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  <row Id="1" PostTypeId="1" AcceptedAnswerId="5" CreationDate="2010-11-02T18:59:01.140" Score="19" ViewCount="482" Body="&lt;p&gt;I often hear about subatomic particles having a property called &quot;spin&quot; but also that it doesn't actually relate to spinning about an axis like you would think. Which particles have spin? What does spin mean if not an actual spinning motion?&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-11-30T19:25:00.320" LastActivityDate="2010-11-30T19:25:00.320" Title="What is spin as it relates to subatomic particles?" Tags="&lt;quantum-mechanics&gt;&lt;spin&gt;&lt;subatomic&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="3" />
  <row Id="2" PostTypeId="1" AcceptedAnswerId="5704" CreationDate="2010-11-02T19:09:48.680" Score="14" ViewCount="586" Body="&lt;p&gt;How would you explain string theory to non physicists such as myself? I'm specially interested on how plausible is it and what is needed to successfully prove it?&lt;/p&gt;&#xA;" OwnerUserId="23" LastEditorUserId="23" LastEditorDisplayName="" LastEditDate="2011-02-24T01:58:12.180" LastActivityDate="2011-03-10T09:27:17.140" Title="What is your simplest explanation of the string theory?" Tags="&lt;general-relativity&gt;&lt;string-theory&gt;&lt;particle-physics&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="4" />
  <row Id="3" PostTypeId="1" AcceptedAnswerId="180" CreationDate="2010-11-02T19:11:05.857" Score="16" ViewCount="376" Body="&lt;p&gt;This is a question that has been posted at many different forums, I thought maybe someone here would have a better or more conceptual answer than I have seen before:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do physicists care about representations of Lie groups? For myself, when I think about a representation that means there is some sort of group acting on a vector space, what is the vector space that this Lie group is acting on? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it that certain things have to be invariant under a group action?&#xA;maybe this is a dumb question, but i thought it might be a good start...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify, I am specifically thinking of the symmetry groups that people think about in relation to the standard model. I do not care why it might be a certain group, but more how we see the group acting, what is it acting on? etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;math kid&lt;/p&gt;&#xA;" OwnerUserId="40" LastEditorUserId="40" LastEditorDisplayName="" LastEditDate="2010-11-03T14:39:08.327" LastActivityDate="2010-11-03T23:01:41.460" Title="Lie theory, Representations and particle physics" Tags="&lt;particle-physics&gt;&lt;standard-model&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="4" />
  <row Id="4" PostTypeId="2" ParentId="1" CreationDate="2010-11-02T19:12:26.543" Score="4" ViewCount="" Body="&lt;p&gt;Imagine going to the rest frame of a massive particle.  In this frame, there is rotational symmetry, which means that the Lie algebra of rotations acts on the wave function.  So the wave function is a vector in a &lt;em&gt;representation&lt;/em&gt; of Lie(SO(3)) = Lie(SU(2)).  &quot;Spin&quot; is the label of precisely which representation this is.  Note that while SO(3) and SU(2) share a Lie algebra, they are different as groups, and it is a fact of life (&quot;the connection between spin and statistics&quot;) that some particles -- fermions, with half-integral spin -- transform under representations of SU(2) while others -- bosons, with integral spin -- transform under SO(3).&lt;/p&gt;&#xA;" OwnerUserId="20" LastActivityDate="2010-11-02T19:12:26.543" CommentCount="3" />
  <row Id="5" PostTypeId="2" ParentId="1" CreationDate="2010-11-02T19:15:20.813" Score="13" ViewCount="" Body="&lt;p&gt;Spin is a technical term specifically referring to &lt;em&gt;intrinsic angular momentum&lt;/em&gt; of particles. It means a very specific thing in quantum/particle physics. (Physicists often borrow loosely related everyday words and give them a very precise physical/mathematical definition.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since truly fundamental particles (e.g. electrons) are point entities, i.e. have no true size in space, it does not make sense to consider them 'spinning' in the common sense, yet they still possess their own angular momenta. Note however, that like many quantum states (fundamental variables of systems in quantum mechanics,) spin is &lt;em&gt;quantised&lt;/em&gt;; i.e. it can only take one of a set of discrete values. Specifically, the allowed values of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Spin_%28physics%29#Spin_quantum_number&quot; rel=&quot;nofollow&quot;&gt;spin quantum number&lt;/a&gt; &lt;code&gt;s&lt;/code&gt; are non-negative multiples of 1/2. The actual spin momentum (denoted &lt;code&gt;S&lt;/code&gt;) is a multiple of Planck's constant, and is given by $S = \sqrt{s (s + 1)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to composite particles (e.g. nuclei, atoms), spin is actually fairly easy to deal with. Like normal (orbital) angular momentum, it adds up linearly. Hence a proton, made of three constituent quarks, has overall spin 1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're curious as to how this (initially rather strange) concept of spin was discovered, I suggest reading about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stern-Gerlach_Experiment&quot; rel=&quot;nofollow&quot;&gt;Stern-Gerlach experiment&lt;/a&gt; of the 1920s. It was later put into the theoretical framework of quantum mechanics by Schrodinger and Pauli.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2010-11-03T13:20:27.003" LastActivityDate="2010-11-03T13:20:27.003" CommentCount="13" />
  <row Id="6" PostTypeId="2" ParentId="1" CreationDate="2010-11-02T19:18:47.330" Score="1" ViewCount="" Body="&lt;p&gt;If a little historical perspective would help, I recommend &lt;a href=&quot;http://www.nature.com/milestones/milespin/timeline.html&quot; rel=&quot;nofollow&quot;&gt;this series&lt;/a&gt; by Nature on the effect that studying &quot;spin&quot; has had on the world of particle physics (and vice-versa).&lt;/p&gt;&#xA;" OwnerUserId="36" LastActivityDate="2010-11-02T19:18:47.330" />
  <row Id="7" PostTypeId="1" AcceptedAnswerId="46" CreationDate="2010-11-02T19:26:38.137" Score="5" ViewCount="301" Body="&lt;p&gt;What are the main problems that we need to solve to prove Laplace's determinism correct and overcome the Uncertainty principle?&lt;/p&gt;&#xA;" OwnerUserId="23" LastEditorUserId="23" LastEditorDisplayName="" LastEditDate="2010-11-02T21:17:54.973" LastActivityDate="2010-11-07T10:56:02.237" Title="Will Determinism be ever possible?" Tags="&lt;uncertainty-principle&gt;&lt;determinism&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="1" />
  <row Id="8" PostTypeId="2" ParentId="3" CreationDate="2010-11-02T19:27:34.463" Score="4" ViewCount="" Body="&lt;p&gt;The vector space that is being acted on typically is a Hilbert space of states in quantum mechanics; very roughly, there's a basis of this vector space which is in one-to-one correspondence with the set of possibilities for a physical system.  The simplest example to try to get your head around is that of the spin 1/2 particle (2 dim representation of SU(2)), which is explained in any introductory quantum mechanics book.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-02T19:27:34.463" CommentCount="6" />
  <row Id="9" PostTypeId="1" CreationDate="2010-11-02T19:28:09.013" Score="16" ViewCount="641" Body="&lt;p&gt;Hamilton's principle states that a dynamic system always follows a path such that action integral is stationary (that is maximum or minimum).&#xA;Why should action integral be stationary? On what basis did Hamilton stated this principle?&lt;/p&gt;&#xA;" OwnerUserId="48" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-01-18T20:38:56.783" LastActivityDate="2011-01-18T22:17:43.910" Title="Hamilton's Principle" Tags="&lt;quantum-mechanics&gt;&lt;mechanics&gt;&lt;hamiltonian-mechanics&gt;&lt;variational-principle&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="5" />
  <row Id="10" PostTypeId="2" ParentId="2" CreationDate="2010-11-02T19:36:09.670" Score="1" ViewCount="" Body="&lt;p&gt;Some of the leaders in the field are Brian Greene and Michio Kaku.  Both have made some Sci channel or Nova series that seemed  appropriate for non-physicists.&lt;br&gt;&#xA;I haven't been following the latest in physics for a while, but I thought that M-Theory had supplanted string theory some time ago.  If so, then Kaku would be the guy.  Greene's series and book &quot;The Elegant Universe&quot; also goes into some Quantum mechanics, which might help as well.&lt;br&gt;&#xA;These are really the only two players I know of that simplify string theory.  Does anyone else have some ideas?&lt;/p&gt;&#xA;" OwnerUserId="49" LastActivityDate="2010-11-02T19:36:09.670" CommentCount="5" />
  <row Id="11" PostTypeId="1" CreationDate="2010-11-02T19:39:28.490" Score="0" ViewCount="142" Body="&lt;p&gt;How can I carry out left hand rule of motors?&lt;/p&gt;&#xA;" OwnerUserId="58" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2010-11-08T02:54:31.453" LastActivityDate="2010-11-08T02:54:31.453" ClosedDate="2010-11-02T22:00:04.677" Title="Left hand rule - how can I carry out it?" Tags="&lt;electromagnetism&gt;" CommentCount="2" />
  <row Id="12" PostTypeId="2" ParentId="7" CreationDate="2010-11-02T19:44:32.050" Score="-2" ViewCount="" Body="&lt;h1&gt;1)  Borrow one of Star Trek's &quot;Heisenberg Compensators&quot;.&lt;/h1&gt;&#xA;&#xA;&lt;h1&gt;2)  There is no barrier as there is no need to combine the two.  What specific barriers do you mean?&lt;/h1&gt;&#xA;" OwnerUserId="49" LastActivityDate="2010-11-02T19:44:32.050" CommentCount="3" />
  <row Id="13" PostTypeId="1" CreationDate="2010-11-02T19:46:46.477" Score="5" ViewCount="319" Body="&lt;p&gt;I've been using the term &quot;sound&quot; all my life, but I really have no clue as to what sound exactly is or how it is created. What is sound? How is it produced? Can it be measured?&lt;/p&gt;&#xA;" OwnerUserId="54" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2011-04-27T09:25:03.957" LastActivityDate="2011-04-27T09:25:03.957" Title="What is sound and how is it produced?" Tags="&lt;sound&gt;&lt;definition&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="14" PostTypeId="2" ParentId="9" CreationDate="2010-11-02T19:50:19.220" Score="4" ViewCount="" Body="&lt;p&gt;As you can see from the image below, you want the variation of the action integral to be a minimum, therefore $\displaystyle \frac{\delta S}{\delta q}$ must be $0$. Otherwise, you are not taking the true path between $q_{t_{1}}$ and $q_{t_{2}}$ but a slightly longer path. However, even following $\delta S=0$, as you know, you might end up with another extremum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/qJWaK.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following the link from j.c., you can find &lt;a href=&quot;http://www.emis.de/classics/Hamilton/GenMeth.pdf&quot; rel=&quot;nofollow&quot;&gt;On a General Method on Dynamics&lt;/a&gt;, which probably answers your question regarding Hamilton's reasoning. I haven't read it but almost surely it is worthwhile.&lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="18" LastEditorDisplayName="" LastEditDate="2010-11-03T15:52:15.833" LastActivityDate="2010-11-03T15:52:15.833" />
  <row Id="15" PostTypeId="1" AcceptedAnswerId="48" CreationDate="2010-11-02T19:51:03.887" Score="16" ViewCount="1682" Body="&lt;p&gt;I know that there's big controversy between two groups of physicists:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;those who support string theory (most of them, I think)&lt;/li&gt;&#xA;&lt;li&gt;and those who oppose it.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;One of the arguments of the second group is that there's no way to disprove the correctness of the string theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is if there's any defined experiment that would disprove string theory?&lt;/p&gt;&#xA;" OwnerUserId="57" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-14T00:46:11.987" LastActivityDate="2011-02-14T00:46:11.987" Title="What experiment would disprove string theory?" Tags="&lt;string-theory&gt;" AnswerCount="9" CommentCount="6" FavoriteCount="5" />
  <row Id="16" PostTypeId="2" ParentId="13" CreationDate="2010-11-02T19:55:07.630" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What it sound?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sound is nothing but a mechanical wave. The human ear can detect mechanical waves from 20 Hz to 20,000 Hz.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How is it produced?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is produced when energy transformation occurs. For eg. beating a drum - mechanical energy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can it be measured?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, it can be measured by the frequency of the wave.&lt;/p&gt;&#xA;" OwnerUserId="48" LastEditorUserId="48" LastEditorDisplayName="" LastEditDate="2010-11-03T06:16:50.360" LastActivityDate="2010-11-03T06:16:50.360" CommentCount="2" />
  <row Id="17" PostTypeId="1" AcceptedAnswerId="22" CreationDate="2010-11-02T20:00:49.790" Score="13" ViewCount="434" Body="&lt;p&gt;Why the sky is blue during the day, red during sunrise/set and black during the night?&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-11-06T13:12:46.880" LastActivityDate="2010-12-15T04:49:33.667" Title="Why does the sky change color?" Tags="&lt;optics&gt;&lt;geophysics&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="18" PostTypeId="2" ParentId="7" CreationDate="2010-11-02T20:03:01.050" Score="3" ViewCount="" Body="&lt;p&gt;Experiments such as the double-slit, EPR and Bell inequality experiments show that quantum mechanics is not just a computational tool, but a genuine property of nature. In this sense, Heisenberg's Uncertainty Principle cannot be overcome and there will be no reversal to determinism.&lt;/p&gt;&#xA;" OwnerUserId="47" LastActivityDate="2010-11-02T20:03:01.050" CommentCount="3" />
  <row Id="19" PostTypeId="1" AcceptedAnswerId="45" CreationDate="2010-11-02T20:03:34.887" Score="2" ViewCount="258" Body="&lt;p&gt;Physicist ofter refer to the energy of collisions between different particles. My questions is how is that energy calculated? Is that kinetic energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, related to this question, I know that the aim is to have higher and higher energy collisions (e.g to test for Higgs Boson). My understanding is that to have higher energy you can either accelerate them more or use particles with higher mass. Is that correct?&lt;/p&gt;&#xA;" OwnerUserId="57" LastEditorUserId="7" LastEditorDisplayName="" LastEditDate="2010-11-02T21:03:58.760" LastActivityDate="2011-04-29T16:03:43.513" Title="How's the energy of particle collisions calculated?" Tags="&lt;particle-physics&gt;&lt;high-energy-physics&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="20" PostTypeId="2" ParentId="13" CreationDate="2010-11-02T20:04:33.713" Score="5" ViewCount="" Body="&lt;p&gt;In general usage, &quot;sound&quot; refers to our perception of the vibrations of particles (atoms, molecules) in some medium, typically air or water, though sound waves can travel through any medium.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vibrations are produced whenever objects cause the particles in the medium to oscillate, e.g. clapping your hands together, beating a drum, or yelling.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sound can be measured the way it is heard - just as air molecules begin to vibrate in the presence of a vibrating object - the air molecules can cause other things to vibrate (such as our eardrums, or a part of a microphone) and we can quantify the level of sound by measuring the motion of the detector in any number of ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a deeper level, sound waves are the &quot;Goldstone modes&quot; corresponding to broken translational symmetry of the surrounding medium.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me try to bring this a little more down to earth.  Think about a parcel of air, large enough to contain a lot of air molecules, but not so large that there are sizable air currents swirling within.  We can think of this in an ideal situation as being a fluid of uniform density.  This fluid has translational symmetry - if I translate it a little in any direction, it looks more or less the same &quot;in the bulk&quot;.  Translational symmetry is a continuous symmetry, so I can squeeze the air in a spatially-varying way.  It turns out that if I do this in a wavelike pattern with a long wavelength, this is an excitation that costs very little energy, and hence will be important to the physics.  These waves are precisely sound waves.  This probably doesn't make much sense as I've written it, so I may try to edit this a bit later on if anyone is actually interested in this point of view.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-02T20:04:33.713" />
  <row Id="21" PostTypeId="1" AcceptedAnswerId="684" CreationDate="2010-11-02T20:06:21.360" Score="2" ViewCount="166" Body="&lt;p&gt;Where is the Monte Carlo method used in physics?&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-11-02T20:41:47.273" LastActivityDate="2010-11-13T01:38:40.830" Title="Monte Carlo use" Tags="&lt;simulation&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="22" PostTypeId="2" ParentId="17" CreationDate="2010-11-02T20:09:04.537" Score="17" ViewCount="" Body="&lt;p&gt;The keywords here are &lt;a href=&quot;http://en.wikipedia.org/wiki/Rayleigh_scattering&quot; rel=&quot;nofollow&quot;&gt;Rayleigh scattering&lt;/a&gt;.  See also &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffuse_sky_radiation&quot; rel=&quot;nofollow&quot;&gt;diffuse sky radiation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But much more simply, it has to do with the way that sunlight interacts with air molecules.  Blue light is scattered more than red light, so during the day when we look at parts of the sky that are away from the sun, we see more blue than red.  During sunset or sunrise, most of the light from the sun comes towards the earth at a sharp angle, so now the blue light is mostly scattered away, and we see mostly red light.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-02T20:09:04.537" />
  <row Id="23" PostTypeId="2" ParentId="13" CreationDate="2010-11-02T20:10:50.757" Score="2" ViewCount="" Body="&lt;p&gt;Sound is air molecules hitting each other, bouncing back, then again hitting each other, and so on. This back-and-forth movement is called vibration. Sound is produced when something like a drum vibrates, and starts the aforementioned bouncing of air molecules. Sound is picked up inside your ears by eardrums, which are like reverse-drums, in that they vibrate with the air molecules. This vibration is then converted to biochemical signals which travel to your brain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a Mickey Mouse visualization here on Youtube:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=qUcI81Ysarg&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=qUcI81Ysarg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="47" LastActivityDate="2010-11-02T20:10:50.757" />
  <row Id="24" PostTypeId="1" AcceptedAnswerId="524" CreationDate="2010-11-02T20:11:14.907" Score="6" ViewCount="781" Body="&lt;p&gt;I think it's clear enough that if you turn your bicycle's steering wheel left, while moving, and you don't lean left, the bike will fall over (to the right) as you turn. I figure this is because the bike's momentum keeps it moving in the direction you were going, and since your wheels have friction against the ground, the top of the bike moves forward relative to the bottom of the wheels. The top of the bike going north while the bottom of the wheels go northwest will understandable cause you to topple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to counteract this and keep you from falling over, leaning into the turn is necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is there also a causal relationship -- that leaning will &lt;strong&gt;cause&lt;/strong&gt; the bike to start to turn? If I start leaning left, I will turn left... but maybe that's because I know that if I don't turn the steering wheel left, the bike will fall over (to the left). I experimented with unruly turns of the steering wheel when I was a kid, and got my scrapes and bruises. Now that I'm a cautious and sedate adult I'm not anxious to experiment that way. :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I also want to ask why airplanes bank into a turn... they don't have the same issues as a bike, i.e. the bottom part has no special friction against the ground. But that would probably make the question too broad.)&lt;/p&gt;&#xA;" OwnerUserId="70" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2010-11-10T18:34:30.180" LastActivityDate="2010-11-24T10:13:58.537" Title="Does leaning (banking) help cause turning on a bike?" Tags="&lt;momentum&gt;&lt;stability&gt;&lt;bicycles&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="2" />
  <row Id="25" PostTypeId="2" ParentId="21" CreationDate="2010-11-02T20:11:28.417" Score="4" ViewCount="" Body="&lt;p&gt;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Monte_Carlo_method_in_statistical_physics&quot; rel=&quot;nofollow&quot;&gt;statistical mechanics&lt;/a&gt;, the Monte Carlo method is used to sample the state space of a statistical system in order to compute physical quantities of interest by averaging.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-02T20:11:28.417" />
  <row Id="26" PostTypeId="1" CreationDate="2010-11-02T20:12:32.630" Score="2" ViewCount="169" Body="&lt;p&gt;I am wondering if someone could provide me with a formula that would tell me at what velocity a projectile can be launched from something using an electromagnetic field. The idea is much like a rail gun or Gauss rifle, but not exactly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just looking to find a formula to determine the speed of the projectile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you,&#xA;Michael Vanderpool&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2010-11-08T02:54:49.893" LastActivityDate="2011-05-30T02:10:03.313" Title="Velocity of Object from electromagnetic field" Tags="&lt;electromagnetism&gt;&lt;projectile&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="27" PostTypeId="1" AcceptedAnswerId="30" CreationDate="2010-11-02T20:17:48.947" Score="16" ViewCount="524" Body="&lt;p&gt;We've learned that the wave function of a particle collapses when we measure a particle's location. If it is found, it becomes more probable to find it a again in the same area, and if not the probability to finding it in the place that was checked decreases dramatically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is about the definition of measurement. What makes a measurement different from any other interaction between two particles (gravity and EM fields for example)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, almost every particle interacts with any other particle, so shouldn't there be constant collapse of the wave function all the time? If this happens we're right back in classical mechanics, aren't we?&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-03T17:54:12.637" LastActivityDate="2011-02-20T13:27:56.247" Title="What is the difference between a measurement and any other interaction in quantum mechanics?" Tags="&lt;quantum-mechanics&gt;&lt;epistemology&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="9" />
  <row Id="28" PostTypeId="2" ParentId="19" CreationDate="2010-11-02T20:18:01.623" Score="0" ViewCount="" Body="&lt;p&gt;It is worth remembering that in at velocities close to c. A particle kinetic energy is inter-wind with its rest mass. So the actual energy equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E = \sqrt{p^2c^2+m^2c^4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the energy of a collision is the sum of the above energy for the two colliding particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is why you construct improved accelerators. As it is the only way to tune up the energy. To the extend of my knowledge we don't how to tune up mass.&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="62" LastEditorDisplayName="" LastEditDate="2010-11-08T07:13:46.200" LastActivityDate="2010-11-08T07:13:46.200" CommentCount="6" />
  <row Id="29" PostTypeId="1" AcceptedAnswerId="33" CreationDate="2010-11-02T20:19:42.540" Score="4" ViewCount="254" Body="&lt;p&gt;I recently encountered a puzzle where a person drove 120 miles at 40mph, then drove back the same 120 miles at 60mph.  The average of the speeds is (40mph+60mph)/2 = 50mph, so the total trip time should be 240mi/50mph = 4.8 hours.  But the trip actually took 5 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is that, and what is the correct way to calculate average speed?&lt;/p&gt;&#xA;" OwnerUserId="77" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-02T20:41:36.290" LastActivityDate="2010-11-03T08:25:11.803" Title="How to calculate average speed" Tags="&lt;speed&gt;&lt;kinematics&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="30" PostTypeId="2" ParentId="27" CreationDate="2010-11-02T20:21:45.097" Score="13" ViewCount="" Body="&lt;p&gt;What you describe in your question is the &quot;Copenhagen interpretation&quot; of quantum mechanics.  There are more nuanced views of this nowadays that don't treat &quot;measurements&quot; quite so asymmetrically, see e.g. sources that talk about decoherence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend watching the classic lecture &lt;a href=&quot;http://media.physics.harvard.edu/video/index.php?id=SidneyColeman_QMIYF.flv&quot; rel=&quot;nofollow&quot;&gt;&quot;Quantum Mechanics in your face&quot; by Sidney Coleman&lt;/a&gt; for a nice take on this sort of thing.  &lt;/p&gt;&#xA;" OwnerUserId="42" LastEditorUserId="42" LastEditorDisplayName="" LastEditDate="2010-11-02T20:37:44.157" LastActivityDate="2010-11-02T20:37:44.157" CommentCount="3" />
  <row Id="31" PostTypeId="1" AcceptedAnswerId="44" CreationDate="2010-11-02T20:22:33.677" Score="4" ViewCount="303" Body="&lt;p&gt;What is Einstein's theory of Special relativity, in terms a lay person can follow?&lt;/p&gt;&#xA;" OwnerUserId="77" LastEditorUserId="7" LastEditorDisplayName="" LastEditDate="2010-11-02T20:56:39.803" LastActivityDate="2011-01-17T00:45:37.377" Title="Lay explanation of the theory of relativity?" Tags="&lt;special-relativity&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="32" PostTypeId="1" AcceptedAnswerId="43" CreationDate="2010-11-02T20:25:08.490" Score="7" ViewCount="300" Body="&lt;p&gt;There is a common myth that water flowing out from a sink should rotate in direction governed by on which hemisphere we are; this is shown false in many household experiments, but how to show it theoretically?  &lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-11-22T18:13:39.927" LastActivityDate="2011-02-05T14:05:02.777" Title="Resolving water sink problem" Tags="&lt;fluid-dynamics&gt;&lt;coriolis-force&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="33" PostTypeId="2" ParentId="29" CreationDate="2010-11-02T20:25:31.397" Score="11" ViewCount="" Body="&lt;p&gt;The reason is because the time taken for the two trips are different, so the average speed is not simply (v1 + v2)/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We should go back to the definition. The average speed is always (total length) &amp;divide; (total time). In your case, the total time can be calculated as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\text{time}_1 &amp;amp;= \frac{120 \mathrm{miles}}{40 \mathrm{mph}} \\&#xA;\text{time}_2 &amp;amp;= \frac{120 \mathrm{miles}}{60 \mathrm{mph}}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the total time is $120\mathrm{miles} \times \left(\frac1{40\mathrm{mph}} + \frac1{60\mathrm{mph}}\right)$. The average speed is therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\text{average speed} &amp;amp;= \frac{2 \times 120\mathrm{miles}}{120\mathrm{miles} \times \left(\frac1{40\mathrm{mph}} + \frac1{60\mathrm{mph}}\right)} \\&#xA;&amp;amp;= \frac{2 }{ \left(\frac1{40\mathrm{mph}} + \frac1{60\mathrm{mph}}\right)} \\&#xA;&amp;amp;= 48 \mathrm{mph}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, when the &lt;em&gt;length&lt;/em&gt; of the trips are the same, the average speed will be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_mean&quot; rel=&quot;nofollow&quot;&gt;harmonic mean&lt;/a&gt; of the respective speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \text{average speed} = \frac2{\frac1{v_1} + \frac1{v_2}} $$&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-11-03T07:02:06.097" LastActivityDate="2010-11-03T07:02:06.097" CommentCount="1" />
  <row Id="34" PostTypeId="2" ParentId="29" CreationDate="2010-11-02T20:26:36.210" Score="2" ViewCount="" Body="&lt;p&gt;The difficulty is that since the trip at 40mph takes longer, you spend more time going 40mph than you do going 60mph, so the average speed is weighted more heavily towards 40 mph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When calculating average speeds for fixed distances, it is better think of everything in minutes per mile rather than miles per hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;60 miles per hour is 1 minute per mile, while 40 miles per hour is 1.5 minutes per mile.  Since we travel the same number of miles at each speed, we can now take the mean of these two figures. That's 1.25 minutes per mile on average.  For 240 miles total, 240miles*1.25minutes/mile = 300 minutes = 5 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method is called finding the &quot;harmonic mean&quot; of the speeds.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-02T20:26:36.210" />
  <row Id="35" PostTypeId="1" AcceptedAnswerId="40" CreationDate="2010-11-02T20:26:45.977" Score="7" ViewCount="589" Body="&lt;p&gt;If I separate two magnets who's opposite poles are facing, I am adding energy. If I let go of the magnets, presumably the energy that I added is used to move the magnets together again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if I start with two separated magnets (with like poles facing), as I move them together, they repel each other. They must be using energy to counteract the force that I'm applying. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does this energy come from? &lt;/p&gt;&#xA;" OwnerUserId="75" LastActivityDate="2011-01-17T07:36:10.233" Title="Where do magnets get the energy to repel?" Tags="&lt;magnets&gt;&lt;energy-conservation&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="36" PostTypeId="2" ParentId="29" CreationDate="2010-11-02T20:27:39.997" Score="3" ViewCount="" Body="&lt;p&gt;Average Speed = Total Distance/ Total time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;t1 = 120/40 = 3 hrs&#xA;t2 = 120/60 = 2 hrs&#xA;&#xA;total time = 5 hrs&#xA;total distance = 240 miles&#xA;&#xA;average speed = 240/5 = 48mph&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="48" LastActivityDate="2010-11-02T20:27:39.997" />
  <row Id="37" PostTypeId="1" CreationDate="2010-11-02T20:32:03.353" Score="4" ViewCount="189" Body="&lt;p&gt;Physicists studying the grounds of physics and some mathematicians often come to a theories which are like a general-relativity, but do not coincide with it. Often these theories contradict the real world. What should one do in order to check one of such theories? Which papers/books should one read in order to be able to understand about some mathematical theories, that &quot;the mathematical theory is nice, but this is not our world&quot;?&lt;/p&gt;&#xA;" OwnerUserId="76" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-25T19:53:17.403" LastActivityDate="2010-11-25T19:53:17.403" Title="How to check Einstein-like equations on their correspondence with the real world?" Tags="&lt;general-relativity&gt;&lt;string-theory&gt;&lt;theory-of-everything&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="38" PostTypeId="2" ParentId="21" CreationDate="2010-11-02T20:34:51.497" Score="2" ViewCount="" Body="&lt;p&gt;For example, a Markov Chain Monte Carlo (MCMC) method was used in the WMAP [1] papers, the most important measurment/papers of modern CMB [2] cosmology, one of the pillars of the Big Bang Theory of the Universe. MCMC was used to explore and fit the parameters of the theory with the WMAP (and other) measurements. You can download the technical papers at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://lambda.gsfc.nasa.gov/product/map/current/map_bibliography.cfm&quot; rel=&quot;nofollow&quot;&gt;http://lambda.gsfc.nasa.gov/product/map/current/map_bibliography.cfm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Wilkinson Microwave Anisotropy Probe&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Cosmic Microwave Background&lt;/p&gt;&#xA;" OwnerUserId="47" LastActivityDate="2010-11-02T20:34:51.497" />
  <row Id="39" PostTypeId="2" ParentId="35" CreationDate="2010-11-02T20:36:14.597" Score="3" ViewCount="" Body="&lt;p&gt;As is said in a comment, the reasoning in the first paragraph is correct but the one in the second paragraph is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you apply a force on something without &quot;moving&quot; the work is null and there is no energy exchange involved (this is not the same thing than doing that with your muscles, but that's another story :p). ( Work =  integral[a to b] of F dot dx ; so Work = 0 if there is not &quot;circulation&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the magnets do not need any energy to statically counteract the force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you do move the magnets, then you need to give some energy. This energy is stored in the system because you cause a variation of magnetic flux: magnet 1 moving induce a variation of flux seen by magnet 2, and this will change the state of magnet 2, increasing its potential energy.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-02T21:23:14.240" LastActivityDate="2010-11-02T21:23:14.240" CommentCount="2" />
  <row Id="40" PostTypeId="2" ParentId="35" CreationDate="2010-11-02T20:39:35.933" Score="9" ViewCount="" Body="&lt;p&gt;Magnetic field in this case (a set of magnets in space, no relativity involved) is conservative, which means it has a potential -- each positional configuration of charges (or dipoles in this case) has its fixed energy which does not depend on history or momenta of charges. So, the work you put or get from displacing them is just exchanged with the potential energy of the field, which means no energy is created or destroyed, just stored.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-11-02T21:26:56.547" LastActivityDate="2010-11-02T21:26:56.547" CommentCount="4" />
  <row Id="41" PostTypeId="1" AcceptedAnswerId="2109" CreationDate="2010-11-02T20:42:22.140" Score="4" ViewCount="339" Body="&lt;p&gt;There have been recent results in mathematics regarding Conformal field theories and topological field theories. I am curious about the reaction to such results in the physics community. So I guess a starting place for a question would be something like: Did Atiyah's axiomatization of TQFT capture what physicists feel are important in QFT's? How many more parameters are there to what a physicist would call a QFT or FT in general?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps I should mention that mathematicians (the ones who dont know any phsyics) think about various Field Theories as things that assign invariants to manifolds that should be sensitive to certain data (depending on the word proceeding field). A lot has been done when you only care about the basic topology of the Manifold you are evalutating your field theory at. Have mathematicians gotten rid of too much for physicists to care? What types of things do physicists want a Field theory to keep track of, what kind of structure on the manifold that is?&lt;/p&gt;&#xA;" OwnerUserId="40" LastActivityDate="2011-01-14T14:07:43.863" Title="Impressions of Topological field theories in mathematics" Tags="&lt;quantum-field-theory&gt;&lt;mathematical-physics&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="42" PostTypeId="2" ParentId="9" CreationDate="2010-11-02T20:46:17.863" Score="8" ViewCount="" Body="&lt;p&gt;The notes from week 1 of John Baez's course in Lagrangian mechanics (&lt;a href=&quot;http://math.ucr.edu/home/baez/classical/#lagrangian&quot; rel=&quot;nofollow&quot;&gt;http://math.ucr.edu/home/baez/classical/#lagrangian&lt;/a&gt;) give some insight into the motivations for action principles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is that least action might be considered an extension of the principle of virtual work.  When an object is in equilibrium, it takes zero work to make an arbitrary small displacement on it.  I.E. the dot product of any small displacement vector and the force is zero (in this case because the force itself is zero).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When an object is accelerating, if we add in an &quot;inertial force&quot; equal to -m*a, then a small, arbitrary, time-dependent displacement from the objects true trajectory would again have zero dot product with F-ma, the true force and inertial force added.   This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(F-ma)\cdot \delta q(t) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, a few calculations found in the notes lead to the stationary action integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Baez discusses D'Alembert more than Hamilton, but either way it's an interesting look into the origins of the idea.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-02T20:46:17.863" CommentCount="1" />
  <row Id="43" PostTypeId="2" ParentId="32" CreationDate="2010-11-02T20:49:09.577" Score="5" ViewCount="" Body="&lt;p&gt;The calculation of the Coriolis force is dependent on latitude:&lt;br&gt;&#xA;$F = m a$  &#xA;where $a = 2 \Omega sin(lat)$, with $\Omega$ being the Earth's angular velocity&lt;br&gt;&#xA;$m$ is the mass of the object in question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth's angular velocity is (about) $7.29 \times 10^{-5}$ rad/sec&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for a sink with a couple gallons of water in it at 45 degrees north... the Coriolis force is about $7.57 \times 2 \times 7.29 \times 10^{-5} = 1.10 \times 10^{-3}$ N. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interesting reading about this &lt;a href=&quot;http://www.meyersix.org/the_mysterious_coriolis_effect.htm&quot; rel=&quot;nofollow&quot;&gt;over this way&lt;/a&gt;, and (of course) &lt;a href=&quot;http://en.wikipedia.org/wiki/Coriolis_effect&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="36" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-22T22:10:56.427" LastActivityDate="2010-11-22T22:10:56.427" CommentCount="1" />
  <row Id="44" PostTypeId="2" ParentId="31" CreationDate="2010-11-02T20:51:29.503" Score="9" ViewCount="" Body="&lt;p&gt;Special Relativity derives from two basic ideas:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The speed of light (in a vacuum) is always c.&lt;/li&gt;&#xA;&lt;li&gt;The laws of physics are the same in all inertial reference frames (basically, points of view that aren't accelerating, that is, they obey Newton's Laws.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With these two points and a little math, various proven conclusions may be derived:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Time Dilation&lt;/strong&gt;: When something moves fast relative to something else, time for the faster moving body slows down. It's not an illusion of time slowing down, it's the real thing: individual atoms that make up the body operate slower, chemical reactions function slower, and biological processes (aging) occur slower. From the perspective of the faster moving body, its time progresses at the usual pace.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Length Contraction&lt;/strong&gt;: Objects moving fast relative to other objects shrink along the line of the direction they're moving.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Relativistic Simultaneity&lt;/strong&gt;: There's no such thing as simultaneous events: because time is attached to the observer, different people could witness 2 events happening in different order. The exception to this is &quot;casually-related&quot; events which are events where event A is the cause of event B.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Mass-Energy&lt;/strong&gt;: The math goes into describing the mass of bodies at rest and how that mass changes as the bodies move. As bodies speed up they get &quot;heavier.&quot; Nothing can travel faster than light (and nothing with mass can travel AT the speed of light) because any massive body would reach infinite &quot;relative mass&quot; at that speed. You can derive E=mc^2 and fission/fusion from this.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is a very quick summary of the basic points and principles.&lt;/p&gt;&#xA;" OwnerUserId="79" LastActivityDate="2010-11-02T20:51:29.503" CommentCount="3" />
  <row Id="45" PostTypeId="2" ParentId="19" CreationDate="2010-11-02T20:53:01.300" Score="2" ViewCount="" Body="&lt;p&gt;I think your question is divided in two parts.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When talking about energy, in the field of HEP or accelerator physics we can talk about &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;total energy&lt;/li&gt;&#xA;&lt;li&gt;kinetic energy&lt;/li&gt;&#xA;&lt;li&gt;momentum&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As for elementary particles relativistic effects manifest themselves almost all the time, you need to use a relativistic form for the energy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = \sqrt{p^2c^2+m^2c^4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of a bare $E = p^2/2m$ (as in classical mechanics, where $p = m v$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This relation as two parts: one depending on the momentum and one (constant) given by the mast of the particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be also noted that for ultra-relativistic cases, where $E \gg E_0$, we have $E = p c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually for low energy applications, like linear accelerators or low energy experiments we talk about the kinetic energy, which is $E_k = \sqrt{p^2c^2+m^2c^4} - mc^2$. For example if you talk about protons of 160 MeV obviously it is kinetic energy, as the rest mass of the proton is rhougly 1 GeV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For higher energy application usually you can make the ultra-relativistic approximation and you then talk about total energy (in eV) or about the momentum in $\mathrm{eV}/c$; taking $c=1$, both are numerically equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you are not sure about which approximation you can take, it is better to explain which one you take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Example:&lt;/em&gt; For a proton in the LHC with a momentum of 3.5 TeV/c you can calculate its total energy which is ... &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How is this energy &quot;calculated&quot;? (I assume you meant &quot;experimentally&quot; or something like that.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In HEP physics we use what we call electron-volt as a unit of energy. A particle of unit charge will have an energy of 1 eV if it descends from rest a potential difference of 1 V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example, when you accelerate protons in the LHC, if you have cavities giving you 10 MV, the particle will gain 10 MeV every turns.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="82" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-29T16:03:43.513" LastActivityDate="2011-04-29T16:03:43.513" CommentCount="8" />
  <row Id="46" PostTypeId="2" ParentId="7" CreationDate="2010-11-02T20:54:20.337" Score="9" ViewCount="" Body="&lt;p&gt;Laplace's determinism is not physically correct over long periods of time.   That is, it neglects chaos/&quot;sensitive dependence on initial conditions&quot;/exponential growth of microscopic perturbations already in Newtonian dynamics, which was seriously thought about only in the 20th century.   Being true, this also will not be overcome.   Stochasticity enters some classical dynamical paths with time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is subtlety here.    In classical mechanics, or the evolution of the wave function, there is a kind of microdeterminism, so that what occurs in the next instant is fully determined by what occurred up until that point.    It is in the longer time evolution of a chaotic system that stochasticity creeps in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, Lapalace said&#xA;&quot;We ought to regard the present state of the universe as the effect of its antecedent state and as the cause of the state that is to follow.&quot;   This part remains true in chaotic classical mechanics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, he then continued &quot;An intelligence knowing all the forces acting in nature at a given instant, as well as the momentary positions of all things in the universe, would be able to comprehend in one single formula the motions of the largest bodies as well as the lightest atoms in the world, provided that its intellect were sufficiently powerful to subject all data to analysis; to it nothing would be uncertain, the future as well as the past would be present to its eyes. The perfection that the human mind has been able to give to astronomy affords but a feeble outline of such an intelligence. (Laplace 1820)&quot;  This is the part that classical chaos invalidates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also read &lt;a href=&quot;http://plato.stanford.edu/entries/determinism-causal/&quot; rel=&quot;nofollow&quot;&gt;http://plato.stanford.edu/entries/determinism-causal/&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there are also questions whether, in light of general relativity, black holes, etc, we can even speak of a &quot;state of the universe as a whole&quot;    There may not be such a god's eye view altogether.   These issues need a philosophy forum, however.  &lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-11-06T18:32:03.297" LastActivityDate="2010-11-06T18:32:03.297" CommentCount="2" />
  <row Id="47" PostTypeId="2" ParentId="31" CreationDate="2010-11-02T21:08:52.787" Score="0" ViewCount="" Body="&lt;p&gt;Einstein's Special Theory of Relativity cannot be of much significance to a lay person because he has no use for such knowledge. However, the common idea that this topic is extremely complicated motivates the layman to learn it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic concepts of this theory are actually somewhat simple. In short, when an object is moving a certain velocity $u$, a few seemingly unusual phenomena happen to it. If a bar of length $L_0$ moves in the direction of its length, its new length will appear to be $L_0 \sqrt{1 - u^2 / c^2}$ from the our viewpoint on the ground, where $c$ is the speed of light. If that formula means nothing to you, simply take note that the length decreases as the velocity increases. Of course, from the bar's viewpoint, it will appear that &lt;em&gt;our&lt;/em&gt; length has changed by that factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since time is proportional to distance traveled at a constant velocity, the length contraction can show that the &quot;rate of time&quot; also changes when an object is moving. If a clock is traveling through space close to the speed of light, it will tick significantly slower than it did at rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those are the basics, but phenomena such as mass-energy equivalence and nuclear binding energy can be derived from these concepts and experiments conducted in the early 20th century.&lt;/p&gt;&#xA;" OwnerUserId="69" LastEditorUserId="69" LastEditorDisplayName="" LastEditDate="2010-11-04T02:21:59.290" LastActivityDate="2010-11-04T02:21:59.290" CommentCount="1" />
  <row Id="48" PostTypeId="2" ParentId="15" CreationDate="2010-11-02T21:08:57.003" Score="2" ViewCount="" Body="&lt;p&gt;String theorist should come with a proposal for an experiment, make some predictions about the results of the experiment and then we should check with real results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a theory cannot come with any, then it will disprove itself little by little ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that with string theory this is extremely difficult to do, and string theorist have year in front of them to go in that direction; but if in 100 years we are still at the same status, then it would be a proof that string theory is unfruitful...&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-11-02T21:08:57.003" CommentCount="12" />
  <row Id="49" PostTypeId="1" AcceptedAnswerId="512" CreationDate="2010-11-02T21:09:23.880" Score="6" ViewCount="137" Body="&lt;p&gt;What should be a properties of a body so a contact with it can be detected by a capacitive touchscreen or a touchpad? &lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="48" LastEditorDisplayName="" LastEditDate="2010-11-02T21:39:43.183" LastActivityDate="2010-11-10T16:59:58.837" Title="What is a capacitive screen sensing?" Tags="&lt;sensor&gt;&lt;applied-physics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="50" PostTypeId="2" ParentId="35" CreationDate="2010-11-02T21:11:34.347" Score="2" ViewCount="" Body="&lt;p&gt;As others have pointed out, the question is somewhat misworded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct question is probably just &quot;Why do magnets repel?&quot;, which can be traced back to the question of why electromagnetism (EM) is a fundamental force of nature. Like gravity, it's one of the four known fundamental forces of nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the use of the word &quot;energy&quot;. Energy is the same thing as work. Work = Force x distance. If there is a 10N [Newton] force acting on a ball, and you move the ball by 1m [meter], you have done 10N x 1M = 10Nn = 10J [Joule] of work (provided the force is constant). So, if you move the magnetic poles against each other, you do work (against the EM field). Then, if you attach the magnets to a spring and let go, the springs will compress. The 10J of work you invested is now stored in the springs! Then, if you remove the magnets and put a bullet on the spring, and let go, you can use the 10J of energy to launch the bullet (work is converted to kinetic energy), and so on =)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens when you push two magnets against each other, but then just hold them in place? Per the above definition of Work = Force x distance, you're not doing work, because distance = 0. But clearly your muscles are straining, and you're burning calories (which is just another unit of energy or work, like Joules), so what's going on? From a physics textbook point of view, no work is being done. This is similar to placing an apple on a table. The table is countering the force of gravity, like your muscles in the magnet example. The table has no muscles, because it is made of wood, and it won't collapse under the weight of the apple. In reality, the stiffness of the table/wood is provided by the electromagentic interactions of the atoms making up the table, so in the end it is the EM force counteracting the gravitational force. In the original magnet/muscle example, your arms are not made up of wood, so your body is doing biochemical/mechanical work to stiffen the cells in your arms by burning the food you ate. (The fact that this process is not perfect means that heat is a by product of this process, that's why your muscles get hot and you start sweating.)&lt;/p&gt;&#xA;" OwnerUserId="47" LastActivityDate="2010-11-02T21:11:34.347" CommentCount="1" />
  <row Id="51" PostTypeId="2" ParentId="37" CreationDate="2010-11-02T21:13:59.320" Score="2" ViewCount="" Body="&lt;p&gt;Physical theories are rarely constructed as a purely theoretical exercise - they usually are based on empirical data which is fitted a mathematical solution. The physical theory then tries to explain the data and corresponding math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basis of scientific theory is that it must be verified by repeatable tests and data. Sometimes theories that explain some data well are disproven when applied to other situations - and sometimes amendments are made in order to account for the difference. That is the nature of every scientific theory - it is accepted until real world data says it is wrong. There is no way to know whether any theory truly describes our world or not - it holds until it no longer matches observations.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditorDisplayName="" LastEditDate="2010-11-02T21:24:02.463" LastActivityDate="2010-11-02T21:24:02.463" />
  <row Id="52" PostTypeId="1" AcceptedAnswerId="55" CreationDate="2010-11-02T21:18:10.360" Score="3" ViewCount="290" Body="&lt;p&gt;A few years ago I went to a museum, where there was a board with 2 bar magnets, on a pole each so they could rotate.&#xA;If you rotated them so the lined up with the same poles (N) facing each other, they'd repel spinning until the south poles lined up, then they'd repel and spin so the north poles lined up and so. This was without any extra help, once you lined them up they would just keep spinning by their selves. How exactly is this possible that the 2 magnets kept producing kinetic energy?&lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2010-11-02T21:32:29.757" Title="How do 2 magnets spin by themselves if positioned precisely?" Tags="&lt;magnets&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="53" PostTypeId="2" ParentId="2" CreationDate="2010-11-02T21:20:10.313" Score="3" ViewCount="" Body="&lt;p&gt;Greene's book has already been mentioned, but Nova's &lt;a href=&quot;http://www.pbs.org/wgbh/nova/physics/elegant-universe-einstein.html&quot; rel=&quot;nofollow&quot;&gt;The Elegant Universe&lt;/a&gt; is a somewhat informative 3-hour video series. Obviously, to truly understand the theory's progress, years of mathematics and physics must be studied, preferably at a university.&lt;/p&gt;&#xA;" OwnerUserId="69" LastActivityDate="2010-11-02T21:20:10.313" />
  <row Id="54" PostTypeId="2" ParentId="7" CreationDate="2010-11-02T21:27:59.867" Score="2" ViewCount="" Body="&lt;p&gt;Determinism as more to worry about stochastic and chaotic processes than with the Heisenberg's uncertainty principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a simple proof, you can easily consider a purely classical system and observe that you don't have determinism as Laplace had hopped.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-11-02T21:27:59.867" />
  <row Id="55" PostTypeId="2" ParentId="52" CreationDate="2010-11-02T21:32:29.757" Score="4" ViewCount="" Body="&lt;p&gt;They are not producing energy; their movement is a result of the work you've put during setting same poles to face each other. They will eventually stop, since this energy is dissipated by friction on their fixings and at some point their momentum will decrease to the point it will be not enough to push the system through energy pick when the same poles are nearby. After this, they will relax to their energy optimum, so probably into configuration when different poles are facing each other.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-02T21:32:29.757" CommentCount="8" />
  <row Id="56" PostTypeId="2" ParentId="15" CreationDate="2010-11-02T21:46:26.483" Score="7" ViewCount="" Body="&lt;p&gt;String theory, was constructed with the idea that at low energies it should reduce to the quantum mechanical and particle world that we see every day. This is analogous to the correspondence principle in quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some sense, any experiment that discredits quantum mechanics will cause a serious re-examination of string theory, however, as many experimentalists will joke, a theorist will always find a way to fix his theory to match the observations. In any case, quantum theory seems very well supported and is unlikely that it will be discredited any time soon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Direct tests of string theory, however,  will have to wait until we can probe much higher energies.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2010-11-02T21:46:26.483" CommentCount="1" />
  <row Id="57" PostTypeId="2" ParentId="21" CreationDate="2010-11-02T21:46:38.563" Score="2" ViewCount="" Body="&lt;p&gt;Another important use of Monte-Carlo is in nuclear engineering: basically for two distinct things: nuclear reactor core calculations and radiation shielding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first use Monte-Carlo to obtain precise estimations of the reactivity of the core (and other parameters) and the second allow to calculate, for example, the thickness of a shielding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally this is related to particle-matter interaction : this is a process where each interaction is quantified by its cross-section which then gives a probability for the interaction to occur. This is thus a very good field to apply MC methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is in medical physics, for the calculation of the deposited dose on a patient.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-11-02T21:46:38.563" />
  <row Id="58" PostTypeId="2" ParentId="26" CreationDate="2010-11-02T21:51:45.043" Score="3" ViewCount="" Body="&lt;p&gt;As KennyTM stated, use the conservation of energy. Say for example you have a constant electric field E to accelerate your particle of charge q and mass m, this will mean that the electrical energy E d q accumulated over the distance d will convert entirely into kinetic energy $\frac12 m v^2$ (assuming you stay non-relativistic, otherwise it's $\sqrt{m^2c^4+p^2c^2}$). Just solve the equation for v (or the momentum p in the relativistic case). Things get a bit more complicated for a non-static EM field.&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2010-11-02T21:51:45.043" />
  <row Id="59" PostTypeId="2" ParentId="26" CreationDate="2010-11-02T21:52:00.430" Score="1" ViewCount="" Body="&lt;p&gt;First, from your comment: a magnetic field will never accelerate anything (I am more specifically thinking about a charge &quot;particle&quot; as this is trivial for a golf ball) (&lt;strong&gt;Edit&lt;/strong&gt;: see the comments bellow about that statement).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnetic field is always perpendicular to the trajectory, thus giving no contribution in a change of the kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To accelerate a particle, you need an electric field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the way to proceed is the following: if you know the electric field and the path, you can integrate E dot dx and multiply by the charge. This will give you an energy difference, and this energy is added as kinetic energy, thus from the kinetic energy, if you are non relativistic you can use $E_k = \frac{1}{2} m v^2$ and make a difference.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-04T23:15:33.007" LastActivityDate="2010-11-04T23:15:33.007" CommentCount="2" />
  <row Id="61" PostTypeId="2" ParentId="24" CreationDate="2010-11-02T21:55:50.267" Score="3" ViewCount="" Body="&lt;p&gt;If you just lean without holding the steering wheel, that will automatically turn in order to conserve the total angular momentum. If you hold it tight however, this is not possible and you will crash.&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2010-11-02T21:55:50.267" CommentCount="5" />
  <row Id="62" PostTypeId="1" AcceptedAnswerId="70" CreationDate="2010-11-02T21:55:55.790" Score="7" ViewCount="272" Body="&lt;p&gt;The LHC in Geneva is a circular accelerator, 27 km long, why is it like that ?&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-03T08:03:25.263" LastActivityDate="2010-11-03T08:03:25.263" Title="Why is the LHC circular and 27km long?" Tags="&lt;accelerator-physics&gt;&lt;large-hadron-collider&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="63" PostTypeId="2" ParentId="7" CreationDate="2010-11-02T21:58:50.950" Score="4" ViewCount="" Body="&lt;p&gt;Laplace's determinism is fine if you do no longer consider x(t) as the function you want to deterministically describe but accept the fact that only the wave function $\psi(x,t)$ is deterministic - and even that only when you neglect QFT effects.&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2010-11-02T21:58:50.950" />
  <row Id="64" PostTypeId="2" ParentId="62" CreationDate="2010-11-02T21:59:39.680" Score="1" ViewCount="" Body="&lt;p&gt;To accelerate particles to very high energies, you need a long path, because the force is limited (just as in case of an airplane runway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You run them in circles to have it confined, or you would have to build something the size of a solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are, however, what is called Linear Accelerators, like SLAC. These are used to accelerate electrons, which are too light to be run in circles. They radiate a lot of energy when pushed sideways.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-11-02T21:59:39.680" CommentCount="1" />
  <row Id="65" PostTypeId="2" ParentId="62" CreationDate="2010-11-02T22:07:36.967" Score="2" ViewCount="" Body="&lt;p&gt;Cyclotrons, while generally economic because of the reduced space and cost to build (compared to linear accelerators), suffer from two notable problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the LHC is in fact a &lt;a href=&quot;http://en.wikipedia.org/wiki/Synchotron&quot; rel=&quot;nofollow&quot;&gt;synchotron&lt;/a&gt;, which is notably improved over a cyclotron for relativistic (high energy) particles. The following principles still apply pretty well, however.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The magnetic field required to impart the centripetal acceleration on the particle to keep it in orbit is &lt;em&gt;inversely proportional&lt;/em&gt; to the radius of orbit. Creating stronger magnetic fields requires larger and much more costly magnets. Maximising the radius of orbit helps reduce the required strength of magnetic field.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Maxwell's equations of electromagnetism indicate that an accelerating charge loses energy by emitting radiation. In fact, if you do the maths, this introduces the limit on the maximum velocity/energy of any accelerated particle. The larger the radius, the lower the centripetal force/acceleration, hence the lower the rate of energy loss.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To illustrate point 1, the force on a charged particle is given as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = Bqv$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where B is the magnetic field strength. And the centripetal force on a charge is given as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = \frac{mv^2}{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where r is the radius of the particle accelerator tunnel from the centre of the orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equating the two gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B = \frac{mv}{qr} = \frac{m\omega}{q}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, increasing the radius $r$ requires a lower $B$ (magnetic field).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point 2 is slightly more involved to show, as one has to apply Maxwell's equations and calculate the energy flux emitted. Hopefully you get the point however.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2010-11-02T22:17:32.003" LastActivityDate="2010-11-02T22:17:32.003" CommentCount="8" />
  <row Id="66" PostTypeId="2" ParentId="62" CreationDate="2010-11-02T22:09:13.113" Score="11" ViewCount="" Body="&lt;p&gt;When you accelerate charged particles, they lose energy by emitting photons (a process called &quot;Bremsstrahlung&quot; or &quot;braking radiation&quot;).  This is a nuisance in particle accelerators, because (1) you want to impart as much energy as possible to the particles being accelerated (that's the point!) and this is a loss; and (2) the bremsstrahlung can be in the form of harmful (to people and machines) ionizing radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stronger magnets are also necessary to produce the tighter turns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acceleration in the beam direction is, of course, unavoidable.  But any time you have motion along a curved path, you also need acceleration perpendicular to the direction of motion in order to curve the trajectory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to minimize this acceleration perpendicular to the direction of motion, it is desirable to have the accelerator be as straight as possible.  If you are building a machine that accelerates particles along a closed path, this means you want to make the radius as big as possible.  (I assume that, if you do the math, it turns out that a circular path (with uniform curvature) is better than an elliptical one, or one with long straightaways followed by tight turns.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative is to build a linear accelerator, which simply accelerates particles in a straight line.  The Stanford Linear Accelerator (SLAC) is one such accelerator; and the International Linear Collider is in the planning phases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the LHC 27 km in circumference?  Because they are re-using the tunnels from the  Large Electron–Positron Collider (LEP), which was 27 km around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why was LEP 27 km around?  It was almost certainly a balance between the science goals and the money available (longer tunnels = more expensive).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out the Superconducting Super Collider (SSC) on Wikipedia, the doomed American successor to LEP.  It was going to be 87 km in circumference.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2010-11-02T22:09:13.113" CommentCount="5" />
  <row Id="67" PostTypeId="2" ParentId="41" CreationDate="2010-11-02T22:13:04.010" Score="4" ViewCount="" Body="&lt;p&gt;I think that, as always, there is a significant disconnect between the math and physics camps.  However, physicists probably realize that the topological partition function of the B-model at genus g is not easy to calculate and doesn't have a good mathematical definition, in the holomorphic limit or not.  That is, they would be likely to agree that mathematicians like Costello have done valuable work in trying to nail down this elusive quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physicists working in topological field theory are pretty comfortable with the Atiyah-Segal axioms and with theories of various depths/levels.  Kapustin, for example, has spoken widely about these points (categories and 2-categories of branes).&lt;/p&gt;&#xA;" OwnerUserId="20" LastActivityDate="2010-11-02T22:13:04.010" CommentCount="1" />
  <row Id="68" PostTypeId="1" AcceptedAnswerId="82" CreationDate="2010-11-02T22:21:41.930" Score="8" ViewCount="283" Body="&lt;p&gt;Our physics teacher showed the class a really interesting demonstration. He used two polarised filters in opposite orientations, then he took some antistatic tape and stretched it under the two plates. The resulting image was projected on the wall using an overhead projector unit. Under stress the originally clear looking tape (as it looked between the polarised filters) turned all sorts of weird colours, and apparently different colours correlate to different levels of stress. What causes this odd effect?&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-05T02:35:25.583" LastActivityDate="2010-12-05T02:35:25.583" Title="What causes polarised materials to change colour under stress?" Tags="&lt;light&gt;&lt;polarization&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="69" PostTypeId="2" ParentId="27" CreationDate="2010-11-02T22:22:52.903" Score="-4" ViewCount="" Body="&lt;p&gt;The short answer is that measurement and interactions are two different animals in Quantum Mechanics. In reality measurements are performed using one of the fundamental interactions (usually EM), but this does not enter the framework of QM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The long answer is that you will not receive a satisfying answer to your question. First, because physicists don't know the answer, and second, because physicists don't care.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physics is concerned with understanding nature, in so far as making predictions regarding measurements. If we have a theory of what happens between measurements (things like Lagrangians and forces) and a theory of measurements (a postulate in Quantum Mechanics that wave functions collapse, plus the probabilistic interpretation of the abs.square of the wave function), and this framework works to the desired accuracy, then the philosophical implications of trying to unite the two are of no interest to physicists, unless they bring about a deeper understanding of nature, in so far as making more precise or more general predictions regarding measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, the line of questioning you pose has been investigated ever since the advent of Quantum Mechanics, but to my knowledge, nothing ever came of it regarding unification of forces and measurements (&quot;don't know&quot;), so the mainstream has lost interest a long time ago (&quot;don't care&quot;). (As an interesting side-note, one important result to come out of related type of inquiry is the Bell inequality.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if this answer seems negative. To quote David Fermin [corrected] regarding the philosophical issues regarding Quantum Mechanics, the pragmatic thing to do is to &quot;shut up and calculate!&quot; =)&lt;/p&gt;&#xA;" OwnerUserId="47" LastEditorUserId="47" LastEditorDisplayName="" LastEditDate="2010-11-09T13:29:16.740" LastActivityDate="2010-11-09T13:29:16.740" CommentCount="3" />
  <row Id="70" PostTypeId="2" ParentId="62" CreationDate="2010-11-02T22:23:43.287" Score="2" ViewCount="" Body="&lt;p&gt;The LHC is a synchrotron, that is, a accelerator with a magnetic field confining the orbit on a circular path and using RF accelerating cavities to accelerate the particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The voltage provided by the cavities is limited (the order of MV) and thus a linear accelerator cannot achieve such high energies (of the order of the TeV) (although some projects of TeV linear collider are in development, CLIC and the ILC) because it would be extremely long. The idea is thus to have a circular path, the particle going through the cavities at each turn and gaining a small amount of energy each turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To have this circular path, we use a magnetic field, it does not accelerate the particles, but it provides a force perpendicular to the motion, thus allowing to bend the trajectory and to obtain a circular orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does it have to be that long ? A fundamental relation for the synchrotrons is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p = q B r&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where p is the particle momentum, q is the charge of the particle, B is the magnetic field and r is the radius of curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can then see that to have a high momentum (and energy) we need a high magnetic field and a large radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the LHC, the magnetic field is already at the limit of what a superconducting magnet can achieve (almost 8.5 T).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LHC then needs a very large tunnel. For that it reuses the tunnel of the LEP which was also a synchrotron, but for electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case the size of the tunnel is not really given by the same reasoning. We need to take into account the synchrotron radiation: any accelerated charge radiates energy in the form of a EM radiation: &quot;light&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the amount of radiation goes with the inverse of the fourth power of the mass, the electron being very light they emit a large amount of radiation. For protons, this effects is almost negligible, that's why it does count for the LHC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for LEP (and LEP gave it's tunnel to the LHC) this was the main limitation to the achieved energy. And to obtain a high energy, the larger the tunnel the better, because the amount of radiation decreases with the bending angle of the dipole magnets, meaning that a large radius leads to lower radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the size of precisely 27 km was chosen for geographic consideration: the tunnel is between the Jura Mountains and the Leman lake, this implies strict constraints in the civil engineering.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-11-02T22:23:43.287" CommentCount="1" />
  <row Id="71" PostTypeId="1" AcceptedAnswerId="285" CreationDate="2010-11-02T22:25:49.123" Score="9" ViewCount="215" Body="&lt;p&gt;In laser resonators, higher order modes (i.e. TEM01, etc) accumulate phase faster than the fundamental TEM00 mode.  This extra phase is called Gouy phase.  What is an intuitive explanation of this effect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gouy predicted and then experimentally verified the existence of this effect long before the existence of lasers.  How did he do it, and what motivated him to think about it?&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2010-11-10T00:49:21.097" Title="What is an intuitive explanation of Gouy phase?" Tags="&lt;optics&gt;&lt;laser&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="4" />
  <row Id="72" PostTypeId="1" AcceptedAnswerId="77" CreationDate="2010-11-02T22:32:13.517" Score="2" ViewCount="103" Body="&lt;p&gt;Why are protons used in cancer therapy ? Is there any advantages compared to classics X-rays treatment or electron treatment ?&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-07T14:12:20.933" LastActivityDate="2011-04-07T14:12:20.933" Title="Proton therapy in cancer treatment" Tags="&lt;proton&gt;&lt;x-ray&gt;&lt;medical-physics&gt;" AnswerCount="1" />
  <row Id="73" PostTypeId="1" CreationDate="2010-11-02T22:37:38.660" Score="9" ViewCount="230" Body="&lt;p&gt;As a mathematician working the area of representation of Quantum groups, I am constantly thinking about solutions of the Yang-Baxter equation. In particular, trigonometric solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often research grants in this area cite this as an &quot;application&quot; of their research. This being said, many mathematicians(definitely including myself) don't know why these solutions are important. So, I wonder;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What exactly do physicists do with solutions to the Yang-Baxter Equation once they have them?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="46" LastActivityDate="2010-11-11T23:58:30.617" Title="How do physicists use solutions to the Yang-Baxter Equation?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" />
  <row Id="75" PostTypeId="1" AcceptedAnswerId="2851" CreationDate="2010-11-02T22:41:44.347" Score="2" ViewCount="190" Body="&lt;p&gt;I'm trying to figure out how to remember that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;hardness: how resistant it is to deformation&lt;/li&gt;&#xA;&lt;li&gt;toughness: how resistant it is to brittle failures&lt;/li&gt;&#xA;&lt;li&gt;stress: force on a surface area&lt;/li&gt;&#xA;&lt;li&gt;strength: ability to withstand stress without failure&lt;/li&gt;&#xA;&lt;li&gt;strain: measurement of deformation of a material&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of a mnemonic or easy way? I only know these from googling them, and I'm finding it tricky to remember them.&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-03T16:38:45.657" LastActivityDate="2011-01-14T05:34:11.240" Title="Mnemonics to remember various properties of materials" Tags="&lt;mnemonic&gt;&lt;material-science&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="76" PostTypeId="1" CreationDate="2010-11-02T22:42:47.327" Score="3" ViewCount="225" Body="&lt;p&gt;Every SO has it, Why not here! I'll go first&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many radio astronomers does it take to change a light bulb?&#xA;None. They are not interested in that short-wave stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many astronomers does it take to change a light bulb?&#xA;None, astronomers prefer the dark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many physicists does it take to change a light bulb?&#xA;Eleven. One to do it and ten to co-author the paper. &lt;/p&gt;&#xA;" OwnerUserId="48" LastActivityDate="2010-11-02T22:42:47.327" CommunityOwnedDate="2010-11-02T23:30:32.947" ClosedDate="2010-11-02T23:30:54.880" Title="Physics Humor - jokes and banter" Tags="&lt;humor&gt;" CommentCount="9" FavoriteCount="1" />
  <row Id="77" PostTypeId="2" ParentId="72" CreationDate="2010-11-02T22:46:27.527" Score="4" ViewCount="" Body="&lt;p&gt;The goal of such a treatment is to induce damages in the cells of the tumor by mean of ionizing radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These radiations can be X-rays (photons), electron, proton or things like carbon ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is: if you try to irradiate a tumor, you first have to go through normal tissues and the risk is to damage them also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photons will transfer energy probabilistically at each collision and the X-ray beam will follow a law like exp[-x] where x is the penetration depth. You then give more energy around the tumor than to the tumor itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main advantage of using protons is the &quot;Bragg peak&quot;: the way protons interacts with matter, they will give much more energy at a certain depth (th so called Bragg peak), if you tune the energy of the beam, and other parameters, so that the Bragg peak coincide with the tumor, then you have a big advantage.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-11-02T22:46:27.527" CommentCount="1" />
  <row Id="78" PostTypeId="1" AcceptedAnswerId="80" CreationDate="2010-11-02T22:46:52.640" Score="8" ViewCount="555" Body="&lt;p&gt;I can understand why 2 protons will repel each other, because they're both positive. But there isn't a neutral charge is there? So why do neutrons repel (do they even or Have I been misinformed?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT the reason I ask is because I've just been learning about neutron stars and how the neutrons are forced(as in they repel) together according to my teacher (he's a great teacher btw what I just said doesn't make it seem so though) So I wondered why they have to be forced by gravity and not just pushed. &lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-02-18T16:21:54.370" LastActivityDate="2011-02-18T16:21:54.370" Title="Why do neutrons repel each other?" Tags="&lt;particle-physics&gt;&lt;neutron&gt;" AnswerCount="7" CommentCount="1" FavoriteCount="1" />
  <row Id="79" PostTypeId="1" AcceptedAnswerId="690" CreationDate="2010-11-02T22:49:04.177" Score="7" ViewCount="343" Body="&lt;p&gt;You can get two photons entangled, and send them off in different directions; this is what happens in EPR experiments. Is the entanglement then somehow affected if one puts a thick slab of EM shielding material between the entangled photons? Have such experiments been made?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to EPR experiments measurements of the entangled states are at odds with SR, so based on that I'd assume the answer is &quot;no&quot;/&quot;don't know&quot;, but any citations would be appreciated!&lt;/p&gt;&#xA;" OwnerUserId="47" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-12T18:35:03.753" LastActivityDate="2010-12-04T06:27:23.957" Title="Is quantum entanglement mediated by an interaction?" Tags="&lt;quantum-mechanics&gt;&lt;entanglement&gt;&lt;epr-experiment&gt;" AnswerCount="7" />
  <row Id="80" PostTypeId="2" ParentId="78" CreationDate="2010-11-02T22:50:07.817" Score="10" ViewCount="" Body="&lt;p&gt;Neutrons have spin 1/2 and therefore obey the pauli exclusion principle, meaning two neutrons cannot occupy the same space at the same time.  When two neutrons' wavefunctions overlap, they feel a strong repulsive force.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Exchange_interaction&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Exchange_interaction&lt;/a&gt; .&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2010-11-02T22:56:01.210" LastActivityDate="2010-11-02T22:56:01.210" CommentCount="5" />
  <row Id="81" PostTypeId="2" ParentId="78" CreationDate="2010-11-02T22:50:45.693" Score="0" ViewCount="" Body="&lt;p&gt;As they are not charged, they only interact through their magnetic moment (which at very low energy could lead to kind of &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_force&quot; rel=&quot;nofollow&quot;&gt;Van der Waals forces&lt;/a&gt;, the strong force and the weak force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try to create a nucleon-nucleon system and that you do some QM calculations using an Hamiltonian taking into account the nuclear force, the electric force and the spin 1/2 of the nucleons you will have different contribution to the total energy, some being positive, other negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in the proton-proton case you cannot have a bound state because the electric repulsion is stronger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also have a contribution that depend on the [projection of the] spin of the nucleons. If they are equal the contribution lower the energy of the system and you can have a bound state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the case of the deuton (a proton-neutron bound stage, very weak).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the neutron-neutron state, both being of spin 1/2, you cannot have the same projection, and thus you cannot obtain a bound state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you interpret that as &quot;they repel each other&quot; then you have your explanation.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-02T22:56:58.213" LastActivityDate="2010-11-02T22:56:58.213" CommentCount="5" />
  <row Id="82" PostTypeId="2" ParentId="68" CreationDate="2010-11-02T22:50:57.147" Score="2" ViewCount="" Body="&lt;p&gt;Stress implies changing the thickness, therefore the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bragg_reflection&quot; rel=&quot;nofollow&quot;&gt;Bragg reflection&lt;/a&gt; change. Or shorter: &lt;a href=&quot;http://en.wikipedia.org/wiki/Interference_%28wave_propagation%29&quot; rel=&quot;nofollow&quot;&gt;Interference&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt; Also, as sigoldberg1 &lt;a href=&quot;http://physics.stackexchange.com/questions/68/what-causes-polarised-materials-to-change-colour-under-stress/253#253&quot;&gt;states&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Birefringence&quot; rel=&quot;nofollow&quot;&gt;Birefringence&lt;/a&gt; might occur, too&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2010-11-08T08:13:33.693" LastActivityDate="2010-11-08T08:13:33.693" CommentCount="3" />
  <row Id="83" PostTypeId="1" AcceptedAnswerId="282" CreationDate="2010-11-02T22:53:40.340" Score="4" ViewCount="159" Body="&lt;p&gt;Ordinary laser light has equal uncertainty in phase and amplitude.  When an otherwise perfect laser beam is incident onto a photodetector, the uncertainty in photon number will produce shot noise with poisson statistics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, laser light may be transformed into a 'squeezed state', where the uncertainty is no longer equally divided between the two quadratures, resulting in a reduction of shot noise.  How is this done?&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2010-11-05T17:18:30.797" Title="How is squeezed light produced?" Tags="&lt;optics&gt;&lt;quantum-optics&gt;" AnswerCount="2" />
  <row Id="84" PostTypeId="2" ParentId="78" CreationDate="2010-11-02T22:57:42.537" Score="1" ViewCount="" Body="&lt;p&gt;Neutrons consist of &lt;a href=&quot;http://en.wikipedia.org/wiki/Quarks&quot; rel=&quot;nofollow&quot;&gt;quarks&lt;/a&gt; that are electrically charged, so when two neutrons get close enough to each other the higher electrical multipole moments will become relevant and cause repelling. But also note the magnetic moment and the strong force Cedric mentioned plus the Pauli exclusion mentioned by nibot.&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2010-11-03T06:05:28.423" LastActivityDate="2010-11-03T06:05:28.423" CommentCount="4" />
  <row Id="85" PostTypeId="1" AcceptedAnswerId="87" CreationDate="2010-11-02T23:06:09.733" Score="4" ViewCount="226" Body="&lt;p&gt;Neutron mass: 1.008664 u&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proton mass: 1.007276 u&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why the discrepancy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a related note, how does one go about measuring the mass of a neutron or proton, anyway?&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-05T01:12:18.087" LastActivityDate="2010-12-01T17:42:15.303" Title="What's with the very slightly larger mass of the neutron compared to the proton?" Tags="&lt;particle-physics&gt;&lt;mass&gt;&lt;neutron&gt;&lt;proton&gt;" AnswerCount="3" />
  <row Id="86" PostTypeId="2" ParentId="79" CreationDate="2010-11-02T23:06:17.970" Score="3" ViewCount="" Body="&lt;p&gt;I think you answered your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;According to EPR experiments measurements of the entangled states are at odds with SR&quot;: if you mean that we cannot consider that the result of a measurement made on one entangled particle &quot;propagates&quot; to another one because this propagation would violate SR principles, you have to rules out a interaction in the sense of &quot;strong, weak, ... interaction&quot;, ie an interaction not in violation with SR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, we do not need to have such an interaction as it is directly explained by the principles of quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is like imagining that an &quot;interaction&quot; teaches QM to the particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main outcome of the treatments of the EPR paradox is to make hidden variables theories irrelevant, so basically &quot;quantum mechanics&quot; wins and we don't need other explanations.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-02T23:25:58.803" LastActivityDate="2010-11-02T23:25:58.803" CommentCount="2" />
  <row Id="87" PostTypeId="2" ParentId="85" CreationDate="2010-11-02T23:08:28.433" Score="8" ViewCount="" Body="&lt;p&gt;Masses and coupling between quarks are free parameters in the standard model, so there is not real explanation to that fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the measurment: you can have a look at this &lt;a href=&quot;http://en.wikipedia.org/wiki/Penning_trap&quot; rel=&quot;nofollow&quot;&gt;wikipedia article about Penning traps&lt;/a&gt; which are devices used for precision measurements for nucleus. Through the cyclotron frequency (Larmor factor) we can obtain the mass of the particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &quot;A neutron is a proton + an electron&quot; is a common answer to this question, but it is a totally invalid reasoning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both protons and neutrons are made of three quarks. The mass of the quarks is not known with enough precision, and even more important (and that's a why for the masses of the quarks), the interaction between them is responsible for the mass value to a much larger extend.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-02T23:15:43.980" LastActivityDate="2010-11-02T23:15:43.980" CommentCount="6" />
  <row Id="88" PostTypeId="2" ParentId="79" CreationDate="2010-11-02T23:15:49.237" Score="1" ViewCount="" Body="&lt;p&gt;Pretty sure that EPR does not state entanglement is at odds with SR or if it does it is incorrect. The point of the EPR paper was that the consequences of entanglement were so strange they could not be real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experimental evidence however supports entanglement and has never shown any violation of SR.&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2010-11-02T23:15:49.237" CommentCount="1" />
  <row Id="89" PostTypeId="1" AcceptedAnswerId="112" CreationDate="2010-11-02T23:17:01.137" Score="5" ViewCount="516" Body="&lt;p&gt;By Law of Gravity, I am referring to&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The magnitude of the force of gravity is proportional to the product of&#xA;the mass of the two objects and inversely proportional to their distance squared.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Does this law of attraction still hold under General Relativity's Tensor Equations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really know enough about mathematics to be able to solve any of Einstein's field equations, but does Newton's basic law of the magnitude of attraction still hold?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they are only approximations, what causes them to differ?&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-05T09:48:29.793" LastActivityDate="2010-12-05T09:48:29.793" Title="Is Newton's Law of Gravity consistent under General Relativity?" Tags="&lt;general-relativity&gt;&lt;gravity&gt;&lt;newtonian-gravity&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" />
  <row Id="90" PostTypeId="1" AcceptedAnswerId="136" CreationDate="2010-11-02T23:18:19.560" Score="4" ViewCount="362" Body="&lt;p&gt;I'm assing that what I've been told is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can only detect pulsars if their beam(s) of electromagnet radiation is directed at Earth. &#xA;That pulsars are the same as neutron stars only they emit beams of EM radiation out of their magnetic poles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So isn't possible that neutron stars emit em radiation in the same fashion as pulsars just  not in the right direction for us to detect it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(if you think this is off topic please discuss it in the existing meta topic about astronomy: &lt;a href=&quot;http://meta.physics.stackexchange.com/questions/18/astronomy-astrophyiscs-related-question-here&quot;&gt;http://meta.physics.stackexchange.com/questions/18/astronomy-astrophyiscs-related-question-here&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-12T22:46:21.820" LastActivityDate="2011-04-14T04:34:42.370" Title="Is it possible that all neutron stars are actually pulsars?" Tags="&lt;astrophysics&gt;&lt;neutron-stars&gt;&lt;pulsars&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="92" PostTypeId="2" ParentId="78" CreationDate="2010-11-02T23:26:45.540" Score="-3" ViewCount="" Body="&lt;p&gt;Ah, given your edit, I think you are really asking, why are neutron stars so much denser than other objects (regular stars, planets, etc)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity acts to clump stuff together in the universe.  Objects like stars and planets are formed and held together by gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason planets and stars aren't compressed down to an infinitesimal point by gravity is because the stuff they are made of eventually begins to resist further compaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normal matter is made out of atoms, which consist of a tiny nucleus surrounded by a cloud of electrons (which themselves have no physical size at all).  It's really the electron cloud that makes things seem solid.  The electrons in the atoms in my feet repel the electrons in the floor, so I don't fall through it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But despite this repulsion, most of the space taken up by atoms is empty space.  It is fair to say that the Earth, made of atoms, is also mostly empty space.  It just seems solid because of the electron repulsion.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutron stars, however, are just made out of neutrons, which can nestle up right next to each other.  There is none of this empty space inside the cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially all of the mass of an atom is in the nucleus (the protons and neutrons).  Neutron stars can be vastly, vastly denser than stars and planets made out of atoms, because they lack the empty space inside the electron clouds.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2010-11-02T23:26:45.540" CommentCount="2" />
  <row Id="93" PostTypeId="1" AcceptedAnswerId="120" CreationDate="2010-11-02T23:31:02.713" Score="22" ViewCount="598" Body="&lt;p&gt;Coulomb's Law states that the fall-off of the strength of the electrostatic force is inversely proportional to the distance squared of the charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gauss's law implies that a the total flux through a surface completely enclosing a charge is proportional to the total amount of charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we imagine a two-dimensional world of people who knew Gauss's law, they would imagine a surface completely enclosing a charge as a flat circle around the charge.  Integrating the flux, they would find that the electrostatic force &lt;em&gt;should&lt;/em&gt; be inversely proportional to the distance of the charges, if Gauss's law were true in a two-dimensional world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if they observed a 1/r^2 fall-off, this implies a two-dimensional world is not all there is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this argument correct?  Does the 1/r^2 fall-off imply that there are only three spatial dimensions we live in?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to make sure this is right before I tell this to my friends and they laugh at me.&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-18T02:45:42.287" LastActivityDate="2010-11-18T02:45:42.287" Title="Does Coulomb's Law, with Gauss's Law, imply the existence of only three spatial dimensions?" Tags="&lt;electrostatics&gt;&lt;coulombs-law&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="6" />
  <row Id="94" PostTypeId="1" AcceptedAnswerId="233" CreationDate="2010-11-02T23:32:47.993" Score="6" ViewCount="115" Body="&lt;p&gt;Measuring the polarization of a laser beam is a simple enough task if the polarization is the same everywhere. You can even buy commercial polarimeters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you go about it if the light beam has different polarizations in different parts of the transverse plane? One example is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Radial_polarization&quot; rel=&quot;nofollow&quot;&gt;radially polarized&lt;/a&gt; beam. More generally, is there a good technique for sampling the local polarization (which might be linear, elliptical, or circular, anywhere on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Poincare_sphere&quot; rel=&quot;nofollow&quot;&gt;Poincaré sphere&lt;/a&gt;) in one transverse plane?&lt;/p&gt;&#xA;" OwnerUserId="107" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-05T02:34:49.093" LastActivityDate="2010-12-05T02:34:49.093" Title="What ways are there to measure the local polarization of a laser beam?" Tags="&lt;optics&gt;&lt;experimental-physics&gt;&lt;laser&gt;&lt;polarization&gt;" AnswerCount="2" />
  <row Id="95" PostTypeId="1" CreationDate="2010-11-02T23:33:24.377" Score="3" ViewCount="161" Body="&lt;p&gt;The current &quot;standard model&quot; or concordance model of cosmology is &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambda-CDM_model&quot; rel=&quot;nofollow&quot;&gt;Lambda CDM&lt;/a&gt; which includes late time acceleration due to a cosmological constant, cold dark matter as the missing matter component and an inflationary period at very early times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What alternative theories are also consistent with current observational data? &lt;/p&gt;&#xA;" OwnerUserId="110" LastActivityDate="2010-11-14T16:23:36.590" Title="What current alternatives are there to the standard concordance model of cosmology?" Tags="&lt;cosmology&gt;&lt;cosmological-inflation&gt;&lt;dark-matter&gt;&lt;dark-energy&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="96" PostTypeId="2" ParentId="93" CreationDate="2010-11-02T23:36:17.480" Score="5" ViewCount="" Body="&lt;p&gt;I would say yes !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually some theories explaining quantum gravity use also this reasoning: gravity is a very weak interaction at a quantum level because it &quot;leaks&quot; into other dimensions, not observable at our scale, but that are present at this scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mathematical tools are different, but if you just think about gauss's law you can imagine one explanation why additional dimensions are present in these theories.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-11-02T23:36:17.480" />
  <row Id="97" PostTypeId="2" ParentId="93" CreationDate="2010-11-02T23:42:59.560" Score="3" ViewCount="" Body="&lt;p&gt;Loosely speaking, (super)string theory considers additional spatial dimensions that are &quot;wrapped up&quot; (have unusual topologies of high curvature, I believe). Now it is of course complete speculation, but if these dimensions do exist, electromagnetism would not spread out much into those dimensions, hence it would appear as if there are only three dimensions still (to a very good approximation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saying that, your argument is more or less sound (though far from bulletproof). It certainly suggests we don't live in a 2D world, and that any possible extra dimensions are comparatively very small!&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2010-11-03T02:29:16.347" LastActivityDate="2010-11-03T02:29:16.347" />
  <row Id="98" PostTypeId="1" AcceptedAnswerId="99" CreationDate="2010-11-02T23:52:07.323" Score="1" ViewCount="83" Body="&lt;p&gt;For a given isotope, one can obtain the binding energy using the semi-empirical mass formula. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;img src=&quot;http://i.stack.imgur.com/U4is7.png&quot; alt=&quot;From Codecogs&quot;&gt; has a binding energy of 1782.8 MeV. From this information, how can the likelihood of the isotopes stability be found?&lt;/p&gt;&#xA;" OwnerUserId="112" LastActivityDate="2010-11-03T00:48:26.140" Title="Obtaining isotope stability" Tags="&lt;isotope&gt;&lt;binding-energy&gt;&lt;nuclear-physics&gt;" AnswerCount="1" />
  <row Id="99" PostTypeId="2" ParentId="98" CreationDate="2010-11-03T00:03:28.897" Score="5" ViewCount="" Body="&lt;p&gt;From the binding energy given experimentally, using precise QM calculations or using a given formula, one should first check for &quot;stability in particles&quot;, if the binding is negative, you will of course not have stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the next thing, if you have a formula, is to check for each type of stability. For example, to check for stability against a given fission, calculate the binding energy of the fragments, obtain the new energy and compare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check for, let's say a beta minus, replace the nucleus (A,Z) in the formula by (A,Z+1), obtain the binding energy, the new total energy (you can safely neglect the mass of the neutrino and even the one of the electron in most cases) and compare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your specific example, this is a bit tricky because a large variety of decay channels are potentially allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to proceed is to look at a binding energy per nucleon or mass excess against A diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ZYH9n.jpg&quot; alt=&quot;binding energy&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see that 237Np is far on the right as a binding energy per nucleon smaller that the most stable elements like 56Fe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can then conclude that 237Np can potentially decay to a more stable element to increase it's binding energy (although these decay, that can be alpha decay can have excessively small probability and a time constant excessively long).&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-03T00:48:26.140" LastActivityDate="2010-11-03T00:48:26.140" CommentCount="3" />
  <row Id="100" PostTypeId="1" AcceptedAnswerId="108" CreationDate="2010-11-03T00:05:24.680" Score="2" ViewCount="99" Body="&lt;p&gt;How is it possible to calculate the energy liberated by a fission process ? For example in the fission of a $_92^235U$ nucleus capturing a neutron ?&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-03T00:38:17.090" LastActivityDate="2010-11-03T00:54:51.450" Title="How to obtain the fission energy" Tags="&lt;nuclear-physics&gt;&lt;nuclear-engineering&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="101" PostTypeId="1" AcceptedAnswerId="229" CreationDate="2010-11-03T00:08:59.077" Score="5" ViewCount="133" Body="&lt;p&gt;The Tevatron is a proton-antiproton collider; it collides a beam of protons against a beam of antiproton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can understand how we obtain the protons, but for the antiprotons ? How are they produced ?&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-04T19:00:09.467" LastActivityDate="2010-11-04T19:00:09.467" Title="Production of antiproton at the Tevatron" Tags="&lt;high-energy-physics&gt;&lt;accelerator-physics&gt;&lt;tevatron&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="102" PostTypeId="1" AcceptedAnswerId="128" CreationDate="2010-11-03T00:15:24.310" Score="4" ViewCount="679" Body="&lt;p&gt;For every force there is an equal force in the opposite direction on another body, correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when the Suns gravity acts on Earth where is the opposite and equal force? I also have the same question for centripetal force in a planets orbit. &lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-05T09:49:03.113" LastActivityDate="2011-04-09T19:47:43.490" Title="Is gravity a force and if so what is its opposite?" Tags="&lt;gravity&gt;&lt;newtonian-gravity&gt;" AnswerCount="7" CommentCount="2" FavoriteCount="1" />
  <row Id="103" PostTypeId="2" ParentId="102" CreationDate="2010-11-03T00:20:36.770" Score="7" ViewCount="" Body="&lt;p&gt;The earth feels a force towards the sun.  The sun feels an 'equal and opposite' force towards the earth.  In fact, the earth does not rotate around the sun; instead, the sun and the earth (if you are considering only those two bodies) orbit around their center of mass.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2010-11-03T00:20:36.770" CommentCount="5" />
  <row Id="104" PostTypeId="2" ParentId="102" CreationDate="2010-11-03T00:21:57.203" Score="4" ViewCount="" Body="&lt;p&gt;Sun and Earth interact through gravity. Sun exert a gravitational force on Earth and ... Earth exert and equal and opposite force on Sun  (or is it the opposite ?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Opposite here does not mean &quot;sci-fi anti gravity force repelling objects&quot; but opposite as in &quot;vector of opposite direction and equal magnitude&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The centrifugal force is not a force, but a pseudo force that is introduced because the referential frame is not inertial. In the referential frame where you need to introduce the centifugal force, the &quot;opposite reaction&quot; is the centripedal force itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve the Earth around the Sun problem, first consider that the sun as an infinite mass and is thus without motion. The sun exert a force on Earth, and in some frame of reference you need to introduce an additional pseudo-force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational field &quot;generated&quot; by Earth is weaker, but the forces are equal (in magnitude).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_{earth} = \frac{G m_{earth}}{r^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_{sun} = \frac{G m_{sun}}{r^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = E_{earth} * m_{sun} = E_{sun} * m_{earth}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Another edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the title of your question &quot;is gravity a force&quot;: Gravity is one of the four fundamental interaction of nature (strong and weak interactions, and electromagnetism). No complete treatment of this interaction exists at a quantum level, so from a classical point of view, yes gravity is a force, associated with a conservative [gravitational] field, etc.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-03T17:07:20.370" LastActivityDate="2010-11-03T17:07:20.370" CommentCount="1" />
  <row Id="105" PostTypeId="2" ParentId="102" CreationDate="2010-11-03T00:22:36.580" Score="15" ViewCount="" Body="&lt;p&gt;The opposite is that Earth attracts the sun with the same exact force!&lt;br&gt;&#xA;Thinking smaller, the Earth attracts you and you attract the earth with forces that are equal in magnitude and direction, but of opposite sense. Of course, by Newton second's law, that same force will have a much greater effect on you than on the Earth. The same applies to the Earth+Sun combination!&lt;/p&gt;&#xA;" OwnerUserId="116" LastActivityDate="2010-11-03T00:22:36.580" CommentCount="4" />
  <row Id="106" PostTypeId="2" ParentId="102" CreationDate="2010-11-03T00:24:30.197" Score="0" ViewCount="" Body="&lt;p&gt;In this simplified picture, gravity of the Sun is a centripetal force for Earth's orbital motion; centrifugal force that equalizes it comes from the angular momentum of Earth.&#xA;If Earth didn't have an angular momentum (was, say, created near the sun without relative velocity, at least in the angular direction), it would just fall on the Sun.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-03T00:24:30.197" />
  <row Id="107" PostTypeId="1" CreationDate="2010-11-03T00:27:06.343" Score="1" ViewCount="88" Body="&lt;p&gt;I've been asked to make some data available in NetCDF that's currently in FITS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there's the &lt;a href=&quot;http://cdf.gsfc.nasa.gov/html/dtws.html&quot; rel=&quot;nofollow&quot;&gt;translators available from the CDF group&lt;/a&gt;, but you have to go FITS to CDF then CDF to NetCDF, and I was hoping to do it in one step, as there's generally less problems introduced.  (although, I admit that I haven't tried doing a round trip from FITS to NetCDF and back to FITS to see which headers manage to survive the process)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(and the data is solar intensity data, so it's solar / space physics data, and I'm really not sure where else I can even ask this question, or if it'll be closed immediately as the community sorts out what's considered on/off topic)&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-11-03T00:27:06.343" ClosedDate="2010-12-08T20:09:39.343" Title="Conversion from FITS to NetCDF?" Tags="&lt;data-conversion&gt;" CommentCount="5" />
  <row Id="108" PostTypeId="2" ParentId="100" CreationDate="2010-11-03T00:35:39.267" Score="4" ViewCount="" Body="&lt;p&gt;When 235U captures a neutron, it forms the compound state 236U*, the &lt;em&gt;excitation energy&lt;/em&gt; is given by&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;E_ex = [m(236U*)-m(236U)]c^2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To find the energy of the compound state, the mass energies of 235U and n can be used - assuming that the neutron will be thermalised for this reaction to occur and thus it's kinetic energy is negligible. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;m(236U*) = m(235U)+m(n) = (235.043924u + 1.008665u) = 236.052589u&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thus:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;E_ex = (236.052589u - 236.045563u)931.502MeV/u = 6.5MeV&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This causes a fission event to occur, which can produce over twenty different products, one example is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;235U+n--&amp;gt;236U*--&amp;gt;93Rb+141CS+2n&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The energy liberated is defined by the &lt;em&gt;Q&lt;/em&gt; value: the initial mass energy minus the final mass energy&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Q = (m_initial - m_final)c^2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Using the masses of 93Rb (92.92172u) and 141Cs (140.91949u)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Q = (236.052589u-(92.92172u+140.91949u+(2*1.008665u)))*931.502MeV/u = 180.76 MeV&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="112" LastEditorUserId="112" LastEditorDisplayName="" LastEditDate="2010-11-03T00:54:51.450" LastActivityDate="2010-11-03T00:54:51.450" CommentCount="3" />
  <row Id="109" PostTypeId="2" ParentId="94" CreationDate="2010-11-03T01:00:13.833" Score="2" ViewCount="" Body="&lt;p&gt;Maybe raster scanning the beam in the polarimeters. Or using the same techniques used to measure a single polarization beam but with a camera as a detector.&lt;/p&gt;&#xA;" OwnerUserId="62" LastActivityDate="2010-11-03T01:00:13.833" />
  <row Id="110" PostTypeId="2" ParentId="9" CreationDate="2010-11-03T01:06:46.777" Score="3" ViewCount="" Body="&lt;p&gt;I generally tell the story that the action principle is another way of getting at the same differential equations -- so at the level of mechanics, the two are equivalent.  However, when it comes to quantum field theory, the description in terms of path integrals over the exponentiated action is essential when considering instanton effects.  So eventually one finds that the formulation in terms of actions is more fundamental, and more physically sound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But still, people don't have a &quot;feel&quot; for action the way they have a feel for energy.&lt;/p&gt;&#xA;" OwnerUserId="20" LastActivityDate="2010-11-03T01:06:46.777" />
  <row Id="111" PostTypeId="1" AcceptedAnswerId="116" CreationDate="2010-11-03T01:13:57.047" Score="6" ViewCount="486" Body="&lt;p&gt;I am looking for a book about &quot;advanced&quot; classical mechanics. By advanced I mean a book considering directly Lagrangian and Hamiltonian formulation, and also providing a firm basis in the geometrical consideration related to these to formalism (like tangent bundle, cotangent bundle, 1-form, 2-form, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521636361&quot; rel=&quot;nofollow&quot;&gt;this book from Saletan and Jose&lt;/a&gt;, but I would like to go into more details about the [symplectic] geometrical and mathematical foundations of classical mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional note: A chapter about relativistic Hamiltonian dynamics would be a good thing.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-04T19:02:15.690" LastActivityDate="2011-07-07T14:24:21.893" CommunityOwnedDate="2011-07-07T13:41:37.800" Title="Book about classical mechanics" Tags="&lt;mechanics&gt;&lt;hamiltonian-mechanics&gt;&lt;classical-mechanics&gt;&lt;lagrangian-mechanics&gt;&lt;books&gt;" AnswerCount="9" CommentCount="0" FavoriteCount="6" />
  <row Id="112" PostTypeId="2" ParentId="89" CreationDate="2010-11-03T01:14:56.533" Score="5" ViewCount="" Body="&lt;p&gt;Yes, in the appropriate limit.  Roughly, the study of geodesic motion in the Schwarzschild solution (which is radially symmetric) reduces to Newtonian gravity at sufficiently large distances and slow speeds.  To see how this works exactly, one must look more specifically at the equations.&lt;/p&gt;&#xA;" OwnerUserId="20" LastActivityDate="2010-11-03T01:14:56.533" CommentCount="4" />
  <row Id="113" PostTypeId="2" ParentId="111" CreationDate="2010-11-03T01:17:28.593" Score="2" ViewCount="" Body="&lt;p&gt;Maybe no symplectic geometry or forms here, but this&#xA;book has a LOT to offer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.people.fas.harvard.edu/~djmorin/book.html&quot; rel=&quot;nofollow&quot;&gt;http://www.people.fas.harvard.edu/~djmorin/book.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20" LastActivityDate="2010-11-03T01:17:28.593" CommunityOwnedDate="2011-07-07T13:41:37.800" CommentCount="2" />
  <row Id="114" PostTypeId="2" ParentId="24" CreationDate="2010-11-03T01:17:40.223" Score="0" ViewCount="" Body="&lt;p&gt;What happens if you lean on bike while trying to ride straight?&lt;br&gt;&#xA;It happens that you turn on same side you lean.&lt;br&gt;&#xA;Why, because you have to generate centrifugal force and lean back up if you do not want to fall down. So if you lean left you have to make left turn in order to generate force which will turn you straight position.&lt;br&gt;&#xA;If you turn too much left you will eventually fall on right because of too much centrifugal force kicking you other side.&lt;/p&gt;&#xA;" OwnerUserId="118" LastActivityDate="2010-11-03T01:17:40.223" CommentCount="1" />
  <row Id="115" PostTypeId="1" AcceptedAnswerId="612" CreationDate="2010-11-03T01:34:55.803" Score="4" ViewCount="236" Body="&lt;p&gt;I have an Hamiltonian problem whose 2D phase space exhibit islands of stability (elliptic fixed points).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can calculate the area of these islands in some cases, but for other cases I would like to use Mathematica (or anything else) to compute it numerically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phase space looks like that :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/gCjb0.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a contour plot make with Mathematica. Could anyone with some knowledge of Mathematica provide a way to achieve this ?&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="134" LastEditorDisplayName="" LastEditDate="2010-11-03T10:30:09.853" LastActivityDate="2010-11-12T02:21:45.140" Title="Mathematica to help for an Hamiltonian problem" Tags="&lt;hamiltonian-mechanics&gt;&lt;numerics&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="116" PostTypeId="2" ParentId="111" CreationDate="2010-11-03T01:56:44.590" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://mitpress.mit.edu/sicm/&quot; rel=&quot;nofollow&quot;&gt;Structure and Interpretation of Classical Mechanics&lt;/a&gt; (&lt;a href=&quot;http://mitpress.mit.edu/sicm/book-Z-H-4.html&quot; rel=&quot;nofollow&quot;&gt;table of contents&lt;/a&gt;) certainly deserves mention. It might not have as much differential geometry as you'd like, though they have a followup article titled &lt;a href=&quot;http://groups.csail.mit.edu/mac/users/wisdom/AIM-2005-003.pdf&quot; rel=&quot;nofollow&quot;&gt;Functional Differential Geometry&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2010-11-03T02:02:52.787" LastActivityDate="2010-11-03T02:02:52.787" CommunityOwnedDate="2011-07-07T13:41:37.800" />
  <row Id="117" PostTypeId="2" ParentId="93" CreationDate="2010-11-03T02:14:03.533" Score="-2" ViewCount="" Body="&lt;p&gt;I think it is correct your appreciation &lt;strong&gt;if&lt;/strong&gt; Gauss law holds in a two dimensional world, then the electrostatic force should be inversely proportional to the distance between charges. However, I'm not at all convinced that Gauss law could be true in a two-dimensional world because &#xA;$F_{q}=k \displaystyle \frac{qq'(r-r')}{|r-r'|^{3}}$ is a consequence of a 3-dimensional space and since we derive Gauss law from such a force law (to be precise, from its electric field), we can not assume the validity of Gauss law independently from a 3-dimensional space. &lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="18" LastEditorDisplayName="" LastEditDate="2010-11-03T19:46:55.007" LastActivityDate="2010-11-03T19:46:55.007" CommentCount="4" />
  <row Id="118" PostTypeId="2" ParentId="111" CreationDate="2010-11-03T02:39:47.943" Score="3" ViewCount="" Body="&lt;p&gt;Pure beaten gold, any edition, paperback.    It never leaves you.   Lagrangian approach.&lt;br&gt;&#xA;See the reviews on Amazon &#xA;L D Landau (Author), E.M. Lifshitz (Author)&#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0750628960&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Mechanics-Third-Course-Theoretical-Physics/dp/0750628960&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-11-03T02:39:47.943" CommunityOwnedDate="2011-07-07T13:41:37.800" CommentCount="1" />
  <row Id="119" PostTypeId="2" ParentId="9" CreationDate="2010-11-03T03:32:09.443" Score="8" ViewCount="" Body="&lt;p&gt;There is also Feynman's approach, i.e. least action is true classically just because it is true quantum mechanically, and classical physics is best considered as an approximation to the underlying quantum approach.   See &lt;a href=&quot;http://www.worldscibooks.com/physics/5852.html&quot; rel=&quot;nofollow&quot;&gt;http://www.worldscibooks.com/physics/5852.html&lt;/a&gt;  or &lt;a href=&quot;http://www.eftaylor.com/pub/call_action.html&quot; rel=&quot;nofollow&quot;&gt;http://www.eftaylor.com/pub/call_action.htm&lt;/a&gt;l .  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the whole thing is summarized in a nutshell in &#xA;Richard P. Feynman, The Feynman Lectures on Physics (Addison–Wesley, Reading, MA, 1964), Vol. II, Chap. 19. (I think, please correct me if I'm wrong here).  The fundamental idea is that the action integral defines the quantum mechanical amplitude for the position of the particle, and the amplitude is stable to interference effects (--&gt;has nonzero probability of occurrence) only at extrema or saddle points of the action integral.   The particle really does explore all alternative paths probabilistically.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You likely want to read Feynman's Lectures on Physics anyway, so you might as well start now.  :-)&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-11-03T04:13:20.937" LastActivityDate="2010-11-03T04:13:20.937" CommentCount="1" />
  <row Id="120" PostTypeId="2" ParentId="93" CreationDate="2010-11-03T05:35:51.833" Score="11" ViewCount="" Body="&lt;p&gt;Yes, absolutely. In fact, Gauss's law is generally considered to be the fundamental law, and Coulomb's law is simply a consequence of it (and of the Lorentz force law).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can actually simulate a 2D world by using a line charge instead of a point charge, and taking a cross section perpendicular to the line. In this case, you find that the force (or electric field) is proportional to 1/r, not 1/r^2, so Gauss's law is still perfectly valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the same conclusion can be made from experiments performed in graphene sheets and the like, which are even better simulations of a true 2D universe, but I don't know of a specific reference to cite for that.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-03T23:15:02.760" LastActivityDate="2010-11-03T23:15:02.760" CommentCount="20" />
  <row Id="121" PostTypeId="2" ParentId="83" CreationDate="2010-11-03T06:15:43.947" Score="1" ViewCount="" Body="&lt;p&gt;Squeezed light can be generated from light in a coherent state or vacuum state by using certain optical nonlinear interactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, an optical parametric amplifier with a vacuum input can generate a squeezed vacuum with a reduction in the noise of one quadrature components by the order of 10 dB. A lower degree of squeezing in bright amplitude-squeezed light can under some circumstances be obtained with frequency doubling. Squeezing can also arise from atom-light interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&#xA;&lt;a href=&quot;http://www.squeezed-light.de/body.html#generation&quot; rel=&quot;nofollow&quot;&gt;http://www.squeezed-light.de/body.html#generation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="48" LastActivityDate="2010-11-03T06:15:43.947" />
  <row Id="122" PostTypeId="1" AcceptedAnswerId="125" CreationDate="2010-11-03T06:23:17.300" Score="7" ViewCount="343" Body="&lt;p&gt;I learned today in class that photons and light are quantized.  I also remember that electric charge is quantized as well.  I was thinking about these implications, and I was wondering if mass was similarly quantized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if we describe certain finite irreducible masses x, y, z, etc., then all masses are integer multiples of these irreducible masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or do masses exist along a continuum, as charge and light were thought to exist on before the discovery of photons and electrons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm only referring to invariant/rest mass)&lt;/p&gt;&#xA;" OwnerUserId="71" LastActivityDate="2010-11-03T07:09:33.360" Title="Is mass quantized?" Tags="&lt;particle-physics&gt;&lt;quantum-mechanics&gt;&lt;mass&gt;&lt;atoms&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="123" PostTypeId="1" AcceptedAnswerId="2961" CreationDate="2010-11-03T07:00:58.390" Score="11" ViewCount="256" Body="&lt;p&gt;Trained as a pure mathematician, I see claims about the mass of a galaxy and other such huge measurements that are arrived at experimentally, and I just have to scratch my head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is a bit of a vague question--but does anyone have a good resource for something like &quot;measurement in astrophysics&quot; or an introductory history of how astrophysicists figured out how to make measurements of this kind?&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2011-01-15T18:12:46.067" Title="How does one measure the mass of a galaxy? And other such large quantities" Tags="&lt;astrophysics&gt;&lt;measurement&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="124" PostTypeId="2" ParentId="122" CreationDate="2010-11-03T07:08:10.333" Score="0" ViewCount="" Body="&lt;p&gt;Rest mass of elementary particles is not quantized: in the standard model, the masses are free parameters of the theory; they must be measured and introduced in the model experimentally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the mass of, say the Hydrogen atom is given by the mass of its constituent (proton and electron whose mass are given) minus the binding energy which is quantized.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-11-03T07:08:10.333" />
  <row Id="125" PostTypeId="2" ParentId="122" CreationDate="2010-11-03T07:09:33.360" Score="7" ViewCount="" Body="&lt;p&gt;As far as anyone knows, no, rest mass is not quantized. In fact, generally speaking, we (meaning people/scientists in general) don't know of any pattern to the masses of the fundamental particles (quarks and leptons), nor do we have any reasonable explanation for why the masses have the values they do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're talking about composite particles, though, much of their mass comes from the binding energy between the constituents, and not from the masses of the constituents themselves. For instance, only a small part of the mass of the proton comes from the masses of its quarks. Most of it is energy stored in the strong force that holds the quarks together - in other words, the energy of the gluons. I don't know the exact details of the calculations (which are quite complicated), but those binding energies may be quantized, so the effective mass of the proton may appear to be quantized.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-03T07:09:33.360" CommentCount="5" />
  <row Id="126" PostTypeId="2" ParentId="123" CreationDate="2010-11-03T07:15:13.807" Score="19" ViewCount="" Body="&lt;p&gt;One common way of making these measurements is gravitational lensing. Basically, astronomers look at some distant object which is located directly behind the galaxy in question. Since the galaxy is so massive, it bends the light from the more distant object around it, so we see an image of the object displaced by some angle from where it actually is in the sky. For a distant object in the right position, we can see multiple images, one from light deflected to the left and one from light deflected to the right. Measuring the angular separation between them allows you to compute the angle by which the light was bent, and in turn to determine the mass of the galaxy required to produce that deflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For galaxies that are less massive but closer, close enough to resolve the spectra of individual stars, we can measure the Doppler shifts of the spectra of stars on the advancing side of the galaxy and on the receding side, and taking the difference gives (twice) the tangential speed of the stars at that radius, which is related to the amount of mass contained within that radius. So measuring the spectra of stars at the edge of the galaxy, or in e.g. a globular cluster that orbits the galaxy, can give you the total mass contained in the galaxy.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-03T07:15:13.807" CommentCount="1" />
  <row Id="127" PostTypeId="1" CreationDate="2010-11-03T07:15:24.197" Score="0" ViewCount="283" Body="&lt;p&gt;First, I want to say upfront that this question need not dissolve into arguments and discussion. This question can and should have a correct answer, please don't respond with your opinions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GNUplot is very pervasive in Physics research. Many of the plots appearing in things like PRL and JPB are made in GNUplot. Why is this the case. There are much more modern tools for doing these types of graphs, and from my uninformed position, it appears that this would be easier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One obvious and relevant first point to make is that GNUplot is free and opensource. I respect this, but do not expect that this is the primary reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am hoping for answers that specify what GNUplot can do that can be achieved in other programs efficiently, say Mathematica. If there are none of these, and the reason is simply tradition/resistance to change, that is also fine, but I expect there must be some specific tasks one wishes to perform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="46" LastActivityDate="2010-11-03T08:31:03.573" ClosedDate="2010-11-03T15:33:58.447" Title="Why is GNUplot so pervasive in Physics when there are much more modern tools?" Tags="&lt;software&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="128" PostTypeId="2" ParentId="102" CreationDate="2010-11-03T07:18:32.080" Score="13" ViewCount="" Body="&lt;p&gt;As many others said, the Sun feels the same force towards Earth as the Earth feels towards the sun. That is your equal and opposite force. In practice though the &quot;visible&quot; effects of a force can be deduced through Newton's first law, i.e. ${\bf F} = m{\bf a}$. In other words, you need to divide the force by the mass of the body to determine the net effect on the body itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\bf F_s} = {\bf F_e}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\bf F_s} = m_s {\bf a_s}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\bf F_e} = m_e {\bf a_e}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_s {\bf a_s} = m_e {\bf a_e}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\bf a_s} = {\bf a_s} \frac{m_e}{m_s}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the last term is $3 \cdot 10^{-6}$! This means that the force that the Earth enacts on the sun is basically doing nothing to the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of seeing this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = \frac{G m_s m_e}{r^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_s = \frac{F}{m_s} = \frac{G m_e}{r^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_e = \frac{F}{m_e} = \frac{G m_s}{r^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{a_s}{a_e} = \frac{m_e}{m_s} = 3 \cdot 10^{-6}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, the same big difference in effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the centripetal force, it is still the same force. Gravity provides a centripetal force which is what keeps Earth in orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth pointing out that the mass that acts as the charge for gravity, known as &lt;em&gt;gravitational mass&lt;/em&gt; is not, a priori, the same mass that appears in Newtons's law, known as &lt;em&gt;inertial mass&lt;/em&gt;. On the other hand it is a fact of nature that they have the same value, and as such we may use a single symbol $m$, instead of two, $m_i$ and $m_g$. This is an underlying, unspoken assumption in the derivation above. This is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weak_equivalence_principle#Tests_of_the_weak_equivalence_principle&quot; rel=&quot;nofollow&quot;&gt;weak equivalence principle&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-12-05T16:24:39.923" LastActivityDate="2010-12-05T16:24:39.923" CommentCount="3" />
  <row Id="129" PostTypeId="1" AcceptedAnswerId="131" CreationDate="2010-11-03T07:23:53.470" Score="13" ViewCount="750" Body="&lt;p&gt;Also, if the Higgs Boson is not found with the LHC, it would be successfully proved as non-existant? or we would just wait for an experiment with higher energy ranges?&lt;/p&gt;&#xA;" OwnerUserId="23" LastActivityDate="2011-03-17T11:03:36.903" Title="What would be the implications to the Standard Model if the Higgs Boson isn't found with the LHC?" Tags="&lt;standard-model&gt;&lt;large-hadron-collider&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="130" PostTypeId="2" ParentId="95" CreationDate="2010-11-03T07:24:57.420" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.google.co.uk/search?q=Penrose+Conformal+Cyclic+Cosmology&quot; rel=&quot;nofollow&quot;&gt;Penrose's Conformal Cyclic Cosmology‎&lt;/a&gt; is such an example. There is late time acceleration due to Lambda, but no inflationary period in the beginning. The late universe is equated to the beginning (of the next big bang cycle) through a conformal factor which maintains the physics.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-03T07:24:57.420" CommentCount="2" />
  <row Id="131" PostTypeId="2" ParentId="129" CreationDate="2010-11-03T07:36:18.657" Score="10" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Higgs_boson&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; actually has a very nice graphic with this information (which roughly agrees with what I remember hearing from people &quot;in the know&quot;):&#xA;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/4/44/HiggsGraph_07-26-2010_hr.svg/500px-HiggsGraph_07-26-2010_hr.svg.png&quot; alt=&quot;Image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that there are both lower and upper bounds on the mass of the Higgs boson. The LHC should be able to cover pretty much the entire range that has not yet been searched, so if it doesn't find the Higgs, we can be fairly confident that something is wrong with the Standard Model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the question is, what could be wrong? Well, there are various possibilities. At the simple end, it's possible that there is more than one Higgs boson. The simplest possible model has only one Higgs boson, and for obvious reasons that's the model that many people are hoping is correct, but it's perfectly possible that there could be a &lt;em&gt;multiplet&lt;/em&gt; of several Higgs particles instead. If there is more than one, I'm not sure how exactly that would change the lower and upper bounds on the mass range, but I believe that there is some possibility that if there is a Higgs multiplet, all the particles could have higher masses than we would be able to detect. (I used to know more about this but it's been a little while)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the other extreme, it could be that the whole theoretical framework of the Standard Model is incorrect. That seems pretty unlikely, since pretty much every prediction the SM has made has turned out to be spot on (except for the presence of the Higgs, of course, but that's still an open question). There are definitely alternate theories waiting in the wings that will be receiving quite a bit more attention if the Higgs is not found.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-03T07:36:18.657" CommentCount="1" />
  <row Id="132" PostTypeId="2" ParentId="29" CreationDate="2010-11-03T08:25:11.803" Score="0" ViewCount="" Body="&lt;p&gt;In order to calculate the average speed you have to weight with the time of the different part of the trip and not with the distance covered in the same parts!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the basic formula u hate to use is &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;v_{ave}=S_{tot}/T{tot}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;if ur trip is divided in two parts S1 covere at speed V1 and S2 covered at speed V2 &lt;strong&gt;what you &lt;em&gt;can't&lt;/em&gt; do is&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;V_{ave}=(V1*S1+V2*S2)/(S1+S2)&lt;/code&gt; &#xA;[that is actually what you did with your: (40mph+60mph)/2 = 50mph, since in your example S1=S2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;whereas what u &lt;em&gt;can&lt;/em&gt; do is&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;V_{ave}=(V1*T1+V2*T2)/(T1+T2)&lt;/code&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;that given your input can be written as &lt;code&gt;(S1+S2)/(S1/V1+S2/V2)&lt;/code&gt; and that is indeed equal to &lt;code&gt;(S1+S2)/(T1+T2)&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2010-11-03T08:25:11.803" />
  <row Id="133" PostTypeId="2" ParentId="127" CreationDate="2010-11-03T08:31:03.573" Score="4" ViewCount="" Body="&lt;p&gt;I agree with  David and his disagreement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My experience and personal opinion is that GNUPlot is simpler and quicker for the real time analysis and then u already have the graph in gnuplot.. why would u bother to change it ;)&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2010-11-03T08:31:03.573" CommentCount="2" />
  <row Id="134" PostTypeId="1" AcceptedAnswerId="137" CreationDate="2010-11-03T09:36:15.970" Score="4" ViewCount="162" Body="&lt;p&gt;Deuteron (&lt;sup&gt;2&lt;/sup&gt;H) is composed of a neutron (spin-1/2) and a proton (spin-1/2), with a total spin of 1, which is a boson. Therefore, it is possible for two deuterons to occupy the same quantum state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the protons and neutrons inside are fermions, so they proton/neutron shouldn't be able the share states due to the exclusion principle. Why it does not happen?&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-11-03T11:12:10.003" Title="Deuterons can share quantum states, but why it is fine for the protons and neutrons inside?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="135" PostTypeId="2" ParentId="89" CreationDate="2010-11-03T09:55:03.283" Score="1" ViewCount="" Body="&lt;p&gt;The main problem here is this: Newton gives us formulas for a force, or a field, if you like. Einstein gives us more generic equations from which to derive gravitational formulas.&#xA;In this context, one must first find a solution to Einstein's equations. This is represented by a formula. This formula is what might, or may not, be approximately equal to Newton's laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This said, as answered elsewhere, there is one solution which is very similar to Newton's. It's a very important solution which describes the field in free space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find more about this formula -- in lingo it's a metric, here:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_metric&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Schwarzschild_metric&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that they are approximations fundamentally arises from different factros: the fact that they are invariant laws under a number of transformations, but mostly special relativity concerns - in other words, no action at a distance - is a big one.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-03T09:55:03.283" />
  <row Id="136" PostTypeId="2" ParentId="90" CreationDate="2010-11-03T10:50:46.967" Score="7" ViewCount="" Body="&lt;p&gt;Pulsars are a label we apply to neutron stars that have been observed to &quot;pulse&quot; radio and x-ray emissions. Although all pulsars are neutron stars, not all pulsars are the same. There are three distinct classes of pulsars are currently known: rotation-powered, where the loss of rotational energy of the star provides the power; accretion-powered pulsars, where the gravitational potential energy of accreted matter is the power source; and magnetars, where the decay of an extremely strong magnetic field provides the electromagnetic power. Recent observations with the Fermi Space Telescope has discovered a subclass of rotationally-powered pulsars that emit only gamma rays rather than in X-rays. Only 18 examples of this new class of pulsar are known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While each of these classes of pulsar and the physics underlying them are quite different, the behaviour as seen from Earth is quite similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since pulsars appear to pulse because they rotate, and it's impossible for the the initial stellar collapse which forms a neutron star not to add angular momentum on a core element during its gravitational collapse phase, it's a given that all neutron stars rotate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, neutron star rotation does slow down over time. So non-rotating neutron stars are at least possible. Hence not all neutron stars will necessarily be pulsars, but most will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However practically, the definition of a pulsar is a &quot;neutron star where we observe pulsations&quot; rather than a distinct type of behaviour. So the answer is of necessity somewhat ambiguous.&lt;/p&gt;&#xA;" OwnerUserId="134" LastActivityDate="2010-11-03T10:50:46.967" CommentCount="4" />
  <row Id="137" PostTypeId="2" ParentId="134" CreationDate="2010-11-03T10:57:46.963" Score="5" ViewCount="" Body="&lt;p&gt;The exclusion principle is for identical fermions; but as proton and neutron are different they can occupy the same quantum state in the deuteron: that's the key point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why you can have a bound state proton-neutron (the deuteron) and no neutron-neutron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at my answer &lt;a href=&quot;http://physics.stackexchange.com/questions/78/why-do-neutrons-repel-each-other/81#81&quot;&gt;to this question&lt;/a&gt; for more details. The main problem is that to have a bound state, you need a positive contribution in the exchange integral, requiring the nucleus to have parallel spins. This is possible in the case of the deuteron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you also mean: &quot;if two deuterons are in the same state, which is allowed because they are boson, then the proton/neutron composing these deuterons should also be in the same state&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the exclusion principle does not work like that. If you consider the state vector of the deuteron you cannot after &quot;go down one level&quot; if you consider the deuteron as a boson and consider the individual nucleus. That's because the stage vector of the deuteron is not the one of a neutron + the one of a proton; it is a linear combination of both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try to construct a global state composed of two neutrons and two protons then you will see that you cannot have a bound state where the spins are all parallel. You will have to construct a state where the spins of the protons are anti parallel and the same for the neutrons. At the end you will have a nice state vector the Helium, but that's a more complicated story.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-03T11:12:10.003" LastActivityDate="2010-11-03T11:12:10.003" />
  <row Id="138" PostTypeId="2" ParentId="134" CreationDate="2010-11-03T10:59:37.957" Score="0" ViewCount="" Body="&lt;p&gt;This one state for deuteron is not a single state for both protons and neutrons.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-03T10:59:37.957" />
  <row Id="139" PostTypeId="1" AcceptedAnswerId="1087" CreationDate="2010-11-03T11:25:08.777" Score="5" ViewCount="344" Body="&lt;p&gt;Everyone knows it is close to $c$, but how close? What are the recent results?&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-01-16T08:44:19.813" LastActivityDate="2011-01-16T08:44:19.813" Title="Speed of neutrinos" Tags="&lt;particle-physics&gt;&lt;neutrino&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="140" PostTypeId="2" ParentId="139" CreationDate="2010-11-03T11:31:18.607" Score="8" ViewCount="" Body="&lt;p&gt;It is technically impossible to measure the speed of such a particle directly; and it all depend on &quot;which&quot; neutrino you are talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed is related to the momentum and the momentum to the energy. So you can have a neutrino of some MeV of total energy, another one of some GeV, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in any cases, the answer will be &quot;very very close&quot; to c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all cases neutrinos can be considered as ultra relativistic: their total energy is much higher than their rest energy: $E_t &gt;&gt; m_0 c^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, this relation holds: $E = p c$ where $ p = \gamma m_0 v$. The problem is that $m_0$ is not know with a high precision, it is of the order of the eV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do the math inserting a plausible energy for a solar neutrino, or for a neutrino coming from a high energy collision process, taking any reasonable value for the rest mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;To fix the ideas, the rest mass can be something of the order of 0.1 eV and a typical solar neutrino can be of the order of 10 MeV. That's a $\gamma = 10^8$ !&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-03T11:52:46.037" LastActivityDate="2010-11-03T11:52:46.037" CommentCount="1" />
  <row Id="141" PostTypeId="1" CreationDate="2010-11-03T12:18:08.250" Score="3" ViewCount="70" Body="&lt;p&gt;I'm interested in the process that goes from, for example, huge data sets of temperature, pressure, precipitation etc readings to a model of the atmosphere. I'd like to know about how much relies on a priori knowledge of Navier-Stokes, and how much curve fitting is done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some sort of general overview of the process (ideally a fairly accessible one) would be perfect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a reference, specifically. I know more or less the science from talking to people who do this stuff, but I would like to have some summary article or book chapter overview I can refer to in written work.&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="136" LastEditorDisplayName="" LastEditDate="2010-11-06T14:22:45.580" LastActivityDate="2010-11-06T14:22:45.580" Title="Reference request: statistical inference with real world (chaotic) data, e.g. climate/weather data" Tags="&lt;fluid-dynamics&gt;&lt;atmospheric-science&gt;&lt;climate-science&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="142" PostTypeId="1" AcceptedAnswerId="152" CreationDate="2010-11-03T12:39:37.883" Score="9" ViewCount="315" Body="&lt;p&gt;I would like to have a good understanding of what is happening when you add salt to boiling water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that the boiling point will be higher, thus lengthening the process (obtaining boiling water), but at the same time, the dissolved salt reduce the polarization effect of the water molecules on the heat capacity, thus shortening the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this competition between these two effects real ? Is it something else ?&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2010-11-03T12:56:52.193" LastActivityDate="2010-12-30T15:34:53.583" Title="Boiling water and salt" Tags="&lt;applied-physics&gt;&lt;physical-chemistry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="143" PostTypeId="2" ParentId="111" CreationDate="2010-11-03T12:46:53.823" Score="3" ViewCount="" Body="&lt;p&gt;My favourite for pure classical mechanics is generally the book by &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0321188977&quot; rel=&quot;nofollow&quot;&gt;Goldstein&lt;/a&gt; which includes the Lagrangian and Hamiltonian methods (although I'm not sure about symplectic geometrical and mathematical foundations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a firm look at curved spacetime manifolds (for vectors, one forms, tangent bundles etc.) I would recommend &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0805387323&quot; rel=&quot;nofollow&quot;&gt;Carroll's Spacetime and Geometry&lt;/a&gt; but it deals with the mathematical backing for General Relativity, which is classical mechanics, but on a curved spacetime.&lt;/p&gt;&#xA;" OwnerUserId="133" LastActivityDate="2010-11-03T12:46:53.823" CommunityOwnedDate="2011-07-07T13:41:37.800" CommentCount="1" />
  <row Id="144" PostTypeId="2" ParentId="27" CreationDate="2010-11-03T13:29:36.837" Score="3" ViewCount="" Body="&lt;p&gt;Much of how you answer this question comes down to your view of the wavefunction or state. If you think that the quantum state is a state of reality (that is, an ontic state), then you must either reproduce the predictions of orthodox (Copenhagen) QM without the measurement postulate or you must explain why nature provides two forms of evolution. The former view is basically the Many Worlds Interpretation, which I feel a great degree of attraction to, as it postulates only unitary evolution, and explains measurement as being an emergent, rather than fundamental, effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you hold that the wavefunction is a state of knowledge (epistemic) about some other underlying ontic state, then measurement collapse represents not a true evolution, but a discontinuous change in your knowledge about a system. Alternative formulations of quantum mechanics, such as Bohmian mechanics, explain this in a mathematically rigorous way, but that some find unsatisfying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of these approaches (and the many more I didn't mention) suggests where to look for the next physical theory, and so the question should eventually be experimentally decidable. For now, though, we must rely on mathematics, physical intuition and rational argument.&lt;/p&gt;&#xA;" OwnerUserId="141" LastActivityDate="2010-11-03T13:29:36.837" CommentCount="2" />
  <row Id="145" PostTypeId="1" AcceptedAnswerId="151" CreationDate="2010-11-03T13:44:38.893" Score="1" ViewCount="215" Body="&lt;p&gt;I was wondering if anyone put together a law to describe the rising temperature of the water coming out of a tap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The setup is fairly simple: there's a water tank at temperature T, a metal tube of length L connected to it and a tap at the end where temperature is measured. The water flows at P l/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the metal tube is at room temperature initially, what law describes the temperature of the water at any instant? What is the limit temperature of the water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-03T15:30:06.187" Title="Law for tap water temperature" Tags="&lt;applied-physics&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="146" PostTypeId="2" ParentId="145" CreationDate="2010-11-03T13:56:23.810" Score="0" ViewCount="" Body="&lt;p&gt;The answer is going to depend on the heat transfer coefficient between the tube and the surrounding room (unless you specify that the tube is held at constant temperature), the heat transfer coefficient between the tube and water, the outside &amp;amp; inside diameter of the tube, and the length of the tube. This is a moderately involved heat transfer problem, unless additional constraints are provided to simplify it.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An excellent resource for understanding the mathematics behind this sort of heat transfer problem can be found over &lt;a href=&quot;http://www.nzifst.org.nz/unitoperations/httrapps1.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. It's very similar to the solution posted above by @Cedric, but may be a little easier for some to follow.&lt;/p&gt;&#xA;" OwnerUserId="36" LastEditorUserId="36" LastEditorDisplayName="" LastEditDate="2010-11-03T15:17:07.407" LastActivityDate="2010-11-03T15:17:07.407" CommentCount="1" />
  <row Id="147" PostTypeId="1" AcceptedAnswerId="149" CreationDate="2010-11-03T13:59:20.010" Score="5" ViewCount="209" Body="&lt;p&gt;The answers in this question: &lt;a href=&quot;http://physics.stackexchange.com/questions/1/what-is-spin-as-it-relates-to-subatomic-particles&quot;&gt;What is spin as it relates to subatomic particles?&lt;/a&gt;&#xA;do not address some particular questions regarding the concept of spin:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How  are some useful ways to &lt;em&gt;imagine&lt;/em&gt; a particle without dimensions - like an electron - to spin?&lt;br&gt;&#xA;How  are some useful ways to &lt;em&gt;imagine&lt;/em&gt; a particle with spin 1/2 to make a 360° turn without returning to it's original position (the wave function transforms as: $\Psi \rightarrow -\Psi$).&lt;br&gt;&#xA;When spin is not a classical property of elementary particles, is it a purely relativistic property, a purely quantum-mechanical property or a mixture of both?&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-01T07:35:32.210" LastActivityDate="2011-02-10T13:30:48.493" Title="What are some useful ways to imagine the concept of spin as it relates to subatomic particles?" Tags="&lt;quantum-mechanics&gt;&lt;spin&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="148" PostTypeId="2" ParentId="111" CreationDate="2010-11-03T14:07:48.730" Score="9" ViewCount="" Body="&lt;p&gt;I can't believe nobody's mentioned Arnol'd's book &quot;Mathematical Methods for Classical Mechanics&quot; - it covers everything you ask for in the first paragraph quite elegantly (though sometimes somewhat tersely).&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-03T14:07:48.730" CommunityOwnedDate="2011-07-07T13:41:37.800" CommentCount="4" />
  <row Id="149" PostTypeId="2" ParentId="147" CreationDate="2010-11-03T14:11:11.810" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;How should one imagine a particle without dimensions - like an electron - to spin?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't. If you want to imagine, then you think classically and it is just a particle spinning... Thinking like that doesn't give you any other insight of what spin really is (an intrinsic angular momentum, behaving like an [orbital] angular momentum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How should one imagine a particle with spin 1/2 to make a 360° turn without returning to it's original position (the wave function transforms as: Ψ→−Ψ)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just imagine it ... no big deal. Again, classically this is not possible, but quantically it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When spin is not a classical property of elementary particles, is it a purely relativistic property, a purely quantum-mechanical property or a mixture of both?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spin of elementary particle is a pure quantum mechanical effect. &lt;strong&gt;Edit:&lt;/strong&gt; See @j.c. comment. Relativity also plays a role.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other interpretation/calculation requires things like commutator, symmetry properties and group theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parallel between &quot;real spinning&quot; and &quot;spin&quot; (which is just a name) comes from the fact that the spin operator needed to account for properties of elementary particles behaves (= has the same definition, based on commutators) like orbital angular momentum operator. This again comes from symmetry properties of ... nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal of quantum physics is to provide a way to calculate properties. If you want to calculate or go deeper in the problem, then you don't need this classical interpretation.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-03T14:17:46.170" LastActivityDate="2010-11-03T14:17:46.170" CommentCount="8" />
  <row Id="151" PostTypeId="2" ParentId="145" CreationDate="2010-11-03T14:50:14.863" Score="4" ViewCount="" Body="&lt;p&gt;We can consider the following model: a tube of constant temperature $T_e$ of lenght L, radius $r$ where water is flowing uniformly at a speed $v$ (that you can obtain from your flow $P$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;slice&quot; of water travels an interval $dx$ in a duration $dt = \frac{dx}{v}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tube will contribute to the &quot;heating&quot; of the water by $\frac{dQ}{dt} = (T-T_e) k 2 \pi r dx$ where $k$ is the conductivity and where we use a very simple model (in particular for the radius, we do not distinguish external and internal radii).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During this interval the temperature $T(x)$ of the water will vary by $dT = -\frac{dQ}{c \rho dV}$ where $C$ is the heat capacity at constant pressure of water, and where $dV = 2 \pi r dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replacing we have $\frac{dT}{T-T_e}=-\frac{k}{\rho C v} dx$ whose solution, if the temperature in the tank (ie x = 0) is $T_t$ :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(x) = (T_t - T_e) e^{(-\alpha x)}+T_e$ where $\alpha = \frac{k}{\rho C v}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the lenght of the tube you have the temperature at the tap.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-03T15:30:06.187" LastActivityDate="2010-11-03T15:30:06.187" CommentCount="9" />
  <row Id="152" PostTypeId="2" ParentId="142" CreationDate="2010-11-03T15:01:02.880" Score="1" ViewCount="" Body="&lt;p&gt;I think the dominant effect might actually be the fact that the salt you add might not be at boiling temperature.  But this is just based on the fact that &lt;a href=&quot;http://en.wikipedia.org/wiki/Boiling-point_elevation#Uses&quot; rel=&quot;nofollow&quot;&gt;the boiling-point elevation due to salt in water is actually quite low&lt;/a&gt; for typical amounts of salt used in cooking, say.  I'm not too familiar with the second effect you mention though.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-03T15:01:02.880" />
  <row Id="153" PostTypeId="2" ParentId="75" CreationDate="2010-11-03T15:09:43.840" Score="2" ViewCount="" Body="&lt;p&gt;I would suggest learning about some of the formulas relating them - that way you're not just memorizing things but actually have some grasp of what goes into them. In particular, I only really know about stress and strain, and it's because I think of them as being the analogue in linear elasticity theory of &quot;force&quot; and &quot;displacement&quot; in Hooke's law. &lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-03T15:09:43.840" />
  <row Id="154" PostTypeId="2" ParentId="49" CreationDate="2010-11-03T15:14:11.793" Score="2" ViewCount="" Body="&lt;p&gt;According to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Capacitive_sensing&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt;, capacitative sensors detect things with conductive or dielectric properties; so to me, that means these sensors detect a contrast in the electrical properties between your finger, say, and the air around it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just tested this out on my touchpad with a metal ruler and if I pushed down, I got a little bit of response.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-03T15:14:11.793" CommentCount="4" />
  <row Id="155" PostTypeId="2" ParentId="78" CreationDate="2010-11-03T15:17:06.673" Score="2" ViewCount="" Body="&lt;p&gt;I think this may be a summary of other answers, but there are a couple things going on here. First, neutrons are electrically neutral, so an obvious force is the Van der Waals force. However, due to quantum mechanics and the Pauli exclusion principle (as noted above), neutrons' wave functions cannot (ish) overlap and so they are subject to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Degenerate_matter#Neutron_degeneracy&quot; rel=&quot;nofollow&quot;&gt;Neutron Degeneracy Pressure&lt;/a&gt;. I think if you understand these two concepts, then you should have a fairly good grasp on what is going on between neutrons in a neutron star.&lt;/p&gt;&#xA;" OwnerUserId="40" LastActivityDate="2010-11-03T15:17:06.673" />
  <row Id="156" PostTypeId="1" AcceptedAnswerId="158" CreationDate="2010-11-03T16:28:45.677" Score="7" ViewCount="141" Body="&lt;p&gt;Although I've never seen it myself, I hear the northern lights are a sight to be seen! I know they're related to the Earth's magnetic field but I don't know much more about them. What is the physical phenomenon that creates the northern lights?&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-11-04T12:16:29.730" LastActivityDate="2010-11-04T14:49:41.693" Title="How are the northern lights produced?" Tags="&lt;astrophysics&gt;&lt;atmosphere&gt;&lt;sun&gt;" AnswerCount="2" />
  <row Id="157" PostTypeId="2" ParentId="141" CreationDate="2010-11-03T16:35:47.510" Score="3" ViewCount="" Body="&lt;p&gt;Climate modelling is a giant science of its own, and the proportion of CFD/statistics depends on the particular model.&lt;br&gt;&#xA;In general, what models do is first a simple (often uncompressable) large-scale CFD to advect the scalar fields and then apply a bunch of subrutines simulating small scale and more complex processes, like radiation transfers, heat transfers, cloud dynamics, precipitation, evatranspiration and interactions with land/ocean. Those are mainly governed by some phenomenological theories, based either on some subscale simulations, approximated theories or purely statistical models. The proper construction of that step depends on the purpose of a model and has more to do with art than science (i.e. balancing accuracy and computational time). &lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-03T16:35:47.510" CommentCount="1" />
  <row Id="158" PostTypeId="2" ParentId="156" CreationDate="2010-11-03T16:52:11.147" Score="7" ViewCount="" Body="&lt;p&gt;The ionized particles from mainly solar wind are caught and trapped by Earth magnetic field, which behaves like a magnetic bottle. (The region in which ions are trapped is called Van Allen radiation belts.) This trap is weaker in the polar regions, and there the ions are mainly released into the denser parts of atmosphere. There they collide with air particles (mostly N$_2$ and O$_2$) causing their fluorescence seen as the northern lights.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-03T16:52:11.147" CommentCount="4" />
  <row Id="159" PostTypeId="1" AcceptedAnswerId="162" CreationDate="2010-11-03T17:13:48.773" Score="0" ViewCount="112" Body="&lt;p&gt;In &lt;a href=&quot;http://physics.stackexchange.com/questions/102/is-gravity-a-force-and-if-so-what-is-its-opposite&quot;&gt;this question&lt;/a&gt; I asked about gravity and in the answers it came up that the magnitude is equal (of the gravity acting on the Sun and the of the gravity acting on the Earth)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does magnitude simply mean it's strength?&lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2010-11-09T22:32:10.333" Title="A force's magnitude" Tags="&lt;astrophysics&gt;&lt;forces&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="160" PostTypeId="1" AcceptedAnswerId="238" CreationDate="2010-11-03T17:16:50.347" Score="12" ViewCount="492" Body="&lt;p&gt;I'm trying to explain elementary mechanics - without the benefits of calculus or even algebra - and struggling. I'd like to find reasonable ways to demonstrate Newton's laws, minimally, and possibly continuing though basic projectile motion, although that may be too much to hope for.&lt;/p&gt;&#xA;" OwnerUserId="36" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-11-18T18:02:00.953" LastActivityDate="2011-01-14T16:19:48.123" CommunityOwnedDate="2010-11-10T18:22:11.807" Title="What are good mechanics experiments for 10 year olds?" Tags="&lt;mechanics&gt;&lt;experimental-physics&gt;&lt;education&gt;&lt;experiment&gt;&lt;home-experiment&gt;" AnswerCount="9" CommentCount="6" FavoriteCount="1" />
  <row Id="161" PostTypeId="2" ParentId="159" CreationDate="2010-11-03T17:17:52.670" Score="6" ViewCount="" Body="&lt;p&gt;Sure - a force has a magnitude (amount) and direction (vector).&lt;/p&gt;&#xA;" OwnerUserId="36" LastActivityDate="2010-11-03T17:17:52.670" />
  <row Id="162" PostTypeId="2" ParentId="159" CreationDate="2010-11-03T17:20:02.617" Score="1" ViewCount="" Body="&lt;p&gt;A force is a normal vector and a vector is characterized by its magnitude ( its norm, &quot;lenght&quot; in graphical representation ) and direction.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-11-03T17:20:02.617" CommentCount="2" />
  <row Id="163" PostTypeId="2" ParentId="159" CreationDate="2010-11-03T17:40:49.470" Score="1" ViewCount="" Body="&lt;p&gt;In the case of centripetal forces, like gravity, force is always directed towards the center of mass - it's a radial force. We can therefore study most of its properties by calculating its strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, with gravity it is typical to use polar coordinates (e.g., in 2D, use angle and radius instead of $x$ and $y$). This leads to a gravitational force which only has a radial component. In this particular case the magnitude of the force vector is the same as the radial component, so it's really easy to calculate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the general case, the magnitude is the length of the force vector calculated by the square root of the dot product of the vector by itself. In classical coordinates $(x,y)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|F| = \sqrt{(F_x^2 + F_y^2)}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $F_x$ and $F_y$ are the $x$ and $y$ components of the force vector ${\bf F}$.&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-09T22:32:10.333" LastActivityDate="2010-11-09T22:32:10.333" />
  <row Id="164" PostTypeId="2" ParentId="160" CreationDate="2010-11-03T17:40:55.417" Score="1" ViewCount="" Body="&lt;p&gt;Just some quick ideas.. the question is fairly intriguing: i'm gonna think about it again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First law of Newton: i think that those dry ice containers that mimics the motion without friction can be found quite cheaply and can be very useful:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If set in motion it will keep going&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If connected to a dynamometer from one side and an eletric mini4WD in the other side can demonstrate (with no math at all.. just common sense) that an object stay still not only if no forces are acting upon it but also if the sum of the force actin on an object is zero  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Second Law: using the same electic mini 4WD as before, measure witha  dynamometer the force of the of the motor (u already did that in the previous experiment ;) ) and then vary the mass of the car. It will be quite evident (always thank to the common sense) that the heaviest car will accelerate less than the other..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third law: take two object of similar but not equal mass and compress a spring between them: when you will release (yeah at the same time would be better:P ) them both of them will move in opposite direction.. that can be quite trickier to explain but the experimental evidence shuold be quite visible!&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2010-11-03T17:40:55.417" CommunityOwnedDate="2010-11-10T18:22:11.807" />
  <row Id="165" PostTypeId="2" ParentId="160" CreationDate="2010-11-03T17:43:54.163" Score="1" ViewCount="" Body="&lt;p&gt;Measuring the acceleration of gravity with a pendulum seems to be a good candidate. You don't have to explain the maths. It's the first physics experiment I did, but I was older. On the other hand, it could work with a good teacher.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-03T17:43:54.163" CommunityOwnedDate="2010-11-10T18:22:11.807" CommentCount="1" />
  <row Id="166" PostTypeId="1" AcceptedAnswerId="170" CreationDate="2010-11-03T17:49:55.693" Score="3" ViewCount="315" Body="&lt;p&gt;Energy production is one of the burning issues for humankind. There has been some talk about future energy technologies including Fusion, Anti-matter annihilation and Zero-point-energy (from most to least plausible). I'm interested in hearing what people know about developments in the field what they think will be the next real breakthrough in energy technologies.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditorDisplayName="" LastEditDate="2010-11-03T20:45:13.980" LastActivityDate="2011-08-11T15:03:54.610" Title="What new breakthrough energy technologies are close to being realized?" Tags="&lt;fusion&gt;&lt;renewable-energy&gt;" AnswerCount="3" CommentCount="12" />
  <row Id="167" PostTypeId="1" AcceptedAnswerId="337" CreationDate="2010-11-03T17:57:06.623" Score="4" ViewCount="264" Body="&lt;p&gt;The source of Earth's magnetic field is a dynamo driven by convection current in the molten core.  Using some basic physics principles (Maxwell's equations, fluid mechanics equations), properties of Earth (mass, radius, composition, temperature gradient, angular velocity), and properties of materials (conductivity and viscosity of molten iron) or other relevant facts, is it possible to estimate the strength of the field to order of magnitude (about one gauss)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Descriptions I've seen of the geodynamo all refer to extensive numerical computation on a computer, but can we get a rough idea with simple estimation?&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2010-11-08T02:53:56.790" LastActivityDate="2010-11-08T13:38:25.090" Title="How can we make an order-of-magnitude estimate of the strength of Earth's magnetic field?" Tags="&lt;forces&gt;&lt;order-of-magnitude&gt;&lt;magnetohydrodynamics&gt;&lt;electromagnetism&gt;&lt;estimation&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="168" PostTypeId="1" AcceptedAnswerId="181" CreationDate="2010-11-03T18:06:46.643" Score="5" ViewCount="149" Body="&lt;p&gt;A process is reversible if and only if it's always at equilibrium during the process.  Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have heard several specific example of this, such as adding weight gradually to a piston to compress the air inside reversibly, by why should it be true in general?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;Here is something that would firmly convince me of this:&#xA;Suppose I have a reversible process that is not always in equilibrium.  Describe a mechanism for exploiting this process to create a perpetual motion machine.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-04T05:29:34.817" LastActivityDate="2010-11-04T05:29:34.817" Title="Intuitively, why is a reversible process one in which the system is always at equilibrium?" Tags="&lt;thermodynamics&gt;&lt;reversibility&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="169" PostTypeId="2" ParentId="139" CreationDate="2010-11-03T18:31:39.743" Score="6" ViewCount="" Body="&lt;p&gt;Experimental measurements of neutrino speed have been performed. But thus far, they have all found that the neutrinos were going so close to the speed of light that no difference was detectable within the precision of the experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple measurements are described here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutrino#Speed&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Neutrino#Speed&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="150" LastActivityDate="2010-11-03T18:31:39.743" />
  <row Id="170" PostTypeId="2" ParentId="166" CreationDate="2010-11-03T18:54:31.777" Score="7" ViewCount="" Body="&lt;p&gt;The next serious advance that is not an speculative/fringe idea is most likely to be &lt;strong&gt;fusion power&lt;/strong&gt;. Harnessing the power of nuclear fusion has long been a goal for energy production since the first hydrogen bomb was created in the 1950s. Creating controlled fusion, rather than the chaotic variety has proven a rather challenging task here on Earth however. (The sun does it quite easily largely thanks to its huge gravitational field.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of particular note is the &lt;a href=&quot;http://www.iter.org/&quot; rel=&quot;nofollow&quot;&gt;ITER project&lt;/a&gt;, currently ongoing in Southern France. It is an international collaborative project with the goal of creating a sustainable fusion reactor that produces net energy. (You can read information on the site for the specific target goals/criteria). This is essentially the last step in the 'proof of concept' stage of sustainable nuclear fusion for providing energy, and is to many looking quite promising. It is however only a scientific experiment still, and nuclear fusion power plants are a bit further off in the future, even given success of ITER.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the proposed successor to ITER is &lt;a href=&quot;http://en.wikipedia.org/wiki/DEMO&quot; rel=&quot;nofollow&quot;&gt;DEMO&lt;/a&gt;, a reactor that aims to produce over an order of magnitude more power. Once testing is complete, the idea is to turn it into the world's first fusion power plant. With any luck, this may just be the breakthrough humanity has been waiting for. It will certainly be a revolutionary event whenever it occurs, and likely render fossil fuels totally obsolete.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2010-11-03T18:54:31.777" CommentCount="9" />
  <row Id="173" PostTypeId="1" AcceptedAnswerId="177" CreationDate="2010-11-03T20:37:43.370" Score="3" ViewCount="245" Body="&lt;p&gt;I would like to know if acceleration is an absolute quantity, and if so why?&lt;/p&gt;&#xA;" OwnerUserId="48" LastActivityDate="2010-11-03T23:18:04.710" Title="Is acceleration an absolute quantity?" Tags="&lt;newtonian-mechanics&gt;&lt;acceleration&gt;" AnswerCount="5" FavoriteCount="2" />
  <row Id="175" PostTypeId="2" ParentId="173" CreationDate="2010-11-03T20:54:25.697" Score="0" ViewCount="" Body="&lt;p&gt;Acceleration is relative to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Frame_of_reference#Observational_frames_of_reference&quot; rel=&quot;nofollow&quot;&gt;point of reference&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2010-11-03T20:54:25.697" CommentCount="0" />
  <row Id="176" PostTypeId="2" ParentId="168" CreationDate="2010-11-03T20:56:21.770" Score="0" ViewCount="" Body="&lt;p&gt;Because a reversible process must not increase the entropy in the system. Any change due to lack of equilibrium will lead to an increase in entropy, and thus irreversibility. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can visualize this by thinking of two bottles, one full of gas and one with vacuum. If the bottles were to be connected, atoms would randomly migrate between the two bottles leading to a system with the gas divided between the containers with equal pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the disorder of the system has increased, because the atoms that were all sorted in one bottle initially are now subdivided in two partitions of a much larger volume. In other words, the lack of equilibrium in pressure leads to an uniformization of the same, and to an irreversible increase in entropy.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-03T20:56:21.770" CommentCount="4" />
  <row Id="177" PostTypeId="2" ParentId="173" CreationDate="2010-11-03T20:57:39.023" Score="3" ViewCount="" Body="&lt;p&gt;In standard Newtonian mechanics, acceleration is indeed considered to be an &lt;em&gt;absolute quantity&lt;/em&gt;, in that it is not determined relative to any &lt;em&gt;inertial&lt;/em&gt; frame of reference (constant velocity). This fact follows directly from the principal that forces are the same everywhere, independent of observer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you're doing classical mechanics in an accelerating reference frame, then you introduce a fictitious force, and accelerations are not absolute with respect to an &quot;inertial frame&quot; or other accelerating reference frames - though this is less often considered, perhaps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that the same statement applies to Einstein's Special Relativity. (I don't really understand enough General Relativity to comment, but I suspect it says no, and instead considers other more fundamental things, such as space-time geodesics.)&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2010-11-03T20:57:39.023" CommentCount="5" />
  <row Id="178" PostTypeId="2" ParentId="173" CreationDate="2010-11-03T20:58:09.697" Score="0" ViewCount="" Body="&lt;p&gt;Acceleration will be the same in any two frames that are moving with constant speed with respect to each other (and may also be rotated and translated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you consider two frames that have relative rotation or acceleration,  the acceleration of an object will be different in the two frames.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-03T20:58:09.697" />
  <row Id="179" PostTypeId="2" ParentId="173" CreationDate="2010-11-03T21:09:24.700" Score="2" ViewCount="" Body="&lt;p&gt;Absolutely not. An observer in free fall and an observer in zero gravity both experience and observe no acceleration in their frame of relevance. One, however, is actually in an accelerating frame of reference.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-03T21:09:24.700" CommentCount="2" />
  <row Id="180" PostTypeId="2" ParentId="3" CreationDate="2010-11-03T21:11:52.053" Score="5" ViewCount="" Body="&lt;p&gt;Let me give a try.&#xA;When we construct a theory, we suspect that the objects it deals with can be rather complicated. It is natural that we want to find the simplest «building blocks» which the complicated objects are made of. If our theory were absolutely arbitrary, we won't be able to classify these simple building blocks at all. Fortunately, when constructing theories we note that the lagrangian we specify and the vacuum state have certain symmetries. Once we noted it, then it is pure math to show that the simple objects in our theory should be classified according to representations of the symmetry group of the lagrangian and the vacuum state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there are some symmetries which are obvious to us, which we perceive (like invariance under the Poincare group), and there are some symmetries which we &lt;strong&gt;invent&lt;/strong&gt; (like non-abelian gauge symmetries). In the latter case we know that, by construction, all the macroscopic states (including the vacuum state) must be invariant under this new internal symmetry group. This gives us a short-cut to the assertion that the simple object in our theory must be classified according to the representations of the new group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what concerns the specific question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;so the fundamental particle is acting on the quantum states?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When we say that a particle or a field is in representation R of group G, we do not mean that the particles are associated with matrices of representation R acting on something else. We rather mean that the particle can be written in terms of &lt;strong&gt;eigenstates&lt;/strong&gt; of matrices representing operators in R. So, it is the symmetry group transformations that act on the particles.&lt;/p&gt;&#xA;" OwnerUserId="59" LastEditorUserId="59" LastEditorDisplayName="" LastEditDate="2010-11-03T23:01:41.460" LastActivityDate="2010-11-03T23:01:41.460" />
  <row Id="181" PostTypeId="2" ParentId="168" CreationDate="2010-11-03T21:29:10.653" Score="5" ViewCount="" Body="&lt;p&gt;I guess the simplest answer is just to carefully read you own words again. A reversible process is the one that can be made flow backwards. It is intuitive to think that it can be made flow backwards &lt;strong&gt;at any time&lt;/strong&gt; we wish. But if the system were in a non-equilibrium state, one would need to wait a bit until it goes to equilibrium before trying to drive it back. So, it does not satisfy our desire to have the system under the control at any time.&lt;/p&gt;&#xA;" OwnerUserId="59" LastActivityDate="2010-11-03T21:29:10.653" CommentCount="2" />
  <row Id="182" PostTypeId="2" ParentId="167" CreationDate="2010-11-03T22:06:41.493" Score="-1" ViewCount="" Body="&lt;p&gt;One answer might be &lt;a href=&quot;http://en.wikipedia.org/wiki/Benford%27s_law&quot; rel=&quot;nofollow&quot;&gt;Benford's law&lt;/a&gt; which state that the probability of lediding digit being 1 among natural constants is approx 30%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How to interpret Benford's law:&lt;/strong&gt;&lt;br&gt;&#xA;Any constant and measurment of &quot;new&quot; phenomen can be in any range. Ok max magnitude for known universe is approx 10^60, but still. &#xA;Picking random number from this range is not evenly distributed but logarithmic distributed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;example: We pick random number from 1 to 10^10&lt;br&gt;&#xA;There is equal probability to pick number from 1 to 10^5 and from 10^5 to 10^10, this way we can prove that there is most likely to pick number that begins whit 1 and least likely whit 9. We can check this rule just by looking on &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aLogLogPlot_of_Line.GIF&quot; rel=&quot;nofollow&quot;&gt;logarithmic scale&lt;/a&gt; or doing some math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measuring of earth field, gravity or something else obeys this law and this is the reason that is most likely that you will measure 1*10^n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other reason&lt;/strong&gt; that field is 1 gauss is that 1 is just magnitude/order of field. (by the way wikipedia article about this topic says that it is between 0.3 and 0.6 gauss) &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Earth%27s_magnetic_field#Field_characteristics&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Earth's_magnetic_field#Field_characteristics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="118" LastEditorUserId="118" LastEditorDisplayName="" LastEditDate="2010-11-05T14:34:10.250" LastActivityDate="2010-11-05T14:34:10.250" CommentCount="10" />
  <row Id="183" PostTypeId="2" ParentId="173" CreationDate="2010-11-03T23:18:04.710" Score="0" ViewCount="" Body="&lt;p&gt;I've finally figured it out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let's define precisely what it means for some quantity to be absolute or relative. In the context in question, it has to do with whether a quantity is absolute (that is, has the same value) or relative (that is, has different values) when measured by two inertial observers moving with respect to one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, first we need to define what an inertial observer is: it's an observer for which Newton's laws are applicable without having to resort to adding fictitious forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, so now we have two observers, Alice and Bob, both of which are inertial. They both observe the motion of some object. Let the index 1 correspond to quantities measured in A's reference frame and the index 2 correspond to quantities measured in B's reference frame. The position of the object is clearly a relative concept, since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;r₂ = r₁ + u t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where u is the velocity of Bob with respect to Alice, and is constant since they're both inertial observers). Note that the time, t, is the same for both observers, as it must be according to Newtonian Mechanics. The object position is a relative concept because r₂ ≠ r₁.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, take the time-derivative of both sides and we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v₂ = v₁ + u&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that is, the velocity of the object with respect to one observer is different than the velocity of the same object with respect to the other observer. Hence, velocity is a relative quantity in Newtonian Mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, take the time-derivative of both sides once again, and we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a₂ = a₁&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(since u is constant). Thus, the acceleration of the object is the same in both reference frames. Acceleration, therefore, is absolute in Newtonian Mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we take into account the theory of relativity, then time flows at different rates for different inertial observers and the result above for the acceleration is no longer true. &lt;/p&gt;&#xA;" OwnerUserId="48" LastActivityDate="2010-11-03T23:18:04.710" CommentCount="2" />
  <row Id="184" PostTypeId="2" ParentId="160" CreationDate="2010-11-03T23:25:37.333" Score="3" ViewCount="" Body="&lt;p&gt;When you're driving, make them notice that if you're going at a constant speed, no matter how fast, they don't &quot;feel&quot; it. Yet, when accelerating from 0-40 they'll be pushed in their seats.&lt;/p&gt;&#xA;" OwnerUserId="155" LastActivityDate="2010-11-03T23:25:37.333" CommunityOwnedDate="2010-11-10T18:22:11.807" />
  <row Id="185" PostTypeId="1" AcceptedAnswerId="195" CreationDate="2010-11-04T00:10:22.777" Score="20" ViewCount="875" Body="&lt;p&gt;As I understand the Higg's boson can be discovered by the LHC because the collisions are done at an energy that is high enough to produce it and because the luminosity will be high enough also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is needed to claim a real &quot;discovery&quot; ? I guess there is not one event saying &quot;hey, that's an Higg's boson&quot; ... I also guess that this was the same kind of situation when the top quark was discovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does it work ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a nice introduction to the subject on &lt;a href=&quot;http://cms.web.cern.ch/cms/Physics/HuntingHiggs/CMS.html&quot; rel=&quot;nofollow&quot;&gt;this page of the CMS experiment&lt;/a&gt;, and the various ways to detect it, for example through the following process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/hTjQC.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-12T20:29:23.563" LastActivityDate="2011-04-12T20:29:23.563" Title="What is needed to claim the discovery of the  Higgs boson?" Tags="&lt;particle-physics&gt;&lt;experimental-physics&gt;&lt;standard-model&gt;&lt;high-energy-physics&gt;&lt;higgs-boson&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="5" />
  <row Id="186" PostTypeId="1" AcceptedAnswerId="189" CreationDate="2010-11-04T00:53:26.137" Score="1" ViewCount="174" Body="&lt;p&gt;We have to analyze this video:&#xA;&lt;a href=&quot;http://www.youtube.com/watch?v=Ty-vH42H_7k&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=Ty-vH42H_7k&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Givens:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) An applied net torque due to the wind on the windmill is equal to 1500 N*m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Each (of the 3) propeller props weighs approximately 45 Kg and has a Moment of Inertial equal to about 65% of that of a rod of the same mass being spun about its end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) This torque is applied for approximately 45 seconds before the explosion, prior to which the windmill was at rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: What was the angular acceleration caused by the torque?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's my attempt at it:&#xA;T=Ia (a is alpha)&#xA;T=ML^2/3 * a * .65 (due to the whole 65% thing. Actually not sure if I should put 3*M for each propeller)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so this is where I get stuck. I'm not given L, so I'm not sure how to work around this. I could also use T=1/2*MR^2*a, but then I dont know R.&lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2010-11-04T01:36:56.683" Title="Finding angular acceleration from torque" Tags="&lt;mechanics&gt;&lt;acceleration&gt;&lt;homework&gt;&lt;rotational-dynamics&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="187" PostTypeId="2" ParentId="160" CreationDate="2010-11-04T00:54:33.140" Score="3" ViewCount="" Body="&lt;p&gt;I am always fond of the experiment of &quot;shooting&quot; a rubber ball and dropping the same ball, and seeing that both of them will hit the ground at the same time, showing that the vertical and horizontal components of velocity are independent.&lt;/p&gt;&#xA;" OwnerUserId="71" LastActivityDate="2010-11-04T00:54:33.140" CommunityOwnedDate="2010-11-10T18:22:11.807" />
  <row Id="188" PostTypeId="2" ParentId="185" CreationDate="2010-11-04T01:00:59.280" Score="7" ViewCount="" Body="&lt;p&gt;I'm not a particle physicist, but here's a little overview from what I understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Colliding particles, in particular hadrons (being composed of three quarks and a field of gluons), are quite capable of generating very &quot;messy&quot; collisions, the more so at higher energies. By &quot;messy&quot; I mean that the variations on the possible results of the collision (the number of different Feynmann diagrams) is pretty large. Of course, certain outcomes have much higher outcomes than others, and the probabilities of such can be estimated in quantum field theory. In any case, there can be decays into all sorts of fundamental particles (with different charges, spins, masses, etc.), and then further decays, and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Verifying that a particle (here the Higgs boson) is really what the theory predicts takes lots of runs of the experiment, and is largely just a complex probability game. Given that the properties of the Higgs aren't exactly known, what particle physicists are looking for is &lt;em&gt;signs of a missing particle&lt;/em&gt;. i.e. A violation of some conservation law in the collision (typically energy or momentum). This is a good hint that there's an unknown particle that hasn't been accounted for. (For example, neutrinos are far too weakly interacting to be detected directly, and were first discovered by noticing some small energy was missing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment, all we really know from previous experimental data is that the lower bound for the Higgs boson mass is 115 GeV/c² - pretty high, but theoretically well within the range of the LHC. This helps us a bit in knowing where to look, but at the end of the day, it's smashing together protons over and over again that rather crudely gives you the final discovery!&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="1008" LastEditorDisplayName="" LastEditDate="2011-02-11T01:45:25.433" LastActivityDate="2011-02-11T01:45:25.433" CommentCount="3" />
  <row Id="189" PostTypeId="2" ParentId="186" CreationDate="2010-11-04T01:36:56.683" Score="1" ViewCount="" Body="&lt;p&gt;The first thing I would point out to you is that $T = \frac{1}{2} MR^2 \alpha$ is really just $T = I\alpha$, with a particular choice of $I$. Is that choice appropriate for this problem? (Ask yourself the same thing any other time you consider using $T = \frac{1}{2} MR^2\alpha$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, note that the moments of inertia of different parts of the windmill do add up to produce the total, just like with mass. You can't just use the moment of inertia of one propeller prop, you have to calculate the total moment of inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, consider this: what information can you get from the video, that could supplement the 3 &quot;givens&quot;? There's no length scale in the video, so you can't measure the length of a prop directly, but there is a time scale. What can you do with that?&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-04T01:36:56.683" CommentCount="2" />
  <row Id="190" PostTypeId="2" ParentId="27" CreationDate="2010-11-04T01:49:23.687" Score="6" ViewCount="" Body="&lt;p&gt;Interactions merely involve a correlation developing.  For example, if an electron is put through a Stern-Gerlach apparatus, a correlation develops between the distance travelled in the x direction and the distance deviated in the y direction.   It is reversible. The measurement which occurs when the particle hits the photographic plate is irreversible.   It is associated with irreversible dissipation, i.e. entropy generation.&lt;br&gt;&#xA;This approximation can itself be dissected further, but it gets very tricky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A really good (1983) book is by Wheeler and Zurek, &quot;The quantum theory of measurement&quot; available as a djvu file at &lt;a href=&quot;http://www.4shared.com/get/vw66Qp70/Wheeler_JA_Zurek_WH__eds__Quan.html&quot; rel=&quot;nofollow&quot;&gt;http://www.4shared.com/get/vw66Qp70/Wheeler_JA_Zurek_WH_&lt;em&gt;eds&lt;/em&gt;_Quan.html&lt;/a&gt;  (8 MB, wait 30 sec for the download).    [Now if I can only figure out how to work a reader for a Mac ...]&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-11-04T02:35:09.447" LastActivityDate="2010-11-04T02:35:09.447" CommentCount="2" />
  <row Id="191" PostTypeId="1" CreationDate="2010-11-04T03:18:57.803" Score="1" ViewCount="157" Body="&lt;p&gt;Why do a magnetic core saturates ?&lt;/p&gt;&#xA;" OwnerUserId="157" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2010-11-08T02:55:17.443" LastActivityDate="2010-11-08T02:55:17.443" Title="What is the mechanism of magnetic core saturation?" Tags="&lt;electromagnetism&gt;&lt;magnetism&gt;" AnswerCount="1" />
  <row Id="192" PostTypeId="2" ParentId="191" CreationDate="2010-11-04T04:13:41.080" Score="1" ViewCount="" Body="&lt;p&gt;from wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Saturation_%28magnetic%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Saturation_%28magnetic%29&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ferromagnetic materials like iron that show saturation are composed of microscopic regions called magnetic domains that act like tiny permanent magnets. Before an external magnetic field is applied to the material, the domains are oriented in random directions. Their tiny magnetic fields point in random directions and cancel each other out, so the material has no overall net magnetic field. When an external magnetizing field H is applied to the material, it penetrates the material and aligns the domains, causing their tiny magnetic fields to turn and align parallel to the external field, adding together to create a large magnetic field which extends out from the material. This is called magnetization. The stronger the external magnetic field, the more the domains align. Saturation occurs when practically all the domains are lined up, so further increases in applied field can't cause further alignment of the domains. This is a simplified account; a more complete explanation can be found in Ferromagnetism.&lt;/p&gt;&#xA;" OwnerUserId="62" LastActivityDate="2010-11-04T04:13:41.080" CommentCount="2" />
  <row Id="193" PostTypeId="1" CreationDate="2010-11-04T04:57:45.843" Score="13" ViewCount="946" Body="&lt;p&gt;What are the best textbooks to read for the mathematical background you need for modern physics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some subjects off the top of my head that probably need covering:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Manifolds&lt;/li&gt;&#xA;&lt;li&gt;Lie theory&lt;/li&gt;&#xA;&lt;li&gt;Algebraic topology&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-02T06:37:23.470" LastActivityDate="2011-06-02T06:37:23.470" CommunityOwnedDate="2010-11-10T18:20:54.740" Title="Best books for mathematical background?" Tags="&lt;mathematics&gt;&lt;education&gt;&lt;books&gt;&lt;reference-request&gt;&lt;big-list&gt;" AnswerCount="9" CommentCount="10" FavoriteCount="10" />
  <row Id="194" PostTypeId="2" ParentId="31" CreationDate="2010-11-04T05:11:23.820" Score="2" ViewCount="" Body="&lt;p&gt;An important point about relativity is that it is not quite like the physics you learn in an introductory physics class.  There, you learn about Newton's Laws or Snell's Law or Lenz's Law, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those are all laws that tell things how to act; they tell mass how to respond to force, or light how to bend, or currents which way to run.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relativity is different in that it provides a set of meta-laws, or laws that the other laws of physics must obey.  It doesn't directly tell things what to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classic example is Maxwell's Equations.  These are laws that tell charged particles and electromagnetic fields how to act.  It turns out that these laws obey a certain mathematical criterion, called &quot;Lorentz Invariance&quot;, that is required by relativity.  So Maxwell's equations are good relativistic laws.  They obey the meta-laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, Newton's laws (of motion) are not good relativistic laws.  They don't obey the meta-laws.  So in relativity, we need a slightly new set of laws to describe how mass responds to force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for what the meta-laws are, they were outlined by Nick Gotch as &quot;basic ideas&quot; above.  Those basic ideas turn out to be equivalent to Lorentz invariance.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-04T05:11:23.820" CommentCount="1" />
  <row Id="195" PostTypeId="2" ParentId="185" CreationDate="2010-11-04T05:16:32.427" Score="14" ViewCount="" Body="&lt;p&gt;&lt;em&gt;NOTE: I recommend reading &lt;a href=&quot;http://physics.stackexchange.com/questions/185/what-is-needed-to-claim-the-discovery-of-the-higgs-boson/188#188&quot;&gt;Noldorin's answer&lt;/a&gt; first, for useful background information, and &lt;a href=&quot;http://physics.stackexchange.com/questions/185/what-is-needed-to-claim-the-discovery-of-the-higgs-boson/284#284&quot;&gt;Matt's answer&lt;/a&gt; afterward if you want additional detail&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noldorin is right that there isn't a single event that you can look at and identify a Higgs boson. In fact, unless the theories are drastically wrong, the Higgs particle is unstable and it has an exceedingly short lifetime - so short that it won't even make it out of the empty space inside the detector! Even at the speed of light, it can only travel a microscopic distance before it decays into other particles. (If I can find some numeric predictions I'll edit that information in.) So we won't be able to detect a Higgs boson &lt;em&gt;directly&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What scientists will be looking for are particular patterns of known particles that are signatures of Higgs decay. For example, the standard model predicts that a Higgs boson could decay into two Z bosons, which in turn decay into a muon and antimuon each. So if physicists see that a particular collision produces two muons and two antimuons, among other particles, there's a chance that somewhere in the mess of particles produced in that collision, there was a Higgs boson. This is just one example, of course; there are many other sets of particles that the Higgs could decay into, and the large detectors at the LHC are designed to look for all of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, Higgs decay is not the only thing that could produce two muon-antimuon pairs, and the same is true for other possible decay products. So just seeing the expected decay products is not a sure sign of a Higgs detection. The real evidence is going to come from the results of many collisions (billions or trillions), accumulated over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each possible set of decay products, you can plot the fraction of collisions in which those decay products are produced (or rather, the scattering cross section, a related quantity) against the total energy of the particles coming into the collision. If the Higgs is real, you'll see a spike, called a &lt;em&gt;resonance&lt;/em&gt;, in the graph at the energy corresponding to the mass of the Higgs particle. It'll look something like this plot, which was produced for the Z boson (which has a mass of only 91 GeV):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://blogs.uslhc.us/wp-content/uploads/2010/05/Zres1.png&quot; alt=&quot;Z boson resonance plot&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image is from &lt;a href=&quot;http://blogs.uslhc.us/the-z-boson-and-resonances&quot; rel=&quot;nofollow&quot;&gt;http://blogs.uslhc.us/the-z-boson-and-resonances&lt;/a&gt;, which is actually a pretty good read.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, to sum up: the main signature of the Higgs boson, like other unstable particles, will be this resonance peak that appears in a graph produced by aggregating data from many billions or trillions of collisions. Hopefully this makes it a bit clearer why there's going to be a lot of detailed analysis involved before we get any clear detection or non-detection of the Higgs particle.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-05T23:09:14.757" LastActivityDate="2010-11-05T23:09:14.757" />
  <row Id="196" PostTypeId="1" CreationDate="2010-11-04T05:57:42.347" Score="3" ViewCount="214" Body="&lt;p&gt;I've always heard people saying, &quot;Faster than light information transfer can't happen because it would violate causality!  Effects can precede their causes!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm trying to think of a situation where this would happen.  I'm sure it has something to do with relativity or something like that.  I sort of understand that people going faster perceive time slower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone help illuminate this for me by showing a scenario where causality is clearly violated due to FTL information transfer?&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T07:47:28.513" LastActivityDate="2011-02-03T07:47:28.513" Title="What are some scenarios where FTL information transfer would violate causality?" Tags="&lt;special-relativity&gt;&lt;time-travel&gt;&lt;faster-than-light&gt;&lt;causality&gt;" AnswerCount="3" />
  <row Id="197" PostTypeId="1" AcceptedAnswerId="200" CreationDate="2010-11-04T06:04:58.330" Score="9" ViewCount="373" Body="&lt;p&gt;I hope this is the right word to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, these forces seem kind of fanciful (except for General Relativity and Gravity, which have a geometric interpretation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, how do two charged particles &lt;strong&gt;know&lt;/strong&gt; that they are to move apart from each other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do they communicate with each other somehow through some means?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard some people tell me that they bounce together messenger photons.  So does one electron receive a messenger photon, go, &quot;Oh hey, I should move in the direction opposite of where this came from, due to the data in it&quot;, and then move?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aren't photons also associated with energy, as well?  Does this type of mediation imply that electrons give off energy in order to exert force on other electrons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every electron is repelled by every other electron in the universe, right?  How does it know where to send its force mediators?  Does it just know what direction to point it in?  Does it simply send it in all directions in a continuum?  Does that mean it's always giving off photons/energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just not sure how to view &quot;how&quot; it is that electrons know they are to move away from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These questions have always bugged me when studying forces.  I'm sure the Standard Model has something to shed some light on it.&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-18T02:44:31.400" LastActivityDate="2011-01-18T16:10:38.653" Title="How are forces &quot;mediated&quot;?" Tags="&lt;electromagnetism&gt;&lt;photons&gt;&lt;carrier-particles&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="3" />
  <row Id="198" PostTypeId="1" CreationDate="2010-11-04T06:14:56.820" Score="7" ViewCount="184" Body="&lt;p&gt;I have a particle system of seven protons and seven (or sometimes eight) neutrons (each formed by their appropriate quarks, etc.) bound together in a state that can be macroscopically described as a nucleus.  If relevant, there are also about seven electrons that are bound to this arrangement.  These particle systems are usually found in pairs, bound to eachother.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Macroscopically, this can be modeled as the elemental Nitrogen ($N_2$), and in other disciplines (such as chemistry), it is treated as a basic unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that at a certain level of thermal energy, this system of elementary particles exist inert and packed together in what can be macroscopically described as a &quot;liquid&quot;.  We know that this is this temperature is about 77.36 Kelvin (measured experimentally) at the most.  Any higher and they start repelling each other and behave as a macroscopic gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way, from simply analyzing the particles that make up this system (the quarks making up fourteen protons and 14-16 neutrons, the electrons) and their interactions due to the current model of particles (is this The Standard Model?), to find this temperature 77.36 Kelvin?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we &quot;derive&quot; 77.36 K from only knowing the particles and their interactions with each other, in the strong nuclear force and electromagnetic force and weak nuclear force?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, what is this derivation?&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="71" LastEditorDisplayName="" LastEditDate="2010-11-04T07:34:26.843" LastActivityDate="2011-01-03T21:05:09.190" Title="Do current models of particle physics explain the chemical properties of elements/compounds?" Tags="&lt;particle-physics&gt;&lt;standard-model&gt;&lt;models&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="199" PostTypeId="2" ParentId="196" CreationDate="2010-11-04T06:28:00.227" Score="5" ViewCount="" Body="&lt;p&gt;Suppose you and I have a conversation from a long distance away.  We're at rest with respect to each other and communicate much faster than light.  I say &quot;How are you&quot;, and you wait a short time and say, &quot;I'm fine thanks.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From our point of view, you were responding to my question.  However, from a reference frame moving from me to you at relativistic speed, your clock is significantly ahead of mine (a relativistic effect).  This means that although you thought you received the message shortly after I sent it, in this frame you didn't.  You actually received the message at an earlier time (before I sent it), but you thought it was later because your clock is ahead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your and my point of view, the order of events is&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I say &quot;How are you?&quot;&lt;/li&gt;&#xA;&lt;li&gt;You hear me say &quot;How are you?&quot;&lt;/li&gt;&#xA;&lt;li&gt;You pause a short time.&lt;/li&gt;&#xA;&lt;li&gt;You say, &quot;I'm fine thanks.&quot;&lt;/li&gt;&#xA;&lt;li&gt;I hear you say, &quot;I'm fine thanks.&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;From the frame moving from me to you, the order of events is&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You hear me say &quot;How are you?&quot;&lt;/li&gt;&#xA;&lt;li&gt;You pause a short time.&lt;/li&gt;&#xA;&lt;li&gt;You say &quot;I'm fine thanks.&quot;&lt;/li&gt;&#xA;&lt;li&gt;I say, &quot;How are you?&quot;&lt;/li&gt;&#xA;&lt;li&gt;I hear you say, &quot;I'm fine thanks.&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The fact that the order of events changes between reference frames is simply part of relativity, with or without faster-than-light communication.  However, it seems strange in this scenario because you are responding to me.  Presumably, if I had said, &quot;Where are my car keys?&quot;, you would have chosen a different response than &quot;I'm fine thanks.&quot;  How then is it possible that you responded to my greeting before I uttered it, at least in some frame?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if this &quot;violates causality&quot;, but it's unintuitive.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-04T06:28:00.227" CommentCount="2" />
  <row Id="200" PostTypeId="2" ParentId="197" CreationDate="2010-11-04T06:49:45.187" Score="12" ViewCount="" Body="&lt;p&gt;Biref answer: Read only the &lt;strong&gt;bold&lt;/strong&gt; part (and ignore grammar then).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer you already mentioned lies in &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_field_theory&quot; rel=&quot;nofollow&quot;&gt;Quantum Field Theory&lt;/a&gt; (QFT). But to fully understand it, you must &lt;strong&gt;give up&lt;/strong&gt; a particle as a &lt;strong&gt;point-like thing that is well-localized&lt;/strong&gt;. There is &lt;strong&gt;one Quantum Field per sort of particle&lt;/strong&gt;, e.g. &lt;em&gt;the&lt;/em&gt; electron field for &lt;em&gt;all&lt;/em&gt; electrons, and &lt;em&gt;the&lt;/em&gt; photon field for all photons. (The fact that there is a single field for all electrons also results in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pauli_exclusion_principle&quot; rel=&quot;nofollow&quot;&gt;Pauli exclusion principle&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you consider &lt;strong&gt;a particle is&lt;/strong&gt; basically just &lt;strong&gt;a&lt;/strong&gt; local &lt;strong&gt;peak in the&lt;/strong&gt; respective particle &lt;strong&gt;field&lt;/strong&gt;, but one cannot even say &quot;This peak corresponds to electron A, this one to B&quot;. Now QFT, more specifically &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_electrodynamics&quot; rel=&quot;nofollow&quot;&gt;Quantum Electrodynamcis&lt;/a&gt; (&lt;strong&gt;QED&lt;/strong&gt;), describes the local &lt;strong&gt;interaction between the electron field and the photon field&lt;/strong&gt;. But since the fields have a dynamic, &lt;strong&gt;a local change&lt;/strong&gt; induced in the photon field by the electron field &lt;strong&gt;will propagate&lt;/strong&gt; with the speed of light (flat space assumed) and interact with the electron field in another place, thus creating the &lt;strong&gt;impression&lt;/strong&gt; &quot;Electron A emitted a photon that told electron B to interact electromagnetically&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's similar for the other interactions, there's a gluon field for the strong interaction (&lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_chromodynamics&quot; rel=&quot;nofollow&quot;&gt;Quantum Chromodynamics&lt;/a&gt;), and for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electroweak_interaction&quot; rel=&quot;nofollow&quot;&gt;electroweak interaction&lt;/a&gt; there's kind of a combination of the photon field and the weak-interaction-bosons.&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2010-11-04T06:49:45.187" CommentCount="5" />
  <row Id="201" PostTypeId="2" ParentId="197" CreationDate="2010-11-04T07:15:00.813" Score="6" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;For example, how do two charged particles know that they are to move apart from each other? Do they communicate with each other somehow through some means?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, specifically the electromagnetic field. To give a simplistic view, a charged particle produces an electric field to indicate its presence and a magnetic field to indicate its motion. Any disturbance in these fields propagates outward at the speed of light. Another charged particle some distance away can react to the fields; if it &quot;feels&quot; the presence of an electric field, it will move in response.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I've heard some people tell me that they bounce together messenger photons. So does one electron receive a messenger photon, go, &quot;Oh hey, I should move in the direction opposite of where this came from, due to the data in it&quot;, and then move?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The idea of the messenger photon is really just an analogy. Personally I don't think it's a very good one, but we don't really have anything better. The thing is, even though disturbances in the EM field propagate like waves, when a particle reacts to such a disturbance, it acts as though another particle collided with it. In order to make this fit with our intuition, we're forced to invent the idea that there are particles, or &lt;em&gt;quanta&lt;/em&gt;, associated with the electromagnetic field, and we call them photons.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Aren't photons also associated with energy, as well? Does this type of mediation imply that electrons give off energy in order to exert force on other electrons?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yep. To make sense of this, you really have to consider the whole system of &lt;em&gt;both&lt;/em&gt; electrons, as well as the EM field itself. Each electron &quot;feels&quot; the electromagnetic field produced by the other, so both electrons' motions will change in such a way that the total energy is conserved.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Every electron is repelled by every other electron in the universe, right? How does it know where to send its force mediators? Does it just know what direction to point it in? Does it simply send it in all directions in a continuum? Does that mean it's always giving off photons/energy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is one of those cases in which it makes more sense to think of disturbances in the EM field as waves, which can certainly be radiated in all directions. However, the electron only disturbs the field when its motion (or quantum state, rather) &lt;em&gt;changes&lt;/em&gt;. An electron stuck in a single quantum state, such as an atomic orbital, won't be radiating any energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is that I would advise you not to take the idea of &quot;messenger photons&quot; too literally. It's just a model that works in some situations but not in others.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-04T07:15:00.813" />
  <row Id="202" PostTypeId="1" AcceptedAnswerId="203" CreationDate="2010-11-04T07:25:26.233" Score="12" ViewCount="378" Body="&lt;p&gt;I just came from a class on Fourier Transformations as applied to signal processing and sound.  It all seems pretty abstract to me, so I was wondering if there were any physical systems that would behave like a Fourier transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if given a wave, a purely physical process that would &quot;return&quot; the Fourier transform in some meaningful way.  Like, you gave it a sound wave and you would see, &quot;Oh, there are a lot of components of frequency 1kHz...a few of frequency 10kHz...some of 500Hz...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen things happening where, if you put sand on a speaker, the sand would start to form patterns on the speakers that are related to the dominant wavelengths/fundamental frequencies of the sound.  Is this some sort of natural, physical fourier transform?&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-19T02:43:24.453" LastActivityDate="2010-12-12T21:50:17.207" Title="Fourier transformation in nature/natural physics?" Tags="&lt;sound&gt;&lt;fourier-transform&gt;&lt;nature&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="5" />
  <row Id="203" PostTypeId="2" ParentId="202" CreationDate="2010-11-04T07:34:39.140" Score="16" ViewCount="" Body="&lt;p&gt;Your ear is an effective Fourier transformer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An ear contains many small hair cells.  The hair cells differ in length, tension, and thickness, and therefore respond to different frequencies.  Different hair cells are mechanically linked to ion channels in different neurons, so different neurons in the brain get activated depending on the Fourier transform of the sound you're hearing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A piano is a Fourier analyzer for a similar reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A prism or diffraction grating would be a Fourier analyzer for light.  It spreads out light of different frequencies, allowing us to analyze how much of each frequency is present in a given source.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-04T07:34:39.140" CommentCount="4" />
  <row Id="204" PostTypeId="2" ParentId="198" CreationDate="2010-11-04T07:47:17.867" Score="10" ViewCount="" Body="&lt;p&gt;Theoretically, yes it should be possible to derive the boiling point of diatomic nitrogen from fundamental forces. In fact, you don't even need to involve the strong force or weak force (or the strong nuclear force, which is sort of different). The strong forces bind the quarks together into nucleons and the nucleons together into nuclei, but they have essentially no effect on distance scales much larger than that of an atomic nucleus. So, for purposes of calculating the boiling point of nitrogen, you can treat the nucleus as basically a point charge. The only force that is relevant to calculating a boiling point is the electromagnetic force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the bad news: even something as simple as calculating the energy levels of helium, with 2 electrons, is impossible to do analytically. To analyze the behavior of even just those two electrons (and He nucleus) in detail, you need to use either perturbation theory or a numerical simulation, or both. And of course, the complexity increases with the number of particles, so simulating the 14 electrons and 2 nuclei of a nitrogen molecule is absurdly complicated. Perhaps it's been done, but I'm not a condensed matter physicist so I wouldn't know where to look for a reference. Maybe someone else can provide you with that information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to calculate the boiling point of nitrogen, I believe the main effect that you'd take into account would be the &lt;a href=&quot;http://en.wikipedia.org/wiki/London_dispersion_force&quot; rel=&quot;nofollow&quot;&gt;instantaneous dipole interaction&lt;/a&gt;. According to the Wikipedia article, it gives an interaction energy in terms of the polarizabilities and ionization energies of the molecules. Those are the quantities that you would have to extract from your simulation and/or perturbative calculation of the dynamics of the nitrogen molecule, if you wanted to calculate the effect from first principles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_{AB}^{\rm disp} \approx -{3 \alpha^A \alpha^B I_A I_B\over 4(I_A + I_B)} R^{-6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(that formula is actually for monatomic noble gases, it may not apply to diatomic molecules)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you get the interaction energy as a function of intermolecular separation $R$, you would then have to do either another numerical simulation, or a rather complicated calculation, to show that a large pool of nitrogen molecules subject to the given intermolecular force undergoes a phase transition at 77.36K (at standard pressure, I assume). There are various thermodynamic models you could use, some more accurate than others, but of course the more accuracy you want, the more computation power you'll need. I suspect that in order to get within a few degrees of the actual temperature, you would need to do something more computationally intensive than would be possible by hand.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-03T21:05:09.190" LastActivityDate="2011-01-03T21:05:09.190" CommentCount="4" />
  <row Id="205" PostTypeId="2" ParentId="202" CreationDate="2010-11-04T08:22:11.060" Score="0" ViewCount="" Body="&lt;p&gt;In imaging microscopy the 2D-Fourier Trasform is a useful tool to analyse the images and improve the S/N ratio..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i had your same opinion at the beginning but then my boss, who actually is a FTT entusiast, was able to convince me to the actual usefulness of the technique!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://xkcd.com/26/&quot; rel=&quot;nofollow&quot;&gt;http://xkcd.com/26/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2010-11-04T08:22:11.060" CommentCount="2" />
  <row Id="206" PostTypeId="2" ParentId="160" CreationDate="2010-11-04T08:37:23.740" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Galileo%27s_Leaning_Tower_of_Pisa_experiment&quot; rel=&quot;nofollow&quot;&gt;Galileo's Leaning Tower of Pisa experiment&lt;/a&gt;&lt;/strong&gt; &lt;em&gt;because&lt;/em&gt; &lt;strong&gt;&lt;code&gt;it is sufficiently counter intuitive&lt;/code&gt;&lt;/strong&gt;. &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-11-04T08:37:23.740" CommunityOwnedDate="2010-11-10T18:22:11.807" />
  <row Id="208" PostTypeId="1" AcceptedAnswerId="213" CreationDate="2010-11-04T09:18:02.863" Score="4" ViewCount="463" Body="&lt;p&gt;M = mass of the sun&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m = mass of the earth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;r = distance between the earth and the sun&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun is converting mass into energy by nuclear fusion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = \frac{GMm}{r^2} = \frac{mv^2}{r} \rightarrow r = \frac{GM}{v^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta E = \Delta M c^2 = (M_{t} - M_{t+\Delta t}) c^2 \rightarrow \Delta M = \Delta E / c^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rightarrow \frac{\Delta r}{\Delta t} = \frac{G}{v^2 c^2}.\frac{\Delta E}{\Delta t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sun radiates $3.9 × 10^{26} W = \Delta E/\Delta t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Velocity of the earth $v = 29.8k m/s$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is nothing that is stopping the earth from moving with the same velocity so for centripetal force to balance gravitational force $r$ must change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $r$ increasing? ($\Delta r/ \Delta t = 3.26070717 × 10^{-10} m/s $)&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-02-05T12:59:37.223" LastActivityDate="2011-05-03T14:17:15.340" Title="Is the distance between the sun and the earth increasing?" Tags="&lt;classical-mechanics&gt;&lt;mechanics&gt;&lt;orbital-motion&gt;" AnswerCount="10" CommentCount="5" FavoriteCount="1" />
  <row Id="210" PostTypeId="2" ParentId="208" CreationDate="2010-11-04T09:44:16.457" Score="3" ViewCount="" Body="&lt;p&gt;I think your reasoning is correct, but the values involved are very small.&#xA;In one year r will increase by 1mm, so in 1 billion years it will have increased by  1000 km or by about0.01%&lt;/p&gt;&#xA;" OwnerUserId="57" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-04T18:09:43.597" LastActivityDate="2010-11-04T18:09:43.597" CommentCount="2" />
  <row Id="211" PostTypeId="2" ParentId="208" CreationDate="2010-11-04T09:50:16.890" Score="1" ViewCount="" Body="&lt;p&gt;That is basically correct, however that change is not very significant. The orbit of the planets in the solar system is chaotic over long periods of time (2 - 230 million years according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Stability_of_the_solar_system#Predictability&quot; rel=&quot;nofollow&quot;&gt;this wikipedia entry&lt;/a&gt;), and this effect is relatively minor. Other causes for change of orbit include gravitational pull from other planets, collisions with asteroids, solar wind and other variables.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2010-11-04T09:50:16.890" />
  <row Id="212" PostTypeId="2" ParentId="208" CreationDate="2010-11-04T09:50:42.423" Score="2" ViewCount="" Body="&lt;p&gt;Such a small magnitude makes this process negliable among other factors; indeed this result literally means that you don't need to care about this process until you build extreamly accurate theory of the full solar system dynamics. To the extent that is probably unreachable due to deterministic chaos.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-04T09:50:42.423" />
  <row Id="213" PostTypeId="2" ParentId="208" CreationDate="2010-11-04T10:37:01.557" Score="6" ViewCount="" Body="&lt;p&gt;I think the reasoning has an error.  It assumes $v$ is constant, but instead we ought to assume the angular momentum is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By dimensional analysis that leads to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r \propto \frac{L^2}{GM}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so as $M$ decreases, $r$ increases (the original post had $r \propto M$, not $r \propto 1/M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, assuming a circular orbit seems dubious.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the other commenters said, this effect is minute.  A significant effect on the orbit of the moon around the earth is tidal evolution, which does actually push the moon further away.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbit_of_the_Moon#Tidal_evolution&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Orbit_of_the_Moon#Tidal_evolution&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-04T10:42:38.340" LastActivityDate="2010-11-04T10:42:38.340" CommentCount="2" />
  <row Id="214" PostTypeId="1" AcceptedAnswerId="225" CreationDate="2010-11-04T12:00:50.860" Score="5" ViewCount="470" Body="&lt;p&gt;Imagine you're trapped inside a free falling elevator. Would you decrease your impact impulse by jumping during the fall? When?&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-11-04T12:14:02.823" LastActivityDate="2011-06-09T19:46:31.223" Title="Would it help if you jump inside a free falling elevator?" Tags="&lt;acceleration&gt;" AnswerCount="6" CommentCount="5" />
  <row Id="215" PostTypeId="2" ParentId="214" CreationDate="2010-11-04T12:04:09.250" Score="2" ViewCount="" Body="&lt;p&gt;If you jumped just before impact, your speed towards the bottom of the elevator shaft would go down a little bit.  But consider that the elevator falls tens of meters, while you jump about one meter.  Your jumping ability is quite small and probably won't make a noticeable difference&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-04T12:04:09.250" CommentCount="4" />
  <row Id="216" PostTypeId="2" ParentId="214" CreationDate="2010-11-04T12:05:19.450" Score="1" ViewCount="" Body="&lt;p&gt;No. You'd reduce your impact velocity, but only with a negligible small fraction.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-04T12:05:19.450" />
  <row Id="217" PostTypeId="1" AcceptedAnswerId="221" CreationDate="2010-11-04T12:10:22.390" Score="16" ViewCount="1121" Body="&lt;p&gt;We all know that there is no &quot;sucking&quot;, only pushing. So how are cooked spaghetti pushed into your mouth? The air pressure applies orthogonal on the spaghetti surface. Where does the component directed into your mouth come from?&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-11-15T14:01:46.207" LastActivityDate="2010-11-15T19:54:03.150" Title="Why can you &quot;suck in&quot; cooked spaghetti?" Tags="&lt;pressure&gt;&lt;air&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="4" />
  <row Id="218" PostTypeId="1" CreationDate="2010-11-04T12:20:38.137" Score="5" ViewCount="240" Body="&lt;p&gt;Similar to &lt;a href=&quot;http://physics.stackexchange.com/questions/160/what-are-good-mechanics-experiments-for-10-year-olds&quot;&gt;this question&lt;/a&gt;: what experiements do you think are impressive for layman adults?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I for example like how you can simulate an Einstein Ring with the bottom of a wine glass.&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-11-19T11:46:25.760" LastActivityDate="2010-11-19T11:46:25.760" Title="What are insightful/impressive experiments for adults" Tags="&lt;experiment&gt;&lt;home-experiment&gt;" AnswerCount="2" FavoriteCount="4" />
  <row Id="219" PostTypeId="2" ParentId="218" CreationDate="2010-11-04T13:00:51.143" Score="4" ViewCount="" Body="&lt;p&gt;it depends on the how much structure/money you have to &quot;invest&quot;. I always find a unvaluable amount of ideas watching the lessons of &lt;a href=&quot;http://web.mit.edu/physics/people/faculty/lewin_walter.html&quot; rel=&quot;nofollow&quot;&gt;W. Lewin from MIT&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for each lesson he present a very scenographic but simple experiment to prove with some evidence the theory..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My personal favorite il the pendulum used to prove the conservation of energy! very entertaining!&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-04T18:59:36.193" LastActivityDate="2010-11-04T18:59:36.193" CommentCount="1" />
  <row Id="220" PostTypeId="2" ParentId="217" CreationDate="2010-11-04T13:08:59.730" Score="7" ViewCount="" Body="&lt;p&gt;the component directed into your mouth comes form the different pressure between outside and inside the mouth!&#xA;If u create difference in pressure of $\Delta P$ the force pushing the spaghetti in will be $\Delta P \cdot S$ where $S$ is the section of the spaghetto (or spaghetti.. depending how hungry you are;) )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if for example u create inside your mouth a void with a 10% efficiecy the difference in pressure will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta P\approx 1atm - 0.1atm= 10^4N/m^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if the section of your single spaghetto is $1mm^2$ the force pushing the food in your mouth will be 10 N: almost a kilogram!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you won't even need to be as efficient as 10% ;)&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2010-11-04T13:08:59.730" CommentCount="4" />
  <row Id="221" PostTypeId="2" ParentId="217" CreationDate="2010-11-04T13:24:01.467" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Update/clarification:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you perform the sucking action, a pressure difference is clearly created and maintained by your lungs between the surrounding air and the air inside your mouth. An important point to notice here is that the mouth must not too far open (a bit lets it work still), else the pressure gradient between inside and outside of the mouth cannot be maintained by the expanding lungs over any significant time period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, observe that there is a normal force on the surface of the spaghetti due to (some arbitrary) air pressure. As I have just explained, there is a pressure gradient between the inside and outside of the mouth, directed inwards, hence there is a driving force inwards. Combining the two effects gives a net force directed at some angle to the surface of the spaghetti. Hence, it is an angled force, with some component pointing inwards towards the mouth, that pushes it in.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2010-11-04T22:50:16.853" LastActivityDate="2010-11-04T22:50:16.853" CommentCount="9" />
  <row Id="222" PostTypeId="1" CreationDate="2010-11-04T13:51:29.063" Score="7" ViewCount="510" Body="&lt;p&gt;What is some good &lt;strong&gt;free&lt;/strong&gt; software for doing physics calculations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm mainly interested in symbolic computation (something like Mathematica, but free).&lt;/p&gt;&#xA;" OwnerUserId="150" LastActivityDate="2011-06-25T19:00:34.353" CommunityOwnedDate="2010-12-24T23:45:29.253" Title="Software for physics calculations" Tags="&lt;software&gt;&lt;computation&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="3" />
  <row Id="223" PostTypeId="2" ParentId="222" CreationDate="2010-11-04T14:06:41.340" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.sagemath.org&quot; rel=&quot;nofollow&quot;&gt;Sage&lt;/a&gt; is a &lt;a href=&quot;http://python.org&quot; rel=&quot;nofollow&quot;&gt;Python&lt;/a&gt; based system (including &lt;a href=&quot;http://www.scipy.org&quot; rel=&quot;nofollow&quot;&gt;Numpy and Scipy&lt;/a&gt;) which includes a symbolic computation module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the Sage &lt;a href=&quot;http://www.sagemath.org&quot; rel=&quot;nofollow&quot;&gt;homepage&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sage is a free open-source mathematics software system licensed under the GPL. It combines the power of many existing open-source packages into a common Python-based interface.&#xA;  Mission: Creating a viable free open source alternative to Magma, Maple, Mathematica and Matlab. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="110" LastActivityDate="2010-11-04T14:06:41.340" CommunityOwnedDate="2010-12-24T23:45:29.253" CommentCount="3" />
  <row Id="224" PostTypeId="2" ParentId="222" CreationDate="2010-11-04T14:15:10.113" Score="14" ViewCount="" Body="&lt;p&gt;Some software I have used or has been recommended to me for physics-related work:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.wolframalpha.com&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha&lt;/a&gt; -- when I don't have Maple around, I use it for simple symbolic calculations&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://maxima.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;Maxima&lt;/a&gt; -- free open source alternative to Maple/Mathematica &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sagemath.org/&quot; rel=&quot;nofollow&quot;&gt;Sage&lt;/a&gt; -- quite an interesting open source symbolic/numerical package, you can try it online at &lt;a href=&quot;http://www.sagenb.org&quot; rel=&quot;nofollow&quot;&gt;sagenb.org&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.scilab.org/&quot; rel=&quot;nofollow&quot;&gt;Scilab&lt;/a&gt;/&lt;a href=&quot;http://www.gnu.org/software/octave/&quot; rel=&quot;nofollow&quot;&gt;GNU Octave&lt;/a&gt; -- alternatives to Matlab, best if you want to any numerical computations&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.r-project.org/&quot; rel=&quot;nofollow&quot;&gt;R&lt;/a&gt; -- powerful programming language used for statistical analysis&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.openoffice.org/product/calc.html&quot; rel=&quot;nofollow&quot;&gt;OpenOffice Calc&lt;/a&gt; -- I'll put it here for completeness as Excel and its variants seem to be the software of choice for most experimental calculations.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15" LastActivityDate="2010-11-04T14:15:10.113" CommunityOwnedDate="2010-12-24T23:45:29.253" CommentCount="3" />
  <row Id="225" PostTypeId="2" ParentId="214" CreationDate="2010-11-04T14:29:24.650" Score="8" ViewCount="" Body="&lt;p&gt;As an addition to already posted answers and while realising that experiments on Mythbusters don't really have the required rigour of physics experiments, the Mythbusters have &lt;a href=&quot;http://mythbustersresults.com/episode17&quot; rel=&quot;nofollow&quot;&gt;tested this theory&lt;/a&gt; and concluded that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The jumping power of a human being cannot cancel out the falling velocity of the elevator. The best speculative advice from an elevator expert would be to lie on the elevator floor instead of jumping. Adam and Jamie speculated the attendant survived because the tight elevator shaft created an air cushion. This together with spring action from slack elevator cable could have slowed the car to survivable speeds.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(This myth is fueled by the story of an elevator attendant found alive but badly injured in an elevator car that had fallen down a shaft in the Empire State Building after a B-25 Medium Bomber crashed into it in 1945.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15" LastActivityDate="2010-11-04T14:29:24.650" CommentCount="1" />
  <row Id="226" PostTypeId="2" ParentId="156" CreationDate="2010-11-04T14:49:41.693" Score="2" ViewCount="" Body="&lt;p&gt;The Solar wind flows past the Earth, interacting with the Earth's magnetic field, the magnetosphere. The boundary between the two, the magnetopause, is buffetted as the Solar wind isn't constant. Blobs of plasma can pass by, the upshot being that the magnetic fields get wound up, tangled if you like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been found that magnetic fields can reconnect, this magnetic reconnection, can release an amount of energy heating the plasma. This is flows along the magnetic fields lines into the polar regions, colliding with the atmosphere and causing the lovely aurore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been a while since I did my degree project on it. My memory may be fading, and I haven't kept up with geophysics for a while.&lt;/p&gt;&#xA;" OwnerUserId="89" LastActivityDate="2010-11-04T14:49:41.693" />
  <row Id="227" PostTypeId="2" ParentId="202" CreationDate="2010-11-04T15:07:41.383" Score="5" ViewCount="" Body="&lt;p&gt;This relates to one of the coolest examples ever - lenses and how you see.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very, very, roughly, when light interacts with a macroscopic object the result is the Fourier transform of the shape of the object, contained in the scattered light.   A lens basically computes the Fourier transform of some of the scattered light.  The Fourier transform is its own inverse, i.e. applied twice we get the identity.   So using the lens in your eye, your retina sees the shape of the object!   Same for cameras, obviously.  I'll try to find some references and edit later to include them.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-11-04T15:07:41.383" />
  <row Id="228" PostTypeId="1" CreationDate="2010-11-04T15:33:35.040" Score="1" ViewCount="167" Body="&lt;p&gt;I'm working on a graphic simulation (just for fun, for an open-source screensaver) of an &lt;a href=&quot;http://en.wikipedia.org/wiki/Aerotrim&quot; rel=&quot;nofollow&quot;&gt;Aerotrim&lt;/a&gt; - a &quot;human gyroscope&quot;, one of those exercise/training machines with a human in the middle, perched on a ring that swivels on an axis with respect to another ring, which swivels on an axis perpendicular to the first axis, on another ring. (AFAICT there could be 2 or 3 such axes that turn, each at right angles to the next. If I can simulate 2 axes / 3 rings, that would be sufficient. A general solution for n rings would be lovely. :-) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simulating the thing turning, with each axis rotating at a constant rate, is no problem. But it would &quot;feel&quot; more real if I could apply some realistic acceleration from the interaction between the rings and the mass of the rider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &lt;a href=&quot;http://www.youtube.com/results?search_query=aerotrim&amp;amp;page=&amp;amp;utm_source=opensearch&quot; rel=&quot;nofollow&quot;&gt;Youtube videos&lt;/a&gt;, you sometimes see a bystander push on one of the rings, to help get the rider going. Not only does that ring accelerate, but others start spinning too. And the pushed ring does not accelerate smoothly, but undergoes resistance from the other rings, apparently transferring acceleration to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not up on angular momentum, torque, and all that, so answers will need to take my ignorance into account, though I'm obviously willing to learn a few things as necessary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't feel like it's necessary to take into account the asymmetry of the person's mass - modeling the person as a point mass in the center of the rings should be fine, I think. In other words I don't plan on modeling the way that the rider accelerates the rings by leaning one way or another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking that in order to make things a little more interesting, I would have the program occasionally apply a force like a hand pushing on one of the rings. I can figure out what the torque on that ring would be if it were independent of the others, but I don't know to model a set of 3 rings connected on axes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts on how I could model the interaction between the rings, each affecting the others? Simplifications are fine. My intent is not to discover new phenomena through accurate physical modeling, but to create a visual display that &lt;em&gt;looks&lt;/em&gt; more realistic than just rings turning at constant rates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Part of my uncertainty is, how much does the Aerotrim really behave like a gyroscope? The gyroscope's main properties are focused on the mass in the middle spinning fast on its axis, and therefore its axis tends not to move. This is obviously different from the aerotrim, where the person is often not rotating much around the inner axis, and the inner axis is certainly not staying still. I wonder if other factors, like the conservation of angular momentum of the &lt;em&gt;outer&lt;/em&gt; rings, plays more of a part in the Aerotrim that is negligible with gyroscopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it &lt;em&gt;is&lt;/em&gt; essentially just a gyroscope... I've looked at the &quot;fundamental equation&quot; at &lt;a href=&quot;http://en.wikipedia.org/wiki/Gyroscope#Properties&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Gyroscope#Properties&lt;/a&gt;, but as a non-physicist I find it difficult to imagine how to turn that into code computing what happens when a force is applied. Pointers to example code for simulation of gyroscopes would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lars&lt;/p&gt;&#xA;" OwnerUserId="70" LastEditorUserId="70" LastEditorDisplayName="" LastEditDate="2010-11-04T15:52:45.780" LastActivityDate="2010-11-04T20:16:57.623" Title="acceleration of rings in aerotrim (human gyroscope)" Tags="&lt;acceleration&gt;&lt;torque&gt;&lt;angular-momentum&gt;&lt;gyroscopes&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="229" PostTypeId="2" ParentId="101" CreationDate="2010-11-04T16:23:27.707" Score="7" ViewCount="" Body="&lt;p&gt;The only way to produce antiprotons would be to collide some particles together and then filter out just the antiprotons (as there are no stable particles that we know of containing antiprotons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fermilab produces the protons and antiprotons in a couple of stages. It starts with negatively ionised hydrogen and accelerates it using a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cockcroft%E2%80%93Walton_generator&quot; rel=&quot;nofollow&quot;&gt;Cockcroft-Walton&lt;/a&gt; accelerator to 750keV, by basically cycling the ions through a potential difference multiple times. These ions are then fed into a linear accelerator that uses an oscillating electroc field to accelerate them to 400 MeV. The electrons are then stripped from the ions using a carbon foil, ending up with 400MeV protons. These then go into the Booster which accelerates them to 8GeV. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the proton production stage and following this, the 8GeV protons are directed to the first ring of the Tevatron (the Main Injector) where they can be accelerated up to 150GeV. Once the protons reach 120GeV, some are collided with a nickel target, producing a shower of particles. The antiprotons are filtered and directed back to the Main Injector, which accelerates both the protons and antiprotons to 150GeV. The two beams are now fed into the main accelerator (the Tevatron) which accelerates them to 1TeV and crosses them at the detector site, producing the shower of particles that is detected and analysed :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.fnal.gov/pub/inquiring/physics/accelerators/chainaccel.html&quot; rel=&quot;nofollow&quot;&gt;Fermilab's accelerators&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.fnal.gov/pub/science/accelerator/&quot; rel=&quot;nofollow&quot;&gt;Component details&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15" LastEditorUserId="15" LastEditorDisplayName="" LastEditDate="2010-11-04T16:35:25.650" LastActivityDate="2010-11-04T16:35:25.650" CommentCount="3" />
  <row Id="230" PostTypeId="2" ParentId="202" CreationDate="2010-11-04T16:42:56.810" Score="13" ViewCount="" Body="&lt;p&gt;I see that two examples in optics have been mentioned, a diffraction grating by Mark Eichenlaub, and a lens by sigoldberg1. I would like to elaborate a bit, because there is a subtle difference between the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the one hand, a diffraction grating separates out light of different frequencies, i.e. colors, transforming them into different positions. This is analogous to how the 1-D Fourier transform works on a sound or electrical signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, a lens takes the 2-D Fourier transform of a &lt;em&gt;monochromatic&lt;/em&gt; light beam. Monochromatic means there is only one frequency or color. You might wonder why the Fourier transform isn't a single peak, if there's only one frequency in the light. That's because a lens transforms &lt;em&gt;spatial frequencies&lt;/em&gt; (roughly equivalent to angles, but I will explain them below) into positions. Mathematically, this is the same Fourier transform, but it transforms an entirely different property of the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It took me the longest time to understand exactly what spatial frequencies were. I finally got it one day, reading chapter 4, &quot;Fourier Optics&quot;, of Saleh and Teich's &lt;em&gt;Fundamentals of Photonics&lt;/em&gt;, which I whole-heartedly recommend if you don't understand my explanation here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You probably learned in your class that you can write any signal as a Fourier series, which is the sum of frequency components, each with their own amplitude and phase. The Fourier transform is a sort of continuous version of that. Well, you can express any monochromatic light beam as the sum of many plane waves traveling at different angles, all with the same frequency, but each with their own amplitude and phase. Is this starting to sound familiar? Each of these plane waves is a &lt;em&gt;spatial frequency&lt;/em&gt;. Just as with the step from Fourier series to Fourier transform, you can make the step from discrete spatial frequencies to 2-D Fourier transform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A positive lens focuses each of these spatial frequencies to a separate point. For example, here are illustrations of two different plane waves being focused by a lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/VqqtF.png&quot; alt=&quot;Plane wave hitting a lens face on&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a plane wave hits a lens face on, the lens focuses it to a point on the optical axis, at the focal distance of the lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/dYKxl.png&quot; alt=&quot;Plane wave hitting a lens at an angle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when the plane wave is incident at an angle, the focus is still at the focal distance, but displaced from the optical axis. This is how a lens transforms &quot;angle&quot; to position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/MgbXE.png&quot; alt=&quot;1-D Fourier transform using a lens&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you have a more complicated beam made up of many spatial frequencies (which, as I said, are just plane waves), they are all focused to separate points in the same plane on the other side of the lens, one focal distance away. This is why we say it computes the 2-D Fourier transform of a monochromatic light beam. In fact, in the 1970s, when computers weren't so fast, people actually experimented with using lenses to calculate Fourier transforms instantaneously!&lt;/p&gt;&#xA;" OwnerUserId="107" LastEditorUserId="107" LastEditorDisplayName="" LastEditDate="2010-11-04T17:39:30.500" LastActivityDate="2010-11-04T17:39:30.500" CommentCount="4" />
  <row Id="231" PostTypeId="2" ParentId="193" CreationDate="2010-11-04T19:06:47.060" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://groups.csail.mit.edu/mac/users/wisdom/AIM-2005-003.pdf&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; you have a book from Sussman and Wisdom from MIT about functional differential geometry covering your first query about &quot;manifolds&quot;.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-11-04T19:06:47.060" CommunityOwnedDate="2010-11-10T18:20:54.740" />
  <row Id="232" PostTypeId="2" ParentId="228" CreationDate="2010-11-04T20:16:57.623" Score="1" ViewCount="" Body="&lt;p&gt;Partial answer, but still. Aerotrim has almost nothing to do with gyroscopes, except that the gyroes are usually mounted in gimbal set similar to aerotrim. &lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-04T20:16:57.623" CommentCount="2" />
  <row Id="233" PostTypeId="2" ParentId="94" CreationDate="2010-11-04T20:18:46.267" Score="5" ViewCount="" Body="&lt;p&gt;The usual way linear polarisation is measured is by shining polarised light onto a polarising filter, rotating that filter and then using &lt;a href=&quot;http://en.wikipedia.org/wiki/Polarizer#Malus.27_law_and_other_properties&quot; rel=&quot;nofollow&quot;&gt;Malus' law&lt;/a&gt; to fit the data to a $I_0 cos^2(\theta_{beam} - \theta_{polariser})$ shape. By finding the angular position of the intensity peak we can infer the angle of polarisation of the incoming beam. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, assume we shine a beam of a certain spread in the transverse plane, having different polarisations everywhere. If we shine this beam onto a polariser, we will get a pattern of intensities $I(x, y) = I_0 cos^2(\theta_{beam}(x, y) - \theta_{polariser})$. We can find the values of $\theta_{beam}(x, y)$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Calibration -- rotate $\theta_{polariser}$ and calculate $\max_{x, y} (\theta_{beam}(x, y))$, the peak intensity. The maximum of the peak intensity over all values of $\theta_{polariser}$ will give you $I_0$. &lt;/li&gt;&#xA;&lt;li&gt;Fix $\theta_{polariser}$ to the value that gives the maximum peak intensity. Now you know that you have aligned your polariser with one of the polarisations present in the beam. Then, the ratio $I(x,y) / I_0$ will give you $cos^2(\theta_{beam}(x, y) - \theta_{polariser})$, from which $\theta_{beam}(x, y)$ can be inferred.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This method works for polarisation that is constant in time. For other types of polarisation, you can always take a snapshot over a short period of time that has roughly constant polarisation. Or, if the polarisation is circular, you can use a &lt;a href=&quot;http://www.rp-photonics.com/waveplates.html&quot; rel=&quot;nofollow&quot;&gt;half waveplate&lt;/a&gt; to convert it to linear.&lt;/p&gt;&#xA;" OwnerUserId="15" LastActivityDate="2010-11-04T20:18:46.267" CommentCount="1" />
  <row Id="234" PostTypeId="1" CreationDate="2010-11-04T20:47:05.523" Score="4" ViewCount="693" Body="&lt;p&gt;&lt;strong&gt;Note: I will expand this question with more specific points when I have my own internet connection and more time (we're moving in, so I'm at a friend's house).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is broad, involved, and to some degree subjective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I started out as a physics-only student, but eventually decided to add a mathematics major. I am greatly interested in mathematics; the typical curriculum required for physics students is not deep or thorough enough; mathematics is more general (that means work!); and it only requires a few more classes. Naturally, I enjoy &lt;a href=&quot;http://math.stackexchange.com/&quot;&gt;mathematics&lt;/a&gt; immensely.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question asks mainly of undergraduate-level study, but feel free to discuss graduate-level study if you like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please &lt;em&gt;do not&lt;/em&gt; rush your answer or try to be comprehensive. I realize the StackOverflow model rewards quick answers, but I would rather wait for a thoughtful, thorough (on a point) answer than get a fast, cluttered one. (As you probably know, &lt;a href=&quot;http://www.bethanyroberts.com/writing_quotes.htm#on%20revision&quot; rel=&quot;nofollow&quot;&gt;revision&lt;/a&gt; produces clear, useful writing; and a properly-done comprehensive answer would take more than a reasonable amount of time and effort.) If you think an overview is necessary, that is fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a question this large, I think the best thing to do is focus on a specific area in each answer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; To Sklivvz, Cedric, Noldorin and everyone else: I had to run off before I could finish, but I wanted to say I knew I would regret this; I was cranky and not thinking clearly, mainly from not eating enough during the day. I am sorry for my sharp responses and for not waiting for my reaction to pass. I apologize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Re: Curricula:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that I am not asking about choosing your own curriculum in college or university. I did not explicitly say that, but several people believed that was my meaning. I will ask more specific questions later, but the main idea is how a physics student should study mathematics (on his or her own, but also by choosing courses if available) to be a competent mathematician with a view to studying physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I merely mentioned adding a mathematics major to illustrate my conclusion that physics student need a deeper mathematical grounding than they typically receive.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now I have to run off again.&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-01T01:49:26.937" LastActivityDate="2011-02-01T01:49:26.937" Title="How should a physics student study mathematics?" Tags="&lt;mathematics&gt;&lt;education&gt;" AnswerCount="3" CommentCount="24" FavoriteCount="7" />
  <row Id="235" PostTypeId="2" ParentId="193" CreationDate="2010-11-04T22:07:04.713" Score="5" ViewCount="" Body="&lt;p&gt;Sean Carroll's &lt;a href=&quot;http://preposterousuniverse.com/grnotes/&quot; rel=&quot;nofollow&quot;&gt;Lecture Notes on General Relativity&lt;/a&gt; contain a superb introduction to the mathematics of GR (differential geometry on Riemann manifolds).  These also also published in modified form in his book, Spacetime and Geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spivak's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0805390219&quot; rel=&quot;nofollow&quot;&gt;Calculus on Manifolds&lt;/a&gt; is a gem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bishop's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486640396&quot; rel=&quot;nofollow&quot;&gt;Tensor Analysis on Manifolds&lt;/a&gt; is a great introduction to the subject, and published by Dover, is very cheap (less than $10 on amazon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Georgi's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0738202339&quot; rel=&quot;nofollow&quot;&gt;Lie Algebras In Particle Physics&lt;/a&gt; is enjoyable and fast-paced, but probably skips around too much to be used as an adequate first exposure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shutz's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521298873&quot; rel=&quot;nofollow&quot;&gt;Geometrical Methods of Mathematical Physics&lt;/a&gt; and &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521887054&quot; rel=&quot;nofollow&quot;&gt;A First Course in General Relativity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite it's incredibly pompous title, Penrose's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0679776311&quot; rel=&quot;nofollow&quot;&gt;The Road to Reality: A Complete Guide to the Laws of the Universe&lt;/a&gt; provides an enjoyable high-level view of a vast expanse of mathematical physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned by Cedric, I am a huge fan of Sussman and Wisdom's Structure and Interpretation of Classical Mechanics and the associated Functional Differential Geometry memo.  The citations in those publications will also point to towards a lot of good material and there's more goodies if you dig around in the source code.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2010-11-04T22:07:04.713" CommunityOwnedDate="2010-11-10T18:20:54.740" CommentCount="1" />
  <row Id="236" PostTypeId="1" AcceptedAnswerId="242" CreationDate="2010-11-04T22:14:19.137" Score="8" ViewCount="316" Body="&lt;p&gt;I've just recently heard of &lt;a href=&quot;http://en.wikipedia.org/wiki/Positronium&quot; rel=&quot;nofollow&quot;&gt;Positronium&lt;/a&gt;, an &quot;element&quot; with interesting properties formed by an electron and positron, and I was shocked to hear that physicists were actually working with this element, even if for a very short lifetime. I was always under the impression that matter and antimatter annihilated when they came even remotely close to each other, which is apparently not the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do these two particles combine to form an element if they're oppositely charged and roughly the same mass? What kind of interactions could possibly take place before they're pulled together and annihilated?&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2010-11-06T15:49:27.317" Title="How does Positronium exist?" Tags="&lt;nuclear-physics&gt;" AnswerCount="4" />
  <row Id="237" PostTypeId="1" AcceptedAnswerId="255" CreationDate="2010-11-04T22:18:25.970" Score="5" ViewCount="422" Body="&lt;p&gt;I have an approximately basketball-sized non-hollow piece of aluminum sitting in my house that is of irregular shape.  I need to find the volume of it for a very legitimate yet irrelevant reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the best way I can do this?  In fact, what are &lt;strong&gt;all&lt;/strong&gt; the ways I could feasibly do this without going to a lab? (I don't live near any labs)&lt;/p&gt;&#xA;" OwnerUserId="71" LastActivityDate="2011-05-18T01:04:25.920" Title="Finding the volume of this irregular shape I have" Tags="&lt;mass&gt;&lt;measurement&gt;&lt;volume&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="238" PostTypeId="2" ParentId="160" CreationDate="2010-11-04T22:21:28.517" Score="2" ViewCount="" Body="&lt;p&gt;I remember enjoying the &lt;a href=&quot;http://meritbadge.org/wiki/index.php/Scientist&quot; rel=&quot;nofollow&quot;&gt;Webelos Scientist Activity Badge&lt;/a&gt; (part of the Boy Scouts program applicable to 10-year-olds).  (Here's a &lt;a href=&quot;http://www.boyscouttrail.com/external_frame.asp?goto=/webelos/w2/otherresources/webelos-scientist-ideas.htm&quot; rel=&quot;nofollow&quot;&gt;list of ideas&lt;/a&gt;.)  Don't need to get too fancy, and a lot of these make nice little projects, i.e. 'build a hot-air balloon' is a great problem-solving exercise (think about what shapes you can cut out from tissue-paper and glue together to make a sphere; hold over a toaster to fill it with hot air).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water rockets, model airplanes/gliders (even paper), potato cannon are also fun activities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;How do airplanes fly&quot; is a good question to think about, and refuting popular but &lt;a href=&quot;http://www.grc.nasa.gov/WWW/K-12/airplane/wrong1.html&quot; rel=&quot;nofollow&quot;&gt;incorrect&lt;/a&gt; ideas is a great critical thinking exercise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this age, it's probably more important to make science fun than to be totally pedantic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of good conceptual points to ponder that could provide great topics for discussion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do tides work?  Why are there high tides simultaneously on opposite sides of the earth? (Hint: the earth doesn't &quot;shield&quot; gravity!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do rockets stay in orbit? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does a mirror reverse left and right but not up and down?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does it take to balance a see-saw?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make a Mobius strip.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2010-11-04T22:21:28.517" CommunityOwnedDate="2010-11-10T18:22:11.807" />
  <row Id="239" PostTypeId="1" AcceptedAnswerId="250" CreationDate="2010-11-04T22:21:57.830" Score="4" ViewCount="413" Body="&lt;p&gt;I used to think that the shape of an object determines its ability to float (boat-shaped objects are more likely to float, and spheres tend to sink).  But my friend, who is fond of making me look stupid, took me to the local lake showed me a sphere that floated and a boat-shaped object made out of iron that sunk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it based on the mass?  I'm not sure that is possible, because I've seen really heavy things (like airplane carriers) float, and really light things (like my friend's iron boat) sink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What property of certain objects allow them to float, if any?&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="172" LastEditorDisplayName="" LastEditDate="2010-11-08T03:24:52.660" LastActivityDate="2010-11-08T03:24:52.660" Title="What property of objects allow them to float?" Tags="&lt;buoyancy&gt;" AnswerCount="7" CommentCount="3" FavoriteCount="1" />
  <row Id="240" PostTypeId="2" ParentId="236" CreationDate="2010-11-04T22:22:22.040" Score="5" ViewCount="" Body="&lt;p&gt;That's because they are oppositely charged that they can form a bound state: even classically you can understand that: oppositely charged charges attract each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it is true that a particle and its antiparticle can annihilate each other, they first have to interact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Positronium is a purely electromagnetic bound state: the positron and the electron will form a bound state by electromagnetic interaction (no strong interaction as they are leptons, and the weak interaction does not play a role to form the bound state).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have the same mass, but it is not a real problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum mechanically this problem is treated exactly the same way as the textbook example of the hydrogen atom. You first separate the centre of mass from the problem, but here as they have the same mass this cannot be neglected in the final result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you calculate the interaction of one particle with the centre of mass (in the case of the H atom, this is unambiguously the interaction of the electron with the proton, but here it is one of the two lepton with the centre of mass which is in the middle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should also be noted that even if the bound state is stable from that point of view, the positronium will eventually annihilate because the two wave function will overlap and thus these two anti-particles can interact and annihilate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Positronium can be formed in a variety of ways, one example, where you can create positronium in your bathroom is to have an element which is $\beta^+$ unstable. After this decay, a positron is emitted. It can then interact with the very large number of electron present in the matter and they can form a bound state: the positronium.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-11-04T22:22:22.040" />
  <row Id="241" PostTypeId="2" ParentId="237" CreationDate="2010-11-04T22:23:58.937" Score="7" ViewCount="" Body="&lt;p&gt;Fill a vessel(that can accommodate the irregular shape) with water brim to brim.&#xA;And put that shape into gently into the vessel.Measure the volume of water spilled.&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-11-04T22:23:58.937" CommentCount="1" />
  <row Id="242" PostTypeId="2" ParentId="236" CreationDate="2010-11-04T22:30:19.533" Score="9" ViewCount="" Body="&lt;p&gt;As you've noticed, it's not automatically true that a particle and its antiparticle will annihilate each other when they get close to each other. In fact, no interaction between particles is really certain to happen. Quantum mechanics (and at a higher level, quantum field theory) tells you that all these interactions happen with certain probabilities. So for instance, when a particle and its antiparticle come into close proximity, there is only a chance that they will interact within any given amount of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the longer the particles remain together, the greater the probability that they will interact and annihilate each other. This is responsible for the 142 ns lifetime of positronium as reported in the Wikipedia article: the probability of annihilation increases with time in such a way that the &lt;em&gt;average&lt;/em&gt; lifetime of an &quot;atom&quot; of positronium is 142 ns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Cedric said, as long as the positron and electron don't annihilate each other (and remember, there is only a limited chance of that happening in any given time), they can interact in much the same way as any other charged particles, such as the proton and electron. Being bound together by the electromagnetic interaction, as in a hydrogen atom or a positronium &quot;atom,&quot; is just one example.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-04T22:30:19.533" CommentCount="4" />
  <row Id="243" PostTypeId="2" ParentId="239" CreationDate="2010-11-04T22:33:37.973" Score="2" ViewCount="" Body="&lt;p&gt;The only parameter that matters is the density: the ratio between the weight and the volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the density is higher than the surrounding medium (let's say water, whose density 1000 kg/m³) is will sink and it will float if the density is lower (you have the special where they are both equal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objects float due to Archimedes' principle: basically, and under some hypothesis, there is a force, directed upward, equal in magnitude to the weight of the volume of displaced water (the volume of water that the immersed part of the body occupies). This weight is equal to g * V * density . The other force acting on the object is its own weight: g * V * density object .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then recover the rule about densities.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-11-04T22:33:37.973" />
  <row Id="244" PostTypeId="2" ParentId="239" CreationDate="2010-11-04T22:34:01.827" Score="1" ViewCount="" Body="&lt;p&gt;In the most common cases, it's based on density. An object that is denser than the liquid will sink, and one that is less dense will float.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To go into a little more detail: when an object is placed in a liquid, the liquid exerts an upward force on it, called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Buoyancy&quot; rel=&quot;nofollow&quot;&gt;buoyant force&lt;/a&gt;. This force is equal in magnitude to the weight of the liquid displaced by the object - that is, the amount of liquid that would fill the submerged volume of the object. (I can provide more mathematical detail if you like, or just look at the Wikipedia article linked)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose the object is completely submerged.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If the object is denser than the liquid, the weight of the liquid that would fill the object's volume will be less than the weight of the object. So the buoyant force, acting upward, is not strong enough to counteract the object's weight, and the object sinks.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the liquid is denser than the object, the weight of the liquid that would fill the object's volume is more than the weight of the object, and the buoyant force is strong enough to overpower the weight, lifting the object up. But as the object rises, eventually it starts to rise out of the liquid, its submerged volume becomes less, and in turn the buoyant force becomes less. At some point, the buoyant force will drop to be equal to the weight of the object, and then the object will stop rising - it'll just float there, at that level.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-04T22:42:05.417" LastActivityDate="2010-11-04T22:42:05.417" CommentCount="3" />
  <row Id="245" PostTypeId="2" ParentId="239" CreationDate="2010-11-04T22:39:22.437" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;DENSITY&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is because of densities of the object that is floating and the liquid in which it is floating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an object have density lower than a fluid it will float otherwise it will sink.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Density of entire object [mass / volume] should be&#xA;  taken into account and not merely the&#xA;  density of material it is made up of.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A ship made up of iron floats in sea because density of ship i.e. mass of ship/ volume of ship is less than that of water. Here though density of iron is more than water hollowness of ship makes its volume large hence &lt;strong&gt;density of &quot;ship&quot;&lt;/strong&gt; is lower than that of water. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In dead sea you and I can float.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In mercury iron nails float.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-11-04T22:45:03.717" LastActivityDate="2010-11-04T22:45:03.717" CommentCount="2" />
  <row Id="246" PostTypeId="2" ParentId="222" CreationDate="2010-11-04T22:40:58.243" Score="6" ViewCount="" Body="&lt;p&gt;It is probably worth your while to buy Mathematica, Maple, or Matlab, depending on your needs.  I wish it weren't so, but this is one area in which the commercial tools are still vastly better than their free counterparts.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are a student, you can buy these at fairly afforable prices.  &lt;a href=&quot;https://webstore.maplesoft.com/product.aspx?id=246&quot; rel=&quot;nofollow&quot;&gt;Maple 14 Student Edition&lt;/a&gt; is only $99. &lt;a href=&quot;http://www.wolfram.com/products/student/mathforstudents/licenses.html&quot; rel=&quot;nofollow&quot;&gt;Mathematica for Students&lt;/a&gt; is \$140, and Matlab/Simulink is \$99 for students.  It is also possible that your school or department already has a site license, allowing you to obtain and use this software for no additional cost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For symbolic calculations, you want either Mathematica or Maple, with Maple being more user-friendly, and Mathematica being more prevalent (in my experience) in actual research environments.  Matlab's focus is on numerical calculations.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2010-11-04T22:40:58.243" CommunityOwnedDate="2010-12-24T23:45:29.253" CommentCount="1" />
  <row Id="247" PostTypeId="2" ParentId="239" CreationDate="2010-11-04T22:49:47.763" Score="1" ViewCount="" Body="&lt;p&gt;An object floats when it receives enough push from the water to compensate its weight. The push is equal to the weight of the water displaced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words: objects with a particularly good shape for moving water away (like boat-shaped objects) are allowed to weigh more and still float; objects with a bad shape (like spheres) are allowed to weigh less and still float.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-04T22:49:47.763" />
  <row Id="248" PostTypeId="2" ParentId="239" CreationDate="2010-11-04T22:53:07.197" Score="1" ViewCount="" Body="&lt;p&gt;To solve any mechanical problem like this, you need only consider the forces acting on an object.  In this case, there are two relevant forces on the boat: its weight (pulling it down), and the buoyant force (pushing it up).  The buoyant force on a boat is equal to the &lt;em&gt;weight of the water it displaces&lt;/em&gt; (see: &lt;a href=&quot;http://en.wikipedia.org/wiki/Buoyancy&quot; rel=&quot;nofollow&quot;&gt;Archimedes principle&lt;/a&gt;).  As the boat sinks into the water further, more and more water is displaced, until the weight of the water displaced equals the weight of the boat and equilibrium is achieved, and the boat floats at that level.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2010-11-04T22:53:07.197" />
  <row Id="249" PostTypeId="2" ParentId="239" CreationDate="2010-11-04T22:54:43.597" Score="3" ViewCount="" Body="&lt;p&gt;An object floats if its upthrust (buoyancy) is in equilibrium with its downwards gravitational force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words (as stated by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Upthrust&quot; rel=&quot;nofollow&quot;&gt;wiki page&lt;/a&gt;),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_net = 0 = mg - \rho_f V_{disp} g$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where all the constants are pretty self-explanatory.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly then, the properties of the object that determine whether/how it floats are its mass and volume. More specifically, it is the relationship between the two; the density of the object (considering the &lt;em&gt;enclosed&lt;/em&gt; volume, i.e. that which water can't enter), must be lower than the density of water $\rho_f$ for it to float.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that is the &lt;em&gt;density function&lt;/em&gt; (how the density of the object varies spatially) that determines what  and the &lt;em&gt;shape&lt;/em&gt; of the object that determines exactly how the object floats (e.g. the water line). This is not necessary just to determine if it floats, however.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2010-11-04T22:59:53.410" LastActivityDate="2010-11-04T22:59:53.410" />
  <row Id="250" PostTypeId="2" ParentId="239" CreationDate="2010-11-04T22:59:21.527" Score="5" ViewCount="" Body="&lt;p&gt;Actually, the answer is a bit more subtle than just density. The principle that is behind floating objects is &lt;a href=&quot;http://en.wikipedia.org/wiki/Buoyancy#Archimedes.27_principle&quot; rel=&quot;nofollow&quot;&gt;Archimedes' principle&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A fluid (liquid or gas) exerts a buoyant force, opposite apparent gravity (i.e. gravity + acceleration of fluid) on an immersed object that is equal to the weight of the displaced fluid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, if you have an object fully immersed in a fluid, the total force it feels is given by (positive sign means down):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$F = gravity + buoyancy = \rho_{object} V g - \rho_{fluid} V g = (\rho_{object} - \rho_{fluid}) V g$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, if the average density of the object is lower than that of the water, it floats. If the object is partially immersed, to calculate the buoyant force you have to consider just the immersed volume and its average density: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$F = \rho_{object} V g -  \rho_{fluid} V_{immersed} g$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that when I was talking about density, I was talking about the &lt;em&gt;average density&lt;/em&gt; of the object. That is its &lt;em&gt;total&lt;/em&gt; mass divided by its volume. Thus, a ship, even if it is made out of high-density iron it is full of air. That air will lower the average density, as it will increase the volume considerably while keeping the weight almost constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to understand this better you can give the following problem a try :) &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the height an ice cube of side L floats in water?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15" LastActivityDate="2010-11-04T22:59:21.527" />
  <row Id="251" PostTypeId="2" ParentId="202" CreationDate="2010-11-04T23:11:24.147" Score="6" ViewCount="" Body="&lt;p&gt;Remember the &lt;a href=&quot;http://en.wikipedia.org/wiki/Double-slit_experiment&quot; rel=&quot;nofollow&quot;&gt;double slit experiment&lt;/a&gt;?  &lt;em&gt;The interference pattern is the Fourier transform of the hole(s)&lt;/em&gt;.  This boggled my mind when I first learned it.  In the limit where the screen is far from the mask, the rays of light actually &lt;em&gt;physically compute the Fourier transform&lt;/em&gt; (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Fraunhofer_diffraction&quot; rel=&quot;nofollow&quot;&gt;Fraunhofer diffraction&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2010-11-04T23:11:24.147" CommentCount="2" />
  <row Id="252" PostTypeId="1" AcceptedAnswerId="264" CreationDate="2010-11-04T23:17:38.907" Score="7" ViewCount="426" Body="&lt;p&gt;(This is a simple question, with likely a rather involved answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What are the primary obstacles to solve the many-body problem in quantum mechanics?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, if we have a Hamiltonian for a number of interdependent particles, why is solving for the time-independent wavefunction so hard? Is the problem essentially just mathematical, or are there physical issues too? The many-body problem of Newtonian mechanics (for example gravitational bodies) seems to be very difficult, with no solution for $n &gt; 3$. Is the quantum mechanical case easier or more difficult, or both in some respects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In relation to this, what sort of approximations/approaches are typically used to solve a system composed of many bodies in arbitrary states? (We do of course have perturbation theory which is sometimes useful, though not in the case of high coupling/interaction. Density functional theory, for example, applies well to solids, but what about arbitrary systems?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, is it theoretically and/or practically impossible to simulate high-order phenomena such as chemical reactions and biological functions precisely using Schrodinger's quantum mechanics, over even QFT (quantum field theory)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: this question is largely intended for seeding, though I'm curious about answers beyond what I already know too!)&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2010-11-04T23:33:00.533" LastActivityDate="2010-11-29T13:41:41.207" Title="The Many Body problem" Tags="&lt;quantum-mechanics&gt;&lt;schrodinger-equation&gt;&lt;many-body&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="253" PostTypeId="2" ParentId="68" CreationDate="2010-11-04T23:18:13.837" Score="4" ViewCount="" Body="&lt;p&gt;More likely that the stress induced strain, i.e. stretching of the tape.   When the tape stretches, the polymer molecules in it tend to orient or also stretch.   This leads to a third layer of polarizing material in the middle, or even birefringence, which makes the colors between the crossed polarizing filters you saw on the screen.    At least that's the way I remember it.&lt;br&gt;&#xA;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Birefringence&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Birefringence&lt;/a&gt; , specifically the subsection on stress induced birefringence    See also &lt;a href=&quot;http://www.youtube.com/watch?v=BEClYQbuG7U&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=BEClYQbuG7U&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-11-05T05:20:03.160" LastActivityDate="2010-11-05T05:20:03.160" CommentCount="1" />
  <row Id="254" PostTypeId="2" ParentId="160" CreationDate="2010-11-04T23:24:09.220" Score="0" ViewCount="" Body="&lt;p&gt;Pythagorus's discovery on the length of strings, tension weights, sounds made.   Not too dynamic, however.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-11-04T23:24:09.220" CommunityOwnedDate="2010-11-10T18:22:11.807" />
  <row Id="255" PostTypeId="2" ParentId="237" CreationDate="2010-11-04T23:31:22.353" Score="15" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://hubpages.com/hub/The-origin-of-Eureka&quot; rel=&quot;nofollow&quot;&gt;Eureka!&lt;/a&gt; As Archimedes said, according to legend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, &quot;TheMachineCharmer's&quot; answer is feasible, but I would recommend &lt;strong&gt;recording the change in the volume of water&lt;/strong&gt; instead (if you need an accurate measurement), because (1) it could be difficult to measure the volume of the spilled water, and (2) it is also a little less accurate to do so. (Some water will be left on the sides of the first container, and inserting the object into the filled container, catching all of the spillover while making sure the water level doesn't drop below the brim, could be difficult.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only need a rough idea, the other way is fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need a more accurate measurement, try one of these:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Get the volume of a container by filling it with water (e.g. from a graduated vessel). Empty the container, then place the object in it. Fill it with water again, measuring how much water you added. Subtract this number from the volume of the container to find the volume of your object. (The order in which you do these doesn't matter, of course.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This one could be a little harder, because you need a large graduated vessel. Fill the empty vessel with roughly enough water to submerge the object. Put the object in and record the change in the water volume.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-11-04T23:31:22.353" CommentCount="2" />
  <row Id="256" PostTypeId="2" ParentId="160" CreationDate="2010-11-04T23:33:01.890" Score="1" ViewCount="" Body="&lt;p&gt;Tell a story.   Start say with the Aristotelian notion of why an arrow flies, you recall, wind circulating from the tip to the tail or something.   Then try to come up with experiments as to why that might be false, e.g. slow movements, movements in space, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at crystals.  Grow some.   Weird, isn't it?   How could that be?    Atomic theory.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-11-04T23:38:38.867" LastActivityDate="2010-11-04T23:38:38.867" CommunityOwnedDate="2010-11-10T18:22:11.807" />
  <row Id="257" PostTypeId="1" AcceptedAnswerId="283" CreationDate="2010-11-04T23:42:30.020" Score="5" ViewCount="180" Body="&lt;p&gt;Particle colliders like the LHC or the Tevatron use a complex accelerator chain to have particles at a given energy before being accelerated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://public.web.cern.ch/public/en/Research/AccelComplex-en.html&quot; rel=&quot;nofollow&quot;&gt;The CERN accelerator complex to inject in the LHC&lt;/a&gt;: protons are first accelerated by a LINAC, then by a &quot;booster&quot; synchrotron, then by a larger synchrotron, and finally by a very large synchrotron where they reach 450 GeV/c.&lt;/li&gt;&#xA;&lt;li&gt;As was discussed in &lt;a href=&quot;http://physics.stackexchange.com/questions/101/production-of-antiproton-at-the-tevatron/229#229&quot;&gt;link text&lt;/a&gt; , the &lt;a href=&quot;http://www.fnal.gov/pub/inquiring/physics/accelerators/chainaccel.html&quot; rel=&quot;nofollow&quot;&gt;chain for the Tevatron&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They both require &quot;injector&quot; accelerator(s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/PzikJ.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is simple: why ? Why can't we have a single accelerator with a source and the possibility to accelerate the particles to the top energy.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note: this is also a seeding question, but I am curious how complete and clear an explanation to that can be provided.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-05T11:27:46.133" LastActivityDate="2010-11-11T15:59:10.073" Title="Particle colliders: why do they need an accelerator chain" Tags="&lt;accelerator-physics&gt;&lt;large-hadron-collider&gt;&lt;tevatron&gt;&lt;cern&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="258" PostTypeId="1" AcceptedAnswerId="259" CreationDate="2010-11-05T00:28:45.073" Score="5" ViewCount="187" Body="&lt;p&gt;Dominoes, when placed upright, remain that way.  Sometimes, even if you tip them a little bit, they will go back to their upright position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you tip them too far, they will fall over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After trying this with many different sized/shaped dominoes and some textbooks, I've noticed that this angle of &quot;maximum tippage&quot; varies for each one, depending on its dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taller dominoes seem to have a lower maximum tippage angle.  Dominoes with wider bases seem to have a higher maximum tippage angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are other factors are involved?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to compute this maximum tippage angle, given the height of the domino, and the width of its base, and any other factors that might be involved?  If so, what is this relationship (mathematically)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Dominoes start acting weirdly when their weight/density is unevenly distributed, so for this question, assume that dominoes are of constant density)&lt;/p&gt;&#xA;" OwnerUserId="71" LastActivityDate="2010-11-05T04:05:33.007" Title="What determines the (minimum) angle at which a domino falls over?" Tags="&lt;mechanics&gt;&lt;equilibrium&gt;&lt;statics&gt;&lt;stability&gt;" AnswerCount="2" />
  <row Id="259" PostTypeId="2" ParentId="258" CreationDate="2010-11-05T00:41:03.683" Score="14" ViewCount="" Body="&lt;p&gt;To make it fall you need a torque. This torque is provided by the weight force acting on the center of mass of the object and by the offset between the center of mass and the edge of the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine your domino standing upright then tilt it. You are moving the center of mass. When the center of mass (blue) is on the right of the edge (red) then you have a torque, represented by the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/4DEhR.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torque is $\tau = m g D$ so to make it fall you need $D$ greater than zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the domino (of uniform constant density) as base of width L the center of mass is located at L/2. For an height H, the center of mass is located at height H/2. That his other sketch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Ep24k.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking this one in the limit case where $D=0$ you obtain the last sketch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving the trigonometry you obtain  $\alpha = Atan \frac{L/2}{H/2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle of the domino with respect to the &quot;table&quot; is $90-\alpha$ degrees.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-05T00:56:18.083" LastActivityDate="2010-11-05T00:56:18.083" CommentCount="5" />
  <row Id="260" PostTypeId="2" ParentId="252" CreationDate="2010-11-05T01:19:20.700" Score="5" ViewCount="" Body="&lt;p&gt;The answer is fairly simple -- classical N-body problem has its solution in $6N$ 1D functions of time, quantum N-body problem has its solution in one complex function, but $3N$-dimensional (not counting spin and similar stuff). Then, there is no wonder why one can find analytical solutions only for trivial problems or at least make $N$ huge and escape into statistical mechanics. And yes, this is only the problem of mathematical complexity here.&lt;br&gt;&#xA;From modelling point of view exact solving also seems hopeless, with only memory complexity of  $\mathcal{O}(K^{3N})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the rest of the answer I will restrict myself to quantum chemistry/material science, since this is the most exploited region -- this means we are now talking about atoms. First of all, atoms have small and very heavy nuclei, which thus can be treated as almost stationary sources of electrostatic potential; this reduces the problem to electrons only (Born-Oppenheimer approx.). Now, there are two main routes to follow: Hartree-Fock or Density Functional Theory.&lt;br&gt;&#xA;In HF, one roughly represents the many-body weavefunction as a combination of some standard base functions -- then one can optimize their contributions to get minimal energy, yet using extended Hamiltonian to adjust the effects of such approximation. &#xA;In DFT, one encouraged by Hohenberg-Kohn theorems reduces the many body weavefunction to electron probability density field (3-dimensional), and accordingly Shroedinger equation terms into density functionals (and there approximations are applied). Next, it can be either solved as this 3D field or in Kohn-Sham way, which is pretty much Hartree-Fock for DFT (one represents density with base functions).&#xA;People sometimes are making something analytical here, but those are mostly theories made to support computational approaches. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally your last question: those approximate methods (but still &lt;em&gt;ab initio&lt;/em&gt; -- there are no experimental parameters there) do predict things like chemical reactions, various spectra and other measurable quantities; accuracy is problematic though. Biology is mostly out of reach because of a time scale; at least there are hybrid methods able to mix for instance the classical simulation of protein motion with quantum simulation of the binding site when it is squeezed enough so something quantum like enzymatic reaction can take place.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-05T01:19:20.700" CommentCount="3" />
  <row Id="261" PostTypeId="2" ParentId="85" CreationDate="2010-11-05T02:01:09.437" Score="6" ViewCount="" Body="&lt;p&gt;A proton is made of two up quarks and a down quark, whereas a neutron is made of two down quarks and an up quark. The quark masses contribute very little of the actual mass of the proton and neutron, which mostly arises from energy associated with the strong interactions among the quarks. Still, they do contribute a small fraction, and the down quark is slightly heavier than the up quark, which approximately accounts for the difference. (The masses of the up and down quarks, because they're so small, are not extremely well-measured, and detailed calculations of what the proton and neutron masses would be for given quark masses are difficult. So it's hard to be quantitative about the answer in a precise way.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read a bit about the state of the art in calculations of the proton mass here: &lt;a href=&quot;http://news.sciencemag.org/sciencenow/2008/11/21-02.html&quot; rel=&quot;nofollow&quot;&gt;http://news.sciencemag.org/sciencenow/2008/11/21-02.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="19" LastEditorDisplayName="" LastEditDate="2010-11-05T02:07:29.627" LastActivityDate="2010-11-05T02:07:29.627" CommentCount="1" />
  <row Id="262" PostTypeId="2" ParentId="93" CreationDate="2010-11-05T02:29:45.250" Score="3" ViewCount="" Body="&lt;p&gt;It's more the other way around, I would say. Gauss's law, together with the fact that we live in a world with 3 spatial dimensions, requires that the force between charges falls off as 1/r^2. But there are perfectly consistent analogues of electrostatics in worlds with 2 or more spatial dimensions, which each have their own ``Coulomb's law&quot; -- with a &lt;em&gt;different&lt;/em&gt; falloff of force with distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More to the point, it's a lot more obvious that we live in a world with 3 spatial dimensions (look around!) than it is that the force between charges has an inverse-square law. So empirically, as well as theoretically, the number of spatial dimensions is more fundamental than the force law.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-11-05T02:29:45.250" />
  <row Id="263" PostTypeId="2" ParentId="258" CreationDate="2010-11-05T04:05:33.007" Score="2" ViewCount="" Body="&lt;p&gt;The direction that the domino falls is determined by the location its center of mass. It falls to the left or right depending on whether the center of mass is to the left or right of the bottom-most edge. If the center of mass is precisely above the edge, then it balances on that edge in an unstable equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="150" LastActivityDate="2010-11-05T04:05:33.007" />
  <row Id="264" PostTypeId="2" ParentId="252" CreationDate="2010-11-05T05:05:42.753" Score="7" ViewCount="" Body="&lt;p&gt;First let me start by saying that the $N$-body problem in classical mechanics is not computationally difficult to approximate a solution to. It is simply that in general there is not a closed form analytic solution, which is why we must rely on numerics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For quantum mechanics, however, the problem is much harder. This is because in quantum mechanics, the state space required to represent the system must be able to represent all possible superpositions of particles. While the number of orthogonal states is exponential in the size of the system, each has an associated phase and amplitude, which even with the most coarse grain discretization will lead to a double exponential in the number of possible states required to represent it. Thus in quantum systems you need $O(2^{2^n})$ variables to reasonable approximate any possible state of the system, versus only $O(2^n)$ required to represent an analogous classical system. Since we can represent $2^m$ states with $m$ bits, to represent the classical state space we need only $O(n)$ bits, versus $O(2^n)$ bits required to directly represent the quantum system. This is why it is believed to be impossible to simulate a quantum computer in polynomial time, but Newtonian physics can be simulated in polynomial time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating ground states is even harder than simulating the systems. Indeed, in general finding the ground state of a classical Hamiltonian is &lt;a href=&quot;http://qwiki.stanford.edu/index.php/Complexity_Zoo%3aN#np&quot; rel=&quot;nofollow&quot;&gt;NP-complete&lt;/a&gt;, while finding the ground state of a quantum Hamiltonian is &lt;a href=&quot;http://qwiki.stanford.edu/index.php/Complexity_Zoo%3aQ#qma&quot; rel=&quot;nofollow&quot;&gt;QMA-complete&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditorDisplayName="" LastEditDate="2010-11-05T05:47:25.020" LastActivityDate="2010-11-05T05:47:25.020" CommentCount="4" />
  <row Id="265" PostTypeId="2" ParentId="234" CreationDate="2010-11-05T05:27:18.133" Score="3" ViewCount="" Body="&lt;p&gt;Obviously this is not a comprehensive list, and my aim is simply to give you a pointer to the basic material you need to cover early on. As you progress you may become more specialised and your field may have particular mathematical techniques and formalisms which are particular to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much of the mathematics used in physics is continuous. This ranges from the elementary calculus used to solve simple Newtonian systems to the differential geometry used in general relativity. With this in mind, it is generally necessary to cover calculus in depth, real and complex analysis, fourier analysis, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, many physical transformation have very nice group structures, and so covering basic group theory is a very good idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, strong linear algebra is a prerequisite for many of the techniques used in the other areas I have mentioned above, and is also extremely important in the matrix formulation of quantum mechanics. Finding ground states of discrete systems (for example spin networks) means finding the minimum eigenvalue and corresponding eigenvector of the Hamiltonian.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2010-11-05T05:27:18.133" />
  <row Id="266" PostTypeId="2" ParentId="252" CreationDate="2010-11-05T06:38:07.520" Score="2" ViewCount="" Body="&lt;p&gt;In addition to what mbq &lt;a href=&quot;http://physics.stackexchange.com/questions/252/the-many-body-problem/260#260&quot;&gt;said&lt;/a&gt;, it might be interesting to know that things get really funny in relativistic quantum mechanics, that is using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Klein%E2%80%93Gordon_equation&quot; rel=&quot;nofollow&quot;&gt;Klein-Gordon&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_equation&quot; rel=&quot;nofollow&quot;&gt;Dirac equation&lt;/a&gt; (but without the &quot;second&quot; quantization of &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_field_theory&quot; rel=&quot;nofollow&quot;&gt;Quantum Field Theory&lt;/a&gt;). There, there's &lt;em&gt;one&lt;/em&gt; wave function per particle &lt;em&gt;sort&lt;/em&gt;, so no matter how many particles of one kind you consider, the only thing that changes is the field itself. You only get more degrees of freedom by actually adding another kind of particles. Of course, since Fermions require &lt;a href=&quot;http://en.wikipedia.org/wiki/Spinor&quot; rel=&quot;nofollow&quot;&gt;Spinors&lt;/a&gt;, you may end up with other computational issues then...&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2010-11-05T06:38:07.520" CommentCount="7" />
  <row Id="267" PostTypeId="2" ParentId="68" CreationDate="2010-11-05T07:49:20.843" Score="5" ViewCount="" Body="&lt;p&gt;I agree with the answer given by &lt;a href=&quot;http://physics.stackexchange.com/users/78/sigoldberg1&quot;&gt;sigoldberg1&lt;/a&gt; -- it is most probably &lt;a href=&quot;http://www.rp-photonics.com/birefringence.html&quot; rel=&quot;nofollow&quot;&gt;stress induced birefringence&lt;/a&gt;. Bragg scattering would not (at least to first order) change the polarisation of the incoming light beam, thus making the crossed polarisers pretty useless. Also, changes in colour due to Bragg scattering would be observed on &lt;em&gt;reflection&lt;/em&gt; and would be angle dependent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What actually happens in experiments of the type shows in &lt;a href=&quot;http://www.youtube.com/watch?v=BEClYQbuG7U&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt; is the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first polariser is used to produce linearly polarised light. &lt;/li&gt;&#xA;&lt;li&gt;The second polariser is rotated at 90 degrees with respect to the first, and thus in the absence of any material, no light will pass through (from &lt;a href=&quot;http://en.wikipedia.org/wiki/Polarizer#Malus.27_law_and_other_properties&quot; rel=&quot;nofollow&quot;&gt;Malus' law&lt;/a&gt;, $cos^2(90^\circ) = 0$&lt;/li&gt;&#xA;&lt;li&gt;However, if the material is &lt;a href=&quot;http://www.rp-photonics.com/birefringence.html&quot; rel=&quot;nofollow&quot;&gt;birefringent&lt;/a&gt; (the refractive index depends on polarisation) the polarisation angle of light passing through will be rotated, and this rotation will depend on the wavelength of the light. &lt;/li&gt;&#xA;&lt;li&gt;Thus, the final polarisation angle will be different than 90 degrees, and that difference will depend on the wavelength of the light, giving the patterns you see in such an experiment.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Stress induced birefringence is of technical importance due to the fact that &quot;similar effects occur in bent optical fibers, and also due to thermal effects in laser crystals, which can lead to depolarization loss&quot; (&lt;a href=&quot;http://www.rp-photonics.com/birefringence.html&quot; rel=&quot;nofollow&quot;&gt;ref&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="15" LastActivityDate="2010-11-05T07:49:20.843" />
  <row Id="268" PostTypeId="1" AcceptedAnswerId="275" CreationDate="2010-11-05T08:54:35.513" Score="7" ViewCount="255" Body="&lt;p&gt;Each spring enormous amounts of water rise up in trees and other vegetation. What causes this stream upwards? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I was under the impression that capillary action is a key factor: the original question therefore was: what are the fundamental forces involved in capillary action?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can perhaps anybody make an estimation of the amount of gravity energy that is involved in this process earth-wide each season?&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-21T01:24:51.230" LastActivityDate="2011-07-21T01:24:51.230" Title="What physical forces pull/press water upwards in vegetation?" Tags="&lt;gravity&gt;&lt;forces&gt;&lt;biophysics&gt;&lt;capillary-action&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="269" PostTypeId="2" ParentId="218" CreationDate="2010-11-05T09:38:07.803" Score="2" ViewCount="" Body="&lt;p&gt;Well known is the publication of 1890-1893 &lt;a href=&quot;http://www.fulltable.com/tomtit/tt10.htm&quot; rel=&quot;nofollow&quot;&gt;La Science Amusante par Tom Tit&lt;/a&gt;, 100 magical experiments to set up &quot;science in play&quot; in your living room, by Arthur Good. There are many modern variants in this tradition of amazing science experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experiments designed for children probalby are impressive for layman adults too!&lt;br&gt;&#xA;Visit the famous science musea like &lt;a href=&quot;http://www.tomtit.se/&quot; rel=&quot;nofollow&quot;&gt;Tom Tits Experiment&lt;/a&gt; in Stockholm Sweden.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-05T09:38:07.803" />
  <row Id="270" PostTypeId="1" CreationDate="2010-11-05T11:02:47.797" Score="5" ViewCount="409" Body="&lt;p&gt;The &lt;a href=&quot;http://www.straightdope.com/columns/read/1886/does-black-clothing-keep-you-cooler&quot; rel=&quot;nofollow&quot;&gt;Straight Dope&lt;/a&gt; ran an explanation of why nomads often wear black clothing - it absorbs heat better from the body. On the other hand, white clothing reflects sunlight better. Is it possible to get the best of both worlds and wear clothing that is black on the inside and white on the outside?&lt;/p&gt;&#xA;" OwnerUserId="119" LastEditorUserId="165" LastEditorDisplayName="" LastEditDate="2010-11-05T16:34:22.993" LastActivityDate="2010-12-08T05:21:02.393" Title="Would wearing clothing that is black on the inside and white on the outside keep you cooler?" Tags="&lt;heat&gt;&lt;radiation&gt;&lt;thermal-radiation&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="271" PostTypeId="1" CreationDate="2010-11-05T11:04:16.160" Score="12" ViewCount="1053" Body="&lt;p&gt;One topic which was covered in university, but which I never understood, is how a spinning top &quot;magically&quot; resists the force of gravity. The conservation of energy explanations make sense, but I don't believe that they provide as much insight as a mechanical explanation would.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/top.html&quot; rel=&quot;nofollow&quot;&gt;hyperphysics link&lt;/a&gt; Cedric provided looks similar to a diagram that I saw in my physics textbook. This diagram illustrates precession nicely, but doesn't explain why the top doesn't fall. Since the angular acceleration is always tangential, I would expect that the top should spiral outwards until it falls to the ground. However, the diagram seems to indicate that the top should be precessing in a circle, not a spiral. Another reason I am not satisfied with this explanation is that the calculation is apparently limited to situations where: &quot;the spin angular velocity w is much greater than the precession angular velocity wP&quot;. The calculation gives no explanation of why this is not the case.&lt;/p&gt;&#xA;" OwnerUserId="119" LastEditorUserId="119" LastEditorDisplayName="" LastEditDate="2010-11-12T21:37:28.420" LastActivityDate="2010-12-14T05:03:55.430" Title="Why don't spinning tops fall over?" Tags="&lt;mechanics&gt;&lt;classical-mechanics&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="3" />
  <row Id="272" PostTypeId="1" AcceptedAnswerId="276" CreationDate="2010-11-05T11:12:58.393" Score="5" ViewCount="367" Body="&lt;p&gt;I think that relativity and quantum mechanics would provide some good examples.&lt;/p&gt;&#xA;" OwnerUserId="119" LastActivityDate="2010-11-06T06:50:56.127" CommunityOwnedDate="2010-11-08T14:46:50.100" ClosedDate="2010-11-06T07:54:48.893" Title="What is the most counter-intuitive result in physics?" Tags="&lt;experimental-physics&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="273" PostTypeId="2" ParentId="271" CreationDate="2010-11-05T11:22:38.383" Score="6" ViewCount="" Body="&lt;p&gt;When it is spinning its angular momentum is quite high. By conservation of angular momentum the spinning top is then more stable against small torques like the action of gravity on the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angular momentum of the top is $J = I \omega$ where $I$ is the inertia tensor and $\omega$ is the Darboux vector, whose magnitude is proportional to the rotational speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a detailed discussion &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/top.html&quot; rel=&quot;nofollow&quot;&gt;on this page of Hyperphysics&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-05T11:45:47.473" LastActivityDate="2010-11-05T11:45:47.473" CommentCount="11" />
  <row Id="274" PostTypeId="2" ParentId="271" CreationDate="2010-11-05T11:41:16.890" Score="1" ViewCount="" Body="&lt;p&gt;when the mass is spinning it has an angular momentum pointing in a direction perpendicular to the plane it's spinning on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angular momentum has to be conservate: i.e. has to keep pointing in the plane-perpendicular direction. As cedric said, the gravity, works for the axis of the spinng mass to fall horizontally on the plane: if this happens also the angular momentum as to torque! and this is not convenient from an energy conservation point of view..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then u can consider that the magnitude of the angular momentum is proportional to the spinning speed: so as the spinning velocity gets higher it gets, for lack of a better word, &quot;easier&quot; for the top to resist the gravity..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If u try to spin a top on an inclined plane you will need to spin it faster to obtain the same &quot;resistance to gravity&quot;!&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2010-11-05T11:41:16.890" />
  <row Id="275" PostTypeId="2" ParentId="268" CreationDate="2010-11-05T11:51:58.950" Score="8" ViewCount="" Body="&lt;p&gt;Water is transported through xylem tissue, which reassemble just a passive bundle of pipes. They are narrow enough to provide quite a huge capillary effect, but this is not a process of transport because it converges fast to equilibrium water levels and stops.&lt;br&gt;&#xA;The flow is powered by two other processes; first and most important is evaporation of water from leafs (transpiration pull) and the second is a osmotic pressure in roots (root pressure).  The first one is a passive process, powered by the difference of water potentials in air and soil; the second is active, plant consumes its own energy to produce ion gradient causing osmotic pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I won't give you the accurate numbers in the second part; to give some imagination, about 90% of water absorbed from soil is transpired into atmosphere, so it is near transpiration rates. A single tree growing in humid continental climate zone (most of Europe and USA) transpires yearly something like 100 tons of water.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-05T11:51:58.950" CommentCount="6" />
  <row Id="276" PostTypeId="2" ParentId="272" CreationDate="2010-11-05T12:50:17.473" Score="3" ViewCount="" Body="&lt;p&gt;Laminar flow reversibility; gets straight from equations, but it is hard to believe it works in reality. Yet: &lt;a href=&quot;http://www.youtube.com/watch?v=p08_KlTKP50&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=p08_KlTKP50&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-05T12:50:17.473" CommunityOwnedDate="2010-11-08T14:46:50.100" CommentCount="3" />
  <row Id="277" PostTypeId="2" ParentId="198" CreationDate="2010-11-05T14:36:45.843" Score="8" ViewCount="" Body="&lt;p&gt;Apologies in advance if the first part of this comes off a bit argumentative, but I think there is an important point about physical theory that should be made.  This point is also implicit in David Zaslavsky's answer as well.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Rant on effective theories&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Actually trying to calculate macroscopic properties like &quot;chemistry&quot; from fundamental theories like QCD and QED, or string theory, etc. seems to me to be missing the point of much of modern physics.  In particular, there's enough of a separation of length, energy, and time scales that the &quot;physics&quot; that goes into the boiling point of Nitrogen is almost completely disjoint from the physics of quarks and and the nucleus.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, if to do any physics we had to start from the very bottom every time, science in a form usable and comprehensible by humans could not exist!  One of the deepest ideas of the 20th century is the notion of &quot;effective field theory and renormalization&quot; which to me makes clear why we can treat all the effects coming from high-energy physics as &quot;just&quot; couplings going into our vastly more useful phenomenological theories, i.e. all we get out of particle physics is some numbers such as the mass of various particles, or effective interaction potentials such as the Coulomb interaction, etc. and these numbers are just as useful if you measure them from doing experiments as if you were to calculate them.  (And as calculating many of these (with some notable exceptions) involves intractable many-body problems, at this point in time it's actually much more accurate to measure them).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Breakdown of scales&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In this particular case of determining boiling points, let me lay out what I think the relevant scales are from the bottom up and perhaps real experts can come by and correct me.  &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;QCD and &quot;below&quot;&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;First of all, effects from nuclear physics and below are totally negligible, as pointed out by David Zaslavsky already.  So we can just treat the nuclei (and electrons) as single particles and take the masses as something given from measurement.  If you wanted to calculate these masses, it's my impression that there are pretty good models for the masses of nuclei if you have the masses of protons and neutrons.  However, the masses of protons and neutrons have not been (directly) calculated to any reasonable accuracy up until fairly recently, if at all.  And these models assume certain values for quark masses, which I think nobody really knows how to calculate at all.  In any case, you could certainly dig deeper and deeper with respect to calculating the masses of things, but other effects won't be important for the boiling point of nitrogen.  &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;QED and quantum mechanics&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Second, higher corrections from QED and indeed, even quantum effects between molecules will probably be ignorable too, since for Nitrogen at its boiling temperature, the molecules will be far away from quantum coherence (I'm less sure of this, but a quick check by estimating the de Broglie wavelength seems to work out; note that this will not be true for liquid Helium or Hydrogen).  What this means is that you can probably get away with treating the Nitrogen molecules as classical (dumbbell-shaped) dipoles with an effective pair interaction coming, as David Zaslavsky pointed out, from a single-molecule quantum ab initio calculation.  I'm not sure if current techniques allow for this to be done with any reasonable accuracy, but with 14 electrons, it is a fairly serious problem not too far away from the current frontiers in the field of quantum chemistry.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Classical molecular dynamics&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Now we get to scales which are to me, more interesting, and where the physics of boiling finally appears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As David Zaslavsky and mbq have pointed out, boiling point is a thermodynamic property, thus it is something which only exists for a macroscopic number of molecules.  That said, one won't need to attempt a simulation with 10^23 Nitrogen molecules to get a reasonable estimate; one of the triumphs of statistical mechanics is an understanding of how to estimate &quot;finite-size effects&quot;; however, it should be emphasized that this is yet another many-body problem (read as, impossible to solve analytically, requires difficult computer simulation to get right).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So finally, to get the boiling point, you could model Nitrogen molecules as some kind of pair-interacting dipolar gas in a classical molecular dynamics simulation at fixed pressure, say, and tune the temperature up until the molecules in your simulation begin to behave in a gaslike way rather than liquidlike (this &quot;behavior&quot; can be made more precise with the notion of pair correlation functions, for instance).  Note a few things here.  The greatest error in your estimation will be coming from your ability to put in a bunch of particles here in a controlled fashion in your simulation.  This is going to vastly outweigh any error from the steps &quot;further up&quot; other than having a good effective potential.  This explains why scientists who actually do molecular dynamics simulation don't spend that much time worrying about QCD and particle physics.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Places to start looking and references&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;OK, that all written, I've looked around for some references that would be useful on molecular dynamics simulation of liquid nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.springerlink.com/content/72175162x0458032/&quot; rel=&quot;nofollow&quot;&gt;Dominique Levesque and Jean Jacques Weis write in the 1992 book The Monte Carlo Method in Condensed Matter Physics&lt;/a&gt; a chapter &quot;Recent progress in the simulation of classical fluids&quot; which goes over the main computational techniques commonly used now.  This contains a short section on Nitrogen, as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1016/S0378-3812%2803%2900057-8&quot; rel=&quot;nofollow&quot;&gt;Javier Carrero-Mantilla and Mario Llano-Restrepo, Fluid Phase Equilibria, Volume 208, Issues 1-2, 15 June 2003, Pages 155-169&lt;/a&gt; writes on simulation of liquid Nitrogen and binary mixtures with some other simple liquids.  I've attached a figure from the paper which shows the liquid-vapor coexistence curve as a function of density, from which you could read off boiling point if you wanted.  The agreement is pretty amazing, and they do use the strategy I outlined above - simulate Nitrogen as a classical fluid with a tuned intermolecular potential; their source for the potential is apparently &lt;a href=&quot;http://www.ingentaconnect.com/content/tandf/tmph/1980/00000041/00000006/art00011&quot; rel=&quot;nofollow&quot;&gt;C.S. Murthy, K. Singer, M.L. Klein and I.R. McDonald, Mol. Phys. 41 (1980), pp. 1387–1399.&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/b5tB6.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could dig around more for newer references, but if you're really interested, you can use google scholar and some of the search terms like &quot;ab initio&quot;, &quot;liquid-vapor coexistence&quot;, &quot;Liquid nitrogen&quot;, &quot;molecular dynamics&quot; to find more...&lt;/p&gt;&#xA;" OwnerUserId="42" LastEditorUserId="42" LastEditorDisplayName="" LastEditDate="2010-11-05T15:01:53.753" LastActivityDate="2010-11-05T15:01:53.753" />
  <row Id="278" PostTypeId="2" ParentId="68" CreationDate="2010-11-05T14:59:44.323" Score="0" ViewCount="" Body="&lt;p&gt;This is a little off-topic, in that the answers to the question by sigoldberg1 and Flaviu Cipcigan address the relevant physics in plenty of detail already, but I can't resist linking to this video which shows a related phenomenon: beautiful swirls of colors in liquid crystals viewed under crossed-polarizers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the principle is close to what is described by Flaviu Cipcigan, but in the systems of liquid crystals showed in the movie, the stresses arise from the interaction between flow and temperature gradients in the liquid, and instead of birefringence, it's the fact that the molecular orientations themselves are rotating and changing in response to the stresses, and thus letting different amounts of light of different colors through the two polarizers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=vlgJ5Jk30zk&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=vlgJ5Jk30zk&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-05T14:59:44.323" />
  <row Id="279" PostTypeId="2" ParentId="270" CreationDate="2010-11-05T16:33:13.133" Score="1" ViewCount="" Body="&lt;p&gt;Since both absorption and radiation depend on color in the same way, this would decrease the energy that is radiated off into the surroundings. So if radiation were the only way to lose heat color wouldn't matter, and in the long term you'd have the same temperature as the incoming light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since nomad's don't live in vacuum there are other ways of heat transportation. What color is best depends on which forms of energy transportation dominate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I guess that you lose more heat to other sources than radiation, while a large part of the incoming heat comes from light, making them white on the outside and black on the inside could make sense. But there are complicating factors such as wind, vaporization of sweat,...&lt;/p&gt;&#xA;" OwnerUserId="165" LastActivityDate="2010-11-05T16:33:13.133" />
  <row Id="280" PostTypeId="2" ParentId="214" CreationDate="2010-11-05T16:56:27.763" Score="2" ViewCount="" Body="&lt;p&gt;From the question of simple reduction of velocity, the answer's already been given (yes, but not enough to make a significant difference) ... but there's one other issue at play here -- how the forces are transfered to the body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're standing upright, it'd all be transfered through your legs;  as Flaviu mentioned, laying down so the force is spread across a larger area would be a better option to this.  But, if you could manage to jump at just the right time, and you knew how to take a fall (bend your knees, roll into it, etc.), it might be possible to spread the force over a greater time and distance, therefore reducing the impulse, and thus the actual damage to your body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I don't think the chances of timing it correctly would be very good, so it wouldn't be particularly advisable.  You'd have to weigh the risk &amp;amp; benefit of this strategy vs. just lying down.&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-11-05T16:56:27.763" CommentCount="3" />
  <row Id="281" PostTypeId="2" ParentId="197" CreationDate="2010-11-05T17:17:14.930" Score="3" ViewCount="" Body="&lt;p&gt;The short answer in quantum field theories is &quot;by the exchange of virtual particles&quot;.  Look up &quot;virtual particles&quot; on Wikipedia to get a sense of it.  The ancillary questions you asked about the energy budget, and the derivation of Coulomb's law are very much the right ones.  They should all be answered there.   For example, energy is conserved except for momentary fluctuations due to the uncertainty principle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no separate &quot;messenger particles&quot;, maybe the person who used that phrase misspoke.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For electromagnetism the relevant virtual particles are photons, for the strong force &quot;gluons&quot;, for the weak force the W' and Z, all of which have been found.  All of these are instances of &quot;gauge bosons&quot;.  Again, see Wikipedia for a basic reference.   The inverse square law for EM is related to the zero rest mass of the photon, the short range, and exponential decrease with distance of the strong and weak forces is related to the non zero rest mass of their exchanged virtual particles.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity is a special case, since there is as yet no quantum theory of gravity.   Stay tuned.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-11-06T05:34:21.097" LastActivityDate="2010-11-06T05:34:21.097" />
  <row Id="282" PostTypeId="2" ParentId="83" CreationDate="2010-11-05T17:18:30.797" Score="7" ViewCount="" Body="&lt;p&gt;Squeezing of laser light generally involves a non-linear interaction, where the nature of the interaction depends on the intensity of the light that is present. An easy to understand example is frequency doubling, which takes two photons from a pump laser, and sends out one photon of twice the frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can think of the input beam as a stream of photons with some fluctuation in the &quot;spacing&quot; of the photons along the beam. That is, on average you will receive, say, one photon per some unit of time, but sometimes you get two, and sometimes none.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you send this beam into a nonlinear crystal to do frequency doubling, the doubling will occur only in those instants when you get two photons in one unit of time. In that case, the two photons are removed from the original beam, and produce one photon in the frequency-doubled output beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the transmitted light left behind in the input beam, you will find lower fluctuations in the intensity, because all of the two-photon instants have been removed. Thus, the transmitted beam is &quot;amplitude-squeezed.&quot; It's not quite as obvious that the frequency-doubled beam has lower intensity fluctuations, but it, too is amplitude squeezed, because you get photons only at the times when you had two photons in the original beam, and it's exceedingly unlikely that you would get two of those in very close succession (or four photons from the original beam in one instant). So you have a lower intensity in the doubled beam, and also lower fluctuations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for example, your input beam might give the following sequence of photon numbers in one-unit time steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1112010112110120&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The input beam after the doubling crystal will look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1110010110110100&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the doubled output beam will look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0001000001000010&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of those have lower fluctuations than the initial state.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-05T17:18:30.797" />
  <row Id="283" PostTypeId="2" ParentId="257" CreationDate="2010-11-05T18:42:54.483" Score="8" ViewCount="" Body="&lt;p&gt;First of all, the scheme of the CERN accelerator complex you posted contains not only that single chain which brings the protons to the LHC, but also several other chains which are used for many lower energy experiments conducted in parallel at CERN. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let's focus on the LHC accelerator chain: why do we need several successive accelerators instead of a single one? The answer is very intuitive from the engineering point of view: technologically, it is much easier to construct several devices specialized for different but &lt;strong&gt;limited ranges&lt;/strong&gt; of some physical parameter than to build a single device that exhibits excellent performance in the &lt;strong&gt;entire vast range&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example in transportation: cars are good to travel at lower speed, airplanes are good for traveling fast. It is exceedingly difficult to build a vehicle that would have equally good performance in the entire range of speeds from 1 km/h to 1000 km/h. Another example is given by professional acoustic systems which usually include several speakers optimized for different frequency ranges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thing for the accelerators. Here the key point is not the energy of the protons &lt;em&gt;per se&lt;/em&gt;, but the quality of the beam at a given energy. Accelerating protons is (technically) a minor issue; the major issue is to keep the beam safe and well-behaved. If you dig a bit into the CERN website, you'll soon realize that these accelerators have quite distinct features and each of them was optimized for its own energy range and its own purposes. Some of them serve to accumulate particles, other are specialized in breaking the beams into bunches. At each stage you need to cool the beams down, dump oscillations, etc., and this is done in a different way at different energies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, you could think of an accelerator that would occupy the LHC ring and would take the protons at a very low energy (say, 1 MeV) and accelerate them up to multi-TeV. Since the protons must always circulate in the same ring, the magnetic field must be adjustable (with very high precision and very high spacial homogeneity) from few microtesla to several tesla, in the range of six orders of magnitude. The beam monitoring system must also be adapted to accurately measure the proton currents differing by 6 orders of magnitude. The similar requirements are placed on the beam steering, orbit correcting and focusing magnets, on the kicker magnets, and on many other components of the accelerator. In short, although technically possible, it would be just too difficult and too expensive. A sequence of several accelerators (which existed before the LHC anyway) is a preferred solution.&lt;/p&gt;&#xA;" OwnerUserId="59" LastActivityDate="2010-11-05T18:42:54.483" CommentCount="1" />
  <row Id="284" PostTypeId="2" ParentId="185" CreationDate="2010-11-05T21:35:37.237" Score="13" ViewCount="" Body="&lt;p&gt;David Zaslavsky's answer goes most of the way toward answering the question &quot;what is needed to claim the discovery of a new particle?&quot;, but to claim discovery of &lt;em&gt;the&lt;/em&gt; Higgs boson, one needs much more. The first sign that the particle really is the predicted Higgs will be the decay modes it shows up in. The Standard Model makes a specific prediction for the rate of different types of Higgs events as a function of its mass. It could decay to two Z bosons, each decaying to two muons, as David Zaslavsky explained, but only if it is heavy enough. If it's lighter, it would decay mostly to bottom and anti-bottom quarks, which are quite difficult to see because ordinary strong interactions produce them much more often. It would also decay rarely to two photons, but this is a distinctive enough signal that it's relatively easy to study. The rate and type of events that are seen would be checked against predictions of the Standard Model for the observed mass. For instance, if a signal showed up involving 4 muons, but at a mass of only 100 GeV, we would know that we weren't seeing the Higgs, but something stranger and not predicted. Or if a 160 GeV particle were found to frequently decay to two photons, we would know that it isn't the Higgs, which at that mass would decay mostly to a pair of Z or W bosons. So there are a number of consistency checks to make between the mass and decay modes. Still, much more is needed to really say that it is the Higgs. For one thing, it should be a scalar, i.e. a particle without spin. The spin can be tested by looking at the angular separation among decay products. The Higgs also interacts with known particles in very specific ways; as many tests of this as possible would be performed. Even if the particle is not the Standard Model Higgs, it still might be a type of Higgs boson in an extended theory, such as the supersymmetric Standard Model. The road from discovering &lt;em&gt;some&lt;/em&gt; new particle to having a completely convincing theoretical account of what the particle is could potentially be a long one, and might even involve other colliders in the future (like the proposed International Linear Collider).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For now, of course, we're all just hoping for &lt;em&gt;something&lt;/em&gt; to be discovered; once this happens, the work of pinning down precisely what the discovery means will be exciting and will continue for years.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-11-05T21:35:37.237" CommentCount="1" />
  <row Id="285" PostTypeId="2" ParentId="71" CreationDate="2010-11-05T21:46:59.323" Score="4" ViewCount="" Body="&lt;p&gt;Several sources link to this paper: S. Feng, H. G. Winful, &lt;a href=&quot;http://www.physics.princeton.edu/~mcdonald/examples/optics/feng_ol_26_485_01.pdf&quot; rel=&quot;nofollow&quot; title=&quot;Physical origin of the Gouy phase shift&quot;&gt;Physical origin of the Gouy phase shift&lt;/a&gt;, Optics Letters, 26, 485 (2001), which tries to give an intuitive explanation of the Gouy phase. Briefly, the point is that convergent waves going through the focus have finite spatial extent in the transverse plane. The uncertainty relation induces then some distribution over the transverse and consequently longitudinal wave vectors. It is claimed that the net effect of this distribution over wave vectors is an overall phase shift, which is larger for higher modes. However to see that one really needs to look into the formulas.&lt;/p&gt;&#xA;" OwnerUserId="59" LastActivityDate="2010-11-05T21:46:59.323" />
  <row Id="286" PostTypeId="1" AcceptedAnswerId="293" CreationDate="2010-11-05T23:17:22.260" Score="5" ViewCount="224" Body="&lt;p&gt;Does anyone have an intuitive explanation of why this is the case? &lt;/p&gt;&#xA;" OwnerUserId="119" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-06T03:26:18.373" LastActivityDate="2010-11-06T05:51:44.660" Title="How is it that angular velocities are vectors, while rotations aren't?" Tags="&lt;rotational-dynamics&gt;&lt;angular-velocity&gt;&lt;vectors&gt;" AnswerCount="3" CommentCount="14" />
  <row Id="287" PostTypeId="2" ParentId="208" CreationDate="2010-11-06T02:16:06.950" Score="1" ViewCount="" Body="&lt;p&gt;Right now?  No, it's decreasing, until we get to perihelion (91.4M miles, near the January 3rd), and then it'll start increasing again 'til aphelion (94.5M miles, near July 4th).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As Mark Eichenlaub pointed out -- &lt;a href=&quot;http://en.wikipedia.org/wiki/Earth%27s_orbit&quot; rel=&quot;nofollow&quot;&gt;Earth's orbit is not circular&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-11-06T02:16:06.950" />
  <row Id="288" PostTypeId="2" ParentId="286" CreationDate="2010-11-06T02:17:17.870" Score="5" ViewCount="" Body="&lt;p&gt;Defining properties of vectors are that you can add them and multiply them by constants. These both make sense for angular velocities. On the other hand, adding rotations doesn't make sense. What you can do with two rotations is compose them: first rotate one way, then rotate another. This operation doesn't look like addition of any sort. For one thing, it doesn't commute. Rotating something 30 degrees around the x axis, then 60 degrees around the y axis, is not the same thing as doing those two operations in the opposite order. (If you've never done so, pick up an object and try!) So the mathematical operation that corresponds to rotations has to be something that is able to express noncommutativity. Matrices work very naturally for this; for two matrices A and B, it isn't true in general that AB = BA.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-11-06T02:17:17.870" CommentCount="2" />
  <row Id="289" PostTypeId="2" ParentId="286" CreationDate="2010-11-06T02:50:26.407" Score="5" ViewCount="" Body="&lt;p&gt;There are actually several different ways to interpret that question, depending on what you mean by &quot;vector&quot; and &quot;rotation&quot;. But here's a sense that I've often wondered about myself: in introductory physics, the velocity vector is defined as the time derivative of the position vector (relative to some fixed point). Why is the same not true of angular velocity - that is, why isn't there an &quot;angular position vector&quot; which angular velocity can be the derivative of?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a matter of fact, sometimes there is. Think about this simple case: pick a single, fixed rotation axis and only consider rotations about that axis. (2D rotations, if you prefer to think of it that way) You'd pick a certain orientation to be the &quot;origin,&quot; and you could actually define an angular position vector, pointing along the rotational axis, with length equal to the amount of rotation relative to that &quot;origin&quot; orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose your object's angular position is changing over time. You can take the derivative of the angular position vector, and hopefully you can see that what you'd get is just the good old angular velocity. No problem there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we live in a 3D world (relativity notwithstanding), so what happens when you try to generalize that model to 3 dimensions? That's where you run into problems. As an example, take the object from the last paragraph, which was rotating around one particular axis - say, the $\hat{z}$ axis. Now suppose it changes its motion so that it starts rotating around a different axis, maybe the $\hat{x}$ axis. How will you represent its orientation now?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be tempted to use an &quot;angular position vector&quot; pointing in the $\hat{z}$ direction, whose length represents the amount of rotation around the $\hat{z}$ axis, and another &quot;angular position vector&quot; pointing in the $\hat{x}$ direction, whose length represents the amount of rotation around the $\hat{x}$ axis. After all, that works for position. But it doesn't work for angular position. The reason why is that &lt;strong&gt;rotations do not commute&lt;/strong&gt;, to use the technical lingo. What that means is that if you apply rotation A to an object and follow it up by rotation B which is around a different axis, you get a different result than if you apply rotation B followed by rotation A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This little issue gets in the way if you try to combine your two $x$- and $z$- angular position vectors into one overall angular position vector. Presumably, you would write this overall vector as $(\theta_x, 0, \theta_z)$ (the zero might represent the amount of rotation around the $\hat{y}$ axis). That vector would represent the sum of $\theta_x$ times the unit rotation around the $\hat{x}$ direction and $\theta_z$ times the unit rotation around the $\hat{z}$ direction. But it's missing a critical piece of information: which of those rotations was performed first? If you gave that vector to your physicist friend, he would be unable to reproduce the orientation of the object because he doesn't know whether to perform the $x$-rotation or the $z$-rotation first. Sure, &lt;em&gt;you&lt;/em&gt; may know that the object was rotated in the $z$-direction first, but that information needs to be contained in the vector for it to be of any use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of the last paragraph is that there's no way to sensibly create linear combinations of these &quot;angular position vectors.&quot; And that pretty much ruins their usefulness, because the ability to be linearly combined is absolutely fundamental to the definition of a vector, and it underlies a &lt;em&gt;lot&lt;/em&gt; of the analytical techniques we use in physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, in this view, the reason matrices work to represent rotations is that matrices offer you an additional operation, multiplication, which you can use to combine them. It happens that matrix multiplication, for certain matrices (3x3 antisymmetric with determinant 1), has the same properties as composing rotations; most notably, it's also noncommutative. Multiplying matrix A by matrix B can give you a different result from multiplying matrix B by matrix A.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-06T02:58:16.773" LastActivityDate="2010-11-06T02:58:16.773" />
  <row Id="290" PostTypeId="1" AcceptedAnswerId="295" CreationDate="2010-11-06T03:23:18.563" Score="36" ViewCount="1323" Body="&lt;p&gt;What aerodynamic effects actually contribute to producing the lift on an airplane?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there's a common belief that lift comes from the Bernoulli effect, where air moving over the wings is at reduced pressure because it's forced to travel further than air flowing under the wings. But I also know that this is wrong, or at best a minor contribution to the actual lift. The thing is, none of the many sources I've seen that discredit the Bernoulli effect explain what's actually going on, so I'm left wondering. Why do airplanes actually fly? Is this something that can be explained or summarized at a level appropriate for someone who isn't trained in fluid dynamics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Links to further reading for more detail would also be much appreciated)&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-24T20:45:41.730" Title="What really allows airplanes to fly?" Tags="&lt;fluid-dynamics&gt;&lt;aerodynamics&gt;&lt;airplanes&gt;&lt;lift&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="14" />
  <row Id="291" PostTypeId="2" ParentId="290" CreationDate="2010-11-06T05:29:58.817" Score="17" ViewCount="" Body="&lt;p&gt;Since you asked for an explanation appropriate to an non-specialized audience, maybe this will do: &lt;a href=&quot;http://home.comcast.net/~clipper-108/lift.pdf&quot; rel=&quot;nofollow&quot;&gt;A Physical Description of Flight&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2010-11-06T05:29:58.817" CommentCount="2" />
  <row Id="292" PostTypeId="2" ParentId="268" CreationDate="2010-11-06T05:43:39.587" Score="3" ViewCount="" Body="&lt;p&gt;As pointed out above, it is actually other forces, not capillary action that gets water/sap to the top of tall trees.&lt;br&gt;&#xA;This makes your title off point.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notwithstanding that misunderstanding, in capillary action the fundamental force is surface tension and wetting forces on the side of the tube.   It can raise water in plants only about 1 meter high.    See &lt;a href=&quot;http://en.wikipedia.org/wiki/Capillary_action&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Capillary_action&lt;/a&gt; for an explanation and a formula.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-11-06T05:43:39.587" CommentCount="1" />
  <row Id="293" PostTypeId="2" ParentId="286" CreationDate="2010-11-06T05:51:44.660" Score="7" ViewCount="" Body="&lt;p&gt;This is a note on why angular velocities are vectors, to complement Matt and David's excellent explanations of why rotations are not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we say something has a certain angular velocity $\vec{\omega_1}$, we mean that each part of the thing has a position-dependent velocity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec{v_1}(\vec{r}) = \vec{\omega_1} \times \vec{r}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We might consider another one of these motions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec{v_2}(\vec{r}) = \vec{\omega_2} \times \vec{r}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and wonder what happens when we add them.  We get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec{v_1}(\vec{r}) + \vec{v_2}(\vec{r}) =  \vec{\omega_1} \times \vec{r} + \vec{\omega_2} \times \vec{r}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross product is linear, so this is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\vec{v_1} + \vec{v_2})(\vec{r}) = (\vec{\omega_1} + \vec{\omega_2}) \times \vec{r}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so it makes fine sense to add angular velocities by vector addition.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-06T05:51:44.660" CommentCount="3" />
  <row Id="294" PostTypeId="2" ParentId="272" CreationDate="2010-11-06T06:50:56.127" Score="1" ViewCount="" Body="&lt;p&gt;If you mean things I'm still scratching my head about, starting from the largest scale, it might be that we, life, the earth, the universe, (in Douglas Adam's words, &quot;everything&quot;) is here at all.   Different levels of the &quot;Why something rather than nothing question&quot; I suppose, but that's already too defined.  We have only developed theories e.g. probability theory on branching trees, quantum chemistry, cosmology in the last 200 years to even partially explain the phenomenology.   If someone told you the story from afar, your initial reaction would have to be &quot;No way!&quot;  You could say it's not physics (no math and way too broad), but it's definitely physical phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For something smaller, quantum phenomena, e.g. quantization of spin axis.   Is it as Zeilinger suggests ultimately because the electron spin carries only one bit of information?&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-11-06T06:50:56.127" CommunityOwnedDate="2010-11-08T14:46:50.100" />
  <row Id="295" PostTypeId="2" ParentId="290" CreationDate="2010-11-06T08:28:26.600" Score="23" ViewCount="" Body="&lt;p&gt;A short summary of the &lt;a href=&quot;http://home.comcast.net/~clipper-108/lift.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; mentioned in another answer and &lt;a href=&quot;http://science.howstuffworks.com/transport/flight/modern/airplane.htm&quot; rel=&quot;nofollow&quot;&gt;another good site&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically &lt;strong&gt;planes fly because they push enough air downwards&lt;/strong&gt; and receive an upwards lift thanks to Newton's third law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They do so in a variety of manners, but the most significant contributions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;The angle of attack&lt;/strong&gt; of the wings, which uses drag to push the air down. This is typical during take off (think of airplanes going upwards with the nose up) and landing (flaps). This is also how planes fly upside down.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;The asymmetrical shape of the wings&lt;/strong&gt; that directs the air passing over them downwards instead of straight behind. This allows planes to fly level to the ground without having a permanent angle on the wings.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-06T11:26:09.173" LastActivityDate="2010-11-06T11:26:09.173" CommentCount="6" />
  <row Id="296" PostTypeId="1" AcceptedAnswerId="655" CreationDate="2010-11-06T10:22:06.087" Score="9" ViewCount="644" Body="&lt;p&gt;In high school I was taught energy was conserved. Then I learned that nuclear reactions allow energy to be converted into mass. Then I also heard that apparently energy can spontaneously appear in quantum mechanics. So, are there any other caveats with the conservation of energy?&lt;/p&gt;&#xA;" OwnerUserId="119" LastActivityDate="2011-03-05T15:27:38.310" Title="Is energy really conserved?" Tags="&lt;energy-conservation&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="297" PostTypeId="2" ParentId="296" CreationDate="2010-11-06T10:37:11.727" Score="4" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Then I learned that nuclear reactions allow energy to be converted into mass.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That would be the opposite and in any case, mass is energy (and energy is mass), so converting one into the other does conserve energy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Then I also heard that apparently energy can spontaneously appear in quantum mechanics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a very short time, given by Heisenberg uncertainty principle. And that's not a violation of the conservation of energy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So, are there any other caveats with the conservation of energy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why &quot;other&quot; ?  There's not any problem with the conservation of energy.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-11-06T10:37:11.727" CommentCount="3" />
  <row Id="298" PostTypeId="2" ParentId="296" CreationDate="2010-11-06T10:42:07.543" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Energy is always conserved without any caveat.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the advent of special relativity, mass and energy are considered equivalent. In other words, they are represented by a vectorial quantity called energy-momentum vector. Before relativity there were separate laws which have been unified.&#xA;It is a very fundamental law that is connected to some basic empirical properties of the universe, like the fact that the laws of physics do not change over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy cannot spontaneously appear in quantum mechanics -- however it cannot be precisely measured and this allows for energy fluctuations. The important difference is that although the total amount of energy can change, this is for very brief amount of time, after which the original quantity is restored. So the energy fluctuation can be considered virtual. You don't get energy out of nothing and energy is still conserved.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-06T10:42:07.543" CommentCount="10" />
  <row Id="299" PostTypeId="2" ParentId="236" CreationDate="2010-11-06T11:44:59.513" Score="5" ViewCount="" Body="&lt;p&gt;Just to add. Not only does the positronium exist, it can also interact with matter and allows you to do some interesting physics. For example, in a recent paper &lt;em&gt;S. Mariazzi, P. Bettotti, R.S. Brusa,&lt;/em&gt; &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v104/i24/e243401&quot; rel=&quot;nofollow&quot;&gt;Positronium Cooling and Emission in Vacuum from Nanochannels at Cryogenic Temperature&lt;/a&gt;, &lt;em&gt;Phys. Rev. Lett. 104, 243401 (2010)&lt;/em&gt; positronium created by deposition of positrons on a nanostructured surface was cooled down by collision with walls of nanochannels and thermalized(!) at about 150K. Here is a citation from the abstract of that paper:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;High formation yield and a meaningful cooled fraction of positronium below room temperature were obtained by implanting positrons in a silicon target in which well-controlled oxidized nanochannels (5–8 nm in diameter) perpendicular to the surface were produced. We show that by implanting positrons at 7 keV in the target held at 150 K, about 27% of positrons form positronium that escapes into the vacuum. Around 9% of the escaped positronium is cooled by collision with the walls of nanochannels and is emitted with a Maxwellian beam at 150 K.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="59" LastActivityDate="2010-11-06T11:44:59.513" CommentCount="1" />
  <row Id="300" PostTypeId="1" AcceptedAnswerId="561" CreationDate="2010-11-06T14:27:17.670" Score="6" ViewCount="98" Body="&lt;p&gt;Global Circulation Models typically have grids of 100-300km on a side. There are obviously lots of atmospheric processes that happen at smaller scales than this. Convection, cloud formation, the effect of mountains...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are these processes built in to the model?&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2010-11-11T14:13:11.080" Title="How do subgrid-scale parametrisations in climate models work?" Tags="&lt;fluid-dynamics&gt;&lt;simulation&gt;&lt;climate-science&gt;" AnswerCount="1" />
  <row Id="301" PostTypeId="2" ParentId="17" CreationDate="2010-11-06T15:19:55.847" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://mitworld.mit.edu/video/74&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;You will find this lecture by Walter Lewin interesting.&lt;/strong&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-11-06T15:19:55.847" />
  <row Id="302" PostTypeId="2" ParentId="236" CreationDate="2010-11-06T15:49:27.317" Score="2" ViewCount="" Body="&lt;p&gt;For a slightly different spin on more or less the same thing that's already been said: The idea that an electron coming sort of close to a positron will inevitably annihilate with is comes from the same sort of misconception that leads people to think that objects coming sort of close to a black hole will inevitably fall in. Neither one is inevitable, because the effect of both forces is just to pull the objects toward one another, which may or may not lead to a collision that consumes them, depending on the details of the motion of the particles before they begin interacting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be able to claim that a positron and an electron that start off perfectly at rest would annihilate without ever forming a bound state, but that's a completely unrealistic situation for a bunch of reasons, including the uncertainty principle. If they start out far apart from each other with some initial velocity, though, their fate will depend on the exact arrangement of the initial conditions. Forming a stable bound state most likely requires a third particle, as well, to conserve energy and momentum.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-06T15:49:27.317" />
  <row Id="303" PostTypeId="1" CreationDate="2010-11-06T17:31:09.677" Score="3" ViewCount="302" Body="&lt;p&gt;When I disturb a body of water, what causes the familiar &quot;water moving&quot; sound?&lt;/p&gt;&#xA;" OwnerUserId="169" LastActivityDate="2010-11-13T18:48:11.593" Title="Why does water make a sound when it is disturbed?" Tags="&lt;sound&gt;&lt;water&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="304" PostTypeId="1" AcceptedAnswerId="2437" CreationDate="2010-11-06T20:01:21.020" Score="13" ViewCount="376" Body="&lt;p&gt;It's bonfire weekend here in the UK, and everybody is intent on shooting fireworks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering, what determines the pitch of the sound produced by an explosion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sound has clearly a dominant mid-range pitch, around a few hundred hertz, and bigger fireworks also have a very low, but large component at around 20-50 hertz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What determines this pitch? The amount of explosive energy? The shape of the vessel? Something else?&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-07T10:31:30.153" LastActivityDate="2011-08-16T21:56:01.220" Title="What determines the pitch of an explosion?" Tags="&lt;acoustics&gt;&lt;explosions&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="5" />
  <row Id="305" PostTypeId="1" AcceptedAnswerId="307" CreationDate="2010-11-06T20:59:24.377" Score="-2" ViewCount="134" Body="&lt;p&gt;The number of electrons emitted because of the photoelectric effect &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;rises with the frequency of the radiation&lt;/li&gt;&#xA;&lt;li&gt;&lt;strike&gt;is not influenced by the intensity of the radiation&lt;/strike&gt; (As Chad points out, this is wrong, the electron's &lt;em&gt;energy&lt;/em&gt; isn't influenced by the radiation's intensity.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Why was this not expected when it was discovered?&lt;/p&gt;&#xA;" OwnerUserId="140" LastEditorUserId="140" LastEditorDisplayName="" LastEditDate="2010-11-07T06:33:59.700" LastActivityDate="2010-11-07T06:33:59.700" Title="Why did people expect the number of electrons emitted because of the photoelectric effect to rise with the radiation's intensity?" Tags="&lt;radiation&gt;&lt;photoelectric-effect&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="306" PostTypeId="2" ParentId="305" CreationDate="2010-11-06T21:13:36.453" Score="1" ViewCount="" Body="&lt;p&gt;Classically, light was thought to only have the characteristics of wave. The energy of a wave depends solely on the intensity not frequency. To knock out an electron one expects supplying enough energy would make it escape from the nucleon's attraction, so people won't expect frequency affects the photoelectric effect.&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-11-06T21:19:08.130" LastActivityDate="2010-11-06T21:19:08.130" />
  <row Id="307" PostTypeId="2" ParentId="305" CreationDate="2010-11-06T21:16:16.637" Score="2" ViewCount="" Body="&lt;p&gt;You're mistaken in your history. The number of electrons emitted does, in fact, increase with the intensity of the radiation. Higher intensity means more photons, which means more chances to knock an electron loose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the &lt;em&gt;energy&lt;/em&gt; of the emitted electrons that people expected to increase that doesn't. In a classical wave picture, an increased intensity means a higher amplitude of the wave, which means more dramatic &quot;shaking&quot; of the electrons, and should thus lead to electrons being ejected with higher energy. In the quantum model, the emission is caused by a single photon, and the energy is determined by the frequency of that photon. Thus, the energy depends on the frequency but not the intensity.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-06T21:16:16.637" />
  <row Id="308" PostTypeId="2" ParentId="304" CreationDate="2010-11-06T21:24:24.677" Score="4" ViewCount="" Body="&lt;p&gt;If I had to guess at an answer, I would guess that it has something to do with the size of the blast. Lower frequency means a longer wavelength, which will be excited more effectively by a blast with a bigger displacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see the sort of thing I mean by looking at a piezoelectric transducer driving a speaker. For a fixed peak-to-peak voltage, the sound produced will be louder at higher frequencies, where the limited displacement of the speaker is closer to the wavelength of the sound wave. Speakers that are driven by PZT's tend to have some electronics to boost the amplitude of lower-frequency signals for this reason (with old-school magnetic coil speakers, you get this for free, and the displacement naturally ends up being larger for a slower oscillation). This is also why woofers for a stereo system tend to be larger-- the greater size more effectively excites low-frequency, long wavelength modes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(We discovered this by accident one day when I was in grad school when we were doing an experiment involving a mirror on a PZT, and spent an hour or so trying to figure it out before one of the older staff scientists wandered in and explained it. It sticks in my memory as a result.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could be wrong, but if I had to guess, this would be my guess.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-12-30T11:47:52.520" CommentCount="1" />
  <row Id="309" PostTypeId="2" ParentId="305" CreationDate="2010-11-06T22:24:00.217" Score="1" ViewCount="" Body="&lt;p&gt;The energy of a photon depends on the frequency: $E=h \nu$. This is not a classical result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So classically: To extract electrons from a material you need a minimal energy. This can be provided by a minimal intensity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But:&lt;/strong&gt; from a quantum mechanical point of view this reasoning breaks down. That's the discovery of the photoelectric effect (and its correct explanation): to extract an electron you need a minimal energy that can be provided by photons above a frequency (and thus energy) threshold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the energy of the incident photon is higher than the &quot;extraction work&quot; of the material (the energy needed to extract the electron) this additional energy is transformed in kinetic energy of the [free] electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore an increase in the intensity of light whose frequency is bellow the threshold cannot help as in the classical case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Increasing the intensity of the light means a higher number of photon (thus more total power but I does not mean more energy per photon), this can allow to extract more electrons (&quot;one by one&quot;).&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-11-06T22:24:00.217" />
  <row Id="310" PostTypeId="2" ParentId="290" CreationDate="2010-11-07T00:34:27.950" Score="9" ViewCount="" Body="&lt;p&gt;From &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0070362408&quot; rel=&quot;nofollow&quot;&gt;Stick and Rudder&lt;/a&gt;&lt;/em&gt; by Wolfgang Langewiesche, page 9, published 1944:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The main fact of all heavier-than-air&#xA;  flight is this: &lt;em&gt;the wing keeps the&#xA;  airplane up by pushing the air down&lt;/em&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It shoves the air down with its bottom&#xA;  surface, and it pulls the air down&#xA;  with its top surface; the latter&#xA;  action is the more important. But the&#xA;  really important thing to understand&#xA;  is that the wing, in whatever fashion,&#xA;  makes the air go down. In exerting a&#xA;  downward force upon the air, the wing&#xA;  receives an upward counterforce--by&#xA;  the same principle, known as Newton's&#xA;  law of action and reaction, which&#xA;  makes a gun recoil as it shoves the&#xA;  bullet out forward; and which makes&#xA;  the nozzle of a fire hose press&#xA;  backward heavily against the fireman&#xA;  as it shoots out a stream of water&#xA;  forward. Air is heavy; sea-level air&#xA;  weights about 2 pounds per cubic yard;&#xA;  thus, as your wings give a downward&#xA;  push to a cubic yard after cubic yard&#xA;  of that heavy stuff, they get upward&#xA;  reactions that are equally hefty.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;That's what keeps an airplane up.&#xA;  Newton's law says that, if the wing&#xA;  pushes the air down, the air must push&#xA;  the wing up. It also puts the same&#xA;  thing the other way 'round: if the&#xA;  wing is to hold the airplane up in the&#xA;  fluid, ever-yielding air, it can do so&#xA;  only by pushing the air down. All the&#xA;  fancy physics of Bernoulli's Theorem,&#xA;  all the highbrow math of the&#xA;  circulation theory, all the diagrams&#xA;  showing the airflow on a wing--all&#xA;  that is only an elaboration and more&#xA;  detailed description of just how&#xA;  Newton's law fulfills itself--for&#xA;  instance, the rather interesting but&#xA;  (for the pilot) really quite useless&#xA;  observation that the wing does most of&#xA;  its downwashing work by suction, with&#xA;  its top surface. ...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Thus, if you will forget some of this&#xA;  excessive erudition, a wing becomes&#xA;  much easier to understand; it is in&#xA;  the last analysis nothing but an air&#xA;  deflector. It is an inclined plane,&#xA;  cleverly curved, to be sure, and&#xA;  elaborately streamlined, but still&#xA;  essentially an inclined plane. That's,&#xA;  after all, why that whole fascinating&#xA;  contraption of ours is called an&#xA;  air-plane.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2010-11-07T00:39:30.517" LastActivityDate="2010-11-07T00:39:30.517" CommentCount="1" />
  <row Id="311" PostTypeId="2" ParentId="252" CreationDate="2010-11-07T01:04:54.400" Score="0" ViewCount="" Body="&lt;p&gt;The many-body equation is immensely difficult to study, both classically and quantum-mechanically.  The late John Pople, of Northwestern University, won a Nobel Prize in 1998 for his numerical models of wave functions of atoms, developing a theoretical basis for their chemical properties.  Here is a link:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://nobelprize.org/nobel_prizes/chemistry/laureates/1998/&quot; rel=&quot;nofollow&quot;&gt;http://nobelprize.org/nobel_prizes/chemistry/laureates/1998/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20" LastActivityDate="2010-11-07T01:04:54.400" CommentCount="1" />
  <row Id="312" PostTypeId="1" AcceptedAnswerId="313" CreationDate="2010-11-07T09:45:12.620" Score="4" ViewCount="221" Body="&lt;p&gt;I know classical physics, quantum mechanics, special relativity, basic nuclear physics. &#xA;I would like to get into some particle physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to get into that higgs boson, lepton, quark things :D &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where to start? Any books, websites, video lectures? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematical rigor required! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks &lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-11-07T10:57:03.330" LastActivityDate="2010-11-07T10:57:03.330" Title="Particle physics getting started" Tags="&lt;particle-physics&gt;&lt;books&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="313" PostTypeId="2" ParentId="312" CreationDate="2010-11-07T09:57:31.100" Score="5" ViewCount="" Body="&lt;p&gt;I would definitely recommend &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3527406018&quot; rel=&quot;nofollow&quot;&gt;David Griffiths' book on particle physics&lt;/a&gt;. I don't have my copy with me right now, but as I recall, the book explains what the different particles of the Standard Model are, as well as the various properties of particles that are important in modern particle physics. It also introduces the basics of quantum field theory, just enough to allow you to calculate cross sections and decay rates for various reactions. Toward the end, it shows you the basic ideas behind spontaneous symmetry breaking and the Higgs mechanism, which shows you where this prediction of the Higgs boson comes from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to get into more mathematical detail, another book I could recommend is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471887412&quot; rel=&quot;nofollow&quot;&gt;Halzen and Martin&lt;/a&gt;. It dates back to 1984 but the physics is still basically correct. I've found that that book takes a lot more effort to work through - that is, you actually have to slow down and think about what you're reading, and work through some of the math, but as long as you put the time in, the understanding you gain is well worth it.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-07T09:57:31.100" CommentCount="2" />
  <row Id="314" PostTypeId="2" ParentId="7" CreationDate="2010-11-07T10:56:02.237" Score="0" ViewCount="" Body="&lt;p&gt;There are no problems we can solve to overcome the uncertainty principle. We can't overcome it. Because it is like 1 + 1 = 2. Its mathematical fact and has got nothing to do with physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General uncertainty relation between two operators $\hat{A},\hat{B}$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma_A \sigma_B \geq \frac{|\langle [ \hat{A}  \hat{B}] \rangle|}{2} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when you plug in position $\hat{x}$ and momentum $\hat{p}$ operator in this equation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma_x \sigma_p \geq \frac{\hbar}{2}$&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-11-07T10:56:02.237" CommentCount="1" />
  <row Id="315" PostTypeId="2" ParentId="303" CreationDate="2010-11-07T14:15:23.057" Score="1" ViewCount="" Body="&lt;p&gt;As your question is very general, I can suggest a general answer: when a water wave is hitting a wall for example, you can &quot;trap&quot; an air bubble between the wave and the walls. This bubble can be compressed, the pressure will be higher and when the water moves, this bubble &quot;explode&quot; emitting a sound (which is nothing else than a pressure wave).&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-11-07T14:15:23.057" CommentCount="1" />
  <row Id="316" PostTypeId="2" ParentId="303" CreationDate="2010-11-07T14:57:07.600" Score="13" ViewCount="" Body="&lt;p&gt;If you throw a stone in a large body of water, this is what you should hear:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a high pitched slapping sound when the stone makes contact with the water. This is due to the air between the stone and the water being pushed out, as well as the surface ripples&lt;/li&gt;&#xA;&lt;li&gt;a low pitched sound growing in pitch (like &quot;Doo-eeeee&quot; ;-) due to the stone sinking and leaving an air &quot;hole&quot; in the water in its trail; this is then filled gradually with water which makes the sound raise in pitch&lt;/li&gt;&#xA;&lt;li&gt;a bubbling sound due to air bubbles trapped under water by the stone, emerging.&lt;/li&gt;&#xA;&lt;li&gt;a random noise due to waves colliding on the surface of the water.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-07T14:57:07.600" CommentCount="1" />
  <row Id="317" PostTypeId="1" AcceptedAnswerId="320" CreationDate="2010-11-07T15:29:27.027" Score="10" ViewCount="504" Body="&lt;p&gt;Quantum Mechanics is very successful in determining the overall statistical distribution of many measurements of the same process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it is completely clueless in determining the outcome of a single measurement. It can only describe it as having a &quot;random&quot; outcome within the predicted distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does this randomness come from? Has physics &quot;given up&quot; on the existence of microscopic physical laws by saying that single measurements are not bound to a physical law?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note: repeating the same measurement over and over with the same apparatus makes the successive measurements non-independent, statistically speaking. There could be a hidden &quot;stateful&quot; mechanism influencing the results. Has any study of fundamental QM features been performed taking this into account? What was the outcome?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: since 2 out of 3 questions seem to me not to answer my original question, maybe a clarification on the question itself will improve the quality of the page :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is about why single measurements have the values they have. Out of the, say, 1000 measure that make a successful QM experiment, why do the single measurements happen in that particular order?&#xA;Why does the wave function collapse to a specific eigenvalue and not another? It's undeniable that this collapse (or projection) happens. Is this random? What is the source of this randomness?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words: what is the mechanism of choice?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit 2: More in particular you can refer to chapter 29 of &quot;The road to reality&quot; by Penrose, and with special interest page 809 where the Everett interpretation is discussed - including why it is, if not wrong, quite incomplete.&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-02-18T22:26:26.817" LastActivityDate="2011-02-19T07:23:34.293" Title="Why can't the outcome of a QM measurement be calculated a-priori?" Tags="&lt;quantum-mechanics&gt;&lt;epistemology&gt;&lt;measurement-problem&gt;" AnswerCount="9" FavoriteCount="4" />
  <row Id="318" PostTypeId="2" ParentId="317" CreationDate="2010-11-07T15:52:25.373" Score="10" ViewCount="" Body="&lt;p&gt;First of all, let me start out by pointing out to you that there have been experimental violations of &lt;a href=&quot;http://en.wikipedia.org/wiki/Bell%27s_inequalities&quot; rel=&quot;nofollow&quot;&gt;Bell's inequalities&lt;/a&gt;. This provides damning evidence against hidden variable models of quantum mechanics, and thus essentially proves that the random outcomes are an essential feature of quantum mechanics. If the outcomes of measurements in every basis were predetermined, we should not be able to violate Bell's inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way of seeing why this is in fact a reasonable state of affairs is to consider Schroedinger's cat. Evolution of closed quantum systems is unitary, and hence entirely deterministic. For example, in the case of the cat, at some point in time we have a state of the system which is a superposition of (atom decayed and cat dead) and (atom undecayed and cat alive) with equal amplitude for each. This far quantum mechanics predicts the exact state of the system. We need to consider what happens when we open the box and look at the cat. When we do this, the system should then be in a superposition of (atom decayed, cat dead, box open, you aware cat is dead) and (atom undecayed, cat alive, box closed, you aware cat is alive). Clearly as time goes on the two branches of the wave function diverge further as the consequences of whether the cat is alive or dead propagate out into the world, and as a result no interference is likely possible. Thus there are two branches of the wave function with different configurations of the world. If you believe the &lt;a href=&quot;http://en.wikipedia.org/wiki/Everett_interpretation&quot; rel=&quot;nofollow&quot;&gt;Everett interpretation&lt;/a&gt; of quantum mechanics then both branches continue to exist indefinitely. Clearly our thinking depends on whether we have seen the cat alive or dead so that we ourselves are in a state (seen cat dead and aware we have seen the cat dead and not seen the cat alive) or (seen cat alive and aware we have seen the cat alive and not seen the cat dead). Thus even if we exist in a superposition we are only aware of a classical outcome to the experiment. Quantum mechanics allows us to calculate the exact wavefunction which is the outcome of the experiment, however, it cannot tell us a priori which branch we will find ourselves aware of after the experiment. This isn't really a shortcoming of the mathematical framework, but rather of our inability to perceive ourselves in anything other than classical states.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2010-11-07T15:52:25.373" CommentCount="15" />
  <row Id="319" PostTypeId="2" ParentId="317" CreationDate="2010-11-07T16:04:34.433" Score="0" ViewCount="" Body="&lt;p&gt;Quantum mechanics is inherently (and was developed as) as non-deterministic (stochastic) theory. The answer to your question lies in one of the postulates of quantum mechanics (see &lt;a href=&quot;http://physics.stackexchange.com/questions/317/why-cant-the-outcome-of-a-qm-measurement-be-calculated-a-priori&quot;&gt;this page&lt;/a&gt; for a full description;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In any measurement of the observable associated with operator $\hat{A}$, the only values that will ever be observed are the eigenvalues $a$, which satisfy the eigenvalue equation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since we know that the modulus squared of the wavefunction corresponds to the probability of a certain physical variable taking a given value, this wavefunction can be expanded on its eigenstates, and we see that each measurement result has an associated probability.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2010-11-07T16:04:34.433" CommentCount="6" />
  <row Id="320" PostTypeId="2" ParentId="317" CreationDate="2010-11-07T16:07:16.277" Score="6" ViewCount="" Body="&lt;p&gt;The short answer is that we do not know why the world is this way.   There might eventually be theories which explain this, rather than the current ones which simply take it as axiomatic.    Maybe these future theories will relate to what we currently call the holographic principle, for example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the apparently partially related fact of the quantization of elementary phenomena, e.g. that the measured spin of an elementary particle always is measured in integer or half integer values.   We also do not know why the world is this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we try to unify these two, the essential statistical aspect of quantum phenomena and the quantization of the phenomena themselves, the beginnings of a new theory start to emerge.   See papers by Tomasz Paterek, Borivoje Dakic, Caslav Brukner, Anton Zeilinger, and others for details .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/0804.1423&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0804.1423&lt;/a&gt;  and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.univie.ac.at/qfp/publications3/pdffiles/Paterek_Logical%20independence%20and%20quantum%20randomness.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.univie.ac.at/qfp/publications3/pdffiles/Paterek_Logical%20independence%20and%20quantum%20randomness.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;beginning with Zeilinger's (1999)&#xA;&lt;a href=&quot;http://www.springerlink.com/content/jt342534x711542g/&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/jt342534x711542g/&lt;/a&gt;, also online free &lt;a href=&quot;http://www.quantum.at/fileadmin/quantum/documents/A_Foundational_Principle_for_Quantum_Mechanics.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These papers present phenomenological (preliminary) theories in which logical propositions about elementary phenomena somehow can only carry 1 or a few bits of information. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for asking this question.    It was a pleasure to find these papers.   &lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-11-07T18:27:33.400" LastActivityDate="2010-11-07T18:27:33.400" CommentCount="1" />
  <row Id="321" PostTypeId="1" AcceptedAnswerId="322" CreationDate="2010-11-07T17:10:19.520" Score="2" ViewCount="350" Body="&lt;p&gt;My physics teacher says that centripetal force is caused by gravity. I'm not entirely sure how this works? How can force cause another in space (ie where there's nothing). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My astronomy teacher says that gravity is (note: not like) a 3D blanket and when you put mass on it, the mass causes a dip/dent in the blanket and so if you put another object with less mass it will roll down the dip onto the bigger mass. Is this true and is this what causes the centripetal force. &lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2010-11-13T15:37:29.417" Title="Planet orbits: what's the difference between gravity and centripetal force?" Tags="&lt;astrophysics&gt;&lt;forces&gt;&lt;astronomy&gt;" AnswerCount="4" />
  <row Id="322" PostTypeId="2" ParentId="321" CreationDate="2010-11-07T17:19:19.433" Score="3" ViewCount="" Body="&lt;p&gt;Simple answer: gravity &lt;em&gt;is&lt;/em&gt; a centripetal force, and can be envisaged clearly as such in Newtonian mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Centripetal&lt;/em&gt; just means a force that is &quot;radially inwards&quot; (&quot;directed towards the centre&quot;). The electric force, for example, is also clearly centripetal. (It's slightly harder to define &quot;centripetal&quot; for the magnetic force.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your astronomy teacher is referring to &lt;a href=&quot;http://en.wikipedia.org/wiki/General_relativity&quot; rel=&quot;nofollow&quot;&gt;Einstein's theory of general relativity&lt;/a&gt;. His description is loosely an overview  of the topology (fabric) of space-time and how it interacts with matter/energy - the manifold is however 4-dimensional, not 3D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, test particles (particles which do not really disturb the gravitational field) in general relativity follow a &lt;a href=&quot;http://en.wikipedia.org/wiki/Geodesic&quot; rel=&quot;nofollow&quot;&gt;geodesic&lt;/a&gt;. This is effectively a generalisation of a straight line (shortest route) of normal Euclidian space to the curved space of GR, and may be seen as the source of centripetal force in Newtonian physics.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2010-11-07T17:25:58.717" LastActivityDate="2010-11-07T17:25:58.717" CommentCount="2" />
  <row Id="323" PostTypeId="2" ParentId="321" CreationDate="2010-11-07T17:20:17.203" Score="2" ViewCount="" Body="&lt;p&gt;A centripetal force is a force directed towards the centre. It's just a characterisation of an existing force. &quot;Centripetal&quot; means &quot;towards the centre&quot; in Greek.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in the solar system, the sun exerts a gravitational force towards itself, and it is a centripetal force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your other question: how does gravity work? According to general relativity, energetic or massive objects distort space, so that other objects passing through the distorted space do not go straight, but bend their trajectory. From their point of view, they experience a force (gravity), and consequentially an acceleration which changes their trajectory.&#xA;So according to general relativity the medium through which gravity acts is actually the distortion of space.&#xA;This is what your astronomy teacher called &quot;a 3D blanket&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, according to Quantum Field Theory, forces are mediated by appropriate particles actually moving through space. Gravity would be carried by particles known as gravitons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both these theories, there is no action-at-a-distance, so body don't exert forces on one another instantaneously, but there's always something &quot;in the middle&quot; that carries or represents the force.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-07T17:20:17.203" CommentCount="5" />
  <row Id="324" PostTypeId="2" ParentId="321" CreationDate="2010-11-07T19:29:03.853" Score="0" ViewCount="" Body="&lt;p&gt;Gravity is a force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity is directed towards the center of the orbit i.e. the sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That makes gravity the centripetal force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a ball attached to a string and you are holding the other end of the string and moving your hand in such a way that the ball is in circular motion. Then tension in the string is centripetal force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, &#xA;ball = earth &lt;/p&gt;&#xA;&#xA;&lt;p&gt;you = sun&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tension in the string = gravity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope it helps. I have no idea about general relativity. &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-11-07T19:29:03.853" CommentCount="2" />
  <row Id="325" PostTypeId="2" ParentId="303" CreationDate="2010-11-07T19:35:54.777" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Sound is a mechanical wave that is an&#xA;  &lt;strong&gt;oscillation of pressure&lt;/strong&gt; transmitted&#xA;  through a solid, liquid, or gas,&#xA;  composed of frequencies within the&#xA;  range of hearing and of a level&#xA;  sufficiently strong to be heard, or&#xA;  the sensation stimulated in organs of&#xA;  hearing by such vibrations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Something(can be stone/ pebbles/ hands/ water etc.) interacting with water creates &quot;oscillation of pressure&quot;. Nature of &quot;oscillation of pressure&quot; depends on both water and interacting substance. And mostly familiar things interact with water thus giving rise to the characteristic sounds. &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-11-07T19:35:54.777" />
  <row Id="326" PostTypeId="1" AcceptedAnswerId="327" CreationDate="2010-11-07T21:28:23.650" Score="1" ViewCount="190" Body="&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/X4Ht1.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refer to number 6. This is the one I'm stuck on. So angular momentum is conserved right, so initial angular momentum is equal to final angular momentum. Initial is 7.87 so final must be 7.87, right? And so average torque is just change in angular momentum / change in time, so 0/7=0. What am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-07T22:05:34.710" LastActivityDate="2010-11-29T08:11:24.603" Title="Angular Momentum and Average Torque" Tags="&lt;mechanics&gt;&lt;classical-mechanics&gt;&lt;homework&gt;&lt;angular-momentum&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="327" PostTypeId="2" ParentId="326" CreationDate="2010-11-07T21:37:31.733" Score="2" ViewCount="" Body="&lt;p&gt;The angular momentum of the rod is 0 at the beginning because it is not rotating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would proceed like that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;by conservation of angular momentum, calculate the final rotational speed of the rod&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;with that given, calculate the final angular momentum of the rod&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You have that the torque gives the variation (with time) of the angular momentum. So if the torque is constant you just have &quot;torque = angular momentum / $\Delta t$&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I can be more specific if you want. Tell me where you find a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Apparently the steps are done in the previous questions, so this should just be a &quot;put everything together question&quot;.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-11-07T21:37:31.733" CommentCount="8" />
  <row Id="328" PostTypeId="1" AcceptedAnswerId="330" CreationDate="2010-11-07T22:20:17.770" Score="1" ViewCount="96" Body="&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/9ak7P.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm stuck on number 5. The answers to the first 4 are correct, but I dont know how to set up number 5. Any idea that I would have would require me having some kind of time information, but thats not given. Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2010-11-07T22:26:44.683" Title="Angular Momentum and Force" Tags="&lt;mechanics&gt;&lt;classical-mechanics&gt;&lt;homework&gt;&lt;angular-momentum&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="329" PostTypeId="2" ParentId="328" CreationDate="2010-11-07T22:24:10.930" Score="1" ViewCount="" Body="&lt;p&gt;Think about these:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the shape of the path the person takes?&lt;/li&gt;&#xA;&lt;li&gt;What force is needed to keep the person traveling along that path?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-07T22:24:10.930" />
  <row Id="330" PostTypeId="2" ParentId="328" CreationDate="2010-11-07T22:26:44.683" Score="2" ViewCount="" Body="&lt;p&gt;I'm not sure I totally understand the question, but I can make an educated guess. Firstly, I envisage a person sitting on the merry-go round with their body pointed tangentially to the circle, in which cases &lt;em&gt;no&lt;/em&gt; force &lt;em&gt;in that direction&lt;/em&gt; is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the person is free to move radially (can slip inwards/outwards), then a force is required to counteract the centripetal force. In other words,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{holding} = F_{centripetal}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the appropriate equation for centripetal force in terms of your known variables, and you can then plug in the values to get the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P.S. Don't look at my edit unless you want the answer, best you try yourself first!)&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2010-11-07T22:26:44.683" CommentCount="6" />
  <row Id="331" PostTypeId="1" AcceptedAnswerId="332" CreationDate="2010-11-07T22:42:04.093" Score="0" ViewCount="429" Body="&lt;p&gt;I'm trying to find the period of precession for a gyroscope. Now I was able to find the angular precession rate, which was 1.132 rad/s, but I have no idea how to convert this to a 'period', and google didnt find me anything useful. How do I do this?&lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-07T22:55:34.943" LastActivityDate="2011-04-29T08:28:14.847" Title="Period of Precession" Tags="&lt;mechanics&gt;&lt;angular-momentum&gt;&lt;homework&gt;&lt;precession&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="332" PostTypeId="2" ParentId="331" CreationDate="2010-11-07T22:46:04.410" Score="1" ViewCount="" Body="&lt;p&gt;If your rate is $\omega_p$ given in rad/s then the period is $\frac{2 \pi}{\omega_p}$ with units of seconds.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-11-07T22:46:04.410" CommentCount="14" />
  <row Id="333" PostTypeId="2" ParentId="331" CreationDate="2010-11-07T22:47:06.437" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Angular velocity is related to angular frequency by a constant factor. Period and frequency also have a simple (inverse) relation. Combine the two.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2010-11-07T22:47:06.437" CommentCount="4" />
  <row Id="335" PostTypeId="1" AcceptedAnswerId="336" CreationDate="2010-11-08T00:26:04.460" Score="4" ViewCount="446" Body="&lt;p&gt;My question is basically what exactly is electricity? I've simply been told before that it's a flow of electrons, but this seems too basic and doesn't show that electricity is instant.  What I mean is turning a switch has no delay between that and a light coming on. Is it really instantaneous? Or is it just so fast that we don't notice it?&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="3109" LastEditorDisplayName="" LastEditDate="2011-07-12T23:17:39.890" LastActivityDate="2011-07-12T23:17:39.890" Title="Is electricity instantaneous?" Tags="&lt;electricity&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="3" />
  <row Id="336" PostTypeId="2" ParentId="335" CreationDate="2010-11-08T01:09:45.120" Score="15" ViewCount="" Body="&lt;p&gt;It's just so fast you don't notice it. You won't see the effect of the travel time in something like turning on a light, because your eyes aren't fast enough to register the delay, but if you do even moderately precise experiments involving signal transmission and look at it on an oscilloscope, you will find that the travel time is easily measurable. The speed of signal propagation is close to that of light, or about a foot per nanosecond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It's worth noting that this is not the speed of electrons moving through the wires, which is dramatically slower. The signal is a disturbance that propagates more rapidly than the drift velocity of electrons in a conductor.)&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-08T01:09:45.120" CommentCount="7" />
  <row Id="337" PostTypeId="2" ParentId="167" CreationDate="2010-11-08T02:22:36.327" Score="1" ViewCount="" Body="&lt;p&gt;Try looking at the figures in a 2005 simulation by Takahashi et.al. in &lt;em&gt;Science&lt;/em&gt; magazine, that at least show recurring reversals at &lt;a href=&quot;http://www.sciencemag.org/cgi/content/full/309/5733/459?cookietest=yes&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencemag.org/cgi/content/full/309/5733/459?cookietest=yes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the viscosity, structure, and heat generation of the core are all to some degree unknown, and that the process may depend upon parametric amplification and even parametric resonance, this seems a pretty good beginning.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2010-11-08T02:52:06.663" LastActivityDate="2010-11-08T02:52:06.663" CommentCount="2" />
  <row Id="338" PostTypeId="2" ParentId="335" CreationDate="2010-11-08T02:57:01.413" Score="2" ViewCount="" Body="&lt;p&gt;The speed of conduction of electricity (usually voltage) in a conducting wire is usually determined by the time it takes for the current to saturate the capacitance of the wire or circuit.   For example, this is why local processes on the same computer chip run much faster than the same process offloaded to a separate chip, i.e. the circuits have much lower capacitance.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this is completely distinct from the universal maximal propagation velocity of any causal interaction in the universe, as seen by any observer, which also happens to be the speed of electromagnetic propagation in a vacuum, also called &quot;c&quot;. &lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-11-08T02:57:01.413" />
  <row Id="339" PostTypeId="1" AcceptedAnswerId="345" CreationDate="2010-11-08T03:07:41.340" Score="2" ViewCount="422" Body="&lt;p&gt;When light rays reflect off a boundary between two materials with different indices of refraction, a lot of the sources I've seen (recently) don't discuss the relation between the amplitude (or equivalently, intensity) of the transmitted/reflected rays and the original ray. Mostly they just discuss the phase difference induced by the reflection, for instance to calculate thin film interference effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/W8fL3.png&quot; alt=&quot;reflection/refraction diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to calculate the &lt;a href=&quot;http://en.wikipedia.org/wiki/Transmission_coefficient&quot; rel=&quot;nofollow&quot;&gt;transmission coefficient&lt;/a&gt; $T$ and &lt;a href=&quot;http://en.wikipedia.org/wiki/Reflection_coefficient&quot; rel=&quot;nofollow&quot;&gt;reflection coefficient&lt;/a&gt; $R$ based on other optical properties of the materials, such as the index of refraction? Or do they need to be looked up from a reference table?&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-08T05:36:30.887" LastActivityDate="2010-11-09T06:07:50.457" Title="How can I determine transmission/reflection coefficients for light?" Tags="&lt;optics&gt;&lt;reflection&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="340" PostTypeId="1" AcceptedAnswerId="342" CreationDate="2010-11-08T03:23:42.877" Score="6" ViewCount="607" Body="&lt;p&gt;Hotels usually install peepholes in their doors so that a person inside a suite can see who is at the door without having to open it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that there should be a convex lens within the peephole so that the person inside can easily see the person outside, but what I'm wondering about is the trick such that the person outside can only see blurred shapes if he tries to look though the peephole. Surely, there is more to it than specially-arranged convex lenses!&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-11-08T04:02:33.307" Title="The principle behind door peepholes" Tags="&lt;optics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="341" PostTypeId="1" AcceptedAnswerId="352" CreationDate="2010-11-08T03:50:18.870" Score="7" ViewCount="406" Body="&lt;p&gt;Not sure if this is a 'real' question, but what is the relation between physics and computer science? A lot of physicists are also computer scientists and vice versa. My professor has a PhD in Physics but is a Computer Science professor. Whats the relation? Why is this so common?&lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2010-11-08T13:53:21.037" Title="Physics and Computer Science" Tags="&lt;software&gt;&lt;computation&gt;" AnswerCount="6" CommentCount="4" FavoriteCount="2" />
  <row Id="342" PostTypeId="2" ParentId="340" CreationDate="2010-11-08T04:02:33.307" Score="9" ViewCount="" Body="&lt;p&gt;I'm not an expert in peepholes specifically, but as far as I know, there are a couple of factors that give a peephole its peculiar properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Peephole&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, the hole is fitted with a fisheye lens. This is a lens that, in one direction, takes a very wide field of view (up to a full half-sphere) and focuses it into a small area. This diagram should give you a rough idea (although this is certainly not exactly accurate to reality; maybe I'll do a better one later):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Y4cJx.png&quot; alt=&quot;fisheye lens ray diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lens is mounted such that it gathers light rays from all angles on the outside of the door, and focuses them on a small pupil-sized region on the inside of the door, so that you (being on the inside) can see whatever is on the outside, through a wide range of angles. This also has the effect of concentrating the light from a large area on the outside into a small area on the inside, thus increasing the intensity (power per unit area), thus making the image brighter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the same process also works in reverse, and that's why the peephole makes it difficult for people to see into your room. If your eye, on the inside, can see a wide range of angles from the outside, a person on the outside can only see a small range of angles from the inside. So they wind up getting a magnified view of a small part of your eye, basically. Also, just as the peephole makes the image outside appear brighter to you, it makes you on the inside appear dimmer to a person on the outside, by spreading the light coming out over a large angular area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other factor that tends to help, at least for houses during the daytime, is that there's a lot more light outside than there is inside, so someone attempting to peer in through your peephole also has to deal with reflection of the exterior light getting in the way. (This particular effect also works with windows)&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-08T04:02:33.307" />
  <row Id="343" PostTypeId="2" ParentId="339" CreationDate="2010-11-08T04:04:27.010" Score="3" ViewCount="" Body="&lt;p&gt;The equations that gives the transmission and reflection coefficients are called Fresnel equations.&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fresnel_equations&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fresnel_equations&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="62" LastActivityDate="2010-11-08T04:04:27.010" CommentCount="1" />
  <row Id="344" PostTypeId="2" ParentId="341" CreationDate="2010-11-08T04:30:57.880" Score="0" ViewCount="" Body="&lt;p&gt;A lot of physical modeling can be done in computer programs. Physics is based on creating mathematical pictures of the way the universe works. Physics really builds on the logic that is integral to computer science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this blog post sums it up nicely when it says, &quot;Without algorithms there would be no Physics! Physics is built on the fundamental assumption that we can model the world using algorithms. Computer Science is the most fundamental natural science.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.daniel-lemire.com/blog/archives/2009/10/05/what-is-more-fundamental-physics-or-computer-science/&quot; rel=&quot;nofollow&quot;&gt;Blog post on Physics and Computer Science&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="123" LastActivityDate="2010-11-08T04:30:57.880" CommentCount="4" />
  <row Id="345" PostTypeId="2" ParentId="339" CreationDate="2010-11-08T04:51:40.923" Score="4" ViewCount="" Body="&lt;p&gt;In addition to Fresnel equations, and in response to your question regarding the &quot;... relation between the amplitude of the transmitted/reflected rays and the original ray&quot;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_{\parallel}=\frac{2n_{1}\cos\theta_{i}}{n_{2}\cos\theta_{i}+n_{1}\cos\theta_{t}}A_{\parallel}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_{\perp}=\frac{2n_{1}\cos\theta_{i}}{n_{1}\cos\theta_{i}+n_{2}\cos\theta_{t}}A_{\perp}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_{\parallel}=\frac{n_{2}\cos\theta_{i}-n_{1}\cos\theta_{t}}{n_{2}\cos\theta_{i}+n_{1}\cos\theta_{t}}A_{\parallel}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_{\perp}=\frac{n_{1}\cos\theta_{i}-n_{2}\cos\theta_{t}}{n_{1}\cos\theta_{i}+n_{2}\cos\theta_{t}}A_{\perp}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A_{\parallel}$ and $A_{\perp}$ is the parallel and perpendicular component of the amplitude of the electric field for the incident wave, respectively. Accordingly for the $T$ (transmitted wave) and $R$ (reflected wave). I think the notation is straightforward to understand. This set of equations are also called Fresnel equations (there are three or four representations).&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2010-11-08T04:51:40.923" CommentCount="3" />
  <row Id="346" PostTypeId="2" ParentId="341" CreationDate="2010-11-08T05:02:56.607" Score="0" ViewCount="" Body="&lt;p&gt;A lot of things are common.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Physical processes can be thought of as a computation.&lt;/li&gt;&#xA;&lt;li&gt;Both fields exploit mathematics in very interesting way that is why it appeals to same people.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-11-08T07:25:04.580" LastActivityDate="2010-11-08T07:25:04.580" />
  <row Id="347" PostTypeId="1" CreationDate="2010-11-08T06:00:56.337" Score="8" ViewCount="228" Body="&lt;p&gt;Fish like electric eels and torpedoes have specially designed nerve cells that allow them to discharge hundreds of volts of electricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, while pure water is usually nonconductive, the dissolved salts and other stuff in both sea and fresh water allow them to be conductive. If an electric fish is able to use its electricity to stun enemies or prey, how come the fish itself is unaffected?&lt;/p&gt;&#xA;" OwnerUserId="172" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-11-08T06:13:21.683" LastActivityDate="2010-11-08T06:39:54.137" Title="Why don't electric fish shock themselves?" Tags="&lt;electricity&gt;&lt;biology&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="348" PostTypeId="2" ParentId="347" CreationDate="2010-11-08T06:12:19.093" Score="11" ViewCount="" Body="&lt;p&gt;I &lt;a href=&quot;http://www.google.com/search?q=Why+don%27t+electric+fish+shock+themselves&quot; rel=&quot;nofollow&quot;&gt;checked&lt;/a&gt; to make sure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simple answer is that electric eels insulate their critical tissues with a layer of fat below the skin, preventing the shock from traveling through their body as the &quot;path of least resistance&quot;. I may update with visuals and details if I can find good ones.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-11-08T06:12:19.093" />
  <row Id="349" PostTypeId="2" ParentId="335" CreationDate="2010-11-08T06:22:12.113" Score="6" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;NO. Electricity is NOT instantaneous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Light has finite velocity &lt;code&gt;3 x 10^8 m/s&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing can travel faster than light. Means everything has finite velocity less than light. Nothing travels instantaneously.&#xA;Hence electrons have finite velocity too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For answers to questions like&lt;br&gt;&#xA;How fast does electricity travels?  &lt;a href=&quot;http://en.wikipedia.org/wiki/Speed_of_electricity&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;here&lt;/strong&gt;&lt;/a&gt;&lt;br&gt;&#xA;Why does it appears to travel so fast? &lt;a href=&quot;http://www.radioelectronicschool.net/files/downloads/howfast.pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;here&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-11-08T06:22:12.113" CommentCount="2" />
  <row Id="350" PostTypeId="2" ParentId="347" CreationDate="2010-11-08T06:39:54.137" Score="16" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/uA1JA.png&quot; alt=&quot;alt text&quot;&gt;&#xA;Suppose current entering into this parallel circuit is $10A$&lt;br&gt;&#xA;then almost all the current flows through poor small fish's body &lt;/p&gt;&#xA;&#xA;&lt;p&gt;current through poor small fish's body = $\frac{1M}{1M+1} \approx  10A $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is probably the large picture but I am just guessing. Hope its correct. &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-11-08T06:39:54.137" CommentCount="1" />
  <row Id="351" PostTypeId="2" ParentId="341" CreationDate="2010-11-08T06:54:07.437" Score="4" ViewCount="" Body="&lt;p&gt;From my reasoning and knowledge of one CS professor who has a PhD in astronomy:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Above all, &lt;strong&gt;the answer depends on your definition of what a &quot;computer scientist&quot; is.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What do you mean by &quot;computer scientist&quot;? Someone who does research in a computer science department? Or does perhaps artificial intelligence, algorithm development, or grid computing for a commercial company?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Or do you mean people who study the same thing a &quot;pure&quot; computer scientist might, but in a different setting (perhaps biology or physics)a &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The last part is ambiguous: You probably mean &quot;Why is this [physics doctorates doing computer science] so common?&quot;, but it could also mean &quot;Why are CS professors with degrees in physics so common?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The computer science professor I mentioned above described learning the cost of bugs in a program: They (the astronomy graduate students) would write a complex program, a Fortran simulation of the sun's core, and print it on punched cards. They would then take the shoe-box of cards (all in a specific order!) to the computer desk, which was located in a &quot;bunker&quot; underneath the main building of the university. Perhaps a few days later, they would come back to receive the output of their queued program. If there was anything wrong with the output (perhaps it halted on an error, what a shame), they would have to go through their entire program to figure out what went wrong. Thus, they learned to meticulously debug their program on paper (really the only way, at the time, I suppose, since they could only use the mainframe) and &quot;run&quot; the program in their head.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;However, this professor (technically a lecturer) &quot;only&quot; teaches: first- and second-semester Java, a web development class, a course on building computers (sigh, I think you used to get to keep your build), and an introductory databases class. So, if we are talking about professors (loosely), their background will probably determine what kind of classes they teach.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditorDisplayName="" LastEditDate="2010-11-08T07:07:47.517" LastActivityDate="2010-11-08T07:07:47.517" CommentCount="4" />
  <row Id="352" PostTypeId="2" ParentId="341" CreationDate="2010-11-08T07:23:29.973" Score="9" ViewCount="" Body="&lt;p&gt;First off, physics tends to provide a very good background for people who move on to study problems in other areas, which is perhaps why there is a lot of cross-over to computer science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are also a number of areas at the interface of computer science and physics which attract people from both sides:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Computer hardware (which is generally based on semiconductor physics).&lt;/li&gt;&#xA;&lt;li&gt;Large scale simulations&lt;/li&gt;&#xA;&lt;li&gt;Physics of computation (Quantum computing, reversible computing, etc.)&lt;/li&gt;&#xA;&lt;li&gt;Theoretical computer science&#xA;etc.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Of these, perhaps the last one (TCS) seems the most surprising. However, in recent years, there has been significant success in applying ideas from thermodynamics and statistical mechanics to problems in computational complexity. An example of this would be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Simulated_annealing&quot; rel=&quot;nofollow&quot;&gt;simulated annealing&lt;/a&gt; algorithm which works extremely well for optimization problems, as well as work done on phase transitions in &lt;a href=&quot;http://en.wikipedia.org/wiki/3SAT#3-satisfiability&quot; rel=&quot;nofollow&quot;&gt;3SAT&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2010-11-08T07:23:29.973" />
  <row Id="353" PostTypeId="1" AcceptedAnswerId="354" CreationDate="2010-11-08T07:30:11.813" Score="5" ViewCount="270" Body="&lt;p&gt;The reasons for choosing length, mass, time, temperature, and amount as base quantities look (at least to me) obvious. What I'm puzzling about is why current (as opposed to resistance, electromotive force, etc.) and luminous intensity (as opposed to illuminance, emittance, etc.) were chosen to be base quantities. Does it have something to do with them being the easiest to measure?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2011-06-14T20:01:37.473" Title="Why were the SI base quantities chosen as such?" Tags="&lt;electricity&gt;&lt;history&gt;&lt;si-units&gt;" AnswerCount="6" CommentCount="2" />
  <row Id="354" PostTypeId="2" ParentId="353" CreationDate="2010-11-08T08:28:41.380" Score="6" ViewCount="" Body="&lt;p&gt;Historical issues I suppose; indeed current definition of ampere is rather stupid (force between two cables in vacuum) in light of the fact it could be done with number of elementary charges per second.&lt;br&gt;&#xA;Candela is even worse, because it involves properties of average human eye (so called &quot;luminosity function&quot;) -- so in principle it changes instantaneously as people get birth, die and their eyes age (not to mention various eye/brain fractures and treatments).&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-08T08:28:41.380" />
  <row Id="355" PostTypeId="2" ParentId="335" CreationDate="2010-11-08T08:39:37.763" Score="3" ViewCount="" Body="&lt;p&gt;From the point of view of simple models, one can depict &quot;little particles&quot; called electrons - huge amounts of them - moving slightly through a lattice of atoms because of an external cause that created a force field in the material. The movement of the particles is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Drift_velocity&quot; rel=&quot;nofollow&quot;&gt;drift velocity&lt;/a&gt; really quite slow. E.g. in a cupper wire of 1 mm diameter and a current of 3 amperes, the drift velocity of electrons is approx. 1.0 m/hour (for the calculation see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Drift_velocity#Numerical_example&quot; rel=&quot;nofollow&quot;&gt;numerical example&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The velocity of the electric signal through the wire is with the speed of light however. To get the idea, one can depict a tube filled with marbles. When you insert a marble at  one end, &quot;instantly&quot; another marble falls out at the other end, however the marbles themselves only shift over a small distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an explanation from the point of view of a particular (simple) model. Things become more complicated, more difficult or impossible to depict in the context of other models, e.g. pure quantum mechanics. In a pure quantum mechanical model individual hard core electrons do not exist and the explanation becomes purely mathematical.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-08T10:12:22.313" LastActivityDate="2010-11-08T10:12:22.313" CommentCount="2" />
  <row Id="356" PostTypeId="1" AcceptedAnswerId="421" CreationDate="2010-11-08T10:11:31.953" Score="10" ViewCount="686" Body="&lt;p&gt;How can I measure the mass of the earth at home? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How was the mass of the earth first measured?&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-11-18T17:58:14.263" LastActivityDate="2010-11-18T17:58:14.263" Title="How can I measure the mass of the earth at home?" Tags="&lt;classical-mechanics&gt;&lt;mechanics&gt;&lt;experimental-physics&gt;&lt;education&gt;&lt;home-experiment&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="4" />
  <row Id="357" PostTypeId="1" CreationDate="2010-11-08T10:12:37.333" Score="15" ViewCount="1198" Body="&lt;p&gt;Are there any experiments I can do to derive the speed of light with only common household tools?&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-11-18T17:53:43.660" LastActivityDate="2011-07-23T17:58:13.530" Title="Home experiments to derive the speed of light?" Tags="&lt;electromagnetism&gt;&lt;measurement&gt;&lt;experiment&gt;&lt;speed-of-light&gt;&lt;home-experiment&gt;" AnswerCount="13" CommentCount="7" FavoriteCount="8" />
  <row Id="358" PostTypeId="2" ParentId="357" CreationDate="2010-11-08T10:17:25.940" Score="7" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=9O2Keu6o3i0&quot; rel=&quot;nofollow&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;All you need is Chocolate and Microwave&lt;/strong&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-11-08T10:17:25.940" CommentCount="13" />
  <row Id="359" PostTypeId="2" ParentId="357" CreationDate="2010-11-08T10:21:16.140" Score="7" ViewCount="" Body="&lt;p&gt;There is a trick I have heard about before but never tried. The basic idea is to put a mars bar in a microwave oven for a short amount of time. First you remove the turntable, so the chocolate bar stays stationary. Then you turn the microwave on just long enough for the chocolate to start to melt. It should melt at the nodes of the standing field. You simply measure the distance between the nodes, and multiply by the frequency of the microwave oven to obtain the speed of light. There is a YouTube demonstration (by a kid) &lt;a href=&quot;http://www.youtube.com/watch?v=9O2Keu6o3i0&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2010-11-08T10:21:16.140" CommentCount="11" />
  <row Id="360" PostTypeId="2" ParentId="356" CreationDate="2010-11-08T10:51:18.287" Score="11" ViewCount="" Body="&lt;p&gt;You can make estimates of the earth's mass $M_e$ by estimating it's average density $\rho$ and using the formula $M_e = \rho \cdot V$, where of course $V= \frac{4}{3} \pi R^3$, so you have to know the earth's radius $R$. This method is rather a gamble because you don't know the earth's average density (suppose the earth would be hollow?). Newton made such an estimate himself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton's law of gravitation $F_{grav} = G \cdot \frac{M_e \cdot m}{R^2} = m \cdot g$ describes the attraction between a mass $m$ on the earth's surface and the earth. While $m$, $g$, so $F_{grav}$ and $R$ can be determined, there are two unknowns: $M_e$ and $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one can only determine the mass of the earth together with this constant $G$: the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_constant&quot; rel=&quot;nofollow&quot;&gt;gravitational constant&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot measure the mass of the earth at home, unless you have access to a Cavendish torsion balance to measure the gravitational constant. The Cavendish experiment is described in &lt;a href=&quot;http://en.wikipedia.org/wiki/Cavendish_experiment&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia article&lt;/a&gt;. Approx. 100 years after Newton the first experiment was done to measure $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of $G \cdot M_e$ can be measured very accurately (3,9860042·1014 m³/s²).&lt;br&gt;&#xA;The value of $G$ can be measured less accurately (6,674·10-11 m³/kg·s²) so that's the bottleneck for the determination of the earth's mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earth's mass is not constant, it is estimated that every day 40 tons of dust from meteorites hit it. This is totally neglectible for a mass of 5,9736 ∙ 10^24 kg.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-10T18:16:15.037" LastActivityDate="2010-11-10T18:16:15.037" />
  <row Id="361" PostTypeId="1" AcceptedAnswerId="362" CreationDate="2010-11-08T12:01:50.757" Score="6" ViewCount="146" Body="&lt;p&gt;Take the following gedankenexperiment in which two astronauts meet each other again and again in a perfectly symmetrical setting - a hyperspherical (3-manifold) universe in which the 3 dimensions are curved into the 4. dimension so that they can travel without acceleration in straight opposite directions and yet meet each other time after time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the one hand this situation is perfectly symmetrical - even in terms of homotopy and winding number. On the other hand the Lorentz invariance should break down according to GRT, so that one frame is prefered - but which one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So the question is: Who will be older? And why?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And even if there is one prefered inertial frame - the frame of the other astronaut should be identical with respect to all relevant parameters so that both get older at the same rate. Which again seems to be a violation of SRT in which the other twin seems to be getting older faster/slower...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should one find out what the prefered frame is when everything is symmetrical - even in terms of GRT?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if we are back to a situation with a prefered frame: what is the difference to the classical Galilean transform? Don't we get all the problems back that seemed to be solved by RT - e.g. the speed limit of light, because if there was a prefered frame you should be allowed to classically add velocities and therefore also get speeds bigger than c ?!? (I know SRT is only a local theory but I don't understand why the global prefered frame should not 'override' the local one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone please enlighten me (please in a not too technical way because otherwise I wouldn't understand!)&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2010-11-08T12:09:32.267" Title="Symmetrical twin paradox" Tags="&lt;general-relativity&gt;&lt;special-relativity&gt;&lt;paradox&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="362" PostTypeId="2" ParentId="361" CreationDate="2010-11-08T12:09:32.267" Score="5" ViewCount="" Body="&lt;p&gt;Your question is addressed in the following paper:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0101014&quot; rel=&quot;nofollow&quot;&gt;The twin paradox in compact spaces&lt;/a&gt;&lt;br/&gt;&#xA;  Authors: John D. Barrow, Janna Levin&lt;br/&gt;&#xA;  Phys.Rev. A63 (2001) 044104&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Abstract: Twins travelling at constant relative velocity will each see the other's time dilate leading to the apparent paradox that each twin believes the other ages more slowly. In a finite space, the twins can both be on inertial, periodic orbits so that they have the opportunity to compare their ages when their paths cross. As we show, they will agree on their respective ages and avoid the paradox. The resolution relies on the selection of a preferred frame singled out by the topology of the space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="47" LastActivityDate="2010-11-08T12:09:32.267" />
  <row Id="363" PostTypeId="1" AcceptedAnswerId="376" CreationDate="2010-11-08T12:10:44.927" Score="7" ViewCount="334" Body="&lt;p&gt;What are some good books, videos, websites for getting started with general relativity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically rigorous preferred! &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-11-12T16:33:04.903" Title="Getting started general relativity" Tags="&lt;general-relativity&gt;&lt;books&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="364" PostTypeId="1" CreationDate="2010-11-08T12:33:05.677" Score="6" ViewCount="318" Body="&lt;p&gt;How different would the effects of gravity be if the planet we're on is in the shape of a torus (doughnut-shaped)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an (approximately) spherical planet, it's slightly clear that objects would tend to be gravitationally attracted towards the center. However a torus would have a hole in its center, and I'm not sure if the attraction towards the center still applies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, could a person on such a planet walk in the vicinity of the hole without falling off?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Additional:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar question, but now consider a planet in the shape of a Möbius strip. Not only do you have to contend with the hole, but with the &quot;kink&quot; as well. Can a person stand up on the kink?&lt;/p&gt;&#xA;" OwnerUserId="172" LastEditorUserId="172" LastEditorDisplayName="" LastEditDate="2010-11-08T14:58:01.513" LastActivityDate="2010-11-08T21:35:21.087" Title="Gravity on a doughnut-shaped/Möbius planet" Tags="&lt;gravity&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="365" PostTypeId="2" ParentId="364" CreationDate="2010-11-08T12:58:12.363" Score="1" ViewCount="" Body="&lt;p&gt;I am doing this out of my head, so hopefully it's correct, but please double check.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When you are sufficiently away from the object, the laws will reduce to the classic point mass solution, F = GMmr^-2.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When you are on the plane of the torus hole, furthermore, contributions from the &quot;up&quot; and &quot;down&quot; directions will cancel out. The torus can be considered as a disk with a hole. In this scenario, the gravity will depend only on the amount of mass in the disk defined by the radius from the observer to the centre of of the torus. It will decrease linearly (I think) from the value on the external border (which should be the similar to the one defined at point 1. and the value zero on the inner border).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Inside the torus on the plane of the hole, gravity should be zero.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In other points you actually need to integrate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-08T12:58:12.363" CommentCount="1" />
  <row Id="366" PostTypeId="2" ParentId="341" CreationDate="2010-11-08T13:26:57.800" Score="1" ViewCount="" Body="&lt;p&gt;I have an M.Sc. in CS and an M.Sc. in Physics (astro), from my experience the intersections usually originate in (at least) two areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, in many fields of Physics, experiments and simulation produce large quantity of data, eg. SDSS (dataset is many TBs), Pan-Starrs (will produce TB/day) in astrophysics, large colliders such as LHC in particle physics (&lt;a href=&quot;http://highscalability.com/blog/2010/9/16/how-can-the-large-hadron-collider-withstand-one-petabyte-of.html&quot; rel=&quot;nofollow&quot;&gt;producing 1PB/s&lt;/a&gt;). When dealing with the storage and processing of such large datasets, physicists get involved with CS and some of us enjoy it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, there's a field called &quot;complex systems&quot;, such as spin glasses or stock markets, which can be modeled effectively on computers, hence the computer program becomes the experiment itself. Complex systems is an exciting field for physicists because fairly simple systems (eg. a pendelum) can produce very complex behaviour, they are relatively cheap to research (PCs are cheap, software is free), and they can be found in many fields of physics.&lt;/p&gt;&#xA;" OwnerUserId="47" LastActivityDate="2010-11-08T13:26:57.800" />
  <row Id="367" PostTypeId="2" ParentId="167" CreationDate="2010-11-08T13:38:25.090" Score="4" ViewCount="" Body="&lt;p&gt;Live on earth is protected from solar wind by the earth's magnetic field. Charged particles from the sun (mostly) penetrate the earth's atmosphere with great velocity. These particles can be trapped by a magnetic field to follow &lt;a href=&quot;http://en.wikipedia.org/wiki/Lorentz_force&quot; rel=&quot;nofollow&quot;&gt;circular path's around the magnetic field lines&lt;/a&gt;, thereby losing their energy due to collisions or &lt;a href=&quot;http://en.wikipedia.org/wiki/Bremsstrahlung&quot; rel=&quot;nofollow&quot;&gt;bremstrahlung&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From first principles we can try to make an estimate of the strength of the magnetic field required to trap charged particles arriving with great velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with a lorentz force and a circular movement we have: $Bqv = m\frac{v^2}{r}$, so&#xA;$B= \frac{mv}{qr}$.&lt;br&gt;&#xA;$v$ is the velocity of the particle, approx. light velocity, order of magnitude $10^8$ m/s. $q$ is the charge of the particle, elementary unit order of magnitude $10^{-19}$ C. $m$ is the mass of the particle, approx. proton mass, order of magnitude $10^{-27}$ kg. $r$ is the radius wherein the particle has to be trapped, at the most 10 km (height of the atmosphere) $10^4$ m. This gives $B \sim 10^{-4}$ T $= 1$ Gauss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll have to appreciate the intelligent design ...&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-08T13:38:25.090" CommentCount="2" />
  <row Id="368" PostTypeId="2" ParentId="364" CreationDate="2010-11-08T13:49:26.183" Score="6" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/GravField.htm&quot; rel=&quot;nofollow&quot;&gt;Gravitational Field from a Ring of Mass&lt;/a&gt;&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;http://www.mathpages.com/home/kmath402/kmath402.htm&quot; rel=&quot;nofollow&quot;&gt;Gravity of a Torus&lt;/a&gt;&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;http://cnx.org/content/m15104/latest/&quot; rel=&quot;nofollow&quot;&gt;Gravitation field due to rigid bodies&lt;/a&gt;&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ringworld&quot; rel=&quot;nofollow&quot;&gt;Ringworld :)&lt;/a&gt;&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-11-08T13:49:26.183" />
  <row Id="369" PostTypeId="2" ParentId="341" CreationDate="2010-11-08T13:53:21.037" Score="3" ViewCount="" Body="&lt;p&gt;I think the main reason why this is so common is that many people who are of the tenured professor age now (50-60) were in graduate school before most colleges offered a Ph.D. in computer science.  So back then, people who were interested in theoretical computer science got their doctorate in Mathematics, and people who were interested in applied computer science got their degrees in a field of the natural sciences.  Since physics offers the most readily answerable questions computationally, particularly with the computational power available in the 1950s and 1960s, many applied computer scientists ended up in physics programs.  When schools began to offer full bachelor's, master's, and doctoral programs in computer science, these people returned to their first love.  (I gathered most of this from talking to a CS professor I had in undergrad who followed exactly the same path you describe.)  I'm pretty sure there are a lot more physics Ph.D.s in computer science professorships than vice versa (even though I'm sure there are exceptions, and I'll get several comments pointing this out).  I realize there is a lot of overlap between the two fields, but if you want to know the real reason what there are so many &lt;em&gt;professors&lt;/em&gt; in CS from physics, this is why.&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2010-11-08T13:53:21.037" />
  <row Id="370" PostTypeId="1" AcceptedAnswerId="374" CreationDate="2010-11-08T14:20:02.450" Score="2" ViewCount="261" Body="&lt;p&gt;I read a few days ago that in the LHC temperatures of billions of degrees were achieved. I'm curious to know what does it really mean such a temperature? The concept of temperature is easy to grasp when the numbers are familiar, such as 100 C, but when it's in the range of millions it's difficult to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you can provide some explanation of how the temperature is estimated it would be very helpful.&lt;/p&gt;&#xA;" OwnerUserId="57" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-08T16:22:32.353" LastActivityDate="2010-11-08T16:22:32.353" Title="What does it mean a temperature of billions of degrees?" Tags="&lt;high-energy-physics&gt;&lt;large-hadron-collider&gt;&lt;temperature&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="371" PostTypeId="1" AcceptedAnswerId="375" CreationDate="2010-11-08T14:41:35.107" Score="4" ViewCount="191" Body="&lt;p&gt;Recently I was surprised to discover that no exact solution for the position of a planet as a function of time exists. I am referring to the two-body problem in a gravitational field where Newtons law of gravity holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well known are proofs that the planet will move in an ellipse, Keplers laws can be derived fairly easy etc., but for the exact position of the planet on the ellipse as a function of time, no formula exists, only numerical approximations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;br&gt;&#xA;Can anybody elaborate on the deeper reason(s) that this relatively simple case cannot be solved?&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-08T16:21:30.543" LastActivityDate="2010-12-21T20:49:47.547" Title="How about an exact solution for the position of a planet as a function of time?" Tags="&lt;mathematical-physics&gt;&lt;newtonian-gravity&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="372" PostTypeId="2" ParentId="370" CreationDate="2010-11-08T14:58:21.107" Score="-1" ViewCount="" Body="&lt;p&gt;Physics is about ideas and about experience, so when a certain number is impossible to perceive, the concept of understanding itself must be taken into account,&#xA;i.e. we enter the realm of philosophy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem does not only occur with temperature. What is a distance of 10^16 m? What does it mean to go back in time 10^15 years and from there 10^-3 sec forward?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such numbers - as temperature in the range of millions degrees - emerge from calculations where experimental values have been inserted in a formula. The result (e.g. 10^6 degree Celsius) is not a reality in the sense of a perception by a human.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;understanding&lt;/em&gt; lies in the ability to tackle the mathematical side, the philosophical problem of reality is not taken into account.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-08T14:58:21.107" CommentCount="1" />
  <row Id="373" PostTypeId="2" ParentId="370" CreationDate="2010-11-08T15:14:00.147" Score="0" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;We can't really understand weight beyond a few tons(or whatever you can lift ;) and less than a few grams.&lt;/li&gt;&#xA;&lt;li&gt;Our eyes can perceive only the visible spectrum.&lt;/li&gt;&#xA;&lt;li&gt;Our ears can't hear frequencies only within certain range.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Similarly we can't feel temperature beyond some limit.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These limitations are imposed on our organs by evolution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Its just too hot to be explained intuitively. We need some math ;-)&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-11-08T15:14:00.147" CommentCount="1" />
  <row Id="374" PostTypeId="2" ParentId="370" CreationDate="2010-11-08T15:24:11.143" Score="5" ViewCount="" Body="&lt;p&gt;You posed two distinct questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;how is temperature defined as a&#xA;physical quantity? &lt;/li&gt;&#xA;&lt;li&gt;how is temperature measured in these&#xA;circumstances?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For the first question, temperature is defined as a thermodynamic quantity relating the change of entropy and the change of internal energy of a system. This is not very intuitive, I agree, but it is THE definition, and it make physicists sure they are talking about the same well-defined quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A somewhat less correct but a much more intuitive definition of temperature is the amount of energy of the chaotic motion per particle. If your particles move chaotically very fast, near the speed of light, so that energy per particle is very large, you temperature is large as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the second question, physicists measure temperature of heavy-ion collisions indirectly, on the basis of several characteristics they observe in their detectors. The simplest way is by detecting of energetic photons and fitting them to thermal spectrum; another way is by studying the geometry of flow of hundreds of particles produced in the collision and fitting them to some models.&lt;/p&gt;&#xA;" OwnerUserId="59" LastActivityDate="2010-11-08T15:24:11.143" CommentCount="6" />
  <row Id="375" PostTypeId="2" ParentId="371" CreationDate="2010-11-08T15:26:23.543" Score="11" ViewCount="" Body="&lt;p&gt;This is not that there is no exact solution, only the exact solutions for $x(t)$ and $y(t)$ use elliptic functions. The problem whether elliptic functions (which are defined by inverse of some integrals) are &quot;good&quot; functions is a bit philosophical one; one can on one hand state that sine is not a real function because one must integrate or sum a infinite series to calculate it, and on the other that even Lorentz attractor solution can be called three Lorentz chaotic functions with 4 parameters $a$, $b$, $c$ and $t$ and tabularized. &lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-21T20:49:47.547" LastActivityDate="2010-12-21T20:49:47.547" CommentCount="8" />
  <row Id="376" PostTypeId="2" ParentId="363" CreationDate="2010-11-08T16:17:37.413" Score="15" ViewCount="" Body="&lt;p&gt;I can only recommend textbooks because that's what I've used, but here are some suggestions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0805386629&quot; rel=&quot;nofollow&quot;&gt;Gravity: An Introduction To General Relativity&lt;/a&gt; by &lt;strong&gt;James Hartle&lt;/strong&gt; is reasonably good as an introduction, although in order to make the content accessible, he does skip over a lot of mathematical detail. For your purposes, you might consider reading the first few chapters just to get the &quot;big picture&quot; if you find other books to be a bit too much at first.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521887054&quot; rel=&quot;nofollow&quot;&gt;A First Course in General Relativity&lt;/a&gt; by &lt;strong&gt;Bernard Schutz&lt;/strong&gt; is one that I've heard similar things about, but I haven't read it myself.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0805387323&quot; rel=&quot;nofollow&quot;&gt;Spacetime and Geometry: An Introduction to General Relativity&lt;/a&gt; by &lt;strong&gt;Sean Carroll&lt;/strong&gt; is one that I've used a bit, and which goes into a slightly higher level of mathematical detail than Hartle. It introduces the basics of differential geometry and uses them to discuss the formulation of tensors, connections, and the metric (and then of course it goes on into the theory itself and applications).&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0226870332&quot; rel=&quot;nofollow&quot;&gt;General Relativity&lt;/a&gt; by &lt;strong&gt;Robert M. Wald&lt;/strong&gt; is a classic, though I'm a little embarrassed to admit that I haven't read much of it. From what I know, though, there's certainly no shortage of mathematical detail, and it derives/explains certain principles in different ways from other books, so it can either be a good reference on its own (if you're up for the detail) or a good companion to whatever else you're reading.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0716703440&quot; rel=&quot;nofollow&quot;&gt;Gravitation&lt;/a&gt; by &lt;strong&gt;Charles Misner, Kip Thorne, and John Wheeler&lt;/strong&gt;, is pretty much the authoritative reference on general relativity (to the extent that one exists). It discusses many aspects and applications of the theory in far more mathematical and logical detail than any other book I've seen. (Consequently, it's very thick.) I would recommend having a copy of this around as a reference to go to about specific topics, when you have questions about the explanations in other books, but it's not the kind of thing you'd sit down and read large chunks of at once.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471925675&quot; rel=&quot;nofollow&quot;&gt;Gravitation and Cosmology: Principles and Applications of the General Theory of Relativity&lt;/a&gt; by &lt;strong&gt;Steven Weinberg&lt;/strong&gt; is another one that I've read a bit of. Honestly I find it a bit hard to follow - just like some of Weinberg's other books, actually - since he gets into such detailed explanations, and it's easy to get bogged down in trying to understand the details and forget about the main point of the argument. Still, this might be another one to go to if you're wondering about the details omitted by other books. This is not as comprehensive as the Misner/Thorne/Wheeler book, though.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-08T16:17:37.413" CommentCount="3" />
  <row Id="377" PostTypeId="2" ParentId="364" CreationDate="2010-11-08T16:22:35.140" Score="0" ViewCount="" Body="&lt;p&gt;On the torus:&lt;br&gt;&#xA;Walking towards the inner side of the torus one becomes lighter, because one has gravitational pull under one's feet (which is stronger because it is nearer) and gravitational pull above one's head making you lighter. The outer side of the torus is the side where people are heavier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the Möbius strip:&lt;br&gt;&#xA;The gravitational pull varies like in the case of the torus, one becomes lighter and then heavier again while making a world trip.&#xA;Walking from one side to the &quot;other&quot; side [parenthisis because it is the same side] via the thickness of the strip - assuming that is accessible, otherwise one has to fly to the &quot;other&quot; side - the situation is reversed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;a href=&quot;http://physics.stackexchange.com/questions/364/gravity-on-a-doughnut-shaped-mobius-planet/365#365&quot;&gt;Sklivvz&lt;/a&gt; notes:&lt;br&gt;&#xA;In the center of the torus or Möbius strip all gravity is cancelled.&lt;br&gt;&#xA;At a great distance the form of the object may be neglected, it can be treated as a point mass. &lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-08T21:35:21.087" LastActivityDate="2010-11-08T21:35:21.087" CommentCount="2" />
  <row Id="378" PostTypeId="2" ParentId="357" CreationDate="2010-11-08T19:18:59.170" Score="7" ViewCount="" Body="&lt;p&gt;With a clock and a telescope you could repeat &lt;a href=&quot;http://en.wikipedia.org/wiki/R%C3%B8mer%27s_determination_of_the_speed_of_light&quot; rel=&quot;nofollow&quot; rel=&quot;nofollow&quot;&gt;Rømer's determination of the speed of light&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2010-11-08T19:18:59.170" CommentCount="3" />
  <row Id="379" PostTypeId="1" AcceptedAnswerId="446" CreationDate="2010-11-08T19:45:31.210" Score="3" ViewCount="138" Body="&lt;p&gt;In quantum optics, a perfect absorber of light is said to emit the &quot;vacuum field&quot;.  In practice, any beam dump will be at finite temperature, so it will emit blackbody radiation.  How do these fields compare?  Is there some critical frequency (for a given temperature) above which the vacuum field dominates?&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2010-11-10T00:37:14.450" Title="In optics, how does the vacuum state compare to thermal radiation?" Tags="&lt;optics&gt;&lt;quantum-optics&gt;" AnswerCount="2" />
  <row Id="380" PostTypeId="2" ParentId="339" CreationDate="2010-11-08T20:30:03.840" Score="3" ViewCount="" Body="&lt;p&gt;This was intended as a comment, but for the sake of clarity, I'd better use an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding to the case $\mu \neq 1$, we can start using the following set of equations, which are derived from the Maxwell equations and after applying boundary conditions that demand that across the boundary the tangential components of $E$ and $H$ should be continuous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos\theta_{i}(A_{\parallel}-R_{\parallel})=\cos\theta_{t}T_{\parallel}$$&#xA;$$A_{\perp}+R_{\perp}=T_{\perp}$$&#xA;$$\sqrt{\frac{\epsilon_{1}}{\mu_{1}}}\cos\theta_{i}(A_{\perp}-R_{\perp})=\sqrt{\frac{\epsilon_{2}}{\mu_{2}}}\cos\theta_{t}T_{\perp}$$&#xA;$$\sqrt{\frac{\epsilon_{1}}{\mu_{1}}}(A_{\parallel}+R_{\parallel}=\sqrt{\frac{\epsilon_{2}}{\mu_{2}}}T_{\parallel}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, adding together the first and fourth equation, you obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_{\parallel}=\frac{2\cos\theta_{i}\sqrt{\epsilon_{1}\mu_{2}}}{\cos\theta_{t}\sqrt{\mu_{2}\epsilon_{1}}+\sqrt{\epsilon_{2}\mu_{1}}\cos\theta_{i}}A_{\parallel}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding the second and third equation, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_{\perp}=\frac{2 \sqrt{\mu_{2}\epsilon_{1}}\cos\theta_{i}}{\sqrt{\epsilon_{2}\mu_{1}}\cos\theta_{t}+\cos\theta_{i}\sqrt{\mu_{2}\epsilon_{1}}}A_{\perp}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accordingly for $R_{\parallel}$ and $R_{\perp}$ (in which we have to substitute the value we already found for $T_{\parallel}$ and $T_{\perp}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_{\parallel}=\frac{\sqrt{\epsilon_{2}\mu_{1}}\cos\theta_{i}-\sqrt{\epsilon_{1}\mu_{2}}\cos\theta_{t}}{\sqrt{\epsilon_{2}\mu_{1}}\cos\theta_{i}+\sqrt{\epsilon_{1}\mu_{2}}\cos\theta_{t}}A_{\parallel}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_{\perp}=\frac{\sqrt{\epsilon_{1}\mu_{2}}\cos\theta_{i}-\sqrt{\epsilon_{2}\mu_{1}}\cos\theta_{t}}{\sqrt{\epsilon_{1}\mu_{2}}\cos\theta_{i}+\sqrt{\epsilon_{2}\mu_{1}}\cos\theta_{t}}A_{\perp}$$&lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="18" LastEditorDisplayName="" LastEditDate="2010-11-09T06:07:50.457" LastActivityDate="2010-11-09T06:07:50.457" CommentCount="3" />
  <row Id="381" PostTypeId="2" ParentId="353" CreationDate="2010-11-08T20:34:45.387" Score="4" ViewCount="" Body="&lt;p&gt;Of course, history has its own way of determining things, but I suspect that the greatest reason for the choices of the base SI units was that they were &lt;em&gt;easily measurable&lt;/em&gt; at the time. Other quantities such as resistance, magnetic, density, power, are (or certainly were at the time) much more difficult to measure precisely. In this sense, the choices were fairly reasonable.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2010-11-08T20:34:45.387" />
  <row Id="382" PostTypeId="2" ParentId="317" CreationDate="2010-11-08T22:19:08.283" Score="1" ViewCount="" Body="&lt;p&gt;The mechanism of choice in one particular instant of a quantum-mechanical experiment is unknown in all physics today - it's just that this fact for many physicists is too uncomfortable to accept or to admit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Einstein couldn't accept it, Bohr and Feynmann admitted it though. The question leads us to the never ending &lt;a href=&quot;http://en.wikipedia.org/wiki/Bohr%E2%80%93Einstein_debates&quot; rel=&quot;nofollow&quot;&gt;Bohr–Einstein debates&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fundamental fact at the heart of physics is that wave-functions $\psi$ cannot be measured, only their absolute square $|\psi|^2$. Pure logic forces us to admit that statements about wave functions are not statements about a perceivable reality. Quantum theory is a theory that explains how to calculate the outcome of a measurement, it cannot tell us what happened before or during the measurement because a wave function is an idea, an inaccessible hypothesis .&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-08T22:19:08.283" />
  <row Id="383" PostTypeId="1" AcceptedAnswerId="393" CreationDate="2010-11-08T22:31:31.397" Score="5" ViewCount="560" Body="&lt;p&gt;I was wondering how easily these two pseudo-forces can be derived mathematically in order to exhibit a clear physical meaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you proceed?&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-11-09T01:22:46.823" LastActivityDate="2010-11-09T03:14:19.610" Title="Derivation of the centrifugal and coriolis force" Tags="&lt;classical-mechanics&gt;&lt;coriolis-force&gt;&lt;centrifugal-force&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" />
  <row Id="384" PostTypeId="1" AcceptedAnswerId="948" CreationDate="2010-11-08T22:38:02.693" Score="9" ViewCount="397" Body="&lt;p&gt;Impedance is a concept that shows up in any area of physics concerning waves.  In transmission lines, impedance is the ratio of voltage to current.  In optics, index of refraction plays a role similar to impedance.  Mechanical impedance is the ratio of force to velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is a general definition of impedance?&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What are some examples of &quot;impedance matching&quot; other than in electrical transmission lines?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="62" LastEditorDisplayName="" LastEditDate="2010-11-09T04:28:03.357" LastActivityDate="2010-11-21T06:27:48.627" Title="What is a general definition of impedance?" Tags="&lt;optics&gt;&lt;electrical-engineering&gt;&lt;waves&gt;" AnswerCount="5" FavoriteCount="4" />
  <row Id="385" PostTypeId="2" ParentId="384" CreationDate="2010-11-08T22:49:38.453" Score="0" ViewCount="" Body="&lt;p&gt;Impedance is general simply refers to the &quot;amount of impeding (or block)&quot;. It's specific meanings in the various branches of physics (mechanical, electrical, wave impedance) varies, but are all loosely based on the concept of the everyday word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good example of impedance matching I always think of relates to audio systems. When you want to chain a number of amplifiers in stages, you usually want to match the output impedance of one with the input impedance of the other in order to maximise power transfer.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2010-11-08T22:49:38.453" CommentCount="13" />
  <row Id="386" PostTypeId="1" CreationDate="2010-11-08T22:50:11.163" Score="1" ViewCount="282" Body="&lt;p&gt;I read today (&lt;a href=&quot;http://en.wikipedia.org/wiki/Continuous_Fourier_transform#Eigenfunctions&quot; rel=&quot;nofollow&quot;&gt;ref&lt;/a&gt;) that the Continuous Fourier Transform has four eigenvalues: +1, +i, -1, and -i.   Associated with each eigenvalue is a space of eigenfunctions: functions which retain their form after undergoing the Fourier transform.  Perhaps the best known example is the Gaussian: the Fourier transform of a Gaussian is again Gaussian.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more general example is the Hermite-Gauss functions (Gaussian multiplied by Hermite polynomial).  These are also eigenfunctions of the Fourier transform, which is why TEMxx modes (with Hermite-Gauss transverse profiles) are stable modes of laser beam propagation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads me to wonder a number of things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How do I show that the Fourier transform has just these four eigenvalues?&lt;/li&gt;&#xA;&lt;li&gt;Given an arbitrary function, how do I find the projections onto the four eigenspaces?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="29" LastActivityDate="2010-11-09T09:35:56.943" ClosedDate="2010-11-09T10:43:17.827" Title="How do I compute the eigenfunctions of the Fourier Transform?" Tags="&lt;fourier-transform&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="387" PostTypeId="1" CreationDate="2010-11-08T22:50:36.140" Score="10" ViewCount="735" Body="&lt;p&gt;It is well known that quantum mechanics and (special and/or general) relativity do not fit well. I am wondering whether it is possible to make a list of contradictions or problems between them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. relativity theory uses a space-time &lt;em&gt;continuum&lt;/em&gt;, while quantum theory uses &lt;em&gt;discrete&lt;/em&gt; states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not merely looking for a solution or rebuttal of such opposites, more for a survey of the field out of interest.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-22T23:21:05.550" LastActivityDate="2011-03-11T02:54:01.163" Title="A list of inconveniences between quantum mechanics and relativity?" Tags="&lt;quantum-mechanics&gt;&lt;general-relativity&gt;&lt;special-relativity&gt;&lt;quantum-gravity&gt;" AnswerCount="5" CommentCount="16" FavoriteCount="4" />
  <row Id="388" PostTypeId="1" AcceptedAnswerId="390" CreationDate="2010-11-08T23:03:07.997" Score="4" ViewCount="437" Body="&lt;p&gt;Some high-energy experiments (RHIC, LHC) use ion-ion collisions instead of proton-proton collisions. Although the total center-of-mass energy is indeed higher than p-p collisions, it might happen that the total energy &lt;em&gt;per nucleon&lt;/em&gt; is actually lower. What are the advantages of using ion-ion collisions (e.g. gold-gold or lead-lead) instead of proton-proton collisions, considering the same accelerator?&lt;/p&gt;&#xA;" OwnerUserId="103" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-08T23:43:03.223" LastActivityDate="2011-05-19T16:41:55.427" Title="Advantages of high-energy heavy-ion collisions over proton-proton collisions?" Tags="&lt;particle-physics&gt;&lt;high-energy-physics&gt;&lt;large-hadron-collider&gt;&lt;collision&gt;&lt;heavy-ion&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="389" PostTypeId="1" AcceptedAnswerId="459" CreationDate="2010-11-08T23:07:27.523" Score="4" ViewCount="257" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_sea&quot; rel=&quot;nofollow&quot;&gt;From Wikipedia&lt;/a&gt;: [The Dirac sea is a theoretical model of the vacuum as an infinite sea of particles with negative energy. It was first postulated by the British physicist Paul Dirac in 1930 to explain the anomalous negative-energy quantum states predicted by the Dirac equation for relativistic electrons. The positron, the antimatter counterpart of the electron, was originally conceived of as a hole in the Dirac sea, well before its experimental discovery in 1932.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Dirac's idea is completely correct in the context of solid state physics, where the valence band in a solid can be regarded as a &quot;sea&quot; of electrons. Holes in this sea indeed occur, and are extremely important for understanding the effects of semiconductors, though they are never referred to as &quot;positrons&quot;. Unlike in particle physics, there is an underlying positive charge — the charge of the ionic lattice — that cancels out the electric charge of the sea.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It always confused me to think of holes as positive charge carriers in semi-conductors as &lt;strong&gt;not being real&lt;/strong&gt;: real electrons move from one lattice-position to another lattice-position, which effectively &lt;em&gt;looks like&lt;/em&gt; a positive hole in the lattice that is moving in the other direction, but in reality a real electron moves, the hole is kind of an &quot;illusion&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand the positrons are always introduced as &lt;strong&gt;real&lt;/strong&gt; hard-core particles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quotes from the Wikipedia article make me unsure: how should I look upon these phenomena? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: holes in a Dirac sea give rise to real pos. entities in one case and to unreal pos. entities in another - how can we distinguish, is it a matter of formalism?&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-09T17:58:47.213" LastActivityDate="2011-06-07T23:49:19.763" Title="Positrons versus holes as positive charge carriers" Tags="&lt;solid-state-physics&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="390" PostTypeId="2" ParentId="388" CreationDate="2010-11-08T23:17:19.390" Score="5" ViewCount="" Body="&lt;p&gt;As a clarification the energy per nucleon is always lower: for example, currently in the LHC the proton top energy is 3.5 TeV. Now the Pb energy is 3.5 TeV times Z so the energy per nucleon is 3.5*Z/A and A is greater than Z for every nucleus (except the proton where it is equal to one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the goal of ion-ion collision is not to increase the total energy or the energy per nucleon: it is to obtain a different type of collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted than in a proton-proton collision, the energy involved in the real collision process is variable: each quark and gluon carry a fraction of the energy of the proton, and hard collision involve a collision between a quark/gluon of one proton against a quark/gluon of the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of ion-ion collision you have the same process: the energy is shared by the protons/neutrons and they can have different energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal of such collision is also to obtain a volume (bigger than in a p-p collision) with a very high energy density. In such a volume, a &quot;state of matter&quot; called quark-gluon-plasma is believed to be possibly created. The study of this QGP is one of the main goal of the ALICE experiment at the LHC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few references:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://public.web.cern.ch/public/en/lhc/ALICE-en.html&quot; rel=&quot;nofollow&quot;&gt;The ALICE experiment at CERN&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://op-webtools.web.cern.ch/op-webtools/Vistar/vistars.php?usr=LHC1&quot; rel=&quot;nofollow&quot;&gt;Live control screen of the LHC now running as a Pb-Pb collider at 574 TeV in the center of mass&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="82" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-19T16:41:55.427" LastActivityDate="2011-05-19T16:41:55.427" CommentCount="3" />
  <row Id="391" PostTypeId="1" AcceptedAnswerId="395" CreationDate="2010-11-08T23:18:59.520" Score="7" ViewCount="1526" Body="&lt;p&gt;Or: can it be proved that anti-matter definitely is nót matter going backwards in time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Antimatter&quot; rel=&quot;nofollow&quot;&gt;From wikipedia&lt;/a&gt;:&#xA;[There is considerable speculation as to why the observable universe is apparently almost entirely matter ... the apparent asymmetry of matter and antimatter in the visible universe is one of the greatest unsolved problems in physics.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Couldn't the concept of reversal in time contribute to the explanation of this problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measurements that reveal reversal in time must be rare and exceptional, just like anti-matter.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-01T07:35:15.953" LastActivityDate="2011-07-20T18:02:23.150" Title="Is anti-matter matter going backwards in time?" Tags="&lt;particle-physics&gt;&lt;antimatter&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="392" PostTypeId="2" ParentId="389" CreationDate="2010-11-08T23:24:45.563" Score="3" ViewCount="" Body="&lt;p&gt;In either case the particles remain virtual until promoted to real by conversion of energy via pair production.    The Feynman diagrams seem quite similar whether they are describing  mev processes with thermal phonons as gauge bosons (holes in solid states) or Mev processes with photons (positrons).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course with holes in the solid state, promotion to real free electrons (n type semiconductor) or free holes (p type) comes from the thermal environment, so the excess charge carriers are assumed real from the outset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thats what I recall from long ago, anyway. &lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-11-10T06:18:20.757" LastActivityDate="2010-11-10T06:18:20.757" CommentCount="2" />
  <row Id="393" PostTypeId="2" ParentId="383" CreationDate="2010-11-08T23:44:07.127" Score="9" ViewCount="" Body="&lt;p&gt;Ok, here is my (hopefully rigorous) demonstration of the origin of these forces here, from  first principles. I've tried to be pretty clear what's happening with the maths. Bear with me, it's a bit lengthy!&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Angular velocity vector&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Let us start with the principal equation defining angular velocity in three dimensions,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{\vec{r}} = \vec{\omega} \times \vec{r} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This can be derived roughly by considering a centripetal force acting on a particle. Note that this equation applies symmetrically in inertial and rotating reference frames.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that we can in fact generalise this statement in terms of $r$ for an arbitrary vector $a$ that is known to be fixed in the rotating body.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Transformation between inertial and rotating frames&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Now consider a vector $a$, which we can write in Cartesian coordinates (fixed within the body) as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{a} = a_x \vec{i} + a_y \vec{j} + a_z \vec{k} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Newtonian mechanics, scalar quantities must be invariant for any given choice of frame, so we can say&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left.\frac{da_x}{dt}\right|_I = \left.\frac{da_x}{dt}\right|_R$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $I$ indicates the value is for the inertial frame, and $R$ that the value is for the rotating frame. Equivalent statements apply for $a_y$ and $a_z$, of course. Hence, any transformation of $a$ between frames must be due to changes in the unit vectors of the basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now by the product rule,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left.\frac{d\vec{a}}{dt}\right|_I = \frac{d}{dt} \left( a_x \vec{i} + a_y \vec{j} + a_z \vec{k} \right) \&#xA;= \left( \frac{da_x}{dt} \vec{i} + \frac{da_y}{dt} \vec{j} + \frac{da_z}{dt} \vec{k} \right) + \left( a_x \frac{d\vec{i}}{dt} + a_y \frac{d\vec{j}}{dt} + a_z \frac{d\vec{k}}{dt} \right) .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the previous equation for angular velocity, we then have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left.\frac{d\vec{a}}{dt}\right|_I = \left( \frac{da_x}{dt} \vec{i} + \frac{da_y}{dt} \vec{j} + \frac{da_z}{dt} \vec{k} \right) + \left( a_x \vec{\omega} \times \vec{i} + a_y \vec{\omega} \times \vec{j} + a_z \vec{\omega} \times \vec{k} \right) \&#xA;= \left.\frac{d\vec{a}}{dt}\right|_R + \vec{\omega} \times \vec{a} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider a position vector on the surface of a rotating body. We can write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{v}_I = \left.\frac{d\vec{r}}{dt}\right|_I = \left.\frac{d\vec{r}}{dt}\right|_R + \vec{\omega} \times \vec{r} ,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and similarly for $\vec{a} = \vec{v}_I$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left.\frac{d^2\vec{r}}{dt^2}\right|_I = \left( \left.\frac{d}{dt}\right|_R + \vec{\omega} \times \right)^2 \vec{r} \&#xA;= \left.\frac{d^2\vec{r}}{dt^2}\right|_R + 2\vec{\omega} \times \left.\frac{d\vec{r}}{dt}\right|_R + \vec{\omega} \times (\vec{\omega} \times \vec{r}) .$$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Forces on body in rotating frame&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Now consider a force acting on an object at position $\vec{r}$ (for example, gravity). Newton's third law states&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{F} = m \left.\frac{d^2\vec{r}}{dt^2}\right|_I .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so substituting this into the previous equation for $\left.\frac{d^2\vec{r}}{dt^2}\right|_I$ and rearranging we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{F}_{net} = m \left.\frac{d^2\vec{r}}{dt^2}\right|_R = \vec{F} - 2m \vec{\omega} \times \vec{v}_R - m \vec{\omega} \times (\vec{\omega} \times \vec{r})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= \vec{F} - 2m \vec{\omega} \times \vec{v}_R + m \vec{\omega}^2 \vec{r}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here we have it. The second term on the right is the &lt;strong&gt;coriolis force&lt;/strong&gt;, and the third term is the &lt;strong&gt;centrifugal force&lt;/strong&gt; (clealry pointing away from the centre of rotation). Any interpretation of the coriolis and centrifugal forces then follow naturally from this single important equation.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2010-11-09T03:14:19.610" LastActivityDate="2010-11-09T03:14:19.610" CommentCount="15" />
  <row Id="394" PostTypeId="2" ParentId="391" CreationDate="2010-11-08T23:48:49.020" Score="4" ViewCount="" Body="&lt;p&gt;This refers to Feynman's 1949 theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.upscale.utoronto.ca/PVB/Harrison/AntiMatter/AntiMatter.html&quot; rel=&quot;nofollow&quot;&gt;http://www.upscale.utoronto.ca/PVB/Harrison/AntiMatter/AntiMatter.html&lt;/a&gt;&lt;a href=&quot;http://www.upscale.utoronto.ca/PVB/Harrison/AntiMatter/AntiMatter.html&quot; rel=&quot;nofollow&quot;&gt;link text&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there:&#xA;&quot;Feynman's Theory of Antimatter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1949 Richard Feynman devised another theory of antimatter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spacetime diagram for pair production and annihilation appears to the right. An electron is travelling along from the lower right, interacts with some light energy and starts travelling backwards in time. An electron travelling backwards in time is what we call a positron. In the diagram, the electron travelling backwards in time interacts with some other light energy and starts travelling forwards in time again. Note that throughout, there is only one electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A friend of mine finds the image of an electron travelling backwards in time, interpreted by us as a positron, to be scary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feynman in his original paper proposing this theory wrote:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;It is as though a bombardier flying low over a road suddenly sees three roads and it is only when two of them come together and disappear again that he realizes that he has simply passed over a long switchback in a single road.&quot; (Physical Review 76, (1949), 749.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that Feynman's theory is yet another echo of the fact, noted above, that a negatively charged object moving from left to right in a magnetic field has the same curvature as a positive object moving from right to left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feynman's theory is mathematically equivalent to Dirac's, although the interpretations are quite different. Which formalism a physicist uses when dealing with antimatter is usually a matter of which form has the simplest structure for the particular problem being solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in Feynman's theory, there is no pair production or annihilation. Instead the electron is just interacting with electromagnetic radiation, i.e. light. Thus the whole process is just another aspect of the fact that accelerating electric charges radiate electric and magnetic fields; here the radiation process is sufficiently violent to reverse the direction of the electron's travel in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nambu commented on Feynman's theory in 1950:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The time itself loses sense as the indicator of the development of phenomena; there are particles which flow down as well as up the stream of time; the eventual creation and annihilation of pairs that may occur now and then is no creation or annihilation, but only a change of direction of moving particles, from past to future, or from future to past.&quot; (Progress in Theoretical Physics 5, (1950) 82).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About Formally Equivalent Descriptions ....&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you mix in another very interesting problem, namely the origin of the apparent matter antimatter asymmetry in the observable universe (observed absence of annihilation radiation except in special circumstances) and point out that it may be related to a very very hard problem indeed, namely the origin of time asymmetry.   One problem at a time, please.  Maybe separate questions, but the answers will likely be more or less over your head since, to the extent that they are even partially understood, they are still being figured out. &lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-11-08T23:56:35.973" LastActivityDate="2010-11-08T23:56:35.973" CommentCount="2" />
  <row Id="395" PostTypeId="2" ParentId="391" CreationDate="2010-11-08T23:51:25.353" Score="10" ViewCount="" Body="&lt;p&gt;To the best of my knowledge, most physicists don't believe that antimatter is &lt;em&gt;actually&lt;/em&gt; matter moving backwards in time. It's not even entirely clear what would it really mean to move backwards in time, from the popular viewpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm remembering correctly, this idea all comes from a story that probably originated with Richard Feynman. At the time, one of the big puzzles of physics was why all instances of a particular elementary particle (all electrons, for example) are apparently identical. Feynman had a very hand-wavy idea that all electrons could in fact be the same electron, just bouncing back and forth between the beginning of time and the end. As far as I know, that idea never developed into anything mathematically grounded, but it did inspire Feynman and others to calculate what the properties of an electron moving backwards in time would be, in a certain precise sense that emerges from quantum field theory. What they came up with was a particle that matched the known properties of the positron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to give you a rough idea of what it means for a particle to &quot;move backwards in time&quot; in the technical sense: in quantum field theory, particles carry with them amounts of various conserved quantities as they move. These quantities may include energy, momentum, electric charge, &quot;flavor,&quot; and others. As the particles move, these conserved quantities produce &quot;currents,&quot; which have a direction based on the motion and sign of the conserved quantity. If you apply the time reversal operator (which is a purely mathematical concept, not something that actually reverses time), you reverse the direction of the current flow, which is equivalent to reversing the sign of the conserved quantity, thus (roughly speaking) turning the particle into its antiparticle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider electric current: it arises from the movement of electric charge, and the direction of the current is a product of the direction of motion of the charge and the sign of the charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec{I} = q\vec{v}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Positive charge moving left ($+q\times -v$) is equivalent to negative charge moving right ($-q\times +v$). If you have a current of electrons moving to the right, and you apply the time reversal operator, it converts the rightward velocity to leftward velocity ($-q\times -v$). But you would get the exact same result by instead converting the electrons into positrons and letting them continue to move to the right ($+q\times +v$); either way, you wind up with the net positive charge flow moving to the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, optional reading if you're interested: there is a very basic (though hard to prove) theorem in quantum field theory, the TCP theorem, that says that if you apply the three operations of time reversal, charge conjugation (switch particles and antiparticles), and parity inversion (mirroring space), the result should be exactly equivalent to what you started with. We know from experimental data that, under certain exotic circumstances, the combination of charge conjugation and parity inversion does &lt;em&gt;not&lt;/em&gt; leave all physical processes unchanged, which means that the same must be true of time reversal: &lt;em&gt;physics is&lt;/em&gt; not &lt;em&gt;time-reversal invariant&lt;/em&gt;. Of course, since we can't &lt;em&gt;actually&lt;/em&gt; reverse time, we can't test in exactly what manner this is true.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-08T23:51:25.353" CommentCount="4" />
  <row Id="396" PostTypeId="2" ParentId="19" CreationDate="2010-11-08T23:58:45.327" Score="1" ViewCount="" Body="&lt;p&gt;In experimental high energy physics, the beam energy is well known. For example, if you have a beam of protons, you &lt;em&gt;know&lt;/em&gt; the energy because the engineers controlling the accelerator make sure that the beam is very well collimated, and goes on track, etc. If it weren't so, the beam would hit the pipe walls and you'd lose it... There are also instruments along the beam pipe which measure the current, so all that is used to control the total energy of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, experimental particle physicists &lt;em&gt;seldomly look into individual collisions&lt;/em&gt; &amp;mdash; because there are millions of collisions per second, with millions of electronic channels, the pile up of collision events is not negligible (typically 10-20 events per &quot;frame&quot;). It's simply too complicated (and error prone) to look at individual events. To make stand out the potentially interesting events from the manifestly uninteresting background events (those too common and already studied, such as low-mass-particle decays), they make specific cuts in quantities that they &lt;em&gt;know&lt;/em&gt; (from numerical simulations) will exclude manifestly uninteresting events, and in the end you have events which are likely to be the &quot;type&quot; you're looking for. On example is this: if you only accept reconstructed tracks which have a linear momentum higher than a certain value, you exclude a lot of particles (&quot;good&quot; and &quot;bad&quot;) in the direction of the collision (beam-wise) but those who are scattered perpendicularly to that direction (i.e. away from the beam), &lt;em&gt;and&lt;/em&gt; have a high energy, are likely to be interesting (as expected from numerical simulations, i.e. &quot;Monte Carlo&quot; as it is called). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line (to make this short):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They know very well what goes in, but they know not too well what goes out for &lt;em&gt;individual events&lt;/em&gt;. When you only count high-energy particles (i.e. those who don't bend very much under the detector's strong magnetic field) and you start &lt;em&gt;superimposing&lt;/em&gt; (piling up!!) all the events which have a couple (in about 100 or so) of &quot;promissing&quot; particle tracks, they start accumulating around the &quot;correct value&quot;. That's how they know that &quot;when two particles collided, a heavy one was created momentarily and it decayed into lighter ones&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can have a clue about how piling up individual events might give the approximate good answer from this illustration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a glass of sand and you let it drop slowly into the floor. You then ask a friend to come into the room and tell him to make an estimation of where in the room you dropped the sand from. That should be easy. He might even make a more or less good prediction of &lt;em&gt;how high&lt;/em&gt; you dropped the sand by how much it spreads on the floor (it should spread wider if dropped higher). Even if your friend knows exactly how much sand was dropped (from knowing how much sand was in the glass), he only has a &quot;good enough&quot; hint at what happened (where in the room, how high) when the sand was dropped from the glass.&lt;/p&gt;&#xA;" OwnerUserId="103" LastActivityDate="2010-11-08T23:58:45.327" CommentCount="1" />
  <row Id="397" PostTypeId="2" ParentId="387" CreationDate="2010-11-09T00:11:07.190" Score="-1" ViewCount="" Body="&lt;p&gt;Quantum mechanics takes place on flat space or on manifolds which can be continuously deformed into flat space time for relativistic Quantum Field Theories on curved spaces.   Now consider a point particle of the Planck mass, which is so concentrated in its mass that it forms a black hole.   The whole formalism more or less fails.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-11-09T00:11:07.190" CommentCount="1" />
  <row Id="398" PostTypeId="2" ParentId="379" CreationDate="2010-11-09T02:49:29.097" Score="2" ViewCount="" Body="&lt;p&gt;The vacuum state is not at all like a thermal distribution. The &quot;vacuum field&quot; is just the zero-point fluctuations of the various modes of the electromagnetic field, with each mode containing half of the fundamental energy for that mode. A thermal distribution has the modes populated with a number of photons following a Planck distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the frequencies that most quantum optics experiments work with, this makes almost no difference. A black body at room temperature has the peak of its black-body spectrum at just under 10 microns, which is a photon energy of about 0.13 eV. Most quantum optics experiments are done at optical or near-infrared frequencies-- the fundamental mode of a YAG laser at 1.06 microns is a common wavelength used, and that's got a photon energy of 1.17 eV. The photon occupation number for that mode would be a factor of e^{-9} lower than the peak value (about four orders of magnitude), which is pretty insignificant. And a lot of quantum-optics experiments use visible light sources, with a frequency almost twice as high as the YAG light, with even less radiation in the thermal modes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an issue for some experiments, particularly those in the microwave regime where the black-body contribution at room temperature is significant. People who do quantum optics experiments in the microwave regime generally do them at cryogenic temperatures (liquid nitrogen or lower) for just this reason, and that's enough to eliminate the problem.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-09T02:49:29.097" />
  <row Id="399" PostTypeId="2" ParentId="387" CreationDate="2010-11-09T03:13:16.117" Score="20" ViewCount="" Body="&lt;p&gt;There are zero contradictions between quantum mechanics and special relativity; quantum field theory is the framework that unifies them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General relativity also works perfectly well as a low-energy effective quantum field theory. For questions like the low-energy scattering of photons and gravitons, for instance, the Standard Model coupled to general relativity is a perfectly good theory. It only breaks down when you ask questions involving invariants of order the Planck scale, where it fails to be predictive; this is the problem of &quot;nonrenormalizability.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonrenormalizability itself is no big deal; the Fermi theory of weak interactions was nonrenormalizable, but now we know how to complete it into a quantum theory involving W and Z bosons that is consistent at higher energies. So nonrenormalizability doesn't necessarily point to a contradiction in the theory; it merely means the theory is incomplete. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity is more subtle, though: the real problem is not so much nonrenormalizability as  high-energy behavior inconsistent with local quantum field theory. In quantum mechanics, if you want to probe physics at short distances, you can scatter particles at high energies. (You can think of this as being due to Heisenberg's uncertainty principle, if you like, or just about properties of Fourier transforms where making localized wave packets requires the use of high frequencies.) By doing ever-higher-energy scattering experiments, you learn about physics at ever-shorter-length scales. (This is why we build the LHC to study physics at the attometer length scale.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With gravity, this high-energy/short-distance correspondence breaks down. If you could collide two particles with center-of-mass energy much larger than the Planck scale, then when they collide their wave packets would contain more than the Planck energy localized in a Planck-length-sized region. This creates a black hole. If you scatter them at even &lt;em&gt;higher&lt;/em&gt; energy, you would make an even &lt;em&gt;bigger&lt;/em&gt; black hole, because the Schwarzschild radius grows with mass. So the harder you try to study shorter distances, the worse off you are: you make black holes that are bigger and bigger and swallow up ever-larger distances. No matter what completes general relativity to solve the renormalizability problem, the physics of &lt;em&gt;large&lt;/em&gt; black holes will be dominated by the Einstein action, so we can make this statement even without knowing the full details of quantum gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tells us that quantum gravity, at very high energies, is &lt;em&gt;not&lt;/em&gt; a quantum field theory in the traditional sense. It's a stranger theory, which probably involves a subtle sort of nonlocality that is relevant for situations like black hole horizons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of this is really a &lt;em&gt;contradiction&lt;/em&gt; between general relativity and quantum mechanics. For instance, string theory is a quantum mechanical theory that includes general relativity as a low-energy limit. What it does mean is that quantum field theory, the framework we use to understand all non-gravitational forces, is not sufficient for understanding gravity. Black holes lead to subtle issues that are still not fully understood.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-11-09T03:13:16.117" CommentCount="3" />
  <row Id="400" PostTypeId="2" ParentId="384" CreationDate="2010-11-09T03:24:23.983" Score="1" ViewCount="" Body="&lt;p&gt;Uh... I liked this question. And a quick look at the Google or Google scholar did not give me much. So is written here more what I grasp people understand about impedance. Like Noldorin was saying I take impedance as how much is impeding power to be transferred from one place to another. In a homogeneous medium power is transferred as wave unimpeded but when it find an interface of another medium then power transfer won't be so straight forward.&#xA;The point is at the two mediums dynamic equations of the two place does no match. Say, electromagnetically one medium has the Maxwell equations of $(\mu_1,\epsilon_1)$ and the other medium has the equations with $(\mu_2,\epsilon_2)$&#xA;At the interface the countour conditions needs to be satisfied&#xA;\begin{equation}\left(\vec{D_2}-\vec{D_1}\right)\cdot\hat{n}=0\end{equation}&#xA;\begin{equation}\left(\vec{B_2}-\vec{B_1}\right)\cdot\hat{n}=0\end{equation}&#xA;\begin{equation}\left(\vec{E_2}-\vec{E_1}\right)\times\hat{n}=0\end{equation}&#xA;\begin{equation}\left(\vec{H_2}-\vec{H_1}\right)\times\hat{n}=0\end{equation}&#xA;where $\hat{n}$ is the normal of the interface. Using the above conditions to connect the solutions of wave equations at each medium, will give rise to the Reflections and Transmission Coefficients. In the case of Maxwell's equations&#xA;\begin{equation}R=\frac{Z_2-Z_1}{Z_2+Z_1}\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a one dimensional wave, with standard solutions&#xA;\begin{equation}f_i\left(x,t\right)=A e^{k_i x-\omega t}+B e^{k_i x+\omega t}\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the conditions that needs to be satisfied are the equality of displacement and the its derivative&#xA;\begin{equation}f_2(x_0,t)=f_1(x_0,t)\end{equation}&#xA;\begin{equation}f'_2(x_0,t)=f'_1(x_0,t)\end{equation}&#xA;where $x_0$ is the position of the interface. Using these conditions with the solutions of the wave equations we will get.&#xA;\begin{equation}R=\frac{k_2-k_1}{k_2+k_1}\end{equation}&#xA;So in this case the impedance will be characterized by $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So generalizing what you want is find what is the physical conditions that needs to be matched at the interface and use it to connect the solutions of each side of the interface, and figure how you get the reflections coefficients, which should look something like the reflection coefficients above. The terms should be your impedance for that system. Note that the definition of your impedance in that system is not quite unique. As in the previous example you could normalize the $k_i$ by some arbitrary $k_0$ and call this the impedance.&#xA;\begin{equation}z_i=\frac{k_i}{k_0}\end{equation}&#xA;Check the literature to find what is the exact definition for the system you are interested in. I suppose you already went to Wikipedia for &lt;a href=&quot;http://en.wikipedia.org/wiki/Impedance_matching&quot; rel=&quot;nofollow&quot;&gt;other examples&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="62" LastEditorDisplayName="" LastEditDate="2010-11-09T13:40:10.663" LastActivityDate="2010-11-09T13:40:10.663" />
  <row Id="401" PostTypeId="1" CreationDate="2010-11-09T05:27:15.270" Score="23" ViewCount="1457" Body="&lt;p&gt;People undoubtedly use a variety of programs to draw diagrams for physics, but I am not familiar with many of them.  I usually hand-draw things in &lt;a href=&quot;http://www.gimp.org/&quot; rel=&quot;nofollow&quot; title=&quot;GIMP&quot;&gt;GIMP&lt;/a&gt;.  GIMP is powerful in some regards, but it's time consuming to do things like draw circles or arrows because I make them from more primitive tools.  It is also difficult to be precise.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know some people use LaTeX, but am not sure how versatile or easy it is.  The only other tools I know are Microsoft Paint and the tools built into Microsoft Office.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which tools are commonly used?  What are their good and bad points (features, ease of use, portability, etc.)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a tool with flexibility and minimal learning curve/development time.  I would like to hand-draw and to drag-and-drop pre-made shapes, but I also want to specify the locations of curves and shapes with equations when I want better precision. Some minimal programming functionality would also be nice (i.e. the ability to run through a loop that draws a series of lines with a varying parameter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please recommend a few pieces of software if they're good for different situations.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-07-20T19:57:59.497" CommunityOwnedDate="2010-11-10T18:21:08.203" Title="What software programs are used to draw physics diagrams, and what are their relative merits?" Tags="&lt;software&gt;" AnswerCount="12" CommentCount="5" FavoriteCount="13" />
  <row Id="402" PostTypeId="2" ParentId="401" CreationDate="2010-11-09T05:43:37.553" Score="2" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://math.stackexchange.com/q/1985/705&quot;&gt;Software for drawing geometry diagrams&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Sometimes I use &lt;a href=&quot;http://www.sagemath.org/tour-graphics.html&quot; rel=&quot;nofollow&quot;&gt;SAGE&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-11-09T07:40:23.850" LastActivityDate="2010-11-09T07:40:23.850" CommunityOwnedDate="2010-11-10T18:21:08.203" CommentCount="2" />
  <row Id="403" PostTypeId="2" ParentId="401" CreationDate="2010-11-09T06:46:01.390" Score="10" ViewCount="" Body="&lt;p&gt;I've had good experiences with &lt;a href=&quot;http://inkscape.org/&quot; rel=&quot;nofollow&quot;&gt;Inkscape&lt;/a&gt;.  It has a GUI interface, but allows you to enter coordinates directly if you want, and it's scriptable.  There is a plug-in that allows you to enter LaTeX directly (for labels and such).  The downside is that it is very much still in development, so sometimes you find that a feature you want is not completely implemented yet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, here is a poster I made last week, entirely within Inkscape:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/XyXm0.png&quot; alt=&quot;Poster made with Inkscape&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a Mac and don't mind spending some money ($100), I've heard good things about &lt;a href=&quot;http://www.omnigroup.com/products/omnigraffle/&quot; rel=&quot;nofollow&quot;&gt;OmniGraffle&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-01-17T16:31:15.690" LastActivityDate="2011-01-17T16:31:15.690" CommunityOwnedDate="2010-11-10T18:21:08.203" CommentCount="6" />
  <row Id="404" PostTypeId="2" ParentId="401" CreationDate="2010-11-09T08:01:19.813" Score="4" ViewCount="" Body="&lt;p&gt;I would try matplotlib, but first check here &lt;a href=&quot;http://matplotlib.sourceforge.net/gallery.html&quot; rel=&quot;nofollow&quot;&gt;http://matplotlib.sourceforge.net/gallery.html&lt;/a&gt; and decide if these pictures satisfies your needs. Also click some picture and inspect source code.&lt;/p&gt;&#xA;" OwnerUserId="118" LastActivityDate="2010-11-09T08:01:19.813" CommunityOwnedDate="2010-11-10T18:21:08.203" CommentCount="2" />
  <row Id="405" PostTypeId="2" ParentId="386" CreationDate="2010-11-09T09:35:56.943" Score="0" ViewCount="" Body="&lt;p&gt;You'd have better luck on math.stackexchange.com, but here's a short outline of how to attack your problem.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;By definition, to calculate the eigenvalues of the Fourier transform, solve the equation &#xA;$$\mathcal{F}{f(x)} = \lambda f(x)$$&#xA;or&#xA;$$\int_{-\infty}^\infty f(x) e^{-2\pi i X x} dx - \lambda f(x) = 0$$&lt;/li&gt;&#xA;&lt;li&gt;To decompose a function into Hermite-Gaussian functions $HG_n(x)$, i.e. find $c_n$ in&#xA;$$f(x) = \sum_n c_n HG_n(x),$$&#xA;take the inner product of $f(x)$ with each Hermite-Gauss function:&#xA;$$c_n = \int_{-\infty}^\infty f(x)\overline{HG_n(x)} dx$$&#xA;where $\bar{x}$ denotes the complex conjugate of $x$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Disclaimer: this is all from memory, so please inform me of the inevitable mistakes, and I'll be glad to correct them.&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2010-11-09T09:35:56.943" />
  <row Id="407" PostTypeId="2" ParentId="401" CreationDate="2010-11-09T10:16:55.550" Score="9" ViewCount="" Body="&lt;p&gt;I'm learning &lt;a href=&quot;http://sourceforge.net/projects/pgf/&quot; rel=&quot;nofollow&quot;&gt;TikZ&lt;/a&gt; (a drawing package for LaTeX) as we speak. It's good for two-dimensional line drawings, the syntax for specifying shapes and curves is extremely versatile, but the learning curve is steeper than LaTeX even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a superb &lt;a href=&quot;http://www.texample.net/tikz/examples/all/&quot; rel=&quot;nofollow&quot;&gt;gallery of TikZ examples&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="107" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-01-17T17:30:37.777" LastActivityDate="2011-01-17T17:30:37.777" CommunityOwnedDate="2010-11-10T18:21:08.203" CommentCount="8" />
  <row Id="408" PostTypeId="1" AcceptedAnswerId="416" CreationDate="2010-11-09T10:26:25.820" Score="2" ViewCount="329" Body="&lt;p&gt;Are there any methods of direct detection for black holes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not referring to gravitational lensing, or measuring the orbits of a star in a binary pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way of directly 'seeing' them? &lt;/p&gt;&#xA;" OwnerUserId="133" LastActivityDate="2011-04-04T16:04:22.643" Title="Can astronomers directly detect black holes?" Tags="&lt;general-relativity&gt;&lt;experimental-physics&gt;&lt;astronomy&gt;" AnswerCount="7" CommentCount="3" />
  <row Id="409" PostTypeId="1" AcceptedAnswerId="438" CreationDate="2010-11-09T10:34:28.043" Score="4" ViewCount="295" Body="&lt;p&gt;How can astronomers say, we know there are black holes at the centre of each galaxy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What methods of indirect detection are there to know where and how big a black hole is?&lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="133" LastEditorDisplayName="" LastEditDate="2011-03-28T11:44:55.823" LastActivityDate="2011-04-12T08:21:24.447" Title="What methods can astronomers use to find a black hole?" Tags="&lt;general-relativity&gt;&lt;experimental-physics&gt;&lt;astronomy&gt;" AnswerCount="6" />
  <row Id="411" PostTypeId="2" ParentId="409" CreationDate="2010-11-09T10:38:59.643" Score="3" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.science20.com/scientific_tapestry/blog/black_hole_detection_methods&quot; rel=&quot;nofollow&quot;&gt;Methods of black hole detection&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://askville.amazon.com/current-method-detecting-black-holes/AnswerViewer.do?requestId=6179744&quot; rel=&quot;nofollow&quot;&gt;There are several ways&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-11-09T10:38:59.643" />
  <row Id="412" PostTypeId="2" ParentId="391" CreationDate="2010-11-09T10:47:27.830" Score="4" ViewCount="" Body="&lt;p&gt;There is one technical inaccuracy in saying that antimatter moves back in time (whatever it might mean). In quantum field theory we get positive energy solutions (usual particles) and negative energy solutions. Negative energy solutions behave in time as if they were propagating backward in time. But they are &lt;strong&gt;not&lt;/strong&gt; the antiparticles, they are just the &quot;negative-energy particles&quot;. Antiparticles are positive energy solutions, and they are obtained by acting with charge conjugation operator on the negative-energy solutions. So, antiparticles move forward in time, as usual particles.&lt;/p&gt;&#xA;" OwnerUserId="59" LastActivityDate="2010-11-09T10:47:27.830" CommentCount="2" />
  <row Id="413" PostTypeId="2" ParentId="356" CreationDate="2010-11-09T11:14:06.807" Score="2" ViewCount="" Body="&lt;p&gt;As Gerard stated you can't directly measure the mass of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can measure the acceleration of gravity at the Earth's surface using a simple pendulum.&#xA;You need to know two things:&#xA;The length of the pendulum, which can be accurately and easily measured with a ruler.&#xA;The period of the oscillation, which can be measured by taking a stop-watch and timing a number of small swings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then use the formula: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g = \frac{4 \pi ^2 L}{T^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where L is the length and T is the period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is then related to the mass by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M = \frac{g R^2}{G}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where R is the radius of the earth and G is Newtons gravitational constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You still need to know these other quantities, as Gerard said. I don't know how you would find R or G at home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Formatted the Formulae using Latex.&lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="133" LastEditorDisplayName="" LastEditDate="2010-11-10T15:24:48.863" LastActivityDate="2010-11-10T15:24:48.863" CommentCount="2" />
  <row Id="414" PostTypeId="1" AcceptedAnswerId="480" CreationDate="2010-11-09T11:45:38.263" Score="22" ViewCount="991" Body="&lt;p&gt;As a Graduate Mathematics student, my interests lies in Number theory. I am curious to know if Number theory has any connections or applications to physics. I have never even heard of any applications of Number theory to physics. I have heard Applications of linear algebra and analysis to many branches of physics, but not number theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Waiting forward in receiving interesting answers!&lt;/p&gt;&#xA;" OwnerUserId="159" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-10T14:35:26.763" LastActivityDate="2011-07-10T14:35:26.763" Title="Number theory in Physics" Tags="&lt;mathematical-physics&gt;&lt;mathematics&gt;&lt;big-list&gt;" AnswerCount="10" CommentCount="1" FavoriteCount="12" />
  <row Id="415" PostTypeId="2" ParentId="414" CreationDate="2010-11-09T12:26:24.313" Score="6" ViewCount="" Body="&lt;p&gt;Here's a journal link (full disclosure:  I'm on the editorial board).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.intlpress.com/CNTP/&quot; rel=&quot;nofollow&quot;&gt;http://www.intlpress.com/CNTP/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20" LastActivityDate="2010-11-09T12:26:24.313" CommentCount="1" />
  <row Id="416" PostTypeId="2" ParentId="408" CreationDate="2010-11-09T12:34:57.487" Score="8" ViewCount="" Body="&lt;p&gt;A colleague in astronomy had a student a few years ago who did a calculation about the possibility of primordial black holes, created in the Big Bang. If the size of these was just right, they could be evaporating into nothing due to Hawking radiation right &quot;now&quot; (scare quotes because this would necessarily include distant black holes that evaporated many years ago, whose light is just reaching us now). The last burst of Hawking radiation for these would look basically like a faint gamma-ray burst, in which case it ought to be directly detectable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure of the current status of this-- their preliminary result was, if I remember correctly, that you might be able to test this by measuring the probability distribution for gamma-ray bursts of the appropriate size and duration, but we didn't have any telescopes capable of picking them up at the time. I'm not sure if that's changed or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, that would give you a direct way to detect black holes of a certain size, though they wouldn't be around after the detection, so it might not really fit the spirit of the question...&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-09T12:34:57.487" CommentCount="4" />
  <row Id="417" PostTypeId="2" ParentId="414" CreationDate="2010-11-09T13:55:53.893" Score="2" ViewCount="" Body="&lt;p&gt;I recall being stunned (they can do that?) by the mere mention of the technique called &quot;Zeta Function Regularization&quot;, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Renormalization&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Renormalization&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Zeta_function_regularization&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Zeta_function_regularization&lt;/a&gt; ,&#xA;to sum divergent series like zeta(-n) to get finite results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that was the limit.  Then I heard about p-adic strings, complex dimensions, and more.  See e.g. &lt;a href=&quot;http://inc.web.ihes.fr/prepub/PREPRINTS/2008/M/M-08-42.pdf&quot; rel=&quot;nofollow&quot;&gt;http://inc.web.ihes.fr/prepub/PREPRINTS/2008/M/M-08-42.pdf&lt;/a&gt; .&lt;br&gt;&#xA; The ideas are certainly very beautiful.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-11-09T13:55:53.893" />
  <row Id="418" PostTypeId="2" ParentId="408" CreationDate="2010-11-09T16:59:27.307" Score="5" ViewCount="" Body="&lt;p&gt;As far as anyone knows, with the exception of the Hawking radiation that Chad mentioned, black holes don't emit anything that we could use to detect them directly. And Hawking radiation from most black holes is so weak that it'd be impossible to detect from any significant distance, except for the final burst of radiation when it finally evaporates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the straightforward answer would be: generally no, black holes can't be detected directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, any black hole that is close enough to another star will have an accretion disk, and that can certainly be detected because it gives off a &lt;em&gt;lot&lt;/em&gt; of X-rays and gamma rays.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-09T16:59:27.307" />
  <row Id="419" PostTypeId="1" AcceptedAnswerId="5139" CreationDate="2010-11-09T17:36:11.747" Score="1" ViewCount="288" Body="&lt;p&gt;I'm not sure if it's relevant, but I'm talking about a situation where a particle is moving in an electro-magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand, if we see the term $\nabla \cdot \vec{v}$ or $\nabla \times \vec{v}$ we can say it is equal to 0, because the velocity vector is independent of the location vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this: Is this always true, or only in specific settings (particle in an electro-magnetic field for example)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is always true: What about a state where an electron's velocity changes with respect to x? Then they can't be independent, can they?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, are the following always true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\partial v_{x}/\partial x = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\partial v_{y}/\partial x = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\partial v_{z}/\partial x = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\partial v_{x}/\partial y = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\partial v_{y}/\partial y = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\partial v_{z}/\partial y = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\partial v_{x}/\partial z = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\partial v_{y}/\partial z = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\partial v_{z}/\partial z = 0$&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2010-11-10T02:02:58.273" LastActivityDate="2011-02-14T01:50:38.963" Title="Why do we automatically assume that the velocity vector, v, and location vector, r, are independent?" Tags="&lt;vectors&gt;" AnswerCount="5" CommentCount="9" />
  <row Id="420" PostTypeId="1" AcceptedAnswerId="422" CreationDate="2010-11-09T17:49:50.370" Score="7" ViewCount="376" Body="&lt;p&gt;Let's say we have a magnet stuck to a metal bar, suspended above the ground. If I attach a paperclip to the magnet, where is the energy to hold the paperclip coming from (against the force of gravity), and for how long will the paperclip remain there - will it remain there forever?&lt;/p&gt;&#xA;" OwnerUserId="104" LastActivityDate="2010-11-09T22:02:41.620" Title="Would a magnet attract a paperclip indefinitely?" Tags="&lt;electromagnetism&gt;&lt;magnets&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="421" PostTypeId="2" ParentId="356" CreationDate="2010-11-09T17:57:10.487" Score="16" ViewCount="" Body="&lt;p&gt;Yes we/you can.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recall seeing a famous video of a homemade version of the Cavendish torsion balance experiment from the early 1960's, made I think for the PSSC high school course.    Basically, the physicist hung a torsion balance from a high ceiling by a long (&gt;10 m?) piece of computer data tape (chosen because it would not stretch).   He carefully minimized air currents.  The torsion masses were two .5 kg bottles of water on a wooden bar (no magnetic interference).   Mass, in the form of boxes of sand, say 20kg was piled around on the floor as static mass and then reversed in position with respect to the suspended masses.   There was a clear plastic box around the balance (with a hole in its top for the suspending tape to pass through) also to minimize the effect of air currents, since the lateral force on each bottle is about G*m1*m2/r^2 = (6.7e-11)*0.5kg*20kg/(0.1m)^2 N ~ 6.7e-8 N, i.e. a lateral force on each bottle equivalent to that generated by a weight of about 7 micrograms, about that of a 1 mm^3 grain of sand.   This is visible to us because the long arms of the torsion balance convert this small force into a torque on the suspending filament, and the restoring torque is itself very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found an Italian dubbed version of the video on Youtube.  See &lt;a href=&quot;http://www.youtube.com/watch?v=uUGpF3h3RaM&amp;amp;feature=related&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=uUGpF3h3RaM&amp;amp;feature=related&lt;/a&gt; and a slightly longer version at &lt;a href=&quot;http://www.youtube.com/watch?v=V4hWMLjfe_M&amp;amp;feature=related&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=V4hWMLjfe_M&amp;amp;feature=related&lt;/a&gt;.   I believe the demonstrator was Prof. Jerrold Zacharias from MIT and the PSSC staff.  If anyone can point me to the original undubbed black and white film loop, I'd appreciate it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looked really crude but qualitatively it worked.    The mirror moved upon reversal of the mass positions.   Yeah, experimental physics!!    Calculate it out.  Use your laser pointer.   Glue mirrors.  Calibrate.  Give it as an experiment in class.   Make a (music?) video.  Put it on Youtube and embed it here.   Social physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found some other do it your self experimenters with crude equipment, experimental tips (try fishing line) and different masses.&#xA;See &lt;a href=&quot;http://funcall.blogspot.com/2009/04/lets-do-twist.html&quot; rel=&quot;nofollow&quot;&gt;http://funcall.blogspot.com/2009/04/lets-do-twist.html&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.hep.fsu.edu/~wahl/phy3802/expinfo/cavendish/cenco_grav.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.hep.fsu.edu/~wahl/phy3802/expinfo/cavendish/cenco_grav.pdf&lt;/a&gt;&#xA;and &lt;a href=&quot;http://www.fourmilab.ch/gravitation/foobar/&quot; rel=&quot;nofollow&quot;&gt;http://www.fourmilab.ch/gravitation/foobar/&lt;/a&gt;, which uses a ladder, some cobblestones, monofilament fishing line and has videos.   For the experiment in this last reference, you don't need mirrors, since you can see the balance masses move directly because their excursion is so large.   See also &lt;a href=&quot;http://www.youtube.com/watch?v=euvWU-4_B5Y&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=euvWU-4_B5Y&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all these experiments there is no calibration of the restoring force of the twisted filament (which Cavendish did from the free torsion period of the balance), the balance beam of one appears to be styrofoam, (so I would worry about subtle charge effects), and the beam hits the support of the fixed masses so that it bounces and we do not see the harmonic angular acceleration we might expect.  This last problem is apparently well known to amateur experimenters in this field.&#xA;Another exposition and video is at &lt;a href=&quot;http://www.juliantrubin.com/bigten/cavendishg.html&quot; rel=&quot;nofollow&quot;&gt;http://www.juliantrubin.com/bigten/cavendishg.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best summary and historical exposition I found is at &lt;a href=&quot;http://wapedia.mobi/en/Torsion_bar_experiment&quot; rel=&quot;nofollow&quot;&gt;http://wapedia.mobi/en/Torsion_bar_experiment&lt;/a&gt; .   I did not realize that the experiment was originally designed by John Michell, a contemporary, whose designs and apparatus passed to Cavendish upon his death.  See &lt;a href=&quot;http://wapedia.mobi/en/John_Michell&quot; rel=&quot;nofollow&quot;&gt;http://wapedia.mobi/en/John_Michell&lt;/a&gt;.  Newton had considered the deviation from vertical that a stationary pendulum would have near a terrestrial mountain in the Principia (1686).   Although he considered the deviation too small to measure, it was measured 52 years later at Chimborazo, Ecuador in 1738, which was the first experiment showing that the Earth was not hollow, apparently a live hypothesis at the time.   The same experiment was repeated in Scotland in 1774.   See &lt;a href=&quot;http://wapedia.mobi/en/Schiehallion_experiment&quot; rel=&quot;nofollow&quot;&gt;http://wapedia.mobi/en/Schiehallion_experiment&lt;/a&gt; .  Mitchell devised the torsion balance experiment in 1783, and started construction of a torsion balance.   Cavendish did his experiment in 1797-1798.   To me this is all quite inspiring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Editorial (I'll move this positive rant to meta soon) - given the obviously widely varied audience on this site, I would very much like to see more questions like this one relating to amateur or home experiments.   The analysis of the data and possible sources of errors in these experiments is often subtle, and is very instructive.    To have real physicists and other clever students publicly criticize some aspect of an experiment provides something that many students may never get otherwise.   The social network framework will help many newcomers from different countries learn what real science is in a way that yet another dose of imperfectly understood theory never will.   And it's fun too.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-11-17T07:04:22.347" LastActivityDate="2010-11-17T07:04:22.347" CommunityOwnedDate="2010-11-11T13:29:14.623" CommentCount="3" />
  <row Id="422" PostTypeId="2" ParentId="420" CreationDate="2010-11-09T18:01:47.923" Score="11" ViewCount="" Body="&lt;p&gt;When the magnet attracts and moves the paperclip, moving it in the gravitational field of the earth, the energy comes from the potential energy that [in that case] we can associate to the magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By attracting the paperclip you increase the gravitational potential energy of the paperclip but you reduce the one it has due to the magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You do not need energy to hold the paperclip. As an analogy: when you hold something in your hands, you feel like you are &quot;working&quot; even in a static case; but in mechanics, work = exchange of energy is defined as $W = \int \vec{F} d\vec{s}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the situation is stationary, no energy exchange is involved, no force does no work at all, so it can stay forever.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-09T20:35:24.710" LastActivityDate="2010-11-09T20:35:24.710" CommentCount="9" />
  <row Id="424" PostTypeId="2" ParentId="384" CreationDate="2010-11-09T19:36:12.327" Score="2" ViewCount="" Body="&lt;p&gt;A good example of impedance matching outside the world of transmission lines is &lt;a href=&quot;http://en.wikipedia.org/wiki/Horn_%28acoustic%29&quot; rel=&quot;nofollow&quot;&gt;acoustic horns&lt;/a&gt;, which match acoustic impedance between a sound source (like a vibrating string or reed) and the air.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-11-09T19:36:12.327" />
  <row Id="425" PostTypeId="1" CreationDate="2010-11-09T19:41:12.453" Score="6" ViewCount="245" Body="&lt;p&gt;All objects above absolute zero emit radiation due to random collisions between the atoms they are made of. The spectrum of radiation emitted varies according to the temperature of the object, I believe because of the random speeds at which atoms hit each other, producing photons of random energies/wavelengths that follow a probability distribution.  The energy and frequencies increase as the temperature increases. This is called thermal radiation, related to black-body radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All electrical components create small noise currents due to the random thermal motion of electrons they are made of. The current/voltage level increases as the temperature increases, with a constant white spectrum. This is called thermal noise or Johnson noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the relationship between these? Are they created by the same process? Why are their spectra different?&lt;/p&gt;&#xA;" OwnerUserId="176" LastEditorUserId="176" LastEditorDisplayName="" LastEditDate="2011-04-08T14:49:35.350" LastActivityDate="2011-04-08T14:49:35.350" Title="What’s the relationship between thermal radiation and Johnson thermal noise?" Tags="&lt;electricity&gt;&lt;electromagnetic-radiation&gt;&lt;temperature&gt;&lt;thermal-radiation&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="426" PostTypeId="1" AcceptedAnswerId="440" CreationDate="2010-11-09T19:42:14.100" Score="2" ViewCount="175" Body="&lt;p&gt;Can gamma rays of high enough energy entering our planet's atmosphere reach the surface (50% probability)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, in other words, is there a window for extremely high-energy gamma rays like for the visible spectrum and radio?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This figure, from &lt;a href=&quot;http://imagine.gsfc.nasa.gov/docs/science/know_l1/emspectrum.html&quot; rel=&quot;nofollow&quot;&gt;Electromagnetic Spectrum&lt;/a&gt;, shows that the penetration increases with increasing energy, but it seems to level out at about 25 km altitude:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://imagine.gsfc.nasa.gov/Images/introduction/emsurface.gif&quot; rel=&quot;nofollow&quot;&gt;http://imagine.gsfc.nasa.gov/Images/introduction/emsurface.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no units on the X-axis, and thus it does not show the energy for the highest energy gamma rays for this figure. This does not rule out a window at even higher energies.&lt;/p&gt;&#xA;" OwnerUserId="174" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-09T20:05:09.393" LastActivityDate="2010-11-09T20:48:47.870" Title="Can extra-solar gamma rays reach the Earth's surface?" Tags="&lt;electromagnetism&gt;&lt;radiation&gt;&lt;air&gt;&lt;atmosphere&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="427" PostTypeId="1" AcceptedAnswerId="428" CreationDate="2010-11-09T19:47:43.983" Score="8" ViewCount="464" Body="&lt;p&gt;General relativity tells us that what we perceive as gravity is curvature of space-time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand (as I understand it) gravity can be understood as a force between objects which are exchanging (hypothetical) virtual particles called gravitons, similar to the way electromagnetic forces are due to objects exchanging virtual photons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least at first glance, the two concepts seem mutually exclusive. Is there a description of gravity which includes both, or is this contradiction one of the problems in combining GR with quantum mechanics?&lt;/p&gt;&#xA;" OwnerUserId="17" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-09T20:04:11.460" LastActivityDate="2010-11-11T10:02:14.260" Title="Nature of gravity: gravitons, curvature of space-time or both?" Tags="&lt;quantum-mechanics&gt;&lt;general-relativity&gt;&lt;gravity&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="4" />
  <row Id="428" PostTypeId="2" ParentId="427" CreationDate="2010-11-09T19:55:50.603" Score="6" ViewCount="" Body="&lt;p&gt;Well, consider this: the same thing happens with electromagnetic forces. We can describe them as particles responding to the presence of electric and magnetic fields, or we can describe them as resulting from the exchange of virtual photons. Those views seem similarly incompatible, but nevertheless both theories (classical electrodynamics and quantum electrodynamics, respectively) give excellent predictions. We can't really say that one is more &quot;right&quot; than another; we just have to accept them both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation with gravity is pretty much a direct analogy to electromagnetism. We can describe gravity as particles responding to the presence of spacetime curvature, or we can describe them as resulting from the exchange of virtual gravitons. As with EM, these views would correspond to classical gravity and quantum gravity, respectively. But the difference is that, although general relativity fills the role of the classical theory, we don't have a good quantum theory of gravity yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't say that the field/particle duality is one of the problems that impedes the combination of quantum mechanics with GR. After all, we had no problem getting around the dual descriptions of electromagnetism. It's just the peculiar details of quantum gravity that make it a difficult theory to develop.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-09T20:02:41.517" LastActivityDate="2010-11-09T20:02:41.517" CommentCount="6" />
  <row Id="429" PostTypeId="1" CreationDate="2010-11-09T20:02:09.707" Score="4" ViewCount="225" Body="&lt;p&gt;In an optics lab, where all optical beams pretty much reside in a plane, it is fairly simple to describe (linear) polarizations as vertical or horizontal (or s and p).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we start talking about light coming from any angle (say, from any sky position), we now need a basis of polarization states &lt;em&gt;at every sky position&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that this requires a vector field on the sphere; and we know that no continuous, nonzero vector field on the sphere exists (the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hairy_ball_theorem&quot; rel=&quot;nofollow&quot;&gt;hairy ball theorem&lt;/a&gt;).  One might try to use parallel transport to move a basis of polarization vectors from one position to another, but of course this is path-dependent and also leads to inconsistencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this resolved?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it make sense to compare the polarizations of light coming from two different directions?&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2010-11-19T01:19:48.830" Title="How can we describe the polarization (of light) coming from an arbitrary angle?" Tags="&lt;optics&gt;&lt;polarization&gt;&lt;vector-fields&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="430" PostTypeId="2" ParentId="429" CreationDate="2010-11-09T20:13:25.403" Score="1" ViewCount="" Body="&lt;p&gt;Sure, you can just express the polarization vector in terms of a Cartesian coordinate system, or any coordinate system you like. This requires that the polarization vectors are constrained to be perpendicular to the light's momentum, $\vec{k}\cdot\vec{\epsilon}=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I quite understand the basis (no pun intended) of your objection, though. Is there a particular physical situation you have in mind in which this problem would arise?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically what's important for light coming from a wide range of angles, e.g. with sunglasses, is how much of it is horizontally polarized and how much is vertically polarized, so you can just split the polarization vector into a vertical component and a component in the 2D horizontal plane.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-09T20:13:25.403" CommentCount="3" />
  <row Id="431" PostTypeId="2" ParentId="401" CreationDate="2010-11-09T20:22:41.977" Score="2" ViewCount="" Body="&lt;p&gt;To start with, for scientific drawing usually vector graphics is more suitable - scalable, convenient to modify and produce less bulky files.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simple general-purpose graphics I use &lt;a href=&quot;http://www.openoffice.org/&quot; rel=&quot;nofollow&quot;&gt;OpenOffice.org Draw&lt;/a&gt; (I prefer it to Incscape).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For abstract diagrams there is &lt;a href=&quot;http://www.yworks.com/en/products_yed_about.html&quot; rel=&quot;nofollow&quot;&gt;yEd - Graph Editor&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both are free, for Win/Linux/MacOSX, easy to learn and can export to vector graphics and pdf.&lt;/p&gt;&#xA;" OwnerUserId="184" LastEditorUserId="184" LastEditorDisplayName="" LastEditDate="2010-11-09T22:56:57.133" LastActivityDate="2010-11-09T22:56:57.133" CommunityOwnedDate="2010-11-10T18:21:08.203" />
  <row Id="432" PostTypeId="2" ParentId="426" CreationDate="2010-11-09T20:23:02.583" Score="2" ViewCount="" Body="&lt;p&gt;I found a reference through Google Books, &lt;em&gt;Very high energy gamma-ray astronomy&lt;/em&gt; by Trevor Weekes, which says that the atmosphere is essentially opaque to high-energy gamma rays, equivalent to a meter-thick wall of lead. We are able to do gamma-ray astronomy with ground-based telescopes by detecting the decay products of the gamma rays' interactions with atmospheric particles, but the photons themselves never (well, essentially never) reach the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'd like to look up the book yourself, the information I found is at the beginning of chapter 2.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-09T20:23:02.583" CommentCount="4" />
  <row Id="433" PostTypeId="1" CreationDate="2010-11-09T20:32:29.217" Score="0" ViewCount="219" Body="&lt;p&gt;In interferometry (specifically, in the domain of &lt;a href=&quot;http://en.wikipedia.org/wiki/Fabry_Perot&quot; rel=&quot;nofollow&quot;&gt;Fabry-Perot cavities&lt;/a&gt;), the function $$f(\phi) = \frac{1}{1 + F \sin^2 \phi}$$ , which describes the shape of the resonant structure of the cavity, is often called the &quot;Airy function&quot; (for instance, in &lt;a href=&quot;http://scienceworld.wolfram.com/physics/AiryFunction.html&quot; rel=&quot;nofollow&quot;&gt;Wolfram Mathworld&lt;/a&gt;).  However, it is obviously quite different from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Airy_function&quot; rel=&quot;nofollow&quot;&gt;special functions Ai(x)&lt;/a&gt; that usually go by that name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This function resembles probability density function of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wrapped_Cauchy_distribution&quot; rel=&quot;nofollow&quot;&gt;wrapped Cauchy distribution&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How did it get the name &quot;Airy function&quot;?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard that Fabry and Perot gave it this name in one of their original papers (maybe &lt;a href=&quot;http://hal.archives-ouvertes.fr/docs/00/23/96/30/PDF/ajp-jphystap_1892_1_313_0.pdf&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;? PDF, in French, which I can't read), in honor of (the same) George Biddell Airy who had earlier considered similar interferometers.  It would be great if someone could help ferret out the first reference to that function by this name.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-03-15T22:59:45.427" LastActivityDate="2011-03-15T22:59:45.427" Title="In interferometry, what is the origin of the name &quot;Airy function&quot;?" Tags="&lt;optics&gt;&lt;history&gt;&lt;interferometry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="434" PostTypeId="2" ParentId="427" CreationDate="2010-11-09T20:34:53.903" Score="3" ViewCount="" Body="&lt;p&gt;Although the analogy between gravity and electromagnetism made by David is fine and self-suggesting, it must be cautiously added that there is no proof that gravity must look like exchange of gravitons at the microscopic level. We actually do not know what the microscopic picture of gravity is, and it might turn out to be very different from the familiar description in terms of particle-carriers of that force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, earlier this year there was a &lt;a href=&quot;http://arxiv.org/abs/arXiv:1001.0785&quot; rel=&quot;nofollow&quot;&gt;preprint by Erik Verlinde&lt;/a&gt; suggesting that gravity might be an &lt;strong&gt;entropic force&lt;/strong&gt;. It this is true, gravitons do not appear in this picture at all. This preprint is being actively discussed (more that 100 citations this year). However it also must be said that Verlinde's suggestion still remains a suggestion, not a theory, as it relies on some murky heuristic arguments, not a solid mathematical theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; the commenter below correctly points out that regardless of the microscopic theory the large-wavelength gravitational waves exist in any case and they can be quantized giving rise to gravitons. So, I guess my caution was misleading. &lt;/p&gt;&#xA;" OwnerUserId="59" LastEditorUserId="59" LastEditorDisplayName="" LastEditDate="2010-11-09T21:53:05.940" LastActivityDate="2010-11-09T21:53:05.940" CommentCount="2" />
  <row Id="437" PostTypeId="1" AcceptedAnswerId="521" CreationDate="2010-11-09T20:38:46.200" Score="7" ViewCount="678" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation&quot; rel=&quot;nofollow&quot;&gt;Schrödinger equation&lt;/a&gt; describes the quantum mechanics of a single massive non-relativistic particle. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_equation&quot; rel=&quot;nofollow&quot;&gt;Dirac equation&lt;/a&gt; governs a single massive relativistic spin-½ particle. The photon is a massless, relativistic spin-1 particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the equivalent equation giving the quantum mechanics of a single photon?&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-11T15:47:04.747" LastActivityDate="2011-06-01T04:55:34.117" Title="What equation describes the wavefunction of a single photon?" Tags="&lt;electromagnetism&gt;&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="1" />
  <row Id="438" PostTypeId="2" ParentId="409" CreationDate="2010-11-09T20:47:33.723" Score="12" ViewCount="" Body="&lt;p&gt;There are three main feasible ways of detecting a black hole:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Gravitational lensing: the strong gravitational attraction of a black hole bends space time and the light coming from nearby stars (nearby in the sense of being in the same are in our sky) is bent inwards. There are a few well known distorsion types due to gravity, but mainly we can see galaxies, which are more or less elliptical, bent into pancake shapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/vwZuR.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Accretion disks and jets: as the black hole &quot;sucks in&quot; dust and other similar matter from nearby space, the matter is accelerated at relativistic velocities and it emits x-rays as it goes to die inside the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Q9VzY.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Stars orbiting black holdes: if a star is orbiting a black hole, it will appear to be orbiting empty space (since we can't basically see a black hole directly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/PmX7R.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Other ways, like Hawking radiation, are only theoretically possible for now  -we could maybe be able to see old mini black holes &quot;popping&quot; but it's not really clear how that would happen exactly and none has been seen so far.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-09T20:47:33.723" CommentCount="1" />
  <row Id="439" PostTypeId="1" CreationDate="2010-11-09T20:47:51.813" Score="1" ViewCount="196" Body="&lt;p&gt;I recently read about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kessler_syndrome&quot; rel=&quot;nofollow&quot;&gt;Kessler Syndrome&lt;/a&gt; and am thinking about writing a story set in a world suffering from it. In the interest of realism, I am curious about the secondary effects which would be produced by this kind of scenario.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;About what distance would the debris cloud reside? How deep would it be?&lt;/li&gt;&#xA;&lt;li&gt;Would the debris cloud destroy satellites? Some satellites? All satellites? Could you build a satellite which would survive it?&lt;/li&gt;&#xA;&lt;li&gt;What would be the effects on the ground? Would there be frequent meteor showers? Would the meteor showers cause damage? Would the sun or stars be obscured?&lt;/li&gt;&#xA;&lt;li&gt;How long would the cloud last? How could it be manually eliminated?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I understand that there are a huge number of variables here, and that many of the answers depend upon how the syndrome got its start. I'm just curious about the best-case and worst-case scenarios. For the best-case scenario, imagine that the &lt;a href=&quot;http://en.wikipedia.org/wiki/International_space_station&quot; rel=&quot;nofollow&quot;&gt;International Space Station&lt;/a&gt; explodes. For worst-case, imagine that &lt;a href=&quot;http://en.wikipedia.org/wiki/433_Eros&quot; rel=&quot;nofollow&quot;&gt;433 Eros&lt;/a&gt; is in geosynchronous orbit and explodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also understand that this is a subjective question. I am trying to make it a &lt;a href=&quot;http://blog.stackoverflow.com/2010/09/good-subjective-bad-subjective/&quot; rel=&quot;nofollow&quot;&gt;good subjective question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="195" LastActivityDate="2010-11-15T16:53:14.610" Title="What would be the real-world implications of the Kessler Syndrome?" Tags="&lt;astrophysics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="440" PostTypeId="2" ParentId="426" CreationDate="2010-11-09T20:48:47.870" Score="3" ViewCount="" Body="&lt;p&gt;Propagation of very energetic photons through medium looks like this. The photon enter the medium, at some point it scatters off an electron or creates an electron-positron pair in a coulomb field, the initial energy distributed between the two daughter particles. Each of these particles then &quot;splits&quot; again, and so on, and as a result an electromagnetic shower develops. The typical length at which this &quot;doubling&quot; occurs is called the radiation length. This length is calculated (and measured) for many materials; it is a property of a medium and does not depend on the energy of the initial photon. For air at normal pressure and temperature is it of the order of hundreds of meters (and it is increases with pressure drop). So, it means that the initial extremely energetic photon will hardly make it into the troposphere, not to mention ground.&lt;/p&gt;&#xA;" OwnerUserId="59" LastActivityDate="2010-11-09T20:48:47.870" />
  <row Id="441" PostTypeId="2" ParentId="437" CreationDate="2010-11-09T20:51:01.967" Score="3" ViewCount="" Body="&lt;p&gt;The maxwell equations, just like in classical electrodynamics. You'll need to use quantum field theory to work with them though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Relativistic_wave_equations&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Relativistic_wave_equations&lt;/a&gt;&lt;br/&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_electrodynamics&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Quantum_electrodynamics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="165" LastActivityDate="2010-11-09T20:51:01.967" />
  <row Id="442" PostTypeId="1" AcceptedAnswerId="448" CreationDate="2010-11-09T20:54:37.543" Score="3" ViewCount="238" Body="&lt;p&gt;This isn't a homework problem. I'm asking cause I'm wondering how strong a person would have to be to pull it off. Their arms would be extended, both hands gripping the staff at one end, and the staff is facing directly away from them, parallel to the floor, going through their 'z-axis' as it were (with x being left-right and y being top-bottom).&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-09T21:57:22.177" LastActivityDate="2010-11-10T03:57:52.183" Title="How much force would be necessary to hold a 6-foot steel staff weighing 20 lbs from one end of it, parallel to the floor?" Tags="&lt;forces&gt;&lt;torque&gt;&lt;statics&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="444" PostTypeId="2" ParentId="442" CreationDate="2010-11-09T21:08:51.313" Score="4" ViewCount="" Body="&lt;p&gt;It's a 20-lbs staff.  20-lbs net force are required to hold it up, regardless of its orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just apply this force to one end of the staff, though, there would be a net torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, you need to use your hands to apply two forces to the staff.  One force, exerted by the near hand on the very end of the staff, should be down.  The other, exerted by the far hand a few inches down the staff, should be up.  You can imagine a see-saw.  The far hand is the pivot of the see-saw, which pushes up on the board.  The near hand is a very heavy person sitting very near the pivot, and the pole itself is a very light person sitting far from the pivot.  (The weight of the pole itself can be considered to be concentrated at  the half-way point.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three forces on the pole and hence two distances between them.  The ratio of the forces at the extremes needs to be the inverse of the ratio of the distances.  That is, if one force is three times further out from the pivot, it should be one-third as large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say the distance between your hands is four inches, or 1/3 a foot.  The distance to the middle of the pole is about 3 feet, or about 10 times as much.  So you need to exert a force of about 200 pounds down with your near hand and 220 pounds up with your far hand.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-09T21:18:04.270" LastActivityDate="2010-11-09T21:18:04.270" CommentCount="3" />
  <row Id="445" PostTypeId="2" ParentId="442" CreationDate="2010-11-09T21:14:22.663" Score="0" ViewCount="" Body="&lt;p&gt;Since you cannot hold the staff at exactly the end, your hand (or rather, your finger on the longer side of the staff) would become your fulcrum and your question becomes a first class lever problem with the weight of the long side becoming the force and the resistance the small part against your hand, in equilibrium.&#xA;Roughly, something like...&#xA;F*lever arm = R*Resistance Arm... you'll have to get one of the smarter guys here to calculate the force on the lever arm.. I think you might have to integrate over the length since the force is not acting along a single point on the lever.&lt;/p&gt;&#xA;" OwnerUserId="49" LastActivityDate="2010-11-09T21:14:22.663" />
  <row Id="446" PostTypeId="2" ParentId="379" CreationDate="2010-11-09T21:16:39.130" Score="6" ViewCount="" Body="&lt;p&gt;The vacuum state is the thermal state for $T=0K$. How to compare if a state is close enough to the vacuum state? By counting photons (for vacuum it is zero). The occupation for photons is given by Bose-Einstein distribution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n = \frac{1}{\exp( E/(kT)) - 1},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $E$ is the photon energy ($E = \hbar \omega = h \nu$) and $k$ is the Boltzmann constant. For room temperature ($T\approx 300K \Rightarrow kT \approx 0.025eV$) and visible light ($E\approx 2.5eV$) it gives&#xA;$$n\approx 10^{-44}$$&#xA;that is, very very few (and in practice - the vacuum state).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.qis.ucalgary.ca/quantech/wiggalery.php&quot; rel=&quot;nofollow&quot;&gt;a gallery of Wigner functions&lt;/a&gt; (the so-called Winger function illustrates fluctuation of electrical field around 0; note that even fore $T=0$ there are some fluctuations (zero-point energy fluctuations), but for $T&gt;0$ there are higher).&lt;/p&gt;&#xA;" OwnerUserId="184" LastEditorUserId="184" LastEditorDisplayName="" LastEditDate="2010-11-10T00:37:14.450" LastActivityDate="2010-11-10T00:37:14.450" />
  <row Id="447" PostTypeId="2" ParentId="442" CreationDate="2010-11-09T21:19:52.763" Score="1" ViewCount="" Body="&lt;pre&gt;&lt;code&gt; ==================================================&#xA;&#xA; |&amp;lt;---d---&amp;gt;|&amp;lt;----(L/2-d)---&amp;gt;|&amp;lt;--------L/2---------&amp;gt;&#xA; |                          |&#xA; |                          V&#xA; |                         mg&#xA; |&#xA; V&#xA; F&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;$ F = \frac{mg(L/2 -d )}{d}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need to support the rod at distance 'd' using your hand at the end where you are holding it.(&lt;a href=&quot;http://physics.stackexchange.com/questions/442/how-much-force-would-be-necessary-to-hold-a-6-foot-steel-staff-weighing-20-lbs-fr/445#445&quot;&gt;look here&lt;/a&gt;) i.e. 'd' can't be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m - mass of the rod&lt;br&gt;&#xA;l - length of the rod&lt;br&gt;&#xA;F - force required&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-11-09T22:03:03.757" LastActivityDate="2010-11-09T22:03:03.757" />
  <row Id="448" PostTypeId="2" ParentId="442" CreationDate="2010-11-09T21:22:21.573" Score="2" ViewCount="" Body="&lt;p&gt;If both of the person's hands were &lt;em&gt;really&lt;/em&gt; at one end of the staff, it wouldn't be possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that, in order for the staff not to move, the net torque on it has to be zero. With only one plane of rotation, the torque due to a given force can be defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tau = rF\sin\theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $r$ is the distance from a chosen &quot;center of rotation&quot; to the point where the force is applied, and $F$ is the magnitude of the force. Now, suppose you choose the point where the staff is being held as the center of rotation. In that case, the hands exert zero torque since $r=0$ for them, but the force of gravity acts on the center of the staff, 3 feet away, so it exerts a nonzero torque. Thus the total net torque is nonzero, and the staff will drop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you take into account that a person's hands have some finite size, and thus they can't be holding the staff &lt;em&gt;exactly&lt;/em&gt; at its end, then it's possible to hold it up. The force required will depend on where the hands are positioned, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/oppg1.png&quot; alt=&quot;force diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $d$ is the width of the person's hand, and suppose that the force exerted by each hand is applied exactly in the center of that hand. (This last assumption isn't true but it's probably good enough for a basic explanation) Then you have three forces being exerted on the staff: gravity acting through its center, the force from one hand acting at a distance $d/2$ from one end, and the force from the other hand acting at a distance $3d/2$ from that same end. In order for the staff to be stationary, all the forces have to cancel out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-mg + F_1 + F_2 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and all the torques have to cancel out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-mg\times\frac{L}{2} + F_1\times\frac{d}{2} + F_2\times\frac{3d}{2} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($L$ is the length of the staff) Solving this system of equations gives you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_1 = \frac{mg}{2d}(3d-L)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_2 = \frac{mg}{2d}(L-d)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging in the given numbers with a reasonable guess for hand width, I get 770 N (170 lbf), directed downward, for the first force (closer to the end of the staff), and 860 N (190 lbf), directed upward, for the second force. So we're basically talking about the ability to exert 200 lbf with each hand, although you could rely on the structural integrity of your arm to help you out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that, since one of your hands is pulling upward and the other is pushing downward, the &lt;em&gt;net&lt;/em&gt; force is 90 N, or 20 lbf, upwards - exactly what is needed to support the weight of the staff.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-10T03:53:13.303" LastActivityDate="2010-11-10T03:53:13.303" CommentCount="2" />
  <row Id="449" PostTypeId="2" ParentId="414" CreationDate="2010-11-09T21:54:07.197" Score="4" ViewCount="" Body="&lt;p&gt;A semi-silly idea that I've read about is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Primon_gas&quot; rel=&quot;nofollow&quot;&gt;Primon gas&lt;/a&gt;, a model where the Riemann zeta function arises as the partition function of a quantum statistical mechanical system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More seriously, take a look at the papers of &lt;a href=&quot;http://arxiv.org/abs/hep-th/0201036&quot; rel=&quot;nofollow&quot;&gt;Yuri Manin and Matilde Marcolli on the hep-th arxiv&lt;/a&gt;, which attempt to connect the holographic principle to arithmetic geometry.  I think there's a lot of hope that the techniques in physics inspired by quantum field theory and string theory might have applications to various branches of mathematics including number theory (for this sort of thing, I can't do better than point you to &lt;a href=&quot;http://math.ucr.edu/home/baez/TWF.html&quot; rel=&quot;nofollow&quot;&gt;the writings of John Baez&lt;/a&gt;) -- I am not as aware of applications of number theory to the kind of physics that can be tested experimentally (though I'd love to be corrected).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One unrelated example -- Freeman Dyson has made vague speculations on quasicrystals and the Riemann hypothesis, you can read about it along with some entertaining history in &lt;a href=&quot;http://www.ams.org/notices/200902/rtx090200212p.pdf&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-09T21:54:07.197" />
  <row Id="450" PostTypeId="2" ParentId="420" CreationDate="2010-11-09T22:02:41.620" Score="0" ViewCount="" Body="&lt;p&gt;The paperclip is not moving relative to earth - this means that no energy is being spent by magnet to hold it, so it can hold the paperclip as long as the magnet has magnetic properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a contrast, when you hold something with your hands, energy is being spent by your muscles not to hold the object, but to remain contracted against gravity. This is because natural state of muscle cells is to be stretched or contracted as gravity tells them to, not against it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By analogy, electromagnet spends energy not to hold things, but to to have magnetic properties, because its natural state is not magnetic. So natural magnet does not spend any energy, since its natural state is being magnetic.&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2010-11-09T22:02:41.620" CommentCount="1" />
  <row Id="451" PostTypeId="2" ParentId="356" CreationDate="2010-11-09T22:04:20.943" Score="5" ViewCount="" Body="&lt;p&gt;You can't measure the mass of Earth directly, as others have stated. You can calculate it knowing:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The value of $g$, the gravitational acceleration (approximately $9.8 m/s$)&lt;/li&gt;&#xA;&lt;li&gt;The value of $R_e$, the radius of the Earth (approximately $6378.1 km$)&lt;/li&gt;&#xA;&lt;li&gt;The value of $G$, the gravitational constant (approximately $6.67×10^{-11} N m^2/kg^2$)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and solving the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$mg = \frac{GM_em}{R_e^2}$ or $M_e = \frac{g R_e^2}{G}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;to measure $g$ you can &lt;a href=&quot;http://en.wikipedia.org/wiki/Pendulum#Gravity_measurement&quot; rel=&quot;nofollow&quot;&gt;use a pendulum&lt;/a&gt; - this &lt;strong&gt;can&lt;/strong&gt; be done at home.&lt;/li&gt;&#xA;&lt;li&gt;to measure $R_e$ the simplest experiment is &lt;a href=&quot;http://en.wikipedia.org/wiki/Eratosthenes#Eratosthenes.27_measurement_of_the_earth.27s_circumference&quot; rel=&quot;nofollow&quot;&gt;Eratosthenes' experiment&lt;/a&gt; - this &lt;strong&gt;cannot&lt;/strong&gt; be done at home&lt;/li&gt;&#xA;&lt;li&gt;to measure $G$ you need to use a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cavendish_experiment&quot; rel=&quot;nofollow&quot;&gt;Cavendish balance&lt;/a&gt; - which &lt;strong&gt;cannot&lt;/strong&gt; be done at home because it's a notoriously difficult experiment (the constant is really small, requires custom apparatus, a very long time, etc.).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-09T22:10:06.583" LastActivityDate="2010-11-09T22:10:06.583" CommentCount="2" />
  <row Id="453" PostTypeId="2" ParentId="425" CreationDate="2010-11-09T22:40:25.673" Score="5" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Black_body&quot; rel=&quot;nofollow&quot;&gt;Blackbody radiation&lt;/a&gt; is an idealized description of &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_radiation&quot; rel=&quot;nofollow&quot;&gt;thermal radiation&lt;/a&gt; of a substance which is in &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_equilibrium&quot; rel=&quot;nofollow&quot;&gt;thermal equilibrium&lt;/a&gt; with the photon field.  Your description in the question, which equates thermal radiation and blackbody radiation is therefore not quite accurate.  Indeed, blackbody radiation is quite simple compared to thermal radiation in general -- blackbody radiation doesn't depend at all on material properties, all it depends on is the temperature and fundamental constants of the electromagnetic field.  Why is this?  Philosophically, when two things are in thermal equilibrium, we can &quot;set an equals sign&quot; between many of their properties.  Therefore, when an object is in thermal equilibrium with the field of photons, to understand the radiation it emits, we only need properties arising from the statistical mechanics and spectrum of photons in three dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Johnson%E2%80%93Nyquist_noise&quot; rel=&quot;nofollow&quot;&gt;Johnson noise&lt;/a&gt; also is independent of the properties of the material, and as it also has to do with electromagnetic fields, one might expect it to be related to blackbody radiation inside the conductor.  This is indeed the case, but now we must concern ourselves with photons in one dimension, since the typical setting for Johnson noise is in a wire, rather than free space!  This explains the difference in the formulas.  Derivations of Johnson noise in this context can be found &lt;a href=&quot;http://www.claysturner.com/dsp/Johnson-Nyquist%20Noise.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://socrates.berkeley.edu/~phylabs/bsc/LV_Programs/Lab_9/Johnson%20Noise%20Docs/Phys122_Johnson_Noise.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (just from a google search of &quot;Johnson noise&quot; and &quot;blackbody&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read a discussion of how to think about Johnson noise in terms of blackbody radiation in &lt;a href=&quot;http://www.eng.yale.edu/rslab/internal/Papers/dickepaper.pdf&quot; rel=&quot;nofollow&quot;&gt;this classic 1946 paper by Robert Dicke&lt;/a&gt;.  The physical point made there is that an antenna receiving blackbody radiation at temperature $T$ and a resistor at temperature $T$ experiencing Johnson noise must have equal power.  The difference in the forms of the power spectra is apparently due to the frequency dependence of the antenna's detection pattern.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-09T22:40:25.673" />
  <row Id="454" PostTypeId="1" CreationDate="2010-11-09T22:47:23.933" Score="2" ViewCount="92" Body="&lt;p&gt;How can you use the relaxation method to model negative dielectrics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relaxation method is usually used to model electrostatics problems but negative dielectrics are only see in dynamic systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my knowledge at frequency(f)=0 no materials act like they have a negative dielectric properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The university I attend offers a numerical electrodynamics course, every year as part of a homework they ask for a a negative dielectric be put into the relaxation method simulation that the students have programmed through the course. A negative dielectric slab in a parallel plate would be an example. I assume even though all of the voltages are static in time domain, that a f!=0 system is being simulated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am interested in is what systems can be modelled in this fashion? How do I correlate the results of the simulation to a physical system. Or a example of a physical system that can be simulated in the above fashion.&lt;/p&gt;&#xA;" OwnerUserId="50" LastEditorUserId="50" LastEditorDisplayName="" LastEditDate="2010-11-10T00:19:34.817" LastActivityDate="2010-11-10T18:06:01.140" Title="Using Relaxation Method to Model Negative Dielectrics in an electric field?" Tags="&lt;electromagnetism&gt;&lt;simulation&gt;&lt;computational-physics&gt;&lt;dielectric&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="455" PostTypeId="1" AcceptedAnswerId="456" CreationDate="2010-11-09T23:02:57.423" Score="2" ViewCount="243" Body="&lt;p&gt;When an object is in free fall, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a(t) = g - \frac{c}{m}v(t)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $g$ is acceleration due to gravity, $m$ is the mass of the object, and $c$ is the coefficient of air resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one get the distance traveled after t seconds? I tried integrating it, giving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v(t) = gt - \frac{c}{3m}s(t)^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s(t) = \sqrt[3]{\frac{3m}{c}(gt - v(t))}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is a function defined in terms of its derivative. How would I find a better definition of s(t)?&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-09T23:45:37.080" LastActivityDate="2010-12-14T02:03:36.130" Title="Distance travelled in free-fall" Tags="&lt;fluid-dynamics&gt;&lt;kinematics&gt;&lt;aerodynamics&gt;" AnswerCount="3" />
  <row Id="456" PostTypeId="2" ParentId="455" CreationDate="2010-11-09T23:23:12.063" Score="4" ViewCount="" Body="&lt;p&gt;You're right about the equation of motion for an object in free fall with air resistance (well, almost right: your $c$ is not the usual definition of the drag coefficient), but when you integrate it, you don't go from $v^2$ to $s^3/3$. That only works when the thing being squared is actually the variable of integration: $\int t^2\mathrm{d}t = t^3/3$, but $\int f(t)^2\mathrm{d}t \neq f(t)^3/3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To properly solve the equation, you'll need to start by finding speed as a function of time. You can write the equation as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}v}{\mathrm{d}t} = g - \frac{c}{m}v^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a separable differential equation, so you can put everything involving the independent variable $t$ on one side and everything involving the dependent variable $v$ on the other side,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}v}{g - \frac{c}{m}v^2} = \mathrm{d}t$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be integrated over $t$, giving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t = \int_{v(0)}^{v(t)}\frac{\mathrm{d}v}{g - \frac{c}{m}v^2} = \sqrt{\frac{m}{cg}}\tanh^{-1}\biggl(\sqrt{\frac{c}{mg}}v\biggr)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(assuming $v(0) = 0$). Then you can solve this for velocity,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}s}{\mathrm{d}t} = v = \sqrt{\frac{mg}{c}}\tanh\biggl(\sqrt{\frac{cg}{m}}t\biggr)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is another separable equation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{s(0)}^{s(t)}\mathrm{d}s = \int_0^t\sqrt{\frac{mg}{c}}\tanh\biggl(\sqrt{\frac{cg}{m}}t\biggr)\mathrm{d}t$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result of that integration is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s(t) = s(0) + \frac{m}{c}\log\cosh\biggl(\sqrt{\frac{cg}{m}}t\biggr)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've written a &lt;a href=&quot;http://www.ellipsix.net/blog/2010/04/calculating-terminal-speed.html&quot; rel=&quot;nofollow&quot;&gt;blog post&lt;/a&gt; about a (possibly) interesting &quot;application&quot; of this calculation. The math above is basically a summary of part of that post.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-14T01:03:54.427" LastActivityDate="2010-12-14T01:03:54.427" CommentCount="2" />
  <row Id="457" PostTypeId="2" ParentId="455" CreationDate="2010-11-09T23:23:33.173" Score="2" ViewCount="" Body="&lt;p&gt;The equation you're looking for is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m\frac{dv}{dt}=\frac{1}{2}\rho C_{D} A v^{2}-mg$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $C_{D}$ is the drag coefficient and $A$ is the cross-sectional area of the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the vertical position as a function of time, you can see the solution in &lt;a href=&quot;http://en.wikipedia.org/wiki/Free_fall#Uniform_gravitational_field_with_air_resistance&quot; rel=&quot;nofollow&quot;&gt;Free Fall&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, your calculation is not quite right. You have $s(t)$ (that is, a function which depends on the time) but you're integrating as if $s(t)$ were $t$ (which is not necessarily true).&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2010-11-09T23:23:33.173" CommentCount="0" />
  <row Id="458" PostTypeId="2" ParentId="71" CreationDate="2010-11-09T23:36:12.263" Score="0" ViewCount="" Body="&lt;p&gt;A naive explanation says that a beam after its focal point is inverted, not only in the sense of its spatial distribution but also in the sense of the direction of the electrical field vector (minus sign = adding $\pi$  to the phase). It's perfectly compatible with the fact why even beam profiles change the phase by $\pi$ and odd do not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this explanations says nothing about behaviour of the phase near the focal point. &lt;/p&gt;&#xA;" OwnerUserId="184" LastEditorUserId="184" LastEditorDisplayName="" LastEditDate="2010-11-10T00:49:21.097" LastActivityDate="2010-11-10T00:49:21.097" />
  <row Id="459" PostTypeId="2" ParentId="389" CreationDate="2010-11-09T23:39:12.650" Score="5" ViewCount="" Body="&lt;p&gt;I want to come back to this answer in a bit and expand a bit more, but it seems to me that the notion that you are reaching for is that of &lt;a href=&quot;http://en.wikipedia.org/wiki/Quasiparticle&quot; rel=&quot;nofollow&quot;&gt;quasiparticles&lt;/a&gt;, and I want to argue that these might be best understood in the context of effective field theories. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly: the electrons and holes in solid state physics are not actually necessarily true single particles, but actually collective excitations of the material (you can think of an electron quasiparticle as being an electron plus some minor deformation of the underlying lattice plus effects due to interactions with other electrons; the picture for holes is that of an empty electron quasiparticle state).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the same is true for electrons and positrons in vacuum (though the energy scales are much different)!  The mass of the electron has contributions from a cloud of virtual electron-positron pairs (this is an effect called mass renormalization).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a bit confusing -- it turns out that e.g. talking about an electron as a single particle that doesn't interact with itself is a classical notion that breaks down when quantum field theory comes into the picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, quantum field theories (i.e. the best physical pictures that we currently have) are actually agnostic in a certain strong sense as to whether an electron is a single particle, or a particle surrounded by a cloud, or whether a hole is a true particle or a quasiparticle, etc.   What do I mean by this?  Well, the mathematical models that currently describe the systems best do not talk about particles at all, they talk about excitations in quantum fields, and for interacting quantum fields, it turns out that the coupling constants that go into the Lagrangians for these field theories will get (in general divergent!) corrections from self-interactions.  Luckily, we now understand that this is not something to worry about --  there's a technique called &quot;renormalization&quot; which fixes the mathematical divergences, and there's a philosophy of &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Effective_field_theory&quot; rel=&quot;nofollow&quot;&gt;effective field theories&lt;/a&gt;&quot; which explains how to think about these cancellations.  Roughly speaking, depending on what energy scale you want to work in, you might interpret the electron as being a single particle, or a fuzzy cloud, but with regards to calculating measurable quantities, it doesn't really matter.  For a better take on this, I like the first few sections of Howard Georgi's review &quot;&lt;a href=&quot;http://www.people.fas.harvard.edu/~hgeorgi/review.pdf&quot; rel=&quot;nofollow&quot;&gt;Effective field theory&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if this doesn't make much sense at the moment.  I hope I've given you a few key words to look up.  Perhaps someone who understands all this a bit better can come by and explain it in easier terms, or I'll come back and edit it in a few days.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-09T23:39:12.650" CommentCount="1" />
  <row Id="460" PostTypeId="2" ParentId="389" CreationDate="2010-11-09T23:41:00.637" Score="1" ViewCount="" Body="&lt;p&gt;Holes in semi-conductors are considered to be &lt;a href=&quot;http://en.wikipedia.org/wiki/Quasiparticle&quot; rel=&quot;nofollow&quot;&gt;quasiparticles.&lt;/a&gt; A quasiparticle is a group of particles that end up acting and having many of the properties of a single particle for a period of time. They are normally considered to have a lifetime tau and decay exponentially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are positrons and electrons that we obverse quasiparticles too? QPs with a very long decay time?  That goes beyond our current level of physics.  I have not heard any evidence that they are but the conjecture has been considered.&lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2010-11-09T23:41:00.637" CommentCount="1" />
  <row Id="461" PostTypeId="1" AcceptedAnswerId="477" CreationDate="2010-11-10T00:02:28.050" Score="2" ViewCount="216" Body="&lt;p&gt;A black hole is so dense that a sphere around it called the event horizon has a greater escape velocity than the speed of light, making it black. So why do astronomers think that there is anything weird (or lack of anything Inc space) inside the event horizon. Why isn't simple the limit to where light can escape and in the middle of event horizon (which physically isnt a surface) is just a hyper dense ball of the matter that's been sucked in and can't escape just like light. Why is it thought that the laws of physics don't exist in the event horizon?&lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2010-11-15T00:14:28.717" Title="Why is it thought that normal physics doesn't exist inside the event horizon of a black hole?" Tags="&lt;astrophysics&gt;&lt;laws-of-physics&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="462" PostTypeId="2" ParentId="437" CreationDate="2010-11-10T00:30:37.547" Score="3" ViewCount="" Body="&lt;p&gt;The general concept of quantum mechanics is that particles are waves. On of hand-waveing &quot;derivations&quot; of quantum mechanics is assumption that phase of particles behaves in the same way as phase of light $\exp( i \vec{k}\cdot \vec{x} - iE t / \hbar)$ (see &lt;em&gt;Feynman Lectues on Physics&lt;/em&gt;, Volume 3, Chapter 7-2).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For light that is monochromatic (or almost monochromatic), just take Maxwell Equations plus add assumption that one photon can't be partially absorbed. Most of the time it suffices to use the paraxial approximation, or even - plane wave approximation. It works for standard quantum mechanics setups like&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Elitzur-Vaidman_bomb-tester&quot; rel=&quot;nofollow&quot;&gt;Elitzur–Vaidman bomb-tester&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For nonmonochronatic light its much more complicated. More on nature of quantum mechanics of one photon:&#xA;Iwo Bialynicki-Birula, &lt;a href=&quot;http://www.cft.edu.pl/~birula/publ/APPPwf.pdf&quot; rel=&quot;nofollow&quot;&gt;On the Wave Function of the Photon&lt;/a&gt;, Acta Physica Polonica 86, 97-116 (1994).&lt;/p&gt;&#xA;" OwnerUserId="184" LastEditorUserId="184" LastEditorDisplayName="" LastEditDate="2010-11-10T00:38:16.430" LastActivityDate="2010-11-10T00:38:16.430" />
  <row Id="463" PostTypeId="1" AcceptedAnswerId="479" CreationDate="2010-11-10T00:31:55.720" Score="5" ViewCount="445" Body="&lt;p&gt;Can any body give a concise of the most fundamental assumptions of quantum mechanics in plain english ?&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2010-11-10T05:54:01.383" Title="List of the most fundamental assumptions of quantum mechanics?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="7" />
  <row Id="464" PostTypeId="2" ParentId="419" CreationDate="2010-11-10T00:44:13.540" Score="1" ViewCount="" Body="&lt;p&gt;I'm still not sure quite what you're after, but the following statements might be helpful/relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a conservative vector field $\vec{F}$, the curl is always zero, i.e. $\nabla \times \vec{F} = \vec{0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The divergence of the curl of any vector field is always zero, i.e. $\nabla \cdot \nabla \times \vec{F} = 0$.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2010-11-10T00:44:13.540" CommentCount="1" />
  <row Id="465" PostTypeId="2" ParentId="463" CreationDate="2010-11-10T00:51:03.837" Score="1" ViewCount="" Body="&lt;p&gt;I don't really understand exactly what you mean with assumptions, but I guess you ask about the physical nature and behavior of the quantistic world. I am going to be very inaccurate in the following writing, but I need some space to get things understood. What you are going to read is a harsh simplification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first assumption is the fact that particles at the quantistic level don't have a position, like real-life objects, they don't even have a speed. Well, they sort of have it, but in reality they are &quot;fuzzy&quot;. Fuzzy in the sense that you cannot really assign a position to them, or a speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be fair, you perfectly know a real-world case scenario where you have to give up the concepts of speed and position for an entity. Take a guitar, and pluck a string. Where is the &quot;vibration&quot; ? Can you point at one specific point in the string where the &quot;vibration&quot; is, or how fast this vibration is traveling. I guess you would say that the vibration is &quot;on the string&quot; or &quot;in the space between one end of the string, and the other hand&quot;. However, there are parts of the string that are vibrating more, and parts that are not vibrating at all, so you could rightfully say that a larger part of the &quot;vibration&quot; is in the center of the string, a smaller part is on the lateral parts close to the ends, and no vibration is at the very ends. In some sense, you will experience higher &quot;presence&quot; of the vibration in the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With electrons, it's exactly the same. Think &quot;electron&quot; as &quot;vibration&quot; in the previous example. Soon, you will realize that you cannot really claim anything about the position of an electron, for the fact that has wave nature. You are forced to see the electron not as a charged ball rolling here and there, but as a diffuse entity totally similar to the concept of &quot;vibration&quot;. As a result, you cannot say anything about the position of an electron, but you can claim that there are zones in the space where the electron has &quot;higher presence&quot; and zones that have &quot;smaller presence&quot;. These zones are probabilistic in nature, and this probability is directly obtained by a mathematical description called the wavefunction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wavefunction mainly depends on the external potential, meaning that the presence of charges, such as protons and other electrons, will affect the perceived potential and will have an effect on the probability distribution in space of an electron. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second assumption is the mapping between physical quantities (such as energy, or momentum) and quantistic operators. Take for example the momentum of an everyday object: it is given by its mass times its speed. In the quantistic world, you don't really have the concept of position, hence you don't have the concept of speed, hence you have to reinterpret the whole thing in terms of probability (rememeber the string ?), which means that what you have about a quantistic object is its wavefunction, and you have to extract the information about the momentum from this wavefunction. How do you do it ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, there's a dogma in quantum mechanics, that if you apply a magic &quot;operator&quot; to the wavefunction, it gives you the quantity you want. There are (simple) rules to generate these operators, and you can apply them to a wavefunction to query the momentum, or the total energy, the kinetic energy and so on.&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="217" LastEditorDisplayName="" LastEditDate="2010-11-10T01:02:40.253" LastActivityDate="2010-11-10T01:02:40.253" CommentCount="4" />
  <row Id="466" PostTypeId="1" AcceptedAnswerId="474" CreationDate="2010-11-10T00:52:20.423" Score="8" ViewCount="188" Body="&lt;p&gt;So light travels slower in glass (for example) than in a vacuum.&#xA;What causes light to slow down? Or: How does it slow down?&#xA;If light passes through the medium, is it not essentially traveling in the &quot;vacuum&quot; between the atoms?&lt;/p&gt;&#xA;" OwnerUserId="207" LastActivityDate="2010-11-10T04:12:13.433" Title="What is the mechanism behind the slowdown of light in a transparent medium?" Tags="&lt;electromagnetism&gt;" AnswerCount="4" />
  <row Id="467" PostTypeId="2" ParentId="419" CreationDate="2010-11-10T00:53:01.123" Score="1" ViewCount="" Body="&lt;p&gt;I don't know whether the second sentence makes any sense. Let's omit it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electron in a constant homogeneous magnetic field moves on a circle with a constant value of velocity (in this meaning the velocity is independent of the location) but with a changing velocity vector. When we add an electric field, it will become only a bit more complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is, the velocity vector is constant in space (independent of the location) only for very specific settings, when a body moves along a straight line.&lt;/p&gt;&#xA;" OwnerUserId="219" LastActivityDate="2010-11-10T00:53:01.123" />
  <row Id="468" PostTypeId="2" ParentId="466" CreationDate="2010-11-10T01:13:41.770" Score="0" ViewCount="" Body="&lt;p&gt;Photons interact with matter all the time when passing through a medium. Only particular frequencies are resonant with the electronic (or rotovibronic) levels in the molecules and get to promote an excitation (thus originating colors, for example), but for most part they are just an electromagnetic disturbance which still has an effect. This interaction slows the propagation of the photon down.&lt;/p&gt;&#xA;" OwnerUserId="217" LastActivityDate="2010-11-10T01:13:41.770" />
  <row Id="469" PostTypeId="2" ParentId="353" CreationDate="2010-11-10T01:21:20.943" Score="0" ViewCount="" Body="&lt;p&gt;We see that candela is very well-defined, without using properties of, for example, a human eye. Let me remind the definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency $540 \cdot 10^{12}$ hertz and that has a radiant intensity in that direction of $\frac{1}{683}$ watt per steradian.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ampere of course cannot be defined as a number of elementary charges per second, because without definition of ampere we cannot define a value of an elementary charge! The definitions is below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to $2 \cdot 10^{−7}$ newton per metre of length.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is important, both definitions don't use any prototypes and &lt;strong&gt;define clear ways how to reproduce these units inside a lab&lt;/strong&gt;. Such experiments should &lt;strong&gt;always give the same results&lt;/strong&gt; anywhere in the world - and they do. The definitions are &lt;strong&gt;simple and unambiguous&lt;/strong&gt;. In SI, only the definition of a kilogram uses a prototype, because the definition of mass still causes problems to physicists.&lt;/p&gt;&#xA;" OwnerUserId="219" LastActivityDate="2010-11-10T01:21:20.943" CommentCount="6" />
  <row Id="470" PostTypeId="2" ParentId="466" CreationDate="2010-11-10T01:29:16.313" Score="0" ViewCount="" Body="&lt;p&gt;Thinking of light as a wave, you can see that when a wave pass through a medium, its wavelength is perturbed a little (in the same way as you would expect a change when a wave is produced in the water and pass through an obstacle). Therefore, given the relation $\lambda = v f$, it is obvious that the velocity needs to change. &lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="18" LastEditorDisplayName="" LastEditDate="2010-11-10T02:05:42.893" LastActivityDate="2010-11-10T02:05:42.893" />
  <row Id="471" PostTypeId="2" ParentId="463" CreationDate="2010-11-10T01:53:17.600" Score="12" ViewCount="" Body="&lt;p&gt;In a rather concise manner, Shankar describes four postulates of nonrelativistic quantum mechanics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I&lt;/strong&gt;. The state of the particle is represented by a vector $|\Psi(t)\rangle$ in a Hilbert space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;II&lt;/strong&gt;. The independent variables $x$ and $p$ of classical mechanics are represented by Hermitian operators $X$ and $P$ with the following matrix elements in the eigenbasis of $X$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle x|X|x'\rangle = x \delta(x-x')$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle x|P|x' \rangle = -i\hbar \delta^{'}(x-x')$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The operators corresponding to dependent variable $\omega(x,p)$ are given Hermitian operators&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Omega(X,P)=\omega(x\rightarrow X,p \rightarrow P)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;III&lt;/strong&gt;. If the particle is in a state $|\Psi\rangle$, measurement of the variable (corresponding to) $\Omega$ will yield one of the eigenvalues $\omega$ with probability $P(\omega)\propto |\langle \omega|\Psi \rangle|^{2}$. The state of the system will change from $|\Psi \rangle$ to $|\omega \rangle$ as a result of the measurement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;IV&lt;/strong&gt;. The state vector $|\Psi(t) \rangle$ obeys the Schrödinger equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$i\hbar \frac{d}{dt}|\Psi(t)\rangle=H|\Psi(t)\rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $H(X,P)=\mathscr{H}(x\rightarrow X, p\rightarrow P)$ is the quantum Hamiltonian operator and $\mathscr{H}$ is the Hamiltonian for the corresponding classical problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, Shankar discusses the postulates and the differences between quantum mechanics and classical mechanics, &lt;strong&gt;hopefully in plain english&lt;/strong&gt;. Probably you want to take a look at that book: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0306447908&quot; rel=&quot;nofollow&quot;&gt;Principles of Quantum Mechanics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are, of course, other sets of equivalent postulates.  &lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="18" LastEditorDisplayName="" LastEditDate="2010-11-10T02:04:09.287" LastActivityDate="2010-11-10T02:04:09.287" CommentCount="5" />
  <row Id="472" PostTypeId="2" ParentId="463" CreationDate="2010-11-10T01:57:10.423" Score="3" ViewCount="" Body="&lt;p&gt;Here are some ways of looking at it for an absolute beginner.   It is only a start.  It is not the way you will finally look at it, but there is nothing actually incorrect here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, plain English just won't do, you are going to need some math.  Do you know anything about complex numbers?   If not, learn a little more before you begin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, in quantum mechanics, we don't work with probabilities, you know, numbers between 0 and 1 that represent the likelihood of something happening.   Instead we work with the square root of probabilities, called &quot;probability amplitudes&quot;, which are complex numbers.   When we are done, we convert these back to probabilities.   This is a fundamental change; in a way the complex number amplitudes are just as important as the probabilities.  The logic is therefore different from what you are used to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, in quantum mechanics, we don't imagine that we know all about objects in between measurements, rather we know that we don't.   Therefore we only focus on what we can actually measure, and possible results of a particular measurement.   This is because some of the assumed imagined properties of objects in between measurements contradict one another, but we can get a theory where we can always work with the measurements themselves in a consistent way.    Any measurement (meter reading, etc.) or property in between measurements has only a probability amplitude of yielding a given result, and the property only co-arises with the measurement itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fourth, if we have several mutually exclusive possibilities, the probability amplitude of either one or the other happening is just the sum of the probability amplitudes of each happening.    Note that this does not mean that the probabilities themselves add, the way we are used to classically.   This is new and different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we add up the probabilities (not the amplitudes this time but their &quot;squares&quot;) of all possible mutually exclusive measurements, gotten by &quot;squaring&quot; the probability amplitudes of each mutually exclusive possibilities, we always get 1 as usual.   This is called unitarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not the whole picture, but it should help somewhat.&#xA;As a reference, try looking at three &lt;a href=&quot;http://vega.org.uk/video/subseries/8&quot; rel=&quot;nofollow&quot;&gt;videos of some lectures that Feynman once gave at Cornell&lt;/a&gt;.   That should help more.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-11-10T04:03:35.897" LastActivityDate="2010-11-10T04:03:35.897" CommentCount="1" />
  <row Id="473" PostTypeId="2" ParentId="461" CreationDate="2010-11-10T02:12:52.053" Score="1" ViewCount="" Body="&lt;p&gt;For example, for starters, outside a Schwarzschild black hole horizon, particles can move in any direction, but time only goes one way. i.e. forward.    Inside the horizon, particles can only move inward toward the central singularity, i.e. one way, but time can go either forward or backwards.   This is a result of the &quot;light cones tipping over&quot;.    That's different. &lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-11-10T02:12:52.053" />
  <row Id="474" PostTypeId="2" ParentId="466" CreationDate="2010-11-10T02:23:25.783" Score="9" ViewCount="" Body="&lt;p&gt;The easiest way to get the exact behavior is from thinking about light as a classical wave interacting with the atoms in the solid material. As long as you're far away from any of the resonant frequencies of the relevant atoms, this picture isn't too bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can think each of the atoms as being like a little dipole, consisting of some positive and some negative charge that is driven back and forth by the off-resonant light field. Being an assemblage of charges that are accelerating due to the driving field, these dipoles will radiate, producing waves at the same frequency as the driving field, but slightly out of phase with it (because a dipole being driven at a frequency other than its resonance frequency will be slightly out of phase with the driving field). The total light field in the material will be the sum of the driving light field and the field produced by the oscillating dipoles. If you go through a little bit of math, you find that this gives you a beam in the same direction as the original beam-- the waves going out to the sides will mostly interfere destructively with each other-- with the same frequency but with a slight delay compared to the driving field. This delay registers as a slowing of the speed of the wave passing through the medium. The exact amount of the delay depends on the particulars of the material, such as the exact resonant frequencies of the atoms in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as you're not too close to one of the resonant frequencies, this gives you a really good approximation of the effect (and &quot;too close&quot; here is a pretty narrow range). It works well enough that most people who deal with this stuff stay with this kind of picture, rather than talking in terms of photons. The basic idea of treating the atoms like little dipoles is a variant of &quot;Huygens's Principle,&quot; by the way, which is a general technique for thinking about how waves behave.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-10T02:23:25.783" CommentCount="1" />
  <row Id="475" PostTypeId="2" ParentId="442" CreationDate="2010-11-10T03:57:52.183" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm asking cause I'm wondering how strong a person would have to be to pull it off. Their arms would be extended, both hands gripping the staff at one end, and the staff is facing directly away from them, parallel to the floor, going through their 'z-axis' as it were (with x being left-right and y being top-bottom).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ridiculously strong. I've done this with a shorter, lighter bamboo staff.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do this occasionally at my martial arts dojo, using a 1 meter, 1 kg staff, as a conditioning exercise. The physiology is very interesting, and the physics of the joints, bones, and supporting muscles goes far beyond the raw torque. Within a minute or two, you start feeling truly astonishing levels of pain and burn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm actually skeptical that any normal human could succeed with a 2 meter, 9 kg staff for a significant length of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, I'll try to remember this, and do the calculations next time we do this exercise as a distraction from the agony.&lt;/p&gt;&#xA;" OwnerUserId="224" LastActivityDate="2010-11-10T03:57:52.183" CommentCount="2" />
  <row Id="476" PostTypeId="2" ParentId="466" CreationDate="2010-11-10T04:12:13.433" Score="2" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/y8Dmw.gif&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;&lt;strong&gt;&lt;a href=&quot;http://www.physicsclassroom.com/class/refrn/u14l1d.cfm&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;&lt;/strong&gt; is a nice explanation.&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-11-10T04:12:13.433" />
  <row Id="477" PostTypeId="2" ParentId="461" CreationDate="2010-11-10T05:08:36.757" Score="2" ViewCount="" Body="&lt;p&gt;For large enough black holes, space is still weakly curved at the event horizon, so of course we should expect that normal physics still exists there. An infalling observer wouldn't experience anything out of the ordinary when crossing an event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is true is that for an outside observer, it's impossible to probe what's happening inside the event horizon of a black hole. (The best you can do is wait a long time and collect the outgoing Hawking radiation.) So from the point of view of such an observer, you can't really tell the difference between living in a world where spacetime keeps going across the horizon, or living in a world where space just ends there and some radiation emerges. This might be the sort of idea your teacher was getting at. You might want to look up &quot;black hole complementarity&quot; to learn more.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-11-10T05:08:36.757" CommentCount="2" />
  <row Id="478" PostTypeId="1" CreationDate="2010-11-10T05:09:18.123" Score="2" ViewCount="362" Body="&lt;p&gt;A magnetic field strength drops-off quickly as the distance from a magnet increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to use electromagnetic fields to create a magnetic field at a location. For example, if there are strong electromagnetic fields intersecting at a location from many strong transmitters, could a constructive interference be created which creates a local magnetic field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The engineering idea is that if you create an array of radio transmitters at a strong strength working together, could there be a interference pattern at a point far from the transmitters (maybe 100 meters away) where the fields produce a magnetic field at that location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this could be done as a EM field is composed of a electric field and a magnetic field, shouldn't there be a way to make the magnetic field strong at a location in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A positive consequence of this would be that a magnetic placed at that location 100 meters away could be moved by the field. Also, if the field was alternating, rather than steady, then a electromagnet at that location could be alternating, and the force on the electromagnet could provide motive force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess since I have never heard of this, it cannot be done. But then again, it was once said that levitating a normal magnet could not be done, then someone did it with a spinning magnet which levitates above another magnet, now it is a popular toy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--- Update&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hi, I would like to clarify, I wish to move a physical object, a small magnet at a distance, not to transmit power to the object. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="226" LastEditorUserId="226" LastEditorDisplayName="" LastEditDate="2010-11-10T13:58:10.683" LastActivityDate="2010-11-27T04:19:25.077" Title="Is it possible to project a magnetic field at a location in space?" Tags="&lt;electromagnetism&gt;&lt;magnets&gt;" AnswerCount="4" FavoriteCount="3" />
  <row Id="479" PostTypeId="2" ParentId="463" CreationDate="2010-11-10T05:54:01.383" Score="5" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://scienceblogs.com/principles/2010/01/seven_essential_elements_of_qu.php&quot; rel=&quot;nofollow&quot;&gt;Uncertain Principles: 7 essential elements of QM&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry but I can't be made plainer than that in English. &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-11-10T05:54:01.383" CommentCount="1" />
  <row Id="480" PostTypeId="2" ParentId="414" CreationDate="2010-11-10T06:04:56.500" Score="15" ViewCount="" Body="&lt;p&gt;I'm not sure i'll be able to post all the links i'd like to (not enough 'reputation points' yet), but i'll try to point to the major refs i know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matilde Marcolli has a nice paper entitled &quot;Number Theory in Physics&quot; (you can find the PDF online and also in the CiteSeerX website: just Google for the title ;-) explaining the several places in Physics where Number Theory shows up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Tangentially, there's a paper by Christopher Deninger entitled &quot;Some analogies between number theory and dynamical systems on foliated spaces&quot; that may open some windows in this theme: after all, Local Systems are in the basis of much of modern Physics (bundle formulations, etc).]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a website called &quot;number theory and physics archive&quot; — &lt;a href=&quot;http://empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/physics.htm&quot; rel=&quot;nofollow&quot;&gt;http://empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/physics.htm&lt;/a&gt; — that contains a vast collection of links to works in this interface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atiyah just gave a talk (last week) at the Simons Center Inaugural Conference, talking about the recent interplay between Physics and Math. And he capped his talk speculating about the connection between Quantum Gravity and the Riemann Hypothesis. He was supposed to give a talk at the IAS on this last topic, but it was canceled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To finish it off, let me bring the Langlands Duality to the table: it's related to Modular Forms and, a such, Number Theory. (Cavalier version: Think of the QFT Path Integral as having a Möbius symmetry with respect to the coupling constants in the Lagrangian.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that out of the way, i think the better angle to see the connection between Number Theory and Physics is to think about the physics problem in a different way: think of the critical points in the Potential and what they mean in Phase Space (Hamiltonian and/or Geodesic flow: Jacobi converted one into another; think of Jacobi fields in Differential Geometry), think about how this plays out in QFT, think about Moduli Spaces and its connection to the above. This is sort of how I view this framework… ;-)&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2010-11-10T06:04:56.500" CommentCount="4" />
  <row Id="481" PostTypeId="2" ParentId="478" CreationDate="2010-11-10T06:06:16.013" Score="0" ViewCount="" Body="&lt;p&gt;Sure, it's possible. That's how sets of cell phone radio towers work. In a typical setup, several of these towers are placed in a line, some distance outside of a town, such that the line is roughly perpendicular to the direction to the town. The placement of the towers is designed so that the signals reinforce each other as much as possible only in the town. That way, most of the cell towers' transmitted power gets directed toward the area where most people are actually using their cell phones, and you don't waste a whole lot of power by transmitting into open wilderness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the cell towers, though, the only things that really get pushed around by the EM waves are electrons in people's cell phones. Theoretically, there's no reason the same concept couldn't be used to move something larger, like a magnet, but you'd need &lt;em&gt;vastly&lt;/em&gt; more powerful transmitters - perhaps more powerful than anyone knows how to build with current technology (depending on how large of a magnet you're trying to move, and how far away it is). Still, it's just an engineering problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a side note: it's generally the electric field that's responsible for moving things around, not the magnetic field. Magnetic fields themselves don't transfer energy, they only change the direction of existing motion. But the two kinds of fields mix when you change your reference frame, so it's kind of a semi-arbitrary distinction anyway.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-10T06:06:16.013" CommentCount="1" />
  <row Id="482" PostTypeId="1" CreationDate="2010-11-10T07:51:36.360" Score="5" ViewCount="144" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Potts_model&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; writes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In statistical mechanics, the Potts&#xA;  model, a generalization of the Ising&#xA;  model, is a model of interacting spins&#xA;  on a crystalline lattice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From combinatorics conferences and seminars, I know that the Potts model has something to do with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chromatic_polynomial&quot; rel=&quot;nofollow&quot;&gt;chromatic polynomial&lt;/a&gt;, but it's not clear to me where it arises (and for which graph it's the chromatic polynomial of).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question:  What does the chromatic polynomial have to do with the Potts model?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="199" LastEditorUserId="42" LastEditorDisplayName="" LastEditDate="2010-11-10T18:06:43.480" LastActivityDate="2010-11-10T23:31:11.473" Title="What does the chromatic polynomial have to do with the Potts model?" Tags="&lt;statistical-mechanics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="483" PostTypeId="2" ParentId="478" CreationDate="2010-11-10T08:01:39.353" Score="1" ViewCount="" Body="&lt;p&gt;This can certainly be done, as &lt;a href=&quot;http://en.wikipedia.org/wiki/Inductive_charging&quot; rel=&quot;nofollow&quot;&gt;induction chargers&lt;/a&gt; work pretty much that way. That said, this technique is not very efficient and it is hard to scale at large distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For large distances, one of the most popular ideas is using &lt;a href=&quot;http://en.wikipedia.org/wiki/Microwave_power_transmission#Microwave_power_transmission&quot; rel=&quot;nofollow&quot;&gt;microwave transmission&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a &lt;a href=&quot;http://en.wikipedia.org/wiki/Wireless_energy_transfer&quot; rel=&quot;nofollow&quot;&gt;very good article&lt;/a&gt; over at wikipedia with a round-up of all these technologies with pros and cons.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-10T08:01:39.353" CommentCount="3" />
  <row Id="484" PostTypeId="1" AcceptedAnswerId="1367" CreationDate="2010-11-10T08:14:50.070" Score="5" ViewCount="278" Body="&lt;p&gt;I have a question about cosmology. At popular level people explain the time of decoupling of matter and radiation as the moment when temperature falls enough for nuclei and electrons to recombine into atoms. People say that the Universe became &amp;ldquo;transparent&amp;rdquo;. The photon cross section by an electro-neutral system is smaller than Thomson's one, i.e., technically it means that the mean free path became very large or even infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is another mechanism for the mean free path to become infinite. It is because during its expansion the universe becomes less dense. Let me explain it by example. Imagine that you are in the forest where the diameter of tree is $a$, and mean distance between trees is $b$. What is the mean diameter of the observed area? It is proportional to $b^{2}/a$, I suppose. Now let's imagine that our forest is in the expanding universe, i.e., $b$ grows (linearly, for example) with time while $a$ remains constant. Then at some moment &amp;ldquo;the diameter of the observed area&amp;rdquo; starts to grow faster than the speed of light, i.e., becomes infinite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It implies that the recombination is not necessary for the decoupling. Did the recombination start earlier than the moment I described above? Or both mechanisms (recombination and density fall) are equally important for the decoupling?&lt;/p&gt;&#xA;" OwnerUserId="225" LastActivityDate="2011-03-01T23:09:24.700" Title="Nature of decoupling of matter and radiation" Tags="&lt;cosmology&gt;" AnswerCount="4" />
  <row Id="485" PostTypeId="2" ParentId="433" CreationDate="2010-11-10T09:15:24.707" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt; this is the wrong Airy function, answer is only kept for discussion&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;According to the &lt;a href=&quot;http://de.wikipedia.org/wiki/Airy-Funktion&quot; rel=&quot;nofollow&quot;&gt;german wikipedia entry on Airy function&lt;/a&gt; (-&gt; &lt;a href=&quot;http://translate.google.com/translate?hl=de&amp;amp;sl=de&amp;amp;tl=en&amp;amp;u=http%3A%2F%2Fde.wikipedia.org%2Fwiki%2FAiry-Funktion&quot; rel=&quot;nofollow&quot;&gt;google translation&lt;/a&gt;), it is named after the british Astronomer &lt;a href=&quot;http://en.wikipedia.org/wiki/George_Biddell_Airy&quot; rel=&quot;nofollow&quot;&gt;George Biddell Airy&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;update&lt;/strong&gt; &lt;a href=&quot;http://wordiq.com/definition/Airy_function&quot; rel=&quot;nofollow&quot;&gt;http://wordiq.com/definition/Airy_function&lt;/a&gt; cites Airy (1838). On the intensity of light in the neighbourhood of a caustic. Transactions of the Cambridge Philosophical Society, 6, 379—402., which may simply be the first publication with these functions thus justifying the naming, but I haven't read it and can only guess. I found it on the archive after some searching: &lt;a href=&quot;http://www.archive.org/details/transactionsofca06camb&quot; rel=&quot;nofollow&quot;&gt;http://www.archive.org/details/transactionsofca06camb&lt;/a&gt; (page 379 ff)&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2010-11-15T08:26:13.987" LastActivityDate="2010-11-15T08:26:13.987" CommentCount="4" />
  <row Id="486" PostTypeId="1" AcceptedAnswerId="489" CreationDate="2010-11-10T09:26:06.553" Score="1" ViewCount="139" Body="&lt;p&gt;In school-level tasks, when (almost) all substances are linear, homogeneous and isotropic, we have $D=\epsilon E$, $H=B/\mu$ and thus Maxwell &quot;in material&quot; equations (1) say how $E$ and $B$ depend on time given known dependence of $\rho$ (free charge density) and $j$ (free current density). Here they are in CGS unit system:&#xA;$$\left\{\begin{aligned}&#xA;\text{div} D=4\pi\rho\\&#xA;\text{div} B=0\\&#xA;\text{rot} E = -\frac{1}{c}\frac{\partial B}{\partial t}\\&#xA;\text{rot} H = \frac{4\pi}{c}j+\frac{1}{c}\frac{\partial D}{\partial t}&#xA;\end{aligned}\right.$$&#xA;Also we know continuity equation $\partial \rho/\partial t + \text{div} j=0$. But this is not enough to determine, how j will change over time or in statical case, how $j$ is distributed in the conductor. What are other equations for $j$? Are there any for some &quot;ideal case&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I don't know actually, is the following task correct or under determined:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Electric current I flows along infinite cylindrical conductor. Inner radius is $r$, outer is $R$, magnetic constants of all substances are given ($\mu_1,\mu_2, \mu_3$ from inside out). Find magnetic field ($B$ and $H$) and current distribution in a conductor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The question:&lt;/strong&gt; Is there any &quot;standard&quot; equations for $j$? Particularly, is the task above well-determined?&lt;/p&gt;&#xA;" OwnerUserId="76" LastActivityDate="2010-11-10T10:39:21.013" Title="Maxwell equations: how to know the behavior of charge and current?" Tags="&lt;electromagnetism&gt;&lt;electricity&gt;" AnswerCount="2" />
  <row Id="487" PostTypeId="2" ParentId="486" CreationDate="2010-11-10T09:34:44.027" Score="1" ViewCount="" Body="&lt;p&gt;In general, &lt;a href=&quot;http://en.wikipedia.org/wiki/Ohm%27s_law&quot; rel=&quot;nofollow&quot;&gt;Ohm's law&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\vec j = \sigma \vec E$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is used, where $\sigma$ is the conductivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example however, you'll use $j = I / A$ where $A$ is the cross section of the conductor, in general electric current through an area $A$ is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I := \iint_{A} \vec j\cdot d\vec n.$$&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2010-11-10T09:41:04.157" LastActivityDate="2010-11-10T09:41:04.157" CommentCount="3" />
  <row Id="488" PostTypeId="2" ParentId="419" CreationDate="2010-11-10T10:26:32.203" Score="2" ViewCount="" Body="&lt;p&gt;Just to show that the question makes little sense, let's take two of your equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 . $\partial v_{x}/\partial x = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a particle moving along the x axis. Now imagine a force accelerating along the x axis. The equation is not true.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\partial v_{y}/\partial x = 0$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Imagine a particle moving in a plane, let's with a 45 degrees angle with respect to the x and y axis (a diagonal). Not imagine a force (let's say gravity) accelerating it in the y-direction. The particle continues to be free in the x direction. Again the equation is wrong.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another example: consider the movement of a particle attached to an infinite wire (so a 1 dimensional problem). We define what we call the &quot;velocity-phase-space&quot; which is 2 dimensional. Imagine it as a plane where one axis is the position and the other is the speed of the particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the trajectory of the particle will be a curve in that place parametrized by a free parameter t: at each instant the particle has a given location and a given speed. So even if &quot;speed&quot; is a concept defined in every point of the plane, the speed of the particle takes only a set of values which is a 1-dimensional path defined in the velocity-phase-space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you take derivatives, you therefore have to follow the path $q(t), \dot{q}(t)$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A last comment: Two things can be dependent even if in some cases it appear they are not. In theory you can write $y=f(x)$ so that $y$ depends on $x$, then you consider the example $f(x) = a$ and you say &quot;look $y$ does not depend on $x$&quot;, OK ... but only in that case.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-11-10T10:26:32.203" CommentCount="3" />
  <row Id="489" PostTypeId="2" ParentId="486" CreationDate="2010-11-10T10:39:21.013" Score="2" ViewCount="" Body="&lt;p&gt;Maxwell equation need to be accompanied by Lorentz equation&#xA;$$\vec{a} = \frac{q}{m} \vec{E} + \frac{q}{m} \vec{v} \times \vec{B}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As there are different masses for charged particles, one cannot form one general equation for behaviour of charge (or current).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In complex scenarios (like in a conductor), there are effective relations like&#xA;$$\vec{j} = \sigma \vec{E}$$&#xA;(see post by Tobias Kienzler). In case of magnetic field in a conductor, current 'generates' electrical field not only parallel, but also perpendicular, to its vector. See - Hall effect. &lt;/p&gt;&#xA;" OwnerUserId="184" LastActivityDate="2010-11-10T10:39:21.013" />
  <row Id="490" PostTypeId="1" AcceptedAnswerId="499" CreationDate="2010-11-10T10:45:57.583" Score="0" ViewCount="417" Body="&lt;p&gt;Ive been given the 4 equations of motions   The fourth being:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s=ut+\frac{1}{2}at^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If rearranged it forms the quadratic equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$at^2+2ut-2s=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that means that t has 2 values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will one of them always be negative. So only one value is realistically possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And how could you rearrange it to get t on it's own? Would that be using the quadratic formula?&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="4" LastEditorDisplayName="" LastEditDate="2010-11-21T14:13:59.563" LastActivityDate="2011-03-18T17:02:25.020" Title="4th equation of motion" Tags="&lt;mechanics&gt;&lt;homework&gt;" AnswerCount="3" CommentCount="10" FavoriteCount="1" />
  <row Id="491" PostTypeId="2" ParentId="490" CreationDate="2010-11-10T10:56:58.137" Score="1" ViewCount="" Body="&lt;p&gt;You are correct about how to solve for &lt;em&gt;t&lt;/em&gt;, it's a quadratic equation of the form $at^2 + bt + c = 0$, so the solution is&#xA;$$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your first question, solving for &lt;em&gt;t&lt;/em&gt; means you are looking for the exact time at which the position of the object is &lt;em&gt;s&lt;/em&gt;. The velocity at time $t=0$ is &lt;em&gt;u&lt;/em&gt;, but the acceleration &lt;em&gt;a&lt;/em&gt; is constant, so that means that &lt;em&gt;before&lt;/em&gt; time $t=0$, the velocity was &lt;em&gt;less&lt;/em&gt; than &lt;em&gt;u&lt;/em&gt; and was accelerating up to &lt;em&gt;u&lt;/em&gt;. If you look far enough back in the past, the velocity must have been negative, and so the object would have passed through the point &lt;em&gt;s&lt;/em&gt; sometime. Therefore, both solutions are valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when we pose a problem like this, it is usually implicit that we are only interested in what happens starting from $t=0$, unless we specifically mention otherwise. So you can usually ignore the negative solution.&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2010-11-10T10:56:58.137" />
  <row Id="493" PostTypeId="1" AcceptedAnswerId="498" CreationDate="2010-11-10T11:12:31.533" Score="0" ViewCount="151" Body="&lt;p&gt;Consider a monochromatic plane wave (I am using bold to represent vectors)&#xA;$$ \mathbf{E}(\mathbf{r},t) = \mathbf{E}_0(\mathbf{r})e^{i(\mathbf{k} \cdot \mathbf{r} - \omega t)}, $$&#xA;$$ \mathbf{B}(\mathbf{r},t) = \mathbf{B}_0(\mathbf{r})e^{i(\mathbf{k} \cdot \mathbf{r} - \omega t)}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few ways to simplify this notation. We can use the &lt;em&gt;complex field&lt;/em&gt;&#xA;$$ \tilde{\mathbf{E}}(\mathbf{r},t) = \tilde{\mathbf{E}}_0 e^{i(\mathbf{k} \cdot \mathbf{r} - \omega t)} $$&#xA;to represent both the electric and magnetic field, where the real part is the electric and the imaginary part is proportional to the magnetic. Often it is useful to just deal with the &lt;em&gt;complex amplitude&lt;/em&gt; ($\tilde{\mathbf{E}}_0$) when adding or manipulating fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when you want to coherently add two waves with the same frequency but different propagation directions, you need to take the spatial variation into account, although you can still leave off the time variation. So you are dealing with this quantity:&#xA;$$ \tilde{\mathbf{E}}_0 e^{i\mathbf{k} \cdot \mathbf{r}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what is this quantity called? I've been thinking &lt;em&gt;time-averaged complex field&lt;/em&gt;, but then again, it's not really time-averaged, is it? &lt;em&gt;Time-independent&lt;/em&gt;? Also, what is its notation? $\langle\tilde{\mathbf{E}}\rangle$?&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2010-11-10T13:11:51.130" Title="Notation of plane waves" Tags="&lt;notation&gt;&lt;electrodynamics&gt;&lt;plane-wave&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="494" PostTypeId="1" CreationDate="2010-11-10T11:53:11.903" Score="3" ViewCount="210" Body="&lt;p&gt;Where can I find some examples of cross sections calculations in QFT done step-by-step? Those integrals are a little horror. For example - a simple scalar+scalar -&gt; scalar+scalar at the tree level in a theory scalar $\phi^4$ ?&lt;/p&gt;&#xA;" OwnerUserId="241" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-20T06:25:38.280" LastActivityDate="2011-07-20T06:25:38.280" Title="Quantum Field Theory cross sections integrals" Tags="&lt;quantum-field-theory&gt;&lt;particle-physics&gt;&lt;books&gt;&lt;reference-request&gt;&lt;cross-section&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="495" PostTypeId="2" ParentId="484" CreationDate="2010-11-10T12:05:06.380" Score="1" ViewCount="" Body="&lt;p&gt;I'm not a cosmologist, but I find you scaling argument a little off in that a) you scale the finite distance between trees, but want to keep the finite radii of your trees constant and b) mean free paths are usually inversely proportional to the cross sections, which scale with a^2 as well. Scaling both a and b with the same factor gives me a constant mean free path.&lt;/p&gt;&#xA;" OwnerUserId="243" LastActivityDate="2010-11-10T12:05:06.380" CommentCount="1" />
  <row Id="496" PostTypeId="2" ParentId="493" CreationDate="2010-11-10T12:13:51.800" Score="1" ViewCount="" Body="&lt;p&gt;I'd call it the initial complex field since it's {E(t=0)}. No reference though.&lt;/p&gt;&#xA;" OwnerUserId="243" LastActivityDate="2010-11-10T12:13:51.800" />
  <row Id="497" PostTypeId="2" ParentId="257" CreationDate="2010-11-10T12:20:20.967" Score="1" ViewCount="" Body="&lt;p&gt;Also, usually: History. Most of the machines in the LHC accelerator chain (well, besides the LINACs, who have always been mainly used as initial stages for the PS) once were cutting edge research tools. When building the next generation design, the old accelerator is already there and usually well understood, and so it is natural to include it in the design.&lt;/p&gt;&#xA;" OwnerUserId="243" LastEditorUserId="243" LastEditorDisplayName="" LastEditDate="2010-11-11T15:59:10.073" LastActivityDate="2010-11-11T15:59:10.073" />
  <row Id="498" PostTypeId="2" ParentId="493" CreationDate="2010-11-10T12:32:57.583" Score="2" ViewCount="" Body="&lt;p&gt;Stationary field or monochromatic field. Yes, basically that is the field including the $e^{i\omega t}$ term, but even when it is omitted one still knows what is meant.&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2010-11-10T12:32:57.583" CommentCount="6" />
  <row Id="499" PostTypeId="2" ParentId="490" CreationDate="2010-11-10T12:39:45.707" Score="6" ViewCount="" Body="&lt;p&gt;The equation describes parabolic motion, if $a\neq 0$ is a non-zero constant acceleration, which I will assume from now on. If you think about it, your solution provides an answer to the question: at what time does the object is in the position $s$? [A note on notation: Traditionally, the letter $s$ denotes distance (I guess from the German word &quot;Strecke&quot;), which by definition is a non-negative quantity, but your formula makes more sense, if we interpret $s$ as a position $x$, which can also be negative.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{2}at^2+ut-x=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t_1 = \frac{-u-\sqrt{D}}{a}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t_2 = \frac{-u+\sqrt{D}}{a}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $D := u^2+2ax$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's think about it for a moment, and see what answers we get by varying $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case $D&amp;lt;0$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discriminant is negative, there are no solutions, therefore at no time your object will have that position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case $D=0 \Leftrightarrow x=-\frac{u^2}{2a}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discriminant is zero, there is only one solution which is the &quot;top&quot; (&quot;bottom&quot;) point reached by the object,if $a&amp;lt;0$ ($a&gt;0$), respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case $D&amp;gt;0$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discriminant is positive and there are two solutions. This means that the object will reach that position twice, once going &quot;up&quot; and once going &quot;down&quot; the parabola.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now will one of them always be negative? Not necessarily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case $ax &amp;gt; 0$: One positive and one negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case $ax &amp;lt; 0$ and $\frac{u}{a}&amp;gt;0$: Two negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case $ax &amp;lt; 0$ and $\frac{u}{a}&amp;lt;0$: Two positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case $x = 0$ and $\frac{u}{a}&amp;gt;0$: One zero and one negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case $x = 0$ and $\frac{u}{a}&amp;lt;0$: One zero and one positive.&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-03-18T08:03:16.573" LastActivityDate="2011-03-18T08:03:16.573" />
  <row Id="500" PostTypeId="2" ParentId="494" CreationDate="2010-11-10T12:49:54.357" Score="3" ViewCount="" Body="&lt;p&gt;Take a look at M.Srednicki, &quot;Quantum field theory&quot;. He is doing it in a very detailed manner.&lt;/p&gt;&#xA;" OwnerUserId="59" LastActivityDate="2010-11-10T12:49:54.357" CommentCount="2" />
  <row Id="501" PostTypeId="1" CreationDate="2010-11-10T12:57:31.590" Score="6" ViewCount="491" Body="&lt;p&gt;Consider the problem of three bodies two of which having mass &lt;em&gt;M&lt;/em&gt;, one of them having mass &lt;em&gt;m&lt;/em&gt;. Body &lt;em&gt;m&lt;/em&gt; is in the middle between the other two, coupled to them by two equal linear springs in rest. Now fix the two bodies &lt;em&gt;M&lt;/em&gt; and move body &lt;em&gt;m&lt;/em&gt; for a small amount perpendicular to the connection line. Now let loose the three of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is trivial when we assume &lt;em&gt;m&lt;/em&gt; to be finite and &lt;em&gt;M&lt;/em&gt; to be infinite (or vice versa), or when the spring constant is 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the problem is so symmetric, that there might be hope to get a closed or approximate solution even for finite &lt;em&gt;m&lt;/em&gt; &gt;&gt; &lt;em&gt;M&lt;/em&gt;, &lt;em&gt;M&lt;/em&gt; &gt;&gt; &lt;em&gt;m&lt;/em&gt;, or even &lt;em&gt;M&lt;/em&gt; = &lt;em&gt;m&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="229" LastActivityDate="2010-12-21T19:45:21.210" Title="Simple three-body-problem?" Tags="&lt;classical-mechanics&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="1" />
  <row Id="503" PostTypeId="1" CreationDate="2010-11-10T13:26:09.143" Score="6" ViewCount="525" Body="&lt;p&gt;What classical open problems are there in special relativity, including questions about non-inertial motion in flat-spacetime, but excluding questions about quantum theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answers can include things which were open problems until recently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;New things in special relativity (and non-gravitational relativity) still seem to be discovered relatively recently:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance the book &quot;A broader view of relativity&quot; by Jong-Ping Hsu, Leonardo Hsu, contains formulas for transformations between some types of accelerated frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Abraham Ungar derived &lt;a href=&quot;http://en.wikipedia.org/wiki/Gyrovector_space&quot; rel=&quot;nofollow&quot;&gt;formulas&lt;/a&gt; for things like proper velocity composition, and for Lorentz transformation composition in terms of velocity and rotation parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it would be good to have a list of things that we still haven't worked out, or have only been worked out recently, to challenge the assumption that special relativity is all figured out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Frame-dragging&quot; rel=&quot;nofollow&quot;&gt;linear frame-dragging&lt;/a&gt; it also seems that the usual saying that GR tends to SR in the limit of negligble gravity should be stated as GR tends to SR for negligble gravity and low acceleration, so any calculations concerning the half-way house of non-inertial relativity between SR and GR would only be valid for low acceleration. Perhaps this should be the definition of non-inertially extended special relativity: acceleration without linear frame-dragging.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I +1ed the answer on self-force because it was an interesting general issue, but once we are talking about fields and particles then we are getting away from the question which is really about calculations in the theory of special relativity itself. Once we talk about fields and particles then I guess there are lots of problems and an accurate treatment requires quantum theories which this question is not about.&lt;/p&gt;&#xA;" OwnerUserId="244" LastEditorUserId="244" LastEditorDisplayName="" LastEditDate="2010-12-08T22:52:44.703" LastActivityDate="2011-08-19T17:00:42.267" Title="Open problems in special relativity, and non-inertial motion in flat-spacetime, or things that have been discovered recently" Tags="&lt;special-relativity&gt;" AnswerCount="7" CommentCount="1" FavoriteCount="3" />
  <row Id="504" PostTypeId="2" ParentId="503" CreationDate="2010-11-10T13:34:06.263" Score="-1" ViewCount="" Body="&lt;p&gt;Are there any? AFAIK classical special relativity is very closed theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And - why are you asking? &lt;/p&gt;&#xA;" OwnerUserId="184" LastActivityDate="2010-11-10T13:34:06.263" CommentCount="2" />
  <row Id="505" PostTypeId="2" ParentId="501" CreationDate="2010-11-10T15:08:29.007" Score="1" ViewCount="" Body="&lt;p&gt;Let $\vec{r_1},\vec{r_3}$ be position vectors of the bodies with mass $M$.&lt;br&gt;&#xA;Let $\vec{r_2}$ be position vector of the body body in the middle with mass $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L$ is the length of spring at rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M\ddot{\vec{r_1}} = k(|\vec{r_2} - \vec{r_1}| - L) \frac{\vec{r_2} - \vec{r_1}}{|\vec{r_2} - \vec{p_1}|} = \vec{F_{12}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M\ddot{\vec{r_3}} = k(|\vec{r_2} - \vec{r_3}| - L) \frac{\vec{r_2} - \vec{r_3}}{|\vec{r_2} - \vec{r_3}|}  = \vec{F_{32}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m\ddot{\vec{r_2}} = -(\vec{F_{12}} + \vec{F_{32}})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to solve this.&#xA;Take a look at &lt;a href=&quot;https://cs.nyu.edu/~dzorin/numcomp08/springs_notes.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://www.ph.biu.ac.il/~rapaport/java-apps/spring.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-11-11T11:13:26.713" LastActivityDate="2010-11-11T11:13:26.713" CommentCount="3" />
  <row Id="506" PostTypeId="1" CreationDate="2010-11-10T15:51:37.623" Score="9" ViewCount="728" Body="&lt;p&gt;Why does the bike fall when its speed is very low or close to zero and is balanced when going with a high speed?&lt;/p&gt;&#xA;" OwnerUserId="248" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2010-11-10T18:33:17.450" LastActivityDate="2011-04-15T20:30:17.287" Title="Why doesn't the bike fall if going with a high speed?" Tags="&lt;classical-mechanics&gt;&lt;bicycles&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="2" />
  <row Id="507" PostTypeId="1" AcceptedAnswerId="520" CreationDate="2010-11-10T16:16:21.193" Score="3" ViewCount="220" Body="&lt;p&gt;I have an old, non power steered car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the car is stationary, the steering wheel is really tough to move, but as soon as I gain very little (0.5 kph or so) speed, the resistance is dramaticly decreased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But at higher speeds, if I wish to make a full steer, the steering wheel will be resisting again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is going on with the wheel friction?&lt;/p&gt;&#xA;" OwnerUserId="250" LastActivityDate="2010-11-11T12:52:28.917" Title="Tire speed dependent friction" Tags="&lt;applied-physics&gt;" AnswerCount="3" />
  <row Id="508" PostTypeId="1" CreationDate="2010-11-10T16:23:39.967" Score="4" ViewCount="248" Body="&lt;p&gt;What are the details of how neutrons are produced as a result of &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_ray&quot; rel=&quot;nofollow&quot;&gt;cosmic ray particles&lt;/a&gt; hitting our planet's atmosphere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, what is the pathway that creates the highest number of neutrons from cosmic ray particles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article &quot;&lt;a href=&quot;http://www.statemaster.com/encyclopedia/Single_event-upset&quot; rel=&quot;nofollow&quot;&gt;Single event upset&lt;/a&gt;&quot; states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An SEU happens due to cosmic particles which collide with atoms in the atmosphere, creating cascades or showers of neutrons and protons. At deep sub-micrometre geometries, this affects semiconductor devices at sea level.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it does not describe the mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia article on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutron&quot; rel=&quot;nofollow&quot;&gt;neutron&lt;/a&gt; states something similar:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cosmic radiation interacting with the Earth's atmosphere continuously generates neutrons that can be detected at the surface.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What are the processes, in detail?&lt;/p&gt;&#xA;" OwnerUserId="174" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-10T16:46:20.320" LastActivityDate="2011-08-17T06:44:27.170" Title="How are neutrons produced from cosmic ray particles?" Tags="&lt;particle-physics&gt;&lt;neutron&gt;&lt;cosmic-rays&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="509" PostTypeId="2" ParentId="506" CreationDate="2010-11-10T16:42:41.977" Score="0" ViewCount="" Body="&lt;p&gt;The basic concept (at least, as I've heard it) is angular momentum. As a bike wheel turns, it has an amount of angular momentum proportional to its rotational speed, associated with the plane of rotation of the wheel. This makes it act basically like a gyroscope: it &quot;resists&quot; any change in the amount or direction of that angular momentum, in the same sense that mass &quot;resists&quot; any change in the amount or direction of its velocity. This basically slows down the tipping of the bike to the point where you are able to prevent it by pushing down on the opposite side pedal.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-10T16:42:41.977" CommentCount="5" />
  <row Id="510" PostTypeId="2" ParentId="506" CreationDate="2010-11-10T16:43:04.077" Score="-2" ViewCount="" Body="&lt;p&gt;The answer is &quot;Centrifugal force&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biger your speed is the biger this force is too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can notice that when you steer left you make your bike's slope on left side. &#xA;And Contrifugal force don't let you fall (when your steer angle is constant at the end bike will make a circle). Then when you make your bike steer more left your bike returns to the balance because you increase that force (it comes from equation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When your speed is smaller contrifugal force is smaller and bike is harder to steer so you can fall easier.&lt;/p&gt;&#xA;" OwnerUserId="58" LastActivityDate="2010-11-10T16:43:04.077" CommentCount="4" />
  <row Id="512" PostTypeId="2" ParentId="49" CreationDate="2010-11-10T16:59:58.837" Score="1" ViewCount="" Body="&lt;p&gt;This reference from Wikipedia seems like a good place to start.&#xA;See &lt;a href=&quot;http://www.lionprecision.com/tech-library/technotes/cap-0020-sensor-theory.html&quot; rel=&quot;nofollow&quot;&gt;http://www.lionprecision.com/tech-library/technotes/cap-0020-sensor-theory.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-11-10T16:59:58.837" />
  <row Id="513" PostTypeId="2" ParentId="506" CreationDate="2010-11-10T17:32:36.613" Score="0" ViewCount="" Body="&lt;p&gt;oneat's answer is correct (I would have commented but I think I'm going to need more space)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a vector (line) starting at your center of gravity. The line represents all the forces acting on you. When you are standing still, the direction of the line is straight down (gravity is the single force present).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To not fall over whens standing still on a bike, you have to keep the point where the line intersects the ground (let's call it point &lt;strong&gt;A&lt;/strong&gt;), between the two wheels of the bike. If you don't, you'll start tipping over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When standing still, the only way to affect that point is to move your center of gravity which you do by shifting your weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's say you're moving. If you're moving in a straight line, at a constant speed, everything is the same, the only force acting on you is gravity. But if you change direction, you get centrifugal force (as oneat correctly pointed out), the same as what you get when you make a sharp turn in a car moving at speed. The value of that force is proportional with your speed, your weight and the speed of the turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This centrifugal force is added to the gravity, and changes the direction of the resulting force acting on you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember point &lt;strong&gt;A&lt;/strong&gt;? If you're riding your bike and it starts to lean to the right, point A starts to move to the right and the bikes leans even more, and so on. But, you instinctively know to turn your bike to the right. This causes a centrifugal force, pointing left, to appear). If point &lt;strong&gt;A&lt;/strong&gt; is still between your wheels then you're fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're moving slowly, the centrifugal force is small, so you have to take the turn more sharply to compensate. If you're moving fast, you might only need to nudge your bike a little to compensate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's explained in more detail &lt;a href=&quot;http://en.wikipedia.org/wiki/Bicycle_and_motorcycle_dynamics&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. (I actually thought of looking it up in wikipedia only after writing this answer, I don't have time to read the article now, hopefully I'm not too wrong)&lt;/p&gt;&#xA;" OwnerUserId="260" LastActivityDate="2010-11-10T17:32:36.613" CommentCount="3" />
  <row Id="514" PostTypeId="2" ParentId="506" CreationDate="2010-11-10T17:47:08.087" Score="10" ViewCount="" Body="&lt;p&gt;The surprising answer is that the stability of the modern bicycle has little or nothing to do with centrifugal force or gyroscopes or any of that.  Look up &quot;bicycle stability&quot; on Google.   Experiments show that the sloped angle of the front fork is very important, e.g. If the fork pointed backwards it is very difficult to stay upright at any speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At higher speeds a very slight turn of the handles moves the bicycle under the center of gravity of the rider quicker, so that the dynamical stability is improved.     As usual experiment corrects theory here.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the answer of Tristan at &lt;a href=&quot;http://physics.stackexchange.com/questions/24/does-leaning-banking-help-cause-turning-on-a-bike/524#524&quot;&gt;Does leaning (banking) help cause turning on a bike?&lt;/a&gt; as well for an even better answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the comment of nibot below for a reference to an actual definitive article.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-11-15T21:56:00.297" LastActivityDate="2010-11-15T21:56:00.297" />
  <row Id="515" PostTypeId="2" ParentId="501" CreationDate="2010-11-10T17:55:40.260" Score="2" ViewCount="" Body="&lt;p&gt;If you restrict the motion of the bodies to 1D, then the problem is pretty easy -- you can just solve it by changing to &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_mode&quot; rel=&quot;nofollow&quot;&gt;normal mode coordinates&lt;/a&gt;. These are collective coordinates, each normal mode coordinate being, in general, a linear combination of the coordinates of all the particles in the system. This transforms the three coupled oscillators (which means energy having terms of the form $x_i x_j$) to three uncoupled oscillators (energy only depending on $q_i^2$), each having a different frequency. The trick is actually similar to completing the square in the total energy of the system, to bring it to a $\sum_i \frac{1}{2} m_i \omega_i^2 q_i^2$ form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can solve exactly for these simple harmonic oscillators and then the general solution are going to be a linear combination of these normal modes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if the particles are allowed to move in all three dimensions, the system should be chaotic. Your initial conditions do restrict the motion in 1D, so this shouldn't be a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that solutions (albeit approximate) for these kind of problems exist for even $10^{23}$ particles, for example if you consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Debye_model&quot; rel=&quot;nofollow&quot;&gt;Debye model&lt;/a&gt;, which models the vibration of solids as a lattice of coupled (quantum) harmonic oscillators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&#xA;Expanding on &lt;a href=&quot;http://physics.stackexchange.com/users/229/hans-stricker&quot;&gt;Hans'&lt;/a&gt; comment, even you choose to represent your problem in a 2D coordinate system, the initial conditions and motion will restrict it to 1D. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that, let's consider taking our x axis to be the line joining the three particles and fix the origin of the coordinate system on one of the mass $M$ particles. The initial conditions can be written as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$x_1(0) = 0$, $m_1 = M$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$x_2(0) = L + d$, $m_2 = m$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$x_3(0) = 2 L$, $m_3 = M$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$y_1(0) = y_2(0) = y_3(0) = 0$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where $d$ is the displacement of particle 2 along the line joining 1 and 3 (the $x$ axis) and $L$ is the natural length of the spring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As initially all particles are at rest, and all forces are along the $x$ axis, the $y$ coordinate of either particle cannot change and the situation remains identical for all subsequent times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, this is an idealised model and in real life you never get the particles to perfectly move in 1 dimension. This is because the general 2D motion of 3 particles connected by springs is chaotic -- a small change in the initial conditions (let's say $y_2 = \epsilon$, which is small) will lead to a large change in the displacements at time $t$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, there is a &lt;a href=&quot;http://en.wikipedia.org/wiki/N-body_problem#Sundman.27s_theorem_for_the_3-body_problem&quot; rel=&quot;nofollow&quot;&gt;series solution&lt;/a&gt; for the three body problem, though I am not sure about the exact assumptions that govern it and if it is applicable to the case of three particles linked by springs free to move in 2D. In the case of all initial velocities zero, this solution wouldn't apply as it &quot;is convergent for all real t, except initial data which correspond to &lt;em&gt;zero angular momentum&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;" OwnerUserId="15" LastEditorUserId="15" LastEditorDisplayName="" LastEditDate="2010-11-11T14:13:30.843" LastActivityDate="2010-11-11T14:13:30.843" CommentCount="2" />
  <row Id="516" PostTypeId="2" ParentId="482" CreationDate="2010-11-10T17:56:05.503" Score="5" ViewCount="" Body="&lt;p&gt;The relationship between the chromatic polynomial and the Potts model is a special case of the relationship between the Tutte polynomial and the random cluster model of Fortuin and Kastelyn.  There's a very tiny bit about this in &lt;a href=&quot;http://en.wikipedia.org/wiki/Tutte_polynomial#Definitions&quot; rel=&quot;nofollow&quot;&gt;the wikipedia page on the Tutte polynomial&lt;/a&gt;, but there's &lt;a href=&quot;http://books.google.com/books?id=SbZKSZ-1qrwC&amp;amp;pg=PA342&amp;amp;lpg=PA342&amp;amp;dq=statistical+mechanics+tutte&amp;amp;source=bl&amp;amp;ots=NzBEf6i4u_&amp;amp;sig=p3ou8AaZYDM6PeVCu1defABrECA&amp;amp;hl=en&amp;amp;ei=yNnaTMTxEobGlQf4mMG9CQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CCoQ6AEwAg#v=onepage&amp;amp;q=statistical%20mechanics%20tutte&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;a section about this&lt;/a&gt; in Bollobàs's &quot;Modern Graph Theory&quot; and plenty more articles that pop up under a google search for &quot;Tutte statistical mechanics&quot; including &quot;&lt;a href=&quot;http://arxiv.org/abs/0804.2468&quot; rel=&quot;nofollow&quot;&gt;A little statistical mechanics for the graph theorist&lt;/a&gt;&quot; which I've always wanted to finish reading...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, it seems there's a little bit of confusion in your question about what the Potts model &quot;is&quot;.  It doesn't quite make sense to ask &quot;for which graph is the Potts model the chromatic polynomial&quot;.  The Potts model is a probabilistic model (roughly, a measure on random vertex &quot;colorings&quot;, i.e. assign a color (often called a spin value) to each vertex) and it may be defined on any graph, where the parameters like temperature or applied field change the measure on the set of vertex colorings of the graph, e.g. at low temperatures, colorings with the same color on neighboring vertices are preferred.  As typically studied in physics, the Potts model (and other such models) are put on periodic graphs (often called &quot;lattices&quot; in physics), such as the square lattice graph, triangular lattice graph, etc.  The chromatic and Tutte polynomials show up when you compute the &quot;partition function&quot; associated to the model, which is a quantity like a generating function... this paragraph is a bit rough, so I'd refer to the other references.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if you have any questions.  I could talk about this stuff all day.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-10T17:56:05.503" />
  <row Id="517" PostTypeId="2" ParentId="454" CreationDate="2010-11-10T18:06:01.140" Score="1" ViewCount="" Body="&lt;p&gt;I would ask your instructor for clarification, but I'd bet that they just want you to model some hypothetical &quot;static negative index material&quot;, even though none are known to exist.  I don't know how to apply relaxation to the full time-dependent Maxwell's equations...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For examples of applications of negative index materials, a recently popular idea is &quot;cloaking&quot;, which you can read about &lt;a href=&quot;http://physics.aps.org/articles/v2/17&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  In general the website &quot;Physics&quot; has had a few articles on recent research on negative index materials, for instance &lt;a href=&quot;http://physics.aps.org/articles/v2/3&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://physics.aps.org/articles/v2/91&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-10T18:06:01.140" CommentCount="1" />
  <row Id="518" PostTypeId="2" ParentId="501" CreationDate="2010-11-10T18:51:57.793" Score="4" ViewCount="" Body="&lt;p&gt;This can be solved without much calculation.  First notice that by conservation of momentum, the momentum will always be zero, and the center of mass won't move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, ask what the forces are at the moment when we let all the masses go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The little mass, (let's say it was pulled to the right by a distance $\Delta x$) experiences a force to the left, coming half from each spring.  The force is $2k\Delta x$.  By Newton's third law, the big masses feel equal and opposite forces to the right of size $k \Delta x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The forces on the two big masses are the same because the springs are compressed/extended the same amount. As the little mass moves, the compression/expansion of the two springs remains the same.  Thus, the two big masses always move together.  The little mass is forced to move opposite the two big masses to keep the total momentum of the system zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motion of the system is that the two big masses oscillate one way while the little mass oscillates the opposite direction, all with the same frequency.  It looks a little bit like a ping-pong ball bouncing back and forth between two large paddles, with the paddles moving in unison, but always coming forward to smack the ping pong ball when it gets close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amplitude of the big mass oscillation is $m/2M$ times the amplitude of the small mass (from the condition that the center of mass is stationary).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The frequency is $\sqrt{k_{eff}/M}$, with $k_{eff}$ an &quot;effective spring constant&quot; felt by the big mass.  When the big mass moves 1cm, it does not contract the spring 1cm, because the little mass is moving the opposite direction and contracting the spring even more.  Instead, the spring contracts $1cm*(1+2M/m)$, so the effective spring constant is $k_{eff} = k(1+2M/m)$ and the frequency of oscillation is $\sqrt{k(1+2M/m)/M}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an example of a &quot;normal mode&quot; as mentioned by Flaviu.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-10T18:51:57.793" CommentCount="4" />
  <row Id="519" PostTypeId="1" CreationDate="2010-11-10T19:10:10.857" Score="3" ViewCount="179" Body="&lt;p&gt;In what part of the spectrum is it radiating? In the infrared, in the microwave? Or is not radiating anymore at all? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In russian:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Чему сейчас равна температура поверхности и ядра нейтронной звезды, которая образовалась 12 миллиардов лет назад?&#xA;В каком диапазоне она сейчас излучает? В инфракрасном, микроволновом? Или не излучает вообще?&lt;/p&gt;&#xA;" OwnerUserId="238" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2010-11-11T19:03:38.827" LastActivityDate="2010-11-12T21:37:27.607" Title="What is the temperature of the surface and core of a neutron star formed 12 billion years ago now equal to?" Tags="&lt;astrophysics&gt;&lt;neutron-stars&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" />
  <row Id="520" PostTypeId="2" ParentId="507" CreationDate="2010-11-10T19:26:35.423" Score="3" ViewCount="" Body="&lt;p&gt;If the car is stationary, then you're rotating the tires about a vertical axis, so you're rotating the contact patch of the tire relative to the road, and there is a lot of friction coming from those two surfaces sliding across each other. (In the center of the contact patch, there's no friction because there's no relative motion, but on the edges of the contact patch, there's a lot of friction because the rotation of the tire moves those surfaces directly across each other.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the car is driving in a straight line, then there's no friction because there's no relative motion between the contact patch and the road (no slippage). The tire is rotating about a perfectly horizontal axis, so the part of the tire just in front of the contact patch is about to press down onto the road and the part just behind is lifting off, but no surfaces are sliding across each other. (There is still energy loss, but it comes from deformation of the tire and the road, not from friction between surfaces. See [Rolling resistance].&lt;a href=&quot;http://en.wikipedia.org/wiki/Rolling_resistance&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the complicated part. If the car is moving forward at a moderate speed and the wheels are rolling as usual, but you're also turning the steering wheel so the car is following a curved path, then in order to think about the friction we have to think about the axis of the NET rotation of the tire. The forward motion of the car gives it a lot of rotation about a horizontal axis. If we use the right-hand rule then this angular velocity vector points to the left. But the steering motion is also rotating the tire, much more slowly, about a vertical axis. For example, if you're turning the car to the left, then this gives a small angular velocity vector pointing up. To find the NET angular velocity, we should add together those two angular velocity vectors, which gives a vector pointing mostly to the left, but slightly up. This is the axis of the NET rotation of the tire at an instant in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this axis is much closer to horizontal than it is to vertical, the tire is mostly rolling forward and the difference in speed between the left and right sides of the contact patch (which causes the relative motion between surfaces, hence friction) is much less than it is when the car is stationary.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2010-11-10T19:26:35.423" CommentCount="1" />
  <row Id="521" PostTypeId="2" ParentId="437" CreationDate="2010-11-10T20:00:36.563" Score="8" ViewCount="" Body="&lt;p&gt;There is no quantum mechanics of a photon, only a quantum field theory of electromagnetic radiation. The reason is that photons are never non-relativistic and they can be freely emitted and absorbed, hence no photon number conservation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, there exists a direction of research where people try to reinterpret certain quantities of electromagnetic field in terms of the photon wave function, see for example &lt;a href=&quot;http://www.cft.edu.pl/~birula/publ/APPPwf.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="59" LastActivityDate="2010-11-10T20:00:36.563" CommentCount="2" />
  <row Id="522" PostTypeId="2" ParentId="24" CreationDate="2010-11-10T20:18:51.137" Score="1" ViewCount="" Body="&lt;p&gt;I disagree with the answer, which currently has the most votes, stating that the normal force changes as the bike leans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider this experiment:  A hollow block of metal shaped like a text book with mass similar to that of the average physics student + bike.  The spine (binding) of the book is rounded, similar to the rounded surface of the bike tire.  Place the block with rounded side down on a hard smooth surface in a vacuum with low friction (similar to the friction a rolling bike tire experiences).  Tilt the book the slightest bit to the right and give it a shove, perfectly applying force in only one direction (forward), accelerating the block to 50 kph, and allowing the block to slide on its own after the initial shove.  What's to stop the block from falling to the right?  Not much...just the friction of rolling the spine of the block on the surface.  It's going to fall and it is not going to turn.&lt;/p&gt;&#xA;" OwnerUserId="268" LastEditorUserId="268" LastEditorDisplayName="" LastEditDate="2010-11-11T12:22:08.503" LastActivityDate="2010-11-11T12:22:08.503" CommentCount="1" />
  <row Id="523" PostTypeId="1" CreationDate="2010-11-10T20:21:55.270" Score="7" ViewCount="322" Body="&lt;p&gt;Let's have two orthogonal n-particle quantum states: $|\psi \rangle$ and $|\phi \rangle$. In theory it is always possible to make an unambiguous measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, things get complicated when one restricts oneself to a certain class of measurements. With so called LOCC (Local Operations and Classical Communication, that is, we have to measure particles separately, but we are allowed to communicate and to have dependence for measurements on the outcomes of previous measurement) still it is possible to unambiguously distinguish any two states (see: Walgate et al., &lt;em&gt;Local Distinguishability of Multipartite Orthogonal Quantum States&lt;/em&gt;, Phys. Rev. Lett. 85, 23: 4972-4975 (2000) &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0007098&quot; rel=&quot;nofollow&quot;&gt;arXiv:quant-ph/0007098&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With fixed local operations (and thus classical communications only after all measurements are done) sometimes we can't unambiguously distinguish between $|\psi \rangle$ and $|\phi \rangle$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is there any simple argument why?&lt;/li&gt;&#xA;&lt;li&gt;Are there any simple criteria which says which orthogonal states can be unambiguously distinguished with local measurements and communication only after them?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="184" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-11-19T15:11:35.417" LastActivityDate="2010-11-29T13:57:51.490" Title="Unambiguous distinguishing of quantum states by local measurement" Tags="&lt;quantum-mechanics&gt;&lt;quantum-optics&gt;&lt;measurement&gt;&lt;quantum-information&gt;" AnswerCount="2" FavoriteCount="5" />
  <row Id="524" PostTypeId="2" ParentId="24" CreationDate="2010-11-10T21:13:47.193" Score="13" ViewCount="" Body="&lt;p&gt;The simple answer is that the angle between the front fork and the vertical causes the force from the ground to create a moment about the axis of rotation that turns the wheel in that direction.  This has nothing to do with actually riding the bike, and it will happen even if the bike is stationary.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, if you project the axis of (steering) rotation all the way through the wheel, top to bottom, it will not be coincident with the point of contact with the ground.  When the bike leans over, the upward (normal) force from the ground is not in the same plane as the axis of rotation, which causes a moment about that axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the bike begins to turn, the frictional component of the contact force will cause the force to go back into the same plane as the axis of rotation, which causes the wheel to hold its position steady.&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2010-11-10T21:13:47.193" CommentCount="4" />
  <row Id="525" PostTypeId="2" ParentId="401" CreationDate="2010-11-10T21:17:43.293" Score="2" ViewCount="" Body="&lt;p&gt;For primitive drawings, I am a big fan of XFig.  The UI is a little clunky, but it can save to dozens of graphics formats and creates figures that are downright trivial to include in a LaTeX document.  The biggest thing for me is that the file format is text-based, so it is completely possible to script more complicated drawings.&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2010-11-10T21:17:43.293" CommunityOwnedDate="2010-11-10T21:17:43.293" />
  <row Id="527" PostTypeId="1" CreationDate="2010-11-10T22:14:34.047" Score="3" ViewCount="492" Body="&lt;p&gt;My anecdotal observations (which could be incorrect, they're totally unscientific) indicate that it takes almost as long to evenly heat a big bowl of soup in a microwave as it does to heat it on the stove - 4-5 minutes in my most recent example. Yet a plate of meat or vegetables is reheated in 30 seconds or less. &lt;/p&gt;&#xA;" OwnerUserId="277" LastActivityDate="2011-01-25T02:59:56.117" Title="Why does evenly heating soup with a microwave take so long?" Tags="&lt;electromagnetism&gt;&lt;heat&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="528" PostTypeId="2" ParentId="508" CreationDate="2010-11-10T22:19:38.227" Score="4" ViewCount="" Body="&lt;p&gt;Between what I've learned about cosmic rays and what I can find online (example: &lt;a href=&quot;http://www.fisica.unlp.edu.ar/~veiga/experiments.html&quot; rel=&quot;nofollow&quot;&gt;http://www.fisica.unlp.edu.ar/~veiga/experiments.html&lt;/a&gt;), it seems that the primary source of neutrons in cosmic ray showers is the disintegration of the atomic nuclei that are struck by the cosmic ray or its decay products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you may know, cosmic rays enter the atmosphere with enormous amounts of energy, so when one strikes a nucleus, it's going to disrupt the nuclear structure. The net effect could be anything from just kicking out a single particle to shattering the entire nucleus into several pieces, as in nuclear fission - it depends on the energy of the cosmic ray and the type of atom/molecule it hits. If the cosmic ray is especially high-energy, the products of the reaction will also have high energies, and they can interact with other nuclei, repeating the same process. A wide variety of particles are produced in these interactions, including neutrons, pions, kaons, and other hadrons.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-10T22:19:38.227" />
  <row Id="529" PostTypeId="2" ParentId="508" CreationDate="2010-11-10T22:59:17.893" Score="6" ViewCount="" Body="&lt;p&gt;OK, here is something concrete and quantitative, &quot;&lt;a href=&quot;http://web.mst.edu/~umrr/cf074.pdf&quot; rel=&quot;nofollow&quot;&gt;Guidelines for predicting single-event upsets in neutron environments&lt;/a&gt;&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Neutrons in the atmosphere result from cosmic-ray spallation interactions with nitrogen and oxygen nuclei. A typical reaction is a 1 GeV proton fragmenting a nitrogen necleus into lighter charged particles and simultaneoously emitting a couple of neutrons. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_ray_spallation&quot; rel=&quot;nofollow&quot;&gt;Cosmic ray spallation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From &quot;&lt;a href=&quot;http://www.spaceweather.ac.cn/publication/jgrs/2006/pdf/2006JD007150.pdf&quot; rel=&quot;nofollow&quot;&gt;Cosmic ray induced ionization in the atmosphere: Full modeling and practical applications&lt;/a&gt;&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... full numerical model,&#xA;  which computes the cosmic ray induced ionization in the&#xA;  entire atmosphere, from the ground level up to the strato-&#xA;  sphere, all over the Globe. The model computations repro-&#xA;  duce actual measurements of the atmospheric ionization in&#xA;  the full range of parameters, from equatorial to polar regions&#xA;  and from the solar minimum to solar maximum. A detailed&#xA;  numerical recipe is given in section 2.5 together with the&#xA;  precalculated tabulated ionization yield function (Tables 1&#xA;  and 2). Using this method, one can easily compute the CRII&#xA;  for any desired location and conditions, instead of using,&#xA;  e.g., a neutron monitor count rate as a proxy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It does not describe it, but the &lt;a href=&quot;http://www-ik.fzk.de/corsika&quot; rel=&quot;nofollow&quot;&gt;CORSIKA (Cosmic Ray Simulations for Kascade) simulation tool&lt;/a&gt; is used.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wiktionary.org/wiki/Google-fu&quot; rel=&quot;nofollow&quot;&gt;Google-fu&lt;/a&gt;: &lt;code&gt;quantitative neutrons cosmic rays atmosphere&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="174" LastEditorUserId="174" LastEditorDisplayName="" LastEditDate="2010-11-10T23:27:34.520" LastActivityDate="2010-11-10T23:27:34.520" CommentCount="1" />
  <row Id="530" PostTypeId="1" AcceptedAnswerId="534" CreationDate="2010-11-10T23:03:54.083" Score="13" ViewCount="719" Body="&lt;p&gt;Knowing black is supposed to be the &quot;color&quot; (I don't want to get into the color/hue/shade debate, please) that absorbs light. how does one manage to have shiny black surfaces? I know about &quot;gloss black&quot; versus &quot;matte black&quot; finishes, but shouldn't the light passing through the gloss (if they didn't pass through the gloss, you wouldn't see the black, right?) be absorbed by the underlying black object? Then there are black gemstones like jet and opal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do black objects shine?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-11-17T01:20:16.433" Title="Why can you have shiny black objects?" Tags="&lt;optics&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="531" PostTypeId="2" ParentId="482" CreationDate="2010-11-10T23:13:52.587" Score="3" ViewCount="" Body="&lt;p&gt;There are a couple of references that i think are pretty decent about this subject:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://books.google.com/books?id=T3tTZS3xpsMC&quot; rel=&quot;nofollow&quot;&gt;Potts Model and Related Problems in Statistical Mechanics&lt;/a&gt; (see chapter 1.5);&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://link.aps.org/doi/10.1103/RevModPhys.54.235&quot; rel=&quot;nofollow&quot;&gt;The Potts Model&lt;/a&gt; (see also, &lt;a href=&quot;http://link.aps.org/doi/10.1103/RevModPhys.64.1099&quot; rel=&quot;nofollow&quot;&gt;Knot Theory and Statistical Mechanics&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These two refs talk about the chromatic polynomial: roughly speaking, you can have the number of states in the Potts model be analogous to a certain number of colors, and the chromatic polynomial emerges from the Partition Function of the system (after a suitable change of variables).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A key concept in this game is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Renormalization_group#Block_spin_renormalization_group&quot; rel=&quot;nofollow&quot;&gt;block variables&lt;/a&gt;, that is, you simply walk through your lattice looking for same-state sites and binding adjacent ones together. After performing this operation in the whole lattice, you end up with &quot;blocks&quot; of &quot;same state&quot; sites, which you just define as your &quot;new variables&quot; and call them &quot;block variables&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you can only do this because, in principle, you have a finite system and a finite sum defining your Partition Function: in this sense, &quot;block variables&quot; are nothing more than a simple re-arrangement of the sum defining the Partition Function. Then, your original Partition Function,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathcal{Z} = \sum_{{s}} e^{-\beta\, \sum_{\langle i\, j\rangle} \delta_{s_i,\, s_j}} = \sum_{{s}} \prod_{\langle i\, j\rangle} e^{-\beta\, \delta_{s_i,\, s_j}} \; ; $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be re-written as,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathcal{Z} = \sum_{{s}} \prod_{\langle i\, j\rangle} (1 + v\, \delta_{s_i,\, s_j}) \; ; $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $ v = e^{-\beta} - 1$; and you can derive the chromatic polynomial from this last expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to think about this problem is as follows: think you have an initial and a final state, i.e., an initial lattice configuration and a final one, where the configuration is a given &quot;spin&quot; (&quot;state&quot;) distribution or, analogously, a given &quot;color&quot; distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think of your &quot;colors&quot; as &quot;billiard balls&quot;, you can think of the process of going from the initial to the final state as a sequence of scatterings among all sites of your lattice. This way you can build an oriented graph from the initial state descending to the final one and, in some sense, the &quot;extremization&quot; (minimization) of the Partition Function is a sorting algorithm to run through this graph in an &quot;efficient&quot; way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I hope this gives you a flavor of how things work.&lt;/p&gt;&#xA;" OwnerUserId="142" LastEditorUserId="142" LastEditorDisplayName="" LastEditDate="2010-11-10T23:31:11.473" LastActivityDate="2010-11-10T23:31:11.473" />
  <row Id="532" PostTypeId="2" ParentId="494" CreationDate="2010-11-10T23:29:52.600" Score="8" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://physics.weber.edu/schroeder/qftbook.html&quot; rel=&quot;nofollow&quot;&gt;Peskin and Schroeder&lt;/a&gt; tends to be the book used in most introductory QFT courses, so you'll definitely find all things there done in a pretty detailed way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Warren Siegel has an online book which is also pretty good, &lt;a href=&quot;http://insti.physics.sunysb.edu/~siegel/errata.html&quot; rel=&quot;nofollow&quot;&gt;Fields&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2010-11-10T23:29:52.600" CommentCount="1" />
  <row Id="533" PostTypeId="2" ParentId="429" CreationDate="2010-11-10T23:30:20.717" Score="4" ViewCount="" Body="&lt;p&gt;You don't need a vector field on the sphere - you just need vectors. Vectors don't have any intrinsic location, just a direction and a magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The polarization of light is independent of the propagation direction of the light. Let's examine this with a simple experiment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ZQAVq.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider an ideal plane-wave laser beam, beam 1, propagating in the &lt;em&gt;z&lt;/em&gt;-direction and striking a screen some distance away. Another beam with exactly the same intensity and wavelength, beam 2, hits the screen at the same point, but from a different angle. Both beams are vertically polarized -- that is to say, their &lt;em&gt;E&lt;/em&gt;-vectors point in the &lt;em&gt;x&lt;/em&gt;-direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the point where these beams hit the screen, you will observe interference fringes, which will go all the way down to zero intensity at their darkest points. This is because the &lt;em&gt;x&lt;/em&gt;-polarized components of the electric field interfere with each other. Besides the &lt;em&gt;x&lt;/em&gt;-components, there are no &lt;em&gt;y&lt;/em&gt; or &lt;em&gt;z&lt;/em&gt;-components to interfere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/0kUTT.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the same situation again, but with both beams horizontally polarized. That is to say, the &lt;em&gt;E&lt;/em&gt;-field of each beam points at right angles to the beam's propagation direction, and lies in the &lt;em&gt;yz&lt;/em&gt;-plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the polarizations are not the same, even though they are both known as &quot;&lt;em&gt;p&lt;/em&gt;-polarized&quot;. (At least I hope so -- I can never remember which is &lt;em&gt;s&lt;/em&gt; and which is &lt;em&gt;p&lt;/em&gt;.) Beam 1's &lt;em&gt;E&lt;/em&gt;-field points purely in the &lt;em&gt;y&lt;/em&gt;-direction, but beam 2's &lt;em&gt;E&lt;/em&gt;-field has both a &lt;em&gt;y&lt;/em&gt; and a &lt;em&gt;z&lt;/em&gt;-component, as I've drawn on the beam in the illustration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the point where these beams hit the screen, the &lt;em&gt;y&lt;/em&gt;-components of the electric fields interfere, once again producing interference fringes. However, even though both beams' amplitudes are the same, the &lt;em&gt;y&lt;/em&gt;-components are not equal, so the dark parts of the fringes don't go all the way down to zero intensity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you see, polarization vectors are polarization vectors, no matter which direction they're coming from. Equally valid, you could define your coordinate axes according to the propagation direction of beam 2, so that beam 1 came in at an angle instead, and still arrive at the same result. Specifically, this means that you don't need to worry about parallel transport. &lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2010-11-10T23:30:20.717" CommentCount="4" />
  <row Id="534" PostTypeId="2" ParentId="530" CreationDate="2010-11-10T23:38:06.200" Score="11" ViewCount="" Body="&lt;p&gt;At any surface (at least one which has a different index of refraction from air) some light is reflected, depending on the angle at which the light hits the surface and the polarization of the light; the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fresnel_equations&quot; rel=&quot;nofollow&quot;&gt;Fresnel equations&lt;/a&gt; will tell you what fraction of your light is reflected and what fraction is transmitted. When you see a black object &quot;shine&quot;, you are seeing the reflected light. But since the object is black, all the &quot;transmitted&quot; light is simply absorbed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between a matte black and a gloss black finish is one of index of refraction, I guess, and possibly of rough/smoothness.&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2010-11-10T23:38:06.200" />
  <row Id="535" PostTypeId="1" AcceptedAnswerId="543" CreationDate="2010-11-10T23:59:43.297" Score="7" ViewCount="868" Body="&lt;p&gt;As Wikipedia says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...the kinetic energy of a non-rotating object of mass $m$ traveling at a speed $v$ is $mv^2/2$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why does this not increase linearly with velocity? Why does it take so much more energy to go from 1 m/s to 2 m/s than it does to go from 0 m/s to 1 m/s?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition is wrong here, please help it out!&lt;/p&gt;&#xA;" OwnerUserId="285" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-01-03T10:30:17.390" LastActivityDate="2011-03-18T07:11:57.743" Title="Why does kinetic energy increase quadratically, not linearly, with velocity?" Tags="&lt;energy&gt;&lt;kinematics&gt;" AnswerCount="6" CommentCount="7" />
  <row Id="536" PostTypeId="2" ParentId="527" CreationDate="2010-11-11T00:16:48.943" Score="5" ViewCount="" Body="&lt;p&gt;That's because water has a higher heat capacity (the amount of energy it takes per kilogram to increase the temperature by 1 degree) than meat or vegetables and since a bowl of soup typically weighs more than a carrot or a piece of chicken, more mass needs to be heated, which takes more energy and thus more time. While meat and veggies do cotain a significant amount of water, a bowl of soup is essentially all water.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2010-11-11T00:16:48.943" CommentCount="2" />
  <row Id="537" PostTypeId="2" ParentId="535" CreationDate="2010-11-11T00:19:09.357" Score="1" ViewCount="" Body="&lt;p&gt;One way to look at this question of yours is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E(v) = \frac{m v^2}{2} \; . $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we multiply the velocity by a certain quantity, i.e., if we scale the velocity, we get the following,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E(\lambda v) = \frac{m (\lambda v)^2}{2} = \lambda^2  \frac{m v^2}{2} = \lambda^2 E(v)\; . $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if you scale your velocity by a factor of $\lambda$, your Energy is scaled by a factor of $\lambda^2$ — this should answer your question (just plug in the numbers).&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2010-11-11T00:19:09.357" CommentCount="3" />
  <row Id="538" PostTypeId="2" ParentId="535" CreationDate="2010-11-11T00:47:36.780" Score="2" ViewCount="" Body="&lt;p&gt;The only real physical reason (which is not really a fully satisfying answer) is that $E \sim v^2$ is what experiments tell us. For example, gravitational potential energy on the Earth's surface is proportional to height, and if you drop an object, you can measure that the height it falls is proportional to the &lt;em&gt;square&lt;/em&gt; of its speed. Thus, if energy is to be conserved, the kinetic energy has to be proportional to $v^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you could question why gravitational potential energy is proportional to height, and once that was resolved, question why some other kind of energy is proportional to something else, and so on. At some point it becomes a philosophical question. The bottom line is, defining kinetic energy to be proportional to the square of the speed has turned out to make a useful theory. That's why we do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, you could always say that if it were linear in velocity, it would be called momentum ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. It may be worth mentioning that kinetic energy is not &lt;em&gt;exactly&lt;/em&gt; proportional to $v^2$. Special relativity gives us the following formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K = mc^2\left(1 - 1/\sqrt{1 - v^2/c^2}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For low speeds, this is essentially equal to $mv^2/2$.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-11T00:47:36.780" CommentCount="1" />
  <row Id="539" PostTypeId="2" ParentId="535" CreationDate="2010-11-11T01:25:15.467" Score="3" ViewCount="" Body="&lt;p&gt;In comes down to definitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Momentum is defined as $p = mv$. Momentum grows linearly with velocity making momentum a quantity that is intuitive to understand (the more momentum the harder an object is to stop). Kinetic energy is a less intuitive quantity associated with an object in motion. KE is assigned such that the instantaneous change in the KE yields the momentum of that object at any given time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dKE}{dv} = p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A separate question one might ask is why do we care about this quantity? The answer is that in  a system with no friction, the sum of the kinetic and potential energies of an object is conserved:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d(KE + PE)}{dt} = 0 $&lt;/p&gt;&#xA;" OwnerUserId="173" LastEditorUserId="173" LastEditorDisplayName="" LastEditDate="2010-11-11T01:30:28.563" LastActivityDate="2010-11-11T01:30:28.563" CommentCount="3" />
  <row Id="540" PostTypeId="2" ParentId="535" CreationDate="2010-11-11T01:48:59.427" Score="4" ViewCount="" Body="&lt;p&gt;As Piotr suggested, accepting the definition of work $W=\mathbf{F}\cdot d\mathbf{x}$, it &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_energy#Derivation&quot; rel=&quot;nofollow&quot;&gt;follows&lt;/a&gt; that the kinetic energy increases quadratically. Why? Because the force and the infinitesimal interval depend linearly on the velocity. Therefore, it is natural to think that if you multiply both quantities, you need to end up with something like $K v^{2}$, where $K$ is an 'arbitrary' constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A much more interesting question is why the Lagrangian depends on the velocity squared. Given the homogeneity of space, it can not contain explicitly $\mathbf{r}$ and given the homogeneity of time, it can not depend on the time. Also, since space is isotropic, the Lagrangian can not contain the velocity $\mathbf{v}$. Therefore, the next simplest choice should be that the Lagrangian must contain the velocity squared. I do think that the Lagrangian is more fundamental in nature than the other quantities, however, its derivation involves the definition of work or equivalently, energy. So probably you won't buy the idea that this last explanation is the true cause of having the kinetic energy increasing quadratically, although, I think it is much more satisfactory than the first explanation.&lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="18" LastEditorDisplayName="" LastEditDate="2010-11-11T02:45:55.263" LastActivityDate="2010-11-11T02:45:55.263" CommentCount="1" />
  <row Id="541" PostTypeId="1" AcceptedAnswerId="569" CreationDate="2010-11-11T01:51:04.993" Score="3" ViewCount="766" Body="&lt;p&gt;I was recently wondering what would happen if the force sliding two surfaces against each other were somehow weaker than kinetic friction but stronger than static friction. Since the sliding force is greater than the maximum force of static friction, $F &gt; f_s = \mu_s F_N$, it seems that the surfaces should slide. But on the other hand, if the force of kinetic friction is greater than the applied force, there'll be a net force $\mu_k F_N - F$ acting &lt;em&gt;against&lt;/em&gt; the motion, suggesting that the surfaces should move opposite to the direction they're being pushed! That doesn't make sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only logical resolution I can think of is that the coefficient of static friction can never be less than the coefficient of kinetic friction. Am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-11T13:00:41.933" Title="Can the coefficient of static friction be less than that of kinetic friction?" Tags="&lt;forces&gt;&lt;newtonian-mechanics&gt;&lt;friction&gt;" AnswerCount="4" />
  <row Id="542" PostTypeId="1" AcceptedAnswerId="547" CreationDate="2010-11-11T01:58:41.103" Score="3" ViewCount="161" Body="&lt;p&gt;My books uses the argument that the multiplicities of a few macrostates in a macroscopic object take up an extraordinarily large share of all possible microstates, such that even over the entire lifetime of the universe, if each microstate had an equal chance of being accessed, big fluctuations away from equilibrium would never occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question to this statistical proof is that, in the real world, is there really an infinitesimal but nonzero chance that macroscopic systems could access some of the more unlikely macrostates (e.g. heat transfer from a cold object to a hot object)?&lt;/p&gt;&#xA;" OwnerUserId="202" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-11T05:51:41.980" LastActivityDate="2011-02-13T03:52:34.797" Title="Chance of objects going against greater entropy" Tags="&lt;statistical-mechanics&gt;&lt;entropy&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="543" PostTypeId="2" ParentId="535" CreationDate="2010-11-11T02:19:52.450" Score="14" ViewCount="" Body="&lt;p&gt;The question is especially relevant from a didactical point of view because one has to learn to distingish between energy (work) and momentum (quantity of motion). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinematic property that is proportional to $v$ is nowadays called momentum, it is the &quot;quantity of motion&quot; residing in a moving object, it's definition is $p:= mv$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;change&lt;/em&gt; of momentum is proportional to the impulse: impulse is the product of a force $F$ and the timespan $\Delta t$ it is applied. This relation is also known as the second law of Newton: $F \Delta t = \Delta p$ or $F dt = dp$. When one substitutes $mv$ for $p$ one gets it's more common form: $F= m \frac{\Delta v}{\Delta t} = ma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for an intuitive explanation that an object with double velocity has four times as much kinetic energy.&lt;br&gt;&#xA;Say A has velocity $v$ and B is an identical object with velocity $2v$.&lt;br&gt;&#xA;B has a double quantity of motion (momentum) - that's were your intuition is correct!&lt;br&gt;&#xA;Now we apply a constant force $F$ to slow both objects down to standstill. From  $F \Delta t = \Delta p$ it follows that the time $\Delta t$ needed for B to slow down is twice as much (we apply the same force to A and B). Therefore the braking distance of B will be a factor of 4 bigger then the braking distance of A (its starting velocity, and therefore also its mean velocity, being twice as much, and its time $\Delta t$ being twice as much, so the distance, $s = \bar{v}\Delta t$, increases 2x2=4 times).&lt;br&gt;&#xA;The work $W$ needed to slow down A and B is calculated as the product of the force and the braking distance $W=Fs$, so this is also four times as much. The kinetic energy is defined as this amount of work, so there we are. &lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-18T07:11:57.743" LastActivityDate="2011-03-18T07:11:57.743" CommentCount="2" />
  <row Id="544" PostTypeId="2" ParentId="541" CreationDate="2010-11-11T02:43:16.917" Score="0" ViewCount="" Body="&lt;p&gt;The kinetic friction is &lt;em&gt;usually&lt;/em&gt; not greater than the applied force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You start moving an object which increases the static force to prevent any motion. However, you keep applying more force until you reach a maximum value for the static friction. Then, the object begins to move. However, the kinetic friction (which is produced when the object moves) is less than the applied force. Furthermore, the kinetic friction doesn't depend on the applied force, therefore, you don't need to apply more force to keep moving the object (actually, you need less force).&lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="18" LastEditorDisplayName="" LastEditDate="2010-11-11T06:37:19.827" LastActivityDate="2010-11-11T06:37:19.827" CommentCount="5" />
  <row Id="545" PostTypeId="2" ParentId="530" CreationDate="2010-11-11T05:55:12.473" Score="0" ViewCount="" Body="&lt;p&gt;Reflection of radiowaves is the result of electrical currents or dielectric charge/discharge induced on the surface of conductors, dielectrics and everything in between. For visible light the model is still valid, but may be more affected by atomic/molecular properties of surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So black glossy dielectric partially reflects (horizontally) polarized white light and absorbs the rest of energy. The proportion depends on angle. Black matte dielectric or white matte conductor (say PLatinum with porous surface) has more complex path with more than 1 changes of light path, so more energy is absorbed. The purely metal (white) conductor with very complex surface can look like completely matte black. &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-11T05:55:12.473" />
  <row Id="546" PostTypeId="2" ParentId="507" CreationDate="2010-11-11T06:00:35.403" Score="0" ViewCount="" Body="&lt;p&gt;On high speed the wheels behave like a gyro. So you feel the resistance proportional to rotational speed and wheels mass.&lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-11T06:00:35.403" />
  <row Id="547" PostTypeId="2" ParentId="542" CreationDate="2010-11-11T06:07:08.830" Score="9" ViewCount="" Body="&lt;p&gt;I guess so - I mean, as far as I know, there's no law of physics that strictly prohibits those &quot;exotic&quot; states from being realized. As long as the state exists and can be reached by some path from the &quot;center&quot; of the state space where the likely states are, there should be a nonzero (not even infinitesimal, really) probability of accessing it. But for a typical system, that probability is really, &lt;em&gt;really&lt;/em&gt;, &lt;strong&gt;really&lt;/strong&gt; small. So small that it's impossible to intuitively comprehend just how unlikely such an event is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is, a lot of people aren't used to dealing with even moderately large or small numbers. If you confront them with a probability like $10^{-10^{23}}$, they often fail to put the smallness of that value in perspective, and instead focus on the fact that it's not strictly equal to zero. From there they may start coming up with all sorts of nonsensical ideas about walking through walls and spontaneous combustion (the weird kind) and the like. So physicists usually find it easier to just say the probability is zero - and in fact, for any purpose other than a rigorous mathematical proof, it might as well be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry about the rant, I know most people are actually relatively sensible about these things, but it bothers me that the crazy ones seem to get all the attention despite being wrong.)&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-11T06:07:08.830" CommentCount="1" />
  <row Id="548" PostTypeId="2" ParentId="530" CreationDate="2010-11-11T06:07:17.237" Score="1" ViewCount="" Body="&lt;p&gt;A perfectly black body is not something you see on your daily life, so when you see a &quot;black object&quot; it's actually a almost black object (black enough, for our perception) but it is reflecting some light, wich we are able to perceive as specular.&lt;/p&gt;&#xA;" OwnerUserId="301" LastActivityDate="2010-11-11T06:07:17.237" />
  <row Id="549" PostTypeId="2" ParentId="535" CreationDate="2010-11-11T06:29:33.677" Score="0" ViewCount="" Body="&lt;p&gt;For every relatively equal (in percents) increase of the speed, the applied force must be present over increasingly (quadratically) long travel distance. F=m*a. At the same time force*distance=work, where work=energy. &lt;/p&gt;&#xA;" OwnerUserId="299" LastActivityDate="2010-11-11T06:29:33.677" />
  <row Id="550" PostTypeId="1" AcceptedAnswerId="553" CreationDate="2010-11-11T07:41:25.607" Score="4" ViewCount="338" Body="&lt;p&gt;I have an electric stove, and when I turn it on and turn off the lights, I notice the stove glowing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as I turn down the temperature, it eventually goes away completely.  Is there a cut-off point for glowing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What actually is giving off the light?  Does the heat itself give off the light, or the metal?&lt;/p&gt;&#xA;" OwnerUserId="71" LastActivityDate="2010-12-12T21:42:28.730" Title="What causes hot things to glow, and at what temperature?" Tags="&lt;light&gt;&lt;heat&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="551" PostTypeId="2" ParentId="317" CreationDate="2010-11-11T07:58:32.653" Score="0" ViewCount="" Body="&lt;p&gt;I accidentally found a partial answer to my question on &lt;a href=&quot;http://arxiv.org/abs/1011.1657&quot; rel=&quot;nofollow&quot;&gt;arxiv&lt;/a&gt;.&#xA;It's still completely theoretical, but the answer can be summarized like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is assumed that some form of event horizon exists on a microscopic level. This event horizon prevents some information from escaping. With these premises, a QFT-like theory can be developed by observing the event horizon. The source of the randomness is the fact that information, as seen from outside the event horizon, is incomplete - the assumption being that randomness is the opposite of information.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As the field enters the Rinder horizon for the observer R, the observer shall not get information about future configurations of φ any more and all what the observer can expect about φ evolution beyond the horizon is a probabilistic distribution P[φ] of φ beyond the horizon. Already known information about φ acts as constraints for the distribution. I suggested that this ignorance is the origin of quantum randomness. Physics in the F wedge should reflect the ignorance of the observer in the R wedge, if information is fundamental.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-11T07:58:32.653" CommentCount="3" />
  <row Id="552" PostTypeId="2" ParentId="541" CreationDate="2010-11-11T08:05:31.533" Score="2" ViewCount="" Body="&lt;p&gt;This answer is speculative - not based on my experience with friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Logically, there's no reason kinetic friction has to be velocity-independent.  You could have kinetic friction that increases with velocity.  That way, if you push on something with more force than static friction, the thing would accelerate up to some certain velocity at which kinetic friction equaled the applied force, and then accelerate no more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that speed were very slow, you could say that kinetic friction is greater than static friction for all normally-encountered speeds without a paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as you pointed out, kinetic friction would have to be less than or equal to static friction for speeds right next to zero.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-11T08:05:31.533" />
  <row Id="553" PostTypeId="2" ParentId="550" CreationDate="2010-11-11T09:19:50.977" Score="7" ViewCount="" Body="&lt;p&gt;It is due to thermal radiation. Bodies with temperature above absolute zero emits radiation. If frequency of the radiation is in the visual range the body &quot;glows&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When the electrons in the atom are excited, for example by being heated, the additional energy pushes the electrons to higher energy orbits. When the electrons fall back down and leave the excited state, energy is re-emitted in the form of a photon.&#xA;  &lt;a href=&quot;http://en.wikipedia.org/wiki/Emission_spectrum&quot; rel=&quot;nofollow&quot;&gt;-wikipedia(emission spectrum)&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This explains that heat itself is not giving off the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Planck%27s_law&quot; rel=&quot;nofollow&quot;&gt;Spectral energy density as function of of wavelength and temperature&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u(\lambda,T) = \frac{8\pi hc}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k T}}-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you integrate $u(\lambda,T)$ wrt $\lambda$ from $\lambda = 380nm$ to $\lambda = 750nm$ you will find that there is always some radiation in visual range. But it is very very small at room temperature(T = 300K) and hence undetectable to naked eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get cut off temperature($T_c$) for glow. You need to solve &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_0^{T_c} \int_{\lambda = 380nm}^{\lambda=780nm} u(\lambda,T)d\lambda dT$ = minimum  required to be detected by human eye.&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-11-11T10:18:17.423" LastActivityDate="2010-11-11T10:18:17.423" CommentCount="5" />
  <row Id="554" PostTypeId="2" ParentId="550" CreationDate="2010-11-11T09:44:54.063" Score="1" ViewCount="" Body="&lt;p&gt;If you do the approximation of a black body (and it is not a bad approximation), you can measure the temperature of a body with the help of the spectrum of an object. &#xA;The key is the Wien's displacement law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Wiens_displacement_law&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Wiens_displacement_law&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the way a lot of thermometers work.&lt;/p&gt;&#xA;" OwnerUserId="306" LastActivityDate="2010-11-11T09:44:54.063" />
  <row Id="555" PostTypeId="2" ParentId="427" CreationDate="2010-11-11T10:02:14.260" Score="4" ViewCount="" Body="&lt;p&gt;Here is just a small remark. It is possible to give a strict mathematical proof about the equivalence of these two pictures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just start with the three (semi-experimental) facts: Lorentz invariance, $1/r$ long-range tail of gravitational force and its one-way action (attraction only) and the fact that the bending of light almost doesn't depend on its frequency and polarization, then you will find that these facts are compatible (in the large distance limit) only with the massless helicity $\pm2$ particle exchange. After that, it has been proved that special relativity and analytic properties of scattering amplitude lead to the equivalence principle [1,2]. This theorem is a pure analog of Gell-Mann-Low-Goldberger soft photon theorem, which claims that the power expansion of the amplitude of photon scattering by a hadron (with respect to photon frequency) does not depend on the spin or internal structure of the hadron (up to the second order). By considering multigraviton scattering amplitudes one can prove that the all local vertices for soft gravitons correspond to the expansion of the Einstein action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It means that the exchange of helicity $\pm2$ massless particle unavoidably leads to the classical general relativity (the opposite statement is trivial).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This program was initiated by Steven Weinberg [1,2] and finished by Deser and Boulware [3]. You can find the complete consideration in their paper [3] with the title &amp;ldquo;&lt;strong&gt;Classical general relativity derived form quantum gravity&lt;/strong&gt;&amp;rdquo;. This paper is a real masterpiece of clear physical explanation of this problem.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;References&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;[1] S. Weinberg, &lt;em&gt;Photons and gravitons in S-matrix theory: derivation of charge conservation and equality of gravitational and inertial mass&lt;/em&gt;, Phys. Rev. B135 (1964) 1049.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] S. Weinberg, &lt;em&gt;Photons and gravitons in perturbation theory: derivation of Maxwell’s and Einstein’s equations&lt;/em&gt;, Phys. Rev. B138 (1965) 988.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3] D. G. Boulware, S. Deser, &lt;em&gt;Classical general relativity derived form quantum gravity&lt;/em&gt;, Ann. Phys. 89 (1975) 193.&lt;/p&gt;&#xA;" OwnerUserId="225" LastActivityDate="2010-11-11T10:02:14.260" CommentCount="4" />
  <row Id="556" PostTypeId="1" AcceptedAnswerId="559" CreationDate="2010-11-11T10:04:47.860" Score="16" ViewCount="876" Body="&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/IKFiu.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize this isn't possible, but I can't see why not, especially if you change the model a little bit so that the balls simply travel through a tube of water on the way up, rather than exactly this model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please be clear and detailed.  I've heard explanations like &quot;the balls wouldn't move&quot; but that doesn't do it for me - I really do not see why the balls on the right would not be pulled/pushed up, and the rest of the chain wouldn't continue on.&lt;/p&gt;&#xA;" OwnerUserId="308" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2010-12-20T00:27:10.650" LastActivityDate="2010-12-20T00:27:10.650" Title="What is the fallacy in this infinite motion machine?" Tags="&lt;paradox&gt;&lt;buoyancy&gt;&lt;hydrodynamics&gt;&lt;perpetual-motion&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="558" PostTypeId="2" ParentId="556" CreationDate="2010-11-11T10:18:24.243" Score="6" ViewCount="" Body="&lt;p&gt;The energy needed to submerge a ball is equal to the energy gain from other ball to emerge from the water on the other side, so any waste on friction drives the process impossible. &lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-11T10:18:24.243" CommentCount="0" />
  <row Id="559" PostTypeId="2" ParentId="556" CreationDate="2010-11-11T10:19:12.900" Score="29" ViewCount="" Body="&lt;p&gt;The balls are entering the water well below the surface.  The pressure there is much higher than at the surface.  The work needed to push the balls into the water at this depth cancels  the work gained when they float back up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can ignore the gravitational force on the balls since gravity pulls down as much as up as you traverse the loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, if the balls enter the water at depth $d$, the pressure is $g \rho d$ with $g$ gravitational acceleration and $\rho$ the density of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The work done to submerge the balls is then the pressure times their volume, or $W_{ball} = g \rho V d$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force upwards on the ball is the weight of the water they displace which is $g \rho V$, and the work the water does on the balls is this force times the distance up that they travel, or $W_{water} = g \rho V d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The work the ball does on the water is the same as the work the water does on the ball.  No free energy.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-11T10:41:58.730" LastActivityDate="2010-11-11T10:41:58.730" CommentCount="1" />
  <row Id="560" PostTypeId="1" AcceptedAnswerId="1887" CreationDate="2010-11-11T10:22:13.270" Score="6" ViewCount="406" Body="&lt;p&gt;I am a part time physics enthusiast and I seldom wonder about the intergalactic space.  First, it is my perception that all(almost all) the objects in the universe are organized in the forms of galaxies; or in other words, the universe is a collection of billions and billions of galaxies which are spaced far apart.  Does anything actually exist in the empty space between galaxies?  If yes, how did it reach there?&lt;/p&gt;&#xA;" OwnerUserId="309" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-13T09:43:03.487" LastActivityDate="2010-12-13T21:42:24.287" Title="Does anything exist in the intergalactic space?" Tags="&lt;universe&gt;" AnswerCount="6" CommentCount="6" FavoriteCount="2" />
  <row Id="561" PostTypeId="2" ParentId="300" CreationDate="2010-11-11T11:26:48.413" Score="4" ViewCount="" Body="&lt;p&gt;In general small scale motions (like convection and formation of local eddies) in geophysical fluid dynamics are treated as turbulence, that is a regime characterized by chaotic motions and rapid, quasi random variations of pressure, temperature and velocity. Those random processes cannot be neglected &lt;em&gt;in boundary layers&lt;/em&gt; (layers of flow close to bounding surfaces atmosphere-soil, atmpsphere-ocean) because they averagely involve a flux of momentum and heat from the atmosphere to the soil or to the ocean (or viceversa). In geophysics most notable boundary layers are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&gt; The atmospheric boundary layer, that is the bottom part of the atmosphere, about 1000 m thick, in contact with soil and sea surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&gt; The oceanic boundary layer, that is the top layer of the sea, about 10-100 m thick, close to the boundary with the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hence a climatic large scale model ivolving boundary layers (for example models describing wind driven oceanic circulation) must take into account turbulence &lt;em&gt;average&lt;/em&gt; effects: indeed &lt;em&gt;parameterizing&lt;/em&gt; small scale phenomena means &quot;taking their average effects into account&quot;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, &lt;em&gt;far from the boundary layers&lt;/em&gt;, turbulence can be neglected. For example global circulation models describing motions of high atmosphere usually neglect turbulence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest way to take into account the average effects of turbulence is to introduce in the equations of dynamics terms that represent the average friction, for example that described by drag equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fd = - ρ Cd |U| u&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where ρ is fluid density, |U| is velocity scale, u is velocity. Fd is called drag force, and is a force by surface unit. It represents the average friction exerted by the atmosphere on the surface. Cd is called &quot;drag coefficient&quot;, and can be estimated through experimental observations. Its value can be different in different situations. Drag equation is an empirical relation, and can be deduced by purely dimensional consideretions, like Reynolds number. In particular we can find, using Buckingham theorem, that Cd depends &lt;em&gt;only&lt;/em&gt; on Reynolds number.&lt;/p&gt;&#xA;" OwnerUserId="312" LastEditorUserId="312" LastEditorDisplayName="" LastEditDate="2010-11-11T14:13:11.080" LastActivityDate="2010-11-11T14:13:11.080" CommunityOwnedDate="2010-11-11T14:15:14.700" />
  <row Id="562" PostTypeId="2" ParentId="193" CreationDate="2010-11-11T11:52:36.510" Score="2" ViewCount="" Body="&lt;p&gt;For a general approach to the maths involved in both classical and quantum physics, one of my favourite books is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-&quot;Mathematics of classical and quantum physics&quot;, Byron &amp;amp; Fuller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the more geometrical side, besides the already mentioned books, you can try:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-&quot;The geometry of physics. An introduction&quot;, Theodore Frankel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, as a general reference, the usual text is Arfken's &quot;Mathematical methods for physicists&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, IMHO, if you want to thoroughly understand the mathematical tools of physics, you should use &quot;Methods of Theoretical Physics&quot;, by Morse &amp;amp; Feshbach. It is an old book, but essential if you want to understand Jackson's Classical Electrodynamics or Messiah's Quantum Mechanics.&lt;/p&gt;&#xA;" OwnerUserId="306" LastActivityDate="2010-11-11T11:52:36.510" CommunityOwnedDate="2010-11-11T11:52:36.510" />
  <row Id="563" PostTypeId="2" ParentId="363" CreationDate="2010-11-11T11:58:50.170" Score="2" ViewCount="" Body="&lt;p&gt;I think D'Inverno's &quot;Introducing Einstein's Relativity&quot; is a good text for a rigorous primer in GR. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following link could be useful for you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.desy.de/user/projects/Physics/Administrivia/rel_booklist.html&quot; rel=&quot;nofollow&quot;&gt;http://www.desy.de/user/projects/Physics/Administrivia/rel_booklist.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To have fun while reading these books, you can enjoy &quot;The Einstein Theory of Relativity: A trip to the fourth dimension&quot;, by Lillian Lieber.&lt;/p&gt;&#xA;" OwnerUserId="306" LastActivityDate="2010-11-11T11:58:50.170" CommentCount="1" />
  <row Id="564" PostTypeId="2" ParentId="24" CreationDate="2010-11-11T12:33:45.243" Score="1" ViewCount="" Body="&lt;p&gt;if u just lean the steering wheel will turn in order to conservate the angular momentum!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;didn't u ever try to ride your bike not holding the steereing wheel with your hands? &#xA;if u try and u want to be a pro, not moving just on a straight line, leaning your weight left or right is exactely how you will curve your trajectory or you will miss the street-hole in your path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;try it! i guarantee no hospital bill will be needed: i do it on regular basis when for example i drink or stretch my back during long bicycle rides!&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditorDisplayName="" LastEditDate="2010-11-13T16:07:39.283" LastActivityDate="2010-11-13T16:07:39.283" CommentCount="3" />
  <row Id="565" PostTypeId="2" ParentId="541" CreationDate="2010-11-11T12:35:11.380" Score="1" ViewCount="" Body="&lt;p&gt;With the simple model of friction, using coefficients of static and kinetic friction, you are right - the coefficient of kinetic friction cannot be higher than the coefficient of static friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In real world, the phenomenon of friction may be much more complex. Your reasoning points out that with the velocity very close to zero kinetic friction cannot be much higher than static friction. To be more accurate, if the velocity is infinitely small, the kinetic friction may be higher than static one, but then it may only be infinitely small! An example is air friction, which is zero when a body doesn't move and increases with velocity. In a simple model, for small velocities, air friction is just proportional to velocity.&lt;/p&gt;&#xA;" OwnerUserId="219" LastActivityDate="2010-11-11T12:35:11.380" CommentCount="1" />
  <row Id="566" PostTypeId="1" AcceptedAnswerId="2008" CreationDate="2010-11-11T12:38:04.347" Score="2" ViewCount="143" Body="&lt;p&gt;Keeping the average power constant, why some materials are more eager to be damaged by pulsed laser with respect of C.W. lasers, or viceversa?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When i talk about pulsed lasers i think for examples of duty cycles in the order of $10^5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example optical elements (such as a vortex phase plate for donut-shaping the beam) have different tolleration regimes regarding the incident power not simple dependent on the average power but also on the peak power for pulsed beam.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-17T00:07:56.580" LastActivityDate="2010-12-17T00:07:56.580" Title="Material resistency to lasers beam" Tags="&lt;optics&gt;&lt;laser&gt;&lt;material-science&gt;&lt;non-linear-optics&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="567" PostTypeId="1" AcceptedAnswerId="587" CreationDate="2010-11-11T12:43:21.107" Score="3" ViewCount="261" Body="&lt;ul&gt;&#xA;&lt;li&gt;What makes it burn, what kind of&#xA;fusion/decay is happening there ?&lt;/li&gt;&#xA;&lt;li&gt;Another question is, what suppose to&#xA;happen with neutron star on long run ? What if it cools, then what&#xA;degenerated matter looks like after&#xA;it cools ? will the gravitational&#xA;equilibrium be ruined after some burn&#xA;time ? how it explodes if it can&#xA;explode at all ?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2010-11-11T13:25:31.387" LastActivityDate="2010-11-14T17:35:40.660" Title="How neutron stars burn ? Is it decay or fusion or something else?" Tags="&lt;astrophysics&gt;&lt;neutron-stars&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="568" PostTypeId="2" ParentId="507" CreationDate="2010-11-11T12:52:28.917" Score="0" ViewCount="" Body="&lt;p&gt;when you stay still you have to win the friction caused by the contact between your tires and the street in order to turn your wheels. &#xA;If, for example, you try to inflate your wheels more (increasing the air pressure inside), you will reduce the contact surface, and the stearing wheel will be less hard! (but don't drive with your tires too inflated, please!!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the contrary, when your moving the contact surface (and hence the source of the friction) is dramatically reduced and that's the reason why the steering wheel becomes bearable when moving at slow speeds. &#xA;When you keep increasing your velocity, the tires will rotate on their axis with higher velocity, thus incrementing their angular momentum that has the tendency to conservate itself! (that's also the reason the &lt;a href=&quot;http://physics.stackexchange.com/questions/271/why-dont-spinning-tops-fall-over&quot;&gt;spinning tops work&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2010-11-11T12:52:28.917" />
  <row Id="569" PostTypeId="2" ParentId="541" CreationDate="2010-11-11T13:00:41.933" Score="8" ViewCount="" Body="&lt;p&gt;The problem with this question is that static friction and kinetic friction are not fundamental forces in any way-- they're purely phenomenological names used to explain observed behavior. &quot;Static friction&quot; is a term we use to describe the observed fact that it usually takes more force to set an object into motion than it takes to keep it moving once you've got it started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, with that in mind, ask yourself how you could measure the relative sizes of static and kinetic friction. If the coefficient of static friction is greater than the coefficient of kinetic friction, this is an easy thing to do: once you overcome the static friction, the frictional force drops. So, you pull on an object with a force sensor, and measure the maximum force required before it gets moving, then once it's in motion, the frictional force decreases, and you measure how much force you need to apply to maintain a constant velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would it mean to have kinetic friction be greater than static friction? Well, it would mean that the force required to &lt;em&gt;keep&lt;/em&gt; an object in motion would be greater than the force required to &lt;em&gt;start&lt;/em&gt; it in motion. Which would require the force to go &lt;em&gt;up&lt;/em&gt; at the instant the object started moving. But that doesn't make any sense, experimentally-- what you would see in that case is just that the force would increase up to the level required to keep the object in motion, as if the coefficients of static and kinetic friction were exactly equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, common sense tells us that the coefficient of static friction can never be less than the coefficient of kinetic friction. Having greater kinetic than static friction just doesn't make any sense in terms of the phenomena being described.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As an aside, the static/kinetic coefficient model is actually pretty lousy. It works as a way to set up problems forcing students to deal with the vector nature of forces, and allows some simple qualitative explanations of observed phenomena, but if you have ever tried to devise a lab doing quantitative measurements of friction, it's a mess.)&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-11T13:00:41.933" />
  <row Id="570" PostTypeId="1" AcceptedAnswerId="572" CreationDate="2010-11-11T13:06:40.527" Score="1" ViewCount="87" Body="&lt;p&gt;I'd like to know (my knowledge at physics are realy poor) what would happen if I've got a generator connected to another generator and both of them generating electricity for a house or whatever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G1 is connected to G2 and the plug. So it gives electricity to both of them making G2 to &quot;load electricity&quot; and plug usable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When G1's energy is finished, G2 starts using the &quot;loaded electricity&quot; making G1 to &quot;load electricity&quot; and still the plug usable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Woudn't that make an infinite electricity system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diego&lt;/p&gt;&#xA;" OwnerUserId="315" LastEditorUserId="315" LastEditorDisplayName="" LastEditDate="2010-11-11T13:35:33.217" LastActivityDate="2010-11-11T13:41:59.383" Title="2 Electricity generator connected" Tags="&lt;energy-conservation&gt;&lt;energy&gt;" AnswerCount="1" />
  <row Id="571" PostTypeId="1" CreationDate="2010-11-11T13:28:33.800" Score="7" ViewCount="380" Body="&lt;p&gt;I have always wondered why salt is white, water is clear and gold is, well, gold. What determines the color of a substance? Does it have something to do with the electrons? And is it possible to predict the color of a substance by looking at the formula?&lt;/p&gt;&#xA;" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-11T13:46:57.963" LastActivityDate="2010-11-11T23:48:32.580" Title="What determines the color of a pure substance and is it possible to predict it?" Tags="&lt;material-science&gt;&lt;solid-state-physics&gt;&lt;chemical-compounds&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="572" PostTypeId="2" ParentId="570" CreationDate="2010-11-11T13:30:51.317" Score="2" ViewCount="" Body="&lt;p&gt;I assume with energy you mean electricity, and your generators sound like rechargeable batteries to me. In that case you have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Series_and_parallel_circuits#Cells_and_batteries&quot; rel=&quot;nofollow&quot;&gt;series connection&lt;/a&gt;, which will simply double the voltage available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, this wouldn't generate infinite energy. Whatever you do, each generator needs fuel that contains chemical energy. In the best (=impossible) case it would be entirely converted into electrical energy. Whether you temporarily store a part of G1's energy in G2 or not does not change anything about the fact that the total energy cannot exceed the chemical energy of your fuel (unless of course there are other influences, e.g. a solar panel and the very energetic sun which won't run out of fuel for some billion years...)&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2010-11-11T13:41:59.383" LastActivityDate="2010-11-11T13:41:59.383" CommentCount="2" />
  <row Id="573" PostTypeId="2" ParentId="560" CreationDate="2010-11-11T13:56:18.627" Score="2" ViewCount="" Body="&lt;p&gt;A simple picture is the following: stars are organized in &lt;strong&gt;galaxies&lt;/strong&gt; of various types (like elliptical, spiral, ...). These galaxies are organized in &lt;strong&gt;groups&lt;/strong&gt; which are local aggregates of galaxies (our galaxies is in the so-called &quot;local group&quot; of around 40-50 galaxies). The largest structure is a &lt;strong&gt;cluster&lt;/strong&gt; of galaxies: the groups are bound by gravitational interaction in a cluster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This classification is not really precise but still. A logical conclusion is that the &quot;empty&quot; space in between these cluster is quasi-completely empty, the density of particles being much much lower than in a galaxy (in average) and is uniform: if it was not uniform, the particles would start to gravitationally interact and start to form other structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that's a valid reasoning for, say, hydrogen gas (atoms). The conclusion as pointed by @Mark Eichenlaub is then that it can be considered empty and that it is often called &quot;the void&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course the space cannot be considered empty: the photons (light) traveling from one place to another are as real as other particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step in the discussion would involve things like dark energy, but that's a different story.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-11-11T13:56:18.627" CommentCount="1" />
  <row Id="574" PostTypeId="2" ParentId="571" CreationDate="2010-11-11T13:59:44.137" Score="0" ViewCount="" Body="&lt;p&gt;The light is just a kind of electromagnetic radiation. You can see it, but it is the same phenomena as X rays. So, when this electromagnetic radiation, light, arrives at a material, the electrons of this material are driven by light. It depends on how are arranged these electrons, their energy... the way they will be driven. &#xA;Thus, electrons, as charged particles, in a movement which is imposed by light, begin to radiate themselves. And this new radiation is the characteristic light reflected by the material, with a well-known frequency (a range of frequencies actually).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some simple cases it is possible to predict exactly the way the material will reflect (re-radiate) light, and to know the color. &lt;/p&gt;&#xA;" OwnerUserId="306" LastActivityDate="2010-11-11T13:59:44.137" />
  <row Id="575" PostTypeId="1" AcceptedAnswerId="576" CreationDate="2010-11-11T14:19:27.107" Score="3" ViewCount="369" Body="&lt;p&gt;Comparing the simples form of the forces of both phenomena: the law of Newton for gravitation  $V\propto \frac{1}{r}$, and the Coulomb law for electrostatics $V\propto \frac{1}{r}$, one might think that if one can be extended relativistically to a curvature in space-time, that the other one would lend itself to a similar description. Is this so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any useful thoughts and/or suggestions!&lt;/p&gt;&#xA;" OwnerUserId="317" LastEditorUserId="2151" LastEditorDisplayName="" LastEditDate="2011-03-01T13:11:56.920" LastActivityDate="2011-03-01T13:11:56.920" Title="Is it possible/correct to describe electromagnetism using curved space(-time)?" Tags="&lt;electromagnetism&gt;&lt;general-relativity&gt;&lt;gravity&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="576" PostTypeId="2" ParentId="575" CreationDate="2010-11-11T14:41:32.910" Score="3" ViewCount="" Body="&lt;p&gt;I don't really follow the text underneath the title, but the answer to this&#xA;question is most certainly, yes.  (For example, we witness electromagnetic&#xA;waves traveling through curved space.)  Here is how electromagnetism is&#xA;described mathematically (I fear this answer is slightly beyond the level of the&#xA;questioner -- sorry -- but perhaps not of other readers):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maxwell's equations are best expressed in terms of the field strength&#xA;tensor.  That tensor is the curvature of a connection on a circle bundle&#xA;(over spacetime).  The connection is the four-vector electromagnetic&#xA;potential, in physics terms.  In this set up, two of Maxwell's equations&#xA;are automatic (for example, saying that the magnetic field is locally the curl&#xA;of a 3-vector is saying that it is divergence-free).  The other two are&#xA;equations for the &quot;divergence&quot; of this tensor.  This way of phrasing&#xA;the problem makes sense in any metric, i.e. on curved spacetimes.&lt;/p&gt;&#xA;" OwnerUserId="20" LastActivityDate="2010-11-11T14:41:32.910" CommentCount="7" />
  <row Id="578" PostTypeId="2" ParentId="571" CreationDate="2010-11-11T15:20:00.977" Score="4" ViewCount="" Body="&lt;p&gt;It's important to note that &lt;a href=&quot;http://en.wikipedia.org/wiki/Color&quot; rel=&quot;nofollow&quot;&gt;color&lt;/a&gt; is only defined given knowledge of three things, see especially &lt;a href=&quot;http://en.wikipedia.org/wiki/Color#Color_of_objects&quot; rel=&quot;nofollow&quot;&gt;this section of the wikipedia article which has a nice discussion&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Light absorbing / reflecting properties of a material&#xA;2) Properties of the light source&#xA;3) Properties of the human perception of light&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, 1 is determined by the interaction of the electrons in the material with varying frequencies of light, see e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Absorption_spectroscopy&quot; rel=&quot;nofollow&quot;&gt;absorption spectrum&lt;/a&gt; for one thing that goes into it, but there are other properties like reflection that I won't go into.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 is determined by the spectrum of the light that goes in.  For instance, things will have a different perceived color under moonlight or sunlight or under mercury lamps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, 3 is mainly determined by properties of the human color receptors in the eye, but also there are perceptual issues coming from how the brain processes visual information; you can sometimes see this in various optical illusions.  Note that this too can vary from person to person, indeed some people are color blind or have color receptors for 4 basic colors, not just the usual 3, see e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrachromat&quot; rel=&quot;nofollow&quot;&gt;tetrachromat&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, the wikipedia page on color goes much more into detail on all this.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-11T15:20:00.977" />
  <row Id="579" PostTypeId="2" ParentId="566" CreationDate="2010-11-11T15:25:17.607" Score="0" ViewCount="" Body="&lt;p&gt;One thing that ought to matter is how much laser light gets reflected versus how much gets absorbed and transmitted.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know much about this, but my naïve guess is that materials get damaged by lasers primarily because they are heated by the intensity of the light, and then they melt (or burn!).   Therefore thermal conductivity or response properties of the material will come into play as well.  For instance, if heat conducts well in a material, the local heating of the material by the laser will be able to diffuse to a larger region.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-11T15:25:17.607" CommentCount="3" />
  <row Id="580" PostTypeId="2" ParentId="560" CreationDate="2010-11-11T15:27:41.463" Score="1" ViewCount="" Body="&lt;p&gt;Try this link for a semipopular view of the overall picture.&#xA;&lt;a href=&quot;http://www.nasa.gov/multimedia/videogallery/index.html&quot; rel=&quot;nofollow&quot;&gt;http://www.nasa.gov/multimedia/videogallery/index.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-11-11T15:27:41.463" />
  <row Id="581" PostTypeId="2" ParentId="193" CreationDate="2010-11-11T15:31:34.567" Score="2" ViewCount="" Body="&lt;p&gt;The last book I read on &quot;background in math for physicists&quot; was &quot;Mathematics for Physics&quot; by Stone and Goldbart, and I enjoyed it quite a bit.  (Since then I've tended to hit the pure math books, but that's a different story).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even better, &lt;a href=&quot;http://webusers.physics.illinois.edu/~goldbart/PG_MS_MfP.htm&quot; rel=&quot;nofollow&quot;&gt;a version of the book&lt;/a&gt; is available online at Paul Goldbart's webpage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a list of topics:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* Calculus of Variations&#xA;* Function Spaces&#xA;* Linear Ordinary Differential Equations&#xA;* Linear Differential Operators&#xA;* Green Functions&#xA;* Partial Differential Equations&#xA;* The Mathematics of Real Waves&#xA;* Special Functions&#xA;* Integral Equations&#xA;* Vectors and Tensors&#xA;* Differential Calculus on Manifolds&#xA;* Integration on Manifolds&#xA;* An Introduction to Differential Topology&#xA;* Groups and Group Representations&#xA;* Lie Groups&#xA;* The Geometry of Fibre Bundles&#xA;* Complex Analysis I&#xA;* Complex Analysis II&#xA;* Special Functions and Complex Variables&#xA;      o Appendix A: Linear Algebra Review&#xA;      o Appendix B: Fourier Series and Integrals &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-11T15:31:34.567" CommunityOwnedDate="2010-11-11T15:31:34.567" />
  <row Id="582" PostTypeId="2" ParentId="575" CreationDate="2010-11-11T15:57:12.387" Score="7" ViewCount="" Body="&lt;p&gt;I'm not sure if this is exactly what you're getting at, but there have been attempts to derive electromagnetism as a consequence of general relativity. The most famous one is &lt;a href=&quot;http://en.wikipedia.org/wiki/Kaluza%E2%80%93Klein_theory&quot; rel=&quot;nofollow&quot;&gt;Kaluza-Klein theory&lt;/a&gt;, published in 1921. The theory is basically just general relativity in a 5-dimensional spacetime, where one of the dimensions is &quot;curled up&quot; (compact). It turns out that the additional equations of motion obtained from the compact dimension are equivalent to the Maxwell equations for electromagnetism (in the relativistic form Eric described).&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-11T15:57:12.387" CommentCount="1" />
  <row Id="583" PostTypeId="2" ParentId="527" CreationDate="2010-11-11T16:27:02.160" Score="6" ViewCount="" Body="&lt;p&gt;What makes your microwave &quot;better&quot; than a stove at heating things is that is rather efficient at getting the heat &lt;em&gt;into your food&lt;/em&gt; - the wave field permeates the food and resonates with dielectric molecules, which then distribute their rotational energy to their surroundings as heat. This does not rely greatly on heat conductivity inside the food since it is heated locally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a stove, all the heat comes from the burner below and somehow has to get to your food - first by transfer from the hot gas/plate to the vessel, and then from that vessel to food. This process is highly inefficient for food that is not in good thermal contact with the vessel - you get a very hot pot and warm air in the room. For water or your soup on the other hand, the thermal contact with the vessel is excellent and almost all the heat that is transferred to the vessel actually ends up in your food.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, the fact that the times for microwave and stovetop heating come out close is mainly an artifact of the fact that your stove outputs much more power than the microwave - a 9000 BTU/hr burner is equivalent to about 2.6 kW, while your usual microwave outputs less than half of that) , but is a lot less efficient. For non-liquid foods, the stove is even less efficient while the efficiency of the microwave doesn't change much.&lt;/p&gt;&#xA;" OwnerUserId="243" LastEditorUserId="243" LastEditorDisplayName="" LastEditDate="2010-11-12T07:44:35.220" LastActivityDate="2010-11-12T07:44:35.220" CommentCount="2" />
  <row Id="584" PostTypeId="1" CreationDate="2010-11-11T16:28:58.683" Score="4" ViewCount="205" Body="&lt;p&gt;A causal set is a poset which is reflexive, antisymmetric, transitive and locally finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a motivation, there is a programme to model spacetime as fundamentally discrete, with causal sets providing the underlying structure. Typically this is constructed by sprinkling (a poisson process) existing spacetime with elements, endowing these elements with an ordering given by causal cones, and removing spacetime. Volume is then given by a counting metric, which with the causal structure is enough to build a geometry. By the poisson nature of this process, the distribution is Lorentz-invariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This only makes sense in nature if the causal set is manifold-like, by which we mean it can be faithfully embedded into a manifold, such that the element count gives volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precisely when is a causal set manifold-like: what are the necessary conditions for the existence of such an embedding? (Are there interesting sufficient conditions?) Do they have natural interpretations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;[This should be tagged quantum-gravity I think.]&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="236" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-11T17:08:43.047" LastActivityDate="2011-04-17T20:32:35.453" Title="What is necessary for a causal set to be manifold-like?" Tags="&lt;mathematical-physics&gt;&lt;quantum-gravity&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="585" PostTypeId="2" ParentId="571" CreationDate="2010-11-11T16:29:01.457" Score="9" ViewCount="" Body="&lt;p&gt;Yes, it is possible to predict the color of a substance, but it is, in some cases, very complex. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The color of a substance is decided at various levels. The most &quot;trivial&quot; level is the molecule in gas phase. You have a molecule, all by itself, and when you send some white light on it you provide all the colors of the spectrum. The electronic configuration of the molecule is such that it &quot;prefers&quot; specific light frequencies (hence, specific light colors). This preference is due to the electronic transitions between a ground state and an excited state, which is then &quot;quenched&quot; as heat. As a consequence, the molecule absorb some colors, thus subtracting them from the white light. What you see is the complementary color. If the molecule absorb blue, you get red. If it absorb yellow, you get violet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electronic transitions, however, are not the only responsible for absorbing light. A molecule can absorb by excitation of rotation and vibration (meaning that the molecule spins faster, or vibrates more). One case is water. Water appears as transparent, but in reality it's slightly blue. The reason is that you are exciting it vibrationally (to be exact, there's an absorption in the red which is an &quot;overtone&quot; of a vibration which actually absorb in the infrared). As a result, a minimal amount of red is subtracted from white light and water ends up being slightly blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecules however are not alone in the universe. They can come close, and eventually have other molecules around, either of the same species, or of other species, such as those of a solvent. No reactions are involved, just the proximity of other molecules, with their protons and electrons. The presence of nearby partners alters the electronic setup of the molecule, and as a consequence, a slight variation of the absorption is produced, leading to a change of color, either as a shift towards blue or red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you have anything that can change the structure of the molecule through chemical reaction. Take tea, put some lemon into it. Its color becomes lighter, because you are increasing the acidity of the water, and the colored substances in the tea receive the H+ protons, creating a change in the molecular electronic distribution. The result is a different color. This effect can be dramatic: from blue to red, from transparent to purple. These are the so-called pH indicators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you have the crystals. When you have ordered atoms, they can absorb light by virtue of their ordered crystalline structure. Note that the starting atom or molecule, by itself, may absorb nothing in the visible, so it would appear &quot;transparent&quot;, or white. Nevertheless, it's by virtue of the highly ordered crystalline structure that, in the end, what you handle with your hands absorbs light, and thus has a color. This, at the quantum level, has to do with band structure and Bloch wavefunctions. The same fact explains semiconductors and conductivity of metals. How the atoms (or molecules) are arranged in space, and how many you have makes a difference in the final color. You say gold has gold color, but if you take a small cluster (say, 100) of gold atoms, what you see is red, not gold-colored. As a side note to crystals, you have impurities. Take aluminium oxide, it is transparent. Add some Chromium Iron and Titanium and it becomes Ruby, which is red, or sapphire, which is blue, depending on the crystal structure, and the relative quantities of these impurities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you have how the substance is structured at the macroscopic level. Take a smooth platinum electrode. It's platinum color. Make it sponge-like (by making very tiny bubbles and pits) to increase the surface area and it appears black as coal. This is because the light is scattered and absorbed completely, leading to a black color. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Potentially, you have additional effects perturbing matter-light interaction. What is the color of a CD ? is it silver ? is it &quot;rainbow&quot; ? What about the color of a oil slick on the road in a rainy day ? What about the color of &lt;a href=&quot;http://en.wikipedia.org/wiki/Opal&quot; rel=&quot;nofollow&quot;&gt;opal&lt;/a&gt;, or of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Tiger_eye&quot; rel=&quot;nofollow&quot;&gt;tiger eye&lt;/a&gt; ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you see, color is a very particular property, and while you can make an educated guess (in particular through quantum mechanics techniques) it's not always easy to know what color a given substance can have. This is just the tip of the iceberg. You have many other phenomena (such as how much light penetrates into the substance, or what imperfections are present) which affects both the color and the reflective properties of a substance. Ice is transparent, but if it's full of bubbles it's white. Plastic looks like plastic, and metal looks like metal, because how the light is scattered and absorbed changes the way light is reflected back to the viewer. This however, does not affect only color, but the general material texture. Pure spectroscopic color for white, non-polarized light is probably what you are referring to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer directly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What determines the color of a substance?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;many different things.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does it have something to do with the electrons?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, in some cases, electrons are &quot;excited&quot;, leading to absorption of some light frequencies, but it's not the only factor.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And is it possible to predict the color of a substance by looking at the formula?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In general, no. But you can, from personal experience, know if a substance is colored or not, and more or less in which part of the spectrum it will absorb. To have a more accurate answer, you need a course in computational chemistry, because you need a program, a computational method, and a good dose of skills to get it out. In any case, the required input is just the xyz coordinates of the atoms and their atomic numbers.&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="217" LastEditorDisplayName="" LastEditDate="2010-11-11T23:48:32.580" LastActivityDate="2010-11-11T23:48:32.580" />
  <row Id="586" PostTypeId="2" ParentId="584" CreationDate="2010-11-11T16:40:51.100" Score="0" ViewCount="" Body="&lt;p&gt;There are some papers that try and construct this embedding, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0207103&quot; rel=&quot;nofollow&quot;&gt;Manifold dimension of a causal set: Tests in conformally flat spacetimes&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Causal_sets#Further_reading&quot; rel=&quot;nofollow&quot;&gt;Manifoldness :: Further Reading (Wikipedia: Causal Sets)&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2010-11-11T16:40:51.100" />
  <row Id="587" PostTypeId="2" ParentId="567" CreationDate="2010-11-11T16:42:04.157" Score="4" ViewCount="" Body="&lt;p&gt;I think &lt;a href=&quot;http://www.astro.umd.edu/~miller/nstar.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; is a good almost-popular introduction to neutron stars, the processes expected to occur therein and their evolution.&lt;/p&gt;&#xA;" OwnerUserId="243" LastActivityDate="2010-11-11T16:42:04.157" CommentCount="2" />
  <row Id="588" PostTypeId="2" ParentId="560" CreationDate="2010-11-11T16:43:42.633" Score="6" ViewCount="" Body="&lt;p&gt;As others have said, it's almost empty, but not quite, as there are gas particles and so on floating around.  As &lt;a href=&quot;http://en.wikipedia.org/wiki/Intergalactic_medium#Intergalactic&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Generally free of dust and debris,&#xA;  intergalactic space is very close to a&#xA;  total vacuum. The space between galaxy&#xA;  clusters, called the voids, is&#xA;  probably nearly empty. Some theories&#xA;  put the average density of the&#xA;  Universe as the equivalent of one&#xA;  hydrogen atom per cubic meter.&#xA;  The density of the universe, however,&#xA;  is clearly not uniform; it ranges from&#xA;  relatively high density in galaxies&#xA;  (including very high density in&#xA;  structures within galaxies, such as&#xA;  planets, stars, and black holes) to&#xA;  conditions in vast voids that have&#xA;  much lower density than the universe's&#xA;  average.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And that's only if you consider empty to mean void of matter - there's also electromagnetic waves permeating most (all?) of space.  And when you get down to the subatomic level, quantum mechanics ensures that particles are constantly popping into and out of existence as well, even in 'empty' space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for how the matter got there, well aside from the normal ways (being shot out of exploding stars and so on), don't forget that before it all started expanding, all of the matter was in the same place anyway, so the particles in intergalactic space haven't necessarily travelled anywhere to get there.  They could have simply stayed where they were while particles around them got gravitationally drawn into nearby clumps of matter/galaxies.&lt;/p&gt;&#xA;" OwnerUserId="327" LastActivityDate="2010-11-11T16:43:42.633" />
  <row Id="589" PostTypeId="1" CreationDate="2010-11-11T16:57:34.707" Score="13" ViewCount="228" Body="&lt;p&gt;Question says it all. Ripping a piece of tape along the width is tough, stretches the tape, messes it up, etc, but if you put the tiniest nick at the top, it rips without any problems. Why is that?&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="42" LastEditorDisplayName="" LastEditDate="2010-11-11T17:44:27.577" LastActivityDate="2010-11-11T18:12:21.780" Title="Why is tape so hard to rip when it is not broken, but so easy once there is the tiniest cut in it?" Tags="&lt;material-science&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" />
  <row Id="590" PostTypeId="2" ParentId="589" CreationDate="2010-11-11T18:12:21.780" Score="12" ViewCount="" Body="&lt;p&gt;This is a problem in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fracture_mechanics&quot; rel=&quot;nofollow&quot;&gt;theory of cracks&lt;/a&gt;, but let me try to give an intuitive discussion on the level of linear elasticity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a rectangular sheet of material with two opposite ends (which I'll call East and West) being pulled apart slowly.  The stress in the material due to the tension from the boundary conditions will be symmetric across the North South axis by symmetry, and is intuitively &quot;spread out&quot; on the rectangle (though the distribution will not be uniform, of course).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though I won't do it, this stress distribution can be calculated in linear elasticity theory, as part of the theory of plates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if we cut a notch in say the North edge of the rectangle, the symmetry of the material is broken and we should expect the stress distribution to be asymmetrical as well.  A calculation similar to above (but harder due to the funnier geometry) in linear elasticity will probably show that the stress will become focused near the notch.  The wikipedia article above seems to claim that the stress near a very sharp notch will actually come out infinite!  (That is, linear elasticity theory breaks down)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously the actual tearing process goes outside of linear elasticity and into plasticity theory somewhat, but I think what happens can be described as follows.  The high concentration of stress near the notch means that when the material gives, it will begin to tear at the &quot;corner&quot; of the notch.  Then this tearing causes the notch to get larger, and weakens more nearby material at that corner, which then becomes torn, etc.  This is a bit like a positive feedback loop.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't quite relate it to the question you're asking yet, but I remember it's possible to do some cute dimensional analysis to show that cracks below a certain size in a material tend to shrink, and those above the threshold size tend to grow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me just add that the shapes of tears in thin sheets has been a recently popular topic in the field of &quot;extreme mechanics&quot; in soft matter physics.  See for instance &lt;a href=&quot;http://proxy.library.upenn.edu:3532/abstract/PRL/v95/i2/e025502&quot; rel=&quot;nofollow&quot;&gt;this recent article by Audoly, Reis and Roman&lt;/a&gt; and citing references.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-11T18:12:21.780" />
  <row Id="591" PostTypeId="2" ParentId="567" CreationDate="2010-11-11T18:42:47.267" Score="9" ViewCount="" Body="&lt;p&gt;With normal stars, the &quot;burning&quot;, that is to say the fusion reaction, produces a pressure that counteracts the pull of gravity to keep the star from collapsing.  But with neutron stars, the protons and electrons in the star have combined into neutrons*.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Pauli exclusion principle causes the neutrons to resist further compression.  That is, the neutrons, being identical fermions, can't all be put in the same state.  So to get them closer and closer together you have to go into higher and higher energy states.  Thus, there is an energy cost in compressing the star, and this results in a sort of pressure called &quot;degeneracy pressure&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is this pressure that stabilizes the neutron star against collapse (assuming it doesn't have enough mass to overcome this pressure and become a black hole).  So they don't need to &quot;burn&quot; to maintain their stability, and so far as I know, they don't.  At least not in the sense of a normal star where you have atomic nuclei fusing.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Note: Neutrons aren't &lt;em&gt;made&lt;/em&gt; of protons and elections, but this transformation can happen by means of the weak nuclear force.  Normally neutrons aren't stable outside of the atomic nucleus -- instead the transformation would go the other way and a free neutron would decay into a proton and electron (there's also an anti-electron neutrino produced).  But under the intense gravitational pressure in a collapsed star, the neutrons are stable, which allows us to end up with neutron stars.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; This is of course a very approximate picture.  The link posted by Thomas Thernel has much more detail.  One good point to emphasize is that, as you might expect, the density is greater at the center of a neutron star than at its outskirts, so the star won't &lt;em&gt;really&lt;/em&gt; be all neutrons... you'll have more neutrons closer to the center, and more ordinary atomic nuclei further out.  Apparently some interesting sorts of structures can form from the remaining nuclei, even at the point where it's 90-95% neutrons.&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorUserId="11" LastEditorDisplayName="" LastEditDate="2010-11-11T18:58:29.930" LastActivityDate="2010-11-11T18:58:29.930" CommentCount="1" />
  <row Id="592" PostTypeId="2" ParentId="527" CreationDate="2010-11-11T20:30:00.733" Score="1" ViewCount="" Body="&lt;p&gt;The wavelength of the microwaves inside your microwave oven is somewhere around 3 cm, or a little more than an inch.  When the oven is on, the microwaves inside its cavity tend to form standing waves.  This means that there are nodes in the electromagnetic waves every 1.5 cm or so, which are places within the oven at which the amplitude of the waves is always zero.  Since the energy of the wave at a given point is proportional to the square of its amplitude, these nodes correspond to points at which the oven is not delivering any energy to the food.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, the existence of the nodes allows for a neat experiment that can be performed to measure the speed of light using marshmallows in a microwave.  See, for example, &lt;a href=&quot;http://orbitingfrog.com/blog/2008/05/13/measure-the-speed-of-light-using-your-microwave/&quot; rel=&quot;nofollow&quot;&gt;this website&lt;/a&gt;.  Briefly, the bottom of a microwave oven (without a spinning platter) is covered in marshmallows and the oven is turned on.  Some marshmallows start to melt, while others are unaffected -- corresponding to the locations of the nodes.  Measuring the distance between them can tell you the wavelength of the radiation which, combined with the stated frequency of the oven, lets you calculate an experimental value for the speed of light!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The existence of these nodes is the reason that most microwave ovens have a rotating platter: the food gets moved through the nodes, but also through the regions that deliver heating energy to it, and on average all of the food is supposed to heat.  In practice though it does often seem, at least anecdotally, that some parts of the food just receive a higher dose of energy than others.  This can lead to soup that is hot in some spots, and still cold in others -- a problem that can be somewhat alleviated by stirring the soup periodically, encouraging the soup to reach thermodynamic equilibrium more quickly.&lt;/p&gt;&#xA;" OwnerUserId="38" LastActivityDate="2010-11-11T20:30:00.733" CommentCount="2" />
  <row Id="594" PostTypeId="1" CreationDate="2010-11-11T20:51:33.617" Score="3" ViewCount="165" Body="&lt;p&gt;In a basic friction problem with Block A sliding on top of Block B, the direction of the friction force is usually explained as being simply the opposite of the direction of motion. So if Block A is sliding to the right, the friction force is pointing to the left. But this reasoning implicitly assumes that we are calculating friction force from the reference frame of Block B. What if we instead look at the problem from the reference frame of Block A? To Block A, it looks as if Block B is sliding to the left, so an observer on Block A would say that there is a friction force which, to oppose the direction of motion, points to the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems counterintuitive and probably wrong for the direction of friction force to depend on reference frame like this. Where is the flaw in the reasoning above? Are the two reference frames described above not exactly equivalent in a way that leads to the force changing directions?&lt;/p&gt;&#xA;" OwnerUserId="340" LastActivityDate="2010-11-11T20:56:13.847" Title="Does the friction force change directions with a change in reference frame?" Tags="&lt;mechanics&gt;&lt;friction&gt;" AnswerCount="1" />
  <row Id="595" PostTypeId="2" ParentId="594" CreationDate="2010-11-11T20:56:13.847" Score="6" ViewCount="" Body="&lt;p&gt;Everything in your  first paragraph is correct.  Friction opposes the relative motion of the blocks.  The thing that's tripping you up is that there are two friction forces.  One is the force of Block A on Block B and the other is the force of Block B on Block A.  These forces point opposite direction (and are equally size, by Newton's third law).&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-11T20:56:13.847" CommentCount="3" />
  <row Id="596" PostTypeId="2" ParentId="357" CreationDate="2010-11-11T22:06:53.270" Score="0" ViewCount="" Body="&lt;p&gt;Didn't the amateur radio guys (Hams) once launch a ballon like reflecting satellite?   Is it still in orbit?   Even at a few hundred kms, the delay would be in the ms.  Maybe the ISS is partly reflective.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-11-11T22:06:53.270" CommentCount="5" />
  <row Id="598" PostTypeId="1" AcceptedAnswerId="599" CreationDate="2010-11-11T22:12:19.927" Score="2" ViewCount="162" Body="&lt;p&gt;I need some books to learn the basis of linear operator theory and the spectral theory with, if it's possible, physics application to quantum mechanics. Can somebody help me?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-12T13:39:20.780" Title="Books for linear operator and spectral theory" Tags="&lt;books&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="599" PostTypeId="2" ParentId="598" CreationDate="2010-11-11T22:21:16.820" Score="3" ViewCount="" Body="&lt;p&gt;I think a good, and classic, reference for your case is the following,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://books.google.com/books?id=nZmpQgAACAAJ&quot; rel=&quot;nofollow&quot;&gt;Introductory functional analysis with applications&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The very last chapter of Kreyszig deals with Quantum Mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, once you've learned how to &quot;translate&quot; the language of Functional Analysis into that of Quantum Mechanics, you can go to more advanced texts in specific topics.&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2010-11-11T22:21:16.820" CommentCount="1" />
  <row Id="600" PostTypeId="2" ParentId="523" CreationDate="2010-11-11T22:44:45.560" Score="3" ViewCount="" Body="&lt;p&gt;To keep things simple, let's talk about two-qubit states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single qubit could have an orthonormal basis $\{|0\rangle, |1\rangle\}$.  But it could also have a different orthonormal basis $\{|+\rangle,|-\rangle\}$, where $$|+\rangle = \large(\normalsize|0\rangle \small+\normalsize |1\rangle\large)\normalsize / \sqrt{2}$$ $$|-\rangle = \large(\normalsize|0\rangle \small-\normalsize |1\rangle\large)\normalsize / \sqrt{2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No suppose the two states you're trying to distinguish are: $$|\psi\rangle = \large(\normalsize|0\rangle|0\rangle \small+\normalsize |1\rangle|+\rangle\large)\normalsize / \sqrt{2}$$ and $$|\phi\rangle = \large(\normalsize|0\rangle|1\rangle \small+\normalsize |1\rangle|-\rangle\large)\normalsize / \sqrt{2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the first qubit in each term is Alice's and the second qubit in each term is Bob's.&lt;br&gt;&lt;br&gt;Let's say Alice does a measurement in the $\{|0\rangle, |1\rangle\}$ basis.  After the measurement the state will be projected into a new state, but &lt;i&gt;which&lt;/i&gt; new state depends on the result of Alice's measurement.&lt;br&gt;&lt;br&gt;If Alice observes state $|0\rangle$: $$|\psi\rangle \rightarrow |0\rangle|0\rangle$$ $$|\phi\rangle \rightarrow |0\rangle|1\rangle$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Alice observes state $|1\rangle$:$$|\psi\rangle \rightarrow |1\rangle|+\rangle$$ $$|\phi\rangle \rightarrow |1\rangle|-\rangle$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Bob &lt;i&gt;knows&lt;/i&gt; Alice observes $|0\rangle$, then he can distinguish the two states by making a measurement in the $\{|0\rangle,|1\rangle\}$ basis.  Likewise, if Bob &lt;i&gt;knows&lt;/i&gt; Alice observes $|1\rangle$, then he can distinguish the two states by making a measurement in the $\{|+\rangle,|-\rangle\}$ basis.&lt;br&gt;&lt;br&gt;&lt;b&gt;But if Bob doesn't know the result of Alice's measurement until &lt;i&gt;after&lt;/i&gt; he makes his measurement, then he doesn't know &lt;i&gt;which&lt;/i&gt; measurement to make.&lt;/b&gt;&lt;br&gt;&lt;br&gt;If the true state is $|1\rangle|+\rangle$ but Bob guesses it's either $|0\rangle|0\rangle$ or $|0\rangle|1\rangle$, then he's going to make a measurement in the $\{|0\rangle,|1\rangle\}$ basis. This means he's equally likely to observe $|0\rangle$ or $|1\rangle$, since the true state of his qubit, $|+\rangle$, is an equal superposition of $|0\rangle$ and $|1\rangle$.  Note that even if the true state of his qubit were $|-\rangle$ (the only other possibility after Alice's measurement), Bob would still have been equally likely to observe $|0\rangle$ or $|1\rangle$.&lt;br&gt;&lt;br&gt;So after comparing results, Alice and Bob realize that Bob's measurement tells them nothing, and moreover the result of Alice's measurement by itself does nothing to distinguish the two initial states.&lt;br&gt;&lt;br&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;br&gt;&lt;br&gt;I'm going to try to give a more general rule for which states Alice and Bob can distinguish with local measurements only (no communication until afterwards.)&lt;br&gt;&lt;br&gt;If Alice and Bob aren't going to communicate until their measurements are done, then the combined measurement is effectively chosen before hand.  As Frédéric Grosshans points out in his answer, this measurement will be a projection into the eigenbasis of $A \otimes B$, where $A$ is a local observable that Alice can measure and $B$ is a local observable that Bob can measure.&lt;br&gt;&lt;br&gt;If $S_A$ is the set of eigenstates of $A$, and $S_B$ is the set of eigenstates of $B$, then the set of eigenstates of $A \otimes B$ is the direct product (the Cartesian product) of $S_A$ and $S_B$.&lt;br&gt;&lt;br&gt;As an example, if $S_A = S_B = \{|0\rangle, |1\rangle\}$, Alice and Bob will only distinguish $\{|0\rangle|0\rangle, |0\rangle|1\rangle, |1\rangle|0\rangle, |1\rangle|1\rangle\}$.  If $S_A = \{|0\rangle,|1\rangle\}$ and $S_B = \{|+\rangle,|-\rangle\}$, Alice and Bob will only distinguish $\{|0\rangle|+\rangle, |0\rangle|-\rangle, |1\rangle|+\rangle, |1\rangle|-\rangle\}$.&lt;br&gt;&lt;br&gt;If you have a set like my two entangled states above, or Frédéric Grosshans' four states $\{|0\rangle|0\rangle, |0\rangle|1\rangle, |1\rangle|+\rangle, |1\rangle|-\rangle\}$, then there's no choice of $A$ and $B$ that will suffice, because these states aren't in the direct product of the eigenstates of two local measurements.&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorUserId="11" LastEditorDisplayName="" LastEditDate="2010-11-29T13:57:51.490" LastActivityDate="2010-11-29T13:57:51.490" CommentCount="9" />
  <row Id="601" PostTypeId="1" CreationDate="2010-11-11T23:08:43.390" Score="6" ViewCount="183" Body="&lt;p&gt;What's the difference between something being labeled a &quot;nano-particle&quot; or it being called a &quot;molecule&quot;?&lt;/p&gt;&#xA;" OwnerUserId="344" LastEditorUserId="42" LastEditorDisplayName="" LastEditDate="2010-11-11T23:27:42.653" LastActivityDate="2010-12-05T14:58:39.560" Title="Nano-particle or Molecule?" Tags="&lt;solid-state-physics&gt;&lt;chemical-compounds&gt;&lt;nanoscience&gt;" AnswerCount="5" FavoriteCount="2" />
  <row Id="602" PostTypeId="2" ParentId="601" CreationDate="2010-11-11T23:13:19.140" Score="5" ViewCount="" Body="&lt;p&gt;It has to do with size. A molecule is in the range of picometers. A carbon-carbon bond is 154 picometers, so you expect most molecules to fall within the range 100-1000 picometers = 0.1 to 1 nanometers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A big molecule, such as a ribosome, or DNA, falls in the range of 10-1000 nanometers. Everything that falls in this range is &quot;nano&quot; by very nature. Nano-particle can refer, for example, to a cluster of gold atoms in the nanometer range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fascination about nanotech is the convergence between the bottom-up approach of synthesis chemists do (such as building large molecules from smaller ones) vs. the top-down approach physicists do (such as making smaller and smaller transistors on a chip). The encounter point is in the realm of nanotechnology, a realm where nature makes all its wonder happen since billion of years, and we are just starting to move the first steps.&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="217" LastEditorDisplayName="" LastEditDate="2010-11-11T23:20:08.627" LastActivityDate="2010-11-11T23:20:08.627" CommentCount="1" />
  <row Id="603" PostTypeId="1" AcceptedAnswerId="608" CreationDate="2010-11-11T23:21:55.747" Score="1" ViewCount="155" Body="&lt;p&gt;We may choose a non-rotating earth as our reference frame and ask ourselves: how about the planetary and stellar motions. A star at a distance of 10 million light years would turn around the earth in 24h with a velocity of 10^18 m/s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A friend once told me that actually articles have been published delving into this problem, e.g. to prove that fictitious forces emerge from the choice of such a bizar reference frame that ensure that the earth is still (somewhat) flattened at it's poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Does anybody know of such a publication?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I know that even such speeds of 10^18 m/s are not in contradiction with relativity because a limiting velocity only exists for exchange of information, which apparantly does not occur.&lt;br&gt;&#xA;Still: could anybody explain why such bizar velocities are allowed?&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2011-02-06T23:55:22.097" Title="Chosing a reference frame in which the earth is at rest and doesn't rotate" Tags="&lt;speed-of-light&gt;&lt;relative-motion&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="604" PostTypeId="2" ParentId="601" CreationDate="2010-11-11T23:25:46.927" Score="5" ViewCount="" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Nanoparticle&quot; rel=&quot;nofollow&quot;&gt;nanoparticle&lt;/a&gt; as typically used in nanotechnology refers to a particle with diameter on the order of 1-100 nanometers, or $10^{-9}$ to $10^{-7}$ meters.  However, it's not just a matter of size.  These particles are not typically &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Molecule&quot; rel=&quot;nofollow&quot;&gt;molecular&lt;/a&gt;&quot; in the sense that they are not stoichiometric units made out of atoms held together by covalent bonds.  Indeed, most nanoparticles are made out of very tiny pieces of crystals or metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stefano Borini gives the rough size of molecules, for a bit more detail, atoms can vary from 50-500 picometers, or $5\times10^{-11}$ to $5\times10^{-10}$ meters (a picometer is 1000 times smaller than a nanometer!) in diameter; small molecules (2-10 atoms) typically are on the order of 1-100 angstroms, or 100 picometers or so.  For instance, benzene (C6H6) has a side length of 140 picometers.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-11T23:25:46.927" />
  <row Id="605" PostTypeId="2" ParentId="598" CreationDate="2010-11-11T23:27:22.920" Score="3" ViewCount="" Body="&lt;p&gt;There are literally hundreds of introductory books on linear operators and their use in quantum mechanics. Finding the right one for you can be tricky. Let me recommend two that have been successful enough to be republished as Dover paperbacks. They're both by Thomas F. Jordan and are specifically oriented towards quantum mechanics applications. The most elementary one (perhaps too elementary) is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Quantum Mechanics in Simple Matrix Form&quot; (1986) Dover (2005) ISBN 0-486-44530-5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other one, significantly more sophisticated, but still very accessible is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Linear Operators for Quantum Mechanics&quot;  (1997) Dover (2006) ISBN 0-486-45329-4&lt;/p&gt;&#xA;" OwnerUserId="345" LastActivityDate="2010-11-11T23:27:22.920" />
  <row Id="606" PostTypeId="2" ParentId="603" CreationDate="2010-11-11T23:49:20.140" Score="2" ViewCount="" Body="&lt;p&gt;1) In this reference frame it is obvious that Earth is flattened at the poles.  There is a centrifugal force pushing out away from the axis of rotation with magnitude $\omega^2 r$, where $r$ is the distance from the axis.  If you're at the poles, $r = 0$ and you aren't pushed out at all.  If you're at the equator $r = R_{earth}$ and you're pushed out a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) (Not a direct answer) Rotation in special relativity is tricky.  For example, the ratio of the circumference of a disk to its diameter is not $\pi$.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Ehrenfest_paradox&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Ehrenfest_paradox&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-11T23:49:20.140" CommentCount="2" />
  <row Id="607" PostTypeId="2" ParentId="73" CreationDate="2010-11-11T23:58:30.617" Score="3" ViewCount="" Body="&lt;p&gt;Ah. Finally a topic I know something about !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many places in physics where the YB equation pops up. I can think of two at the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. Exactly solvable lattice models&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. Quantum Computation (QC)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the second application I find most exciting, so I'll focus on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The canonical reference (IMHO) on the link between the YB equation and QC is the wonderful paper by Lomonaco and Kauffmann (LK04) &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0401090&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/quant-ph/0401090&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In topological quantum computation, the hope is to be able to perform unitary operations on qubits by moving them around each other. A typical arena is a 2D electron gas, where our qubits are the quasiparticles of the system. In 2D when we exchange two objects we get a richer symmetry group than in 3D, where we get the permutation group whose eigenvalues $ \pm 1$ correspond to the case of bosons and fermions respectively. However, in 2D this symmetry group is enlarged to the braid group - one can exchange two objects by moving them around such that their worldlines &quot;braid&quot; around each other. This braiding cannot be eliminated by deforming the trajectories, because we don't have the third dimension to utilize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow to cut a long story short, the YBE can be shown diagramatically as a relationship between three particles under exchange (see fig. 1 on pg. 8 of above ref.). What LK04 then show is that solutions of the YBE are unitary matrices which are universal for quantum computation. In much the same way that any classical binary circuit can be built out of NAND gates along, any quantum circuit can be built out of a set of universal quantum gates.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-11T23:58:30.617" CommentCount="1" />
  <row Id="608" PostTypeId="2" ParentId="603" CreationDate="2010-11-12T00:01:38.013" Score="6" ViewCount="" Body="&lt;p&gt;Velocities in General Relativity can only be compared at a point, where local tangent planes coincide.  Talking about the velocities of far-away stars in any sort of absolute sense is an empty question.  Saying 'the coordinate velocity of Andromeda is 10^huge m/s' is, in a sense, not a statement about physics, but rather about your coordinate system.  In order to get a meaningful prediction, you would have to devise an experiment whereby you compare the two velocities--say, andromeda sends the earth a light signal at a preassigned 100 Hz.  An Earth-based observer then measures the redshift for the light signal, and then uses that to decide their relative velocities.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-11-12T00:01:38.013" CommentCount="3" />
  <row Id="609" PostTypeId="2" ParentId="601" CreationDate="2010-11-12T00:03:16.430" Score="8" ViewCount="" Body="&lt;p&gt;It's not simply a matter of size. Generally nanoparticles are a few nm to 100 nm and most molecules are smaller. But, for example, single chain of a high molecular weight polymer or a DNA molecule can easily be much larger than 100 nm which would put it outside the conventional &quot;nano&quot; range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distinction is somewhat fuzzy. Is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Fullerene&quot; rel=&quot;nofollow&quot;&gt;fullerene&lt;/a&gt; a molecule or a nanoparticle? &lt;a href=&quot;http://www.sigmaaldrich.com/materials-science/nanomaterials/Silsesquioxanes.html&quot; rel=&quot;nofollow&quot;&gt;silsesquioxanes&lt;/a&gt;? Why is polystyrene a &quot;molecule&quot; and a carbon nanotube a &quot;nanoparticle&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this &quot;gray area&quot; it seems to come down to &lt;strong&gt;how it behaves, and how it's used&lt;/strong&gt;. If the thing behaves in a similar (qualitatively) way as larger particles, and can be used in a similar way, you usually call it a nanoparticle. If the thing sticks together to make a solid or a liquid material, you usually call it a molecule. &lt;/p&gt;&#xA;" OwnerUserId="17" LastActivityDate="2010-11-12T00:03:16.430" CommentCount="2" />
  <row Id="610" PostTypeId="1" AcceptedAnswerId="611" CreationDate="2010-11-12T00:28:58.360" Score="4" ViewCount="225" Body="&lt;p&gt;A cylinder whose cross section is represented below is placed on an inclined plane. I would like to determine the maximum slope of the inclined plane so that the cylinder does not roll. The mass centre (CM) of the cylinder is at a distance r from the central axis. The cylinder consists of a cylindrical shell with mass $m_1$ and a smaller cylinder with mass $m_2$ placed away from the axis and rigidly attached to the larger cylinder. What is the influence of friction? Is it possible to establish the law of the movement? I think that the piece may roll upwards until it stops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure was copied from &lt;em&gt;Projecto Ciência na Bagagem&lt;/em&gt; -- &lt;a href=&quot;http://ead.liberato.com.br/~mitza/rt_cb_20.pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Cilindro desobediente&lt;/em&gt;&lt;/a&gt; &#xA;&lt;img src=&quot;http://i.stack.imgur.com/1ih2z.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: Depending on the initial conditions is it possible to find the highest point the cylinder rolls to, before stopping?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT2: From Institute and Museum of the History of Science -- &lt;a href=&quot;http://brunelleschi.imss.fi.it/museum/esim.asp?c=417020&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Cylinder on inclined plane&lt;/em&gt;&lt;/a&gt; [another cylinder]&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;When placed on the inclined plane,&#xA;  [another] cylinder  tends to&#xA;  roll upward, coming to a halt at a&#xA;  well-determined position.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="251" LastEditorUserId="251" LastEditorDisplayName="" LastEditDate="2010-11-12T18:48:19.640" LastActivityDate="2011-03-11T07:22:36.127" Title="Equilibrium and movement of a cylinder with asymmetric mass centre on an inclined plane" Tags="&lt;mechanics&gt;&lt;classical-mechanics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="611" PostTypeId="2" ParentId="610" CreationDate="2010-11-12T01:17:02.933" Score="5" ViewCount="" Body="&lt;p&gt;The effect of friction is to make the cylinder roll down the ramp rather than slide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find an equilibrium angle, use virtual work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\phi$ changes by a small amount $d\phi$, as the cylinder rolls, then everything goes down a little (neglecting at first the small interior cylinder's upward movement) because you're moving down the ramp.  You move $R d\phi$ down the ramp, and lose elevation $\sin \Phi R d\phi$.  The total work done by gravity is $(m_1 + m_2) g \sin\Phi R d\phi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the interior cylinder rises with respect to the center of the big cylinder by an amount $(R - R_2) d\phi$.  The work done by gravity on the little cylinder is $g m_2 (R-R_2) d\phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equilibrium is achieved when these are equal, so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_2 (R - R_2) = (m_1 + m_2) R \sin\Phi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin\Phi = \frac{m_2(R-R_2)}{(m_1+m_2)R} = \frac{r}{R}$&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-03-11T07:22:36.127" LastActivityDate="2011-03-11T07:22:36.127" CommentCount="4" />
  <row Id="612" PostTypeId="2" ParentId="115" CreationDate="2010-11-12T02:15:34.303" Score="3" ViewCount="" Body="&lt;p&gt;There is rather nice function in Mathematica 7, which allows one to integrate over an arbitrary complicated region. It is &lt;strong&gt;Boole&lt;/strong&gt;:[True,False]$\to${1,0}. Below is just an example taken from Mathematica Documentation Center. If you have a 2D area defined by the inequality $4 x^4-4 x^2+y^2\leq 0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/9hlVb.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can integrate any function $f(x,y)$ over this domain as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Integrate[f[x,y] Boole[y^2 - 4 x^2 + 4 x^4 &amp;lt;= 0], {x, -Infinity,Infinity}, &#xA;{y, -Infinity,Infinity}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For example, if $f(x,y)$ is unity then it gives you the total volume of the integration domain:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;In[1]:= Integrate[Boole[y^2 - 4 x^2 + 4 x^4 &amp;lt;= 0], {x, -Infinity,Infinity}, &#xA;{y, -Infinity,Infinity}]&#xA;Out[1]= 8/3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In fact, you can use any condition you want, including that is determining your islands of stability. Numerical integration is also possible:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;In[1]:= NIntegrate[Boole[y^2 - 4 x^2 + 4 x^4 &amp;lt;= 0], {x, -Infinity,Infinity}, &#xA;{y, -Infinity,Infinity}]&#xA;Out[1]= 2.66667&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="225" LastEditorUserId="225" LastEditorDisplayName="" LastEditDate="2010-11-12T02:21:45.140" LastActivityDate="2010-11-12T02:21:45.140" CommentCount="2" />
  <row Id="613" PostTypeId="1" CreationDate="2010-11-12T03:07:27.480" Score="2" ViewCount="282" Body="&lt;p&gt;Fastest known method for computing Potts model partition function (Bedini and Jacobsen's &lt;a href=&quot;http://arxiv.org/abs/1003.4847&quot; rel=&quot;nofollow&quot;&gt;&quot;A tree-decomposed transfer matrix for computing exact Potts model partition functions for arbitrary graphs, with applications to planar graph colourings&quot;&lt;/a&gt;) uses a &quot;tree decomposition&quot; of the square lattice. How can this method be extended to computing two-point correlation function? An obvious approach is to re-run it for every pair of vertices fixed to a particular value, but that ends up with a lot of redundant computation, is there a more efficient way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example splitting scheme for grid&#xA;&lt;img src=&quot;http://yaroslavvb.com/upload/grid-decompose.png&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="331" LastEditorDisplayName="" LastEditDate="2010-11-18T04:29:56.610" LastActivityDate="2010-11-18T04:29:56.610" Title="Two-point correlation function for planar Potts model" Tags="&lt;statistical-mechanics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="614" PostTypeId="1" CreationDate="2010-11-12T07:16:03.527" Score="1" ViewCount="241" Body="&lt;p&gt;[I edited the question according to Mark's and Grisha's answers.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider two point-like particles of equal mass colliding centrally in 2D. The final directions of the momenta of the two particles are not determined by conservation of energy and momentum: they can go anywhere.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Under which circumstances the direction of the two particles &lt;em&gt;is&lt;/em&gt; determined?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="229" LastEditorUserId="229" LastEditorDisplayName="" LastEditDate="2010-11-12T16:17:43.343" LastActivityDate="2010-11-13T04:48:22.357" Title="Where do atoms go after collision?" Tags="&lt;classical-mechanics&gt;&lt;paradox&gt;&lt;scattering&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="615" PostTypeId="1" AcceptedAnswerId="654" CreationDate="2010-11-12T07:23:28.107" Score="4" ViewCount="470" Body="&lt;p&gt;Corrective glasses are usually intended to help focus light on your retina.  Supposing I had good vision already, but simply wanted more light, could I make glasses that would send more light into my eye without magnifying, focusing, or distorting the image?  (Ignore chromatic aberration.)&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-06-05T21:40:31.800" Title="Is it possible to make glasses that make everything brighter, but do not magnify or focus?" Tags="&lt;optics&gt;&lt;geometric-optics&gt;" AnswerCount="8" FavoriteCount="1" />
  <row Id="616" PostTypeId="2" ParentId="615" CreationDate="2010-11-12T07:32:06.507" Score="3" ViewCount="" Body="&lt;p&gt;No, not using optics; you'd need some sort of active amplification. Without magnifying, focusing, or distorting the image, the glasses would need to be flat. But consider the energy in the light entering the front face and leaving the rear face of the lenses - what you want would require there to be more coming out than going in, which violates energy conservation. The extra energy has to come from somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question caused a lot more discussion than I expected. I would like to try to explain my answer a little further, because people seem to have misunderstood me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me say first, I want to consider the case of using only lenses and similar devices. In j.c.'s answer we have already established that what you ask is possible using amplification, frequency conversion, or thermal imaging -- all of which are used in night vision / image intensification technology. Browsing Wikipedia I found out that some animals have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Tapetum_lucidum&quot; rel=&quot;nofollow&quot;&gt;tapetum lucidum&lt;/a&gt;, a reflecting membrane behind the retina which turns around any light that made it through the retina without getting absorbed, allowing it to pass through the retinal cells once more and be absorbed on the rebound. This increases the efficiency of the retina, which is also a way to do what you ask, but I'm discounting it since it is not available to us humans without resorting to surgery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now down to business. We want to create corrective glasses that make your vision brighter, but with no&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;magnifying&lt;/li&gt;&#xA;&lt;li&gt;focusing&lt;/li&gt;&#xA;&lt;li&gt;or distortion.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This means, considering the following diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/77EaW.png&quot; alt=&quot;a wavefront entering a mystery &amp;quot;?&amp;quot; box and emerging the same&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The optical system we want to create is represented by the &quot;?&quot; box. The wavefront on the left has to be the same as the wavefront on the right, only brighter. (Except for the effects of propagation across the distance that the box occupies, but let's ignore that.) Notice that I have drawn the wavefronts as squiggly lines with arrows all over them. This is because the images that our eyes observe are not the same as the point sources we deal with in undergraduate optics, they are made up of many different rays traveling in different directions and with different phases (i.e. running a little ahead of or behind each other.) So it's not enough to just capture &quot;more light&quot; into the system, it has to be in the right place, traveling at the proper angle, with the right amount of delay compared to the other rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now look at my crude drawing of an eye looking at a triangle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/YbT6L.jpg&quot; alt=&quot;an eye regarding a triangle, with some light rays from the triangle entering the eye and some missing it&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have drawn the outermost rays of the outermost points of the triangle that enter the eye. I have also drawn some rays that go off in different directions that don't enter the eye. To make the image brighter, you would have to ensure that more of the rays that diverge from a single point on the triangle, also converge on the same point on the retina. But how? If the eye's iris and lens were infinitely large, then they would naturally converge there. But we don't have huge eyes, so we would have to use external lenses to achieve that. There is no way, though, to build a system of lenses that does that, without changing the rays that are already on the right track. You can't refract some of the rays and not others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what I definitely want to make clear is that the idea of &quot;if the glasses can capture more light, then it will brighten the image&quot; is a partial misconception. I say 'partial' because what coneslayer says is true: binoculars with a bigger aperture will produce a brighter image. However, this is only true if the brightness was already reduced due to a smaller aperture! Think of it this way: the &lt;em&gt;absence&lt;/em&gt; of binoculars is equivalent to 1x binoculars with an infinitely large aperture!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, the brightness of the image on your retina is limited by the natural aperture of your eye: your pupil. This is why your pupils get bigger in the dark. And you can't make your pupils any bigger than you already do instinctively, at least not without resorting to surgery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, I would like to thank everyone who contributed here for making it such a lively discussion. This forced me to think more carefully about the problem than I did at first, and get all my arguments in order.&lt;/p&gt;&#xA;" OwnerUserId="107" LastEditorUserId="107" LastEditorDisplayName="" LastEditDate="2010-11-13T22:34:19.797" LastActivityDate="2010-11-13T22:34:19.797" CommentCount="10" />
  <row Id="617" PostTypeId="2" ParentId="614" CreationDate="2010-11-12T07:34:39.727" Score="0" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In a frame of reference where one of the particles is initially at rest, the final direction of its momentum is determined by conservation of energy and momentum. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This statement is incorrect. Both particles can have transverse components of momentum equal in absolute value but into opposite directions. Just consider slightly non-central collision of two billiard balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[update]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy-momentum conservation law determines only so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinematics&quot; rel=&quot;nofollow&quot;&gt;kinematics&lt;/a&gt; of the collision. The directions of final particles is the question of &lt;a href=&quot;http://en.wikipedia.org/wiki/Dynamics_%28mechanics%29&quot; rel=&quot;nofollow&quot;&gt;dynamics&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine the directions of final particles you should know the equations of motion at each moment of time (more generally, the interaction law). For classical mechanics it is &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion&quot; rel=&quot;nofollow&quot;&gt;Newton's laws of motion&lt;/a&gt;. Solving these equations gives you particle's trajectories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summarizing, there is no general prescription to determine the angular distribution of colliding particles, it depends on details how they interact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, if you don't know the interaction law, the measuring of angular distribution is a good method to obtain some information about it. This problem is usually called as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Inverse_scattering_problem&quot; rel=&quot;nofollow&quot;&gt;inverse scattering problem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="225" LastEditorUserId="225" LastEditorDisplayName="" LastEditDate="2010-11-12T10:20:14.040" LastActivityDate="2010-11-12T10:20:14.040" CommentCount="8" />
  <row Id="618" PostTypeId="2" ParentId="614" CreationDate="2010-11-12T07:56:26.767" Score="2" ViewCount="" Body="&lt;p&gt;(Edited to reflect updated question).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution has a single parameter, which we can take to be the angle the impacting particle goes after the collision.  A particular solution is determined only after this parameter is given.  Usually we would talk about the collision of, for example, two spheres, in which case this parameter could be related to whether the spheres are  colliding head on or glancing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in the frame where one particle is initially moving and the other is not, one solution is for the moving particle to come to a halt and for the stationary particle to take up all its motion.  Another solution is for both particles to fly off from the impact site at 45 degree angles to the original motion, each with $\sqrt{2}/2$ the initial velocity.  Both of these scenarios conserve momentum and kinetic energy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many other solutions.  In this frame, the conservation of momentum says&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec{p} = \vec{p_1} + \vec{p_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\vec{p}$ the momentum before the collision and $\vec{p_1}$ and $\vec{p_2}$ the momenta after.  Squaring,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p^2 = p_1^2 + 2\vec{p_1}\cdot\vec{p_2} + p_2^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kinetic energy is proportional to $p^2$, so conservation of kinetic energy gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p^2 = p_1^2 + p_2^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These last two combine to give&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec{p_1}\cdot\vec{p_2} = 0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the momenta point at a 90-degree angle.  However, any one particle can choose any angle within 90 degrees of the initial direction of motion to go after the collision.  Once this direction is chosen, the direction of motion of the other particle, as well as their momenta, are fixed.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-12T08:39:41.800" LastActivityDate="2010-11-12T08:39:41.800" CommentCount="1" />
  <row Id="619" PostTypeId="1" AcceptedAnswerId="623" CreationDate="2010-11-12T08:06:36.197" Score="1" ViewCount="188" Body="&lt;p&gt;It seems all we're capable of accelerating currently are atomic particles. Why can't we, say, accelerate a clock to relativistic speeds?&lt;/p&gt;&#xA;" OwnerUserId="202" LastActivityDate="2010-11-12T14:12:16.450" Title="Why is it so hard to accelerate macroscopic objects?" Tags="&lt;general-relativity&gt;&lt;accelerator-physics&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="620" PostTypeId="2" ParentId="601" CreationDate="2010-11-12T08:07:30.607" Score="1" ViewCount="" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecule&quot; rel=&quot;nofollow&quot;&gt;molecule&lt;/a&gt; is the collection of atoms all connected by covalent bonds, while a &lt;a href=&quot;http://en.wikipedia.org/wiki/Nanoparticle&quot; rel=&quot;nofollow&quot;&gt;nanoparticle&lt;/a&gt; is any kind of particle&lt;sup&gt;1&lt;/sup&gt; with a size in the nanometer scale, that is (usually) between 1-999 nm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nanoparticle&quot; rel=&quot;nofollow&quot;&gt;Although the size of most molecules would fit into the above outline, individual molecules are usually not referred to as nanoparticles.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1)&lt;/sup&gt;For example crystals and amorphous bulks&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2010-11-12T08:07:30.607" />
  <row Id="621" PostTypeId="2" ParentId="619" CreationDate="2010-11-12T08:13:34.967" Score="3" ViewCount="" Body="&lt;p&gt;How about Newton's law $F=ma$? The force needed to accelerate 1 kg is 27 orders of magnitude ($10^{27}$) bigger then to accelerate a proton.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-12T08:13:34.967" />
  <row Id="622" PostTypeId="1" CreationDate="2010-11-12T08:29:13.930" Score="3" ViewCount="242" Body="&lt;p&gt;While at lunch with my lab group we got into discussion of the different types of scattering phenomenon that we encounter in everyday life and physical experiments. We ended up listing about a dozen or so types of scattering that we see constantly, but I left unfulfilled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to compile a list of scattering phenomenon and a brief description. At the very least this will be helpful for those taking the physics GRE in the near future, otherwise its a good exercise for anyone interested in physics, even those in professional physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll Begin to give you the basic jist of what I'm looking for, feel free to elaborate on the description as much as you want. An equation is useful if it's easily expressed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Compton scattering&lt;/strong&gt;: Scattering of a photon off of a particle, photon imparts momentum to the particle and loses some of its energy, the change in wavelength is described by &#xA;$$\Delta_{\lambda} = \frac{h}{mc}(1-\cos{\theta})$$&#xA;It is independent of the frequency of the incoming photon and inversely proportional to the mass of the particle scattered from. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Raleigh Scattering&lt;/strong&gt;: Elastic Scattering of photons off of molecules and other (micro and macro) particles, photons do not lose energy but change direction. Has a strong dependence on frequency and is the reason the sky is blue (and why you can see laser pointer beams in the dark). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I don't have the reputation to create tags yet, and there are no tags relevant to this question (big-list scattering) please feel free to retag as necessary &lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2010-11-12T09:11:50.383" LastActivityDate="2010-11-14T03:39:56.760" CommunityOwnedDate="2010-11-18T19:14:25.597" Title="List of Scattering Phenomena" Tags="&lt;particle-physics&gt;&lt;mechanics&gt;&lt;scattering&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="623" PostTypeId="2" ParentId="619" CreationDate="2010-11-12T08:44:03.670" Score="7" ViewCount="" Body="&lt;p&gt;The obvious answer is their mass, accelerating a subatomic particle to relativistic velocity takes many orders of magnitude less energy than accelerating a macroscopic particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More subtly, we can only accelerate subatomic particles that we can manipulate with electromagnetic fields, nobody has a proposed viable experiments to study relativistic neutron or neutrino reactions because we really have no way to manipulate them. What this means is that we need relatively light, charged, particles that we can accelerate to relativistic velocity in a reasonable space with reasonable amount of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which brings us to our next point. Modern accelerators use what are called &quot;RF cavities&quot; to accelerate charged particles like protons and electrons, these are effectively giant pulsed capacitors with the particle flying in the space between the plates. We are governed by material science and the permiativity of the vacuum and have a limit to the maxium voltage we can put across the plates, and hence the maximum energy we can impart to a particle per unit length of the cavity. For modern accelerators this is on the order of 1 MeV per cm. 1 MeV on an electron goes a long way (which is why your TV works). 1 MeV on a baseball barely does anything. If the LHC seems big, an accelerator to collide two baseballs with the same velocity would be wider than the solar system (using modern technology). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More realistically though, we accelerate particles so we can study the physics during collisions of two particles. When we collide electrons and positrons for instance, nearly 100% of the energy carried by the particle is deposited into the reaction if they collide head on, its this fact that lets us see the creation of new and exotic particles following the collision (And why LEP began seeing Z bosons as soon as they crossed the 180GeV threshold). When we move to the LHC for example, we are colliding two protons- which, roughly speaking are sacks of consisting of 3 quarks. And the momentum (and therefore the energy) of the proton is more or less evenly distributed among the three quarks. So the energy available to us in the collision is actually about 6 times less than an eqiuivalent electron-positron collider. Imagine doing this with a baseball, which is composed of trillions of trillions of particles. The energy that each particle carries is a tiny fraction of the total energy of the baseball, therefore every single collision event only has a tiny fraction of the total energy of the system, and in the end we get no useful reaction out of a collision that could not have just as easily been accomplished by putting two televisions face to face.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2010-11-12T08:44:03.670" CommentCount="2" />
  <row Id="624" PostTypeId="2" ParentId="615" CreationDate="2010-11-12T08:56:12.833" Score="5" ViewCount="" Body="&lt;p&gt;Expanding a bit on what ptomato said, I think there's a problem with the second law of thermodynamics regardless of any problems with the first law.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the universe consisted of two things: a black body the shape of eyeballs and the microwave background.  Then the black body would come into thermal equilibrium with the microwave background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we introduce these glasses that make things brighter, placing them in front of the black body.  Suddenly the black body has more light coming on to it.  It absorbs all this light, so it has to heat up.  We made heat flow from a colder place (the microwave background) to a hotter place (the black body) by a passive optical phenomenon.  That violates the second law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;It turns out this principle is called &quot;conservation of brightness&quot; or &quot;conservation of surface brightness&quot; and is well-known in optics.  See,  for example, &lt;a href=&quot;http://www.cv.nrao.edu/course/astr534/Brightness.html&quot; rel=&quot;nofollow&quot;&gt;http://www.cv.nrao.edu/course/astr534/Brightness.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-29T15:08:37.650" LastActivityDate="2010-11-29T15:08:37.650" />
  <row Id="625" PostTypeId="1" CreationDate="2010-11-12T08:57:27.627" Score="12" ViewCount="606" Body="&lt;ul&gt;&#xA;&lt;li&gt;Classical mechanics: $t\mapsto \vec x(t)$, the world is described by particle trajectories $\vec x(t)$ or $x^\mu(\lambda)$, i.e. the Hilbert vector is the particle coordinate function $\vec x$ (or $x^\mu$), which is then projected into the space parametrized by the &quot;coordinate&quot; time $t$ or the relativistic parameter $\lambda$ (which is not necessarily monotonous in $t$).&lt;br&gt;&#xA;Interpretation: For each parameter value, the coordinate of a particle is described.&lt;br&gt;&#xA;Deterministic: The particle position itself&lt;/li&gt;&#xA;&lt;li&gt;Quantum mechanics: $x^\mu\mapsto\psi(x^\mu)$, (sometimes called &quot;the &lt;em&gt;first quantization&lt;/em&gt;&quot;) yields &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_mechanics&quot; rel=&quot;nofollow&quot;&gt;Quantum mechanics&lt;/a&gt;, where the Hilbert vector is the wave function (being a field) $|\Psi\rangle$ that is for example projected into coordinate space so the parameters are $(\vec x,t)$ or $x^\mu$.&lt;br&gt;&#xA;Interpretation: For each coordinate, the quantum field describes the charge density (or the probability of measuring the particle at that position if you stick with the non-relativistic theory).&lt;br&gt;&#xA;Deterministic: The wave function&lt;br&gt;&#xA;Non-deterministic: The particle position&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_field_theory&quot; rel=&quot;nofollow&quot;&gt;Quantum Field Theory&lt;/a&gt;: $\psi(x^\mu)\mapsto \Phi[\psi]$, (called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_field_theory#Second_quantization&quot; rel=&quot;nofollow&quot;&gt;second quantization&lt;/a&gt; despite the fact that now the wave field is quantized, not the coordinates for a second time) basically yields a functional $\Phi$ as Hilbert vector projected into quantum field space parametrized by the wave functions $\psi(x^\mu)$.&lt;br&gt;&#xA;Interpretation: For each possible wave function, the (to my knowledge nameless) $\Phi$ describes something like the probability of that wave function to occur (sorry, I don't know how to formulate this better, it's not really a probability). One effect is for example particle generation, thus the notion &quot;particle&quot; is fishy now&lt;br&gt;&#xA;Deterministic: The functional $\Phi$  &#xA;Non-deterministic: The wave function $\psi$ and the &quot;particle&quot; position&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, could there be a third quantization $\Phi[\psi(x^\mu)] \mapsto \xi\{\Phi\}$? What would it mean? And what about fourth, fifth, ... quantization? Or is second quantization something ultimate?&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2010-11-15T12:16:17.110" LastActivityDate="2011-01-19T01:02:19.700" Title="Is a &quot;third quantization&quot; possible?" Tags="&lt;quantum-field-theory&gt;&lt;quantization&gt;&lt;second-quantization&gt;" AnswerCount="8" FavoriteCount="3" />
  <row Id="626" PostTypeId="2" ParentId="619" CreationDate="2010-11-12T08:59:49.807" Score="1" ViewCount="" Body="&lt;p&gt;It's not particularly hard to accelerate anything, it just requires an energy source and time. &#xA;The problems are merely practical:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;To go really fast you require a large vacuum. So in the case of particles we have a 27km underground tunnel (LHC) and in the case of space probes we have space itself.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the case of probes, there's the necessity of making them self propelled, so a lot of fuel must be accelerated too, requiring even more energy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Given the costs, we tend to accelerate stuff as little as possible. In the case of space probe, once they reach the escape velocity of the solar system, nothing more is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fastest large object is currently Voyager 1, running at $5.8\times10^{-5}c$ or 0.006% of the speed of light.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-12T08:59:49.807" />
  <row Id="627" PostTypeId="2" ParentId="622" CreationDate="2010-11-12T09:14:28.167" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mie_theory&quot; rel=&quot;nofollow&quot;&gt;Mie scattering&lt;/a&gt;: The exact Theory to which Raleigh Scattering is the small particle approximation.&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2010-11-12T09:14:28.167" CommunityOwnedDate="2010-11-18T19:14:25.597" />
  <row Id="628" PostTypeId="2" ParentId="615" CreationDate="2010-11-12T09:26:43.740" Score="1" ViewCount="" Body="&lt;p&gt;I'm assuming that brighter means more light power hitting your retina. Focusing light increases the &lt;em&gt;Intensity&lt;/em&gt; of light (Power per unit area), but there is no passive way to increase the light power that hits your retina passively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to avoid an appeal to thermodynamics and instead try to reason this to you somewhat intuitively. The light power that hits your eyes, is essentially the amount of photons that are hit your retina. There are ways to increase the energy of each individual photon (there are special crystals that can double or even triple the frequency of light), but always at the expense of the number of photons (i.e. the power stays at most the same as it was before doubling). You are essentially asking for a way to &lt;em&gt;passively&lt;/em&gt; increase the number of photons hitting your eye. The simplest way is to merely turn up the brightness of the light in the room. There is no magic trick you can do with a piece of glass that will spit out more photons than you originally had. Let alone one that will increase the amount of information that you have available to you. &lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2010-11-12T09:26:43.740" CommentCount="1" />
  <row Id="629" PostTypeId="1" AcceptedAnswerId="6107" CreationDate="2010-11-12T10:01:54.560" Score="16" ViewCount="1123" Body="&lt;p&gt;When you're in a train and it slows down, you experience the push forward from the deceleration which is no surprise since the force one experiences results from good old $F=m a$. However, the moment the train stops one is apparently pulled backwards. But is that a real physical effect or just the result from leaning backwards to compensate the deceleration and that force suddenly stopping?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So far the answers basically agree that there are two spring forces involved, for one thing oneself as already guessed by me and for the other the vehicle itself as first suggested in &lt;a href=&quot;http://physics.stackexchange.com/questions/629/why-does-one-experience-a-short-pull-in-the-wrong-direction-when-a-vehicle-stops/638#638&quot;&gt;Robert's answer&lt;/a&gt;. Also, as &lt;a href=&quot;http://physics.stackexchange.com/questions/629/why-does-one-experience-a-short-pull-in-the-wrong-direction-when-a-vehicle-stops/670#670&quot;&gt;Gerard suggested&lt;/a&gt; the release of the brakes and some other friction effects might play a role. So let's be more precise with the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which effect &lt;strong&gt;dominates&lt;/strong&gt; the wrong-pull-effect? And thus, who can reduce it most:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;the traveler&lt;/li&gt;&#xA;  &lt;li&gt;the driver&lt;/li&gt;&#xA;  &lt;li&gt;the vehicle designer?&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt; Let's make this more interesting: I'm setting up a bounty of &lt;s&gt;50&lt;/s&gt; 100 (see edit below) for devising an experiment to explain this effect or at least prove my explanation right/wrong, and by the end of this month I'll award a second bounty of &lt;s&gt;200&lt;/s&gt; 150 for what I subjectively judge to be the best answer describing either:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;an accomplished experiment (some video or reproducibility should be included)&lt;/li&gt;&#xA;&lt;li&gt;a numerical simulation&lt;/li&gt;&#xA;&lt;li&gt;a rigorous theoretical description&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;update&lt;/strong&gt; since I like both the suggestions of &lt;a href=&quot;http://physics.stackexchange.com/questions/629/why-does-one-experience-a-short-pull-in-the-wrong-direction-when-a-vehicle-stops/5052#5052&quot;&gt;QH7&lt;/a&gt; and &lt;a href=&quot;http://physics.stackexchange.com/questions/629/why-does-one-experience-a-short-pull-in-the-wrong-direction-when-a-vehicle-stops/4926#4926&quot;&gt;Georg&lt;/a&gt;, I decided to put up a second bounty of 50 (thus reducing the second bounty to 150 however)&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-02-23T15:52:36.960" LastActivityDate="2011-02-28T17:02:18.063" Title="Why does one experience a short pull in the wrong direction when a vehicle stops?" Tags="&lt;classical-mechanics&gt;" AnswerCount="16" CommentCount="7" FavoriteCount="4" />
  <row Id="630" PostTypeId="1" AcceptedAnswerId="1064" CreationDate="2010-11-12T10:47:31.417" Score="3" ViewCount="372" Body="&lt;p&gt;It seems that if the coherence length of a laser is big enough, it is possible to observe a (moving) interference picture by combining them. Is it true? How fast should photo-detectors be for observing of the interference of beams from two of the &quot;best available&quot; lasers? What is the coherence length of the best-available laser? More specifically, does there exist any (optical single-wavelength) laser with coherence length exceeding 500 meters?&lt;/p&gt;&#xA;" OwnerUserId="76" LastEditorUserId="42" LastEditorDisplayName="" LastEditDate="2010-11-12T14:39:54.700" LastActivityDate="2010-11-18T14:31:46.883" Title="Is it possible to observe interference from 2 independent optical lasers?" Tags="&lt;optics&gt;&lt;laser&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="631" PostTypeId="2" ParentId="615" CreationDate="2010-11-12T11:18:42.007" Score="3" ViewCount="" Body="&lt;p&gt;You want an &lt;a href=&quot;http://en.wikipedia.org/wiki/Optical_amplifier&quot; rel=&quot;nofollow&quot;&gt;optical amplifier&lt;/a&gt;. It uses stimulated emission to amplify the light, like in laser, but without the feedback loop. Generally used for amplifying fiber optics signal to greater than 200 km. Making wearable glasses with this wouldn't be easy. Inverting electron population of a piece of glass with enough gain to be useful. As ptomato noted it is an active device, therefore in needs a power source to feed it. But I don't think your are concerned about this.&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="62" LastEditorDisplayName="" LastEditDate="2010-11-12T13:37:11.297" LastActivityDate="2010-11-12T13:37:11.297" CommentCount="2" />
  <row Id="632" PostTypeId="2" ParentId="630" CreationDate="2010-11-12T11:44:12.267" Score="3" ViewCount="" Body="&lt;p&gt;personally i work with two pulsed laser source and i managed to use one as master and the other one as slaver succeding in this way to have a perfect sincronization in wavelenght,  phase and pulse frequency.. i've never tried to see interference path between them (I use these just for microscopy purpose) but as Tobias was saying I don't see why that should not work as long as a stable sincronyzation is kept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really don't know about the 500m coherence length.. &lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2010-11-12T11:44:12.267" CommentCount="1" />
  <row Id="633" PostTypeId="2" ParentId="566" CreationDate="2010-11-12T11:47:54.103" Score="2" ViewCount="" Body="&lt;p&gt;Not being an exhausting list. Some phenomena that allows for this to happens is non-linear effects. A short pulse allow a higher intensity for a small period to time which enhance non-linear effects such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Two-photon_absorption&quot; rel=&quot;nofollow&quot;&gt;two-photon absorption&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Self-focusing&quot; rel=&quot;nofollow&quot;&gt;self-focusing&lt;/a&gt;. Two photon absorption will make a material absorb light which it wouldn't if light was low intensity. Self-focusing would make the light focus more which would decrease area which the heat can be diffused, or increase intensity for two or more photon absorption. The self-defocusing is possible too which would allow for decrease damage in the material if pulsed.&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="62" LastEditorDisplayName="" LastEditDate="2010-11-15T13:43:33.867" LastActivityDate="2010-11-15T13:43:33.867" CommentCount="2" />
  <row Id="634" PostTypeId="1" AcceptedAnswerId="689" CreationDate="2010-11-12T11:56:28.333" Score="6" ViewCount="425" Body="&lt;p&gt;As I understand, Dirac equation was first interpreted as a wave equation following the ideas of non relativistic quantum mechanics, but this lead to different problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation was then reinterpreted as a field equation and it is now a crucial part of quantum field theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: could you provide me a reference (paper, book) that explains this evolution, including the different historical steps, etc. ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a good knowledge of QM and I studying field theories, but I would like to have a clearer view on this historical evolution.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="42" LastEditorDisplayName="" LastEditDate="2010-11-12T14:39:01.253" LastActivityDate="2011-01-16T16:51:28.540" Title="Evolution in the interpretation of the Dirac equation" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;history&gt;&lt;dirac-equation&gt;" AnswerCount="6" FavoriteCount="2" />
  <row Id="635" PostTypeId="2" ParentId="634" CreationDate="2010-11-12T12:09:43.153" Score="5" ViewCount="" Body="&lt;p&gt;I think that the first volume of the series &quot;The Quantum Theory of Fields&quot;, by Steven Weinberg, is a good text to understand the origin of Dirac equation, QFT, and all these kind of topics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe Weinberg's books are not the best for a first course in QFT (or in General Relativity, he has also a great book on this topic), but his great coverage and unique point of view make them very worth reading.&lt;/p&gt;&#xA;" OwnerUserId="306" LastActivityDate="2010-11-12T12:09:43.153" />
  <row Id="636" PostTypeId="2" ParentId="625" CreationDate="2010-11-12T12:17:51.880" Score="2" ViewCount="" Body="&lt;p&gt;Wow, that's a very good question. Unfortunately, I can't write down a question, because a haven't got one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, I tried to found something related to third quantization in arxiv, and surprisingly (or not so surprisingly), you can find some papers related to this new step. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to name a few:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0606021&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/gr-qc/0606021&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/9212044&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9212044&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really hope that someone can get a full answer here.&lt;/p&gt;&#xA;" OwnerUserId="306" LastActivityDate="2010-11-12T12:17:51.880" CommentCount="2" />
  <row Id="637" PostTypeId="2" ParentId="630" CreationDate="2010-11-12T12:20:58.967" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.home.agilent.com/agilent/product.jspx?pn=Q85026A&quot; rel=&quot;nofollow&quot;&gt;Detectors as fast as 50 GHz&lt;/a&gt; can be easily bought (if you have the money :P). This means that if the difference in the frequency of the lasers is smaller then 50 GHz or wavelength difference is smaller than 60 pm then you can detect the &lt;a href=&quot;http://en.wikipedia.org/wiki/Beat_%28acoustics%29&quot; rel=&quot;nofollow&quot;&gt;beating&lt;/a&gt; using these fast detectors. This wavelength difference can be achieved (sorry but I am not in the mood of finding a papers reporting said achievement), you can even phase-lock the lasers (but then I would not call them independent anymore)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coherence length of 500 m means &lt;a href=&quot;http://en.wikipedia.org/wiki/Coherence_time&quot; rel=&quot;nofollow&quot;&gt;coherent time&lt;/a&gt; of $t=500/c = 1.7 \mu s$ which means a bandwidth of $\Delta\nu = 600 kHz$, which can be achieved by &lt;a href=&quot;http://www.rp-photonics.com/narrow_linewidth_lasers.html&quot; rel=&quot;nofollow&quot;&gt;external-cavity&lt;/a&gt; diode lasers.&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="62" LastEditorDisplayName="" LastEditDate="2010-11-12T13:01:56.480" LastActivityDate="2010-11-12T13:01:56.480" />
  <row Id="638" PostTypeId="2" ParentId="629" CreationDate="2010-11-12T12:39:32.903" Score="3" ViewCount="" Body="&lt;p&gt;I think since the acceleration of the motion of the vehicle towards voyage direction is always &amp;lt;= zero for the slowing down process, the explanation has to be searched somewhere else since this cannot explain your &quot;kickback&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my opinion it might be that the slowing down acts on the vehicle also as angular acceleration such that it gets tilted a little (front &quot;into the ground&quot;, back &quot;into air&quot;). At the point when the vehicle stops, this tilt will vanish (might be described by aperiodic oscillation). This movement, call it &quot;up-whipping&quot; will again have a turning point causing the kick into the seat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, maybe :)&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2010-11-12T15:53:02.163" LastActivityDate="2010-11-12T15:53:02.163" CommentCount="3" />
  <row Id="639" PostTypeId="1" AcceptedAnswerId="1340" CreationDate="2010-11-12T12:50:05.657" Score="3" ViewCount="302" Body="&lt;p&gt;Under some atmospheric stability condition, over flat terrain, it has been observed for a while that the ratio between wind speed at height $h_1$ above the earth and the wind speed at height $h_0$ is $\log\frac{h_1}{h^*}/\log\frac{h_0}{h^*}$ where $h^*$ is related to the terrain (called roughness length). (see for example &lt;a href=&quot;http://en.wikipedia.org/wiki/Log_wind_profile&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Log_wind_profile&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the theories (with some details or references please) that explain this rule. Please put only your prefered theory (and hence one per post).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance &lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-12T17:26:05.963" LastActivityDate="2010-11-27T11:13:53.917" Title="logarithmic wind speed profile" Tags="&lt;fluid-dynamics&gt;&lt;aerodynamics&gt;&lt;atmosphere&gt;&lt;atmospheric-science&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="640" PostTypeId="2" ParentId="598" CreationDate="2010-11-12T13:39:20.780" Score="2" ViewCount="" Body="&lt;p&gt;For a fast and useful introduction, you can read the second chapter of the book&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Quantum Mechanics, Vol. I&quot;, by Claude Cohen-Tannoudji et al.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will find here almost all you need to study quantum physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a deeper treatment, you can try,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Functional Analysis. Volume I.&quot;, by Reed &amp;amp; Simon. This is the first volume of the collection Methods of Modern Mathematical Physics.&lt;/p&gt;&#xA;" OwnerUserId="306" LastActivityDate="2010-11-12T13:39:20.780" />
  <row Id="641" PostTypeId="2" ParentId="603" CreationDate="2010-11-12T13:45:47.263" Score="0" ViewCount="" Body="&lt;p&gt;Planetary motions in such a frame would be depicted properly by the deferent an epicycle model of Ptolemy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Deferent_and_epicycle&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Deferent_and_epicycle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question is actually answered by Ptolemy. For more reference, look up details in the Almagest! :-)&lt;/p&gt;&#xA;" OwnerUserId="306" LastActivityDate="2010-11-12T13:45:47.263" CommentCount="2" />
  <row Id="642" PostTypeId="2" ParentId="615" CreationDate="2010-11-12T13:55:56.763" Score="5" ViewCount="" Body="&lt;p&gt;Others have discussed things around this, but doesn't &lt;a href=&quot;http://en.wikipedia.org/wiki/Night_vision&quot; rel=&quot;nofollow&quot;&gt;night vision&lt;/a&gt; qualify as an answer to &quot;glasses that make everything brighter&quot; ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 3 main categories discussed.  Most likely you're only interested in the first category.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Image intensification&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Image intensification technologies work on the principle of magnifying the amount of received photons from various natural sources such as starlight or moonlight. Examples of such technologies include night glasses and low light cameras.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Active illumination&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Active illumination technologies work on the principle of coupling imaging intensification technology with an active source of illumination in the near infrared (NIR) or shortwave infrared (SWIR) band. Examples of such technologies include low light cameras.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Thermal imaging&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Thermal imaging technologies work by detecting the temperature difference between the background and the foreground objects. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For more info on the first category, the article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Night_vision_device&quot; rel=&quot;nofollow&quot;&gt;Night vision device&lt;/a&gt; discusses current technologies.  The article &lt;a href=&quot;http://en.wikipedia.org/wiki/Image_intensifier&quot; rel=&quot;nofollow&quot;&gt;image intensifier&lt;/a&gt; goes even further.  Apparently these primarily use &lt;a href=&quot;http://en.wikipedia.org/wiki/Photocathode&quot; rel=&quot;nofollow&quot;&gt;photocathodes&lt;/a&gt; to intensify images.  They are a bit different from the optical amplifiers discussed by Bernardo Kyotoku in that they rely on the photoelectric effect.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-12T13:55:56.763" CommentCount="3" />
  <row Id="644" PostTypeId="2" ParentId="619" CreationDate="2010-11-12T14:01:08.087" Score="0" ViewCount="" Body="&lt;p&gt;The following formula is very self-explained:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m=\frac{m_{0}}{\sqrt{1-v^{2}/c^{2}}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as an object go faster, its mass increases, and it is more difficult to accelerate it. Mass tends to infinity when you approach to the speed of light. Thus, only massless particles can go so fast!&lt;/p&gt;&#xA;" OwnerUserId="306" LastEditorUserId="306" LastEditorDisplayName="" LastEditDate="2010-11-12T14:12:16.450" LastActivityDate="2010-11-12T14:12:16.450" CommentCount="2" />
  <row Id="645" PostTypeId="2" ParentId="630" CreationDate="2010-11-12T14:24:51.783" Score="2" ViewCount="" Body="&lt;p&gt;I'm assuming from the phrasing of the question that you would like to see a spatial interference pattern-- bright and dark fringes-- not something in the time domain. As noted earlier, you can easily beat two independent lasers together and get a signal that pulses at the difference between the two. You can also do coincidence counting experiments using photons derived from two different lasers and show that they interfere with each other-- I believe that's the &quot;Hong-Ou-Mandel&quot; experiment, but I could have the name wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If what you want to see is a pattern of interference fringes in space, you need a detector that is both fast and position-sensitive, as the interference pattern will shift around rapidly in time. You would need to take a &quot;snapshot&quot; of the pattern at some instant, before it has time to shift position and wash out the fringes from an earlier time. So, what kind of speed are we talking about?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, the pattern between the lasers would be steady for a time on the order of the coherence time of the lasers. A good rule of thumb for finding the coherence time for a fairly ordinary laser is that the coherence length of the laser is roughly equal to the length of the cavity. A typical gas laser of the sort you would readily find on the shelves of your local physics department stockroom has a cavity that is roughly one foot long, which means a coherence time of the time required for light to travel one foot, which is one nanosecond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you would want to be able to pick up the pattern in less than a nanosecond, which means your detector needs to be able to handle count rates of at least one gigahertz. That's not too difficult to manage-- you can get photodiodes with a bandwidth of 50-60GHz pretty easily. In order to see a spatial pattern, though, you really want a linear array of these at the very least, and a CCD camera would be even better. You also need to be able to pick up multiple photons in that span, in order to be able to see the pattern, so you can clearly resolve the difference between bright and dark fringes-- I would say you probably want at least 100 photons/pixel in the bright fringes to be able to get decent fringe contrast. This is doable with fairly basic lasers-- a few milliwatts in the red range of wavelengths is about 10^15 photons/s, or about 10^6 photons/ns, so you can spread that out over a few thousand pixels and still be safe. And, of course, you need a low level of &quot;dark counts&quot; in your detector, so it can readily tell the difference between a bright spot with 100 photons and a dark spot with none.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a pretty challenging set of detector requirements. You want, essentially, a CCD detector with single-photon sensitivity and a bandwidth of 100 GHz. I don't think you're going to find that just lying around. Even the fast CCD detectors people use for real-time monitoring of BEC experiments and the like have a frame readout time of a millisecond or so, and those aren't single-photon sensitive. Single-photon detectors tend to be single avalanche photodiodes, which are easily damaged by photon count rates of a few tens of kilohertz. You might be able to construct a detector with these specs, but it would be an extremely difficult problem, and not something that would be worth doing just to see a spatial interference pattern between two independent lasers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can bump the coherent time of the lasers up by a factor of 1000 (which is not a trivial undertaking) then it becomes a little easier to do, but it's still very difficult-- you're looking for a CCD with a frame readout time of a microsecond or so, able to handle photon counts in the megahertz range. Which is still a very hard problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is probably why it hasn't been done, at least not that I'm aware of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; The closest to the sort of thing you're talking about is the &lt;a href=&quot;http://prola.aps.org/abstract/PR/v159/i5/p1084_1&quot; rel=&quot;nofollow&quot;&gt;Pfleegor-Mandel experiment&lt;/a&gt;, which used a statistical technique to show that there were spatial fringes in the interference pattern of two independent lasers. It's a long way from a direct observation of the pattern, though.&lt;/p&gt;&#xA;" OwnerUserId="166" LastEditorUserId="166" LastEditorDisplayName="" LastEditDate="2010-11-12T16:01:28.903" LastActivityDate="2010-11-12T16:01:28.903" />
  <row Id="646" PostTypeId="2" ParentId="622" CreationDate="2010-11-12T14:25:06.210" Score="1" ViewCount="" Body="&lt;p&gt;Scattering, interpreted broadly is just &lt;strong&gt;&quot;probe with wave or particle&quot; and &quot;measure response&quot;&lt;/strong&gt; and thus I claim that it is the principle behind most if not all experimental techniques in physics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the scattering techniques I give below are &quot;just&quot; applications of Rayleigh scattering, but I hope you can see that the simplicity of Rayleigh scattering is precisely what makes them useful experimental tools!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X-ray and neutron scattering experiments are used to find the structure of complicated materials, for instance.  The keyword here is &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bragg%27s_law&quot; rel=&quot;nofollow&quot;&gt;Bragg's law&lt;/a&gt;&quot; in the case of crystal lattices. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A popular technique to probe dynamics is &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dynamic_light_scattering&quot; rel=&quot;nofollow&quot;&gt;Dynamic light scattering&lt;/a&gt;&quot;.  The principle here is that scattering can detect fluctuations due to Brownian motion in complex fluids, and this information is encoded in the autocorrelation of the measured intensity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With scattering experiments, you might think you're screwed if your probe scatters multiple times before it comes back to your detector (because then how will you be able to interpret the data).  However, &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffusing-wave_spectroscopy&quot; rel=&quot;nofollow&quot;&gt;Diffusing wave spectroscopy&lt;/a&gt; is a very clever technique that applies in the presence of multiple scattering events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The field of &lt;a href=&quot;http://en.wikipedia.org/wiki/Acoustics&quot; rel=&quot;nofollow&quot;&gt;acoustics&lt;/a&gt; is all about scattering of sound waves.  In materials there's a lot of interest in studying phonons - these are also probed via scattering of various types.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the generalized interpretation of scattering, &lt;strong&gt;conductivity / resistivity measurements&lt;/strong&gt; in hard condensed matter also become scattering experiments.  There you probe with a electromagnetic wave by causing the voltage in a wire to your sample to oscillate, and you measure the response in the form of electron current.  Examples of recently popular materials studied this way: &lt;a href=&quot;http://en.wikipedia.org/wiki/High-temperature_superconductivity&quot; rel=&quot;nofollow&quot;&gt;high-temperature superconductors&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_Hall_effect&quot; rel=&quot;nofollow&quot;&gt;integer&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Fractional_quantum_Hall_effect&quot; rel=&quot;nofollow&quot;&gt;fractional&lt;/a&gt; quantum Hall devices, &lt;a href=&quot;http://en.wikipedia.org/wiki/Graphene&quot; rel=&quot;nofollow&quot;&gt;graphene&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Topological_insulator&quot; rel=&quot;nofollow&quot;&gt;topological insulators&lt;/a&gt;, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And obviously, from this point of view, particle colliders and detectors do a kind of scattering experiment as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The list goes on and on.  Many, if not most things studied in physics are &lt;strong&gt;waves&lt;/strong&gt; of some sort, and thus scattering processes are very relevant to experiments.  One prominent non-example is most of astrophysics, where we have no chance of actually probing the stars ourselves and can only observe what they do on their own.  You might argue that astrophysics is therefore just &quot;observational&quot; and not &quot;experimental&quot;.&lt;/p&gt;&#xA;" OwnerUserId="42" LastEditorUserId="42" LastEditorDisplayName="" LastEditDate="2010-11-12T14:42:29.723" LastActivityDate="2010-11-12T14:42:29.723" CommunityOwnedDate="2010-11-18T19:14:25.597" />
  <row Id="647" PostTypeId="1" AcceptedAnswerId="708" CreationDate="2010-11-12T14:43:25.543" Score="4" ViewCount="215" Body="&lt;p&gt;The difference between kinematics and dynamics that Grisha pointed to in his &lt;a href=&quot;http://physics.stackexchange.com/questions/614/where-do-particles-go-after-collision/617#617&quot;&gt;answer&lt;/a&gt; comes up in a similar but slightly more complicated case: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider (in 1D) two point-like particles connected by a spring (&quot;the molecule&quot;), the right of them being hit by a third particle (&quot;the bullet&quot;) coming from the right with initial velocity $v_0$. Let the molecule be in rest initially. Let all three particles have equal mass $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the collision took place the bullet will move to the right with velocity $v_b$, the molecule will move to the left with velocity $v_m$ and oscillate with frequency $\omega$ and amplitude $\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total energy of the molecule and its total momentum after the collision are then (I hope I didn't make any serious errors):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E = \frac{m}{2}(v_1^2 + v_2^2) = m (v_m^2 + \alpha^2\omega^2)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p = 2mv_m$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conservation laws yield:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_m = \frac{1}{3}(v_0 + \sqrt{v_0^2 - 3\alpha^2\omega^2})$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_b = v_0 - 2v_m$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we know that $v_m \geq \frac{1}{2}v_0$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 \leq \alpha \leq \frac{v_0}{2\omega}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e., the range in which the amplitude $\alpha$ can vary depends on the spring constant $k$ (which is associated with the frequency by $\omega \propto \sqrt{k/m}$): if the spring constant tends to $\infty$ the amplitude must tend to 0 and the molecule must behave approximately like a rigid rod of mass $2m$, as would have been expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the amplitude $\alpha$ is &lt;em&gt;not&lt;/em&gt; determined by the conservation laws, only its range of possible values. (This is analoguous to the case of two colliding point-like particles, see answers to &lt;a href=&quot;http://physics.stackexchange.com/questions/614/where-do-particles-go-after-collision&quot;&gt;Where do particles go after collision?&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But:&lt;/strong&gt; At least I would see it as unphysical if the molecule would &lt;em&gt;not&lt;/em&gt; oscillate after the collision for finite spring constants $k&amp;gt;0$, which is nevertheless allowed by the conservation laws. I cannot imagine an &quot;inner structure&quot; or a micro-process taking place during the collision which would produce this behaviour. What results could inverse scattering experiments yield?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How am I to think about this mind-boggling&#xA;  puzzle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[Postscript]&lt;/strong&gt; In the case of two colliding point-like particles it is easy to specify the dynamics in a most natural way, at least in the case when the two collide with diametrical velocities: &quot;for symmetry reasons&quot; their velocities after the collision shall be in the same/opposite direction. (For real point-like particles without inner structure that's the only law/rule that makes sense.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now the puzzle continues: Which &quot;natural&quot; law could determine the solution of the molecule-bullet-problem?&lt;/p&gt;&#xA;" OwnerUserId="229" LastEditorUserId="229" LastEditorDisplayName="" LastEditDate="2010-11-13T11:23:03.843" LastActivityDate="2011-05-08T17:24:36.067" Title="How do molecules vibrate after collision?" Tags="&lt;classical-mechanics&gt;&lt;paradox&gt;&lt;scattering&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" />
  <row Id="648" PostTypeId="2" ParentId="629" CreationDate="2010-11-12T14:45:58.037" Score="5" ViewCount="" Body="&lt;p&gt;I think @Robert's answer about 'up-whipping' might explain some of the effect in some cases, but I doubt it's noticeable in trains or other things with limited or no suspension travel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm fairly certain it's mainly down to the 'leaning back' effect (the Kienzler effect??).  Take a marble with you next time you travel by train, you can probably devise an experiment to find out!&lt;/p&gt;&#xA;" OwnerUserId="327" LastActivityDate="2010-11-12T14:45:58.037" CommentCount="1" />
  <row Id="650" PostTypeId="1" AcceptedAnswerId="657" CreationDate="2010-11-12T15:44:54.350" Score="2" ViewCount="272" Body="&lt;p&gt;If the physical world we know is made of matter stretched in time and space and some forces affecting it, what would be an Idea, pure Information or something similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does those things don't exist in a physical model or did I just miss something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes where to arrange it into the physical model? Is an idea just matter without time and Space?&lt;/p&gt;&#xA;" OwnerUserId="337" LastActivityDate="2010-11-12T17:02:00.340" ClosedDate="2010-11-13T10:28:18.993" Title="What is an &quot;Idea&quot; in terms of time space and matter?" Tags="&lt;energy&gt;&lt;space&gt;" AnswerCount="2" CommentCount="12" FavoriteCount="1" />
  <row Id="651" PostTypeId="1" AcceptedAnswerId="669" CreationDate="2010-11-12T15:50:46.747" Score="2" ViewCount="212" Body="&lt;p&gt;If metal moves in a magnetic field It creates a current so when the ISS or any other satellite/object charged so that if you connected a light/ put it in a circuit/ it would light up? also why doesn't this have an effect on electronics on earth. &lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="4" LastEditorDisplayName="" LastEditDate="2010-11-20T21:37:35.407" LastActivityDate="2010-11-20T21:37:35.407" Title="Why isn't the ISS electrically charged?" Tags="&lt;magnets&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="652" PostTypeId="2" ParentId="584" CreationDate="2010-11-12T16:27:25.313" Score="3" ViewCount="" Body="&lt;p&gt;In particular the following reference attempts to construct a useful necessary condition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S. Major, D.P. Rideout, S. Surya, Stable Homology as an Indicator of Manifoldlikeness in Causal Set Theory, arXiv:0902.0434 (Continuum topology and homology)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know of any sufficient conditions, beyond giving an explicit embedding.  Such a thing would be great!&lt;/p&gt;&#xA;" LastActivityDate="2010-11-12T16:27:25.313" CommentCount="1" />
  <row Id="653" PostTypeId="2" ParentId="363" CreationDate="2010-11-12T16:33:04.903" Score="1" ViewCount="" Body="&lt;p&gt;For the mathematically inclined there is also:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sachs, Wu, &lt;strong&gt;General Relativity for Mathematicians&lt;/strong&gt; (Springer Graduate Texts in Mathematics)&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2010-11-12T16:33:04.903" />
  <row Id="654" PostTypeId="2" ParentId="615" CreationDate="2010-11-12T16:36:33.907" Score="7" ViewCount="" Body="&lt;p&gt;First, let me try to clarify the question, because not everyone seems to get it (&quot;where would the photons come from?&quot;). If I'm buying binoculars, I might choose from 8x24, 8x36, or 8x50 binoculars. The first number is the magnification; 8x for all the examples. The second number is the size of the objective in millimeters, 24 to 50 millimeters in the example. All else being equal, the larger objectives will gather more light, and produce a brighter image. The 8x24 and 8x50 binoculars will deliver the same 8x magnification, but the 8x50 pair will be brighter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's reasonable to ask if you could make 1x binoculars that have a big objective that brightens the scene in front of you. There's no conservation of energy issue here; the larger front objective would gather more light. The problem is that there's a relationship between the magnification and the sizes of the entrance and exit pupils:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;magnification = entrance_pupil_size / exit_pupil_size&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you create a binocular with an entrance pupil (objective, basically) that's larger than your eye's pupil, so that it can gather more light than your eye's pupil, then at 1x magnification, the exit pupil is going to be just as big as the objective lens. And since that's bigger than your eye's pupil, the &quot;extra&quot; light is going to run into your iris instead of going through the pupil, and you will gain no advantage from it.&lt;/p&gt;&#xA;" OwnerUserId="292" LastActivityDate="2010-11-12T16:36:33.907" CommentCount="5" />
  <row Id="655" PostTypeId="2" ParentId="296" CreationDate="2010-11-12T16:36:50.590" Score="15" ViewCount="" Body="&lt;p&gt;The topic of &quot;Energy Conservation&quot; really depends on the particular &quot;theory&quot;, paradigm, that you're considering — and it can vary quite a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good hammer to use to hit this nail is &lt;a href=&quot;http://en.wikipedia.org/wiki/Noether%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Nöther's Theorem&lt;/a&gt;: see, e.g., how it's applied in &lt;a href=&quot;http://en.wikipedia.org/wiki/Noether%27s_theorem#Example_1%3a_Conservation_of_energy&quot; rel=&quot;nofollow&quot;&gt;Classical Mechanics&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same principle can be applied to all other theories in Physics, from Thermodynamics and Statistical Mechanics all the way up to General Relativity and Quantum Field Theory (and Gauge Theories).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the lesson to learn is that Energy is only conserved if there's translational time symmetry in the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which brings us to General Relativity: in several interesting cases in GR, it's simply impossible to properly define a &quot;time&quot; direction! Technically speaking, this would imply a certain global property (called &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Globally_hyperbolic&quot; rel=&quot;nofollow&quot;&gt;global hyperbolicity&lt;/a&gt;&quot;) which not all 4-dimensional spacetimes have. So, in general, Energy is &lt;strong&gt;not&lt;/strong&gt; conserved in GR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for quantum effects, Energy is conserved in Quantum Field Theory (which is a superset of Quantum Mechanics, so to speak): although it's true that there can be fluctuations, these are bounded by the &quot;uncertainty principle&quot;, and do not affect the application of Nöther's Theorem in QFT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the bottom line is that, even though energy is &lt;em&gt;not&lt;/em&gt; conserved always, we can always understand what this non-conservation mean via Nöther's Theorem. ;-)&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2010-11-12T16:36:50.590" CommentCount="2" />
  <row Id="656" PostTypeId="2" ParentId="79" CreationDate="2010-11-12T16:44:37.700" Score="9" ViewCount="" Body="&lt;p&gt;The original goal of the EPR paper was to show that quantum mechanics is incomplete. Hence, that extra variables have to be added to complete it, contrary to what Cedric claims. The goal of EPR is to show that either nature is non-local (and thus in conflict with SR) either quantum mechanics is incomplete. Since Einstein was not ready to abandon locality and SR, he concluded that quantum mechanics is incomplete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later however, John Bell would show that quantum mechanics is in fact non-local. To do that, he first devised an inequality that would have to be satisfied by any local physical theory. Then, he showed that this inequality is violated for certain entangled states, thereby proving that quantum mechanics is non-local. In the 70's Alain Aspect then made an experiment to check if Bell's inequalities were violated in nature or not. There have been many similar experiments since then and they all point to nature being non-local and quantum mechanics being a good description of this non-locality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there are possible loopholes in the experiments, which I won't discuss here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can also object that quantum mechanics described by Schrödinger's equation is not Lorentz-invariant, so we should not expect quantum mechanics to agree with SR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about equations which are Lorentz-invariant? Dirac equations, Klein-Gordon, etc... That's where it gets difficult. We know that a correct description of these equations requires quantum field theories. But we only manage to treat the field theories perturbatively. Other approaches are numerical or very limited. So I don't know of any detailed study within the context of quantum field theory of entanglement and Bell inequalities. But I hope someone can come in with more information about these. My knowledge is limited in these areas.&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2010-11-13T08:23:20.483" LastActivityDate="2010-11-13T08:23:20.483" />
  <row Id="657" PostTypeId="2" ParentId="650" CreationDate="2010-11-12T16:46:05.040" Score="2" ViewCount="" Body="&lt;p&gt;An &quot;idea&quot; is something that happens in the confines of your &lt;em&gt;mind&lt;/em&gt;: two particles colliding don't have ideas, they simply behave following the so-called &quot;Laws of Physics&quot;, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this sense, an idea, or, more generally, &lt;em&gt;creativity&lt;/em&gt; itself, is something that only &quot;exists&quot; (if that's an appropriate term in this case) in your &quot;mind&quot;, in this &quot;emergent phenomena&quot; driven by the inner working of your brain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this sense, i'd classify this question as a &quot;neuroscience&quot; one... ;-)&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2010-11-12T16:46:05.040" CommentCount="8" />
  <row Id="659" PostTypeId="2" ParentId="634" CreationDate="2010-11-12T17:00:42.073" Score="3" ViewCount="" Body="&lt;p&gt;@Cedric: It's true that the notion of &quot;fields&quot; (as we have today in QFT) is something that wasn't completely developed at that time. But i never heard that Dirac's Eq was understood as a &quot;wave equation&quot;: he was trying to explicitly describe particles (electrons) — besides, the notion that &quot;particles&quot; and &quot;waves&quot; were connected already existed at that time, in the form of von Neumann's &quot;transformation theory&quot; (read: Fourier Transform + distributions). On to of that, people knew that &quot;Matrix Mechanics&quot; (à la Heisenberg) and &quot;Wave Mechanics&quot; (à la Schrödinger) were two sides of the same coin (one being particle-like and the other one being wave-like).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, before anything else, it'd be nice to corroborate this story of yours. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, let me give you a couple of references that may shed some light:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://press.princeton.edu/titles/5609.html&quot; rel=&quot;nofollow&quot;&gt;An Interpretive Introduction to Quantum Field Theory&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dl.dropbox.com/u/122003/bonolis2004.pdf&quot; rel=&quot;nofollow&quot;&gt;From the Rise of the Group Concept to the Stormy Onset of Group Theory in the New Quantum Mechanics (PDF, 739Kb)&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2010-11-12T17:00:42.073" CommentCount="1" />
  <row Id="660" PostTypeId="2" ParentId="650" CreationDate="2010-11-12T17:02:00.340" Score="2" ViewCount="" Body="&lt;p&gt;Information must always be represented by the state of some physical system.  The idea in your head exists somewhere within the biological structures and electrochemical state of your brain, just as information in the computer is represented by the presence or absence of electric charge in the billions of bits of RAM.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2010-11-12T17:02:00.340" />
  <row Id="661" PostTypeId="1" AcceptedAnswerId="667" CreationDate="2010-11-12T17:24:44.000" Score="8" ViewCount="447" Body="&lt;p&gt;If someone has short or long sight, is it possible to tune image on a computer monitor in such way, that a person could see it sharp as if they were wearing glasses? If not, will 3d monitor make it possible?&lt;/p&gt;&#xA;" OwnerUserId="213" LastActivityDate="2011-04-11T09:01:12.717" Title="Is it possible to blur an image in such way that a person with sight problems could see it sharp?" Tags="&lt;optics&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="662" PostTypeId="2" ParentId="27" CreationDate="2010-11-12T17:25:09.767" Score="0" ViewCount="" Body="&lt;p&gt;This is a question that philosophers of physics try to answer now, not physicists (even if most of the times the border is not really sharp). So if you are looking for a more detailed discussion (and ressources), you should have a look at this article of the Stanford Encyclopedia of Philosophy: &lt;a href=&quot;http://plato.stanford.edu/entries/qt-measurement/&quot; rel=&quot;nofollow&quot;&gt;http://plato.stanford.edu/entries/qt-measurement/&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-11-12T17:25:09.767" CommentCount="1" />
  <row Id="663" PostTypeId="2" ParentId="651" CreationDate="2010-11-12T17:25:40.157" Score="5" ViewCount="" Body="&lt;p&gt;Well, it probably &lt;em&gt;is&lt;/em&gt; charged, as it continuously exposed to all sorts of radiation (including charged particles) from the sun and elsewhere.  The fact that it is made of metal and moving through a magnetic field is not a source of charging, however.  The principle of &lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_induction&quot; rel=&quot;nofollow&quot;&gt;magnetic induction&lt;/a&gt; says that a conductor moving through a magnetic field will develop a &lt;em&gt;voltage&lt;/em&gt; across it.  A &lt;em&gt;changing&lt;/em&gt; magnetic field will induce &lt;a href=&quot;http://en.wikipedia.org/wiki/Eddy_current&quot; rel=&quot;nofollow&quot;&gt;eddy currents&lt;/a&gt; in the metal shell of the spacecraft which will result in some heating and will also produce magnetic fields.  Managing the effects of radiation, charging, and torques introduced from magnetic fields are all important parts of spacecraft engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you hook up giant wires to the ISS to make it into a giant generator as it spins through the earth's magnetic field?  Maybe, though the back-action would cause the orbit to decay faster than it usually does.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2010-11-12T17:25:40.157" CommentCount="1" />
  <row Id="664" PostTypeId="2" ParentId="271" CreationDate="2010-11-12T17:27:11.047" Score="1" ViewCount="" Body="&lt;p&gt;All the explanations given involve conservation of angular momentum, which is perfectly correct, but I feel that people with no thorough background in physics and mathematics will be left unsatisfied with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to explain conservation of angular momentum in terms which would be understandable to the layman? It's a pedagogical problem I have given a lot of thought and I have yet to find a satisfactory answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely, you need to start from something, some basic axiom that the person will be willing to accept at face value. I thought about using either the law of action and reaction or conservation of momentum. I think these are relatively easy to describe &quot;pictorially&quot;. But going from these to angular momentum using a vector product, is a mathematical procedure I'm not sure I can explain to someone who doesn't know anything about math. So, this should be circumvented in some way by a nice visual example again to make things clearer and I haven't found one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anybody having ideas?&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2010-11-12T17:27:11.047" CommentCount="9" />
  <row Id="665" PostTypeId="2" ParentId="661" CreationDate="2010-11-12T17:55:28.793" Score="2" ViewCount="" Body="&lt;p&gt;No, if you are someone who needs glasses the effect without them would be like a blur effect which can not be reversed the same strongly blurred picture can not be rendered to its original detailed focus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking a picture without focusing the lens would have the same effect. You can neither create an image that will be sharp with an unfocused lens nor reconstruct the detailed image taken with an unfocused lens.&lt;/p&gt;&#xA;" OwnerUserId="337" LastActivityDate="2010-11-12T17:55:28.793" CommentCount="1" />
  <row Id="666" PostTypeId="2" ParentId="661" CreationDate="2010-11-12T18:17:43.343" Score="-1" ViewCount="" Body="&lt;p&gt;Well, I think that it is possible to create a &quot;blurred&quot; image from a real one in order to be correct for someone with short/long sight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is &quot;only&quot; an optical problem, and, as long as we can apply the paraxial approximation, it is a linear system, very suitable to a computer program. We know the laws that focus the rays in one point, so we can disturb these rays so that they arrive at a different point, and it would be possible to &quot;tune&quot; this point to be correct to every people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that the point is that, from a correct and well-defined image, we can do a lot of things with it. If we have a blurred image, and we want to make it correct, it will be more difficult, because now we haven't got all the information as in the first case.&lt;/p&gt;&#xA;" OwnerUserId="306" LastActivityDate="2010-11-12T18:17:43.343" CommentCount="1" />
  <row Id="667" PostTypeId="2" ParentId="661" CreationDate="2010-11-12T18:18:00.230" Score="8" ViewCount="" Body="&lt;p&gt;Let's take a simple original picture to look at - just two nearby dots on a white background.  If you have bad vision, the dots look blurred.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way good vision works is to ensure that all the light hitting any particular small area of your retina comes from the same direction in front of you.  Conversely, all the light coming from one direction hits one specific spot on your retina.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have bad vision, the light from a locus of nearby directions all hits on the same part of your retina, and the light from a particular direction is smeared out over an area on your retina.  Hence, blurred vision is an averaging effect.  When you look at the dots, you'll see them smear out into each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might try to compensate for this by making a &quot;counter-blurred&quot; image where the source dots are smaller, but if the original dots are close enough that light from the center of one dot is spilling over to overlap light from the center of the second dot, making the dots smaller won't fix that problem.  Hence, the dots will always appear blurred.  You can't create the impression that the original has for someone with good vision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A photograph is really just a bunch of nearby dots, and so the same problem applies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know about the 3D monitor, though.  I suppose if it can control the direction of light coming off it, it could be modified to focus the light some and create a sharp image for someone with blurred vision.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-04-11T09:01:12.717" LastActivityDate="2011-04-11T09:01:12.717" CommentCount="2" />
  <row Id="668" PostTypeId="2" ParentId="661" CreationDate="2010-11-12T18:28:12.387" Score="7" ViewCount="" Body="&lt;p&gt;Not on regular monitor screen. The technology necessary to achieve such effect would be holographic display, holographic in the sense of wavefront synthesis. Although this device would be a 3D display, not all 3D display are holographic. You would need technologies such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Spatial_light_modulator&quot; rel=&quot;nofollow&quot;&gt;spatial light modulator&lt;/a&gt;. Which only exists as low specs laboratory devices.&lt;/p&gt;&#xA;" OwnerUserId="62" LastActivityDate="2010-11-12T18:28:12.387" />
  <row Id="669" PostTypeId="2" ParentId="651" CreationDate="2010-11-12T18:29:50.710" Score="5" ViewCount="" Body="&lt;p&gt;Charging, as nibot said, happens because the &quot;balance of charge&quot; is altered. If you have a neutral ISS, and you assume that no electrons are added or removed, the ISS will stay neutral. This can of course change, and the ISS can charge, if electrons are removed or added due to the exposure to ionizing radiations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you are right to say that if a metal moves in a magnetic field, you obtain a charge separation, and consequently a voltage. The same concept is used in any dynamo to produce electricity. There was indeed an experiment using this concept:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electrodynamic_tether&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Electrodynamic_tether&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tether_satellite&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Tether_satellite&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It was deployed twice with little success, but it did produce a (small) voltage, and therefore create a current.&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="217" LastEditorDisplayName="" LastEditDate="2010-11-12T18:39:26.960" LastActivityDate="2010-11-12T18:39:26.960" />
  <row Id="670" PostTypeId="2" ParentId="629" CreationDate="2010-11-12T20:29:15.840" Score="1" ViewCount="" Body="&lt;p&gt;Perhaps it can be explained like this:&lt;br&gt;&#xA;At a certain moment the train pilot releases the brake system. At that point the heavy mass of the train is released from a rather powerfull constraint and it's wheels can move freely (well not completely free of course). When the wheels turn slightly forwards you experience a force in the other direction. The mass of the train being quite large and the release of the brakes quite sudden, this force is stronger then the previous carefull deceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the short pull wouldn't be in a &quot;wrong&quot; direction.&lt;br&gt;&#xA;I should be easy to verify in reality whether this explanation is correct.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-12T20:45:22.593" LastActivityDate="2010-11-12T20:45:22.593" CommentCount="3" />
  <row Id="671" PostTypeId="2" ParentId="519" CreationDate="2010-11-12T21:37:27.607" Score="1" ViewCount="" Body="&lt;p&gt;You'll have to wait for a real physicist to jump in for a numerical answer. As far as I know, it doesn't burn but is simply radiating the energy it started out with. I suspect the&#xA;thermal conductivity is quite high. If you assume the thermal conductivity is so high that there is essentially no temperature difference between the photosphere and the core you should be able to use the Stefan Boltzmann law to determine the rate of heat loss. But even then you'd need to know the heat capacity of matter in that form, so any physicists willing to jump in with actual theory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I assume heat capacity is independent of temperature, the cooling equation (after throwing out all the constants) would look like:&#xA;  dT/dt=-T*&lt;em&gt;4&#xA;That has a solution, (again throwing out constants) like T=t&lt;/em&gt;*(-1/3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There could be other sources of energy, initial magnetic field, rotation, gravitational&#xA;(it would probably shrink at least somewhat as it cools), and it there is any matter nearby&#xA;the potential for stuff to fall onto it. {Thats about as far as a dare venture, again hopefully an astrophysicist will jump in}&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2010-11-12T21:37:27.607" />
  <row Id="672" PostTypeId="1" AcceptedAnswerId="688" CreationDate="2010-11-12T21:37:28.973" Score="3" ViewCount="235" Body="&lt;p&gt;Hi there i am wondering, air is making sound propagate. So in vacuum there is no sound. but what is the relation of pressure to sound volume? is it linear?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if i have a source of sound and lets say i am in a room with air pressure 0.5 of the pressure outside would the sound be 0.5 lighter? or does air pressure have no impact on the sound volume?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a comprehensible formula for that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks a lot!&lt;/p&gt;&#xA;" LastActivityDate="2010-11-13T02:41:54.460" Title="What is the relation of sound propagation to air pressure?" Tags="&lt;sound&gt;&lt;pressure&gt;&lt;acoustics&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="673" PostTypeId="2" ParentId="672" CreationDate="2010-11-12T21:44:44.097" Score="1" ViewCount="" Body="&lt;p&gt;Propagation speed in an ideal gas varies &lt;a href=&quot;http://en.wikipedia.org/wiki/Speed_of_sound&quot; rel=&quot;nofollow&quot;&gt;as the square root of pressure&lt;/a&gt;.  It also depends on the adiabatic index, the density of the gas, and the temperature of the gas.&lt;/p&gt;&#xA;" OwnerUserId="379" LastActivityDate="2010-11-12T21:44:44.097" />
  <row Id="675" PostTypeId="2" ParentId="672" CreationDate="2010-11-12T21:50:52.560" Score="1" ViewCount="" Body="&lt;p&gt;The volume of sound is measured in decibel (dB), which are a measure of air pressure.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sound pressure level (SPL) or sound level Lp is a logarithmic measure of the effective sound pressure of a sound relative to a reference value. It is measured in decibels (dB) above a standard reference level.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sound_pressure#Sound_pressure_level&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sound_pressure#Sound_pressure_level&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;em&gt;undistorted&lt;/em&gt; sound cannot be carried by more pressure than it's available. In other words, at 1 atm, a sound can only reach 191dB, which correspond to 1 atm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over that, we are basically talking about shockwaves.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-12T21:50:52.560" CommentCount="1" />
  <row Id="676" PostTypeId="1" AcceptedAnswerId="682" CreationDate="2010-11-12T22:02:06.553" Score="3" ViewCount="312" Body="&lt;p&gt;Or in proton-electron collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To destroy is to turn  into other particles, not baryons.&#xA;In context of the baryon asymmetry.&lt;/p&gt;&#xA;" OwnerUserId="238" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-01-16T08:19:16.543" LastActivityDate="2011-01-16T08:19:16.543" Title="Is it possible to destroy proton in proton-proton collision?" Tags="&lt;particle-physics&gt;&lt;experimental-physics&gt;&lt;accelerator-physics&gt;&lt;proton&gt;&lt;barion-asymmetry&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="677" PostTypeId="2" ParentId="676" CreationDate="2010-11-12T22:09:17.653" Score="2" ViewCount="" Body="&lt;p&gt;If changing the protons into something else counts as &quot;destroying&quot; it, then yes, &lt;a href=&quot;http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction&quot; rel=&quot;nofollow&quot;&gt;this is what keeps stars burning&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, two protons can interact and form a deuteron, a positron and an antineutrino and some energy.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-12T22:09:17.653" />
  <row Id="678" PostTypeId="2" ParentId="676" CreationDate="2010-11-12T22:09:49.610" Score="-1" ViewCount="" Body="&lt;p&gt;You can't destroy a particle. Without involving more complex concepts such as colour, momentum and energy must be conserved, and this implies that you cannot destroy particles. You can produce new particles or radiation by colliding protons (or neutrons...), but, in the sense that they explode and disappear, it is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, you can get muons, gamma-rays, or, more interesting, you could be able to &quot;see&quot; the quarks inside the protons for a moment.&lt;/p&gt;&#xA;" OwnerUserId="306" LastActivityDate="2010-11-12T22:09:49.610" CommentCount="2" />
  <row Id="679" PostTypeId="1" AcceptedAnswerId="762" CreationDate="2010-11-12T22:21:33.623" Score="10" ViewCount="711" Body="&lt;p&gt;I'm thinking of following some kind of education in physics online. I have a master degree in Computer Science and have reasonable good knowledge in physics. I would like a program of 1-2 years and I'm more interested in particle physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any good online program that offer something similar?&lt;/p&gt;&#xA;" OwnerUserId="57" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-25T21:02:13.070" LastActivityDate="2011-08-28T20:57:09.830" Title="Learning physics online?" Tags="&lt;particle-physics&gt;&lt;soft-question&gt;&lt;education&gt;&lt;reference-request&gt;&lt;big-list&gt;" AnswerCount="11" CommentCount="6" FavoriteCount="9" />
  <row Id="680" PostTypeId="2" ParentId="679" CreationDate="2010-11-12T22:33:26.263" Score="6" ViewCount="" Body="&lt;p&gt;List of recommended books and notes from the beginner level to advanced:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.phys.uu.nl/~thooft/theorist.html&quot; rel=&quot;nofollow&quot;&gt;How to Become a Good Theoretical Physicist&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar list on mathematics, parts of it may be useful for physics too:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hbpms.blogspot.com/2008/05/content-and-about.html&quot; rel=&quot;nofollow&quot;&gt;How to Become a Pure Mathematician (or Statistician)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MIT OCW offers complete courses in &lt;a href=&quot;http://ocw.mit.edu/courses/ocw-scholar/&quot; rel=&quot;nofollow&quot;&gt;Classical Mechanics, Electricity and Magnetism and others&lt;/a&gt;, designed for independent students. &lt;a href=&quot;http://ocw.mit.edu/courses/physics/&quot; rel=&quot;nofollow&quot;&gt;Tens of other physics courses are available&lt;/a&gt;, but may not be complete.&lt;/p&gt;&#xA;" OwnerUserId="306" LastEditorUserId="3728" LastEditorDisplayName="" LastEditDate="2011-05-25T19:41:52.547" LastActivityDate="2011-05-25T19:41:52.547" />
  <row Id="681" PostTypeId="2" ParentId="676" CreationDate="2010-11-12T23:04:42.523" Score="5" ViewCount="" Body="&lt;p&gt;Most proton-proton collisions will be elastic: throw in two protons and two protons will come out, deflected at some angle. But the more interesting collisions are those where individual constituents of the proton (quarks, antiquarks, or gluons) interact. For instance, all the interesting high-energy proton-proton collisions at the LHC are really collisions of two quarks, or two gluons, or a gluon and a quark (or similar combinations involving antiquarks) coming from the two protons. The results of this &quot;hard&quot; collision often come out at a large angle away from the proton beam, while the &quot;remnants&quot; of the proton that weren't directly involved in the collision sail off down the beamline in roughly the same direction the proton was originally going. It's probably reasonable to say that the original proton was &quot;destroyed&quot; in this process, although some large fraction of its energy and its constituents keep moving in the same direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the word &quot;destroyed&quot; is a little fuzzy. The protons don't just disappear. There are a few constraints on the final result of the collision process: it must conserve momentum, energy, and baryon number. A proton has baryon number +1, as does a neutron and various heavier &quot;hyperons,&quot; whereas an antiproton has baryon number -1. The momentum, energy, and baryon number can all be divided up in complicated ways. So you might argue that there's a sense in which you didn't &quot;destroy&quot; the protons, since at the end you still have to have a total of 2 baryons. But in general, the proton remnants moving down the beamline &lt;em&gt;could&lt;/em&gt; join up with antiquarks and make mesons with no net baryon number, while baryons could form from the hard collision and move away at a large angle. In that case I wouldn't say there's any sense in which the original protons maintained their &quot;identity,&quot; and I think it would be reasonable to say they were destroyed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All things considered, though, it's probably best to not think in terms of the word &quot;destroy&quot; that you chose at all. Maybe the way to visualize it is like this: most proton-proton collisions are like two billiard balls that bounce off each other in an elastic way. But the collisions that are of the most interest at a collider are more like cases where two billiard balls hit each other and small pieces of each shear off and fly out at a big angle, but the bulk of each ball keeps moving in roughly the direction it started from.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-11-12T23:04:42.523" CommentCount="1" />
  <row Id="682" PostTypeId="2" ParentId="676" CreationDate="2010-11-12T23:42:09.007" Score="1" ViewCount="" Body="&lt;p&gt;The way you used baryon made me wonder if you are unaware of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Baryon_number&quot; rel=&quot;nofollow&quot;&gt;baryon nearly conservation law&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the link:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The baryon number is nearly conserved&#xA;  in all the interactions of the&#xA;  Standard Model. 'Conserved' means that&#xA;  the sum of the baryon number of all&#xA;  incoming particles is the same as the&#xA;  sum of the baryon numbers of all&#xA;  particles resulting from the reaction.&#xA;  An exception is the chiral anomaly.&#xA;  However, sphalerons are not all that&#xA;  common. Electroweak sphalerons can&#xA;  only change the baryon number by 3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="62" LastActivityDate="2010-11-12T23:42:09.007" CommentCount="1" />
  <row Id="683" PostTypeId="2" ParentId="271" CreationDate="2010-11-13T01:20:47.873" Score="1" ViewCount="" Body="&lt;p&gt;The point is that conservation principles are not generally intuitive. For example, why should energy be conserved? One must have a grip of the dynamics involved in order to understand them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the precession of the spinning top doesn't have to do with the conservation of angular momentum. It has to do with the strange nature of torque and its interaction with angular momentum. When a force acts on a spinning top, it excerpts a torque perpendicular to the plane defined by the axis of the top and the direction of gravity, which is a vertical plane. That direction is horizontal. &#xA;On the other hand, the torque is the rate of change of angular momentum. That means that the direction of the torque is the direction towards which the vector of angular momentum changes. Thus, since the torque is horizontal and perpendicular to the angular momentum, it can only change the direction of angular momentum along the horizontal direction and not towards the ground. That means that the vector of angular momentum has its back on the ground, at the point that the tip of the top touches the ground, and its head is performing a circle on a plane that is parallel to the ground. That motion is the precession of the spinning top.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I think that the reason for assuming a much faster rotation than precession for the top, is to simplify the calculations and consider the top as a gyroscope. &lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2010-11-13T01:20:47.873" />
  <row Id="684" PostTypeId="2" ParentId="21" CreationDate="2010-11-13T01:38:40.830" Score="5" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Monte_Carlo_method&quot; rel=&quot;nofollow&quot;&gt;Monte Carlo&lt;/a&gt; is a particular numerical technique heavily used in Physics, mainly when one needs to &quot;brute force&quot; the calculation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Virtually, all areas of Physics can make use of simulations that include some sort of Monte Carlo code, from the designing of detectors in particle physics (think Fermilab and LHC: &lt;a href=&quot;http://geant4.cern.ch/&quot; rel=&quot;nofollow&quot;&gt;Geant4&lt;/a&gt;), passing by out-of-equilibrium statistical mechanics (search for meta-stable stationary states), all the way to &lt;a href=&quot;http://en.wikipedia.org/wiki/Lattice_QCD&quot; rel=&quot;nofollow&quot;&gt;Lattice QCD&lt;/a&gt; and Climate Modeling!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the techniques used to simulate Quantum Field Theories are heavily based in Monte Carlo and are flexible enough to be of value in several other branches of Physics (from Statistical Mechanics to Climate Modeling).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can think in the following terms: in Physics, many problems are integro-differential in nature, i.e., they can be expressed either as a Differential equation or as an Integral one. Solving Diff Eqs can be tricky (think in terms of variable mesh sizes, etc: hyperbolic and parabolic diff eqs can make use of quite different mesh choices) and, sometimes, it pays to cast the problem as an Integral eq and use Monte Carlo to brute force the calculation of the answer, given that MC can be a very effective integration tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays, the basic tool-belt of a computational physicist includes Monte Carlo and LAPACK (ATLAS, BLAS) in a couple of different languages (C/C++, FORTRAN, python, etc): these tools, combined with an effective modeling and coding of the problem, can be as sharp as the state-of-the-art requires (think parallel processing, GPU, CUDA, etc).&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2010-11-13T01:38:40.830" />
  <row Id="685" PostTypeId="2" ParentId="79" CreationDate="2010-11-13T01:48:25.407" Score="1" ViewCount="" Body="&lt;p&gt;The answer to the question depends a bit on what is meant by &quot;mediated&quot;. A composite quantum system composed of two or more quantum subsystems can be in a quantum state in which the subsystems are entangled from the beginning, i.e., from the initial state. If the composite system evolves without any interaction among the subsystems, then the form and degree of entanglement between them will not change. If there is interaction, the entanglement will, usually, change in form and/or degree. In particular, if the initial state is not entangled, then subsequent interaction between the subsystems will entangle them. But this fact, per se, doesn't depend on the kind of interaction. Any interaction, whatsoever, between the subsystems will, at least for awhile, entangle the initially non-entangled subsystems. One might well regard this as &quot;mediation&quot; of entanglement by interaction. But physicists, generally, do not think of entanglement, per se, as a dynamical feature of quantum mechanics. Rather, it is regarded as a kinematical/structural possibility of quantum states for composite systems which can be modified by the presence of interactions but is not ultimately, due to interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me add that entangled states are the most common states, by far. The unentangled states of composite systems (which are just the so-called product states) are much rarer by comparison.     &lt;/p&gt;&#xA;" OwnerUserId="345" LastActivityDate="2010-11-13T01:48:25.407" />
  <row Id="686" PostTypeId="2" ParentId="661" CreationDate="2010-11-13T02:00:03.410" Score="1" ViewCount="" Body="&lt;p&gt;The light source with this properties exists in technology named &quot;Adaptive Optics&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is yes, if your monitor (not existing on market yet) has controllable phase source for each pixel on screen (like phased array antenna), and even more. Each pixel actually needs multiples of phase values, depending on angle of look. So for megapixel display you may need gigapixel of phasing elements which follow and compensate phase exactly to errors on the ray trace to location of retina cell of each individual eyeball, looking at screen. &lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-13T02:07:21.563" LastActivityDate="2010-11-13T02:07:21.563" />
  <row Id="687" PostTypeId="2" ParentId="79" CreationDate="2010-11-13T02:23:51.790" Score="1" ViewCount="" Body="&lt;p&gt;Conclusion&#xA;1.  Definitely not a causal interaction - see Alain Aspect delayed choice experiments&#xA;2.  Rather the amplitudes of causally separated particles are merely remain correlated due to past common origin events&#xA;3.  See Smerlak and Rovelli at &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0604064&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/quant-ph/0604064&lt;/a&gt; .&lt;br&gt;&#xA;See also Rovelli at &lt;a href=&quot;http://arxiv.org/abs/quant-ph/9609002&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/quant-ph/9609002&lt;/a&gt; for a coherent point of view, but it's somewhat subtle.  &lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-11-13T02:23:51.790" />
  <row Id="688" PostTypeId="2" ParentId="672" CreationDate="2010-11-13T02:41:54.460" Score="3" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Sound_intensity&quot; rel=&quot;nofollow&quot;&gt;sound intensity&lt;/a&gt; is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = \xi^2 \omega^2 c \rho$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\xi$ is the particle displacement, $\omega$ the frequency, $c$ the speed of sound and $\rho$ the density of the medium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're in a room where the pressure is lower (and so the density of air is also lower). Say that your sound source (string, speaker or whatever) is vibrating with the same amplitude and frequency, so $\xi$ and $\omega$ don't change. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Speed_of_sound#Dependence_on_the_properties_of_the_medium&quot; rel=&quot;nofollow&quot;&gt;speed of sound&lt;/a&gt; is roughly independent of the pressure. So your &lt;strong&gt;sound intensity will be roughly proportional to the density of the air, or proportional to the pressure&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17" LastActivityDate="2010-11-13T02:41:54.460" CommentCount="3" />
  <row Id="689" PostTypeId="2" ParentId="634" CreationDate="2010-11-13T02:52:35.387" Score="5" ViewCount="" Body="&lt;p&gt;For the details of the physics involved in the two ways of interpreting the Dirac wave equation I recommend chapters XI and XII of Dirac's &quot;Principles of Quantum Mechanics&quot; 4th edition, and chapters XX and XXI of Messiah's &quot;Quantum Mechanics&quot;, vol. II. For the more historical details I recommend chapters 5 and 6 of Crease and Mann's &quot;The Second Creation&quot;, chapters 13 and 15 of Pais' &quot;Inward Bound&quot; and the first chapter of Schweber's &quot;QED And The Men Who Made It&quot;.&lt;/p&gt;&#xA;" OwnerUserId="345" LastActivityDate="2010-11-13T02:52:35.387" CommentCount="1" />
  <row Id="690" PostTypeId="2" ParentId="79" CreationDate="2010-11-13T03:06:09.493" Score="8" ViewCount="" Body="&lt;p&gt;The standard test for whether two things are really entangled with one another in the spooky-action-at-a-distance sense of the EPR picture is to see whether measurements of the states of the two particles violate one of the Bell inequalities, meaning that the correlation between the states is stronger than can be explained by any local hidden variable theory. This has been done with lots of systems having significant separation between the particles-- as long ago as 1982, Alain Aspect's group did a test with time-varying detectors that were separated by 40 feet or so, and the results were something like nine standard deviations away from the LHV limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More recently, Chris Monroe's group at the University of Maryland has done experiments where they entangle the states of two ions in two different ion traps, and showed Bell violation by something like 3.5 standard deviations. I &lt;a href=&quot;http://scienceblogs.com/principles/2009/07/entanglement_by_accident.php&quot; rel=&quot;nofollow&quot;&gt;wrote this up on the blog a while back&lt;/a&gt;, and the post includes links to the relevant papers. I'm not sure whether there's a complete lack of a straight-line path between the ions, but they're in completely separate vacuum chambers (mostly stainless steel), so I think it's fairly likely to meet the requirements of the question.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-13T03:06:09.493" CommentCount="1" />
  <row Id="691" PostTypeId="1" CreationDate="2010-11-13T03:15:35.493" Score="2" ViewCount="249" Body="&lt;p&gt;If you look up while you are falling into a black hole you see the universe blue shifted, that is, you see the universe moving quickly forward in time compared to your local time. Since this effect increases as you get closer to the singularity, a star aimed at you could blast you with a billion years of radiation in maybe a second. Of course, no star will be perfectly aligned for billions of years. On the other hand, there is lots of cosmic background radiation coming at you from everywhere in the sky and it will be blue shifted too. So, how bad is it? Would this radiation cook you before you hit the singularity?&lt;/p&gt;&#xA;" LastActivityDate="2010-12-15T20:16:31.597" Title="Which gets you first, the black hole singularity or the cosmic background radiation?" Tags="&lt;astrophysics&gt;&lt;cosmology&gt;" AnswerCount="3" FavoriteCount="0" />
  <row Id="693" PostTypeId="2" ParentId="615" CreationDate="2010-11-13T03:51:08.680" Score="0" ViewCount="" Body="&lt;p&gt;Suppose you just start with an enormous magnifying glass (convex lens) and project the (inverted?)image onto a very fine ground glass intermediate screen.  That would be plenty bright and small.  It Seems to work with my reading glasses projecting an image onto a piece of paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think something like this (with complex lenses separated by a ground glass screen) is sometimes used in the motion picture industry.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-11-13T04:02:12.773" LastActivityDate="2010-11-13T04:02:12.773" CommentCount="2" />
  <row Id="694" PostTypeId="2" ParentId="95" CreationDate="2010-11-13T03:52:16.960" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What alternative theories are also consistent with current observational data? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is an alternative line of thought according to which inhomogeneities in large-scale structure, known as voids, are responsible for the &quot;perceived&quot; or &quot;apparent&quot; acceleration on large time scales. This viewpoint was first advocated by &lt;a href=&quot;http://arxiv.org/abs/astro-ph/9907206&quot; rel=&quot;nofollow&quot;&gt;Celerier&lt;/a&gt; and &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0602478&quot; rel=&quot;nofollow&quot;&gt;Inoue and Silk&lt;/a&gt;, among others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A numerical comparison of WMAP3 data with LCDM and non-Lambda model is done &lt;a href=&quot;http://arxiv.org/abs/0712.0370&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. It is found that the likelihood of the non-Lambda model is comparable to that of the LCDM one. However such a fit requires a very low Hubble rate ( $ \sim 0.40 - 0.45 h^{-1} $ ) compared to the widely accepted $ \sim 0.73 h^{-1} $, and also requires a significant amount of negative curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These shortcomings aside, a very recent work by &lt;a href=&quot;http://arxiv.org/abs/1011.2046&quot; rel=&quot;nofollow&quot;&gt;Hael Collins&lt;/a&gt; allows an analytic approach to the problem, albeit in a simplified setting. To quote from the abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This example provides an illustration of a universe where the inhomogeneities can affect its average expansion rate; and its simplicity allows a condition to be derived that tells when their presence should begin to become important. &lt;em&gt;Since the averages of the non-uniform parts of the metric and the matter density grow faster than their uniform parts, the average expansion rate accelerates with the advent of the era governed by the inhomogeneities.&lt;/em&gt; (emphasis mine).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So as they say, you free to &quot;worship at the church of your choice&quot; :-D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Update: Nov. 14, 2010]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... but if these (&lt;a href=&quot;http://arxiv.org/abs/0709.0732&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0702082&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;) references to papers by David Wiltshire have anything to say about the question of &quot;apparent&quot; vs. &quot;actual&quot; cosmic acceleration, it would appear that many in the cosmology community have been worshipping the false god of &quot;dark energy&quot; !&lt;/p&gt;&#xA;" LastActivityDate="2010-11-14T08:31:13.053" />
  <row Id="695" PostTypeId="2" ParentId="691" CreationDate="2010-11-13T04:30:03.240" Score="0" ViewCount="" Body="&lt;p&gt;Time dilation and the blue shifting of incoming radiation go to infinity as you approach the event horizon, and you'll cross that before you get near the singularity. This means that, as you said, you'll be fried by blue-shifted radiation from the rest of the universe before your vaporized remains cross the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;To clarify my answer, from the point of view of an observer approaching the event horizon of a black hole, down-falling radiation from a distant source will be blue-shifted. This blue-shifting goes to infinity as the observer approaches the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, from his point of view, the regions of the universe far from the black hole will be accelerated in time. The rest of the universe sees him slow down, and he sees the rest of the universe speed up. This time dilation effect also goes to infinity as he approaches the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not really correct to say that an observer won't know that he is crossing an event horizon. The size of a black hole only affects tidal forces; they are small at the event horizon of a large black hole. But other effects are still important.&lt;/p&gt;&#xA;" OwnerUserId="150" LastEditorUserId="150" LastEditorDisplayName="" LastEditDate="2010-12-15T20:16:31.597" LastActivityDate="2010-12-15T20:16:31.597" CommentCount="3" />
  <row Id="696" PostTypeId="2" ParentId="614" CreationDate="2010-11-13T04:48:22.357" Score="0" ViewCount="" Body="&lt;p&gt;Look up the phrase &quot;impact parameter&quot;. Perhaps in combination with &quot;Rutherford scattering&quot;.   Your basic geometric model is not exactly right.   Once you see it, everything will work out quantitatively, both in classical and quantum mechanics, using different other ideas, of course.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-11-13T04:48:22.357" />
  <row Id="697" PostTypeId="2" ParentId="625" CreationDate="2010-11-13T09:05:44.373" Score="3" ViewCount="" Body="&lt;p&gt;As far as I know -- &quot;second quantization&quot; is just a deprecated term, used traditionaly.&#xA;See e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Second_quantization#Second_quantization%3a_field_theory&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2010-11-13T09:05:44.373" CommentCount="2" />
  <row Id="698" PostTypeId="2" ParentId="196" CreationDate="2010-11-13T09:06:04.437" Score="1" ViewCount="" Body="&lt;p&gt;Basically, this is because simultaneity is not an invariant notion in SR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You probably know the classical example of two lightnings striking at the same time but different spots with respect to an observer who is at rest. But for an observer moving towards one of the lightnings, the lightning he is moving towards will have struck first. To an observer moving in the opposite direction, it will be the lightning he's moving towards. So each observer arranges the events in a different temporal sequence. This is true for space-like separated events, to use the technical jargon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not problematic because space-like separated events should not be causally related... unless there is FTL communication/travel/whatever... That's when the funny things happen, when Jack responds to the phone call before Judy ever made it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've always wondered if the show writers of Star Trek, or any other SF series using FTL com, thought about it. For all the time travel stories these shows have, I can't remember a story that exploited this particular effect.&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2010-11-13T09:37:44.283" LastActivityDate="2010-11-13T09:37:44.283" />
  <row Id="699" PostTypeId="2" ParentId="196" CreationDate="2010-11-13T09:30:13.943" Score="0" ViewCount="" Body="&lt;p&gt;Let's assume you are stationary and are observing a spaceship flying by at a velocity $v$. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Case $v \lt c$: If the spaceship flies at relativistic speeds, you will notice their clocks going slower, but everything would happen in the same order as if they were stationary with respect to you. The causal relationships between events is preserved, so is the second law of thermodynamics.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Case $v = c$: If the spaceship flies at the speed of light, you will notice their clocks stopped. Nothing is moving, so, in a limit sense, the causal relationships between events is preserved, so is the second law of thermodynamics.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Case $v \gt c$: If the spaceship flies faster than light, you will notice their clocks going backwards. All the causal relationships are reversed and the second law of thermodynamics is violated.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-13T09:30:13.943" />
  <row Id="700" PostTypeId="2" ParentId="408" CreationDate="2010-11-13T09:38:40.023" Score="2" ViewCount="" Body="&lt;p&gt;There is some work on the subject by professor Narayan. As David said, a black hole will usually have an accretion disk that radiates a lot of energy. That energy comes from the gravitational potential energy. As the material of the disk falls towards the central object, it transforms its potential energy to radiation while the material is on the disk and interacts with the rest of the disk. On the final part of the infall of the material to the central object, where the infalling material gets separated from the interior edge of the disk, the gravitational potential energy is transformed to radiation only if there is a central object with a surface. If the central object is a black hole, then there should be a lot of energy missing from the center of the accretion disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is again, not a direct detection, in the sense that you don't see the black hole itself. The only thing that a black hole can emit is Hawking radiation, but that radiation is relevant only for non-astrophysical black holes (collapsed stars or centers of galaxies), like the primordial black holes that Chad mentioned (which are small) or mini black holes that could be created in an accelerator or from cosmic radiation.&lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2010-11-13T09:38:40.023" />
  <row Id="701" PostTypeId="2" ParentId="78" CreationDate="2010-11-13T10:09:47.477" Score="10" ViewCount="" Body="&lt;p&gt;Neutrons (and protons) being spin 1/2 fermions, must fit antisymmetric wavefunctions.  This &quot;wavefunction&quot; doesn't always involve waves, though.  For nucleons - the generic term for neutron or proton - this  wavefunction for the pair is a produce of (1) a spatial part, (2) a spin part, and (3) an isospin part.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The isospin part is a clever way to describe charge possibilities of otherwise identical particles. We regard neutrons and protons as being in a sense identical.  Just as a spin 1/2 particle can be &quot;up&quot; or &quot;down&quot; along some chosen axis, so is an isospin 1/2 particle can be &quot;up&quot; or &quot;down&quot; along an abstract mathematical axis - it's exactly the same SU(2) math as spin - but it plays out in physical reality as charge.   For nucleons it's not +1/2 and -1/2 charge but with an offset, so we have +1 (proton) and 0 (neutron).   This idea is from Heisenberg in 1932.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, how can the overall wavefunction of a pair of particles be antisymmetric?  There are three factors - right away we can imagine three possibilites: any one factor being antisymmetric with the other two symmetric.   We could also have all three antisymmetric at the same time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An antisymmetric spatial wavefunction would have a node, like an atomic p orbital, like the electric potential around a dipole antenna.  This is a higher energy state than a simple spherical blog, a Gaussian.   Given the range of internuclear forces, this nodal antisymmetric wavefunction has more energy than if the two nucleons just stayed apart. This is a matter of radial or angular kinetic energy has to be either &quot;zero&quot; or some quantized value that exceeds &quot;escape velocity&quot;  So forget that part of the system wavefunction being antisymmetric.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, we don't have separate spatial wavefunctions for the two nucleons - whatever one does, the partner does the exact opposite, like a two-body celestial mechanics problem.  They orbit a common barycenter.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spin part could be antisymmetric.  This is a bit tricky.  If particle #1 is up and #2 is down, we can write &quot;UD&quot;.  There is also &quot;DU&quot;.  We form the spin part of the wavefunction for the pair as UD-DU.  We could instead choose UD+DU but note this is symmetric. So are UU and DD.  Just how UD-DU differs from UD+DU may mystify beginners in quantum mechanics, but it's important, and it's how physical matter works whether we humans like it or not.  (You might also see where the 'u' and 'd' quarks got their names.  The quark idea came along years after isospin.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither D nor U is really a wave or a function; they're at most just rows and columns in matrices if you must represent them in familiar math.  Otherwise quantum physicists just deal with these symbolically.  Still the jargon is &quot;wavefunction&quot; - we silly humans and our primitive scientific language!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same math applies for isospin.   But the physics differs.   We've concluded that the spatial part of the system wavefunction must be symmetric, so it's up to either the spin part or the isospin part to be antisymmetric.  But not both!   If spins are symmetric - they're parallel.  This is experimentally the case - the Deuteron (obtain by distilling &quot;heavy water&quot; from water) - so we deduce that the isospin part is antisymmetric.  That is, we must have one isospin &quot;up&quot; and one isospin &quot;down&quot; - neutron and a proton, not two neutrons or two protons.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just why must the spins of the two nucleons be parallel?  The strong force holding them together - the exchange of pions, kaons and other mesons - works better in that case.  To explain that takes deeper analysis than I can go into here.   When the spins are antiparallel, there's not enough force to keep the nucleons together.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be the case though, were to try pushing two neutrons together.  They'd be both isospin &quot;up&quot; therefore symmetric isospin part of the wavefunction, therefore requiring an antisymmetric spin part, which leads to the pions and their buddies not getting as good a grip on the neutrons, which drift off going their separate ways.&lt;/p&gt;&#xA;" OwnerUserId="353" LastActivityDate="2010-11-13T10:09:47.477" CommentCount="2" />
  <row Id="702" PostTypeId="2" ParentId="625" CreationDate="2010-11-13T10:16:29.613" Score="-4" ViewCount="" Body="&lt;p&gt;At the cost of attracting the ire of the &lt;em&gt;hardcore&lt;/em&gt; physicists, if I can rephrase your question in &quot;mere mortal&quot; terms:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;According to classical physics: $x^\mu$ describes reality&lt;/li&gt;&#xA;&lt;li&gt;According to quantum mechanics: $\psi$ describes a &quot;meta-reality&quot; which allows us to describe the result of large numbers of measurements, but not single ones.&lt;/li&gt;&#xA;&lt;li&gt;According to quantum field theory: $\Phi$ describes a &quot;meta-meta-reality&quot; which allows us to describe the &quot;meta-realities&quot; that allow us to describe the result of large numbers of measurements.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At each step, the theory moves away from the physical world, and becomes, more and more, a collection of mathematical algorithms - in the sense that it describes something that might as well not exist in the &quot;real world&quot; (the &quot;meta-&quot; realities). The &quot;meta-&quot; realities cannot be directly measured, so I think it's reasonable to question whether they are real at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this sense, your question is more philosophical or epistemological than physical. Surely, mathematically you can quantise as many times as you like. Our ability to calculate predictions and the detachment from physical reality, though, diminish the predictive power of these theories and surely the theories tend to increase our understanding of reality much less at each step. It becomes harder and harder to defend them as real &quot;physical laws&quot;.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-13T10:16:29.613" CommentCount="5" />
  <row Id="703" PostTypeId="1" CreationDate="2010-11-13T10:28:05.627" Score="-1" ViewCount="260" Body="&lt;p&gt;If gravity exists quantumly it must take up space, have frequency, velocity, etc. Perhaps gravity is too fine a frquency for detection.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-14T18:27:15.003" Title="Is gravity a vector or tensor function and does gravity have velocity or is it instantaneous?" Tags="&lt;gravity&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="704" PostTypeId="1" AcceptedAnswerId="713" CreationDate="2010-11-13T10:29:49.087" Score="3" ViewCount="269" Body="&lt;p&gt;If you read the article &quot;More Is Different&quot;, by &lt;a href=&quot;http://en.wikipedia.org/wiki/Philip_Warren_Anderson&quot; rel=&quot;nofollow&quot;&gt;P.W. Anderson&lt;/a&gt; (Science, 4 August 1972), you will find a deep question: are the physical laws dependent of the size of the system under study?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, we can ask ourselves, are the description of a hundred of atoms more than simply one hundred times the description of one atom alone? Of course we have interactions, but, are these interactions dependent of the number of particles implied?&lt;/p&gt;&#xA;" OwnerUserId="306" LastEditorUserId="306" LastEditorDisplayName="" LastEditDate="2010-11-13T10:46:44.627" LastActivityDate="2010-11-13T17:47:24.780" Title="Are the physical laws scale-dependent?" Tags="&lt;mechanics&gt;&lt;quantum-mechanics&gt;&lt;statistical-mechanics&gt;" AnswerCount="4" FavoriteCount="4" />
  <row Id="705" PostTypeId="2" ParentId="703" CreationDate="2010-11-13T10:35:37.600" Score="1" ViewCount="" Body="&lt;p&gt;According to general relativity, gravity is an apparent effect of the distortion of space-time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both quantum field theory and general relativity describe some of the properties you mentioned:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In QFT it is possible to describe particles known as gravitons that mediate gravity&lt;/li&gt;&#xA;&lt;li&gt;In GR it is predicted that gravitational effects could spread in gravitational waves&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are some experiments being performed to detect gravitational waves through &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_wave_detector&quot; rel=&quot;nofollow&quot;&gt;gravitational wave detectors&lt;/a&gt;. Although the experiments are very difficult due to the minuscule effects expected, the frequencies are not the problem.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-13T10:35:37.600" />
  <row Id="706" PostTypeId="2" ParentId="703" CreationDate="2010-11-13T10:40:07.313" Score="0" ViewCount="" Body="&lt;p&gt;The speed of gravitational interactions is well-known, in the classical GR context, but must be the same in a quantum theory of gravity. This speed is c, the speed of light. The hypotethical particle of gravitational interaction is called graviton. It is a massless particle (so it can go at the speed of light) and it should have integer spin (as all bossons, which carries forces).&lt;/p&gt;&#xA;" OwnerUserId="306" LastActivityDate="2010-11-13T10:40:07.313" />
  <row Id="707" PostTypeId="2" ParentId="704" CreationDate="2010-11-13T10:48:49.350" Score="2" ViewCount="" Body="&lt;p&gt;It's a very interesting question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly, QM is scale dependent, as the planck-level world described is completely separated from the human scale world and can only be accessed through a disruptive measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, there must be a crossover point at which the quantum description, as described by the wave function $\psi$ collapses in a measured value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General Relativity is not scale dependent. However, gravity is very weak and it's very hard to perform experiments about GR at a microscopic level, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other theories also tend to describe the very small, most of them, like string theory, do tend to reduce to QM in some appropriate limit and hence inherit its scale dependency.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-13T10:48:49.350" CommentCount="9" />
  <row Id="708" PostTypeId="2" ParentId="647" CreationDate="2010-11-13T11:58:52.737" Score="1" ViewCount="" Body="&lt;p&gt;I'm not entirely sure that this is what you're looking for, but here's a Lagrangian take on the system. I choose coordinates so that $x_1(0)=0,x_2(0)=0$, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L=m/2(\dot x_1^2 + \dot x_2^2) - k/2 (x_1-x_2)^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easier to solve if we go to coordinates $$X=1/2(x_1+x_2), \Delta x = 1/2 (x_1-x_2)$$ so that $$L = m(\dot X^2 + \Delta x^2) - 2k\Delta x^2$$. This gives us the following Euler-Lagrange equations -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dt}\left(2m\dot X\right) = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dt}\left(2m\dot\Delta X\right) = -4k\Delta x$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you'd expect, the center-of-mass motion is constant and the second equation describes an oscillating $\Delta X$. What determines the amplitude of your oscillation here is the initial condition - the bullet strikes one end of the molecule and transfers momentum $p_0$there, so that you get initial conditions $$x_1=0,x_2=0,\dot x_1 = p_0/m, \dot x_2 = 0$$. If we parametrize the oscillation as $\Delta x = \alpha \sin(\omega t + \phi)$, at $t=0$ we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta x(0) = \alpha \sin(\phi) = 0, \Delta \dot x = \alpha \omega \cos(\phi) = p_0/2m $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives us &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi=0, \alpha = \frac{p_0}{2\omega m}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So of course $\alpha$ is completely determined in this picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing with the model in your question, I would say you mistreat the collision by assuming that the bullet collides with the entire molecule as a rigid unit, while in fact the collision is between the bullet and one atom, both of equal mass, so that $v_b$ will be zero, $v_m$ becomes exactly $v_0/2$ and you can calculate your $\alpha$ from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering a little about the validity of my approach when going to infinite $k$ since that SHOULD change the collision process to be 'with the entire system' and I'd like to see a kind of $k$ dependence in $v_b$, but I can't see how to properly treat this at the moment.&lt;/p&gt;&#xA;" OwnerUserId="243" LastEditorUserId="243" LastEditorDisplayName="" LastEditDate="2010-11-13T17:44:01.250" LastActivityDate="2010-11-13T17:44:01.250" CommunityOwnedDate="2010-11-13T11:58:52.737" CommentCount="2" />
  <row Id="709" PostTypeId="2" ParentId="89" CreationDate="2010-11-13T12:53:53.407" Score="1" ViewCount="" Body="&lt;p&gt;Newton's Law of Gravity is consistent under General Relativity at high speed too :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets consider Newton equation of energy conservation for free fall from the infinity with initial speed of object equal to zero:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\large {mc^2=E-\frac{GMm}{R}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\large {mc^2=E-\frac{R_{g*}}{R}\;mc^2}$  where $\large {R_{g*}=GM/c^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\large {E=mc^2\left(1+\frac{R_{g*}}{R}\right)=mc^2\left(\frac{R+R_{g*}}{R}\right)}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\large {mc^2=E\;\frac{R}{R+R_{g*}}=E\left(1-\frac{R_{g*}}{R+R_{g*}}\right)}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and as the result&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bf\large {mc^2=E-\frac{GM}{R+R_{g*}}\;\frac{E}{c^2}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bf\large {mc^2=E-\frac{GMm}{R}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the resulting equation energy ($E/c^2$) is attracted, not  mass ($m$).&#xA;That's why gravitational redshift  is the same in  Newton Gravity and in General Relativity (for $R&gt;&gt;R_g$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slight modification of Newton equation describes radial movement of an object &lt;strong&gt;at any speed&lt;/strong&gt; with different initial conditions in the same way as General Relativity. Not only free fall from infinity with initial speed equal to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bf\large {E_1\left(1-\frac{GM}{c^2(R_1+R_{gm}+R_{gM})}\right)=E_2\left(1-\frac{GM}{c^2(R_2+R_{gm}+R_{gM})}\right)}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it has no any singularity! So I like it :)&lt;/p&gt;&#xA;" OwnerUserId="238" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2010-11-18T19:21:38.650" LastActivityDate="2010-11-18T19:21:38.650" CommentCount="13" />
  <row Id="710" PostTypeId="2" ParentId="625" CreationDate="2010-11-13T13:33:33.097" Score="2" ViewCount="" Body="&lt;p&gt;I agree with Kostya that these names are deprecated and, in this sense, should be avoided (A. Zee's, &quot;QFT in a Nutshell&quot;,  book makes this point pretty straightforwardly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you think of the process of &quot;quantization&quot; as a &lt;em&gt;functor&lt;/em&gt;, you get to Baez's constructions. But, note that the objects being acted upon by this 'quantization functor' get progressively different from what you may be expecting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example that comes to mind is the quantization of &lt;a href=&quot;http://en.wikipedia.org/wiki/Bundle_gerbes&quot; rel=&quot;nofollow&quot;&gt;gerbes&lt;/a&gt;, which does make an appearance in high energy physics (see section 3 of &lt;a href=&quot;http://arxiv.org/abs/0905.2720&quot; rel=&quot;nofollow&quot;&gt;Geometric Langlands From Six Dimensions&lt;/a&gt;). But these objects are very non-intuitive from the Physics point-of-view: you don't even get an &lt;a href=&quot;http://mathoverflow.net/questions/38825/what-kind-of-lagrangians-can-we-have/38828#38828&quot; rel=&quot;nofollow&quot;&gt;Action&lt;/a&gt; associated to this construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, at this point, the furthest we've moved in this direction is String Field Theory. But, in some sense, &quot;quantization&quot; is still a mystery…&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2010-11-13T13:33:33.097" />
  <row Id="711" PostTypeId="2" ParentId="704" CreationDate="2010-11-13T13:46:31.303" Score="0" ViewCount="" Body="&lt;p&gt;As Sklivivvz already said, &lt;em&gt;most&lt;/em&gt; Quantum Mechanical effects only show effectively on a &quot;quantum scale&quot;, which is dependent on the variables involved. Although this scale is again dependent on the system itself, and so makes the reasoning moot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean scale dependent as in &quot;the speed of light may well be different on a &lt;em&gt;very&lt;/em&gt; small scale&quot;, or &quot;the planck constant may be different on very large scale&quot;, I cannot answer as I do not know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I know is this: most physical laws as shown in textbooks are approximations in well-defined areas (e.g. the law of Newton for (not too) small r, the particle physics DGLAP evolution equations for their domain of relevance in kinetic space, etc...) To this effect, these laws in physics are scale dependent. The much more general derivations are (as far as I understand) not scale dependent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example to &quot;prove&quot; the above statements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum effects in macroscopic systems: Fe as a conductor can be described as a Fermi sea of electrons, which is a very quantized state and obeys quantum mechanical laws on a macroscopic level. Properties like specific heat and conduction are direct consequences of the quantum mechanical description of this system, and to that effect, the macroscopic result is somewhat &quot;the sum of quantum states&quot;. This is also exactly what statistical physics is based on: sums of states, and this is exactly what connects the &quot;quantum world&quot; with what we can observe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I misunderstood your question or make claims you think are false, please say so.&lt;/p&gt;&#xA;" OwnerUserId="317" LastActivityDate="2010-11-13T13:46:31.303" />
  <row Id="712" PostTypeId="2" ParentId="679" CreationDate="2010-11-13T13:53:17.017" Score="-1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.khanacademy.org/#Physics&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="309" LastActivityDate="2010-11-13T13:53:17.017" />
  <row Id="713" PostTypeId="2" ParentId="704" CreationDate="2010-11-13T14:16:22.713" Score="9" ViewCount="" Body="&lt;p&gt;It is important to digest appropriately Anderson's comments about scaling. In Physics, when one talks of &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Scale_invariance&quot; rel=&quot;nofollow&quot;&gt;scaling phenomena&lt;/a&gt;&quot;, what's &lt;em&gt;really&lt;/em&gt; being talked about are these two things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Renormalization_group&quot; rel=&quot;nofollow&quot;&gt;Renormalization Group&lt;/a&gt;; &amp;amp;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/nucl-th/0510023&quot; rel=&quot;nofollow&quot;&gt;Effective Field Theory&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And, as i mentioned above, &lt;a href=&quot;http://en.wikipedia.org/wiki/Conformal_symmetry&quot; rel=&quot;nofollow&quot;&gt;conformal symmetry&lt;/a&gt; plays a leading role in all of this discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly, the bottom line is something like this: every physical theory has its domain of validity, ie, its &quot;laws&quot; are only valid within certain &quot;conditions&quot;, which we usually express in terms of an Energy Scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, e.g., GR is valid in certain regimes that we call &quot;relativistic&quot;: if you're too slow compared to the speed-of-light, Newtonian gravity is a very good approximation. In this sense, Newtonian gravity is an &quot;effective description&quot; of GR (in the appropriate energy scale).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true for Quantum Field Theories: you can start with a given description at a given energy (or length) scale and, as you change your scale, either increasing or decreasing the energy of the phenomena involved, you'll be lead to different theories, in order to describe the new, effective, phenomena that you'll see. For instance, you can describe the world in terms of protons and neutrons or in terms of quarks and gluons — the only change is in the energy scale used and, as such, in the &quot;effective theory&quot; you'll be using to describe the ingredients your experiments measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the concepts really behind Anderson's argument. In fact, when he says &quot;more is different&quot;, he's alluding to a concept called &lt;a href=&quot;http://en.wikipedia.org/wiki/Emergence&quot; rel=&quot;nofollow&quot;&gt;emergent phenomena&lt;/a&gt;, which is basically described by the notion of Effective Field Theory i mentioned above. Here's a picture: you can describe a proton in terms of quarks and gluons, but it's very hard to describe a whole nucleus in terms of quarks and gluons — essentially because there are so many of them, that calculations become virtually impossible. So, what people do is to compute the Effective Field Theory of quarks and gluons, and use it instead to describe the whole nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar thing can be seen in Statistical Mechanics, when one observes that the behavior of  a collection of particles is very different from that of a single particle — this is the prototypical &quot;more is different&quot;: the physical properties of the collection of particles are not mirrored by the individual properties of each particle — this is &quot;emergence&quot;, and that's why we use Effective Field Theories to describe the collection of particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's really astounding is that, using Renormalization Group techniques, we can actually compute Effective Field Theories for several different energy scales! 8-)&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2010-11-13T14:16:22.713" CommentCount="3" />
  <row Id="714" PostTypeId="2" ParentId="408" CreationDate="2010-11-13T15:07:46.060" Score="3" ViewCount="" Body="&lt;p&gt;If you were close enough, or the black hole big enough, beams of light passing near it&#xA;would distort the image of objects more distant than the BH, also some of this light would &#xA;disappear down the BH's event horizon. So yes, if you had a big enough telescope you could see&#xA;it. But the size of BHs relative the their probable distance means this telescope won't be&#xA;around for many many years. In a similar vein, wouldn't looking at a BH against the cosmic&#xA;background radiation do the same trick? You'd see Hawking radiation at a much lower&#xA;temperature than the CMB (unless the BH is very small). But, again given the actual &#xA;size/distance of BHs, the needed resolution is far away from being achievable.&lt;/p&gt;&#xA;" OwnerUserId="391" LastEditorUserId="965" LastEditorDisplayName="" LastEditDate="2011-04-03T22:43:27.923" LastActivityDate="2011-04-03T22:43:27.923" CommentCount="3" />
  <row Id="715" PostTypeId="2" ParentId="317" CreationDate="2010-11-13T15:14:32.283" Score="1" ViewCount="" Body="&lt;p&gt;You might want to read about Bohmian mechanics. Bohmian mechanics is perfectly deterministic. The reason randomness appears is explained in the same way as the appearance of randomness in thermodynamic equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's some further reading with links to several papers at the bottom of the page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://plato.stanford.edu/entries/qm-bohm/#qr&quot; rel=&quot;nofollow&quot;&gt;http://plato.stanford.edu/entries/qm-bohm/#qr&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2010-11-13T15:14:32.283" CommentCount="4" />
  <row Id="716" PostTypeId="1" AcceptedAnswerId="719" CreationDate="2010-11-13T15:36:32.037" Score="0" ViewCount="196" Body="&lt;p&gt;The relativistic energy-momentum equation is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;E^2 = (pc)^2 + (mc^2)^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, we have pc = Ev/c, so we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E = mc^2/(1-v^2/c^2)^1/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, accelerating a proton to near the speed if light, I get the following results for the energy of proton:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.990000000000000   c =     0.0000000011    J =      0.01    TeV &lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.999000000000000   c =     0.0000000034    J =      0.02    TeV &lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.999900000000000   c =     0.0000000106    J =      0.07    TeV &lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.999990000000000   c =     0.0000000336    J =      0.21    TeV &lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.999999000000000   c =     0.0000001063    J =      0.66    TeV &lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.999999900000000   c =     0.0000003361    J =      2.10    TeV &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;0.999999990000000 c =     0.0000010630    J =      6.64    TeV&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.999999999000000   c =     0.0000033614    J =      20.98   TeV &lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.999999999900000   c =     0.0000106298    J =      66.35   TeV &lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.999999999990000   c =     0.0000336143    J =      209.83      TeV &lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.999999999999000   c =     0.0001062989    J =      663.54      TeV &lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.999999999999900   c =     0.0003360908    J =      2,097.94    TeV &lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.999999999999990   c =     0.0010634026    J =      6,637.97    TeV &lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.999999999999999   c =     0.0033627744    J =      20,991.10   TeV &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the LHC is accelerating protons to 7TeV it means they're traveling with a speed of 0.99999999c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is everything above correct?&lt;/p&gt;&#xA;" OwnerUserId="57" LastActivityDate="2010-11-13T16:03:23.893" Title="Relativistic speed/energy relation. Is this correct?" Tags="&lt;particle-physics&gt;&lt;special-relativity&gt;" AnswerCount="1" />
  <row Id="717" PostTypeId="2" ParentId="321" CreationDate="2010-11-13T15:37:29.417" Score="0" ViewCount="" Body="&lt;p&gt;cetripital force only exists when you have prescribed motion (due to constraints). Think of a roller coaster car riding on a rail. To keep the car on the rail and tangetial to it's direction a force and moment need to be applied to the car. When the path is circular we call the cetripetal force. In fact, with any path, instanteously it is said to be following a circle and therefore there is always an instanteneous centripetal force (unless in free fall).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planets and things in orbit do not have a prescribed motion, but are following the free fall path whichever way they need to go. Gauss called this the principle of least action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-11-13T15:37:29.417" />
  <row Id="718" PostTypeId="2" ParentId="639" CreationDate="2010-11-13T15:49:53.207" Score="1" ViewCount="" Body="&lt;p&gt;My &lt;em&gt;guess&lt;/em&gt; is this is a property of turbulent boundary layers over rough surfaces. You may be able to associate the roughness and windspeed to a &lt;a href=&quot;http://en.wikipedia.org/wiki/Strouhal_number&quot; rel=&quot;nofollow&quot;&gt;Strouhal&lt;/a&gt; number, and together with the Raynolds number give you at what situation you have this $\log()$ law be dominant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for a derivation of this behavior, I am hoping someone with fluids experience can actually answer your question.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-11-13T15:49:53.207" />
  <row Id="719" PostTypeId="2" ParentId="716" CreationDate="2010-11-13T15:55:47.100" Score="1" ViewCount="" Body="&lt;p&gt;Yes you are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the rest mass of a particle is $m$ and the total energy is $E$, then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E = \gamma mc^2 = \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}, $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac vc = \sqrt{ 1 - \left( \frac{mc^2}E \right)^2 } \approx 1 - \frac12 \left( \frac{mc^2}E \right)^2 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proton rest mass is 938 MeV, so at 7 TeV, the proton's speed is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1 - \frac vc = \frac12 \left( \frac{938\times10^6}{7\times10^{12}} \right)^2 = 9 \times 10^{-9} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;meaning v ~ 0.999 999 991 c&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-11-13T16:03:23.893" LastActivityDate="2010-11-13T16:03:23.893" />
  <row Id="720" PostTypeId="2" ParentId="24" CreationDate="2010-11-13T16:09:32.263" Score="0" ViewCount="" Body="&lt;p&gt;Yes. A lean induces a side force that makes the bike follow a circular path.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-11-13T16:09:32.263" CommentCount="2" />
  <row Id="721" PostTypeId="1" AcceptedAnswerId="724" CreationDate="2010-11-13T17:38:07.253" Score="4" ViewCount="301" Body="&lt;p&gt;My question is silly formulated, but I want to know if there is some sensible physical question buried in it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose an exact copy of our universe is made, but where spatial distances and sizes are twice as large relative to ours.&#xA;Would this universe evolve and function just as ours?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since mass is proportional to volume which is distance cubed,&#xA;but strength of a rope is only distance squared so it would not work in proportion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that a universe with our physical laws that evloves like ours can only have 1 certain size ?&#xA;Or would the physical laws scale in proportion so it evolves the same ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can it only have 1 size, what is it relative to ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, suppose another exact copy of our universe is made, but where everything happends twice as fast relative to ours, would it evolve in the same way as ours?&#xA;(We can suppose if needed that our universe is deterministic)&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-01-21T10:18:02.830" Title="If the size of universe doubled" Tags="&lt;astrophysics&gt;&lt;mathematical-physics&gt;&lt;measurement-problem&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" />
  <row Id="722" PostTypeId="2" ParentId="704" CreationDate="2010-11-13T17:47:24.780" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;The question depends on what one's definition of &quot;physical law&quot; is.&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of the point of Anderson's article is to argue that strict reductionism is not what scientists do in practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More explicitly, one caricature of pure reductionism is that only an explanation starting from the very bottom, i.e. involving strings and quarks, etc. counts as physical law, whereas Anderson would argue more that any reasonably quantitative workable theory of phenomena based at any scale is a physical law.  I really doubt that any working physicists really hold the former position anymore, in part due to one of the great advances of post 1950's theoretical physics, the effective field theory and renormalization group philosophy that Daniel so nicely explained in his answer.  We don't have to go that deep to see some illustrations of this though.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;examples&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Let me give some easy examples and raise some discussion questions to help you decide what qualifies as a physical law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the first emergent insight that one learns in physics is the notion of &quot;center of mass&quot;, that is, the idea that in many cases you can treat an object made out of $10^{27}$ atoms as a single point particle!  Though it's not pointed out when you're a freshman, what's going on here is exactly the same as happens in all other applications of emergence - you ignore all the internal degrees of freedom of your sliding block, for instance, because they are much higher energy than the stuff that you want to talk about.  Does the fact that we ignore all that information make the explanation of the acceleration of a block any less of a &quot;physical law&quot;?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Already we see a generic fact about emergence -- at some point the explanation breaks down.  For instance, there's no way from treating a block as a point particle to be able to understand the sound it makes when it hits the ground, though we can get a very good description of its motion (in a vacuum, say)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two discoveries at the start of the 20th century showed that the entire edifice of Newton's laws and Galilean relativity is an emergent phenomenon.  Note that this understanding doesn't invalidate most applications of Newton's laws -- that the limit $c\rightarrow\infty$ and $\hbar\rightarrow0$ simplifies to something that high school students can calculate with is really amazing, isn't it?  But, does the fact that they are &quot;just&quot; some limiting case mean that Newton's laws and Galilean relativity are not &quot;physical laws&quot;?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also turn these questions around and imagine ourselves in a universe where QM and SR were discovered first.  In that case, if people then discovered the classical limits, would people think of them as less &quot;physical&quot;?  &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;conclusion&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;One could and should spend some time thinking about what our physical principles are and what &quot;emergent&quot; assumptions and approximations go into them; breaking down the assumption that mechanics ended with Newton was the work of the 1st half of the 20th century.  On the other hand, Anderson points out that it's also useful to go in the opposite direction, in some ways, this is where the big problems in physics that I'm interested in are; to start with many-body descriptions which are impossible to calculate with and not very illuminating, and pick out the simplifications and insights that emerge!  Whether you call it a physical law or not is sort of besides the point, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are some of the things that come up when I think a bit about your question.  &lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-13T17:47:24.780" />
  <row Id="723" PostTypeId="1" CreationDate="2010-11-13T17:58:14.233" Score="5" ViewCount="360" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Polywell&quot; rel=&quot;nofollow&quot;&gt;Polywell&lt;/a&gt; is a proposed new type of fusion reactor, which is designed to use magnetic fields to overcome the problems with the Elmore-Tuck-Watson fusor.  I'm trying to understand exactly how it works.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An Elmore-Tuck-Watson fusor ... consists of a vacuum chamber containing a negatively charged outer grid (which may be the chamber) and a positively charged inner grid. Electrons are injected into the system and accelerated toward the inner grid. Most of the time, the electrons pass through the grid, through the core, and through the inner grid again, which then decelerates them and reaccelerates them inward wherein they return through the core. As they pass repeatedly through the core, they generate a negatively charged zone, a potential well, which is called a virtual cathode.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt; So this device uses a constant high voltage applied to the electrodes to produce a region of negative electric charge in the center.  How is this possible?  Electrons repel each other, so creating a region in empty space in which electrons are more dense than their surroundings would seem to require a constant energy input.  Without any energy input, they'll just fly apart to a minimum energy equilibrium where the charge density is equal everywhere.  Applying a voltage to electrodes with no current flowing between them does not use up any energy, however.  So where is the energy to hold the clump of electrons together coming from?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Like the Elmore-Tuck-Watson (ETW) fusor, the polywell confines positive ions through their attraction to the negative potential well which is created by the electrons that are held inside a positively charged grid. However, to avoid the losses related to the electrons striking the grid, the Polywell uses magnetic fields to shield the grid. The magnetic fields are configured in a way that adds to the confinement of the electrons so that there are many more electrons inside the core than outside.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2:&lt;/strong&gt; So the magnetic fields generated by the grid would supposedly prevent the electrons from hitting the grid.  The positively charged inner grid is attractive to electrons, but it's also magnetized, and somehow the magnetic fields repel electrons away from the grid?  But how can unchanging/steady-state magnets confine (or repel) electrons?  A bar magnet doesn't create any kind of movement or displacement of charge.&lt;/p&gt;&#xA;" OwnerUserId="176" LastEditorUserId="176" LastEditorDisplayName="" LastEditDate="2011-04-08T14:28:08.753" LastActivityDate="2011-04-12T13:04:13.000" Title="Help fill in my understanding of the Polywell fusion reactor" Tags="&lt;electromagnetism&gt;&lt;electricity&gt;&lt;electrodynamics&gt;&lt;fusion&gt;&lt;magnetohydrodynamics&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="724" PostTypeId="2" ParentId="721" CreationDate="2010-11-13T18:01:32.763" Score="5" ViewCount="" Body="&lt;p&gt;If you think for a moment about how lengths and speeds in our universe are set (that is, independent of how we choose to measure them, by meters or seconds or whatever), you'll see that these must ultimately come from different ratios of fundamental constants.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know and would be very surprised if there's a way to change these ratios so that all lengthscales or all timescales would change, since so much of what we observe actually comes from very complicated interacting systems acting at different timescales and lengthscales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is probably no, there is no sensible physical question behind what you've asked, at least the way you've asked it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could ask of course, what would happen if some (dimensionless) physical constant was doubled or halved, and then we could chat some more.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do I emphasize dimensionless here?  What if you asked &quot;What happens if the speed of light was doubled?&quot;  Well, because the speed of light is what sets our time and length scales, we would observe no changes at all!  In other words, think about what you can compare the speed of light to that doesn't depend on the speed of light itself.  Just saying the numerical quantity changes is meaningless because the units we use to measure it rely (via a possibly long chain of dependencies) on the speed of light itself!&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-13T18:01:32.763" CommentCount="3" />
  <row Id="725" PostTypeId="2" ParentId="550" CreationDate="2010-11-13T18:47:42.183" Score="3" ViewCount="" Body="&lt;p&gt;I would like to emphasize a point mentioned in the first answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classical black body radiation has to do with a gas of photons in thermal equilibrium. The photons interact with each other and that interaction gives the usual Planck spectrum for their energies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside stars, there are photons produced which interact with the stellar matter and they both come to a thermal equilibrium. When the photons escape the star, they have a black body spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you take some amount of matter (an electric stove) and heat it, there is the question of where do the photons come from. In low temperatures, you don't have enough thermal energy to excite electrons in atoms. So, where do the photons come from? The answer is that they must come from the excitation of the background electromagnetic field. Even if you tried to isolate the heated matter from every electromagnetic field, you could never block out the vacuum fluctuations. &lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2010-11-13T18:47:42.183" />
  <row Id="726" PostTypeId="2" ParentId="303" CreationDate="2010-11-13T18:48:11.593" Score="3" ViewCount="" Body="&lt;p&gt;The major source of sound when water is disturbed is the creation of bubbles, which oscillate at a resonant frequency underwater.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ffden-2.phys.uaf.edu/311_fall2004.web.dir/Ryan_Rankin/bubble%20resonance.htm&quot; rel=&quot;nofollow&quot;&gt;Bubble Resonance&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.annualreviews.org/doi/abs/10.1146/annurev.fl.25.010193.003045&quot; rel=&quot;nofollow&quot;&gt;The Impact of Drops on Liquid Surfaces and the Underwater Noise of Rain&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-11-13T18:48:11.593" />
  <row Id="727" PostTypeId="1" AcceptedAnswerId="732" CreationDate="2010-11-13T18:55:39.760" Score="0" ViewCount="101" Body="&lt;p&gt;Suppose I have a hill that goes up and down unevenly.  It's frictionless, and I want to slide a point mass down the hill.  I am interested in the path it takes.  (By &quot;path&quot; I mean the trail it leaves behind, not the time-dependent trajectory.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can find the equations of motion of the point mass and, from  there, find the path by finding the entire time-dependent motion.  Is there a way to find the path directly, given the initial velocity of the particle?  For example, does the correct path extremize something?&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-13T19:05:11.817" LastActivityDate="2010-11-13T21:12:46.043" Title="What condition is fulfilled by the path of a mass sliding down a lumpy hill?" Tags="&lt;classical-mechanics&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="728" PostTypeId="1" AcceptedAnswerId="819" CreationDate="2010-11-13T19:00:42.380" Score="15" ViewCount="1038" Body="&lt;p&gt;Suppose I have a lumpy hill.  In a first experiment, the hill is frictionless and I let a ball slide down, starting from rest.  I watch the path it takes (the time-independent trail it follows).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the next experiment, the hill stays the same shape, but the ball now rolls without slipping down the hill.  Assume there is no deformation of the ball or of the hill, no micro-sliding of the contact surface, and no other forms of &lt;a href=&quot;http://en.wikipedia.org/wiki/Rolling_resistance&quot; rel=&quot;nofollow&quot; title=&quot;rolling resistance&quot;&gt;rolling resistance&lt;/a&gt;. Energy is conserved.  I release the ball from rest at same point on the hill.  I track the path the rolling ball takes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the rolling ball follow the same path as the sliding ball, but more slowly, or does it sometimes follow a different path?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: There may be some &quot;trick&quot; answers.  For example, if the hill curves significantly on scales similar to  the radius of the ball, the ball could get &quot;wedged in&quot; somewhere, or two parts of the ball could contact the hill at the same time.  Let's assume that geometrically, the shape of the hill is such that it is only possible to have the ball contact the hill at one point, and that the ball always contacts the hill at a single point (i.e. it never flies off).&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-15T22:38:00.283" LastActivityDate="2011-04-25T01:57:19.990" Title="Will a ball slide down a lumpy hill over the same path it rolls down the hill?" Tags="&lt;classical-mechanics&gt;" AnswerCount="10" CommentCount="2" FavoriteCount="3" />
  <row Id="729" PostTypeId="2" ParentId="727" CreationDate="2010-11-13T19:14:20.703" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strike&gt;I think you're simply looking for the shortest path between the starting point and the arrival point under the constraint that the path should be on the surface.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as Mark pointed out, this can't be right. But what is still correct is the principle of least action: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=\int \; \left[ \frac{1}{2}m v^2 - mgz - \lambda (z-f(x,y)) \right] \; dt $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I chose a function of the form $z=f(x,y)$ to describe the landscape, but more general forms are possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extremizing this action leads to the following equations of motion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;m\frac{d^2 x}{dt^2} &amp;amp; = \lambda \frac{\partial f}{\partial x} \\\\&#xD;&#xA;&#xD;&#xA;m\frac{d^2 y}{dt^2} &amp;amp; = \lambda \frac{\partial f}{\partial y} \\\\&#xD;&#xA;&#xD;&#xA;m\frac{d^2 z}{dt^2} &amp;amp; = -mg -\lambda \\\\&#xD;&#xA;&#xD;&#xA;z=f(x,y)&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2010-11-13T21:12:46.043" LastActivityDate="2010-11-13T21:12:46.043" CommentCount="8" />
  <row Id="730" PostTypeId="2" ParentId="208" CreationDate="2010-11-13T19:16:14.920" Score="2" ViewCount="" Body="&lt;p&gt;The Sun is also losing mass due to the solar wind. Again the fraction of mass lost is very small compared to the mass of the Sun, so the effect is very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are these relevant papers that I think you will find interesting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0511138&quot; rel=&quot;nofollow&quot;&gt;Orbital effects of Sun's mass loss and the Earth's fate&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/0907.2469&quot; rel=&quot;nofollow&quot;&gt;Astrometric Solar-System Anomalies&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2010-11-13T19:16:14.920" />
  <row Id="731" PostTypeId="2" ParentId="270" CreationDate="2010-11-13T19:16:15.057" Score="4" ViewCount="" Body="&lt;p&gt;If I recall correctly, the color of a surface doesn't reliably indicate the emissivity.  The color in the visible spectrum is more of a side effect than anything.  Most emissions around body temperature or room temperature happen in the infrared, not the visible, and that's not reliably indicated by visible color:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/A0UZn.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The emissions of the sun, on the other hand, occur mostly in the visible region, which might explain the common perception that white reflects and black absorbs.  In the case of sunlight, this might be the dominant effect.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For example, white paint is quoted as having an absorptivity of 0.16, while having an &lt;a href=&quot;http://en.wikipedia.org/wiki/Emissivity&quot; rel=&quot;nofollow&quot;&gt;emissivity&lt;/a&gt; of 0.93. This is because the absorptivity is averaged with weighting for the solar spectrum, while the emissivity is weighted for the emission of the paint itself at normal ambient temperatures. ... The white paint will serve as a very good insulator against solar radiation, because it is very reflective of the solar radiation, and although it therefore emits poorly in the solar band, its temperature will be around room temperature, and it will emit whatever radiation it has absorbed in the infrared, where its emission coefficient is high. &lt;a href=&quot;http://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="176" LastEditorUserId="176" LastEditorDisplayName="" LastEditDate="2010-12-08T05:21:02.393" LastActivityDate="2010-12-08T05:21:02.393" />
  <row Id="732" PostTypeId="2" ParentId="727" CreationDate="2010-11-13T19:28:08.310" Score="1" ViewCount="" Body="&lt;p&gt;I think what Raskolnikov says is incorrect (I'd post this as a comment but can't because of low reputation). What he says would be correct in the absence of gravity, so it is just about finding geodesics on some manifold given by your surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your problem is harder, though, because what you are interested in are not geodesics, but trajectories that correspond to non-zero force acting on the particle. I am not sure what more could I write about this, because your question is too general: it is equivalent to solving equations of motion on (almost) arbitrary two-dimensional manifold with (almost) arbitrary potential.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-11-13T19:28:08.310" CommentCount="5" />
  <row Id="733" PostTypeId="2" ParentId="208" CreationDate="2010-11-13T19:29:20.077" Score="2" ViewCount="" Body="&lt;p&gt;The distance is increasing due to friction from the tides. Besides the moon, the sun has a tidal component for the earths oceans, and when that crashes into continents the energy absorbed comes from the potential energy of the sun-earth system. I am not sure how this compares to the distance loss due to the mass and energy radiation of the sun you mentioned.&lt;/p&gt;&#xA;" OwnerUserId="392" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2010-11-18T05:04:00.623" LastActivityDate="2010-11-18T05:04:00.623" CommentCount="2" />
  <row Id="735" PostTypeId="2" ParentId="728" CreationDate="2010-11-13T19:41:11.607" Score="4" ViewCount="" Body="&lt;p&gt;In one dimension, the answer is that the path must be the same, since the only possible thing that could go wrong is that the rolling ball might not make it over a hill that the sliding ball does (or vice versa).  However, this will never happen due to conservation of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried for a bit to put the general case into Lagrangian dynamics (there the question is formulated as whether the projections of $\vec{q}(t)$ will be the same, given two different Lagrangians) but then I remembered that a ball rolling without slipping is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Nonholonomic_system&quot; rel=&quot;nofollow&quot;&gt;non-holonomic system&lt;/a&gt; which makes the problem non-integrable, and I gave up.  Maybe I stopped too soon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect there is a clever solution to the problem; my intuition is that the potential energy term depends only on position and this implies that the trajectories will also be independent of the speed of the ball too.  I'll think a bit more about this.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-13T19:41:11.607" CommentCount="5" />
  <row Id="736" PostTypeId="2" ParentId="727" CreationDate="2010-11-13T20:37:16.937" Score="0" ViewCount="" Body="&lt;p&gt;If you know the path $\vec{r}(t)$ then you know the acceleration $\vec{a}=\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}$ and from that you get the force needed. $\vec{F}=m\cdot\vec{a}$ .This is also known as a forward dynamics problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the hill shape $\vec{r}(s)$ where $s$ is a parameter and the&#xA;derivatives $\vec{r}&amp;#39;=\frac{\mathrm{d}\vec{r}}{\mathrm{d}s}$, $\vec{r}&amp;#39;&amp;#39;=\frac{\mathrm{d}^{2}\vec{r}}{\mathrm{d}s^{2}}$  &#xA;fully describe the hill properties. To follow the hill, the particle&#xA;velocity has to be tangent to the hill $\vec{v}=\upsilon\vec{r}&amp;#39;$  &#xA;and the acceleration $\vec{a}=\dot{\upsilon}\vec{r}&amp;#39;+\upsilon^{2}\vec{r}&amp;#39;&amp;#39;$.&lt;br&gt;&#xA;The equations of motion are $\sum\vec{F}=m\,\vec{a}$ and the reaction&#xA;forces have to provide no work (or power) making them normal to the&#xA;path. Why? $\vec{F}_{\mathrm{reaction}}\cdot\vec{v}=\vec{F}_{\mathrm{reaction}}\cdot\upsilon\vec{r}&amp;#39;=0$  only if $\vec{F}_{\mathrm{reaction}}$ is perpendicular to $\vec{r}&amp;#39;$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can define the instanteneous tangential and normal unit vectors&#xA;at each point $s$ then the equations of motion are decomposed nicely&#xA;along those directions allowing to solve for the tangential acceleration&#xA;$\dot{\upsilon}$ and the reaction force $F_{\mathrm{reaction}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not fully understanding your question of, given a shape (path) what is the path without the equations of motion? You either know the path and get the forces, or you know the forces and get the path.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-11-13T20:37:16.937" CommentCount="1" />
  <row Id="737" PostTypeId="2" ParentId="721" CreationDate="2010-11-13T20:52:25.243" Score="1" ViewCount="" Body="&lt;p&gt;If the universe was flat (which is false), the size would be determined by the age and the speed of light. In other words, it would be a sphere of 13.7 billion light years in diameter.&#xA;But the speed of light is merely a physical constant which depends on our choice of units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It wouldn't be twice as large without being twice as old, and obviously this is a big difference :-) In other words, a universe twice as large with the same age, cannot have a big bang, or if you were to measure the age of the universe it would appear to be twice as much as we measure it.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-13T20:52:25.243" />
  <row Id="738" PostTypeId="2" ParentId="728" CreationDate="2010-11-13T20:58:58.747" Score="1" ViewCount="" Body="&lt;p&gt;This certainly isn't a rigorous answer, but it should be 'almost certainly not':&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pointwise, gravity will pull with a force in the direction of the gradient of the height function projected onto hill's surface, while friction will push in a direction opposing the motion of the ball.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the hill is lumpy, the direction of this projection of the hill's geometric gradient, and thus, the direction of the gravitational force, will vary with position in a way different from the variation in the direction of the ball's path.  Thus the direction of the net force on the ball will depend on whether or not you have the no-slip condition 'turned on'.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-11-13T20:58:58.747" CommentCount="4" />
  <row Id="739" PostTypeId="2" ParentId="728" CreationDate="2010-11-13T21:03:35.373" Score="0" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It the original trajectory, as seen from above, is straight, then friction will not curve it because it's always parallel and opposite to velocity, and velocity is always parallel to the trajectory. The only difference is that the ball will have less energy and it will not go the same distance. Sooner or later it will stop due to friction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the original trajectory, as seen from above, is curved, then the trajectory will be changed because in a curve velocity and trajectory are not parallel. If you need an example think about a ball going around a bowl. Without friction, the ball will keep on going on a stable &quot;orbit&quot; around the bowl. With friction it will spiral down until it stops.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edited due to a possible different interpretation of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the only effect of friction is to make the ball roll. The ball will acquire moment of inertia that will oppose any change in direction. If the trajectory was initially straight, nothing should change. If the trajectory was curved, it would follow a different one.&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-13T21:37:40.257" LastActivityDate="2010-11-13T21:37:40.257" CommentCount="9" />
  <row Id="740" PostTypeId="2" ParentId="721" CreationDate="2010-11-13T21:09:09.300" Score="4" ViewCount="" Body="&lt;p&gt;There is no absolute scale to the universe unless you 'turn on' quantum mechanics and general relativity simultaneously.  If you do this, then dimensional analysis can tell you that there is a way to use $\hbar$, $c$ and $G$ to construct a unique set of units--the Planck set of units--you can do it yourself pretty easily just by multiplying the three constants together, raised to some power, and trying to get meters, kilograms or whatever out.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, conceivably, we could tell if were in this other, 'distance squared' universe, by simply doing experiments to measure these three constants, and then calculating the Planck length.  If it is different than our Planck length, then it must be some other universe, where we chose a different scale for our distances, or one of the forces is different, or whatever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that answer what you were getting at?&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-11-13T21:09:09.300" CommentCount="2" />
  <row Id="741" PostTypeId="1" CreationDate="2010-11-13T21:16:03.690" Score="7" ViewCount="521" Body="&lt;p&gt;I would like to know more about &lt;a href=&quot;http://en.wikipedia.org/wiki/Ehresmann_connection&quot; rel=&quot;nofollow&quot;&gt;Ehresmann connections&lt;/a&gt; in vector bundles and how they relate to the electromagnetic field and the electron in quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt; The Schrödinger equation for a free electron is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{(-i\hbar\nabla)^2}{2m} \psi = i\hbar\partial_t \psi $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to write down the Schrödinger equation for an electron in an electromagnetic field given by the vector potential $A=(c\phi,\mathbf{A})$, we simply replace the momentum and time operator with the following operators&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rcl}&#xD;&#xA;  -i\hbar\nabla &amp;amp;\mapsto&amp;amp; D_i = -i\hbar\nabla + e\mathbf{A} \\&#xD;&#xA;   i\hbar\partial_t &amp;amp;\mapsto&amp;amp; D_0 = i\hbar\partial_t - e\phi&#xD;&#xA;\end{array}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have heard that this represents a &quot;covariant derivative&quot;, and I would like to know more about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My questions&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;(Delegated to &lt;a href=&quot;http://physics.stackexchange.com/questions/768/notation-for-sections-of-vector-bundles&quot;&gt;Notation for Sections of Vector Bundles&lt;/a&gt;.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I have heard that a connection is a &quot;Lie-algebra-valued one-form&quot;. How can I visualize that? Why does it take values in the Lie-algebra of $U(1)$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since a connection is a one-form, how can I apply it to a section $\psi$? I mean, a one-form eats vectors, but I have a section here? What is $D_\mu \psi(x^\mu)$, is it a section, too?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I apologize for my apparent confusion, which is of course the reason for my questions.&lt;/p&gt;&#xA;" OwnerUserId="188" LastEditorUserId="188" LastEditorDisplayName="" LastEditDate="2010-11-14T13:59:55.317" LastActivityDate="2010-12-16T09:45:09.077" Title="Electromagnetic Field as a Connection in a Vector Bundle" Tags="&lt;quantum-mechanics&gt;&lt;electromagnetism&gt;&lt;quantum-field-theory&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="6" />
  <row Id="742" PostTypeId="2" ParentId="484" CreationDate="2010-11-13T21:17:11.633" Score="1" ViewCount="" Body="&lt;p&gt;The only thing I can think of for the 'diameter of the observed area' is the inter-atom distance within the IGM, and for recombination to not take place, I would assume that the IGM in this model would still be hot enough to be a plasma.  A naïve answer would be that this is unlikely, considering that we're still not at infinite redshifts with respect to neighboring galaxies, much less the IGM between us and the neighboring galaxies.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-11-13T21:17:11.633" />
  <row Id="743" PostTypeId="1" CreationDate="2010-11-13T21:27:40.173" Score="15" ViewCount="647" Body="&lt;p&gt;What papers/books/reviews can you suggest to learn what renormalization &quot;really&quot; is? Standard QFT textbooks are usually computation-heavy and provide little physical insight in this respect - after my QFT course, I was left with the impression that renormalization is just a technical, somewhat arbitrary trick (justified by experience) to get rid of divergences. However, the appearance of renormalization in other fields of physics (renormalization group approach in statistical physics etc.), where its necessity and effectiveness have, more or less, clear physical meaning, suggests a general concept beyond mere &quot;shut up and calculate&quot; ad hoc gadget. I'm esp. interested in texts providing some unifying insight about renormalization in QFT, statistical physics or pure mathematics.&lt;/p&gt;&#xA;" OwnerUserId="307" LastActivityDate="2011-08-22T15:21:57.010" Title="Suggested reading for renormalization (not only in QFT)" Tags="&lt;quantum-field-theory&gt;" AnswerCount="12" CommentCount="2" FavoriteCount="7" />
  <row Id="744" PostTypeId="2" ParentId="743" CreationDate="2010-11-13T21:36:35.203" Score="2" ViewCount="" Body="&lt;p&gt;A really good textbook on QFT, in a new and exciting approach is &quot;Quantum Field Theory in a Nutshell&quot;, by Anthony Zee. It is a not so technical book in QFT, and with a deep insight in physics.&lt;/p&gt;&#xA;" OwnerUserId="306" LastActivityDate="2010-11-13T21:36:35.203" />
  <row Id="745" PostTypeId="2" ParentId="743" CreationDate="2010-11-13T21:45:30.217" Score="9" ViewCount="" Body="&lt;p&gt;Renormalization is &lt;em&gt;absolutely not&lt;/em&gt; just a technical trick, it's a key part of understanding effective field theory and why we can compute &lt;em&gt;anything&lt;/em&gt; without knowing the final microscopic theory of all physics. One good online source that explains a nice physical example is Joe Polchinski's &lt;a href=&quot;http://arxiv.org/abs/hep-th/9210046&quot; rel=&quot;nofollow&quot;&gt;&quot;Effective field theory and the Fermi surface&quot;&lt;/a&gt; (and you can also look up the references therein). Also, pretty much any modern field theory textbook will explain the modern Wilsonian point of view on effective field theory. Some recent books that try to emphasize physical insight and not just calculation are those by Zee and by Banks.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-11-13T21:45:30.217" />
  <row Id="746" PostTypeId="2" ParentId="721" CreationDate="2010-11-13T21:46:44.313" Score="0" ViewCount="" Body="&lt;p&gt;From what I know of cosmology, we are only able to see a small (and shrinking) portion of a much bigger universe. The only real measures of size (I think) are age (from the hubble constant) and density. If we wait for the age to roughly double (the dark energy expansion means we may get there a bit sooner), we will be at your proposed state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course one diference is the effect of time. Already at our present age, the universe is in a sort of old age, more than 90% of all the stars that will ever be have already been born. The galxacies are running low on the gas from which new stars are created. So your older less dense universe should be considerably less energetic on average than what we see around us.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-11-13T21:46:44.313" />
  <row Id="747" PostTypeId="2" ParentId="743" CreationDate="2010-11-13T22:04:16.917" Score="4" ViewCount="" Body="&lt;p&gt;There are several books that do this, from &lt;a href=&quot;http://books.google.com/books?id=nAXncL7_KrQC&quot; rel=&quot;nofollow&quot;&gt;Renormalization: an introduction&lt;/a&gt; and &lt;a href=&quot;http://books.google.com/books?id=60wX3xfdGRAC&quot; rel=&quot;nofollow&quot;&gt;Renormalization: an introduction to renormalization, the renormalization group, and the operator-product expansion&lt;/a&gt;, to &lt;a href=&quot;http://books.google.com/books?id=N8DBpTzBCJYC&quot; rel=&quot;nofollow&quot;&gt;Quantum field theory and critical phenomena&lt;/a&gt; and &lt;a href=&quot;http://books.google.com/books?id=mJku0CuJhVAC&quot; rel=&quot;nofollow&quot;&gt;Renormalization methods: a guide for beginners&lt;/a&gt;; or the more classics &lt;a href=&quot;http://books.google.com/books?id=Wt804S9FjyAC&quot; rel=&quot;nofollow&quot;&gt;Scaling and renormalization in statistical physics&lt;/a&gt; and &lt;a href=&quot;http://books.google.com/books?id=h8bvAAAAMAAJ&quot; rel=&quot;nofollow&quot;&gt;Finite quantum electrodynamics: the causal approach&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps…&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2010-11-13T22:04:16.917" CommentCount="3" />
  <row Id="749" PostTypeId="2" ParentId="647" CreationDate="2010-11-13T22:19:54.237" Score="0" ViewCount="" Body="&lt;p&gt;I believe, Thomas' answer is correct, even when I didn't think of the bullet colliding with the entire molecule as a &lt;em&gt;rigid&lt;/em&gt; unit, literally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The critical point is, why the target takes the momentum of the bullet instantly and completely (stopping the latter instantly), even when antagonized by the spring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's because the force of the spring is finite, while the force-by-collision is infinite. This state of affairs changes only - but drastically then -, when the force of the spring - i.e. $k$ - is infinite, too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or when we consider a smooth potential between the bullet and the target like $V(\Delta x) \propto \beta\text{e}^{-\beta \Delta x^2}$ or - more &quot;realistically&quot; something like $V(\Delta x) \propto \frac{1}{\beta \Delta x}$ - with $\beta \rightarrow \infty$ giving &quot;hard sphere collision&quot;. &lt;/p&gt;&#xA;" OwnerUserId="229" LastEditorUserId="229" LastEditorDisplayName="" LastEditDate="2010-11-13T22:41:50.200" LastActivityDate="2010-11-13T22:41:50.200" />
  <row Id="750" PostTypeId="2" ParentId="743" CreationDate="2010-11-13T22:34:23.620" Score="2" ViewCount="" Body="&lt;p&gt;Other people provided lots of references, so I'll just state what I think about the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are familiar with statistical physics part of the renormalization, you should already have a good grasp also on QFT renormalization (even if you don't yet know it!). The moral is the same here: divergences arise because our picture is only effective and, more generally, the theory doesn't account for all the realistic effects of the nature (like measurement).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The UV divergences appear because of infinite interaction energies and they point at the fact that the theory might be incomplete (i.e. it's just an approximation of some better underlying theory), so we are not really allowed to take infinite energy limit without somehow modifying our theory to accommodate for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about IR divergence? Well, this limit again cannot be taken if you think about it a little but the reason is different from the UV case. IR limit lets you count with arbitrarily small energies. But is this really physical? What about our measurement? Are we really able to measure arbitrarily small energies? Well, of course not. But QFT doesn't know anything about our measuring devices, so it's not surprising that you again have to account for this by hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another novel point of renormalization is a running coupling. And again, this arises precisely because we started to notice that coupling constants aren't really constants when one thinks more deeply about what it constitutes to measure something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the whole point of renormalization can be stated pretty concisely: it arises because we realized how ignorant we were. Both of ignoring the fact that QFT is not the ultimate theory of everything and also ignoring the subject of measuring.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-11-13T22:34:23.620" CommentCount="5" />
  <row Id="751" PostTypeId="2" ParentId="728" CreationDate="2010-11-13T23:09:15.477" Score="5" ViewCount="" Body="&lt;p&gt;A ball rolling down the hill will go slower than a ball sliding down the hill, because of its angular momentum. So if it rolls down the hill and into a banked curve, it will take the curve lower on the bank, and thus have a different path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: as the other answers show, this is in fact incorrect, for the reason that I suggested in my comment.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditorDisplayName="" LastEditDate="2011-02-06T02:38:54.373" LastActivityDate="2011-02-06T02:38:54.373" CommentCount="15" />
  <row Id="752" PostTypeId="2" ParentId="741" CreationDate="2010-11-14T01:04:25.170" Score="7" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Not really sure about this question. Are you perhaps just asking about notation? You can choose whatever you like. But usually you choose one coordinate system and just work in that. It's certainly a non-issue if you are working in flat spacetime: there you have nice global coordinates for everything.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First sentence is correct. For arbitrary Lie group $G$ with Lie algebra $\mathfrak{g}$ you might get a so called &lt;a href=&quot;http://en.wikipedia.org/wiki/G-structure&quot; rel=&quot;nofollow&quot;&gt;G-structure&lt;/a&gt; (e.g. $O(n)$-structure for Riemannian manifolds) and you can define a connection on that. It's rather heavy-duty mathematics, but at the end of the day you obtain a $\mathfrak{g} $-valued (more precisely, it takes values in the adjoint representation of $\mathfrak{g} $) one-form $ \omega $ and covariant derivative $D_{\mu}$ such as the one you've written.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In general I visualize connections like this: you put in a vector and the connection gives you back an element of the Lie algebra that is a generator of some transformation in the Lie group. E.g. on a Riemannian manifold you insert into the connection the direction you want to go to and you'll get back (very roughly said) information on how much the space is curved in that direction. More preciselly, if you integrate $\oint_{\gamma} \exp(\omega(\dot{\gamma}(t))) dt$ you obtain some element $T \in G$ that tells you how any vector gets parallel-transported along the closed curve $\gamma$ (this is called holonomy).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As for the last question of why $ U(1) $: well, because it's electromagnetism. Various groups will give you various interactions (e.g. $SU(3)$ gives you QCD). These groups arise because the theories contain something called gauge symmetry. I am sure you know that Maxwell equations, when written out in $A$ and $\phi$, are invariant to certain transformations. More concretely and clearly, as ${\bf F} = d{\bf A}$, where ${\bf F}$ is the electromagnetic tensor and ${\bf A} $ is the four-potential. In this form it's obvious that equations for $ {\bf F}$ don't change upon the transformation ${\bf A&amp;#39;} = {\bf A} + d\chi$. Now, let us step back to $\psi$ and note that if you want to make the theory locally invariant (why would you do that? because it's nice to have local properties instead of global ones. And the theory is clearly invariant with respect to global phase change, so let's just try this) with respect to the change of phase, you'll have to introduce new degree of freedom ${\bf A}$ that transforms precisely like the four-potential in Maxwell theory! Actually, what we just did is that we recovered photons. So, to conclude: if you choose $ U(1)$ as a group of symmetries, electromagnetism pops out. But note that to incorporate all of this consistently, you need to work in the framework of quantum field theory because Schrödinger's equation clearly isn't relativistically invariant, which is a feature we'd surely like to have in the theory of electromagnetism.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This can be a little confusing, especially in the physical literature and I doubt I can really make it clearer. Probably this will just confuse you even more but here goes: there exist various spaces of sections and $ G$-structures and representations of both $ G $ and $ \mathfrak{g}$ that one should really discern between but which are identified in physics. To make this rigorous would take too much space, so I suggest you look at some books on gauge theory. I'll just tell you that vector (more precisely vector &lt;em&gt;field&lt;/em&gt;) is also a section and vice-versa. But note that another two terms are getting mixed together here: abstract vector as a concept from linear algebra (this is our $ \psi $: you know it's this kind of vector because it has no spacetime index $\mu $) and vector as an element of Minkowski spacetime, say $V^{\mu}$. As for the $ D_{\mu} $, it acts (by extension) also on tensor algebra of the space of sections $ \psi $ (in this simple case where $\psi $ has no index, the tensor algebra is isomorphic to normal tangent tensor algebra) and gives you a one-form living in this tensor algebra, so yeah: it is a section, but in a totally different (altough isomorphic) space!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;No need to apologize. These things are pretty hard and it takes a lot of time to grok all of it and I guess lot of the physicists would just hand-wave most of the mathematical content away and go ahead to calculate something. Which is possible here because $ U(1) $ is only one-dimensional, its Lie algebra is one-dimensional, and so is $ \psi  $. So you just work with numbers all the time and there arises no need for abstract concepts. Except that this doesn't help you one bit when you'll try to generalize this (either to QCD with non-abelian group $ SU(3) $, or to the curved spacetime), or even if you try to think about some concepts that are rather basic from the point of view of geometry (e.g. try thinking about what is the meaning of the above-mentioned holonomy with respect to $\gamma $ in the case of $ U(1 )  $). So it's good you're trying to understand the core principles already.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt; in regard to Greg's question in the comments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure I understand you completely but I got a feeling (probably wrong) that you are mixing up various notions of invariance here. There are at least two notions of physical invariance (under the action of Lorentz group and under the action of $U(1)$) and also a notion of invariance in the sense of being &quot;coordinate-free&quot;. Now, if my feeling is correct, you are asking for an analogue of $v = v^{\mu}{{\rm \partial} \over {\rm \partial} x^{\mu}}$ for $\psi$. These two things are indeed very similar, but in a little disguised way. More precisely: $v$ is a section of a tangent bundle $TM$ and we are decomposing it with respect to some section of the canonical tangent frame bundle $FM$, which also carries a natural action of the group $GL_k({\mathbb{R}})$ (the action is a local change of basis and $k$ is the rank of $TM$). In other words, we have a $G$-structure here and this is where &quot;coordinate-free&quot; invariance comes from. The situation is similar with $\psi$: it is a section of a vector bundle $\pi: V \to M$ which carries an $U(1)$-structure. At this point it should also be clear where the difference between the two cases is: in the former you have two bundles $TM$ and $FM$ while in the latter there is only $\pi: V \to M$. So it doesn't really make sense to ask for $\psi$ to be any more invariant than it already is: you have nothing with respect to which you could decompose it. So instead of thinking about $\psi$ as an analogue of section of $TM$, think of it instead as an analogue of a section of $FM$.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-14T14:07:36.280" LastActivityDate="2010-11-14T14:07:36.280" CommentCount="5" />
  <row Id="753" PostTypeId="2" ParentId="741" CreationDate="2010-11-14T02:41:11.863" Score="10" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;Here's a short(-ish) answer.  A vector bundle is a family of vector spaces over a manifold.  The vector spaces can have bases.  The manifold can have coordinates.  The two concepts are not a priori related (now for the bundle of &lt;em&gt;tangent&lt;/em&gt; spaces, a coordinate change happens to induce a change of basis; this fact often sows confusion).  Once you pick a basis for your vector space, you &lt;em&gt;do&lt;/em&gt; define a vector by its components -- but someone else may be describing the &lt;em&gt;same&lt;/em&gt; vector in a different basis.  To translate to physics:  change of basis = gauge transformation.  In the case of a charged particle, the wave function is the component of a one-vector section; in a new basis, this number changes by a non-zero complex number (which can vary from point to point).  Again, the wave function is a section, and a section means one vector for each point in the manifold.  How do we differentiate a function which takes values in &lt;em&gt;different&lt;/em&gt; vector spaces over different points?  We need a way of &lt;em&gt;connecting&lt;/em&gt; the vector spaces.  The connection does this; pragmatically, it is just a rule for doing this differentiation.  Of course, differentiation will look different in different vector spaces, so the &lt;em&gt;form&lt;/em&gt; of the connection will depend on the basis and change under gauge transformations (just as the form of a linear transformation changes under change of bases).  That's what various messy formulas about how things &quot;transform&quot; are trying to tell you.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="20" LastActivityDate="2010-11-14T02:41:11.863" CommentCount="6" />
  <row Id="754" PostTypeId="2" ParentId="208" CreationDate="2010-11-14T02:55:31.920" Score="0" ViewCount="" Body="&lt;p&gt;I think even currently mass loss due to the solar wind is orders of magnitude greater than mass loss due to $E=mc^2$. But even so it isn't much. Ideally you would have enough&#xA; mass loss, so that a planets radius would increase at just that rate that kept the stars&#xA; luminosity divided by r squared constant, i.e. if that were the case the planet would stay &#xA;within the habitable zone, even as the star brightened due to stellar evolution. Alas the&#xA; solar wind is much too weak to accomplish the task. But we should be moving further out in&#xA; any case.&lt;/p&gt;&#xA;" OwnerUserId="391" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-12-04T15:55:08.753" LastActivityDate="2010-12-04T15:55:08.753" CommentCount="2" />
  <row Id="755" PostTypeId="1" CreationDate="2010-11-14T02:56:20.010" Score="7" ViewCount="205" Body="&lt;p&gt;Interstellar space is pretty empty but there a small number of of atoms (mostly hydrogen?) floating around. How many atoms per light year would a photon encounter while traversing interstellar space?&lt;/p&gt;&#xA;" OwnerUserId="207" LastActivityDate="2010-11-14T06:28:19.250" Title="How many atoms per light year does light encounter when traversing interstellar space?" Tags="&lt;electromagnetism&gt;&lt;astrophysics&gt;&lt;cosmology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="756" PostTypeId="1" AcceptedAnswerId="761" CreationDate="2010-11-14T03:02:29.403" Score="8" ViewCount="673" Body="&lt;p&gt;To date, what is the most mathematically precise formulation of the AdS/CFT correspondence, and what are the most robust tests of the conjecture?&lt;/p&gt;&#xA;" OwnerUserId="20" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-14T05:15:20.513" LastActivityDate="2011-01-14T13:38:51.447" Title="Mathematics of AdS/CFT" Tags="&lt;high-energy-physics&gt;&lt;string-theory&gt;&lt;ads-cft&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="5" />
  <row Id="757" PostTypeId="2" ParentId="723" CreationDate="2010-11-14T03:18:27.917" Score="3" ViewCount="" Body="&lt;p&gt;A charged particle moving in a magnetic field experiences a force f= q*VxB. This force is&#xA;perpendicular to both the mag field and the direction of motion. If it is a uniform magnetic field charged particles move in circles, and the frequency is called the cyclotron frequency.&#xA;You can essentially decompose the velocity into a component parrallel to the mag field, and a component perpendicular. The later is like an orbital velocity, the former (parallel to the field) velocity is unaffected. So your electrons would turn sideways rather than go straight at the wires.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in a real plasma, the charged particles will occasionally collide, and new orbits established after every collision. I think (but its been decades for me) as a result they will drift in the direction of the electric field anyway. I don't know if this happens in a fast enough time span to seriously affect the device. But in oder for the fusion device to work, the hydrogen (or Helium-3) nuclei have to be colliding with each other. So the system will not&#xA;be lossless. I presume the scientists in charge have worked this all out.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-11-14T03:18:27.917" CommentCount="1" />
  <row Id="758" PostTypeId="2" ParentId="622" CreationDate="2010-11-14T03:39:56.760" Score="1" ViewCount="" Body="&lt;p&gt;The inverse process of Compton scattering (&lt;strong&gt;Inverse Compton&lt;/strong&gt;) is a process where a relativistic charged particle scatters a low energy photon and boosts its energy by an amount proportional to the square of the Lorentz factor of the particle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\epsilon_1=4\gamma^2\epsilon_0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process is very important in Astrophysics. It is one of the basic mechanism that produce gamma radiation in AGNs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An other process that could be considered as a scattering process is the &lt;strong&gt;gravitational deflection of light&lt;/strong&gt; from the Sun or any other massive object. That process can be described as if the space-time is a medium of variable refractive index&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n=\sqrt{1+\frac{2a^2GM}{c^2r^3}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where alpha is the minimum distance of the light beam from the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, in &lt;strong&gt;quantum mechanics&lt;/strong&gt;, one can consider the scattering of particles by a potential. The description is given with the help of the wavefunction, where the particles are described as waves which are transmitted across, or reflected by the potential.&lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2010-11-14T03:39:56.760" CommunityOwnedDate="2010-11-14T03:39:56.760" />
  <row Id="759" PostTypeId="1" CreationDate="2010-11-14T04:31:31.593" Score="-7" ViewCount="649" Body="&lt;p&gt;I was reading about interesting article &lt;a href=&quot;http://www.donnee.com/universe-as-computer.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; which suggests that our universe is a big computer simulation and the proof of it is a Quantum Physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know quantum physics tries to  provide some explanation about unpredictable behavior of universe at atomic scale but could not believe that thing has anyway relates to pixels of the universe ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can it be true that behavior of universe at atomic level coming from some super computer ?&lt;/p&gt;&#xA;" OwnerUserId="146" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-04T22:18:27.520" LastActivityDate="2011-07-29T04:16:28.157" Title="Does Quantum Physics really suggests this universe as a computer simulation?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="1" />
  <row Id="760" PostTypeId="2" ParentId="759" CreationDate="2010-11-14T05:14:28.363" Score="17" ViewCount="" Body="&lt;p&gt;No, quantum physics suggests no such thing. As Mark pointed out in his comment, the &quot;article&quot; you linked to is not an article at all; as far as I can tell, it's a promotional page for a science fiction novel. As actual science, it would be completely bogus.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-14T05:14:28.363" />
  <row Id="761" PostTypeId="2" ParentId="756" CreationDate="2010-11-14T05:24:10.750" Score="11" ViewCount="" Body="&lt;p&gt;I'm not sure what you have in mind with &quot;mathematically precise&quot;; as with almost anything involving quantum field theory or string theory, there's no rigorous definition of the theories involved in the duality. But, if you grant me their existence, I would say the sharpest statement is still the early formulation by &lt;a href=&quot;http://arxiv.org/abs/hep-th/9802109&quot; rel=&quot;nofollow&quot;&gt;Gubser, Klebanov, and Polyakov&lt;/a&gt; and by &lt;a href=&quot;http://arxiv.org/abs/hep-th/9802150&quot; rel=&quot;nofollow&quot;&gt;Witten&lt;/a&gt;, i.e. that the partition function of the CFT in the presence of external sources for single-trace operators is the same as that for string theory in AdS with boundary conditions determined by the sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most detailed computational checks of the correspondence are probably those that use integrability to compute anomalous dimensions of operators over the full range from weak to strong coupling. I'm not an expert on this, but I'll point you to &lt;a href=&quot;http://arxiv.org/abs/0901.3753&quot; rel=&quot;nofollow&quot;&gt;one fairly recent paper&lt;/a&gt; that contains some of the major references to get you started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a more global perspective, though, gauge/gravity duality extends well beyond the original case of ${\cal N}=4$ SYM and $AdS_5 \times S^5$, to any theory that meets the two requirements of having a large-$N$ expansion and a large 't Hooft coupling. The important ideas, again, were mostly there in the very early papers, but I would say they've been put on a somewhat more solid footing. The key point is that the bulk theory is tractable in the case when only a few fields are involved and curvatures are weak. This condition, translated to a statement about the dual field theory, is that most of the single-trace operators acquire very large anomalous dimensions (which is natural in very strongly coupled theories). Recently there has been some progress in formulating a bottom-up argument from the opposite direction, i.e. starting with the assumption that a large-$N$ conformal field theory includes few low-dimensional single-trace operators and arguing that this implies the existence of a bulk dual theory. See &lt;a href=&quot;http://arxiv.org/abs/0907.0151&quot; rel=&quot;nofollow&quot;&gt;this paper of Heemskerk et al.&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if any of this is what you would think of as &quot;mathematics&quot;....&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-11-14T05:24:10.750" CommentCount="4" />
  <row Id="762" PostTypeId="2" ParentId="679" CreationDate="2010-11-14T05:53:27.327" Score="7" ViewCount="" Body="&lt;p&gt;Since you say you're mostly interested in learning physics, not in a degree, there are lots of options. You say you have good knowledge of physics; if you want to learn particle physics, you need a good undergrad physics background plus some additional knowledge. I tend to think that reading textbooks is still the best way to learn the basics; if you want to learn particle physics, you might start with the book by Griffiths. If you find that you don't have enough background knowledge for it, then you would need to read other, more basic textbooks. Once you pass some threshold of background knowledge, though, you can learn a lot online. You could try searching &lt;a href=&quot;http://arxiv.org&quot; rel=&quot;nofollow&quot;&gt;arxiv.org&lt;/a&gt; for introductory lectures on various topics. In particle physics, for instance, you might search for the word &quot;TASI&quot;, which is a summer school for graduate students to learn more about the field (the lectures are often written up and posted online). Most of this will assume you already know quantum field theory, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get a prepublication draft of Mark Srednicki's QFT textbook &lt;a href=&quot;http://www.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf&quot; rel=&quot;nofollow&quot;&gt;from his website&lt;/a&gt; (PDF file).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Perimeter Institute has a &lt;a href=&quot;http://www.perimeterscholars.org/home.html&quot; rel=&quot;nofollow&quot;&gt;one-year Master's degree program&lt;/a&gt; and videotapes all of the lectures. You can watch the videos from their archive.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-11-14T05:53:27.327" CommentCount="1" />
  <row Id="763" PostTypeId="1" AcceptedAnswerId="996" CreationDate="2010-11-14T06:17:57.923" Score="-4" ViewCount="131" Body="&lt;p&gt;I have 2 equation describing the alternating amperage i1 and i2. I need to get amount of these amperages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My equations:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;i1= 10 sin (ωt+30)&#xA;&#xA;i2= 20 sin (ωt-50) &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How can i make it? I forgot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="402" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-11-14T08:24:17.427" LastActivityDate="2010-11-17T13:16:34.673" Title="Amount of 2 amperage" Tags="&lt;electricity&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="764" PostTypeId="2" ParentId="755" CreationDate="2010-11-14T06:28:19.250" Score="12" ViewCount="" Body="&lt;p&gt;The number of scattering events (&quot;encounters&quot;) per unit path length is equal to the density of scatterers times the scattering cross section (summed over all species of scatterers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density of particles (mostly hydrogen atoms) in the interstellar medium (ISM) is on the order of 1/cm^3. If we assume that the photon scattering cross section of a hydrogen atom is the square of a common atomic length scale, 1 angstrom^2, then the number of scattering events per unit path length is 10^-14 per meter, or ~100 per light year, give or take a factor of 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take this with a grain of salt, however! The scattering cross section for photons is strongly energy-dependent. For visible photons, say, the cross section is greatly increased if the photon is sufficiently near an atomic transition energy. The hydrogen atoms &quot;look much larger&quot; to these photons, so there are much more collisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lower in the EM spectrum, you run into the fact that the ISM is actually a plasma (it contains ionized hydrogen and free electrons). I believe the plasma frequency is on the order of 1 kHz (strongly depends on temperature...), so radio waves significantly below 1 kHz cannot propagate through the ISM at all. They scatter from the plasma as a whole, rather than individual particles, so the ISM is actually opaque to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the real answer is &quot;it entirely depends on what energy photons you're talking about&quot;.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2010-11-14T06:28:19.250" CommentCount="2" />
  <row Id="765" PostTypeId="2" ParentId="759" CreationDate="2010-11-14T08:09:34.147" Score="6" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;The thing you are reading is NOT an article(that you should trust).&lt;/li&gt;&#xA;&lt;li&gt;Even if the universe were a simulation there is NO such proof in QM.&lt;/li&gt;&#xA;&lt;li&gt;Though physical processes can be thought of as computations it does not imply existence of &quot;super computer&quot;. &lt;/li&gt;&#xA;&lt;li&gt;Also, even if such a &quot;super computer&quot; exists it must be a part of some universe. That universe must also be simulated by some &quot;super-super-computer&quot; and so on. So, its basically &lt;a href=&quot;http://en.wikipedia.org/wiki/Turtles_all_the_way_down&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;turtles all the way down&lt;/strong&gt;&lt;/a&gt;. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-11-14T08:09:34.147" />
  <row Id="766" PostTypeId="2" ParentId="759" CreationDate="2010-11-14T08:59:21.717" Score="4" ViewCount="" Body="&lt;p&gt;While your reference does not quite meet the standards for scientific publishing ;), your question itself addresses a very profound notion. I disagree with the appraisal of any such notion of a &lt;em&gt;computational universe&lt;/em&gt; as being dead on arrival. Some of the best minds in physics - Wheeler, Lloyd, Wolfram, Schmidhuber and others - have been strong advocates of this paradigm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[This is not an attempt on my part to start, or indulge in any kind of war of words, but a simple statement of my opinion. Take it or leave it.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me frame the question in a slightly less controversial form than the one you put it in Xinus:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does it make sense to address the motions and interactions of elementary particles - working within the standard Hilbert space framework of quantum mechanics - from a computational perspective?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When put this way the question appears a little less shocking. I was planning to write more here but there is a wonderfully detailed wikipedia [entry][http://en.wikipedia.org/wiki/Digital_physics] on this very issue which can answer your query in a much more complete way than I could attempt here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ps: btw the &lt;a href=&quot;http://www.donnee.com/chapters.htm&quot; rel=&quot;nofollow&quot;&gt;novel&lt;/a&gt; in question by Jim Wills looks very intriguing and happens to be free to read online. So thanks for that !&lt;/p&gt;&#xA;" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-14T12:48:54.273" LastActivityDate="2010-11-14T12:48:54.273" CommentCount="4" />
  <row Id="767" PostTypeId="2" ParentId="679" CreationDate="2010-11-14T11:40:20.747" Score="9" ViewCount="" Body="&lt;p&gt;Stanford has posted a bunch of theoretical physics courses by top physicist &lt;a href=&quot;http://en.wikipedia.org/wiki/Leonard_Susskind&quot; rel=&quot;nofollow&quot;&gt;Leonard Susskind&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/view_play_list?p=189C0DCE90CB6D81&quot; rel=&quot;nofollow&quot;&gt;Classical Mechanics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://youtube.com/view_play_list?p=84C10A9CB1D13841&quot; rel=&quot;nofollow&quot;&gt;Quantum Mechanics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://youtube.com/view_play_list?p=CCD6C043FEC59772&quot; rel=&quot;nofollow&quot;&gt;Special Relativity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/user/StanfordUniversity#grid/user/6C8BDEEBA6BDC78D&quot; rel=&quot;nofollow&quot;&gt;General Relativity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=32wIKaLkvc4&amp;amp;feature=PlayList&amp;amp;p=888811AA667C942F&amp;amp;index=0&amp;amp;playnext=1&quot; rel=&quot;nofollow&quot;&gt;Cosmology&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/stanford#g/c/B72416C707D85AB0&quot; rel=&quot;nofollow&quot;&gt;Statistical Mechanics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a physicist but I still watched all these, just because Susskind is such a great teacher of physics. This is as good as it gets!&lt;/p&gt;&#xA;" OwnerUserId="47" LastActivityDate="2010-11-14T11:40:20.747" CommentCount="1" />
  <row Id="768" PostTypeId="1" AcceptedAnswerId="769" CreationDate="2010-11-14T13:58:07.020" Score="4" ViewCount="216" Body="&lt;p&gt;(Reformulation of part 1 of &lt;a href=&quot;http://physics.stackexchange.com/questions/741/electromagnetic-field-as-a-connection-in-a-vector-bundle&quot;&gt;Electromagnetic Field as a Connection in a Vector Bundle&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a good notation for sections of vector bundles that is both invariant &lt;em&gt;and&lt;/em&gt; references bundle coordinates. Is there a standard notation for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum mechanics, the wave function $\psi(x,t)$ of an electron is usually introduced as a function $\psi : M \to \mathbb{C}$ where $M$ is the space-time, usually $M=\mathbb{R}^3\times\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when modeling the electron in an electromagnetic field, it is best to think of $\psi(x,t)$ as a section in a $U(1)$-vector bundle $\pi : P \to M$. Actually, $\psi(x,t)$ *itself* is not a section, it's just the image of a section in one particular local trivialization $\pi^{-1}(U) \cong U\times\mathbb{C}$ of the vector bundle. In a different local trivialization (= a different gauge), the image will be $e^{i\chi(x,t)}\psi(x,t)$ with a different phase factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I feel uncomfortable with this notation. Namely, I would prefer an invariant notation, like for the tangent bundle. For a section $\vec v$ of the tangent bundle (= a vector field), I can write $\vec v = v^\mu \frac{\partial}{\partial x^\mu}$. This expression mentions the &lt;em&gt;coordinates&lt;/em&gt; $v^\mu$ in a particular coordinate system, but it is also &lt;em&gt;invariant&lt;/em&gt;, because I also write down the basis vector $\frac{\partial}{\partial x^\mu}$ of the coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The great benefit of the vector notation is that it automatically deals with coordinate changes: $\frac{\partial}{\partial x^\mu} = \frac{\partial}{\partial y^\nu}\frac{\partial y^\nu}{\partial x^\mu}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a notation for sections of vector bundles that is similar to the notation $\vec v = v^\mu \frac{\partial}{\partial x^\mu}$ for the tangent bundle? What does it look like for our particular example $\psi$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no, what are the usual/standard notations for this? How do they keep track of the bundle coordinates?&lt;/p&gt;&#xA;" OwnerUserId="188" LastEditorUserId="188" LastEditorDisplayName="" LastEditDate="2010-11-14T14:06:39.620" LastActivityDate="2010-12-15T22:20:25.600" Title="Notation for Sections of Vector Bundles" Tags="&lt;quantum-mechanics&gt;&lt;electromagnetism&gt;&lt;quantum-field-theory&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="769" PostTypeId="2" ParentId="768" CreationDate="2010-11-14T14:32:06.067" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I realized that what I've written wasn't really correct, so let me change the text a little. I'll mark the additions by &lt;em&gt;italics&lt;/em&gt;, so that the old text stays as a reference.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I gave a (partial) answer to this in the update of my answer to your &lt;a href=&quot;http://physics.stackexchange.com/questions/741/electromagnetic-field-as-a-connection-in-a-vector-bundle/752#752&quot;&gt;previous question &lt;/a&gt; so let me copy&amp;amp;paste that answer (with some modifications):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to the first question is no &lt;em&gt;but for a different reason than I stated in the above link and in the text below!&lt;/em&gt;. There is no such notation and to see why, we first have to understand where &quot;coordinate-free&quot; vector invariance comes from. The $v$ in you question is a section of a tangent bundle $TM$ and we are decomposing it with respect to some section of the canonical tangent frame bundle $FM$, which also carries a natural action of the group $GL_k({\mathbb{R}})$ (the action is a local change of basis and $k$ is the rank of $TM$). In other words, we have a $GL_k({\mathbb{R}})$-structure here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation is superficially similar with $\psi$: it is a section of a vector bundle $\pi: V \to M$ which carries an $U(1)$-structure. At this point it should also be clear where the difference between the two cases is: in the former you have two bundles $TM$ and $FM$ while in the latter there is only $\pi: V \to M$. So it doesn't really make sense to ask for $\psi$ to be any more invariant than it already is: you have nothing with respect to which you could decompose it. So instead of thinking about $\psi$ as an analogue of section of $TM$, think of it instead as an analogue of a section of $FM$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I made a mistake in the above reasoning because in the case of one-dimensional $U(1)$ the concepts of $V$ and $F(V)$ (associated frame bundle) coincide. So you also have two bundles in the second case. But the difference comes from the fact that $TM$ is a very special vector bundle: it's structure comes from the manifold $M$ itself, whereas $V$ is an extrinsic structure. So you certainly cannot get a decomposition with respect to coordinate derivates on $M$ as is the case of $TM$.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the second question: in gauge theories one usually fixes gauge before-hand (think of Lorenz or Coulomb gauge) and work in that forever. You don't really get anything interesting here by working in some &quot;gauge-free&quot; way (or at least I don't know about it). So, these things really aren't an issue at least until the point when you come across QFT and start wondering how to account for all this huge gauge freedom. And there it actually &lt;em&gt;is&lt;/em&gt; a big problem that must be dealt with and it &lt;em&gt;can&lt;/em&gt; be dealt with in various ways (including gauge-fixing). But none of this is relevant for you at this point, I guess.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-14T18:32:27.590" LastActivityDate="2010-11-14T18:32:27.590" CommentCount="3" />
  <row Id="770" PostTypeId="1" CreationDate="2010-11-14T15:29:34.897" Score="6" ViewCount="1325" Body="&lt;p&gt;Don't worry... I won't actually do it, just would like to know how.&lt;/p&gt;&#xA;" OwnerUserId="409" LastActivityDate="2011-01-20T17:17:26.223" CommunityOwnedDate="2011-01-07T14:13:00.227" ClosedDate="2011-01-20T18:06:18.233" Title="What is the most efficient way to destroy the universe?" Tags="&lt;cosmology&gt;" AnswerCount="11" CommentCount="15" FavoriteCount="3" />
  <row Id="772" PostTypeId="2" ParentId="95" CreationDate="2010-11-14T16:23:36.590" Score="0" ViewCount="" Body="&lt;p&gt;Prof. Alex Mayer has proposed a new cosmology that matches observables and which does not require dark energy. Google &quot;Alex Mayer&quot; (his website is the first hit) and then first review the &quot;An Introduction to the New Cosmology&quot; talk he gave in September. You can also download a preview of his new book.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-14T16:23:36.590" />
  <row Id="773" PostTypeId="2" ParentId="408" CreationDate="2010-11-14T16:50:10.230" Score="4" ViewCount="" Body="&lt;p&gt;Are you aware of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cygnus_X-1&quot; rel=&quot;nofollow&quot;&gt;Cygnus X-1&lt;/a&gt; x-ray source? This might give you an example of &quot;seeing&quot; a black hole. This is an example of an accretion disk x-ray source, discussed in other answers here.&lt;/p&gt;&#xA;" OwnerUserId="411" LastActivityDate="2010-11-14T16:50:10.230" />
  <row Id="774" PostTypeId="2" ParentId="523" CreationDate="2010-11-14T16:52:51.540" Score="2" ViewCount="" Body="&lt;p&gt;(Edited to correct TeX typo)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I complete Tim Goodman's answer in answer to get something more systematic. A local measurement has to be written as a tensorial product of two observables A⊗B. And it can only distinguish (with probability 1) its eigenstates. The states |ϕ⟩ and |ψ⟩ of Tim's example cannot be written as eigenstates of a tensorial product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this does not correspond exactly to states which cannot be written as tensorial product of states. For example, if let's use the following 4 states :&#xA;$|\psi_0\rangle=|00\rangle$ , $|\psi_1\rangle=|1+\rangle$,&lt;br&gt;&#xA;$|\phi_0\rangle=|01\rangle$ , $|\phi_1\rangle=|1-\rangle$ and let's try to distinguish the ψs from the ϕs.Tim's arguments are still valid, even if each the 4 states is a product state and is orthogonal to all the 3 others. Furthermore, each pair of state is locally distinguishable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think locally distinguishable subspaces has something to do with to the direct sum of locally orthogonal subspaces, but I don't exactly know how to write it.&lt;/p&gt;&#xA;" OwnerUserId="373" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-11-15T10:14:57.730" LastActivityDate="2010-11-15T10:14:57.730" CommentCount="4" />
  <row Id="775" PostTypeId="2" ParentId="408" CreationDate="2010-11-14T16:54:32.693" Score="2" ViewCount="" Body="&lt;p&gt;If a black hole is illuminated along our line of site by a source behind us, it might be possible to see its shadow against a bright background.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-11-14T16:54:32.693" CommentCount="1" />
  <row Id="776" PostTypeId="1" CreationDate="2010-11-14T16:54:55.627" Score="9" ViewCount="609" Body="&lt;p&gt;Our soler system itself contains two candidate &quot;Earths&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is Jupiter's moon Europa and another is Saturn's moon Titan. Both of them have the problem of having at low temperature as Sun's heat cannot reach them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jupiter is made up mostly of Hydrogen so is our Sun. Can we ignite Jupiter so that it will produce enough heat to warm these two earth like planets/satellites ? What can be consequences of any of such attempt ?&lt;/p&gt;&#xA;" OwnerUserId="146" LastActivityDate="2011-03-04T03:45:35.790" Title="Can Jupiter be ignited?" Tags="&lt;astronomy&gt;" AnswerCount="6" CommentCount="8" />
  <row Id="777" PostTypeId="2" ParentId="703" CreationDate="2010-11-14T17:05:30.173" Score="1" ViewCount="" Body="&lt;p&gt;Be careful (as the other two answers here are) to distinguish between the speed of DISTURBANCES in the gravitational field, and the 'speed of gravity'.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The former is a well-posed, meaningful thing.  The latter is not--the Earth's gravitational field, for instance, is mostly static--we just feel the same ol' gravity all the time.  So us Earthlings talking about the 'speed' of Earth's gravity doesn't mean much.  Now if a black hole hit the sun, and we started noticing changes in our gravitational field, we'd be able to talk about how quickly the gravitational news of that collision reached us.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-11-14T17:05:30.173" />
  <row Id="778" PostTypeId="2" ParentId="776" CreationDate="2010-11-14T17:15:32.237" Score="5" ViewCount="" Body="&lt;p&gt;No, Jupiter is simply not massive enough to sustain nuclear fusion (and effectively become a star).               &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Off the top of my head, I remember the statistic that it needs to be roughly 10 times bigger in order to generate the necessary gravitational forces at the core to initiate fusion and become a star.             &lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-03-03T18:01:45.143" LastActivityDate="2011-03-03T18:01:45.143" CommentCount="4" />
  <row Id="779" PostTypeId="2" ParentId="770" CreationDate="2010-11-14T17:21:58.747" Score="1" ViewCount="" Body="&lt;p&gt;I think that only an Universe can destroy an Universe (there's a lot of energy implied in doing so). If there is enough mass in our Universe, maybe its end could be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Big_Crunch&quot; rel=&quot;nofollow&quot;&gt;Big Crunch&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="306" LastActivityDate="2010-11-14T17:21:58.747" CommunityOwnedDate="2011-01-07T14:13:00.227" />
  <row Id="780" PostTypeId="2" ParentId="567" CreationDate="2010-11-14T17:35:40.660" Score="2" ViewCount="" Body="&lt;p&gt;Tim Goodman's answer is right, and I just want to add a couple of things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though Neutron stars do not &quot;burn&quot; anything, for the most part, they DO shine, thanks to the fact that they are the remnants of the core of a star, and thus, when they are born, are roughly as hot as the center of a star.  They will thus shine in the same way that a hot poker shines when you stick it in a fire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, neutron stars have a maximum possible mass, dictated by the rules of general relativity and the mass of a neutron.  If a neutron star has extra mass beyond this dumped on its surface (say, because it is orbiting an ordinary star, and that star accretes mass onto the NS), they can explode in a violent way.  To my understanding, this has not been observed, but it has been known to happen in White Dwarfs, which are held together by a very similar principle to neutron stars.  The end result is a supernova, and then a black hole.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-11-14T17:35:40.660" CommentCount="2" />
  <row Id="781" PostTypeId="2" ParentId="770" CreationDate="2010-11-14T17:37:29.177" Score="0" ViewCount="" Body="&lt;p&gt;Inverse the forces in nature is the fastest way to destroy it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Make &quot;Weak force&quot; strong&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Make &quot;Strong force&quot; weak&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First one will increase radiation and will decay matter to energy instantaneously and second one will explode atom nucleus.&lt;/p&gt;&#xA;" OwnerUserId="146" LastActivityDate="2010-11-14T17:37:29.177" CommunityOwnedDate="2011-01-07T14:13:00.227" CommentCount="2" />
  <row Id="782" PostTypeId="1" AcceptedAnswerId="827" CreationDate="2010-11-14T17:45:03.010" Score="3" ViewCount="369" Body="&lt;p&gt;I'm trying to calculate the initial launch angles and velocity of a projectile (atmosphere's effects can be neglected), assuming that I know the lat/lon coordinates of both the launch and the destination points. Can you recommend any software libraries or reading materials that could help? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-12T22:48:13.330" LastActivityDate="2010-12-12T22:48:13.330" Title="How to calculate a ballistic trajectory for a suborbital flight?" Tags="&lt;newtonian-mechanics&gt;&lt;orbital-motion&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="783" PostTypeId="1" AcceptedAnswerId="785" CreationDate="2010-11-14T18:06:27.640" Score="5" ViewCount="525" Body="&lt;p&gt;I'm fascinated by the fundamental questions raised by the Double Slit Experiment at the quantum level. I found this &lt;a href=&quot;http://www.youtube.com/watch?v=DfPeprQ7oGc&quot; rel=&quot;nofollow&quot;&gt;&quot;Dr Quantum&quot; video clip&lt;/a&gt; which seems like a great explanation. But is it scientifically accurate?&lt;/p&gt;&#xA;" OwnerUserId="414" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-03-13T20:17:01.410" LastActivityDate="2011-04-16T23:31:06.903" Title="Is Dr Quantum's Double Slit Experiment video scientifically accurate?" Tags="&lt;quantum-mechanics&gt;&lt;experimental-physics&gt;&lt;double-slit-experiment&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="2" />
  <row Id="784" PostTypeId="2" ParentId="770" CreationDate="2010-11-14T18:10:52.310" Score="17" ViewCount="" Body="&lt;p&gt;The physical process that might best be described as &quot;destroying the universe&quot; would be vacuum decay. Our entire universe might be in a &lt;a href=&quot;https://secure.wikimedia.org/wikipedia/en/wiki/Metastability&quot; rel=&quot;nofollow&quot;&gt;metastable&lt;/a&gt; state, much like a supercooled liquid, and at some point quantum or thermal fluctuations could send some region of the universe into a lower ground state. If the region that tunnels to this ground state is large enough, it will expand at the speed of light, swallowing up the rest of the universe and converting it all to the lower state. This type of vacuum decay, in theories involving gravity, was calculated in a famous paper by &lt;a href=&quot;http://prd.aps.org/abstract/PRD/v21/i12/p3305_1&quot; rel=&quot;nofollow&quot;&gt;Coleman and de Luccia&lt;/a&gt;. Anyone with the relevant background knowledge should read the paper, which like much of Coleman's work is beautifully clear and comprehensible. One passage in the paper is particularly famous (though it arguably shouldn't be taken too seriously now, given some of the assumptions it involves, including that our vacuum has exactly zero cosmological constant):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The possibility that we are living in a false vacuum has never been a cheering one to contemplate. Vacuum decay is the ultimate ecological catastrophe; in the new vacuum there are new constants of nature; after vacuum decay, not only is life as we know it impossible, so is chemistry as we know it. However, one could always draw stoic comfort from the possibility that perhaps in the course of time the new vacuum would sustain, if not life as we know it, at least some structures capable of knowing joy. This possibility has now been eliminated.&quot;&lt;br/&gt;&amp;mdash; Sidney Coleman&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we live in a metastable vacuum (and, due to the observed positive cosmological constant, most physicists working on high-energy physics, cosmology, or gravity believe that we must, though there might be some unknown loopholes), then the universe will eventually decay. Luckily, the production of critical bubbles is quite rare, allowing us the time we needed to evolve and live in this metastable universe before the inevitable catastrophe. [&lt;strong&gt;Edited for clarification&lt;/strong&gt; due to Marek's comment: it's only &quot;inevitable&quot; if our universe really is metastable, which despite strong suggestive arguments we don't definitively know.] You would have to wait around for billions of years to witness the end of the universe. If you're more ambitious (let's hope not), you could try to cause the decay by &lt;em&gt;producing&lt;/em&gt; a region of the true vacuum, thus catalyzing the process instead of letting it happen naturally. A number of crackpots have tried to start a panic over accelerator experiments by suggesting that they can do this. However, higher-energy collisions happen in nature, and we're still here, so this is nothing to worry about. (There are more quantitative versions of this argument, of course. The risk really is effectively zero.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vacuum decay processes I've mentioned don't necessarily &quot;destroy&quot; the universe, they just radically alter it by leaving it in a completely different state. There is one type of vacuum decay, discovered by Witten and known as the &lt;a href=&quot;http://dx.doi.org/10.1016/0550-3213%2882%2990007-4&quot; rel=&quot;nofollow&quot;&gt;bubble of nothing&lt;/a&gt; instability, which is more radical and leaves behind nothing, not even a new vacuum state.&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="19" LastEditorDisplayName="" LastEditDate="2010-11-14T19:43:37.987" LastActivityDate="2010-11-14T19:43:37.987" CommunityOwnedDate="2011-01-07T14:13:00.227" CommentCount="8" />
  <row Id="785" PostTypeId="2" ParentId="783" CreationDate="2010-11-14T18:15:31.070" Score="10" ViewCount="" Body="&lt;p&gt;A bad thing about the video is how they explained the part where you try to observe which slit the electron goes through. They made it sound more mysterious than it really is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we have to ask ourselves is: what does it mean to observe an electron? What does it mean to observe anything? If we want to look at something, we need light. We see things because light is reflected off objects and our eyes collect this light which is then interpreted by our brains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want to see which slit the electron goes through, we shine light upon it, but this fundamentally alters the experiment. Small particles are very sensitive to perturbations and shining light on an electron is a big perturbation. Now, and this is technical, the Heisenberg uncertainty relation tells you how much the electron's path will be perturbed by the light. The path is more perturbed as the energy of the photon is greater, but to determine the position of the electron accurately, you need high energy according to Heisenberg. High energy means perturbing the electron a lot and as a consequence destroy the interference pattern. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you might want to give up on accuracy to avoid perturbing the path of the electron too much, but if you do that, the Heisenberg relation will show you that you have to diminish the energy of the photon so much that you will not be able to locate the electron anymore. The interference pattern on the other hand will reappear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More details can be found in the &lt;a href=&quot;http://student.fizika.org/~jsisko/Knjige/Opca%20Fizika/Feynman%20Lectures%20on%20Physics/Vol%203%20Ch%2001%20-%20Quantum%20Behavior.pdf&quot; rel=&quot;nofollow&quot;&gt;Feynman Lectures, Volume 3, Chapter 1&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2010-11-14T18:15:31.070" CommentCount="1" />
  <row Id="786" PostTypeId="2" ParentId="770" CreationDate="2010-11-14T18:26:58.453" Score="7" ViewCount="" Body="&lt;p&gt;We are only humans. We don't know how to destroy the universe.&lt;/p&gt;&#xA;" OwnerUserId="150" LastActivityDate="2010-11-14T18:26:58.453" CommunityOwnedDate="2011-01-07T14:13:00.227" />
  <row Id="787" PostTypeId="2" ParentId="703" CreationDate="2010-11-14T18:27:15.003" Score="1" ViewCount="" Body="&lt;p&gt;As for the first question: gravitation is a tensor interaction of rank 2 (for more advanced readers: traceless metric transforms under the (1,1) representation of the Lorentz group. On quantum level this gives spin 2 particles called gravitons). This is really the same thing as with electromagnetism. There you have a vector interaction because field potential $A_{\mu}$ is a vector. (and the same comments can be made about Lorentz group and spin; in particular, photons are spin 1 particles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not really correct to ask for velocity of gravitation and two distinct things can be meant by this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the speed of expansion of universe: this can be a lot larger than a speed of light. For more on this see any book on cosmology (or google) and expansion (in particular inflation).&lt;/li&gt;&#xA;&lt;li&gt;the speed of gravitational waves, i.e. disturbances in the background metric. These waves propagate at the speed of light.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now your questions are actually interesting when asked together. Because the answer to the second one (propagation of waves at the speed of light) means that the corresponding particles are massless and this implies that they have number of degrees of freedom smaller by one than what you would originally expect, because longitudinal modes aren't allowed for massless particles. E.g. with electromagnetism, this corresponds to the fact that you get only two polarizations (right- and left-moving).&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-11-14T18:27:15.003" />
  <row Id="788" PostTypeId="2" ParentId="234" CreationDate="2010-11-14T19:11:14.437" Score="0" ViewCount="" Body="&lt;p&gt;The question is way too broad. Different areas of physics requires different level (and area) of maths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One general list is here:&#xA;&lt;a href=&quot;http://www.phys.uu.nl/~thooft/theorist.html&quot; rel=&quot;nofollow&quot;&gt;Gerard ’t Hooft, Theoretical Physics as a Challenge&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also one approach is to learn maths when you encounter it in physics (*), making sure each time you learn a bit more than solely to understand (*).&lt;/p&gt;&#xA;" OwnerUserId="184" LastActivityDate="2010-11-14T19:11:14.437" CommentCount="1" />
  <row Id="789" PostTypeId="1" CreationDate="2010-11-14T19:19:34.720" Score="3" ViewCount="308" Body="&lt;p&gt;Lets see the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^- \mu^- \to e^- \pi^- \nu_\mu \;\;\;\;\;\;\;\;\;\;\; {(1)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose, that this reaction occurs as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^- \mu^- \to e^- \mu^- \pi^+ \pi^- \to e^- \pi^- \nu_\mu$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible at energy less than 2*140 MeV?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is for the analogous proton-muon reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p^+ \mu^+ \to p^+ \pi^+ \bar{\nu}_\mu \;\;\;\;\;\;\;\;\;\;\;{(2)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once more reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p^+ p^+ \to p^+ p^+ \pi^- \pi^+ \to p^+ n \pi^+ \;\;\;\;\;\;\;\;\;\;\;{(3)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the experimental data?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;P.S.&lt;/strong&gt; This question is important enough. The main solar reaction is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p^+ p^+ \to d^+ e^+ \nu_e $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this reaction occurs as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p^+ p^+ \to p^+ p^+ e^- e^+ \to p^+ n \nu_e e^+ \to d^+ e^+ \nu_e$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then  it would require the energy of more than 2*0.511 MeV to take place. &#xA;Cross section of this reaction will be much less, so the main solar reaction should be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p^+ p^+ e^- \to d^+ \nu_e $&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 20.02.11&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why I think so?&#xA;I suppose that new particles are created in pairs particle-antiparticle.&#xA;So the reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^- e^- \to e^- e^- \pi^- \pi^+ $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;requires the energy of more than 2*140 MeV to take place&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well as the reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^- e^- \to e^- e^- \pi^- e^+ \nu_e $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;from symmetry considerations, since there is a decay&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\pi^+ \to  e^+ \nu_e$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the result is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^- e^- \to e^- e^- \pi^- e^+ \nu_e \to e^- \pi^- \nu_e $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;contrary to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^- e^- \to e^- \nu_e W^- \to e^- \pi^-  \nu_e$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;with the minimum reaction energy of 140 MeV  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is valid for the reactions (1) (2) (3)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the &lt;strong&gt;experimental data&lt;/strong&gt; on minimum energy of these reactions?&lt;/p&gt;&#xA;" OwnerUserId="238" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-02-23T18:20:43.257" LastActivityDate="2011-02-23T18:20:43.257" CommunityOwnedDate="2011-02-20T12:34:56.737" Title="Energy of the electron-muon reaction" Tags="&lt;particle-physics&gt;&lt;astrophysics&gt;&lt;experimental-physics&gt;&lt;sun&gt;&lt;weak-force&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="1" />
  <row Id="790" PostTypeId="2" ParentId="756" CreationDate="2010-11-14T19:24:59.553" Score="1" ViewCount="" Body="&lt;p&gt;The most robust experimental test of AdS/CFT so far is the measurement of the viscosity of the quark-gluon plasma formed in the aftermath of collisions between heavy atoms. There is a nice paper on arXiv about &lt;a href=&quot;http://arxiv.org/abs/0805.3579&quot; rel=&quot;nofollow&quot;&gt;AdS-CFT and the RHIC fireball&lt;/a&gt;, where RHIC is the relativistic heavy ion collider at Brookhaven. And another nice review paper on the underlying physics by &lt;a href=&quot;http://arxiv.org/abs/0704.0240&quot; rel=&quot;nofollow&quot;&gt;Son and Starinets&lt;/a&gt;. You might also find stimulating Subir Sachdev's &lt;a href=&quot;http://arxiv.org/abs/1010.0682&quot; rel=&quot;nofollow&quot;&gt;work&lt;/a&gt; relating AdS/CFT to problems in condensed matter.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                              All the best,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" LastActivityDate="2010-11-14T19:24:59.553" />
  <row Id="791" PostTypeId="1" AcceptedAnswerId="794" CreationDate="2010-11-14T19:35:11.250" Score="3" ViewCount="157" Body="&lt;p&gt;First off, I swear this is not homework. I'm doing some practice problems because I got an exam coming up. I'm stuck on this one:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/w7KMz.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured I would use energy conservation for this problem. So since the thing is not moving initially, I tried doing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$mgh=1/2 I\omega^2+1/2 mv^2$, but that doesnt give me the right answer. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2010-11-14T20:09:20.850" Title="Question about moment of inertia and velocity" Tags="&lt;mechanics&gt;&lt;classical-mechanics&gt;&lt;rotational-dynamics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="792" PostTypeId="2" ParentId="789" CreationDate="2010-11-14T19:52:35.630" Score="6" ViewCount="" Body="&lt;p&gt;There is no need for any intermediate state involving multiple pions. The leading Feynman diagram has a photon exchanged between the electron and muon, and the muon radiating a (very off-shell) W that decays into quarks that hadronize to a pion. The only constraint is that the initial energy is larger than the sum of the masses of the final state particles. (The pion in the final state, of course, is unstable and will soon decay back to a muon and antineutrino.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you don't want to think in terms of the W and quarks, you can just consider that there's an effective pion&amp;ndash;muon&amp;ndash;neutrino vertex which is what allows the pion to decay, and the muon is using this interaction to turn into a pion and neutrino.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, for any process you can consider, the minimum energy needed is just the sum of the masses of the final-state particles.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-11-14T19:52:35.630" CommentCount="5" />
  <row Id="793" PostTypeId="2" ParentId="791" CreationDate="2010-11-14T20:03:29.700" Score="1" ViewCount="" Body="&lt;p&gt;Don't forget the potential energy of m2 which is at height 2h when m1 hits the ground.&lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2010-11-14T20:03:29.700" CommentCount="3" />
  <row Id="794" PostTypeId="2" ParentId="791" CreationDate="2010-11-14T20:09:20.850" Score="5" ViewCount="" Body="&lt;p&gt;The basic setup is correct, conservation of energy might be the quickest way to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(m_1 -m_2) g h = \frac 1 2 I \omega^2 + \frac 1 2 (m_1+m_2) v^2, I=\frac{MR^2}{2}, \omega = v/R$$ gives me one of your options as the result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two $m$ in your formula seem to refer to different quantities.&lt;/p&gt;&#xA;" OwnerUserId="243" LastActivityDate="2010-11-14T20:09:20.850" CommentCount="4" />
  <row Id="798" PostTypeId="1" AcceptedAnswerId="808" CreationDate="2010-11-14T21:16:57.567" Score="4" ViewCount="274" Body="&lt;p&gt;I really hope someone will take a quick look at the following, I would just love to better understand it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This exercise is from Arnold's &quot;Mathematical Methods of Classical Mechanics&quot;, p. 97 in the chapter on d'Alemberts principle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rod of weight P, tilted at an angle of 60° to the plane of a table, begins to fall with initial velocity zero. Find the constraint force of the table at the initial moment, considering the table as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) absolutely smooth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) absolutely rough&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In the first case, the holonomic constraint holds the end of the rod on the plane of the table, and in the second case, at a given point.)&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Mq3d0.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I must admit, I am pretty unsure on how to do calculations using this &quot;fancy&quot; mathematical kind of physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, I'm lost with (a). But I'll have a go at (b):&#xA;As far as I understood, d'Alemberts principle states, that if $M$ is a constraining manifold, $x(q)$ is a curve in $M$ and $\xi$ is a vector perpendicular to $T_xM$, then $x$ satisfies Lagrange's equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{dt} \frac{\partial L}{\partial \dot q} = \frac{\partial L}{\partial q} \qquad L = \frac{{\dot x}^2}{2} - U(x)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;iff for the following inner product, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(m \ddot x + \frac{\partial U}{\partial x}, \xi \right) = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess that's about right so far? The constraint would in this case essentially be $\mathbb S^1$, since the rod moves on a circe around the point in contact with the table. Can we assume that all the rod's mass is at the center of mass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would we then have $U(x) = -gx_2$ in this case (where $x_2$ is the vertical component of $x$)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, then $\partial U / \partial x = -ge_2$. Where $e_1, e_2$ are the horizontal and vertical unit vectors, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the inital moment, we have $x = \cos(60&amp;#176;) e_1 + \sin(60&amp;#176;) e_2$, so by d'Alembert's principle we must have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(m \ddot x + \frac{\partial U}{\partial x}, \cos(60&amp;#176;) e_1 + \sin(60&amp;#176;) e_2 \right) = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or written differently&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m \ddot {x_1} \cos(60&amp;#176;) + m \ddot{x_2} \sin(60&amp;#176;) - \sin(60&amp;#176;)g = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So: Is this correct so far? Or am I way off? Is (a) handled any differently up to this point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for reading (and hopefully thanks for your helpful reply)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone could recommend a good problem book (with solutions), in which this kind of mathematical approach is used (I don't know if this is how physicists would actually compute stuff??), I would greatly appreciate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kind regards,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sam&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditorDisplayName="" LastEditDate="2010-11-15T04:46:43.007" LastActivityDate="2010-11-15T04:46:43.007" Title="Constraint force on a rod" Tags="&lt;classical-mechanics&gt;&lt;lagrangian-mechanics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="799" PostTypeId="1" CreationDate="2010-11-14T21:54:36.877" Score="6" ViewCount="295" Body="&lt;p&gt;Say I have a series of tubes (not the internet) looking like this, where &lt;code&gt;w&lt;/code&gt; represents water:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;|    |      | |&#xA;|    |      | |&#xA;|wwww|      |w|&#xA;|wwww|      |w|&#xA;|wwww+------+w|&#xA;|wwwwwwwwwwwww|&#xA;+-------------+&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Why is it that if I put some water in either side, the water level changes on the other side until they equalize? Why does this work even if I put water on the thinner side? And why is it that when you tilt the whole contraption, the water level is still even, although slanted?&lt;/p&gt;&#xA;" OwnerUserId="196" LastActivityDate="2010-11-15T10:07:13.977" Title="Why does the water level equalize in a series of tubes?" Tags="&lt;fluid-dynamics&gt;&lt;pressure&gt;&lt;water&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="800" PostTypeId="2" ParentId="799" CreationDate="2010-11-14T22:17:01.223" Score="2" ViewCount="" Body="&lt;p&gt;In a liquid like water, the pressure acts in an isotropic way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, imagine a slice of water in the middle tube; what are the forces acting on this slide ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force exerted by the pressure on the left side, and the one on the right side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one on the left depends and the height of the water column in the left pipe. The one on the right depends on the height in the right pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want equilibrium, both have to be equal. Therefore the heights have to be equal.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;About the pressure: pressure has dimension of force divided by surface, in common units: $N/m^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The column of water on the left pipe exert a force, due to its weight (gravity) that is $g \rho S h$, where $\rho$ is the volumic mass, S is the cross section of the pipe and h is the height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the pressure is $g \rho h$, thus independent of the cross section of the pipe. This force (gravitational) acts downward, but the fluid make it acts in an isotropic way, thus being directed from left to right on the slice of water (see above).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A good schematic explanation is available in &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/pflu.html&quot; rel=&quot;nofollow&quot;&gt;hyperphysics&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Altough the diameter of the left pipe is bigger, the force exerted on the &quot;slice&quot; of water is not higher because the pressure on a given infinitesimal volume depends only on the height of the column of water above it. Imagine two simple straight vertical pipes filled with water and with equal height, one with a large diameter and the other with a smaller. It is true that the force on the bottom of the big one is higher, but the pressure will be the same, because the force acts on a larger surface.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-14T23:12:27.650" LastActivityDate="2010-11-14T23:12:27.650" CommentCount="4" />
  <row Id="801" PostTypeId="2" ParentId="799" CreationDate="2010-11-14T22:23:07.487" Score="1" ViewCount="" Body="&lt;p&gt;The short answer is: the system (after long enough time passes) wants to attain a configuration which is stable (we say the system is in equilibrium).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now would the system be stable if one side had higher column of water than the other? Of course not, because the pressure at the bottom would be different between the two sides, introducing some forces. It is these forces which lower the height of the higher column and vice versa. Now, if the system were without any friction this would result in harmonic oscillations with water going up left column, then right, then left, ad infinitum. But because of friction, energy is lost in the form of heat when water moves and it pretty quickly attains a stable configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, I suppose you should be able to observe few oscillations if you make the height difference really big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slanted case is completely the same. Applying above intuition about stability, it should be clear that the surface of the water must always be perpendicular to the gradient of the gravitational potential and the surface must be a level surface of the gravitational potential (otherwise you introduce some nonequilibrium forces). In homogeneous potential this results in a plane perpendicular to the direction to the center of the Earth.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-11-14T22:23:07.487" />
  <row Id="802" PostTypeId="2" ParentId="789" CreationDate="2010-11-14T22:26:07.597" Score="0" ViewCount="" Body="&lt;p&gt;To complement Matt's answer, we just compare the total energy in both sides:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_e$ = 0.5 MeV, $M_\mu = 106 MeV$, $M_\pi = 140 MeV, the mass of the neutrino can be neglected so we obtain that &lt;strong&gt;kinematically&lt;/strong&gt; this process is possible only if we give 34 MeV in the form of kinetic energy.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-11-14T22:26:07.597" />
  <row Id="803" PostTypeId="2" ParentId="776" CreationDate="2010-11-14T22:34:36.263" Score="11" ViewCount="" Body="&lt;p&gt;Jupiter's mass is too small to produce nuclear fusion. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Jupiter would need to be about 75 times as massive to fuse hydrogen and become a star &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Jupiter&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Jupiter&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This wikipedia page explains the detailed requirements of nuclear fusion: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_fusion&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Nuclear_fusion&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="259" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-03-03T21:25:26.307" LastActivityDate="2011-03-03T21:25:26.307" CommentCount="1" />
  <row Id="804" PostTypeId="2" ParentId="634" CreationDate="2010-11-14T22:59:22.043" Score="0" ViewCount="" Body="&lt;p&gt;I recommend J.D. Bjorken, S.D. Drell, &lt;em&gt;Relativistic Quantum Mechanics&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this book can provide smooth introduction into Dirac equation - the book is almost dedicated to it. At the same time I discourage starting with Quantum Field Theory before learning Dirac equation - the conceptual step may be way to difficult.&lt;/p&gt;&#xA;" OwnerUserId="184" LastActivityDate="2010-11-14T22:59:22.043" />
  <row Id="805" PostTypeId="2" ParentId="776" CreationDate="2010-11-15T00:11:20.147" Score="3" ViewCount="" Body="&lt;p&gt;A star is not burning in the typical sense of burning that we have on earth. It's fusion hydrogen into helium. And it does this because it has so much mass that gravity is trying to collapse the star but the immense pressure means fusion takes place (creating an outward pressure to counteract the gravity pulling inwards creating a bance which must be maintained to keep the sun from collapsing inwards or exploding. ) so a star is defined by it's mass. You have brown dwarfs that are almost massive enough to start fusion. &lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2010-11-15T00:11:20.147" />
  <row Id="806" PostTypeId="2" ParentId="798" CreationDate="2010-11-15T00:13:42.530" Score="2" ViewCount="" Body="&lt;p&gt;I will not address your first question as I am not really sure. I've read what you've written and given enough time I'd be probably be able to understand whether the derivation is correct, but from top of my head I can't and, more importantly (and this answers your other question): this is not how physicists think about these problems (at least I don't know anyone who does), because once you learn Lagrange's formalism there's no need to go back to this strange principle that, as far as I know, is only used as a motivation on the road to better formalisms (Lagrange's and Hamilton's).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, here's what I'd do:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In (b), as you correctly state, the endpoint (or any other point for that matter) of the rod will be constrained to $S^1$. Now this gives us a nice way to parametrize the problem: the angle $\varphi$ between the rod and the table (this is all part of the Lagrange formalism. If you haven't yet learned it, I strongly advise you do). Also let us denote by $r$ the distance between the center of mass and the point of contact with the table. We write out the Lagrangian for the center of mass (assuming the rod is homogeneous)&#xA;$$L(\varphi, \dot{\varphi}) = T(\varphi, \dot{\varphi}) - U(\varphi) = m{r^2\dot{\varphi}^2 \over 2} - mgr\sin(\varphi)$$&#xD;&#xA;Now just apply Lagrange's equations to obtain&#xA;$$mr^2\ddot{\varphi} = - mgr\cos(\varphi)$$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For (a) we should assume smooth table, which means no friction, which means the rod will slide. It is clear that it will be confined to the half-plane perpendicular to the table in which the rod initially lies. We can parametrize this half-plane by $p$: the position of the point of contact of the rod with the table and $\phi$ as before. Again using Lagrange formalism one obtains&#xA;$$L(p, \varphi, \dot{p}, \dot{\varphi}) = {m \over 2}(\dot{p}^2 -2r\dot{p}\sin(\phi)\dot{\phi} + r^2\dot{\varphi}^2 ) - mgr\sin(\varphi)$$&#xD;&#xA;This is not a particularly nice Lagrangian and one has to wonder whether there are some better coordinates in which the problem simplifies (as in the case (b) above).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-15T00:38:14.703" LastActivityDate="2010-11-15T00:38:14.703" CommentCount="1" />
  <row Id="807" PostTypeId="2" ParentId="461" CreationDate="2010-11-15T00:14:28.717" Score="2" ViewCount="" Body="&lt;p&gt;If you are inside the event horizon of a spinning black hole, there are closed timelike curves--paths through spacetime which, when you progress them into the future, they end up in their past.  These curves clearly violate the principle of causality, and thus, the region inside the black hole can be considered unphysical in a sense.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, of course, the spacetime singularity also lives inside of the black hole's horizon.  Classically, it is a point or ring of infinite density at which the curvature of spacetime is infinite.  Spacetime histories intersecting the singularity must end if you believe in classical mechanics.  So, there is clearly some new physics there--most relativity researchers believe in something called the Cosmic Censorship conjecture, which says that nearly all* physically reasonable solutions of Einstein's equations hide their black hole singularities behind a horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*There are some known exact solutions where physically reasonable matter distributions collapse to horizonless singularities, but they are believed to form a set of measure zero in the space of matter distributions.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-11-15T00:14:28.717" />
  <row Id="808" PostTypeId="2" ParentId="798" CreationDate="2010-11-15T00:20:50.513" Score="1" ViewCount="" Body="&lt;p&gt;The first way to consider the problem, as we have a simple rigid body is to consider two points on it, say one end (the one that will touch the table first) and the center of mass (or the other end).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The embedding configuration space is thus $R^2 \times R^2$ (the three coordinates of the end of the rod and the three coordinates of the other) if we consider that the rod can only move in the vertical plane including a line on the table (I hope this is the actual question ...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The action of the surface lead to the [holonomic] constraint equation that can be written $f(x,y) = 0$ which is the one of a surface (here a curve) in the configuration space $R^2$. In addition to this constraint equation we have the Newton's equation which is $m \frac{d^2}{dt^2} \vec{r} = \vec{F}+\vec{N}$ (where $N$ is the force of constraint. This is for one end. For the other (or for the center of mass) you have two similar equations, with the constraint being that the other end (or the center of mass) is at a given distance to the other end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that view, we can consider the points as being free and we have 4 Newtonian equations and 2 constraint equations thus reducing the degrees of freedom to 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also consider that the constraints do not work and that they are perpendicular to the plane of the table: $\vec{N} = \lambda \vec{\nabla}f$ where $f$ is the one of the constraint equation. Newton's equation can be rewritten $\langle m \frac{d^2}{dt^2} \vec{r} - \vec{F}, \vec{\xi} \rangle = 0$ where $\xi$ is a tangent vector defined as $\langle \vec{\xi} ,\vec{\nabla} f \rangle= 0$ ** (1) **.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is d'Alembert principle. As the components of $\vec{xi}$ are constrained by (1), this lead to 2N (newton) - 2 (constraints) independant relations.  The next step is to choose a system of coordinates where (1) is automatically satisfied: these are the generalized coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rod is modeled when in addition to specifying the coordinate of the end that is on the plane you also specify the angle that the rod makes with the surface: we can choose to independent generalized coordinates: $x$ and $\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motion of the system (now the system is not free anymore) will take place in the Manifold $M$ where $x$ and $\theta$ are independent coordinates. These $(x, \theta)$ are usually written $q = (q_1, q_2)$. This manifold is the direct product of $R$ and $S^1$ (the angles take values in a 1-torus, a segment whose ends are identified).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can then be derived that the motion in these coordinates follow Lagrange's equations. The Lagrangian is this case as the natural $L(x, \theta, \dot{x}, \dot{\theta})=T-V$ form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For your case (b)&lt;/strong&gt; this problem is simplified: x is fixed and you should only care about $\theta$, you have $L = \frac{m}{2} l^2 \dot{\theta}^2 - l m g  sin\theta$ where $l$ is the distance  between one end and the center of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Lagrange's equation you obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m l \ddot{\theta} = -m g cos\theta$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that you can obtain the complete motion of the rod.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For the case (a)&lt;/strong&gt; you have to consider a more complete Lagrangian:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L = \frac{m}{2}( l^2 \dot{\theta}^2 - 2 l \dot{x} sin\theta \dot{\theta} + \dot{x}^2)- l m g  sin\theta$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Lagrangian lead to a more complicated motion; if I did the calculation correctly we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ddot{\theta} = - \frac{g}{l} cos\theta$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ddot{x} = l sin \theta \ddot{\theta}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To obtain the constraint, you can solve this and then you go back to Newton's equations. In the case (b) you have in addition to force $\dot{x} = \ddot{x} = 0$ to obtain the tangential constraint.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-15T01:45:31.813" LastActivityDate="2010-11-15T01:45:31.813" CommentCount="6" />
  <row Id="809" PostTypeId="2" ParentId="776" CreationDate="2010-11-15T01:36:55.923" Score="7" ViewCount="" Body="&lt;p&gt;There is a minor change to yyahn's answer. The isotopes Deuterium, and Lithium7 which are present in small amounts -left over from the big bang, can fuse at lower mass than pure hydrogen burning. The estimate is at around 65 mass Jupiter Lithium fuses with hydrogen to form two helium nuclei. Brown dwarfs as light as 13 mass Jupiter can ignite Deuterium fusion. So we&#xA;seem to have some weak forms of fusion (on low abundance fuels) that can happen for some range of brown dwarf masses. See wikipedia on &lt;a href=&quot;http://en.wikipedia.org/wiki/Brown_dwarf&quot; rel=&quot;nofollow&quot;&gt;brown dwarfs&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="391" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-03-03T13:52:43.673" LastActivityDate="2011-03-03T13:52:43.673" />
  <row Id="810" PostTypeId="1" AcceptedAnswerId="826" CreationDate="2010-11-15T01:47:10.507" Score="2" ViewCount="185" Body="&lt;p&gt;I have heard the terms up and down conversion in photonics/photovoltaics articles. What do the terms mean?&lt;/p&gt;&#xA;" OwnerUserId="293" LastActivityDate="2010-11-15T11:21:54.583" Title="What is up and down conversion in photonics?" Tags="&lt;photons&gt;" AnswerCount="2" />
  <row Id="811" PostTypeId="2" ParentId="799" CreationDate="2010-11-15T01:49:52.983" Score="4" ViewCount="" Body="&lt;p&gt;Another way to think of it is to take the rule that the water wants to minimize its potential energy.  (This is a basic principle of statics.)  The potential energy is just the average height of all the water, times its weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the water level were higher on the right, we could skim a little water off the top there and dump it on the left.  The water we moved would go down some while the rest of the water would stay at the same height, so the average level would go down.  Hence, it wasn't at a minimum before.  Only if the water level is the same everywhere is there no way to skim a little from one spot, dump it somewhere else, and reduce the potential energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Caveat: see comments on this answer for a little more explanation&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-15T10:07:13.977" LastActivityDate="2010-11-15T10:07:13.977" CommentCount="5" />
  <row Id="812" PostTypeId="2" ParentId="679" CreationDate="2010-11-15T02:56:14.837" Score="-1" ViewCount="" Body="&lt;p&gt;Kahn Academy has a bunch of great online lessons - in the Physics section - too many to count - proibibly about a hundred or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.khanacademy.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.khanacademy.org/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="423" LastActivityDate="2010-11-15T02:56:14.837" />
  <row Id="813" PostTypeId="2" ParentId="27" CreationDate="2010-11-15T03:39:47.950" Score="0" ViewCount="" Body="&lt;p&gt;It is the case that all measurements proceed via the exploitation of the natural interactions that we understand theoretically. But once the measurement is completed and the result in hand, the QM analysis of the subsequent evolution of just those systems that yielded that particular result can no longer employ the original state function (which allows for all the different possible results), but must then employ just that part of the original state function that corresponds to the particular result. This 'sudden' change in the state function used is called state function collapse. Many physicists regard this change as corresponding to nothing more than the change in the experimenter's knowledge once the result is in hand. This is the epistemological interpretation of the state function. But many regard the change as also reflecting a genuine physical change in the state of those systems that came through the measurement yielding the particular result. This is the ontological interpretation of the state function and it has many variations. Still many others hold an ontological interpretation of the state function while denying that collapse happens at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These latter views, which also have many versions, have given rise to various interpretations of and/or alternatives to QM that go by names such as Pilot wave, deBroglie-Bohm, Modal interpretations, Relative state, Many Worlds, Many Minds, Consistent Histories, Decoherence theoretic, Information theoretic etc. Collectively these are all called NO-Collapse theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The champions of real, physical collapse have also been at work creating alternative theories of their own that replace the collapse postulate by evolutions that generate collapse dynamically. These theories go by the names of their authors, Ghirardi-Rimini-Weber-Pearl, Karolahazy, Penrose, Gisin, Percival, etc. Collectively these are the Collapse theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty in deciding among these many and still proliferating alternatives is due to the incredible success of standard QM. All the alternatives must, at least, reproduce the corroborated results of QM while possibly allowing for deviations in, as yet, untested waters. Some of them offer no deviations from QM, whatsoever! So deciding between them &#xA;and QM must be a matter of philosophy or aesthetics. In any case, the days of the hegemony of the Copenhagen interpretation, if they ever really existed, are gone forever.      &lt;/p&gt;&#xA;" OwnerUserId="345" LastActivityDate="2010-11-15T03:39:47.950" CommentCount="1" />
  <row Id="814" PostTypeId="1" AcceptedAnswerId="878" CreationDate="2010-11-15T05:33:53.873" Score="11" ViewCount="355" Body="&lt;p&gt;One of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tests_of_general_relativity#Classical_tests&quot; rel=&quot;nofollow&quot;&gt;three classic tests&lt;/a&gt; of general relativity is the calculation of the precession of the perihelion of Mercury's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This precession rate had been precisely measured using data collected since the 1600's, and it was later found that Newton's theory of gravity predicts a value that differs from the observed value. That difference, which I am calling the &lt;em&gt;anomalous precession&lt;/em&gt;, was estimated to be about 43 arcseconds per century in Einstein's time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have heard that general relativity predicts an additional correction that is almost exactly sufficient to account for that 43&quot;/century difference, but I've never seen that calculation done, at least not correctly. Can anyone supply the details?&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-16T04:04:11.900" Title="How do you calculate the anomalous precession of Mercury?" Tags="&lt;general-relativity&gt;&lt;astrophysics&gt;&lt;precession&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="4" />
  <row Id="815" PostTypeId="1" AcceptedAnswerId="4003" CreationDate="2010-11-15T07:26:15.083" Score="6" ViewCount="157" Body="&lt;p&gt;In a recent (and to me, very enlightening) paper, entitled &quot;Relational EPR&quot;, &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0604064&quot; rel=&quot;nofollow&quot;&gt;Smerlak and Rovelli&lt;/a&gt; present a way of thinking about EPR which relies upon Rovelli's previously published work on relational quantum mechanics (see &lt;a href=&quot;http://arxiv.org/abs/quant-ph/9609002&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/quant-ph/9609002&lt;/a&gt; ). In relational quantum mechanics, there is no non-locality, but the definition of when an event occurs is weakened from Einstein's strict definition and instead is localized to each observer-measurement apparatus, including subsequent observers. There are (informal) coherence assumptions to ensure the consistency of reports from different subsequent observers (all possible friends of Wigner).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this seems very similar to various results in modern categorical mathematics. Is there a standard mathematical structure which well describes the structure of the space of localized measurements which Rovelli has envisioned? I know of Isham's work on topos theory and quantum mechanics, but I think he is aiming at something a little different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS I first asked this on mathunderflow, then on mathoverflow but to no avail, I am therefore reposting it here.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2011-01-27T16:29:52.540" Title="What categorical mathematical structure(s) best describe the space of “localized events” in “relational quantum mechanics”?" Tags="&lt;quantum-mechanics&gt;&lt;category&gt;&lt;theory&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="816" PostTypeId="2" ParentId="814" CreationDate="2010-11-15T07:39:00.320" Score="1" ViewCount="" Body="&lt;p&gt;Try &lt;a href=&quot;http://www.mathpages.com/rr/s6-02/6-02.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.mathpages.com/rr/s6-02/6-02.htm&lt;/a&gt; .  Caveat, I haven't looked at it carefully yet.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a detailed discussion at &lt;a href=&quot;http://wapedia.mobi/en/Two-body_problem_in_general_relativity?t=3.&quot; rel=&quot;nofollow&quot;&gt;http://wapedia.mobi/en/Two-body_problem_in_general_relativity?t=3.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-11-15T14:12:51.673" LastActivityDate="2010-11-15T14:12:51.673" />
  <row Id="817" PostTypeId="1" AcceptedAnswerId="835" CreationDate="2010-11-15T09:17:36.937" Score="2" ViewCount="369" Body="&lt;p&gt;This thought gave rise to some new questions in my mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the consequences for:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How would it affect duality i.e. particle, wave property of photons?&lt;/li&gt;&#xA;&lt;li&gt;How does this statement affect the information theoretical aspect (entropy) of the universe?  &lt;strong&gt;Update:&lt;/strong&gt; Given a volume V of space, is the entropy (maximum information that can be store) in this volume changed when this statement is applied?&lt;/li&gt;&#xA;&lt;li&gt;How is a black hole affect by this statement? &lt;strong&gt;Update&lt;/strong&gt; How is entropy changed inside the black hole?&lt;/li&gt;&#xA;&lt;li&gt;Could one consequence be that the universe is hologram, since the construction isn’t continues?&lt;/li&gt;&#xA;&lt;li&gt;Would the smallest quantified space be planck's constant? Is there an equivalent constant for time?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I hope to get some of your feedbacks regarding this statement.&lt;/p&gt;&#xA;" OwnerUserId="425" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-12T22:48:55.763" LastActivityDate="2011-03-29T16:22:14.550" Title="What are some approaches to discrete space-time used in modern physics?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;string-theory&gt;&lt;spacetime&gt;" AnswerCount="3" CommentCount="11" FavoriteCount="1" />
  <row Id="818" PostTypeId="2" ParentId="814" CreationDate="2010-11-15T09:35:22.273" Score="3" ViewCount="" Body="&lt;p&gt;Well, it goes like this: consider Schwarzschild metric and testing particle (this is Mercury) with energy $E$ and momentum $L$. Because you have enough integrals of motion, the equations basically solve themselves and you obtain &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_geodesics#Effective_radial_potential_energy&quot; rel=&quot;nofollow&quot;&gt;an effective potential&lt;/a&gt; that contains basic Newton potential, the centrifugal potential and a correction term. Then you apply &lt;a href=&quot;http://en.wikipedia.org/wiki/Binet_equation&quot; rel=&quot;nofollow&quot;&gt;Binet equation&lt;/a&gt; and you are left with some differential equation that is not easy to solve, but essentially is an equation for conic section (as in the classical case) plus some correction terms. So you make some approximation (based on the parameters of the problem) and are left with &quot;conic section&quot; that precesses a little.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I wonder: how much more precise do you want the above argument to be made? Do you want a complete derivation, or just to clarify some confusing point? Also, how much are you familiar with GTR? I am asking just so that I know at what level should I be explaining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, see the wikipedia pages, it looks like they have some derivation there (although I did not check it and you can't always trust wikipedia).&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-15T14:44:17.430" LastActivityDate="2010-11-15T14:44:17.430" CommentCount="4" />
  <row Id="819" PostTypeId="2" ParentId="728" CreationDate="2010-11-15T10:10:59.963" Score="7" ViewCount="" Body="&lt;p&gt;I think the answer is that yes, they do follow the same path.  I'll try to show this by doing some kinematics, then simply calculating the accelerations.  I could use someone to check it over carefully.  I have had to use some heuristics in a couple of places because my vocabulary for discussing this problem is limited.  I don't see fundamental flaws in the physics, but please point them out (as well as typos) if you see them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;$m$ = mass of ball&lt;/li&gt;&#xA;&lt;li&gt;$I$ = moment of inertia of ball&lt;/li&gt;&#xA;&lt;li&gt;$R$ = radius of ball&lt;/li&gt;&#xA;&lt;li&gt;$\vec{R}$ = vector pointing from the center of the ball to the point of contact with the hill&lt;/li&gt;&#xA;&lt;li&gt;$\vec{g}$ = acceleration due to gravity&lt;/li&gt;&#xA;&lt;li&gt;$\vec{g}_h$ = acceleration due to gravity projected into the plane of the hill&lt;/li&gt;&#xA;&lt;li&gt;$\vec{\omega}$ = angular velocity of the ball&lt;/li&gt;&#xA;&lt;li&gt;$\vec{v}$ = velocity of the ball&lt;/li&gt;&#xA;&lt;li&gt;$\vec{\tau}$ = torque on the ball&lt;/li&gt;&#xA;&lt;li&gt;$\vec{f}$ = force of friction exerted on the ball by the hill&lt;/li&gt;&#xA;&lt;li&gt;$\vec{a}$ = acceleration of the ball&lt;/li&gt;&#xA;&lt;li&gt;$\vec{a}_h$ = acceleration of the ball projected in the plane of the hill&lt;/li&gt;&#xA;&lt;li&gt;subscript &quot;r&quot; on the above refers to rolling ball&lt;/li&gt;&#xA;&lt;li&gt;subscript &quot;s&quot; refers to sliding ball&lt;/li&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy is conserved in both cases, so the two balls must have the same kinetic energy as they pass through any given point.  The kinetic energy of the sliding ball is $1/2 mv_s^2$, and the kinetic energy of the rolling ball is $1/2(mv_r^2 + I\omega^2)$.  Using $|\omega| = |v|/R$, some algebra yields $(v_s/v_r)^2 = 1+I/mR^2$.  This tells us that if the rolling ball follows the same path as the sliding ball, it does so by doing everything slower by this set amount.  Let's let $v_r/v_s = c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to know whether the balls move on the same path.  They will do this if they always point the same direction when they get to the same place.  If we allow their velocities to be functions of position, they follow the same path if $\vec{v_r}(\vec{x}) = c\vec{v_s}(\vec{x})$.  We know this will hold for the velocities' magnitudes, so this equation is expressing the condition that the velocities' directions are the same.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's introduce a parameters $\lambda_r$, $\lambda_s$ along the balls' paths.  If they will follow the same path, the $\lambda$'s can be identical.  Make $\lambda$ equal to the distance the ball travels.  Differentiating both sides of the previous condition with respect to $\lambda$ and applying the chain rule gives $d\vec{v_r}/dt*dt/d\lambda = c d\vec{v_s}/dt*dt/d\lambda$.  These expressions are accelerations and velocities, and so this can be rewritten $\vec{a_r}/|v_r| = c\vec{a_s}/|v_s|$.  The definition of $c$ allows us to write this as $\vec{a_r} = c^2\vec{a_s}$ or $\vec{a_s} = (1 + I/mR^2)\vec{a_r}$.  As long as this condition holds, the assumption that the $\lambda$'s are the same is justified, the velocities of the balls will change the same way as they move through space, and the paths will be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we'll look at the component of acceleration normal to the hill.  Imagine that at some point on the hill both balls are moving in the same direction.  Then we take a circle that is tangent to the path the balls are following right at that point, like a loop coming up out of the hill.  The acceleration of the balls as they follow their paths at this point is just their centripetal acceleration going around this circle, and that's $v^2/r$.  Since the ratio of the velocities is $c$, the ratio of the centripetal accelerations is $c^2$, which is exactly the ratio specified in the condition that the accelerations must satisfy.  This means the projections of the accelerations normal to the hill &quot;work out&quot; in terms of the balls following the same path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we need the acceleration of the sliding ball projected into the hill, $\vec{a_{sh}}$.  There is only one force that has any component in the hill, and that's gravity.  So we have $\vec{a_{sh}} = \vec{g_h}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rolling ball has two forces that contribute in the plane of the hill - gravity and friction.  This gives $\vec{a_{rh}} = \vec{g_h} + \vec{f}/m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find $\vec{f}$, we'll make two requirements of it.  One is what's above - that $\vec{f}$ is what's needed to add to gravity to make the acceleration.  Another is that $\vec{f}$ contributes a torque about the center of mass of the ball.  This torque can be determined kinematically though, using only knowledge of $\vec{a}$.  There will be only one friction force capable of satisfying both conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torque due to friction is $\vec{\tau} = \vec{R}\times\vec{f} = I d\vec{\omega}/dt$.  To find another expression involving $d\vec{\omega}/dt$, note that the bottom of the ball is moving backwards relative to the center of mass, so $\vec{\omega} \times \vec{R} = -\vec{v_r}$.  Differentiate with respect to time.  $d\vec{R}/dt\times\vec{\omega} + \vec{R}\times d\vec{\omega}/dt = \vec{a_r}$.  Plugging in for $d\vec{\omega}/dt$ gives $d\vec{R}/dt\times\vec{\omega} + \vec{R} \times (1/I* \vec{R} \times \vec{f}) = \vec{a_r}$.  We are only interested in the component of $\vec{a_r}$ in the plane of the hill.  The term $d\vec{R}/dt\times\vec{\omega}$ is normal to the hill because both vectors being crossed are in the plane of the hill.  The term $\vec{R} \times (1/I *\vec{R} \times \vec{f})$ is completely in the plane of the hill because it's perpendicular to $\vec{R}$.  A vector identity lets us simplify this to $1/I(\vec{R}(\vec{R}\cdot\vec{f}) - \vec{f}(\vec{R}\cdot\vec{R})) = \vec{a_{rh}}$.  Use that $\vec{R} \cdot \vec{f} = 0$ because they are perpendicular, so $\vec{f} R^2/I = \vec{a_{rh}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have two expressions for $\vec{a_{rh}}$, so equate them.  $\vec{f}R^2/I = \vec{g_h} + \vec{f}/m$.  This gives $\vec{f} = -\vec{g_h}+mI/(R^2m+I)$.  Plugging this back into the equation for acceleration gives $\vec{a_{rh}} = \vec{g_h}(1 - I/(R^2m+I))$.  Comparing to the sliding ball, $\vec{a_{sh}} = (1 + I/mR^2)\vec{a_{rh}}$.  This is the condition on the accelerations we sought to demonstrate.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-15T10:34:44.867" LastActivityDate="2010-11-15T10:34:44.867" CommentCount="25" />
  <row Id="820" PostTypeId="2" ParentId="817" CreationDate="2010-11-15T10:34:30.713" Score="1" ViewCount="" Body="&lt;p&gt;OK, I found a review article that might be useful to you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cdsweb.cern.ch/record/704227&quot; rel=&quot;nofollow&quot;&gt;http://cdsweb.cern.ch/record/704227&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I quote the abstract here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We review some modern theories about the structure of space and time, in particular those related to discrete space and time. Following an epistemological method we start from theories which discuss discrete space and time as a mathematical tool to solve physical models. Antother theories look for physical content of the discrete structure of space and time, based in relational theories of space and time which are derived from the relations of some fundamental entities. Finally we present some philosophical positions who try to find the ontological foundation of the relational theories os space and time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hope this is the kind of thing you were looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Woops, this was misleading, the abstract is in english, but the paper is in spanish. The references in the article are still useful though.&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2010-11-15T10:34:30.713" />
  <row Id="821" PostTypeId="2" ParentId="728" CreationDate="2010-11-15T10:45:11.010" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; the example below doesn't work (as a counterexample). I realized that thanks to the conservation of the angular momentum in the direction perpendicular to the slope the motion decouples also in the second case. Now I am starting to believe that the result holds in general and am going to try to find a simple conceptual proof of this. I think Mark's argument is correct but it looks like an accidental mathematical result that is not at all obvious physically. In any case: thanks for the interesting food for thought!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Ok, here's a particularly easy example where trajectories (probably) differ. Just consider an inclined plane and suppose that the initial velocity of the ball is perpendicular to the slope. In the case where the ball is just sliding, the momentum in the perpendicular direction will be conserved. In particular, these two directions (perpendicular and along the slope) can be treated independently and you obtain a parabola as the trajectory. But  when the ball is rotating the equations of motion are quite complicated (I will try to solve them later to complete the argument) and it would surprise me hugely if the result was again parabola (not to mention the same one).&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-16T23:22:40.593" LastActivityDate="2010-11-16T23:22:40.593" CommentCount="1" />
  <row Id="822" PostTypeId="1" CreationDate="2010-11-15T10:56:22.267" Score="20" ViewCount="1179" Body="&lt;p&gt;This may seem like a slightly trite question, but it is one that has long intrigued me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I formally learned classical (Newtonian) mechanics, it has often struck me that angular momentum (and generally rotational dynamics) can be fully derived from normal (linear) momentum and dynamics. Simply by considering circular motion of a point mass and introducing new quantities, it seems one can describe and explain angular momentum fully without any new postulates. In this sense, I am lead to believe only ordinary momentum and dynamics are fundamental to mechanics, with rotational stuff effectively being a corollary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then at a later point I learned quantum mechanics. Alright, so orbital angular momentum does not really disturb my picture of the origin/fundamentality, but when we consider the concept of &lt;em&gt;spin&lt;/em&gt;, this introduces a problem in this proposed (philosophical) understanding. Spin is apparently intrinsic angular momentum; that is, it applies to a point particle. Something can possess angular momentum that is not actually moving/rotating - a concept that does not exist in classical mechanics! Does this imply that angular momentum is in fact a fundamental quantity, intrinsic to the universe in some sense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It somewhat bothers me that that fundamental particles such as electrons and quarks can possess their own angular momentum (spin), when otherwise angular momentum/rotational dynamics would fall out quite naturally from normal (linear) mechanics. There are of course some fringe theories that propose that even these so-called fundamental particles are composite, but at the moment physicists widely accept the concept of intrinsic angular momentum. In any case, can this dilemma be resolved, or do we simply have to extend our framework of fundamental quantities?&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2011-01-28T13:15:04.190" Title="Is Angular Momentum truly fundamental?" Tags="&lt;classical-mechanics&gt;&lt;angular-momentum&gt;" AnswerCount="8" FavoriteCount="10" />
  <row Id="823" PostTypeId="2" ParentId="625" CreationDate="2010-11-15T11:06:09.633" Score="2" ViewCount="" Body="&lt;p&gt;One more answer against &amp;ldquo;second quntization&amp;rdquo;, because I think it is a good demonstration of how a lame notation can obscure a physical meaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The first statement is&lt;/strong&gt;: there is no second quantization. For example, here is citation from Steven Weinberg's book &amp;ldquo;&lt;a href=&quot;http://www.scribd.com/doc/3082871&quot; rel=&quot;nofollow&quot;&gt;The Quantum  Theory of Fields&lt;/a&gt;&amp;rdquo; Vol.I:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It would  be a good thing if the&#xA;  misleading expression &amp;lsquo;second&#xA;  quantization&amp;rsquo; were  permanently&#xA;  retired.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[I would even say that there is no quantization at all, as a procedure to pass from classical theory to quantum one, because (for example) quantum mechanics of single particle is more fundamental than the classical mechanics, therefore you can derive all &amp;ldquo;classical&amp;rdquo; results from QM but not vice versa. But I understand that it is a too speculative answer.]  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a procedure called &amp;ldquo;canonical quantization&amp;rdquo;, which is used to construct a quantum theory for a classical system which has Hamiltonian dynamics, or more generally, to construct a quantum theory which has a certain classical limit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, if by the &amp;ldquo;canonical quantization&amp;rdquo; of a Hamiltonian system with finite number of degrees of freedom (classical mechanics) you imply quantum mechanics (QM) with fixed number of particles, then quantum field theory (QFT) is the &amp;ldquo;canonical quantization&amp;rdquo; of a classical Hamiltonian system with infinite number of degrees of freedom - classical field theory, &lt;em&gt;not quantum mechanics&lt;/em&gt;. For such procedure, there is no difference between quantization of the electro-magnetic field modes and quantization of vibrational modes of the surface of the droplet of superfluid helium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more citation from Weinberg's book:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The  wave fields $\phi$, $\varphi$,&#xA;  etc, are not probability amplitudes at&#xA;  all...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is useful to keep in mind the following analogy: the coordinates are the &amp;ldquo;classical configuration&amp;rdquo; of a particle. QM wave function $\psi(x)$ corresponds to the &amp;ldquo;smearing&amp;rdquo; of a quantum particle over all possible &amp;ldquo;classical configurations&amp;rdquo;. QFT wave function $\Psi(A)$ corresponds to &amp;ldquo;smearing&amp;rdquo; of a quantum field over all possible configurations of a classical field $A$. Operator $\hat{A}$ corresponds to the observable $A$ in the same way as observable $x$ is represented by Hermitian operators $\hat{x}$ in QM.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The second statement is&lt;/strong&gt;: &amp;ldquo;canonical quantization&amp;rdquo; is irrelevant in the context of fundamental theory. QFT is the only way to marry quantum mechanics to special relativity and can be contracted without a reference to any &quot;classical crutches&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;: There is not any sequence of &amp;ldquo;quantizations&amp;rdquo; (1st, 2nd,.. nth).&lt;/p&gt;&#xA;" OwnerUserId="225" LastActivityDate="2010-11-15T11:06:09.633" CommentCount="6" />
  <row Id="824" PostTypeId="2" ParentId="810" CreationDate="2010-11-15T11:12:15.473" Score="4" ViewCount="" Body="&lt;p&gt;In the field of quantum optics &quot;&lt;a href=&quot;http://omnis.if.ufrj.br/~phsr/PHSR/PDC.htm&quot; rel=&quot;nofollow&quot;&gt;parametric down-conversion&lt;/a&gt;&quot; is a process where an intense pump laser shines an optical non-linear crystal and strongly correlated or entangled photon pairs are generated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an optical non-linear crystal the polarization response of the material to the optical field is non-linear, which typically occurs at high intensity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The photons of the pump laser have energy $\hbar \omega_p$ and momentum $\vec{k_p}$ which must be equal to the sum of the energy and sum of momentum for the produced photon pair: $\hbar \omega_p = \hbar \omega_1 + \hbar \omega_2$ and $\vec{k_p} =\vec{ k_1} + \vec{k_2}$. Conservation of the parameters energy and momentum lead to strong correlations or entanglement between the produced pairs. I believe the name &lt;em&gt;parametric&lt;/em&gt; indicates the fact that certain parameters are entangled whereas other parameters are not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process is not very efficient but the experimental setup is not very difficult and acts as source of entangled photon-pairs in many experiments on the foundations of quantum mechanics addressing the EPR paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parametric up conversion probably is the reverse process where entangled photon pairs form one new photon - I am not sure however.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2010-11-15T11:21:45.960" LastActivityDate="2010-11-15T11:21:45.960" CommentCount="1" />
  <row Id="825" PostTypeId="2" ParentId="822" CreationDate="2010-11-15T11:20:58.550" Score="2" ViewCount="" Body="&lt;p&gt;In classical mechanics, the fundamental entities change according to the framework you opt for. If you do classical Newtonian mechanics, I'd say the fundamental entities are postions and velocities. All the others can be derived from them and the dynamics of particles are described in terms of functions of these (forces are functions of time, positions and velocities).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you go to Hamiltonian mechanics, then positions and momenta become fundamental. And the Hamiltonian can be expressed as a function of these and possibly time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, in classical mechanics, angular momentum is always a derived quantity, because it is always an orbital angular momentum, never an intrinsic angular momentum. Even when you have an object rotating on an own axis, this can be understood as the particles constituting the object executing an orbital motion. Of course, you can write Hamiltonians which depend on the angular momentum of the top, but these are higher level descriptions, the angular momentum of the top could still be decomposed in principle into the orbital angular momenta of its constituents. This would not be a very practical approach to problem-solving of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, as you say, a fundamental intrinsic angular momentum is a novelty in quantum mechanics. The way it enters the equations is usually through the multiple-valuedness of the wave function. Say a spin 1/2 particle has to be described by two independent component wave functions (there could be more components, but these would not be independent). I don't know of any way around this. This is a basic fact of how nature works and it is related to the representations of the symmetry group of space-time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the symmetry group of space-time is basically the same in quantum and in classical physics, I don't see however why it should not be possible to describe particles with intrinsic momenta in classical mechanics. I think it is certainly possible in principle. The question is, is it useful? Since all our elementary particles have to be described at the quantum level, what use is a classical theory of particles with intrinsic momenta? Except in the sense of tackling problems like the top by simplification or something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: As a matter of fact, classical field theories do have spin. Think of the Maxwell equations for instance.&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2010-11-15T11:33:32.530" LastActivityDate="2010-11-15T11:33:32.530" CommentCount="10" />
  <row Id="826" PostTypeId="2" ParentId="810" CreationDate="2010-11-15T11:21:54.583" Score="4" ViewCount="" Body="&lt;p&gt;I complete Gerard's answer: up-conversion is the reverse process, where two low frequency photons are converted to a single high-frequency photon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically  down- and up-conversion correspond to a frequency conversion of the photons through a nonlinear interaction. The up/down term correspond to the &quot;direction&quot; of the frequency change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to add that the &quot;parametric&quot; term has nothing to do with entanglement, but with the fact that the χ⁽²⁾ nonlinearity used is analogue to a classical parametric oscillator &lt;a href=&quot;http://en.wikipedia.org/wiki/Parametric_oscillator&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Parametric_oscillator&lt;/a&gt; .&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2010-11-15T11:21:54.583" CommentCount="1" />
  <row Id="827" PostTypeId="2" ParentId="782" CreationDate="2010-11-15T11:25:41.910" Score="6" ViewCount="" Body="&lt;p&gt;If you don't consider atmospheric drag and other dissipative effects, assuming that the Earth is non-rotating, and ignore general relativity (which is a lot of assumptions to make, and later I can try to edit this to comment on how they might be relaxed), a ballistic trajectory is just a segment of an orbit. So finding the launch parameters just entails finding an  orbit that intersects both the launch point and the destination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbital trajectories are given by the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_o = \frac{\rho}{1 + \epsilon\cos\phi}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\rho$ is a length parameter related to the size of the orbit and $\epsilon$ is the eccentricity. At the points where the orbit intersects the Earth's surface, you'll have $r_o$ equaling the radius of the Earth at that point; let's say $r_o = R_L$ for the launch point and $r_o = R_D$ for the destination. This gives you the following two equations,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_L = \frac{\rho}{1 + \epsilon\cos\phi_L}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_D = \frac{\rho}{1 + \epsilon\cos\phi_D}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;along with the constraint that $\phi_D - \phi_L$ (or vice-versa, depending on how you define coordinates) has to be equal to the angular separation $\Delta$ between the source and destination. That angular separation can be calculated using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Haversine_formula&quot; rel=&quot;nofollow&quot;&gt;law of haversines&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This system of equations can be solved for the orbital parameters as follows, assuming $R_D \neq R_L$ (see below for the other case),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\epsilon = \frac{R_L - R_D}{R_D\cos(\phi_L + \Delta) - R_L\cos\phi_L}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho = \frac{R_D R_L [\cos(\phi_L + \Delta) - \cos\phi_L]}{R_D\cos(\phi_L + \Delta) - R_L\cos\phi_L}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that they are dependent on $\phi_L$; this is because only the &lt;em&gt;difference&lt;/em&gt; $\phi_D - \phi_L = \Delta$ is physically relevant, so you're free to choose the actual &lt;em&gt;values&lt;/em&gt; to be whatever you like. Your choice of $\phi_L$ will influence the shape of the trajectory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having calculated $\epsilon$ and $\rho$, you can determine the launch angle $\alpha$ above the horizon by calculating the slope of the orbit at the launch point:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan\alpha = \left.\frac{1}{r_o}\frac{\mathrm{d}r_o}{\mathrm{d}\phi}\right|_{\phi_L} = \frac{\epsilon\sin\phi_L}{1 + \epsilon\cos\phi_L} = \frac{(R_L - R_D)\sin\phi_L}{R_D[\cos(\phi_L + \Delta) - \cos\phi_L]}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine the speed, you can use the fact that (according to my notes) the total energy of the projectile is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = \frac{GMm(\epsilon^2 - 1)}{2\rho}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is equal to the sum of kinetic and potential energies, $\frac{1}{2}mv^2 - \frac{GMm}{r}$. ($m$ is the mass of the projectile, $M$ is that of the Earth) Plugging in $r = R_L$, I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = \sqrt{\frac{2GM}{R_L} + \frac{GM(\epsilon^2 - 1)}{\rho}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the bottom line is that you plug $R_D$, $R_L$, $\Delta$, and some choice of $\phi_L$ into the formulas for $v$ and $\alpha$ to get the launch parameters.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I mentioned that the procedure above hits a snag if $R_D = R_L$. You'd wind up calculating $\epsilon = 0$, which on a spherical planet entails rolling your projectile along the ground ;-) which doesn't make sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $R_D = R_L$, you can go back to the orbital equations and find that $\cos\phi_D = \cos\phi_L$. (Alternatively, this could come from the condition that the denominator of $\epsilon$ also be zero.) The only way to satisfy this is by setting $\phi_L = -\phi_D = \Delta/2$. This gives you the condition&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho = R_L\left(1 + \epsilon\cos\frac{\Delta}{2}\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, you have a degree of freedom: you can choose any value of $\epsilon$, and plugging into this equation will give you the corresponding value of $\rho$. Once you have those, you can use the same procedures to calculate the launch angle and speed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan\alpha = \left.\frac{1}{r_o}\frac{\mathrm{d}r_o}{\mathrm{d}\phi}\right|_{\phi_L} = \frac{\epsilon\sin\frac{\Delta}{2}}{1 + \epsilon\cos\frac{\Delta}{2}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = \sqrt{\frac{2GM}{R_L} + \frac{GM(\epsilon^2 - 1)}{R_L(1 + \epsilon\cos\frac{\Delta}{2})}}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-16T07:00:10.357" LastActivityDate="2010-11-16T07:00:10.357" CommentCount="2" />
  <row Id="828" PostTypeId="2" ParentId="822" CreationDate="2010-11-15T11:33:43.513" Score="16" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; As David pointed out, it's better to distinguish between generic &lt;em&gt;angular momentum&lt;/em&gt; and &lt;em&gt;orbital angular momentum&lt;/em&gt;. The first concept is more general and includes spin while the second one is (as the name suggests) just about orbiting. There is also the concept of &lt;em&gt;total angular momentum&lt;/em&gt; which is the quantity that is really conserved in systems with rotational symmetry. But in the absence of spin it coincides with &lt;em&gt;orbital angular momentum&lt;/em&gt;. This is the situation I analyze in the first paragraph.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Angular momentum &lt;strong&gt;is&lt;/strong&gt; fundamental. Why? &lt;a href=&quot;http://en.wikipedia.org/wiki/Noether%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Noether's theorem&lt;/a&gt; tells us that the symmetry of the system (in this case space-time) leads to the conservation of some quantity (momentum for translation, orbital angular momentum for rotation). Now, as it happens, Euclidean space is both translation and rotation invariant in compatible manner, so these concepts are related and it can appear that you can derive one from the other. But there might exist space-time that is translation but not rotation invariant and vice versa. In such a space-time you wouldn't get a relation between orbital angular momentum and momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to address the spin. Again, it is a result of some symmetry. But in this case the symmetry arises because of &lt;a href=&quot;http://en.wikipedia.org/wiki/Wigner%27s_classification&quot; rel=&quot;nofollow&quot;&gt;Wigner's correspondence&lt;/a&gt; between particles and irreducible representations of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Poincare_group&quot; rel=&quot;nofollow&quot;&gt;Poincaré group&lt;/a&gt; which is the symmetry group of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Minkowski_space&quot; rel=&quot;nofollow&quot;&gt;Minkowski space-time&lt;/a&gt;. This correspondence tells us that massive particles are classified by their mass and spin. But spin is not orbital angular momentum! The spin corresponds to group $Spin(3) \cong SU(2)$ which is a double cover of $SO(3)$ (rotational symmetry of three-dimensional Euclidean space). So this is a completely different concept that is only superficially similar and can't really be directly compared with orbital angular momentum. One way to see this is that spin can be a half-integer, but orbital angular momentum must always be an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to summarize:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;orbital angular momentum&lt;/em&gt; is a classical concept that arises in any space-time with rotational symmetry.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;spin&lt;/em&gt; is a concept that comes from quantum field theory built on the Minkowski space-time. The same concept also works for classical field theory, but there we don't have a clear correspondence with particles, so I omitted this case.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addition for the curious&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Eric has pointed out, there is more than just a superficial similarity between orbital angular momentum and spin. To illustrate the connection, it's useful to consider the question of how particle's properties transform under the change of coordinates (recall that conservation of total angular momentum arises because of the invariance to the change of coordinates that corresponds to rotation). Let us proceed in a little bit more generality and consider any transformation $\Lambda$ from the Lorentz group. Let us have a field $V^a(x^{\mu})$ that transforms in matrix representation ${S^a}_b (\Lambda)$ of the Lorentz group. Thanks to Wigner we know this corresponds to some particle; e.g. it could be scalar (like Higgs), bispinor (like electron) or vector (like Z boson). Its transformation properties under the element ${\Lambda^{\mu}}_{\nu}$ are then determined by (using Einstein summation convention)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ V&amp;#39;^a ({\Lambda^{\mu}}_{\nu} x^{\nu}) = {S^a}_b(\Lambda) V^b (x^{\mu}) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this one can at least intuitively see the relation between the properties of the space-time ($\Lambda$) and the particle ($S$). To return to the original question: $\Lambda$ contains information about the orbital angular momentum and $S$ contains information about the spin. So the two are connected but not in a trivial way. In particular, I don't think it's very useful to imagine spin as the actual &lt;em&gt;spinning&lt;/em&gt; of the particle (contrary to the terminology). But of course anyone is free to imagine whatever they feel helps them grasp the theory better.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-15T22:40:59.440" LastActivityDate="2010-11-15T22:40:59.440" CommentCount="23" />
  <row Id="829" PostTypeId="2" ParentId="822" CreationDate="2010-11-15T11:42:49.367" Score="6" ViewCount="" Body="&lt;p&gt;In the field of classical mechanics almost always angular momentum is derived form linear momemtum. This actually might be the problem because it is also possible to do it the other way around: linear momentum is a limiting case of angular momentum where the radius of rotation becomes infinite. In this view the split between rotational and linear &lt;em&gt;vanishes&lt;/em&gt; - the new concept that is introduced is: &lt;em&gt;infinity&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a new idea of mine, it has been established in the 19th century. By using projective geometry one can integrate linear and angular kinematics and dynamics in one framework (i.e. a translation is a rotation around an infinite axis; a pure moment is a force along an infinite line of action). Keywords: Felix Klein, linear complexes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another issue is intrinsic angular momentum. I could say: study the fundamentals, the principles, and the maths, and eventually you'll get a holistic picture, but that's not what I believe. I think we are in need of some kind of &lt;em&gt;geometrical&lt;/em&gt; electron model that permits us to depict intrinsical angular momentum. &lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-15T11:42:49.367" CommentCount="1" />
  <row Id="830" PostTypeId="1" AcceptedAnswerId="832" CreationDate="2010-11-15T12:01:33.207" Score="0" ViewCount="321" Body="&lt;p&gt;Suppose you have a solid ball on a horizontal table.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the direction of friction force when the ball I pushed horizontally and &lt;strong&gt;starts rolling&lt;/strong&gt;?&lt;/li&gt;&#xA;&lt;li&gt;Why is the direction of friction as it is? &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Which forces acts at the contact point between delta time t0 to t1? (If we divide friction force in sub forces)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;V=1Vx m/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fx=?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="425" LastActivityDate="2010-11-15T12:17:00.667" Title="What is the direction of the friction force on a rolling ball?" Tags="&lt;mechanics&gt;&lt;newtonian-mechanics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="831" PostTypeId="2" ParentId="613" CreationDate="2010-11-15T12:13:01.560" Score="4" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First, what we clam is that our method is the best for arbitrary planar graphs, due to the fact that the treewidth for a planar graph of size N scales a O(N^1/2). For the particular case of the square lattice, our method is not different from the traditional TM, since a strip of width L has a tree decomposition of width L+1 which is a path decomposition.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Second, in the Fortuin Kasteleyn (FK) representation you can't really fix vertices to any particular state since the degrees of freedom sit on the bonds. Nevertheless it's known that the two point function in the state representation equals the probability of the two vertices to be connected in a FK cluster. So what one should do, perhaps, is to give a special mark to connectivity states touching the two vertices.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Andrea Bedini&lt;/p&gt;&#xA;" OwnerUserId="426" LastActivityDate="2010-11-15T12:13:01.560" CommentCount="1" />
  <row Id="832" PostTypeId="2" ParentId="830" CreationDate="2010-11-15T12:13:33.820" Score="2" ViewCount="" Body="&lt;p&gt;First consider there is no friction. The point of contact between the ball and the table moves with the direction of the global motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now introduce friction: you have kinematic friction slowing down this point thus make the ball roll due to the induced torque. You will have a motion in between the cases of pure sliding and pure rolling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the direction of the friction force is obvious (by definition of the friction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you do the things at the limit case, you will have a pure rolling. In that case the point of contact has zero instantaneous velocity and if the motion is horizontal, with constant and angular and linear motion, you don't need any friction, if you had friction, this would induce a torque and the angular momentum will change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you introduce acceleration or a non horizontal surface: in that case you have static friction: the point cannot move forward, friction is directed opposite to the &quot;accelerated&quot; direction, you introduce a torque.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-11-15T12:13:33.820" />
  <row Id="834" PostTypeId="2" ParentId="217" CreationDate="2010-11-15T12:52:31.273" Score="3" ViewCount="" Body="&lt;p&gt;Since I'm not entirely content with the answers to date, here's my take - everyone seems to agree on the basics of forces generated by pressure differentials. If you took a rigid uncooked spaghetto with a cross section of $A$, the case is quite clear - on the cylinder's base in your mouth, a force of $p_{in} A$ is trying to push the spaghetto out, and on the other end, a force of $p_{out} A$ pushes it inwards. The forces generated by the pressures on the sides even out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It becomes more confusing when referring to cooked spaghetti since a) the cooked spaghetto won't stick straight out from your mouth and b) we intuitively don't want it to transmit forces because &quot;you can't push on a string&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, these complications do not change the underlying principle. Imagine cutting the spaghetto right outside your mouth while you keep sucking on the inner part. Clearly, the free spaghetto wouldn't feel any net force from air pressure after the cut and just float away/fall down. By cutting, you have changed the total picture by adding $p_{out} A$ orthogonal to the cut surface, but removed the push of $p_{in} A$, so before there must have been a net force on this surface before if there is none afterwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that it's not the qusestion, but in this case it is much simpler to look at the basic thermodynamics - the system is basically a microcanonical ensemble, with $E = p_{out} V_{out} + p_{in} V_{in}$ under the constraint that $V_{in} + V_{out} = \mathrm{const}$, and if $p_{in} &amp;lt; p_{out}$, the minimum energy (and thus equilibrium) state is clearly that with $V_{out}=0$.&lt;/p&gt;&#xA;" OwnerUserId="243" LastEditorUserId="243" LastEditorDisplayName="" LastEditDate="2010-11-15T13:00:01.097" LastActivityDate="2010-11-15T13:00:01.097" />
  <row Id="835" PostTypeId="2" ParentId="817" CreationDate="2010-11-15T14:01:54.627" Score="5" ViewCount="" Body="&lt;p&gt;Let's try and make things more precise, step-by-step.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There's no such thing as &quot;particle-wave duality&quot;: the name-of-the-game is &quot;Quantum Field Theory&quot;. This paradoxical notion of a possible &quot;duality&quot; only happens when you don't use the appropriate framework to describe your Physics. Therefore, it makes no sense to speculate on what would happen if spacetime were quantized/discrete: in this scenario, the question would be: &lt;em&gt;&quot;Would a quantized/discrete spacetime affect Quantum Field Theory?&quot;&lt;/em&gt; And the answer to this question is &quot;No.&quot; The reason being that different physical theories have &lt;strong&gt;different&lt;/strong&gt; domains of validity, given by the characteristic energy of the phenomena they describe.&lt;/li&gt;&#xA;&lt;li&gt;What &lt;em&gt;is&lt;/em&gt; the &quot;information theoretical aspect of the universe&quot;?! This is not even appropriately defined, let alone &quot;well defined&quot;.&lt;/li&gt;&#xA;&lt;li&gt;The black hole is the stereotypical object in a quantum gravity theory. So, when you quantize spacetime, you should look at black holes to see what happens. We already know that black holes have Entropy. So, the very first question should be: What does your particular quantization scheme yields for black hole Entropy? The current state-of-the-art, as far as i know, is that all different schemes of quantization of spacetime yield a reasonable answer to this question.&lt;/li&gt;&#xA;&lt;li&gt;This question, again, is not even appropriately defined, let alone &quot;well defined&quot;. Holography has a very precise and well defined meaning in Physics, which is not related to the hologram in a credit card, for example. So, holography does play a role in quantum gravity, the more famous statement being that of AdS/CFT. But, as it stands, your question does not have meaning.&lt;/li&gt;&#xA;&lt;li&gt;This has already been stablished a long time ago: if you quantize spacetime, the smallest unit of spacetime is given in terms of &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck_units&quot; rel=&quot;nofollow&quot;&gt;Natural units&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="142" LastActivityDate="2010-11-15T14:01:54.627" CommentCount="17" />
  <row Id="836" PostTypeId="2" ParentId="815" CreationDate="2010-11-15T14:27:20.967" Score="0" ViewCount="" Body="&lt;p&gt;There exists a notion of &lt;em&gt;locality&lt;/em&gt; in Physics, broadly given by a WKB-style approximation done to the Path Integral: if you do a steepest-descent approximation to the Path Integral, the critical points of the Action will give you the regions that mostly contribute to the Path Integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it's possible to generalize this notion to gauge theories… and the name of the game is &lt;a href=&quot;http://arxiv.org/abs/hep-th/9608068&quot; rel=&quot;nofollow&quot;&gt;equivariant localization&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to categorify this notion of localization à la &lt;a href=&quot;http://arxiv.org/abs/0806.1079&quot; rel=&quot;nofollow&quot;&gt;AQFT from n-functorial QFT&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure this answers your question, in the sense that i don't know how much these concepts are related to &quot;relational quantum mechanics&quot;. But, maybe this can help…&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2010-11-15T14:27:20.967" CommentCount="1" />
  <row Id="837" PostTypeId="2" ParentId="782" CreationDate="2010-11-15T14:31:19.793" Score="1" ViewCount="" Body="&lt;p&gt;This book may be what you're looking for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486651134&quot; rel=&quot;nofollow&quot;&gt;Introduction to Space Dynamics&lt;/a&gt; by William Tyrrell Thomson&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amazon also recommends:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486600610&quot; rel=&quot;nofollow&quot;&gt;Fundamentals of Astrodynamics&lt;/a&gt; by Roger R. Bate Paperback&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486646874&quot; rel=&quot;nofollow&quot;&gt;An Introduction to Celestial Mechanics&lt;/a&gt; by Forest Ray Moulton&lt;/p&gt;&#xA;" OwnerUserId="47" LastActivityDate="2010-11-15T14:31:19.793" />
  <row Id="838" PostTypeId="1" AcceptedAnswerId="939" CreationDate="2010-11-15T15:29:45.587" Score="3" ViewCount="286" Body="&lt;p&gt;Without any a priori knowledge of the mass, speed, distance, and size of local celestial bodies (aside from Earth's size), what can I calculate and how from my &quot;backyard&quot; through observation?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Edit:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;My goal is to create a high-school level physics project that students can conduct over the course of a year/term as both a physics problem mixed with a bit of astrophysics history.  Core concepts would include observation (within reason).  This isn't meant to be extremely precise work (within an order of magnitude would be excellent), but to demonstrate the methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a bunch of knowns, it's trivial to find out one unknown, but I am unsure of how to start from square one like the early physicists and astronomers did in times past.&lt;/p&gt;&#xA;" OwnerUserId="431" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-24T21:27:32.190" LastActivityDate="2011-03-24T21:27:32.190" Title="How would I calculate celestial body characteristics without a priori knowledge?" Tags="&lt;astrophysics&gt;&lt;astronomy&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="839" PostTypeId="2" ParentId="679" CreationDate="2010-11-15T15:47:13.783" Score="0" ViewCount="" Body="&lt;p&gt;Here are a few that don't appear to have been mentioned:&#xA;[EDIT: I will list one per answer, as the spam filter won't allow me to post more than one link at a time.]&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I second MIT's &lt;a href=&quot;http://ocw.mit.edu&quot; rel=&quot;nofollow&quot;&gt;OCW&lt;/a&gt; as the very best resource for self-learners.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="380" LastActivityDate="2010-11-15T15:47:13.783" />
  <row Id="840" PostTypeId="2" ParentId="679" CreationDate="2010-11-15T15:51:40.240" Score="1" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://nptel.iitm.ac.in/courses.php&quot; rel=&quot;nofollow&quot;&gt;NPTEL&lt;/a&gt; has a bunch of undergrad review level physics and math courses in the Basic Courses category.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="380" LastActivityDate="2010-11-15T15:51:40.240" />
  <row Id="841" PostTypeId="2" ParentId="679" CreationDate="2010-11-15T15:56:33.263" Score="1" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ictp.tv/diploma/&quot; rel=&quot;nofollow&quot;&gt;ICTP&lt;/a&gt; has a lot of high-undergrad and graduate-level courses, including condensed matter and HEP.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="380" LastActivityDate="2010-11-15T15:56:33.263" />
  <row Id="842" PostTypeId="2" ParentId="679" CreationDate="2010-11-15T16:00:15.460" Score="1" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.uccs.edu/~math/vidarchive.html&quot; rel=&quot;nofollow&quot;&gt;UCCS&lt;/a&gt; has a large collection of online math courses - some very useful.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Good luck! (And sorry the links had to be spread across multiple answers.)&lt;/p&gt;&#xA;" OwnerUserId="380" LastActivityDate="2010-11-15T16:00:15.460" />
  <row Id="843" PostTypeId="1" AcceptedAnswerId="845" CreationDate="2010-11-15T16:05:51.687" Score="2" ViewCount="272" Body="&lt;p&gt;New Scientist article: &lt;a href=&quot;http://www.newscientist.com/article/dn19723-summon-a-demon-to-turn-information-into-energy.html&quot; rel=&quot;nofollow&quot;&gt;Summon a 'demon' to turn information into energy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed of light c converts between space and time and also appears in e=mc^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maxwell's Demon can turn information supplied into a system into energy, which suggests there is a constant 'd' such that e=i*f(d) where i is information and f is a function of d.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If d appears in the formula relating energy and information like c relates energy and mass, but c is a conversion factor bewteen space and time, then what is d a conversion factor between?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What dimensions does d have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than &quot;Maxwell's demon constant&quot;, is there another name for d?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is there not a 'd' after all, instead 'c' appears once again in the information-energy equivalence fomula e=i*f(c)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is there not a constant here at all and the situation is more complicated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Living things have evolved to exploit all kinds of opportunities - even quantum effects in the form of photosynthesis. Living things have senses to extract information from the environment, but do living cells utilise any demon effects to turn any of this information into energy ?&lt;/p&gt;&#xA;" OwnerUserId="244" LastActivityDate="2010-11-16T10:07:28.943" Title="Maxwell's Demon Constant (information-energy equivalence)" Tags="&lt;energy&gt;" AnswerCount="2" />
  <row Id="844" PostTypeId="2" ParentId="815" CreationDate="2010-11-15T16:11:15.957" Score="2" ViewCount="" Body="&lt;p&gt;To respond to Marek's comment: as a reader familiar with physics and categories, I can say why I am unable to even approach a response.  The question cites some papers, but is not readable on its own (not &quot;self-contained&quot;).  We only learn is that there is some nonlocal theory out there which obeys some consistency conditions.  I am sure that this would seriously impair the possibility of help from Math Overflow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than trying to guess at the theories being asked about, then, maybe it's a good idea to say a word about Segal's axioms and S-matrix theory.  Picture a quantum field theory on (IN)xR, where IN is a spatial manifold.  Imagine that the incoming spacelike hypersurface of spacetime is a disjoint union of IN_i.  To each component IN_i we assign a Hilbert space H_i, so an incoming state is in a tensor product of the H_i.  The scattering of these states leads to an outgoing state in OUT_j, where OUT_j are the components of the outgoing spacelike hypersurface.  This map from IN states to OUT state is called the S-matrix.  &quot;Factorizability&quot; of the S-matrix says that when spacetime is made from &quot;gluing&quot; together two spacetimes along a common OUT/IN, the S-matrix map is the appropriate tensor product of maps.  (This does correspond to a locality assumption, roughly that the path integral has a continuous integrand called the action.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In two dimensions, where compact one-manifolds are very simple (always circles), this business simplifies greatly -- especially if your quantum field theory is independent of the metric on spactime -- and the &quot;pair-of-pants&quot; surface provides a multiplication map on the Hilbert space (two go in, one comes out),  the notion of an algebra.  Allowing four one-manifolds with boundary leads to the notion of a category (an algebra is a category with one object, for what it's worth).&lt;/p&gt;&#xA;" OwnerUserId="20" LastActivityDate="2010-11-15T16:11:15.957" />
  <row Id="845" PostTypeId="2" ParentId="843" CreationDate="2010-11-15T16:18:37.727" Score="3" ViewCount="" Body="&lt;p&gt;You're looking for this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Entropy_in_thermodynamics_and_information_theory#Information_is_physical&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Entropy_in_thermodynamics_and_information_theory#Information_is_physical&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, one bit of information allows an amount of work equal to $kT\ln2$. Where $k$ is Boltzmann's constant and $T$ is the thermodynamic temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen the article in Nature and I think it is horrible. They do a very bad service in explaining what's happening. The way they do it, it sounds like they invented a perpetuum mobile, which is not true. How was the information uncovered in the first place? That involved energy. And as thermodynamics will have it, more energy than the information provided you to perform work with the bead.&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2010-11-15T16:18:37.727" CommentCount="4" />
  <row Id="846" PostTypeId="2" ParentId="439" CreationDate="2010-11-15T16:53:14.610" Score="1" ViewCount="" Body="&lt;p&gt;If you read Kessler explanation at &lt;a href=&quot;http://webpages.charter.net/dkessler/files/KesSym.html&quot; rel=&quot;nofollow&quot;&gt;http://webpages.charter.net/dkessler/files/KesSym.html&lt;/a&gt; , he predicts that the dangerous region will be between 800 km and 1000 km. And that you do not deed a special event to start it. Just continue to launch satellites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the ISS is not so dangerous, because it is in low earth orbit, and the debris will quickly (a few years) fall down sue to atmospheric friction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not think that the geosynchronous orbit is a big problem, because it is only the equator of a big sphere (36000 km radius).&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2010-11-15T16:53:14.610" />
  <row Id="847" PostTypeId="1" AcceptedAnswerId="854" CreationDate="2010-11-15T17:02:26.190" Score="4" ViewCount="361" Body="&lt;p&gt;&lt;br&gt;&#xA;I am wondering if there is a device for measuring wavelengths of things. As semiconductors emit infrared radiation and I would like to know what is the wavelength it is so I can check to see if its going to be interfering with my project.&lt;br&gt;&#xA;Thanks in Advance,&lt;br&gt;&#xA;Dean&lt;/p&gt;&#xA;" OwnerUserId="433" LastActivityDate="2010-11-24T22:41:09.867" Title="Way to measure wavelength of infrared radiation?" Tags="&lt;electromagnetic-radiation&gt;" AnswerCount="3" />
  <row Id="848" PostTypeId="2" ParentId="847" CreationDate="2010-11-15T17:08:54.237" Score="0" ViewCount="" Body="&lt;p&gt;You are looking for a lambdameter. There are various ways to build one, but you can also buy one, e.g. &lt;a href=&quot;http://www.photonicsonline.com/product.mvc/Lambdameter-LRL-005-0001&quot; rel=&quot;nofollow&quot;&gt;http://www.photonicsonline.com/product.mvc/Lambdameter-LRL-005-0001&lt;/a&gt; . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are often spectrometers, but, if you are analysing light with a narrow spectrum, an interferometric system gives you the best precision. The idea is to build an interferometer with a moving mirror, and counting the fringes during the movement gives you the wavelength if you know the amplitude of the movement. Of course, the best way to know the amplitude of the movement is to count the fringes of a known radiation (usually given by HeNe laser)&lt;/p&gt;&#xA;" OwnerUserId="373" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-11-24T22:41:09.867" LastActivityDate="2010-11-24T22:41:09.867" />
  <row Id="849" PostTypeId="1" AcceptedAnswerId="857" CreationDate="2010-11-15T17:21:20.217" Score="7" ViewCount="122" Body="&lt;p&gt;The title says it. An explanation for only 2 qubits would already be interesting, since I already have difficulties to find the 5 MUBs in this simple case.&lt;/p&gt;&#xA;" OwnerUserId="373" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-11-17T17:51:19.427" LastActivityDate="2010-11-17T17:51:19.427" Title="Is there a simple way to express the 2ⁿ+1 mutually unbiased bases for n qubits?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-information&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="850" PostTypeId="1" AcceptedAnswerId="852" CreationDate="2010-11-15T17:27:05.563" Score="2" ViewCount="357" Body="&lt;p&gt;What exactly is a field theory? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we classify theories as field theories and non field theories? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;After reading the answers I am under the impression that almost every theory is a field theory. Is this correct?&lt;/li&gt;&#xA;&lt;li&gt;What is difference between QM and QFT?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-11-15T19:02:57.863" LastActivityDate="2010-11-15T19:55:52.840" Title="What does a Field Theory mean?" Tags="&lt;theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="851" PostTypeId="2" ParentId="850" CreationDate="2010-11-15T17:57:46.777" Score="9" ViewCount="" Body="&lt;p&gt;A field theory is a physical description of reality in which the fundamental entities are fields, i.e. objects having no definite spatial location but a certain value or intensity at each place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of fields are the temperature in a room, for each location in the room, a temperature can be specified, although in most cases temperature will be pretty uniform, unless for instance if you just turned on a heater, then there will be a temperature gradient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational field in Newtonian mechanics is a description of what the force of attraction on a test particle is as generated by a large mass. This field is vector-valued.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example of a vector-valued field is the velocity field in a fluid. It gives the velocity of each infinitesimal piece of fluid at some instant t. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electromagnetic field is specified by giving the value of an antisymmetric rank-2 tensor at each space-time location.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The novelty of quantum mechanics with respect to classical mechanics is that it has to incorporate the discreteness of the action. That's what we call quantizing. In particular, this means that energy will be quantized under certain circumstances (namely due to some boundary conditions or potentials limiting the amount of possible states). In classical mechanics, systems typically have a finite amount of degrees of freedom. For instance, the 1D harmonic oscillator has two degrees of freedom, the position and the momentum of the oscillator. In quantum mechanics, the energy of the oscillator (which is a combination of position and momentum) becomes quantized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum field theory takes this one step further, instead of quantizing systems with a finite amount of degrees of freedom, it tackles systems with an infinite amount, in other words fields. The way non-interacting fields are quantized is reminiscent of the way the harmonic oscillator is quantized except you now have an infinite amount of oscillators. This brings with it a lot of technical complications.  &lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2010-11-15T19:55:52.840" LastActivityDate="2010-11-15T19:55:52.840" CommentCount="1" />
  <row Id="852" PostTypeId="2" ParentId="850" CreationDate="2010-11-15T18:10:04.483" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt; to address new questions.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The answer to this question is no. At least if you take the question purely formally. Only theories such as classical field theory, quantum field theory and continuum mechanics are field theories (you generally recognize them by having continuous degrees of freedom; also they usually have the word &lt;em&gt;field&lt;/em&gt; in the title :-)). But physically, lots of different theories may be equivalent, or may be approximations of some other theory, so there are many connections among them (this is the point I was trying to illustrate, but maybe I overemphasized it).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Difference between QM and QFT is essentially the same as between classical mechanics and classical field theory. In the mechanics you have just a few particles (or more generally, small number of degrees of freedom), while fields have an infinite number of degrees of freedom. Naturally, field theories are a lot harder than the corresponding mechanics. But there is a connection I already mentioned: you can see what happens when you let the number of particles grow arbitrarily large. This system will then essentially behave as a field theory. So in a sense, we can say that field theory is a large $N$ (number of degrees of freedom) limit of the corresponding mechanical theory. Of course, this view is very simplified, but I don't want to get too technical here.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Field theory is a theory that studies fields. Now what is a field? I suppose everyone should be familiar with at least some of them, e.g. gravitational or electromagnetic (EM) field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, how do you recognize that object is a field? Well, essentially, you look at how complicated the object is. To make this more precise: main objects of study of classical mechanics are point particles. All you need to keep track of them is just few parameters (position, velocity). On the other hand, consider the EM field: you need to keep track of the data (electric and magnetic field vector) in every point of the universe, so there is infinitely many parameters of this system! This is what I meant by system being large: you need a lot of data to describe it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it might seem that something is amiss. You &lt;strong&gt;do&lt;/strong&gt; need a lot of data to describe real objects (just think of how many atoms there are in the grain of sand). So are ordinary objects fields? Yes and no, both answers are correct depending on your point of view. If you consider a massive object as essentially being described by few parameters (like center of mass velocity and moment of inertia) and completely ignore all information about atoms then it's clearly not a field. Nevertheless, at the microscopic level, atoms wiggle around and even the grain of sand really is as complicated object as any EM field (not to mention that atoms themselves produce EM field), so it's certainly correct to call them that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let us see where our definition of field takes us. Let's talk about quantum mechanics for a while. What about two quantum particles? Is it a field? Well, clearly not. What about three? Still not. And what if we keep adding particles so that there will be a huge number of them? Well, it turns out that we'll get a quantum field! This is precisely the correspondence between e.g. photons and quantum EM field. You can either look at EM field as being described by vector of electric and magnetic field at every point as in the classical case, or you can instead reorganize your data so that you keep track of what kind of photons you have. It's useful to carry both pictures in head and use the more appropriate one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a subject of continuum mechanics. There you can also start with particles (describing atoms in some real object, e.g. water) and because there are so many of them, you can again reorganize your data, consider the object as being essentially continuous (which real objects surely are at least unless you look at them with a microscope), and instead describe them by parameters such as pressure and temperature at every point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize: the field theory is essentially about dealing with large objects. However, when we are looking at the problem with particle hat on, we usually don't say it's a field. For instance, when describing real objects as consisting of atoms, we are usually talking about statistical mechanics, or condensed matter physics. Only when we move to the realm of continuum mechanics, we say that there are fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is much more to be said on the topic but this post got already too long so I'll stop here. If you have any questions, ask away!&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-15T19:23:41.420" LastActivityDate="2010-11-15T19:23:41.420" CommentCount="3" />
  <row Id="854" PostTypeId="2" ParentId="847" CreationDate="2010-11-15T18:35:11.587" Score="5" ViewCount="" Body="&lt;p&gt;You probably want to know the amount of light of each different wavelength (a.k.a. light spectrum) the semiconductor is emitting. The device you use for this is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectrometer&quot; rel=&quot;nofollow&quot;&gt;spectrometer&lt;/a&gt; or &lt;a href=&quot;http://www.us.anritsu.com/products/Test-and-Measurement-Solutions/Optical-Solutions/Optical-Spectrum-Analyzers_ARPC2022ARPC.aspx&quot; rel=&quot;nofollow&quot;&gt;optical spectral analyzer&lt;/a&gt;. What semiconductors are you using? Unless you are using an uncommon semiconductor, you can probably find the emission spectrum online.&lt;/p&gt;&#xA;" OwnerUserId="62" LastActivityDate="2010-11-15T18:35:11.587" />
  <row Id="855" PostTypeId="2" ParentId="849" CreationDate="2010-11-15T18:35:19.900" Score="1" ViewCount="" Body="&lt;p&gt;I don't know anything about the topic but just by searching the internet I found &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.11.2635&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; which claims to prove that $2^n+1$ MUBs always exist for $n$ qubits and moreover it says in the abstract that the proof is constructive. Check it out.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-11-15T18:35:19.900" />
  <row Id="856" PostTypeId="2" ParentId="847" CreationDate="2010-11-15T18:57:25.357" Score="3" ViewCount="" Body="&lt;p&gt;Do you want to know the distribution over a broad range of wavelengths, or do you have a single semiconductor device emitting at a single wavelength that you need to measure? And do you know the approximate wavelength? &quot;Infrared&quot; covers a pretty broad range, from wavelengths that are not-quite-visible to wavelengths that are practically microwaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a single source, the best way to measure the wavelength is using interference. If you've got an easy way to see the light, you can just sent it through a diffraction grating of known spacing, and measure the spacing between the bright spots. Or, if you want to be more precise, you can use a Michelson interferometer (which is what's in a lot of commercial wavemeters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're worried about thermal emission over a broad range of infrared wavelengths, then you need a spectrometer of some sort. A grating monochromator world be the simplest thing, and probably work over the largest range of wavelengths. If you care about near-infrared (shorter than about 2 microns), there's &lt;a href=&quot;http://www.oceanoptics.com/Products/nirquest.asp&quot; rel=&quot;nofollow&quot;&gt;this commercial unit&lt;/a&gt;, which I haven't used, but the other spectrometers we have from them are pretty good.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-15T18:57:25.357" />
  <row Id="857" PostTypeId="2" ParentId="849" CreationDate="2010-11-15T19:10:25.033" Score="7" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Sorry I misread 2&lt;sup&gt;n&lt;/sup&gt;+1 as the number of dimensions.  It becomes more systematic then. The case N=4 can be found in &lt;a href=&quot;http://en.wikipedia.org/wiki/Mutually_unbiased_bases#Example_for_d_.3D_4&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mutually_unbiased_bases#Example_for_d_.3D_4&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general construction can be found in [4]. Let's consider the case n=2. Firstly, use the Galois ring $GR(4,n=2) = \mathbb Z_4[x] / \langle h(x)\rangle$ where $h(x)=x^2+x+1$ is an irreducible polynomial of the field (i.e. its root cannot be represented in $\mathbb Z_4$. Let $\xi$ be that root. We define the Teichmüller set $T_n = \{0,1,\xi,\xi^2,\dotsc\}$, where&#xA;$$\begin{aligned}&#xD;&#xA;1 &amp;amp;= 1 \\&#xD;&#xA;\xi &amp;amp;= \xi \\&#xD;&#xA;\xi^2 &amp;amp;= -\xi-1 = 3\xi+3 \\&#xD;&#xA;\xi^3 &amp;amp;= 3\xi^2 + 3\xi = -3 = 1 \text{  (cycle formed)}&#xD;&#xA;\end{aligned}$$&#xD;&#xA;so $T_2 = \{0,1,\xi,3\xi+3\}$. Now, the basis are given by&#xA;$$ v_{a,b} = \frac1{\sqrt{2^n}} \exp\left(\frac{2\pi i}4 \mathrm{tr}((a+2b)x)\right)_{x\in T_n} $$&#xD;&#xA;where $a,b\in T_n$, the trace is defined as $\mathrm{tr}(x)=\sum_{k=0}^{n-1}\sigma^k(x)$ where $\sigma(a+2b)=a^2+2b^2$. In our case, $\mathrm{tr}(a+2b)=a+2b+a^2+2b^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance,&#xA;$$\begin{aligned}&#xD;&#xA;v_{1,\xi} &amp;amp;= \frac12 \exp \left(\frac{2\pi i}4 \mathrm{tr} ( (1+2\xi) \{ 0, 1, \xi, 3\xi+3 \} ) \right) \\&#xD;&#xA;&amp;amp;= \frac12 \exp \left(\frac{2\pi i}4 \mathrm{tr} ( \{0, 1+2\xi, \xi+2\xi^2, 9\xi+3+6\xi^2 \} )  \right) \\&#xD;&#xA;&amp;amp;= \frac12 \exp \left(\frac{2\pi i}4 \mathrm{tr} ( \{0, 1+2\xi, 3\xi+2, 3\xi+1 \} )  \right) \\&#xD;&#xA;&amp;amp;= \frac12 \exp \left(\frac{2\pi i}4 \mathrm{tr} ( \{0, 1+2\xi, \xi+2(3\xi+3), 3\xi+3+2 \} )  \right) \\&#xD;&#xA;&amp;amp;= \frac12 \exp \left(\frac{2\pi i}4 \{0, 1+2\xi+1+2\xi^2, \xi+2(3\xi+3)+\xi^2+2(3\xi+3)^2, (3\xi+3)+2+(3\xi+3)^2+2 \} \right) \\&#xD;&#xA;&amp;amp;= \frac12 \exp \left(\frac{2\pi i}4 \{0, 0, 1, 3 \} \right) \\&#xD;&#xA;&amp;amp;= \frac12 \{1, 1, i, -i \}&#xD;&#xA;\end{aligned}$$&#xD;&#xA;You could then compute the other 15 combinations to get the MUBs (after including the standard basis). (Please ask on math.SE or MathOverflow for the details of Galois ring.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The N=5 case is kept below for those who are interested.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As far as I can check, the MUBs for N not a power of prime is in general unknown. In fact, even for N = 6 only 3 MUBs are found, and it's still a conjecture mathematically if there the 4th set doesn't exist[1,2].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since not all 2&lt;sup&gt;n&lt;/sup&gt;+1 are primes, I doubt if it is possible to analytically find MUBs for all n. But for N=5, we could use the &quot;Unitary operators method using Galois fields&quot; as explained in Wikipedia[1,3].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ X = \sum_0^4 |k+1\rangle\langle k| = \begin{pmatrix}&#xD;&#xA;0&amp;amp;1&amp;amp;0&amp;amp;0&amp;amp;0\\&#xD;&#xA;0&amp;amp;0&amp;amp;1&amp;amp;0&amp;amp;0\\&#xD;&#xA;0&amp;amp;0&amp;amp;0&amp;amp;1&amp;amp;0\\&#xD;&#xA;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;1\\&#xD;&#xA;1&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0 \end{pmatrix} $$&#xD;&#xA;$$ Z = \sum_0^4 \omega^k|k\rangle\langle k| = \begin{pmatrix}&#xD;&#xA;1&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;0\\&#xD;&#xA;0&amp;amp;\omega&amp;amp;0&amp;amp;0&amp;amp;0\\&#xD;&#xA;0&amp;amp;0&amp;amp;\omega^2&amp;amp;0&amp;amp;0\\&#xD;&#xA;0&amp;amp;0&amp;amp;0&amp;amp;\omega^3&amp;amp;0\\&#xD;&#xA;0&amp;amp;0&amp;amp;0&amp;amp;0&amp;amp;\omega^4 \end{pmatrix} $$&#xD;&#xA;where $\omega = e^{2\pi i/5}$. Then we compute the eigenvectors of the 6 matrices&#xA;$$X, Z, XZ, XZ^2, XZ^3, XZ^4$$&#xD;&#xA;to get the MUBs (each row represents a base):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{aligned} &#xD;&#xA;Z &amp;amp;\to \begin{bmatrix} 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA; 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 \\&#xD;&#xA; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1\end{bmatrix} \\&#xD;&#xA;X &amp;amp;\to \frac1{\sqrt5}\begin{bmatrix} &#xD;&#xA; \omega ^2 &amp;amp; \omega ^3 &amp;amp; \omega ^4 &amp;amp; 1 &amp;amp; \omega  \\&#xD;&#xA; \omega ^3 &amp;amp; \omega ^2 &amp;amp; \omega  &amp;amp; 1 &amp;amp; \omega ^4 \\&#xD;&#xA; -\omega ^2 &amp;amp; -\omega ^4 &amp;amp; -\omega  &amp;amp; -\omega ^3 &amp;amp; -1 \\&#xD;&#xA; -\omega ^3 &amp;amp; -\omega  &amp;amp; -\omega ^4 &amp;amp; -\omega ^2 &amp;amp; -1 \\&#xD;&#xA; -1 &amp;amp; -1 &amp;amp; -1 &amp;amp; -1 &amp;amp; -1&#xD;&#xA;\end{bmatrix}\\&#xD;&#xA;XZ &amp;amp;\to \frac1{\sqrt5}\begin{bmatrix}&#xD;&#xA; \omega ^4 &amp;amp; 1 &amp;amp; 1 &amp;amp; \omega ^4 &amp;amp; \omega ^2 \\&#xD;&#xA; 1 &amp;amp; 1 &amp;amp; \omega ^4 &amp;amp; \omega ^2 &amp;amp; \omega ^4 \\&#xD;&#xA; \omega ^2 &amp;amp; \omega ^4 &amp;amp; 1 &amp;amp; 1 &amp;amp; \omega ^4 \\&#xD;&#xA; 1 &amp;amp; \omega ^3 &amp;amp; 1 &amp;amp; \omega  &amp;amp; \omega  \\&#xD;&#xA; \omega  &amp;amp; 1 &amp;amp; \omega ^3 &amp;amp; 1 &amp;amp; \omega &#xD;&#xA;\end{bmatrix} \\&#xD;&#xA;XZ^2 &amp;amp;\to \frac1{\sqrt5}\begin{bmatrix} \omega ^4 &amp;amp; \omega  &amp;amp; \omega  &amp;amp; \omega ^4 &amp;amp; 1 \\&#xD;&#xA; 1 &amp;amp; \omega ^3 &amp;amp; \omega ^4 &amp;amp; \omega ^3 &amp;amp; 1 \\&#xD;&#xA; \omega ^4 &amp;amp; 1 &amp;amp; \omega ^4 &amp;amp; \omega  &amp;amp; \omega  \\&#xD;&#xA; 1 &amp;amp; 1 &amp;amp; \omega ^3 &amp;amp; \omega ^4 &amp;amp; \omega ^3 \\&#xD;&#xA; \omega ^3 &amp;amp; 1 &amp;amp; 1 &amp;amp; \omega ^3 &amp;amp; \omega ^4 \\&#xD;&#xA; \omega  &amp;amp; 1 &amp;amp; \omega ^2 &amp;amp; \omega ^2 &amp;amp; 1&#xD;&#xA;\end{bmatrix}\\&#xD;&#xA;XZ^3 &amp;amp;\to \frac1{\sqrt5}\begin{bmatrix}&#xD;&#xA; 1 &amp;amp; \omega ^2 &amp;amp; \omega  &amp;amp; \omega ^2 &amp;amp; 1 \\&#xD;&#xA; \omega  &amp;amp; 1 &amp;amp; \omega  &amp;amp; \omega ^4 &amp;amp; \omega ^4 \\&#xD;&#xA; 1 &amp;amp; 1 &amp;amp; \omega ^2 &amp;amp; \omega  &amp;amp; \omega ^2 \\&#xD;&#xA; \omega ^2 &amp;amp; 1 &amp;amp; 1 &amp;amp; \omega ^2 &amp;amp; \omega  \\&#xD;&#xA; \omega ^4 &amp;amp; 1 &amp;amp; \omega ^3 &amp;amp; \omega ^3 &amp;amp; 1&#xD;&#xA;\end{bmatrix}\\&#xD;&#xA;XZ^4 &amp;amp;\to \frac1{\sqrt5}\begin{bmatrix}&#xD;&#xA;  \omega  &amp;amp; 1 &amp;amp; 1 &amp;amp; \omega  &amp;amp; \omega ^3 \\&#xD;&#xA; 1 &amp;amp; 1 &amp;amp; \omega  &amp;amp; \omega ^3 &amp;amp; \omega  \\&#xD;&#xA; \omega ^3 &amp;amp; \omega  &amp;amp; 1 &amp;amp; 1 &amp;amp; \omega  \\&#xD;&#xA; 1 &amp;amp; \omega ^2 &amp;amp; 1 &amp;amp; \omega ^4 &amp;amp; \omega ^4 \\&#xD;&#xA; \omega ^4 &amp;amp; 1 &amp;amp; \omega ^2 &amp;amp; 1 &amp;amp; \omega ^4&#xD;&#xA;\end{bmatrix}&#xD;&#xA;\end{aligned}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ref:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;[1]: &lt;a href=&quot;http://en.wikipedia.org/wiki/Mutually_unbiased_bases&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mutually_unbiased_bases&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;[2]: Stephen Brierley, Stefan Weigert. &lt;em&gt;Constructing Mutually Unbiased Bases in Dimension Six&lt;/em&gt;. &lt;a href=&quot;http://arxiv.org/abs/0901.4051&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0901.4051&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;[3]: Somshubhro Bandyopadhyay, P. Oscar Boykin, Vwani Roychowdhury, Farrokh Vatan. &lt;em&gt;A new proof for the existence of mutually unbiased bases&lt;/em&gt;. &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0103162&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/quant-ph/0103162&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;[4] Andreas Klappenecker, Martin Rötteler, &lt;em&gt;Constructions of Mutually Unbiased Bases&lt;/em&gt;. &lt;a href=&quot;http://xxx.lanl.gov/abs/quant-ph/0309120&quot; rel=&quot;nofollow&quot;&gt;http://xxx.lanl.gov/abs/quant-ph/0309120&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-11-15T23:11:40.613" LastActivityDate="2010-11-15T23:11:40.613" CommentCount="2" />
  <row Id="859" PostTypeId="2" ParentId="217" CreationDate="2010-11-15T19:54:03.150" Score="5" ViewCount="" Body="&lt;p&gt;I think he is envisaging the spaghetti as a long thin cylinder. Air pressure can only exert a force normal to the surface, so the &quot;push&quot; has to come from the end of the noodle, which &#xA;would seem to be too far away to transmit stress through such a squishy medium. Mechanical engineers sometimes decompose stress into pressure (a component uniform if all directions), and deviatoric, whatever is left to fill out the stress tensor. The later deviatoric stress would indeed not transmit well&#xA;through the spaghetti. But as long as the medium is incompressible (like water) the pressure&#xA;will be easily transmitted. This would also apply to compressible spaghetti, but the volume must change with a change of pressure.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2010-11-15T19:54:03.150" />
  <row Id="860" PostTypeId="2" ParentId="629" CreationDate="2010-11-15T20:15:38.790" Score="3" ViewCount="" Body="&lt;p&gt;I believe this has to do with Jerk. Jerk is the derivative of acceleration. The breaks of the vehicle will provide a nearly constant force against the motion of the car. Since F=MA, as your vehicle slows at a constant rate of negative acceleration. Once the car stops, the force (and consequently your acceleration) go to zero very quickly. This results in a high amount of Jerk, and this is what you're feeling.&lt;/p&gt;&#xA;" OwnerUserId="435" LastActivityDate="2010-11-15T20:15:38.790" />
  <row Id="861" PostTypeId="2" ParentId="789" CreationDate="2010-11-15T21:19:06.577" Score="1" ViewCount="" Body="&lt;p&gt;The &quot;main solar reaction&quot; you are talking about,  $p^+ p^+ \to d^+ e^+ \nu_e $ is the first step of the &lt;strong&gt;pp-chain&lt;/strong&gt; allowing the fusion of 4 protons to form Helium. This &quot;pp-chain&quot; has different &quot;branches&quot; (possible reactions) but all of them start with the process you mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process is exoenergetic and releases 0.420 MeV even if deuton has a very low binding energy the reaction is possible because of the small difference in mass between the proton and the neutron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a process involving the weak interaction: one of the proton &quot;decay&quot; ($\beta^+$ process) into a neutron, a positron and a neutrino.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now about your second equation with an intermediate state where an electron-positron pair appear: you are making a mistake because you take the difference between the two sides, but you artificially introduced this electron-positron pair: this has to be taken into account:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You wrote&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p^+ + p^+ \rightarrow p + p + e^- + e^+$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but that's wrong, it should be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p + p + \textrm{“some kinetic energy allowing the production of the pair = 0.511 MeV”} \rightarrow p + p + e^- + e^+$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and now, it should be right.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="1656" LastEditorDisplayName="" LastEditDate="2011-02-21T00:06:34.887" LastActivityDate="2011-02-21T00:06:34.887" CommentCount="1" />
  <row Id="862" PostTypeId="1" CreationDate="2010-11-15T22:00:09.287" Score="3" ViewCount="390" Body="&lt;p&gt;I am asking this question completely out of curiosity. The other day, my roommate, by mistake, used 'Light year' as a unit of time instead of distance. When I corrected him (pedantic, much), he said the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Units are relative. And according to Fourier Transforms, units can be changed so Light year is a unit of time.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That got me thinking and I read up Fourier Transforms on wikipedia but couldn't find anything about using a unit in one domain as a unit for another measurement. I do agree that units (particularly, base units are relative. eg: the meter), but does his statement make any sense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT&#xA;Thank you everyone for all the answers. It isn't so much to in it in or prove a point as it is to understand the concept better. Anyways this is his response after I showed him this thread. Any comments would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His response:&#xA;Nevermind, for the first time I accept I was wrong. BUT using lightyears to measure time is possible. My example didn't make sense bacause I was wrong when I meantioned that I'm still measuring dist. If you have a signal in time domain and ...take the FT, I get a signal which DOES NOT HAVE to be in frequency domain. Clarify this to the guy who posted last. Now the new signal is in a domain defined by me and so is its units. This signal although not equal to the original signal, still represents that if ya take an inverse FT. So, the idea of time will still be there. Now coming back to our case: lightyears here is not the lightyears you are used to read when dealing with distance. It represents time.&lt;/p&gt;&#xA;" OwnerUserId="436" LastEditorUserId="436" LastEditorDisplayName="" LastEditDate="2010-11-16T04:35:04.560" LastActivityDate="2010-11-16T14:21:07.200" Title="Does this statement make any sense?" Tags="&lt;si-units&gt;&lt;fourier-transform&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="863" PostTypeId="2" ParentId="862" CreationDate="2010-11-15T22:06:44.510" Score="4" ViewCount="" Body="&lt;p&gt;No, the definition of light year is &quot;the distance that light travels in a year in vacuum&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.google.co.uk/search?sourceid=chrome&amp;amp;ie=UTF-8&amp;amp;q=define+light+year&quot; rel=&quot;nofollow&quot;&gt;http://www.google.co.uk/search?sourceid=chrome&amp;amp;ie=UTF-8&amp;amp;q=define+light+year&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short of redefining English, the statement is incorrect!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By changing the Units, you would simply change the &lt;em&gt;numeric value&lt;/em&gt; of a l.y., but not its &lt;em&gt;definition&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-15T22:06:44.510" />
  <row Id="864" PostTypeId="2" ParentId="862" CreationDate="2010-11-15T22:09:22.153" Score="10" ViewCount="" Body="&lt;p&gt;This doesn't make much sense: light year is in any case a unit of distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is common is to use &quot;reduced units&quot;, for examples units where $c=1$ (speed of light) or $h=2\pi$. But in these cases the opposite would happen: you would say &quot;year&quot; to mean a distance. Or for example you say &quot;has a mass of xyz MeV&quot; instead of &quot;$MeV/c^2$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;About the Fourier transform: this allow to go from the so-called &quot;time domain&quot; (even if &quot;time&quot; is not always the usual time) to the &quot;frequency domain&quot; involving ... frequencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as you can see this cannot change the definition of light-year.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-11-15T22:09:22.153" CommentCount="3" />
  <row Id="865" PostTypeId="2" ParentId="862" CreationDate="2010-11-15T22:09:57.723" Score="2" ViewCount="" Body="&lt;p&gt;He's BS-ing you to the extent that an explanation of the fourier transform is not necessary to point this out (&quot;units are relative&quot; - what like length and time are the same thing? then all unit definitions would be circular as length is defined in terms of time at present. is mass also length? can you give my height, weight, and age in one number and define each as the same number in different units? =) ). I'll be specific though, anyway. The fourier transform of a scalar number is simply not defined (no such thing as a fourier transform of a distance, for instance). Typically fourier transforms give you a function of frequency for every function of time or of space. Under certain conditions, where there is a relationship between spatial frequency and temporal frequency (like waves propagating in a nondispersive medium here wavelength*frequency = speed of waves) you can then get a function of time for any function of space using a fourier transform, indirectly. But this is a special case, involves functions, not numbers, and this fourier transform of the original function is NOT THE SAME as the original function, so it's irrelevant to your original argument. So your friend is screwing with you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a good one &lt;/p&gt;&#xA;" OwnerUserId="437" LastActivityDate="2010-11-15T22:09:57.723" />
  <row Id="866" PostTypeId="1" AcceptedAnswerId="868" CreationDate="2010-11-15T22:16:07.667" Score="2" ViewCount="203" Body="&lt;p&gt;I beg your pardon in advance if this question is &lt;em&gt;naive&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Quantum Mechanics, discrete values of measurements occur &lt;del&gt;only&lt;/del&gt; in relation to &lt;em&gt;bound&lt;/em&gt; states. This is because of the general solution for the time-independent Shroedinger's equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, in order to quantise space-time, do we need to make space-time compact? How would a &lt;em&gt;compact&lt;/em&gt; space-time look like? Would we simply need close it by including $\infty$? Is that physical (infinity doesn't sound very physical to me)?&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-15T22:44:49.313" LastActivityDate="2010-11-15T22:44:49.313" Title="Possibly naive question about quantized space-time" Tags="&lt;cosmology&gt;&lt;quantum-gravity&gt;&lt;quantization&gt;" AnswerCount="1" />
  <row Id="867" PostTypeId="1" AcceptedAnswerId="872" CreationDate="2010-11-15T22:17:58.353" Score="-1" ViewCount="439" Body="&lt;p&gt;In which real case scenarios a convolution or deconvolution operation is useful ?&lt;/p&gt;&#xA;" OwnerUserId="217" LastActivityDate="2010-11-15T23:50:31.903" Title="Meaning and application of convolution or deconvolution in physical sciences" Tags="&lt;mathematics&gt;" AnswerCount="4" CommentCount="10" />
  <row Id="868" PostTypeId="2" ParentId="866" CreationDate="2010-11-15T22:40:59.650" Score="5" ViewCount="" Body="&lt;p&gt;Discrete eigenvalues for measurable observables in QM occur more broadly than just for bound states. For example angular momentum, whether orbital or spin, is always restricted to discrete eigenvalues for all systems, free and unbound as well as bound systems. It is the case, however, that discrete spectra is 'associated' with compact geometries. Thus the angular momentum operators are the generators of rotations and the group of rotations is a compact group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this stage of our understanding, one can not make a final statement about possible discrete quantization of space-time since we have no established/corroborated theory that entails such quantization. However, recent research in the Loop Quantum Gravity research program looking for a quantum theory of gravity, has led to the result that if the LQG approach holds, then the boundary surface areas and the internal volumes of compact space-like regions of space-time must have discrete spectra.&lt;/p&gt;&#xA;" OwnerUserId="345" LastActivityDate="2010-11-15T22:40:59.650" />
  <row Id="870" PostTypeId="2" ParentId="867" CreationDate="2010-11-15T23:05:21.063" Score="0" ViewCount="" Body="&lt;p&gt;Solving the Fresnel diffraction integral, to find the amplitude and phase of waves near a source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Fresnel_diffraction#Convolution&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fresnel_diffraction#Convolution&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2010-11-15T23:05:21.063" />
  <row Id="871" PostTypeId="2" ParentId="867" CreationDate="2010-11-15T23:08:31.867" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Convolution#Applications&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Convolution#Applications&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="184" LastActivityDate="2010-11-15T23:08:31.867" CommentCount="3" />
  <row Id="872" PostTypeId="2" ParentId="867" CreationDate="2010-11-15T23:12:45.383" Score="1" ViewCount="" Body="&lt;p&gt;Any real instrument will have some &lt;a href=&quot;http://en.wikipedia.org/wiki/Impulse_response&quot; rel=&quot;nofollow&quot;&gt;impulse response&lt;/a&gt;.  The measured signal is the convolution of the source signal with this impulse response.  For example, if you aim a telescope at a point source, you will see not a point source but the point source convolved with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Point_spread_function&quot; rel=&quot;nofollow&quot;&gt;point spread function&lt;/a&gt; (2D impulse response) of the telescope.  Some kind of (usually approximate) &lt;a href=&quot;http://en.wikipedia.org/wiki/Deconvolution&quot; rel=&quot;nofollow&quot;&gt;deconvolution&lt;/a&gt; is applied to correct this and better estimate the source signal.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2010-11-15T23:12:45.383" />
  <row Id="873" PostTypeId="2" ParentId="867" CreationDate="2010-11-15T23:50:31.903" Score="1" ViewCount="" Body="&lt;p&gt;This is a very general topic to discuss, (I'm not sure there's a right answer here), but I think it's very much worth pointing out the convolution theorem. It's not a &quot;practical&quot; application per se, but it's a very useful method, since Fourier analysis enters physics in so many areas. Anywhere where you encounter &lt;em&gt;spectrum analysis&lt;/em&gt;, convolutions will quite possibly come into play somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Convolution_theorem&quot; rel=&quot;nofollow&quot;&gt;convolution theorem&lt;/a&gt; states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{F}\{f*g\} = \mathcal{F}\{f\} \cdot \mathcal{F}\{g\}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{F}\{f \cdot g\}= \mathcal{F}\{f\}*\mathcal{F}\{g\}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathcal{F}$ represents the Fourier transform operator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are remarkably useful identities when performing Fourier transforms. Since Fourier transforms can in fact be used for solving some differential equations, they have a notable application there too.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2010-11-15T23:50:31.903" />
  <row Id="874" PostTypeId="1" CreationDate="2010-11-15T23:59:17.513" Score="0" ViewCount="294" Body="&lt;p&gt;Here is the description of experiment which uses Mach–Zehnder interferometer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mach-Zehnder&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mach-Zehnder&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://departments.colgate.edu/physics/research/Photon/root/ajpbs02.pdf&quot; rel=&quot;nofollow&quot;&gt;http://departments.colgate.edu/physics/research/Photon/root/ajpbs02.pdf&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(see section &quot;C. The experiments: The photon interferes with itself&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I don't think I can fully understand the nature of what's happening in this experiment&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could possibly someone give an explanation preferably in simple terms or point to another source?&lt;/p&gt;&#xA;" OwnerUserId="438" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-16T04:13:28.143" LastActivityDate="2010-11-16T04:13:28.143" Title="Explanation of the experiment with Mach–Zehnder interferometer (self-interference of a single photon)" Tags="&lt;quantum-mechanics&gt;&lt;quantum-optics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="875" PostTypeId="2" ParentId="838" CreationDate="2010-11-16T00:15:32.713" Score="0" ViewCount="" Body="&lt;p&gt;There's a surprising amount you can get from very basic observations; basically just looking at how bright certain stars are over time, that is, measuring their &lt;a href=&quot;http://en.wikipedia.org/wiki/Light_curve&quot; rel=&quot;nofollow&quot;&gt;light curves&lt;/a&gt;.  I would look into the literature of &lt;a href=&quot;http://en.wikipedia.org/wiki/Variable_star&quot; rel=&quot;nofollow&quot;&gt;variable stars&lt;/a&gt;.  In particular, (if you're in the United States) the &lt;a href=&quot;http://www.aavso.org/&quot; rel=&quot;nofollow&quot;&gt;AAVSO&lt;/a&gt; is an organization of amateur observers that may have resources on their website for people interested in these things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me sketch one prototypical example.  There is a famous class of stars called &lt;a href=&quot;http://en.wikipedia.org/wiki/Cepheid_variable&quot; rel=&quot;nofollow&quot;&gt;Cepheid variables&lt;/a&gt; with a well-known relationship between the frequency of oscillation and their absolute magnitude.  By comparing the absolute magnitude to the apparent magnitude, you can actually determine the distance to these stars!  People have used these sorts of observations to measure the distances of nearby galaxies, and is an important step in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_distance_ladder&quot; rel=&quot;nofollow&quot;&gt;cosmic distance ladder&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, many of these relationships between period and brightness are purely empirical -- I don't know that we currently understand stellar astrophysics well enough to derive these relationships from scratch.  But it's been a while since I've read about these things.  If you're more interested in this side of things, I'd look for books on stellar astrophysics; though the background you'll need is fairly broad (nuclear physics, statistical mechanics and thermodynamics, knowledge of radiative processes)...&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-16T00:15:32.713" CommentCount="3" />
  <row Id="876" PostTypeId="2" ParentId="234" CreationDate="2010-11-16T02:34:14.973" Score="3" ViewCount="" Body="&lt;p&gt;I feel very strongly about this question.  I believe that for an experimentalist, it's fine to not go very deeply into advanced mathematics whatsoever.  Mostly experimentalists need to understand one particular experiment at a time extremely well, and there are so many skills an experimentalist needs to focus all of their time/energy on developing as students.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe experimentalists should derive their physical intuition from lots of time spent in the lab, whereas theoreticians should develop their physical intuition from a sense of &quot;mathematical beauty&quot; in the spirit of Dirac.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoreticians, in my opinion, should study mathematics like math majors, almost forgetting about physics for a time; this is the point I feel so strongly about.  The thing is that math is such a big subject, and once you have the road map of what is important for theoretical physics; then it really takes years of study to learn all the mathematics.  I think it's so bad how many physics proffessors, who are themselves experimentalists, teach math improperly to young theoretician's.  I personally had to unlearn many of the things I thought I knew about math, once I took a course based on Rudin's &quot;Principle's of Analysis&quot;.&lt;/p&gt;&#xA;" OwnerUserId="442" LastActivityDate="2010-11-16T02:34:14.973" />
  <row Id="877" PostTypeId="2" ParentId="874" CreationDate="2010-11-16T03:28:50.353" Score="4" ViewCount="" Body="&lt;p&gt;A very nice explanation &lt;a href=&quot;http://www.upscale.utoronto.ca/PVB/Harrison/MachZehnder/MachZehnder.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't forget:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When a light ray is incident on a surface and the material on the other side of the surface has a higher index of refraction... then the reflected light ray is shifted in its phase by exactly one half a wavelength.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18" LastActivityDate="2010-11-16T03:28:50.353" CommentCount="2" />
  <row Id="878" PostTypeId="2" ParentId="814" CreationDate="2010-11-16T04:04:11.900" Score="5" ViewCount="" Body="&lt;p&gt;A very detailed computation with a comparison between the classical and the relativistic solution: &lt;a href=&quot;http://www.math.washington.edu/~morrow/papers/Genrel.pdf&quot; rel=&quot;nofollow&quot;&gt;The Precession of Mercury’s Perihelion&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2010-11-16T04:04:11.900" CommentCount="4" />
  <row Id="879" PostTypeId="1" AcceptedAnswerId="1003" CreationDate="2010-11-16T04:05:53.910" Score="7" ViewCount="276" Body="&lt;p&gt;Seems like all the simpler Lie algebras have a use in one or another branch of theoretical physics.  Even the exceptional E8 comes up in string theory.  But G2?   I've always wondered about that one.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know only of one false alarm in the 1960s or 1970s before SU(3) quark theory was understood, some physicists tried to fit mesons into a G2 representation.&lt;/p&gt;&#xA;" OwnerUserId="353" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-16T05:04:04.517" LastActivityDate="2010-11-17T15:16:08.443" Title="Is the G2 Lie algebra useful for anything?" Tags="&lt;mathematical-physics&gt;&lt;theory&gt;&lt;group-theory&gt;&lt;lie-algebra&gt;&lt;group-representations&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" />
  <row Id="880" PostTypeId="1" AcceptedAnswerId="1654" CreationDate="2010-11-16T04:16:33.230" Score="2" ViewCount="468" Body="&lt;p&gt;Also the side question is how many Joules is one photon (any between 450-660nm).&#xA;Thank you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I am asking because I want to estimate how much thermal energy should be dissipated by LED when part of known energy is emitted as light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Got an answer from Robert. Thank you.&#xA;So those 6500 Lumen 100Watt white LED arrays emit about 9 watt of energy in photons alone and rest 91W goes into heat. Not bad.&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="299" LastEditorDisplayName="" LastEditDate="2010-11-16T04:42:30.767" LastActivityDate="2010-12-05T17:11:45.483" Title="How many photons per second is one Lumen?" Tags="&lt;photons&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="1" />
  <row Id="881" PostTypeId="2" ParentId="879" CreationDate="2010-11-16T04:32:50.867" Score="8" ViewCount="" Body="&lt;p&gt;I don't know if these rise to the level of &quot;useful,&quot; but:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Yang-Mills theory with gauge group $G_2$ is interesting because $G_2$ has trivial center. So people simulate it on a lattice, try to understand in what sense it might be confining, how string tensions scale, if it has a deconfinement phase transition, and so on. The idea is that looking at a group with no center provides an interesting window into which phenomena in gauge theories rely crucially on the existence of a center and which do not. One recent paper (selected more or less at random from a search; I don't know this literature well enough to make useful suggestions) is &lt;a href=&quot;http://arxiv.org/abs/0807.1390&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;M-theory compactified on seven-dimensional manifolds of $G_2$ holonomy gives rise to four-dimensional theories with ${\cal N} = 1$ supersymmetry. I don't know the earliest references (probably this knowledge goes back to early work on supergravity before M-theory), but one place to look might be &lt;a href=&quot;http://arxiv.org/abs/hep-th/0107177&quot; rel=&quot;nofollow&quot;&gt;this paper of Atiyah and Witten&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="19" LastActivityDate="2010-11-16T04:32:50.867" />
  <row Id="882" PostTypeId="1" CreationDate="2010-11-16T04:38:28.817" Score="4" ViewCount="78" Body="&lt;p&gt;How to estimate the amount of water condensing from air on a surface, given the air's temperature and relative humidity and how they change over time, the surface temperature, material's thermal properites, roughness and whatever else needs to be given about the air and surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For my purpose, we may assume the surface starts off dry, but the more general situation would be as interesting.  If enough water condenses, it'll form drops and run off - can we account for that?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there some fairly simple formulas or rules of thumb?  High accuracy is not needed; I'll be happy to get grams per sq meter per second (or whatever) to within a factor of two.  (What if we wanted higher accuracy?)&lt;/p&gt;&#xA;" OwnerUserId="353" LastActivityDate="2010-11-16T08:45:10.887" Title="How to estimate condensation from air?" Tags="&lt;thermodynamics&gt;&lt;thermal-radiation&gt;&lt;water&gt;&lt;estimation&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="883" PostTypeId="2" ParentId="37" CreationDate="2010-11-16T05:01:32.953" Score="3" ViewCount="" Body="&lt;p&gt;Considering all the possible low-order corrections to Newtonian motion that there might be to a particle moving in a gravitation field of any kind,  or rather, in comparing relative motions of neighboring particles, there are defined some paramaters according to the Parameterized Post Newtonian (PPN) formalism.  This is described nicely in the famous book by Misner Thorne &amp;amp; Wheeler.  (My other three books on GTR don't mention PPN.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GTR can be analyzed in terms of PPN parameters - beta=1, eta=0, etc.  Alternative theories work out to having different PPN values.   Any experimental observations, after being calibrated and curves fitted to and so on, can be judged in terms of PPN parameters, at least some of them, or combinations of some of them.   So it's a matter of seeing what fits within error bars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Parameterized_post-Newtonian_formalism&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Parameterized_post-Newtonian_formalism&lt;/a&gt; tells more detail.&lt;/p&gt;&#xA;" OwnerUserId="353" LastActivityDate="2010-11-16T05:01:32.953" />
  <row Id="884" PostTypeId="1" CreationDate="2010-11-16T05:24:31.580" Score="12" ViewCount="1195" Body="&lt;p&gt;Inspired by &lt;a href=&quot;http://physics.stackexchange.com/questions/234/how-should-a-physics-student-study-mathematics&quot;&gt;How should a physics student study mathematics?&lt;/a&gt; and in the same vein as &lt;a href=&quot;http://physics.stackexchange.com/questions/193/best-books-for-mathematical-background&quot;&gt;Best books for mathematical background?&lt;/a&gt;, although in a more general fashion, I'd like to know if anyone is interested in doing a list of the books 'par excellence' for a physicist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In spite of the frivolous nature of this post, I think it can be a valuable resource.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0750628960&quot; rel=&quot;nofollow&quot;&gt;Course of Theoretical Physics&lt;/a&gt; - &lt;strong&gt;L.D. Landau, E.M. Lifshitz.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0805370021&quot; rel=&quot;nofollow&quot;&gt;Mathematical Methods of Physics&lt;/a&gt; - &lt;strong&gt;Mathews, Walker&lt;/strong&gt;. Very nice chapter on complex variables and evaluation of integrals, presenting must-know tricks to solve non-trivial problems. Also contains an introduction to groups and group representations with physical applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/048667164X&quot; rel=&quot;nofollow&quot;&gt;Mathematics of Classical and Quantum Physics&lt;/a&gt; - &lt;strong&gt;Byron and Fuller.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471010901&quot; rel=&quot;nofollow&quot;&gt;Topics in Algebra&lt;/a&gt; - &lt;strong&gt;I. N. Herstein&lt;/strong&gt;. Extremely well written, introduce basic concepts in groups, rings, vector spaces, fields and linear transformations. Concepts are motivated and a nice set of problems accompany each chapter (some of them quite challenging).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0126546568&quot; rel=&quot;nofollow&quot;&gt;Partial Differential Equations in Physics&lt;/a&gt; - &lt;strong&gt;Arnold Sommerfeld&lt;/strong&gt;. Although a bit dated, very clear explanations. First chapter on Fourier Series is enlightening. The ratio interesting information/page is extremely large. Contains discussions on types of differential equations, integral equations, boundary value problems, special functions and eigenfunctions.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="18" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-18T15:03:23.763" LastActivityDate="2011-07-18T15:03:23.763" CommunityOwnedDate="2010-11-16T19:26:35.393" Title="Books that every physicist should read" Tags="&lt;soft-question&gt;&lt;books&gt;&lt;reference-request&gt;&lt;big-list&gt;" AnswerCount="10" CommentCount="9" FavoriteCount="9" />
  <row Id="885" PostTypeId="1" CreationDate="2010-11-16T05:50:57.240" Score="11" ViewCount="884" Body="&lt;p&gt;How does it make sense to vary the position and the velocity independently?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Velocity is the derivative of position, so how can you treat them as independent variables? Doesn't every physics student ask this question when he learns calculus of variations? Does anybody ever answer this question? Ever? If so, please educate me.&lt;/p&gt;&#xA;" OwnerUserId="150" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-19T02:59:09.483" LastActivityDate="2011-06-08T15:12:57.833" Title="Why does calculus of variations work?" Tags="&lt;mathematical-physics&gt;&lt;lagrangian-mechanics&gt;&lt;variational-principle&gt;" AnswerCount="5" CommentCount="9" FavoriteCount="7" />
  <row Id="886" PostTypeId="1" AcceptedAnswerId="1739" CreationDate="2010-11-16T06:49:25.190" Score="10" ViewCount="320" Body="&lt;p&gt;My question is about the article &lt;a href=&quot;http://www.sciencemag.org/cgi/content/full/299/5614/1865&quot; rel=&quot;nofollow&quot;&gt;Swimming in Spacetime&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My gut reaction on first reading it was &quot;this violates conservation of momentum, doesn't it?&quot;. I now realize, however, that this doesn't allow something to change its momentum; it only allows something to move (change position) without ever having a nonzero momentum. Since this is relativity, there is no simple relationship between momentum and velocity like p = mv, so this is all well and good. An object can move, with a constant momentum of zero, by changing its shape in a nontrivial cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, now I'm thinking about a different conservation law and I can't see how &quot;swimming through spacetime&quot; is possible without violating it. The conserved quantity I'm thinking of is &lt;a href=&quot;http://math.ucr.edu/home/baez/boosts.html&quot; rel=&quot;nofollow&quot;&gt;the Noether charge associated with Lorentz boosts&lt;/a&gt;, which is basically x - (p/E)t, that is, the position of the center of mass projected back to time t=0. If p = 0, then the conserved quantity is simply x, the position of the center of mass. This obviously contradicts the whole swimming idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's going on here? Is swimming through spacetime only possible if the spacetime is curved in some way that breaks the symmetry under Lorentz boosts? Or is there some error in my reasoning?&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-04-06T08:51:39.057" Title="Swimming in Spacetime - apparent conserved quantity violation" Tags="&lt;general-relativity&gt;&lt;paradox&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" />
  <row Id="887" PostTypeId="1" CreationDate="2010-11-16T07:32:10.670" Score="10" ViewCount="427" Body="&lt;p&gt;Cellular automata provide interesting models of physics: Google Scholar gives more than 25,000 results when searching for &lt;a href=&quot;http://scholar.google.com/scholar?q=%22cellular+automata%22+physics&amp;amp;hl=en&amp;amp;btnG=Search&amp;amp;as_sdt=2001&amp;amp;as_sdtp=on&quot; rel=&quot;nofollow&quot;&gt;&quot;cellular automata&quot; physics&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google Scholar still gives more than 2.000 results when searching for &lt;a href=&quot;http://scholar.google.com/scholar?hl=en&amp;amp;q=%22quantum+cellular+automata%22&amp;amp;btnG=Search&amp;amp;as_sdt=2000&amp;amp;as_ylo=&amp;amp;as_vis=0&quot; rel=&quot;nofollow&quot;&gt;&quot;quantum cellular automata&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it gives only &lt;a href=&quot;http://books.google.com/books?hl=en&amp;amp;lr=&amp;amp;id=sab-9fGPCEcC&amp;amp;oi=fnd&amp;amp;pg=PA1&amp;amp;dq=%22relativistic+cellular+automata%22&amp;amp;ots=rG9eoNgPYh&amp;amp;sig=ZNQpI-0B-0S2358zX9To-h4NtdI#v=onepage&amp;amp;q=%22relativistic%20cellular%20automata%22&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt; (one!) result when searching for &lt;a href=&quot;http://scholar.google.com/scholar?hl=en&amp;amp;q=%22relativistic+cellular+automata%22&amp;amp;btnG=Search&amp;amp;as_sdt=2000&amp;amp;as_ylo=&amp;amp;as_vis=0&quot; rel=&quot;nofollow&quot;&gt;&quot;relativistic cellular automata&quot;&lt;/a&gt;, i.e. cellular automata with a (discrete) Minkoswki space-time instead of an Euclidean one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can this be understood? &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Why does the concept of QCA seem more&#xA;  promising than that of RCA?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Are there conceptual or technical barriers for a thorough treatment of RCA?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="229" LastEditorUserId="229" LastEditorDisplayName="" LastEditDate="2010-11-16T08:39:07.033" LastActivityDate="2011-01-30T03:34:08.527" Title="Relativistic Cellular Automata" Tags="&lt;special-relativity&gt;&lt;models&gt;" AnswerCount="5" FavoriteCount="4" />
  <row Id="888" PostTypeId="1" CreationDate="2010-11-16T08:32:30.313" Score="0" ViewCount="72" Body="&lt;p&gt;who of you in this forum uses R (http://www.r-project.org/) with the multicore, SNOW or CUDA packages, so for advanced calculations that need more power than a workstation CPU? On which hardware do you compute these scripts? At home/ at work or do you have data center access somewhere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The background of these questions is the following: I am currently writing my M.Sc. thesis about R and High-Performance-Computing and need a strong knowledge about who actually uses R. I read that R had 1 million users in 2008, bu thats more or less the only user statistics I could find on this topic - so I hope for your answers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely Heinrich&lt;/p&gt;&#xA;" OwnerUserId="446" LastActivityDate="2010-11-16T08:32:30.313" ClosedDate="2010-11-16T15:47:52.897" Title="Who uses R with multicore, SNOW or CUDA package?" Tags="&lt;computation&gt;" CommentCount="3" />
  <row Id="889" PostTypeId="2" ParentId="882" CreationDate="2010-11-16T08:45:10.887" Score="1" ViewCount="" Body="&lt;p&gt;So you want to know how much water a certain surface adsorbs. This is really dependent on the surface material/conditions. Check &lt;a href=&quot;http://en.wikipedia.org/wiki/Adsorption&quot; rel=&quot;nofollow&quot;&gt;adsorption&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Relative_humidity&quot; rel=&quot;nofollow&quot;&gt;relative humidity&lt;/a&gt; on Wikipedia. To where I have analyzed, it seems that there is about enough information in the two articles. I am not a specialist on the subject so I might be missing some important factor.&lt;/p&gt;&#xA;" OwnerUserId="62" LastActivityDate="2010-11-16T08:45:10.887" />
  <row Id="890" PostTypeId="2" ParentId="886" CreationDate="2010-11-16T08:59:38.183" Score="1" ViewCount="" Body="&lt;p&gt;Well, I hope that my primitive understanding of GR makes for a good non-expert explanation... In GR, the Lorentz group symmetries are generally only valid locally, that is for a given space-time point. If you want to translate a vector to another space-time point, you need to do a parallel transport, which usually introduces correction terms depending on the curvature&lt;/p&gt;&#xA;" OwnerUserId="243" LastActivityDate="2010-11-16T08:59:38.183" CommentCount="3" />
  <row Id="891" PostTypeId="2" ParentId="885" CreationDate="2010-11-16T09:01:59.053" Score="19" ViewCount="" Body="&lt;p&gt;Unlike your question suggests, it is &lt;em&gt;not&lt;/em&gt; true that velocity is varied independently of position. A variation of position $q \mapsto q + \delta q$ induces a variation of velocity $\partial_t q \mapsto \partial_t q + \partial_t (\delta q)$ as you would expect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing that may seem strange is that $q$ and $\partial_t q$ are treated as independent variables of the Lagrangian $L(q,\partial_t q)$. But this is not surprising; after all, if you ask &quot;what is the kinetic energy of a particle?&quot;, then it is not enough to know the position of the particle, you also have to know its velocity to answer that question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put differently, you can choose position and velocity independently as &lt;em&gt;initial conditions&lt;/em&gt;, that's why the Lagrangian function treats them as independent; but the calculus of variation does &lt;em&gt;not vary them independently&lt;/em&gt;, a variation in position induces a fitting variation in velocity.&lt;/p&gt;&#xA;" OwnerUserId="188" LastActivityDate="2010-11-16T09:01:59.053" />
  <row Id="892" PostTypeId="2" ParentId="884" CreationDate="2010-11-16T09:03:05.917" Score="11" ViewCount="" Body="&lt;p&gt;There is a lot of good books in CM, QM, EM... but what every physicist should read, undoubtly, are &lt;a href=&quot;http://www.google.com/search?tbs=bks:1&amp;amp;tbo=1&amp;amp;q=The+Feynman+Lectures+on+Physics&amp;amp;btnG=Search+Books#q=The+Feynman+Lectures+on+Physics&amp;amp;hl=en&amp;amp;tbo=1&amp;amp;tbs=bks:1,bkt%3ab,bkv%3aa,vw%3ag,qdr%3at&amp;amp;source=lnt&amp;amp;sa=X&amp;amp;ei=mufiTP2vEpC1hAfW_ozPDQ&amp;amp;ved=0CBgQpwU&amp;amp;fp=7a3f99f36efcd1b2&quot; rel=&quot;nofollow&quot;&gt;The Feynman Lectures on Physics&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="306" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-16T20:21:14.957" LastActivityDate="2010-11-16T20:21:14.957" CommunityOwnedDate="2010-11-16T19:26:35.393" />
  <row Id="893" PostTypeId="1" CreationDate="2010-11-16T09:25:08.897" Score="12" ViewCount="669" Body="&lt;p&gt;I heard about &lt;a href=&quot;http://en.wikipedia.org/wiki/Yakov_Perelman#Books&quot; rel=&quot;nofollow&quot;&gt;Yakov Perelman&lt;/a&gt; and his books. I just finished reading his two volumes of &lt;strong&gt;Physics for Entertainment&lt;/strong&gt;. What a delightful read! What a splendid author. This is the exact book I've been searching for. I can use it to develop interest for science (math &amp;amp; physics) in my students.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His math books:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mathematics Can Be Fun&lt;/li&gt;&#xA;&lt;li&gt;Figures for Fun&lt;/li&gt;&#xA;&lt;li&gt;Arithmetic for entertainment&lt;/li&gt;&#xA;&lt;li&gt;Geometry for Entertainment&lt;/li&gt;&#xA;&lt;li&gt;Lively Mathematics&lt;/li&gt;&#xA;&lt;li&gt;Fun with Maths &amp;amp; Physics&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;His physics books:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Physics for Entertainment (1913)&lt;/li&gt;&#xA;&lt;li&gt;Physics Everywhere&lt;/li&gt;&#xA;&lt;li&gt;Mechanics for entertainment&lt;/li&gt;&#xA;&lt;li&gt;Astronomy for entertainment&lt;/li&gt;&#xA;&lt;li&gt;Tricks and Amusements&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I want to get all the above books. Because books from author like this cannot be disappointing. But unfortunately not all of them are available. :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also read another amazing book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/069111966X&quot; rel=&quot;nofollow&quot;&gt;How to Solve It: A New Aspect of Mathematical Method by G.Polya&lt;/a&gt;. This books actually teaches you &lt;em&gt;how to think&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the similar lines if you have any book suggestions (with very practical problems &amp;amp; case studies) for physics &amp;amp; Math (Please don't differentiate between math &amp;amp; physics here. If I can develop interest in one of the subject he'll gain interest in other.) please contribute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cross Post:&lt;/strong&gt; &lt;a href=&quot;http://math.stackexchange.com/questions/10543/books-that-develop-interest-critical-thinking-among-high-school-students&quot;&gt;http://math.stackexchange.com/questions/10543/books-that-develop-interest-critical-thinking-among-high-school-students&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="447" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-01T12:28:47.023" LastActivityDate="2011-02-05T03:24:12.173" CommunityOwnedDate="2010-11-18T19:13:02.650" Title="Books that develop interest &amp; critical thinking among high school students" Tags="&lt;soft-question&gt;&lt;books&gt;" AnswerCount="7" CommentCount="10" FavoriteCount="5" />
  <row Id="894" PostTypeId="2" ParentId="893" CreationDate="2010-11-16T09:38:50.800" Score="2" ViewCount="" Body="&lt;p&gt;try &#xA;&lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0062731009&quot; rel=&quot;nofollow&quot;&gt;The Cartoon Guide to Physics&lt;/a&gt;&lt;/em&gt; by Larry Gonick&lt;/p&gt;&#xA;&#xA;&lt;p&gt;funny and smart!&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="447" LastEditorDisplayName="" LastEditDate="2010-11-21T00:03:54.473" LastActivityDate="2010-11-21T00:03:54.473" CommunityOwnedDate="2010-11-18T19:13:02.650" CommentCount="1" />
  <row Id="895" PostTypeId="2" ParentId="893" CreationDate="2010-11-16T09:39:12.343" Score="5" ViewCount="" Body="&lt;p&gt;I saw that some people wanted to learn physics after they saw&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://research.microsoft.com/apps/tools/tuva/index.html&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Project TUVA lectures by Richard Feynman&lt;/strong&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://videolectures.net/mit801f99_lewin_lec10/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;10th classic classical mechanics lecture by Walter Lewin&lt;/strong&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Surely you are joking Mr. Feynman book by Richard Feynman&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To get people interested you need to show them that &lt;strong&gt;doing physics is COOL&lt;/strong&gt; and is FUN. &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-11-16T09:39:12.343" CommunityOwnedDate="2010-11-18T19:13:02.650" CommentCount="12" />
  <row Id="896" PostTypeId="2" ParentId="887" CreationDate="2010-11-16T09:48:53.867" Score="6" ViewCount="" Body="&lt;p&gt;Check out Mark Smith's PhD thesis titled &lt;a href=&quot;http://dspace.mit.edu/handle/1721.1/33501&quot; rel=&quot;nofollow&quot;&gt;Cellular automata methods in mathematical physics&lt;/a&gt;, specifically Chapter 4: Lorentz Invariance in Cellular Automata.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion part of the chapter:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Symmetry is an important aspect of physical laws, and it is therefore desirable to identify analogous symmetry in CA rules. Furthermore, the most important symmetry &#xA;  groups in physics are the Lorentz group and its relatives. While there is a substantial &#xA;  difference between the manifest existence of a preferred frame in CA and the lack &#xA;  of a preferred frame demanded by special relativity, there are still some interesting &#xA;  connections. In particular, CA have a well-defined speed of light which imposes a &#xA;  causal structure on their evolution, much as a Minkowski metric imposes a causal &#xA;  structure on spacetime. To the extent that these structures can be made to coincide &#xA;  between the CA and continuum cases, it makes sense to look for Lorentz invariant &#xA;  CA. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The diffusion of massless particles in one spatial dimension provides a good example of a Lorentz invariant process that can be expressed in alternative mathematical &#xA;  forms. A corresponding set of linear partial differential equations can be derived with &#xA;  a simple transport argument and then shown to be Lorentz invariant. A CA formulation of the process is also Lorentz invariant in the limit of low particle density and &#xA;  small lattice spacing. The equations can be solved with standard techniques, and the &#xA;  analytic solution provides a check on the results of the simulation. Generalization to &#xA;  higher dimensions seems to be difficult because of anisotropy of CA lattices, though &#xA;  it is still plausible that symmetry may emerge in complex, high-density systems. The &#xA;  model and analyses presented here can be used as a benchmark for further studies of &#xA;  symmetry in physical laws using CA. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="47" LastActivityDate="2010-11-16T09:48:53.867" CommentCount="1" />
  <row Id="897" PostTypeId="2" ParentId="884" CreationDate="2010-11-16T09:56:06.730" Score="3" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0198520115&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;The Principles of Quantum Mechanics by P A M Dirac&lt;/strong&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0198517440&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Newton's Principia for the Common Reader by S Chandrasekhar&lt;/strong&gt;&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-11-16T09:56:06.730" CommunityOwnedDate="2010-11-16T19:26:35.393" CommentCount="2" />
  <row Id="898" PostTypeId="2" ParentId="838" CreationDate="2010-11-16T10:04:47.627" Score="1" ViewCount="" Body="&lt;p&gt;Possibly the easiest one, but often overlooked: earth itself. How would you determine it's size? There are a number of fairly easy ways to measure it within an oder of magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next up is the moon. Obviously there's the earth's shadow, and we've just determined that radius. So, how big is the moon in comparison? How heavy would it be? (It's rock, you might assume the same density) What is it's rotation period? Its orbit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big advantage of the moon is that it's quite easy to find and observe, even under less than stellar atmospheric conditions.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2010-11-16T10:04:47.627" />
  <row Id="899" PostTypeId="2" ParentId="843" CreationDate="2010-11-16T10:07:28.943" Score="1" ViewCount="" Body="&lt;p&gt;To supplement, the answer for your final question. As Raskolnikov pointed out, this article lies about the fact that this process can be a source of energy, and indeed all known biological mechanisms of gathering energy satisfy energy conservation; photosynthesis gathers the energy from light and various kinds of respiration from chemical energy of various substances available in environment (that mostly also originated in photosynthesis, with exception of some esoteric ecosystems).&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-16T10:07:28.943" CommentCount="1" />
  <row Id="900" PostTypeId="1" AcceptedAnswerId="903" CreationDate="2010-11-16T11:57:05.633" Score="4" ViewCount="202" Body="&lt;p&gt;Why do frozen objects have a higher propensity to snap/break cleanly than non-frozen/warmer ones?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-16T17:17:49.330" Title="Why do frozen objects have a higher propensity to snap/break cleanly than non-frozen/warmer ones?" Tags="&lt;temperature&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="901" PostTypeId="2" ParentId="357" CreationDate="2010-11-16T11:58:45.913" Score="18" ViewCount="" Body="&lt;p&gt;I don't know if it qualify as home experiment, but you can use the internet to get access to thousands of kilometres of optical fibres for free. It allows you to measure the speed of light in the fibres, which is &lt;em&gt;c/n&lt;/em&gt;, where &lt;em&gt;n&lt;/em&gt; is the refractive index of glass, &lt;em&gt;i.e.&lt;/em&gt; 1.5. This corresponds to 2×10⁸ m·s⁻¹. Using ping, you measure a roundtrip time, that is it should correspond to 100 km/ms of roundtrip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Paris, I ping the website of Columbia, in New-York, I have&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;fred@sanduleak2:~$ ping www.columbia.edu&#xA;PING www.columbia.akadns.net (128.59.48.24) 56(84) bytes of data.&#xA;64 bytes from www-csm.cc.columbia.edu (128.59.48.24): icmp_req=1 ttl=113 time=125 ms&#xA;64 bytes from www-csm.cc.columbia.edu (128.59.48.24): icmp_req=2 ttl=113 time=116 ms&#xA;....&#xA;64 bytes from www-csm.cc.columbia.edu (128.59.48.24): icmp_req=16 ttl=113 time=112 ms&#xA;^C&#xA;--- www.columbia.akadns.net ping statistics ---&#xA;17 packets transmitted, 16 received, 5% packet loss, time 16023ms&#xA;rtt min/avg/max/mdev = 108.585/118.151/132.156/7.728 ms&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The minimum roundtrip time is 108 ms, which would correspond to 10800 km instead of 5839 km. Off by a factor of 2, but the correct order of magnitude, due to delays. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one looks more precisely the traject of my packets to new-york with tracepath&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;fred@sanduleak2:~$ tracepath www.columbia.edu&#xA;&#xA; 1:  sanduleak2                                            0.266ms pmtu 1500&#xA; ....  &#xA; 3:  pioneer.ens-cachan.fr                                 1.072ms &#xA; ....&#xA; 6:  vl172-orsay-rtr-021.noc.renater.fr                   28.747ms asymm  9 &#xA; 7:  te0-1-0-5-paris1-rtr-001.noc.renater.fr              20.931ms &#xA; 8:  renater.rt1.par.fr.geant2.net                        30.307ms asymm  9 &#xA; 9:  so-3-0-0.rt1.lon.uk.geant2.net                       33.780ms asymm 10 &#xA;10:  so-2-0-0.rt1.ams.nl.geant2.net                       36.570ms asymm 11 &#xA;11:  xe-2-3-0.102.rtr.newy32aoa.net.internet2.edu        127.394ms asymm 12  &#xA;12:  nyc-7600-internet2-newy.nysernet.net                128.238ms &#xA;13:  columbia.nyc-7600.nysernet.net                      135.948ms &#xA;14:  ....&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We see that the packets travel around (Paris, London, Amsterdam) and cross the Atlantic between Amsterdam (10) and New-York (11) in 127-37=90 ms (roundtrip). This still gives us a 9000 km distance, way too long. I don't know if it is due to the cable trajectory, electronic delays, to small sampling by tracepath or an error on my calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related to this ping delay, you have the funny &lt;a href=&quot;http://www.ibiblio.org/harris/500milemail.html&quot; rel=&quot;nofollow&quot;&gt;500 miles bug&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another in-the-lab experiment using cheap material and computers is in the arXiv paper &lt;a href=&quot;http://arxiv.org/abs/physics/0201053&quot; rel=&quot;nofollow&quot;&gt;speed of light measurement using ping&lt;/a&gt;. However, their measurement is indirect (they measure the propagation inside CAT5 cables), but it should also be doable with optical fibres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited to add&lt;/strong&gt;: My idea of using tracepath probably comes from &lt;a href=&quot;http://arxiv.org/abs/physics/0208087&quot; rel=&quot;nofollow&quot;&gt;Measuring the Earth with Traceroute&lt;/a&gt;. In this paper they are more lucky than I was (only 20% slower, instead of 100% !)&lt;/p&gt;&#xA;" OwnerUserId="373" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-11-17T17:25:58.153" LastActivityDate="2010-11-17T17:25:58.153" CommentCount="11" />
  <row Id="902" PostTypeId="2" ParentId="838" CreationDate="2010-11-16T12:04:52.080" Score="0" ViewCount="" Body="&lt;p&gt;You might try talking to the guy who wrote: &lt;a href=&quot;http://www.fallingapplescience.com/neglected&quot; rel=&quot;nofollow&quot;&gt;http://www.fallingapplescience.com/neglected&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He's apparently working on the very same problem of how to teach basic astronomy from the ground up without a priori knowledge.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-16T12:04:52.080" />
  <row Id="903" PostTypeId="2" ParentId="900" CreationDate="2010-11-16T12:13:18.950" Score="2" ViewCount="" Body="&lt;p&gt;First of all, it is not always the case -- the most known example is meat frozen in liquid nitrogen. In general, I think there are many different processes involved; temperature change may lead to small fractures in solid materials, in organic objects we have formation of ice from water inside cells what both damages the structure and lowers elasticity, there may be more general phase transitions involved, like for instance crystal structure changes of cocoa butter in chocolate.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-16T12:13:18.950" />
  <row Id="904" PostTypeId="2" ParentId="862" CreationDate="2010-11-16T12:17:42.723" Score="3" ViewCount="" Body="&lt;p&gt;If you FT from time domain, you end up in frequency domain, not in space domain. Of course you can use the relativistic interpretation by dividing a lightyear by a factor of c to end up with - oh, a year! But instead of that FT explanation one could of course just say &quot;it's a figure of speech as a synonym to half an eternity&quot;...&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2010-11-16T12:17:42.723" />
  <row Id="905" PostTypeId="1" CreationDate="2010-11-16T12:39:22.573" Score="8" ViewCount="550" Body="&lt;p&gt;It seems odd that entropy is usually only defined for a system in a single 'slice' of time or spacelike region.  Can one define the entropy of a system defined by a 4d region of spacetime, in such a way that yields a codimension one definition which agrees with the usual one when the codimension one slice is spacelike?&lt;/p&gt;&#xA;" OwnerUserId="457" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-05T12:44:38.023" LastActivityDate="2011-02-24T19:10:04.270" Title="Is there a four-dimensional definition of entropy?" Tags="&lt;special-relativity&gt;&lt;statistical-mechanics&gt;&lt;entropy&gt;" AnswerCount="7" CommentCount="5" FavoriteCount="2" />
  <row Id="906" PostTypeId="2" ParentId="905" CreationDate="2010-11-16T12:52:57.537" Score="7" ViewCount="" Body="&lt;p&gt;You are thinking about Boltzmann's definition of entropy, I guess?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Boltzmann's definition, entropy is just the logarithm of the amount of possible states associated with certain macroscopic variables. In its generality, therefore, it doesn't seem to me to exclude the possibility of counting states with different time coordinates. Or in your more general context, on different time-slices. The question is, what does this correspond to? Does it make sense to do that? You would have to specify the time-development of the macroscopic variables and count the number of microscopic trajectories compatible with those macroscopic trajectories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a matter of fact, there exist so-called dynamical entropies. In a heuristic sense, what they do is counting the density of phase-space trajectories of a system, whereas Boltzmann entropy just counts the amount of accessible states under certain macroscopic constraints. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Kolmogorov%E2%80%93Sinai_entropy#Measure-theoretic_entropy&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Kolmogorov%E2%80%93Sinai_entropy#Measure-theoretic_entropy&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2010-11-16T12:52:57.537" />
  <row Id="907" PostTypeId="1" CreationDate="2010-11-16T13:09:28.923" Score="3" ViewCount="213" Body="&lt;p&gt;In &lt;a href=&quot;http://physics.stackexchange.com/questions/9/hamiltonian-principle&quot;&gt;this question&lt;/a&gt; it is discussed why by Hamilton's principle the action integral must be stationary. Most examples deal with the case that the action integral is minimal: this makes sense - we all follow the path with the least resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a stationary point can be a maximal or minimal extremum or even a point of inflexion (rising or falling). Can anybody give examples from theory or practice where the action integral takes on a maximal extremum or a point of inflexion?&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-16T23:53:41.050" Title="Type of stationary point in Hamilton's principle" Tags="&lt;hamiltonian-mechanics&gt;&lt;variational-principle&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="908" PostTypeId="1" CreationDate="2010-11-16T13:54:05.617" Score="-5" ViewCount="220" Body="&lt;p&gt;Why is there something rather than nothing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And before you answer &quot;God did it,&quot; consider Carl Sagan's reply &quot;then how did God come about?&quot;&lt;/p&gt;&#xA;" OwnerUserId="460" LastActivityDate="2010-11-16T14:07:59.413" ClosedDate="2010-11-16T15:48:51.773" Title="Why is there something rather than nothing?" Tags="&lt;cosmology&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="909" PostTypeId="2" ParentId="862" CreationDate="2010-11-16T13:54:31.840" Score="5" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;His response: Nevermind, for the first&#xA;  time I accept I was wrong. BUT using&#xA;  lightyears to measure time is&#xA;  possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's possible if you rape physics, definitions, and just organize things so that you actually, in the end, obtain time from dimensional analysis. The idea would go along this line of reasoning:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I know a man walks 5 km/h, hence&lt;/li&gt;&#xA;&lt;li&gt;space is a unit of time because I can say 10 km is equivalent to two hours, hence&lt;/li&gt;&#xA;&lt;li&gt;I can measure time by specifying a distance. Let's meet in 5 kilometers at the pub 200 meters from here.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But that's not a unit definition. it's just a deep misrepresentation of concepts: using this approach, you can define everything in terms of anything else, assuming there's a relationship among them. you could define time in terms of weight of apples a man can move from the tree to a box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the whole setup is pretty circular in definition. You define time as space traveled by something at a constant speed for a defined time, so in the end your pulled you own bootstraps.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My example didn't make sense&#xA;  bacause I was wrong when I meantioned&#xA;  that I'm still measuring dist. If you&#xA;  have a signal in time domain and&#xA;  ...take the FT, I get a signal which&#xA;  DOES NOT HAVE to be in frequency&#xA;  domain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A Fourier transform is nothing but finding the coefficients of a linear combination of plane waves. When you find these coefficients, you can express the original function as the linear combination of these coefficients and a plane wave. If you note, there's a unit relationship between the domain before the FT and after. seconds -&gt; seconds^(-1) = Hz. So even if you want to do fourier decomposition of a space-based periodic or aperiodic system, the resulting domain will be in meters^(-1), which is eventually a &lt;a href=&quot;http://en.wikipedia.org/wiki/Wave_number&quot; rel=&quot;nofollow&quot;&gt;wave number&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Clarify this to the guy who&#xA;  posted last. Now the new signal is in&#xA;  a domain defined by me and so is its&#xA;  units.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nope, it just turns out from the dimensional analysis that this is not the case. Clearly, you can always transform your frobbles units into Hertz through an ad-hoc transformation you invent (see the man-apples above) but that would still not change the final dimensional analysis of your FT, and you would, in any case, introduce an arbitrary constant (in our case above, the walking speed) which, in the end, likely produces a circular definition.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This signal although not equal&#xA;  to the original signal, still&#xA;  represents that if ya take an inverse&#xA;  FT. So, the idea of time will still be&#xA;  there. Now coming back to our case:&#xA;  lightyears here is not the lightyears&#xA;  you are used to read when dealing with&#xA;  distance. It represents time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lightyears is a measure of distance. It is a product of two well defined constants: the speed of light and a well defined amount of time. Simple dimensional analysis tells you it's a distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: non-tech note. there's nothing wrong not to know things. Tell your friend it's not my intention to mock him. There's, however, something wrong to pretend to be right through misunderstood justification. It looks like he is mature enough to understand he is wrong, which is the spirit we should all live with. I hope my answers clarifies his doubts. Note that I could be wrong myself. I don't know the technicalities of standard unit definitions, and my exposition could be patched by someone who knows more than me on this field. The approximation I present here is good enough for the purpose of explanation, but it is wrong nevertheless when we go down to the gritty details, and I am ready to accept criticism on this regard. This is how we progress.&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="217" LastEditorDisplayName="" LastEditDate="2010-11-16T14:21:07.200" LastActivityDate="2010-11-16T14:21:07.200" CommentCount="1" />
  <row Id="910" PostTypeId="2" ParentId="908" CreationDate="2010-11-16T14:03:15.940" Score="0" ViewCount="" Body="&lt;p&gt;The question is unanswerable.  If I tell you &quot;because P&quot;, then you can simply ask what caused P.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-16T14:03:15.940" />
  <row Id="911" PostTypeId="2" ParentId="908" CreationDate="2010-11-16T14:07:59.413" Score="0" ViewCount="" Body="&lt;p&gt;Prove to me there is something!  Maybe existence is a fluctuation of nothingness.&lt;/p&gt;&#xA;" OwnerUserId="463" LastActivityDate="2010-11-16T14:07:59.413" />
  <row Id="912" PostTypeId="2" ParentId="905" CreationDate="2010-11-16T14:08:46.463" Score="4" ViewCount="" Body="&lt;p&gt;As far as know, entropy works in systems with Hamiltonian dynamics, that is, when there is explicit dependence on time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In classical mechanics (where position and momentum depend on time), there is Boltzmann entropy $S = k_b \ln \Omega$ ($\Omega$ - 'number' of states). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In (non-relativistic) quantum mechanics (where wavefunction depend on time), there is von Neumann Entropy $S = -k_b \langle \rho \ln \rho \rangle $ ($\rho$ - density matrix).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though in general there is information-theoretic quantity Shannon entropy $S = \sum_i p_i \ln p_i$ ($p_i$ are probabilities that the system is in the $i$-th state). Maybe in Quantum Field Theory there is some kind of 4-d entropy, but I am not sure. Anyway, the fundamental property 'entropy is non-decreasing function of time' has any meaning only if $S$ is a function of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a broader discussion see great review/didactic paper:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A Wehrl, &lt;a href=&quot;http://rmp.aps.org/abstract/RMP/v50/i2/p221_1&quot; rel=&quot;nofollow&quot;&gt;General properties of entropy&lt;/a&gt;, Rev. Mod. Phys. 50, 221–260 (1978)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="184" LastEditorUserId="184" LastEditorDisplayName="" LastEditDate="2010-11-16T14:19:39.980" LastActivityDate="2010-11-16T14:19:39.980" CommentCount="1" />
  <row Id="913" PostTypeId="2" ParentId="905" CreationDate="2010-11-16T14:15:47.083" Score="0" ViewCount="" Body="&lt;p&gt;Entropy doesn't need a four dimensional definition. It is a scalar amount, since the number of states is just a number. These states could be parts of a n-dimensional phase-space, but it doesn't change entropy. &lt;/p&gt;&#xA;" OwnerUserId="306" LastActivityDate="2010-11-16T14:15:47.083" />
  <row Id="915" PostTypeId="2" ParentId="893" CreationDate="2010-11-16T14:34:06.627" Score="0" ViewCount="" Body="&lt;p&gt;If it do not need to be a book, I recommend&#xA;&lt;a href=&quot;http://www.scientificamerican.com/department.cfm?id=skeptic&quot; rel=&quot;nofollow&quot;&gt;Michael Shermer's Skeptic Column&lt;/a&gt; in &lt;a href=&quot;http://www.scientificamerican.com&quot; rel=&quot;nofollow&quot;&gt;Scientific American Magazine&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="184" LastActivityDate="2010-11-16T14:34:06.627" CommunityOwnedDate="2010-11-18T19:13:02.650" />
  <row Id="916" PostTypeId="1" AcceptedAnswerId="960" CreationDate="2010-11-16T14:48:50.857" Score="9" ViewCount="543" Body="&lt;h1&gt;Problem&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;If you had a long bar floating in space, what would be the compressive force at the centre of the bar, due to the self-weight of both ends?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diagram - what is the force at point X in the middle of the bar?:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;----------------------L---------------------&amp;gt;, total mass M&#xA;=======================X======================   &amp;lt;- the bar&#xA;                 F---&amp;gt; X &amp;lt;---F&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h1&gt;Summary&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;You should be able to simplify by cutting the bar into pieces, but that gives a different answer depending on how many pieces you use (see below). So the simplification must be wrong - but why?&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;My approach&lt;/h1&gt;&#xA;&#xA;&lt;h3&gt;Split bar in two&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;So, one approximation would be to cut the bar in half - two pieces of length L/2, mass M/2:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         (M/2)&amp;lt;-------L/2-------&amp;gt;(M/2)&#xA;           #1          X           #2   &amp;lt;- bar approximated as blobs #1 and #2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Force at X is G(M1.M2)/(R^2) = G (M/2)^2 / (L/2)^2 = G M^2 / L^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or Fx / (G. M^2 / L^2) = &lt;strong&gt;1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is that really valid? If so, shouldn't you get the same answer if you split the bar into four pieces?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Split bar into four&lt;/h3&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   (M/4)&amp;lt;-L/4-&amp;gt;(M/4)&amp;lt;-L/4-&amp;gt;(M/4)&amp;lt;-L/4-&amp;gt;(M/4)&#xA;     #1          #2    X     #3          #4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;My assumption is that the force at X is the sum of the attractions of each blob on the left to every blob on the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force at X = #1&amp;lt;&gt;#3 + #1&amp;lt;&gt;#4 + #2&amp;lt;&gt;#3 + #2&amp;lt;&gt;#4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;('&amp;lt;&gt;' being force between blobs #x and #y).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fx / (G.M^2 / L^2) = (2/4)^-2 + (3/4)^-2 + (1/4)^-2 + (2/4)^-2 = &lt;strong&gt;1.61&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is bigger than the previous result (1.61 vs 1).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Split bar into six&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Similarly, if you split into 6 blobs, the total force comes out as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fx / (G.M^2 / L^2) = (3/6)^-2+(4/6)^-2+(5/6)^-2 + (2/6)^-2+(3/6)^-2+(4/6)^-2 + (1/6)^-2+(2/6)^-2+(3/6)^-2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fx / (G.M^2 / L^2) = &lt;strong&gt;2.00&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;So what's wrong with my approach? And what is the real answer?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;So it seems the more pieces we split the bar into, the larger the result gets. There's clearly something wrong with my assumptions! - but what?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be very glad if someone here could explain this. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&#xA;As Peter Shor pointed out, my calculations had some dodgy algebra and I'd calculated $$L^2/M^2$$ values rather than $$M^2/L^2$$. I've now corrected that - the value still increases as you divide into more masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll do a bit more work with more divisions and see if this leads to convergence or not.&lt;/p&gt;&#xA;" OwnerUserId="474" LastEditorUserId="474" LastEditorDisplayName="" LastEditDate="2010-11-19T18:05:54.393" LastActivityDate="2010-11-19T18:05:54.393" Title="Imagine a long bar floating in space. What force does it exert on itself in the middle due to gravity?" Tags="&lt;classical-mechanics&gt;&lt;gravity&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="2" />
  <row Id="917" PostTypeId="1" AcceptedAnswerId="981" CreationDate="2010-11-16T14:49:59.750" Score="15" ViewCount="755" Body="&lt;p&gt;I'm swimming in the ocean and there's a thunderstorm. Lightning bolts hit ships around me. Should I get out of the water?&lt;/p&gt;&#xA;" OwnerUserId="473" LastActivityDate="2011-08-26T15:59:27.283" Title="Lightning strikes the Ocean I'm swimming in - what happens?" Tags="&lt;electricity&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="2" />
  <row Id="921" PostTypeId="2" ParentId="691" CreationDate="2010-11-16T15:28:49.950" Score="5" ViewCount="" Body="&lt;p&gt;That is incorrect description of what happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the black hole is small, the gravitational forces pull you apart before you get anywhere near the horizon. On the other hand, if the black hole is large, you won't even notice anything as you cross the horizon, until you hit the singularity and &lt;em&gt;than&lt;/em&gt; you'll be pulled apart. Remember, from your point of view you are just free-falling, the event horizon only has meaning for the outside observers. The stars will look somewhat different during your fall because of the holes gravitational lensing effect, but no fiery death by cosmic radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the outside observer's point of view this is all different: they will see you getting closer and closer to the horizon but never reaching it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some more detailed explanation about this (and many other things about black holes) is for example here:&#xA;&lt;a href=&quot;http://cosmology.berkeley.edu/Education/BHfaq.html#q3&quot; rel=&quot;nofollow&quot;&gt;http://cosmology.berkeley.edu/Education/BHfaq.html#q3&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="472" LastActivityDate="2010-11-16T15:28:49.950" CommentCount="2" />
  <row Id="922" PostTypeId="2" ParentId="917" CreationDate="2010-11-16T15:39:00.817" Score="0" ViewCount="" Body="&lt;p&gt;Most probably the current just spreads in all directions and weakens quite fast (at least like $r^{-2}$, not counting resistance), so I don't think the hazard is much (in magnitude) larger than on land in similar conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: In what I found in Internet salty water has only 10 times better conductivity than wet soil; yet on land the wet soil layer lays on insulating layer of dry soil, so the current is directed to rather &quot;flood&quot; than penetrate. &lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-11-16T15:48:57.473" LastActivityDate="2010-11-16T15:48:57.473" />
  <row Id="924" PostTypeId="1" AcceptedAnswerId="965" CreationDate="2010-11-16T16:17:10.407" Score="0" ViewCount="349" Body="&lt;p&gt;Years ago there was some hype around &lt;a href=&quot;http://en.wikipedia.org/wiki/Heim_theory&quot; rel=&quot;nofollow&quot;&gt;a theory that would supposedly allow for FTL&lt;/a&gt;. But I have since heard nothing of this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is anything happening with the theory? How would a magnetic field allow us alter spacial dimensions?&lt;/p&gt;&#xA;" OwnerUserId="486" LastActivityDate="2010-11-17T00:14:36.323" ClosedDate="2010-11-18T10:53:30.917" Title="Heim Theory and FTL" Tags="&lt;electromagnetism&gt;&lt;theory&gt;&lt;faster-than-light&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="925" PostTypeId="1" AcceptedAnswerId="949" CreationDate="2010-11-16T16:26:04.680" Score="2" ViewCount="243" Body="&lt;p&gt;An alkali atom has a single outer electron that interacts with incoming photons of the right wavelength (for alkalies it's in the visible &amp;amp; IR range). If there is an external magnetic field, the electron has a well defined quantization axis and the the incoming light can be separated into three components: linearly polarized and left/right-handed circular polarized parts. The interaction strength between the different atomic energy levels (levels described by different quantum numbers) and each of the polarization components is relatively straightforward to calculate (with the appropriate Clebsch-Gordan coefficients and 6-j symbols).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is then the situation if there is no external magnetic field to provide the quantization axis? Is &quot;polarization&quot; meaningful from the atoms point of view? How one sets out to calculate the interaction strength for the different polarizations in the laboratory frame?&lt;/p&gt;&#xA;" OwnerUserId="472" LastActivityDate="2010-11-16T21:16:17.823" Title="Alkali atom - photon interaction in zero magnetic field" Tags="&lt;quantum-mechanics&gt;&lt;electromagnetism&gt;&lt;quantum-optics&gt;" AnswerCount="2" />
  <row Id="927" PostTypeId="2" ParentId="917" CreationDate="2010-11-16T16:35:16.357" Score="1" ViewCount="" Body="&lt;p&gt;either: it comes down very close to you, so the lightning will probably go through your head and fry you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or: it comes down some distance away and dissipates rather quickly. salt water is probably even more conductive than your body, so the current might even flow right around you. I doubt it would make much of a difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the danger from thunderstorms while you're in the water is really that your head is the most elevated thing in a large area, not the conductivity of water.&lt;/p&gt;&#xA;" OwnerUserId="489" LastActivityDate="2010-11-16T16:35:16.357" />
  <row Id="928" PostTypeId="2" ParentId="893" CreationDate="2010-11-16T16:40:36.710" Score="3" ViewCount="" Body="&lt;p&gt;Back when I was in my final two years of high school, there are a handful of books (the first two with co-incidentally similar titles!) that I remember reading and enjoying:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0132366789&quot; rel=&quot;nofollow&quot;&gt;The Theory of Almost Everything&lt;/a&gt; by &lt;em&gt;Robert Oerter&lt;/em&gt; - a great in-depth popular science book on 20th century physics and the goals of unification.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0552997048&quot; rel=&quot;nofollow&quot;&gt;A Short History if Nearly Everything&lt;/a&gt; by &lt;em&gt;Bill Bryson&lt;/em&gt; - very much a popular science book (covering many aspects of science from physics to geology to taxonomy). As from any Bill Bryson book, you can expect a lot of humour, storytelling, and generally an entertaining read. The science content is less than some, but still worthwhile.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0140289208&quot; rel=&quot;nofollow&quot;&gt;Godel, Escher, Bach&lt;/a&gt; (GEB) by &lt;em&gt;Douglas Hofstadter&lt;/em&gt; - this book is probably famous enough not to need an introduction. I actually read it in my earlier university years, but it's even more suitable for a keen final-year high-school student I'd think. It explores the very nature of the consciousness, thought, complexity, and beauty - and gives mind-opening insights fields as mathematical logic, music, art, AI, and physics.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13" LastActivityDate="2010-11-16T16:40:36.710" CommunityOwnedDate="2010-11-18T19:13:02.650" />
  <row Id="929" PostTypeId="2" ParentId="916" CreationDate="2010-11-16T16:43:08.140" Score="4" ViewCount="" Body="&lt;p&gt;As other people answered, the force is a vector quantity, so the net force at the center is indeed zero. But it seems what you have in mind is actually the pressure (stress) at that particular point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, regarding your question. What you are computing is&#xA;$$ \sum_{\substack{i\in\text{first part}\\j\in\text{second part}}} \mathbf F_{ij} $$&#xD;&#xA;the total force acting on each particle on the first part due to the second part. This is not equivalent to &quot;force at X&quot;. But anyway, but why do you get infinity? Because &lt;em&gt;this is the answer&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we compute the gravitational field due to the 2nd part ($\lambda=M/L$):&#xA;$$ g(h) = G\lambda \int_0^{L/2} \frac{1}{(z+h)^2}dz = \frac{G\lambda L}{h(2h+L)}, $$&#xD;&#xA;we find that the gravitational force scales as $\frac1h$. Now suppose you split the first part into $N$ blobs. The position of the $i$-th blob closest to the center will be $h_i=\frac{Li}{2N}$, its mass will be $\frac{\lambda L}{2N}$, so together the contribution of force of each blob will be $F_i = m_ig(h_i) \sim \frac{1/N}{i/N} = \frac1i$, and the &lt;em&gt;total&lt;/em&gt; force is $\sim\sum_{i=1}^N\frac1i\sim\ln N$ which diverges logarithmically to $\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sounds unphysical, but this is because the setting itself is unphysical — you can't have a mass with no &quot;width&quot; but finite mass, this makes the density infinite, and this causes the infinite force as the two part are right next to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's makes sense when the bar has size. So assume that bar is a cylinder of radius $\epsilon \to 0$. Also assume the bar's density is $\rho = M/\pi\epsilon^2L$. The gravity field along the z direction is then&#xA;$$ g(h) = G\rho \int_0^{L/2} \int_0^{2\pi} \int_0^{\epsilon} \frac{h+z}{((h+z)^2+r^2)^{3/2}} rdrd\phi dz = G\rho\pi  \left(2 \sqrt{h^2+\epsilon ^2}-\sqrt{(2 h+L)^2+4 \epsilon ^2}+L\right), $$&#xD;&#xA;and the force is now bounded above by a constant $G\rho\pi(2\epsilon+L-\sqrt{4\epsilon^2+L^2})$ for varying $h$, and the total force by partitioning the 1st part would be $\sim \sum_{i=1}^N\frac{\text{constant}}N = \text{constant}$ as you would expect. (Nevertheless, this constant will diverge with $1/\epsilon$ as $\epsilon\to0$.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;But the result above has nothing to do with the pressure. At precisely the center, there is no force, and the magnitude of force scales as $r$ in its immediate neighborhood. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's mark a small square membrane of size $a\times a\times dz$ with negligible mass $a^2\rho dz$ at position $z\to0$ above the center of the bar to record the pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The membrane will experience a &lt;em&gt;net&lt;/em&gt; push from above. Since the membrane is small, we could assume it is a uniform force. The total force that pushes the membrane downwards is&#xA;$$\begin{aligned} F &amp;amp;= a^2\rho dz g_{\text{total}}(z) \\&#xD;&#xA;&amp;amp;= a^2\rho dz\cdot G\rho\pi \left(\sqrt{(L+2z)^2+4\epsilon^2}-\sqrt{(L-2z)^2+4\epsilon^2}\right) \\&#xD;&#xA;&amp;amp;\sim  \frac{4a^2z\rho dz\cdot G\rho L\pi }{\sqrt{L^2+4\epsilon^2}}&#xD;&#xA; \end{aligned}$$&#xD;&#xA;and thus the pressure is &#xA;$$ dP = -\frac F{a^2} = -\frac{4zG\rho^2 L\pi dz }{\sqrt{L^2+4\epsilon^2}} \sim 4zdz \frac{GM^2}{L^2\epsilon^4} $$&#xD;&#xA;$$ \implies P \sim -z^2 \frac{GM^2}{2L^2\epsilon^4} $$&#xD;&#xA;note that the $z$ is still here. We see that the pressure is &lt;em&gt;finite&lt;/em&gt; and maximal when we measure it &lt;em&gt;precisely at the center&lt;/em&gt; (and decreases away from it). But this will diverge if we let $\epsilon\to0$.&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-11-17T09:51:23.797" LastActivityDate="2010-11-17T09:51:23.797" CommentCount="13" />
  <row Id="930" PostTypeId="2" ParentId="917" CreationDate="2010-11-16T17:01:50.667" Score="5" ViewCount="" Body="&lt;p&gt;One crude way to look at this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there are $N$ wires each with resistance $R$ connected in parallel and common potential difference across the wires is $V$. So current through each wire is $I = \frac{V}{NR}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume hypothetical wire formed by sea water of length $L$ and cross sectional area $S$. Number of such wires in the hemisphere of radius $L$ is approximately $\frac{2\pi L^2}{S}$. Resistance of such a wire will be $\frac{\rho L}{S} $ where $\rho$ is resistivity of sea water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of such wires that can be connected to your body (with area $A$) is $\frac{A}{S}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So approximate current that your body will face is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = \frac{\frac{A}{S}V}{\frac{2\pi L^2}{S} \frac{\rho L}{S}} = \frac{AVS}{2\pi\rho L^3}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assuming $L=100, \rho=0.25, A=1, V=100M, S=10^{-2}, \rho=0.25$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = 0.6 Amp$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most important thing is $I \propto A$ and $I \propto \frac{1}{L^3}$. &lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-11-17T07:31:48.377" LastActivityDate="2010-11-17T07:31:48.377" CommentCount="2" />
  <row Id="931" PostTypeId="1" CreationDate="2010-11-16T17:14:17.223" Score="5" ViewCount="364" Body="&lt;p&gt;Recently a &lt;a href=&quot;http://www.fasterthanthewind.org/2010/09/well-we-exceeded-our-goal.html&quot; rel=&quot;nofollow&quot;&gt;group&lt;/a&gt; set a record for sailing a wind-powered land vehicle directly down wind, and a speed faster than wind speed. Wikipedia has a &lt;a href=&quot;http://en.wikipedia.org/wiki/Sailing_faster_than_the_wind#Sailing_dead_downwind_faster_than_the_wind&quot; rel=&quot;nofollow&quot;&gt;page&lt;/a&gt; talking about it, but it doesn't explain exactly how it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can imagine some configurations, but I don't get how a system could be set up to push forward with a tailwind (when the cart is moving slower than the wind) and push forward with a head wind (when the cart is moving faster than the wind).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone help explain how this is possible?&lt;/p&gt;&#xA;" OwnerUserId="488" LastEditorUserId="488" LastEditorDisplayName="" LastEditDate="2010-11-16T19:29:00.913" LastActivityDate="2010-11-17T05:46:16.867" Title="How to sail downwind faster than the wind?" Tags="&lt;mechanics&gt;&lt;forces&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="4" />
  <row Id="932" PostTypeId="1" CreationDate="2010-11-16T17:17:33.533" Score="1" ViewCount="296" Body="&lt;p&gt;like 110, 220 volts e.t.c?&lt;/p&gt;&#xA;" LastActivityDate="2010-11-16T17:49:36.920" ClosedDate="2010-11-16T20:32:04.390" Title="why is the electric power voltages multiples of 11?" Tags="&lt;electricity&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="933" PostTypeId="2" ParentId="900" CreationDate="2010-11-16T17:17:49.330" Score="3" ViewCount="" Body="&lt;p&gt;Depends on the kind of object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A really &quot;hard&quot; object (i.e. a crystal) will break just as easy when it's warm, but most other materials are flexible and/or soft, so they give way before breaking. Cooling them lowers that flexibility, so e.g. a cold string of rubber will snap much earlier, but a string of crystal will have snapped when stretched that far more or less regardless of it's temperature.&lt;/p&gt;&#xA;" OwnerUserId="489" LastActivityDate="2010-11-16T17:17:49.330" />
  <row Id="934" PostTypeId="2" ParentId="931" CreationDate="2010-11-16T17:21:46.077" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://terrytao.wordpress.com/2009/03/23/sailing-into-the-wind-or-faster-than-the-wind/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;By Terry Tao.&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sailing_faster_than_the_wind&quot; rel=&quot;nofollow&quot;&gt;Sailing faster than the wind&lt;/a&gt; is possible.&#xA;Good explanation @ &lt;a href=&quot;http://www.phys.unsw.edu.au/~jw/sailing.html&quot; rel=&quot;nofollow&quot;&gt;The physics of sailing&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-11-16T19:14:32.080" LastActivityDate="2010-11-16T19:14:32.080" CommunityOwnedDate="2010-11-17T18:27:01.813" CommentCount="6" />
  <row Id="935" PostTypeId="2" ParentId="884" CreationDate="2010-11-16T17:21:59.950" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0679776311&quot; rel=&quot;nofollow&quot;&gt;Roger Penrose: The Road to Reality: A Complete Guide to the Laws of the Universe&lt;/a&gt;, before they start studying. If that doesn't discourage you you're up to the task.&lt;/p&gt;&#xA;" OwnerUserId="489" LastActivityDate="2010-11-16T17:21:59.950" CommunityOwnedDate="2010-11-16T19:26:35.393" CommentCount="3" />
  <row Id="936" PostTypeId="2" ParentId="932" CreationDate="2010-11-16T17:37:05.720" Score="0" ViewCount="" Body="&lt;p&gt;I guess voltages have the values that they have for historical reasons.&lt;br&gt;&#xA; I learned this from &lt;a href=&quot;http://uk.answers.yahoo.com/question/index?qid=20100221075203AANuZ1O&quot; rel=&quot;nofollow&quot;&gt;yahoo answers :)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-11-16T17:37:05.720" />
  <row Id="937" PostTypeId="1" AcceptedAnswerId="943" CreationDate="2010-11-16T17:39:59.213" Score="36" ViewCount="1969" Body="&lt;p&gt;My understanding is that light can not escape from within a black hole (within the event horizon).  I've also heard that information cannot propagate faster than the speed of light.  It would seem to me that the gravitational attraction caused by a black hole carries information about the amount of mass within the black hole.  So, how does this information escape?  Looking at it from a particle point of view: do the gravitons (should they exist) travel faster than the photons?&lt;/p&gt;&#xA;" OwnerUserId="493" LastEditorUserId="2352" LastEditorDisplayName="" LastEditDate="2011-03-02T12:02:47.337" LastActivityDate="2011-03-02T12:02:47.337" Title="How does gravity escape a black hole?" Tags="&lt;general-relativity&gt;&lt;gravity&gt;&lt;black-holes&gt;&lt;speed-of-light&gt;" AnswerCount="14" CommentCount="13" FavoriteCount="7" />
  <row Id="939" PostTypeId="2" ParentId="838" CreationDate="2010-11-16T17:47:02.913" Score="2" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;calculate the size of the earth e.g. through vertical sticks (what &lt;a href=&quot;http://en.wikipedia.org/wiki/Eratosthenes&quot; rel=&quot;nofollow&quot;&gt;Erathostenes&lt;/a&gt; did) (or just assume it, this requires travelling)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;measure the acceleration of gravity at the earth's surface and calculate the earth's mass (this is cheating by assuming newton's law of gravity, which was developed astronomically)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;assume the moon's mass is small and calculate its distance from its orbital period. confirm the orbit is roughly circular by measuring the moons position at a specific time of day over 1 month (or so).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;in principle the sun's distance can then be determined by measuring how much of the moon is lit on the night exactly between new and full moon, thus measuring the angles of the sun-moon-earth triangle (greeks did it, but this is probably pretty hard to get right)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;sun's and moon's diameter can be determined through their viewing angle&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;sun's mass can be calculated through the earth's orbital period&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="489" LastActivityDate="2010-11-16T17:47:02.913" CommentCount="2" />
  <row Id="940" PostTypeId="2" ParentId="932" CreationDate="2010-11-16T17:49:36.920" Score="0" ViewCount="" Body="&lt;p&gt;well actually, whereas the normal voltage is 220V, high voltage is 360V and 480V (at least here in italy)..&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2010-11-16T17:49:36.920" />
  <row Id="941" PostTypeId="2" ParentId="937" CreationDate="2010-11-16T17:53:21.317" Score="2" ViewCount="" Body="&lt;p&gt;Gravitation doesn't work the way light does (which is why quantum gravity is &lt;em&gt;hard&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A massive body &quot;dents&quot; space and time, so that, figuratively speaking, light has a hard time running uphill. But the hill itself (i.e. the curved spacetime) has to be there in the first place.&lt;/p&gt;&#xA;" OwnerUserId="489" LastActivityDate="2010-11-16T17:53:21.317" CommentCount="2" />
  <row Id="942" PostTypeId="2" ParentId="937" CreationDate="2010-11-16T17:53:32.133" Score="1" ViewCount="" Body="&lt;p&gt;The simple answer is that, in every practical circumstance, gravity is considered not a particle but as a warping of spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use the old bowling-ball-on-a-bed example, the bowling ball creates a depression in the bed that pulls other objects towards it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, gravity is geometric, and not limited by transmission of particles.&lt;/p&gt;&#xA;" OwnerUserId="71" LastActivityDate="2010-11-16T17:53:32.133" CommentCount="1" />
  <row Id="943" PostTypeId="2" ParentId="937" CreationDate="2010-11-16T17:55:40.360" Score="26" ViewCount="" Body="&lt;p&gt;Well, the information doesn't have to escape from inside the horizon, because it is not inside. The information is on the horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to see that is from the fact that from the perspective of an observer outside the horizon of a black hole, nothing ever crosses the horizon. It asymptotically gets to the horizon in infinite time (as it is measured from the perspective of an observer at infinity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An other way to see that is the fact that from the boundary conditions on the horizon you can get all the information you need to describe the space-time outside but that is something more technical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, since classical GR is a geometrical theory and not a quantum field theory*, gravitons is not the appropriate way to describe it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*To clarify this point, GR can admit a description in the framework of gauge theories like the theory of electromagnetism. But even though electromagnetism can admit a second quantization (and be described as a QFT), GR can't.&lt;/p&gt;&#xA;" OwnerUserId="383" LastEditorUserId="383" LastEditorDisplayName="" LastEditDate="2010-11-19T11:09:37.723" LastActivityDate="2010-11-19T11:09:37.723" CommentCount="8" />
  <row Id="944" PostTypeId="1" CreationDate="2010-11-16T17:59:34.267" Score="3" ViewCount="272" Body="&lt;p&gt;In classical mechanics, we had Newton's law of gravity $F \propto \frac{Mm}{r^2}$.  Because of this, all laws of classical electrostatics applied to classical gravity if we assumed that all charges attracted each other due to Coulomb's law being analogous.  We can &quot;tweak&quot; classical electrostatics to fit gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In modern physics, does the reverse work?  Can we &quot;tweak&quot; General Relativity to accurately describe electrostatics or even electromagnetism?&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="71" LastEditorDisplayName="" LastEditDate="2010-11-17T21:20:02.210" LastActivityDate="2010-11-17T21:20:02.210" Title="Noticing that Newtonian gravity and electrostatics are equivalent, is there also a relationship between the general relativity and electrodynamics?" Tags="&lt;electromagnetism&gt;&lt;general-relativity&gt;&lt;gravity&gt;&lt;electrodynamics&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="2" />
  <row Id="945" PostTypeId="2" ParentId="931" CreationDate="2010-11-16T18:05:03.360" Score="-1" ViewCount="" Body="&lt;p&gt;The reason is in the mechanism of the force that drives the sailboat. The mechanism is not the pressure that the wind excerpts on the sail, but the lift that is created by the wind flow around the sail. So it is the Bernoulli principle that is powering the sailboats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: Sorry, I didn't notice that you said downwind...&lt;/p&gt;&#xA;" OwnerUserId="383" LastEditorUserId="383" LastEditorDisplayName="" LastEditDate="2010-11-16T18:15:27.653" LastActivityDate="2010-11-16T18:15:27.653" CommentCount="1" />
  <row Id="946" PostTypeId="1" AcceptedAnswerId="1113" CreationDate="2010-11-16T18:09:38.317" Score="6" ViewCount="60" Body="&lt;p&gt;Quite a technical question!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have measured the Point Spread Function of 100nm fluorescent breads with my Olympus scanning head. I'm two-photon exciting the beads with a wavelength of 800nm and focused in the sample with a 100x with N.A.: 1.4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory suggests me the resolution of such a system to be:&#xA;$$d=\frac{0.7\cdot\lambda}{N.A.}\approx 400nm$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that value should be equal (with at most a 18% correction) to the FWHM of the 2D gaussian I obtain on the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But from my analysis of the images I obtain a FWHM close to 2um. Now surely the formula is for an &lt;em&gt;ideal&lt;/em&gt; optical setup but a factor five seems to me quite strange! Is that possible to obtain such a result, in the case of &lt;strong&gt;very not ideal&lt;/strong&gt; optical elements, or should I look for some sort of problem with the acquisition sw that tells me the pixel dimension of the images? &lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2010-11-19T18:19:23.177" Title="PSF Measurements In Fluorescence Imaging" Tags="&lt;laser&gt;&lt;microscopy&gt;&lt;non-linear-optics&gt;&lt;imaging&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="947" PostTypeId="2" ParentId="937" CreationDate="2010-11-16T18:11:55.470" Score="1" ViewCount="" Body="&lt;p&gt;The black hole does &quot;leak&quot; information, but it is not due to &quot;gravitions, but in the form of the Hawking radiation. It has its basis in quantum mechanics, and is a thermal sort of radiation with extremely low rate. This also means that the black hole is slowly evaporates, but on a time scale that is comparable to the age of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The origin of this radiation can be described in a little bit hand-waving way as such: due to quantum fluctuations, there's particle-antiparticle pair creation going on in the vacuum. If such a pair-creating happens on the horizon, one of the pair can fall into the black hole while the other can escape. To preserve the total energy (since the vacuum fluctuations are around 0) with a particle now flying away, its fallen pair has to have a negative energy from the black hole's point of view, thus it is effectively losing mass. The outside observer perceives this whole process as &quot;evaporation&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This radiation has a distribution as described by a &quot;temperature&quot;, which is inversely proportional to the black hole's mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Might want to check out &lt;a href=&quot;http://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hawking_radiation&lt;/a&gt; and other sources for more details...&lt;/p&gt;&#xA;" OwnerUserId="472" LastActivityDate="2010-11-16T18:11:55.470" />
  <row Id="948" PostTypeId="2" ParentId="384" CreationDate="2010-11-16T18:37:03.237" Score="4" ViewCount="" Body="&lt;p&gt;I found a general, qualitative answer in David Blackstock's book &lt;em&gt;Physical Acoustics&lt;/em&gt;, on page 46:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Impedance is often described as the ratio of a &quot;push&quot; variable $q_p$ (such as voltage or pressure) to a corresponding &quot;flow&quot; variable $q_f$ (such as current or particle velocity).  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I also received &lt;a href=&quot;http://www.quora.com/What-is-a-nice-general-definition-of-impedance-that-applies-to-all-wave-phenomena&quot; rel=&quot;nofollow&quot;&gt;a nice answer&lt;/a&gt; to this question on another Q&amp;amp;A site which expands a bit on this qualitative statement with a quantitative one.  In particular, this answer makes the point that impedance is the ratio (transfer function) of a force applied &lt;em&gt;at a particular point&lt;/em&gt; to the velocity at that point.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose what I am looking for next is an intuitive explanation of the phenomenon of impedance &lt;em&gt;matching&lt;/em&gt; and maximum power transfer.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2010-11-16T18:45:51.887" LastActivityDate="2010-11-16T18:45:51.887" />
  <row Id="949" PostTypeId="2" ParentId="925" CreationDate="2010-11-16T18:41:56.210" Score="3" ViewCount="" Body="&lt;p&gt;Basically, if there is no magnetic field, you are free to chose any quantization axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory, the quantization axis choice can be arbitrary, various axis choice corresponding to different &quot;coordinate choices&quot; in the Hilbert space. However, when you have a magnetic field, the state corresponding to axis different than the magnetic field direction are not eigenstates of the energy, and choosing the &quot;right&quot; axis makes the maths simpler. When you have no field, the different states are degenerate, and every choice is as simple as the others.&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2010-11-16T18:41:56.210" />
  <row Id="950" PostTypeId="2" ParentId="944" CreationDate="2010-11-16T18:42:12.790" Score="3" ViewCount="" Body="&lt;p&gt;First a note of caution: what you are talking about is electro-&lt;strong&gt;statics&lt;/strong&gt;, not dynamics. The reason electrostatics and Newtonian gravitation are similar is that this is really the only possible law that has certain nice properties in 3 dimensions (e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%27s_law&quot; rel=&quot;nofollow&quot;&gt;Gauss' law&lt;/a&gt;). This is of course no definitive proof (the actual proof being based on that these theories are similar approximations of some better theories) but it might give you at least some intuition into the matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you appreciate the point that there might be currents and electromagnetic waves, you must instead consider electrodynamics and then you surely can't &quot;tweak&quot; it to fit Newtonian gravity anymore. Going further, even if you consider gravitational field (as studied in general theory of relativity (GTR)) and electromagnetic field, these concepts are pretty different.&#xA;So in order to give any answers I have to take your question quite liberally and just tell you what are the similarities between the two field theories and whether they can be modeled together. And for this some answers can indeed be given:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitomagnetism&quot; rel=&quot;nofollow&quot;&gt;Gravitomagnetism&lt;/a&gt; is a linear approximation of the GTR that is equivalent to electrodynamics. This is probably the closest answer to your question there is because in your case you approximate electrodynamics with electrostatics and this is then equivalent with Newtonian gravitation. So here you go the other way although on a much higher level of field theories.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can consider electrodynamics on the curved background having an effective model for both of them at once. They are completely compatible and in some regards similar (e.g. by having waves propagating at the speed of light).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is something called &lt;a href=&quot;http://en.wikipedia.org/wiki/Kaluza-Klein_theory&quot; rel=&quot;nofollow&quot;&gt;Kaluza-Klein theory&lt;/a&gt; where you study 5-dimensional space-time in which one dimension models electrodynamics. And, in fact, this turns out to be equivalent to the second point but with an extra field called &lt;a href=&quot;http://en.wikipedia.org/wiki/Radion&quot; rel=&quot;nofollow&quot;&gt;radion&lt;/a&gt;. This is is a baby version of the modern stuff treated in string theory and similar research areas.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Both theories are an example of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauge_theory&quot; rel=&quot;nofollow&quot;&gt;gauge theory&lt;/a&gt; although general relativity is quite special in this regard.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-16T20:38:05.933" LastActivityDate="2010-11-16T20:38:05.933" CommentCount="1" />
  <row Id="951" PostTypeId="2" ParentId="944" CreationDate="2010-11-16T19:10:31.523" Score="17" ViewCount="" Body="&lt;p&gt;The parallel between Gravity and E&amp;amp;M is that both forces are mediated by massless particles, the graviton and the photon, respectively. This, in the end of the day, is the reason why both classical theories look similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, when you &lt;em&gt;really&lt;/em&gt; study what's going on behind the scene, you learn that Gravity is more appropriately described by General Relativity (GR) and ElectroMagnetism is better described by Quantum ElectroDynamics (QED).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resemblance of these two theories, one may say, rests in the fact that both are described by the same mathematical framework: a &lt;a href=&quot;http://en.wikipedia.org/wiki/Principal_bundle&quot; rel=&quot;nofollow&quot;&gt;principle bundle&lt;/a&gt;. In GR's (ie, gravity) case this bundle is a Tangent Bundle (or an $SO(3,1)$-bundle) and in QED's (ie, E&amp;amp;M) case it's a $U(1)$-bundle. The geometric structure is the same, what changes is the &quot;gauge group&quot;, the object that describes the symmetries of each theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under this new sense, then, your question could be posed this way: &lt;em&gt;&quot;Is there a way to modify geometry in order to incorporate both of the symmetries of these two theories?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this question was attacked by &lt;a href=&quot;http://en.wikipedia.org/wiki/Hermann_Weyl&quot; rel=&quot;nofollow&quot;&gt;Hermann Weyl&lt;/a&gt; in his book &lt;a href=&quot;http://books.google.com/books?id=KCgZAQAAIAAJ&quot; rel=&quot;nofollow&quot;&gt;Space, Time and Matter&lt;/a&gt;, giving birth to what we now call &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauge_theory&quot; rel=&quot;nofollow&quot;&gt;Gauge Theory&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it turns out, Weyl's observations amounts to a slight change on what symmetries we use to describe Gravity: rather than only using $SO(3,1)$, Weyl used a different group of symmetries, called &lt;a href=&quot;http://en.wikipedia.org/wiki/Conformal_symmetry&quot; rel=&quot;nofollow&quot;&gt;Conformal&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Einstein later showed, it turns out that if you try and describe Gravity and E&amp;amp;M using this generalized group of symmetries (under this new geometrical framework of principle bundles) you do not get the appropriate radiation rates for atoms, ie, atoms which we know to be stable (they don't spontaneously decay radioactively) would not be so under Weyl's proposal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this blow, this notion of &lt;em&gt;unifying&lt;/em&gt; Gravity and E&amp;amp;M via a generalization of the geometry (&lt;a href=&quot;http://en.wikipedia.org/wiki/Connection_(principal_bundle)&quot; rel=&quot;nofollow&quot;&gt;principal bundles&lt;/a&gt;) that describes both of them, was put aside: it's virtually impossible to get stable atoms (stability of matter) this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, people tried a slightly different construction: they posited that spacetime was 5-dimensional (rather than 4-dimensional, as we see everyday) and constructed something called &#xA;a &lt;a href=&quot;http://en.wikipedia.org/wiki/Kaluza_Klein&quot; rel=&quot;nofollow&quot;&gt;Kaluza-Klein&lt;/a&gt; theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, rather than encode the E&amp;amp;M symmetries by changing the geometry via the use of the Conformal Group, they changed it by increasing its dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this proposal has its own drawbacks, for instance, the sore thumb that is &lt;em&gt;supradimensionality&lt;/em&gt;, ie, the fact that spacetime is assumed to be 5-dimensional (rather than 4-dim) — there are other technicalities, but let's leave those for later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom-line is that it's proven very hard to describe gravity together with the other forces of Nature. In fact, we can describe the Strong Force, the Wear Force and ElectroMagnetism all together: this is called the &quot;Standard Model of Particle Physics&quot;. But we cannot incorporate gravity in this description, despite decades of trying.&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2010-11-16T19:10:31.523" CommentCount="2" />
  <row Id="952" PostTypeId="2" ParentId="542" CreationDate="2010-11-16T19:10:41.007" Score="3" ViewCount="" Body="&lt;p&gt;You can get a quantitative estimation of the relative probability of 2 macrostates of different entropies $S$ from the $S=k\ln\Omega$ formula, where $k=1.38\times 10^{-23}\ \mathrm{J}/\mathrm{K}$ is the Boltzmann constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $P\propto\Omega\propto e^{-S/k}$ . This means that $\frac{P_1}{P_2}=e^{-\frac{S_1-S_2}{k}}$.The presence of $k$ as a denominator of the exponent ensures that the probability is small as soon as the entropy difference is bigger than a few times $k$, and completely negligible when the entropy difference is as small as $1000k\sim10^{-20}$ J/K. When you allow $S_1-S_2$ to take any reasonable macroscopic value , the you have the insanely small probabilities David Zaslavsky spoke about.&lt;/p&gt;&#xA;" OwnerUserId="373" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-13T03:52:34.797" LastActivityDate="2011-02-13T03:52:34.797" />
  <row Id="953" PostTypeId="2" ParentId="937" CreationDate="2010-11-16T19:23:26.780" Score="3" ViewCount="" Body="&lt;p&gt;I think the best explanation that can be given is this: you have to discern between statical and dynamical properties of the space-time. What do I mean by that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, there are certain space-times that are static. This is for example the case of the prototypical &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_metric&quot; rel=&quot;nofollow&quot;&gt;black-hole solution&lt;/a&gt; of GTR. Now, this space-time exists a priori (by definition of static: it always was there and always will be), so the gravity doesn't really need to propagate. As GTR tells us gravity is only an illusion left on us by the curved space-time. So there is no paradox here: black holes appear to be gravitating (as in producing some force and being dynamical) but in fact they are completely static and no propagation of information is needed. In reality we know that black-holes are not completely static but this is a correct first approximation to that picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to address the dynamical part, two different things can be meant by this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Actual global change of space-time as can be seen e.g. in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Metric_expansion_of_space&quot; rel=&quot;nofollow&quot;&gt;expansion of the universe&lt;/a&gt;. This expansion need not obey the speed of light but this is in no contradiction with any known law. In particular you cannot send any superluminal signals. In fact, opposite is true: by too quick an expansion parts of universe might go too far away for even their light to ever reach us. They will get causally disconnected from our sector of space-time and to us it will appear as if it never existed. So it shouldn't be surprising that no information can be communicated.&lt;/li&gt;&#xA;&lt;li&gt;Gravitational waves, which is a just fancy name for the disturbances in the underlying space-time. They obey the speed of light and the corresponding quantum particles are called gravitons. Now these waves/particles indeed wouldn't be able escape from underneath the horizon (in the precisely the same way as any other particle, except for &lt;a href=&quot;http://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow&quot;&gt;Hawking radiation&lt;/a&gt;, but this is a special quantum effect).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-16T19:30:48.913" LastActivityDate="2010-11-16T19:30:48.913" />
  <row Id="954" PostTypeId="2" ParentId="937" CreationDate="2010-11-16T19:32:11.380" Score="10" ViewCount="" Body="&lt;p&gt;Let's get something out of the way: let's agree not to bring gravitons into this answer. The rationale is simple: when you talk about gravitons you imply a whole lot of things about quantum phenomena, none of which is really necessary to answer your main question. In any case, gravitons propagate with the very same speed as photons: the speed of light, $c$. This way we can focus simply in Classical GR, ie, the Differential Geometry of Spacetime: this is more than enough to address your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this setting, GR is a theory that says how much curvature a space &quot;suffers&quot; given a certain amount of mass (or energy, cf &lt;a href=&quot;http://en.wikipedia.org/wiki/Stress-energy_tensor&quot; rel=&quot;nofollow&quot;&gt;Stress-Energy Tensor&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_hole&quot; rel=&quot;nofollow&quot;&gt;Black Hole&lt;/a&gt; is a region of spacetime that has such an intense curvature that it &quot;pinches out&quot; a certain region of spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this sense, it's not too bad to understand what's going on: if you can measure the curvature of spacetime, you can definitely tell whether or not you're moving towards a region of increasing curvature (ie, towards a block hole).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is exactly what's done: one measures the curvature of spacetime and that's enough: at some point, the curvature is so intense that the light-cones are &quot;flipped&quot;. At that exact point, you define the &lt;a href=&quot;http://en.wikipedia.org/wiki/Event_horizon&quot; rel=&quot;nofollow&quot;&gt;Event Horizon&lt;/a&gt;, ie, that region of spacetime where causality is affected by the curvature of spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how you make a map of spacetime and can chart black holes. Given that curvature is proportional to gravitational attraction, this sequence of ideas completely addresses your doubt: you don't have anything coming out of the black hole, nor anything like that. All you need is to chart the curvature of spacetime, measuring what happens to your light-cone structure. Then, you find your Event Horizon and, thus, your black hole. This way you got all the information you need, without having anything coming out of the black hole.&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2010-11-16T19:32:11.380" CommentCount="15" />
  <row Id="955" PostTypeId="2" ParentId="907" CreationDate="2010-11-16T19:44:27.170" Score="2" ViewCount="" Body="&lt;p&gt;If you look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Instanton&quot; rel=&quot;nofollow&quot;&gt;(anti-)self-duality&lt;/a&gt;, you get exactly inflection points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you can think in terms of the &lt;a href=&quot;http://en.wikipedia.org/wiki/WKB_approximation&quot; rel=&quot;nofollow&quot;&gt;WKB-approximation&lt;/a&gt; of Feynman Path Integrals, and classify your extrema accordingly. Behind this, there's a whole story about Morse Theory and manifold classification/decomposition, etc... but, the bottom line is that you can think of these WKBs of the Path Integral as the tool to classify the extrema of your Action.&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2010-11-16T19:44:27.170" />
  <row Id="956" PostTypeId="2" ParentId="907" CreationDate="2010-11-16T20:01:19.880" Score="0" ViewCount="" Body="&lt;p&gt;This isn't an action from mechanics, but in gravitational lensing we look for stationary points of the time travel of light.  GR basically tells us that light travels at different speeds depending on the gravitational potential.  (This is distant light, not local right here in our lab.)  If light from a distant quasar travels through a galaxy before reaching us, it experiences a varying gravitational potential depending on the mass distribution of the galaxy and the path the light takes.  Different paths for the light then take different amounts of time.  Looking in the sky, we can associate a travel time with each angle in the sky.  Then we make a 2D map of travel time versus point in the sky. &#xA;Here's an example from a quick search (the source is &lt;a href=&quot;http://iopscience.iop.org/1538-3881/125/6/2769/202407.text.html&quot; rel=&quot;nofollow&quot; title=&quot;here&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ivM3i.gif&quot; alt=&quot;contour diagram for gravitational lensing&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lines connect parts of the sky that have equal travel time.  The different panels show qualitatively different types of gravitational lensing systems.  If you look at the upper-right panel, you see five dots, which are five distinct images of a distant quasar.  The ones at intersections of contour lines are saddle points.  The others are either local minima or maxima of travel time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference in the time it takes light to reach Earth from different images can be dramatic - weeks or even months!&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-16T20:01:19.880" />
  <row Id="957" PostTypeId="2" ParentId="884" CreationDate="2010-11-16T20:33:12.283" Score="8" ViewCount="" Body="&lt;p&gt;I think everyone needs to read Feynman's &lt;a href=&quot;http://en.wikipedia.org/wiki/QED%3a_The_Strange_Theory_of_Light_and_Matter&quot; rel=&quot;nofollow&quot;&gt;QED: The Strange Theory of Light and Matter&lt;/a&gt;. While there are many great books on QFT, this one shows you the inner workings of the microscopic world like none other. It also gets bonus points for being accessible to basically everyone.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-11-16T20:33:12.283" CommunityOwnedDate="2010-11-16T20:33:12.283" CommentCount="1" />
  <row Id="958" PostTypeId="2" ParentId="925" CreationDate="2010-11-16T21:16:17.823" Score="2" ViewCount="" Body="&lt;p&gt;The choice of the axes is arbitrary, but if you're dealing with light incident from a well-defined direction, that naturally picks out a preferred axis. In the zero field case, you would take the direction of the incoming light beam as your quantization axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The polarization of the light is meaningful even without an external field, because photons carry angular momentum that depends on the polarization. That's why the different polarizations drive different transitions in the atoms: absorbing right-hand circular polarization increases the angular momentum of the electron by one unit (in addition to raising it to a higher energy state), and absorbing left-hand circular polarization decreases the angular momentum of the electron by one unit, with the angular momentum being directed along the direction of propagation of the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're dealing with light incident from multiple different directions, well, that's kind of a mess. Each individual interaction would have its own quantization axis, and you would need to do the appropriate projection of the state for the next interactions. And, of course, unless extreme care is taken to shield out stray magnetic fields, whatever small local field you may have will pick out a preferred direction, which also complicates matters. This is why people doing experiments where they really care about the distribution over Zeeman sublevels tend to apply a known field, just to make life easier.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-16T21:16:17.823" />
  <row Id="960" PostTypeId="2" ParentId="916" CreationDate="2010-11-16T22:10:04.947" Score="6" ViewCount="" Body="&lt;p&gt;The reason you encountered higher and higher pressure at the center of the rod as you cut it into more pieces is that you were essentially approximating an integral, but the integral diverges (&quot;is infinity&quot; colloquially).  When the rod has zero thickness, but still has mass, the density of the matter is infinite, and this leads to infinitely strong gravitational forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer this question, we'll imagine the cylinder has some small, finite radius $R$.  We want to find the force between the two halves of the cylinder.  We'll let one half just sit stationary in space.  It will create a gravitational potential.  Then we'll grab the other half and pull it away to some distance $d$.  The gravitational potential energy is a function of $d$.  The force between the two halves of the cylinder is the derivative of the gravitational potential energy with respect to $d$ when $d=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem described above is too hard.  It is quite difficult to calculate the gravitational potential of a cylinder at an arbitrary point.  The gravitational potential of a point mass is just $-Gm/r$, but for a cylinder that extends out in three dimensions, we need to replace $m$ with the density $\rho$ and then integrate over the mass of the entire cylinder.  The expression for $r$, the distance from an arbitrary point outside the cylinder to a point inside it, is not very tractable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, at a point on the axis of the cylinder, the gravitational potential is more accessible due to the extra symmetry.  If we set up cylindrical coordinates with the axis of the cylinder along the z-axis, and then integrate over the bottom half of the cylinder, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V(z) = \int_{z&amp;#39;=0}^{-L/2}\int_{r=0}^{R}\int_{\theta=0}^{2\Pi} \frac{G\rho}{\sqrt{(z-z&amp;#39;)^2+r^2}} r\textrm{d}\theta\textrm{d}r\textrm{d}z&amp;#39;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and doing the integral of $\theta$ it's&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V(z) = 2\pi G\rho\int_{z&amp;#39;=0}^{-L/2}\int_{r=0}^{R} \frac{1}{\sqrt{(z-z&amp;#39;)^2+r^2}} r\textrm{d}r\textrm{d}z&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This allows us to make an approximation.  Although the half of the cylinder we use to calculate the potential must have finite width, we can calculate the potential energy by assuming that the other half of the cylinder is located perfectly along the axis.  As long as the radius of the cylinder is very small compared to the length, this is a valid approximation.  So the potential energy comes from integrating the previous expression for $V$ along the $z$-axis for the length of the cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't actually want the potential energy, but the derivative of the potential energy.  So we imagine moving the top half up the cylinder up a little bit $dz$, and ask how the potential energy changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving the entire top half of the cylinder up by $dz$ is equivalent to taking a piece of thickness $dz$ and slicing it off the bottom and moving it to the top.  So we really just need to find the difference in the potential between the top and bottom of the top half of the cylinder and multiply by the mass-per-unit-length of the cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force between the two halves of the cylinder is $\frac{M}{L}[V(L/2) - V(0)]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That still leaves two integrals to evaluate.  $V(L/2)$ is easy, because it's far away from the half of the cylinder providing the gravitational potential (compared to $R$).  That lets us approximate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V(L/2) = \frac{-GM}{L} \int_{-L/2}^0 \frac{1}{L/2-x}dx = -\frac{GM}{L}\ln 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral for $V(0)$ is trickier, so I put it in Mathematica and got&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V(0) = -\frac{GM}{L}\textrm{arcsinh}\left(\frac{L}{2R}\right)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the regime we are interested in ($R$ small compared to $L$) the $\sinh(x)$ is just $e^x/2$, so this simplifies to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V(0) = -\frac{GM}{L} \ln\left(\frac{L}{R}\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives a final answer for the force&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = \frac{M^2G}{L^2}\ln\left(\frac{L}{2R}\right)$&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-19T06:52:16.877" LastActivityDate="2010-11-19T06:52:16.877" CommentCount="3" />
  <row Id="961" PostTypeId="1" AcceptedAnswerId="973" CreationDate="2010-11-16T23:50:16.207" Score="6" ViewCount="479" Body="&lt;p&gt;Lighter nuclei liberate energy when undergoing fusion, heavier nuclei when undergoing fission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is it about the nucleus of an Iron atom that makes it so stable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively: Iron has the greatest nuclear binding energy - but why?&lt;/p&gt;&#xA;" OwnerUserId="453" LastActivityDate="2010-11-17T11:19:39.433" Title="Why is the nucleus of an Iron atom so stable?" Tags="&lt;nuclear-physics&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="962" PostTypeId="2" ParentId="907" CreationDate="2010-11-16T23:53:41.050" Score="1" ViewCount="" Body="&lt;p&gt;If you have in mind a mechanical system and if you are really referring to the Hamilton's principle (which tells how to find the actual trajectory the system follows between the &lt;em&gt;fixed&lt;/em&gt; initial and final times at &lt;em&gt;fixed&lt;/em&gt; initial conditions), then I doubt you can find such an example. The reason is that a mechanical system contains the kinetic term in its lagrangian, which is unbounded from above. That is, you can always construct a trajectory $\{x(t), {\dot x}(t)\}$ with as large action as you wish; so there is no maximum, not even a local one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In his answer above, Mark refers to an optical example that sort of maximizes, not minimizes the action. The example can be reformulated in mechanical form using small balls rolling on a surface instead of light rays, but I don't think it really refers to the Hamilton's principle. In this example you send light rays (or small balls) from a fixed initial point into &lt;em&gt;different&lt;/em&gt; directions and search for caustics in balls' trajectories. That is, you vary the initial condition (the direction of initial velocity) and at the same time you assume that each trajectory (for each choice of the velocity) is already known. That's not what the Hamilton's principle is about (there you fix all the initial conditions and solve for the trajectory).&lt;/p&gt;&#xA;" OwnerUserId="59" LastActivityDate="2010-11-16T23:53:41.050" />
  <row Id="963" PostTypeId="1" AcceptedAnswerId="968" CreationDate="2010-11-16T23:56:57.440" Score="0" ViewCount="309" Body="&lt;p&gt;According to general relativity, an observer not in the close proximity of a black hole, observing a mass fall into that black hole, will never see that mass cross event horizon(it will gradually fade but never actually pass through the event horizon) as it requires an infinite amount of time to do so. So, effectively, for the observer, the mass of the object falling in will never reach the middle of the event horizon, or the so called &quot;singularity&quot;, as it will require more-than-infinite amount of time to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then why do we even define such a thing as a singularity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am alright with the singularities in space-time curvature, but not with a finite amount of mass being compressed into zero volume, not because it is silly, because it would never happen in a finite time for a casual observer outside of black hole.&lt;/p&gt;&#xA;" OwnerUserId="504" LastActivityDate="2011-02-07T14:14:07.790" Title="Why do we define such a thing as singularity?" Tags="&lt;general-relativity&gt;&lt;gravity&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="964" PostTypeId="2" ParentId="917" CreationDate="2010-11-17T00:04:20.243" Score="0" ViewCount="" Body="&lt;p&gt;While moving across a conductance, electricity tends to follow the more conductive(i.e. less resistive) path. In the salt-water case, it will, depending on the distance from you, mostly ignore you. However, if you are sufficiently close, since a lightning bolt has such a huge energy output, it can still be enough to fry you even though it mostly ignores you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would feel uncomfortable swimming in a thunderstorm and you should too.&lt;/p&gt;&#xA;" OwnerUserId="504" LastActivityDate="2010-11-17T00:04:20.243" />
  <row Id="965" PostTypeId="2" ParentId="924" CreationDate="2010-11-17T00:14:36.323" Score="2" ViewCount="" Body="&lt;p&gt;As far as I know, you can not alter space-time curvature with any realistic amount of magnetic field(you'd need a planck energy per planck length to rip the fabric, which is equivalent to a black hole anyway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for FTL, it brings up so many problems that it should not really exist in a universe obeying &quot;casuality&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About Heim Theory,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some scientists are so concerned with addressing some specific problems in physics that their resultant theory is in contrary with even every-day phenomena like F=ma or x=vt. They, of course, do not do this just to mess around, it is done to lead some other theories to a reasonable final theory. However, as far as I know, it was not Heim himself that tried to apply FTL travel to Heim theory, it was some other physicist who found a way to implement Heim theory in a way for it to allow FTL travel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Heim theory is not really considered a valuable candidate for replacement of GR&amp;amp;QM.&#xA;-The fact that it proposes FTL does not really mean much for us since it is so lackluster in so many aspects that you can't really trust it.&#xA;-A stronger magnetic field would not really allow for much alteration in any spatial dimension. It would just tear you apart because of diamagnetism of water and strip the info out of your credit cards at a longer distance. Incredibly strong magnetic fields exist near magnetars and FTL phenomena would have been observed near magnetars if there were such a condition.&lt;/p&gt;&#xA;" OwnerUserId="504" LastActivityDate="2010-11-17T00:14:36.323" />
  <row Id="966" PostTypeId="2" ParentId="530" CreationDate="2010-11-17T01:20:16.433" Score="0" ViewCount="" Body="&lt;p&gt;As ptomato points out, even a &quot;very black&quot; object reflects &lt;em&gt;some&lt;/em&gt; light. If the surface is rough (at a small scale), the light is reflected off in all directions, and the object appears, matte. If the surface is smooth, then all of the reflected light will bounce in the same way, so that you can see the reflection of the light source; this reflection will be easier to see against the black surface than it would be on a similar white (and shiny) object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my model is correct, then shiny black should look shiny under direct light, but shiny black and matte black should look very similar under indirect light (in monotonous surroundings).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, matte black objects are shiny when wet, so if my model is correct then that shine must be from the liquid surface. So I predict that the color of the shine will be the color of the liquid. For example, if a white light shines on a matte maroon object, if you wet it with water the reflections will be white, but if you polish it the reflections will be maroon.&lt;/p&gt;&#xA;" OwnerUserId="390" LastActivityDate="2010-11-17T01:20:16.433" />
  <row Id="968" PostTypeId="2" ParentId="963" CreationDate="2010-11-17T01:35:05.117" Score="3" ViewCount="" Body="&lt;p&gt;While the distant observer will never observe the infalling observer crossing the event horizon, the infalling observer will intersect the horizon in a finite, calculable amount of local time.  So, understanding the singularity is important for understanding the fate of that observer, certainly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, if you believe that Hawking radiation is real, then the picture isn't quite so stark--the radius of the black hole will decrease, and it will be possible to see some of the collapsing matter at late times.  Understanding the singularity is critical to understanding what exactly happens when the radius of the black hole becomes very small (say, a radius of one planck length).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, in addition to black hole singularities, you have big bang singularities, which are also a consequence of a general theorem in relativity.  Understanding what happens near the big bang singularity (or what quantum effects may cause the singularity not to happen, and perhaps trigger something like inflation) is one of the central questions in cosmology.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-11-17T01:35:05.117" CommentCount="40" />
  <row Id="969" PostTypeId="2" ParentId="937" CreationDate="2010-11-17T01:44:11.147" Score="5" ViewCount="" Body="&lt;p&gt;The problem here is a misunderstanding of what a particle is in QFT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A particle is an $\bf{excitation}$ of a field, not the field itself.  In QED, if you set up a static central charge, and leave it there a very long time, it sets up a field $E=k{q \over r^2}$.  No photons.  When another charge enters that region, it feels that force.  Now, that second charge WILL scatter and accelerate, and THERE, you will have a $e^{-}-&amp;gt;e^{-}+\gamma$ reaction due to that acceleration, (classically, the waves created by having a disturbance in the EM field) but you will not have a photon exchange with the central charge, at least until it feels the field set up by our first charge, which will happen at some later time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider the black hole.  It is a static solution of einstein's equations, sitting there happily.  When it is intruded upon by a test mass, it already has set up its field.  So, when something scatters off of it, it moves along the field set up by the black hole.  Now, it will accelerate, and perhaps, &quot;radiate a graviton&quot;, but the black hole will only feel that after the test particle's radiation field enters the black hole horizon, which it may do freely.  But nowhere in this process, does a particle leave the black hole horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example of why the naïve notion of all forces coming from a Feynman diagram with two pairs of legs is the Higgs boson--the entire universe is immersed in a nonzero Higgs field.  But we only talk about the 'creation' of Higgs 'particles' when we disturb the Higgs field enough to create ripples in the Higgs field--Higgs waves.  THOSE are the Higgs particles we're looking for in the LHC.  You don't need ripples in the gravitational field to explain why a planet orbits a black hole.  You just need the field to have a certain distribution.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2010-11-18T22:22:10.967" LastActivityDate="2010-11-18T22:22:10.967" CommentCount="0" />
  <row Id="970" PostTypeId="1" AcceptedAnswerId="971" CreationDate="2010-11-17T01:49:49.447" Score="9" ViewCount="345" Body="&lt;p&gt;This question has been bothering me for a while. I have a crude hypothesis...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, an observer falling into a black hole will cross the event horizon at some specific future (proper) time in, and that it will not be a traumatic event if the black hole is big enough (e.g. tidal forces will be quite mild).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the observer will see the universe above &quot;speed up&quot;, and can see any future date arrive at a distant point &lt;em&gt;before crossing the event horizon&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, black holes evaporate, which may lead to some caveats about the previous two statements (which do not take evaporation into account).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So suppose we have a large black hole, destined to evaporate and vanish in the the year 10&lt;sup&gt;50&lt;/sup&gt; AD. And suppose I jump into it, equipped with a telescope that lets me observe the Earth. Before I reach the event horizon I will see 10&lt;sup&gt;50&lt;/sup&gt; AD arrive on earth. At that point I will see astronomers on earth waving flags to indicate that they have seen the black hole vanish. So if I look &quot;down&quot; I will see empty space with no black hole looming. So where am I? If I'm just adrift in space, am I in a cloud of all the other objects that ever fell into the hole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for my crude hypothesis: as I fall, and the hole gets smaller, and the curvature near the horizon gets more acute, I'll be racked by tidal forces and blasted by Hawking radiation. Any extended body I happen to have will be disintegrated, so &quot;I&quot; will survive only if I'm an indestructible point, and the cloud of such particles is what astronomers see as the final flash of Hawking radiation. Is this even close to plausible?&lt;/p&gt;&#xA;" OwnerUserId="390" LastActivityDate="2011-02-28T12:16:04.293" Title="If I fall into an evaporating black hole, where do I end up?" Tags="&lt;general-relativity&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="4" />
  <row Id="971" PostTypeId="2" ParentId="970" CreationDate="2010-11-17T02:03:59.820" Score="6" ViewCount="" Body="&lt;p&gt;A few things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)Just because an observer crossing the event horizon doesn't necessarily feel ill effects AT THE TIME OF CROSSING the horizon, it doesn't mean that they won't inevitably end up at the singularity, where there will be plenty of ill effects--all timelike curves that cross the horizon end up at the singularity in a finite amount of proper time.  For a particle falling into a non-spinning black hole, it's actually the same amount of proper time that it would take to fall into a Newtonian point mass.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) You have to be very careful about what you mean by 'horizon' in the case of a black hole that eventually evaporates.  There are several definitions of 'horizon', and depending on how you resolve the singularity, and upon how the hole evaporates these different definitions can differ in meaning--the most common difference is the apparent horizon-a 'point at which, for this given time, you can't go back', and the event horizon--'the point at which, you MUST end up at the singularity'.  It might be possible that your evaporating black hole spacetime may have an apparent horizon but no event horizon, for instance.  In that case, the whole paradox goes away.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) A careful answer of this requires the careful drawing of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Penrose_diagram&quot; rel=&quot;nofollow&quot;&gt;Penrose-Carter diagram&lt;/a&gt; of the relevant spacetime.  If you managed to tweak it somehow so that you fell in, blasted your rockets for long enough to outlive the recontraction of the horizon, the short answer is that you wouldn't receive all of the information about all of the future, just that determined by the &quot;null past&quot; of the horizon--you would find out about all of the lightlike and timelike rays that fell into the horizon, but not those that would head toward the spot where the horizon used to be at times later than when the horizon was there.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-11-17T02:03:59.820" CommentCount="8" />
  <row Id="972" PostTypeId="2" ParentId="937" CreationDate="2010-11-17T02:20:05.113" Score="2" ViewCount="" Body="&lt;p&gt;The various theories - QED, GTR, classical electromagnetism, quantum loop gravity, etc - are all different ways to describe nature.  Nature is what it is; theories all have defects.  So far as saying whether gravity resembles electromagnetism in some way or not, is just blowing warm air  about how humans think and not saying anything substantial about physical reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what if we don't have a full grasp of quantum gravity?  Gravitons are a sensible concept, and a key part in some unified (or semi-unified) field theories.  It might get tricky because unlike other quantum particles, gravitons are a part of the curvature of spacetime and the relations of nearby lightcones, as they fly through said spacetime. We can sort of ignorer that for now.   The question is good, and can be answered in terms of quantum theory and gravitons.  We just don't know, given the existing state  of physics knowledge, how far we can push the idea.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When charged particle attract or repel, the force is due to  virtual photons.  Photons &lt;em&gt;like&lt;/em&gt; to travel at  the universal speed &lt;i&gt;c&lt;/i&gt;, but they  don't have to.  Heisenberg says so! You can break the laws of conservation of energy and  momentum as much as you like, but the more you deviate, the shorter the time span and smaller the bit of space in which you violate these laws.  For the virtual photons connecting two charged particles, they've got the room  between the two particles, and a time span matching that at lightspeed.  These not running waves with a well-defined wavelength, period or phase velocity.  This ill-defined velocity can be faster than c or less equally well.  In QED, the photon propagator - the wavefunction giving the probability amplitude of a virtual photon connecting (x1, t1) to (x2, t2) is nonzero everywhere - inside and outside the past and future light cones, though becoming unlimited in magnitude on the light cones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So gravitons, if they are that  much like photons, can exist just fine outside the horizon and inside.   They are, in a rough sense, as big as the space between the black hole and whatever is orbiting or falling into it.  Don't  picture them as little energy  pellets flying from the black  hole center (singularity or whatever) - even with Heisenberg's indulgence, it's just not a matter of small particles  trying to get  through the horizon the wrong way.   A graviton is probably already on both sides!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more satisfying answer, I suspect it takes knowing the math,Fourier transforms, Riemann tensors and all that.  &lt;/p&gt;&#xA;" OwnerUserId="353" LastActivityDate="2010-11-17T02:20:05.113" />
  <row Id="973" PostTypeId="2" ParentId="961" CreationDate="2010-11-17T02:20:11.350" Score="18" ViewCount="" Body="&lt;p&gt;It all comes down to a balance between a number of different physical interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The binding energy of a nucleus is commonly described with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Semi-empirical_mass_formula&quot; rel=&quot;nofollow&quot;&gt;semiempirical mass formula&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(A, Z) = a_V A - a_S A^{2/3} - a_C \frac{Z(Z-1)}{A^{1/3}} - a_A \frac{(A-2Z)^2}{A} + \delta(A,Z)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A = Z + N$ is the total number of nucleons, $Z$ the number of protons, and $N$ the number of neutrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The different contributions have physical explanation as:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$a_V$ : volume term, the bigger the volume the more nucleons interact with each other through the strong interaction, the more they attract each other&lt;/li&gt;&#xA;&lt;li&gt;$a_S$ : surface term, similar to the surface tension, some energy stored in there, reducing the binding interaction&lt;/li&gt;&#xA;&lt;li&gt;$a_C$ : the Coulomb repulsion of the protons within the nucleus&lt;/li&gt;&#xA;&lt;li&gt;$a_A$ : asymmetry term, rooted in the Pauli exclusion principle. Basically if there are more of one type of nucleon (generally of neutrons) then the overall energy is larger than needs to be thus decreasing the binding energy (note: $A-2Z = Z - N$)&lt;/li&gt;&#xA;&lt;li&gt;$\delta$ : pairing term, depends on whether there are even or odd number of nucleons altogether and even or odd number of protons/neutrons. In empirical description usually modeled as a continuous variable $a_P/A^{1/2}$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is of the expression for the &lt;strong&gt;total&lt;/strong&gt; binding energy, what is interesting is the binding energy &lt;strong&gt;per nucleus&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(A, Z)/A \approx a_V - a_S \frac{1}{A^{1/3}} - a_C \frac{Z(Z-1)}{A^{4/3}} - a_A \frac{(A-2Z)^2}{A^2} + a_P \frac{1}{A^{3/2}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see which nucleus (what value of $A$) is the most stable one has to find for which $A$ is this function maximal. At this point $Z$ is arbitrary but we should chose a physically meaningful value. From theoretical point of view a good choice is the $Z$ that gives the highest binding energy for a given $A$ (the most stable isotope), for which we need to solve solve $\frac{\partial (E/A)}{\partial Z} = 0$. The results is $Z_{stable}(A) \approx \frac{1}{2}\frac{A}{1+A^{2/3} \frac{a_C}{4 a_A}}$. After putting back the $Z_{stable}(A)$ into $E(A, Z)/A$ one can maximize the function value to get the &quot;optimal number&quot; of nucleons for the most stable element. Depending on the empirically determined values of $a_S, a_C, a_A, a_P$ the maximum will occur in the area $A \approx 58 \ldots 63$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interpretation of this result is something like this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;for &lt;strong&gt;small atoms&lt;/strong&gt; (small $A$) the biggest contribution is the &lt;strong&gt;surface term&lt;/strong&gt; (they have a large surface-to-volume ratio), and they want increase the number of nucleons to reduce it - hence you have fusion&lt;/li&gt;&#xA;&lt;li&gt;for &lt;strong&gt;large atoms&lt;/strong&gt; (large $A$) the &lt;strong&gt;Coulomb term&lt;/strong&gt; increases because more protons mean more repulsion between them, and also, to keep everything together more neutrons are needed (thus $N \gg Z$ which makes the &lt;strong&gt;asymmetry term&lt;/strong&gt; larger as well. By ejecting some nucleons (alpha decay), or converting between neutrons and protons (beta decay) the nucleus can reduce this terms.&lt;/li&gt;&#xA;&lt;li&gt;optimally bound $A$ (and $Z$) happens when these two groups of competing contributions balance each others out.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="472" LastActivityDate="2010-11-17T02:20:11.350" CommentCount="3" />
  <row Id="974" PostTypeId="2" ParentId="886" CreationDate="2010-11-17T02:42:32.203" Score="2" ViewCount="" Body="&lt;p&gt;Hard to tell exactly what the scenario is from that article.  From what is shown, my guess is that the swimmer is doing work in deforming the object, which then moves the object.  Then, the object having moved, the swimmer deforms the object BACK.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this cycle would create zero work classically, in the case of relativity, you are now at a point where the gravitational potential has a different value, and therefore, the work you do to restore the object has been &quot;redshifted&quot; to a different value.  In essence, the 'swimming' scheme converts gravitational potential energy into kinetic energy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that might not be quite what they're doing in this article.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-11-17T02:42:32.203" />
  <row Id="975" PostTypeId="2" ParentId="937" CreationDate="2010-11-17T02:45:27.443" Score="18" ViewCount="" Body="&lt;p&gt;There are some good answers here already but I hope this is a nice short summary:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electromagnetic radiation cannot escape a black hole, because it travels at the speed of light. Similarly, &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_wave&quot; rel=&quot;nofollow&quot;&gt;gravitational &lt;strong&gt;radiation&lt;/strong&gt;&lt;/a&gt; cannot escape a black hole either, because it too travels at the speed of light. If gravitational radiation could escape, you could theoretically use it to send a signal from the inside of the black hole to the outside, which is forbidden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black hole, however, &lt;strong&gt;can&lt;/strong&gt; have an electric charge, which means there is an electric field around it. This is not a paradox because a static electric field is different from electromagnetic radiation. Similarly, a black hole has a mass, so it has a gravitational field around it. This is not a paradox either because a gravitational field is different from gravitational radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say the gravitational field carries information about the amount of mass (actually energy) inside, but that does not give a way for someone inside to send a signal to the outside, because to do so they would have to create or destroy energy, which is impossible. Thus there is no paradox.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2010-11-17T02:45:27.443" CommentCount="3" />
  <row Id="976" PostTypeId="1" CreationDate="2010-11-17T02:49:33.763" Score="1" ViewCount="159" Body="&lt;p&gt;Among common household appliances, things one can make from stuff in the garage  or hardware store, and reasonably safe, e.g. within reach  of hobbyists and high school kids entering a science fair, what is the highest speed of motion of any bit of matter one can obtain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bit of matter, or surface points of some object, should be macroscopic - big enough to see, time the motion  of somehow, make collisions with other bits of matter.  Electrons aren't big enough!   Fluids are fine too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Guns, dynamite would be too dangerous.  The tips of a fan blade or shutter edges in a camera are fine.   Using a drill to make something spin real fast might be okay if all mad  scientists involved are careful.  &lt;/p&gt;&#xA;" OwnerUserId="353" LastEditorUserId="353" LastEditorDisplayName="" LastEditDate="2010-11-17T03:51:22.643" LastActivityDate="2010-11-17T03:51:22.643" ClosedDate="2010-11-18T15:18:35.080" Title="What's the highest speed one can obtain safely at home?" Tags="&lt;experiment&gt;&lt;speed&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="977" PostTypeId="2" ParentId="905" CreationDate="2010-11-17T03:03:28.443" Score="3" ViewCount="" Body="&lt;p&gt;Do you mean the entropy flux density 4-vector, and its 4 divergence, along with their projections onto the plane orthogonal to the energy transport?   See e.g. &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1989A%26A...211..476O&amp;amp;db_key=AST&amp;amp;page_ind=0&amp;amp;data_type=GIF&amp;amp;type=SCREEN_VIEW&amp;amp;classic=YES&quot; rel=&quot;nofollow&quot;&gt;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1989A%26A...211..476O&amp;amp;db_key=AST&amp;amp;page_ind=0&amp;amp;data_type=GIF&amp;amp;type=SCREEN_VIEW&amp;amp;classic=YES&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-11-17T03:03:28.443" />
  <row Id="978" PostTypeId="2" ParentId="976" CreationDate="2010-11-17T03:19:49.327" Score="3" ViewCount="" Body="&lt;p&gt;I'm going to assume that by &quot;any bit of matter&quot; you did not intend to include bits of matter of molecular or atomic size. For if you allow such small objects to be in the running then the higher velocity molecules in the air or in any steam you generate by heating water to boiling (the average of the squared speed is proportional to the absolute temperature) or in the glowing gas molecules comprising a candle flame would be moving at a good clip. But if we exclude such motions and just accept macroscopic or mesoscopic motions, then a reasonable candidate is the speed of the tip of an adroitly snapped rope or whip which exceeds the speed of sound when it produces the characteristic 'Crack!'.&lt;/p&gt;&#xA;" OwnerUserId="345" LastActivityDate="2010-11-17T03:19:49.327" />
  <row Id="979" PostTypeId="2" ParentId="478" CreationDate="2010-11-17T04:52:33.933" Score="2" ViewCount="" Body="&lt;p&gt;Nope, this method is not feasible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are right that the EM field is &quot;Electro&quot;+&quot;Magnetic&quot;, but these contributions are oscillating in time. Every interference they produce will oscillate in time as well, even if the peak magnitude of these oscillation can (theoretically) be high (that is the constructive interference you mentioned).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you just take the magnetic field vector of your $i$th beam at the position of your (pointlike) object:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_i(t) = B_{0i} \sin(2 \pi \nu_i t + \phi_i)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a $n$ waves that constructively interfere, which have to have the same frequency ($\nu_i = \nu$), and the the total magnetic field will be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B(t) = (\sum_{i=1}^n B_i) \sin(2 \pi \nu t) = n B \sin(2 \pi \nu t)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;with the final, very simplified step being true if all the field magnitudes are the same. Thus your field will still have the oscillation at the original frequency (eg. radio frequency (~ tens-hundreds of MHz) or laser beam (hundreds of THz). No macroscopic object would react to this high frequency perturbation due to inertia. And even if they would, the resulting motion wouldn't be a push or a pull, but oscillation on the time scale of $1/\nu$. That's why in general, motion is induced with stationary or slowly moving fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... and this haven't even taken into account a lot of practical limitations to this scheme.&lt;/p&gt;&#xA;" OwnerUserId="472" LastActivityDate="2010-11-17T04:52:33.933" CommentCount="7" />
  <row Id="980" PostTypeId="2" ParentId="575" CreationDate="2010-11-17T05:04:43.987" Score="1" ViewCount="" Body="&lt;p&gt;Everyone else here is right, but there is also an unmentioned complication to using curvature to explain electromagnetic forces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use curvature to explain gravity in an attempt to explain why intertial and gravitational masses are the same--why a bowling ball falls the same rate as a penny.  Einstein's answer was that it's because they're just travelling in 'straight' lines in the same spacetime, and aren't feeling any forces at all--they're just trying to move in the straightest line they can.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But with E&amp;amp;M, not all charges feel the same force in a given electric field--in fact, chargless particles feel no force at all!  So, if we're going to explain E&amp;amp;M using curvature, we're going to need to explain why different particles have different charges.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-11-17T05:04:43.987" />
  <row Id="981" PostTypeId="2" ParentId="917" CreationDate="2010-11-17T05:34:27.887" Score="3" ViewCount="" Body="&lt;p&gt;In fresh water what makes lightening so dangerous to a swimmer is that most of the current travels on the surface of the water, so rather then getting a $1/r^2$ falloff in current density, you see a $1/r$ falloff. Obviously eventually it will be conducted down into the mass of the water, but this takes a many meters. In salt water, this should happen much quicker. I'm not sure how the conductivity of the inside of your body compares to seawater. Even if it is less, some current would still flow through you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For normal dry skin, it takes considerable voltage to penetrate the skin (maybe a hundred volts), wet your skin with saltwater and you'll conduct electricity quite well! As a teenager playing with chemistry and water, that happened to me once, 12 volts AC and ionic solutions made for a pretty nasty shock. Normally 12 volts won't penetrate the skin, so I was unrealistically confident!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a spark generator that makes roughly 20KV sparks (from a capacitor), discharge it into water, and you see surface sparks spread from the point of entry in all directions.&lt;/p&gt;&#xA;" OwnerUserId="391" LastEditorUserId="98" LastEditorDisplayName="" LastEditDate="2011-08-26T15:59:27.283" LastActivityDate="2011-08-26T15:59:27.283" CommentCount="2" />
  <row Id="982" PostTypeId="2" ParentId="931" CreationDate="2010-11-17T05:46:16.867" Score="2" ViewCount="" Body="&lt;p&gt;It does sound counterintuitive. The actual device used gears attached to a wind turbine. Even though the turbine was moving faster than the wind, the thrust (IIRC) was such as to slow down the wind field. So energy was being extracted from the wind field to turn wheels via the right gear ratio. In any case energy is removed from the wind field so no laws of physics were violated. Traveling at angles to the wind is easier to visualize. In fact wind turbine blades travel at right angles to the wind at multiples of the wind speed. But that is different. This was a really clever device, and unless shown how it works (I don't remember the details) most physicists think &quot;no way&quot;. I suspect it might have to be pushed to faster than wind speed to get it going. Without wheels attached to the solid ground it wouldn't work, its basically increasing the wind field coupling to the solid earth.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-11-17T05:46:16.867" CommentCount="2" />
  <row Id="983" PostTypeId="1" AcceptedAnswerId="991" CreationDate="2010-11-17T06:10:57.973" Score="3" ViewCount="193" Body="&lt;p&gt;This (see &lt;a href=&quot;http://www.youtube.com/watch?v=TMCf7SNUb-Q&amp;amp;feature=related&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=TMCf7SNUb-Q&amp;amp;feature=related&lt;/a&gt;) is so cool, I wanted to show it.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a pro forma question for this forum, any idea what is going on with the breaking of larger, presumably vortex related bubble rings and reconnection making the smaller rings?   The dolphins appear to be making smaller vortices by suddenly moving their snouts through a ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a simpler level, why do the rings get larger?   Any other good questions come to mind?&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="67" LastEditorDisplayName="" LastEditDate="2010-11-18T19:11:52.433" LastActivityDate="2010-11-18T19:11:52.433" Title="Why do those dolphin bubble rings I saw in a video behave the way they do?" Tags="&lt;hydrodynamics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="984" PostTypeId="2" ParentId="961" CreationDate="2010-11-17T06:18:59.413" Score="1" ViewCount="" Body="&lt;p&gt;In some sense the nucleus of a Helium (He-4) is more stable, than  the nucleus of an Iron.&#xA;About 20 MeV is required to detach  any particle from the nucleus of a Helium.&#xA;But only about 10 MeV is enough to detach a nucleon from the nucleus of an Iron.&lt;/p&gt;&#xA;" OwnerUserId="238" LastActivityDate="2010-11-17T06:18:59.413" CommentCount="3" />
  <row Id="985" PostTypeId="2" ParentId="885" CreationDate="2010-11-17T06:19:39.013" Score="5" ViewCount="" Body="&lt;p&gt;Considering what Greg Graviton wrote, I'll write out the derivation and see if I can make sense of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S = \int_{t_1}^{t_2} L(q, \dot q, t) dt $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where S is the action and L the Lagrangian. We vary the path and find the extremum of the action:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \delta S = \int_{t_1}^{t_2} ({\partial L \over \partial q}\delta q + {\partial L \over \partial \dot q}\delta \dot q) dt = 0. $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, q and $\dot q$ are varied independently. But then in the next step we use this identity,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \delta \dot q = {d \over dt} \delta q. $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is where the relationship between q and $\dot q$ enters the picture. I think that what is happening here is that q and $\dot q$ are treated as independent initially, but then the independence is removed by the identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \delta S = \int_{t_1}^{t_2} ({\partial L \over \partial q}\delta q + {\partial L \over \partial \dot q}{d \over dt} \delta q) dt = 0 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then follows the rest of the derivation. We integrate the second term by parts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \delta S = \lbrack {\partial L \over \partial \dot q}\delta q\rbrack_{t_1}^{t_2} + \int_{t_1}^{t_2} ({\partial L \over \partial q} - {d \over dt}{\partial L \over \partial \dot q})\delta q dt = 0, $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the bracketed expression is zero because the endpoints are held fixed. And then we can pull out the Euler-Lagrange equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ {\partial L \over \partial q} - {d \over dt}{\partial L \over \partial \dot q} = 0. $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it makes more sense to me. You start by treating the variables as independent but then remove the independence by imposing a condition &lt;em&gt;during&lt;/em&gt; the derivation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that makes sense. I expect in general other problems can be treated the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I copied the above equations from &lt;em&gt;Mechanics&lt;/em&gt; by Landau and Lifshitz.)&lt;/p&gt;&#xA;" OwnerUserId="150" LastEditorUserId="150" LastEditorDisplayName="" LastEditDate="2010-11-17T06:25:16.030" LastActivityDate="2010-11-17T06:25:16.030" CommentCount="7" />
  <row Id="986" PostTypeId="2" ParentId="884" CreationDate="2010-11-17T07:34:28.977" Score="9" ViewCount="" Body="&lt;p&gt;One formative book for me was Ed Purcell's &lt;a href=&quot;http://goo.gl/JjqHs&quot; rel=&quot;nofollow&quot;&gt;Electricity and Magnetism&lt;/a&gt;.  Purcell was my early education in thinking like a physicist.  It introduced me to thought experiments, simple models, and the usefulness of new mathematical tools.  It's mathematically very clear, and physically insightful.  The problems are extremely rich.  It manages a huge deal of physics without much unnecessary computation.  When I post replies to physics questions on this board, I sometimes wonder how Purcell would handle it - setting out the physical principles first, carrying out the calculations as clearly and succinctly as possible, and using physical insight for shortcuts and simplifications.  I doubt I will ever live up to his model, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My entire freshman class read Purcell during second quarter.  In conversation later, I heard over and over from people that after reading chapter 5, &quot;The Fields Of Moving Charges&quot;, they were so awestruck by this 25-page illustrated introduction to relativistic E&amp;amp;M that when they finished, they stared at the wall for about ten minutes, then read the entire thing over again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also from the Berkeley physics series, Frank Crawford's &lt;a href=&quot;http://books.google.com/books?id=JLURAQAAIAAJ&amp;amp;q=crawford+waves&amp;amp;dq=crawford+waves&amp;amp;hl=en&amp;amp;ei=LYPjTIGPDYf6sAPXiaVm&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCUQ6AEwAA&quot; rel=&quot;nofollow&quot;&gt;Waves&lt;/a&gt; is charming, delving into many interesting wave phenomena with very nice explanations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;James Nearing's &lt;a href=&quot;http://www.physics.miami.edu/~nearing/mathmethods/&quot; rel=&quot;nofollow&quot;&gt;free book, Mathematical Methods for Physics&lt;/a&gt; on undergraduate math methods deserves more attention that it gets.  It's clear, insightful, and packed with good problems.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-17T07:34:28.977" CommunityOwnedDate="2010-11-17T07:34:28.977" CommentCount="2" />
  <row Id="987" PostTypeId="2" ParentId="884" CreationDate="2010-11-17T08:20:11.657" Score="2" ViewCount="" Body="&lt;p&gt;There are many books which are very beneficial for every physicist, and it is difficult to judge which ones really make up the must-read list. But in any case the top books in that list, in my opinion, should be not technical ones but the two autobiographic books by Feynman: &quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0393316041&quot; rel=&quot;nofollow&quot;&gt;Surely, you are joking Mr. Feynman&lt;/a&gt;&quot; and &quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0393320928&quot; rel=&quot;nofollow&quot;&gt;What do you care hat other people think&lt;/a&gt;&quot;. These books are about how a physicist sees the world and how much he enjoys seeing it from that perspective.&lt;/p&gt;&#xA;" OwnerUserId="59" LastEditorUserId="429" LastEditorDisplayName="" LastEditDate="2011-07-18T14:32:15.267" LastActivityDate="2011-07-18T14:32:15.267" CommunityOwnedDate="2010-11-17T08:20:11.657" CommentCount="2" />
  <row Id="989" PostTypeId="2" ParentId="887" CreationDate="2010-11-17T09:40:58.017" Score="3" ViewCount="" Body="&lt;p&gt;I asked this very same question at mathoverflow, too (do the policies of PSE have anything against this?), and got one further answer, which I leave to your attention:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/46212/relativistic-cellular-automata/46227#46227&quot; rel=&quot;nofollow&quot;&gt;There are no &quot;non-trivial&quot; finite&#xA;  sub-groups of $O(3,1)$&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="229" LastEditorUserId="229" LastEditorDisplayName="" LastEditDate="2010-11-17T13:51:31.717" LastActivityDate="2010-11-17T13:51:31.717" />
  <row Id="990" PostTypeId="2" ParentId="946" CreationDate="2010-11-17T09:58:04.927" Score="1" ViewCount="" Body="&lt;p&gt;You have many things which can go wrong when you try to go down to the wavelength limit, without resorting to &lt;strong&gt;Very not ideal&lt;/strong&gt; optical element : non-Gaussian beam, wrong wavelength, spherical aberration, etc. I fear there is no easy answer to your question.&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2010-11-17T09:58:04.927" CommentCount="1" />
  <row Id="991" PostTypeId="2" ParentId="983" CreationDate="2010-11-17T10:10:59.637" Score="7" ViewCount="" Body="&lt;p&gt;Heh, I usually show this movie during my presentations.&lt;br&gt;&#xA;First of all, this has nothing to do with surface tension. Those are just vortex rings that are only visualised by dolphins by filling their cores with air (you can see this happening about 0:12).&lt;br&gt;&#xA;At first the rings grow to optimize vortex core size, but the most increase is made by the fact that dolphins make water flow through the ring (you can see they kinda push the rings through the water). The speed of each point on a vortex ring in localised induction approximation is $\dot{\vec{s}}=\frac{\beta}{R}\hat{\mathbf{b}}$, where $\hat{\mathbf{b}}$ is a binormal vector of the ring, $\beta$ is constant and $R$ is the diameter of the ring; so when the dolphin pushes the ring its effective speed decreases and so its diameter must increase. Not counting friction, free ring holds its diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The making of small rings is due to the fact that dolphins distort the ring making it reconnect with itself. Two rings are created in the process, small and big, yet the big degenerates in eddies created by the dolphin swimming past it to play with the smaller one.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-11-17T10:17:05.703" LastActivityDate="2010-11-17T10:17:05.703" CommentCount="3" />
  <row Id="993" PostTypeId="2" ParentId="961" CreationDate="2010-11-17T11:19:39.433" Score="6" ViewCount="" Body="&lt;p&gt;The bonding of nuclei is dominated by 2 main forces - the strong nuclear force, and the electromagnetic force. The strong nuclear force is much stronger than the electromagnetic force, but acts over much shorter distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For small nuclei (eg hydrogen and helium), if you're able to add more nucleons, they are likely to stick due to the attraction of the strong force. This is why smaller nuclei tend to fuse together. Sticking the particles together results in a lower-energy configuration so it is more stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For larger nuclei, the size of the nucleus means particles on one side don't feel much strong force attraction from particles on the other side, but they still feel electromagnetic repulsion (if they are charged, ie protons). This means that larger nuclei are less stable, and can form lower-energy configurations by splitting into smaller parts (fission).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron is at the middle point in terms of nucleus size, where either adding or removing particles would result in a higher-energy configuration, and so it is regarded as the most stable nucleus.&lt;/p&gt;&#xA;" OwnerUserId="474" LastActivityDate="2010-11-17T11:19:39.433" CommentCount="1" />
  <row Id="994" PostTypeId="1" CreationDate="2010-11-17T12:00:13.247" Score="-4" ViewCount="161" Body="&lt;p&gt;By &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Real_space&quot; rel=&quot;nofollow&quot;&gt;real space&lt;/a&gt;&quot; I mean a space in which physical particles move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a color sphere and let a bunch of objects &quot;move&quot; on its surface. &quot;Move&quot; means &quot;change colors&quot;. Let there be some rules governing the changes of color, maybe giving rise to something like inertness or collisions. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Under which conditions will we be willing or&#xA;  forced to think of the movements as&#xA;  real movements in real space?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/zhVN1.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="229" LastEditorUserId="229" LastEditorDisplayName="" LastEditDate="2010-11-17T14:40:33.250" LastActivityDate="2010-11-17T20:38:31.217" Title="What makes a space a real space?" Tags="&lt;classical-mechanics&gt;&lt;space&gt;" AnswerCount="1" CommentCount="20" />
  <row Id="995" PostTypeId="1" AcceptedAnswerId="1022" CreationDate="2010-11-17T12:32:58.727" Score="5" ViewCount="143" Body="&lt;p&gt;I want to simulate an impact between two bodies according to gravity, and eventually considering other forces to stick matter together. I'd like to use python to do this, but I am open to alternatives. What kind of tools or libraries can I use to perform this task ?&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="217" LastEditorDisplayName="" LastEditDate="2010-11-18T23:39:12.663" LastActivityDate="2010-11-18T23:39:12.663" Title="Simulate a physical impact of objects made of finite, small elements" Tags="&lt;gravity&gt;&lt;simulation&gt;&lt;computational-physics&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="996" PostTypeId="2" ParentId="763" CreationDate="2010-11-17T12:34:30.597" Score="1" ViewCount="" Body="&lt;p&gt;I am going to assume that you ask R.M.S. current and will answer that. If you want some other form of answer just clarify the question a bit more and I'll adjust the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the RMS(Root-Mean-Square) value of a current, you do the following calculation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{\frac{1}{T} \int_0^T i(t) dt }$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where T is the period of the signal and t is the variable of integration.&lt;/p&gt;&#xA;" OwnerUserId="504" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-17T13:16:34.673" LastActivityDate="2010-11-17T13:16:34.673" CommentCount="1" />
  <row Id="997" PostTypeId="2" ParentId="995" CreationDate="2010-11-17T12:45:06.017" Score="0" ViewCount="" Body="&lt;p&gt;Check out the site of &lt;a href=&quot;http://physbam.stanford.edu/~fedkiw/&quot; rel=&quot;nofollow&quot;&gt;Ron Fedkiw&lt;/a&gt;; it is a good starting point with comprehensive set of keywords.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-17T12:45:06.017" />
  <row Id="998" PostTypeId="2" ParentId="995" CreationDate="2010-11-17T12:47:51.023" Score="0" ViewCount="" Body="&lt;p&gt;it depends on what kind of simulation you trying to build:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if your simulation has the purpose build a simulative model, that, for example, avoids the experimental noise, maybe with a complex dynamics algorithm and so on, i think C or C++ are the best choices..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If on the other hand you want to create a quick simulation with graphical output and built-in analysis tools (maybe even for didactic purpose), python is your choice! in this case I suggest you check out the &lt;a href=&quot;http://www.enthought.com/products/epd.php&quot; rel=&quot;nofollow&quot;&gt;Enthought Python Distribution&lt;/a&gt;.. for accademic use it is freeware and it has a built-in release of scipy. &lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2010-11-17T12:47:51.023" CommentCount="1" />
  <row Id="999" PostTypeId="1" AcceptedAnswerId="1001" CreationDate="2010-11-17T13:40:28.137" Score="6" ViewCount="497" Body="&lt;p&gt;I've noticed that whenever I turn the lamp off in my room at night, the lightbulb seems to continue to glow for a minute or so after that.  It's not bright though; the only way I even notice it is if the room is dark. So why does it keep glowing?  &lt;/p&gt;&#xA;" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-18T00:26:11.280" LastActivityDate="2010-12-20T10:48:10.723" Title="Why do lightbulbs continue to glow after the light is turned off?" Tags="&lt;thermodynamics&gt;&lt;thermal-radiation&gt;&lt;electromagnetic-radiation&gt;&lt;household-physics&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="1000" PostTypeId="2" ParentId="999" CreationDate="2010-11-17T13:50:06.943" Score="4" ViewCount="" Body="&lt;p&gt;It is still hot enough to emit light.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-17T13:50:06.943" CommentCount="1" />
  <row Id="1001" PostTypeId="2" ParentId="999" CreationDate="2010-11-17T13:52:19.423" Score="13" ViewCount="" Body="&lt;p&gt;If it's an incandescent bulb, it's because the whole operating principle of the bulb is based on getting the filament really hot, hot enough to glow. When you cut off the current, it stops heating the filament, so it cools down fairly rapidly, but there may be enough residual heat for a faint glow lasting a little while afterwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's a CFL bulb or an ordinary fluorescent bulb, the &quot;white&quot; color we see is produced by a fluorescent coating on the glass of the bulb, which converts some of the invisible ultraviolet light produced by the excited gas atoms into visible light. When you cut off the current, you stop exciting the gas atoms and thus stop producing new ultraviolet light, but there may be enough residual excitation in the material to keep glowing for a little while afterwards. This would be similar to the way glow-in-the-dark  materials work-- those absorb UV light when they're in bright light, exciting atoms and molecules in the material, which then slowly emit visible photons.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-17T13:52:19.423" />
  <row Id="1003" PostTypeId="2" ParentId="879" CreationDate="2010-11-17T14:55:47.083" Score="8" ViewCount="" Body="&lt;p&gt;Yes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;G2 shows up often, starting with atomic physics (perhaps Racah is the first; see R. E. Behrends, J. Dreitlein, C. Fronsdal, and B. W. Lee, “Simple groups and strong interaction symmetries,” Rev. Mod. Phys. 34, 1 (1962).). You will find some refences in my 1976 Phys rev paper on &lt;a href=&quot;http://ChaosBook.org/GroupTheory/refs&quot; rel=&quot;nofollow&quot;&gt;cns.physics.gatech.edu/GroupTheory/refs&lt;/a&gt; . I have whole folder of physics G2 papers, but now I see I did not bother to enter G2 history into www.birdtracks.eu. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nobody's perfect. Sorry&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Predrag (for responses, email to dasgroup [snail] gatech.edu, I sometimes look at those. Pure accident I saw this question...)&lt;/p&gt;&#xA;" OwnerUserId="516" LastEditorUserId="516" LastEditorDisplayName="" LastEditDate="2010-11-17T15:16:08.443" LastActivityDate="2010-11-17T15:16:08.443" CommentCount="2" />
  <row Id="1005" PostTypeId="2" ParentId="887" CreationDate="2010-11-17T15:37:27.277" Score="4" ViewCount="" Body="&lt;p&gt;In cellular automata I do know there is explicit dependence on step/time. In quantum mechanics (and other  many other theories) it is natural to write local evolution with respect to time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the contrary, in 'pure' relativity, time is not that different from position. And thus there is no such natural interpretation like 'the next step is the next time'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are relativity-based equations (e.g. Dirac Equation, Maxwell Equation) in which time can be taken to be an independent variable. And for sure there are more papers on the topic than one, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Iwo Bialynicki-Birula, &lt;a href=&quot;http://www.cft.edu.pl/~birula/publ/lattice.pdf&quot; rel=&quot;nofollow&quot;&gt;Weyl, Dirac, and Maxwell equations on a lattice as unitary cellular automata&lt;/a&gt;, Phys. Rev. D 49, 6920–6927 (1994)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Furthermore, some relativistic aspects are easily implemented, like $c=\hbox{(pixel size)}/\hbox{(time step)}$. One cellular automaton-like thing is so-called Feynman Checkerboard, which bases on the assumption that every particle &lt;strong&gt;always&lt;/strong&gt; travels at $c$ but also sometimes gets bounced (it turns that effective mass depends on the amplitude of bouncing).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Feynman_checkerboard&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Feynman_checkerboard&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Feynman and Hibbs, Quantum Mechanics and Path Integrals, New York: McGraw-Hill, Problem 2-6, pp. 34-36, 1965 (the original textbook, when it appears as a problem for the reader)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="184" LastActivityDate="2010-11-17T15:37:27.277" CommentCount="1" />
  <row Id="1006" PostTypeId="2" ParentId="759" CreationDate="2010-11-17T15:53:06.760" Score="10" ViewCount="" Body="&lt;p&gt;I'm the author of the novel mentioned above and I'd like to clarify. Neither I nor my novel claim the universe is a computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The operating principle of the novel's universe is that all things are One through balance. The protagonist, Paul, is asked if his universe is a computer. He answers that in a balanced quantum universe this would be impossible. Why? Here's a direct quote from the page cited above: &lt;a href=&quot;http://donnee.com/universe-as-computer.htm&quot; rel=&quot;nofollow&quot;&gt;http://donnee.com/universe-as-computer.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Calculation is a process, or time-dependent function. A balanced quantum universe is a holistic, non-local phenomenon. There is no time and no process. Balance simply is. Balance requires no computation. So, my initial response—that the universe is a quantum computer—is correct only from a classical, local perspective. From a holistic, non-local quantum perspective, the answer is: No, the universe is not a computer.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enjoy.&lt;/p&gt;&#xA;" OwnerUserId="517" LastActivityDate="2010-11-17T15:53:06.760" CommentCount="2" />
  <row Id="1007" PostTypeId="2" ParentId="357" CreationDate="2010-11-17T16:56:17.150" Score="0" ViewCount="" Body="&lt;p&gt;What about a doppler shift method?   A doppler speed radar or lidar gun might have all the necessary components, for its logic to be reversed.  Here's one for instance, ready for disassembly.&#xA;&lt;a href=&quot;http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&amp;amp;item=300374815766&amp;amp;rvr_id=169891150704&amp;amp;crlp=1_263602_304642&amp;amp;UA=M&quot; rel=&quot;nofollow&quot;&gt;http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&amp;amp;item=300374815766&amp;amp;rvr_id=169891150704&amp;amp;crlp=1_263602_304642&amp;amp;UA=M&lt;/a&gt;*S%3F&amp;amp;GUID=0537e92612c0a06456359f45ffd1174f&amp;amp;itemid=300374815766&amp;amp;ff4=263602_304642#ht_2332wt_979&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-11-17T17:02:31.920" LastActivityDate="2010-11-17T17:02:31.920" />
  <row Id="1008" PostTypeId="1" AcceptedAnswerId="2955" CreationDate="2010-11-17T17:23:26.860" Score="11" ViewCount="500" Body="&lt;p&gt;There's more gravitational force in our galaxy (and others) than can be explained by counting stars. So why not lots of dark planetery systems (ie without stars) ? Why must we assume some undiscovered and unexplained form of matter ?&lt;/p&gt;&#xA;" OwnerUserId="495" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-15T03:27:36.050" LastActivityDate="2011-01-15T16:03:00.700" Title="Why isn't dark matter just matter?" Tags="&lt;astrophysics&gt;&lt;cosmology&gt;&lt;dark-matter&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="1" />
  <row Id="1009" PostTypeId="2" ParentId="1008" CreationDate="2010-11-17T17:33:42.640" Score="4" ViewCount="" Body="&lt;p&gt;All possible dim/dwarf/died stars, interstellar medium(gas,dust,molecular cloud etc.), have been considered. But they are not enough. The rest part can only be something we never know before. We are expecting some unknown new particles.&lt;/p&gt;&#xA;" OwnerUserId="518" LastActivityDate="2010-11-17T17:33:42.640" />
  <row Id="1010" PostTypeId="2" ParentId="1008" CreationDate="2010-11-17T17:36:19.277" Score="4" ViewCount="" Body="&lt;p&gt;The possibility of large dark objects made of normal baryonic matter has been considered.  These are called &lt;a href=&quot;http://en.wikipedia.org/wiki/MACHOs&quot; rel=&quot;nofollow&quot;&gt;MACHOs&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are various reasons to think that most of the dark matter can't be in the form of MACHOs.  From the above wikipedia article (which links to some relevant journal articles):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;The Big Bang as it is currently understood simply couldn't produce enough baryons without causing major problems in the observed elemental abundances,&lt;a href=&quot;http://arxiv.org/abs/astro-ph/9904291&quot; rel=&quot;nofollow&quot;&gt;[6]&lt;/a&gt; including the abundance of deuterium.&lt;a href=&quot;http://arxiv.org/abs/astro-ph/9504082&quot; rel=&quot;nofollow&quot;&gt;[7]&lt;/a&gt; Furthermore, separate observations of baryon acoustic oscillations, both in the cosmic microwave background and large-scale structure of galaxies, set limits on the total baryon-to-total matter ratio. These observations show that a large fraction of non-baryonic matter is necessary regardless of the presence or absence of MACHOs.&lt;/blockquote&gt;&#xA;" OwnerUserId="11" LastActivityDate="2010-11-17T17:36:19.277" CommentCount="1" />
  <row Id="1011" PostTypeId="2" ParentId="89" CreationDate="2010-11-17T17:40:33.260" Score="2" ViewCount="" Body="&lt;p&gt;Eric's answer is not really correct (or at least not complete). For instance, it doesn't tell you anything about the motion of two comparably heavy bodies (and indeed this problem is very hard in GR, in stark contrast to the Newtonian case). So let me make his statements a bit more precise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct approach is to treat the Newtonian gravity as a perturbation of the flat &lt;a href=&quot;http://en.wikipedia.org/wiki/Minkowski_space&quot; rel=&quot;nofollow&quot;&gt;Minkowski space-time&lt;/a&gt;. One writes $g = \eta + h$ for the metric of this space-time ($\eta$ being Minkowski metric and $h$ being the perturbation that encodes curvature of the space-time) and &lt;a href=&quot;http://en.wikipedia.org/wiki/Linearized_gravity#Linearised_Einstein_field_equations&quot; rel=&quot;nofollow&quot;&gt;linearize&lt;/a&gt; the theory in $h$. By doing this one actually obtains a lot more than just Newtonian gravity, namely &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitomagnetism&quot; rel=&quot;nofollow&quot;&gt;gravitomagnetism&lt;/a&gt;, in which one can also investigate dynamical properties of the space-time not included in the Newtonian picture. In particular the propagation of gravitational waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to recover Newtonian gravity we have to make one more approximation. Just realize that Newtonian gravity is not relativistic, i.e. it violates finite speed of light. But if we assume that $h$ changes only slowly and make calculations we will find out that the perturbation metric $h$ encodes the Newtonian field potential $\Phi$ and that the space-time is curved in precisely the way to reproduce the Newtonian gravity. Or rather (from the modern perspective): Newtonian picture is indeed a correct low-speed, almost-flat description of GR.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-11-17T17:40:33.260" />
  <row Id="1012" PostTypeId="2" ParentId="1008" CreationDate="2010-11-17T17:46:38.483" Score="6" ViewCount="" Body="&lt;p&gt;As above answers have mentioned most of the ordinary matter has been considered as candidates and we are fairly certain that there has to be some sort of &quot;dark&quot; matter at work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, we take on the phenomenon of gravitational lensing. A very famous example is the Bullet cluster where you can clearly observe the effects of a compact mass acting as an optical lens. A mass of such magnitude can not conveniently be a cluster of brown dwarves and most black holes we observe have many objects orbiting and a particle jet accompanying. Besides, a black hole of small magnitude would not be enough to explain the magnitude of the gravitational lensing in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, much more complicated computations made with general relativity principles need a lot more mass to account for the current shape and angular speed of galaxies. It may be easy to say that GR is wrong(which you shouldn't say to the face of a physicist), even though GR is incomplete in the sense that it is not a theory of everything, it still explains most of the gravitational phenomena pretty well. Besides, the fact that there is gravitational lensing means that there really IS some sort of mass or gravitational pull(or more like space time curvature anomaly) in certain parts of our galaxy, and the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though dark matter seems so mysterious, we can guess most of its properties from our &quot;lack of knowledge&quot;. Firstly, since it is &quot;invisible&quot; in all spectrum of light, we can assume that it does not interact via electromagnetic force. Since it is a fountain of gravitational force, we can say that it, surprise surprise, interacts via gravitational force. Calculations for weak&amp;amp;strong forces are pretty complicated and very indirect so I'm just going to say that most of the currently proposed dark matters interact via strong and weak forces as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A planetary system without a star is only possible if the star died out to be a neutron star, black hole, or a white dwarf, all of which are detectable in some way.&lt;/p&gt;&#xA;" OwnerUserId="504" LastActivityDate="2010-11-17T17:46:38.483" CommentCount="3" />
  <row Id="1013" PostTypeId="2" ParentId="357" CreationDate="2010-11-17T17:55:16.320" Score="7" ViewCount="" Body="&lt;p&gt;You might also want to try the rotating mirror method, of Léon Foucault. It is detailed &lt;a href=&quot;http://www.pas.rochester.edu/~pavone/particle-www/teachers/demonstrations/FoucaultDemonstration.htm&quot; rel=&quot;nofollow&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.njsas.org/projects/speed_of_light/cache/2/lightspeedformal.htm&quot; rel=&quot;nofollow&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. The only difficult part is the rotating mirror, but it could probably be done with a drill.&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2010-11-17T17:55:16.320" CommentCount="4" />
  <row Id="1014" PostTypeId="1" AcceptedAnswerId="1016" CreationDate="2010-11-17T19:42:17.023" Score="5" ViewCount="188" Body="&lt;p&gt;Premise: the LHC is obviously mapping unseen territory in high energies, and therefore it's always possible to imagine &lt;em&gt;far out&lt;/em&gt; results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Excluding completely unexpected outcomes - is the LHC performing any experiment that could help with string theory or m-theory? For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;direct super-strings or m-theory predictions to be tested or confuted&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;but also&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;measurements that would help &quot;shape&quot; string/m-theory into something more concretely testable or practical than the current blurry incarnation?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="66" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-11-17T19:55:03.333" LastActivityDate="2010-11-17T20:25:16.900" Title="How relevant is LHC to quantum gravity?" Tags="&lt;experimental-physics&gt;&lt;large-hadron-collider&gt;&lt;cern&gt;&lt;super-strings&gt;" AnswerCount="2" />
  <row Id="1015" PostTypeId="2" ParentId="1014" CreationDate="2010-11-17T20:00:39.207" Score="2" ViewCount="" Body="&lt;p&gt;This &lt;a href=&quot;http://cerncourier.com/cws/article/cern/28895&quot; rel=&quot;nofollow&quot;&gt;article from CERN Courier&lt;/a&gt; is about string-theory and experimental tests of it, and about LHC. It is not really technical but it also links to various references and articles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for more technical info, you can browse &lt;a href=&quot;http://indico.cern.ch/categoryDisplay.py?categId=104&quot; rel=&quot;nofollow&quot;&gt;the list of String Theory Seminar of the TH department at CERN&lt;/a&gt; and search for &quot;LHC&quot;; very interesting.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-11-17T20:00:39.207" />
  <row Id="1016" PostTypeId="2" ParentId="1014" CreationDate="2010-11-17T20:25:16.900" Score="9" ViewCount="" Body="&lt;p&gt;Well, there's no reason to believe in supersymmetry, beyond some theoretical niceness to it, so if they see THAT at the LHC, then string theory gets a big boost, as there is no way other than supersymmetry to produce fermions in string theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing that might be relevant to quantum gravity is that if there are large extra dimensions (as in, large compared to the Planck length, but smaller than detectable by things like the Cavendish experiment).  If that is the case, then the 'fundamental' gravitational constant may be much larger than Newton's constant (they differ by a factor of the volume of the large extra dimensions), and quantum gravitational effects would be accessible at the LHC.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-11-17T20:25:16.900" CommentCount="2" />
  <row Id="1017" PostTypeId="2" ParentId="994" CreationDate="2010-11-17T20:38:31.217" Score="1" ViewCount="" Body="&lt;p&gt;The answer to your question is that the way we perceive reality is governed by physical laws as well. In your coloured sphere model, the coloured sphere people would perceive the changing of colour as we perceive movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, it is quite customary in physics to think using completely abstract representations of space, and only ultimately reduce the equations to real space. In other words there are &lt;a href=&quot;http://en.wikipedia.org/wiki/Phase_space&quot; rel=&quot;nofollow&quot;&gt;representations of the universe which do not resemble real space&lt;/a&gt;, but are equally as valid.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-17T20:38:31.217" />
  <row Id="1018" PostTypeId="1" AcceptedAnswerId="1021" CreationDate="2010-11-17T21:49:53.793" Score="4" ViewCount="964" Body="&lt;p&gt;I am Math Grad student with a little bit of interest in physics. Recently i looked into the wikipedia page for &lt;a href=&quot;http://en.wikipedia.org/wiki/Simple_harmonic_motion&quot; rel=&quot;nofollow&quot;&gt;Simple Harmonic Motion&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Guess, I am too bad at physics to understand it. Considering me as a &lt;em&gt;layman&lt;/em&gt;  how would one explain:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What a Simple Harmonic motion is? And why does this motion have such an equation $$x(t)= A \cos(\omega{t} + \varphi)$$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;associated with it? Can anyone give examples of where &lt;strong&gt;S.H.M.&lt;/strong&gt; is tangible in Nature. &lt;/p&gt;&#xA;" OwnerUserId="159" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-22T22:16:56.857" LastActivityDate="2011-01-23T19:20:24.220" Title="Explanation: Simple Harmonic Motion" Tags="&lt;harmonic-oscillator&gt;" AnswerCount="9" FavoriteCount="1" />
  <row Id="1019" PostTypeId="1" AcceptedAnswerId="1027" CreationDate="2010-11-17T21:59:59.027" Score="30" ViewCount="3125" Body="&lt;p&gt;Well, in Mathematics there are somethings, which appear true but they aren't true. Naive students often get fooled by these results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me consider a very simple example. As a child one learns this formula $$(a+b)^{2} =a^{2}+ 2 \cdot a \cdot b + b^{2}$$ But as one mature's he applies this same formula for Matrices. That is given any two $n \times n$ square matrices, one believes that this result is true: $$(A+B)^{2} = A^{2} + 2 \cdot A \cdot B +B^{2}$$ But eventually this is false as Matrices aren't necessarily commutative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know whether there any such things happening with physics students as well. My motivation came from the following MO thread, which many of you might take a look into:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="159" LastEditorUserId="103" LastEditorDisplayName="" LastEditDate="2011-08-25T22:29:51.033" LastActivityDate="2011-08-26T08:36:23.683" CommunityOwnedDate="2010-11-26T21:47:08.610" Title="Common false beliefs in Physics" Tags="&lt;soft-question&gt;" AnswerCount="45" CommentCount="8" FavoriteCount="19" />
  <row Id="1020" PostTypeId="2" ParentId="1018" CreationDate="2010-11-17T22:05:01.590" Score="4" ViewCount="" Body="&lt;p&gt;Simple harmonic motion (SMH) describes the behavior of systems characterized by a equilibrium point and a restoring &quot;force&quot; (in some generalized sense) proportional to the displacement from the equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example system&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A simple mechanical system with this behavior is a mass on a spring (which we will consider in one dimension for ease). There is some point where the mass is stable: it can be left alone with no external forces on it and will not accelerate. That's the equilibrium, call it $x_0$. If you move the mass from that point, the spring exerts a restoring force $F = -k(x-x_0)$. Here $k$ is a property of the spring called the &quot;spring constant&quot;; a stiff spring hs a high spring constant and a weak spring has a low value for $k$. Lets consider the special case where $x_0 = 0$ (with no loss of generality, but just to keep the number symbols to a minimum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the acceleration of the mass at any point in time is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a = \frac{d^2x}{dt^2} = \frac{F}{m} = -\frac{kx}{m} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a second order linear differential equation, the solutions to which are of the form you exhibit above (where $A$ and $\phi$) take the part of the two constants of integration needed to make the solution agree with an arbitrary set of boundary conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's prove it by explicit substitution:&#xA;$$x(t)=A \cos(\omega t + \varphi )$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v(t) = \frac{d}{dt} x(t) = A \sin(\omega t + \varphi ) \omega $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a(t) = \frac{d}{dt} v(t) = \frac{d^2}{dt^2} x(t) = -A \cos(\omega t + \varphi ) \omega^2 $$&#xD;&#xA;so substituting back in we get&#xA;$$&#xD;&#xA;-A \cos(\omega t + \varphi ) \omega^2 = \frac{F}{m} = -\frac{k x}{m} = -\frac{k A}{m} \cos(\omega t + \varphi )&#xD;&#xA;$$&#xD;&#xA;which implies that &#xA;$$ \omega^2 = \frac{k}{m} $$&#xD;&#xA;which relates the spring constant and the mass of the object to the frequency of oscillation.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;General discussion&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The thing that is important about SHM is that these restoring &quot;forces&quot; linearly proportional to &quot;displacement&quot; (which we allow that there may be generalized meanings for both &quot;force&quot; and &quot;displacement&quot;) are &lt;em&gt;very&lt;/em&gt; common in the universe. So a great many phenomena may be described this way (and even more may be if we limit ourselves to small perturbations).&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2010-11-17T22:35:35.740" LastActivityDate="2010-11-17T22:35:35.740" CommentCount="3" />
  <row Id="1021" PostTypeId="2" ParentId="1018" CreationDate="2010-11-17T22:19:39.400" Score="6" ViewCount="" Body="&lt;p&gt;This is all about potential; it is common that a particle movement is described by a following ODE:&lt;br&gt;&#xA;$m\ddot{\vec{x}}=-\nabla V(\vec{x})$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $V$ is some function; usually one is interested in minima of $V$ (they correspond to some stable equilibrium states). Now, however complex $V$ generally is, its minima locally looks pretty much like some quadratic forms, and so the common assumption that $V(x)=Ax^2$... this makes the last equation simplify to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ddot{x}=-\omega^2x$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with solution in harmonic oscillations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The common analogy of this is a ball in a paraboloid dish resembling potential shape; it oscillates near the bottom.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-17T22:19:39.400" CommentCount="12" />
  <row Id="1022" PostTypeId="2" ParentId="995" CreationDate="2010-11-17T22:27:02.123" Score="3" ViewCount="" Body="&lt;p&gt;I recently did something like this, in order to simulate a system of two masses connected by a spring. Those masses lay horizontally on a frictionless plane. One of these masses got an initial impulse and thereafter the system was left alone. While the entire system (the controid to be precies) moves with constant velocity, the two masses are oscillating, while moving forward. Here is a short ASCII drawing of the system &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; Initial Impulse     ______              ______ &#xA; ----&amp;gt;               | m1 |/\/\/\/\/\/\/\| m2 |&#xA;_____________________|____|______________|____|______________________&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After writing down the differential equations, I wrote a small python programm simulating the problem. This programm relies on the method of small steps (also called the Eueler Method). Here is the correspondig wikipedia article:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Euler_method&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Euler_method&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I implemented this alogorithm for the problem described above and plotted the results using gnuplot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gnuplot.info (I am only allowed to add one hyperlink, so please add www)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you are free to use any tool you like for this purpose. Here comes the sourcecode of my small programm:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/usr/bin/python&#xA;import os&#xA;&#xA;steps = 100000&#xA;time = 100.&#xA;&#xA;# Initial conditions&#xA;D = 0.9&#xA;m1 = 1.2&#xA;m2 = 0.4&#xA;v1 = 1.3&#xA;v2 = 0.&#xA;x1 = 0.&#xA;x2 = 1.&#xA;l = 1.&#xA;&#xA;#Since I also tried to implement other algorithmus i specify which one to use&#xA;Euler = 1&#xA;&#xA;&#xA;#Euler &#xA;&#xA;if Euler == 1:&#xA;    timesteps = time / steps&#xA;    # Open the files for writing the results to&#xA;    f = open('results_x1', 'w')&#xA;    f2 = open('results_x2', 'w')&#xA;    f3 = open('results_com', 'w')&#xA;&#xA;    # The real calculation   &#xA;    for i in range(0,steps):&#xA;        x1 = x1 + (D * (x2 - x1 -l) / m1)* (timesteps**2) + v1 * timesteps&#xA;        x2 = x2 - (D * (x2 - x1 -l) / m2)* (timesteps**2) + v2 * timesteps&#xA;        v1 = v1 + (D * (x2 - x1 -l) / m1)* (timesteps)&#xA;        v2 = v2 - (D * (x2 - x1 -l) / m2)* (timesteps)  &#xA;        f.write(str(i*timesteps) + &quot; &quot; + str(x1) + &quot;\n&quot;)&#xA;        f2.write(str(i*timesteps) + &quot; &quot; + str(x2) + &quot;\n&quot;)&#xA;        f3.write(str(i*timesteps) + &quot; &quot; + str((x1*m1 + x2*m2)/(m1+m2)) + &quot;\n&quot;)&#xA;&#xA;&#xA;    f.close()&#xA;    f2.close()&#xA;    f3.close()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Of course there are better alogorithmus than the euler one, but this one is definitly the easiest to implement (I failed implementing more advanced algorithms ;-)). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So these are the steps you should probably follow:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Write down the differential equations for you problem &lt;/li&gt;&#xA;&lt;li&gt;Understand the Euler Method&lt;/li&gt;&#xA;&lt;li&gt;Take my code as a reference point and modify it for your problem&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I know that this is quite an extensive topic and that my answer is therefore just superficial. Just tell what you want to know more about, and I will try to add corresponding comments ;-)&lt;/p&gt;&#xA;" OwnerUserId="393" LastActivityDate="2010-11-17T22:27:02.123" CommentCount="2" />
  <row Id="1023" PostTypeId="2" ParentId="1018" CreationDate="2010-11-17T22:33:36.007" Score="1" ViewCount="" Body="&lt;p&gt;Simple harmonic motions are designated by that name because the time dependence of the coordinate being described is given by a constant multiple of a sine or cosine function of a multiple of the time, i.e., a harmonic function in the sense of being the kind of function employed in the harmonic, or Fourier analysis of a periodic, but otherwise arbitrary motion in terms of sinusoidal functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The importance of simple harmonic motion stems, first of all, from its pervasive presence throughout macroscopic nature as the characteristic response to small disturbances from equilibrium which produce restoring forces proportional to the disturbance. Beyond that, and as indicated in the previous paragraph, almost all varieties of motion are analyzable as superpositions of simple harmonic motions of varying frequencies and amplitudes. Such analysis is just what we call Fourier analysis of the motion.  &lt;/p&gt;&#xA;" OwnerUserId="345" LastActivityDate="2010-11-17T22:33:36.007" CommentCount="1" />
  <row Id="1024" PostTypeId="2" ParentId="1019" CreationDate="2010-11-17T22:33:38.843" Score="7" ViewCount="" Body="&lt;p&gt;Historically the concept of &lt;em&gt;absolute velocity&lt;/em&gt; was commonly believed until the time of &lt;a href=&quot;http://en.wikipedia.org/wiki/Galileo&quot; rel=&quot;nofollow&quot;&gt;Galileo Galilei&lt;/a&gt; in the early 17th century. As a naive child, before studying basic physics, it is surprisingly easy to believe in this even these days!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of absolute velocity states that all velocities are fixed with respect to an absolute frame of reference. Galileo showed that velocity is relative to your frame of reference, a principle known as &lt;em&gt;Galilean relativity&lt;/em&gt;. This was later fully quantified by &lt;a href=&quot;http://en.wikipedia.org/wiki/Isaac_Newton&quot; rel=&quot;nofollow&quot;&gt;Isaac Newton&lt;/a&gt;, who also proposed that acceleration is invariant with respect to inertial frames.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-08-14T21:05:08.883" LastActivityDate="2011-08-14T21:05:08.883" CommunityOwnedDate="2010-11-26T21:47:08.610" CommentCount="2" />
  <row Id="1025" PostTypeId="2" ParentId="1018" CreationDate="2010-11-17T22:38:29.843" Score="0" ViewCount="" Body="&lt;p&gt;I was once trying to push a large log off a hiking trail trail.  The log had been there a while and had sunk into the dirt, so to push it out I would have to push hard enough to  force it up out of the depression.  I kicked at it, but couldn't kick it hard enough to get it out.  I was with a couple other people, so we all kicked it, but again we couldn't get it out.  What did happen was that it rolled up a little way out of its hole, then rolled back towards us and  climbed a small way out of its hole in the opposite direction.  Finally it rolled back the way we originally kicked it again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We soon caught on to the idea of kicking the log, waiting for it to roll back towards us, then roll the other way, and kick it again while it was already rolling the direction we wanted it to go.  With a little effort, this strategy finally let us roll the log out of its depression and off to the side of the trail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The log was exhibiting simple harmonic motion - it shook back and forth at a certain frequency, oscillating around the spot it would stay at rest.  Our kicks were timed to coincide with the period of the log's oscillations.  This is called achieving &quot;resonance&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The log was not a perfect simple harmonic oscillator because it would only shake back and forth a few times before settling down again.  Ideally, a  simple harmonic oscillator will shake back and forth indefinitely.  Further, it can shake more or less vigorously (different amplitude of the sine wave), but it will always have the same period and move sinusoidally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many mechanical systems are close to simple harmonic oscillators.  A pendulum in a clock is a good example.  If you take a rectangular or vertical cylindrical ice cube floating in water and tap it very gently, it will oscillate up and down in harmonic motion with a period of around a half a second.  The water itself is a harmonic oscillator, too.  If you give it a short shove across the table, you'll see the entire surface of the water shake back and forth in a characteristic way called the &quot;sloshing mode&quot;.  Electrical circuits with an inductor and a capacitor make extremely good simple harmonic oscillators.  A quartz crystal is a good one, too.  Since it oscillates with a definite period, it makes for a good clock.  Older clocks use a different simple harmonic oscillator, a wound up spring that shakes back and forth.  If a roller coaster were to roll back and forth at the bottom of a hill, it would be a simple harmonic oscillator, too.  If you dug a hole from the north pole to the south pole and dropped something in, gravity would swing it back and forth through the Earth in simple harmonic oscillation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple harmonic oscillators can be used to describe molecular or atomic systems.  You can discover a lot about the way that light interacts with matter simply by pretending that an electron attached to an atom is a simple harmonic oscillator, and molecules like CO2 vibrate like simple harmonic oscillators as well.  Even you are a simple harmonic oscillator.  Spread your feet, make yourself rigid, and rock back and forth from foot to foot, staying stiff so that the foot not supporting your weight actually comes off the ground.  You'll find that as long as you don't go too far, there's a certain frequency that your body wants to rock at, and that this frequency changes when you change the distance between your feet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many caveats to all this.  These systems are subject to friction, so they run down eventually, unlike an ideal simple harmonic oscillator.  Most of them are simple harmonic oscillators only for small amplitudes.  If you push the entire ice cube under water, it doesn't obey a sine wave oscillation any more.  The roller coaster, even if frictionless, is only a simple harmonic oscillator if the shape of the track is a special shape called a brachistochrone.  The hole through the Earth only results in simple harmonic motion under the false assumption that the density of Earth is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, these models are extremely useful at explaining natural phenomena.  A large body of water like a lake or bay is a simple harmonic oscillator in the same way as the glass of water pushed across the table to excite the sloshing mode.  But the period of the sloshing mode gets longer and longer as the body of water gets bigger.  For the Bay of Fundy, the sloshing mode is very close to twelve hours.  Since the tides come once every twelve hours, the tides act in a way very similar to the way my friends and I kicked the heavy log out of its hole.  The tides build up very strongly, and tides there can be ten meters!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineers will have to worry about this resonance phenomenon all the time.  The most famous example is the Tacoma Narrows suspension bridge, which started shaking wildly and finally collapsed on a windy day when the turbulent eddies of the wind were in resonance with the bridge.  If you're building a building, you don't want it to undergo simple harmonic motion at the same frequency as earthquake waves, and you don't want the suspension on a car to have the same frequency as some sort of regular dip in the road, for example.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-20T05:06:04.643" LastActivityDate="2010-11-20T05:06:04.643" />
  <row Id="1026" PostTypeId="1" AcceptedAnswerId="1028" CreationDate="2010-11-17T22:41:25.123" Score="0" ViewCount="179" Body="&lt;p&gt;The weighing scales in my kitchen are currently showing 0 Kg. However, there is a column of air between the scales and the ceiling that is presumably exerting pressure on them. If I were to place those scales in a vacuum, would those scales show a negative weight?&lt;/p&gt;&#xA;" OwnerUserId="526" LastActivityDate="2010-11-19T05:43:48.797" Title="Would weighing scales show a negative weight if placed in a vacuum?" Tags="&lt;pressure&gt;&lt;atmosphere&gt;" AnswerCount="2" />
  <row Id="1027" PostTypeId="2" ParentId="1019" CreationDate="2010-11-17T22:44:03.137" Score="23" ViewCount="" Body="&lt;p&gt;A mistake that I often come across and that is so easy to make: people somehow have a visceral belief that heavy objects fall faster than light ones. Setting aside of course problems of air resistance, this is obviously false, but it seems to be so counterintuitive and I think it is somehow tied to our intuitive understanding of mass as inertia. Since higher mass means higher inertia. People understand this intuitively as it takes more force to push a fat guy than a thin one. But they don't see that gravity is a force proportional to mass, so that more inertia is paralleled by more gravity as well. With as a result the same gravitational acceleration for light and heavy bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've even noticed the mistake being made by professional physicists in colloquial conversations.&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2010-11-17T22:56:31.937" LastActivityDate="2010-11-17T22:56:31.937" CommunityOwnedDate="2010-11-26T21:47:08.610" CommentCount="12" />
  <row Id="1028" PostTypeId="2" ParentId="1026" CreationDate="2010-11-17T22:47:12.147" Score="7" ViewCount="" Body="&lt;p&gt;They would show only a very tiny &lt;del&gt;negative&lt;/del&gt; positive weight.  A typical kitchen scale has air above and below the weighing pan, and the pressure is very nearly the same because the thickness of the weighing pan is millimeters and the height of the atmosphere is a million times that.  The net force of the pressure is quite small - equal to the weight of the air it displaces, and air is about 10^3 or 10^4 times less dense than typical solid things.  So if your pan weights a few hundred grams, the scale would read a few tenths of a gram in a vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: When there's atmosphere, there's a buoyant force pointing up.  When we lose atmosphere, there's more force down and the tared scale now reads positive.  Originally I got the buoyant force correct but the reading on the scale backwards.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-19T05:43:48.797" LastActivityDate="2010-11-19T05:43:48.797" CommentCount="26" />
  <row Id="1029" PostTypeId="2" ParentId="1026" CreationDate="2010-11-17T22:59:16.947" Score="3" ViewCount="" Body="&lt;p&gt;It would vary from weighing scale to weighing scale. In my experience, most of the scales subtract the weight of the &quot;tare&quot;, which is the thing that you put your &quot;thing to be weighed on&quot;. The scales which use this system will show a noticeable amount of negative weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the types which do not subtract the amount of dare, rather compare it with some &quot;example&quot; weight will not be affected much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some scales are more sophisticated and completely ignore the initial dare by some methods. They, also, would not exhibit much difference in any gravity or pressure change.&lt;/p&gt;&#xA;" OwnerUserId="504" LastEditorUserId="504" LastEditorDisplayName="" LastEditDate="2010-11-18T17:32:31.193" LastActivityDate="2010-11-18T17:32:31.193" CommentCount="2" />
  <row Id="1030" PostTypeId="1" AcceptedAnswerId="1034" CreationDate="2010-11-17T23:01:27.767" Score="5" ViewCount="300" Body="&lt;p&gt;The theory of Inflation explains the apparent consistency of the universe by proposing that  the early universe grew exponentially for a 1E-36 seconds. Isn't a simpler explanation that the universe is just older and so the homogeneousness comes from a slower more steady growth? Is there any evidence that rules out a slow growing universe and supports Inflation theory?&lt;/p&gt;&#xA;" OwnerUserId="526" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-19T02:49:34.997" LastActivityDate="2011-08-11T03:55:05.377" Title="What is the evidence for Inflation of the early universe?" Tags="&lt;general-relativity&gt;&lt;cosmology&gt;&lt;cosmological-inflation&gt;&lt;experimental-evidence&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="1031" PostTypeId="2" ParentId="1019" CreationDate="2010-11-17T23:05:38.380" Score="3" ViewCount="" Body="&lt;p&gt;While the other answers are absolutely correct, they are very subtle in the way they appear. However one very large false-belief in electrical science is that current is taken to be &quot;flowing&quot; from the + terminal to the - terminal in DC current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it doesn't really matter which way you choose since you are dealing with close to light speed current, electrons are actually moving from - terminal to + terminal(do not get me wrong electrons are NOT moving anywhere near light speed while drifting. They are moving in the order of centimeters per second.). Hence the current actually moves from - to +.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it would be interesting to note that not a soul in the professional area of electrical science considers the current from - to +, as it would cause inconsistency with his/her colleagues.&lt;/p&gt;&#xA;" OwnerUserId="504" LastEditorUserId="504" LastEditorDisplayName="" LastEditDate="2010-11-17T23:19:15.143" LastActivityDate="2010-11-17T23:19:15.143" CommunityOwnedDate="2010-11-26T21:47:08.610" CommentCount="5" />
  <row Id="1032" PostTypeId="2" ParentId="1019" CreationDate="2010-11-17T23:10:03.370" Score="3" ViewCount="" Body="&lt;p&gt;One widespread belief (I think due to popular books such as Hawking's) is that GTR can &lt;strong&gt;never ever, ever&lt;/strong&gt; be quantized and you always obtain infinities and blah blah. Well, it can, in many situations and in many theories. What is actually meant is that GTR is not a renormalizable quantum field theory in the naive way. But this specification is never explicitly pointed out so people get a false impression that quantum gravity is something completely out of the realms of current physics. Well, surprise, surprise, it's not. We can quantize many gravitational effects (such as waves), we understand that black holes have entropy, we understand they produce Hawking radiation and eventually disappear, etc. And assuming that string theory is correct we can predict whole deal more about it.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-11-17T23:10:03.370" CommunityOwnedDate="2010-11-26T21:47:08.610" CommentCount="3" />
  <row Id="1033" PostTypeId="2" ParentId="1019" CreationDate="2010-11-17T23:13:53.273" Score="13" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Quantum mechanics is way too strange so it can't possibly be a correct description of the real world. Right?&lt;/em&gt; I think nothing else needs to be said about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or maybe on a second thought, some more concrete beliefs and their solution are in order:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The physical world has to be deterministic (it doesn't).&lt;/li&gt;&#xA;&lt;li&gt;Every possible question that can occur to you must have a precise answer by measurement (we observe only what we can, not what we want to).&lt;/li&gt;&#xA;&lt;li&gt;The collapse of the wavefunction is in contradiction with finite speed of light (no information is being transmitted).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-17T23:23:23.393" LastActivityDate="2010-11-17T23:23:23.393" CommunityOwnedDate="2010-11-26T21:47:08.610" CommentCount="11" />
  <row Id="1034" PostTypeId="2" ParentId="1030" CreationDate="2010-11-17T23:16:07.330" Score="2" ViewCount="" Body="&lt;p&gt;I can not go into much detail here but let me say that exponential growth brings many things that we see around us right now: absence of magnetic monopoles, a homogeneous universe in which no section is a &quot;preferred&quot; section i.e. has more matter density, and many more observable quantities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually after a brief search I found a wiki article stating most of the things i said above and much, much more: &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_inflation#Observational_status&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Cosmic_inflation#Observational_status&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="504" LastActivityDate="2010-11-17T23:16:07.330" CommentCount="1" />
  <row Id="1035" PostTypeId="2" ParentId="1030" CreationDate="2010-11-17T23:44:41.893" Score="0" ViewCount="" Body="&lt;p&gt;The main (and original) reason for the proposal of inflationary theory was the &lt;a href=&quot;http://en.wikipedia.org/wiki/Horizon_problem&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;horizon problem&lt;/strong&gt;&lt;/a&gt;. That is, the fact the the universe is so incredibly homogeneous and isotropic despite the fact that some parts of the universe are apparently too far away to have exchanged energy. Inflation in the early universe is a powerful explanation for this intriguing observation. Also of note is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Flatness_problem&quot; rel=&quot;nofollow&quot;&gt;flatness problem&lt;/a&gt;, which inflation also helps address.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are of course various other theories to explaining this problems, not in any way related to inflation, such as the varying speed of light (VSL) theory. These are however under active research and still not widely accepted.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2010-11-17T23:44:41.893" />
  <row Id="1036" PostTypeId="2" ParentId="1008" CreationDate="2010-11-18T00:11:20.130" Score="3" ViewCount="" Body="&lt;p&gt;All the matter that we &lt;em&gt;do&lt;/em&gt; know to exist (called Baryonic matter) emits &lt;em&gt;some&lt;/em&gt; kind of electromagnetic radiation at &lt;em&gt;some&lt;/em&gt; frequency. Sometimes it's measured in infrared radiation, because matter, no matter how cold, will still radiate some amount of heat. To the best of our knowledge, it's not actually possible to cool any matter to absolute zero, and it's certainly not happening naturally. I believe the coolest matter known to exist in the universe is around 3 degrees Kelvin. Other wavelengths can determine the exact nature of the matter in question, and its temperature. For example, ionized oxygen glows in visible light at a certain wavelength - that's one of the things that make emission nebulae visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, after measuring all that matter in other galaxies, we've noticed that there's simply not enough mass to keep them from falling apart (namely, stars within galaxies are moving faster than the escape velocity of the gravitational pull of the center of mass of the galaxy) - especially when we've tried to make computer models of galaxies. That was the first clue that there was something going on that we didn't fully understand. As others have described, other methods of determining the total mass of a galaxy have shown similar results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, there's only one possible explanation for this phenomenon: there must be some kind of matter (and all matter has mass) that we &lt;em&gt;can't&lt;/em&gt; detect. In fact, the amount of mass that can't be accounted for in this way is about 95% of the mass of any given galaxy. Of course, that's a pretty big deal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, we've been trying to work out this very large discrepancy since 1933. The difference between now and then is that accuracy of our measurements of velocity of stars in other galaxies has improved, as well as our ability to measure other phenomena happening within galaxies and galaxy clusters. The more accurate we've become in our measurements, the more this glaring discrepancy has become evident.&lt;/p&gt;&#xA;" OwnerUserId="529" LastActivityDate="2010-11-18T00:11:20.130" />
  <row Id="1037" PostTypeId="1" AcceptedAnswerId="1040" CreationDate="2010-11-18T00:49:18.020" Score="2" ViewCount="283" Body="&lt;p&gt;I understand how telescope, microscope and glasses work.&lt;br&gt;&#xA;But how do contact lenses work?&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-11-22T15:43:04.213" Title="How do contact lenses work?" Tags="&lt;optics&gt;&lt;geometric-optics&gt;" AnswerCount="3" />
  <row Id="1038" PostTypeId="2" ParentId="123" CreationDate="2010-11-18T00:50:00.267" Score="0" ViewCount="" Body="&lt;p&gt;The real fun part is where you try to determine the mass of the visible matter in the galaxy. :)&lt;/p&gt;&#xA;" OwnerUserId="529" LastActivityDate="2010-11-18T00:50:00.267" />
  <row Id="1039" PostTypeId="1" AcceptedAnswerId="1090" CreationDate="2010-11-18T00:55:42.133" Score="5" ViewCount="338" Body="&lt;p&gt;If you see through small enough aperture, you can see things without glasses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this trick work?&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2011-06-16T22:43:45.560" LastActivityDate="2011-06-16T22:43:45.560" Title="How &quot;How to See Without Your Glasses&quot; works?" Tags="&lt;optics&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="1040" PostTypeId="2" ParentId="1037" CreationDate="2010-11-18T02:40:16.147" Score="3" ViewCount="" Body="&lt;p&gt;Your question is really &quot;how does the human eye work?&quot;, since the contact lens is designed to adjust the optics of the human biological lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/X3oTy.png&quot; alt=&quot;lens in the human eye&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image &lt;a href=&quot;http://en.wikipedia.org/wiki/Lens_%28anatomy%29&quot; rel=&quot;nofollow&quot;&gt;from the wikipedia article on the anatomical lens&lt;/a&gt; shows how the lens focuses incoming light from the left onto the retina (right).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For people who are &lt;a href=&quot;http://en.wikipedia.org/wiki/Myopia&quot; rel=&quot;nofollow&quot;&gt;nearsighted&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperopia&quot; rel=&quot;nofollow&quot;&gt;farsighted&lt;/a&gt;, the light coming into the eye does not end up in focus.  See this picture for the case of nearsightedness:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/QKbb8.gif&quot; alt=&quot;nearsightedness&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should imagine the &lt;a href=&quot;http://en.wikipedia.org/wiki/Contact_lens&quot; rel=&quot;nofollow&quot;&gt;contact lens&lt;/a&gt; being placed over the cornea (left surface) and causing the rays to adjust so that the image ends up in focus.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-18T02:40:16.147" CommentCount="2" />
  <row Id="1041" PostTypeId="2" ParentId="1018" CreationDate="2010-11-18T03:19:49.967" Score="2" ViewCount="" Body="&lt;p&gt;This echoes some of the other responses, but simple harmonic motion is &lt;em&gt;universal&lt;/em&gt; in the following sense.  A system displaced from equilibrium will exhibit some force (possibly zero) as a function of the displacement.  &quot;Equilibrium&quot; means that if x is the displacement coordinate, then the response force F(x) obeys F(0)=0.  So if we Taylor expand F in x we get F(x) = 0 + F'(0)x + F''(0)x^2/2 + ..., where here x can be a multi-coordinate, F vector-valued, with F' a Jacobian, etc.  The point is that for small displacements, the linear term is a good approximation (it is nonzero generically, but not always) and the equations take the form d^2x^i/dt^2 = \sum_j (dF^i/dx^j)(0) x^j.  This linear system is reducible to copies of simple harmonic motion if we can diagonalize the matrix F'(0) (e.g., if the force is conservative this is symmetric, a Hessian, and therefore diagonalizable).  This same reasoning works in infinite dimensions, formally, so in quantum field theory.  If you want an everyday example, kick a wall.  It will shake.  This is why.&lt;/p&gt;&#xA;" OwnerUserId="20" LastActivityDate="2010-11-18T03:19:49.967" />
  <row Id="1042" PostTypeId="1" AcceptedAnswerId="1045" CreationDate="2010-11-18T03:24:40.090" Score="3" ViewCount="445" Body="&lt;p&gt;In the photoelectric effect, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Work_function&quot; rel=&quot;nofollow&quot;&gt;work function&lt;/a&gt; is the minimum amount of energy (per photon) needed to eject an electron from the surface of a metal. Is it possible to calculate this energy from the atomic properties of the metal (atomic number, atomic mass, electron configuration, etc.), and the properties of the lattice structure, in a way that doesn't require a complex computational model? (If so, how?) Or, if not, is there a simple formula in terms of the metal's atomic/lattice properties that can approximate it to within a few percent?&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-18T19:06:16.637" LastActivityDate="2010-11-18T19:06:16.637" Title="How would I calculate the work function of a metal?" Tags="&lt;solid-state-physics&gt;&lt;photoelectric-effect&gt;" AnswerCount="2" />
  <row Id="1043" PostTypeId="2" ParentId="1018" CreationDate="2010-11-18T04:26:41.123" Score="1" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/1lqca.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a particle in circular motion with angular velocity $\omega$ and radius of the circle $A$. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Projection of the particle on diagonal&#xA;  of the circle is in SHM.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suppose the diagonal of the circle is on X-axis and center of the circle is $(0,0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that equation of position of the projection at time $t$ is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi$ angle at time $t=0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x(t) = Acos(\omega t + \phi)$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-11-18T04:51:35.143" LastActivityDate="2010-11-18T04:51:35.143" />
  <row Id="1044" PostTypeId="2" ParentId="1042" CreationDate="2010-11-18T04:34:55.267" Score="0" ViewCount="" Body="&lt;p&gt;In order to calculate a work function dependent on properties of the metal, the traditional method is to calculate the electronic current density $j_{z}$ by using the Richardson-Dushman equation for the thermionic emission which is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$j_{z}=BT^{2}e^{-\frac{W}{K_{B}T}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $W$ is the work function of the metal, $T$ is the thermodynamic temperature of the metal, $B$ is a constant calculated as $1.20173 X 10^{6} \text{A} m^{-2}K^{-2}$. See more information about the constant $B$ in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermionic_emission#Richardson.27s_Law&quot; rel=&quot;nofollow&quot;&gt;thermionic emission&lt;/a&gt; article in Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then plotting $\ln\bigg|\displaystyle \frac{j_{z}}{T^{2}}\bigg|$ against $\displaystyle\frac{1}{K_{B}T}$ produces a straight line with slope $-W$. That's the way in which most of the work functions are calculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electron distribution is considered in the derivation of the Richardson-Dushman equation by using the Fermi-Dirac statistics.&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2010-11-18T04:34:55.267" CommentCount="6" />
  <row Id="1045" PostTypeId="2" ParentId="1042" CreationDate="2010-11-18T04:45:58.580" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;You can't just get it from the atomic properties, the electronic properties of a metal are dominated by &quot;solid state&quot;-type considerations&lt;/strong&gt;, for instance, the fact that electrons live in a band structure rather than something more akin to the usual discrete levels that one learns about in QM 1.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thankfully, Ashcroft and Mermin's classic book has a long discussion on the work function in chapter 18.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their formula is $W=-\epsilon_F+W_s$, where $\epsilon_F$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermi_energy&quot; rel=&quot;nofollow&quot;&gt;Fermi energy&lt;/a&gt;, a quantity determined by the density of electrons and the properties of the crystal lattice of the metal; you can work out reasonable approximations to this for alkali metals by using the free electron approximation.  $W_s$ is a quantity related to surface effects; for this term Ashcroft and Mermin give a model with a dipole moment per unit area of $P$, so that $W_s=-4\pi e P$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure whether you can really get &quot;within a few percent&quot; with such crude techniques, but it's certainly something that's calculable.  In particular, getting a good approximation comes down to two things, 1) getting a handle on the band structure of the metal so that you can calculate $\epsilon_F$ accurately 2) having a good model for the surface of the metal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a wrinkle here about how $\epsilon_F$ is defined.  One can't just use the usual expression $\epsilon_F=\hbar^2k_F^2/2m$ as one needs to add in a term corresponding to the electrostatic energy of the ions, something like the Madelung constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, &lt;strong&gt;I recommend the book of Ashcroft and Mermin&lt;/strong&gt; for this (and any other just-beyond-basic questions you might have about thinking about electrons in metals and semiconductors).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just out of curiosity, I dug into the literature a little bit to see what researchers have done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1971, &lt;a href=&quot;http://prb.aps.org/abstract/PRB/v3/i4/p1215_1&quot; rel=&quot;nofollow&quot;&gt;Lang and Kohn&lt;/a&gt; were able to get work functions of simple metals to about 5% and noble metals to 15%.  I think it would be possible to reproduce these calculations today quite easily.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more recent paper (2005) by &lt;a href=&quot;http://dx.doi.org/10.1016/j.susc.2005.12.008&quot; rel=&quot;nofollow&quot;&gt;Da Silva, Stampfl, and Scheffler&lt;/a&gt; uses all-electron first principles calculations, as opposed to the more empirical fitted pseudopotential calculation that Lang and Kohn used.  My impression from skimming the data is that they only get marginal agreement with experiment; in particular, I noticed that the experimental values seem to have quite a bit of variance in them, consistent with the fact that work function is strongly dependent on surface properties, which are hard to get consistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for Aluminum along the 1 1 1 crystal plane, their calculations yield 4.21eV, compared to experimental values of 4.48, 4.24 and 4.33.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By comparison, Lang and Kohn's 1979 calculation gave 4.05 eV and they quoted an experimental value of 4.19 eV.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently the last big review on calculating work functions of metals is this one by &lt;a href=&quot;http://www.springerlink.com/content/p1174q76124k8584/&quot; rel=&quot;nofollow&quot;&gt;Hözl et al&lt;/a&gt; from 1979.  I didn't read it though.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-18T04:45:58.580" CommentCount="1" />
  <row Id="1046" PostTypeId="2" ParentId="1018" CreationDate="2010-11-18T05:17:14.257" Score="0" ViewCount="" Body="&lt;p&gt;Anywhere there is a restoring force which is linear with some kind of pertrubation from equilibrium, harmonic motion occurs. Most dynamic systems ehibiting either elastic forces, or some kind of conservative field (like gravity) have a tendency to exhibit harmonic motion. I guess you can view it as the natural dance of potential and kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example we used in high school was a cork floating on the surface of water vertically. Starting from the position where all is in equilibrium, a small dip of the cork increases the boyancy linearly, and a small rise of the cork decreases. Since the sum of the forces are zero at equilibrium then changes in boyancy are viewed as an additional opposing force to motion that is linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A practical application is the measurement of mass moment of inertia of an axisymmetric shape. You suspend the shape from a thin elastic wire and twist it slightly. The resulting oscilation back and forth is S.H.M. and its period is used to calculate what the MMOI around the axis of rotation is. There are companies out there that part of quality assurance for manufactured parts, they run samples through this &quot;simple&quot; test to check the expected performance in the field.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-11-18T05:17:14.257" />
  <row Id="1047" PostTypeId="2" ParentId="937" CreationDate="2010-11-18T05:26:42.887" Score="1" ViewCount="" Body="&lt;p&gt;Isn't there a rule that comes out of the field equations for gravity that the gravity felt in a closed surface (like a shpere) is proportional to the total mass enclosed by the surface? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this is just too simplistic of a view for gravity, I will admit to it.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-11-18T05:26:42.887" />
  <row Id="1048" PostTypeId="1" AcceptedAnswerId="12781" CreationDate="2010-11-18T05:39:42.043" Score="16" ViewCount="721" Body="&lt;p&gt;Suppose in the milliseconds after the big bang the cosmic egg had aquired some &lt;strong&gt;large&lt;/strong&gt; angular momentum. As it expanded, keeping the momentum constant (not external forces) the rate of rotation would have slowed down, but it would never reach zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What implications would that have to measurments to distant supernova and CMB radiation? Do we have any experimental data that definitely rules out such as scenario? And to what confidence level?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;A &lt;a href=&quot;http://news.discovery.com/space/do-we-live-in-a-spinning-universe-110708.html#mkcpgn=rssnws1&quot; rel=&quot;nofollow&quot;&gt;Recent Article&lt;/a&gt; suggests that the universe might indeed be spinning as a whole. Anyone care to poke holes at it?&lt;/p&gt;&#xA;" OwnerUserId="392" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2011-07-10T05:25:29.777" LastActivityDate="2011-07-26T01:59:09.297" Title="What if the universe is rotating as a whole?" Tags="&lt;general-relativity&gt;&lt;cosmology&gt;&lt;angular-momentum&gt;&lt;measurement&gt;" AnswerCount="5" CommentCount="9" FavoriteCount="1" />
  <row Id="1049" PostTypeId="2" ParentId="1037" CreationDate="2010-11-18T05:54:41.480" Score="2" ViewCount="" Body="&lt;p&gt;The thing that confused me at first about contact lenses was that they didn't look like &quot;normal&quot; lenses, in which each side of them curved in the opposite direction to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand them, contact lenses are Meniscus lenses - wherein the sides are curved in the same direction as each other, but more curved on one of the sides. You can see some examples &lt;a href=&quot;http://en.wikipedia.org/wiki/Meniscus_lens#Types_of_simple_lenses&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="158" LastActivityDate="2010-11-18T05:54:41.480" />
  <row Id="1050" PostTypeId="2" ParentId="387" CreationDate="2010-11-18T07:31:58.940" Score="2" ViewCount="" Body="&lt;p&gt;Matt Reece gives a good answer, but one additional area of tension that seems worth mentioning is the problem of time.  The role of time in quantum theory is quite different from general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See here for a review of some of the issues involved: &lt;a href=&quot;http://arxiv.org/abs/gr-qc/9210011&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/gr-qc/9210011&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11" LastActivityDate="2010-11-18T07:31:58.940" CommentCount="1" />
  <row Id="1051" PostTypeId="1" AcceptedAnswerId="1052" CreationDate="2010-11-18T08:25:55.393" Score="0" ViewCount="108" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/843/maxwells-demon-constant-information-energy-equivalence&quot;&gt;Maxwell&amp;#39;s Demon Constant (information-energy equivalence)&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I was reading:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.scientificamerican.com/article.cfm?id=demonic-device-converts-inform&quot; rel=&quot;nofollow&quot;&gt;Demonic device converts information to energy : Experiment inspired by a paradox tempts a bead uphill.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its good to see conservation of energy is violated. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know more about it. What other resources are available (that doesn't involve too much math). Especially, I want to see that beed experiment. Are there any demonstrations available on net?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I did learn more about maxwell's daemon from this:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/yrLNz.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="447" LastEditorUserId="447" LastEditorDisplayName="" LastEditDate="2010-11-18T13:51:06.903" LastActivityDate="2010-11-18T13:51:06.903" ClosedDate="2010-11-22T07:14:34.293" Title="conversion of information to energy" Tags="&lt;energy-conservation&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="1052" PostTypeId="2" ParentId="1051" CreationDate="2010-11-18T08:41:41.100" Score="8" ViewCount="" Body="&lt;p&gt;I'm sorry to crush your joy, but conservation of energy is not violated. However, Scientific American does rape science, that I can assure you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just ask yourself: how did they get the information in the first place? Didn't that require energy? How much? You'll see that the answer to these questions rules out any violation of the laws of thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saying that energy conservation is violated in this case is about as deep as saying that it is violated in the extraction of energy from petroleum. &lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2010-11-18T08:41:41.100" CommentCount="4" />
  <row Id="1054" PostTypeId="1" CreationDate="2010-11-18T10:32:59.860" Score="1" ViewCount="160" Body="&lt;p&gt;What do you guys think of &lt;a href=&quot;http://www.dilbert.com/blog/entry/what_is_the_universe_made_of/&quot; rel=&quot;nofollow&quot;&gt;Scott Adams' theory&lt;/a&gt; that the Universe might be made of recursions, created by observation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find it fascinating but have no strong opinion yet if it makes sense.&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="145" LastEditorDisplayName="" LastEditDate="2010-11-18T11:09:42.947" LastActivityDate="2010-11-18T11:47:32.853" ClosedDate="2010-11-18T16:59:02.380" Title="What if the Universe is made of recursions?" Tags="&lt;universe&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="1055" PostTypeId="2" ParentId="1048" CreationDate="2010-11-18T11:32:54.250" Score="1" ViewCount="" Body="&lt;p&gt;If you say that universe is rotating &quot;as a whole&quot; you'd have to define a rotation axis. A rotation axis would cause some problems, like violating a homogeneous and isotropic universe, as proposed by cosmic inflation model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, you can not say that the universe is rotating at the same angular speed everywhere with respect to the rotation axis. This would cause faster than light linear velocities, although I simply used classical mechanics to calculate, maybe a relativistic model is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The existence of such an angular momentum might have some implications though. For example, one could say, if the theory &quot;Big Bounce&quot;(http://en.wikipedia.org/wiki/Big_bounce) is correct, one could say the origin of the entire universe is a black hole with an angular momentum, that is, a rotating black hole with possibly no electrical charge.&lt;/p&gt;&#xA;" OwnerUserId="504" LastActivityDate="2010-11-18T11:32:54.250" CommentCount="3" />
  <row Id="1056" PostTypeId="2" ParentId="1054" CreationDate="2010-11-18T11:47:32.853" Score="1" ViewCount="" Body="&lt;p&gt;Although that particular blog is mostly philosophy, there are some ideas which are scientifically relevant in there. However he, as many other people who are not very deeply interested in physics, is making a mistake of trying to make analogies to classical world while examining the microcosmos, like questioning what an elementary particle is made of. It, of course, sounds absurd when considered from a macro point of view, however the particles in question have such sophisticated behavior that they are almost virtual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To inspect &lt;em&gt;what&lt;/em&gt; a certain particle like electron is &lt;em&gt;made of&lt;/em&gt; can be a bit out of reach for humanity at the moment, and possibly in the near future. As far as our observations go, electron is not made out of anything but itself, but string theory proposes that it is actually made of tiny strings of energy, vibrating at certain frequencies and properties to create different forms of particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give a very confusing example, what exactly is an electron made of? Well, an electron orbiting an atom is made of probability waves. Yep, not electromagnetic, not sound, not seismic, but these are very real probability waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coming to your main question, I think this is much more a philosophy than science as it has no scientific basis and no provability in the foreseeable future. Most importantly, it does not address to any phenomena we see around us. Accepting the existence of such a theory would be a bit hypocritical as we do not question either existence or non-existence of God in science since &quot;God&quot; is not a necessary fact to question while observing the universe.&lt;/p&gt;&#xA;" OwnerUserId="504" LastActivityDate="2010-11-18T11:47:32.853" CommentCount="1" />
  <row Id="1057" PostTypeId="2" ParentId="1039" CreationDate="2010-11-18T11:56:43.520" Score="6" ViewCount="" Body="&lt;p&gt;Basically, what your eye does is take an average value of all the light that comes into your eye, and present it as the final image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider a wide open eye. This eye receives information from all angles possible, and the larger angle the incoming angle has with respect to the normal line of the eye, the more refraction it will go through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When your eye does not function well, like myopia, your lens will not be able to refract properly. And every light that needs a refraction will undergo some deviation from its correct path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you do when squinting, or the tactic shown in Wikihow, is basically eliminating the larger angle light rays, while only receiving those that require little to none refraction to reach the optic disc in your retina. &lt;/p&gt;&#xA;" OwnerUserId="504" LastActivityDate="2010-11-18T11:56:43.520" CommentCount="1" />
  <row Id="1058" PostTypeId="1" CreationDate="2010-11-18T12:00:00.317" Score="2" ViewCount="121" Body="&lt;p&gt;&lt;sup&gt;[Remark: I admit that my first attempt on &lt;a href=&quot;http://physics.stackexchange.com/questions/994/what-makes-a-space-a-real-space&quot;&gt;What makes a space a real space?&lt;/a&gt; was rather ill-posed and led to some confusion. Sorry for that, but please give me a second try. Part of the confusion arose from the ununsual term &quot;real space&quot;. The far better term would have been &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Physical_space&quot; rel=&quot;nofollow&quot;&gt;physical space&lt;/a&gt;&quot; which I hope to be unambiguous.]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following let a &quot;mechanical system&quot; be a system of $n$ spatial objects moving in physical space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider you are given a function $q:\mathbb{R} \rightarrow \mathcal{M}^n$ with $\mathcal{M}$ a mathematical space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When would you assume that $q$ describes the evolution of a mechanical system?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would, if I found or was given a function $\mathcal{L}(x,\dot{x},t)$ such that the Lagrange equations hold for all times $t$ when inserting $q(t)$ for $x$ and $\dot{q}(t)$ for $\dot{x}$, and which I can consider physically plausible. In this case I would interpret $\mathcal{M}$ as physical space, maybe distorted by generalized coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would &lt;em&gt;not&lt;/em&gt;, if there is provably no such function. In this case $q$ describes the evolution of a general dynamical system, but not a mechanical one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there mathematically definable conditions on $\mathcal{L}$ for being physically plausible, or possible, let's say? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Besides, maybe, that $\mathcal{L}$ must be of the form $\mathcal{T} - \mathcal{V}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can anyone give an example of a non-mechanical system that &lt;em&gt;behaves&lt;/em&gt; like a mechanical one (in the narrow sense above)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One answer to this question comes immediately in mind: a system that is &lt;em&gt;designed&lt;/em&gt; to behave like a mechanical one, e.g. a computer running a simulation. I'd like to exclude such cases if I could, and restrict the question to &quot;natural&quot; systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What about possible worlds with a physical space like ours but in which the Lagrange equations do not hold?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="229" LastEditorUserId="229" LastEditorDisplayName="" LastEditDate="2010-11-18T14:58:56.820" LastActivityDate="2010-11-18T17:13:46.463" Title="Distinguishing mechanical systems from general dynamical systems" Tags="&lt;classical-mechanics&gt;&lt;space&gt;&lt;lagrangian-mechanics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="1059" PostTypeId="2" ParentId="1048" CreationDate="2010-11-18T12:08:41.213" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mach%27s_principle&quot; rel=&quot;nofollow&quot;&gt;Mach's principle&lt;/a&gt; says that inertia arises from the mass of everything else in the universe, and so if the universe were rotating, it would be dragging us around in a circle with it, and because of that we wouldn't notice it. This frame-dragging actually happens, and is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lense%E2%80%93Thirring_precession&quot; rel=&quot;nofollow&quot;&gt;Lense-Thirring effect&lt;/a&gt;. If you were inside a massive rotating sphere, your frame of reference would be rotating with respect the rest of the universe outside the massive rotating sphere. However, I'm not sure whether the Lense-Thirring effect actually implies Mach's principle. Maybe somebody who knows more could help.&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2010-11-18T12:08:41.213" CommentCount="3" />
  <row Id="1060" PostTypeId="1" AcceptedAnswerId="1062" CreationDate="2010-11-18T13:11:21.970" Score="7" ViewCount="324" Body="&lt;p&gt;A superconducting wire has no electrical resistance and as such it does not heat up when current passes through it. Non-superconducting wires can be damaged by too much current, because they get too hot and melt. But since current is the flow of electrons, surely unlimited electrons can't flow?&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="42" LastEditorDisplayName="" LastEditDate="2010-11-18T18:42:35.267" LastActivityDate="2010-11-18T19:58:19.023" Title="Can a superconducting wire conduct unlimited current?" Tags="&lt;solid-state-physics&gt;&lt;superconductivity&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" />
  <row Id="1061" PostTypeId="1" AcceptedAnswerId="1066" CreationDate="2010-11-18T13:20:41.843" Score="9" ViewCount="649" Body="&lt;p&gt;I am reading &lt;em&gt;A Brief History of Time&lt;/em&gt; by Stephen Hawking, and in it he mentions that without compensating for relativity, GPS devices would be out by miles. Why is this? (I am not sure which relativity he means as I am several chapters ahead now and the question just came to me.)&lt;/p&gt;&#xA;" OwnerUserId="104" LastActivityDate="2010-11-30T21:17:41.397" Title="Why does GPS depend on relativity?" Tags="&lt;general-relativity&gt;&lt;special-relativity&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" />
  <row Id="1062" PostTypeId="2" ParentId="1060" CreationDate="2010-11-18T13:47:58.460" Score="9" ViewCount="" Body="&lt;p&gt;There is a critical current density for every superconductor where the superconductor acts as an ordinary conductor and a voltage difference can be measured between its ends.&lt;/p&gt;&#xA;" OwnerUserId="504" LastActivityDate="2010-11-18T13:47:58.460" CommentCount="4" />
  <row Id="1063" PostTypeId="2" ParentId="1061" CreationDate="2010-11-18T13:51:57.817" Score="2" ViewCount="" Body="&lt;p&gt;The effect of gravitational time dilation can even be measured if you go from the surface of the earth to an orbit around the earth. Therefore, as GPS satellites measure the time it's messages take to reach you and come back, it is important to account for the real time that the signal takes to reach the target.&lt;/p&gt;&#xA;" OwnerUserId="504" LastActivityDate="2010-11-18T13:51:57.817" CommentCount="6" />
  <row Id="1064" PostTypeId="2" ParentId="630" CreationDate="2010-11-18T14:31:46.883" Score="4" ViewCount="" Body="&lt;p&gt;This paper seems relevant to your question:&#xA;&lt;a href=&quot;http://ajp.aapt.org/resource/1/ajpias/v68/i3/p245_s1&quot; rel=&quot;nofollow&quot;&gt;http://ajp.aapt.org/resource/1/ajpias/v68/i3/p245_s1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm reading the abstract correctly, the answers to your questions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q: It seems that if the coherence length of a laser is big enough, it is possible to observe a (moving) interference picture by combining them. Is it true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: Yes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q: How fast should photo-detectors be for observing of the interference of beams from two of the &quot;best available&quot; lasers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: 1 millisecond or faster&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q: What is the coherence length of the best-available laser?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: More than 300 km&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q: More specifically, does there exist any (optical single-wavelength) laser with coherence length exceeding 500 meters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: Yes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The abstract in question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Interference fringes produced by a pair of intracavity stabilized diode laser beams, each impinging separately on one aperture of a double slit, are recorded on a linear charge-coupled device array. The peculiar result of the experiment is that the fringe system is found to persist for a time of the order of 1 ms and loses contrast for longer integration times. This implies that the individual linewidths of the two beams from the stabilized lasers are narrower than 1 kHz and that the average drift rates of the central peaks are far less than 0.1 MHz/s. The device was built within the advanced undergraduate electronics laboratory of the department of physics and represents a considerable improvement over previous demonstration apparatuses used to detect interference fringes from independent lasers.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting 1986 review of interference from independent sources:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rev. Mod. Phys. 58, 209–231 (1986) &quot;Interference between independent photons&quot;&lt;/p&gt;&#xA;" OwnerUserId="539" LastActivityDate="2010-11-18T14:31:46.883" />
  <row Id="1065" PostTypeId="2" ParentId="1061" CreationDate="2010-11-18T14:40:30.363" Score="5" ViewCount="" Body="&lt;p&gt;You can find out about this in great detail in the excellent summary over here: &lt;a href=&quot;http://metaresearch.org/cosmology/gps-relativity.asp&quot; rel=&quot;nofollow&quot;&gt;What the Global Positioning System Tells Us about Relativity?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a nutshell:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;General Relativity&lt;/strong&gt; predicts that clocks go slower in a higher gravitational field. That is the clock aboard the GPS satellites &quot;clicks&quot; faster than the clock down on Earth.&lt;/li&gt;&#xA;&lt;li&gt;Also, &lt;strong&gt;Special Relativity&lt;/strong&gt; predicts that a moving clock is slower than the stationary one. So this effect will slow the clock compared to the one down on Earth.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As you see, &lt;em&gt;in this case&lt;/em&gt; the two effects are acting in opposite direction but their magnitude is not equal, thus don't cancel each other out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you find out your position by comparing the time signal from a number of satellites. They are at different distance from you and it then takes different time for the signal to reach you. Thus the signal of &quot;Satellite A says right now it is 22:31:12&quot; will be different from what you'll hear Satellite B at the &lt;em&gt;same moment&lt;/em&gt;). From the time difference of the signal and knowing the satellites positions (your GPS knows that) you can triangulate your position on the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one does not compensate for the different clock speeds, the distance measurement would be wrong and the position estimation could be hundreds or thousands of meters or more off, making the GPS system essentially useless.&lt;/p&gt;&#xA;" OwnerUserId="472" LastActivityDate="2010-11-18T14:40:30.363" />
  <row Id="1066" PostTypeId="2" ParentId="1061" CreationDate="2010-11-18T14:53:04.923" Score="15" ViewCount="" Body="&lt;p&gt;Error margin for position predicted by GPS is $15m$. So GPS system must keep time with accuracy of at least $15m/c$ which is roughly $50ns$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $50ns$ error in timekeeping corresponds to $15m$ error in distance prediction.&lt;br&gt;&#xA;Hence, for $38\mu s$ error in timekeeping corresponds to $11km$ error in distance prediction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we do not apply corrections using GR to GPS then $38\mu s$ error in timekeeping is introduced PER DAY.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can check it yourself by using following formulas&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_1 = \frac{T_0}{\sqrt{1-\frac{v^2}{c^2}}}$ ...clock runs relatively slower if it is moving at high velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_2 = \frac{T_0}{\sqrt{1-\frac{2GM}{c^2 R}}}$ ...clock runs relatively faster because of weak gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_1$ = 7 microseconds/day&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_2$ = 45 microseconds/day&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_2 - T_1$ = 38 microseconds/day&lt;/p&gt;&#xA;&#xA;&lt;p&gt;use values given in &lt;a href=&quot;http://www.crownedanarchist.com/emc2/applications_of_relativity_in_gps.doc&quot; rel=&quot;nofollow&quot;&gt;this very good article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for equations refer to &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/gratim.html#c5&quot; rel=&quot;nofollow&quot;&gt;hyperphysics&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Stephen Hawking is right! :-)&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-11-18T15:02:07.517" LastActivityDate="2010-11-18T15:02:07.517" CommentCount="1" />
  <row Id="1067" PostTypeId="1" AcceptedAnswerId="1070" CreationDate="2010-11-18T15:19:13.580" Score="3" ViewCount="324" Body="&lt;p&gt;I don't really know much about Quantum mechanics, but would like to know one simple fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The state function $\Psi(r, t)|$ whose magnitude gives the probability density of the position of the particle and the magnitude of its ($\Psi(r, t)$) fourier transform gives probability density of its momentum. Is there any rule that these state functions are smooth (possess infinite order derivatives everywhere) (derivatives of all orders exist)?&lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="540" LastEditorDisplayName="" LastEditDate="2010-11-18T16:14:13.337" LastActivityDate="2011-03-01T03:36:49.240" Title="Is it guaranteed that wavefunction is well behaved everywhere?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="2" />
  <row Id="1069" PostTypeId="2" ParentId="1060" CreationDate="2010-11-18T16:01:00.823" Score="13" ViewCount="" Body="&lt;p&gt;The problem with trying to pass huge amounts of current through a superconductor is that any flowing current creates a magnetic field circling around it (&lt;a href=&quot;http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_circuital_law&quot; rel=&quot;nofollow&quot;&gt;Ampere's law&lt;/a&gt;). A superconductor also expels all magnetic fields from inside itself (&lt;a href=&quot;http://en.wikipedia.org/wiki/Meissner_effect&quot; rel=&quot;nofollow&quot;&gt;Meissner effect&lt;/a&gt;), so what you get is a lot of magnetic field lines all bunched up just outside the surface of the superconductor, which makes a strong magnetic field there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnetic fields kill superconductivity. Every superconductor has a &quot;critical field&quot; at which it stops superconducting (which actually depends strongly on temperature). So at some point the magnetic field produced by the current will be too much for the superconductor to handle, and it will &quot;quench&quot;, which means it abruptly stops superconducting, and then the energy in the flowing current is very rapidly turned into heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A superconducting magnet quench can produce a sizeable explosion!&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2010-11-18T16:01:00.823" CommentCount="6" />
  <row Id="1070" PostTypeId="2" ParentId="1067" CreationDate="2010-11-18T16:15:09.250" Score="8" ViewCount="" Body="&lt;p&gt;The only general requirement on the state function for a single, spinless, quantum particle (quanton) in a physically realistic state is that the state function be square integrable, i.e., the integral of its absolute value squared over all space be finite. Non-square integrable state functions are used for many purposes, but they are all idealizations that do not, individually, represent realistic states. If the state function is also to belong to the domain of definition of the Hamiltonian, then, in non-relativistic QM, the state function must be spatially differentiable to second order as well. State functions which are square integrable but not second order differentiable do not satisfy the Schroedinger equation. But their time evolution is still determined by continuity considerations since the second order differentiable state functions are everywhere dense in the state space, i.e., Hilbert space.  &lt;/p&gt;&#xA;" OwnerUserId="345" LastActivityDate="2010-11-18T16:15:09.250" CommentCount="4" />
  <row Id="1071" PostTypeId="2" ParentId="1067" CreationDate="2010-11-18T16:23:17.137" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/qm.html#c4&quot; rel=&quot;nofollow&quot;&gt;Some of the conditions&lt;/a&gt; for wavefunctions $\Psi(x)$, for all elements $x$ of a subset of $\mathbb{R^{d}}$ (in the hyperphysics link, they use $x \in \mathbb{R}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.- Must be a solution of the Schrodinger equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.- Must be normalizable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.- Must be a continuous function of $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.- The slope of the function in $x$ must be continuous, that is, $\displaystyle \frac{\partial \Psi(x)}{\partial x}$ must be continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The property of being square-integrable is included in the condition 2.&lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="18" LastEditorDisplayName="" LastEditDate="2010-11-19T02:08:04.530" LastActivityDate="2010-11-19T02:08:04.530" CommentCount="31" />
  <row Id="1072" PostTypeId="2" ParentId="1058" CreationDate="2010-11-18T17:13:46.463" Score="2" ViewCount="" Body="&lt;p&gt;The initial definition given here of a mechanical system;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In the following let a &quot;mechanical system&quot; be a system of n spatial objects moving in physical space.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is much broader than the restriction to a 'standard' Lagrangian framework would allow. By 'standard' I mean a Lagrangian depending only on q and its first time derivative, q', as well as, possibly, time itself. If one generalises the Lagrangian framework to allow dependence on higher order time derivatives, with the corresponding change in the Lagrangian equations of motion, then one comes closer to allowing for the variety permitted in the initial definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the rest of the question suggests that systems governed by a 'standard' Lagrangian are the real focus of interest here. The restriction of L to the form, T - V, queried in the 2nd question is devoid of content unless, as usual, one restricts the appearance of the time derivative, q', to T. Usually (but not always), such T's depend on q' via quadratic forms like q'Mq', where M is a matrix/tensor depending only on q.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for examples of systems evolving in accordance with equations of motion that would follow from such a Lagrangian formalism, but not be mechanical systems as per the initial definition, there may well be examples in the mathematical models used by economists and some social scientists to describe evolution of market variables or population parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one allows the exponent, n, of the mathematical space, M, in the initial definition to be infinite (corresponding to an infinite number of degrees of freedom) then the evolution of fields, such as the electromagnetic field, would provide examples. The fields take there values at all the points of physical space, but those values evolve in field-space, not physical space, so they are not mechanical according to the initial definition.   &lt;/p&gt;&#xA;" OwnerUserId="345" LastActivityDate="2010-11-18T17:13:46.463" CommentCount="4" />
  <row Id="1073" PostTypeId="1" AcceptedAnswerId="1086" CreationDate="2010-11-18T17:25:16.843" Score="9" ViewCount="395" Body="&lt;p&gt;First, I'll state some background that lead me to the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking about quantization of space-time on and off for a long time but I never really looked into it any deeper (mainly because I am not yet quite fluent in string theory). But the recent discussion about &lt;a href=&quot;http://physics.stackexchange.com/questions/937/how-does-gravity-escape-a-black-hole/953#953&quot;&gt;propagation of information in space-time&lt;/a&gt; got me thinking about the topic again. Combined with the fact that it has been more or less decided that &lt;a href=&quot;http://meta.physics.stackexchange.com/questions/94/level-of-questions-what-do-we-do-now&quot;&gt;we should also ask graduate level questions here&lt;/a&gt;, I decided I should give it a go.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;So, first some of my (certainly very naive) thoughts.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's no problem to quantize gravitational waves on a curved background. They don't differ much from any other particles we know. But what if we want the background itself to change in response to movement of the matter and quantize these processes? Then I would imagine that space-time itself is built from tiny particles (call them &lt;em&gt;space-timeons&lt;/em&gt;) that interact by means of exchanging gravitons. I drawed this conclusion from an analogy with how solid matter is built from atoms in a lattice and interact by means of exchanging phonons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I am aware that the above picture is completely naive but I think it must in some sense also be correct. That is, if there exists any reasonable notion of quantum gravity, it &lt;em&gt;has to&lt;/em&gt; look something like that (at least on the level of QFT).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, having come this far I decided I should not stop. So let's move one step further and assume string theory is correct description of the nature. Then all the particles above are actually strings. So I decided that space-time must probably arise as condensation of huge number of strings. Now does this make any sense at all? To make it more precise (and also to ask something in case it doesn't make a sense), I have two questions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;In what way does the classical space-time arise as a limit in the string theory? Try to give a clear, conceptual description of this process. I don't mind some equations, but I want mainly ideas.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Is there a good introduction into this subject? If so, what is it? If not, where else can I learn about this stuff?&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Regarding 2., I would prefer something not too advanced. But to give you an idea of what kind of literature might be appropriate for my level: I already now something about classical strings, how to quantize them (in different quantization schemes) and some bits about the role of CFT on the worldsheets. Also I have a qualitative overview of different types of string theories and also a little quantitative knowledge about moduli spaces of Calabi-Yau manifolds and their properties.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-11-19T21:38:34.940" Title="How does classical GR concept of space-time emerge from string theory?" Tags="&lt;string-theory&gt;&lt;quantization&gt;&lt;spacetime&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="4" />
  <row Id="1074" PostTypeId="1" AcceptedAnswerId="1078" CreationDate="2010-11-18T18:10:37.593" Score="4" ViewCount="343" Body="&lt;pre&gt;&lt;code&gt;                    Wire A&#xA;                  -----R-----    &#xA;                 /           \&#xA;------I---&amp;gt;-[p]--             ---------I--&amp;gt;&#xA;                 \           /&#xA;                  -----R-----    &#xA;                    Wire B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Wire A and B both have resistance $R$.&lt;br&gt;&#xA;So current $\frac{I}{2}$ flows through each of the wires.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How come electrons know which wire path to choose? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am thinking is if we consider electrons at point P are in state&#xA;$\frac{|A\rangle + |B\rangle}{\sqrt{2}}$ and at the junction some kind of measurement occurs which leaves half of the electrons in state $|A\rangle$ and other half in $|B\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please explain how exactly this works. Please point out why argument above is right or wrong (or at all relevant) in this situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;How can one derive basic laws for series and parallel ckts from more basic principles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! :-)&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-11-18T20:40:04.463" LastActivityDate="2011-06-04T22:04:13.130" Title="How do electrons choose their path?" Tags="&lt;quantum-mechanics&gt;&lt;electricity&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="1075" PostTypeId="1" AcceptedAnswerId="1079" CreationDate="2010-11-18T18:12:21.553" Score="10" ViewCount="504" Body="&lt;p&gt;How to get an approximation of Avogadro or Boltzmann constant through experimental means accessible by an hobbyist ?&lt;/p&gt;&#xA;" OwnerUserId="373" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-11-23T10:55:07.950" LastActivityDate="2011-04-18T21:08:13.053" Title="Home experiment to estimate Avogadro's number?" Tags="&lt;thermodynamics&gt;&lt;statistical-mechanics&gt;&lt;home-experiment&gt;&lt;mole&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="3" />
  <row Id="1076" PostTypeId="2" ParentId="1074" CreationDate="2010-11-18T18:19:37.550" Score="4" ViewCount="" Body="&lt;p&gt;Think of it as a queue of people trying to enter a building with two entrances, both the same size and length. As people push each other, and as the people on the front move, queue proceeds, and statistically on equal amount in both entrances provided they are identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same principle applies for different R's, R and 2R, where one entrance is smaller than the other one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your argument. It is correct &lt;strong&gt;mathematically&lt;/strong&gt;, in my opinion.&lt;/p&gt;&#xA;" OwnerUserId="504" LastEditorUserId="504" LastEditorDisplayName="" LastEditDate="2010-11-18T19:33:15.407" LastActivityDate="2010-11-18T19:33:15.407" CommentCount="5" />
  <row Id="1077" PostTypeId="1" CreationDate="2010-11-18T18:24:03.537" Score="10" ViewCount="542" Body="&lt;p&gt;There is this famous example about the order difference between gravitational force and EM force. All the gravitational force of Earth is just countered by the electromagnetic force between the electrons on the surface of my feet and the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But also wave-function of electrons(fermions) do not overlap each other due to Pauli exclusion principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So which one is the true reason of me not flopping inside the earth forever? Is it only one of them(my guess is Pauli exclusion) or is it both?&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-07T15:24:14.147" Title="How can I stand on the ground? EM or/and Pauli?" Tags="&lt;electromagnetism&gt;&lt;gravity&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="6" />
  <row Id="1078" PostTypeId="2" ParentId="1074" CreationDate="2010-11-18T18:28:17.390" Score="4" ViewCount="" Body="&lt;p&gt;To speak of quantum state, the electrons should be coherent, have a non-negliglible de Broglie wavelength. On the other hand, a non-zero resistance imply a nonnegligible dissipation and is sure to break any coherence. Your pure quantum state is then quickly turned into a classical probabilistic mixture, and your electron behaves exactly like a water drop in a river which separates itself into to branch in front of an island: its trajectory depends on its position inside the cable, relative to impurity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more quantitative, the quantum effect can roughly be seen on a length of the scale Λ, where Λ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_de_Broglie_wavelength&quot; rel=&quot;nofollow&quot;&gt;thermal de Broglie wavelength&lt;/a&gt;, given by :&#xA;$$&#xD;&#xA;   \Lambda&#xD;&#xA;   =&#xD;&#xA;   \frac{h}{\sqrt{2\pi mkT}}&#xD;&#xA;   \simeq 10^{-11}\mathrm{ m} &#xD;&#xA;$$&#xD;&#xA;where the numerical application is for an electron at room temperature. The quantum effects are therefore already negligible at the nanometric scale.&lt;/p&gt;&#xA;" OwnerUserId="373" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-11-19T11:29:49.627" LastActivityDate="2010-11-19T11:29:49.627" />
  <row Id="1079" PostTypeId="2" ParentId="1075" CreationDate="2010-11-18T18:30:53.330" Score="6" ViewCount="" Body="&lt;p&gt;Your best bet is to try to replicate the experiments of Perrin who first measured Avogadro's constant.  This is a common lab in &quot;Advanced Lab&quot; courses in undergraduate or graduate courses, so you can probably find writeups and such via google.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principle is to observe &lt;a href=&quot;http://en.wikipedia.org/wiki/Brownian_motion&quot; rel=&quot;nofollow&quot;&gt;Brownian motion&lt;/a&gt; under a microscope and measure the diffusion constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Einstein's theory for Brownian motion relates the diffusion constant of a spherical particle to the temperature via the &lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein_relation_%28kinetic_theory%29&quot; rel=&quot;nofollow&quot;&gt;Einstein-Stokes law&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D=\frac{k_BT}{6\pi\eta r}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $D$ is the diffusion constant, $T$ is the temperature, $\eta$ is the viscosity, and $r$ is the radius of a small spherical particle.  All of these properties should be measurable at least crudely with home equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I did this in my lab course was as follows.  We had access to a microscope with a CCD camera which allowed digital video recording, as well as samples of monodisperse polystyrene particles (which are commercially available, and labeled with their size).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suspend the particles in water (the viscosity of water is well known) at room temperature and place on a slide (better, put a thermocouple in your sample).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take video of the particles Brownian motion, and then using something like &lt;a href=&quot;http://www.physics.emory.edu/~weeks/idl/&quot; rel=&quot;nofollow&quot;&gt;John Crocker, David Grier, and Eric Weeks's celebrated particle tracking code&lt;/a&gt; extract 2D (or maybe 3D?) particle trajectories (i.e. $x(t)$, $y(t)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now plot the mean squared displacement of particles versus time.  The slope of this curve is the diffusion constant, which then yields an estimate for $k_B$ via Stokes-Einstein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To recover Avogadro's constant, you need the ideal gas constant $R$, which is measured through independent means; typically via macroscopic thermodynamic experiments which probe the slope in $pV=nRT$.&lt;/p&gt;&#xA;" OwnerUserId="42" LastEditorUserId="42" LastEditorDisplayName="" LastEditDate="2010-11-18T18:36:46.050" LastActivityDate="2010-11-18T18:36:46.050" />
  <row Id="1080" PostTypeId="1" CreationDate="2010-11-18T18:36:38.357" Score="7" ViewCount="455" Body="&lt;p&gt;When I first learned about the four fundamental forces of nature, I assumed that they were just the only four kind of interactions there were. But after learning a little field theory, there are many other kinds of couplings, even in the standard model. So why isn't the Yukawa Higgs coupling considered a fifth fundamental force? To be a fundamental force, does there need to be a gauge boson mediating it?&lt;/p&gt;&#xA;" OwnerUserId="542" LastActivityDate="2010-11-18T22:38:00.540" Title="Why isn't Higgs coupling considered a fifth fundamental force?" Tags="&lt;particle-physics&gt;&lt;experimental-physics&gt;&lt;quantum-field-theory&gt;&lt;forces&gt;&lt;higgs-boson&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="1081" PostTypeId="2" ParentId="1077" CreationDate="2010-11-18T18:43:12.543" Score="4" ViewCount="" Body="&lt;p&gt;OK, I'll make a complete revision of my original reply since it was quite sloppy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I originally confused two issues that are nevertheless related, I confused stability of matter and the impenetrability of matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, it must be clear that the two questions are related. If I have two chunks of matter of the same type on top of one another, one can't imagine that the explanation for the fact that these chunks don't &quot;fall through&quot; each other would be unrelated to the explanation of why we do not fall through the ground. So, in ultimate analysis, the question is tied to the question of stability of matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there are several steps in the problem. To explain the stability of matter, one has to explain why atoms are stable (and before that why nuclei are stable), then one has to explain why agregates of atoms like solids or liquids can be stable, i.e. why bulk matter is stable. The stability of bulk matter will then serve as the basis to explain why &quot;we can stand on the ground&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with the last step and assuming we already know about the stability of bulk matter, we can imagine that when we exert a pressure on stable bulk matter, we can expect by what it means to be in a stable equilibrium, that the piece of matter would exert an opposing pressure, trying to restore itself to its most stable configuration, provided stresses are not too great. So solving the problem of stability of bulk matter will help us understand what the nature of the restoring force will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as is well known, electromagnetic forces can not be the sole explanation. There exist no stable equilibria when there are only electric charges interacting electromagnetically. I won't go through the proof here, but it is accessible to undergraduates, it can be found in the Feynman Lectures, book 2, chapter 5. It's an application of Gauss' law in the static case. The dynamic case only complicates matters in the wrong direction. As we know, accelerated charges radiate energy away, thus an electron orbiting a nucleus would soon fall inward if nothing prevented it, to take the classic example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enters Elliott Lieb and his paper '&lt;a href=&quot;http://128.112.100.2/~kirkmcd/examples/QM/lieb_rmp_48_553_76.pdf&quot; rel=&quot;nofollow&quot;&gt;The stability of matter&lt;/a&gt;' that can be easily found online. So, I'll quote a lot from there. It reviews a lot of the results in the field of mathematical physics of the problem of stability of matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what does Lieb essentially say about stability of atoms: that it is a consequence of a principle introduced by Sobolev. Sobolev's inequality states in a mathematically precise form that if one tries to compress the wavefunction anywhere, the kinetic energy will increase. It's a kind of stronger version of the HUP. (Note that at this point, Lieb does not use Pauli's exclusion principle. This is to be expected, take a hydrogen atom, it is stable, since there's only one electron, Pauli's exclusion principle can't be invoked here to explain its stability.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, Lieb goes on to explain the stability of bulk matter by using Sobolev's inequality again. But this time around, he extends the inequality, and takes into account the fact that matter is made up of fermions. So, the Pauli exclusion principle is indeed used. So, again a lower bound for the kinetic energy is found, the interesting thing is that this lower bound is proportional to $N^{5/3}$ where $N$ is the number of fermions. If the particles were not fermions, the proportionality would have been $N$, which we can see by using the previous bound for 1 atom and multiplying by the number of atoms. So it is really the Pauli exclusion principle that contributes the factor $N^{2/3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lieb goes then to show that this factor is crucial. He uses Thomas-Fermi theory as a relevant approximation of the behaviour of bulk matter to demonstrate this. This is were the analysis becomes very intricate. I have no time to summarize it in more detail. So I'll just say that some theorems about the nature of TF theory are derived, these are then combined in the end to show that the minimum energy or ground state energy of the system is bounded from below. A numerical value for this bound is derived which is $-23Ry/$particle, $(1Ry \approx 13.6 eV)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important take-away message is though that Fermi-statistics or the Pauli exclusion principle is indeed essential to explain the stability of bulk matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Lieb's paper, there is an extra chapter which tackles the question of why matter doesn't explode, instead of implode. The interesting thing is that pure EM is sufficient to answer this question.&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2010-12-04T13:57:53.723" LastActivityDate="2010-12-04T13:57:53.723" CommentCount="12" />
  <row Id="1083" PostTypeId="2" ParentId="1060" CreationDate="2010-11-18T19:58:19.023" Score="3" ViewCount="" Body="&lt;p&gt;In addition to the fundamental limits noted by others there is the practical engineering issue of dealing with the increasing heating at the point where the superconductor joins the normally conducting parts of your circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Get it wrong, and &lt;em&gt;&quot;Hello, Quench City!&quot;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2010-11-18T19:58:19.023" />
  <row Id="1084" PostTypeId="2" ParentId="1075" CreationDate="2010-11-18T20:12:06.607" Score="4" ViewCount="" Body="&lt;p&gt;Well, addressing only $N_A$, and allowing that it only gets you halfway there: Carl Sagan did a bit on &lt;em&gt;Cosmos&lt;/em&gt; for estimating the &lt;em&gt;size&lt;/em&gt; of a oil molecule. Put a known amount (both volume and mass) of cooking oil on a calm body of water and wait for the slick to diffuse to it's maximum contiguous area. Estimate the area.{*} Divide the original volume by that area and you have a rough value for the linear size scale of the molecule and thus how much volume it occupies. (We're neglecting issues of aspect ratio and polarity here so this is only a order of magnitude value.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you have the molecular mass (from a mass spectrometer or some such) you're got all you need.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;{*} A challenging step, but I'd use a photograph with some know length reference for scale and a &lt;a href=&quot;http://en.wikipedia.org/wiki/Planimeter&quot; rel=&quot;nofollow&quot;&gt;planimeter&lt;/a&gt;. If you're not familiar with these instruments get ready to geek out. Very cool.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2010-11-18T20:12:06.607" CommentCount="5" />
  <row Id="1085" PostTypeId="2" ParentId="1080" CreationDate="2010-11-18T20:28:54.977" Score="3" ViewCount="" Body="&lt;p&gt;The Higgs exchange between matter particles can certainly be called a force. Whether it can be viewed a &lt;em&gt;fundamental&lt;/em&gt; force, is a matter of taste. But there is one important distinction between the force due to the Higgs exchange and the usual fundamental interactions. The strong and electroweak interactions are described as gauge interactions. It means that they are not put by hand but arise automatically when you require that the matter fields be invariant under local certain internal transformations (phase rotations, color transformations, etc.). In contrast to that, the &quot;Higgs force&quot; is put into the model by hand, as its presence is not driven by any symmetry consideration. You can equally nicely consider a model with zero Yukawa coupling.&lt;/p&gt;&#xA;" OwnerUserId="59" LastActivityDate="2010-11-18T20:28:54.977" CommentCount="3" />
  <row Id="1086" PostTypeId="2" ParentId="1073" CreationDate="2010-11-18T20:39:59.970" Score="8" ViewCount="" Body="&lt;p&gt;First, you are right in that non-Minkowski solutions to string theory, in which the gravitational field is macroscopic, it should be thought of as a condensate of a huge number of gravitons (which are one of the spacetime particles associated to a degree of freedom of the string).  (Aside:  a point particle, corresponding to quantum field theory, has no internal degrees of freedom; the different particles come simply from different labels attached to ponits.  A string has many degrees of freedom, each of which corresponds to a particle in the spacetime interpretation of string theory, i.e. the effective field theory.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To your question (1):  certainly there is no great organizing principle of string theory (yet).  One practical principle is that the 2-dimensional (quantum) field theory which describes the fluctuations of the string worldsheet should be &lt;em&gt;conformal&lt;/em&gt;, i.e. independent of local scale invariance of the metric.  This allows us to integrate over all metrics on Riemann surfaces &lt;em&gt;only up to diffeomorphisms and scalings&lt;/em&gt;, which is to say only up to a finite number of degrees of freedom.  That's an integral we can do.  (Were we able to integrate over all metrics in a way that is sensible within quantum field theory, we would already have been able to quantize gravity.)  Now, scale invariance imposes constraints on the background spacetime fields used to construct the 2d action (such as the metric, which determines the energy of the map from the worldsheet of the string).  These constraints reduce to Einstein's equations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's not a very fundamental derivation, but formulating string theory in a way which is independent of the starting point (&quot;background independence&quot;) is notoriously tricky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2):  This goes under the name &quot;strings in background fields,&quot; and can be found in Volume 1 of Green, Schwarz and Witten.&lt;/p&gt;&#xA;" OwnerUserId="20" LastEditorUserId="20" LastEditorDisplayName="" LastEditDate="2010-11-19T21:38:34.940" LastActivityDate="2010-11-19T21:38:34.940" CommentCount="3" />
  <row Id="1087" PostTypeId="2" ParentId="139" CreationDate="2010-11-18T21:45:53.330" Score="4" ViewCount="" Body="&lt;p&gt;Your question is equivalent to asking what the absolute mass of the neutrinos is, and the answer is currently unknown. We do have decent values for the differences of the squared masses for all three required neutrino states (one pair separated by about $7.7 \times 10^{-5}\text{ eV}^2$, and another one standing off from them by about $\pm 2.4 \times 10^{-3}\text{ eV}^2$ (but note the sign ambiguity a problem known as the mass hierarchy question)). This puts a lower bound on the mass of the &lt;em&gt;most&lt;/em&gt; massive state at about $0.05\text{ eV}$, but puts no non-trivial bound on the lowest neutrino mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supernova neutrinos may eventually be able to answer this by time-of-flight, but it depends on the theoretical understanding of exactly what is going on inside the exploding star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is some data from SN 1987a, but it is not of sufficient quality to provide even a rough answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best we can say at this time is that they are known to be quite small, indeed. (And that is &lt;em&gt;another&lt;/em&gt; problem to keep the theorists busy...)&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-01-15T19:13:58.493" LastActivityDate="2011-01-15T19:13:58.493" />
  <row Id="1088" PostTypeId="2" ParentId="1080" CreationDate="2010-11-18T22:38:00.540" Score="3" ViewCount="" Body="&lt;p&gt;Forces and interactions are similar, so maybe for the purposes of this discussion we should define force as &quot;interaction with a massless particle,&quot; as this gives the possibility of macroscopic-range &quot;forces&quot; (barring impeding effects such as confinement or symmetry breaking).  After all, gravity is a massless but non-Yang-Mills force (mediated by spin-2, not spin-1, particle).  There are some constraints on which spins (representations) of massless particles are possible in quantum field theory, but I'm not familiar enough to try to cite them off-hand, sorry.&lt;/p&gt;&#xA;" OwnerUserId="20" LastActivityDate="2010-11-18T22:38:00.540" />
  <row Id="1089" PostTypeId="2" ParentId="916" CreationDate="2010-11-18T22:48:34.333" Score="2" ViewCount="" Body="&lt;p&gt;For one thing, you've made a careless algebra mistake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're computing $(L/M)^2$ instead of $(M/L)^2$.&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2010-11-18T22:48:34.333" CommentCount="2" />
  <row Id="1090" PostTypeId="2" ParentId="1039" CreationDate="2010-11-18T23:02:27.070" Score="14" ViewCount="" Body="&lt;p&gt;For someone with perfect vision, the lens in their eye focuses a point source in their field of vision to a point source on their retina. For someone with less than perfect vision, the focus lies either before or behind the retina, resulting in a large spot on the retina, instead of a point. See the top illustration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/5upTV.png&quot; alt=&quot;Nearsighted eye with and without aperture&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you squint, or roll your fingers into a tube in front of your eye, you're placing an aperture in the way of the light. See the bottom illustration. As the diagram shows, the focus stays the same distance away from the retina, but since the angles are smaller, the spot on the retina is also smaller. A smaller spot is closer to a point, and therefore seems &quot;sharper&quot; - even though the vision is just as bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Camera_obscura&quot; rel=&quot;nofollow&quot;&gt;camera obscura&lt;/a&gt; works on a similar principle.&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2010-11-18T23:02:27.070" CommentCount="2" />
  <row Id="1091" PostTypeId="1" AcceptedAnswerId="1093" CreationDate="2010-11-19T00:03:38.763" Score="1" ViewCount="70" Body="&lt;p&gt;What are the current explanations and models for the formation of a central bar in some galaxies, and what is the difference that can produce a barred galaxy instead of a spiral one ?&lt;/p&gt;&#xA;" OwnerUserId="217" LastActivityDate="2010-11-19T03:36:30.047" Title="Proposed model for existence of bar in galaxies" Tags="&lt;astrophysics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="1092" PostTypeId="2" ParentId="429" CreationDate="2010-11-19T01:19:48.830" Score="1" ViewCount="" Body="&lt;p&gt;I think what you are looking for are known as &quot;&lt;a href=&quot;http://www.atnf.csiro.au/computing/software/atca_aips/node11.html&quot; rel=&quot;nofollow&quot;&gt;Stokes Parameters&lt;/a&gt;&quot;. There is also a very nice description of the polarization tensor in Sec. 9.10 of Mukhanov's &quot;Physical Foundations of Cosmology&quot;. With these tools you can compare radiation patterns from different directions and what have you. Hope that helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers,&lt;/p&gt;&#xA;" LastActivityDate="2010-11-19T01:19:48.830" />
  <row Id="1093" PostTypeId="2" ParentId="1091" CreationDate="2010-11-19T03:23:26.207" Score="4" ViewCount="" Body="&lt;p&gt;It's my impression from skimming the literature just now that much of the current activity in this field is based on N-body simulations.  It's hard for me as an outsider to evaluate the analytical scenarios that are being tested, but I think that the key idea is that certain &lt;a href=&quot;http://en.wikipedia.org/wiki/Lindblad_resonance&quot; rel=&quot;nofollow&quot;&gt;Lindblad resonances&lt;/a&gt; which excite &lt;a href=&quot;http://en.wikipedia.org/wiki/Density_wave_theory&quot; rel=&quot;nofollow&quot;&gt;density waves&lt;/a&gt; are the main instabilities that lead to bar formation.  See also section (4) of &lt;a href=&quot;http://www.astro.virginia.edu/class/whittle/astr553/Topic06/Lecture_6.html&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt;, which gives a broad overview of stellar dynamics in galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, there are several guesses out there.  See &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0208252&quot; rel=&quot;nofollow&quot;&gt;this paper by Steinmetz&lt;/a&gt; for an overview of the understanding of galaxy shapes in general in 2002.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key citation in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Barred_spiral_galaxy&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt; linked to by Cem is &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0206273&quot; rel=&quot;nofollow&quot;&gt;a 2002 paper by Bournard and Combes (free on the arxiv)&lt;/a&gt; which describes gravitational N-body simulations which follow processes of bar formation and re-formation.  They find a strong dependence on the frequency of bar formation on the distribution of mass in the galaxy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A galactic disk may either keep a single bar or get several successive bars. By simulating several disks, we study the dependence on physical parameters. We find that the two determining parameters are the bulge to disk and halo to disk masses ratios. Light bulge galaxies maintain their bar, while bars are destructed and regenerated in heavy bulge galaxies. The halo mass has also an effect on the disk evolution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1016/S1384-1076%2802%2900102-1&quot; rel=&quot;nofollow&quot;&gt;Another paper from 2002 by Steinmetz and Navarro&lt;/a&gt; does a different simulation and observes that a bar is formed in their model galaxy due to the tidal forces from a satellite galaxy.  Here's a related image from their paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/hPPTc.jpg&quot; alt=&quot;Steinmetz and Navarro&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Caption: Fig. 4.  The tidal triggering of bar instability by a satellite. At z=1.6 the disk develops a well defined, long-lasting bar pattern as a result of tidal forcing by the satellite shown in the z=1.2 panel. The satellite is finally disrupted at z=1.18 after 7 pericentric passages. The bar pattern, however, survives for several Gyr, as shown in the picture. Stars less than 1.5 Gyr old are shown in blue, older stars in red. Note that the bar is most prominent in the young component. Horizontal bars in each panel are 5 (physical) kpc long. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="42" LastEditorUserId="42" LastEditorDisplayName="" LastEditDate="2010-11-19T03:36:30.047" LastActivityDate="2010-11-19T03:36:30.047" />
  <row Id="1094" PostTypeId="1" CreationDate="2010-11-19T03:29:47.077" Score="3" ViewCount="258" Body="&lt;p&gt;how I can create hindrances to radio waves?&lt;/p&gt;&#xA;" OwnerUserId="547" LastEditorUserId="42" LastEditorDisplayName="" LastEditDate="2010-11-19T03:34:09.097" LastActivityDate="2011-02-05T21:47:56.790" Title="how I can create hindrances to radio waves?" Tags="&lt;electromagnetic-radiation&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="1096" PostTypeId="1" AcceptedAnswerId="1127" CreationDate="2010-11-19T04:13:32.960" Score="-3" ViewCount="160" Body="&lt;p&gt;How could one explain the collision of two continuous mass distributions in view of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;gravitation (Newtonanian and General relativity) ?&lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="540" LastEditorDisplayName="" LastEditDate="2010-11-20T18:29:28.853" LastActivityDate="2010-11-20T18:29:28.853" ClosedDate="2010-11-22T08:20:54.490" Title="Gravity and Collision of two continuous mass distributions" Tags="&lt;general-relativity&gt;&lt;mechanics&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="1098" PostTypeId="2" ParentId="1096" CreationDate="2010-11-19T04:55:26.490" Score="0" ViewCount="" Body="&lt;p&gt;In General Relativity, a point mass is a black hole.  The collision of two black holes is a very complicated problem that has to be solved using numeric supercomputer simulation.  The short story is that they end up coalescing to a single black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some complications from this answer that could arise from spin-orbit coupling, but I don't think that's what you're asking about.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2010-11-19T16:15:06.967" LastActivityDate="2010-11-19T16:15:06.967" CommentCount="2" />
  <row Id="1099" PostTypeId="1" AcceptedAnswerId="1104" CreationDate="2010-11-19T05:20:16.840" Score="-1" ViewCount="227" Body="&lt;p&gt;A 1.0 mW laser ($\lambda$ = 590 nm) shines on a cesium photocathode (ϕ = 1.95 eV). Assume an efficiency of $10^{-5}$ for producing photoelectrons (that is, 1 photoelectron is produced for every 10^5 incident photons) and determine the photoelectric current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the way I tackled this problem was the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we know &#xA;$$KE_{max} = hf - \phi$$&#xD;&#xA;$$KE_{max} = 0.1180 eV$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now I dont know how to finish? I know the current is a rate of time but there is no time given here? Also what is the stopping potential? I did a little bit of research online and found this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_{0} = \frac{W}{q}$$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$KE_{max} = eV_{0} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried equation them but I dont know what to solve for (i solved for e which was the wrong answer).&lt;/p&gt;&#xA;" OwnerUserId="551" LastEditorUserId="504" LastEditorDisplayName="" LastEditDate="2010-11-19T12:24:40.493" LastActivityDate="2010-11-19T13:39:54.223" Title="photoelectric effect" Tags="&lt;quantum-mechanics&gt;&lt;homework&gt;&lt;photoelectric-effect&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="1101" PostTypeId="2" ParentId="1099" CreationDate="2010-11-19T07:24:43.617" Score="2" ViewCount="" Body="&lt;p&gt;Note the power units given for the laser intensity. Power is energy transferred or transformed per unit &lt;em&gt;time&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how many photons per second, and from that how many photo-electrons?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;And an interesting question to ask yourself then is if the metal is electrically isolated, can this go one forever, and if not why and how does it stop? What would you need to know to give a quantitative answer?&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2010-11-19T07:24:43.617" />
  <row Id="1102" PostTypeId="1" AcceptedAnswerId="1109" CreationDate="2010-11-19T08:01:00.573" Score="9" ViewCount="355" Body="&lt;p&gt;Let's imagine the following situation:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At an initial moment t=0 a large water drop with diameter for example D=10 cm is placed in deep space (Say an astronaut is experimenting). Let's the initial temperature of the drop be moderate $T_0=283 K$ and the drop itself is at rest at t=0. What will happen with the drop? Maybe it will decompose into smaller droplets while boiling? Or maybe it is going to flash-freeze and the ice shell is forming? Or maybe something else?&lt;/p&gt;&#xA;" OwnerUserId="552" LastActivityDate="2010-12-03T15:11:21.070" Title="A water drop in vacuum" Tags="&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="1103" PostTypeId="2" ParentId="1077" CreationDate="2010-11-19T08:05:58.120" Score="-1" ViewCount="" Body="&lt;p&gt;The interaction between the nucleus (the core of the atoms) and the electrons is electromagnetic, but the Pauli exclusion principle keeps them from falling into the core. This is why matter has volume, and why different objects cannot occupy the same point in space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find more details about the Pauli exclusion principle &lt;a href=&quot;http://www.fisica.aquila.infn.it/homepages/cavanna/SSIS_DidFisIII_web/MeccQuantNonRel/6_Pauli_Principle.pdf&quot; rel=&quot;nofollow&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is the interaction between atoms that make up the different structures in nature, the rocks and the whole Earth (&lt;strong&gt;solids&lt;/strong&gt;) beneath you, for instance. In these structures the atoms are bound together, mostly by electromagnetic interactions, and to pass some other atoms between them you have to invest some energy against that bound. For certain objects this energy is a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of crystals. Our puny human fingers are way too weak to &quot;pass between&quot; the crystal atoms. If we build some strong machines, those &lt;em&gt;can&lt;/em&gt; pass between the atoms in a crystal - that's called cutting. Like when the silicon crystal is sliced in a computer chip factory. Needs a lot of energy...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there other objects that are much less bound - you can e.g. pass between the molecules of water, it's called swimming (could be swimming in other &lt;strong&gt;liquids&lt;/strong&gt; too). Those molecules are not interacting with each other much. You still cannot make your atoms occupy the same space as the H or O atoms in the water, and that's because of Pauli. And of course, you are walking around on the surface of the Earth - passing between the atoms of the air, because atoms and molecules in &lt;strong&gt;gasses&lt;/strong&gt; are not bound to each other at all, you just have to move them out of your way (Pauli) but no bounding (EM) to work agains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this case you mentioned, the bounding force between the atoms that make up the material beneath you is stronger than the gravitational force pushing you and the ground together, thus not enough to break those bounds and let you &quot;slide down&quot;. &lt;/p&gt;&#xA;" OwnerUserId="472" LastActivityDate="2010-11-19T08:05:58.120" CommentCount="1" />
  <row Id="1104" PostTypeId="2" ParentId="1099" CreationDate="2010-11-19T08:12:31.967" Score="1" ViewCount="" Body="&lt;p&gt;Describe the process at a high level:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The laser produces a certain number of photons per second $N_{\gamma}$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;The photons free a certain number of electrons per second $N_e$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;The free electrons per second give you the current $I$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N_{\gamma}= \frac{P_{laser}}{E_{\gamma}}$, where $E_{\gamma} = h\nu = \frac{hc}{\lambda}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $N_{\gamma}= \frac{\lambda}{hc}P_{laser}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N_e = \eta N_{\gamma}$, where $\eta$ is the efficiency&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $N_e = \frac{\eta \lambda}{hc}P_{laser}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = N_e q = \frac{q \eta \lambda}{hc}P_{laser}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the current is proportional to the intensity (power) of the laser as expected:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = kP_{laser}$, where $k=\frac{q\eta\lambda}{hc}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last thing to check is the cutoff point, in other words if the energy of a photon is enough to overcome the binding energy of the electrons. This you already did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See here &lt;a href=&quot;http://physics.info/photoelectric/&quot; rel=&quot;nofollow&quot;&gt;http://physics.info/photoelectric/&lt;/a&gt;, in particular where it says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The rate (n/t) at which photoelectrons (with charge e) are emitted from a photoemissive surface can be determined from the photoelectric current (I)...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-19T13:39:54.223" LastActivityDate="2010-11-19T13:39:54.223" CommentCount="3" />
  <row Id="1105" PostTypeId="2" ParentId="1099" CreationDate="2010-11-19T10:38:43.447" Score="1" ViewCount="" Body="&lt;p&gt;The first thing you have to calculate is the number of photon incident on the photocathode per unit time!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That you can calculate by:&#xA;$P_{laser}=E_{inc}/dt=[J/sec]$&#xD;&#xA;since every photon carries an energy equal to $h\nu=h\cdot c/\lambda=3.37\cdot 10^{-19} J$ (greater than the work power of course, if not what the heck are we doing here??) you can obtain:&#xA;$$\frac{P_{laser}}{E_{photon}}=\frac{n}{dt}$$&#xD;&#xA;being $n/dt$ the photon/second flux.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You do the math and you find: $n/dt\approx3\cdot 10^{11}$ photon per second.. (check that!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The efficiency is $10^{-5}$ so that means that you produce $\approx 3\cdot 10^6$ electon per second: and that is you photoelectric current that you can convert in your favorite unit.&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2010-11-19T10:38:43.447" />
  <row Id="1106" PostTypeId="1" AcceptedAnswerId="1108" CreationDate="2010-11-19T11:52:11.357" Score="3" ViewCount="237" Body="&lt;p&gt;A friend of mine claims to have been able to surf the Internet without fuss on a Wi-Fi connection while performing NMR on samples he was analyzing. I would have thought the strong magnets needed for this would have washed out Wi-Fi signals due to the radio waves emitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How plausible is his claim?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, to be more general, what kind of (electro?)magnet would it take to interfere with mobile devices?&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-11-19T12:04:37.663" Title="Wi-Fi in the presence of very strong magnets?" Tags="&lt;electromagnetism&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="1107" PostTypeId="2" ParentId="1019" CreationDate="2010-11-19T11:53:15.050" Score="1" ViewCount="" Body="&lt;p&gt;A common misconception is that in a double slit experiment, the electrons or the photons go through both slits at the same time and interfere with themselves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing of the sort ever happens and that idea doesn’t come from quantum mechanics. It is just meaningless to ask the question (from which slit did the particle pass?) in the context of the particular experiment and even more meaningless to say that it went through both. The particle always passes from the one or the other slit if you setup an experiment that asks the question and in the case that you shoot the particles one at a time you always get one hit on the screen and not an interference of portions of the particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that the misconception has its roots in the wave analogy of the wave function description of quantum mechanics, where you must have a wave passing from both slits in order to have interference on the other side. Of course that picture doesn’t translate to having a particle passing from both slits, but it only states that there exists interference between the wave functions of the two independent events that translates to a distribution of probability for finding a particle on a particular point on the screen. That is a statement of &lt;a href=&quot;http://en.wikipedia.org/wiki/Born_rule&quot; rel=&quot;nofollow&quot;&gt;Born’s interpretation&lt;/a&gt; that has been recently tested with a &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0811/0811.2068v1.pdf&quot; rel=&quot;nofollow&quot;&gt;triple slit experiment&lt;/a&gt;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could ask at this point, “Ok, I get it for the electrons, but light behaves like a wave in everyday experience. What happens in that case?”, and the answer is that light behaves like a wave only when you have a large flux of photons and you can go to the continuous field approximation. It is then that the wave-like properties become apparent both for photons and electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; &lt;a href=&quot;http://bloggingheads.tv/diavlogs/13487&quot; rel=&quot;nofollow&quot;&gt;A very interesting video on Quantum Mechanics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update2:&lt;/strong&gt; &lt;a href=&quot;http://media.physics.harvard.edu/video/index.php?id=SidneyColeman_QMIYF.flv&quot; rel=&quot;nofollow&quot;&gt;QM in your face&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update3:&lt;/strong&gt; The Feynman Lectures on Physics vol3: Scattering from a crystal (neutrons).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... &lt;em&gt;Let’s review the physics of this experiment. If we could, in principle, distinguish the alternative final states (even though you do not bother to do so), the total, final probability is obtained by calculating the probability for each state (not the amplitude) and then adding them together. If you cannot distinguish the final states even in principle, then the probability amplitudes must be summed before taking the absolute square to find the actual probability. The thing you should notice particularly is that if you were to try to represent the neutron by a wave alone, you would get the same kind of distribution for the scattering of a down spinning neutron as for an up spinning neutron. You would have to say that the “wave” would come from all the different atoms and interfere just as for the up spinning one with the same amplitude. But we know that this is not the way it works. &lt;strong&gt;So as we stated earlier, we must be careful not to attribute too much reality to the waves in space. They are useful for certain problems but not for all&lt;/strong&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="383" LastEditorUserId="383" LastEditorDisplayName="" LastEditDate="2010-12-14T20:57:16.707" LastActivityDate="2010-12-14T20:57:16.707" CommunityOwnedDate="2010-11-26T21:47:08.610" CommentCount="33" />
  <row Id="1108" PostTypeId="2" ParentId="1106" CreationDate="2010-11-19T12:04:37.663" Score="8" ViewCount="" Body="&lt;p&gt;In NMR, the strong magnets set the frequency of the nuclear resonance, using the constant magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically the resonance radio waves are around the MHz frequencies whereas Wi-Fi is around 2.5GHz. When the frequencies are different, they don't disrupt each other's signals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electro-magnets wouldn't interfere, as it is a constant field produced rather than an elecromagnetic wave, and even then it would have to be at the same frequency.&lt;/p&gt;&#xA;" OwnerUserId="133" LastActivityDate="2010-11-19T12:04:37.663" CommentCount="1" />
  <row Id="1109" PostTypeId="2" ParentId="1102" CreationDate="2010-11-19T12:54:13.747" Score="10" ViewCount="" Body="&lt;p&gt;I think what would happen is that any water molecule with enough energy to escape the surface tension would escape. Because there is no air to provide the water molecule with a way of turning round and going back in, it would permanently leave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the highest energy molecules would selectively evaporate, lowering the average energy of the remaining water. This process is known as evaporative cooling, where you selectively remove the most energetic molecules (and yes, it's what happens when you blow on a hot drink).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the size, shape and method of getting it into the vacuum at that temperature (in a jar, take the jar away, blasted out a tube with gas) it may or may not freeze on its way to total evaporation. This is assuming zero pressure in outer space and close enough to zero temperature. If you look at the pressure / temperature phase diagram for water (&lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aPhase-diag2.svg&quot; rel=&quot;nofollow&quot;&gt;wikipedia page of generic phase diagram&lt;/a&gt;) it depends if the rapid drop in external pressure causes the evaporative cooling to drop the internal temperature fast enough to briefly solidify on it's way to pure gas. Carbon dioxide will go from solid straight to gas at atmospheric pressure, (smoke machines) and, as the pressure is so low, if the water did freeze solid, it wouldn't be for long, as it would continue evaporating in such a strong vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Fixed spelling typos.&lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="133" LastEditorDisplayName="" LastEditDate="2010-11-19T13:33:45.240" LastActivityDate="2010-11-19T13:33:45.240" />
  <row Id="1110" PostTypeId="1" CreationDate="2010-11-19T14:19:38.503" Score="6" ViewCount="188" Body="&lt;p&gt;Given a quantum state of $n$ qubits, and being restricted to linear optics (that is, the output annihilation operators are linear combinations of the input annihilation operators):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Which states are accessible, that is can be made from $|\psi \rangle$ with certainty? (Are there any simple criteria?)&lt;/li&gt;&#xA;&lt;li&gt;If we want to get (any) $|\phi \rangle$ from $|\psi \rangle$ and we allow post-selection, are there any known bounds on the success rate? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If it is going to simplify anything, one may assume that we consider states having exactly one photon in each rail:&#xA;$\langle \psi | \hat{a}_{\updownarrow i}^{\dagger 2} \hat{a}_{\updownarrow i}^{2} | \psi \rangle&#xD;&#xA;= \langle \psi | \hat{a}_{\mathord{\leftrightarrow} i}^{\dagger 2} \hat{a}_{\mathord{\leftrightarrow}i}^{2} | \psi \rangle&#xD;&#xA;= \langle \psi | \hat{a}_{\mathord{\leftrightarrow} i}^{\dagger} \hat{a}_{\updownarrow i}^{\dagger}\hat{a}_{\mathord{\leftrightarrow}i} \hat{a}_{\updownarrow i} | \psi \rangle&#xD;&#xA;= 0$.&lt;/p&gt;&#xA;" OwnerUserId="184" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-11-19T15:09:56.793" LastActivityDate="2010-11-21T13:44:53.670" Title="Which qubit states are accessible with linear optics operations?" Tags="&lt;quantum-optics&gt;&lt;quantum-information&gt;&lt;reseach-level&gt;&lt;quantum-computation&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="1111" PostTypeId="1" AcceptedAnswerId="1116" CreationDate="2010-11-19T17:27:11.870" Score="7" ViewCount="471" Body="&lt;p&gt;When a particle spins in the same direction as its momentum, it has right helicity, and left helicity otherwise. Neutrinos, however, have some kind of inherent helicity called chirality. But they can have either helicity. How is chirality different from helicity?&lt;/p&gt;&#xA;" OwnerUserId="556" LastActivityDate="2011-05-16T14:01:12.447" Title="What's the difference between helicity and chirality?" Tags="&lt;particle-physics&gt;&lt;quantum-field-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="1112" PostTypeId="1" CreationDate="2010-11-19T17:55:15.093" Score="6" ViewCount="221" Body="&lt;p&gt;In 1985, Harmuth wrote that Maxwell's equations are incompatible with causality, and overcame the problem by adding a term for magnetic dipole currents, and as a consequence the problem of infinite zero-point energy and renormalization disappears. At least according to Harmuth's book:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculus of finite differences in quantum electrodynamics&#xA; By Henning F. Harmuth, Beate Meffert&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The foreword is readable at &lt;a href=&quot;http://books.google.com/books?id=EQFENquY6fwC&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=EQFENquY6fwC&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are Harmuth's modifications generally accepted by the physics community as a more accurate description of reality than the unmodified equations ? &lt;/p&gt;&#xA;" OwnerUserId="244" LastActivityDate="2010-11-20T17:31:52.010" Title="Modified Maxwell's equations" Tags="&lt;electromagnetism&gt;" AnswerCount="2" />
  <row Id="1113" PostTypeId="2" ParentId="946" CreationDate="2010-11-19T18:19:23.177" Score="1" ViewCount="" Body="&lt;p&gt;ok i've found the solution to my problem.. first of all i had a factor 2 discrepancy due to the pixel dimension and, more important from an optical point of view, the laser beam was underfilling the entrance pupil of the objective: i.e. the focusing was worse!&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2010-11-19T18:19:23.177" />
  <row Id="1114" PostTypeId="2" ParentId="1112" CreationDate="2010-11-19T19:13:59.163" Score="11" ViewCount="" Body="&lt;p&gt;For what it's worth, throughout my career as a professor of theoretical physics specializing in high energy physics and the foundations of Lorentz covariant quantum theory, I never heard of Harmuth or his modification of Maxwell's equations until reading this question and the indicated foreword to his book. Now it's hard to keep track of everything going on in science, even within ones specialty. But I think it's safe to say that Harmuth's theory is not accepted by the physics community as an improvement on Maxwell's equations. There have been proposed modifications that have received extensive attention and study, such as Born-Infeld electrodynamics, and ever since the development of the Electro-Weak unified theory our understanding of the status of the EM field has been altered. But no proposals for modifying Maxwell's equations, per se, have been accepted as established improvements.&lt;/p&gt;&#xA;" OwnerUserId="345" LastActivityDate="2010-11-19T19:13:59.163" />
  <row Id="1115" PostTypeId="1" CreationDate="2010-11-19T19:38:18.837" Score="5" ViewCount="224" Body="&lt;p&gt;It's always possible to do that decomposition? I'm asking it because I know Helmholtz theorem that says a field on R3 that vanishes at infinity (r-&gt;+oo) can be decomposed in one way univocally in a gradient and a curl.&#xA;But i know for example that a costant field E on R3 is a gradient (not univocally definied): E(x+y+z+cost)...And the electric field is -gradG+ dA/dt, where A can be (Coulomb Gauge) free-divergence.&#xA;So, it's always possible to do the decomposition of a (regular, of course) field on R3 in two fields,  free-curl and free-divergence? &#xA;And on a limted domain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for the attention&#xA;SB&lt;/p&gt;&#xA;" OwnerUserId="558" LastEditorUserId="558" LastEditorDisplayName="" LastEditDate="2010-11-19T20:02:47.447" LastActivityDate="2011-03-20T11:49:03.640" Title="Decomposition of a vectorial field in free-curl and free-divergence fields" Tags="&lt;mathematics&gt;&lt;electrodynamics&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="1116" PostTypeId="2" ParentId="1111" CreationDate="2010-11-19T19:49:40.950" Score="4" ViewCount="" Body="&lt;p&gt;At first glance, chirality and helicity seem to have no relationship to each other. Helicity, as you said, is whether the spin is aligned or anti aligned with the momentum. Chirality is like your left hand versus your right hand. Its just a property that makes them different than each other, but in a way that is reversed through a mirror imaging - your left hand looks just like your right hand if you look at it in a mirror and vice-versa.&#xA;If you do out the math though, you find out that they are linked. Helicity is not an inherent property of a particle because of relativity. Suppose you have some massive particle with spin. In one frame the momentum could be aligned with the spin, but you could just boost to a frame where the momentum was pointing the other direction (boost meaning looking from a frame moving with respect to the original frame). But if the particle is massless, it will travel at the speed of light, and so you can't boost past it. So you can't flip its helicity by changing frames. In this case, if it is &quot;chiral right-handed&quot;, it will have right-handed helicity. If it is &quot;chiral left-handed&quot;, it will have left-handed helicity.&#xA;So chirality in the end has something to do with the natural helicity in the massless limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that chirality is not just a property of neutrinos. It is important for neutrinos because it is not known whether both chiralities exist. It is possible that only left-handed neutrinos (and only right-handed antineutrinos) exist.&lt;/p&gt;&#xA;" OwnerUserId="542" LastActivityDate="2010-11-19T19:49:40.950" CommentCount="1" />
  <row Id="1118" PostTypeId="2" ParentId="1115" CreationDate="2010-11-19T20:56:14.293" Score="6" ViewCount="" Body="&lt;p&gt;As long as the field can be Fourier transformed,&#xA;$$\tilde{\mathbf F}(\mathbf k) = \frac1{(2\pi)^{3/2}} \iiint e^{-i\mathbf k\cdot\mathbf r} \mathbf F(\mathbf r) d^3\mathbf r, $$&#xD;&#xA;we can separate $\tilde{\mathbf F}$ into the longitudinal and traverse parts&#xA;$$ \tilde{\mathbf F}(\mathbf k) = \tilde{\mathbf F}_\parallel(\mathbf k) + \tilde{\mathbf F}_\perp(\mathbf k)$$&#xD;&#xA;where&#xA;$$ \tilde{\mathbf F}_\parallel(\mathbf k) = \hat{\mathbf k} (\hat{\mathbf k}\cdot\tilde{\mathbf F}(\mathbf k)).$$&#xD;&#xA;This makes&#xA;$$ \mathbf k\cdot\tilde{\mathbf F}_\perp(\mathbf k) = 0, \quad \mathbf k\times\tilde{\mathbf F}_\parallel(\mathbf k) = 0,$$ which is equivalent to ($\mathbf k \mapsto -i\nabla$)&#xA;$$ \nabla\cdot{\mathbf F}_\perp(\mathbf r) = 0, \quad \nabla\times{\mathbf F}_\parallel(\mathbf r) = 0,$$&#xD;&#xA;when performing the inverse Fourier transform (again requiring it works). Thus shows $\mathbf F$ is split into a divergence-free part $\mathbf F_\perp$ and curl-free part $\mathbf F_\parallel$.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-11-19T20:56:14.293" CommentCount="5" />
  <row Id="1119" PostTypeId="2" ParentId="1019" CreationDate="2010-11-19T20:57:22.393" Score="7" ViewCount="" Body="&lt;p&gt;As a tutor, I frequently have conversations like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;So we worked out that if when I toss my pen up at 2m/s, it will go 20cm high.  How high will it go if I toss it up at 4m/s?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;40 cm.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Well, okay, let's check by working through that equation again...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[We find out that the answer is 80 cm]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;So when I throw it up twice as fast, it goes four times as high, because it takes twice as long to get to the top, but is also going twice as fast.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Okay.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Now what if I throw it up three times as fast?  How many times as high will it go?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Six.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't a misconception about kinetic energy, so much as a lack of comprehension about what scaling is.  When students are missing this concept, almost all of physics is more difficult to discuss.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-19T20:57:22.393" CommunityOwnedDate="2010-11-26T21:47:08.610" CommentCount="5" />
  <row Id="1120" PostTypeId="2" ParentId="1075" CreationDate="2010-11-19T21:07:33.760" Score="4" ViewCount="" Body="&lt;p&gt;Electrolysis.  Run a current through a weak acid, and measure the current going in and coming out.  Hydrogen ions in the acid will capture electrons and bond to each other to form hydrogen gas.  If you accept a measured value for the charge of an electron, you can find the number of hydrogen molecules liberated.  Then measure the volume of hydrogen formed at STP.  Avogadro's number is the number of molecules in a gas at STP with a volume of 2.2*10^-2 m^3.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-19T21:07:33.760" CommentCount="3" />
  <row Id="1121" PostTypeId="1" CreationDate="2010-11-19T21:09:37.690" Score="6" ViewCount="153" Body="&lt;p&gt;Electromagnetic induction is the production of voltage across a conductor moving through a magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since an orbiting satellite is passing through the Earth's magnetic field would a voltage be induced in a conductor inside the satellite?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would expect this voltage to be incredibly small but if true where is this energy &quot;coming from&quot;?&lt;/p&gt;&#xA;" OwnerUserId="561" LastActivityDate="2011-01-25T02:28:56.933" Title="Satellite Power" Tags="&lt;electromagnetism&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="1122" PostTypeId="2" ParentId="1019" CreationDate="2010-11-19T21:20:51.557" Score="15" ViewCount="" Body="&lt;p&gt;I would say that for most people, the quadratic scaling of kinetic energy with speed is somewhat of a mystery. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;People don't understand how if you go twice as fast, a car accident actually four times as energetic, hence the high number of reckless drivers and deadly accidents.&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-19T21:28:51.833" LastActivityDate="2010-11-19T21:28:51.833" CommunityOwnedDate="2010-11-26T21:47:08.610" CommentCount="5" />
  <row Id="1123" PostTypeId="2" ParentId="1121" CreationDate="2010-11-19T21:21:36.390" Score="5" ViewCount="" Body="&lt;p&gt;From the kinetic energy of the satellite -- this will make it eventually fall on the Earth surface.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-19T21:21:36.390" CommentCount="3" />
  <row Id="1124" PostTypeId="2" ParentId="1019" CreationDate="2010-11-19T21:28:17.963" Score="12" ViewCount="" Body="&lt;p&gt;A couple of space and sci-fi derived misconceptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An orbiting satellite needs propulsion and that orbiting is different from free fall.&lt;/li&gt;&#xA;&lt;li&gt;You can actually see a laser beam in free space! I've seen this in my experimental physics class a few years back, before keychain lasers were common. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="66" LastEditorUserId="431" LastEditorDisplayName="" LastEditDate="2010-12-05T01:36:11.450" LastActivityDate="2010-12-05T01:36:11.450" CommunityOwnedDate="2010-11-26T21:47:08.610" CommentCount="3" />
  <row Id="1126" PostTypeId="1" AcceptedAnswerId="1130" CreationDate="2010-11-19T23:36:45.733" Score="0" ViewCount="284" Body="&lt;p&gt;Is it possible to throw say a tennis ball at $1ms^{-1}$ for 20m and then be able to throw the same ball at $2ms^{-1}$ for 20m at the same angle? That is throwing the ball in a traditional curve withe surface between the landing point and the take off point being flat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Logic screams that this is not possible if so what is the relation between the speed and the distance (I mean what says how big the distance will be depending on how fast the ball is thrown).&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-11-22T20:38:43.930" LastActivityDate="2010-12-17T18:35:45.690" Title="Is it possible to throw an object faster but over the same distance?" Tags="&lt;forces&gt;&lt;motion&gt;" AnswerCount="6" CommentCount="1" />
  <row Id="1127" PostTypeId="2" ParentId="1096" CreationDate="2010-11-19T23:44:20.320" Score="0" ViewCount="" Body="&lt;p&gt;In the context of classical mechanics, your question is probably ill-formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since point masses have no physical extent, the gravitational force increases without bound as &lt;code&gt;r&lt;/code&gt; approached zero, which it will indeed do because they'll only collide when they're superposed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Infinite force on a finite mass implies infinite acceleration (F = ma) which implies infinite velocity, which is inconsistent with special relativity.&lt;/p&gt;&#xA;" OwnerUserId="379" LastActivityDate="2010-11-19T23:44:20.320" />
  <row Id="1128" PostTypeId="1" AcceptedAnswerId="1133" CreationDate="2010-11-19T23:51:48.087" Score="5" ViewCount="259" Body="&lt;p&gt;The escape velocity of earth is roughly $11 kms^{-1}$. However, what if a long ladder was built extending out of earth's atmosphere and considerably more. Then if something was to climb up at much less than the escape velocity, what would happen when it reached the end?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what if the object that climbed the &quot;ladder&quot; then fired some kind of thruster/rocket and was going fast enough so that it orbited the earth. It would mean less energy required to get into orbit?&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-20T00:03:24.337" LastActivityDate="2010-11-30T21:11:48.627" Title="Escape velocity" Tags="&lt;astrophysics&gt;" AnswerCount="6" CommentCount="0" />
  <row Id="1129" PostTypeId="2" ParentId="1126" CreationDate="2010-11-19T23:54:18.537" Score="1" ViewCount="" Body="&lt;p&gt;It's possible to have two angles with the same initial velocity that carry the ball the same distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html&quot; rel=&quot;nofollow&quot;&gt;You can play here.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't run through the calculations for your particular example, but you can do that easily enough yourself if you want.&lt;/p&gt;&#xA;" OwnerUserId="379" LastActivityDate="2010-11-19T23:54:18.537" />
  <row Id="1130" PostTypeId="2" ParentId="1126" CreationDate="2010-11-19T23:55:39.630" Score="4" ViewCount="" Body="&lt;p&gt;The relationship is between speed, distance, and the angle thrown.  The distance the ball travels before coming back to the same height is further if you throw it faster, but less if you throw it at a lower angle (up to about 45 degrees).  By adjusting both the speed and angle of your throw, you can have two throws that go the same distance at different speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, neither 2m/s nor 1m/s is fast enough to throw something 20m.  There is a maximum distance you can throw for a given velocity.  This distance depends on the square of the velocity, because if you throw something faster at a given angle it will be up in the air for more time and travel faster during that time.  It also depends inversely on gravity's strength, because if gravity becomes stronger the time in the air is less.  $g$ is about 10m/s^2, so for a throw of 2m/s $v^2/g = .4m$ is nowhere close to the desired 20m throw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The full expression for the distance traveled, neglecting air resistance, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sin(2\theta)\frac{v^2}{g}$&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-20T05:49:24.157" LastActivityDate="2010-11-20T05:49:24.157" CommentCount="2" />
  <row Id="1131" PostTypeId="2" ParentId="1128" CreationDate="2010-11-20T00:00:36.273" Score="1" ViewCount="" Body="&lt;p&gt;Don't know how tall the ladder is, but it would get easier and easier to climb the ladder as you went up, because the gravititational force goes down as distance from the earth's center of mass increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're already up at some orbit height, you need only to thrust along your intended orbit until you reach the orbital velocity.  No need to thrust to get up to altitude since you're already there.&lt;/p&gt;&#xA;" OwnerUserId="379" LastActivityDate="2010-11-20T00:00:36.273" CommentCount="1" />
  <row Id="1132" PostTypeId="2" ParentId="1128" CreationDate="2010-11-20T00:10:01.977" Score="5" ViewCount="" Body="&lt;p&gt;If your ladder were the height of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Geostationary_orbit&quot; rel=&quot;nofollow&quot;&gt;geostationary orbit&lt;/a&gt;, which is about 6 times the radius of the Earth, then when you got to the end you could step off and be in orbit.  If it were lower, you would need to add energy to obtain a circular orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea you're talking about is essentially a &lt;a href=&quot;http://en.wikipedia.org/wiki/Space_elevator&quot; rel=&quot;nofollow&quot;&gt;space elevator&lt;/a&gt;.  If it were feasible, it would be a far more efficient means of getting to orbit than chemical rockets.  Unfortunately, it is not feasible at this time for a variety of engineering reasons.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-20T00:10:01.977" CommentCount="4" />
  <row Id="1133" PostTypeId="2" ParentId="1128" CreationDate="2010-11-20T00:18:20.663" Score="6" ViewCount="" Body="&lt;p&gt;&quot;Escape velocity&quot; is really just a measure of the kinetic energy an object near the surface of the Earth would need to have to start with in order to just run out of energy at the point where it was infinitely far from Earth, having converted all of its initial kinetic energy to gravitational potential energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you built a giant ladder or a space elevator or whatever, the total energy required to get to orbit is exactly the same, it just comes in a less spectacular form. Rather than burning a rocket the whole way, you would be doing a slower conversion of energy into gravitational potential energy-- electricity running a motor to winch the rocket up to the top of a space elevator, or chemical energy from food as you climbed a bazillion stairs to get there, or whatever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A space elevator would be an attractive way to get a rocket into orbit, or away from the Earth because it reduces the amount of rocket fuel you need to use to get there, replacing it with some other source that is more convenient (and less explosive) to work with. But you still need the same total amount of energy to get your payload into orbit.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-20T00:18:20.663" CommentCount="5" />
  <row Id="1134" PostTypeId="1" CreationDate="2010-11-20T00:33:10.390" Score="4" ViewCount="245" Body="&lt;p&gt;Is the &quot;event horizon&quot; of a black hole potentially violable?  Black holes are commonly described as being unidirectional (matter / energy goes in, but doesn't come out), but is the event horizon of a black hole potentially deformable?  That is, in the case of two black holes colliding (let's say), is there not a point when the effective gravitational field between the two black holes &quot;cancel out&quot;, thereby causing a potential &quot;escape path&quot; for light traveling perpendicular to the axis the two black holes are on?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a more complicated case, consider two black holes of similar mass orbiting each other in a stable configuration; since no matter or energy actually pass the event horizon of a black hole, would the area between the black holes be a potential zone for escape of energy?  (I'm assuming minimal frame dragging effects here.)  And since the black holes would need to rotate at a very high rate of speed to avoid actually colliding, would that not restrict the emission of energy to be along the axis of rotation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit of clarification here; the primary question here is whether or not the gravitational force vectors of the black holes can properly cancel each other out at an appropriate point between them.  If that's the case, then (in a static system) it can be easily determined that there is a plane through that point perpendicular to the axis along which the two black holes exist where the gravitational force vectors are deeply reduced (not zero, of course, except at the precise point where the masses perfectly balance; however, reduced significantly based upon the inverse distance from said axis).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an (again static) two black hole system, orbiting material could enter a &quot;figure eight&quot; orbit; much of this material would pass through this &quot;nodal point&quot; (or very nearby it).  The high speeds and volumes of the material would indicate a large number of collisions; some resultant (high-energy) ejecta from these collisions would result in some (small) quantity of ejecta traveling along the plane of reduced gravitational force.  Now consider the (more dynamically stable) case of stably orbiting black holes; the plane of reduced gravitational force is itself rotating (about the axis of rotation of the black holes); this turns the &quot;escape plane&quot; (not entirely just a plane, it having some (small) quantity of thickness) into a line (again, not entirely just a line, but a very thin cone).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary point is that that the cancellation of force vectors effectively makes for a &quot;dimple&quot; in the event horizon of each of the black holes; that is, the existence of another strong gravitational attractor in close proximity deforms the event horizon of a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, is there anything which invalidates this analysis?  As mentioned before, frame dragging is a concern; is there anything else which would be a concern?  (I suspect the frame dragging concern would actually be resolved by the mutual forced frame dragging, i.e. the holes' spins would &quot;synchronize&quot; because of the mutual frame drag effect.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is a possibility, then does this not imply another means of black hole mass &quot;evaporation&quot; other than Hawking radiation?  (And, in fact, wouldn't it imply the type of &quot;polar jet&quot; that has been observed?)&lt;/p&gt;&#xA;" OwnerUserId="543" LastEditorUserId="543" LastEditorDisplayName="" LastEditDate="2010-11-20T05:13:02.570" LastActivityDate="2011-02-07T00:07:18.957" Title="Event Horizon violability?" Tags="&lt;gravity&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="1135" PostTypeId="1" AcceptedAnswerId="1136" CreationDate="2010-11-20T00:59:02.127" Score="5" ViewCount="816" Body="&lt;p&gt;I have noticed that authors in the literature sometimes divide characteristics of some phenomenon into &quot;kinematics&quot; and &quot;dynamics&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I first encountered this in Jackson's E&amp;amp;M book, where, in section 7.3 of the third edition, he writes, on the reflection and refraction of waves at a plane interface:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Kinematic properties:&#xA;  (a) Angle of reflection equals angle of incidence&#xA;  (b) Snell's law&lt;/li&gt;&#xA;  &lt;li&gt;Dynamic properties&#xA;  (a) Intensities of reflected and refracted radiation&#xA;  (b) Phase changes and polarization&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But this is by no means the only example.  A quick Google search reveals &quot;dynamic and kinematic viscosity,&quot; &quot;kinematic and dynamic performance,&quot;  &quot;fully dynamic and kinematic voronoi diagrams,&quot; &quot;kinematic and reduced-dynamic precise orbit determination,&quot; and many other occurrences of this distinction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the real distinction between &lt;em&gt;kinematic&lt;/em&gt; and &lt;em&gt;dynamic&lt;/em&gt;?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-01-26T17:10:48.123" Title="What is the difference between &quot;kinematics&quot; and &quot;dynamics&quot;?" Tags="&lt;kinematics&gt;" AnswerCount="7" CommentCount="2" FavoriteCount="3" />
  <row Id="1136" PostTypeId="2" ParentId="1135" CreationDate="2010-11-20T01:27:43.583" Score="9" ViewCount="" Body="&lt;p&gt;In classical mechanics (which is where I usually encounter it), &quot;kinematics&quot; generally refers to the study of properties of motion-- position, velocity, acceleration, etc.-- without any consideration of why those quantities have the values they do. &quot;Dynamics&quot; means a study of the rules governing the interactions of these particles, which allow you to determine why the quantities have the values they do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, for example, problems involving motion with constant acceleration (&quot;A car starts from rest and accelerates at 4m/s/s. How long does it take to cover 100m?&quot;) are classified as kinematics, while problems involving forces (&quot;A 100g mass is attached to a spring with a spring constant of 10 N/m and hangs vertically from a support. How much does the spring stretch?&quot;) are classified as &quot;dynamics.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's kind of an operational definition, at least.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-20T01:27:43.583" CommentCount="2" />
  <row Id="1137" PostTypeId="2" ParentId="1134" CreationDate="2010-11-20T01:35:19.503" Score="1" ViewCount="" Body="&lt;p&gt;Well, by &lt;em&gt;definition&lt;/em&gt;, an event horizon is a surface such that any event inside cannot affect an outside observer. (I think I've heard it defined as the boundary of a region in spacetime excluded from the causal past of future null infinity.) So the simple answer to your question has to be no, the event horizon cannot be &quot;violated,&quot; otherwise it would not actually be an event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think you're getting at something interesting by considering the &quot;cancellation&quot; of gravitational fields between two black holes. I guess the question is, if you take an isolated black hole and combine it with another one to form a binary BH, are there any points that would be inside the event horizon of the single BH that are outside the horizon of the binary? That I'm not sure about, at least not without doing some analysis of spacetime around a black hole collision (and I don't think that's an analysis I would know how to do in detail). Besides, you'd have to define some physically meaningful procedure for identifying spacetime points around the original black hole with spacetime points around the binary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that the answer to this is also &quot;no,&quot; though; that is, every point that is inside the event horizon of a black hole cannot be placed outside an event horizon by the addition of additional matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you had two orbiting black holes with a region between them that was not inside the event horizon, energy (in the form of light, for example) should certainly be able to escape from that region. Again, this stems from the definition of the event horizon: if light had no way to escape, then the region would be inside the horizon.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-20T01:42:49.967" LastActivityDate="2010-11-20T01:42:49.967" CommentCount="3" />
  <row Id="1138" PostTypeId="2" ParentId="1134" CreationDate="2010-11-20T01:51:33.783" Score="1" ViewCount="" Body="&lt;p&gt;In the case of &lt;strong&gt;colliding black holes&lt;/strong&gt; there is no line where the gravitational force acts the opposite direction and cancels out: only one point. There is a line where the gravitational force is &lt;em&gt;equal magnitude&lt;/em&gt;, but the force vectors add up to a non-zero value. Thus there is no escape route. Also, when the two colliding black holes are so close that the event horizon overlaps, they are going to merge in a very short time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;THis has bearing on your second question too. In the case of &lt;strong&gt;orbitin black holes&lt;/strong&gt; there are the usual &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrangian_point&quot; rel=&quot;nofollow&quot;&gt;Lagrangian points&lt;/a&gt; of the orbit, In a nutshell, if two very massive bodies are on a stable orbit around each other, there are 5 points where you could place objects with a very small mass (compared to the orbiting bodies) so that they stay there relatively to the other two because the gravitational pull of the two &quot;big ones&quot; kinda cancel each other out, together with the orbiting. But this also means, that there is no escape route: if the object leaves the Lagrangian point, it will start to orbit the two bodies or likely fall into one of them in the case of black holes. These orbits are also not possible &lt;em&gt;within&lt;/em&gt; the event horizon, then the black holes will just merge.&lt;/p&gt;&#xA;" OwnerUserId="472" LastActivityDate="2010-11-20T01:51:33.783" CommentCount="4" />
  <row Id="1139" PostTypeId="2" ParentId="1134" CreationDate="2010-11-20T01:55:31.173" Score="2" ViewCount="" Body="&lt;p&gt;Well, if I'm being pedantic, your question is self-contradicting.  The mathematical definition of a black hole is determined by tracing back the points that escape to infinity at late times, and then taking their boundary--thus, the event horizon is PRECISELY those points that cannot escape to infinity.  But that's a boring answer that doesn't tell you anything about physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A less boring answer states that there is a very strong theorem that says that the area of a black hole is necessarily nondecreasing (so long as you have ordinary matter and no quantum mechanics), and as a corollary, you can prove that the null generators* of the hole either are destined to sit on the black hole's horizon forever, or to fall into the hole.  They cannot escape.  And since these light rays can't escape, then neither can particulate matter nor rays farther into the hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*The null generators are the light beams that are the boundary between the outgoing light rays that can freely leave the hole outside the horizon, and the ones forced to fall in that like inside the horizon.  They simply float in space, neither able to go out or go in.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-11-20T01:55:31.173" CommentCount="2" />
  <row Id="1140" PostTypeId="2" ParentId="1134" CreationDate="2010-11-20T02:20:11.353" Score="1" ViewCount="" Body="&lt;p&gt;I'm not sure I understood your questions properly, but the black hole region of a (strongly asymptotically predictable*) spacetime is a well-defined topological subspace (the complement of the causal past of future null infinity). You have a foliation of spacetime by Cauchy surfaces, each one corresponding to an &quot;instant of time&quot;. Then a black hole in time T is just a connected component of the black hole region, restricted to the corresponding Cauchy surface, and its boundary is the event horizon. It is therefore clear what it means to be inside or outside the black hole, and those notions are mutually exclusive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, classically (i.e., in the context of general relativity) it is impossible to escape the event horizon by its own definition; no causal curve starting inside of it can get to future null/timelike infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You asked if the event horizon can be &quot;deformed&quot;. I'm not sure what you mean, but there are very general theorems which restrict the way black holes and event horizons behave and evolve. For instance, the event horizon must be topologically a 2-sphere; A black hole cannot bifurcate into 2 differente black holes, nor &quot;evaporate&quot; (note that Hawking radiation is a quantum effect, so it does not belong in this context). Finally, the event horizon area of all black holes never decreases in time. These are precise theorems of topology/Lorentzian geometry which are true in any physically reasonable spacetime. Of course, I'm not stating then precisely here, there are hypothesis about &quot;energy conditions&quot; and about the asymptotic behaviour of spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see all these results stated and proved in Hawking, Ellis - The large scale structure of spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*this is a technical condition for the definition to make sense.&lt;/p&gt;&#xA;" OwnerUserId="565" LastActivityDate="2010-11-20T02:20:11.353" CommentCount="2" />
  <row Id="1141" PostTypeId="2" ParentId="1018" CreationDate="2010-11-20T04:24:08.223" Score="3" ViewCount="" Body="&lt;p&gt;I guess this thread shows everyone has their own tastes when it comes to this topic!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, it's called harmonic motion because sine and cosine are the elementary harmonic functions.  Recall that in general a harmonic function is a solution of Laplace's equation (which shows up everywhere in physics), and in we initially study $sin(x+vt)=sin(kx+\omega t)$ since these are the building blocks of solutions of the wave equation.  In QM, the principle of superposition (from ODE's) takes on an entirely different, physical meaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also obvious that the equation of harmonic motion is the projection of $e^{i\omega t}$ onto the real axis, which is a standard trick to derive this equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now an important example of harmonic oscillators in nature are the atoms in a solid.  This is basic 1905 Einstein, in 1905 Einstein showed that atoms in a solid can be treated as quantum harmonic oscillators, and that this explains the behavior of specific heat at low temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another important example is that the EM field can be considered as three space where each point is a quantum harmonic oscillator.  In fact, the EM field is quantized by considering the Fourier components of the field to be creation and inhalation operators for photons of a given frequency $\omega$, which are exactly alike to the raising and lowering operators of the QM harmonic oscillator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final standard remark is that almost all small oscillations in nature can be approximated by $F=-kx$, by a Taylor expansion of the force $F(x)$ to first order in $x$.&lt;/p&gt;&#xA;" OwnerUserId="442" LastEditorUserId="442" LastEditorDisplayName="" LastEditDate="2010-11-20T04:35:13.193" LastActivityDate="2010-11-20T04:35:13.193" />
  <row Id="1142" PostTypeId="1" AcceptedAnswerId="1162" CreationDate="2010-11-20T04:24:22.043" Score="3" ViewCount="215" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Coefficient_of_restitution&quot; rel=&quot;nofollow&quot;&gt;coefficient of restitution&lt;/a&gt; characterizes a collision in one dimension by relating the initial and final speeds of the particles involved,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_R = -\frac{v_{2f} - v_{1f}}{v_{2i} - v_{1i}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a 2D collision, the velocity can be split into components parallel and perpendicular to the &quot;plane of collision&quot; (the plane tangent to the two objects' surfaces at their contact point), and the equation above applies to the perpendicular components of the velocities only. One could write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix}v_{2f\shortparallel} - v_{1f\shortparallel} \\ v_{2f\perp} - v_{1f\perp}\end{pmatrix} = \begin{pmatrix}1 &amp;amp; 0 \\ 0 &amp;amp; -C_R\end{pmatrix}\begin{pmatrix}v_{2i\shortparallel} - v_{1i\shortparallel} \\ v_{2i\perp} - v_{1i\perp}\end{pmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is it useful (i.e. does it produce a more accurate description of realistic 2D collisions) to generalize that matrix? Perhaps by allowing the top left element to be unequal to 1, or allowing nonzero off-diagonal elements? Or is there some nonlinear relation that works better?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As usual, if you know of any relevant published research, references would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-21T00:28:59.843" LastActivityDate="2010-11-21T00:28:59.843" Title="Is there a 2D generalization of the coefficient of restitution?" Tags="&lt;kinematics&gt;&lt;collision&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="1143" PostTypeId="1" AcceptedAnswerId="1180" CreationDate="2010-11-20T04:58:48.267" Score="11" ViewCount="405" Body="&lt;p&gt;Suppose a particle has 50% probability of being at location $A$, and 50% probability being at location $B$ (see double slit experiment). According to QM the particle is at both $A$ and $B$ at the same time, so is there a force of gravity between the two particle superpositions? Is there &lt;em&gt;self&lt;/em&gt;-gravity when a wave-function reaches over a finite distance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot seem to wrap my head around this. Is the gravity a proportional fraction of the entire mass based on the probabilities. How do you combine a wavefunction with &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%27_law_for_gravity&quot; rel=&quot;nofollow&quot;&gt;Gauss' law of gravity&lt;/a&gt;? I have being trying to think about self-gravity for a long time now.&lt;/p&gt;&#xA;" OwnerUserId="392" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-03-13T20:17:26.837" LastActivityDate="2011-03-13T20:17:26.837" Title="Particle wavefunction and gravity" Tags="&lt;quantum-mechanics&gt;&lt;gravity&gt;&lt;double-slit-experiment&gt;" AnswerCount="6" FavoriteCount="4" />
  <row Id="1144" PostTypeId="2" ParentId="1135" CreationDate="2010-11-20T05:11:57.000" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Statics&lt;/strong&gt;: Study of forces in equilibrium without consideration of changes over time.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Kinematics&lt;/strong&gt;: Study of motions (position, velocity, acceleration)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Kineto-statics&lt;/strong&gt;: Study of forces in equilibrium, with the addition of motion related forces (like inertia forces via &lt;a href=&quot;http://en.wikipedia.org/wiki/D%27Alembert%27s_principle&quot; rel=&quot;nofollow&quot;&gt;D'Alembert's principe&lt;/a&gt;) one instant at the time. Results from one time frame do not affect the results on the next time frame.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Dynamics&lt;/strong&gt;: Full consideration of time varying phenomena in the interaction between motions, forces and material properties. Typically there is an time-integration process where results from one time frame effect the results on the next time frame.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As far as the source if kinematic and dynamic viscocity, I am not sure, and I have wondered this myself. Maybe it stems from the test methods used to measure each property.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-11-20T05:11:57.000" />
  <row Id="1145" PostTypeId="2" ParentId="1135" CreationDate="2010-11-20T05:13:39.010" Score="2" ViewCount="" Body="&lt;p&gt;I'd say that kinematics has to do with the space of all possible configurations of a system at one time, for example what restrictions are placed on those by conservation laws. Dynamics has to do with how configurations change as a function of time. As always, the way the term is actually used may depend on the person.&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2010-11-20T05:13:39.010" />
  <row Id="1146" PostTypeId="2" ParentId="1142" CreationDate="2010-11-20T05:21:31.340" Score="0" ViewCount="" Body="&lt;p&gt;What happens during a true collision is that the contact forces are split into normal and tangential components. The normal forces arise from the elastic compression and expansion of the material near the surface, and the tangential from friction, stiction or viscocity in the slip direction. Those are two different completely effects and trying to group them together into one modeling construct will result in loss of detail. It is an gross approximation to just say there is % loss of energy in the normal direction due to hysteresis losses on the contact deflection, that lumping that with an artificial % loss of energy in the tangential direction due to friction is  going to result in completely unrealistic results in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend spliting the problem up in normal and tangential components and handling them each with their own modeling techniques. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-11-20T05:21:31.340" CommentCount="1" />
  <row Id="1147" PostTypeId="2" ParentId="1128" CreationDate="2010-11-20T05:29:47.713" Score="1" ViewCount="" Body="&lt;p&gt;When you reach the end of the ladder you will fall back to earth, unless you have reached geo-synchronous orbit. Just compare the energy expended climbing with that required to reach orbit (which is more that the energy to just escape gravity vertically). If the ladder is longer than geo-synchronous orbit then you are no longer expending energy climbing, but on the contrary the earth is giving you energy. If you let go of the ladder you will fly out in space overcoming gravity.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-11-20T05:29:47.713" CommentCount="2" />
  <row Id="1148" PostTypeId="2" ParentId="1143" CreationDate="2010-11-20T05:42:24.830" Score="4" ViewCount="" Body="&lt;p&gt;I'm fairly sure it's not correct to say that the particle is at both A and B at the same time. If it interacts with something at A, then it's at A, not at B. And vice-versa. I believe the production of a gravitational field would be one such interaction (although perhaps we might need a quantum theory of gravity to be truly sure), so when you detect the gravitational field produced by the particle, it will appear to be &quot;emanating&quot; from either A or B, but not both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would mean that the particle can't interact with itself, since if it exists at point A to be &quot;emitting&quot; the gravitational field, it can't &lt;em&gt;also&lt;/em&gt; exist at point B to be reacting to the gravitational field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the same question could apply to electromagnetic self-interaction of a charged particle. But for that case we do have a theory that should explain what happens, namely quantum electrodynamics. Perhaps someone else can explain that case in detail, or if I can figure out something, I'll edit it in here.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-20T05:42:24.830" CommentCount="9" />
  <row Id="1149" PostTypeId="2" ParentId="1126" CreationDate="2010-11-20T05:45:41.170" Score="2" ViewCount="" Body="&lt;p&gt;The total distance traveled by a ball launched with velocity $v$ at an angle $\theta$ from the horizontal to reach the save vertical level is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta x=v^2\sin(2\theta)/g $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That comes from&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{align*}&#xD;&#xA;\Delta x &amp;amp; =(v\,\cos\theta)\, t\\&#xD;&#xA;\Delta y &amp;amp; =(v\,\sin\theta)\, t-\frac{1}{2}\, g\, t^{2}\end{align*} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;solved for $t$ when $\Delta y = 0$ as $t=2\, (v\,\sin\theta)/g$ and ignoring the trivial solution of $t=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is &lt;strong&gt;no&lt;/strong&gt; because for the same $\Delta x$ when $v$ changes so does $\theta$ has to change.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-11-20T05:45:41.170" />
  <row Id="1150" PostTypeId="2" ParentId="1019" CreationDate="2010-11-20T06:37:00.977" Score="28" ViewCount="" Body="&lt;p&gt;Amazingly, Wikipedia has an article titled &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_common_misconceptions&quot; rel=&quot;nofollow&quot;&gt;List of common misconceptions&lt;/a&gt;&quot;.  There is a (short) section dedicated to &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_common_misconceptions#Physics&quot; rel=&quot;nofollow&quot;&gt;Physics&lt;/a&gt;, which mentions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The role of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Water_vortex#Draining_in_bathtubs_and_toilets&quot; rel=&quot;nofollow&quot;&gt;Coriolis effect&lt;/a&gt; in bathtubs and sink drains&lt;/li&gt;&#xA;&lt;li&gt;The role of angular momentum in bicycle stability&lt;/li&gt;&#xA;&lt;li&gt;The &quot;equal time&quot; fallacy in explaining the lift developed by an airfoil&lt;/li&gt;&#xA;&lt;li&gt;Glass isn't actually a high viscosity fluid&lt;/li&gt;&#xA;&lt;li&gt;Composition of air&lt;/li&gt;&#xA;&lt;li&gt;&quot;Lightning never strikes twice&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_common_misconceptions#Astronomy&quot; rel=&quot;nofollow&quot;&gt;Astronomy&lt;/a&gt; section has some good ones too:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When a star collapses into a black hole, its gravitational pull does not actually increase.&lt;/li&gt;&#xA;&lt;li&gt;Meteorites are not actually hot when they land; usually they are cold.  (I would add: the heating of meteors is more due to the compression of the air in front of them than to 'friction with the air' as commonly believed.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some that I would add:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Once something is in orbit it is free from Earth's gravity.&quot;  Even educated people get tripped up on this one; the internet is rife with people suggesting we just &quot;nudge&quot; the International Space Station into lunar orbit.  At a much more basic level of misunderstanding, there is the idea that astronauts are &quot;weightless&quot; because they are far away from the earth.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;There is a high tide on the opposite side of the earth from the moon/sun because the earth 'shields' the ocean from the gravitational pull.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="29" LastEditorUserId="1155" LastEditorDisplayName="" LastEditDate="2011-01-08T09:10:57.200" LastActivityDate="2011-01-08T09:10:57.200" CommunityOwnedDate="2010-11-26T21:47:08.610" CommentCount="6" />
  <row Id="1151" PostTypeId="1" AcceptedAnswerId="1152" CreationDate="2010-11-20T06:56:02.403" Score="8" ViewCount="220" Body="&lt;p&gt;If I throw a ball straight up, it deflects slightly to the west due to Coriolis forces.  If instead I watch a bubble float up in water, is the bubble deflected west, east, or neither?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the bubble also moves west, but am not sure.  My reasoning is that the air in the bubble must feel a Coriolis force to the west because it is rising.  Further, as the bubble rises, the water around it moves, and the net motion of the water is down.  Water moving down experiences a Coriolis deflection to the east.  Bubbles move the opposite direction of water, so if the water is moving east, then the bubble should move west.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;I don't have any reasonable-sounding counter-arguments, but I'm not completely convinced by my argument, either.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit&lt;/b&gt;: One reason I'm a unsure about this problem is that a bubble does not get thrown outward by a centrifugal force.  Imagine a plastic ball filled with water and spinning fast.  A bubble is sitting in the water in the equatorial place, half way to the edge of the ball.  There's centrifugal force out towards the wall, but the bubble does not move that way.  The bubble moves up the pressure gradient.  The water gets thrown to the wall, and thus there is higher pressure at the wall than at the center.  The bubble moves towards the center.  If a bubble moves counter to the centrifugal force, I should be careful before claiming it moves with the Coriolis force.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-20T22:42:06.383" LastActivityDate="2010-11-23T17:04:17.417" Title="Which direction will Coriolis forces deflect a bubble?" Tags="&lt;classical-mechanics&gt;&lt;fluid-dynamics&gt;&lt;rotational-dynamics&gt;&lt;coriolis-force&gt;" AnswerCount="3" />
  <row Id="1152" PostTypeId="2" ParentId="1151" CreationDate="2010-11-20T09:03:24.733" Score="2" ViewCount="" Body="&lt;p&gt;It does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To convince yourself, remember that rising hot air does experience a Coriolis force, so I am quite sure that your bubble does too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, think of what the Coriolis acceleration is - it is an apparent acceleration due to the fact that you, the observer, are in a rotating reference frame, so your definition of &quot;straight up&quot; is actually a curve. When the bubble goes up you see it as having a curved trajectory - but this has to do with your rotating definition of straight up. Therefore this acceleration applies to it.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-20T09:03:24.733" CommentCount="1" />
  <row Id="1153" PostTypeId="2" ParentId="1151" CreationDate="2010-11-20T09:26:05.887" Score="0" ViewCount="" Body="&lt;p&gt;I think you are right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As i know Coriolis force is not so significant on a small scale  to curve a bubble in bottle of Pepsi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if the bubble will go to the west or not because i depends how these fluid interact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Mechanically i should go.&lt;/p&gt;&#xA;" OwnerUserId="567" LastEditorUserId="567" LastEditorDisplayName="" LastEditDate="2010-11-20T11:41:02.380" LastActivityDate="2010-11-20T11:41:02.380" CommentCount="1" />
  <row Id="1154" PostTypeId="2" ParentId="1128" CreationDate="2010-11-20T09:27:54.777" Score="2" ViewCount="" Body="&lt;p&gt;Energy required to escape from the earth's gravity is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{GMm}{R} = mgR$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you may pay it in installments, you may pay it slowly but you may not pay less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Escape velocity from point at height $h$ above the earth's surface is $V_{eh} = \sqrt{2g(R-h)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case I:&lt;/strong&gt; Launching rocket from the surface of the earth i.e. h = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total energy = Energy required to launch &#xA;             = $\frac{mv_{e0}^2}{2} = \frac{2mg(R+0)}{2} = mgR $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case II:&lt;/strong&gt; Launching rocket from height $h$ above the surface of the earth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total energy = Energy required to climb upto height $h$ + Energy required to launch&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$mgh + \frac{mv_{eh}^2}{2} = mgh + \frac{2mg(R-h)}{2} = mgR$ &lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-20T09:32:20.537" LastActivityDate="2010-11-20T09:32:20.537" CommentCount="2" />
  <row Id="1155" PostTypeId="1" AcceptedAnswerId="1156" CreationDate="2010-11-20T09:34:00.793" Score="6" ViewCount="258" Body="&lt;p&gt;I would like to ask a stupid question here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a body 'b' moving downward with a velocity v in a rotating frame of reference with angular velocity w, and w and v not being parallel and anti parallel. We all know that the body 'b' experiences a Coriolis force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it gets deflected in its trajectory then according to newtons third law  shouldn't it push some body with equal force?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what if the fame of reference was moon where there is no atmosphere? will the body be deflected?&lt;/p&gt;&#xA;" OwnerUserId="567" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-11-22T20:45:15.537" LastActivityDate="2010-11-22T20:45:15.537" Title="Coriolis force and Newton's third law" Tags="&lt;classical-mechanics&gt;&lt;mechanics&gt;" AnswerCount="1" />
  <row Id="1156" PostTypeId="2" ParentId="1155" CreationDate="2010-11-20T09:44:11.000" Score="6" ViewCount="" Body="&lt;p&gt;Nothing feels a reaction force to the Coriolis force because Newton's laws only apply in an inertial frame.  If we imagine a frame with some sort of wild, random acceleration, then in that frame at a given moment it would look like, in a room with completely stationary tables and chairs and things, everything was simultaneously accelerating the same direction, and there would be huge &quot;fictitious forces&quot; needed to explain this.  There are no third-law pairs going on, though.  The fictitious forces were necessary to make $F=ma$ work out in this accelerating frame, but by introducing them we sacrificed the third law (at least in regards to those kinematic forces).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine dropping a ball off a tower.  It gets deflected east.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we analyze this scenario from a frame rotating along with the Earth, there is no conflict with Newton's third law because it does not hold for the Coriolis force or other kinematic forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we analyze the scenario from an inertial frame in which the Earth is rotating, the trajectory simply is not deflected.  The ball lands east of where it was dropped because its velocity in a tangential direction, which comes from the rotation of the Earth, stayed the same, but as it fell its angular velocity $v/r$ increased because $r$ decreased.  That led it to &quot;deflect east&quot; when it was really moving in a perfectly ordinary parabola.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-20T09:44:11.000" CommentCount="1" />
  <row Id="1157" PostTypeId="1" AcceptedAnswerId="1169" CreationDate="2010-11-20T10:32:43.673" Score="4" ViewCount="396" Body="&lt;p&gt;If I have an over-damped mechanical system that is excited with a sinusoidal motion. That sinusoidal motion starts with a determined frequency then increases frequency over time.&#xA;Of course, it is known that there will be a phase shift between the driving force and the motion of the hanging mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how to figure out phase lag of mass motion in relation to driving force?&lt;/p&gt;&#xA;" OwnerUserId="568" LastActivityDate="2010-11-21T05:30:44.330" Title="How to determine phase angle for a sinusoidal motion?" Tags="&lt;harmonic-oscillator&gt;" AnswerCount="3" />
  <row Id="1158" PostTypeId="2" ParentId="1157" CreationDate="2010-11-20T11:06:54.033" Score="1" ViewCount="" Body="&lt;p&gt;You can see this for yourself by swinging a pendulum or waving a flexible curtain rod. If you swing it slower than the resonance frequency, the mass (or the other end of the curtain rod) just follows your hand, so the phase lag is zero. If you swing faster than the resonance frequency, then the mass does the opposite of what your hand does, so the phase lag is 180 degrees. As you approach the resonance frequency from slow to fast, the phase starts going from 0 to 180, and the amplitude increases a lot, hitting a maximum at the resonance frequency and decreasing afterwards. Swinging at exactly the resonance frequency using only your hand is pretty much impossible, but if you could, you would see that the phase lag was 90 degrees - halfway in between 0 and 180.&lt;/p&gt;&#xA;" OwnerUserId="107" LastEditorUserId="107" LastEditorDisplayName="" LastEditDate="2010-11-20T19:34:55.787" LastActivityDate="2010-11-20T19:34:55.787" CommentCount="1" />
  <row Id="1159" PostTypeId="1" AcceptedAnswerId="1274" CreationDate="2010-11-20T11:14:59.427" Score="0" ViewCount="326" Body="&lt;p&gt;My original question was in an effort to understand the electrical analogy to Markov chains, which is explained in Snell's article.  There are some neat parallels that involve taking a Markov chain and considering the edge weights to be wire conductances.  Unfortunately, after reading the article, I realize that they specifically say that their analogy is unrealistic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how is electrical energy transmitted?&lt;/p&gt;&#xA;" OwnerUserId="569" LastEditorUserId="569" LastEditorDisplayName="" LastEditDate="2010-11-24T23:18:27.993" LastActivityDate="2010-11-24T23:18:27.993" Title="How is electrical energy transmitted?" Tags="&lt;electricity&gt;&lt;electrodynamics&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="1" />
  <row Id="1160" PostTypeId="2" ParentId="1159" CreationDate="2010-11-20T12:30:27.630" Score="2" ViewCount="" Body="&lt;p&gt;uhm.. if i understood your doubt correctly, you want a macroscopic analogy of what happens with resistance and circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, imagine a great concert in a stadium. hundred of thousand of people want to get in at the same time and they push so that, if materially possible, they would all entre the stadium at the same time (and this push is analog to the voltage drops i.e. the energy for unit of charge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you imagine the gate of the stadium, the wider the gate the higher the flux of people going in will be, i.e. the higher the current of people entering the stadium. The gate is analogous to the resistance: wider gate means lower resistance higher current (if the push of the crowd, i.e. the voltage drop, is kept constant).&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="28" LastEditorDisplayName="" LastEditDate="2010-11-24T14:23:50.477" LastActivityDate="2010-11-24T14:23:50.477" />
  <row Id="1161" PostTypeId="2" ParentId="1135" CreationDate="2010-11-20T14:05:43.760" Score="2" ViewCount="" Body="&lt;p&gt;Since everybody already gave nice replies to this question, I'll give a more pragmatic answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't worry about it. It is an arbitrary distinction made by humans. Nature doesn't care if some phenomenon can be described/explained purely from kinematic considerations or not. It's not a fundamental distinction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it is a useful distinction. I'm sure you know the distinction somehow implicitly when you solve problems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me give an example in mechanics: you swing a pendulum in a vertical plane, you swing sufficiently fast so that the trajectory is a circle. What is the tension in the pendulum when it passes in the lowest point of the circle. The tension is a dynamical quantity, because it is a force. Now, when you solve the problem, you don't write down the full equation of Newton and solve them. You use the kinematic information you have about the trajectory: it's a circle, in the lowest part of the trajectory there is no tangential acceleration, so the acceleration is directed radially inwards and is $v^2/r$. From this you can find the tension by using purely kinematic considerations and never solving $\vec{F}=m\vec{a}$ as a differential equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess you understood that in physics we do this all the time. If we didn't, many problems would be impossible to tackle without resorting to extensive computer simulations all the time. In most problems we consider, we already have some idea of the kinematics, which permits to reduce the space of acceptable solutions. Sometimes so drastically (but that is only for the simplest problems) that we can solve them by purely kinematic considerations.&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2010-11-20T14:05:43.760" />
  <row Id="1162" PostTypeId="2" ParentId="1142" CreationDate="2010-11-20T14:09:24.797" Score="1" ViewCount="" Body="&lt;p&gt;To be simple, consider an object hitting a heavy plane. If there is coefficient of friction $C_F$ (and restitution $C_R$) between the bodies, parallel velocity is modified as well:&#xA;$\Delta p_{\parallel} = F_{\parallel} \Delta t = -C_F F_{\perp} \Delta t = -C_F \Delta &#xD;&#xA;p_{\perp} = -C_F(1+C_R) p_{\perp}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus obtaining&#xA;$$&#xD;&#xA;\left[&#xD;&#xA;\begin{array}{c}&#xD;&#xA;v_{\parallel}^{out} \\&#xD;&#xA;v_{\perp}^{out}&#xD;&#xA;\end{array}&#xD;&#xA;\right]&#xD;&#xA;=&#xD;&#xA;\underbrace{&#xD;&#xA;\left[&#xD;&#xA;\begin{array}{cc}&#xD;&#xA;1 &amp;amp; - C_F(1+C_R) \\&#xD;&#xA;0 &amp;amp; - C_R&#xD;&#xA;\end{array}&#xD;&#xA;\right]}_{A}&#xD;&#xA;&#xD;&#xA;\left[&#xD;&#xA;\begin{array}{c}&#xD;&#xA;v_{\parallel}^{in} \\&#xD;&#xA;v_{\perp}^{in}&#xD;&#xA;\end{array}&#xD;&#xA;\right].&#xD;&#xA;$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that it works as long as $v_{\parallel}^{in}\geq C_F(1+C_R) v_{\perp}^{in}$.&#xA;If there were a general (working for all input velocities) matrix $A$, it should allow dissipation, but not creation, of energy i.e. $||A||_2\leq 1$ (the condition is not fulfilled in the above example). I means that either the matrix need to have other entries as well or that the problem intrinsically needs restriction of the initial conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, bear in mind coefficient of restitution is only a effective parameter (as it has been already mentioned by jalexiou).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, it may be a nice experimental problem for an undergraduate student to find the all coefficients (and check for which conditions they work). I am curious of the results.&lt;/p&gt;&#xA;" OwnerUserId="184" LastEditorUserId="184" LastEditorDisplayName="" LastEditDate="2010-11-20T15:31:37.707" LastActivityDate="2010-11-20T15:31:37.707" CommentCount="1" />
  <row Id="1163" PostTypeId="2" ParentId="1143" CreationDate="2010-11-20T16:55:58.703" Score="1" ViewCount="" Body="&lt;p&gt;Let's first recognize that the gravitational force is 10^{-42} times weaker than the electrostatic force.  Now pause to ingest that number.  The earth is only about 10^{23} times more massive than we are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is not a practical concern, but one of fundamental physical theory.  We don't have a good theory in which we can analyze dynamical gravitational and quantum-mechanical processes.  Hawking radiation is perhaps the best blend of the two, but it takes place in an extremely large gravitational field, where the discrepancy between the forces is much less.  (Within string theory, which is a consistent quantum theory which includes gravity, the best quantum-gravity calculations regard the number of microstates of a black hole, i.e. its entropy.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To guess that an incoming state's gravitational self-interaction will affect its evolution -- and that this effect will be different for superpositions of localized states -- but I have no idea what would actually happen.  I'm sure that people more connected with experiments will have a better feel, but a fundamental understanding is currently beyond our ken.&lt;/p&gt;&#xA;" OwnerUserId="20" LastEditorUserId="20" LastEditorDisplayName="" LastEditDate="2010-11-20T20:56:59.990" LastActivityDate="2010-11-20T20:56:59.990" CommentCount="3" />
  <row Id="1164" PostTypeId="2" ParentId="1112" CreationDate="2010-11-20T17:31:52.010" Score="2" ViewCount="" Body="&lt;p&gt;I have done two projects when I was a graduate student on magnetic monopoles. It seems to me that Harmuth's equations are just the same as those for electromagnetism with magnetic monopoles, which has been studied by Dirac before Harmuth. But I may be missing something here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, while magnetic monopoles are not part of mainstream physics in the sense that they have never been detected and that they don't show up in the Standard Model, they are an important part in many of the theories that try to go beyond the Standard Model.&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2010-11-20T17:31:52.010" />
  <row Id="1165" PostTypeId="1" AcceptedAnswerId="1167" CreationDate="2010-11-20T18:54:06.640" Score="13" ViewCount="233" Body="&lt;p&gt;Ordinary matter and antimatter have the same physical properties when it comes to, for example, spectroscopy. Hydrogen and antihydrogen atoms produce the same spectroscopy when excited, and adsorb the same frequencies. The charge does not make a difference in the potential (regardless if it's generated by a proton or an antiproton) nor in how the positron behaves in this potential (being its mass equal to the mass of an electron)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can astronomy evaluate if a far galaxy is made of matter or antimatter, given that from the spectroscopy point of view, they behave in the same way? In other words, how do we know that an asymmetry exists between matter and antimatter in the universe? &lt;/p&gt;&#xA;" OwnerUserId="217" LastActivityDate="2010-11-29T18:39:15.497" Title="Experimental observation of matter/antimatter in the universe" Tags="&lt;astrophysics&gt;&lt;spectroscopy&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="1166" PostTypeId="2" ParentId="1143" CreationDate="2010-11-20T19:23:28.593" Score="2" ViewCount="" Body="&lt;p&gt;This is similar to what my old boss called the &quot;Lafyatis Problem,&quot; because it was first posed to him by Greg Lafyatis at Ohio State, which involved light emission from a particle in a &quot;Schroedinger cat&quot; state. The question is, if you have an atom in a superposition of two position states, and it emits light, should you expect to see an interference pattern in the emitted light due to the light being emitted by the two different possible locations of the atom? This got debated periodically, and as I recall the consensus answer that was settled on was that there wouldn't be any interference, but it was never fully settled, and kept being brought up from time to time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that Dave Wineland's group did experiments in which they put trapped ions into a &quot;cat&quot; state, a superposition of two positions in the trap (Science 272, 1131 (1996)). They didn't deal with this issue experimentally, but I think it's what prompted the discussion, and might be a place to start looking for information.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-20T19:23:28.593" />
  <row Id="1167" PostTypeId="2" ParentId="1165" CreationDate="2010-11-20T19:31:25.590" Score="18" ViewCount="" Body="&lt;p&gt;To be a little pedantic, nobody has yet done precision spectroscopy of antihydrogen, though the recent success in trapping it at CERN (all over the news this week, &lt;a href=&quot;http://dx.doi.org/10.1038/nature09610&quot; rel=&quot;nofollow&quot;&gt;paper here&lt;/a&gt;) is an early step toward that. It's possible that there are small differences in the spectrum of antihydrogen and hydrogen, though these differences can't be all that large, or they would be reflected in the interactions of antiprotons and positrons with ordinary matter in ways that would've shown up in other experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it (and I am not an astronomer) the primary evidence for a lack of vast amounts of antimatter out there in the universe is a lack of radiation from the annihiliation. We're very confident that our local neighborhood is matter, not antimatter, which means that if there were an anti-galaxy somewhere, there would also need to be a boundary region between the normal matter and antimatter areas. At that boundary region, there would be a constant stream of particle-antiparticle annihilations, which produce gamma rays of a very particular energy. We don't see any such region when we look out at the universe, though, which strongly suggests that there aren't any anti-galaxies running around out there.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-20T19:31:25.590" CommentCount="4" />
  <row Id="1168" PostTypeId="2" ParentId="15" CreationDate="2010-11-20T19:56:58.993" Score="1" ViewCount="" Body="&lt;p&gt;As suggested by other posters, the key question is energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At very high energy levels, approaching some of the quantum gravity 'limits', the string-like  nature of fundamental particles would become increasingly apparent. (In terms of an experiment, at a high enough energy level, for instance, there would likely be new specific 'resonances' of the material that could be identified.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also find &lt;a href=&quot;http://www.wired.com/wiredscience/2010/09/stringy-quantum/&quot; rel=&quot;nofollow&quot;&gt;this article interesting&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="574" LastEditorUserId="574" LastEditorDisplayName="" LastEditDate="2010-11-20T21:44:01.293" LastActivityDate="2010-11-20T21:44:01.293" />
  <row Id="1169" PostTypeId="2" ParentId="1157" CreationDate="2010-11-21T00:22:29.113" Score="4" ViewCount="" Body="&lt;p&gt;A damped harmonic oscillator with a sinusoidal driving force is represented by the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ddot{x} + \gamma\dot{x} + \omega_0^2x = \frac{F_D \sin(\omega_D t)}{m}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\gamma = b/m$ ($b$ is the damping coefficient, $b=F/v$) and $\omega_0^2 = k/m$ is the resonant frequency of the oscillator. The particular solution to this equation can be determined by taking the imaginary part of the solution to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ddot{x} + \gamma\dot{x} + \omega_0^2x = \frac{F_D}{m}e^{i\omega_D t}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you assume* the solution takes the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x(t) = A e^{i(\omega_D t + \phi)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and plug that in, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-A \omega_D^2 + \omega_0^2 A = \frac{F_D}{m}\cos(\phi)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\gamma\omega_D A = \frac{F_D}{m}\sin(\phi)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for the phase difference gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan\phi = \frac{\gamma\omega_D}{\omega_0^2 - \omega_D^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This depends on the frequency of the driving force and the resonant frequency of the oscillator, but not on the amplitude of the driving force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can express this in terms of the dimensionless variable $x = \omega_D / \omega_0$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan\phi = \frac{\gamma}{\omega_0}\frac{x}{1 - x^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if you graph it,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/9D3Su.gif&quot; alt=&quot;plot of phi versus x&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(graph generated by Wolfram Alpha) you'll see how the response of the oscillator jumps from leading to lagging when $\omega_D = \omega_0$ (at $x=1$), that is, when the driving and resonant frequencies are equal.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;*The same solution can be obtained from Fourier decomposition without making this assumption.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-21T04:01:10.237" LastActivityDate="2010-11-21T04:01:10.237" />
  <row Id="1170" PostTypeId="2" ParentId="1048" CreationDate="2010-11-21T03:03:48.330" Score="9" ViewCount="" Body="&lt;p&gt;I think in that case the universe will be homogeneous but not isotropic. Such geometries were classified by Bianchi, and they have parameters that are constrained be experiment (not my area of expertise, so I am not sure to what degree). As far as I know, there is no indication that angular momentum is needed to explain any cosmological observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, I don't think general relativity (or any other current theory) obeys Mach's principal, as usually phrased it requires highly non-local effects. Frame dragging may be in the spirit of Mach's principle, but it follows from perfectly local interactions as encoded in GR.&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2010-11-21T03:03:48.330" />
  <row Id="1171" PostTypeId="1" AcceptedAnswerId="1173" CreationDate="2010-11-21T03:22:07.750" Score="0" ViewCount="131" Body="&lt;p&gt;Assume that there are only well behaved functions as mass distributions, and there are no other forces except gravitation. Is it than possible to create an arrangement where a variation of a certain quantity (could be mass density or gravitational field or momentum) has a resonance?&lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-11-22T20:41:40.920" LastActivityDate="2010-11-22T20:41:40.920" Title="Resonance in a gravitational field?" Tags="&lt;gravity&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="1172" PostTypeId="1" CreationDate="2010-11-21T04:03:40.977" Score="2" ViewCount="159" Body="&lt;p&gt;I have tried to learn it myself but I was unable to understand. Can you please explain it in detaails with simple example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="577" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2010-11-21T21:24:23.887" LastActivityDate="2010-11-21T21:24:23.887" Title="How will you fix the direction of the lines of force due to circular current?" Tags="&lt;experimental-physics&gt;&lt;homework&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="1173" PostTypeId="2" ParentId="1171" CreationDate="2010-11-21T04:35:06.520" Score="2" ViewCount="" Body="&lt;p&gt;A matter distribution with a sinsuoidally varying monopole or dipole moment will only produce variations in the gravitational field within the matter distribution.  If the quadrupole (or a higher multipole moment) varies sinsuoidally, you will produce gravitational waves in a way very analogous to how electromagnetic waves are produced, with the amplitude differing by just a few numeric constants.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't expect the back-reaction of the wave to amplify the variation that created the wave in the first place, though.  Is this what you mean by 'a resonance'?&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-11-21T04:35:06.520" CommentCount="7" />
  <row Id="1174" PostTypeId="1" AcceptedAnswerId="1176" CreationDate="2010-11-21T05:19:50.397" Score="1" ViewCount="115" Body="&lt;p&gt;In a gravitational field, should the mass distributions always behave well?&lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-11-22T20:40:58.860" LastActivityDate="2010-11-22T20:40:58.860" Title="Mass distribution driven by gravitational field" Tags="&lt;gravity&gt;" AnswerCount="1" />
  <row Id="1175" PostTypeId="2" ParentId="1157" CreationDate="2010-11-21T05:30:44.330" Score="1" ViewCount="" Body="&lt;p&gt;David has the correct answer, with the addition that if $\gamma$ is not known is can be calculated if the system is driven at the natural frequency and the amplitude is measured (you mentioned this is an overdamped system). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A_{x=1}=F_D/(\gamma\,m\,\omega_0) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \gamma=F_D/(A_{x=1}\,m\,\omega_0) $$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-11-21T05:30:44.330" />
  <row Id="1176" PostTypeId="2" ParentId="1174" CreationDate="2010-11-21T05:33:04.130" Score="3" ViewCount="" Body="&lt;p&gt;No.  There are numerous well-known mass distributions that start out non-singular, and which collapse to form a black-hole after a finite amount of time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black hole solutions, however, have a region of infinite density and infinite spacetime curvature, so they are not at all well-behaved in any ordinary sense.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-11-21T05:33:04.130" CommentCount="6" />
  <row Id="1177" PostTypeId="2" ParentId="384" CreationDate="2010-11-21T06:06:26.843" Score="0" ViewCount="" Body="&lt;p&gt;Mechanical impedance matching does have an application in electrical transmission lines (or any elastic cable/structure vibration) because it helps describe how much of the wave gets through a discontinuity, and how much is being reflected. Mechanical impedance is force over velocity and along the cable it is equal to tension over wave speed. The discontinuity may be either an elastic support (with damping properties), or a change unit weight of the cable (bigger cross section) which changes the wave speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a cable with tension $T$ and unit weight $w$ (weight per length) the wave speed is $c=\sqrt{T/w}$ and the impedance is $z=\frac{T}{c}=\sqrt{T\,w}$. The practical application of this is in the design of stockbridge dampers that go on cables and structures to absorb aeolian vibrations. There is a theory of matching impedances and measuring the &quot;Inverse Standing Wave Ratio&quot; in testing to check the efficiency of the damper. With a good match most of the wave energy does not get reflected back into the cable once it reaches the end near the tower. There is a IEEE standard covering all this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you missed it, &lt;em&gt;mechanical impedance&lt;/em&gt; is force over speed. Maybe someone something similar for the electrical, or acoustical impedance.&lt;/p&gt;&#xA;" OwnerUserId="392" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2010-11-21T06:27:48.627" LastActivityDate="2010-11-21T06:27:48.627" />
  <row Id="1178" PostTypeId="2" ParentId="1172" CreationDate="2010-11-21T07:08:55.293" Score="6" ViewCount="" Body="&lt;p&gt;By &lt;a href=&quot;http://en.wikipedia.org/wiki/Line_of_force&quot; rel=&quot;nofollow&quot;&gt;lines of force&lt;/a&gt;, do you mean the magnetic field? If so, you could determine the direction of the B-field by right-hand  rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://frank.mtsu.edu/~phys2020/Lectures/L12-L18/L17/Current_Loops/Fig17.73.JPG&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that a current element flowing up will create a B-field circulating out of the screen on the wire's left, and into the screen on the right's right. This direction is chosen by convention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/3/3e/Manoderecha.svg/337px-Manoderecha.svg.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wrap the wire into a loop (making it flows counterclockwise), we find that the B-field will point into the screen outside of the loop,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://frank.mtsu.edu/~phys2020/Lectures/L12-L18/L17/Current_Loops/Fig17.72a.JPG&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and out of the screen inside the loop&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://frank.mtsu.edu/~phys2020/Lectures/L12-L18/L17/Current_Loops/Fig17.72b.JPG&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing up all elements (Biot-Savart law) we will find that the B-field should look like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://frank.mtsu.edu/~phys2020/Lectures/L12-L18/L17/Current_Loops/a_Fig17.07.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Picture source: &lt;a href=&quot;http://frank.mtsu.edu/~phys2020/Lectures/L12-L18/L17/Current_Loops/current_loops.html&quot; rel=&quot;nofollow&quot;&gt;http://frank.mtsu.edu/~phys2020/Lectures/L12-L18/L17/Current_Loops/current_loops.html&lt;/a&gt;; Wikipedia)&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-11-21T07:08:55.293" />
  <row Id="1179" PostTypeId="2" ParentId="1172" CreationDate="2010-11-21T07:09:44.017" Score="2" ViewCount="" Body="&lt;p&gt;A circular current, running counter-clockwise when viewed from above, induces a magnetic field whose field lines point mostly up through the center of the loop and then curl around to point down outside the loop.  A clockwise current's magnetic field lines point the opposite direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a picture from &lt;a href=&quot;http://lempel.pagesperso-orange.fr/rappels_physique_uk.htm&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; source&#xA;&lt;img src=&quot;http://i.stack.imgur.com/o9xJx.gif&quot; alt=&quot;magnetic field lines&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-21T07:09:44.017" />
  <row Id="1180" PostTypeId="2" ParentId="1143" CreationDate="2010-11-21T09:52:38.663" Score="6" ViewCount="" Body="&lt;p&gt;There is some work by Roger Penrose on the subject. The papers title is, &quot;&lt;a href=&quot;http://www.sciencetimes.com.cn/upload/blog/file/2010/8/201081019170575880.pdf&quot; rel=&quot;nofollow&quot;&gt;On Gravity's Role in Quantum State Reduction&lt;/a&gt;&quot;, and it discuses how the interaction of two states that have different mass distributions with spacetime can cause the wavefunction to collapse in the one state or the other. There is also a following paper that discuses the same thing in Newtonian gravity, &quot;&lt;a href=&quot;http://iopscience.iop.org/0264-9381/15/9/019&quot; rel=&quot;nofollow&quot;&gt;Spherically-symmetric solutions of the Schrödinger-Newton equations&lt;/a&gt;&quot; (and there is also &lt;a href=&quot;https://www.worldscibooks.com/etextbook/4977/4977_chap01.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; that you could have a look).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one thing that I should point out that is also pointed out by David. In a situation as the one described in the question (double slit experiment), the particle is not at two different places at the same time and interacts with it self. It is the two states (wavefunctions) that interact to give you the interference. &lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2010-11-21T09:52:38.663" CommentCount="2" />
  <row Id="1181" PostTypeId="2" ParentId="1143" CreationDate="2010-11-21T11:06:35.953" Score="1" ViewCount="" Body="&lt;p&gt;We don't know, really.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking in Quantum Mechanical terms, you should proceed like this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The setup of the experiment is represented in quantised form in Schrödinger's equation (it gets more complex with QED and QFT, but the principle is pretty much the same). This includes the physical setup (screens, etc.) and all the fields you wish to account for. &lt;strong&gt;This is where you should include gravity&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The solution of the equation gives you a function which ultimately describes your experiment. You apply specific operators which correspond to measurements and obtain a series of possible outcomes and the probabilities of each.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The thing of note here is that when you add gravity initially, its effects are taken into account globally (in Schrödinger's equation). So this would take into account all the effects you are talking about. What doesn't work in your line of thinking is that you account for gravity locally (i.e. with particle at point A or B or both), whereas, in a QM world you can't do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the problem is that obviously we don't have a good and valid representation of gravity in QM terms.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-21T11:06:35.953" />
  <row Id="1182" PostTypeId="2" ParentId="1110" CreationDate="2010-11-21T13:44:53.670" Score="10" ViewCount="" Body="&lt;p&gt;Thanks for the clarification. Your question makes sense to me now. I'm not really going to be able to answer it. In general, if you start with a photon number state, and put it through linear optics, I believe the state you get looks like a big, ugly mess if you try to write it down in any reasonable basis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think you'll be able to get most quantum states without postselection. An argument for this is: suppose you could get all quantum states. Then you could get coherent states. Reversing the circuit, you would then be able to get your start state $|{\psi}\rangle$ by starting with some coherent state and using linear optics. But if you start with coherent  states, and use linear optics, all you'll get is coherent states. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe a better path is to investigate is what happens when you do a small linear optics transformation, and then project back into your DFS. What you would then want are a set of transformations which keep you in the DFS (using the quantum Zeno effect) but where the quantum Zeno effect doesn't end up keeping your original state exactly the same. I don't know whether such linear optics transformations exist.&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2010-11-21T13:44:53.670" CommentCount="1" />
  <row Id="1184" PostTypeId="1" CreationDate="2010-11-21T16:19:07.163" Score="3" ViewCount="240" Body="&lt;p&gt;For a school project, I'm trying to make an automated flight planner, considering mainly the differences in flight paths according to wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I've heard that flying with/against the wind affects airspeed. But I really don't know any specific(even empirical) laws on determining by how much it does so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's more, I've also heard about gas savings through these effects. That I would have pretty much no idea of how to calculate but that's less important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically if anyone can point me in the right direction to figure out how to figure out solid speed gains, I'd be grateful, even moreso if I can find out the theoretical origins of such laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The only thing I found that I think is close to what I want is &lt;a href=&quot;http://en.wikipedia.org/wiki/E6B&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, but I have no clue what the laws derive from)&lt;/p&gt;&#xA;" OwnerUserId="579" LastActivityDate="2011-03-16T10:52:44.573" Title="How do I figure out the effects of wind on flight?" Tags="&lt;applied-physics&gt;&lt;aerodynamics&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="1185" PostTypeId="2" ParentId="1184" CreationDate="2010-11-21T18:29:49.987" Score="3" ViewCount="" Body="&lt;p&gt;Well, the formulae in the Wikipedia page you reference are simple consequences of this model:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Your aircraft instruments measure speed relative to the surrounding air (air speed)&lt;/li&gt;&#xA;&lt;li&gt;The surrounding air moves with respect to the ground (wind speed)&lt;/li&gt;&#xA;&lt;li&gt;The speed of the aircraft vs the ground will be the vector sum of both these speeds.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To first order (and I'm pretty sure that should suffice for your school project) the resulting fuel savings ensue through the fact that you only need to expend fuel to maintain air speed - basically, if the wind carries the surrounding air in your direction of travel, you have to travel shorter distances &quot;through the air&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mathematics underlying this consists of vector addition in the simple case and maybe addition of vector fields/line integrals in these vector fields for more complex cases.&lt;/p&gt;&#xA;" OwnerUserId="243" LastActivityDate="2010-11-21T18:29:49.987" />
  <row Id="1186" PostTypeId="2" ParentId="1184" CreationDate="2010-11-21T18:33:30.563" Score="0" ViewCount="" Body="&lt;p&gt;Basically what you can precisely calculate are just simple effects based on the fact that the velocity of a plane in ground reference frame is a vector sum of plane's velocity in air reference frame and the velocity of air in ground's reference frame (=wind) -- this lead to some corrections in direction (very handy in GPS era, though), speed and time of flight (and so fuel issues), but this is kindergarten physics.&#xA;Of course wind modifies the flight dynamics too, but this on the other hand cannot be precisely calculated because hydrodynamics is just too complex. This gap is partially filled by some phenomenological laws, the rest is guided by &quot;piloting skills&quot; of the pilot.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-21T18:33:30.563" />
  <row Id="1188" PostTypeId="2" ParentId="1143" CreationDate="2010-11-21T20:37:27.017" Score="4" ViewCount="" Body="&lt;p&gt;First a general comment - everything in the world is described by either classical or quantum fields. Point particles are a fiction, sometimes useful, sometimes not. Starting with classical field theories like Maxwell equations or general relativity, you find that you are forced to forget about point sources and repalce them by continuous charge or mass distribution, otherwise you get all kinds of nonsense (non-locality, acausality, etc. etc.). One of the reasons for that is the infinite self-force or self-energy problem that crops up already at the classical level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can approximate a continuous distribution of matter by a &quot;particle&quot; if it obeys certain conditions, roughly speaking it has to be localized and weakly interacting. By &quot;localized&quot; I mean that all relevant observable quantities (expectation values of operators) are localized. This is not the situation you describe - the wavefunction is approximately localized (with two centres) but it is not observable. Relevant observable quantities like the expectation value of currents will not necessarily be localized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what you are asking in effect is the self-force for a particular distribution of mass (or charge). There is an answer for that, but since you are asking a question that had to do with short distance physics, the quantum mechanics of the gravitational (or electromagnetic) field comes into it. Probably not enough space-time to elaborate on this here.&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2010-11-21T20:37:27.017" />
  <row Id="1189" PostTypeId="2" ParentId="1184" CreationDate="2010-11-21T21:08:53.167" Score="3" ViewCount="" Body="&lt;p&gt;For your purposes, you can simply assume that the airplane travels at a constant velocity through the air mass.  &quot;Wind&quot; simply means that this air mass is moving with respect to the ground.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the airplane's velocity through the air is (a vector) v, and the air velocity with respect to the ground is a vector w, then the airplane's velocity with respect to the ground is simply v+w.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the same problem as swimming in a river with a current.  If you are swimming directly upstream (or downstream), then the velocity of river water simply subtracts (or adds) to your velocity.   If, on the other hand, you want to swim directly across, without floating downstream with the current, you have to angle your swim slightly upstream so that the component of your velocity in the direction of the river flow cancels out the river velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In flying, this correction is called the &quot;wind correction angle&quot;; it corrects for the crosswind component of the wind, in order to fly a straight path (relative to the ground) from point A to point B.  If you google for &quot;cross-country flight planning&quot; you will find lots of materials about how this is done in practice.  The old-fashioned way is to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/E6B&quot; rel=&quot;nofollow&quot;&gt;E6B circular slide-rule&lt;/a&gt;, which you mentioned.  The math is actually easier than that device/page might have you believe.  The advantage of the E6B is that you can use it with one hand while you are flying an airplane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you assume that the wind is constant everywhere, then the problem is fairly trivial.  If you assume that the wind is different from place to place and at different altitudes, and that the airplane's engine's efficiency is different at different altitudes, then the problem is more complex/interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the wind really matters is when the airplane is transitioning from the ground to the air (and thus their relative velocities really matter!).  Search youtube for &quot;cross-wind landing&quot; for some dramatic examples.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2010-11-21T21:23:36.897" LastActivityDate="2010-11-21T21:23:36.897" />
  <row Id="1190" PostTypeId="1" AcceptedAnswerId="1191" CreationDate="2010-11-21T21:47:24.863" Score="11" ViewCount="280" Body="&lt;p&gt;In general what will holding an anti-hydrogen atom for more than a 1/10th of second allow scientists to discover? &#xA;Specifically, given that they can hold one for &amp;lt;1/10th of a second, what would they discover that have not previously been able to determine. Or if not known, what have they been able to discover to date?&lt;/p&gt;&#xA;" OwnerUserId="582" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-22T05:48:41.840" LastActivityDate="2010-11-22T05:48:41.840" Title="In general what will holding an anti-hydrogen atom for more than a 1/10th of second allow scientists to discover?" Tags="&lt;particle-physics&gt;&lt;cern&gt;&lt;antimatter&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="1191" PostTypeId="2" ParentId="1190" CreationDate="2010-11-21T22:20:24.167" Score="11" ViewCount="" Body="&lt;p&gt;The ultimate goal is to be able to do precision spectroscopy of antihydrogen, to make sure that the energy states are the same as in ordinary matter. If there are differences between the energy levels of ordinary hydrogen and antihydrogen, that would violate &quot;CP&quot; symmetry, which says that if you change the sign of all the charges in some system, and invert the parity, every interaction should be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that CP violation occurs in nature-- it's been observed in kaon decay, among other things-- and it's related to the observed asymmetry between matter and antimatter in the universe. The known sources of CP violation are not enough to explain the matter-antimatter imbalance in the universe, though, so there have to be other forms of it out there that have yet to be discovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what we known about the interactions of matter and antimatter, any differences in the antihydrogen states would have to be very small, but laser spectroscopy can be used to do measurements of astonishing precision-- there are single-ion atomic clocks that are good to a few parts in $10^{18}$. Having the target atoms trapped for only a tenth of a second complicates matters, but a group at Argonne National Lab did spectroscopic measurements of the charge radius of unstable helium isotopes that don't last very much longer than that, so it's a good step toward the goal of doing spectroscopy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing that people talk about testing with anti-atoms is the behavior of gravity. Again, you need to have trapped neutral atoms for this, because electrostatic forces are thirty-some orders of magnitude stronger than gravity. That will also require extreme precision, and many more atoms than have been trapped to date, but the recent experiments are a good start, and the remaining issues are mostly technical, not fundamental.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-21T22:20:24.167" CommentCount="5" />
  <row Id="1192" PostTypeId="1" AcceptedAnswerId="1197" CreationDate="2010-11-21T23:30:41.127" Score="2" ViewCount="245" Body="&lt;p&gt;I heard Carl Sagan talking about the Universe 15 Billion years ago, and the Big Bang.  He made the statement that it was the biggest explosion of all time (at first I thought this a subtle pun).  This leads me to my question.  What would time have been like at +1 &quot;moment&quot; after the big bang?  What I'm trying to ask is, and I hate to say it because I'm afraid I'll sound foolish, did time flow at the same rate?  Wouldn't all that mass in one place have distorted space/time (and why didn't it &quot;rip&quot; it)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were inside of that mass with a stop watch, I'm guessing I wouldn't have been able to measure a difference because time would have effected me the same way as it effected all the other space/mass in the area.  I'm guessing I would have to have something inside (the initial Big Bang mass) and something outside measuring time and see if there was a difference (intuitively this feels weird to consider, could I actually place something &quot;outside&quot; the &quot;Big Bang mass&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe I've said too much, or made the question too complicated.  I apologize if this is the case.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Update&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A black hole is a lot of Mass collapsed into a small space.  I believe that as mass increases time dilation increases.  I remember hearing that if you fell into a black hole, you'd never experience the last second of your life...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is true of black holes, how did time pass in the mass/energy that is responsible for the big bang?  As the big bang occurred, did time speed up with the expansion of the universe?  I'm trying to explain what I'm getting at by asking more questions related to what I was asking.  I'm trying to understand what time itself looked like.  As I understand it time prevents everything from happening at once.  If time was a line, were the ends smashed together into a point before the Big Bang ?  Maybe as mass/energy expanded the &quot;time line&quot; expanded too?&lt;/p&gt;&#xA;" OwnerUserId="573" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2010-11-22T03:33:54.633" LastActivityDate="2010-11-22T09:32:10.673" Title="Big bang, and time" Tags="&lt;cosmology&gt;&lt;big-bang&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="1" />
  <row Id="1193" PostTypeId="1" CreationDate="2010-11-22T02:51:56.277" Score="10" ViewCount="1150" Body="&lt;p&gt;I was reading somewhere about a really cheap way of travelling: using balloons to get ourselves away from the surface of the earth. The idea held that because the earth rotates, we should be able to land in a different place after some time. As we all know, this doesn't happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone said that the reason why this doesn't happen is because the atmosphere (air, clouds etc.) also revolves around the earth (with the same angular velocity as the earth's rotation). Since we are also part of the atmosphere, our position doesn't change relatively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, I'm not convinced with that answer. Why does the atmosphere rotate along with earth?  Gravitational force is towards the centre of the earth, but I don't see how it's making the atmosphere rotate.&lt;/p&gt;&#xA;" OwnerUserId="447" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-12T16:42:53.360" LastActivityDate="2011-04-12T16:42:53.360" Title="Why does the atmosphere rotate along with the earth?" Tags="&lt;fluid-dynamics&gt;&lt;newtonian-mechanics&gt;&lt;atmosphere&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="3" />
  <row Id="1194" PostTypeId="2" ParentId="1193" CreationDate="2010-11-22T03:05:42.463" Score="10" ViewCount="" Body="&lt;p&gt;Friction AKA wind resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must have tried to stand in a strong wind or stuck you hand out the window of a traveling vehicle. From that you can feel the force that moving air exerts on objects in its way, and by Newton's law of reaction things in the way exert an equal force tending to move the air up to speed with the ground near it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if the atmosphere started at rest, it would have long since been brought up to speed.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-22T03:27:21.253" LastActivityDate="2010-11-22T03:27:21.253" />
  <row Id="1195" PostTypeId="2" ParentId="1193" CreationDate="2010-11-22T03:18:48.340" Score="15" ViewCount="" Body="&lt;p&gt;The atmosphere rotates along with the Earth for the same reason you do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force isn't needed to make something go.  That's a basic law of physics - that a thing that's moving will just keep moving if there's no force on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force is needed either to make something change its speed, or to make its motion point in a new direction.  A force can do both or just one of these.  Most forces do both, but a force that pushes in the exactly the same direction you're already going only changes your speed, and does not change your direction.  A force that pushes at a right angle to the direction you're already going only changes your direction, and does not add any speed.  A force at &quot;10 o'clock&quot;, for example, will change both your speed and your direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you stand still on Earth, you continue going the same speed, but your direction changes; between day and night you move opposite directions.  So the forces on you must be at a right angle to your direction of motion.  Indeed, they are.  Your motion is from west to east along the surface of the Earth, and the force of gravity pulls you down towards the center of the Earth - the force and your motion are at right angles.  Similarly for the atmosphere.  It is moving along with the Earth, and moving at a constant speed.  It does not need anything to push it along with the Earth.  Since only its direction of motion is changing, it only needs a force at a right angle to its motion, the same as you, and the force that does the job is again gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's not the whole picture, because the amount that your direction of motion changes depends on how strong the right-angle force is.  It turns out gravity is much too strong for how much our direction of motion changes as the Earth spins.  There must be some other force on us and on the atmosphere canceling out most of the gravity.  There is.  For me it's the force of the chair on my butt.  For the atmosphere, it's the air pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So gravity doesn't &quot;make the air rotate&quot;.  The air is already going, and gravity simply changes its direction to pull it in a circle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be wondering why the air doesn't just sit there and have the Earth spin underneath it.  One answer to that is that from our point of view that would mean incredibly strong wind all the time.  That wind would run into stuff and eventually get slowed down to zero (that's from our point of view - the air would &quot;speed up&quot; to our speed of rotation from a point of view out in space watching everything happen).  Even the air high up would eventually rotate with the Earth because although it can't slam into mountains or buildings and get stopped from blowing, it can essentially &quot;slam into&quot; the air beneath it due to friction in the air.  (This is a little redundant with dmckee's answer; I was half way done when he beat me to the punch)&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-03-11T06:51:56.340" LastActivityDate="2011-03-11T06:51:56.340" CommentCount="1" />
  <row Id="1196" PostTypeId="2" ParentId="1184" CreationDate="2010-11-22T03:31:17.213" Score="0" ViewCount="" Body="&lt;p&gt;Wind speed affects ground speed, not air speed. The airplanes fly at a specified IAS (indicated air speed). Add the IAS with the wind speed in the direction of travel and you get the ground speed. Using simple algebra&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ u_{ground} = u_{IAS} + u_{wind} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to travel a distance $S$ with head wind takes $t_{A\rightarrow B}=\frac{S}{u_{IAS}-u_{wind}}$, but to return with tail wind $t_{B\rightarrow A}=\frac{S}{u_{IAS}+u_{wind}}$. Obviously $t_{A\rightarrow B} &amp;gt; t_{B\rightarrow A}$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-11-22T03:31:17.213" />
  <row Id="1197" PostTypeId="2" ParentId="1192" CreationDate="2010-11-22T04:12:18.527" Score="2" ViewCount="" Body="&lt;p&gt;Here is a sense in which this can be answered a bit unambiguously--it is a known effect that gravitational fields both dilate time, by a factor $\sqrt{1-\frac{2\,G\,M}{c^{2}\,r}}$ and redshift light waves by that same factor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also known that cosmological effects redshift gravitational waves.  This time, it is done by a factor of $a(t)$, the so-called 'radius of the universe'.  For example, the cosmic microwave background radiation was believed to have been radiated from a surface whose temperature (and therefore, emitted wavelength) is roughly equivalent to the surface of a hot star.  It is a matter of simple algebra to find a value of $\frac{M}{r}$ for which the two effects are roughly equivalent, and, if you wish, you can think of this as describing a &quot;different rate of time flow.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my knowledge, there really isn't a useful reason to do this, though.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-11-22T04:12:18.527" CommentCount="2" />
  <row Id="1198" PostTypeId="2" ParentId="1193" CreationDate="2010-11-22T06:53:32.353" Score="9" ViewCount="" Body="&lt;p&gt;There are at least two reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the air layer adjacent to the Earth surface is dragged with it (being at rest with it).&lt;/li&gt;&#xA;&lt;li&gt;air &lt;a href=&quot;http://en.wikipedia.org/wiki/Viscosity&quot; rel=&quot;nofollow&quot;&gt;viscosity&lt;/a&gt; -- it could be thought as a friction between different air layers. Upper layers are carried along by underlying layers.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If the air were to stop suddenly it would result in &lt;a href=&quot;http://www.wolframalpha.com/input/?i=earth+diameter+in+km+divided+by+24+hours&quot; rel=&quot;nofollow&quot;&gt;~500 km/h wind speed&lt;/a&gt;. For comparison Hurricane Katrina's winds were &lt;a href=&quot;http://www.nasa.gov/vision/earth/lookingatearth/h2005_katrina.html&quot; rel=&quot;nofollow&quot;&gt;~100 km/h&lt;/a&gt;. It tells us that the air viscosity is sufficiently large that other effects (e.g., weather related) do not dominate its motion relative to the Earth surface.&lt;/p&gt;&#xA;" OwnerUserId="405" LastActivityDate="2010-11-22T06:53:32.353" />
  <row Id="1199" PostTypeId="2" ParentId="893" CreationDate="2010-11-22T07:20:11.063" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.fanfiction.net/s/5782108/1/Harry_Potter_and_the_Methods_of_Rationality&quot; rel=&quot;nofollow&quot;&gt;Harry Potter and the Methods of Rationality&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's &lt;a href=&quot;http://i.stack.imgur.com/WS5aw.jpg&quot; rel=&quot;nofollow&quot;&gt;a fan art&lt;/a&gt; that gives a taste of the kind of critical thinking being present in the book (it is very very slightly NSFW so I didn't include it inline).&lt;/p&gt;&#xA;" OwnerUserId="405" LastActivityDate="2010-11-22T07:20:11.063" CommunityOwnedDate="2010-11-22T07:20:11.063" />
  <row Id="1200" PostTypeId="2" ParentId="1192" CreationDate="2010-11-22T09:32:10.673" Score="2" ViewCount="" Body="&lt;p&gt;In the context of FRW cosmology, there is no difference in the rate of time between the epochs of the evolution of the universe. You can see that from the form of the line element&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ds^2=-dt^2+a(t)^2\gamma_{ij}dx^idx^j.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is a result of the symmetries that you assume for the matter distribution (homogeneous, isotropic) and the choice of observers that you make. So the observers that follow the expansion of the Universe, which are the galaxies more or less, perceive the same time wherever and whenever they are. The cosmological time is the proper time of all the comoving observers, as it is evident from the line element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a Schwarzschild metric and static observers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ds^2=-(1-\frac{2M}{r})dt^2+(1-\frac{2M}{r})^{-1}dr^2+r^2d\Omega^2,$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is the factor in front of dt that makes the difference and you have different time rates for observers at different positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one more point. Someone mentions the redshift and the perceived difference of the rate of time for faraway objects. That would appear to contradict what I am saying, but it isn't. The redshift effect is an observer symmetric effect. Like in the case of SR where you have two inertial observers with different velocities and each of them thinks that the others time runs slower, when both of them actually experience proper time. That is very different from the case of the static observers near a gravitating object, where there is no such symmetry. The clock of the observer that is at bigger r runs faster than the clock of the one that is at smaller r.&lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2010-11-22T09:32:10.673" CommentCount="17" />
  <row Id="1201" PostTypeId="1" AcceptedAnswerId="1207" CreationDate="2010-11-22T12:59:14.677" Score="6" ViewCount="385" Body="&lt;p&gt;We live in a world where almost all physical phenomena is non-linear, while the description of microscopic phenomena is based on quantum mechanics wich is linear by definition. What do you think the points of connection between the two descriptions?&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2011-07-30T15:04:10.477" ClosedDate="2010-11-22T19:04:59.287" Title="Linearity of quantum mechanics" Tags="&lt;quantum-mechanics&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="1202" PostTypeId="2" ParentId="1201" CreationDate="2010-11-22T13:18:11.443" Score="0" ViewCount="" Body="&lt;p&gt;There is other &quot;domain of linearity&quot;; $\ddot{x}=-x$ is a linear equation with solutions nonlinear in time.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-22T13:18:11.443" CommentCount="6" />
  <row Id="1203" PostTypeId="1" AcceptedAnswerId="1214" CreationDate="2010-11-22T13:34:35.267" Score="5" ViewCount="268" Body="&lt;p&gt;Could any form of matter teleportation every really happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, what form would it take? Some science-fiction talks about literally &lt;em&gt;sending&lt;/em&gt; the atoms from one location to another; others talk of creating a duplicate and destroying the original.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question originally added &lt;a href=&quot;http://area51.stackexchange.com/proposals/23848/theoretical-physics/23955#23955&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; there are a couple of comments asking me to specify what I mean by &lt;strong&gt;teleportation&lt;/strong&gt;. I'm talking about stuff of Star Trek - taking an object or person and transporting them almost instantly across great distances (thousands of KMs) by:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;breaking them up at the atomic (or smaller) level&lt;/li&gt;&#xA;&lt;li&gt;transporting those atomic components (or the &lt;em&gt;information&lt;/em&gt;, &quot;recipe&quot; to recreate) and&lt;/li&gt;&#xA;&lt;li&gt;(re-)assembling at the new location.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="414" LastEditorUserId="414" LastEditorDisplayName="" LastEditDate="2010-11-22T21:13:40.810" LastActivityDate="2010-11-22T21:13:40.810" Title="Could teleportation ever really happen?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="1204" PostTypeId="2" ParentId="1203" CreationDate="2010-11-22T13:51:26.670" Score="4" ViewCount="" Body="&lt;p&gt;What do you mean by &quot;teleportation&quot;? In the strictest sense, you will never be able to make an object disappear from one position and reappear in another instantaneously. That would create all sorts of problems with causality in relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's nothing particularly wrong with the idea of disassembling an object in one position, sending the information about how it was put together to another position at the speed of light, and reassembling the object at that location from raw materials already in place. That's effectively like teleportation in a lot of ways, but doesn't violate any rules of physics. Using the techniques of quantum teleportation, you can even make sure that the distant copy is created in exactly the same quantum state as the original, should that be important.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-22T13:51:26.670" CommentCount="5" />
  <row Id="1205" PostTypeId="1" AcceptedAnswerId="1225" CreationDate="2010-11-22T13:56:05.730" Score="10" ViewCount="411" Body="&lt;p&gt;When a wind blow through sharp edge, say, edge of a paper, you can see the vibration of the paper and hear the sound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this type of oscillation, it should be a damped oscillation with external driving force $\mathcal{F}(x)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ddot{x}+2\lambda\dot{x}+\omega_{0}^{2}x=\mathcal{F}(x,t)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with the driving force in the form of sinusoidal $\mathcal{F}(x,t)\sim F\sin(\omega t)$ (It should be a 3D version though). However, if the air flow is really steady, the external force should be constant so there is no oscillation at all from this equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple guess is that the motion of the paper influence the streamline of air flow so that the pressure gradient provides the exact sinusoidal driving force. This mechanism is reasonable because the propagation of sound wave has the oscillation relationship between the displacement and the pressure. A question here is how the air flow sustain the energy loss due to the damping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also not easy to think of the initial condition as well as external force for these oscillation. So &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What exactly is the physical mechanism to generate sound when a 'steady' air flowing through a sharp edge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Another similar question is how the driving force act in the resonance in musical instrumental such as pipe? A 'steady' flow of air is also provided somewhere to the pipe, but air oscillates inside the pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: From the answers, people suggest few mechanism for different situations. Few more questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) For static object, Kármán vortex street can form behind the object. So is the sound frequency the same as frequency of vortex generation, or the same as the expansion of the vortex? Sound is a spherical wave propagating outward, so identify the sound frequency should locate the point of generation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Where is sound generated in the situation (1)? Is the sound frequency the same as the frequency of the vibrating paper?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fluid mechanics is a difficult topics, but there are approximation for special cases trying to explain the underlying mechanism, such as the flag fluttering cited by j.c.&lt;/p&gt;&#xA;" OwnerUserId="562" LastEditorUserId="562" LastEditorDisplayName="" LastEditDate="2010-11-24T04:21:06.723" LastActivityDate="2010-11-24T04:21:06.723" Title="How does a steady wind flow generate sound?" Tags="&lt;fluid-dynamics&gt;&lt;pressure&gt;&lt;sound&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="3" />
  <row Id="1206" PostTypeId="2" ParentId="1201" CreationDate="2010-11-22T13:58:13.110" Score="3" ViewCount="" Body="&lt;p&gt;Assuming you mean &quot;linear&quot; in the mathematical sense of &quot;the sum of two solutions to the relevant equation is also a solution,&quot; there's no particular reason why macroscopic objects are inherently non-linear. In fact, there is a large body of work in the quantum foundations community on ways to have macroscopic objects behave in a linear manner but &lt;em&gt;look&lt;/em&gt; non-linear. That's the whole point of things like the Many-Worlds interpretation of quantum mechanics, and the research into decoherence by people like W. Zurek. There may be a scale above which it is impractical to &lt;em&gt;see&lt;/em&gt; superposition states, but that doesn't mean that they can't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's not what you mean, then I don't know how to answer you.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-22T13:58:13.110" />
  <row Id="1207" PostTypeId="2" ParentId="1201" CreationDate="2010-11-22T14:44:10.403" Score="2" ViewCount="" Body="&lt;p&gt;Linear in the quantum mechanics has nothing to do with its complexity. A two-state spin can be described by a simple 2-by-2 matrix; however, 30 interacting spin, in general, must be described by a 1 billion by 1 billion matrix. It grows exponentially as the number of spins increases, for $10^{23}$ spin, you may need a matrix of size $2^{10^{23}}$. It is not easy to understand and not simple in most sense. If you learn some statistical mechanics, you will know that this number is large enough to have new emergent phenomenon.&lt;/p&gt;&#xA;" OwnerUserId="562" LastEditorUserId="4466" LastEditorDisplayName="" LastEditDate="2011-07-30T15:04:10.477" LastActivityDate="2011-07-30T15:04:10.477" CommentCount="3" />
  <row Id="1208" PostTypeId="2" ParentId="1205" CreationDate="2010-11-22T14:59:04.550" Score="2" ViewCount="" Body="&lt;p&gt;It is always about a resonance between the vibrating element and a pressure mod in some resonator (in case of sharp edge, the edge itself oscillates).&lt;br&gt;&#xA;I believe the oscillations itself are produced by the turbulent vortices forming &lt;a href=&quot;http://en.wikipedia.org/wiki/K%C3%A1rm%C3%A1n_vortex_street&quot; rel=&quot;nofollow&quot;&gt;Kármán street&lt;/a&gt; behind the reed or edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I have even found a reference: &lt;a href=&quot;http://arxiv.org/abs/physics/0008053v1&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/physics/0008053v1&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT2: In detail, the process starts with static edge in laminar flow; speed of flow increases and finally the Reynolds number exceeds critical value in which streamlines start to disengage from edge first forming vortices in Karman street (this is called Strouhal instability) and then a totally random turbulent flow. The vortices interact with the edge &quot;poking&quot; it quite randomly (Karman street has some frequency, but I think it does matter only for some narrow cases); now the process is driven by resonances. The edge has some resonance frequencies, the air around can also have them if enclosed in some container, in case of some instruments the musician's lips are also involved -- all those for some kind of mechanical filter that amplifies certain frequencies (instruments) or ranges of frequencies (random setups). Finally, those mechanic vibrations induce pressure weaves in air that we hear as a sound, again either clear as in case of instruments or a noise as in case of trees, roofs and other stuff.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-11-22T20:32:29.100" LastActivityDate="2010-11-22T20:32:29.100" CommentCount="2" />
  <row Id="1209" PostTypeId="2" ParentId="1201" CreationDate="2010-11-22T15:35:01.253" Score="5" ViewCount="" Body="&lt;p&gt;There is an all too common misconception that because the Schrodinger equation is linear, non-linear phenomena (like chaos) are only classical. The wavefunction does obey a linear equation, the Schrodinger equation, but it is not directly related to observable physics. Observables quantities, like expectation values of operators, obey non-linear equations. In fact, many times the same equations as their classical counterparts, with small corrections.&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2010-11-22T15:35:01.253" />
  <row Id="1210" PostTypeId="2" ParentId="1037" CreationDate="2010-11-22T15:43:04.213" Score="0" ViewCount="" Body="&lt;p&gt;Contact lenses work in the same way as glasses, by adding or subtracting wavefront curvature.&#xA;This basically adjusts where the focal point of light entering the eye is, glasses and contact lenses are designed to adjust it so that the focal point of the light lies on the retina.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main differences I can see between glasses and contact lenses, is that the contact lens faces are both curving outwards like this (( rather than this () or this )( like traditional glasses.&#xA;This is so that the lens can rest comfortably on the outside of the eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we remember the laws of optics, we remember that the light is only affected by the changes in refractive indices, in the boundaries between air and glass for example.&#xA;In normal glasses the two surfaces are air to glass and glass to air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contact lenses we have again two surfaces, but it's air to contact lens then contact lens to eye.&#xA;Note I say contact lens not glass as contact lenses are made from a variety of materials.&#xA;The focal point adjustment in a contact lens happens mostly at the outer boundary - the one between air and the contact lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the other surface is curved the 'wrong' way (for some types of lens requirements) means that it's effect must be calculated and compensated for by the outer surface.&lt;/p&gt;&#xA;" OwnerUserId="133" LastActivityDate="2010-11-22T15:43:04.213" />
  <row Id="1211" PostTypeId="2" ParentId="1205" CreationDate="2010-11-22T16:01:11.897" Score="3" ViewCount="" Body="&lt;p&gt;The full description of viscous fluid flow (i.e. the &lt;a href=&quot;http://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations&quot; rel=&quot;nofollow&quot;&gt;Navier-Stokes equations&lt;/a&gt;) is non-linear and can be sensitively dependent on initial conditions. What this means in practical terms it that you can't always count on your intuition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The evolution of the &lt;a href=&quot;http://physics.stackexchange.com/questions/1205/how-does-a-steady-wind-flow-generate-sound/1208#1208&quot;&gt;Kármán vortex streets linked by mbq&lt;/a&gt; from &lt;a href=&quot;http://en.wikipedia.org/wiki/Laminar_flow&quot; rel=&quot;nofollow&quot;&gt;laminar flow&lt;/a&gt; around an obstacle are are a classic demonstration. (And easy enough for a determined middle school student to create in the garage for a science project, though things are likely to get wet...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any way, once the wind starts doing non-linear things, it can generate periodic stresses, and from that you get the whistling or humming noise we all know and love. Add a resonant cavity and you can amplify nice pure tones which is the short-short version of how this works in wind instruments.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2010-11-22T16:01:11.897" CommentCount="2" />
  <row Id="1213" PostTypeId="1" AcceptedAnswerId="1288" CreationDate="2010-11-22T16:35:42.123" Score="5" ViewCount="224" Body="&lt;p&gt;Is there a precise mathematical derivation for the correspondence principle for which I can replace $E \rightarrow i \hbar \frac{\partial}{\partial t}$ and $p \rightarrow -i \hbar \nabla$?&lt;/p&gt;&#xA;" OwnerUserId="592" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-24T23:09:49.510" LastActivityDate="2010-11-25T04:28:04.027" Title="Correspondence principle" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="2" />
  <row Id="1214" PostTypeId="2" ParentId="1203" CreationDate="2010-11-22T20:27:14.940" Score="4" ViewCount="" Body="&lt;p&gt;I don't think it's possible in the way you mean it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Option 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Decompose, move, rebuild&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Decomposition is possible as long as we can consider the matter classical. If any quantum state is relevant, it will be automatically destroyed in the process of measuring the state of each atom in order to be able to rebuild later. This will lead to a large increase in entropy (which is relevant in the last step).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The movement of a stream of atoms is possible, but not simple at all, especially if you need to focus it in a spot. The energy required, though is probably very similar to the energy required to transport the matter classically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rebuilding is possible (given enough technology, that is), but it would be enormously expensive in terms of energy because of the need of reversing the increase of entropy in the first step, locally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all, it's probably possible, but much more expensive in terms of energy than simply moving the body, so it's not clear to me how useful it would be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, were the matter being transported a living being, there are a number of ethical questions involved: would the person rebuilt be still the same person as the one killed by the first step of the process?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Option 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Use &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_teleportation&quot; rel=&quot;nofollow&quot;&gt;quantum teleportation&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, as far as I know, requires a quantum entanglement between the mass and a perfect copy of the mass. Although this is probably possible for simple systems, I am not really sure it represents the kind of teleportation you are looking for. It's a teleportation of quantum states, not of matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Option 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Abuse Quantum Mechanics&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is far off, and there's nothing suggesting that it's possible, but if we found a way of interfering radically with the probabilities of quantum mechanical measurements, we could leverage the indeterminacy in the body's position to make it move. It really depends on us understanding better how the wave function actually collapses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Option 4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Abuse General Relativity&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also far off, but if we were to open a stable &lt;a href=&quot;http://en.wikipedia.org/wiki/Wormhole&quot; rel=&quot;nofollow&quot;&gt;wormhole&lt;/a&gt; between two positions then it should be possible to step from one place to another instantly. This &lt;em&gt;merely&lt;/em&gt; requires using currently unknown forms of exotic matter to generate the holes, because all the wormholes that we know of, in GR, are actually of really small diameter and also apparently a not very healthy place to be because of extreme gravity, radiation and so on.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A couple of good resources (divulgative but physically accurate):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0141030909&quot; rel=&quot;nofollow&quot;&gt;Physics of the Impossible: A Scientific Exploration of the World of Phasers, Force Fields, Teleportation and Time Travel&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0006550428&quot; rel=&quot;nofollow&quot;&gt;The Physics of Star Trek&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-22T20:33:17.400" LastActivityDate="2010-11-22T20:33:17.400" CommentCount="6" />
  <row Id="1215" PostTypeId="2" ParentId="1126" CreationDate="2010-11-22T21:03:57.013" Score="0" ViewCount="" Body="&lt;p&gt;If you throw a ball with an angle $\theta$ and initial velocity $v_1$, the ball will hit the ground at a distance given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d_1 = \frac{v_1^2}{g}sin(2\theta)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if you throw it again with a different velocity $v_2$, the ball will hit the ground at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d_2 = \frac{v_2^2}{g}sin(2\theta)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are asking is if there is such a $\theta$ for which $d_1 = d_2$ while $v_1 \neq v_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d_1 = d_2 \implies \frac{v_1^2}{g}sin(2\theta) = \frac{v_2^2}{g}sin(2\theta)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_1^2 sin(2\theta) = v_2^2 sin(2\theta)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $sin(2\theta) \neq 0$ this leads to $v_1^2 = v_2^2$ which is never true for positive velocities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$sin(2\theta) = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this gives the following solutions: $0, \frac{\pi}{2}, \pi...$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if the the angle is 0 or 180 degrees, then the ball will hit the ground instantaneously in both cases. If the angle is 90 degrees, the ball will simply go up and fall back on the same spot in both cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: this does not take into account air resistance, variations in the acceleration of gravity due to distance and the rotation of the Earth.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-22T21:03:57.013" />
  <row Id="1216" PostTypeId="1" AcceptedAnswerId="1218" CreationDate="2010-11-22T21:35:13.243" Score="16" ViewCount="833" Body="&lt;p&gt;Sorry if this question is naive. It is just a curiosity that I have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there theoretical or experimental reasons why gravity should not be an emergent property of nature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume a &lt;em&gt;standard model&lt;/em&gt; view of the world in the very small. Is it possible that gravity only applies to systems with a scale large enough to encompass very large numbers of particles as an emergent property?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all: the standard model works very well without gravity; general relativity (and gravity in general) has only been measured at distances on the millimeter scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could gravity emerge? For example, it could be that space-time only gets curved by systems which have measurable properties, or only gets curved by average values. In other words that the stress-energy tensor has a minimum scale by which it varies.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit to explain a bit better what I'm thinking of.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;We would not have a proper quantum gravity as such. I.e. no unified theory that contains QM and GR at the same time. &lt;/li&gt;&#xA;&lt;li&gt;We could have a &quot;small&quot; (possibly semi-classical) glue theory that only needs to explain how the two theories cross over: &#xA;&lt;ul&gt;&#xA;&lt;li&gt;the conditions and mechanism of wave packet reduction (or the other corresponding phenomena in other QM interpretations, like universe branching or decoherence or whatnot)&lt;/li&gt;&#xA;&lt;li&gt;how this is correlated to curvature - how GM phenomena arise at this transition point.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there theoretical or experimental reasons why such a reasoning is fundamentally incorrect?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-09T14:54:29.327" LastActivityDate="2011-03-17T01:23:43.700" Title="Could gravity be an emergent property of nature?" Tags="&lt;general-relativity&gt;&lt;gravity&gt;&lt;cosmology&gt;&lt;standard-model&gt;" AnswerCount="5" CommentCount="9" FavoriteCount="4" />
  <row Id="1217" PostTypeId="1" CreationDate="2010-11-22T21:38:47.007" Score="8" ViewCount="392" Body="&lt;p&gt;For example, why is the semileptonic $B$ decay $B \to X\ell\nu$ inclusive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't find any definition of these frequently used terms, strange.&lt;/p&gt;&#xA;" OwnerUserId="598" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-29T17:14:20.743" LastActivityDate="2011-08-29T17:14:20.743" Title="What's the difference between inclusive and exclusive decays?" Tags="&lt;high-energy-physics&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="1218" PostTypeId="2" ParentId="1216" CreationDate="2010-11-22T21:49:24.463" Score="3" ViewCount="" Body="&lt;p&gt;I'm not an expert in gravity, however, this is what I know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a hypothesis about gravity being &lt;a href=&quot;http://en.wikipedia.org/wiki/Entropic_gravity&quot; rel=&quot;nofollow&quot;&gt;an entropic property&lt;/a&gt;. The paper from Verlinde is available at &lt;a href=&quot;http://arxiv.org/abs/1001.0785&quot; rel=&quot;nofollow&quot;&gt;arXiv&lt;/a&gt;. That said, I would be surprised for this to be true. The reason is simple. As you probably know, entropy is an emergent property out of statistical probability. If you have non-interacting, adimensional particles into one half of a box, with the other half empty and separated by a valve, it's probability, thus entropy, that drives the transformation. If you look at it from the energetic point of view, the energy is exactly the same before and after the transformation. This works nicely for statistical distribution, but when you have to explain why things are attracted to each other statistically, it's much harder. From the probabilistic point of view, it would be the opposite: the more degrees of freedom your particles have, the more entropy they have. A clump has less degrees of freedom, hence has less entropy, meaning that, in a closed system, the existence of gravity is baffling. This is out of my speculation, and I think I am wrong. The paper seems to be a pleasure to read, but I haven't had the chance to go through it.&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="217" LastEditorDisplayName="" LastEditDate="2010-11-29T23:42:25.633" LastActivityDate="2010-11-29T23:42:25.633" CommentCount="2" />
  <row Id="1219" PostTypeId="2" ParentId="1217" CreationDate="2010-11-22T22:02:45.833" Score="6" ViewCount="" Body="&lt;h2&gt;An experimental take&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exclusive&lt;/strong&gt; implies that you have measured the energy and momenta of &lt;em&gt;all&lt;/em&gt; the products  (well, with an exception I'll discuss below). &lt;strong&gt;Inclusive&lt;/strong&gt; means that you may have left some of the products unmeasured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This applies to scattering processes as well as decays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some things to note:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Exclusive measurements allow you to nail down &lt;em&gt;one&lt;/em&gt;, well defined physics process, while inclusive measurements may tell you about a collection of processes&lt;/li&gt;&#xA;&lt;li&gt;It is generally difficult to measure neutral particles&lt;/li&gt;&#xA;&lt;li&gt;If there are more than a couple of products it begins to require a lot of instrumentation to reliably collect them all and (crucially) to know how well you have done so&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the process you are asking about the neutrino is necessarily unobserved rendering the measurement inclusive, further an $X$ in the final state is often used to indicate unmeasured and unspecified stuff (i.e. it means the measurement is inclusive by design).&#xA;Here unspecified includes case in high acceptance instruments where you consider all events with the specified products: those for which we know $X$ is empty those for which $X$ is non-empty and well characterized, and those for which $X$ is ill-characterized.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Theoretical view&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I'm less sure of how theorist use these terms, but I believe there is a parallel. Something like: &lt;strong&gt;exclusive&lt;/strong&gt; means one and only one process, while &lt;strong&gt;inclusive&lt;/strong&gt; means all processes that include the specified products.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Convergence of theory and experiment&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Of course, we haven't really learned anything until we get theory and experiment together, which is sometimes traumatic for both communities. Still &lt;em&gt;exclusive&lt;/em&gt; measurements and calculations are clearing talking about the same thing, and &lt;em&gt;inclusivity&lt;/em&gt; can be made to agree with some care in building the experiment and assembling the theoretical results.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Experimenters cheating on exclusivity&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Sometimes in nuclear physics we talk about scattering measurements as &lt;em&gt;exclusive&lt;/em&gt; when there is an unmeasured, heavy, recoiling nucleus involved. The assumption being that that it carries a small fraction of the total energy and momentum involved and can be neglected, though there is some risk from this if the nucleus is left in a highly excited state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular my dissertation project was on $A(e,e&amp;#39;p)$ reaction (elastic scattering of protons out of a stationary nuclear target where the beam was characterized and both the proton and outgoing electron were observed), and we assumed that the remnant nucleus was left largely undisturbed and recoiling with a momentum opposite the Fermi motion of the target proton.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2010-11-23T22:39:12.577" LastActivityDate="2010-11-23T22:39:12.577" CommentCount="7" />
  <row Id="1220" PostTypeId="1" AcceptedAnswerId="1221" CreationDate="2010-11-23T01:12:22.473" Score="3" ViewCount="528" Body="&lt;p&gt;If two ends of a rope are pulled with forces of equal magnitude and opposite direction, the tension at the center of the rope must be zero. True or false?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is false. I chose true though and I'm not understanding why. Forces act at the center of mass of the object, so if there are two forces of equal and opposite magnitude, then they should cancel out resulting in zero tension, no?&lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2010-11-24T05:11:33.683" Title="Rope tension question" Tags="&lt;classical-mechanics&gt;&lt;forces&gt;" AnswerCount="4" />
  <row Id="1221" PostTypeId="2" ParentId="1220" CreationDate="2010-11-23T01:42:21.537" Score="16" ViewCount="" Body="&lt;p&gt;The tension of the rope is the shared magnitude of the two forces. Imagine cutting the rope at a point and inserting a spring scale in its place. The reading will show the tension. A rope with zero tension would be hanging loosely or laying on the ground, neglecting the rope's mass.&lt;/p&gt;&#xA;" OwnerUserId="69" LastEditorUserId="69" LastEditorDisplayName="" LastEditDate="2010-11-23T22:45:17.427" LastActivityDate="2010-11-23T22:45:17.427" CommentCount="3" />
  <row Id="1222" PostTypeId="1" AcceptedAnswerId="1287" CreationDate="2010-11-23T01:55:55.747" Score="3" ViewCount="308" Body="&lt;p&gt;So, the canned explanation that I always hear about why the white hole solution of the extended Schwarzschild solution is non-physical is that &quot;The matter distribution cuts off the white hole solution.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I look at Oppenheimer-Snyder collapse, however, the cosmological solution that I paste to in the dust's interior is just the closed pressureless model of the Robertson-Walker metric, which is perfectly time-symmetric about the point of maximal expansion.  If I look at this naïvely, then, the fluid in the Oppenheimer-Snyder solution originated in a singularity, which then disintegrates, to form an expanded fluid, which reaches some maximal extent, and then recollapses to form a singularity in the future, after some finite time.  That past singularity sure seems like a white hole to me.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I misinterpreting here?  Or is it just a pathology of the Oppenheimer-Snyder construction that it is over-simple, and more physically realistic matter models WOULD cut off the past singularity?&lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2010-11-23T02:30:40.677" LastActivityDate="2011-01-18T14:43:49.463" Title="White Holes and Time-Reversed Oppenheimer-Snyder collapse" Tags="&lt;general-relativity&gt;&lt;gravitational-collapse&gt;&lt;black-holes&gt;" AnswerCount="4" />
  <row Id="1223" PostTypeId="1" AcceptedAnswerId="1284" CreationDate="2010-11-23T02:19:04.650" Score="5" ViewCount="90" Body="&lt;p&gt;There are by now (late 2010) many tests of GTR.  GTR is holding up quite well!  But, surely, we do not yet have it cornered in every possible way.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of the PPN parameters used to characterize low-order deviations of any theory of gravity from Newtionian ideals,  where are the loosest screws?   Which PPN parameter is least constrained by experiment?   What kinds of experiments would help constrain it?  &lt;/p&gt;&#xA;" OwnerUserId="353" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-24T23:26:18.717" LastActivityDate="2010-11-24T23:26:18.717" Title="Which PPN parameters are least well determined?" Tags="&lt;general-relativity&gt;&lt;experimental-physics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="1224" PostTypeId="2" ParentId="1216" CreationDate="2010-11-23T02:52:14.390" Score="4" ViewCount="" Body="&lt;p&gt;Isn't the answer to the question of the title widely believed to be &quot;yes&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you believe that searching for what high-energy theorists call a &quot;theory of everything&quot; is a valuable and worthwhile enterprise, then you probably also believe that &lt;strong&gt;gravity as we currently understand it&lt;/strong&gt; (General Relativity, say) &lt;strong&gt;&quot;emerges&quot; from some deeper theory&lt;/strong&gt; (in the effective field theory sense) &lt;strong&gt;which unifies it with all other known fundamental forces&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course nobody yet knows for sure what that theory is, but I'm told &lt;strong&gt;certain flavors of string theory&lt;/strong&gt; are the most viable candidates as of 2010.  You can find some indication towards how gravity emerges from string theory in &lt;a href=&quot;http://physics.stackexchange.com/questions/1073/how-does-classical-gr-concept-of-space-time-emerge-from-string-theory/1086#1086&quot;&gt;the first few sentences of this answer by Eric Zaslow&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps Eric Zaslow or some other expert can give more details at the level of saying, for instance, how Einstein's equations arise from string theory (I would ask this as a question on this site, except that I know I could find the answer in any book on string theory if I cared enough to look).  I'm told that it has something to do with the renormalization group equations of the conformal field theory on the worldsheet, but I'm afraid I can't reproduce or explain that argument any further for you here.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-23T02:52:14.390" CommentCount="6" />
  <row Id="1225" PostTypeId="2" ParentId="1205" CreationDate="2010-11-23T03:08:03.180" Score="2" ViewCount="" Body="&lt;p&gt;There are really two parts to your question.  &lt;strong&gt;First, how does the wind affect the motion of the paper, and how does that motion then couple to the behavior of the wind?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the case of a flag fluttering due to the wind (which may or may not be applicable depending on what you have in mind), the physics of the instability leading to fluttering has been worked out in a &lt;a href=&quot;http://www.pnas.org/content/102/6/1829.full&quot; rel=&quot;nofollow&quot;&gt;fairly celebrated paper of 2005 by Argentina and Mahadevan&lt;/a&gt;.  They argue that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] in a particular limit corresponding to a low-density fluid flowing over a soft high-density flag, the flapping instability is akin to a resonance between the mode of oscillation of a rigid pivoted airfoil in a flow and a hinged-free elastic plate vibrating in its lowest mode. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suspect this paper and probably some of the papers that cite it would be the right place to look.  As you can see, this part of the question that you've asked is of fairly high interest in current research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second, how does this motion generate sound?&lt;/strong&gt;  This is also an interesting question but I know less about this.  I would suppose that some resonant vibrational modes of the sheet of paper that you're asking about (those involved in the flapping modes described by Argentina and Mahadevan) generate the sound.  But I don't know very much about sound generation or acoustics.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-23T03:08:03.180" CommentCount="2" />
  <row Id="1226" PostTypeId="2" ParentId="1094" CreationDate="2010-11-23T05:18:51.250" Score="1" ViewCount="" Body="&lt;p&gt;A large sheet or volume of matter that contains plenty of freely moveable charge carriers.  Examples: metal, which contains electrons not assigned to particular atoms, but free to move;  plasmas which consist of electrons and positive ions free to move in space; or  a cloud of charged dust particles.   If you just want to slow it down not stop the radio waves, any dielectric material will do, the the index of refraction can vary greatly with frequency.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have enough of this material to block the line of sight between all points of the emitter and all points of the sensor/observer, and more beyond that to reduce radiation sneaking around due to diffraction.   To say much more, the question needs to be more specific.&lt;/p&gt;&#xA;" OwnerUserId="353" LastActivityDate="2010-11-23T05:18:51.250" />
  <row Id="1227" PostTypeId="1" AcceptedAnswerId="1228" CreationDate="2010-11-23T05:45:05.720" Score="1" ViewCount="244" Body="&lt;p&gt;I have a formula which is $\text{G-force} = \frac{v\omega}{9.8}$, where $v$ is speed and $\omega$ is the angular velocity. I've seen on the internet that G-force is actually $\text{acceleration}/9.8$. I'm confused as to which formula is correct. For simulating the motion of particle taking a turn, would omega simply be velocity divided by radius of turn? Assuming Cartesian coordinates.&lt;br&gt;&#xA;Another funny thing I noticed is that while simulating particle motion, a 7G turn showed up as an almost straight line (while using a constant turn motion model) with a velocity of 900m/s and time interval of 1second. Am I simulating wrong or is my use of the first equation wrong?&lt;/p&gt;&#xA;" OwnerUserId="604" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-11-23T15:07:38.357" LastActivityDate="2010-11-23T15:07:38.357" Title="Calculation of the G-force" Tags="&lt;simulation&gt;&lt;angular-velocity&gt;&lt;centrifugal-force&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="1228" PostTypeId="2" ParentId="1227" CreationDate="2010-11-23T06:14:20.843" Score="3" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/G_force&quot; rel=&quot;nofollow&quot;&gt;g force&lt;/a&gt; is a unit of acceleration. 1 g is equal to 9.80665 m s&lt;sup&gt;-2&lt;/sup&gt;. So the correct formula is&#xA;$$ \text{G force} = \frac{\text{Acceleration in m s}^{-2}}{9.8}. $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when describing uniform circular motion (i.e. $\boldsymbol\omega$ is constant) in free space, the only acceleration felt by the person rotating (in their frame of reference) is the centrifugal acceleration, which is exactly &#xA;$$ a = \frac{v^2}r = v\omega = \omega^2 r,$$&#xD;&#xA;so the first expression is also correct for centrifugal acceleration &lt;em&gt;of uniform circular motion&lt;/em&gt;. (If the motion is not a uniform circular motion, only $a = \omega^2 r$ can be used to describe the centrifugal acceleration.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I don't know how do you get the 7 g.)&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-11-23T06:22:03.647" LastActivityDate="2010-11-23T06:22:03.647" CommentCount="9" />
  <row Id="1229" PostTypeId="1" CreationDate="2010-11-23T06:24:03.533" Score="5" ViewCount="192" Body="&lt;p&gt;We use Right Hand Rule in calculating Torque not because that's the direction torque is pointing in the real, physical world, but because it's a convenient way to indicate the &quot;sign&quot; of the rotation and its axis, and as long as we pick one coordinate system and stick to it, everything works out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think (I might be incorrect) we use the Right Hand Rule in Electromagnetism arbitrarily, but it happens to be very convenient only because of the right combination of Benjamin Franklin's arbitrary choice for designating current flow and the way our magnetic field was aligned at the time of the convention choice for North.  But essentially, it is arbitrary -- if Benjamin Franklin had chosen current flow to match the flow of electrons, we'd either use another choice of axes or say that the magnetic field is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm saying is that I don't think that nature itself is inherently &quot;right handed&quot; in this situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any case/situation in physics where the Right Hand Rule and choice of coordinate axes is not arbitrary, but rather inherent in nature itself?  How so?&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="71" LastEditorDisplayName="" LastEditDate="2010-11-23T07:30:19.263" LastActivityDate="2010-11-23T19:17:52.180" Title="Is there any situation in Physics where the Right Hand Rule is not arbitrary?" Tags="&lt;chirality&gt;&lt;coordinate-systems&gt;&lt;parity&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="1230" PostTypeId="1" CreationDate="2010-11-23T06:31:55.727" Score="5" ViewCount="401" Body="&lt;p&gt;This might be a question to ask in a Chemistry site, but because there is a lot of talk about redefining many units of measurements in terms of Avogadro's Number / the Mole, I was wondering why we don't just redefine the Mole to infinite precision, since it is basically inherently an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might be the only unit/physical constant that &lt;em&gt;can&lt;/em&gt; be defined to infinite precision.  The only unit that is an &quot;integer&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess it's not really a physical constant, per se, in that it is not a property of nature.  But then we can easily create a definition of the Kilogram that doesn't change over time as exactly something-something moles worth of Carbon-12.&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-11-29T16:46:56.337" LastActivityDate="2010-12-21T14:30:21.547" Title="What's keeping us from simply redefining Avogadro's Number / the Mole as a definite integer?" Tags="&lt;si-units&gt;&lt;mole&gt;&lt;physical-constants&gt;&lt;metrology&gt;" AnswerCount="5" CommentCount="7" />
  <row Id="1231" PostTypeId="2" ParentId="1220" CreationDate="2010-11-23T07:04:29.900" Score="10" ViewCount="" Body="&lt;p&gt;Here's a slightly more mathematical answer to go with Vortico's excellent physical answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tension is not a force, and doesn't have a magnitude and direction.  This is confusing, because we frequently talk about the &quot;force of tension&quot;, which is a different thing from the tension itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tension is actually an example of a mathematical object called a &quot;rank-two tensor&quot;.  A rank two tensor is a function whose input and output are both vectors.  The input is a direction, and the output is the force exerted on the piece of rope lying in that direction.  So for a vertical rope, you can take a unit vector pointing down as the input to the tension tensor, and it gives back a force of, say, 20N up.  That would mean that at any point in the rope, the rope is pulling up on the parts beneath it with 20N of force.  Then if you input a unit vector point up, you'll get back a force of 20N down.  Any part of the rope pulls down on whatever is above it (more rope, or maybe a ceiling) with 20N of force.  We then say that &quot;the tension in the rope is 20N&quot;, but what we mean is actually the entire story above about inputting directions and getting out forces.  Every little piece of rope has two 20N forces on it from the surrounding rope, and also exerts two 20N forces on the surrounding rope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a little pedantic, since the tension is just a number and the whole bit about the tensor does not seem very important.  It becomes more important when we talk about surface tension, perhaps in a metal sheet.  In that case we could imagine that the tension really is different in different directions.  Physically, if you cut a slit in the sheet, and then measured how much force it takes to hold the slit closed, you could find that the force required (per unit length of the slit) depends on the slit's direction.  You would need the full power of the rank-two tensor to describe that tension.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-23T07:04:29.900" CommentCount="7" />
  <row Id="1233" PostTypeId="2" ParentId="1230" CreationDate="2010-11-23T07:36:15.563" Score="9" ViewCount="" Body="&lt;p&gt;The problem is that you want your unit definitions to be realizable - so specifying &quot;1 mol is &lt;em&gt;long number&lt;/em&gt; molecules, 1 gram is 1/12 of the mass of one mol of $C_{12}$&quot; is nice for your thought process, but as long as there is no practical way to count molecules at such scales to a precision of better than $10^{-9}$ (which I think is the precision of the kilogram standard), there is no operational advantage to it.&lt;/p&gt;&#xA;" OwnerUserId="243" LastEditorUserId="243" LastEditorDisplayName="" LastEditDate="2010-11-23T13:36:02.027" LastActivityDate="2010-11-23T13:36:02.027" />
  <row Id="1234" PostTypeId="2" ParentId="1229" CreationDate="2010-11-23T07:43:39.997" Score="6" ViewCount="" Body="&lt;p&gt;No, it really is arbitrary. The reason we use the right hand rule &lt;em&gt;today&lt;/em&gt; (although it may have been chosen for different reasons of convenience in the past) is simply that our coordinate system of choice is right-handed. Mathematically, this means that we define the directions of the axes so that you have to use the right-hand rule to evaluate this cross product:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hat{x}\times\hat{y} = \hat{z}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and everything flows from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be just as valid to define a left-handed coordinate system, in which the $\hat{z}$-axis points the other way. In that coordinate system, you would use the left-hand rule to evaluate cross products, but physics would still work the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the experiment that discovered parity violation, the researchers found that muons were emitted parallel (or antiparallel, I forget which - but it was one and not the other) to the angular momentum of decaying cobalt nuclei. If physics had been developed with a left-handed coordinate system, I think they would have found the opposite result, since angular momentum is something that we use the right-hand rule to define - but for exactly that reason, there's no &lt;em&gt;fundamental&lt;/em&gt; distinction between a &quot;parallel&quot; result and an &quot;antiparallel&quot; result. The direction of angular momentum, and thus the word we use for the result of the cobalt decay experiment, is an artifact of our coordinate system.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-23T08:02:44.227" LastActivityDate="2010-11-23T08:02:44.227" />
  <row Id="1235" PostTypeId="1" AcceptedAnswerId="1242" CreationDate="2010-11-23T08:08:07.367" Score="23" ViewCount="667" Body="&lt;p&gt;One can obtain the solution to a $2$-Body problem analytically. However, I understand that obtaining a general solution to a &lt;a href=&quot;http://en.wikipedia.org/wiki/N-body_problem&quot; rel=&quot;nofollow&quot;&gt;$N$-body problem&lt;/a&gt; is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a proof somewhere that shows this impossibility?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I am looking to prove or disprove the below statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;there exists a power series that that&#xA;  solve this problem, for all the terms&#xA;  in the series and the summation of the&#xA;  series must converge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="605" LastEditorUserId="605" LastEditorDisplayName="" LastEditDate="2010-11-24T13:02:59.470" LastActivityDate="2011-08-09T04:46:05.577" Title="The Impossibility of Solving $N$-Body Problem" Tags="&lt;classical-mechanics&gt;" AnswerCount="2" CommentCount="10" FavoriteCount="2" />
  <row Id="1236" PostTypeId="1" AcceptedAnswerId="1237" CreationDate="2010-11-23T08:51:46.560" Score="4" ViewCount="103" Body="&lt;p&gt;Inert gases are chemically neutral. They don't react with anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, How were they discovered?&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-11-23T11:27:33.557" Title="How were inert gases discovered?" Tags="&lt;inert-gases&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="1237" PostTypeId="2" ParentId="1236" CreationDate="2010-11-23T09:55:26.000" Score="6" ViewCount="" Body="&lt;p&gt;First of all, this is not true that noble gases does not form any compounds -- it can be done with some chemical tricks, usually using fluorine and some hell conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, you don't need any chemistry to detect a new element -- helium was for instance first spotted in spectrum of sunligh. The isolation can also by made on physical means only; the most efficient idea is to cool down air isolating each new fraction that turns into liquid, but there are dozens other.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-23T09:55:26.000" CommentCount="1" />
  <row Id="1238" PostTypeId="2" ParentId="1230" CreationDate="2010-11-23T10:14:32.323" Score="3" ViewCount="" Body="&lt;p&gt;Basically, you are proposing to redefine the kilogram, and your approach has been proposed and recently (in october 2010) abandoned ( &lt;a href=&quot;http://en.wikipedia.org/wiki/Kilogram#Carbon-12&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Kilogram#Carbon-12&lt;/a&gt; ). &#xA;I think the reason why the Watt-balance approach has been preferred for the future definition of the kilogramme was mainly technological : it is more precise and would allow more practical realization of the kilogramme.&lt;/p&gt;&#xA;" OwnerUserId="373" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-11-23T10:20:51.367" LastActivityDate="2010-11-23T10:20:51.367" />
  <row Id="1239" PostTypeId="2" ParentId="1236" CreationDate="2010-11-23T11:27:33.557" Score="5" ViewCount="" Body="&lt;p&gt;The history is summed up in &lt;a href=&quot;http://en.wikipedia.org/wiki/Noble_gas#History&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Noble_gas#History&lt;/a&gt; . The concept of noble gas emerged from the discovery of argon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As said by mpq, the first to be seen spectroscopically was Helium. Then Argon was detected as a component of the air less reactive than nitrogen (http://en.wikipedia.org/wiki/Argon#History ).&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2010-11-23T11:27:33.557" />
  <row Id="1240" PostTypeId="2" ParentId="1230" CreationDate="2010-11-23T11:30:22.710" Score="12" ViewCount="" Body="&lt;p&gt;There was a proposal in 2006 trying to define N&lt;sub&gt;A&lt;/sub&gt; as an exact number[&lt;a href=&quot;http://arxiv.org/abs/physics/0612087&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;,&lt;a href=&quot;http://www.americanscientist.org/issues/pub/2007/2/an-exact-value-for-avogadros-number/3&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;]:&#xA;$$ N_A^* = 84\;446\;888^3 = 6.022\;141\;410\;704\;090\;840\;990\;72 \times 10^{23} $$&#xD;&#xA;the problem? This value is incorrect, as the currently most accurate result is[&lt;a href=&quot;http://arxiv.org/abs/1010.2317&quot; rel=&quot;nofollow&quot;&gt;3&lt;/a&gt;] &#xA;$$ N_A = 6.022\;140\;84(18) \times 10^{23} $$&#xD;&#xA;i.e. $N_A^*$ is now 3 s.d. away from $N_A$. As I have commented, if we randomly pick a number within the current error bound and call it $N_A$, we risk the problem that a better experiment for the old definition will invalidate the proposed value. To be safe about the validity of that number, you need to produce an equally accurate experiment to show that it is actually valid (like the 299792458 m/s in the definition of meter, and 9192631770 Hz in the definition of second.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the rationale for redefinition of SI base unit always involve that the current one isn't accurate enough or hard to realize: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.bipm.org/en/CGPM/db/13/1/&quot; rel=&quot;nofollow&quot;&gt;second (1967)&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the definition ... is inadequate for the present needs of metrology&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.bipm.org/en/CGPM/db/11/6/&quot; rel=&quot;nofollow&quot;&gt;meter (1960)&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the international Prototype does not define the metre with an accuracy adequate for the present needs of metrology, &lt;/li&gt;&#xA;&lt;li&gt;it is moreover desirable to adopt a natural and indestructible standard,&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.bipm.org/en/CGPM/db/17/1/&quot; rel=&quot;nofollow&quot;&gt;meter (1983)&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the present definition does not allow a sufficiently precise realization of the metre for all requirements&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.bipm.org/en/CGPM/db/16/3/&quot; rel=&quot;nofollow&quot;&gt;candela (1979)&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the time has come to give the candela a definition that will allow an improvement in both the ease of realization and the precision of photometric standards, ...&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Do the current experiments which reduces to N&lt;sub&gt;A&lt;/sub&gt; = 12 gram of carbon-12 atoms not accurate enough or hard to realize? I don't think so; 9 significant figures are already very accurate. However, the redefinition of mole would be &lt;a href=&quot;http://www.bipm.org/en/CGPM/db/23/12/&quot; rel=&quot;nofollow&quot;&gt;put on board on 2011&lt;/a&gt; (24th CGPM). One proposal is to define[&lt;a href=&quot;http://iopscience.iop.org/0026-1394/43/3/006/&quot; rel=&quot;nofollow&quot;&gt;4&lt;/a&gt;]&#xA;$$ N_A \overset{\underset{\mathrm{def}}{}}{=} 6.022\;141\;5 \times 10^{23} \mathrm{mol}^{-1}, $$ to decouple kilogram from the definition of mole. So if this path is taken, the only thing that keep us from defining it as a definite number &lt;em&gt;to 10 significant figures&lt;/em&gt; is because &quot;the conference haven't started yet&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But infinite precision? That would be a long way before we can reach and need that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ref:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Ronald Fox and Theodore Hill, &lt;em&gt;A Proposed Exact Integer Value for Avogadro's Number&lt;/em&gt;. &lt;a href=&quot;http://arxiv.org/abs/physics/0612087&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/physics/0612087&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Ronald Fox and Theodore Hill, &lt;em&gt;An Exact Value for Avogadro's Number&lt;/em&gt;. &lt;a href=&quot;http://www.americanscientist.org/issues/pub/2007/2/an-exact-value-for-avogadros-number/3&quot; rel=&quot;nofollow&quot;&gt;http://www.americanscientist.org/issues/pub/2007/2/an-exact-value-for-avogadros-number/3&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;B. Andreas, Y. Azuma, G. Bartl, et. al., &lt;em&gt;An accurate determination of the Avogadro constant by counting the atoms in a 28Si crystal&lt;/em&gt; &lt;a href=&quot;http://arxiv.org/abs/1010.2317&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1010.2317&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Ian M Mills, Peter J Mohr, Terry J Quinn, et. al., &lt;em&gt;Redefinition of the kilogram, ampere, kelvin and mole: a proposed approach to implementing CIPM recommendation 1 (CI-2005)&lt;/em&gt;. &lt;a href=&quot;http://iopscience.iop.org/0026-1394/43/3/006&quot; rel=&quot;nofollow&quot;&gt;http://iopscience.iop.org/0026-1394/43/3/006&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-11-23T11:30:22.710" />
  <row Id="1241" PostTypeId="1" CreationDate="2010-11-23T12:58:03.367" Score="0" ViewCount="152" Body="&lt;p&gt;I have a set of samples that represents a waveform. This waveform resembles a frequency modulated sinusoidal wave (only it is not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to invert this waveform or &lt;strike&gt;shift it by $2\pi$&lt;/strike&gt; shift it by $\pi$. of course taking the cosine of samples as they are without preprocessing is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What should I do to achieve this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="568" LastEditorUserId="568" LastEditorDisplayName="" LastEditDate="2010-11-23T13:37:10.963" LastActivityDate="2011-03-15T04:20:11.113" Title="How to get the cosine of a waveform?" Tags="&lt;electricity&gt;&lt;waves&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="1242" PostTypeId="2" ParentId="1235" CreationDate="2010-11-23T13:12:37.387" Score="21" ViewCount="" Body="&lt;p&gt;While the N-body Problem is chaotic, a convergent  expansion exists.&#xA;The 3-Body expansion was found by Sundman in 1912, and the full N-body problem in 1991 by wang. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, These expansions are pretty much useless for real problems( millions of terms are required for even short times); you're much better off with a numerical integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The history of the 3-Body problem is in itself pretty interesting stuff. Check out June Barrow-Green's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821803670&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; which include a pretty good analysis of all the relevant physics, along with a ripping tale.&lt;/p&gt;&#xA;" OwnerUserId="606" LastEditorUserId="606" LastEditorDisplayName="" LastEditDate="2010-11-23T13:21:18.320" LastActivityDate="2010-11-23T13:21:18.320" CommentCount="10" />
  <row Id="1243" PostTypeId="1" CreationDate="2010-11-23T13:29:42.090" Score="8" ViewCount="160" Body="&lt;p&gt;The ground state for  a quantum Hall system on a torus with fractional filling factor can be classified by the Chern number, which is why the Hall conductance is quantized. Is there another method or classification one can use to distinguish states?&lt;/p&gt;&#xA;" OwnerUserId="606" LastEditorUserId="42" LastEditorDisplayName="" LastEditDate="2010-11-23T15:42:22.617" LastActivityDate="2010-11-24T19:36:55.083" Title="Is there a method for differentiating fractional quantum Hall states aside from finding Chern numbers?" Tags="&lt;quantum-mechanics&gt;&lt;solid-state-physics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="1244" PostTypeId="2" ParentId="1241" CreationDate="2010-11-23T14:00:28.760" Score="1" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;For each sample I calculated the angle $\theta = i * 2\pi$&lt;/li&gt;&#xA;&lt;li&gt;Then I added $\pi$ to $\theta$ while calculating sine and cosine components.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="568" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-15T04:20:11.113" LastActivityDate="2011-03-15T04:20:11.113" CommentCount="1" />
  <row Id="1245" PostTypeId="2" ParentId="1241" CreationDate="2010-11-23T14:59:18.023" Score="2" ViewCount="" Body="&lt;p&gt;Based on what you write in the comments, perhaps you can just calculate the DC component by taking the average, subtract that, flip it over the X axis, and add the DC component back.&lt;/p&gt;&#xA;" OwnerUserId="47" LastActivityDate="2010-11-23T14:59:18.023" CommentCount="3" />
  <row Id="1246" PostTypeId="2" ParentId="1151" CreationDate="2010-11-23T17:04:17.417" Score="0" ViewCount="" Body="&lt;p&gt;Here is one case in which a bubble will move against the Coriolis force, although it's not exactly the same case as addressed in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.inwit.com/inwit/writings/coriolisforce.html&quot; rel=&quot;nofollow&quot;&gt;This reference&lt;/a&gt; claims that Coriolis forces used to come into play in ship navigation by deflecting the bubble in a level, but doesn't say which way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the bubble in a level.  If you are in the northern hemisphere and are moving further north, the Coriolis force is to the west.  The Coriolis force acts on the air in the level's bubble, but also on the water in the level.  The force is proportional to mass, so by the equivalence principle we can think of it as a small gravitational force pointing west.  A bubble in water moves opposite gravity, so the bubble moves east, against the Coriolis force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between this scenario and a bubble simply floating up in water is that in the level, the bubble and water move in the same direction, but not so for a bubble floating up.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-23T17:04:17.417" CommentCount="2" />
  <row Id="1247" PostTypeId="1" CreationDate="2010-11-23T17:15:46.630" Score="2" ViewCount="256" Body="&lt;p&gt;It's comparatively easy (cum grano salis) to grasp the following concepts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Euclidean space-time &lt;em&gt;(continous space and continuous time)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;classical mechanics &lt;em&gt;(discretely distributed matter in continous space and continuous time)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Minkowskian space &lt;em&gt;(continously intermingled space and time)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;special relativistic mechanics &lt;em&gt;(discretely distributed matter in continously intermingled space and time)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;classical electrodynamics&lt;/li&gt;&#xA;&lt;li&gt;classical quantum mechanics &lt;em&gt;(discrete energies, continuously distributed matter in continous space and continuous time)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;quantum electrodynamics&lt;/li&gt;&#xA;&lt;li&gt;general relativity &lt;em&gt;(continously intermingled space-time and matter)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Accordingly, there are lots of introductory texts and text-books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also easy to grasp&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;numerical simulations &lt;em&gt;(on artificially - and mostly unphysically - discretized spaces, times, and space-times)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;cellular automata &lt;em&gt;(on unphysical regular spatial grids)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It's definitely hard to grasp (for somehow graspable reasons)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;quantum gravity&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I do not know whether there are empirical evidences for a discrete space-time or only theoretical desiderata, anyhow I cannot figure a discrete space and/or time out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it so hard to introduce and explain the concept of a physical discrete space-time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are there no easy to understand&#xA;introductory texts or text-books on definitions,&#xA;concepts, models, pros and cons of&#xA;discrete space, time, and - finally -&#xA;space-time? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Respectively: &lt;em&gt;Where are they?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the reasons for this maybe related to the reasons why quantum gravity is so hard to grasp?&lt;/p&gt;&#xA;" OwnerUserId="229" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-23T21:50:05.543" LastActivityDate="2010-11-23T21:50:05.543" Title="Introductions to discrete space-time" Tags="&lt;soft-question&gt;&lt;quantum-gravity&gt;&lt;spacetime&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="1" />
  <row Id="1248" PostTypeId="2" ParentId="1229" CreationDate="2010-11-23T19:17:52.180" Score="1" ViewCount="" Body="&lt;p&gt;I read in a book recently a theoretical conversation between an alien and and an earthling where the earthling was trying to explain which hand is the &lt;em&gt;right&lt;/em&gt; hand (without any visual contact) and which the &lt;em&gt;left&lt;/em&gt;. The alien would have to perform experiments to determine a reference left or right. It turns out after a whole chapter showing how there isn't anything in nature inheritly differentiating between left and right, there was an experiment involving beta decay and the spin of neutrinos that came out. I don't remeber the details, but apparently deep in the bowels of the standard model in the universe there is sense of left and right, at least when it comes to subatomic particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if I could find this book again in my library ... :-)&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-11-23T19:17:52.180" CommentCount="1" />
  <row Id="1249" PostTypeId="2" ParentId="1205" CreationDate="2010-11-23T19:41:09.510" Score="1" ViewCount="" Body="&lt;p&gt;This is a similar problem to the karman vortices developing around cables/pipes when wind blows around them. When each vortex sheds and releases from the boundary layer there is a reactive aerodynamic &quot;lift&quot; developed which might move the structure. When the structure returns it forces the next vortex to shed aplifying the effect. It is like pushing a pendulum. If you push on the right time, even a tiny force can have a large effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Search cable galloping, aeolian vibrations and structural wind noise. Note that with cables the wind might excite the 160-th basic harmonic of the system, and not the first few as you might expect. Thats because the internal turbulence in wind resonates based on the Strouhal number is it might be of the order of 40 Hz-160 Hz.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-11-23T19:41:09.510" CommentCount="2" />
  <row Id="1250" PostTypeId="1" AcceptedAnswerId="1253" CreationDate="2010-11-23T19:52:22.680" Score="3" ViewCount="155" Body="&lt;p&gt;If B is magnetic field and E electric Field, B=rotA, E= -gradV+dA/dt.There is Gauge's invariance, for the trasformation A'-&gt;A+gradL/dt, V'-&gt;V-dL/dt. Now, we can write:&#xA;-Coulomb Gauge : the choice of a L that implies rotA=0&#xA;-Lorenz Gauge : the choice of a L that implies gradA+1/c^2 dV/dt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm trying to demonstrate that MATHEMATICALLY it's always possible to find a L that satisfies CG or LG.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would someone help me :-) ?   &lt;/p&gt;&#xA;" OwnerUserId="558" LastActivityDate="2010-11-23T20:42:11.250" Title="Are gauge choices in electrodynamics really always possible?" Tags="&lt;mathematics&gt;&lt;electrodynamics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" />
  <row Id="1251" PostTypeId="1" CreationDate="2010-11-23T20:18:34.687" Score="2" ViewCount="262" Body="&lt;p&gt;This is a follow-up question to &lt;a href=&quot;http://physics.stackexchange.com/questions/1247/introductions-to-discrete-space-time&quot;&gt;Introductions to discrete space-time&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this line of thought hopeless?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Classical mechanics can be understood&#xA;  as the limit of relativistic mechanics&#xA;  $RM_c$ for $c \rightarrow \infty$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Classical mechanics can be understood&#xA;  as the limit of quantum mechanics&#xA;  $QM_h$ for $h \rightarrow 0$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As a limit of which discrete geometry&#xA;  $\Gamma_\lambda$ can classical&#xA;  mechanics be understood &#xA;  for $\lambda \rightarrow 0$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="229" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-23T21:22:40.867" LastActivityDate="2010-11-25T23:20:01.307" Title="Why is it hopeless to view differential geometry as the limit of a discrete geometry?" Tags="&lt;quantum-mechanics&gt;&lt;classical-mechanics&gt;&lt;special-relativity&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="1252" PostTypeId="2" ParentId="1250" CreationDate="2010-11-23T20:31:27.470" Score="0" ViewCount="" Body="&lt;p&gt;I'm not sure i understood your question. Look, all E&amp;amp;M Lagrangians have gauge freedom &quot;built in&quot;, in the sense that you can re-write the $E$ and $B$ fields and the Lagrangian won't change. Therefore, it's always possible for you to make a choice of gauge, you always have that freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a slightly tangential note, remember what &lt;a href=&quot;http://en.wikipedia.org/wiki/Helmholtz_decomposition&quot; rel=&quot;nofollow&quot;&gt;Helmholtz theorem&lt;/a&gt; has to say: you can always decompose a &quot;well behaved&quot; vector field into a sum of a &lt;em&gt;curl&lt;/em&gt; and a &lt;em&gt;grad&lt;/em&gt; part. And this is exactly what you're doing to the $E$ and $B$ fields in Maxwell's eqs. So, now your question becomes: what happens when you apply your gauge transformation in this context? That is, what is $\nabla\times\nabla$ and $\nabla\cdot\nabla$? What does this imply for $A$ (what equation you get for the vector-potential)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should get you going...&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2010-11-23T20:31:27.470" CommentCount="1" />
  <row Id="1253" PostTypeId="2" ParentId="1250" CreationDate="2010-11-23T20:34:42.597" Score="3" ViewCount="" Body="&lt;p&gt;The correct &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauge_fixing#Gauge_freedom&quot; rel=&quot;nofollow&quot;&gt;Gauge transformation formula&lt;/a&gt; should be&#xA;$$\begin{aligned} \mathbf A &amp;amp;\mapsto \mathbf A + \nabla \lambda \\&#xD;&#xA;\mathbf V &amp;amp;\mapsto V - \frac{\partial\lambda}{\partial t}, \end{aligned} $$&#xD;&#xA;not something with &quot;gradL/dt&quot;. The Coulomb gauge requires $\nabla\cdot\mathbf A=0$, not &quot;rotA = 0&quot;. The Lorenz gauge requires $\nabla\cdot\mathbf A + \frac1{c^2}\frac{\partial V}{\partial t}=0$, not &quot;gradA+1/c^2 dV/dt&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Coulomb gauge can be chosen by solving the Poisson equation&#xA;$$ \nabla^2 \lambda = -\nabla\cdot\mathbf A$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lorenz gauge can be chosen by solving the inhomogeneous wave equation&#xA;$$ \nabla^2 \lambda - \frac1{c^2}\frac{\partial^2\lambda}{\partial t^2} = -\nabla\cdot\mathbf A+\frac1{c^2}\frac{\partial V}{\partial t}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Substitute the transformed potentials into the conditions to get the PDEs)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Existence of solutions of these PDEs are guaranteed as long as the source terms (stuff on the RHS) are &quot;well-behaved&quot; (e.g. $\nabla\cdot\mathbf A$ should grow slower than $1/r$ in the Poisson equation)&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-11-23T20:42:11.250" LastActivityDate="2010-11-23T20:42:11.250" CommentCount="1" />
  <row Id="1254" PostTypeId="2" ParentId="1251" CreationDate="2010-11-23T20:34:48.993" Score="0" ViewCount="" Body="&lt;p&gt;To begin with, what &lt;em&gt;is&lt;/em&gt; $\lambda$? And why would this particular limit be the relevant one (and, say, not $\lambda\rightarrow\infty$)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, this notion that $\hbar\rightarrow 0$ recovers Classical Mechanics (from Quantum Mechanics) is not quite correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, i think this question is a bit too vague...&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2010-11-23T20:34:48.993" CommentCount="14" />
  <row Id="1255" PostTypeId="2" ParentId="1247" CreationDate="2010-11-23T20:44:51.717" Score="5" ViewCount="" Body="&lt;p&gt;I think you're looking for something like &lt;a href=&quot;http://en.wikipedia.org/wiki/Regge_calculus&quot; rel=&quot;nofollow&quot;&gt;Regge Calculus&lt;/a&gt; (there are plenty of extra refs in this link, and many more in &lt;a href=&quot;http://scienceworld.wolfram.com/physics/ReggeCalculus.html&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also check this article, &lt;a href=&quot;http://arxiv.org/abs/gr-qc/9701052&quot; rel=&quot;nofollow&quot;&gt;Quantum Gravity and Regge Calculus&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, &lt;a href=&quot;http://ccom.ucsd.edu/~astern/&quot; rel=&quot;nofollow&quot;&gt;this guy&lt;/a&gt; has a whole research area in this topic (discrete spaces, etc)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there's &lt;strong&gt;plenty&lt;/strong&gt; of stuff out there... you just have to look. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;Edited&lt;/strong&gt;) PS: In fact, here's the name of the game: &lt;a href=&quot;http://en.wikipedia.org/wiki/Discrete_differential_geometry&quot; rel=&quot;nofollow&quot;&gt;Discrete Differential Geometry&lt;/a&gt;. Google away...&lt;/p&gt;&#xA;" OwnerUserId="142" LastEditorUserId="142" LastEditorDisplayName="" LastEditDate="2010-11-23T21:02:04.103" LastActivityDate="2010-11-23T21:02:04.103" CommentCount="2" />
  <row Id="1256" PostTypeId="2" ParentId="1217" CreationDate="2010-11-23T22:18:36.503" Score="3" ViewCount="" Body="&lt;p&gt;Some remarks on theoretical issues involved in inclusive/exclusive processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, in the inclusive vs. exclusive vs. semi-inclusive classification, we care only about final &lt;em&gt;hadrons&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, exclusive processes are much more difficult to calculate than inclusive ones. When you calculate the cross section of an inclusive process, it is usually sufficient to calculate it at the quark/gluon level. You know that these quarks and gluons will eventually hadronize into some hadrons, but in this calculation you don't care how exactly this will happen. So, we effectively integrate over the full final phase space of hadrons by setting it equal to the phase space of quarks/gluons, which you know how to calculate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you move from fully inclusive to less inclusive processes (jet production, semi-inclusive processes, exclusive processes), you are trying to gain some insight into the structure of the hadronic final state. And the more details you want to see, the more difficult is the calculation. It is not sufficient anymore to calculate production of quarks/gluons. You now need to specify their probability to hadronize into a given hadron or set of hadrons (e.g. via fragmentation functions). You need to consider differential distributions w.r.t. more kinematic variables than before. You often need generalization of partonic distributions, but factorization theorems become much harder to prove for less exclusive processes. All this makes exclusive processes difficult to calculate, at least at high energies, when quark and gluon degrees of freedom are fully important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same time, exclusive processes often give you a deeper insight into the structure of hadrons than the fully inclusive ones. For example, inclusive deep-inelastic scattering (DIS) is sensitive only to very few partonic densities, even when polarization degree of freedom is taken into account. Semi-inclusive and exclusive processes in DIS (such as SIDIS, various diffractive processes, DVCS = deeply virtual compton scattering) are sensitive to many new partonic distributions, which reflect more subtler aspects of the proton's structure. That's why they are studied despite being so difficult.&lt;/p&gt;&#xA;" OwnerUserId="59" LastActivityDate="2010-11-23T22:18:36.503" />
  <row Id="1257" PostTypeId="1" AcceptedAnswerId="1258" CreationDate="2010-11-24T00:04:57.423" Score="22" ViewCount="920" Body="&lt;p&gt;What is the reason? &#xA;Is it caused by their narrow shape, the soft material, walking vibration or something else? &lt;/p&gt;&#xA;" OwnerUserId="518" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-11-24T11:46:05.200" LastActivityDate="2010-11-28T19:38:19.853" Title="Why do earphone wires always get tangled up in pocket?" Tags="&lt;experimental-physics&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="7" />
  <row Id="1258" PostTypeId="2" ParentId="1257" CreationDate="2010-11-24T00:24:57.310" Score="26" ViewCount="" Body="&lt;p&gt;I have to go, but I will leave you with this for now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In March 2009, a group at UCSD published a paper titled &lt;a href=&quot;http://physics.ucsd.edu/~des/DESPublications.htm&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;&quot;Spontaneous knotting of an agitated string&quot;&lt;/strong&gt;&lt;/a&gt;. It is under reference number 30 on Doug Smith's site (linked). Below the paper are links to newspaper articles that discuss it and surely provide a summary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a paper titled &lt;strong&gt;&quot;The Biophysics of Knotting&quot;&lt;/strong&gt; (reference #37), but that is about biophysical examples of knotting (DNA, umbilicus, etc.).&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2010-11-24T00:24:57.310" CommentCount="9" />
  <row Id="1259" PostTypeId="2" ParentId="1220" CreationDate="2010-11-24T00:29:58.537" Score="0" ViewCount="" Body="&lt;p&gt;The &quot;tensor&quot; is simply a 3by3 matrix, which when multiplies by the normal vector gives the force. And you can argue that in a box with a uniform (stress) tensor has to have a finite rotational acceleration, so if you take the limit&#xA;shrinking the box to size zero, you discover the stress tensor must be a symmetric matrix.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2010-11-24T00:29:58.537" CommentCount="1" />
  <row Id="1260" PostTypeId="2" ParentId="1251" CreationDate="2010-11-24T02:10:41.437" Score="2" ViewCount="" Body="&lt;p&gt;Not exactly sure what you are asking. For a cubic lattice, the limit as the lattice spacing goes to 0 recovers all of classical physics, and, if you want to discretize time, then as $\Delta{}t$ approaches 0, then finite difference equations become differentials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, Low energy sound waves in a crystal don't see the discreteness, and it can be treated as a continuous medium.     &lt;/p&gt;&#xA;" OwnerUserId="606" LastActivityDate="2010-11-24T02:10:41.437" />
  <row Id="1261" PostTypeId="2" ParentId="1220" CreationDate="2010-11-24T05:11:33.683" Score="1" ViewCount="" Body="&lt;p&gt;Free Body Diagram (&lt;a href=&quot;http://en.wikipedia.org/wiki/Free_body_diagram&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Free_body_diagram&lt;/a&gt;) is your friend. Cut the rope at any point and replace the missing piece with the internal tension force and see how and it changes along the cable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider gravity then the rope forms a catenary shape and the tensions is minimum at the sag point (middle) and peak at the ends.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-11-24T05:11:33.683" />
  <row Id="1263" PostTypeId="1" AcceptedAnswerId="1501" CreationDate="2010-11-24T10:48:54.307" Score="9" ViewCount="537" Body="&lt;p&gt;I saw this &lt;a href=&quot;http://www.ted.com/talks/woody_norris_invents_amazing_things.html&quot; rel=&quot;nofollow&quot;&gt;TED talk&lt;/a&gt; and I am curious as to how the sound is focused on the general level. Can anyone explain this or does anyone have any good articles?&lt;/p&gt;&#xA;" OwnerUserId="119" LastActivityDate="2010-12-08T15:32:41.640" Title="How can you focus sound?" Tags="&lt;sound&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="1264" PostTypeId="2" ParentId="1263" CreationDate="2010-11-24T11:04:08.863" Score="2" ViewCount="" Body="&lt;p&gt;On the general level, you focus sound the same way you focus light -- either by reflecting it from a parabolic surface, or letting it pass through an acoustic lens. An acoustic lens is just like an optical lens in that it consists of a material with a different propagation speed of sound, with varying thickness. See the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Acoustic_mirror&quot; rel=&quot;nofollow&quot;&gt;acoustic mirrors&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2010-11-24T11:04:08.863" CommentCount="2" />
  <row Id="1265" PostTypeId="2" ParentId="1263" CreationDate="2010-11-24T12:42:11.313" Score="2" ViewCount="" Body="&lt;p&gt;Sound is a type of wave, so it has all the wave properties similar to other wave such as light wave. For light wave, you can use a len to focus the light. A len has higher refractive index, or lower light speed than the environment. The same is true for sound wave, so what you need is to make a high refractive region [1].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The air surrounding us can be approximated by the ideal gas, so the speed of sound is [2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c=\sqrt{\gamma\frac{P}{\rho}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&#xA;$\gamma$ is the adiabatic index,&#xA;$p$ is pressure of the air,&#xA;$\rho$ is density of the air&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, we want to create a region with high refractive, or equivalently low sound speed. There are few way to achieve this, one is to decrease the pressure, another way is to decrease the temperature (by the ideal gas law $PV=NRT$). However, in both cases, you either need a hard container or a refrigerator near it to keep it cold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, increasing the density can be easily done by using a heavy gas such as carbon dioxide. You just need to fill the gas in a balloon and it can act as a very simple acoustic len. Note that the size of the balloon or other container must be large compared with the wavelength. There are also other methods to focus sound without using lens. [3]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As said before, the same mechanism can be applied for other wave, for example, a water wave. In a shallow water tank, adding a len shape obstacle at the bottom can converge water wave because water wave move slowly at the shallow region. This experiment can be easily preformed in oneself's home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/sound/refrac.html&quot; rel=&quot;nofollow&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/sound/refrac.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] &lt;a href=&quot;http://en.wikipedia.org/wiki/Speed_of_sound#Speed_in_ideal_gases_and_in_air&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Speed_of_sound#Speed_in_ideal_gases_and_in_air&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3] &lt;a href=&quot;http://focus.aps.org/story/v14/st3&quot; rel=&quot;nofollow&quot;&gt;http://focus.aps.org/story/v14/st3&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="562" LastEditorUserId="562" LastEditorDisplayName="" LastEditDate="2010-11-24T12:56:33.323" LastActivityDate="2010-11-24T12:56:33.323" CommentCount="1" />
  <row Id="1266" PostTypeId="1" AcceptedAnswerId="1272" CreationDate="2010-11-24T12:44:16.477" Score="8" ViewCount="266" Body="&lt;p&gt;Question shortly: &#xA;How far would a hockey puck slide in two different cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The puck is sliding (translation) on ice and spinning on its flat surface.  &lt;/li&gt;&#xA;&lt;li&gt;The puck is sliding on ice without spinning. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Other conditions are the same in both cases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplifications: no air resistance, the coefficient of friction is const while sliding(spinning)&lt;br&gt;&#xA;Let's formalize the problem:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a disc with diameter D = 8 cm, mass m=170 g, the coefficient of friction on ice let be  $\mu=0.02$  &#xA;Initial values:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$v_0=10 \frac{m}{s}$ and $\omega_0=100 \frac{rad}{s}$  &lt;/li&gt;&#xA;&lt;li&gt;$v_0=10 \frac{m}{s}$ and $\omega_0=0 \frac{rad}{s}$  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Obviously the part 2) is trivial. I am working currently on part 1). This seems to be very difficult. Instead of a solid disc i took a thin ring at first. But even in that case the solution seems to be an integro-differential equation. If you have some ideas let us know.&lt;/p&gt;&#xA;" OwnerUserId="552" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-11-24T16:43:28.000" LastActivityDate="2010-11-24T23:28:19.377" Title="A problem inspired by the ice hockey game" Tags="&lt;mechanics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="1267" PostTypeId="1" AcceptedAnswerId="1271" CreationDate="2010-11-24T12:49:55.893" Score="2" ViewCount="233" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/312/particle-physics-getting-started&quot;&gt;Particle physics getting started&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I'm looking for a book that introduces the building blocks refer to the standard model for a course in nuclear physics. There are good books on this?&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2010-11-25T09:44:55.313" ClosedDate="2011-07-11T18:26:12.947" Title="Book for particles physics" Tags="&lt;books&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="1268" PostTypeId="1" CreationDate="2010-11-24T12:59:19.030" Score="1" ViewCount="887" Body="&lt;p&gt;In his book &quot;Einstein's mistakes&quot; H. C. Ohanian, the author, holds that Einstein delivered 7 proofs for $E=mc^2$ in his life that all were in some way incorrect. This despite the fact that correct proves had been published and mistakes in his proofs were sometimes pointed out to him.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first proof e.g. contains a circular line of thought in that it falsely assumes special relativity to be compatible with rigid bodies.&lt;br&gt;&#xA;EDIT: Reference: stated by V. Icke, prof. theoretical astronomy at Leiden University in his (Dutch) book 'Niks relatief': &quot;Einstein made an error in his calculation of 1906&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you think Ohanian's statement is true or was/is he biased in his opinion?&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-13T07:31:20.513" LastActivityDate="2011-08-15T03:47:59.630" Title="Did Einstein prove $E=mc^2$ correctly?" Tags="&lt;special-relativity&gt;&lt;history&gt;" AnswerCount="6" CommentCount="7" FavoriteCount="2" />
  <row Id="1269" PostTypeId="2" ParentId="1268" CreationDate="2010-11-24T13:22:55.477" Score="12" ViewCount="" Body="&lt;p&gt;I will exaggerate a bit, but in physics, proof in the sense of mathematical proof is irrelevant. Even if all of Einstein's deductions of the formula were wrong, it still turns out that empirical evidence supports $E=mc^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, without the exaggeration, mathematical deduction is important in physical theories because it shows us how conclusions and principles hang together. This can be important when elaborating further theories. Imagine, for the sake of argument, it turns out that the relativity principle is not correct, yet $E=mc^2$. Since usually we deduce the latter from the former, there is something interesting here, it would mean that $E=mc^2$ is more fundamental than the principle of relativity. Special relativity arises itself from this kind of considerations. Realizing that the invariance of the speed of light is more fundamental than the invariance of time, the latter being only approximately true at very low speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I still wasn't able to read what Ohanian is saying in particular, but it is no secret that Einstein was not a great mathematician. For instance, if it was not for the help of his friend Marcel Grossmann, Einstein might never have been able to develop the theory of general relativity. From his intuition about the equivalence principle in 1905 to the actual GR in 1915, he had to toil for 10 years with non-Euclidean geometry. &lt;strike&gt;In the meantime, he nearly got overtaken by David Hilbert.&lt;/strike&gt; (See Marek's comment.) &lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2010-11-26T11:41:23.387" LastActivityDate="2010-11-26T11:41:23.387" CommentCount="18" />
  <row Id="1270" PostTypeId="2" ParentId="1263" CreationDate="2010-11-24T13:30:11.917" Score="3" ViewCount="" Body="&lt;p&gt;These ‘audio spotlights’ work by emitting ultrasound at two different frequencies; it is the short wavelength of the ultrasound that causes the beam to be so directed. The two waves interfere and produce sum and different tones at frequencies of $f_1+f_2$ and $f_1-f_2$; if the ultrasound frequencies are, say, $f_1=45\,$kHz and $f_2=44\,$kHz, the difference tone will be at $1$kHz which is in the audible range for humans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I knew some people who looked at using audio spotlights for noise control some time ago, but (IIRC) the general consensus is that since you're subjecting your victims with very large amounts of ultrasonic noise (greater than 100dB), these devices are probably not too safe for continuous use. (Or, at least, their safety was in no way assured.)&lt;/p&gt;&#xA;" OwnerUserId="420" LastActivityDate="2010-11-24T13:30:11.917" />
  <row Id="1271" PostTypeId="2" ParentId="1267" CreationDate="2010-11-24T13:49:52.143" Score="2" ViewCount="" Body="&lt;p&gt;I'd suggest &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471603864&quot; rel=&quot;nofollow&quot;&gt;Griffiths&lt;/a&gt; and/or &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201624605&quot; rel=&quot;nofollow&quot;&gt;Kane&lt;/a&gt;. Read the reviews to see whether this is the kind of books you are after.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a pretty nice (if quite popular) Flip Tanedo's &lt;a href=&quot;http://blogs.uslhc.us/lets-draw-feynman-diagams&quot; rel=&quot;nofollow&quot;&gt;blog series&lt;/a&gt; on the topic. If you like that, I also recommend the rest of his posts $-$ they're pretty cool.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-11-24T13:49:52.143" />
  <row Id="1272" PostTypeId="2" ParentId="1266" CreationDate="2010-11-24T14:06:09.447" Score="14" ViewCount="" Body="&lt;p&gt;Your problem is deeper than it might seem at the first glance. Apart from numbers, it involves a remarkable phenomenon, which was studied in detail only a few years ago in the paper &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v90/i24/e248302&quot; rel=&quot;nofollow&quot;&gt;Phys. Rev. Lett. 90, 248302 (2003)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the puck is an infinitely thin disc, and it is sent across the ice sliding and rotating. What will stop first, translational or rotational motion? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is they will stop &lt;strong&gt;simultaneously&lt;/strong&gt;, and this does not depend on the initial translational and angular velocities. The origin is the intrinsic friction-mediated coupling between rotation and sliding. If rotation is too fast for a given speed $v$, the translational friction will be very small. And vice versa, if rotation is too slow, the rotational deceleration is small. Thus, there exists a &quot;magic&quot; value of $\epsilon=v/R\omega$ towards which every initial condition is attracted. In the paper cited above this magic value was found to be approximately 0.653. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This value fully characterizes the asymptotic motion of the puck and allows you to calculate the friction force distribution, deceleration and finally the path (which is a straight line in this approximation). In fact, a part of that was already found in the paper cited above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this concerns only an infinitely thin disc, for which the pressure distribution is equal everywhere under the puck. With a thick disc you have more pressure under the front part of the puck, which additionally complicates the problem (still, see some discussion in the paper). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if you change the shape of the pack (as you in fact did by taking a ring instead of a disc), the magic value of $\epsilon$ also changes. See another paper that discusses this: &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v95/i26/e264303&quot; rel=&quot;nofollow&quot;&gt;Phys. Rev. Lett. 95, 264303 (2005)&lt;/a&gt;. See also &lt;a href=&quot;http://ajp.aapt.org/resource/1/ajpias/v74/i6/p554_s1&quot; rel=&quot;nofollow&quot;&gt;this comment&lt;/a&gt; to &lt;a href=&quot;http://ajp.aapt.org/resource/1/ajpias/v53/i12/p1149_s1&quot; rel=&quot;nofollow&quot;&gt;this old paper on puck motion on the ice&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="59" LastActivityDate="2010-11-24T14:06:09.447" CommentCount="1" />
  <row Id="1273" PostTypeId="1" CreationDate="2010-11-24T15:29:37.970" Score="1" ViewCount="252" Body="&lt;p&gt;So instead of assuming that the velocity $c$ is a maximal velocity, proving that while assuming $E=mc^2$.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-25T22:35:34.683" Title="Can special relativity be deduced from $E=mc^2$?" Tags="&lt;special-relativity&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="1274" PostTypeId="2" ParentId="1159" CreationDate="2010-11-24T15:47:54.407" Score="2" ViewCount="" Body="&lt;p&gt;While the answer to the question in the title is simply 'yes', I assume that your question is: 'how do individual electrons move in electrical circuits'?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/335/is-electricity-instantaneous/355#355&quot;&gt;My answer to this question&lt;/a&gt; should explain that - it uses a simple but correct model for individual electrons versus macroscopic current.&#xA;The relationship between macroscopic current and microscopic drift velocity is a linear relation. The same holds for the relationship between macroscopic resistance and microscopic resistance.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2010-11-24T15:47:54.407" CommentCount="1" />
  <row Id="1275" PostTypeId="2" ParentId="1273" CreationDate="2010-11-24T15:52:29.900" Score="3" ViewCount="" Body="&lt;p&gt;These two concepts are not related at all, so it's not possible to deduce anything either way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Special relativity in its bare form talks only about space-time, i.e. geometry. You don't have any energy or mass there without first somehow postulating what it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the reverse direction: you can't really deduce anything about SR from $E = mc^2$ because $E$ and $m$ could be anything (in particular you don't know whether $E$ is a component of a four-momentum $p$ or something completely different). And even if you knew what $E$ and $m$ was that would still give you no clue about what $c$ is (and that it should be constant in every inertial frame).&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-11-24T15:52:29.900" CommentCount="4" />
  <row Id="1276" PostTypeId="2" ParentId="1273" CreationDate="2010-11-24T16:24:08.637" Score="4" ViewCount="" Body="&lt;p&gt;Usually, the cornerstone of Special Relativity is the constancy of the speed-of-light: form this fact alone, much of Special Relativity can be &quot;deduced&quot;. (For a very interesting article about this point, see &lt;a href=&quot;http://dx.doi.org/10.1119/1.13917&quot; rel=&quot;nofollow&quot;&gt;Relativity without light&lt;/a&gt;, by David Mermin.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, this formula for energy is just an approximation of the fully relativistic expression, cf &lt;a href=&quot;http://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence&quot; rel=&quot;nofollow&quot;&gt;Mass-Energy Equivalence&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2010-11-24T16:24:08.637" CommentCount="13" />
  <row Id="1277" PostTypeId="2" ParentId="1268" CreationDate="2010-11-24T17:02:47.043" Score="0" ViewCount="" Body="&lt;p&gt;I remember reading Einstein's original paper on this, and it seemed to be argued pretty clearly. I believe he considers a scenario involving the emission and absorption of photons, and uses the length/time dilation factors to get an expression for energy, which hey then takes to the classical (Newtonian) limit and equates with $\frac{1}{2}mv^2$ to show the relation.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-24T20:30:02.793" LastActivityDate="2010-11-24T20:30:02.793" CommentCount="7" />
  <row Id="1278" PostTypeId="2" ParentId="1243" CreationDate="2010-11-24T19:36:55.083" Score="2" ViewCount="" Body="&lt;p&gt;In your title, you ask about &lt;strong&gt;fractional QHE&lt;/strong&gt; but the Chern number description as far as I know holds for &lt;strong&gt;integer QHE&lt;/strong&gt;.  I don't know that much about FQHE, but let me say a bit about IQHE which I understand a bit better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The famous paper of &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v49/i6/p405_1&quot; rel=&quot;nofollow&quot;&gt;TKNN&lt;/a&gt; established that the Hall conductivity at integer filling is proportional to a topological invariant associated to the band structure of 2 dimensional Hamiltonians.  This topological invariant, the Chern number, is an integer which tells us how the band structure &quot;twists&quot; over the Brillouin zone (this is the &quot;torus&quot; in your question) (more formally, the Chern number classifies the complex vector bundle associated to the band Hamiltonian).  Remember for now that this is a property of the band structure, which comes out of a description where electron-electron interactions are neglected, i.e. we are not dealing with a &quot;strongly-correlated system&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side comment: considering a sort of $Z_2$ equivariant (roughly, time-reversal invariant) version of this topological invariant led to the current extremely hot topic of &lt;a href=&quot;http://arxiv.org/abs/1002.3895&quot; rel=&quot;nofollow&quot;&gt;3D topological insulators&lt;/a&gt;, as initiated by &lt;a href=&quot;http://prb.aps.org/abstract/PRB/v76/i4/e045302&quot; rel=&quot;nofollow&quot;&gt;Fu and Kane&lt;/a&gt; (among others).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fractional QHE does not admit a single-particle description -- that is, you can't understand the properties from band theory, as in IQHE -- it is a phase of electron behavior which results from &lt;strong&gt;interactions&lt;/strong&gt;.  Thus, I don't think the Chern number description carries over easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The FQHE admits a &lt;strong&gt;Chern-Simons Landau-Ginzburg&lt;/strong&gt; description, which I've been reading a bit about in &lt;a href=&quot;http://so5.stanford.edu/Research/Projects/Zhang1992A.pdf&quot; rel=&quot;nofollow&quot;&gt;this 1992 review by Shou-Cheng Zhang&lt;/a&gt;.  The Chern-Simons term in this field theory should not be confused with Chern numbers!  (I'm not sure if that's what you're doing in the question, but I want to make that clear.)  The notions are related mathematically, but I believe the physics here is distinct.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were just asking about IQHE, the insight of TKNN that the IQHE states are classified by a topological invariant probably rules out other independent descriptions.  I might be confused as to your intent, but it seems unlikely that there could be a useful description of the IQHE states which doesn't use topology (topological invariant = stable to perturbations, which leads to the amazing plateaus, after all), and the topological situation is really pretty well understood at this point in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please let me know if anything is unclear or if I've said something wrong.  I too am just a learner in this field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I might come back and add some stuff about FQHE if I ever get around to understanding it better.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-24T19:36:55.083" CommentCount="4" />
  <row Id="1279" PostTypeId="1" AcceptedAnswerId="1282" CreationDate="2010-11-24T20:28:55.820" Score="5" ViewCount="240" Body="&lt;p&gt;I was always intrigued by the phenomena that govern the returning back  of a boomerang to the thrower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if it is dependent on various factors such as velocity, aerodynamics, wind resistance and so on I was wondering, there must be some concise explanation to it?&lt;/p&gt;&#xA;" OwnerUserId="619" LastEditorUserId="619" LastEditorDisplayName="" LastEditDate="2010-11-29T14:33:08.853" LastActivityDate="2010-11-29T14:33:08.853" Title="Why exactly does a boomerang return back to the thrower?" Tags="&lt;fluid-dynamics&gt;&lt;angular-momentum&gt;&lt;aerodynamics&gt;&lt;lift&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="1280" PostTypeId="2" ParentId="1230" CreationDate="2010-11-24T20:38:52.267" Score="0" ViewCount="" Body="&lt;p&gt;No one can prove with 100% certainty that there are will be no benefits. In every discipline acquiring a higher degree of precision is advancement. Unless there is an uncertainty principal for a more precise definition, then the idea of striving for a higher precision should be encouraged. The fact that no one at the moment knows how to achieve a higher precision makes the task a perfect Phd project! &lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2010-11-24T20:38:52.267" />
  <row Id="1282" PostTypeId="2" ParentId="1279" CreationDate="2010-11-24T20:45:43.113" Score="11" ViewCount="" Body="&lt;p&gt;Fundamentally, a boomerang has two arms that spin. One arm spins in the same direction of flight and the other spins away from the direction of flight. For this reason, there's a tilt force on the boomerang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/OPT17.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since the boomerang is spinning it has angular momentum. Therefore the tilt force generates precession which is pretty much what makes the boomerang come back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/uNxqC.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://plus.maths.org/issue7/features/boomerangs/index.html&quot; rel=&quot;nofollow&quot;&gt;Full explanation here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-24T20:45:43.113" CommentCount="1" />
  <row Id="1283" PostTypeId="2" ParentId="1268" CreationDate="2010-11-24T21:37:36.720" Score="3" ViewCount="" Body="&lt;p&gt;In special relativity, it's only accelerating bodies which are not allowed to be rigid. Non-accelerating bodies don't have any forces on them, so there is no obstacle to their retaining the same shape. I haven't checked, but I believe Einstein's intuitive derivation of relativity didn't involve any accelerating bodies.&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2010-11-24T21:37:36.720" CommentCount="5" />
  <row Id="1284" PostTypeId="2" ParentId="1223" CreationDate="2010-11-24T23:03:33.727" Score="1" ViewCount="" Body="&lt;p&gt;From the &lt;a href=&quot;http://en.wikipedia.org/wiki/Parameterized_post-Newtonian_formalism#Accuracy_from_experimental_tests&quot; rel=&quot;nofollow&quot;&gt;articles&lt;/a&gt; &lt;a href=&quot;http://relativity.livingreviews.org/Articles/lrr-2006-3/&quot; rel=&quot;nofollow&quot;&gt;linked&lt;/a&gt; in the comments:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Parameter  Bound        Effects                             Experiment&#xA;---------  -----        -------                             ----------&#xA;γ − 1      2.3 x 10^-5  Time delay, Light deflection        Cassini tracking&#xA;β − 1      2.3 x 10^-4  Nordtvedt effect, Perihelion shift  Nordtvedt effect&#xA;ξ          0.001        Earth tides                         Gravimeter data&#xA;α1         10^-4        Orbit polarization                  Lunar laser ranging&#xA;α2         4 x 10^-7    Spin precession                     Sun axis' alignment&#xA;                                                            with ecliptic&#xA;α3         4 x 10^-20   Self-acceleration                   Pulsar spin-down&#xA;                                                            statistics&#xA;ζ1         0.02         -                                   Combined PPN bounds&#xA;ζ2         4 x 10^-5    Binary pulsar acceleration          PSR 1913+16&#xA;ζ3         10^-8        Newton's 3rd law                    Lunar acceleration&#xA;ζ4         0.006        -                                   Kreuzer experiment&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where the parameters are so defined:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The parameters $\gamma$ and $\beta$ are the usual Eddington–Robertson–Schiff parameters used to describe the “classical” tests of GR, and are in some sense the most important; they are the only non-zero parameters in GR and scalar-tensor gravity. The parameter $\xi$ is non-zero in any theory of gravity that predicts preferred-location effects such as a galaxy-induced anisotropy in the local gravitational constant GL (also called “Whitehead” effects); $\alpha_1$, $\alpha_2$, $\alpha_3$ measure whether or not the theory predicts post-Newtonian preferred-frame effects; $\zeta_1$, $\zeta_2$, $\zeta_3$, $\zeta_4$, measure whether or not the theory predicts violations of global conservation laws for total momentum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And these are the GR-predicted values:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;γ = β = 1 and α1 = α2 = α3 = ζ1 = ζ2 = ζ3 = ζ4 = ξ = 0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-24T23:03:33.727" />
  <row Id="1285" PostTypeId="2" ParentId="1266" CreationDate="2010-11-24T23:28:19.377" Score="2" ViewCount="" Body="&lt;p&gt;This is a tricky question, one that I encountered in a similar situation a few years ago when working on a similar simulation (for a curling game, not hockey, but fundamentally similar).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few things jump out at you immediately about the problem; the other poster is correct that the translational and rotational motion indeed must stop simultaneously.  My approach to this was a bit more intuitive than his described answer though; the reason is simply that due to the motion of one or the other, the transition from kinetic to static friction is never made independently of the angular or translational velocity, but made simultaneously.  This is true of both ideal and more complex simulations.  (Note, however, that that does not mean that the angular velocity cannot become zero, particularly in situations where the coefficient of friction is not constant at all points on the traveling surface; however, it the angular velocity will not remain precisely zero while the translational velocity is non-zero.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the distance the &quot;puck&quot; will travel when rotating or not...  consider the situation where the translational velocity is zero, and the rotational velocity is non-zero.  Assuming constant kinetic (and static) friction at all contact points, the puck will not move; therefore, when translational energy is added to the system via a force vector, there is no disturbing force to perturb the original energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, this seems not to be precisely correct in the indication that the angular force will not affect the translational force, as the case of curling shows quite clearly; the &quot;rock&quot; (equivalent to the puck) will turn (or curl, hence the name of the game) in the direction of the rotation of the leading edge of the rock.  I don't have precise equations for this, but intuition indicates that the complete path traveled by a puck (rock) imparted with an angular velocity is the same distance as a rock (puck) without an angular velocity, but that the puck (rock) with the angular velocity travels a curved path (the total length of which is equal to the distance the unrotating rockpuck would travel in a straight line).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the reasons why a curling rock curls... I'm uncertain, but you've reminded me to ask a question about that.  :-)&lt;/p&gt;&#xA;" OwnerUserId="543" LastActivityDate="2010-11-24T23:28:19.377" CommentCount="1" />
  <row Id="1286" PostTypeId="1" CreationDate="2010-11-24T23:40:09.683" Score="7" ViewCount="280" Body="&lt;p&gt;In the game of curling, players &quot;curl&quot; a granite &quot;rock&quot; (of precise size and roughly a flattened cylinder) down a &quot;sheet&quot; of ice towards a target; the &quot;rock&quot; will curve in its path in the direction of the motion of the leading edge.  The question here is, what is the mechanism for the curling of the rock?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard frictional physics indicates that angular velocity should not have any effect on translational vector; while moving, the bottom surface of the rock is subject to kinetic friction.  Due to frictional deceleration, the leading edge of the rock exerts (slightly) more force on the ice than the trailing edge; this would seem to indicate that the rock would curve in the opposite direction from what it does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuition would indicate that the reason for the curling motion would be due to kinetic friction being modified by velocity; that is, along the &quot;left&quot; side of the rock (assuming a rock moving away from the observer) for a rock rotating counterclockwise, the velocity along the ground is slower than for the &quot;right&quot; side; if kinetic friction is decreased as velocity increases, this would be a good explanation for the motion.  However, everything I've seen seems to indicate that the coefficient of kinetic friction is constant; is this correct?  If so, what is the reason for the (well-observed) phenomenon?&lt;/p&gt;&#xA;" OwnerUserId="543" LastActivityDate="2011-04-04T14:09:45.343" Title="Why does a curling rock curl?" Tags="&lt;angular-momentum&gt;&lt;friction&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="1287" PostTypeId="2" ParentId="1222" CreationDate="2010-11-25T00:21:34.703" Score="3" ViewCount="" Body="&lt;p&gt;The point is that the configuration that you are describing as the time-reversed Oppenheimer-Snyder collapse would need very specific initial conditions to occur. A collapse process on the other hand is very generic. You can have a collapse with any number of initial configurations. In that sense a collapse process is more physical than the reverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is like dropping a needle and having it land on its tip. There are many ways to drop it and have it lying down on its side, but there is only one that you can drop it and have it land and stay balanced on its tip. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a mathematical possibility but a physical improbability.&lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2010-11-25T00:21:34.703" CommentCount="19" />
  <row Id="1288" PostTypeId="2" ParentId="1213" CreationDate="2010-11-25T04:28:04.027" Score="5" ViewCount="" Body="&lt;p&gt;There are a few different ways to establish this correspondence, eg, using Lie groups or Fourier Transform. But, in the end of the day, the notion that this takes one from Classical to Quantum Mechanics is nothing but an 'axiom'. So, in this sense, it's a bit weird to use the word &quot;derive&quot; the principle: this 'principle' is used as one of the &lt;em&gt;axioms&lt;/em&gt; that &lt;em&gt;defines&lt;/em&gt; Quantum Mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, von Neumann was probably the one to first formalize this construction in terms of what he called &quot;Transformation Theory&quot;, which is the theory of &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform#Functional_relationships&quot; rel=&quot;nofollow&quot;&gt;Fourier Transforms&lt;/a&gt; for distributions (generalized functions).&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2010-11-25T04:28:04.027" />
  <row Id="1289" PostTypeId="1" AcceptedAnswerId="1291" CreationDate="2010-11-25T06:34:55.373" Score="6" ViewCount="373" Body="&lt;p&gt;I have had a question since childhood.  Why do we always get circular waves (ripples) in water even when we throw irregularly shaped object in it? &lt;/p&gt;&#xA;" OwnerUserId="309" LastEditorUserId="309" LastEditorDisplayName="" LastEditDate="2010-11-25T07:56:45.287" LastActivityDate="2010-11-25T19:43:38.210" Title="Waves in water always circular" Tags="&lt;water&gt;&lt;waves&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="1290" PostTypeId="2" ParentId="1289" CreationDate="2010-11-25T07:59:11.090" Score="1" ViewCount="" Body="&lt;p&gt;I don't think that water ripples are circular in nature because of the size of the object is small compared to ripples because you can even create perfect ripple with kitchen knife.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is circular in nature because the transmission of disturbance from a source event takes place in a wave fashion in a 2-D medium. And intensity of wave is distributed equally in all directions of medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you know wave motion all particles in ripple execute a harmonic motion, they remain in phase i.e. particles at equal distance from center remain at equal height, there by creating a circle.&lt;/p&gt;&#xA;" OwnerUserId="567" LastEditorUserId="567" LastEditorDisplayName="" LastEditDate="2010-11-25T08:15:21.663" LastActivityDate="2010-11-25T08:15:21.663" />
  <row Id="1291" PostTypeId="2" ParentId="1289" CreationDate="2010-11-25T08:04:57.510" Score="9" ViewCount="" Body="&lt;p&gt;Actually the ripples are not circular at all. See photo below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/fWFL6.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a long stick will generate a straight water front on from its sides and circular waves from its edges. Something similar to a rectangle where the two short sides are replaced by semi-circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the waves spread, the straight front will retain its length, whereas the circular sides will grow in bigger and bigger circles, hence the impression that on a large body of water the waves end up being circular - they are not, but very close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that an irregular object generates &quot;circular&quot; ripples is therefore this: as the waves propagate, &lt;a href=&quot;http://en.wikipedia.org/wiki/Huygens%27_principle&quot; rel=&quot;nofollow&quot;&gt;the irregularities are maintained&lt;/a&gt; but &lt;em&gt;spread&lt;/em&gt; across a larger and larger circular wave front.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very good example of this phenomenon is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_microwave_background_radiation&quot; rel=&quot;nofollow&quot;&gt;Cosmic Microwave Background (CMB)&lt;/a&gt; where electromagnetic waves from the Big Bang are measured after having spread for 13.7 billion years. Although the CMB is really, really smooth - because of the &quot;circular ripple&quot; effect, if you like, we can still measure small irregularities, which we think are due to the &quot;irregular shape&quot; of the Big Bang at a certain time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/m4eAa.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-25T19:43:38.210" LastActivityDate="2010-11-25T19:43:38.210" CommentCount="2" />
  <row Id="1292" PostTypeId="1" CreationDate="2010-11-25T08:26:35.970" Score="0" ViewCount="145" Body="&lt;p&gt;Is there any example of a transmission of energy in a medium that does not show wave nature? &lt;/p&gt;&#xA;" OwnerUserId="567" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-11-25T09:23:13.247" LastActivityDate="2010-11-25T12:51:41.387" ClosedDate="2010-12-14T15:34:04.847" Title="Transmission of energy in medium and wave nature (periodicity)?" Tags="&lt;mechanics&gt;&lt;waves&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="1293" PostTypeId="2" ParentId="1292" CreationDate="2010-11-25T08:34:46.830" Score="4" ViewCount="" Body="&lt;p&gt;Yes, for example heat energy is transferred through three different means:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_radiation&quot; rel=&quot;nofollow&quot;&gt;Radiation&lt;/a&gt; (which is wave like)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_conduction&quot; rel=&quot;nofollow&quot;&gt;Conduction&lt;/a&gt; (which is not wave like)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Convection&quot; rel=&quot;nofollow&quot;&gt;Convection&lt;/a&gt; (which is not wave like)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of course, according to the Standard Model, everything has wave-like properties, so in a way there's no escape from waves, but I believe that your question is best answered with classical phenomena.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-25T08:34:46.830" CommentCount="4" />
  <row Id="1294" PostTypeId="1" AcceptedAnswerId="1297" CreationDate="2010-11-25T08:49:10.420" Score="2" ViewCount="361" Body="&lt;p&gt;While i was in high-school i learn't the &lt;strong&gt;Doppler's Effect&lt;/strong&gt; which if i remember correctly is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Apparent change in the frequency of sound caused due the relative motion between the sound and the observer.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This phenomenon seems obvious, but what i would like to know is, what use does &lt;strong&gt;Doppler Effect&lt;/strong&gt; have in real life. Why is it useful?&lt;/p&gt;&#xA;" OwnerUserId="159" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-11-25T19:09:27.543" LastActivityDate="2011-07-22T16:16:01.157" Title="Doppler's effect use" Tags="&lt;sound&gt;&lt;waves&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="2" />
  <row Id="1295" PostTypeId="2" ParentId="1294" CreationDate="2010-11-25T09:32:09.290" Score="1" ViewCount="" Body="&lt;p&gt;I haven't used it much, but you can use it in games development to implement 3D sound.&lt;/p&gt;&#xA;" OwnerUserId="627" LastActivityDate="2010-11-25T09:32:09.290" CommentCount="2" />
  <row Id="1296" PostTypeId="2" ParentId="1267" CreationDate="2010-11-25T09:35:25.083" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540366830&quot; rel=&quot;nofollow&quot;&gt;Particles and Nuclei&lt;/a&gt; is an excellent book, without derivations, but gets important ideas and results across. Then, a good but very short (at ~300 pages) introduction to the Standard Model is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521852498&quot; rel=&quot;nofollow&quot;&gt;An Introduction to the Standard Model of Particle Physics&lt;/a&gt;. Then there is Zee's outstanding &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691140340&quot; rel=&quot;nofollow&quot;&gt;Quantum Field Theory in a Nutshell&lt;/a&gt;, it has short, readable chapters, although it's almost certainly useless in terms of studying for tests for you course unless the prof. specifically uses this book. Good luck!&lt;/p&gt;&#xA;" OwnerUserId="47" LastEditorUserId="47" LastEditorDisplayName="" LastEditDate="2010-11-25T09:44:55.313" LastActivityDate="2010-11-25T09:44:55.313" />
  <row Id="1297" PostTypeId="2" ParentId="1294" CreationDate="2010-11-25T09:53:45.527" Score="5" ViewCount="" Body="&lt;p&gt;It can be used to measure speed - that's how police radar guns and speed cameras work. Radar waves from the gun/ camera are reflected off the moving vehicle, and the wavelength is shifted according to the speed of the vehicle relative to the gun/ camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In astrophysics, looking at light from distant galaxies, we notice that certain characteristics of the light are shifted in wavelength due to the Doppler effect. This is known as red-shift, as we notice the light is mostly shifted to the longer-wavelength (red) end of the spectrum. This tells us that distant galaxies are moving away from us, which is the primary piece of information that led to the development of the Big Bang theory.&lt;/p&gt;&#xA;" OwnerUserId="474" LastActivityDate="2010-11-25T09:53:45.527" CommentCount="1" />
  <row Id="1299" PostTypeId="2" ParentId="1292" CreationDate="2010-11-25T12:51:41.387" Score="3" ViewCount="" Body="&lt;p&gt;Three basic kinds of energy transfer are usually mentioned. However I don't quite agree with Sklivvz's answer so here is my take on this issue.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'll start with &lt;strong&gt;radiation&lt;/strong&gt; because it is the most fundamental of the three and so most easily answered. It is governed by Maxwell equations which are equivalent to classical wave-equations. So all radiative effects are manifestly wave-like in nature. Nevertheless, it has to be mentioned that by using various optical materials one can obtain also dissipative solutions and lots of non-linear dispersive effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could say that &lt;strong&gt;conduction&lt;/strong&gt; doesn't have wave-like properties because it is governed by parabolic heat equation. However, this equation surely isn't correct because it allows for infinite speed of propagation. There exists also relativistic heat equation which admits wave-like solutions (as one would expect for any relativistic system).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for &lt;strong&gt;convection&lt;/strong&gt;, this is usually governed by some kind of fluid equation from continuum mechanics (like Navier-Stokes) and I suppose that also here one could easily obtain wave-like solution (for specific system and initial conditions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to give an answer to your question: yes, there exist many situations where energy is transferred in a non-wave-like fashion (like exponential decay) but I believe it's not correct to say that any kind of above three energy transfers is or isn't wave-like in general. It depends on precise situation one is dealing with and the level of approximation one is using (like taking relativity into account).&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-11-25T12:51:41.387" CommentCount="2" />
  <row Id="1300" PostTypeId="2" ParentId="1294" CreationDate="2010-11-25T15:33:51.513" Score="3" ViewCount="" Body="&lt;p&gt;I haven't worked with it myself, but an important application of the Doppler effect seems to be to &lt;a href=&quot;http://en.wikipedia.org/wiki/Medical_ultrasonography#Doppler_sonography&quot; rel=&quot;nofollow&quot;&gt;use ultrasound to detect the flow direction of blood&lt;/a&gt;. Apparently you can do some really awesome heart diagnostics stuff by using Doppler ultrasound to map flow around valves and the such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, and many weather radar systems use the Doppler effect to determine the movement of weather systems (actually precipitation patterns, I guess?).&lt;/p&gt;&#xA;" OwnerUserId="636" LastActivityDate="2010-11-25T15:33:51.513" />
  <row Id="1301" PostTypeId="2" ParentId="1286" CreationDate="2010-11-25T17:10:29.657" Score="2" ViewCount="" Body="&lt;p&gt;The coefficient of friction for a curling stone is not constant - motion on ice is different from motion on an unchanging solid surface, thanks to melting, which in turn is proportional to most of the factors that affect friction (e.g. contact area, velocity). So, exactly as you suspect, friction decreases as velocity increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the changing CofF, the stone curls more strongly at the end of its trajectory (as it gets slower) than at the beginning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, as Mark C says, the team can affect the trajectory after delivery by sweeping - the entire purpose of the sweepers is to melt or at least smooth the ice ahead of the stone (or not, as required), which tends to straighten and lengthen the trajectory by reducing the effect of friction on the leading edge of the stone.&lt;/p&gt;&#xA;" OwnerUserId="637" LastActivityDate="2010-11-25T17:10:29.657" CommentCount="3" />
  <row Id="1302" PostTypeId="1" AcceptedAnswerId="1306" CreationDate="2010-11-25T17:32:57.650" Score="5" ViewCount="438" Body="&lt;p&gt;If you shake a soda bottle before opening it, and then open it, you get the fizz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the compressed CO&lt;sub&gt;2&lt;/sub&gt; releasing to the atmosphere which is at comparatively low pressure value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two questions (related)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Why does shaking a bottle make the compresses gases &lt;em&gt;un&lt;/em&gt;-dissolve?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why can't the gases dissolve back into the liquid?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="230" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-01T00:13:44.757" LastActivityDate="2011-05-30T16:41:38.707" Title="Why can't CO2 mix back with the liquid after a soda bottle has been shaken?" Tags="&lt;pressure&gt;&lt;atmosphere&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="1303" PostTypeId="2" ParentId="1302" CreationDate="2010-11-25T18:20:39.110" Score="1" ViewCount="" Body="&lt;p&gt;For the &lt;em&gt;first question&lt;/em&gt;, I'll give an analogy that I hope illustrates quite well what is happening here. Imagine you have a ball on a floor. Now an earthquake comes. If its not too destructive, the ball will just roll and maybe bounce but after the earthquake is over, it will still be on the floor. Now imagine that you would put that ball on your desk (so you use some work to overcome gravity). What will happen after the earthquake? Well, it's pretty certain that the ball will fall to the floor and all your work will go uselessly away in the form of heat (that is produces when ball hits the ground) and the entropy of the system increases (as it should by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Second_law_of_thermodynamics&quot; rel=&quot;nofollow&quot;&gt;Second Law of Thermodynamics&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it should be pretty obvious that if you replace ball in the above with $CO_2$ bubble, work to overcome gravity by work needed to compress the bubble and earthquake by the shaking, you have precisely the same situation. So the system is in an unstable equilibrium and any big enough perturbation will bring it to the stable equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your &lt;em&gt;second question&lt;/em&gt; is either trivial or wrong, depending on how you meant it. If you meant it in the way &lt;em&gt;why doesn't $CO_2$ compress itself back&lt;/em&gt; then the reason is the same as why balls don't hop up on your desk when left to themselves. They don't have energy for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you meant it as just dissolving, then as a matter of fact, they do. Both water and $CO_2$ molecules are both in the volume of the liquid and the air. For example, there is always some water vapor above the surface of the water. The exact concentration depends on the volume and pressure in the bottle, but except for that it is constant and non-zero. But the concentrations in the stable equilibrium case are very different from concentrations in the unstable (compressed) case and it is always preferable for the system to increase its entropy thereby moving from the unstable to the stable equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-11-25T18:20:39.110" />
  <row Id="1304" PostTypeId="2" ParentId="1294" CreationDate="2010-11-25T19:06:52.300" Score="2" ViewCount="" Body="&lt;p&gt;Others have answered about real-life applications of Doppler effect, but you have many others. For example, in quantum optics, it is at the heart of &lt;a href=&quot;http://en.wikipedia.org/wiki/Doppler_cooling&quot; rel=&quot;nofollow&quot;&gt;Doppler cooling&lt;/a&gt;, which is used to cool down some dilute atom clouds down to a few hundred microkelvins, and then use them for some fundamental physics experiment. If you really want to have a &lt;em&gt;practical&lt;/em&gt; application, then this technique is used since the 1990's to improve the precision of &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_clock&quot; rel=&quot;nofollow&quot;&gt;atomic clocks&lt;/a&gt;, because the movement of the atoms induce a Doppler-shift of the frequency used as reference for this clocks. It's a way to fight Doppler-effect with Doppler-effect ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to use the Doppler shift to create on each a force which is proportional to its speed of each atom in order to slow it down. This cools the atomic cloud down because the temperature is proportional to the average kinetic energy. This kind of speed dependence is achieved using detuned lasers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you shine a laser to an atom with a frequency tuned to an electronic transition, the atom will absorb photons from the laser beam and feel a strong force. If the laser is detuned towards the red, it can be &quot;retuned&quot; through a Doppler shift when the atom moves towards the beam. As a consequence, the more speed the atom has towards the laser, the more it &quot;feels&quot; the laser, and the stronger the force is. Putting 2 lasers in each direction of the space (i.e. six laser) allows then, in the volume where the laser overlap, to create a force proportional to the speed of the atom. Every atom which enters this zone is slowed down to a very small speed, hence the name &quot;optical molasse&quot; given to these set-ups.&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2010-11-25T19:06:52.300" />
  <row Id="1305" PostTypeId="1" AcceptedAnswerId="1313" CreationDate="2010-11-25T19:14:10.300" Score="3" ViewCount="166" Body="&lt;p&gt;I've simplified this down by quite a lot but as far as I understand it, magnetic fields on the Sun's surface twist together and when it gets all a bit too much they release energy.&lt;br&gt;&#xA; My question is why can't we twist magnetic fields here on Earth to produce energy?&lt;/p&gt;&#xA;" OwnerUserId="635" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-27T07:34:58.810" LastActivityDate="2010-11-27T07:34:58.810" Title="Producing energy from magnetic flux loops" Tags="&lt;electromagnetism&gt;&lt;magnetohydrodynamics&gt;&lt;sun&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="1306" PostTypeId="2" ParentId="1302" CreationDate="2010-11-25T19:43:11.207" Score="5" ViewCount="" Body="&lt;p&gt;This is what I think that happens. The bottles with the soda have the liquid inside under pressure. Under these conditions you have an amount of CO2 dissolved in the liquid. If you shake it then it will make some bubbles but not a lot and it will return to the previous state. If you don't disturb the liquid and open the bottle, then the pressure will drop to atmospheric levels. The dissolved CO2 in the liquid is in a grater concentration that the liquid can hold for that pressure. But there are no evaporation seeds for the bubbles to form. It is a similar thing with the superheated water, where you have crossed the threshold for a phase transition, but you need impurities to initiate the process in the liquid. If you disturb the liquid it will produce bubbles. If you first shake the bottle and then open it, then the agitated liquid will form bubbles of CO2 with the pressure drop. Now, once you have reduced the pressure on the liquids surface, there is no way to put the CO2 back in to the liquid, except for increasing the pressure again.&lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2010-11-25T19:43:11.207" CommentCount="1" />
  <row Id="1307" PostTypeId="1" AcceptedAnswerId="1309" CreationDate="2010-11-25T21:50:18.897" Score="6" ViewCount="482" Body="&lt;p&gt;While the speed of light in vacuum is a universal constant (c), the speed at which light propagates in other materials/mediums may be less than c. This is obviously suggested by the fact that different materials(especially in the case of transparent ones) have a particular &lt;a href=&quot;http://en.wikipedia.org/wiki/Refractive_index&quot; rel=&quot;nofollow&quot;&gt;Refractive Index&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But surely, matter or even photons can be accelerated beyond this speed in a medium?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, what are the effects?&lt;/p&gt;&#xA;" OwnerUserId="619" LastEditorUserId="619" LastEditorDisplayName="" LastEditDate="2010-12-05T19:05:09.240" LastActivityDate="2010-12-05T21:56:23.043" Title="What happens if light/particles exceeded the speed of light for a particular medium" Tags="&lt;speed-of-light&gt;&lt;cherenkov-radiation&gt;" AnswerCount="2" CommentCount="17" FavoriteCount="2" />
  <row Id="1308" PostTypeId="2" ParentId="1307" CreationDate="2010-11-25T22:08:10.557" Score="11" ViewCount="" Body="&lt;p&gt;There is a phenomenon somewhat analogous to an &quot;optical boom&quot; know as &lt;a href=&quot;http://en.wikipedia.org/wiki/Cherenkov_radiation&quot; rel=&quot;nofollow&quot;&gt;Cherenkov radiation&lt;/a&gt; which occurs when charged particles pass through a medium at a speed greater than the speed of light for that particular medium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This produces a beautiful blue glow often seen in pictures of the cores of nuclear reactors. See the wikipedia page for pictures&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not entirely sure if this answers your question but it may be of interest to you.&lt;/p&gt;&#xA;" OwnerUserId="632" LastEditorUserId="632" LastEditorDisplayName="" LastEditDate="2010-11-25T22:16:34.523" LastActivityDate="2010-11-25T22:16:34.523" CommentCount="3" />
  <row Id="1309" PostTypeId="2" ParentId="1307" CreationDate="2010-11-25T22:12:34.567" Score="12" ViewCount="" Body="&lt;p&gt;Speed of light is indeed lower when light propagates through materials (as opposed to vacuum). This doesn't mean that individual photons go slower but rather that the apparent speed of light pulse is lower due to interactions with atoms of the material. So in this case it is possible for some objects to go &quot;faster than light&quot; and indeed very similar effect to sonic boom, called &lt;a href=&quot;http://en.wikipedia.org/wiki/Cherenkov_radiation&quot; rel=&quot;nofollow&quot;&gt;Cherenkov radiation&lt;/a&gt;, appears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for most materials the apparent speed of light is still huge (of the order of speed of light in vacuum) so you need very energetic particles to generate Cherenkov radiation. So this effect is mainly relevant for high-energy particle physics, astrophysics and nuclear physics.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-05T21:56:23.043" LastActivityDate="2010-12-05T21:56:23.043" CommentCount="2" />
  <row Id="1310" PostTypeId="2" ParentId="1273" CreationDate="2010-11-25T22:28:09.850" Score="-1" ViewCount="" Body="&lt;p&gt;Here is a quick result that might interest you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's begin from the equation of relativistic energy in terms of the rest mass, $m_0$, $E=\gamma m_oc^2$, then by inserting $\gamma = \sqrt{\frac{1}{1-\frac{v^2}{c^2}}}$, we find:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{6cm}E^2-E^2\frac{v^2}{c^2}=(m_0)^2c^4$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{6cm}E^2=(E_0)^2+p^2c^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the equation that relates relativistic energy and momentum.  So maybe the best we can say is if you start from the equations $E^2=(E_0)^2+p^2c^2$, and $E=\gamma m_oc^2$, we can derive the Lorentz contraction. &lt;/p&gt;&#xA;" OwnerUserId="442" LastEditorUserId="442" LastEditorDisplayName="" LastEditDate="2010-11-25T22:35:34.683" LastActivityDate="2010-11-25T22:35:34.683" CommentCount="1" />
  <row Id="1311" PostTypeId="2" ParentId="1251" CreationDate="2010-11-25T23:05:41.330" Score="3" ViewCount="" Body="&lt;p&gt;I think you might be interested in the following important distinction: relativistic mechanics can be built up upon it's own first principles and then derives classical mechanics as a limiting case; and although QM reduces to classical mechanics in a limiting case, QM actual requires classical mechanics for it's very formulation, and thus QM occupies a peculiar place among physical theories.  See Landau, volume 3, chapter 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I will attempt to answer your question, even though in my opinion this question should be asked at math.stack exchange.  In the broadest sense, a continuous manifold IS the limit of a discrete set of points, just like the set of real numbers is the limit set of the set of fractions.  But after viewing your previous post that you linked to, I would like you to consider the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cardinality of a discrete set of points is the cardinality of the integers, and the cardinality of the set of points on a continuous manifold is at least that of the continuum.  There are certainly many consequences of this fact which distinguish the two cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another thing you should bear in mind, the FT defines functions by a countable set of coefficients, but FT's only converge up to a set of measure zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In QM the FT plays a very important role, and I think the discrete nature of QM is not related to a discrete space time necessarily, but rather to the mathematics of continuous symmetries, i.e. Lie groups.  This explains for example the quantum numbers associated with angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Lie group is a group which is also a continuous manifold, and thus we see that already, a discrete set of points can never be a Lie group, because a discrete set of points is by definition not continuous.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly computer methods which make use of discrete approximation have produced good results, but the thing about computer models is you have to be so careful all the time to make sure you are not introducing an error by choosing a bad parametrization for your discrete &quot;mesh&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take a look at the case of fluid mechanics simulations.  It is well known, and should be intuitively clear, that for the case of turbulent flows, the density of the mesh at regions of intense turbulence is going to have a strong influence on the results of your computer simulation.  There are so many subtle things that are involved with mathematical convergence for these types of situations.  It turns out, that there are lots of situations where you just can't simulate the turbulence because the chaotic nature of the turbulence is so sensitive to the initial conditions and the mesh density, that it's very hard to get your simulation to compile (converge) in any meaningful way.  I think this example of simulating fluid mechanics really shows the limitations of using discrete methods to approximate solutions of certain problems.&lt;/p&gt;&#xA;" OwnerUserId="442" LastEditorUserId="442" LastEditorDisplayName="" LastEditDate="2010-11-25T23:20:01.307" LastActivityDate="2010-11-25T23:20:01.307" />
  <row Id="1312" PostTypeId="1" CreationDate="2010-11-25T23:36:06.780" Score="-1" ViewCount="330" Body="&lt;p&gt;If I'm moving through MY reference frame at the speed of light, isn't time still passing by normally for me?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help me think about this fourth dimension- space-time.  I want to intuitively understand it.  I understand some thought-experiments (Einstein's train experiment, for example, and why events occur slower/later for people that are farther away or moving from the event because of the infinitesimal amounts of time it takes for light to travel to them).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question I could ask here is what caused you to understand space-time when you finally got it (if you didn't immediately get it).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-13T07:31:02.340" LastActivityDate="2011-01-13T07:31:02.340" ClosedDate="2010-11-26T17:24:54.720" Title="Why does moving through space at the speed of light automatically limit us from moving through time?" Tags="&lt;speed-of-light&gt;&lt;spacetime&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="1313" PostTypeId="2" ParentId="1305" CreationDate="2010-11-25T23:36:40.640" Score="6" ViewCount="" Body="&lt;p&gt;We have conservation of energy to contend with. The magnetic energy being released in the upper layers of the suns atmosphere originated in turbulent convection lower down. So in effect the upper layers of the sun (upper here meaning roughly from the photosphere and downward) are acting as a heat engine putting some energy into the suns magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can't &quot;twist&quot; magnetic field lines without providing the energy to do so. It would&#xA;be possible to generate electricity from changes in the earth's magnetic field caused by&#xA;the impingement of solar wind born magnetic energy on the earths magnetic field. Any change&#xA;in the total flux of magnetic field going through a current loop, generates a voltage. In fact&#xA;during geomagnetic storms there is a danger that currents induced in power transmission lines could cause serious damage to th electrical grid system. But this is not a practical way to generate energy, but rather a hazard that our infrastructure has to contend with.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-11-25T23:36:40.640" />
  <row Id="1314" PostTypeId="2" ParentId="1294" CreationDate="2010-11-26T00:34:53.563" Score="4" ViewCount="" Body="&lt;p&gt;The main &quot;uses&quot; of the Doppler effect are, as other people have already mentioned, in the measurement of velocities. In addition to police radar, it's used in weather radar. That's why local news stations (at least in the US) are forever touting their &quot;Doppler &lt;em&gt;N&lt;/em&gt;&quot; (for large values of &lt;em&gt;N&lt;/em&gt;) weather system-- those systems use the Doppler effect to measure the speed of water droplets in clouds, which provides additional information about their composition and nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can also confirm that they use the Doppler effect in ultrasound imaging of the heart, to measure and map out the flow of blood in different regions. Before my daughter was born, they thought she might be at risk for heart problems, so we had a pediatric cardiologist look at it. They have a nifty color-coded display showing blood moving toward the imager as blue, and away from it as red. (That was the second-best part of the whole thing, after the doctor giving her a clean bill of health...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My own research area is in laser cooling, where we use the Doppler effect to ensure that atoms only interact with light when they're moving toward the laser. If you tune the laser frequency slightly below the frequency the atoms want to absorb, a stationary atom will not interact with the laser, but an atom moving toward the laser will see the frequency shifted up, closer to the resonant frequency for the atom, and will be more likely to absorb light. When an atom absorbs a photon of light, it gets a small &quot;kick&quot; in the direction the photon was headed. If the atom is moving in the opposite direction to the photon (that is, toward the laser), then the &quot;kick&quot; will act to slow the atom down. With the proper arrangement of lasers, you can slow atoms down in any direction they try to move, thus reducing their overall velocity. Since temperature is a measure of the average kinetic energy of the atoms making up a sample, slower velocities mean lower temperatures. You can easily reach microkelvin temperatures this way-- a few millionths of a degree (C) above absolute zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a lot of complications along the way-- the Doppler effect is both the key to laser cooling and a problem that needs to be overcome in slowing atoms from room temperature-- but that's the basic idea. There's more information, and nifty Java applet video games, at the &lt;a href=&quot;http://www.colorado.edu/physics/2000/bec/lascool1.html&quot; rel=&quot;nofollow&quot;&gt;Physics 2000 site at the University of Colorado&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Laser cooled atoms are the starting point for all manner of experiments in atomic and molecular physics, and quantum optics. The most important technological application at the moment is in the area of atomic clocks, with the best cesium atomic clocks in the world using laser cooled atoms. At various times, there have been suggestions to put atomic clocks in space, potentially as a next generation GPS system, which would improve the already impressive precision of GPS navigation. I'm not sure what the current status of those ideas is, though.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-26T00:34:53.563" />
  <row Id="1315" PostTypeId="2" ParentId="1312" CreationDate="2010-11-26T00:45:00.467" Score="1" ViewCount="" Body="&lt;p&gt;As noted in comments, your question is problematic because in your own frame of reference, your velocity would be zero. You will never see yourself moving at the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a bunch of different ways to see that the speed of light is a limit, among them the fact that the relativistic velocity transformations prevent any object that is seen to be moving at a speed less than that of light from being seen at light speed by any other observer, the fact that an infinite amount of energy would be required to accelerate an object to light speed, and the fact that any object moving faster than the speed of light would create problems with causality, with effects happening before the things that caused them according to at least some observers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two best general books on Special Relativity that I've seen (and I've been reading a bunch of them as research for the book I'm writing) are David Mermin's &lt;a href=&quot;http://press.princeton.edu/titles/8112.html&quot; rel=&quot;nofollow&quot;&gt;It's About Time&lt;/a&gt; and Tatsu Takeuchi's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521141001&quot; rel=&quot;nofollow&quot;&gt;An Illustrated Guide to Relativity&lt;/a&gt;. Both are aimed at non-scientists, but explain the underlying ideas in a really clear manner that can provide some additional insight even for physicists.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-26T00:45:00.467" />
  <row Id="1316" PostTypeId="2" ParentId="1312" CreationDate="2010-11-26T03:55:18.007" Score="0" ViewCount="" Body="&lt;p&gt;Here is my method for intuitively understanding space-time. Imagine you are holding a clock in front of you. At any given moment, you have the choice to allow the clock to tick at the correct pace (from your viewpoint) by holding the clock still, stop the clock by moving it at the speed of light, or somewhere in between. The &quot;rate of observed time&quot; is given by the Lorentz factor in all three cases.&lt;/p&gt;&#xA;" OwnerUserId="69" LastActivityDate="2010-11-26T03:55:18.007" />
  <row Id="1317" PostTypeId="1" CreationDate="2010-11-26T05:15:07.327" Score="3" ViewCount="449" Body="&lt;p&gt;It is generally assumed that pushing a cart is more easier than pulling one. But why?&lt;br&gt;&#xA;Is there any difference in terms of force required to achieve the same amount of displacement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it just a human perception?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do almost all automobiles transfer torque to the back axle. But then, why do trains have engines in the front?&lt;/p&gt;&#xA;" OwnerUserId="619" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-11-26T09:41:42.880" LastActivityDate="2011-08-30T08:15:37.030" Title="Is pushing actually easier than pulling?" Tags="&lt;classical-mechanics&gt;&lt;mechanics&gt;&lt;forces&gt;&lt;newtonian-mechanics&gt;&lt;momentum&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="1318" PostTypeId="2" ParentId="1312" CreationDate="2010-11-26T07:46:37.000" Score="1" ViewCount="" Body="&lt;p&gt;I can answer both questions at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to realize that space-time is not merely space plus time. It's a unified concept, in which we create extra mathematical structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, you may want to take a look at the definition of distance between events:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And start to ask yourself questions, like: what does $ds$ represent? What happens to $ds$ if an object is still with respect to the observer? What if the object is moving at the speed of light? What is represented by the locus of points defined by $ds=0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://en.wikipedia.org/wiki/Minkowski_space&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Minkowski_space&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-26T07:46:37.000" />
  <row Id="1319" PostTypeId="2" ParentId="506" CreationDate="2010-11-26T08:09:49.157" Score="0" ViewCount="" Body="&lt;p&gt;Lets assume that the bike is stationary. &#xA;If the bike is leaning (not perfectly vertical upright) then bike would surely fall off sooner or later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if the bike is moving, the bike would not fall even if it's leaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its because the bike might seem as it is moving straight forward( as circle or curve is big it looks straight), but it isn't. Its moving in a circular motion (depends on your velocity).&#xA;I you have high velocity then you are moving in a bigger radius (and note that you always lean on bike on circular motion to balance the torque produced by centrifugal acceleration by gravity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let me present an example. We all have rolled circular objects like circular lid of jar.&#xA;When roll it, it goes straight forward but when it loses its velocity it begins to go round in a small radius with high leaning angle(from vertical upright) which is clearly visible. As you have been already explained, it's centrifugal force that does the work.&lt;/p&gt;&#xA;" OwnerUserId="567" LastEditorUserId="567" LastEditorDisplayName="" LastEditDate="2010-11-28T05:05:11.283" LastActivityDate="2010-11-28T05:05:11.283" />
  <row Id="1320" PostTypeId="1" AcceptedAnswerId="1322" CreationDate="2010-11-26T08:49:39.310" Score="5" ViewCount="178" Body="&lt;p&gt;If I break one bulb on a string of Christmas lights, or the filament goes out, then all of the rest seem to stop working.  Why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I've also seen the opposite happen in some newer lights -- when one breaks, the others stay on.  What makes this one different than the others?&lt;/p&gt;&#xA;" OwnerUserId="71" LastActivityDate="2010-11-26T09:40:47.247" Title="Why does breaking one bulb on a string of Christmas Lights cause all the rest to break in some lights?" Tags="&lt;electric-circuits&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="1321" PostTypeId="2" ParentId="1317" CreationDate="2010-11-26T09:17:59.347" Score="10" ViewCount="" Body="&lt;p&gt;The force required to accelerate an object of a given mass by a given amount will always be constant ($F=ma$). The difference between pushing and pulling is that humans are built in a particular way - our muscles, joints and tendons won't make different actions cost the same in terms of our perceived effort, or our energy expenditure (although the useful working energy transferred to the cart will be same).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not any kind of expert on human kinematics, but I expect the reason that pushing feels easier is because the point of contact (hands or shoulders) can be braced up against the cart, allowing you to use your big driving muscles (quadriceps/gluteal muscles) much more efficiently. When you pull, you've got to try to keep your arms rigid to transfer more of your driving energy into the cart - that means you're burning more energy just in your arm muscles than when you push.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the axle receiving the torque of a vehicular engine, I'm pretty sure it's far more to do with steering mechanics and manoeuvrability than any kind of energy transfer efficiency.&lt;/p&gt;&#xA;" OwnerUserId="637" LastEditorUserId="637" LastEditorDisplayName="" LastEditDate="2010-11-26T10:48:07.587" LastActivityDate="2010-11-26T10:48:07.587" />
  <row Id="1322" PostTypeId="2" ParentId="1320" CreationDate="2010-11-26T09:28:00.973" Score="9" ViewCount="" Body="&lt;p&gt;They are in series circuit, so breaking one bulb breaks the circuit itself:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/yQJok.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-11-26T09:40:47.247" LastActivityDate="2010-11-26T09:40:47.247" CommentCount="1" />
  <row Id="1323" PostTypeId="1" AcceptedAnswerId="1358" CreationDate="2010-11-26T09:44:06.653" Score="4" ViewCount="306" Body="&lt;p&gt;What is the Madelung transformation and how is it used?&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2011-07-08T14:31:14.650" Title="Madelung transformation" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" />
  <row Id="1324" PostTypeId="1" AcceptedAnswerId="1333" CreationDate="2010-11-26T11:24:49.157" Score="10" ViewCount="533" Body="&lt;p&gt;While it is quite common to use piecewise constant functions to describe reality, e.g. the optical properties of a layered system, or the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermi%E2%80%93Dirac_statistics&quot; rel=&quot;nofollow&quot;&gt;Fermi–Dirac statistics&lt;/a&gt; at (the impossible to reach exactly) $T=0$, I wonder if in a fundamental theory such as QFT some statement on the analyticity of the fields can be made/assumed/proven/refuted?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take for example the Klein-Gordon equation. Even if you start with the non-analytical Delta distribution, after infinitesimal time the field will smooth out to an analytical function. (Yeah I know, that is one of the problems of relativistic quantum &lt;em&gt;mechanics&lt;/em&gt; and why QFT is &quot;truer&quot;, but intuitively I don't assume path integrals to behave otherwise but smoothing, too).&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2011-04-23T19:40:46.990" Title="Is the world $C^\infty$?" Tags="&lt;quantum-field-theory&gt;&lt;analyticity&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="1325" PostTypeId="2" ParentId="1324" CreationDate="2010-11-26T12:20:29.967" Score="1" ViewCount="" Body="&lt;p&gt;What quantities supposed to be $C^\infty$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if it answers you question, but AFAIK smooth functions are nice and useful tool to describe many aspects of the physical world. However, I don't see why they should be considered as fundamental in any sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to QFT, even there you often encounter Dirac delta (and you can't get rid of it easily).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One professor from my department when asked if all physical dependences are continuous answered &quot;Yes - and even more - with discrete domain&quot; (as you will never make an infinite number of measurements).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, in my opinion there may be more specific (and purposeful) questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If for a given theory such-and-such dependences are continuous/analytic/smooth/(other nice property)?&lt;/li&gt;&#xA;&lt;li&gt;If in practice one can restrict oneself to using only smooth functions, resulting in approximation error below measurement error?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="184" LastEditorUserId="184" LastEditorDisplayName="" LastEditDate="2010-11-26T12:47:35.340" LastActivityDate="2010-11-26T12:47:35.340" CommentCount="1" />
  <row Id="1326" PostTypeId="1" CreationDate="2010-11-26T16:24:50.893" Score="12" ViewCount="412" Body="&lt;p&gt;I know it has to do with electricity flowing and generating a magnetic field, but I would like a thorough explanation (with perhaps a picture). In particular:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is in the ear piece?&#xA;&lt;li&gt;Why do they repel as opposed to attract?&#xA;&lt;li&gt;How come my ear buds don't seem to do it?&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="649" LastActivityDate="2011-08-15T18:32:55.993" Title="Why do earphone pieces repel each other when music is on?" Tags="&lt;electromagnetism&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="4" />
  <row Id="1327" PostTypeId="1" AcceptedAnswerId="1337" CreationDate="2010-11-26T16:33:14.757" Score="7" ViewCount="475" Body="&lt;p&gt;If all distances are constantly increasing, as Hubble's law say, then lots of potential energies of form ~$\frac{1}{r}$ changes, so how is the total energy of the Universe conserved with Hubble's expansion?&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-26T18:53:40.413" LastActivityDate="2011-01-16T16:51:44.407" Title="Hubble's law and conservation of energy" Tags="&lt;astrophysics&gt;&lt;mathematical-physics&gt;&lt;spacetime&gt;" AnswerCount="5" FavoriteCount="3" />
  <row Id="1328" PostTypeId="2" ParentId="1327" CreationDate="2010-11-26T16:52:10.270" Score="10" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Update 1:&lt;/strong&gt; I edited the answer a little to make it more accessible.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 2:&lt;/strong&gt; Based on the discussion with Sklivvz I'll elaborate a bit more why the concept of conservation of energy is problematic in General Relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First one has to note that conservation of some quantity means that the quantity is constant through the evolution of the system in time. But what is time? Already Special Relativity tells us that every observer carries his own local notion of time. General Relativity complicates this notion a great deal further. So it doesn't even make sense to talk about constancy of some quantity in time unless we specify what time we mean! To make sense of this one usually restricts to small volume (so that time has almost same meaning for each point of the volume) or else one has a good notion of what time is globally. This second point is fortunately true in our universe (and also in flat Minkowski space-time) because it can be described quite well by some &lt;a href=&quot;http://en.wikipedia.org/wiki/FRW&quot; rel=&quot;nofollow&quot;&gt;FRLW solutions&lt;/a&gt; of Einstein equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the general notion of time complicates things but can be dealt with. The worse problem is that in General Relativity it is really hard to say what &lt;strong&gt;gravitational energy&lt;/strong&gt; (i.e. energy stored as curvature of space-time) is. It turns out that different observers won't really agree on this (so the concept is not &lt;a href=&quot;http://en.wikipedia.org/wiki/General_covariance&quot; rel=&quot;nofollow&quot;&gt;covariant&lt;/a&gt;) and except for some special situations one can't say anything useful.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Short answer: conservation of energy is not a fundamental law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Noether%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Noether's theorem&lt;/a&gt; tells us that whenever there is some symmetry in physical laws, one obtains some conserved quantity. For translations in space one obtains conserved momentum. For rotations one gets &lt;a href=&quot;http://physics.stackexchange.com/questions/822/is-angular-momentum-truly-fundamental/828#828&quot;&gt;angular momentum&lt;/a&gt;. And for translations in time, one gets the law of conservation of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that conservation of energy only holds for systems that are governed by laws which are invariant in time, i.e. &lt;em&gt;static&lt;/em&gt;. Most systems one encounters in daily life are of this sort (even friction, when looked upon closely, conserves energy; the missing energy is just transformed into kinetic energy of atoms as heat). But this is only the consequence of living in an idealized static &lt;a href=&quot;http://en.wikipedia.org/wiki/Minkowski_space&quot; rel=&quot;nofollow&quot;&gt;Minkowski space-time&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moment one leaves this nice static place and considers dynamical universe, one has to dispose of the simple concept of conservation of energy. One can consider various notions of &lt;a href=&quot;http://en.wikipedia.org/wiki/Mass_in_general_relativity&quot; rel=&quot;nofollow&quot;&gt;energy&lt;/a&gt; in General Relativity but these concepts are pretty messy and one is pretty much forced to talk only about energy locally in some small volume. But you can't really say anything about universe as a whole.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; To elaborate on a small volume in the above (as asked by Sklivvz in comments), one can often limit himself to a volume where the system is approximately isolated from the rest of the universe. E.g. solar system is to a good degree isolated and the energy would be conserved if it wasn't for incoming (and outgoing) particles (like light) and bodies (like asteroids).&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-27T11:43:18.597" LastActivityDate="2010-11-27T11:43:18.597" CommentCount="17" />
  <row Id="1329" PostTypeId="2" ParentId="1326" CreationDate="2010-11-26T16:58:15.153" Score="3" ViewCount="" Body="&lt;p&gt;Because the loudspeakers in your earphones (or at least @Marek's) contain some permanent magnets. To make a long story short, the magnetic field produced by the current is used to move the permanent magnet according to the music you're hearing. Then, the details depend on the specific technology your earphones use, as you can check with &lt;a href=&quot;http://en.wikipedia.org/wiki/Earphone#Technology&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Earphone#Technology&lt;/a&gt; (with diagrams)&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2010-11-26T16:58:15.153" CommentCount="1" />
  <row Id="1330" PostTypeId="2" ParentId="1327" CreationDate="2010-11-26T17:53:52.267" Score="3" ViewCount="" Body="&lt;p&gt;There are a few answers to this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Conservation of energy in relativity is only a local law.  Energy is conserved in a local reference frame, but it is not conserved globally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The cosmological constant has a negative pressure associated with it, so as the universe expands, the matter in the universe has work done on it by the cosmological constant, which then causes more expansion, which causes more work done, etc.  So conservation of energy is still there, it's just enforced by the novel thermodynamics of a cosmological fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Marek's answer--a completely coordinate-free definition of energy doesn't exist for Big Bang Cosmology, and so this isn't a well-defined question.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-26T18:50:42.333" LastActivityDate="2010-11-26T18:50:42.333" />
  <row Id="1331" PostTypeId="2" ParentId="1324" CreationDate="2010-11-26T20:55:25.813" Score="3" ViewCount="" Body="&lt;p&gt;I am not even sure if the world is $C^{0}$. The concept of uncountability in &quot;real&quot; world is still hard for me to digest. I am happy to deal with uncountability in pure mathematics but I am not sure if it is the case in the &quot;real&quot; world. It might be possible to reformulate all of physics in terms of discrete and not continuous.&#xA;One such attempt is Discrete Philosophy though I don't know how much of this is true and how much is bullshit. &quot;http://en.wikipedia.org/wiki/Digital_philosophy&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be possible to reformulate them in terms of some fundamental quantities and assume that these quantities cannot be sub divided further. For instance discretize space in terms of say Planck's length and time in terms of say Planck's time and so on.&lt;/p&gt;&#xA;" OwnerUserId="403" LastEditorUserId="403" LastEditorDisplayName="" LastEditDate="2010-11-26T21:09:46.213" LastActivityDate="2010-11-26T21:09:46.213" CommentCount="3" />
  <row Id="1332" PostTypeId="2" ParentId="1326" CreationDate="2010-11-26T21:38:49.957" Score="14" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;What is in the ear piece?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally there is a small magnet, a loop of coil and a diaphragm. The current going through the coil is modulated so that it makes the coil move in the magnetic field and the diaphragm, which is attached to it. The movement of the diaphragm generates the sound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why do they repel as opposed to attract?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the two magnets are oriented the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A longer and more interesting answer is the following. Most modern music has monaural lower frequencies. This is because the kick drum and the bass in rock and pop are positioned in the centre of the audio spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the lower frequencies are the ones that need more electromagnetic energy, because they need to push more air - the wavelength is longer - and they are in-phase between the two ear pieces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in practice, when you listen to music, you feel more force pushing outwards. If you were to listen to music with out of phase basses, you would feel less force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How come my ear buds don't seem to do it?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a number of reasons.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;they are smaller, therefore there are smaller forces involved&lt;/li&gt;&#xA;&lt;li&gt;the diaphragms are smaller and the earbuds are therefore built to have far less low frequency response &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="66" LastEditorUserId="176" LastEditorDisplayName="" LastEditDate="2011-08-15T18:32:55.993" LastActivityDate="2011-08-15T18:32:55.993" CommentCount="3" />
  <row Id="1333" PostTypeId="2" ParentId="1324" CreationDate="2010-11-26T22:43:16.277" Score="5" ViewCount="" Body="&lt;p&gt;This is a really interesting, but equally beguiling, question.&#xA;Shock waves are discontinuities that develop in solutions of the wave equation.&#xA;Phase transitions (of various kinds) are non-continuities in thermodynamics, but as thermodynamics is a study of aggregate quantitites, one might argue that the microscopic system is still continuous.  However, the Higgs mechanism is an analogue in quantum field theory, where continuity is a bit harder to see.  It is likely that smoothness is simply a convenience of our mathematical models (as was mentioned above).  It is also possible that smooth spacetime is some aggregate/thermodynamic approximation of discrete microstates of spacetime -- but our model of &lt;em&gt;that&lt;/em&gt; discrete system will probably be described by the mathematics of continuous functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(p.s.:  Nonanalyticity is somehow akin to free will:  our future is not determined by all time-derivatives of our past!)&lt;/p&gt;&#xA;" OwnerUserId="20" LastActivityDate="2010-11-26T22:43:16.277" CommentCount="3" />
  <row Id="1334" PostTypeId="1" AcceptedAnswerId="1351" CreationDate="2010-11-27T02:52:32.257" Score="2" ViewCount="100" Body="&lt;p&gt;In the elasticity theory, you can derive a wave equation from the fundamental equation of motion for an elastic linear homogeneous isotropic medium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho \partial^2_t \overline{u} = \mu \nabla^2 \overline{u} + (\mu+\lambda) \nabla(\nabla \cdot \overline{u})$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the seismology tradition, you introduce scalar and vector potentials for the P and S components of the displacements, derive wave equations for them and use them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in electrodynamics you can derive from Maxwell's equations the wave equations for Fields and for Potentials; but there you use the potentials because they compose a quadrivector. In geophysics what's the convenience of it?&lt;/p&gt;&#xA;" OwnerUserId="558" LastEditorUserId="42" LastEditorDisplayName="" LastEditDate="2010-11-27T17:16:38.013" LastActivityDate="2010-12-02T08:44:31.040" Title="Utility of displacements potentials in geophysics" Tags="&lt;electrodynamics&gt;&lt;waves&gt;&lt;geophysics&gt;&lt;elasticity&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="1335" PostTypeId="2" ParentId="478" CreationDate="2010-11-27T04:19:25.077" Score="1" ViewCount="" Body="&lt;p&gt;This is possible, it's just very very difficult to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People regularly do this with the electric field of light to move dielectric particles (insulators) in the lab, and the technique is known as &quot;optical tweezers&quot;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Optical_tweezers&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Optical_tweezers&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason you don't want to try this with magnetic particles is that the magnetic field of light is much weaker than the electric field, or rather that it doesn't interact very much with most magnetic materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Greg's answer above is half correct - The fact that these fields oscillate in time means that the applied force would oscillate as well, however, the field gradient is exploited instead to make optical tweezers work.&lt;/p&gt;&#xA;" OwnerUserId="651" LastActivityDate="2010-11-27T04:19:25.077" />
  <row Id="1336" PostTypeId="1" AcceptedAnswerId="1339" CreationDate="2010-11-27T05:34:06.867" Score="4" ViewCount="297" Body="&lt;p&gt;I recently read that the &lt;a href=&quot;http://www.livescience.com/environment/050825_earthcore.html&quot; rel=&quot;nofollow&quot;&gt;earth's core rotates faster than the surface.&lt;/a&gt;&lt;br&gt;&#xA;Well, firstly, it's easier to digest the concept of planetary bodies, stars, galaxies in rotation and/or orbital motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, what makes a planet's core rotate? And in the earth's case, faster than its surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, I am aware that the core's rotation is what lends the earth its magnetic field but.. what keeps it going in the first place? &lt;/p&gt;&#xA;" OwnerUserId="619" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-28T11:06:14.320" LastActivityDate="2010-11-28T11:06:14.320" Title="What sustains the rotation of earth's core (faster than surface)?" Tags="&lt;classical-mechanics&gt;&lt;magnetohydrodynamics&gt;&lt;geophysics&gt;" AnswerCount="4" />
  <row Id="1337" PostTypeId="2" ParentId="1327" CreationDate="2010-11-27T09:07:12.590" Score="5" ViewCount="" Body="&lt;p&gt;@kalle43,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is such a thing as a law of conservation of energy. And IMHO it is a fundamental law of nature - though its statement admittedly becomes murky in a general relativistic or quantum setting. This remains true no matter what background metric describes your spacetime or whether you are talking about equilibrium or non-equilibrium processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In physics we describe systems, and processes which affect those systems, by understanding the conserved quantities of the system in question. So to answer your question about energy conservation in the context of Hubble expansion, we first have to identify the system we are dealing with and its dynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the presence of homogenous and isotropic matter with stress-energy tensor:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T_{\mu\nu} = diag(-\rho+\kappa\Lambda,p-\kappa\Lambda,p-\kappa\Lambda,p-\kappa\Lambda) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\rho$ and $p$ are the density and pressure of our matter distribution; $\Lambda$ is the cosmological constant and $\kappa = 1/8\pi G$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the ansatz of of a homogenous and isotropic metric $ g = a(t)^2 diag(-1,1,1,1) $, Einstein's equations $ G_{\mu\nu} = \kappa T_{\mu\nu}$ yield the two Friedmann equations. They are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ H^2 = \left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G}{3} \rho + \frac{\Lambda}{3} - \frac{k}{a^2} $ (First  Friedmann Eq.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{\ddot a}{a} = -\frac{4\pi G}{3}(\rho + 3p) + \frac{\Lambda}{3} $ (Second Friedmann Eq.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $k \in {-1,0,1} $ determines whether the spacetime is open ($ k = -1 $), flat ($k=0$) or closed ($k=+1$). For our purposes we can set $k$ to zero. Also for our purposes $p=0$ in the second equation. $ H = \dot a/a $ is the Hubble parameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now taking the time-derivative of the first equation gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2 H \dot H = 2 \left( \frac{\dot a}{a} \right) \left( \frac{\ddot a}{a} - \frac{\dot a^2}{a^2} \right) = \frac{d}{dt}\left( \frac{\kappa}{3} \rho + \frac{\Lambda}{3} -  \frac{k}{a^2} \right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting the r.h.s of the 2nd Friedmann Equation in the lhs of the above expression, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d}{dt}\left( \kappa \rho + \Lambda - \frac{3k}{a^2} \right) = - 3 H \left( \rho + p - \frac{k}{a^2} \right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the statement of energy conservation for our system. For $ H &amp;gt; 0 $ ($H &amp;lt; 0$) the energy density (the r.h.s.) in a given volume is a decreasing (increasing) quantity. This in line with our intuitive expectation regarding an expanding (contracting) cosmology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies for the long-winded answer. You could also have found this material on the relevant wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;nofollow&quot;&gt;page&lt;/a&gt;. I wanted to make my answer self-contained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the Universe as we observe it consists of bound systems - solar systems, galaxies, galaxy clusters - which are seemingly not affected by cosmic expansion. For these cases one needs to do a &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0001056&quot; rel=&quot;nofollow&quot;&gt;little more work&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers,&lt;/p&gt;&#xA;" LastActivityDate="2010-11-27T09:07:12.590" CommentCount="5" />
  <row Id="1338" PostTypeId="2" ParentId="1336" CreationDate="2010-11-27T10:25:03.677" Score="1" ViewCount="" Body="&lt;p&gt;Precisely the inner core is rotating a bit faster than the planet surface.  The outer core, which separates it from the mantle and crust, is a liquid with quite a complex dynamics --mostly probably rotation, convection plus creation of Earth magnetic field. While this all is governed by magnetohydrodynamics it very hard even to simulate the phenomenon, and the obvious lack of experimental data is not helping. In a light of this there is no wonder it does not directly transfers angular speed from inner core to the surface.&lt;br&gt;&#xA;The question what is the origin is even harder; most probably the whole energy of Earth interior is derived from the times of Solar System creation, but as I wrote before how this energy manifests is a  highly nonlinear and difficult to predict.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-27T10:25:03.677" />
  <row Id="1339" PostTypeId="2" ParentId="1336" CreationDate="2010-11-27T11:07:08.023" Score="5" ViewCount="" Body="&lt;p&gt;The same thing that makes the surface rotate: the fact that it always rotated. Because angular momentum is (almost) conserved our planet has no other option than to always rotate. &lt;em&gt;(Of course in reality this picture is complicated by interactions with rest of the universe, but those are just small corrections)&lt;/em&gt;. You can ask then why did it rotate in the first place and this has to do with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Formation_and_evolution_of_the_Solar_System&quot; rel=&quot;nofollow&quot;&gt;origin of the Solar system&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the main question is not what makes Earth rotate but what layers of Earth rotate at what speed. If Earth were a solid body then it would rotate with same angular speed everywhere. But the interior of the Earth is liquid, so the picture is quite complicated. Only thing that is obvious is that every part of Earth must rotate (at some speed or other) because of friction: if one layer (say crust) were rotating and other (say mantle) weren't then after a little while it would start to rotate anyway because the atoms of the moving layer would drag the atoms of the static layer with them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To say anything more than this one would have to consider precise materials and fluid equations in the interior of Earth. Hopefully someone else will come along and complete the picture.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-27T11:26:59.560" LastActivityDate="2010-11-27T11:26:59.560" />
  <row Id="1340" PostTypeId="2" ParentId="639" CreationDate="2010-11-27T11:13:53.917" Score="4" ViewCount="" Body="&lt;p&gt;Logarithmic profile for wind speed regards the bottom part of atmospheric boundary layer (say, about the bottom 100 m, on a boundary layer about 1000 m high). It can be deducted doing some non obvious but reasonable assumptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) Vertical flux of horizontal momentum due to turbulence must be uniform in the lowest part of the atmosphere. Let's consider a reference frame where the average velocity $\overline{u}$ is directed along x axis. Let's decompose velocity in its average and random (turbulent) parts, according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Reynolds_decomposition&quot; rel=&quot;nofollow&quot;&gt;Reynolds decomposition&lt;/a&gt;: x component of velocity is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u = \overline{u} + u&amp;#39;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vertical component is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$w = w&amp;#39;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\overline{u&amp;#39;} = 0$, and $\overline{w&amp;#39;} = 0$, but in general $\overline{u&amp;#39;w&amp;#39;} \neq 0$: u' and w' are covariant.&#xA;Vertical flux of horizontal momentum is given by $\overline{u&amp;#39;w&amp;#39;}$. Thus the first assumption can be expressed as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: $\overline{u&amp;#39;w&amp;#39;} = constant$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) Prandtl hypothesis: random part of horizontal velocity u' is proportional to vertical wind shear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: $u&amp;#39; = l&amp;#39; \frac{\partial \overline{u}}{\partial z}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where l' is the &quot;mixing length&quot;: we can suppose that an air particle maintains its original horizontal speed during its random motion for a length l', before mixing with the surrounding air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C) Vertical length scale of turbulent eddies is comparable to their horizontal length scale, thus the random part of vertical velocity is of the same order as the horizontal one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3: $w&amp;#39; \approx l&amp;#39; \frac{\partial \overline{u}}{\partial z}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using expressions 2 and 3 in 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4: $\overline{u&amp;#39;w&amp;#39;} = \overline{l&amp;#39;^2} \left(\frac{\partial \overline{u}}{\partial z}\right)^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D) At the bottom part of the atmosphere the absolute value of mixing length l' is proportional to high z: this is reasonable because random motion are limited at the bottom by the earth surface. The hypothesis is: $(|l&amp;#39;| = kz)$ where k is Kàrmàn constant. Substituting l' in expression 4 we obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5: $\overline{u&amp;#39;w&amp;#39;} = (kz)^2 \left(\frac{\partial \overline{u}}{\partial z}\right)^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extracting the square root of 4 and separating the variables we obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6: $d\overline{u} = \frac{\sqrt{\overline{u&amp;#39;w&amp;#39;}}}{k} \frac{dz}{z}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating we obtain the logarithmic profile:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta \overline{u} = \frac{\sqrt{\overline{u&amp;#39;w&amp;#39;}}}{k} \log{\frac{z}{z_0}}$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2010-11-27T11:13:53.917" CommentCount="2" />
  <row Id="1341" PostTypeId="2" ParentId="1336" CreationDate="2010-11-27T11:14:47.590" Score="1" ViewCount="" Body="&lt;p&gt;According to our current understanding of the formation of planets, they are created from dust which originates in a previous supernova (or other large) explosion. This is called a nebula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nebular_hypothesis&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Nebular_hypothesis&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this nebula contracts into planets, there are generally two possible cases: that the dust that is eventually going to form a planet is spinning, overall, or not. The most general and likely case is that it is spinning in some form. There is no overall reason why it shouldn't, short of some &quot;magical&quot; combination of canceling factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens when this spinning dust cloud contracts under gravity? It spins faster (think of an ice skater spinning), due to the conservation of angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, you end up with planets with a relatively fast spinning core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what keeps this spin going? Fundamentally, conservation of angular momentum. Nothing is keeping it going, and it's actually spinning down. It should eventually stop - in million or billions of years. That's how much momentum is stored in it :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the crust spin slower than the core? For two reasons: since things spin faster as they contract, you naturally end up with a variation of speed between the outer layers and the inner core; secondly, the core is actually dragging the crust around, but the Earth is not really a solid so the drag cannot be thought of as a rigid motion. &lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-27T21:01:14.580" LastActivityDate="2010-11-27T21:01:14.580" CommentCount="7" />
  <row Id="1342" PostTypeId="2" ParentId="1327" CreationDate="2010-11-27T11:48:15.590" Score="2" ViewCount="" Body="&lt;p&gt;Expansion is found between galaxies and thought as a uniform law in the universe. But it doesn't mean that everything is expanding, especially not for gravitational bounded systems (&lt;a href=&quot;http://nedwww.ipac.caltech.edu/level5/Peacock/frames.html&quot; rel=&quot;nofollow&quot;&gt;Peacock &quot;Cosmological physics&quot; 3.3&lt;/a&gt; ).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;THE NATURE OF THE EXPANSION An inability to see that the expansion is locally just kinematical also lies at the root of perhaps the worst misconception about the big bang. Many semi-popular accounts of cosmology contain statements to the effect that ``space itself is swelling up'' in causing the galaxies to separate. This seems to imply that all objects are being stretched by some mysterious force: are we to infer that humans who survived for a Hubble time would find themselves to be roughly four metres tall? Certainly not. Apart from anything else, this would be a profoundly anti-relativistic notion, since relativity teaches us that properties of objects in local inertial frames are independent of the global properties of spacetime. If we understand that objects separate now only because they have done so in the past, there need be no confusion. A pair of massless objects set up at rest with respect to each other in a uniform model will show no tendency to separate (in fact, the gravitational force of the mass lying between them will cause an inward relative acceleration). In the common elementary demonstration of the expansion by means of inflating a balloon, galaxies should be represented by glued-on coins, not ink drawings (which will spuriously expand with the universe). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So conservation of energy can still be used to study the evolution of solar system, Galactic system, galaxy clusters etc. But the WHOLE universe is something different. What we observed is only a part of it, we can not simply take it as the whole vision. Since the age of the universe is limited (13.7Gyr), there may be much more objects outside our sight. Then all the conservation laws of a closed system can not be used.&lt;/p&gt;&#xA;" OwnerUserId="518" LastEditorUserId="518" LastEditorDisplayName="" LastEditDate="2010-11-27T13:24:08.403" LastActivityDate="2010-11-27T13:24:08.403" />
  <row Id="1343" PostTypeId="1" CreationDate="2010-11-27T11:51:25.403" Score="24" ViewCount="1323" Body="&lt;p&gt;Why does dust stick to rotating fan propeller?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, most people (including I) think of the dust will not stick to rotating fan propellers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for the great explanations. I am still waiting for the &quot;better&quot; one if any.&lt;/p&gt;&#xA;" OwnerUserId="652" LastEditorUserId="652" LastEditorDisplayName="" LastEditDate="2011-07-10T04:50:21.887" LastActivityDate="2011-07-10T04:50:21.887" Title="Why does dust stick to rotating fan propeller?" Tags="&lt;fluid-dynamics&gt;&lt;rotational-dynamics&gt;&lt;electrostatics&gt;" AnswerCount="9" CommentCount="4" FavoriteCount="7" />
  <row Id="1344" PostTypeId="2" ParentId="1343" CreationDate="2010-11-27T12:14:19.717" Score="6" ViewCount="" Body="&lt;p&gt;&lt;em&gt;First taking only speed of the fan into the account.&lt;/em&gt; If the fan rotates slowly then the situation is obviously not very much different from if it weren't rotating at all. The centrifugal force on the dust particles is not big enough to throw them away off the fan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, there's static electricity that has to be taken into the account. It's perfectly possible that some residual charge is generated on the fan (this depends very much on what is the fan made off) and as the dust particles are often charged the ones with correct polarity will get attracted to the fan. In this case they would stick even if fan were rotating very fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you can test whether the second option is realized in your case by touching the fan to discharge it and it should &quot;kick&quot; you a little. Or if you don't like that you can bring some charged object close to the fan and see whether it is affected.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-11-27T12:14:19.717" />
  <row Id="1345" PostTypeId="1" AcceptedAnswerId="1346" CreationDate="2010-11-27T12:40:11.890" Score="4" ViewCount="353" Body="&lt;p&gt;If the light velocity is a vector quantity, why vector addition cannot be applied to it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or the light velocity is not a vector quantity?&lt;/p&gt;&#xA;" OwnerUserId="652" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-11-27T18:05:04.353" LastActivityDate="2011-02-06T23:10:45.510" Title="If the light velocity is a vector quantity, why vector addition cannot be applied to it?" Tags="&lt;special-relativity&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="1346" PostTypeId="2" ParentId="1345" CreationDate="2010-11-27T12:55:02.897" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Update 2:&lt;/strong&gt;  It might also be useful add the relation to usual Galilean physics. This will probably only make sense after reading &lt;em&gt;Update 1&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that one can parametrize boosts in (1+1) by $[\cosh(\eta), \sinh(\eta)]$. As it so happens this equals to $[\gamma, \gamma {v \over c}]$. The classical physics correspond to the asymptotic case $c \to \infty$ (i.e. no limit on the speed of light). So this reduces to $\gamma \to 1$, ${v \over c} \to 0$ and that means $\eta \to 0$. So in Galilean case all boosts degenerate to trivial transformation and this is why time and space separate and addition of velocities starts to work.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 1:&lt;/strong&gt; having seen KennyTM's strange answer I decided to add some notes to this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, there exists a concept of &lt;strong&gt;rapidity&lt;/strong&gt;. This is a natural variable for parametrization of &lt;a href=&quot;http://en.wikipedia.org/wiki/Lorentz_boost#Lorentz_transformation_for_frames_in_standard_configuration&quot; rel=&quot;nofollow&quot;&gt;boosts&lt;/a&gt;. First consider a circle. Why circle? Because it has to do with rotations and rotations are very similar to boosts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Circle is an object given by an equation $x^2 + y^2 = 1$. You can decide that you'll parametrize (part of) it by coordinates $[x, \sqrt{1 - x^2}]$. Now what if you want to rotate the circle by some angle? Can you figure out how will the parametrization change? Maybe you can but I assure you this is not pretty. But there is a nicer way. Let's try parametrizing the circle by an angle $\phi$ so that it would become set of points $[\cos(\phi), \sin(\phi)]$ and now the rotation by angle $\psi$ corresponds to parametrization $[\cos(\phi + \psi), \sin(\phi + \psi)]$. So our parameter is additive! You won't find any better parametrization of the circle than this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, so we understand circles a little better now. But as already said, rotation in two space dimensions is almost the same thing as boost in (1+1) space-time dimensions. In the same way that rotations (around origin) preserve circles, boosts preserve hyperbolas. So instead of working with $\sin$ and $\cos$ you'll work with hyperbolic functions $\sinh$ and $\cosh$ and instead of $\phi$ you'll obtain &lt;strong&gt;rapidity&lt;/strong&gt; $\eta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this only works this nicely in (1+1)-dimensions. In (3+1) you'll have many more interesting effects (similarly to like rotations are strange beasts in 3 dimensions as opposed in 2). But it's still true that like the general 3-dimensional rotations are nicely parametrized by an axis and rotation angle, the boosts are nicely parametrized by the direction of boost and rapidity. So if you perform two rotations about the same axis it's the same as rotation by a sum of the angles and if you perform two boosts in one direction it's the same as doing a boost with added rapidities.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'll assume you're talking about &lt;a href=&quot;http://en.wikipedia.org/wiki/Galilean_invariance&quot; rel=&quot;nofollow&quot;&gt;Galilean principle of relativity&lt;/a&gt; whereby the velocities transform by pure addition. This concept breaks down when the speeds one is dealing with are too large. Speed of light is an extreme case of such speed. Then one has to use Special Relativity and instead consider &lt;a href=&quot;http://en.wikipedia.org/wiki/Four-vector&quot; rel=&quot;nofollow&quot;&gt;four-vectors&lt;/a&gt; transforming by &lt;a href=&quot;http://en.wikipedia.org/wiki/Lorentz_transformation&quot; rel=&quot;nofollow&quot;&gt;Lorentz transformation&lt;/a&gt;. Now, this transformations preserve the &lt;em&gt;Minkowski length&lt;/em&gt; of four-vectors (in the same way that rotations preserve length of usual vectors).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that velocity of light corresponds to &lt;strong&gt;zero&lt;/strong&gt; &lt;em&gt;Minkowski length&lt;/em&gt; and so light moves at the speed of light in every inertial frame. This is the famous Einstein's postulate.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-30T21:00:09.820" LastActivityDate="2010-11-30T21:00:09.820" CommentCount="4" />
  <row Id="1347" PostTypeId="2" ParentId="1336" CreationDate="2010-11-27T16:05:31.290" Score="4" ViewCount="" Body="&lt;p&gt;The earth's rotation is slowing down, due primarily to tidal interaction with the moon (transfering angular momentum to the moon). This frictional loss occurs near the surface. On this basis, if the rotational coupling of the inner core to the mantle is imperfect, it would be expected to lag with respect to the slowdown. I think precessional changes (cycle length roughly 23000 years) are probably more important for generating rotational mistmatches. Then it all gets mixed up with/by&#xA;thermal convection and magnetic fields as well.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-11-27T16:05:31.290" CommentCount="2" />
  <row Id="1349" PostTypeId="2" ParentId="1343" CreationDate="2010-11-27T17:08:40.450" Score="2" ViewCount="" Body="&lt;p&gt;I'm guessing that a very thin layer of oil covers the blade, maybe because of proximity to the oiled bearings, making it a little sticky.   The blade would pick up excess dust because it moves through more air than if it were stationary.    According to this hypothesis, static electricity is not involved.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-11-27T17:08:40.450" CommentCount="2" />
  <row Id="1350" PostTypeId="2" ParentId="1343" CreationDate="2010-11-27T17:10:28.257" Score="18" ViewCount="" Body="&lt;p&gt;The reason is that you have a boundary layer on the surface of the blade of the fan. On the frame of the blade (the blade moves with some velocity, but at the frame of the blade the air moves) the boundary layer starts from the surface of the blade where the fluids velocity is zero and as you move away from the blade, the velocity increases up to the value of the velocity of the blade (you can call that the undisturbed velocity of the flow). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you have some fine dust, it actually doesn't feel much wind and it can't be blown away. Static electricity could be an other factor, but you can see that on metallic propellers also.&lt;/p&gt;&#xA;" OwnerUserId="383" LastEditorUserId="383" LastEditorDisplayName="" LastEditDate="2010-11-27T17:17:03.820" LastActivityDate="2010-11-27T17:17:03.820" CommentCount="4" />
  <row Id="1351" PostTypeId="2" ParentId="1334" CreationDate="2010-11-27T17:28:36.993" Score="2" ViewCount="" Body="&lt;p&gt;I believe potentials in Maxwell's equations were introduced originally to make solving equations simpler -- it wasn't until a bit later that the Lorentz symmetry was noticed. Similar potentials are introduced in 2D hydrodynamics as well, see e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Stream_function&quot; rel=&quot;nofollow&quot;&gt;stream function&lt;/a&gt;.  So I would say that the reason for introducing these potentials originally was always to simplify the algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, whenever one has some nice trick like this, there's a deeper reason.  Here the &quot;deeper reason&quot; for introducing various potentials has to do with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Helmholtz_decomposition&quot; rel=&quot;nofollow&quot;&gt;Helmholtz decomposition&lt;/a&gt; (even deeper, coming from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hodge_decomposition#Hodge_decomposition&quot; rel=&quot;nofollow&quot;&gt;Hodge decomposition&lt;/a&gt;), which guarantees that vector fields can be written as a sum of a curl-free vector field and a divergence-free vector field.  This decomposition &quot;plays nicely&quot; with wave equations in the sense that there are natural wave modes, transverse (divergence-free) and longitudinal (curl-free) which propagate independently.  See the discussion of &lt;a href=&quot;http://en.wikipedia.org/wiki/Helmholtz_decomposition#Longitudinal_and_transverse_fields&quot; rel=&quot;nofollow&quot;&gt;transverse and longitudinal&lt;/a&gt;, i.e. the application of the Helmholtz decomposition to a Fourier transform of a vector field, which obviously is highly applicable to the P (longitudinal) and S (transverse) waves in seismology.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, the idea of the Helmholtz decomposition is a bit like breaking a vector into independent components, and then calculating what happens to each separately, except now one is working with vector fields.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-27T17:28:36.993" />
  <row Id="1352" PostTypeId="2" ParentId="1343" CreationDate="2010-11-27T18:00:19.977" Score="11" ViewCount="" Body="&lt;p&gt;What about this hypothesis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dust sticks everywhere, but since the propeller cuts through a lot of air, it meets more dust particles. Thus, more dust sticks to the propeller than elsewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Evidence&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I (Mark) took photos my the fan my room to support Damien's hypothesis.  The first photo is of the leading edge of the fan blade, which impacts a lot of air, and the second photo is of the trailing edge of the same fan blade.  I've never cleaned this fan.  The leading edge is covered in a thick, 3-5mm layer of dust, while the trailing edge is almost clean.&#xA;&lt;img src=&quot;http://i.stack.imgur.com/M5yUG.jpg&quot; alt=&quot;leading edge of fan blade covered in dust&quot;&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/r7YOb.jpg&quot; alt=&quot;trailing edge of fan blade nearly clean&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="657" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-12-07T06:22:27.063" LastActivityDate="2010-12-07T06:22:27.063" CommentCount="4" />
  <row Id="1353" PostTypeId="1" CreationDate="2010-11-27T18:21:12.473" Score="10" ViewCount="334" Body="&lt;p&gt;First, let me state that I'm a lot less experienced with physics than most people here. Quantum mechanics was as far as I got and that was about 9 years ago, with no use in the meantime. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of people seem to think that the act of observing the universe we are changing the universe. This is the standard woo hand waving produced by people who don't understand the physics -- thank you Fritjof Capra. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reading about quantum physics out there on the internet, I find this idea is propagated far too often. Often enough that I question whether my understanding is accurate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does observing a particle always mean hitting it with something to collapse the waveform? Is there any other type of observation? What do physicists mean when they say &quot;observe&quot;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related, doesn't get to the question of what observation is: &lt;a href=&quot;http://physics.stackexchange.com/questions/27/what-is-the-difference-between-a-measurement-and-any-other-interaction-in-quantum&quot;&gt;What is the difference between a measurement and any other interaction in quantum mechanics?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="656" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-11-28T18:03:56.343" LastActivityDate="2010-11-29T04:40:18.263" Title="Is there a difference between observing a particle and hitting it with another particle?" Tags="&lt;quantum-mechanics&gt;&lt;measurement&gt;&lt;epistemology&gt;&lt;measurement-problem&gt;" AnswerCount="5" FavoriteCount="4" />
  <row Id="1354" PostTypeId="1" AcceptedAnswerId="1356" CreationDate="2010-11-27T18:25:55.297" Score="2" ViewCount="356" Body="&lt;p&gt;What are the best examples in practical life&#xA;of an energy-entropy competition which favors entropy over energy?&#xA;My initial thought is a clogged drain -- too unlikely for the&#xA;hair/spaghetti to align itself along the pipe -- but this is probably&#xA;far from an optimal example.  Curious to see what you got.  Thanks.&lt;/p&gt;&#xA;" OwnerUserId="20" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-28T11:07:24.363" LastActivityDate="2010-12-25T00:27:46.450" Title="Best example of energy-entropy competition?" Tags="&lt;soft-question&gt;&lt;entropy&gt;" AnswerCount="8" CommentCount="4" />
  <row Id="1355" PostTypeId="2" ParentId="1353" CreationDate="2010-11-27T18:43:19.417" Score="1" ViewCount="" Body="&lt;p&gt;The thing is there are a variety of different opinions that, since they can not be distinguished by experiment, are around and used by different people to interpret experiments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conventional view of quantum mechanics, although it has eroded over time, is that a sharp disctinction has to be made between the classical and the quantum. The apparatus has to be described classicaly, while the quantum describes the measurement results of the experiment. Von Neumann has then tried to show that the distinction need not be sharp and that you can include the apparatus in the quantum description, but it then has to be observed itself by another apparatus which has to be described classically. Wigner argued that this regression of the quantum/classical divide can be carried up to the mind, hence why there is a lot of woo latching onto these ideas, because they seem to justify the importance of the human mind over anything else in the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other approaches have argued that there is no distinction between classical and quantum, at least. One is the Many Worlds Interpretation, which states that a superposition of states is representing actually realized states but in different universes. Another is the Bohmian or pilotwave interpretation, which states that the wave equation is describing a wave that guides particles. An extra equation is then supplemented to the Schrödinger equation to show how this guiding happens. In both theories, there is no need to speak about measurement, at least, not in any deeper sense than what would be done in classical physics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Interpretations_of_quantum_mechanics&quot; rel=&quot;nofollow&quot;&gt;Here's a non-exhaustive list of interpretations of quantum mechanics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, within the context of the Copenhagen interpretation and the von Neumann/Wigner paradigm, the answer to the title question would be yes, there is a difference between measurement and hitting with another particle. Within the context of Bohmian or MWI interpretations, the answer would be no.&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2010-11-27T20:11:29.883" LastActivityDate="2010-11-27T20:11:29.883" />
  <row Id="1356" PostTypeId="2" ParentId="1354" CreationDate="2010-11-27T20:27:08.873" Score="4" ViewCount="" Body="&lt;p&gt;One example I know is so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Granular_convection&quot; rel=&quot;nofollow&quot;&gt;Brazil nut effect&lt;/a&gt;. When you place balls in a container of two different sizes and shake it, the larger ones will go up (even if they are denser than the smaller ones). So the final energy of system after introducing noise is clearly greater than the initial. I believe that the phenomena needs to be entropy-driven. However, I don't know a proof. &lt;/p&gt;&#xA;" OwnerUserId="184" LastActivityDate="2010-11-27T20:27:08.873" CommentCount="3" />
  <row Id="1357" PostTypeId="2" ParentId="1354" CreationDate="2010-11-27T20:36:22.553" Score="6" ViewCount="" Body="&lt;p&gt;The air in this room.  Because of gravity, the lowest energy configuration is clearly one in which all the molecules lie on the floor.  But from the entropic point of view they should be exploring all of their phase space and bouncing around the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to say also that examples such as the clogged drain and so on should be viewed purely as analogies.  Entropy is a real physical quantity that can be calculated.  As an actual physics problem, the calculation of the entropy of the drain + hair + water system would be a monster and there is no telling what the result will be! Is it even in equilibrium?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Brazil nut effect cited above is a non-equilibrium effect, so although it is interesting I don't think it necessarily means anything about energy-entropy competition.  The balance of energy and entropy happens when a system (at constant temperature for example) reaches equilibrium, minimizing its free energy.  The shaken Brazil nuts are a non-equilibrium problem, and thus are not minimizing their free energy! But very interesting nevertheless, all the more so since it is non-equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="266" LastActivityDate="2010-11-27T20:36:22.553" CommentCount="10" />
  <row Id="1358" PostTypeId="2" ParentId="1323" CreationDate="2010-11-27T20:40:49.340" Score="4" ViewCount="" Body="&lt;p&gt;I googled &lt;em&gt;&quot;Madelung Transformation&quot;&lt;/em&gt; in German and found the following answer in &lt;a href=&quot;http://www.wvsgym.de/faecher/Physik/langfassung_jufo_08.pdf&quot; rel=&quot;nofollow&quot;&gt;an article about the theory of super conductivity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Schrödinger equation is an axiom of quantum physics that is difficult to interpret i.a. because of its usage of complex numbers:&#xA;$i\hbar \frac{\partial}{\partial t}\Psi = \mathcal{H} \Psi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Kopenhagen interpretation $\Psi\Psi^*$ is known to describe the probability distribution of a particle. Erwin Madelung tried (already in 1926) to understand the nature of the Schrödinger equation from a different point of view. He substituted $\Psi = ae^{i\phi}$ and split the Schrödinger equation in its real part and its imaginary part (first and second Madelung equation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $\Psi\Psi^* = a^2$, the second Madelung equation (see above reference) becomes a continuity equation of the probability density $a^2$. This equation can thus be interpreted in a more physical way than the Schrödinger equation and &lt;em&gt;that's the point&lt;/em&gt;. Similar arguments hold for the first equation but these arguments are more involved (see above reference).&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-07-08T14:31:14.650" LastActivityDate="2011-07-08T14:31:14.650" CommentCount="6" />
  <row Id="1359" PostTypeId="2" ParentId="1354" CreationDate="2010-11-27T21:30:35.093" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Black_body_radiation&quot; rel=&quot;nofollow&quot;&gt;Blackbody radiation&lt;/a&gt;&lt;/strong&gt;: anything hotter than its environment radiates energy thus increasing the entropy of the universe. Entropy wins :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Sun&lt;/strong&gt; :-) The Sun's energy does not increase the Earth's total energy!&#xA;In fact, the Earth radiates almost exactly the same amount of energy as it receives from the Sun. What we &lt;em&gt;really&lt;/em&gt; gain from the Sun is that we use the sun rays' low entropy to power life on Earth, and the Earth radiates high entropy microwaves in the night.&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-28T10:03:58.383" LastActivityDate="2010-11-28T10:03:58.383" CommentCount="15" />
  <row Id="1360" PostTypeId="2" ParentId="1354" CreationDate="2010-11-27T21:36:25.763" Score="0" ViewCount="" Body="&lt;p&gt;Read as comment to Piotr's answer (not enough points to comment).  I believe that &quot;Brazil nut simulations&quot; have been done with &quot;Brazil nuts&quot; that are denser (i.e. lead), as well as larger, than the other nuts, and they still rise to the top. I think they do that in this article:&#xA;&lt;a href=&quot;http://prl.aps.org/abstract/PRL/v58/i10/p1038_1&quot; rel=&quot;nofollow&quot;&gt;http://prl.aps.org/abstract/PRL/v58/i10/p1038_1&lt;/a&gt;&#xA;Packing is of course important, since it keeps the lead nuts from falling to the bottom!&lt;/p&gt;&#xA;" OwnerUserId="266" LastActivityDate="2010-11-27T21:36:25.763" />
  <row Id="1361" PostTypeId="1" AcceptedAnswerId="1364" CreationDate="2010-11-27T21:38:05.727" Score="14" ViewCount="372" Body="&lt;p&gt;Scattering of light by light does not occur in the solution of Maxwell's equations (since they are linear and EM waves obey superposition), but it is a prediction of QED (the most significant Feynman diagrams have a closed loop of four electron propagators).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has scattering of light by light in vacuum ever been observed, for photons of any energy? If not, how close are we to such an experiment?&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-01-15T22:44:39.247" Title="Scattering of light by light: experimental status" Tags="&lt;quantum-field-theory&gt;&lt;light&gt;&lt;scattering&gt;&lt;experimental-evidence&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="1362" PostTypeId="2" ParentId="1361" CreationDate="2010-11-27T21:40:35.910" Score="4" ViewCount="" Body="&lt;p&gt;I've a vague memory that light-on-light scattering puts an upper limit on the energy of photons that will propagate over cosmic distance scales. So one probe is very high energy gamma astronomy.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2010-11-27T21:40:35.910" CommentCount="3" />
  <row Id="1363" PostTypeId="2" ParentId="1354" CreationDate="2010-11-27T21:45:30.177" Score="0" ViewCount="" Body="&lt;p&gt;General question/comment (again, not enough points).  What exactly (quantitatively) is meant when we say that &quot;energy-entropy competition favors entropy&quot;? In the appropriate ensemble, entropy and energy are always competing.  In equilibrium a situation is reached that balances them exactly.  Who is to say who &quot;won&quot;? If instead we are talking about a non-equilibrium situation, then what you are asking is for a process where entropy increases.  Then there are plenty of examples - see for example Sklivvz's answer.  Similarly, take any example of heat conduction from a hot object to a cold one, etc.&lt;/p&gt;&#xA;" OwnerUserId="266" LastActivityDate="2010-11-27T21:45:30.177" />
  <row Id="1364" PostTypeId="2" ParentId="1361" CreationDate="2010-11-27T22:05:21.397" Score="9" ViewCount="" Body="&lt;p&gt;This was demonstrated by &quot;&lt;a href=&quot;http://www.slac.stanford.edu/exp/e144/e144.html&quot; rel=&quot;nofollow&quot;&gt;Experiment 144&lt;/a&gt;&quot; at SLAC in 1997.  Here is a &lt;a href=&quot;http://www.slac.stanford.edu/exp/e144/papers.html&quot; rel=&quot;nofollow&quot;&gt;list of publications&lt;/a&gt; from that project, for instance &quot;&lt;a href=&quot;http://prl.aps.org/abstract/PRL/v79/i9/p1626_1&quot; rel=&quot;nofollow&quot;&gt;Positron Production in Multiphoton Light-by-Light Scattering&lt;/a&gt;&quot;, whose abstract reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A signal of 106±14 positrons above background has been observed in collisions of a low-emittance 46.6 GeV electron beam with terawatt pulses from a Nd:glass laser at 527 nm wavelength in an experiment at the Final Focus Test Beam at SLAC. The positrons are interpreted as arising from a two-step process in which laser photons are backscattered to GeV energies by the electron beam followed by a collision between the high-energy photon and several laser photons to produce an electron-positron pair. &lt;strong&gt;These results are the first laboratory evidence for inelastic light-by-light scattering involving only real photons&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2010-11-27T22:12:54.747" LastActivityDate="2010-11-27T22:12:54.747" CommentCount="3" />
  <row Id="1365" PostTypeId="2" ParentId="1324" CreationDate="2010-11-27T23:10:56.973" Score="1" ViewCount="" Body="&lt;p&gt;[Some very nice answers by Eric, Sivaram and Piotr above. Here's my take!]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short answer: NO !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The notion of $C^\infty$ is a mathematical aberration that was conjured up to help smooth (pun intended) discussions in real analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, remember, you asked &quot;Is the &lt;strong&gt;world&lt;/strong&gt; $C^\infty$?&quot;. By &quot;world&quot; I take it to mean the physical world around us, our notions of which are based on what we can observe. A &lt;strong&gt;physical observable&lt;/strong&gt; which is infinitely differentiable, would require an infinite number of measurements to determine the value of that observable in a given region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the consensus is emerging that information is the underlying substrate of the Universe (in the various forms of the holographic principle), it becomes even more urgent to reject a notion of $C^\infty$ observables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note how I have stressed the words &quot;physical observables&quot; rather than functions or mathematical entities that are used as intermediaries to compute any measured quantity. This is in harmony with Eric's statement that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is also possible that smooth spacetime is some aggregate/thermodynamic approximation of discrete microstates of spacetime -- but our model of that discrete system will probably be described by the mathematics of continuous functions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" LastActivityDate="2010-11-27T23:10:56.973" CommentCount="2" />
  <row Id="1366" PostTypeId="2" ParentId="1353" CreationDate="2010-11-27T23:35:29.353" Score="3" ViewCount="" Body="&lt;p&gt;&quot;Collapse the waveform&quot; is a loaded term, that would not be agreed to by all physicists. There are a great many &quot;no-collapse&quot; interpretations out there in which there is no special role for measurement that directly alters the wavefunction. There are also collapse-type interpretations in which the collapse happens more or less spontaneously, as in Roger Penrose's theory whereby gravitational effects cause any superposition above a certain mass threshold to collapse incredibly quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a practical matter, it's hard to think of a measurement technique that couldn't be described as hitting one particle with another. Most quantum optical experiments rely on scattering light off an atom in order to detect the state of the atom, many charged-particle experiments involve running the particles into a surface or a wire in order to detect it, and so on. I think that the solid-state qubit experiment done by people like Rob Schoelkopf at Yale would probably count as an exception, because I believe they use a SQUID to detect the state of their artificial atoms via magnetic fields. If you want to get really picky, though, you could probably consider that a particle interaction as well, though, in some QED sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even there, though, the act of measurement does not leave the initial system unchanged. While there would not be general agreement with the specific phrasing &quot;observation changes the universe,&quot; the idea that quantum systems behave differently after a measurement is central to the theory, and can't be avoided.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-27T23:35:29.353" />
  <row Id="1367" PostTypeId="2" ParentId="484" CreationDate="2010-11-28T04:12:07.987" Score="2" ViewCount="" Body="&lt;p&gt;You are right in that you don't need recombination to decouple and that Hubble expansion is an important part of decoupling.  Decoupling basically happens when the interaction rate drops to the Hubble expansion: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$  \Gamma = H$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in terms of the cross section: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Gamma = n_X &amp;lt;\sigma v&amp;gt;$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n_X$ is the number density, $\sigma$ is the interaction cross section, and $v$ is the speed of the particle.  So, if there is expansion the cross section need not vanish for decoupling.  Also, even if the free electron number density didn't drop precipitously due to recombination, the photons would decouple eventually.  See Modern Cosmology by Dodelson pg. 73 for example (available in Google books).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dodelson states the free electron fraction drops to about 2% to get decoupling, so in some sense recombination had to proceed partially (though not to completion) to get decoupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this answers your question.&lt;/p&gt;&#xA;" LastActivityDate="2010-11-28T04:12:07.987" />
  <row Id="1368" PostTypeId="1" AcceptedAnswerId="1370" CreationDate="2010-11-28T04:17:47.130" Score="10" ViewCount="267" Body="&lt;p&gt;If the velocity is a relative quantity, will it make inconsistent equations when applying it to the conservation of energy equations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the train moving at $V$ relative to ground, there is an object moving at $v$ relative to the frame in the same direction the frame moves. Observer on the ground calculates the object kinetic energy as $\frac{1}{2}m(v+V)^2$. However, another observer on the frame calculates the energy as $\frac{1}{2}mv^2$. When each of these equations is plugged into the conservation of energy, they will result in 2 distinct results (I think).&lt;/p&gt;&#xA;" OwnerUserId="652" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-28T11:03:32.573" LastActivityDate="2010-11-28T23:04:38.010" Title="Is kinetic energy a relative quantity? Will it make inconsistent equations when applying it to the conservation of energy equations?" Tags="&lt;classical-mechanics&gt;&lt;special-relativity&gt;&lt;energy-conservation&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="1369" PostTypeId="2" ParentId="1368" CreationDate="2010-11-28T04:31:07.133" Score="1" ViewCount="" Body="&lt;p&gt;You should remain in one frame of reference when applying the law of conservation of energy. Then you should be fine.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2010-11-28T04:31:07.133" CommentCount="1" />
  <row Id="1370" PostTypeId="2" ParentId="1368" CreationDate="2010-11-28T05:08:25.027" Score="13" ViewCount="" Body="&lt;p&gt;Yes, kinetic energy is a relative quantity. As you might guess, this means that when you're using energy conservation, you have to stay within a single frame of reference; all that energy conservation tells you is that the amount of energy as measured in any &lt;em&gt;one&lt;/em&gt; frame stays the same over time. You can't meaningfully compare the amount of energy measured in frame A (e.g. the ground) to the amount of energy measured in frame B (e.g. the train).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you can &lt;em&gt;convert&lt;/em&gt; an amount of kinetic energy measured in one frame to another frame, if you know their relative velocity. If you're working at low speeds, the easy (approximate) way to do this is to just calculate the relative velocity, as you did. So if the train observer measures a kinetic energy $K = \frac{1}{2}mv^2$, the ground observer will measure a kinetic energy of $\frac{1}{2}m(v + V)^2$, or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K + \sqrt{2Km}V + \frac{1}{2}mV^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(in one dimension).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you get up to higher speeds, or you want an exact expression, you'll have to use the relativistic definition of energy. In special relativity, the kinetic energy is given by the difference between the total energy and the &quot;rest energy,&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K = E - mc^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to figure out the transformation rule is to use the fact that the total energy is part of a four-vector, along with the relativistic momentum,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix}E/c \\ p\end{pmatrix} = \begin{pmatrix}\gamma_v mc \\ \gamma_v mv\end{pmatrix}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\gamma_v = 1/\sqrt{1 - v^2/c^2}$. This four-vector transforms under the Lorentz transformation as you shift from one reference frame to another,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix}E/c \\ p\end{pmatrix}_\text{ground} = \begin{pmatrix}\gamma &amp;amp; \gamma\beta \\ \gamma\beta &amp;amp; \gamma\end{pmatrix}\begin{pmatrix}E/c \\ p\end{pmatrix}_\text{train}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $\beta = V/c$ and $\gamma = 1/\sqrt{1 - \beta^2}$), so the energy as observed from the ground would be given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_\text{ground} = \gamma(E_\text{train} + \beta c p_\text{train})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinetic energy is obtained by subtracting $mc^2$ from the total energy, so you'd get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\text{ground} = \gamma(E_\text{train} + \beta c p_\text{train}) - mc^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which works out to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\text{ground} = \gamma K_\text{train} + (\gamma - 1) mc^2 + \gamma\beta c p_\text{train}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $K$ is the relativistic kinetic energy and $p$ is the relativistic momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted it in terms of energy alone:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\text{ground} = \gamma K_\text{train} + (\gamma - 1) mc^2 + \gamma\beta\sqrt{K_\text{train}^2 + 2 mc^2 K_\text{train}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might start to notice a similarity to the non-relativistic expression above ($K + \sqrt{2Km}V + \frac{1}{2}mV^2$), and indeed, if you plug in some approximations that are valid at low speeds ($\gamma \approx 1$, $\gamma - 1 \approx V^2/c^2$, $K_\text{train} \approx \frac{1}{2}mv^2 \ll mc^2$), you will recover exactly that expression.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-28T23:04:38.010" LastActivityDate="2010-11-28T23:04:38.010" />
  <row Id="1371" PostTypeId="2" ParentId="1354" CreationDate="2010-11-28T06:39:10.803" Score="1" ViewCount="" Body="&lt;p&gt;Well, as a physicist working mainly in the field of statistical physics, I obviously have to mention Ising model. In this case the model is actually tractable (at least in two dimensions) and can tell us a whole lot about (not just) energy-entropy battle. It's obvious that the ground state (for ferromagnetic case) is all spins pointing one way (say up). Now, if you point some spins the other way then you are losing in terms of energy (something like number of neighbors times number of &lt;em&gt;wrong&lt;/em&gt; spins) but actually you gain hugely in entropy (because of translational invariance of lattice models). This argument can be made very precise when working with polymer model (which is isomorphic to Ising model) and considering low-temperature cluster expansion.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am sorry, but I am not really able to provide nice references. Wikipedia articles are pretty bad. Maybe I should spend some time bringing them up to the current knowledge (that is to say, knowledge since like 1970s) about cluster expansion. For now, if anyone is interested, just read the &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.cmp/1104114796&quot; rel=&quot;nofollow&quot;&gt;basic paper&lt;/a&gt; on the topic.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-11-28T06:39:10.803" />
  <row Id="1372" PostTypeId="1" CreationDate="2010-11-28T08:26:06.043" Score="3" ViewCount="351" Body="&lt;p&gt;Let's assume that there is nothing in the universe except Earth. If the Earth rotates on its axis as it does, then would we experience the effects of rotational motion like centrifugal force and Coriolis force?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The meaning of my question is: is Earth rotating relative to space?&lt;/p&gt;&#xA;" OwnerUserId="567" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-11-28T11:02:04.587" LastActivityDate="2010-11-30T16:12:54.777" Title="Is rotational motion relative to space?" Tags="&lt;classical-mechanics&gt;&lt;astronomy&gt;" AnswerCount="5" CommentCount="0" />
  <row Id="1373" PostTypeId="2" ParentId="1372" CreationDate="2010-11-28T10:01:08.680" Score="2" ViewCount="" Body="&lt;p&gt;The Earth is rotating relative to its axis. So, yes, the effects would be there - &lt;a href=&quot;http://en.wikipedia.org/wiki/Foucault_pendulum&quot; rel=&quot;nofollow&quot;&gt;Focault's pendulum&lt;/a&gt; would still precede, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_rotation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Absolute_rotation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-28T10:01:08.680" />
  <row Id="1374" PostTypeId="2" ParentId="1372" CreationDate="2010-11-28T11:21:55.787" Score="-2" ViewCount="" Body="&lt;p&gt;The statement &quot;The Earth is rotating relative to its axis&quot; is circular, what defines the axis is the rotation, and without anything else in universe to observe the rotation of earth by, is earth really rotating? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The assumption of a universe with only earth in it should imply that the space and time and anything else are also bounded to the surface of the earth. Now in such a universe Earth is  rotating respective to what?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If space is necessary as a medium for earth to rotate in, in a universe that there is nothing except earth, is space also part of that universe? What about dark matter, dark energy and anything else that could be the cause of all the &quot;rotational effects&quot; in a the normal universe? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and according to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_rotation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Absolute_rotation&lt;/a&gt; there is no reason to conclude the effects will still be there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whoever down voted without commenting as to why, does have any logical reason?&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2010-11-28T11:27:24.487" LastActivityDate="2010-11-28T11:27:24.487" CommentCount="18" />
  <row Id="1375" PostTypeId="1" CreationDate="2010-11-28T12:04:34.207" Score="-3" ViewCount="103" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Four-dimensionalism#Presentism_and_eternalism&quot; rel=&quot;nofollow&quot;&gt;Four-dimensionalism&lt;/a&gt; claims that the universe is basically one huge space-time worm and that everything exists at once (however you want to say that since &quot;internal time&quot; is then just another coordinate in this worm and I'm talking about the look from the outside).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OTOH, all processes in the universe seem to follow an energy economic principle (&lt;a href=&quot;http://en.wikipedia.org/wiki/Second_law_of_thermodynamics&quot; rel=&quot;nofollow&quot;&gt;second law of thermodynamics&lt;/a&gt;), that is, everything tries to reach a state of lowest energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm wondering if that isn't enough to refute the idea of 4D-ism: For it to be the correct theory about space-time, you'd need a lot more matter (at least) because every point in time must be materialized as a &quot;solid&quot; state. That means for everything can happen within what we perceive as &quot;one second&quot;, one state must be encoded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum processes happen at attoseconds (10^-18s), so you need &lt;em&gt;at least&lt;/em&gt; 10^18 states just to encode one second of the 4D worm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my argument is that, just to prove the theory, you're &quot;wasting&quot; insane amounts of whatever the universe is made of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that correct or am I missing something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: English is not my mother language, so I may have use the wrong terms but I hope you get the idea.&lt;/p&gt;&#xA;" OwnerUserId="662" LastActivityDate="2010-11-28T14:57:35.633" Title="Four-dimensionalism vs energy economy" Tags="&lt;spacetime&gt;&lt;universe&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="1376" PostTypeId="2" ParentId="1353" CreationDate="2010-11-28T12:24:59.417" Score="0" ViewCount="" Body="&lt;p&gt;How do you know the output of an experiment? You &quot;observe&quot; it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you actually do that? Well, with your eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does that work? Photons emitted or reflected by a surface hit special molecules in your eyes. This leads to a signal which is transmitted into the rest of your brain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the actual observation eventually happens by exchanging photons. Now you want to observe really small particles. Particles that are so small that interaction with a photon actually changes them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which begs the question whether you can examine something like an electron &lt;em&gt;without&lt;/em&gt; hitting it with a photon or anything that might change it's state. This is really hard. So in most cases, observing something does influence it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a couple of tricks like observing something that is by itself influenced by the particle you want to measure (say the electric field of a moving electron could influence another molecule and you could do your measurement on the molecule).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the end, all these processes are based on resonance. If you want to measure anything, you must create a resonance of some kind and that always means two-way interaction at some level.&lt;/p&gt;&#xA;" OwnerUserId="662" LastActivityDate="2010-11-28T12:24:59.417" />
  <row Id="1377" PostTypeId="1" AcceptedAnswerId="1378" CreationDate="2010-11-28T13:11:03.977" Score="0" ViewCount="99" Body="&lt;p&gt;We have a model of electricity which says electrons flow from one place to another. We have a model of optics which says that photons go from one place to another. As I understand, there is currently research for a similar model to explain gravity (the hypothetical &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Graviton&quot; rel=&quot;nofollow&quot;&gt;graviton&lt;/a&gt;&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a similar approach to kinetic energy? That is a theory which is based on small information packages traveling from one atom to another to change it's velocity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Background: I'm researching material for a hypothetical space drive for aliens in a SciFi story that I'm writing which is based on direct kinetic energy manipulation (DKEM). The idea is that instead of throwing stuff around, they directly manipulate the kinetic energy of the matter that makes up the space ship.&lt;/p&gt;&#xA;" OwnerUserId="662" LastActivityDate="2010-11-28T13:57:55.933" ClosedDate="2010-11-28T22:10:39.377" Title="Is there a theory about kinetic energy &quot;particles&quot;?" Tags="&lt;energy&gt;&lt;kinematics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="1378" PostTypeId="2" ParentId="1377" CreationDate="2010-11-28T13:57:55.933" Score="3" ViewCount="" Body="&lt;p&gt;I think you are confusing forces with energy. Forces are mediated by particles, photons, gravitons, etc. The particles interact with the normal constituents of matter (typically electrons, but also nucleons) and change their energy. Thus a force is &quot;felt&quot; by the particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: a photon hits a free electron, the electron changes its velocity and potentially its speed and kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this sense the &quot;small packages&quot; you are looking for are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Force_carrier&quot; rel=&quot;nofollow&quot;&gt;force carrying particles&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;photons (electro-magnetic force)&lt;/li&gt;&#xA;&lt;li&gt;gluons (strong force)&lt;/li&gt;&#xA;&lt;li&gt;W/Z particles (weak force)&lt;/li&gt;&#xA;&lt;li&gt;gravitons (gravity)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are a few more mentioned in the linked article.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-28T13:57:55.933" CommentCount="1" />
  <row Id="1379" PostTypeId="2" ParentId="1343" CreationDate="2010-11-28T14:07:54.520" Score="6" ViewCount="" Body="&lt;p&gt;Wind doesn't actually touch the surface. You can see the same effect on a car: even if you move at speeds beyond 70mph, the dust doesn't get blown away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look closely, there is a boundary layer between the matter of the fan and the air around the fan. When you get closer to the fan blades, the air starts to move with the fan (the blade pulls it along), so air very close to the blade doesn't move (much) relative to the blade itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, this is true when you add matter to the blade (like dust). In this case, the friction of the air is less than the adhesion of the dust to the blade, so the dust sticks to the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On my fans, I find a lot of dust and short strands on the edge that cuts the air. Here, the air flow presses the strands to the blade (parts of it on either side of the edge). This way, the fan actively collects dust. Again, the force of the air pressing the strands against the blade (plus the friction between dust and blade) is much stronger than the centrifugal force which might pull them sideways. Since the strands cling to the surface, the blade isn't strong enough to cut them, so they stay where they are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This relation is true for all fan speeds, so the dust always ever gets more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple countermeasure is a coarse net on the side where the fan sucks the air in. Most dust strands will get caught in the net and you can easily wash them away every few weeks or collect them with a vacuum cleaner.&lt;/p&gt;&#xA;" OwnerUserId="662" LastActivityDate="2010-11-28T14:07:54.520" />
  <row Id="1380" PostTypeId="1" AcceptedAnswerId="1389" CreationDate="2010-11-28T14:38:30.923" Score="0" ViewCount="148" Body="&lt;p&gt;This is a thought experiment, so please don't treat it too harsh :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short: If we could isolate two places A and B in the universe from all and any interaction with the surroundings, is there a physical law which states &quot;if something is dropped in place A, it has to stay there&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long version: Let's assume that the energy of the whole universe is fixed. Let's further assume that it is (by some trick) possible to completely isolate a box of 1m^3, say in the center of a planet (all gravitational and centrifugal forces cancel themselves out), the mass of the planet shields against radiation and we use a trick to shield against neutrinos or we ignore them since the rarely interact with matter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does an object behave when it has no interaction with the rest of the universe whatsoever? If I put an object in a box described above and I have several such boxes, would it matter in which box the object is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a law which says &quot;even if no one knows, the object &lt;strong&gt;has&lt;/strong&gt; to stay where it is&quot;? Or is that just our expectation based on everyday experience?&lt;/p&gt;&#xA;" OwnerUserId="662" LastActivityDate="2011-02-20T18:07:18.067" Title="Teleportation by quantum isolation" Tags="&lt;quantum-mechanics&gt;" AnswerCount="3" />
  <row Id="1381" PostTypeId="2" ParentId="1375" CreationDate="2010-11-28T14:57:35.633" Score="1" ViewCount="" Body="&lt;p&gt;In both cases you need a very large infinity of states to describe the universe. Whether you treat &lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_space&quot; rel=&quot;nofollow&quot;&gt;time as a special coordinate&lt;/a&gt;, and thus use standard mechanics with trajectories, velocities etc., or if you treat time as just one of the four coordinates in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Spacetime&quot; rel=&quot;nofollow&quot;&gt;special four dimensional manifold&lt;/a&gt;, these coordinates are assumed to vary continuously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between the two representations are the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In 3d you have the concept of a place, and of a time-evolution; in 4d you have the concept of an event, or series of events.&lt;/li&gt;&#xA;&lt;li&gt;In 3d time and space are absolute, in 4d are bound to an observer and freely mix, in general relativity they are curved, even absolutely (e.g. the whole universe might not be flat).&lt;/li&gt;&#xA;&lt;li&gt;In 3d you have one &quot;copy&quot; of matter that evolves through time; in 4d you have one &quot;copy&quot; of matter as well, which &quot;lives&quot; along a 4d curve, describing its position at each time (depending on the observer).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, both theories are just as economical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you may be interested by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Many-worlds_interpretation&quot; rel=&quot;nofollow&quot;&gt;many-world hypothesis&lt;/a&gt;. That theory is much more expensive as a (at least partial) copy of the universe is created each time a measurement is made. However, though, nothing changes - energy only tends to settle to local minima and not necessarily absolute ones: each copy of the universe has then its minima to tend to, and nothing is out of place.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-28T14:57:35.633" CommentCount="3" />
  <row Id="1382" PostTypeId="2" ParentId="1380" CreationDate="2010-11-28T15:04:22.147" Score="2" ViewCount="" Body="&lt;p&gt;Of course: &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion&quot; rel=&quot;nofollow&quot;&gt;Newton's first law&lt;/a&gt;!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Every body remains in a state of rest or uniform motion (constant velocity) unless it is acted upon by an external unbalanced force. This means that in the absence of a non-zero net force, the center of mass of a body either remains at rest, or moves at a constant speed in a straight line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-11-28T15:04:22.147" CommentCount="3" />
  <row Id="1383" PostTypeId="1" AcceptedAnswerId="1387" CreationDate="2010-11-28T15:10:48.737" Score="4" ViewCount="296" Body="&lt;p&gt;If the metre is now defined as the distance light travels in vacuum in &lt;code&gt;1⁄299,792,458th&lt;/code&gt; of a second and the speed of light is accepted to be &lt;code&gt;299,792,458 m/s&lt;/code&gt;, doesn't this seem like the chicken and egg problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember reading somewhere (in the context of uncertainity principles)..&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;..the more precisely one property is measured, the less precisely the other can be measured&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, how then do physicists claim to have accurately measured the speed of light? Does this mean the definition of the metre is dependent on our ability to accurately measure the speed of light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all, to determine speed (distance traveled/time taken) you must first choose some standards of distance and time, and so different choices can give different answers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about so many other factors that affect these measurements. I do not claim to understand the theories of relativity entirely but what about the chosen frame of reference? Spacetime curvature? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that our measurements are only relative and not absolute?&lt;/p&gt;&#xA;" OwnerUserId="619" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-11-29T16:45:16.260" LastActivityDate="2011-01-16T05:30:58.923" Title="Measuring the speed of light and defining the metre - absolute or relative?" Tags="&lt;experimental-physics&gt;&lt;measurement&gt;&lt;spacetime&gt;&lt;speed-of-light&gt;&lt;metrology&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="1384" PostTypeId="2" ParentId="1380" CreationDate="2010-11-28T15:22:50.390" Score="1" ViewCount="" Body="&lt;p&gt;How does an object behave when is has no interaction with the rest of the universe whatsoever? Like every object you ever studied in a physics class. Interaction with the rest of the universe is the hard part that we always abstract away so we can isolate some particular interaction of interest. A perfectly isolated system would always behave in perfect accordance with the rules governing whatever sort of object it was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum terms (since you've got &quot;quantum&quot; in the question title), what you're describing is an infinite square well. That is, the potential barrier for the system in question is of infinite height, perfectly isolating the system from everything else. The wavefunction at the edges of the box would be exactly zero, and the wavefunction everywhere outside the box would be zero. This is the only system in quantum physics that does not have some probability (however infinitesimal) of turning up at some different location. This is, of course, a textbook idealization and not anything you could actually make in reality.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-28T15:22:50.390" />
  <row Id="1385" PostTypeId="2" ParentId="1383" CreationDate="2010-11-28T15:30:40.523" Score="3" ViewCount="" Body="&lt;p&gt;Basically, what happened is that we managed to measure time very accurately with the advent of atomic clocks. Einstein's theory of relativity predicts, or rather posits, that the speed of light is an invariant. So measuring time accurately implies we can measure distances accurately too. We only have to check that light is indeed an invariant and there are also easily reproducible and precise lab techniques to check that, for instance by interferometry. This allows then to define the meter. The number 299,792,458 is then chosen so to make the meter agree with the old definition which used a standard rod. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that in practice, the meter is often determined by measuring wavelengths of certain atomic transitions, read further on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Metre&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Metre&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2010-11-28T15:43:07.153" LastActivityDate="2010-11-28T15:43:07.153" />
  <row Id="1386" PostTypeId="1" AcceptedAnswerId="1399" CreationDate="2010-11-28T15:31:29.303" Score="4" ViewCount="119" Body="&lt;p&gt;Suppose we now consider a lattice of spin, say Ising model, and the phase transition at the critical temperature $T_c$. There are few scaling laws describe the regime around the critical temperature such as correlation length $\xi$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\xi\sim|T-T_{c}|{}^{-\nu}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\nu$ is the critical exponent. This equation describes the divergence of the correlation length as the temperature approaching $T_c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it gives no direct physical picture of the spins near the $T_c$ because we need to compute the summation over each pair of spins to find the correlation. So the question is: If we really look at the lattice, what is the most obvious feature and scaling that we can obverse?&lt;/p&gt;&#xA;" OwnerUserId="562" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-28T23:13:28.957" LastActivityDate="2010-11-28T23:13:28.957" Title="What happens for the spins around the phase transition" Tags="&lt;statistical-mechanics&gt;&lt;phase-transition&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" />
  <row Id="1387" PostTypeId="2" ParentId="1383" CreationDate="2010-11-28T15:32:21.653" Score="9" ViewCount="" Body="&lt;p&gt;There are three relevant quantities involved here: the length of a meter, the duration of one second, and the speed of light. You only need to absolutely measure one of them, after which the other two can be defined in terms of the one that is measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For technological reasons, we have chosen to make the measured reference quantity the length of one second, which is defined in terms of the number of oscillations of radiation associated with the transition between the hyperfine ground states in cesium (specifically, it's 9,192,631,770 oscillations of that light). This is basically because there are experimental techniques that allow incredibly precise measurements of the frequency of radiation, at a level that really can't be matched by length or speed measurements. (The best frequency measurements in the world use trapped aluminum ions as the &quot;clock,&quot; and are good to something like one part in $10^{18}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having defined the second in terms of some physically measurable quantity, we are then free to define the speed of light as having some particular value in meters/second, and then define the meter in terms of the distance traveled by light in one second. The size of a meter is merely a matter of convention, not anything fixed in the physical world, so as long as we have anchored the second to something fundamental, we can make the meter be whatever we want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The particular values of the meter and the speed of light that we choose are based on older measurements using a meter defined in terms of the circumference of the Earth. We've chosen to keep that value, because it would be a hassle to make a wholesale change.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-28T15:32:21.653" CommentCount="2" />
  <row Id="1388" PostTypeId="2" ParentId="1386" CreationDate="2010-11-28T15:42:27.450" Score="3" ViewCount="" Body="&lt;p&gt;The correlation length &lt;em&gt;is&lt;/em&gt; something that you can see from the spins (looking at the lattice).  You can determine it from a sequence of snapshots of an Ising model simulation.  Visually, as the critical point is reached, you will see that the typical size of clusters having the same magnetization (up or down) becomes larger and larger, and this size is none other than the correlation length.  This seems to me to be a very &quot;direct physical picture.&quot;&lt;/p&gt;&#xA;" OwnerUserId="266" LastActivityDate="2010-11-28T15:42:27.450" CommentCount="2" />
  <row Id="1389" PostTypeId="2" ParentId="1380" CreationDate="2010-11-28T15:42:37.363" Score="4" ViewCount="" Body="&lt;p&gt;Classically Sklivvz's answer would be correct. But in quantum world the story is not quite over. In the following I'll talk about particles (because for them quantum effects are more apparent) but it would also be true for bigger objects (although the bigger the object the more improbable the &quot;teleportation&quot; would be).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;First, from the point of view of Quantum Mechanics&lt;/em&gt; and for a while assuming that you can really cancel all forces on the particle you are essentially investigating a double-well potential. There will always be some &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_tunnelling&quot; rel=&quot;nofollow&quot;&gt;tunelling&lt;/a&gt; between two boxes. If the boxes are too far away then it will be very improbable (where this improbability increases exponentially with distance) but surely possible that if you look into second box after some time, you'll find your particle there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in reality the picture is complicated by &lt;em&gt;Quantum Field Theory&lt;/em&gt;. First, it stops to make sense to talk about particles because they are &lt;a href=&quot;http://en.wikipedia.org/wiki/Identical_particles&quot; rel=&quot;nofollow&quot;&gt;indistinguishable&lt;/a&gt; and also they can be created out of nothing. Related to this is the fact that you can't ever dispose of all interaction because vacuum itself is very lively place! There are particles created and annihilated all the time and this has observable effects on any object (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Casimir_effect&quot; rel=&quot;nofollow&quot;&gt;Casimir effect&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you would need you object to be big enough so that it becomes distinguishable (with high probability). But as already stated the bigger the object, the lower the probability that it leaves the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion, as you might have guessed beforehand, is that with extremely high probability nothing interesting will happen at all.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-11-28T15:42:37.363" CommentCount="3" />
  <row Id="1390" PostTypeId="2" ParentId="1353" CreationDate="2010-11-28T15:55:47.637" Score="4" ViewCount="" Body="&lt;p&gt;Assuming that the incoming &quot;first&quot; particle is prepared in a pure state, interaction with another particle does seem necessary.  Such an interaction might simply be the spontaneous emission of a photon or other particle by the original incoming particle, however.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Most importantly, such an interaction is not itself sufficient.&lt;/em&gt; For a measurement event to occur (wave function collapse in the Von Neumann formalism)  we must also &quot;physically lose track&quot; of the some of the information of the interacting particle after the interaction has taken place, so that we must replace the entangled state description of the second particle after the interaction with a probabilistic mixture of such states, forcing a description of the first particle after the interaction in terms of a real valued probability density matrix rather than as the complex valued pure state amplitude we started with. This change of description automatically includes an increase in entropy, which also occurs physically.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless the second, interacting, particle either escapes the apparatus or interacts with a third particle which so escapes, i.e. &quot;interacts with the environment&quot;, no measurement has yet occurred, the entire interaction is in principle reversible, and the complex amplitude description remains appropriate.   Measurement requires &quot;loss&quot; (via decoherence) of the entangling information by further entangling with the environment and dissipation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The escaping third particle is often an emitted photon or phonon.  See the reference in the linked answer &lt;a href=&quot;http://physics.stackexchange.com/questions/27/what-is-the-difference-between-a-measurement-and-any-other-interaction-in-quantum/190#190&quot;&gt;What is the difference between a measurement and any other interaction in quantum mechanics?&lt;/a&gt;, particularly the 1939 article by London and Bauer (but avoid their metaphysics) for details.  More recently, see  &lt;a href=&quot;http://books.google.com/books?id=1YO9tQ4mFY8C&amp;amp;pg=PA101&amp;amp;lpg=PA101&amp;amp;dq=london+bauer+quantum+measurement&amp;amp;source=bl&amp;amp;ots=o2BWVWuglD&amp;amp;sig=hJRwr_SVYZk3mfVE4mu9hjl2RAc&amp;amp;hl=en&amp;amp;ei=qHryTOyBPcH78Ab-5oGyDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;sqi=2&amp;amp;ved=0CB0Q6AEwAQ&quot; rel=&quot;nofollow&quot;&gt;this book on quantum measurement theory&lt;/a&gt;, particularly page 102 referring to the view of Zeh.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may have noticed that some ambiguity remains in this descriptipn.   This has been analyzed in great detail and resolved by Zurek, but it gets a little tricky.   See e.g. &lt;a href=&quot;http://arxiv.org/abs/1001.3419&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1001.3419&lt;/a&gt; and references therein.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-11-28T17:22:39.203" LastActivityDate="2010-11-28T17:22:39.203" />
  <row Id="1391" PostTypeId="2" ParentId="1343" CreationDate="2010-11-28T16:25:19.287" Score="1" ViewCount="" Body="&lt;p&gt;Here are some of my observations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If a fan is not used, the sticking of dust is at the same rate as other objects.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If a fan is used, even the rotation is slow, the dust can still stick much faster than un-used fan.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The dust not only stick at the propeller, but also the cover bar behind and in front of it. It can give a thick layer of dust at both place.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is (plastic ?) coating around the fan, so not metal is direct expose there.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Dust accumulation is a bit faster near the kitchen. The fan surface also has many oil.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hence, I think that the most important reason of high sticking rate should be the high throughput of the air. Dust will not move toward the fan itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the reason of sticking, I think the it should be due to the electrostatic charge on the dust. It is similar to the dust removal in power plant. Also, dust can adhere at most common surface such as table, wall and plastic, so whether the fan has charge should not be important factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both electrostatic charge and oil should increase the sticking rate. After the initial sticking, the rough surface should allow easier sticking for later dust, so I guess the accumulation rate speed up.&lt;/p&gt;&#xA;" OwnerUserId="562" LastActivityDate="2010-11-28T16:25:19.287" />
  <row Id="1392" PostTypeId="1" AcceptedAnswerId="1398" CreationDate="2010-11-28T17:58:45.923" Score="10" ViewCount="144" Body="&lt;p&gt;Sometimes I may come up with an equation in mind, so I want to search for the related material. It may be the case that I learn it before but forget the name, or, there is no name for the equation yet. In this case, I may be able to recall a reference book. Searching in Internet can be a fast way though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are cases that I get some idea and equations (maybe a modified one) myself. So I want to know whether there are any other people working on it before. Because of the vagueness, it is difficult to search it by keywords because I do not know the 'name' of this idea: it is either too board or no results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can try to search google scholar, arxiv and maybe the prola, but there is no support of equation pattern matching. For example, entering ∇⋅V in google give you no useful results (it is better if you input the 'divergence of potential').&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any good way to search papers in this case? It is even better that I can use the combination of equations and keywords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Another reason is that Mathematician should have done some deep analysis on the related mathematical topics. Yet it is difficult to know their results because there is a gap between the terms used in physics and mathematics. It would be really useful if we can find and learn their results.&lt;/p&gt;&#xA;" OwnerUserId="562" LastEditorUserId="562" LastEditorDisplayName="" LastEditDate="2010-11-28T18:32:25.617" LastActivityDate="2010-12-10T19:38:55.073" Title="Searching books and papers with equations" Tags="&lt;books&gt;&lt;theory&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="3" />
  <row Id="1393" PostTypeId="2" ParentId="1302" CreationDate="2010-11-28T18:34:34.743" Score="1" ViewCount="" Body="&lt;p&gt;Water can dissolve a certain (low) quantity of CO2. To create this &quot;refreshing&quot; effect (that is when all the nerves in your gullet fire because they have the impression to being dissolved by an acid :-)), a lot more CO2 is dissolved in the soda.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the liquid can't keep so much CO2. To keep the CO2 there for a certain time (say a couple of months), the bottle is under pressure. This way, the CO2 molecules will spread evenly in the liquid. That only works because the liquid can dissolve some CO2 in the first place. It wouldn't work for something that doesn't dissolve (rock, oil, ...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A liquid can't be compressed, so the volume of the water doesn't change when you open the bottle. You will get a little bit of fizz from the small amount of air on top of the liquid. But most of it will come from the CO2 which isn't kept in the liquid by the pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does CO2 dissolve? It actually forms a (very weak) molecular bond with the water. This way, all the molecules can reach a lower energy state (the water forms &quot;pouches&quot; for the CO2 molecules and they &quot;rest&quot; in there unless disturbed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is very fragile because the water doesn't really want to dissolve so much CO2 -- it just has to because of the pressure under which this happens. Now when you share the bottle, you disturb this fragile balance. The CO2 molecules move and start to form tiny gas clouds that tend to absorb each other (because they like each other more than the water): You get bubbles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't open the bottle and wait for a few days, the water will dissolve the CO2 again. This is slow for many reasons. First of all, the water doesn't really want to. Second, no one mixes the water, so the part near the surface has a high saturation and that disperses very slowly through an undisturbed liquid. Mixing the water would be bad, even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the production plant, a nozzle is inserted into the bottle and the CO2 is released near the center of the liquid. If the bottle wasn't sealed, the CO2 would quickly evaporate. But since it is sealed, the CO2 spreads evenly in the liquid (so actually the water dissolves the CO2 &lt;em&gt;bubbles&lt;/em&gt;, not the CO2 itself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the reason why you only get a small fizz when you buy a bottle since it has been shaken a lot before it arrived at your place (moving to a storage, on a train/truck, loading/unloading, stacking onto a shelve, carrying it home, etc).&lt;/p&gt;&#xA;" OwnerUserId="662" LastActivityDate="2010-11-28T18:34:34.743" />
  <row Id="1394" PostTypeId="2" ParentId="1257" CreationDate="2010-11-28T18:50:26.557" Score="4" ViewCount="" Body="&lt;p&gt;Because they are too long for the confined space. Reasoning: A very short string can't clot (say 1mm). So it needs a certain length before it can start clotting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they could be placed neatly (say big pocket), they wouldn't clot because the ends are too far apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They must be flexible. A steel bar doesn't clot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need certain elements to be just right:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Long enough&lt;/li&gt;&#xA;&lt;li&gt;Confined space&lt;/li&gt;&#xA;&lt;li&gt;Flexible&lt;/li&gt;&#xA;&lt;li&gt;Sticky surface (finger/skin oil, soft material, etc).&lt;/li&gt;&#xA;&lt;li&gt;Ends must be free to move&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In the usual case, the string will be stuffed into the pocket without much consideration, so it will actually start somewhat clotted. If the ends aren't fixed somewhere, they will easily weave with the rest of the string. What makes matters worse is that the ends aren't thin. So you can't untangle everything by simple pulling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest way to avoid clotting is to fixate the two ends. In my case, I have a clip to pin the volume control to my clothes. If you fixate the two ends with that clip, they don't clot.&lt;/p&gt;&#xA;" OwnerUserId="662" LastActivityDate="2010-11-28T18:50:26.557" />
  <row Id="1395" PostTypeId="2" ParentId="1257" CreationDate="2010-11-28T19:38:19.853" Score="1" ViewCount="" Body="&lt;p&gt;Research into this topic won the 2008 (or was it 2009) IgNobel price in physics where it was quantified just how complex these nodes can get.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2010-11-28T19:38:19.853" CommentCount="1" />
  <row Id="1398" PostTypeId="2" ParentId="1392" CreationDate="2010-11-28T21:04:53.693" Score="8" ViewCount="" Body="&lt;p&gt;You can perform $\LaTeX$ search - that is, write formula in &lt;a href=&quot;http://web.ift.uib.no/Teori/KURS/WRK/TeX/symALL.html&quot; rel=&quot;nofollow&quot;&gt;LaTeX&lt;/a&gt; in an appropriate search engine:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.latexsearch.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.latexsearch.com/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, as one can type the same expression in different ways and with different symbols, I never used it it practice. (Anyone did?) &lt;/p&gt;&#xA;" OwnerUserId="184" LastActivityDate="2010-11-28T21:04:53.693" CommentCount="5" />
  <row Id="1399" PostTypeId="2" ParentId="1386" CreationDate="2010-11-28T21:30:39.100" Score="5" ViewCount="" Body="&lt;h2&gt;Generalities on Conformal Invariance&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In two dimensions, a lot is known / conjectured about statistical models at criticality.  For instance, at $T_c$, the spin configuration that you see will not only be self-similar (what others here have been calling &quot;fractal&quot;) but actually fully conformally invariant (in the continum limit); that is, &lt;strong&gt;the probability distributions (&quot;ensembles&quot;) are invariant not only under scale transformations, but indeed, under any transformation which preserves angles&lt;/strong&gt;.  (I think conformal invariance is expected in models with &quot;short-range&quot; interactions at criticality in any dimension, but it's only well understood in 2 dimensions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me go a bit more into what &quot;conformal invariance&quot; means.  Consider the following &quot;thought experiment&quot;, which you might try with a computer simulation.  First, intersect a disk with a square lattice with side length $\epsilon$ with $\epsilon$ quite small.  On this finite graph, draw a random spin configuration for the Ising model at criticality.  Call this ensemble &quot;A&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, perform a &lt;a href=&quot;http://en.wikipedia.org/wiki/Conformal_map&quot; rel=&quot;nofollow&quot;&gt;conformal mapping&lt;/a&gt; of a disk onto whatever (simply connected) shape you like, say a triangle or a square (this is guaranteed by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemann_mapping_theorem&quot; rel=&quot;nofollow&quot;&gt;Riemann mapping theorem&lt;/a&gt;).  Approximate this shape with a square lattice with side $\epsilon$ and draw a random critical Ising spin configuration for this.  Call this ensemble &quot;B&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The claim is that if you perform a conformal transformation of ensemble &quot;A&quot; to get A', the typical picture from A' will be identical to the typical picture from B (this is a statement that the distribution of spin configurations will be nearly identical, if $\epsilon$ is small).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's some illustrations from &lt;a href=&quot;http://www.unige.ch/~smirnov/slides/index.html&quot; rel=&quot;nofollow&quot;&gt;slides of Stanislav Smirnov&lt;/a&gt; (who recently won a Fields medal for related work!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/xdlNJ.png&quot; alt=&quot;Ising model&quot;&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/ex8UQ.png&quot; alt=&quot;random walk&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the left is a critical Ising model configuration under a conformal transformation; on the right is a random walk under a conformal transformation.  You have to kind of use your imagination to see that (other than finite lattice effects) the pictures would be similar to those if you just ran a Ising model or random walk in the funny shapes below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has recently become a hot topic in mathematics, due to the breakthroughs in &lt;a href=&quot;http://en.wikipedia.org/wiki/Schramm%E2%80%93Loewner_evolution&quot; rel=&quot;nofollow&quot;&gt;Schramm-Loewner evolution&lt;/a&gt;, though it has been studied a bit longer in physics using the less rigorous techniques of &lt;a href=&quot;http://en.wikipedia.org/wiki/Conformal_field_theory&quot; rel=&quot;nofollow&quot;&gt;conformal field theory&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;A specific &quot;geometrical observable&quot;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I've been kind of vague above about what conformal invariance means for individual &quot;pictures&quot; of spins (well, most of the claims are about probability distributions).  You seem to be dissatisfied with various claims about correlation functions; I also feel that these are often a bit hard to grasp geometrically.  There is one solidly geometric thing that one can say about certain lattice models at criticality, and this is about the &lt;strong&gt;boundaries between spin clusters&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boundaries between up and down spins in the Ising model at criticality form an ensemble of closed loops in the plane.  This loop ensemble is believed to be related to SLE $\kappa=3$ curves; this is a very particular (conformally invariant) ensemble of random planar curves.  SLE curves are parametrized by a real number $\kappa\geq0$, when $\kappa=0$ one just has straight lines, as $\kappa$ gets larger, the curve gets more &quot;wriggly&quot;, the fractal dimension of these curves is equal to $1+\kappa/8$ for $\kappa&amp;lt;8$ (thus in the continuum limit, boundaries between Ising spin clusters are expected to be 11/8 = 1.375 dimensional!).  For $\kappa\geq8$, the curves actually become plane-filling!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the things I talked about above do not hold except when you are exactly at $T=T_c$.  For the Ising model, it is sometimes hard to generate spin configurations at $T=T_c$ due to critical slowing down of Markov chains which are used, but there are many other systems such as critical percolation which are easier to study.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;References&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;One good place to start reading about conformal invariance in lattice models for a theoretical physicist is Malte Henkel's book &quot;Conformal Invariance and Critical Phenomena&quot;.  Of course if you're serious, you'll eventually turn to the big yellow book by DiFrancesco et al on conformal field theory...  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good intro to SLE for physicists is &lt;a href=&quot;http://arxiv.org/abs/math-ph/0607046&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt; by Ilya Gruzberg.  Probably it won't be long before some books combining the SLE approach with the CFT approach start coming out.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2010-11-28T21:30:39.100" CommentCount="5" />
  <row Id="1400" PostTypeId="2" ParentId="111" CreationDate="2010-11-28T22:31:26.380" Score="1" ViewCount="" Body="&lt;p&gt;As already mentioned the standard introductory books in hamiltonian geometrical (point-) mechanics are Foundations of Mechanics by Abraham and Marsden and Arnolds Mathematical Methods of Classical Mechanics. Another standard book is: &quot;Classical Mathematical Physics&quot; by Walter Thirring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also have a look at &quot;Symmetry in mechanics: a gentle, modern introduction&quot; by Stefanie Frank Singer, which bridges the gap between standard college courses on classical point mechanics and books like those mentioned above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another interesting book for geometrical hamilton mechanics is &quot;Introduction to Symmetry and Mechanics&quot; by J. Marsden, which gives a nice introductory overview to that topic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more examples from a geometrical point of view you may also consult &quot;Global aspects of classical integrable systems&quot; by Cushman and Bates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One sould also mention &quot;Mathematical Aspects of Classical and Celestical Mechanics&quot; by Arnold, Kozlov and Neishtadt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a less advanced (and less rigorous approach) with very much examples you may have a look at the german book &quot;Klassische Mechanik&quot; by F. Kuypers.&lt;/p&gt;&#xA;" OwnerUserId="667" LastActivityDate="2010-11-28T22:31:26.380" CommunityOwnedDate="2011-07-07T13:41:37.800" />
  <row Id="1401" PostTypeId="2" ParentId="1372" CreationDate="2010-11-28T22:51:42.523" Score="4" ViewCount="" Body="&lt;p&gt;The same &lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_rotation&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; everyone else is citing is a decent reference on this. Basically, we don't know, and probably never will, because we can't put an object in an otherwise empty universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you could, though. So we've got a planet in an otherwise empty universe. To test the hypothesis of absolute rotation, you could do various experiments on the planet's surface to measure the fictitious forces arising from being in a rotating reference frame. For example, you could set up &lt;a href=&quot;http://en.wikipedia.org/wiki/Foucault_pendulum&quot; rel=&quot;nofollow&quot;&gt;Foucault's pendulum&lt;/a&gt; at various locations on the planet and measure the precession rate at each location. (I think you'd need at least 3 locations) From those results you could determine the planet's rotation axis and rotational velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Newtonian viewpoint holds that yes, rotation is relative to space. If this view is correct, and the isolated planet were rotating relative to space, you would see your pendulums (pendula?) precessing at a nonzero rate, and you could solve for the planet's rotational velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the Einstein/Mach viewpoint holds that rotation is not relative to space, but rather is defined relative to the matter in the universe. If this viewpoint is correct, you would never see any precession of the pendulums because the bulk of the matter in this experimental universe is the planet itself, so it basically &lt;em&gt;defines&lt;/em&gt; the frame of zero rotation. In our universe, of course, there is a much larger distribution of matter to define a nonrotating rotational reference frame. Mathematically, this results from a phenomenon in GR known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Frame_dragging&quot; rel=&quot;nofollow&quot;&gt;frame dragging&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Newtonian/absolute view has the advantage of being kind of intuitive, but it does require that space defines some sort of absolute rotational reference frame. Given that we know all linear motion is relative, it seems odd (to me, and others) that rotational motion could be absolute. In addition, if rotation could be absolute, for any nonzero rotational velocity, a large enough distribution of matter in the universe would require the outer objects to be moving at the speed of light relative to a nonrotating reference frame. This could conceivably be allowed, it would just mean that no matter could be boosted into that nonrotating reference frame, but again, it seems odd. The Einstein/Mach view has the advantage that it makes this &quot;faster-than-light rotation&quot; extremely unlikely as a consequence of the structure of the theory alone.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-28T22:51:42.523" CommentCount="9" />
  <row Id="1402" PostTypeId="1" AcceptedAnswerId="1403" CreationDate="2010-11-28T22:57:52.977" Score="7" ViewCount="610" Body="&lt;p&gt;It might seem common sense that when we split a magnet we get 2 magnets with their own N-S poles. But somehow, I find it hard to accept this fact.(Which i now know is stated by Gauss's Law)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have had this doubt ever since reading about the quantum-field-theory and I know I might sound crazy but is it really impossible to separate the poles of a magnet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about black holes? They attract matter and everything else only inwards into them, right?  So then, are they the only such case/exception?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I being naive? Is there some proof/explanation that can give closure to this question of an independently existing pole?&lt;/p&gt;&#xA;" OwnerUserId="619" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-28T23:12:48.660" LastActivityDate="2011-01-22T13:08:02.213" Title="Is it possible to separate the poles of a magnet?" Tags="&lt;electromagnetism&gt;&lt;quantum-field-theory&gt;&lt;magnets&gt;&lt;magnetic-monopoles&gt;" AnswerCount="7" />
  <row Id="1403" PostTypeId="2" ParentId="1402" CreationDate="2010-11-28T23:12:21.017" Score="9" ViewCount="" Body="&lt;p&gt;Well, in order for this splitting to be possible, the magnet would have to be made of two &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_monopole&quot; rel=&quot;nofollow&quot;&gt;magnetic monopoles&lt;/a&gt; (like charged particles, but with &quot;magnetic charge&quot; instead of electric charge) bound together. No known magnet is actually constructed this way; all real magnets that have been studied are made of either little current loops, or particles that have a spin magnetic moment (and these basically act like little current loops).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's still an open question whether or not magnetic monopoles exist. Some theories predict that they should, but most have nothing to say about it either way. I am not aware of any theories that prohibit the existence of these monopoles. Quantum field theory in general falls in the second category; that is, there is nothing inherent in QFT that requires magnetic monopoles to exist or not exist.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-28T23:12:21.017" CommentCount="6" />
  <row Id="1404" PostTypeId="2" ParentId="1402" CreationDate="2010-11-29T00:17:51.600" Score="3" ViewCount="" Body="&lt;p&gt;I suspect your problem is you may want to think about it rhetorically. Magnetic poles are really just a mental shortcut useful to provide a bit of intuition to something that is inherently just math. We don't have physical entities called mag poles, we have a magnetic field, and it works as if it were generated by currents (and maybe spin, which may or may not work like moving charge (a current)). So cut your magnet, and you have two similar, but shorter pieces, and a local concentration of field lines is usually called a &quot;pole&quot;, and polarity refers to the signed value of the magnetic field normal to the surface.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-11-29T00:17:51.600" />
  <row Id="1405" PostTypeId="1" AcceptedAnswerId="1407" CreationDate="2010-11-29T00:27:21.913" Score="2" ViewCount="224" Body="&lt;p&gt;I understand that the 2nd law of thermodynamics roughly states that, if you have a body (or a gas in a chamber) that is hot at one end and cold on the other, the heat will always flow from the hot to the cold part and to get the opposite effect, one has to put energy in from the outside (through a machine or something).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I don't understand why this fact cannot be explained just through probabilities (of the velocities of the gas molecules, say). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would seem to me that it is simply very, very, very unlikely that faster moving molecules all end up in (approximately) one spot at any time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But from all the fuzz about the 2nd law, I'm led to believe that there has to be more behind it than probability. So where am I wrong? Why is the second law beyond probability? How is the 2nd law tested? (so that one can rule out simple probability?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ps.: I haven't yet had a course on probability theory. So my understanding of it is limited.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2010-11-29T02:17:39.563" Title="2nd Law of Thermodynamics" Tags="&lt;thermodynamics&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="1406" PostTypeId="2" ParentId="1402" CreationDate="2010-11-29T00:30:45.557" Score="11" ViewCount="" Body="&lt;p&gt;This is basically how a magnet's atoms look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/1O2L2.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so, when you split it into two, you do not change anything but the length of the magnet. As you can see the North poles(Black sides or &quot;K&quot; as it's in a different language) face north and south poles face south in each part of the magnet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;IF&lt;/em&gt; there are such things as monopoles, then it is possible to define a magnetic charge which would allow us to separate a unit magnet into two monopoles. However there are no confirmations for magnetic monopoles so we'll have to accept magnets as a whole for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for black holes, their attraction is gravitational, and not very different from the gravity earth applies on you.&lt;/p&gt;&#xA;" OwnerUserId="504" LastActivityDate="2010-11-29T00:30:45.557" />
  <row Id="1407" PostTypeId="2" ParentId="1405" CreationDate="2010-11-29T00:35:52.747" Score="4" ViewCount="" Body="&lt;p&gt;I assure you, it is all probability AND statistics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well you see when you say &quot;a gas is at 300 Kelvin&quot;, then it does not mean all the molecules in the gas are &quot;at 300 Kelvin&quot;, it rather states an average. It represents the total behavior of the gas compared to anything at any other temperature. So there are actually molecules with more kinetic energy, and those with less kinetic energy, interacting with each other and the environment they are in, however due to the massive number of collisions and interactions, the result with highest probability(i.e. same temperature box&amp;amp;gas stays at the same temperature) is the macroscopic outcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probability, combined with statistics is a very powerful tool to represent macroscopic nature, and I suggest you to take probabilistic maths and statistical thermodynamics courses to further investigate such issues.&lt;/p&gt;&#xA;" OwnerUserId="504" LastActivityDate="2010-11-29T00:35:52.747" CommentCount="1" />
  <row Id="1408" PostTypeId="2" ParentId="1405" CreationDate="2010-11-29T00:45:17.530" Score="2" ViewCount="" Body="&lt;p&gt;It's very, very unlikely that heat should flow from cold to hot, for values of &quot;very very unlikely&quot; involving extremely large powers of ten. We're talking monkeys-with-typewriters-writing-Shakespeare kind of unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a very good, largely nonmathematical discussion of the second law and entropy in Sean Carroll's &lt;em&gt;&lt;a href=&quot;http://preposterousuniverse.com/eternitytohere/&quot; rel=&quot;nofollow&quot;&gt;From Eternity to Here&lt;/a&gt;&lt;/em&gt;, which discusses entropy in detail in the context of the &quot;arrow of time.&quot; It's a very good treatment of the subject, and well worth reading.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-11-29T00:45:17.530" CommentCount="4" />
  <row Id="1409" PostTypeId="2" ParentId="1405" CreationDate="2010-11-29T01:21:20.323" Score="1" ViewCount="" Body="&lt;p&gt;I belief that when the 2nd law of thermodynamics is stated carefully in the right way it becomes a total no-brainer, i.e., just plain obvious. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only concept you have to know about is that of macro- and micro-states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A micro-state is a state that characterizes position/velocity of each and every molecule in your gas, or, more general, the exact state of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A macro state is a collection of many micro-states that share a common characteristic. For example, there are many ways you can arrange the molecules of a gas and still have the same volume, temperature and pressure. The number of micro-states that make up a given macro-state is called &lt;em&gt;multiplicity&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2nd law of thermodynamics simply states that over the course of time, multiplicity will increase: Macro-states with more micro-states are more probable than macro-states with fewer micro-states.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2010-11-29T01:21:20.323" />
  <row Id="1410" PostTypeId="1" AcceptedAnswerId="1411" CreationDate="2010-11-29T01:27:52.457" Score="1" ViewCount="914" Body="&lt;p&gt;I'm in a classical mechanics class now. On our exams, most questions are quantitative. And in general, besides the theory part, all physics problems just require you to gather formulas, manipulate them a little bit by using other formulas, and come up with an appropriate value. In the question, you are given all the information you need. For our exams, they even give us a formula sheet, with every formula and equation we can possibly need. But even after all this, physics remains to be one of the most challenging subject for many students, including myself. I just dont get it. It seems like it should be so easy. Grab a formula or two, and manipulate it until all the given quantities can fit into these formulas. But yet, even after all this, its not easy to do, but it &lt;em&gt;should&lt;/em&gt; be. Sorry if I'm not making any sense. I'm a student, and most of you already know this stuff, so try thinking as a newbie to physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Although this seems like a general study question, its actually a very specific question. I'm a really good student, but I think I've met my match with physics. It seems you just cant 'memorize' things in this class, like you can with any other class. You really just have to understand it. And lately, I've just been so discouraged to study since no matter how hard I study, I do really bad on the tricky exams. My study routine goes a little something like this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Watch pre-lecture video that explains the current chapter, and take notes on that.&lt;/li&gt;&#xA;&lt;li&gt;Do the weekly homework, which itself is pretty challenging.&lt;/li&gt;&#xA;&lt;li&gt;To study for the exams, I do a BUNCH of practice problems from previous semester exams that they supply us with. I literally go over and do 5 old exams, one question at a time, and I understand everything perfectly-atleast I think I do, until the REAL exam comes around, and I do absolutely terrible. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I dont know what else to do at this point :(&lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="41" LastEditorDisplayName="" LastEditDate="2010-11-29T02:10:08.630" LastActivityDate="2010-11-29T20:45:47.617" ClosedDate="2010-11-29T21:15:47.810" Title="Why is Physics so hard?" Tags="&lt;soft-question&gt;" AnswerCount="10" CommentCount="19" FavoriteCount="1" />
  <row Id="1411" PostTypeId="2" ParentId="1410" CreationDate="2010-11-29T01:35:51.127" Score="11" ViewCount="" Body="&lt;p&gt;The problem is: There are so many formulas which you can combine, rearrange and manipulate in oh so many ways that just knowing the formulas doesn't get you that far. What you need is intuition and understanding as to which formulas relate to your problem and how you should relate them to get what you want. This intuition can only be built through experience, i.e. numerous hours of problem solving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why it's so important you do all the problem sets you can find, because only then will you become really familiar with the formulas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of it like learning to play chess: You can learn the rules in an afternoon, they aren't that hard. But this teaches you nothing about how to be a good chess player. That takes years of practice.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2010-11-29T01:35:51.127" CommentCount="2" />
  <row Id="1412" PostTypeId="2" ParentId="1410" CreationDate="2010-11-29T01:42:49.187" Score="1" ViewCount="" Body="&lt;p&gt;Physics is about the ability to predict the outcome of scenarios. Either with formulas, simulation, or good old curve fitting. The underlying principles a few basic ones, but to use them you required to understand higher lever math, so to make it &quot;easy&quot;, basic physics classes boil it down to a &lt;em&gt;cookbook&lt;/em&gt; method. Add one of this, two of this, mix and voila here is the result. That is not how physics supposed to be, and it is a pitty you are stuck in such a situation. My suggestion is to bear with it, until you discover some of the commonalities by yourself and find your own way of making sense of the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hang in there, it gets only beter from here!&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-11-29T01:42:49.187" CommentCount="1" />
  <row Id="1413" PostTypeId="2" ParentId="1410" CreationDate="2010-11-29T01:56:26.267" Score="1" ViewCount="" Body="&lt;p&gt;I don't know how helpful this group will be, we are probably a selfselected subgroup whose natural intuitions were such that physics was easy (at least up until some advanced level, when the math got challenging). A lot has to do with&#xA;understanding the material, and being very good at problem solving. And I've always felt the statement about the worst teachers being those who didn't have to struggle to learn the material, has a lot of truth to it. I think being good at undrgrad level physics has a very high correlation with being good at math word problems, so if you don't get a satisfactory answer here (on how to make it much easier), I'd ask the math group the same question about word problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best learning technique for me when I was learning it was to speedread the chapter in the text book, then take out lots of paper and derive all the formulas just taught -the speedreading was to get the gist of where you are trying to go to and a feeling for how to get there. Also you should have mentally noted any nonobvious tricks needed to get the math to work out. Once you can do that, your understanding should be high enough to enable you to work through problems that need the understanding of the theory, but for which a cookbook like approach (simply plug in equation XXX for problems of type YYY, almost never works). If&#xA;you have a really good understanding of equation XXX, how it was arrived at etc., you have a good chance of being able to use it when it is appropriate, (at least if you are creative about problem solving).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the biggest thing about physics, is its mostly about a mode of thinking, and secondarily about the detailed material. Watch how those who are good at it descibe how they solve problems, hopefully some of their thinking style can be learned.&lt;/p&gt;&#xA;" OwnerUserId="391" LastEditorUserId="391" LastEditorDisplayName="" LastEditDate="2010-11-29T02:09:40.143" LastActivityDate="2010-11-29T02:09:40.143" />
  <row Id="1414" PostTypeId="2" ParentId="1405" CreationDate="2010-11-29T02:17:39.563" Score="1" ViewCount="" Body="&lt;p&gt;I also found that with modern computers with about a days effort, I could write a 2D kinetic particle simulation code. Collide two particles, and conservation of energy and momentum leaves only a single undetrmined quantity, which I simply set by using a random number generator. Throw in a few hundred or thousand particles, and you can see how the system evolves, and what sort of statistical properties emerge. This allows someone who isn't really good at math, but good at programming to design numerical experiments. I postulate that if these tools had been available when the subject was first researched that if would have developed much quicker.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-11-29T02:17:39.563" />
  <row Id="1415" PostTypeId="2" ParentId="1353" CreationDate="2010-11-29T04:40:18.263" Score="0" ViewCount="" Body="&lt;p&gt;This question reminds me of the Zen koan:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the sound of one hand?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.io.com/~snewton/zen/onehand.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; site (which also elaborates on the koan) if:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;while walking, standing, sitting, and reclining, you proceed straightforwardly without interruption in the study of this koan, you will suddenly pluck out the karmic root of birth and death and break down the cave of ignorance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;... and/or understand quantum mechanics :-) Ok ... maybe not quite well enough for a degree, but you get my point ;)&lt;/p&gt;&#xA;" LastActivityDate="2010-11-29T04:40:18.263" />
  <row Id="1416" PostTypeId="2" ParentId="1410" CreationDate="2010-11-29T04:48:30.647" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I dont know what else to do at this point :(&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Party less? Just kidding. Well @fprime, the first thing to realize is that physics is really friggin' hard and it seems to me that once you're over that hump you might actually end up enjoying it. However, loving the subject does not imply that you will never have night terrors where you wake up screaming &quot;Its a real job I tell you, its a real job&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you think you have what it takes then hold your nose and dive in. Six months and no social life later you might start finding physics a lot easier and also a lot more different than what you expect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One good place for beginners seeking inspiration is the set of lectures by Feynman. Good luck !&lt;/p&gt;&#xA;" LastActivityDate="2010-11-29T04:48:30.647" CommentCount="3" />
  <row Id="1417" PostTypeId="1" AcceptedAnswerId="1420" CreationDate="2010-11-29T05:05:49.233" Score="2" ViewCount="282" Body="&lt;p&gt;I don't have any idea of general relativity but intend to learn. Is it a good idea to study general relativity in two dimensions (time and single spatial dimension) in the begining to get good idea on the subject? If it is, then please give some references for such a treatment. &lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-11-29T12:10:46.423" LastActivityDate="2010-12-22T19:10:09.813" Title="General relativity (gravitation) in time and one spatial dimension" Tags="&lt;general-relativity&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="1418" PostTypeId="2" ParentId="1417" CreationDate="2010-11-29T05:19:29.430" Score="2" ViewCount="" Body="&lt;p&gt;No, not really. One of the neat things about GR is that the same theory works in any number of dimensions. In other words, when studying GR, most of what you learn is independent of the dimensionality of spacetime, so there is little to be gained from limiting yourself to 1+1D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it can certainly be instructive to &lt;em&gt;apply&lt;/em&gt; the theory in 1+1D spacetime or other dimensionalities, to calculate how gravity or electromagnetism would work in &quot;flatland,&quot; for example. But those sorts of calculations are interesting mainly because of the results you get. The procedure to apply GR in different dimensionalities is essentially the same as in 3+1D.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-29T05:19:29.430" CommentCount="4" />
  <row Id="1419" PostTypeId="1" AcceptedAnswerId="1425" CreationDate="2010-11-29T06:15:50.823" Score="2" ViewCount="233" Body="&lt;p&gt;&lt;strong&gt;This question is Edited after recieving comments.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the definition of momentum when a mass distribution $\rho(r,t)$ is given?&lt;br&gt;&#xA;Assuming a particle as a point mass we know the definition of momentum as $p = mv$.&lt;br&gt;&#xA; I need a definition where it is assumed that point masses are not present.&lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-11-29T12:04:35.773" LastActivityDate="2010-12-03T12:57:58.340" Title="What is the definition of momentum when a mass distribution $\rho(r,t)$ is given?" Tags="&lt;newtonian-mechanics&gt;&lt;lagrangian-mechanics&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="1420" PostTypeId="2" ParentId="1417" CreationDate="2010-11-29T06:23:05.670" Score="6" ViewCount="" Body="&lt;p&gt;Depends on your reference.  One thing that should be said is that 1+1 relativity theory is trivially simple in several ways.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the Riemann Curvature tensor $R_{abcd}$ is determined completely by one component, meaning that Einstein's equation reduces to a single PDE.  Since this is generally a tensor equation involving a somewhat large number of coupled PDE's, you will likely not get appropriate intuition for the theory.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, there is a conformal invariance$^{1}$ built into the 1+1 Einstein's equations that further makes the theory very easy to solve, since we're now dealing with a single PDE (plus a matter equation of state) with conformal invariance that lets us control the input o that PDE.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two complications will give a very bad intuition for how Einstein's equations work--in particular, they reduce Einstein's equation to an equation depending only on topological invariants, and not upon local properties of the theory.  There can be no radiation in 1+1 gravity, for example.   In 3+1 dimensions, where neither of the above simplifications apply, the qualitative behaviour of the theory is very different.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, a guided study of particular Minkowskian two-spaces can help one understand full Relativity theory, but that is something best done for you by an expert.  Please just start with Schutz's book if you're just trying to teach yourself Relativity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^{1}$conformal invariance means I can map the metric $g_{ab}$ into $\bar g_{ab} = \phi g_{ab}$ and have $\bar g_{ab}$ satisfy the same equation that $g_{ab}$ did.  Einstein's equations in 1+1 dimensions are unable to distinguish between the metric and a rescaling of the metric.&lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2010-11-30T03:49:43.750" LastActivityDate="2010-11-30T03:49:43.750" CommentCount="1" />
  <row Id="1421" PostTypeId="1" AcceptedAnswerId="1422" CreationDate="2010-11-29T06:32:10.413" Score="0" ViewCount="406" Body="&lt;p&gt;This measurement comes up a lot in my E&amp;amp;M class, in regards to inductance and inductors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there really no conventional term for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, is there some historical reason for this omission?&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="71" LastEditorDisplayName="" LastEditDate="2010-11-29T10:05:43.903" LastActivityDate="2010-12-01T17:29:30.533" Title="Is there a name for the derivative of current with respect to time, or the second derivative of charge with respect to time?" Tags="&lt;electromagnetism&gt;&lt;nomenclature&gt;" AnswerCount="1" />
  <row Id="1422" PostTypeId="2" ParentId="1421" CreationDate="2010-11-29T07:13:45.023" Score="2" ViewCount="" Body="&lt;p&gt;Nope, not as far as I know. It's just &quot;rate of change of current&quot; or something like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose it's possible someone has given it a name in some paper or textbook, but if so, it's not widely used.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-29T07:13:45.023" CommentCount="1" />
  <row Id="1423" PostTypeId="2" ParentId="326" CreationDate="2010-11-29T07:55:42.677" Score="0" ViewCount="" Body="&lt;p&gt;I take it that the idea for part 6 is that a non-rotating rectangular bar is dropped onto the rotating disk, but initially does not notate with it (why not?).  Then (not by friction or eany other simple coupling I can imagine where the torque on the bar would typically be proportional to the angular velocity difference, but by some other unstated mechanism,) the rod is angularly accelerated with constant torque (weird no?), somehow coupling to and using the rotational energy of the disk, and that frictional energy losses can nevertheless be neglected.    No wonder our student was confused.   This is a rather unphysical or unusual situation, no?   What might be a mechanism to do this?   Does anyone else find this difficult to visualize as an actual situation?   To me, at a minimum, little more clarity and detail in the initial description of the problem might have been helpful for our student's initial visualization.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-11-29T08:11:24.603" LastActivityDate="2010-11-29T08:11:24.603" CommentCount="1" />
  <row Id="1424" PostTypeId="2" ParentId="1410" CreationDate="2010-11-29T08:27:32.133" Score="1" ViewCount="" Body="&lt;p&gt;The relevant concepts involved may not have been properly explained to you in sufficient detail for you to manipulate them easily.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you considered using a book with very clear definitions and many worked examples, like Schaum's outline series?   See &lt;a href=&quot;http://www.amazon.com/s/ref=nb_sb_ss_i_6_15?url=search-alias%3Dstripbooks&amp;amp;field-keywords=schaums+physics&amp;amp;sprefix=schaums+physics&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/s/ref=nb_sb_ss_i_6_15?url=search-alias%3Dstripbooks&amp;amp;field-keywords=schaums+physics&amp;amp;sprefix=schaums+physics&lt;/a&gt;.   They get surprisingly good reviews.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone recommend very good online lectures, preferably videos?&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-11-29T08:27:32.133" CommentCount="1" />
  <row Id="1425" PostTypeId="2" ParentId="1419" CreationDate="2010-11-29T10:01:19.123" Score="5" ViewCount="" Body="&lt;p&gt;If you have multiple masses the total momentum is&#xA;$$ p = \sum_i m_i v_i $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a continuum distribution then you can proceed like follows.&#xA;You divide the continuum into small boxes where each part of the box has approximately same mass and velocity (this assumes some kind of smoothness of the distribution). Then you can obtain whole momentum with the above formula. Now letting the box sizes go to zero you obtain integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ p(t) = \int \rho(r, t) v(r, t) dr $$&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-11-29T10:01:19.123" CommentCount="14" />
  <row Id="1427" PostTypeId="2" ParentId="1410" CreationDate="2010-11-29T11:08:45.083" Score="4" ViewCount="" Body="&lt;p&gt;There is an answer to your question in the form of a physics paper : &lt;a href=&quot;http://arxiv.org/abs/1005.0005&quot; rel=&quot;nofollow&quot;&gt;Determining dynamical equations is hard&lt;/a&gt; by Toby Cubitt, Jens Eisert and Michael Wolf. Basically, of you define physics as determining dynamical equations &lt;em&gt;—the laws of physics—&lt;/em&gt; from some observations &lt;em&gt;—the experiments—&lt;/em&gt;, they prove that this problem is NP-complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, physics is not only hard, it is NP-hard !&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2010-11-29T11:08:45.083" />
  <row Id="1428" PostTypeId="1" AcceptedAnswerId="1435" CreationDate="2010-11-29T12:16:18.810" Score="0" ViewCount="133" Body="&lt;p&gt;Let $\rho(r,t)$ and $v(r,t)$ be mass and velocity distributions. Given $\rho(r,0)$ and $v(r,0)$ (initial conditions) what is the differential equation that describes the evolution of $\rho(r,t)$ and $v(r,t)$ in space and time, assuming Newtonian gravitation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may assume $r$ to be one dimensional, if it makes writing the equations convenient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think $v(r,t)$ should be replaced with $p(r,t) = \rho(r,t)v(r,t)$ and name it as momentum distribution to avoid some technical faults.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that to completely characterize an inviscid fluid flow, Euler equations are &#xA; needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The law of conservation of mass is already given in the answer. A component of pressure&#xA; is missing in the equation related to conservation of momentum.(first equation). Lastly a&#xA; third equation corresponding to law of conservation of energy should be introduced.(3rd&#xA; one of the euler equations of inviscid fluid flow).For the third equation we assume that&#xA; the internal energy per unit volume of the fluid is zero.&lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="540" LastEditorDisplayName="" LastEditDate="2010-11-30T12:49:40.723" LastActivityDate="2010-11-30T12:49:40.723" Title="Evolution of mass and velocity distributions under newtonian gravitation" Tags="&lt;newtonian-gravity&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="1429" PostTypeId="2" ParentId="1410" CreationDate="2010-11-29T12:37:32.107" Score="1" ViewCount="" Body="&lt;p&gt;This is perhaps a bit besides the point and not really answering your question, but I'll mention it anyway. For me at least, the difference between &quot;how can I possibly memorize all of this?&quot; and true understanding often boils down to an &lt;em&gt;interest&lt;/em&gt; and curiosity. I realize classes have a tendency to kill these things (even when done right, because you're now doing something because you have to), but I think it's worth the effort to try and rekindle your true interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, being told to &quot;be more interested&quot; is hardly help, and I'm sorry for that. I just know it tends to make the difference between being miserable cramming things into my head and working hard learning to understand something. When you learn something that explains a thing you've been wondering about, you no longer have to memorize. You feel in some sense immersed in your newfound knowledge, and you &lt;em&gt;know&lt;/em&gt; it instead of just remembering it. I think this applies to most sciences (I've definitely experienced it in both physics, mathematics, computer science and electronics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my real advice is this: There will always be courses that don't interest you. For those you'll just have to cram it in. But at least the majority of courses can hopefully be viewed with from a different angle where they seem more like a way of answering a question or curiosity you have than a list of things you must know for the exam.&lt;/p&gt;&#xA;" OwnerUserId="636" LastActivityDate="2010-11-29T12:37:32.107" />
  <row Id="1430" PostTypeId="1" AcceptedAnswerId="1432" CreationDate="2010-11-29T13:04:36.580" Score="0" ViewCount="250" Body="&lt;p&gt;I was watching a documentary entitled &lt;a href=&quot;http://www.bbc.co.uk/bbcfour/documentaries/features/atom.shtml&quot; rel=&quot;nofollow&quot;&gt;&quot;The Atom&quot;&lt;/a&gt; and one of the statements made was that Atoms behave differently when we look at them. I wasn't too sure about the reasoning behind this and i'm hoping someone could explain how or why this happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not too knowledgeable about the field of Physics but the theory of Atoms is rather interesting.&lt;/p&gt;&#xA;" OwnerUserId="672" LastActivityDate="2010-11-30T13:09:57.500" Title="Appearance of atoms" Tags="&lt;quantum-mechanics&gt;&lt;atoms&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="1431" PostTypeId="2" ParentId="1410" CreationDate="2010-11-29T13:07:59.490" Score="3" ViewCount="" Body="&lt;p&gt;&lt;b&gt;You can't just memorize it, you have to understand it&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I tell you that an electron has been fired through an electric field towards an oscillating platform, and I want to know the time after which it will hit the platform, and what minimum initial energy I need. To answer that, you need to have a clear model in your head of what an electron is, how it interacts with electric fields, the idea that a particle follows a path defined by an equation of motion, the thought that the oscillation of the platform will change both the time taken and the minimum initial energy needed to reach the platform. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Learn the appropriate theory&lt;/li&gt;&#xA;&lt;li&gt;Form a mental model of that theory&lt;/li&gt;&#xA;&lt;li&gt;Using your mental model see, in your mind, what you think might happen in this specific circumstance&lt;/li&gt;&#xA;&lt;li&gt;Use the equations to construct some equations that reflect your intuition about what will happen.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You see the process of starting from the general (e.g. electromagnetism theory) to the specific (e.g. motion of an electron through an electric field). When doing research in Physics, you tend to do this to get an intuitive expectation, or prediction, of what will happen, and then you do an experiment to test it. Experimental Physics often works in the opposite direction, starting from some specific experimental results, and coming up with a general theory to tie them together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you haven't got the understanding, the intuitive mental model that captures all the rules (theory) that you know, then you will struggle to see your way to the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you were right when you said &quot;You really just have to understand it.&quot;. &lt;/p&gt;&#xA;" OwnerUserId="670" LastActivityDate="2010-11-29T13:07:59.490" />
  <row Id="1432" PostTypeId="2" ParentId="1430" CreationDate="2010-11-29T13:26:41.110" Score="5" ViewCount="" Body="&lt;p&gt;The big change in understanding that happened with Quantum Physics is the idea that the universe is random, rather than clockwork, at its lowest level. &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The double slit experiment&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The classic example is that if I fire a particle (e.g. an atom) at board with 2 slits in it, the particle could go through slit A, or slit B, before hitting a plate at the back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out, however, that the particle behaves like a wave, and goes through both at the same time, and you see an interference pattern on the plate at the back. But, if I put a measuring device on one of the slits to find out whether it really goes through both, it stops behaving like a wave, and switches to behaving like a particle instead, going through one slit or the other  - no interference pattern.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;It changes if you watch it&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;What seems to be happening here is that the atom encounters the 2 slits and has a certain probability of going through either slit. But since nothing (like a detector) is forcing it to reveal which one it went through, it continues as though it went through both. When it hits the plate at the back, it is forced to reveal its position, so it randomly picks a position to be in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when you put the detector next to one slit, the particle has to reveal its position earlier, so the possibilities are different, and we get different behaviour and a different pattern on the back plate.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Consciousness&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Often people use this to suggest that it's because we 'look' at the particle that its behaviour changes. This is rubbish. The difference appears when you put a detector there, not when you, as a human, look at the result. If you put the detector in place and then refused to view the result, it wouldn't change the result.&lt;/p&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;p&gt;So what Prof Jim was saying is that when we measure things like the position of an atom, we unavoidably change its behaviour, because we force it to decide on an answer, rather than carrying on as it was.&lt;/p&gt;&#xA;" OwnerUserId="670" LastEditorUserId="670" LastEditorDisplayName="" LastEditDate="2010-11-30T13:09:57.500" LastActivityDate="2010-11-30T13:09:57.500" CommentCount="1" />
  <row Id="1433" PostTypeId="2" ParentId="252" CreationDate="2010-11-29T13:41:41.207" Score="1" ViewCount="" Body="&lt;p&gt;On a more abstract level, the problem is linearity versus non-linearity. It's straightforward to solve a number of linear equations, and they always yield an analytic answer. However, non-linear equations produce chaotic behaviour, which cannot be generalised in most cases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, the 3-body Newtonian problem involves 2C3 = 3 non-linear equations; the nonlinearity comes from the r&lt;sup&gt;2&lt;/sup&gt; relationship. And 3 non-linear relations are the minimum requirement for a chaotic system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, quantum mechanics involves a large number of non-linear equations - given a set of 3 electrons, each will repel the others via a non-linear relation, and with even more complexity than the Newtonian problem where all things are known and determinable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the simple answer is that the problem is mathematics that can't be solved for the general case, which result from the physics, and that the quantum case is indeed worse than the classical one.&lt;/p&gt;&#xA;" OwnerUserId="670" LastActivityDate="2010-11-29T13:41:41.207" />
  <row Id="1434" PostTypeId="2" ParentId="1410" CreationDate="2010-11-29T14:20:07.030" Score="1" ViewCount="" Body="&lt;p&gt;The way I see it, there are two aspects to the practice of science, there is a constructive one and a destructive one. Let me try to expand on that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The destructive one is the easiest to explain and usually the one on which a fair amount of time is spent. It is basically the skeptical part of science, the one where you test one idea against another. There are various tools at your disposition to do this: experiment, checking for mathematical/logical coherence of the ideas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constructive or creative aspect is the hardest part. We don't have a systematic approach to it, basically, all we do is just try. The reason some people are more successful than others at doing this is experience, they can immediately select the potentially successful ideas from a larger set of ideas, without going through all the steps of the destructive part. Or they just are willing to spend so much more time on working on the subject, giving up on many other activities, which of course forges experience. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize: critique is always relatively easy, coming up with good ideas is the hard part. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, one might say that these two aspects are not necessarily exclusive to physics, which is true. An artist has to go through the same motions, but the criteria are different. Nature is a harsh mistress, and that's why physics is so difficult. We can come up with lots of ideas, but nature doesn't care about how creative we are, if the idea doesn't work it doesn't work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd say sciences in general are difficult because of this. If you didn't encounter the problem in other sciences yet, it's because you are still at a more elementary study level in them as compared to physics.&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2010-11-29T14:20:07.030" />
  <row Id="1435" PostTypeId="2" ParentId="1428" CreationDate="2010-11-29T14:21:57.437" Score="4" ViewCount="" Body="&lt;p&gt;Newton's law of gravitation says that to get the acceleration at a given point, you take any distant bit of mass and make a vector pointing towards it with length proportional to the mass and inversely proportional to the square of the distance to it.  Integrate that expression over the entire mass distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial{\vec{v}}(\vec{r},t)}{\partial t} + [\vec{v}(\vec{r},t)\cdot\nabla]\vec{v}(\vec{r},t) = G \int_{\textrm{space}}\frac{\rho(\vec{r&amp;#39;},t)}{|(\vec{r&amp;#39;}-\vec{r})|^3}(\vec{r&amp;#39;}-\vec{r}) \textrm{d}V(\vec{r&amp;#39;})$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_constant&quot; rel=&quot;nofollow&quot;&gt;Newton's gravitational constant&lt;/a&gt;, and $\textrm{d}V(\vec{r&amp;#39;})$ is a volume element at the location $\vec{r&amp;#39;}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, in terms of $\vec{p}$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\textrm{d}\vec{p}}{\textrm{d}t} = G\rho\int_{\textrm{space}}\frac{\rho&amp;#39;}{|\vec{r&amp;#39;}-\vec{r}|^3}(\vec{r&amp;#39;}-\vec{r}) \textrm{d}V(\vec{r}&amp;#39;)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conservation of mass requires that the net flow of mass into a region of space result in an increase in the density there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial\rho(\vec{r},t)}{\partial t} = \nabla \cdot \vec{p}(\vec{r},t)$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-29T20:35:09.380" LastActivityDate="2010-11-29T20:35:09.380" CommentCount="20" />
  <row Id="1436" PostTypeId="2" ParentId="1428" CreationDate="2010-11-29T15:42:44.940" Score="0" ViewCount="" Body="&lt;p&gt;Gauss law of gravity is what you want &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%27_law_for_gravity&quot; rel=&quot;nofollow&quot;&gt;wikipeadia link&lt;/a&gt;. This will give you $\vec{g}$ at each location.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-11-29T15:42:44.940" />
  <row Id="1437" PostTypeId="2" ParentId="1430" CreationDate="2010-11-29T16:03:24.080" Score="2" ViewCount="" Body="&lt;p&gt;First, a caveat: There are several interpretations of quantum mechanics -- that is, different ways of thinking about what's happening that nevertheless give the same predictions for what we can measure in experiments.  How best to explain measurement is somewhat dependent on your choice of interpretation.  For this answer I'll stick with something like the traditional Copenhagen interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, &quot;behave differently when we look at them&quot; is sort of a shorthand.  It would be more accurate to say that the state of the atom changes when you measure it.  We're not necessarily &quot;looking at&quot; the atom at all -- for instance, sometimes you have multiple atoms in an entangled state where performing a measurement on one may change the state of &lt;em&gt;both&lt;/em&gt; atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, putting aside these details, the basic idea is this.  In quantum mechanics, atoms are described by a wave equation (the Schrodinger equation), and waves can be in a superposition.  Think of a sound wave, which may contain multiple frequencies (that is, multiple pitches).  But whereas with a classical wave (like sound) we'd perform a frequency measurement and find a little bit of each frequency component, with quantum mechanics our measurements give a definite answer.  So even if the wave is a superposition of frequency X and frequency Y, your frequency measurement gives you just X, or it gives you just Y.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of these possible outcomes has some probability of occurring, and this probability depends on how much of each frequency we started with.  And &lt;em&gt;after&lt;/em&gt; the measurement, if you only measured frequency X, then you are really left with a state that only contains frequency X, even if that's not what you started with.  The measurement changes the state of the system being measured, and does it in a way we can't predict ahead of time; we can only say what the &lt;em&gt;probability&lt;/em&gt; of getting that result would be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used &quot;frequency&quot; to maintain my analogy with classical waves, but this can be &lt;em&gt;any&lt;/em&gt; observable (that is, any measurable property of the atom).  The position of the atom, the momentum of the atom, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, there is a different rule, the uncertainty principle, that says the particle can't be in a state with a definite momentum and a state with a definite position at the same time.  So measuring the position will change the state of the particle to one with a single value of position (at least if I pretend we could measure it with perfect precision), and measuring the momentum will change it to a &lt;em&gt;different&lt;/em&gt; state, one with a single value of momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People often confuse the effect of measurement with the uncertainty principle, but note that they are two separate things.  With measurement, we say &quot;Measuring property P puts the particle in a state where it has one value of P, not a state that's a superposition of states with two different P values.&quot;  The uncertainty principle says &quot;There's no state which both has only one value of position and only one value of momentum&quot;.  (In fact it says more than that, it says something specific about how the range of momentum values grows the more precisely we narrow down the position value and vice versa, but I'm trying to keep things simple here.)&lt;/p&gt;&#xA;" OwnerUserId="11" LastActivityDate="2010-11-29T16:03:24.080" CommentCount="1" />
  <row Id="1438" PostTypeId="1" AcceptedAnswerId="1449" CreationDate="2010-11-29T16:42:43.640" Score="12" ViewCount="495" Body="&lt;p&gt;Is there an easy way to understand and/or visualize the reciprocal lattice of a two or three dimensional solid-state lattice?  What is the significance of the reciprocal lattice, and why do solid state physicists express things in terms of the reciprocal lattice rather than the real-space lattice?&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2011-01-03T05:47:40.070" Title="Reciprocal Lattices" Tags="&lt;solid-state-physics&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" />
  <row Id="1439" PostTypeId="2" ParentId="1383" CreationDate="2010-11-29T16:54:01.260" Score="1" ViewCount="" Body="&lt;p&gt;I just complete @Chad Orzel's response by point : you seem to be worried on the effect of chosen frame of reference and space-time curvature. One of the reasons to redefine the meter in term of speed of light is because the speed of light is independent of the frame of reference and the space time curvature, while distance and time are not. That the whole point of Einstein relativity theories.   &lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2010-11-29T16:54:01.260" />
  <row Id="1440" PostTypeId="2" ParentId="1438" CreationDate="2010-11-29T16:58:17.363" Score="10" ViewCount="" Body="&lt;p&gt;The significance of the reciprocal lattice is tied with diffraction of waves on a crystal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we determine the crystalstructure of some material? We usually do by bombarding a tiny piece of crystal with X-rays or neutrons or another type of wave of appropriate wavelength and properties. We then look at the diffraction pattern. Now, to make a long story short, the pattern that appears will essentially be a pattern in the reciprocal space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/X-ray_crystallography&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/X-ray_crystallography&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2010-11-29T17:08:59.973" LastActivityDate="2010-11-29T17:08:59.973" CommentCount="1" />
  <row Id="1441" PostTypeId="1" CreationDate="2010-11-29T17:20:08.007" Score="9" ViewCount="381" Body="&lt;p&gt;It's easy to find the &quot;magic&quot; numbers of neutrons on the diagrams of alpha-decay energy: 82, 126, 152, 162. Such  &quot;magic&quot; nuclei should be more stable than their neighbors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;IMG src=&quot;http://elementy.ru/blogs/u/i/2598.gif&quot;&gt;&#xA;&lt;IMG src=&quot;http://elementy.ru/blogs/u/i/2599.gif&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why some  nuclei with &quot;magic&quot; numbers of neutrons have a half-life less than their    neighbor isotopes with odd numbers of neutrons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples for  &quot;magic&quot; number 126:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A half-life of &quot;magic&quot; Po-210 is &lt;strong&gt;138 days&lt;/strong&gt;,   whereas a half-life of neighbor isotope Po-209 is &lt;strong&gt;102 years&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A half-life of &quot;magic&quot; Ra-214 is &lt;strong&gt;2.46 sec&lt;/strong&gt;, whereas a half-life of neighbor isotope Ra-213 is &lt;strong&gt;2.74 min&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples for  &quot;magic&quot; number 152:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A half-life of &quot;magic&quot; Cm-248 is &lt;strong&gt;348 thousand years&lt;/strong&gt;,  whereas a half-life of neighbor isotope Cm-247 is &lt;strong&gt;16 million years&lt;/strong&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A half-life of &quot;magic&quot; Cf-250 is &lt;strong&gt;13 years&lt;/strong&gt;,  whereas a half-life of neighbor isotope Cf-249 is &lt;strong&gt;351 years&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.  &lt;a href=&quot;http://www.nndc.bnl.gov/masses/mass.mas03&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt; of the diagrams data&lt;/p&gt;&#xA;" OwnerUserId="238" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2010-12-09T17:44:15.463" LastActivityDate="2010-12-11T20:27:27.533" Title="Why some nuclei with &quot;magic&quot; numbers of neutrons have a half-life less than their neighbor isotopes?" Tags="&lt;nuclear-physics&gt;&lt;isotope&gt;&lt;radiation&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="1442" PostTypeId="1" AcceptedAnswerId="1444" CreationDate="2010-11-29T17:20:20.630" Score="5" ViewCount="287" Body="&lt;p&gt;Assume the time standard clocks and any memories about the time standard are destroyed. Can we recover the time standard again exactly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recovering the time standard again means we can determine the date  and time&#xA;that are exactly the same as the destroyed time standard clocks will show (if they are not destroyed). &lt;/p&gt;&#xA;" OwnerUserId="652" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-11-29T18:00:11.077" LastActivityDate="2011-01-18T14:05:20.853" Title="If time standard clocks and any memories about the time standard are destroyed, can we recover the time standard again?" Tags="&lt;measurement&gt;&lt;astronomy&gt;&lt;metrology&gt;&lt;time&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="1443" PostTypeId="2" ParentId="1438" CreationDate="2010-11-29T17:37:59.433" Score="8" ViewCount="" Body="&lt;p&gt;To understand why reciprocal space is important it is perhaps useful to illustrate related concept of frequency space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anytime one is analyzing waves (be it sound waves, EM waves or any other kind) one can exploit the translational symmetry of the laws of motion. The dual quantity to position is momentum (for massless fields like EM waves this also corresponds to frequency) and because of the said symmetry everything becomes whole lot easier. Instead of working with general wave (which can be a pretty difficult beasts) one instead works with monochromatic (i.e. single frequency) waves. For this special kind of waves the differential equations simplify just to algebraic equations, so the problem becomes easily tractable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the method explained in the previous paragraph is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_analysis&quot; rel=&quot;nofollow&quot;&gt;Fourier analysis&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;Fourier transform&lt;/a&gt; and is very general. Put simply, anytime you have some nice symmetry you can use Fourier analysis to move to the dual space where the problem will simplify greatly. When applied to lattices (which have a lot of translational symmetry) one obtains the concept of a reciprocal lattice.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mathematical note:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From mathematical point of view one exploits that the system is essentially described by integrable functions on some &lt;a href=&quot;http://en.wikipedia.org/wiki/Locally_compact_group&quot; rel=&quot;nofollow&quot;&gt;locally compact abelian group&lt;/a&gt; $G$. By &lt;a href=&quot;http://en.wikipedia.org/wiki/Peter%2dWeyl_theorem&quot; rel=&quot;nofollow&quot;&gt;Peter-Weyl theorem&lt;/a&gt; we know that the space of such functions is parametrized by irreducible representations of $G$ which form a dual Pontryagin group $\hat{G}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example when working with periodic functions one is actually working with the functions defined on the circle $S^1$. Now the irreps of $S^1$ are parametrized precisely by integers. So in this case we get the decompositions of a periodic function $$f(x) = \sum_{k \in \mathbb{Z}} f_k \rho_k(x)$$ where $f_k$ are the Fourier modes of $f(x)$ and $\rho_k(x) = \exp(-i k x)$ are irreps of $S^1$. In general the irreps of an LCA group are given by some exponential so this explains the ubiquity of exponentials in physics.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-11-29T17:45:06.843" LastActivityDate="2010-11-29T17:45:06.843" CommentCount="6" />
  <row Id="1444" PostTypeId="2" ParentId="1442" CreationDate="2010-11-29T17:44:04.150" Score="11" ViewCount="" Body="&lt;p&gt;It depends on your definition of &quot;any memories&quot;. If you don't remember what a second is, there is no solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remember the &quot;old&quot; definition (a day has 24 h on 60 minutes, each of it is 60 s), and live on Earth not to far from now, you can rebuild an approximate time standard. If you remember the modern definition, &lt;em&gt;i.e.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(the duration of 9,192,631,770 periods&#xA;  of the radiation corresponding to the&#xA;  transition between the two hyperfine&#xA;  levels of the ground state of the&#xA;  caesium 133 atom.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;you can reconstruct it as good as we can now, even if you are in a spaceship far away from Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would give you a second duration, from which you can define a time-scale. The next-step is to synchronize your new time-scal with the old time-scale. This is basically what historian do when they translate Egyptian or Mayan date in our modern system of datation. To have something precise, you basically need some record of an astronomical event. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best analogue I can see is the work discussed &lt;a href=&quot;http://www.ucolick.org/~sla/leapsecs/deltat.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.ucolick.org/~sla/leapsecs/dutc.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, where the difference between the solar time and atomic time is extrapolated in the past, before the atomic clock era. There are 10 s error bars around 1700, and 3h error bar for the year 1000 BC. &lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2010-11-29T17:44:04.150" CommentCount="1" />
  <row Id="1445" PostTypeId="1" AcceptedAnswerId="1451" CreationDate="2010-11-29T17:48:46.777" Score="0" ViewCount="68" Body="&lt;p&gt;In the context of this &lt;a href=&quot;http://physics.stackexchange.com/questions/1428/a-question-on-evolution-of-mass-and-velocity-distributions-rhor-t-and-v&quot;&gt;question&lt;/a&gt; should mass distribution $\rho(r,t)$ and momentum distribution $p(r,t)$ be well behaved ? By 'well behaved' it is meant that derivatives of all orders exist everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see any reason why they should be well behaved. For the equations (given in the answer (by Mark Eichenlaub) to the above mentioned question) to be consistent the conditions required are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\vec{p}$ should be continuously differentiable in order that $\nabla\cdot\vec{p}$ exists.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\rho(r,t)$ should be partially differentiable wrt $t$ i.e., $\frac{\partial\rho(\vec{r},t)}{\partial t}$ exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please give strong reasons as to why it is required that $\rho(r,t)$ and $\vec{p}(r,t)$ to be always well behaved ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="540" LastEditorUserId="540" LastEditorDisplayName="" LastEditDate="2010-11-29T18:29:36.720" LastActivityDate="2010-11-29T19:55:29.290" Title="Behaviour of mass and momentum distributions under Newtonian Gravity" Tags="&lt;newtonian-gravity&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="1446" PostTypeId="2" ParentId="1445" CreationDate="2010-11-29T17:58:41.283" Score="1" ViewCount="" Body="&lt;p&gt;They are not necessary well behaved. But if they're not you will use distributions instead of functions. For example, if $\rho$ is not continuous, $\partial\rho/\partial t$ will behave like a Dirac distribution. The maths are just more complex, rarely bring new insight about the physics, that's why physicists often assume the functions to be regular enough.&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2010-11-29T17:58:41.283" CommentCount="1" />
  <row Id="1447" PostTypeId="2" ParentId="1165" CreationDate="2010-11-29T18:39:15.497" Score="3" ViewCount="" Body="&lt;p&gt;Matter and anti-matter behave in the same way with respect to electromagnetic interactions, so we could not distinguish the two by electromagnetic observations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was mentioned in the other reply that annihilation from the contact of matter with antimatter would produce a signature gamma radiation that would be easily observable. That is true, but things are not that simple. It has been calculated that matter and antimatter could be kept separated by a type of &lt;a href=&quot;http://en.wikipedia.org/wiki/Leidenfrost_effect&quot; rel=&quot;nofollow&quot;&gt;Leidenfrost&lt;/a&gt; layer between them that would be supported by a relatively slow rate of annihilation (Collective effects in diffuse matter-antimatter plasma: &lt;a href=&quot;http://www.springerlink.com/content/v33083853m7020j6/&quot; rel=&quot;nofollow&quot;&gt;I&lt;/a&gt;, &lt;a href=&quot;http://www.springerlink.com/content/p47730l28xr37548/&quot; rel=&quot;nofollow&quot;&gt;II&lt;/a&gt;). That rate could be so slow and the layer so small, that it could be practically unobservable in cosmological distances. I am not trying to support a theory that would claim that there are great amounts of antimatter in the universe, but there are some who do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from all that, one way to distinguish between the two could be &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutrino#Supernova_neutrinos&quot; rel=&quot;nofollow&quot;&gt;the detection of neutrinos&lt;/a&gt;. In a supernova of an antimatter star for example, the neutrinos that we would get would not be the same as the ones from a normal matter supernova, I suppose.&lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2010-11-29T18:39:15.497" />
  <row Id="1449" PostTypeId="2" ParentId="1438" CreationDate="2010-11-29T19:42:50.123" Score="3" ViewCount="" Body="&lt;p&gt;Its questions like this one that keep me coming back to this site !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your first question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there an easy way to understand and/or visualize the reciprocal lattice of a two or three dimensional solid-state lattice?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;YES ! The reciprocal lattice is simply the dual of the original lattice. And the dual lattice has a simple visual algorithm.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Given a lattice $L$, for each unit cell of $L$ find the point corresponding to that cell's &quot;center of mass&quot; (see below).&lt;/li&gt;&#xA;&lt;li&gt;Connect each such &quot;center of mass&quot; to its nearest neighbors.&lt;/li&gt;&#xA;&lt;li&gt;The resulting lattice is the &lt;strong&gt;dual&lt;/strong&gt; of $L$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To find center of mass of unit cell (we consider 2d case, generalizes to arbitrary dimension):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Draw the perpendicular bisectors of the edges which bound the unit cell.&lt;/li&gt;&#xA;&lt;li&gt;For regular lattices these lines should intersect at a single point in the interior of the cell. This point is the &quot;center of mass&quot; of the cell.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Performing these simple steps you find that the dual of a square lattice is also a square lattice, and that the triangular and hexagonal lattices are each others duals ! You can see a nice illustration of this fact &lt;a href=&quot;http://web.me.com/paulscott.info/polyhedra/compounds/duals_comps.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your second question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the significance of the reciprocal lattice, and why do solid state physicists express things in terms of the reciprocal lattice rather than the real-space lattice?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As mentioned by others this has to do with fourier transforms. In solid-state physics we want to understand the excitations (waveforms) that a certain material, whose structure is given by some lattice $L$, can support. For a lattice only certain momenta are allowed due to its discrete structure. These allowed momenta correspond to the vertices of the dual lattice! For more see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Reciprocal_lattice&quot; rel=&quot;nofollow&quot;&gt;wikipedia page&lt;/a&gt; or check out the first couple of chapters of little Kittel or Ashcroft and Mermin.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                                Cheers,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; This to clarify some doubts about my answer @wsc has expressed in the comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, it is incorrect that reciprocal lattice vectors in 3D have dimensions $1/L^2$. Consider a 3D lattice with basis vectors $\{a_i\}$. The reciprocal lattice has basis vectors given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ b_i = \frac{1}{2V} \epsilon_i{}^{jk} \, a_j a_k $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in index notation, with summation convention. A more familiar way to write this is in vector notation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbf{b}_i = 2\pi \frac{\mathbf{a}_j \times \mathbf{a}_k}{\mathbf{a}_i \cdot (\mathbf{a}_j \times \mathbf{a}_k)} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $(i,j,k)$ are cyclic permutations of $(1,2,3)$. We can see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dim[\mathbf{b}_i] = \frac{\dim[\mathbf{a}]^2}{\dim[\mathbf{a}]^3} = \frac{1}{L} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and in terms of the lattice spacing $a$, $\vert\mathbf{b}\vert \sim \frac{1}{a}$. In fact, this is a basic fact true in any dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also understand the normalization of the reciprocal lattice vectors by the factor $\mathbf{a}_i \cdot (\mathbf{a}_j \times \mathbf{a}_k)$ as being nothing more than $V$ - the volume of the unit cell. Why? So that the transformation between the lattice and reciprocal lattice vector spaces is invertible and the methods of Fourier analysis can be put to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all regular lattices &lt;strong&gt;AFAIK&lt;/strong&gt; the &quot;dual&quot; and &quot;reciprocal&quot; lattices are identical. For irregular lattices - with defects and disorder - this correspondence would possibly break down.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-03T05:47:40.070" CommentCount="10" />
  <row Id="1451" PostTypeId="2" ParentId="1445" CreationDate="2010-11-29T19:55:29.290" Score="2" ViewCount="" Body="&lt;p&gt;Will $\rho$ always behave well? No, not at all.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is physics, not mathematics -- if something is wrong with solutions, the explanation is simple: too bad for equations, probably approximations behind them lack some process or the mathematical description starts to be invalid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, it is obvious that a cloud with no angular momentum will be collapsing into one point, yet this is not a problem -- it is just &quot;physical&quot; that there will be some repulsion that would start to be significant in high density regions invalidating this simplified model.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-29T19:55:29.290" CommentCount="16" />
  <row Id="1452" PostTypeId="2" ParentId="1410" CreationDate="2010-11-29T20:45:47.617" Score="2" ViewCount="" Body="&lt;p&gt;Formulas are (merely) a means of describing relations between physical quantities. You should understand what a formula expresses, the physical concept (e.g. energy conservation) behind it. It's often useful if you can find an everyday analogy for it (e.g. 'money conservation': income equals expenses plus savings). Also, before applying a given formula you must check whether the assumptions made when deriving it are fulfilled in a given problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately it's not clear why despite the learning effort you seem not to perform so well at the exams. Try studying together with other students (from whom you can learn). Or try asking questions here. And try to find out where your exam answers differ from the correct answers.&lt;/p&gt;&#xA;" OwnerUserId="671" LastActivityDate="2010-11-29T20:45:47.617" />
  <row Id="1453" PostTypeId="1" CreationDate="2010-11-29T20:49:35.867" Score="30" ViewCount="2740" Body="&lt;p&gt;We all know that some theoretical ideas lack experimental evidence while in other cases there's a lack of a suitable theory for known phenomena and established facts and concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what problem in physics, according to you, deserves a mention? &lt;strong&gt;And why you think solving that particular problem is of utmost importance and/or how far-reaching its effects/repercussions would be.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One unsolved problem per post.&lt;/p&gt;&#xA;" OwnerUserId="619" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-19T22:58:46.953" LastActivityDate="2011-06-04T17:18:20.393" CommunityOwnedDate="2011-01-22T18:11:53.713" Title="Great unsolved physics problems" Tags="&lt;experimental-physics&gt;&lt;soft-question&gt;&lt;theory&gt;" AnswerCount="32" CommentCount="6" FavoriteCount="13" />
  <row Id="1454" PostTypeId="2" ParentId="1453" CreationDate="2010-11-29T21:14:30.000" Score="13" ViewCount="" Body="&lt;p&gt;One fundamental problem in mathematical physics that has to my knowledge not been resolved, but I might be mistaken since I haven't been working in this particular field for already 5 years and 5 years is a long time, is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In macroscopic bodies, there are several laws existing describing transport phenomena like heat conduction. They are described by precise mathematical laws which are well established experimentally. A well-known one is Fourier's law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c\frac{\partial}{\partial t}T(\vec{r},t)=\nabla \cdot \left(\kappa \nabla T(\vec{r},t)\right) \; ,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $T(\vec{r},t)$ is the local temperature, $c=c(T)$ the specific heat and $\kappa=\kappa(T)$ the heat conductivity. There is however no rigorous mathematical derivation of this equation for any classical or quantum model with a Hamiltonian microscopic evolution. This problem is related to the question wether deterministic microscopic systems can fully explain the behaviour of macroscopic matter. (Remember &lt;a href=&quot;http://physics.stackexchange.com/questions/1405/2nd-law-of-thermodynamics&quot;&gt;this question&lt;/a&gt;, well basically, we have no rigorous proof that it's all just probability applied on huge amounts of microscopic particles. At least not for out of equilibrium processes, which transport phenomena are.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I don't expect anyone but mathematical physicists invested in this particular field to loose any sleep over it. We have a good heuristic understanding of how things work and there's nobody seriously doubting that the explanation of macroscopic phenomena lies in understanding the underlying microscopic phenomena (except maybe some crackpots and fringe physicists). Still, a mathematical derivation could bring us a deeper understanding of why it works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Green-Kubo relation gives $\beta V \int_0^\infty d\tau \left\langle J(0)J(\tau) \right\rangle$ for the transport coefficients. The time autocorrelation function can be computed using the closed time path formalism.&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-23T07:18:45.020" LastActivityDate="2011-01-23T07:18:45.020" CommunityOwnedDate="2011-01-22T18:11:53.713" CommentCount="7" />
  <row Id="1455" PostTypeId="2" ParentId="1453" CreationDate="2010-11-29T21:15:52.093" Score="22" ViewCount="" Body="&lt;p&gt;Q: Are there magnetic monopoles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why: If there are, Maxwell's equations become more symmetric and, more important, it immediately explains why charge is quantized. If there aren't, then the interesting question might be: Why?&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2010-11-29T21:15:52.093" CommunityOwnedDate="2011-01-22T18:11:53.713" CommentCount="2" />
  <row Id="1456" PostTypeId="2" ParentId="1453" CreationDate="2010-11-29T21:34:34.250" Score="7" ViewCount="" Body="&lt;p&gt;I think one important problem (at least from the point of mathematical physics) is to establish the Standard Model as a mathematically complete and consistent quantum field&#xA;theory. This is related to: &lt;a href=&quot;http://www.claymath.org/millennium/Yang-Mills_Theory/&quot; rel=&quot;nofollow&quot;&gt;http://www.claymath.org/millennium/Yang-Mills_Theory/&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that the Standard Model of particle physics deals with fundamental structures of matter and is one of the most successful models in physics in terms of accuracy of predictions. However it seems to consist of a bunch of unjustified rules, people argue using undefined objects etc. Up to now there is no conceptual clear, mathematical and logical consistent description of the Standard Model available. So I think it would give us a quite deeper understanding how nature works if we had such a consistent formualtion of the Standard Model.&lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorUserId="667" LastEditorDisplayName="" LastEditDate="2010-11-29T21:43:22.570" LastActivityDate="2010-11-29T21:43:22.570" CommunityOwnedDate="2011-01-22T18:11:53.713" CommentCount="4" />
  <row Id="1457" PostTypeId="2" ParentId="1453" CreationDate="2010-11-29T22:48:38.213" Score="21" ViewCount="" Body="&lt;p&gt;A general existence and uniqueness theorem for the Navier-Stokes equations.  The lack of one is probably the largest open problem in classical mechanics, considering that the study of fluid dynamics is based on the Navier-Stokes equations and generalizations thereof.  If there were an equation of state/initial data set where the solution was either non-existent or non-unique, even if such a thing were not experimentally realizable (for instance, the &quot;good&quot; initial data formed a dense set in the space of possible initial data), it would still fundamentally change how we look at the evolution of classica systems.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-11-29T22:48:38.213" CommunityOwnedDate="2011-01-22T18:11:53.713" CommentCount="3" />
  <row Id="1458" PostTypeId="2" ParentId="1428" CreationDate="2010-11-29T23:05:49.580" Score="1" ViewCount="" Body="&lt;p&gt;This problem is insoluble (since it is not well-posed) without a matter equation of state (and, unless you assume $T\approx 0$, a thermal equation of state).  If you specify a non-dust equation of state, it ends up having the Navier-Stokes equations embedded in it.  What do you need this for?  The above answers discuss some special cases (most specifically, static distributions) well, but this is still an open problem in the general case.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2010-11-30T02:57:59.353" LastActivityDate="2010-11-30T02:57:59.353" CommentCount="13" />
  <row Id="1459" PostTypeId="2" ParentId="1453" CreationDate="2010-11-30T01:25:38.680" Score="17" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Is supersymmetry (SUSY) a symmetry of our world?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has to do with many outstanding problems in modern physics, like &lt;a href=&quot;http://en.wikipedia.org/wiki/Hierarchy_problem&quot; rel=&quot;nofollow&quot;&gt;hierarchy problem&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Dark_matter&quot; rel=&quot;nofollow&quot;&gt;nature of dark matter&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmological_constant#Cosmological_constant_problem&quot; rel=&quot;nofollow&quot;&gt;cosmological constant problem&lt;/a&gt; and others. If true, it explains some of them in one stroke and for others it provides clues as to what to do next. But note that it would also bring many other questions (e.g. precisely what SUSY model is realized in nature and how exactly is spontaneous breaking of SUSY realized). It also points a way to quantum gravity (via supergravity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, this symmetry is very appealing theoretically and many people are quite sure it is indeed correct despite not yet being observed &lt;em&gt;(so, in this regard it is quite similar to Higgs boson)&lt;/em&gt;. Now, not only is the resolution of this problem (in either way) very interesting but LHC should actually be able to give the answer in next few years and this makes it all the more exciting.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-11-30T01:25:38.680" CommunityOwnedDate="2011-01-22T18:11:53.713" CommentCount="7" />
  <row Id="1460" PostTypeId="2" ParentId="1453" CreationDate="2010-11-30T03:22:46.167" Score="4" ViewCount="" Body="&lt;p&gt;I would like to know how to compute the charge of an electron from first principles.  This would likely have major implications that would depend on the form of the solution.&lt;/p&gt;&#xA;" OwnerUserId="442" LastActivityDate="2010-11-30T03:22:46.167" CommunityOwnedDate="2011-01-22T18:11:53.713" CommentCount="2" />
  <row Id="1461" PostTypeId="2" ParentId="1442" CreationDate="2010-11-30T05:56:54.370" Score="3" ViewCount="" Body="&lt;p&gt;&quot;Imagine for a moment what would happen if, just as a practical joke, someone found a way to stop all atomic clocks, just for a short time.  This would cause such a tremendous disturbance in world affairs that the loss of TAI would be a totally insignificant matter!  Furthermore, when it came to setting it up again, the phase of TAI could be retrieved to within a few tenths of a microsecond by observations of rapidly rotating pulsars...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- Claude Audoin &amp;amp; Bernard Guinot,&#xA;p. 252, sec. 7.3.1 of &quot;The Measurement of Time: Time, Frequency and the Atomic Clock&quot;,&#xA;Cambridge University Press, 2001&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2010-11-30T05:56:54.370" CommentCount="2" />
  <row Id="1462" PostTypeId="1" AcceptedAnswerId="1472" CreationDate="2010-11-30T08:51:20.900" Score="2" ViewCount="328" Body="&lt;p&gt;what is the relation between mass density $\rho$ and pressure $P$ for a perfect fluid ? &lt;/p&gt;&#xA;" OwnerUserId="540" LastActivityDate="2010-12-01T02:05:21.007" Title="Relation between density and pressure for a perfect fluid" Tags="&lt;newtonian-gravity&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="1463" PostTypeId="2" ParentId="1462" CreationDate="2010-11-30T09:20:58.313" Score="1" ViewCount="" Body="&lt;p&gt;I see you are at the moment interested in fluid dynamics and its motion under an external force.&#xA;For the given question, you are searching the Euler equations which should be introduced in any standard fluid dynamics textbook and most of the books on continuum mechanics. I suggest you to pick your favorite one and try to understand from there.&#xA;The community will be glad to help you in any related questions.&#xA;Sincerely,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2010-11-30T09:20:58.313" CommentCount="1" />
  <row Id="1464" PostTypeId="2" ParentId="1462" CreationDate="2010-11-30T09:36:06.183" Score="0" ViewCount="" Body="&lt;p&gt;Yet-to-be-postulated. Perfect fluid is a set of assumptions that does not touch compressibility (ok, it brings some limitation to some very esoteric cases, but this does not change the meritum).   &lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-11-30T09:36:06.183" />
  <row Id="1465" PostTypeId="2" ParentId="1462" CreationDate="2010-11-30T11:23:59.127" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Perfect_fluid&quot; rel=&quot;nofollow&quot;&gt;This page&lt;/a&gt; gives the energy-momentum tensor of a perfect fluid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T^{\mu\nu} = (\rho + p) \, U^\mu U^\nu + p \, \eta^{\mu\nu}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing the energy and momentum densities, it is then trivial to relate them to mass density in either the classical or relativistic cases.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2010-11-30T11:23:59.127" />
  <row Id="1466" PostTypeId="2" ParentId="1372" CreationDate="2010-11-30T14:12:25.903" Score="2" ViewCount="" Body="&lt;p&gt;thank you for your question. Indeed, you are pointing in the directions of two big unifications made in physics:&#xA; Linear motion and rest by Newton and acceleration by Einstein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is convenient for you, I will devide your question into two parts concerning Newtonian and Einsteinian physics which will be related to who is observing the rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, if you are fixed to a point on the surface of the earth which is rotating, you will be able to measure it by means of fictitious forces acting on you, also in a non-relativistic Newtonian setup. But the interesting question is what happens if you are somewhere in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, assuming the earth as axisymmetric, Newtonian physics will tell you that there will be no difference to a non-rotating earth in the gravitational field. You will have to consider general relativity to answer this question. It turns out, that rotating bodies curl spacetime and you can even assign an angular momentum to it. By measuring this effect, which might be very hard, you would be able experimentally verify earth's rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS.: For further information, see e.g. &lt;a href=&quot;http://arxiv.org/abs/0902.1859&quot; rel=&quot;nofollow&quot;&gt;On the multipole moments of a rigidly rotating fluid body&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PPS.: @David: If I understand your argumentation correctly, you state that earth's rotation would define the zero-rotation reference-frame. I must admit that this might not be correct. Think of the Kerr spacetime for a non-vanishing rotation of a black hole. This is a similar situation, you cannot find a coordinate system with vanishing angular momentum of the spacetime as a whole.&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2010-11-30T15:19:42.513" LastActivityDate="2010-11-30T15:19:42.513" />
  <row Id="1467" PostTypeId="1" AcceptedAnswerId="1502" CreationDate="2010-11-30T15:10:40.720" Score="7" ViewCount="297" Body="&lt;p&gt;If a baseball is dropped on a trampoline, the point under the object will move a certain distance downward before starting to travel upward again.  If a bowling ball is dropped, it will deform further downwards.  What is the nature of the relationship between the magnitude of this deformation and the object's mass?   Linear?  Square?  etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;I would like to add that the heart of what I'm asking is along the lines of this:  &quot;If a small child is jumping on a trampoline and the trampoline depresses 25% towards the ground, would an  adult who weighs slighly less than four times as much be safe from depressing it all the way to the ground?&quot;&lt;/p&gt;&#xA;" OwnerUserId="682" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-11-30T21:43:02.490" LastActivityDate="2011-07-31T10:49:12.737" Title="How far does a trampoline vertically deform based on the mass of the object?" Tags="&lt;classical-mechanics&gt;&lt;forces&gt;&lt;spring&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" />
  <row Id="1468" PostTypeId="2" ParentId="1372" CreationDate="2010-11-30T16:12:54.777" Score="0" ViewCount="" Body="&lt;p&gt;From a popular physics standpoint, a good source that constantly discusses the history of the debate about this problem is the popular book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0375727205&quot; rel=&quot;nofollow&quot;&gt;The Fabric of the Cosmos&lt;/a&gt; by Brian Greene.   The answer to this problem has been different in different eras since the time of Newton, who I believe first proposed this question in the form of a person spinning in an empty universe and whether or not their arms splay outward.  Greene seems to think that string theory has something to do with the answer, though I'm not so convinced he's right.  A lot of pseudoscientific philosophy goes into these types of questions, which is why I think it's best addressed in a popular book rather than in a scientific context.  Anyway, there are lots of different angles and approaches to this problem, and this book is a good read if you're interested in knowing what a lot of people think about this problem throughout history, among other things.&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2010-11-30T16:12:54.777" />
  <row Id="1469" PostTypeId="2" ParentId="1441" CreationDate="2010-11-30T16:39:37.530" Score="0" ViewCount="" Body="&lt;p&gt;Have you considered the problem in light of &lt;a href=&quot;http://en.wikipedia.org/wiki/Semi-empirical_mass_formula&quot; rel=&quot;nofollow&quot;&gt;the semi-empirical mass formula&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m = Zm_p + Nm_n - \frac{E_B}{c^2} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the binging energy is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E_B = a_VA  - a_SA^{2/3} - a_C\frac{Z(Z-1)}{A^{1/3}} - A_A\frac{(A - 2Z)^2}{A} +\delta(A,Z)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and all the coefficients are experimentally tuned values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are looking for is a pair where the mass of the post decay nucleus is &quot;much&quot; lighter than the pre-decay nucleus. And the important aspects are the difference in the nucleon masses, and the changes in the Coulomb ($a_C$) and Asymmetry ($a_A$) terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wikipedia article I liked has numeric values. You'll note that this kind of consideration suggests the island and rock of stability.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2010-11-30T19:12:50.910" LastActivityDate="2010-11-30T19:12:50.910" CommentCount="2" />
  <row Id="1470" PostTypeId="2" ParentId="1467" CreationDate="2010-11-30T17:22:05.700" Score="0" ViewCount="" Body="&lt;p&gt;It is not a simple relationship. A trampoline will have various natural modes, which are shapes of deformation that correspond to different frequencies. Any dynamic collision, will excite many of these modes, which will continue until they gradulay damp out. For a very low frequency applied force,for example how far will the baseball depress the trampoline if it just sits on it for a long time, (i.e. a static situation), the answer should be a linear function of the ball's weight (or nearly so), but for dynamic processes all sorts of transient dynamics come into play to complicate things.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2010-11-30T17:22:05.700" CommentCount="5" />
  <row Id="1471" PostTypeId="2" ParentId="1467" CreationDate="2010-11-30T17:47:58.897" Score="0" ViewCount="" Body="&lt;p&gt;Interesting! I don't have time to investigate this in detail, and it's been a while since Itouched mechanical physics, but it sounds to me that a half-decent first approximation to the static case would be to consider the trampoline as a mesh of perfects springs. Somebody please correct my hunch, but: In the limit of infinitely finely woven springs, a point mass interacts with the mesh as though it were instead suspended by a string of 2*2=4 (2 springs per dimension) times the spring constant, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for a suitably regular non-point object, let's say a cube, neighboring springs in the interior of the contact surface (here the base square) between the object and the trampoline won't matter (right?). Thus only the circumference of the object should matter, and depress the trampoline like a continuum of point masses... so we should get a depression increasing linearly with the object's mass and decreasing linearly with the side lengths of its base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, as has been mentioned, what's really needed here is a good/realistic model for an elastic membrane.&lt;/p&gt;&#xA;" OwnerUserId="636" LastActivityDate="2010-11-30T17:47:58.897" CommentCount="2" />
  <row Id="1472" PostTypeId="2" ParentId="1462" CreationDate="2010-11-30T18:36:35.620" Score="4" ViewCount="" Body="&lt;p&gt;There is no restriction.  The simplest choice is $P = \kappa \rho$, where $\kappa$ is a real constant, and that is often used to get simple results.  $\kappa=0$ is dust, $\kappa=-1$ is cosmological constant, and $\kappa=\frac{1}{3}$ is a gas of photons.  A proper analysis also needs an equation describing the flow of internal energy and its relationship to temperature.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a concrete example, note that $PV=NkT$ can be rewritten as $P = \frac{N}{V}kT = \frac{\rho}{m_{p}}kT$, where $m_{p}$ is the mass of the particle making up the ideal gas.  Thus, if the fluid is held at a constant temperature, we get $\kappa = \frac{kT}{m_{p}}$.  If the fluid is not at a constant temperature, we need to use the First Law of Thermodynamics and knowledge of the process (isoentropic, adiabatic, etc) the system is undergoing to get a relationship between the density and pressure.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2010-12-01T02:05:21.007" LastActivityDate="2010-12-01T02:05:21.007" CommentCount="9" />
  <row Id="1473" PostTypeId="2" ParentId="1402" CreationDate="2010-11-30T20:53:00.023" Score="1" ViewCount="" Body="&lt;p&gt;No, because found magnet is not composed of poles. Lacking magnetic charges, the creator had to cheat use currents of electric charges. These currents make two poles at once, and these are all magnets found in nature (and stores) today.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-11-30T20:53:00.023" CommentCount="3" />
  <row Id="1474" PostTypeId="2" ParentId="1343" CreationDate="2010-11-30T20:58:03.167" Score="5" ViewCount="" Body="&lt;p&gt;The short answer is that there's no wind near the blade. This is called &lt;a href=&quot;http://en.wikipedia.org/wiki/No-slip_condition&quot; rel=&quot;nofollow&quot;&gt;no-slip condition&lt;/a&gt; in hydrodynamics of viscous fluids. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Concession] It is actually more than that. There's minor van der waals sticking which contributes to this otherwise purely hydrodynamic phenomenon.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-11-30T20:58:03.167" />
  <row Id="1475" PostTypeId="2" ParentId="1135" CreationDate="2010-11-30T21:03:01.727" Score="1" ViewCount="" Body="&lt;p&gt;The difference is: &lt;strong&gt;consideration of forces&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many cases, forces themselves are related to location (lookup entries potential, potential energy, hamiltonian) so dynamics would allow you to predict system's behavior from given initial state.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-11-30T21:03:01.727" />
  <row Id="1477" PostTypeId="2" ParentId="1128" CreationDate="2010-11-30T21:11:48.627" Score="1" ViewCount="" Body="&lt;p&gt;Short answer: climbing such a ladder &lt;strong&gt;is&lt;/strong&gt; acceleration. This is because the ladder is stuck into spinning earth, and you get higher speed by going up to higher radius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of spinning, escape velocity does depend on height. If you are static at infinity, you already have escaped.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-11-30T21:11:48.627" CommentCount="2" />
  <row Id="1478" PostTypeId="2" ParentId="1061" CreationDate="2010-11-30T21:17:41.397" Score="1" ViewCount="" Body="&lt;p&gt;After dealing with GPS algorithms for significant part of my lifetime, I'd say in one word: GPS is a precise device. It is really state of the art to determine location from timing (aka DTOA, differential time of arrival), catching electromagnetic waves which are so damn fast. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That has to account for many subtle effects which might be insignificant in simple life - such as athmospheric disturbances, relativistic effects, ridiculously hard electronics design issues, etc.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-11-30T21:17:41.397" />
  <row Id="1479" PostTypeId="1" AcceptedAnswerId="1483" CreationDate="2010-11-30T21:22:17.613" Score="8" ViewCount="535" Body="&lt;p&gt;I've watched a video from the American National History Museum entitled &lt;a href=&quot;http://www.youtube.com/watch?v=17jymDn0W6U&quot; rel=&quot;nofollow&quot;&gt;The Known Universe&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The video shows a continuous animation zooming out from earth to the entire known universe. It claims to accurately display every star and galaxy mapped so far. At one point in this video [3:00 - 3:15 minutes] it displays this text:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The galaxies we have mapped so far.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The empty areas where we have yet to&#xA;  map.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At this point, the shape of the &quot;universe&quot; is roughly hourglass, with earth at it's centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/TEQFs.jpg&quot; alt=&quot;The Known Universe - &amp;quot;The galaxies we have mapped so far&amp;quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble understanding what this represents and why they shape is hourglass. Is this simply because we have chosen to map in those directions first? Is there a reason astronomers would choose to map in this pattern, or is this something more fundamental about the shape of the universe? Or is it to do with the speed of light reaching us from these distant galaxies?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing on from the hourglass pattern, the cosmic microwave background radiation is represented as a sphere and labelled &quot;Our cosmic horizon in space and time&quot;. This doesn't help clear anything up. If we can map CMB in all directions, why have we only mapped galaxies in this hourglass shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/qZnhm.jpg&quot; alt=&quot;The Known Universe - &amp;quot;Out cosmic horizon in space and time&amp;quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="692" LastEditorUserId="692" LastEditorDisplayName="" LastEditDate="2010-11-30T21:43:00.210" LastActivityDate="2011-01-15T17:50:16.483" Title="Why is the mapped universe shaped like an hourglass?" Tags="&lt;astrophysics&gt;&lt;astronomy&gt;&lt;universe&gt;" AnswerCount="6" FavoriteCount="1" />
  <row Id="1480" PostTypeId="2" ParentId="1479" CreationDate="2010-11-30T21:27:04.007" Score="7" ViewCount="" Body="&lt;p&gt;Someone will correct me if I'm wrong, but I think the explanation is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our planet is part of the Solar system, which is itself rotating around the center of the Milky Way in one of its spiral arms. Whenever we want to look at deep space in the direction of these arms, we have to look through thick layers of dust and other stars from the different spiral arms our line of sight will cross. So basically, our own galaxy is impeding our view in certain directions. So the cones you see are basically the up and down direction with respect to the galactic plane.&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2010-11-30T21:27:04.007" CommentCount="3" />
  <row Id="1482" PostTypeId="2" ParentId="1019" CreationDate="2010-11-30T21:34:24.277" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;&quot;Long hair grows slower&quot;&lt;/strong&gt; is due to biological effects&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this is a purely mathematical phenomenon. The bigger average length, the more decrease is caused by each hair fall. This leads to a differential equation $$\frac{dL}{dt} = K - \alpha L$$ that has a solutions which decay exponentially to an equilibrium at $$K/\alpha$$. &lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-11-30T21:34:24.277" CommunityOwnedDate="2010-11-30T21:34:24.277" CommentCount="13" />
  <row Id="1483" PostTypeId="2" ParentId="1479" CreationDate="2010-11-30T21:37:32.050" Score="16" ViewCount="" Body="&lt;p&gt;First of all, the universe is most certainly &lt;strong&gt;not&lt;/strong&gt; shaped like an hourglass. It simply looks that way because the gas and dust in the plane of our galaxy obstruct our view of anything outside the galaxy in those directions. So we can only see other galaxies (and similarly distant objects) by pointing telescopes at some angle to the galactic plane. That gives the &quot;hourglass&quot; shape: it's simply because those are the only directions we can see in. In reality, we have every reason to think galaxies are distributed more or less uniformly, once you look at a large enough scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The video description doesn't cite its sources, but I suspect that (some/most of) the information on the distant galaxies comes from the &lt;a href=&quot;http://www.sdss.org&quot; rel=&quot;nofollow&quot;&gt;Sloan Digital Sky Survey&lt;/a&gt;, which is AFAIK the most comprehensive survey of objects in the universe outside our own cluster of galaxies. You might want to check out the information on their website if you're interested in this stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as long as I'm citing sources, the latest CMB data comes from the &lt;a href=&quot;http://map.gsfc.nasa.gov/&quot; rel=&quot;nofollow&quot;&gt;WMAP project&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-11-30T21:37:32.050" CommentCount="5" />
  <row Id="1484" PostTypeId="2" ParentId="1019" CreationDate="2010-11-30T21:40:49.260" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;&quot;Burning coal for heating is more efficient than electrical due to thermodynamical losses at coal power plant&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is false, even though it is true that converting electrical power to heat would be even worse. The correct way to heat with a given amount of heat source is this: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;burn it at high temperature&lt;/li&gt;&#xA;&lt;li&gt;make work with heat machine between T&lt;sub&gt;high&lt;/sub&gt; and the environment&lt;/li&gt;&#xA;&lt;li&gt;use the work to power an air conditioner to bring heat from T&lt;sub&gt;environment&lt;/sub&gt; into T&lt;sub&gt;room&lt;/sub&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This gives efficiency more than 1 (more heat brought into room than heat produced by burning), and net cooling of the surrounding environment.&lt;/p&gt;&#xA;" OwnerUserId="100" LastEditorUserId="176" LastEditorDisplayName="" LastEditDate="2011-06-07T02:18:34.133" LastActivityDate="2011-06-07T02:18:34.133" CommunityOwnedDate="2010-11-30T21:40:49.260" CommentCount="3" />
  <row Id="1485" PostTypeId="2" ParentId="629" CreationDate="2010-11-30T21:53:36.687" Score="3" ViewCount="" Body="&lt;p&gt;You are pulled wrong way &lt;strong&gt;by your own hands&lt;/strong&gt; and legs, which were stressed to keep you decelerating together with the vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the vehicle suddenly stopped accelerating (this is how friction works: it is opposite to speed, and then speed suddenly reaches zero) then this tension keeps pulling you until you react and reconfigure.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-11-30T21:53:36.687" CommentCount="5" />
  <row Id="1486" PostTypeId="1" AcceptedAnswerId="1499" CreationDate="2010-11-30T21:57:11.013" Score="2" ViewCount="232" Body="&lt;p&gt;From what I remember, one of the first steps in finding the equations of motion for an orbiting body is to argue that the body's motion has to be restricted to a plane, because the central force has no component perpendicular to the plane defined by the initial position and velocity vectors. (Or equivalently, because the direction of the total angular momentum has to be constant, since there is no torque.) This means that you can choose coordinates such that $\dot\phi = 0$, which makes the third term of this general Lagrangian vanish:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = \frac{1}{2}m\dot r^2 + \frac{1}{2}mr^2\dot\phi^2 + \frac{1}{2}mr^2\dot\theta^2\sin^2\phi - U(r)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, this is done in &lt;a href=&quot;http://www.worldforge.org/project/newsletters/July2002/LagrangianP3&quot; rel=&quot;nofollow&quot;&gt;this derivation&lt;/a&gt; around equation (11). (Note: in this post I use the Zwillinger convention for spherical coordinates listed on &lt;a href=&quot;http://mathworld.wolfram.com/SphericalCoordinates.html&quot; rel=&quot;nofollow&quot;&gt;MathWorld&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I could be wrong, but I think there should be a total of three conserved quantities in this problem. If I use the argument in the previous paragraph to impose a coordinate system in which $\dot\phi = 0$, then I can easily identify the energy as the Noether invariant corresponding to time translation invariance, and the angular momentum as the Noether invariant corresponding to rotational invariance around the $z$ axis ($\theta\to\theta + \epsilon$). But I can't think of a third one, and I suspect that the restriction $\dot\phi = 0$ (by the reasoning of the first paragraph) eliminates that third conserved quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what if I &lt;em&gt;don't&lt;/em&gt; use the physical argument from the first paragraph, so that I'm &lt;em&gt;not&lt;/em&gt; limited to $\dot\phi = 0$? From the general Lagrangian I've written above, if my intuition is right, I should be able to obtain three conserved quantities from Noether's theorem &lt;em&gt;alone&lt;/em&gt;. Of course, the energy is one, and that's still easy to compute; another one should be angular momentum, although that no longer corresponds to $\theta\to\theta + \epsilon$ but to some more complicated transformation. So I guess my question has two parts,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Prerequisite:&lt;/strong&gt; What is the symmetry transformation that generates angular momentum when $\dot\phi \neq 0$?&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Main question:&lt;/strong&gt; Is there another spatial symmetry that generates another conserved quantity? If so, what is it?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;P.S. I do know there's basically no practical value to this question, since you can always choose coordinates such that $\dot\phi = 0$, but I'm just curious.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-03T00:37:13.803" LastActivityDate="2010-12-03T00:37:13.803" Title="Is there a conserved quantity that enforces planar orbits in central force motion?" Tags="&lt;lagrangian-mechanics&gt;&lt;orbital-motion&gt;&lt;noethers-theorem&gt;" AnswerCount="2" CommentCount="12" />
  <row Id="1487" PostTypeId="2" ParentId="270" CreationDate="2010-11-30T22:04:02.570" Score="1" ViewCount="" Body="&lt;p&gt;Since outside interacts with the radiation both ways, it would not help. But, there's something that would:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to have clothes which are Black (totally absorbing) in visible light and White in Thermal-infrared wavelengths (totally reflecting). This color is called Red (well, to some extent: if you shift the spectrum into visible terms).&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-11-30T22:04:02.570" />
  <row Id="1488" PostTypeId="2" ParentId="1479" CreationDate="2010-11-30T22:05:08.703" Score="4" ViewCount="" Body="&lt;p&gt;The intuitive reason why there is a black strip in the area corresponding to where Milky Way lies is explained in the other answers but this is far from being the end of story.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, &lt;em&gt;why doesn't it show also on WMAP picture?&lt;/em&gt; Well, the short answer is &lt;strong&gt;it does&lt;/strong&gt;. But what they do (and what hasn't been answered in other questions) is that they scan different wavelengths of light and that gives them enough information (by using various spectral methods) to &lt;a href=&quot;http://en.wikipedia.org/wiki/Wilkinson_Microwave_Anisotropy_Probe#Foreground_radiation_subtraction&quot; rel=&quot;nofollow&quot;&gt;subtract the effects of Milky Way&lt;/a&gt; (and also other objects that obstruct the view).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's obvious that one can't really say that we don't &lt;strong&gt;see&lt;/strong&gt; through Milky Way. We don't see only in the narrow sense of &quot;see&quot; as in &lt;em&gt;seeing in visible spectrum of light&lt;/em&gt; but obviously this is not physical at all. Visible light isn't any special just because we, humans, happen to have eyes that register it. Astronomers observe the universe at very different wavelengths (like X-ray radiation, gamma bursts, IR, UV and others). And even this would be too narrow. They observe all kind of particles other then light. Neutrinos are one particular type that is obviously not very much bothered by our galaxy.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-11-30T22:05:08.703" />
  <row Id="1489" PostTypeId="2" ParentId="1019" CreationDate="2010-11-30T22:17:44.270" Score="5" ViewCount="" Body="&lt;p&gt;I think one common false belief is that a &lt;a href=&quot;http://en.wikipedia.org/wiki/Crookes_radiometer&quot; rel=&quot;nofollow&quot;&gt;light mill&lt;/a&gt; rotates because photons deposit more momentum on the shiny side (where they are reflected) than on the black side (where they are absorbed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find it quite astonishing to see that many people think so despite the fact that a light mill spins to the opposide direction than predicted by that explanation.&lt;/p&gt;&#xA;" OwnerUserId="667" LastActivityDate="2010-11-30T22:17:44.270" CommunityOwnedDate="2010-11-30T22:17:44.270" CommentCount="2" />
  <row Id="1490" PostTypeId="2" ParentId="1486" CreationDate="2010-11-30T22:28:43.120" Score="4" ViewCount="" Body="&lt;p&gt;The plane of motion at any given instant is the space spanned by the orbiting body's position vector and velocity vector.  To get a normal to this plane, take the cross product of those two vectors.  The normal to the plane equally can be used to define what plane we're talking about because a given vector is only normal to one plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross product of position and velocity is just a constant times the angular momentum, so the conservation of angular momentum implies that the plane of motion doesn't change (and is stronger).&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-30T22:28:43.120" CommentCount="1" />
  <row Id="1491" PostTypeId="1" CreationDate="2010-11-30T22:39:07.040" Score="4" ViewCount="81" Body="&lt;p&gt;As the simplest example, consider a set of two qubits where the reduced density matrix of each qubit is known. If the two qubits are not entangled, the overall state would be given by the tensor product of the one qubit states. More generally, I could write a set of contraints on the elements of a two-qubit density matrix to guarantee the appropriate reduced description.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there is a way to do this more elegantly and systematically for arbitrary bi-partite quantum systems? I'm particularly interested in systems where one of the Hilbert spaces is infinite dimensional, such as a spin 1/2 particle in a harmonic oscillator.&lt;/p&gt;&#xA;" OwnerUserId="694" LastActivityDate="2010-11-30T23:56:34.747" Title="Can I parameterize the state of a quantum system given reduced density matrices describing its subparts?" Tags="&lt;quantum-information&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="1492" PostTypeId="2" ParentId="1479" CreationDate="2010-11-30T22:52:39.437" Score="4" ViewCount="" Body="&lt;p&gt;As it has been pointed out, the Milky Way is to blame for the hourglass shape of the maps of the universe. I would like though to point out for the case of the CMB and the WMAP images that have the contribution of the Milky Way removed, that there seem to be some open issues. There is a very nice lecture given at PI on the subject of CMB anomalies inferred from WMAP, that discuses the possibility that they could have something to do with the way that the signal from the Milky Way is subtracted. The lecture is:&#xA;&lt;a href=&quot;http://www.pirsa.org/index.php?p=media&amp;amp;url=http://streamer.perimeterinstitute.ca/Flash/9cd6f9d2-a6bc-48c8-b94e-fbcb0f1c2c4a/viewer.html&amp;amp;pirsa=08100080&amp;amp;type=Flash%20Presentation&quot; rel=&quot;nofollow&quot;&gt;If the CMB is right, it is inconsistent with standard inflationary Lambda CDM (2008, Glenn Starkman)&lt;/a&gt; &#xA;and a relevant paper:&#xA;&lt;a href=&quot;http://arxiv.org/abs/1004.5602&quot; rel=&quot;nofollow&quot;&gt;Large-angle anomalies in the CMB&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2010-11-30T22:52:39.437" />
  <row Id="1493" PostTypeId="1" AcceptedAnswerId="1494" CreationDate="2010-11-30T23:01:02.160" Score="7" ViewCount="325" Body="&lt;p&gt;How efficient is an electric heater? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess: greater than 95%. Possibly even 99%. I say this because most energy is converted into heat; some is converted into light and kinetic energy, and possibly other forms of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone other opinions? (This is not homework. I am just curious and I'm having a discussion with a friend who says an electric heater is horribly inefficient, less than 5%.)&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-11-30T23:15:26.767" LastActivityDate="2010-12-01T09:27:38.277" Title="How efficient is an electric heater?" Tags="&lt;thermodynamics&gt;&lt;energy&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="1494" PostTypeId="2" ParentId="1493" CreationDate="2010-11-30T23:30:56.057" Score="19" ViewCount="" Body="&lt;p&gt;It depends on what you mean by efficiency.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you want to heat your house.  An electric heater like you're considering would do this by converting electrical energy directly into heat.  Pretty much all the electrical energy does get converted to heat, as you suggest.  The energy used to get a certain amount of heat into the house is simply equal to that amount of heat.  In that sense, the electric heater is 100% efficient, since energy not directly turned into heat will be turned into heat soon.  That isn't a very useful way of thinking about efficiency, though, because any form of energy in your house will probably decay into heat energy pretty quickly.  Your computer, television, and refrigerator are 100% efficient at heating your house from this point of view, because although they do things other than generate heat, the energy they use to do those things becomes heat in short order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, a heat pump would heat your house by taking heat from the outside and moving it inside.  The energy it needs to do this depends on the outside and inside temperatures.  If the temperatures inside and outside are $T_i$ and $T_o$, an ideal heat pump (i.e. a &lt;a href=&quot;http://en.wikipedia.org/wiki/Carnot_heat_engine&quot; rel=&quot;nofollow&quot;&gt;Carnot engine&lt;/a&gt;) would require &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1-\frac{T_o}{T_i})*dH$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joules of work energy to move $dH$ Joules of heat energy from outside to inside (if the outside temperature were greater, this number is negative, meaning the heat pump can extract energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The efficiency of the electric heater, compared to the idealized heat pump, is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1-\frac{T_o}{T_i}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for given inside and outside temperatures.  When the inside and outside temperatures are the same, the electric heater is zero percent efficient.  If it's 0C outside and 25C inside, the electric heater is about 8% efficient.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-11-30T23:30:56.057" CommentCount="7" />
  <row Id="1495" PostTypeId="2" ParentId="1491" CreationDate="2010-11-30T23:56:34.747" Score="2" ViewCount="" Body="&lt;p&gt;Density matrices often admit an interesting geometric interpretations when you map them to the space of generalized Bloch vectors, see for example the book I. Bengtsson, K. Życzkowski, &lt;a href=&quot;http://books.google.com/books?id=aA4vXMbuOTUC&amp;amp;lpg=PA416&amp;amp;ots=IaBET8vZuG&amp;amp;dq=geometry%20of%20quantum%20states&amp;amp;pg=PA416#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Geometry of quantum states&lt;/a&gt;, 2006. I won't be surprised if it turns out that the result has something to do with the coset space $SU(2N)/[SU(N)\times SU(N)]$, where N is the dimensionality of the Hilbert space of a single qubit.&lt;/p&gt;&#xA;" OwnerUserId="59" LastActivityDate="2010-11-30T23:56:34.747" />
  <row Id="1496" PostTypeId="2" ParentId="1493" CreationDate="2010-11-30T23:59:42.757" Score="5" ViewCount="" Body="&lt;p&gt;It was a good answer Mark. Of course by drawing a lot of current some Joule heating will happen outside the house as well, in the transmission lines and transformers especially. So the efficiency will get lower depending upon where you draw the (electrical box). Some of the energy from the TV and refrigerator will also escape from the house before being degraded to heat (harmonics in&#xA;the electric lines, noise, and light from the picture tube (I'd bet these loses are well under one percent)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case air sourced heat pumps are far less efficient than the theoretical limit. The figure I've seen is that a quality heat pump might give you about 3.5 times as much heat as the electrical power input. Not bad compared to the alternatives, but a very long way from Carnot efficiency. Using real (im)perfect working materials is really a drag. &lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2010-11-30T23:59:42.757" CommentCount="1" />
  <row Id="1498" PostTypeId="2" ParentId="1479" CreationDate="2010-12-01T00:44:46.483" Score="3" ViewCount="" Body="&lt;p&gt;The visualisation shows an 'hour glass' shape because you're looking at it from above. The actual layout of the points is like two pie pieces put point to point. The visualisation is of the distribution of galaxies in the universe at large distances, and has either come from the Sloan digital sky survey (SDSS) or the 2 degree field galaxy redshift survey (2dfGRS).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making these plots requires a large amount of telescope time, as for each galaxy you must measure the spectra of the light it emits accurately enough to get the redshift, which is a measure of how distant it is. The surveys thus don't cover the whole sky, instead they are done as stripes, which leads to the pie shape when projected through the universe. In order to observe the structure that forms from the galaxy distribution you want a continuous sample which goes over a large angular range, which the stripe gives in one direction. You also want to make sure that you aren't sampling a special region of sky, hence there's another sample taken on the other side of the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, the shape of the sample is purely due to the survey format, the actual structure of galaxies will fill a sphere. &lt;/p&gt;&#xA;" OwnerUserId="695" LastActivityDate="2010-12-01T00:44:46.483" />
  <row Id="1499" PostTypeId="2" ParentId="1486" CreationDate="2010-12-01T01:13:14.883" Score="7" ViewCount="" Body="&lt;p&gt;I think you make some quite confusing statements, so let me be a little bit too explicit.&#xA;First, for any system whose laws don't depend explicitly on time one obtains conserved energy as an integral of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Central force systems are invariant under the action of $SO(3)$. This is so because both kinetic and potential energy are scalars. Noether's theorem then tells us that the generators of this action must be conserved. Now, $SO(3)$ is three-dimensional and it can be shown that the corresponding generators are e.g. $L_x$, $L_y$, $L_z$ components of angular momentum vector $\mathbf L$. These generators correspond to infinitesimal rotations around $x$, $y$ and $z$ axis respectively. You can check that your Lagrangian is indeed invariant with respect to these transformations (this calculation is a bit tedious though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have in total four conserved quantities $E$ and $\mathbf L$. Now, important fact to note is that setting $\dot{\phi} = 0$ amounts to redefining coordinates and taking $\phi$ as an azimuthal parameter. So instead let's just work with the condition $\theta = \pi/2$. This indeed eliminates two quantities (not just one as you thought) because this restricts our configuration variety to a plane and we remain only with action of $SO(2)$ group (the surviving generator being $L_z$). What is being implicitly used is the fact that $\mathbf L$ is orthogonal to this plane and so the fact that $L_y$ and $L_z$ is conserved is automatically satisfied because they are both zero.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-01T01:13:14.883" CommentCount="1" />
  <row Id="1500" PostTypeId="2" ParentId="1019" CreationDate="2010-12-01T01:17:06.240" Score="-5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Something cannot come from nothing&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it can: In quantum field theories, the vacuum is &lt;em&gt;not empty&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2010-12-01T01:17:06.240" CommunityOwnedDate="2010-12-01T01:17:06.240" CommentCount="3" />
  <row Id="1501" PostTypeId="2" ParentId="1263" CreationDate="2010-12-01T04:44:57.167" Score="7" ViewCount="" Body="&lt;p&gt;I don't think anyone here has really answered your question.  In this case, the sound is &quot;focused&quot; using &lt;a href=&quot;http://en.wikipedia.org/wiki/Phased_array&quot; rel=&quot;nofollow&quot;&gt;phased arrays&lt;/a&gt;.  The face of the audio spotlight has multiple transducers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/3xSRA.jpg&quot; alt=&quot;audio spotlight close-up&quot;&gt; &lt;a href=&quot;http://www.flickr.com/photos/chris_carter_/2756363738/&quot; rel=&quot;nofollow&quot;&gt;Flickr&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same signal is output from each of them, but delayed slightly by different amounts, so that the wavefronts all reach the same point in front of the device at the same time.  This &quot;virtual focus&quot; is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Beamforming&quot; rel=&quot;nofollow&quot;&gt;beamforming&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/FyHW8.jpg&quot; alt=&quot;phased array beam steering and focusing&quot;&gt; &lt;a href=&quot;http://www.ob-ultrasound.net/somers.html&quot; rel=&quot;nofollow&quot;&gt;ref&lt;/a&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/PQm44.gif&quot; alt=&quot;phased-array principles&quot;&gt; &lt;a href=&quot;http://www.bercli.net/documentation/article_principles.htm&quot; rel=&quot;nofollow&quot;&gt;ref&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how modern radars focus their beams, too.  Instead of spinning a satellite dish around, they have lots of little elements that don't move, but the signals are delayed to produce different beam shapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/QQOhd.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;img src=&quot;http://i.stack.imgur.com/TasoC.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="176" LastEditorUserId="176" LastEditorDisplayName="" LastEditDate="2010-12-08T15:32:41.640" LastActivityDate="2010-12-08T15:32:41.640" CommentCount="4" />
  <row Id="1502" PostTypeId="2" ParentId="1467" CreationDate="2010-12-01T04:59:49.060" Score="3" ViewCount="" Body="&lt;p&gt;It depends also on the shape of the object. If you assume the trampoline is circular, and the object is much smaller (like a point mass) then you can start developing the equations. You have to know the initial tension of the trampoline, and also assume the material non-elastic but supsended by perfect strings in a radial direction (with known stiffness). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some math the static deflection (with pre-tension) obeys the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{W}{k\, R}=\tan\theta+\left(\frac{F_{0}}{k\, R}-1\right)\sin\theta $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $R$ is the radius of the trampoline then the dip is $H=R\,\tan\theta$ and so $\theta$ is the angle from horizontal that the cone makes. For any given angle $\theta$ the trampoline supports weight $W$ (given above) given total stiffness of $k$ and pre-tension of $F_{0}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the above will give you the weight $W$ it will support given a dip $H$. It is the reciprocal of what you want, but it is solvable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the trampoline has $N$ linear springs each with stiffness rate of $k_i$ then the total stiffness (springs in parallel) is $k=k_i\,N$. To define the pre-tension $F_{0}$ assume that the free radius of the trampoline surface is $R$, but the springs are located at $R_0$ then the pre-tension is $F_{0}=k\,(R_0-R)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: A trampoline of 12 feet in diameter needs $F_{0}=100$ lbs total of pulling to string into a 12.5 foot ring. The stiffness is $k=\frac{100}{6}$ in pounds per inch. To dip the trampoline by 5 feet $\tan(\theta)=\frac{H}{R}=\frac{5\times12}{12\times12}$. Plug these into the above and you should get $W=115.4$ lbs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I am going to confuse some people because I am treating radial quantities such as stiffness and loads as linear, but it works out (just use cylindrical coordinates).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Approximation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Small angle approximation (weight &amp;lt; 10% pre-tension, cone angle &amp;lt; 6°)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W = F_{0}\,\frac{H}{R}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: Using the same numbers as above a $W=10$ lbs weight will dip $H = (12\times12)\frac{W}{100} = 14.4$ inches. The full solution above gives $13.1$ inches&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Theory&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The deformed shape of the trampoline is a perfect cone. The distance from the center to where the springs start in the deformed state is always equal to $R$. The extension of the springs tension is then $F=F_{0}+k\,\left(\frac{R}{\cos\theta}-R\right)$ which needs to be balanced by the weight as $W = F\,\sin\theta$. The pieces come together to make the equation shown above.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Update&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Solution is corrected for the fact the radial distance is constant, not the surface area of the trampoline. As the trampoline deforms its perimeter crumples and folds on itself like a napkin when lifted from the center.&lt;/p&gt;&#xA;" OwnerUserId="392" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2011-07-31T10:49:12.737" LastActivityDate="2011-07-31T10:49:12.737" CommentCount="4" />
  <row Id="1503" PostTypeId="1" AcceptedAnswerId="1506" CreationDate="2010-12-01T09:16:47.230" Score="1" ViewCount="209" Body="&lt;p&gt;When a charged particle moves in an electric field, the field performs work on the particle. Thus, the energy of the field decreases, turning into kinetic energy of the particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the magnetic field of a permanent magnetic similarly lose energy and perform work when moving a conductor with a current?&lt;/p&gt;&#xA;" OwnerUserId="690" LastEditorUserId="690" LastEditorDisplayName="" LastEditDate="2010-12-11T18:04:30.973" LastActivityDate="2010-12-11T18:04:30.973" Title="Does the energy of a magnetic field decrease when it moves a conductor carrying a current?" Tags="&lt;electromagnetism&gt;&lt;energy&gt;&lt;magnets&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="1504" PostTypeId="2" ParentId="1493" CreationDate="2010-12-01T09:27:38.277" Score="6" ViewCount="" Body="&lt;p&gt;The efficiency is 100%, which is considered a poor heating efficiency. It is 100% because all light, motion, etc produced by the heater at some point becomes dissipated to heat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same heating efficiency have all other completely-in-door devices: &lt;strong&gt;heating with a TV set is as efficient as an electric heater&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is a better way to heat. A typical air conditioner can heat your house by 3 joules consuming only one joule of electric energy, making it &lt;strong&gt;more than 300%&lt;/strong&gt; efficient. This is done by actually cooling the outside world.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-12-01T09:27:38.277" />
  <row Id="1505" PostTypeId="2" ParentId="1019" CreationDate="2010-12-01T09:50:13.927" Score="4" ViewCount="" Body="&lt;p&gt;Some sport instructors would tell you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Running on a treadmill is easier because you only have to jump, while on the street you also push forward&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggested then that running in a train should be the easiest, by this line of thought. However, it is true that starting to run (or, accelerating) is indeed easier. Or the air resistance, unless there's a slowest forward wind. Or, the randomly changing slope. Also, lack of the air conditioner. That pretty much summarizes the difference.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-12-01T09:50:13.927" CommunityOwnedDate="2010-12-01T09:50:13.927" CommentCount="3" />
  <row Id="1506" PostTypeId="2" ParentId="1503" CreationDate="2010-12-01T10:45:39.143" Score="2" ViewCount="" Body="&lt;p&gt;Magnetic forces do not perform work because they are always perpendicular to the motion of the charged particle they act on.  However, it is possible to transform the energy stored in a magnetic field into the bulk motion of a conductor carrying a current.  I'll give one heuristic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider two long wires running parallel.  Give them currents running in the same direction, and hold the wires still for a time that's long compared to the distance between them divided by the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there is a magnetic field from the first wire at the location of the second and vice versa.  The wires attract each other.  Once released, they experience an acceleration towards each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this analysis only holds up until the first moment the wires are released.  After that, they are moving towards each other.  This motion of the wires entails a motion of the charge they carry - a current.  We will need to account for this current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say the wires run horizontally on your computer monitor, and the current is to the right.  Then the top wire is pulled down, and its current is now mostly to the right and a little bit down.  The magnetic field it feels is pointing towards you, out of the monitor.  Thus, the force on this top wire points mostly down, but it also points a little bit backwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the force on the bottom wire is mostly up, but also a little bit backwards.  The backwards components of the magnetic forces act to decrease the currents in the wires.  If  the currents in the wire decrease, the magnetic field gets weaker.  In this way energy from the magnetic fields gets converted into the bulk motion of the wires carrying the currents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnetic forces did not actually do any of the work here, though.  In the absence of any wire, the charged particles in the magnetic field would like to move in circles.  Instead, electrical forces between the particles and the wire they're trapped in forced the particles to stay inside the wire instead.  These electrical interactions between the charged particles and the rest of the wire did the work that accelerated the wires towards each other.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-12-01T10:45:39.143" />
  <row Id="1507" PostTypeId="1" AcceptedAnswerId="1851" CreationDate="2010-12-01T13:07:17.953" Score="9" ViewCount="493" Body="&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; I provided an answer of my own (reflecting the things I discovered since I asked the question). But there is still lot to be added. I'd love to hear about other people's opinions on the solutions and relations among them. In particular &lt;strong&gt;short, intuitive descriptions of the methods used&lt;/strong&gt;. Come on, the bounty awaits ;-)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, this might look like a question into the history of Ising model but actually it's about physics. In particular I want to learn about all the approaches to Ising model that have in some way or other relation to Onsager's solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I am asking three questions at once but they are all very related so I thought it's better to put them under one umbrella. If you think it would be better to split please let me know.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;When reading articles and listening to lectures one often encounters so called &lt;a href=&quot;http://en.wikipedia.org/wiki/Ising_model#Two_dimensions_.E2.80.93_Onsager.27s_solution&quot; rel=&quot;nofollow&quot;&gt;Onsager's solution&lt;/a&gt;. This is obviously very famous, a first case of a complete solution of a microscopic system that exhibits phase transition.&#xA;So it is pretty surprising that each time I hear about it, the derivation is (at least ostensibly) completely different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more precise and give some examples:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The most common approach seems to be through computation of eigenvalues of some transfer matrix.&lt;/li&gt;&#xA;&lt;li&gt;There are few approaches through Peierl's contour model. This can then be reformulated in terms of a model of cycles on edges and then one can either proceed by cluster expansion or again by some matrix calculations.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The solutions differ in what type of matrix they use and also whether or not they employ Fourier transform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my questions (or rather requests) are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Try to give another example of an approach that might be called &lt;em&gt;Onsager's solution&lt;/em&gt; (you can also include variations of the ones I already mentioned).&lt;/li&gt;&#xA;  &lt;li&gt;Are all of these approaches really that different? Argue why or why not some (or better yet, &lt;strong&gt;all&lt;/strong&gt;) of them could happen to be equivalent.&lt;/li&gt;&#xA;  &lt;li&gt;What approach did Onsager actually take in his original paper. In other words, which of the numerous &lt;em&gt;Onsager's solutions&lt;/em&gt; is actually &lt;strong&gt;&lt;em&gt;the&lt;/em&gt; Onsager's solution&lt;/strong&gt;.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For 3.: I looked at the paper for a bit and I am a little perplexed. On the one hand it looks like it might be related to transfer matrix but on the other hand it talks about quaternion algebras. Now that might be just a quirk of Onsager's approach to 4x4 matrices that pop basically in every other solution but I'll need some time to understand it; so any help is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-08T13:48:43.927" LastActivityDate="2010-12-12T17:05:36.957" Title="How many Onsager's solutions are there?" Tags="&lt;statistical-mechanics&gt;&lt;ising-model&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="3" />
  <row Id="1508" PostTypeId="1" AcceptedAnswerId="1511" CreationDate="2010-12-01T13:35:17.940" Score="1" ViewCount="125" Body="&lt;p&gt;I've read that QM operates in a Hilbert space (where the state functions live). I dont know if its meaningful to ask such a question, what are the answers to an anlogous questions on GR and newtonian gravity?&lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-01T15:06:34.967" LastActivityDate="2010-12-01T15:24:50.280" Title="What is an analog to QM's Hilbert space in GR?" Tags="&lt;theory&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="1509" PostTypeId="2" ParentId="1508" CreationDate="2010-12-01T14:42:49.560" Score="0" ViewCount="" Body="&lt;p&gt;First, I'll assume that you're talking about quantization. To understand how to quantize GR it is absolutely necessary to give an account (however sketchy) of the approach used to quantize simpler systems.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Classical mechanics&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This is a procedure whereby one transfers from the classical point of view (Newtonian mechanics or equivalently Lagrangian or Hamiltonian mechanics) to the quantum point of view. Now, there are some general prescriptions how one can quantize classical mechanical systems. The most common one is that one replaces the phase space by Hilbert space, functions on phase space by operators on the Hilbert space and &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson_bracket&quot; rel=&quot;nofollow&quot;&gt;Poisson bracket&lt;/a&gt; of functions by commutator of the operators.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Field theory&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The previous paragraph was only dealing with mechanics, i.e. case where there are only few degrees of freedom. But GR is a field theory (of gravitational field) and is actually a kind of gauge theory (but a little special at that). One has to first learn how to quantize classical fields and then gauge fields. To do that you can replace the (infinite-dimensional) phase space of the field by (very large) Hilbert space and produce an analogue of Poisson brackets called &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_bracket&quot; rel=&quot;nofollow&quot;&gt;Dirac bracket&lt;/a&gt; which you then replace by commutators. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(The second very common approach to quantization is via &lt;a href=&quot;http://en.wikipedia.org/wiki/Path_integral_formulation&quot; rel=&quot;nofollow&quot;&gt;path integral&lt;/a&gt; for which you don't need any operators but I won't elaborate on that here because it is a huge area that would take us far away off the topic of your question)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then to quantize a gauge theory with its own huge gauge symmetry one has to carry out a very nontrivial discussion about the structure of these Dirac brackets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(There also exist other approaches to this but none of them is particularly easy for a beginner. If you're interested see &lt;a href=&quot;http://en.wikipedia.org/wiki/Faddeev%2dPopov_ghost&quot; rel=&quot;nofollow&quot;&gt;Faddeev-Popov ghosts&lt;/a&gt; in path integral gauge quantization and &lt;a href=&quot;http://en.wikipedia.org/wiki/BRST_quantization&quot; rel=&quot;nofollow&quot;&gt;BRST quantization&lt;/a&gt;)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Gravitation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now, the thing is that GR (as a field theory) is hard to quantize. I.e. if you repeat the above approach for GR, you'll find out that your quantum theory doesn't make sense (because it is not &lt;a href=&quot;http://en.wikipedia.org/wiki/Renormalization#Renormalizability&quot; rel=&quot;nofollow&quot;&gt;renormalizable&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This suggests that something more than naive approach is needed. And there are actually lots of them. For one thing, one can quantize gravity in certain special dimensions (like 2+1) if one generalizes GR a little (this was done by Witten in '80s). There are also various reformulations that relate quantum gravity and QFT (like &lt;a href=&quot;http://en.wikipedia.org/wiki/AdS/CFT_correspondence&quot; rel=&quot;nofollow&quot;&gt;AdS/CFT correspondence&lt;/a&gt;). There is also &lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix_string_theory&quot; rel=&quot;nofollow&quot;&gt;matrix string theory&lt;/a&gt; that shows duality between matrix quantum mechanics and GR (as pointed out to me by Matt in &lt;a href=&quot;http://physics.stackexchange.com/questions/1073/how-does-classical-gr-concept-of-space-time-emerge-from-string-theory&quot;&gt;this question of mine&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, quantization of GR is very hard. There are many theories and as of yet there is no experimental evidence that would let us know which one is the correct one.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-01T14:50:19.823" LastActivityDate="2010-12-01T14:50:19.823" CommentCount="2" />
  <row Id="1510" PostTypeId="2" ParentId="629" CreationDate="2010-12-01T15:06:55.713" Score="6" ViewCount="" Body="&lt;p&gt;We have two effects going on. One is that the person is countering a constant acceleration, but when it rapidly stops there is nothing to stop the counterforce (lean) from creating acceleration of the person. The train is not a rigidbody, but in fact is a deformable body, so your intuition which assumes it posses infinite stiffness is wrong.. A deformable body&#xA;under stress (to transmit the deceleration from the brakes to the distributed mass of the train car), is also under strain (i.e. the body of the train experiences some deformation). The instant the braking force is removed, these stresses and strains remain, and the train car is in a nonequlibrium state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try modeling it as mass on a spring attached to a rigid body. Subject the rigid body to a constant acceleration, and let the spring/mass come to its equilibrium position. Then eliminate the acceleration. The spring is not in its equilibrium position, and the mass will see acceleration.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-12-01T15:06:55.713" CommentCount="2" />
  <row Id="1511" PostTypeId="2" ParentId="1508" CreationDate="2010-12-01T15:09:52.047" Score="2" ViewCount="" Body="&lt;p&gt;I interpreted your question differently, more like a mathematics question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Quantum Mechanics, we basically have an equation, the Schrödinger equation, which is a differential equation on the space of square-integrable complex valued functions. This space is a Hilbert space, which means that it is a vector space, and it also has a nice topological structure, basically all Cauchy-sequences of vectors converge in that space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Newtonian mechanics, the equations are defined on phase space, which is basically a $6N$-dimensional space, $N$ is the total number of particles, on which coordinates for a point consist of the positions and momenta of each particle you want to describe. The solution of the equations induces a flow on this phase space. The structure of phase space is usually that of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Symplectic_manifold&quot; rel=&quot;nofollow&quot;&gt;symplectic manifold&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In General Relativity, the equations are Einstein's field equations. They link the Riemann tensor to the energy-momentum tensor. They are difficult to solve in the sense that they are nonlinear and you have to specify an energy-momentum tensor, but this tensor will also depend on the geometry of space-time, thus the Riemann tensor. So you have to solve in one go for the geometry and energy-matter distribution. In practice, many simplifying assumptions will be made. But the &quot;space&quot; of solutions is the space of geometries and energy-matter distributions compatible with the field equations.&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2010-12-01T15:24:50.280" LastActivityDate="2010-12-01T15:24:50.280" CommentCount="8" />
  <row Id="1513" PostTypeId="1" AcceptedAnswerId="1514" CreationDate="2010-12-01T15:17:56.067" Score="7" ViewCount="419" Body="&lt;p&gt;I have a digital watch, rated to go underwater to 100m. When it is underwater it can be read normally, up until you reach a certain angle, then suddenly, it becomes almost like a mirror, reflecting all the light that hits it, almost perfectly (you cannot read the digits, the entire surface becomes reflective.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen this happen with other waterproof watches too, so I don't think it's unique to mine or the specific model. I'm wondering what causes this? My physics teacher was stumped when I told him about this (we're doing lenses and imaging in physics right now.) I think it has something to do with internal refraction. I haven't been able to measure the angle it becomes reflective accurately, I estimate about 30-40 degrees. Near this critical point, it can be half and half reflective, where only half becomes a mirror, but it's always either reflective or not - never in between being reflective and non-reflective.&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-12-02T16:10:54.627" LastActivityDate="2010-12-02T16:10:54.627" Title="Why does my watch act like a mirror under water?" Tags="&lt;optics&gt;&lt;light&gt;&lt;mirror&gt;" AnswerCount="1" />
  <row Id="1514" PostTypeId="2" ParentId="1513" CreationDate="2010-12-01T15:35:22.403" Score="10" ViewCount="" Body="&lt;p&gt;What you are observing is total internal reflection.  Snell's law tells you that, for a ray transmitting through a surface $n_{1}\sin\theta_{1} = n_{2}\sin\theta_{2}$, where $\theta$ represents the angle of reflection from the surface, $n$ represents the index of refraction of the substance in question, and the labels 1 and 2 represent the source medium and the destination medium.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n_{1}&amp;gt;n_{2}$, as would be the case for light leaving water ($n\approx 1.33$)and entering the air ($n\approx 1$)inside of your watch, simple algebra will tell you that there is a range of $\theta_{1}$ at which you will find that Snell's law predicts $\sin \theta_{2}&amp;gt;1$.  For this range of angles, since you can't solve for $\theta_{2}$, light cannot be transmitted, and must be reflected.  So the watch looks like a mirror.  In fact, if you flip over, and look at the surface of the water, you will find a portion of the surface of the water looks like a mirror, too!&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-12-01T15:35:22.403" CommentCount="10" />
  <row Id="1515" PostTypeId="2" ParentId="1019" CreationDate="2010-12-01T15:37:19.847" Score="10" ViewCount="" Body="&lt;p&gt;Some I have heard:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Heavier objects fall faster (this is just plain wrong.) However, bigger and smaller objects in a typical Earth environment would fall at different rates due to air resistance, but actual mass has no influence.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Two cars colliding at 60 mph is the same as one car colliding into a wall at 120 mph. I think MythBusters did something on this.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Electrons travel really fast around a circuit: actually, they travel really slow but it is similar to Newton's cradle in that a small movement in one ball can transfer the energy to the last one almost instaneously.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The laws of thermodynamics have been broken by some guy in a garage with some magnets. Well, no, they still seem to be intact, and a lot of modern science depends on them!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;That an oscilloscope trace travels faster than the speed of light on expensive high speed analog scopes (~1-2GHz.) This is not quite true: although the beam may sweep the surface of the CRT faster than &lt;em&gt;c&lt;/em&gt; (due to the relatively small movement at the neck of the CRT), the trace cannot communicate information faster than &lt;em&gt;c&lt;/em&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;More related to chemistry, but the fact that water can have &quot;memory&quot; and all that homeopathic nonsense that con arti^H^H^H^H^H^H^H^H homeopaths spurt out.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="104" LastActivityDate="2010-12-01T15:37:19.847" CommunityOwnedDate="2010-12-01T15:37:19.847" CommentCount="9" />
  <row Id="1516" PostTypeId="2" ParentId="1467" CreationDate="2010-12-01T17:06:48.683" Score="2" ViewCount="" Body="&lt;p&gt;Since @jalaxiou provided a solution for the static case looking at the balance of forces, I will try to provide a solution for the dynamic case from energetic considerations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ǹotation and energies&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will first restrict my self to a 1D trampoline, &lt;em&gt;i.e.&lt;/em&gt; a rope with 2 springs of stiffness $k$. The length of the string + the springs without pretention is $2R_0$, ant the diameter of the trampoline is $2R&amp;gt;2R_0$. When a point-mass in the middle of the trampoline is at a depth $H$ below the initial level, each spring is elongated by $\sqrt{H^2+R^2}-R_0$. The elastic potential energy of the 2 springs is therefore :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_{\text{elastic}}=k(\sqrt{H^2+R^2}-R_0)^2 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational potential energy of the mass is &#xA;$$E_{\text{gravitational}}=-mgH.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Static solution&lt;/strong&gt;&#xA;The static solution corresponds to the minimum value of $E_{\text{gravitational}}+E_{\text{elastic}}$ obtained by derivation i*H*. However I get a fourth order equation which is not trivial to solve analytically, except for small $\frac{H}{R}$, where $H\simeq\frac{mg}{2k(1-R_0/R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dynamic solution&lt;/strong&gt;&#xA;For the dynamic solution, you add the kinetic energy $E_{\text{kinetic}}=\tfrac12mv^2$ of the mass into the game. If you suppose that there is no friction (yo have an ideal trampoline), the total energy $E=E_{\text{elastic}}+E_{\text{gravitational}}+E_{\text{kinetic}}$ is conserved during the motion. If the motion is vertical, the kinetic energy is 0 at the top and the bottom of the movement, since it is by definition the point where the vertical speed is 0. If you make jump of height $H_J$ above the trampoline, the energy at the top of the jump is $mgH_J+k(R-R_0)^2$. comparing thin energy with the one at the bottom of the movement, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{gather}&#xA;  mgH_J+k(R-R_0)^2=-mgH+k(\sqrt{H^2+R^2}-R_0)^2 \\&#xA;  mg(H+H_J)=k(H^2-2R_0(\sqrt{R^2+H^2}-R))&#xA;\end{gather}&#xA;whch can also me converted to a fourth order polynomial in $H$, with probably ugly but analytical solutions. In the limit of small deformation $H\ll R$, we have&#xA;\begin{gather}&#xA;  mg(H+H_J)\simeq k(H^2-2R_0\frac{H^2}{2R})=k(1-\frac{R_0}R)H^2&#xA;\end{gather}&#xA;which can be solved for H and gives:&#xA;$$&#xD;&#xA;  H=\frac{mg}{2k(1-\frac{R_0}{R})}\left(1+\sqrt{1+\frac{4k}{mg}(1-\frac{R_0}{R})H_J}\right).&#xD;&#xA;$$&#xD;&#xA;This ugly expression simplifies for small jumps, when $H_J\ll\frac{mg}{4k(1-R_0/R)}$ into&#xA;$$ H\simeq\frac{mg}{k(1-\frac{R_0}{R})}+H_J $$&#xD;&#xA;If you look at the total amplitude of the movement, between $z=+H_J$ and $z=-H$, you see oscillations centered around the static equilibrium postion which is expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, if you want a circular trampoline, with $n$ spring instead of a rope with 2 springs, you just have to replace $k$ by $\frac n2 k$ in the equations above by symmetry.&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2010-12-01T17:06:48.683" />
  <row Id="1518" PostTypeId="1" AcceptedAnswerId="1594" CreationDate="2010-12-01T17:32:14.507" Score="2" ViewCount="130" Body="&lt;p&gt;Do someone knows the units of the spectra provided &lt;a href=&quot;http://spectral.joensuu.fi/index.php?page=database&amp;amp;info=artlights#1&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; ? It seems obvious enough that it's said nowhere, but even Wikipedia and other sites are quite blurry on this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is it power (watts), radiance(watts/m²/str), something else ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks !&lt;/p&gt;&#xA;" OwnerUserId="702" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-01T17:45:06.707" LastActivityDate="2010-12-03T16:22:41.440" Title="Light emission spectrum units" Tags="&lt;light&gt;&lt;si-units&gt;&lt;spectroscopy&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="1519" PostTypeId="2" ParentId="85" CreationDate="2010-12-01T17:42:15.303" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://physicsworld.com/cws/article/news/36756&quot; rel=&quot;nofollow&quot;&gt;Proton and neutron masses calculated from first principles&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="238" LastActivityDate="2010-12-01T17:42:15.303" CommentCount="1" />
  <row Id="1520" PostTypeId="2" ParentId="1419" CreationDate="2010-12-01T17:44:49.500" Score="1" ViewCount="" Body="&lt;p&gt;Counter-question:&#xA;What is the definition of &lt;strong&gt;mass&lt;/strong&gt;, when a mass distribution given?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strange question, isn't it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you have the mass distribution you got to have the momentum &lt;strong&gt;distribution&lt;/strong&gt; as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is simply ρ*v. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2010-12-01T17:44:49.500" />
  <row Id="1521" PostTypeId="2" ParentId="1518" CreationDate="2010-12-01T18:15:26.363" Score="3" ViewCount="" Body="&lt;p&gt;Usually, the spectrum is in arbitrary unit, proportional to the power, the radiance or whatever. The information the spectrum provides is the relative radiance of various wavelengths which is contained in the spectrum. If you want the radiance of one wavelength band, you simply multiply the relevant band of a normalized spectrum (of integral 1) by the total radiance of your source.&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2010-12-01T18:15:26.363" CommentCount="1" />
  <row Id="1522" PostTypeId="2" ParentId="1518" CreationDate="2010-12-01T18:30:22.887" Score="0" ViewCount="" Body="&lt;p&gt;Most spectrometers work by spatially separating light into its component frequencies by means of a diffraction grating, and measuring that with a CCD array. That means measuring the intensity*, or power per unit area, the units of which are watts per square meter. However, there is usually an unknown scaling factor involved, so don't expect the values to actually be in W/m². I would guess that the values in these files are normalized so that a value of 1 is the intensity at which the corresponding CCD pixel saturates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Strictly speaking, you are measuring the intensity present in one range of frequencies, but it's simpler and usually justified to just treat it as one frequency.&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2010-12-01T18:30:22.887" />
  <row Id="1523" PostTypeId="1" AcceptedAnswerId="1525" CreationDate="2010-12-01T18:31:12.890" Score="8" ViewCount="340" Body="&lt;p&gt;I am looking for some literature on the Ising model, but I'm having a hard time doing so. All the documentation I seem to find is way over my knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you direct me to some documentation on it that can be parsed by my puny undergrad brain? If the answer is negative, can you explain it right there, on the answer form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a bunch,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Francisco&lt;/p&gt;&#xA;" OwnerUserId="704" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-12-01T18:33:17.253" LastActivityDate="2010-12-01T19:56:01.880" Title="Ising model for dummies" Tags="&lt;statistical-mechanics&gt;&lt;ising-model&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="5" />
  <row Id="1524" PostTypeId="1" AcceptedAnswerId="1526" CreationDate="2010-12-01T18:37:59.240" Score="2" ViewCount="375" Body="&lt;p&gt;How can we do this computation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\iiint_{R^3} \frac{e^{ik&amp;#39;r}}{r} e^{ik_1x+k_2y+k_3z}dx dy dz$ &#xA;where $r=\sqrt{x^2+y^2+z^2}$ ? I think we must use distributions...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physically, it's equivalent to find wave vectors k distribution and to write a spherical wave as sum of plane waves...&#xA;I know the formula for the inverse problem: write a plane wave as sum of spherical waves...The solution in this case is a serie of spherical harmonics and shperical bessel functions...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone can help me please? :-)&lt;/p&gt;&#xA;" OwnerUserId="558" LastEditorUserId="558" LastEditorDisplayName="" LastEditDate="2010-12-01T20:54:38.643" LastActivityDate="2010-12-01T20:54:38.643" Title="Spherical wave as sum of plane waves" Tags="&lt;mathematical-physics&gt;&lt;mathematics&gt;&lt;waves&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="0" />
  <row Id="1525" PostTypeId="2" ParentId="1523" CreationDate="2010-12-01T19:01:21.103" Score="12" ViewCount="" Body="&lt;p&gt;The Ising model is a model, originally developed to describe ferromagnetism, but subsequently extended to more problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, it is an interaction model for spins. Imagine you have a system which is a collection of $N$ spins. Each spin $S_i$ has two possible states $+1$ or $-1$. Here you can imagine already a possible extension to more states. You can also imagine a different interpretation to the spins: $-1$ is a box containing no gas particle, $+1$ is a box containing a gas particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm getting ahead of myself. Let's stick to spins for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is to define the energy of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E= - \sum_i h_i S_i - \sum_{i \neq j} J_{ij} S_i S_j$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term can be interpreted as the contribution to the energy of the interaction of a spin with a local magnetic field. If the magnetic field is the same for all spins, then $h_i=h$ for all $i$. You see that aligning with the field will means a lower energy than going against the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second term represents interactions between spins within the system. If $J_{ij}&amp;gt;0$ spins which align will contribute negatively to the energy of the system, thus lowering the total energy. If $J_{ij}&amp;lt;0$ then anti-alignment will contribute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I still haven't specified is how the spins are structured. By choosing the coefficients $J_{ij}$ appropriately, I can introduce this structure. Suppose I want a 1-dimensional system, which is the one Ising originally solved. Then, you have a countable infinity of spins arranged along the real line with equal spacing. Ising imposed an interaction only allong neighboring spins. So spin $S_n$ can interact with spin $S_{n-1}$ and spin $S_{n+1}$. The energy formula I gave above becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E= - J \sum_{n} S_n S_{n+1} \; ,$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;if all spins can interact equally strongly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to each configuration of the system corresponds a certain energy. In statistical mechanics, we know that the probability of a certain configuration is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(\{S\}) \sim e^{-E(\{S\})/kT}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $T$ is the temperature. Or, if we compute the partition sum &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z=\sum_{\{S\}} e^{-E(\{S\})/kT}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can deduce the complete equilibrium thermodynamic properties of the system. In particular, we can see if there are phase transitions for certain values of the parameters. (There is none in the 1D case, which at the time, combined with the lack of attention his model was getting, made Ising abandon physics.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there are many more generalizations of this model. You can also make dynamical versions of the model where you are not just interested in the equilibrium configurations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's &lt;a href=&quot;http://en.wikipedia.org/wiki/Ising_model&quot; rel=&quot;nofollow&quot;&gt;the wikipedia page for Ising models&lt;/a&gt;. &#xA;Some references:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;H.E. Stanley 'Introduction to Phase Transitions and Critical Phenomena' Clarendon Press Oxford &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;J.M. Yeomans 'Statistical Mechanics of Phase Transitions' Clarendon&#xA;Press Oxford &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;R.J. Baxter 'Exactly Solved Models in Statistical Mechanics' Academic New York &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2010-12-01T19:38:34.257" LastActivityDate="2010-12-01T19:38:34.257" CommentCount="6" />
  <row Id="1526" PostTypeId="2" ParentId="1524" CreationDate="2010-12-01T19:31:21.067" Score="5" ViewCount="" Body="&lt;p&gt;From your description, I believe you want to find the Fourier transform of &#xA;$$ f(\mathbf r) = \frac{e^{ik&amp;#39;r}}r, $$&#xD;&#xA;and the wave can be recovered from the linear superposition of plane waves identified by &lt;strong&gt;k&lt;/strong&gt;&#xA;$$ f(\mathbf r) = \frac1{(2\pi)^{3/2}}\iiint \mathcal F[f](\mathbf k)e^{i\mathbf k\cdot\mathbf r} d^3 \mathbf k. $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spherical wave have spherical symmetry, so what you should do is to perform the integration in spherical coordinates instead of Cartesian. WLOG, assume &lt;strong&gt;k&lt;/strong&gt; is along the &lt;em&gt;z&lt;/em&gt; axis, thus&#xA;$$\begin{aligned}&#xD;&#xA;\mathcal F[f](k\hat{\mathbf z}) &amp;amp;= \frac1{(2\pi)^{3/2}} \iiint \frac{e^{ik&amp;#39;r}}r e^{-i\mathbf k\cdot \mathbf r} d^3\mathbf r \\&#xD;&#xA;&amp;amp;= \frac1{(2\pi)^{3/2}} \iiint \frac{e^{ik&amp;#39;r}}r e^{-ikr\cos\theta} r^2 \sin\theta dr d\theta d\phi \\&#xD;&#xA;&amp;amp;= \frac1{(2\pi)^{1/2}} \int_0^\infty \left(re^{ik&amp;#39;r} \int_0^{\pi} e^{-ikr\cos\theta} \sin\theta d\theta\right) dr \\&#xD;&#xA;&amp;amp;= \frac1{(2\pi)^{1/2}}  \int_0^\infty r e^{ik&amp;#39;r} \frac{2 \sin kr}{kr} dr \\&#xD;&#xA;&amp;amp;= \frac1k\sqrt{\frac2\pi}  \int_0^\infty e^{ik&amp;#39;r} \sin kr dr \\&#xD;&#xA;&amp;amp;= \sqrt{\frac2\pi}\frac1{k^2 - k&amp;#39;^2}&#xD;&#xA;\end{aligned}$$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-12-01T19:31:21.067" CommentCount="6" />
  <row Id="1527" PostTypeId="2" ParentId="1523" CreationDate="2010-12-01T19:56:01.880" Score="4" ViewCount="" Body="&lt;p&gt;Probably the best book in the field of spin models is the classic book by &lt;strong&gt;Baxter &quot;Exactly Solved Models in Statistical Mechanics&quot;&lt;/strong&gt;, it is very well written and contains both the basics and some more advanced topics.&lt;/p&gt;&#xA;" OwnerUserId="154" LastActivityDate="2010-12-01T19:56:01.880" CommentCount="2" />
  <row Id="1528" PostTypeId="1" AcceptedAnswerId="1529" CreationDate="2010-12-01T20:35:29.993" Score="9" ViewCount="396" Body="&lt;p&gt;There are six types of quarks, known as &lt;em&gt;flavors&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why where these types called flavors? &lt;/li&gt;&#xA;&lt;li&gt;Why do the flavors have such odd names (&lt;em&gt;up&lt;/em&gt;, &lt;em&gt;down&lt;/em&gt;, &lt;em&gt;charm&lt;/em&gt;, &lt;em&gt;strange&lt;/em&gt;, &lt;em&gt;top&lt;/em&gt;, and &lt;em&gt;bottom&lt;/em&gt;)?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="246" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T14:31:27.437" LastActivityDate="2011-02-03T14:31:27.437" Title="Why are quark types known as flavors?" Tags="&lt;particle-physics&gt;&lt;standard-model&gt;&lt;definition&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="1529" PostTypeId="2" ParentId="1528" CreationDate="2010-12-01T20:55:43.637" Score="15" ViewCount="" Body="&lt;p&gt;It is important to remember that words like this get used in spit-balling sessions and then stick. You have to think of a couple of guy sitting by a black board, coffee in hand saying something like&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;OK, OK! SO &lt;em&gt;that&lt;/em&gt; doesn't work. But what if we assume these things come in three flavors and ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's just a word made up on the spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, I think that &quot;up&quot; and &quot;down&quot; came from an analogy with spin. &quot;Strange&quot; interaction were so called because they violated rules of thumb for other known interactions...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that things got out of hand.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Side note: &quot;flavor&quot; is also used to distinguish the generation of leptons (and will no doubt be applied to any sterile neutrinos that make an appearance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side note$^2$: For a long time there were competing names for the third generation flavors. I'm a little sad that &quot;top&quot; and &quot;bottom&quot; won out against &quot;truth&quot; and &quot;beauty&quot;.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2010-12-02T02:11:07.220" LastActivityDate="2010-12-02T02:11:07.220" CommentCount="5" />
  <row Id="1530" PostTypeId="1" AcceptedAnswerId="1553" CreationDate="2010-12-01T21:01:12.027" Score="4" ViewCount="458" Body="&lt;p&gt;Is there a series of transmutations through nuclear decay that will result in the stable gold isotope ${}^{197}\mathrm{Au}$ ? How long will the process take? &lt;/p&gt;&#xA;" OwnerUserId="246" LastActivityDate="2011-01-17T11:43:23.153" Title="Is it possible to obtain gold through nuclear decay?" Tags="&lt;nuclear-physics&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="1531" PostTypeId="2" ParentId="1530" CreationDate="2010-12-01T21:07:46.697" Score="3" ViewCount="" Body="&lt;p&gt;Natural gold exists, so the answer to the first part of your question is unambiguously &lt;em&gt;&quot;Yes&quot;&lt;/em&gt;. 'Cause all those heavy elements get made by transmutation in supernovae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't answer the time scale thing because I haven't a table of the isotopes in front of me right now.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Checking with &lt;a href=&quot;http://ie.lbl.gov/education/isotopes.htm&quot; rel=&quot;nofollow&quot;&gt;http://ie.lbl.gov/education/isotopes.htm&lt;/a&gt; I find that $^{197}$Pt has a beta decay branching fraction of about 3% and a halflife of about 95 minutes, and $^{197}$Ir decays by beta with 5 minute halflife and $^{197}$Os decays by beta with a 64 hour halflife...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, you can chase this as far as you care to.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2010-12-01T21:40:36.563" LastActivityDate="2010-12-01T21:40:36.563" CommentCount="3" />
  <row Id="1532" PostTypeId="2" ParentId="1528" CreationDate="2010-12-01T21:52:34.943" Score="5" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There are six flavors of quarks.&#xA;  &quot;Flavors&quot; just means different kinds.&#xA;  The two lightest are called up and&#xA;  down.     &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The third quark is called strange. It&#xA;  was named after the &quot;strangely&quot; long&#xA;  lifetime of the K particle, the first&#xA;  composite particle found to contain&#xA;  this quark.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The fourth quark type, the charm&#xA;  quark, was named on a whim. It was&#xA;  discovered in 1974 almost&#xA;  simultaneously at both the Stanford&#xA;  Linear Accelerator Center (SLAC) and&#xA;  at Brookhaven National Laboratory.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The fifth and sixth quarks were&#xA;  sometimes called truth and beauty in&#xA;  the past, but even physicists thought&#xA;  that was too cute.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The bottom quark was first discovered&#xA;  at Fermi National Lab (Fermilab) in&#xA;  1977, in a composite particle called&#xA;  Upsilon ($\Upsilon$).     &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The top quark was discovered last,&#xA;  also at Fermilab, in 1995. It is the&#xA;  most massive quark. It had been&#xA;  predicted for a long time but had&#xA;  never been observed successfully until&#xA;  then.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Copied shamelessly from &lt;strong&gt;&lt;a href=&quot;http://particleadventure.org/quarknaming.html&quot; rel=&quot;nofollow&quot;&gt;How did quarks get their silly names?&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-12-05T13:58:49.987" LastActivityDate="2010-12-05T13:58:49.987" CommentCount="1" />
  <row Id="1533" PostTypeId="1" AcceptedAnswerId="1549" CreationDate="2010-12-01T23:08:04.113" Score="3" ViewCount="101" Body="&lt;p&gt;A lot of work has been done recently on electron dynamics using attosecond pump-probe techniques; for instance in &lt;a href=&quot;http://www.pnas.org/content/103/18/6793.abstract&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;.  In this particular paper, the authors photoionized the neutral tetrapeptide $\mathrm{TrpLeu}_3$ to the cation and observed the repopulation from the HOMO-1 to the HOMO (highest-occupied molecular orbital) of the cation in a time-resolved manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my question:  Since Heisenberg's uncertainty principle guarantees that the natural linewidth of a energy measurement will be greater than or equal to the inverse of the state's lifetime times $\hbar$, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta E \geq \frac{\hbar}{\Delta t}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;could you just measure the natural linewidth of the photons emitted by the transition between the transient state and the final state and deduce the lifetime of the transient state?  Or would the broadening be too large for subfemtosecond states to accurately determine the natural linewidth?  Or, would other effects such as Doppler broadening be too large to get a reasonable measurement of the natural linewidth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:  I'm a long way out of my field of specialty (computational chemistry) in this question, so I apologize in advance if this makes no sense whatsoever (please tell me why, though, in your answer if this is the case).  Also, obviously this would only yield an upper bound for the lifetime; my question is more along the lines of if it would yield an upper bound worth obtaining. &lt;/p&gt;&#xA;" OwnerUserId="144" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-12-02T11:09:16.320" LastActivityDate="2010-12-02T14:24:33.063" Title="Is it possible to determine timescales of electron dynamics from the natural linewidth of an electronic transition?" Tags="&lt;quantum-mechanics&gt;&lt;spectroscopy&gt;&lt;uncertainty-principle&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="1534" PostTypeId="1" CreationDate="2010-12-02T00:19:49.803" Score="3" ViewCount="258" Body="&lt;p&gt;A few quark related questions (I don't knowmuch about them other than there are 2 flavours concerning protons and neutrons)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are there 3 quarks in a proton or neutron? Why not 2 or 4?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an upper limit to the size of an atom, before gravity starts combining protons and neutrons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard in here that quarks don't make up all the mass of a neutron/proton, if so what does?&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2010-12-02T01:55:57.300" LastActivityDate="2010-12-02T01:55:57.300" Title="Why are there 3 quarks in proton?" Tags="&lt;standard-model&gt;&lt;subatomic&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="1535" PostTypeId="2" ParentId="1534" CreationDate="2010-12-02T00:49:44.913" Score="5" ViewCount="" Body="&lt;h2&gt;Why three quarks?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In very simple terms bound states of quarks (&lt;a href=&quot;http://en.wikipedia.org/wiki/Hadron&quot; rel=&quot;nofollow&quot;&gt;hadrons&lt;/a&gt;) have to be &lt;a href=&quot;http://en.wikipedia.org/wiki/Color_confinement&quot; rel=&quot;nofollow&quot;&gt;color neutral&lt;/a&gt; so that means either color quark + anticolor antiquark (&lt;a href=&quot;http://en.wikipedia.org/wiki/Meson&quot; rel=&quot;nofollow&quot;&gt;mesons&lt;/a&gt;) or three quarks carrying R, G and B color charge respectively (&lt;a href=&quot;http://en.wikipedia.org/wiki/Baryon&quot; rel=&quot;nofollow&quot;&gt;baryons&lt;/a&gt;). &lt;em&gt;(&lt;strong&gt;Note:&lt;/strong&gt; There should also exist exotic particles like tetraquarks and pentaquarks but these haven't been observed yet and there is &lt;a href=&quot;http://en.wikipedia.org/wiki/Quark%2dgluon_plasma&quot; rel=&quot;nofollow&quot;&gt;quark-gluon plasma&lt;/a&gt; which was observed. But none of this exotic matter can play role of a proton)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it turns out that the most stable of all those hadronic particles is proton (which happens to be baryon). Everything else decays to other particles sooner or later (usually very soon) and can't possibly make up the stable matter around us. Neutron makes an exception to this because when it is bound in nucleus it becomes stable (well, not quite, radioactive beta decay can still occur).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that even if some meson particle were stable, mesons are still bosons. So they wouldn't obey Pauli's exclusion principle and it would probably be impossible to build anything like nucleus from them. One very likely needs fermions for that and that means baryons and three quarks.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Size of an atom&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This has nothing to do with gravity. Big atoms are unstable purely for nuclear reasons (i.e. their nuclei decay extremely fast via radioactivity). Gravitational considerations are only important for huge objects (typically stars and their afterlife products like white dwarfs and neutron stars).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Mass of a proton&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Since Einstein we know that $E=mc^2$, i.e. energy is mass. So anytime there is some interaction you have to account for its binding energy when computing mass of a composite object. Mass of a proton comes almost exclusively from strong interactions that bind the three quarks together.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-02T01:00:57.270" LastActivityDate="2010-12-02T01:00:57.270" CommentCount="4" />
  <row Id="1536" PostTypeId="1" AcceptedAnswerId="1538" CreationDate="2010-12-02T02:15:31.693" Score="4" ViewCount="414" Body="&lt;p&gt;&lt;em&gt;Disclaimer: I am not a physicist; I am a geometer (and a student!) trying to learn some physics.  Please be gentle.  Thanks!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When solving the Schrödinger equation for a particle in a spherical potential, it seems common to separate variables into angular and radial components.  The angular evolution can then be expressed in terms of eigenfunctions of the Laplace-Beltrami operator $\Delta$ on the sphere, i.e., the spherical harmonics.  (It is my understanding that these eigenfunctions or eigenstates also have some physical significance, namely that eigenfunctions with the same eigenvalue correspond to states of equal energy.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When solving the Dirac equation (again with a spherical potential) you'd expect a similar story: separate into angular and radial components and write the angular evolution in terms of the eigenfunctions of the Riemannian Dirac operator $D$ on the sphere.  And, you'd expect these eigenfunctions would have a similar physical interpretation to the non-relativistic case (after all, the only thing we changed was the energy-momentum relationship).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the references I'm finding on the Dirac equation with central potential write solutions in terms of the &lt;em&gt;spherical spinors&lt;/em&gt; $\Omega$, which are themselves simple functions of the spherical harmonics $Y_l^m$.  This situtation seems odd to me because, although eigenfunctions of $D$ are also eigenfunctions of $\Delta$, the opposite is not true.  In particular, $D$ will have both positive and negative eigenvalues, and so eigenspaces with equal value but opposite sign get &quot;mixed&quot; when we square $D$ (recall that on, say, Euclidean $R^3$, $D^2=\Delta$).  I'm not sure about the physical interpretation, though, because I don't understand the physical meaning of eigenfunctions of the Dirac operator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some more concrete questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;what do eigenfunctions of the $D$ represent physically?&lt;/li&gt;&#xA;&lt;li&gt;why are the spherical harmonics used for separation of variables rather than eigenfunctions of $D$?&lt;/li&gt;&#xA;&lt;li&gt;alternatively, are there cases where eigenfunctions of $D$ are used to solve Dirac's equation?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Pedagogical references are appreciated.  Thanks!&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="708" LastEditorDisplayName="" LastEditDate="2010-12-02T18:39:04.437" LastActivityDate="2010-12-02T18:39:04.437" Title="Separation of variables, eigenfunctions of the Dirac operator" Tags="&lt;quantum-mechanics&gt;&lt;mathematical-physics&gt;&lt;spin&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="1537" PostTypeId="2" ParentId="723" CreationDate="2010-12-02T02:15:33.240" Score="2" ViewCount="" Body="&lt;p&gt;The idea is that the magnetic fields contain the electrons in the center. Yes there will be losses, but the containment factor is the key to current research. Is it enough to build a sufficient population of electrons to create a large enough negative potential well to in turn attract ions (fusion fuel) at a high enough velocity and density that they will collide and fuse as they accellerate towards the center of the volume from opposing directions. The current state of the research states that confinement is proven. The step in progress is the next after confinement, and that is scaling. In the current project, the magnetic field is being increased by a factor of 8 (.1T to .8T) to verify the predicted containment scaling. If the scaling is proven, the next step would be to build a unit at the predicted breakeven size, which is about ten times the currently tested diameter. They will increase the magnet torus from 30cm to 3Meters (or 15cm radius to 1.5Meter radius, as is commonly discussed). The primary loss in a fusor is collisions with the central grid. The polywell eliminates this grid by use of the magnetic fields.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-02T02:15:33.240" CommentCount="1" />
  <row Id="1538" PostTypeId="2" ParentId="1536" CreationDate="2010-12-02T03:11:08.797" Score="4" ViewCount="" Body="&lt;p&gt;There might be a small bit of confusion in the question here.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Skip all this stuff below until the next bracketed comment.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First recall that so(3) has representations that integrate to representations of SU(2) rather than SO(3), and these are the &quot;half-integral&quot; spin.  Likewise for so(3,1) and SU(2)xSU(2) and SO(3,1), respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for a manifold with Lorentzian metric, a representation of so(3,1) indicates what kind of particle/object/tensor/spinor you're talking about.  The representation tells you how to transform the object/section when you change coordinates.  (N.B.:  since we're talking about bundles associated to the frame bundle, changing coordinates induces a change of trivialization of the bundle.  Ordinarily, the two concepts are indpendent.)  More specifically, tensors correspond to &quot;integral spin&quot; representations, e.g. the 4-dimensional representation is vectors, while the 6-dimensional anti-symmetric representation describes two-forms (such as field strengths).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So spinors are sections of bundles which correspond to representations of so(3,1) which integrate to a representation of SU(2)xSU(2).  Practically speaking, in order to covariantly differentiate such an object, you can do the following.  The Levi-Civita connection already gives you an element of so(3,1) (a matrix) for each tangent index.  Plus you have a representation of so(3,1), so you act with this element of so(3,1) by your representation.  This is what the gamma matrices are about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Oy, this is getting too long.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here's the thing:  the spinor representations decompose into different irreducible components, and the Dirac operator maps one representation (positive chirality) into another (negative chirality)!  That is, it maps sections of one spinor bundle into sections of another.  You can of course look at the Dirac operator on the sum of these bundles, but eigenvectors do not have an evident physical interpretation (they are of mixed chirality).&#xA;In flat space, the square of the Dirac operator is a multiple of the identity endomorphism on the bundle, so eigenvalues make perfect sense and can be written in terms of functions times the (global) basis elements for spinors.&lt;/p&gt;&#xA;" OwnerUserId="20" LastActivityDate="2010-12-02T03:11:08.797" CommentCount="3" />
  <row Id="1539" PostTypeId="2" ParentId="1536" CreationDate="2010-12-02T04:46:23.740" Score="0" ViewCount="" Body="&lt;p&gt;I think the essential physics is simply this:  The conserved quantity associated with the spherical symmetry of the system is the angular momentum vector.  In the case of the scalar particle described by the Schrödinger equation, this is unambiguous.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one progresses to the Dirac operator, however, you are now describing a particle with intrinsic spin.  Now, the conserved quantity is $\vec J = \vec S + \vec L$.  In general, this means that you would expect some sort of spin-orbit coupling in the system, breaking the degeneracy between different values of the z component of spin seen in the Schrödinger-equation based solution for the central force problem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've never seen anyone do this by a direct application of the Dirac equation.  Typically, the approach is to solve this using the Schrödinger equation, and then treat the spin-orbit coupling as a small perturbation on top of that, and spin as another quantum number that shows up whose explanation is just a handwavey QFT.  But all of this information would have to be embedded in the Dirac equation.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-12-02T04:46:23.740" CommentCount="5" />
  <row Id="1540" PostTypeId="1" CreationDate="2010-12-02T06:15:09.823" Score="1" ViewCount="241" Body="&lt;p&gt;My Question was like this and i have realised few things and still have some doubts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a book in which a paragraph goes like this&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now, $\dot\phi$, $\dot \theta$,&#xA;  $\dot\psi$ are respectively the&#xA;  angular speeds about the space z-axis,&#xA;  line of node and the body z-axis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know how Euler's angles of rotation are even connected with the angular velocity. Since it's my assumption that the the space set of axis remains unchanged, and the body set of axes rotate with the the axis of rotation. According to my assumption, the angular velocity vector components should remain constant w.r.t the space set of axes (not equal but unchanged) and the angular velocity vector is zero w.r.t the body set of axes because the body set of axes is rotating with the axis so the angular velocity is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if  the body set of axes were stationary and the rigid body is rotating, would that mean the the components would be connected to the Euler's angle of rotation anyway? I think that Euler's angle are just angles of rotation that transforms the space set of axes into body set of axes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I also don't understand what does this 'line of node mean'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have come to realize that in Euler's rotation, The space axis is rotated about space Z axis, new space X-axis and,  body Z axis (which is aligned by the new space X axis rotation). Since there is rotation, there is angular speed, and the rotation are $ \phi , \theta ,and \psi $, then obviously the the angular speeds are  $\dot\phi$, $\dot \theta$,&#xA; $\dot\psi$ and the line of node is the new space X-axis from space Z-axis rotation. And there is no rigid body involved. But has this angular velocity got something to do with the rotation of rigid body? Like stability of spinning top?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know but I hope i am right.&lt;/p&gt;&#xA;" OwnerUserId="567" LastEditorUserId="567" LastEditorDisplayName="" LastEditDate="2010-12-03T05:06:58.217" LastActivityDate="2010-12-03T05:06:58.217" Title="Relation of angular speed of a rigid body to Euler's Angles" Tags="&lt;classical-mechanics&gt;&lt;rotational-dynamics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="1541" PostTypeId="1" CreationDate="2010-12-02T07:29:53.613" Score="-1" ViewCount="230" Body="&lt;p&gt;I heard that an &lt;a href=&quot;http://apple.stackexchange.com/questions/4555/iphone-dropped-from-the-top-on-my-desktop/4558#4558&quot;&gt;iphone falling on carpet would not cause damage to it&lt;/a&gt;, both internal and external.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any physical explanation for this?&lt;/p&gt;&#xA;" OwnerUserId="605" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-02T08:49:44.907" LastActivityDate="2010-12-02T17:39:39.533" ClosedDate="2010-12-03T01:18:51.133" Title="An iPhone falling on carpet is fine, is it true?" Tags="&lt;everyday-life&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="1542" PostTypeId="2" ParentId="1541" CreationDate="2010-12-02T08:18:51.647" Score="4" ViewCount="" Body="&lt;p&gt;During a fall, the phone's potential energy ($mgh$) is converted into kinetic energy ($\frac{1}{2}mv^2$). When the phone hits the floor, it stops falling, so the kinetic energy becomes zero. That energy has to go somewhere. Some of it will be used to exert a force on the phone. If the floor is hard, then the phone gets that energy at the moment of impact, and that will probably be enough to break it. If the floor is a carpet, which is compressible, then the phone will compress it and the energy will be stored as potential energy in the &quot;springiness&quot; of the carpet, and released gradually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you drop the phone from high enough, say the top of a crane, onto a carpet, then it will still break. That's because there was more potential energy to start with, because $h$ was 100 meters instead of 1. If the phone hits the carpet and compresses it as much as possible, but still has enough kinetic energy left to break on impact, then it certainly will break.&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2010-12-02T08:18:51.647" />
  <row Id="1543" PostTypeId="2" ParentId="1334" CreationDate="2010-12-02T08:44:31.040" Score="1" ViewCount="" Body="&lt;p&gt;Also, adding to what j.c. states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a complicated constitutive relation between the electric field and the electric displacement field, e.g. nonlinear and in terms of a convolution (no instantanious reaction of the medium), it might be very complicated to find a representation in terms of potentials obaying a wave equation using a non-perturbative ansatz.&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2010-12-02T08:44:31.040" />
  <row Id="1544" PostTypeId="1" AcceptedAnswerId="1976" CreationDate="2010-12-02T09:30:43.433" Score="6" ViewCount="412" Body="&lt;p&gt;the widely used approach to nonlinear optics is a &lt;strong&gt;Taylor expansion&lt;/strong&gt; of the dielectric displacement field $\mathbf{D} = \epsilon_0\cdot\mathbf{E} + \mathbf{P}$ in a Fourier representation of the polarization $\mathbf{P}$ in terms of the dielectric susceptibility $\mathcal{X}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{P} = \epsilon_0\cdot(\mathcal{X}^{(1)}(\mathbf{E}) + \mathcal{X}^{(2)}(\mathbf{E},\mathbf{E}) + \dots)$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This expansion &lt;strong&gt;does not work&lt;/strong&gt; anymore if the excitation field has components close to the resonance of the medium. Then, one has to take the whole quantum mechanical situation into account by e.g. describing light/matter interaction by a two-level Hamiltonian. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this approach is certainly &lt;strong&gt;not the most general one&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Intrinsically nonlinear formulations of electrodynamics&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;So, what kind of nonlinear formulations of electrodynamics given in a Lagrangian formulation are there? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One known ansatz is the &lt;strong&gt;Born-Infeld model&lt;/strong&gt; as pointed out by Raskolnikov. There, the Lagrangian density is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathcal{L} = b^2\cdot \left[ \sqrt{-\det (g_{\mu \nu})} - \sqrt{-\det(g_{\mu \nu} + F_{\mu \nu}/b)} \right]$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the theory has some nice features as for example a maximum energy density and its relation to gauge fields in string theory. But as I see it, this model is an intrinsically nonlinear model for the free-space field itself and not usefull for describing nonlinear matter interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same holds for an ansatz of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathcal{L} = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} + \lambda\cdot\left( F^{\mu\nu}F_{\mu\nu} \right)^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;proposed by &lt;a href=&quot;http://www.springerlink.com/content/759u2jh547pk107m/&quot; rel=&quot;nofollow&quot;&gt;Mahzoon and Riazi&lt;/a&gt;. Of course, describing the system in &lt;strong&gt;Quantum Electrodynamics&lt;/strong&gt; is intrinsically nonlinear and ... to my mind way to complicated for a macroscopical description for nonlinear optics. The &lt;strong&gt;question&lt;/strong&gt; is: Can we still get a nice formulation of the theory, say, as a mean field theory via an effective Lagrangian?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think a suitable ansatz could be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathcal{L} = -\frac{1}{4}M^{\mu\nu}F_{\mu\nu}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $M$ now accounts for the matter reaction and depends in a nonlinear way on $\mathbf{E}$ and $\mathbf{B}$, say&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M^{\mu\nu} = T^{\mu\nu\alpha\beta}F_{\alpha\beta}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where now $T$ is a nonlinear function of the field strength and might obey certain symmetries. The equation $T = T\left( F \right)$ remains unknown and depends on the material.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Metric vs. $T$ approach&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As pointed out by &lt;strong&gt;space_cadet&lt;/strong&gt;, one might ask the question why the nonlinearity is not better suited in the metric itself. I think this is a matter of taste. My point is that explicitly changing the metric might imply a non-stationary spacetime in which a &lt;a href=&quot;http://math.stackexchange.com/questions/13902/does-a-fourier-transformation-on-a-pseudo-riemannian-manifold-make-sense&quot;&gt;Fourier transformation might not be well defined&lt;/a&gt;. It might be totally sufficient to treat spacetime as Lorentzian manifold.&lt;br&gt;&#xA;Also, we might need a simple spacetime structure later on to explain the material interaction since the polarization $\mathbf{P}$ depends on the matter response generally in terms of an integration over the past, say  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{P}(t) = \int_{-\infty}^{t}R\left[\mathbf{E}\right](\tau )d\tau$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $R$ beeing some nonlinear response function(al) related to $T^{\mu\nu\alpha\beta}$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Examples for $T$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;To illustrate the idea of $T$, here are some examples.&lt;br&gt;&#xA;For &lt;strong&gt;free space&lt;/strong&gt;, $T$ it is given by $T^{\mu\nu\alpha\beta} = g^{\mu\alpha}g^{\nu\beta}$ resulting in the free-space Lagrangian $\mathcal{L} = -\frac{1}{4}T^{\mu\nu\alpha\beta}F_{\alpha\beta}F_{\mu\nu} = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$&#xD;&#xA;The Lagrangian of Mahzoon and Riazi can be reconstructed by&lt;br&gt;&#xA;$T^{\mu\nu\alpha\beta} = \left( 1 + \lambda F^{\gamma\delta}F_{\gamma\delta} \right)\cdot g^{\mu\alpha}g^{\nu\beta}$.&lt;br&gt;&#xA;One might be able to derive a Kerr nonlinearity using this Lagrangian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is anyone familiar in a description of nonlinear optics/electrodynamics in terms of a gauge field theory or something similar to the thoughts outlined here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Comments on the first Bounty&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I want to &lt;strong&gt;thank&lt;/strong&gt; everyone actively participating in the discussion, especially &lt;strong&gt;Greg Graviton&lt;/strong&gt;, &lt;strong&gt;Marek&lt;/strong&gt;, &lt;strong&gt;Raskolnikov&lt;/strong&gt;, &lt;strong&gt;space_cadet&lt;/strong&gt; and &lt;strong&gt;Willie Wong&lt;/strong&gt;. I am enjoying the discussion relating to this question and thankfull for all the nice leads you gave. I decided to give the &lt;strong&gt;bounty&lt;/strong&gt; to Willie since he gave the thread a new direction introducing the &lt;strong&gt;material manifold&lt;/strong&gt; to us.&lt;br&gt;&#xA;For now, I have to reconsider all the ideas and I hope I can come up with a new revision of the question that should be formulated in a clearer way as it is at the moment.&lt;br&gt;&#xA;So, thank you again for your contributions and feel welcome to share new insights.  &lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2010-12-17T16:12:10.403" LastActivityDate="2010-12-17T16:12:10.403" Title="Nonlinear optics as gauge theory" Tags="&lt;optics&gt;&lt;electrodynamics&gt;&lt;quantum-optics&gt;&lt;non-linear-optics&gt;&lt;gauge-theory&gt;" AnswerCount="4" CommentCount="24" FavoriteCount="2" />
  <row Id="1545" PostTypeId="1" AcceptedAnswerId="1548" CreationDate="2010-12-02T12:10:37.283" Score="3" ViewCount="168" Body="&lt;p&gt;I haven't understood this thing: Physics is invariant for CPT trasform...But the Heat or diffusive equation $\nabla^2 T=\partial_t T$ is not invariant for time reversal...but it's P invariant..So CPT simmetry would be violed...&#xA;What I haven't undestood?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="558" LastActivityDate="2010-12-02T12:34:34.033" Title="CPT and heat equation" Tags="&lt;quantum-field-theory&gt;&lt;heat&gt;" AnswerCount="2" />
  <row Id="1546" PostTypeId="2" ParentId="1545" CreationDate="2010-12-02T12:29:27.913" Score="3" ViewCount="" Body="&lt;p&gt;CPT theorem is not a theorem for all of physics but only for a quantum field theory (QFT). Also CPT invariance doesn't mean that QFT is necessarily invariant with respect to any of C, P and T (or PT, TC and CP, which is the same by CPT theorem) transform. Indeed, all of these symmetries are violated by weak interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, even if the macroscopic laws were completely correct it wouldn't mean that they need to preserve microscopic laws. E.g. most of the microscopic physics is time symmetric (except for small violation by the weak interaction) but second law of thermodynamics (which is universally true for any macroscopic system just by means of logic and statistics) tells you that entropy has to increase with time. We can say that the huge number of particles &lt;strong&gt;breaks&lt;/strong&gt; the microscopical time-symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the heat equation essentially captures dynamics of this time asymmetry of the second law. It tells you that temperatures eventually even out and that is an irreversible process that increases entropy.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-02T12:29:27.913" CommentCount="2" />
  <row Id="1547" PostTypeId="2" ParentId="1540" CreationDate="2010-12-02T12:30:54.257" Score="1" ViewCount="" Body="&lt;p&gt;If we express the change in anglular velocity as $\Delta\vec\omega$ in local coordinates, with for example the angles $\phi$, $\theta$ and $\psi$ rotations about the $Z$, $X$ and $Z$ respectively then the answer is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta\vec{\omega}=\dot{\phi}\hat{k}+\mathrm{{Rot}(\hat{k},\phi})\left(\dot{\theta}\hat{i}+\mathrm{{Rot}(\hat{i},\theta})\left(\dot{\psi}\hat{k}\right)\right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;strong&gt;Rot&lt;/strong&gt;(axis,angle) is a 3x3 rotation matrix. We apply the angular rotation components in sequence on the local coordinate axes this way. Also $\hat{i}=(1,0,0)$, $\hat{j}=(0,1,0)$ and $\hat{k}=(0,0,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Line of node must be the common normal between the two $z$-axes. We typically denote that as the $x$-axis (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Denavit-Hartenberg_Parameters&quot; rel=&quot;nofollow&quot;&gt;Denavit-Hartenberg&lt;/a&gt; notation).&lt;/p&gt;&#xA;" OwnerUserId="392" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2010-12-02T22:41:20.883" LastActivityDate="2010-12-02T22:41:20.883" CommentCount="1" />
  <row Id="1548" PostTypeId="2" ParentId="1545" CreationDate="2010-12-02T12:34:34.033" Score="10" ViewCount="" Body="&lt;p&gt;The heat equation is a macroscopic equation. It describes the flow of heat from hot objects to cold ones. Of course it can not be time-reversible, since the opposite movement never happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, I say 'of course' but you actually have stumbled on something important. As you say, the fundamental laws of nature should be CPT invariant, or at least we expect them to be. The reason the heat equation is not CPT invariant is that it is not a fundamental law, but a macroscopic law emerging from the microscopic laws governing the motions of elementary particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is however a problem here, how does this time asymmetry arise from microscopic laws that are themselves time reversal invariant? The answer to that is given by statistical mechanics. While the microscopic laws are time-reversible (I'll focus on T, and leave CP aside), not all states are equally likely with respect to certain choices of the macroscopic variables. There are more configurations of particles corresponding to a room filled with air than with a room where all the air would be concentrated in one corner. It is this asymmetry that forms the basis of all explanations in statistical mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that clears things up a bit.&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2010-12-02T12:34:34.033" CommentCount="6" />
  <row Id="1549" PostTypeId="2" ParentId="1533" CreationDate="2010-12-02T14:24:33.063" Score="3" ViewCount="" Body="&lt;p&gt;I don't know a great deal about attosecond spectroscopy, but my inclination is to say that it would be extremely difficult to get a lifetime from a linewidth in this manner. The energy width due to a femtosecond lifetime of the intermediate state would be about 2/3 of an eV, which is a sizable fraction of the energy of a photon of the base laser wavelength (usually they start with something like a YAG or Ti:Sapph, so in the 800-1000nm kind of range, which is between 1 and 1.5 eV photon energy). That means you'd be looking for a spread of frequencies that was almost as big as the excitation frequency, and it's kind of a hassle to extract a linewidth from that sort of thing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the real attraction of the attosecond pump-probe sort of experiments: when you've got that kind of timing resolution, you don't need to try to back time dependence out of broad-band spectroscopic features. You can just measure it directly by stroboscopically following the transition. They can watch electrons redistributing themselves in the molecules directly, by doing an initial excitation, then probing the state at a series of short time steps after the excitation. They can basically make movies of the evolution of the electron wavefunction directly, which is just awesome.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-12-02T14:24:33.063" CommentCount="3" />
  <row Id="1550" PostTypeId="1" CreationDate="2010-12-02T14:49:27.000" Score="0" ViewCount="93" Body="&lt;p&gt;A few questions about a magnet and a paperclip:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What do you call a material that attracts another material via magnetism? (i.e. the magnet)&lt;/li&gt;&#xA;&lt;li&gt;What do you call the material that is attracted in #1? (i.e. the paperclip)&lt;/li&gt;&#xA;&lt;li&gt;What properties make #1 (the magnet) and #2 (the paperclip) get along so well? Electron configurations?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These are pretty elementary...please be gentle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I'm looking for basic answers that might be found in a level 200 course.&lt;/p&gt;&#xA;" OwnerUserId="253" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-02T15:13:01.157" LastActivityDate="2010-12-06T14:37:04.427" Title="Magnetism-Related Terminology" Tags="&lt;magnets&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="1551" PostTypeId="2" ParentId="1550" CreationDate="2010-12-02T15:11:51.517" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;well, a &lt;em&gt;magnet&lt;/em&gt;... If its magnetic properties are permanent it is called a &lt;em&gt;permanent magnet&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;again just magnet or if you want to be more precise a magnetizable material. This material need not be a permanent magnet on its own but when external magnetic field is applied its internal magnetic configuration realigns with the external field and this makes the material exhibit its own magnetic field. When the external field is turned off, material may lose its magnetic properties (depending on precise material and temperature). You can see for yourself that if you magnetize the paperclip, it itself starts to act like a permanent magnet and you can attract other paperclips. But after a little while its magnetic field will fade away because the ordered internal configuration will slowly be disturbed by random thermal fluctuations.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When put together, permanent magnet's field magnetizes the paperclip (as explained above). The rest is just a simple magnetic interaction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-02T15:11:51.517" CommentCount="6" />
  <row Id="1552" PostTypeId="2" ParentId="1544" CreationDate="2010-12-02T15:42:52.190" Score="4" ViewCount="" Body="&lt;p&gt;Nonlinear is a buzzword used to cover anything that is not linear. Depending on what kind of nonlinearity is involved, and thus what kind of material, there could be one symmetry or another, or there could be no symmetry at all. For instance, in superconductors, gauge symmetry is broken and photons behave as if they have acquired a mass. The result is that magnetic fields have limited penetration in the superconductor. And I think this is still described by linear equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know of one gauge-invariant theory that is non-linear, this model is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Born%E2%80%93Infeld_model&quot; rel=&quot;nofollow&quot;&gt;Born-Infeld model&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2010-12-02T15:42:52.190" CommentCount="1" />
  <row Id="1553" PostTypeId="2" ParentId="1530" CreationDate="2010-12-02T16:39:09.657" Score="3" ViewCount="" Body="&lt;p&gt;I guess you are really looking for this wikipedia page : &lt;a href=&quot;http://en.wikipedia.org/wiki/Synthesis_of_noble_metals#Gold&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Synthesis_of_noble_metals#Gold&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, there are gold synthesis technique, but they apparently all need some external energy (either γ-ray or neutron capture) and are not restricted to nuclear decay. One of them has for intermediate step the nuclear decay ¹⁹⁷Hg →　¹⁹⁷Au + e⁺ with a 2 days half life. The unstable ¹⁹⁷Hg is obtained from a stable Hg-isotope by γ-ray irradiation (¹⁹⁸Hg + γ → ¹⁹⁷Hg + n).&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2010-12-02T16:39:09.657" />
  <row Id="1554" PostTypeId="2" ParentId="1541" CreationDate="2010-12-02T17:39:39.533" Score="0" ViewCount="" Body="&lt;p&gt;Manufacturers routinely simulate these sorts of accidents. I would be surprised if it&#xA;wasn't engineered to survive a fall from desktop height.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2010-12-02T17:39:39.533" />
  <row Id="1555" PostTypeId="1" CreationDate="2010-12-02T17:39:46.687" Score="1" ViewCount="108" Body="&lt;p&gt;After seeing this image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mynasadata.larc.nasa.gov/images/EM_Spectrum3-new.jpg&quot; rel=&quot;nofollow&quot;&gt;http://mynasadata.larc.nasa.gov/images/EM_Spectrum3-new.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And reading this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The long wavelength limit is the size of the universe itself, while it is thought that the short wavelength limit is in the vicinity of the Planck length, although in principle the spectrum is infinite and continuous.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question is, how did the EMS evolve? Was it weak at first, i.e. at Planck time, which allowed certain reactions to modify, shape or add properties to it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What really defined the position and order of everything that is the EMS? E.g. colors. What defined or added something as colors to be inserted in that specific range of 400nm to 700nm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this is not to philosophical and thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="3037" LastActivityDate="2010-12-02T20:10:40.763" ClosedDate="2010-12-02T20:41:48.890" Title="Creation of the Electromagnetic Spectrum" Tags="&lt;electromagnetism&gt;&lt;waves&gt;&lt;electromagnetic-radiation&gt;&lt;radiation&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="1556" PostTypeId="2" ParentId="1019" CreationDate="2010-12-02T18:01:54.793" Score="7" ViewCount="" Body="&lt;p&gt;If you somehow manage to BREAK a law of physics, the universe will vanish!&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2010-12-02T18:01:54.793" CommunityOwnedDate="2010-12-02T18:01:54.793" CommentCount="2" />
  <row Id="1557" PostTypeId="1" AcceptedAnswerId="1558" CreationDate="2010-12-02T18:50:45.907" Score="4" ViewCount="607" Body="&lt;p&gt;I'm in a freshmen level physics class now, so I dont know much, but something I heard today intrigued me. My TA was talking about how at the research facility he worked at, they were able to accelerate some certain particle to &quot;99.99% the speed of light&quot;. I said why not 100%, and I didnt quite understand his explanation, but he said it wasnt possible. This confused me. Since the speed of light is a finite number, why can we go so close to its speed but not quite?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I read all the answers, and I &lt;em&gt;think&lt;/em&gt; I'm sort of understanding it. Another silly question though: If we are getting this particle to 99.99% the speed of light by giving it some sort of finite acceleration, and increasing it more and more, why cant we increase it just a &lt;em&gt;little more&lt;/em&gt;? Sorry I know this is a silly question. I totally accept the fact we cant reach 100%, but I'm just trying to break it down. If we've gotten so close by giving it larger and larger acceleration every time, why cant we just supply it with more acceleration? And how much of a difference is there between 99.99% the speed of light, and &lt;em&gt;the&lt;/em&gt; speed of light? (I'm not really sure if &quot;difference&quot; is a good word to use, but hopefully you get what I'm asking).&lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="41" LastEditorDisplayName="" LastEditDate="2010-12-02T19:52:39.710" LastActivityDate="2011-01-28T23:35:29.807" Title="Accelerating particles to the speed of light" Tags="&lt;particle-physics&gt;&lt;speed-of-light&gt;" AnswerCount="8" CommentCount="2" FavoriteCount="2" />
  <row Id="1558" PostTypeId="2" ParentId="1557" CreationDate="2010-12-02T19:14:48.150" Score="9" ViewCount="" Body="&lt;p&gt;By special relativity, the energy needed to accelerate a particle (with mass) grow super-quadratically when the speed is close to &lt;em&gt;c&lt;/em&gt;, and is &amp;infin; when it is &lt;em&gt;c&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E = \gamma mc^2 = \frac{mc^2}{\sqrt{1 - (\text{“percent of speed of light”})^2}} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you can't supply infinite energy to the particle, it is not possible to get to 100% &lt;em&gt;c&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Suppose you have got an electron (m = 9.1 &amp;times; 10&lt;sup&gt;-31&lt;/sup&gt; kg) to 99.99% of speed of light. This is equivalent to providing 36 MeV of kinetic energy. Now suppose you accelerate &quot;a little more&quot; by providing yet another 36 MeV of energy. You will find this this only boosts the electron to 99.9975% &lt;em&gt;c&lt;/em&gt;. Say you accelerate &quot;a lot more&quot; by providing 36,000,000 MeV instead of 36 MeV. That will still make you reach 99.99999999999999% &lt;em&gt;c&lt;/em&gt; instead of 100%. The energy increase explodes as you approach &lt;em&gt;c&lt;/em&gt;, and your input will exhaust eventually no matter how large it is. The difference between 99.99% and 100% is infinite amount of energy.&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-12-02T22:46:17.587" LastActivityDate="2010-12-02T22:46:17.587" CommentCount="4" />
  <row Id="1559" PostTypeId="2" ParentId="1557" CreationDate="2010-12-02T19:15:01.100" Score="4" ViewCount="" Body="&lt;p&gt;You have to understand special relativity. It's basically because newtonian mechanics breaks down at speeds close to the speed of light and &lt;code&gt;F=ma&lt;/code&gt; is false. It's basically because your mass isn't constant, it varies based on your speed. And as you approach &lt;code&gt;c&lt;/code&gt;, your mass has to approach infinity and thus you'll need infinite force to move accelerate from &lt;code&gt;c - delta(v)&lt;/code&gt; to &lt;code&gt;c&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just a basic overview, I'm sure someone will come with a much more detailed overview. But you can take a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Special_relativity&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Special_relativity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically the part about relativistic mechanics.&lt;/p&gt;&#xA;" OwnerUserId="149" LastActivityDate="2010-12-02T19:15:01.100" CommentCount="2" />
  <row Id="1560" PostTypeId="2" ParentId="1557" CreationDate="2010-12-02T19:15:36.123" Score="2" ViewCount="" Body="&lt;p&gt;It is a direct consequence of the theory of special relativity that no &lt;em&gt;massive&lt;/em&gt; massive particle can travel at the speed of light. (And every &lt;em&gt;massless&lt;/em&gt; particle &lt;em&gt;must&lt;/em&gt; travel at the speed of light.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can consider the impossibility of accelerating a particle to precisely speed &lt;code&gt;c&lt;/code&gt; in one of several ways, but the most obvious is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hypothetical massive particle traveling at speed &lt;code&gt;c&lt;/code&gt; would have &lt;em&gt;infinite&lt;/em&gt; mass (or mass-energy). Singularities are bad! (Or if you want, it would require an infinite force/amount of energy to accelerate the particle to &lt;code&gt;c&lt;/code&gt;, approaching the limit.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side note: If you're a freshman in physics, you will most likely be studying basic special relativity quite soon. Everything should be a lot clearer after taking such a course.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2010-12-02T19:15:36.123" />
  <row Id="1561" PostTypeId="2" ParentId="1557" CreationDate="2010-12-02T19:20:49.303" Score="4" ViewCount="" Body="&lt;p&gt;There are (at least) two explanations, kinematical and dynamical.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Dynamics&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;When you want to make an object accelerate you have to use energy to produce force on the object. The force is $F = ma$ &lt;em&gt;(this equation is not really correct in SR but it suffices for our purposes)&lt;/em&gt; Now the point of SR is that the mass $m$ that the object seems to be having when it is moving with respect to you is not constant. It goes like $m = m_0 \gamma(v)$ where $m_0$ is the objects invariant mass (as seen from its own rest frame) and $\gamma(v)$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lorentz_factor&quot; rel=&quot;nofollow&quot;&gt;Lorentz factor&lt;/a&gt;. Now $\gamma(v) \to \infty$ as $v \to c$. So this means that the (apparent or relativistic) mass of the object becomes arbitrarily large and you would need an infinite amount of energy to get to the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Kinematics&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;From the kinematical point of view it all boils down to relativistic concept of velocity. In SR when you want to change particle's speed you have to &lt;a href=&quot;http://en.wikipedia.org/wiki/Lorentz_boost#Matrix_form&quot; rel=&quot;nofollow&quot;&gt;boost&lt;/a&gt; it. This is described by a certain &lt;a href=&quot;http://en.wikipedia.org/wiki/Lorentz_transformation&quot; rel=&quot;nofollow&quot;&gt;Lorentz transformation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now its useful to move to the dual point of view. Instead of saying that you boost the particle you can just change your reference frame in the opposite way. So instead of giving the particle speed $v$ in direction $\mathbf x$ you will look at the particle at rest from a reference frame that has speed $v$ in direction $-\mathbf x$. This transformation is also described by a Lorentz transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now every Lorentz transformation preserves the relations $v &amp;lt; c$, $v = c$ and $v &amp;gt; c$ (the middle one is actually Einstein's postulate on invariance of speed of light in every inertial frame). That means that if your velocity is less than speed of light it will be so in any reference frame. And also that if some particle was once going slower than speed of light, it will always do so.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-02T19:20:49.303" CommentCount="12" />
  <row Id="1562" PostTypeId="2" ParentId="1555" CreationDate="2010-12-02T19:30:37.197" Score="3" ViewCount="" Body="&lt;p&gt;I think saying that the EMS evolved isn't really an accurate description. The EMS simply is. The upper limit (the size of the universe) is because a wave can't be bigger than the system that contains it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing defines the position or order. It's simply a number line, from small to large. There's nothing inherently special to the visible range. It's simply the range that humans evolved to perceive. We might just have easily evolved to perceive the infrared spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're putting too much weight on what it's called. All of that is just human-centric thinking. A wavelength of &lt;strike&gt;400&lt;/strike&gt;700nm doesn't care that it looks red.&lt;/p&gt;&#xA;" OwnerUserId="149" LastEditorUserId="149" LastEditorDisplayName="" LastEditDate="2010-12-02T20:10:40.763" LastActivityDate="2010-12-02T20:10:40.763" CommentCount="2" />
  <row Id="1563" PostTypeId="1" CreationDate="2010-12-02T19:30:44.387" Score="0" ViewCount="378" Body="&lt;p&gt;Here's a mechanical/mathematical problem I hope that I could get help with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to solve the differential equations of double pendulum using Matlab's ODE45. I have a system of two differential equations and two initial conditions. I know how to solve simple system but substituting the dependent variables. But this is different. Let's say I do this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x(1) = x(2) &#xA;x(2) = ... (1) &#xA;x(3) = x(4)&#xA;x(4) = ... (2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now (1) will contain x'(4) and (2) will contain x'(2). That is, the second derivatives that I have no way of substituting make up a part of the other differential equations. How can I solve this??&lt;/p&gt;&#xA;" OwnerUserId="714" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2010-12-02T23:09:46.260" LastActivityDate="2010-12-02T23:09:46.260" ClosedDate="2010-12-02T21:28:53.330" Title="Double pendulum, written as a system of differential equations. How?" Tags="&lt;classical-mechanics&gt;&lt;simulation&gt;" CommentCount="3" />
  <row Id="1565" PostTypeId="2" ParentId="1555" CreationDate="2010-12-02T20:08:01.477" Score="2" ViewCount="" Body="&lt;p&gt;The EMS is a reflection of the properties and laws of the universe. The light that you see starts from the emission of a fusion reaction, filtered by molecules in the atmosphere which themselves are selected by the construction and mass of the planet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So just after the big bang, the light would have been mostly very high energy, and short wavelength, as the types of matter that produce longer wavelength EM (molecules, polymers) didn't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the universe expanded and matter formed, fusion produced heavier atoms and enabled the larger molecules, and the length scales at work allowed resonances at longer wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, though, that light is emitted in a continuous spectrum as black body radiation, from all objects with a temperature above zero. So once anything could vibrate and thus have a temperature, it would emit such a spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Largely, the universe began at the very short, hot end of the spectrum, and will eventually die at the long, cold end, unless some sort of crunch reverses it. Even the beyond-gamma radiation of the big bang itself is now microwave background radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So goes the current theory, anyway.&lt;/p&gt;&#xA;" OwnerUserId="670" LastActivityDate="2010-12-02T20:08:01.477" CommentCount="1" />
  <row Id="1566" PostTypeId="1" CreationDate="2010-12-02T21:05:28.990" Score="9" ViewCount="438" Body="&lt;p&gt;Which experiments prove atomic theory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sub-atomic theories:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;atoms have: nuclei; electrons; protons; and neutrons.  &lt;/li&gt;&#xA;&lt;li&gt;That the number of electrons atoms have determines their relationship with other atoms.  &lt;/li&gt;&#xA;&lt;li&gt;That the atom is the smallest elemental unit of matter. &#xA; a. meaning: that we can't continue to divide atoms into anything smaller and have them      retain the characteristics of the parent element&lt;/li&gt;&#xA;&lt;li&gt;That everything is made of atoms.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These sub-theories might spur more thoughts of individual experiments that prove individual sub-atomic theories (my guess is more was able to be proven after more experiments followed).&lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="642" LastEditorDisplayName="" LastEditDate="2010-12-04T02:34:41.550" LastActivityDate="2011-05-24T01:41:18.147" CommunityOwnedDate="2010-12-04T15:44:03.543" Title="Which experiments prove atomic theory?" Tags="&lt;atoms&gt;" AnswerCount="4" CommentCount="12" FavoriteCount="2" />
  <row Id="1567" PostTypeId="2" ParentId="1566" CreationDate="2010-12-02T22:15:23.490" Score="6" ViewCount="" Body="&lt;p&gt;I would say that one experiment that demonstrates the atomic nature of things is the &lt;a href=&quot;http://en.wikipedia.org/wiki/%C3%9Cber_die_von_der_molekularkinetischen_Theorie_der_W%C3%A4rme_geforderte_Bewegung_von_in_ruhenden_Fl%C3%BCssigkeiten_suspendierten_Teilchen&quot; rel=&quot;nofollow&quot;&gt;observation of Brownian motion&lt;/a&gt;. But it is not the experiment itself that convinces that things are made of atoms, rather its theoretical explanation given by Einstein in one of his 1905 papers (actually Einsteins work for his PhD was on the subject of atomic theory and there are several publications in the period 1903-1905). Of course there is also the observation of Rayleigh who calculated Avogadro’s number by the distance from which he could make out the figure of Mount Everest, assuming that light is scattered by atoms and that is why far away objects look fuzzy (1,2). Also scattering experiments demonstrated the atomic nature of things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Rayleigh, On the transmission of light through an atmosphere containing small&#xA;particles in suspension, in Scientific Papers by Lord Rayleigh Vol. 4, pp. 247–405,&#xA;New York: Dover, 1899/1964.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) &lt;a href=&quot;http://iopscience.iop.org/0143-0807/26/1/018/&quot; rel=&quot;nofollow&quot;&gt;P. Pesic, Eur. J. Phys. 26, 183 (2005)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) &lt;a href=&quot;http://dx.doi.org/10.1016/j.endeavour.2007.05.003&quot; rel=&quot;nofollow&quot;&gt;Patterson, G. Jean Perrin and the triumph of the atomic doctrine (2007) Endeavour, 31 (2), pp. 50-53&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="383" LastEditorUserId="383" LastEditorDisplayName="" LastEditDate="2010-12-02T23:29:57.480" LastActivityDate="2010-12-02T23:29:57.480" CommunityOwnedDate="2010-12-04T15:44:03.543" CommentCount="8" />
  <row Id="1569" PostTypeId="2" ParentId="1566" CreationDate="2010-12-02T23:44:29.630" Score="2" ViewCount="" Body="&lt;p&gt;The history of atoms is definitely intertwined with quantum mechanics. There are many features of the quantum theory that make atomic nature of our world apparent. But here I'd like to state an earlier result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thomson's 1897 &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_theory#Discovery_of_subatomic_particles&quot; rel=&quot;nofollow&quot;&gt;discovery of the electron&lt;/a&gt; not only showed that atoms exist but also that they have substructure.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-02T23:44:29.630" CommunityOwnedDate="2010-12-04T15:44:03.543" />
  <row Id="1570" PostTypeId="1" AcceptedAnswerId="1579" CreationDate="2010-12-02T23:53:32.867" Score="1" ViewCount="331" Body="&lt;p&gt;Hey, &#xA;trying to finish an assignment but having some trouble with it. I will show all my work. The topic is on wave/particle dualty, uncertainty principle (second year modern physics course).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the de Broglie wavelength of a 5.7 MeV α particle emitted from an atomic nucleus whose diameter is approximately $1.6 \times 10^{-14} m$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I know the de broglie wavelength is $\lambda = h / p$. But we dont know p in this case, however we do know the energy. So to relate energy and momentum I use the following formula: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E^2 = p^2c^2+m^2c^4$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using $m = 6.644 \times 10^{-27}$ and using $ E = 9.13\times 10^{-13}$ gives me a complex solution. (Check WolframAlpha)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I've tried leaving the energy in terms of eV and MeV and they give me real solutions however the answer is not correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is 6.02 fentometers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone have any advice? &lt;/p&gt;&#xA;" OwnerUserId="551" LastActivityDate="2010-12-03T09:40:36.510" Title="Calculating de Broglie wavelength" Tags="&lt;uncertainty-principle&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="1571" PostTypeId="2" ParentId="1570" CreationDate="2010-12-03T00:09:54.720" Score="4" ViewCount="" Body="&lt;p&gt;Normally I'd knock on you for simply asking a homework question, but in this case you're dealing with a poorly-worded problem.  The rest mass of an alpha particle is 3.7 GeV; you can't have an alpha particle with less energy than its rest mass.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 5.7 MeV is supposed to be interpreted as the kinetic energy of the alpha particle, not its total energy.  I guess you can take it from there.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-12-03T00:09:54.720" CommentCount="12" />
  <row Id="1572" PostTypeId="1" AcceptedAnswerId="1589" CreationDate="2010-12-03T01:28:19.317" Score="3" ViewCount="109" Body="&lt;p&gt;I've been reading about nanotubes lately, and I keep seeing the $ (n,m) $ notation. How does this describe a nanotube's structure? How do I determine which is $n$ and which is $m$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm familiar with matrix notation referring to rows and columns, but I couldn't connect it with the nanotube structure which is, albeit predictable, not quite like a row-column grid.&lt;/p&gt;&#xA;" OwnerUserId="293" LastActivityDate="2010-12-03T14:31:28.260" Title="How to determine the (n,m) dimensions of a carbon nanotube?" Tags="&lt;nanoscience&gt;" AnswerCount="2" />
  <row Id="1573" PostTypeId="2" ParentId="1572" CreationDate="2010-12-03T02:43:01.277" Score="4" ViewCount="" Body="&lt;p&gt;A carbon nanotube can be seen as a sheet of graphene that is &quot;rolled up&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, graphene is a two-dimensional lattice and hence has two lattice vectors, $\vec{a}_1$ and $\vec{a}_2$. (If you are unfamiliar with lattice vectors let me know and I will expand on this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers $(n,m)$ simply state that your tube is obtained from taking one atom of the sheet and rolling it onto that atom that is at located $n \vec{a}_1 + m \vec{a}_2$ away from your original atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Graphene is a &lt;em&gt;tridiagonal&lt;/em&gt; lattice with two atoms per unit cell.&#xA;&lt;img src=&quot;http://i.stack.imgur.com/iyDKW.jpg&quot; alt=&quot;Left: Lattice structure of graphene, right: Brillouin zone (not of interest here)&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditorDisplayName="" LastEditDate="2010-12-03T05:02:52.583" LastActivityDate="2010-12-03T05:02:52.583" CommentCount="3" />
  <row Id="1574" PostTypeId="1" CreationDate="2010-12-03T06:03:49.250" Score="10" ViewCount="225" Body="&lt;p&gt;I've been told that, from Maxwell's equations, one can find that the propagation of change in the Electromagnetic Field travels at a speed $\frac{1}{\sqrt{\mu_0 \epsilon_0}}$ (the values of which can be empirically found, and, when plugged into the expression, yield the empirically found speed of light)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really not sure how I would go about finding $v = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$ simply from Maxwell's equations in the following form, in SI units --&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla \cdot \mathbf{E} = \frac {\rho} {\epsilon_0}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla \cdot \mathbf{B} = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla \times \mathbf{B} = \mu_0 \left( \mathbf{J} + \epsilon_0 \frac{\partial \mathbf{E}} {\partial t} \right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is what I believe true? (that the speed of propagation is derivable from Maxwell's Equations)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, what else is needed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, can you provide a clear and cogent derivation?&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="71" LastEditorDisplayName="" LastEditDate="2010-12-03T10:07:27.310" LastActivityDate="2010-12-03T21:38:14.020" Title="Deriving the speed of the propagation of a change in the Electromagnetic Field from Maxwell's Equations" Tags="&lt;electromagnetism&gt;&lt;speed-of-light&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="1575" PostTypeId="2" ParentId="1566" CreationDate="2010-12-03T06:31:53.463" Score="4" ViewCount="" Body="&lt;p&gt;I once heard Uhlenbeck give a lecture on this to high school students over the Christmas break at the Rockefeller Univ. years ago.   He recounted a published argument he attributed to Einstein around 1905 (I think), which was that atoms were real if you could count the number of them/mole (Avogadro's number) many different independent ways, and you always got more or less the same answer.   So Brownian motion, gas law, counting with an atomic force microscope, X-ray diffraction, spreading oil film, and many other possibilities would all count as subarguments to the main argument, i.e. that atoms were real.   If anyone knows the reference(s), I would appreciate them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was a live question at the time.  For example Mach, who died in 1916, was apparently an atomic skeptic.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-12-03T16:24:05.403" LastActivityDate="2010-12-03T16:24:05.403" CommunityOwnedDate="2010-12-04T15:44:03.543" CommentCount="3" />
  <row Id="1577" PostTypeId="2" ParentId="1574" CreationDate="2010-12-03T07:56:46.433" Score="11" ViewCount="" Body="&lt;p&gt;Although this is a standard derivation, you frequently don't see it in introductory electromagnetism courses, maybe because those courses shy away from the heavy use of vector calculus.  Here's the usual approach.  We'll find a &lt;a href=&quot;http://en.wikipedia.org/wiki/Wave_equation&quot; rel=&quot;nofollow&quot;&gt;wave equation&lt;/a&gt; from Maxwell's equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla \times \vec{E} = -\frac{\partial\vec{B}}{\partial t}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a partial derivative of both sides with respect to time.  The curl operator has no partial with respect to time, so this becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla \times \frac{\partial\vec{E}}{\partial t} = -\frac{\partial^2\vec{B}}{\partial t^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's another of Maxwell's equations that tells us about $\partial\vec{E}/\partial t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla \times \vec{B} = \mu_0\epsilon_0\frac{\partial \vec{E}}{\partial t}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve this for $\partial\vec{E}/\partial t$ and plug into the previous expression to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla \times \frac{(\nabla \times \vec{B})}{\mu_0\epsilon_0} = -\frac{\partial^2 \vec{B}}{\partial t^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_calculus_identities#Curl_of_the_curl&quot; rel=&quot;nofollow&quot;&gt;curl of curl identity&lt;/a&gt; lets us rewrite this as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{\mu_0 \epsilon_0}\left(\nabla(\nabla \cdot \vec{B}) + \nabla^2\vec{B}\right) = -\frac{\partial^2 \vec{B}}{\partial t^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the divergence of the magnetic field is zero, so kill that term, and rearrange to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{\mu_0 \epsilon_0}\nabla^2\vec{B} + \frac{\partial^2 \vec{B}}{\partial t^2} = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the wave equation we're seeking.  One solution is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec{B} = B_0 e^{i (\vec{x}\cdot\vec{k} - \omega t) }$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This represents a plane wave traveling in the direction of the vector $\vec{k}$ with frequency $\omega$ and phase velocity $v = \omega/|\vec{k}|$.  In order to be a solution, this equation needs to have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\omega^2}{k^2} = \frac{1}{\mu_0\epsilon_0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, setting $v = 1/\sqrt{\mu_0\epsilon_0}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\omega}{k} = v$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dispersion_relation&quot; rel=&quot;nofollow&quot;&gt;dispersion relation&lt;/a&gt;.  The speed that electromagnetic signals travel is given by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_velocity&quot; rel=&quot;nofollow&quot;&gt;group velocity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d\omega}{d k} = v$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So electromagnetic signals in a vacuum travel at speed $c = 1/\sqrt{\mu_0\epsilon_0}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit&lt;/b&gt;&#xA;You can follow the same steps to derive the wave equation for $\vec{E}$, but you will have to assume you're in free space, i.e. $\rho = 0$.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-12-03T21:38:14.020" LastActivityDate="2010-12-03T21:38:14.020" CommentCount="9" />
  <row Id="1578" PostTypeId="2" ParentId="1019" CreationDate="2010-12-03T08:57:08.340" Score="13" ViewCount="" Body="&lt;p&gt;&quot;Summer is when the earth is closest to the sun, and winter is when it's furthest away.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's true that the earth's orbit is slightly elliptical, but the effect of this, as far as seasons, is very small. For one thing, this wouldn't explain why the sun rises and sets at different times in different seasons, and if this were true, the whole planet would have summer at the same time.&lt;/p&gt;&#xA;" OwnerUserId="127" LastActivityDate="2010-12-03T08:57:08.340" CommunityOwnedDate="2010-12-03T08:57:08.340" CommentCount="2" />
  <row Id="1579" PostTypeId="2" ParentId="1570" CreationDate="2010-12-03T09:40:36.510" Score="0" ViewCount="" Body="&lt;p&gt;When the kinetic energy is given (5.7 MeV) and the mass is known, calculating momentum is very straightforward. I would like to point to the part of the assignment that you missed which is the nucleus diameter. It tells you something about the position of the particle and thus prevents you from knowing the momentum (and hence the de Broglie wavelength) all too well.&lt;/p&gt;&#xA;" OwnerUserId="701" LastActivityDate="2010-12-03T09:40:36.510" CommentCount="2" />
  <row Id="1580" PostTypeId="2" ParentId="1019" CreationDate="2010-12-03T09:41:39.970" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Why do I have to learn this law when they change it every few years?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has to do with a fact that some (actually, a lot of) people believe that the progress in physics is done in form of &lt;strong&gt;revolution&lt;/strong&gt; and (in particular) that one day we might find laws that will contradict everything we knew up till then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, if one looks closely on history of physics, it should become apparent that progress was always just evolutionary. Even when some idea needed a revolution in the way people &lt;em&gt;think&lt;/em&gt; (as with SR and QM) it always turned out to be just a generalization of our previous ideas (so both SR and QM have nice classical limits which coincide with Newtonian mechanics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Barring the useless philosophical views (like we might live in Matrix or we don't know whether the sun will rise tomorrow for sure) it's pretty certain that our universe is a comprehensible place and our theories are just better and better approximations to the reality. So it will &lt;strong&gt;always&lt;/strong&gt; be useful to learn Newtonian mechanics, even one million years from now.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-03T09:41:39.970" CommunityOwnedDate="2010-12-03T09:41:39.970" />
  <row Id="1581" PostTypeId="2" ParentId="1019" CreationDate="2010-12-03T09:48:37.143" Score="8" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;You need something more to get something more&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is about emergent macroscopic properties of microscopical laws. Some people can't understand that statistics is powerful enough to make seemingly random heap of molecules suddenly show macroscopic properties like &lt;em&gt;being solid&lt;/em&gt; or &lt;em&gt;being magnetic&lt;/em&gt; and they think some hand of god is required to make this happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best illustration of a contradiction to this principle are living organisms. They consist of nothing else than few physical laws all the way down and large number of molecules. All that was needed was statistics and natural selection.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I will agree though that it is quite amazing all of our nature can spontaneously emerge just from particles given enough space and time.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-03T09:48:37.143" CommunityOwnedDate="2010-12-03T09:48:37.143" CommentCount="2" />
  <row Id="1582" PostTypeId="2" ParentId="1574" CreationDate="2010-12-03T10:21:39.080" Score="5" ViewCount="" Body="&lt;p&gt;Mark's answer is correct but it is way too long and hides the punchline. So let me show a shorter derivation using more advanced math. Not too advanced though, just tensor formalism in &lt;a href=&quot;http://en.wikipedia.org/wiki/Minkowski_space&quot; rel=&quot;nofollow&quot;&gt;Minkowski space-time&lt;/a&gt; for Special Relativity and &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_form&quot; rel=&quot;nofollow&quot;&gt;differential forms&lt;/a&gt;. You will need all of this sooner of later, so it should be useful to learn (at least a little) about it already. This answer would be just a few lines if you already knew the formalism but it will be a little longer because I'll try to teach you also about the formalism.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You probably know that &lt;a href=&quot;http://en.wikipedia.org/wiki/Lorentz_transformation&quot; rel=&quot;nofollow&quot;&gt;Lorentz transformations&lt;/a&gt; mix $\mathbf E$ and $\mathbf B$. So they are not really independent and it turns out that they are just a part of rank 2 antisymmetric 4-dimensional tensor (this really means $4 \times 4$ antisymmetric matrix) $\mathbf F$. Now, it should be at least dimensionally clear that such a matrix has 6 independent components which precisely coincides with 3+3 degrees of freedom of $\mathbf E$ and $\mathbf B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should probably also know that both $\mathbf E$ and $\mathbf B$ can be expressed in terms of potentials. In our formalism it translates into ${\mathbf F} = {\rm d} {\mathbf A}$ where ${\rm d}$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Exterior_derivative&quot; rel=&quot;nofollow&quot;&gt;exterior derivative&lt;/a&gt; $\mathbf A$ is the four-potential that combines scalar $\phi$ and three-vector $\mathbf A$ potentials you should already know and love.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it turns out that Maxwell equations in vacuum have really simple form in this formalism&#xA;$${\rm d}{\mathbf F} = 0$$&#xD;&#xA;$${\rm \delta}{\mathbf F} = 0$$&#xD;&#xA;with $\delta$ being the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hodge_dual#The_codifferential&quot; rel=&quot;nofollow&quot;&gt;codifferential&lt;/a&gt; which is dual to $\rm d$. The first equation actually tells us that four-potential exists (because ${\rm d}^2 = 0)$ and the second one is the actual evolutionary equation that would contain four-current $\mathbf j$ if we weren't in vacuum.&#xA;Now whenever we have a solution to these equations, they will also solve&#xA;$$\square {\mathbf F} = ({\rm d\delta + \delta d}) {\mathbf F} = 0$$&#xD;&#xA;But this $\square$ is precisely &lt;a href=&quot;http://en.wikipedia.org/wiki/D%27Alembert_operator&quot; rel=&quot;nofollow&quot;&gt;d'Alembert wave operator&lt;/a&gt; and so indeed $\mathbf F$ propagates at the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference:&lt;/strong&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Covariant_formulation_of_classical_electromagnetism&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; on the covariant or four-vector formalism&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-03T21:19:04.187" LastActivityDate="2010-12-03T21:19:04.187" CommentCount="11" />
  <row Id="1583" PostTypeId="2" ParentId="1019" CreationDate="2010-12-03T12:34:42.877" Score="10" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Classical mechanics is boring and mostly solved&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...especially in case of fluid dynamics (-;&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-12-03T12:34:42.877" CommunityOwnedDate="2010-12-03T12:34:42.877" CommentCount="2" />
  <row Id="1584" PostTypeId="2" ParentId="1419" CreationDate="2010-12-03T12:57:58.340" Score="0" ViewCount="" Body="&lt;p&gt;Motion of a non-point object can be described as a combination of translation (which only depends on the total mass of the object) and rotation (which depends on the mass distribution). Rotational momentum is defined as $\mathbf{L}= I \boldsymbol{\omega}$ where $I$ is a moment of inertia $I = \int_V \rho(\mathbf{r})\,d(\mathbf{r})^2 \, \mathrm{d}V\!(\mathbf{r})$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the mass distribution defines the &lt;strong&gt;rotational&lt;/strong&gt; momentum of a system.&lt;/p&gt;&#xA;" OwnerUserId="701" LastActivityDate="2010-12-03T12:57:58.340" />
  <row Id="1585" PostTypeId="2" ParentId="1126" CreationDate="2010-12-03T13:24:09.197" Score="2" ViewCount="" Body="&lt;p&gt;In reality you could do that by throwing curved balls. You curve the slower ball in one direction so it travels further and you curve the faster one in the opposite direction. You can see that in table tennis.&lt;/p&gt;&#xA;" OwnerUserId="701" LastActivityDate="2010-12-03T13:24:09.197" CommentCount="2" />
  <row Id="1586" PostTypeId="1" CreationDate="2010-12-03T13:32:35.817" Score="3" ViewCount="150" Body="&lt;p&gt;(Probably related to &lt;a href=&quot;http://physics.stackexchange.com/questions/721&quot;&gt;this one&lt;/a&gt;, and probably should be CW.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very long time ago, I had the good fortune to read George Gamow's excellent series of &lt;em&gt;Mr. Tompkins&lt;/em&gt; books. That introduced me to the idea of a world where the usual physical constants (e.g. the speed of light and Planck's constant) were changed such that &quot;paradoxical&quot; effects became apparent in the macroworld.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My memory is hazy now, but I do recall the concepts of relativity (e.g. dilation) becoming more pronounced when the speed of light is reduced to &quot;human-sized&quot; speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this vein, I ask this: assuming all other physical constants being fixed, what exactly can be expected to happen if (&lt;em&gt;physical constant of your choice&lt;/em&gt;) is increased/or decreased?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One physical constant per answer, please.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-12-04T12:06:33.410" Title="What if physical constants were increased or decreased?" Tags="&lt;physical-constants&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="1587" PostTypeId="2" ParentId="1102" CreationDate="2010-12-03T14:14:28.757" Score="3" ViewCount="" Body="&lt;p&gt;The water drop will immideately start to &lt;em&gt;boil&lt;/em&gt; because below $0.006 atm$ it boils even at $273 K$. However, boiling in vacuum is something special since there is no buoyancy that would force bubbles to leave the droplet. I think that the process will look similar to the one shown in &lt;a href=&quot;http://www.youtube.com/watch?v=bgC-ocnTTto&quot; rel=&quot;nofollow&quot; title=&quot;Alka-Seltzer added to spherical water drop in microgravity&quot;&gt;this beautiful video&lt;/a&gt; and the droplet will soon end up as a ball of ice foam.&lt;/p&gt;&#xA;" OwnerUserId="701" LastActivityDate="2010-12-03T14:14:28.757" />
  <row Id="1588" PostTypeId="2" ParentId="1077" CreationDate="2010-12-03T14:26:08.410" Score="-1" ViewCount="" Body="&lt;p&gt;My first answer contained a very sloppy statement that EM interaction is irrelevant because of neutrality of matter which had to be corrected. I consider interaction of two neutral atoms and leave out the question of why some matter is solid and other is not.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Higher-order EM interaction does exist between atoms. What is maybe counterintuitive, it is an &lt;em&gt;attractive&lt;/em&gt; interaction called &lt;strong&gt;van der Waals force&lt;/strong&gt;. &lt;/li&gt;&#xA;&lt;li&gt;Electrons repel each other but manage to stay together even if there are 50 of them in an atom. We could imagine moving two hydrogen atoms together until they (almost) form a helium atom. Electric repulsion is not going to become infinite, the total energy would be in fact lower. Electric repulsion does not prevent one atom going through another.&lt;/li&gt;&#xA;&lt;li&gt;However, two electrons in the same state just can not exist close to each other (&lt;strong&gt;Pauli exclusion&lt;/strong&gt;) and the atomic orbitals would have to distort to avoid overlap. There is no special force that does that so eventually it comes back to an electromagnetic interaction but the fact is that without Pauli exclusion this interaction would not happen.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I should also mention the name of Lennard-Jones whose &lt;a href=&quot;http://en.wikipedia.org/wiki/Lennard-Jones_potential&quot; rel=&quot;nofollow&quot; title=&quot;potential&quot;&gt;potential&lt;/a&gt; describes an interaction between two atoms. Note the $r^{-12}$ term for the short-range Pauli repulsion.&lt;/p&gt;&#xA;" OwnerUserId="701" LastEditorUserId="701" LastEditorDisplayName="" LastEditDate="2010-12-07T15:24:14.147" LastActivityDate="2010-12-07T15:24:14.147" CommentCount="11" />
  <row Id="1589" PostTypeId="2" ParentId="1572" CreationDate="2010-12-03T14:31:28.260" Score="3" ViewCount="" Body="&lt;p&gt;In the picure below, you can see how the row-column grid correspond to the graphene structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/LJRLx.png&quot; alt=&quot;Graphene structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To have a (n,m) nanotube, you &quot;just&quot; have to roll your graphene sheet so that the (0,0) hexagon coincides with the (n,m) hexagon. Of course, it is much easier said than done !&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2010-12-03T14:31:28.260" CommentCount="2" />
  <row Id="1590" PostTypeId="2" ParentId="1102" CreationDate="2010-12-03T15:06:04.143" Score="-1" ViewCount="" Body="&lt;p&gt;I'm not to sure it would boil, because the effects of surface tension would raise the  internal pressure above zero. Also in order to have boiling, you have to overcome surface  tension. Surface tension basically means that the pressure inside a microbubble is greater  than the fluid pressure by an amount that is proportional to the inverse of its diameter. So&lt;br&gt;&#xA;you need to have seed bubbles of a certain size inorder to get internal bubbles to grow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it would cool by a combination of radiation (Stefan-Boltzmann) and evaporation. It wouldn't flash freeze, but over perhaps a hour or less it would start to freeze. So we would end up with a nearly spherical ice ball. It is easy to compute the cooling rate from radiation, but I'm not sure of the evaporative contribution to the cooling rate.&lt;/p&gt;&#xA;" OwnerUserId="391" LastEditorUserId="391" LastEditorDisplayName="" LastEditDate="2010-12-03T15:11:21.070" LastActivityDate="2010-12-03T15:11:21.070" />
  <row Id="1591" PostTypeId="2" ParentId="1077" CreationDate="2010-12-03T15:50:07.500" Score="3" ViewCount="" Body="&lt;p&gt;Ok, I'll bite the bullet and get the downvotes, but my answer is EM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? Well, you can't stand on water and you can't stand on air. The Pauli principle applies to those cases but doesn't make matter solid. It's the (fundamentally) cristalline structure that makes an object solid enought to stand on. This is yes, related to QM (what is not) but certainly EM in nature.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Marek asked me to specify a bit more and add some meat to my answer. Fair enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question itself is not a well-posed one. Electro-magnetism is quantum mechanical in nature, and it's basically impossible to speak about atoms without speaking about Pauli's principle and EM. Actually one could say that Pauli's principle is at the basis of chemistry, together with EM. So in this sense, what is left to answer? There can't be an atom without either of the two. In this sense the answer is BOTH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I choose therefore to interpret the question differently: is the exclusion principle acting between the soles of my shoes and the ground, that which is holding me up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the reason why we can stand on ground is undoubtedly chemical in nature - there are chemical bonds between atoms and molecules in the ground that make it solid (as opposed to, say, liquid). Frozen water is governed by EM, by the fact that the molecule is electrically charged, and not by chemical covalent bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this sense, for me, EM is probably more important than Pauli's into making something solid (and not liquid or gaseous).&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-12-03T21:00:55.007" LastActivityDate="2010-12-03T21:00:55.007" CommentCount="4" />
  <row Id="1592" PostTypeId="2" ParentId="1586" CreationDate="2010-12-03T15:53:30.137" Score="0" ViewCount="" Body="&lt;p&gt;Increasing the value of $\hbar$ significantly would be pretty interesting, as matter has a wavelength proportional to it.  Walking through a doorway might become a new experience as you get diffracted into a wall.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is all backwards thinking.  The physical constants calibrate our mathematical models of the universe, like $\lambda = h/p$, not the other way around.  So it is better to say that if you got diffracted every time you walked through a doorway, you would need a large $\hbar$ to model that.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, since constants like $\hbar$ have units, their value is somewhat arbitrary anyway.  Which brings us to dimensionless constants...  here's a good one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{\rm{grav}} / F_{\rm{em}} \propto \frac{\frac{G m m}{r^2}}{\frac{K q q }{ r^2}} \propto \frac{G}{K}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the ratio of the gravitational to electromagnetic force for a unit charge and mass at a unit distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently the electromagnetic force kicks gravity's butt, but I wonder what life would be like if that weren't the case.      &lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2010-12-03T15:53:30.137" CommentCount="7" />
  <row Id="1593" PostTypeId="2" ParentId="1019" CreationDate="2010-12-03T16:20:46.743" Score="6" ViewCount="" Body="&lt;p&gt;The most common misconceptions are about &lt;strong&gt;gravity&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(1) Gravity turns off at space-shuttle orbit distance because the astronauts are weightless&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity is at about 80% strength compared to the surface of the Earth.  The astronauts are weightless because the shuttle is in free-fall (orbit).  If there was no gravity, the shuttle could not orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(2) Gravity is generated by the spinning Earth.  If the Earth stopped spinning, gravity would turn off.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity is generated by virtue of the Earth's mass and the mass of the object; the two exert a mutual pull.  There are some smaller effects associated with the spinning Earth (e.g. the Coriolis effect), but gravity would still work fine if the Earth stopped spinning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another great misconception about the force of impact between a truck and a small car in a collision:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(3) The truck exerts a greater force of impact on the car, than the car on the truck&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the damage can be certainly unequal, the forces are.&lt;/p&gt;&#xA;" OwnerUserId="722" LastActivityDate="2010-12-03T16:20:46.743" CommunityOwnedDate="2010-12-03T16:20:46.743" CommentCount="1" />
  <row Id="1594" PostTypeId="2" ParentId="1518" CreationDate="2010-12-03T16:22:41.440" Score="0" ViewCount="" Body="&lt;p&gt;From the author : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data is in W/sr/m2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jussi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, radiance. Thanks anyway :)&lt;/p&gt;&#xA;" OwnerUserId="702" LastActivityDate="2010-12-03T16:22:41.440" CommentCount="2" />
  <row Id="1595" PostTypeId="1" AcceptedAnswerId="1596" CreationDate="2010-12-03T17:48:02.630" Score="2" ViewCount="194" Body="&lt;p&gt;Gravitational attraction is given by $\frac{GMm}{r^2}$ while attraction due to electric charge is given by $\frac{q_1 q_2}{r^2}$. Why does gravity need a constant while electric charge doesn't? Because we've picked &quot;the right&quot; units for charge so that the constant is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my vague understanding of &quot;natural units&quot;. My question is: what would natural units for mass look like, and why don't we use them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(related point: it seems weird that what a kilogram is is still decided by a lump of metal in Paris, while other SI units are defined by reproduceable standards...)&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-03T19:04:35.233" LastActivityDate="2010-12-03T19:16:38.777" Title="&quot;Natural units&quot; of mass" Tags="&lt;si-units&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1596" PostTypeId="2" ParentId="1595" CreationDate="2010-12-03T18:24:59.333" Score="8" ViewCount="" Body="&lt;p&gt;Well, that equation for the force due to electric charges is only true for a very special choice for the unit of the electric charge.  Typically, you would write down Coulomb's law as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k\frac{q_{1}q_{2}}{r^{2}}$, where $k$ is a constant of proportionality chosen to make the units work out.  IN the SI system, the unit of charge is the Coulomb (C) and the value of $k$ is approximately $8.99 \cdot 10^{9} \frac{N\cdot m^{2}}{C^{2}}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you choose to express your charge in Gauss units, defining one Gauss to be numerically equal to $\sqrt{8.99\cdot 10^{9}} \,\,C$, then the value of the Coulomb constant becomes 1 in this new system of units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you could do the same thing for gravity, but your unit of mass would be really small: $\sqrt{6.67 \cdot 10^{-11}} \,\, kg$.  It would also mess up the definitions for any unit that had dimensionality of mass, like the Newton, Joule, etc.  But you could do that.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there is a scheme called &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck_units&quot; rel=&quot;nofollow&quot;&gt;Planck Units&lt;/a&gt; that attempts to set as many fundamental constants of nature as possible equal to one.    &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-12-03T18:24:59.333" CommentCount="2" />
  <row Id="1598" PostTypeId="2" ParentId="1586" CreationDate="2010-12-03T20:16:17.383" Score="0" ViewCount="" Body="&lt;p&gt;See Smolin's book &quot;The Life of the Cosmos&quot;, &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Life_of_the_Cosmos&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://books.google.com/books?id=EF5jZ_eev-cC&amp;amp;printsec=frontcover&amp;amp;dq=smolin+The+Life+Cosmos&amp;amp;source=bl&amp;amp;ots=pICurxIw2S&amp;amp;sig=yeuUzglCG92kKkNClBiEkpy77ak&amp;amp;hl=en&amp;amp;ei=yk_5TP2_HcX7lweJvdjhBw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=5&amp;amp;ved=0CDwQ6AEwBA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, wherein he suggests common descent and Darwinian selection amongst multiverses for peturbed physical constants maximizing black hole formation in that universe.   His most recent paper on arxiv on this topic is &lt;a href=&quot;http://arxiv.org/abs/hep-th/0612185&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0612185&lt;/a&gt;, where he argues that this is still a live hypothesis, which has already survived several experimental tests.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-12-03T20:22:04.073" LastActivityDate="2010-12-03T20:22:04.073" />
  <row Id="1599" PostTypeId="1" AcceptedAnswerId="1600" CreationDate="2010-12-03T22:17:20.443" Score="6" ViewCount="200" Body="&lt;p&gt;the internet has changed science drastically. Not only in terms of distributing knowledge e.g. via online encyclopedias as wikipedia and freely available sources of publications as arXiv but also as a tool for communication through email and lively discussions in forums, newsgroups and communities like StackExchange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scientific success nowadays greatly depends on the ability to build up teams with different scientific background and experience. To my mind, the ideal researcher has to seek for people helping him to tackle parts of his problems in which he is not an expert instead of spending months on something, someone else could have been done in days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, can't &lt;strong&gt;we&lt;/strong&gt; be this ideal researcher, or: Can a community on a webpage do research by openly presenting ideas and working them out? Or is this not possible due to &quot;stealing&quot; of ideas, lacking institutional structures and the like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you think?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-03T22:42:27.450" LastActivityDate="2010-12-03T22:42:27.450" Title="Can a web community write papers?" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="4" />
  <row Id="1600" PostTypeId="2" ParentId="1599" CreationDate="2010-12-03T22:20:02.847" Score="5" ViewCount="" Body="&lt;p&gt;It's possible: see for instance the success of the various polymath projects&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Polymath_project#Polymath_Project&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Polymath_project#Polymath_Project&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="396" LastActivityDate="2010-12-03T22:20:02.847" CommentCount="7" />
  <row Id="1601" PostTypeId="1" AcceptedAnswerId="1604" CreationDate="2010-12-03T22:24:14.060" Score="12" ViewCount="590" Body="&lt;p&gt;I am a math grad student who would like to learn some classical mechanics. The caveat is I am not to interested in the standard coordinate approach. I can't help but think of the fields that arise in physics are actually sections of vector bundles (or maybe principal bundles) and would love an approach to classical mechanics or what have you that took advantage of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is there a text book you would recommend that phrases the constructions in classical mechanics via bundles without an appeal to transition functions?  &lt;/li&gt;&#xA;&lt;li&gt;What are the drawbacks to this approach other than the fact that it makes computations less doable? (if it does that)  &lt;/li&gt;&#xA;&lt;li&gt;Are there benefits to thinking about things this way, ie would it be of benefit to someone attempting to learn this material to do it this way?  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;thanks for your time,&#xA;sean&lt;/p&gt;&#xA;" OwnerUserId="40" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-03T22:50:32.753" LastActivityDate="2011-04-01T14:03:03.780" Title="Classical mechanics without coordinates book" Tags="&lt;classical-mechanics&gt;&lt;mathematical-physics&gt;&lt;books&gt;&lt;mathematics&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="8" />
  <row Id="1602" PostTypeId="2" ParentId="1599" CreationDate="2010-12-03T22:30:33.257" Score="4" ViewCount="" Body="&lt;p&gt;Not quite an answer but in the same vein:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many nuclear and high energy collaboration write papers in the name of the collaboration (and for that matter coordinate mostly over the intertubes). That is you'll see a paper by &lt;em&gt;&quot;the KamLAND collaboration&quot;&lt;/em&gt;. A list of the participants broken out by time can generally then be found in the collaboration's web space. This prevents the need for multiple page author lists.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2010-12-03T22:30:33.257" CommentCount="2" />
  <row Id="1603" PostTypeId="1" CreationDate="2010-12-03T22:58:22.453" Score="8" ViewCount="624" Body="&lt;p&gt;I have always wondered about applications of Algebraic Topology to Physics, seeing as am I studying algebraic topology and physics is cool and pretty. My initial thoughts would be that since most invariants and constructions in algebraic topology can not tell the difference between a line and a point and $\mathbb{R}^4$ so how could we get anything physically useful?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course we know this is wrong. Or at least I am told it is wrong since several people tell me that both are used.I would love to see some examples of applications of topology or algebraic topology to getting actual results or concepts clarified in physics. One example I always here is &quot;K-theory is the proper receptacle for charge&quot; and maybe someone could start by elaborating on that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sure there are other common examples I am missing. I won't make this CW unless someone has a really compelling reason why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks,&#xA;sean&lt;/p&gt;&#xA;" OwnerUserId="40" LastActivityDate="2011-01-20T02:37:13.593" Title="Applications of Algebraic Topology to physics" Tags="&lt;particle-physics&gt;&lt;mathematical-physics&gt;&lt;mathematics&gt;&lt;string-theory&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="4" />
  <row Id="1604" PostTypeId="2" ParentId="1601" CreationDate="2010-12-03T23:16:08.747" Score="7" ViewCount="" Body="&lt;h2&gt;1.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I am in love with Fecko's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521845076&quot; rel=&quot;nofollow&quot;&gt;Differential Geometry and Lie Groups for Physicists&lt;/a&gt;. Despite not being just about mechanics (but rather about more or less all rudimentary modern theoretical physics) it discusses both Lagrangian and Hamiltonian formalism. It also provides countless exercises (with nice hints) so that you can really get a feel for the matter.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;2.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I can't think of any major drawbacks. Of course, if the problem has no symmetry you sometimes have no other choice than to go back to some coordinates and solve numerically. But this is probably non-issue for you because I suppose you first want to understand physical problems with some structure.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;3.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There are countless benefits. To list just few of them.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;relation to symmetries and conserved quantities becomes obvious. Noether's theorem in Hamiltonian formalism is so amazingly simple statement (Hamiltonian is constant for symmetry flow if an only if the generator of the symmetry is constant for Hamiltonian flow) that one has to wonder where all the long-winded coordinate calculations went.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Not only are the calculations short, one also gains valuable geometrical insights e.g. about the flow of the configuration on the manifold.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It's a beautiful formalism.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I don't know about others but whenever I have to calculate in coordinates I become nervous. I can compute the results but after few pages when most of the quantities mysteriously cancel, you don't really know why what you derived is true. So then you go back to geometry and lo and behold, the derivation is just few lines and obvious. Of course I am exaggerating now but that's the way I feel.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It's the basis for all of modern physics. If the above four points were true in classical mechanics, they are even more true when dealing with things like gauge theories (and that is where the full beauty and power of mathematics comes out).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-04T00:37:07.787" LastActivityDate="2010-12-04T00:37:07.787" CommentCount="2" />
  <row Id="1605" PostTypeId="2" ParentId="1603" CreationDate="2010-12-04T00:34:51.673" Score="5" ViewCount="" Body="&lt;p&gt;First a warning: I don't know much about either algebraic topology or its uses of physics but I know of some places so hopefully you'll find this useful.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Topological defects in space&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The standard (but very nice) example is &lt;a href=&quot;http://en.wikipedia.org/wiki/Aharonov%2dBohm_effect#Magnetic_solenoid_effect&quot; rel=&quot;nofollow&quot;&gt;Aharonov-Bohm effect&lt;/a&gt; which considers a solenoid and a charged particle. Idealizing the situation let the solenoid be infinite so that you'll obtain ${\mathbb R}^3$ with a line removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the particle is charged it transforms under the $U(1)$ gauge theory. More precisely, its phase will be parallel-transported along its path. If the path encloses the solenoid then the phase will be nontrivial whereas if it doesn't enclose it, the phase will be zero. This is because $$\phi \propto \oint_{\partial S} {\mathbf A} \cdot d{\mathbf x}  = \int_S \nabla \times {\mathbf A} \cdot d{\mathbf S} = \int_S {\mathbf B}\cdot d{\mathbf S}$$ and note that $\mathbf B$ vanishes outside the solenoid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The punchline is that because of the above argument the phase factor is a topological invariant for paths that go between some two fixed points. So this will produce an interference between topologically distinguishable paths (which might have a different phase factor).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Instantons&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;One place where homotopy pops up are &lt;a href=&quot;http://en.wikipedia.org/wiki/Instanton#Quantum_field_theory&quot; rel=&quot;nofollow&quot;&gt;Instantons&lt;/a&gt; in gauge theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, if you consider a &lt;a href=&quot;http://en.wikipedia.org/wiki/Yang%2dMills_theory&quot; rel=&quot;nofollow&quot;&gt;Yang-mills theory&lt;/a&gt; in ${\mathbb R}^4$ (so this means Euclidean time) and you want the solution (which is a connection) to have a finite energy then its curvature has to vanish at infinity. This allows you to restrict your attention to $S^3$ (this is where the term instanton comes from; it is localized) and this is where homotopy enters to tell you about topologically inequivalent ways the field can wrap around $S^3$. Things like these are really big in modern physics (both QCD and string theory) because instantons give you a way to talk about non-perturbative phenomena in QFT. But I am afraid I can't really tell you anything more than this. &lt;em&gt;(I hope I'll get to study these things more myself).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;TQFT&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Last point (which I know nearly nothing about) concerns &lt;a href=&quot;http://en.wikipedia.org/wiki/Topological_quantum_field_theory&quot; rel=&quot;nofollow&quot;&gt;Topological Quantum Field Theory&lt;/a&gt; like &lt;a href=&quot;http://en.wikipedia.org/wiki/Chern%2dSimons_theory&quot; rel=&quot;nofollow&quot;&gt;Chern-Simons theory&lt;/a&gt;. These again arise in string theory (as does all of modern mathematics). And again, I am sorry I cannot tell you more than this yet.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-04T00:40:33.323" LastActivityDate="2010-12-04T00:40:33.323" />
  <row Id="1606" PostTypeId="1" AcceptedAnswerId="1617" CreationDate="2010-12-04T00:58:25.703" Score="2" ViewCount="99" Body="&lt;p&gt;I've seen several popular reports of a new count of low-mass stars in elliptical galaxies (&lt;a href=&quot;http://www.usatoday.com/tech/science/space/2010-12-01-dwarf-stars_N.htm&quot; rel=&quot;nofollow&quot;&gt;here's one&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Pursuant to several correct comments I've changed the title to agree with the actual report which is that the recount concerns elliptical galaxies---and I don't know where I got the notion that it concerned dwarf galaxies---but I am leaving my comments belo intact as they represent the way I was thinking before I was corrected. Note that we are in fact talking about relatively few very massive galaxies instead of many very light ones, but the questions are largely unchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first instinct was to dismiss it as mostly interesting to those who specialize in galactic dynamics, but then it occurred to me that there must be a &lt;em&gt;lot&lt;/em&gt; of those galaxies, and I began to wonder about the baryonic-matter/dark-matter/dark-energy balance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that this makes little difference to the matter/dark energy part of the equation because the total matter fraction is derived from large scale measurements of cluster dynamics. But even if I am right about the matter/dark-energy thing, that leaves the question of baryonic vs. dark matter fraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone shed some light on this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, links to pre-prints or journal articles related to this measurement would be welcome.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2010-12-04T02:55:41.370" LastActivityDate="2010-12-04T15:44:53.177" Title="Does the recent re-count of stars in elliptical gallaxies affect our understanding of the universal mass balance?" Tags="&lt;cosmology&gt;&lt;galaxy&gt;&lt;stellar-population&gt;&lt;mass-balance&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="1607" PostTypeId="2" ParentId="1606" CreationDate="2010-12-04T02:47:42.797" Score="1" ViewCount="" Body="&lt;p&gt;I guess I shouldn't try more than a one liner in a &quot;comment&quot;, it looks like (enter) causes a new comment....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, it is an interesting question. My reading was that the unexpectedly high abundance was for eliptical galaxies, which are usually the largest ones. So the significance may be higher than you assume. Supposedly the Milky Way (which is a spiral galaxy) doesn't show the same superabundance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case stellar populations generally follow a powerlaw mass distribution function, with the lowest&#xA;mass objects dominating both the numbers and the mass. It has generally been assumed that there&#xA;is a low mass cutoff. I think this follows from star formation theory, which says low mass stars take too long to form (their birth clouds would be disrupted). Obviously our understanding of star birth is inadaquate. I wonder about the abundance of brown dwarfs. Perhaps the total amount of mass in dark clumps of normal matter is greater than currently assumed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think that the case for exotic dark matter to not exist has any support however. Theories of cosmology including the clumping of matter and formation of galaxy clusters do seem to require dark matter. But an increase in the amount of normal matter would be interesting. &lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-12-04T02:47:42.797" CommentCount="3" />
  <row Id="1608" PostTypeId="2" ParentId="1603" CreationDate="2010-12-04T02:55:19.993" Score="1" ViewCount="" Body="&lt;p&gt;All of Marek's examples are good.  (I had to write a new answer due to space limitations on comments.)  Instantons are probably the single best place to explore this relationship.  Maxwell's equations in vacuo read dF = 0 and d*F = 0, where F is the field strength tensor.  The charge of a particle (by Gauss' law) can be obtained by calculating the integral of *F, on a surrounding sphere, while the magnetic charge (always zero, as we have not yet observed magnetic monopoles reliably) is the integral of F.  Now if you're studying algebraic topology, F is the Chern form of the connection defined by the gauge field (vector potential), namely it represents the first Chern class of this bundle.  This is the prime example of how a characteristic class -- which measures the topological type of the bundle -- appears in physics as a quantum number, magnetic charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, in quantum field theory we are instructed to integrate over ALL connections, including those for different topological types of bundles -- so the configuration space has different components.  The minimum (Euclidean) energy configurations in these different components are called &quot;instantons.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many other examples which involve mildly exotic theories of physics.&lt;/p&gt;&#xA;" OwnerUserId="20" LastActivityDate="2010-12-04T02:55:19.993" />
  <row Id="1609" PostTypeId="2" ParentId="1601" CreationDate="2010-12-04T03:07:26.413" Score="5" ViewCount="" Body="&lt;p&gt;The story is not very long.  A symplectic manifold is a manifold $X$ with a nondegenerate, closed two-form, $\omega$.  Given this, from a function $h$ (&quot;Hamiltonian&quot;) we can construct a vector field $v$ by $\omega(v,-) = dh$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flow by $v$ defines the classical trajectories.  (Note we have not used a metric on $X$, i.e. $v$ is NOT the gradient of $h$.)  In fact, $h$ remains constant (conserved) in the flow, since $v(h) = \omega(v,v) = 0.$  Also note ${\mathcal L}_v \omega = 0$, meaning $\omega$ is conserved in the flow -- in particular, so is the Liouville phase $\omega^{\wedge n}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example, take $X$ to be the cotangent space $T^*({\mathbb R}^n)$ with coordinates $x$ (position) and $y$ (momentum), with $\omega = {\rm d}x \wedge {\rm d}y$.  Take $h = y^2/2 + V(x)$, i.e. KE + PE.  Then $v = y {\partial \over \partial x} - V&amp;#39; {\partial \over \partial y}$, so the flow equations are $\dot{x} = y$, $\dot{y} = -V&amp;#39;$, or $\ddot{x} = -V&amp;#39;$, Newton's law.&lt;/p&gt;&#xA;" OwnerUserId="20" LastEditorUserId="20" LastEditorDisplayName="" LastEditDate="2010-12-05T03:07:12.490" LastActivityDate="2010-12-05T03:07:12.490" CommentCount="2" />
  <row Id="1610" PostTypeId="1" AcceptedAnswerId="1611" CreationDate="2010-12-04T03:49:35.907" Score="0" ViewCount="315" Body="&lt;p&gt;so I've been at this for about 3 - 4 hours now. It is an homework assignment (well part of a question which i've already completed). We did not learn this in class. All work is shown below. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An atom in an excited state of $4.9 eV$ emits a photon and ends up in the ground state. The lifetime of the excited state is $1.2 \times 10^{-13} s$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(b) What is the spectral line width (in wavelength) of the photon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So lets look at what I have done so far. I have done the following: &#xA;$$\Delta E \Delta t = \frac{\hbar}{2} $$ but &#xA;$$E = h f$$ so&#xA;$$\Delta f = \frac{1}{4\pi \Delta t}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but if I take $\Delta f$ and convert it into wavelength using $\lambda f = c $ then it gives me the wrong answer. I've tried MANY variations of the above formulas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct answer is $0.142 nm $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone give me a hint?&lt;/p&gt;&#xA;" OwnerUserId="551" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-04T11:01:20.473" LastActivityDate="2010-12-04T11:01:20.473" Title="Spectral Line Width and Uncertainty principle" Tags="&lt;homework&gt;&lt;uncertainty-principle&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="1611" PostTypeId="2" ParentId="1610" CreationDate="2010-12-04T04:53:36.150" Score="3" ViewCount="" Body="&lt;p&gt;Hint: Your problem is in the &quot;take $\Delta f$ an convert it to wavelength using $\lambda f = c$&quot; part.  The equation $\lambda f = c$ does not imply $\Delta \lambda \Delta f = c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: Rather it implies,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda f = c$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda = \frac{c}{f} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now differentiate:&#xA;$d\lambda = -c\frac{df}{f^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$df \ll f$ so treating the $\Delta f$ as a differential works fine.  If this was not the case, you'd want to integrate from $f_{min}$ to $f_{max}$ (not given in the problem, just how you'd have to do it otherwise).  This formula gives you the right answer.  The sign opposite signs just indicate the higher frequency corresponds to the lower wavelength and is ignored in the final answer.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-04T04:53:36.150" />
  <row Id="1612" PostTypeId="2" ParentId="1610" CreationDate="2010-12-04T04:54:42.530" Score="2" ViewCount="" Body="&lt;p&gt;Okay so a buddy helped me out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You had to use the following formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta \lambda = hc \( \frac{1}{E_1} - \frac{1}{E_2} \) $$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="551" LastActivityDate="2010-12-04T04:54:42.530" />
  <row Id="1613" PostTypeId="2" ParentId="1077" CreationDate="2010-12-04T05:53:39.947" Score="2" ViewCount="" Body="&lt;p&gt;Holy moly, there are a lot of confused parts of answers out there.   Here is a way to begin to sort out the different related principles, physically. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The question currently asks &quot;EM or/and Pauli?&quot;.    Short answer: Neither, although it is true that electromagnetism is the force involved (rather than the strong or weak nuclear forces, the only other candidates), and it is true that the size of atoms is determined by the uncertainty and exclusion principles, as it were.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Neglect or remove gravity from the situation.   Then the question really becomes, &quot;Why are solids, at the earth's crust say (but it doesn't matter), solid, i.e. why do they resist strain, to the quantitative degree that they do?&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We can idealize the problem.   Why do single crystals resist strain?   We replace rocks and dirt by interlaced microcrystals, or mixtures bound by friction.   In other words, why are crystals rigid?   Why can't we walk on water until it freezes?   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What are the energy scales of the problem?   Well the typical force to be considered will be the force required to disrupt the crystal lattice.  This is about the thermal energy at melting, = k*T(melting).  So for water (since I like to think in electron volts) about 0.025 ev/molecule, for rocks about 10 times as much.  This is because the water molecules in an ice crystal are bound by hydrogen bonds with strength about 0.1 e.v. and atoms in quartz are bound by covalent bonds of about 1 ev.   So yes, electromagnetism is the involved force.&#xA;To be specific, the force on the bottom of a shoe, normally about (1kg*9.8 m/sec^2)/cm^2 = 2 x 10 -15 kg m s-2/(the area corresponding to a molecule of water on the surface of the crystal)  times the distance such a force must move the molecule (maybe 10-10 m?) = 2 x 10^-25 J or (using 6.25 x 10^18 ev/joule) 1.2 x 10-6 ev gets compared to kT as above. So the force of the shoe walking on ice supplies only about 4.8 x 10-5 of the required energy to melt or deform a crystal of ice.  So we don't sink through solid ice. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Here we learn an important lesson.   In physics, to talk about &quot;a cause&quot; is to secretly talk about the about the calculations and equations underlying computing the magnitude of the phenomenon.   So, Weisskopf, (see below) points out that the same equations with different numbers, this time using the pressure developed at the bottom of a mountain actually does reach the energy scale required to deform or melt crystals of rock (say quartz), and therefore computes the maximal height of mountains on earth (or any planet), using only a few fundamental constants.   This then carries over.   To compute the maximal height of mountains on white dwarf stars (fanciful, I think), or importantly on the surface of neutron stars (real) use the same principles, but now invoke the exclusion principal, or other forces, since these are now relevant to the energy scales involved.&#xA;On a more mundane scale and as a check, if we concentrate one's weight on a much smaller area like the blade of an ice skate, to yield a higher force, we indeed can deform ice crystals, which is said to be the reason for the low friction we experience when ice skating.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Back to earth, shoes.   At these very small energies, compared to any atomic energy scale, to say nothing of the scales involved in the weak or strong interactions, we can should just consider atoms as single units, not as separate electrons or nuclei.   Quantum mechanics is required to allow us to neglect all the higher energy degrees of freedom which have been &quot;frozen out&quot; at the low (300K) average energy of the molecules involved.  Weisskopf has a &lt;a href=&quot;http://books.google.com/books?id=te1LFEIXxjcC&amp;amp;printsec=frontcover&amp;amp;dq=inauthor%3a%22Victor+F.+Weisskopf%22&amp;amp;hl=en&amp;amp;ei=mNL5TOOPAcWqlAfX4cmRDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CDAQ6AEwAQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;good exposition&lt;/a&gt; of this in his popular book &quot;Knowledge and Wonder&quot;, see esp chap. 7 on &quot;The quantum ladder&quot;  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We are left with a more well defined problem:  why do liquids freeze and become solid, rigid crystalline shapes, and why are crystals rigid in the first place?   Rephrased: why does lowering the temperature a little bit, lead to &quot;freezing out&quot; of the translational degrees of freedom of individual molecules or atoms, with the simultaneous aquisition of a single rigid global translational phase (lattice) for the positions of all the molecules/atoms in a crystal.   In other words, why, after the phase transition, this change in the microscopic symmetry of the material, is there rigidity? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This also suggests the answer to the original question.  In order for strain to become large, molecules need to break from the global phase and leave their position in the crystal lattice, and the lattice resists this change &lt;em&gt;cooperatively&lt;/em&gt;, i.e. many molecules must get disrupted, so the energy required is large, considerably larger than the energy available from the tiny pressures involved from the pressure from your weight on the sole of your shoe.   There is more, but I'll stop here for now.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;P.S.   All of the Weisskopf popular physics books are terrific&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-12-04T19:10:19.437" LastActivityDate="2010-12-04T19:10:19.437" CommunityOwnedDate="2010-12-04T19:13:52.633" CommentCount="2" />
  <row Id="1614" PostTypeId="2" ParentId="79" CreationDate="2010-12-04T06:27:23.957" Score="1" ViewCount="" Body="&lt;p&gt;There is no effect of the one measurement event upon the other.  It is not until the results from both measurements are brought together for comparison, and accumulated statistically, that it gets interesting.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only interactions relevant to the entanglement are at the source, when the singlet spin system falls apart into two spin one particles (or whatever exactly it is you're doing) and again when  the measurements are correlated in one place.  The latter is not often mentioned in QM, but only in discussions of the philosophy of QM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Too many writers assume something nonlocal is going on, or some superluminal effect occurs.  The classic papers on the Bell experiments typically state that one (at least) of these has to go:  locality, causality, realism.  It's an open discussion even today in 2010, which one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I like to toss out causality, and understand things according to Cramer's Transactional Interpretation -- &lt;a href=&quot;http://mist.npl.washington.edu/npl/int_rep/tiqm/TI_toc.html&quot; rel=&quot;nofollow&quot;&gt;http://mist.npl.washington.edu/npl/int_rep/tiqm/TI_toc.html&lt;/a&gt; -- but at the end of the day I really don't know any better than anyone esle.&lt;/p&gt;&#xA;" OwnerUserId="353" LastActivityDate="2010-12-04T06:27:23.957" />
  <row Id="1615" PostTypeId="1" AcceptedAnswerId="1616" CreationDate="2010-12-04T08:34:54.350" Score="0" ViewCount="113" Body="&lt;p&gt;How to start with a perfect fluid concept and reach (by approximations through certain mathematically well defined assumptions) to the concept of particle ? Here newtonian gravitation is being assumed. The state equation of matter can be assumed to be that of dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume any initial mass distribution $\rho(r,t=0)$ and momentum distribution $p(r,t=o)$ and state equation of dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't intend to throw away the fluid concept,but still want to arrive at a particle as some form of a mass distribution.One example could be $\rho_p(r,t)$ which has a compact support in r describing a particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume there are a number of such particles as described in the earlier edit part of the question as $\rho_p(\vec{r}-\vec{r_i},t=0)$ for $i = 1,2,3,...$. distributed arbitrarily in space such that  no two are overlapping. This could be considered as $\rho(r,t=0) = \displaystyle\sum\limits_{i=0}^n \rho_p(\vec{r}-\vec{r_i},t=0)$. Similarily for momentum distibution at $t=0$ as $\vec{p}(r,t=0) = \displaystyle\sum\limits_{i=0}^n \vec{p_p}(\vec{r}-\vec{r_i},t=0)$. What would eventually happen to such a system under its own newtonian gravity. Just out of curiosity what would happen if GR is assumed.&lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="540" LastEditorDisplayName="" LastEditDate="2010-12-04T14:41:41.900" LastActivityDate="2010-12-04T14:41:41.900" Title="Fluid to particles under newtonian gravity" Tags="&lt;newtonian-gravity&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="1616" PostTypeId="2" ParentId="1615" CreationDate="2010-12-04T10:17:08.437" Score="1" ViewCount="" Body="&lt;p&gt;If I understand your question correctly you want to get from Gauss' gravitational law to Newton's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \nabla \cdot \textbf{g} = -4 \pi G \rho$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating both sides over a volume $V$ enclosing a mass $M$ and having surface $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{V} \! \nabla \cdot \textbf{g } \mathrm{d}\textbf{V} = -4 \pi G \int_{V} \! \rho\ \mathrm{d}\textbf{V}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{V} \! \nabla \cdot \textbf{g } \mathrm{d}\textbf{V} = -4 \pi G M$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Divergence_theorem&quot; rel=&quot;nofollow&quot;&gt;divergence theorem&lt;/a&gt;: $\int_{V} \! \nabla \cdot \textbf{g } \mathrm{d}\textbf{V} = \oint_{\partial{V}}  \textbf{g} \cdot \mathrm{d}\textbf{S}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\oint_{\partial{V}} \! \textbf{g} \cdot \mathrm{d}\textbf{S} = -4\pi GM $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that $\textbf{g}$ is constant over $\mathrm{d}\textbf{S}$, and that the force is centripetal we can write &#xA;$$\textbf{g} = g(r){\textbf{e}_r}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\textbf{e}_r$ are the unit vectors of the acceleration that point towards the center on the surface $S$ and g(r) gives us the magnitude of the acceleration at radius $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our equation then becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(r)\oint_{\partial{V}} \! \textbf{e}_r \cdot \mathrm{d}\textbf{S} = -4\pi GM $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(r) 4\pi r^2 = -4\pi GM $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(r) = -\frac{GM}{r^2} $$&#xD;&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-12-04T10:48:26.507" LastActivityDate="2010-12-04T10:48:26.507" CommentCount="12" />
  <row Id="1617" PostTypeId="2" ParentId="1606" CreationDate="2010-12-04T11:50:43.800" Score="3" ViewCount="" Body="&lt;p&gt;The fraction of baryonic matter to dark matter is not deduced only from galactic dynamics. It is also derived from big bang nucleosynthesis and from the higher multipole acoustic peaks in the CMB spectrum. I would say that the element abundance is a far more important indicator of the fraction between baryonic and dark matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pdg.lbl.gov/2006/reviews/bigbangnucrpp.pdf&quot; rel=&quot;nofollow&quot;&gt;Big-Bang nucleosynthesis&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretical overview of Cosmic Microwave Background anisotropy: &lt;a href=&quot;http://nedwww.ipac.caltech.edu/level5/Sept07/Wright/Wright2.html&quot; rel=&quot;nofollow&quot;&gt;1.2. Results&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="383" LastEditorUserId="383" LastEditorDisplayName="" LastEditDate="2010-12-04T15:44:53.177" LastActivityDate="2010-12-04T15:44:53.177" CommentCount="2" />
  <row Id="1618" PostTypeId="2" ParentId="1586" CreationDate="2010-12-04T12:06:33.410" Score="1" ViewCount="" Body="&lt;p&gt;I think you can find one of the answers &lt;a href=&quot;http://www.youtube.com/watch?v=JQnHTKZBTI4&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="498" LastActivityDate="2010-12-04T12:06:33.410" />
  <row Id="1619" PostTypeId="1" CreationDate="2010-12-04T14:26:30.063" Score="0" ViewCount="36" Body="&lt;p&gt;I need to build a device to measure the impedance in AC, at various frequency (100Hz - 100KHz), with the quadrupole method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, I have quite a good understanding of the principle behind the this kind of measurements. What I have some problem with is the construction of the setup i'm gonna need beside the 4-pin probe itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess I'm gonna need a AC current generator (right?) but what is necessary to acquire the data and be able to plot them both in a Bode representation (|Z| vs. freq and PhaseAngle vs. freq) and in Nyquist representation (-Im(Z) vs. Re(Z)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is really a little be outside my field of study so please, kindly elaborate :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thx&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2010-12-04T16:02:09.870" ClosedDate="2010-12-10T22:07:34.407" Title="Impedance Measurements in AC at Various Frequency" Tags="&lt;electromagnetism&gt;&lt;electricity&gt;&lt;electronics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="1620" PostTypeId="2" ParentId="1557" CreationDate="2010-12-04T14:47:49.757" Score="0" ViewCount="" Body="&lt;p&gt;I'm going to try for a qualitative, equation-free version of the answer..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you push on an object, you increase its momentum, which is the product of the object's mass and its speed. When you push on an object that is at rest, i.e., not already moving relative to you, the change in the object's momentum is realized almost entirely through a change in the speed component. This is what gives us the 'common sense' that if you push something a little bit harder, it will go a little bit faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as the object's speed approaches the speed of light, the effect of applying force to the object is altered. Rather than the object's speed increasing, its mass begins to increase instead. So when the object's apparent speed is for example 99.99% of light, if you push it a little harder, while it does speed up slightly, it mainly just &lt;em&gt;becomes a little bit heavier&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This changeover from effect-on-speed to effect-on-mass happens gradually (not all at once!), and there are equations in the other answers that describe it quantitatively. At everyday speed scales, the change-in-mass effect is practically immeasurable, so it seems counter-intuitive, but put particles in an accelerator, and it becomes observable fact.&lt;/p&gt;&#xA;" OwnerUserId="717" LastActivityDate="2010-12-04T14:47:49.757" />
  <row Id="1621" PostTypeId="1" AcceptedAnswerId="1622" CreationDate="2010-12-04T14:54:39.257" Score="0" ViewCount="163" Body="&lt;p&gt;I am not aware of GR, but due to curiosity i have a question in my mind. Please let me know if it is inappropriate to ask here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is about singularity. I am under the assumption that singularity means in mathematical terms equivalent to a &lt;a href=&quot;http://en.wikipedia.org/wiki/Discontinuity_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;discontinuity&lt;/a&gt; in a function. My question is what type of discontinuity are the ones corresponding to &lt;a href=&quot;http://en.wikipedia.org/wiki/Penrose-Hawking_singularity_theorems&quot; rel=&quot;nofollow&quot;&gt;Penrose-Hawking singularity theorems&lt;/a&gt; and also to the the &lt;a href=&quot;http://en.wikipedia.org/wiki/Naked_singularity&quot; rel=&quot;nofollow&quot;&gt;naked singularity&lt;/a&gt; ?&#xA;by type i mean&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;removable (left and right limits exist and are equal &#xA;         but not equal to the function value at that point).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a jump type discontinuity (left and right limits exists and are &#xA;       not equal to each other).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Oscillating discontinuity (as described as described in the above given link)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="540" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-12-04T18:41:19.077" LastActivityDate="2010-12-06T04:43:35.767" Title="A Question on Singularity" Tags="&lt;general-relativity&gt;&lt;gravity&gt;" AnswerCount="3" />
  <row Id="1622" PostTypeId="2" ParentId="1621" CreationDate="2010-12-04T15:10:50.513" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Depending on one's definition of the discontinuity (whether one allows talking about the continuity of the function at a point where the function is undefined), pole might also be considered an infinite discontinuity.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Singularity is certainly not equivalent to a discontinuity (not even mathematically). It can also mean &lt;a href=&quot;http://en.wikipedia.org/wiki/Pole_%28complex_analysis%29&quot; rel=&quot;nofollow&quot;&gt;pole&lt;/a&gt; and this is the usual meaning it has in GR. I.e. the curvature of the space-time blows up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_metric#Singularities_and_black_holes&quot; rel=&quot;nofollow&quot;&gt;Schwarzschild metric&lt;/a&gt;'s curvature diverges at the origin as may be checked by looking at &lt;a href=&quot;http://en.wikipedia.org/wiki/Kretschmann_scalar&quot; rel=&quot;nofollow&quot;&gt;Kretschmann invariant&lt;/a&gt;. There might also exist coordinate singularities but these are unphysical and can be removed by a better choice of coordinates.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; What does one mean by curvature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic object associated with the metric tensor is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Connection_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;connection&lt;/a&gt; and for that one can define a &lt;a href=&quot;http://en.wikipedia.org/wiki/Curvature_form&quot; rel=&quot;nofollow&quot;&gt;curvature form&lt;/a&gt;. Now this is quite complicated object and it is not quite clear how to detect singularities in it. One easy way is to define &lt;a href=&quot;http://en.wikipedia.org/wiki/Curvature_invariant_%28general_relativity%29&quot; rel=&quot;nofollow&quot;&gt;invariants&lt;/a&gt; (such as the above-mentioned Kretschmann invariant) associated with this form.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-05T20:13:42.607" LastActivityDate="2010-12-05T20:13:42.607" CommentCount="7" />
  <row Id="1623" PostTypeId="2" ParentId="1019" CreationDate="2010-12-04T15:22:53.923" Score="3" ViewCount="" Body="&lt;p&gt;the belief that cartoon-universe rules apply to falling objects:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as embodied by the statement that, &quot;if you are trapped in a falling elevator, you can avoid destruction by jumping at the last moment, so that when the elevator hits, you are in the air, and only fall the last inch or two.&quot;&lt;/p&gt;&#xA;" OwnerUserId="717" LastActivityDate="2010-12-04T15:22:53.923" CommunityOwnedDate="2010-12-04T15:22:53.923" CommentCount="6" />
  <row Id="1624" PostTypeId="2" ParentId="1619" CreationDate="2010-12-04T16:02:09.870" Score="0" ViewCount="" Body="&lt;p&gt;I am not familiar with impedance measurement using the quadrupole method. But with the information you have given I can make some guesses. The number of signal say the how many channels you need to measure. From the frequency band you said you are interested according to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem&quot; rel=&quot;nofollow&quot;&gt;sampling theorem&lt;/a&gt;, your oscilloscope sampling rate needs to be at least 200 KHz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computer sound cards have a digitizer of 44 KHz, if measuring up to 20 kHz would do and you need only 2 channels that should be enough. It can generate the 20 kHz signal too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise usb oscilloscope, should do the job. Supposing that you have the signal generator as you talked. And I am supposing you can code your way up to make the whole acquisition using your computer.&#xA;BTW, have any online reference to the quadrupole method you are talking about?&lt;/p&gt;&#xA;" OwnerUserId="62" LastActivityDate="2010-12-04T16:02:09.870" />
  <row Id="1625" PostTypeId="2" ParentId="1019" CreationDate="2010-12-04T16:24:53.347" Score="3" ViewCount="" Body="&lt;p&gt;It really bugs me. I've even heard from some quantum mechanics lecturers that they think quantum entanglement implies faster than light communication!&lt;/p&gt;&#xA;" OwnerUserId="304" LastActivityDate="2010-12-04T16:24:53.347" CommunityOwnedDate="2010-12-04T16:24:53.347" CommentCount="6" />
  <row Id="1626" PostTypeId="1" AcceptedAnswerId="1632" CreationDate="2010-12-04T17:28:35.833" Score="2" ViewCount="796" Body="&lt;p&gt;Please, take this question seriously, because this is a real problem to me. I have a door in my flat. A closet door, to be specific. And there is a problem with it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the summer everything is ok, the standard wood door opens and closes as predicted, but in the winter time it drives me crazy! It just won't close! Seems that it doesn't fit where it belongs. Damn door just doesn't allow me to welcome guests and take a cup of mulled wine with them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is, what physical processes are standing behind that unfitting? I believe that sun and gravity may be involved, but I don't really understand it. So, please, help. If I'll know the problem, I'll figure out a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S Maybe it is important that I live in &lt;a href=&quot;http://en.wikipedia.org/wiki/Kazan&quot; rel=&quot;nofollow&quot;&gt;Kazan', Russia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="729" LastActivityDate="2011-02-11T15:34:47.547" Title="Why won't my door close in the winter?" Tags="&lt;everyday-life&gt;" AnswerCount="7" CommentCount="1" />
  <row Id="1627" PostTypeId="2" ParentId="1626" CreationDate="2010-12-04T17:35:25.863" Score="1" ViewCount="" Body="&lt;p&gt;Seasonality suggest it has something to do with &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_expansion&quot; rel=&quot;nofollow&quot;&gt;thermal expansion&lt;/a&gt;; yet if you seek for solution, it is better to ask on &lt;a href=&quot;http://diy.stackexchange.com&quot;&gt;http://diy.stackexchange.com&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-12-04T17:35:25.863" />
  <row Id="1628" PostTypeId="1" AcceptedAnswerId="1629" CreationDate="2010-12-04T18:42:20.593" Score="1" ViewCount="291" Body="&lt;p&gt;Is the curvature of space-time a smooth function everywhere (except at black holes) in view of General relativity. By 'smooth' it is meant that it possesses derivatives of all order at a given point. &lt;/p&gt;&#xA;" OwnerUserId="540" LastActivityDate="2011-02-12T02:13:38.293" Title="Is the curvature of space-time a smooth function everywhere ? (except at black holes)" Tags="&lt;general-relativity&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="1629" PostTypeId="2" ParentId="1628" CreationDate="2010-12-04T19:04:52.907" Score="3" ViewCount="" Body="&lt;p&gt;No, not at the boundary of a solid object like a planet.  There's a step function in the stres-energy tensor, and so you'll have a step function in the Riemann tensor.&lt;/p&gt;&#xA;" OwnerUserId="353" LastActivityDate="2010-12-04T19:04:52.907" CommentCount="8" />
  <row Id="1630" PostTypeId="2" ParentId="1019" CreationDate="2010-12-04T19:18:37.193" Score="7" ViewCount="" Body="&lt;p&gt;I hear from time to time people highly educated and skilled in Physics (unlike those believing that heavy objects fall faster than light ones) making the following claim:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... quantum mechanics indicates that certain physical quantities can take only a countable set of discrete values. Consequently, many current approaches to foundational questions in physics and cosmology advocate novel discrete or 'digital' pictures of nature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://fqxi.org/community/essay&quot; rel=&quot;nofollow&quot;&gt;&quot;Is Reality Digital or Analog&quot; essay contest at FQXi&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discrete spectra of some quantum observables do not imply/suggest that nature, in particular spacetime, is fundamentally discrete. The spectrum of a continuous operator acting on Hilbert spaces [which is a topological (vector) space, hence is continuous], often has a discrete part. This has nothing to do with spacetime being discrete. If it will (eventually) turn out to be discrete, it will be for other reasons.&lt;/p&gt;&#xA;" OwnerUserId="730" LastEditorUserId="730" LastEditorDisplayName="" LastEditDate="2010-12-15T20:13:38.037" LastActivityDate="2010-12-15T20:13:38.037" CommunityOwnedDate="2010-12-04T19:18:37.193" CommentCount="1" />
  <row Id="1631" PostTypeId="1" AcceptedAnswerId="1635" CreationDate="2010-12-04T19:19:52.187" Score="0" ViewCount="446" Body="&lt;p&gt;Total number of subatomic particles in the universe. Are they finite ? assuming any of GR or QM or even ST.&lt;/p&gt;&#xA;" OwnerUserId="540" LastActivityDate="2010-12-04T21:32:20.303" Title="Total number of subatomic particles in the universe. Are they finite ? assuming any of GR or QM or even ST" Tags="&lt;quantum-mechanics&gt;&lt;string-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="1632" PostTypeId="2" ParentId="1626" CreationDate="2010-12-04T19:21:53.363" Score="7" ViewCount="" Body="&lt;p&gt;I would say it has to do with humidity, since it is a wooden door. It probably gets too humid during the winter and the wood expands. If it had to do with temperature, it would be the opposite effect (it would expand during the summer when it is hot).&lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2010-12-04T19:21:53.363" CommentCount="5" />
  <row Id="1633" PostTypeId="2" ParentId="1631" CreationDate="2010-12-04T19:25:00.927" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://pdglive.lbl.gov/listings1.brl?exp=Y&quot; rel=&quot;nofollow&quot;&gt;PDGLive&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="238" LastActivityDate="2010-12-04T19:25:00.927" CommentCount="2" />
  <row Id="1634" PostTypeId="1" AcceptedAnswerId="1644" CreationDate="2010-12-04T19:44:07.690" Score="1" ViewCount="297" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;1.The magnitude of the total force acting on a ball rolling without&#xA;  slipping down a ramp is greater than&#xA;  the magnitude of the total force&#xA;  acting on the same ball if it slides&#xA;  down the ramp without friction. True&#xA;  or false?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I chose true for this one, as I figured that if our force has to oppose friction, then it must be greater than that force without friction. However the correct answer is false.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2.The magnitude of the velocity of an object must change if the magnitude of&#xA;  its acceleration is a constant. True&#xA;  or false?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I chose true for this one too. If acceleration is constant, then velocity is linear. This turned out to be false as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain to me why I was wrong for both of these?&lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2010-12-06T04:04:59.140" Title="Velocity and acceleration question" Tags="&lt;acceleration&gt;&lt;velocity&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="1635" PostTypeId="2" ParentId="1631" CreationDate="2010-12-04T20:04:52.223" Score="3" ViewCount="" Body="&lt;p&gt;It's possible to make an estimate of the amount of baryons in the &lt;strong&gt;observable&lt;/strong&gt; universe. But it's more difficult to make an estimate of anything else. It gets particularly more difficult when you consider things like photons, because they can pop in and out of existance, i.e. the number of photons is not constant. Actually, that is true for all elementary particles, since they are considered as excitations of quantum fields in most modern physics theories. Thus, the number of particles is not constant. But the heavier the particle, the less likely it will pop in existance. And if it's too heavy, it'll decay in lighter, more stable particles. So maybe you'll have something a couple of orders bigger than the amount of baryons, but probably not much larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, there is dark matter, of which we don't really know much. So, I have no idea if an estimate has been attempted of the amount of dark matter particles. Any estimate will be highly dependent on the theory we have for these particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In his book The Emperor's New Mind, Penrose estimates the number of baryons in the observable universe to be of the order of $10^{80}$. &lt;a href=&quot;http://www.physicsforums.com/showthread.php?t=325578&quot; rel=&quot;nofollow&quot;&gt;This seems to confirm it.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2010-12-04T21:32:20.303" LastActivityDate="2010-12-04T21:32:20.303" CommentCount="2" />
  <row Id="1636" PostTypeId="2" ParentId="1634" CreationDate="2010-12-04T20:45:38.127" Score="0" ViewCount="" Body="&lt;p&gt;I'll just give you (very obvious) hints because I don't want to solve the problem completely for you (it's better to solve it on your own).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint for 1:&lt;/strong&gt; just think what kind of forces are there in both cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint for 2:&lt;/strong&gt; think about all the possible values the constant might have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hint for 2 for a different solution:&lt;/strong&gt; think in more dimensions.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-04T20:45:38.127" CommentCount="5" />
  <row Id="1637" PostTypeId="1" AcceptedAnswerId="3787" CreationDate="2010-12-04T21:55:28.437" Score="0" ViewCount="246" Body="&lt;p&gt;I've ran into conflicting information on how to calculate the Gibbs free energy of fuels during combustion per unit mass, volume and mole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sample solution for hydrogen would be really appreciated!&lt;/p&gt;&#xA;" OwnerUserId="734" LastActivityDate="2011-01-24T20:33:52.323" Title="how to calculate gibbs free energy per unit mass, per unit volume, and per mole?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="1638" PostTypeId="1" CreationDate="2010-12-04T22:12:40.377" Score="1" ViewCount="89" Body="&lt;p&gt;What would I need to do to create water droplets of a consistent size and shape over a distance of 3 to 10 feet?  Are there any special requirements for the nozzle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I use pure water or should I have some kind of additive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any special requirements on the valve, etc, that is produced?&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="735" LastEditorDisplayName="" LastEditDate="2010-12-05T01:49:30.190" LastActivityDate="2011-01-04T12:32:05.363" ClosedDate="2011-01-04T19:40:57.660" Title="What is required to create water droplets of a consistent size and shape over a distance of 3 to 10 feet?" Tags="&lt;hydrodynamics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="1639" PostTypeId="1" AcceptedAnswerId="1640" CreationDate="2010-12-04T22:13:27.743" Score="7" ViewCount="470" Body="&lt;p&gt;Clearly there will be differences like air resistance; I'm not interested in that. It seems like you're working against gravity when you're actually running in a way that you're not if you're on a treadmill, but on the other hand it seems like one should be able to take a piece of the treadmill's belt as an inertial reference point. What's going on here?&lt;/p&gt;&#xA;" LastActivityDate="2010-12-12T15:47:57.493" Title="What's the difference between running up a hill and running up an inclined treadmill?" Tags="&lt;mechanics&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="1640" PostTypeId="2" ParentId="1639" CreationDate="2010-12-04T22:47:28.523" Score="8" ViewCount="" Body="&lt;p&gt;For me it is axiomatic that machine miles are easier than real miles, but let's analyze the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the runner maintains a constant velocity up the hill, or remains stationary in the frame of the gym on the treadmill. In both cases the runner's acceleration is zero, so we know that her legs must provide a constant force with upward magnitude $mg$, and the they have to do this against a surface passing by at an angle $\theta$ below the horizontal and moving with a velocity $v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinematics in the runners frame of reference look the same. This is not the cause of the difference in perceived difficulty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have always assumed that the difference in difficulty was two fold:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Wind resistance is not really negligible.&lt;/li&gt;&#xA;&lt;li&gt;The treadmill presents a very uniform reliable surface and the runner need not lift her legs as high to insure non-tripping progress.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also modern treadmill are designed to be relatively easy on the knees, and the accomplish this by having a slightly springy feeling which presumably returns some energy to the runner.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2010-12-04T22:56:07.567" LastActivityDate="2010-12-04T22:56:07.567" CommentCount="8" />
  <row Id="1641" PostTypeId="2" ParentId="1639" CreationDate="2010-12-04T22:47:33.797" Score="4" ViewCount="" Body="&lt;p&gt;The word 'difference' may be ambiguous, but let's look at the situation from several points of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy balance: Indeed, your potential energy does increase in case 1 and not in case 2. Muscles clearly perform the same work, so the energy must go somewhere? Yes, to the electric grid. The treadmill device's engine, to maintain constant velocity, will consume less electrical power to do so (or might even push energy back into the grid, in case of an efficient motor) because your legs are actually pulling it now downwards. If you do the math, you see it exactly compensates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Muscle work: work, in thermodynamical sense, is not just F*dx. One has to take a machine and consider all interfaces. For example, a spring or a muscle have two ends, and dx in the formula is actually the difference between two paths. Muscle expansion/contraction will be the same, and so is the force. Therefore, they are doing the same work. This work is, the amount of chemical internal energy stored within the muscle converted to mechanical work.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-12-04T22:47:33.797" />
  <row Id="1642" PostTypeId="1" CreationDate="2010-12-04T22:49:40.440" Score="2" ViewCount="160" Body="&lt;p&gt;I want to create a stream of water that emits only a droplet of water, waits a few milliseconds, and then continues.  The important thing is that I need to create a visible gap between drops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the desire to have droplets created with a consistent size and shape (asked in &lt;a href=&quot;http://physics.stackexchange.com/questions/1638/what-is-required-to-create-water-droplets-of-a-consistent-size-and-shape-over-a-d&quot;&gt;this&lt;/a&gt; question), how would one go about creating a stream analogous to a morose-code-stream of droplets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[Edit]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm considering having a solenoid valve connected to a tube pointed downward, where capillary action holds the water in place.  The solenoid releases the amount equal to one droplet of water.  What I'm having trouble with is making the drops look semi-uniform as they fall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an artistic rendering of what I'm trying to accomplish:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/2uYXm.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="735" LastEditorDisplayName="" LastEditDate="2010-12-05T01:59:38.983" LastActivityDate="2010-12-05T01:59:38.983" Title="Low pressure, pulsating, series of water droplets" Tags="&lt;hydrodynamics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="1643" PostTypeId="2" ParentId="1642" CreationDate="2010-12-05T00:32:56.273" Score="1" ViewCount="" Body="&lt;p&gt;I guess you can do it with &quot;brute force&quot;, like pushing water through a narrow nozzle by applying very short, hard and precise hits on a piston behind the nozzle. But that seems to be a rather technologically challenging device for a &quot;do-it-yourself&quot; lab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simpler solution you could try is just to have a steady water flow through a nozzle. A cylindrical column of water is intrinsically unstable due to capillary forces and starts breaking into droplets when set free. This breaking begins by generation of ripples of a certain wavelength (proportional to the radius of the column) which corresponds to the most unstable mode. So, if you let this instability develop by shooting the water jet at some distance, then you'll have at the end a more or less periodic sequence of droplets of roughly the same size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this instability needs some time to develop, and a big difficulty here is to suppress other instabilities which unavoidably begin to develop in parallel. You could perhaps do this by applying sound with a well calculated frequency to the tip the nozzle. The sound should selectively excite your preferred oscillation mode so that the desired instability sets off immediately after the water leaves the nozzle, while other instabilities don't have enough time to develop. However there is no guarantee that it will really work :)&lt;/p&gt;&#xA;" OwnerUserId="59" LastActivityDate="2010-12-05T00:32:56.273" CommentCount="3" />
  <row Id="1644" PostTypeId="2" ParentId="1634" CreationDate="2010-12-05T01:19:53.893" Score="2" ViewCount="" Body="&lt;p&gt;Mark I liked his approach very much, by he stating his reasoning I had enough information to figure out where the flaws on his rationale were. Which for he to do it by himself would be much more difficult. The things I did not like in his question were, the vague title question, and that two question were asked in a single thread. fprime, to set an example,  would split te two questions and reformulate it. I will put the proper answer after.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second question you could have asked more directly. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the magnitude of the velocity of&#xA;  an object must change if the magnitude&#xA;  of its acceleration is a constant?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then put your answer and rationale below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets go for the answers. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It seems to me that you got the concept of &quot;total force acting on an object&quot; wrong. When it says force acting it does not mean that we are going to apply a force to it. I it means, list all the forces that are applied on it and sum everything vectorially. In this case there is only the weight, normal and friction force. Check the picture below, and try to do the vectorial sum. Remember that in one of the cases the friction force is absent, so just imagine the picture without it. Did I get it right? I mean does the answer make sense now?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/4LsbJ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;That is a very common misconception of acceleration, that I struggled myself with when I was learning it. Problem lies on the missmatch of the popular definition of acceleration and the formal definition of acceleration.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The definition you are working with is probably.&#xA;&quot;Acceleration is the rate by which the &lt;em&gt;magnitude of the&lt;/em&gt; velocity changes&quot;.&#xA;While the formal definition of acceleration is&#xA;&quot;Acceleration is the rate by which the velocity is changes&quot;.&#xA;Now because velocity has the direction property it can change its direction without changing its magnitude. Changing direction IS a change in velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We actually use the wrong definition all around in our daily life. When we say do not accelerate while turning in a curve. We mean do not step the gas. But by the formal definition, it is impossible to turn without accelerating. So when we hear this phrase we know the person is not using the formal acceleration concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you must understand that in the question the formal definition is to be assumed. For it is this definition that is used in professional setting. So take care with it, all physics and engineering books and articles assume this the formal definition of acceleration.&lt;/p&gt;&#xA;" OwnerUserId="62" LastActivityDate="2010-12-05T01:19:53.893" CommentCount="2" />
  <row Id="1645" PostTypeId="2" ParentId="1441" CreationDate="2010-12-05T02:05:04.713" Score="1" ViewCount="" Body="&lt;p&gt;You're assuming that nuclei with exactly a magic number of neutrons are more stable than all their non-magic neighbors in the chart of nuclides, but there's no reason to think that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a nucleus has a magic number of neutrons, that means one shell is completely full, and the next shell is empty. Therefore the next neutron you add (going to magic+1) will have significantly greater single-particle energy, so that nucleus should be less stable than the magic one. This is born out by both of your examples for the well-known magic number 126:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Po-211 (N = 126+1): 0.5 seconds&lt;/li&gt;&#xA;&lt;li&gt;Ra-215 (N = 126+1): 1.5 milliseconds&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, if you remove one neutron, then nothing changes about the other (magic−1) neutrons; they are still in the same shells as in the magic case. No extra stability is caused by having a completely full shell as opposed to an almost full one. (In this respect, the word &quot;magic&quot; is misleading.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's another way to think of it: In alpha decay (which is the main decay branch for all of your examples), two neutrons are removed from the nucleus. It will always be the two most energetic neutrons that are removed (those in the highest shells). In a magic nucleus, the lower shell is completely full, so two neutrons are removed from that lower shell. In a magic+1 nucleus, there is a lone neutron in the upper shell, so when two neutrons are removed, one comes from the upper shell and one from the lower shell. Since one comes from the upper shell, more energy is released in the alpha decay and in general the half-life will be shorter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in a magic−1 nucleus, there are no neutrons in the upper shell, so both neutrons come from the lower shell. This is the same situation as the magic case, so there's no reason to expect the decay energies or half-lives to be drastically different. (Of course they won't be exactly the same, but the differences come from other, more subtle effects.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, 152 is not a &quot;canonical&quot; or &quot;universal&quot; magic number. It shows up on the plot you have there of some specific elements, but if you look at other elements, the gap between the shells occurs at a different place. 126 is a universal magic number, but at 152 the situation is more complicated because changing the neutron/proton ratio also shifts the shells relative to each other. This is why the thing I said about magic+1 always being less stable than magic doesn't hold for one of those. Nuclear structure is really complicated.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2010-12-05T02:05:04.713" CommentCount="8" />
  <row Id="1646" PostTypeId="1" AcceptedAnswerId="2941" CreationDate="2010-12-05T02:42:28.453" Score="11" ViewCount="304" Body="&lt;p&gt;This is obviously a &quot;fun&quot; question, but I'm sure it still has valid physics in it, so bear with me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How great of a magnetic field would you need to transmute other elements into iron/nickel, if that's in fact possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnetic field near a neutron star, roughly of order 10^10 tesla, is so strong that it distorts atomic orbitals into thin &quot;cigar shapes&quot;. (The cyclotron energy becomes greater than the Coulomb energy.) Certainly if a solid crystal were placed in such a field it would become very anisotropic, and at some field strength the lattice constant in the direction transverse to the field could become small enough for nuclear fusion rates between the nuclei to become non-negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How high do we need to crank up the field before the nuclei all equilibrate to the absolute energy minimum of iron and nickel in, say, a matter of hours or days?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: From &lt;a href=&quot;http://dx.doi.org/10.1086/152986&quot; rel=&quot;nofollow&quot;&gt;http://dx.doi.org/10.1086/152986&lt;/a&gt; it appears that matter in strong magnetic fields forms into strongly-bound 1D chains along the field lines, which are only weakly bound to each other, and the parallel and transverse lattice constants are actually comparable.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditorDisplayName="" LastEditDate="2010-12-05T19:13:07.313" LastActivityDate="2011-01-15T14:00:46.813" Title="Turned to steel in the great magnetic field" Tags="&lt;nuclear-physics&gt;&lt;solid-state-physics&gt;&lt;atoms&gt;&lt;neutron-stars&gt;&lt;order-of-magnitude&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="1647" PostTypeId="1" AcceptedAnswerId="1649" CreationDate="2010-12-05T04:22:25.887" Score="1" ViewCount="330" Body="&lt;p&gt;I am preparing to write my paper about particle acceleration recently, and i need some general knowledge to get my feet wet about it.&lt;br&gt;&#xA;May you please give me an introduction of what particle acceleration is, the concept, the basic mechanism of particle accelerator, the general principle, or please refer me a link of particle acceleration. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards.&lt;/p&gt;&#xA;" OwnerUserId="738" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-12-05T10:59:18.650" LastActivityDate="2010-12-05T19:39:20.223" Title="An Introduction to particle acceleration and particle accelerator" Tags="&lt;particle-physics&gt;&lt;accelerator-physics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="1648" PostTypeId="2" ParentId="1647" CreationDate="2010-12-05T09:06:09.480" Score="1" ViewCount="" Body="&lt;p&gt;You can also read this book if you're interested, I think it's great for an introduction:&#xA;E. J. N. Wilson, ‘&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0198508298&quot; rel=&quot;nofollow&quot;&gt;An introduction to Particle Accelerators&lt;/a&gt;’, OXFORD, 2001.&lt;/p&gt;&#xA;" OwnerUserId="592" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-12-05T19:39:20.223" LastActivityDate="2010-12-05T19:39:20.223" />
  <row Id="1649" PostTypeId="2" ParentId="1647" CreationDate="2010-12-05T11:05:10.727" Score="1" ViewCount="" Body="&lt;p&gt;If you're writing a paper I don't think it makes much sense to try to describe such a general topic here. Instead I can give you a few good references or you might ask more specific questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two books that are considered in the field to be excellent references:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540490434&quot; rel=&quot;nofollow&quot;&gt;Wiedemann, Particle Accelerator Physics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/9812562001&quot; rel=&quot;nofollow&quot;&gt;Lee, Accelerator Physics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For a good basic introduction you have the book of Wilson: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0198508298&quot; rel=&quot;nofollow&quot;&gt;An introduction to particle accelerators&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also have a look at the proceedings of the &lt;a href=&quot;http://cdsweb.cern.ch/search?ln=fr&amp;amp;cc=CERN+Yellow+Reports&amp;amp;sc=1&amp;amp;p=CAS&amp;amp;f=&amp;amp;action_search=Recherche&quot; rel=&quot;nofollow&quot;&gt;CERN Accelerator School (CAS)&lt;/a&gt;. You can search for beginner level, intermediate or advanced. The 2005-004 is a good one to start.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2010-12-05T11:05:10.727" />
  <row Id="1650" PostTypeId="2" ParentId="1638" CreationDate="2010-12-05T11:43:53.767" Score="1" ViewCount="" Body="&lt;p&gt;I think that you only need an appropriate nozzle, called an atomizer. It is basically a very fine nozzle which uses the &lt;a href=&quot;http://en.wikipedia.org/wiki/Venturi_effect&quot; rel=&quot;nofollow&quot;&gt;Venturi effect&lt;/a&gt; to create a very think cloud of droplets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are commercial companies selling specific nozzles by droplet size, distance and distribution (they are typically used in factories).&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-12-05T11:43:53.767" />
  <row Id="1652" PostTypeId="2" ParentId="601" CreationDate="2010-12-05T14:58:39.560" Score="0" ViewCount="" Body="&lt;p&gt;A nanopaticle is collection of atoms or molecules that is less than 100 nm in any dimension and has a special and controllable function according to the molecular blocks of it. &lt;/p&gt;&#xA;" LastActivityDate="2010-12-05T14:58:39.560" />
  <row Id="1654" PostTypeId="2" ParentId="880" CreationDate="2010-12-05T17:11:45.483" Score="5" ViewCount="" Body="&lt;p&gt;Approximately $10^{15}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.wolframalpha.com/input/?i=How+many+photons+per+second+is+one+Lumen&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Photon flux of 540 nm light from the mechanical equivalent of light and the integrated spectral sensitivity of the human eye:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$3.8&amp;#215;10^{15}\ photons/s$  (photons per second)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$6.3&amp;#215;10^{-9}\ mol/s$  (moles of photons per second)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also see &lt;a href=&quot;http://books.google.com/books?id=rCBuW7u6qhsC&amp;amp;pg=PA60&amp;amp;lpg=PA60&amp;amp;dq=lumen+%22photons+per+second%22&amp;amp;source=bl&amp;amp;ots=LI_IGUcrPh&amp;amp;sig=B1ZLJjnm4QzW7fjCt3Ku4sPnirc&amp;amp;hl=en&amp;amp;ei=CwfiTPrFH4mphAfvkfGmDQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=10&amp;amp;ved=0CFYQ6AEwCQ#v=onepage&amp;amp;q=lumen%20%22photons%20per%20second%22&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this reference&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: this summarises Robert's answer in the question comments and is set to CW.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-12-05T17:11:45.483" CommunityOwnedDate="2010-12-05T17:11:45.483" />
  <row Id="1655" PostTypeId="2" ParentId="102" CreationDate="2010-12-05T17:22:48.337" Score="1" ViewCount="" Body="&lt;p&gt;Just a caveat that is left out of the answers above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton's third law is only approximately true.  If a system radiates (either electromagnetically or gravitationally), then the net momentum of the system can change, and you will find that $F_{12}\neq F_{21}$ if the radiation is not uniform in all directions.  The reason for this is simple enough: radiation carries energy and momentum away from the system.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Earth and the sun, this really doesn't matter much, because the gravitational radiation of the system is so close to zero that we wouldn't even know how to begin to search for the effect.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for things like colliding black holes, the effect can become very significant.  Two orbiting black holes with zero center of mass velocity can end up with more than the escape velocity needed to leave the center of the galaxy, for example.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-12-05T17:22:48.337" />
  <row Id="1656" PostTypeId="2" ParentId="905" CreationDate="2010-12-05T19:13:54.597" Score="0" ViewCount="" Body="&lt;p&gt;Well, if one knows the entropy in the microcanonical ensemble, then one can find the internal energy.  But we already know that the energy of a system depends on the reference frame in which it is observed.  Thus, it is more appropriate to think of entropy in the same terms one thinks of energy--the 0-component of a four-vector, rather than as a 4-scalar.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-12-05T19:13:54.597" CommentCount="2" />
  <row Id="1657" PostTypeId="5" CreationDate="2010-12-05T21:17:14.717" Score="0" ViewCount="0" Body="&lt;p&gt;&lt;strong&gt;What does classical mechanics entail?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classical mechanics can be subdivided based on various criteria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Historically they were divided into&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/tagged/kinematics&quot;&gt;kinematics&lt;/a&gt; studies concepts of position, velocity and acceleration, but no forces.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/tagged/statics&quot;&gt;statics&lt;/a&gt; investigating equilibrium properties.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;dynamics&lt;/em&gt; studying the actual effects of forces on bodies.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Depending on a mathematical formalism, they can be divided into&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/tagged/newtonian-mechanics&quot;&gt;Newtonian mechanics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/tagged/lagrangian-mechanics&quot;&gt;Lagrangian mechanics&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/tagged/hamiltonian-mechanics&quot;&gt;Hamiltonian mechanics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Depending on a precise object of study&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/tagged/fluid-dynamics&quot;&gt;Fluid dynamics&lt;/a&gt; studies motion of classical fluids.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/tagged/statistical-mechanics&quot;&gt;Statistical mechanics&lt;/a&gt; studies systems composed of huge number of particles by means of statistics and probability theory.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-05T21:17:14.717" LastActivityDate="2010-12-05T21:17:14.717" />
  <row Id="1658" PostTypeId="4" CreationDate="2010-12-05T21:17:14.717" Score="0" ViewCount="0" Body="A part of physics studying the motion of bodies subjected to forces. It does so in a classical setting, i.e. not quantum and not special relativistic." OwnerUserId="-1" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-05T21:17:14.717" LastActivityDate="2010-12-05T21:17:14.717" />
  <row Id="1659" PostTypeId="5" CreationDate="2010-12-05T21:27:39.983" Score="0" ViewCount="0" Body="" OwnerUserId="-1" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-05T21:27:39.983" LastActivityDate="2010-12-05T21:27:39.983" />
  <row Id="1660" PostTypeId="4" CreationDate="2010-12-05T21:27:39.983" Score="0" ViewCount="0" Body="Quantum mechanics describes the microscopic properties of nature in a regime where classical mechanics no longer applies. It explains phenomena such as wave-particle duality, quantization of energy and uncertainty principle." OwnerUserId="-1" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-05T21:27:39.983" LastActivityDate="2010-12-05T21:27:39.983" />
  <row Id="1661" PostTypeId="2" ParentId="1019" CreationDate="2010-12-05T21:45:12.690" Score="4" ViewCount="" Body="&lt;p&gt;If you are riding a bicycle and you turn the front wheel to the left then the bicycle will steer to the left.&lt;/p&gt;&#xA;" OwnerUserId="207" LastActivityDate="2010-12-05T21:45:12.690" CommunityOwnedDate="2010-12-05T21:45:12.690" CommentCount="1" />
  <row Id="1662" PostTypeId="2" ParentId="1639" CreationDate="2010-12-05T22:36:42.840" Score="-1" ViewCount="" Body="&lt;p&gt;(Running up a treadmill) = (expend energy to keep feet moving at a constant speed) + (other effects)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Running up a hill) = (expend energy to keep feet moving at a constant speed) + (energy to lift center of gravity by hill height) + (other effects)&lt;/p&gt;&#xA;" OwnerUserId="392" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2010-12-06T18:19:32.593" LastActivityDate="2010-12-06T18:19:32.593" CommentCount="4" />
  <row Id="1663" PostTypeId="1" AcceptedAnswerId="1664" CreationDate="2010-12-05T23:05:54.177" Score="4" ViewCount="257" Body="&lt;p&gt;I've seen videos of people in space (on ISS) who squeeze a bottle or something and liquid comes out, it then separates into smaller balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this surely it should stay pretty much together because theres no gravity from the Earth so the liquid should attract itself?&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-06T08:55:55.003" LastActivityDate="2010-12-06T08:55:55.003" Title="Why do liquids separate in space?" Tags="&lt;gravity&gt;&lt;fluid-dynamics&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="1" />
  <row Id="1664" PostTypeId="2" ParentId="1663" CreationDate="2010-12-05T23:16:22.877" Score="7" ViewCount="" Body="&lt;p&gt;Because liquids, water in particular have a high &lt;a href=&quot;http://en.wikipedia.org/wiki/Surface_tension&quot; rel=&quot;nofollow&quot;&gt;surface tension&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this reason blobs of water tend to become spherical. Now, given that it starts as an elongated stream (say because it's pushed out of a bottle), the stream breaks up in different pseudo-spherical bubbles;  if an astronaut were to pour water very, very slowly and carefully he could create &lt;a href=&quot;http://www.youtube.com/watch?feature=player_embedded&amp;amp;v=qxk3VqZZBsw#!&quot; rel=&quot;nofollow&quot;&gt;a single spherical blob&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept is not dissimilar to drops, but without the gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ol10p.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-12-05T23:22:08.173" LastActivityDate="2010-12-05T23:22:08.173" />
  <row Id="1665" PostTypeId="2" ParentId="1663" CreationDate="2010-12-05T23:19:49.603" Score="1" ViewCount="" Body="&lt;p&gt;Well, I haven't seen your videos so I don't know what precise conditions were present but here's my guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It depends on how precisely they squeezed the bottle. It's true that if you had a ball of liquid in space on its own then it would remain together (except for slowly evaporating) because of surface tension. But if you let it out of the bottle with non-zero speed then there might be various instabilities and smaller balls can form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it all depends on the actual speed the liquid is leaving the bottle and the precise shape of the bottle (or at least its neck). Just consider showering with a hose -- lots of droplets form there.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-05T23:19:49.603" />
  <row Id="1666" PostTypeId="2" ParentId="1663" CreationDate="2010-12-05T23:24:42.560" Score="4" ViewCount="" Body="&lt;p&gt;A cylindrical column of a liquid is intrinsically unstable due to surface tension. This is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Plateau%E2%80%93Rayleigh_instability&quot; rel=&quot;nofollow&quot;&gt;Plateau-Rayleigh instability&lt;/a&gt;. Check out &lt;a href=&quot;http://www.princeton.edu/~stonelab/Teaching/Oren%20Breslouer%20559%20Final%20Report.pdf&quot; rel=&quot;nofollow&quot;&gt;this pdf&lt;/a&gt; for some pictures and equations. When you squeeze a liquid out of a bottle, it initially has a somewhat elongated shape, and the instability develops. &lt;/p&gt;&#xA;" OwnerUserId="59" LastActivityDate="2010-12-05T23:24:42.560" />
  <row Id="1667" PostTypeId="2" ParentId="1663" CreationDate="2010-12-05T23:27:32.017" Score="2" ViewCount="" Body="&lt;p&gt;Consider a spherical drop of liquid floating weightlessly. Now apply some mechanical energy unevenly to it. You have the mechanical energy of internal fluid flows competing against surface tension, which seeks to restore it to the lowest energy spherical state. But&#xA;surface tension if not all that strong, if the internal energies within the drop are too large,&#xA;they can cause it to break into two of more droplets. The real key to getting large drops, is to launch it with minimal internal (fliud) energy.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-12-05T23:27:32.017" />
  <row Id="1668" PostTypeId="2" ParentId="905" CreationDate="2010-12-05T23:43:31.620" Score="0" ViewCount="" Body="&lt;p&gt;See the wiki page for &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_hole_thermodynamics&quot; rel=&quot;nofollow&quot;&gt;Black Hole Entropy&lt;/a&gt;.  If I recall correctly from my GR course, the area of a certain hyper-surface is always increasing in one direction of time.&lt;/p&gt;&#xA;" OwnerUserId="442" LastActivityDate="2010-12-05T23:43:31.620" CommentCount="1" />
  <row Id="1669" PostTypeId="2" ParentId="1628" CreationDate="2010-12-05T23:54:43.470" Score="-1" ViewCount="" Body="&lt;p&gt;I actually believe the answer to this question is yes, and the reason is that a fundamental assumption of GR is that the gravitational field is constant over small regions of space time.  Another way of phrasing this is to say the GR is only valid in the limit of geometrical optics, which is the limit for which the wavelength of light is small compared with the length scale of the problem under consideration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very hard to do any calculations in GR, and the 2 well known cases (black holes and cosmology) both involve smooth functions for all intents and purposes.  Actually I think it's generally accepted that almost all functions that arise in physics are analytic (i.e. can be represented by power series), which except for some pathological examples means the functions are smooth.&lt;/p&gt;&#xA;" OwnerUserId="442" LastEditorUserId="442" LastEditorDisplayName="" LastEditDate="2010-12-06T00:01:43.310" LastActivityDate="2010-12-06T00:01:43.310" CommentCount="11" />
  <row Id="1670" PostTypeId="2" ParentId="1634" CreationDate="2010-12-06T04:04:59.140" Score="0" ViewCount="" Body="&lt;p&gt;I am going to comment on the first question only. The sum of the forces is equal to acceleration, so I guess this poorly worded question boils down to which ball is accelerating more: a) With friction and angular acceleration, or b) Without friction and zero angular acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From an energy perspective you can suspect that with friction some of the work done goes towards rotating the ball and thus will be less left to move it linearly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The math also supports this argument if you work out all the forces/moments, as well required rotational motion for each scenario.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-12-06T04:04:59.140" />
  <row Id="1671" PostTypeId="2" ParentId="1621" CreationDate="2010-12-06T04:09:42.020" Score="0" ViewCount="" Body="&lt;p&gt;I suspect mathematically singularity means $\frac{1}{0}\rightarrow\infty$. The size goes to zero and the math blows up. In fact, the size does not have to be zero for singularity to exist. If you think of trying to force a wave to exist in a sub-harmonic wavelength (size), it also makes the math blow up.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-12-06T04:09:42.020" />
  <row Id="1672" PostTypeId="2" ParentId="1621" CreationDate="2010-12-06T04:43:35.767" Score="1" ViewCount="" Body="&lt;p&gt;The formal definition is 'a point through which a geodesic cannot be extended'.  It can be a point where the curvature is infinite, as said above, but the concept is a little more general--it is a point at which there is no smooth way to extend a path into the future.  Formally, what the singularity theorems show is that, when the curvature obeys certain conditions, and you have positive energy, then certain geodesics must end in a finite proper time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the contrast with Minkowski space, where a particle moving in a straight line is free to continue moving in a straight line forever, always tracing out more and more proper time.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-12-06T04:43:35.767" CommentCount="1" />
  <row Id="1673" PostTypeId="1" AcceptedAnswerId="1678" CreationDate="2010-12-06T05:50:46.407" Score="4" ViewCount="332" Body="&lt;p&gt;Coulomb force in SI is&#xA;$ F = \frac{Q1*Q2}{4\pi\varepsilon  R^{2}} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while in CGS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ F = \frac{Q1*Q2}{R^{2}} $&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;why is it? I mean doesn't it any make difference in dimension? since $  \varepsilon $ itself has dimention. And similarly for magnetic force also.&lt;/p&gt;&#xA;" OwnerUserId="567" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-12-06T07:45:22.640" LastActivityDate="2010-12-06T08:39:55.963" Title="coulomb force in SI and cgs" Tags="&lt;electromagnetism&gt;&lt;si-units&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="1675" PostTypeId="2" ParentId="1673" CreationDate="2010-12-06T07:26:13.020" Score="1" ViewCount="" Body="&lt;p&gt;It is true because all the quantities in the expression change dimension, not only force, but also charge and distance. But not only that: there are different variations of CGS system. The one you wrote is &lt;a href=&quot;http://en.wikipedia.org/wiki/Centimetre_gram_second_system_of_units&quot; rel=&quot;nofollow&quot;&gt;Electrostatic CGS&lt;/a&gt;, and it chooses the units of charge so that $\frac{1}{4\pi\epsilon_0}=1$.&lt;/p&gt;&#xA;" OwnerUserId="154" LastActivityDate="2010-12-06T07:26:13.020" />
  <row Id="1676" PostTypeId="2" ParentId="1663" CreationDate="2010-12-06T07:33:05.477" Score="0" ViewCount="" Body="&lt;p&gt;The liquid does not stay in one piece, because there is an excess kinetic energy it gains when coming out of bottle, this energy is in the end transformed into the surface tension energy of the liquid drops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more question is as follows: why the liquid drops in space, while attracting to one another, do not assemble into a single drop? This is because the gravitational force among them is far to small for that to happen.&lt;/p&gt;&#xA;" OwnerUserId="154" LastActivityDate="2010-12-06T07:33:05.477" />
  <row Id="1677" PostTypeId="2" ParentId="1673" CreationDate="2010-12-06T07:41:27.557" Score="5" ViewCount="" Body="&lt;p&gt;Yes the dimension is different. In SI the current (A) a base unit independent from length (m), mass (kg) and time (s) because we choose to, but in &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_units&quot; rel=&quot;nofollow&quot;&gt;CGS Gaussian unit&lt;/a&gt; this is not (1 unit of current = 1 g&lt;sup&gt;1/2&lt;/sup&gt; cm&lt;sup&gt;3/2&lt;/sup&gt; s&lt;sup&gt;-2&lt;/sup&gt;), by setting $\epsilon_{0,SI} = \frac1{4\pi}$. This also leads to some perhaps unintuitive results, like the unit capacitance in CGS Gaussian is &quot;centimeter&quot;, and resistivity is &quot;second&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unit system is actually arbitrary. We could, for example, make every physical observable a unitless number by setting $c = \hbar = k_B = \frac1{4\pi\epsilon_0} = G = 1$, and we get the Planck's natural unit, although those numbers become inconveniently large or small. This is also the reason of introducing ampere (and volt and ohm) as separate units, as that &quot;unit of current&quot;'s size is impractically small in everyday life (1 unit of CGS current is 3 &amp;times; 10&lt;sup&gt;-8&lt;/sup&gt; A).&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-12-06T07:41:27.557" />
  <row Id="1678" PostTypeId="2" ParentId="1673" CreationDate="2010-12-06T08:10:53.760" Score="7" ViewCount="" Body="&lt;p&gt;&lt;b&gt;Short Answer&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You've hit upon the quirk that the SI and CGS systems not only measure electric charge with different units, but also assign them different dimensionality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In SI, the Ampere is a &lt;a href=&quot;http://en.wikipedia.org/wiki/SI_base_unit&quot; rel=&quot;nofollow&quot;&gt;base unit&lt;/a&gt;.  Amperes are not made out of anything else - they are primitive, like meters, kilograms, and seconds.  One Ampere is one Coulomb per second, so the unit of electric charge, the Coulomb, has dimension of Amperes*seconds (and is equal to one Ampere-second).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When there's an equation that has Amperes or Coulombs on one side, and something without those units (like force) on the other side, the equation will always have a constant that has the correct units to balance things out.  That job is done by $\mu_0$ and $\epsilon_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In CGS, charge is measured in &lt;a href=&quot;http://en.wikipedia.org/wiki/Centimetre_gram_second_system_of_units#Electrostatic_units_.28ESU.29&quot; rel=&quot;nofollow&quot;&gt;esu&lt;/a&gt;, but this is a derived unit.  Charge is considered to be made up out of length, mass, and time the way, for example, angular momentum is.  Charge has dimensions &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[M]^{1/2}[L]^{3/2}[S]^{-1}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and one esu is equal to the square root of a dyne-cm^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Longer Answer&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In SI, the we start with meters, kilograms, and seconds.  Then we set $\mu_0$ to be $4\pi 10^{-7}\textrm{ohm-sec/m}$.  Now take two wires  very long compared to the distance between them, and run the same current through them.  Make the distance between them $d$.  A section of length $l$ of the wires will feel a force of attraction $F$, which depends on the square of the current.  The Ampere is defined such that when the current is measured in Amperes, we find that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = \frac{\mu_0 I^2 l}{2\pi d}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when $F$ is measured in Newtons, $d$ and $l$ in meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This definition requires that you can create the same current in each wire, that you can accurately measure the force per unit length, and that the force is perfectly proportional to the square of the current.  In truth it won't be because the wires aren't infinitely long and perfectly straight and parallel, but some equivalent operational definition might be used in practice.  (The fact that the definition is at all possible is a test of the physical hypothesis of the proportionality.)  The important point is that the Ampere becomes a new basic unit.  (Technically, what's been said so far would not define Amperes, but give us a relationship between Amperes and Ohms.  We could hash that out by looking at the units of V=IR, for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Coulomb used in Coulomb's law is then defined as one Ampere-second.  This would seem to allow us to experimentally measure $\epsilon_0$ because it is now the only unknown in Coulomb's law.  Originally, that was correct, but now we have defined the speed of light to be c = 2.99792458 * 10^8 meters/second, and so we're constrained by $c = 1/\sqrt{\mu_0\epsilon_0}$.  This forces $\epsilon_0$ to be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\epsilon_0 = \frac{1}{4\pi*8.9875517853681764}\frac{\textrm{sec}}{\textrm{ohm-m}}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;meaning that we could also define the Coulomb directly from Coulomb's law - take two charged bodies, measure the force between them, and define the Coulomb to be the unit of charge such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = \frac{Q_1Q_2}{4\pi\epsilon_0 R^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $R$ in meters and $F$ in Newtons.  The experimental observation that these two definitions of the Coulomb (one as the Ampere-second and one directly from Coulomb's law) agree then becomes a test of the physical theory of electromagnetism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In CGS units, charge is measured in esu, a derived unit defined by Coulomb's law as you've written it.  One esu is the charge such that two charged bodies feel the Coulomb force&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = \frac{Q_1Q_2}{R^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $F$ in dynes and $R$ in centimeters.  If both charges are one esu, this gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1\textrm{dyne} = \frac{1 \textrm{esu}^2}{1\textrm{cm}^2}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1\textrm{esu} = 1\textrm{cm}\sqrt{\textrm{dyne}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;finally, since the dyne is a derived unit, this could be written in terms of base units as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1\textrm{esu} = \sqrt{\frac{\textrm{g-cm^3}}{\textrm{s}^2}}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, there is one less base unit involved in electromagnetic formulas when using CGS.  Also, there is no need for the constant $\epsilon_0$ used in SI.  There is also no need for $\mu_0$.  Instead, things like Maxwell's equation explicitly include the speed of light $c$.  You can see a side-by-side comparison of basic equations of electromagnetism in SI and CGS units &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_units#Maxwell.27s_equations&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting CGS centimeters and grams to SI meters and kilograms, then equating the two expressions for the force in Coulomb's law, we find the conversion that one Coulomb is the same amount of charge as 2.99792458 * 10^9 esu.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;b&gt;Reference&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer is a summary of an appendix to Purcell's &lt;i&gt;Electricity and Magnetism&lt;/i&gt;&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-12-06T08:39:55.963" LastActivityDate="2010-12-06T08:39:55.963" CommentCount="2" />
  <row Id="1679" PostTypeId="2" ParentId="1663" CreationDate="2010-12-06T08:26:34.227" Score="0" ViewCount="" Body="&lt;p&gt;Given a stream or column of liquid, it &quot;wants&quot; to find some local minimum potential energy state. Less energy is required to &quot;pinch off&quot; a number of smaller spheres than to bring all of the water together in one large sphere, so that's what happens.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-12-06T08:26:34.227" />
  <row Id="1680" PostTypeId="1" AcceptedAnswerId="1682" CreationDate="2010-12-06T11:27:23.477" Score="1" ViewCount="93" Body="&lt;p&gt;If the captured object do not have tangential velocity, it's just the free-fall time. But when it has, it may take longer time to fall in, right ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function should be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ddot{r} = -GM/r^2 + (v_0r_0/r)^2 / r = -GM/r^2 + v_0^2r_0^2 / r^3$  , &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where v_0 is the initial tangential velocity . After one integration, it becomes &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{r}^2/2=GM(1/r-1/r_0)-v_0^2r_0^2(1/r^2-1/r_0^2)/2 $ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to deal with it. But I guess there is a analytic solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone knows something about it ?&lt;/p&gt;&#xA;" OwnerUserId="518" LastEditorUserId="518" LastEditorDisplayName="" LastEditDate="2010-12-06T11:34:59.993" LastActivityDate="2010-12-06T12:18:08.307" Title="How long does it take a object captured by a star falling to the center?" Tags="&lt;classical-mechanics&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="1681" PostTypeId="1" AcceptedAnswerId="1724" CreationDate="2010-12-06T11:39:53.880" Score="4" ViewCount="147" Body="&lt;p&gt;I've come across black holes thermodynamics multiple times recently (both at this site and elsewhere) and some things started bugging me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one thing, first law bothers me a little. It is a reflection of the law of conservation of energy. This is fine when the space-time is stationary (as in Kerr solution) and is consistent with system being in a thermal equilibrium (so that thermodynamics apply at all). But what about more general systems of black holes?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What are the assumptions on the system of black holes for it to be in thermal equilibrium so that laws of BH thermodynamics can apply?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; the reason I am asking is that I heard that the laws should also be correct for system of multiple BH (so that their total event horizon area is increasing, for example). But I cannot wrap my head about &lt;strong&gt;how&lt;/strong&gt; might the system of BH be in thermal equilibrium. I mean, they would be moving around, generating gravitational waves that carry energy away (violating the first law) all the time. Right?&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-01-07T03:46:46.260" Title="What kind of systems of black holes satisfy the laws of black hole thermodynamics?" Tags="&lt;general-relativity&gt;&lt;thermodynamics&gt;&lt;black-holes&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="1682" PostTypeId="2" ParentId="1680" CreationDate="2010-12-06T12:06:06.253" Score="3" ViewCount="" Body="&lt;p&gt;We are talking just Newtonian gravity here. You should probably know that the orbit in this case is just a conic section. Depending on the initial angular momentum and energy you can compute the closest point to the star on the orbit. The object will fall into star if and only if this point is inside the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, supposing you already determined that the object will fall into the star, you can solve for the point of crossing star's surface (this is just geometry, intersection of the conic section and the circle) and then solving for the time of arrival into that point. All of this is simplified by noting that you only need to know the $r$-coordinate which is simply the radius of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this all boils down to finding a solution to your equation. You can do that just by taking a square root, &lt;a href=&quot;http://en.wikipedia.org/wiki/Separation_of_variables&quot; rel=&quot;nofollow&quot;&gt;separating variables&lt;/a&gt; and computing the integral. That will give you dependence of $t = t(r)$. So then just plug in the radius of the star and you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the problem is reduced to finding integrals. Try some software for that (because the integral doesn't look easy) like &lt;a href=&quot;http://integrals.wolfram.com/index.jsp&quot; rel=&quot;nofollow&quot;&gt;Wolfram Mathematica Integrator&lt;/a&gt;. If you have any problems with this I suggest you ask what to do next at math.SE to get better answers.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-06T12:18:08.307" LastActivityDate="2010-12-06T12:18:08.307" CommentCount="1" />
  <row Id="1683" PostTypeId="1" AcceptedAnswerId="1856" CreationDate="2010-12-06T12:10:54.640" Score="24" ViewCount="1487" Body="&lt;p&gt;Some time ago I came across a problem which might be of interest to the physics.se, I think. The problem sounds like a homework problem, but I think it is not trivial (i am still thinking about it):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a rail tank wagon filled with liquid, say water.&lt;img src=&quot;http://i.stack.imgur.com/nunmv.jpg&quot; alt=&quot;wagon&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that at  some moment $t=0$, a nozzle is opened at left side of the tank at the bottom. The water jet from the nozzle is directed &lt;strong&gt;vertically down&lt;/strong&gt;. Question:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the final velocity of the rail tank wagon after emptying?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplifications and assumptions:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rail tracks lie horizontally, there is no rolling (air) friction, the speed of the water jet from the nozzle is subject to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Torricelli%27s_law&quot; rel=&quot;nofollow&quot;&gt;Torricelli's law&lt;/a&gt;, the horizontal cross-section of the tank is a constant, the water surface inside the tank remains horizontal.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Data given:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M$ (mass of the wagon without water)&lt;br&gt;&#xA;$m$ (initial mass of the water)&lt;br&gt;&#xA;$S$ (horizontal cross-section of the tank)&lt;br&gt;&#xA;$S\gg s$ (cross sectional area of the nozzle)&lt;br&gt;&#xA;$\rho$ (density of the water)&lt;br&gt;&#xA;$l$ (horizontal distance from the nozzle to the centre of the mass of the wagon with water)&lt;br&gt;&#xA;$g$ (gravitational acceleration)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thinking at the moment is whether dimensional methods can shed light on a way to the solution. One thing is obvious: If $l=0$ then the wagon will not move at all.&lt;/p&gt;&#xA;" OwnerUserId="552" LastEditorUserId="887" LastEditorDisplayName="" LastEditDate="2011-01-28T22:28:53.093" LastActivityDate="2011-01-30T17:55:31.860" Title="Mechanics around a rail tank wagon" Tags="&lt;mechanics&gt;&lt;fluid-dynamics&gt;" AnswerCount="10" CommentCount="33" FavoriteCount="10" />
  <row Id="1684" PostTypeId="2" ParentId="1683" CreationDate="2010-12-06T13:27:34.100" Score="-2" ViewCount="" Body="&lt;p&gt;Clearly the water going out of the nozzle does not contribute any horizontal momentum change. Initially the wagon is still and the water flows downward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only reason why the wagon could move is that there is a force acting on the right size of the nozzle as the water hits it and its direction is turned towards the floor, thus exerting a force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let's think about this. How can we calculate this force? The force is equal to the pressure of the water times the vertical cross section area of the nozzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the water pressure is the same on the side with the nozzle and the side without. The force on the left side of the nozzle is compensated by an equal force on the right side of the wagon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/UjIks.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The forces on the left are exactly canceled by the ones on the right. F3 is canceled out by -F3 acting on the left side of the tap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there were no left side of the tap we would have a net horizontal force (the wagon would be propelled by recoil), but having a left side keeps the wagon still.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's clear to me that there will be, in real life, second order effects like unbalances in the density of the water which could make the wagon oscillate or move. But the question clearly states that the water remains horizontal (therefore undisturbed) and that Torricelli's law applies. This only happens when the outward flow is so slow that any inhomogeneities in density are second order effects and the water can be treated as to always have a laminar flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case the system is analogous as standing on a frictionless surface. Short of throwing something outwards, one wouldn't be able to propel. Throwing something downwards wouldn't help.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address Mark's and Marek's concern about the conservation of momentum, I can say this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the water, internally, initially falls down and gains vertical momentum&lt;/li&gt;&#xA;&lt;li&gt;at some point it will necessarily turn left. The momentum will not change in magnitude, but in direction: from down to left. This creates a reaction force on the bottom and on the right side.&lt;/li&gt;&#xA;&lt;li&gt;at the final point (the nozzle): the water will turn down again, from left to down. This creates a reaction force on the top of the nozzle and on the left of the nozzle&lt;/li&gt;&#xA;&lt;li&gt;since the water flows vertically w.r.t. wagon, it has zero horizontal momentum at the exit point&lt;/li&gt;&#xA;&lt;li&gt;this &lt;em&gt;constraint&lt;/em&gt; implies that the left hand force and the right hand force compensate.&lt;/li&gt;&#xA;&lt;li&gt;instead, there will be a torque. I have not calculated this, but depending on the length of the tube, this torque could eventually make the wagon tilt (if the weight of the tube remains negligible). Normally, though, the torque will not have a movement effect, it would merely move the center of mass towards the right.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To understand this a bit better:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Imagine the same problem &lt;strong&gt;without the nozzle&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Water flows freely left, horizontally.&lt;/li&gt;&#xA;&lt;li&gt;The water flows out with a speed of $v(t)=\sqrt{2gh(t)}$ and a &lt;strong&gt;horizontal&lt;/strong&gt; momentum that can be calculated via the parameter $s$ and $v(t)$, and a &lt;strong&gt;vertical&lt;/strong&gt; momentum of zero&lt;/li&gt;&#xA;&lt;li&gt;The wagon's horizontal momentum momentum changes by the same amount, opposite sign&lt;/li&gt;&#xA;&lt;li&gt;The wagon recoils right&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Re-imagine the original problem, &lt;strong&gt;with the nozzle&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Water flows freely downwards, vertically&lt;/li&gt;&#xA;&lt;li&gt;The water flows out with a speed of $v(t)=\sqrt{2gh(t)}$ and a &lt;strong&gt;vertical&lt;/strong&gt; momentum that can be calculated via the parameter $s$ and $v(t)$ and a &lt;strong&gt;horizontal&lt;/strong&gt; momentum of zero&lt;/li&gt;&#xA;&lt;li&gt;The wagon's horizontal momentum changes by the same amount, &lt;strong&gt;which is zero&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;The wagon stays still&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-12-10T15:55:05.770" LastActivityDate="2010-12-10T15:55:05.770" CommentCount="97" />
  <row Id="1685" PostTypeId="1" AcceptedAnswerId="1698" CreationDate="2010-12-06T14:02:02.907" Score="5" ViewCount="744" Body="&lt;p&gt;How can I derive the Euler-Lagrange equations valid in the field of special relativity? Specifically, consider a scalar field.&lt;/p&gt;&#xA;" OwnerUserId="592" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-06T21:22:12.363" LastActivityDate="2011-08-11T18:25:57.757" Title="The Euler-Lagrange equation in special relativity" Tags="&lt;quantum-field-theory&gt;&lt;special-relativity&gt;&lt;lagrangian-mechanics&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="1686" PostTypeId="1" AcceptedAnswerId="1696" CreationDate="2010-12-06T14:19:43.410" Score="3" ViewCount="491" Body="&lt;p&gt;I'm reading &lt;em&gt;Nano: The Essentials&lt;/em&gt; by T. Pradeep and I came upon this statement in the section explaining the basics of scanning electron microscopy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, the equation breaks down when the electron velocity approaches the speed of light as mass increases. At such velocities, one needs to do relativistic correction to the mass so that it becomes...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We all know about the famous theory of relativity, but I couldn't quite grasp the &quot;why&quot; of its concepts yet. This might shed new light on what I already know about time slowing down for me if I move faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the mass of an object increase when its speed approaches that of light?&lt;/p&gt;&#xA;" OwnerUserId="293" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-06T17:11:36.147" LastActivityDate="2010-12-06T22:38:12.217" Title="Why does the mass of an object increase when its speed approaches that of light?" Tags="&lt;special-relativity&gt;&lt;speed-of-light&gt;&lt;mass&gt;&lt;velocity&gt;" AnswerCount="7" CommentCount="0" FavoriteCount="2" />
  <row Id="1687" PostTypeId="2" ParentId="1550" CreationDate="2010-12-06T14:37:04.427" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;Magnet&lt;/li&gt;&#xA;&lt;li&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Ferromagnetism&quot; rel=&quot;nofollow&quot;&gt;ferromagnetic&lt;/a&gt; material,more rarely a &lt;a href=&quot;http://en.wikipedia.org/wiki/Paramagnetism&quot; rel=&quot;nofollow&quot;&gt;paramagnetic&lt;/a&gt; one. Other materials can&#xA; be diamagnetic, antiferromagnetic. You can see &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetism&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; for a review&lt;/li&gt;&#xA;&lt;li&gt;Actually, they go along so well because they interact the same way with the magnetic field, which mediates the interaction. As you guessed, it is related to their electronic configuration and how they interact inside the material. This interaction defines the nature of the magnetism of the material. If you have a diamagnetic material, for example, it is (slightly) repelled by a magnet. If you have a strong enough magnet, you can levitate a diamagnetic object (&lt;a href=&quot;http://en.wikipedia.org/wiki/Diamagnetism#Diamagnetic_levitation&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;e.g.&lt;/em&gt; a frog&lt;/a&gt;).  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="373" LastActivityDate="2010-12-06T14:37:04.427" />
  <row Id="1688" PostTypeId="2" ParentId="1686" CreationDate="2010-12-06T14:44:52.787" Score="1" ViewCount="" Body="&lt;p&gt;Fundamentally, mass and energy are the same thing. They are two &quot;points of view&quot; of the same reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the &quot;point of view&quot; (inertial frame) of an electron, its mass does not increase, its speed is always zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the &quot;point of view&quot; (inertial frame) of a stationary observer, the electron has a very high kinetic energy (some in the mass term and some in the speed term)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the &quot;point of view&quot; (inertial frame) of a moving observer, the electron has a different kinetic energy (some in the mass term and some in the speed term)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so on.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-12-06T14:44:52.787" />
  <row Id="1689" PostTypeId="2" ParentId="1683" CreationDate="2010-12-06T15:06:50.683" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;My answer below is wrong: it doesn't take into account the momentum of water leaving the cart once it has started moving.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, by conservation of the horizontal momentum in the absence of any horizontal force, the speed of the wagon at the end will be 0.  However, the position of the centre of mass of the (Wagon+Water) system should also be conserved, so the wagon will move slowly to the right during the process, which can probably be linked to a pressure difference inside the tank. But it will stop by the time the Wagon is empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real question is therefore not the final speed, but the final displacement. Let &lt;em&gt;x&lt;/em&gt; be the current position of the Wagon's centre of mass. When a mass -dµ of water goes through the nozzle, its centre of mass is displaced by &lt;em&gt;l&lt;/em&gt; to the left, and the centre of mass of the wagon is displaced by -l·dµ/(µ+M) to the right, where &lt;em&gt;µ&lt;/em&gt; is the remaining mass of water inside the wagon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating this gives&#xA;$$\Delta x=-l\int_m^0\frac{\mathrm d\mu}{\mu+M}=l\ln\frac{m+M}M $$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if the wagon moves initially to a (non-relativistic !) speed, the previous analysis stays true in the moving reference frame. The speed will not change, but the wagon will have a &lt;em&gt;Δx&lt;/em&gt; advance compared to a wagon with the same initial speed, but a closed nozzle   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited to correct a sign error*&lt;em&gt;strong text&lt;/em&gt;*&lt;/p&gt;&#xA;" OwnerUserId="373" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-12-08T09:19:16.730" LastActivityDate="2010-12-08T09:19:16.730" CommentCount="21" />
  <row Id="1690" PostTypeId="1" CreationDate="2010-12-06T15:37:17.690" Score="6" ViewCount="757" Body="&lt;p&gt;Although I doubt somewhat whether this question is really appropriate for this site, I hope it gets answered anyways. I guess, what I'm wondering is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How does one get to work as a theoretical physicist and - probably more importantly - what do theoretical physicist actually do all day long?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How are theoretical physicists distinguishable from mathematicians? Does a physicists day look very different from that of a mathematician?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I have a great interest in physics, but I'm not really much interested in doing experiments: Would it be advisable to do my bachelor in mathematics and try to get into theoretical physics later on?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a real chance of getting into research afterwards? (not that any kind of answer to this question would ever stop me from trying...)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Well, I hope this question is acceptable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think 1) might for example be answered by giving a link to a blog of a working theoretical physicist, who gives some insight into his or her everyday life, or some kind of an essay on the topic. Of course any other kind of answer is greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kind regards,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sam&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit after several answers:&lt;/strong&gt; Thanks a lot for all the responses! I found it very interesting and helpful to to get some input from you guys.&#xA;Although the opinions seem to differ a bit, one can definitely see many overlaps, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do still have some time do decide, and will definitely look at some books suggested here, visit some lectures and try to get a feel for what it would be like to do either physics or maths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks again for your effort! :-)&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditorDisplayName="" LastEditDate="2010-12-08T09:58:58.037" LastActivityDate="2011-08-07T20:45:29.630" Title="Theoretical Physics - How to?" Tags="&lt;soft-question&gt;" AnswerCount="9" CommentCount="6" FavoriteCount="9" />
  <row Id="1691" PostTypeId="2" ParentId="1686" CreationDate="2010-12-06T15:48:12.937" Score="2" ViewCount="" Body="&lt;p&gt;There is a point of view, that under the term &quot;the mass&quot; one must mean &quot;the &lt;strong&gt;rest&lt;/strong&gt; mass&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that point of view there is obviously no dependence of the (rest) mass on the speed of an object. And, therefore, the mass of an object does not increase when its speed increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct (from that point of view) way to talk about the fenomenon is to say that &lt;em&gt;with increase of the speed of an object you need more and more energy in order to make it move faster&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of cource there is no fundamental controversy between thsi point of view and that of many books and articles. But the usage of the concept of &quot;realtivistic mass&quot; makes things much more complicated, even if it was introduced in pursuit of simplicity. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2010-12-06T15:48:12.937" CommentCount="4" />
  <row Id="1692" PostTypeId="2" ParentId="1690" CreationDate="2010-12-06T16:00:17.197" Score="0" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;This I don't know. I think that everyone have his own way of doing research... But you'll need to work hard anyway.&lt;/li&gt;&#xA;&lt;li&gt;I don't think there is a clear distinction now. There is some &quot;gradient&quot; between those.&lt;/li&gt;&#xA;&lt;li&gt;I think that there is a reason beyond teaching students experiment even if they would be theorticians. But matematicians keep coming to physics as if they don't care about my opinion. &lt;/li&gt;&#xA;&lt;li&gt;Everyone I know who tried it -- made it. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="386" LastActivityDate="2010-12-06T16:00:17.197" CommentCount="3" />
  <row Id="1693" PostTypeId="2" ParentId="1690" CreationDate="2010-12-06T16:14:42.730" Score="6" ViewCount="" Body="&lt;p&gt;If you want to work as a theoretical physicist, it would be advisable to get a little bit of grounding in experimental physics anyway. So my answer to #3 is, if you want to get into theoretical physics, get a Bachelor's in physics, not mathematics, and take at least one or two experimental courses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I work at a university where students often do shoddy work in the undergraduate experimental courses, because &quot;I came here to study theoretical physics and I'm not interested in experiments.&quot; What they don't realize is that most of our professors in theoretical physics are of the opinion that to be an excellent theoretician, you primarily need to be a well-rounded physicist, with both theoretical and experimental skills. These students run into trouble when they're looking for internships or final projects, because no professor will accept them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I edited this answer, because I didn't mean to imply that you had to be an excellent experimenter to be a good theoretician. If you don't enjoy experimenting, then you don't enjoy experimenting. Just don't dismiss it or ignore it altogether.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Saturday Morning Breakfast Cereal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/2AH00.gif&quot; alt=&quot;Saturday Morning Breakfast Cereal&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="107" LastEditorUserId="107" LastEditorDisplayName="" LastEditDate="2010-12-07T00:34:48.243" LastActivityDate="2010-12-07T00:34:48.243" CommentCount="1" />
  <row Id="1694" PostTypeId="2" ParentId="1686" CreationDate="2010-12-06T16:32:37.040" Score="6" ViewCount="" Body="&lt;p&gt;If you want to intuitively see why the mass increases, consider the following.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Firstly, nothing can travel faster than the speed of light (this is the premise on which Special Relativity is based)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Secondly, applying a force to an object will increase its kinetic energy (assuming the force acts in the same direction as the object's motion)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since kinetic energy $K.E.$ = $0.5 * m * v^2$, if $v$ is limited to $c$, then as $v$ approaches $c$ the only way for $K.E.$ to increase is for $m$ to increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't a fully mathematical answer, but may help you to intuit why the mass increases.&lt;/p&gt;&#xA;" OwnerUserId="474" LastActivityDate="2010-12-06T16:32:37.040" CommentCount="3" />
  <row Id="1695" PostTypeId="2" ParentId="1690" CreationDate="2010-12-06T16:38:23.823" Score="3" ViewCount="" Body="&lt;p&gt;This was originally a comment under ptomato's answer but got too long and became point 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I am still just a student but I guess what you do is you mostly read papers and study books and try to work on some problems (both your thesis and things from physics.SE ;-) )...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;...which is probably the same thing mathematicians do, only the area of research and methods used differ. But I don't think there's a clear distinction. You have people like Witten who won Field's medal but is clearly a theoretical physicist. Now what's that about? :-)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I don't think being good experimenter is important at all. Sure, it helps. So does good teaching skills, programming skills, communication skills, algebraic geometry skills, ... But none of that is really necessary for theoretical physics. All that is needed is that person likes science and math and has inquisitive mind that is curious about the ways nature works.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Certainly. I know both of physics majors who got into math research and math majors who got into physics research. I think the required skills greatly overlap. The main difference is level of rigor and amount of physical intuition. You can adapt both to the environment you are working in. Of course, it might be excruciating for physicist to be forced to write formal proof where he just &lt;em&gt;sees&lt;/em&gt; the answer, or for mathematician to watch hand-wavy derivations in physics.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Personally, I am more on the math side and I hated experiments. So we'll see if I can succeed in being a good theorist :-)&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-06T16:38:23.823" CommentCount="3" />
  <row Id="1696" PostTypeId="2" ParentId="1686" CreationDate="2010-12-06T16:45:41.350" Score="8" ViewCount="" Body="&lt;p&gt;The complete relevant text in &lt;a href=&quot;http://books.google.com.hk/books?id=gPRPYyfZJvkC&amp;amp;pg=PA21&amp;amp;lpg=PA21&amp;amp;dq=%22However,+the+equation+breaks+down+when+the+electron+velocity+approaches+the+speed+of+light+as+mass+increases.%22&amp;amp;source=bl&amp;amp;ots=Kq_ZtVxXBw&amp;amp;sig=q5zcvkELqDY0XoK74OtDU-Syiq4&amp;amp;hl=zh-TW&amp;amp;ei=uA_9TKfmM4WycLimrfQO&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBcQ6AEwAA#v=onepage&amp;amp;q=%22However%2C%20the%20equation%20breaks%20down%20when%20the%20electron%20velocity%20approaches%20the%20speed%20of%20light%20as%20mass%20increases.%22&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;the book&lt;/a&gt; is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The de Broglie wave equation relates the velocity of the electron with its wavelength, &lt;em&gt;&amp;lambda;&lt;/em&gt; = &lt;em&gt;h&lt;/em&gt;/&lt;i&gt;mv&lt;/i&gt; ... However, the equation breaks down when the electron velocity approaches the speed of light as mass increases. ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually, the &lt;a href=&quot;http://en.wikipedia.org/wiki/De_Broglie_wavelength#The_de_Broglie_relations&quot; rel=&quot;nofollow&quot;&gt;de Broglie wavelength&lt;/a&gt; should be&#xA;$$ \lambda = \frac hp, $$&#xD;&#xA;where &lt;em&gt;p&lt;/em&gt; is the momentum. While &lt;em&gt;p&lt;/em&gt; = &lt;em&gt;mv&lt;/em&gt; in classical mechanics, in special relativity the actual relation is&#xA;$$ \mathbf p = \gamma m \mathbf v = \frac{m\mathbf v}{\sqrt{1-\frac{v^2}{c^2}}} $$&#xD;&#xA;where &lt;em&gt;m&lt;/em&gt; is the rest mass. If we still need to make the equation &lt;em&gt;p&lt;/em&gt; = &lt;em&gt;mv&lt;/em&gt; correct, we introduce the concept of &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Relativistic_mass#The_relativistic_mass_concept&quot; rel=&quot;nofollow&quot;&gt;relativistic mass&lt;/a&gt;&quot; &lt;em&gt;M&lt;/em&gt; = &lt;em&gt;&amp;gamma;m&lt;/em&gt; which increases with &lt;em&gt;v&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-12-06T16:45:41.350" CommentCount="1" />
  <row Id="1697" PostTypeId="2" ParentId="1690" CreationDate="2010-12-06T17:11:02.200" Score="0" ViewCount="" Body="&lt;p&gt;Answer to 2: In mathematics, truth is based on logic and deductive reasoning from axioms and definitions.  In physics, the only truth comes from the results of experiments.  Furthermore, in physics there can be no true rigor, because in principle it will never be possible to input the initial conditions for a macroscopic number of particles and simulate their motion, or solve the large number of differential equations which arise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to be a theorist, you really have to be the best of the best to succeed at research.  Experimentalist have an easier time getting jobs because they are expected to bring in a lot of grant money (the administration often takes a sizable portion of this money, which is why they love to hire experimentalists), whereas theorists generally don't get nearly as much grant money and therefore need to really be special to set themselves apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretical physicists are often doing elaborate and detailed calculations with fairly low level mathematics, in an attempt to explain the results of some recent experiment.  Mathematician's also do long calculations, but they mostly are trying to prove things.  It is fairly rare for a theorist to predict a phenomenon before it is discovered in the lab, it is much more common that experimental discoveries are made without being predicted in advance (and that theoretical predictions don't pan out); whereas mathematicians &quot;discover&quot; or &quot;invent&quot; things all the time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorists are the companions of experimenters, they support them in many ways; not only by carrying out calculations and giving them numbers, but also by thinking carefully about the physical mechanisms/principles of everything which is going on in the experiment; to perhaps discover an insight which the experimenter has not thought of.&lt;/p&gt;&#xA;" OwnerUserId="442" LastEditorUserId="442" LastEditorDisplayName="" LastEditDate="2010-12-06T20:44:39.773" LastActivityDate="2010-12-06T20:44:39.773" CommentCount="10" />
  <row Id="1698" PostTypeId="2" ParentId="1685" CreationDate="2010-12-06T17:20:06.940" Score="7" ViewCount="" Body="&lt;h2&gt;General approach&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;First recall that Euler-Lagrange equations are conditions for the vanishing of the variation of action $S$. For a scalar field $\Phi$ with Lagrangian density $\mathcal L$ on some open subset U we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S[\Phi] = \int_U {\mathcal L}(\Phi(x), \partial^{\mu}\Phi(x)) {\rm d}^4 x$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a variation of the field in direction $\chi$ and compute&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S[\Phi + \varepsilon \chi] = \int_M {\mathcal L}(\Phi(x) + \varepsilon \chi(x), \partial^{\mu}(\Phi(x) + \varepsilon \chi(x))) {\rm d}^4 x$$&#xD;&#xA;Then using Taylor expansion&#xA;$$S[\Phi + \varepsilon \chi] - S[\Phi] =&#xD;&#xA;\int_U \left[ \varepsilon \chi(x) {\partial{\mathcal L} \over \partial \Phi}(\Phi(x), \partial^{\mu}\Phi(x)) +&#xD;&#xA; \varepsilon (\partial^{\mu} \chi(x)) {\partial{\mathcal L} \over \partial (\partial^{\mu} \Phi)}(\Phi(x), \partial^{\mu}\Phi(x)) + O(\varepsilon^2) \right] {\rm d}^4 x$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using integration by parts on the second term (assuming $\chi$ vanishes on $\partial U$), diving by $\varepsilon$ on both sides and letting $\varepsilon \to 0$ this becomes a variation in direction $\chi$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta S [\Phi][\chi] = \int_U \chi(x) \left[  {\partial{\mathcal L} \over \partial \Phi}(\Phi(x), \partial^{\mu}\Phi(x)) -&#xD;&#xA;\partial^{\mu}\left( {\partial{\mathcal L} \over \partial (\partial^{\mu} \Phi)}(\Phi(x), \partial^{\mu}\Phi(x))\right) \right] {\rm d}^4 x$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By requiring variations in all directions equal zero we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ {\partial{\mathcal L} \over \partial \Phi} -&#xD;&#xA;\partial^{\mu}\left( {\partial{\mathcal L} \over \partial (\partial^{\mu} \Phi)}\right) = 0 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(arguments the same as always, so omitted).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Massive scalar field example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Consider Lagrangian density $${\mathcal L} = {1 \over 2}\eta_{\mu \nu} \partial^{\mu} \Phi \partial^{\nu} \Phi - {1 \over 2} m^2 \Phi^2 $$&#xD;&#xA;By using E-L equations we have just derived we obtain Klein-Gordon equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \eta_{\mu \nu} \partial^{\mu} \partial^{\nu} \Phi + m^2 \Phi^2 = \square \Phi + m^2 \Phi^2 = 0$$&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-06T17:20:06.940" CommentCount="6" />
  <row Id="1699" PostTypeId="2" ParentId="1603" CreationDate="2010-12-06T17:22:47.327" Score="1" ViewCount="" Body="&lt;p&gt;I'm looking for similar examples. The given examples are good -- I just want to add one more, that I recently discovered: &lt;a href=&quot;http://en.wikipedia.org/wiki/Topological_insulator&quot; rel=&quot;nofollow&quot;&gt;Topological insulators&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2010-12-06T17:22:47.327" CommentCount="2" />
  <row Id="1701" PostTypeId="2" ParentId="1690" CreationDate="2010-12-06T17:43:25.693" Score="3" ViewCount="" Body="&lt;p&gt;First you must ask yourself what excites you.  Is it watching events happen?  Comparing what happens to what is predicted to happen?  Making the prediction?  Understanding the prediction?  Making a model for hypotehtical predictions?  Understanding what constitutes a model?  None of the above?  Unless you're just in the business for the money, your research interests should drive your education as much as possible.  Otherwise, why bother?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my case, what excites me is the &lt;em&gt;connection&lt;/em&gt; between the concrete and the abstract, which is why I always straddled the math-physics fence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, the general pattern of any researcher is the same.  Make observations.  Ask questions.  Compare with previous questions and snwers.  Figure out how to find answers, when none exists.  The &lt;em&gt;methods&lt;/em&gt; are very different, of course.  It is true that a theoretician and a mathematician have similar daily patterns, and even similar methods (talking, reading, writing e-mail, working things out on a chalkboard/tablet/whatever).  The nature of the research is different, but there's lots of overlap.  When people pose the &quot;what are you?&quot; question to me, I try to punt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Note added:  Back now to finish.  I cut my answer short earlier.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main difference between the fields is not so much one of rigor, though that is an obvious one.  The more relevant difference is the questions that drive research.  In mathematics, one tries to understand the structure behind constructions of objects.  In theoretical physics, one tries to understand the structures behind physical models and physical phenomena -- and these can often &lt;em&gt;employ&lt;/em&gt; or &lt;em&gt;reveal&lt;/em&gt; mathematical structures.  Sometimes, the fields work in parallel, with both camps trying to get to the bottom of things.  &quot;Bottom&quot; in physics means understanding how hypothetical physical calculations (e.g., correlation functions, metrics) are affected.  &quot;Bottom&quot; in mathematics might mean a definition and/or a relation to existing structures.&lt;/p&gt;&#xA;" OwnerUserId="20" LastEditorUserId="20" LastEditorDisplayName="" LastEditDate="2010-12-06T18:24:31.120" LastActivityDate="2010-12-06T18:24:31.120" CommentCount="4" />
  <row Id="1702" PostTypeId="2" ParentId="1690" CreationDate="2010-12-06T19:18:01.700" Score="0" ViewCount="" Body="&lt;p&gt;&lt;em&gt;1. How does one get to work as a theoretical physicist&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go to a good university and get a firm foundation in math and physics.  Go to graduate school in theoretical physics.   Then apply for post-doc and faculty positions; repeat until you get a faculty position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;3. Would it be advisable to do my bachelor in mathematics and try to get into theoretical physics later on?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do whichever you find yourself liking better.  A background in math could be excellent preparation for theoretical physics.  You will either do &quot;mainly physics with a lot of math&quot; or &quot;mainly math with a lot of physics.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;4. Is there a real chance of getting into research afterwards?&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you're good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good way to start (right now) is to get yourself a copy of Landau and Lifshitz's &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0750628960&quot; rel=&quot;nofollow&quot;&gt;Mechanics&lt;/a&gt;&lt;/em&gt; and work through it.  See where that takes you.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2010-12-06T19:28:05.717" LastActivityDate="2010-12-06T19:28:05.717" />
  <row Id="1704" PostTypeId="2" ParentId="1690" CreationDate="2010-12-06T19:36:49.873" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://math.ucr.edu/home/baez/&quot; rel=&quot;nofollow&quot;&gt;John Baez&lt;/a&gt; addresses some of these questions on his page &quot;&lt;a href=&quot;http://math.ucr.edu/home/baez/advice.html&quot; rel=&quot;nofollow&quot;&gt;Advice for the Young Scientist&lt;/a&gt;&quot;.  I think it is more oriented towards, for example, someone deciding between grad-schools or later appointments rather than choosing between majors at the university, but it may still be useful to you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Math or Physics?&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This is for people who are torn between a research career in math and one in physics.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Nobody can stop you from learning and thinking about both math and physics - you should go on doing both! The real issue is whether you want to work in a math department or a physics department. It's possible to switch from one discipline to another after grad school, but it's not easy, since departments prefer to hire people with an appropriate degree. So, it's wise to decide which job suits you best before you apply for grad school - if not sooner.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;To decide, you need to know how these two jobs differ. For this, it's&#xA;  best to talk to as many mathematicians&#xA;  and physicists as you can, and find&#xA;  out what their jobs are like. Talk to&#xA;  your professors! It's also great to go&#xA;  to some conferences - there's often&#xA;  money for students to attend&#xA;  conferences. I can only summarize:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Mathematicians get promoted by&#xA;  publishing in math journals;&#xA;  physicists by publishing in physics&#xA;  journals. Read both kinds of journals&#xA;  and see which you can best imagine&#xA;  yourself publishing in. Spend time in&#xA;  a good library and browse. I spent my&#xA;  whole undergraduate career doing this!&#xA;  There are some journals at the&#xA;  boundary of math and physics, like&#xA;  Advances in Theoretical and&#xA;  Mathematical Physics and&#xA;  Communications in Mathematical&#xA;  Physics. Look at these. But also look&#xA;  at a bunch of journals that are&#xA;  full-fledged physics, like Physical&#xA;  Review A, B, C, D, and E, and Physical&#xA;  Review Letters, or solidly&#xA;  mathematical, like Annals of&#xA;  Mathematics or the Transactions of the&#xA;  American Mathematical Society. You'll&#xA;  see these come from different worlds!&#xA;  Mathematicians typically write in the&#xA;  definition/theorem/proof style, while&#xA;  physicists write shorter papers, and&#xA;  more of them, often packed with&#xA;  formulas, but usually leaving out all&#xA;  the details of calculations.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Physics is a faster, looser, more&#xA;  energetic discipline. You are also&#xA;  evaluated more heavily on how much&#xA;  grant money you can pull in.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Here's another way to put it: do you&#xA;  like things to be clearly stated in a&#xA;  rigorous way, or do you like to use&#xA;  your physical intuition to get to the&#xA;  answers?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He also has a page titled &quot;&lt;a href=&quot;http://math.ucr.edu/home/baez/books.html&quot; rel=&quot;nofollow&quot;&gt;How to Learn Math and Physics&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2010-12-06T19:42:48.777" LastActivityDate="2010-12-06T19:42:48.777" CommentCount="1" />
  <row Id="1705" PostTypeId="2" ParentId="1686" CreationDate="2010-12-06T19:40:41.210" Score="10" ViewCount="" Body="&lt;p&gt;The mass (the &lt;strong&gt;true mass&lt;/strong&gt; which physicists actually deal with when they calculate something concerning relativistic particles) &lt;strong&gt;does not change with velocity&lt;/strong&gt;. The mass (the true mass!) is an intrinsic property of a body, and it does not depends on the observer's frame of reference. I strongly suggest to read &lt;a href=&quot;http://arxiv.org/abs/hep-ph/0602037&quot; rel=&quot;nofollow&quot;&gt;this popular article&lt;/a&gt; by Lev Okun, where he calls the concept of relativistic mass a &quot;pedagogical virus&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What actually changes at relativistic speeds is the &lt;strong&gt;dynamical law&lt;/strong&gt; that relates momentum and energy depend with the velocity (which was already written). Let me put it this way: trying to ascribe the modification of the dynamical law to a changing mass is the same as trying to explain non-Euclidean geometry by redefining $\pi$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why this law changes is the correct question, and it is discussed in the answers here. &lt;/p&gt;&#xA;" OwnerUserId="59" LastActivityDate="2010-12-06T19:40:41.210" CommentCount="1" />
  <row Id="1706" PostTypeId="1" AcceptedAnswerId="1755" CreationDate="2010-12-06T21:19:22.867" Score="4" ViewCount="252" Body="&lt;p&gt;I guess the title says it all: how could/would you experimentally test whether our universe is truly &lt;a href=&quot;http://en.wikipedia.org/wiki/T-symmetry&quot; rel=&quot;nofollow&quot;&gt;time reversal invariant&lt;/a&gt;, without relying on the &lt;a href=&quot;http://en.wikipedia.org/wiki/CPT_symmetry&quot; rel=&quot;nofollow&quot;&gt;CPT theorem&lt;/a&gt;? What experiments have been proposed to check this? Have any of them been performed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that there are &lt;em&gt;indirect&lt;/em&gt; tests of time reversal invariance by observing &lt;a href=&quot;http://en.wikipedia.org/wiki/CP_violation&quot; rel=&quot;nofollow&quot;&gt;CP violation&lt;/a&gt;, in the decay $K_L \to 2\pi$ for example. Then if you assume that the necessary conditions for CPT symmetry are satisfied in our universe, that means there must be time reversal symmetry violation as well. But I'm curious about ways to test time reversal invariance without relying on CPT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, how could we distinguish between the standard model, which predicts T violation, and some hypothetical other theory that matches current experimental results as well as the SM, but in which CPT symmetry does not hold?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. This may be more of a research-level question, but I'm hoping it won't be &lt;em&gt;too&lt;/em&gt; obscure for this site ;-)&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-06T22:13:00.413" LastActivityDate="2010-12-08T21:00:45.577" Title="What sort of experiment would directly test time reversal invariance?" Tags="&lt;quantum-field-theory&gt;&lt;time-reversal&gt;&lt;symmetry&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="1707" PostTypeId="2" ParentId="1193" CreationDate="2010-12-06T21:43:02.447" Score="2" ViewCount="" Body="&lt;p&gt;Because of friction.&lt;/p&gt;&#xA;" OwnerUserId="154" LastActivityDate="2010-12-06T21:43:02.447" />
  <row Id="1708" PostTypeId="2" ParentId="1690" CreationDate="2010-12-06T21:53:36.803" Score="0" ViewCount="" Body="&lt;p&gt;I can see theoretical physicists at work every day, and I've also seen experimentalists... My own field is computer simulations. To my understanding the answers to your questions can be like this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What theoretical physicists do all the day long? Calculate and think.&lt;/li&gt;&#xA;&lt;li&gt;Physics vs Math? Physics is more intuitive than mathematics, and there is more freedom in physics than in mathematics. There is also much less abstraction in physics.&lt;/li&gt;&#xA;&lt;li&gt;Maybe yes, maybe no. This strongly depends on what exactly kind of physics you want to do.&lt;/li&gt;&#xA;&lt;li&gt;Yes, of course. But one has to work really hard for this to happen. Usually everything depends on the level of one's publications. And - in my opinion - if one loves physics he has a REALLY good chance of having good achievements and, as a consequence, publications.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="154" LastActivityDate="2010-12-06T21:53:36.803" />
  <row Id="1709" PostTypeId="2" ParentId="1686" CreationDate="2010-12-06T21:58:07.497" Score="0" ViewCount="" Body="&lt;p&gt;I don't really think we really know &quot;why&quot;, we only know that it does increase. The answer to the questions like &quot;why the laws of nature are the way they are?&quot; is beyond the scope of physics, it's a matter of philosophy.&lt;/p&gt;&#xA;" OwnerUserId="154" LastActivityDate="2010-12-06T21:58:07.497" CommentCount="1" />
  <row Id="1710" PostTypeId="2" ParentId="1686" CreationDate="2010-12-06T22:38:12.217" Score="0" ViewCount="" Body="&lt;p&gt;In special relativity the actual invariant is the magnitude of the covariant &lt;a href=&quot;http://en.wikipedia.org/wiki/Four-momentum&quot; rel=&quot;nofollow&quot;&gt;energy momentum 4-vector&lt;/a&gt; (E/c, p_x,p_y,p_z), not the apparent mass itself.  See also the section on &quot;momentum in 4 Dimensions&quot;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Special_relativity#Momentum_in_4D&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  The apparent mass in a moving frame is just a projection.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-12-06T22:38:12.217" />
  <row Id="1711" PostTypeId="2" ParentId="1706" CreationDate="2010-12-06T23:16:44.793" Score="1" ViewCount="" Body="&lt;p&gt;The C, P and T symmetries came from the equations due to Schrodinger, Klein-Gordon and Dirac. These partial differential equations depend on time and position. By changing the sign of time and position, these equations remained unchanged, so here is the origin of P and T. The C symmetry resulted originally from Dirac's equation. Since these equations are very good at describing the behaviors of particles, we can safely say that C, P and T came from observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the discovery of the weak interaction, C and P turned out to be violated, and the hope was that their combination CP is still preserved. After the discovery of the violation of CP, the hope moved to CPT. So, the experimental data said that each of C, P, T is respected when no weak interaction takes place, otherwise we should be happy with the combined CPT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I think that the answer is that the experimental data told us that the universe is T-invariant, then when the weak interaction was involved, it told us that CP is violated, and from CPT that T is violated. And CPT is not specific to the standard model, &lt;a href=&quot;http://en.wikipedia.org/wiki/CPT_symmetry#Derivation&quot; rel=&quot;nofollow&quot;&gt;it results from the Lorentz invariance and the fact that the energy is bounded for below&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A presumable CPT violation would imply the violation of the Lorentz invariance (http://en.wikipedia.org/wiki/CPT_symmetry#CPT_violation). So, if an alternative theory wants to distinguish itself by experiment, it should predict violations of the Lorentz symmetry. The theory itself will say the regimes in which this violation occurs (I bet that it will be at the Plank scale!). In this case, probably the Lorentz violation will imply with ease the CPT violation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find it very unlikely a violation of the Lorentz invariance, because this principle is (probably) the most ubiquitous principle in Physics, but who knows...&lt;/p&gt;&#xA;" OwnerUserId="730" LastActivityDate="2010-12-06T23:16:44.793" CommentCount="2" />
  <row Id="1712" PostTypeId="2" ParentId="1690" CreationDate="2010-12-07T00:24:31.707" Score="2" ViewCount="" Body="&lt;p&gt;Good question. I did my diploma in Theoretical Particle Physics and therefore worked one year as a theoretical pysicist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To your questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The short answer: Find the error in your calculations. One guy at our institute once said: &quot;Theoretical physicists spend half of their time in finding algebraic sign errors.&quot; This is to a degree true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer: It depends on the field you are working at. Actually Theoretical Physics starts at the point the calculations become too complicated for experimental physics. So for the particle physics this is &quot;Phenomenology&quot;. (hep-ph on arxiv.org -- don't be surprised if you don't understand anything there) These are the guys that calculate the - theoretical - cross sections for experiments that are tested at the LHC and other colliders. So this is in the end squaring Feynman diagrams, integrating them and analyzing the results. For some problems you can do that completely with pen and paper for other problems it makes sense to do 95% of that work on the computer. Depends on the research group and what they are working on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand there is this completely theoretical physics area. Of course they also need to make sure at some point that their work corresponds with the experiments but they mostly concentrate on &quot;model building&quot;. (hep-th on arxiv.org) For instance: We are seeing that Standard Model Higgs mechanism is probably not the way nature implemented the Electro-Weak-Force. Well, why don't we check out what happens if we have 2 Higgs fields instead of one? Or what happens if we change the symmetry group? Sounds interesting but you also pay a price: Most stuff you do is really abstract and will probably not be the way nature has implemented things. :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there is also Mathematical Physics (math-ph on arxiv), but that is usually done in a Mathematics department. Theoretical physcists usually don't have the time to check every mathematical detail. In particular nobody hasn't been able yet to formalize Quantum Field Theory (= Theoretical Particle Physics) like it has been done with non-relativistic Quantum Mechanics or even Newtonian Mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Physicists don't do mathematics in such a formal way. If they we're that detailed as mathematicians are, there wouldn't be able to do actual physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physicists approach is: Let's do this calculation, it looks formally ok and makes sense. If our cross-checks are not consistent, let's check again and pay more attention to the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll probably want to ask a mathematician how his work day looks like. (stackexchange...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Me not either. And no, please don't do the same mistake as I did by concentrating to much on the mathematics. Except you have enough time to learn all the important stuff about physics you want to work later on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make yourself clear: Physics and Mathematics are completely different subjects. Physics is a Natural science. Are you interested in that? Really? Study physics. Of course you need to do some experiments, they are physicists' source of information. But in the end it matters that you are informed about the important experiments that matter to your work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you interested in abstract concepts and the physics doesn't matter to you? Study mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are in between, consider studying Mathematical physics. But make sure your university has a mathematical physics department. I recommend you to talk to students or professors to get some first hand information. Mathematical physics is also really interesting. Take dynamical systems for instance or google for Wightman axioms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Yes of course. Is there a chance to become a professor? Of course, but you need to be focussed and make smart decisions. The process differs from country to country. About Germany I can say it is easy to get into research. But it's not easy to become a professor. Make yourself clear (in one or two years) whether you are a research guy or whether you are just interested about the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again I recommend you to visit a university and ask as many questions as possible to the mathematicians and physicians there. Most universities have regular events for students interested in studying at their university. At least in Germany there are a lot of student's organizations you can talk to and a lot of mathematics and physics departments have professors that are specialized in helping you with your questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I want to be more precise about the rigor. Theoretical physicists need to be as rigor as mathematicians. &lt;em&gt;But&lt;/em&gt;: Their rigor is focussed on the physics not on the mathematics.&lt;/p&gt;&#xA;" OwnerUserId="744" LastEditorUserId="4521" LastEditorDisplayName="" LastEditDate="2011-08-07T20:45:29.630" LastActivityDate="2011-08-07T20:45:29.630" />
  <row Id="1714" PostTypeId="2" ParentId="1706" CreationDate="2010-12-07T01:39:05.107" Score="7" ViewCount="" Body="&lt;p&gt;There are numerous research groups engaged in a search for an electric dipole moment of the electron, which, if it exists, would violate time-reversal symmetry. You can see this because any dipole moment the electron might have would need to be parallel to the spin (or anti-parallel). When you reverse time, the spin necessarily flips, but the electric dipole moment would not change, so the relative orientation of the two would change. That's the best example of a T-violating phenomenon that I know of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the risk of unseemly self-promotion, I wrote an &lt;a href=&quot;http://physicsworld.com/cws/article/print/41093&quot; rel=&quot;nofollow&quot;&gt;article on edm searches for Physics World&lt;/a&gt; last year. You need to register to read the whole thing, but it's free.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-12-07T01:39:05.107" CommentCount="1" />
  <row Id="1715" PostTypeId="2" ParentId="1077" CreationDate="2010-12-07T04:03:48.627" Score="-1" ViewCount="" Body="&lt;p&gt;Coulomb repulsion including the inverse square dependence has been experimentally verified down to nuclear lengths (by Rutherford's gold foil experiment) and even the fm (femtometer) scale (Breton V et al 1991 Phys. Rev. Lett. 66 572–5).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get 1000 N of force from a mol of electrons (seems like a reasonable order of magnitude for a contact area) we need to get them $ r = \sqrt{10^{23}k e^2 / 10^3} \approx 10^{-4} $m apart.   This shows that the Coulomb repulsion is more than sufficient to hold us up before we get even close to the quantum mechanical regime. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More importantly, the Pauli Exclusion Principle is not a force.  It simply says two fermions, in this case electrons, cannot occupy the same quantum state.  It doesn't say that fermions necessarily push apart if you were to get them very close to the same quantum state (whatever that means), nor does it say that fermionic wavefunctions cannot occupy the same space (as in, x,y,z).  I think if you look at the charge distribution in orbitals of any molecule or atom, you will find plenty of non-negligible spatial overlap.  &lt;/p&gt;&#xA;" OwnerUserId="722" LastEditorUserId="722" LastEditorDisplayName="" LastEditDate="2010-12-07T04:38:01.200" LastActivityDate="2010-12-07T04:38:01.200" CommentCount="7" />
  <row Id="1716" PostTypeId="1" AcceptedAnswerId="1718" CreationDate="2010-12-07T04:39:44.247" Score="2" ViewCount="81" Body="&lt;p&gt;I know from the book that electrons will be kicked out from the metal plate if the light of appropriate wavelength is exposed to the metal plate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My mental model says if we let the light expose the plate in a long enough interval, the plate will have no electrons anymore. So what happens on the plate? Does it still looks like a plate as the original? Or it become lighter due to loss of electrons?&lt;/p&gt;&#xA;" OwnerUserId="652" LastEditorUserId="652" LastEditorDisplayName="" LastEditDate="2010-12-07T04:47:05.973" LastActivityDate="2010-12-07T04:59:20.313" Title="In photo-electric experiment, if the light is exposed to a metal plate for a long enough interval, does the plate become lighter?" Tags="&lt;photoelectric-effect&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="1717" PostTypeId="1" CreationDate="2010-12-07T04:58:39.757" Score="3" ViewCount="358" Body="&lt;p&gt;The speed of sound is the rate that disturbances in air propagate through it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to have a wind that itself is moving at supersonic speeds relative to stationary winds around it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or perhaps a fluid flowing through a pipe at a speed greater than the speed of sound within it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, what would it &quot;look&quot; like?  What kind of phenomenon would occur?&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="71" LastEditorDisplayName="" LastEditDate="2010-12-07T05:15:09.313" LastActivityDate="2011-02-24T15:37:23.570" Title="Is it possible to accelerate air to supersonic speeds? What would it look like?" Tags="&lt;fluid-dynamics&gt;&lt;sound&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="1718" PostTypeId="2" ParentId="1716" CreationDate="2010-12-07T04:59:20.313" Score="3" ViewCount="" Body="&lt;p&gt;Well, it will become lighter with each electron removed, but unless you do it in the vacuum, it will get its electrons back from the environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, unless we are talking about &lt;em&gt;really&lt;/em&gt; high energies, only the weakly bound valence electrons will be removed. The work to remove the electron, $W_a$, becomes higher the more electrons have left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, now if we suppose that you have somehow managed to remove &lt;em&gt;all&lt;/em&gt; of the valence electrons, then sure, the plate will behave differently, as the optical and electrical properties are related to the electronic structure. If you have kicked out all (quasi-)free electrons, it might not be such a good conductor any more, for example.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2010-12-07T04:59:20.313" CommentCount="5" />
  <row Id="1719" PostTypeId="2" ParentId="1717" CreationDate="2010-12-07T05:10:10.820" Score="0" ViewCount="" Body="&lt;p&gt;I do not see why would would see anything. You will hear a noise however. But due to a lack of propulsion, the moving air will very quickly lose its speed and dissipate to subsonic speed.&lt;/p&gt;&#xA;" OwnerUserId="747" LastActivityDate="2010-12-07T05:10:10.820" />
  <row Id="1720" PostTypeId="1" AcceptedAnswerId="1727" CreationDate="2010-12-07T05:15:01.923" Score="7" ViewCount="330" Body="&lt;p&gt;Okay, some textbooks I came across, and a homework assignment I had to do several years ago, suggested that the reason we can skate on ice is the peculiar $p(T)$-curve of the ice-water boundary. The reasoning is that due to the high pressure the skates put on the ice, it will melt at temperaturs below $273 K$ and thus provide a thin film of liquid on which we can skate. It was then mentioned as fun fact that you could ice-skate on a planet with lakes of frozen dioxide because that gas has the $p(T)$-curve the other way round.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My calculations at that time told me that this was, pardon my french, bollocks. The pressure wasn't nearly high enough to lower the melting point to even something like $-0.5$ degrees Celsius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose it is some other mechanism, probably related to the crystal structure of ice, but I'd really appreciate if someone more knowledgeable could tell something about it. &lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-01-17T18:32:11.927" Title="Ice skating, how does it really work?" Tags="&lt;thermodynamics&gt;&lt;ice&gt;&lt;sports&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1721" PostTypeId="2" ParentId="1717" CreationDate="2010-12-07T05:33:10.240" Score="2" ViewCount="" Body="&lt;p&gt;It's possible and in fact easy to do: take a tube filed with air and move a piston down it faster than the speed of sound in the surrounding air (doing this is just a matter of having enough force available). The air exiting the tube will be moving faster then the speed of sound in the air around it. OTOH it will be at a much higher pressure and as a result the speed of sound in the air exiting the tube will be higher than the speed of the air its self.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to see this in action, you need only watch a rife be fired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also of interest is the &lt;a href=&quot;http://en.wikipedia.org/wiki/De_laval_nozzle&quot; rel=&quot;nofollow&quot;&gt;De Laval nozzle.&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2010-12-07T05:33:10.240" />
  <row Id="1722" PostTypeId="2" ParentId="1717" CreationDate="2010-12-07T06:15:49.203" Score="8" ViewCount="" Body="&lt;p&gt;The facility to accelerate wind to supersonic speed is called wind tunnel. Here is the pamphlet of an hypersonic (Mach 7 and 8) wind tunnel in Japan. It is explained how it works and you can even see some photos of hypersonic flow.&#xA;&lt;a href=&quot;http://daedalus.k.u-tokyo.ac.jp/wt/WTpamphE.pdf&quot; rel=&quot;nofollow&quot;&gt;http://daedalus.k.u-tokyo.ac.jp/wt/WTpamphE.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="157" LastActivityDate="2010-12-07T06:15:49.203" />
  <row Id="1723" PostTypeId="2" ParentId="503" CreationDate="2010-12-07T08:53:37.283" Score="5" ViewCount="" Body="&lt;p&gt;I can think of the problem of a self-force on an accelerating charged particle as one that still needs a full solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another important question in classical relativity is that of the motion of extended bodies in curved spacetime. Turns out that because the motion of an &lt;strong&gt;extended&lt;/strong&gt; body in a curved space appears differently to different observers, such an object can actually translate (&quot;swim&quot;) through space by periodic distortion of its shape &lt;a href=&quot;http://groups.csail.mit.edu/mac/users/wisdom/swimming.pdf&quot; rel=&quot;nofollow&quot;&gt;[reference]&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other open classical problems in relativity but again, as Piotr, said its hard to think of any in flat space.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-07T08:53:37.283" CommentCount="6" />
  <row Id="1724" PostTypeId="2" ParentId="1681" CreationDate="2010-12-07T09:34:28.170" Score="6" ViewCount="" Body="&lt;p&gt;The first law is not violated if stated properly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ dM = \frac{\kappa}{2 \pi}dA + \Omega dJ + \Phi dQ $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(reference: &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_hole_thermodynamics&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\kappa,\Omega,J,\Phi,Q$ are the surface gravity,angular velocity, angular momentum, electric potential and charge of the black hole, respectively. Compare this to the usual expression for the first law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ dE = TdS + PdV + \mu dN $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can (heuristically) make the identifications $ T = \frac{\kappa}{2\pi}$, $ S = A/4 $, $ \mu  = \Phi $ and $ N = Q $. The first two of these are well-established. Hawking showed that the temperature of the black hole is proportional to its surface gravity ($T \propto \kappa$) and Bekenstein showed that its entropy should be proportional to its area ($S=A/4$). The third and fourth equalities ($\mu = \Phi$  and $ N = Q $) can be understood if we think of the black hole as an aggregate of N particle with unit charge. Adding another charged particle to this ensemble of $N$ particles, with total charge $Q$, will cost an amount of work given by $\Phi dQ$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the case of a single black hole, one can use the framework of dynamical horizons developed by Ashtekar, Badri Krishnan, Sean Hayward and others [refs &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0308033&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0207080&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;]. Turns out the laws of black hole entropy can be extended to &lt;strong&gt;completely dynamical&lt;/strong&gt; black holes with a well-defined expressions for the first and second laws in terms of fluxes through the dynamical horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of a dynamical horizon is in terms of the expansion of the inward pointing, null normal vector field on the 2+1 d boundary of a 3+1 d region. I can't think of the detailed expressions of the top of my head, but you can find them in the above reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't give a concrete answer for the case of multiple black holes, but I would think you could extend the dynamical horizon framework to that case - probably not without some serious effort though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, there will be no violation of energy conservation. The sum of the energy and momentum carried off by gravitational waves (and detected by an observer at infinity) and the change in the black hole's energy and momentum (again w.r.t such an asymptotic observer) will remain a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps !&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Corrected proportionality constants for $T$ and $S$. Thanks @Jeff!&lt;/p&gt;&#xA;" LastActivityDate="2011-01-07T03:46:46.260" CommentCount="12" />
  <row Id="1725" PostTypeId="1" CreationDate="2010-12-07T09:59:02.630" Score="5" ViewCount="499" Body="&lt;p&gt;Almost every solid state physics textbook says crystal momentum is not really physical momentum. For example, phonons always carry crystal momentum but they do not cause a translation of the sample at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I learned that in indirect-band-gap semiconductors, we need phonons to provide the crystal momentum transfer to make happen electron transitions between the top of the valance band and the bottom of the conduction band. Along with absorbing or emitting photons, of course. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photons do carry physical momentum. For the purpose of momentum conservation, it seems that phonons do carry physical momentum as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we explain this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;========================================&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it more specifically, I drew a graph to tell the story:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/syVah.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;K (capital) is crystal momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For such transition, photon provides most of the energy transfer (and a little momentum transfer hk, k in lower case), phonon provides most of the momentum transfer (and a little energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar graphs can be found in most solid state physics textbooks. The picture tells me, either the photon participating in the transition carries crystal momentum, which value is equal to physical moemntum hk, or the crystal momentum itself is a kind of physical momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, one can prove that a phonon does not carry physical momentum (here I quote Kittel's &quot;Introduction to Solid State Physics&quot;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/AohrD.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how do we explain the momentum transfer in the electron transition aforementioned?&lt;/p&gt;&#xA;" OwnerUserId="749" LastEditorUserId="67" LastEditorDisplayName="" LastEditDate="2010-12-10T15:17:08.017" LastActivityDate="2010-12-10T15:17:08.017" Title="Is crystal momentum really momentum?" Tags="&lt;solid-state-physics&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="1" />
  <row Id="1726" PostTypeId="2" ParentId="1683" CreationDate="2010-12-07T11:03:36.097" Score="1" ViewCount="" Body="&lt;p&gt;&lt;b&gt;Qualitative Answer&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the cart exhibits an extremely surprising behavior.  The cart begins by sitting still on the track.  The hole is to the left of the center.  When the nozzle is opened, water in the cart begins a net flow to the left.  The cart, conserving momentum, picks up a velocity to the right.  In a steady state, the flow of water would be constant and the cart would move at constant velocity.  However, as the flow rate begins to decrease, the velocity of the cart decreases.  Eventually, the cart comes to a standstill, then actually reverses directions, moving to the left before the last water falls out.  When the last of the water is gone, the cart is coasting to the left.  The center of mass of the system never moves, because as the center of mass of the cart moves, the center of mass of the water moves oppositely.  Momentum is also conserved, because as the cart picks up momentum, the water picks up opposite momentum.  If the water also slides after hitting the track, by the end of the process the water will have a net motion somewhat to the right to compensate the motion to the left of the cart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Quantitative Answer&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the cart move at a speed $v$ to the right, and the water move at an average speed $w$ to the right.  In general, $v \neq w$ because the water's center of mass is moving relative to the cart.  The hole is at $l$.  If the hole is on the left then $l$ is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The velocity of the water relative to the cart is $w-v$.  This velocity comes from the fact that the water, if it were to continue as it is now, would all move from the center of the cart to the hole, a distance $l$, in a time $m/f$, with $f$ the mass flow rate.  Thus the kinematic relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w-v = \frac{lf}{m}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we want to conserve momentum.  This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dt}(Mv + mw) = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking this derivative, we have to keep in mind that $M$ and $m$ are changing because water is flowing out of the cart.  $m$ is decreasing at the rate $f$, and $M$ is increasing at the rate $f$ when we think of $M$ as the total mass moving at speed $v$ rather than the mass of the cart.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M\dot{v} + m\dot{w} + f(v-w) = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physically, the first two terms represent the force on the cart and the force on the water in the cart.  The last term represents the force on the water entering the nozzle.  Water entering the nozzle goes from $w$ to $v$, thus experiencing acceleration.  We have an earlier expression for $v-w$, so plug it in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M\dot{v}+m\dot{w} = \frac{lf^2}{m}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to solve for $\dot{v}$.  To do this, take the time derivative of the kinematic equation for $w-v$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{w} - \dot{v} = \frac{l\dot{f}}{m} + \frac{lf^2}{m^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These last two equations simplify to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{v} = \frac{-l\dot{f}}{M+m}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the flow rate is constant, there is no acceleration.  This is plausible because we can imagine watching in a center-of-mass frame where the cart moves to the right and the water moves to the left.  The water entering the nozzle feels an acceleration, but the water in the cart is also accelerating, and in the opposite direction.  (The water in the cart is accelerating because there is less and less of it, so on average it must move faster to deliver the correct flow rate from the center of the cart to the nozzle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right when we release the nozzle, the flow rate very quickly jumps up, and so the cart quickly picks up speed, too.  $m$ is essentially constant over the course of this acceleration, so the cart jumps up to a speed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = -\frac{lf}{M+m}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m$ were to remain constant, we would find that this relation continues to hold, so that when the water stops flowing, the cart also stops.  However, $m$ is not constant; it decreases.  When the flow slows to a stop, the acceleration of the cart is now larger because $m$ is smaller.  Hence, by the time all the water has left the cart, it is actually moving to the left.  This is surprising but necessary - the water is mostly moving to the right because the cart initially moved to the right.  The cart must wind up moving left when all is said and done to compensate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we suppose the flow rate is constant the entire time, except abruptly beginning and ending (an assumption not in the original problem, which is qualitatively similar but more work to calculate), the final velocity of the cart is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_f = \frac{lfm}{M(M+m)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water is all flowing at the speed the cart originally jumped to,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w_f = -\frac{lf}{M+m}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so we see that momentum is conserved.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-12-07T13:33:00.927" LastActivityDate="2010-12-07T13:33:00.927" CommentCount="42" />
  <row Id="1727" PostTypeId="2" ParentId="1720" CreationDate="2010-12-07T11:16:10.087" Score="6" ViewCount="" Body="&lt;p&gt;Yup, this is true that the pressure is too small, but the true explanation is not justified yet. Nevertheless the common sense is that there is a lubricating film of water or at least anomalous ice.&#xA;For an overview, see: &lt;a href=&quot;http://lptms.u-psud.fr/membres/trizac/Ens/L3FIP/Ice.pdf&quot; rel=&quot;nofollow&quot;&gt;http://lptms.u-psud.fr/membres/trizac/Ens/L3FIP/Ice.pdf&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-12-07T11:16:10.087" />
  <row Id="1728" PostTypeId="2" ParentId="1725" CreationDate="2010-12-07T11:34:33.527" Score="0" ViewCount="" Body="&lt;p&gt;I'm not an expert, but I dont see any contradiction between following statements.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Crystal momentum is not really a physical momentum.&lt;/li&gt;&#xA;&lt;li&gt;Phonons carry crystal momentum.&lt;/li&gt;&#xA;&lt;li&gt;Phonons do carry physical momentum as well.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="386" LastActivityDate="2010-12-07T11:34:33.527" CommentCount="2" />
  <row Id="1730" PostTypeId="2" ParentId="1725" CreationDate="2010-12-07T14:47:08.543" Score="0" ViewCount="" Body="&lt;p&gt;Cristal momentum is not a real momentum because this is not an eingevalue of the hamiltonian. Infact you define the cristal momentum according to the periodicity of the bravais lattice. Then the sistem is invariant only under discrete traslation and then k is not a good quantic number.&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2010-12-07T14:47:08.543" CommentCount="5" />
  <row Id="1731" PostTypeId="2" ParentId="1717" CreationDate="2010-12-07T14:50:25.320" Score="0" ViewCount="" Body="&lt;p&gt;Isn't this what bullets (traveling at over the speed of sound) do to air inside a gun when it's fired?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as what it looks like, everone I've shot a gun at hasn't been in good enough shape to tell me.&lt;/p&gt;&#xA;" OwnerUserId="752" LastActivityDate="2010-12-07T14:50:25.320" CommentCount="1" />
  <row Id="1732" PostTypeId="2" ParentId="503" CreationDate="2010-12-07T15:29:55.247" Score="2" ViewCount="" Body="&lt;p&gt;Would you consider something like finding a classical solution to the Yang-mills equation of motion a problem in special relativity?  Because I think that's still open for non-$U(1)$ symmetry groups.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-12-07T15:29:55.247" CommentCount="7" />
  <row Id="1733" PostTypeId="2" ParentId="1441" CreationDate="2010-12-07T15:38:49.353" Score="1" ViewCount="" Body="&lt;p&gt;As @Keenan rightly pointed out, the main decay branch for all of my examples is alpha-decay. Alpha particles consist of two protons and two neutrons. And it's easy to remove two neutrons from a completely full (&quot;magic&quot;) neutron shell than from a &quot;magic-1&quot; neutron shell. Because  a &quot;magic-1&quot; neutron shell is more neutron-deficit than a &quot;magic&quot; one.&#xA;Maybe that is the answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, &quot;magic&quot; nuclei  No(264,102) and Rf(266,104) are also beta-stable, so they should have  a long enough  half-life. But  &quot;magic-1&quot; nuclei No(263,102) and Rf(265,104) should be even more stable. &lt;/p&gt;&#xA;" OwnerUserId="238" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2010-12-11T20:27:27.533" LastActivityDate="2010-12-11T20:27:27.533" />
  <row Id="1734" PostTypeId="1" AcceptedAnswerId="1735" CreationDate="2010-12-07T16:13:57.690" Score="1" ViewCount="97" Body="&lt;p&gt;Suppose you wanted to do a time-resolved experiment with a molecular beam traveling at, say, 300 m/s involving a mobile excitation (pump) laser that scans across the length of the molecular beam and a stationary detector.  So the time-resolution of the experiment depends on how precisely the position of the excitation laser can be fine-tuned,  e.g. if the laser can be tuned in the millimeter range, then the time resolution would be in the microsecond time scale.  (Don't worry about the path length of the beam to the detector.  It's reasonable to imagine that one would be interested in obtaining ns accuracy for a system in which nothing happens for several ms).  Also suppose that the velocity distribution of the particles in the beam have a sharp velocity distribution, such that ∆v/v≈0.01.  My question is, then, to what degree of accuracy can the excitation laser's position be fine-tuned using modern optics?  Obviously the intensity, wavelength, and type of laser will have some effect on the answer, but I'm interested in a ballpark figure anyway.  The obvious follow-up question is how, experimentally, does one tune the laser to that level of accuracy?  Would relativistic effects start to be an issue at some point?  And is the velocity distribution (or, for that matter, the measurement of the velocity) actually going to end up being the limiting factor on time resolution in the end?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for your answers.&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2010-12-07T16:31:24.803" Title="How much can a laser's position be fine-tuned?" Tags="&lt;optics&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1735" PostTypeId="2" ParentId="1734" CreationDate="2010-12-07T16:31:24.803" Score="4" ViewCount="" Body="&lt;p&gt;The resolution would ultimately depend on the size of the laser beam where it crosses the molecular beam. The smallest size you'll be able to obtain with the beam is on the order of the optical wavelength (there are some numerical factors that depend on the exact geometry, but if you're just looking for a ballpark figure, that will give you a lower bound). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accuracy you can get with the physical position pretty much depends on how much money you're willing to spend. Small fractions of a millimeter are pretty trivial, and can be done with fairly ordinary optical mounts-- people doing cold-atom experiments routinely manage to hit micron-scale clouds of atoms with micron-sized laser beams just using optical mounts from ThorLabs. (It takes some patience, but it's not a major technical challenge.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're operating in the infinite money limit, you can get commercial position systems that will give you nanometer-ish accuracy over a fairly wide range of positions-- I think I've seen claims of being able to position some object within 100nm over a span measured in meters. That'll cost you big money, though, and is probably complete overkill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, I suspect that the velocity spread and beam size will be the ultimate limiting factors. If your laser beam is 600 nm wide and the speed is 300 m/s, that corresponds to an uncertainty of a couple of nanoseconds, which is a distance uncertainty of a few hundred nm. Positioning accuracy much better than a micron isn't going to make much difference in that case.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-12-07T16:31:24.803" CommentCount="1" />
  <row Id="1736" PostTypeId="2" ParentId="1343" CreationDate="2010-12-07T16:33:43.513" Score="2" ViewCount="" Body="&lt;p&gt;Dust sticks almost anywhere. Almost? Yes almost, because yet we don't have no super perfect nano-surfaces. The point is that every macroscopic surface is raspy and not perfectly smooth. Therefore very small dust can easily stick. This very small dust makes the surface even more raspy, making it easy for the heavier dust to arrive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the near future we might have fan propellors that have no easy visible dust sticking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course it is the electro-magnetic force that lets the dust stick. Make yourself clear how very, very small pieces of dust look like and you understand how it can easily stick on a non-smooth surface by means of the Coulomb-Force. In the end: Most things you see are electrically charged, at least if you &quot;zoom in&quot; a lot. Looking from far away the charges are effectively neutralized. Of course a super perfect nano surface even looks very good is you &quot;zoom in&quot;.&lt;/p&gt;&#xA;" OwnerUserId="744" LastActivityDate="2010-12-07T16:33:43.513" />
  <row Id="1737" PostTypeId="1" AcceptedAnswerId="1963" CreationDate="2010-12-07T18:13:18.743" Score="3" ViewCount="196" Body="&lt;p&gt;What are phonons, rotons, and maxons, and what does their dispersion curve have to do with superfluidity?  I understand that they are quasiparticles, but I'm not entirely sure what that implies.  Are these concepts purely notational and/or mathematical?  Or are these three quasiparticles fundamentally different from each other?&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2010-12-15T18:00:01.323" Title="Phonons, rotons, and maxons" Tags="&lt;spectroscopy&gt;&lt;quasiparticles&gt;&lt;superfluidity&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" />
  <row Id="1739" PostTypeId="2" ParentId="886" CreationDate="2010-12-07T23:00:45.883" Score="6" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What's going on here? Is swimming through spacetime only possible if the spacetime is curved in some way that breaks the symmetry under Lorentz boosts? Or is there some error in my reasoning?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is precisely the case. No error in your reasoning. In the case of a curved spacetime the &quot;center of mass&quot; of an &lt;strong&gt;extended&lt;/strong&gt; body is no longer well-defined w.r.t external - i.e. located in an asymptotically flat region - observers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to &quot;swim&quot; through spacetime one exploits the inhomogeneities of the gravitational field. The presence of these inhomogeneities breaks local Lorentz symmetry which is necessary for the mechanism to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular the scale of the swimmer and the inhomogeneities should be comparable. This is one reason why, at present, the construction of an &lt;em&gt;actual&lt;/em&gt; swimmer is far beyond our technological means.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; For those interested on extended body effects in GR there is are classic papers by Dixon. More recently Abraham Harte has done some amazing work along these lines &lt;a href=&quot;http://arxiv.org/abs/0706.2909v2&quot; rel=&quot;nofollow&quot;&gt;Extended-body effects in cosmological spacetimes&lt;/a&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2011-04-06T08:51:39.057" CommentCount="1" />
  <row Id="1740" PostTypeId="2" ParentId="1725" CreationDate="2010-12-07T23:55:56.450" Score="0" ViewCount="" Body="&lt;p&gt;Crystal Momentum can be seen as &quot;less than momentum&quot;, that holds a part of the information from what real momentum usually tells, whether or not the other part makes any sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once wave function values is of interest only on lattice sites, for example, as in case of phonons, what matters is how phase changes from one site to another and not elsewhere in space. In such case, 2pi jumps are meaningless, so it is known &quot;up to addition of a reciprocal latice vector&quot;&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-12-07T23:55:56.450" />
  <row Id="1741" PostTypeId="1" CreationDate="2010-12-08T00:37:48.847" Score="1" ViewCount="127" Body="&lt;p&gt;I'm looking for something that will generate scatter plots comparing different properties of isotopes. Ideally I'd like some web page that lets me select axis and click go but a CSV file with lost of properties would work.&lt;/p&gt;&#xA;" OwnerUserId="754" LastEditorUserId="754" LastEditorDisplayName="" LastEditDate="2010-12-08T15:31:20.190" LastActivityDate="2010-12-08T20:15:20.617" Title="Isotope properties plotting tool?" Tags="&lt;nuclear-physics&gt;&lt;software&gt;&lt;isotope&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="1742" PostTypeId="2" ParentId="1741" CreationDate="2010-12-08T01:59:49.230" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://ie.lbl.gov/education/isotopes.htm&quot; rel=&quot;nofollow&quot;&gt;Exploring the Table of Isotopes&lt;/a&gt; is my go-to site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few others out there, but when I moved from my last computer those were some of the bookmarks I &lt;em&gt;didn't&lt;/em&gt; bring, 'cause I never used them... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You haven't specified what you want in your CVS file. I think this site only offers transition energies and branching fractions by radiation type. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For searching a whole database of transistions for lines of [energy range]$\times$[A or Z range]$\times$[halflife range] use &lt;a href=&quot;http://nucleardata.nuclear.lu.se/NuclearData/toi/&quot; rel=&quot;nofollow&quot;&gt;The Lund/LBNL Nuclear Data Search&lt;/a&gt;. Very useful if you are thinking of building a novel calibration source or some thing like that. This site also has stopping powers and other good stuff (go up a level from the link...). Another go-to resource.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2010-12-08T02:10:31.080" LastActivityDate="2010-12-08T02:10:31.080" CommentCount="1" />
  <row Id="1743" PostTypeId="1" CreationDate="2010-12-08T03:01:04.087" Score="2" ViewCount="227" Body="&lt;p&gt;I have recently become curious about modeling the repulsion of everyday objects in contact with one another.  By repulsion I mean as you attempt to walk through a wall, the pain in your nose suddenly alerts you to the fact that it won't be possible.  I've come up with this exercise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the mutual pressure exerted by two dipole sheets of finite extent, each area $A$, separated by a distance $D$?  Assume $D \gg$ d where the dipole moment $p = ed$ and $e = 1.6\times10^{-19}$.  There are $N$ dipoles per unit area.  Make an order of magnitude estimate for $N$ to represent a density one would find in ordinary matter.  Make a numerical plot for $10^{-10} &amp;lt; D &amp;lt; 10^{-3}$ meters.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a crude electrostatic model of two electrically neutral materials brought close together.  The oppositely oriented dipole sheets are generated by the electronic cloud repulsion and deformation as the two interfaces approach.  It is meant to serve as a quantitative estimate of the &lt;em&gt;maximum&lt;/em&gt; distance at which Coulomb repulsion could possibly stop one body from passing through another.  Interface physics is pretty rich, but I wonder how this model does in predicting such pressures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone has related references please post them, especially any experimental studies.&lt;/p&gt;&#xA;" OwnerUserId="722" LastEditorUserId="722" LastEditorDisplayName="" LastEditDate="2010-12-08T22:04:59.037" LastActivityDate="2010-12-17T17:13:42.883" Title="What is the pressure between two electric dipole sheets of finite extent?" Tags="&lt;electromagnetism&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="1744" PostTypeId="1" AcceptedAnswerId="1746" CreationDate="2010-12-08T05:46:52.623" Score="7" ViewCount="416" Body="&lt;p&gt;Tesla patented a device for gathering energy from light, using the photoelectric effect. (&lt;a href=&quot;http://www.google.com/patents/about?id=YitoAAAAEBAJ&quot; rel=&quot;nofollow&quot;&gt;US 685,957 - Apparatus for the Utilization of Radiant Energy&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/cSGJ1.png&quot; alt=&quot;Telsa's photoelectric effect generator&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically just a sheet of &quot;highly polished or amalgamated&quot; metal (top) connected to a capacitor (the ⊔⊓ shapes) connected to the Earth.  (The other stuff on the right of the image is a load intermittently driven when the charge on the capacitor rises high enough.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this actually work?  Why or why not?  Does it need to be in a vacuum?  Would the plate stay charged after electrons are thrown off, or would they be attracted and drift back to it and neutralize it?  How does the capacitor to earth make this better than a plate hanging in space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The patent demonstrates that he doesn't understand the cause of the photoelectric effect (&quot;sources of such radiant energy throw off with great velocity minute particles of matter which are strongly electrified, and therefore capable of charging an electrical conductor&quot;), but he understands the results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Incidentally, I have tried building this with a piece of aluminum foil, electrolytic cap, and wire strong enough to stick a few inches into the ground.  I didn't see any voltage with a multimeter in bright sunlight.)&lt;/p&gt;&#xA;" OwnerUserId="176" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-13T10:04:57.973" LastActivityDate="2011-07-13T10:04:57.973" Title="Does Tesla's photoelectric &quot;solar cell&quot; really work?" Tags="&lt;electricity&gt;&lt;photoelectric-effect&gt;&lt;renewable-energy&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" />
  <row Id="1745" PostTypeId="2" ParentId="1343" CreationDate="2010-12-08T08:33:16.033" Score="2" ViewCount="" Body="&lt;p&gt;since it seems like participating in bounty questions is something like our group sport, I could not resist to also think about the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answers given according to the vanishing velocity on the propeller are unlikely to explain the phenomenon. Dust particles, even though they are very thin, have a three dimensional shape. So even if there is no velocity at zero distance (which is the mentioned boundary condition), there will be a relative movement along the dust particle (Remember that turbulence takes place on every scale for the Navier-Stokes equations) which to my opinion acts as a pulling force on the particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my mind, the problem relates to an electrostatic one. The propeller is very likely to have some charge due to his movement through air and the ongoing friction. The (might even be very small) charge induces a dipole moment on the dust particles nearby which I guess might be treated as dielectric balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This dipole moment will then attract the dust particles which as a result touch the surface. If now everything would be metal, there would be immediate (full) charge balance and the attractive force would vanish. But under the assumption of a dielectric medium, some dipole moment stays, so does the attractive force and the dust on the propeller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Greetings&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I just saw that the electrostatic argument already came up. Nevertheless, I hope my explanations are still usefull.&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2010-12-08T08:46:09.387" LastActivityDate="2010-12-08T08:46:09.387" />
  <row Id="1746" PostTypeId="2" ParentId="1744" CreationDate="2010-12-08T08:44:45.767" Score="8" ViewCount="" Body="&lt;p&gt;The energy needed to remove an electron from a solid is called  the &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Work_function&quot; rel=&quot;nofollow&quot;&gt;work function&lt;/a&gt;&lt;/strong&gt;. For most metals you would need UV photons (300 nm for Aluminium) that rarely reach the Earth's surface. Visible light can eject electrons from alkali metals, but the &lt;strong&gt;quantum yield&lt;/strong&gt; (the probability of electron emission per incident photon) for pure metals is low (probably less than 1%). Materials like CsTe that are used in photocathodes have efficiency up to 40% (at certain wavelength) but they are expensive and difficult to handle in open air.&#xA;Silicon solar cells also utilize photoelectric effect and compared to metals they are efficient and inexpensive.&lt;/p&gt;&#xA;" OwnerUserId="701" LastEditorUserId="701" LastEditorDisplayName="" LastEditDate="2010-12-08T09:43:16.443" LastActivityDate="2010-12-08T09:43:16.443" CommentCount="13" />
  <row Id="1747" PostTypeId="1" AcceptedAnswerId="1753" CreationDate="2010-12-08T09:01:44.593" Score="3" ViewCount="291" Body="&lt;p&gt;Consider a conducting wire of 1M and 1000KM. Now if we connect a battery and a bulb to both these wires. Bulb glows instantaneously its because (my guess:) electric filed travels from positive terminal of battery to negative terminal through wire almost instantaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, My question is there any speed for this electric field? If yes, what is it? Can I induce any delay in in glowing a bulb by similar expermient?&lt;/p&gt;&#xA;" OwnerUserId="447" LastActivityDate="2010-12-08T14:56:01.487" Title="Is there a travelling speed of for electric field? If yes, what is it?" Tags="&lt;electricity&gt;&lt;electrodynamics&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="1748" PostTypeId="2" ParentId="1747" CreationDate="2010-12-08T09:06:13.837" Score="1" ViewCount="" Body="&lt;p&gt;Electric field is mediated by photons, photons are light, information of a change in the electric field thus travel at lightspeed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesnt mean your bulb is gonna light that fast. For that to happend there has to be electrons moving through it, these travel much slower through the wire, of the order 0.01 cm/s IIRC. However since the photons travel at lightspeed with information that the electric field has changed the electrons near the bulb start to move once the photons has reached them.&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2010-12-08T09:31:41.420" LastActivityDate="2010-12-08T09:31:41.420" CommentCount="5" />
  <row Id="1749" PostTypeId="2" ParentId="1747" CreationDate="2010-12-08T09:55:41.067" Score="3" ViewCount="" Body="&lt;p&gt;Yes, the relativity tell you that no information can travel faster than speed of light. That is the disturbance of electric field can only travel slower than the speed of light. The typically speed is around few tens percentage of speed of light for a copper wire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You find the response almost instantaneous because you use only a short wire, say 1 m. In this case, it responses with a time delay of the order $10^{-8}$ s in which you certainly cannot feel any time delay because you cannot tell difference for events shorter than $10^{-1}$ s as a human.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to introduce delay, the easiest way is to use a longer wire. By wrapping a long wire, say, 10 km, but you still cannot feel any different since the delay is still around $10^{-4}$ s. However, you may use device to measure this delay. You cannot use a very long wire, say 1000 km, because the electric field will be vanishing small for you to do the experiment.&lt;/p&gt;&#xA;" OwnerUserId="562" LastActivityDate="2010-12-08T09:55:41.067" />
  <row Id="1750" PostTypeId="2" ParentId="1725" CreationDate="2010-12-08T11:12:57.693" Score="0" ViewCount="" Body="&lt;p&gt;I don't see why the crystal should move as a whole. A first argument is the one given by Marek in the comments of the question. But that's also the very principle of a wave: transfer of energy/momentum and no transfer of matter; in this case the ions will vibrate around their equilibrium position in the lattice and the modes of this vibration if a phonon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering this &quot;real momentum&quot;-ness whatever it means this looks strange to me: consider the absorption of a photon (carrying momentum and energy) by the crystal. This process can happen due to the creation of a phonon, or more generally a bunch of them, creating a wave packet. This wave packet will propagate in the crystal (seen as a medium, from the point of view of the phonon this is equivalent to the vacuum from the point of view of the photon) and conserve the momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally I think this question is not really interesting from a physical point of view, it is the same as asking if the phonons are real particles or not. I will depend on what you call &quot;real&quot; and following a pragmatic physicist way we call them particles because they have properties that more &quot;naturally real&quot; particles have and they can be treated in the same way...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also find &lt;a href=&quot;http://i.stack.imgur.com/AohrD.jpg&quot; rel=&quot;nofollow&quot;&gt;that proof&lt;/a&gt; a little odd. A real propagating phonon (or plasmon, magnon, whatever) must be seen as a wave packet. In that case the proof given will show that this packet does carry momentum.&lt;/p&gt;&#xA;" OwnerUserId="82" LastEditorUserId="82" LastEditorDisplayName="" LastEditDate="2010-12-08T11:19:00.823" LastActivityDate="2010-12-08T11:19:00.823" />
  <row Id="1751" PostTypeId="2" ParentId="1725" CreationDate="2010-12-08T12:09:27.407" Score="2" ViewCount="" Body="&lt;p&gt;First (as already pointed out in my comments) crystal momentum is in every regard the same as usual momentum except the fact that it takes values only in Brillouin zone (as a consequence of discrete symmetry of lattice; or more precisely its continuum limit). So the answer to your question is: for most purposes crystal momentum is a real momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the term &lt;em&gt;crystal momentum&lt;/em&gt; is being used here in two different meanings and that is probably where confusion arises. Your citation uses it as a &lt;strong&gt;total momentum&lt;/strong&gt; of the crystal. This is obviously zero for phonons (which are just harmonic modes of the material) because on average the atoms of the crystal don't move (they just oscillate around stable positions). And that is precisely why nobody uses the term in this way (and I don't understand why your book does).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But locally energy and momentum are still being transfered (jumping from one atom to the next as they interact). So in fact, phonon is a wave that propagates in the material in some direction and carries some energy. Obviously this is a very physical wave with physical energy and physical momentum. It is this latter momentum which is usually referred to as &lt;strong&gt;crystal momentum&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-08T12:09:27.407" />
  <row Id="1752" PostTypeId="2" ParentId="1507" CreationDate="2010-12-08T13:44:38.860" Score="4" ViewCount="" Body="&lt;p&gt;Seeing as no one is trying to give an answer, I'll take a stab at it myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shortly after writing this question I learned (in this &lt;a href=&quot;http://physics.stackexchange.com/questions/1523/ising-model-for-dummies/1525#1525&quot;&gt;cute answer of Raskolnikov's&lt;/a&gt;) about Baxter's wonderful book on exact solutions in statistical mechanics. Slowly but surely I realized that Ising model has been solved so many times by some many different methods by virtually every famous physicist (I'll list some of the solutions later ) that it became clear that my question is hopefully inadequate and only reflects my huge ignorance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make up for that, I started reading papers. The Onsagar's paper itself came out in 1944. In 1949 there appeared Bruria Kaufman's paper where she notes that the transfer matrix can be interpreted as $2^n$-dimensional representation of $2n$-dimensional rotations. So she introduces spinor analysis (e.g. Pauli and Dirac matrices) and goes ahead to solve the problem. I must say I am in love with this approach (okay, you got me, I &lt;strong&gt;am&lt;/strong&gt; a group person).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1952 Kac and Ward used a purely combinatorial method of some polygons (which I don't yet quite understand, but it probably has to do with Peierl's contours). Other papers note the duality with free fermionic field. Or note that Ising is just a special case of Random Cluster Model; or a dimer model. These papers carry names (in no particular order) such as Potts, Ward, Kac, Kasteleyn, Yang, Baxter, Fisher, Montroll and others. It's quite obvious that it will take me some time to understand (or indeed, even read) all those papers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I took a different road and used &lt;a href=&quot;http://www.google.cz/search?q=ising+onsager+kaufman+ward+potts+hurst+green+montroll+kac+kasteleyn+yang&quot; rel=&quot;nofollow&quot;&gt;asked google&lt;/a&gt;. Querying all the names above at once returns precious gems:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.scholarpedia.org/article/Ising_model%3a_exact_results&quot; rel=&quot;nofollow&quot;&gt;Amazing article&lt;/a&gt; over at Scholarpedia. It contains historical treatment, main methods of solution, references to the papers I mentioned and much much more.&lt;/li&gt;&#xA;&lt;li&gt;paper &lt;a href=&quot;http://rmp.aps.org/abstract/RMP/v39/i4/p883_1&quot; rel=&quot;nofollow&quot;&gt;History of Lenz-Ising model&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;paper &lt;a href=&quot;http://www.google.cz/url?sa=t&amp;amp;source=web&amp;amp;cd=3&amp;amp;ved=0CB0QFjAC&amp;amp;url=http%3A%2F%2Fpsroc.phys.ntu.edu.tw%2Fcjp%2Fdownload.php%3Fd%3D1%26pid%3D1025&amp;amp;rct=j&amp;amp;q=ising%20onsager%20kaufman%20ward%20potts%20hurst%20green%20montroll%20kac%20kasteleyn%20yang%20%20fisher%20vdovichenko%20baxter%20mccoy&amp;amp;ei=a4r_TK7oIND2sgbTsPzyDg&amp;amp;usg=AFQjCNHHRsxMuhHpGcTO2UVMgBJxUqv20A&amp;amp;sig2=3Bc2Vr2xDKBxEQATcQ1GLA&amp;amp;cad=rja&quot; rel=&quot;nofollow&quot;&gt;Spontaneous magnetization of the Ising model&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-08T13:44:38.860" CommentCount="9" />
  <row Id="1753" PostTypeId="2" ParentId="1747" CreationDate="2010-12-08T14:56:01.487" Score="5" ViewCount="" Body="&lt;p&gt;There are two different questions at work here, that you've kind of mashed together. The first question is &quot;What is the speed at which a change in the electric field propagates?&quot; The answer to that is the speed of light. In QED terms, the electromagnetic interaction that we see as the electric field is mediated by photons, so any change in an established field (say, due to shifting the position of the charge creating the field) won't be felt by a distant object until enough time has passed for a photon from the source to make it to the observation point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second question is &quot;What is the speed of propagation of electric current?&quot; This speed is slower than the speed of light, but still on about that order of magnitude-- the exact value depends a little on the arrangement of wires and so on, but you won't be far off if you assume that electrical signals propagate down a cable at the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This relates to electric field in that the charge moving through a circuit to light a light bulb has to be driven by some electric field, so you can reasonably ask how that field is established, and how much time it takes. Qualitatively, the necessary field is established by excess charge on the surface of the wires, with the surface charge being generally positive near the positive terminal of a battery and generally negative near the negative terminal, and dropping off smoothly from one to the other so that the electric field is more or less piecewise constant (that is, the field is the same everywhere inside a wire, and the field is the same everywhere inside a resistor, but the two field values are not the same). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the circuit is first connected, there is a rapid redistribution of the charge on the surface of the wires which establishes the surface charge gradients that drive the steady-state current that will eventually do whatever it is you want it to do. The time required to establish the gradients and settle in to the steady-state condition is very fast, most likely on the order of nanoseconds for a normal circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a good discussion of the business of how, exactly, charges get moved around to drive a current in the textbook that we use for our introductory classes, &lt;a href=&quot;http://matterandinteractions.org/&quot; rel=&quot;nofollow&quot;&gt;Matter and Interactions&lt;/a&gt;, by Chabay and Sherwood. It doesn't go into enough detail to let you calculate the relevant times directly, but it lays out the basic science pretty well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It's a textbook for a first-year introductory physics class, so it sweeps a lot of condensed matter physics under the metaphorical rug-- there's no discussion of band structure or surface modes, or any of that. It's fairly solid conceptually, though, at least according to colleagues who know more about those fields than I do.)&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-12-08T14:56:01.487" />
  <row Id="1754" PostTypeId="2" ParentId="1741" CreationDate="2010-12-08T20:15:20.617" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.nndc.bnl.gov/chart/&quot; rel=&quot;nofollow&quot;&gt;Chart of Nuclides&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or like &lt;a href=&quot;http://atom.kaeri.re.kr/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="238" LastActivityDate="2010-12-08T20:15:20.617" />
  <row Id="1755" PostTypeId="2" ParentId="1706" CreationDate="2010-12-08T21:00:45.577" Score="2" ViewCount="" Body="&lt;p&gt;At research level, you might be interested in the PDG review on conservation laws:&#xA;&lt;a href=&quot;http://pdg.lbl.gov/2009/reviews/rpp2009-rev-conservation-laws.pdf&quot; rel=&quot;nofollow&quot;&gt;http://pdg.lbl.gov/2009/reviews/rpp2009-rev-conservation-laws.pdf&lt;/a&gt;&#xA;Also, the review about CPT invariance gives information about tests of CPT violation in neutral kaons, but I can apparently only post one hyperlink (reputation too low).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that CP violation itself is still an active area of research (particularly at Belle and LHCb), since we don't know definitively how many systems display it and whether there's a deeper explanation for it.&lt;/p&gt;&#xA;" OwnerUserId="762" LastActivityDate="2010-12-08T21:00:45.577" CommentCount="1" />
  <row Id="1756" PostTypeId="1" AcceptedAnswerId="1759" CreationDate="2010-12-08T22:38:11.380" Score="5" ViewCount="374" Body="&lt;p&gt;as you might know, the Standard Model of physics can be seen as a $U(1)\times SU(2)\times SU(3)$ gauge theory where each symmetry group accounts for different force fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The behaviour for the force field of a &quot;point&quot; charge in the most simple cases of this model, the electromagnetic interaction expressed as an abelian (elements commute) $U(1)$ theory, is well known and falls of as $r^{-2}$ and is proportional to the total charge of the source. One states that this force is long range as it only falls off polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the system for the other interactions, the weak and strong ones, is much more complicated since the underlying groups are not abelian which makes the problem intrinsically nonlinear as can be seen from the Yang-Mills equations, $F = DA, DF = 0$ and its dual counterpart. In contrast to electromagnetism, the interaction range falls off rather quickly and different potentials are known to describe different phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can one see directly (e.g. from the non-Abelian character of the group) that the decay of the force field must be faster than for electrodynamics? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2010-12-09T08:40:02.417" LastActivityDate="2011-01-16T21:07:25.150" Title="Interaction ranges in the Standard Model - Electrodynamics vs QCD" Tags="&lt;quantum-field-theory&gt;&lt;electrodynamics&gt;&lt;theory&gt;&lt;gauge-theory&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="2" />
  <row Id="1757" PostTypeId="2" ParentId="1756" CreationDate="2010-12-08T23:21:05.463" Score="5" ViewCount="" Body="&lt;p&gt;The picture is a little more complicated. The decay of the weak interaction has to do with whether or not propagators are massive, not with abelianness. The decay of strong force has to do with &lt;a href=&quot;http://en.wikipedia.org/wiki/Color_confinement&quot; rel=&quot;nofollow&quot;&gt;confinement&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Spontaneous symmetry breaking&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The group for electroweak theory is $SU(2) \times U(1)$. There are four gauge bosons here. At high enough energies they are all massless (and therefore provide long-range interactions) and transform among each other. But in nature we can observe weak interactions (which means they have massive propagators that decay quickly). This is because the above group is &lt;a href=&quot;http://en.wikipedia.org/wiki/Spontaneous_symmetry_breaking&quot; rel=&quot;nofollow&quot;&gt;spontaneously broken&lt;/a&gt; at low enough energies by &lt;a href=&quot;http://en.wikipedia.org/wiki/Higgs_mechanism&quot; rel=&quot;nofollow&quot;&gt;Higgs mechanism&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the most simple picture, there are four scalar Higgs fields. Three of them couple to the (originally massless) gauge bosons and you obtain massive $W^-, W^+, Z$. These form $SU(2)$. But note that this is a different $SU(2)$ than the original one (i.e. it also contains part of the fourth gauge boson from the original $U(1)$. One of the Higgs fields remains (this is the one people look for at LHC). You also obtain photon, which is massless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Note that this model is not a consequence of some particular theory. It was built with all the observations in mind and this is the most natural way to do it).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the weak bosons are very heavy and decay quickly. They are only present in the actual interactions as virtual particles.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Confinement&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The other part of the standard model is $SU(3)$. Gluons are massless particles and so they propagate at the speed of light. The difference with respect to $U(1)$ case is that the field theory is not free (except at high energies where the theory possesses an &lt;a href=&quot;http://en.wikipedia.org/wiki/Asymptotic_freedom&quot; rel=&quot;nofollow&quot;&gt;asymptotic freedom&lt;/a&gt;) and so the model has complicated dynamics also if no charges are present. But here renormalization comes into play. It tells you that as the relevant energy you are dealing with gets too low, the strong coupling diverges. So you can't separate quarks. If you try, you'll just create &lt;a href=&quot;http://en.wikipedia.org/wiki/Jet_%28particle_physics%29&quot; rel=&quot;nofollow&quot;&gt;jets of hadrons&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Nuclear interactions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;So when you are talking about decay of strong force field, you must in fact be talking about the nuclear force between nucleons (protons and neutrons). This is an effective interaction which mediated by &lt;a href=&quot;http://en.wikipedia.org/wiki/Pion&quot; rel=&quot;nofollow&quot;&gt;pion&lt;/a&gt; $\pi$. Pions are massive and described by &lt;a href=&quot;http://en.wikipedia.org/wiki/Yukawa_potential&quot; rel=&quot;nofollow&quot;&gt;Yukawa potential&lt;/a&gt; which indeed decays exponentially.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-08T23:36:56.097" LastActivityDate="2010-12-08T23:36:56.097" CommentCount="9" />
  <row Id="1759" PostTypeId="2" ParentId="1756" CreationDate="2010-12-09T02:28:54.203" Score="3" ViewCount="" Body="&lt;p&gt;This is a very nice question. There is indeed a simple way to see that a non-abelian theory will be of shorter range than an abelian one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The action of a gauge theory, generically contains terms of the form $ Tr[F F] $ or $ Tr[F \star F] $, where $ F $ is the curvature or field strength of the gauge connection. For an abelian connection $A_\mu$, the field strength is of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F_{\mu\nu} = \partial_{[\mu}A_{\nu]} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the $ [ \dots ] $ represents antisymmetrization over the indices within the brackets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently the $F^2$ type terms in the action are of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F^2 \sim (\partial A) (\partial A) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a non-abelian connection $A_\mu^I$, where $I$ is now an index in the lie algebra of some non-abelian group, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F^I_{\mu\nu} = \partial_{[\mu}A^I_{\nu]} + f^I_{JK}[A^J_\mu,A^K_\nu] $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $ f_{IJK} $ are the &lt;em&gt;structure constants&lt;/em&gt; of the group in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently the $ \mathcal{O}(F^2) $ terms in the action now contain terms of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (\partial A) A^2 \textrm{ and } A^4 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are self-interaction terms which will, in general, endow the connection $A^I_\mu$ with a mass - in a suitable symmetry broken phase of the theory. And a massive gauge particle leads to a short range (and/or confining) interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the gist of it. There are likely other ways to approach the problem, but this is the one I'm most familiar with.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In response to some comments I'd like to quote the following line from the Jaffe-Witten &lt;a href=&quot;http://www.claymath.org/millennium/Yang-Mills_Theory/yangmills.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; introducing the Yang-Mills problem as part of the Clay math prize:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot; ... One view of the mass gap in Yang–Mills theory suggests that it could arise from the quartic potential $(A \wedge A)^2$ in the action, where $ F = dA + g A \wedge A $, see [11], &#xA;  and may be tied to curvature in the space of connections, see [44].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reference [11] cited in the line above is a &lt;a href=&quot;http://adsabs.harvard.edu/abs/1981NuPhB.188..479F&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; by Feynman where he studies SU(2) gauge theory in 2+1 dimensions and concludes the gauge invariance dictates the presence of a mass gap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can argue about fixed points and phases and whatnot at different temperatures. But unless you have something that beats Jaffe, Witten and Feynman I guess it is safe to conclude that @robert's intuitive guess that &lt;em&gt;the non-linear nature of non-abelian gauge theory is responsible for its short-range/massive/confining character&lt;/em&gt; is right on target.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-09T04:41:18.027" CommunityOwnedDate="2010-12-09T02:28:54.203" CommentCount="8" />
  <row Id="1760" PostTypeId="2" ParentId="1756" CreationDate="2010-12-09T02:55:27.750" Score="2" ViewCount="" Body="&lt;p&gt;The non-linearities and the short range nature if the force are completely independent. You can have Abelian theories which are short ranged (look up the Abelian Higgs model), and non-linear interacting theories that are long ranged (e.g gravity). The same theory can have more than one phase (depending on temperature and other control parameter) in which the forces are long ranged or not, it all depends on the details.&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2010-12-09T02:55:27.750" CommentCount="10" />
  <row Id="1762" PostTypeId="1" CreationDate="2010-12-09T04:50:49.247" Score="3" ViewCount="141" Body="&lt;p&gt;I am not sure if I'm clear with my question or not. Usually people give example of rocket science, meaning space physics, to refer to something which is very complex and esoteric. Could you give some examples, relatively, defining the level and type of physics required in dfferent areas?&lt;/p&gt;&#xA;" OwnerUserId="766" LastActivityDate="2010-12-09T05:28:59.127" ClosedDate="2010-12-09T09:14:51.123" Title="Which field of area is based on most complex physics (relatively speaking)?" Tags="&lt;education&gt;&lt;epistemology&gt;" CommentCount="2" />
  <row Id="1763" PostTypeId="1" AcceptedAnswerId="1764" CreationDate="2010-12-09T06:15:40.887" Score="0" ViewCount="482" Body="&lt;p&gt;Consider an electron in an infinite square well. The expectation values of momentum and angular momentum are all zero for energy eigenstates. An electron transition is accompanied by the emission or absorption of photons. And we know the momentum of a photon is hk and the angluar momentum (spin) is 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The momentum and angular momentum should be conserved in the transition process. Does that mean all electron transitions between energy levels are prohibited in such system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If such transitions are prohibited, how do we explain light-emitting quantum dots?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;===============================&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the answers and comments below. Now I realize a quantum dot is more like an atom than a hollow box. But if we consider the theoretical ideal square well, there seems to be no simultaneous eigenstates of both energy and momentum - at least I can't figure it out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a 2D quare well. The energey eigenstates can be shown as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Particle_in_a_box#Higher-dimensional_boxes&quot; rel=&quot;nofollow&quot; title=&quot;2D box&quot;&gt;Energy eigenstates in a 2D Box&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently they are not momentum eigenstates. Since these states form a complete set, we can set a momentum eigenstate to be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|k_x\rangle = \sum_{n_x,n_y} C_{n_x,n_y} |n_x,n_y\rangle$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where C's are constants, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_x |k_x\rangle = \hbar k_x |k_x\rangle$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep going on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_x |k_x\rangle = -i\hbar\nabla_x \sum_{n_x,n_y} C_{n_x,n_y} |n_x,n_y\rangle = -i\hbar\nabla_x \left( C_{1,1} \sin(k_{1}x)\sin(k_{1}y)+\cdots\right)$&#xD;&#xA;$= -i\hbar\left( C_{1,1} k_1 \cos(k_1 x)\sin(k_1 y)+\cdots\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=???\; \hbar k_x \left( C_{1,1} \sin(k_{1}x)\sin(k_{1}y)+\cdots\right) = \hbar k_x |k_x\rangle$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know any way to transform cos*sin's to sin*sin's without messing them up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything wrong with my calculation?&lt;/p&gt;&#xA;" OwnerUserId="749" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2010-12-11T16:36:25.763" LastActivityDate="2010-12-11T16:36:25.763" Title="Electron transitions in an infinite square well" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="1764" PostTypeId="2" ParentId="1763" CreationDate="2010-12-09T08:35:37.803" Score="4" ViewCount="" Body="&lt;p&gt;In an ideal 3-D square well the electron's movement along the three axes is independent. This is not true for a real quantum dot which has a spherical shape (close to it) and is thus more similar to hydrogen atom. As soon as there is a coupling between movement along $x$,$y$ and $z$ the electron will start orbiting around the quantum dot and will posess an angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a linear conjugated molecule can be considered as a 1-D square well. Linear carbon chains have been discovered in the interstellar space and also produced in the lab. To interact with a photon such molecule has to have a  &lt;a href=&quot;http://en.wikipedia.org/wiki/Transition_dipole_moment&quot; rel=&quot;nofollow&quot;&gt;transition dipole moment&lt;/a&gt; which means that the electron cloud should &lt;em&gt;move&lt;/em&gt;. If we don't consider $p$-orbitals (that would add a second dimension), only a non-symmetric molecule like $C_6N$ can have a nonzero transition moment and such molecules are in fact very strong light absorbers.&lt;/p&gt;&#xA;" OwnerUserId="701" LastActivityDate="2010-12-09T08:35:37.803" />
  <row Id="1765" PostTypeId="2" ParentId="1683" CreationDate="2010-12-09T11:21:48.760" Score="2" ViewCount="" Body="&lt;p&gt;Here is my attempt. I went to a somewhat different path than kalle43 and this is a little easier i think.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x(t)$ be the coordinate of the nozzle  at time $t$. Consider an infinitesimal mass of water $dm$ departing the nozzle at time $\tau$ : $$dm=-m&amp;#39;(\tau)d\tau$$ Here $m&amp;#39;(t)$ denotes the time derivative of the mass of water inside the tank.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x(\tau)$ be the horizontal coordinate of $dm$ at time $\tau$. Then at time $t&amp;gt;\tau$ the horizontal coordinate of $dm$ will be: $$x(\tau)+(t-\tau)x&amp;#39;(\tau)$$ Here $x&amp;#39;(t)$ denotes the time derivative of the coordinate of the nozzle  at time $t$ or simply velocity of the wagon.&lt;br&gt;&#xA;Now sum $x_idm_i$ (static moment of mass) over all infinitesimal particles emitted from the nozzle within the time period $(0...t)$  will be expressed by the integral:&lt;br&gt;&#xA;$$-\int_0^t [x(\tau)+(t-\tau)x&amp;#39;(\tau)]m&amp;#39;(\tau)d\tau$$ The following step is to get static moment of mass of the wagon with water inside it. This is a simple:$$[l+x(t)][M+m(t)]$$ Now the static moment of mass of the whole system(the wagon with water + emitted water) is expressed as the sum of  last two expressions:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\int_0^t [x(\tau)+(t-\tau)x&amp;#39;(\tau)]m&amp;#39;(\tau)d\tau+[l+x(t)][M+m(t)]=pt+c$$ $p=const$ and $c=const$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you ask what means $pt+c$.This becomes clear when we differentiate the last equation with respect to $t$:  $$-\int_0^t x&amp;#39;(\tau)m&amp;#39;(\tau)d\tau- x(t)m&amp;#39;(t)+x&amp;#39;(t)[M+m(t)]+m&amp;#39;(t)[l+x(t)]=p$$ $p=const$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This result represents the  horizontal momentum of the whole system(the wagon with water + emitted water)&lt;/strong&gt;. &lt;strong&gt;This must be conserved&lt;/strong&gt;. So $c$ is simply integration constant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the most important part follows:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the initial moment $t=0$. At this moment let the coordinate of the nozzle be zero:$(x(0)=0)$ as well as the initial velocity of the wagon:$(x'(0)=0)$. Then the momentum equation gives:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$lm&amp;#39;(0)=p=const$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can we conclude from this result? First, before the opening of the nozzle the momentum of the whole system(wagon+water inside it) is definitely zero. But after opening, at $t=0$ the momentum remains zero only if $m&amp;#39;(0)=0$. Otherwise it suddenly becomes different from zero. &lt;strong&gt;And this last is realized in the given problem.&lt;/strong&gt; The momentum of the whole system(the wagon with water + emitted water) becomes different from zero and the wagon starts to move in one direction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if $m&amp;#39;(0)=0$ then &lt;strong&gt;Mark Eichenlaub's scenario&lt;/strong&gt; will start, i think.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's differentiate the momentum equation with respect to $t$ to get the equation of motion of the wagon: $$[M+m(t)]x&amp;#39;&amp;#39;(t)=-lm&amp;#39;&amp;#39;(t)$$ Actually, I was shocked that the equation turned out to be as simple. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I drifted from Torricelli's law and added an example which confirms quantitatively Mark Eichenlaub's qualitative answer. This shows also that the law of conservation of energy is irrelevant  in this problem. &#xA;Only the mass change of the wagon does matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I picked a function $m(t)$ such that $m&amp;#39;(0)=0$. So there is no need to worry about any instantaneous jump at $t=0$ and the horizontal momentum remains zero.&#xA;$$m(t)=\frac{m}{2}\left(1+cos\frac{\pi{t}}{T}\right);0\leq t\leq T$$ and the equation of motion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[M+m(t)]x&amp;#39;&amp;#39;(t)=-lm&amp;#39;&amp;#39;(t)$$ The solution of the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{x}(t)=\frac{l\pi^2}{T}\left(\frac{t}{T}-\frac{2 }{\pi}\frac{\eta+1}{\sqrt{{2\eta+1}}}\arctan\frac{tan\frac{\pi{t}}{2T}}{\sqrt{2\eta+1}}\right)$$ where $\eta=\frac{m}{2M}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This solution  follows closely by the behavior Mark gave. Final velocity is directed to the left $(v_f&amp;lt;0)$ and is given by the expression:&#xA;$$v_f=\frac{l\pi^2}{T}\left(1-\frac{\eta+1}{\sqrt{{2\eta+1}}}\right)$$&lt;/p&gt;&#xA;" OwnerUserId="552" LastEditorUserId="887" LastEditorDisplayName="" LastEditDate="2011-01-30T17:55:31.860" LastActivityDate="2011-01-30T17:55:31.860" CommentCount="41" />
  <row Id="1766" PostTypeId="1" AcceptedAnswerId="1767" CreationDate="2010-12-09T13:25:41.763" Score="2" ViewCount="245" Body="&lt;p&gt;Is well known that to calculate the probability of transition in the scattering processes, as a first approximation, we use the Fermi golden rule. This rule is obtained considering the initial system in an eigenstate of the unperturbed Hamiltonian $H_0$ and considering the time-dependent perturbation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V(t)=\begin{cases} V(\underline{r}),&amp;amp; \mbox{if } t_i&amp;lt;t&amp;lt;t_f \\ 0, &amp;amp; \mbox{otherwise} \end{cases}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $t_f-t_i$ is the interval of interaction, and then considering the limit for $t_i \rightarrow -\infty$, $t_f \rightarrow + \infty$.&#xA;Now, what is the range of validity of this approximation?&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2010-12-09T14:11:26.667" Title="Fermi's Golden Rule" Tags="&lt;quantum-mechanics&gt;&lt;nuclear-physics&gt;&lt;scattering&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="1767" PostTypeId="2" ParentId="1766" CreationDate="2010-12-09T14:11:26.667" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fermi%27s_golden_rule&quot; rel=&quot;nofollow&quot;&gt;Here:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...the assumption that the time of the measurement is much larger than the time needed for the transition...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="386" LastActivityDate="2010-12-09T14:11:26.667" />
  <row Id="1768" PostTypeId="1" CreationDate="2010-12-09T15:17:25.213" Score="3" ViewCount="333" Body="&lt;p&gt;I am in no way experienced in the Physics field so this question may seem a bit silly but i'd appreciate an answer :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why doesn't air freeze?&lt;/p&gt;&#xA;" OwnerUserId="672" LastActivityDate="2010-12-12T00:12:47.230" Title="Why doesn't air freeze?" Tags="&lt;temperature&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="0" />
  <row Id="1769" PostTypeId="2" ParentId="1768" CreationDate="2010-12-09T15:40:56.177" Score="13" ViewCount="" Body="&lt;p&gt;Air is composed of several different gases, but the most important ones are &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrogen&quot; rel=&quot;nofollow&quot;&gt;nitrogen&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Oxygen&quot; rel=&quot;nofollow&quot;&gt;oxygen&lt;/a&gt;. Since these components have very low melting points, you won't see air freeze any time soon on Earth, but on the other hand, liquid nitrogen is produced routinely for all kinds of purposes (like for physics students to toy with during boring lab work).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is now reduced to why nitrogen and oxygen are present in gaseous forms at temperatures we are accustomed to and why they have such low melting points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the first question, temperature on Earth is what it is thanks to the Sun's light that warms us, the Earth's own geothermal heat, but also because of the greenhouse properties of a lot of gases in our air. (So air has a &quot;hand&quot; in warming itself if you want. :p)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second, the metling points of nitrogen and oxygen are so low because the molecules that make up these gases are so tiny and loosely bound. It takes little energy to excite them, therefore at &quot;room temperatures&quot; they are extremely mobile and are basically moving around in straight lines until they bump on each other or on anything else. The characteristic behaviour of molecules in liquids and solids is much less mobile.&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2010-12-09T15:40:56.177" CommentCount="3" />
  <row Id="1770" PostTypeId="2" ParentId="1768" CreationDate="2010-12-09T16:28:04.360" Score="4" ViewCount="" Body="&lt;p&gt;At the normal pressure, 99.9% of air (nitrogen, oxygen and argon) will solidify in 55K (where the oxygen does). Below about 15K also hydrogen (0.000524% of normalized air) solidifies leaving only helium which will be a liquid up to 0K (yet will change into a superfluid about 2.17K).&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-12-09T16:28:04.360" />
  <row Id="1771" PostTypeId="2" ParentId="1763" CreationDate="2010-12-09T18:17:07.480" Score="2" ViewCount="" Body="&lt;p&gt;If you're thinking about a 1-D square well, it doesn't really make sense to worry about angular momentum, since angular momentum necessarily requires multiple dimensions. When it comes to the linear momentum, there may be some subtlety regarding the meaning of expectation values. The expectation value of momentum ($\langle p \rangle $)  is zero for a square-well state, but that does not mean that the particle is not moving. The expectation value of momentum squared ($\langle p^2 \rangle$) is non-zero. The conceptual explanation of this is that the particle is always in motion, but is equally likely to be found moving in either direction. Thus, when you average the momentum of many different realizations to get the expectation value, the momentum averages to zero, but when you remove the directional dependence by squaring the momentum, you get a non-zero value that corresponds to the speed of the particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might allow enough wiggle room to let a photon absorption/emission conserve both energy and momentum, in that the momentum squared increases as you increase the energy state (as it must, the kinetic energy being $\frac{p^2}{2m}$). I'm not sure the numbers really work out, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll also note that when we talk about transitions in atoms at the same sort of level as used in talking about the 1-D infinite square well, we generally ignore the photon momentum-- we talk about electrons moving from one orbital to another, but the selection rules we find are based only on conservation of energy and conservation of angular momentum. The linear momentum carried by the photon is transferred to the atom as a whole, which is the basis for laser cooling, but this is generally neglected at the intro quantum physics level. So, it's not unprecedented for linear momentum to be ignored at this sort of crude level of approximation.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-12-09T18:17:07.480" />
  <row Id="1772" PostTypeId="1" CreationDate="2010-12-09T18:21:48.660" Score="1" ViewCount="170" Body="&lt;p&gt;It seems like everything in the universe is in motion, at least relative to some other object.  That leads me to believe that all objects are in motion.  But how do we measure motion when we are talking about galaxies, is there any reference point that's not moving?&lt;/p&gt;&#xA;" OwnerUserId="771" LastActivityDate="2010-12-10T03:00:50.107" Title="Are all objects in motion?" Tags="&lt;motion&gt;" AnswerCount="4" />
  <row Id="1773" PostTypeId="2" ParentId="1772" CreationDate="2010-12-09T18:51:25.527" Score="0" ViewCount="" Body="&lt;p&gt;Well, all objects are in motion with respect to some other object and none of them are superior frames of reference due to Special Relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you may think of the expansion of the universe. Everything in the universe that is far enough is moving away from every observer in the universe in a macroscopic scale. To clarify what I mean, if an object is far enough from you, you can easily detect that it is moving away from you using redshifting and you would see that everything distant enough is moving away from you without any exceptions, although &quot;distant enough&quot; is dependent on some other factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Motion of galaxies are measured with respect to another galaxy, or if it's the rotation you're talking about it's measured with respect to the rotation axis.&lt;/p&gt;&#xA;" OwnerUserId="504" LastActivityDate="2010-12-09T18:51:25.527" CommentCount="2" />
  <row Id="1774" PostTypeId="2" ParentId="1772" CreationDate="2010-12-09T19:04:33.570" Score="3" ViewCount="" Body="&lt;p&gt;You are correct: everything is in motion (or not) based on the reference frame. Motion is a relative concept, so you are never &quot;moving&quot; but only &quot;moving with respect to something&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find a good basic primer here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Principle_of_relativity&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Principle_of_relativity&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-12-09T19:04:33.570" CommentCount="1" />
  <row Id="1775" PostTypeId="1" AcceptedAnswerId="1778" CreationDate="2010-12-09T23:21:08.810" Score="8" ViewCount="651" Body="&lt;p&gt;If temperature makes particles vibrate faster, and movement is limited by the speed of light, then temperature must be limited as well I would assume. Why there is no limits?&lt;/p&gt;&#xA;" OwnerUserId="213" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-17T19:33:58.843" LastActivityDate="2011-03-17T19:33:58.843" Title="Why is there no absolute maximum temperature?" Tags="&lt;special-relativity&gt;&lt;temperature&gt;" AnswerCount="7" FavoriteCount="4" />
  <row Id="1776" PostTypeId="2" ParentId="1772" CreationDate="2010-12-09T23:28:27.460" Score="0" ViewCount="" Body="&lt;p&gt;But, are all inertial frames really equal? Place two objects (or reference frames if you prefer) close together, but with a high relative velocity. Do they see the same cosmic background radiation? Does the one moving relative to the other see a doplar effect in their measurements of the CMB, and in redshifts of distant objects? If the answer isn't no (i.e. we can detect velocity wrt the CMB), then there is some prefered velocity for each point in spacetime.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2010-12-09T23:28:27.460" />
  <row Id="1777" PostTypeId="2" ParentId="1775" CreationDate="2010-12-09T23:40:03.490" Score="4" ViewCount="" Body="&lt;p&gt;The speed of light is an upper limit for the speed of a massive object, but there is no upper bound on the kinetic energy of an object. In fact, that's why the speed of light is an upper limit (one of many reasons, anyway)-- an object moving at the speed of light would have infinite kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature is a measure of the average kinetic energy of particles in a sample. Since kinetic energy does not have an upper limit, temperature does not have an absolute maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In equations, the kinetic energy is:&#xA;$K=(\gamma - 1)mc^2 = (\frac{1}{\sqrt{1-v^2/c^2}}-1)mc^2$&#xD;&#xA;which becomes infinitely large as v gets very close to the speed of light c.)&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-12-09T23:40:03.490" />
  <row Id="1778" PostTypeId="2" ParentId="1775" CreationDate="2010-12-09T23:40:47.227" Score="12" ViewCount="" Body="&lt;p&gt;I think the problem here is that you're being vague about the limits Special Relativity impose. Let's get this clarified by being a bit more precise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;velocity&lt;/em&gt; of any particle is of course limited by the speed of light &lt;em&gt;c&lt;/em&gt;. However, the theory of Special Relativity does not imply any limit on &lt;em&gt;energy&lt;/em&gt;. In fact, as energy of a massive particle tends towards infinity, its velocity tends toward the speed of light. Specifically,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = \text{rest mass energy} + \text{kinetic energy} = \gamma mc^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\gamma = 1/\sqrt{1-(u/c)^2}$. Clearly, for any energy and thus any gamma, $u$ is still bounded from above by $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that microscopic (internal) energy relates to macroscopic temperature by a constant factor (on the order of the Boltzmann constant), hence temperature of particles, like energy, has no real limit.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2010-12-09T23:40:47.227" CommentCount="6" />
  <row Id="1779" PostTypeId="1" AcceptedAnswerId="1795" CreationDate="2010-12-10T02:08:01.257" Score="11" ViewCount="382" Body="&lt;p&gt;I am a graduate student in experimental physics, currently writing up my dissertation and beginning to apply to post-doc positions.  While it is very easy to find out about open positions in my particular sub-field (via various mailing lists), it is more difficult to find out about  potentially interesting positions in other sub-fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is a good way to identify potentially interesting post-doc position openings outside of one's sub-field?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-04-21T16:36:00.817" Title="What is an effective way to search for post-doc positions in physics?" Tags="&lt;soft-question&gt;&lt;physics-careers&gt;&lt;grad-school&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="9" />
  <row Id="1780" PostTypeId="2" ParentId="1779" CreationDate="2010-12-10T02:10:40.567" Score="3" ViewCount="" Body="&lt;p&gt;In addition to searching for job postings, simply emailing research groups where you would like to work is surprisingly effective.  Even if they do not have an advertised opening, they might make one for you if they want to hire you.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-04-21T16:00:46.313" LastActivityDate="2011-04-21T16:00:46.313" CommentCount="1" />
  <row Id="1782" PostTypeId="2" ParentId="1779" CreationDate="2010-12-10T02:45:49.867" Score="4" ViewCount="" Body="&lt;p&gt;If you are in nuclear or particle physics or a related field you use &lt;a href=&quot;http://www.slac.stanford.edu/spires/jobs/&quot; rel=&quot;nofollow&quot;&gt;Job Search page on Spires&lt;/a&gt; (and not just for postdocs either). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course Physics Today and network like crazy. People who already know your work are likely to believe in you. People who know you adviser or current boss are also good marks.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2010-12-10T02:45:49.867" />
  <row Id="1783" PostTypeId="2" ParentId="1775" CreationDate="2010-12-10T02:54:02.233" Score="3" ViewCount="" Body="&lt;p&gt;While special relativity does not, &lt;em&gt;a priori&lt;/em&gt;, place any constraints on the maximum temperature a system can attain, the situation changes when we consider the quark-gluon plasma - a stage you will eventually reach if you heat up any hadronic matter sufficiently. Rolf Hagedorn realized that for hadronic matter there exists a maximum temperature above which the partition function of the system is not well-defined. In other words you can only heat up &lt;strong&gt;hadronic matter&lt;/strong&gt; to a maximum given by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hagedorn_temperature&quot; rel=&quot;nofollow&quot;&gt;Hagedorn temperature&lt;/a&gt; $T_H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since hadronic-matter constitutes the vast majority of the matter we interact with (excluding dark matter and dark energy), in some sense $T_H$ is the maximum temperature that ordinary matter can attain, though this is by no means the end of the story ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, even with special relativity alone, one can see that when the temperature of a gas of particles becomes comparable to the rest energy of the particles in question, any attempt to increase the temperature beyond that point will only lead to pair creation. This was, vaguely speaking, the reasoning behind Hagedorn's work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also find this &lt;a href=&quot;http://www.pbs.org/wgbh/nova/zero/hot.html&quot; rel=&quot;nofollow&quot;&gt;Nova column&lt;/a&gt; on the Hagedorn phase enlightening.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                              Cheers,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" LastActivityDate="2010-12-10T02:54:02.233" CommentCount="2" />
  <row Id="1784" PostTypeId="2" ParentId="1779" CreationDate="2010-12-10T02:55:37.243" Score="4" ViewCount="" Body="&lt;p&gt;best site for astronomy/astrophysics:&#xA;&lt;a href=&quot;http://members.aas.org/JobReg/JobRegister.cfm&quot; rel=&quot;nofollow&quot;&gt;AAS&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the famous &lt;a href=&quot;http://cdm.berkeley.edu/doku.php?id=astrophysicsjobs&quot; rel=&quot;nofollow&quot;&gt;Rumor Mill&lt;/a&gt; about those jobs&lt;/p&gt;&#xA;" OwnerUserId="775" LastEditorUserId="775" LastEditorDisplayName="" LastEditDate="2010-12-10T03:02:06.093" LastActivityDate="2010-12-10T03:02:06.093" CommentCount="2" />
  <row Id="1785" PostTypeId="2" ParentId="1772" CreationDate="2010-12-10T03:00:50.107" Score="0" ViewCount="" Body="&lt;p&gt;The CMB definitely gives a preferred inertial frame at any point in the universe. However, this gets a little trickier when you include GR, and then if you are sitting in/near a deep potential well (e.g., a cluster of galaxies), the CMB may never appear isotropic, regardless of your velocity&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2010-12-10T03:00:50.107" CommentCount="1" />
  <row Id="1786" PostTypeId="2" ParentId="1775" CreationDate="2010-12-10T04:08:22.620" Score="2" ViewCount="" Body="&lt;p&gt;We have two reasons for there not being a limit. As everyother commenter here has said, SR does not limit the energy per particle. Actually energy per degree of freedom would be a more precise statement. In any case temperature does not equate directly to the energy per particle DOF, but rather to the staistical probabilities, namely that the relative probabilities of a particular state being occupied is proportional to e(- deltaE/kT ). (Even that only applies to the low density limit, fermions are limited to one particle per allowable state, so in some high density low temperature limits (solid state, and degenerate state (some stellar interiors, white dwarfs etc.)) the lowest energy states are almost fully occupied. But, in any case, temperature applies to the probability distribution of the occupation of states with different energies, average energy per particle is just the normalized integral of this density time energy.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-12-10T04:08:22.620" />
  <row Id="1787" PostTypeId="1" CreationDate="2010-12-10T05:55:01.617" Score="17" ViewCount="665" Body="&lt;p&gt;General relativity says that spacetime is a Lorentzian 4-manifold $M$ whose metric satisfies Einstein's field equations.  I have two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What topological restrictions do Einstein's equations put on the manifold.  For instance, the existence of a Lorentz metric implies some topological things, like the euler characteristic vanishing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are there any experiments being done or even any hypothetical experiments that can give information on the topology?  E.g. is there a group of graduate students out there trying to contract loops to discover the fundamental group of the universe?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="37" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-12-10T12:00:29.590" LastActivityDate="2010-12-13T23:36:57.773" Title="What is known about the topological structure of spacetime?" Tags="&lt;general-relativity&gt;" AnswerCount="6" FavoriteCount="5" />
  <row Id="1789" PostTypeId="1" AcceptedAnswerId="1790" CreationDate="2010-12-10T08:05:06.613" Score="1" ViewCount="136" Body="&lt;p&gt;If the induced emf in a circuit is negative, and current from this emf is the emf over the resistance, what happens to the negative sign in the induced emf when solving for the current? Surely there's no such thing as &quot;negative&quot; current?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if we have a constant magnetic field that is pointing down, and the area of a loop is increasing (so that the flux is positive), then the emf will be negative. But if the magnetic field is pointing down, then since the emf opposes a change, it will point up and the current will go counterclockwise. So is this what the negative current is accounting for?&lt;/p&gt;&#xA;" OwnerUserId="202" LastEditorUserId="202" LastEditorDisplayName="" LastEditDate="2010-12-10T08:14:29.083" LastActivityDate="2010-12-10T08:32:10.777" Title="Current from induced emf" Tags="&lt;electromagnetism&gt;&lt;electricity&gt;&lt;electric-circuits&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="1790" PostTypeId="2" ParentId="1789" CreationDate="2010-12-10T08:07:28.857" Score="5" ViewCount="" Body="&lt;p&gt;Negative current just means it flows opposite to the chosen direction.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Take your example, you have a loop with increasing area and downward pointing B-field&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x   x  x  x  x  x  x  x  x&#xA;  +-------------|-----------  ^&#xA;x | x  x  x  x  |  x  x  x    |     y&#xA;  |             | ---&amp;gt;        h     ^&#xA;x | x  x  x  x  |  x  x  x    |     |&#xA;  +-------------|-----------  v     o---&amp;gt; x&#xA;x   x  x  x  x  x  x  x  x         z&#xA;&#xA;  &amp;lt;--- w(t) ----&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Therefore, the flux is $\Phi = hw(t)\mathbf B\cdot \hat{\mathbf z} = -Bhw(t)$ which is &lt;em&gt;negative&lt;/em&gt;. Since the area is increasing, $dw/dt &amp;gt; 0$, so $d\Phi/dt = -Bh\,dw/dt &amp;lt; 0$ is also &lt;em&gt;negative&lt;/em&gt;. The emf $\mathcal E = -d\Phi/dt &amp;gt; 0$ is thus &lt;em&gt;positive&lt;/em&gt;, and the induced current flows counterclockwise as expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sign of flux doesn't just depend on the area, but also the direction of the B-field relative to the surface normal.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As suggested by Mark in the comments, it would be easier to determine the sign using Lenz's law.&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-12-10T08:32:10.777" LastActivityDate="2010-12-10T08:32:10.777" CommentCount="1" />
  <row Id="1791" PostTypeId="1" AcceptedAnswerId="1793" CreationDate="2010-12-10T08:08:56.953" Score="2" ViewCount="187" Body="&lt;p&gt;At least from their names, it seems motional emf is induced, so what's the difference?&lt;/p&gt;&#xA;" OwnerUserId="202" LastActivityDate="2011-02-20T19:47:19.007" Title="What is the difference between induced and motional emf?" Tags="&lt;electricity&gt;" AnswerCount="1" />
  <row Id="1792" PostTypeId="2" ParentId="1787" CreationDate="2010-12-10T08:15:58.103" Score="7" ViewCount="" Body="&lt;p&gt;That's a &lt;strong&gt;great&lt;/strong&gt; question ! What you are asking about is one of the missing links between classical and quantum gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On their own, the Einstein equations are local field equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ G_{\mu\nu} = 8 \pi G T_{\mu\nu} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and do not contain any topological information. At the level of the action principle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S_{eh} = \int_\mathcal{M} d^4 x \sqrt{-g} \mathbf{R} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the term we generally include is the Ricci scalar $ \mathbf{R} = Tr[ R_{\mu\nu} ] $, which depends only on the first and second derivatives of the metric and is, again, a local quantity. So the action does not tell us about topology either, unless you're in two dimensions, where the Euler characteristic is given by the integral of the ricci scalar:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int d^2 x \mathcal{R} = \chi $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(modulo some numerical factors). So gravity in 2 dimensions is &lt;strong&gt;entirely&lt;/strong&gt; topological. This is in contrast to the 4D case where the Einstein-Hilbert action appears to contain no topological information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should cover your first question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All is not lost, however. One can &lt;strong&gt;add&lt;/strong&gt; topological degrees of freedom to 4D gravity by the addition of terms corresponding to various topological invariants (Chern-Simons, Nieh-Yan and Pontryagin). For instance, the Chern-Simons contribution to the action looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S_{cs} = \int d^4 x {}^\star R \, R $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $ R \equiv R_{abcd} $ is the Riemann tensor and $ {}^\star R_{abcd} = 1/2 \epsilon_{ab}{}^{ij} R_{cd\,ij} $ is its dual. Here is a very nice paper by &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0308071&quot; rel=&quot;nofollow&quot;&gt;Jackiw and Pi&lt;/a&gt; for the details of this construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's plenty more to be said about topology and general relativity. Your question only scratches the surface. But there's a goldmine underneath ! I'll let someone else tackle your second question. Short answer is &quot;yes&quot;.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-10T08:15:58.103" CommentCount="2" />
  <row Id="1793" PostTypeId="2" ParentId="1791" CreationDate="2010-12-10T08:27:23.137" Score="3" ViewCount="" Body="&lt;p&gt;Physically, they're the same, but historically this was not understood until special relativity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a loop of wire and a bar magnet.  In one experiment, we move the magnet towards the loop of wire.  The flux through the loop increases. The loop now has a current, meaning there is an electric field pointing around the loop.  This is induced EMF.  Induced EMF is calculated using the change in flux through a stationary loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a second experiment, we hold the magnet still and move the loop towards it.  We start the same distance away and move the same speed, but change the role of stationary and moving agents.  Now, $\partial B/\partial t = 0$ because the magnetic field is not changing, so there is no induction.  Instead, there is a Lorentz force on the charge carriers in the loop because they are moving through a magnetic field.  The induced current works out to be exactly the same as before, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physically, the same thing is happening, but the mathematical description is different depending on whether you view it from the rest frame of the magnet or the rest frame of the loop.  One of the most important physics papers ever, Einstein's &lt;a href=&quot;http://www.fourmilab.ch/etexts/einstein/specrel/www/&quot; rel=&quot;nofollow&quot;&gt;&quot;On the Electrodynamics of Moving Bodies&quot;&lt;/a&gt;, begins by pointing out this strange coincidence.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;It is known that Maxwell's electrodynamics--as usually understood at the present time--when applied to moving bodies, leads to asymmetries which do not appear to be inherent in the phenomena. Take, for example, the reciprocal electrodynamic action of a magnet and a conductor. The observable phenomenon here depends only on the relative motion of the conductor and the magnet, whereas the customary view draws a sharp distinction between the two cases in which either the one or the other of these bodies is in motion. For if the magnet is in motion and the conductor at rest, there arises in the neighbourhood of the magnet an electric field with a certain definite energy, producing a current at the places where parts of the conductor are situated. But if the magnet is stationary and the conductor in motion, no electric field arises in the neighbourhood of the magnet. In the conductor, however, we find an electromotive force, to which in itself there is no corresponding energy, but which gives rise--assuming equality of relative motion in the two cases discussed--to electric currents of the same path and intensity as those produced by the electric forces in the former case.&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Special relativity allows us to understand this problem by describing how different observers perceive magnetic and electric fields when they are in motion relative to each other.  If you walk towards me down a hallway, we will measure different electric and magnetic fields in the hallway, even when you're right next to me as you walk past. Once we understand how to transform the magnetic field of the bar magnet, we see that induced and motional emf are just transformed versions of the same thing, accurately described by special relativity.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-12-10T08:40:40.307" LastActivityDate="2010-12-10T08:40:40.307" CommentCount="3" />
  <row Id="1794" PostTypeId="2" ParentId="1787" CreationDate="2010-12-10T08:28:02.740" Score="2" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Einstein equations describe local structure of the space-time. They contain no global or topological information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I heard that some restricitons on the &lt;em&gt;scale&lt;/em&gt; of topology can be derived from curvature of the Universe if the curvature is negative. (Something like &quot;scale = integer multiple of 1/curvature&quot;.)  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Well, if our space has non-trivial topology, then light rays will &quot;wrap around&quot; our universe multiple times and you'll be able to see the same (similar) copies of galaxies. I heard of people searching for such similarities without success. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also nontrivial topology must result in some correlation in CMB -- no such correlations were found (yet?) either.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="386" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-12-10T12:02:20.350" LastActivityDate="2010-12-10T12:02:20.350" CommentCount="3" />
  <row Id="1795" PostTypeId="2" ParentId="1779" CreationDate="2010-12-10T10:15:36.790" Score="5" ViewCount="" Body="&lt;p&gt;I think this is a proper question to be answered with wiki.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Resources For Job seeker&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;General:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://jobs.phds.org/&quot; rel=&quot;nofollow&quot;&gt;Jobs for PhD&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://academicjobsonline.org/&quot; rel=&quot;nofollow&quot;&gt;Academic jobs&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Physics:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.slac.stanford.edu/spires/jobs/&quot; rel=&quot;nofollow&quot;&gt;SPIRES&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://careers.aps.org/jobs/&quot; rel=&quot;nofollow&quot;&gt;American Physical Society (APS)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Astrophysics:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://members.aas.org/JobReg/JobRegister.cfm&quot; rel=&quot;nofollow&quot;&gt;American Astronomical Society (AAS)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://eas.unige.ch/display_job.jsp&quot; rel=&quot;nofollow&quot;&gt;European Astronomical Society (EAS)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.esa.int/hr/vn.htm&quot; rel=&quot;nofollow&quot;&gt;European Space Agency (ESA)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://jobs.eso.org/&quot; rel=&quot;nofollow&quot;&gt;European Southern Observatory (ESO)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Gravitational waves:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://gwave.wikispot.org/Jobs&quot; rel=&quot;nofollow&quot;&gt;Gravitational wave detection jobs wiki&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="518" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-04-21T15:52:02.217" LastActivityDate="2011-04-21T15:52:02.217" CommunityOwnedDate="2010-12-10T10:15:36.790" />
  <row Id="1796" PostTypeId="1" CreationDate="2010-12-10T10:24:08.483" Score="5" ViewCount="119" Body="&lt;p&gt;Where do you check and put them usually?&#xA;Let's make a wiki ~&lt;/p&gt;&#xA;" OwnerUserId="518" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-10T11:04:55.717" LastActivityDate="2011-01-14T08:56:30.457" Title="Where can we find information of International Physics/Astrophysics conferences?" Tags="&lt;astrophysics&gt;&lt;soft-question&gt;&lt;physics-careers&gt;" AnswerCount="5" FavoriteCount="4" />
  <row Id="1797" PostTypeId="1" CreationDate="2010-12-10T11:02:04.200" Score="2" ViewCount="124" Body="&lt;p&gt;Many experiments, such as optical, superconductivity, etc, use the samples that involve rare earth metals and transition metals. Why are they used that often. Is the main reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;They have the required electronic structure in its d- and f- orbital so that we can create the required spin-spin interaction.&lt;/li&gt;&#xA;&lt;li&gt;They provide the required structure of the crystal&lt;/li&gt;&#xA;&lt;li&gt;They provide the required energy band gap&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Any elaboration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Here the samples I mean is the usual YBCO, or some papers, say &lt;a href=&quot;http://arxiv.org/abs/1012.2102&quot; rel=&quot;nofollow&quot;&gt;Gd5Si2Ge2&lt;/a&gt;, that I do not have any idea. I am also curious about why they decide to investigate a particular compound in the first place?&lt;/p&gt;&#xA;" OwnerUserId="562" LastEditorUserId="562" LastEditorDisplayName="" LastEditDate="2010-12-10T13:35:32.653" LastActivityDate="2010-12-10T19:33:22.620" Title="Experimental samples with rare earth metal" Tags="&lt;solid-state-physics&gt;&lt;experiment&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="1798" PostTypeId="2" ParentId="1796" CreationDate="2010-12-10T11:05:43.120" Score="0" ViewCount="" Body="&lt;p&gt;If you mean conference with international people, I think &lt;a href=&quot;http://www.aps.org/meetings/march/&quot; rel=&quot;nofollow&quot;&gt;March meeting&lt;/a&gt; should be one of them.&lt;/p&gt;&#xA;" OwnerUserId="562" LastActivityDate="2010-12-10T11:05:43.120" />
  <row Id="1799" PostTypeId="1" AcceptedAnswerId="2082" CreationDate="2010-12-10T15:06:34.387" Score="8" ViewCount="707" Body="&lt;p&gt;I would be very happy if someone could take a stab at conveying what conformal blocks are and how they are used in conformal field theory (CFT). I'm finally getting the glimmerings of understanding from reading &lt;a href=&quot;http://www.physics.rutgers.edu/~gmoore/MooreReadNonabelions.pdf&quot; rel=&quot;nofollow&quot;&gt;Moore and Read's&lt;/a&gt; wonderful paper. But I think/hope this site has people who can explain the notions involved in a simpler and more intuitive manner.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: Here is a simple example, taken from pg 8 of the reference cited above ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a 2D CFT we have correlation functions of fields $ \phi_i(z,\bar z) $, (where $ z = x+\imath y$) at various points on the complex plane. The n-point correlation function can be expanded as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left \langle \prod_{a=1}^n \phi_{i_a}(z_a,\bar z_a) \right \rangle = \sum_p | F_{p\; i_{1} \dots i_n}(z_{1} \dots z_n)|^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $p$ labels members of a basis of functions $ F_{p\; i_1 \dots i_n}(z_{1} \dots z_n) $ which span a vector space for each n-tuple $(z_{1} \dots z_n)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These functions $F_p$ are known as conformal blocks, and appear to give a &quot;fourier&quot; decomposition of the correlation functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I've gathered so far. If someone could elaborate with more examples that would be wonderful !&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: It is proving very difficult to decide which answer is the &quot;correct&quot; one. I will give it a few more days. Perhaps the situation will change !&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The &quot;correct&quot; answer goes to (drum-roll): David Zavlasky. Well they are all great answers. I chose David's for the extra five points because his is the simplest, IMHO. He also mentions the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cross-ratio&quot; rel=&quot;nofollow&quot;&gt;&quot;cross-ratio&quot;&lt;/a&gt; which is a building block of CFT.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-24T21:54:15.493" Title="A pedestrian explanation of conformal blocks" Tags="&lt;mathematical-physics&gt;&lt;condensed-matter&gt;&lt;conformal-field-theory&gt;" AnswerCount="4" CommentCount="23" FavoriteCount="4" />
  <row Id="1800" PostTypeId="2" ParentId="1683" CreationDate="2010-12-10T15:23:51.127" Score="5" ViewCount="" Body="&lt;p&gt;OK, that is my second tentative to solve this problem. I think I have a solution this time, thank to the discussion of others in that thread. The solution is $v_{\text{final}}=\sqrt{2gh(0)}\frac{ls}{h(0)S}(1-\frac{\pi}{2})$ if $m\gg M$. This corresponds to a few millimetres per second towards the left for a wagon full of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is how I've derived it :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order not to neglect not negligible contributions, I will pose the problem for a cart of a quite arbitrary shape, before restricting it to our cart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$S(z)$ : section of the cart at altitude $z$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$h(t)$ : height of water at time $t$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$l(z)$ : abscissa of the centre of mass (CoM)  of the slice of water at altitude $z$&#xD;&#xA; - $M$ : Mass of the empty cart&lt;/li&gt;&#xA;&lt;li&gt;$m = \int_0^h(0) dz S(z) \rho$ : initial mass of water&lt;/li&gt;&#xA;&lt;li&gt;$\mu(t)$ : remaining mass of water at time $t$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$f(t)=-d&amp;#181;/dt &amp;gt; 0$ is the mass flow of water&lt;/li&gt;&#xA;&lt;li&gt;$v_v(z,t) &amp;lt; 0$ : vertical speed of the water slice at altitude $z$ &lt;/li&gt;&#xA;&lt;li&gt;$v_h(z,t)$ : horizontal speed of its CoM.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the case of the cart, we will have :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$S(z)$ is constant above the nozzle. Let $\delta+\epsilon$ be the nozzle height. We then have $S(z)=S$ for $z&amp;gt;\delta+\epsilon$. For numerical appplications, we'll suppose a $3\times3\times10$ m³ cart, with $S=30$ m².&lt;/li&gt;&#xA;&lt;li&gt;The last part of the nozzle is a pipe of height $\delta\ll h(0)$. In this pipe $S(z&amp;lt;\delta)= s\ll S$. If the output has a 10 cm side, $s=1O^{-2}$ m².&lt;/li&gt;&#xA;&lt;li&gt;$h(0) = 3$ m&lt;/li&gt;&#xA;&lt;li&gt;Above the nozzle, the CoM of the water is fixed at $l(z&amp;gt;\delta+\epsilon)=0$, while in the lower part, $l(z&amp;lt;\delta)=-l$, wher $l=5$ m.&lt;/li&gt;&#xA;&lt;li&gt;I'll assume $M=10^4$ kg, but I've no idea whether it's realistic.&lt;/li&gt;&#xA;&lt;li&gt;$\rho = 10^3$ kg·m⁻³&lt;/li&gt;&#xA;&lt;li&gt;$m=\rho S h(0) =$ 9·10⁴ kg&lt;/li&gt;&#xA;&lt;li&gt;$g=10 m&amp;#183;s⁻&amp;#178;$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Vertical movement of water&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following, we will assume that the horizontal acceleration $a$ of the cart stays $a\ll g$ during the movement. A nonzero acceleration would induce correction terms proportional to $\frac{a^2}{g^2}$, and we will check that this hypothesis is consistent later. This assumption allows us to neglect any motion of the cart when looking at the movement of water in the cart referential, and then compute $f(t)$, $h(t)$ and $\mu(t)$. We will then use the resulting f this computation to find the horizontal movement of the cart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The incompressibility of water allows us to write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(t)=-\rho S(z,t) v_v(z,t) =\rho S(h(t)) \frac{dh}{dt} =- \rho s v_v(0,t)  \quad(*)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bernoulli, at altitude $h$ and $0$ gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{gather}&#xA;\left(\frac{dh}{dt}\right)^2 +  2gh = (v_v(0,t))^2 \\ &#xA;2gh=(dh/dt)² (\frac{S(h)²}{S(0)}² -1)&#xA;\end{gather}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case, except in the nozzle, $\frac{S(h)^2}{S(0)^2}=\frac{S^2}{s^2}\simeq 10^7$. We will therefore neglect the $-1$ in the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation has the following solution :&#xA;$$ h(t)=h(0)(1-t/t_m)^2 \text{ for } t\in[0, t_m]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $h(t&amp;gt;t_m)=0$, with $t_m=\frac Ss \sqrt{2h(0)/g}$. Here $t_m=3\cdot 10^3 \sqrt(6/10) \sim 2000$ s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have then $\mu(t)=m (1-t/t_m)^2$ and $f(t)=f(0)(1-t/t_m)$ with $f(0)=\rho s \sqrt{2gh(0)}\sim=10^{-2+3}\sqrt{60}\sim80$ kg·s⁻¹.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conservation of the horizontal momentum&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now comes the interesting part of the problem, the horizontal movement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Momenta will be computed in the cart referential ($P^{CR}$) and in the rail referential ($P^RR$).If you look at the water inside the cart, its momentum will be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P^{CR}_{\text{water}}=\rho\int_{0}^{h(t)}dz S(z) v_h(z,t)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $v_h(z,t)= dl/dz v_v(z,t)$.  From that and the expression $(*)$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P^{CR}_{\text{water}}=- f(t) \int_{0}{h(t)}dz  dl/dz= f(t) (l(0)-l(h(t))).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going back to the more physical rail-refrential, we have then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P^{RR}_{\text{water}}=&amp;#181;(t)v(t) + f(t) (l(0)-l(h(t)))$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also have, for the cart,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P^{RR}_{\text{cart}}=M v(t)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As stated in other answers (but not my previous one :-(), one should not forget the momentum of the water which has left the cart in previous time :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P^{RR}_{\text{leaked water}}=\int_0^t d\tau f(\tau) v(\tau)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing these term, together with the momentum conservation, we have :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0=P^{RR}_{\text{total}}=(M+\mu(t))v(t) + f(t) (l(0)-l(h(t))) + \int_0^t d\tau f(\tau) v(\tau) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example when the cart is empty, $f(t)=0$, $\mu(t)=0$ and the above equations becomes :&#xA;$$ 0=P^{RR}_{\text{total}}=Mv_{\text{final}} + \int_0^t d\tau f(\tau) v(\tau) $$&#xD;&#xA;The cart can have a final nonzero speed, if its momentum is compensated by the net momentum of the water having left the cart. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiating the momentum conservation relatively to $t$, we obtain,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0=(M+&amp;#181;(t))\frac{dv}{dt} - f(t) v(t) + \frac{df}{dt}(l(0)-l(h(t))) - f(t) \frac{dh}{dt} \frac{dl}{dz} + f(t) v(t)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation can be simplified into&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{dv}{dt}=\frac{1}{M+\mu(t)}\left[\frac{df}{dt}[l(h(t))-l(0)] - \frac{dl}{dz}\frac{f(t)^2}{\rho S(h(t))}\right] $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing $f(t)$ as per the previous section allows us to integrate this equation, at least numerically, for any cart. In the following, we solve the equation for our cart geometry, distinguishing three steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 1: opening the nozzle&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the nozzle is quickly opened at $t=0$, the cart is full and $\mu=m$ is constant. the equation we have to solve is then&#xA;$$\frac{dv}{dt}=\frac{1}{M+m}\frac{df}{dt}l-0 $$&#xD;&#xA;from which we easily deduce&#xA;$$\Delta v = \frac{l\Delta f}{M+m}=\frac{lf(0)}{M+m}.$$&#xD;&#xA;With the numerical values above, this corresponds to a speed of 4 mm·s⁻¹. This movement of the cart compensate the internal acceleration of the water inside the cart towards the nozzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As wee will see later, this abrupt speed change is the biggest acceleration taken by the cart. If the nozzle is opened in one second, which is still quickly enough to keep $\mu=m$ approximation valid, the horizontal acceleration $a$ is still small : $\frac{a}{g}=4\cdot10^{-4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2: Emptying the cart above the nozzle&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above the nozzle, we have a constant $l(h)=0$ and the differential equation is&#xA;$$\frac{dv}{dt}=\frac{l}{M+\mu(t)}\frac{df}{dt}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the cart is emptied with a constant $f(t)$, it does not accelerate nor slow down, until the f(t) is cut. In that moment the back action is the same in a reverse direction, but with a lower mass. (M instead of M+m). We end therefore with a net speed towards the left of value $lf(1/(M+m)-1/M)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the more general case where f slowly decrease to 0, $df/dt &amp;lt;0$, implying a slow down, and indeed a reversal of the speed, since the total mass $M+&amp;#181;(t)$ decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we plug into the above equation the values we have for $f(t)$ and $\mu(t)$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dv}{dt}=\frac{lf(0)}{t_m(M+m(1-t/t_m)^2)}=-g\frac{ls^2m}{h(0)S^2M}\frac1{1+\frac mM(1-t/t_m)^2}$$&#xD;&#xA;which can be analytically integrated using $\int dt/(1+t^2)= \arctan t$. We have then&#xA;$$v(t)-v(0)=-\frac{ls}{h(0)S}\sqrt{2gh(0)}\left[\arctan\sqrt{\frac mM} - \arctan\left(\frac{t_m-t}{t_m}\sqrt{\frac mM}\right)\right]$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have then &#xA;$$v(t_m)=v(0)-\frac{ls}{hS}\sqrt{2gh(0)}\arctan\sqrt{\frac mM}$$&#xD;&#xA;In the limit $m\gg M$, where the mass of water is larger than the cart mass, $\arctan\sqrt{m/M}\simeq\pi/2$ and $v(0)=\sqrt{2gh(0)}\frac{ls}{h(0)S}$, so that :&#xA;$$v(t_m)=\sqrt{2gh(0)}\frac{ls}{h(0)S}(1-\frac{\pi}{2})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;step 3: Showing that the nozzle has no influence, so long at it is small&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem of the nozzle is the zone where $\frac{dl}{dz}$ is not small. let say that this zone is of height $\epsilon$, above a vertical pipe of height $\delta$, with $\epsilon\ll\delta\ll h(0)$. I have the intuition that the problem is not so dangerous, since the $\propto l/\epsilon$ derivative will be only relevant for a time proportional to $\epsilon$, and the small amount of water involved should keep the corrective term small. But I have nothing more rigorous yet :-(&lt;/p&gt;&#xA;" OwnerUserId="373" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-12-10T16:32:52.203" LastActivityDate="2010-12-10T16:32:52.203" CommentCount="28" />
  <row Id="1801" PostTypeId="1" AcceptedAnswerId="1833" CreationDate="2010-12-10T18:27:55.907" Score="5" ViewCount="297" Body="&lt;p&gt;If space itself is expanding, then why would it have any effect on matter (separates distant galaxies)? &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Space is &quot;nothing&quot;, and if &quot;nothing&quot; becomes bigger &quot;nothing&quot; it's still a &quot;nothing&quot; that shouldn't interact with matter in any way (it doesn't have mass, energy, etc).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Gravity doesn't have a cutoff distance afaik, so even the most distant galaxies should be attracted to each other. Gravity force would be very very tiny, but it would still dominate &quot;nothing&quot; from space expansion.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lets take inertia from Big Bang into account. Inertia would be the primary force that moves galaxies away comparing to their tiny gravity and even more tiny, if any, force of our &quot;nothing&quot; that is still expanding inbetween. Wouldn't expansion decelerate if driven mostly by inertia?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="213" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-01-17T19:33:36.370" LastActivityDate="2011-03-31T17:54:58.270" Title="Why space expansion affects matter?" Tags="&lt;general-relativity&gt;&lt;gravity&gt;&lt;space-expansion&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" />
  <row Id="1802" PostTypeId="2" ParentId="1801" CreationDate="2010-12-10T19:01:36.650" Score="4" ViewCount="" Body="&lt;h2&gt;Space is &quot;nothing&quot;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Not really. General relativity tells us that space-time is an actual living entity that responds to what's matter doing and matter responds to the way space-time is curved. Or as &lt;a href=&quot;http://en.wikipedia.org/wiki/John_Archibald_Wheeler&quot; rel=&quot;nofollow&quot;&gt;John Wheeler&lt;/a&gt; put it: &lt;em&gt;&quot;matter tells Spacetime how to curve, and Spacetime tells matter how to move.&quot;&lt;/em&gt; For an introduction into general relativity, you can start at this &lt;a href=&quot;http://en.wikipedia.org/wiki/Introduction_to_general_relativity&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Gravity doesn't have a cutoff distance&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You are talking just classical Newtonian gravity. But this isn't how gravity works on big scales. On big scales gravitation is just an appearance caused by the fact that the space-time is curved. If it is curved in a certain way then the appearance is that the objects attract each other. Like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/LWTI6.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But on huge scales these local, attractive, deformations are completely negligible. When you take whole space-time into account, it is curved differently (more onto this in the last section). And the important fact is that the space-time expands (like when you blow into a balloon). Because it expands, the distances between any two points are getting bigger and it seems like everything is getting away from you. So that gravitation is actually a repulsive force on big scales (or at least, it can be; again see the last section for a clarification).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Wouldn't the expansion decelerate?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There are many solutions of Eistein's equations. Some of them are stationary, some of them are expanding (either at accelerated rate or decelerated rate), some of them revert from expansion to retraction. Now, which of these solutions is correct depends on the precise conditions at (or more precisely shortly after) the big bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Important part of the present theoretical understanding of the space-time expansion is that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmological_constant&quot; rel=&quot;nofollow&quot;&gt;cosmological constant&lt;/a&gt; is non-zero and this is what accelerates the expansion. The complete picture about the matter content of our universe and its relation to the expansion of the universe is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambda%2dCDM_model&quot; rel=&quot;nofollow&quot;&gt;Lambda-CDM&lt;/a&gt; (Lambda is an alias for the cosmological constant and CDM is a cold dark matter, which makes up most of the matter in the universe). Now, cosmological constant is also called &lt;a href=&quot;http://en.wikipedia.org/wiki/Dark_energy&quot; rel=&quot;nofollow&quot;&gt;dark energy&lt;/a&gt;. But currently nothing much is known about these matters and major conceptual problems in modern theoretical physics and cosmology have to do with the nature of the cosmological constant.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-10T19:21:28.217" LastActivityDate="2010-12-10T19:21:28.217" CommentCount="4" />
  <row Id="1803" PostTypeId="2" ParentId="1797" CreationDate="2010-12-10T19:33:22.620" Score="1" ViewCount="" Body="&lt;p&gt;It's the open d- and f-shells that make for very interesting physics, because it can introduce effects that dramatically depend on doping. Look up &quot;Strongly Correlated Materials&quot; for an overview.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2010-12-10T19:33:22.620" CommentCount="2" />
  <row Id="1804" PostTypeId="2" ParentId="1392" CreationDate="2010-12-10T19:38:55.073" Score="3" ViewCount="" Body="&lt;p&gt;There was a similar question at &lt;a href=&quot;http://mathoverflow.net/questions/46417/strategies-for-digging-through-literature&quot; rel=&quot;nofollow&quot;&gt;Mathoverflow&lt;/a&gt;. I think it contains some useful references and discussion so it's definitely worth checking out.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Usually I would post this as a comment under the question but probably nobody would notice it anymore. And seeing that the other answer is also quite short I hope this is fine.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-10T19:38:55.073" />
  <row Id="1805" PostTypeId="2" ParentId="1801" CreationDate="2010-12-10T22:56:20.073" Score="1" ViewCount="" Body="&lt;p&gt;I actually think you've been misled. Usually (in my experience) when people say that space is expanding, what they mean is that the &lt;em&gt;universe&lt;/em&gt; is expanding - but that just means that the objects in the universe are getting further apart (or, in the case of a continuous mass/energy distribution, that the density is decreasing with time). This is what the cosmological scale factor $a(\tau)$ is for: it describes the change in the distance between two objects (like galaxies) whose motion is subject to large-scale interactions only. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r(t_1) = \frac{a(t_1)}{a(t_2)}r(t_2)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, because $a(\tau)$ is part of the metric, which specifies the distortion (or &quot;curvature&quot;) of spacetime, it's easy to think that this scale factor would characterize the expansion of space itself. But I think that's the wrong interpretation, or at least a confusing interpretation. The scale factor really just relates to measurable distances between objects. So instead of trying to figure out what it means for space to expand, just think about things getting further apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving on, you're right to say that, if things behave the way we would intuitively expect them to, the expansion of the universe &lt;em&gt;should&lt;/em&gt; be slowing down due to gravitational attraction. But the best experimental observations that I am aware of show that the opposite is true: the expansion is speeding up. Unless you're prepared to argue that the experiments were performed or interpreted incorrectly, that means that something nonintuitive must be going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's still a lot of debate about what exactly could be making the expansion of the universe speed up. Whatever it is, cosmologists are calling it dark energy (back in Einstein's day they called it the &quot;cosmological constant&quot;), but nobody has a satisfactory explanation of why this dark energy exists, or what its properties might be, other than the fact that it makes the universe's expansion speed up. This is one of the biggest open questions in modern cosmology.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-12-10T22:56:20.073" CommentCount="17" />
  <row Id="1806" PostTypeId="2" ParentId="1787" CreationDate="2010-12-11T03:45:56.653" Score="6" ViewCount="" Body="&lt;p&gt;I don't know the answer, but your intuition is right on -- the fact that the &lt;em&gt;equations&lt;/em&gt; are local doesn't mean that there cannot be a constraint on the topology of a global solution.  For example, in Euclidean signature, $R_{ij} = g_{ij}$ immediately implies that the scalar curvature is positive, which in turn leads to topological constraints.  If the four-manifold is Einstein &lt;em&gt;and&lt;/em&gt; complex, then it must be a del Pezzo surface (highly constrained).  I don't know much about Lorentzian signature, but I know that the PDE's are a whole different beast.  I have seen a few results about classification of possible holonomy groups of Lorentzian Einstein manifolds, but I don't know anything global (I actually don't know anything at all).&lt;/p&gt;&#xA;" OwnerUserId="20" LastActivityDate="2010-12-11T03:45:56.653" />
  <row Id="1807" PostTypeId="2" ParentId="1787" CreationDate="2010-12-11T04:01:16.263" Score="2" ViewCount="" Body="&lt;p&gt;These are two independent questions, one mathematical, and one about observations.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What constraints do the Einstein equations imply about the global structure of space and/or spacetime? I don't know the general answer, my impression is that not as much as is known about Lorentzian manifolds as about Euclidean manifolds. Furthermore, there is no reason to suspect the space/spacetime is singularity-free (at the very least we know of many black holes in the universe), and I doubt much can be said about the global structure of any manifold if you allow for singularities.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;About observational physics: the only observable I can think of that is sensitive to global structure is the low multipoles of the CMB, and every now and again there are papers on the subject, to explain anomalies in such multipoles (e.g. stories about football-shaped universe). Alas, cosmic variance limits how seriously you can take such observations and models aimed to explain them.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="566" LastActivityDate="2010-12-11T04:01:16.263" />
  <row Id="1808" PostTypeId="2" ParentId="357" CreationDate="2010-12-11T04:08:44.503" Score="7" ViewCount="" Body="&lt;p&gt;You could find a capacitor and read of its capacitance, alternately build one and measure it, and measure its dimensions. Now you can get a good estimate on the permitivity of vacuum, epsilon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are possibly other intricate ways to measure this number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed of light is then given by a relation involving another number, the vacuum permeability, µ , which needs no measuring as it is defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This relation can be derived from Maxwell's equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c=\frac{1}{\sqrt{\varepsilon&amp;#181;}}$&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-07-23T17:58:13.530" LastActivityDate="2011-07-23T17:58:13.530" CommentCount="1" />
  <row Id="1809" PostTypeId="2" ParentId="1683" CreationDate="2010-12-11T04:25:10.863" Score="-2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;A quantitative answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three main conservation laws of fluid mechanics are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Conservation of mass&lt;/li&gt;&#xA;&lt;li&gt;Conservation of momentum&lt;/li&gt;&#xA;&lt;li&gt;Conservation of energy&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fluid_dynamics#Equations_of_fluid_dynamics&quot; rel=&quot;nofollow&quot;&gt;Reference&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between the time $t$ and $t+\mathrm{d}t$ a mass of water $\mathrm{d}m(t)$ escapes through the nozzle. The mass escapes at a speed governed by Torricelli's law - obtained through 1. and 3.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v(t) = \sqrt{2gh(t)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The direction of the water is determined by the inclination of the nozzle $\theta$ which we may generalize to vary from $0$ radians (horizontal, pointing left) to $\frac{\pi}{2}$ (vertical, pointing down).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{v}(t) = -v(t) \pmatrix{ \sin \theta \\ \cos \theta }$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The momentum of the water flowing out is determined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbf{p}(t) = m(t) \mathbf{v}(t) = -m(t)v(t) \pmatrix{ sin \theta \\ cos \theta }$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbf{p}(t) = p(t) \pmatrix{ sin \theta \\ cos \theta }$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the fluid is incompressible and mass is conserved, the mass flowing out corresponds to an equivalent decrease in the amount of water from the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm{d}m(t) = \rho S \mathrm{d}h(t)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But also, the water will flow at a speed $v(t)$ at the nozzle, so the water that escapes is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm{d}m(t) = \rho s v(t)\mathrm{d}t$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S \mathrm{d}h(t) = s v(t)\mathrm{d}t$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm{d}h(t) = \frac{s}{S}v(t)\mathrm{d}t$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging in the equation for $v(t)$ and introducing $\sigma=\frac{s}{S}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm{d}h(t) = -\sigma \sqrt{2g h(t)} \mathrm{d}t$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;solving this first-order nonlinear ordinary differential equation and using $h_0 = h(t=0)$ and $v_0 = v(t=0) = \sqrt{2gh_0}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h(t) = \frac{1}{2}g\sigma^2t^2 - v_0\sigma t+h_0 \approx h_0 - v_0\sigma t$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This lets us find $v(t)$, $m(t)$ and $\mathbf{p}(t)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v(t) \approx -\sqrt{2gh_0 - 2gv_0\sigma t}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}m}{\mathrm{d}t} = \rho s v(t) \approx - \rho s \sqrt{2gh_0 - 2gv_0\sigma t}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is solved by the (approximate) solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m(t) \approx m(t) = C +\frac{2 \rho s \sqrt{2g} (h_0-\sigma v_0 t)^{\frac{3}{2}}}{3 \sigma v_0}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: an analytical solution exists, but it's really ugly&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate $C$ we must use the condition that when all the water is gone, $m(t) = 0$. To do so we can solve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0=h(t)\approx h_0 - v_0\sigma t \implies t_f \approx \frac{h_0}{v_0 \sigma}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 = m(t = t_f) = C +\frac{2 \rho s \sqrt{2g} (h_0-\sigma v_0 \frac{h_0}{v_0 \sigma})^{\frac{3}{2}}}{3 \sigma v_0}\implies C=0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m(t) \approx \frac{2 \rho s \sqrt{2g} (h_0-\sigma v_0 t)^{\frac{3}{2}}}{3 \sigma v_0} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;finally, the magnitude of the linear momentum is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ p(t) = m(t)v(t) \approx -\frac{2 \rho s \sqrt{2g} (h_0-\sigma v_0 t)^{\frac{3}{2}}}{3 \sigma v_0} \sqrt{2gh_0 - 2gv_0\sigma t}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's see the effect of the two components of $\mathbf{p}$. The &lt;strong&gt;horizontal&lt;/strong&gt; component propels the wagon by reaction; the &lt;strong&gt;vertical&lt;/strong&gt; component creates a torque that pushes the center of mass to the right - note that the flow pushes the center of mass to the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\theta = 0$, all the linear momentum is horizontal. There is no torque, the wagon will move by reaction and the center of mass doesn't move because the water flowing out and the wagon move in opposite directions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_{wagon} = p(t)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\theta = \frac{\pi}{2}$ all the linear momentum is vertical. There will be a torque but no horizontal movement, as there is no horizontal momentum. This implies that the contributions to the center of mass by the water flowing out and the torque must cancel out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if $\theta$ has a middle value, a compositions of the two behaviours will occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In regards to the problem, $\theta = \frac{\pi}{2}$ and therefore the wagon will not move&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-12-11T04:25:10.863" CommentCount="14" />
  <row Id="1811" PostTypeId="2" ParentId="357" CreationDate="2010-12-11T07:15:56.670" Score="6" ViewCount="" Body="&lt;p&gt;I can't think of a way to do it with &quot;common household tools&quot; but if you have an oscilloscope, a laser diode, a couple of photo-sensors, a beam splitter, you can do it.  All of these things are readily available from science supply/hobby stores online, but not usually in most homes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set up the laser diode to hit the beam splitter and be split into two beams.  Set up the two beams so that they hit two photo-sensors, but make one of the photo-sensors exactly &lt;strong&gt;twice&lt;/strong&gt; the distance from the beam splitter as the other.  This will create two separate paths for the light, one twice as long as the other.  Run the output of the photo-diodes into two channels of the oscilloscope.  Switch on the laser diode, and you should see two pulses on the o-scope, one from each of the two laser diodes.  The difference between them is the time it takes the light beam to travel the distance of the &lt;strong&gt;difference&lt;/strong&gt; in the two paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason to do it this way is accuracy - if you only had one beam, and your photo-diode took, say, 1 microsecond longer to turn on than what was in the documentation, or your laser were slow to turn on, then you would get very inaccurate results.  But with two beams, those errors cancel each other out, and so all you're left with is the time of the light.&lt;/p&gt;&#xA;" OwnerUserId="792" LastActivityDate="2010-12-11T07:15:56.670" CommentCount="3" />
  <row Id="1812" PostTypeId="2" ParentId="1796" CreationDate="2010-12-11T09:42:50.743" Score="1" ViewCount="" Body="&lt;p&gt;I would say that the best place to look for conferences is the site of &lt;a href=&quot;http://physicsworld.com/cws/events&quot; rel=&quot;nofollow&quot;&gt;physics world and particularly the events page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2010-12-11T09:42:50.743" />
  <row Id="1813" PostTypeId="1" CreationDate="2010-12-11T11:45:22.530" Score="4" ViewCount="100" Body="&lt;p&gt;How do I get rid of the static charge I get from my microfiber couches? Every time I get up and touch something it zaps me...&lt;br&gt;&#xA;I'm mostly worried about my laptop getting fried.&lt;/p&gt;&#xA;" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-09T13:21:24.690" LastActivityDate="2011-01-09T13:21:24.690" Title="Couch shocks everything!" Tags="&lt;electrostatics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="1814" PostTypeId="2" ParentId="1813" CreationDate="2010-12-11T11:56:26.777" Score="6" ViewCount="" Body="&lt;p&gt;The reason why you get shocked is the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;As you move on the couch some static charge accumulates on your clothes and the couch&lt;/li&gt;&#xA;&lt;li&gt;As you stand up, the charges are separated, increasing the electric potential (the voltage) between you and the couch&lt;/li&gt;&#xA;&lt;li&gt;As soon as this voltage is high enough, a spark occurs&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How to avoid it:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Wear clothes that do not favour this static accumulation: for example cotton&lt;/li&gt;&#xA;&lt;li&gt;Alternatively, while you stand up, touch something conductive and grounded (or at worst, keep one hand on the sofa). This will allow you to discharge as the voltage increases, without shocks.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-12-11T11:56:26.777" CommentCount="1" />
  <row Id="1815" PostTypeId="2" ParentId="1787" CreationDate="2010-12-11T12:22:36.083" Score="1" ViewCount="" Body="&lt;p&gt;On the experiments and topology question, there is some work on the subject by Glenn Starkman et al. In their work, they search for structures in the CMB that would indicate some particular topology for the universe. &lt;a href=&quot;http://www.pirsa.org/index.php?p=media&amp;amp;url=http://streamer.perimeterinstitute.ca/Flash/9cd6f9d2-a6bc-48c8-b94e-fbcb0f1c2c4a/viewer.html&amp;amp;pirsa=08100080&amp;amp;type=Flash%20Presentation&quot; rel=&quot;nofollow&quot;&gt;There is a very nice lecture given in PI on the subject as well as other issues that have to do with CMB&lt;/a&gt;. To give you a spoiler on the lecture, they haven't found anything in large angle correlations.&lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2010-12-11T12:22:36.083" />
  <row Id="1816" PostTypeId="1" AcceptedAnswerId="1914" CreationDate="2010-12-11T14:56:19.230" Score="3" ViewCount="295" Body="&lt;p&gt;In the novel &lt;a href=&quot;http://en.wikipedia.org/wiki/Dragon%27s_Egg&quot; rel=&quot;nofollow&quot;&gt;Dragon's Egg&lt;/a&gt;, the human crew use one asteroid to swing other asteroids in place to counter the gravity of the neutron star. I understood that it was similar to a gravity sling shot, but I wasn't able to fully get how the crew were able to move the smaller asteroids in place using the big one. Can anyone explain that further?&lt;/p&gt;&#xA;" OwnerUserId="797" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-11T18:30:46.600" LastActivityDate="2010-12-14T20:38:45.493" Title="Orbital mechanics of Dragon's Egg" Tags="&lt;gravity&gt;&lt;orbital-motion&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="1817" PostTypeId="2" ParentId="1796" CreationDate="2010-12-11T15:31:56.017" Score="1" ViewCount="" Body="&lt;p&gt;There is a number of sites which lists conferences, e.g.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.conference-service.com/conferences/index.html&quot; rel=&quot;nofollow&quot;&gt;Conference Listing in Mathematics, Physics and Chemistry - Mandl&lt;/a&gt; (this one is good and aesthetic; once I attended a conference I had found there)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Associations/societies list their conferences&lt;/li&gt;&#xA;&lt;li&gt;Some research groups homepages have list on conferences in their field.&lt;/li&gt;&#xA;&lt;li&gt;Facebook, your faculty homepage, etc&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(However, none of the current solutions is perfect. I have one idea what to do... but let's wait - or contact me.)&lt;/p&gt;&#xA;" OwnerUserId="184" LastActivityDate="2010-12-11T15:31:56.017" />
  <row Id="1818" PostTypeId="2" ParentId="1222" CreationDate="2010-12-11T16:03:33.700" Score="1" ViewCount="" Body="&lt;p&gt;So, I brought this up at my research group meeting this week.  Turns out my initial guess &lt;b&gt;was&lt;/b&gt; right--the past development of the Oppenheimer-Snyder spacetime &lt;b&gt;does&lt;/b&gt; contain a white hole.  When people say that the matter distribution cuts off the white hole, what they generally assume is that the spacetime does not contain a moment of time symmetry, and that the matter distribution is something like a star, that goes into the distant past.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-12-11T16:03:33.700" />
  <row Id="1819" PostTypeId="1" AcceptedAnswerId="1827" CreationDate="2010-12-11T16:24:30.163" Score="3" ViewCount="337" Body="&lt;p&gt;What does an atom radiate: a wave packet or a single photon?&lt;/p&gt;&#xA;" OwnerUserId="690" LastEditorUserId="690" LastEditorDisplayName="" LastEditDate="2010-12-11T18:07:17.803" LastActivityDate="2010-12-11T20:14:53.213" Title="What does an atom radiate: a wave packet or a single photon?" Tags="&lt;quantum-mechanics&gt;&lt;electromagnetism&gt;&lt;optics&gt;&lt;quantum-field-theory&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="2" />
  <row Id="1820" PostTypeId="2" ParentId="1756" CreationDate="2010-12-11T16:24:36.230" Score="-2" ViewCount="" Body="&lt;p&gt;I well appreciated the &lt;strong&gt;lively discussion&lt;/strong&gt; arising to the question. I am truly interested in grasping an idea of the standard model and understanding it on a level such that I somehow know what is going on in QFT research e.g. at CERN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know if it is a no-go to answer one's own question but I want to give it a try here to see if I somehow got the ideas of the people contributing, namely &lt;strong&gt;Marek&lt;/strong&gt;, &lt;strong&gt;space-cadet&lt;/strong&gt; and &lt;strong&gt;Moshe&lt;/strong&gt; (thank you!). I might also want to add some additional information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following &lt;a href=&quot;http://www.claymath.org/millennium/Yang-Mills_Theory/yangmills.pdf&quot; rel=&quot;nofollow&quot;&gt;Jaffe &amp;amp; Witten&lt;/a&gt;, Quantum Chromodynamics (QCD, the $SU(3)$ &quot;part&quot; of the Standard Model) has to fulfill three properties to be successfully describing the strong force. One of these is the so-called &lt;strong&gt;mass gap&lt;/strong&gt; which means that every excitation of the vacuum must have a truly positive energy $E &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For QCD, the mass-gap is responsible for &lt;strong&gt;massive gauge Bosons&lt;/strong&gt; and therefore &lt;strong&gt;short interaction ranges&lt;/strong&gt;. So what about the other theories? My thought was that the non-abelian structure of $SU(2)\times U(1)$ and $SU(3)$ somehow directly relates to short interaction ranges. A nice argument was given by &lt;strong&gt;space-cadet&lt;/strong&gt; who points out that the $(A\wedge A)^2$ term in the Langrangian can be interpreted as the term responsible for the mass gap (again from the Jaffe &amp;amp; Witten paper).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And indeed, a recent &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0807/0807.2179v5.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; by &lt;a href=&quot;http://marcofrasca.wordpress.com/2008/07/15/classical-yang-mills-theory-and-mass-gap/&quot; rel=&quot;nofollow&quot;&gt;Frasca&lt;/a&gt; shows that for classical Yang-Mills theory &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;mass gap&lt;/strong&gt; is simply a dynamical effect arising from the &lt;strong&gt;self-interaction term&lt;/strong&gt; of the equations of motion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So there is evidence that $A\wedge A \neq 0$ might be an explanation of short interaction ranges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining question is if the non-Abelian character can be &lt;strong&gt;the (only) reason&lt;/strong&gt; for massive gauge particles or if &lt;strong&gt;other effects&lt;/strong&gt; such as asymptotic freedom and confinement are important to understand different ranges of inderaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;abelian Higgs model&lt;/strong&gt; (thanks to &lt;strong&gt;Moshe&lt;/strong&gt;) is an example where &lt;strong&gt;symmetry breaking&lt;/strong&gt; accounts for massive gauge particles. Also, there might be examples of long-range non-Abelian gauge theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To &lt;strong&gt;sum up&lt;/strong&gt;, I would state that the situation is indeed complicated as &lt;strong&gt;Marek&lt;/strong&gt; pointed out. But for some theories, the interaction range can be explained by the (non-)Abelian character of the underlying group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2010-12-11T16:24:36.230" CommentCount="8" />
  <row Id="1821" PostTypeId="2" ParentId="357" CreationDate="2010-12-11T17:17:06.150" Score="1" ViewCount="" Body="&lt;p&gt;Those laser tape-measures operate in an interesting way, that relies on the speed of light to determine distance. So conversely, if you have a known distance, then with the same equipment you should be able to estimate c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the tape measures do is modulate the intensity of the outgoing laser according to the intensity of the reflected light. It's basically an oscillator whose frequency depends on the optical propagation delay. The commercial products use the resulting frequency to determine a distance to display.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can get at the oscillator output, and set up to measure a known distance, you should be able to estimate c as the frequency in Hz times the round-trip distance in meters.&lt;/p&gt;&#xA;" OwnerUserId="717" LastActivityDate="2010-12-11T17:17:06.150" />
  <row Id="1822" PostTypeId="1" AcceptedAnswerId="1824" CreationDate="2010-12-11T17:59:52.460" Score="1" ViewCount="156" Body="&lt;p&gt;Suppose there are N non-interacting classical particles in a box, so their state can be described by the $\{\mathbf{x}_i(t), \mathbf{p}_i(t) \}$. If the particles are initially at the left of the box, they can eventually occupy the whole box according to the Newtons law. In this case, we call the gas expand and this process is irreversible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, Newtons laws also have time-reversal symmetry, so we should be able to construct an initial condition such that particles occupy the whole box (i.e. not all particles in the left chamber) will all move to the left chamber. Here are the simple questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How to select the initial condition $\{\mathbf{x}_i(0), \mathbf{p}_i(0) \}$ if you can solve the set of equation of motion.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why the gas expansion is irreversible even though you can select the initial condition above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="562" LastActivityDate="2010-12-11T21:30:40.437" Title="Irreversible expansion and time reversal symmetry" Tags="&lt;thermodynamics&gt;&lt;statistical-mechanics&gt;&lt;newtonian-mechanics&gt;&lt;reversibility&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="1823" PostTypeId="2" ParentId="1819" CreationDate="2010-12-11T18:16:46.623" Score="3" ViewCount="" Body="&lt;p&gt;When with &quot;radiate&quot; you mean the radiation emitted from an excited state of an atom decaying into the ground state, then you have a single photon, but its frequency is not known in advance because of the spectral broadening of the transition. &lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2010-12-11T18:16:46.623" />
  <row Id="1824" PostTypeId="2" ParentId="1822" CreationDate="2010-12-11T18:30:58.740" Score="4" ViewCount="" Body="&lt;p&gt;1/ OK, let's start from an initial condition where all the particles are made to fit a tiny little corner of the room and their initial velocities are chosen randomly, according to a Maxwell-Boltzmann distribution for instance. As we let the system evolve, the gas will expand, that is true because it corresponds to the behaviour the Maxwell-Boltzmann predicts, that of an ideal gas near equilibrium and therefore the gas will try to fill the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose at some moment we stop the evolution of the system, for instance after the gas has filled the entire room. Now, if we reverse all the velocities of the particles but do not change their positions, it is clear that by construction, we have found a solution that will evolve backwards to the corner where the gas was originally. So, it is indeed possible to find an initial configuration and in fact several initial configurations that satisfy your requirement. (But it takes a Maxwell Demon to realize them in practice.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2/ Now, why doesn't that invalidate the macroscopic irreversibility of the system? Well, we should look at how many microstates corresponding to the macrostate &quot;box entirely filled&quot; do go back to the corner. Say the box has volume $V$. Now, assuming we have $N$ particles, the total number of microscopic configurations compatible with the box filled will be proportional to $V^N$. (I am omitting the velocities in this analysis, they will only make the analysis less simple to follow.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many of these configurations go back to the corner? Well, to compute that, remember that we can easily construct the solutions which go back to the corner by taking the configurations that start from the corner and expand to fill the entire room and then reverse the velocities. So, let's say we call the volume of the corner $W$. So the amount of trajectories that go back to the corner will be proportional to $W^N$. So, the fraction of trajectories that go back to the corner among the trajectories that fill the entire volume will be $W^N/V^N$. Since $W/V&amp;lt;1$ and $N$ is a large number, of the order of the Avogadro number, you can see why you never actually observe these trajectories in real life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, if you make your system small enough, less than 10 particles, and your corner is a half of the room, then the probability is about $1/2^{10} \approx 0.001$, I'd say it is worth the wait. ;p&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OK, I wanted to add something to this explanation. The apparant paradox that is proposed in the OP is originally attributed to &lt;a href=&quot;http://en.wikipedia.org/wiki/Loschmidt%27s_paradox&quot; rel=&quot;nofollow&quot;&gt;Loschmidt&lt;/a&gt;. The paradox Marek is talking about which is related to the ergodicity of the system is due to &lt;a href=&quot;http://www-users.math.umd.edu/~lvrmr/History/Recurrence.html&quot; rel=&quot;nofollow&quot;&gt;Zermelo&lt;/a&gt;, they are different paradoxes and require different answers.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2010-12-11T21:30:40.437" LastActivityDate="2010-12-11T21:30:40.437" CommentCount="9" />
  <row Id="1825" PostTypeId="2" ParentId="1822" CreationDate="2010-12-11T19:14:48.797" Score="5" ViewCount="" Body="&lt;p&gt;The short answer is that it is not technically irreversible.  If you wait some huge amount of time, the gas will de-expand, as per the &lt;a href=&quot;http://en.wikipedia.org/wiki/Poincar%C3%A9_recurrence_theorem&quot; rel=&quot;nofollow&quot;&gt;Poincaré recurrence theorem&lt;/a&gt;.  The problem with this is that the amount of time for a system to evolve from a low-entropy configuration to a high entropy configuration is very small, while you would be waiting several ages of the universe for the process to reverse itself.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note how the recent &lt;a href=&quot;http://www.nature.com/news/2010/101114/full/news.2010.606.html&quot; rel=&quot;nofollow&quot;&gt;Maxwell Demon&lt;/a&gt; experiment has exploited the subatomic reversibility of macroscopically irreversible processes to generate useable work. (Ignore the overstatements about 'converting information to energy').  &lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2010-12-11T21:18:18.830" LastActivityDate="2010-12-11T21:18:18.830" CommentCount="2" />
  <row Id="1826" PostTypeId="2" ParentId="1816" CreationDate="2010-12-11T20:06:44.353" Score="0" ViewCount="" Body="&lt;p&gt;Not an answer but to give context:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are eight asteroids: 2 large and 6 smaller ones. All the asteroids have been collapsed by injecting their core with magnetic monopoles that makes them collapse to white dwarf density. The large ones were originally 250km in diameter and collapsed to 100m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The larger asteroids are on highly elliptical orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.co.uk/books?id=YataAAAAMAAJ&amp;amp;dq=isbn:034543529X&amp;amp;q=six+asteroids#search_anchor&quot; rel=&quot;nofollow&quot;&gt;http://books.google.co.uk/books?id=YataAAAAMAAJ&amp;amp;dq=isbn:034543529X&amp;amp;q=six+asteroids#search_anchor&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-12-12T04:20:19.747" LastActivityDate="2010-12-12T04:20:19.747" CommentCount="20" />
  <row Id="1827" PostTypeId="2" ParentId="1819" CreationDate="2010-12-11T20:14:53.213" Score="10" ViewCount="" Body="&lt;p&gt;An excited atom radiates off a &lt;strong&gt;wave packet&lt;/strong&gt;, that is, an oscillatory pattern of EM field which is effectively localized in space and time. It certainly does not emit the strictly monochromatic plane wave (which is infinite in space and time), nor a pointlike object (a pointlike distribution of a field includes Fourier components with arbitrary large energy). One can even speak of the &quot;shape&quot; of this wave packet: it typically has a sharp front and an exponentially decaying tail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A separate question is &lt;strong&gt;what is the photon?&lt;/strong&gt; I guess there is no universal answer; the actual thing that we call &quot;the photon&quot; depends on the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we quantize electromagnetic field, &quot;the photon&quot; means the excitation of the EM field with a fixed energy and momentum. (To be more precise, we introduce a regulator function that decays off at infinity, but then we consider everything in the limit when the regulator function tends towards identical unity. So, for all purposes we still can think of the monochromatic excitation). But when we discuss the normal modes of EM field with non-trivial boundary conditions, for example in a resonator inside a laser, the photons although being monochromatic are not spatially infinite plane waves anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we talk about emission and absorption of light quanta, which always proceeds in more or less localized wave packets, we usually call these wave packets the photons. In certain cases, for example in astrophysics, we can even speak of the time when a photon was emitted (although this is not a momentary act!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these subtle differences become even more delicate when discussing neutrino oscillation. There has been and still is a lot of confusion which originates from unjustified assumptions of how a neutrino is produced and detected (wave packets vs. plane waves, the moment and the position when the neutrino is emitted, etc.). Luckily, the EM field do not mix (at least in vacuum).&lt;/p&gt;&#xA;" OwnerUserId="59" LastActivityDate="2010-12-11T20:14:53.213" />
  <row Id="1828" PostTypeId="2" ParentId="1816" CreationDate="2010-12-11T20:38:08.063" Score="0" ViewCount="" Body="&lt;p&gt;When a small spacecraft slingshots around a large moon or planet, it can greatly increase its speed at the expense of very slightly decreasing the speed of the larger body. Given a method by which to move a larger body, you could very simply cause it to move smaller bodies by traveling near enough to attract them. More complicated would be moving it in such a way that the smaller body was given a specific increase in absolute velocity, which would be the &quot;reverse slingshot&quot;.&lt;/p&gt;&#xA;" OwnerUserId="218" LastActivityDate="2010-12-11T20:38:08.063" CommentCount="1" />
  <row Id="1829" PostTypeId="2" ParentId="1507" CreationDate="2010-12-11T20:55:55.097" Score="4" ViewCount="" Body="&lt;p&gt;I am about halfway the most important part of Onsager's paper, so I'll try to summarize what I've understood so far, I'll edit later when I have more to say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Onsager starts by using the 1D model to illustrate his methodology and fix some notations, so I'm gonna follow him but I'll use some more &quot;modern&quot; notations.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 1D Ising model, only neighbouring spins interact, therefore, the energy of interactions is represented by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E=-J\mu^{(k)}\mu^{(k-1)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $J$ is the interaction strength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The partition function is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z = \sum_{\mu^{(1)},\ldots,\mu^{(N)}=\pm 1} e^{-\sum_k J\mu^{(k)}\mu^{(k-1)}/kT}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Onsager notes that the exponential can be seen as a matrix component:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle \mu^{(k-1)}| V | \mu^{(k)} \rangle = e^{-J\mu^{(k)}\mu^{(k-1)}/kT}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The partition sum becomes the trace of a matrix product in this notation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z = \sum_{\mu^{(1)},\mu^{(N)}=\pm 1} \langle \mu^{(1)}| V^{N-1} | \mu^{(N)} \rangle$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for large powers $N$ of $V$, the largest eigenvalue will dominate. In this case, $V$ is just a $2\times 2$ matrix and the largest eigenvalue is $2\cosh(J/kT)=2\cosh(H)$, introducing $H=J/kT$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now, to construct the 2D Ising model, Onsager proposes to build it by adding a 1D chain to another 1D chain, and then repeat the procedure to obtain the full 2D model.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, he notes that the energy of the newly added chain $\mu$ will depend on the chain $\mu&amp;#39;$ to which it is added as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = -\sum_{j=1}^n J \mu_j \mu&amp;#39;_j $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we exponentiate this to go to the partition formula, we get the $n$th power of the matrixwe defined previously, so using notation that Onsager introduced there&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ V_1 = (2 \sinh(2H))^{n/2} \exp(H^{*}B)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $H^{*}=\tanh^{-1}(e^{-2H})$ and $B=\sum_j C_j$ with $C_j$ the matrix operator that works on a chain as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_j |\mu_1,\ldots,\mu_j,\ldots,\mu_n \rangle = |\mu_1,\ldots,-\mu_j,\ldots,\mu_n \rangle $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, to account for the energy contribution from spins within a chain, he notes that the total energy is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = -J&amp;#39; \sum_{j=1}^n \mu_j\mu_{j+1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;adding periodicity, that is the $n$th atom is a neighbor to the 1st. Also note that the interaction strength should not be equal to the interchain interaction strength. He introduces new matrix operators $s_j$ which act on a chain as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s_j|\mu_1,\ldots,\mu_j,\ldots,\mu_n \rangle = \mu_j |\mu_1,\ldots,\mu_j,\ldots,\mu_n \rangle $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and in this way constructs a matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_2 = \exp(H&amp;#39;A) = \exp(H&amp;#39;\sum_j s_j s_{j+1})$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the 2D model can be constructed by adding a chain through application of $V_1$ and then define the internal interactions by using $V_2$. So one gets the following chain of operations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cdots V_2 V_1 V_2 V_1 V_2 V_1 V_2 V_1 V_2 V_1$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is thus clear that the matrix to be analyzed in our 2D model is $V=V_2 V_1$. This is our new eigenvalue problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda | \mu_1,\ldots,\mu_n \rangle = \exp(H&amp;#39;\sum_j s_j s_{j+1}) \sum_{\mu&amp;#39;_1,\ldots,\mu&amp;#39;_n=\pm 1} \exp(H\sum_j \mu_j \mu&amp;#39;_{j})| \mu&amp;#39;_1,\ldots,\mu&amp;#39;_n \rangle$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now, the quaternions come into play. Onsager notes that the operators $s_j$ and $C_j$ he constructed form a quaternion algebra.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the basis elements $(1,s_j,C_j,s_jC_j)$ generate the quaternions and since for different $j$'s the operators commute, we have a tensor product of quaternions, thus a quaternion algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- To be continued --&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2010-12-11T20:55:55.097" CommentCount="6" />
  <row Id="1830" PostTypeId="1" CreationDate="2010-12-11T20:56:13.803" Score="0" ViewCount="100" Body="&lt;p&gt;If we have 1D wave equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial^2 \psi}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 \psi}{\partial t^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we say that it's always possible to decompose the generic solution $f(x+ct)+g(x-ct)$ using Fourier Transforms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{2\pi} \int_R f(\omega)e^{i\omega t}d\omega$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we have two variables, $x$ and $t$, how can we use the FT with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_R f(\omega)e^{i(k(\omega)x\mp\omega t)}d\omega \; ?$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have two different types of evolution, $+ct$ and $-ct$... &#xA;I have seen situations with no dispersion law, where we consider only the $x+ct$ wave and use the point $x=0$ to find the spectrum of waves $f(\omega)$, but can someone explain me the general method? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2D or 3D I haven't any idea about how I can do all this. Someone can explain me this too?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much!&lt;/p&gt;&#xA;" OwnerUserId="558" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2010-12-11T21:30:14.040" LastActivityDate="2010-12-11T21:30:14.040" ClosedDate="2010-12-12T01:24:27.577" Title="Fourier analysis for waves" Tags="&lt;mathematics&gt;&lt;waves&gt;&lt;plane-wave&gt;" CommentCount="2" FavoriteCount="2" />
  <row Id="1831" PostTypeId="2" ParentId="1822" CreationDate="2010-12-11T21:11:00.610" Score="1" ViewCount="" Body="&lt;p&gt;Trivially, we could set the initial conditions so that particles all have velocities pointing directly towards one corner with speeds proportional to their distances from the corner.  I don't think this is what you were looking for, though, since it doesn't look like the typical  configurations of state space you'd expect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing to keep in mind (in addition to Jerry and Rasklonikov's answers) is that any realistic macroscopic system is not in going to be perfectly isolated.  Even if you could reach in and adjust the positions and velocities of every particle, the particles would interact with the edges of the box somewhat randomly because the box is presumably sitting in some sort of heat bath.  These unpredictable interactions would mess up any carefully-calculated dance of the particles.  Even if you set up the initial conditions correctly,  in practice you still wouldn't get the particles to go back to the corner.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-12-11T21:11:00.610" />
  <row Id="1832" PostTypeId="2" ParentId="1756" CreationDate="2010-12-11T22:25:50.677" Score="-2" ViewCount="" Body="&lt;p&gt;@Moshe: As you may know, when people are not able to manage quantum field theories in some limits tend to form a lot of prejudices about how something should come out. This is the situation about mass gap in Yang-Mills theory. You can read &lt;a href=&quot;http://marcofrasca.wordpress.com/2010/12/01/current-status-of-yang-mills-mass-gap-question/&quot; rel=&quot;nofollow&quot;&gt;this post&lt;/a&gt; to have a correct idea about the current situation on this question and, of course, if you are able to find some unsatisfactory points in that papers, published in respectful journals and with an intervention of Terry Tao, I will be happy to hear from you. On the other side, the paper you are discussing is old, unpublished and overcome by 0907.4053 [math-ph] that will appear in the Journal of Nonlinear Mathematical Physics.&lt;/p&gt;&#xA;" OwnerUserId="799" LastActivityDate="2010-12-11T22:25:50.677" CommentCount="4" />
  <row Id="1833" PostTypeId="2" ParentId="1801" CreationDate="2010-12-11T23:17:41.040" Score="3" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Space actually has energy&lt;/strong&gt;, vacuum energy. It has been shown by various experiments and can be explained by Quantum Mechanics. So more space means more energy. Although it probably can't be used to do work, it does act to increase expansion. Check out the &lt;a href=&quot;http://en.wikipedia.org/wiki/Vacuum_energy&quot; rel=&quot;nofollow&quot;&gt;wikipedia page&lt;/a&gt; for more info.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Although galaxies are massive, they are far away and thus the resulting acceleration towards each other is weak.  &lt;strong&gt;If the space between galaxies is expanding at a faster rate than their mutual atraction; galaxies will move apart.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a football player running from one endzone toward the other as fast as he could. However the distance between them endzones doubles every 4 seconds.  The endzones are not moving; the space between them is growing. There is no way the player could hope to reach his goal.  In a short time he would not be able to see the other end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scary thing is this not only goes for the field but everything.  The player himself would be ripped apart as his own body parts moved farther and farther apart.  If space was expanding fast enough the attractive forces keeping black holes or even atoms together would not be enough. This what is commonly referred to as the Big Rip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luckily the expansion is slow enough that only &quot;distant&quot; galaxies are moving away from each other. Gravitational forces in a solar system or a galaxy are more than enough to withstand expansion.  The larger the scale of the system the more expansion has to play.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;When we talk about expansion we are not talking about objects moving away because of their great velocities against a static backdrop.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What your describing is like marbles on a grid.  The marbles get farther away due to their velocities relative to the grid pointing in opposite directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead expansion is like the grid growing in scale. This effects the distances between the marbles independently of their velocities; which is why we observe all far away objects as moving away from us. If it wasn't for expansion we would expect a more random distribution.  It is only at smaller scales where expansion is less of a factor that we can see objects moving toward us such as Andromeda's Galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are not in any sense moving away from the center of the universe.  The idea of the Big Bang and Inflation is that not only every thing, matter and energy; but everywhere was contained in the Big Bang.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="800" LastEditorUserId="800" LastEditorDisplayName="" LastEditDate="2010-12-21T15:24:12.997" LastActivityDate="2010-12-21T15:24:12.997" CommentCount="3" />
  <row Id="1834" PostTypeId="2" ParentId="1768" CreationDate="2010-12-12T00:12:47.230" Score="3" ViewCount="" Body="&lt;p&gt;Air does freeze just at temperatures and pressures we don't often experience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an entire industry around producing and distilling liquid air.  You take air and compress it.  As a result the air increases in temperature; the air is allowed to cool. Then the air is expanded by venting it into a new chamber. The result is a much colder gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through cycles of compression, cooling and venting we can get air to condense into a liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If liquid nitrogen is then placed in a vacuum chamber, it will freeze.&lt;/p&gt;&#xA;" OwnerUserId="800" LastActivityDate="2010-12-12T00:12:47.230" CommentCount="2" />
  <row Id="1835" PostTypeId="2" ParentId="1544" CreationDate="2010-12-12T01:51:30.583" Score="3" ViewCount="" Body="&lt;p&gt;You have been asking some seriously interesting questions! Here's my take on this one ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say this about the Born-Infeld action:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But as I see it, this model is an intrinsically nonlinear model for the free-space field itself and not useful for describing nonlinear matter interaction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure exactly what you mean by &quot;free-space&quot; field. I take it that you're referring to $ F_{\mu\nu} $. Well there is no reason why one cannot define an $ F_{\mu\nu} $ for waves propagating non-linearly, within a medium or in a vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matter-light interaction can be specified (at least in part if not wholly) by the form of $ g_{\mu\nu} $. Now bear with me for a minute. I'm not referring to the metric generated by some kind of matter. The metric in question does not, &lt;em&gt;a priori&lt;/em&gt;, satisfy the Einstein equations. It is instead the &lt;strong&gt;effective&lt;/strong&gt; metric experienced by the light-rays propagating within the given material. See these excellent papers by Ulf Leonhardt and Thomas Philbin &lt;a href=&quot;http://arxiv.org/abs/cond-mat/0607418&quot; rel=&quot;nofollow&quot;&gt;[1]&lt;/a&gt;,&lt;a href=&quot;http://arxiv.org/abs/0805.4778&quot; rel=&quot;nofollow&quot;&gt;[2]&lt;/a&gt; for more details on this notion. In brief the off-diagonal components $ g_{ij}$ (where $ (i,j \in \{1,2,3\}\,\, i \neq j) $ encode the susceptibility tensor and the diagonal components $ g_{0i} $ determine the mixing between the electric and magnetic components of the wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the lagrangian density for the matter-light interaction you posit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathcal{L}_{int} \propto M^{\mu\nu} F_{\mu\nu} = T^{\mu\nu\alpha\beta} F_{\alpha\beta} F_{\mu\nu} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for flat space (or no-medium) $ T^{\mu\nu\alpha\beta} = g^{\mu\nu}g^{\alpha\beta} $, this term reduces to $ F^{\mu\nu} F_{\mu\nu}$ which is nothing more than the Maxwell term ! On the face of it this gives us nothing new, unless we adopt the route outlined above and use the metric $g_{\mu\nu}$ to encode the optical properties of the medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another line of thought which exploits this notion of the metric to allow one to speak of an analogy between optical processes and the big-bang is the phenomenal work of Igor Smolyaninov &lt;a href=&quot;http://arxiv.org/abs/0908.2407&quot; rel=&quot;nofollow&quot;&gt;[3]&lt;/a&gt;. This paper was accepted by PRL btw, so its nothing to sneeze at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the above line of reasoning is not fatally flawed, and that one can encode the effects of the medium in the metric, it seems that either the Maxwell or the Born-Infeld action are  perfectly good &lt;em&gt;candidates&lt;/em&gt; of gauge-invariant actions for your purposes.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                                Cheers,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: &lt;strong&gt;Non-linearity redux&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @Raskolnikov pointed out, the identification of the components $g_{ab}$ with the optical susceptibilities of a material, does not give us a &lt;em&gt;nonlinear&lt;/em&gt; material. For that, you have to have a dependence of the susceptibilities on the field strengths themselves. So you have a feedback mechanism $ \mathbf{g} \rightarrow \mathbf{F} \rightarrow \mathbf{g} $ and &lt;em&gt;therefore&lt;/em&gt; the non-linearity ! Therefore in general, as @robert has been trying to convey to me without success, $\mathbf{g}$ should in general be a function of $\mathbf{F}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then you start treading dangerously close to the speculation that somehow the eventual picture (for the fully non-linear case) might be somehow general relativistic. That is a very tempting idea, but I leave that for another time.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-12T10:18:18.037" CommentCount="20" />
  <row Id="1836" PostTypeId="1" AcceptedAnswerId="1838" CreationDate="2010-12-12T02:01:54.143" Score="4" ViewCount="434" Body="&lt;p&gt;I think that something is invisible if its isolated particles are smaller than the wavelength of visible light. Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is air invisible? What about other gases and fumes which are visible?&lt;/p&gt;&#xA;" OwnerUserId="293" LastActivityDate="2011-03-19T19:01:47.427" Title="Why is air invisible?" Tags="&lt;light&gt;&lt;air&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" />
  <row Id="1837" PostTypeId="2" ParentId="1836" CreationDate="2010-12-12T02:18:40.943" Score="0" ViewCount="" Body="&lt;p&gt;The air has purely real rifraction indix $\sim 1$, no dispersive, &quot;like vacuum&quot;. So there is not absorption, and the air is transparent for visible wavelenght, like X rays for metals over $plasm frequency$.&lt;/p&gt;&#xA;" OwnerUserId="558" LastActivityDate="2010-12-12T02:18:40.943" CommentCount="1" />
  <row Id="1838" PostTypeId="2" ParentId="1836" CreationDate="2010-12-12T02:20:15.417" Score="11" ViewCount="" Body="&lt;p&gt;I think the pithy answer is that our eyes adapted to see the subset of the electromagnetic spectrum where air has no absorption peaks.  If we saw in different frequency ranges, then air would scatter the light we saw, and our eyes would be less useful.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-12-12T02:20:15.417" CommentCount="9" />
  <row Id="1839" PostTypeId="1" AcceptedAnswerId="1840" CreationDate="2010-12-12T03:49:14.497" Score="0" ViewCount="88" Body="&lt;p&gt;I was thinking about how they say those sails on top of some dinosaurs helped regulate their body temperature. If a dinosaur didn't have that sail, would it &lt;em&gt;really&lt;/em&gt; make any difference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you heated up two towels (large ones) to 50 degrees Celsius, and spread one out, and shaped the other one into a ball, which one would cool faster?&lt;/p&gt;&#xA;" OwnerUserId="125" LastActivityDate="2010-12-12T04:45:34.667" Title="Does a towel that's spread out cool faster than one that isn't?" Tags="&lt;temperature&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="1840" PostTypeId="2" ParentId="1839" CreationDate="2010-12-12T04:45:34.667" Score="5" ViewCount="" Body="&lt;p&gt;Heat loss is largely proportional to surface area, so your spreadout towel more more efficiently cool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About those dinosaurs. It is know that elephants large ears do get used for cooling. With a good blood supply those sails certainly could have dissipated at lot of heat. That doesn't prove that that was their primary purpose however.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-12-12T04:45:34.667" CommentCount="1" />
  <row Id="1841" PostTypeId="2" ParentId="1836" CreationDate="2010-12-12T05:16:13.917" Score="7" ViewCount="" Body="&lt;p&gt;I don't think one can just state that particle size smaller than the wavelength of light implies no interaction with light. You have to look at the quantum mechanical modes of the atoms/molecules. If they have modes with frequencies in the wvaelength range you are interested in, then you will get interaction/absorption. I think that even clean air does have nonzero absorption in the visible. Have a large enough column density along a beam of light, and there&#xA;will be some absorption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also do note, that the speed of light is slower in air than in a vacuum, so the air does have some effect. You do get effects of refraction, such as mirages, and heat waves seen looking across convection over a hor surface. Also if you point a telescope at a star, you&#xA;see the mess the atmosphere makes of the image.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-12-12T05:16:13.917" />
  <row Id="1842" PostTypeId="2" ParentId="1507" CreationDate="2010-12-12T08:54:54.037" Score="2" ViewCount="" Body="&lt;p&gt;Not to state the obvious, but it seems the information at the &lt;a href=&quot;http://www.scholarpedia.org/article/Ising_model%3a_exact_results#Methods_of_Solution&quot; rel=&quot;nofollow&quot;&gt;scholaropaedia&lt;/a&gt; article which @marek mentioned in his answer, is more comprehensive than any answer I or anyone else is likely to come up with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quote this article &quot;there are &lt;strong&gt;five different methods&lt;/strong&gt; which have been used to compute the free energy of the Ising model&quot;. For details best check out the link above. Anything more I add will just be repetition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the bounty, it should go to Barry McCoy - the author of the scholaropaedia article ;)&lt;/p&gt;&#xA;" LastActivityDate="2010-12-12T08:54:54.037" CommentCount="2" />
  <row Id="1843" PostTypeId="1" AcceptedAnswerId="1850" CreationDate="2010-12-12T11:04:31.047" Score="9" ViewCount="275" Body="&lt;p&gt;The simple picture of Hawking radiation is that a pair-antiparticle pair is produced near the event horizon, then one falls into the black hole while the other escapes. Suppose the particles are quarks-antiquarks, which experience quark confinement thanks to QCD. If one of them is swallowed by the black hole, its partner is left alone. Eventually the quark gains enough energy and turns into a hadronic jet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my line of thinking correct? If yes, is it (or QCD in general) taken into account when calculating Hawking radiation?&lt;/p&gt;&#xA;" OwnerUserId="47" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-12T11:10:00.303" LastActivityDate="2010-12-31T07:29:08.260" Title="Hawking radiation and quark confinement" Tags="&lt;high-energy-physics&gt;&lt;quantum-gravity&gt;&lt;hawking-radiation&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="1844" PostTypeId="2" ParentId="1843" CreationDate="2010-12-12T11:41:07.387" Score="2" ViewCount="" Body="&lt;p&gt;I think your reasoning should be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as I understand the Hawking radiation (i.e. not much) the effect is derived independently of the microscopic theory. You just assume that particle (which is to escape from around the horizon) needs to accelerate (a lot) in order to overcome black hole's gravity and then notice that accelerating observer will register a black body radiation. This is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Unruh_temperature#Unruh_temperature&quot; rel=&quot;nofollow&quot;&gt;Unruh effect&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you can certainly try to derive the effect from the first (i.e. QFT) principles. E.g. &lt;a href=&quot;http://www.springerlink.com/content/g04t647066763623/&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; claims that it is indeed possible to derive the results from the &lt;em&gt;free&lt;/em&gt; QFT. But for realistic QFT (like QCD) this doesn't seem technically feasible and the results you obtain (i.e. Hawking temperature) can depend on the precise theory used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-ph/0612151&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; talks about similar effect but instead from tunneling from under the black hole horizon it discusses tunneling out of the confinement horizon (this is called white hole because confinement means white hadrons; not to be confused with white holes from GR).&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-12T11:41:07.387" CommentCount="1" />
  <row Id="1845" PostTypeId="1" CreationDate="2010-12-12T13:06:31.457" Score="11" ViewCount="433" Body="&lt;p&gt;Are there any nice experiments using wireles LAN access points or routers or mobile phones to demonstrate physical features of electromagnetic fields, especially em-waves? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely I am looking for experiments which use one of the devices above as the main tool and some materials that can be purchased at a reasonable cost from, say, a hardware store or electronic supply store only. I do not impose any restrictions on the cognitive level the experiments aim at; undergraduate level is fine as well as graduate level in em-theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please use separate answers for different experiments.&lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorUserId="667" LastEditorDisplayName="" LastEditDate="2010-12-17T22:05:35.730" LastActivityDate="2010-12-20T17:54:35.597" CommunityOwnedDate="2010-12-17T13:31:18.583" Title="Home experiments using wireless LAN or mobile phones about electromagnetism" Tags="&lt;electromagnetism&gt;&lt;experimental-physics&gt;&lt;electrodynamics&gt;&lt;electromagnetic-radiation&gt;&lt;home-experiment&gt;" AnswerCount="5" CommentCount="10" FavoriteCount="6" />
  <row Id="1846" PostTypeId="1" CreationDate="2010-12-12T14:42:39.533" Score="3" ViewCount="176" Body="&lt;p&gt;How much of classical mechanics can be modelled with &lt;a href=&quot;http://en.wikipedia.org/wiki/Soliton&quot; rel=&quot;nofollow&quot;&gt;solitons&lt;/a&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am aware of is that single solitons behave in a way like free particles: they move along as stable entities with constant velocity. But:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can solitons interact with an external field?&lt;/li&gt;&#xA;&lt;li&gt;Can solitons collide?&lt;/li&gt;&#xA;&lt;li&gt;Can solitons interact with each other by general (central) forces?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That means: Are there PDEs with solutions that can be interpreted as &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a particle in an external field &lt;/li&gt;&#xA;&lt;li&gt;two colliding particles&lt;/li&gt;&#xA;&lt;li&gt;two generically interacting particles &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;obeying appropriate laws of classical mechanics? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would the mass/momentum of a soliton come into play?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pointers to references would be very welcome!&lt;/p&gt;&#xA;" OwnerUserId="229" LastActivityDate="2010-12-13T08:42:28.590" Title="Soliton mechanics" Tags="&lt;classical-mechanics&gt;&lt;mechanics&gt;" AnswerCount="1" />
  <row Id="1847" PostTypeId="1" CreationDate="2010-12-12T15:01:19.907" Score="4" ViewCount="290" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/8441/what-is-a-complete-book-for-quantum-field-theory&quot;&gt;What is a complete book for quantum field theory?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;At the moment I am studying &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Piron: Foundations of Quantum Physics,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jauch: Foundations of Quantum Mechanics, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ludwig: Foundations of Quantum Mechanics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of them discuss nonrelativistic quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is, if there are corresponding approaches to quantum field theory (in particular aiming to the standard model of particle physics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; (in response to noldorin's comment): I think if one knows the books mentioned above, the question is not very vague. Take for example Ludwigs approch. For instance it is written with the aim in mind to provide clear foundations of nonrelativistic quantum mechanics which match Ludwigs epistemological theory about physics (&quot;A new foundation of Physical Theories&quot;). It is axiomatic, mathematical sound, emphasises the idea of preperation and registration procedures, uses the mathematical language of lattice theory etc. However since I am studying this book at the moment I might miss some important points in Ludwigs approach. So I ask a bit vague if there is an approach to QFT which corresponts &lt;strong&gt;in your view&lt;/strong&gt; to Ludwigs approach to nonrelativistic quantum mechanics in &lt;strong&gt;the most essential points (from your point of view)&lt;/strong&gt; (in style, epistemological background, mathematical language etc.). Perhaps there are students of Ludwig which transfered and developed his ideas for QFT as well, I don't know.&lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-01T07:35:54.597" LastActivityDate="2011-02-01T07:35:54.597" ClosedDate="2011-04-11T18:11:02.563" Title="Jauch, Piron, Ludwig -&gt; QFT?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;particle-physics&gt;&lt;mathematical-physics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="1848" PostTypeId="2" ParentId="1846" CreationDate="2010-12-12T15:47:14.150" Score="7" ViewCount="" Body="&lt;p&gt;I think your question might be reformulated in some way. The general approach to solitons is to derive a differential equation with respect to a given potential. Then, the solution to this equation might have soliton solution, which of course interact with the external field, can collide etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think this may not cover what you are thinking off. Let me give you a quick derivation of Sine-Gordon equation arising from a model of coupled pendulums. This might help to specify what you want to know.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Sin Gordon Equation: elastic coupled pendulums&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The kinetic energy of $N$ pendulums with mass $m$ and length $l$ is given by&lt;br&gt;&#xA;$T = \sum_{i=1}^N\frac{m}{2}l^2\dot{\varphi}_i^2$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The potential energy is a little more complicated:&#xA;$U = \sum_{i=1}^Nmgl\left( 1 - \cos\varphi_i\right) + \sum_{i=1}^{N-1}\frac{D}{2}\left( \varphi_{i+1}-\varphi_{i}\right)^2$&lt;br&gt;&#xA;where $D$ accounts for the strength of the band and $g$ for gravitation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lagrange function is then given by $L = T - U$ and we find the equations of motion using Lagrangian mechanics via the famous&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{dt}\frac{\partial L}{\partial\dot{\varphi}_i}-\frac{\partial L}{\partial\varphi_i}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;leading to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{D}{mgl}\left( \varphi_{i+1} - 2\varphi_{i} + \varphi_{i-1}\right) - \frac{l}{g}\partial_{tt}\varphi_i=\sin\varphi_i$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, by letting $N\rightarrow\infty$ and rescaling of coordinates we arrive at the famous &lt;a href=&quot;http://en.wikipedia.org/wiki/Sine%E2%80%93Gordon_equation&quot; rel=&quot;nofollow&quot;&gt;Sine-Gordon-equation&lt;/a&gt; for the continuous system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\partial_{\rho\rho}\varphi - \partial_{\tau\tau}\varphi = \sin\varphi$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation has mono- or multisoliton solutions which are nicely described in the Wikipedia article mentioned earlier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that for the derivation of the equation of motion we needed to have the potential already at hand, it does not work the other way round as far as I know. The solutions are very interesting and widely used throughout physics to explain &quot;non-dispersive&quot; water waves, scattering of particles and so on.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Some further remarks&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For a much deeper insights you might have a look at the book &quot;Solitonen - Nichtlineare Strukturen&quot; by R. Meinel written in German (and judging from your name you might understand) or research articles like &lt;a href=&quot;http://arxiv.org/PS_cache/hep-th/pdf/0008/0008108v3.pdf&quot; rel=&quot;nofollow&quot;&gt;Non-commutative soliton scattering&lt;/a&gt; by Lindström et al.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know there does not exist a strict kind of approach that will tell you if a system has soliton solutions or not; the search for such solutions is a science on its own. An expert for solitons in optical systems is O. Egorov, some of his papers on the subject can be found &lt;a href=&quot;http://scholar.google.de/scholar?q=oleg+egorov+soliton&quot; rel=&quot;nofollow&quot;&gt;on google scholar&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2010-12-13T08:42:28.590" LastActivityDate="2010-12-13T08:42:28.590" CommentCount="8" />
  <row Id="1849" PostTypeId="2" ParentId="1639" CreationDate="2010-12-12T15:47:57.493" Score="1" ViewCount="" Body="&lt;p&gt;Assume that the hill and the treadmill have the same angle of elevation (are inclined identically), and that two identical persons A and B are running on them at the same speed $v$. Here the speed of the person B running on the treadmill is obviously zero with respect to the ground, but we will consider the speed of the treadmill's belt to be $-v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume that B and the treadmill are now in a truck which is running up the hill parallel to A, and with the same speed $v$ as A. The truck should be arranged so that B and the treadmill are not inclined while the truck is running up the hill. By our hypothesis, the speed of the upper part of the belt is zero with respect to the ground. This will not affect the effort done by the runner B, because the truck is moving with constant velocity. That is, there are no extra forces caused by inertia, because the truck doesn't accelerate, decelerate or change direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By looking now at both runners A and B we see that they are moving parallel with the same speed. They can even do the same moves in synchrony. The angle elevation is the same for both of them too. But B may even not realize that he is climbing a hill, he may think that he is in a room with no windows, which doesn't move. The conditions are identical. So, there is no difference between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If our intuition still saids that the treadmill guy is burning less calories, let's imagine that the road on which the person A is running up the hill is a very long treadmill belt. Imagine that underneath the belt there is a treadmill which is doing two things: is moving with the same speed $v$ towards the top of the hill, and the upper side of the belt is moving backwards with $-v$, so that the belt appear to be fixed with respect to the ground. From the outside, the belt doesn't move (and for runner A too). Now, it should be clear that there is no difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above I assumed that there is no wind, the treadmill and the runners are moving with uniform speed, there is no difference between the runners. I also assumed that gravity is not weaker towards the top of the hill.&lt;/p&gt;&#xA;" OwnerUserId="730" LastActivityDate="2010-12-12T15:47:57.493" CommentCount="1" />
  <row Id="1850" PostTypeId="2" ParentId="1843" CreationDate="2010-12-12T15:55:57.500" Score="5" ViewCount="" Body="&lt;p&gt;Actually working out a rigorous prediction for Hawking radiation, you need to solve the equations for the relevant QFT semi-classically--treating the metric of spacetime as a substitute for the Minkowski metric usually used in QFT. This is similar to deriving the quantum mechanics of the Hydrogen atom without paying attention to the fact that photons exist and the electromagnetic field is quantized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, when you work out the Hawking effect for a scalar field, or a fermion or whatever, what you do is start with the &lt;a href=&quot;http://prime-spot.de/Bilder/BR/schwarzschild_l.jpg&quot; rel=&quot;nofollow&quot;&gt;Penrose-Carter diagram for the Schwarzschild spacetime&lt;/a&gt;.  You define a set of linearized plane wave solutions at past infinity, and then use your curved-space equations of motion for the field to evolve forward at time,and then measure what projection onto a set of linearized plane wave solutions that you get at future infinity.  Hawking's result in his original paper was that even if you define the &quot;in&quot; state at past null infinity to be the vacuum state (i.e., the state with no particles), then you STILL have a blackbody distribution at future null infinity--thus, we interpret the black hole as radiating.  Note how none of this relies upon virtual pair creation, and how it ties pretty directly into ordinary QFT.  The bit about pair creation mostly serves to give a nice physical picture as to &lt;b&gt;why&lt;/b&gt; there could be radiation at future null infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there are a few reasons why this procedure will not work for quarks.  Marek points out the clearest of them--there is no &quot;in&quot; or &quot;out&quot; state that describes a set of free, unbound quarks or gluons.  The best we can do is approximate such a state at high energy.  So the semi-classical formalism breaks down on some level.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second reason that this won't work out is more astrophysical than anything.  Black holes are known to likely have a very small physical charge, because if they were to acquire a charge of any size, they would have a very large electromagnetic field that would polarize everything around them, suck in oppositely charged ions, and neutralize themselves.  If, by some chance, a black hole did gain some sort of color charge by creating an oppositely colored particle near it, the force on that particle would be so great that it would not be able to escape to infinity, anyway, and the black hole would remain uncolored, and we wouldn't see free quarks escaping to infinity, so the effect you describe shouldn't be possible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're asking whether or not we would see bound states of quarks escaping, however, the answer to &lt;b&gt;that&lt;/b&gt; question is almost certainly yes.  A pion or a proton is a perfectly good state to do semiclassical analysis with, and (I don't know this for sure, but I'm pretty sure that someone has looked at this before, and there isn't a plausible counterargument that I can think of) that we would get perfectly sensible blackbody radiation out of using them as a test particle, so long as the mass of the black hole is low enough (and therefore, its temperature high enough) to not allow the production of such massive particles.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2010-12-12T16:45:53.773" LastActivityDate="2010-12-12T16:45:53.773" CommentCount="2" />
  <row Id="1851" PostTypeId="2" ParentId="1507" CreationDate="2010-12-12T17:05:36.957" Score="3" ViewCount="" Body="&lt;p&gt;I wish I could do your question justice, but I will content myself with a remark on the connection between two of the solution methods mentioned in Barry McCoy's article, namely Baxter's commuting transfer matrix method, and Onsager's original algebraic approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a certain sense, these methods have to be considered distinct since Baxter's method is applicable to a vast family of additional models, whereas Onsager's method applies only to Ising and closely related models.  A related fact is that, while the free energy and order parameter have been computed for a great many two-dimensional models, only for Ising are the correlation functions completely understood.  (They can be written in terms of simple determinants.)  Among solvable two-dimensional models, Ising appears to be very special.  It lies at the intersection of many infinite families  of models.  Although all solvable lattice models have lots of unexpected structure - in particular, they have infinitely many conserved quantities - Ising is even more special.  Onsager's original method of solution exploited some of this special structure - in particular, the direct-product structure of the transfer matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Baxter's commuting transfer matrix method does not exploit this special structure, it can be used to solve the many other models that don't have it.  His method uses the Yang-Baxter relation to establish that the transfer matrices commute for different  values of the spectral parameter (which, in the Ising model, parametrizes the difference between horizontal and vertical coupling strengths).  Since the eigenvectors must therefore be independent of the spectral parameter, one can derive functional relations for the eigenvalues, which can then be solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Onsager's method was expanded upon by Dolan and Grady, who showed that a certain set of commutation relations implies the existence of an infinite set of conservation laws.  In the 1980s, a solvable n-state generalization of the Ising model, known as the superintegrable chiral Potts model, was discovered that satisfies Dolan and Grady's conditions and, as a consequence, has transfer matrices with the same direct-product structure that Onsager exploited in 1944.  Interestingly, the superintegrable chiral Potts model corresponds to a special point in a one-parameter family of solvable models, the integrable chiral Potts models.  The latter are solvable by Baxter's method, but can be solved by Onsager's method only at the superintegrable point.  There seems to be a lot of work going on currently on the correlation functions of the superintegrable chiral Potts model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other solution methods that Barry McCoy mentions in his Scholarpedia article - Kaufman's free fermions, the combinatorial method, Baxter and Enting's 399th solution - also seem to make use of the particular structure of the Ising model.  In this sense, they are more akin to Onsager's original method than to Baxter's commuting-transfer-matrix method.  As you have already suggested, there may be some equivalences among them, but I would have to give this more study before commenting further.&lt;/p&gt;&#xA;" OwnerUserId="793" LastActivityDate="2010-12-12T17:05:36.957" CommentCount="2" />
  <row Id="1852" PostTypeId="1" AcceptedAnswerId="1866" CreationDate="2010-12-12T17:18:04.803" Score="5" ViewCount="321" Body="&lt;p&gt;There have been a few &lt;a href=&quot;http://www.crunchgear.com/2010/12/10/navys-mach-7-railgun-fires-a-round-at-33000-megajoules/&quot; rel=&quot;nofollow&quot;&gt;articles about the Navy's new Mach 7 33 Megajoule railgun.&lt;/a&gt; As a physics teacher, I have a couple of questions about this, and was hoping for some help. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is the kinetic energy of the projectile 33 MJ, or is that the energy that was delivered from the capacitor bank that was used to fire the projectile, so that some of it would have been lost in resistive heating, etc. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The photos of the projectile show a cloud of hot gas behind the projectile? What is this? superheated air? Plasma from the electrical breakdown inside the railgun? What causes this? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="807" LastActivityDate="2010-12-16T00:12:25.367" Title="Navy's new Mach 7 Rail Gun Photo" Tags="&lt;electromagnetism&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="1853" PostTypeId="2" ParentId="1847" CreationDate="2010-12-12T17:56:35.120" Score="3" ViewCount="" Body="&lt;p&gt;There are three QFT books that I would recommend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anthony Zee's &lt;a href=&quot;http://press.princeton.edu/titles/7573.html&quot; rel=&quot;nofollow&quot;&gt;Quantum Field Theory in a Nutshell&lt;/a&gt; is a very good book written for advanced undergraduates/beginning grad students.  It is much more physics-y than math-y in focus, with everything motivated as carefully as possible.  While the book is definitely quantitative, he puts a bigger priority on learning the conceptual underpinnings and the &lt;b&gt;why&lt;/b&gt; of QFT than he does on either mathematical rigor or heavy calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peskin and Schroder's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201503972&quot; rel=&quot;nofollow&quot;&gt;An Introduction to Quantum Feild Theory&lt;/a&gt; is a bit more advanced than Zee's book, and much less physically motivated.  There is a much larger focus on calculating cross-sections in this book, though.  It's also a bit older, so some newer developments aren't discussed.  It's very clearly written for those more interested in phenomonlogy than fundamental theory, too.  But you'll finish this book knowing how to do QFT, certainly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then, there's Stephen Weinberg's three part &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521550017&quot; rel=&quot;nofollow&quot;&gt;quantum field theory&lt;/a&gt; series.  These books are quite mathematically rigorous, and also quite difficult to read.  You'll understand the underlying math if you start with this book (and it sounds like that's what you want from your question and comment), but teaching yourself QFT out of this at a first pass would be like trying to learn E&amp;amp;M out of Jackson before having heard of an electric field.  There is a lot of insight and careful mathematical development that you can only find in these books, though.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-12-12T17:56:35.120" CommentCount="2" />
  <row Id="1854" PostTypeId="2" ParentId="1845" CreationDate="2010-12-12T18:12:26.457" Score="5" ViewCount="" Body="&lt;p&gt;Simplest thing coming to my mind -- wrapping your mobile (or anything that is able to receive a signal) in some aluminium foil (or in anything made of conductive material).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then your reciever won't be able to receive the signal. &quot;'Cause conductors block e.m. waves.&quot;&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2010-12-12T18:12:26.457" CommunityOwnedDate="2010-12-17T13:31:18.583" CommentCount="2" />
  <row Id="1855" PostTypeId="2" ParentId="1601" CreationDate="2010-12-12T20:25:54.657" Score="0" ViewCount="" Body="&lt;p&gt;For Newtonian geometric mechanics you may have a look at Oliva's Geometric mechanics book. For the Hamiltonian view my answer to &lt;a href=&quot;http://physics.stackexchange.com/questions/111/book-about-classical-mechanics&quot;&gt;this&lt;/a&gt; question may be usefull.&lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorUserId="667" LastEditorDisplayName="" LastEditDate="2011-03-27T10:50:58.510" LastActivityDate="2011-03-27T10:50:58.510" CommentCount="0" />
  <row Id="1856" PostTypeId="2" ParentId="1683" CreationDate="2010-12-12T20:34:15.837" Score="10" ViewCount="" Body="&lt;p&gt;Interesting problem. I think my approach and answer is very close to other posted solutions. I also added a possible scenario. The basic summary is it is the change in the average momentum of the water in the wagon that causes the wagon to move.  Requiring the water to distribute it self evenly in the wagon causes this relation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;average momentum of water in the wagon = $l\times$ mass flow out of wagon&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In cases where the wagon has been and forever shall expel water at a constant rate, the wagon stands still.&lt;/strong&gt; Imagine it being refilled from above its center of mass. You can actually do this same problem with an empty cart being filled from above instead of emptying below. With $l$ being the horizontal point from the wagon's center of mass at which the water falls down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The wagon does move if&lt;/strong&gt; there is some fluctuation in the mass flow out of the wagon either by abrupt starts/stops or by running out of water.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Variables&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$t_{c}\to$ time when wagon runs dry&lt;/li&gt;&#xA;&lt;li&gt;$l\to$ distance from center of mass of wagon to nozzle, positive $l$ implies nozzle is on the right side of the wagon&lt;/li&gt;&#xA;&lt;li&gt;$x(t)\to$ center of mass of wagon&lt;/li&gt;&#xA;&lt;li&gt;$x_{cm}(t)\to$ center of mass of everything&lt;/li&gt;&#xA;&lt;li&gt;$h(t)\to$ height of water in the container&lt;/li&gt;&#xA;&lt;li&gt;$m(t)\to$total mass of the wagon including any water it holds&lt;/li&gt;&#xA;&lt;li&gt;$m_{w}\to$ mass of initial water&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$m_{c}\to$ mass of the wagon; the c is for the critical point of $m(t)$ when all the water is gone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Originally c was for container but it makes sense $m(t_{c})=m_c$&#xD;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Frame of Reference&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$x(0)=0$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;$\dot{x}(0)=0$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Drainage&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to side step the issue of initial conditions for now. I'm going to treat the system as if the nozzle was always open and water has always been running. Only concerned with how a container with a constant cross section, S, would drain.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Torricelli's Law : Mass Flow =$-\dot{m}(t)$ : Mass of System&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$v(t)=\sqrt{2 g h(t)}$$&#xD;&#xA;$$-\dot{m}(t)=\rho s v(t)$$&#xD;&#xA;$$m(t)=\rho S h(t) + m_{c}$$&#xD;&#xA;Combine to eliminate $m(t)$ and $v(t)$&#xD;&#xA;$$\frac{\dot{h}}{\sqrt{h(t)}}=-\frac{s}{S}\sqrt{2 g}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to the differential equation:&#xA;$$h(t)=h(0){\left(1-t\sqrt{\frac{g {s}^{2}}{2 {S}^{2} h(0)}}\right)}^{2}$$&#xD;&#xA;$$h(t)=h(0){\left(1-\frac{t}{t_{c}}\right)}^{2}$$&#xD;&#xA;where $t_{c}=\sqrt{\frac{2 {S}^{2} h(0)}{g {s}^{2}}}$ and $h(t&amp;gt;t_c)=0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from there we get $m(t)$:&#xA;$$m(t)=\rho S h(0) {(1-\frac{t}{t_{c}})}^{2} + m_{c}$$&#xD;&#xA;$$m(t)=m_{w} {(1-\frac{t}{t_{c}})}^{2} + m_{c}$$&#xD;&#xA;and for $m(t&amp;gt;t_{c})$ is simply $m_{c}$, the mass of the wagon&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Center of Mass&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to find the center of mass we will account for all of it. At $t=0$, $x_{cm}(0)=x(0)$=0 since all the mass is in the wagon and we assumed equally distributed.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Wagon and its contents&#xA;$$m(t)x(t)$$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;Water that has left the wagon&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If water leaves the the wagon at $t=\tau$, then it will have speed $\dot{x}(\tau)$. Therefore its location is $f(t,\tau)$:&#xA;$$f(t,\tau) = l+x(\tau)+\dot{x}(\tau)(t-\tau)$$&#xD;&#xA;Then we just integrate to get their contributions.  We get their infinitesimal masses from our mass flow:&#xA;$$\int_0^t f(t,\tau) [-\dot{m}(\tau)]d\tau$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Combine&#xA;$$m(0)x_{cm}(t)=m(t)x(t)-\int_0^t f(t,\tau)\dot{m}(\tau)d\tau$$&#xD;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Differentiating gives us:&#xA;$$m(0)\dot{x_{cm}}(t)=\dot{m}(t)x(t)+m(t)\dot{x}(t)-f(t,t)\dot{m}(t)-\int_0^t \frac{df(t,\tau)}{dt}\dot{m}(\tau)d\tau$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplifying:&#xA;$$f(t,t)=x(t)+ l$$&#xD;&#xA;$$\frac{df(t,\tau)}{dt}=\dot{x}(\tau)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integration by parts:&#xA;$$\int_0^t\dot{m}(\tau)\dot{x}(\tau)d\tau=m(t)\dot{x}(t)-\int_0^tm(\tau)\ddot{x}(\tau)d\tau$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repalce:&#xA;$$m(0)\dot{x_{cm}}(t)=\dot{m}(t)x(t)+m(t)\dot{x}(t)-\dot{m}(t)(x(t)+ l)-m(t)\dot{x}(t)+\int_0^tm(\tau)\ddot{x}(\tau)d\tau$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation - In order these terms stand for:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;mass dissapearing from wagon at the center of mass&lt;/li&gt;&#xA;&lt;li&gt;momentum of wagon and its contents&lt;/li&gt;&#xA;&lt;li&gt;mass appearing outside of wagon at the nozzle&lt;/li&gt;&#xA;&lt;li&gt;last two terms account for momentum of water outside of the wagon&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Combining the first and third terms gives us the average momentum the water in the wagon must have to maintain its even distribution horizontally in the container.  They are not evidence for instantaneous dissapearance from the center and reappearance at the nozzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Result:&#xA;$$m(0)\dot{x_{cm}}(t)=-\dot{m}(t) l+\int_0^tm(\tau)\ddot{x}(\tau)d\tau$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&#xA;$$m(t)=m_{w} {(1-\frac{t}{t_{c}})}^{2} + m_{c}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Wagon w/ Brakes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this scenario, the wagon has been losing water before $t=0$.  However the force of the brakes keeps $\dot{x}(t)=0$. At $t=0$ the brakes are released and it is allowed to move.  This avoids any instantaneous jump in velocity by the wagon.  It also allows $x_{cm}$ to be a non-zero constant after $t=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $t=0$:&#xA;$$m(0)\dot{x_{cm}}(0)=-\dot{m}(0) l+\int_0^0m(\tau)\ddot{x}(\tau)d\tau$$&#xD;&#xA;$$m(0)\dot{x_{cm}}(0)=-\dot{m}(0) l$$&#xD;&#xA;$$\dot{x_{cm}}(0)=-\frac{\dot{m}(0)}{m(0)} l$$&#xD;&#xA;$$\dot{x_{cm}}(0)=\frac{2 l m_w}{t_c m(0)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $t&amp;gt;0$ there is no force from the brakes:&#xA;$$\ddot{x_{cm}}(t\ge0)=0$$&#xD;&#xA;$$\dot{x_{cm}}(t\ge0)=\frac{2 l m_w}{t_c m(0)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words in this situation at $t=0$ the momentum of the whole system matches that of the water in side the wagon. The only question now is as time evolves how is that momentum transfered to the wagon and water leaving the moving wagon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiate the system's momentum:&#xA;$$m(0)\ddot{x_{cm}}(t)=-\ddot{m}(t) l+\frac{d}{d t}\int_0^tm(\tau)\ddot{x}(\tau)d\tau$$&#xD;&#xA;$$0=-\ddot{m}(t)l+m(t)\ddot{x}(t)$$&#xD;&#xA;$$\ddot{x}(t)=\frac{\ddot{m}(t)l}{m(t)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Physical Considerations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we have a simple system as long as $\ddot{m}(t)$ is continuous.    The physical explanation is that if we abruptly closed the nozzle the water in the wagon does not come to an immediate stop relative to the wagon.  It sloshes around and after a certain relaxation time redistributes its momentum to the system as a whole. Similarly with the quick turn on, the water in the container can't just gain an average momentum to match $-\dot{m}(t)l$. Again there must be some relaxation time for the water to hit that equilibrium where it can evenly distribute itself in the wagon.  It is not that these situations are impossible but that my equations would not take into account these relaxation times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My situation just avoids that. The water in the wagon has already hit some equilibrium before $t=0$.  Also having the water move under its own weight provides a slow turn off.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Velocity of Wagon&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining the results from previous sections:&#xA;$$\ddot{x}(t)=\frac{2\frac{m_w}{{t_c}^2}l}{m_{w} {(1-\frac{t}{t_{c}})}^{2} + m_{c}}$$&#xD;&#xA;$$\ddot{x}(t)=\frac{2 l m_w}{{t_c}^2 m_c}{\left[\frac{m_w}{m_c}{(1-\frac{t}{t_c})}^{2}+1\right]}^{-1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\frac{du}{1+u^2}=\arctan(u)$$&#xD;&#xA;$$u=\sqrt{\frac{m_w}{m_c}}(1-\frac{t}{t_c})$$&#xD;&#xA;$$\dot{x}(t)=-\frac{2 l}{t_c}\sqrt{\frac{m_w}{m_c}}\int\frac{du}{1+u^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{x}(t)=\frac{2 l}{t_c}\sqrt{\frac{m_w}{m_c}}\left[\arctan\sqrt{\frac{m_w}{m_c}}-arctan\sqrt{\frac{m_w}{m_c}}\left(1-\frac{t}{t_c}\right)\right]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Extremely Heavy Wagon&lt;/strong&gt;: $\sqrt{\frac{m_w}{m_c}}\ll1$&#xD;&#xA;$$\arctan(x)\to x-\frac{1}{3}x^3$$&#xD;&#xA;$$\dot{x}(t_c)=\frac{2 l m_w}{t_c m_c}$$&#xD;&#xA;$$\dot{x_{cm}}(t\ge0)=\frac{2 l m_w}{t_c m(0)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes physical sense.  The wagon's final momentum is just about equal to our initial momentum. The higher order terms would account for the momentum that the dispensed water has.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Regular Wagon&lt;/strong&gt;:  $\sqrt{\frac{m_w}{m_c}}\gg1$&#xD;&#xA;$$\arctan(x)\to \frac{\pi}{2}$$&#xD;&#xA;$$\dot{x}(t_c)=\frac{\pi l}{t_c}\sqrt{\frac{m_w}{m_c}}$$&#xD;&#xA;$$\dot{x_{cm}}(t\ge0)=\frac{2 l m_w}{t_c m(0)}$$&#xD;&#xA;$$p_{cm}(t\ge0)=\frac{2}{\pi}\sqrt{\frac{m_w}{m_c}}p(t)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This case has the wagon with a significantly smaller portion of the systems momentum.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="800" LastEditorUserId="800" LastEditorDisplayName="" LastEditDate="2010-12-12T22:56:07.633" LastActivityDate="2010-12-12T22:56:07.633" CommentCount="11" />
  <row Id="1857" PostTypeId="2" ParentId="1845" CreationDate="2010-12-12T20:58:34.523" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Mapping EM Waves from Dipole Antennae&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Apparatus&lt;/strong&gt;: a wireless adapter with movable antennae, an OS with a wireless signal strength indicator (XP shows the number of bars for instance - or you could write a program for it). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Procedure&lt;/strong&gt;: place apparatus at different locations w.r.t the wireless router and measure the signal intensity as a function of location and the orientation of the receiving antennae &lt;strong&gt;relative to&lt;/strong&gt; the orientation of the emitting antennae (on the router).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Analysis&lt;/strong&gt;: From form of intensity function determine signal strenth loss as a function of distance and &lt;strong&gt;relative&lt;/strong&gt; orientation between the two antennae. Repeat in different locations and environments - indoors, outdoors, etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" LastActivityDate="2010-12-12T20:58:34.523" CommunityOwnedDate="2010-12-17T13:31:18.583" CommentCount="3" />
  <row Id="1858" PostTypeId="1" CreationDate="2010-12-12T21:16:27.570" Score="9" ViewCount="479" Body="&lt;p&gt;I'm trying to get motivated in learning the Atiyah-Singer index theorem.  In most places I read about it, e.g. wikipedia, it is mentioned that the theorem is important in theoretical physics.  So my question is, what are some examples of these applications?&lt;/p&gt;&#xA;" OwnerUserId="37" LastActivityDate="2010-12-14T22:05:36.053" Title="Where is the Atiyah-Singer index theorem used in physics?" Tags="&lt;mathematical-physics&gt;" AnswerCount="4" FavoriteCount="7" />
  <row Id="1859" PostTypeId="2" ParentId="550" CreationDate="2010-12-12T21:42:28.730" Score="0" ViewCount="" Body="&lt;p&gt;So the light given off from an incandecent light bulb is the same type (photonic) as the light given off from an LED (just a different frequency makup).&lt;/p&gt;&#xA;" LastActivityDate="2010-12-12T21:42:28.730" CommentCount="1" />
  <row Id="1860" PostTypeId="2" ParentId="202" CreationDate="2010-12-12T21:50:17.207" Score="1" ViewCount="" Body="&lt;p&gt;I want to add something to the speaker issue and show that the &lt;strong&gt;Fourier transformation&lt;/strong&gt; is &lt;strong&gt;not enough&lt;/strong&gt; to explain it (as maybe stated implicitly by the answers provided so far).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a first approximation you can describe the &lt;strong&gt;eigenfunctions&lt;/strong&gt; of a plate subject to some mechanical (e.g. acoustic excitation) by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Helmholtz_equation&quot; rel=&quot;nofollow&quot;&gt;Helmholtz equation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta\psi(x,y) + \mathbf{k}^2\psi(x,y) = 0$ for  ${x,y}\in \Omega = {(x,y)\in(-L_x...L_x,-L_y...L_y)}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solutions to this equation are plane waves if, and now comes the point that was not discussed so far, you have some rectangular &lt;strong&gt;boundary&lt;/strong&gt; with Dirichlet conditions, say&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\psi \equiv 0$ for $(x,y) \in \partial\Omega$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you really have (the m'th) resonances/eigenfrequencies at some $k_x*L_x = m$ and analogously for $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But&lt;/strong&gt; if the boundaries are somehow different, you get other eigenfunctions like &lt;a href=&quot;http://en.wikipedia.org/wiki/Bessel_function&quot; rel=&quot;nofollow&quot;&gt;Bessel Functions&lt;/a&gt; for a spherical system. The situation gets totally messi if you have a shape that is not integrable like for &lt;a href=&quot;http://en.wikipedia.org/wiki/Dynamical_billiards&quot; rel=&quot;nofollow&quot;&gt;dynamical billiards&lt;/a&gt;. Then you can observe &lt;strong&gt;chaotic&lt;/strong&gt; eigenfunctions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to sum up, it is not enough to know the Fourier transformation to explain such wave phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2010-12-12T21:50:17.207" />
  <row Id="1861" PostTypeId="2" ParentId="1601" CreationDate="2010-12-12T21:57:20.423" Score="2" ViewCount="" Body="&lt;p&gt;To answer 3: it really depends on why you want to learn this material.  To me, the modern view is important because it is very elegant and generalizable (you can &quot;do&quot; classical mechanics on any Poisson manifold).  It also leads to very interesting mathematics.  For example, the evolution of an observable is given by $f&amp;#39; = \{H,f\}$ where $H$ is the Hamiltonian function and $\{,\}$ is the Poisson bracket.  In the Heisenberg picture of quantum mechanics, an observable (represented by an operator $A$) evolves according to $A&amp;#39; = [H,A]$ where $H$ is the quantum mechanical Hamiltonian and $[,]$ is the Lie bracket (commutator).  This resemblance has led to things like deformation theory and quantization.  &lt;/p&gt;&#xA;" OwnerUserId="37" LastActivityDate="2010-12-12T21:57:20.423" />
  <row Id="1862" PostTypeId="1" CreationDate="2010-12-12T21:58:11.463" Score="2" ViewCount="185" Body="&lt;p&gt;I am studying social networks in terms of graph theory and linear algebra. I know that physicists have published and worked alot in this field. This causes me to assume that there are sub-fields in physics which overlap in the essence of their problems with those of small world networks. Which natural phenomenon exhibit these kind of features similar to small world networks?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know so maybe I can look at those problems to get inspiration that can be taken to social network theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best,&lt;/p&gt;&#xA;" OwnerUserId="813" LastActivityDate="2010-12-14T12:22:49.187" Title="which areas in physics overlap with those of social network theory for the analysis of the graphs?" Tags="&lt;mathematical-physics&gt;&lt;mathematics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="1863" PostTypeId="2" ParentId="1858" CreationDate="2010-12-12T22:02:58.687" Score="4" ViewCount="" Body="&lt;p&gt;First let me explain what the &lt;strong&gt;index&lt;/strong&gt; in question refers to. If the math gets too full of jargon let me know in the comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In physics we are often interested in the spectrum of various operators on some manifolds we care about. Eg: the Dirac operator in 3+1 spacetime. In particular the low-energy long distance physics is contained in the zero modes (ground states).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what the &quot;index&quot; measures, for the Dirac operator $D$ and a given manifold $M$, is the difference between the number of left-handed zero modes and the number of right-handed zero modes. More technically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ ind\,D = dim\,ker\,D - dim\,ker\,D^{+} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $D$ is the operator in question; $ker\,D$ is the kernel of $D$ - the set of states which are annihilated by $D$; and $ker\,D^{+}$ is the kernel of its adjoint. Then, as you can see, $ind\,D$ counts the difference between the dimensionalities of these two spaces. This number depends only on the topology of $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, the ASI theorem relates the topology of a manifold $M$ to the zero modes or ground states of a differential operator $D$ acting on $M$. This is obviously information of relevance to physicists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps someone else can elaborate more on the physical aspects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best reference for this and other mathematical physics topics, in my opinion, is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0750306068&quot; rel=&quot;nofollow&quot;&gt;Nakahara&lt;/a&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-12T22:02:58.687" CommunityOwnedDate="2010-12-12T22:02:58.687" />
  <row Id="1864" PostTypeId="2" ParentId="1683" CreationDate="2010-12-12T23:12:22.697" Score="-1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Short version:&lt;/strong&gt; movement inside the closed system cannot accelerate it. Zero horizontal speed at exit means zero speed at t-&gt;infinity. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;More detailed version:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me transfer the problem to a simpler one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have an open wagon with me standng on one side of it holding a heavy box. Now I will start running towards the other side of the wagon. This will cause the wagon to move in the opposite direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a certain point I will have to decelarate so that I stop at the other side of the wagon. This will create eqaual force to accelerating thus compensating any speed that developed during the accelerating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position of the wagon will be changed so that the center of mass will not have moved. The speed will be equal to starting speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I drop the box straight down. (I will use a bit of force to simulate the water pressure, &#xA; but that is not important) Speed is zero, wagon moved box is down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's say I have multiplied, have negligible weight and the box is a molecule of water. The final speed will be certainly zero again. The question is, what the displacement of the wagon will be. I have two answers and cannot choose either:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The centre of mass has to be kept the same (horizontally, gravitation can move it vertically down). This determines the final position of the wagon. &lt;/li&gt;&#xA;&lt;li&gt;The final displacement is speed integrated over time. Now for each molecule that will start moving left, there will be one stopping at the nozzle. This would compensate the forces in real time keeping the speed at zero and so the displacement. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Please correct me if my analogy is wrong at some point and try to answer the question about final displacement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit - more explanations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the wagon moves during the process it's true that the water will have a momentum relative to rail and it will travel at the same speed as the wagon. That means there will be no net force from this water coming down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a very long tube open on both sides filled with water. If you put this tube vertically in a homogeneous gravitation field the water will flow (fall) out of it. If it moves at a constant speed the water would behave the same relative to the tube. The outside observer would see a tube moving to the side and a column of water moving down and to the side (at the same speed, so it would stay under the tube all the time). The same goes for the water from the nozzle: it will always have the same horizontal speed as the wagon at the point of leaving thus having no effect whatsoever on its movement. This is true disregarding the speed of the wagon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said this the only forces affecting the whole water-wagon system are those caused by the internal movement of water. On this frictionless rail you can change the wagon's position from inside only at the cost of regrouping the stuff inside (changing the mass distribution through the system). Someone (let's say a lobster) walking on a wagon (of zero weigh for simplification) on a frictionless rail cannot move relative to the rail. It is the same as if this lobster was trying to walk on frictionless ice: there would be no reactive force to move him. Looking at the lobster on the zero-weight wagon we would see a lobster walking, though not moving, and a wagon moving under him. As the only mass in this system is the lobster, the centre of mass would not move. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Returning to the water - after opening the nozzle the water starts moving to the left and because there was no speed at t=0 there had to be some acceleration. than the water is gradually moved towards the left end of the wagon where it loses its horizontal speed and leaves the wagon at zero horizontal speed. While stopping the decelerating will compensate any forces (and speed) created during the acceleration. Whether this is going on at zero or non zero speed relative to the rail has no influence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we have no external force in the horizontal direction, there centre of mass has to stay unmoved (which requires the wagon to move). At the same time the zero momentum of the water train system has to be preserved so unless the water leaves the train with non-zero horizontal speed relative to the wagon the wagon cannot end up with non-zero horizontal speed relative to the water expelled. &lt;/p&gt;&#xA;" OwnerUserId="816" LastEditorUserId="816" LastEditorDisplayName="" LastEditDate="2010-12-13T08:58:41.963" LastActivityDate="2010-12-13T08:58:41.963" CommentCount="15" />
  <row Id="1865" PostTypeId="2" ParentId="1858" CreationDate="2010-12-13T00:31:21.100" Score="4" ViewCount="" Body="&lt;p&gt;The equations of motion, or the equations of instantons, or solitons, or Einstein's equations, or just about any equations in physics, are differential equations.  In many cases, we are interested in the &lt;em&gt;space of solutions&lt;/em&gt; of a differential equation.  If we write the total (possibly nonlinear) differential equation of interest as $L(u) = 0,$ we can &lt;em&gt;linearize&lt;/em&gt; near a solution $u_0,$ i.e. write $u = u_0 + v$ and expand $L(u_0 + v) = 0 + L&amp;#39;|_{u_0}(v) + ... =: D(v)$ to construct a linear equation $D(v)=0$ in the displacement $v.$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A linear differential equation is like a matrix equation.  Recall that an $n\times m$ matrix $M$ is a map from $R^n$ to $R^m$, and $dim(ker(M)) - dim(ker(M^*)) = n-m,$ independent of the particular matrix (or linear transformation, more generally).  This number is called the &quot;index.&quot;  In infinite dimensions, these numbers are not generally finite, but often (especially for elliptic differential equations) they are, and depend only on certain &quot;global&quot; information about the spaces on which they act.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The index theorem tells you what the index of a linear differential operator ($D,$ above) is.  You can use it to calculate the &lt;em&gt;dimension&lt;/em&gt; of the space of solutions to the equation $L(u)=0.$  (When the solution space is a manifold [another story], the dimension is the dimension of the tangent space, which the equation $D(v)=0$ describes.)  It does &lt;em&gt;not&lt;/em&gt; tell you &lt;em&gt;what&lt;/em&gt; the actual space of solutions is.  That's a hard, nonlinear question.&lt;/p&gt;&#xA;" OwnerUserId="20" LastActivityDate="2010-12-13T00:31:21.100" CommentCount="6" />
  <row Id="1866" PostTypeId="2" ParentId="1852" CreationDate="2010-12-13T02:21:02.590" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;33 MJ is the electrical energy. I think the projectile is about 1kg, so the efficiency is about 10%, not so bad.&lt;/li&gt;&#xA;&lt;li&gt;Plasma from electrical breakdown, which then gets accelerated the same way the projectile does, with ExB force&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="775" LastActivityDate="2010-12-13T02:21:02.590" CommentCount="2" />
  <row Id="1867" PostTypeId="2" ParentId="409" CreationDate="2010-12-13T02:31:21.347" Score="3" ViewCount="" Body="&lt;p&gt;For the record, gravitational lensing has never been used to detect or even observe a known black hole. The above spectacular image of lensed galaxies is due to a galaxy cluster which makes the lens, and is roughly a million times more massive than the most massive black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all likelihood, we will detect black holes with gravitational waves (e.g., LIGO or LISA) before strong lensing. However &lt;em&gt;microlensing&lt;/em&gt; may very soon be used to detect isolated BHs in our galaxy, along the line of site to individual stars in the galactic bulge.&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2010-12-13T02:31:21.347" CommentCount="1" />
  <row Id="1868" PostTypeId="2" ParentId="560" CreationDate="2010-12-13T02:35:17.023" Score="2" ViewCount="" Body="&lt;p&gt;At higher redshifts, we actually see huge clouds of neutral Hydrogen (atomic, not molecular) in the form of absorption lines from distant quasars, called the &quot;Lyman alpha forest&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These clouds may eventually form galaxies and stars, but they are currently just gas, an not particularly low density either. &lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2010-12-13T02:35:17.023" />
  <row Id="1869" PostTypeId="1" CreationDate="2010-12-13T02:46:44.200" Score="8" ViewCount="666" Body="&lt;p&gt;It's a truism that in science it is at least as important to state the reasons why a theory or idea might be wrong, as to state the reasons why it is might be correct.  For example, early renormalization methods made experimental predictions that were very accurate, but the inventor's famously worried for decades about the conceptual problems of divergences; and many of their papers/articles/books on the subject include a long discussion about how acutely worried they were that their method for handling divergences was deeply flawed somehow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intention with this post is not to engage any user in a debate about whether string theory is true or not.  I think anyone who takes string theory seriously should be happy to acknowledge as clearly as possible all the potential theoretical issues/problems with the theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore I would like to ask, what are the major theoretical problems which would lead a reasonable person to be worried that string theory is not an accurate theory?  I am excluding the fact that string theory has made no (major) predictions that have been confirmed, I am only interested in conceptual and theoretical ideas which might discredit the theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please keep all answers to one issue per post.&lt;/p&gt;&#xA;" OwnerUserId="442" LastEditorUserId="442" LastEditorDisplayName="" LastEditDate="2010-12-13T03:03:01.633" LastActivityDate="2011-01-25T15:38:55.950" Title="What reasons are there to suspect string theory may be an incorrect theory?" Tags="&lt;string-theory&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="2" />
  <row Id="1870" PostTypeId="2" ParentId="560" CreationDate="2010-12-13T04:37:51.480" Score="2" ViewCount="" Body="&lt;p&gt;There certainly is gas within galaxy clusters. It tends to have very high temperature, circa a hundred million K, and is detectable by diffuse x-ray emission. Of course we have the elusive dark matter, which is probably more diffusely spread out than luminous matter. So maybe your question should be modified to, &quot;is the space between galaxy clusters a vacuum?&quot;. I suspect the answer to that is also no, although the density is clearly very low.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-12-13T04:37:51.480" />
  <row Id="1872" PostTypeId="2" ParentId="1869" CreationDate="2010-12-13T06:16:46.597" Score="2" ViewCount="" Body="&lt;p&gt;If I understand what you're looking for correctly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One issue is that string theory predicts the number of spacetime dimensions to be 10 (or 11 in M-theory), larger than the 4 we observe. Of course, it's possible to explain that away by postulating that 6 (or 7) of the dimensions form a compact manifold with characteristic scale much smaller than we are able to detect, but as far as I know, nobody knows whether that's a realistic configuration. The predictions of the theory would depend on the particular topology of the compact manifold, and it's not clear whether there is any choice of topology that makes the theory correspond to reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And even if there is, it'd still be an open question why nature has &quot;chosen&quot; that particular topology, out of all the possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;N.B. I haven't studied this in full detail so perhaps I've gotten some of the details wrong, in which case hopefully someone more knowledgeable will correct me.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-12-13T06:16:46.597" CommentCount="4" />
  <row Id="1873" PostTypeId="1" AcceptedAnswerId="1874" CreationDate="2010-12-13T09:19:45.943" Score="6" ViewCount="414" Body="&lt;p&gt;I realize that path integral techniques can be applied to quantitative finance such as option value calculation. But I don't quite understand how this is done. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to explain this to me in qualitative and quantitative terms?&lt;/p&gt;&#xA;" OwnerUserId="605" LastEditorUserId="605" LastEditorDisplayName="" LastEditDate="2010-12-15T01:09:56.890" LastActivityDate="2010-12-16T10:14:31.070" Title="The Concepts of Path Integral in Quantitative Finance" Tags="&lt;path-integral&gt;" AnswerCount="1" CommentCount="22" FavoriteCount="2" />
  <row Id="1874" PostTypeId="2" ParentId="1873" CreationDate="2010-12-13T12:33:50.407" Score="11" ViewCount="" Body="&lt;p&gt;The fundamental equation which serves as the basis for the path-integral formulation of finance and many physical problems is the Chapman-Kolmogorov equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p(X_f|X_i)=\int p(X_f|X_k)p(X_k|X_i) dX_k $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is analogous to the following equation for amplitudes in quantum mechanics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle X_f|X_i \rangle=\int \langle X_f|X_k\rangle\langle X_k|X_i\rangle dX_k $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's right, it's the same form, but the interpretation of the basic entities changes. In the former, they are probability densities and thus real and positive, in the latter they are probability amplitudes and thus complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The class of physical problems that can be tackled with the first type of equation are called Markov processes, their characteristic is that the state of the system depends only on its previous state. Despite its seeming limitedness, this comprises many phenomena since any process with a long but finite memory can be mapped onto a Markov process provided the state space is enlarged appropriately. On the other hand, the second equation is pretty natural and general in quantum mechanics. It is basically stating that the unity operator can always be decomposed into a, possibly overcomplete, sum of pure states &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbb{I}=\int |X_k\rangle\langle X_k| dX_k \; .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, constructing a path integral is done by slashing up the path from $X_i$ to $X_k$ into ever smaller components. Let's suppose that the endpoints are fixed, then we might assume that to go from one endpoint to another, the system has to go through paths $(X_i,X_1(t_1),X_2(t_2),\ldots,X_n(t_n),X_f)$. This leads to the following integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p(X_f|X_i)=\int\cdots\int \prod_{k=0}^n p(X_{k+1}(t_{k+1})|X_k(t_k)) \prod_{k=1}^n dX_k(t_k) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where I put $X_0(t_0)=X_i$ and $X_{n+1}(t_{n+1})=X_f$. The tricky part is now to see if the limit can be defined meaningfully. This can be very problematic, especially in the quantum case. Ironically, the cases that are used for finance and statistical mechanics are often much more well-behaved. This is again related to one integral being over complex numbers and the other over real numbers, but it's not the only reason. Up till now, I have not been specific about the kind of system I want to study, this will play an important role as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's take an option which is a financial security of which the price is dependent on the price of the underlying stock and time. So we can write $O(X,t)$ for the price of the option and we'll assume the underlying stock follows a geometric brownian motion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dX}{X}=\mu dt + \sigma dW$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $W$ represents a Wiener process with increments $dW$ having mean zero and variance $dt$. Also assume that the pay-off of the option at the expiration time $T$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$O(X_T,T)=F(X_T)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $F$ a given function of the terminal stock price.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, &lt;a href=&quot;http://en.wikipedia.org/wiki/Black%E2%80%93Scholes&quot; rel=&quot;nofollow&quot;&gt;Fisher Black and Myron Scholes&lt;/a&gt; have shown that the option, under the 'no arbitrage' assumption, satisfies the following PDE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial O}{\partial t} + \frac{1}{2}\sigma^2X^2\frac{\partial^2 O}{\partial X^2} + r X \frac{\partial O}{\partial X} - rO = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in which $r$ is the risk free interest rate. If instead of the geometric brownian motion variable $X$, I reformulate this into $x=\ln X$ which is an arithmetic brownian motion variable, I can reformulate the equation as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial O}{\partial t} + \frac{1}{2}\sigma^2\frac{\partial^2 O}{\partial x^2} + (r-\frac{\sigma^2}{2}) \frac{\partial O}{\partial x} - rO = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is nothing else but a special case of the PDE's that can be solved by using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Feynman-Kac_formula&quot; rel=&quot;nofollow&quot;&gt;Feynman-Kac formula&lt;/a&gt;, which includes also the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fokker%E2%80%93Planck_equation&quot; rel=&quot;nofollow&quot;&gt;Fokker-Planck equation&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Smoluchowski_equation&quot; rel=&quot;nofollow&quot;&gt;Smoluchowski equation&lt;/a&gt;, both related to the description of diffusion processes in physics. In the diffusion problem, O is to be interpreted as a distribution of velocities of the particle (Fokker-Planck) or of the positions of the particle (Smoluchowski). That's how we relate to what I introduced above. Also note that the Schrödinger equation in quantum mechanics is very similar in form, except you'll get complex coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Feynman-Kac formula tells us that the solution to the PDE is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$O(X,t) = e^{-r(T-t)}\mathbb{E}\left[ F(X_T)|X(t)=X \right]$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is this expectation value that will now be represented as a pathintegral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$O(X,t) = e^{-r(T-t)}\int_{-\infty}^{+\infty}\left(\int_{x(t)=x}^{x(T)=x_T} F(e^{x_T}) e^{A_{BS}(x(t&amp;#39;))} \mathcal{D}x(t&amp;#39;)\right) dx_T$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{BS}(x(t&amp;#39;))=\int_t^{T} \frac{1}{2\sigma^2}\left(\frac{dx(t&amp;#39;)}{dt&amp;#39;}-\mu\right)^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the action functional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this path integral can be built is the same explained before, here it is possible to split the conditional expectation ever further in smaller intervals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rcl}\mathbb{E}\left[ F(e^{x_T})|x(t)=x \right] &amp;amp; = &amp;amp; \int_{-\infty}^{+\infty} F(e^{x_T}) p(x_T|x(t)=x) dx_T \\&#xD;&#xA;&amp;amp; = &amp;amp; \int_{-\infty}^{+\infty} F(e^{x_T}) \int_{\tilde{x}(t)=x}^{\tilde{x}(T)=x_T} p(x_T|\tilde{x}(\tilde{t})) p(\tilde{x}(\tilde{t})|x(t)=x) d\tilde{x}(\tilde{t}) dx_T \end{array}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the conditional probabilities satisfying the PDE for the arithmetic brownian motion as noticed above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll stop here for now, but I refer to the &lt;a href=&quot;http://www.springerlink.com/content/k0u754r084j3t757/&quot; rel=&quot;nofollow&quot;&gt;following article&lt;/a&gt; for further details.&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2010-12-16T10:14:31.070" LastActivityDate="2010-12-16T10:14:31.070" CommentCount="10" />
  <row Id="1875" PostTypeId="1" AcceptedAnswerId="1890" CreationDate="2010-12-13T12:53:57.763" Score="0" ViewCount="223" Body="&lt;p&gt;Here is a link to a &lt;a href=&quot;http://www.youtube.com/watch?feature=player_embedded&amp;amp;v=0JOgTKigxGs&quot; rel=&quot;nofollow&quot;&gt;video of a self-balencing unicycle&lt;/a&gt;. I'm curious as to how this works. In particular:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What provides the torque to keep the rider from falling over? I know it's somehow provided by the motor, but this isn't an external torque (if we define the unicycle as our system), and I assume we need an external torque to change the angular momentum of the unicycle. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you were to try to ride this thing on ice, I assume that it could not prevent the rider from falling over. Is this correct? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What would the free body diagrams look like for this unicycle when it is in the act of falling over? When it is righting itself? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="807" LastActivityDate="2010-12-16T08:53:30.827" Title="Self Balancing Unicycle" Tags="&lt;torque&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="1876" PostTypeId="1" AcceptedAnswerId="1885" CreationDate="2010-12-13T13:14:15.237" Score="6" ViewCount="353" Body="&lt;p&gt;Can you tell me how graph theory comes into physics, and the concept of small world graphs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(inspired to ask from comment from sean tilson in): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/1862/which-areas-in-physics-overlap-with-those-of-social-network-theory-for-the-analys&quot;&gt;which areas in physics overlap with those of social network theory for the analysis of the graphs?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best,&lt;/p&gt;&#xA;" OwnerUserId="813" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-12-13T13:33:12.673" LastActivityDate="2010-12-13T21:54:39.557" Title="Why are physicists interested in graph theory?" Tags="&lt;graph-theory&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="3" />
  <row Id="1877" PostTypeId="2" ParentId="691" CreationDate="2010-12-13T14:09:43.363" Score="1" ViewCount="" Body="&lt;p&gt;I agree with Greg. There is no infinite blueshift, because you as an observer are also getting blue-shifted. In fact, due to the tidal forces, the radiation coming just behind you actually appears redshifted (although not infinitely so).&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2010-12-13T14:09:43.363" />
  <row Id="1878" PostTypeId="2" ParentId="1876" CreationDate="2010-12-13T14:20:00.383" Score="6" ViewCount="" Body="&lt;p&gt;Richard Feynman reformulated quantum mechanics (and quantum field theory) in terms of a path integral, meaning that in order to find the likelihood of some process occurring, you take a kind of weighted average over all potential trajectories.  The weighting function is the exponentiated &quot;action,&quot; $\exp(iS/\hbar).$ and the dominant contribution comes from paths which extremize this function, i.e. classical trajectories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically -- almost always -- this integral is too hard for anyone to do (let alone define rigorously), so Feynman developed a perturbation theory, an expansion in terms of graphs.  The nature of the graphs depends on the interactions and coupling constants of your model -- that is, on the action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An (oversimplified) example:  Suppose you only had one degree of freedom, x, and the action is $S_0(x) = i\hbar x^2/2$, so that $\exp(iS_0/\hbar) = \exp(-x^2/2).$  (You can ignore $\hbar$ in this example.)  Then the path integral is $\int \exp(-x^2/2) dx$ and equals $\sqrt{\pi}$.  However, if we add a cubic &quot;interaction&quot; term, so $S = S_0 - i\hbar a x^3$ then we can expand $\int \exp(iS/\hbar)$ in powers of $a,$ the first nonzero contribution being $a^2 \int (x^3)^2 \exp(-x^2/2) dx,$ which you can do exactly.  (The term linear in $a$ is zero because the three powers of $x$ can't be paired up [&quot;contracted&quot;].)  The graph for this term has two vertices (the two powers of $a$), each with three edges attached (the three powers of $x$ in the interaction term).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So graphs are ubiquitous in QFT!&lt;/p&gt;&#xA;" OwnerUserId="20" LastEditorUserId="20" LastEditorDisplayName="" LastEditDate="2010-12-13T16:33:36.520" LastActivityDate="2010-12-13T16:33:36.520" />
  <row Id="1879" PostTypeId="2" ParentId="1876" CreationDate="2010-12-13T14:22:52.220" Score="1" ViewCount="" Body="&lt;p&gt;One context in which graphs can be useful in physics is in the discrete representation of spacetime in quantum gravity, where events are represented by the nodes of a type of poset (partially ordered set) called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Causal_sets&quot; rel=&quot;nofollow&quot;&gt;causet&lt;/a&gt; and causal relationships are represented by the edges. This is particularly suited to a graph-theoretic interpretation, since posets can be intuitively visualized as &lt;a href=&quot;http://en.wikipedia.org/wiki/Directed_acyclic_graph&quot; rel=&quot;nofollow&quot;&gt;DAG&lt;/a&gt;s.&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="354" LastEditorDisplayName="" LastEditDate="2010-12-13T14:47:44.967" LastActivityDate="2010-12-13T14:47:44.967" />
  <row Id="1880" PostTypeId="2" ParentId="1876" CreationDate="2010-12-13T16:34:48.350" Score="3" ViewCount="" Body="&lt;p&gt;Not sure this is what you are aiming at, but &lt;strong&gt;graphs are ubiquitous in statistical physics!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They really crop up all over the place. To give you some ideas from the top of my head:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Probability&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Classical statistical physics is built on the concept of microstates. Each microstate has a certain energy $E$ gets a probability $P = {1 \over Z} exp(-\beta E)$ where $\beta$ is a parameter depending on the temperature of the system and $$Z = \sum_{microstates} exp(-\beta E)$$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Partition_function_%28statistical_mechanics%29&quot; rel=&quot;nofollow&quot;&gt;partition function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you make this assignment of probabilities you can ask for things like average energy of the system and fluctuations and lots of other interesting stuff. And all of that stuff can be computed easily if you can somehow carry out the summation and determine the partition function. That is, the problem can often (at least in the discrete case) be reduced to counting, which means &lt;em&gt;combinatorics&lt;/em&gt;. And often the combinatorial problem has to do with &lt;em&gt;graph theory&lt;/em&gt; (either directly or by means of some clever duality).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Crystals&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Statistical physics often investigates lattices. This is because they model crystals or other ordered forms of matter. These are very special graphs that possess translational symmetries (and more generally also some rotational and reflectional symmetries; think about hexagonal lattice). Once again you can define energy for microstates as in the first example and proceed to formulate the probabilistic problem. But it can be observed that lots of methods that work for investigation of the system on the lattice can actually be generalized to arbitrary graphs.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Arbitrary graphs&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Perhaps nicest connection (and one positively stunning if you haven't heard about it before) is between correlation function of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_free_field&quot; rel=&quot;nofollow&quot;&gt;Gaussian free field&lt;/a&gt; on the graph and random walks on the same graph (see e.g. &lt;a href=&quot;http://tcsmath.wordpress.com/2010/12/09/open-question-cover-times-and-the-gaussian-free-field/&quot; rel=&quot;nofollow&quot;&gt;this recent blog&lt;/a&gt; on the topic). This is a discrete version of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Path_integral_formulation&quot; rel=&quot;nofollow&quot;&gt;Feynman path integral&lt;/a&gt; which gives you probability amplitudes of particle getting from one place to another in terms of summing over every path between the two points.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Polymer model&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;One more model I'd like to mention is the Polymer model. The idea is that you have some objects, called &lt;em&gt;polymers&lt;/em&gt;, that usually live on some kind of lattice (imagine e.g. cycles on the edges of the hexagonal lattice). Now the requirement is that these objects do not occupy the same space of the lattice (i.e. they don't intersect). This idea can be rigorously captured by the means of a huge infinite-dimensional graph where vertices are all of the possible polymers with edges between any two of them that are not compatible (that is, when they intersect on the original lattice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks like a hard problem but actually it can be investigated by the means of cluster expansion. To give (an oversimplified) idea: it all stems from the simple but very useful combinatorial identity&#xA;$$ \exp(\sum_i x_i) = \sum_N {1 \over N!} (\sum_i x_i)^N = \sum_N \sum_{i_1 + \cdots + i_N = N} \prod_k{x_k^{i_k} \over i_k!}$$&#xD;&#xA;Using this one can transform the partition function $Z$ (which is an ugly sum over all possible subgraphs of the original huge graph) into an exponential of sum over just &lt;strong&gt;connected&lt;/strong&gt; subgraphs of the polymer graph. Now connected subgraphs are much nicer objects than arbitrary subgraphs and one indeed obtains nice results by expanding the logarithm of the partition function in this way.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-13T16:41:23.370" LastActivityDate="2010-12-13T16:41:23.370" CommentCount="7" />
  <row Id="1881" PostTypeId="2" ParentId="1845" CreationDate="2010-12-13T18:01:40.073" Score="3" ViewCount="" Body="&lt;p&gt;Show that a metal dish pantenna increases Wi-Fi reception  &lt;a href=&quot;http://www.instructables.com/id/Wifi-Signal-Strainer-WokFi/&quot; rel=&quot;nofollow&quot;&gt;http://www.instructables.com/id/Wifi-Signal-Strainer-WokFi/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2010-12-13T18:01:40.073" CommunityOwnedDate="2010-12-17T13:31:18.583" />
  <row Id="1884" PostTypeId="2" ParentId="1876" CreationDate="2010-12-13T19:52:58.380" Score="0" ViewCount="" Body="&lt;p&gt;Graph theory is very useful in design and analysis of electronic circuits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very useful in designing various control systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Signal-flow_graph&quot; rel=&quot;nofollow&quot;&gt;Signal Flow Graphs&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Mason%27s_rule&quot; rel=&quot;nofollow&quot;&gt;Meson's Rule&lt;/a&gt; make your life a lot easier while trying to find transfer functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, while solving differential equations numerically Graph Theory is used for mesh generation.&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-12-13T19:52:58.380" />
  <row Id="1885" PostTypeId="2" ParentId="1876" CreationDate="2010-12-13T20:16:47.043" Score="3" ViewCount="" Body="&lt;p&gt;I don't know why physicists are interested in &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_network&quot; rel=&quot;nofollow&quot;&gt;complex network theory&lt;/a&gt;, but well, whenever you can create a physical model describing some behavior you could call it &quot;physics&quot; (&lt;a href=&quot;http://en.wikipedia.org/wiki/Econophysics&quot; rel=&quot;nofollow&quot;&gt;econophysics&lt;/a&gt;, sociophysics, etc), and this is likely the reason why they study complex network. I will just answer the 2nd part of the question — the concept of &lt;a href=&quot;http://en.wikipedia.org/wiki/Small-world_network&quot; rel=&quot;nofollow&quot;&gt;small-world network&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Short average distance&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The small-world phenomena is best known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Six_degrees_of_separation&quot; rel=&quot;nofollow&quot;&gt;six degree of separation&lt;/a&gt;, i.e. two persons are related to each other by at most 6 steps in a network of human relationship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this network, people are represented by &lt;strong&gt;nodes&lt;/strong&gt;, and if two people knows each other directly we create a &lt;strong&gt;link&lt;/strong&gt; between the two nodes. If there are two nodes without a direct link, we could possibly take other routes. The smallest number of links need to be traversed is called the &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Distance_%28graph_theory%29&quot; rel=&quot;nofollow&quot;&gt;distance&lt;/a&gt;&lt;/strong&gt; (a.k.a. path length) between the two nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small-world network (with size &lt;em&gt;N&lt;/em&gt;)  needs to satisfy a condition, that the &lt;strong&gt;average distance&lt;/strong&gt; $\langle\ell\rangle$ over all possible node pairs is &quot;short&quot;, i.e.&#xA;$ \langle\ell\rangle \sim \log N $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many natural and artificial complex networks, like the human relationship network above, neural network, metabolic networks etc. have short average distance, which is why this property is being studied, because physicists wanted to have a model that describes most of these complex networks. The &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93R%C3%A9nyi_model&quot; rel=&quot;nofollow&quot;&gt;Erdős-Rényi model&lt;/a&gt;&lt;/strong&gt; — which is essentially a network that the links are constructed with some constant probability — has the short average distance property.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;High clustering coefficient&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The ER model was one of the first models used to describe real-world complex networks, but was soon found to be insufficient because it doesn't have another important property — a &quot;high&quot; &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Clustering_coefficient&quot; rel=&quot;nofollow&quot;&gt;clustering coefficient&lt;/a&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (local) clustering coefficient was introduced by Duncan J. Watts and Steven Strogatz in 1998 as a measure of how well nodes are &quot;clustered&quot; together locally. It is defined as&#xA;$$ c_{v} = \frac{\text{number of triangles containing }v}{\frac{k_v(k_v-1)}2} $$ where $k_v$ is the number of links connected to $v$, i.e. its &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Degree_%28graph_theory%29&quot; rel=&quot;nofollow&quot;&gt;degree&lt;/a&gt;&lt;/strong&gt;. The local clustering coefficient is actually a ratio of &quot;neighbors of $v$ that knows each other&quot;, and &quot;maximum number of potential neighbors of $v$ that can know each other&quot;. For example, in this graph:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/SalWt.png&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;the clustering coefficient of the central green node is&#xA;$$ c = \frac{\text{number of triangles}}{\frac{k(k-1)}2} = \frac{4}{\frac{5(5-1)}2} = 0.4. $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the local clustering coefficient is 0, then all your friends don't know each other, which is not true in the social network. The expected behavior is a large clustering coefficient, where groups of your friends know each other (thus forming triangles, i.e. A knows B, B knows C, C knows A). This property is where the ER model breaks down — its clustering coefficient is close to zero with the same number of nodes and links with a real network.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a network has &lt;em&gt;both&lt;/em&gt; short average distance and high average local clustering coefficient, we call it a &lt;strong&gt;small-world network&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;And beyond&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Watts_and_Strogatz_model&quot; rel=&quot;nofollow&quot;&gt;Watts-Strogatz model&lt;/a&gt;&lt;/strong&gt; was invented to address the small-world property. However, it was soon determined that even the WS model is not good enough as it is not &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Scale-free_network&quot; rel=&quot;nofollow&quot;&gt;scale-free&lt;/a&gt;&lt;/strong&gt;.  And then the &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Barab%C3%A1si%E2%80%93Albert_model&quot; rel=&quot;nofollow&quot;&gt;Barabási-Albert model&lt;/a&gt;&lt;/strong&gt; was created to describe why real-world networks both small-world and scale-free, although it also cannot explain other properties like the clustering coefficient distribution, hierarchical structure, etc, and of course more and more sophisticated models are proposed as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, these properties are studied to construct a universal model that describes all (if not possible, &quot;most&quot;) real-world complex networks, and use it to test and improve behaviors such as error and attack tolerance, evolution dynamics etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't fear a lot of mathematics, the 2002 review paper &lt;em&gt;&lt;a href=&quot;http://arxiv.org/pdf/cond-mat/0106096&quot; rel=&quot;nofollow&quot;&gt;Statistical mechanics of complex networks&lt;/a&gt;&lt;/em&gt; by Réka Albert and Albert-László Barabási is a (IMO) must-read classic for anyone beginning to study complex networks. All of the above can be found in this paper.&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-13T21:54:39.557" LastActivityDate="2010-12-13T21:54:39.557" CommentCount="4" />
  <row Id="1887" PostTypeId="2" ParentId="560" CreationDate="2010-12-13T21:42:24.287" Score="4" ViewCount="" Body="&lt;p&gt;I might add some further notes to the actual material things existing in intergalactic space. One might wonder but the notion that there is &lt;strong&gt;space&lt;/strong&gt; is already stating that there is more than nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It implies that there is at least vacuum which is a pretty interesting thing on its own. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Classical harmonic oscillator&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Maybe you know that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_oscillator&quot; rel=&quot;nofollow&quot;&gt;harmonic oscillator&lt;/a&gt; has energy levels&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_n = \hbar \omega \left( n + \frac{1}{2}\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and an astonishing result is that the lowest energy state is $E_0 = \frac{1}{2}\hbar\omega &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Quantum electrodynamical oscillator&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Coming back to the vacuum, the situation is somewhat comparable. Considering Heisenberg's Principle of Uncertainty in its energy-time form,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta{t}\cdot\Delta{E} \geq \hbar$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can see already that a state of a quantum system with &lt;strong&gt;definite zero energy&lt;/strong&gt; for all times &lt;strong&gt;cannot exist&lt;/strong&gt;, even though the expectation value might vanish.&lt;br&gt;&#xA;Going more into detail, we see that the operator of the vector potential fullfills the wave equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta{A_l} - \frac{1}{c^2}\partial_{tt}A_l = 0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and a Helmholtz equation if one puts $\partial_{tt}\rightarrow{-\omega^2}$. This equation is usually tackled by separation of variables and after some math we arrive at a Hamilton&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H = \frac{1}{2}\sum_{\lambda}\left({p^2_\lambda+\omega_\lambda^2\lambda{q^2_\lambda}}\right)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where now $\lambda$ accounts for some mode index. And here comes the magic. This is a description equation for &lt;strong&gt;harmonic oscillators&lt;/strong&gt;! But here we run into a &lt;strong&gt;conceptional difficulty&lt;/strong&gt;. The vacuum energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_{vac} = \frac{1}{2}\sum_\lambda{\hbar\omega_\lambda}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is &lt;strong&gt;infinitely large&lt;/strong&gt; since there are infinitely many modes of the vacuum. But this is not very physical, so most of the time for calculations you just &quot;leave out&quot; this part.  &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Implications of a vacuum energy&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;In the case of different separated domains where you are able to allow a different  different number of modes (e.g. via metal plates), this energy will be different for those domains resulting in a force which is the famous &lt;a href=&quot;http://en.wikipedia.org/wiki/Casimir_effect&quot; rel=&quot;nofollow&quot;&gt;Casimir effect&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But vacuum energy has other implications. One hope it that it might some day explain the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmological_constant&quot; rel=&quot;nofollow&quot;&gt;cosmological constant&lt;/a&gt; in terms of a unified field theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I hope, I could convince you that &quot;empty&quot; might be much more one would expect :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2010-12-13T21:42:24.287" CommentCount="8" />
  <row Id="1888" PostTypeId="1" CreationDate="2010-12-13T22:23:49.120" Score="2" ViewCount="277" Body="&lt;p&gt;Hydrogen is the lightest element, so it's cable of lifting the most weight in out atmosphere (probably not the best terminology there, but you get the picture)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would hot hydrogen (in the same sense as hot air) be able to lift even more mass?&#xA;Would a higher or lower density of hydrogen in a ballon lift more?&#xA;If you could have a balloon which had nothing in it (it was a vacuum inside) would that lift more than a hydrogen balloon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically what are the physics to balloons and lifting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(really not sure what to tag this, so if someone else could that'd be great)&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2010-12-13T22:27:53.657" LastActivityDate="2010-12-14T01:38:51.600" Title="Balloons and lifting gases" Tags="&lt;forces&gt;&lt;buoyancy&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="1889" PostTypeId="2" ParentId="1888" CreationDate="2010-12-13T22:30:49.397" Score="5" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Would hot hydrogen (in the same sense as hot air) be able to lift even more mass? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. Though I suppose the fire danger goes up, and you certainly can't use a propane burner to warm it...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would a higher or lower density of hydrogen in a ballon lift more? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lower density always means higher buoyancy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you could have a balloon which had nothing in it (it was a vacuum inside) would that lift more than a hydrogen balloon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, and this has been proposed in various ways in science fiction literature. The engineering challenge is finding a away to confine the vacuum that is as light as a gas bag so that you don't loose the advantage to extra weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general a volume $V$ of material of density $\rho$ immersed in a fluid of density $\rho_f$ experiences a buoyant force of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F_b = gV\rho_f $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and a weight of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W = -gV\rho $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the available lifting force is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F_l = gV(\rho_f - \rho) .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the object is floating at the surface of a liquid the buoyant force is modified to reflect the volume of liquid displaced $F_b = g V_d \rho_f$ where $V_d$ is enough to cover the weight of the floating object.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2010-12-13T22:30:49.397" CommentCount="10" />
  <row Id="1890" PostTypeId="2" ParentId="1875" CreationDate="2010-12-13T22:32:35.153" Score="2" ViewCount="" Body="&lt;p&gt;Basically it works on the following principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a gyroscope used to determine what is the current angle between the vertical and the seat of the monocycle. This can be done because the gyroscope will tend to maintain the same position no matter what the orientation of the monocycle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gyroscope is connected to the engine, so that when it detects torque towards the front, the engine will generate more torque oriented backwards correspondingly (spin up); when it detects torque towards the back, the engine will spin down accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The engine torque mentioned is generated by increasing (or the decreasing) the angular velocity of the wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this has two purposes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It corrects any imbalances of the passenger and keeps the wheel directed up.&lt;/li&gt;&#xA;&lt;li&gt;The corrections have a net effect of speeding up or slowing down the linear motion (i.e. how fast the monocycle goes), so it's an effective way of controlling the monocycle. If you want to brake you lean back.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="66" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2010-12-14T00:06:05.323" LastActivityDate="2010-12-14T00:06:05.323" CommentCount="2" />
  <row Id="1891" PostTypeId="2" ParentId="1787" CreationDate="2010-12-13T23:36:57.773" Score="4" ViewCount="" Body="&lt;p&gt;Just one additional point that I haven't seen mentioned above: &lt;em&gt;if the space-time has non-trivial fundamental group, it won't be seen by an observer at infinity&lt;/em&gt;. This is the content of the &lt;strong&gt;Topological Censorship Theorem&lt;/strong&gt;. The implication is that for an asymptotically flat space-time, any interesting topology will be hidden behind the event-horizon. The proof of the theorem is rather surprisingly simple: it is more or less a direct extension of Penrose's singularity theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Friedman, J. L.; Schleich, K. &amp;amp; Witt, D. M. Topological censorship Phys. Rev. Lett., American Physical Society, 1993, 71, 1486-1489&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Schleich, K. &amp;amp; Witt, D. M. Singularities from the Topology and Differentiable Structure of Asymptotically Flat Spacetimes &lt;a href=&quot;http://arxiv.org/abs/1006.2890&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1006.2890&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galloway, G. J.. &lt;a href=&quot;http://iopscience.iop.org/0264-9381/12/10/002/fulltext&quot; rel=&quot;nofollow&quot;&gt;On the topology of the domain of outer communication&lt;/a&gt;. Class. Quantum Grav. 12 No 10 (October 1995) L99 (3pp)&lt;/p&gt;&#xA;" OwnerUserId="829" LastActivityDate="2010-12-13T23:36:57.773" CommentCount="4" />
  <row Id="1892" PostTypeId="2" ParentId="1888" CreationDate="2010-12-14T01:38:51.600" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/1888/balloons-and-lifting-gases/1889#1889&quot;&gt;dmckee's answer&lt;/a&gt; is a great not-too-technical description of buoyancy. Read that first. But in case you're interested, I thought I would go into some more detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The buoyant force on a submerged object (e.g. a balloon submerged in air) is equal to the weight of the displaced fluid,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_b = \rho_f g V$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as dmckee said. The physical origin of this force is actually the pressure difference between the top and bottom surfaces of the floating object. Pressure in a fluid at a certain height is related to the depth of the fluid above that height by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(z_2) - P(z_1) = \rho_f g (z_2 - z_1)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that is, density of fluid times gravitational acceleration times height difference. If you have a rectangular box whose top and bottom surfaces are flat, then it's pretty easy to calculate the buoyant force as the pressure differential times the area of those surfaces,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_b = (\Delta P)(A) = \rho_f g \Delta z A = \rho_f g V$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an irregular shape, you'll have to do an integral of some sort. For example, I once wrote a &lt;a href=&quot;http://www.ellipsix.net/blog/2009/11/buoyancy-part-1.html&quot; rel=&quot;nofollow&quot;&gt;blog post&lt;/a&gt; which discusses, in part, deriving the buoyant force (and weight) from the minimization of potential energy, and that method can be easier to apply to irregular objects. (There are also a couple of interesting applications, even if you don't care about the math.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also take into account variations in density (or gravitational acceleration) over the size of the balloon by doing an integral. But, according to the &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aUs_standard_atmosphere_model.png&quot; rel=&quot;nofollow&quot;&gt;US standard atmosphere model&lt;/a&gt;, the density of the atmosphere takes about $20\ \mathrm{km}$ to drop off to near zero, which corresponds to a fraction of a percent change over the height of a typical hot air balloon (a few tens of meters). That fraction of a percent is generally negligible, so you're pretty safe just using a single value for the density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you can't neglect differences in density between vastly different altitudes. Remember that the buoyant force on the balloon is equal to the weight of the amount of fluid displaced. As you go higher, the density of the air drops, which means the balloon displaces a lower mass of air. Therefore, as the balloon rises, the buoyant force drops. Eventually it reaches a height at which the buoyant force exactly balances out the weight of the balloon (and basket), and the balloon levitates at that level. As has been said in the comments, for a controlled balloon, the operator can adjust the level by either heating the gas inside the balloon (thus making it expand and displace more air) or by letting some gas out (thus making the balloon contract and displace less air).&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-12-14T01:38:51.600" CommentCount="3" />
  <row Id="1893" PostTypeId="2" ParentId="455" CreationDate="2010-12-14T02:03:36.130" Score="0" ViewCount="" Body="&lt;p&gt;When accelereration is given as a function of velocity only then the following apply&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x(u)=\int\frac{u}{a(u)}\,\mathrm{d}u+K_{1} $$ and $$ t(u)=\int\frac{1}{a(u)}\,\mathrm{d}u+K_{2} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the rest is solved similarly to David Z (above). &lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-12-14T02:03:36.130" />
  <row Id="1894" PostTypeId="1" AcceptedAnswerId="1896" CreationDate="2010-12-14T04:39:55.020" Score="13" ViewCount="469" Body="&lt;p&gt;Coming from a mathematical background, I'm trying to get a handle on the path integral formulation of quantum mechanics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Feynman, if you want to figure out the probability amplitude for a particle moving from one point to another, you 1) figure out the contribution from every possible path it could take, then 2) &quot;sum up&quot; all the contributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually when you want to &quot;sum up&quot; an infinite number of things, you do so by putting a measure on the space of things, from which a notion of integration arises. However the function space of paths is not just infinite, it's &lt;em&gt;extremely infinite&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the path-space has a notion of dimension, it would be infinite-dimensional (eg, viewed as a submanifold of $C([0,t] , {\mathbb R}^n))$. For any reasonable notion of distance, every ball will fail to be compact. It's hard to see how one could reasonably define a measure over a space like this - Lebesgue-like measures are certainly out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The books I've seen basically forgo defining what anything is, and instead present a method to do calculations involving &quot;zig-zag&quot; paths and renormalization. Apparently this gives the right answer experimentally, but it seem extremely contrived (what if you approximate the paths a different way, how do you know you will get the same answer?). Is there a more rigorous way to define Feynman path integrals in terms of function spaces and measures?&lt;/p&gt;&#xA;" OwnerUserId="821" LastEditorUserId="821" LastEditorDisplayName="" LastEditDate="2010-12-14T09:38:22.613" LastActivityDate="2011-08-09T15:53:45.567" Title="Path integral vs. measure on infinite dimensional space" Tags="&lt;quantum-mechanics&gt;&lt;path-integral&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="3" />
  <row Id="1895" PostTypeId="2" ParentId="271" CreationDate="2010-12-14T05:03:55.430" Score="1" ViewCount="" Body="&lt;p&gt;This is a rather old topic, but I felt I might have what you're looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In response to some of the answers, you write:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since the angular acceleration is always tangential, I would expect that the top should spiral outwards until it falls to the ground.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Absolutely, that is what you &lt;em&gt;should&lt;/em&gt; expect to happen. And it does ... momentarily. The final solution is a little more involved than just being uniform rotation around the vertical axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to understand this, imagine that you take a spinning top which you have just set down at time $t = t_0 $ on the ground. Now, what happens in the next instant is &lt;strong&gt;exactly&lt;/strong&gt; what you intuitively expect - the top begins to fall under gravity's influence and $ \phi $ (see &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/top.html&quot; rel=&quot;nofollow&quot;&gt;figure&lt;/a&gt; for notation) starts to increase going from $ \phi \rightarrow \phi + \delta \phi $ at time $t_1$. Consequently the angular momentum $ \mathbf{L} $ of the top changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is similar to what happens in the 2nd figure on the hyperphysics page, where $\delta \mathbf{L}$ is in the direction of $ \delta \theta $, only now $ \delta \mathbf{L} $ in the direction $ \delta \phi $ and lies in the plane containing the longitudinal axis $L_A$ of the top and the central vertical axis $V_A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Increasing $\phi$ lowers the center of mass of the top and thus its potential energy by an amount $ -\delta U $. Assuming energy conservation, this translates to an increase in the kinetic energy $\delta K$ of the top. Since the top is constrained to have zero linear momentum, this $ \delta K$ contributes entirely to the top's rotational energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind, however, that the top is now rotating around &lt;em&gt;two&lt;/em&gt; different axes. One component is the original spinning motion around its own longitudinal axes and the other is the rotation induced by gravity around the direction $N_A$ &lt;em&gt;normal to&lt;/em&gt; the plane containing $L_A$ and $V_A$. Therefore, the $\delta K$ must be appropriately portioned between these two motions. Let's see how this happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moment of inertia of the top ($I_A$) around the axis $L_A$ is clearly less than that ($I_V$) around the axis $N_A$. This is true for all but the most oddly shaped tops. Convince yourself that this is the case. In circuits more current flows through paths with lower resistance. Likewise in mechanics more energy is transferred to the component with lesser inertia. Thus the greater portion of $\delta K$ will go to increasing the angular momentum of the top around its longitudinal axis $L_A$ by some amount $\delta L&amp;#39;$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, conservation of angular momentum requires that there be a torque corresponding to this increase. The effect of this &lt;em&gt;induced&lt;/em&gt; torque is to cause the falling top to start swinging back upwards. In this way, instead of a spiral, the tip of the top traces out something like a cycloid as it precesses around the central axis.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, the diagram seems to indicate that the top should be precessing in a circle, not a spiral.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The circular trajectory is an idealization only achieved in the limit that $\omega_s / \omega_p \rightarrow \infty$, where $\omega_s$ is the spin angular velocity and $\omega_p$ is the precession angular velocity. Any top with realistic values of $\omega_s$ and $\omega_p$ will have finite &quot;wobble&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I would not have known of this rather elaborate dynamics if not for one of Feynman's lecture volumes (Part I, I think) where this question is considered in great detail!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The above write-up is a little on the hand-wavy side and there probably are errors in my reasoning. For the full kahuna look up the Feynman lectures !&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                          Cheers,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" LastActivityDate="2010-12-14T05:03:55.430" CommentCount="3" />
  <row Id="1896" PostTypeId="2" ParentId="1894" CreationDate="2010-12-14T08:45:44.870" Score="6" ViewCount="" Body="&lt;p&gt;Path integral is indeed very problematic on its own. But there are ways to &lt;em&gt;almost&lt;/em&gt; capturing it rigorously.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Wiener process&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;One way is to start with &lt;a href=&quot;http://en.wikipedia.org/wiki/Abstract_Wiener_space&quot; rel=&quot;nofollow&quot;&gt;Abstract Wiener space&lt;/a&gt; that can be built out of the Hamiltonian and carries a canonical Wiener measure. This is the usual measure describing properties of the random walk. Now to arrive at path integral one has to accept the existence of &quot;infinite-dimensional &lt;a href=&quot;http://en.wikipedia.org/wiki/Wick_rotation&quot; rel=&quot;nofollow&quot;&gt;Wick rotation&lt;/a&gt;&quot; and analytically continue Wiener measure to the complex plane (and every time this is done a probabilist dies somewhere).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the usual connection between statistical physics (which is a nice, well-defined real theory) at inverse temperature $\beta$ in (N+1,0) space-time dimensions and evolution of the quantum system in (N, 1) dimensions for time $t = -i \hbar \beta$ that is used all over the physics but almost never justified. Although in some cases it was actually possible to prove that &lt;a href=&quot;http://en.wikipedia.org/wiki/Wightman_axioms&quot; rel=&quot;nofollow&quot;&gt;Wightman QFT&lt;/a&gt; theory is indeed a Wick rotation of some Euclidean QFT (note that quantum mechanics is also a special case of QFT in (0, 1) space-time dimensions).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Intermezzo&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a good place to point out that while path integral is problematic in QM, whole lot of different issues enter with more space dimensions. One has to deal with operator valued distributions and there is no good way to multiply them (which is what physicist absolutely need to do). There are various &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_field_theory#Axiomatic_approaches&quot; rel=&quot;nofollow&quot;&gt;axiomatic approaches&lt;/a&gt; to get a handle on this and they in fact do look very nice. Except that it's very hard to actually find a theory that satisfies these axioms. In particular, none of our present day theories of Standard model have been rigorously defined.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Anyway, to make the Wick rotation a bit clearer, recall that Schrödinger equation is a kind of diffusion equation but for an introduction of complex numbers. And then just come back to the beginning and note that diffusion equation is macroscopic equation that captures the mean behavior of the random walk. &lt;em&gt;(But this is not to say that path integral in any way depends on the Schrödingerian, non-relativistic physics)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Others&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There were other approaches to define the path-integral rigorously. They propose some set of axioms that path-integral has to obey and continue from there. To my knowledge (but I'd like to be wrong), all of these approaches are too constraining (they don't describe most of physically interesting situations). But if you'd like I can dig up some references.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-14T08:45:44.870" CommentCount="2" />
  <row Id="1897" PostTypeId="2" ParentId="1894" CreationDate="2010-12-14T09:58:57.690" Score="3" ViewCount="" Body="&lt;p&gt;The answer is: forget about it. :-) Currently, there is no satisfying mathematical formalization of the path integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coming from a mathematics background myself, I was equally dismayed at this state of affairs. But I have come to terms with it, mainly due to the following historical observation: for several centuries, &lt;em&gt;infinitesimal quantities&lt;/em&gt; did not have a satisfying formalization, but that didn't stop mathematicians from understanding and using them. We now have the blessing of Weierstraß' epsilons and deltas, but it is also a curse, since the infinitesimal quantities disappeared as well (outside of non-standard analysis). I would say that the path integral is a similar situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you accept the path integral as a &quot;figure of speech&quot;, then there are ways to formalize it to some extend. Namely, you can interpret it as an &quot;integration&quot; of the &lt;em&gt;propagator&lt;/em&gt;, much like the exponential function is the solution of the differential equation $\dot y = Ay$. The &lt;em&gt;propagator&lt;/em&gt; is the probability amplitude&#xA;$$ U(r,t; r_0,t_0) $$&#xD;&#xA;of finding a particle at place and time $r,t$ when it originally started at place and time $r_0,t_0$. It is the general solution to the Schrödinger equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ i\hbar \frac{\partial}{\partial t}U(r,t; r_0,t_0) = \hat H(r,t) U(r,t; r_0,t_0), \quad U(r,t_0; r_0,t_0) = \delta(r-r_0) .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, pick a time $t_1$ that lies between $t$ and $t_0$. The interpretation of the propagator as probability amplitude makes it clear that you can also obtain it by integrating over all intermediate positions $r_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ U(r,t; r_0,t_0) = \int dr_1 U(r,t; r_1,t_1)U(r_1,t_1; r_0,t_0) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you repeat that procedure and divide the time interval $[t,t_0]$ into infinitely many parts, thus integrating over all possible intermediate positions, you will obtain the path integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More details on this construction can be found in &lt;a href=&quot;http://books.google.com/books?id=EYbIQQAACAAJ&quot; rel=&quot;nofollow&quot;&gt;Altland, Simons. Condensed Matter Field Theory.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="188" LastActivityDate="2010-12-14T09:58:57.690" CommentCount="5" />
  <row Id="1898" PostTypeId="1" AcceptedAnswerId="1899" CreationDate="2010-12-14T10:14:07.810" Score="10" ViewCount="612" Body="&lt;p&gt;Please correct me if I'm wrong, but I believe that photons slow down when travelling through glass. Does this mean they gain mass? Otherwise, what happens to extra kinetic energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I understand now that the apparent slowing down is due to electron interactions, does the glass gain weight due to light travelling through it?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="832" LastEditorDisplayName="" LastEditDate="2010-12-14T16:18:25.357" LastActivityDate="2011-01-14T13:18:01.910" Title="Do photons gain mass when they travel through glass?" Tags="&lt;energy&gt;&lt;photons&gt;" AnswerCount="4" FavoriteCount="3" />
  <row Id="1899" PostTypeId="2" ParentId="1898" CreationDate="2010-12-14T10:24:13.247" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; thanks to all the commentators. Before I mixed together both scattering and absorption of light. I tried to update the answer to more accurately describe what is actually going on &lt;em&gt;down there&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt; I will only consider interaction with molecules of the material here. More advanced stuff like interaction with the lattice of crystals or interaction with free electrons in metals would need a separate discussion.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What happens is that when photon enters the matter it has a non-zero probability to scatter on atoms of the material. In QED (quantum electrodynamics) this process is realized by summing over all possible ways the photon can interact with electrons of the material. Simplest way is that the photon is absorbed by electron, thereby increasing its energy (but this is not an excitation to a precise energetic level; any energy will do) and after a little while electron emits a different photon. As correctly pointed out by Tobias, if there are more photons with the same energy and momentum nearby, the emitted photon will tend to have same characteristics. This is because photons are &lt;a href=&quot;http://en.wikipedia.org/wiki/Boson&quot; rel=&quot;nofollow&quot;&gt;bosons&lt;/a&gt; and bosons like to occupy same states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, all of these processes contribute to the final &lt;em&gt;scattering amplitude&lt;/em&gt;. This is a complex number describing both the apparent slow-down of the photons on the matter and also absorption of light in the matter. Its value depends on the precise way molecule looks, what energy levels do electrons occupy and so on. In any case, you can (at least in principle) reduce all that complexity of single atom into a number that tells you the index of refraction and the coefficient of absorption. Note that this number will also depend on the energy of the incoming photon, giving dispersion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want to find the actual time it will take photon (note that here the word photon is used liberally as it might get absorbed and re-emitted) to travel through the material, we are again encouraged to sum over all the possible trajectories and this means over all the possible scatterings on all of the atoms. One possible trajectory is that photon doesn't interact with anything. This is a dominant one that would be correct in vacuum. But now there is also a possibility that photon will scatter on some atoms (usually just one of them though, because the scattering probability is small) and this will modify the final amplitude. If there is no absorption, the only effect will be that it will &quot;take the photon longer time to travel through material&quot;. If there is also absorption the probability of the photon to pass through the material will lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, quantum theory is just probabilistic in nature and what this means is that if you let lots of photons go through the material then they will, in general, take some scattering on the atoms. So it can be said (and it is great deal correct) that matter's electrons &quot;trap&quot; the incoming light, making it propagate slower.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-14T13:09:15.353" LastActivityDate="2010-12-14T13:09:15.353" CommentCount="12" />
  <row Id="1900" PostTypeId="2" ParentId="1898" CreationDate="2010-12-14T12:15:47.837" Score="0" ViewCount="" Body="&lt;p&gt;Transmission of light through glass has nothing to do with electron excitation and that is precisely why glass is transparent. In fact, incoming electromagnetic wave polarizes the medium that re-emits the radiation. Theoretically it could be re-emitted in any direction but it can be shown that different &lt;em&gt;wavelets&lt;/em&gt; (small parts of the wave) will interfere positively only in the initial direction of light. How difficult it is to polarize a certain medium is characterized by its &lt;em&gt;polarizability&lt;/em&gt; which is directly linked to the refractive index.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to the question about the mass of the photon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photon momentum is defined as $\textbf{p}=\hbar\textbf{k}$. It can be shown that momenta of incident(i) and transmitted(t) photons are related as $$n_{ti}=\frac{p_t}{p_i}$$ where &lt;em&gt;n&lt;/em&gt; is the refractive index. This means that for $n_{ti}&amp;gt;1$, $p_t&amp;gt;p_i$ so the momentum of the photon actually increases which can be attributed to an increase of the photon's effective mass (see F.R. Tangherlini, &quot;On Snell's law and the Gravitational Deflection of Light&quot;, &lt;em&gt;Am. J. Phys.&lt;/em&gt; &lt;strong&gt;36&lt;/strong&gt;, 1001 (1968). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: The argument whether a photon's momentum in medium is $n$ times smaller of $n$ times larger is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Abraham%2DMinkowski_controversy&quot; rel=&quot;nofollow&quot;&gt;Abraham–Minkowski controversy&lt;/a&gt; and there are strong evidences for both definitions. &lt;/p&gt;&#xA;" OwnerUserId="701" LastEditorUserId="701" LastEditorDisplayName="" LastEditDate="2010-12-16T08:07:20.663" LastActivityDate="2010-12-16T08:07:20.663" CommentCount="7" />
  <row Id="1901" PostTypeId="1" AcceptedAnswerId="2866" CreationDate="2010-12-14T12:20:51.040" Score="9" ViewCount="576" Body="&lt;p&gt;I understand only a little of general relativity, but that's why I'm here! :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the hypothetical situation of some extra-terrestrial intelligence pushing all the mass in the universe, every last subatomic particle, into a giant black hole. What is the surface area of such an object? For the sake of argument, let's keep this very idealized. Dark energy, whatever it is, can be captured and thrown into the SMBH; assume there's no Hawking radiation causing it to slowly evaporate, etc. &lt;/p&gt;&#xA;" OwnerUserId="756" LastActivityDate="2011-01-15T17:42:48.983" Title="If the universe were compressed into a super massive black hole, how big would it be?" Tags="&lt;cosmology&gt;&lt;spacetime&gt;&lt;mass&gt;&lt;black-holes&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" />
  <row Id="1902" PostTypeId="2" ParentId="1862" CreationDate="2010-12-14T12:22:49.187" Score="1" ViewCount="" Body="&lt;p&gt;If there is a subfield in physics that studying network, the closest one must be the statistical mechanics, and the second should be the soft condensed matter physics.&#xA;In my opinion, there are three main branches of network study.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Part 1: Study of interaction in network&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For this area of study, the focus is the equilibrium state and the statistical mechanics properties.&#xA;Strong correlation between elements can often give rise to emergent behaviors that do not appear in the underlying system.&#xA;If there are no interaction, these phenomena cannot occurs because the resulting system is only the sum of individual part.&#xA;Therefore, the network of interacting elements is important, and indeed, the precise structure of the network can determine whether the emergence phenomenon exist or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given problem in statistical physics, we can write down the Hamiltonian of the system,&#xA;say, Ising model&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathcal{H}=-\sum_{i,j} A_{ij} J_{ij}\sigma_{i}\sigma_{j}-h\sum_{i}\sigma_{i}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the $\sigma_{i}$ represents state (spin here) of a node and the $A_{ij}$ is the adjacency matrix of the network.&#xA;Note that $A_{ij}$ can have the topology of simple square lattice, or a complex network with non-local linkage.&#xA;Using this Hamiltonian, we can write down the corresponding partition function $Z = \sum_i {exp(-\beta \mathcal{E_i})}$,&#xA;which allows us to find the set of standard macroscopic properties.&#xA;In statistical mechanics, this model is interesting because it is the first model showing the phase transition and the critical phenomenon.&#xA;This model also show different behaviour when the topology is a complex network.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within all properties, the most important one is the ground state and its energy&#xA;because their roles in physics and they are always defined for all system.&#xA;Furthermore, there is another reason for non-physics problem: the relationship with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Optimization_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;optimization problem&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is obvious that finding the ground state is an optimization problem because we need to find a state that has minimum energy.&#xA;Hence, it is related to, say, the minimizing of overall error in &lt;a href=&quot;http://en.wikipedia.org/wiki/Supervised_learning&quot; rel=&quot;nofollow&quot;&gt;supervised learning&lt;/a&gt; or maximizing the some type of cost function in financial market.&#xA;An example of optimization problem is the &lt;a href=&quot;http://arxiv.org/abs/condmat/0603718&quot; rel=&quot;nofollow&quot;&gt;community detection&lt;/a&gt; of network,&#xA;which classify each state as one community by minimizing the Hamiltonian. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, most &lt;a href=&quot;http://en.wikipedia.org/wiki/Np-complete&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;NP-complete&lt;/strong&gt;&lt;/a&gt; problem also define an optimization problem.&#xA;And some of them can be mapped (polynomial time reduction) directly to problems in the statistical mechanics,&#xA;such as the problem of finding ground state of antiferromagnetic Ising model and MAXCUT problem in graph theory.&#xA;(see review &lt;a href=&quot;http://arxiv.org/abs/0705.0010&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Critical phenomena in complex networks&lt;/strong&gt;&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Returning to the question of why the network topology can significantly change the problem, here is an example:&#xA;An antiferromagnetic Ising model on a square lattice (a bipartite network) has a simple unique ground state such that all nodes has spin opposite to its neighbor.&#xA;However, if we consider the model defined on a triangular lattice, the ground state is nondegenerate and the state can change easily because of the frustration of the neighboring spin. For this type of topology or other complex network, the finding of ground state a highly non-trivial task.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea in statistical mechanics can also be apply to other problems.&#xA;The idea of entropy (complexity) can possibly explain why a problem is hard to solve sometimes, but easy in other case when network topology is changed. The idea of temperature in statistical mechanics also benefit the study of randomized algorithm for optimization problems. The most obvious one is the Monte Carlo algorithm for finding ground state. In real world, if the truly optimized solution cannot be found, suboptimal solution is usually acceptable. The energy landscape of the problem can usually give an insight of the problem itself and an estimate of true ground state and the accessibility of the ground state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under construction...&lt;/p&gt;&#xA;" OwnerUserId="562" LastActivityDate="2010-12-14T12:22:49.187" CommentCount="2" />
  <row Id="1903" PostTypeId="2" ParentId="1898" CreationDate="2010-12-14T14:25:11.067" Score="8" ViewCount="" Body="&lt;h2&gt;Fundamental approach&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Marek's answer is summarised shortly as &quot;no&quot;. It is based on the most &quot;fundamental&quot; concepts of physics -- you got fundamental quantum particles -- photons, electrons and some others. And these particles interact with each other producing all the world around us. The properties of the particles, like their mass, charge, e.t.c. doesn't change whatever you do with them. And, therefore, the mass of the photon is always zero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This appoach is very intuitive and of course the answer is correct... &#xA;But one can look at the same problem from different perspective, obtaining a different answer:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Quasiparticles approach&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Those fundamental particles are just excitations of the vacuum -- the universal medium for everything around us. We like to talk about particles, because they are &quot;free&quot; -- they are flying freely in the vacuum, rarely interacting with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now instead of vacuum we consider another &quot;not so universal&quot; medium -- a glass. Like everything else the glass is made of the mentioned fundamental particles.  It turns out that one wouldn't want to talk about the &lt;em&gt;fundamental&lt;/em&gt; photon inside a glass -- it always interacting with stuff in the matter: it scatters, got absorbed, got re-emitted e.t.c. In other words it is not &quot;free&quot;. It is much easier to consider a &lt;strong&gt;quasiparticle&lt;/strong&gt;, which is &quot;nearly a photon&quot;. A quasiparticle is an excitation of the glassy medium. And it behaves like it is &quot;free&quot; in the glass -- it is flying freely in the glass, rarely interacting with other quasiparticles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that point of view the answer to the question is &quot;yes&quot; -- inside the glass the quasiparticle called &quot;photon&quot; has some mass, while in the vacuum the fundamental particle called &quot;photon&quot; hasn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This  second point of view is much more elaborate and takes more effort to understand, but I think that it is more &quot;flexible&quot; and allows you to understand such things as renormalization, effective field theories, quarks and hadron structure and QCD, thermal field theory, e.t.c.&#xA;After all, the thing we now call &quot;the fundamental vacuum&quot; can be just &quot;a glass&quot; made of something more fundamental. &lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2010-12-14T15:25:18.050" LastActivityDate="2010-12-14T15:25:18.050" CommentCount="2" />
  <row Id="1904" PostTypeId="2" ParentId="1901" CreationDate="2010-12-14T14:45:31.233" Score="9" ViewCount="" Body="&lt;h2&gt;Calculation&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;First I went &lt;a href=&quot;http://en.wikipedia.org/wiki/Observable_universe#Mass&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, getting the number 3.14×10^54 kg.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Under &quot;the size of a black hole&quot; I understand the &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow&quot;&gt;Schwarzschild radius&lt;/a&gt;. Which is $r=\frac{2 G M}{c^2}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then I just did the math:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=2G%2A(3.14E54+kg)%2Fc%5E2&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=2G%2A(3.14E54+kg)%2Fc%5E2&lt;/a&gt;&lt;br&gt;&#xA;Obtaining ~5 diameters of observable Universe...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;The Problem&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I felt really uneasy about the result. Indeed: I took the mass of &lt;strong&gt;observable Universe&lt;/strong&gt; and obtained 5 times radius of the &lt;strong&gt;observable Universe&lt;/strong&gt;. &lt;em&gt;But how on Earth the observable Universe is smaller than the black hole you can make from it?!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To elaborate, let's also calculate how the &quot;density&quot; of the black hole depends on its mass:&lt;br&gt;&#xA;$\rho(M) = \frac{3M}{4\pi r^3}=\frac{3Mc^6}{4\pi\cdot 8 G^3 M^3} = \frac{3 c^6}{32\pi G^3}\cdot\frac{1}{M^2}$&lt;br&gt;&#xA;We got that with the growth of mass of the black hole its &quot;density&quot; decreases. At really high masses it gets so disperse that it is even gets smaller than the density of the Universe as we see it. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Interpretation and the answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now, how do I understand this.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_hole&quot; rel=&quot;nofollow&quot;&gt;black hole&lt;/a&gt; is &quot;a &lt;strong&gt;region of space&lt;/strong&gt; from which nothing, not even light, can escape&quot;. So, in order to make sense of the question we got to imagine some (presumably, flat) empty space. And then we are putting all the matter we gathered into it. It turns out, that even if we do not &quot;compress&quot; this stuff (we can even make it $\simeq5^3=125$ times more disperse), the gravitation attraction of the system would be strong enough for &quot;nothing, not even light, could escape&quot; from the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as far as I understand the situation now, answer to the question is: aliens don't need to do anything with the Universe in order to make a black hole (in a sense) of it.&lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-12-14T20:40:32.250" LastActivityDate="2010-12-14T20:40:32.250" CommentCount="8" />
  <row Id="1905" PostTypeId="2" ParentId="1858" CreationDate="2010-12-14T16:22:04.243" Score="2" ViewCount="" Body="&lt;p&gt;In the case of a Dirac operator, the index is the (signed) excess dimension of the space of vacuum modes of one chirality w/r/t the other: i.e., the number of anomalous “ghost” states in a chiral field theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anomalies arise when the classical/quantum symmetry correspondence breaks down under renormalization (a global anomaly could be responsible for quark mass in QCD; resolving the local chiral anomaly in the SM accounts for quarks and leptons; resolving it in superstring theory fixes the gauge group [to either SO(32) or E8 x E8], and the resolution of a conformal anomaly fixes the dimension of spacetime and the fermion content). When trying to turn string theory into actual physics, one asks &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can it explain three generations of chiral fermions? &lt;/li&gt;&#xA;&lt;li&gt;Can it explain the experimental results on proton decay?&lt;/li&gt;&#xA;&lt;li&gt;Can it explain the smallness of the electron mass? &lt;/li&gt;&#xA;&lt;li&gt;Can it explain [things about the cosmological constant]?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and AST helps to answer these questions.&lt;/p&gt;&#xA;" OwnerUserId="711" LastActivityDate="2010-12-14T16:22:04.243" />
  <row Id="1906" PostTypeId="1" AcceptedAnswerId="2802" CreationDate="2010-12-14T16:45:04.717" Score="4" ViewCount="1367" Body="&lt;p&gt;I'm trying to understand the Kubo Formula for the electrical conductivity in the context of the Quantum Hall Effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is that several papers, for instance the famous &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v49/i6/p405_1&quot; rel=&quot;nofollow&quot;&gt;TKNN (1982)&lt;/a&gt; paper, or an elaboration by &lt;a href=&quot;http://linkinghub.elsevier.com/retrieve/pii/0003491685901484&quot; rel=&quot;nofollow&quot;&gt;Kohmoto (1984)&lt;/a&gt;, write the diagonal entries of the conductivity tensor in the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sigma_{xy}(\omega \to 0) = \frac{ie^2}{\hbar} \sum_{E^a &amp;lt; E_F &amp;lt; E^b} \frac{\langle a|v_x|b \rangle \langle b|v_y|a \rangle - \langle a|v_y|b \rangle \langle b|v_x|a \rangle}{(E^a - E^b)^2} .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the static limit $\omega\to 0$ and low temperature $T\to 0$. The sum goes over all eigenstates $|a\rangle$ and $|b\rangle$ of the single-particle Hamiltonian. $E_F$ is the Fermi energy. $v_x$ and $v_y$ are the single-particle velocity operators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, these papers don't derive this equation, which is unfortunate because the Kubo formula is usually not presented in this form. I have found (and succeeded in rederiving) the following variation instead &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sigma_{xy}(\omega+i\eta) = \frac{-ie^2}{V(\omega + i\eta)} \sum_{a,b} f(E^a) \left( \frac{\langle a|v_x|b \rangle \langle b|v_y|a \rangle}{\hbar\omega + i\eta + E^a - E^b} + \frac{\langle a|v_y|b \rangle \langle b|v_x|a \rangle}{-\hbar\omega - i\eta + E^a - E^b} \right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is formula (13.37) from &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0030839939&quot; rel=&quot;nofollow&quot;&gt;Ashcroft, Mermin&lt;/a&gt;, though they don't actually prove it. $f(E)$ is the Fermi distribution. A nice derivation is given in &lt;a href=&quot;http://www.springer.com/materials/book/978-3-540-74789-5&quot; rel=&quot;nofollow&quot;&gt;Czycholl&lt;/a&gt; (german).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my question is, obviously&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How to derive the first formula from the second?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can see that the first equation arises as the linear term when writing the sum as a power series in $\omega$, but why doesn't the constant term diverge?&lt;/p&gt;&#xA;" OwnerUserId="188" LastEditorUserId="188" LastEditorDisplayName="" LastEditDate="2010-12-14T19:28:40.753" LastActivityDate="2011-01-13T21:02:52.430" Title="Kubo Formula for Quantum Hall Effect" Tags="&lt;electromagnetism&gt;&lt;second-quantization&gt;&lt;condensed-matter&gt;&lt;many-body&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" />
  <row Id="1907" PostTypeId="1" AcceptedAnswerId="1954" CreationDate="2010-12-14T19:05:33.983" Score="6" ViewCount="596" Body="&lt;p&gt;I realize that this is more of an astronomy question, but I figured this would be an appropriate place for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the moon's phases are determined by its position in orbit relative to the sun. (See: &lt;a href=&quot;http://www.extremescience.com/zoom/index.php/space/35-space-science/77-about-the-moon&quot; rel=&quot;nofollow&quot;&gt;http://www.extremescience.com/zoom/index.php/space/35-space-science/77-about-the-moon&lt;/a&gt;). The &quot;shadow&quot; is not cast by the Earth (a common misconception-this is actually a lunar eclipse), but by the moon's body itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would appear that, in order for the moon to be a new moon, it would have to be somewhere between the Earth and the Sun. However, it would seem that, if we were looking at a new moon, we'd necessarily be looking at the Sun as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why, then, can we sometimes see a new moon at night? Why doesn't it vanish at night for a half-month every month, between its last/first quarters? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &lt;a href=&quot;http://www.dubaiastronomy.com/Portals/0/all%20images/Planets/moon/new%20moon.JPG&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a picture of a new moon at nighttime, for visualization purposes (and so nobody has to comb thru Twilight pictures).&lt;/p&gt;&#xA;" OwnerUserId="834" LastEditorUserId="834" LastEditorDisplayName="" LastEditDate="2010-12-15T14:52:53.010" LastActivityDate="2010-12-15T15:22:31.577" Title="Why CAN we see the new moon at night?" Tags="&lt;astronomy&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="1" />
  <row Id="1908" PostTypeId="1" AcceptedAnswerId="1925" CreationDate="2010-12-14T19:26:58.180" Score="1" ViewCount="106" Body="&lt;p&gt;I've read about the idea that buckyballs and other nanostructures could be used to hold drugs and things until they reach certain places in the body and then get released.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I was wondering, if you created a buckyball in air, so that some molecules that are in air (such as oxygen and nitrogen) were inside the buckyball. if you then put the buckyball in a vacuum, would the air be able to escape, or would it be trapped in the buckyball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And vice versa, a buckyball created in a vacuum put into the atmosphere, would any molecules fit through into the buckyball?)&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-14T23:30:09.887" LastActivityDate="2010-12-14T23:30:09.887" Title="Buckyballs in vacuum" Tags="&lt;chemical-compounds&gt;&lt;condensed-matter&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="1909" PostTypeId="1" AcceptedAnswerId="1921" CreationDate="2010-12-14T19:52:04.157" Score="4" ViewCount="428" Body="&lt;p&gt;This was discussed in an answer to a &lt;a href=&quot;http://physics.stackexchange.com/questions/1898/do-photons-gain-mass-when-they-travel-through-glass&quot;&gt;related question&lt;/a&gt; but I think that it deserves a separate and, hopefully, more clear answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a single photon ($\lambda$=532 nm) traveling through a plate of perfect glass with a refractive index $n=1.5$. We know that it does not change its direction or other characteristics in any particular way and propagating 1 cm through such glass is equivalent to 1.5 cm of vacuum. Apparently, the photon interacts with glass, but what is the physical nature of this interaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's not consider high-order effects like Rayleigh/Raman scattering.&lt;/p&gt;&#xA;" OwnerUserId="701" LastActivityDate="2010-12-14T21:56:40.477" Title="How does a photon travel through glass?" Tags="&lt;optics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="1910" PostTypeId="2" ParentId="1907" CreationDate="2010-12-14T19:55:10.287" Score="7" ViewCount="" Body="&lt;p&gt;I don't get the question. What do you mean that you shouldn’t be able to see a new moon at night. You never see a new moon at night. That phase lasts a couple of days. After that there is a crescent moon visible after sunset. The moon continues to move eastwards (its position at sunset) and after about two weeks you have a full moon.&lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2010-12-14T19:55:10.287" CommentCount="1" />
  <row Id="1911" PostTypeId="2" ParentId="1907" CreationDate="2010-12-14T20:11:51.670" Score="4" ViewCount="" Body="&lt;p&gt;The new moon isn't directly inline with the sun -if it was it would be a solar eclipse. So the new moon is only near to the sun (within several degrees), if the moon is several degrees behind the setting sun, the sun sets while this &quot;new moon&quot; is still slightly above the horizon. Roughly everyday after the new moon the moonset is about an hour later. By the&#xA;time it gets to be a quarter moon moonset comes near midnight.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2010-12-14T20:11:51.670" />
  <row Id="1912" PostTypeId="2" ParentId="1907" CreationDate="2010-12-14T20:13:56.567" Score="1" ViewCount="" Body="&lt;p&gt;I'll just add that during the &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_eclipse&quot; rel=&quot;nofollow&quot;&gt;solar eclipse&lt;/a&gt; you can actually see &quot;brand new&quot; moon.&lt;br&gt;&#xA;Backlit. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2010-12-14T20:13:56.567" />
  <row Id="1913" PostTypeId="2" ParentId="1907" CreationDate="2010-12-14T20:28:34.860" Score="4" ViewCount="" Body="&lt;p&gt;Imagine for a moment that you are standing on the Earth facing surface of the moon at new moon. What do you see in the sky? You see the Earth and in &lt;em&gt;its&lt;/em&gt; full phase. The Earth has 13 times the area of the moon, and three times the albedo: that sucker is &lt;em&gt;bright&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: the &quot;dark&quot; side of the moon is illuminated by earth-shine, and on a clear dark night you can see it. Of course, you have to manage this just at full dark or just before dawn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB: If you live in the city you will have to take a drive into the country and wait for your eyes to adjust.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2010-12-14T20:28:34.860" CommentCount="3" />
  <row Id="1914" PostTypeId="2" ParentId="1816" CreationDate="2010-12-14T20:38:45.493" Score="1" ViewCount="" Body="&lt;p&gt;I know nothing of this book, but I do know a little about N-body gravitational interactions. When N &gt;= 3, you can do just about anything you want with a little propulsion, but it may take a very long time. This has been proposed by NASA (good approximate for &lt;em&gt;hard&lt;/em&gt; SF) as a way of sending probes to the outer solar system:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Interplanetary_Transport_Network&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Interplanetary_Transport_Network&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the stability questions, it is known that you can put 7 or 8 equal-mass objects on a stable circular orbit around their center of mass:&#xA;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1988A%26A...205..309S&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/1988A%26A...205..309S&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After introducing an external mass and its accompanying tidal forces, I imagine there is still some stability regime, which could be pretty close to the NS, especially if the &quot;asteroids&quot; have WD density, their constellation could be on the order of 1km, while the NS radius is probably closer to 15km.&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2010-12-14T20:38:45.493" />
  <row Id="1915" PostTypeId="1" AcceptedAnswerId="1922" CreationDate="2010-12-14T20:56:34.207" Score="5" ViewCount="373" Body="&lt;p&gt;This question is about cosmology and general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the difference between the universe and the observable universe. What I am not really clear about is what is meant when I read that the universe is infinite.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does it have infinite mass or is it&#xA;dishomogeneous? &lt;/li&gt;&#xA;&lt;li&gt;How can the universe&#xA;transition from being finite near the&#xA;big bang and infinite 14 billion&#xA;years later? Or would an infinite&#xA;universe not necessarily have a big&#xA;bang at all?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-08-05T04:52:34.360" Title="What does it mean that the universe is &quot;infinite&quot;?" Tags="&lt;general-relativity&gt;&lt;cosmology&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="1916" PostTypeId="1" AcceptedAnswerId="1930" CreationDate="2010-12-14T21:05:45.967" Score="9" ViewCount="555" Body="&lt;p&gt;I think we all heard general statements like &quot;once big enough star burns out there is nothing to prevent the gravitational collapse ending in a black hole&quot;. But I can't remember even seeing the process described precisely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the deal: at first there is a nice object, a star. Stars can be nicely modeled by general relativity, nuclear physics and statistical physics combined and very much is known about these models and it can be observed whether they agree with things like light and neutrino fluxes, surface temperature and probably also lot of other stuff I know nothing about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the collapse we are left with another nice object, a black hole. We know that &lt;a href=&quot;http://en.wikipedia.org/wiki/No-hair_theorem&quot; rel=&quot;nofollow&quot;&gt;black holes have no hair&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: what happens in-between? More precisely, between the time when all of the nuclear material has been burned out (and if possible ignore effects like reheating of the star after big enough compression) and the time where there is nothing more than just a black hole.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;Give a description of what happens during the collapse?&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;How does the star &quot;lose its hair&quot;?&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Can the actual collapse be solved analytically?&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;At what point is singularity created?&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; &lt;em&gt;I don't want to know what an outside observer will see. Instead, I'd like to find out what an individual part of the dead star will &quot;feel&quot; when a black hole is about to form near it. In other words, I want a complete solution (ideally analytical, but numerical would be also completely fine)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feel free to assume anything that makes your life easier. Spherical symmetry is definitely fine. Also, if for any reason the questions don't make sense (like Cauchy problem is ill-defined in the presence of the singularity) feel free to interpret them in a way that make them sensible (e.g. assume that black hole is built from D-branes).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Also, I have a feeling that what I intended as a simple question at first ended up being pretty complex. If you think it should be split into smaller (and therefore more manageable and answerable) parts, let me know.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-14T22:19:55.960" LastActivityDate="2011-02-28T13:22:08.607" Title="How precisely does a star collapse into a black hole?" Tags="&lt;general-relativity&gt;&lt;black-holes&gt;" AnswerCount="7" CommentCount="19" FavoriteCount="5" />
  <row Id="1917" PostTypeId="2" ParentId="1916" CreationDate="2010-12-14T21:18:43.883" Score="0" ViewCount="" Body="&lt;p&gt;I'm kinda out of my depth here, but I think the balding issue can be understood in cartoon terms as an effect of the time dilation of events nearing the event horizon. From outside all the material appears to stack-up against the horizon and things appear to stop happening. If there are no dynamics there can only be static effects: i.e. electrostatics and gravitation.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2010-12-14T21:18:43.883" CommentCount="1" />
  <row Id="1918" PostTypeId="1" CreationDate="2010-12-14T21:41:08.423" Score="10" ViewCount="178" Body="&lt;p&gt;I've been wondering exactly why the elements are distributed the way they are on earth. The heavier elements have their origins in the centers of stars, or in supernova. After the death of the stars, you end up with a dust cloud containing the heavier elements. Later, planets form out of these along with new stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the heavier elements were randomly distributed in these explosions, how do we end up with special deposits of minerals on earth? We have mines and special areas where the concentrations of different materials are higher. For example, we have iron, gold, nickel and uranium mines. Why aren't these materials uniformly distributed? Why isn't the earth composed of a substance which is a uniform mixture of all the heavy elements? Why does it seem there is sometimes a preference for like materials to clump together?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nickel and iron mostly make up the earth's core, which I guess makes sense in terms of them being fairly heavy while comparatively abundant relative to heavier elements, but what about everything near the surface that we observe?&lt;/p&gt;&#xA;" OwnerUserId="197" LastActivityDate="2010-12-14T23:41:53.063" Title="Why the distribution of elements on earth?" Tags="&lt;geophysics&gt;&lt;gravitational-collapse&gt;&lt;stellar-population&gt;&lt;physical-chemistry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="1920" PostTypeId="2" ParentId="1544" CreationDate="2010-12-14T21:50:15.773" Score="2" ViewCount="" Body="&lt;p&gt;In a condensed matter field theory course, I learned the following: &lt;em&gt;microscopically&lt;/em&gt;, the Lagrangian for the electromagnetic field looks like it is supposed to, coupling minimally to the particle coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ L = \sum_i\left( \frac m2 (p_i-\frac ec \mathbf A(r_i))^2 - e\Phi(r_i) + \dots \right) .$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a &lt;em&gt;macroscopic&lt;/em&gt; level, however, after getting rid of all the individual particle degrees of freedom via the grand canonical ensemble, new behavior may emerge. Namely, the &lt;em&gt;effective Lagrangian&lt;/em&gt; for the electromagnetic field in the body may look very different from a linear one. For example, the effective action for the e.m. field in a superconductor is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S_{\text{eff}}[\mathbf A] = \frac\beta2 \int d^3r \mathbf A^\perp(r) \left(-\frac 1{\mu_0}\nabla^2 + \frac {n_s}m \right)\mathbf A^\perp(r)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mu_0$ is the vacuum permeability, $n_s$ the superfluid density, $m$ the electron mass and $\mathbf A^\perp$ is the perpendicular component of the gauge field, defined in Fourier space as $\mathbf A^\perp(q) = \mathbf A(q) - q(q\cdot \mathbf A(q))/q^2$. The difference to the vacuum action is the additional &quot;mass term&quot; $n_s/m$, which causes the Meissner effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose that you are asking for the most general form that such effective actions may have? I don't have an answer, but I don't see why a most general form should actually exist in the first place.&lt;/p&gt;&#xA;" OwnerUserId="188" LastActivityDate="2010-12-14T21:50:15.773" CommentCount="6" />
  <row Id="1921" PostTypeId="2" ParentId="1909" CreationDate="2010-12-14T21:56:40.477" Score="7" ViewCount="" Body="&lt;p&gt;This is a tricky question to answer, because in many ways it doesn't make sense to talk about a definite path followed by a single photon. Quantum mechanics is inherently probabilistic, so all we can really talk about are the probabilities of various outcomes over many repeated experiments with identically prepared initial states. All we can measure is something like the average travel time for a large number of single photons passing through a block of glass one after the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transmission of light through a medium is easier to explain in a classical sense, where you think of the light beam as a wave that drives oscillations in the atomic dipoles making up the material. Each atom then re-radiates its own waves at the same frequency, but slightly out of phase. The sum of the initial wave and the re-radiated wave is a wave that lags behind the incoming wave a little bit, which explains the reduced speed. A beam of light entering a block of material tends to continue in the same direction because forward scattered light from any individual atom tends to interfere constructively with forward scattered light from other atoms in the material, while light scattered off to the sides mostly interferes destructively, and cancels out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carrying this picture over to the quantum regime, you would say that a single photon entering the material will potentially be absorbed and re-emitted by each of the atoms making up the first layer of the material. Since we cannot directly measure which atom did the absorbing, though, we treat the situation mathematically as a superposition of all the possible outcomes, namely, each of the atoms absorbing then re-emitting the photon. Then, when we come to the next layer of the material, we first need to add up all the wavefunctions corresponding to all the possible absorptions and re-emissions, and when we do that, we find that just as in the classical wave case, the most likely result is for the photon to continue on in the same direction it was originally headed. Then we repeat the process for all the atoms in the second layer, and the third, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any given layer, the probability of being absorbed then re-emitted by any individual atom is pretty tiny, but there are vast numbers of atoms in a typical solid, so the odds are that the photon will be absorbed and re-emitted during the passage through the glass are very good. Thus, on average, the photon will be delayed relative to one that passes through an equal length of vacuum, giving rise to the lower observed transmission velocity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it's not possible to observe the exact path taken by any photon-- that is, which specific atoms it scattered from-- and if we attempted to make such a measurement, it would change the path of the photon to such a degree as to be completely useless. Thus, when we talk about the transmission of a single photon through a refractive material, we assign the photon a velocity that is the average velocity determined from many realizations of the single photon experiment, and go from there.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-12-14T21:56:40.477" CommentCount="6" />
  <row Id="1922" PostTypeId="2" ParentId="1915" CreationDate="2010-12-14T22:03:44.620" Score="5" ViewCount="" Body="&lt;p&gt;Basically, I think the idea that the universe is infinite comes from considerations of the large-scale curvature of spacetime. In particular, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann-Lema%C3%AEtre-Robertson-Walker&quot; rel=&quot;nofollow&quot;&gt;FLRW cosmological model&lt;/a&gt; predicts a certain critical density of matter and energy which would make spacetime &quot;flat&quot; (in the sense that it would have the Minkowski metric on large scales). If the actual density is greater than that density, then spacetime is &quot;positively curved,&quot; which implies that it is also &lt;em&gt;bounded&lt;/em&gt; - that is, that there is a certain maximum distance between any two spacetime points. (I don't know the details of how you get from positive curvature to being bounded) However, if the actual density is not greater than that critical density, there is no bound, which means that for any distance $d$, you could find two points in the universe that are at least that far away. I think that's what it means to be infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, the observations done to date, paired with current theoretical models, are inconclusive as to whether the actual density of matter and energy in the universe is greater than or less than (or exactly equal to) the critical density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the universe is in fact infinite in this sense, it still could have had a big bang. The FLRW metric includes a scale factor $a(\tau)$ which characterizes the relative scale of the universe at different times. Specifically, the distance between two objects (due only to the change in scale, i.e. ignore all interactions between the objects) at different times $t_1$ and $t_2$ satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d(t_1)}{a(t_1)} = \frac{d(t_2)}{a(t_2)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, it seems that the universe is expanding, so $a(\tau)$ is getting larger. But if you imagine running that expansion in reverse, eventually you'd get back to a &quot;time&quot; where $a(\tau) = 0$, and at that time all objects would be in the same position, no matter whether space was infinite or not. That's what we call the Big Bang.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-05T04:52:34.360" LastActivityDate="2011-08-05T04:52:34.360" CommentCount="5" />
  <row Id="1923" PostTypeId="2" ParentId="1858" CreationDate="2010-12-14T22:05:36.053" Score="8" ViewCount="" Body="&lt;p&gt;Eric and others have given good answers as to why one expects the index theorem to arise in various physical systems. One of the earliest and most important applications is 't Hooft's resolution of the $U(1)$ problem. This refers to the lack of a ninth pseudo-Goldstone boson (like the pions and Kaons) in QCD that one would naively expect from chiral symmetry breaking. There are two parts to the resolution. The first is the fact that the chiral $U(1)$ is anomalous. The second is the realization that there are configurations of finite action (instantons) which contribute to correlation functions involving the divergence of the $U(1)$ axial current. The analysis relies heavily on the index theorem for the Dirac operator coupled to the $SU(3)$ gauge field of QCD. For a more complete explanation see S. Coleman's Erice lectures&#xA;&quot;The uses of instantons.&quot; There are also important applications to S-duality of $N=4$ SYM which involve the index theorem for the Dirac operator on monopole moduli spaces.&lt;/p&gt;&#xA;" OwnerUserId="838" LastActivityDate="2010-12-14T22:05:36.053" CommentCount="4" />
  <row Id="1924" PostTypeId="2" ParentId="1916" CreationDate="2010-12-14T22:11:20.850" Score="1" ViewCount="" Body="&lt;p&gt;I will try to do an intuitive explanation as I do not possess the mathematical knowledge for any further analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not think, for an outside observer, singularity would form in a definite time. Rather, when the density inside is enough for an event horizon to form, i.e., the radius is smaller than the Schwarzschild radius, a black hole is said to be formed, not its singularity though. Because, for an outside observer, time comes to a complete stop at the event horizon, and only if the event horizon gets smaller, can the region inside the initial event horizon make sense for the outside observer. So if you believe that Hawking Radiation exists, then in a finite amount of time, the black hole will slowly evaporate until it's event horizon is nothing more than a singularity, then evaporate completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So rather than an actual forming of a singularity at a certain time for an outside observer, the mass that is &quot;queued up&quot; gradually condenses into a singularity, while also evaporating via hawking radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the losing of the hair, once the event horizon forms, the hair of the materials at the event horizon or inside are lost because there are two possible fates(actually one final ultimate fate) for a particle until the forming of the singularity that I mentioned: it will be emitted as energy due to hawking radiation, or it will be the part of the last singularity which will also evaporate due to hawking radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is actually similar to a question that I had asked and couldn't explain what I meant very clearly.&lt;/p&gt;&#xA;" OwnerUserId="504" LastActivityDate="2010-12-14T22:11:20.850" CommentCount="1" />
  <row Id="1925" PostTypeId="2" ParentId="1908" CreationDate="2010-12-14T22:13:35.620" Score="6" ViewCount="" Body="&lt;p&gt;It is just that the buckyball's faces are not holes as in popular view; there is pretty much electron probability density there forming quite a strong barrier. Also the size of the faces is comparable to the size of a, for instance, oxygen molecule, so I'm pretty sure that it is impossible for a buckyball to release a molecule just because of one atmosphere pressure difference; yet I don't really think that anything can be put inside without a targeted, careful procedure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: A picture of C60, with, I hope, van der Waals surface. This shows that this structure is indeed pretty dense.&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/WH9vw.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-14T23:29:17.750" LastActivityDate="2010-12-14T23:29:17.750" />
  <row Id="1926" PostTypeId="2" ParentId="1683" CreationDate="2010-12-14T22:41:10.517" Score="1" ViewCount="" Body="&lt;p&gt;This answer presents an analogy that I hope will clarify how it is possible that 1) the wagon moves 2) the wagon winds up with a net velocity at the end of the problem.  This isn't a direct answer - it's intended as supporting conceptual material (so I've marked it community wiki).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Setup&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Throughout this answer, all velocities and all momenta are calculated solely in the reference frame of the rail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that the tank does not have water in it.  Instead it has a gun that shoots clay lumps. The gun is mounted at the middle.  It can shoot any size clay lump at any speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a hole in the wagon floor.  For convenience, the hole is all the way at the left side of the wagon.  If the gun shoots a lump of clay to the left, the gun, which is rigidly attached to the rest of the wagon, will recoil some.  The lump will fly towards the left side of the wagon and collide with the left wall completely inelastically.  Then it will fall down through  the hole in the floor and exit the wagon with exactly the same horizontal speed (if any) as the wagon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;First experiment&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tank starts out stationary with a lump of mass $m$ in the gun.  It shoots the lump at speed $v$.  The lump is moving to the left; $v$ is negative.  The momentum of the lump is $mv$.  Let the recoil speed of the wagon be $w_0$.  By conservation of momentum, $mv + Mw_0 = 0$.  Therefore, the cart recoils, moving at speed &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w_0 = -v*m/M$$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is to the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, the lump collides with the left wall.  At this point the lump and wagon must move at some new, mutual speed after the collision.  Call that $w_f$.  Conservation of momentum implies $w_f = 0$ and the wagon has come to a dead stop.  The lump  falls through the hole straight down and the wagon sits still for the remainder of eternity.  It is displaced from its original position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Second Experiment&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tank starts out with two lumps of clay in the gun, each of mass $m/2$.  The gun shoots one lump at speed $v$ as before.  Conservation of momentum gives $m/2*v + (M+m/2)*w_0 = 0$, or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w_0 = -v\frac{m}{2(M+m/2)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we wait until the moment when that lump hits the left wall.  At precisely that moment, we fire the next lump, also at speed $v$.  We make the acceleration profiles of the two lumps exactly equal in magnitude and opposite in sign.  This way, the forces on the two lumps must be equal.  Those forces come from the rigid body of the gun and wagon combined.  Hence, the gun/wagon feels no net force and no acceleration during this process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first lump is now comoving with the wagon at speed $w_0$.  It falls through the hole moving at that speed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, the second lump collides with the wagon.  The second lump and the wagon come to some mutual velocity $w_f$.  Conservation of momentum gives $mw_0/2 + (M + m/2)w_f = 0$, or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w_f = -w_0 \frac{m}{2(M+m/2)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or substituting in for $w_0$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w_f = v \left(\frac{m}{2(M+m/2)}\right)^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second lump falls out of the wagon and moves at speed $w_f$, and the wagon coasts at speed $w_f$ from then on.  $w_f$ is proportional to $v$ and has the same sign.  The wagon is moving to the left at the end of the process.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-12-15T09:18:35.057" LastActivityDate="2010-12-15T09:18:35.057" CommunityOwnedDate="2010-12-14T22:41:10.517" CommentCount="4" />
  <row Id="1927" PostTypeId="1" CreationDate="2010-12-14T22:44:44.543" Score="3" ViewCount="240" Body="&lt;p&gt;I am thinking of a (greatly simplified) computer simulation of a universe that followed something like Newtonian rules.  Inside the simulation are A.I.s that are made from those same rules, and can only use those rules interact the world around them.  Would there be some fundamental limits on what those A.I.s could work out about their universe, like their own version of an uncertainty principle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for phrasing this question in such a convoluted way.  If anyone recognises what I am asking, and can point me in the right direction that would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="840" LastActivityDate="2010-12-16T08:41:42.417" Title="Would something like the uncertainty principle arise even if the universe was built on something like Newtonian mechanics?" Tags="&lt;uncertainty-principle&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="1" />
  <row Id="1928" PostTypeId="2" ParentId="1918" CreationDate="2010-12-14T22:56:43.210" Score="5" ViewCount="" Body="&lt;p&gt;It is an interesting question, I will take a stab at it. Hopefully my answer will be generic enough to remain plausible.&#xA;  First of all the ISM (Interstellar Medium) which collapsed to form the sun would not likely have been well mixed. It would be some sort of turbulently mixed mixture of recent SN remnants, and the molecular cloud with which it presumably collided to set off star formation. So probably this was a bit of a mess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then in the solar accretion disk from which the plantissimals form we have temperature gradients. In the higher temperature regions volatile elements don't condense so we get some&#xA;differentiation based upon where the material was in the disk when it condensed. Then the planets are builtup via collision of lots of smaller planetissimals, coming from different regions. The larger&#xA;bodies will melt and undergo differentiation due to internal gravity. The iron nickle and elements heavily attracted to it go to the core, and the lighter oxidized material form a rocky mantle. Stuff further out, contains a lot of lighter volatile elements, like ice and frozen CO2 etc. In any case all this stuff continues to bombard the planet, and indeed at a reduced flux still does. Note we have several different classes of meteorites with differing composition. So the most recent additions won't have been mixed into the planets interior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various geologic processes which are driven by the earth as a heat engine (primordial heat plus radioactive decay) does drive flows of stuff like magma (partial melts of parent materials), and hydrothermal activity which dissolves and deposits stuff as it circulates, and samples differnt temperatures and pressures. We also&#xA;have various surface processes, including biological that move stuff around, and sometimes&#xA;locally concentrate some elements. Because of the energy flows in the system (driven by internal heat and/or solar heating), we have the potential for local concentrations to happen.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2010-12-14T22:56:43.210" CommentCount="0" />
  <row Id="1929" PostTypeId="2" ParentId="1927" CreationDate="2010-12-14T23:02:38.850" Score="2" ViewCount="" Body="&lt;p&gt;Well, Newtonian mechanics are abandoned in our world in some scales because it fails to explain certain phenomena. Which means that there are some stuff in work other than Newtonian mechanics, such as the properties of atoms, electrons, spectra, time, relative motion, limit of speed...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the universe that the AIs that you proposed live in is a stable one(because electrons would spiral down to the core and annihilate in classical mechanics), and the energy was not quantized(i.e. h=0), uncertainty principle would no longer be valid. Wave-particle duality would not exist either as the de Broglie wavelengths of all particles would be 0. Similarly if c were equal to infinity, time dilation and length contraction would not exist in special relativity. To model gravity you would probably use springs between each and every object in the universe, so curvature of space time would disappear, hence the time dilation as well. Basically every single quantum and relativity based phenomena would not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me note that I do not know if a stable universe capable of sustaining protons and electrons is possible solely by Newtonian mechanics without manipulating fundamental constants.&lt;/p&gt;&#xA;" OwnerUserId="504" LastEditorUserId="504" LastEditorDisplayName="" LastEditDate="2010-12-14T23:07:50.903" LastActivityDate="2010-12-14T23:07:50.903" />
  <row Id="1930" PostTypeId="2" ParentId="1916" CreationDate="2010-12-14T23:41:15.017" Score="6" ViewCount="" Body="&lt;p&gt;The solution for this problem for a dust equation of state and spherical symmetry is known as the Oppenheimer-Snyder solution. You model the interior of the distribution as a FRW universe with positive spatial curvature, zero pressure, and zero cosmological constant. You model the exterior of the solution as the Schwarzschild solution cut off at a time-dependent radius. So long as the matter distribution is dust, the thing satisfies all of the junction conditions you need. See Poisson's relativity book or MTW.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more general solution requires numerics.  But one thing we can say for sure is that there is no need for the black hole to shed its 'hair' in the case of spherical symmetry--the radial dependence of the solution will just compress into the singularity eventually, or scatter out to infinity.  Birchoff's theorem tells us every spherically symmetric vacuum solution must be the Schwarzschild solution (perhaps with an electrostatic charge, which is technically not vacuum).  This is related to the fact that there can be no monopole radiation in relativity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the general case for this problem is very likely chaotic.  Already, if the equation of state of the matter is that of a classical, spherically symmetric, Klein-Gordon field, which is a relatively simple generalization, the system exhibits a (link is a large postscript file)&lt;a href=&quot;http://laplace.physics.ubc.ca/People/petryk/phd.ps&quot; rel=&quot;nofollow&quot;&gt;second-order phase transition&lt;/a&gt;, a result found by Matt Choptuik, and related to the settling of the Hawking naked singularity bet.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2010-12-15T17:38:26.093" LastActivityDate="2010-12-15T17:38:26.093" CommentCount="1" />
  <row Id="1931" PostTypeId="2" ParentId="1918" CreationDate="2010-12-14T23:41:53.063" Score="1" ViewCount="" Body="&lt;p&gt;The answer is also a bit vague here, because it happens differently to each ore/element; yet this is not that the deposits were created among with Earth, rather than they were created with a mix of geological and sometimes biological processes that gathered some material from larger area, purified and transported it and finally gathered in the deposit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See this wikipedia article for more details: &lt;a href=&quot;http://en.wikipedia.org/wiki/Ore_genesis&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Ore_genesis&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-12-14T23:41:53.063" />
  <row Id="1932" PostTypeId="2" ParentId="1927" CreationDate="2010-12-14T23:47:58.710" Score="3" ViewCount="" Body="&lt;p&gt;Newtonian mechanics is pretty good in producing deterministic chaos which is much more important factor in every-day uncertainty than the uncertainty principle.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-12-14T23:47:58.710" />
  <row Id="1933" PostTypeId="2" ParentId="1019" CreationDate="2010-12-14T23:56:45.477" Score="-3" ViewCount="" Body="&lt;p&gt;This is tongue-in-cheek!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That an emptying tank with a nozzle pointing downwards would actually experience a force!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(see &lt;a href=&quot;http://arxiv.org/pdf/physics/0312087v3&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/physics/0312087v3&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;:-)&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-12-14T23:56:45.477" CommunityOwnedDate="2010-12-14T23:56:45.477" />
  <row Id="1934" PostTypeId="2" ParentId="1927" CreationDate="2010-12-15T01:15:33.423" Score="3" ViewCount="" Body="&lt;p&gt;Classical chaos will always limit the ability to predict the future, no matter how good the AI, because there will always be some finite uncertainty in the measurement of the initial conditions. A remarkable number of seemingly simple systems are, in fact, sensitively dependent on the initial conditions, and will become unpredictable after a short time. There's a bit in Iain Banks's &lt;em&gt;The State of the Art&lt;/em&gt;, one of his stories about the Culture, an interstellar civilization ruled by super-intelligent AI's where one of the Minds who run the Culture says &quot;I’m the smartest thing for a hundred light year radius, and by a factor of about a million … but even I can’t predict where a snooker ball’s going to end up after more than six collisions.&quot; While I can't vouch for the exact accuracy of the snooker reference, the general idea is sound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This uncertainty is not at all like quantum uncertainty a la the Heisenberg Uncertainty Principle, though. The Uncertainty Principle is, fundamentally, a consequence of the fact that quantum objects have both particle and wave properties, and any attempt to have both of those things at the same time will necessarily involve some uncertainty in both position and momentum (or any other pair of conjugate observables). A purely classical universe would not have that feature, but would have some uncertainty due to finite measurement capabilities.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-12-15T01:15:33.423" />
  <row Id="1935" PostTypeId="2" ParentId="1019" CreationDate="2010-12-15T01:25:02.307" Score="12" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;That if an object is moving, there must be a force propelling it in that direction.&lt;/strong&gt; &#xA;Students very commonly think that forces cause objects to have velocity, rather than the fact that forces cause objects to &lt;em&gt;change&lt;/em&gt; velocity. &lt;/p&gt;&#xA;" OwnerUserId="807" LastActivityDate="2010-12-15T01:25:02.307" CommunityOwnedDate="2010-12-15T01:25:02.307" CommentCount="1" />
  <row Id="1936" PostTypeId="2" ParentId="1354" CreationDate="2010-12-15T03:06:11.487" Score="3" ViewCount="" Body="&lt;p&gt;One of the nicest examples I know is the Kosterlitz-Thouless phase transition in the XY model. What is cool is that the transition is driven by the condensation of vortices which have an energy that diverges logarithmically with the size of the system. You would think they couldn't contribute at all because of this, but it turns out their entropy also diverges in the same way, so the free energy $F=E-TS \propto (c-k_BT) log(R)$ where $c$ is a parameter and $R$ is the size of the system. At sufficiently large $T$ the entropy terms wins and the system undergoes a transition through formation of vortices. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;p.s. after rereading the question I realize my answer does not involve &quot;practical life&quot; but I'll leave it anyway.&lt;/p&gt;&#xA;" OwnerUserId="838" LastActivityDate="2010-12-15T03:06:11.487" />
  <row Id="1937" PostTypeId="2" ParentId="17" CreationDate="2010-12-15T04:49:33.667" Score="2" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Rayleigh_scattering&quot; rel=&quot;nofollow&quot;&gt;Rayleigh scattering&lt;/a&gt; is one element of the solution, as &lt;a href=&quot;http://physics.stackexchange.com/users/42/j-c&quot;&gt;j.c.&lt;/a&gt; explained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when there is dust in the air (for instance after a volcano sends huge quantities of tiny rock particles into the sky, like what happened with &lt;a href=&quot;http://en.wikipedia.org/wiki/Eyjafjallaj%C3%B6kull&quot; rel=&quot;nofollow&quot;&gt;Eyjafjallajökull&lt;/a&gt;, or because of sand or smoke particles), you might have noticed that the sunsets happen to be more colourful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is caused by another related light scattering effect, known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tyndall_effect&quot; rel=&quot;nofollow&quot;&gt;Tyndall effect&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="843" LastActivityDate="2010-12-15T04:49:33.667" />
  <row Id="1938" PostTypeId="2" ParentId="1916" CreationDate="2010-12-15T05:21:12.560" Score="0" ViewCount="" Body="&lt;p&gt;I would hazard a guess that aside from time dilation effects, in many case it is probably a slow process. If the mass of the core is marginal for forming a BH, you should initially get a neutron star. As the NS cools, accumulates more material, and/or sheds angular momentum it&#xA;can cross a threshold for collapse to a BH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that in many SN, you get enough angular momentum, so that you get some sort of accretion disklike thing, which again must shed angular momentum before it gets small/dense enough to collapse into the BH. And even then I suspect you might have a phase where there is a smallish BH surrounded by a large heavy accretion disk, which takes its time getting swallowed up. I'm not so sure if any of that happens with ultramassive stars (i.e &gt; 150 times a massive as the sun), these might directly collapse. But, the more common run of the mill heavystars might not take a direct collapse trajectory, but instead spend a period of time losing various forms of energy (thermal, rotational, magnetic) before they can collapse all the way. This energy loss probably results in gamma ray bursts. &lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-12-15T05:21:12.560" CommentCount="6" />
  <row Id="1939" PostTypeId="1" AcceptedAnswerId="1941" CreationDate="2010-12-15T07:16:43.190" Score="2" ViewCount="199" Body="&lt;p&gt;Ok, so my boss is trying to make a car turntable. In essence, he has a two boards that sit atop a rotating ring. He wants to put two wheels at the end of each board (8 wheels total). He thinks that you can angle wheels properly (2d) such that they can be flat (axis is parallel to the floor) and the whole contraption move smoothly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at &lt;a href=&quot;http://www.flickr.com/photos/evancarroll/5263073898/&quot; rel=&quot;nofollow&quot;&gt;this picture&lt;/a&gt; (Edit: copied below - Mark)&#xA;&lt;img src=&quot;http://i.stack.imgur.com/YujYw.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, so the question, will this design move smoothly? My guess is &quot;no.&quot; I reason that if the wheels are unbanked then they will move in a straight line when driven, and the ring will realign them periodically to its circular path (probably causing a skip at the wheel as the stress is released). He thinks it will just work.. Please use technical terms with very easy layman definitions so I can reference wikipedia and texts and learn more about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, for the utter lack of useful physics terms.&lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-12-15T07:25:23.263" LastActivityDate="2010-12-15T16:21:50.590" Title="Physics of a fixed wheel moving on a flat surface held at distance from a center point" Tags="&lt;classical-mechanics&gt;&lt;soft-question&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="1940" PostTypeId="2" ParentId="1019" CreationDate="2010-12-15T07:19:42.517" Score="0" ViewCount="" Body="&lt;p&gt;Here's a false belief. Everything your physics teacher says is true. Because physics teachers would never mislead you over how many states of matter there are or how lift is generated on a wing.&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-12-15T07:19:42.517" CommunityOwnedDate="2010-12-15T07:19:42.517" />
  <row Id="1941" PostTypeId="2" ParentId="1939" CreationDate="2010-12-15T09:13:17.397" Score="1" ViewCount="" Body="&lt;p&gt;I think you are essentially correct that the device does not run smoothly, but the importance of this effect will depend on things like the thickness of the wheels and how well-built the table is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will try to answer the following question: &quot;What is the minimum force we must pull with on the outside rim to make the table spin?&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this calculation we'll pretend that the construction is essentially perfect - the wheels are super-smooth, perfectly aligned, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the turntable is spinning, any given wheel is moving in a circle.  An ideal wheel contacts the ground beneath it with just the very bottom of the wheel.  This bottom portion is a line segment whose width is the width of the wheel.  Let's call that the &quot;contact line&quot; and make its width $d$ (I want to use $w$ later for the weight).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we painted the rim of one of the wheels.  Then as it went around in the circle, it would paint the ground beneath it with a ring whose thickness is $d$.  Let's call the radius of that ring $R$.  This is the same as the distance from the center of the device to the wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resistance to turning arises when the bottom of the wheel slips, or moves relative to the ground.  For a wheel going in a straight line, this slipping could ideally be zero - the bottom of the wheel would not be moving at all relative to the ground (while the top of the wheel would be moving at double speed to make up for it.)  When the wheel moves in a circle, though, there must be some slipping because different parts of the wheel's contact line move different distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we go around one full circle, we want to know the total amount of slipping that occurs.  We'll say that the very center of the contact line doesn't slip at all.  The outside and inside edges move in larger and smaller circles respectively, so their total slipping distance is $d\pi$ each (in opposite directions, but that doesn't matter much).  The average slipping distance across the wheel is $d \pi/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next we want to know what force friction with the  ground exerts on slipping wheels.  This is proportional to the weight of the turntable, $w$, and to an unknown coefficient of friction with the ground $\mu$.  $\mu$ might be around one half or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total work done is the force $w\mu$ times the slipping distance $2 d \pi$, so the work needed to rotate the ring one full rotation is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W = 2w\mu d \pi$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The number of wheels doesn't matter, because if we add more wheels, the weight on each wheel goes down proportionately.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This work is also equal to the force we exert on the ring times the distance around, which in this case is $2 \pi R$.  This means we can set&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2 \pi R F = 2 w \mu d \pi$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and simplify to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = w \mu \frac{d}{R}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wider the wheels, the heavier the table, the higher the friction, and the smaller the table, the harder it will be to turn the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we want to know whether this matters.  Is this effect comparable in size to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rolling_resistance&quot; rel=&quot;nofollow&quot;&gt;rolling resistance&lt;/a&gt; we expect with wheels anyway?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia article linked above says that a typical car tire has a rolling resistance of $0.01$.  Perhaps a well-constructed turn table could achieve the same.  The comparable statistic from the sliding of the wheels is $\mu d/R$.  A reasonable ratio of the wheel thickness to the radius is $1/10$, so if $\mu \approx 1/2$ we get that the resistance due to slipping is around $0.05$, perhaps five times as big as the rolling resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The punch line is this:  if the cart is exquisitely well-constructed and has very thick wheels, then the fact that the wheels don't align perfectly will be the dominant cause of resistance.  If the cart is poorly-constructed, with sticky wheels, or if the wheels are very thin, then the slightly-misaligned wheels don't matter much.  In my best guess for an average scenario, the extra resistance added by slipping wheels is, to an order of magnitude, about five times the ordinary rolling resistance, and the spinning turntable will be significantly harder to spin than a simple cart constructed the same way would be to push in a straight line.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-12-15T09:13:17.397" CommentCount="5" />
  <row Id="1942" PostTypeId="1" AcceptedAnswerId="1947" CreationDate="2010-12-15T09:17:29.037" Score="2" ViewCount="164" Body="&lt;p&gt;Suppose a cube of mass $m$ enters a loop-de-loop of radius $r$, travelling halfway around and reaching the top. At the top, it has velocity $v$, and the second half of the loop is frictionless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I determine whether or not it will continue around the loop, or fall off?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the notes I took during class I can check to see if $v^2/r &amp;lt; g$, but I don't understand why that works. Can someone explain this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I determine if the cube stays on the loop if the second half of the loop isn't frictionless?&lt;/p&gt;&#xA;" OwnerUserId="768" LastEditorUserId="768" LastEditorDisplayName="" LastEditDate="2010-12-15T10:19:47.843" LastActivityDate="2010-12-15T10:26:34.710" Title="Does the cube stay on the loop-de-loop?" Tags="&lt;classical-mechanics&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="1943" PostTypeId="1" CreationDate="2010-12-15T09:19:15.930" Score="2" ViewCount="351" Body="&lt;p&gt;If I am sitting on a skateboard and travel downhill, will the velocity depend at all on the size of the wheel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The skateboard is only powered by gravity. There are a lot of variables to consider. Given the best conditions, where the surface of the road is smooth, and the tires are roughly equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard answers that larger wheels are faster because they offer more springiness, and larger wheels are slower because they deform more and thus have higher rolling resistance. Larger wheels are faster because they roll over imperfections in the road better than smaller wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious from a theoretical standpoint, does the size of the wheel matter? That is perhaps, ignoring things like rolling resistance. And also from a practical standpoint. &lt;/p&gt;&#xA;" LastActivityDate="2010-12-15T10:58:45.007" Title="If a vehicle is rolling down a hill, will its speed depend on the size of the wheel?" Tags="&lt;classical-mechanics&gt;&lt;applied-physics&gt;" AnswerCount="2" />
  <row Id="1944" PostTypeId="2" ParentId="1453" CreationDate="2010-12-15T09:58:03.783" Score="12" ViewCount="" Body="&lt;p&gt;Since no one mentioned &lt;a href=&quot;http://en.wikipedia.org/wiki/Color_confinement&quot; rel=&quot;nofollow&quot;&gt;the confinement&lt;/a&gt;:&lt;br&gt;&#xA;&lt;em&gt;&quot;no analytic proof exists that quantum chromodynamics should be confining&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Briefly: it turns out that one cannot isolate color-charged particles. Intuitively one can understand it -- gluons, being the carriers of strong interaction, carry the color-charge themselves. So they are &quot;screening&quot; the color-charge of a carrier. And when one tries to, say, separate two quarks, the &quot;gluon tube&quot; appears between them. Which leads to production of new quarks, and those new quarks combine with the quarks being separated, producing colorless states in the end.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up to now there is no analytic calculation, that supports this picture.&lt;br&gt;&#xA;I think that the solution of that would be a great achievement.&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2010-12-15T09:58:03.783" CommunityOwnedDate="2011-01-22T18:11:53.713" />
  <row Id="1945" PostTypeId="2" ParentId="1943" CreationDate="2010-12-15T10:14:44.733" Score="4" ViewCount="" Body="&lt;p&gt;From a purely theoretical standpoint, the radius of the wheel doesn't matter, but heavy wheels are slow.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an idealized scenario, the skateboard conserves energy.  This means that its total energy when it gets to the bottom of the hill is the same regardless of how it goes down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a wheel rolls down the hill, it picks up kinetic energy.  Some of that goes into its translational motion, while some goes in its rotation.  The energy in the rotation is essentially wasted from the point of view of going fast.  A solid cylinder or disk, for example, will move $1/\sqrt{1.5} = 0.81$ times as fast at the bottom of a hill it has rolled down as it would go if it slid down without rolling (and without friction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have heavy wheels compared to the weight of the skateboard and rider, then you suffer most of this slowdown.  If you have light wheels, the energy of the wheels hardly matters and you can approach the ideal sliding speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next we want to know if this matters.  On &lt;a href=&quot;http://en.wikipedia.org/wiki/Rolling_resistance&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; I found that rolling resistances can be about $0.01$.  When you roll down an incline of angle $\theta$, you lose about $0.01 \cot\theta$ of the energy you pick up to friction.  For now set rolling resistance equal to $c$ instead of the number $0.01$.  For cylindrical wheels of total mass $m$ and total wheels and rider of mass $M$, a fraction $m/3M$ is used in rotational rather than translational kinetic energy.  Thus, the rotational energy stored in the wheels becomes important when, roughly speaking,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m/M &amp;gt; 3 c \cot\theta$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question we wanted to answer was not when rolling resistance becomes important, but what size wheels are faster.  So imagine that $c$ is a function of $R$, the wheel radius, and that $m = \lambda R^2$, saying that we'll consider wheels of the same density and thickness, but different radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we differentiate both sides of the previous expression with respect to $R$, we get a condition for the extra rotational energy stored in the wheel size to start being a bad trade off for any improvement in rolling resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\lambda R/M &amp;gt; - 3 c&amp;#39;(R) \cot\theta$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ R &amp;gt; -\frac{3 M c&amp;#39;(R) \cot\theta}{2\lambda}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When this inequality is satisfied, making the wheels large will slow you down.  Otherwise larger wheels are better.  Note that if $c&amp;#39;(R)$ is zero or positive, larger wheels are always worse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I can't think of good ways to estimate $c&amp;#39;(R)$ without experimentation.  One thing I can think of is this (it's highly speculative):  A skateboard has pretty hard wheels that probably don't deform much under the weight of a rider, but skateboards are treated roughly and may get some grime in their bearings.  So I would guess that friction in the bearing could be the most important factor in rolling resistance for a skateboard.  I don't have much experience with them, but I think that if I take a skateboard wheel and spin it while holding it up in the air, it won't spin and spin for a minute or more like a  bicycle wheel will.  Larger wheels mean fewer rotations and less motion in the bearing, so this would give $c(R) = \alpha/R$ for some constant $\alpha$.  This guess would give&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ R &amp;gt; \frac{3 M \alpha \cot\theta}{2\lambda R^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R &amp;gt; \left(\frac{3 M \alpha \cot\theta}{2\lambda}\right)^{1/3}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I let the density of the wheels be $1g/cm^3$, the mass of the rider by $75 kg$, the slope $10 ^\circ$, the width of the wheels $2 cm$, and the rolling resistance $0.01$ when the wheels are $5cm$ tall, I get that larger wheels slow you down when $R &amp;gt; 17 cm$.  That would indicate that larger wheels are actually better in this case up to a pretty big size wheel (for a skateboard), but take it with a large grain of salt.  There are lots of unjustified assumptions in there.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-12-15T10:50:10.813" LastActivityDate="2010-12-15T10:50:10.813" />
  <row Id="1946" PostTypeId="2" ParentId="1942" CreationDate="2010-12-15T10:20:34.200" Score="1" ViewCount="" Body="&lt;p&gt;Try finding the normal force between the loop and the block.  Ask yourself what normal forces are physically possible.  Is there some condition on the normal force?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the second half of the loop is not frictionless, you would have a fairly complicated calculation on your hands to find the full motion.  You'd need to find the energy dissipated due to friction, which in turn depends on the normal force.  However, my intuition is saying that if the block gets up to the top of the loop and isn't going to fall off at the top, it isn't going to fall off anywhere in the second half of the loop, no matter how high the friction coefficient.  You might want to check that with some math if you're interested.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-12-15T10:20:34.200" CommentCount="2" />
  <row Id="1947" PostTypeId="2" ParentId="1942" CreationDate="2010-12-15T10:26:34.710" Score="2" ViewCount="" Body="&lt;p&gt;If the second half of the loop is frictionless, this is just a simple application of Newton's second law. In order to stay on the loop, the cube needs to move in a circle of radius $r$, but no smaller, which means it has to be subject to a net inward force of $mv^2/r$, but no larger. Gravity always provides a force of $mg$, so if $g &amp;gt; v^2/r$, the inward force will be too large for the block to stay on the loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is friction on the second half of the loop, you could use the same idea in principle, but it becomes a lot messier because you need to check the force at every point on the second half, not just the top. Consider a point at angle $\theta$ around the loop, where $\theta = 0$ corresponds to the very top. If you draw a free-body diagram, you'll see that there are two forces acting perpendicular to the loop (gravity and the normal force) and two components acting parallel to the loop (gravity and friction). Write out Newton's second law in each case and you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$mg\cos\theta + F_N(\theta) = m r\dot\theta^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(perpendicular) and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$mg\sin\theta - \mu F_N(\theta) = mr\ddot\theta$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(parallel). Notice that I've written the centripetal acceleration ($r\dot\theta^2$) and tangential acceleration ($r\ddot\theta$) in polar coordinates for convenience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a system of coupled differential equations that you would need to solve to find the normal force at each angle. If $F_N(\theta) &amp;lt; 0$ for any $\theta$, the block will fall off at that point. Of course, those are rather complicated equations, and I'm pretty sure you could only solve them numerically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also try using the Euler-Lagrange equation and Lagrange multipliers for the case with friction. I don't have time to write that up now but I'll come back to this as soon as I can and fill it in.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-12-15T10:26:34.710" CommentCount="3" />
  <row Id="1948" PostTypeId="2" ParentId="1943" CreationDate="2010-12-15T10:53:16.680" Score="5" ViewCount="" Body="&lt;p&gt;Well, you can't directly compare speeds, because the vehicle has different speeds at different times (or equivalently, at different positions). But you can compare the &lt;em&gt;accelerations&lt;/em&gt;. It turns out that if you look at the theoretically simplest effect, moment of inertia, the acceleration does not depend on the wheel size, but it does depend on the &lt;em&gt;shape&lt;/em&gt; of the wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose a wheel has a moment of inertia $I$, which for a solid cylinder would be $\frac{1}{2}mr^2$ ($m$ is the wheel's mass and $r$ is its radius). The kinetic energy of the wheel is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose you have $N$ of these wheels mounted to a chassis of mass $M$. The kinetic energy of the entire ensemble is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K = \frac{1}{2}(M + Nm)v^2 + \frac{1}{2}NI\omega^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the wheels roll without slipping, then $v = r\omega$ and that becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K = \frac{1}{2}\left(M + Nm + \frac{NI}{r^2}\right)v^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if this whole thing is rolling down a slope of inclination $\theta$ (where $\theta = 0$ corresponds to horizontal), the total energy is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = \frac{1}{2}\left(M + Nm + \frac{NI}{r^2}\right)v^2 - (M + Nm)gd\sin\theta = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $d$ is the distance traveled along the slope from the point at which the vehicle started rolling. If you take the time derivative of this and do a little algebra, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a = \frac{(M + Nm)g\sin\theta}{M + Nm + \frac{NI}{r^2}}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point it's worth noting that moment of inertia for any kind of rotationally symmetric object (like a wheel) is given by a formula of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = bmr^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $b$ is some number. For instance, $b = \frac{1}{2}$ for the solid cylinder I mentioned above. For a hollow wheel (no hubcaps or anything) you'd have $b = 1$, for a sphere $b = \frac{2}{5}$, and so on. So the above formula reduces to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a = \frac{(M + Nm)g\sin\theta}{M + Nm(1 + b)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that it doesn't depend on the size of the wheel ($r$) anymore. But it does depend on that constant $b$, which is determined by the wheel's shape. A wheel with a larger value of $b$ will undergo less acceleration.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-15T10:58:45.007" LastActivityDate="2010-12-15T10:58:45.007" CommentCount="1" />
  <row Id="1949" PostTypeId="1" AcceptedAnswerId="1951" CreationDate="2010-12-15T10:53:21.933" Score="6" ViewCount="296" Body="&lt;p&gt;This is a quantitative question. The problem is inspired by this event:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On August 5, 2010, an enormous chunk of ice, roughly 97 square miles (251 square kilometers) in size, broke off the Petermann Glacier along the northwestern coast of Greenland. The Petermann Glacier lost about one-quarter of its 70-kilometer- (40-mile-) long floating ice shelf, said researchers who analyzed the satellite data at the University of Delaware.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine an iceberg that is moving freely in the ocean. Given that the temperature of the surrounding water is $T = 4$ Celsius and the temperature $T=0$ Celsius is evenly distributed throughout the volume of the iceberg estimate how long does it take the iceberg to melt completely in the ocean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will find the mass of the iceberg from the event description. Average thickness of the chunc is estimated about $500$ m. For simplify we suppose that the iceberg is spherical during the melting.&lt;/p&gt;&#xA;" OwnerUserId="552" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-15T13:34:55.460" LastActivityDate="2010-12-15T18:28:15.170" Title="How long does it take an iceberg to melt in the ocean?" Tags="&lt;thermodynamics&gt;&lt;geophysics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="1950" PostTypeId="2" ParentId="741" CreationDate="2010-12-15T11:05:11.983" Score="3" ViewCount="" Body="&lt;p&gt;I can fully understand your confusion since it is natural that you feel overwhelmed by this new viewpoint on the theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers given by &lt;strong&gt;Eric&lt;/strong&gt; and &lt;strong&gt;Marek&lt;/strong&gt; are just fine and I will not directly talk about principal bundles, local trivialization and the like. I want to present a very intuitive approach here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest that you go one or two steps back and try to understand the notion of a covariant derivative in classical differential geometry. There, the covariant derivative $D$ assures that if you derive some quantity $F$ on a manifold, say some surface, this new quantity $DF$ will also lie &quot;on the manifold&quot; (actually, something related to it like the tangent space).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following example will hopefully illustrate the issue what it means that something has to &quot;stay on the manifold&quot;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Mass-point on a surface&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Ok, lets do the most simple example one could think of, the motion of a free mass point on a surface in Newtonian mechanics. As you know, the Lagrangian in this case is just the kinetic energy,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = T = \frac{m}{2}\mathbf{v}^2$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what is now $\mathbf{v}^2$? We have to assume that in every point, the velocity is tangential to the surface. Then, we know that $\mathbf{v}^2 = g_{ab}\dot{x}^a\dot{x}^b$ where we sum over the indices and the surface is described by some metric $g_{ab}(x)dx^adx^b$ and we have replaced the velocity by the time derivative of the position of the particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the solution of the system we will need two terms. First of all, we want to calculate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial{L}}{\partial{x^k}} = \frac{m}{2}\partial_k{g_{ab}(x)}\dot{x}^a\dot{x}^b$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;second,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dt}\frac{\partial{L}}{\partial{\dot{x}^k}} = \frac{d}{dt}\frac{m}{2}g_{ab}(x) \left( \delta^a_k\dot{x}^b + \dot{x}^a\delta^b_k \right) = m\frac{d}{dt}g_{kb}(x)\dot{x}^b$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since $g$ is symmetric. Now,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m\frac{d}{dt}g_{kb}(x)\dot{x}^b = m\left( \partial_lg_{kb}(x)\dot{x}^l\dot{x}^b + g_{kb}(x)\ddot{x}^b\right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the equations of motion are given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dt}\frac{\partial{L}}{\partial{\dot{x}^k}} - \frac{\partial{L}}{\partial{x^k}} = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and dropping m, re-using the symmetry in $g$ and renaming some indices, we arrive at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g_{kb}\ddot{x}^b+\frac{1}{2}\left( \partial_a g_{kb} + \partial_b g_{ka} - \partial_k g_{ab} \right)\dot{x}^a\dot{x}^b = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is exactly by applying $g^{ik}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ddot{x}^i + \Gamma^i_{ab}\dot{x}^a\dot{x}^b = 0$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the &lt;a href=&quot;http://en.wikipedia.org/wiki/Christoffel_symbols&quot; rel=&quot;nofollow&quot;&gt;Christoffel symbols&lt;/a&gt; can directly seen as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma^i_{ab} = \frac{1}{2}g^{ik}\left( \partial_a g_{kb} + \partial_b g_{ka} - \partial_k g_{ab} \right)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I hope I did not miscalculate anything. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Relation to electrodynamics&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now this equation of motion is already magic since it is precisely the equation of motion for a testparticle in &lt;strong&gt;general relativity&lt;/strong&gt;. But what about (other) gauge field theories?&lt;br&gt;&#xA;Here, the &lt;strong&gt;curvature&lt;/strong&gt; is not defined with respect to the manifold directly but to a group somehow &quot;attached&quot; to it. That's why it will have some group indices but one can drop this if the Lie Algebra of the group is one dimensional as in the case of electrodynamics. There, our curvature is $F_{\mu\nu}$ but we could also state $F^a_{\mu b\nu}$ where now $a$ and $b$ are indices of the group. This looks much more familliar to the curvature of general relativity, $R^{\mu}_{\nu\alpha\beta}$ where all indices correspond to the manifold, in some sense, the tangent space is the group of general relativity, roughly speaking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For historical reasons, the Christoffel symbols that somehow catch (not invariantly!) the, force, acting on the particle because of the curvature are in gauge theories called gauge fields $A$ and rescaled,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\beta A^a_{\mu b}dx^{\mu}&amp;quot; = &amp;quot;\Gamma^a_{\mu b}dx^\mu $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with again group indices $a$ and $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you derive something on the manifold, you will always have to define that derivative with respect to the Christoffel Symbols to stay &quot;in the manifold&quot;. On the other hand, the derivation will also have to respect the group character, one could say the result has to stay &quot;in the group&quot;. This will be realized by a covariant derivative and here the &quot;Christoffel symbols&quot; are called gauge fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2010-12-16T09:45:09.077" LastActivityDate="2010-12-16T09:45:09.077" CommentCount="6" />
  <row Id="1951" PostTypeId="2" ParentId="1949" CreationDate="2010-12-15T11:27:45.947" Score="8" ViewCount="" Body="&lt;p&gt;Heat leaves through the surface of the berg $\propto r^2$, but the heat required to melt depends on mass $\propto r^3$.  Thus, $t \propto r$.  Also assume $t \propto 1/\Delta T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took a cube with $r \approx 2 cm$ and put in in my water bottle $\Delta T = 20 C$.  It melted in $5$ minutes.  Let $r_{berg} = (8 km * 8 km * 250m)^{1/3} = 2*10^3 m$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t_{berg} = 3 min 10^5*20/4 = 1.5 * 10^6 min = 3 years$&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-12-15T11:27:45.947" CommentCount="8" />
  <row Id="1952" PostTypeId="1" AcceptedAnswerId="1973" CreationDate="2010-12-15T13:37:04.897" Score="6" ViewCount="166" Body="&lt;p&gt;I'm preparing a presentation on &lt;a href=&quot;http://en.wikipedia.org/wiki/Spin_ice&quot; rel=&quot;nofollow&quot;&gt;Spin-Ice&lt;/a&gt;, but something's been bugging me for a while. On the Wikipedia page for &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometrical_frustration&quot; rel=&quot;nofollow&quot;&gt;Geometrical Frustration&lt;/a&gt;, it says the following about easy spins on a tetrahedron with ferromagnetic interactions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are three different equivalent arrangements with two spins out and two in, so the ground state is three-fold degenerate.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/7CRYL.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just don't understand why. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We're considering easy spins with ferromagnetic interactions, so any ground state needs 2 spins pointing towards the center (in) and 2 spins pointing away from the center (out). &lt;/li&gt;&#xA;&lt;li&gt;Given a tetrahedron,  I can think of 6 different ways to distribute 2 spins in and 2 spins out on the vertexes. $(iioo,ioio,oiio,iooi,oioi,ooii)$&#xD;&lt;/li&gt;&#xA;&lt;li&gt;You can rotate any of the above states to reproduce another, which might mean they are not different. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But then, why does wiki state there are three &lt;em&gt;different&lt;/em&gt; arrengements?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; Each of the above configurations has a total magnetic moment in a different direction. We can place our $(x,y,z)$ axis so that each one of the first three cases has a total moment in the positive direction of one of the axis, and each of the last three has a total moment in the negative direction of one of the axis.&lt;/p&gt;&#xA;" OwnerUserId="724" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2010-12-15T22:30:39.967" LastActivityDate="2010-12-16T04:10:35.530" Title="Why is the ground state of the ferromagnetic tetrahedron threefold degenerate?" Tags="&lt;electromagnetism&gt;&lt;spin&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="1953" PostTypeId="2" ParentId="1916" CreationDate="2010-12-15T14:02:44.840" Score="1" ViewCount="" Body="&lt;p&gt;The &quot;hair&quot; is lost via gravitational radiation. This is also known as quasi-normal ringdown, as the BH vibrates at different frequencies much like a drum (maybe a &quot;gong&quot; is better analogy). Any charge on the black hole will simple get shorted out by free charges in the surrounding plasma, on a very short time scale.&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2010-12-15T14:02:44.840" CommentCount="4" />
  <row Id="1954" PostTypeId="2" ParentId="1907" CreationDate="2010-12-15T14:44:06.510" Score="5" ViewCount="" Body="&lt;p&gt;Well, I'd like to say that you are almost there. The key point of this question is to know that usually illustrations are just showing the relative positions but not with the real ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/bimrR.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the size and distance of the moon is the same as such pictures show, it will much harder to find when it lies at the same side of the sun. Because to see it, the smallest position angle from the sun is $\alpha = (R_E+R_M)/D_{EM}$ .Then it will vanish more days every month as you say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the real situation is like this:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/RAQlk.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the real distance is so far, even the moon is quite close to the sun, people besides the Day Night Terminator can still find it. An estimate can be given in this way:&#xA;$\alpha = (R_E+R_M)/D_{EM} = (6471+3476)/384400 = 0.0259 $ rad.&#xA; So there a only $ 2 \alpha \times 180/\pi \times 28/360 = 0.23 $ days we can not saw it from the earth. Due to the strong light of the sun, it's longer but still within one day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: This is not a physics question at all. But here may be the best place to put it before we have an astronomy part. Click &lt;a href=&quot;http://area51.stackexchange.com/proposals/51/astronomy&quot;&gt;here&lt;/a&gt; to help it reach Beta ~&lt;/p&gt;&#xA;" OwnerUserId="518" LastEditorUserId="518" LastEditorDisplayName="" LastEditDate="2010-12-15T15:22:31.577" LastActivityDate="2010-12-15T15:22:31.577" CommentCount="2" />
  <row Id="1955" PostTypeId="2" ParentId="1875" CreationDate="2010-12-15T15:52:55.667" Score="1" ViewCount="" Body="&lt;p&gt;Sklivvz has the mechanics down. To answer your questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You're right, if the system were only the rider and the unicycle, there would be nothing to prevent it from falling over. However, there's another component of the system that you're missing: &lt;b&gt;the ground.&lt;/b&gt;&lt;/li&gt;&#xA;&lt;li&gt;Yes. If the wheel were unable to &quot;push against&quot; the ground, as on a frictionless surface, there would be nothing preventing the rider falling over.&lt;/li&gt;&#xA;&lt;li&gt;The FBD for falling over would show the force of gravity on the rider/unicycle system's COG &#xA;pointing down, and the force of the ground on the wheel pointing up, creating a moment couple. (Note that if there was no ground in the system, there would be no moment, and hence no falling.)&lt;/li&gt;&#xA;&lt;li&gt;The FBD for self-righting would show the same system, with an additional force (friction) tangential to the wheel on the ground. This force is caused by the wheel's motion in the same direction as the falling.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You could also draw FBD's for the unicycle itself, with the force of the rider bearing down on the seat and the weight of the unicycle on the COG, but unless you need to account for the rider bending back &amp;amp; forth and waving his arms wildly, it's safe to assume that the rider/unicycle are a single body.&lt;/p&gt;&#xA;" OwnerUserId="834" LastActivityDate="2010-12-15T15:52:55.667" />
  <row Id="1956" PostTypeId="2" ParentId="1939" CreationDate="2010-12-15T16:21:50.590" Score="1" ViewCount="" Body="&lt;p&gt;I'm going to give a sort of meta-answer to this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sounds like a hokey design for a turntable. Having multiple bearings for the same rotation motion is begging for alignment problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your best bet is probably to use a single large &lt;a href=&quot;http://en.wikipedia.org/wiki/Tapered_roller_bearing&quot; rel=&quot;nofollow&quot;&gt;tapered-roller bearing&lt;/a&gt; in the base of ring, and make sure the &quot;boards&quot; are rigid enough to not cause any deflection loading when they're loaded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a 25,000 N pickup truck and a safety factor of 4, you're probably looking at around $150 for a bearing that will handle whatever car you plunk onto it, no wheels necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you decide to go the wheels route anyways (probably a bit cheaper), use &lt;a href=&quot;http://www.mcmaster.com/#casters&quot; rel=&quot;nofollow&quot;&gt;casters&lt;/a&gt;, and you won't have to worry about wheel alignment.&lt;/p&gt;&#xA;" OwnerUserId="834" LastActivityDate="2010-12-15T16:21:50.590" />
  <row Id="1957" PostTypeId="1" AcceptedAnswerId="1958" CreationDate="2010-12-15T16:48:45.450" Score="10" ViewCount="497" Body="&lt;p&gt;Question sounds stupid but I was taught that something which reflects all colors in light is white. Same is the function of mirror, it also reflects all light. What is the difference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;br&gt;&#xA;But what if the white object is purely plain and does not scatter the light. Will it be a mirror?&lt;/p&gt;&#xA;" OwnerUserId="847" LastEditorUserId="847" LastEditorDisplayName="" LastEditDate="2010-12-16T09:57:37.003" LastActivityDate="2010-12-16T09:57:37.003" Title="What is the difference between a white object and a mirror?" Tags="&lt;light&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="1958" PostTypeId="2" ParentId="1957" CreationDate="2010-12-15T17:01:36.470" Score="17" ViewCount="" Body="&lt;p&gt;The difference is the direction the light is emitted in. Mirrors 'bounce' light in a predictable direction, white objects scatter light.&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-12-15T17:01:36.470" CommentCount="6" />
  <row Id="1959" PostTypeId="2" ParentId="1957" CreationDate="2010-12-15T17:02:09.640" Score="9" ViewCount="" Body="&lt;p&gt;A white object reflects the light in all the directions, independently of the original direction. It is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffuse_reflection&quot; rel=&quot;nofollow&quot;&gt;diffuse reflection&lt;/a&gt;. If you shine a beam of light onto a white surface, it is scattered in all the directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, a mirror reflect the light symmetrically to the input direction, with no scattering. This is what allows us to see &quot;through&quot; the mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit to add:&lt;/em&gt; &quot;Normal&quot; coloured object all have a diffuse reflection, their colour being given by the mix of wavelength they reflect. To better understand the difference between the two kind of reflections, you can also read the related question : &lt;a href=&quot;http://physics.stackexchange.com/questions/530/why-can-you-have-shiny-black-objects&quot;&gt;&quot;Why can you have shiny black objects?&quot;&lt;/a&gt; (and the answers !)&lt;/p&gt;&#xA;" OwnerUserId="373" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2010-12-15T17:17:49.690" LastActivityDate="2010-12-15T17:17:49.690" CommentCount="4" />
  <row Id="1960" PostTypeId="2" ParentId="1957" CreationDate="2010-12-15T17:08:53.433" Score="6" ViewCount="" Body="&lt;p&gt;Other answers are correct except that they forget to mention why white objects behave differently than mirrors. Well, generic white objects have very rough bumpy surfaces. This causes &lt;a href=&quot;http://en.wikipedia.org/wiki/Scattering&quot; rel=&quot;nofollow&quot;&gt;scattering&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, mirror has a very even and polished surface so that the simple law for &lt;a href=&quot;http://en.wikipedia.org/wiki/Specular_reflection&quot; rel=&quot;nofollow&quot;&gt;specular reflection&lt;/a&gt; of incoming rays holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you can also obtain any kind of roughness in between. Actually, you can produce a mirror by starting with a generic white object and polishing it.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-15T17:08:53.433" CommentCount="4" />
  <row Id="1961" PostTypeId="1" CreationDate="2010-12-15T17:20:09.487" Score="3" ViewCount="439" Body="&lt;p&gt;Does the transmission medium affect the speed of a signal? For instance does light traveling through a fiber cable get a bit from A =&gt; B faster than copper can transmit a bit the over the same distance?&lt;/p&gt;&#xA;" OwnerUserId="848" LastActivityDate="2011-04-02T16:52:54.093" Title="Is a signal traveling through fiber faster than a signal traveling through copper?" Tags="&lt;electricity&gt;&lt;speed&gt;" AnswerCount="3" CommentCount="14" FavoriteCount="1" />
  <row Id="1962" PostTypeId="2" ParentId="1961" CreationDate="2010-12-15T17:27:17.057" Score="5" ViewCount="" Body="&lt;p&gt;The medium &lt;em&gt;does&lt;/em&gt; affect the speed of propagation, but I don't have detailed information on the technologies currently in use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rule of thumb is about 2/3 of the speed in vacuum for either coax cable or fiber. No idea for twisted pair.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2010-12-15T17:27:17.057" CommentCount="1" />
  <row Id="1963" PostTypeId="2" ParentId="1737" CreationDate="2010-12-15T18:00:01.323" Score="2" ViewCount="" Body="&lt;p&gt;Superfluid Helium-4 has a very well studied excitation structure -- at very low momenta, there is a low energy excitation, the phonon, that corresponds to a periodic density fluctuation in the superfluid with well defined wave-number and an energy $E = c \hbar k$ (c being the speed of sound in the superfluid). Though others might quibble with me over vocabulary, I prefer to call phonons &quot;collective excitations&quot; and reserve &quot;quasiparticle&quot; for excitations that correspond to renormalized single-particle excitations (like an electron in a Fermi Liquid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In either case though, what is meant is simply that the excitation is long-lived, or, by the uncertainty principle, that it has a sharply defined energy, and here the phonon does. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This linear relation breaks down at higher momenta, where the $E,k$ curve turns down, then turns back up. Near the local maximum, there are sharp excitations with an inverse-parabolic dispersion. These are the maxons. Near the local minimum we have a sharp excitation with a parabolic dispersion, and these are the rotons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Historically rotons were introduced by Landau, with a guess for their dispersion ($E=\Delta+(p-p_0)^2/2m$), not merely as a mathematical device, but because a superfluid described only by the phonon dispersion fails to capture the actual thermodynamics observed in Helium-4. On the other hand, Landau wasn't concerned about a qualitative, microscopic picture of what rotons &lt;em&gt;are,&lt;/em&gt; in the sense that we know what phonons mean for the local density of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I am actually not too sure what rotons are in that same microscopic sense, and the same for maxons, although it is apparent that a gas of phonons and a gas of rotons are very different things. On the other hand, this is somewhat paradoxical since all three of these excitations are just different parts of the same dispersion curve, so to say they are &lt;em&gt;fundamentally different&lt;/em&gt; is a phrase I wouldn't use without a great deal of caution...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, knowing all of the excitations present in superfluid helium is very important for calculating its thermodynamic properties, etc., however &lt;em&gt;I think&lt;/em&gt; (and hopefully someone will correct me if I'm wrong!) the only excitation that is &lt;em&gt;intrinsic&lt;/em&gt; to superfluidity is the phonon mode - this is because the superfluid state breaks a continuous gauge symmetry of the normal fluid, and thus phonons are the massless Goldstone modes of the superfluid state.&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2010-12-15T18:00:01.323" CommentCount="3" />
  <row Id="1964" PostTypeId="2" ParentId="1949" CreationDate="2010-12-15T18:28:15.170" Score="5" ViewCount="" Body="&lt;p&gt;I don't think its very easy to answer. This is a large tabular berg, so its width is many times its height. The key will be fluid flow of warm seawater to its underside. Without this flow, the sea underneath the berg would be depleted of heat and melting would stop. The transport of warmer water is probably determined by a combination of ocean currents and storm driven waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note: 4C is a max (not min) density for water, and thus would sink. Sea water freezes&#xA;at roughly 28F (-2C), so if it is embedded in water that is at the feezing point, the heat transfer would be from the iceberg into the sea. {But actually an antarctic iceberg probably&#xA;has an internal temperature of -20 to -40C}. It probably has to be transported north into warmer ocean water before the melting really gets going.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2010-12-15T18:28:15.170" />
  <row Id="1965" PostTypeId="2" ParentId="1961" CreationDate="2010-12-15T20:05:02.770" Score="5" ViewCount="" Body="&lt;p&gt;Well, definitely yes, i think. Depending on properties of the cable as well as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Relative_permittivity&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Relative permittivity&lt;/strong&gt;&lt;/a&gt;(also measured as dielectric constant) of the material surrounding the copper is a factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: The velocity factor also called &lt;strong&gt;wave propagation speed&lt;/strong&gt; or velocity of propagation (VoP or vP), of a transmission medium is the speed at which a wavefront (of an acoustic signal, for example, or an electromagnetic signal, a radio signal, a light pulse in a fibre channel or a change of the electrical voltage on a copper wire) passes through the medium, relative to the speed of light. For optical signals, &lt;strong&gt;refractive index&lt;/strong&gt; is a similar quantity.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Twisted pair copper cabling are known to transmit signals at speeds anywhere between 40% to 70% of that of light. So clearly, there are a lot of variables that come into play here that vary with material used, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cable#Electrical_cable_types&quot; rel=&quot;nofollow&quot;&gt;methods of cabling&lt;/a&gt; etc. See this &lt;a href=&quot;http://en.wikipedia.org/wiki/Wave_propagation_speed#Typical_velocity_factors&quot; rel=&quot;nofollow&quot;&gt;table&lt;/a&gt; for reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your question, a quick wikipedia lookup for &lt;a href=&quot;http://en.wikipedia.org/wiki/Optical_fibers#Light_scattering&quot; rel=&quot;nofollow&quot;&gt;fiber optic cables&lt;/a&gt; reveals this..&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The typical value for the cladding of an optical fiber is 1.46. The core value is typically 1.48. The larger the index of refraction, the slower light travels in that medium. From this information, a good rule of thumb is that signal using optical fiber for communication will travel at around 200 million meters per second.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which at first glance, might seem like faster but does not necessarily mean &quot;always faster&quot; than any/all types of copper cables.&lt;/p&gt;&#xA;" OwnerUserId="619" LastEditorUserId="619" LastEditorDisplayName="" LastEditDate="2010-12-16T02:36:04.327" LastActivityDate="2010-12-16T02:36:04.327" />
  <row Id="1966" PostTypeId="2" ParentId="1453" CreationDate="2010-12-15T21:42:19.343" Score="14" ViewCount="" Body="&lt;p&gt;A completely general method for overcoming the fermion sign problem in quantum monte carlo sampling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it is, any special circumstances in which we can guarantee the weights of a fermion path integral to be nonnegative are highly prized, but in general, in spatial dimensions greater than d = 1 we just can't accurately do QMC simulations for fermions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequences? Well, in addition to making my life (exponentially!) easier, it would also prove P=NP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(see, e.g.: Phys. Rev. Lett. 94, 170201)&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2010-12-15T21:42:19.343" CommunityOwnedDate="2011-01-22T18:11:53.713" CommentCount="4" />
  <row Id="1967" PostTypeId="2" ParentId="768" CreationDate="2010-12-15T22:20:25.600" Score="2" ViewCount="" Body="&lt;p&gt;Sorry I just noticed your comment on a previous question where you stated you would ask a separate question.  Here is a response.  (Sorry for any overlap with marek's.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as in order to talk about vectors in an n-dimensional vector space as n-tuples of numbers, you need to first choose basis, in order to talk about sections of vector bundles in a concrete way, you must choose a &quot;frame&quot; $\{e_a\}$ (which is just a fancy way of saying a family of basis vectors/sections).  Then the notation is exactly as before, a section $s$ looks like $s^a e_a$ (summed over $a$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example of a wave function for a charged particle, the vector bundle is one (complex) dimensional, so the single basis vector $e$ is usually dropped from the notation.  But it should be there, morally, as you note.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gauge transformation by $\chi$ amounts to changing your basis vector $e$ by multipliying it by $\exp(-i\chi)$, so the (invariant) vector $\Psi = \psi e$ has coordinate $\psi \exp(i\chi)$ in the new basis.  The gauge field $A_\mu$ is a 1x1 matrix, and has to be changed by adding $\exp(i\chi)\partial_\mu \exp(-i\chi) = -i\partial_\mu \chi$.  That way $D_\mu \Psi = \partial_\mu \psi + iqA_\mu \psi$ makes invariant sense, where $q$ is the charge (or representation, telling how $A$ acts on $\psi$ -- i.e. by multiplying with $q$ in front).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helped.&lt;/p&gt;&#xA;" OwnerUserId="20" LastActivityDate="2010-12-15T22:20:25.600" CommentCount="4" />
  <row Id="1968" PostTypeId="1" AcceptedAnswerId="2016" CreationDate="2010-12-15T22:38:27.830" Score="3" ViewCount="120" Body="&lt;p&gt;Summary: when the Moon is x degrees below the horizon, it interferes &#xA;w/ stargazing the same as astronomical twilight would. What is x (as a &#xA;function of the Moon's phase)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define civil, nautical, and astronomical twilight as when the sun &#xA;is 0-6, 6-12, and 12-18 degrees below the horizon respectively. This &#xA;corresponds roughly to what most people call twilight, the ability to &#xA;distinguish a horizon at sea, and the ability to see 6th magnitude &#xA;stars at the zenith. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, even when below the horizon, the Moon shines brightly enough &#xA;to interfere w/ stargazing. What are the equivalent twilight angles &#xA;for the Moon? I realize that even the full Moon overhead isn't bright &#xA;enough for civil twilight, so my real interest is in astronomical &#xA;twilight. Of course, this will vary greatly w/ the Moon's phase, and &#xA;slightly with Moon's distance. &lt;/p&gt;&#xA;" OwnerUserId="854" LastActivityDate="2010-12-17T11:04:26.007" Title="Lunar twilight and sixth magnitude stars" Tags="&lt;light&gt;&lt;astronomy&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="1969" PostTypeId="2" ParentId="1019" CreationDate="2010-12-15T22:41:44.120" Score="2" ViewCount="" Body="&lt;p&gt;That the PI Team is the best group to design and implement their own data systems, rather than bring in people experienced with IT security, data modeling for reuse by other groups, and other data informatics issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related misconception: that their data is so different from other data that a data system must be designed from the ground up for each new experiment.&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-12-15T22:41:44.120" CommunityOwnedDate="2010-12-15T22:41:44.120" />
  <row Id="1970" PostTypeId="1" CreationDate="2010-12-15T23:39:34.930" Score="0" ViewCount="75" Body="&lt;p&gt;Is there a formula that gives me the instantaneous change in temperature under ideal circumstances? Details: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a cloudless day, temperature is affected by two major things(?): &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;While the Sun is up, incoming solar radiation, which increases the temperature at a known rate(?) depending on the sun's angle. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Heat loss to &quot;outer space&quot;. &quot;Outer space&quot; is colder than the ground, so heat is always flowing away from the ground into outer space. The hotter it is on the ground, the faster the heat flows away. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is there a semi-accurate mathematical function that models these temperature changes for a given location on a given cloudless day? &lt;/p&gt;&#xA;" OwnerUserId="854" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-16T02:02:57.463" LastActivityDate="2010-12-16T03:12:48.240" Title="Computing delta temperature based on radiation and heat loss" Tags="&lt;temperature&gt;&lt;climate-science&gt;&lt;models&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="1971" PostTypeId="1" CreationDate="2010-12-16T00:08:23.980" Score="3" ViewCount="239" Body="&lt;p&gt;Summary: What &quot;well-known&quot; and short parametrized mathematical &#xA;function describes daily and hourly temperature for a given location? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the mean daily temperature graph for a given location, &#xA;it looks like a sine wave with a period of one year. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the hourly temperature for a given day for a given location &#xA;also looks like a sine wave with a period of one day. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, closer inspection (Fourier analysis) shows that they're &#xA;not. There are fairly strong components of frequency 2/year and 3/year &#xA;for the daily temperature, and the hourly temperature also has strong &#xA;non-single-period terms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a parametrized function that reasonably describes the daily &#xA;mean temperature and (a separate function) the hourly mean &#xA;temperature? The parameters would be location-based. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize I can keep taking more Fourier terms to increase accuracy, &#xA;but I was hoping for something more elegant. For example, maybe the &#xA;graph is a parametrized version of sin^2(x) or some other &#xA;&quot;well-known&quot; function. &lt;/p&gt;&#xA;" OwnerUserId="854" LastActivityDate="2011-02-24T18:52:58.927" Title="Approximating mean daily and hourly temperature beyond Fourier series" Tags="&lt;temperature&gt;&lt;climate-science&gt;&lt;models&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="1972" PostTypeId="2" ParentId="1852" CreationDate="2010-12-16T00:12:25.367" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;33 MJ is probably the projectile kinetic energy. The stored electrical energy required is this value, divided by the efficiency.  Since this is a military project, the actual efficiency value is probably sensitive, and therefore not released.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The fireball behind the projectile may be burning vaporized metal from the projectile and rails.  In a railgun, the current flows from one rail, through the projectile, and back into the other rail. The projectile makes a sliding electrical contact with the rails as it gets launched, and this contact carries an electrical current of tens of thousands of amps.  This kind of current through any sort of moving or sliding high-current electrical contact is probably going to burn off a little metal from both the projectile and the rails.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="857" LastActivityDate="2010-12-16T00:12:25.367" />
  <row Id="1973" PostTypeId="2" ParentId="1952" CreationDate="2010-12-16T00:22:33.177" Score="3" ViewCount="" Body="&lt;p&gt;What you have to take into account here is the discrete rotational symmetries of the tetrahedron. For instance let us write the state of the tetrahedron as $\mid i_1, i_2, i_3, i_4\rangle $ where $i_k$ is the spin on the $k^{\textrm{th}}$ vertex. The state in the figure you show above can then be written as $\mid o,i,i,o \rangle$ (with $o$ and $i$ meaning &quot;outward pointing&quot; and &quot;inward pointing&quot; respectively).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the absence of any anisotropies which break the rotational symmetry, the state $\mid i_3, i_1, i_2, i_4\rangle $ can be obtained from the state $\mid i_1, i_2, i_3, i_4\rangle $ by rotating the tetradhedron by $2\pi/3$ around the axis passing through the 4th vertex ($v_4$) and the center of triangle $\Delta_{123}$, i.e.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mid i,o,i,o \rangle = \hat R_4(2\pi/3) \mid o,i,i,o \rangle $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\hat R_i (\theta)$ is the operator for rotations by $\theta$ around the $i^\textrm{th}$ axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;alternatively you can also obtain $\mid i,o,i,o \rangle$ by performing a reflection across the axis passing through $v_3$ and bisecting the edge ($e_{12}$) between $v_1$ and $v_2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mid i,o,i,o \rangle = \hat S_{123} \mid o,i,i,o \rangle $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\hat S_{ijk} $ is the generator of reflections through the axis passing through $v_k$ and bisecting the edge ($e_{ij}$). Similarly we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mid i,i,o,o \rangle = \hat R_4(4\pi/3) \mid o,i,i,o \rangle $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, w.r.t these discrete symmetries the six-states you mention are not independent. We must take suitable linear combinations of these states to obtain a set of independent basis vectors which are invariant under the action of these symmetries. When you do this correctly the six states will reduce to three states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mid \Psi_4 \rangle = \frac{1}{\sqrt{3}}\left(\mid v_1, v_2, v_3, v_4\rangle + \mid v_3, v_1, v_2, v_4\rangle + \mid v_2, v_3, v_1, v_4\rangle \right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and likewise for $ \mid \Psi_3 \rangle $ and $ \mid \Psi_2 \rangle $. There are only &lt;em&gt;three&lt;/em&gt; such states, and not four (we have four triangles), because the fourth state (in this case $\mid \Psi_1 \rangle$ ) can be written as a linear sum of the other three !&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                        Cheers,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Following a suggestion by @bruce, just want to clarify that each $ \mid \Psi_i \rangle $ is invariant only under the action of the permutation group on the triangle &lt;em&gt;dual&lt;/em&gt; (opposite) to the vertex $v_i$. This is a subgroup of the full symmetry group of the tetrahedron.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-16T04:10:35.530" CommunityOwnedDate="2010-12-16T00:22:33.177" CommentCount="8" />
  <row Id="1974" PostTypeId="2" ParentId="1952" CreationDate="2010-12-16T01:00:27.037" Score="0" ViewCount="" Body="&lt;p&gt;It also bugs me that the description of the triangular lattice in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometrical_frustration&quot; rel=&quot;nofollow&quot;&gt;Geometric frustration&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The third spin cannot simultaneously minimize its interactions with both of the other two. Thus the ground state is twofold degenerate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within all eight state, except all up and all down, the ground state is should be six folds degenerate, instead of two folds. I guess something wrong here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, it is uncommon to consider the rotational symmetry of a lattice because the counting cannot be scaled for large $N$. So I guess they may consider the following symmetry:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$ s_i \rightarrow -s_i $$&#xD;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you consider the Hamiltonian of a antiferromagnetic Ising model or Heisenberg model:&#xA;$$H=J\sum_{i,j}\mathbf{s}_{i}\cdot\mathbf{s}_{j}$$ &#xA;This Hamiltonian is invariant under the above symmetry, so every state $\{s_i\}$ has the corresponding state $\{-s_i\}$. This symmetry should correspond to the $i\leftrightarrow o$ inversion in your counting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a reason why the above symmetry is considered. Considering the magnetization of a system:&#xA;$$M=\sum_i s_i$$&#xD;&#xA;However, because of the symmetry above, the ensemble average of this value is always 0 for all spin model with no external magnetic field. A remedy to this situation is to add a small external magnetic field to break the symmetry. Another one might be considering only half of the ensemble under the symmetry...&lt;/p&gt;&#xA;" OwnerUserId="562" LastEditorUserId="562" LastEditorDisplayName="" LastEditDate="2010-12-16T01:10:21.830" LastActivityDate="2010-12-16T01:10:21.830" CommentCount="5" />
  <row Id="1975" PostTypeId="1" CreationDate="2010-12-16T01:39:04.323" Score="5" ViewCount="267" Body="&lt;p&gt;For a given Hamiltonian with spin interaction, say Ising model&#xA;$$H=-J\sum_{i,j} s_i s_j$$&#xD;&#xA;in which there are no external magnetic field. The Hamiltonian is invariant under transformation $s_i \rightarrow -s_i$, so there are always two spin states with exactly same energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the magnetization $M = \sum_i s_i$, we can take the ensemble average&#xA;$$\left\langle M\right\rangle =\sum_{\{s_{i}\}}M\exp(-\beta E)$$&#xD;&#xA;and the result should be $\left\langle M\right\rangle  = 0$ because the two states with all spin flipped will exactly cancel each other. There is an argument for this in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ising_model#Peierls_droplets&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question: how is this situation handled for finite and infinite lattice? How can they obtain the non-zero magnetization for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ising_model#Two_dimensions_.E2.80.93_Onsager.27s_solution&quot; rel=&quot;nofollow&quot;&gt;2D Ising model&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M=\left(1-\left[\sinh\left(\log(1+\sqrt{2})\frac{T_{c}}{T}\right)\right]^{-4}\right)^{1/8}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some information&lt;/strong&gt;: for a 1D Ising model with external magnetic field, one can solve the Hamiltonian&#xA;$$H=-J\sum_{i,j} s_i s_j - \mu B \sum_i s_i$$&#xD;&#xA;and obtains the magnetization as&#xA;$$\left\langle M\right\rangle =\frac{N\mu\sinh(\beta\mu B)}{\left[\exp(-4\beta J)+\sinh^{2}(\beta\mu B)\right]^{1/2}}$$&#xD;&#xA;It gives the result $\left\langle M\right\rangle  \rightarrow 0$ when $B \rightarrow 0$ for any temperature and this match the definition of the magnetization above. However, it gives $\left\langle M\right\rangle  \rightarrow N\mu$ when we take the limit of $T \rightarrow 0$. It suggests the ordering of limit is important, but we still get $\left\langle M\right\rangle  = 0$ when there is no external magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reminder&lt;/strong&gt;: There are some precautions needed to care when you run computer simulation using the definition of magnetization above directly, otherwise, you will always get 0. These methods are similar to create a spontaneous symmetry breaking manually, for the Ising model, the following may be used:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Use the $\left\langle |M|\right\rangle$ instead&lt;/li&gt;&#xA;&lt;li&gt;Fix the state of one spin so that the system will close to either $M=+1$ or $M=-1$ at low temperature.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In general, the finite size scaling should be used because we are likely interested in the thermodynamic limit of the system.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;:&#xA;Visualization should explain the problem better. Here is the figure of the canonical distribution as a function of magnetization $M$ and temperature $T$ for 3D Ising model ($L = 10$).&#xA;&lt;img src=&quot;http://i.stack.imgur.com/jsf5R.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a fixed temperature $T \lesssim T_c$, there are two symmetric peaks with opposite magnetization. If we blindly use the $\left\langle M\right\rangle$ defined before, we will get $\left\langle M\right\rangle = 0$. So how do we deal with this situation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a finite system, there is a finite probability that the transition between two peaks can occur. However, the &quot;valley&quot; between two peaks will become deeper and deeper when the size of the system $L$ increase. When $L \rightarrow \infty$, the transition probability tends to zero and those two configuration space should be separated. Note that the &quot;flat mountain&quot; in the figure at $L = 10$ will also become a very very sharp peak when $L \rightarrow \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One method discussed in the answers below is to consider the average of $M$ for separate configuration spaces. This seems reasonable for infinite system, but becomes a problem for finite system. Another problem raised here is that how to find each separated configuration spaces?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Thanks people try to give the answers to this question. In the following discussion, Kostya gives the typical treatment of spontaneous symmetry breaking. Marek discusses the ensemble average below and above the critical temperature. Greg Graviton gives an analogue for the real space spontaneous symmetry breaking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone can explain better to the problem of configuration space and how to take the average for Ising model, other spin models or general case, you are welcome to leave answer here.&lt;/p&gt;&#xA;" OwnerUserId="562" LastEditorUserId="562" LastEditorDisplayName="" LastEditDate="2010-12-17T11:53:40.157" LastActivityDate="2011-05-07T12:26:37.513" Title="Zero magnetization of spin model without external magnetic field" Tags="&lt;statistical-mechanics&gt;&lt;ising-model&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="2" />
  <row Id="1976" PostTypeId="2" ParentId="1544" CreationDate="2010-12-16T02:30:41.203" Score="4" ViewCount="" Body="&lt;p&gt;Just a few random thoughts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is something important in your observation that the Born-Infeld model is essentially a free-space model. It is known to Boillat and Plebanski (separately in 1970) that the Born-Infeld model is the only model of electromagnetism (as a connection on a $U(1)$ vector bundle) that satisfies the following conditions&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Covariance under Lorentz transformations&lt;/li&gt;&#xA;&lt;li&gt;Reduces to Maxwell's equation in the small-field strength limit&lt;/li&gt;&#xA;&lt;li&gt;$U(1)$ gauge symmetry&lt;/li&gt;&#xA;&lt;li&gt;Integrable energy density for a point-charge&lt;/li&gt;&#xA;&lt;li&gt;No birefringence (speed of light independent of polarization). &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(the linear Maxwell system fails condition 4.) (See Michael Kiessling, &quot;Electromagnetic field theory without divergence problems&quot;, J. Stat. Phys. (2004) doi:10.1023/B:JOSS.0000037250.72634.2a for an exposition on this and related issues.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since you are interested in nonlinear optics inside a material, instead of in vacuum, I think conditions 1 and 5 can safely be dropped. (Though you may want to keep 5 as a matter of course.) Condition 4 is intuitively pleasing, but maybe not &lt;em&gt;too&lt;/em&gt; important, at least not until you have some candidate theories in mind that you want to distinguish. Condition 3 you must keep. Condition 2, on the other hand, really depends on what kind of material you have in mind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, a small suggestion: personally I think it is better to, from the get-go, write your proposed Lagrangian as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ L = T^{abcd} F_{ab}F_{cd} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;instead of $M^{ab}F_{cd}$. I think it is generally preferable to consider Lagrangian field theories of at least quadratic dependence on the field variables. A pure linear term suggests to me an external potential which I don't think should be built into the theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want something like condition 2, but with a dielectric constant or such, then you must have that $T^{abcd}$ admit a Taylor expansion looking something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T^{abcd} = \tilde{g}^{ac}\tilde{g}^{bd} + O(|F|) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\tilde{g}$ is some effective metric for the material. Birefringence, however, you don't have to insert in explicitly: most likely a generic (linear or nonlinear) $T^{abcd}$ you write down will have birefringence; it is only when you try to rule it out that you will bring in some constraints. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting thing is to consider what it means to have an analogous notion to condition 1. In the free-space case, condition 1 implies that the Lagrangian should only be a function of the Lorentz invariant $B^2 - E^2$ (in natural units) and of the pseudo-scalar invariant $B\cdot E$. In terms of the Faraday tensor these two invariants are $F^{ab}F_{ab}$ and $F^{ab}{}^*F_{ab}$ respectively, where ${}^*$ denote the Hodge dual. The determination of the &lt;em&gt;linear&lt;/em&gt; part of your theory (of electromagnetic waves in a material) is essentially by what you will use to replace condition 1. If you assume your material is isotropic and homogeneous, then some similar sort of scalar + pseudo-scalar invariants is probably a good bet. &lt;/p&gt;&#xA;" OwnerUserId="829" LastActivityDate="2010-12-16T02:30:41.203" CommentCount="8" />
  <row Id="1977" PostTypeId="2" ParentId="1970" CreationDate="2010-12-16T03:12:48.240" Score="2" ViewCount="" Body="&lt;p&gt;The radiative portion is actually pretty straightforward. For the absorption of sunlight&#xA;you multiply the strength of the sun (usually taken as 1000 watt/meter**2 at sea level by&#xA;1 minus the albedo, and multiply by the cosine of the angle. For IR, you need to know the&#xA;emissivity (but for most surfaces .95 to 1 can be assumed), and use the Stefan Bolzmann law. You also have downgoing IR radiation from the sky, this flux will depend upon the temperature structure of the atmosphere and the atmospheric humidity. I have an IR thermometer, and typical clear sky temperatures are rougly -20-30F during the summer, and -60F during the winter. This is typically 130watts (at -60f) and 200watts at -20f, so it is significant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat transport to the atmosphere is more difficult, as transport through a thin boundary layer will dominate. Although typically well more than half of any excess flux will be accounted for by the net IR flux.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-12-16T03:12:48.240" />
  <row Id="1978" PostTypeId="2" ParentId="1927" CreationDate="2010-12-16T04:12:03.050" Score="1" ViewCount="" Body="&lt;p&gt;Well, let's clear up some issues. contrary to the answers above I think you mean working out the uncertainty relationship (UR) itself as opposed to predicting the future. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the case first of all you shouldn't expect your simulated intelligence to come up with a version of quantum uncertainty relation since you're limiting your system to the classical laws. Then even if you mean some sort of classical chaos then it would be tricky. There is no physical limitation to prevent your intelligence to come up with an UR. However, computationally it has to be capable of abstraction, categorisation and a rule-based reasoning. My intuition is some techniques such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Adaptive_resonance_theory&quot; rel=&quot;nofollow&quot;&gt;adaptive resonance theory&lt;/a&gt; and a kind of evolutionary computation (http://en.wikipedia.org/wiki/Evolutionary_computation) plus the introduction of some noise can work in principle. The reason for introducing noise is that this is something you gonna need to get to the state of complexity you need in your system which comes out of the evolutionary computation.&lt;/p&gt;&#xA;" OwnerUserId="304" LastActivityDate="2010-12-16T04:12:03.050" CommentCount="1" />
  <row Id="1979" PostTypeId="2" ParentId="1927" CreationDate="2010-12-16T07:22:07.867" Score="1" ViewCount="" Body="&lt;p&gt;Some sources of uncertainties I could think of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-They could never determine what physical laws governs their universe with 100% certainty. If an experiment has the same outcome a googleplex times, you still cannot know the same outcome will occur the next time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Computational problems, we dont even have a solution to the 3-body problem in CM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-They would have to interact with their universe to measure it, how to account for what degree one at the same time impacts it ? &#xA;Question: Sometimes the source of UP in QM is explained by the fact one cannot measure a system without interacting with it, but this same applies to calssical mechanics, so what is different?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-If the space they live in is continuous, it seems impossible to measure position etc exactly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Also as you mentioned, a complete model of the universe, within the universe, would include the model, which includes the model of the model etc. This leads to many problems..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-In the case one overcomes all these, Gödel's incompleteness theorem would ultimately rule over all.&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2010-12-16T08:41:42.417" LastActivityDate="2010-12-16T08:41:42.417" />
  <row Id="1980" PostTypeId="2" ParentId="1453" CreationDate="2010-12-16T08:04:53.923" Score="18" ViewCount="" Body="&lt;p&gt;There is not (as far as I know) a satisfactory microscopic description of high-T&lt;sub&gt;c&lt;/sub&gt; superconductors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classical (s-wave) superconductors are described by &lt;a href=&quot;http://en.wikipedia.org/wiki/BCS_theory&quot; rel=&quot;nofollow&quot;&gt;BCS theory&lt;/a&gt;, which qualitatively assembles charge-carriers into phonon-mediated Cooper pairs, and successfully predicts many macroscopic phenomena like the Meissner effect.  This theory has been known for over 50 years.  The behavior of high-T&lt;sub&gt;c&lt;/sub&gt; superconductors seems to be a different animal, with substantially more complexity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were to have a good grasp of the microscopic theory, this would be a major insight into the behavior of certain highly-correlated electronic systems, and one could perhaps build on such an understanding to analyze more complicated systems.  On a more applications-oriented level, it is conceivable that we would have an easier time cooking up better-performing or more industrially economical superconductors.&lt;/p&gt;&#xA;" OwnerUserId="761" LastActivityDate="2010-12-16T08:04:53.923" CommunityOwnedDate="2011-01-22T18:11:53.713" />
  <row Id="1981" PostTypeId="2" ParentId="1875" CreationDate="2010-12-16T08:53:30.827" Score="1" ViewCount="" Body="&lt;p&gt;It is in principle possible for a unicycle to prevent the rider from falling over, even on a frictionless surface, by spinning up the (hopefully massive) wheel enough for the resulting torque to pull the axis under the rider.  Naturally, locomotion on a frictionless surface has its own set of problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the rider is falling sideways, the unicycle would need an additional motor to twist the wheel in an appropriate direction relative to the seat to compensate.  Such twisting could be a potential problem for people whose legs aren't tentacles.&lt;/p&gt;&#xA;" OwnerUserId="761" LastActivityDate="2010-12-16T08:53:30.827" />
  <row Id="1982" PostTypeId="1" AcceptedAnswerId="1983" CreationDate="2010-12-16T09:38:40.990" Score="10" ViewCount="277" Body="&lt;p&gt;Slightly biological, hopefully physical enough to be answered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose a magenta hue is represented by a mix of red and blue pigment. This is all very well for a creature with red and blue photoreceptors, but suppose it was seen by a creature which had a magenta-sensitive receptor, but no red or blue. Would the colour appear the same (insomuch as a qualititave concept can appear the same, I suppose I mean 'Would they interpret it as colour of the same wavelength?')?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crux of my question is, do the particular bands in which photoreceptors are activated affect vision of additive, as opposed to pure, hues?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, completely off-topic, but if there happens to be a biologist around, do animals on the whole have similar photoreceptors, or are they placed largely randomly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;Wyatt&lt;/p&gt;&#xA;" OwnerUserId="860" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-17T10:37:36.217" LastActivityDate="2011-02-04T07:56:21.577" Title="Is colour, as represented using primary colours, accurate only to humans?" Tags="&lt;biophysics&gt;&lt;color&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" />
  <row Id="1983" PostTypeId="2" ParentId="1982" CreationDate="2010-12-16T10:15:16.933" Score="15" ViewCount="" Body="&lt;p&gt;No, they will not appear the same. Humans have three color receptors so any possible color for us is just three numbers in RGB space. However, electromagnetic spectrum is continuous and there is an infinite number of spectra that would produce the same RGB stimulus. That is why you perceive the this page as white although it is in fact a combination of R,G,B tuned for a human eye. A creature with other set of photoreceptors would not see this page as white. Actually, white is also subjective, see how many settings for white balance your digital camera has, but suppose that we set it to 'daylight' and consider the continuous spectrum of sun as white. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another interesting point: magenta hue cannot be represented by any single wavelength, check out this famous horseshoe diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/b4HvK.png&quot; alt=&quot;CIE 1931 Color Space&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So magenta is even more subjective than green or red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, &lt;s&gt;most&lt;/s&gt; some animals don't perceive colors at all. There are some who have only two receptors and others who have four. &lt;a href=&quot;http://en.wikipedia.org/wiki/Mantis_shrimp&quot; rel=&quot;nofollow&quot;&gt;Mantis shrimp&lt;/a&gt; has 16 different photoreceptors!&lt;/p&gt;&#xA;" OwnerUserId="701" LastEditorUserId="701" LastEditorDisplayName="" LastEditDate="2011-02-04T07:56:21.577" LastActivityDate="2011-02-04T07:56:21.577" CommentCount="10" />
  <row Id="1984" PostTypeId="1" CreationDate="2010-12-16T10:30:17.663" Score="10" ViewCount="462" Body="&lt;p&gt;I read the definition of Work as $W = F \times d$ (Work = (Force) $\times$ (Distance)).&lt;br&gt;&#xA;If a book is there on the table, no work is done as no distance is covered. If I hold up a book in my hand and my arm is stretched, if no work is being done, where is my energy going?&lt;/p&gt;&#xA;" OwnerUserId="847" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2010-12-16T16:38:12.997" LastActivityDate="2010-12-16T16:38:12.997" Title="Why does holding something up cost energy while no work is being done?" Tags="&lt;energy&gt;&lt;biophysics&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="1" />
  <row Id="1985" PostTypeId="2" ParentId="1984" CreationDate="2010-12-16T11:48:50.573" Score="1" ViewCount="" Body="&lt;p&gt;When a physicist talks about work, they are using the word in the technical sense of the equation you quote. To a biologist, though, work might be defined as energy expended to carry out a task. In your example, your arm will not naturally stay in the position described. Your body (mostly your muscles) must expend energy to hold your arm (and the book) in a set position, unsupported by anything but your own physiology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, by the biologist's definition, your muscles are doing work to hold up the book and your arm (muscle fibers are contracting and relaxing based on a host of chemical processes at the cellular level). But by the physicist's technical definition, no work is being done.&lt;/p&gt;&#xA;" OwnerUserId="861" LastActivityDate="2010-12-16T11:48:50.573" CommentCount="2" />
  <row Id="1986" PostTypeId="2" ParentId="1984" CreationDate="2010-12-16T11:52:33.783" Score="6" ViewCount="" Body="&lt;p&gt;This is about how your muscles work -- the're an ensemble of small elements that, triggered by a signal from nerves, use chemical energy to go from less energetical long state to more energetical short one. Yet, this obviously is not permanent and there is spontaneous come back, that must be compensated by another trigger.&#xA;This way there are numerous streches and releases that in sum gives small oscillations that create macroscopic work on the weight.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-12-16T11:52:33.783" CommentCount="1" />
  <row Id="1987" PostTypeId="2" ParentId="1984" CreationDate="2010-12-16T11:53:15.883" Score="15" ViewCount="" Body="&lt;p&gt;While you do spend some body energy to keep the book lifted, it's important to differentiate it from physical effort. They are connected but are not the same. Physical effort depends not only on how &lt;strong&gt;much&lt;/strong&gt; energy is spent, but also on &lt;strong&gt;how&lt;/strong&gt; energy is spent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Holding a book in a stretched arm requires &lt;strong&gt;a lot&lt;/strong&gt; of physical effort, but it doesn't take that much energy. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In the &lt;strong&gt;ideal&lt;/strong&gt; case, &lt;em&gt;if you manage to hold your arm perfectly steady, and your muscle cells managed to stay contracted without requiring energy input&lt;/em&gt;, there wouldn't be any energy spent at all because there wouldn't be any distance moved.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On &lt;strong&gt;real&lt;/strong&gt; scenarios, however, you do spend (chemical) energy stored within your body, but &lt;em&gt;where&lt;/em&gt; is it spent? It is spent on a cellular level. The only way your muscle cells can stay contracted is by receiving electrical impulses. This electrical energy comes from the chemical energy contained within your molecules. What happens on a cellular level is that the cells contract and relax very quickly, thus resulting in a macroscopic contracted muscle. Since the cells are moving and applying a force, they are doing work. Since they are not efficient machines, they are also releasing a lot of heat. So, finally, the Chemical energy stored within your body is released by the cell as both work and heat.* &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Both on the ideal and the real scenarios we are talking about the physical definition of energy. On your consideration, you ignore the movement of muscle cells, so you're considering the ideal case. A careful analysis of the &lt;em&gt;real&lt;/em&gt; case leads to the conclusion that work is done and heat is released, even though the arm itself isn't moving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* &lt;em&gt;Ultimately, the work done by the cells is actually done on other cells, which eventually dissipates into heat due to friction and non-elasticity. So all the energy you spend is invested in keeping the muscle tension and eventually dissipated as heat.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="724" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2010-12-16T16:17:17.933" LastActivityDate="2010-12-16T16:17:17.933" CommentCount="5" />
  <row Id="1989" PostTypeId="1" AcceptedAnswerId="1990" CreationDate="2010-12-16T14:07:32.497" Score="2" ViewCount="129" Body="&lt;p&gt;What is terminal velocity? I've heard the term especially when the Discovery Channel is covering something about sky diving. Also, it is commonly known that HALO (Hi-Altitude, Lo-Opening) infantry reach &lt;em&gt;terminal velocity&lt;/em&gt; before their chutes open.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the terminal velocity be different for one individual weighing 180 pounds versus an individual weighing 250 pounds?&lt;/p&gt;&#xA;" OwnerUserId="862" LastActivityDate="2010-12-16T15:10:45.170" Title="What is terminal velocity?" Tags="&lt;gravity&gt;&lt;velocity&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="1990" PostTypeId="2" ParentId="1989" CreationDate="2010-12-16T14:33:40.857" Score="7" ViewCount="" Body="&lt;p&gt;Terminal velocity is the maximum velocity that you can reach during free-fall. If you imagine yourself falling in gravity, and ignore air resistance, you would fall with acceleration $g$, and your velocity would grow unbounded (well, until special relativity takes over). This effect is independent of your mass, since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = ma = mg \Rightarrow a = g$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where terminal velocity arises is that air resistance is a &lt;em&gt;velocity dependent&lt;/em&gt; force acting against your free fall. If we had, for example, a drag force of $F_D=KAv^2$ ($K$ is just a constant to make all the units work out, and depends on the properties of the fluid you're falling through, and $A$ is your surface area along the direction of motion) then the terminal velocity is the velocity at which the forces cancel (i.e., no more acceleration, so the velocity becomes constant):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = 0 = mg - KAv_t^2 \Rightarrow v_t=\sqrt{mg/KA}$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we see that a more massive object can in fact have a larger terminal velocity.&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2010-12-16T14:33:40.857" CommentCount="4" />
  <row Id="1991" PostTypeId="2" ParentId="1989" CreationDate="2010-12-16T14:36:01.090" Score="3" ViewCount="" Body="&lt;p&gt;You can find a good article here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Terminal_velocity&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Terminal_velocity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context you provide, terminal velocity is the maximum speed that an object in free fall reaches in the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When an object is falling, or in free fall, there are two forces that determine whether it will accelerate downwards or not:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;gravity (trying to accelerate the body downwards)&lt;/li&gt;&#xA;&lt;li&gt;air friction (trying to push the body upwards)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Initially, as the body is not moving, there is no air drag, and the object starts falling due to gravity.&#xA;Now, as the object speeds up, the gravity contribution remains constant, whereas the drag increases with the speed of the object.&#xA;Finally a point is reached where the drag is so much that the object does not accelerate anymore. Velocity stays constant and it is called terminal velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value for it is proportional to $\sqrt{m}$ so clearly objects of different weights have, in general different terminal velocities (heavier objects having higher values), but there are also other factors to account for, like how aerodynamic the object is. A sphere has higher terminal velocity than a sheet of metal of the same mass.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-12-16T14:36:01.090" />
  <row Id="1992" PostTypeId="2" ParentId="1984" CreationDate="2010-12-16T14:45:49.657" Score="-5" ViewCount="" Body="&lt;p&gt;Energy is being expended maintaining it in position. Earth's gravity is applying a force downwards, the book is being accelerated down gravitation force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A force is being applied to the hand and arm which must be resisted and thus energy expended. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arm and book are not a closed system.&lt;/p&gt;&#xA;" OwnerUserId="785" LastEditorUserId="785" LastEditorDisplayName="" LastEditDate="2010-12-16T15:20:48.787" LastActivityDate="2010-12-16T15:20:48.787" CommentCount="10" />
  <row Id="1993" PostTypeId="2" ParentId="1984" CreationDate="2010-12-16T14:52:25.360" Score="2" ViewCount="" Body="&lt;p&gt;The reason is that you need to &lt;strong&gt;spend energy to keep muscle stretched&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing you need know is that the work $W=F \Delta x$ is the energy transfer between objects. Hence, there are no work done on the book when it is put on the table because there are no movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When your arm &lt;a href=&quot;http://en.wikipedia.org/wiki/Muscle&quot; rel=&quot;nofollow&quot;&gt;muscle&lt;/a&gt; is stretched, however, it consumes energy continuously to keep this state so you feel tire very fast. This energy comes from the chemical energy in your body and most of them are converted into heat and lost to the surrounding. In this situation, no energy is transferred to the book, so no work is done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can feel the different energy consumption when your arm is stretched in different angle. A particular case is that you put the book on your leg when you sit on a chair so your muscle is relaxed and the energy spent is less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also a &lt;strong&gt;special type of muscle, &lt;a href=&quot;http://en.wikipedia.org/wiki/Smooth_muscle#Advances_and_current_research_in_contraction_and_relaxation&quot; rel=&quot;nofollow&quot;&gt;smooth muscle&lt;/a&gt;, requires very little energy to keep its state&lt;/strong&gt; so that it can always keep it stretched and you won't get tire: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Tonic smooth muscle contracts and relaxes slowly and exhibits force maintenance such as vascular smooth muscle. Force maintenance is the maintaining of a contraction for a prolonged time with little energy utilization.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="562" LastEditorUserId="562" LastEditorDisplayName="" LastEditDate="2010-12-16T16:13:12.857" LastActivityDate="2010-12-16T16:13:12.857" CommentCount="1" />
  <row Id="1994" PostTypeId="2" ParentId="1984" CreationDate="2010-12-16T15:05:44.113" Score="3" ViewCount="" Body="&lt;p&gt;Perhaps an analogy is in order. Lets hold up the book by using an electromagnet (say we put a piece if steel under it ). If the coils were made of superconducting material it would take no energy input to maintain the position/field strength. But if we use ordinary wire, ohmic loses within&#xA;the coil must be made up for by externally supplied electrical energy.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-12-16T15:05:44.113" />
  <row Id="1995" PostTypeId="2" ParentId="1989" CreationDate="2010-12-16T15:10:45.170" Score="1" ViewCount="" Body="&lt;p&gt;If the falling body is non-spherical, then the drag will be dependent upon the bodies&#xA;orientation. Skydivers exploit this to fly(fall) in formation, assume a higher drag configuration to slow down, or a lower drag configuation to speed up.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-12-16T15:10:45.170" />
  <row Id="1997" PostTypeId="2" ParentId="1975" CreationDate="2010-12-16T15:37:02.230" Score="4" ViewCount="" Body="&lt;p&gt;Maybe I didn't get the question, but the whole point of the discussion of Ising model is &lt;strong&gt;not&lt;/strong&gt; getting zero magnetization. This is a topic of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Spontaneous_symmetry_breaking&quot; rel=&quot;nofollow&quot;&gt;spontaneous symmetry breaking&lt;/a&gt; (further &quot;SSB&quot;). The symmetry you are talking about is broken spontaneously leading to nonzero magnetization. I must admit that I never got into details of Onsager solution, but I know that there are three approaches to this very important concept of modern physics.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Limit of small &quot;magnetic field&quot;.&lt;/strong&gt;&lt;br&gt;&#xA;You can consider a situation, when you got your system in a magnetic field $H$ first -- in that case there is no mentioned symmetry, and the magnetization gets along the magnetic field. Then you turn the magnetic field off.&lt;br&gt;&#xA;The point is that the limits $H\to\infty$ and $N\to\infty$ do not commute, so if you first take the $N\to\infty$ limit and then $H\to\infty$ you will end up with spontaneous magnetization.&lt;br&gt;&#xA;As far as I know -- this is the oldest approach to SSB and it is repeated in many statistical physics textbooks.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Boundary condition.&lt;/strong&gt;&lt;br&gt;&#xA;That so trivial that it is boring -- you just &quot;put in&quot; the preferred magnetization by setting appropriate boundary conditions for your system. Just fix the direction of your spins at the boundary or &quot;at infinity&quot;.&#xA;As far as I know -- this is the approach adopted in mathematical physics analysis of SSB.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;No transition in your lifetime&lt;/strong&gt;&lt;br&gt;&#xA;That is the most &quot;useless&quot; and the most beautiful in my opinion. There is always a probability for your system to go to the state with opposite direction of magnetization due to some fluctuation with really small probability. So the magnetization is still zero if you average over infinite time. But those transitions are so improbable that you will definitely not see one even if you will wait for &quot;billions of zillions&quot; of years.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="386" LastActivityDate="2010-12-16T15:37:02.230" CommentCount="3" />
  <row Id="1998" PostTypeId="1" CreationDate="2010-12-16T16:32:58.183" Score="1" ViewCount="143" Body="&lt;h2&gt;&lt;strong&gt;Preface&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The definition of &lt;strong&gt;average speed&lt;/strong&gt; of an object is defined by the distance travelled over time:&#xA;$$v_{avg} = \frac{x_2 - x_1}{t_2 - t1}$$&#xD;&#xA;The interpretation of speed is that when you multiply speed with the time interval, you should get the distance you travelled at this interval. However, it does not measure the local variation so a better definition is the &lt;strong&gt;instantaneous speed&lt;/strong&gt;:&#xA;$$v = \lim_{\Delta t\rightarrow0} \frac{\Delta x}{\Delta t}$$&#xD;&#xA;The above are very standard stuff for introduction mechanics course.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Average speed with equal segments&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now, if we have speed for consecutive paths with equal length, beginner students often  commit an error. They may use the following:&#xA;$$v_{avg} = \frac{1}{2}(v_1 + v_2)$$&#xD;&#xA;as average speed between the start point and ending point, which is WRONG in general. The correct way to obtain the effective average speed is &#xA;$$v_{avg}=\left(\frac{1}{N}\sum_{i}^{n}v_{i}^{-1}\right)^{-1}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Mean of varying speed measurements for the same path&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now, suppose that there is a situation that the distance between a starting point and an ending point is fixed. Now, there are experiments to measure the time spent by the traveller (or vehicle, or particle) moving along this path multiple time. Because the time it takes is always different, so we can obtain a list of speed $v_1, v_2, v_3, ..., v_n$ that differ with each other. Presumably there is a corresponding speed distribution $\mathcal{P}(v)$ for the speed travelling along this path. The &lt;strong&gt;mean of the speed distribution&lt;/strong&gt; is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\langle v\right\rangle = \int_0^\infty \mathcal{P}(v) dv  \approx \frac{1}{N} \sum_i^n v_i$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that it is different from the average speed that defined at the first part. It is the mean of the speed distribution function and it is similar to the &quot;wrong $v_{avg}$&quot; discussed in the previous part. Please &lt;strong&gt;do not mix it up&lt;/strong&gt; with the average speed defined above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question: I am seeking an interpretation of this quantity $\left\langle v\right\rangle$, that is, I want to know the &lt;em&gt;situations/problems&lt;/em&gt; that this idea of $\left\langle v\right\rangle$ can be applied.&lt;/p&gt;&#xA;" OwnerUserId="562" LastEditorUserId="562" LastEditorDisplayName="" LastEditDate="2010-12-16T19:13:52.663" LastActivityDate="2011-01-18T00:23:49.257" Title="Application of mean of speed distribution" Tags="&lt;speed&gt;" AnswerCount="2" CommentCount="13" />
  <row Id="1999" PostTypeId="2" ParentId="1975" CreationDate="2010-12-16T17:37:53.777" Score="3" ViewCount="" Body="&lt;p&gt;Let me ask a seemingly naive question first: what is actually a statistical ensemble? Mathematically, it's a prescription of an assignment of probability to every microstate. So this is a probability measure. But we have to be careful with measures defined on infinite systems. You certainly can't blindly work with $\exp(-\beta E)$ factors as energy will generally be infinite for an infinite system. But fortunately in the Ising model case (and lots of other spin models), it turns out that there exists a very nice &lt;a href=&quot;http://en.wikipedia.org/wiki/Gibbs_measure&quot; rel=&quot;nofollow&quot;&gt;Gibbs measure&lt;/a&gt;. I am not going to derive all the results on Gibbs measures here (see e.g. &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0899254624&quot; rel=&quot;nofollow&quot;&gt;Georgii&lt;/a&gt;'s bible) but basically one consistently prescribes how the measure should behave in all &lt;em&gt;finite&lt;/em&gt; regions when subjected to some boundary conditions. The logic being that in finite regions everything behaves nicely (and energy is finite) and we can recover infinite-volume thermodynamic behavior by some limiting procedures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you are talking about &lt;strong&gt;the&lt;/strong&gt; ensemble average, you are supposing there is only one such ensemble and therefore only one Gibbs measure for a given model. It turns out that this assumption fails. There can exist more than one measure and this is actually equivalent with an appearance of a first-order phase transition. In other words, you have only one measure (or one phase) for $\beta \leq \beta_c$. For $\beta &amp;gt; \beta_c$ you will always have two (equally good) measures to choose from and actually you can also take their convex combination but it turns out only extreme points of the convex set of all Gibbs measures are what we would call a phase. The reason you obtained zero in your average is precisely because you've picked such a combination $1/2 (\mu^{\beta}_+ + \mu^{\beta}_-)$. But this is not a macroscopic phase but rather a description of your (complete) uncertainty that you don't yet know whether system is in $+$ state or the $-$ state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second-order phase transition (the one that corresponds to SSB here) is associated with &quot;merging&quot; of the Gibbs measures $$\lim_{\beta \to \beta_c} \mu^{\beta}_+ = \lim_{\beta \to \beta_c} \mu^{\beta}_- = \mu^{\beta_c}$$&#xA;Or perhaps more natural point of view is the &quot;splitting&quot; of the measure as you decrease the temperature, so that system has to &quot;decide&quot; which of the low-temperature phases it will pick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt; what I implicitily assumed above was that the measures are shift-invariant (homogeneous, if you will). If this condition is dropped there exist many more measures (associated with phases) which describe an interfaces. E.g. consider an Ising system in a square with $+$ boundary condition on the bottom of the box and $-$ on the top. In a thermodynamic limit one obtains a system which is mostly $+$ on the bottom, mostly $-$ on the top and with a fluctuating interface in the middle.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-05-07T12:26:37.513" LastActivityDate="2011-05-07T12:26:37.513" CommentCount="6" />
  <row Id="2000" PostTypeId="1" AcceptedAnswerId="2020" CreationDate="2010-12-16T17:59:14.147" Score="1" ViewCount="83" Body="&lt;p&gt;I've been trying to understand the various concepts behind x-ray optics compared to standard visible/IR optics like mirrors and such.  However, the x-ray mirrors I've been finding typically have optimal angles of reflectance on the order of a few milliradians, particularly for hard x-rays.  I think I mostly understand the reasons for this (having something to do with Bragg diffraction?), but, say, for absorption experiments with low sample density, it may be necessary to use something analogous to a multipass absorption cell.  Since x-rays can't be reflected at anywhere near 90° with high efficiency, can a multipass absorption cell be constructed for x-rays (say, with fiber optics or something else)?  Has anyone done so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiple Google searches for such a thing have turned up nothing, but maybe I'm just searching for the wrong thing.&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2010-12-17T13:45:49.527" Title="Can a multipass x-ray absorption cell be constructed?" Tags="&lt;optics&gt;&lt;spectroscopy&gt;&lt;x-ray&gt;" AnswerCount="2" />
  <row Id="2001" PostTypeId="1" AcceptedAnswerId="2002" CreationDate="2010-12-16T18:40:27.773" Score="1" ViewCount="67" Body="&lt;p&gt;When we derive refraction and reflection laws for a generical plane wave on a surface, we say that reflected and refracted must be in phase with the incident wave. Why a medium cannot do a sphased reaction beside incident wave? Sorry for my trivial question.&lt;/p&gt;&#xA;" OwnerUserId="558" LastEditorUserId="107" LastEditorDisplayName="" LastEditDate="2010-12-16T18:51:13.227" LastActivityDate="2010-12-16T18:51:13.227" Title="Reflected and refracted wave sphased" Tags="&lt;optics&gt;&lt;waves&gt;&lt;plane-wave&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="2002" PostTypeId="2" ParentId="2001" CreationDate="2010-12-16T18:50:55.130" Score="2" ViewCount="" Body="&lt;p&gt;It is possible. Take a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Phase_conjugate_mirror#Optical_phase_conjugation&quot; rel=&quot;nofollow&quot;&gt;phase conjugated mirror&lt;/a&gt; on Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, phase shifts do happen on reflection from flat surfaces -- all the time. It's called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Goos-Hanchen_effect&quot; rel=&quot;nofollow&quot;&gt;Goos-Hänchen effect&lt;/a&gt;. It's just very very small, so that's probably why you ignore it in your derivation.&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2010-12-16T18:50:55.130" CommentCount="2" />
  <row Id="2004" PostTypeId="2" ParentId="1975" CreationDate="2010-12-16T19:16:44.297" Score="3" ViewCount="" Body="&lt;p&gt;If I understand you correctly, you seem to be asking the fundamental question of how on earth broken symmetries are possible. After all, there is an apparent contradiction: on one hand, the thermal average $\langle M \rangle= \sum_{\{s_i\}} M \exp(-\beta E)$ must vanish due to symmetry, while on the other hand, physicists claim that the 2D Ising model gives rise to a nonzero magnetization. How can that be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following thought experiment will provide an instructive answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a table with an upside-down pendulum attached to it. We assume that the head of the pendulum can only move from left to right or the other way round until it hits the table, i.e. it is confined to one dimension and its position can be described by a single angle $\phi\in[-\pi,\pi]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/3qRKi.png&quot; alt=&quot;Pendulum, upside-down&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The potential energy of the pendulum is given by $E(\phi)=-A\cos \phi$. Clearly, the whole situation is symmetric with respect to a reflection $\phi \mapsto -\phi$; the potential energy is invariant under this reflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this symmetry of the energy does &lt;em&gt;not&lt;/em&gt; imply that the &lt;em&gt;stable states&lt;/em&gt; are symmetric themselves. Namely, the energy has &lt;em&gt;two, asymmetric&lt;/em&gt; minima at $\phi=\pi$ and $\phi=-\pi$ respectively (the pendulum is stopped by the table); the stable states are no longer symmetric themselves. Of course, the symmetry is not lost; if you have calculated one stable state, you can apply the reflection and get the other, but the point is that there are several stable states now. The energy is symmetric, but its minima are not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, imagine the following experiment: the table with the pendulum is located in one room and you are located in a different room. A friend of yours puts the pendulum exactly in the $\phi=0$ position (which is unstable) and then lets go. Since you are in another room, you can't see what is going to happen next in the thermodynamic system consisting of the pendulum and your friend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: before you go into the other room, what is the average position, the &quot;ensemble average&quot; of the pendulum? Well, it will fall to either the left or the right with equal probability, so after a short amount of time in which the pendulum hits the table and thermal equilibrium is restored, the average will be…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle \phi \rangle = \frac12 \pi + \frac12 (-\pi) = 0 \quad!$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you now run into the other room expecting the pendulum to be at position $\phi=0$, dangerously balancing on top of the table, you will be disappointed to learn that it has fallen to either the left $\phi=\pi$ or to the right $\phi=-\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What has happened? The point is that the thermal average is indeed zero, but the situation is so unstable that as you soon as you acquire a teensy bit of information, like a short glimpse into the other room, or knowledge of the position of &lt;em&gt;just one single molecule&lt;/em&gt; of the pendulum, the situation will split in two and you will observe one of two different outcomes. The outcomes have equal probability, that's why you get an ensemble average of zero, either one could have happened. But they occupy quite different parts of the configuration space, and you suddenly get a nonzero angle or nonzero magnetization. Compare this to an ideal gas where getting information about the position of a single molecule has a vanishing impact on the situation at large. For the pendulum, knowing the position of just one molecule changes everything, and that's why the thermal average does not give an adequate description of what you will measure.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Update&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;To elaborate on the last paragraph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to understand what the thermal average $\langle \phi \rangle$ does. A very good explanation is provided in E.T. Jaynes' article &lt;a href=&quot;http://bayes.wustl.edu/etj/articles/ccarnot.pdf&quot; rel=&quot;nofollow&quot;&gt;The Evolution of Carnot's Principle&lt;/a&gt;. Namely, the thermal average is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the average over the probability distribution that maximizes the &quot;number of ways in which the system can be realized&quot; (= entropy), subject to the constraint that the average energy has a fixed value $\langle E \rangle = E_0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, if we only know the average energy of the system and are ignorant about everything else, the maximum entropy principle gives a probability distribution with which we can calculate other quantities, like the average magnetization $\langle M \rangle$. For our pendulum, this reasoning makes sense: if we are in the other room, the &lt;em&gt;best prediction&lt;/em&gt; about the position of the pendulum that we can make is indeed $\langle \phi \rangle = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we go into the other room and look at the pendulum, we acquire slightly more information than just the average energy, for instance the position of a single molecule of the pendulum. The point is that the probability distribution is peaked around $\phi=\pi$ and $\phi=-\pi$, and the little information we acquire is enough to see that the system is located at one of the peaks. In other words, lifting our ignorance just a little bit gives rise to a marked change in prediction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare this to an ideal gas, where learning of the position of a single molecule does not change anything about the probability distribution at large and won't change our prediction of macroscopic quantities.&lt;/p&gt;&#xA;" OwnerUserId="188" LastEditorUserId="188" LastEditorDisplayName="" LastEditDate="2010-12-17T09:45:06.013" LastActivityDate="2010-12-17T09:45:06.013" CommentCount="19" />
  <row Id="2005" PostTypeId="2" ParentId="1998" CreationDate="2010-12-16T21:07:31.190" Score="2" ViewCount="" Body="&lt;p&gt;Suppose an object travels through a fluid at low Reynolds number.  Then the drag force on it is proportional to its velocity.  In an talk called &lt;a href=&quot;http://jila.colorado.edu/perkinsgroup/Purcell_life_at_low_reynolds_number.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Life at Low Reynolds Number&quot;&lt;/a&gt;, Ed Purcell wrote about bacteria swimming at low Reynolds number,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA; For these animals inertia is totally irrelevant. We know that F=ma, &#xA;but they could scarcely care less. I'll show you a picture of the real animals in a bit but we are going to be taking about objects &#xA;which are the order of a micron in size... In water where the &#xA;kinematic viscosity is $10^{-2}$ cm/sec these things move around with a typical speed of $30$ micron/sec. If I have to push that animal &#xA;to move it, and suddenly I stop pushing, how far will it coast before it slows down? The answer is, about $0.1$ angstrom. And it takes it about $0.6$ microsec to slow down. I think this makes it clear what &#xA;low Reynolds number means. Inertial plays no role whatsoever.&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, if you're a bacterium, then energy you need to use to take a stroll can be calculated solely from the energy needed to overcome drag.  The drag is proportional to velocity, so the energy needed to travel a certain distance is proportional to the velocity you travel at while doing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mean of the speed distribution, defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle v \rangle = \int_{path} v(x) dx$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is proportional to the energy expended per unit distance traveled.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-12-16T21:07:31.190" CommentCount="4" />
  <row Id="2006" PostTypeId="1" AcceptedAnswerId="2007" CreationDate="2010-12-16T22:07:41.800" Score="-4" ViewCount="143" Body="&lt;p&gt;In theory, dark energy be used to generate electricity. Given its apparent abundance, this could provide rather a lot (perhaps too much) clean energy.  Has any research been conducted  on this or any other practical applications for dark energy? (other than repelling galaxies ;)&lt;/p&gt;&#xA;" OwnerUserId="868" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-21T20:37:35.743" LastActivityDate="2010-12-21T20:37:35.743" Title="Dark energy converted to electricity" Tags="&lt;dark-energy&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="2007" PostTypeId="2" ParentId="2006" CreationDate="2010-12-16T22:10:34.450" Score="1" ViewCount="" Body="&lt;p&gt;Dark energy is not really an energy in the common sense and there absolutely is no project to produce electricity via dark energy. Dark energy is most probably a mistake in our calculations. Even if it is not so, the nature of dark energy is currently so far from our grasp that there is no way, even theoretically, to generate energy from dark energy yet.&lt;/p&gt;&#xA;" OwnerUserId="504" LastActivityDate="2010-12-16T22:10:34.450" CommentCount="7" />
  <row Id="2008" PostTypeId="2" ParentId="566" CreationDate="2010-12-16T23:31:33.343" Score="5" ViewCount="" Body="&lt;p&gt;Typically, a laser will damage an optical surface in one of two ways. The first is just what you would expect: the laser heats the material up until something bad happens. The second is also pretty simple, but less common because (AFAIK) it is really only a problem with very short pulses (on the order of femtoseconds). In this case a small but rapidly changing mechanical stress is produced by the beam, either as an effect of heating or the electrostrictive effect. The resulting shock is strong enough to damage the optical surface, usually by producing microscopic chips or causing optical coatings to de-laminate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the important issue here, which should answer your question, is that average power doesn't really mean much in the context of optical damage. It can be an issue in cases where heat dissipation over long time scales becomes a problem, but in cases where the laser is pulsed, damage mechanics are primarily determined by the peak power and the pulse width.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make analogy: I have never been shot with a gun. If I get shot in the head tomorrow, my lifetime average bullet flux will be very low, just one bullet in 26 years. Yet I'll still be dead.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2010-12-16T23:31:33.343" />
  <row Id="2009" PostTypeId="2" ParentId="2000" CreationDate="2010-12-17T00:40:35.627" Score="2" ViewCount="" Body="&lt;p&gt;I guess the idea here would be to make something like a ring cavity for ordinary light, where the light is bounced around a closed loop. Which is something you see done in dye lasers, and has probably been used to do laser spectroscopy at some point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with this is that if your deflection angle is a few milliradians, then you would require thousands of bounces to come all the way back around to where you started. Since the reflection efficiency isn't perfect, you would lose most of your x-rays by the time you managed to get them back in line for a second pass. Even if the reflectivity was 99.9% (which it isn't in the x-ray band), you'd only have about 37% of the light left after 1000 bounces.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-12-17T00:40:35.627" CommentCount="2" />
  <row Id="2010" PostTypeId="1" CreationDate="2010-12-17T01:20:01.267" Score="1" ViewCount="120" Body="&lt;p&gt;Whenever we measure something, it is usually inexact.  For example, the mass of a baseball will never be measured &lt;em&gt;exactly&lt;/em&gt; on a scale in any unit of measurement besides &quot;mass in baseballs that are currently being measured&quot; and rational multiples thereof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any non-arbitrary physical quantities that are &quot;inherently&quot; rational?  That is, quantities that can be expressed &quot;exactly&quot; in a non-trivial, fundamental unit of measurement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, there is &lt;strong&gt;total charge&lt;/strong&gt;, because charge can always be represented as an integer multiple of the charge of an electron.  And to an extent, photons/light can be described this way too.  Are there any others?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, are there any inherently &quot;algebraic&quot; quantities that are not inherently rational?  That is, quantities that can be expressed closed-form as the solution to an algebraic equation in a non-trivial unit of measurement, but not as an &quot;exact&quot; ratio to that unit of measurement?&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="71" LastEditorDisplayName="" LastEditDate="2010-12-17T01:35:09.657" LastActivityDate="2010-12-17T05:02:36.963" Title="Are there any quantities in the physical world that are inherently rational/algebraic?" Tags="&lt;quantization&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="2011" PostTypeId="2" ParentId="2010" CreationDate="2010-12-17T02:50:41.823" Score="4" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there any non-arbitrary physical quantities that are &quot;inherently&quot; rational?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Absolutely! The conductivity plateaus in the integer and fractional quantum hall effects correspond to conductivities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sigma = \nu e^2/h $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\nu$ take on values which are integers (obviously) for the IQHE and fractions for the FQHE. So if you fix the &quot;non-trivial, fundamental unit of measurement&quot;, as you put it, for conductance to be $e^2/h$ then the hall conductivities for the IQHE and FQHE are &quot;inherently rational&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy levels of the Hydrogen atom also satisfy these criteria. The energy of the $n^\textrm{th}$ orbital is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E_n = - \frac{E_r}{n^2} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take the Rydberg energy $E_r$ to be a &quot;fundamental unit of measurement&quot; then $E_n$ again satisfy the criteria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some might argue that $E_r$ is not fundamental enough. However, the quantum of hall conductance $e^2/h$ certainly is as fundamental as it gets. There are plenty of other examples such as, as mentioned by you and others, charge and spin. I'd like to add magnetic flux to that list. Again, from the quantum hall effect, where $p$ quanta of magnetic flux $\phi_0$ pair up with $q$ electrons to give you a composite fermion with an anyonic exchange phase $pq$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, (almost?) any observable that corresponds to a topological invariant has eigenvalues which are integer or rational.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, are there any inherently &quot;algebraic&quot; quantities that are not inherently rational?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Interestingly, that appears to be a harder question to answer. So I'll let someone handle this part ;-)&lt;/p&gt;&#xA;" LastActivityDate="2010-12-17T02:50:41.823" CommentCount="11" />
  <row Id="2013" PostTypeId="2" ParentId="1743" CreationDate="2010-12-17T06:14:20.257" Score="2" ViewCount="" Body="&lt;p&gt;I'll take a shot at it with a few assumptions. Since you're talking about contact between macroscopic objects, and the area of the sheets represents the area of contact, we can assume $R&amp;gt;&amp;gt;D&amp;gt;&amp;gt;d$. As has been mentioned in the comments, we cannot extrapolate to infinite sheets or there won't be an electric field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method here will be to keep the first sheet fixed, and estimate the potential energy of the second sheet as a function of $D$. The derivative $\frac{dU}{dD}$ will give us the force suffered by it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Placing the fixed sheet on the x-y plane, with its center on the origin, we have to start by calculating its electric field. We'll restrict ourselves to calculating the electric field it generates on a point that lies on the $z$ axis, a distance $D$ from its center (justified later). For that, we'll assume the sheet is disk-shaped, and treat this dipole disk as a pair of charged disks separated by a distance $d$ that coincide with the planes $z=\pm d/2$. The electric field generated by a charged disk is easy to find on textbooks (and easy to integrate as well), being&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E(z) = \frac{\sigma}{2\epsilon_0}\left(1 - \frac{z}{\sqrt{R^2+z^2}}\right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rewriting it to suit our dipole:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E_T(z) = E(z+d&amp;#39;) - E(z-d&amp;#39;) = \frac{\sigma}{2\epsilon_0}\left(- \frac{z+d&amp;#39;}{\sqrt{R^2+(z+d&amp;#39;)^2}} + \frac{z-d&amp;#39;}{\sqrt{R^2+(z-d&amp;#39;)^2}}\right),$$&#xD;&#xA;where $d&amp;#39;=d/2$ and $\sigma = eN$. We could use a couple approximations on the above equation, but they're unnecessary right now. The point is that we know what is the electric field that the fixed sheet generates on the center of the second sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll now have to use the approximations. This result is the electric field that's felt by a single dipole on the second sheet. If the sheets were infinite, symmetry would imply that all dipoles on the second sheet are under the same (constant) electric field, and the potential energy per unit area of the second sheet would be $U=\vec{E}_T\cdot\vec{p}\,N = E_T(D)p\,N$. Since $R&amp;gt;&amp;gt;D$, the border effects won't be too relevant, and we'll assume that $E_T$ is constant along the entire second sheet (much like one does inside capacitors). That leads to the total potential energy of the second sheet&#xA;$$U(D)=\frac{(pN)^2}{2\epsilon_0d}\left(- \frac{D+d&amp;#39;}{\sqrt{R^2+(D+d&amp;#39;)^2}} + \frac{D-d&amp;#39;}{\sqrt{R^2+(D-d&amp;#39;)^2}}\right)\pi R^2,$$&#xD;&#xA;as well as the potential energy per unit area&#xA;$$u(D)=\frac{(pN)^2}{2\epsilon_0d}\left(- \frac{D+d&amp;#39;}{\sqrt{R^2+(D+d&amp;#39;)^2}} + \frac{D-d&amp;#39;}{\sqrt{R^2+(D-d&amp;#39;)^2}}\right).$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the first derivative with respect to $D$, will give us the force and the pressure (respectively) that the fixed sheet applies on the second sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here go the answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F=\frac{\pi(RpN)^2}{2\epsilon_0d}(G_- - G_+)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P=\frac{(pN)^2}{2\epsilon_0d}(G_- - G_+)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G_\pm = (R^2 + (D\pm d/2)^2)^{-.5} - (D\pm d/2)^2(R^2 + (D\pm d/2)^2)^{-1.5}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I can tell, these are the Force and Pressure that the question asks for. Below are a couple of plots for $G_--G_+$ (measured in $1/m$), just multiply their $y$ values by $\frac{(pN)^2}{2\epsilon_0d}$ to get the pressure. The $x$ axis is $D$ measured in $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This plot uses $R=1mm$. &lt;strong&gt;Green line is $d=1nm$, Red line is $d=10nm$.&lt;/strong&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/0B8Pj.jpg&quot; alt=&quot;Plot 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This plot uses $R=1cm$. &lt;strong&gt;Green line is $d=1nm$, Red line is $d=10nm$.&lt;/strong&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/h5poh.jpg&quot; alt=&quot;Plot 2&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="724" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2010-12-17T17:13:42.883" LastActivityDate="2010-12-17T17:13:42.883" CommentCount="5" />
  <row Id="2014" PostTypeId="1" CreationDate="2010-12-17T06:48:27.953" Score="8" ViewCount="453" Body="&lt;p&gt;The correlation functions found in Barouch and McCoy's paper (PRA 3, 2137 (1971)) for the XX spin chain use a method which uses Wick's theorem. For the zz correlation function, this gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\langle \sigma_l^z \sigma_{l+R}^z \rangle = \langle \sigma_l^z \rangle^2 - G_R^2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where for $R=1$, $G_1 = -\langle \sigma_l^x \sigma_{l+1}^x+ \sigma_l^y \sigma_{l+1}^y \rangle/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I calculate $\langle \sigma_l^z \sigma_{l+1}^z \rangle$ both explicitly and using the equation above for 8 qubits, I get different answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is Wick's theorem still valid for 8 qubits, which means I've just made a mistake? Or is it valid only in the thermodynamic limit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your replies everyone. @lcv However, I haven't used the analytical diagonalisation for this - I have simply used Mathematica to diagonalise the 8 qubit chain numerically after substituting arbitrary values for the coupling strength, magnetic field and temperature. Hence it can't be an error in the diagonalisation. It is the thermal average I have calculated, that is $\langle \sigma^z_l \rangle=tr(\rho \sigma^z_l )$ where $\rho=e^{−H/T}/tr(e^{−H/T})$ and T is temperature. But in doing this, I find that $\langle \sigma^z_l \sigma^z_{l+R} \rangle \neq \langle \sigma^z_l \rangle^2 - G_1^2$ where I've defined $G_1$ above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit2&lt;/strong&gt; (@marek @lcv @Fitzsimons @Luboš) I'm going to try to clarify - The open XX Hamiltonian in a magnetic field is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;H=-\frac{J}{2}\sum_{l=1}^{N-1} (\sigma^x_l \sigma^x_{l+1} + \sigma^y_l \sigma^y_{l+1})- B \sum_{l=1}^N \sigma^z_l &#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Mathematica, I have defined the Pauli spin matrices, then the Hamiltonian for 8 qubits. I then put in values for $J$, $B$ and $T$, and calculate the thermal density matrix, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\rho = \frac{e^{-H/T}}{tr(e^{-H/T})}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now I have numerical density matrix. I then calculate $\langle \sigma^z_l \sigma_{l+1}^z \rangle=tr(\rho \sigma^z_l \sigma_{l+1}^z )$ using the definitions of the Pauli spin matrices and $\rho$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next I calculate $\langle \sigma_l^z \sigma_{l+R}^z \rangle$ using the result from Wick's theorem which gives $\langle \sigma_l^z \rangle^2 - G_R^2$ where for $R=1$, $G_1 = -\langle \sigma_l^x \sigma_{l+1}^x+ \sigma_l^y \sigma_{l+1}^y \rangle/2$. I again use the Pauli spin matrices I defined and the same numerical $\rho$ to calculate them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I get a different (numerical) answer for each of these.&lt;/p&gt;&#xA;" OwnerUserId="873" LastEditorUserId="873" LastEditorDisplayName="" LastEditDate="2011-02-25T16:17:06.643" LastActivityDate="2011-02-25T16:17:06.643" Title="To calculate the correlation functions of an XX spin chain, Wick's theorem is used. But is it valid for a chain of any size?" Tags="&lt;quantum-mechanics&gt;&lt;statistical-mechanics&gt;&lt;spin-chains&gt;" AnswerCount="3" CommentCount="11" FavoriteCount="1" />
  <row Id="2015" PostTypeId="1" CreationDate="2010-12-17T10:01:24.917" Score="10" ViewCount="293" Body="&lt;h3&gt;Dear Community,&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;recently, &lt;a href=&quot;http://en.wikipedia.org/wiki/Transformation_optics&quot; rel=&quot;nofollow&quot;&gt;Transformation optics&lt;/a&gt; celebrates some sort of scientific revival due to its (possible) applications for cloaking, see e.g. &lt;a href=&quot;http://www.sciencemag.org/content/323/5910/110.abstract&quot; rel=&quot;nofollow&quot;&gt;Broadband Invisibility by Non-Euclidean Cloaking&lt;/a&gt; by Leonhard and Tyc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/users/829/willie-wong&quot;&gt;Willie Wong&lt;/a&gt; explained to me in &lt;a href=&quot;http://physics.stackexchange.com/questions/1544/nonlinear-optics-as-gauge-theory&quot;&gt;another thread&lt;/a&gt; how a so-called &lt;strong&gt;material manifold&lt;/strong&gt; could be used to include the effects of nonlinear optics as some sort of effective metric $\tilde{g}$. Indeed, this should not only work in that case, the easiest example of such an approach should be &lt;strong&gt;transformation optics&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bevore explaining what is meant with such a construction, I will formulate the &lt;strong&gt;question&lt;/strong&gt;:&lt;br&gt;&#xA;How can one understand transformation optics in terms of the material manifold?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The material manifold&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The following explanations are directly linked to Willie's comments in the mentioned thread. Since I will formulate them on my own, mistakes might occur, which are in turn only related to me. Willie further gives the references &lt;a href=&quot;http://archive.numdam.org/ARCHIVE/AIHPA/AIHPA_1998__69_3/AIHPA_1998__69_3_275_0/AIHPA_1998__69_3_275_0.pdf&quot; rel=&quot;nofollow&quot;&gt;Relativistic and nonrelativistic elastodynamics with small shear strains&lt;/a&gt; by Tahvildar-Zahdeh for a mathematical introduction and Beig &amp;amp; Schmidt's &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0211054&quot; rel=&quot;nofollow&quot;&gt;Relativistic Elasticity&lt;/a&gt; which is a very clear work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a spacetime $(M,g)$ and a timelike killing vector field $\xi$, $g(\xi, \xi) = 0$ and a three-dimensional Riemannian &lt;strong&gt;material manifold&lt;/strong&gt; $(N,h)$, the &lt;strong&gt;material map&lt;/strong&gt; $\phi{:} M\rightarrow N$ for which $d\phi{(\xi)} = 0$ is an equivalence class mapping trajectories of point particles on $M$ onto one point in $N$. One can further define a metric $\tilde{g}$ as the pull-back of $h$, $\tilde{g}_{(x)}(X,Y) = h_{(\phi{(x)})}(d\phi{(X)}, d\phi{(Y)})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my &lt;strong&gt;question&lt;/strong&gt; could be &lt;strong&gt;recast&lt;/strong&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can $\tilde{g}$ be used to explain transformation optics?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edith adds some &lt;strong&gt;further thoughts&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Special case: Geometrical optics&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Geometrical_optics&quot; rel=&quot;nofollow&quot;&gt;Geometrical optics&lt;/a&gt; is widely used when wave-phenomena like &lt;a href=&quot;http://en.wikipedia.org/wiki/Interference_%28wave_propagation%29&quot; rel=&quot;nofollow&quot;&gt;interference&lt;/a&gt; do not play a major role. In this case we approach the question from relativity. I will scetch my thoughts in the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Minkowsky spacetime&lt;/strong&gt; is described by the metric $\eta = \eta_{\mu\nu}dx^\mu dx^\nu$. Fixing coordinates, we can always bring it into the form&#xA;$$\eta = -c^2 dt^2 + d\mathbf{r}^2$$&#xA;where I explicitly kept the &lt;strong&gt;speed of light&lt;/strong&gt; $c$ and differ between space and time.&lt;br&gt;&#xA;&lt;strong&gt;Light rays&lt;/strong&gt; can be described by &lt;strong&gt;null geodesics&lt;/strong&gt; $v$ for which $\eta(v,v) = 0$, hence&#xA;$$\left( v^t \right)^2\eta_{tt} = \frac1{c^2} v^i v^j \eta_{ij}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interpreting the &lt;strong&gt;speed of light&lt;/strong&gt; not as a constant but as a &lt;strong&gt;function&lt;/strong&gt;, $\frac1{c^2} = \epsilon_0\mu_0 \epsilon_r(x^\mu) \mu_r(x^\mu)$ (isotropic quantities assumed here, not tensorial ones - generaliziation is obvious but would blow up formulas here), we can directly interpret&#xA;$$\tilde{g} = \epsilon_0\mu_0 \epsilon_r(x^\mu) \mu_r(x^\mu)\eta_{ij}dx^i dx^j$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, as before, $i$ and $j$ correspond to space-coordinates only. Since $\epsilon_r$ and $\mu_r$ are functions, $\tilde{g}$ will correspond to a curved three dimensional space. Furthermore, the &lt;strong&gt;material manifold&lt;/strong&gt; can be identified as&#xA;$$(d\phi)_i (d\phi)_j = \epsilon_r \mu_r \eta_{ij}$$&#xA;and has (under the given assumption of isotropy) the form of a &lt;a href=&quot;http://mathworld.wolfram.com/ConformalTransformation.html&quot; rel=&quot;nofollow&quot;&gt;conformal transformation&lt;/a&gt;, also discussed in &lt;a href=&quot;http://physics.stackexchange.com/q/4847/358&quot;&gt;another question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure about &lt;a href=&quot;http://en.wikipedia.org/wiki/Refraction&quot; rel=&quot;nofollow&quot;&gt;refraction&lt;/a&gt; at jumps of $\epsilon_r$ and $\mu_r$ at this point - the formulation given makes only sense for &lt;strong&gt;differentiable&lt;/strong&gt; quantities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what about &lt;strong&gt;full electrodynamics&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2011-03-02T09:25:40.430" LastActivityDate="2011-03-02T09:25:40.430" Title="Formulation of Transformation optics using a Material Manifold" Tags="&lt;optics&gt;&lt;electrodynamics&gt;&lt;differential-geometry&gt;" CommentCount="4" FavoriteCount="4" />
  <row Id="2016" PostTypeId="2" ParentId="1968" CreationDate="2010-12-17T11:04:26.007" Score="3" ViewCount="" Body="&lt;p&gt;It was really difficult to find any mentioning of the &quot;lunar twilight&quot; in astronomical literature. But I managed to find one in an old book by G.V.Rosenberg called &quot;General picture of twilight phenomena&quot;. The book is in russian, but there is not too much about the moon. So I'll just translate all the relevant stuff. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all there is this plot for the &lt;strong&gt;solar&lt;/strong&gt; twilight (I've translated it):&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/6o024.gif&quot; alt=&quot;Translated from russian&quot;&gt;&lt;br&gt;&#xA;Where E is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Illuminance&quot; rel=&quot;nofollow&quot;&gt;illuminance&lt;/a&gt; of a horizontal surface in &lt;a href=&quot;http://en.wikipedia.org/wiki/Lux&quot; rel=&quot;nofollow&quot;&gt;luxes&lt;/a&gt;. And lg -- is a logarithm base 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then there is a small chapter:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Lunar twilight&lt;/strong&gt;&lt;br&gt;&#xA;  Lunar twilight is similar to solar twilight, but it is much more bleak. The illumination from the moon varies from $10^{-9}$ (new moon) to $2\cdot 10^{-8}$ (full moon) from the illumination from the sun at the same point in the sky. Therefore the illumination from the full moon high in the sky corresponds approximately to the middle of nautical twilight. And lunar twilight ends virtually when the moon gets under the horizon. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="386" LastActivityDate="2010-12-17T11:04:26.007" CommentCount="2" />
  <row Id="2017" PostTypeId="1" AcceptedAnswerId="2019" CreationDate="2010-12-17T12:16:00.943" Score="2" ViewCount="180" Body="&lt;p&gt;Every time we have to stop the car, it is costing us extra money, because we have to then accelerate to full speed again. I would like to know how much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to simplify the situation, we can make these assumptions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;we disregard cost of car maintenance (brake pads wearing out etc.). This is focused entirely on fuel economy.&lt;/li&gt;&#xA;&lt;li&gt;A car is always traveling at 60km/h when it needs to stop&lt;/li&gt;&#xA;&lt;li&gt;The car's fuel economy is 8L/100kms when traveling at 60km&lt;/li&gt;&#xA;&lt;li&gt;The cost of fuel is $1.25 per litre.&lt;/li&gt;&#xA;&lt;li&gt;The car weighs 1000kg&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I will be happy with either a definitive answer, an explanation of how to arrive at the answer, or an explanation of why there is not enough information in the question to come up with the answer.&lt;/p&gt;&#xA;" OwnerUserId="406" LastEditorUserId="406" LastEditorDisplayName="" LastEditDate="2010-12-17T12:39:31.773" LastActivityDate="2010-12-17T13:07:56.460" Title="How much does it cost every time your car stops for a traffic light?" Tags="&lt;energy&gt;&lt;velocity&gt;&lt;motion&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="2018" PostTypeId="2" ParentId="2017" CreationDate="2010-12-17T12:33:12.823" Score="0" ViewCount="" Body="&lt;p&gt;I believe more information is needed to formulate an answer. Either you'd need the potential energy per liter of the gas and the necessary information to determine how much energy is lost in breaking (in which case, assuming no other loss of energy for simplicity's sake, you can work it out based on pure energy considerations) or you need more detail regarding the acceleration and deceleration phases of motion (in which case you can use forces and work).&lt;/p&gt;&#xA;" OwnerUserId="861" LastActivityDate="2010-12-17T12:33:12.823" CommentCount="2" />
  <row Id="2019" PostTypeId="2" ParentId="2017" CreationDate="2010-12-17T13:07:56.460" Score="5" ViewCount="" Body="&lt;p&gt;Here is a very basic estimation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinetic energy of a 1000 kg car moving at 60 km/h is $$E=\frac{mv^{2}}{2}=\frac{1000kg(16.7m/s)^{2}}{2}=138.9 kJ$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat of gasoline combustion is 47 MJ/kg = 35000 kJ/litre. Assuming 10% efficiency of the car's engine, you would need to burn $$\frac{138.9 kJ}{0.1\cdot35000kJ/l}=0.04 litre$$ of gasoline to accelerate the car to a given speed which would cost you 5 cents.&lt;/p&gt;&#xA;" OwnerUserId="701" LastActivityDate="2010-12-17T13:07:56.460" CommentCount="1" />
  <row Id="2020" PostTypeId="2" ParentId="2000" CreationDate="2010-12-17T13:45:49.527" Score="1" ViewCount="" Body="&lt;p&gt;Maybe you don't need a multipass cell after all. Check out &lt;a href=&quot;http://www.nature.com/nphys/journal/v6/n3/abs/nphys1506.html&quot; rel=&quot;nofollow&quot;&gt;this recent paper&lt;/a&gt; in Nature Physics which reports experimental results on very high (above 90%) reflection off diamond even at the normal incidence(!) for hard X-rays (energies up to 20-30 keV).&lt;/p&gt;&#xA;" OwnerUserId="59" LastActivityDate="2010-12-17T13:45:49.527" />
  <row Id="2021" PostTypeId="1" CreationDate="2010-12-17T14:35:31.377" Score="-2" ViewCount="194" Body="&lt;p&gt;I would like to know in which Derived International System Unit a Brix value could be converted to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;UPDATE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is Brix the SI units to measure the sugar content of an aqueous solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that the good SI derived measurement units would be g/l. Is it true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="874" LastEditorUserId="874" LastEditorDisplayName="" LastEditDate="2010-12-28T15:43:27.613" LastActivityDate="2010-12-28T15:43:27.613" ClosedDate="2010-12-17T18:34:03.413" Title="What is the SI Derived Unit of Brix?" Tags="&lt;si-units&gt;&lt;data-conversion&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="2022" PostTypeId="2" ParentId="2021" CreationDate="2010-12-17T14:55:01.810" Score="0" ViewCount="" Body="&lt;p&gt;Its a dimensionless quantity. Its a ratio of grams to grams.&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2010-12-17T14:55:01.810" CommentCount="1" />
  <row Id="2023" PostTypeId="1" CreationDate="2010-12-17T17:46:43.667" Score="3" ViewCount="269" Body="&lt;p&gt;I need &quot;educated&quot; reasons whether it is safe to use any wireless device during a lightning storm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most people said don't use it but they cannot explain why.&lt;/p&gt;&#xA;" OwnerUserId="652" LastActivityDate="2010-12-19T02:34:25.957" Title="Is it safe to use any wireless device during a lightning storm?" Tags="&lt;electromagnetism&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="2024" PostTypeId="2" ParentId="2023" CreationDate="2010-12-17T18:32:41.590" Score="8" ViewCount="" Body="&lt;p&gt;Most probably yes; wireless devices are not grounded, so they are not lighting rods of any kind as it is frequently assumed. There are some theories that cell phones somehow attracts lightnings by the field they produce, but the theory behind is weak. Experimental evaluation is very hard, since lightning hits are quite rare, such events are guided by numerous parameters and there are lots of other, hard to evaluate factors.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-12-17T18:32:41.590" CommentCount="1" />
  <row Id="2025" PostTypeId="2" ParentId="1126" CreationDate="2010-12-17T18:35:45.690" Score="0" ViewCount="" Body="&lt;p&gt;I suppose you could reproduce the exact trajectory (i.e. same angle and distance) by changing gravity, or more physically realistic, using an electric charge on the ball and uniform background E field.&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2010-12-17T18:35:45.690" />
  <row Id="2026" PostTypeId="1" CreationDate="2010-12-17T21:16:45.627" Score="12" ViewCount="482" Body="&lt;p&gt;I'm going to be teaching a course on gauge/gravity duality (aka AdS/CFT) in the winter.&#xA;The focus will be on applications in particle theory including $N=4$ SYM, the viscosity/entropy bound, and aspects of large $N$ QCD. I would like to include at least one application to condensed matter physics. There has been a lot of work on this over the last couple of years but I haven't followed it very closely. Can anybody suggest a nice application that is simple enough to work through in a few lectures for students who know QFT and GR and will have had some exposure to AdS/CFT?&lt;/p&gt;&#xA;" OwnerUserId="838" LastActivityDate="2011-01-14T13:32:36.013" Title="Condensed matter application of AdS/CFT" Tags="&lt;condensed-matter&gt;&lt;ads-cft&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="5" />
  <row Id="2027" PostTypeId="1" AcceptedAnswerId="2044" CreationDate="2010-12-17T21:28:25.347" Score="3" ViewCount="228" Body="&lt;p&gt;Consider a spherical rigid stone rotating with angular velocity $\omega$ being dropped vertically onto a horizontal rigid surface with the coefficient of friction $\mu$. Can the stone roll on the surface? If it can, what type of the motion is it? Constant velocity, constant accelaration, or varying acceleration? Ignore air resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality if we drop a rotating ball, it will roll on the surface. But I am not sure whether it is because of the deformation of the ball at the surface of contact.&lt;/p&gt;&#xA;" OwnerUserId="749" LastEditorUserId="749" LastEditorDisplayName="" LastEditDate="2010-12-17T21:43:53.197" LastActivityDate="2010-12-20T12:19:30.910" Title="Rolling stone on a frictional surface" Tags="&lt;classical-mechanics&gt;&lt;mechanics&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="2028" PostTypeId="2" ParentId="2026" CreationDate="2010-12-17T23:49:42.247" Score="2" ViewCount="" Body="&lt;p&gt;I would suggest looking at Sachdev's &quot;&lt;a href=&quot;http://arxiv.org/abs/arXiv:1002.2947&quot; rel=&quot;nofollow&quot;&gt;Condensed Matter and AdS/CFT&lt;/a&gt;&quot; and McGreevy's &quot;&lt;a href=&quot;http://arxiv.org/abs/0909.0518&quot; rel=&quot;nofollow&quot;&gt;Holographic duality with a view toward many-body physics&lt;/a&gt;&quot; for an overall perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to condensed matter, AdS/CFT is generally used to calculate transport coefficients for systems near their critical point such as in the &lt;strong&gt;antiferromagnetic (Neel state) - dimer gas crossover&lt;/strong&gt; or for high-Tc superconductivity to understand the transition between psuedogap and superconducting phases. Again for a more precise statement of the various cases I would look at these and other references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their is also a &lt;a href=&quot;http://motls.blogspot.com/2007/02/adscft-and-condensed-matter-physics.html&quot; rel=&quot;nofollow&quot;&gt;nice post&lt;/a&gt; on Lubos Motl's blog regarding Sachdev's work.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                             Cheers,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" LastActivityDate="2010-12-17T23:49:42.247" />
  <row Id="2029" PostTypeId="1" AcceptedAnswerId="6540" CreationDate="2010-12-18T02:18:29.193" Score="6" ViewCount="564" Body="&lt;p&gt;I know that it's possible to make an object levitate using an electromagnet to hold it up. &#xA;But is it also possible to do this with regular magnets? Is there a special kind of magnet I need in order to have one powerful enough to hold an object up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm asking because I have this idea in mind where I want to make a decorative item that is levitating an inch or so above its container.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/30uUp.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A&lt;/strong&gt; is the object I want to levitate.&lt;br&gt;&#xA;&lt;strong&gt;B&lt;/strong&gt; is the container.&lt;br&gt;&#xA;&lt;strong&gt;C&lt;/strong&gt; shows the magnetic field that is repelling &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; from each other achieving the levitation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of this would be as small as a food plate, maybe even smaller. &lt;strong&gt;B&lt;/strong&gt; is barely a pound or two in weight.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that possible?  &lt;/p&gt;&#xA;" OwnerUserId="877" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-19T05:47:01.883" LastActivityDate="2011-03-15T18:32:11.187" Title="Can I levitate an object without using an electromagnet?" Tags="&lt;electromagnetism&gt;&lt;magnets&gt;" AnswerCount="3" CommentCount="10" FavoriteCount="1" />
  <row Id="2030" PostTypeId="2" ParentId="2027" CreationDate="2010-12-18T02:30:33.577" Score="0" ViewCount="" Body="&lt;p&gt;It can and will roll (and/or slide) if its inital axis of rotation does not lie in the vertical.&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2010-12-18T04:55:09.543" LastActivityDate="2010-12-18T04:55:09.543" />
  <row Id="2031" PostTypeId="2" ParentId="2029" CreationDate="2010-12-18T03:12:18.203" Score="-1" ViewCount="" Body="&lt;p&gt;yes, you can.  Not easily though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My favorite is the Meisner effect.  It would require a superconductor for B, which would most likely require liquid nitrogen.  Basically, the levitating magnet (A) would create an current in the superconductor (which has &quot;zero&quot; resistance).  The current would then go for as long as the superconductor superconducts, creating a magnetic field, which would repulse A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more explanation, check out wikipedia.  Basically, a  I have no idea how much this would cost.  Although, the liquid nitrogen needed would most likely make it impractical.  By the way, there are no room temp superconductors currently known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;more methods are shown here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_levitation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Magnetic_levitation&lt;/a&gt;.  Some methods shown there do in fact use electricity, but some don't.  Check out the spin-stabilized levitation.  Very cool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last would be to build a bowl of magnets.  If you arrange it right, you should be able to get it stable.&lt;/p&gt;&#xA;" OwnerUserId="879" LastActivityDate="2010-12-18T03:12:18.203" CommentCount="4" />
  <row Id="2032" PostTypeId="1" AcceptedAnswerId="2042" CreationDate="2010-12-18T04:36:27.587" Score="6" ViewCount="375" Body="&lt;p&gt;If the earth's gravity exerts a net downward gravitational force on all air molecules, how come the molecules don't eventually lose their momentum and all settle down? How is the atmosphere is still miles thick after billions of years?&lt;/p&gt;&#xA;" OwnerUserId="202" LastActivityDate="2010-12-19T08:25:47.923" Title="How come gas molecules don't settle down?" Tags="&lt;gravity&gt;&lt;atmosphere&gt;" AnswerCount="5" />
  <row Id="2033" PostTypeId="2" ParentId="2032" CreationDate="2010-12-18T04:39:56.207" Score="3" ViewCount="" Body="&lt;p&gt;This is because the sun is supplying them with kinetic energy all the time, which you know is temperature. Then the ideal gas law states p*V=nRT, so the volume keeps pretty much constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure is causes by the electrostatic repulsion between the molecules pushing them apart, and the gravity pulling them towarsd earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another important reason they dont 'settle down' is that temperature is the lowest form of energy, this means that what you actually ask for cannot happen in any closed system, the temperature energy can only go towards heating something else, the core of the atmosphere is in this sense not a perfect analogy, because some of its energy will be &quot;lost&quot; to heating the earth and the oceans, and most will be radiated into the universe.&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2010-12-18T08:22:51.593" LastActivityDate="2010-12-18T08:22:51.593" CommentCount="5" />
  <row Id="2034" PostTypeId="2" ParentId="2032" CreationDate="2010-12-18T04:43:11.313" Score="0" ViewCount="" Body="&lt;p&gt;The sun keeps the earth warm. Without it, the atmosphere would freeze solid.&lt;/p&gt;&#xA;" OwnerUserId="150" LastActivityDate="2010-12-18T04:43:11.313" CommentCount="6" />
  <row Id="2035" PostTypeId="2" ParentId="2029" CreationDate="2010-12-18T05:00:54.810" Score="3" ViewCount="" Body="&lt;p&gt;Earnshaw's theorem proves that using only static ferromagnetism (like refridgerator magnets, which have a very slowly changing magnetic field) it is impossible to stably levitate against gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Earnshaw%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Earnshaw%27s_theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diamagnetic materials might be used to levitate objects.&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2011-03-09T07:10:22.740" LastActivityDate="2011-03-09T07:10:22.740" CommentCount="6" />
  <row Id="2036" PostTypeId="1" AcceptedAnswerId="2037" CreationDate="2010-12-18T06:37:10.243" Score="5" ViewCount="251" Body="&lt;p&gt;Something I have read multiple times that I've never intuitively understood is that &quot;heavier&quot; particles are harder to detect than &quot;lighter&quot; ones...   For example, I quote from Stephen Hawking's &quot;The Grand Design&quot; in relation to supersymmetry:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But various calculations that physicists have performed indicate that the partner particles corresponding to the particles we observe ought to be a thousand times as massive as a proton, if not even heavier. That is too heavy for such particles to have been seen in any experiments to date, but there is hope that such particles will eventually be created in the Large Hadron Collider in Geneva.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could someone please explain, in simple terms, why heavy particles are harder to detect?  Intuitively (to me, a non-physicist) it would seem that it should be the other way around, because a particle with more mass should interact more strongly with other matter.&lt;/p&gt;&#xA;" OwnerUserId="880" LastActivityDate="2010-12-19T16:49:25.527" Title="Why are &quot;heavier&quot; particles harder to detect than &quot;lighter&quot; ones?" Tags="&lt;particle-physics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="2037" PostTypeId="2" ParentId="2036" CreationDate="2010-12-18T06:49:33.193" Score="10" ViewCount="" Body="&lt;p&gt;They are harder to detect for mainly two reasons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, because they decay very quickly into lighter particles. Infact the heaviest elementary particle we know of ( top quark ) decay's so rapidly it is theoretically and technically impossible to measure it in any other way than indirectly through its decay products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This poses another problem as to what statistical and theoretical basis one can with some certainty claim detection of such particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem is that of production, since they have more mass, they require much more energy to produce, thus larger and more expensive accelerators.&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2010-12-18T06:49:33.193" CommentCount="8" />
  <row Id="2038" PostTypeId="2" ParentId="1845" CreationDate="2010-12-18T07:20:22.310" Score="1" ViewCount="" Body="&lt;p&gt;You could construct a Faraday cage and see that if you put a mobile phone inside you don't get any signal.&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2010-12-18T07:20:22.310" CommunityOwnedDate="2010-12-18T07:20:22.310" />
  <row Id="2039" PostTypeId="2" ParentId="2032" CreationDate="2010-12-18T07:55:48.177" Score="5" ViewCount="" Body="&lt;p&gt;The key ingredient is temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were zero then all the air would indeed just fall down to the ground (actually, this is a simplification I'll address later). As you increase the temperature the atoms of the ground will start to wiggle more and they'll start to kick the air molecules giving them non-zero average height. So the atmosphere would move a little off the ground. The bigger the temperature is the higher the atmosphere will reach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt; there are number of assumptions above that simplify the picture. They are not that important but I want to provide a complete picture:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Even at the zero temperature the molecules would wiggle a little because of quantum mechanics&lt;/li&gt;&#xA;&lt;li&gt;the atmosphere would freeze at some point (like 50K) so under that temperature it would just lie on the ground&lt;/li&gt;&#xA;&lt;li&gt;I assumed that the ground and the atmosphere have the same temperature because they are in the thermal equilibrium; in reality their temperatures can differ a little because of additional slow heat-transfer processes&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To leave you with something a little bit more concrete (although quite simplified), statistical mechanics tells us that probability $p$ that some portion of the system will occupy a given energy $E$ level is $p \propto \exp({-E \over kT})$ where $k$ is a Boltzmann constant (note that you have to normalize this probability to 1 by summing over all energy levels; that's why it's only proportional, $\propto$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the atmosphere of ideal gas where molecules have only potential energy (kinetic energy is not important here because the molecules don't interact with each other in the ideal gas) we have $E = mgh$. This gives us $p \propto \exp({-mgh \over kT})$. You can see that atmosphere &quot;wants&quot; to get lower ($p$ is bigger for $h$ lower) but the temperature &quot;forces&quot; it to go up. Play with the equation for different values of $T$ and $g$ (gravity strength) to see what happens.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-18T07:55:48.177" CommentCount="8" />
  <row Id="2040" PostTypeId="2" ParentId="2032" CreationDate="2010-12-18T09:13:33.080" Score="5" ViewCount="" Body="&lt;p&gt;The other answers are correct but to understand them you have to get an idea of how much thermal energy does an average molecule have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Maxwell-Boltzmann distribution, the most probable speed of an air (say, nitrogen) molecule at room temperature is $v_p = \sqrt { \frac{2kT}{m} } = 422 m/s$. Without collisions with other molecules it can travel upwards $h=\frac{v_{0}^{2}}{2g}$ = 9 kilometers before the gravity stops it and pulls back to Earth.&#xA;Basically, potential energy of molecules in gravitational field is too small compared to their kinetic energy to keep them low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Still, gravity is the reason why we &lt;em&gt;have&lt;/em&gt; atmosphere after billions of years.&lt;/p&gt;&#xA;" OwnerUserId="701" LastEditorUserId="701" LastEditorDisplayName="" LastEditDate="2010-12-19T08:25:47.923" LastActivityDate="2010-12-19T08:25:47.923" CommentCount="4" />
  <row Id="2041" PostTypeId="1" AcceptedAnswerId="2049" CreationDate="2010-12-18T09:52:58.607" Score="8" ViewCount="475" Body="&lt;p&gt;How is the following classical optics phenomenon explained in quantum electrodynamics?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Reflection and Refraction&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Are they simply due to photons being absorbed and re-emitted? How do we get to Snell's law, for example, in that case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Split by request: See the other part of this question &lt;a href=&quot;http://physics.stackexchange.com/questions/2055/how-are-classical-optics-phenomena-explained-in-qed-color&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-12-19T10:43:59.650" LastActivityDate="2010-12-19T10:43:59.650" Title="How are classical optics phenomena explained in QED (Snell's law)?" Tags="&lt;optics&gt;&lt;quantum-electrodynamics&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="4" />
  <row Id="2042" PostTypeId="2" ParentId="2032" CreationDate="2010-12-18T10:07:40.253" Score="1" ViewCount="" Body="&lt;p&gt;First of all, gravity does continually accelerate the air molecules. I don't see how this could make them lose their momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the net effect of gravity on the atmosphere? Simple, gravity prevents the atmosphere from flying off in space, and instead it keeps it comfortably wrapped around our planet!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why the atmosphere is still thick after billions of years is because you have two net effects on the air molecules, gravity, which keeps it as close as possible to the ground, and inertia, who has the opposite net effect. So as long as the molecules do not slow down they &quot;orbit&quot; our planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is the same as why is the moon orbiting the Earth after billions of years. There's a balance between the kinetic energy of the moon and the gravitational potential energy - or a balance between gravity and inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other answers give you a summary explanation of what determines the temperature of the atmosphere and hence its molecules' average velocity. The reality is way more complicated as the temperature of the atmosphere is not constant with height and you have to take into account many more factors like varying pressure, convection and so on. Modelling the Earth's atmosphere accurately is very complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion the basic mechanisms are outlined above. I hope they answer your question.&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-12-18T10:15:48.003" LastActivityDate="2010-12-18T10:15:48.003" CommentCount="20" />
  <row Id="2043" PostTypeId="1" AcceptedAnswerId="2920" CreationDate="2010-12-18T10:58:37.830" Score="13" ViewCount="541" Body="&lt;p&gt;At the moment I am studying operator algebras from a mathematical point of view. Up to now I have read and heard of many remarks and side notes that von Neumann algebras ($W^*$ algebras) are important in quantum physics. However I didn't see where they actually occur and why they are important. So my question is, where they occur and what's exactly the point why they are important.&lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorUserId="667" LastEditorDisplayName="" LastEditDate="2010-12-18T11:12:27.220" LastActivityDate="2011-01-24T10:34:17.363" Title="Why are von Neumann Algebras important in quantum physics?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;mathematical-physics&gt;&lt;quantum-information&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="2044" PostTypeId="2" ParentId="2027" CreationDate="2010-12-18T13:30:34.193" Score="2" ViewCount="" Body="&lt;p&gt;I'll assume that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The stone has horizontal axis of rotation.&lt;/li&gt;&#xA;&lt;li&gt;The objects in question are perfectly rigid.&lt;/li&gt;&#xA;&lt;li&gt;The collision is perfectly inelastic.&lt;/li&gt;&#xA;&lt;li&gt;The static and dynamic coefficients of friction are equal.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In this case, you will get a vertical delta impulse on the stone, and up to $\mu$ of that impulse is converted to an instantaneous jump in horizontal speed.  If the angular momentum of the stone is not exhausted by that impulse, it will skid for a while, producing constant acceleration until it grips the surface.  If the angular momentum is exhausted, then the stone will just roll.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relaxing one or more of the assumptions may yield more interesting behavior.  For example, under certain conditions, rubber balls will switch the direction of rotation with each bounce.&lt;/p&gt;&#xA;" OwnerUserId="761" LastActivityDate="2010-12-18T13:30:34.193" />
  <row Id="2045" PostTypeId="1" AcceptedAnswerId="2050" CreationDate="2010-12-18T13:39:06.463" Score="7" ViewCount="464" Body="&lt;p&gt;Surely group theory is a very handy tool in the problems dealing with symmetry. But is there any application for ring theory in physics? If not, what's this that makes rings not applicable in physics problems?&lt;/p&gt;&#xA;" OwnerUserId="304" LastActivityDate="2010-12-18T17:46:58.103" Title="Ring theory in physics" Tags="&lt;mathematical-physics&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="2046" PostTypeId="2" ParentId="2045" CreationDate="2010-12-18T15:23:28.453" Score="7" ViewCount="" Body="&lt;p&gt;I don't know any important applications of rings in undergraduate level physics, but there are plenty of applications in the study of supersymmetric field theories. For example, take an $N=2$ supersymmetric $\sigma$-model with whose target space is a Kahler manifold $X$. One can define a set of twisted supercharges $Q$ (the exact form depends on the twisting) and there is a ring structure on the cohomology of $Q$. Classically this is the same as the cohomology ring of $X$, but quantum mechanically there are corrections from instantons. The chiral ring of the Q's is often called the quantum cohomology of $X$. The chiral ring structure also played an important role in the discovery of mirror symmetry. The lectures by E. Witten in &quot;Quantum fields and strings: a course for mathematicians, Volume 2&quot; are a good reference on this for the mathematically inclined. If there are some more down to earth examples where rings are important in physics I'd also be interested to hear about them.&lt;/p&gt;&#xA;" OwnerUserId="838" LastActivityDate="2010-12-18T15:23:28.453" />
  <row Id="2047" PostTypeId="2" ParentId="2041" CreationDate="2010-12-18T15:51:04.487" Score="2" ViewCount="" Body="&lt;p&gt;It really desires a long discussion. You may be interested in the book &quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691024170&quot; rel=&quot;nofollow&quot;&gt;QED: The Strange theory of light and matter&lt;/a&gt;&quot; written by Richard Feynman (or the corresponding &lt;a href=&quot;http://vega.org.uk/video/subseries/8&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt;), which gives a comprehensive introduction with almost no number and formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the solution of Schrödinger equation of hydrogen atom. The energy level is discrete, so its absorption spectrum is also discrete. In this case, only few colors can be seen. However, in solid, the atom are interacting strongly with each other and the resulting absorption spectrum can be very complicated. This interaction depends strongly with its structure and the outside electrons. The temperature can play an essential role for the structural change and a phase transition can occur and so does color change. I think there is no easy explanation for the exact absorption spectrum, or color, for a material without doing complicated calculation.&lt;/p&gt;&#xA;" OwnerUserId="562" LastActivityDate="2010-12-18T15:51:04.487" />
  <row Id="2048" PostTypeId="2" ParentId="2045" CreationDate="2010-12-18T16:21:45.053" Score="2" ViewCount="" Body="&lt;p&gt;In geometry, spaces are often characterized by their ring of continuous (or smooth) R or C valued functions.  This is the basic philosophy of algebraic geometry, noncommutative geometry, and deformation theory, which all have applications to physics (the latter two being almost exclusively motivated by quantum mechanics).  For example any compact Hausdorff space can be constructed just from its ring of continuous functions so one could say that a space is a ring.  This ring is always commutative since we multiply pointwise and R and C are commutative.  The viewpoint of noncommutative geometry is to study noncommutative rings as if they were the ring of continuous functions of some &quot;noncommutative&quot; space.  This is motivated by quantum mechanics where the algebra of observables is noncommutative.&lt;/p&gt;&#xA;" OwnerUserId="37" LastActivityDate="2010-12-18T16:21:45.053" />
  <row Id="2049" PostTypeId="2" ParentId="2041" CreationDate="2010-12-18T17:33:23.567" Score="5" ViewCount="" Body="&lt;p&gt;Hwlau is correct about the book but the answer actually isn't that long so I think I can try to mention some basic points.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Path integral&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;One approach to quantum theory called path integral tells you that you have to sum &lt;a href=&quot;http://en.wikipedia.org/wiki/Probability_amplitude&quot; rel=&quot;nofollow&quot;&gt;probability amplitudes&lt;/a&gt; (I'll assume that you have at least some idea of what probability amplitude is; QED can't really be explained without this minimal level of knowledge)) over all possible paths that the particle can take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for photons probability amplitude of a given path is $\exp(i K L)$ where $K$ is some constant and $L$ is a length of the path (note that this is very simplified picture but I don't want to get too technical so this is fine for now). The basic point is that you can imagine that amplitude as a unit vector in the complex plane. So when doing a path integral you are adding lots of short arrows (this terminology is of course due to Feynman). In general for any given trajectory I can find many shorter and longer paths so this will give us a nonconstructive interference (you will be adding lots of arrows that point in random directions). But there can exist some special paths which are either longest or shortest (in other words, extremal) and these will give you constructive interference. This is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat%27s_principle&quot; rel=&quot;nofollow&quot;&gt;Fermat's principle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Fermat's principle&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;So much for the preparation and now to answer your question. We will proceed in two steps. First we will give classical answer using Fermat's principle and then we will need to address other issues that will arise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's illustrate this first on a problem of light traveling between points $A$ and $B$ in free space. You can find lots of paths between them but if it won't be the shortest one it won't actually contribute to the path integral for the reasons given above. The only one that will is the shortest one so this recovers the fact that light travels in straight lines. The same answer can be recovered for reflection. For refraction you will have to take into account that the constant $K$ mentioned above depends on the index of refraction (at least classically; we will explain how it arises from microscopic principles later). But again you can arrive at Snell's law using just Fermat's principle.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;QED&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now to address actual microscopic questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, index of refraction arises because  &lt;a href=&quot;http://physics.stackexchange.com/questions/1898/do-photons-gain-mass-when-they-travel-through-glass&quot;&gt;light travels slower&lt;/a&gt; in materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what about reflection? Well, we are actually getting to the roots of the QED so it's about time we introduced interactions. Amazingly, there is actually only one interaction: electron absorbs photon. This interaction again gets a probability amplitude and you have to take this into account when computing the path integral. So let's see what we can say about a photon that goes from $A$ then hits a mirror and then goes to $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We already know that the photon travels in straight lines both between $A$ and the mirror and between mirror and $B$. What can happen in between? Well, the complete picture is of course complicated: photon can get absorbed by an electron then it will be re-emitted (note that even if we are talking about &lt;strong&gt;the&lt;/strong&gt; photon here, the emitted photon is actually distinct from the original one; but it doesn't matter that much) then it can travel for some time inside the material get absorbed by another electron, re-emitted again and finally fly back to $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make the picture simpler we will just consider the case that the material is a 100% real mirror (if it were e.g. glass you would actually get multiple reflections from all of the layers inside the material, most of which would destructively interfere and you'd be left with reflections from front and back surface of the glass; obviously, I would have to make this already long answer twice longer :-)). For mirrors there is only one major contribution and that is that the photon gets scattered (absorbed and re-emitted) directly on the surface layer of electrons of the mirror and then flies back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Quiz question:&lt;/strong&gt; and what about process that the photon flies to the mirror and then changes its mind and flies back to $B$ without interacting with any electrons; this is surely a possible trajectory we have to take into account. Is this an important contribution to the path integral or not?&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-18T17:33:23.567" CommentCount="4" />
  <row Id="2050" PostTypeId="2" ParentId="2045" CreationDate="2010-12-18T17:46:58.103" Score="8" ViewCount="" Body="&lt;p&gt;This really comes down to the question of how broadly you define ring theory.  Special types of rings appear all over the place in physics, but often their focused study is given a more specialized name.  The term &quot;ring theory&quot; is sometimes used to indicate the specific study of rings as a general class, and under that interpretation, the discipline seems to be closer to logic and set theory than questions of current physical relevance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, rings show up in the following contexts (this list is not comprehensive):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The representation theory of a finite or compact group $G$ can be studied from the optic of ring theory, since representations are modules over a (suitably topologized in the infinite case) group ring $\mathbb{C}[G]$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Algebras of operators, such as C* algebras and von Neumann algebras, are rings.  The ring of functions on a manifold is a commutative C* algebra, and geometric quantization is done by deforming the product structure on the ring of functions on a symplectic manifold (e.g., taking the ring of functions on the cotangent bundle of a configuration space to the ring of differential operators on the space).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Algebraic varieties such as Calabi-Yau varieties and Riemann surfaces show up in some string theory and conformal field theory papers.  They are patched together using commutative rings of functions on open sets.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cohomology and K-theory of a topological space form (graded-)commutative rings.  I am told that string theorists sometimes view these rings as places where certain charges live.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you like vertex algebras, their representation theory is captured by the module theory of a current algebra, which is a big ring.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm told that perturbative renormalization can be placed into Wightman's axiomatic framework in a mathematically rigorous fashion, if the framework is suitably generalized by replacing $\mathbb{C}$ with a formal power series ring in one ore more variables, such as $\mathbb{C}[[\lambda]]$, where the variables are the coupling constants, assumed to be infinitesimal.  Objects like the Hilbert space of states are replaced with modules over this ring equipped with a sesquilinear form.  Borcherds has a &lt;a href=&quot;http://arxiv.org/abs/1008.0129&quot; rel=&quot;nofollow&quot;&gt;recent preprint&lt;/a&gt; establishing this.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="761" LastActivityDate="2010-12-18T17:46:58.103" />
  <row Id="2051" PostTypeId="1" AcceptedAnswerId="2053" CreationDate="2010-12-18T22:17:35.197" Score="16" ViewCount="542" Body="&lt;p&gt;In the &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_Model&quot; rel=&quot;nofollow&quot;&gt;standard model&lt;/a&gt; of particle physics, there are three generations of quarks (up/down, strange/charm, and top/bottom), along with three generations of leptons (electron, muon, and tau).  All of these particles have been observed experimentally, and we don't seem to have seen anything new along these lines.  A priori, this doesn't eliminate the possibility of a fourth generation, but the physicists I've spoken to do not think additional generations are likely.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What sort of theoretical or experimental reasons do we have for this limitation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One reason I heard from my officemate is that we haven't seen new neutrinos.  Neutrinos seem to be light enough that if another generation's neutrino is too heavy to be detected, then the corresponding quarks would be massive enough that new physics might interfere with their existence.  This suggests the question: is there a general rule relating neutrino masses to quark masses, or would an exceptionally heavy neutrino just look bizarre but otherwise be okay with our current state of knowledge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason I've heard involves the Yukawa coupling between quarks and the Higgs field.  Apparently, if quark masses get much beyond the top quark mass, the coupling gets strong enough that QCD fails to accurately describe the resulting theory.  My wild guess is that this really means perturbative expansions in Feynman diagrams don't even pretend to converge, but that it may not necessarily eliminate alternative techniques like lattice QCD (about which I know nothing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional reasons would be greatly appreciated, and any words or references (the more mathy the better) that would help to illuminate the previous paragraphs would be nice.&lt;/p&gt;&#xA;" OwnerUserId="761" LastActivityDate="2010-12-18T23:10:37.913" Title="Why do we think there are only three generations of fundamental particles?" Tags="&lt;particle-physics&gt;&lt;standard-model&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="2052" PostTypeId="2" ParentId="2051" CreationDate="2010-12-18T23:03:00.413" Score="2" ViewCount="" Body="&lt;p&gt;It is a good question, and I got someone to explain it to me in grad school and have since forgotten, so I ran off to the PDG website to read.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find in the chapter on neutrino oscillations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The data on the invisible decay width of the $Z_0$-boson is compatible with only 3 light ﬂavour neutrinos coupled to $Z_0$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and references D. Karlen in RPP2010 which addresses the question in the lepton sector.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2010-12-18T23:03:00.413" CommentCount="1" />
  <row Id="2053" PostTypeId="2" ParentId="2051" CreationDate="2010-12-18T23:10:37.913" Score="11" ViewCount="" Body="&lt;p&gt;There are very good experimental limits on light neutrinos that have the same electroweak couplings as the neutrinos in the first 3 generations from the measured width of the $Z$ boson. Here light means $m_\nu &amp;lt; m_Z/2$. Note this does not involve direct detection of neutrinos, it is an indirect measurement based on the calculation of the $Z$ width given the number of light neutrinos. Here's the PDG citation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pdg.lbl.gov/2010/listings/rpp2010-list-number-neutrino-types.pdf&quot; rel=&quot;nofollow&quot;&gt;http://pdg.lbl.gov/2010/listings/rpp2010-list-number-neutrino-types.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a cosmological bound on the number of neutrino generations coming from production of Helium during big-bang nucleosynthesis. This is discussed in &quot;The Early Universe&quot; by Kolb and Turner although I am sure there are now more up to date reviews. This bound is around 3 or 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no direct relationship between quark and neutrino masses, although you can derive&#xA;possible relations by embedding the Standard Model in various GUTS such as those based on&#xA;$SO(10)$ or $E_6$. The most straightforward explanation in such models of why neutrinos are light is called the see-saw mechanism &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Seesaw_mechanism&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Seesaw_mechanism&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and leads to neutrinos masses $m_\nu \sim m_q^2/M$ where $M$ is some large mass scale on the order of $10^{11} ~GeV$ associated with the vacuum expectation value of some Higgs field that plays a role in breaking the GUT symmetry down to $SU(3) \times SU(2) \times U(1)$. If the same mechanism is at play for additional generations one would expect the neutrinos to be lighter than $M_Z$ even if the quarks are quite heavy. &#xA;Also, as you mentioned, if you try to make fourth or higher generations very heavy you have to increase the Yukawa coupling to the point that you are outside the range of perturbation  theory. These are rough theoretical explanations and the full story is much more complicated  but the combination of the excellent experimental limits, cosmological bounds and theoretical expectations makes most people skeptical of further generations. Sorry this wasn't mathier.&lt;/p&gt;&#xA;" OwnerUserId="838" LastActivityDate="2010-12-18T23:10:37.913" CommentCount="1" />
  <row Id="2054" PostTypeId="2" ParentId="2023" CreationDate="2010-12-19T02:34:25.957" Score="3" ViewCount="" Body="&lt;p&gt;A modern wireless electrical device should be perfectly safe, all it is is putting a metal piece near your ear, basically giving you the same chance of being hit by lightning as any other time. Older (Or more accurately, larger) technologies with larger antennas might work as a lightning rod, and thus give you a somewhat higher chance of being hit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically lightning is a low frequency electric signal, very broad band, while most cell phones are exactly the opposite. &lt;/p&gt;&#xA;" OwnerUserId="871" LastActivityDate="2010-12-19T02:34:25.957" />
  <row Id="2055" PostTypeId="1" AcceptedAnswerId="2058" CreationDate="2010-12-19T10:42:47.030" Score="4" ViewCount="164" Body="&lt;p&gt;How is the following classical optics phenomenon explained in quantum electrodynamics?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Color&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;According to Schroedinger's model of the atom, only particular colors are emitted depending on the type of atom and the state of its electrons. Then, how it is possible that the same type of matter changes color in relation to temperature? For example, water is transparent with a light blue tint, but snow is white (which means, pretty much, that photons have no particular color in the visible spectrum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Split by request: See the other part of this question &lt;a href=&quot;http://physics.stackexchange.com/questions/2041/how-are-classical-optics-phenomena-explained-in-qed-snells-law&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-12-19T12:41:51.800" Title="How are classical optics phenomena explained in QED (color)?" Tags="&lt;optics&gt;&lt;quantum-electrodynamics&gt;" AnswerCount="1" />
  <row Id="2056" PostTypeId="1" CreationDate="2010-12-19T12:26:14.240" Score="0" ViewCount="72" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/2023/is-it-safe-to-use-any-wireless-device-during-a-lightning-storm&quot;&gt;Is it safe to use any wireless device during a lightning storm?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;We often hear the tale that a person gets stuck by lighting when he is unfortunate enough to use a cell phone outdoor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why is this so?&lt;/p&gt;&#xA;" OwnerUserId="605" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-19T14:02:32.543" LastActivityDate="2010-12-19T14:02:32.543" ClosedDate="2010-12-19T13:52:42.203" Title="Why is it that using cell phone can invite lighting strike?" Tags="&lt;lightning&gt;" AnswerCount="0" CommentCount="5" />
  <row Id="2058" PostTypeId="2" ParentId="2055" CreationDate="2010-12-19T12:41:51.800" Score="3" ViewCount="" Body="&lt;p&gt;It's important to distinguish different phases of the material. While it's true that water and snow consists of the same building blocks of $H_2 O$, those individual blocks are actually not as important as the resulting material. You can build a storage room and pyramids just out of bricks (at least in principle).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Gases&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let's start with the simplest case. Simplest in the sense that there is not much going on, just molecules flying around. Let us just assume ideal gas where molecules don't interact with each other. Then to explain everything it suffices to look at just one molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The color of the molecule of course arises because of its absorption spectrum. This in turn depends on the molecule's energy levels. For atoms these are pretty nice discrete levels. For more complicated molecules you also have rotational and vibrational degrees of freedom to take into account and besides discrete levels you'll also see continuous strips that consist of very fine energy levels corresponding to that. For illustration, see the wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Spin_isomers_of_hydrogen&quot; rel=&quot;nofollow&quot;&gt;$H_2$ hydrogen&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, if you are somehow able to obtain that energy spectrum you can then investigate the macroscopic properties by the means of the usual Boltzmann statistics &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat W = Z^{-1} \exp(-{\beta \hat H}) = Z^{-1} \sum_n\exp(- \beta E_n) {\hat P}_n $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $Z = {\rm Tr} \exp(-{\beta \hat H})$ being the partition function, $\beta$ the inverse temperature and $P_n$ projector on $n$-th energy level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this you can see that as you increase temperature the energy level distribution will change to occupy higher levels and this in turn will change the probability of individual absorption processes between certain levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the only place where you need to talk about QED is the connection between absorption spectrum and energy levels. Recall that in quantum mechanics energy levels are &lt;strong&gt;stable&lt;/strong&gt;. They don't change, so there would be neither emission nor absorption. To resolve this apparent paradox we have to recall that we forgot to quantize the electromagnetic field. If you take this into account then atom's excited energy levels are no longer stable because of the fluctuations of electromagnetic field. Or in particle terms: because the number of particles is not conserved and it's easy to create photon out of nothing. Of course, energy is conserved so only photon's corresponding to the energy difference are possible. And the same can be said for absorption.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Liquids and solids&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For now I won't talk about these phases because the answer is both becoming long and because optical properties of solids would constitute a &lt;a href=&quot;http://books.google.com/books?id=K9YJ950kBDsC&quot; rel=&quot;nofollow&quot;&gt;book of its own&lt;/a&gt;. As for liquids, I have a feeling that the same analysis as for gases should hold except it's of course no longer sufficient to talk about individual molecules because of non-negligible interactions. But I guess starting from ideal gas's statistics and just switching temperature should give a reasonable first approximation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Remark:&lt;/em&gt; Looking back on the answer, I didn't really explain your question completely. But I guess it's because it's quite hard to encompass everything that is going on. Maybe another split and specialization (e.g. to color of solids) of the question would be in order? :-)&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-19T12:41:51.800" />
  <row Id="2059" PostTypeId="1" AcceptedAnswerId="2068" CreationDate="2010-12-19T13:06:41.430" Score="3" ViewCount="538" Body="&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_theory&quot; rel=&quot;nofollow&quot;&gt;Wikipedia entry on Kinetic Theory&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The temperature of an ideal monatomic gas is a measure of the average kinetic energy of its atoms. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now if I remove all the particles from the box shown below will the temperature be zero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/3E0Ak.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-12-19T13:22:20.720" LastActivityDate="2010-12-19T20:08:58.817" Title="Is temperature in vacuum zero?" Tags="&lt;statistical-mechanics&gt;&lt;temperature&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="2060" PostTypeId="2" ParentId="2059" CreationDate="2010-12-19T13:19:53.150" Score="4" ViewCount="" Body="&lt;p&gt;According to the above definiton the answer is not defined mathematically, the average kinetic energy is  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle 1/N \sum_{i=1}^N m_i*v_i^2/2$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we plug in N=0 we see that it blows up  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we had some other definiton or the gas was not quantized maybe we could take the limit as N-&gt;0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;This answers your question if you define a thermometer as something that measures temperature, and you have defined temperature as above.&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2010-12-19T13:25:26.730" LastActivityDate="2010-12-19T13:25:26.730" CommentCount="8" />
  <row Id="2061" PostTypeId="2" ParentId="2059" CreationDate="2010-12-19T13:25:24.293" Score="5" ViewCount="" Body="&lt;p&gt;The notion of temperature doesn't make any sense in complete vacuum (meaning the absence of all objects). It only makes sense as a description of how much some objects wiggle around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To discuss the thermometer problem, one first needs to know what it means to measure temperature. &quot;You just insert thermometer wait a little and you're done&quot;, I can hear people say. Well, not quite. What happens microscopically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the simplest case, if you want to measure temperature of a sample you need to attach thermometer to it. Molecules of these objects will interact and eventually will come into thermal equilibrium. Thermometer then has some calibration that tells you that so-and-so temperature corresponds to so-and-so much wiggling of its molecules. Well, it should be obvious that for this to work, concept of thermal equilibrium is essential. But you won't get thermal equilibrium if there are very few molecules of the sample. In particular zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that the contact of surfaces is not the only way to attain thermal equilibrium. Any heat transfer process will do and that means any interaction. So you can try to measure the temperature e.g. by electromagnetic radiation. Well, if you insert such a thermometer into completely empty box then depending on the box's properties the electromagnetic radiation will either leave completely and thermometer will show zero or the box would trap radiation and the box would no longer be empty (it would contain photons). In any case, what you are measuring now is not temperature of the vacuum but rather the EM transmission properties of the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize: problem of measurement is not a trivial one and it has actually lead physicists to great many discoveries. Noting that you can't simultaneously measure position and momentum gave rise to quantum mechanics. Noting that couplings of our elementary particle theories depend on the energy you input into measurement gave rise to renormalization and better understanding of quantum field theories as a whole. So it's always important to think about what are you actually measuring microscopically.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, let me talk about some related things for a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that box full of gas again. As you'd lower temperature of the walls the molecules would transfer their energy to the walls and become slower. Now, you can imagine that by doing this for a long time you'll eventually reach zero temperature and all the movement will stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, this is not possible because you'd need infinite time to reach that temperature. And even if you had that time, you have to take uncertainty principle (you can't know the position of an object absolutely precisely) into account. Actually, cooling is a big field of physics in itself and entails various extremely sophisticated techniques that are very close to 0K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that in reality there is no such thing as vacuum (again in the above sense) because of quantum fluctuations.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-19T13:44:37.480" LastActivityDate="2010-12-19T13:44:37.480" CommentCount="9" />
  <row Id="2062" PostTypeId="2" ParentId="2059" CreationDate="2010-12-19T13:41:09.260" Score="2" ViewCount="" Body="&lt;p&gt;Even if you remove the particles, there will be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_radiation&quot; rel=&quot;nofollow&quot;&gt;thermal radiation&lt;/a&gt; coming from the borders. This way a thermometer placed inside will eventually show the temperature of borders.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-19T14:11:26.797" LastActivityDate="2010-12-19T14:11:26.797" CommentCount="9" />
  <row Id="2063" PostTypeId="2" ParentId="2036" CreationDate="2010-12-19T16:49:25.527" Score="2" ViewCount="" Body="&lt;p&gt;Hawking's phrase &quot;seen in any experiments&quot; really means the following. First you have to produce the heavy particle. Ordinary matter, consisting of electrons, protons and neutrons (the latter two themselves bound states of quarks), does not contain these new heavy particles, so they must be produced in high-energy collisions. By Einstein's famous formula $E=mc^2$, the production of very heavy particles (say with mass in the $TeV/c^2$ range) requires energies of order a $TeV$. Hence the LHC. Then there is the separate problem of detecting the particles once you have produced them. Your expectation that particles with more mass interact more strongly with other particles is true for gravity, but it is not true for the strong, weak or electromagnetic interactions which are relevant for the detection of elementary particles (the gravitational interaction is much too weak for these purposes). The strength of electromagnetic interactions depends on the electric charge and there are analogous &quot;charges&quot; for the weak and strong interactions. Detecting the particles is extremely complicated and depends on their mass, their lifetime, whether they have strong, weak or electromagnetic interactions or not, and what kinds of Standard Model background effects might have the same experimental signal as the particle you are trying to detect. This link from the LHC website has some information about particle detectors:&#xA;&lt;a href=&quot;http://public.web.cern.ch/public/en/LHC/LHCExperiments-en.html&quot; rel=&quot;nofollow&quot;&gt;http://public.web.cern.ch/public/en/LHC/LHCExperiments-en.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="838" LastActivityDate="2010-12-19T16:49:25.527" CommentCount="1" />
  <row Id="2064" PostTypeId="1" AcceptedAnswerId="6137" CreationDate="2010-12-19T16:49:59.847" Score="4" ViewCount="288" Body="&lt;p&gt;Anyone have good mnemonics for remembering standard packets of data in physics?&#xA;Any field within physics would be welcomed.  Examples of such &quot;packets&quot;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;data in the standard model of particle physics&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;charges/masses of common subatomic particles and light atoms&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;cosmological data -- numbers, masses, luminosities&#xA;(I recall &quot;Oh Be A Fine Girl/Guy Kiss Me Right Now Smack&quot; for star types)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;thermodynamics&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;material properties&#xA;(I note also &lt;a href=&quot;http://physics.stackexchange.com/questions/75/mnemonics-to-remember-various-properties-of-materials&quot;&gt;Mnemonics to remember various properties of materials&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="20" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-19T18:11:44.653" LastActivityDate="2011-03-01T12:29:47.960" CommunityOwnedDate="2010-12-19T18:11:54.420" Title="Mnemonics for remembering physical data" Tags="&lt;nomenclature&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="2065" PostTypeId="1" CreationDate="2010-12-19T16:51:50.030" Score="1" ViewCount="165" Body="&lt;p&gt;Would an object like a wooden bed interfere or block the signal coming from a 802.11 wireless router?&lt;/p&gt;&#xA;" OwnerUserId="771" LastEditorUserId="771" LastEditorDisplayName="" LastEditDate="2010-12-19T23:04:12.280" LastActivityDate="2011-05-05T15:16:54.450" Title="Do non-metal objects reduce the signal strength of a computer wireless network device?" Tags="&lt;waves&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="2066" PostTypeId="1" CreationDate="2010-12-19T18:14:46.200" Score="6" ViewCount="625" Body="&lt;p&gt;We found that water with salt, sugar, or baking soda dissolved in it cools faster than pure water.&lt;br&gt;&#xA;Water has a very high specific heat; how do these solutes lower it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We heated a beaker (300ml) of water to 90° C and let it cool, checking the temperature every 5 minutes.  We repeated the experiment adding a tablespoon of salt.  At each 5 minute interval, the temperature was higher for pure water than for salt water.  Same result with baking soda and sugar.&lt;/p&gt;&#xA;" OwnerUserId="889" LastEditorUserId="889" LastEditorDisplayName="" LastEditDate="2010-12-19T22:36:25.573" LastActivityDate="2010-12-20T21:30:15.603" Title="Why does adding solutes to pure water lower the the specific heat?" Tags="&lt;thermodynamics&gt;&lt;water&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="2068" PostTypeId="2" ParentId="2059" CreationDate="2010-12-19T19:02:28.327" Score="10" ViewCount="" Body="&lt;p&gt;There's no temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we use the following definition &quot;temperature is the average&#xA;kinetic energy of the particles&quot;. Then no particles - no temperature.&#xA;As the first sight this answer doesn't seem to be good enough, but if you want&#xA;to calculate &quot;average spin&quot; or &quot;average charge&quot; those parameters will have no sense&#xA;if there's no particles to calculate data on.&lt;/p&gt;&#xA;" OwnerUserId="890" LastEditorUserId="890" LastEditorDisplayName="" LastEditDate="2010-12-19T20:08:58.817" LastActivityDate="2010-12-19T20:08:58.817" CommentCount="4" />
  <row Id="2069" PostTypeId="1" CreationDate="2010-12-19T21:58:54.793" Score="0" ViewCount="390" Body="&lt;p&gt;Let's look at the measurement problem in the orthodox interpretation of quantum mechanics as an inconsistency between inner and outer treatment of the measurement apparatus. You can always push your boundaries of treating the evolution of your system as unitary further and further. You can say OK, the universe as a whole is evolving unitarily (let's not worry about information loss in a blackhole). So it's up to me to consider the boundary to see the evolution of my system and apparatus together or just my system. And I should be able to work out the reduced density matrix of my system equally in every treatment unambiguously! However, If you treat the apparatus externally, the evolution of the system would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a|\uparrow\rangle + b|\downarrow\rangle \to |\uparrow\rangle$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with probability $|a|^2$ or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a|\uparrow\rangle + b|\downarrow\rangle \to |\downarrow\rangle$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with probability $|b|^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas, an internal treatment of the apparatus would give&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\uparrow\rangle\otimes|\text{ready}\rangle\to U\bigl(|\uparrow\rangle\otimes|\text{ready}\rangle\bigr) = |\uparrow\rangle\otimes|\text{up}\rangle$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\downarrow\rangle\otimes|\text{ready}\rangle\to U\bigl(|\downarrow\rangle\otimes|\text{ready}\rangle\bigr) = |\uparrow\rangle\otimes|\text{down}\rangle$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $U$ a linear operator, $U(a|\psi\rangle + b|\phi\rangle) = aU|\psi\rangle + bU|\phi\rangle$, which evolves&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bigl(a|\uparrow\rangle + b|\downarrow\rangle\bigr)\otimes|\text{ready}\rangle$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U\bigl[a|\uparrow\rangle\otimes|\text{ready}\rangle + b|\downarrow\rangle\otimes|\text{ready}\rangle\bigr] =a|\uparrow\rangle\otimes|\text{up}\rangle + b|\downarrow\rangle\otimes|\text{down}\rangle$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, pushing the boundary after the measuring apparatus gives a difference physics. This could be viewed as a problem with measurement in orthodox quantum mechanics (as opposed to realist or operational strategies to solve the measurement problem) &#xA;But I was thinking it's not really an inconsistency within a theory. It's just an inconsistency between two different choices of the internal-external boundaries! I'm not asking about the role of decoherence theory. It sounds to me like the measurement problem wasn't really a problem in the first place! Am I right about that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;update:&lt;/strong&gt; It has been pointed out that the question is not clear enough yet. Here is my last attempt: It's believed that for an adequate postulates for quantum mechanics, the inner and outer treatment of measuring apparatus shouldn't affect the physics of the system. Which for the orthodox interpretation of quantum mechanics does. For instance in the Bohm's model this has been resolved by denial of representational completeness. And in Operational interpretation it's bypassed by avoiding talking about physical state of the system. Here the question is &lt;strong&gt;Are we really allowed to change the boundaries?&lt;/strong&gt; Because if you don't believe you can, the problem will never appear in the first place.&#xA;I hope that explains what I'm asking. Because I don't think I can make it more clear :-)&lt;/p&gt;&#xA;" OwnerUserId="304" LastEditorUserId="304" LastEditorDisplayName="" LastEditDate="2010-12-20T08:52:04.443" LastActivityDate="2011-08-11T05:59:47.623" Title="Measurement problem in the orthodox interpretation" Tags="&lt;quantum-mechanics&gt;&lt;measurement-problem&gt;" AnswerCount="4" CommentCount="12" FavoriteCount="1" />
  <row Id="2070" PostTypeId="2" ParentId="2065" CreationDate="2010-12-19T22:15:34.363" Score="3" ViewCount="" Body="&lt;p&gt;The question is about physics, but not in terms easily answerable by a physicist. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the bed made of (spring mattresses have lots of metal inside)? &lt;/li&gt;&#xA;&lt;li&gt;At what frequency is the source of EM waves? &lt;/li&gt;&#xA;&lt;li&gt;Where are you measuring the waves w.r.t. the bed and router?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;etc.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In general, yes there will be some dampening, depending on the factors above.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-12-19T22:15:34.363" CommentCount="3" />
  <row Id="2071" PostTypeId="1" CreationDate="2010-12-19T22:43:11.793" Score="6" ViewCount="199" Body="&lt;p&gt;'tis the season as they say!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me obvious that it's better to drive in high gear on snowy roads to reduce the torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are completely opposite advices being given on different sites:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;weather.com says &quot;&lt;a href=&quot;http://www.weather.com/activities/driving/drivingsafety/drivingsafetytips/snow.html&quot; rel=&quot;nofollow&quot;&gt;Use low gears to keep traction&lt;/a&gt;&quot;&lt;/li&gt;&#xA;&lt;li&gt;bbc.co.uk says &quot;&lt;a href=&quot;http://news.bbc.co.uk/1/hi/8443690.stm&quot; rel=&quot;nofollow&quot;&gt;Stay in a higher gear for better control&lt;/a&gt;&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Who is right?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; trying to make this a bit more physically answerable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use a simplified model of the car:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the torque of the engine is directly controlled through a smooth limited function ($0\leq T\leq T_{max}$ ) whose time derivative is also limited.&lt;/li&gt;&#xA;&lt;li&gt;torque is transmitted through a choice of gears to the wheels (ratios are givens)&lt;/li&gt;&#xA;&lt;li&gt;the wheels need not slip on the road ($\mu$ is given, circumference is given)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the relationship between gear ratio, torque function and car acceleration?&lt;/li&gt;&#xA;&lt;li&gt;Is it true that there is a preferred gear ratio (given a set) that always maximises acceleration, independent of the choice of function? If so, which?&lt;/li&gt;&#xA;&lt;li&gt;What is the best choice of function and gear to maximise acceleration?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-12-29T11:57:46.290" LastActivityDate="2011-07-03T02:10:07.453" Title="Driving on snowy roads" Tags="&lt;mechanics&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="2072" PostTypeId="1" AcceptedAnswerId="2073" CreationDate="2010-12-20T00:11:12.577" Score="13" ViewCount="868" Body="&lt;p&gt;I searched and couldn't find it on the site, so here it is (quoted to the letter):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;On this infinite grid of ideal one-ohm resistors, what's the equivalent resistance between the two marked nodes?&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;img src=&quot;http://i.stack.imgur.com/gUKoJ.png&quot; alt=&quot;Nerd Sniping&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://xkcd.com/356/&quot; rel=&quot;nofollow&quot;&gt;With a link to the source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not really sure if there is an answer for this question. However, given my lack of expertise with basic electronics, it could even be an easy one.&lt;/p&gt;&#xA;" OwnerUserId="724" LastActivityDate="2010-12-20T00:38:12.877" Title="On this infinite grid of resistors, what's the equivalent resistance?" Tags="&lt;electronics&gt;&lt;resistors&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="5" />
  <row Id="2073" PostTypeId="2" ParentId="2072" CreationDate="2010-12-20T00:38:12.877" Score="11" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://xkcd.com/356/&quot; rel=&quot;nofollow&quot;&gt;Nerd Sniping&lt;/a&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is $\frac{4}{\pi} - \frac{1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple explanation: &lt;a href=&quot;http://www.mbeckler.org/resistor_grid/&quot; rel=&quot;nofollow&quot;&gt;http://www.mbeckler.org/resistor_grid/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematical derivation: &lt;a href=&quot;http://www.mathpages.com/home/kmath668/kmath668.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.mathpages.com/home/kmath668/kmath668.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-12-20T00:38:12.877" CommentCount="13" />
  <row Id="2074" PostTypeId="2" ParentId="2071" CreationDate="2010-12-20T01:48:23.623" Score="3" ViewCount="" Body="&lt;p&gt;I used to be really really good at this(but now I live in California and my skill ha gone totally stale). I think it depends on the sort of snow we are talking about. If it is packed or ice then the ability to control torque -especially when starting from a dead stop is important. In&#xA;really deep snow, then torque (and maintaining momentum is the key). I used to drive my old Z71 pickup in as much a 3/4 meter of unplowed (uphill too), and compound low plus a heavy foot was essentail for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the real key to safety is stopping. That requires practice. And keeping in mind that&#xA;you want the wheels in rolling contact, i.e. you want to keep any braking/steering forces below the value where they breakfree and can slide sideways. Being able to feather both&#xA;the accelerator and the braking in response to whether the wheels have traction or not is key. The feel for that can vary greatly by vehicle model. My old landcruiser was terrible tha way, you became aware of a skid by looking out the window, whereas with my even older suburu my brake foot somehow knew if the tires had traction.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-12-20T01:48:23.623" CommentCount="2" />
  <row Id="2075" PostTypeId="2" ParentId="2069" CreationDate="2010-12-20T03:18:35.030" Score="1" ViewCount="" Body="&lt;p&gt;@Sina is making a perfectly good point with this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement is this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;The measurement problem is not really a problem, because the result of a measurement is not only dependent on the system being &quot;measured&quot;, but also on the observer doing the &quot;measuring&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To actually realize this dichotomy between an &quot;internal&quot; and &quot;external&quot; observer, the size of the observer's Hilbert space, given by its dimension $\dim(H_O)$, must be comparable to $\dim(H_S)$ - the dimension of the Hilbert space corresponding to the system under observation. Instead, what we generally encounter is $\dim(H_O) \gg \dim(H_S)$ as is the case for, say, an apparatus with a vacuum chamber and other paraphernalia which is being used to study an atomic scale sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the apparatus is not described by the three states $\{|\text{ready}\rangle, |\text{up}\rangle, |\text{down}\rangle\}$, but by the large (infinite?) family of states $\{|\text{ready};\alpha\rangle, |\text{up};\alpha\rangle, |\text{down};\alpha\rangle\}$ where $\alpha$ parametrizes the &quot;helper&quot; degrees of freedom of the apparatus which are not directly involved in generating the final output, but are nevertheless present in any interaction. Examples of these d.o.f are the states of the electrons in the wiring which transmits data between the apparatus and the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we consider the apparatus to be &quot;internal&quot; then the Hilbert space of the total &quot;system+apparatus&quot; is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H_{S+O} = H_S \otimes H_O $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which has as its basis vectors &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \{ \left|\uparrow\right\rangle|\text{ready};\alpha\rangle,&#xD;&#xA;      \left|\uparrow\right\rangle|\text{up};\alpha\rangle,&#xD;&#xA;      \left|\uparrow\right\rangle|\text{down}; \alpha\rangle;&#xD;&#xA;      \left|\downarrow\right\rangle|\text{ready};\alpha\rangle,&#xD;&#xA;      \left|\downarrow\right\rangle|\text{up};\alpha\rangle,&#xD;&#xA;      \left|\downarrow\right\rangle|\text{down};\alpha\rangle  \} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the states of the form $\left|\uparrow\right\rangle|\text{down};\alpha\rangle$ and $\left|\downarrow\right\rangle|\text{up};\alpha\rangle$ which are &quot;counterfactual&quot; (and you neglected to mention in the question).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point I wave my hands and say when an external &lt;em&gt;super-observer&lt;/em&gt; looks at the states of the system described by $H_{S+O}$, the conterfactual states of the type mentioned above will interfere destructively due to the presence of the numerous $\alpha$ d.o.f; leaving only the &quot;consistent&quot; states of the type $\left|\uparrow\right\rangle|\text{up};\alpha\rangle$  and $ \left|\downarrow\right\rangle|\text{down};\alpha\rangle$ as the ones with non-negligible amplitudes. So in such cases, their is no contradiction between what the super-observer sees and whatever output the apparatus yields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is when the observer and observed systems become comparable in size that we run into all kinds of problems. As far as I know no apparatus has yet been constructed which is described by the same number of d.o.f as the system it is supposed to measure. But we are rapidly approaching that limit with nanotechnology and then this measurement dichotomy will have to dealt with head on.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I hope this answer makes sense. However, such questions always lie in treacherous territory. So if I've made some tautological error which invalidates everything I have said, please point it out !&lt;/p&gt;&#xA;" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-12-20T06:33:40.390" LastActivityDate="2010-12-20T06:33:40.390" CommentCount="12" />
  <row Id="2076" PostTypeId="2" ParentId="2066" CreationDate="2010-12-20T03:33:47.097" Score="3" ViewCount="" Body="&lt;p&gt;The obvious answer for at least part of this simply concerns the new substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water has a fairly high specific heat.  It is greater than that of sugar, salt, baking soda, etc.  The specific heat of the combination (solution) of these two is somewhere between that of either one alone (probably a weighted average by mass) simply by merit of the temperature change occurring to both substances rather than just one&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I suspect this answer is incomplete.  There could be another phenomena in play to explain the cooling differences, perhaps associated with how the solute changes the temperature of phase changes (ie higher boiling pt, lower freezing)&lt;/p&gt;&#xA;" OwnerUserId="893" LastActivityDate="2010-12-20T03:33:47.097" CommentCount="4" />
  <row Id="2077" PostTypeId="1" AcceptedAnswerId="2078" CreationDate="2010-12-20T03:56:13.550" Score="9" ViewCount="315" Body="&lt;p&gt;It is said that Klein-Gordon equation is a relativistic version of the Schrodinger equation. In Schrodinger equation, it is straightforward to include potential energy. But for K-G eqn things seem to be more complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it specifically, how can one find the energy eigenfunctions of a particle in a finite square well via K-G eqn?&lt;/p&gt;&#xA;" OwnerUserId="749" LastEditorUserId="749" LastEditorDisplayName="" LastEditDate="2010-12-20T04:05:44.947" LastActivityDate="2011-04-19T08:58:39.130" Title="Klein–Gordon equation" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="2078" PostTypeId="2" ParentId="2077" CreationDate="2010-12-20T04:34:12.123" Score="11" ViewCount="" Body="&lt;p&gt;K-G is a relativistic equation. So if you want to couple it to external fields you have to use a vector potential $A_\mu$ which will modify the partial-derivative, so that the Klein-Gordon equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left( \partial^\mu \partial_\mu + m^2 \right)\phi = 0 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left ( D^\mu D_\mu + m^2 \right) \phi = 0 $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $ D_\mu = \partial_\mu + \imath A_\mu $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it remains to figure out what $A_\mu$ will correspond to a given potential - finite square well or otherwise. For this you will likely have to fix a gauge before you can identify the potential as such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A square well potential will correspond to $ A_\mu = (V(x), 0,0,0)$, where $ V(x) = V_0 \,\forall |x| &amp;gt; \pm a $ and $V(x) = 0, x \in [-a,a] $. The resulting behavior is a little counter-intuitive, yielding something known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Klein_paradox&quot; rel=&quot;nofollow&quot;&gt;Klein paradox&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further details see ch. 4 for this excellent &lt;a href=&quot;http://soar.wichita.edu/dspace/bitstream/10057/1170/1/t07046.pdf&quot; rel=&quot;nofollow&quot;&gt;thesis&lt;/a&gt; on this problem by D. Schaffer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: &lt;a href=&quot;http://wiki.physics.fsu.edu/wiki/index.php/Klein-Gordon_equation&quot; rel=&quot;nofollow&quot;&gt;This page&lt;/a&gt; also has excellent information along these lines.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-20T04:34:12.123" CommentCount="5" />
  <row Id="2080" PostTypeId="2" ParentId="2027" CreationDate="2010-12-20T08:14:50.700" Score="0" ViewCount="" Body="&lt;p&gt;The sphere is initially going to accelerate in the horizontal direction as it transferes the rotational speed into linear motion, by slipping. This will end when the sphere is going to start rolling with speed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \omega_{roll} = \frac{I}{I+m\,R^2}\,\Omega $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Omega$ is the initial rot. speed, $m$ the mass, $I$ the mass moment of inertia and $R$ the radius of the sphere. Once the sphere is rolling its velocity is going to be constant. This will happen at a time $t_c$ after the impact. You can find the time by plotting the horizontal speed $\dot{x}$ and the rot. speed $\omega$ as a function of time and note when $\dot{x}=\omega\,R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you assume the (vetrical) impact velocity to be $v$, the coefficient of friction $\mu$ and the contact to be fairly stiff (fast), but also critically damped (with no rebound), I get the following equations for the rotational speed as a function of time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \omega = \frac{\mu\,m\,R}{I}\left(v\,\left(\exp({-\beta\,t})-1\right)-g\,t\right)+\Omega $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linear speed is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dot{x} = \mu\left(v\,\left(1-\exp({-\beta\,t})\right)+g\,t\right) $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\beta = \sqrt{k/m} $ is the contact frequency (sqrt. stiffness over mass), and it is a huge number making the contact last milliseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve for all this I start from the vertical penetration (approach) of the part as it makes contact and comes to rest at a value less than zero due to gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ y = \frac{v}{\beta}\exp({-\beta\,t})-\frac{g}{\beta^2} $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above comes from the spring damper contact model&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ddot{y} = -\frac{k}{m} y - \frac{d}{m} \dot{y} - g $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with damping $d=2\sqrt{k\,m}$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peace!&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-12-20T08:14:50.700" CommentCount="2" />
  <row Id="2081" PostTypeId="2" ParentId="2066" CreationDate="2010-12-20T08:26:22.983" Score="2" ViewCount="" Body="&lt;p&gt;If your description of the experiment is accurate then the result you got is unexpected. It is true that specific heat capacity of salt solution (per &lt;em&gt;mass&lt;/em&gt; unit) is lower than of pure water, you can estimate it as $$C_{p} = wC_{p}^{salt}+(1-w)C_p^{H_2O}$$ where &lt;em&gt;w&lt;/em&gt; is the mass fraction of salt. However, as you describe it, you didn't keep the mass constant but increased it by adding some salt to a fixed volume of water so your total heat capacity should be the sum of the heat capacity of water (which is the same for all samples) and that of salt, sugar or baking soda.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;em&gt;w&lt;/em&gt; in your experiment was around 0.04, the effect you were measuring was quite small and could easily be smaller then the experimental error. This error consists of the accuracy of measuring volume of water, of measuring temperature, of timing. The easiest way to reduce these errors is to repeat each experiment several times in random order and see if the results are consistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: I found &lt;a href=&quot;http://www.genchem.com/SABrochure/Fig0205.html&quot; rel=&quot;nofollow&quot;&gt;a plot&lt;/a&gt; of specific heat of soda solutions and I calculated heat capacity for two cases: a) 300 ml of pure water and b) 300 ml of water + 12.5 g of soda = 312.5 g of 4% soda solution. The heat capacity of pure water sample would be 1254 J/K and that of water with soda 1276 J/K - as I expected, it is higher but the difference is less than 2%.&lt;/p&gt;&#xA;" OwnerUserId="701" LastEditorUserId="701" LastEditorDisplayName="" LastEditDate="2010-12-20T21:30:15.603" LastActivityDate="2010-12-20T21:30:15.603" CommentCount="6" />
  <row Id="2082" PostTypeId="2" ParentId="1799" CreationDate="2010-12-20T09:34:46.937" Score="6" ViewCount="" Body="&lt;p&gt;I did a bit of reading about this, and it turns out that conformal blocks are actually quite relevant to my research! So I figured it was worth the time to investigate in some more detail. I've never studied conformal field theory formally, but I hope I'm not writing anything outright wrong here. (I lost my first draft and had to reconstruct it, which is why it's taken so long)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In conformal field theory, it's common to represent coordinates on a two-dimensional space by using complex numbers, so $\vec{r} = (x,y)$ becomes $\rho = x + iy$. In this notation, the theory is invariant under the action of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Mobius_transformation&quot; rel=&quot;nofollow&quot;&gt;Möbius transformation&lt;/a&gt; (a.k.a. conformal transformation),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho \to \frac{a\rho + b}{c\rho + d}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in which $a$, $b$, $c$, and $d$ are complex constants that satisfy $ad - bc \neq 0$. The transformation has three complex degrees of freedom - in other words, if you specify three initial points and three final points on the complex plane, there is a unique Möbius transformation that maps those three initial points to the three final points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So any function of four coordinates on the plane, for example a four-point correlation function of quantum fields,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G_4 = \langle \phi_1(\rho_1,\rho_1^*) \phi_2(\rho_2,\rho_2^*) \phi_3(\rho_3,\rho_3^*) \phi_4(\rho_4,\rho_4^*) \rangle$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has only one real degree of freedom, after you factor out the gauge freedoms corresponding to the Möbius transformation. In other words, you can map any three of those coordinates on to three fixed reference points (for example $0$, $1$, and $\infty$), and you're left with a function of only one variable, something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = \frac{(\rho_4 - \rho_2)(\rho_3 - \rho_1)}{(\rho_4 - \rho_1)(\rho_3 - \rho_2)}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This opens the door to write $G_4$ as a simple function of this one ratio (at least, simpler than a function of four independent coordinates).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The particular part of CFT in which conformal blocks are applied (as far as I can tell; I'm starting to get a little out of my depth here) has to do with Virasoro algebras. Specifically, the way the individual fields $\phi_i$ transform under a conformal transformation is described by the group defined by the Virasoro algebra. The four-point function $G_4$ can be written as a sum of contributions from different representations of the group,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G_4(\rho_1,\rho_2,\rho_3,\rho_4) = \sum_l G_l f(D_l, d_i, C, x) f(D_l, d_i, C, x^*)$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $l$ indexes the different representations; $C$ is a constant (the &quot;central charge&quot; of the Virasoro algebra); and $d_i$ and $D_l$ are anomalous dimensions of the external fields and the internal field respectively. The function $f$ is called a conformal block. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/IuXt7.png&quot; alt=&quot;Feynman diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f$ is useful because it can be calculated (in principle or in practice, I'm not sure which) using only information about a single representation of the Virasoro group. It can be expressed as a series in $x$ of a known form, the coefficients of which depend on the structure of the group.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Further Reading&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Belavin A. &lt;strong&gt;Infinite conformal symmetry in two-dimensional quantum field theory.&lt;/strong&gt; &lt;em&gt;Nuclear Physics B&lt;/em&gt;. 1984;241(2):333-380. Available at: &lt;a href=&quot;http://dx.doi.org/10.1016/0550-3213(84)90052-X&quot; rel=&quot;nofollow&quot;&gt;http://dx.doi.org/10.1016/0550-3213(84)90052-X&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Zamolodchikov AB. &lt;strong&gt;Conformal symmetry in two dimensions: an explicit recurrence formula for the conformal partial wave amplitude.&lt;/strong&gt; &lt;em&gt;Communications in Mathematical Physics (1965-1997)&lt;/em&gt;. 1984;96(3):419-422. Available at: &lt;a href=&quot;http://projecteuclid.org/euclid.cmp/1103941860&quot; rel=&quot;nofollow&quot;&gt;http://projecteuclid.org/euclid.cmp/1103941860&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Zamolodchikov AB. &lt;strong&gt;Conformal symmetry in two-dimensional space: Recursion representation of conformal block&lt;/strong&gt;. &lt;em&gt;Theoretical and Mathematical Physics&lt;/em&gt;. 1987;73(1):1088-1093. Available at: &lt;a href=&quot;http://www.springerlink.com/content/khq7730604681676/&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/khq7730604681676/&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;and of course DiFrancesco et al's book.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-12-20T09:34:46.937" CommentCount="12" />
  <row Id="2084" PostTypeId="1" CreationDate="2010-12-20T10:33:27.300" Score="4" ViewCount="210" Body="&lt;p&gt;I am looking for reference papers and/or books on the principles of LASER. I've understood that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Gould, R. Gordon (1959). &quot;The LASER, Light Amplification by Stimulated Emission of Radiation&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;was one of the principal papers published by the disputed inventor himself. What are other good sources?&lt;/p&gt;&#xA;" OwnerUserId="250" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-27T06:56:28.607" LastActivityDate="2011-07-27T06:56:28.607" Title="Good (citable) paper on Laser Physics" Tags="&lt;books&gt;&lt;reference-request&gt;&lt;laser&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="2085" PostTypeId="2" ParentId="999" CreationDate="2010-12-20T10:48:10.723" Score="2" ViewCount="" Body="&lt;p&gt;The filament will cool very rapidly after current stops flowing and I would say will only account for a second or two of your perceived after glow. If you are noticing a longer duration of glow (you state a minute or so) that is probably retinal in cause &lt;a href=&quot;http://en.wikipedia.org/wiki/Afterimage&quot; rel=&quot;nofollow&quot;&gt;retinal afterimage&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="896" LastActivityDate="2010-12-20T10:48:10.723" />
  <row Id="2086" PostTypeId="2" ParentId="2027" CreationDate="2010-12-20T12:19:30.910" Score="2" ViewCount="" Body="&lt;p&gt;I will assume that the collision to the surface is perfectly elastic and $\vec{\omega}$ is parallel to the surface. Let's denote the collision time by $\tau$ and the vertical velocity of the ball at the moment of collision  by $v_y$. $\tau$ is supposed to be infinitesimally short. Now we will concentrate in what happens during collision. Let's denote the force exerted by sliding friction while colliding by $F_\mu$.(due to the rotation of the ball) Then $$F_\mu=\mu\frac{\Delta P}{\tau}$$ where $\Delta P=2mv_y$ is the change in vertical momentum of the ball. This force obviously accelerates the ball in horizontal direction. So after the collision the velocity of the ball has a horizontal component&#xA;$$v_x=\frac{F_\mu\tau}{m}=2\mu v_y$$ and it bounce up at an angle $\alpha$ to the horizontal&lt;br&gt;&#xA;$$\alpha=arctan\frac{v_y}{v_x}=arctan\frac{1}{2\mu}$$However this solution is valid only if$\mu&amp;lt;\frac{1}{5}$ because in case of $\mu&amp;gt;\frac{1}{5}$ the rotational motion of the ball ends before the end of the collision.&lt;/p&gt;&#xA;" OwnerUserId="552" LastActivityDate="2010-12-20T12:19:30.910" />
  <row Id="2087" PostTypeId="2" ParentId="2084" CreationDate="2010-12-20T12:40:25.243" Score="1" ViewCount="" Body="&lt;p&gt;The Springer book linked &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387017690&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; helped me a lot during the writing on my master thesis on femptosecond-pulsed lasers...try it!&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="1616" LastEditorDisplayName="" LastEditDate="2011-02-28T01:44:49.017" LastActivityDate="2011-02-28T01:44:49.017" CommentCount="1" />
  <row Id="2088" PostTypeId="1" AcceptedAnswerId="2094" CreationDate="2010-12-20T13:03:10.267" Score="4" ViewCount="149" Body="&lt;p&gt;Suppose I tie one end of a rope to my ceiling and the other end to a spot on my floor directly underneath it.  Because the rope has some mass, the tension varies along the rope, from highest at the ceiling to lowest at the floor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a wave packet begins propagating down the rope, will its shape change?  If so, it is possible to calculate the shape of a Gaussian wave packet as it travels down the rope?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, it seems to me that the wave packet will change shape for two reasons.  First, the portions higher up the rope have higher tension, and therefore higher speed.  They will &quot;catch up&quot; to portions of the wave packet further down.  Second, because the tension is changing, the wave equation now has a term related to the first derivative of the displacement of the rope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, I tried assuming that the displacement of the rope is only horizontal and that the slope of the rope is never far from vertical.  I made $y$ a coordinate measuring up from the floor and $x$ a coordinate to the right.  Letting the tension be $T(y) = T_0+\lambda g y$, with $\lambda$ the mass per unit length, I got the wave equation (&lt;b&gt;edit:&lt;/b&gt; important typo corrected)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial^2{x}}{\partial t^2} = g \frac{\partial x}{\partial y} + \frac{T_0+\lambda g y}{\lambda} \frac{\partial^2 x}{\partial y^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I don't know what to do with it.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-12-21T00:58:19.873" LastActivityDate="2010-12-21T12:46:35.063" Title="Will a wave packet undergo dispersion when traveling down a hanging rope?" Tags="&lt;mechanics&gt;&lt;waves&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="2089" PostTypeId="1" AcceptedAnswerId="2093" CreationDate="2010-12-20T13:14:27.937" Score="5" ViewCount="197" Body="&lt;p&gt;Closely related to &lt;a href=&quot;http://physics.stackexchange.com/questions/2088/will-a-wave-packet-undergo-dispersion-when-traveling-down-a-hanging-rope&quot;&gt;this question on traveling waves on a hanging rope&lt;/a&gt;, I would also like to know what the normal modes are on a rope that hangs vertically, fixed at both ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tension in the rope increases with height, so I expect that the wavelength of standing waves decreases as we move up, and nodes are  clustered somewhat more towards the top of the rope than the bottom.  Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wave equation I found was (&lt;b&gt;note&lt;/b&gt;: important typo corrected)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial^2{x}}{\partial t^2} = g \frac{\partial x}{\partial y} + \frac{T_0+\lambda g y}{\lambda} \frac{\partial^2 x}{\partial y^2}$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so mathematically, my question is about how to find the eigenfunctions of this equation with the boundary conditions $x(0) = x(h) = 0$ with $h$ the height of the ceiling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I was able to &lt;a href=&quot;http://adsabs.harvard.edu/abs/1998PhTea..36..108G&quot; rel=&quot;nofollow&quot;&gt;locate a reference&lt;/a&gt;, but don't have access to it.)&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-12-21T00:56:49.703" LastActivityDate="2010-12-21T12:38:56.790" Title="What are the normal modes of a vertical rope?" Tags="&lt;mechanics&gt;&lt;waves&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="2090" PostTypeId="2" ParentId="2069" CreationDate="2010-12-20T13:21:30.980" Score="2" ViewCount="" Body="&lt;p&gt;I think you've just rediscovered the Many-Worlds interpretation. All that Many-Worlds really says is that when a system in a superposition state is measured, the state of the measurement apparatus (and associated observers, etc.) becomes entangled with the state of the system . The system continues to be in a superposition of all possible states, it's just a somewhat more complicated superposition, including the measurement apparatus and all that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contrary to what a lot of popular treatments lead people to believe, Many-Worlds does not involve the creation of an entire separate universe worth of matter for each of the possible measurement outcomes. There is just a single wavefunction describing the single universe worth of matter that we have, which becomes more and more complicated as time goes on. We only see a single branch because decoherence in the form of unmeasured interactions with a larger environment destroys our ability to detect any influence of these different branches on one another. This makes the different branches effectively separate &quot;universes,&quot; because the piece of an observer in one branch is entangled with the state of the system in that branch, and cannot detect any influence of the other branches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your internal/ external phrasing is different than I'm used to hearing, but the math you've written down looks exactly like the way you would formulate a Many-Worlds type description of the measurement process. If that's not what you're doing, the distinction isn't clear to me.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-12-20T13:21:30.980" CommentCount="6" />
  <row Id="2091" PostTypeId="2" ParentId="2088" CreationDate="2010-12-20T13:34:56.227" Score="0" ViewCount="" Body="&lt;p&gt;I don't think it's possible to get an analytical solution to that, but I could be wrong. Conceptually, what you have is a situation where the speed of the wave varies as you move along the rope, as the square root of the local tension ($v\propto \sqrt{T(y)}$). This is analogous to things like water waves moving up a beach (where the speed of the wave depends on the depth), so that might be a place to look for tips on the math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This scenario is also sort of similar to the case of the Schrodinger equation in a linear potential (a ball bouncing vertically under the influence of gravity, say), though the Schrodinger equation is only first-order in time, so that's another place you might look for solutions.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-12-20T13:34:56.227" />
  <row Id="2092" PostTypeId="2" ParentId="1799" CreationDate="2010-12-20T13:39:57.393" Score="6" ViewCount="" Body="&lt;p&gt;Conformal field theory is the theory of scale invariance (or large-order behavior) in two dimensions.  Scaling means dependence on angles only.  In 2d, group of angle-preserving (conformal) transformations is infinite-dimensional, and in fact there are only a finite number of degrees of freedom in a 2d metric after conformal transformations and diffeomorphisms.  (The degrees of freedom are the moduli space of Riemann surfaces.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fields in a theory with conformal symmetry must give representations of this symmetry algebra, and such representations are labeled by a quantum number called conformal dimension or weight.  The transformations themselves are holomorphic changes of coordinates (z --&gt; f(z) and they are generated by the Lie algebra of holomorphic vector fields $L_n := -z^{n+1}\partial_z$ and their complex conjugates.  You can calculate this algebra:  $[L_n,L_m] = (n-m)L_{m+n}$ which is called the Virasoro algebra.  (There are two of these, one with z and one with z-bar.)  Quantum mechanically, this algebra can be corrected by the &lt;code&gt;conformal anomaly' parametrized by the&lt;/code&gt;central charge' (&quot;central&quot; because the extra term commutes with all others).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now just as in a rotation-invariant theory, if you want to know how a solution looks after a rotation you only need to know which representation the state lies in, in a conformal theory if you want to change coordinates infinitesimally you only need to know the conformal weights of the fields.  But such transformations are infinitesimal coordinate changes, so this gives a differential equation that the correlator must obey.  Everything in the theory can be written in terms of solutions to these differential equations -- these are called `conformal blocks.'  (There are solutions in z-bar, too.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method is detailed in the classic work of Belavin, Polyakov and Zamolodchikov (NPB 241 (1988) p. 333) (another pioneer is Knizhnik).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p.s.  String theory is all about 2d field theories and their dependence on the moduli of Riemann surfaces.  The condition that the conformal theory be anomaly-free is the most common way of deriving dimension formulas in string theory.&lt;/p&gt;&#xA;" OwnerUserId="20" LastActivityDate="2010-12-20T13:39:57.393" CommentCount="7" />
  <row Id="2093" PostTypeId="2" ParentId="2089" CreationDate="2010-12-20T13:55:45.067" Score="5" ViewCount="" Body="&lt;p&gt;I want to try to give a solution to your equation using a &lt;strong&gt;separation ansatz&lt;/strong&gt; of the form&lt;br&gt;&#xA;$$x(y,t) = T(t)\cdot Y(y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We find&#xA;$$\frac{T&amp;#39;&amp;#39;}{T} = \frac{1}{Y}\left( gY&amp;#39; + \left(\frac{T_0 + \lambda g y}{\lambda}\right) Y&amp;#39;&amp;#39; \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both sides do depend on different variables, hence the given equation must equal a constant, say $-\omega_n^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see directly that the fundamental system for $T$ can be stated as oscillations (if this notation is convenient for you):&#xA;$$eig(T) = \{ \sum_n \left[ e^{\mathrm{i}\omega_n t}, e^{-\mathrm{i}\omega_n t} \right]\}$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;New solution for Y&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The differential equation for $Y$ now takes the form&#xA;$$ \frac{\omega_n^2}{g} Y + Y&amp;#39; + \left(\frac{T_0}{\lambda} + y\right) Y&amp;#39;&amp;#39; = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and it is obvious that one should first of all make a substitution of the form&#xA;$$z = \frac{T_0}{\lambda} + y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such that we find with $Y(y) = Z(z)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\omega_n^2}{g} Z + Z&amp;#39; + z Z&amp;#39;&amp;#39; = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution to this equation are indeed &lt;strong&gt;Bessel functions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$eig(Z) = \{J_0\left(2\omega_n \sqrt{z/g} \right) , Y_0\left( 2\omega_n \sqrt{z/g}\right) \}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, finally&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x{(t,y)}=\sum_n e^{\{+,-\}\mathrm{i}\omega_n t}\left[ a_n J_0 \left( 2\omega_n \sqrt{\frac{T_0 + \lambda y}{\lambda g}} \right) + b_n Y_0 \left( 2\omega_n \sqrt{\frac{T_0 + \lambda y}{\lambda g}} \right) \right]$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall solution is then given due to the starting and boundary conditions you specified. Since this is a discussion on its own, I will leave this for the &lt;a href=&quot;http://physics.stackexchange.com/questions/2088/will-a-wave-packet-undergo-dispersion-when-traveling-down-a-hanging-rope&quot;&gt;other thread&lt;/a&gt; :)&lt;br&gt;&#xA;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2010-12-21T12:38:56.790" LastActivityDate="2010-12-21T12:38:56.790" CommentCount="7" />
  <row Id="2094" PostTypeId="2" ParentId="2088" CreationDate="2010-12-20T14:04:47.863" Score="2" ViewCount="" Body="&lt;p&gt;I tried to calculate the solution to your problem in &lt;a href=&quot;http://physics.stackexchange.com/questions/2089/what-are-the-normal-modes-of-a-vertical-rope/2093#2093&quot;&gt;your related question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I did not miscalculate, the dispersion relation should be implicitly given by the boundary conditions at $y=0$ and $y=h$. The solution was&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x{(t,y)}=\sum_n e^{\{+,-\}\mathrm{i}\omega_n t}\left[ a_n J_0 \left( 2\omega_n \sqrt{\frac{T_0 + \lambda y}{\lambda g}} \right) + b_n Y_0 \left( 2\omega_n \sqrt{\frac{T_0 + \lambda y}{\lambda g}} \right) \right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but it is very unhandy. Performing a coordinate transformation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho = \frac{T_0 + \lambda y}{\lambda g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can drop the $Y_0$ part of the solution since it is not limited at the origin and $J_0$ is already zero there.&lt;br&gt;&#xA;We thus find that the dispersion relation can be stated as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\omega_n \cdot \rho{(y = h)} = \mathrm{&amp;quot;root\ of\ }J_0&amp;quot;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2010-12-21T12:46:35.063" LastActivityDate="2010-12-21T12:46:35.063" CommentCount="3" />
  <row Id="2095" PostTypeId="1" AcceptedAnswerId="2096" CreationDate="2010-12-20T15:12:28.580" Score="3" ViewCount="273" Body="&lt;p&gt;A while ago someone proposed the following thought experiment to me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A horse attached to a cart is resting on a horizontal road. If the horse attempts to move by pulling the cart, according to the 3rd Newton's Law, the cart will exert a force equal in magnitude and opposite in direction, cancelling each other out and thus the horse and the cart &lt;em&gt;should&lt;/em&gt; not move. And yet it moves (pun intended ;) Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I never got a satisfactory answer, my guess is that the answer lies in the frames of reference involved: horse-cart and horse-road. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="892" LastActivityDate="2010-12-20T18:37:01.607" Title="Why does the cart move?" Tags="&lt;classical-mechanics&gt;&lt;mechanics&gt;&lt;newtonian-mechanics&gt;" AnswerCount="3" />
  <row Id="2096" PostTypeId="2" ParentId="2095" CreationDate="2010-12-20T15:21:53.647" Score="9" ViewCount="" Body="&lt;p&gt;The opposing force always work on a different body, thats why they dont cancel. For instance the moon pulls the earth and the earth pulls the moon with same force. Also, the horse dont move by pulling the cart, but by pulling the earth.&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2010-12-20T15:30:50.330" LastActivityDate="2010-12-20T15:30:50.330" CommentCount="2" />
  <row Id="2097" PostTypeId="2" ParentId="2095" CreationDate="2010-12-20T17:44:51.207" Score="4" ViewCount="" Body="&lt;p&gt;The horse also exerts a force on the road, and the road reacts on the horse. To compute the movement of the horse, you need to add the 2 forces acting on the horse (by the road and by the cart).&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2010-12-20T17:44:51.207" CommentCount="2" />
  <row Id="2098" PostTypeId="2" ParentId="1845" CreationDate="2010-12-20T17:54:35.597" Score="0" ViewCount="" Body="&lt;p&gt;You can construct a &lt;a href=&quot;http://en.wikipedia.org/wiki/Slotted_waveguide&quot; rel=&quot;nofollow&quot;&gt;slotted waveguide antenna&lt;/a&gt; for a wifi router. I guess that you can construct 2 of them, a working one with the slots at the right points, were the microwave power is at its maximum, and one with the slots at the wrong point, with the slots shifted by $\lambda/2$ to be at the point where the microwave power it minimal. The fact that the good one works and the other don't can be seen as a way to check the wavelength of the wifi signal. Maybe the best obtion would be to build one with &quot;sliding slots&quot;, in order to go from one situation to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You probably can play similar tricks with other wavelength dependent antenna design.&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2010-12-20T17:54:35.597" CommunityOwnedDate="2010-12-20T17:54:35.597" />
  <row Id="2099" PostTypeId="2" ParentId="1799" CreationDate="2010-12-20T18:10:46.927" Score="7" ViewCount="" Body="&lt;p&gt;Now that we have a physicist's perspective, I don't feel too bad outlining conformal blocks from a mathematician's point of view.  Presumably there is a dictionary connecting the two worlds, but I don't understand physics well enough to say coherent sentences about it.  I apologize in advance for any confusion - this is not a very pedestrian topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll approach conformal blocks from the standpoint of conformal vertex algebras, which typically appear in mathematics as algebraic structures that you can use to prove theorems in representation theory.  Vertex algebras are vector spaces $V$ equipped with a &quot;multiplication with singularities&quot; $V \otimes V \to V((z))$ that encodes a best effort at multiplying quantum fields (which are sometimes called &quot;operator-valued distributions&quot;).  Left multiplication by an element $u$ yields a formal power series $\sum_{n \in \mathbb{Z}} u_n z^{-n-1}$ whose coefficients are operators.  To make a vertex algebra conformal is to choose a distinguished vector $\omega$ whose corresponding operators generate an action of the Virasoro algebra, which is a central extension of the complexified Lie algebra of polynomial vector fields on the circle.  You don't lose much conceptually by thinking of Virasoro as the tangent space of the group $Diff(S^1)$ at the identity, but there is a &quot;nonzero central charge&quot; anomaly in play that can make the central extension necessary.  The circle shows up here because it is the boundary of a puncture where we will insert a field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding of the physical interpretation is the following incomplete and possibly incorrect picture: Inside a 2D conformal field theory, there is an algebra of (say, left-moving) chiral symmetries, and this is precisely the information captured by the conformal vertex algebra.  The space of states in the theory decomposes into a set of &quot;sectors&quot; which are modules of the vertex algebra.  If we choose a Riemann surface (which is a sphere in most textbooks), and attach states from various sectors to a set of distinct points, we should get a set of amplitudes, which are values of chiral correlation functions attached to these input data.  I have heard that there is some way to pass from the chiral stuff to the conformal field theory proper, where the ambiguity in the correlators disappears and one gets honest correlation functions, but I haven't seen it in the math literature.  In any case, conformal blocks live inside this machine - given sectors attached to points on a Riemann surface, a conformal block is a gadget that eats choices of states in those sectors, and outputs values of correlation functions in a manner consistent with the chiral symmetries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a sketch of the mathematical construction, due to Edward Frenkel (and described in more detail in his book &lt;em&gt;Vertex Algebras and Algebraic Curves&lt;/em&gt; with David Ben-Zvi): There is a &quot;positive half&quot; of the Virasoro algebra, spanned by generators $-z^n\frac{d}{dz}$ for $n \geq 0$, and it generates the Lie algebra of derivations on the infinitesimal complex disk, and also acts on the conformal vertex algebra $V$.  We can use this action to construct a vector bundle $\mathscr{V}$ with flat connection on our Riemann surface of choice by the Gelfand-Kazhdan &quot;formal geometry&quot; method (which I won't describe).  Given punctures $p_1, \dots, p_n$, one constructs, from the De Rham complex of $\mathscr{V}$, a Lie algebra $L$ that acts naturally on $n$-tuples of $V$-modules.  Given $V$-modules $M_i$ attached at points $p_i$, a conformal block is an $L$-module map from $\bigotimes M_i$ to the trivial module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is in general quite difficult to do any explicit calculations with conformal blocks, because of the amount of geometry involved.  If your Riemann surface has handles, you will have to deal with a choice of complex structure, and if it has a lot of punctures, you have to deal with a complicated configuration space of points.  You typically see tree-level diagrams with 4 inputs, because:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;That is where the bare minimum of geometry appears - since the automorphism group of the complex projective line is triply transitive, the configuration space of four points is a thrice-punctured line (by which I mean a sphere).&lt;/li&gt;&#xA;&lt;li&gt;Depending on the level of detail you seek, it is often all that you need - the spaces of blocks can be assembled by gluing surfaces together out of pants and taking sums over sectors where the sewing happens.  In the complex algebro-geometric picture, this sewing means sticking spheres together transversely at points to get a nodal curve.  One then deforms to get a smooth complex curve, and does a parallel transport along the corresponding path in the moduli space of marked curves.  The four-point configuration is a situation where you have exactly one sewing operation (and the other such situation is a punctured torus, which is important for getting characters).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In fact, when the conformal field theory is suitably well-behaved (read: rational), one gets dimensions of spaces of all conformal blocks from just the dimensions of three-point genus zero blocks, also known as structure constants of the fusion algebra.  One sees this in the Verlinde formula, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think examples of conformal blocks have a certain necessary complexity, but here is an overview of a reasonably simple case that is motivated by the WZW model.  Pick a simple Lie group, like $SU(2)$, and a level $\ell$ (which we can view as a positive integer).  One constructs the vertex algebra and its modules as level $\ell$ integrable representations of the affine Kac-Moody Lie algebra $\hat{\mathfrak{sl}_2}$, which is a central extension of the loop algebra of the complexification of the Lie algebra $\mathfrak{su}_2$.  If we choose a Riemann surface (such as a sphere), and decorate points with just the vacuum module, we get a space of conformal blocks that is the space of global sections of a certain line bundle $L_G^{\otimes \ell}$ on the moduli space of $SU(2)$ bundles on the surface.  Here $L_G$ is the ample generator of the Picard group of the moduli space.&lt;/p&gt;&#xA;" OwnerUserId="761" LastEditorUserId="761" LastEditorDisplayName="" LastEditDate="2010-12-20T19:35:49.893" LastActivityDate="2010-12-20T19:35:49.893" CommentCount="8" />
  <row Id="2100" PostTypeId="1" AcceptedAnswerId="2102" CreationDate="2010-12-20T18:27:34.003" Score="3" ViewCount="212" Body="&lt;p&gt;When you stick ice in a drink, AFAICT (the last physics I took was in high school) two things cool the drink:&lt;ul&gt;&lt;li&gt;The ice, being cooler than the drink, gets heat transferred to it from the drink (Newton's law of cooling). (This continues after it melts, too.)&lt;li&gt;The ice melts, and the cool liquid mixes with the drink, which makes the mixture feel cooler.&lt;/ul&gt;My first question is, to what extent does each of these cool the drink: which has a greater effect, and how much greater?&lt;br&gt;&lt;br&gt;Secondly, how much of the cooling by the first method (heat transfer) is without melting the ice? That is, is there any significant amount of heat transfer to each speck of ice before it's melted, and how much does that cumulatively affect the drink's temperature?&lt;br&gt;&lt;br&gt;I suppose all this will depend on a bunch of variables, like the shape and number of ice cubes and various temperatures and volumes. But any light that can be shed, let's say for &quot;typical&quot; situations, would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="902" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-22T11:38:39.780" LastActivityDate="2010-12-22T11:38:39.780" Title="What cools a drink?" Tags="&lt;fluid-dynamics&gt;&lt;thermodynamics&gt;" AnswerCount="3" />
  <row Id="2101" PostTypeId="2" ParentId="2095" CreationDate="2010-12-20T18:37:01.607" Score="3" ViewCount="" Body="&lt;p&gt;Let us draw free body diagram for the cart. &#xA;It is important to note that reaction force is NOT acting on the cart. And that is why my friend it moves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does not mean that reaction is absent. It is there but its NOT acting on the cart itself. So not all the forces on the cart are balanced. There is net force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/WXtUw.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is common misunderstanding. &lt;a href=&quot;http://en.wikipedia.org/wiki/Reaction_%28physics%29#Examples_of_common_misunderstandings&quot; rel=&quot;nofollow&quot;&gt;See examples here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-12-20T18:37:01.607" />
  <row Id="2102" PostTypeId="2" ParentId="2100" CreationDate="2010-12-20T19:15:21.523" Score="6" ViewCount="" Body="&lt;p&gt;There are three processes to take into account:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The warming of ice towards the melting point if it was originally below $0^{\circ} C$.&lt;/li&gt;&#xA;&lt;li&gt;The melting of ice itself&lt;/li&gt;&#xA;&lt;li&gt;The warming of the resulting water&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The 1. and 3. part is addressed by &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_capacity&quot; rel=&quot;nofollow&quot;&gt;heat capacity&lt;/a&gt; of ice and water respectively and the amount of heat will be directly proportional to temperature difference and weight of the water/ice. The proportionality constant (actually it also depends on the temperature but not very strongly so let's just ignore that) is called specific heat. For &lt;a href=&quot;http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html&quot; rel=&quot;nofollow&quot;&gt;water&lt;/a&gt; it is about twice as large as that of &lt;a href=&quot;http://www.engineeringtoolbox.com/ice-thermal-properties-d_576.html&quot; rel=&quot;nofollow&quot;&gt;ice&lt;/a&gt; at temperatures around $0^{\circ} C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the 2. part, this has to do with &lt;a href=&quot;http://en.wikipedia.org/wiki/Latent_heat&quot; rel=&quot;nofollow&quot;&gt;latent heat&lt;/a&gt;. Simply put, this is an amount of heat you need to change phases without changing temperature. Less simply put, when warming you are just converting the heat into greater wiggling of water molecules around their stable positions in the crystal thereby increasing their temperature. But at the melting point that heat will instead go into breaking chemical bonds between molecules in the ice lattice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, latent heat is really big (you need lots of energy to break those bonds). To get a hang on it: you would need the same amount of heat to warm water from $0^{\circ} C$ to $80^{\circ} C$ as you would need to melt the same amount of ice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, presumably you want your drink cold in the end so that temperature for 3. will be close to $0^{\circ} C$ and also the ice cubes should be pretty warm (no use in producing ice cubes of e.g. $-50^{\circ} C$, right?). This means that these processes won't contribute much cooling. It's fair to say that melting of the ice takes care of everything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt; we can also quickly estimate how much ice you need by neglecting the processes 1. and 3. Say you are starting with a warm drink of $25^{\circ} C$ and you want to get it to $5^{\circ} C$. So, reusing the argument about the $80^{\circ} C$ difference being equivalent to a latent heat of the same mass, we see that you need four times less ice than water to get the job done.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-20T19:15:21.523" CommentCount="1" />
  <row Id="2103" PostTypeId="2" ParentId="2100" CreationDate="2010-12-20T19:20:29.650" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;If you want to cool the drink by heat transfer from the drink to ice then you should not put ice into it. You should surround the glass of liquid with ice. &lt;/li&gt;&#xA;&lt;li&gt;Suppose you pour cold water(melted ice) into the drink then &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;the heat difference between the drink and the mixture is&lt;br&gt;&#xA;$H_h = m_h c_h (T_h - T_m)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;similarly, for cold water&lt;br&gt;&#xA;$H_c = m_c c_c (T_m - T_c)$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $H_h = H_c$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_m = \frac{m_h c_h T_h + m_c c_c T_c}{m_h c_h + m_c c_c}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, whenever $T_m &amp;lt; T_h$ we say resultant mixture is cooled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.newtonsballs.com/CalcPages/FluidMix.htm&quot; rel=&quot;nofollow&quot;&gt;--copied shamelessly from&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-12-20T19:20:29.650" />
  <row Id="2104" PostTypeId="2" ParentId="2100" CreationDate="2010-12-20T19:33:44.483" Score="1" ViewCount="" Body="&lt;p&gt;Suppose we can an ice cube at 0 °C with mass m&lt;sub&gt;i&lt;/sub&gt; kg, and some water at temperature T with mass m&lt;sub&gt;w&lt;/sub&gt; kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt;&lt;br&gt;&#xA;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Latent_heat#Table_of_latent_heats&quot; rel=&quot;nofollow&quot;&gt;specific latent heat&lt;/a&gt; of ice is L&lt;sub&gt;i&lt;/sub&gt; = 334 000 J kg&lt;sup&gt;-1&lt;/sup&gt;, and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Water_%28properties%29#Heat_capacity_and_heats_of_vaporization_and_fusion&quot; rel=&quot;nofollow&quot;&gt;specific heat capacity of water&lt;/a&gt; is 4200 J K&lt;sup&gt;-1&lt;/sup&gt; kg&lt;sup&gt;-1&lt;/sup&gt;. Therefore, to completely melt the ice cube, the water will need to release heat of (334000 m_i) J, i.e. the temperature drop is&#xA;$$334000 m_i = 4200 m_w \Delta T \implies \Delta T = \frac{334000 m_i}{4200 m_w}.$$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/2066/why-does-adding-solutes-to-pure-water-lower-the-the-specific-heat&quot;&gt;As discussed in the previous question&lt;/a&gt;, if the drink is not pure water, the heat capacity can become less than 4200 J K&lt;sup&gt;-1&lt;/sup&gt; kg&lt;sup&gt;-1&lt;/sup&gt;, and this only makes &amp;Delta;T larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.&lt;/strong&gt;&lt;br&gt;&#xA;When the completely melted ice is mixed with water. The final temperature will be the weighted average of the individual mixture after all heat transfer.&#xA;$$ m_w T + m_i 0 = (m_w + m_i) (T - \Delta T) \implies \Delta T = \frac{m_i}{m_w+m_i}T $$&#xD;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the melting of ice, the temperature will not be a constant (T is decreasing), and thus the &amp;Delta;T by this mixing process will also become less. The result thus only gives an &lt;em&gt;upper bound&lt;/em&gt; of temperature decrease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider an example case where your drink is at room temperature 20 °C, the mass of water is 100 g and the mass of ice cube is 10 g. Then the &amp;Delta;T just by melting is 8 C°, and that by mixing is at most 2 C°, this shows the cooling is mainly due to melting.&lt;/p&gt;&#xA;" OwnerUserId="56" LastActivityDate="2010-12-20T19:33:44.483" />
  <row Id="2105" PostTypeId="2" ParentId="1799" CreationDate="2010-12-20T22:13:20.333" Score="7" ViewCount="" Body="&lt;p&gt;There are already nice answers from both a physical and mathematical perspective, explaining the basic idea - given the algebra of holomorphic operators (or equivalently the symmetry algebra) of a CFT, we can write down a collection of equations (the Ward identities) that the partition function of the theory must satisfy on any Riemann surface. The space of solutions of these equations is the space of conformal blocks. If we indeed have a full CFT then &#xA;the partition function will be a particular conformal block. But given any conformal block we can still make sense of correlation functions on the Riemann surface, so can perform much of the field theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a fair amount of mathematical work on extending a chiral algebra to a full CFT, especially in the rational case (as Scott pointed out this is a central focus of the extended oeuvre of Fuchs, Schweigert, Runkel and collaborators). This involves finding modular invariant combination of modules for the chiral algebra, and can be reduced to finding special modules (Frobenius algebra objects in the braided tensor category of modules with some conditions). In the irrational case this theory is really in its infancy -- there's a notion of what branes should be, but there isn't a full structure theory..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think a very illuminating point of view on conformal blocks derives from the idea that a chiral CFT is more like a three-dimensional [topological] quantum field theory than like an honest CFT (and this can be made precise in the rational case, see e.g. the book by Bakalov-Kirillov).&#xA;From this point of view, we have a 3d QFT which makes sense on curved backgrounds (in fact topologically invariant),  so we can assign a Hilbert space of states from quantizing the theory on a Riemann surface times R. This space of states is the space of conformal blocks. More generally we can consider line operators in this three-dimensional theory, which means we can insert operators at points of the Riemann surface times R. These operators correspond to modules for the chiral algebra, and the resulting Hilbert space is the space of conformal blocks with module insertions. If we have a non-rational CFT we don't get a full 3d topological QFT but we can still assign Hilbert spaces to Riemann surfaces or surfaces with module insertions, hence conformal blocks. (In a full-fledged theory these vector spaces would be forced to e finite dimensional by the well-definition of the trace of the Hamiltonian, which is zero in a topological theory).&lt;/p&gt;&#xA;" OwnerUserId="905" LastActivityDate="2010-12-20T22:13:20.333" />
  <row Id="2106" PostTypeId="2" ParentId="2089" CreationDate="2010-12-20T23:26:56.840" Score="2" ViewCount="" Body="&lt;p&gt;The normal modes are related to Bessel functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[I made an error when I first submitted this answer which I am about to correct.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a web page discussing the free-hanging version of the problem: &lt;a href=&quot;http://mysite.du.edu/~jcalvert/math/hchain.htm&quot; rel=&quot;nofollow&quot;&gt;The Hanging Chain&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is the oldest journal reference I can find, though still much later than the first solution of the problem, which was in the 18th century: &lt;a href=&quot;http://dx.doi.org/10.1119/1.1932623&quot; rel=&quot;nofollow&quot;&gt;American Journal of Physics 18:405&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditorDisplayName="" LastEditDate="2010-12-21T01:58:41.013" LastActivityDate="2010-12-21T01:58:41.013" CommentCount="2" />
  <row Id="2107" PostTypeId="1" CreationDate="2010-12-20T23:46:46.510" Score="0" ViewCount="71" Body="&lt;p&gt;does deep frying wood cause it to undergo pyrolysis, such that bio-oil is produced?&#xA;by submerging it under a high smoke point oil, such as avocado oil, would it react with the  oxygen bonded to the oil molecule and combust, or would it turn into bio-oil?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also, please migrate if there's a chemistry stackexchange&lt;/p&gt;&#xA;" OwnerUserId="907" LastActivityDate="2010-12-20T23:46:46.510" ClosedDate="2010-12-21T00:05:10.933" Title="does deep frying wood cause it to undergo pyrolysis?" Tags="&lt;physical-chemistry&gt;" CommentCount="1" />
  <row Id="2109" PostTypeId="2" ParentId="41" CreationDate="2010-12-21T02:17:49.827" Score="4" ViewCount="" Body="&lt;p&gt;The work being done on TQFT by physicists and mathematicians is wonderful, but in no way should you think it somehow captures what is important in QFT to physicists trying to explain the real world. QFT as applied to the real world has particle-like excitations, non-trivial correlation functions depending on the spacetime distance between operators, spontaneous symmetry breaking, non-trivial renormalization group flows and so on and calculations are done in 4d flat Minkoswki space because this is an excellent approximation to the local spacetime metric. TQFT, as the name implies, captures information about the topology of manifolds and would be completely boring on $R^{3,1}$.  You can't use it to study pion scattering, or compute the short distance interaction between quarks, or the production cross-section for the Higgs boson or really anything that particle theorists do with QFT. So the answer to your first question is no. The axioms of TQFT do not capture what physicists feel is important in QFT. It isn't really a question of adding more parameters, they are just very different beasts. To your question of whether mathematicians have gotten rid of too much for physicists to care, the answer is yes, for most physicists. As I said earlier, for all practical purposes on can do particle theory on flat Minkowski space and the only structure that really matters is the causal structure. None of this is meant to denigrate work on TQFT of course.&lt;/p&gt;&#xA;" OwnerUserId="838" LastActivityDate="2010-12-21T02:17:49.827" CommentCount="1" />
  <row Id="2110" PostTypeId="1" CreationDate="2010-12-21T03:59:29.420" Score="14" ViewCount="566" Body="&lt;p&gt;REFORMULATED:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked at the other questions (ie &quot;why does space expansion affect matter&quot;) but can't find the answer I am looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question:&#xA;There is always mention of space expanding when we talk about the speed of galaxies relative to ours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why, if space is expanding, does matter not also expand? If a circle is drawn on balloon (2d plane), and the balloon expands, then the circle also expands. If matter is an object with 3 spatial dimensions, then when those 3 dimensions expand, so should the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that was the case, we wouldn't see the universe as expanding at all, because we would be expanding (spatially) with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a few potential answers for this, which raise their own problems:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Fundamental particles are 'point sized' objects. They cannot expand because they do not have spatial dimension to begin with. The problem with this is that while the particles would not expand, the space between them would, leading to a point where the 3 non-gravity forces would no longer hold matter together due to distance&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Fundamental particles are curled up in additional dimensions ala string theory. These dimensions are not expanding. Same problems as 1, with the added problem of being a bit unsatisfying.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The answer seems to be (from Marek in the previous question) that the gravitational force is so much weaker than the other forces that large (macro) objects move apart, but small (micro) objects stay together. However, this simple explanation seems to imply that expansion of space is a 'force' that can be overcome by a greater one. That doesn't sound right to me.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I think some of the problems in this question verge into metaphysics, but I think from his (?) previous answer, Marek can probably explain the physical side of things a bit more thoroughly. I will leave it there cos anything else I write sounds rambling!&lt;/p&gt;&#xA;" OwnerUserId="909" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-01-23T21:54:17.370" LastActivityDate="2011-02-03T00:40:43.280" Title="Why does space expansion not expand matter?" Tags="&lt;particle-physics&gt;&lt;astrophysics&gt;&lt;spacetime&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="2" />
  <row Id="2111" PostTypeId="1" AcceptedAnswerId="2138" CreationDate="2010-12-21T04:54:44.973" Score="22" ViewCount="1284" Body="&lt;p&gt;To suck water through a straw, you create a partial vacuum in your lungs.  Water rises through the straw until the pressure in the straw at the water level equals atmospheric pressure.  This corresponds to drinking water through a straw about ten meters long at maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By taping several straws together, a friend and I drank through a $3.07m$ straw.  I think we may have had some leaking preventing us going higher.  Also, we were about to empty the red cup into the straw completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/VCLN5.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is about what would happen if Superman were to drink through a straw by creating a complete vacuum in the straw.  The water would rise to ten meters in the steady state, but if he created the vacuum suddenly, would the water's inertia carry it higher?  What would the motion of water up the straw be?  What is the highest height he could drink from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignore thermodynamic effects like evaporation and assume the straw is stationary relative to the water and that there is no friction.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-12-21T10:01:35.043" LastActivityDate="2011-01-31T07:47:26.830" Title="How long a straw could Superman use?" Tags="&lt;fluid-dynamics&gt;&lt;pressure&gt;" AnswerCount="12" CommentCount="4" FavoriteCount="3" />
  <row Id="2112" PostTypeId="1" AcceptedAnswerId="2126" CreationDate="2010-12-21T09:40:32.757" Score="3" ViewCount="223" Body="&lt;p&gt;What is the cross section of the smallest object that can be detected with say &lt;a href=&quot;http://en.wikipedia.org/wiki/Visible_spectrum&quot; rel=&quot;nofollow&quot;&gt;visible light&lt;/a&gt; ($\lambda$ ~380 - 750 nm) or &lt;a href=&quot;http://en.wikipedia.org/wiki/X_band&quot; rel=&quot;nofollow&quot;&gt;X - band radar&lt;/a&gt; ($\lambda$ ~20 - 50 mm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the object need to have one side larger then $\frac{\lambda}{2}$ or something like that to get a reflection?&lt;/p&gt;&#xA;" OwnerUserId="250" LastEditorUserId="250" LastEditorDisplayName="" LastEditDate="2010-12-21T10:49:29.483" LastActivityDate="2010-12-27T22:06:44.957" Title="How large is the smallest object that can be detected at a given wavelength?" Tags="&lt;electromagnetic-radiation&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="2113" PostTypeId="2" ParentId="2111" CreationDate="2010-12-21T09:58:07.210" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;I must say that i was totally wrong! Of course Mark Eichenlaub and Omega Centauri are right!&lt;br&gt;&#xA;My apologies!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="552" LastEditorUserId="552" LastEditorDisplayName="" LastEditDate="2011-01-31T07:46:42.167" LastActivityDate="2011-01-31T07:46:42.167" CommentCount="27" />
  <row Id="2114" PostTypeId="1" CreationDate="2010-12-21T10:19:13.557" Score="1" ViewCount="188" Body="&lt;p&gt;I want to compare the much-talked about Cheeta running prosthesis to a the normal running process in terms of force and energy, but I don't know where to start. How would you start the comparison? A mathematical model would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="910" LastActivityDate="2011-09-01T12:38:49.460" Title="Cheetah prosthesis efficiency" Tags="&lt;mechanics&gt;&lt;mathematical-physics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="2115" PostTypeId="1" AcceptedAnswerId="2120" CreationDate="2010-12-21T12:20:45.230" Score="3" ViewCount="235" Body="&lt;p&gt;Is is possible to set a swing into oscillations without touching the ground? &#xA;This occurred to me while watching the &lt;a href=&quot;http://www.imdb.com/title/tt0383574/&quot; rel=&quot;nofollow&quot;&gt;second pirates movie&lt;/a&gt;. There is a scene where the ship's crew is suspended in a cage from a bridge in between two cliffs. They escape by swinging the cage towards one of the cliff. Is that even possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&#xA;From the answers, it is clear that it is possible to make the swing oscillate. Assuming the model Mark has &lt;a href=&quot;http://physics.stackexchange.com/questions/2115/oscillate-a-swing-without-touching-the-ground/2120#2120&quot;&gt;proposed&lt;/a&gt; would there be limits to much you can swing? Is it possible to quantify it?&lt;/p&gt;&#xA;" OwnerUserId="804" LastEditorUserId="804" LastEditorDisplayName="" LastEditDate="2010-12-22T08:30:42.237" LastActivityDate="2010-12-22T08:30:42.237" Title="Oscillate a swing without touching the ground" Tags="&lt;mechanics&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="2116" PostTypeId="2" ParentId="2026" CreationDate="2010-12-21T12:33:54.720" Score="1" ViewCount="" Body="&lt;p&gt;Jeff, holographic superconductivity/superfluidity is fairly simple conceptually and mechanically. The basic physics is breaking spontaneously an Abelian global symmetry. After you translate that to AdS language, this has to do with the existence of solutions to certain differential equations, with certain boundary conditions. The solutions can be constructed numerically (or in some cases analytically), but even without an explicit solution there are some arguments to demonstrate that such solutions exist. The first paper in the series by Horowitz, Hartnoll and Herzog may be a good one to follow.&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2010-12-21T12:33:54.720" CommentCount="1" />
  <row Id="2117" PostTypeId="2" ParentId="501" CreationDate="2010-12-21T13:02:21.663" Score="5" ViewCount="" Body="&lt;p&gt;So far the 2 highest rated answers appear to be looking at the case where $m$ is displaced in line with $M$ - $m$ - $M$, whereas the question asks what happens when the displacement is perpendicular to the connection line. I'll attempt to answer that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll assume that the rest position of the masses were all along the x-axis. So after the initial displacement, we have $M_1$ and $M_2$ at $(-L, 0)$, $(L, 0)$ respectively, and $m$ at (0, $d_0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By symmetry, $m$ must always remain at $x = 0$, so its position is $(0,d)$. Assuming small displacements, $M_1$ moves about $(-L, 0)$ by small amounts $x, y$, so its position is $(-L + x, y)$. Correspondingly, $M_2$ is always at $(L -x, y)$. Displacements $d, x, y$ are all functions of time $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the forces on the masses, we need to know the angle and extension of the spring at any given time. So $\theta, e$ as functions of $d, x, y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For small displacements, $Lsin\theta = d - y$ and $sin\theta \approx \theta$. So $$\theta = (d-y)/L$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find extension $e$, we start with the triangle $(-L+x, y), (0, y), (0, d)$ which has sides of length $$X = L-x$$ (parallel to x-axis), $$Y = d-y$$ (along y-axis) and $$H = L+e$$ (the hypotenuse).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $Hcos\theta = X$ and $cos\theta \approx 1 - \theta^2/2$ for small $\theta$. This gives $Lcos\theta + e cos\theta = L - x$, expanding to $L - L\frac{\theta^2}{2} + e -e\frac{\theta^2}{2} = L -x$. The $L$s cancel from both sides, and we can also ignore $e\frac{\theta^2}{2}$ as it is of order $small^3$, compared to the other values which are order $small$ ($L\frac{\theta^2}{2}$ is of order $small$ as it is $large * small^2$). This leaves us with $$e = L\frac{\theta^2}{2} - x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Forces&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force in each spring $F = ke$ allows us to write equations of motion for each mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m$ is the simplest, since we know motion is constricted to be along the y-axis by symmetry. Force on $m$ is twice the spring force (2 springs) applied at an angle of $\theta$, and in the opposite direction to the displacement $d$. So $$F_{m,y} = -2 (ke)sin\theta = -2k (\frac{(d-y)^2}{2L} -x) \frac{d-y}{L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us the acceleration of $m$ from $F = ma$, as $$\ddot{d} = -\frac{2k}{m} (\frac{(d-y)^2}{2L} -x) \frac{d-y}{L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, the forces on $M_1$, resolved into $x$ and $y$ components. $$F_{M_1,x} = (ke) cos\theta = k (\frac{(d-y)^2}{2L} -x) (1 - \frac{(d-y)^2}{2L^2})$$ Which gives us $$\ddot{x} = \frac{k}{M} (\frac{(d-y)^2}{2L} -x)(1 - \frac{(d-y)^2}{2L^2})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in the $y$-direction: $$F_{M_1,y} = (ke)sin\theta = k (\frac{(d-y)^2}{2L} -x) \frac{x-y}{L}$$ and so $$\ddot{y} = \frac{k}{M} (\frac{(d-y)^2}{2L} -x) \frac{x-y}{L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The forces acting upon, and motion of, $M_2$ is a reflection of $M_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sadly solving these equations is beyond my mathematical capability - would be very interested to see what the outcome is. The equations are simple enough that it should be trivial to solve numerically - if I have time over the next few days I'll try to write a simulation for it to report back on the behaviour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're going to have the central mass moving up and down, and the end masses moving elliptically (and symmetrically or course). I strongly suspect the motion will be chaotic - it looks similar in terms of degrees of freedom to a double pendulum.&lt;/p&gt;&#xA;" OwnerUserId="474" LastActivityDate="2010-12-21T13:02:21.663" />
  <row Id="2118" PostTypeId="2" ParentId="2115" CreationDate="2010-12-21T13:32:26.977" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;It is, because you are always touching the ground.&lt;/strong&gt; You are sitting on the bench, which is attached to the swing's &lt;em&gt;frame&lt;/em&gt;, which is touching the ground. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By pulling the chains the right way you can displace your centre of mass, which will cause the swing to swing. The reason you can displace your centre of mass is that you+swing are not an isolated system. The swing is attached to the ground, and so you can displace yourself without moving the swing.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you somehow managed to build a swing without attaching it to the ground one of the below would be true:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;it would be attached to something else, and the same thing would happen;&lt;/li&gt;&#xA;&lt;li&gt;it would be free-falling, and so it would not work as a swing (there would be no tension in the rope/chain);&lt;/li&gt;&#xA;&lt;li&gt;it would be on a gravity-less system, and it would be identical to the free-fall case.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; &lt;em&gt;(idea inspired by Mark's answer)&lt;/em&gt;&lt;br&gt;&#xA;The third possibility would be to build a swing that rests on the ground without being attached to it. The swing could be built on top of a huge bowl that rests on the ground. In that case the answer is still &lt;strong&gt;yes, you can&lt;/strong&gt;.   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If there is friction between the swing/bowl and the ground, then it's basically the same case at the beginning of my answer. Having friction would be the same as being attached to the ground. &lt;/li&gt;&#xA;&lt;li&gt;If there is no friction, then you still can. When you pull yourself to the right (by pushing your hands against the swing itself) the swing will go left. That will lead to a complicated movement, which will depend a lot on how your swing rests on the ground.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="724" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2010-12-21T23:04:02.507" LastActivityDate="2010-12-21T23:04:02.507" CommentCount="5" />
  <row Id="2119" PostTypeId="1" AcceptedAnswerId="2128" CreationDate="2010-12-21T13:40:22.190" Score="4" ViewCount="533" Body="&lt;p&gt;Consider an infinite square grid, where each side of a square is a spring following Hooke's law, with spring constant k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the relation between the force and displacement between two points?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they are proportional;&lt;br&gt;&#xA;What is the equivalent spring constant between the origo and the point (x,y) (integers) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&#xA;I also want to know this;&lt;br&gt;&#xA;Suppose you make the springs so small that this can be treated as a continuos sheet.&lt;br&gt;&#xA;At what speed will a wave propagate ?&lt;br&gt;&#xA;(A wave starting as an initial displacement perpendicular to the sheet.)    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given some inital state;&lt;br&gt;&#xA;Is there an equation for the time-evolution of the continuous sheet?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2:&#xA;Suppose there is a mass at every node, and that its (x,y)-coordinates is fixed, it only vibrates out of the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider then that we take the continuous limit, such that we get a 2D membrane of mass density µ. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is the membrane isotropic ?&lt;/li&gt;&#xA;&lt;li&gt;Consider that we used another tiling (like hexagonal), before taking the continuous limit, will this sheet behave the same way?&lt;/li&gt;&#xA;&lt;li&gt;If not, but they are both isotropic, how does one characterize their difference, can they be made to behave the same way by changing the spring constant k?&lt;/li&gt;&#xA;&lt;li&gt;What is the equation of motion for the square sheet with spring constant k?&lt;/li&gt;&#xA;&lt;li&gt;What is the equation of motion for the square sheet if the springs obey a generalized Force law, F=k*x^n, where n is a variable.&lt;/li&gt;&#xA;&lt;li&gt;What is the equation of motion for a 3D cubic grid?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I am particularly interested in answer to 1., 2. and 3.&#xA;I dont expect anyone to answer all these and will also accept an answer which does not explain anything but simply provides a good reference.&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2010-12-22T18:33:59.203" LastActivityDate="2010-12-30T16:47:08.840" Title="Equivalent spring-constant for infinite square grid of springs" Tags="&lt;classical-mechanics&gt;&lt;mechanics&gt;" AnswerCount="3" CommentCount="17" />
  <row Id="2120" PostTypeId="2" ParentId="2115" CreationDate="2010-12-21T13:47:38.880" Score="2" ViewCount="" Body="&lt;p&gt;Yes, you can start swinging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a swing in a two-dimensional world.  The swing consists of a stiff rod hanging down from a ceiling rigidly attached to a bowl with a ball inside.  The rod can swing freely about its connection point to the ceiling, so it's a pendulum.  The bowl is a section of a circle whose radius of curvature equals the length of the rod.  The ball slides frictionlessly inside the bowl, except that it can exert a force on the bowl at any time if it wants to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for choosing this system is that as the bowl swings, the physical space it occupies doesn't change much.  A swinging motion is as if a little bit of the bowl gets lopped off one side and pasted on to the other.  This means that from the ball's point of view, as long as all oscillations remain small, the bowl is essentially stationary, and the ball oscillates like a pendulum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bowl itself also oscillates like a pendulum, but it does so at a different frequency.  If you write down the equations of motion for the system, you find that the ball oscillates with angular frequency $\omega_0 = \sqrt{g/l}$, but the bowl oscillates with angular frequency $\omega_0 \sqrt{\sin\theta_0/\theta_0}$, with $\theta_0$ half the angular size of the bowl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we start from the equilibrium position with the bowl and ball at rest at the bottom of the swing.  If the ball momentarily exerts a force on the bowl, they will begin oscillating in different directions.  Because they have different frequencies, they will eventually both be displaced in the same direction.  When that happens, have the ball latch on to the bowl and cease all motion relative to the bowl.  At this point you have a swing that's displaced from equilibrium - you're swinging.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-12-21T13:47:38.880" CommentCount="2" />
  <row Id="2121" PostTypeId="2" ParentId="1230" CreationDate="2010-12-21T14:30:21.547" Score="2" ViewCount="" Body="&lt;p&gt;Basically, there's no reason why we couldn't redefine the mole as as simple integer number of atoms or molecules. In fact, as other users have mentioned, there's a lot of people who'd like to do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, that's not how chemists actually use the mole in practice. You simply can't count 6×10^23 atoms or molecules, nor do you need to. What is important for chemists is to know (for example) that there are the same number of atoms in 58 grams of iron as in 12 grams of carbon, and so on for all the other elements. It's not important to know exactly what that number is, just that it's the same number, and, for much of the history of chemistry, we had absolutely no idea what the number was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should point out as well that the mole is not the only unit which is an &quot;integer&quot;. If you take the coulomb as a unit of electric charge, that should be equal to an integer number of elementary charges, shouldn't it? Actually, it's not, for historical reasons, and there doesn't seem to be much enthusiasm for making it an integer number of elementary charges either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find my paper discussing this in more technical terms at &lt;a href=&quot;http://precedings.nature.com/documents/5138/version/1&quot; rel=&quot;nofollow&quot;&gt;http://precedings.nature.com/documents/5138/version/1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="915" LastActivityDate="2010-12-21T14:30:21.547" />
  <row Id="2122" PostTypeId="2" ParentId="2115" CreationDate="2010-12-21T15:22:55.527" Score="1" ViewCount="" Body="&lt;p&gt;This reminds me of one of my favorite physics articles, called &quot;Swimming in space-time&quot;, which shows that in any space-time metric with intrinsic curvature, simple periodic oscillations in isolation can lead to translational motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencemag.org/content/299/5614/1865.full&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencemag.org/content/299/5614/1865.full&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0510054&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/gr-qc/0510054&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0612131&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/gr-qc/0612131&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2010-12-21T15:22:55.527" CommentCount="6" />
  <row Id="2123" PostTypeId="2" ParentId="2111" CreationDate="2010-12-21T15:32:28.770" Score="0" ViewCount="" Body="&lt;p&gt;Unless I'm missing something, this is simply the height of a water-based barometer, since it is really the atmospheric pressure that is pushing the water up the straw. at STP, the answer is 33 1/2 feet. If he were sucking Hg up the straw (not recommended for non-superheros!), the height would be ~30 inches.&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2010-12-21T15:32:28.770" CommentCount="2" />
  <row Id="2124" PostTypeId="2" ParentId="501" CreationDate="2010-12-21T15:38:01.900" Score="0" ViewCount="" Body="&lt;p&gt;I happen to have a 2d dynamics simulator that I hoped my game programming kids might expand into a game physics engine. In any case it took about a minute to set up the problem:&#xA;*timestep&#xA;$       dt  ncyc&#xD;&#xA;       .01    25&#xD;&#xA;*nodes&#xD;&#xA;$ node               x               y            mass              vx              vy&#xA;     0             -1.             0.0            1.e0           .0000&#xA;     1             -.0             0.1              .1           .0000&#xA;     2              1.             0.0            1.e0           .0000&#xA;*elements&#xA;$  nel    n1    n2    n3            l0              st            damp&#xA;     0     0     1                 1.0              1.           0.000&#xA;     1     1     2                 1.0              1.           0.000&#xA;*history_nodes&#xA;     0&#xA;     1&#xA;*end&#xA;   You should note that there is some nonlinearity in the system, as the geometric relation between the vertical displacement changes the spring length nonlinearly (except for small oscillations). The bending mode also nonlinearly couples to the strictly horizontal mode, and the frequency of the later mode is several times higher than for the bending mode. So&#xA;you get both modes excited, although mostly the vertical one.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-12-21T15:38:01.900" CommentCount="1" />
  <row Id="2125" PostTypeId="2" ParentId="2112" CreationDate="2010-12-21T15:39:06.940" Score="5" ViewCount="" Body="&lt;p&gt;Seems like in principle, there should be no minimum size. For example, a small hand-held radio (cell phone for those of you in the 21st century) can detect, generate, and scatter EM waves with wavelength much longer than the object itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For optical and X-ray scales, one would probably be limited by quantum or thermal effects, which may in principle be mitigated at very low temperatures. But even so, a single atom is MUCH MUCH smaller than 500nm, and can absorb and emit light at that wavelength.&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2010-12-21T15:39:06.940" CommentCount="4" />
  <row Id="2126" PostTypeId="2" ParentId="2112" CreationDate="2010-12-21T16:51:42.303" Score="0" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Angular_resolution&quot; rel=&quot;nofollow&quot;&gt;Rayleigh criterion&lt;/a&gt; determines the &lt;strike&gt;maximum&lt;/strike&gt; minimum size of objects - in terms of the &lt;strike&gt;maximum&lt;/strike&gt; minimum angular resolution $\theta$ - which can be resolved at a give wavelength $\lambda$ and with a lens of a given diameter $D$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ sin(\theta) = 1.22 \frac{\lambda}{D} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spatial resolution is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta l = 1.22 \frac{f\lambda}{D} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f$ is the focal length. For visible light this gives a limit of $\sim $ 200 nm for the minimum resolvable distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is all based on classical optics. With newer metamaterial based lenses one can do much better than the Rayleigh limit.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-22T21:36:00.190" CommentCount="7" />
  <row Id="2127" PostTypeId="2" ParentId="1603" CreationDate="2010-12-21T17:03:33.823" Score="3" ViewCount="" Body="&lt;p&gt;Marek and Eric have given good answers. I think many particle physicists first encountered homotopy theory in the context of magnetic monopoles. Take the Standard Model gauge group $H=SU(3) \times SU(2) \times U(1)$ and embed it into a GUT gauge group $G$ such as&#xA;$SU(5)$ or $SO(10)$. Assuming there is no accidental degeneracy, the space of minima of the symmetry breaking potential is the coset $G/H$. Static, finite energy configurations must&#xA;approach a point in $G/H$ at spatial infinity and so are classified by $\pi_2(G/H)$ which&#xA;equals $\pi_1(H)$ provided that $\pi_1(G)= 0$. Since $\pi_1(H)=\mathbb{Z}$ there is an integer valued topological charge for these configurations which turns out to be magnetic monopole charge. Sidney Colemans lectures ( &lt;a href=&quot;http://ccdb4fs.kek.jp/cgi-bin/img/allpdf?198211084&quot; rel=&quot;nofollow&quot;&gt;http://ccdb4fs.kek.jp/cgi-bin/img/allpdf?198211084&lt;/a&gt; ) explain this in much more detail. Condensed matter systems have a much wider range of &quot;Higgs&quot; fields (i.e. order parameters) and so have much more interesting and complicated symmetry breaking patterns and a much richer classification of topological defects by homotopy groups. Mermin has a very nice review here: &lt;a href=&quot;http://rmp.aps.org/abstract/RMP/v51/i3/p591_1&quot; rel=&quot;nofollow&quot;&gt;http://rmp.aps.org/abstract/RMP/v51/i3/p591_1&lt;/a&gt; .&lt;/p&gt;&#xA;" OwnerUserId="838" LastActivityDate="2010-12-21T17:03:33.823" />
  <row Id="2128" PostTypeId="2" ParentId="2119" CreationDate="2010-12-21T19:06:42.697" Score="1" ViewCount="" Body="&lt;p&gt;I'll answer only the third one (for now at least); the movement with limit to small vertical oscillations will be governed by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Vibrations_of_a_circular_drum&quot; rel=&quot;nofollow&quot;&gt;drum equation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ddot{s}(x,y)=c^2 \nabla^2 s(x,y)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $s(x,y)$ is a vertical displacement in point $(x,y)$ and $c$ is the weave speed; using dimensional analysis I would say that $c\sim\sqrt{\frac{k}{\sigma}}$, where $\sigma$ is the mass density. Of course everything is getting much more complex with larger amplitudes.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-21T22:46:28.630" LastActivityDate="2010-12-21T22:46:28.630" CommentCount="9" />
  <row Id="2129" PostTypeId="2" ParentId="2119" CreationDate="2010-12-21T19:32:13.013" Score="0" ViewCount="" Body="&lt;p&gt;If on every node of the grid you have a small mass, then you have a model for a two dimensional solid. That would behave like a two dimensional membrane. The equation of motion for every disturbance would be a wave equation. &#xA;In the case of an one-dimensional grid, the wave velocity for such a wave would be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c^2=\frac{kl^2}{m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where l is the distance between two neighbouring masses. In the case of the two dimensional grid, you will probably also have a geometric factor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Regarding the last edit, if you have a tension that characterizes the membrane, then the velocity is the square root of the tension over the mass density.&#xA;So the geometry of the thing would play some part both to the tension and the mass density. That is because if you change the shape of the cell, then you assign different surface for every mass and you also assign a different number of brunches with springs to every node thus changing the effective spring constant. These are my qualitative guesses.&lt;/p&gt;&#xA;" OwnerUserId="383" LastEditorUserId="383" LastEditorDisplayName="" LastEditDate="2010-12-23T00:22:56.263" LastActivityDate="2010-12-23T00:22:56.263" CommentCount="12" />
  <row Id="2130" PostTypeId="2" ParentId="501" CreationDate="2010-12-21T19:45:21.210" Score="0" ViewCount="" Body="&lt;p&gt;Now that I've had time to properly think about it. The problem has multiple zero energy modes, and at least one of these is excited by the proposed initialization. But to cut to the chase:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give the initial configuration masses at x=-1,0,1. This is the equlibrium, so the springs at length=1 have no tension. Now we have three rigid body modes: translation in the X or Y directions. These are zero-energy modes, as the only energy is kinetic which scales as Vsquared, not as V. Similarly we have rotation about the Z axis. This is also a zero energy mode. By symmetry we aren't exiting any of these modes. Then consider the mode represented by&#xA;the center mode moving up, and the end modes moving downwards. An infinitessimal perturbation produces no internal energy, (spring tension is sqrt(1+deltaY**2)-1, which is quadratic in&#xA;deltaY, so this is also zero energy mode. So we could have one endnode rotationg counterclockwise, and one rotating clockwise, with no restoring forces to oppose this motion. Initial problem symmetry does not rule out this mode. We have two modes with only x displacements, a symmetric one, where the center mass is fixed and the end nodes move in and out. We also have a mode where the center particles moves right and left, this modeshape would be antisymmetric in x. We should have a sixth mode (because 3 2D particles have six degrees of freedom), but I can't seem to envisage it (alhough&#xA;an eigenanalysis should reveal it). In any case at least one of these zero energy modes is excited by the initial perturbation. And the restoring forces are nonlinear (the square or great power of the perturbation), so interesting nonlinear behavior will result.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2010-12-21T19:45:21.210" />
  <row Id="2131" PostTypeId="1" AcceptedAnswerId="2135" CreationDate="2010-12-21T20:24:19.293" Score="6" ViewCount="199" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dark_energy&quot; rel=&quot;nofollow&quot;&gt;Dark energy&lt;/a&gt; is introduced as a constant inside Einstein's equations. Its primary purpose, from what I understand, is to make Einstein's equations compatible with the &lt;em&gt;accelerating&lt;/em&gt; expansion of the universe. As a consequence, of the &quot;predictions&quot; of dark energy is the expansion of the universe according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble_expansion&quot; rel=&quot;nofollow&quot;&gt;Hubble's law&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there are numerous experiments that verify this expansion (and its acceleration), and thus indirectly support the dark energy theory. My question is:&lt;br&gt;&#xA;&lt;strong&gt;Are there &lt;em&gt;other&lt;/em&gt; factors that give credit to the existence of dark energy?&lt;/strong&gt;&lt;br&gt;&#xA;Are there any experiments that support this theory, but not only through the verification of Hubble expansion?&lt;/p&gt;&#xA;" OwnerUserId="724" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2010-12-21T20:48:29.070" LastActivityDate="2011-07-10T02:53:07.133" Title="What experiments, other than Hubble Expansion, support the Dark Energy theory?" Tags="&lt;general-relativity&gt;&lt;astrophysics&gt;&lt;experimental-physics&gt;&lt;dark-energy&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="1" />
  <row Id="2132" PostTypeId="2" ParentId="2114" CreationDate="2010-12-21T20:44:32.277" Score="0" ViewCount="" Body="&lt;p&gt;As mentioned in comments, the question isn't very well-defined, but I'll seed some ideas for basis of comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Force&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Put a series of pressure transducers on the sole of a shoe &amp;amp; on the bottom of the Cheeta&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Measure the distribution/duration of force. &lt;/li&gt;&#xA;&lt;li&gt;Compare the amount of force that gets translated into forward momentum.&lt;/li&gt;&#xA;&lt;li&gt;Compare the duration of force, see if one exerts horizontal force longer.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Measure the strength of various muscles on both runners, look for discrepancy in muscles used.&lt;/li&gt;&#xA;&lt;li&gt;Measure displacement of the body of the Cheeta, calculate stress, compare to stress in joints.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Energy&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Compare runner's speeds under different loading conditions (50 lbs. backpack, etc.)&lt;/li&gt;&#xA;&lt;li&gt;Compare runner's speeds up/down hills.&lt;/li&gt;&#xA;&lt;li&gt;Hook runners up to oxygen consumption monitors, measure VO2max,  calculate energy expenditure while running.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Just some ideas, any of which would provide interesting basis for comparison in terms of &quot;force and energy.&quot;&lt;/p&gt;&#xA;" OwnerUserId="834" LastActivityDate="2010-12-21T20:44:32.277" />
  <row Id="2133" PostTypeId="2" ParentId="2111" CreationDate="2010-12-21T20:55:57.700" Score="4" ViewCount="" Body="&lt;p&gt;I went back and took a more careful look at this. I'm still not convinced it's correct, but I'm hoping this is at least better than what I had before.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let $h$ be the height of the column of water inside the straw. As this height rises by an amount $\delta h$, the work done on the column is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta W = P A \delta h$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ is the cross-sectional area. The change in potential energy is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta U = \rho A \delta h g h$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so by conservation of energy,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta K = \delta W - \delta U = \left(P - \rho g h\right) A \delta h$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This excess kinetic energy comes from two contributions: the added mass,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta K_1 = \frac{1}{2}mv^2 = \frac{1}{2}(\rho A \delta h)\dot{h}^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and any change in speed of the column of water,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta K_2 = \frac{1}{2}m(2v\delta v) = (\rho A h)\dot{h} \delta\dot{h}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting it all together, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P A \delta h - \rho A g h \delta h = \frac{1}{2}\rho A \dot{h}^2 \delta h + \rho A h\dot{h}\delta\dot{h}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you assume (or prove) that $P$ is dependent on $h$ through Bernoulli's theorem,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P + \frac{1}{2}\rho\dot{h}^2 = P_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting in (and canceling the common factor of $A$), you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_0 \delta h - \rho g h \delta h = \rho \dot{h}^2 \delta h + \rho h \dot{h}\delta\dot{h}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which at least accounts for the mysterious factor of $\frac{1}{2}$ that appeared in previous versions of my answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I don't think we can simply assume that $\delta h \neq 0$ and divide it out, because if we do that, we get a factor of $\frac{\delta\dot{h}}{\delta h}$ which is undefined at the initial and maximum heights. (Roughly speaking, the variation $\delta h$ is second-order at those points whereas the variation $\delta\dot{h}$ is still first-order.) Instead, I'll divide by $\delta t$, which certainly should not produce any singularities, to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_0 \dot{h} - \rho g h \dot{h} = \rho \dot{h}^3 + \rho h \dot{h}\ddot{h}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the initial and maximum heights, $\dot{h} = 0$, so the equation is trivially satisfied there. But consider the situation when displaced from either initial or maximum height by an arbitrarily small amount, such that $\dot{h}\neq 0$. Here we can cancel out $\dot{h}$ to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_0 - \rho g h = \rho \dot{h}^2 + \rho h \ddot{h}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\dot{h}$ will be infinitesimally small around the maximum height, we can neglect the first term on the right, but not the second. So we're left with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_0 - \rho g h = \rho h \ddot{h}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this agrees with a simple analysis using Newton's second law. (The forces acting on the column of water at its maximum height are the pressure force $P_0 A$ acting upwards and gravity $\rho Ahg$ acting downwards, and the difference is equal to $ma = \rho Ah\ddot{h}$.) So the differential equation passes at least one basic consistency test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, this equation no longer admits the solution $h = \frac{P_0}{\rho g}$. Instead we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h = \frac{P_0}{\rho (g + \ddot{h})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I can't think of a way to determine $\ddot{h}$ at maximum without solving the equation, so for now I'm limited to a numerical solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a quick estimation, I plugged the full differential equation from above into Mathematica's &lt;code&gt;NDSolve&lt;/code&gt; function. With boundary conditions $h(0) = 0$ and $\dot{h}(0) = 0$, it complained about undefined expressions, so I used boundary conditions at a nonzero time, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h(\epsilon_t) = \epsilon_h$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{h}(\epsilon_t) = \epsilon_{\dot{h}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for values of the various $\epsilon$ constants ranging from $10^{-3}$ to $10^{-8}$. In my tests, I get this graph, seemingly independently of the values of $\{\epsilon\}$ or the ratios between them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/HHHwY.png&quot; alt=&quot;Graph of solution&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematica indicates that the graph peaks at $15.5\,\mathrm{m}$, so if this analysis is correct, that would be the maximum height. (FWIW I am still &lt;em&gt;very&lt;/em&gt; suspicious of this calculation though)&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-22T05:37:30.267" LastActivityDate="2010-12-22T05:37:30.267" CommentCount="13" />
  <row Id="2135" PostTypeId="2" ParentId="2131" CreationDate="2010-12-21T21:27:15.080" Score="3" ViewCount="" Body="&lt;p&gt;I think the strongest evidence comes from the CMB fluctuations, namely the location of the first acoustic peak. This gives the overall geometry of the Universe ($\Omega_{tot}=1$; the Universe is flat). Then with a multitude of observations of dark &lt;em&gt;matter&lt;/em&gt; (e.g., galaxy cluster counts, large-scale structure, and weak lensing) to get $\Omega_{matter}=0.3$, we are left with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Omega_{\Lambda}=\Omega_{tot}-\Omega_{matter}=0.7$$&lt;/p&gt;&#xA;" OwnerUserId="775" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-21T22:04:20.130" LastActivityDate="2010-12-21T22:04:20.130" CommentCount="5" />
  <row Id="2136" PostTypeId="2" ParentId="2131" CreationDate="2010-12-21T21:33:34.710" Score="2" ViewCount="" Body="&lt;p&gt;No, there isn't much beyond the acceleration parameter of the universe to support DE.  In fact, if you're willing to abandon homogenity and isotropy, you can even get away &lt;a href=&quot;http://blogs.discovermagazine.com/cosmicvariance/2010/07/29/are-you-the-center-of-the-universe/&quot; rel=&quot;nofollow&quot;&gt;without DE&lt;/a&gt; by choosing a void model, where you replace a fine tuning of the matter distribution with a fine tuning of the dependence of the density on radius from the 'center of the universe'.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-12-21T21:33:34.710" CommentCount="3" />
  <row Id="2137" PostTypeId="2" ParentId="2111" CreationDate="2010-12-21T22:19:21.623" Score="6" ViewCount="" Body="&lt;p&gt;I have an argument that the water in the straw will rise to twice the equilibrium height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;David and Martin's answers consider the system of water in the straw.  I will consider the system of the water in the straw plus the water in the reservoir.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As water goes into the straw, the water level in the reservoir drops, and the atmosphere does work on the system.  If a volume $V$ of water enters the straw, the work done on the system is $PV$, with $P$ the atmospheric pressure.  Assume that the reservoir has a large surface area so that the level the reservoir drops is negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the water is at its peak in the straw, the kinetic energy of the system is zero, so the potential energy is $PV$.  The potential energy is also $\rho g V h/2$.  So the maximum height of the water is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h = \frac{2P}{\rho g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer is different from Martin and David's.  I think this might be because when the water starts moving, the pressure at the entrance to the straw may not be $P$ any more.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-12-22T11:16:59.903" LastActivityDate="2010-12-22T11:16:59.903" CommentCount="9" />
  <row Id="2138" PostTypeId="2" ParentId="2111" CreationDate="2010-12-21T22:20:54.607" Score="9" ViewCount="" Body="&lt;p&gt;I think we can most easily consider the problem from the perspective of energy. For a unit area column the external energy put in equals the volume of the column times the air pressure. The gravitational energy is the mass of water raised times the average height of the water. So&#xA;as we pull the water up, we the water is gaining kinetic energy until the water reaches the static limit (roughly 10meters), but at this point the average water in the tube has only risen by half that amount, so the rest is kinetic energy of (upward) water motion. So the vacuum energy in dimensionless units =h, while the gravitational energy is h**2/2. The solution is h=2.&#xA;When we reach 2 times the static limit (20meters) then the gravitational energy in the water matches the &quot;vacuum&quot; energy we put in, so that would represent the high point of the oscillation. So I think we would get 2times the static limit. The water velocity will be messy to solve for, as the amount of water moving in the column depends upon height, so just look only at net energy as a function of water height... Of course he will only get a sip of water, then the column would start to fall........&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2010-12-21T22:20:54.607" CommentCount="10" />
  <row Id="2139" PostTypeId="2" ParentId="2111" CreationDate="2010-12-21T23:17:00.557" Score="2" ViewCount="" Body="&lt;p&gt;If I follow up on keenan pepper's suggestion, if the water is deep, and especially if you can mess with the topology of the straw you can go to almost unlimited height!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a straw that is stuck very deeply into the ocean. Then coil the straw around at great depth many many times. I then blow very hard (I am superman afterall), and create a huge volume of airfilled straw at great depth. This configuration has a great deal of potential energy, so if we simply stop blowing we have the pressure at the great depth of the bottom of the straw accelerating water into and up the straw. Since by coiling the straw at great depth I can obtain an unlimited ratio of volume of the straw underwater to volume above water, the energy analysis allows me to reach an unlimited height. So the issue becomes if there is some other sort of limit. Can we get cavitation of water trying to enter the straw or something if the velocity gets too high? But, in any case you should be able to get really high, tens or hundreds of times the static limit, by preenergizing the system in this way. In the real world friction will limit how far you can take it.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2010-12-21T23:17:00.557" CommentCount="4" />
  <row Id="2140" PostTypeId="2" ParentId="2110" CreationDate="2010-12-22T00:24:44.010" Score="3" ViewCount="" Body="&lt;p&gt;Let's talk about the balloon first because it provides a pretty good model for the expanding universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's true that if you draw a big circle then it will quickly expand as you blow into the balloon. Actually, the apparent speed with which two of the points on the circle in a distance $D$ of each other would move relative to each other will be $v = H_0 D$ where $H_0$ is the speed the balloon itself is expanding. This simple relation is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble%27s_law&quot; rel=&quot;nofollow&quot;&gt;Hubble's law&lt;/a&gt; and $H_0$ is the famous &lt;em&gt;Hubble constant&lt;/em&gt;. The moral of this story is that the expansion effect is dependent on the distance between objects and really only apparent for the space-time on the biggest scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, this is only part of the full picture because even on small distances objects should expand (just slower). Let us consider galaxies for the moment. According to wikipedia, $H_0 \approx 70\, {\rm km \cdot s^{-1} \cdot {Mpc}^-1}$ so for Milky way which has a diameter of $D \approx 30\, {\rm kPc}$ this would give $v \approx 2\,{\rm km \cdot s^{-1}}$. You can see that the effect is not terribly big but the given enough time, our galaxy should grow. But it doesn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand why, we have to remember that space expansion isn't the only important thing that happens in our universe. There are other forces like electromagnetism. But most importantly, we have forgotten about good old Newtonian gravity that holds big massive objects together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see, when equations of space-time expansion are derived, nothing of the above is taken into account because all of it is negligible on the macroscopic scale. One assumes that universe is a homogenous fluid where microscopic fluid particles are the size of the galaxies (it takes some getting used to to think about &lt;a href=&quot;http://htwins.net/scale/&quot; rel=&quot;nofollow&quot;&gt;galaxies as being microscopic&lt;/a&gt;). So it shouldn't be surprising that this model doesn't tell us anything about the stability of galaxies; not to mention planets, houses or tables. And conversely, when investigating stability of objects you don't really need to account for space-time expansion unless you get to the scale of galaxies and even there the effect isn't that big.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-22T00:24:44.010" CommentCount="9" />
  <row Id="2141" PostTypeId="2" ParentId="2110" CreationDate="2010-12-22T02:03:40.770" Score="3" ViewCount="" Body="&lt;p&gt;This problem has been extensively studied. These are some references that I think answer various aspects of your question. I would say that it is not exactly a solved problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0508052/&quot; rel=&quot;nofollow&quot;&gt;In an expanding universe, what doesn't expand?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1088/0264-9381/16/4/020&quot; rel=&quot;nofollow&quot;&gt;Size of a hydrogen atom in the expanding universe&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://prl.aps.org/abstract/PRL/v12/i15/p435_1&quot; rel=&quot;nofollow&quot;&gt;Evolution of the Solar System and the Expansion of the Universe&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://prl.aps.org/abstract/PRL/v75/i20/p3602_1&quot; rel=&quot;nofollow&quot;&gt;Multiparticle Dynamics in an Expanding Universe&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/astro-ph/9902189&quot; rel=&quot;nofollow&quot;&gt;Life, The Universe, and Nothing: Life and Death in an Ever-Expanding Universe&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; This review (Rev.Mod.Phys.) is free from arXiv, &lt;a href=&quot;http://arxiv.org/abs/0810.2712&quot; rel=&quot;nofollow&quot;&gt;Influence of global cosmological expansion on local dynamics and kinematics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="383" LastEditorUserId="383" LastEditorDisplayName="" LastEditDate="2010-12-22T14:06:44.587" LastActivityDate="2010-12-22T14:06:44.587" CommentCount="1" />
  <row Id="2142" PostTypeId="2" ParentId="2111" CreationDate="2010-12-22T02:51:32.443" Score="0" ViewCount="" Body="&lt;p&gt;If you immerse the staw in 10m of water, while holding the end closed, and then release it, the water will accelerate upwards past the level point and it will overshoot up to a height of about 6m. No sucking needed. Add some suction to this and you can go higher. A square section straw should minimize the friction loss allowing for better &quot;spring back&quot;. Overall you can help the suction, by immersing the straw deeper and deeper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that it takes work do immerse the straw (displacing the water) and that is the energy conveted into ponetial energy that allows the water to rise. Close the end when the water reaches the maximum height and you can measure how high you can reach.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-12-22T02:51:32.443" CommentCount="7" />
  <row Id="2143" PostTypeId="2" ParentId="2119" CreationDate="2010-12-22T03:04:29.150" Score="1" ViewCount="" Body="&lt;p&gt;I stick to the first question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only do small displacements, and the two points are along the same line of springs then the effective spring rate is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ k_{eff} = \frac{k}{N} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $N$ is the number of springs between the points. Why? Well split the problem like this&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(inf)---[k_out]---(A)---[k_in]---(B)---[k_out]---(inf)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where &lt;code&gt;(A)&lt;/code&gt; and &lt;code&gt;(B)&lt;/code&gt; are the two points, and the springs are replaces with the effective springs &lt;code&gt;[k_out]&lt;/code&gt; between the points and infinity, and &lt;code&gt;[k_in]&lt;/code&gt; between the two points. The formula for springs in series is $\frac{1}{k_{eff}} = \frac{1}{k_1}+\frac{1}{k_2}+\ldots$, or $k_{eff}=k/N$ if all the springs have the same rate. So &lt;code&gt;[k_out]&lt;/code&gt; is zero because $N=\infty$ and whats left to consider is only the springs in-between the points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the springs out of the line of the points are un-important for small displacements because they only contribute higher order non-linearities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Completely different equations are needed for the continious sheet. The wave speed has to do with the mass/density of the sheet also, not just the elasticity and the sitffness.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-12-22T03:04:29.150" />
  <row Id="2144" PostTypeId="1" CreationDate="2010-12-22T05:40:01.540" Score="2" ViewCount="177" Body="&lt;p&gt;Can someone please explain what a potential is? Example. velocity potential in ideal flows, acoustic potential (gradient of which gives the particle velocity in a sound wave).&#xA;Whenever I see potential functions complex analysis is applied to compute integrals, complex functions for conformal mapping. I vaguely understand that potential functions are independent of path and in complex plane the integration around a contour is independent of path as well (the integration depends only on the end points, hence in a closed curve say region Z, if function, f is analytic then integration of f over Z is 0). Can you please explain on this?&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2010-12-22T22:20:21.983" Title="Potential functions" Tags="&lt;mathematical-physics&gt;&lt;fluid-dynamics&gt;&lt;acoustics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="2145" PostTypeId="2" ParentId="2115" CreationDate="2010-12-22T05:41:44.520" Score="1" ViewCount="" Body="&lt;p&gt;Yes. It is certainly possible to make a swing oscillate without touching the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important note (particularly with respect to some of the other answers that have been posted) is that I'm talking about periodic changes in the state of the system, not necessarily motion of the center of mass of the whole system. In other words, a swing-set which is not coupled to the ground (I'll say it is on a frictionless surface, for example) can oscillate, although the frame will move opposite the swinging child such that the CoM of the whole system is stationary. For the CoM to move, the swing-set does indeed need some coupling to the ground, be it frictional, rigid, or other, but in either case there may be oscillations and the system can posses non-zero kinetic and potential energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, from a conservation of energy argument, there is no contradiction. Before even considering the system dynamics and its equation of motion it is obvious that if the child on the swing moves around he/she will influence the state of the swing system. If you are having conceptual issues because no force is being exerted on the ground, and it therefore seems that no work is being done, consider that the child represents a bank of stored energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the question is whether the child on the swing can initiate oscillations without touching the ground. Clearly, yes he can. By shifting his CoM, the angle of the swing is forced from its equilibrium, which means the system will now oscillate. For the swing-set on a frictionless surface these oscillations will include the counter-motion of the frame, but they will be oscillations none the less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, once oscillations are initiated, they can be amplified without the child exerting any force along the direction of his motion. Simply by raising and lowering his center of mass along the direction towards the swing's pivot, the oscillations can be amplified via parametric excitation, on which wikipedia has a good page.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2010-12-22T05:41:44.520" />
  <row Id="2146" PostTypeId="2" ParentId="2144" CreationDate="2010-12-22T06:34:18.773" Score="1" ViewCount="" Body="&lt;p&gt;Your question needs improvement, but I'll try to help with an answer. You're not really asking one thing specifically, so I'll try clarify a few points.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The word &quot;potential&quot; doesn't necessarily mean anything it needs to be part of an expression.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Holomorphic_function&quot; rel=&quot;nofollow&quot;&gt;Holomorphic functions&lt;/a&gt;, which you usually study in complex analysis, and conformal mapping is a common way of find the static electric potential for complicated geometries.&lt;/li&gt;&#xA;&lt;li&gt;The contour integration of a vector field that's the gradient of a scalar function (called the potential) does not depend on the shape of the curve, only on its end points. &lt;/li&gt;&#xA;&lt;li&gt;Countour integration of a scalar analytic function over a closed curve on the complex plane also does not depend on the shape of the curve. &lt;/li&gt;&#xA;&lt;li&gt;The velocity vector in fluid dynamics is a vector field. Sometimes, it is convenient to define this vector field as the gradient of a scalar field, which &lt;em&gt;might&lt;/em&gt; be called the velocity potential (I'm not sure).&lt;/li&gt;&#xA;&lt;li&gt;Acoustic waves are waves in the vector field of a fluid, so the same as the above applies.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you wanted a more in depth explanation on one of these, please specify &lt;em&gt;what&lt;/em&gt; you want to know. &lt;/p&gt;&#xA;" OwnerUserId="724" LastActivityDate="2010-12-22T06:34:18.773" />
  <row Id="2147" PostTypeId="1" AcceptedAnswerId="2149" CreationDate="2010-12-22T07:59:59.237" Score="5" ViewCount="176" Body="&lt;p&gt;This is not another question about faster-than-light travel or superluminal communication. I totally appreciate the speed limit capped by physical laws (or theories.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just curious, since there is no limit for phase velocity, is there any way to generate a wave with controllable phase velocity higher than speed of light?&lt;/p&gt;&#xA;" OwnerUserId="749" LastActivityDate="2011-03-09T03:19:20.887" Title="Controllable faster-than-light phase velocity" Tags="&lt;electromagnetism&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="2148" PostTypeId="1" AcceptedAnswerId="2151" CreationDate="2010-12-22T08:56:29.510" Score="3" ViewCount="113" Body="&lt;p&gt;I have always pictured volume element as a small cuboid in with volume $dx dy dz$. however i in curvilinear system, how would the shape of this volume element be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean i spherical polar coordinate system, how the shape of this volume element be visualized (is it a small sphere whose integration give the volume of object or same), or my idea of 3-d cartisan coordinate absolutely wrong.&lt;/p&gt;&#xA;" OwnerUserId="567" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-22T11:42:55.707" LastActivityDate="2010-12-22T11:42:55.707" Title="Shape of volume element in curviliner system" Tags="&lt;mathematical-physics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="2149" PostTypeId="2" ParentId="2147" CreationDate="2010-12-22T09:10:42.850" Score="3" ViewCount="" Body="&lt;p&gt;Yes. The phase velocity $v_\phi$ of light in a medium is equal to $c_0/n$, where $c_0$ is the speed of light in vacuum, and &lt;em&gt;n&lt;/em&gt; is the material's refractive index. The index is usually greater than 1, so normally this doesn't happen; but occasionally it is smaller &lt;em&gt;for certain frequencies of light&lt;/em&gt; (&lt;a href=&quot;http://en.wikipedia.org/wiki/Refractive_index#Speed_of_light&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; gives the examples of X-rays and frequencies near absorption resonances), giving a phase velocity greater than $c_0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to change the refractive index of a material: see for example the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pockels_effect&quot; rel=&quot;nofollow&quot;&gt;Pockels effect&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kerr_effect&quot; rel=&quot;nofollow&quot;&gt;Kerr effect&lt;/a&gt;. I don't know if this means that it would actually be possible to build a material with a &quot;phase velocity knob&quot; though, as suggested in the comment below.&lt;/p&gt;&#xA;" OwnerUserId="107" LastEditorUserId="107" LastEditorDisplayName="" LastEditDate="2010-12-22T11:05:25.677" LastActivityDate="2010-12-22T11:05:25.677" CommentCount="4" />
  <row Id="2150" PostTypeId="2" ParentId="2144" CreationDate="2010-12-22T09:43:06.033" Score="5" ViewCount="" Body="&lt;p&gt;Potential is a special case of a more general construction in differential geometry. Let's start abstractly and we'll get to the potentials again at the end.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Differential forms&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The framework of &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_form&quot; rel=&quot;nofollow&quot;&gt;differential forms&lt;/a&gt; provides a basis for integration on arbitrary manifold. Differential $p$-forms are totally antisymmetric covariant $p$-tensors. What's special about them is that you can define &lt;a href=&quot;http://en.wikipedia.org/wiki/Exterior_derivative&quot; rel=&quot;nofollow&quot;&gt;exterior derivative&lt;/a&gt; taking $p$-form to $(p+1)$-form. Now, $0$-forms are just functions and $1$-forms are like usual vectors (if you have a metric on your manifold, as is the case e.g. in Euclidean space, you can freely move between forms (also called covectors) and vectors).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, exterior derivative on functions gives us a 1-form. Let's say ${\rm d}f =  \alpha$. The important point about the operator $\rm d$ is that it is nilpotent ${\rm d}^2 = 0$. So this means that ${\rm d} \alpha = {\rm d}^2 f = 0$. Important point is that this relation can be reversed (at least on topologically trivial manifolds that don't have any holes in them, and so on -- see &lt;a href=&quot;http://en.wikipedia.org/wiki/Closed_and_exact_differential_forms#Poincar.C3.A9_lemma&quot; rel=&quot;nofollow&quot;&gt;Poincaré lemma&lt;/a&gt; for more information. But as always, Euclidean space is fine): if you have some form $ \alpha$ such that ${\rm d} \alpha = 0$ (we say that the form is &lt;em&gt;closed&lt;/em&gt;) then there will exist another form $\beta$ such that $\alpha = {\rm d} \beta$ (we say that the form is &lt;em&gt;exact&lt;/em&gt;). So this is you potential in a general setting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to understand why potentials are useful in the first we have to talk a bit about integration. It is possible to integrate $n$-forms on $n$-dimensional manifolds (the reason for this is that they have the similar transformation properties to Jacobian of usual integration substitution). So if you have some $p$-form $\alpha$ you can integrate it over some $p$-dimensional subset $U$ of the manifold and this is denoted by $\int_U \alpha$. The punchline is that if $\alpha$ has a potential $\alpha = {\rm d} \beta$ we can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Stokes%27_theorem&quot; rel=&quot;nofollow&quot;&gt;Stokes' theorem&lt;/a&gt; that tells us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_U \alpha = \int_U {\rm d} \beta = \int_{\partial U} \beta $$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\partial U$ is a boundary of the given subset. So that we can transform some integrals into others which can often simplify calculations.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Physics&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;To connect again with your questions: potentials arise from closed forms. The closeness conditions can take various guises in standard vector formalism. The usual one is for conservative forces $\nabla \times {\mathbf F} = 0$. This can be translated to the language of the differential forms as the condition on $\mathbf F$ being closed and so we know that there must be another form, say $\phi$ such that ${\mathbf F} = \nabla \phi$ (notice the identity $\nabla \times \nabla \phi = 0$ --  this is our good old ${\rm d}^2 = 0$ in action again). Because of the Stoke theorem we know that usefulness of the concept of conservative forces stems from the fact that their integral over a closed path doesn't depend on the path (this is a trivial consequence of $U$ having no boundary in that case). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another famous closed form is magnetic induction ${\mathbf B}$ because there are no monopoles (yet): $\nabla \cdot {\mathbf B} = 0$. This gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\mathbf B} = \nabla \times {\mathbf A}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathbf A$ is a vector potential. Again by using Stokes' theorem we can find that flow of $\mathbf B$ through any closed surface is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt; it might seems strange that we integrate vector $\mathbf B$ over a surface which is two-dimensional. This is not how we defined the integration for forms. But $\mathbf B$ is actually a two-form (you can see this from its relation to $\mathbf A$ which is a genuine one-form) and one is exploiting that in three-dimensional space one can identify these with one-forms. This is actually the usual identification of antisymmetric $3\times 3$ matrix with a pseudovector.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Complex analysis&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The complex analysis is very similar (although slightly harder) setting. The complex plane can be regarded as a two-dimensional real manifold so that there are two linearly independent one-forms: &lt;em&gt;holomorphic&lt;/em&gt; and &lt;em&gt;antiholomorphic&lt;/em&gt; forms ${\rm d}z$ and ${\rm d} \bar z$ where $z = x + iy$ and $\bar z = x - iy$. It can be shown that holomorphic functions satisfy $\partial_{\bar z} f = 0$ and that any holomorphic form $h = f {\rm d}z$ is closed so this gives the Cauchy formula &#xA;$$\oint_{\gamma} f {\rm d}z = 0$$ as a special case of Stokes' theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, one can exploit the full power of the complex formalism by using both holomorphic and antiholomorphic functions to encode any information that can be found in a plane which is usually described by functions ${\mathbb R}^2 \to {\mathbb R}^2$ as functions ${\mathbb C} \to {\mathbb C}$. It is again possible to translate all the language of differential forms and potentials into the complex setting.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-22T22:20:21.983" LastActivityDate="2010-12-22T22:20:21.983" CommentCount="2" />
  <row Id="2151" PostTypeId="2" ParentId="2148" CreationDate="2010-12-22T11:23:15.630" Score="6" ViewCount="" Body="&lt;p&gt;the easiest way of picturing the volume element is to start in a point $x\in M$ given in local coordinates $x^\mu$ and go along a (maybe curved) line $x^\mu + dx^\mu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, lets do this! In a two dimensional &lt;a href=&quot;http://en.wikipedia.org/wiki/Polar_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;polar coordinate system&lt;/a&gt; you can start at some point that is given by a radius $r = r_0$ and some angle $\varphi = \varphi_0$. Now, you &lt;strong&gt;draw a small arc segment&lt;/strong&gt; from $(r_0, \varphi_0)$ to $(r_0, \varphi_0 + d\varphi)$ where $d\varphi$ is just a small angle and you hold the radius constant.&lt;br&gt;&#xA;Then, you do the same from  $(r_0, \varphi_0)$ to $(r_0 + dr, \varphi_0)$. Now you mark the point $(r_0 + dr, \varphi_0 + d\varphi)$ and connect this point (in the same manner as you drew the segments before) to the endpoints of your lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, can you calculate the &lt;strong&gt;surface area&lt;/strong&gt; of this little element?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have done this, can you relate your calculation to the &lt;strong&gt;volume form&lt;/strong&gt; $\mathrm{vol} = \sqrt{g} dy^1\wedge\dots\wedge dy^n = r dr \wedge d\varphi$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS.: I left out the &lt;a href=&quot;http://mathworld.wolfram.com/ManifoldOrientation.html&quot; rel=&quot;nofollow&quot;&gt;orientation of the manifold&lt;/a&gt; for clearity.&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2010-12-22T11:23:15.630" CommentCount="9" />
  <row Id="2152" PostTypeId="1" AcceptedAnswerId="2154" CreationDate="2010-12-22T14:46:02.810" Score="4" ViewCount="206" Body="&lt;p&gt;Laser beams are said to have high &lt;a href=&quot;http://en.wikipedia.org/wiki/Laser&quot; rel=&quot;nofollow&quot;&gt;&quot;spacial coherence&quot;&lt;/a&gt;. This means that the beam is highly concentrated even at long distances (low spread).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this be achieved with radio waves (much longer waves) or is it due to laser's &lt;a href=&quot;http://en.wikipedia.org/wiki/Stimulated_emission&quot; rel=&quot;nofollow&quot;&gt; stimulated emission&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="250" LastActivityDate="2010-12-22T20:12:10.413" Title="Can radio waves be formed into a pencil beam?" Tags="&lt;electromagnetic-radiation&gt;&lt;laser&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="2153" PostTypeId="2" ParentId="2147" CreationDate="2010-12-22T15:13:59.220" Score="1" ViewCount="" Body="&lt;p&gt;The highest-profile schemes for manipulating the speed of light to make &quot;slow light&quot; use quantum phenomena like electromagnetically induced transparency to manipulate the index of refraction. The &quot;slow light&quot; effects come not from having a large index, but rather a steep slope in the index as a function of frequency. They produce this by pumping the medium with additional light so that the combination of the two light fields puts the atoms in the medium into a particular quantum superposition. The exact final speed can be tuned by choosing the properties of the pump field appropriately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Fast light&quot; experiments can work in a similar manner-- for example, this paper on &lt;a href=&quot;http://pra.aps.org/abstract/PRA/v63/i5/e053806&quot; rel=&quot;nofollow&quot;&gt;superluminal pulse propagation&lt;/a&gt; (&lt;a href=&quot;http://arxiv.org/abs/physics/0012060&quot; rel=&quot;nofollow&quot;&gt;arxiv version&lt;/a&gt;) uses a cesium vapor pumped into a particular state as the medium through which the pulse propagates faster than the normal speed of light. The resulting speed depends on the parameters of the laser used to excite the vapor, and thus can be tuned by varying those parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These experiments generally deal with the group velocity rather than the phase velocity, as that's the thing that you can measure in an experiment. I'm not sure if that invalidates these as responses to your question, but that's what comes to mind on the topic of varying the speed of light.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-12-22T15:13:59.220" CommentCount="1" />
  <row Id="2154" PostTypeId="2" ParentId="2152" CreationDate="2010-12-22T15:22:35.030" Score="6" ViewCount="" Body="&lt;p&gt;Laser light is spatially &lt;em&gt;and&lt;/em&gt; temporally coherent. The stimulated emission is mainly responsible for the temporal coherence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is yes, you can create an electromagnetic beam that is spatially but not temporally coherent by placing a pinhole close to the source, and then another pinhole in the far field of the first pinhole. This beam will not spread out very much. (But also remember that laser light does spread out.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for RF frequencies, a &quot;pinhole&quot; is probably several meters in diameter. The far field distance is given by this inequality: $L \gg a^2/\lambda$, where &lt;em&gt;L&lt;/em&gt; is the distance, &lt;em&gt;a&lt;/em&gt; is the diameter of the hole, and $\lambda$ is the wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, creating a RF pencil beam is probably not practical. The term &quot;pencil beam&quot; mentioned in the Wikipedia article is explained as being diffraction-limited. The size of a diffraction-limited beam gets larger with, I believe, the square root of the wavelength. It would be more like a gas-pipeline beam than a pencil beam.&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2010-12-22T15:22:35.030" />
  <row Id="2155" PostTypeId="2" ParentId="2152" CreationDate="2010-12-22T15:24:01.567" Score="5" ViewCount="" Body="&lt;p&gt;It depends on how big a pencil you're thinking about. There's no fundamental reason why radio waves can't be collimated in the same sort of way that visible light beams are. In fact, some radar systems send out fairly collimated beams at radio frequencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to make a radio-wave beam that is the same size as a typical laser beam, though, you're out of luck. You can't focus light of whatever wavelength down to a distance much smaller than a wavelength and expect it to stay there for very long. Making a reasonably collimated laser beam with a width of a millimeter or so isn't really a problem because the wavelengths of visible light are in the neighborhood of 500 nm, or about 2000 times smaller than the beam. Radio waves, however, have wavelengths that are measured in centimeters or even meters, and those aren't going to let you make a tight beam a millimeter across. &lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-12-22T15:24:01.567" />
  <row Id="2156" PostTypeId="1" CreationDate="2010-12-22T15:29:32.483" Score="6" ViewCount="152" Body="&lt;p&gt;I remember a while ago my father dropped a glass lid and it smashed. &lt;a href=&quot;http://images.google.com/imgres?imgurl=http://www.webvert-cookware.co.uk/flatglasslid-300.jpg&amp;amp;imgrefurl=http://www.webvert-cookware.co.uk/flatglasslid.htm&amp;amp;usg=__dPS-9DaG0CrN-iOj0zLN8v15L9k=&amp;amp;h=300&amp;amp;w=300&amp;amp;sz=21&amp;amp;hl=no&amp;amp;start=0&amp;amp;sig2=3ETNZ53Jv3YGPd32e5Ng6g&amp;amp;zoom=1&amp;amp;tbnid=3YP3t6rszo9PVM%3a&amp;amp;tbnh=166&amp;amp;tbnw=180&amp;amp;ei=uxgSTdmEOIal8QOVqc2EBw&amp;amp;prev=/images%3Fq%3Dglass%2Blid%26hl%3Dno%26biw%3D1438%26bih%3D815%26gbv%3D2%26tbs%3Disch:1&amp;amp;itbs=1&amp;amp;iact=rc&amp;amp;dur=198&amp;amp;oei=uxgSTdmEOIal8QOVqc2EBw&amp;amp;esq=1&amp;amp;page=1&amp;amp;ndsp=24&amp;amp;ved=1t:429,r:0,s:0&amp;amp;tx=159&amp;amp;ty=107&quot; rel=&quot;nofollow&quot;&gt;It looked something like this.&lt;/a&gt; When that happened, for about 5 minutes afterwards, the glass parts were splitting, kind of like popcorn, and you could hear the sound. I was just wondering why this happened, and the particles didn't just sit quietly in their own original parts?&lt;/p&gt;&#xA;" OwnerUserId="927" LastActivityDate="2010-12-23T06:46:21.393" Title="Why did this glass start popping?" Tags="&lt;energy&gt;&lt;mass&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="2157" PostTypeId="1" AcceptedAnswerId="2161" CreationDate="2010-12-22T15:42:12.870" Score="0" ViewCount="103" Body="&lt;p&gt;I essentially have three questions concerning &lt;a href=&quot;http://en.wikipedia.org/wiki/High-energy_radio-frequency_weapons&quot; rel=&quot;nofollow&quot;&gt;weapons based on EM waves&lt;/a&gt; or more &lt;a href=&quot;http://en.wikipedia.org/wiki/Directed-energy_weapon#Lasers&quot; rel=&quot;nofollow&quot;&gt;generally&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Focusing on the weapons using radio-waves and/or micro-waves, what power do these types of weapons need to radiate to generate the described effects?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How easy or difficult would it be to make your own weapon or to acquire one. As I understand it from the wikipage, only the US army is known to have such weapons, so I suppose it is not something you could easily make unless you have a physics lab at your disposition.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We live in a world where the &quot;amount&quot; of radio-waves and micro-waves that pervade our space is impressive. Isn't there any danger to be expected from these waves, considering that weapons can be built based solely on concentrated pulses of EM waves? Could something apparently as benign as a cell phone or a WiFi connection have the same effects, even though they would require longer exposition? I am aware that &lt;a href=&quot;http://en.wikipedia.org/wiki/Wireless_electronic_devices_and_health&quot; rel=&quot;nofollow&quot;&gt;studies&lt;/a&gt;, also this &lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_radiation_and_health&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; and this &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrical_sensitivity&quot; rel=&quot;nofollow&quot;&gt;one&lt;/a&gt;, have been done, and they seem to be either inconclusive or answering in the negative. Yet, I have this nagging doubt.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I am aware that the question is bordering on questions of medicine and physiology as well as engineering rather than pure physics, but I'm not sure where else to ask.&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2010-12-22T16:08:23.887" LastActivityDate="2010-12-22T20:08:53.397" Title="Electromagnetic weapons: power?" Tags="&lt;electromagnetism&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="2158" PostTypeId="1" AcceptedAnswerId="2165" CreationDate="2010-12-22T17:35:10.410" Score="4" ViewCount="145" Body="&lt;p&gt;I have a 4' hose that is closed at one end and connected to a Airdata Test Set (precise control of pressure) and a high accuracy pressure monitor on the other end with a T and valve.  The valve allows the Airdata Test Set connection to be closed off resulting in a hose connected to the pressure monitor and closed at the other end.  The valve is a high quality needle valve.  The pressure monitor is a a Druck DPI 142.  Half the length of the hose is in a temperature chamber controlled to 70 C.  The Druck connected end is outside the chamber at roughly 22 C.  When the airdata test set is commanded to a pressure of 1300 mb and allowed to settle for 30 seconds or so, then the valve closed, the pressure reported by the Druck drops over 20 minutes or so with a decreasing rate of change.  The airdata test set draws air from the room when operating.  The hose is ~0.190&quot; ID neoprene, Saint-Gobain P/N 06404-15.  The temp chamber, hose, etc, are given 1 hour to thermally stabilize prior to commanding the pressure to 1300 mb. The difference between initial pressure and stable pressure is ~18 mb. Why does the pressure take 20 minutes to stabilize?&lt;/p&gt;&#xA;" OwnerUserId="929" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-22T18:01:44.083" LastActivityDate="2011-01-27T21:35:58.240" Title="Why does measured pressure change over time in closed hose with temperature gradient" Tags="&lt;fluid-dynamics&gt;" AnswerCount="1" />
  <row Id="2159" PostTypeId="2" ParentId="1417" CreationDate="2010-12-22T19:10:09.813" Score="5" ViewCount="" Body="&lt;p&gt;It is true that gravity in $1+1$ dimensions is pretty trivial, but if you couple it to a scalar field in the right way it becomes much richer and turns into a very nice model which allows one to study black holes, black hole formation, and black hole evaporation. I apologize for advertising some of my own work, but one of the first papers on this is C. Callan,&#xA;S. Giddings, JH, and A. Strominger, hep-th/9111056. This CGHS model has been studied in great detail, both analytically and numerically. For example, here is a recent paper that contains a numerical study of black hole evaporation in this model: &lt;a href=&quot;http://arXiv.org/pdf/1012.0077&quot; rel=&quot;nofollow&quot;&gt;http://arXiv.org/pdf/1012.0077&lt;/a&gt;.&#xA;I think one can learn quite a bit about black holes and GR in this model without many of the complications of four dimensions.&lt;/p&gt;&#xA;" OwnerUserId="838" LastActivityDate="2010-12-22T19:10:09.813" CommentCount="1" />
  <row Id="2160" PostTypeId="1" AcceptedAnswerId="11166" CreationDate="2010-12-22T20:02:54.557" Score="3" ViewCount="209" Body="&lt;p&gt;Could you suggest any if it's possible to deposit a layer of SiO2 via PVD?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why doesn't it widely used? Is that correct that I need ~1600C in vacuum?&#xA;Could you suggest any books/papers about that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately cannot tag this properly :-)&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-06-16T07:05:04.403" Title="Physical vapor deposition of SiO2" Tags="&lt;experimental-physics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="2161" PostTypeId="2" ParentId="2157" CreationDate="2010-12-22T20:08:53.397" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Minimal required energy starting from 100J in ns-scale pulses, any kind of microwawe/light will do. It would be Very hard to deliver energy to distance &gt; 10m. In constant emission you would need power in the range 100-1000W+ to see minimal effects.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Russia also had laser weapons, but they are not widely used (basically because conventional weapons are much cheaper with the same effect). Well, you can make 100W CO2 laser at lab or 1000W microwave emitter, but it's not really useful due to short range. Conventional weapons are really much more efficient for the price.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Amount of energy is to low (constant emission &amp;lt;1W). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="930" LastActivityDate="2010-12-22T20:08:53.397" />
  <row Id="2162" PostTypeId="2" ParentId="2152" CreationDate="2010-12-22T20:12:10.413" Score="1" ViewCount="" Body="&lt;p&gt;Well, you just need Maser :-)&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Maser&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Maser&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2010-12-22T20:12:10.413" />
  <row Id="2163" PostTypeId="2" ParentId="2156" CreationDate="2010-12-22T20:15:30.240" Score="4" ViewCount="" Body="&lt;p&gt;Due to internal stress in the material.&#xA;This stress might be there due to fabrication technology or due to heat cycling while usage (less likely).&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2010-12-22T20:15:30.240" />
  <row Id="2164" PostTypeId="1" CreationDate="2010-12-22T20:59:29.567" Score="2" ViewCount="241" Body="&lt;p&gt;I just saw &lt;a href=&quot;http://www.youtube.com/watch?v=fC942wZ0Nv4&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt; which was kind of nifty. What principles govern this? Is it simply that 700 lbs of air pressure are exerted from that little 1 HP blower? What would you have to take into consideration to adopt the same principles to move something like a car on a turntable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is moving something with air more economical than moving it by other means?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just thinking out loud...&lt;/p&gt;&#xA;" OwnerUserId="325" LastActivityDate="2011-07-14T19:18:26.230" Title="What principles does an air glider use?" Tags="&lt;air&gt;&lt;lift&gt;" AnswerCount="2" />
  <row Id="2165" PostTypeId="2" ParentId="2158" CreationDate="2010-12-22T21:01:11.393" Score="1" ViewCount="" Body="&lt;p&gt;Firstly, and forgive me for asking the obvious, are you &lt;em&gt;certain&lt;/em&gt; that there are no leaks anywhere in your setup? I'd suggest getting someone else to check it over in person - it may be something obvious you've overlooked that a fresh pair of eyes would see. Again, apologies if you've already tried this :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that's not it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you increase the pressure in the hose, the air temp in the hose will also increase (work is done on the gas to increase the pressure). As the temp settles back to equilibrium, the pressure decreases slightly according to $pV = NRT$. (I think you say you allow 1 hour for equilibrium &lt;em&gt;prior&lt;/em&gt; to increasing the pressure). &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If this is true, then increasing the pressure again should result in a much smaller pressure drop (since an increase of 18mb will result in much less heating than 1300mb). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are you able to monitor temperatures inside the hose?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Given that the pressure does seem to settle, this is where I'd put my money.&lt;/p&gt;&#xA;" OwnerUserId="474" LastActivityDate="2010-12-22T21:01:11.393" CommentCount="1" />
  <row Id="2166" PostTypeId="1" AcceptedAnswerId="2168" CreationDate="2010-12-22T21:24:13.017" Score="3" ViewCount="525" Body="&lt;p&gt;I read somewhere that according to relativity, black holes and other space related stuff it is possible to jump into past.&lt;br&gt;&#xA;Is it possible for anything to go back in time either continuously or by jumping?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I doesn't fit my mind and seems totally illogical, unreasonable and even stupid thought because if it was possible, we must had someone who came from future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt;&#xA;&lt;a href=&quot;http://news.discovery.com/space/time-travel-impossible-photon-110724.html&quot; rel=&quot;nofollow&quot;&gt;Time Travel Impossible, Say Scientists&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="847" LastEditorUserId="847" LastEditorDisplayName="" LastEditDate="2011-07-28T10:39:59.510" LastActivityDate="2011-07-29T05:01:15.510" Title="Is it possible to go back in time?" Tags="&lt;time-travel&gt;&lt;time&gt;&lt;time-reversal&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="3" />
  <row Id="2167" PostTypeId="1" CreationDate="2010-12-22T21:30:56.423" Score="7" ViewCount="416" Body="&lt;p&gt;There are many videos on youtube in which people arranged magnets in circle and rotated one placing in middle of that circle on a shaft, and the magnet (magnet motor) starts madly and continues its movement.&lt;br&gt;&#xA;Do they really rotate infinitely?&lt;br&gt;&#xA;If so, do they from where they get that extra energy to move so fast and infinitely?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This question is about real world scenario&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE:&#xA;I am unable to find that particular video about which i talked about -- I added them to my favourites, but they are all now deleted :( -- there are hundreds of that that kind of videos. Some example videos links&lt;br&gt;&#xA;&lt;a href=&quot;http://www.youtube.com/watch?v=dT9s33X9D4I&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=dT9s33X9D4I&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.youtube.com/watch?v=yei0NMqUaZ8&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=yei0NMqUaZ8&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.youtube.com/watch?v=5DvPLFX5gwA&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=5DvPLFX5gwA&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.youtube.com/watch?v=i0oUaPZ_wF8&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=i0oUaPZ_wF8&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://peswiki.com/index.php/OS:Screw-Magnet_Motor&quot; rel=&quot;nofollow&quot;&gt;http://peswiki.com/index.php/OS:Screw-Magnet_Motor&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.youtube.com/watch?v=aHVBu77jz4w&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=aHVBu77jz4w&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="847" LastEditorUserId="847" LastEditorDisplayName="" LastEditDate="2010-12-24T13:15:03.243" LastActivityDate="2011-02-01T17:27:03.767" Title="Can magnets rotate infinitely?" Tags="&lt;electromagnetism&gt;&lt;magnets&gt;&lt;perpetual-motion&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="2168" PostTypeId="2" ParentId="2166" CreationDate="2010-12-22T21:38:46.117" Score="3" ViewCount="" Body="&lt;p&gt;It is mathematically possible to create some instances in which an object goes back in time &lt;em&gt;relative to some observer&lt;/em&gt;. For example, simply going faster than light causes such an effect, but of course, speed of light is the limit for any massive object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it is mathematically possible, there are many paradoxes caused by time travel to past, unless you accept existence of some sorts of multiverses, which doesn't really qualify as a time travel to past anyway, rather travel between said multiverses. Mainly, and most importantly, time travel to past violates &quot;casuality&quot;, one of the main principles the universe is thought to have, in which the cause precedes the result, i.e. your father is born before you are born. Through time travel to past, it is possible for the &quot;result&quot; to eliminate its &quot;cause&quot; before it causes the result, that is, you killing your father before you were conceived. This is not the only paradox caused by time travel to past, however this is one of the main ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I think it was Hawking that suggested this, we see no time-travelers around us, pointing out to the fact that humanity will not achieve the technology to go back in time as far as it wants. HOWEVER, it is possible that time travel to past may be discovered, only limited to taking travelers back to the creation of the machine, as some time travel machines work on this principle, namely wormholes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Needless to say, wormholes require extremely exotic conditions to form and maintain, and therefore are very far from being created deliberately, if they even exist.&lt;/p&gt;&#xA;" OwnerUserId="504" LastActivityDate="2010-12-22T21:38:46.117" CommentCount="10" />
  <row Id="2169" PostTypeId="1" CreationDate="2010-12-22T22:27:06.893" Score="0" ViewCount="122" Body="&lt;p&gt;In the classic metaphor, a light beam bends for the same reason that a wagon getting one wheel stuck in the sand does...the wheels travel at uneven speeds, and the wheel on the smoother surface travels faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the key to the wagon scenario is the axle - if the two wheels were not bound, the faster wheel would sail on, heedless of the other wheel's difficulty. So, if the metaphor is useful at all, there is an axle force binding the photos in a beam of light, which causes it to turn when it hits a different medium non-perpendicularly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is this just a bad metaphor for the rubes?&lt;/p&gt;&#xA;" OwnerUserId="935" LastActivityDate="2010-12-22T23:14:14.423" Title="Why does light refract if photons are not bound by an axle?" Tags="&lt;optics&gt;" AnswerCount="2" />
  <row Id="2170" PostTypeId="2" ParentId="2169" CreationDate="2010-12-22T23:12:03.733" Score="0" ViewCount="" Body="&lt;p&gt;First important point is that light always travels in a straight lines in an environment of constant index of refraction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/eejRR.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What creates the dispersion is the fact that the index of refraction depends on the frequency of the light $n = n(f)$. This also means that phase velocity (defined as $v = {c \over n}$) depends on the frequency. You don't see this in the image but you can imagine that in usual materials (e.g. glass) a wavefront will propagate faster along red tracks than along the violet tracks (equivalently, red tracks are less bent than the violet tracks, as can be seen in the picture). So this will make the wavefront bulging up in the red part of the spectrum (the precise shape depending on the precise material and angle of incidence).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's probably just the wavefront the metaphor is talking about. But you can see that wavefront will spread out as it propagates in the material (precisely because there is no &quot;axle&quot; between individual photons). In my opinion it's a pretty bad metaphor except for giving a correct illustration of the relation &quot;denser environment&quot; $\leftrightarrow$ &quot;slower speed&quot;.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-22T23:12:03.733" />
  <row Id="2171" PostTypeId="2" ParentId="2169" CreationDate="2010-12-22T23:14:14.423" Score="1" ViewCount="" Body="&lt;p&gt;Its a metaphor. But any physical analogy can only be taken so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an alternative instead of a wagon we can consider a marching band. In this case the band marches in formation until it reaches the edge of a muddy patch. If the approach was head on, the band would keep marching in formation albeit at a slower speed. If the encounter is at an angle, then the marchers on the patch move slower than the ones not on it - leading to a &quot;bending&quot; in the ranks. In this case there is no &quot;axle&quot; connecting the marchers but they &quot;refract&quot; nevertheless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://physics.bu.edu/~duffy/semester2/c24_refract_model.html&quot; rel=&quot;nofollow&quot;&gt;page&lt;/a&gt; has a beautiful applet illustrating this analogy.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-22T23:14:14.423" />
  <row Id="2172" PostTypeId="1" CreationDate="2010-12-22T23:53:59.300" Score="2" ViewCount="207" Body="&lt;p&gt;I am a physics novice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google tells me that electron microscopes work much like their optical counterparts -- but the analogy falls apart for me when I think about what I'm &quot;viewing.&quot;  Obviously, you can &lt;em&gt;see&lt;/em&gt; light through the lenses, but what is the &quot;image&quot; analog for electron microscopes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it at-all like spraying an invisible shape with bullets and examining where collisions took place?  Like if you shot at an invisible car with a tommy gun and were able to make out bullet holes -- so that the more bullets you shoot the better your image?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, just for completeness, I suspect this implies that the best resolution you can get is the bullet-size, or in this case the size of the electron.  How do you map &quot;objects&quot; or whatever they are considered on that scale if they are smaller than an electron?  Is our perception of how &lt;em&gt;small&lt;/em&gt; we can see limited by this cap?&lt;/p&gt;&#xA;" OwnerUserId="936" LastActivityDate="2010-12-23T05:24:32.960" Title="How does an electron microscope work?" Tags="&lt;electromagnetism&gt;&lt;microscopy&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="2173" PostTypeId="1" CreationDate="2010-12-23T00:52:21.507" Score="0" ViewCount="265" Body="&lt;p&gt;[This question has arisen from a wish to understand an end-of-universe scenario: heat death]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are time and mass intrinsically linked?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, does time &quot;run slower&quot; (whatever that may mean) in a universe with less mass?&#xA;And ultimately, in a universe without any mass, just space and energy (photons), then the universal clock stops ticking. I.e. no further aging takes place??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So does time slow down gradually as baryonic mass breaks down?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the case, would the rate at which time slows down also slow down - because time itself as slowed down - leading to an asymptotic decay of time. I.e. it would take forever for all the particles to decay because each particle takes longer and longer to break down due to time slowing down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers,&#xA;Dave&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: In case you hadn't guessed - I don't have a physics background.&lt;/p&gt;&#xA;" OwnerUserId="937" LastActivityDate="2010-12-23T00:52:21.507" ClosedDate="2010-12-23T04:53:57.177" Title="What's the relationship between mass and time?" Tags="&lt;mass&gt;&lt;universe&gt;&lt;time&gt;" CommentCount="3" />
  <row Id="2174" PostTypeId="2" ParentId="2156" CreationDate="2010-12-23T00:56:22.287" Score="1" ViewCount="" Body="&lt;p&gt;Here is an idea. You could have cracking (fracture) behaving as a critical process. From the dropping of the glass you could have micro-fractures forming. In time and because of the amount of tension (and energy) concentrated in the glass from the fall, these micro-fractures slowly could organize to form big fractures, which would cause the popping sounds you continued to hear for some time after the fall of the glass lid. If it was some kind of crystal that broke I would be more confident that that was the case, but since it is some kind of glass, I am not so sure that it can happen something like that, since glass is amorphous, but I am not sure that it can't either.&lt;/p&gt;&#xA;" OwnerUserId="383" LastEditorUserId="383" LastEditorDisplayName="" LastEditDate="2010-12-23T06:46:21.393" LastActivityDate="2010-12-23T06:46:21.393" />
  <row Id="2175" PostTypeId="1" AcceptedAnswerId="2176" CreationDate="2010-12-23T01:40:14.267" Score="7" ViewCount="698" Body="&lt;p&gt;Is it possible for information (like 1 and 0s) to be transmitted faster than light? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, take a rigid pole of several AU in length. Now say you have a person on each end, and one of them starts pulling and pushing on his/her end. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The person on the opposite end should receive the pushes and pulls instantaneously as no particle is making the full journey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this actually work?&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-20T14:47:22.943" LastActivityDate="2011-05-29T21:06:17.967" Title="Is it possible for information to be transmitted faster than light?" Tags="&lt;special-relativity&gt;&lt;speed-of-light&gt;&lt;faster-than-light&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="1" />
  <row Id="2176" PostTypeId="2" ParentId="2175" CreationDate="2010-12-23T02:00:18.680" Score="19" ViewCount="" Body="&lt;p&gt;The answer is no. The pole would bend/wobble and the effect at the other end would still be delayed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that the force, which binds the atoms of the pole together: The Electro-Magnetic force, needs to be transmitted from one end of the pole to the other; and the transmitter of the EM-force is light, and thus the signal cannot travel faster than the speed of light, and in stead the pole will bend, because close end will have moved, and the far end will not yet have received intelligence of the move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: A simpler reason. &lt;br&gt;&#xA;In order to move the whole pole, you need to move every atom of the pole.&lt;br&gt;&#xA;You might like to think of atoms as next door neighbours If one of them decides to move, he sends out a messenger to all his closest neighbours telling them he is moving. Then they all decide to move as well, so they each send out messengers to to their closest neighbours to let them know they are moving; and so it continues, until the message to move has travelled all the way to the end. No atom will move until he has received the message to do so, and the message won't travel any faster than all the messengers can run; and the messengers can't run faster than the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;/B2S&lt;/p&gt;&#xA;" OwnerUserId="938" LastEditorUserId="938" LastEditorDisplayName="" LastEditDate="2010-12-23T02:45:40.033" LastActivityDate="2010-12-23T02:45:40.033" CommentCount="10" />
  <row Id="2177" PostTypeId="2" ParentId="2166" CreationDate="2010-12-23T02:23:59.730" Score="1" ViewCount="" Body="&lt;p&gt;I think &lt;a href=&quot;http://adsabs.harvard.edu/abs/1988PhRvL..61.1446M&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; is the &quot;gold standard&quot; of time travel articles in the refereed literature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For popular literature, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0393312763&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; is my favorite&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for movies, &lt;a href=&quot;http://www.primermovie.com/story.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; is the best I have ever seen (other than &quot;Back to the Future&quot; of course, but that is only for nostalgic, and not physics reasons)&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2010-12-23T02:23:59.730" />
  <row Id="2178" PostTypeId="1" AcceptedAnswerId="2184" CreationDate="2010-12-23T02:47:04.993" Score="5" ViewCount="312" Body="&lt;p&gt;As I understand it (and admittedly it's a weak grasp), a computer processes information irreversibly (AND gates, for example), and therefore has some minimum entropy increase associated with its computations.  The true entropy increase is much greater, and comes from the conversion of electrical energy to heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How efficient is a typical desktop computer when viewed in this light?  Make any assumptions you find useful about the energy use, computations per second, temperature of the room, etc.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-12-24T19:02:41.640" Title="How efficient is a desktop computer?" Tags="&lt;thermodynamics&gt;&lt;computation&gt;&lt;entropy&gt;&lt;information&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="3" />
  <row Id="2179" PostTypeId="1" AcceptedAnswerId="2180" CreationDate="2010-12-23T03:06:58.620" Score="1" ViewCount="131" Body="&lt;p&gt;Hey guys, &#xA;I have a final tomorrow and I am going over some assignments. One of the questions from the assignment was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A photon having 37 keV scatters from a&#xA;  free electron at rest. What is the&#xA;  maximum energy that the electron can&#xA;  obtain?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, I got the right answer (4.67 KeV) but I can't seem to figure it out now. Maybe I am just too tired. &#xA;Any hints to where I should begin?&lt;/p&gt;&#xA;" OwnerUserId="551" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-23T10:08:14.640" LastActivityDate="2010-12-23T10:08:14.640" Title="Electron/Photon Scattering" Tags="&lt;homework&gt;&lt;scattering&gt;" AnswerCount="1" />
  <row Id="2180" PostTypeId="2" ParentId="2179" CreationDate="2010-12-23T03:20:29.610" Score="3" ViewCount="" Body="&lt;p&gt;There is only one parameter in the system. You have several choices, but you might use the scattering angle, $\Theta$, of the photon in the electrons initial rest frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now conserve 4-momentum, and write the final state energy of the electron as a function of $\Theta$ and maximize.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2010-12-23T03:20:29.610" />
  <row Id="2181" PostTypeId="2" ParentId="2175" CreationDate="2010-12-23T04:52:25.557" Score="3" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible for information (like 1 and 0s) tO be transmitted in anyway faster than light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;No.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Born2Smile said the same thing (which I +1'd) but I figured it's worth repeating for emphasis. It'd be a violation of causality. For some more details on why this is not allowed, in addition to Born2Smile's answer, see &lt;a href=&quot;http://physics.stackexchange.com/questions/196/what-are-some-scenarios-where-ftl-information-transfer-would-violate-causality&quot;&gt;What are some scenarios where FTL information transfer would violate causality? &lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-12-23T04:52:25.557" CommentCount="6" />
  <row Id="2182" PostTypeId="2" ParentId="2172" CreationDate="2010-12-23T05:24:32.960" Score="2" ViewCount="" Body="&lt;p&gt;those answers were all pretty heavy for a physics novice!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer you are looking for is easier than you might think - &#xA;you 'see' light through a microscope's lens because your eyes are great photon detectors. without a detector, a light microscope can't create an image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;an electron microscope (there are various types) simply has other types of detectors. depending on what size scale you are looking on the could variously be detecting reflected electrons (the proper term is backscattered), deflected electrons, emitted light from excited electrons, or even variations in various forces. This is the same for light microscopes, you are simply seeing reflected/emitted light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there are even electron microscopes that rely on quantum tunneling of electrons, and measure the tunneled current, which changes depending on the distance between microscope tip and object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the point is, all they do is measure interactions of the wave and the object, in the same way your eyes do with a normal light microscope. a computer (or your brain) then models what was detected to resolve an image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the only other concept you need is that the resolution of any microscope is proportional to the wavelength used. A smaller wavelength can generate better resolution. So the best you can do with a light microscope is violet light, with a wavelength around 400nm. An electron can be seen as a wave as well, and has a far smaller wavelength, in the range of picometres. Marek's suggestion re:wave-particle duality could help here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been trying to put why small wavelengths mean better resolution into simple terms but it is eluding me. Kinda needs diagrams. Maybe someone else can help?&lt;/p&gt;&#xA;" OwnerUserId="909" LastActivityDate="2010-12-23T05:24:32.960" CommentCount="5" />
  <row Id="2183" PostTypeId="2" ParentId="2167" CreationDate="2010-12-23T08:29:58.850" Score="8" ViewCount="" Body="&lt;p&gt;Anything can rotate infinitely, if there is no friction. Rotating an object with constant velocity uses no energy to maintain. With friction it will loose energy and eventually stop. If you add a static magnetic field to the object, it will gain some finite potential energy, which in turn can be converted to kinetic energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you add a changing magnetic field, you can make it continuously rotate or even accelerate, but this requires energy to maintain.&#xA;This is how regular electromotors work, the battery is an electromagnetic potential, and will give the motor kinetic energy, but eventually the battery will be depleted.&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2011-02-01T16:19:14.977" LastActivityDate="2011-02-01T16:19:14.977" CommentCount="3" />
  <row Id="2184" PostTypeId="2" ParentId="2178" CreationDate="2010-12-23T10:27:10.050" Score="4" ViewCount="" Body="&lt;p&gt;Assuming a typical computer with CPU processing power ~1 GHz. It means that it can generate output byte sequence at ~$10^9$ byte/s, which is about ~$10^{-13}$ J/K in terms of von Neumann entropy. Also, the power consumption of a typical CPU is ~100 W, which gives entropy ~0.3 J/K at room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the (minimum ΔS) / (actual ΔS) ~ $10^{-14}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This calculation is not quite right because it is hard to determine what is the actual output of a computer. In most case, the previous output will be used as input later. The above calculation has also made the assumption that all output is continuously written in some external device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better point of view is that each gates taking two inputs and one output, such as AND, OR, NAND, ...,  must drop one bit to the surrounding as heat. This is the minimum energy $W$ required to process information in a classical computer. In this sense, we may define the efficiency as $e = W/Q$, where $Q$ is the actual heat generation per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The efficiency depends on how many such logical gates that will be used, but I guess it is less than thousand in a typical clock rate, so $e \approx 10^{-11}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It means that our computer is very low efficiency in terms of information processing, but probably good as a heater. This theoretical minimum energy requirement is also hard to verified by experiment because of the high accuracy required. &lt;/p&gt;&#xA;" OwnerUserId="562" LastActivityDate="2010-12-23T10:27:10.050" CommentCount="1" />
  <row Id="2185" PostTypeId="2" ParentId="2178" CreationDate="2010-12-23T12:41:57.270" Score="7" ViewCount="" Body="&lt;p&gt;Some info from ASIC world:&#xA;For example, you processor have 300 mil. transistors, and most of these do some work. But, in order to make for example pure 32-bit add operation you need just about 1000 of them. Others are for caching and passing data back and forth - support functions which are impossible to estimate. So estimations from math side are very hard to make.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern process design already targets for energy per switch, and it's being optimized.&#xA;Unfortunately, the slower speed you need - the more efficient processor runs.&#xA;For example, to get 50% of speed you can spend just about 10% of power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So they are deadly inefficient (and there is still room for 100-10000 times improvement), but estimating CPU as a whole is wrong. You should take into account only minimal computing unit used, like summator, you cannot predict how many switches you would have in support logic which takes 98% of chip area. &lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2010-12-23T12:41:57.270" />
  <row Id="2186" PostTypeId="2" ParentId="2111" CreationDate="2010-12-23T12:45:51.713" Score="1" ViewCount="" Body="&lt;p&gt;I am not fully convinced by this argument, but can't find a flaw in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's analyse a similar experiment, which I believe to be equivalent. Assume that initially, the water is already at the stable level $H=\frac{P_{atm}}{\rho g}$. The vacuum is already present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, by some unimportant means, we lower the water to $h=0$, and then let it go up freely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much energy are we storing in the system by lowering the water? We can find out by calculating the work done. The work is done against pressure and in favour of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W= sP_{atm}H + \int^0_H{m(h)g\ \mathrm{d}h}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting $H=\frac{P_{atm}}{\rho g}$, $m(h)=\rho s h$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W= \frac{sP_{atm}^2}{\rho g} + s\rho g\int^0_H{h\ \mathrm{d}h}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W= \frac{sP_{atm}^2}{\rho g} - \frac{1}{2}s\rho gH^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W= \frac{sP_{atm}^2}{2\rho g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now when the water is released and allowed to rise, all this energy will used to make the water rise. No energy is assumed to be wasted on friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the highest point, all the energy will be converted in gravitational potential energy. This can be expressed through the following formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U(h)=\frac{1}{2}s\rho gh^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, at the top point, $W=U(h)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{sP_{atm}^2}{2\rho g} = \frac{1}{2}s\rho gh^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $h$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h^2=\frac{P_{atm}^2}{g^2\rho^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h = \frac{P_{atm}}{g\rho} = H$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the water will raise up to $H$, which it will reach with zero velocity.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-12-23T12:45:51.713" CommentCount="34" />
  <row Id="2187" PostTypeId="1" AcceptedAnswerId="2192" CreationDate="2010-12-23T13:41:32.720" Score="0" ViewCount="162" Body="&lt;p&gt;How does the optical zoom works..&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Does it brings the light closer to the object... Or&lt;/li&gt;&#xA;&lt;li&gt;Does it just enlarges the picture of the object.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In other words... Suppose I have telescope with 1 light-year zoom capacity, could I see stars 1 year ahead of its light reaching earth...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ps : I'm not physics graduate, but interested in physics. I joined here, to learn more...  Please excuse me if my question is trivial or stupid.&lt;/p&gt;&#xA;" OwnerUserId="923" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-12-23T22:41:10.997" LastActivityDate="2010-12-23T22:42:33.367" Title="What does the optical zoom in telescopes do?" Tags="&lt;optics&gt;&lt;light&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="2188" PostTypeId="1" CreationDate="2010-12-23T14:16:17.700" Score="9" ViewCount="671" Body="&lt;p&gt;Background: I'm in a legal academic discussion about the status of electronic 'goods' and whether they qualify as 'goods' in the same way a chair and a pen do. In this context (and specifically at the exact circumstance under discussion), it matters if electricity is 'tangible'. So far, most authors have blindly assumed electricity to be a flow of electrons, making a literal analogy with water, making statements such as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;information is stored in capacitors in the form of electrons. When a capacitor is filled more than 50% with electrons, it's considered to be 'on' (a bit with value '1').&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the information represented by a certain current (or rather, a series of on/off currents) has mass, because it consists of the electrons that flow through the wire.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a virtual object is tangible because it exists in memory in the form of electrons that are there (or not) in a certain pattern.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now I have a background in informatics but only a basic knowledge of electricity, and as good as none on the fundamental (physics) level. However I still feel this representation is wrong, and that you can't just say that information in a RAM chip has mass because it consists of electrons that are or are not in the capacitors on that chip. I have found hints in that direction on sites such as &lt;a href=&quot;http://amasci.com/miscon/eleca.html#made&quot; rel=&quot;nofollow&quot;&gt;http://amasci.com/miscon/eleca.html#made&lt;/a&gt;, but I can't quite make out what 'electricity' &lt;em&gt;is&lt;/em&gt; and how it relates to &lt;em&gt;current&lt;/em&gt; and &lt;em&gt;potential&lt;/em&gt; and other words that are used interchangeably in these discussions but which are, I think, different things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my questions are (all just different angles of looking at the same underlying concept):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What is 'electricity', really, on a fundamental level; but explained in terms a layman can understand? Is there an analogy with other things that is accurate, unlike the 'flowing water' analogy, which is sufficient for high school level but is a simplification? (at least, I think...)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do 'electricity', an 'electrical current' and an 'electrical charge' have mass, apart from the object they are embodies in? Does the mass of a copper wire change when you put a current through it, because of the electrons coming in and going out?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How do electrons fit into this? Is electricity composed of a bunch of electrons that flow through a mass? I think not, reading the link I gave before, but I don't quite understand what their role is.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Most authors blindly assume that electricity is merely a flow of electrons through mass. In how far and under what assumptions is this correct?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="944" LastActivityDate="2011-04-16T23:58:01.373" Title="Does 'electricity' have mass? Is 'electricity' tangible?" Tags="&lt;electricity&gt;" AnswerCount="9" CommentCount="8" FavoriteCount="2" />
  <row Id="2189" PostTypeId="1" AcceptedAnswerId="2683" CreationDate="2010-12-23T14:26:17.873" Score="6" ViewCount="372" Body="&lt;p&gt;It's known that single spherical lens cannot focus parallel beam of monochromatic light into single point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you suggest how should aspherical lens look like to be able to focus in single point (again, light is monochromatic)? Do we need conic factor, or higher components?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you suggest any decent books on aspherical optics &amp;amp; it's manufacturing?&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="930" LastEditorDisplayName="" LastEditDate="2011-01-09T02:54:28.970" LastActivityDate="2011-01-10T19:12:04.947" Title="Aspherical lenses" Tags="&lt;optics&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="2190" PostTypeId="2" ParentId="2188" CreationDate="2010-12-23T14:54:45.763" Score="4" ViewCount="" Body="&lt;p&gt;Electricity is just a generic term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electric current is a number representing the number of charged particles travelling through a given surface per time. Like how many women walk in the door to the mall in a given day, thats a number, it dont have mass. Electric charge would here be 1 if you are a women, and 0 if you are a man, the door would be the surface thorugh which we want to measure women current.&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2010-12-23T14:54:45.763" CommentCount="1" />
  <row Id="2191" PostTypeId="2" ParentId="2188" CreationDate="2010-12-23T14:55:31.543" Score="2" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;This is directional movement of electrones in the field&lt;/li&gt;&#xA;&lt;li&gt;Electrons have a mass, but when current flows through a wire number of electrons does not change. They just move in the field, number is the same.&lt;/li&gt;&#xA;&lt;li&gt;Yes, it is. Electrons flow easily through metals in the presence of electric field, that's why they are conductive.&lt;/li&gt;&#xA;&lt;li&gt;Well, I think it's true by definition.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The only way you can increase/decrease mass of an object is when you charge metallic object - in this case count of electrons increases/decreases.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2010-12-23T14:55:31.543" CommentCount="4" />
  <row Id="2192" PostTypeId="2" ParentId="2187" CreationDate="2010-12-23T15:04:26.533" Score="4" ViewCount="" Body="&lt;p&gt;The answer is 2. Zoom isnt measured in distance, it is more like a ratio of magnification. So it tells you how much you magnified something, ofcourse for lenses the zoom is slightly more technical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lightyear is a distance, its actually the distance travelled by light in 1 year, so when you see stars 1 ly. away, you see how they appeared 1 year ago.&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2010-12-23T15:15:13.057" LastActivityDate="2010-12-23T15:15:13.057" CommentCount="1" />
  <row Id="2193" PostTypeId="2" ParentId="2188" CreationDate="2010-12-23T15:07:49.713" Score="4" ViewCount="" Body="&lt;p&gt;Electronic goods can be transmitted without exchanging electrons (think of wi-fi networks). They actually don't even need electricity at all, for example vouchers can be sent through normal post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What changes is the configuration of electrons and electromagnetic fields in the electronic device which contains the goods. This is simply a &quot;state&quot;. For example goods stored on your hard drive are merely stored as particular stable configurations of magnetic particles on a platter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in practice, electronic goods are merely numbers. There is no mass associated with them, and no particular exchange of charges. They are configurations, or states.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-12-23T15:07:49.713" CommentCount="2" />
  <row Id="2194" PostTypeId="2" ParentId="2188" CreationDate="2010-12-23T15:08:44.233" Score="7" ViewCount="" Body="&lt;p&gt;Interesting, but I'm don't think you are asking the right questions in the context of &lt;em&gt;law&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that electrons and electricity are completely irrelevant when it comes to the question of &quot;tangible&quot; and &quot;electronic&quot; &quot;goods&quot;. You will obtain a good answer only if you forget about electricity, which just happens to be a &lt;em&gt;convenient physical carrier&lt;/em&gt; of information, and focus on the objects that might or might not be goods like a chair or a pen. A proper question would be this: &quot;Is a &lt;em&gt;newspaper article&lt;/em&gt; a good, like a chair or a pen?&quot;. Specifying whether the newspaper article is written on paper or &quot;on electrons&quot; or on something else is besides the point because the concept of &quot;newspaper article&quot; is entirely independent of the material it is written on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it more pointedly: if you have to know the metaphysical nature of electricity to make a law about newspapers, you're definitely doing it wrong. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keeping in mind that your questions and their answers are completely useless in the context of law, I can now proceed to answer them.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I don't know of any good analogy to electricity that captures it properly. It is like &lt;em&gt;gravitation&lt;/em&gt; in that distant bodies attract each other, except that in electricity, bodies can also repel each other. Furthermore, the gravitational pull of, say, a chair is so small that we don't think of a chair as attracting us gravitationally. This is also why the &lt;em&gt;water analogy&lt;/em&gt; doesn't work so well: water does not attract other water from afar. The &quot;cause&quot; for attraction/repulsion is the &lt;em&gt;electrical charge&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;em&gt;carriers of charge&lt;/em&gt;, like &lt;em&gt;electrons&lt;/em&gt; or &lt;em&gt;ions&lt;/em&gt; (= atoms with electrons missing) do have a mass. You can think of them as tiny, charged balls flying around through space (keep in mind that a copper wire consist of mostly empty space, too). On the other hand, &lt;em&gt;electrical current&lt;/em&gt; does not have a mass, just like a &lt;em&gt;water current&lt;/em&gt; does not have a mass, it simply doesn't make sense. (Both imply a &lt;em&gt;mass current&lt;/em&gt;, however.) Likewise, &lt;em&gt;electricity&lt;/em&gt; is a general term and does not have a mass, just like &quot;law&quot; and &quot;liberty&quot; don't have a mass.&lt;br&gt;&lt;br&gt;The mass of a copper wire is the sum of the masses of its constituents, some of which are electrons. However, the number electrons exiting a copper wire is usually the same as the number of electrons entering the copper wire, so its mass does not change. In any case, the mass of electrons is way too small to make more than a negligible contribution to the total mass of the wire anyway.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Again, &quot;electricity&quot; is a general term. The link you mentioned refers to &lt;em&gt;electric current&lt;/em&gt;, which is the same as &lt;em&gt;flow of charge&lt;/em&gt;. Basically, the link says that electrons are not the only tiny balls that carry a charge. This is indeed the case. It's just that in the common case of metals, electrical current is usually carried by electrons.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This question is not well-posed. Again, electricity is a very general term and encompasses things like &lt;em&gt;electric field&lt;/em&gt;, &lt;em&gt;electric current&lt;/em&gt;, &lt;em&gt;electric charge&lt;/em&gt; etc. For instance, &lt;em&gt;light&lt;/em&gt; is part of electricity as well, because it's an &lt;em&gt;electromagnetic wave&lt;/em&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="188" LastActivityDate="2010-12-23T15:08:44.233" CommentCount="11" />
  <row Id="2195" PostTypeId="2" ParentId="2189" CreationDate="2010-12-23T15:58:42.993" Score="2" ViewCount="" Body="&lt;p&gt;Thorlabs has a little information on aspheric lens design on their &lt;a href=&quot;http://www.thorlabs.de/NewGroupPage9.cfm?ObjectGroup_ID=3810&amp;amp;pn=352140-A&quot; rel=&quot;nofollow&quot;&gt;product page&lt;/a&gt; (scroll down below the product pictures and click on &quot;Lens Formula&quot;). Note that &lt;em&gt;no&lt;/em&gt; traditional lens can focus light down to a single point; the minimum size is subject to the diffraction limit and is in the order of the wavelength.&lt;/p&gt;&#xA;" OwnerUserId="107" LastEditorUserId="107" LastEditorDisplayName="" LastEditDate="2011-01-09T16:32:25.470" LastActivityDate="2011-01-09T16:32:25.470" CommentCount="3" />
  <row Id="2196" PostTypeId="1" AcceptedAnswerId="2198" CreationDate="2010-12-23T15:59:54.977" Score="5" ViewCount="189" Body="&lt;p&gt;Another interesting infinite lattice problem I found while watching a &lt;a href=&quot;http://www.google.co.uk/search?aq=1&amp;amp;oq=stephen+haw&amp;amp;sourceid=chrome&amp;amp;ie=UTF-8&amp;amp;q=stephen+hawking+universe#q=stephen+hawking+universe&amp;amp;hl=en&amp;amp;safe=off&amp;amp;prmd=ivnso&amp;amp;source=univ&amp;amp;tbs=vid:1&amp;amp;tbo=u&amp;amp;ei=znETTc_lHOCAhAeerum2Dg&amp;amp;sa=X&amp;amp;oi=video_result_group&amp;amp;ct=title&amp;amp;resnum=1&amp;amp;ved=0CDsQqwQwAA&amp;amp;fp=ebe8500894cfa2f2&quot; rel=&quot;nofollow&quot;&gt;physics documentary&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine an infinite square lattice of point masses, subject to gravity. The masses involved are all $m$ and the length of each square of the lattice is $l$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the symmetries of the problem the system should be in (unstable) balance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens if a mass is removed to the system? Intuition says that the other masses would be repelled by the hole in a sort of &quot;anti-gravity&quot;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is my intuition correct? &lt;/li&gt;&#xA;&lt;li&gt;Is it possible to derive analytically a formula for this apparent repulsion force?&lt;/li&gt;&#xA;&lt;li&gt;If so, is the &quot;anti-gravity&quot; force expressed by $F=-\frac{Gm^2}{r^2}$, where $r$ is the radial distance of a point mass from the hole?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Video here: &lt;a href=&quot;http://www.youtube.com/watch?v=mPJ3TUAwI74&amp;amp;feature=player_detailpage#t=408s&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=mPJ3TUAwI74&amp;amp;feature=player_detailpage#t=408s&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-12-23T17:22:01.257" LastActivityDate="2010-12-23T20:55:53.117" Title="Anti-gravity in an infinite lattice of point masses" Tags="&lt;classical-mechanics&gt;&lt;gravity&gt;&lt;newtonian-mechanics&gt;&lt;newtonian-gravity&gt;" AnswerCount="2" CommentCount="11" />
  <row Id="2197" PostTypeId="1" AcceptedAnswerId="2450" CreationDate="2010-12-23T16:05:07.423" Score="6" ViewCount="317" Body="&lt;p&gt;If you take a hardboiled egg and put it on a table and start to spin it, if you spin it fast enough it will start to spin in an upright position. What is the angular velocity needed for this transition to occur?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is energy conserved in the transition?&#xA;Must there be some imperfection in the egg for the transition to occur?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume a prolate spheroid egg of constant density.&#xA;If that doesnt work, assume its center of mass is shifted abit, if that dont work, assume an egg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: One is allowed to use simpler shapes of the same symmetry type, to avoid messy integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And most importantly, can you explain without equations, why does the egg prefer to spin in an upright position ?&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2010-12-30T16:53:30.410" LastActivityDate="2011-01-02T16:21:18.280" Title="How fast do you have to spin an egg to have it standing?" Tags="&lt;classical-mechanics&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="2198" PostTypeId="2" ParentId="2196" CreationDate="2010-12-23T16:32:36.217" Score="2" ViewCount="" Body="&lt;p&gt;I think that your initial intutiion is right--before the point particle is removed, you had (an infinite set of) two $\frac{G\,m\,m}{r^{2}}$ forces balancing each other, and then you remove one of them in one element of the set.  So initially, every point particle will feel a force of $\frac{G\,m\,m}{r^{2}}$ away from the hole, where $r$ is the distance to the hole.  An instant after that, however, all of the particles will move, and in fact, will move in such a way that the particles closest to the hole will be closer together than the particles farther from the hole.  The consequence is that the particles would start to clump in a complicated way (that I would expect to depend on the initial spacing, since that determines how much initial potential energy density there is in the system)&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-12-23T16:32:36.217" CommentCount="3" />
  <row Id="2200" PostTypeId="1" CreationDate="2010-12-23T17:19:51.497" Score="1" ViewCount="94" Body="&lt;p&gt;How big could an array of space telescopes acting as an interferometer be ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How big would it have to be to resolve exoplanet surface detail the size of Iceland at a distance of 100 light years ?&lt;/p&gt;&#xA;" OwnerUserId="945" LastActivityDate="2010-12-23T20:55:09.143" Title="Exoplanet surface detail: Limitations on size of space telescope array" Tags="&lt;astronomy&gt;" AnswerCount="1" />
  <row Id="2201" PostTypeId="1" AcceptedAnswerId="2226" CreationDate="2010-12-23T17:20:50.587" Score="3" ViewCount="138" Body="&lt;p&gt;I recently heard about jet quenching concerning data taken by the experiments at the LHC. Apparently it is related to the existence to the quark-gluon plasma. As far as I understood this interpretation is an analogy to hydrodynamics (the jet is &quot;blocked&quot; by the plasma).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to understand that in more details: why is this analogy a &quot;good&quot; one, how far can it go and how do we detect that experimentaly.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2011-09-03T05:44:19.427" Title="What is jet quenching and how far can the hydrodynamic analogy go?" Tags="&lt;experimental-physics&gt;&lt;high-energy-physics&gt;&lt;large-hadron-collider&gt;&lt;heavy-ion&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="2202" PostTypeId="2" ParentId="2196" CreationDate="2010-12-23T18:19:06.837" Score="2" ViewCount="" Body="&lt;p&gt;I assume by square lattice you mean a 3D &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_crystal_system&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;cubic&lt;/em&gt; lattice&lt;/a&gt; because there's no translational symmetry along the $z$-axis for a 2D &lt;em&gt;square&lt;/em&gt; lattice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the masses are located at $(n_x, n_y, n_z)$ where $n_{x,y,z}\in\mathbb Z$. Let's also define the unit of mass and length so that $m=l=1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the total force acted on the mass point at (0, 0, 0) just due to the 1st octant $(x&amp;gt;0,y&amp;gt;0,z&amp;gt;0)$:&#xA;$$\begin{aligned}&#xD;&#xA;\mathbf F_{+++} &amp;amp;= -G \sum_{n_x=1}^\infty \sum_{n_y=1}^\infty \sum_{n_z=1}^\infty &#xD;&#xA;\frac{n_x \hat{\mathbf x} + n_y \hat{\mathbf y} + n_z \hat{\mathbf z}}{(n_x^2+n_y^2+n_z^2)^{3/2}} \\&#xD;&#xA;&amp;amp;= -G \left( \hat{\mathbf x} \sum_{n_x=1}^\infty \sum_{n_y=1}^\infty \sum_{n_z=1}^\infty \frac{n_x}{(n_x^2+n_y^2+n_z^2)^{3/2}} + \dotsb \right),&#xD;&#xA;\end{aligned}$$&#xA;however, the sum actually &lt;em&gt;diverges&lt;/em&gt;, since,&#xA;$$\begin{aligned}\sum_{n_x=1}^\infty \sum_{n_y=1}^\infty \sum_{n_z=1}^\infty \frac{n_x}{(n_x^2+n_y^2+n_z^2)^{3/2}} &amp;amp;\ge  \int_1^\infty \int_1^\infty \int_1^\infty \frac{n_x}{(n_x^2+n_y^2+n_z^2)^{3/2}} dn_x dn_y dn_z \\&#xD;&#xA;&amp;amp;= \int_1^\infty \int_1^\infty \frac1{\sqrt{1+n_y^2+n_z^2}} dn_y dn_z \\ &amp;amp;= \infty, \end{aligned} $$&#xA;so while symmetry suggests that the force at center is &lt;strong&gt;0&lt;/strong&gt;, mathematically it is not well defined.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Of course, if we assume the net force can be well-defined as &lt;strong&gt;0&lt;/strong&gt; (e.g. the gravity actually decays faster than $1/r^3$!), then Points 1 and 3 are correct. When we remove a particle from the lattice, the contribution $-\frac{GMm\hat{\mathbf r}}{r^\alpha}$ will be subtracted from it, so it is as if there is a particle of negative mass $-m$ put to that point. This is because force is additive and gravity is proportional to mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Yeah this is stating the obvious.)&lt;/p&gt;&#xA;" OwnerUserId="56" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-12-23T20:55:53.117" LastActivityDate="2010-12-23T20:55:53.117" CommentCount="8" />
  <row Id="2204" PostTypeId="1" AcceptedAnswerId="2205" CreationDate="2010-12-23T18:47:09.253" Score="5" ViewCount="316" Body="&lt;p&gt;I'm interested in the meaning of the phrase &quot;continuum limit&quot; specifically as it is often used in expressions relating to the ability of a quantum gravity theory to recover GR in the continuum limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe I understand the meaning but want to make sure I am not missing some important part of the precise definition as my intuition may be off and I have not seen it defined anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pointers to a place where it is defined in an online resource would be appreciated. A google search just turned up many references to it being used in papers and articles and such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="948" LastActivityDate="2010-12-25T13:36:46.410" Title="What specifically does the phrase &quot;continuum limit&quot; mean?" Tags="&lt;quantum-gravity&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="2205" PostTypeId="2" ParentId="2204" CreationDate="2010-12-23T19:10:59.923" Score="6" ViewCount="" Body="&lt;p&gt;Usually it relates to discrete models where it means &quot;becoming less discrete&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make this more formal and give a simple example, consider a set $$\Lambda_{\alpha}(M) := \{ x \in {\mathbb R}^d \,|\, x_i = k \alpha, \, 0 \leq k \leq \lfloor {M \over \alpha} \rfloor \}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and imagine there is an edge between nearest neighbors so that part of it looks like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/TYV7x.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this can be thought of as discretization of a $d$-dimensional cube of side $M$ modelling some crystal. Now by &lt;em&gt;continuum limit&lt;/em&gt; one understand the limit $\alpha \to 0$. In this case one would recover the original cube $[0, M]^d$. The reason this is important is that lattice is a much simpler object than a continuous space and we can compute lots of things there directly. In some cases we are then able to carry out limits and recover the original continuous theory.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for your talk about GR, I think there is only one possible interpretation in the context of &lt;a href=&quot;http://en.wikipedia.org/wiki/Loop_quantum_gravity&quot; rel=&quot;nofollow&quot;&gt;Loop Quantum Gravity&lt;/a&gt; which deals with discrete space-time. I don't know much about this theory but I'd assume that in order to recover GR one would need to perform the continuum limit of the space-time spacing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But note that LQG doesn't mean quantum gravity in general. It is just one of proposed theories for quantum gravity and other theories (e.g. string theory) don't assume discrete space-time, so there is no continuum limit to carry out.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; David has made a very good point about the nature of continuum limit, so I decided to elaborate on the matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two semantically different concepts of continuum limit (although mathematically they are the same):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;One can start with a discrete microscopic theory that is already complete (e.g. lattice of crystals; or with space-time in LQG). In this case the semantics of continuum limit is &lt;em&gt;getting rid of microscopic physics&lt;/em&gt;. Once you perform the limit, you obtain a simpler theory.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;But sometimes the complete theory can already be continuous (e.g. Quantum Field Theory) and one first wants to discretize it in order to be able to perform the calculations more easily. In this case when we are shrinking the scale we are actually making the discrete model &lt;em&gt;a better approximation to the original theory&lt;/em&gt; until in continuum limit we recover the original theory completely. So here we also lose the microscopic degrees of freedom but we don't care because the discrete model wasn't physical anyway -- it was just a mathematical tool.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-23T22:36:22.523" LastActivityDate="2010-12-23T22:36:22.523" CommentCount="12" />
  <row Id="2206" PostTypeId="1" AcceptedAnswerId="2209" CreationDate="2010-12-23T19:21:04.553" Score="9" ViewCount="383" Body="&lt;p&gt;What ever happened to &quot;action at a distance&quot; in entangled quantum states, i.e. the Einstein-Rosen-Podulsky paradox? I thought they argued that in principle one could communicate faster than speed of light with entangled, separated states when one wave function gets collapsed. I imagine this paradox has been resolved, but I don't know the reference.&lt;/p&gt;&#xA;" OwnerUserId="775" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-05T11:42:05.060" LastActivityDate="2011-02-05T11:42:05.060" Title="Is there such a thing as &quot;Action at a distance?&quot;" Tags="&lt;quantum-mechanics&gt;&lt;quantum-information&gt;&lt;entanglement&gt;&lt;paradox&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="2207" PostTypeId="1" AcceptedAnswerId="2224" CreationDate="2010-12-23T20:02:25.587" Score="7" ViewCount="373" Body="&lt;p&gt;It's a Christmas time and so I hope I'll be pardoned for asking a question which probably doesn't make much sense :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In standard undergraduate nuclear physics course one learns about models such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Liquid-drop_model&quot; rel=&quot;nofollow&quot;&gt;Liquid Drop model&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_shell_model&quot; rel=&quot;nofollow&quot;&gt;Shell model&lt;/a&gt; that explain some properties of nucleus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as I understand it these models are purely empirical and most importantly &lt;em&gt;incompatible&lt;/em&gt;. Perhaps it's just my weak classical mind but I can't imagine same object being described as a liquid with nucleons being the constituent particles floating freely all around the nucleus and on the other hand a shell model where nucleons occupy discrete energy levels and are separated from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I wonder whether these empirical models are really all we've got or whether there are some more precise models. I guess one can't really compute the shape of the nucleus from the first principles as one can do with hydrogen atom in QM. Especially since first principles here probably means starting with QCD (or at least nucleons exchanging pions, but that is still QFT). But I hope there has been at least some progress since the old empirical models. So we come to my questions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;Do we have a better model for description of a nucleus than the ones mentioned?&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;How would some nuclei (both small and large) qualitatively look in such a better model? &lt;em&gt;Look&lt;/em&gt; here means that whether enough is known so that I could imagine nucleus in the same way as I can imagine an atom (i.e. hard nucleus and electrons orbiting around it on various orbitals).&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;What is the current state of first-principles QCD computations of the nucleus?&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="329" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-12-23T21:23:12.653" LastActivityDate="2010-12-24T18:59:04.323" Title="What does a nucleus look like?" Tags="&lt;quantum-field-theory&gt;&lt;nuclear-physics&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="2208" PostTypeId="2" ParentId="2200" CreationDate="2010-12-23T20:55:09.143" Score="1" ViewCount="" Body="&lt;p&gt;Well, with current or immediate-future technology, let's use &lt;a href=&quot;http://lisa.nasa.gov/&quot; rel=&quot;nofollow&quot;&gt;LISA&lt;/a&gt; as an example of the largest baseline achievable at $5\times 10^6$ km. Then the diffraction limit says $\theta \sim \frac{\lambda}{D}$, where $\theta$ is angular resolution, in this case ($10^7$ cm)/($10^{20}$ cm)=$10^{-13}$ radians. At optical wavelengths, the LISA-type detector would have resolution of $10^{-16}$, so yes, in principle, this seems possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest problem I think I am ignoring is flux: there wouldn't be enough photons coming from exo-Iceland to do proper interferometry, and then there is also the problem of contrast: what is the difference in flux between land and sea features, aside from contrast with host star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With current radio telescopes, a much-quoted factoid is that you could read the date on a dime on the moon, but in practice you never could because it doesn't emit in the radio, and certainly not with that contrast in signal.&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2010-12-23T20:55:09.143" />
  <row Id="2209" PostTypeId="2" ParentId="2206" CreationDate="2010-12-23T21:28:39.737" Score="12" ViewCount="" Body="&lt;p&gt;It's not possible to communicate faster than light using entangled states. All you get out of entanglement is a &lt;em&gt;correlation&lt;/em&gt; between the values of two measurements.; the entanglement doesn't allow you to &lt;em&gt;influence&lt;/em&gt; the value measured at another location in a non-causal way. In other words, the correlation only becomes evident after combining the results from the measurements afterwards, for which you need classical information transfer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider the thought experiment described on the &lt;a href=&quot;http://en.wikipedia.org/wiki/EPR_paradox&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; for the EPR paradox: a neutral pion decays into an electron and a positron, emitting them in opposite directions and with opposite spins. However, the actual value of the spin is undetermined, so with respect to a spin measurement along a chosen axis, the electron and positron are in the state&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\sqrt{2}}\left(|e^+ \uparrow\rangle|e^- \downarrow\rangle + |e^+ \downarrow\rangle|e^- \uparrow\rangle\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you measure the spin of the positron along this chosen axis. If you measure $\uparrow$, then the state will collapse to $|e^+ \uparrow\rangle|e^- \downarrow\rangle$, which determines that the spin of the electron must be $\downarrow$; and vice versa. So if you and the other person (who is measuring the electron spin) get together and compare measurements afterwards, you'll always find that you've made opposite measurements for the spins. But there is no way to &lt;em&gt;control&lt;/em&gt; which value you measure for the spin of the positron, which is what you'd need to do to send information. As long as the other person doesn't know what the result of your measurement is, he can't attach any informational value to either result for his measurement.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-23T21:54:51.890" LastActivityDate="2010-12-23T21:54:51.890" CommentCount="1" />
  <row Id="2210" PostTypeId="2" ParentId="2206" CreationDate="2010-12-23T21:31:18.563" Score="8" ViewCount="" Body="&lt;p&gt;Well, the problem in that paradox is that yes, one of the parties will measure the entangled particle to get the wave function collapsed and yes it will collapse for the other party. However, the other party will still have to measure the thing to learn what it is or has to wait for the initial party to send them a message telling what the wave function has collapsed to. The first method will result in a 50% +x and 50% -x(if it is spin you are measuring), as the wave function that collapsed can collapse to either one of these states. So the fact that the wave function collapsed does not really transfer any usable information to the other side. The second method is capped with the speed of light anyway.&lt;/p&gt;&#xA;" OwnerUserId="504" LastActivityDate="2010-12-23T21:31:18.563" CommentCount="1" />
  <row Id="2211" PostTypeId="2" ParentId="2206" CreationDate="2010-12-23T22:27:31.020" Score="-1" ViewCount="" Body="&lt;p&gt;Being silly, in the vein of the season, if you agreed before hand that if in David's example your colleague could open their christmas present on the other side of the world if their electron was spin up, and the same time they measured their electron you measured your positron as spin down, you now know they have opened their present, faster than the speed of light&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instantaneous data transfer!&lt;/p&gt;&#xA;" OwnerUserId="909" LastActivityDate="2010-12-23T22:27:31.020" CommentCount="8" />
  <row Id="2212" PostTypeId="2" ParentId="2187" CreationDate="2010-12-23T22:36:14.287" Score="4" ViewCount="" Body="&lt;p&gt;the magnification is a property of diverging light. it really needs a diagram, but the further away the detector is from the focal spot, the bigger the image will appear&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this is the best picture I could find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/lGMGP.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the picture is a bit sloppy because it implies the focal spot is in the lens, which is not usually true, but it gives you the concept&lt;/p&gt;&#xA;&#xA;&lt;p&gt;try this yourself with a magnifying glass - the further you hold your eye from the glass (not moving the object) the bigger the object should appear&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the same is true if you move the magnifying glass away from the object, but for a slightly more complex reason about how far the lens can bend the light&lt;/p&gt;&#xA;" OwnerUserId="909" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-12-23T22:42:33.367" LastActivityDate="2010-12-23T22:42:33.367" CommentCount="2" />
  <row Id="2213" PostTypeId="2" ParentId="401" CreationDate="2010-12-23T23:08:59.877" Score="8" ViewCount="" Body="&lt;p&gt;I'll interpret your term &lt;em&gt;diagram&lt;/em&gt; as &quot;any fancy image that captures some physics&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this I can hardly recommend anything else then &lt;a href=&quot;http://en.wikipedia.org/wiki/MetaPost&quot; rel=&quot;nofollow&quot;&gt;MetaPost&lt;/a&gt;. It's on par with TeX in being a little hard to learn but once you do master the basics you won't believe you could have ever used anything else (in particular, GIMP and Inkscape; good analogy here would be to TeX vs. MS Word).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Basic properties&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;it's a (simple) programming language&lt;/li&gt;&#xA;&lt;li&gt;it's vectorial (this should probably go without saying but still)&lt;/li&gt;&#xA;&lt;li&gt;it's primitives are things like points, lines, paths, splines&lt;/li&gt;&#xA;&lt;li&gt;it contains excellent image manipulation facilities; you can say things like &quot;take this image, scale it up by two and rotate it by 60 degrees&quot;&lt;/li&gt;&#xA;&lt;li&gt;you can insert TeX labels&lt;/li&gt;&#xA;&lt;li&gt;it can solve equations; This is a real killer that no one else offers. You can draw two curves $X(t)$ and $Y(t)$ (defined most comfortably as splines) and tell MP to compute their intersection, draw a point there and label it with some text&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Success story&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I used MP to create some polygons on a hexagonal lattice in the context of cluster expansions. There was a huge number of those polygons to draw, so I quickly abandoned all hope of trying to draw them by hand in Inkscape or something similar. True, it would probably be quicker in the end, but I hate manual work; I rather spend much longer learning some programming language and then just code all the work in few minutes. So I put together simple MP program that has converted my input data (vertices and edges of the polygons as just numbers) into beautiful images. For a one night's work and my first time with MP I was more than satisfied.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Goodies&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/xCb9P.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Credit for this amazing picture goes to &lt;a href=&quot;http://kahrstrom.com/mathematics/&quot; rel=&quot;nofollow&quot;&gt;Johan Kåhrström&lt;/a&gt; (go also see more stuff under &lt;em&gt;illustrations&lt;/em&gt; there).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.cs.toronto.edu/~mackay/metapost/&quot; rel=&quot;nofollow&quot;&gt;http://www.cs.toronto.edu/~mackay/metapost/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ursoswald.ch/metapost/tutorial.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ursoswald.ch/metapost/tutorial.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://commons.wikimedia.org/wiki/File:Contravariant_Coordinates.png&quot; rel=&quot;nofollow&quot;&gt;http://commons.wikimedia.org/wiki/File:Contravariant_Coordinates.png&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-01-13T09:21:03.427" LastActivityDate="2011-01-13T09:21:03.427" CommunityOwnedDate="2010-12-23T23:08:59.877" CommentCount="5" />
  <row Id="2215" PostTypeId="2" ParentId="401" CreationDate="2010-12-23T23:41:25.640" Score="3" ViewCount="" Body="&lt;p&gt;Related post in &lt;a href=&quot;http://stackoverflow.com/questions/4005042/language-or-package-for-drawing-geometric-diagrams&quot;&gt;SO&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My personal favorite is &lt;a href=&quot;http://asymptote.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;Asymtpote&lt;/a&gt; which is like MetaPost on steroids. A gallery is &lt;a href=&quot;http://asymptote.sourceforge.net/gallery/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-12-23T23:41:25.640" CommunityOwnedDate="2010-12-23T23:41:25.640" CommentCount="1" />
  <row Id="2216" PostTypeId="2" ParentId="2207" CreationDate="2010-12-24T00:03:22.580" Score="2" ViewCount="" Body="&lt;p&gt;When you get down to these length scales, you have to be careful with words such as &quot;look&quot;. In my mind, the &quot;answer&quot; for this question requires a roundabout in the philosophical area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Look&quot; - what does it mean in physics? Even more relevant: &quot;is&quot;. When a physicists uses the word &quot;is&quot;, s/he is really using a shortcut for saying &quot;behaves as it would be&quot;, that is, it behaves according to some model (this is a realization that took a couple of years for me to reach. Of course, I might change it in the future as I continue my physics intake). This is not only true for quantum physics, but in the macroscopic world, the distinction is often thought of as not as important. No one can deny the following though: a rod is a model for a certain object, and we can prescribe this model to real-life objects and calculate. In some cases, the physical object that we one day called a rod might be called a spring, depending on what we want to do. What &quot;is&quot; the object? In the physicists eyes, it is both a rod and a spring, and there is no incompatibility, as long as one knows what one is doing. What does it &quot;look&quot; like? It depends on the length scale and the application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the quantum world, de Broglie says that the constituents are both particles and waves. Contradiction? No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does a nucleus &quot;look&quot;? Which model is &quot;right&quot;? It depends. How I would intuitively imagine it in my mind is as a continuous domain of a charge (strong/weak/EM) cloud. Is this &quot;correct&quot;? It is one possible model, that works well in some applications. In other applications, it is just a hard sphere/wave/point/oscillator...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry about the flaky amateur-philosophical mumbo-jumbo :-) Question to think about: what does &quot;is&quot; mean for a physicist (or a mathematician, or many other people).&lt;/p&gt;&#xA;" OwnerUserId="949" LastActivityDate="2010-12-24T00:03:22.580" CommentCount="19" />
  <row Id="2217" PostTypeId="2" ParentId="2188" CreationDate="2010-12-24T00:15:46.860" Score="0" ViewCount="" Body="&lt;p&gt;I agree with greg above, that the properties of electricity are irrelevant in law. There is (as far as I know) no good set of precedents/examples to quote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I justify my position is the electrons are not related to the good in any way, other than being a transmission medium. A newspaper article was not a great example, because the paper is always the same, it is attached to the good. Perhaps a closer example is saying a TV broadcast is not made valuable by the radiowaves it was broadcast on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrons required (or not, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Optical_computing&quot; rel=&quot;nofollow&quot;&gt;optical computing&lt;/a&gt;) to form a 'electrical good' are not constant - ie the same electrons do not remain as part of that 'good'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing that never changes, the basis of what that good is, is the coded description of it. The ones and zeros as Sklivvz put it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working out whether electricity is a physical thing is irrelevant, because the 'good' would theoretically be made of &lt;strong&gt;every electron in the world&lt;/strong&gt;, as they could all be used at different times to make those ones and zeros.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of an electrical good as I see it is defined by 2 things - the intellectual property in it, and the accepted ownership of it. Like a famous painting the has been restored, you arent paying for the pigments, or even the brushstrokes.&lt;/p&gt;&#xA;" OwnerUserId="909" LastActivityDate="2010-12-24T00:15:46.860" />
  <row Id="2218" PostTypeId="2" ParentId="2206" CreationDate="2010-12-24T02:40:08.033" Score="4" ViewCount="" Body="&lt;p&gt;Let's be more rigorous. No-signalling has been proven safely and shouldn't be worried about. Nevertheless, you'd notice that the point of EPR paper was to show that if quantum mechanics is considered to be a description of &quot;reality&quot;, then it is &quot;incomplete&quot;. There is an approach, such as in operationalism, to say quantum mechanics isn't meant to be a description of &lt;em&gt;reality&lt;/em&gt;. It's a description of our &lt;em&gt;knowledge of reality&lt;/em&gt;, due to Asher Peres. Another approach is to say we can give an ontological model of quantum theory using contextual hidden variables, such as the one in de Broglie-Bohm model.&#xA;So conclusion: EPR argument hasn't been resolved if you mean it's gone! Because in fact orthodox quantum mechanics isn't a complete description of reality. However, it doesn't mean one can signal faster than light!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some interesting extra information: There is &lt;a href=&quot;http://arxiv.org/pdf/0706.2661v1&quot; rel=&quot;nofollow&quot;&gt;an interesting paper&lt;/a&gt; which kind of analyses Einstein's argument. It bring historical facts that show Einstein didn't like the EPR and wrote another paper with the same title in correspondence with Schrödinger, and the one with Rosen and Podolsky was never reviewed by him. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This quotation is from a letter of Einstein to Schrödinger, dated June 19, 1935:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;“For reasons of language this [paper] was written by Podolsky after many discussions. But still it has not come out as well as I really wanted; on the contrary, the main point was, so to speak, buried by the erudition.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; A source of confusion in my answer has been pointed out by Marek. I'll try to clarify here:&#xA;Scientific realism assumes there is an underlying objective reality which has attributes regardless of them being measured by an observer. &#xA;One can suggest a model which ignores such reality and say &quot;...there is no logical necessity for a realistic worldview to always be obtainable&quot;(Fuchs and Peres, Physics Today &lt;em&gt;53 (3), 70-71.&lt;/em&gt;).&#xA;On the other hand, one can offer an ontological model which in this case it can be located in 3 different category as in figure below:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/XUFsV.jpg&quot; alt=&quot;alt text&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe Einstein had in mind to show that quantum mechanics can't give a picture of type (a). Which was successful. Because, even if there is an underlying reality, quantum states can't sharply specify them by any means. &lt;/p&gt;&#xA;" OwnerUserId="304" LastEditorUserId="304" LastEditorDisplayName="" LastEditDate="2010-12-24T10:59:45.327" LastActivityDate="2010-12-24T10:59:45.327" CommentCount="6" />
  <row Id="2219" PostTypeId="1" CreationDate="2010-12-24T04:21:41.817" Score="2" ViewCount="341" Body="&lt;p&gt;I'm interested in learning physics. I do realize that the subject is large and that it would be easier if I had a specific area of interest. However, I do not. I suppose I want to learn about the fundamentals of it all; the axioms that combine all physics fields. Or, in other words, a high school physics class. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, a book or series of videos would be helpful. I looked over MIT and unfortunately the material wasn't for me. I don't mean to be &quot;picky&quot; so I am not completely ruling out any resource just yet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="951" LastActivityDate="2010-12-24T06:10:37.637" CommunityOwnedDate="2010-12-24T06:59:54.527" Title="Beginner Physics Resources?" Tags="&lt;education&gt;" AnswerCount="6" CommentCount="9" FavoriteCount="3" />
  <row Id="2220" PostTypeId="2" ParentId="2219" CreationDate="2010-12-24T04:31:46.563" Score="6" ViewCount="" Body="&lt;p&gt;I say, start with the &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Feynman_Lectures_on_Physics&quot; rel=&quot;nofollow&quot;&gt;Feynman lectures&lt;/a&gt;. Here's a link to them in &lt;a href=&quot;http://student.fizika.org/~jsisko/Knjige/Opca%20Fizika/Feynman%20Lectures%20on%20Physics/&quot; rel=&quot;nofollow&quot;&gt;PDF form&lt;/a&gt;. You can even &lt;a href=&quot;http://research.microsoft.com/apps/tools/tuva/index.html&quot; rel=&quot;nofollow&quot;&gt;watch them online&lt;/a&gt;. There are also many online physics courses by reputed physicists on Youtube. Some with Leonard Susskind, but I think they are advanced.&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2010-12-24T04:31:46.563" CommunityOwnedDate="2010-12-24T06:59:54.527" CommentCount="5" />
  <row Id="2221" PostTypeId="2" ParentId="2219" CreationDate="2010-12-24T04:51:29.513" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.learner.org/courses/physics/&quot; rel=&quot;nofollow&quot;&gt;Physics for the 21st century&lt;/a&gt;is a new, free, on-line course that explores the most modern development in physics. It includes videos, readings, and interactive labs. It's quite good, and represents a modern reformulation of The Mechanical Universe and Beyond, a series of videos from the mid-eighties that covers all of introductory physics—also &lt;a href=&quot;http://www.learner.org/resources/series42.html&quot; rel=&quot;nofollow&quot;&gt;available online for free.&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="807" LastActivityDate="2010-12-24T04:51:29.513" CommunityOwnedDate="2010-12-24T06:59:54.527" CommentCount="1" />
  <row Id="2222" PostTypeId="2" ParentId="2219" CreationDate="2010-12-24T05:18:32.127" Score="0" ViewCount="" Body="&lt;p&gt;You could try the &lt;a href=&quot;http://www.youtube.com/user/khanacademy#p/c/AD5B880806EBE0A4&quot; rel=&quot;nofollow&quot;&gt;Khan Academy&lt;/a&gt;'s room on Physics. It's been known to be an excellent resource, even to the likes of Bill Gates.&lt;/p&gt;&#xA;" OwnerUserId="871" LastActivityDate="2010-12-24T05:18:32.127" CommunityOwnedDate="2010-12-24T06:59:54.527" />
  <row Id="2223" PostTypeId="2" ParentId="2219" CreationDate="2010-12-24T06:10:37.637" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471320579&quot; rel=&quot;nofollow&quot;&gt;The book by by Resnick and Halliday&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/oMeVL.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to solve as many problems as possible. :) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;May the force be with you! &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-12-24T06:10:37.637" CommunityOwnedDate="2010-12-24T06:59:54.527" />
  <row Id="2224" PostTypeId="2" ParentId="2207" CreationDate="2010-12-24T06:57:45.480" Score="2" ViewCount="" Body="&lt;p&gt;I can't give you a real answer but I figured this may be better than nothing:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I don't know of any better model, although I'm not an expert in nuclear physics so I can't be sure one doesn't exist.&lt;/li&gt;&#xA;&lt;li&gt;I suspect that any better model would not give a nice neat physical picture of the nucleus, the way we have a nice neat physical picture of the atom. In particular, I wouldn't expect to see a model that predicts concentric wavefunctions for nucleons, the way quantum mechanics describes electrons. Protons and neutrons are complex composite particles with many internal (and external) interactions, so their wavefunctions will tend to be much more localized than those of electrons. This is one case where the classical particle model actually is not too bad.&lt;/li&gt;&#xA;&lt;li&gt;From what I've heard (from people who know this business), the current state of the art in lattice QCD computations is calculating the mass of the proton with some accuracy. They're a long way away from being able to actually simulate a whole nucleus.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-12-24T06:57:45.480" CommentCount="1" />
  <row Id="2225" PostTypeId="2" ParentId="2188" CreationDate="2010-12-24T08:03:08.617" Score="2" ViewCount="" Body="&lt;p&gt;At the risk of being redundant, I thought I would add my take on this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;What is 'electricity', really, on a fundamental level; but explained in terms a layman can understand? Is there an analogy with other things that is accurate, unlike the 'flowing water' analogy, which is sufficient for high school level but is a simplification? (at least, I think...)&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First, let me say that in physics, we don't really talk about &quot;electricity&quot; in general, because that word doesn't have a precise definition in physics. As far as we're concerned, usually, it's just the stuff that makes our computers work ;-) But I think if you really pushed a physicist to define electricity, they'd probably come up with something like &quot;energy and/or information transmitted through the movement of electric charge.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the analogy of water might actually not be a bad one, but you have to get a little more general than just a simple flow through a pipe. For instance, one way to transmit information and energy through water is by sending waves from one end to the other. You could imagine dropping stones into a pond on one side of the pond, and a person on the other side could look at the pattern of ripples produced by your stones and read a message from them. Or they could set up a tiny &lt;a href=&quot;http://en.wikipedia.org/wiki/Water_wheel&quot; rel=&quot;nofollow&quot;&gt;water wheel&lt;/a&gt; that would be pushed by the waves, and use it to power some sort of device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's sort of like the way our electricity works, at least the kind we get off the public power grid. The water is analogous to electrons, and the ripples are like alternating current (so named because the direction of motion of the electrons - or water molecules - alternates back and forth as the waves pass by). Note that the water itself barely moves at all, it simply acts as a medium for the waves to propagate through. Similarly, with electricity as we typically use it, the electrons barely move at all, they simply act as a medium for waves.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Do 'electricity', an 'electrical current' and an 'electrical charge' have mass, apart from the object they are embodies in? Does the mass of a copper wire change when you put a current through it, because of the electrons coming in and going out?&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well... &lt;em&gt;technically&lt;/em&gt; you &lt;em&gt;could&lt;/em&gt; say that electricity has mass in the sense that running a current through a wire does add a tiny amount of energy to the wire, and when you measure mass (e.g. on a scale) you're really measuring energy instead. But detecting the difference in energy (mass) produced by an electric current would require a scale that is far more sensitive than any we have on Earth, and perhaps far more sensitive than any that we could ever build.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of it this way: electricity has mass only in the same sense that a person walking weighs more than the same person standing still.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;How do electrons fit into this? Is electricity composed of a bunch of electrons that flow through a mass? I think not, reading the link I gave before, but I don't quite understand what their role is.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Going back to my third paragraph, electrons form the medium that &quot;electrical waves&quot; exist in, just like water is the medium that water waves exist in.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Most authors blindly assume that electricity is merely a flow of electrons through mass. In how far and under what assumptions is this correct?&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not a correct description of the electricity we get from our public power grid. As I've said, that sort of electricity involves alternating current, in which the information is transmitted in waves, and the electrons simply form the medium for those waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is another kind of electricity, direct current, in which the electrons actually do flow through the wire. This is what you get from batteries, for instance, and it also is used in parts of many electronic devices. Direct current would be analogous to water flowing through a pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both alternating and direct current are equally capable of transmitting information and/or energy.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;P.S. For what it's worth, I once wrote a &lt;a href=&quot;http://www.ellipsix.net/blog/2009/04/how-much-does-data-weigh.html&quot; rel=&quot;nofollow&quot;&gt;blog post&lt;/a&gt; addressing a similar question, namely how much the data stored on a hard drive would influence its weight. Hard drives basically use tiny magnets, not electric charges or currents, but the general idea is closely related. Note that the weight difference I calculated in that post (10&lt;sup&gt;-14&lt;/sup&gt; grams) is due to the &lt;em&gt;energy&lt;/em&gt; difference between different ways the hard drive could be filled, specifically between two possible extreme configurations that are unlikely to ever occur in reality.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-12-24T08:03:08.617" />
  <row Id="2226" PostTypeId="2" ParentId="2201" CreationDate="2010-12-24T08:30:33.953" Score="4" ViewCount="" Body="&lt;p&gt;As you probably know, in high-energy hadron collisions, you typically have a quark in one of the incoming hadrons scatter off a quark in the other incoming hadron (deep inelastic scattering). For the brief moment in which the quarks are interacting, they can be considered as free particles to some extent, so they &quot;bounce&quot; off each other and emerge back-to-back along a random axis. Since this typically knocks the quarks out of their original particles, you'll get two jets of daughter hadrons, one for each of the scattered quarks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jet quenching can occur when this process takes place in the presence of a &quot;soup&quot; of many other hadrons. Typically, what happens is that the quark scattering takes place near the edge of the hadron soup. One of the jets will be directed outward, towards lower particle densities, and the other will be directed inward, toward higher particle densities. The latter jet will lose most of its energy due to further collisions and gluon radiation (it gets &quot;quenched&quot;), but the former jet continues out to the detector. So what you see is only one jet of daughter hadrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several years ago I wrote a &lt;a href=&quot;http://www.phys.psu.edu/~dzaslavs/papers/rhicstring.f06.pdf&quot; rel=&quot;nofollow&quot;&gt;research paper&lt;/a&gt; about calculating the drag force on a quark in a quenched jet. Figure 5 on page 12 shows jet quenching in action. (I'll include the figure itself here once I get back to my other computer which has the source graphic on it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogy between jet quenching and hydrodynamics is not exactly in my area of expertise, but from what I've heard, hydrodynamic models (developed for classical liquids) have had some success describing certain properties of the quark-gluon plasma. It has been said that this suggests that the QGP is actually more like a liquid than a plasma, but I'm not sure to what extent that's literally true. You don't need the hydrodynamic analogy to explain how jet quenching is able to occur, though, at least not at a qualitative level.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-09-03T05:44:19.427" LastActivityDate="2011-09-03T05:44:19.427" CommentCount="2" />
  <row Id="2227" PostTypeId="1" CreationDate="2010-12-24T08:57:41.070" Score="16" ViewCount="226" Body="&lt;p&gt;Theory predicts that uniform acceleration leads to experiencing thermal radiation (so called Fulling Davies Unruh radiation), associated with the appearance of an event horizon.    For non uniform but unidirectional acceleration the shape of the experienced radiation changes from thermal to other spectral densities, but also is predicted to exist.    But suppose the acceleration is periodic and oscillatory, i.e. no permanent horizon persists?    In particular, what about the case of harmonic motion, for a full cycle, half a cycle, etc.?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an even simpler related problem that makes the apparent paradox easier to see.   Suppose at proper time t=0, I accelerate at constant acceleration k in the x direction for t0 seconds, presumably experiencing Unruh radiation.  Then I accelerate with acceleration -k, (in the -x direction,) for 2*t0 seconds, seeing more Unruh radiation coming from the opposite direction, and then I finish with  with acceleration +k for the final t0 seconds.   At the end of the 4*t0 proper seconds, I'm back where I started, at rest, without any event horizon.   Was the Unruh radiation I felt when reversing acceleration secretly correlated or entangled with the radiation I initially and finally saw?   Otherwise, from a more macro scale, I didn't actually necessarily move much, and the acceleration event horizon was instantaneous, evanescent and fleeting, so whence arose the Unruh radiation?&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="176" LastEditorDisplayName="" LastEditDate="2010-12-26T08:01:37.593" LastActivityDate="2011-08-18T19:31:19.050" Title="Paradox?: What is the form of radiation experienced by a harmonically accelerated observer?" Tags="&lt;mathematical-physics&gt;&lt;thermal-radiation&gt;&lt;relativity&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="4" />
  <row Id="2228" PostTypeId="1" AcceptedAnswerId="2237" CreationDate="2010-12-24T12:50:33.093" Score="6" ViewCount="317" Body="&lt;p&gt;I heard somewhere that quarks have a property called 'colour' - what does this mean?&lt;/p&gt;&#xA;" OwnerUserId="954" LastActivityDate="2010-12-24T16:15:43.003" Title="What is the definition of colour (the quantum state)?" Tags="&lt;quantum-mechanics&gt;&lt;particle-physics&gt;&lt;color&gt;" AnswerCount="2" FavoriteCount="4" />
  <row Id="2229" PostTypeId="1" AcceptedAnswerId="2233" CreationDate="2010-12-24T13:17:10.747" Score="11" ViewCount="1424" Body="&lt;p&gt;As an explanation of why a large gravitational field, such as a black hole, can bend light, I have heard that light has momentum. This is given as a solution to the problem of only massive objects being affected by gravity. However, momentum is the product of mass and velocity, so, by this definition, massless photons can not have momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can photons have momentum? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this momentum defined (equations)?&lt;/p&gt;&#xA;" OwnerUserId="954" LastActivityDate="2011-08-08T16:59:16.493" Title="If photons have no mass, how can they have momentum?" Tags="&lt;energy&gt;&lt;light&gt;&lt;mass&gt;&lt;photons&gt;&lt;momentum&gt;" AnswerCount="8" CommentCount="4" FavoriteCount="1" />
  <row Id="2230" PostTypeId="1" CreationDate="2010-12-24T13:25:21.753" Score="1" ViewCount="565" Body="&lt;p&gt;I was told that the relative speed rule does not apply to the speed of light. No matter how fast two objects are moving, the speed of light will remain same for both of them.&lt;br&gt;&#xA;How and why is this possible? Also why can't anything travel faster than light?&lt;/p&gt;&#xA;" OwnerUserId="847" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-12-27T07:54:08.103" LastActivityDate="2011-02-06T20:19:01.607" Title="Why and how is the speed of light constant?" Tags="&lt;light&gt;&lt;speed-of-light&gt;" AnswerCount="5" CommentCount="11" FavoriteCount="3" />
  <row Id="2231" PostTypeId="2" ParentId="2229" CreationDate="2010-12-24T13:30:09.863" Score="1" ViewCount="" Body="&lt;p&gt;The reason why the path of photons is bent is that the space in which they travel is distorted. The photons follow the shortest possible path (called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Geodesic&quot; rel=&quot;nofollow&quot;&gt;geodesic&lt;/a&gt;) in bent space. When the space is not bent, or flat, then the shortest possible path is a straight line. When the space is bent with some spherical curvature, the shortest possible path lies actually on an equatorial circumference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, this is in General Relativity. In Newtonian gravitation, photons travel in straight lines.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We can associate a momentum of a photon with the De Broglie's relation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p=\frac{h}{\lambda}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $h$ is Planck's constant and $\lambda$ is the wavelength of the photon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also allows us to associate a mass:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m=p/c=h/(\lambda c)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we plug in this mass into the Newtonian gravitational formula, however, the result is not compatible with what is actually measured by experimentation.&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-12-24T16:12:44.997" LastActivityDate="2010-12-24T16:12:44.997" CommentCount="18" />
  <row Id="2232" PostTypeId="2" ParentId="2230" CreationDate="2010-12-24T13:34:40.393" Score="2" ViewCount="" Body="&lt;p&gt;In actual fact, the relative speed rule does not apply, ever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Velocity-addition_formula&quot; rel=&quot;nofollow&quot;&gt;relativistically correct speed addition rule&lt;/a&gt; is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s=\frac{v+u}{1+\frac{vu}{c^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $\frac{vu}{c^2}$ is close to zero (in other words when the velocities invloved are much less than the speed of light, then the correct formula reduces to the Galilean version $s=u+v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing can be faster than light, fundamentally, because as you accelerate you not only gain speed, but also mass. As you approach the speed of light, the energy given to you by the force causing the acceleration basically contributes more and more to the increase of your mass and less and less to the increase of your speed. It does this precisely so you never reach the speed of light. Instead, massless particles like photons always travel at the speed of light.&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-12-24T13:41:57.533" LastActivityDate="2010-12-24T13:41:57.533" />
  <row Id="2233" PostTypeId="2" ParentId="2229" CreationDate="2010-12-24T14:10:23.017" Score="10" ViewCount="" Body="&lt;p&gt;There are two important concepts here that explain the influence of gravity on light (photons).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The theory of Special Relativity, proved in 1905 (or rather the 2nd paper of that year on the subject) gives an equation for the relativistic energy of a particle;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E^2 = (m_0 c^2)^2 + p^2 c^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m_0$ is the rest mass of the particle (0 in the case of a photon). Hence this reduces to $E = pc$. Einstein also introduced the concept of relativistic mass (and the related mass-energy equivalence) in the same paper; we can then write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m c^2 = pc$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ is the &lt;em&gt;relativistic&lt;/em&gt; mass here, hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = p/c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, a photon &lt;em&gt;does have relativistic mass proportional to its momentum&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/De_Broglie_relation&quot; rel=&quot;nofollow&quot;&gt;De Broglie's relation&lt;/a&gt;, an early result of quantum theory (specifically wave-particle duality), states that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda = h / p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $h$ is simply Planck's constant. This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p = h / \lambda$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hence combining the two results, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = E / c^2 = h / \lambda c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;again, paying attention to the fact that $m$ is &lt;em&gt;relativistic mass&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here we have it: photons have 'mass' inversely proportional to their wavelength! Then simply by Newton's theory of gravity, they have gravitational influence. (To dispel a potential source of confusion, Einstein specifically proved that relativistic mass is an extension/generalisation of Newtonian mass, so we should conceptually be able to treat the two the same.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few different ways of thinking about this phenomenon in any case, but I hope I've provided a fairly straightforward and apparent one. (One could go into general relativity for a full explanation, but I find this the best overview.)&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2010-12-25T16:48:14.470" LastActivityDate="2010-12-25T16:48:14.470" CommentCount="14" />
  <row Id="2234" PostTypeId="2" ParentId="2207" CreationDate="2010-12-24T14:45:43.667" Score="2" ViewCount="" Body="&lt;p&gt;Relevant link - &lt;a href=&quot;http://www.nature.com/news/2010/100824/full/4661034a.html?s=news_rss&quot; rel=&quot;nofollow&quot;&gt;Nucleons in platinum nucleus seem to move in a coordinated fashion&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more link - &lt;a href=&quot;http://arxiv.org/abs/0809.2607&quot; rel=&quot;nofollow&quot;&gt;One-neutron halo nucleus Be-11&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="238" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2010-12-24T18:59:04.323" LastActivityDate="2010-12-24T18:59:04.323" CommentCount="5" />
  <row Id="2235" PostTypeId="2" ParentId="2167" CreationDate="2010-12-24T14:45:58.503" Score="-1" ViewCount="" Body="&lt;p&gt;Looking at the first video -- if magnetism was sharply directional (i.e. a little searchlight of attraction coming out from the tip of the magnet) then this device would work. Unfortunately magnetism is (roughly) an omni-directional field and so doesn't just let go of one screw and start tugging on the next. The field attracts strongest the objects that are closer and attracts more weakly the objects that are farther away. If you make the magnet strong enough to start pulling at the screw-head that is next in line it will be strong enough to pull &lt;em&gt;very&lt;/em&gt; strongly at the screw-head that is passing under it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The person who built this wants a force to be applied to the drum at a tangent to the surface of the drum (kind of like a rope wrapped around it) to spin the drum. But any force applied by the magnet is going to be nearly straight out, which won't induce a tendency to spin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof comes from the fact that the device doesn't self-start. When the magnets are brought close to the drum the drum jiggles a bit as it comes to a static resting place, but the supposed spinning force doesn't happen -- the narrator has to start it spinning.&lt;/p&gt;&#xA;" OwnerUserId="956" LastActivityDate="2010-12-24T14:45:58.503" CommentCount="2" />
  <row Id="2236" PostTypeId="2" ParentId="2228" CreationDate="2010-12-24T15:49:21.240" Score="7" ViewCount="" Body="&lt;p&gt;Let me give a straightforward and basic answer here; perhaps someone else can elaborate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way to think of &lt;em&gt;colour&lt;/em&gt; is something analogous to &lt;em&gt;charge&lt;/em&gt; in electromagnetism. (Indeed, colour is often called &lt;em&gt;colour charge&lt;/em&gt;). It is the fundamental property of particles relating to the strong force, and like electric charge, comes in discrete values (termed red, green, and blue - don't take them literally though). Note however that strong integration works in a far more complex way than electromagnetism (Maxwellian, or even quantum electrodynamics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia article states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The &quot;color&quot; of quarks and gluons is&#xA;  completely unrelated to visual&#xA;  perception of color.[1] Rather, it is&#xA;  a whimsical name for a property that&#xA;  has almost no manifestation at&#xA;  distances above the size of an atomic&#xA;  nucleus. The term color was chosen&#xA;  because the abstract property to which&#xA;  it refers has three aspects, which are&#xA;  analogized to the three primary colors&#xA;  of red, green, and blue.[2] By&#xA;  comparison, the electromagnetic charge&#xA;  has a single aspect, which takes the&#xA;  values positive or negative.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13" LastActivityDate="2010-12-24T15:49:21.240" CommentCount="2" />
  <row Id="2237" PostTypeId="2" ParentId="2228" CreationDate="2010-12-24T16:15:43.003" Score="9" ViewCount="" Body="&lt;p&gt;You heard well. But note that this doesn't have anything to do with real color. The reason that property was so named is because of accidental similarities to how color mix and because certain physicist have a weird sense of humor.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Quantum electrodynamics&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Note that the thing they are talking about is &lt;em&gt;color charge&lt;/em&gt;. So it seems useful to first review properties of normal (electric) charges. Let's talk about electrons for instance. We know from classical physics that charged objects produce electromagnetic field around them and this field in turn affects charged particles. In quantum theory one has to quantize this field. The quantum of electromagnetic field is photon and so quantum theory tells us that charged objects interact electromagnetically by exchanging photons&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www-sldnt.slac.stanford.edu/alr/images/photon_exch.jpg&quot; alt=&quot;Electron - electron interaction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read the image from left to right. Electron radiates photon, this makes it change direction and the other electron absorbs it changing direction too. This is the way quantum theory explain electromagnetic interaction.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Quantum chromodynamics&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt; the discussion below will not be technically precise because of inherent complexity of quantum theory. I'll elaborate on certain points in a separate paragraph after this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the above knowledge under our belts, it's not that hard to describe Quantum chromodynamics (from greek Χρώμα, chroma meaning color) anymore. Instead of electric charge, objects can carry so called color charge. But there is not just one of them, but three: red, green, blue (yes, this is one of the superficial similarities to usual color).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, we need one more ingredient in our theory -- something that would replace photons. Such a particle exists and it is called &lt;em&gt;gluon&lt;/em&gt;. But note that while it was enough to have only one kind of photon (because there was only one electric charge), we need more gluons to mediate various types of interactions (e.g. interaction between red quark and blue quark, which would be mediated by red-antiblue gluon, and so on). So we have nine type of gluons in total, right? Well, there are actually only eight of them and this is one of the technicalities that I'll address later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that quantum chromodynamics predicts that bound states of colored particles have to be &quot;white&quot; meaning that it should e.g. contain three particles, one red, one blue and one green (this is another similarity with the real color mixing). &quot;White&quot; particles containing three quarks are called baryons and you should know at least two of them: protons and neutrons. Actually, it turns out that there is more than one quark type (six actually) and you get various particles by mixing different kinds of them. Two lights quarks are called &lt;em&gt;up&lt;/em&gt; and &lt;em&gt;down&lt;/em&gt;. Proton contains two ups and one down while neutron contains one up and two downs. Okay, let's see how this works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/3/31/Pn_scatter_quarks.png&quot; alt=&quot;Proton - neutron interaction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is pretty complicated diagram because I wasn't able to find anything simpler, but let me try to explain what's going on. The incoming proton's blue quark emits blue-antired gluon and by doing that changes to red and also change direction a little. This gluon is then caught by its red down quark and this changes it to blue and throws it out of the proton. Similar thing happens with the green up quark, the only difference being that in the end &lt;em&gt;antiblue&lt;/em&gt; up is thrown out of the proton. So we have blue-antiblue up/down pair. This is a color neutral particle and it is called &lt;em&gt;pion&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the point of view of the neutron, exactly the same discussion applies, so I hope this is at least a little clear by now. What that picture actually explained is how proton - neutron strong interactions in nucleus are explained from the point of view of quantum chromodynamics.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Technicalities&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;All of the above discussion is subsumed in so called gauge theories. These are theories that contain certain number of charges, $N$ (e.g. $N = 1$ for electromagnetism) and certain number of interaction particles. But to specify their number, we have to first talk about the group of symmetries of the said theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For electromagnetism this group is $U(1)$ which is one-dimensional and so there is only one photon. For weak interactions there are two charges, so called flavors and the group that corresponds to that is $SU(2)$; this one is three-dimensional, so we obtain three mediating particles: $Z, W^{\pm}$. For strong interactions, there are three colors and the group that describes their mixing is $SU(3)$. This one is eight-dimensional, giving us eight gluons. One could also try to use $U(3)$ group, which is nine-dimensional. But this is ruled out by experiment (!). To understand why, let me make precise other statements I greatly simplified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum theory, states are often by superposition of other, more elementary states. It turns out, that when one talks about blue-antired gluon, what is actually meant is the superposition&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${1 \over \sqrt 2} \big( \big|b\big&amp;gt;\big|\bar r\big&amp;gt; + \big|\bar b\big&amp;gt;\big| r\big&amp;gt; \big)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and similarly in other cases. This is because there are certain symmetry conditions imposed on the theory and this state is invariant if we swap $b$ and $r$ (it wouldn't be if it contained only one part); but of course it would be painful to explicitly read out that formula all the time one wants to talk about gluons; therefore the simplifictation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as we mentioned, the actual bound states or particles have to be color neutral. It turns out gluons themselves could create a scalar state&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${1 \over \sqrt 3} \big( \big|b\big&amp;gt;\big|\bar b\big&amp;gt; + &#xD;&#xA;\big|r\big&amp;gt;\big|\bar r\big&amp;gt; + &#xD;&#xA;\big|g\big&amp;gt;\big|\bar g\big&amp;gt; \big)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this state existed in nature then there would exist long-range gluon interaction. But we know it doesn't. So the actual group is smaller, just $SU(3)$ and we are left with eight gluons.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-24T16:15:43.003" CommentCount="5" />
  <row Id="2238" PostTypeId="2" ParentId="2197" CreationDate="2010-12-24T16:46:36.583" Score="2" ViewCount="" Body="&lt;p&gt;I think one suited approach to your problem is to look at the &lt;strong&gt;stability&lt;/strong&gt; of the movement with respect to the energy.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Energy stability analysis&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;rotation energy&lt;/strong&gt; is given by&#xA;$$E_{rot} = \frac{1}{2}\Theta{(\mathbf{\omega},\mathbf{\omega})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Theta$ is the (tensorial) &lt;a href=&quot;http://en.wikipedia.org/wiki/Moment_of_inertia&quot; rel=&quot;nofollow&quot;&gt;moment of inertia&lt;/a&gt; with respect to the (vectorial) rotation $\mathbf{\omega}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of a prolate, homogeneous ellipsoid is given by $M = \frac{4}{3}\rho r_a^2r_b$ where $\rho$ is the mass density and $r_b$ is assumed to be the axis longer (or, say different) than $r_b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, we can write down the &lt;a href=&quot;http://scienceworld.wolfram.com/physics/MomentofInertiaEllipsoid.html&quot; rel=&quot;nofollow&quot;&gt;principal moments of inertia&lt;/a&gt; for this ellipsoid as&#xA;$$\Theta_b = \frac{M}{5}\cdot 2 r_a^2$$&#xA;and&#xA;$$\Theta_a = \frac{M}{5}\cdot (r_a^2 + r_b^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we now chose $\theta$ to be the angle between $\mathbf{\omega}$ and the $z$-axis, we find the rotation energy as&#xA;$$E_{rot} = \frac{M}{10}\left( 2r_a^2 \cos^2\theta \omega^2  + (r_a^2 + r_b^2) \sin^2\theta \omega^2 \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\omega$ now being the scalar value of rotation (I could not make it bold).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we have to turn to the &lt;strong&gt;potential energy&lt;/strong&gt; in the gravitational field. I suppose, this should take the form&#xA;$$E_{pot} = Mg\left( r_b \cos\theta + r_a \sin\theta \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you now analyze the energy &#xA;$$E = E_{rot}+ E_{pot}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with respect to $\theta$ and the parameters $r_a$ and $r_b$, you will find out that $\partial_\theta E$ will vanish at $\theta = 0$ but it will be a &lt;strong&gt;maximum&lt;/strong&gt; (remember, $\theta\geq 0$) assuming $r_b &amp;gt; r_a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means, that the movement will be &lt;strong&gt;unstable&lt;/strong&gt; assuming a rotating prolate body. Or, in other words, you cannot find some $\omega_s$ to have the egg at resting rotation in this model.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Other approaches&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;One further ansatz to calculate a stable rotating egg, one could assume that the rotating egg is not homogeneous. A simple model could be to assume that a small &lt;strong&gt;ring of mass&lt;/strong&gt; is attached to the &quot;smaller belt&quot; leading to&#xA;$$\Theta_b = \frac{M}{5}\cdot 2 r_a^2 + \frac{M_{ring}}{2}r_a^2\quad, \qquad \Theta_a = \frac{M}{5}\cdot (r_a^2 + r_b^2) + \frac{M_{ring}}{2\cdot 2\pi r_a}r_a^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, can you show, when $E$ has a minimum in $\theta = 0$ leading to a &lt;strong&gt;stability condition&lt;/strong&gt; of the form $F(M_{ring}, r_a/r_b; \omega_s) = 0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2010-12-24T20:20:12.800" LastActivityDate="2010-12-24T20:20:12.800" CommentCount="11" />
  <row Id="2239" PostTypeId="1" AcceptedAnswerId="2240" CreationDate="2010-12-24T17:52:28.467" Score="6" ViewCount="367" Body="&lt;p&gt;I've noticed that an airplane appears to have more lift when it's almost touching the ground then it has 100 feet or more in the air. What causes this to occur?&lt;/p&gt;&#xA;" OwnerUserId="871" LastActivityDate="2011-04-13T15:37:26.853" Title="Why does an airplane have more lift near the ground?" Tags="&lt;air&gt;&lt;airplanes&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="2240" PostTypeId="2" ParentId="2239" CreationDate="2010-12-24T18:17:13.343" Score="6" ViewCount="" Body="&lt;p&gt;This is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Ground_effect_in_aircraft&quot; rel=&quot;nofollow&quot;&gt;Ground Effect&lt;/a&gt;.  Not to be confused with flaring, which is a technique used by pilots to gain lift by increasing the angle of attack as airspeed decreases. &lt;br&gt;&#xA;Technicality, you can flare an aircraft at any altitude.  The higher the altitude, the faster the airspeed of which you can flare an aircraft before stalling due to air thinning as altitude increases.&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="959" LastEditorUserId="959" LastEditorDisplayName="" LastEditDate="2011-04-12T19:32:38.840" LastActivityDate="2011-04-12T19:32:38.840" />
  <row Id="2241" PostTypeId="1" CreationDate="2010-12-24T18:37:08.397" Score="1" ViewCount="218" Body="&lt;p&gt;Please excuse my question, as i don't come from a physics background, but i was really wondering.&#xA;Assuming we are in one of numerous universes which all have physical dimensions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the name of the physical space that envelopes all of these universes?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="959" LastEditorUserId="959" LastEditorDisplayName="" LastEditDate="2010-12-24T22:02:15.140" LastActivityDate="2010-12-24T22:02:15.140" ClosedDate="2010-12-24T19:18:01.027" Title="What is the name of the physical space enveloping all universes?" Tags="&lt;astronomy&gt;&lt;universe&gt;&lt;space&gt;" CommentCount="20" />
  <row Id="2242" PostTypeId="2" ParentId="2178" CreationDate="2010-12-24T19:02:41.640" Score="2" ViewCount="" Body="&lt;p&gt;For all practical purposes today the answers above are very informative.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as Marek has pointed out above, your fundamental theoretical model of the thermodynamics of computation, on which you are basing the question is, surprisingly, wrong, as we first began to discover 50 years ago (&lt;a href=&quot;http://strangepaths.com/reversible-computation/2008/01/20/en/&quot; rel=&quot;nofollow&quot;&gt;see refs&lt;/a&gt;. to Landauer Charlie Bennet, Friedkin, others).    Actually, all computations are in principle, dissipation free, except for the dissipation required to overwrite or forget previously stored bits.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classic example is this.   Suppose you want to compute the next to the last binary digit of the zillionth prime or some such.   Then you do so, slowly and reversibly, carefully not overwriting any of the intermediate bits you generate, which requires a lot of space.  Perhaps you even make use of quantum entanglement in the computer.  Then you write the answer, by overwriting (irreversibly forgetting) the single bit of the answer in some (say external) register.   Then you can reverse the original computation, also without any dissipation at all.   You are left having to dissipate only the entropy necessary to overwrite the 1 bit of the recorded answer, because that is the only information you were forced to forget.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the denominator approaches zero, in theory, the theoretical answer to your question is infinity.   There is a tradeoff with space to hold all the intermediate results.  This is a surprise, a shock really, but it shows the power of clear thought.   It is intimately connected with quantum computing, but also has entirely classical models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the right theoretical way to ask your question would be more like, for a particular computation, to be completed in a time t, operating with a limited memory of x bits, what is the necessary dissipation.   I'm no expert, but will try to get more refs.   PS.  The resting brain probably uses about 20 Watts.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-12-24T19:02:41.640" />
  <row Id="2244" PostTypeId="1" AcceptedAnswerId="2252" CreationDate="2010-12-24T19:27:46.953" Score="8" ViewCount="376" Body="&lt;p&gt;Pardon me for my stubborn classical/semiclassical brain. But I bet I am not the only one finding such description confusing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If EM force is caused by the exchange of photons, does that mean only when there are photons exchanged shall there be a force? To my knowledge, once charged particles are placed, the electromagnetic force is always there, uninterruptedly. According to such logic, there has to be a stream of infinite photons to build EM force, and there has to be no interval between one &quot;exchange event&quot; to another. A free light source from an EM field? The scenario is really hard to imagine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For neclei the scenario becomes even odder. The strong interaction between protons is caused by the exchange of massive pions. It sounds like the protons toss a stream of balls to one another to build an attractive force - and the balls should come from nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please correct me if I am wrong: the excitations of photons and pions all come from nothing. So there should be EM force and strong force everywhere, no matter what type of particles out there. Say, even electrical neutral, dipole-free particles can build EM force in-between. And I find no reason such exchanges of particles cannot happen in vacuum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope there will be some decent firmware to refresh my classical brain with newer field language codes. &lt;/p&gt;&#xA;" OwnerUserId="749" LastActivityDate="2010-12-26T07:55:31.493" Title="The exchange of photons gives rise to the electromagnetic force" Tags="&lt;quantum-field-theory&gt;" AnswerCount="3" CommentCount="11" FavoriteCount="0" />
  <row Id="2245" PostTypeId="1" AcceptedAnswerId="2246" CreationDate="2010-12-24T21:19:12.423" Score="7" ViewCount="168" Body="&lt;p&gt;It seems like among the electrical conductors there's a relationship between the ability to conduct heat as well as electricity. Eg: Copper is better than aluminum at conducting both electricity and heat, and silver is better yet at both. Is the reason for this known? Are there materials that are good at conducting electricity, but lousy at conducting heat?&lt;/p&gt;&#xA;" OwnerUserId="961" LastActivityDate="2010-12-25T10:40:22.027" Title="Why are materials that are better at conducting electricity also proportionately better at conducting heat?" Tags="&lt;electricity&gt;&lt;material-science&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="2246" PostTypeId="2" ParentId="2245" CreationDate="2010-12-24T21:36:50.933" Score="6" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_conductivity&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Thermal_conductivity&lt;/a&gt;   In metals, I think it generally has to do with the higher valence band electron mobility, but it gets more interesting elsewhere.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In metals, thermal conductivity approximately tracks electrical conductivity according to the Wiedemann-Franz law, as freely moving valence electrons  transfer not only electric current but also heat energy. However, the general correlation between electrical and thermal conductance does not hold for other materials, due to the increased importance of phonon carriers for heat in non-metals. As shown in the table below, highly electrically conductive silver is less thermally conductive than diamond, which is an electrical insulator.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="78" LastEditorUserId="113" LastEditorDisplayName="" LastEditDate="2010-12-25T01:35:32.360" LastActivityDate="2010-12-25T01:35:32.360" />
  <row Id="2247" PostTypeId="2" ParentId="2244" CreationDate="2010-12-24T21:56:24.620" Score="2" ViewCount="" Body="&lt;p&gt;The original question seemed highly confused IMO.   I wrote a comment to the original question (see above) which I slightly paraphrase as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, so we've sharpened your question. You want to know 1. What are virtual particles and 2. How, in principal does their exchange give rise to real forces, e.g. a) how does the exchange of the virtual gauge bosons of EM (virtual photons) give rise to the EM force, b) how do the exchange of virtual gauge bosons of the strong interaction (virtual gluons) give rise to the strong force, and c) similar for W and Z0 and the weak interaction? You do not want the details, but the basic principal of how exchange of virtual particles can yield a real force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in answer to question #1 What are virtual particles?&#xA;- I suggested he/she start at &lt;a href=&quot;http://en.wikipedia.org/wiki/Virtual_particle&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Virtual_particle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In answer to questions #2. How, in principal does the exchange of virtual gauge bosons give rise to real forces?  For this, see&#xA;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Force_carrier&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Force_carrier&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In answer to questions #2b and 2c, i.e. give some indication at the undergraduate level how these same sort of Feynman diagrams could also apply to the strong and weak interactions, see &#xA;&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/particles/expar.html&quot; rel=&quot;nofollow&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/particles/expar.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a popular reference book, consistent with the apparent level of knowledge of the original poster, I gave &lt;a href=&quot;http://books.google.com/books?id=pgasJfizmocC&amp;amp;pg=PA158&amp;amp;lpg=PA158&amp;amp;dq=virtual+bosons+exchange+forces&amp;amp;source=bl&amp;amp;ots=0r6h5zmNL6&amp;amp;sig=1UsmuWiAZGhsL1R9B7SQGCB4nc4&amp;amp;hl=en&amp;amp;ei=MBcVTbWrPIOglAeS5NS6BQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=7&amp;amp;ved=0CEwQ6AEwBg#v=onepage&amp;amp;q=virtual%20bosons%20exchange%20forces&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this free reference&lt;/a&gt; to &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0750305436&quot; rel=&quot;nofollow&quot;&gt;The Particle Century by Gordon Fraser&lt;/a&gt; for good measure.    IMHO, the original poster is going to be best served by a list of good semi popular books on these issues.  Constructive additional comments to my attempt at an answer might best be served by a list of such books which were thought of highly by the commentator(s), for a) general undergraduates b) physics undergraduates and c) possibly even h.s. students.  &lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-12-26T07:55:31.493" LastActivityDate="2010-12-26T07:55:31.493" CommentCount="2" />
  <row Id="2248" PostTypeId="2" ParentId="2229" CreationDate="2010-12-24T22:24:27.430" Score="-1" ViewCount="" Body="&lt;p&gt;Until a good specialist comes and clean up this mess, I am going to touch in point that was not raised yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Momentum&quot; rel=&quot;nofollow&quot;&gt;Momentum&lt;/a&gt; is the Noether charge of translational invariance. If you consider only massive entities in system &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrangian&quot; rel=&quot;nofollow&quot;&gt;Lagrangian&lt;/a&gt;, it will be noticed that the Lagrangian is not invariant over shift translation unless you add the electromagnetic field Lagrangian.&lt;/p&gt;&#xA;" OwnerUserId="62" LastActivityDate="2010-12-24T22:24:27.430" CommentCount="1" />
  <row Id="2249" PostTypeId="1" CreationDate="2010-12-24T22:55:53.920" Score="8" ViewCount="235" Body="&lt;p&gt;I've been reading this &lt;a href=&quot;http://www.if.ufrgs.br/~magusmao/FIP10604/RefsSemin/GMRreview_2003.pdf&quot; rel=&quot;nofollow&quot;&gt;review&lt;/a&gt; on &lt;a href=&quot;http://en.wikipedia.org/wiki/Giant_magnetoresistance&quot; rel=&quot;nofollow&quot;&gt;Giant Magnetoresistance&lt;/a&gt;, and something about it is bothering me. The basic effect is that, using a special &quot;stack&quot; of layers (alternating between ferromagnetic and not magnetic) you can cause great variations on the electrical resistance of the material by applying a small magnetic field.&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/jJQRA.png&quot; alt=&quot;alt text&quot;&gt;&lt;br&gt;&#xA;The explanation given both on the review and on Wikipedia is the following. The configuration on the left will cause scattering for all spin down electrons that cross it, but will present very low resistance for the spin up electrons (thus a small resistance overall). The one on the right, will scatter all electrons and thus have very large resistance overall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Here's my problem&lt;/strong&gt; (quoting the review)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These experimental results were all obtained for the ’Current In the Plane of the layers’ &#xA;  (CIP) geometry. The current is then parallel to the plane of the layers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can this be? CIP in the above image would mean the current is horizontal, with probes positioned on the left and right of the middle layer. But, if you don't force the current to cross the ferromagnetic layers, why would there be an increase in resistance? I checked with a professor, and he confirmed that the standard method is indeed to apply the probes horizontally, but he couldn't tell why the effect still worked. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One option that came up is that electrons don't follow a straight line from one probe to the other, and thus they would still &quot;flow&quot; through the other layers, but I know that's not the case. Using the resistors analogy seen in the image above, applying a CIP is equivalent to setting the resistors in parallel for each of the spin orientations, and parallel resistors &lt;em&gt;always&lt;/em&gt; have less (or equal) resistance than the weakest resistor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be clear, my question: How can the above explanation be compatible with the CIP case? If it's not, then why does it still work?&lt;/p&gt;&#xA;" OwnerUserId="724" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2010-12-24T23:08:52.963" LastActivityDate="2011-01-29T17:57:00.553" Title="What's the explanation for the Giant Magnetoresistance effect?" Tags="&lt;electromagnetism&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="2250" PostTypeId="1" AcceptedAnswerId="2405" CreationDate="2010-12-24T23:38:23.307" Score="6" ViewCount="301" Body="&lt;p&gt;What is the theoretical difference between the physical elementary interaction that causes an e+ to attract an e− when they exchange a virtual photon? Why is this exchange different from an e-e- scattering (which produces repulsion)?&lt;br&gt;&#xA;This question is related to but more specific than the recently posted &lt;a href=&quot;http://physics.stackexchange.com/questions/2244/the-exchange-of-photons-gives-rise-to-the-electromagnetic-force&quot;&gt;The exchange of photons gives rise to the electromagnetic force&lt;/a&gt;.  It has not been answered there.   It would be good to see the answer somewhere.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2010-12-25T21:29:48.307" LastActivityDate="2011-01-01T03:26:57.683" Title="What is the difference which leads to attraction in e+e- scattering and repulsion in e-e- scattering in QFT" Tags="&lt;quantum-field-theory&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="1" />
  <row Id="2251" PostTypeId="2" ParentId="1354" CreationDate="2010-12-25T00:27:46.450" Score="0" ViewCount="" Body="&lt;p&gt;If you are willing to go down to microscopic scales, a nice example of &quot;entropy winning&quot; is the phenomenon of &lt;em&gt;depletion forces&lt;/em&gt;. Large particles in a suspension of smaller ones feel an effective attractive force, even if the interaction between all particles is just hard-wall. The attractive force arises because the volume available to the smaller particles increases when the larger ones get sufficiently close, and hence their entropy increases. See e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sho Asakura and Fumio Oosawa,&lt;br&gt;&#xA;  &quot;Interaction between particles suspended in solutions of macromolecules&quot;,&lt;br&gt;&#xA;  J. Pol. Sci. 33 (1958) 183-192.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Depletion forces can be measured directly and are quite important for biological systems.&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2010-12-25T00:27:46.450" CommentCount="1" />
  <row Id="2252" PostTypeId="2" ParentId="2244" CreationDate="2010-12-25T02:12:57.073" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; I went over this answer and clarified some parts. Most importantly I expanded the &lt;em&gt;Forces&lt;/em&gt; section to connect better with the question.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I like your reasoning and you actually come to the right conclusions, so congratulations on that! But understanding the relation between forces and particles isn't that simple and in my opinion the best one can do is provide you with the bottom-up description of how one arrives to the notion of force when one starts with particles. So here comes the firmware you wanted. I hope you won't find it too long-winded.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Particle physics&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;So let's start with particle physics. The building blocks are particles and interactions between them. That's all there is to it. Imagine you have a bunch of particles of various types (massive, massless, scalar, vector, charged, color-charged and so on) and at first you could suppose that all kinds of processes between this particles are allowed (e.g. three photons meeting at a point and creating a gluon and a quark; or sever electrons meeting at a point and creating four electrons a photon and three gravitons). Physics could indeed look like this and it would be an incomprehensible mess if it did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately for us, there are few organizing principles that make the particle physics reasonably simple (but not &lt;em&gt;too&lt;/em&gt; simple, mind you!). These principles are known as conservation laws. After having done large number of experiments, we became convinced that electric charged is conserved (the number is the same before and after the experiment). We have also found that momentum is conserved. And lots of other things too. This means that processes such as the ones I mentioned before are already ruled out because they violate some if these laws. Only processes that can survive (very strict) conservation requirements are to be considered possible in a theory that could describe our world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another important principle is that we want our interactions simple. This one is not of experimental nature but it is appealing and in any case, it's easier to start with simpler interactions and only if that doesn't work trying to introduce more complex ones. Again fortunately for us, it turns out basic interactions are very simple. In a given interaction point there is always just a small number of particles. Namely:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;two: particle changing direction&lt;/li&gt;&#xA;&lt;li&gt;three:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;particle absorbing another particle, e.g. $e^- + \gamma \to e^-$&lt;/li&gt;&#xA;&lt;li&gt;or one particle decaying to two other particles $W^- \to e^- + \bar\nu_e$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;four: these ones don't have as nice interpretation as the above ones; but to give an example anyone, one has e.g. two gluons going in and two gluons going out&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So one example of such a simple process is electron absorbing a photon. This violates no conservation law and actually turns out to be the building block of a theory of electromagnetism. Also, the fact that there is a nice theory for this interaction is connected to the fact that the charge is conserved (and in general there is a relation between conservation of quantities and the way we build our theories) but this connection is better left for another question.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Back to the forces&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;So, you are asking yourself what was all that long and boring talk about, aren't you? The main point is: our world (as we currently understand it) is indeed described by all those different species of particles that are omnipresent everywhere and interact by the bizarre interactions allowed by the conservation laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when one wants to understand electromagnetic force all the way down, there is no other way (actually, there is one and I will mention it in the end; but I didn't want to over-complicate the picture) than to imagine huge number of photons flying all around, being absorbed and emitted by charged particles all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's illustrate this on your problem of Coulomb interaction between two electrons. The complete contribution to the force between the two electrons consists of all the possible combination of elementary processes. E.g. first electron emits photon, this then flies to the other electron and gets absorbed, or first electron emits photon, this changes to electron-positron pair which quickly recombine into another photon and this then flies to the second electron and gets absorbed. There is huge number of these processes to take into account but actually the simplest ones contribute the most.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But while we're at Coulomb force, I'd like to mention striking difference to the classical case. There the theory tells you that you have an EM field also when one electron is present. But in quantum theory this wouldn't make sense. The electron would need to emit photons (because this is what corresponds to the field) but they would have nowhere to fly to. Besides, electron would be losing energy and so wouldn't be stable. And there are various other reasons while this is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am getting at is that a single electron doesn't produce any EM field until it meets another charged particle! Actually, this should make sense if you think about it for a while. How do you detect there is an electron if nothing else at all is present? The simple answer is: you're out of luck, you won't detect it. You always need some test particles. So the classical picture of an electrostatic EM field of a point particle describes only what &lt;em&gt;would happen&lt;/em&gt; if another particle would be inserted in that field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above talk is part of the bigger bundle of issues with measurement (and indeed even of the very definition of) the mass, charge and other properties of system in quantum field theory. These issues are resolved by the means of renormalization but let's leave that for another day.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Quantum fields&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Well, turns out all of the above talk about particles (although visually appealing and technically very useful) is just an approximation to the more precise picture of there existing just one quantum field for every particle type and the huge number of particles everywhere corresponding just to sharp local peaks of that field. These fields then interact by the means of quite complex interactions that reduce to the usual particle stuff when once look what those peaks are doing when they come close together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This field view can be quite enlightening for certain topics and quite useless for others. One place where it is actually illuminating is when one is trying to understand to spontaneous appearance of so-called virtual particle-antiparticle pairs. It's not clear where do they appear from as particles. But from the point of view of the field, they are just local excitations. One should imagine quantum field as a sheet that is wiggling around all the time (by the means of inherent quantum wigglage) and from time to time wiggling hugely enough to create a peak that corresponds to the mentioned pair.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-25T23:06:45.153" LastActivityDate="2010-12-25T23:06:45.153" CommentCount="9" />
  <row Id="2253" PostTypeId="2" ParentId="2244" CreationDate="2010-12-25T04:19:32.143" Score="4" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;To my knowledge, once charged particles are placed, the electromagnetic force is always there, uninterruptedly. According to such logic, there has to be a stream of infinite photons to build EM force, and there has to be no interval between one &quot;exchange event&quot; to another.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nope. Any exchange event requires a finite amount of time and energy. So, no, electrons do not experience a &quot;continuous&quot; force. At the quantum level you would not expect processes to be continuous in general. Also you are conflating two different pictures of interaction. In a classical framework yes the em force is &quot;always there&quot;. This is not true in a quantum description. Google &lt;strong&gt;checkerboard model&lt;/strong&gt; to learn about just such a discrete picture of dynamics for elementary particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine two people sitting in cars on a frictionless surface throwing snowballs at each other. Every time person A tosses a ball with momentum $\vec{k}$ he/she gains momentum $-\vec{k}$. When person B catches said snowball, assuming a perfectly elastic catch, he/she will gain momentum $\vec{k}$. At the end of this process, A has gained $-\vec{k}$ momentum and B has gained $\vec{k}$. The total momentum of the system is conserved. To an external observer it appears as if A and B have &quot;repelled&quot; each other. How to extend this analogy to induce &quot;attraction&quot; between objects is not clear to me at this time - but let me assume for the time being that it can be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for the electromagnetic and gravitational fields, the &quot;snowballs&quot; in question - the photon and graviton - have a zero (or vanishingly small) mass. It doesn't take much effort for A and B to throw these massless snowballs leading to the long-range nature of these forces.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Please correct me if I am wrong: the excitations of photons and pions all come from nothing. So there should be EM force and strong force everywhere, no matter what type of particles out there. Say, even electrical neutral, dipole-free particles can build EM force in-between.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wrong. Photon and gravitons do not arise from &quot;nothing&quot;. To use a cliched phrase - &quot;nothing&quot; comes from &quot;nothing&quot;. They are, as @marek mentioned, &lt;em&gt;borrowed&lt;/em&gt; from the ground state or vacuum of the field in question. And yes, &quot;electrical neutral, dipole-free particles&quot; can develop attractive interactions purely due to quantum fluctuations. This is the origin of the Van der Waals force.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For nuclei [sic.] the scenario becomes even odder. The strong interaction between protons is caused by the exchange of massive pions. It sounds like the protons toss a stream of balls to one another to build an attractive force - and the balls should come from nothing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is precisely because the &quot;snowballs&quot; (pions) are massive, that the strong force is short-ranges and confined to a finite region of space - i.e. the nucleus! Once again, the pions don't come from nothing. Protons, neutrons, pions and all other excitations arise from the same vacuum and should rightly be referred to as &quot;quasiparticles&quot; rather than elementary particles. As to what the nature of this vacuum is, we are still trying to figure that out ;)&lt;/p&gt;&#xA;" LastActivityDate="2010-12-25T04:19:32.143" />
  <row Id="2254" PostTypeId="1" CreationDate="2010-12-25T04:30:02.167" Score="11" ViewCount="488" Body="&lt;p&gt;Take a glass of water and piece of toilet paper. If you keep the paper vertical, and touch the surface of the water with the tip of the paper, you can see the water being absorbed and climbing up the paper. This &lt;em&gt;climbing up&lt;/em&gt; is intriguing me. Where is the energy coming from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supposedly, a lot of water molecules are moving up and gaining potential energy. This has to be balanced by something. I can think of a few possibilities, but I can't tell which one it is.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When the water molecules dilute into the paper, they are at a state of lower potential binding energy. Some molecular interaction (van der Waals?) is at lower energy when you have a water+paper solution, compared to a water-only solution. This compensates the gain in gravitational energy.&lt;/li&gt;&#xA;&lt;li&gt;The surface between the paper and the water is at lower pressure than the atmosphere. This causes the water to be pushed into the paper by the atmospheric pressure, up until the point that the column of water above the surface is heavy enough to counterbalance. The potential energy would be a result of work done by the atmosphere.&lt;/li&gt;&#xA;&lt;li&gt;Some water molecules climb up randomly and loose kinetic energy going up, and somehow get &quot;stuck&quot; up there.&lt;/li&gt;&#xA;&lt;li&gt;Something else?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="724" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2010-12-25T05:52:06.073" LastActivityDate="2010-12-27T22:53:00.840" Title="When water climbs up a piece of paper, where is the energy coming from?" Tags="&lt;fluid-dynamics&gt;&lt;capillary-action&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" />
  <row Id="2255" PostTypeId="2" ParentId="2254" CreationDate="2010-12-25T05:03:52.763" Score="3" ViewCount="" Body="&lt;p&gt;The physics behind this is the same &lt;a href=&quot;http://en.wikipedia.org/wiki/Capillary_action&quot; rel=&quot;nofollow&quot;&gt;capillary action&lt;/a&gt; that causes water to move up narrow cylindrical channels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tissue paper is extremely porous, but the pores are sufficiently narrow that the cohesion between water molecules (actually driven by the Coulomb interaction, since water molecules are polar) and adhesion between the water and the surfaces of the pores combine to &quot;lift&quot; the water into the pores. Adhesion is, I think, most likely driven by hydrogen bonding. In the case of a SiO2 glass tube, there will be surface oxygen atoms to bind to, and though I don't know anything about the chemistry of paper, it is organic, so I'm sure there's hanging O-H chains to bond with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of water moving through each pore is actually pretty small, so if you imagine the paper instead as a set of &quot;ideal&quot; cylindrical capillaries, the gravitational potential for each &lt;em&gt;individual column&lt;/em&gt; is very weak, and so the water can rise quite a bit before the gravitational potential beats out the combination of cohesion and adhesion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In the interests of giving a broad answer, I remember in high school doing a neat chemistry experiment with a technique based on this effect, called &lt;a href=&quot;http://en.wikipedia.org/wiki/Paper_chromatography&quot; rel=&quot;nofollow&quot;&gt;paper chromatography&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I've added a bit on the possible microscopic origin of adhesion.&lt;/p&gt;&#xA;" OwnerUserId="791" LastEditorUserId="791" LastEditorDisplayName="" LastEditDate="2010-12-26T02:19:38.520" LastActivityDate="2010-12-26T02:19:38.520" CommentCount="2" />
  <row Id="2256" PostTypeId="1" AcceptedAnswerId="2263" CreationDate="2010-12-25T08:57:50.617" Score="2" ViewCount="149" Body="&lt;p&gt;I am working on a project that requires some amount of number crunching. I require some basic data about the galaxy M82 and our own Milky Way (particularly luminosities). My advisor is a nut and he blurts out numbers from his head that seem to be right. All the more, I would like to be sure. So is there any place where I can find reliable data? (Google isn't helping much here and I also tried &lt;a href=&quot;http://nedwww.ipac.caltech.edu/&quot; rel=&quot;nofollow&quot;&gt;http://nedwww.ipac.caltech.edu/&lt;/a&gt;). Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I am working on gas outflows from galaxies. I am focusing on galaxies which have luminosities beyond their Eddington limit. Thus to compare some calculated results, it would be useful to have the bolometric luminosities of M82 and Milky Way. Hope this clarifies.&lt;/p&gt;&#xA;" OwnerUserId="966" LastEditorUserId="966" LastEditorDisplayName="" LastEditDate="2010-12-25T09:35:49.197" LastActivityDate="2010-12-25T19:15:29.247" Title="Astrophysical Data" Tags="&lt;astrophysics&gt;&lt;astronomy&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="2257" PostTypeId="2" ParentId="2245" CreationDate="2010-12-25T10:40:22.027" Score="2" ViewCount="" Body="&lt;p&gt;This is true only for metals. &lt;strong&gt;Diamond&lt;/strong&gt;, for example, is barely a semiconductor. But it has a better heat conductivity than &lt;strong&gt;any metal&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-12-25T10:40:22.027" />
  <row Id="2258" PostTypeId="2" ParentId="2229" CreationDate="2010-12-25T10:43:13.037" Score="-2" ViewCount="" Body="&lt;p&gt;Of course they have mass. When saying &quot;photons have no mass&quot; in &lt;a href=&quot;http://www.youtube.com/watch?v=j50ZssEojtM&quot; rel=&quot;nofollow&quot;&gt;LHC rap&lt;/a&gt;, they were referring to the rest mass, it just didn't rhyme. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you pack a bunch of photons into your mirror-coated box, it will be heavier, by E/mc^2 as usual)&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2010-12-25T10:43:13.037" CommentCount="3" />
  <row Id="2260" PostTypeId="1" AcceptedAnswerId="2267" CreationDate="2010-12-25T17:16:04.687" Score="4" ViewCount="240" Body="&lt;p&gt;In an astronomy forum that I frequent, I have been having a discussion where the state of quantum gravity research came up. I claimed that Loop Quantum Gravity theories couldn't prove GR in the continuum limit, nor could they compute Mercury's perihelion because they didn't have any matter, at this point. Another poster claimed that one of the first things that any gravity scientist checks for the theory is the theory's prediction for the precession of Mercury's perihelion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for cosmological extensions to GR like, MOND, TeVeS or STVG, I can see where this might make sense. But I can't see how this poster's claim holds for quantum-gravity theories that haven't been able to establish that they reproduce GR in the continuum limit and that don't have matter at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I feel like I'm either missing something important about quantum gravity theories, in general, or the poster's comment only applies to cosmological-scale gravity modifications designed as an alternative to dark matter and not quantum gravity theories. Or perhaps a little of each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How, for instance, would one compute the precession of Mercury's perihelion using CDT, or Rovelli et al.'s spin-foam version of LQG? It may be that the dynamics of any object are implied in the spacetime microstructure, I don't understand the math well enough yet to follow the implications to know if this is true. But how can you compute the orbit of a planet when you require mass and fermions in order to even have the Sun and a planet like Mercury?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I note the very very recent paper of this week, Dec. 21, &lt;a href=&quot;http://arxiv.org/abs/1012.4719&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1012.4719&lt;/a&gt;, where the Marseille LQG group run by Carlo Rovelli claims they can now incorporate Fermions, so my question applies to research prior to this event, not that going forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not even close to an expert, so I'm looking for enough information that I'll be able to respond intelligently to the other poster. I'll point them to the answer I received here so perhaps this will help the marketing for this excellent resource.&lt;/p&gt;&#xA;" OwnerUserId="948" LastActivityDate="2010-12-26T08:18:30.717" Title="Quantum Gravity and Calculations of Mercury's Perihelion" Tags="&lt;general-relativity&gt;&lt;gravity&gt;&lt;quantum-gravity&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="1" />
  <row Id="2261" PostTypeId="2" ParentId="2254" CreationDate="2010-12-25T17:42:05.180" Score="1" ViewCount="" Body="&lt;p&gt;The previous answer tells you why the water moves up but doesn't explain where the energy comes from. In order for water to move up and thereby gain gravitational potential energy, you need to have some energy loss somewhere else to compensate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the energy comes from the random molecular movement of heat which extends the edge of the water itself up the capillary tubes/surfaces. So the water cools as it climbs because of the reduction in energy required to compensate for the reduced gravitational energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of it comes from loss of gravitational energy in the glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This part of the problem can be seen as the same as what happens when you take a test tube and place it in a tub filled with water, fill the tube with water and then invert it and pull the closed end out of the water so that some of the water in the closed end of the tube rises above the water in the rest of the tub. Where does the energy come from for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two sources, not one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) From your arm moving the tube upwards and pulling on the end of the tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) From the water in the tub dropping down in level and thereby leaving less gravitational potential energy for the remaining water in the tub.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider a long tube placed into the tub. That water in the tube will rise to the level of the water outside the tube. If you attach the tube end above the water to a closed valve that leads to a vacuum chamber, the water level will stay the same. If you open the valve, the water level will rise. The vacuum itself supplies no energy and can do no work since there are no molecules or forces involved. What does the work is exchange of potential energy from the gravity of the water in the tub. It pushes on the water in the tube causing it to rise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for capillary action you describe, some of the energy comes from the loss of the potential energy of the water or fluid outside the capillary tube, and some of it comes from the loss of heat in the water and toilet paper.&lt;/p&gt;&#xA;" OwnerUserId="948" LastActivityDate="2010-12-25T17:42:05.180" CommentCount="10" />
  <row Id="2262" PostTypeId="2" ParentId="2239" CreationDate="2010-12-25T17:54:24.000" Score="5" ViewCount="" Body="&lt;p&gt;Ground effect is caused by the increased pressure under the wing because the vortex at the end of the wing which normally just twists behind the wingtips for long distances runs into the ground itself thereby increasing the pressure under the wing as a whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can think of why the vortex forms if you think of the end of the wing as deflecting air downward. Right past the end of the wing, the air is not deflected at all. This makes some of the air deflected down move sideways outward from the wing because the pressure is lower there. This is what creates the spinning action of the vortex. The wing comes through the air and presses some of the air down but the air right next to it is not pressed at all, so there is a downward AND sideways movement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In ground effect the downward effect is increased while the sideways effect is reduced. When you are close to the ground, the air pressure under the wing is higher and at the same time there is less room for the air to flow sideways because of the distance. Big vortices can't form as easily so less of the planes energy goes into spinning air and more into keeping it up.&lt;/p&gt;&#xA;" OwnerUserId="948" LastActivityDate="2010-12-25T17:54:24.000" />
  <row Id="2263" PostTypeId="2" ParentId="2256" CreationDate="2010-12-25T19:15:29.247" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.google.com/search?hl=en&amp;amp;q=bolometric+luminosities+of+M82+&quot; rel=&quot;nofollow&quot;&gt;Google&lt;/a&gt; - &lt;a href=&quot;http://www.stsci.edu/science/starburst/Preprints/moriond99.pdf&quot; rel=&quot;nofollow&quot;&gt;Bolometric luminosity of M82&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/find/all/1/ti%3a+M82/0/1/0/all/0/1&quot; rel=&quot;nofollow&quot;&gt;arxiv&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="238" LastActivityDate="2010-12-25T19:15:29.247" />
  <row Id="2264" PostTypeId="1" AcceptedAnswerId="2868" CreationDate="2010-12-25T19:53:26.703" Score="1" ViewCount="125" Body="&lt;p&gt;Why in the irradiament mulipoles of Lienard-Wiechert's potential we say that electric quadrupole give a contribute of the same order of the magnetic dipole? How can we see it from their equations? And is there a physical reason for this? Sorry for my trivial question.&lt;/p&gt;&#xA;" OwnerUserId="558" LastActivityDate="2011-01-14T11:15:19.833" Title="EM irradiament and multipoles" Tags="&lt;electromagnetism&gt;&lt;waves&gt;&lt;electrodynamics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="2265" PostTypeId="1" CreationDate="2010-12-25T21:28:29.460" Score="-1" ViewCount="199" Body="&lt;p&gt;Say that the airplane is going in 1000 km/h. On the side of the airplane, there is a 10x10 cm window. How much friction would this window cause. For the sake of the calculation, imagine that the airplane is infinitely long. The window is also perfectly level with the rest of the wall, and there is no turbulence on the side of the plane. The outside air is also standing perfectly still. The airplane is travelling at sea level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know it is neglectable compared to the turbulence etc. caused by the form of the airplane, but I really need the answer to the question.&lt;/p&gt;&#xA;" OwnerUserId="904" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-26T21:21:31.440" LastActivityDate="2010-12-31T23:42:37.837" Title="How much friction is caused by a 10x10 cm window on the side of an airplane?" Tags="&lt;air&gt;&lt;friction&gt;&lt;airplanes&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="2266" PostTypeId="2" ParentId="2265" CreationDate="2010-12-26T02:02:08.857" Score="2" ViewCount="" Body="&lt;p&gt;If the window is pefectly flush, so the the surface geometry is the same as without it, why would we expect there to be any effect whatsoever? I think all that matters is geometry and surface roughness. Also note that 1000KM/hour is pretty fast, airliners might travel that fast a stratospheric height, but drag would be a few times higher at sealevel -because the density is maybe 3 times as high as at cruising altitude.&lt;/p&gt;&#xA;" OwnerUserId="391" LastEditorUserId="391" LastEditorDisplayName="" LastEditDate="2010-12-31T23:42:37.837" LastActivityDate="2010-12-31T23:42:37.837" CommentCount="7" />
  <row Id="2267" PostTypeId="2" ParentId="2260" CreationDate="2010-12-26T08:03:43.657" Score="2" ViewCount="" Body="&lt;p&gt;First of all, I think you're correct that Mercury's precession is only useful as a check on classical, large-scale theories of gravity, like MOND, which would actually govern astronomical interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed out in the comments, quantum gravity mainly concerns itself with an entirely different domain, namely Planck-scale distances, times, and densities, since those are the scenarios in which classical GR seems to break down. Accordingly, the typical checks on theories of quantum gravity are things like black hole entropy, and predicting some sort of non-singular behavior for the universe at the Big Bang. Using LQG to calculate a property of a complex real-world system like the solar system is at worst impossible and at best absurdly complicated, if it in fact reduces to classical GR (which I have also heard is yet to be proven).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, a theory of quantum gravity could perhaps handle the &quot;interesting&quot; part of the calculation of planetary precession without incorporating mass directly. All it really needs is a connection, which in LQG is provided by the SU(2) gauge field $A^i_a$ defined on the edges of a graph. Given a graph and a connection, one might be able to compute something akin to a geodesic through the graph, which in the continuum limit would correspond to the orbit - though my knowledge of LQG falls short here, I'm not sure whether that's how you would &lt;em&gt;actually&lt;/em&gt; get from quantum geometry to an orbit. And anyway, as far as I know, in order to get the connection for a given mass distribution (i.e. the Sun) in the first place, you would have to use classical GR to find the Christoffel symbols.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-26T08:18:30.717" LastActivityDate="2010-12-26T08:18:30.717" CommentCount="1" />
  <row Id="2268" PostTypeId="1" AcceptedAnswerId="2282" CreationDate="2010-12-26T08:44:59.457" Score="1" ViewCount="283" Body="&lt;p&gt;This question was inspired by the answer to the question &lt;a href=&quot;http://physics.stackexchange.com/questions/1901/if-the-universe-were-compressed-into-a-super-massive-black-hole-how-big-would-it&quot;&gt;&quot;If the universe were compressed into a super massive black hole, how big would it be&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that we have a matter with a uniform density $\rho$.&#xA;Some mass of this matter may forms a  black hole with the Schwarzschild radius:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\large{R_S=c*\sqrt{\frac{3}{8\pi G\rho}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation is easy to get from&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\large{R_S=\frac{2GM}{c^2}}$ and $\large{\rho=\frac{3M}{4\pi R_S^3}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For  &lt;a href=&quot;http://en.wikipedia.org/wiki/Observable_universe#Mass&quot; rel=&quot;nofollow&quot;&gt;universe density&lt;/a&gt; ($9.3*10^{-27} kg/m^3$) Schwarzschild radius of the black hole is &lt;a href=&quot;http://www.wolframalpha.com/input/?i=c%2a%283%2F%288%2aG%2api%2a9.3%2a10%5E-27kg%2Fm%5E3%29%29%5E1%2F2+&quot; rel=&quot;nofollow&quot;&gt;13.9 billion light-years&lt;/a&gt;. While radius of observable Universe is &lt;a href=&quot;http://en.wikipedia.org/wiki/Observable_universe#Size&quot; rel=&quot;nofollow&quot;&gt;46 billion light-years&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could be located inside such black hole, but we don't observe its singularity and event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why there are no supermassive black holes with the density of the Universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it means, that the whole Universe is infinite and has uniform density?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Relative link - &lt;a href=&quot;http://www.xs4all.nl/~johanw/PhysFAQ/Relativity/BlackHoles/universe.html&quot; rel=&quot;nofollow&quot;&gt;Is the Big Bang a black hole?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="238" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2010-12-27T19:29:09.620" LastActivityDate="2010-12-27T19:29:09.620" Title="Supermassive black holes with the density of the Universe" Tags="&lt;cosmology&gt;&lt;universe&gt;&lt;black-holes&gt;" AnswerCount="1" CommentCount="15" />
  <row Id="2269" PostTypeId="1" AcceptedAnswerId="2270" CreationDate="2010-12-26T13:05:13.277" Score="12" ViewCount="521" Body="&lt;p&gt;We are taught that the universe began as a singularity - an infinitely small and infinitely dense point. At the beginning of time there was a 'Big Bang' or, more accurately, 'Inflation'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main evidence for this is the observation of the red shift of all of the galaxies. This shows us that as time increases, the universe becomes bigger. A logical outcome of this is that going back in time, the universe shrinks. This is then extrapolated back to the beginning of time, where the universe was infinitely small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the only thing that I can see that we know for sure is that the universe used to be smaller. This does not necessarily imply that it used to be infinitely small. How do we know, for example, that the universe doesn't oscillate and that we are in a time where the universe is expanding (and accelerating) and will eventually contract again?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other evidence is there that suggests we started from an singularity?&lt;/p&gt;&#xA;" OwnerUserId="954" LastActivityDate="2011-07-08T09:54:31.770" Title="Does red shift evidence necessarily imply that the universe started from a singularity?" Tags="&lt;universe&gt;&lt;experimental-evidence&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" />
  <row Id="2270" PostTypeId="2" ParentId="2269" CreationDate="2010-12-26T13:48:12.740" Score="16" ViewCount="" Body="&lt;p&gt;There are 3 observations that support the big bang theory, i.e. origin of the universe in a singularity:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The redshift of galaxies, as you already mentioned.&lt;/li&gt;&#xA;&lt;li&gt;The cosmic background radiation.&lt;/li&gt;&#xA;&lt;li&gt;The amounts of different nuclei in the universe, notably the preponderance of light elements like hydrogen and helium.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Each of these alone would probably not be sufficient to support the big bang theory. The redshift of galaxies could be explained by some other theory, some have been suggested by Hoyle and Narlikar in the past. Probably the other two phenomena could be explained independently as well, but it is the conjunction that fits so well with the big bang hypothesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that settle the matter once and for all? Short answer is no. Since these 3 observations have been made and confirmed, more detailed observations have been added to the mix and this has complicated the story for the big band model. But that would take us into a longer post. The current model which is the most widely accepted is the so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Lambda-CDM_model&quot; rel=&quot;nofollow&quot;&gt;Lambda-CDM model&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the problem of the universe starting in a real singularity, instead of a very dense state, this is still an open problem related to a yet to be invented (or completed) theory of quantum gravity. Our current understanding of singularities in General Relativity is going back to the Penrose-Hawking singularity theorems. They are of the kind &quot;Here be dragons!&quot; in that they delineate the conditions for singularities to form and point where our knowledge ends. More can not be done, because a singularity is basically a failure of the theory.&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2010-12-27T09:17:31.957" LastActivityDate="2010-12-27T09:17:31.957" CommentCount="1" />
  <row Id="2271" PostTypeId="2" ParentId="2269" CreationDate="2010-12-26T13:53:45.653" Score="7" ViewCount="" Body="&lt;p&gt;If someone taught you that the expansion of the universe necessarily implies a singularity, then he was wrong. The past singularity is not the only possible initial condition. There are actually several models that predict different scenarios. For example there is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Eternal_inflation&quot; rel=&quot;nofollow&quot;&gt;eternal inflation&lt;/a&gt; scenario, where there is no initial singularity. There is also &lt;a href=&quot;http://en.wikipedia.org/wiki/Neil_Turok&quot; rel=&quot;nofollow&quot;&gt;Turoks&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Ekpyrotic_universe&quot; rel=&quot;nofollow&quot;&gt;ecpyrotic model&lt;/a&gt;, which has big bangs without singularities and there are many other models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the oscillation, things are more straightforward. The dynamics of the universe depend on the mass content of the universe. The mass content is measured by the energy density parameter. If that parameter is over a critical value, then the universe is closed and could oscillate as you say. If the density is smaller than or equal to the critical, then the universe expands for ever. The density of the universe is measured in the cosmic microwave background and it is measured to be almost exactly the critical. So the universe is not going to recollapse. But that doesn't exclude something like the Turok oscillating model.&lt;/p&gt;&#xA;" OwnerUserId="383" LastEditorUserId="383" LastEditorDisplayName="" LastEditDate="2010-12-26T15:04:35.920" LastActivityDate="2010-12-26T15:04:35.920" />
  <row Id="2272" PostTypeId="1" AcceptedAnswerId="2274" CreationDate="2010-12-26T19:04:27.407" Score="0" ViewCount="139" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/2175/faster-than-light-information&quot;&gt;Faster than light information&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Hey there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just want to make clear that I'm new here and I don't know whether this is the right place to ask such a hypothetical question. So if it's not, I'd be glad if someone could tell me where to ask. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you and a friend of yours were very far away from each other, let's say the distance is 1 light-year. Now you're holding a bar which measures exactly 1 light-year and currently touches your friend's shoulder. What happens when you move that bar (in order to poke your friend)? Will there be a delay? And what magnitude will that delay have? A year? Several years?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking forward to your answers. :)&lt;/p&gt;&#xA;" OwnerUserId="977" LastActivityDate="2010-12-26T19:15:34.837" ClosedDate="2010-12-26T21:15:25.670" Title="What happens when I move a very long bar?" Tags="&lt;mechanics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="2273" PostTypeId="2" ParentId="222" CreationDate="2010-12-26T19:14:07.440" Score="5" ViewCount="" Body="&lt;p&gt;I've recently discovered &lt;a href=&quot;http://cadabra.phi-sci.com/&quot; rel=&quot;nofollow&quot;&gt;Cadabra&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A field-theory motivated approach to computer algebra&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm really impressed.&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2010-12-26T19:14:07.440" CommunityOwnedDate="2010-12-26T19:14:07.440" CommentCount="2" />
  <row Id="2274" PostTypeId="2" ParentId="2272" CreationDate="2010-12-26T19:15:34.837" Score="1" ViewCount="" Body="&lt;p&gt;The change would transmit at the P wave velocity (look up elastic waves). This is typically a few kilometers per second for most materials you would be likely to make a bar out of. You can maximize this speed by going for low density at high strength, something like titanium comes to mind. In any case the result would be many thousands of years.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-12-26T19:15:34.837" />
  <row Id="2275" PostTypeId="1" CreationDate="2010-12-26T19:35:23.533" Score="0" ViewCount="120" Body="&lt;p&gt;Say you pick a point on the Earth. The earth spins at some constant $x (I don't know how fast, sorry). Now, that point has a different relative speed to a fixed point on Earth's orbit as it revolves around the Sun. This point &lt;em&gt;also&lt;/em&gt; has a relative speed to the galaxy from a fixed observer, and etc down to an arbitrary point in the Universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you calculate the absolute/actual speed of this point from a theoretical relative point in the Universe traveling at constant speed 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this be considered the absolute speed on the universe that all objects should be compared against?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm not a physics person really, just got into an interesting conversation with a buddy at work)&lt;/p&gt;&#xA;" OwnerUserId="978" LastActivityDate="2010-12-26T19:58:54.007" ClosedDate="2010-12-26T21:09:28.433" Title="Calculating absolute speed of a relative object" Tags="&lt;general-relativity&gt;&lt;astrophysics&gt;&lt;speed&gt;" AnswerCount="3" CommentCount="9" />
  <row Id="2276" PostTypeId="2" ParentId="2275" CreationDate="2010-12-26T19:47:41.733" Score="0" ViewCount="" Body="&lt;p&gt;The problem is: you don't have a point in the Universe traveling at constant speed 0, thus the entire proposal can't be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could define a point (actually, a reference frame, but I don't think you know what it is) in the universe to have constant 0 speed. But then the so called &quot;absolute speed&quot; would just be the speed relative to that frame you picked arbitrarily, and wouldn't be absolute at all.&lt;/p&gt;&#xA;" OwnerUserId="724" LastActivityDate="2010-12-26T19:47:41.733" />
  <row Id="2277" PostTypeId="2" ParentId="2275" CreationDate="2010-12-26T19:54:35.583" Score="1" ViewCount="" Body="&lt;p&gt;The flip side of &lt;a href=&quot;http://physics.stackexchange.com/questions/2275/calculating-absolute-speed-of-a-relative-object/2276#2276&quot;&gt;Bruce's answer&lt;/a&gt; is that &lt;em&gt;any&lt;/em&gt; inertial trajectory that you select is an equally valid candidate for the &lt;em&gt;&quot;theoretical relative point in the Universe traveling at constant speed 0&quot;&lt;/em&gt;. That's the whole point Einstein kept going on about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nor---if the currently favored Big Bang cosmology is correct---can you locate (even in theory) a unique point which can be called the &quot;origin&quot; or &quot;center&quot; of the universe. The origin was &lt;em&gt;everywhere&lt;/em&gt; at the same time.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2010-12-26T19:54:35.583" />
  <row Id="2278" PostTypeId="2" ParentId="2275" CreationDate="2010-12-26T19:58:54.007" Score="1" ViewCount="" Body="&lt;p&gt;There are several problems with this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;There is the assumption that the Earth rotates at a constant speed. This is not true. For example, powerful earthquakes can change the speed of the Earth's rotation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Everything is relative. You can, however, choose your '0' to be whatever you want but then your 'absolute' measurements would be relative to this.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The word 'absolute' suggests to me an exact measurement (I don't know if that was intended) but all measurements have uncertainty, even theoretically, as the act of observing will always affect the system you are observing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://www.youtube.com/watch?v=PmJV8CHIqFc&amp;amp;feature=BF&amp;amp;list=PLF946A3C19A47C6A2&amp;amp;index=1&quot; rel=&quot;nofollow&quot;&gt;introduction to classical mechanics&lt;/a&gt; on YouTube may be helpful for understanding measurements further. &lt;/p&gt;&#xA;" OwnerUserId="954" LastActivityDate="2010-12-26T19:58:54.007" />
  <row Id="2279" PostTypeId="1" AcceptedAnswerId="2286" CreationDate="2010-12-26T22:59:12.953" Score="5" ViewCount="350" Body="&lt;p&gt;Using purely gravitational potential energy, what is the highest efficiency one can achieve with a trebuchet counter-weight type of machine? Efficiency defined here as transformation of potential energy in the counterweight to kinetic energy of the trajectile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: To be more specific, we can use the following idealization:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;no friction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;no air resistance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;no elastic/material losses&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But any of the &quot;standard&quot; trebuchet designs are allowed: simple counterweight, hinged counterweight, vertical counterweight, etc. &lt;/p&gt;&#xA;" OwnerUserId="775" LastEditorUserId="775" LastEditorDisplayName="" LastEditDate="2010-12-27T00:17:16.923" LastActivityDate="2010-12-27T06:29:25.413" Title="What is the maximum efficiency of a trebuchet?" Tags="&lt;classical-mechanics&gt;&lt;projectile&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="2280" PostTypeId="2" ParentId="2188" CreationDate="2010-12-27T01:25:00.870" Score="1" ViewCount="" Body="&lt;p&gt;&quot;What Is &quot;Electricity&quot;?&quot;&#xA;1996 William J. Beaty &#xA;&lt;a href=&quot;http://amasci.com/miscon/whatis.html&quot; rel=&quot;nofollow&quot;&gt;http://amasci.com/miscon/whatis.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;COME ON, WHAT IS ELECTRICITY, REALLY?&quot;&#xA;by Bill Beaty. &#xA;&lt;a href=&quot;http://amasci.com/elect/elefaq1.html#ae&quot; rel=&quot;nofollow&quot;&gt;http://amasci.com/elect/elefaq1.html#ae&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I hear the phrase &quot;electronic goods&quot;, I think of physical, tangible objects such as AM radio receivers, cell phones, MP3 players, LCD wristwatches, the &quot;Simon&quot; electronic game, GPS receivers, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course these physical objects -- what I call &quot;electronic goods&quot; -- are tangible and qualify as &quot;goods&quot; in the same way a chair and a pen do.&#xA;Both these electronic goods and chairs are made of physical materials -- metal, plastic, etc. -- and require some skills and tools to manufacture.&#xA;Both these electronic goods and chairs cost time and effort to acquire the raw materials and convert them into the final shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds like you are also interested in &quot;intangible goods&quot;, aka &quot;digital goods&quot;, which can easily be perfectly copied digitally from any digital media to any other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such goods (songs in MIDI format, songs in MP3 format, software, news articles, photographs in JPG format, documents describing the exact shape of each part of a chair and how to assemble it, etc.)&#xA;are in some sense &quot;the same&quot; whether they are printed with ink on paper, stored as magnetic patterns on a hard drive, stored in computer memory, (extremely temporarily) stored as fluctuations in photon density in space between a satellite in geostationary orbit and an earth station, stored as tiny dimples on a DVD, or stored in any one of a dozen other media.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Information can be stored in paper tape in the form of holes punched in the paper. When the left hole is punched, it's considered to be &quot;on&quot; or value &quot;1&quot;. When the right hole is punched, it's considered to be &quot;off&quot; or &quot;zero.&#xA;(Standard paper tape has a different arrangement).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the paper tape substrate has mass, the information represented by these holes does not have mass, because a spool of paper tape with some digital good stored on it weighs &lt;em&gt;less&lt;/em&gt; than a completely blank (unpunched) paper tape.&#xA;Also, if you start with two identical blank spools, and you put some valuable digital good on one tape, and completely fill the other tape with all-zeros (which has practically no value), the resulting spools weigh practically the same amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A virtual object stored on a spool of paper tape can be considered, in some sense, to be tangible because it is stored in the form of a pattern of holes punched in paper that can be seen and felt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since these digital goods can be so easily translated from one media to another, it doesn't make sense to me to focus on one subtle detail of the physics of one particular media, and assume that subtle detail has any relevance to the digital good -- when that detail is completely different when &quot;the same&quot; digital good is stored in a different media.&lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2010-12-27T01:25:00.870" CommunityOwnedDate="2010-12-27T01:25:00.870" CommentCount="3" />
  <row Id="2281" PostTypeId="1" AcceptedAnswerId="2283" CreationDate="2010-12-27T02:54:58.020" Score="10" ViewCount="1337" Body="&lt;p&gt;Our ability to store data on or in physical media continues to grow, with the maximum amount a data you can store in a given volume increasing exponentially from year to year. Storage devices continue to get smaller and their capacity gets bigger. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can't continue forever, though, I would imagine. &quot;Things&quot; can only get so small; but what about information? How small can a single bit of information be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put another way: given a limited physical space -- say 1 cubic centimeter -- and without assuming more dimensions than we currently have access to, what is the maximum amount of information that can be stored in that space? At what point does the exponential growth of storage density come to such a conclusive and final halt that we have no reason to even attempt to increase it further?&lt;/p&gt;&#xA;" OwnerUserId="985" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-28T00:15:16.707" LastActivityDate="2010-12-28T00:15:16.707" Title="Maximum theoretical data density" Tags="&lt;quantum-mechanics&gt;&lt;information&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="2282" PostTypeId="2" ParentId="2268" CreationDate="2010-12-27T02:57:26.300" Score="1" ViewCount="" Body="&lt;p&gt;You are asking a wrong question. Here is the problem with your reasoning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are assuming a Schwarzschild metric and a homogenous distribution of mass. But the Schwarzschild geometry describes a vacuum spacetime. So you can't use it for a spacetime filled with matter. For a cosmological spacetime filled with matter, like our universe, the suitable metric to use would be something else, like the FRW for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could only use the Schwarzschild spacetime if you assumed a sphere of some uniform density $\rho$ and vacuum outside the radius of the sphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me illustrate how things would work out then. As you can see, a particular density corresponds to a particular $R_s$, lets call it $R_s(\rho)$. So if you had a sphere of matter with a radius $R_1$ grater than $R_s(\rho)$, then you couldn't apply the formula $R_s(\rho)=c\sqrt{\frac{3}{8\pi G \rho}}$. You would have to use the Schwarzschild metric only in the vacuum region outside of the sphere. So you would have then $R_s=\frac{8\pi G\rho R_1^3}{3c^2}$. In order to see how the $R_s$ compares with $R_s(\rho)$, you can replace the density with $\rho=\frac{3c^2}{8\pi G R_s(\rho)}$. So you would get that the Schwarzschild radius for a sphere of uniform density $\rho$ and radius $R_1&amp;gt;R_s(\rho)$ is $R_s=\left(\frac{R_1}{R_s(\rho)}\right)^2R_1$, which is grater than the radius of the sphere. So the sphere is inside its Schwarzschild horizon. If on the other hand, the radius $R_1$ is smaller than $R_s(\rho)$, then the corresponding horizon would have to be inside the sphere. But inside the sphere the Schwarzschild metric doesn't apply. So it isn't necessary that there should be a horizon inside the matter distribution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you apply these to the universe and assume for example that the radius of the visible universe is the radius $R_1$ of the sphere, then you would have a horizon radius (using your numbers) that would be almost 10 times the radius of the observable universe. So, the entire universe would have to be in a black hole of radius of 460 billion light-years. So the assumption that we should see black holes with horizons of radii of 13.9 billion light-years is not correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one assumes the above point of view, one could say that the universe is a white hole that is exploding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that all these are helpful and not confusing.&lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2010-12-27T02:57:26.300" CommentCount="5" />
  <row Id="2283" PostTypeId="2" ParentId="2281" CreationDate="2010-12-27T03:15:52.700" Score="11" ViewCount="" Body="&lt;p&gt;The answer is given by the &lt;a href=&quot;http://arxiv.org/abs/hep-th/9905177&quot; rel=&quot;nofollow&quot;&gt;covariant entropy bound&lt;/a&gt; (CEB) also referred to as the Bousso bound after Raphael Bousso who first suggested it. The CEB sounds very similar to the Holographic principle (HP) in that both relate the dynamics of a system to what happens on its boundary, but the similarity ends there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The HP suggests that the physics (specifically Supergravity or SUGRA) in a d-dimensional spacetime can be mapped to the physics of a conformal field theory living on it d-1 dimensional boundary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CEB is more along the lines of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bekenstein_bound&quot; rel=&quot;nofollow&quot;&gt;Bekenstein bound&lt;/a&gt; which says that the entropy of a black hole is proportional to the area of its horizon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S = \frac{k A}{4} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To cut a long story short the maximum information that you can store in $1 cc = 10^{-6} m^3$ of space is proportional to the area of its boundary. For a uniform spherical volume, that area is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A = V^{2/3} = 10^{-4} m^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the maximum information (# of bits) you can store is approximately given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S \sim \frac{A}{A_{pl}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A_{pl}$ is the planck area $ \sim 10^{-70} m^2 $. For our $ 1 cc $ volume this gives $ S_{max} \sim 10^{66} $ bits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is a rough order-of-magnitude estimate, but it lies in the general ballpark and gives you an idea of the limit that you are talking about. As you can see, we still have decades if not centuries before our technology can saturate this bound !&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                         Cheers,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Thanks to @mark for pointing out that $1 cc = 10^{-6} m^3$ and not $10^{-9} m^3$. Changes final result by three orders of magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;On Entropy and Planck Area&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In response to @david's observations in the comments let me elaborate on two issues.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Planck Area&lt;/strong&gt;: From lqg (and also string theory) we know that geometric observables such as the area and volume are quantized in any theory of gravity. This result is at the kinematical level and is independent of what the actual dynamics are. The quantum of area, as one would expect, is of the order of $\sim l_{pl}^2$ where $l_{pl}$ is the Planck length. In quantum gravity the dynamical entities are precisely these area elements to which one associates a spin-variable $j$, where generally $j = \pm 1/2$ (the lowest rep of SU(2)). Each spin can carry a single &lt;em&gt;qubit&lt;/em&gt; of information. Thus it is natural to associate the planck areas with a single unit of information.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Entropy as a measure of Information&lt;/strong&gt;: There is a great misunderstanding in the physics community regarding the relationship between entropy $S$ - usually described as a measure of &lt;em&gt;disorder&lt;/em&gt; - and useful information $I$ such as that stored on a chip, an abacus or any other device. However they are one and the same. I remember being laughed out of a physics chat room once for saying this so I don't expect anyone to take this at face value.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But think about this for a second (or two). What is entropy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S = k_B \ln(N) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $k_B$ is Boltzmann's constant and $N$ the number of microscopic degrees of freedom of a system. For a gas in a box, for eg, $N$ corresponds to the number of different ways to distribute the molecules in a given volume. If we were able to actually use a gas chamber as an information storage device, then each one of these configurations would correspond to a unit of memory. Or consider a spin-chain with $m$ spins. Each spin can take two (classical) values $\pm 1/2$. Using a spin to represent a bit, we see that a spin-chain of length $m$ can encode $2^m$ different numbers. What is the corresponding entropy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ S \sim \ln(2^m) = m \ln(2) \sim \textrm{number of bits} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since we have identified each spin with a bit (more precisely qubit). Therefore we can safely say that the entropy of a system is proportional to the number of bits required to describe the system and hence to its storage capacity.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-27T05:07:31.257" CommentCount="15" />
  <row Id="2284" PostTypeId="2" ParentId="2269" CreationDate="2010-12-27T04:28:04.380" Score="2" ViewCount="" Body="&lt;p&gt;I grew up with Hoyles steady state universe. The universe was expanding, but continuous creation of matter (light elements to match observed abundance) was postulated so that the universe had a (quasi) steady form (i.e. density of the universe was maintained throughout time). It sounds a bit magical to be sure, but I don't think it could be conclusively cast aside until the CMB was discovered.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-12-27T04:28:04.380" />
  <row Id="2285" PostTypeId="1" CreationDate="2010-12-27T06:00:39.110" Score="5" ViewCount="602" Body="&lt;p&gt;As an example of such a historical myth that we all learned, there is the story that in his confrontation with Cardinal Bellarmine in 1632, Galileo had all the evidence on his side, won the intellectual argument by the weight of that evidence, and was defeated only by the ignorant prejudices the trial judges.   However, historical studies have also advanced as more detailed information was obtained and reflected upon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual situation was much more nuanced.   The version of Copernican theory defended by Galileo assumed 1. a very small solar system, the size of which should clearly be seen by measurements of planetary parallax and 2. circular orbits, not elliptical (although Galileo knew of the work of Kepler, but declined to incorporate it).  Both Copernicus and Galileo had attempted to measure parallax in the solar system, and failed to get a result &gt; 0.  (It was only measured by Bessell in 1838.)   Furthermore, because it used circles for orbits, the Copernican model was actually less accurate than the old Ptolemaic epicycle system of approximations.  In addition, Galileo predicted only a single tide/day, not the actual two.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand his unambiguous observation of Venus in all four phases in 1610 definitely ruled out the Aristotelian model of heavens, since in this model the sphere of the sun was always beyond the sphere of Venus, although the Tychonic, Capellan and Extended Capellan models were still possible, each either with or without a daily rotating Earth. These latter models had the virtue of explaining the phases of Venus without the vice of the 'refutation' of full Copernican heliocentrism’s prediction of stellar parallax.  From &lt;a href=&quot;http://en.wikipedia.org/wiki/Galileo_Galilei&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's article on Galileo&lt;/a&gt;, confirmed by Weinberg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conclusion: on the basis of the evidence available to them, the Church had a pretty strong case on the evidence against Copernican heliocentrism, arguably stronger than Galileo.  Furthermore, on the basis of contemporary documents and doctrines going back to Augustine and the 4th century, if the church had seen conclusive proof in favor of the Copernican system, it was already formally committed to changing its interpretation.   The Church, being engaged in a power struggle during the reformation, would likely have followed conclusive evidence had it been presented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refs: see the historically accurate and nuanced video at &lt;a href=&quot;http://online.itp.ucsb.edu/online/colloq/tutino1/rm/qt.html&quot; rel=&quot;nofollow&quot;&gt;http://online.itp.ucsb.edu/online/colloq/tutino1/rm/qt.html&lt;/a&gt;, for slides see &lt;a href=&quot;http://online.itp.ucsb.edu/online/colloq/tutino1/&quot; rel=&quot;nofollow&quot;&gt;http://online.itp.ucsb.edu/online/colloq/tutino1/&lt;/a&gt;, book at &lt;a href=&quot;http://books.google.com/books?id=k1czB51a8eEC&amp;amp;printsec=frontcover&amp;amp;dq=stefania+tutino&amp;amp;source=bl&amp;amp;ots=aW1M70o-vF&amp;amp;sig=bSwW_PAkCFUvcAnRePm2Qn0wT-s&amp;amp;hl=en&amp;amp;ei=VC0YTZuuMYaKlwen8r3iDA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=2&amp;amp;ved=0CBkQ6AEwATgK#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Google&lt;/a&gt;.  This is not an obscure, idiosyncratic view of the Galileo affair, but rather the generally accepted view of knowledgeable scholars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other famous oversimplified or false myths of this sort do you know and can document and or reference corrections to?&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2011-01-29T04:27:56.947" LastActivityDate="2011-01-30T13:28:22.193" CommunityOwnedDate="2010-12-27T07:02:15.633" Title="Myths in the history of Physics" Tags="&lt;history&gt;" AnswerCount="7" CommentCount="23" FavoriteCount="8" />
  <row Id="2286" PostTypeId="2" ParentId="2279" CreationDate="2010-12-27T06:29:25.413" Score="5" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Trebuchet&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; on trebuchets links to a &lt;a href=&quot;http://www.algobeautytreb.com/trebmath35.pdf&quot; rel=&quot;nofollow&quot;&gt;PDF paper&lt;/a&gt; which discusses exactly this question. It considers several models of varying complexity and finds a maximum &lt;em&gt;range efficiency&lt;/em&gt; of 83% for a 100 pound counterweight, 1 pound projectile, a 5 foot long beam pivoted 1 foot from the point of attachment of the counterweight, and a 3.25 foot long sling. Here range efficiency is defined as the horizontal range of the realistic trebuchet model relative to the range of a &quot;black box model&quot; which is able to completely convert the gravitational potential energy of the counterweight into kinetic energy of the projectile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to find the &lt;em&gt;energy efficiency&lt;/em&gt;, defined as the fraction of the counterweight's gravitational potential energy that actually gets transferred to the projectile, you would need to use the relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\epsilon_R}{\epsilon_E} = 2\sin\alpha\cos\alpha = \sin 2\alpha$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\alpha$ is the angle of release of the projectile above the horizontal. Unfortunately, the paper doesn't give the value of $\alpha$ corresponding to the simulation that produced the maximum efficiency, so I can't give you a specific number without running the simulations myself. (Perhaps I'll do that when I have time; if anyone else gets to it first, feel free to edit the relevant numbers in.)&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-12-27T06:29:25.413" CommentCount="5" />
  <row Id="2288" PostTypeId="2" ParentId="2230" CreationDate="2010-12-27T06:49:03.970" Score="2" ViewCount="" Body="&lt;p&gt;As to part 2 of your question &quot;Also why nothing can be more speedy than light?&quot;, the answer is that it's not just light.  The point is that c is the &lt;em&gt;maximal velocity of any causal, information transmitting interaction in the universe&lt;/em&gt;, mediated by anything travelling forwards in time (see footnote).  Its just that photons, having 0 rest mass, travelling in a vacuum approach that fundamental limit, c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Footnote: Except maybe 'tachyons' - never seen and traveling backwards in time because they go faster than c.   (Note that Norbert Wiener once pointed out that for a causal influence travelling backwards in time, we would experience it as &quot;random&quot;, since it would apparently be an event without an antecedent cause to us).&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-12-27T06:49:03.970" />
  <row Id="2289" PostTypeId="1" CreationDate="2010-12-27T07:07:07.223" Score="4" ViewCount="323" Body="&lt;p&gt;What is the point of having 45 degreed blade in Guillotine? After all, the pressure-Force over Area-blade imposes to the neck is the same either way. &lt;/p&gt;&#xA;" OwnerUserId="987" LastActivityDate="2010-12-27T07:07:07.223" ClosedDate="2010-12-27T12:11:00.640" Title="Why does Guillotine have of 45 degreed blade rather than the one parallel to the ground?" Tags="&lt;mechanics&gt;" CommentCount="16" />
  <row Id="2290" PostTypeId="2" ParentId="2230" CreationDate="2010-12-27T07:42:15.020" Score="5" ViewCount="" Body="&lt;p&gt;The view of most physicists is that asking &quot;How can it be that the speed of light is constant?&quot; is similar to asking &quot;How can it be that things don't always go in the direction of the force on them?&quot; or &quot;How can it be that quantum-mechanical predictions involve probability?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual answer is that these things simply &lt;i&gt;are&lt;/i&gt;.  There is no deeper, more fundamental explanation.  There is some similarity here with the viewpoint you may have learned in studying Euclidean geometry; we need to start with some axioms that we assume to be true, and cannot justify.  Philosophically, these ideas are not precisely the same (mathematical axioms are not subject to experimental test), but the constant speed of light is frequently described as a &quot;postulate&quot; of relativity.  Once we assume it is true, we can work out &lt;a href=&quot;http://en.wikipedia.org/wiki/Introduction_to_special_relativity&quot; rel=&quot;nofollow&quot;&gt;its logical consequences&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not to say that, in physics, postulates stay postulates.  For example, many people are especially concerned about probability in quantum mechanics, and are trying to understand it based on more fundamental ideas (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_decoherence&quot; rel=&quot;nofollow&quot;&gt;decoherence&lt;/a&gt; as one example).  As another example, Newton's laws of motion were originally taken as unprovable postulates, but are now explained via quantum mechanics (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Ehrenfest_theorem&quot; rel=&quot;nofollow&quot;&gt;Ehrenfest's theorem&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this time, the constancy of the speed of light, or more generally the principle of &lt;a href=&quot;http://en.wikipedia.org/wiki/Lorentz_covariance&quot; rel=&quot;nofollow&quot;&gt;Lorentz symmetry&lt;/a&gt;, is not justified by anything considered to be more fundamental.  In fact, the assumption that it is true has been a guiding light to theoretical physicists; quantum field theory was invented by thinking about how quantum mechanics could be made to respect the ideas of relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although we do not have a theoretical justification for the constancy of the speed of light, we do have very accurate experimental tests of the idea.  The most famous is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Michelson%E2%80%93Morley_experiment&quot; rel=&quot;nofollow&quot;&gt;Michelson-Morley experiment&lt;/a&gt;, which measured the relative speed of light in different directions to see if it was affected by the motion of the Earth.  This experiment rejected the hypothesis that the motion of the Earth affects the speed of light.  According to the Wikipedia article I linked, a modern version of this experiment by Hils and Hall concluded that the difference in the speed of light along directions parallel and perpendicular to Earth's motion is less than one part in $5*10^{12}$.  In addition to direct tests of the speed of light, there have also been &lt;a href=&quot;http://www.xs4all.nl/~johanw/PhysFAQ/Relativity/SR/experiments.html&quot; rel=&quot;nofollow&quot;&gt;many other experimental tests of special relativity&lt;/a&gt;.  (I haven't read this last page carefully, but, on flipping through, it looks good.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few caveats worth mentioning.  In general relativity, the speed of light is only constant locally.  This means that the distance between two objects can increase faster than the speed of light, but it is still impossible for light to zip past you at a speed faster than the normal one.  Also, in quantum theory, the speed of light is a statistical property.  A photon may travel slightly slower or faster than light, and only travels at light speed on average.  However, deviations from the speed of light would be probably be too small to observe directly.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2010-12-27T07:42:15.020" />
  <row Id="2291" PostTypeId="1" AcceptedAnswerId="2303" CreationDate="2010-12-27T08:44:57.503" Score="5" ViewCount="448" Body="&lt;p&gt;Suppose I do two experiments to find the triple point of water, one in zero-g and one on Earth.  On Earth, water in the liquid or solid phase has less gravitational potential per unit mass than water in the gas phase.  Therefore, the solid and liquid phases should be favored slightly more on Earth than in zero-g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a back-of-the-envelope calculation, how does the temperature of the triple-point of water depend on the gravitational acceleration and, if necessary, on the mass of water and volume and shape of container?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit&lt;/b&gt;&#xA;Let's say I have a box in zero-g.  The box is one meter on a side.  It has nothing in it but water.  Its temperature and pressure are just right so that it's at the triple point.  All the water and ice and steam are floating around the box because it's zero-g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I turn on gravity.  The liquid water and ice fall to the bottom of the box, but the average height of the steam remains almost half a meter above the bottom of the box.  So when gravity got turned on, the potential energy of the ice and liquid water went down significantly, but the potential energy of the steam didn't.  Doesn't this mean that once gravity is turned on, water molecules would rather be part of the ice or liquid phase so that they can have lower energy?  Wouldn't we no longer be at the triple point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several people have posted saying the answer is &quot;no&quot;.  I don't disbelieve that.  Maybe the answer is just &quot;no&quot;.  I don't understand why the answer is no.  Answers such as &quot;No, because gravity doesn't affect the triple point,&quot; or &quot;No, because the triple point only depends on pressure and temperature&quot; simply restate the answer &quot;no&quot; with more words.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-12-28T01:28:37.110" LastActivityDate="2010-12-28T01:28:37.110" Title="How does the temperature of the triple point of water depend on gravitational acceleration?" Tags="&lt;thermodynamics&gt;&lt;order-of-magnitude&gt;" AnswerCount="5" CommentCount="7" FavoriteCount="1" />
  <row Id="2292" PostTypeId="1" AcceptedAnswerId="2293" CreationDate="2010-12-27T09:01:18.663" Score="5" ViewCount="152" Body="&lt;p&gt;When judging if relativity is important in a given phenomenon, we might examine the number $v/c$, with $v$ a typical velocity of the object.  If this number is near one, relativity is important.  In optics, we examine $\lambda/d$, with $\lambda$ the wavelength of light involved, and $d$ a typical size-scale for the object interacting with light.  If this number is near one, physical optics becomes important.  In fluid mechanics we have the Reynolds number, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we compute a number to tell roughly when a gas will deviate significantly from the ideal gas law?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, you may want to deal only with a monatomic gas, but a more general approach would be welcome.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-12-27T09:07:31.237" LastActivityDate="2010-12-30T05:41:20.197" Title="Is there a number that describes a gas's departure from the ideal gas law?" Tags="&lt;thermodynamics&gt;&lt;order-of-magnitude&gt;" AnswerCount="3" />
  <row Id="2293" PostTypeId="2" ParentId="2292" CreationDate="2010-12-27T09:17:30.763" Score="6" ViewCount="" Body="&lt;p&gt;The simplest extension of the ideal gas law is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_equation&quot; rel=&quot;nofollow&quot;&gt;van der Waals equation of state&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (p + \frac{a&amp;#39;}{v^2})(v - b&amp;#39;) = kT $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a&amp;#39;$ and $b&amp;#39;$ account for the interatomic attraction and the finite particle size respectively. These two parameters are thus candidates for &quot;number(s) that describe deviations from the ideal gas law&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;proper&lt;/em&gt; measure of such deviations are referred to as &lt;a href=&quot;http://en.wikipedia.org/wiki/Virial_coefficient&quot; rel=&quot;nofollow&quot;&gt;virial coefficients&lt;/a&gt; which are parameters in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Virial_expansion&quot; rel=&quot;nofollow&quot;&gt;virial expansion&lt;/a&gt; - a power series expansion of the pressure in terms of the temperature. You can find details at the referenced links or in a stat mech textbook such as Pathria or Huang.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-27T09:17:30.763" CommentCount="5" />
  <row Id="2294" PostTypeId="2" ParentId="2291" CreationDate="2010-12-27T09:42:01.907" Score="2" ViewCount="" Body="&lt;p&gt;There is no difference; phase transitions does not change gravitational potential.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-12-27T09:42:01.907" CommentCount="7" />
  <row Id="2295" PostTypeId="2" ParentId="2291" CreationDate="2010-12-27T10:17:02.273" Score="0" ViewCount="" Body="&lt;p&gt;I don't see how to derive triple point directly, but let me talk about something similar and see where it leads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ideal gas atmosphere you'll find out that $g$ works as an inverse temperature. That is, $0K$ is the same as infinitely strong gravity because both keep the gas at the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real gases exhibit phase transitions, so we should promote ideal gas to something like van der Waals gas. This should not change the above correspondence of $\beta = {1 \over k_B T}$ and $g$ much. Therefore, I would imagine that one would get a phase transition at high $g$ with constant $\beta$. Or in other words, that in the presence of gravitational field, higher temperatures are enough for the phase transition. Or in other words yet, we have another coupling $g$ that tells system to cluster besides the usual thermal coupling $\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this carries over (by a huge and completely unjustified extrapolation) to the triple point of water. I'll try to provide some calculations later.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-27T10:35:27.227" LastActivityDate="2010-12-27T10:35:27.227" CommentCount="6" />
  <row Id="2296" PostTypeId="2" ParentId="2111" CreationDate="2010-12-27T10:46:11.923" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;I must say that i was totally wrong! Of course Mark Eichenlaub and Omega Centauri are right!&lt;br&gt;&#xA;My apologies!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="552" LastEditorUserId="552" LastEditorDisplayName="" LastEditDate="2011-01-31T07:47:26.830" LastActivityDate="2011-01-31T07:47:26.830" CommentCount="8" />
  <row Id="2297" PostTypeId="2" ParentId="2227" CreationDate="2010-12-27T13:53:47.033" Score="1" ViewCount="" Body="&lt;p&gt;I will attempt a qualitative answer to the question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We assume that we have an observer that is performing an oscillatory motion in which we have a repetition of two constant acceleration phases. The trajectory would be the sum of parts of hyperbola, where each part would correspond to a constant acceleration phase ($k$ or $-k$) with the acceleration directed towards the $x=0$. In this motion there will be no event horizon, but there will be several apparent horizons made out of the corresponding Rindler horizons of each part of the orbit that has constant acceleration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a lot of work that supports the fact that Unruh/Hawking-like radiation can be observed from apparent horizons (see &lt;a href=&quot;http://arxiv.org/abs/hep-th/0106111&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, &lt;a href=&quot;http://arxiv.org/abs/0809.1554v2&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://arxiv.org/abs/0812.4217&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for example). So my guess would be that if the essential features that Visser describes exist, then you would see radiation. Thus in this particular case, for each accelerated branch the observer would se thermal radiation corresponding to acceleration $k$ as long as the acceleration phase is long enough to satisfy the condition for slow evolution of the apparent horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am guessing that you can apply the same reasoning for the case of harmonic motion, as long as the period of oscillation is long enough. The spectrum of the radiation would evolve in time and would probably be something like an integrated Planck spectrum over a range of temperatures which would probably be something like a power law.&lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2010-12-27T13:53:47.033" CommentCount="2" />
  <row Id="2298" PostTypeId="2" ParentId="2285" CreationDate="2010-12-27T15:05:16.880" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Maxwell's equations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't so much a myth, as a misnomer! The four equations commonly called Maxwell's equations should in fact be attributed to Gauss, Faraday, and Ampere. Maxwell added the important displacement current term to Ampere's equation, which then allowed him to derive and explain electromagnetic waves - his true achievement. The equations however, were a joint work over almost a century of physics.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2010-12-27T15:05:16.880" CommunityOwnedDate="2010-12-27T15:05:16.880" CommentCount="4" />
  <row Id="2299" PostTypeId="2" ParentId="2291" CreationDate="2010-12-27T16:06:32.757" Score="0" ViewCount="" Body="&lt;p&gt;Back of the envelope calculation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;gravity will manifest itself by increasing the pressure by a factor of $mg/A$ where A is the surface of the container with water at the triple point. So basically the effect is shifting the triple point &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aPhase-diag2.svg&quot; rel=&quot;nofollow&quot;&gt;phase diagram&lt;/a&gt; down. So the triple point would be at the same temperature with a lower pressure.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$p_{measured} = p_{gas} + p_{gravity}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Assuming a regular shape we can say that $p_{gravity}=\rho g h$ where $\rho$ is the density of the gas at our triple point pressure and h is the typical height of a column of gas.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$p_{gravity}=\rho g h$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;to calculate $\rho$ we can use the ideal gas law: &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\rho = \frac{Nm_w}{V} = \frac{p_{gas}m_w}{kT}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;we can now calculate what fraction of pressure is due to gravity&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\frac{p_{gravity}}{p_{gas}}=\rho g h/p_{gas}=\frac{m_w g h}{kT}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;using the following values we can calculate this ratio&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$m_w = 3 \times 10^{-26} \mathrm{kg}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g = 9.81 \mathrm{m/s^{2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h = 1 \mathrm{m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k = 1.38 \times 10^{-23} \mathrm{J/K}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T = 273.16 K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{p_{gravity}}{p_{gas}}=0.000078$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that you need a value for g one thousand times stronger to see a large effect.&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2010-12-27T17:06:56.363" LastActivityDate="2010-12-27T17:06:56.363" CommentCount="6" />
  <row Id="2300" PostTypeId="1" CreationDate="2010-12-27T18:38:33.290" Score="3" ViewCount="183" Body="&lt;p&gt;Could you please help me in this problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 3 independent variables, (T,H,t) , as inputs and one output P  ( I have all data for these inputs and the output, done experimentally measured every hour during one year).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find a formula of this form: P=f(T,H,t)  where t  is the time in hours and it is always in the x -axis (index), T  is temperature, H  is humidity, and P  is power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have all the data, and when I draw them in the same graph during one year, meaning that P , T , and H  vs. hours. I found that the behavior of P  is oscillating, making a sinusoidal shape over the entire year:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www10.0zz0.com/2010/12/27/18/159867792.jpg&quot; rel=&quot;nofollow&quot;&gt;http://www10.0zz0.com/2010/12/27/18/159867792.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if I make a zoom view to this figure, for example from the 2000 th hour of the year to the 3000 th hour, it is clear that it almost has the same shape but it is oscillating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it keeps oscillating and increasing up to the peak point and then it starts decreasing till the end of the year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how can I predict the structure of the formula that relates P  with T , H , and t ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the effect of T,H and t on P .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any approach that you advise me to follow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for this long question and any help from you is highly appreciated. I read many papers but I could not know how to solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-28T00:14:39.343" LastActivityDate="2011-08-25T08:34:30.120" Title="Relating the power to weather variables" Tags="&lt;mathematics&gt;&lt;models&gt;&lt;atmospheric-science&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="2301" PostTypeId="2" ParentId="2285" CreationDate="2010-12-27T18:52:26.500" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;MYTH:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (L)arge (H)adron (C)ollider will destroy the earth. High energy collisions will create a black hole and it will swallow the earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read this in a newspaper. It was also mentioned there that this is myth and physicist are pretty/damn sure that this won't happen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; &lt;em&gt;I don't know enough about LHC and particles to debunk this one. But I am sure that no sane physicist would want to commit suicide.&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-12-28T06:47:04.140" LastActivityDate="2010-12-28T06:47:04.140" CommunityOwnedDate="2010-12-27T18:52:26.500" CommentCount="12" />
  <row Id="2302" PostTypeId="2" ParentId="2300" CreationDate="2010-12-27T19:26:33.053" Score="0" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;You can train some &lt;a href=&quot;http://en.wikipedia.org/wiki/Neural_network&quot; rel=&quot;nofollow&quot;&gt;Neural Networks&lt;/a&gt; with your data and then use them for predictions afterwards.&lt;/li&gt;&#xA;&lt;li&gt;You can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Interpolation&quot; rel=&quot;nofollow&quot;&gt;interpolation&lt;/a&gt; techniques to arrive at approximate formula from your data.&lt;/li&gt;&#xA;&lt;li&gt;One cool idea is to try applying models used in &lt;a href=&quot;http://en.wikipedia.org/wiki/Algorithmic_trading&quot; rel=&quot;nofollow&quot;&gt;Algorithmic Trading&lt;/a&gt; for prediction from data.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;From the graph it looks like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P \propto T$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P \propto \frac{1}{H}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can also be noted that power is not very sensitive to changes in humidity but is highly correlated with changes in temperature. &lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2010-12-27T19:31:33.190" LastActivityDate="2010-12-27T19:31:33.190" CommentCount="5" />
  <row Id="2303" PostTypeId="2" ParentId="2291" CreationDate="2010-12-27T19:52:56.973" Score="9" ViewCount="" Body="&lt;p&gt;This is perhaps similar to what mbq meant, but I will elaborate.&#xA;The T-p phase diagram of water tells us, for a given temperature and pressure, what phase we will get if we have a bunch of that substance.  If I apply different pressures to a bottle of water, I am moving around in the p-direction of the T-p plane.  I am not changing the pressure of the triple point of water, just changing the pressure of that particular bottle of water! Similarly, if a tank of water is in a gravitational field, it affects the pressure.  In fact, it leads to different pressures at different locations of the tank.  It could lead some parts of the tank to freeze, for example.  But it does not in any way change the triple point of water itself, which is an intrinsic property of that substance.  So I would say that the question is ill-posed.  It might be better to ask: what will happen to a tank of water at a given temperature and density if we now apply a gravitational field?&lt;/p&gt;&#xA;" OwnerUserId="266" LastActivityDate="2010-12-27T19:52:56.973" CommentCount="40" />
  <row Id="2304" PostTypeId="2" ParentId="2291" CreationDate="2010-12-27T20:43:02.860" Score="2" ViewCount="" Body="&lt;p&gt;To expand on &lt;a href=&quot;http://physics.stackexchange.com/questions/2291/how-does-the-temperature-of-the-triple-point-of-water-depend-on-gravitational-acc/2294#2294&quot;&gt;mpq&lt;/a&gt; and &lt;a href=&quot;http://physics.stackexchange.com/questions/2291/how-does-the-temperature-of-the-triple-point-of-water-depend-on-gravitational-acc/2303#2303&quot;&gt;Greg's remarks&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A triple point is a unique (pressure,temperature) pair for each material (and where appropriate group of phases) where three phases of matter can coexist in equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are not at the right pressure, there is &lt;em&gt;no&lt;/em&gt; temperature where this occurs, and if you are not at the right temperature there is &lt;em&gt;no&lt;/em&gt; pressure where this occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that sense the question is simply based on a misconception.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Several posters have tried to address the effects of changing gravitation on the locally experienced pressure due to a atmosphere column. That's fine as far as it goes, but it doesn't really go to the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diddling the local gravity will change the pressure naturally experienced due the local atmosphere but &lt;em&gt;will not&lt;/em&gt; affect the parameters of the triple point.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2010-12-27T21:07:50.307" LastActivityDate="2010-12-27T21:07:50.307" CommentCount="1" />
  <row Id="2305" PostTypeId="2" ParentId="2249" CreationDate="2010-12-27T21:22:03.140" Score="0" ViewCount="" Body="&lt;p&gt;One page review for beginners like me at &lt;a href=&quot;http://www.ornl.gov/info/ornlreview/v30n3-4/giant.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.ornl.gov/info/ornlreview/v30n3-4/giant.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-12-27T21:22:03.140" />
  <row Id="2306" PostTypeId="2" ParentId="2112" CreationDate="2010-12-27T22:06:44.957" Score="1" ViewCount="" Body="&lt;p&gt;Detecting an object requires you to get enough of a signal from it that you can discriminate that signal from background noise. This is not necessarily dependent on the size of the object, although it can be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you are trying to detect an object which has very low reflectivity at your given wavelength, you might get a stronger signal if the object is larger. Of course this also depends on the geometry of the object, and how much of the reflected light is reflected towards your detector. Ultimately the deciding factor in whether you can detect an object is the ratio of the signal strength to the noise in your detection system. Engineers typically call this the &quot;Signal to noise ratio&quot; or SNR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Rayleigh limit will describe the minimum distance between two point sources such that an imaging system can distinguish one point from the other, rather than seeing them as a single point. However, this has nothing to do with the fundamental requirement of detecting the presence of an object in the first place. It is quite possible to, for example, detect medium-wavelength infrared light (with a wavelength of 3 to 8 microns) emanating from a pinhole or the tip of an optical fiber which is only a couple microns across. In this case an imaging system would indeed see a spot of light with a radius given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{array}{rl}&#xD;&#xA; \Delta l = 1.22 \frac{f\lambda}{D} &amp;amp;\mbox{for a circular aperture}\\&#xD;&#xA; \Delta l = \frac{f\lambda}{D} &amp;amp;\mbox{for a square aperture}&#xD;&#xA;\end{array}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply by detecting this spot, you would have detected the presence of the object, although you would not be able to say anything about its size or shape, because it is below the resolution of your device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good way to think about this is to get away from the concept of an imaging system. Imagine you've got a lens focusing light onto a photodiode, instead of an imaging detector like the CCD in a camera. With a single photodiode, you can't determine the size or shape of anything. All you can do is detect a signal, or the lack of a signal. This is mostly independent of resolving power or the size of the object being detected.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2010-12-27T22:06:44.957" />
  <row Id="2307" PostTypeId="2" ParentId="1094" CreationDate="2010-12-27T22:37:31.473" Score="1" ViewCount="" Body="&lt;p&gt;Any conductive material will block any EM radiation, including radio waves. However, the thickness of material required to fully block the radiation depends on the wavelength of the radiation and the properties of the material. With a perfect conductor, the radiation is stopped completely by any thickness of the material, but in any real material (except for a superconductor) the radiation is attenuated as it travels through the material. The strength of the radiation will fall off with an exponential decay, and it will do so more quickly in a better conductor. The important number here is called &quot;skin depth&quot; which is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta = \frac{c}{\sqrt{2\pi\sigma\mu\omega}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\sigma$ is the conductivity of the material, $\mu$ is the magnetic permeability of the material, and $\omega$ is the frequency of the radiation. The skin depth is the distance into the material at which the radiation will have fallen to a fraction of $1/e \approx 0.37$ relative to its original intensity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the skin depth will be smaller for a better conductor or higher frequency radiation. For example, for Aluminum, radiation at 60Hz will have a skin depth of over 8 millimeters, while optical radiation around 600 THz will have a skin depth around 3 nanometers! This is why metal blocks visible light so well, yet radio waves easily penetrate into your house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Lectures on Electromagnetism&quot; page 280. Ashok Das; Hindustan Book Agency, New Delhi, 2004.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Skin_effect&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Skin_effect&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="869" LastActivityDate="2010-12-27T22:37:31.473" CommentCount="1" />
  <row Id="2308" PostTypeId="1" CreationDate="2010-12-27T22:47:12.467" Score="3" ViewCount="479" Body="&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Spacetime#Space-like_interval&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When a space-like interval separates&#xA;  two events, not enough time passes&#xA;  between their occurrences for there to&#xA;  exist a causal relationship crossing&#xA;  the spatial distance between the two&#xA;  events at the speed of light or&#xA;  slower. Generally, the events are&#xA;  considered not to occur in each&#xA;  other's future or past. There exists a&#xA;  reference frame such that the two&#xA;  events are observed to occur at the&#xA;  same time, but there is no reference&#xA;  frame in which the two events can&#xA;  occur in the same spatial location.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can someone give me an example of this?&lt;/p&gt;&#xA;" OwnerUserId="993" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-28T00:13:10.483" LastActivityDate="2010-12-28T01:20:34.460" Title="Example of space-like intervals in spacetime" Tags="&lt;special-relativity&gt;&lt;spacetime&gt;&lt;causality&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="2309" PostTypeId="2" ParentId="2254" CreationDate="2010-12-27T22:53:00.840" Score="6" ViewCount="" Body="&lt;p&gt;The surface of any fluid has an associated energy-per-unit-area, known as the surface energy, a.k.a. surface tension. This energy is not a property of the fluid alone, but of the fluid and the medium it is in contact with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case you would have associated surface energies for the water-air interface, e_wa, as well as for the water-paper interface, e_wp. The total energy of the fluid in a configuration is the sum of the potential energy, plus the product of the corresponding surface energies by their respective surface areas, S_wa*e_wa +S_wp*e_wp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you want to look at it in a purely energy balance point of view, the increase in potential energy of the water climbing up the paper is compensated by a reduction of the total surface energy. When capillary action makes things rise, it is because the liquid-solid energy is lower than the liquid-air energy. By wicking into the porous material, the solid-liquid contact area is increased at the expense of the liquid-air one, resulting in an overall reduction of contact energy, which is what drives the rise in potential energy.&lt;/p&gt;&#xA;" OwnerUserId="995" LastActivityDate="2010-12-27T22:53:00.840" CommentCount="3" />
  <row Id="2310" PostTypeId="2" ParentId="2292" CreationDate="2010-12-27T23:06:27.880" Score="0" ViewCount="" Body="&lt;p&gt;I seem to recall it was primarily about heat capacity. A perfect gas has heat capacities of 5/3 or 1 (constant pressure versus volume), because the only energy sinks/sources are translational kinetic energy of the particles (atoms/molecules), and pressure times dV. If these molecules have other degrees of freedom, vibrational or rotational usually, then there are other places for energy (heat) to be stored and the heat capacity increases. This can be made intuitive by thinking where extra heat can go. In an imperfect gas, some of the additional energy doesn't go to translational kinetic energy of the particles (where it contributes to pressure), but instead it goes to other internal (to the particles) degrees of freedom. Even monoatomic gases have some other degrees of freedom (quantum states(orbitals) for the electron(s)), but&#xA;these are usually of high ennough energy as to not be activated at room temperature. For a given material the heat capacity is a function of temperature and pressure, any departure from the values for a perfcet gas, means that an adiabatic change in pressure/volume will have a different curve than for a perfect gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the comment about van der Waals, is valid, the particles (atoms/molecules) do take up volume, so not all of the volume is not available for the other particles to bounce around in. This means, the pressure/volume/temperature relationship is modified by this effect as well. But this effect is proportional to density, so at low pressure the heat capacity will be the dominate cause of departure from a perfect gas. &lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2010-12-27T23:06:27.880" />
  <row Id="2311" PostTypeId="2" ParentId="2308" CreationDate="2010-12-27T23:33:27.913" Score="7" ViewCount="" Body="&lt;p&gt;A very simple example is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two observers A and B are spatially separated (one in London, the other in Cambridge). They are static with respect to each other and can therefore measure the same time. So, we can fix a particular time T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two events $(T,\mathbf{x_{London}})$ and $(T,\mathbf{x_{Cambridge}})$ are spatially separated - one cannot influence the other. Anything happening at $(T,\mathbf{x_{London}})$, for example, can only influence B later on when it crosses its light cone at $T+\frac{||\mathbf{x_{Cambridge}}-\mathbf{x_{London}}||}{c}$.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-12-27T23:33:27.913" />
  <row Id="2312" PostTypeId="2" ParentId="2308" CreationDate="2010-12-27T23:49:07.950" Score="5" ViewCount="" Body="&lt;p&gt;Ultimately the question is about the causal structure of geometry. The local, causal structure of any event $E$ (represented by a point in Minkowski space) is determined by its light-cone, the boundary between events - past and future - that are spacelike or timelike w.r.t $E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.theory.caltech.edu/people/patricia/gifs/glcaus1.gif&quot; alt=&quot;Light-Cone&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this figure the vertical axis is time (t) and the two horizontal axes are spatial (x,y). The points inside the cone are those which are in the causal past and future of the event $E$. The points lying outside the cone cannot have any causal influence on &lt;em&gt;or&lt;/em&gt; cannot be causally affected by $E$. If you pick any point $P =(x,y)$ in this region, then the interval between $E = (0,0)$ and $P$ will be &quot;spacelike&quot;. The interval between any two points in Minkowski space is calculated using the (-,+,+) metric:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ l_{12}^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2 - c^2(t_1 - t_2)^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c$ is the speed of light. Note the minus sign. That is what distinguishes timelike dimensions from spacelike ones. Points lying &lt;em&gt;on&lt;/em&gt; the light-cone itself have zero or &quot;null&quot; separation in this metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, pick any two points $(x_1,y_1,t_1)$ and $(x_2,y_2,t_2)$ such that $l_{12}^2 \gt 0$. These two points have a space-like separation. As for a &lt;strong&gt;concrete example&lt;/strong&gt; of such a pair of events, consider any two clocks in your house with spatial separation $\delta x$ between them, synchronized to within $\delta t$ seconds of each other such that $\delta x \gt c \delta t$. Then everytime the clocks tick, the separation between those two events in spacelike.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; The sign associated with spacelike or timelike intervals is a matter of convention and depends on your choice of metric $(-,+,+)$ or (+,-,-)&lt;/p&gt;&#xA;" LastActivityDate="2010-12-27T23:49:07.950" />
  <row Id="2313" PostTypeId="2" ParentId="2308" CreationDate="2010-12-28T01:20:34.460" Score="2" ViewCount="" Body="&lt;p&gt;Recipe to generate events with space like separation (and be able to prove it):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Set up a radio transmitter at point $A$.&lt;/li&gt;&#xA;&lt;li&gt;At point $B$ and $C$ which are arrange so that A lies at the center point of the segment between them we put radio receivers that will perform some action on receiving a radio pulse from A. We make the distance between them greater than the speed-of-light * however-long-it-takes-the gadget to act.&lt;/li&gt;&#xA;&lt;li&gt;Emit a pulse from the transmitter at $A$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The result is that two events will happen---one each at $B$ and $C$---which appear to be simultaneous as observed from A. Thus they have space-like separation. &lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2010-12-28T01:20:34.460" />
  <row Id="2317" PostTypeId="1" AcceptedAnswerId="2318" CreationDate="2010-12-28T15:17:58.630" Score="4" ViewCount="220" Body="&lt;p&gt;Including air resistance, what is the escape velocity from the surface of the earth for a free-flying trajectile?&lt;/p&gt;&#xA;" OwnerUserId="775" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-28T15:59:16.630" LastActivityDate="2010-12-28T15:59:16.630" Title="Including air resistance, what is the escape velocity from Earth?" Tags="&lt;classical-mechanics&gt;&lt;air&gt;&lt;projectile&gt;" AnswerCount="2" />
  <row Id="2318" PostTypeId="2" ParentId="2317" CreationDate="2010-12-28T15:33:24.020" Score="2" ViewCount="" Body="&lt;p&gt;For the lower atmosphere, where most of the air is, the temp, pressure, and density of air is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T=T_0-Lh$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p=p_0\left(\frac{T}{T_0}\right)^{\frac{gM}{RL}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho = \frac{pM}{RT} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the following constants:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;sea level standard atmospheric pressure p0 = 101325 Pa&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;sea level standard temperature T0 = 288.15 K&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Earth-surface gravitational acceleration g = 9.80665 m/s2.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;temperature lapse rate L = 0.0065 K/m&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;universal gas constant R = 8.31447 J/(mol·K)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;molar mass of dry air M = 0.0289644 kg/mol&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The force due to air resistance can be written:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = -\rho v^2 C_d A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $C_d$ is the coefficient of drag, $v$ is the velocity, and $A$ is the surface area of the projectile. The goal is to get out of the atmosphere (where force of gravity is roughly constante) with the Earth's escape velocity, $11.2$ km/s. For a bullet-shaped 1-kg projectile of steel, $C_d \approx 0.04$ and $A \approx 4\times 10^{-4}$ m$^2$. This leads to a initial velocity of $13.5$ km/s. While not much higher than the vaccuum value, it is still high enough that the bullet would probably vaporize in the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, if one used a sphere instead of a bullet, then $C_d=0.4$, $A=4\times 10^{-3}$ m$^2$, the air resistance is 100 times higher, and thus a much greater velocity is needed. But since the drag scales like $v^2$, this leads to much higher drag. So much so, that even if one could launch the ball at the speed of light (Newtonians only, please!), it still could not make it out of the atmosphere!&lt;/p&gt;&#xA;" OwnerUserId="775" LastEditorUserId="775" LastEditorDisplayName="" LastEditDate="2010-12-28T15:54:25.867" LastActivityDate="2010-12-28T15:54:25.867" CommentCount="4" />
  <row Id="2319" PostTypeId="2" ParentId="2285" CreationDate="2010-12-28T15:34:33.623" Score="3" ViewCount="" Body="&lt;p&gt;&lt;em&gt;All of the physical laws are $C$, $P$ and $T$ symmetric.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is actually not that far from truth, which is why it took so long to notice it doesn't hold. Only weak interactions violate these symmetries. In 1950s it was indeed realized that while then known electromagnetic and strong processes were tested to be $P$-symmetric, weak processes were omitted. And indeed as soon as experiments were conducted (on a certain beta decay), violation has been found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C$-symmetry is violated by the fact that we only observe left-handed neutrinos and right-handed antineutrinos. But $C$-symmetry would imply also right-handed neutrinos and left-handed antineutrinos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for $T$-symmetry, this is the same as $CP$-symmetry, by $TCP$ theorem (and this is no blind belief anymore because this theorem is an immediate consequence of any reasonable quantum field theory). This symmetry was proposed after the discovery of $P$-symmetry and later found to be violated because $K_L$, a long-lived kaon eigenstate, can decay into two pions (which would be forbidden if $CP$-symmetry held). Today, these violations are understood in terms of &lt;a href=&quot;http://en.wikipedia.org/wiki/Cabibbo%2dKobayashi%2dMaskawa_matrix&quot; rel=&quot;nofollow&quot;&gt;CKM matrix&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also our intuition about violations is by now completely reversed: besides $TCP$-symmetry (which has been verified to a very large degree) people became actually suspicious when some system &lt;em&gt;obeys&lt;/em&gt; some additional symmetry. This is best illustrated on &lt;a href=&quot;http://en.wikipedia.org/wiki/CP_violation#CP_violation_in_strong_sector&quot; rel=&quot;nofollow&quot;&gt;strong $CP$ problem&lt;/a&gt; because strong interactions obey this symmetry and there is no a priori reason why they should.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2010-12-28T15:34:33.623" CommunityOwnedDate="2010-12-28T15:34:33.623" CommentCount="4" />
  <row Id="2320" PostTypeId="2" ParentId="2285" CreationDate="2010-12-28T15:42:21.570" Score="6" ViewCount="" Body="&lt;p&gt;I'll add the popular myth that Special relativity was made &quot;in response&quot; to &lt;a href=&quot;http://en.wikipedia.org/wiki/Michelson%E2%80%93Morley_experiment&quot; rel=&quot;nofollow&quot;&gt;Michelson–Morley experiment&lt;/a&gt;. Actually, Einstein never mentioned the experiment in his work, as well as he didn't mention aether.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that new theory must not be constructed &quot;in response&quot; to some new experiment. New theory should embrace all existent experimental evidence. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2010-12-28T15:42:21.570" CommunityOwnedDate="2010-12-28T15:42:21.570" CommentCount="2" />
  <row Id="2321" PostTypeId="2" ParentId="2317" CreationDate="2010-12-28T15:44:30.350" Score="0" ViewCount="" Body="&lt;p&gt;It depends on the altitude, but let's assume sea level. The base escape velocity, assuming a vacuum, is &lt;a href=&quot;http://en.wikipedia.org/wiki/Escape_velocity#List_of_escape_velocities&quot; rel=&quot;nofollow&quot;&gt;11.2 km/s&lt;/a&gt;. From there, it depends on the aerodynamical properties of the object. From &lt;a href=&quot;http://library.thinkquest.org/10568/aerodynamics/forces.html&quot; rel=&quot;nofollow&quot;&gt;Hitchhiker's Guide to Model Rocketry&lt;/a&gt;, we get this formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drag = 1/2 * V ^ 2 * (Air Density) * CD * (Projected Area)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CD, or Coefficient of Drag, is a dimensionless number dependent on the shape of an object. A typical model rocket has a CD of about 0.75, while a high performance, highly streamlined rocket may be as low as 0.4. Smaller fins will reduce the CD, but also cause the stability to go down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm going to assume for simplicities sake that we've got a 0.4 CD rocket. I'll assume a 1 m diameter circular projected area, or pi/4 projected area. That leaves us still needing an adequate Air Density model. The air density model I'm using is pulled from &lt;a href=&quot;http://en.wikipedia.org/wiki/Density_of_air#Altitude&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, and is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/cYdtu.png&quot; alt=&quot;T term&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/UUlAm.png&quot; alt=&quot;p term&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/XAtgM.png&quot; alt=&quot;Density Term&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I might try and model this sometime, but let me just give you the force at sea level, given a normal escape velocity. The drag is 2414 kN. Assuming this object was roughly cylndrical, having a density of 1, and is 10 m long, that would give it a weight of roughly 8000 kg. That would be somewhat less then the affect of gravity on the same object. I'd guess from these numbers that for this object, if you were going around 14-15 km/s, you'd be able to escape the Earth in one shot, but I'd have to crunch the numbers more carefully.&lt;/p&gt;&#xA;" OwnerUserId="871" LastEditorUserId="871" LastEditorDisplayName="" LastEditDate="2010-12-28T15:56:25.080" LastActivityDate="2010-12-28T15:56:25.080" CommentCount="2" />
  <row Id="2322" PostTypeId="1" CreationDate="2010-12-28T16:00:33.587" Score="7" ViewCount="705" Body="&lt;p&gt;I was reading up on Tesla's Wikipedia page last night, and I came across this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When he was 81, Tesla stated he had&#xA;  completed a &quot;dynamic theory of&#xA;  gravity&quot;. He stated that it was&#xA;  &quot;worked out in all details&quot; and that&#xA;  he hoped to soon give it to the&#xA;  world. [75] The theory was never&#xA;  published.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was wondering, has this theory ever been found? &lt;a href=&quot;http://en.wikipedia.org/wiki/Nikola_Tesla#Field_theories&quot; rel=&quot;nofollow&quot;&gt;Here's a link to the Wikipedia article on Tesla&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1000" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T15:00:05.020" LastActivityDate="2011-03-10T02:48:16.940" Title="Tesla's theory of gravity" Tags="&lt;gravity&gt;&lt;theory&gt;&lt;history&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="5" />
  <row Id="2323" PostTypeId="2" ParentId="1453" CreationDate="2010-12-28T17:38:38.640" Score="5" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Is physical world ultimately describable by a renormalizable theory?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most physicists tend to assume that this &lt;em&gt;has&lt;/em&gt; to be so. The truth is that this assumption is just a convenience so we can add a cutoff that will keep the low-energy models safe from interacting with features of the high-energy underlying models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is nothing that will force this; in fact, is not out of the park to believe that most of the (27? 28? can't remember) &quot;free&quot; parameters in the standard model might be predicted by a non-perturbative nonrenormalizable high-energy model&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More importantly though, this very assumption is directly related to the fact that current physical theories have remained largely unfalsifiable in practical terms. So this assumption (may) eventually turn out to be a self-induced dead-end&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-01-20T16:22:27.640" LastActivityDate="2011-01-20T16:22:27.640" CommunityOwnedDate="2011-01-22T18:11:53.713" CommentCount="5" />
  <row Id="2324" PostTypeId="2" ParentId="2322" CreationDate="2010-12-28T17:50:33.410" Score="5" ViewCount="" Body="&lt;p&gt;It was said at the time (1930's) that only 10 people understood General Relativity. Tesla was clearly &lt;strong&gt;not&lt;/strong&gt; one of them! :-)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.tesla.hu/tesla/articles/19311100/page_1.gif&quot; rel=&quot;nofollow&quot;&gt;Faster than light&lt;/a&gt; article&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.tesla.hu/tesla/articles/19311100/index.htm&quot; rel=&quot;nofollow&quot;&gt;Bonkers article&lt;/a&gt; in which he still asserts that the aether exists &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, no, the theory has never been released - but if it was consistent with the various beliefs held by Tesla at the time, it must have been completely wrong.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-12-28T17:50:33.410" CommentCount="1" />
  <row Id="2325" PostTypeId="1" AcceptedAnswerId="2326" CreationDate="2010-12-28T19:25:42.333" Score="2" ViewCount="270" Body="&lt;p&gt;This is my first question on this site so forgive me for the awkward wording of the question. Basically, my question is why does light from, say, a sparkler, seem to remain where it just came from to you EYE? I attached a sample photograph and i understand that cameras work differently in that they're designed to &quot;retain&quot; light so to speak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why does the human eye have a similar, albeit less dramatic effect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ytwcr.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1002" LastActivityDate="2010-12-29T16:58:17.473" Title="Why can you see a light in the location that the light source just left?" Tags="&lt;light&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="2326" PostTypeId="2" ParentId="2325" CreationDate="2010-12-28T19:36:47.243" Score="3" ViewCount="" Body="&lt;p&gt;The human eye is actually quite similar to a camera. It also requires a finite exposure time to build up sufficient photon signal. If the source is moving over that time (typically about 0.1 sec for human eye), it will be blurred out over the exposure.&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2010-12-28T19:36:47.243" CommentCount="2" />
  <row Id="2328" PostTypeId="1" AcceptedAnswerId="2338" CreationDate="2010-12-28T21:59:37.997" Score="8" ViewCount="485" Body="&lt;p&gt;While looking at some exercises in my physics textbook, I came across the following problem which I thought was quite interesting:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;It is possible for the electron beam in a television picture tube to move across&#xA;the screen at a speed faster than the speed of light. Why does this not&#xA;contradict special relativity?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I suspect that it's because the television is in air, and light in air travels slower than light in a vacuum. So I suppose they're saying the the electron could travel faster in air than the speed of light in air, like what causes Cerenkov radiation?&lt;/p&gt;&#xA;" OwnerUserId="202" LastEditorUserId="202" LastEditorDisplayName="" LastEditDate="2010-12-28T22:05:55.803" LastActivityDate="2011-07-17T23:29:30.843" Title="Electrons faster than speed of light" Tags="&lt;special-relativity&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="2" />
  <row Id="2329" PostTypeId="2" ParentId="2328" CreationDate="2010-12-28T22:05:12.780" Score="8" ViewCount="" Body="&lt;p&gt;This is a conflation of phase velocity, and group velocity.&#xA;  The beam can be seen to move from say left to right at higher than c, but no information or particles are traveling that fast. Information is being transmitted from the electron gun to the phosphor at well under the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has nothing to do with the media it is embedded in. The information is going from the electron gun to the screen, not from one location on the screen to another.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2010-12-28T22:05:12.780" CommentCount="4" />
  <row Id="2330" PostTypeId="1" CreationDate="2010-12-28T22:29:28.427" Score="0" ViewCount="87" Body="&lt;p&gt;Recently a &quot;U&quot; shaped scratch appeared on my car's front tire. It looks like someone rolled past a curb and scraped it. This has led me to the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generate a plot (ideally polar) of a scratch pattern on a tire given tire radius and the ground height of a protrusion. Assume the protrusion is a single point and remains stationary while the tire rolls past. Also assume that the tire rolls straight past the protrusion, which scratches the side of the tire (i.e. hubcap, trim, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I wasn't so rusty on this material I would have taken a stab at it myself. So, any help in setting up and solving this problem would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1005" LastEditorUserId="1005" LastEditorDisplayName="" LastEditDate="2011-01-03T15:15:20.233" LastActivityDate="2011-01-03T15:15:20.233" Title="Kinematics of scratching a rolling tire" Tags="&lt;kinematics&gt;" AnswerCount="1" CommentCount="11" />
  <row Id="2331" PostTypeId="2" ParentId="2322" CreationDate="2010-12-28T23:36:13.633" Score="-6" ViewCount="" Body="&lt;p&gt;Tesla was (and according to some still is) a very interesting guy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much of his work is still considered &quot;black science&quot; meaning it is privately held by private stakeholders. It's not available to mainstream science - and may not be for some time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most interesting public work available on the subject has been performed by a husband and wife team Dr. Paulo N. Correa and Alexandra N. Correa who have been very careful to publish every detail of everything they have done to hopefully avoid the fate of others (Like Wilhelm Reich) who were researching in the same field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Correa's recreated Telsa and Reich's experiments and published all their results (much of which is publicly available) and have built operational prototypes to prove the theories on: Gravity and Anti-Gravity, Alternative Energy (from sources believed unknown to mainstream science), and Nano-Biology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information check out their site &lt;a href=&quot;http://www.aetherometry.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.aetherometry.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FYI - Anti-gravity engine designs have been available since the 1920's and the intellectual property can be found at the US Patent Office.&lt;/p&gt;&#xA;" OwnerUserId="1006" LastActivityDate="2010-12-28T23:36:13.633" CommentCount="4" />
  <row Id="2332" PostTypeId="1" AcceptedAnswerId="2333" CreationDate="2010-12-29T00:58:01.163" Score="4" ViewCount="297" Body="&lt;p&gt;Is there an equation I can use to calculate the temperature (as a function of time) of an object which is gaining or losing heat by convection? Or equivalently, the rate of energy transfer from the object to the surrounding fluid (or vice-versa)? It can involve constants representing properties of the object and the surrounding fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_equation&quot; rel=&quot;nofollow&quot;&gt;heat equation&lt;/a&gt; for conduction and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law&quot; rel=&quot;nofollow&quot;&gt;Stefan-Boltzmann law&lt;/a&gt; for radiation, but I can't think of one for convection.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-12-29T03:47:28.213" Title="Is there an equation for convective heat transfer?" Tags="&lt;thermodynamics&gt;&lt;temperature&gt;&lt;heat&gt;&lt;convection&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="2333" PostTypeId="2" ParentId="2332" CreationDate="2010-12-29T01:09:28.897" Score="2" ViewCount="" Body="&lt;p&gt;Nothing as easy from basic principles as for conduction or radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You &lt;em&gt;can&lt;/em&gt; multiple the mean heat carried by the convecting liquid by it's mass rate of flow&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta T = c_p * (T_{in} - T_{out}) * \frac{\Delta V}{\Delta t} \rho \Delta t .$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $T_{in}$ is the mean temperature of the liquid moving toward the sink (cool side) and $T_{out}$ is the temperature of the liquid moving the other way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that you can't get the temperature(s) of the convecting liquid or the flow rate without detailed calculations or measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the generaly case you pretty much have to go to CFD.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2010-12-29T03:47:28.213" LastActivityDate="2010-12-29T03:47:28.213" CommentCount="4" />
  <row Id="2334" PostTypeId="1" AcceptedAnswerId="2339" CreationDate="2010-12-29T02:26:26.090" Score="5" ViewCount="188" Body="&lt;p&gt;At the AGU, I presented a poster on &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010AGUFMIN11B1076H&quot; rel=&quot;nofollow&quot;&gt;vocabulary for discussing data systems&lt;/a&gt;, and someone left a note at my poster stating:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You have a bias here towards observational data.  Need to recognize that a lot of data comes from models and analyses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I completely agree; I didn't mention models and the values that come from them at all, and I had defined 'data' in such a way as it only covered observational data:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Values collected as part of a scientific investigation; may be qualified as ‘science data’. This includes uncalibrated values (raw data), derived values (calibrated data), and other transformations of the values (processed data).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;... but do scientists consider those resultant values to be 'data'?  I've talked to a few scientists over this last week (all in solar or space physics), and everyone of them including those who deal with modeling, were of the option that it was 'model output', but not 'data', (although one made the distinction between the philosphical concept of 'data' being multiple values, but said he didn't consider it to be 'science data' and commented that some of the earth science folks considered the values from their models to be data)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question -- is there a better term to use other than 'model output', and are there fields where 'model output' is considered to be 'data'?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;note&lt;/strong&gt; : there's also the issue of the definitions of raw vs. derived vs. processed data, as there's different definitions used by instrument operators, but just ignore that issue in the definition of 'data' for the time being.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;note 2&lt;/strong&gt; : I originally asked this on meta, and there were suggetions I move it to the main site; I've edited the text slightly due to a response I had yesterday from my surveying the scientists I work with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;note 3&lt;/strong&gt; : to set the context, the original intent was to identify clear or ambiguous terms across science disciplines, but the intended users were the data informatics community (those building data systems for the most part, not necessarily the scientists), but part of the issue is using language that won't offend or confuse the discipline scientists&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-12-29T18:11:41.663" Title="Are the results from models considered &quot;data&quot;?" Tags="&lt;soft-question&gt;" AnswerCount="6" />
  <row Id="2335" PostTypeId="2" ParentId="2334" CreationDate="2010-12-29T02:48:28.800" Score="2" ViewCount="" Body="&lt;p&gt;I'm an experimenter, so I stand pretty strongly with the idea that &quot;data&quot; comes from experiments. But life is rarely that easy and there are some complicating factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particle physics the main complicating factor comes in the form of &lt;strong&gt;lattice QCD&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It works like this: you evaluate the time-dependent, field-theoretical potentials on a lattice (and apply some clever magic due to Peter LePage and others to take care of the high frequency stuff that would otherwise make the calculation too big to handle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exactly what potential you evaluate is a matter of some choice as there are an number of free parameters in the theory. Of course, you also get to choose the situation to be simulated. Now, because this is a QCD calculation there are very few situations where you already know the answer. That's what make the technique useful. But it also means you can not anticipate the outcome of an calculation until you run the code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lattice QCD people sometimes speak of this as being &quot;an experiment&quot; and of the results as &quot;data&quot;. At least they do among themselves and when talking to more traditional theorists. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I accept this as a bit of a gray area in the &quot;data&quot; versus &quot;theory&quot; space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On one hand they don't know the results until they crunch the numbers and then they have to live with them. The results are generally repeatable, and sometimes takes some work to compare to &quot;pure&quot; theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other it is calculation and they can (and do) change the micro-physics assumptions that go into the models.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another gray area in my mind concerns (&quot;real&quot; experimental data) that have been transformed in a model to make them comparable to another data set. (This is separate from analysis, for instance in nuclear or particle physics cross-sections, form-factors and the like will sometimes be &quot;adjusted&quot; by factors known from models to allow them to be compared to results from other experiments done at different kinematics.) Data? Well, mostly, but dependent on the theory, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how much selecting and correcting are you willing to allow before a particular set of data has to be taken with a &quot;if they did all the subsequent calculations right&quot; grain of salt.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2010-12-29T02:57:05.537" LastActivityDate="2010-12-29T02:57:05.537" CommentCount="2" />
  <row Id="2336" PostTypeId="2" ParentId="2330" CreationDate="2010-12-29T03:59:35.707" Score="2" ViewCount="" Body="&lt;p&gt;Start with the reverse problem. What are the coordinates of a point fixed on the wheel. I am going to use the polar notation $r$ and $\varphi$ to locate a point on the wheel based on some reference axis. Supposed the center of the wheel is $(x_c,0)$ where $R$ is the radius of the wheel and $x_c$ is a function of time (linear). The rotation of the wheel is designated by the angle $\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The location of a point on the wheel is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x-x_{c}=r\,\sin(\theta+\varphi)$$&#xA;  $$y=r\,\cos(\theta+\varphi)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now the problem is while $x_{c}$ and $\theta$ are varying what is $r$ and $\varphi$ such that $x$ and $y$ are fixed in space. Note that $\theta\,=\,\frac{x_c}{R}$ for a rolling wheel to make everything a function of the independent variable $x_c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a little math you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r(x_{c})=\sqrt{x_{c}^{2}+y^{2}}$$&#xA;  $$\varphi(x_{c})=\arctan\left(\frac{x_{c}}{y}\right)-\frac{x_{c}}{R}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the assumption(initial condition) that $x=0$ when $x_c=0$. I plot the results for various $\frac{y}{R}$ ranging from $-1\ldots0$ and moving the center of the wheel from $x_c=-R\ldots+R$ and make some cool looking $\gamma$ looking curves. They look like the ribbons people hang to bring awareness to various illnesses. If the tire is profiled such that its width varies with radius, then only part of the shape will imprint leaving only the &lt;code&gt;U&lt;/code&gt; shaped part of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a graph of the shapes using &lt;code&gt;Asymtpote Vector Graphics&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/b2kI3.png&quot; alt=&quot;Curves Shapes&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-12-29T03:59:35.707" CommentCount="1" />
  <row Id="2337" PostTypeId="2" ParentId="2160" CreationDate="2010-12-29T06:11:08.683" Score="3" ViewCount="" Body="&lt;p&gt;SiO2 is very commonly used in optical coatings. A typical example is the production of &lt;a href=&quot;http://en.wikipedia.org/wiki/Dielectric_mirror&quot; rel=&quot;nofollow&quot;&gt;Dielectric Mirrors&lt;/a&gt; which works just like a silver or aluminum mirror, but can achieve very high reflectivity, upwards of 99.9%, and can be made to work only at very specific wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this application SiO2 (called silica, generally) is often used in combination with HaO2 (hafnium oxide, or hafnia). Alternating layers are deposited on a glass substrate in very specific thicknesses to produce the desired optical qualities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not really an expert on optical coatings, but I work closely with people who are, and I don't believe that the temperature needs to be quite as high as 1600C. For typical processes, temperatures are in the range of 900C for the silica source, but much lower for the substrate being coated. More recently, I've heard that we are using some sort of plasma assisted coating technique, which I gather can operate at much lower temperatures, 400C or below. This is because the energetic ions in the plasma knock silica off of the source without much heating, and apparently produces higher quality coatings as well. EDIT: As Georg pointed out, the material being removed from the coating source is not molecular silica, but some other combination of silicon/oxygen compounds. I assume that an SiO2 crystal structure re-forms on the coated optic, but I do not really know for sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Silicon_dioxide&quot; rel=&quot;nofollow&quot;&gt;Silicon Dioxide&lt;/a&gt; is quite thorough, and has good references which should give you a good starting point if you want to do more research. I also found this &lt;a href=&quot;http://www.svc.org/Publications/Books-of-Interest-to-Vacuum-Coaters.cfm&quot; rel=&quot;nofollow&quot;&gt;list of books&lt;/a&gt;, and this &lt;a href=&quot;http://books.google.com/books?id=Qi98H-iTgLEC&amp;amp;printsec=frontcover&amp;amp;hl=en#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; as well, although I cannot personally tell you if they are good books or not.&lt;/p&gt;&#xA;" OwnerUserId="869" LastEditorUserId="869" LastEditorDisplayName="" LastEditDate="2011-03-25T20:48:39.377" LastActivityDate="2011-03-25T20:48:39.377" CommentCount="7" />
  <row Id="2338" PostTypeId="2" ParentId="2328" CreationDate="2010-12-29T06:32:24.733" Score="11" ViewCount="" Body="&lt;p&gt;This is an example of what is sometimes called the &quot;Marquee Effect.&quot; Think of the light bulbs surrounding an old-fashioned movie theater marquee, where the light bulbs turn on in sequence to produce the illusion, from a distance, of a light source which is moving around the the marquee.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no limit on how short the time interval is between one light turning on and the next turning on, so the perceived light source position can move arbitrarily fast, but in fact nothing is actually moving at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the television screen, the phosphors on the screen can be lit in rapid sequence, but the electrons in the beam do not ever need to move at (or even near) the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, there are loads of examples of some imaginary or conceptual &quot;object&quot; moving faster than light, but in all these cases there is nothing actually moving at all. A classic example is the intersection point of two nearly parallel lines, which moves very rapidly as the angle between the lines changes. In this case it is obvious that the moving &quot;object&quot; isn't moving at all, but its still a good example of a case where you can discuss something moving faster than light without there being any violation of physical law.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2010-12-29T06:32:24.733" />
  <row Id="2339" PostTypeId="2" ParentId="2334" CreationDate="2010-12-29T06:39:04.930" Score="2" ViewCount="" Body="&lt;p&gt;Well, to be really pedantic, almost anything is data. Any measurement or observation is data. I suppose I'd define data as any piece of information, regardless of how it was obtained or whether it is valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, I think your question was really getting at what would be considered evidence, in the context of providing support or opposition to a theoretical model. I'd say that the answer is very situation dependent. Using simulation results as an example since it was in your question, I would certainty take simulation results in support of some theory, but only on the condition that the simulation has been very well validated against experiment. The details of how and why the simulation results are favorable to the theory is also certainly important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, all science must be validated by experiment or observation, but that can sometimes be a long journey, and there is definitely a place for other sorts of data along the way.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2010-12-29T06:39:04.930" CommentCount="1" />
  <row Id="2340" PostTypeId="2" ParentId="2334" CreationDate="2010-12-29T10:00:29.340" Score="1" ViewCount="" Body="&lt;p&gt;Data form modelling is also a data; the only thing is that it describes the model, not the real phenomenon. &lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-12-29T10:00:29.340" CommentCount="2" />
  <row Id="2341" PostTypeId="1" AcceptedAnswerId="2342" CreationDate="2010-12-29T12:40:24.887" Score="4" ViewCount="359" Body="&lt;p&gt;This is a question motivated by &lt;a href=&quot;http://en.wikipedia.org/wiki/Angry_Birds&quot; rel=&quot;nofollow&quot;&gt;Angry Birds&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When playing the game, I notice that if the initial velocity is constant, the way to land a bird furthest away from the launching point is by launching the bird at 45 degrees from the ground. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to obtain a derivation and a proof of this?&lt;/p&gt;&#xA;" OwnerUserId="605" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-12-31T05:04:24.003" LastActivityDate="2011-01-04T01:23:57.250" Title="The angle of launching an Angry Bird that will land the bird furthest from the launching point is 45 degrees?" Tags="&lt;newtonian-mechanics&gt;&lt;projectile&gt;" AnswerCount="2" />
  <row Id="2342" PostTypeId="2" ParentId="2341" CreationDate="2010-12-29T13:12:51.073" Score="11" ViewCount="" Body="&lt;p&gt;45 degrees is, in fact, the angle for maximum range for a projectile with no air resistance. In the absence of air resistance, the only force acting is gravity, which causes a constant acceleration of g downwards. this determines the amount of time the particle spends in the air, via the formula for the position of a particle with constant acceleration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y(t) = y(0) + v_y t + \frac{1}{2}a_y t^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting in the relevant parameters (start and end positions both 0, acceleration -g (negative because it's downward)) this becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 = v_y t - \frac{1}{2}g t^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which we solve to get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t = \frac{2v_y}{g}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time then goes into the equation for the horizontal position:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x(t) = x(0) + v_x t + \frac{1}{2}a_x t^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As there's no horizontal force acting, this reduces to just&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x(t) = v_x t = \frac{2v_x v_y}{g}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get this in terms of the angle, we use the fact from trigonometry that for a velocity $v$ at an angle $\theta$ from the horizontal,  the vertical velocity is $v \sin \theta$ and the horizontal velocity is $v \cos \theta$, giving us:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = \frac{2v \cos \theta v\sin \theta}{g} =\frac{v^2}{g}\sin 2\theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has its maximum value for $\theta = \frac{\pi}{4}$, namely, 45 degrees from the horizontal.&lt;/p&gt;&#xA;" OwnerUserId="166" LastEditorUserId="166" LastEditorDisplayName="" LastEditDate="2010-12-29T15:25:17.423" LastActivityDate="2010-12-29T15:25:17.423" CommentCount="12" />
  <row Id="2343" PostTypeId="1" CreationDate="2010-12-29T13:48:15.437" Score="2" ViewCount="716" Body="&lt;p&gt;I know a lot about Theoretical Physicists from &lt;a href=&quot;http://en.wikipedia.org/wiki/Sheldon_Cooper&quot; rel=&quot;nofollow&quot;&gt;Sheldon Cooper&lt;/a&gt; of &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Big_Bang_Theory&quot; rel=&quot;nofollow&quot;&gt;Big Bang Theory&lt;/a&gt;. I know about parallel universes (and I wonder if one codingtales in one of the parallel universes believes if parallel universes do not exist).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry, but this is just for fun.&#xA;But I've real questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Are all the equations that appear on the board and Sheldon tries to solve true?&lt;/li&gt;&#xA;&lt;li&gt;How is a Rocket Scientist different from a Theoretical Physicist?&lt;/li&gt;&#xA;&lt;li&gt;I figure out some errors in the definition of a &quot;Screw&quot; that Sheldon gives. He says he is on an inclined plane wrapped helically around its axis. As far as I know there are no inclined planes with axes. Can you please explain it to me?&lt;/li&gt;&#xA;&lt;li&gt;Any other suggestion that would help me understand Big Bang Theory better would be appreciated.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(And please forgive the tags. I didn't find &quot;Theoretical-Physics&quot; tag. I have to bluntly use some other alternative.)&lt;/p&gt;&#xA;" OwnerUserId="1011" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2010-12-29T20:49:25.067" LastActivityDate="2010-12-29T20:49:25.067" CommunityOwnedDate="2010-12-29T15:49:29.520" ClosedDate="2010-12-29T21:17:02.903" Title="Are all theoretical physicists like Sheldon Cooper?" Tags="&lt;soft-question&gt;" AnswerCount="5" CommentCount="7" />
  <row Id="2344" PostTypeId="2" ParentId="2343" CreationDate="2010-12-29T14:08:42.640" Score="4" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;So far, from the few episodes of Big Bang Theory I've seen, all equations on the boards correspond to real physics problems. Kudos to the makers of the TV series for actually presenting real equations and not some mumbo jumbo. The flip side is that very often, the equations are undergraduate or graduate level, but not really worthy of someone doing PhD research. I don't think Leonard or Sheldon would still be busy calculating the eigenvalues of the hydrogen atom.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The rocket scientist is doing applied science and is thus mostly an engineer. The theoretical scientist is focusing on fundamental science and is often doing so by using mathematics and conceptual thinking rather than experiments. Both do use computer simulations nowadays, and even rocket scientists go back to the drawing board and do maths, but their math are not aimed at understanding the fundamental workings of nature but rather at how to build this or that device.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can't help you here, I don't really get Sheldon's point. Maybe he's talking about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Osculating_plane&quot; rel=&quot;nofollow&quot;&gt;osculating plane&lt;/a&gt;? There are some nice pictures on the page for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Frenet-Serret_formulas&quot; rel=&quot;nofollow&quot;&gt;Frenet-Serret formulas&lt;/a&gt; for the case of a helix.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do you mean understanding &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Big_Bang_Theory&quot; rel=&quot;nofollow&quot;&gt;the TV series&lt;/a&gt; or understanding &lt;a href=&quot;http://en.wikipedia.org/wiki/Big_bang_theory&quot; rel=&quot;nofollow&quot;&gt;the cosmological model&lt;/a&gt;? ;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2010-12-29T14:14:09.280" LastActivityDate="2010-12-29T14:14:09.280" CommunityOwnedDate="2010-12-29T15:49:29.520" CommentCount="2" />
  <row Id="2345" PostTypeId="2" ParentId="2334" CreationDate="2010-12-29T14:59:44.393" Score="1" ViewCount="" Body="&lt;p&gt;This is a politically loaded question, which is probably why you got the response you did at the AGU, where there are people who work on climate change issues. There is a concerted effort in some quarters to cast doubt on the science showing that the Earth is warming, and predicting the effects of climate change. As these predictions rely heavily on simulations, one part of this effort is an attempt to disparage models and simulations as having lesser status, scientifically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the actual status of models and simulations, that varies from (sub)field to (sub)field, more or less in accordance with how difficult it is to interpret experimental or observational data. My own field of experimental Atomic, Molecular, and Optical physics has a fairly clear divide between experiment and theory, largely because the experiments we do are relatively unambiguous: an atom either absorbs light or doesn't, or it's either in this position or that one. We do need simulations to compare to some experiments, but there's never much question that those are theory, and not part of the experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you get to nuclear and particle physics, where the detectors are the size of office buildings, the line gets a little fuzzier. The systems they use to detect and identify the products of collisions between particles are so complicated that it's impossible to interpret what happens without a significant amount of simulation. As a result, experimental nuclear and particle physicists spend a great deal of time generating and analyzing simulated data, in order to account for issues of detector efficiency and so on. I don't think they would call these results &quot;data&quot; per se, but computational simulation is an absolutely essential part of experimental physics in those fields, and those simulations are accorded more status than they would be in AMO physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things get even more complicated when you get to parts of physics that are fundamentally observational rather than experimental. If you're a particle physicist, you can repeat your experiments millions or billions of times, and build up a very good statistical understanding of what happens. If you're an astrophysicist or a geophysicist, you only get one data run-- we have only one observable universe, and only one Earth within it to study. You can't rewind the history of the observable universe and try it again with slightly different input parameters. Unless you do it in a simulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My outsider's understanding of those fields is that simulation and modeling is accorded a much higher status than in my corner of physics, just out of necessity. If you want to use a physical model to explain some geological or astrophysical phenomenon, the only way you can really do it is by running a whole lot of simulations, and showing that the single reality we observe is a plausible result of your models. Correctly interpreting and establishing correspondences between simulations and observation is a subtle and complicated business, and constitutes a huge proportion of the work in those communities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know that many astrophysicists, and even fewer geophysicists, so I don't know the terminology they use. My impression of the astrophysics talks I've seen is that they wouldn't put such simulation results on the same level as observational data or experiments, but then my sample isn't remotely representative. It may well be that there are fields in which model results are deemed &quot;data&quot; in the local jargon.&lt;/p&gt;&#xA;" OwnerUserId="166" LastEditorUserId="166" LastEditorDisplayName="" LastEditDate="2010-12-29T15:09:51.750" LastActivityDate="2010-12-29T15:09:51.750" CommentCount="3" />
  <row Id="2346" PostTypeId="2" ParentId="2343" CreationDate="2010-12-29T15:13:14.183" Score="7" ViewCount="" Body="&lt;p&gt;Well, speaking from my own experience, we are pretty much the same as Hollywood portrays us, but even better looking.&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2010-12-29T15:13:14.183" CommunityOwnedDate="2010-12-29T15:49:29.520" CommentCount="2" />
  <row Id="2347" PostTypeId="1" AcceptedAnswerId="2351" CreationDate="2010-12-29T15:18:16.177" Score="4" ViewCount="814" Body="&lt;p&gt;Inspired by a similar post, I wanted to know what video games, past or present, most accurately employ the laws of physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this is an appropriate question for this site, since the users are far more qualified to answer it than anyone on typical gaming sites, who think they know a lot more about physics than they really do. In my limited experience, most games claiming &quot;realistic physics&quot; are anything but.&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2011-04-11T13:41:09.250" CommunityOwnedDate="2010-12-29T15:51:27.737" Title="Best (most realistic) physics game?" Tags="&lt;simulation&gt;&lt;laws-of-physics&gt;" AnswerCount="12" CommentCount="3" FavoriteCount="1" />
  <row Id="2348" PostTypeId="2" ParentId="2347" CreationDate="2010-12-29T15:20:24.663" Score="3" ViewCount="" Body="&lt;p&gt;I was never able to do rigorous quantitative analysis, but in my memory, &lt;a href=&quot;http://en.wikipedia.org/wiki/Scorched_Earth_%28video_game%29&quot; rel=&quot;nofollow&quot;&gt;Scorched Earth&lt;/a&gt; always seemed very realistic, including proper trajectories, air resistance, and wind.&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2010-12-29T15:20:24.663" CommunityOwnedDate="2010-12-29T15:51:27.737" CommentCount="2" />
  <row Id="2349" PostTypeId="2" ParentId="2343" CreationDate="2010-12-29T15:23:51.900" Score="6" ViewCount="" Body="&lt;p&gt;The equations on the whiteboard are, in fact, drawn from real physics. They have a scientific consultant, David Saltzburg of UCLA, who gave a &lt;a href=&quot;http://meetings.aps.org/Meeting/MAR10/Event/120091&quot; rel=&quot;nofollow&quot;&gt;talk at the March Meeting last year&lt;/a&gt; about his work with the show. Apparently, he faxes them pages of relevant equations that they copy over onto the whiteboard. He also has &lt;a href=&quot;http://thebigblogtheory.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;a blog&lt;/a&gt; where he talks about his work with the show, and the science topics that turn up in each episode.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-12-29T15:23:51.900" CommunityOwnedDate="2010-12-29T15:49:29.520" CommentCount="1" />
  <row Id="2350" PostTypeId="2" ParentId="2347" CreationDate="2010-12-29T15:24:27.407" Score="2" ViewCount="" Body="&lt;p&gt;The original &lt;a href=&quot;http://en.wikipedia.org/wiki/Asteroids_%28video_game%29&quot; rel=&quot;nofollow&quot;&gt;Asteroids&lt;/a&gt; arcade game has a very good model of rocket propelled flight in vacuum micro-gravity (Well, it neglects the matter of the fuel mass-fraction. So assume an unreasonable efficient propulsion system...). Of course, the space metric is toroidal and the collisions model is rather primitive... &lt;strong&gt;Edit:&lt;/strong&gt; Wikipedia tells me that the ship in Asteroids would coast to a halt. I don't recall that, but it seems to kill this assertion. ::sigh::&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lunar_Lander_%28arcade_game%29&quot; rel=&quot;nofollow&quot;&gt;Lunar Lander&lt;/a&gt; game many of us recall from the 8-bit era handled vacuum maneuvering in a non-trivial gravitational environment. Again the collisions model was a bit simpleminded...&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2010-12-29T16:21:32.497" LastActivityDate="2010-12-29T16:21:32.497" CommunityOwnedDate="2010-12-29T15:51:27.737" CommentCount="1" />
  <row Id="2351" PostTypeId="2" ParentId="2347" CreationDate="2010-12-29T15:58:54.053" Score="2" ViewCount="" Body="&lt;p&gt;James McNeill is building an almost-true-physics lunar lander clone (with full rigid body motion, like compressing shock absorbers for example). The game is still work in progress but playable (&lt;a href=&quot;http://sites.google.com/site/mcneja/rocket&quot; rel=&quot;nofollow&quot;&gt;see here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very interesting blog &lt;a href=&quot;http://playtechs.blogspot.com/search/label/dev-diary&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-12-29T15:58:54.053" CommunityOwnedDate="2010-12-29T15:58:54.053" />
  <row Id="2352" PostTypeId="2" ParentId="2347" CreationDate="2010-12-29T16:26:13.827" Score="2" ViewCount="" Body="&lt;p&gt;Most first-person shooter today actually have a decent gravity model. Angry Birds is fairly realistic, for a phone game as well. &lt;/p&gt;&#xA;" OwnerUserId="871" LastActivityDate="2010-12-29T16:26:13.827" CommunityOwnedDate="2010-12-29T16:26:13.827" />
  <row Id="2353" PostTypeId="2" ParentId="2325" CreationDate="2010-12-29T16:45:45.907" Score="1" ViewCount="" Body="&lt;p&gt;The human retina and visual perception system of the brain display in combination what is known as &quot;persistence of vision.&quot; The eye and brain retain a visual image for somewhere between 1/30th and 1/25th of a second, or even longer for especially bright images. See: &lt;a href=&quot;http://en.wikipedia.org/wiki/Persistence_of_vision&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia entry on &lt;em&gt;Persistence of Vision&lt;/em&gt;&lt;/a&gt; and &lt;a href=&quot;http://www.cvc.sunysb.edu/334/perception/perception.htm#pov&quot; rel=&quot;nofollow&quot;&gt;the page on &lt;em&gt;The Nature of Visual Perception&lt;/em&gt; by the Center for Visual Computer at Stonybrook&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Persistence of vision is not a myth&lt;/strong&gt;. If you click on the second link above, you will see an image with 9 yellow circles and one small black dot which demonstrates that persistence of vision is real by showing an afterimage you will be able to detect for yourself. Persistence of vision is directly related to the perception of an afterimage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some attribute persistence of vision only to the retina, and some attribute it to the visual processing system in the brain, specifically to what is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Iconic_memory&quot; rel=&quot;nofollow&quot;&gt;iconic memory&lt;/a&gt;. But no one knows for certain what causes this phenomena at this time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is indeed a myth called the &quot;myth of persistence of vision&quot; which &lt;em&gt;relates to the reasons for the perception of movement&lt;/em&gt; specifically and shows that persistence of vision does not account for the perception of movement in film but that movement is accounted for in film by some combination of the phi phenomenon and beta movement. So the myth is that the very real phenomenon of persistence of vision accounts for movement perception in film which generally only displays 24 frames per second. The myth is not that the idea of perception of vision is itself a myth. See:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://web.archive.org/web/20080526105906/http://www.uca.edu/org/ccsmi/ccsmi/classicwork/Myth+Revisited.htm&quot; rel=&quot;nofollow&quot;&gt;THE MYTH OF PERSISTENCE OF VISION REVISITED By Joseph and Barbara Anderson&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: Even the wikipedia entry says that: &quot;Some scientists nowadays consider the entire theory a myth&quot; which is confusing because if one follows the reference for this statement, one reaches the link above from Joseph and Barbara Anderson which clearly is talking about movement perception only and not the general idea of perception of vision.&lt;/p&gt;&#xA;" OwnerUserId="948" LastEditorUserId="948" LastEditorDisplayName="" LastEditDate="2010-12-29T16:58:17.473" LastActivityDate="2010-12-29T16:58:17.473" />
  <row Id="2354" PostTypeId="2" ParentId="2343" CreationDate="2010-12-29T16:52:23.670" Score="0" ViewCount="" Body="&lt;p&gt;I think it's fair to say there's some similarities with Sheldon, but not including his personality... The ones I know are actually pleasant to be around, although perhaps somewhat into conversations that most people just don't understand...&lt;/p&gt;&#xA;" OwnerUserId="871" LastActivityDate="2010-12-29T16:52:23.670" CommunityOwnedDate="2010-12-29T16:52:23.670" />
  <row Id="2355" PostTypeId="1" AcceptedAnswerId="2357" CreationDate="2010-12-29T17:13:10.610" Score="1" ViewCount="292" Body="&lt;p&gt;Does time stretch all the way back for infinity or was there a point when time appears to start in the universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember reading long ago somewhere that according to one theory time began shortly before the creation of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does time have a starting point of note?&lt;/p&gt;&#xA;" OwnerUserId="771" LastActivityDate="2010-12-29T18:01:11.527" Title="Did time exist before the creation of matter in the universe?" Tags="&lt;time&gt;&lt;big-bang&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="2356" PostTypeId="2" ParentId="2334" CreationDate="2010-12-29T17:41:38.427" Score="1" ViewCount="" Body="&lt;p&gt;A great quote I heard at the American Chemical Society conference last year went something like, &quot;All experimentalists trust each other's results, but not their own; All simulators distrust each other's results, but not their own.&quot;  The reason that happens is that experimentalists rigorously describe their methods, use stock reagents, and even publish the model numbers of their instruments.  It's easy, at least in principle, to replicate their work, so many people trust that they don't have to.  Simulators don't do that, partly because there aren't standardized ways to do it(yet), partly because the community hasn't demanded it(yet), and partly because they don't want to give away their software.  It is difficult, even in principle, to replicate the results produced by a simulator, thus many feel that they can't trust the results.  That's changing -- perhaps when it does more people will regard the results of simulations as &quot;data&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For my part, I'm in computational biophysics, doing simulations of protein folding, and I do regard results produced from my simulations (&quot;&lt;em&gt;in silico&lt;/em&gt;&quot;) as data, and the simulations themselves as experiments.  I try to maintain the view that the data and experiments are &lt;em&gt;about the model&lt;/em&gt;, however, especially when talking to experimentalists, and even more so when talking to people with a medical or biological background (they tend to be somewhat hostile to simulation, in my experience).  If I make claims about the relationship between the model, the data from the model, and the results one would obtain by doing an analogous experiment &lt;em&gt;in vitro&lt;/em&gt; I have to be very careful about validation of the model in as many independent ways as possible, and in careful vetting of the methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, my measurement of secondary structure content of an ensemble of protein structures is not fundamentally different from that of an experimentalist, except that I have a lot more information available and have to throw some of it away to produce an average measurement that is commensurable to the one obtained by the experimentalist using circular dichroism spectroscopy (and some interpretation based on a theoretical model). I'm still not going to claim that my measurement is the correct result for the protein of interest unless I'm very certain that the model and the free parameters used in the relevant simulation were the correct ones to use for the situation, but I would give the experimenter the benefit of the doubt that their results are probably the correct ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, if I was to be &lt;em&gt;very&lt;/em&gt; careful about the word &quot;data&quot;, I would say that the only data produced by my simulations are the coordinates of the atoms themselves.  Everything else is a derived measurement using those coordinates.  Why?  Because I could recreate all the measurements from the coordinates, but I couldn't recreate the coordinates from the measurements.&lt;/p&gt;&#xA;" OwnerUserId="1013" LastActivityDate="2010-12-29T17:41:38.427" CommentCount="1" />
  <row Id="2357" PostTypeId="2" ParentId="2355" CreationDate="2010-12-29T18:01:11.527" Score="2" ViewCount="" Body="&lt;p&gt;In short, we don't know. There are a few indications that time started at the big bang, or at least it had some form of discontinuity. This might be wrong though.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;According to General Relativity, there is no such thing as an absolute time. Time is always relative to an observer, without the universe there would be no corresponding concept of time. All observers within the universe would have their clocks &quot;slowed down&quot; the nearer they are to the big bang (nearer in time). At the big bang point, their clock would stop. This said, we know that GR doesn't apply as-is all the way to the Big Bang.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Some cosmological theories like CCC predict a series of aeons and some form of cyclic universe. These predict a discontinuity (CCC predicts a conformal scale change) of time at the big bang, and at the end of the universe.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As a side note: people tend to have a special fascination with time. For all we know though, time is only relatively special. From a cosmological point of view the discussion is whether space-time existed. We are pretty sure that it was very very small at some point.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2010-12-29T18:01:11.527" CommentCount="3" />
  <row Id="2358" PostTypeId="2" ParentId="2334" CreationDate="2010-12-29T18:11:41.663" Score="2" ViewCount="" Body="&lt;p&gt;Cross posted at Chad's blog:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a climate scientist. Complex numerical modeling of climate system is a giant neighborhood in the climate science community. There are a great many scientists who work almost entirely within the world of numerical climate models. Many of them refer to the output of climate models as &quot;data&quot;. This has been exacerbated by the release of huge archives of climate model output on public websites (the CMIP archive hosted by LLNL) which allows anyone to download and analyze the output from more than 20 climate models. This is accomplished in much the same fashion that large observational groups, such as NASA, share publicly-funded observational data. The CMIP archive has been a great development for opening up the world of climate modeling to any interested party (and to valuable scrutiny by any interested party). But it has, in my opinion, blurred the line between data and simulation output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My opinion is that the outputs of numerical models should not be referred to as data. Our technical language, particularly in the earth sciences, should distinguish between observations of the one true realization of the earth's climate and the representations of possible (but not necessarily likely or even plausible) climate states that the models spit out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of this should be taken as demoting the &quot;status of simulations&quot;. I work with models and observations. They both have their place in climate science.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-29T18:11:41.663" />
  <row Id="2359" PostTypeId="2" ParentId="2343" CreationDate="2010-12-29T18:48:56.827" Score="0" ViewCount="" Body="&lt;p&gt;I have seen every one of the traits displayed by the characters on the show, but spread over far more people. I'd say the most common thing that we do is to use physics in places where it doesn't belong, but who doesn't think about the world in terms of things they know; we just sound way dorkier doing it.&lt;/p&gt;&#xA;" OwnerUserId="411" LastActivityDate="2010-12-29T18:48:56.827" CommunityOwnedDate="2010-12-29T18:48:56.827" />
  <row Id="2360" PostTypeId="2" ParentId="2347" CreationDate="2010-12-29T20:16:28.533" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;POOL&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of pool(/snooker/billiards) video games are pretty realistic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because implementation of physics engines of such games is relatively easy task. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=uwfcHrTRLnw&quot; rel=&quot;nofollow&quot;&gt;Here is one example.&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other games in which some physics is real are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.blitwise.com/ptanks.html&quot; rel=&quot;nofollow&quot;&gt;Pocket Tanks&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="126" LastActivityDate="2010-12-29T20:16:28.533" CommunityOwnedDate="2010-12-29T20:16:28.533" />
  <row Id="2361" PostTypeId="1" AcceptedAnswerId="2367" CreationDate="2010-12-29T20:19:08.047" Score="9" ViewCount="386" Body="&lt;p&gt;I recently became really interested in learning about physics and cosmology, but I still know very little. Hopefully someone with more knowledge will be able to shed some light on my questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are my presumptions (please correct me if I'm mistaken):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$t_{0}$: Big Bang.&lt;/li&gt;&#xA;&lt;li&gt;The Universe expands (and thus cools) sufficiently to allow the formation of atoms.&lt;/li&gt;&#xA;&lt;li&gt;The early Universe mostly consists of Hydrogen and is not uniform in its geometry.&lt;/li&gt;&#xA;&lt;li&gt;Gravity pushes lumps of matter together until their density (temperature) is so high that nuclear fusion kicks off. The first giant stars are born.&lt;/li&gt;&#xA;&lt;li&gt;Some heavy elements are created in these stars and when they explode, sufficient temperatures are reached to form heavier elements still.&lt;/li&gt;&#xA;&lt;li&gt;All of the successive stars and planets are the product of these original stars.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If the amount of Hydrogen is finite in the Universe and each successive generation of stars use up most of their Hydrogen, what is the theoretical maximum number of generations of stars that our Universe can support? A ballpark figure in terms of years is fine too. ;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are there some special physical processes that occur in the Universe on a large enough scale that are able to break up heavy elements (i.e., Helium) into lighter elements (i.e., Hydrogen)? Kind of like a cosmic recycling operation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If we discount the possibility of a Big Crunch, is the Universe pretty much headed into a time when there will be no more stars and only a bunch of black holes, planets, and other debris?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1018" LastEditorUserId="1018" LastEditorDisplayName="" LastEditDate="2010-12-29T23:46:46.337" LastActivityDate="2010-12-30T16:05:00.017" Title="How long will the Universe's hydrogen reserves last for?" Tags="&lt;cosmology&gt;&lt;universe&gt;&lt;big-bang&gt;" AnswerCount="3" CommentCount="14" FavoriteCount="2" />
  <row Id="2362" PostTypeId="2" ParentId="2322" CreationDate="2010-12-29T20:49:45.807" Score="0" ViewCount="" Body="&lt;p&gt;Everything else aside - this question was about where to find Tesla's theory of gravity, and my answer pointed to those that have tried to reconstruct it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's interesting that so far that answer has generated a -6, and the answer saying Telsa doesn't know what he's talking about, so even if the theory was available it must be completely wrong has a +4... on a forum supposedly frequented by people of science. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;warning soap box detected&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Couldn't &quot;science&quot; be considered the systematic observation of natural events and conditions in order to discover facts about them and to formulate laws and principles based on these facts. If that's right, Telsa, Reich and the Correa's are practising good science - even if their theories are non-conventional, controversial or even completely off track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physics is concerned with the energy/matter and motion. In that light, Tesla, Reich and the Correa's are absolutely researching physics. The fact they all had a personal disposition towards philosophy as well, doesn't detract from that, even if it colours their extrapolations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's understandable to want to believe what we &quot;know&quot; is &quot;right,&quot; especially considering the time and money we invest into learning. Therefore when something challenges that knowledge, it's equally understandable we become defensive and dismissive. I hope we can agree that &quot;defensive and dismissive&quot; without a reasoned basis, are certainly not attributes of someone practising good science. Like Feynman said &quot;science is a culture of doubt.&quot; - I daresay this is the reason for the down vote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact remains that all these people have results that are testable, repeatable and have led to the manufacture of operational prototypes that conventional &quot;science&quot; insists are impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to see truly open discussion of alternative, but well practised physics, rather than dogmatic dismissal of things that don't immediately fit our understandings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a physics forum, my friends. Let's try to be objective and not colour our answers with personal baggage... like I've just done here :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;falls off soapbox&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1006" LastEditorUserId="1006" LastEditorDisplayName="" LastEditDate="2010-12-29T20:56:43.163" LastActivityDate="2010-12-29T20:56:43.163" CommentCount="8" />
  <row Id="2363" PostTypeId="1" AcceptedAnswerId="2384" CreationDate="2010-12-29T23:08:15.890" Score="4" ViewCount="171" Body="&lt;p&gt;What is the dominant form of heat transfer between warm water and cold air?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a $100 mg$ drop of water falls through $-40 C$ air, how quickly could it freeze?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it credible that in very cold weather spit freezes in the half a second it takes to reach the ground?  &lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2010-12-29T23:24:01.467" LastActivityDate="2010-12-30T15:59:02.967" Title="How cold does it need to be for spit to freeze before hitting the ground?" Tags="&lt;thermodynamics&gt;&lt;water&gt;&lt;order-of-magnitude&gt;&lt;phase-transition&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="2364" PostTypeId="1" CreationDate="2010-12-29T23:42:43.113" Score="5" ViewCount="410" Body="&lt;p&gt;In the book &quot;Vectors, Tensors, and the Basic Equations of Fluid Mechanics&quot; by Rutherford Aris I read the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the fluid is such that the torques&#xA;  within it arise only as the moments of&#xA;  direct forces we shall call it&#xA;  nonpolar. A polar fluid is one that is&#xA;  capable of transmitting stress couples&#xA;  and being subjected to body torques,&#xA;  as in polyatomic and certain&#xA;  non-Newtonian fluids.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can someone help me understand this? Say by giving me another definition of polar and nonpolar fluids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-30T10:52:56.570" LastActivityDate="2011-05-23T03:09:39.160" Title="Polar vs non-polar fluid" Tags="&lt;fluid-dynamics&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="2365" PostTypeId="2" ParentId="2364" CreationDate="2010-12-30T00:32:23.067" Score="1" ViewCount="" Body="&lt;p&gt;A polar fluid is just a fluid where the constituent molecules have a polarization -- it could be a fluid of molecules that have a magnetic spin moment, or something like H2O where each individual molecule has a nonzero electric dipole -- and at the macroscopic level, as you average over all of the microscopic moments, you get a net polarization for the whole fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm no expert on fluid mechanics, but I imagine the polarization somehow couples to the stress tensor in a way that generates torques in whatever equations of motion the author is interested in.&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2010-12-30T00:32:23.067" CommentCount="7" />
  <row Id="2366" PostTypeId="2" ParentId="2347" CreationDate="2010-12-30T01:25:47.790" Score="1" ViewCount="" Body="&lt;p&gt;Angry Birds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a very nice game with realistic collision physics simulation. &lt;/p&gt;&#xA;" OwnerUserId="605" LastActivityDate="2010-12-30T01:25:47.790" CommunityOwnedDate="2010-12-30T01:25:47.790" CommentCount="2" />
  <row Id="2367" PostTypeId="2" ParentId="2361" CreationDate="2010-12-30T03:48:26.257" Score="1" ViewCount="" Body="&lt;p&gt;Tis, a good question. Two related questions arise from it. The first one is, will the hydrogen all be used up in finite time? The second related one is, will star formation completely stop in finite time? They sound related, but the first result doesn't necessarily imply the same result for the second, or vice versa. I.E. a low but nonzero gas density might possibly not allow further star formation, and maybe we could have no hydrogen, but have other types of gas (or even solid objects) still collect into stellar mass objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know for sure the answers. The rate of star formation (and hydrogen consumption) could decline slowly enough as to never formally reach zero. Or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do know that a lot of gas gets blown out of galaxies by massive stars, supernova, and black hole activity, and becomes intergalactic gas -usually staying within the galaxy cluster. On a long time scale this should eventually fall back into the cluster's galaxies. So I would think the star formation rate would have a very long tail.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-12-30T03:48:26.257" CommentCount="3" />
  <row Id="2368" PostTypeId="1" AcceptedAnswerId="2369" CreationDate="2010-12-30T04:25:44.930" Score="6" ViewCount="199" Body="&lt;p&gt;I recently found out that a PET scan stands for a &lt;a href=&quot;http://en.wikipedia.org/wiki/Positron_emission_tomography&quot; rel=&quot;nofollow&quot;&gt;positron emission tomography&lt;/a&gt;. Are there any other practical uses of antimatter in the present?&lt;/p&gt;&#xA;" OwnerUserId="246" LastActivityDate="2010-12-30T15:31:30.233" CommunityOwnedDate="2011-04-08T20:18:24.827" Title="Practical uses of antimatter in the present" Tags="&lt;applied-physics&gt;&lt;antimatter&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="2369" PostTypeId="2" ParentId="2368" CreationDate="2010-12-30T05:08:56.790" Score="4" ViewCount="" Body="&lt;p&gt;I don't know if the other antiparticles have been exploited, but positrons are used in metallurgical studies. &lt;a href=&quot;http://books.google.com/books?id=-uj80fqU8LEC&amp;amp;pg=PA1633&quot; rel=&quot;nofollow&quot;&gt;Positron annihilation spectroscopy&lt;/a&gt; is one particular application.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-12-30T05:08:56.790" CommunityOwnedDate="2011-04-08T20:18:24.827" />
  <row Id="2370" PostTypeId="2" ParentId="2361" CreationDate="2010-12-30T05:39:15.533" Score="1" ViewCount="" Body="&lt;p&gt;This is really a comment but is too long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simpler, but related question is &quot;How much of the primordial hydrogen has been used up so far?&quot;   or &quot;Can we measure a difference between the percentage of primordial hydrogen (~75% H-1) or the percentage of primordial Helium (~25% He-4) and the percentages today.  Short answer - no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, how about a gradient of helium-4 from the galactic center to the outer arms.   I think the answer is yes but it's hard for me to tell.    Well, can we see a metallicity gradient (all elements heavier than He)?  Yes, we can.  See &lt;a href=&quot;http://nedwww.ipac.caltech.edu/level5/Sept04/Henry/Henry2_1.html&quot; rel=&quot;nofollow&quot;&gt;http://nedwww.ipac.caltech.edu/level5/Sept04/Henry/Henry2_1.html&lt;/a&gt;, fig 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conclusion, you have asked an experimental question in astronomy, and I would like to see astronomical evidence in any answer given.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-12-30T05:39:15.533" />
  <row Id="2371" PostTypeId="2" ParentId="2292" CreationDate="2010-12-30T05:41:20.197" Score="1" ViewCount="" Body="&lt;p&gt;There are quantities called the &lt;em&gt;compressibility factor&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z=\frac{P\bar{V}}{RT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed069p218&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;fugacity&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f=P\int_{P^*}^P \frac{Z-1}{P}\mathrm dP$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that can be considered as measures of departure from ideality. For an ideal gas, $Z=1$ and $f=P$. The fugacity is intimately related to the &quot;chemical potential&quot; (molar Gibbs free energy) of a nonideal gas; books such as those by &lt;a href=&quot;http://books.google.com/books?id=f-bje0-DEYUC&quot; rel=&quot;nofollow&quot;&gt;McQuarrie&lt;/a&gt; and &lt;a href=&quot;http://books.google.com/books?id=HQtRAAAAMAAJ&quot; rel=&quot;nofollow&quot;&gt;Hirschfelder-Curtiss-Bird&lt;/a&gt; should have a more complete discussion on these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll just make the additional note that though the van der Waals equation of state (EOS) is standard, there is a more modern EOS that is not more complicated than van der Waals's and usually gives more accurate property predictions (considering its relatively simple form), the &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/cr60137a013&quot; rel=&quot;nofollow&quot;&gt;Redlich-Kwong equation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P=\frac{RT}{\bar{V}-b}-\frac{a}{\sqrt{T}\bar{V}(\bar{V}+b)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expression behaves surprisingly well in phase equilibria studies, but certainly not as good as more elaborate multiparameter EOSs. Its advantage is that the EOS is easily manipulable. As can ascertained from the original article, the proposers of this EOS were associated with Shell Petroleum, and they have used it for the phase equilibria of petroleum components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, there has been much research on &quot;cubic equations of state&quot; since then; see &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ie020447b&quot; rel=&quot;nofollow&quot;&gt;this survey article&lt;/a&gt; for instance.&lt;/p&gt;&#xA;" OwnerUserId="172" LastActivityDate="2010-12-30T05:41:20.197" CommentCount="1" />
  <row Id="2372" PostTypeId="1" AcceptedAnswerId="2401" CreationDate="2010-12-30T06:20:00.310" Score="7" ViewCount="371" Body="&lt;p&gt;Question sounds quite catchy but real question goes like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's suppose there is a straight rigid bar with height $h$ and center of mass at the middle of height $h/2$. Now if the bar is vertically upright from ground, how long will it take to fall on the ground and what is the equation of motion of the center of mass (Lagrangian)?&lt;/p&gt;&#xA;" OwnerUserId="567" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-30T10:46:01.627" LastActivityDate="2011-02-05T10:03:44.043" Title="How long it will take for a tree to fall on a ground?" Tags="&lt;classical-mechanics&gt;&lt;mechanics&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="1" />
  <row Id="2374" PostTypeId="2" ParentId="2364" CreationDate="2010-12-30T11:10:22.350" Score="1" ViewCount="" Body="&lt;p&gt;It is about the stress tensor; it is almost always assumed that it is symmetric to satisfy angular momentum conservation. Yet, there are some fluids capable of creating rotation from squeezing (like those spintops with pistons) and thus have some antisymmetric part in their stress tensors.&lt;br&gt;&#xA;Aris just calls those fluids polar, what is pretty correct but makes confusion with electromagnetic properties -- I believe that &quot;fluid with non-symmetric stress&quot; or &quot;couple stress&quot; are better keywords.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-30T11:22:11.767" LastActivityDate="2010-12-30T11:22:11.767" CommentCount="4" />
  <row Id="2375" PostTypeId="2" ParentId="743" CreationDate="2010-12-30T11:14:08.253" Score="7" ViewCount="" Body="&lt;p&gt;I have written a pedagogical article about renormalization and renormalization group and I would be happy to have your opinion about it. It is published in American Journal of Physics. You'll find it also on ArXiv: &lt;a href=&quot;http://arxiv.org/abs/hep-th/0212049&quot; rel=&quot;nofollow&quot;&gt;A hint of renormalization&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. Delamotte&lt;/p&gt;&#xA;" OwnerUserId="1028" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2010-12-30T16:50:28.350" LastActivityDate="2010-12-30T16:50:28.350" CommentCount="1" />
  <row Id="2377" PostTypeId="1" AcceptedAnswerId="2380" CreationDate="2010-12-30T13:44:17.630" Score="6" ViewCount="347" Body="&lt;p&gt;Is randomness based on lack of knowledge or behaviour of universe is true random?&lt;/p&gt;&#xA;" OwnerUserId="1029" LastEditorUserId="1029" LastEditorDisplayName="" LastEditDate="2011-01-01T17:30:54.943" LastActivityDate="2011-03-16T15:54:17.467" Title="Is randomness deterministic?" Tags="&lt;determinism&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="1" />
  <row Id="2378" PostTypeId="1" AcceptedAnswerId="2379" CreationDate="2010-12-30T13:54:59.283" Score="0" ViewCount="333" Body="&lt;p&gt;Given what we know about space, time and the movement of galaxies, have we or can we determine what our position is in relation to the projected location of the Big Bang? I've read some introductory papers on the superstructure and galaxy cluster movements, but none of them specifically mentioned space in terms of relative or absolute positions relating to the original position of the Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is, does our current understanding of the structure and workings of the Universe give us a good enough estimate to determine our location relative to the Big Bang or can we never guess at it, since every viewpoint in our universe looks the same in every direction?&lt;/p&gt;&#xA;" OwnerUserId="845" LastActivityDate="2011-02-02T00:48:05.820" Title="What is our location relative to the Big Bang?" Tags="&lt;spacetime&gt;&lt;big-bang&gt;" AnswerCount="4" />
  <row Id="2379" PostTypeId="2" ParentId="2378" CreationDate="2010-12-30T14:01:24.613" Score="3" ViewCount="" Body="&lt;p&gt;Current cosmological theorists suppose that the universe is exactly identical, no matter where it is viewed from, so long as it is viewed at the same time.  At the time of the big bang, the distances between any two given points seems to shrink to zero (or some nonzero value that we supposedly will derive from quantum mechanics).  The conclusion is that the Big Bang happened everywhere, all at once.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also how you get out of the 'was the big bang a black hole?'-type questions:  even though you had large concentrations of matter at times close to the big bang, they were spread out over all space, which is different than just having a clump of matter with finite extent (the second thing &lt;b&gt;would&lt;/b&gt; collapse to a black hole).  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2010-12-30T14:01:24.613" CommentCount="9" />
  <row Id="2380" PostTypeId="2" ParentId="2377" CreationDate="2010-12-30T15:00:48.690" Score="10" ViewCount="" Body="&lt;p&gt;This is a very general question, and can be answered from several perspectives. I shall try to give an overview so you can perhaps research the areas that interest you a bit more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, the most fundamental interpretation of probability (as considered by most mathematicians) is &lt;a href=&quot;http://en.wikipedia.org/wiki/Bayesian_probability&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Bayesian probability&lt;/strong&gt;&lt;/a&gt;. This effectively states that probability measures &lt;em&gt;state of knowledge&lt;/em&gt; of an observer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This view has interesting ties with physics, in particular quantum mechanics. One could consider the random outcome of a QM measurement (wavefunction collapse) from a &lt;a href=&quot;http://en.wikipedia.org/wiki/Frequentist&quot; rel=&quot;nofollow&quot;&gt;frequentist approach&lt;/a&gt;, but it is often more appealing philosophically to consider it as a state of knowledge. (The famous thought experiment of Schrodinger's cat is a good example - until one opens the box, we can only say it is an &quot;alive-dead&quot; cat!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, Bayesian probability does &lt;em&gt;not&lt;/em&gt; explicitly preclude determinism (or non-determinism). Our current understanding of quantum mechanics &lt;em&gt;does&lt;/em&gt; however. In other words, even knowing perfectly the state of a system at a given time, we cannot predict the state of the system at a future time. This most famous upset Albert Einstein, who spent many years of his life looking for a more fundamental deterministic theory - a so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Hidden_variable_theory&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;hidden-variables theory&lt;/strong&gt;&lt;/a&gt;. Since then, however, we have learnt of &lt;a href=&quot;http://en.wikipedia.org/wiki/Bell%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Bell's theorem&lt;/a&gt;, which implies the non-existence of &lt;em&gt;local&lt;/em&gt; hidden variables, suggesting that &lt;em&gt;there is no&lt;/em&gt; more fundamental theory that &quot;explains away&quot; the non-determinism of QM. This is however a very contentious issue, and in any case does not rule out the existence of non-local hidden variable theories - the most famous of which is Bohm's interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, this issue is far from settled, and creates a lot of contention between different groups of physicists as well as philosophers today.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2010-12-30T15:00:48.690" CommentCount="7" />
  <row Id="2381" PostTypeId="2" ParentId="2377" CreationDate="2010-12-30T15:11:54.120" Score="1" ViewCount="" Body="&lt;p&gt;There is a fundamental randomness in the universe but we can often treat things as deterministic. For example, we can accurately predict the path of a projectile provided we know the initial velocity and the gravitational acceleration. However, every measurement has uncertainty due to the accuracy and precision of the instruments used to make the measurements. From these measurements, our predictions also have uncertainty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uncertainty becomes a fundamental problem at extremely small scales. You should read up on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Uncertainty_principle&quot; rel=&quot;nofollow&quot;&gt;Uncertainty Principle&lt;/a&gt; for a detailed explanation of this but I will attempt to put it simply. To make a measurement, you actually have to interact with the object. For example, to see in the dark you may use a torch. This will shine light at objects, which will be scattered and reflected and your eyes will detect the reflected light. Here light interacts with the object you are observing. At a large scale this doesn't change much, but at extremely small scales the energy carried by light is enough to change the system significantly. So the action of observing necessarily implies that you are changing the system so that you can never measure something exactly. It is important to realise that this is a fundamental law of the universe, not just that our equipment is not good enough. I recommend searching for&lt;a href=&quot;http://www.youtube.com/results?search_query=Dr+Quantum&amp;amp;aq=f&quot; rel=&quot;nofollow&quot;&gt; Dr Quantum&lt;/a&gt; videos - it is an animated series that explains these concepts. Due to these limitations, we have to model things like the position of a particle as a probability distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another important thing with regards to determinism is radioactive decay. We can predict very well how much of a radioactive substance will decay in a certain time. It is simply an exponential decay. However, if we extract a single atom, we have no idea when it will decay. This is completely random - the decay of an atom is indeterministic and not at all affected by environmental factors. Again, our models are reduced to probability.&lt;/p&gt;&#xA;" OwnerUserId="954" LastActivityDate="2010-12-30T15:11:54.120" />
  <row Id="2382" PostTypeId="2" ParentId="2368" CreationDate="2010-12-30T15:31:30.233" Score="2" ViewCount="" Body="&lt;p&gt;I don't know if this counts as &quot;practical&quot;, but antiprotons are used in the Tevatron to increase the center-of-mass energy of collisions with protons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the LHC, however, the rest-mass is so much smaller than the Lorentz-boosted mass in the center-of-mass frame, that it basically isn't worth the hassle of using antiprotons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may just be a quirk of historical naming, but anti-neutrinos are typically much more common than neutrinos, since they are formed via beta-decay, which is happening all around us all the time.&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2010-12-30T15:31:30.233" CommunityOwnedDate="2011-04-08T20:18:24.827" CommentCount="1" />
  <row Id="2383" PostTypeId="2" ParentId="142" CreationDate="2010-12-30T15:34:53.583" Score="2" ViewCount="" Body="&lt;p&gt;Okay, first we have the phenomenon: Yes. adding salt increases the boiling point of water, which means that you have to input more energy to get the water to boil, but your egg or pasta will cook faster once you do, because the water will be hotter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there's the why.  The boiling point of a liquid is the temperature at which the vapor pressure of the liquid is the same as the atmospheric pressure above the liquid.  If we can artificially increase the vapor pressure of the liquid, we decrease the boiling temperature.  If we can artificially decrease the vapor pressure of the liquid, we increase the boiling temperature.  So the question has now become: why does the vapor pressure of water decrease when we add salt to it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So imagine a pot of water.  At any given temperature there will be some water molecules in the gas phase above the pot (that's the origin of the vapor pressure), and some in the liquid phase in the pot.  The proportion in the two phases is determined by the interplay of lowering potential energy (by decreasing elevation in gravity, by forming hydrogen bonds, by lining up the polar ends of the molecules, etc.) and increasing the entropy (there's more accessible states in the gas phase, most liquids are incompressible, etc.).  The potential energy part favors the liquid phase, while the entropy part favors the gas phase.  The real requirement here is to minimize the free energy, F = U - TS, with F the free energy, U the potential, T the temperature, and S the entropy.  Since S is paired with the temperature, increasing the temperature increases the impact of the entropy part, which is why the vapor pressure increases as we increase the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we toss in some salt, while keeping the temperature fixed.  The volume fraction of the water decreases, and suddenly there are new accessible states for the water molecules in the liquid phase -- so the vapor pressure decreases.  We keep adding salt and the vapor pressure keeps decreasing.  If we keep going, eventually there's no vapor pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Raoult's law says that the vapor pressure of a solution is proportional to the vapor pressure of the pure solvent (basically that there is a straight line between the pure vapor pressure and zero, when we've buried it in salt).  That's taken as the definition of an ideal solution.  Real solutions have a curved functional form between the two boundary conditions, with the deviations from linearity coming from interactions between the solute (the salt) and the solvent (the water).  Those interactions might be things like breaking up the network of hydrogen bonds in the water, disrupting the polarization arrangement (both of which will favor gas phase), or bonding/pairing up with water molecules (which will favor liquid phase).  At relatively low concentrations of solute the interaction effects are pretty small, so the dependence of vapor pressure on solute concentration remains roughly linear.  The cool observation though is that at most temperatures and for most solvents, it doesn't matter what solute you use (as long as the solute itself doesn't have a vapor pressure), the vapor pressure of the solvent is still decreased by adding solute (which indicates that the entropic contribution is the most important part, and the interactions don't play a big role).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to sum up: for a given concentration of salt dissolved in water, there are more states accessible to the water molecules in the liquid phase than there are in pure water.  So at every water temperature as we pour in energy to make it boil, there will be a lower vapor pressure than there would have been without the salt, and thus we won't get to the boiling point until the water has reached a higher temperature (until we've poured in more energy than we would have had to).  Salt does disrupt the network of hydrogen bonds in the water molecules, but the effect isn't very big at reasonable concentrations of salt, and it's never big enough to counteract the entropic effect.&lt;/p&gt;&#xA;" OwnerUserId="1013" LastActivityDate="2010-12-30T15:34:53.583" />
  <row Id="2384" PostTypeId="2" ParentId="2363" CreationDate="2010-12-30T15:59:02.967" Score="2" ViewCount="" Body="&lt;p&gt;Consider a spherical drop of water, initial temp 40C, radius 3mm, mass 0.1g&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get it down to 0C, you need to remove 4.18 (J/gK) * 0.1 g * 40 K = 17 J&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then, to freeze it solid, you need to remove latent heat of fusion 333 (J/g) * 0.1 g = 33 J&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for a total of 50 J.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat conductivity equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H=\frac{\Delta Q}{\Delta t} = k A\frac{\Delta T}{x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $k$ is the thermal conductivity of water ($0.6 W/m\cdot K$), $A$ is the surface area, and $x$ is the thickness. Take $A=4\pi R^2$ and $x=R=3$mm, and you find that it would take 27 sec to freeze the drop of water in -40C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in practice, the drop will be elongated, increasing $A/x$, and really only the surface layer needs to freeze, possibly eliminating 50-90% of the required latent heat of fusion, so in practice, I think it should be possible to freeze in about a second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a real answer, I think we need to go to Mythbusters!!&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2010-12-30T15:59:02.967" CommentCount="1" />
  <row Id="2385" PostTypeId="2" ParentId="2361" CreationDate="2010-12-30T16:05:00.017" Score="1" ViewCount="" Body="&lt;p&gt;The vast majority of hydrogen in the universe is in hot gas in galaxy clusters, or cold, extremely diffuse atomic hydrogen in the intergalactic medium (Lyman alpha absorbers). Neither population is likely to ever form stars, so I think the safe answer is &quot;forever.&quot;&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2010-12-30T16:05:00.017" CommentCount="1" />
  <row Id="2386" PostTypeId="1" CreationDate="2010-12-30T16:30:52.953" Score="4" ViewCount="147" Body="&lt;p&gt;A big argument by the nitrogen-in-the-tire crowd is that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Nitrogen atoms are bigger and thus less likely to escape the tire, bringing stability to your tire pressure.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If &lt;a href=&quot;http://en.wikipedia.org/wiki/Earth%27s_atmosphere#Composition&quot; rel=&quot;nofollow&quot;&gt;Earth's atmosphere is %78.084 percent nitrogen&lt;/a&gt;, then the non-Nitrogen composition is ~22%. If this 22% is more likely to seep out of the tire, and Nitrogen doesn't seep out, then simply filling the tire up with air will maintain the original 78% N, plus the added nitrogen which would be 78% of the remaining 22% (assuming total seepage).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply, won't a tire that's been filled up 10 times in the course of its normal life already be disproportionately Nitrogen if this argument is true?&lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="325" LastEditorDisplayName="" LastEditDate="2010-12-30T16:38:08.970" LastActivityDate="2010-12-30T20:51:47.503" Title="Will an air-filled tire normally gravitate toward being nitrogen-filled just by refilling with air?" Tags="&lt;air&gt;&lt;pressure&gt;" AnswerCount="2" CommentCount="10" />
  <row Id="2387" PostTypeId="2" ParentId="2378" CreationDate="2010-12-30T16:32:57.537" Score="-1" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Qkrwu.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Big bang happened somewhere in the middle... edges... and everywhere at once.&lt;/p&gt;&#xA;" OwnerUserId="576" LastActivityDate="2010-12-30T16:32:57.537" CommentCount="2" />
  <row Id="2388" PostTypeId="2" ParentId="2372" CreationDate="2010-12-30T18:24:45.980" Score="4" ViewCount="" Body="&lt;p&gt;The answer depends on the initial conditions (initial angle, and initial angular velocity). If it starts vertical, then without an initial angular velocity it would take forever to fall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that as the rotational speed inceases, the bar might lift up from the pivot point due to the centifigual forces. Also once friction is overcome the contact is going to slip. So there are three domains for the solution.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Contact is fixed, forces need to be calculated&lt;/li&gt;&#xA;&lt;li&gt;Contact is sliding, vertical force needs to be calculated&lt;/li&gt;&#xA;&lt;li&gt;No longer in contact, no forces applied (other than gravity).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can try to work out the solutions from the following equations:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Equations Of Motion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F=m\ddot{x} $$&#xA;$$ N=m(\ddot{y}+g) $$&#xA;$$ \frac{H}{2}\left(F\,\cos\theta+N\,\sin\theta\right)=I_{G}\ddot{\theta} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where: $F$ frictional (horiz.) force, $N$ normal contact force, $m$ mass of bar, ($\ddot{x}$, $\ddot{y}$) acceleration of center of gravity, $H$ the total height of the bar, $\theta$ angle of bar from vertical (+=CCW), $I_G$ mass moment of inertia of the bar at the c.g.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Velocity of contact point&#xA;$$ vx_{A}=\dot{x}+\frac{H}{2}\left(\dot{\theta}\cos\theta\right) $$&#xA;$$ vy_{A}=\dot{y}+\frac{H}{2}\left(\dot{\theta}\sin\theta\right) $$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Acceleration of contact point&#xA;$$ ax_{A}=\ddot{x}+\frac{H}{2}\left(\ddot{\theta}\cos\theta-\dot{\theta}^{2}\sin\theta\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ ay_{A}=\ddot{y}+\frac{H}{2}\left(\ddot{\theta}\sin\theta+\dot{\theta}^{2}\cos\theta\right) $$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Friction contdition&#xA;$$ F\leq\mu N $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will find the equation of motion for case &lt;code&gt;1&lt;/code&gt; being&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ddot{\theta}=\frac{g\sin\theta-\frac{H}{2}\dot{\theta}\cos2\theta}{\frac{I_{G}}{mH/2}+\frac{H}{2}\sin2\theta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding the condition where you slip into case &lt;code&gt;2&lt;/code&gt; and then case &lt;code&gt;3&lt;/code&gt; involves monitoring the forces $F$ and $N$ and checking when $F&amp;gt;\mu N$ and when $N&amp;lt;0$.&lt;/p&gt;&#xA;" OwnerUserId="392" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2010-12-30T18:49:52.853" LastActivityDate="2010-12-30T18:49:52.853" CommentCount="2" />
  <row Id="2389" PostTypeId="2" ParentId="2386" CreationDate="2010-12-30T20:23:00.970" Score="7" ViewCount="" Body="&lt;p&gt;This probably depends on the tire structure; in general:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;for hydrodynamic size hole there will be no difference in move of oxygen and nitrogen&lt;/li&gt;&#xA;&lt;li&gt;for a very small holes there will be some sorting due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Effusion&quot; rel=&quot;nofollow&quot;&gt;effusion&lt;/a&gt;, but this is pretty inefficient process since atomic masses of oxygen and nitrogen are similar; what's worse, it will work in the opposite direction -- oxygen molecules are a bit heavier than nitrogen ones, so due to the Graham's law the nitrogen will escape about 1.07 times faster than oxygen.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-12-30T20:23:00.970" CommentCount="1" />
  <row Id="2390" PostTypeId="2" ParentId="2386" CreationDate="2010-12-30T20:51:47.503" Score="2" ViewCount="" Body="&lt;p&gt;I thought the big attraction of nitrogen was its low reactivity. The oxygen can chemically combine with rubber molecules, sort of like conbustion, but far below ignition temperature, and the claim is that that effects the tire lifetime. These oxidation reactions will rates will be much faster at higher temperature, but tires do heat up while driving. I was talked into N2, the last time I got new tires. I don't know if it is was really worth the approx $8 per tire it cost me however (i.e. how much the tire lifetime improves and the slight improvement in vehicle milage is actually worth). So if the O2 molecules are diffusing out faster, but more importantly chemically combining with the rubber, I would expect the O2 concentration to decrease with time. &lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2010-12-30T20:51:47.503" CommentCount="5" />
  <row Id="2391" PostTypeId="2" ParentId="304" CreationDate="2010-12-30T22:44:08.510" Score="0" ViewCount="" Body="&lt;p&gt;You have a number of factors, and I'm not even sure I even know them all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For explosives like fireworks, one issue is pressure -- larger fireworks tend to have thicker shells, and be wound more times.  (in the manufacture of larger fireworks, they wrap them tightly with string to compress them and make them burst further).  So, when the firework finally bursts, the internal pressure is higher, and so you'd end up disturbing a larger volume of air in the same amount of time.  I have no idea how much this is a factor ... I assume you've have to take an identical amount of explosive, in identical volume containers of different strengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, using more explosives would also result in disturbing more air, but I don't know how much of a contributing factor volume is vs. pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other factor is the chemical used in the explosion -- obviously, different ones burn differently, but even in the case of balloons bursting, and we're dealing with similar pressures of similar sizes, helium and hydrogen don't make the same sounds as something filled with compressed &quot;air&quot; ... but I don't know if it's an issue with the molecules being smaller, and so less likely to vibrate other things they pass by (just by odds of there being a collision), or if it's simply an issue of different total mass, and so there's less stored energy at the time of release.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing shape is also a factor, but that's going to be even harder to quantify, as with a shaped charge you're going to have gas moving at different speeds in different directions, which may make the pitch vary with the location of the observer.&lt;/p&gt;&#xA;" OwnerUserId="115" LastActivityDate="2010-12-30T22:44:08.510" />
  <row Id="2392" PostTypeId="1" AcceptedAnswerId="2406" CreationDate="2010-12-30T22:45:56.480" Score="2" ViewCount="268" Body="&lt;p&gt;In 2009, Rowan University released a paper claiming to replicate Blacklight Power's results on energy generation using hydrino states of the hydrogen atom. &lt;a href=&quot;http://www.blacklightpower.com/pdf/BLPIndependentReport.pdf&quot; rel=&quot;nofollow&quot;&gt;The paper&lt;/a&gt; appears to describe the procedure in every detail as far as my untrained eye can tell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone else attempted to replicate these results and did they succeed?&lt;/p&gt;&#xA;" OwnerUserId="786" LastActivityDate="2011-02-20T23:50:14.840" Title="Have the Rowan University &quot;hydrino&quot; findings been replicated elsewhere?" Tags="&lt;experimental-physics&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="2393" PostTypeId="2" ParentId="2378" CreationDate="2010-12-30T23:15:23.313" Score="1" ViewCount="" Body="&lt;p&gt;Think about the big bang like this: if you took every possible trajectory that an object could possibly take through the universe, and traced them all backward in time, they would all eventually come together. That &quot;merging of trajectories&quot; is the big bang; all objects, and indeed all of space, would have been at the same position. So there's no &lt;em&gt;one&lt;/em&gt; point in the universe that can be considered the location of the big bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More quantitatively, the FLRW metric, which is how we characterize the universe as a whole in general relativity, includes a scale factor $a$ which characterizes the relative scale of the universe at different times. Specifically, the distance between two objects (due only to the change in scale) at different times $t_1$ and $t_2$ satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d(t_1)}{a(t_1)} = \frac{d(t_2)}{a(t_2)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, $a$ is getting larger with time. But if you imagine running the universe's expansion in reverse, eventually you'd get back to a &quot;time&quot; when $a = 0$, and at that time all objects would be in the same position. That's the big bang. (Adapted from &lt;a href=&quot;http://physics.stackexchange.com/questions/1915/what-does-it-mean-that-the-universe-is-infinite/1922#1922&quot;&gt;another answer&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2010-12-30T23:15:23.313" />
  <row Id="2394" PostTypeId="1" AcceptedAnswerId="2395" CreationDate="2010-12-31T00:07:05.103" Score="3" ViewCount="93" Body="&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=UJJuq_pcyIQ&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=UJJuq_pcyIQ&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly is going on in the video example? I understand the phenomena occurs because of magnetism but I am trying to figure out the mechanics behind this sculpture. There obviously is a magnet underneath but what is it doing? Is it moving? Is it getting some type of charge? Any insight is great!&lt;/p&gt;&#xA;" OwnerUserId="1035" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2010-12-31T00:39:27.187" LastActivityDate="2010-12-31T16:54:28.530" Title="Dynamic ferrofluid sculptures" Tags="&lt;fluid-dynamics&gt;&lt;magnets&gt;&lt;magnetohydrodynamics&gt;" AnswerCount="2" />
  <row Id="2395" PostTypeId="2" ParentId="2394" CreationDate="2010-12-31T01:56:58.520" Score="0" ViewCount="" Body="&lt;p&gt;I think I can tackle the mechanics aspect. What you can see is an inverted metallic cone with a helical groove running down and around its surface. On zooming in, you the perforations in the cone's surface. There is a pump which pumps the fluid back to the top as it flows down. As the empty cone begins to refill, the fluid also starts to leak out of the sides. That is why you see that flowery pattern emerge from the bottom and travel upwards. Once the fluid reaches the top the cone is filled and at maximum pressure, when the pump is turned off and the cycle begins again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the purely mechanical aspect of this construction is very clever on its own and presumably substituting other fluids would yield different kinds of fountains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The leafy structure is generated due to magnetic fields. I can't say how exactly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: On further reflection it seems that one can have an electromagnetic pump rather than a mechanical one. Imagine a inductor running through the center of the cone. A sinusoidal current would generate a sinusoidal magnetic field within the inductor, which would alternately pull the fluid up the cone and then push it back down.&lt;/p&gt;&#xA;" LastActivityDate="2010-12-31T01:56:58.520" />
  <row Id="2396" PostTypeId="2" ParentId="2285" CreationDate="2010-12-31T03:15:16.583" Score="5" ViewCount="" Body="&lt;p&gt;Mark Eichenlaub asked in a comment 2 days ago- &quot;Also, did Galileo actually drops rocks off that damn tower or not?!!&quot;    As you may now it's iffy, and furthermore the dropping different weights experiment had been done and published several times before Galileo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, from &lt;a href=&quot;http://en.wikipedia.org/wiki/Galileo%27s_Leaning_Tower_of_Pisa_experiment&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, Galileo reasoned as follows from one of the first thought experiments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Imagine two objects, one light and one heavier than the other one, are connected to each other by a string. Drop this system of objects from the top of a tower. If we assume heavier objects do indeed fall faster than lighter ones (and conversely, lighter objects fall slower), the string will soon pull taut as the lighter object retards the fall of the heavier object. But the system considered as a whole is heavier than the heavy object alone, and therefore should fall faster. This contradiction leads one to conclude the assumption is false.&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2010-12-31T03:15:16.583" CommunityOwnedDate="2010-12-31T03:15:16.583" CommentCount="4" />
  <row Id="2398" PostTypeId="1" AcceptedAnswerId="2420" CreationDate="2010-12-31T03:31:42.337" Score="9" ViewCount="207" Body="&lt;p&gt;For an experiment I will hopefully be soon conducting at Johns Hopkins I need two different lenses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first needs to allow all wavelengths above 500 nm to pass (thus a high pass filter) and cut off everything else.&lt;br&gt;&#xA;The second needs to allow all wavelengths below 370 nm to pass (thus a low pass filter) and cut off everything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My knowledge of optics is middling.  I know that good old glass cuts of UV light, but I was hoping for something more specific.  Does anyone know of the theory necessary to &quot;tune&quot; materials to make such filters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Truth be told, I'm an experimentalist, so simply giving me a retail source that has such lenses would get me to where I need to go!  But learning the theory would be nice as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;Sam&lt;/p&gt;&#xA;" OwnerUserId="1037" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-01T18:09:26.340" LastActivityDate="2011-01-01T18:09:26.340" Title="Do high/low pass lenses exist?" Tags="&lt;optics&gt;&lt;experimental-physics&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="2399" PostTypeId="2" ParentId="2398" CreationDate="2010-12-31T03:52:08.827" Score="5" ViewCount="" Body="&lt;p&gt;You may consider to use a &lt;a href=&quot;http://en.wikipedia.org/wiki/Prism_%28optics%29&quot; rel=&quot;nofollow&quot;&gt;prism&lt;/a&gt; to separate the different frequency light (essentially you are performing a &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;Fourier transform&lt;/a&gt;). Then physically blocked those light with wavelength higher than 500 nm and lower than 370 nm. Finally, you merge these light ray together. This will be the device you want and you can freely adjust the frequency range.&lt;/p&gt;&#xA;" OwnerUserId="562" LastActivityDate="2010-12-31T03:52:08.827" CommentCount="6" />
  <row Id="2400" PostTypeId="2" ParentId="1216" CreationDate="2010-12-31T05:45:17.250" Score="2" ViewCount="" Body="&lt;p&gt;You might want to look up the &lt;a href=&quot;http://ccdb4fs.kek.jp/cgi-bin/img/allpdf?198101322&quot; rel=&quot;nofollow&quot;&gt;Weinberg-Witten theorem&lt;/a&gt; which shows that's not possible given certain assumptions. If the original model from which quantum gravity is supposed to emerge is an ordinary Poincar&amp;eacute; covariant quantum field theory over flat nondynamical Minkowski space, they showed it's not possible for massless helicity $\pm 2$ particles to emerge. As a theory of quantum gravity ought to contain gravitons, this appears to rule out such models.&#xA;Of course, these assumptions are questionable. For instance, the theory from which gravity emerges might not be a quantum field theory. This is the case for superstring theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility might be the &quot;fundamental&quot; model isn't Lorentz covariant. However, we still need the low energy effective theory to be approximately Lorentz covariant. In typical condensed matter analog models, different quasiparticles couple to different metrics, and there is no universality to the gravitational couplings, or the speed of light. Unless all the quasiparticles co-emerge together, I don't see any way around this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be a bit hard to come up with the positive energy theorem in an emergent theory of gravity. The positive energy theorem states that the ADM energy of an asymptotically flat spacetime always has to be nonnegative. In an emergent theory, the ADM energy could just as easily be negative for some states. To see this, note first that the ADM energy can be defined locally as the limit as we go to spatial infinity of a locally defined integral over an enclosing spatial surface with one spatial and one time codimension. If we assume the &quot;fundamental&quot; theory is local, this means the now emergent ADM flux also has to be defined locally in terms of the more fundamental fields. As the enclosing boundary becomes larger and larger, its extrinsic curvature becomes closer and closer to zero. If we have a positive ADM flux passing through a plane &amp;mdash; as defined with respect to a choice of normal vector orientation &amp;mdash; a reflection by the plane will give us another state where this ADM flux is now negative. So, we can certainly imagine performing some sort of approximate reflection about the enclosing submanifold on a local patchwork basis, at least for the regions at or around the enclosing surface. We then need to find an interpolation of the resulting state far into the interior, which of course, might not look anything like a reflection at all. But if the fundamental theory also satisfies local independence, that ought to be possible. But the end result of all this construction is a state with negative emergent ADM energy. I know this argument is very handwavy and nonrigorous, but it sounds plausible. But there might be some loopholes. For instance, the fundamental theory might be local, but the emergent large scale excitations &amp;mdash; and hence emergent spacetime &amp;mdash; might be delocalized with respect to the underlying background spacetime. Or the underlying fundamental theory might be inherently nonlocal.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-12T14:36:26.017" LastActivityDate="2011-01-12T14:36:26.017" CommentCount="6" />
  <row Id="2401" PostTypeId="2" ParentId="2372" CreationDate="2010-12-31T06:07:49.583" Score="1" ViewCount="" Body="&lt;p&gt;A falling tree is basically an inverted pendulum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The period of a pendulum of length $h$ for small oscillations is $2\pi \sqrt{h/g}$, with $g$ the acceleration due to gravity, about $10\ m/s$.  For an inverted pendulum near the top of its arc, there is no period, but the quantity $\sqrt{h/g}$ does represent a characteristic time scale for this system.  The tree will take a few of these characteristic times to fall.  $h$ for a tree is an &quot;effective height&quot;, and depends on the mass distribution of the tree.  If all the mass is at the top, $h$ is the height of the tree.  If the tree is uniform, $h$ is $2/3$ the true height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a tree with $h = 40\ m$, the characteristic time is $2\ s$.  For small angles, the angle the tree makes with the vertical will be multiplied by $e$ in this time.  Let's start the tree at $1^{\circ}$ so that it needs to multiply its angle by $90$ to fall.  $\ln(90) = 4.5$ so the tree takes about $9\ s$ to fall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is mathematically an underestimate because the characteristic time increases slightly as the tree falls, but not too much.  Give it a nice round $10\ s$ and you get something that matches &lt;a href=&quot;http://www.youtube.com/watch?v=cClxs3j_I6E&quot; rel=&quot;nofollow&quot;&gt;the first YouTube video&lt;/a&gt; I found.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="174" LastEditorDisplayName="" LastEditDate="2011-02-05T10:03:44.043" LastActivityDate="2011-02-05T10:03:44.043" CommentCount="6" />
  <row Id="2402" PostTypeId="2" ParentId="437" CreationDate="2010-12-31T06:48:59.770" Score="2" ViewCount="" Body="&lt;p&gt;According to Wigner's analysis, the single photon Hilbert space is spanned by a basis parameterized by energy-momenta on the forward light cone boundary, and a helicity of $\pm 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a manifestly Lorentz covariant description in position space has to include a fictitious longitudinal photon with a helicity of 0. This degree of freedom is pure gauge, and decouples. Interestingly enough, the state norm is now positive semidefinite, instead of positive definite, with the transverse modes having positive norm and the longitudinal ones having zero norm.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-06T13:17:29.760" LastActivityDate="2011-01-06T13:17:29.760" CommentCount="1" />
  <row Id="2403" PostTypeId="2" ParentId="387" CreationDate="2010-12-31T07:05:19.630" Score="2" ViewCount="" Body="&lt;p&gt;A superposition of causal structures. More precisely, given two events A and B, they could be in a superposition of being spacelike, null and timelike separated. Quantum field theory is built upon a sharp distinction between localized operators which are spacelike separated from those which aren't. With a superposition of causal structures, such distinctions break down.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2010-12-31T07:05:19.630" />
  <row Id="2404" PostTypeId="2" ParentId="1843" CreationDate="2010-12-31T07:21:05.883" Score="1" ViewCount="" Body="&lt;p&gt;Most calculations of the Hawking effect assume free quantum fields. This assumption breaks down for strongly coupled quantum chromodynamics. As the Hawking temperature is much lower than the QCD deconfinement temperature, there isn't enough energy to hadronize &quot;virtual&quot; quark-antiquark pairs. Instead, the particle just outside the event horizon gets pulled into the black hole by the confining QCD flux tube. It can't escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, for astronomical sized black holes, the Hawking temperature is so low only massless gravitons and photons can be radiated. Even neutrinos are too heavy, never mind quarks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for sufficiently light mini black holes, we can have the creation of hadron pairs, with one hadron escaping, and the other falling into the black hole.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2010-12-31T07:29:08.260" LastActivityDate="2010-12-31T07:29:08.260" />
  <row Id="2405" PostTypeId="2" ParentId="2250" CreationDate="2010-12-31T08:46:01.873" Score="5" ViewCount="" Body="&lt;p&gt;Short answer: Calculate the scattering amplitude (matrix element) for electron-electron or electron-positron scattering. They will differ by a minus sign. This means that the electron electron interaction will have a positive sign in front of the 1/r Coulomb potential which corresponds to a positive (repulsive) force. The same procedure for electron-positron scattering will lead to a negative 1/r Coulomb potential and hence a negative (attractive) force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longer answer: To decide whether any scattering process is attractive or repulsive you calculate the scattering amplitude (matrix element). Then compare your answer to the Born approximation for nonrelativistic QM scattering to see what classical potential your QFT result corresponds to. If the potential increases as you separate the particles the corresponding classical force will be attractive and vice versa. The minus signs in the scattering calculation can arise from many sources (anticommuting fermion fields, vertex rules, propagator terms, etc.). When you choose different incoming or outgoing states, like electron-electron or electron-positron, the calculation results in an extra negative sign. Interestingly, for the Coulomb force, one of the deciding minus signs comes from the photon vector propagator. Any vector propagator will cause like fermion charges to repel and unlike fermion charges to attract. Fermions don't always have to repel other fermions of the same type. If fermions exchange a scalar particle (instead of a photon) they can attract another fermion of the same charge! Something else to consider is that the Coulomb potential 1/r you arrive at is only for first order Feynman diagrams. If you include higher order terms you get corrections to the usual Coulomb potential/force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Using a Born approximation from nonrelativistic QM is an admittedly weird way of going about answering this question. It seems strange that we would take a result from our more fundamental theory (QFT) and then have to translate it into a less fundamental theory like nonrelativistic QM. But by asking questions posed in classical or nonrelativistic language (e.g. is this interaction force attractive or repulsive), maybe we are forcing ourselves to use a classical or nonrelativistic approximation/theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out Peskin pg 125-126 for more on deriving the Coulomb potential/force from QED.&lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="1040" LastEditorDisplayName="" LastEditDate="2011-01-01T03:26:57.683" LastActivityDate="2011-01-01T03:26:57.683" />
  <row Id="2406" PostTypeId="2" ParentId="2392" CreationDate="2010-12-31T14:10:18.527" Score="7" ViewCount="" Body="&lt;p&gt;I am highly skeptical of this result, primarily because the theories promoted by Black Light Power are improbable to the point of being gibberish. The energy states of hydrogen can be calculated exactly, and have been both calculated and measured spectroscopically to extremely high precision, and experiment and theory are in perfect agreement. If the modern understanding of quantum physics (including QED) were incomplete enough to leave room for mysterious lower-energy states in hydrogen, there would've been some indication of this in one of the countless experiments that have been done on hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another good reason to be skeptical of this result is that the report in question seems to have been &quot;released&quot; only via Black Light Power's web site. The only mention of the authors of this report in conjunction with &quot;hydrinos&quot; that Google can find come from Black Light Power. This result has not appeared in any scientific journal known to Google. Or even the Rowan University web site. This is not what I would call a ringing endorsement of the work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the report itself, it is entirely concerned with chemical NMR spectra, and I don't have any first-hand experience with those. I know just enough about the field to know that there can be subtle issues involved with the recording and interpretation of these. I'm more inclined to believe that the mysterious peaks seen in their samples are some NMR artifact than that they are the signature of radically new physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's conceivable, barely, that this really does represent some dramatic new discovery, and has not yet appeared in print because it's working its way through the peer review process, taking a long time because extraordinary claims require extraordinary scrutiny. The principal person behind Black Light Power has been making claims like this since I was in grad school in the 1990's, though, and has yet to produce anything solid. I wouldn't hold my breath waiting for this to appear in a reputable peer-reviewed journal, if I were you.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-12-31T14:10:18.527" CommentCount="1" />
  <row Id="2407" PostTypeId="1" AcceptedAnswerId="2421" CreationDate="2010-12-31T14:18:52.413" Score="11" ViewCount="1023" Body="&lt;p&gt;Of course, assuming your grandmother is not a theoretical physicist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to hear the basics concepts that make LQG tick and the way it relates to the GR. I heard about spin-networks where one assigns Lie groups representations to the edges and intertwining operators to the nodes of the graph but at the moment I have no idea why this concept should be useful (except for a possible similarity with gauge theories and Wilson loops; but I guess this is purely accidental). I also heard that this spin-graph can evolve by means of a spin-foam which, I guess, should be a generalization of a graph to the simplicial complexes but that's where my knowledge ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also read the &lt;a href=&quot;http://en.wikipedia.org/wiki/Loop_quantum_gravity&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt; but I don't find it very enlightening. It gives some motivation for quantizing gravity and lists some problems of LQG but (unless I am blind) it never says what LQG actually is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my questions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Try to give a simple description of fundamentals of Loop Quantum Gravity.&lt;/li&gt;&#xA;  &lt;li&gt;Give some basic results of the theory. Not necessary physical, I just want to know what are implications of the fundamentals I ask for in 1.&lt;/li&gt;&#xA;  &lt;li&gt;Why is this theory interesting physically? In particular, what does it tell us about General Relativity (both about the way it is quantized and the way it is recovered from LQG).&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-02-02T04:48:01.943" Title="Give a description of Loop Quantum Gravity your grandmother could understand" Tags="&lt;quantum-gravity&gt;&lt;loop-quantum-gravity&gt;" AnswerCount="4" CommentCount="9" FavoriteCount="3" />
  <row Id="2408" PostTypeId="1" CreationDate="2010-12-31T14:23:43.530" Score="3" ViewCount="346" Body="&lt;p&gt;Does the force of kinetic friction increase with the relative speed of the objects involved?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have heard and read that the answer is no.  This is counter intuitive, and is a big part of why the &quot;Airplane on a Treadmill&quot; question is so interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What phenomena are at work with kinetic friction, and why does it not increase with relative speed?&lt;/p&gt;&#xA;" OwnerUserId="1042" LastEditorUserId="1042" LastEditorDisplayName="" LastEditDate="2010-12-31T14:34:16.263" LastActivityDate="2010-12-31T15:53:41.057" Title="Does the force of kinetic friction increase with the relative speed of the objects involved? If not, why not?" Tags="&lt;friction&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="2409" PostTypeId="1" AcceptedAnswerId="2673" CreationDate="2010-12-31T15:05:15.913" Score="8" ViewCount="328" Body="&lt;p&gt;A type-I superconductor can expel almost all magnetic flux (below some critical value $H_c$) from its interior when superconducting. Light as we know is an electromagnetic wave. So what would happen if we shine a light deep into a type-I superconductor? Suppose the magnetic field of the light is significantly lower than $H_c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what about type-II superconductors?&lt;/p&gt;&#xA;" OwnerUserId="749" LastActivityDate="2011-01-10T12:40:46.287" Title="Shine a light into a superconductor" Tags="&lt;electromagnetism&gt;&lt;solid-state-physics&gt;&lt;condensed-matter&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="2410" PostTypeId="2" ParentId="2398" CreationDate="2010-12-31T15:08:37.173" Score="3" ViewCount="" Body="&lt;p&gt;What you're looking for isn't a lens so much as a filter. There are filters for whatever purpose you might want out there in the world, and they are relatively inexpensive. Doing a google search on &lt;a href=&quot;http://www.google.com/search?q=Optic+Supplies&amp;amp;ie=utf-8&amp;amp;oe=utf-8&amp;amp;aq=t&amp;amp;rls=org.mozilla%3aen-US%3aofficial&amp;amp;client=firefox-a#sclient=psy&amp;amp;hl=en&amp;amp;client=firefox-a&amp;amp;hs=h01&amp;amp;rls=org.mozilla%3aen-US%3Aofficial&amp;amp;source=hp&amp;amp;q=Optic+Supplies+filters&amp;amp;aq=f&amp;amp;aqi=&amp;amp;aql=&amp;amp;oq=&amp;amp;gs_rfai=&amp;amp;pbx=1&amp;amp;fp=9bef8cda26d1a6ec&quot; rel=&quot;nofollow&quot;&gt;Optics Filters&lt;/a&gt;, I came across a couple of sites that might help you.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.edmundoptics.com/onlinecatalog/browse.cfm?categoryid=41&amp;amp;utm_source=google&amp;amp;utm_medium=ppc&amp;amp;utm_campaign=us&quot; rel=&quot;nofollow&quot;&gt;Edmond's Optics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.opticalfiltersusa.com/home.html&quot; rel=&quot;nofollow&quot;&gt;Optical Filters USA&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="871" LastActivityDate="2010-12-31T15:08:37.173" />
  <row Id="2411" PostTypeId="2" ParentId="2408" CreationDate="2010-12-31T15:18:19.370" Score="3" ViewCount="" Body="&lt;p&gt;At the simple level of approximation where you talk about kinetic friction, it doesn't depend on speed. It's not a great approximation (the coefficients of kinetic friction that you find for materials tend to have huge uncertainties), though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason we use the approximation (other than that it makes for good intro mechanics problems) is that the microscopic physics is pretty complicated. At a very small scale, all objects are somewhat rough (at the atomic scale, if not before), and friction is the result of trying to drag one corrugated surface over another. Larger projections from the surfaces will snag against each other and require some force to dislodge, and the sum of all those microscopic snags and drags is the force we see as friction. As it's impossible to keep track of all those interactions in detail for any reasonable size object, we approximate the total force using the kinetic friction model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kinetic friction has nothing to do with the airplane-on-a-treadmill problem, though. Kinetic friction involves two surfaces sliding across one another. In a situation involving rolling, however, there is no sliding. At the point where the wheels of the airplane come in contact with the surface of the ground (or a treadmill), the wheel surface is not moving relative to the ground. The relative speed of the bit of the wheel touching the ground and the bit of ground that it is touching is always zero, no matter how fast the wheel is moving relative to the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cute way to see this is to place a ruler on top of a couple of soda cans (or other convenient round objects), and roll it some distance along the surface of a table next to another ruler. You'll find that the distance covered by the ruler on top of the cans is double that covered by the center of one of the cans. This happens because the point where the cans touch the table is stationary relative to the table. The centers of the cans move, though, which means that the top part of the rolling can must be moving at twice the speed of the center for the average speed of the can to work out. You can think of the contact point, center, and top being (at some instant) like points connected by a stick that pivots about the contact point-- the point at the far end of the stick will move twice as fast as the point at the halfway mark (obviously, this only holds exactly for the infinitesimal amout of time between when a given bit of can hits the table and when it lifts off again as the roll continues, but it gets you the right idea). Thus, the ruler along the top moves twice as far as the center of the cans.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2010-12-31T15:18:19.370" CommentCount="3" />
  <row Id="2412" PostTypeId="2" ParentId="2408" CreationDate="2010-12-31T15:53:41.057" Score="1" ViewCount="" Body="&lt;p&gt;There are many mechanisms of friction to be considered. &quot;Dry&quot; friction does not depend on the sliding speed, only on the normal force. &quot;Viscous&quot; friction depends on speed linearly, &quot;Elastohydrodynamic (EHD)&quot; friction depends on speed, pressure and film properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which type of friction are you specifically asking for?&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2010-12-31T15:53:41.057" />
  <row Id="2413" PostTypeId="2" ParentId="2392" CreationDate="2010-12-31T16:23:21.743" Score="2" ViewCount="" Body="&lt;p&gt;If it were indeed possible to get energy out of a ground state atom, then by definition, that atom would not be in the ground state. In other words, claiming that you can get energy out of ground state atoms is, in fact, a claim that what look like ordinary hydrogen atoms at rest are actually atoms in some kind of meta-stable excited state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the odds that after 13 billion years, ALL of the hydrogen in the universe is still in this 'false' ground state? Surely there would be enough of the 'true' ground state material floating around that someone would already have noticed it? There would have to be some obvious clues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it seems too good to be true, it probably is, especially where claims of 'new energy sources' are concerned. You might as well stick a water wheel in a flat pond.&lt;/p&gt;&#xA;" OwnerUserId="717" LastActivityDate="2010-12-31T16:23:21.743" CommentCount="2" />
  <row Id="2414" PostTypeId="2" ParentId="2394" CreationDate="2010-12-31T16:54:28.530" Score="0" ViewCount="" Body="&lt;p&gt;IMO this is only obtained with an electromagnet inside this cone structure. The cuts only help to spin the fluid moving up/downwards. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &quot;more verbose&quot; example of such a structure: &lt;a href=&quot;http://www.youtube.com/watch?v=OE2pB1pyZN0&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=OE2pB1pyZN0&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2010-12-31T16:54:28.530" />
  <row Id="2415" PostTypeId="1" AcceptedAnswerId="2416" CreationDate="2010-12-31T19:39:39.470" Score="3" ViewCount="227" Body="&lt;p&gt;From my understanding objects do not fall but are pulled to the earth from gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind, I can't understand why if helium filled balloons are not pulled by gravity then shouldn't they float in place (or drift like objects in space with no gravity) rather than actively move away from the earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is gravity pushing it rather than pulling it? Why is this?&lt;/p&gt;&#xA;" OwnerUserId="947" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2010-12-31T19:56:45.447" LastActivityDate="2010-12-31T22:13:57.830" Title="Why do helium filled balloons move away from the Earth?" Tags="&lt;gravity&gt;&lt;buoyancy&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="2416" PostTypeId="2" ParentId="2415" CreationDate="2010-12-31T19:52:34.700" Score="5" ViewCount="" Body="&lt;p&gt;Helium balloons are pulled by gravity, as are all objects with mass.  The reason they don't fall is that there is another force acting on them, a buoyant force from air pressure that is equal to the weight of the air displaced by the balloon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason you don't float is that the weight of the air you displace is quite a bit less than your weight (a person is more dense than air).  The reason a normal balloon doesn't float is that the weight of the air it displaces is just a little bit less than the weight of the balloon (because it is filled with air, but the rubber of the balloon itself is more dense than the air).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogy you want is to objects floating (or suspended) in water.  Most rocks sink to the bottom, pulled by gravity, because the weight of the water they displace is less than their own weight.  A bowling ball (ironically) is very close to the same density as water, so it will float suspended in mid-water, just like the helium balloon that has leaked a little bit.&lt;/p&gt;&#xA;" OwnerUserId="1013" LastActivityDate="2010-12-31T19:52:34.700" />
  <row Id="2417" PostTypeId="2" ParentId="2415" CreationDate="2010-12-31T22:13:57.830" Score="3" ViewCount="" Body="&lt;p&gt;Just a tad about how bouyancy works. Any fluid in a gravitational field possesses a pressure gradient, (which if the gas/liquid is in equilibrium) counterbalances the effect of gravity. Gravity acting on such a fluid creates this pressure, which is referred to a hydrostatic pressure. To make a long story short, the external pressure (of the air) is greater at the bottom of your ballon than at the top. So the net external pressure on the ballon is imbalanced, providing a net upward force equal to the weight of the displaced air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put your ballon in space (I'll ignore the observation it would pop), there is no pressure, and therefore no imbalance of pressure to create the bouyancy effect.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2010-12-31T22:13:57.830" />
  <row Id="2419" PostTypeId="2" ParentId="770" CreationDate="2011-01-01T00:30:42.390" Score="1" ViewCount="" Body="&lt;p&gt;At CERN's website, there was - for the time being - a page (which I cannot find now, someone please help?) about how the LHC could possibly &quot;destroy the earth, the solar system or the whole universe altogether&quot; with possible speculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These included creating black holes, strange matter that is more stable than the regular one, and a bunch of other ideas which I don't remember. All of this of course rather bogus, because such collisions occur naturally in cosmic radiation at sufficient frequency.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2011-01-01T00:30:42.390" CommunityOwnedDate="2011-01-07T14:13:00.227" CommentCount="1" />
  <row Id="2420" PostTypeId="2" ParentId="2398" CreationDate="2011-01-01T03:26:58.167" Score="8" ViewCount="" Body="&lt;p&gt;There are many ways to do this. Which option you choose depends on what degree of performance you require, and how much money you re willing to spend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, you should understand that, while you could apply a wavelength selective coating to a lens, this would much more commonly and cheaply be done with a wavelength filter separate from the lens itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the cheapest way to do this would be with a glass filter which absorbs short or long wavelengths selectively. There is a wide range of these available from &lt;a href=&quot;http://www.us.schott.com/english/index.html&quot; rel=&quot;nofollow&quot;&gt;Schott Glass&lt;/a&gt;. I use these commonly, and I've never had trouble. Most vendors will be happy to produce custom shapes, thicknesses, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The down side to filter glass is that it doesn't have a particularly sharp cutoff between the bassband and the stopband. For that, you will need a dielectric coating engineered to your specifications. For that, I would look at &lt;a href=&quot;http://www.cvimellesgriot.com/&quot; rel=&quot;nofollow&quot;&gt;CVI&lt;/a&gt; or &lt;a href=&quot;http://www.newport.com/&quot; rel=&quot;nofollow&quot;&gt;Newport&lt;/a&gt;, although there are other vendors out there. There may be something you can use in their catalogs, but custom orders are normal for the optical manufacturing industry, so don't hesitate to call up a sales engineer. In my experience, sales people in this industry are very well educated on their products, or will direct you to an engineer who can tell you exactly what they can produce for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, there are other vendors you could look at, but these are the ones I would go to first. At the very least, looking at their catalogs will give you an idea of what it is you are really looking for.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-01-01T03:26:58.167" CommentCount="4" />
  <row Id="2421" PostTypeId="2" ParentId="2407" CreationDate="2011-01-01T07:53:05.813" Score="12" ViewCount="" Body="&lt;p&gt;@Marek your question is very broad. Replace &quot;lqg&quot; with &quot;string theory&quot; and you can imagine that the answer would be too long to fit here ;&gt;). So if this answer seems short on details, I hope you will understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The program of Loop Quantum Gravity is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The notion of &lt;strike&gt;diffeomorphism invariance&lt;/strike&gt; &lt;em&gt;background independence&lt;/em&gt;, which is central to General Relativity, is considered sacrosanct. In other words this rules out the String Theory based approaches where the target manifold, in which the string is embedded, is generically taken to be flat [Please correct me if I'm wrong.] I'm sure that that is not the only background geometry that has been looked at, but the point is that String Theory is not written in a &lt;strong&gt;manifestly background independent&lt;/strong&gt; manner. LQG aims to fill this gap.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The usual quantization of LQG begins with Dirac's recipe for quantizing systems with constraints. This is because General Relativity is a theory whose Hamiltonian density ($\mathcal{H}_{eh}$), obtained after performing a $3+1$ split of the Einstein-Hilbert action via the ADM procedure [&lt;strong&gt;&lt;a href=&quot;http://grwiki.physics.ncsu.edu/wiki/Hamiltonian_%28ADM%29&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;,&lt;a href=&quot;http://en.wikipedia.org/wiki/ADM_formalism&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;&lt;/strong&gt;], is composed only of constraints, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathcal{H}_{eh} = N^a \mathcal{V}_a + N \mathcal{H} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $N^a$ and $N$ are the lapse and shift vectors respectively which determine the choice of foliation for the $3+1$ split. $\mathcal{V}_a$ and $\mathcal{H}$ are referred to as the vector (or diffeomorphism) constraint and the scalar (or &quot;hamiltonian&quot;) constraint. In the resulting phase space the configuration and momentum variables are identified with the intrinsic metric ($h_{ab}$) of our 3-manifold $M$ and its extrinsic curvature ($k_{ab}$) w.r.t its embedding in the full $3+1$ spacetime, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ {p,q} \rightarrow \{\pi_{ab},q^{ab}\} := \{k_{ab},h^{ab}\} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This procedure is generally referred to as &lt;strong&gt;canonical quantization&lt;/strong&gt;. It can also be shown that $ k_{ab} = \mathcal{L}_t h_{ab} $, where $ \mathcal{L}_t $ is the Lie derivative along the time-like vector normal to $M$. This is just a fancy way of saying that $ k_{ab} = \dot{h}_{ab} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where, in olden days, our progress would come to a halt, because after applying the ADM procedure to the usual EH form of the action, the resulting constraints are complicated &lt;strong&gt;non-polynomial&lt;/strong&gt; expressions in terms of the co-ordinates and momenta. There was little progress in this line until in 1986 $\sim$ 88, Abhay Ashtekar put forth a form of General Relativity where the phase space variables were a canonically transformed version of $ \{k_{ab},h^{ab}\} $ This change is facilitated by writing GR in terms of connection and vielbien (tetrads) $ \{A_{a}^i,e^{a}_i\}$ where $a,b,\cdots$ are our usual spacetime indices and $i,j,\cdots$ take values in a &lt;em&gt;Lie Algebra&lt;/em&gt;. The resulting connection is referred to as the &quot;Ashtekar&quot; or sometimes &quot;Ashtekar-Barbero&quot; connection. The metric is given in terms of the tetrad by :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ h_{ab} = e_a^i e_b^j \eta_{ij} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\eta_{ij}$ is the Minkowski metric $\textrm{diag}(-1,+1,+1,+1)$. After jumping through lots of hoops we obtain a form for the constraints which is polynomial in the co-ordinates and momenta and thus amenable to usual methods of quantization:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathcal{H}_{eha} = N^a_i \mathcal{V}_a^i + N \mathcal{H} + T^i \mathcal{G_i} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, once again, $ \mathcal{V}_a^i $ and $\mathcal{H}$ are the vector and scalar constraints. The explanation of the new, third term is postponed for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Nb:&lt;/strong&gt; Thus far we have made &lt;em&gt;no&lt;/em&gt; modifications to the theoretical structure of GR. The Ashtekar formalism describes the exact same physics as the ADM version. However, the ARS (Ashtekar-Rovelli-Smolin) framework exposes a new symmetry of the metric. The introduction of spinors in quantum mechanics (and the corresponding Dirac equation) allows us to express a scalar field $\phi(x)$ as the &quot;square&quot; of a spinor $ \phi = \Psi^i \Psi_i $. In a similar manner the use of the vierbien allows us to write the metric as a square $ g_{ab} = e_a^i e_b^j \eta_{ij} $. &lt;strong&gt;The transition from the metric to connection variables in GR is analogous to the transition from the Klein-Gordon to the Dirac equation in field theory.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The application of the Dirac quantization procedure for constrained systems shows us that the kinematical Hilbert space, consisting of those states which are annihilated by the quantum version of the constraints, has spin-networks as its elements. All of this is very &lt;strong&gt;rigorous&lt;/strong&gt; and several mathematical technicalities have gradually been resolved over the past two decades.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This answer is already pretty long. It only gives you a taste of things to come. The explanation of the Dirac quantization procedure and spin-networks would be separate answers in themselves. One can give an algorithm for this approach:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Write GR in connection and tetrad variables (in first order form).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Perform $3+1$ decomposition to obtain the Einstein-Hilbert-Ashtekar Hamiltonian $\mathcal{H}_{eha}$ which turns out to be a sum of constraints. Therefore, the action of the quantized version of this Hamiltonian on elements of the physical space of states yields $ \mathcal{H}_{eha} \mid \Psi \rangle = 0 $. (After a &lt;strong&gt;great deal&lt;/strong&gt; of investigation) we find that these states are represented by graphs whose edges are labeled by representations of the gauge group (for GR this is $SU(2)$).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Spin-foams correspond to histories which connect two spin-networks states. On a given spin-network one can perform certain operations on edges and vertices which leave the state in the kinematical Hilbert space. These involve moves which split or join edges and vertices and those which change the connectivity (as in the &quot;star-triangle transformation&quot;). One can &lt;em&gt;formally&lt;/em&gt; view a spin-foam as a succession of states $\{ \mid \Psi(t_i) \rangle \}$ obtained by the repeated action of the scalar constraint $ \mid \Psi(t_1) \rangle \sim \exp{}^{-i\mathcal{H}_{eha}\delta t} \mid \Psi (t_0); \mid \Psi(t_2) \rangle \sim \exp{}^{-i\mathcal{H}_{eha}\delta t} \mid \Psi (t_1) \cdots \rangle $ [&lt;a href=&quot;http://arxiv.org/abs/gr-qc/9612035&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;3&lt;/strong&gt;&lt;/a&gt;].&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The graviton propogator has a robust quantum version in these models. Its long-distance limit yields the $1/r^2$ behavior expected for gravity and an effective coarse-grained action given by the usual one consisting of the Ricci scalar plus terms containing quantum corrections.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There is a great deal of literature to back up everything I've said here, but this is already pretty exhausting so you'll have to take me on my word. Let me know what your Grandma thinks of this answer ;).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" LastActivityDate="2011-01-02T09:42:30.710" CommentCount="19" />
  <row Id="2422" PostTypeId="2" ParentId="2372" CreationDate="2011-01-01T11:01:48.930" Score="2" ViewCount="" Body="&lt;p&gt;To complete the other answers, here's the Lagrangian of the system. Using polar coordinates centred at the base of the rod,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{L}=T-V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{L}=\frac{1}{2}I\dot{\theta^2}-\frac{1}{2}mgh\cos{\theta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the known value for $I$, $\frac{mh^2}{3}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{L}=\frac{1}{6}mh^2\dot{\theta^2}-\frac{1}{2}mgh\cos{\theta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Euler-Lagrange, gives us the equation of motion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{\theta}}=\frac{\partial\mathcal{L}}{\partial\theta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dt}(\frac{1}{3}mh^2\dot{\theta})=\frac{1}{2}mgh\sin{\theta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ddot{\theta}=\frac{3}{2}\frac{g}{h}\sin{\theta}$$&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-01T19:28:45.343" LastActivityDate="2011-01-01T19:28:45.343" CommentCount="2" />
  <row Id="2424" PostTypeId="2" ParentId="2285" CreationDate="2011-01-01T12:42:06.167" Score="3" ViewCount="" Body="&lt;p&gt;That nuclear bombs are fission powered, but thermonuclear bombs are fusion powered. Although thermonuclear bombs do use fusion as a secondary reaction, this is used to produce vast amounts of energy through fission, which produces around 50% of the energy output.&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-30T09:42:04.760" LastActivityDate="2011-01-30T09:42:04.760" CommunityOwnedDate="2011-01-01T12:42:06.167" CommentCount="9" />
  <row Id="2425" PostTypeId="1" AcceptedAnswerId="2427" CreationDate="2011-01-01T15:28:43.473" Score="4" ViewCount="428" Body="&lt;p&gt;In its simplest form, &lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalence_principle&quot; rel=&quot;nofollow&quot;&gt;equivalence principle&lt;/a&gt; states that the inertial mass and the gravitational mass should be the same. This is easy to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why is it so important to the formulation of General Relativity? To be more specific, I don't understand how the gravitational field equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Ote28.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be derived from this principle. &lt;/p&gt;&#xA;" OwnerUserId="605" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-29T16:59:28.937" LastActivityDate="2011-03-29T16:59:28.937" Title="Why is the equivalence principle so important to general relativity?" Tags="&lt;general-relativity&gt;&lt;equivalence-principle&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="2426" PostTypeId="2" ParentId="2425" CreationDate="2011-01-01T15:53:12.980" Score="5" ViewCount="" Body="&lt;p&gt;Equivalence principle states (very roughly) that movement of objects doesn't depend on their mass (so long as they are massive, of course). These important observation is what introduces (pseudo)Riemannian geometry into the theory of gravitation, because it essentially tells us that matter that is not acted on by other forces follows the geodesics of the given manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In total, there are three key ingredients to General Relativity and only one of them relates directly to the Einstein's equations. Let me mention each one of them briefly:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;General covariance&lt;/strong&gt;&#xA;This a requirement that &lt;em&gt;every&lt;/em&gt; reference frame is good for doing physics and brings in the concept of manifold and diffeomorphism invariance.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Equivalence principle&lt;/strong&gt;&#xA;This is a requirement that the manifold be (pseudo)Riemannian and that test particles follow geodesics (a note on this at the end of the answer).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Interaction of matter and space-time in a certain manner and reduction to Newtonian gravity in a classical limit&lt;/strong&gt;&#xA;&lt;em&gt;This&lt;/em&gt; is what gives you Einstein equations.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So it's not really well-posed question to ask what role does equivalence principle play in deriving Einstein's equations. Only correct answer would be: it lets you introduce the concept of (pseudo)Riemannian manifold so that you have some well-defined object to define the equations for in the first place. In particular the Einstein tensor $G_{\mu \nu}$ depends on a &lt;strong&gt;metric tensor&lt;/strong&gt;, so you already have to know there is such a thing as a metric (which is what equivalence principle tells you) even before you start pondering whether there might be such a thing as Einstein's equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt; One can have geodesics on other mathematical structures than just (pseudo)Riemannian manifolds. A manifold with connection is enough to have but there are other reasons people like to have metric tensor around (which induces a special kind of connection by itself).&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-01-01T15:53:12.980" />
  <row Id="2427" PostTypeId="2" ParentId="2425" CreationDate="2011-01-01T16:05:26.797" Score="9" ViewCount="" Body="&lt;p&gt;A derivation of Einstein's equation isn't why the Equivalence principle is central to GR. The reason that the equivalence principle is central to GR is in the fact that you can represent the gravitational field with a metric tensor at all--you can replace a force equation with a geodesic equation for a test mass precisely due to the fact that the geodesic that that test mass follows (or the &quot;acceleration&quot; felt by a Newtonian mass) is independent of the mass of that test$^{1}$ particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equivalence principle, however, only selects out that one can represent gravity with a metric tensor.  There are a great many other so-called &quot;metric theories of gravity&quot; that obey the equivalence principle, but are not general relativity--amongst other things, they will differ in the field equation for the metric tensor, or have extra fields in addition to the metric--the most famous of these is the Brans-Dicke theory, which treats Newton's constant as a scalar field coupled to the metric tensor.  Most alternative metric theories have either been experimentally ruled out, or have had their additional fields constrained to the point where their values are consistent with zero (for instance, Brans-Dicke theory has a parameter $\omega$, and tends to GR if $\frac{1}{\omega}\rightarrow 0$.  Current data says that $\omega &amp;gt; 4000$, or some similar number.).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^{1}$Note that this is generally only true if the mass of the test particle is &quot;small&quot; compared to the local curvature of the spacetime, and if it's motion is slow enough to not produce gravitational radiation comparable to its energy.  Either of these effects will cause the test mass to perturb the background spacetime, and those effects will both be mass dependent and cause the test mass to not follow a geodesic of the background spacetime.  Both of these approximations are true (to great precision, at least) of all of the planets, asteroids and comets orbiting the sun, amongst many other things.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2011-01-01T16:26:57.160" LastActivityDate="2011-01-01T16:26:57.160" CommentCount="1" />
  <row Id="2428" PostTypeId="2" ParentId="2407" CreationDate="2011-01-01T17:47:14.493" Score="5" ViewCount="" Body="&lt;p&gt;Let me make an attempt: LQG is a name for a collection of research programs aimed at quantizing a metric theory of gravity (usually the Einstein-Hilbert action) directly. Those attempts use different variables (starting with Ashtekar's new variables) and different techniques of quantization (e.g path integrals, canonical quantization, loop quantization, etc.) in an attempt to circumvent the problems one encounters when perturbatively quantizing the EH action around flat spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a system with finitely many degrees of freedom, unless you are making some preverse choices, quantizing a classical theory usually gives a quantum theory whose classical limit, in turn, is the theory you started from. Also, most the above quantization procedures (with the possible exception of loop quantization) usually give the same quantum theory, perhaps differing in some minor ways. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this is far from guaranteed in field theories, because of ultra violet divergences. The questions of whether when you &quot;quantize GR&quot; the resulting theory has a classical limit where it turns into classical GR in flat spacetime, or whether all the different quantizations used in LQG are equivalent to each other, those are open questions as far as I can tell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The names of different fields tend to be historical, in this case it has to do with the loop variables employed in constructing the so-called kinematical Hilbert space of LQG. Non -perturbative quantum gravity (or simply quantum gravity) usually refers to more than just LQG, for example including things like causal dynamical triangulations (CDT). All of those approaches have in common the belief that some appropriate approach to quantization will enable us to quantize the metric (or some alias thereof) directly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is in distinction to the approach titled (for similarly obscure historical reasons) &quot;string theory&quot; in which the fundamental degrees of freedom are not the metric (or the connection, or anything else appearing in the low energy description of gravity), and the low energy degrees of freedom are &quot;emergent&quot; in some appropriate sense of that word.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-01-01T19:01:08.180" LastActivityDate="2011-01-01T19:01:08.180" CommentCount="5" />
  <row Id="2429" PostTypeId="1" CreationDate="2011-01-01T19:29:38.103" Score="0" ViewCount="71" Body="&lt;p&gt;I post the question &lt;a href=&quot;http://tex.stackexchange.com/questions/8081/why-doesnt-the-color-of-overlapped-region-between-opaque-red-and-green-objects-b&quot;&gt;here&lt;/a&gt; several minutes ago.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In theory, additive mixing Red color and Green color produces Yellow color.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In my mental model, additive mixing opaque Red and opaque Green will produce opaque Yellow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the following attempt, I got a result that is different from my mental model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WHY?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/nkONV.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="652" LastActivityDate="2011-01-01T20:10:26.067" ClosedDate="2011-01-01T20:40:56.853" Title="Why DOESN'T the color of overlapped region between opaque red and green objects become opaque yellow?" Tags="&lt;color&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="2430" PostTypeId="1" AcceptedAnswerId="2433" CreationDate="2011-01-01T20:06:57.803" Score="1" ViewCount="205" Body="&lt;p&gt;I've always heard that watering plants if the temperature goes a few degrees below freezing will help prevent them from freezing, but I've never quite understood the physics behind it. Can you guys help me to understand this a bit better? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="871" LastEditorUserId="871" LastEditorDisplayName="" LastEditDate="2011-01-01T21:31:23.390" LastActivityDate="2011-01-02T18:08:33.733" Title="How does watering your plants help protect against freezing?" Tags="&lt;heat&gt;&lt;water&gt;&lt;freezing&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="2431" PostTypeId="1" AcceptedAnswerId="2434" CreationDate="2011-01-01T20:09:05.787" Score="3" ViewCount="183" Body="&lt;p&gt;I've heard that turning on faucets to a slight drip will prevent pipes from freezing, but I've never understood why this is the case. Can anyone out there help me to understand? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="871" LastEditorUserId="871" LastEditorDisplayName="" LastEditDate="2011-01-01T21:30:59.320" LastActivityDate="2011-01-01T22:00:39.983" Title="Why does a slight drip of water protect pipes from freezing?" Tags="&lt;water&gt;&lt;heat&gt;&lt;freezing&gt;" AnswerCount="2" />
  <row Id="2432" PostTypeId="2" ParentId="2429" CreationDate="2011-01-01T20:10:26.067" Score="0" ViewCount="" Body="&lt;p&gt;You've darkened the center color more, I think that's the problem. Try lightening it somewhat and I think you'll have the right color.&lt;/p&gt;&#xA;" OwnerUserId="871" LastActivityDate="2011-01-01T20:10:26.067" />
  <row Id="2433" PostTypeId="2" ParentId="2430" CreationDate="2011-01-01T20:44:41.017" Score="3" ViewCount="" Body="&lt;p&gt;If the temperature is not much below freezing, the rate of heat transfer from your plants (and particularly from the earth around their roots) is low, if there is a lot of water present, the high heat of fusion means that it will take a long time to actually freeze much of it. So maybe the plant makes it through the night without too much damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if it doesn't warm up enough the next day the second night will kill them because it starts close to freezing.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-01-01T21:07:32.097" LastActivityDate="2011-01-01T21:07:32.097" />
  <row Id="2434" PostTypeId="2" ParentId="2431" CreationDate="2011-01-01T20:47:44.213" Score="6" ViewCount="" Body="&lt;p&gt;Pipes are damaged when ice forms a complete blockage, and the expansion of water trapped by it puts too much pressure on them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, ice is a pretty good thermal insulator, so once a little ice forms on the inside of the pipe further freezing proceeds slowly. If the water is flowing there will not be enough time for it to freeze between leaving the (relatively warm and protected) underground run and exiting the pipe at the dripping faucet. Thus the pipe never freezes all the way through, no blockage forms and excess pressure is never applied.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-01-01T20:47:44.213" />
  <row Id="2435" PostTypeId="1" AcceptedAnswerId="2472" CreationDate="2011-01-01T21:32:25.303" Score="9" ViewCount="382" Body="&lt;p&gt;This is a second question (in what will probably become a series) in my attempt to understand LQG a little. In the &lt;a href=&quot;http://physics.stackexchange.com/questions/2407/give-a-description-of-loop-quantum-gravity-your-grandmother-could-understand&quot;&gt;previous question&lt;/a&gt; I asked about the general concepts behind LQG to which &lt;strong&gt;space_cadet&lt;/strong&gt; provided a very nice answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me summarize that answer so that I can check that I've understood it properly. In LQG, one works with connections instead of a metric because this greatly simplifies the equations (space_cadet makes an analogy to &quot;taking a square root&quot; of K-G equation to obtain Dirac equation). The connections should determine the geometry of a given 3D manifold which is a space-like slice of our 4D space-time. Then, as is usual in quantizing a system, one proceeds to define &quot;wave-functions&quot; on the configuration space of connections and the space of these functionals on connections should form a Hilbert space. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt; I guess there is more than one Hilbert space present, depending on precisely what space of connections we work with. This will probably have to do with enforcing the usual Einstein constraints and also diffeomorphism constraints. But I'll postpone these technicalities to a later question.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So that's one picture we have about LQG. But when people actually talk about LQG, one always hears about spin-networks and area and volume operators. So how does these objects connect with the space_cadet's answer? Let's start slowly&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;What is a spin-network exactly?&lt;/li&gt;&#xA;  &lt;li&gt;What are the main mathematical properties?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just a reference to the literature will suffice because I realize that the questions (especially the second one) might be quite broad. Although, for once, &lt;a href=&quot;http://en.wikipedia.org/wiki/Spin_network&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt; does a decent job in hinting at answers of both 1. and 2. but it leaves me greatly dissatisfied. In particular, I have no idea what happens at the vertices. Wikipedia says that they should carry intertwining operators. Intertwinors always work on two representations so presumably there is an intertwinor for every pair of edges joining at the vertex? Also, by Schur's lemma, intertwinors of inequivalent irreps are zero, so that usually this notion would be pretty trivial. As you can see, I am really confused about this, so I'd like to hear&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&amp;nbsp;3. What is the significance of vertices and intertwinors for spin-networks?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Okay, having the definitions out of the way, spin-networks should presumably also form a basis of the aforementioned Hilbert space (I am not sure which one Hilbert space it is and what conditions this puts on the spin-networks; if possible, let's postpone this discussion until some later time) so there must be some correspondence between connection functionals and spin-networks.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&amp;nbsp;4. How does this correspondence look precisely? If I have some spin-network, how do I obtain a functional on connections from that?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-02-06T08:53:45.473" Title="What is the role spin-networks play in Loop Quantum Gravity?" Tags="&lt;quantum-gravity&gt;&lt;loop-quantum-gravity&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="2" />
  <row Id="2436" PostTypeId="2" ParentId="2431" CreationDate="2011-01-01T22:00:39.983" Score="2" ViewCount="" Body="&lt;p&gt;I would say it is a simple case of heat transfer. The new water (from the mains) is above freezing (usually by 10C or more), so the flow is transfering heat from the relatively warm input water. I would also say, that the opening through the end of the pipe might act a bit as a pressure relief valve, i.e. some freezing of the contents of a closed pipe means a major increase in pressure, however if you have it slightly open the excess pressure can be relieved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said all that, I've seen how I can't use the garden on a clear morning with a temp less than roughly 40F. Radiational cooling of the hose at night, means some ice is present in the hose, and turning it on (with the hose nozzle present), usually just results in ice jamming up the nozzle. So I wouldn't count on your slow drip to not clog up with ice.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-01-01T22:00:39.983" />
  <row Id="2437" PostTypeId="2" ParentId="304" CreationDate="2011-01-01T22:35:12.710" Score="7" ViewCount="" Body="&lt;p&gt;I will go mostly with Chad's argumentation. Larger equals slower, which should excite relatively more low frequency sound. Also note that the larger the blast (hopefully) the further away the observer is. And air is not a perfectly elastic acoustic medium, some energy is lost, and the higher frequencies attenuate quicker than shorter, so distance will selectively filter out the higher frequencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, note, explosion usually means detonation. A detonation is an exothermic reaction which spreads by the compression (adiabatic) heating from the shockwave, and the chemical energy maintains the shockwave. A shockwave is essentially a highly nonlinear soundwave, and as the overpressure decays with distance from the source is will grade into a soundwave. Fireworks (pyrotechnics) are not explosives, but are the (relatively) slow reaction of chemicals (combustables, and an oxidizer) due to heat. Fireworks may generate shockwaves in air, if the package ruptures at sufficiently high pressure. Likewise volcanic blasts are not detonations, but shockwaves formed by the escape of high pressure gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an explosion is fast compared to the sound frequencies the detector is sensitive to (probably human ears in your case), then we might be able to model the explosion as a delta-function in time. A delta function should equally excite all frequencies, so it should be a simple matter of distance attenuation of sound waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An explosion close to a solid surface, creates an amazing effect I've heard called a mach-stem (although wikipedia does not produce anything useful for this term). In any case, at fixed distance the near surface shock wave is much stronger than the shockwave at height above ground. In essence the ground effect part of the shockwave weakens roughly only as 1/R rather that the 1/R**2 one would expect for a free air spherical blastwave. I don't know what effect this has on the sound spectrum (pitch), but you need to be at a much larger standoff distance from a near groundblast than from a samesized high altitude blast because of it. &lt;/p&gt;&#xA;" OwnerUserId="391" LastEditorUserId="391" LastEditorDisplayName="" LastEditDate="2011-01-02T17:36:29.770" LastActivityDate="2011-01-02T17:36:29.770" CommentCount="2" />
  <row Id="2438" PostTypeId="1" AcceptedAnswerId="2446" CreationDate="2011-01-02T02:33:24.517" Score="5" ViewCount="188" Body="&lt;p&gt;I'd like to keep a container of potable water in my car for when I get stuck, but in the winter it gets cold enough for bottled water to freeze.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't put salt in it, because then it won't be drinkable anymore (or at least, will make my condition worse if I do drink it), and I can't mix it with alcohol in case I get the car unstuck and have to drive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What else could be added to water to sufficiently lower its freezing point without making it undrinkable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Or to keep it in line with physics rather than chemistry, how do I calculate the amount of pressure the water need to be under to prevent it from freezing at a given temperature, say 15-20&lt;sup&gt;o&lt;/sup&gt; F?)&lt;/p&gt;&#xA;" OwnerUserId="961" LastEditorUserId="961" LastEditorDisplayName="" LastEditDate="2011-01-02T19:33:23.820" LastActivityDate="2011-01-02T19:33:23.820" ClosedDate="2011-01-02T23:27:55.277" Title="How can I prevent drinking water from freezing in cold weather?" Tags="&lt;water&gt;&lt;freezing&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="2439" PostTypeId="1" AcceptedAnswerId="2440" CreationDate="2011-01-02T05:23:27.497" Score="0" ViewCount="179" Body="&lt;p&gt;I apologise in advance if this question doesn't appeal to the advanced questions being asked in this Physics forum, but I'm a great fan of the Stack Exchange software and would trust the answers provided here to be more correct than that of Yahoo! Answers etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A car is travelling with a constant speed of 80km/h and passes a stationary motorcycle policeman. The policeman sets off in pursuit, accelerating to 80km/h in 10 seconds reaching a constant speed of 100 km/h after a further 5 seconds. At what time will the policeman catch up with the car?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer in the back of the book is 32.5 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steps/logic I completed/used to solve the equation were:&lt;br&gt;&#xA;- If you let x equal each other, the displacement will be the same, and the time can be solved algebraically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;br&gt;&#xA;$$x=vt$$&#xA;As the car is moving at 80km/h, we want to convert to m/s. 80/3.6 = 22.22m/s&lt;br&gt;&#xA;$$x=22.22t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the policeman, he reaches 22.22m/s in 10 seconds.&lt;br&gt;&#xA;$$\begin{aligned}&#xD;&#xA;x &amp;amp;= \frac12 (u+v) t \\&#xD;&#xA;&amp;amp;= \frac12 \times 22.22 \times 10 \\&#xD;&#xA;&amp;amp;= 111.11 \mathrm m&#xD;&#xA;\end{aligned}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The policeman progresses to travel a further 5 seconds and increases his speed to 100km/h.&lt;br&gt;&#xA;100km/h -&gt; m/s = 100 / 3.6 = 27.78m/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{aligned}&#xD;&#xA;x &amp;amp;= \frac12 (u+v) t \\&#xD;&#xA;x &amp;amp;= \frac12 \times (22.22 + 27.78) \times 5 \\&#xD;&#xA;x &amp;amp;= \frac12 \times 50 \times 5 \\&#xD;&#xA;x &amp;amp;= 250 / 2 \\&#xD;&#xA;x &amp;amp;= 125 \mathrm m&#xD;&#xA;\end{aligned}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By adding these two distances together we get 236.1m. &#xA;So the equation I have is:&lt;br&gt;&#xA;$$22.22t = 27.78t - 236.1  $$&#xA;Which solves to let t = 42.47s which is really wrong.&lt;/p&gt;&#xA;" OwnerUserId="1056" LastEditorUserId="56" LastEditorDisplayName="" LastEditDate="2011-01-02T07:14:34.927" LastActivityDate="2011-01-02T07:14:34.927" Title="Solving straight-line motion question for time" Tags="&lt;homework&gt;&lt;kinematics&gt;&lt;motion&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="2440" PostTypeId="2" ParentId="2439" CreationDate="2011-01-02T05:55:48.827" Score="2" ViewCount="" Body="&lt;p&gt;Your mistake is in the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$22.22t = 27.78t - 236.1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything up to there made good sense, but if the police officer has already traveled 236 meters, you should add that to his distance traveled, not subtract it.  You'll also need to account for the way the police officer only began traveling at full speed 15 seconds into the chase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, it is much easier to do the problem by thinking about the relative speeds.  During the first ten seconds, the car is going 80kph and the police officer is going 40kph on average.  So the police officer loses ground at an average of 40kph for 10 seconds.  We can think of this as 10 seconds' worth of loss, and ask how many seconds' worth of loss the police officer gains as he speeds up further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the next segment, the police officer gains ground at an average of 10kph for 5 seconds.  He's gaining ground 1/4 as fast as he lost it earlier and does it for 5 seconds, so this makes up for 5/4 of a second's worth of loss, leaving 8 3/4 seconds' lost ground remaining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally he gains ground at 20kph until he catches up.  He's gaining here at half the rate he was originally losing ground, so it takes him double the remaining seconds' worth time, or 17.5 seconds, to finish the pursuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method is much simpler to calculate, eliminating many opportunities for errors.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-01-02T05:55:48.827" CommentCount="5" />
  <row Id="2441" PostTypeId="2" ParentId="237" CreationDate="2011-01-02T06:54:38.607" Score="4" ViewCount="" Body="&lt;p&gt;Measure the mass. If you know it is solid aluminum, then you know its density, and you can easily calculate its volume.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-01-02T06:54:38.607" CommentCount="4" />
  <row Id="2442" PostTypeId="1" CreationDate="2011-01-02T12:11:57.847" Score="3" ViewCount="137" Body="&lt;p&gt;I'm looking for primer material on the modelling the physics of liquids. In particular I want to make a small simulation (I'm a programmer by profession) of throwing ink at a board, much like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.pond5.com/stock-footage/670934/ink-splatter.html&quot; rel=&quot;nofollow&quot;&gt;http://www.pond5.com/stock-footage/670934/ink-splatter.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm researching ways to approach the problem before I start banging on the keyboard. But I'm finding it hard to find an article about procedural generation of splatters and pools of liquid. There are a lot of search results for After Effects plugins, but this is of limited use. It could be possible that an open source AE plugin exists out there that I can analyze, but I would prefer to have a good maths based article to draw from. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My math skills are quite basic so I'm fully prepared for a hard slog on this one. I would like to make some sort of road map for myself to get to a level where I can make a simulation such as this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Guidance is much appreciated. &lt;/p&gt;&#xA;" OwnerUserId="1058" LastActivityDate="2011-01-02T13:25:38.317" Title="Primer on Liquid/Ink Splatter Physics" Tags="&lt;fluid-dynamics&gt;&lt;simulation&gt;" AnswerCount="1" />
  <row Id="2443" PostTypeId="2" ParentId="2442" CreationDate="2011-01-02T13:25:38.317" Score="3" ViewCount="" Body="&lt;p&gt;There are two separate physical considerations regarding the video you posted:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The physics of a spherical drop hitting a dry surface&lt;/li&gt;&#xA;&lt;li&gt;Multiple spherical drops hitting the surface at different times&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Once the first is solved, you should be able to simulate the second easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the first point, a &lt;a href=&quot;http://arxiv.org/find/all/1/all%3a+splash/0/1/0/all/0/1&quot; rel=&quot;nofollow&quot;&gt;quick arxiv search&lt;/a&gt; returns the following results:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/pdf/0904.0483&quot; rel=&quot;nofollow&quot;&gt;Impact of a Viscous Liquid Drop&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/pdf/0806.3050&quot; rel=&quot;nofollow&quot;&gt;Rayleigh-Plateau instability causes the crown splash&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/pdf/1005.2771&quot; rel=&quot;nofollow&quot;&gt;Thin Film Formation During Splashing of Viscous Liquids&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/pdf/1006.1588&quot; rel=&quot;nofollow&quot;&gt;How micropatterns and air pressure affect splashing on surfaces&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-01-02T13:25:38.317" CommentCount="3" />
  <row Id="2444" PostTypeId="1" CreationDate="2011-01-02T13:36:16.657" Score="0" ViewCount="197" Body="&lt;p&gt;I have some task to do:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Please write equations in Simulink and&#xA;  simulate the circuit elements&#xA;  connected in series RLC (it was the&#xA;  electrical!). Outputs of the system&#xA;  should be voltage on the capacitor and&#xA;  voltage across the resistor. Recalled&#xA;  a very polite, the equation of state&#xA;  has the form: ˙ x = Ax + Bu y = Cx +&#xA;  Du I only equation in this form will&#xA;  be considered appropriate. For&#xA;  example, a voltage boost circuit and a&#xA;  sinusoidal function Heaviside'a, and&#xA;  their sum. Results accumulate in a&#xA;  place accessible to Matlab. In Matlab&#xA;  to check the superposition principle.&#xA;  Repeat this operation for a non-zero&#xA;  initial conditions. Perform the&#xA;  frequency characteristics for&#xA;  frequencies near the resonant&#xA;  frequency for all combinations of&#xA;  input-output (it would be good to&#xA;  benefit from toolbox'a Control&#xA;  System), although this is not&#xA;  necessary. Frequency can be calculated&#xA;  only for this case and can be&#xA;  calculated based on any knowledge of&#xA;  the matrix A, B, C and D. The latter&#xA;  solution will scored higher.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I want to ask how matrixs A,B,C and D can look to solve this problem ? Who to get two output values from this (voltage on capicitor and resistor) ?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-02T20:30:11.013" ClosedDate="2011-01-02T23:13:37.150" Title="Space state equation for RLC to calc voltage on resistor and capicitor" Tags="&lt;electricity&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="2445" PostTypeId="2" ParentId="2341" CreationDate="2011-01-02T13:44:33.520" Score="2" ViewCount="" Body="&lt;p&gt;Marek asks for a &quot;simpler conceptual argument&quot;; there is one, using the basic observation that the distance travelled is linear in both the horizontal and vertical components of the initial velocity.  It's not hard to fill in the rest, but I'm on a cross-country flight right now so it will need to wait a few hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;Ok, details:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance traveled is given by $d = v_xt$.  One can either show with some algebra (as in &lt;strong&gt;Chad Orzel&lt;/strong&gt;'s answer) or otherwise that the flight time is linear in the vertical component: $t = cv_y$.  Thus, $d$ is proportional to $v_xv_y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we maximize the distance travelled by maximizing $v_xv_y$ subject to the constraint that the magnitude of the velocity vector is fixed: $|v| = c_2$.  There's a bunch of ways to see how to do this; you can use calculus, but I prefer to observe that the level sets of $f(x,y) = xy$ are hyperbolas with asymptotes on the $x$ and $y$ axes, so the maximum must occur when $v_x = v_y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even cuter is to appeal to the fact that both the constraint and the equation to maximize are quadratic, so there is either a unique maximum or the maximum occurs at one or both of the endpoints.  Since the distance travelled at the endpoints is obviously zero, there is a unique maximum.  By symmetry, it must occur when $v_x = v_y$.&lt;/p&gt;&#xA;" OwnerUserId="1061" LastEditorUserId="1061" LastEditorDisplayName="" LastEditDate="2011-01-04T01:23:57.250" LastActivityDate="2011-01-04T01:23:57.250" CommentCount="2" />
  <row Id="2446" PostTypeId="2" ParentId="2438" CreationDate="2011-01-02T14:01:01.197" Score="4" ViewCount="" Body="&lt;p&gt;It may not be the answer you are looking for but I recommend you get a thermos or a well insulated flask. These are what mountaineers use and you do not have to change the chemical composition of water this way.&lt;/p&gt;&#xA;" OwnerUserId="504" LastActivityDate="2011-01-02T14:01:01.197" />
  <row Id="2447" PostTypeId="1" AcceptedAnswerId="2449" CreationDate="2011-01-02T14:30:50.797" Score="9" ViewCount="607" Body="&lt;p&gt;Regarding general relativity:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the physical meaning of the Christoffel symbol ($\Gamma^i_{\ jk}$)?&lt;/li&gt;&#xA;&lt;li&gt;What are the (preferably physical) differences between the Riemann curvature tensor ($R^i_{\ jkl}$), Ricci tensor ($R_{ij}$) and Ricci scalar ($R$)? For example why do the Einstein equations include the Ricci tensor and scalar, but not the Riemann tensor?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To be clear, by &quot;physical meaning&quot; I mean something like - what physical effect do these components generate? Or, they make the GR solutions deviate from Newton because of xxx factor... or something similarly physically intuitive.&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2011-01-02T19:34:54.153" LastActivityDate="2011-03-19T22:48:02.283" Title="What is the physical meaning of the connection and the curvature tensor?" Tags="&lt;general-relativity&gt;&lt;differential-geometry&gt;&lt;tensor-calculus&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="3" />
  <row Id="2448" PostTypeId="2" ParentId="2444" CreationDate="2011-01-02T15:03:08.313" Score="0" ViewCount="" Body="&lt;p&gt;It's fairly obvious that the equations you are describing are the equations governing a &lt;a href=&quot;http://en.wikipedia.org/wiki/State_space_%28controls%29#Linear_systems&quot; rel=&quot;nofollow&quot;&gt;state machine&lt;/a&gt;. I could go through all of the equations to help produce it, but let me give you just a few pointers.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The first equations doesn't care at all about the outputs of the system.&lt;/li&gt;&#xA;&lt;li&gt;Everywhere you have a differential, you need to have a state variable. That would include each of the LC components.&lt;/li&gt;&#xA;&lt;li&gt;You need to relate each of the LC components into something that is a component of other state variables.&lt;/li&gt;&#xA;&lt;li&gt;The second equation relates the state variables to your outputs.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There is a paper relating electrical components to state space &lt;a href=&quot;http://www.digsilent.de/Consulting/Publications/ipst01statespace.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, you might take a look at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as modeling these in Matlab, you can use the ss command, I'll leave you to look up how to use it.&lt;/p&gt;&#xA;" OwnerUserId="871" LastActivityDate="2011-01-02T15:03:08.313" />
  <row Id="2449" PostTypeId="2" ParentId="2447" CreationDate="2011-01-02T15:18:52.197" Score="7" ViewCount="" Body="&lt;p&gt;The simplest way to explain the Christoffel symbol is to look at them in flat space.  Normally, the laplacian of a scalar in three flat dimensions is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla^{a}\nabla_{a}\phi = \frac{\partial^{2}\phi}{\partial x^{2}}+\frac{\partial^{2}\phi}{\partial y^{2}}+\frac{\partial^{2}\phi}{\partial z^{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, that isn't the case if I switch from the $(x,y,z)$ coordinate system to cylindrical coordinates $(r,\theta,z)$.  Now, the laplacian becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla^{a}\nabla_{a}\phi=\frac{\partial^{2}\phi}{\partial r^{2}}+\frac{1}{r^{2}}\left(\frac{\partial^{2}\phi}{\partial \theta^{2}}\right)+\frac{\partial^{2}\phi}{\partial z^{2}}-\frac{1}{r}\left(\frac{\partial\phi}{\partial r}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most important thing to note is the last term above--you now have not only second derivatives of $\phi$, but you also now have a term involving a &lt;b&gt;first&lt;/b&gt; derivative of $\phi$.  This is precisely what a Christoffel symbol does.  In general, the Laplacian operator is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla_{a}\nabla^{a}\phi = g^{ab}\partial_{a}\partial_{b}\phi - g^{ab}\Gamma_{ab}{}^{c}\partial_{c}\phi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of cylindrical coordinates, what the extra term does is encode the fact that the coordinate system isn't homogenous into the derivative operator--surfaces at constant $r$ are much larger far from the origin than they are close to the origin.  In the case of a curved space(time), what the Christoffel symbols do is explain the inhomogenities/curvature/whatever of the space(time) itself.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the curvature tensors--they are contractions of each other.  The Riemann tensor is simply an anticommutator of derivative operators--$R_{abc}{}^{d}\omega_{d} \equiv \nabla_{a}\nabla_{b}\omega_{c} - \nabla_{b}\nabla_{a} \omega_{c}$.  It measures how parallel translation of a vector/one-form differs if you go in direction 1 and then direction 2 or in the opposite order.  The Riemann tensor is an unwieldy thing to work with, however, having four indices.  It turns out that it is antisymmetric on the first two and last two indices, however, so there is in fact only a single contraction (contraction=multiply by the metric tensor and sum over all indices) one can make on it, $g^{ab}R_{acbd}=R_{cd}$, and this defines the Ricci tensor.  The Ricci scalar is just a further contraction of this, $R=g^{ab}R_{ab}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, due to Special Relativity, Einstein already knew that matter had to be represented by a two-index tensor that combined the pressures, currents, and densities of the matter distribution.  This matter distribution, if physically meaningful, should also satisfy a continuity equation: $\nabla_{a}T^{ab}=0$, which basically says that matter is neither created nor destroyed in the distribution, and that the time rate of change in a current is  the gradient of pressure.  When Einstein was writing his field equations down, he wanted some quantity created from the metric tensor that also satisfied this (call it $G^{ab}$) to set equal to $T^{ab}$.  But this means that $\nabla_{a}G^{ab} =0$.  It turns out that there is only one such combination of terms involving first and second derivatives of the metric tensor: $R_{ab} - \frac{1}{2}Rg_{ab} + \Lambda g_{ab}$, where $\Lambda$ is an arbitrary constant.  So, this is what Einstein picked for his field equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $R_{ab}$ has the same number of indicies as the stress-energy tensor.  So, a hand-wavey way of looking at what $R_{ab}$ means is to say that it tells you the &quot;part of the curvature&quot; that derives from the presence of matter.  Where does this leave the remaining components of $R_{abc}{}^{d}$ on which $R_{ab}$ does not depend?  Well, the simplest way (not COMPLETELY correct, but simplest) is to call these the parts of the curvature derived from the dynamics of the gravitational field itself--an empty spacetime containing only gravitational radiation, for example, will satisfy $R_{ab}=0$ but will also have $R_{abc}{}^{d}\neq 0$.  Same for a spacetime containing only a black hole.  These extra components of $R_{abc}{}^{d}$ give you the information about the gravitational dynamics of the spacetime, independent of what matter the spacetime contains.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is getting long, so I'll leave this at that.&lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2011-01-02T15:47:36.137" LastActivityDate="2011-01-02T15:47:36.137" CommentCount="20" />
  <row Id="2450" PostTypeId="2" ParentId="2197" CreationDate="2011-01-02T16:21:18.280" Score="1" ViewCount="" Body="&lt;p&gt;In my oppinion this problem is equivalent to the Tippe_top, with the difference of the egg not turning around 180 degrees but 90 degrees. The underlying physics should be the same, since the center of mass and the geomtric center are not being the same. The egg should be a little bit easier to calculate, since its geometric shape is not that horrible. (However its still way over my head). As you already pointed out, this behaviour is not possible on a frictionless table. For some readings see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tippe_top&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Tippe_top&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/chao-dyn/pdf/9501/9501008v1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/chao-dyn/pdf/9501/9501008v1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jstor.org/pss/2102374&quot; rel=&quot;nofollow&quot;&gt;http://www.jstor.org/pss/2102374&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last two links are quite an extensive discussion about the involved physics. A more basic introduction to this topic is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.physik.uni-augsburg.de/~wobsta/tippetop/literature.shtml.de&quot; rel=&quot;nofollow&quot;&gt;http://www.physik.uni-augsburg.de/~wobsta/tippetop/literature.shtml.de&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunatelly, this website is only available in German. However you might take the above links as as starting point for furhter readings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Greetings&lt;/p&gt;&#xA;" OwnerUserId="393" LastActivityDate="2011-01-02T16:21:18.280" />
  <row Id="2451" PostTypeId="1" AcceptedAnswerId="2474" CreationDate="2011-01-02T17:58:30.137" Score="3" ViewCount="379" Body="&lt;p&gt;I saw this mind boggling result that if the tires don't slip then the work done by an engine to move a car is zero. Why is this true? Moreover, what does this truly mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Sorry about not being clear, but I was talking about an idealised case where air resistance is negligible and the tires are perfectly rigid and there is no internal friction at play.&lt;/p&gt;&#xA;" OwnerUserId="1065" LastEditorUserId="1065" LastEditorDisplayName="" LastEditDate="2011-01-02T19:14:26.027" LastActivityDate="2011-01-05T05:56:11.220" Title="Why does a car engine not do work if the wheels don't slip?" Tags="&lt;classical-mechanics&gt;&lt;rotational-dynamics&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="2" />
  <row Id="2453" PostTypeId="2" ParentId="2430" CreationDate="2011-01-02T18:08:33.733" Score="1" ViewCount="" Body="&lt;p&gt;While this may be true, a better option is actually to spray your plants with water. When water freezes, it releases heat (a little counterintuitive, I know, but that's why you have to put water in a cold place to freeze it - you have to take away heat). So, if your plants have a thin layer of water on them that freezes, it actually helps keep them warmer. Additionally, the layer of ice, being a good insulator, will then help keep the plant warmer through the cold spell.&lt;/p&gt;&#xA;" OwnerUserId="861" LastActivityDate="2011-01-02T18:08:33.733" />
  <row Id="2454" PostTypeId="2" ParentId="2451" CreationDate="2011-01-02T18:17:42.560" Score="0" ViewCount="" Body="&lt;p&gt;It's fairly simple:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The work done $W$ must correspond to a change in the kinetic energy of the car $T=\frac{1}{2}mv^2$. It any work is done, there must therefore be a corresponding acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the car is moving at constant velocity, then there is no work done. The engine counterbalances the air drag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the wheels slip, then the car slows down due to drag. Work is done by air to slow down the car. The engine still performs no work on the car as it cannot possibly alter the speed if wheels slip.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-01-02T18:17:42.560" CommentCount="13" />
  <row Id="2455" PostTypeId="1" AcceptedAnswerId="2479" CreationDate="2011-01-02T18:27:52.060" Score="6" ViewCount="257" Body="&lt;p&gt;What happens to the light [energy] after it enters an eye and hits the rods and cones? I presume the energy becomes electrical, and it must be near 100% perfect, else our eyes would heat up? Or am I missing something. (the thinking behind this question is why haven't we managed to grow solar panels) &lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-03T05:23:44.683" LastActivityDate="2011-02-04T06:07:46.480" Title="What happens to light after it enters an eye" Tags="&lt;light&gt;&lt;energy-conservation&gt;&lt;biophysics&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="2456" PostTypeId="2" ParentId="2447" CreationDate="2011-01-02T18:49:20.483" Score="2" ViewCount="" Body="&lt;p&gt;Note that there is no &lt;em&gt;physical meaning&lt;/em&gt; of Christoffel symbols as they are not tensors. It's always possible to choose local coordinates such that all of $\Gamma$ vanish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But their mathematical meaning is that they form a pseudotensor. Technically, if we have two covariant derivatives $\nabla_1$ and $\nabla_2$ then their difference $\Gamma := \nabla_1 - \nabla_2$ satisfies some nice mathematical properties (namely that it is an ultralocal operator) and so its acting on any object is just local and can be represented by a tensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\nabla_1$ we usually take the covariant derivative we are interested in (e.g. a metric covariant derivative with vanishing torsion induced by some metric tensor $g$). For $\nabla_2$ there are two general (and widely used) choices. One can either use coordinate covariant derivative $\partial$ (which anihilates coordinate vector ${\partial \over \partial x}$ and covector fields ${\rm d} x$ and this gives the usual expression $\nabla = \partial + \Gamma_{Christoffel}$. The other choice (which generalizes the previous one) is a covariant derivative $\bar \partial$ which annihilates some tetrad $e$ (in the previous case we had the tetrad ${\rm d} x$ which is very specific; for general tetrad there need not exist associated coordinates). This leads to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrad_formalism&quot; rel=&quot;nofollow&quot;&gt;tetrad formalism&lt;/a&gt; and one writes $\nabla = \bar \partial + \gamma$ where $\gamma$ are Ricci rotation coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for Riemann tensor, it is once again a tensorial representation of an ultralocal operator, namely the curvature operator $R(u, v)$. This is a black box that takes two vector fields (thought of as a direction) and returns an ultralocal operator that tells you how much the space curves along those directions. More precisely, it tells you what happens with a vector if you parallel transport it along the infinitesimal polygon $0 \to u \to u+v \to v \to [u,v] \to 0$; it can be thought of as a square except that the two fields don't need to close and this is measured by their commutator $[u, v]$. So you can express it as $R(e_a, e_b) e_c = {R_{abc}}^d e_d$ and you'll obtain the usual Riemann tensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, because of the (a)symmetry of the Riemann tensor, two inequivalent contractions are possible. One of them is the trace ${R_{abc}}^c$ and this can be trivially seen to be zero for Riemann tensor derived from Levi-Civita connection (more generally for connections that preserve volume elements). The other contraction, ${R_{abc}}^a$ gives the Ricci tensor. This will be symmetric for Levi-Civita connection (because the trace of the Riemann tensor is zero and because the torsion vanishes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One useful (quite mathematical though) view of the Ricci tensor is as a &quot;laplacian of the metric&quot;, $R_{ij} \sim -{1 \over 2}\Delta g_{ij}$ and by an analogy to heat flows this relates Ricci flows which are a basic tool used in the study of Poincaré conjecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the geometrical meaning of the Ricci tensor is that it measures the deformation of the volume element in normal geodesic coordinates. These are coordinates that you can obtain around any point if you parametrize the neighborhood by geodesic flows. So the Ricci tensor measures how geodesics tend to get denser or sparser around a point in a given direction. Think about how sphere with positive curvature has less volume because its geodesics converge (they are the big circles on the sphere) than a hyperbolic space with negative curvature where geodesics diverge (there are infinitely many straight lines parallel to a given line). In particular, Ricci-flat manifolds (which are the solutions of vacuum Einstein's equations with zero cosmological constant) behave in this regard like the usual Euclidean space. You need to generalize this to Einstein manifolds (which are vacuum solutions with non-zero cosmological constant) to obtain analogues of sphere and hyperbolic space (namely, deSitter and anti-deSitter space).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a lot more to be said on these topics but I hope this will be helpful at least a little to you.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-01-02T18:49:20.483" CommentCount="2" />
  <row Id="2457" PostTypeId="2" ParentId="2451" CreationDate="2011-01-02T19:06:51.980" Score="2" ViewCount="" Body="&lt;p&gt;Unless you are talking about in the very limited sense where at the molecular level some number of molecules of the tire must rub off in order to transfer a force to the car, the statement: &quot;&lt;em&gt;if the tires don't slip then the work done by an engine to move a car is zero&lt;/em&gt;&quot; is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tires slipping in the normal English language sense has nothing to do with energy transfer or work. Where did you read that? You could have a car with gears for wheels running on a geared road that would not slip at all, there would still be work involved in accelerating the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work is defined as the amount of energy transferred by a force acting through a distance. It can also be measured by torque acting through a rotation. The work done by an engine is measured in this way by torque and rotation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You said: &quot;the work done by the engine to move a car.&quot; If a car accelerates slow enough that the tires do not slip, there is still torque being applied to the engine and wheels to cause the acceleration. Even in the case of a car that is moving at a constant speed, &lt;em&gt;in order to overcome the drag of the car moving through the air and the frictional drag of the drivetrain, the engine must apply torque to the transmission&lt;/em&gt;. The amount of energy the engine supplies exactly matches the energy drag of the air and drivetrain for a car that is moving at a constant velocity.&lt;/p&gt;&#xA;" OwnerUserId="948" LastActivityDate="2011-01-02T19:06:51.980" CommentCount="4" />
  <row Id="2458" PostTypeId="2" ParentId="2455" CreationDate="2011-01-02T19:12:58.070" Score="3" ViewCount="" Body="&lt;p&gt;Your reasoning is correct. Our eyes do heat up. The constant flow of blood is likely what carries away the heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elaborate explanation: all biological processes are ultimately thermodynamical processes. Any &quot;operation&quot; - such as a single cycle of ATP production by &lt;a href=&quot;http://en.wikipedia.org/wiki/ATP_synthase&quot; rel=&quot;nofollow&quot;&gt;ATP synthase&lt;/a&gt;, or the conversion of incident photons into metabolic energy (&lt;a href=&quot;http://en.wikipedia.org/wiki/Photosynthetic_efficiency&quot; rel=&quot;nofollow&quot;&gt;photosynthesis&lt;/a&gt;) or into neuronal impulses (vision) - can be thought of as an engine with a certain efficiency $e$. This is traditionally given by the ratio of the amount of &lt;em&gt;useful work&lt;/em&gt; $W$ the engine performs in a single cycle to the total amount of energy $E$ fed into it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ e = \frac{W}{E} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since $E = W + Q$, where $Q$ is the waste heat released into the environment, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ e = 1 - \frac{Q}{E} $$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part of your question as I interpret it is: &quot;&lt;em&gt;What pressures of natural selection lead to the evolution of vision&lt;/em&gt;?&quot;. The simplest answer I can think of is the exposure of photo-reactive compounds to a constant bath of photons (from the Sun) over millions of years lead to the evolution of different mechanisms to exploit this energy, one of those mechanisms being that of vision. Of course, this is not &lt;em&gt;really&lt;/em&gt; an answer. Maybe somebody with greater knowledge of evolutionary biology can shed more light on this aspect.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-02T19:12:58.070" CommentCount="5" />
  <row Id="2459" PostTypeId="2" ParentId="2455" CreationDate="2011-01-02T19:32:25.997" Score="2" ViewCount="" Body="&lt;p&gt;You are correct, that the light becomes an electrical charge from our eyes, and the resulting signal is processed by  a very complex system. The rest of it will heat up our eyes, but realize that the same thing is true of light hitting our body anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The signal generated from our eyes is minimal, I wouldn't use the word solar panel in the slightest to describe the affect at all.&lt;/p&gt;&#xA;" OwnerUserId="871" LastActivityDate="2011-01-02T19:32:25.997" />
  <row Id="2460" PostTypeId="1" CreationDate="2011-01-02T20:24:28.443" Score="1" ViewCount="95" Body="&lt;p&gt;I have a noob question about possible effects of time-travel, supposing it's physically possible, relatively to random and pseudo-random events.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Suppose the human race builds a time-travel machine like the ones we see in science-fiction movies&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;and suppose this machine has the ability to bring a person back of any amount of time, from seconds to centuries, with the possibility of some kind of &lt;em&gt;stealth-landing&lt;/em&gt; (one could &lt;em&gt;appear&lt;/em&gt; in the target time/place without necessarily alerting people, ie. no yet any interferece with events).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I won't make assumptions on the nature of &lt;em&gt;space-time&lt;/em&gt;, especially for what concerns the &lt;em&gt;multiverse&lt;/em&gt; theory and the single timeline theory (like we saw in some movies, one could change the storyline with his actions, note 1). Let's just suppose we can go back in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of human events are dicated by choice of people, and then are &lt;strong&gt;deterministic&lt;/strong&gt;. For example, I might assume if we go back some years and live without interfering with the &lt;em&gt;local history&lt;/em&gt;, we would still see &lt;em&gt;Barack Obama&lt;/em&gt; and &lt;em&gt;Mahmud Ahmadinejad&lt;/em&gt; at their respective political places. If no direct interference is done to storyline, there is no reason to doubt that the political events of the world (not only political of course, that's just an example) repeat unchanged like the time-traveler knew in his time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This because I took as example the political polls, which are dictated by volountary choice of people. Suppose now the time-traveler goes back to Berlin stadium the day of 2006 football World Cup Finals. Remember how it got? Italy won the match by &lt;strong&gt;penalty shoot-outs&lt;/strong&gt; thanks to a &lt;strong&gt;lucky shot&lt;/strong&gt; by defender Fabio Grosso. &lt;strike&gt;It's&lt;/strike&gt; I think it's a common opinion that football's penalty kicks are affected by uncertainity, since the goalkeeper must &lt;em&gt;randomly&lt;/em&gt; decide where to jump to catch the ball.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;First part of question:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Do you think that the time-traveler will see the match as it exactly occurred? When the traveler is at the stadium, looking at the game, the &lt;em&gt;goal-event&lt;/em&gt; didn't happen yet, and since there is a certain probability that the player will score, could it be possible that he fails this other time? Sorry for having made the example of shoot-outs since coin toss is more random than those, but I couldn't find a FIFA final, or any other important football match, won by coin toss after continous draw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, another example, about pseudo-randomness. We all know that computer-generated random numbers are &lt;em&gt;pseudo-random&lt;/em&gt;, ie. perfectly deterministic but merely impossible to predict without deep information on the algorithm and its status. As an example, now, I would like to take two popular lotteries: &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Superenalotto&quot; rel=&quot;nofollow&quot;&gt;Superenalotto&lt;/a&gt;&lt;/strong&gt; and &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Win_For_Life&quot; rel=&quot;nofollow&quot;&gt;Win For Life&lt;/a&gt;&lt;/strong&gt;. At least for the Italian version of WFL, draw is done hourly by a computer (so pseudo-randomness), but the Superenalotto is mechanically drawn.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Second part of the question&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Do you agree that a time-traveler will &lt;strong&gt;surely&lt;/strong&gt; (again, supposing no direct or &lt;strong&gt;indirect&lt;/strong&gt; interference, note 2) observe the same combinations of Win For Life being drawn as he could observe in his original timeline? And, like the football example above, will possibly the time-traveler observe different combinations for Superenalotto since &lt;em&gt;the drawing-event didn't happen yet&lt;/em&gt; at the time he sees the draw in the past timeline?&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Third part of the question&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Supposing my &lt;em&gt;theory&lt;/em&gt; (italicized) is correct, would this give strength to the multiverse theory? So, a time-traveler won't actually go back in time in our space-time, but instead go into another parallel universe, to a certain point in the space-time of that unviverse, &lt;strong&gt;and observe &lt;em&gt;again&lt;/em&gt; all the chain of events&lt;/strong&gt;, affected by deterministic and random events, as they haven't happened yet?&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;What do you think about all this?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Note 1 - example: in Star Trek ENT, 4x01-02, killing Lenin brought Nazis to grow their power&#xA;Note 2: surely if the time-traveler has a chat with the employee that is in charge of powering up WFL servers he could &lt;em&gt;push the button&lt;/em&gt; even a fraction of seconds earlier/later and have the software initialize its random seed to a different number, or any other kind of indirect interference could lead to this&lt;/p&gt;&#xA;" OwnerUserId="1066" LastActivityDate="2011-01-02T20:24:28.443" ClosedDate="2011-01-02T20:32:17.263" Title="Time-travel and random events" Tags="&lt;time-travel&gt;" CommentCount="7" />
  <row Id="2461" PostTypeId="2" ParentId="2444" CreationDate="2011-01-02T20:30:11.013" Score="0" ViewCount="" Body="&lt;p&gt;I think matrix looks:&#xA;A = [0, 1; -1/(L*C), -R/L]&#xA;B = [0; 1/L]&#xA;C = [0; -R]&#xA;D = 0&#xA;But what with initial conditions ? And how to get voltage(on capicitor and resistor) from this ?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-02T20:30:11.013" />
  <row Id="2462" PostTypeId="2" ParentId="2435" CreationDate="2011-01-02T20:55:33.697" Score="1" ViewCount="" Body="&lt;p&gt;I have found the LQG introduction references here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.physicsforums.com/showpost.php?p=2868468&amp;amp;postcount=244&quot; rel=&quot;nofollow&quot;&gt;http://www.physicsforums.com/showpost.php?p=2868468&amp;amp;postcount=244&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to be quite useful. Marcus is a very nice guy, knows most everything going on in the LQG community, has followed all the LQG papers since 2003 or so, and is very helpful. I don't know if he hangs out here or not yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Baez reference marcus lists:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Baez on spinfoams &lt;a href=&quot;http://arxiv.org/abs/gr-qc/9709052&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/gr-qc/9709052&lt;/a&gt; (see especially pages 1-4 and 33,34)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;may be what you want. As marcus points out, it is dated, and the issues Baez points out have been largely covered by more recent research, but the explanations Baez offers are among the easiest to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should further point out that most of the research in LQG has converged on the idea of using spinfoam not spin networks per se. I'm not sure I completely understand the difference yet, but that's my impression.&lt;/p&gt;&#xA;" OwnerUserId="948" LastActivityDate="2011-01-02T20:55:33.697" CommentCount="1" />
  <row Id="2463" PostTypeId="2" ParentId="770" CreationDate="2011-01-02T21:02:58.027" Score="10" ViewCount="" Body="&lt;p&gt;Kill all the turtles on which it rests.&lt;/p&gt;&#xA;" OwnerUserId="948" LastActivityDate="2011-01-02T21:02:58.027" CommunityOwnedDate="2011-01-07T14:13:00.227" CommentCount="3" />
  <row Id="2464" PostTypeId="1" CreationDate="2011-01-02T21:04:22.277" Score="3" ViewCount="259" Body="&lt;p&gt;As per Newton objects with mass attract each other, and per Einstein this is further explained by saying that mass warps space-time. So a massive object makes a &quot;dent&quot; into space-time,&#xA;a gravity well. I have taken to visualizing this as placing a object on a rubber sheet and the&#xA;resulting dent, being the gravity field. So obviously placing two objects&#xA;on the sheet not to far from each other will make the dents  overlap, and the object will roll towards each other.&#xA;BUT for a BLACKHOLE, this is not a dent. It's a cut or rupture in the rubber sheet.&#xA;Furthermore, space-time is constantly falling INTO the blackhole, and everything else&#xA;that exists in space-time, including light.&#xA;So a blackhole is not just a super-massive object, it's really a hole, and how can a hole move?&#xA;How does it react to the gravity pull of a nearby object, when everything just falls thru it?&#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1070" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-03T05:22:14.480" LastActivityDate="2011-01-03T05:22:14.480" Title="How do you explain the observed fact that &quot;black hole&quot; objects move?" Tags="&lt;general-relativity&gt;&lt;black-holes&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="2465" PostTypeId="2" ParentId="2455" CreationDate="2011-01-02T21:20:32.410" Score="0" ViewCount="" Body="&lt;p&gt;&quot;The second part of your question as I interpret it is: &quot;What pressures of natural selection lead to the evolution of vision?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;algae grows at sunny spots. having a means to detect these sunny spots gives a greater&#xA;chance of having more food and thus more offspring.&lt;/p&gt;&#xA;" OwnerUserId="1070" LastActivityDate="2011-01-02T21:20:32.410" />
  <row Id="2466" PostTypeId="2" ParentId="2464" CreationDate="2011-01-02T22:41:33.097" Score="5" ViewCount="" Body="&lt;p&gt;Your mental picture is pretty correct except for the part &quot;space-time is falling into the blackhole&quot;. This is completely wrong. Space-time is not falling anywhere. Consider Schwarzschild's solution. This one is static. So it's obvious that &lt;em&gt;nothing&lt;/em&gt; happens with space-time at all, it just sits there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, you wouldn't be able to distinguish the black hole from an usual object of the same mass just by its gravitational effects on other distant objects alone. For objects that are far away from the BH the good old Newtonian limit will apply and it will attract them as described by Newton's gravitational law. In particular, it will attract another BH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture you had in mind about the attraction of two objects as dents in rubber sheet that get closer is correct. This continues to hold even with black holes except that in this case it is not matter that bends the space-time but instead it's a self-sustaining space-time configuration. So a BH can in fact be thought as a special form of matter.&#xA;Also, never mind that the sheet is ruptured at the singularity because on the outside of the horizon the space-time is perfectly regular and the inside of the horizon is causally disconnected from outside observers anyway (meaning that black hole is indeed black).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Einstein's equations will tell you how this curved space-time, which describes e.g. two (or more) black holes, will evolve into the future. Besides the usual Newtonian scattering you'll get relativistic corrections for their motion, you'll get gravitational waves flying away and other effects. E.g. the two black holes could join into one bigger hole.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-01-02T23:46:36.957" LastActivityDate="2011-01-02T23:46:36.957" CommentCount="8" />
  <row Id="2467" PostTypeId="2" ParentId="2464" CreationDate="2011-01-03T00:48:21.977" Score="4" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How do you explain the observed fact that “blackhole” objects move?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$G_{\mu\nu} = 8\pi T_{\mu\nu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Einstein's equation allows for black holes to move when viewed from an external reference frame. That's all the explanation we need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I mean to say is that you're taking the rubber sheet analogy too far. It doesn't &lt;em&gt;explain&lt;/em&gt; anything. All it's supposed to do is give you a sense of how coordinates can be distorted &lt;em&gt;around&lt;/em&gt; a massive object or any gravity well, like a black hole. It doesn't really work when you want to figure out what's going on at the singularity, because rubber sheets in real life don't have singularities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black hole is not literally a &quot;hole&quot; in spacetime; it's just an infinitely deep gravity well. There's really no fundamental qualitative difference between the gravitational field of a black hole and that of a less dense object (they're both described by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_metric&quot; rel=&quot;nofollow&quot;&gt;Schwarzschild metric&lt;/a&gt;); it's just that the black hole happens to be of a high enough density that it prevents light from escaping within a certain surface. Probably a better (though still imperfect) way to relate it to the rubber sheet analogy is to say that a black hole is not a rupture in the sheet, it's just a dent that's deep enough and small enough so that the slope of the sheet exceeds some critical value. (General relativity predicts that whenever matter forms a black hole, it will all fall inward to a central point, creating a singularity, which could be represented in the rubber sheet analogy as a dent which is infinitely deep at a central point.)&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-01-03T00:48:21.977" CommentCount="3" />
  <row Id="2468" PostTypeId="1" AcceptedAnswerId="2556" CreationDate="2011-01-03T01:02:46.633" Score="2" ViewCount="388" Body="&lt;p&gt;Reading a &lt;a href=&quot;http://physics.stackexchange.com/questions/2460/time-travel-and-random-events-closed&quot;&gt;previous closed question&lt;/a&gt; an interesting variation has come to my mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that time travel to the past was possible:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I wait for an atom to decay and measure the time, $t_{1a}$ &lt;/li&gt;&#xA;&lt;li&gt;I travel back in time at $t_0&amp;lt;t_{1a}$&lt;/li&gt;&#xA;&lt;li&gt;I wait for the same atom to decay and measure the time, $t_{1b}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let's think about the two values, $t_{1a}$ and $t_{1b}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they coincide, then from my point of view the measured $t_1$ time would not be governed by chance (the second time I would know $t_1$ &lt;em&gt;a priori&lt;/em&gt;). It would therefore prove some form of existence of hidden variables. Would this violate any known laws of quantum mechanics? Would this prove the existence of hidden variables (e.g. Bohm's interpretation)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they are different, then what to make of $t_{1a}$ and $t_{1b}$? Which would be the &lt;em&gt;correct&lt;/em&gt; value? To me this doesn't make any sense, but maybe it could be compatible with the multiverse interpretation (I don't know how though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, does this gedankenexperiment:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;...help us select viable QM interpretations and exclude others? Which?&lt;/li&gt;&#xA;&lt;li&gt;...lead us to conclude that backwards information time travel is incompatible with QM?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="66" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-14T17:26:14.710" LastActivityDate="2011-01-14T17:26:14.710" Title="Time travel and nuclear decay" Tags="&lt;quantum-mechanics&gt;&lt;epistemology&gt;&lt;time-travel&gt;&lt;measurement-problem&gt;" AnswerCount="3" CommentCount="30" FavoriteCount="1" />
  <row Id="2469" PostTypeId="1" AcceptedAnswerId="2470" CreationDate="2011-01-03T01:40:13.770" Score="0" ViewCount="157" Body="&lt;p&gt;Some friends and I were at Disneyland this past week.  We spent an hour arguing back and forth whether it is possible for someone to throw an NFL football over the Matterhorn?  I think it is, but no one else really thought it was possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you guys think?  Is this humanly possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Matterhorn is 146 feet (44.5 metres) tall. Not sure how wide the base of the mountain is. It's ok if the football hits the other side of the mountain (as long as it clears the top).&lt;/p&gt;&#xA;" OwnerUserId="1073" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-03T04:40:00.940" LastActivityDate="2011-01-03T04:40:00.940" Title="Can a human throw a football over the Matterhorn at Disneyland?" Tags="&lt;newtonian-mechanics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="2470" PostTypeId="2" ParentId="2469" CreationDate="2011-01-03T01:55:31.287" Score="1" ViewCount="" Body="&lt;p&gt;I love manly arguments!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.youtube.com/watch?v=1FuC3jnm4TE&quot; rel=&quot;nofollow&quot;&gt;this hail mary pass&lt;/a&gt;, it looks like, with initial angle 45deg and velocity $\sim 30$ m/s, the best football pass can go 70 yards in 4 sec hang time. If he had thrown it straight up, the maximum height would have been about 50 yards, just &lt;em&gt;barely&lt;/em&gt; making it. Give him a running start and no 300-lb defenders to worry about, and I say it's possible!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best NFL punts have hang time of 6-8 secs, so they also would have just made it. &lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2011-01-03T01:55:31.287" CommentCount="1" />
  <row Id="2471" PostTypeId="2" ParentId="1438" CreationDate="2011-01-03T03:29:21.747" Score="1" ViewCount="" Body="&lt;p&gt;One problem I had in understanding reciprocal space is the origin in a real space unit cell is at infinity in reciprocal space and vise-versa. How could any real construct be mapped out in reciprocal space in a finite cell that included the origin?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it is important to remember that we tend to deal with &lt;em&gt;waves&lt;/em&gt; in reciprocal space. So for example, when crystallographers use x-rays, neutron diffraction etc. to obtain a pattern in reciprocal space, simply applying the reciprocal conversion (i.e. $2\pi/a$) does not obtain the positions of the atoms, but rather the &lt;em&gt;wavelengths&lt;/em&gt; of a set of waves that will denote the positions of the atoms. A wave with wavenumber zero will have infinite wavelength (i.e. is not a wave [is this correct?])&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is basically saying what the others have said about the reciprocal space being the Fourier transform of the real space - waves do not have position that is so easily defined, but they do have well defined wavelength!&lt;/p&gt;&#xA;" OwnerUserId="1072" LastActivityDate="2011-01-03T03:29:21.747" />
  <row Id="2472" PostTypeId="2" ParentId="2435" CreationDate="2011-01-03T08:14:54.990" Score="4" ViewCount="" Body="&lt;p&gt;So what are spin-networks? Briefly, they are graphs with representations (&quot;spins&quot;) of some gauge group (generally SU(2) or SL(2,C) in LQG) living on each edge. At each non-trivial vertex, one has three or more edges meeting up. What is the simplest purpose of the intertwiner? It is to ensure that angular momentum is conserved at each vertex. For the case of four-valent edge we have four spins: $(j_1,j_2,j_3,j_4)$. There is a simple visual picture of the intertwiner in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture a tetrahedron enclosing the given vertex, such that each edge pierces precisely one face of the tetrahedron. Now, the natural prescription for what happens when a surface is punctured by a spin is to associate the Casimir of that spin $ \mathbf{J}^2 $ with the puncture. The Casimir for spin $j$ has eigenvalues $ j (j+1) $. You can also see these as energy eigenvalues for the quantum rotor model. These eigenvalues are identified with the area associated with a puncture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for the said edges and vertices to correspond to a consistent geometry it is important that certain constraints be satisfied. For instance, for a triangle we require that the edge lengths satisfy the triangle inequality $ a + b \lt c $ and the angles should add up to $ \angle a + \angle b + \angle c = \kappa \pi$, with $\kappa = 1$ if the triangle is embedded in a flat space and $\kappa \ne 1$ denoting the deviation of the space from zero curvature (positively or negatively curved).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a similar manner, for a classical tetrahedron, now it is the sums of the areas of the faces which should satisfy &quot;closure&quot; constraints. For a quantum tetrahedron these constraints translate into relations between the operators $j_i$ which endow the faces with area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for a triangle giving its three edge lengths $(a,b,c)$ completely fixes the angles and there is no more freedom. However, specifying all four areas of a tetrahedron &lt;em&gt;does not&lt;/em&gt; fix all the freedom. The tetrahedron can still be bent and distorted in ways that preserve the closure constraints (not so for a triangle!). These are the physical degrees of freedom that an intertwiner possesses - the various shapes that are consistent with a tetrahedron with face areas given by the spins, or more generally a polyhedron for n-valent edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the key players in this arena include, among others, Laurent Friedel, Eugenio Bianchi, E. Magliaro, C. Perini, F. Conrady, J. Engle, Rovelli, R. Pereira, K. Krasnov and Etera Livine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this provides some intuition for these structures. Also, I should add, that at present I am working on a review article on LQG for and by &quot;the bewildered&quot;. &lt;strong&gt;I reserve the right to use any or all of the contents of my answers to this and other questions on physics.se in said work, with proper acknowledgements to all who contribute with questions and comments&lt;/strong&gt;. This legalese is necessary so nobody comes after me with a bullsh*t plagiarism charge when my article does appear :P&lt;/p&gt;&#xA;" LastActivityDate="2011-01-03T08:14:54.990" CommentCount="7" />
  <row Id="2473" PostTypeId="2" ParentId="2377" CreationDate="2011-01-03T08:43:36.110" Score="-2" ViewCount="" Body="&lt;p&gt;The behaviour of the universe is random. Truly random. But there exists order. Patterns emerge from nothing. That's what Chaos Theory tell us. Its completely unpredictable.&lt;/p&gt;&#xA;" OwnerUserId="298" LastActivityDate="2011-01-03T08:43:36.110" CommentCount="2" />
  <row Id="2474" PostTypeId="2" ParentId="2451" CreationDate="2011-01-03T09:00:26.360" Score="4" ViewCount="" Body="&lt;p&gt;The answer to your question depends on precisely how it is interpreted.  In my opinion, the clearest way of understanding a car driving on the road does in fact have the engine doing work on the car, but it is possible to define the system involved such that this is is not so.  However, under this interpretation, the engine does no work on the car regardless of whether or not the tires are slipping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain this distinction, I'll first give an example of work, then talk a little about work on the driving car, then finally discuss no-slip tires.  I think this is a fairly-tricky issue, so please excuse the length of my response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At Anna's request, let's assume no air drag, no internal friction, and no slipping.  If the car is moving at a constant velocity, then the engine can be turned off, allowing the car to coast.  There is clearly no work done and no change in energy in this scenario, so we'll talk about what happens as the car accelerates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Work_(physics)&quot; rel=&quot;nofollow&quot;&gt;Work&lt;/a&gt;&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In physics, &quot;work&quot; is a way of transferring energy into or out of a system.  We say that a force does work on a system, thus adding energy to it.  Work is defined by the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\textrm{d}W = \vec{F}\cdot \textrm{d}\vec{x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left hand side, $\textrm{d}W$, means &quot;a small amount of work done&quot;.  On the right hand side, $\vec{F}$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Force&quot; rel=&quot;nofollow&quot;&gt;force&lt;/a&gt; that is doing the work, and $\textrm{d}\vec{x}$ is the movement of the part of the system where the force is applied.  This does not particularly look like a change in energy, but it is possible to show, using Newton's laws, that work, as defined above, turns out to be the same as the change in &lt;a href=&quot;http://physics.stackexchange.com/questions/535/why-does-kinetic-energy-increase-quadratically-not-linearly-with-velocity&quot;&gt;kinetic energy&lt;/a&gt; induced by the force.  (see &lt;a href=&quot;http://uccpbank.k12hsn.org/courses/APPhysicsCI/course%20files/readings/lesson7studysheet.pdf&quot; rel=&quot;nofollow&quot;&gt;work-energy theorem&lt;/a&gt;)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to keep in mind for later that although work done on a system induces a change in its kinetic energy, not all changes in kinetic energy are due to work being performed.  For example, if I have a gerbil in a cage, and the gerbil starts running, the cage didn't go anywhere, so $\textrm{d}\vec{x} = 0$ for the system (gerbil + cage), so the work done on the cage/gerbil system is zero.  Nonetheless, it gained kinetic energy.  This is an internal conversion of energy from chemical energy in the gerbil to kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example of work, suppose you put your car in neutral and I stand behind it and push.  If I push with a force of $100$ newtons, I do $1$ joule of work for each centimeter the car moves.  In this example, $\textrm{d}W = 1 \textrm{J}$, $\vec{F} = 100 \textrm{N}$, and $\textrm{d}\vec{x} = 1 \textrm{cm}$.  The &quot;$\cdot$&quot; symbol means the vector dot product.  As long as the force and the motion are in the same direction, this is the same as normal multiplication (but if the force is at an angle to the motion, the work goes down; we won't worry about this in the rest of the answer).  This work is the change in the kinetic energy of the car; the car speeds up some while I push.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important to define the system you're doing work on.  In the example above, I considered the system to be your car.  Then I applied an external force with my body.  In an alternative view, we could define a system that consists of the car and me together.  Then I didn't do any work on the system at all since, by definition, only external forces can do work.  Finally, we might consider the system to be just the back half of the car.  In this case, I'm still the external force and I did the same amount of work, but all that work went into the back half of the car.  The back half of the car then in turn did some work on the front half of the car.  This tells us that when I push on the back of the car, the car compresses a little, creating a force between the front half and back half that's half the size of the force I'm pushing with (since half the work done on the back half goes into the front half, giving those two halves the same change in speed).  This is really true.  If you slice your car in half, then hold the two halves together with some slices of white bread, when I start pushing on the back of the car the white bread will get crushed.  Thus, the internal stresses in a car are different for front-wheel-drive and rear-wheel-drive vehicles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Net Work on the Car&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take our system to be the entire car as it drives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The car might be going at $45 \textrm{mph}$ or $20$ meters/second.  In that case, in one second the car travels $20 \textrm{m}$.  If, in one second, the car goes from $20 \textrm{m}/\textrm{s}$ to $21 \textrm{m}/\textrm{s}$, and the car has a mass of $1000 kg$, then the change in the car's kinetic energy works out to $20,500 \textrm{J}$.  Assuming this is work done by an external force (me pushing, for example, although it's unlikely!) the work done on the car is $20,500 \textrm{J}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;No-Slip Tires&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above analysis is wrong - the assumption that work is done by an external force is incorrect.  The car is accelerating, so there must be an external force on it.  But it turns out that external force does no work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The external force on the car is the force from the road on the tires.  Because the tires are not moving with respect to the road, this force does not do work.  Thus, the road does no work on the car.  This is as it should be.  Everyone knows the road is not making your car go.  The road cannot do work on your car because the road has no energy to give up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, there is a force equal and opposite from your tires to the road, but again this force does no work.  This says that your car does not put any energy into the road.  This allows a car engine to be more efficient than a jet engine.  A jet engine works by pushing air backwards.  This air is moving with respect to the jet, and so the jet does considerable work on the air.  That means that lots of the energy used by the jet engine goes into the air, and is wasted from the viewpoint of making the jet go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I'm pushing the car, my feet don't move while in contact with the road, so again no work is done on us.  In both the case of me pushing (as long as I'm part of the system) and the car driving, the car's increase in kinetic energy is not due to external work, but to a conversion of chemical energy to kinetic energy completely within the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;The Engine&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the engine is inside the car, if the system we consider is the car itself, the engine does no work on the car.  A system can't do work on itself, by definition.  Instead, the engine spins an axle that spins the tires, adding kinetic energy to the car, but not doing work.  This is why it's possible to say the engine does no work, and it is not dependent on whether or not the tires are slipping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the kinetic energy of the car is changing, we might learn something more by redefining the system.  If we define two systems, one being the body of the car minus the engine, and the other being the engine, now work is indeed being done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The engine does work on the car.  It does this by spinning an axle.  To spin the axle, the engine must exert force on the outside edge of the axle, and since the outside edge of the axle is moving, the engine does work.  (As inflector pointed out in his answer, this work is more easily calculated in terms of the torque on the axle as it rotates, but the two descriptions are equivalent).  All the energy the car gains comes through the engine, so in this scenario the engine does all the work to accelerate the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The car also does work on the engine in this case.  The car exerts a force on the engine.  This is a structural force that keeps the engine in place inside the car.  The car is paying back a portion of the energy it gained when the engine did work on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Conclusion&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two different people (in the same reference frame) can observe the same events, but disagree on whether work is done during them, or on how much work is done.  This is because work is a transfer of energy between systems, so that if the two people define their systems differently, they will calculate different values for work done (and for heat exchange).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In light of this ambiguity, it is correct, although incomplete, to state that the engine does no work.  This is not dependent on the no-slip tires, and is an artifact of defining systems such that the engine and car are part of the same system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No slip tires imply that the road does no work on the car, and that the car does no work on the road.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-01-03T09:20:42.303" LastActivityDate="2011-01-03T09:20:42.303" CommentCount="8" />
  <row Id="2475" PostTypeId="1" AcceptedAnswerId="2495" CreationDate="2011-01-03T09:56:31.150" Score="6" ViewCount="162" Body="&lt;p&gt;Does the weak nuclear force play a role (positive or negative) in nuclear binding?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally you only see discussions about weak decay and flavour changing physics, but is there a contribution to nuclear binding when a proton and neutron exchange a $W^\pm$ and thus exchange places? Or do $Z$ exchanges / neutral currents contribute?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the strength of this interaction so small that it's completely ignorable when compared to the nuclear binding due to residual strong interactions?&lt;/p&gt;&#xA;" OwnerUserId="429" LastEditorUserId="429" LastEditorDisplayName="" LastEditDate="2011-01-04T11:05:12.783" LastActivityDate="2011-01-04T11:05:12.783" Title="Weak contribution to nuclear binding" Tags="&lt;particle-physics&gt;&lt;quantum-field-theory&gt;&lt;nuclear-physics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="2476" PostTypeId="1" CreationDate="2011-01-03T12:53:39.287" Score="3" ViewCount="258" Body="&lt;p&gt;Is it possible to counter-act g-force for a jet-pilot, by him putting on a scuba-diving suit&#xA;and filling the cockpit with water? On earth we are constantly pulled down, or accelerated with one g. In this situation, if we put the jet-pilot in a pool, he would neither sink nor float.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we could increase Earth's gravity to say 9 g's, the pilot in the pool would float even more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? Thanks...&lt;/p&gt;&#xA;" OwnerUserId="1070" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2011-01-03T19:54:25.003" LastActivityDate="2011-01-03T19:54:25.003" Title="Counteracting g-force" Tags="&lt;buoyancy&gt;&lt;equivalence-principle&gt;" AnswerCount="2" />
  <row Id="2477" PostTypeId="2" ParentId="2476" CreationDate="2011-01-03T13:09:20.640" Score="3" ViewCount="" Body="&lt;p&gt;If you fill the cockpit with water, the pilot will feel a buoyant force.  Humans have about the same density as water, so ignoring the scuba suit, the pilot will feel a buoyant force about equal to his own weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plane's maneuvers don't change this result much.  By the &lt;a href=&quot;http://physics.stackexchange.com/questions/2425/why-is-the-equivalence-principle-so-important-to-general-relativity&quot;&gt;equivalence principle&lt;/a&gt;, when the plane accelerates, the water in the cockpit and the pilot both get heavier.  The buoyancy changes in the same way the pilot's weight does.  &lt;strike&gt;If he feels weightless before takeoff, he'll continue feeling weightless during maneuvers.&lt;/strike&gt;  If he's supported by buoyancy before takeoff, he'll continue to be supported by buoyancy during maneuvers.  (This assumes that the jerk, or time derivative of acceleration, is low enough that the water can remain in hydrostatic equilibrium during the maneuver.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although basic buoyancy works the same, there will be a pressure change during accelerations.  If the plane is accelerating up at 9g, he'll feel pressure as if he's under water ten times as &lt;strike&gt;deep as he really is&lt;/strike&gt; dense as the water really is.  As Bruce pointed out, this is a big pressure gradient, and the pilot will still realize that the force beneath him pushing up is stronger than the force above pushing down.  He will essentially feel heavy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring compression, changing Earth's gravity won't change whether something sinks or floats.  Something more dense than water sinks.  Something less dense floats.  If we multiply gravity by some constant, an object immersed in water has the net force on it also multiplied by that constant.  So if you have something that floats and has a net force on it of $1N$ upward under normal gravity, if you double the gravity, it will experience a net upward force of $2N$ and have double the acceleration (ignoring drag).&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-01-03T17:41:01.283" LastActivityDate="2011-01-03T17:41:01.283" CommentCount="12" />
  <row Id="2478" PostTypeId="2" ParentId="2476" CreationDate="2011-01-03T13:38:39.890" Score="2" ViewCount="" Body="&lt;p&gt;Your logic is mostly precise, but it won't counteract g-forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(I'll assume the jet flyes straight up, for simplicity, but it doesn't make any difference.)&lt;/em&gt;&#xA;Like you said, if the pilot was originally in equilibrium inside the water and we accelerate the jet at 9g, the buoyancy would get 10 times as strong (9+1). However, the pilot will still feel a force accelerating him. The only difference is that, instead of being pushed by the jet seat, he will be pushed up by the water itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So he will still feel the g-force, only a little differently.&lt;/strong&gt; Instead of the force being concentrated on the area of contact between him and the seat, it will be distributed through the area of contact between him and the water (approximately half of it, to be more precise). But the net force will be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final effect will be that he'll float inside the water. The water is accelerating up with the jet, and he'll accelerate up as well (because of all the extra buoyancy).&lt;br&gt;&#xA;If he were submersed in a heavier fluid to start with, the effect would be the same, except he would float partially above water.&lt;br&gt;&#xA;If he were submersed in a lighter fluid to start with, then he would be able to touch his jet seat, and the upward acceleration would come partially from the seat and partially from the liquid.&lt;/p&gt;&#xA;" OwnerUserId="724" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2011-01-03T15:02:22.330" LastActivityDate="2011-01-03T15:02:22.330" CommentCount="19" />
  <row Id="2479" PostTypeId="2" ParentId="2455" CreationDate="2011-01-03T13:45:56.427" Score="3" ViewCount="" Body="&lt;p&gt;There is some heating that takes place, but the amount is pretty trivial, because there just isn't that much light reaching the back of your eye. A back-of-the-envelope sort of estimate would be to say that the light of the Sun reaching the Earth's surface amounts to about a kilowatt of radiation per square meter. Your pupils have a radius of maybe a millimeter, probably much less in bright sunlight. So, if you're staring directly at the Sun (which, hopefully, you're not really doing), you're getting at most a few milliwatts delivered to the back of your eye. That's not going to tax the temperature regulation systems in your body, given that a living human generates about the same heat as a hundred-watt light bulb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you dramatically increase the amount of light delivered to your eye, say by accidentally catching a high-power pulsed laser in the eye, you can overwhelm the body's ability to carry away the heat, and do real damage. The pulsed-laser lab next to the office where I did my undergrad thesis research had a sign on the door explaining in gruesome detail what would happen if you were to catch a full YAG laser pulse in the eye, which involved the boiling retina basically blasting your eyeball out of your skull. Which is why you wear safety glasses in those labs, and knock before entering any optics lab.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2011-01-03T13:45:56.427" CommentCount="2" />
  <row Id="2480" PostTypeId="1" CreationDate="2011-01-03T13:55:55.827" Score="3" ViewCount="254" Body="&lt;p&gt;What happens if you blow bubbles into a glass of lemonade on the international space station? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you are weightless in orbit, there's no up, down, left nor right. We define down on the Earth surface as the direction things tend to fall, down. On Earth bubbles rise if blown into a glass of lemonade (with a straw and a cover).&lt;/p&gt;&#xA;" OwnerUserId="1070" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-04T01:11:18.070" LastActivityDate="2011-01-04T01:11:18.070" Title="Bubbles in zero gravity" Tags="&lt;gravity&gt;&lt;fluid-dynamics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="2481" PostTypeId="1" CreationDate="2011-01-03T14:28:08.633" Score="12" ViewCount="595" Body="&lt;p&gt;If you could travel to the center of the Earth (or any planet), would you be weightless there?&lt;/p&gt;&#xA;" OwnerUserId="1070" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-03T14:40:20.877" LastActivityDate="2011-03-02T16:24:46.163" Title="Would you be weightless at the center of the Earth?" Tags="&lt;gravity&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="2482" PostTypeId="2" ParentId="2481" CreationDate="2011-01-03T14:33:59.047" Score="17" ViewCount="" Body="&lt;p&gt;Correct. If you split the earth up into spherical shells, then the gravity from the shells &quot;above&quot; you cancels out, and you only feel the shells &quot;below&quot; you. When you are in the middle there is nothing &quot;below&quot; you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refrence from Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%27_law_for_gravity&quot; rel=&quot;nofollow&quot;&gt;Gauss&lt;/a&gt; &amp;amp; &lt;a href=&quot;http://en.wikipedia.org/wiki/Shell_theorem&quot; rel=&quot;nofollow&quot;&gt;Shell Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{I am using some simplistic terms, but I don't want to break out surface integrals and radial flux equations}&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Although the inside of the shell will have zero gravity &lt;em&gt;classically&lt;/em&gt;, it will also have non zero gravity &lt;em&gt;relativistically&lt;/em&gt;. At the perfect center the forces may balance out, yielding an &lt;em&gt;unstable&lt;/em&gt; solution, meaning that a small perturbation in position will result in forces that exaggerate this perturbation.&lt;/p&gt;&#xA;" OwnerUserId="392" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2011-03-02T16:24:46.163" LastActivityDate="2011-03-02T16:24:46.163" CommentCount="12" />
  <row Id="2484" PostTypeId="2" ParentId="2475" CreationDate="2011-01-03T15:22:31.827" Score="0" ViewCount="" Body="&lt;p&gt;The answer is, there is a very small affect. If you take a look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weak_Nuclear_Force&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;, you will note that the weak force is about 10^-13 weaker then the strong force. Therefore it's just not going to have much of an affect. &lt;/p&gt;&#xA;" OwnerUserId="871" LastActivityDate="2011-01-03T15:22:31.827" CommentCount="1" />
  <row Id="2485" PostTypeId="2" ParentId="2480" CreationDate="2011-01-03T16:26:20.967" Score="7" ViewCount="" Body="&lt;p&gt;Here is a YouTube link of an experiment done on board the International Space Station.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=cXsvy2tBJlU&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=cXsvy2tBJlU&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second part of the video, air is injected into a spherical water drop.  As you can see, nothing really dramatic happens.  There is certainly no explosion.  The air bubble adopts a spherical shape to minimize its surface area, but otherwise it sits quite happily inside the water drop. This is probably as close as we can get to answering the lemonade question.  One could easily blow bubbles (or a single bubble) into a floating drop of lemonade.  There are some differences due to the lack of gravity.  When you blow bubbles into a liquid here on earth, they are pulled upward by buoyancy due to gravity.  Thus they do not get very big before detaching from the straw.  In contrast, in zero-g the bubble will stay put (other than being pushed a little by air entering the bubble).  So you will not get the same bubble 'train' as on earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has important implications for drinking soda/pop or sparkling wine/champagne on the Space Station: don't do it! These drinks are supersaturated with carbon dioxide (with respect to our normal atmosphere).  On Earth, we open a bottle of champagne and bubbles nucleate around defects (or fibers) on the surface of the glass, but they quickly rise due to buoyancy.  They grow by capturing more carbon dioxide as they rise, but they escape into the atmosphere before getting very big.  In space, the bubbles would quickly grow without rising! So it would seem that opening the champagne in space would always make it bubble over completely!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some other nice experiments in the YouTube video, including the addition of an effervescent tablet (pill) into a water drop, leading to the growth of gas-filled bubbles.&lt;/p&gt;&#xA;" OwnerUserId="266" LastEditorUserId="266" LastEditorDisplayName="" LastEditDate="2011-01-03T19:12:01.640" LastActivityDate="2011-01-03T19:12:01.640" CommentCount="4" />
  <row Id="2486" PostTypeId="1" CreationDate="2011-01-03T16:35:32.717" Score="-4" ViewCount="192" Body="&lt;p&gt;Here's an proposal on how to get from point A to point B in zero-gravity without using any&#xA;propellant and the question why it wouldn't work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A closed tube, filled with water and a round (solid) object. If you need equations, the volume of enclosed water is the same as the round object, but the round object is 10 times lighter. (imagine a glass with water plus a ping-pong ball).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here's a picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img210.imageshack.us/img210/6640/sprite143.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On earth the round object will float on the water inside the tube (subject to one G).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In zero-gravity the round object has no preferred position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we accelerate the tube in zero-gravity by one G, the situation is the same as on earth, the  round object &quot;floats&quot;. In this example we are accelerating the tube from the left side to the right side. The round object will consider floating to the right AS LONG AS THERE IS acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider adding a pipe to the bottom of the tube and connecting it to the top, a loop.&#xA;Inside the pipe is a small water pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we give this apparatus a push, say one G in a zero-g environment, the round object will move &quot;up&quot;, but now we start the water pump and spray the water on the round object, we try to submerge it. It will resist and impart a impulse on the water. like trying to hose down a air balloon floating on the pool. The pump will feel a resistance and hence the whole apparatus will move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just running the pump at constant speed, same volume of water per second, will do nothing.&#xA;but if we run the pump faster and faster the whole apparatus will start moving:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the amount needs to be geometrical. The point is that we need to keep the apparatus feeling&#xA;an acceleration, since only then will the round-object &quot;float&quot; and resist the incoming water at the top, hence we have something to &quot;push on&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before you blow the &quot;foul whistle&quot;: consider the situation if there were no round floating ball in the apparatus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the pump runs on solar power or pre-charged battery)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(disclaimer: i know standing on a sailboat and blowing into the sail will not get me anywhere, action&amp;lt;-&gt;reaction)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks a bunch Sklivvz for the edit. sometimes the idea just needs to get out, never the mind how it looks like : P&lt;/p&gt;&#xA;" OwnerUserId="1070" LastEditorUserId="67" LastEditorDisplayName="" LastEditDate="2011-01-04T02:45:56.350" LastActivityDate="2011-01-04T09:07:36.157" Title="Moving from point A to point B in zero-gravity" Tags="&lt;mechanics&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="2487" PostTypeId="2" ParentId="2486" CreationDate="2011-01-03T18:47:54.587" Score="1" ViewCount="" Body="&lt;p&gt;Consider the following scenario:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A gun is on the left side of a boat, which&#xA;is very long. &lt;/li&gt;&#xA;&lt;li&gt;The gun shoots, recoil&#xA;makes the boat moves left.&lt;/li&gt;&#xA;&lt;li&gt;The bullet&#xA;hits the right side, the boat stops.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The end point of the process is that the boat has moved, but is at rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this work? It works because there is a time between the shooting and the impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following scenario:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A gun is on the left side of a boat, which is very long.&lt;/li&gt;&#xA;&lt;li&gt;The gun shoots repeatedly with a precise frequency so that at each recoil corresponds an impact at the same time.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The boat will start moving at the first shot (which is unbalanced) and will stop at the last impact (which is also unbalanced).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this work? It works because in the beginning and at the end there is no flying bullet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following scenario:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Instead of a gun there is a pump shooting a continuous stream of water in an empty tube&lt;/li&gt;&#xA;&lt;li&gt;Initially the recoil makes the boat move left&lt;/li&gt;&#xA;&lt;li&gt;Once the tube is full, the impact of the water on the right side balances the recoil on the left side&lt;/li&gt;&#xA;&lt;li&gt;When you turn off the water, there is a final impact period on the right side that makes the boat stops&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The boat will start moving as the tube fills, and it will stop moving as the tube empties&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this work? It works because there is no water in the tube initially and at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, consider the following scenario:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Same setup as the previous, with an always full tube (e.g. with a reservoir which is simultaneously filled and emptied)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The boat will not move.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In conclusion: I believe that your setup should be equivalent to one of these setups - I am not sure which - but from them you can derive whether your idea would work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scenarios are taken from this &lt;a href=&quot;http://arxiv.org/pdf/physics/0312087v3&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-01-03T18:47:54.587" CommentCount="5" />
  <row Id="2488" PostTypeId="2" ParentId="2481" CreationDate="2011-01-03T20:30:24.943" Score="10" ViewCount="" Body="&lt;p&gt;The simplest way to think about it is that there is mass all around you in the center of the Earth so you get an equal gravitational &quot;pull&quot; from all directions. The pulls cancel out so you get no acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one assumes constant density for the Earth (which isn't strictly speaking true but it is close enough for this illustration) the gravitational acceleration drops linearly from 1g at the surface to 0 at the center of the Earth. So you'd get a zero if you stepped on a scale at the center of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more complicated explanation is that acceleration due to gravity is the derivative of the gravitational potential. This potential is a minimum at the center of the Earth and grows quadratically up to the surface. It then continues to increase at a lower rate. Since at the exact center is flat (like the bottom of a valley), the derivative which is a measure of the rate of change is zero, and there is no acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, even though you would be weightless there, the effects of gravity are highest at the center of the Earth. You get more gravitational time dilation, for example, than you do at the surface.&lt;/p&gt;&#xA;" OwnerUserId="948" LastEditorUserId="2329" LastEditorDisplayName="" LastEditDate="2011-03-01T06:23:43.700" LastActivityDate="2011-03-01T06:23:43.700" CommentCount="1" />
  <row Id="2489" PostTypeId="1" AcceptedAnswerId="2492" CreationDate="2011-01-03T20:51:58.747" Score="0" ViewCount="43" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/198/do-current-models-of-particle-physics-explain-the-chemical-properties-of-elements&quot;&gt;Do current models of particle physics explain the chemical properties of elements/compounds?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Is there a rule/pattern, or multiple rules and/or patterns, which describe the properties of an element given the number of protons and neutrons it has. For example a carbon atom which has 6 protons and 6 neutrons, howcan you work out what it will act like, reactivity, if there was a bar of carbon, how could they be arranged and what would its properties be like strength. But the rules would work for any number of protons and neutrons.&lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2011-01-03T20:59:44.900" ClosedDate="2011-01-03T21:03:26.273" Title="Is there a [set of] rules/patterns that apply to elements" Tags="&lt;atoms&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="2490" PostTypeId="1" AcceptedAnswerId="2494" CreationDate="2011-01-03T20:57:46.590" Score="17" ViewCount="570" Body="&lt;p&gt;What is the significance about the bell shape, when its hit at the rim it rings/produces sound better than other shaped objects? If so could anyone explain a little bit on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: From the suggestions in the comments, clarification for the term &quot;sound better&quot;: Sound more effective for the purpose which bells are created for. (Thanks Justin)&lt;/p&gt;&#xA;" OwnerUserId="1077" LastEditorUserId="1077" LastEditorDisplayName="" LastEditDate="2011-01-04T20:12:36.383" LastActivityDate="2011-02-10T15:50:27.897" Title="Why is the bell, well, bell shaped?" Tags="&lt;mechanics&gt;&lt;acoustics&gt;" AnswerCount="4" CommentCount="12" FavoriteCount="1" />
  <row Id="2491" PostTypeId="2" ParentId="2489" CreationDate="2011-01-03T20:57:55.960" Score="0" ViewCount="" Body="&lt;p&gt;There are &lt;em&gt;trends&lt;/em&gt;, which is a large part of how we came to have the periodic table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But nothing in a neat little package. The full answer requires considerable quantum mechanics (and is only done on physics first principles as a stunt, there is a reason chemists stay in business...).&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-01-03T20:57:55.960" />
  <row Id="2492" PostTypeId="2" ParentId="2489" CreationDate="2011-01-03T20:59:44.900" Score="0" ViewCount="" Body="&lt;p&gt;Technically, there is one rule that does it all: The Schroedinger equation for a system of N electrons and M nuclei. If you could solve that, you would know everything there is to know about your element. Unfortunately, you cannot solve it exactly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some rules that will tell you which elements might want to form bonds with which other elements in what ratio. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even for &quot;simple&quot; things like the crystal structure, there is no simple rule: You will always have to do the math to find out that these elements form bcc-crystals and these other elements from fcc-crystals or whatever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally: Elements in the same column of the table tend to be have in a similar fashion, because they have the same number of chemically relevant valence electrons.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-01-03T20:59:44.900" />
  <row Id="2493" PostTypeId="2" ParentId="2490" CreationDate="2011-01-03T21:31:01.267" Score="-7" ViewCount="" Body="&lt;p&gt;uhm .. so you can hangup something inside and have it hit the walls?&#xA;if you want a sound and you hit it from the outside, i doesn't need to be bell shaped.&#xA;consider a &quot;gong&quot;.&#xA;also bells are /  were expensive to manufacture during a certain time and used for religious purposes. it is easier to &quot;cast&quot; a huge sounding object in a bell shape rather then as, a tube or pipe?&#xA;furthermore also,you can hang a bell in a way, as opposed to a &quot;gong&quot;, in a way so it doesn't need an army of men to operate and make much noise. one novice hanging on to the rope will suffice.&lt;/p&gt;&#xA;" OwnerUserId="1070" LastEditorUserId="1070" LastEditorDisplayName="" LastEditDate="2011-01-03T23:01:13.623" LastActivityDate="2011-01-03T23:01:13.623" CommentCount="5" />
  <row Id="2494" PostTypeId="2" ParentId="2490" CreationDate="2011-01-03T23:05:41.813" Score="16" ViewCount="" Body="&lt;p&gt;The bell is typically bell-shaped for two reasons, first because the circle is structurally strong and this allows bells to be struck with greater force than if the shape was flat or had sharp edges which would be more prone to cracking, further the circular shape allows a wave to travel around the bells perimeter so that standing waves can develop around the circumference of the bell. It is the resonance from standing waves that is responsible for the sound of the ringing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And second the bell's shape makes the timbre of the bell more musically pleasing. The reason for the increasing diameter as you go from the top to the bottom of the bell is so that the bell resonates at different frequencies which can be tuned in a large bell so that you have what amounts to a complex musical chord playing when the bell is struck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a given bell might have a resonance at the fundamental, a subharmonic one octave lower, a minor third above, a fifth above, and a full octave above. The different diameter sections of the bell contribute to these different harmonics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bell construction is as much an art as a science. Here is a good online resource that describes the process of creating a large bell:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.msu.edu/~carillon/batmbook/chapter4.htm&quot; rel=&quot;nofollow&quot;&gt;https://www.msu.edu/~carillon/batmbook/chapter4.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also the next chapter which goes deeper into the acoustics of bells:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.msu.edu/~carillon/batmbook/chapter5.htm&quot; rel=&quot;nofollow&quot;&gt;https://www.msu.edu/~carillon/batmbook/chapter5.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="948" LastEditorUserId="948" LastEditorDisplayName="" LastEditDate="2011-01-03T23:50:01.713" LastActivityDate="2011-01-03T23:50:01.713" CommentCount="5" />
  <row Id="2495" PostTypeId="2" ParentId="2475" CreationDate="2011-01-03T23:40:29.263" Score="5" ViewCount="" Body="&lt;p&gt;A back of the envelope calculation suffices, meaning all factors of $2,3, \pi$ etc.&#xA;have been ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The residual strong force is mainly due to pion exchange and can be modeled&#xA;by this plus a short range repulsion due to exchange of $\omega$ mesons. The Yukawa potential from pion exchange is $e^{-m_\pi r}/r$. The weak interactions arise from $W^\pm$ and $Z$ exchange and will give rise to a similar potential with&#xA;$m_\pi$ replaced by $m_Z$ (in the spirit of the calculation I take $m_W \sim m_Z$). At typical nuclear densities of $.16/(fermi)^3$ nuclei are separated by distances which are within a factor of two of $1/m_\pi$. Thus the ratio of the Yukawa potential due to weak exchange to the Yukawa potential due to pion exchange at these densities is roughly $e^{-m_W/m_\pi} \sim e^{-640}$. I've ignored the fact that there are different coupling constants in front of the two potentials, but this difference is irrelevant compared to the factor of $e^{-640}$. So yeah, you can ignore the weak contribution to the binding energy.&lt;/p&gt;&#xA;" OwnerUserId="838" LastActivityDate="2011-01-03T23:40:29.263" CommentCount="2" />
  <row Id="2497" PostTypeId="1" AcceptedAnswerId="2499" CreationDate="2011-01-04T01:23:37.030" Score="8" ViewCount="491" Body="&lt;p&gt;I admit, it's been a few years since I've studied physics, but the following question came to me when I was listening to a talk by Lawrence Krauss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any knowledge of from where matter that exists today originated? I recall that the law of conservation of mass asserts that matter cannot be created nor destroyed, but surely the matter we see today had to be created at some point? Perhaps I am applying this law in the wrong fashion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I ask, is because Krauss mentioned that the elements of organic matter where created in stars, not at the beginning of time (whenever that may have been), but I ask, where did the building blocks for these elements arise? Were they too created in stars? If so, from where did their constituent building blocks come?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please forgive me if this off topic, it is my first post on this particular stackexchange site. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="872" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-04T01:56:04.570" LastActivityDate="2011-01-14T12:51:01.220" Title="Where does matter come from?" Tags="&lt;high-energy-physics&gt;&lt;mass&gt;&lt;big-bang&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="2498" PostTypeId="2" ParentId="2497" CreationDate="2011-01-04T01:38:40.410" Score="8" ViewCount="" Body="&lt;p&gt;The origin of matter is a major problem in modern physics. Sure we have a &lt;em&gt;description&lt;/em&gt; in terms of the Standard Model of particle interactions, but that is simply an empirical framework built upon observation. There is no compelling reason, for instance, why the gauge group of the SM should be $ SU(3)\times SU(2) \times U(1) $ and not any other. Grand Unified Theories (GUT) attempt to provide an answer which seemingly complicates the issue further by suggesting bigger groups $SO(5), SO(10), \ldots$ which can contain all three families of particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certain streams of contemporary research appear to favor an alternative and perhaps simpler approach, one that was favored by Einstein and Wheeler among others - that matter is simply another aspect of geometry. The paper of Einstein and Rosen which is commonly cited as the source of the notion of wormholes, was &lt;em&gt;in fact&lt;/em&gt; an attempt to provide an explanation for particles as topological defects in the vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow, all of this sounds pretty technical. To simplify the notion greatly the viewpoint favored by approaches such as Non-Commutative Geometry, LQG and (IMHO) String Theory is that particles are topological defects in a background geometry. If you think of geometry as a sheet then these defects can be thought of as punctures in this sheet. How such punctures can join up and interact should be governed by simple rules and one would hope these rules will yield the S.M. in a suitable coarse-grained approximation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Changing some wording in 2nd para to something a little less &lt;strong&gt;absolute&lt;/strong&gt; in response to comments by @matt and others.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-05T21:30:01.143" CommentCount="25" />
  <row Id="2499" PostTypeId="2" ParentId="2497" CreationDate="2011-01-04T01:53:29.140" Score="5" ViewCount="" Body="&lt;p&gt;The law of conservation of mass is only valid in the classical limit. More in general, the combination of mass and energy is conserved, as they can be exchanged under certain conditions $E=mc^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In very simplified terms, it is currently understood that, initially, possibly only energy was present. Then this energy started to partially change into mass, forming first &lt;em&gt;quarks&lt;/em&gt; and &lt;em&gt;electrons&lt;/em&gt; (mostly). Then heavier particles. Finally, once the universe cooled down a bit more, the first atoms started to form through the aggregation of the particles. It is calculated that atoms would be split between hydrogen and helium with a 3:1 ratio - these are the two lightest atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was enough to create the first generation of stars, which with fusion and with their dying blast generated the heavier elements that build life. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This explains the currently understood principle of nucleogenesis - if you are instead more interested in how and why energy is changed into mass, and what is the nature of mass, then space_cadet's answer gives you a quick overlook of the hypotheses that are being studied.&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-04T10:48:06.710" LastActivityDate="2011-01-04T10:48:06.710" CommentCount="18" />
  <row Id="2500" PostTypeId="2" ParentId="2468" CreationDate="2011-01-04T03:36:14.657" Score="0" ViewCount="" Body="&lt;p&gt;It depends on what type of universe you think we live in.  There are at least two schools of thought here.  One says that when you go back in time the universe splits into two realities.  The original, and the alternate universe.  So in the alternate universe anything can happen, and the atom has two distinct decay times, one for each universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second school of thought says that all time travel into the past has already happened (that is, you can't change the past, because your travel to the past has already been taken into account in our timeline).  So in that case, you go back, wait around, and get to watch yourself take the first measurement.  It is impossible for you to interact with yourself, because then you would have remembered that you saw your self from the future in the first place.  &lt;/p&gt;&#xA;" OwnerUserId="1082" LastActivityDate="2011-01-04T03:36:14.657" />
  <row Id="2501" PostTypeId="1" AcceptedAnswerId="2504" CreationDate="2011-01-04T04:03:36.270" Score="8" ViewCount="427" Body="&lt;p&gt;Except for Mercury, the planets in the Solar System &lt;a href=&quot;http://en.wikipedia.org/wiki/Table_of_planets_in_the_solar_system#Planets&quot; rel=&quot;nofollow&quot;&gt;have very small eccentricities&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this property special to the Solar System?  &lt;a href=&quot;http://en.wikipedia.org/wiki/Extrasolar_planet#Orbital_parameters&quot; rel=&quot;nofollow&quot;&gt;Wikipedia states&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Most exoplanets with orbital periods of 20 days or less have near-circular orbits of very low eccentricity. That is believed to be due to tidal circularization, an effect in which the gravitational interaction between two bodies gradually reduces their orbital eccentricity. By contrast, most known exoplanets with longer orbital periods have quite eccentric orbits. (As of July 2010, 55% of such exoplanets have eccentricities greater than 0.2 while 17% have eccentricities greater than 0.5.&lt;a href=&quot;http://en.wikipedia.org/wiki/Table_of_planets_in_the_solar_system#Planets&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;) This is not an observational selection effect, since a planet can be detected about equally well regardless of the eccentricity of its orbit. The prevalence of elliptical orbits is a major puzzle, since current theories of planetary formation strongly suggest planets should form with circular (that is, non-eccentric) orbits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is special about the Solar System that orbits of planets here are nearly circular, but elsewhere they are moderately or highly eccentric?&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-02-15T05:58:41.077" Title="Why are the orbits of planets in the Solar System nearly circular?" Tags="&lt;astronomy&gt;&lt;orbital-motion&gt;&lt;solar-system&gt;&lt;celestial-mechanics&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="1" />
  <row Id="2502" PostTypeId="1" AcceptedAnswerId="2512" CreationDate="2011-01-04T04:06:54.100" Score="6" ViewCount="116" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Apsis#The_perihelion_and_aphelion_of_the_Earth&quot; rel=&quot;nofollow&quot;&gt;Earth's perihelion&lt;/a&gt; passed about nine hours ago.  How accurately do we know the moment of closest approach of the Earth to the center of the sun?  How do we make this measurement?&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-01-04T22:34:42.133" Title="How accurately is the moment of perihelion of Earth known, and how is it measured?" Tags="&lt;astronomy&gt;&lt;orbital-motion&gt;&lt;celestial-mechanics&gt;" AnswerCount="2" />
  <row Id="2503" PostTypeId="2" ParentId="2501" CreationDate="2011-01-04T04:23:35.327" Score="2" ViewCount="" Body="&lt;p&gt;This property is likely shared by every other planetary system belonging to the same class as ours. Anthropically speaking, orbits with higher ellipticities will have more extreme environments and thus will be less likely to harbor life. As for the physical reason, I'd guess you could come up with a thermodynamic argument. Intuitively, an elliptical orbit seems to be further from equilibrium than a circular one. A collection of bodies orbiting a star, with high ellipticities initially will then relax to a state where the orbits are less and less elliptical. Of course, this is all heuristic.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; In response to @Marek's and @MSalters comments let me add some clarification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a planet with some eccentricity $e$ interacts with a transient object (asteriod etc.) which process is thermodynamically more likely: that the planet eccentricity &lt;em&gt;increases&lt;/em&gt; or that it &lt;em&gt;decreases&lt;/em&gt; as a result of the interaction? The LRL vector is a conserved quantity but not when the object is in a noisy environment. The interactions between the planets themselves are also greater when they have orbits with high $e$. As they slowly lose $e$ by scattering and damping events, the mutual interaction will decrease as the orbits become more circular.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-04T15:56:04.063" CommentCount="6" />
  <row Id="2504" PostTypeId="2" ParentId="2501" CreationDate="2011-01-04T08:43:20.507" Score="4" ViewCount="" Body="&lt;p&gt;This was previously a comment to space_cadet's answer but became long (down-vote wasn't me though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand space_cadet's talk about unstable orbits. Recall that two-body system with Coulomb interaction has an additional $SO(3)$ symmetry and has a conserved &lt;a href=&quot;http://en.wikipedia.org/wiki/Laplace%2dRunge%2dLenz_vector&quot; rel=&quot;nofollow&quot;&gt;Laplace-Runge-Lenz vector&lt;/a&gt; which preserves the eccentricity. Because interactions between planets themselves are pretty negligible one needs to look for explanation elsewhere. Namely, in the initial conditions of the Solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can imagine slowly rotating big ball of dust. This would collapse to the Sun in the center a disk (because of preservation of angular momentum) with circular orbits and proto-planets would form, collecting the dust on their orbits. Initially those planets were quite close and there were interesting scattering processes happening. The last part of the puzzle is mystery though. If there were still large amount of dust present in the Solar system it would damp the orbits to the point of becoming more circular than they are today. The most popular explanation seems to be that the damping of the eccentricity was mediated by smaller bodies (like asteroids). Read more &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0404240&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-01-04T08:43:20.507" />
  <row Id="2505" PostTypeId="2" ParentId="2486" CreationDate="2011-01-04T09:07:36.157" Score="0" ViewCount="" Body="&lt;p&gt;i think it will not work, simply because the round object that floats can only &quot;carry&quot; so much weight before submerging. it's like a life-vest: it can float one person, but not an entire army. i our case of the above mentioned apparatus, the weight of the tube, pipe, pump and water would rest on the small round object, which probably is too much and it would sink, like a overladen life vest, even if the whole apparatus is experiencing an acceleration.&#xA;-BUT- it could augment the acceleration of a rocket attached to the bottom of the apparatus.&#xA;so only by itself, it probably cannot accelerate and move itself, but if we have a small rocket attached&#xA;that would provide 1 G of acceleration, the apparatus could add another x-amount of acceleration when operated in the manner mentioned.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-04T09:07:36.157" CommentCount="2" />
  <row Id="2506" PostTypeId="1" AcceptedAnswerId="14347" CreationDate="2011-01-04T09:27:06.787" Score="5" ViewCount="226" Body="&lt;p&gt;in recent questions like &quot;&lt;a href=&quot;http://physics.stackexchange.com/questions/2041/how-are-classical-optics-phenomena-explained-in-qed-snells-law&quot;&gt;How are classical optics phenomena explained in QED (Snell's law)?&lt;/a&gt;&quot; and &quot;&lt;a href=&quot;http://physics.stackexchange.com/questions/1898/do-photons-gain-mass-when-they-travel-through-glass&quot;&gt;Do photons gain mass when they travel through glass?&lt;/a&gt;&quot; we could learn something about &lt;strong&gt;effective properties&lt;/strong&gt; of matter interacting with a force field in terms of the &lt;a href=&quot;http://physics.stackexchange.com/questions/1898/do-photons-gain-mass-when-they-travel-through-glass/1899#1899&quot;&gt;path integral&lt;/a&gt; and &lt;a href=&quot;http://physics.stackexchange.com/questions/1898/do-photons-gain-mass-when-they-travel-through-glass/1903#1903&quot;&gt;quasiparticles&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely, both approaches must be equivalent but come from a different philosophy. Widely used is the quasiparticle approach in &lt;a href=&quot;http://en.wikipedia.org/wiki/Solid-state_physics&quot; rel=&quot;nofollow&quot;&gt;solid state physics&lt;/a&gt; e.g. calculating dispersion relations of &lt;a href=&quot;http://en.wikipedia.org/wiki/Phonon&quot; rel=&quot;nofollow&quot;&gt;phonons&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would really like to know if there are &lt;strong&gt;simple examples&lt;/strong&gt; for &lt;strong&gt;explicit calculations&lt;/strong&gt; of the &lt;strong&gt;properties of photon-quasiparticles&lt;/strong&gt; coming from a rigorous approach like a matter description via QED and finding an effective action e.g. using the &lt;strong&gt;Wetterich equation&lt;/strong&gt; (see e.g. &lt;a href=&quot;http://arxiv.org/abs/hep-ph/0611146&quot; rel=&quot;nofollow&quot;&gt;Introduction to the functional RG and applications to gauge theories&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any calculations and/or references would be very nice.&lt;br&gt;&#xA;Thank you in advance, sincerely,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2011-09-05T04:30:44.803" Title="How to calculate the properties of Photon-Quasiparticles" Tags="&lt;solid-state-physics&gt;&lt;quantum-electrodynamics&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="1" />
  <row Id="2507" PostTypeId="1" CreationDate="2011-01-04T10:25:52.647" Score="0" ViewCount="54" Body="&lt;p&gt;Is it true, that a sufficiently light and sturdy canister that contains a vacuum would float if it were in total lighter then the equivalent amount of air it could contain?&#xA;And if wind is caused by denser air flowing towards less dense areas, would the canister float against the wind, towards the area of more dense air?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-04T10:33:19.227" ClosedDate="2011-01-04T11:58:30.857" Title="enclosed vacuum" Tags="&lt;mechanics&gt;" CommentCount="4" />
  <row Id="2508" PostTypeId="1" CreationDate="2011-01-04T11:41:36.153" Score="18" ViewCount="1013" Body="&lt;p&gt;Below is a picture of Giant Water Lily. Scientific Name: Victoria Amazonica. Leaves of some of these could be as big as 3 m diameter and carry a weight of 45kg spread evenly and can support a child. Now the problem:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that a leaf of such flower with a child is floating freely on water. Child crawls along the edge of the leaf until it arrives back to the starting point. In other words, he makes a full circle in the reference frame of the leaf. Question:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the total angle $\theta $ that the leaf turns through in time child crawls? (in the reference frame of water). Assume that the leaf is a large rigid circular disk. Ignore air and water resistance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ftiaronsem's solution is absolutely correct if we assume that the leaf can only freely rotate about its geometrical center. However i was keeping in an eye that the leaf is not connected to the ground and can freely move in any direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Data given:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m$ (mass of child)&lt;br&gt;&#xA;$M$ (mass of the leaf)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Wflmi.jpg&quot; alt=&quot;Giant Water Lily&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="552" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2011-01-14T00:44:14.093" LastActivityDate="2011-02-18T18:09:38.083" Title="Rotation angle of a giant lily when a child crawls on its rim" Tags="&lt;classical-mechanics&gt;&lt;angular-momentum&gt;" AnswerCount="6" CommentCount="13" FavoriteCount="5" />
  <row Id="2509" PostTypeId="2" ParentId="2451" CreationDate="2011-01-04T11:43:01.953" Score="0" ViewCount="" Body="&lt;p&gt;on a sufficiently small planet with roads and sufficient &quot;massive&quot; car in respect to the planet, you spin up the planet while accelerating on its surface?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-04T11:43:01.953" CommentCount="2" />
  <row Id="2510" PostTypeId="2" ParentId="2508" CreationDate="2011-01-04T12:37:24.717" Score="9" ViewCount="" Body="&lt;p&gt;This first part asumes the following two things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The flower is connected to the ground, i.e. by its shaft, so that its center of mass is not moving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The child stops as soon as it reaches it's original position on the flower (Not its original position in the ground reference frame).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Marek correctly pointed out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I\omega_l=mR^2\omega_c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $I$ is the Moment of Inertia of the flower, $w_l$ its angular velocity, $R$ its Radius and where $\omega_c$ ist the angular velocity of the child.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can directly be derived from the conservation of angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By integrating this over the time $t$ the child is crawling, you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I\int_0^t\omega_l=mR^2\int_0^t\omega_c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I\omega_lt=mR^2\omega_c t $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, one might be tempted to think that $ \omega_ct = 2 \pi$. However this is not true (Credits go to Sklivvz), since the flower is also spinning. However we know that $w_ct + w_lt = 2\pi$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I\omega_lt=mR^2(2\pi - \omega_lt) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I\omega_lt + m R^2 \omega_lt =mR^22\pi $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now by knowing that the moment of inertia of a disk is $\frac{M}{2}R^2$ (I can add a prove if you want me to), we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{M}{2}R^2 \omega_lt + m R^2 \omega_lt =mR^22\pi $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{M}{2} \theta + m \theta =m2\pi $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M \theta + 2m \theta =m4\pi $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta=\frac{m4\pi}{M+2m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As discussed in the comments, not only the child has an angular velocity around the center of mass, but the entire system is rotating around the center of mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/K9TWC.jpg&quot; alt=&quot;flower rotating around its center of mass&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my first attempt to solve this general case of the problem (which I have deleted due to it being nonsense), I assumed that the translational part of the flower is covered by applying the parallel axis theorem. However, as Mark pointed out correctly, the parallel axis theorem is not applicable here. For a detailed explanation watch Mark's great video and read through the book chapter he linked in the comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please see Mark's Post for a detailed and correct analysis of this situation.&lt;/p&gt;&#xA;" OwnerUserId="393" LastEditorUserId="393" LastEditorDisplayName="" LastEditDate="2011-01-15T14:32:43.780" LastActivityDate="2011-01-15T14:32:43.780" CommentCount="114" />
  <row Id="2511" PostTypeId="2" ParentId="2501" CreationDate="2011-01-04T16:06:04.123" Score="3" ViewCount="" Body="&lt;p&gt;A short answer is that dissipation (e.g. dust, gas interactions with planetessimals) is good at removing energy from a system, but not angular momentum. Circular orbits have the minimum energy for a given angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For short-period exoplanets, the primary form of dissipation is tidal forces of the star on the planet (similarly, the moon is on a nearly circular orbit around the earth, although it certainly didn't start that way!)&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2011-01-04T16:06:04.123" CommentCount="5" />
  <row Id="2512" PostTypeId="2" ParentId="2502" CreationDate="2011-01-04T16:14:17.613" Score="3" ViewCount="" Body="&lt;p&gt;Two things that we are really good at measuring in astronomy: time and angles. With respect to the sun and other planets, we can measure our relative orientation in the solar system to better than a milliarcsec ($8\times 10^{-10}$ of a circle). A year is $3\times 10^7$ sec, so we know passage of pericenter to about 0.02sec. This precesion can be built up even more by observations over many years (the greeks knew the average length of a month to much better than a second by comparing eclipse times separated by hundreds of years). Of course, all our atomic clocks keep absolute time much better than this, and in practice we know (because of these accurate astronomical measurements) that the earth rotation is not nearly so constant, so we often have to add or subtract leap-seconds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: from radar ranging, we can actually measure our instantaneous location in the solar system to within 3m. With an orbital velocity of about 4.7km/s, that gives pericenter passage to better than a millisecond!&lt;/p&gt;&#xA;" OwnerUserId="775" LastEditorUserId="775" LastEditorDisplayName="" LastEditDate="2011-01-04T22:34:42.133" LastActivityDate="2011-01-04T22:34:42.133" CommentCount="3" />
  <row Id="2513" PostTypeId="2" ParentId="2502" CreationDate="2011-01-04T18:04:17.733" Score="1" ViewCount="" Body="&lt;p&gt;I'm pretty sure the current method of measuring solar-system scale distances is radar ranging.  I don't know enough about the details of this for measuring the Earth-Sun distance.  A first order estimate of the error in measuring the moment of closest approach would have to be something of the order of the 16 minute delay time in sending a signal from Earth and receiving a response on the Earth's surface from the reflection event at the Sun's surface.  There might be subtle problems tied to the fact that the Sun doesn't have a solid surface, but we could certainly do radar ranging with several planets--triangulating the Earth-Sun distance using Mercury and Venus, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and like Jeremy said, we could probably improve this time by using several years' worth of data.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-01-04T18:04:17.733" CommentCount="1" />
  <row Id="2515" PostTypeId="2" ParentId="2490" CreationDate="2011-01-04T22:20:39.563" Score="2" ViewCount="" Body="&lt;p&gt;The correct answer is the following: bells have the shape they have because they are made to be hung in high places. The shape focuses sound downwards and maximises loudness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has a cost. The shape is not favourable to hold a steady pitch, and they sound much worse than tubular bells, which are used in symphony orchestras because of their clear and precise sound, which is obviously a more important requirement in that context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Japanese bells have cylindrical shapes, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, cow bells have yet a different oval-like shape probably because of the ease of production. They don't sound very well at all, they are more percussive than melodic.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As a side note. The argument in another answer about harmony is quite wrong. First of all basically any musical interment produces harmonics. That is the fundamental reason why different instruments have different timbres. So, sure, it would apply to bells too (with the caveat that church bells don't have a constant pitch) but it is not specific at all to bells or to their choice of shape in particular. For example the material with which a bell is wrought is just as important.&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-04T22:56:28.940" LastActivityDate="2011-01-04T22:56:28.940" CommentCount="2" />
  <row Id="2516" PostTypeId="1" AcceptedAnswerId="2518" CreationDate="2011-01-05T01:54:21.210" Score="2" ViewCount="282" Body="&lt;p&gt;I am trying to get a common understanding from these two previous questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/q/1686/246&quot;&gt;Why does the mass of an object increase when its speed approaches that of light?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/q/1307/246&quot;&gt;What happens if light/particles exceeded the speed of light for a particular medium&lt;/a&gt; (sic) &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Does the increase of mass occur only if the particle approaches &lt;em&gt;c&lt;/em&gt; (speed of light in a &lt;em&gt;vacuum&lt;/em&gt;) or if it simply approaches the speed of light in its current medium? For example, does the mass of charged particles increase during &lt;a href=&quot;http://en.wikipedia.org/wiki/Cherenkov_radiation&quot; rel=&quot;nofollow&quot;&gt;Cherenkov radiation&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="246" LastActivityDate="2011-01-05T14:59:41.000" Title="Mass of particle near light speed in a medium" Tags="&lt;special-relativity&gt;&lt;speed-of-light&gt;&lt;mass&gt;&lt;velocity&gt;&lt;cherenkov-radiation&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="2517" PostTypeId="2" ParentId="2516" CreationDate="2011-01-05T03:15:58.070" Score="2" ViewCount="" Body="&lt;p&gt;Special relativity is based on the concept of a single absolute &quot;speed limit&quot; for the universe, from all (inertial) reference frames. That &quot;speed limit&quot; is the constant $c$, which is equivalent to the speed of light in a vacuum. The speed of light in other materials is not fundamental in any way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the simple answer is, the two points you bring up are quite orthogonal. Sure, a particle undergoing Cherenkov radiation is likely to be going a high percentage of the speed of the light, and would thus have increased mass. (Note that the Cherenkov radiation would decrease its energy and thus relativistic mass somewhat, due to the EM interaction with the surrounding molecules of the material). At the end of the day, consider 1. as a relativistic effect and 2. as a electromagnetic one. They're both related of course, but there's no direct connection that I think you're looking for.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2011-01-05T14:42:45.310" LastActivityDate="2011-01-05T14:42:45.310" CommentCount="3" />
  <row Id="2518" PostTypeId="2" ParentId="2516" CreationDate="2011-01-05T03:26:08.967" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;First, your question.&lt;/strong&gt;&lt;br&gt;&#xA; Yes, the increase of mass occurs only when a particle approaches $c$ (speed of light in vacuum). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c$ is fundamental in Special Relativity, not because it is the speed of photons, but because it is &lt;em&gt;the&lt;/em&gt; constant speed in the universe (the only speed invariant to boosts). Just because macroscopic light is transmitted at a lower speed inside a particular medium, that doesn't mean that the fundamental speed of Special Relativity is any different. Even inside mediums where light travels more slowly, all relativistic effects happen when a particle approaches $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Cherenkov radiation (CR) is just an effect related to the speed of light in a medium (and not to $c$), it doesn't have anything to do with mass increase either. Though CR and mass increase can happen simultaneously to a particle, they are independent (the first does &lt;strong&gt;not&lt;/strong&gt; imply the second, and vice-versa).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second, about the increase of mass.&lt;/strong&gt;&lt;br&gt;&#xA;It has been a historical habit to say that a particle's mass increases as $m=\gamma m_0$ when its velocity approaches $c$. That is not very appropriate. While it may seem convenient to define this &lt;em&gt;relativistic mass&lt;/em&gt;, it's not a good habit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, because it's confusing to some people. There are physically intuitive ways to explain to a student why time intervals must stretch and why space intervals have to contract, but there's no way at all to explain why a particle's mass should increase. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, it's also not accurate. The defined &lt;em&gt;relativistic mass&lt;/em&gt; parameter does not sustain the properties you would expect from a mass under close analysis. &lt;em&gt;(I have a reference for this, but the pdf file somehow got corrupted in the last 8 years. I'm looking for a copy.)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="724" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2011-01-05T14:59:41.000" LastActivityDate="2011-01-05T14:59:41.000" CommentCount="2" />
  <row Id="2519" PostTypeId="2" ParentId="2451" CreationDate="2011-01-05T05:56:11.220" Score="0" ViewCount="" Body="&lt;p&gt;In the idealized scenario, you turn off the engine and the car keeps going forever, since the engine is not doing any work.  If the car is completely at rest, then the engine is doing work.  Maybe not in regards to the tires against the road, but somewhere in the system, pistons are moving, and a crank-shaft is turning the wheels.  That initial impulse that gets the car moving happens over some distance, and that is where your work is.  But remember, work only happens when velocity changes (in the ideal scenario).&lt;/p&gt;&#xA;" OwnerUserId="1082" LastActivityDate="2011-01-05T05:56:11.220" />
  <row Id="2520" PostTypeId="2" ParentId="2501" CreationDate="2011-01-05T09:18:55.283" Score="0" ViewCount="" Body="&lt;p&gt;The short answer is tides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe there're more shaping effects for the sun, but at least that's the primary effect shaping moon's orbit to be circular. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is that the tidal torque on the sattelite has a strong dependence on the distance. Secondly, the orbital mechanics motion has an interesting property: if you give a bump to an orbitting body it will still pass thru the same point (but it won't get as far out). This is, qualitatively, how tides slowly shape off outer parts of the orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the effect would be the same for other types of friction strongly depending on the distance, for example friction between the bodies' athmospheres.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2011-01-05T09:18:55.283" CommentCount="1" />
  <row Id="2521" PostTypeId="1" CreationDate="2011-01-05T10:34:10.677" Score="11" ViewCount="370" Body="&lt;p&gt;I was sent here from mathoverflow, hoping for a complete answer to this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;===&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A friend of mine asked me what is the flux of the electric field (or any vector field like &#xA;$$&#xD;&#xA;\vec r=(x,y,z)\mapsto \frac{\vec r}{|r|^3}&#xD;&#xA;$$ where $|r|=(x^2+y^2+z^2)^{1/2}$) through a Mobius strip. It seems to me there are no way to compute it in the &quot;standard&quot; way because the strip is not orientable, but if I think about the fact that such a strip &lt;em&gt;can&lt;/em&gt; indeed be built (for example using a thin metal layer), I also think that an answer &lt;em&gt;must&lt;/em&gt; be mathematically expressible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Searching on wikipedia I found that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mobius_resistor&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mobius_resistor&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Möbius resistor is an electrical component made up of two conductive surfaces separated by a dielectric material, twisted 180° and connected to form a Möbius strip. &lt;em&gt;It provides a resistor which has no residual self-inductance, meaning that it can resist the flow of electricity without causing magnetic interference at the same time.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I relate the highlighted phrase to some known differential geometry (physics, analysis?) theorem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;===&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm convinced that there are no way to apply Gauss law since there are no ways to bound a portion of space with a Mobius strip. Nonetheless I would like to &quot;see&quot; some equations showing me that &quot;[the mobius resistor] can resist the flow of electricity without causing magnetic interference at the same time&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1099" LastActivityDate="2011-01-05T15:31:12.483" Title="Flux through a Mobius strip" Tags="&lt;electromagnetism&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="2522" PostTypeId="2" ParentId="2521" CreationDate="2011-01-05T11:33:04.550" Score="5" ViewCount="" Body="&lt;p&gt;Assuming the material of the strip is a good conductor any charges, and currents, on the strip must end up on the boundary. This is true at least in the limit of small applied e.m.f.s. So its not really Gauss', but Ampere's law you're looking for here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{\partial M} I \cdot dl = \Phi_B $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. the line integral of the current $I$ around the boundary $\partial M$ of the Mobius strip $M$. This quantity is actually a topological invariant of the manifold (in this case $M$) under question. For the Mobius strip it is as can be seen to be zero by starting at any given point on the boundary and evaluating the integral as one moves along the edge for one complete cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/7/7d/M%C3%B6bius_resistor.svg/200px-M%C3%B6bius_resistor.svg.png&quot; alt=&quot;Mobius Strip&quot;&gt;&lt;strike&gt;&lt;strong&gt;Note&lt;/strong&gt;: Some of the arrows are incorrect. Think of the strip as a freeway with the vertical axis running through its center, where cars drive only on the edges. You can divide the edge into parts &quot;upper&quot; and &quot;lower&quot;, then the current should be circulating one way in the &quot;upper&quot; half and the opposite way in the &quot;lower&quot; half, changing directions at the place where the &quot;interchange&quot; is located. Label accordingly.&lt;/strike&gt;&lt;/p&gt;&#xA;" LastActivityDate="2011-01-05T15:31:12.483" CommentCount="14" />
  <row Id="2523" PostTypeId="1" AcceptedAnswerId="2526" CreationDate="2011-01-05T13:03:12.867" Score="4" ViewCount="159" Body="&lt;p&gt;Is there any reason to believe that any measure of loudness (e.g. sound pressure) might have an upper boundary, similar to upper limit (c) of the speed of mass?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-05T16:03:22.933" Title="Is there a limit to loudness?" Tags="&lt;acoustics&gt;" AnswerCount="4" />
  <row Id="2524" PostTypeId="2" ParentId="2523" CreationDate="2011-01-05T13:27:15.257" Score="0" ViewCount="" Body="&lt;p&gt;The only practical limit to sound-pressure might be when the medium were to be compressed into a black hole. Although I do not know about the sound-propagating features of black holes. Long before that the medium would come apart, f.e. into plasma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all, you compare two things: an upper limit of a speed with a power. How much power can you put into a particle to speed it up with &lt;em&gt;c&lt;/em&gt; in mind?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I'd rather not guess what kind of sound-generator would be necessary. One might ask the band &quot;Disaster Area&quot; (https://secure.wikimedia.org/wikipedia/en/wiki/Disaster_Area)&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2011-01-05T13:27:15.257" />
  <row Id="2525" PostTypeId="2" ParentId="2523" CreationDate="2011-01-05T13:29:12.530" Score="1" ViewCount="" Body="&lt;p&gt;Well, the short answer is: there is, when the hydrodynamic approximation (that fluid is composed of small &quot;fluid particles&quot; i which real particles move in the reference frame of the &quot;fluid particle&quot; like in stationary fluid) breaks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upper bound can be approximated with wave amplitude equal to ambient pressure, so that the pressure is going down to 0 in wave minimas (this plus minus corresponds to cavitation); yet this corresponds to loudness of $\infty$dB.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-05T15:07:42.787" LastActivityDate="2011-01-05T15:07:42.787" CommentCount="3" />
  <row Id="2526" PostTypeId="2" ParentId="2523" CreationDate="2011-01-05T13:33:30.417" Score="2" ViewCount="" Body="&lt;p&gt;Let's talk about the sound waves in the air first. Physically they are longitudinal waves of pressure. The bunch of air in one place will get compressed (in comparison with equilibrium state) and after that will expand, compressing the adjacent air and so on the wave propagates. These (single frequency) waves are essentially described by three numbers: the speed of propagation (called speed of sound; it depends on the type of the material and the temperature), the frequency of oscillation (determining the pitch), and the amplitude of oscillation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the amplitude of oscillations which determines the loudness. So you are essentially asking whether there is any limit on amplitude of compression. Well, of course. At high enough pressures, the air would freeze even at normal temperatures. So this is the limit for air. Similar thing would happen with liquids: at certain pressures they would condense into solids. You could continue with phases of matter in this way and applying higher and higher pressures you would eventually end up with a black hole. That would be an ultimate limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course, in reality the limit is set by our engineering capabilities and I doubt it's possible to create sound waves that would be able to freeze air.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-01-05T13:33:30.417" CommentCount="4" />
  <row Id="2527" PostTypeId="2" ParentId="2523" CreationDate="2011-01-05T16:03:22.933" Score="4" ViewCount="" Body="&lt;p&gt;Yes - there is a sound pressure limit for &lt;em&gt;undistorted&lt;/em&gt; sound. Over that limit we have a shock wave. It depends on the environmental pressure, but there is a theoretical limit to loudness which you can &lt;a href=&quot;http://en.wikipedia.org/wiki/Sound_pressure#Examples_of_sound_pressure_and_sound_pressure_levels&quot; rel=&quot;nofollow&quot;&gt;find here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limit is basically equal to the pressure.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Theoretical limit for undistorted sound at 1 atmosphere environmental pressure    101,325 Pa  ~194.094 dB&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The lower limit of audibility is therefore defined as 0 dB, but the upper limit is not as clearly defined. While 1 atm (191 dB) is the largest pressure variation an undistorted sound wave can have in Earth's atmosphere, larger sound waves can be present in other atmospheres, or on Earth in the form of shock waves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-01-05T16:03:22.933" />
  <row Id="2528" PostTypeId="1" CreationDate="2011-01-05T16:10:21.580" Score="7" ViewCount="418" Body="&lt;p&gt;could you please recommend any good texts on algebraic geometry (just over the complex numbers rather than arbitrary fields) and on complex geometry including Kahler manifolds that could serve as an informal introduction to the subject for a theoretical physicist (having in mind the appliactions in physics, e.g. in the string theory). What I want for a moment is to get some informal picture of the subject rather than being dug up into the gory details of the proofs and lost in higher and higher levels of abstraction of commutative algebra and category theory. The texts I have found so far are all rather dry and almost completely lack this informal streak, and all of them are greared towards pure mathematicians, so if there exists something like &quot;Algebraic geometry for physicists&quot; and &quot;Kahler manifolds for physicists&quot; (of course, they would probably have different titles :)), I would greatly appreciate the relevant references. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks in advance.&lt;/p&gt;&#xA;" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-05T19:09:59.753" LastActivityDate="2011-01-10T13:32:04.627" Title="Crash course on algebraic geometry with view to applications in physics" Tags="&lt;mathematical-physics&gt;&lt;mathematics&gt;&lt;books&gt;" AnswerCount="5" FavoriteCount="4" />
  <row Id="2529" PostTypeId="2" ParentId="2528" CreationDate="2011-01-05T16:16:48.547" Score="3" ViewCount="" Body="&lt;p&gt;One text that immediately comes to mind is &quot;lectures on complex geometry&quot; by Philip Candelas, an accessible introduction covering the very basics. I think it is hidden in some Trieste proceedings or some such.&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-01-05T16:16:48.547" CommentCount="2" />
  <row Id="2530" PostTypeId="1" AcceptedAnswerId="2533" CreationDate="2011-01-05T16:49:19.757" Score="8" ViewCount="318" Body="&lt;p&gt;What is the advantage for rockets to &lt;a href=&quot;http://en.wikipedia.org/wiki/Multistage_rocket&quot; rel=&quot;nofollow&quot;&gt;have multiple stages&lt;/a&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wouldn't a single stage with the same amount of fuel weigh less?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; I would like a quantitative answer, if possible :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/lswdV.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-01-05T18:12:47.307" Title="Why do rockets have multiple stages?" Tags="&lt;mechanics&gt;&lt;gravity&gt;&lt;rocket-science&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="2531" PostTypeId="2" ParentId="2530" CreationDate="2011-01-05T17:04:33.807" Score="4" ViewCount="" Body="&lt;p&gt;Launch weight would be lower if you had a fixed fuel load, and only one engine and fuel tank. However the specific impulse applied to the payload would be lower. The problem is that even when the fuel is say 90% exhausted, the rocket is still trying to accelerate the now grossly oversized fuel tank and engine. So the trick is to try to reduce the deadweight (structural mass) as the fuel is consumed. Another compromise system is to have jetisonable external fuel tanks, like the shuttle, which are thrown away once their fuel is consumed.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-01-05T17:04:33.807" CommentCount="4" />
  <row Id="2532" PostTypeId="1" AcceptedAnswerId="2539" CreationDate="2011-01-05T17:33:08.517" Score="9" ViewCount="344" Body="&lt;p&gt;I know that string theory is still under heavy development, and as such it still can't make predictions (or not that many predictions anyways).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it is clear from the number of years this theory has been under development and from the large number of theoretical physicists studying it, that it is considered a good and viable candidate as a quantum gravity theory.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So, what is the evidence that this is true? Why is it considered such a good candidate as the correct quantum gravity theory?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without wanting to sound inflammatory in the least, it has been under heavy development for a very long time and it's still not able to make predictions, for example, or still makes outlandish statements (like extra dimensions) that would require a high amount of experimental evidence to be accepted. So - if so many people believe it is the way to go, there have to be good reasons, right? What are they? &lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-01-06T14:28:57.030" Title="What evidence exists for string theory viability?" Tags="&lt;string-theory&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="2" />
  <row Id="2533" PostTypeId="2" ParentId="2530" CreationDate="2011-01-05T17:53:54.287" Score="4" ViewCount="" Body="&lt;p&gt;As Omega Centauri wrote, it is mainly about removing unused tank mass; for numbers, see Wikipedia article about &lt;a href=&quot;http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation&quot; rel=&quot;nofollow&quot;&gt;Ciołkowski's equation&lt;/a&gt;, especially the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation#Examples&quot; rel=&quot;nofollow&quot;&gt;example&lt;/a&gt; there.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2011-01-05T17:53:54.287" />
  <row Id="2534" PostTypeId="2" ParentId="2530" CreationDate="2011-01-05T17:55:40.647" Score="8" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edited a little now that I better understand your question.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Short&lt;/strong&gt;&lt;br&gt;&#xA;In a multi-stage, the weight of the parts that are dropped along the ride compensates for the fact that the extra engines make it heavier in the beginning. Partially because a rocket's engine isn't that heavy compared to the fuel tank. The engine mostly just ignites and controls the combustion, while the fuel tank needs to be huge.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long&lt;/strong&gt;&lt;br&gt;&#xA;I'll analyse the very optimal case, in which the energy spent is the minimum energy needed to get your rocket in orbit. Real cases go waaay above this and it's not even practical to do energy balance. &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0534408966&quot; rel=&quot;nofollow&quot;&gt;Thornton &amp;amp; Marion&lt;/a&gt; does a great (undergraduate level) analysis using linear momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to launch a single stage rocket, there's a specific size that minimizes the needed energy. The weight of your rocket (just the hull, not the fuel) increases with size $P_R=g\rho V$. And the minimum amount of energy necessary to get your rocket in orbit obviously increases with its weight: $$W_{min} = hP_R + W&amp;#39;,$$ where $W&amp;#39;$ is the work you do to lift the fuel. $W&amp;#39;$ is not important here but, for the record, it goes something like:&#xA;$$W&amp;#39; = g\int_0^h m_{fuel}(z)dz&amp;#39;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the rocket is too small, the little amount of fuel you can fit into it does not contain enough energy to lift the hull. If the rocket is too big, then you're just wasting energy because it's unnecessarily heavy. That means you have an optimal size which minimizes the energy of a single-stage rocket. Once you find it, the amount of fuel associated to that size is the very minimum you'll need to get your rocket on orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider a rocket of that same size, but with 2 stages. Let's assume that it has an extra engine of weight $p_e$. Let's also assume that, when your rocket reaches a height $h/2$ more than $1/3$ of its tank will be empty (which is true). Since that part of my tank is empty, I might as well leave it behind (along with the engine) and drop the dead weight (I'll call it $p_t$). If I drop it at $h/2$ the energy I'll need to reach orbit is&#xA; $$W_{min2} = \frac{h}{2}(P_R + p_e) + \frac{h}{2}(P_R - p_t) + W&amp;#39;_2 = hP_R + W&amp;#39;_2 -  \frac{h}{2}(p_t - p_e).$$&#xA;If we sightly reduce the amount of initial fuel we can make $W&amp;#39;_2\leq W&amp;#39;$. Therefore, $W_{min2}$ is clearly less than $W_{min}$ as long as the engine you had to add is lighter than the tank you dropped.&lt;/p&gt;&#xA;" OwnerUserId="724" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2011-01-05T18:12:47.307" LastActivityDate="2011-01-05T18:12:47.307" CommentCount="1" />
  <row Id="2535" PostTypeId="2" ParentId="2530" CreationDate="2011-01-05T18:01:43.940" Score="3" ViewCount="" Body="&lt;p&gt;The easiest way to think of it is this, imagine all the mass left over when a rocket has burned 85% of it's fuel. The mass of most of the tank and structure is now overkill and waste. It would be nice to be able to jettison that extra mass so that the fuel left can accelerate only the payload.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's what a multi-stage rocket does. It jettisons the mass of initial stages so that the remaining fuel and thrust can accelerate much smaller mass to a much higher velocity than it would have been able to if there was only one stage. Remember acceleration is proportional to mass, so if you can get rid of say 80% of the mass then you can accelerate  the payload 5 times more for the same remaining fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another benefit is that you can use rocket motors that are tuned for different velocities.  In the initial stage you need maximum thrust and the rocket is not moving as fast. In the later stages you want high efficiency motors, not necessarily high thrust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get very high velocities it requires less overall fuel and mass with multiple stages. This comes at the cost of greater complexity and cost.&lt;/p&gt;&#xA;" OwnerUserId="948" LastActivityDate="2011-01-05T18:01:43.940" CommentCount="0" />
  <row Id="2536" PostTypeId="2" ParentId="2528" CreationDate="2011-01-05T18:56:53.290" Score="3" ViewCount="" Body="&lt;p&gt;I am not sure about generic applications to physics but when I hear about algebraic geometry I immediately think of string theory. And when I hear about Kähler manifolds it's pretty hard not to think of Calabi-Yaus :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you don't mind this kind of applications, you could find useful a paper by Brian Greene: &lt;a href=&quot;http://arxiv.org/abs/hep-th/9702155&quot; rel=&quot;nofollow&quot;&gt;String Theory on Calabi-Yau Manifolds&lt;/a&gt;. It contains also general talk about differential geometry and string theory. But most importantly, you'll find there the basics of Kähler and Calabi-Yau manifolds as well as lots of applications like mirror symmetry and investigation of moduli spaces.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-01-05T18:56:53.290" />
  <row Id="2537" PostTypeId="1" AcceptedAnswerId="2540" CreationDate="2011-01-05T19:27:17.193" Score="3" ViewCount="135" Body="&lt;p&gt;As a trade I am a software architect. It does happen that I have to design sites where the servers are located in different parts of the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now an essential parameter to consider is latency: that is the time it takes between when a packet is sent from one server, and when it is received on the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the packet might go through other systems etcetera. However, the hallmark of a good connection is low latency.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question: what is the minimal latency that can be obtained over a coaxial wire (i.e. what is the maximum speed reachable by an electrical impulse over a coaxial wire)?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Question: what is the minimal latency that can be obtained over a fiber optic connection (if different from distance/c)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="66" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2011-01-05T22:22:54.457" LastActivityDate="2011-01-06T02:20:25.313" Title="Physically induced latency in internet connections" Tags="&lt;electromagnetism&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="2538" PostTypeId="1" CreationDate="2011-01-05T20:23:06.260" Score="0" ViewCount="88" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/2175/moving-a-very-long-bar-ftl-information&quot;&gt;Moving A Very Long Bar (FTL Information)&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;The &quot;gedankenexperiment&quot; goes like this:&#xA;There are two teams, each with to scientists.&#xA;Both team members of both teams are distanced &quot;x&quot; meters from each other.&#xA;The goal is to determine, who can transmit &quot;one bit of information&quot; faster to the other&#xA;team member.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Team A, &quot;light-team&quot;, uses a LASER. They turn it ON and Off. The time needed to travel&#xA;from the fist team member to the second is &quot;t-light&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Team B, &quot;stick-team&quot;, uses a very rigid stick. they move the stick, from left to right.&#xA;The time needed for the second team member to register the &quot;poke&quot;, is &quot;t-stick&quot;.&#xA;(we are saying that both member of team B are holding on to the stick).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so will &quot;team light&quot; transmit &quot;one bit&quot; faster or &quot;team stick&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(disclaimer: There's a second part to the question, later) &lt;/p&gt;&#xA;" LastActivityDate="2011-01-05T20:23:06.260" ClosedDate="2011-01-05T21:21:16.267" Title="faster then light?" Tags="&lt;mechanics&gt;" CommentCount="4" />
  <row Id="2539" PostTypeId="2" ParentId="2532" CreationDate="2011-01-05T20:42:45.563" Score="11" ViewCount="" Body="&lt;p&gt;Some random points in support of ST, with no attempt to be systematic or comprehensive. I will not get into a long discussion, if someone does not find this convincing I'd advice them not to work on the subject. I also don't have the time to elaborate or justify the claims below, just take them at face value, or maybe you can ask more specific follow up questions.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;ST incorporated naturally and effortlessly all of the mathematical structure underlying modern particle physics and gravity. It does so many times in surprising and highly non-trival ways, many times barely surviving difficult consistency checks. Certainly anyone with any familiarity with the subject has a strong feeling that you get out much more than you put in.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;ST quantizes gravity, and that form of quantization avoids all the difficulties  that more traditional approaches encounter. This is also surprising: it was developed originally as theory of the strong interactions, and when people discovered it contains quantized gravity they spent years trying to get rid of it, with no success. As a theory of quantum gravity it passes many consistentcy checks, failure of any of them would invalidate the whole framework, for example in providing a microscopic description of a large class of black holes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;ST extends the calculational tools available to us to investigate interesting physical systems, many times the only such calculational techniques available. Again, it does so in novel and unexpected ways. It therefore provides a natural language to extend quantum field theory, to models which include quantized gravity, and (relatedly) models with strong interactions. Many calculations using that language are simpler and more natural than other methods, it seems therefore to be the right language to discuss a large class of physical systems.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;ST contains in principle all the ingredients needed to construct a model for particle physics, though this has proven to be more difficult than originally thought. But, in view of the above, even if it turns out not to provide a model for beyond the standard model physics, it is certainly useful enough for many physicists to decide spending their time stuying it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Of course other people may have their own reasons to find the subject interesting, I'm only speaking for myself.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-01-06T14:28:57.030" LastActivityDate="2011-01-06T14:28:57.030" CommentCount="2" />
  <row Id="2540" PostTypeId="2" ParentId="2537" CreationDate="2011-01-05T23:57:50.670" Score="5" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Wave_propagation_speed&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Wave_propagation_speed&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;79% for coax, 67% of speed of light for fiber.&#xA;So it's 42ms for 10'000km of coax, and 49.7ms for 10'000km of fiber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For roundtrip time multiply by 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reality adds some more latency (up to 1.5-2x) on transcievers, queues on overloaded segments and non-optimal paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: As a side note, I really doubt someone could make 10'000km link at 10Gbit+ via coax :-D Signal amplification is just much easier &amp;amp; cheaper in case of fiber.&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="930" LastEditorDisplayName="" LastEditDate="2011-01-06T02:20:25.313" LastActivityDate="2011-01-06T02:20:25.313" CommentCount="3" />
  <row Id="2541" PostTypeId="2" ParentId="770" CreationDate="2011-01-06T00:03:06.297" Score="1" ViewCount="" Body="&lt;p&gt;Find antiuniverse somewhere and annihilate it with what we have here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes does not work as they evaporate when growing.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-01-06T00:03:06.297" CommunityOwnedDate="2011-01-07T14:13:00.227" CommentCount="2" />
  <row Id="2542" PostTypeId="1" CreationDate="2011-01-06T03:54:16.067" Score="4" ViewCount="289" Body="&lt;p&gt;I don't know much about light-emitting diodes, but I imaging if you had a panel of RGB diodes you could produce any wavelength of color within the visible light spectrum.  However, if I&#xA;also wanted to generate specific wavelengths of UVA or UVB (anywhere from 290 to 400nm), could I also accomplish this using diodes?  Essentially, I am interested in making a small panel of diodes in which I could produce any specific wavelength of visible light, UVA, or UVB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" LastActivityDate="2011-04-25T16:08:39.953" Title="Can light emitting diodes be used to generate UV wavelengths?" Tags="&lt;light&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="2543" PostTypeId="2" ParentId="2542" CreationDate="2011-01-06T04:04:12.610" Score="0" ViewCount="" Body="&lt;p&gt;RGB diodes are actually LED with fluorescent dye which is stimulated and thus mixed to different colours. LED-light itself is nearly monochrome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So unless you find LEDs with dye that fluoresces in UV (I doubt that there are some for the normal market) you'll have to buy some UV-LED for these wavelengths.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2011-01-06T04:04:12.610" />
  <row Id="2544" PostTypeId="2" ParentId="2542" CreationDate="2011-01-06T04:23:23.023" Score="4" ViewCount="" Body="&lt;p&gt;RGB diodes, the kind you find in an LED computer monitor or television, don't actually produce any wavelength of light you want. As you probably know, the diodes come in triplets, one each of red, green, and blue. Each of those individual diodes only emits one specific wavelength of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, by varying the relative intensities of the three colors, it's possible to &lt;em&gt;simulate&lt;/em&gt; the effect of any other wavelength of light on the human eye. Since there are only three types of cones on the retina, for any given wavelength which the eye detects, you can induce the same chemical reaction in the cones as that wavelength naturally would, using only three colors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is, in RGB LEDs there are only three wavelengths ever being produced, and it's only by &quot;fooling&quot; our eye that they create the illusion of the full spectrum. Since human eyes aren't sensitive to UV radiation, the trick wouldn't work for ultraviolet wavelengths. You can't &lt;em&gt;actually&lt;/em&gt; create a diode that will produce any wavelength you want dynamically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there are a wide range of materials that you can put in a diode in order to affect the wavelength of radiation it does produce. From a physics perspective, UV radiation is no different from light except that it has a higher frequency (shorter wavelength), so by choosing the right material, you can in principle create a diode that will get you any pre-chosen UV wavelength you want. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Light-emitting_diode&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article on LEDs&lt;/a&gt; gives a table of various materials grouped by the wavelengths they can produce, and there are several options for UV radiation. (There may be a lower limit on the wavelength obtainable from an LED due to energy considerations, but I'm not familiar with the details of how LEDs work so I couldn't tell you whether that's the case or what wavelength it might be.)&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-01-06T04:23:23.023" CommentCount="1" />
  <row Id="2548" PostTypeId="2" ParentId="2468" CreationDate="2011-01-06T06:33:36.097" Score="2" ViewCount="" Body="&lt;p&gt;Because we can't actually go back in time to test the idea behind this question, what you're asking in some sense boils down to a question of what theory you really believe governs the universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the mathematics of QM, if you run experiments backwards in time (mathematically, of course), you only recover your starting state if you do not disturb the experimental system (with a measurement, for example) before running it backwards. So, implicit in the question is a desire to take a measurement and then undo the measurement by traveling back in time. This is impossible in standard QM because it would allow violation of uncertainty and all sorts of other fundamental underpinnings of the theory. So, the question fundamentally can't be asked in a meaningful way. This is the view I would tend to take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, on the other hand, you believe in, say, the many-worlds interpretation of QM, then traveling back in time probably corresponds to taking a different path down the infinitely bifurcating tree of futures, meaning the question can be asked and the answer will be that your second time measurement may be different from the first. Both are &quot;correct&quot; in that one takes place in one section of the multiverse while the other takes place in a different section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you believe in a hidden variables interpretation, that implies (in the sense that if you hold that view, you probably don't believe in the probabilistic nature of QM) that the decay will happen at a set time. In that case, you'd get the same measurement both times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to answer your numbered questions with this subset of interpretations:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The real experiment would help us select between some interpretations of QM. However, the thought experiment does not, as the result depends upon what assumptions you use, which reduces, in this case, to which interpretation you believe in.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The second paragraph of my answer implies not that backwards information travel is prohibited, but that information about the state of the system is outdated because the system has been altered. In fact, I would personally argue that, based on the reversibility of the mathematics, backwards information travel has occurred in the form of the measurement altering the system. Of course, that's a clear violation of causality, that the effect of the measurement is felt before the measurement is done, but obviously, in this posited situation, causality violation should be expected.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="861" LastActivityDate="2011-01-06T06:33:36.097" />
  <row Id="2549" PostTypeId="2" ParentId="2542" CreationDate="2011-01-06T07:12:39.490" Score="4" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Light-emitting diodes (LEDs) can be&#xA;  manufactured to emit light in the&#xA;  ultraviolet range, although practical&#xA;  LED arrays are very limited below 365&#xA;  nm. LED efficiency at 365 nm is approx&#xA;  5-8%, whereas efficiency at 395 nm is&#xA;  closer to 20%, and power outputs at&#xA;  these longer UV wavelengths are also&#xA;  better.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ultraviolet#Ultraviolet_LEDs&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Ultraviolet#Ultraviolet_LEDs&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are not much options in production currently. Mainly ~395nm is available and that's it. Spectrum is quite wide. You can change wavelength a little by altering current/temperature, but that's tricky. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to look at non-linear crystals too (KTP for example).&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-01-06T07:12:39.490" />
  <row Id="2550" PostTypeId="1" AcceptedAnswerId="2551" CreationDate="2011-01-06T09:34:53.660" Score="0" ViewCount="179" Body="&lt;p&gt;I read an article which tells power consumption by many devices. It say that a desktop computer (computer and monitor) use 400 to 600 watt.&lt;br&gt;&#xA;While when i checked my computer and monitor with meter, it was about 60 + 60 = 120 watt (computer + 17&quot; CRT monitor) after loading windows xp and running an application. The voltage is 220V here.&lt;br&gt;&#xA;Which one is correct? How much power does it consume?&lt;/p&gt;&#xA;" OwnerUserId="847" LastActivityDate="2011-01-06T18:06:49.357" ClosedDate="2011-01-06T22:39:40.997" Title="What is the Power Consumption of a desktop computer?" Tags="&lt;electricity&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="2551" PostTypeId="2" ParentId="2550" CreationDate="2011-01-06T09:48:32.563" Score="2" ViewCount="" Body="&lt;p&gt;I sincerely doubt that a computer could use 400+W under normal circumstances. That is the typical power rating of the transformer powering the computer. So, &lt;em&gt;at peak&lt;/em&gt; the computer could use that much power. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;em&gt;peak consumption&lt;/em&gt; in practice means something like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;CPU fully loaded&lt;/li&gt;&#xA;&lt;li&gt;GPU fully loaded&lt;/li&gt;&#xA;&lt;li&gt;Hard disk at maximum throughput&lt;/li&gt;&#xA;&lt;li&gt;All the USB and Firewire ports supporting powered devices&lt;/li&gt;&#xA;&lt;li&gt;Wi-Fi and Bluetooth turned on and trasmitting&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also, IBM compatible motherboards and power transformers are made to support different hardware configurations. Examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Some motherboards support multiple CPUs&lt;/li&gt;&#xA;&lt;li&gt;All support many hard drives&lt;/li&gt;&#xA;&lt;li&gt;etc...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You can think of it this way as well. Servers normally have two PSUs (for redundancy). Both are plugged in at the same time and both could theoretically support the server by themselves. So if the maximum peak wattage of the server is 500W, you would have 2 500W PSUs plugged in at the same time. However, the consumption of the server will not exceed 500W.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-01-06T09:48:32.563" />
  <row Id="2552" PostTypeId="1" CreationDate="2011-01-06T11:24:02.600" Score="2" ViewCount="274" Body="&lt;p&gt;Mass m is connected to the end of a cord at length R above it's rotational axis (the axis is parallel to the horizon, the position of the mass is perpendicular to the horizon). It is given initial velocity V0 at a direction parallel to the horizon. Initial state is depicted at position A in the image.&lt;br&gt;&#xA;the forces working on the mass are MG from the earth and T the tension of the cord.&lt;br&gt;&#xA;How can i calculate the tension of the cord when the mass is at some angle $\theta$  from its initial position (position B in the image)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img535.imageshack.us/img535/8167/76300324.png&quot; alt=&quot;image&quot;&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what i thought:&lt;br&gt;&#xA;since the mass is moving in a circle then the total force in the radial direction is T - MG*$\cos\theta$=M*(V^2)/R&lt;br&gt;&#xA;and so T = MG*$\cos\theta$+M*(V^2)/R&lt;br&gt;&#xA;but since MG applies acceleration in the tangential direction then V should also be a function of $\theta$ and that is where i kind of got lost. I tried to express V as the integration of MG*$\sin\theta$ but i wasn't sure if that's the right approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd very appreciate an answer accompanied with an explanation.&#xA;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="1111" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-06T12:10:00.887" LastActivityDate="2011-04-07T03:39:42.940" Title="how to calculate cord tension in a vertical circle?" Tags="&lt;homework&gt;&lt;newtonian-mechanics&gt;&lt;rotational-dynamics&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="2553" PostTypeId="2" ParentId="2552" CreationDate="2011-01-06T12:30:46.963" Score="1" ViewCount="" Body="&lt;p&gt;Some hints to get you started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass is always on the circle. The radius is constant. Therefore the tension supplied must always be equal to the centrifugal forces acting on the mass. This is the composition of the rotational centrifugal force and gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{tension} = F_{rot}+F_{grav}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the point of view of the equations this is a pendulum - you will obtain a set of equations of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ddot\theta(t) = k \sin \theta(t)$$&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-01-06T12:30:46.963" CommentCount="2" />
  <row Id="2554" PostTypeId="1" AcceptedAnswerId="2576" CreationDate="2011-01-06T12:41:34.890" Score="10" ViewCount="515" Body="&lt;p&gt;I read a lot about the classical twin paradox recently. What confuses me is that some authors claim that it can be resolved within SRT, others say that you need GRT. Now, what is true (and why)? Thank you!&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-06T16:20:00.827" LastActivityDate="2011-03-04T23:19:59.590" Title="How is the classical twin paradox resolved?" Tags="&lt;general-relativity&gt;&lt;special-relativity&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="3" />
  <row Id="2555" PostTypeId="2" ParentId="2554" CreationDate="2011-01-06T13:11:37.863" Score="-1" ViewCount="" Body="&lt;p&gt;The problem is the symmetry break caused by the traveling twin when he changes his direction. This break makes the two twins distinguishable from each other. Before the traveling one changes his direction, both think of their partner to be the younger one (because the time in your own system is always the fastest possible time). So IMHO this paradox has not much to do with general relativity - special relativity can of course not describe accelerated movements, but what really matters is that the twin reverses, not how he accelerates and so on.&lt;/p&gt;&#xA;" OwnerUserId="1112" LastActivityDate="2011-01-06T13:11:37.863" CommentCount="10" />
  <row Id="2556" PostTypeId="2" ParentId="2468" CreationDate="2011-01-06T13:37:51.730" Score="3" ViewCount="" Body="&lt;p&gt;There's a prescription by &lt;a href=&quot;http://prd.aps.org/abstract/PRD/v44/i10/p3197_1&quot; rel=&quot;nofollow&quot;&gt;Deutsch&lt;/a&gt; for the quantum mechanics of closed timelike curves. It works on the level of density states, instead of Hilbert space states. Given his prescription, he showed that a fixed point solution always exists no matter what the initial conditions are. However, this solution isn't unique in general. Also, pure states can be converted into mixed states. Not only that, the solution isn't even linear in the initial density state. It has been shown by &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0808/0808.2669v1.pdf&quot; rel=&quot;nofollow&quot;&gt;Aaronson and Watrous&lt;/a&gt; that this prescription allows time travelling computers to solve PSPACE-complete problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's another prescription based upon post-selection, as expounded by &lt;a href=&quot;http://arxiv.org/abs/1007.2615&quot; rel=&quot;nofollow&quot;&gt;Seth Lloyd&lt;/a&gt; where you would observe the same decay time. This prescription has the drawback that not every initial condition admits a solution. It has been shown that time-travelling computers can only solve PP-complete problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both prescriptions violate unitarity. In fact, the density matrix evolves nonlinearly in both prescriptions. This nonlinearity has the potential to wreak havoc upon quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between both prescriptions can be seen most clearly with the grandfather paradox. To simplify matters, consider a qubit which can only take on two states, $| 0 \rangle$ and $| 1 \rangle$. Let's also suppose this qubit is sent around on a closed timelike curve and loops back onto itself. During a cycle around the loop, the qubit is flipped. Now classically, there clearly isn't any consistent solution. However, according to Deutsch's prescription, the mixed density state $\begin{pmatrix} \frac{1}{2} &amp;amp; a\\ a &amp;amp; \frac{1}{2} \end{pmatrix}$ where $a$ is a real number between $-\frac{1}{2}$ and $\frac{1}{2}$ is a fixed-point solution. He chooses to interpret this using the many worlds interpretation as follows; say the qubit starts off with a value of $0$ in one world. After a loop, it ends up with a value of $1$ in a &lt;em&gt;different&lt;/em&gt; parallel universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Lloyd's prescription, on the other hand, there is no solution at all!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for the example you presented, both prescriptions will give the same answer, namely the time traveller will observe the same decay time both times around. This is because the nuclear decay is not an integral part of the closed timelike loop. To see this, suppose we have the original prepared unstable particle, plus some experimental apparatus with a clock and a clock pointer which will be set to the time of the nuclear decay. After waiting for some time much longer than the half-life, the clock pointer will end up in the state $\sqrt{k}\int^\infty_0 dt\, e^{-kt/2} | t \rangle$. The whole point is, it doesn't make any difference if the time traveller doesn't have direct access to the unstable particle, but only to the clock pointer, and it doesn't make any difference either if the clock pointer is prepared outside the time machine, i.e. is part of the initial conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it might turn out the correct prescription is none of the above. Or it might also turn out that closed timelike curves are absolutely forbidden in a complete theory of quantum gravity. Yet another possibility might be closed timelike curves are allowed, but a complete theory of quantum gravity somehow manages to preserve unitarity, just as it presumably preserves unitarity in evaporating black holes.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-09T15:16:57.817" LastActivityDate="2011-01-09T15:16:57.817" CommentCount="1" />
  <row Id="2557" PostTypeId="2" ParentId="2508" CreationDate="2011-01-06T13:50:26.893" Score="11" ViewCount="" Body="&lt;p&gt;&lt;b&gt;Note:&lt;/b&gt; to illustrate the motion of the pad more clearly, I made &lt;a href=&quot;http://www.youtube.com/watch?v=hqCjbo-Ua2M&quot; rel=&quot;nofollow&quot;&gt;a video&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the pad have radius $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine looking down on the pad from above.  The baby is on the right, crawling counterclockwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The center of mass of the (baby + pad) system can't move.  In the picture below, the green circle represents the starting position of the pad.  The starting position of the baby is marked with an arrow pointing in the direction of its motion.  $C$ is the center of the pad, pointing in its direction of motion (opposite the baby's by conservation of momentum).  $M$ is the center of mass of the system.  $d = R\frac{m}{m+M}$ is the distance from the center of the pad to the center of mass of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/JC8We.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As time goes on, the baby moves around the edge, and the lily pad moves to stay perfectly opposite the baby.  This ensures that the center of mass doesn't move.  Because the length $d$ is fixed, $C$ must always stay the same distance from $M$, so $C$ must move in a circle around $M$.  The baby moves in another circle of radius $R-d$ so that the distance between the center of the pad and the baby remains $R$.  We can draw in the trajectories of the point $C$ and the baby, and add a radius, like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/mb2qw.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Together, the baby and the center of the pad have only one degree of freedom.  Either of them can choose to be at any given point on their trajectories, but the other one is then forced to be opposite them.  Let's describe their positions by an angle $\theta$ from the horizontal.  To show this, I'll move the baby up a bit and draw in $\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/TBvXw.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lily pad has one more degree of freedom - its rotation.  Let's call its rotation relative to the water (relative to north) $\phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total angular momentum of the system is zero.  There are contributions from the translational angular momentum of the pad and the baby, and the rotational angular momentum of the pad.  This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Md^2\dot{\theta} + m(R-d)^2\dot{\theta} + \frac{MR^2}{2}\dot{\phi} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplifying the algebra using the expression for $d$, then integrating over time and using the initial condition $\phi(0) = \theta(0) = 0$, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi = \frac{-2m}{m+M}\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi$ is what we're after, but we want to know $\phi$ after the baby has crawled far enough to return to its starting point.  The minus sign on $\phi$ indicates that the pad spins counter to the rotation of pad and baby around the center of mass.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find $\phi$, we need to use the information about the total amount the baby crawls.  Introduce a new angle $\alpha$ that represents how much the baby has crawled around the pad from the pad's point of view - we are done when the baby gets to $\alpha = 2\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The true speed of the baby (relative to the water) is $(R-d)\dot{\theta}$.  Another way to calculate this is to find the baby's speed relative to the pad, then add the pad's speed relative to the water.  We don't need to worry about vectors here because all motion is tangent to the baby's trajectory.  This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(R-d)\dot{\theta} = R\dot{\alpha} + R \dot{\phi} - d\dot{\theta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;b&gt;note:&lt;/b&gt; I originally left out the term $R \dot{\phi}$, which led the initial answer to be off) This simplifies to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta = \alpha + \phi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;again using the initial condition $\theta(0) = \alpha(0) = 0$.  We want $\alpha = 2\pi$, so set $\theta = 2\pi + \phi$.  After simplifying, this gives the final answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi = \frac{-4\pi m}{3m+M}$$&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-01-13T03:16:30.520" LastActivityDate="2011-01-13T03:16:30.520" CommentCount="31" />
  <row Id="2558" PostTypeId="1" AcceptedAnswerId="2562" CreationDate="2011-01-06T14:04:14.413" Score="7" ViewCount="351" Body="&lt;p&gt;What are the different death scenarios for a black hole?  I know they can evaporate through Hawking radiation - but is there any other way?  What if you just kept shoveling more and more mass and energy into the black hole?&lt;/p&gt;&#xA;" OwnerUserId="922" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-21T15:58:29.680" LastActivityDate="2011-01-24T04:09:16.093" Title="The final death of a black hole" Tags="&lt;astrophysics&gt;&lt;quantum-gravity&gt;&lt;black-holes&gt;" AnswerCount="6" FavoriteCount="2" />
  <row Id="2559" PostTypeId="1" CreationDate="2011-01-06T14:34:55.513" Score="4" ViewCount="185" Body="&lt;p&gt;Imagine a circuit with a voltage source, a switch and an inductivity all connected in series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the switch is open and there's no current and no magnetic field around. If we close the switch, the potential difference of the voltage source is instantaneously applied to the inductivity. Lenz's Law tells us that the induced voltage from the inductivity will always be such that the change in the magnetic flux is reduced. And we have Faraday's Law which tells us that the induced voltage is equal to the change in the magnetic flux.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the theoretical reason why the change in magnetic flux must be smaller than the applied voltage? Why does for t -&gt; infinity always run a current trough the inductivity as it would be just a resistor? Is it just because the wires have a resistance? What would then happen when taking superconductors instead of wires? Is it the inner resistivity of the voltage source then?&lt;/p&gt;&#xA;" OwnerUserId="1112" LastActivityDate="2011-01-07T22:18:14.640" Title="Self induction: Why is induced voltage smaller than the applied voltage?" Tags="&lt;electromagnetism&gt;" AnswerCount="1" />
  <row Id="2560" PostTypeId="2" ParentId="2554" CreationDate="2011-01-06T14:43:32.663" Score="11" ViewCount="" Body="&lt;p&gt;The &quot;paradox&quot; in the twin paradox is that a naive view of the problem would suggest that the situation ought to be perfectly symmetric: each twin should believe that he or she is really at rest, while the other twin is the one who moves off at high speed then returns. This is incorrect, though, because one of the two necessarily accelerates, which provides a means of distinguishing between the two frames of reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can understand the difference in times entirely from within special relativity, with no need to invoke general relativity. You can easily Google up a dozen different explanations of how it works, but the best I've seen is probably the one in Tatsu Takeuchi's &lt;em&gt;An Illustrated Guide to Relativity&lt;/em&gt;. Sadly, being illustrated and not online, it's a bit beyond my ability to reproduce here, but it's worth a look.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem that needs to be resolved is that both twins should see the other's clock running slow during the trip, but somehow, the moving twin has to end up seeing less time pass than the Earthbound twin. The resolution is, very roughly, that in the moving frame of the twin on the return trip, the departure point is much farther in the past, so the trip has been going on for a longer time. An observer who has always been in that frame (say the captain of the interstellar transport the twin is hitching a ride on) would say that the Earth clock has always been slow, but it's been going for far longer than the twin's internal clock would suggest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The details are somewhat subtle, though, which is why there are dozens of semi-contradictory explanations available via Google. It's important to note, though, that it is &lt;strong&gt;not&lt;/strong&gt; the case that the acceleration causes the difference between the clocks (in which case you might actually need GR to work it out). It's the transitions between frames of reference that leads to the difference, not the acceleration itself, and you can get a twin-paradox-like difference between clocks even when both twins undergo exactly the same acceleration, as shown in this &lt;a href=&quot;http://ajp.aapt.org/resource/1/ajpias/v57/i9/p791_s1?isAuthorized=no&quot; rel=&quot;nofollow&quot;&gt;American Journal of Physics paper&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2011-01-06T14:43:32.663" CommentCount="5" />
  <row Id="2561" PostTypeId="2" ParentId="2558" CreationDate="2011-01-06T14:59:48.500" Score="0" ViewCount="" Body="&lt;p&gt;Well, you could merge it with another black hole, so that is a way of reducing the total number of black holes in the universe, but I'm not sure you would call that &quot;death.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may also be possible to accelerate its evaporation by distorting its surface via tidal interactions with a nearby object, such as a fly-by of an unbound black hole, but all the radiated gravitational wave energy might just come from the tidal energy and not the initial BH mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choptuik and Christodoulou both showed that you could create a naked singularity by the collapse of a very special scalar field, but I don't know what its fate would be.&lt;/p&gt;&#xA;" OwnerUserId="775" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-06T15:55:02.823" LastActivityDate="2011-01-06T15:55:02.823" />
  <row Id="2562" PostTypeId="2" ParentId="2558" CreationDate="2011-01-06T15:10:21.910" Score="4" ViewCount="" Body="&lt;p&gt;Hawking radiation is a very slow process of the black hole losing energy and shrinking. If you counter this by supplying a little bit of matter or energy falling into the black hole you can easily overcome it and sustain the black hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than Hawking radiation I don't think there is any known process for black holes to shrink. The area theorem in classical general relativity states that the area of black hole horizon always increases in any physical process. So at least classically there is no way for black holes to die, or even shrink a little. Hawking radiation evades that because it is a quantum  process (which is also why it is a slow process). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the final stage of the evaporation, I think the honest answer is nobody knows. The logical possibilities are either the black hole shrinks to nothing and disappears, or it leaves behind some long-lived &quot;remnant&quot;. Either one of this possibilities has its strong and weak points, but ultimately you'd need to know more about a quantum theory of gravity to know for sure.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-01-06T16:06:46.547" LastActivityDate="2011-01-06T16:06:46.547" />
  <row Id="2563" PostTypeId="1" AcceptedAnswerId="2565" CreationDate="2011-01-06T17:06:33.207" Score="3" ViewCount="275" Body="&lt;p&gt;Suppose a black hole were to suck up a massive amount of surrounding oxygen and hydrogen.  Would this ignite inside the black hole and turn it back into a star?  &lt;/p&gt;&#xA;" OwnerUserId="1115" LastActivityDate="2011-01-14T12:37:51.180" Title="What happens if a black hole sucks up a massive amount of oxygen and hydrogen?" Tags="&lt;cosmology&gt;" AnswerCount="4" />
  <row Id="2564" PostTypeId="2" ParentId="2528" CreationDate="2011-01-06T17:10:18.480" Score="4" ViewCount="" Body="&lt;p&gt;Griffiths and Harris' &quot;Principles of Algebraic Geometry&quot; (Wiley) is the best for your purposes (read only the parts on Kahler geometry).  The sections on algebraic geometry in &quot;Mirror Symmetry&quot; (Clay/AMS) are essentially a Crib Notes version of that paper and some of the classic CY and special geometry papers referred to above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you should keep in mind going in is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kahler manifolds are complex manifolds with a hermitian inner product on tangent vectors which have a metric that is determined (locally) by a single function.  It is the geometry in which the metric and the complex structure &quot;get along very nicely.&quot;  This simplifies lots of calculations and adds new symmetries.  That's why we know so much about them.&lt;/p&gt;&#xA;" OwnerUserId="20" LastActivityDate="2011-01-06T17:10:18.480" />
  <row Id="2565" PostTypeId="2" ParentId="2563" CreationDate="2011-01-06T17:16:14.130" Score="5" ViewCount="" Body="&lt;p&gt;No. The basic point about black holes is that they are black, i.e. nothing escapes from them. Barring Hawking radiation (a quantum effect) there is no way to destroy a black hole. By throwing things into it, you just make it bigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it's perfectly possible for a star to form inside a black hole (at least in principle). This is because big black holes are pretty nicely behaved and if there would be a proto-star dust sucked into them (meaning they would fell under the horizon), it wouldn't experience anything out of ordinary and could form a star. And perhaps, given enough time, also form a black hole of their own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, life under the horizon of the big black hole is in every way normal except that you can't escape from the inside of the horizon and eventually will hit a singularity and be destroyed (that is, if you won't get destroyed sooner by lots of other effects, like tidal forces).&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-01-06T17:16:14.130" CommentCount="8" />
  <row Id="2566" PostTypeId="2" ParentId="2563" CreationDate="2011-01-06T17:22:26.317" Score="1" ViewCount="" Body="&lt;p&gt;No, it would not. An explosion of any power involves matter which has mass and cannot even reach the speed of light. And even the speed of light is not fast enough to escape a black hole. So the explosion would never escape the black hole if it were to occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no way to turn a black hole into a star. Adding mass just makes it bigger. A little chemical explosion is very little energy compared with the energy of fusion which powers a star. The power of the explosion doesn't matter. Even a supernova inside the event horizon would not turn it into a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that oxygen and hydrogen would burn beyond the event horizon if they were in close enough proximity and temperature, they might even explode. But the byproducts, even the light from the burning or explosion would not escape the black hole's event horizon.&lt;/p&gt;&#xA;" OwnerUserId="948" LastActivityDate="2011-01-06T17:22:26.317" />
  <row Id="2567" PostTypeId="1" CreationDate="2011-01-06T17:37:09.397" Score="5" ViewCount="472" Body="&lt;p&gt;Many light sources like LEDs and lasers only emit a single wavelength of light. Is there a light source that emits all wavelengths of visible light at the same time?&lt;/p&gt;&#xA;" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-15T04:04:06.923" LastActivityDate="2011-01-15T15:05:20.993" Title="Is there a light source that emits all wavelengths of visible light at the same time?" Tags="&lt;light&gt;&lt;spectroscopy&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="2568" PostTypeId="2" ParentId="2563" CreationDate="2011-01-06T17:44:14.920" Score="1" ViewCount="" Body="&lt;p&gt;the answers given by Marek and inflector are perfectly fine. Given &lt;strong&gt;our knowledge&lt;/strong&gt; of general relativity and the spacetimes of black holes, there is no escape, no black hole transformation, not even another supermassive black hole would change the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I am writing is that we should be a little more careful. The point is that the laws of nature we are aware of are reliable on &lt;strong&gt;energy scales&lt;/strong&gt; we can &lt;strong&gt;access/measure&lt;/strong&gt; somehow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analogously to your question we could ask if there is an &quot;amount of&quot; &lt;strong&gt;energy&lt;/strong&gt; we would need to transform a black hole somehow. And the answer is: We just &lt;strong&gt;don't know&lt;/strong&gt;, we are not aware of some &lt;a href=&quot;http://en.wikipedia.org/wiki/Grand_Unified_Theory&quot; rel=&quot;nofollow&quot;&gt;black hole transformation theory&lt;/a&gt; :)&lt;br&gt;&#xA;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2011-01-06T17:44:14.920" CommentCount="2" />
  <row Id="2570" PostTypeId="2" ParentId="2567" CreationDate="2011-01-06T17:49:30.083" Score="4" ViewCount="" Body="&lt;p&gt;Blackbody radiation for a white hot object emits the spectrum from infrared to ultraviolet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See: &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_body&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Black_body&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graphite is a decent approximation of a blackbody radiator. So graphite heated to white hot will emit the full spectrum of visible light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note however, that the spectrum will not be flat. There will be more energy on the ultraviolet end of the scale. See the curves in the graph on the upper right of the Wiki page for more information.&lt;/p&gt;&#xA;" OwnerUserId="948" LastActivityDate="2011-01-06T17:49:30.083" CommentCount="4" />
  <row Id="2571" PostTypeId="2" ParentId="2550" CreationDate="2011-01-06T18:06:49.357" Score="0" ViewCount="" Body="&lt;p&gt;I have a dual socket 8core system on my desk right now, with dual moniters.&#xA;  In near idle mode, with monitiers swicthed off it draws 200watts (measured), even with moniters on (30-40watts apiece), and 8 CPUs pounding away it is only about 360. Most systems are a lot less beefy than that. Some of the hi performance graphics cards may be pretty hungry however.&#xA;   For about $20 you can buy a Kill-a-Watt meter and measure such things yourself.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-01-06T18:06:49.357" />
  <row Id="2572" PostTypeId="2" ParentId="2528" CreationDate="2011-01-06T18:46:29.617" Score="2" ViewCount="" Body="&lt;p&gt;Though not particularly aimed at physicist, you could take a look at &lt;a href=&quot;http://books.google.com/books?id=8zQYxx7PjqsC&amp;amp;lpg=PP1&amp;amp;ots=tQQeavPBzH&amp;amp;dq=an%20invitation%20to%20algebraic%20geometry&amp;amp;pg=PP1#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;An Invitation to Algebraic Geometry&lt;/a&gt; by Smith, Kahanpää, Kekäläinen and Traves. It's very short and avoids many technicalities and proof. Instead it give a bird's-eye view and succeeds well at transmitting some of the basics.&lt;/p&gt;&#xA;" OwnerUserId="128" LastActivityDate="2011-01-06T18:46:29.617" />
  <row Id="2573" PostTypeId="1" CreationDate="2011-01-06T19:28:26.923" Score="2" ViewCount="280" Body="&lt;p&gt;Quite a while ago I read about a series of experiments that basically suggested that a certain kind of particle/atom/(something) were &quot;intelligent&quot; and could appear in two places at once, or essentially could &quot;tell the future&quot; when it came to navigating a &quot;maze&quot; ...I think it might have involved lasers or mirrors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does any one a) know what I'm talking about, and b) have links/further information on it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really don't have much more recollection than that I'm afraid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will probably come across as a rather vague question so my apologies but hopefully someone will know what I'm talking about!&lt;/p&gt;&#xA;" LastActivityDate="2011-03-05T22:40:10.397" ClosedDate="2011-03-05T22:56:02.550" Title="Intelligent(?) Particles" Tags="&lt;particle-physics&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="2574" PostTypeId="2" ParentId="2508" CreationDate="2011-01-06T19:43:04.950" Score="0" ViewCount="" Body="&lt;p&gt;This problem can be solved in two steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Prove that the lily will only spin and not translate&lt;/li&gt;&#xA;&lt;li&gt;Find the spinning velocity and therefore the angle&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The lily only rotates&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be resolved geometrically. The baby must turn right to keep their circular trajectory, with respect to the lily. It will therefore exert a force on the lily equal and opposite to their centripetal force (i.e. the force they use to turn right).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force exerted by the baby is of constant magnitude and direction towards the center of the lily. The reaction force is also of constant magnitude but with direction outside the lily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the net effect of force on the baby as time passes? We can obtain it be subtracting the centripetal forces at small intervals of time. Geometry and common sense show that the net effect is a vector tangential to the trajectory of the baby (obviously).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can think about the net effect of the force on the lily. The force is exactly the same magnitude as the one on the baby, but with opposite direction. We can also see it as the same direction and negative magnitude. So the forces will compose over time in the same way as the baby, but with the opposite sign. So in other words the net effect is that the lily spins backwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As there are no other forces acting on the system, the lily will only spin on its axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calculating the angle&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\theta$ be the angle of rotation of the baby, and $\phi$ the angle of rotation of the lily. Let the ratio of the masses, $\frac{M}{m}=\mu$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the baby always keeps the same distance from the center of the lily, both $\ddot\theta=0=\ddot\phi$ and therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot\theta = \omega_{b}$$&#xA;and&#xA;$$\dot\phi = \omega_{l}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both are constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the frame of reference of the pond, the lily will rotate with angular velocity $\omega_l$ and the baby will rotate with angular velocity $\omega_b+\omega_l$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Angular momentum must be conserved, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\omega_b+\omega_l) m r^2 = -I\omega_{l} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting $I=\frac{Mr^2}{2}$ and solving for $\omega_l$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega_l = -\omega_b\frac{2}{\mu+2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the time that is necessary for a full rotation of the baby is $t_{f}=2\pi/\omega_b$. The angle $\phi$ rotated by the lily is therefore equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi=\omega_l t_f=-\frac{4\pi}{\mu+2}$$&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-01-06T19:43:04.950" CommentCount="13" />
  <row Id="2575" PostTypeId="2" ParentId="2573" CreationDate="2011-01-06T19:58:11.603" Score="3" ViewCount="" Body="&lt;p&gt;I don't know about intelligent, but they have free will. At least if you do.&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Free_will_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Free_will_theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T04:12:04.513" LastActivityDate="2011-03-02T04:12:04.513" CommentCount="2" />
  <row Id="2576" PostTypeId="2" ParentId="2554" CreationDate="2011-01-06T20:38:00.697" Score="7" ViewCount="" Body="&lt;p&gt;To understand this paradox it's best to forget about everything you know (even from SR) because all of that just causes confusion and start with just a few simple concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of them is that the space-time carries a metric that tells you how to measure distance and time. In the case of SR this metric is extremely simple and it's possible to introduce simple $x$, $t$ coordinates (I'll work in 1+1 and $c = 1$) in which space-time interval looks like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ ds^2 = -dt^2 + dx^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's see how this works on this simple doodle I put together&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/m8SF3.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vertical direction is time-like and the horizontal is space-like. E.g. the blue line has &quot;length&quot; $ds_1^2 = -20^2 = -400$ in the square units of the picture (note the minus sign that corresponds to time-like direction) and each of the red lines has length zero (they represent the trajectories of light).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length of the green line is $ds_2^2 = -20^2 + 10^2 = -300$. To compute proper times along those trajectories you can use $d\tau^2 = -ds^2$. We can see that the trip will take the green twin shorter proper time than the blue twin. In other words, green twin will be younger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, any kind of curved path you might imagine between top and bottom will take shorter time than the blue path. This is because time-like geodesics (which are just upward pointing straight lines in Minkowski space) between two points maximize the proper time. Essentially this can be seen to arise because any deviation from the straight line will induce unnecessary space-like contributions to the space-time interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that there was no paradox because we treated the problem as what is really was: computation of proper-time of the general trajectories. Note that this is the only way to approach this kind of problems in GR. In SR that are other approaches because of its homogeneity and flatness and if done carefully, lead to the same results. It's just that people often aren't careful enough and that is what leads to paradoxes. So in my opinion, it's useful to take the lesson from GR here and forget about all those ad-hoc SR calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to give you a taste what a SR calculation might look like: because of globally nice coordinates, people are tempted to describe also distant phenomena (which doesn't really make sense, physics is always only local). So the blue twin might decide to compute the age of the green twin. This will work nicely because it is in the inertial frame of reference, so it'll arrive at the same result we did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the green twin will come to strange conclusions. Both straight lines of its trajectory will work just fine and if it weren't for the turn, the blue twin would need to be younger from the green twin's viewpoint too. So the green twin has to conclude that the fact that blue twin was in a strong gravitational field (which is equivalent to the acceleration that makes green twin turn) makes it &lt;em&gt;older&lt;/em&gt;. This gives a mathematically correct result (if computed carefully), but of course, physically it's a complete nonsense. You just can't expect that your local acceleration has any effect on a distant observer. The point that has to be taken here (and that GR makes clear only too well) is that you should &lt;em&gt;never&lt;/em&gt; try to talk about distant objects.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-01-06T21:58:21.593" LastActivityDate="2011-01-06T21:58:21.593" CommentCount="6" />
  <row Id="2577" PostTypeId="2" ParentId="2558" CreationDate="2011-01-06T21:09:23.043" Score="0" ViewCount="" Body="&lt;p&gt;If you shovel lots of mass and energy into a black hole, you will most likely get a bigger black hole.  It will not get indigestion, and it will not explode due to being full.  You may get accretion and jets if your shoveling is suitably generic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any list of possible &quot;final death&quot; scenarios will depend on your threshold for improbable events, and possibly which theory of quantum gravity you choose to accept.  Hawking radiation appears to be a fairly random process, and there is a very small but nontrivial probability for a black hole to destroy itself with a spontaneous eruption of big chunks of matter and energy.  The far more likely scenario seems to be that the temperature (and hence the flux from the radiation) increases steadily as the mass of the black hole decreases, until there is a very small hot black hole whose behavior can't be described well with semi-classical methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This paragraph is very speculative, and may not reflect expert opinion.)  It seems possible to me that a collapsing body can form an apparent horizon (with respect to distant observers), and not collapse to form a singularity.  The final death may just be that it emits Hawking radiation until the radius crosses some threshold and the horizon disappears to reveal a dense macroscopic object made of matter.&lt;/p&gt;&#xA;" OwnerUserId="761" LastActivityDate="2011-01-06T21:09:23.043" CommentCount="1" />
  <row Id="2578" PostTypeId="2" ParentId="2559" CreationDate="2011-01-06T21:48:14.463" Score="1" ViewCount="" Body="&lt;p&gt;Well there are many questions in your question and I'll try to address them one by one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you may know,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L* di/dt = V (t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one important thing to note about inductors is that unless there is an impulsive excitation(that is, the input is bounded), it's current can not make instantaneous jumps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, when you apply the voltage, it will still have a 0 current, it will have a high slope, but 0 current nonetheless at $0^+$ where $0^+$ denotes the instant right after 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, so using $L* di/dt = V (t)$ and basic first order circuit analysis coupled with a bit of differential equations, you get the current equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$i(t) = V/R *(1 - e^ {-tR/L})$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;R is the total thevenin resistance in the circuit. This includes the internal resistance of the inductor. This is the Zero state response of the inductor and it does not take into account any form of initial condition, however it is not hard to implement that into the formula. As you can see, the current will only reach its final value as t -&gt; infinity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you assume the inductor circuit has zero resistance, it would be a short circuit between the terminals of the voltage source. If you assume there is an output resistance of voltage source, you can separate it from the voltage source and use in the calculation as R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Change in the magnetic flux is basically the self induced EMF. The differential equation basically points out to the fact that most things in the inductor occur in an exponential decay kind-of fashion. I hope this answers at least some of your questions. Let me know in the comments if anything is not clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Just to be clear * is multiplication in my answer, not convolution or something like that.&lt;/p&gt;&#xA;" OwnerUserId="504" LastEditorUserId="504" LastEditorDisplayName="" LastEditDate="2011-01-07T22:18:14.640" LastActivityDate="2011-01-07T22:18:14.640" />
  <row Id="2579" PostTypeId="1" AcceptedAnswerId="2593" CreationDate="2011-01-06T21:51:44.700" Score="5" ViewCount="367" Body="&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I edited the question to reflect Moshe's objections. Please, look at it again.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It's apparently a black hole time around here so I decided to ask a question of my own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a few generic questions about black holes, I am wondering whether string theory is able to provide something beyond the usual semiclassical Hawking radiation talk. Feel free to provide an answer from the standpoint of other theories of quantum gravity but AFAIK none of the other theories has yet come close to dealing with these questions. That's why I focus on string theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's talk about micro black holes. They have extreme temperature, extreme curvature, and I guess they must be exceptional in other senses too. At some point the gravitational description of these objects breaks down and I imagine this kind of black hole could be more properly modeled like a condensate of some stringy stuff. So let's talk about fuzzballs instead of black holes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;What does that microscopic fuzzball model look like?&lt;/li&gt;&#xA;  &lt;li&gt;What does string theory tell us about the evaporation of those fuzzballs? Is the Hawking radiation still the main effect (as for the regular black holes) or do other phenomena take over at some point?&lt;/li&gt;&#xA;  &lt;li&gt;Also feel free to add any other established results regarding black hole decay (as Jeff did with information preservation).&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-01-07T09:10:17.870" LastActivityDate="2011-03-02T04:11:13.517" Title="Does string theory tells us anything concrete about black hole decay?" Tags="&lt;string-theory&gt;&lt;quantum-gravity&gt;" AnswerCount="4" CommentCount="9" FavoriteCount="1" />
  <row Id="2580" PostTypeId="2" ParentId="2573" CreationDate="2011-01-06T21:59:12.443" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/physics/0004047&quot; rel=&quot;nofollow&quot;&gt;Informational interpretation of quantum mechanics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/physics/0509057&quot; rel=&quot;nofollow&quot;&gt;Informational experiments with microparticles and atom&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="238" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-01-06T22:19:00.087" LastActivityDate="2011-01-06T22:19:00.087" />
  <row Id="2581" PostTypeId="2" ParentId="193" CreationDate="2011-01-06T22:14:01.433" Score="0" ViewCount="" Body="&lt;p&gt;I found &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471198269&quot; rel=&quot;nofollow&quot;&gt;Mathematical Methods in the Physical Sciences&lt;/a&gt; by Mary Boas to be a very good broad book covering the basics. You'll need other books obviously but if you are looking for one book for a solid review of the basics, this book is excellent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the chapter titles:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Infinite series, power series &lt;/li&gt;&#xA;&lt;li&gt;Complex numbers &lt;/li&gt;&#xA;&lt;li&gt;Linear algebra &lt;/li&gt;&#xA;&lt;li&gt;Partial differentiation &lt;/li&gt;&#xA;&lt;li&gt;Multiple integrals &lt;/li&gt;&#xA;&lt;li&gt;Vector analysis &lt;/li&gt;&#xA;&lt;li&gt;Fourier series and transforms &lt;/li&gt;&#xA;&lt;li&gt;Ordinary differential equations &lt;/li&gt;&#xA;&lt;li&gt;Calculus of variations &lt;/li&gt;&#xA;&lt;li&gt;Tensor analysis &lt;/li&gt;&#xA;&lt;li&gt;Special functions &lt;/li&gt;&#xA;&lt;li&gt;Series solutions of differential equations, legendre, bessel, hermite, and laguerre functions &lt;/li&gt;&#xA;&lt;li&gt;Partial differential equations &lt;/li&gt;&#xA;&lt;li&gt;Functions of a complex variable &lt;/li&gt;&#xA;&lt;li&gt;Probability and statistics &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I also second &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0679776311&quot; rel=&quot;nofollow&quot;&gt;Roger Penrose's The Road to Reality&lt;/a&gt; as a good book with a broad scope of math with a more theoretical slant.&lt;/p&gt;&#xA;" OwnerUserId="948" LastActivityDate="2011-01-06T22:14:01.433" CommunityOwnedDate="2011-01-06T22:14:01.433" />
  <row Id="2582" PostTypeId="1" AcceptedAnswerId="2590" CreationDate="2011-01-06T22:15:19.883" Score="3" ViewCount="119" Body="&lt;p&gt;I'm guessing this isn't a great physics question, but I just can't find an answer with Google.&#xA;If the galactic plane is perfectly horizontal at what angle and rotation will the ecliptic plane of our Solar System be?&lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2011-01-07T02:21:29.627" Title="The Galactic Plane" Tags="&lt;astrophysics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="2584" PostTypeId="2" ParentId="193" CreationDate="2011-01-06T22:52:29.507" Score="0" ViewCount="" Body="&lt;p&gt;Nice question. I don't know much about either differential geometry or algebraic topology, but having studied groups a little, I think I can provide some references for Lie groups. So here are the books I found useful&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://math.stanford.edu/~ykatznel/samelson/liealgeb.pdf&quot; rel=&quot;nofollow&quot;&gt;Samelson, Notes on Lie Algebras&lt;/a&gt; written in a &lt;em&gt;Definition, Theorem, Proof&lt;/em&gt; style, so it's little hard to grasp (I recommend mutliple rereading) but gives a good overview of the structure, classification (root systems and Dynkin diagrams) and representations (highest weight theory) of Lie algebras.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387900535&quot; rel=&quot;nofollow&quot;&gt;Humphreys, Introduction to Lie Algebras and Representation Theory&lt;/a&gt; less theorem-heavy and more talkative than &lt;em&gt;Samelson&lt;/em&gt; and contains huge number of great exercises.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387974954&quot; rel=&quot;nofollow&quot;&gt;Fulton, Harris, Representation Theory A First Course&lt;/a&gt; discusses more or less everything a physicist needs to know about groups (also mentions some finite groups). Lacks the systematic theorem based approach of the two books above, but boasts great explanations and nice pictures. I'd suggest it as a nice first reading about groups it if weren't for its length.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521663482&quot; rel=&quot;nofollow&quot;&gt;Goodman, Wallach, Representations and Invariants of the Classical Groups&lt;/a&gt; this is an ultimate bible on groups. Authors take an algebraic geometrical approach to the Lie groups (instead of the usual differential geometrical) which makes the book somewhat hard to read for a regular physicist. But besides this the book provides an in-depth look at lots of concrete representations (e.g. tensor representations and connection with symmetric group; this is often omitted elsewhere), discusses highest weight theory at great length, provides a nice introduction to spinors and also mentions branching rules. And lots of other stuff. Definitely recommended.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-01-06T22:52:29.507" CommunityOwnedDate="2011-01-06T22:52:29.507" />
  <row Id="2585" PostTypeId="1" AcceptedAnswerId="2874" CreationDate="2011-01-06T23:05:15.163" Score="5" ViewCount="314" Body="&lt;p&gt;It is true for fermions in the same potential that the total wavefunction of two particles must be antisymmetric with respect to exchange of electrons. Which means the spin wavefunction is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\chi=\frac{1}{\sqrt{2}}[\chi_+ (1)\chi_- (2)-\chi_+ (2)\chi_- (1)] $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which looks very much like the bell state, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\beta_{11}=\frac{1}{\sqrt{2}}[ |01\rangle - |10 \rangle]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, can we call those fermions, entangled states, as long as they are within the potential or there is something fundamentally special about entangled states (e.g. difference in measurement statistics) which makes them more unique?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies if the question is too simple for the level of this website. However, apparently it has made a lot of confusion for many people!&lt;/p&gt;&#xA;" OwnerUserId="304" LastActivityDate="2011-01-14T12:16:19.523" Title="Pauli exclusion principle and Entangled pairs" Tags="&lt;quantum-mechanics&gt;&lt;measurement&gt;&lt;entanglement&gt;" AnswerCount="3" CommentCount="10" />
  <row Id="2586" PostTypeId="1" CreationDate="2011-01-06T23:28:52.287" Score="9" ViewCount="294" Body="&lt;p&gt;On one hand, in the classical electrodynamics polarization of transparent media yields in lowering the speed of light by the factor of $n=\sqrt{\epsilon_r \mu_r}$ (refractive index).&#xA;On the other, in the quantum field theory the vacuum polarization does not decrease the speed of light. The thing it does is increase of strength of electromagnetic interaction. Why is it so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My two guesses are the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In principle the vacuum polarization does decrease speed of light ($\rightarrow 0$) but we implicitly put on-shell renormalization to keep $c$ constant.&lt;/li&gt;&#xA;&lt;li&gt;For some reason the process is completely different from polarization of transparent media.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="184" LastActivityDate="2011-01-16T21:23:08.103" Title="Why the vacuum polarization does not decrease the speed of light?" Tags="&lt;quantum-field-theory&gt;&lt;special-relativity&gt;&lt;speed-of-light&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="2587" PostTypeId="2" ParentId="2586" CreationDate="2011-01-06T23:53:38.230" Score="5" ViewCount="" Body="&lt;p&gt;In transparent media, there is matter present. The refractive index has its origin in the tree level process of $$e^- \gamma \to e^- \to e^- \gamma$$ (with $e^-$ either a part of the atom or free as in metals). There is no need for renormalization here, it's just a standard contribution to the probability amplitude of photon propagation. There are also other processes (interaction with nuclei and whatever loop corrections you can think of) but these are only second (and more) order corrections. In any case, what happens here is that photons gain effective mass via all those interactions, which is the ultimate reason they propagate slower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In vacuum the relevant process is a loop-level one $$\gamma \to e^+e^- \to \gamma$$. You'll obtain an UV divergence here and to take care of it introduce a renormalization of the field which is what that process corresponds to in the QED Lagrangian (the relevant term being $F^2 = (\partial A)^2$. But this is just a kinetic term, not a mass term (which would be something like $m_A^2 A^2$). You could imagine adding such a term to the Lagrangian but that is of course inconsistent with the observation. We &lt;em&gt;know&lt;/em&gt; that photons travel at the speed of light (in the stronger sense of special relativity that is built-in in every QFT).&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-01-07T01:26:04.490" LastActivityDate="2011-01-07T01:26:04.490" CommentCount="2" />
  <row Id="2588" PostTypeId="2" ParentId="2579" CreationDate="2011-01-07T00:29:38.883" Score="5" ViewCount="" Body="&lt;p&gt;I think there is an interesting question in there, but you are not quite there yet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mental picture of black holes in string theory is as collection of many microscopic ingredients in some generic complicated state, which looks for almost all intents and purposes as thermal state made of these constituents. If you think about an analogy to a gas made of many molecules, your questions are analogous to asking what is the smallest possible gas and what is the fate of such a gas. Roughly speaking, when the system becomes smaller the thermodynamic description becomes less accurate, and at some stage you'd have to discuss the dynamics of the individual constituents, none of which is of course described by thermodynamics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, when black holes become smaller, they become highly curved and are described less and less accurately by their geometry alone, ultimately you'd need to discuss the microscopic physics of their constituents. I personally don't expect that individual constituents are usefully described as black holes, or any other classical geometry for that matter, at least not in general. However, the fuzzball program of Mathur's seeks exactly such description, at least partially and under favourable conditions, so who knows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To paraphrase slightly what I am saying is that black holes in GR should be thought of as coarse grained (thermodynamic) description of a complicated system (whose details, amounting to a statistical mechanical description of the same system, are exactly known in string theory in some special cases). When the black hole gets smaller, corresponding to your micro-black holes question, you are starting to talk about a small collection of those constituents. Those objects are no longer black holes, as those are defined by some aspect of their geometry, which is no longer a useful description of the system.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-01-07T02:14:01.670" LastActivityDate="2011-01-07T02:14:01.670" CommentCount="3" />
  <row Id="2589" PostTypeId="2" ParentId="2579" CreationDate="2011-01-07T02:00:43.290" Score="3" ViewCount="" Body="&lt;p&gt;One can argue formally using the AdS/CFT correspondence that black hole formation and evaporation does not destroy information, contrary to earlier claims by Hawking that it must. This was the basis for Hawking's payoff of a bet he made with John Preskill. The bet, and a brief discussion of some of the issues involved can be found here &lt;a href=&quot;http://www.theory.caltech.edu/~preskill/jp_24jul04.html&quot; rel=&quot;nofollow&quot;&gt;http://www.theory.caltech.edu/~preskill/jp_24jul04.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although Hawking claimed to have solved the problem in a press conference, credit is really due to J. Maldacena for AdS/CFT and for his later work using this to argue for the unitarity of BH formation and evaporation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Important caveats are that there is no proof of AdS/CFT (although at this point it is very hard to believe it is not correct) and that the argument really doesn't tell one very explicitly just how the information manages to be encoded in the outgoing Hawking radiation.&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T04:11:13.517" LastActivityDate="2011-03-02T04:11:13.517" CommentCount="3" />
  <row Id="2590" PostTypeId="2" ParentId="2582" CreationDate="2011-01-07T02:21:29.627" Score="1" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/tXOrO.jpg&quot; alt=&quot;Galactic Plane and Solar Orienation&quot;&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/TX8kf.jpg&quot; alt=&quot;Galactic Coordinates and Earth&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer seems to near &lt;a href=&quot;http://cs.astronomy.com/asycs/forums/p/31176/368880.aspx&quot; rel=&quot;nofollow&quot;&gt;60 degrees&lt;/a&gt;. Additional description &lt;a href=&quot;http://stars.astro.illinois.edu/celsph.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found this mathematica description of the ecliptic plane relative to the galactic plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://scienceworld.wolfram.com/astronomy/GalacticCoordinates.html&quot; rel=&quot;nofollow&quot;&gt;Wolfram.com&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-01-07T02:21:29.627" CommentCount="3" />
  <row Id="2591" PostTypeId="2" ParentId="2552" CreationDate="2011-01-07T02:42:27.563" Score="1" ViewCount="" Body="&lt;p&gt;If you know the motion path, then you know what the acceleration of the mass should be at any angle $\theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Position&lt;/strong&gt;&#xA;$$\vec{r}=[\mbox{-}R\,\sin\theta,R\,\cos\theta]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Velocity&lt;/strong&gt;&#xA;$$\vec{v}=[\mbox{-}R\dot{\theta}\,\cos\theta,\mbox{-}R\dot{\theta}\,\sin\theta]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Acceleration&lt;/strong&gt;&#xA;$$\vec{a}=[R\dot{\theta}^{2}\,\sin\theta-R\ddot{\theta}\,\cos\theta,\mbox{-}R\dot{\theta}^{2}\cos\,\theta-R\ddot{\theta}\,\sin\theta]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;Sum of the forces&lt;/strong&gt; acting on the mass are&#xA;$$\sum\vec{F}=[T\,\sin\theta,\mbox{-}T\,\cos\theta]-[0,m\, g]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newtons law of motion is used to solve for the angular acceleration $\ddot{\theta}$ and the tension $T$ (must be positive).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ddot{\theta}=\frac{g}{R}\,\sin\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T=m\,R\,\dot{\theta}^2-m\,g\,\cos\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to get the tension, you not only need the angle $\theta$, but also the angular speed $\dot{\theta}$.&lt;/p&gt;&#xA;" OwnerUserId="392" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2011-01-07T02:47:52.027" LastActivityDate="2011-01-07T02:47:52.027" />
  <row Id="2592" PostTypeId="2" ParentId="193" CreationDate="2011-01-07T02:46:16.327" Score="2" ViewCount="" Body="&lt;p&gt;Physics Reports 66:  Gravitation, Gauge Theories, and Geometry (Eguchi, Gilkey and Hanson).&lt;/p&gt;&#xA;" OwnerUserId="20" LastActivityDate="2011-01-07T02:46:16.327" CommunityOwnedDate="2011-01-07T02:46:16.327" />
  <row Id="2593" PostTypeId="2" ParentId="2579" CreationDate="2011-01-07T02:58:29.537" Score="2" ViewCount="" Body="&lt;p&gt;Moshe's answer gets the main point across. If you're interested in trying to learn something more detailed, here's a thought about where you might start. (I'm not even close to expert on these matters, but this is something that I've found readable and worthwhile.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's an interesting thread of literature including &lt;a href=&quot;http://arxiv.org/abs/hep-th/9309145&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9309145&lt;/a&gt; by Susskind and &lt;a href=&quot;http://arxiv.org/abs/hep-th/9612146&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9612146&lt;/a&gt; by Horowitz and Polchinski, along with other papers that those might lead you to. The idea is that free strings have a certain entropy, in the sense that for a given mass, there are many different combinations of oscillator modes you can turn on to find a string state of that mass. The string energy is proportional to its length, and a typical string of a given energy looks roughly like a random walk of a particular length. Once you turn on a coupling, at some point some of these states will become black holes, because their energy is contained in a region smaller than their Schwarzschild radius. For a given mass, a black hole state also has some entropy. There are various consistency checks you can do on the way these things scale, to see that it's sensible to talk about a sort of smooth transition between string states and black hole states if you fix the energy and vary the coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are more technical and more precise papers in the literature on black hole entropy and microstate counting, but for a non-expert like me this particular thread of literature seems interesting because it relies on parametric scaling laws that are pretty readily comprehensible, and paints a relatively clear picture of the physics.&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="19" LastEditorDisplayName="" LastEditDate="2011-01-07T03:05:45.190" LastActivityDate="2011-01-07T03:05:45.190" CommentCount="1" />
  <row Id="2594" PostTypeId="1" AcceptedAnswerId="2601" CreationDate="2011-01-07T07:30:04.437" Score="9" ViewCount="470" Body="&lt;p&gt;A lot of string-theory questions have been asked here. This one is, hopefully, different in that this inquiry is specific and the expected answer would be more mathematical than philosophical in character.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give some context, I'm familiar with LQG, QFT etc. I'm a beginner in CFT and string-theory. On reading papers on AdS/CFT and String Theory the following sentence often appears: &quot;&lt;strong&gt;The low-energy effective action for $x$ dimensional ST of type $y$ reduces to that of supergravity (SUGRA) in $z$ dimensions with matter fields of types $a$, $b$ and $c$&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $64,000 dollar question any noob, such as me, wants answered is what are the mathematical/technical steps or arguments which lead us from the fundamental action of string-theory (the Polyakov/Nambu-Goto action) to a low-energy effective description. Of course, I might be asking a question for which the only concise answer is: &quot;&lt;em&gt;go read GSW&lt;/em&gt;&quot;. Here's hoping otherwise ;)&lt;/p&gt;&#xA;" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-07T11:03:40.797" LastActivityDate="2011-03-02T04:10:53.423" Title="Steps for going from Polyakov action to low-energy effective action (SUGRA?) in String Theory" Tags="&lt;quantum-field-theory&gt;&lt;string-theory&gt;&lt;effective-action&gt;&lt;supergravity&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="2595" PostTypeId="2" ParentId="2586" CreationDate="2011-01-07T10:15:59.637" Score="-1" ViewCount="" Body="&lt;p&gt;The &quot;vacuum polarization&quot; is just like &quot;&lt;a href=&quot;http://physics.stackexchange.com/questions/1686/why-does-the-mass-of-an-object-increase-when-its-speed-approaches-that-of-light/1691#1691&quot;&gt;relativistic mass&lt;/a&gt;&quot;.&lt;br&gt;&#xA;Both concepts were introduced &quot;to make things simpler&quot;, while they just lead to confusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;vacuum polarization&quot; is an artefact of &lt;a href=&quot;http://en.wikipedia.org/wiki/SI_electromagnetism_units&quot; rel=&quot;nofollow&quot;&gt;SI units&lt;/a&gt;. In SI units magnetic and electric fields are measured in different units. This is very unnatural, since these fields are just components of one tensor. It is like measuring height and length in different units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to this unnaturalness one have to introduce extra dimensional quantities to &quot;convert&quot; between magnetic, electric and other units: vacuum permittivity and permeability. And these quantities have not too much physical sence -- it is like 3.2808399 between meter and foot -- just a matter of convention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there is no realtion of this &quot;vacuum polarization&quot; to the polarization of vacuum in QFT.&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-01-07T10:15:59.637" CommentCount="2" />
  <row Id="2596" PostTypeId="1" AcceptedAnswerId="2603" CreationDate="2011-01-07T12:30:35.767" Score="14" ViewCount="704" Body="&lt;p&gt;shortly after the invention of &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_electrodynamics&quot; rel=&quot;nofollow&quot;&gt;quantum electrodynamics&lt;/a&gt;, one discovered that the theory had some very bad properties. It took twenty years to discover that certain &lt;strong&gt;infinities&lt;/strong&gt; could be overcome by a process called &lt;a href=&quot;http://en.wikipedia.org/wiki/Renormalization&quot; rel=&quot;nofollow&quot;&gt;renormalization&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One might state the physical reason behind this that we are only aware of &lt;strong&gt;effective theories&lt;/strong&gt; which are reliable on certain &lt;strong&gt;scales&lt;/strong&gt; given by more or less fundamental constants. Renormalization tells us how to deal with this situation and only consider effects of a specific range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;technique&lt;/strong&gt; to perform the calculations is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Renormalization_group&quot; rel=&quot;nofollow&quot;&gt;renormalization group&lt;/a&gt;. It is a powerful tool and it is no wonder it is under heavy development since nothing can be calculated without it in quantum field theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per-se, this procedure is not limited to its root and one might ask the &lt;strong&gt;question&lt;/strong&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;How can we use the renormalization group to find effective theories for classical field theories?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I suppose, an &lt;strong&gt;example&lt;/strong&gt; where this has been done very recently can be found in &lt;a href=&quot;http://arxiv.org/abs/1012.0461&quot; rel=&quot;nofollow&quot;&gt;Renormalization Group Analysis of Turbulent Hydrodynamics&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be thankful for any insight, examples etc.&lt;br&gt;&#xA;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS.: The question is naturally linked to &lt;a href=&quot;http://physics.stackexchange.com/questions/2506/how-to-calculate-the-properties-of-photon-quasiparticles&quot;&gt;How to calculate the properties of Photon-Quasiparticles&lt;/a&gt; and in a loose line with &lt;a href=&quot;http://physics.stackexchange.com/questions/1799/a-pedestrian-explanation-of-conformal-blocks&quot;&gt;A pedestrian explanation of conformal blocks&lt;/a&gt;.&lt;br&gt;&#xA;Since I am no expert in the field please advice me if something is not clear or simply wrong.&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2011-01-08T18:01:38.043" LastActivityDate="2011-06-09T19:01:17.173" Title="A pedestrian explanation of Renormalization Groups - from QED to classical field theories" Tags="&lt;quantum-field-theory&gt;&lt;effective-action&gt;&lt;renormalization&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="6" />
  <row Id="2597" PostTypeId="1" AcceptedAnswerId="2609" CreationDate="2011-01-07T13:32:40.143" Score="8" ViewCount="401" Body="&lt;p&gt;I understand that energy conservation is not a rule in general relativity, but I'd like to know under what circumstances it can still be possible. In other words, when is it possible to associate a potential energy to the gravitational field, so that the energy is constant in the evolution of the system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some examples, is there a convenient way to define energy in these scenarios?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Just a system of gravitational waves.&lt;/li&gt;&#xA;&lt;li&gt;A point mass moving in a static (but otherwise arbitrary) space-time. Equivalent (if I'm not mistaken) to a test mass moving in the field of a second much larger mass, the larger mass wouldn't move.&lt;/li&gt;&#xA;&lt;li&gt;Two rotating bodies of similar mass.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Overall, I'm trying to understand what keeps us from associating a potential energy to the metric. When we break the time translation symmetry of a system by introducing an electromagnetic field, we can still conserve energy by defining an electromagnetic potential energy. &lt;strong&gt;Why can't we do the same when we break TT symmetry by making space-time curved?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="724" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-07T15:31:40.017" LastActivityDate="2011-01-20T15:55:25.783" Title="Energy conservation in General Relativity" Tags="&lt;general-relativity&gt;&lt;energy-conservation&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="2598" PostTypeId="1" AcceptedAnswerId="2886" CreationDate="2011-01-07T13:40:17.090" Score="6" ViewCount="586" Body="&lt;p&gt;In a recent discussion about black holes, space_cadet provided me with the following paper of Rovelli: &lt;a href=&quot;http://arxiv.org/abs/gr-qc/9603063&quot; rel=&quot;nofollow&quot;&gt;Black Hole Entropy from Loop Quantum Gravity&lt;/a&gt; which claims to derive the Bekenstein-Hawking formula for the entropy of the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parts of his derivation seem strange to me, so I hope someone will able to clarify them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the computation hangs on the notion of &lt;em&gt;distinguishable&lt;/em&gt; (by an outside observer) states. It's not clear to me how does one decide which of the states are distinguishable and which are not. Indeed, Rovelli mentions a different paper that assumes different condition and derives an incorrect formula. It seems to me that the concept of Rovelli's distinctness was arrived at either accidentally or a posteriori to derive the correct entropy formula.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the concept of distinguishable states discussed somewhere more carefully?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After this assumption is taken, the argument proceeds to count number of ordered partitions of a given number (representing the area of the black hole) and this can easily be seen exponential by combinatorial arguments, leading to the proportionality of the area and entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it turns out that the constant of proportionality is wrong (roughly 12 times smaller than the correct B-H constant). Rovelli says that this is because number of issues were not addressed. The correct computation of area would also need to take the effect of nodes intersecting the horizon. It's not clear to me that addressing this would not spoils the proportionality even further (instead of correcting it). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Has a more proper derivation of the black hole entropy been carried out?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-01-28T14:54:49.270" Title="Discussion of the Rovelli's paper on the black hole entropy in Loop Quantum Gravity" Tags="&lt;quantum-gravity&gt;&lt;loop-quantum-gravity&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="2599" PostTypeId="2" ParentId="2596" CreationDate="2011-01-07T14:12:49.710" Score="4" ViewCount="" Body="&lt;p&gt;First, the full notion of renormalization group, as studied in QFT, is definitely not needed in the classical theory. This is because QFT actually doesn't make sense without a renormalization scheme and for any theory one has to always investigate the flow of couplings towards some fixed points (corresponding to conformal field theories) to check whether the given theory is renormalizable in the first place. So renormalization is an integral part of the QFT in the great contrast with classical theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other place where notion of renormalization group flow is important is condensed matter theory. This is because these flows have fixed points that (when they are non-trivial) correspond to critical points (this is again connected with the mentioned conformal symmetry). The renormalization group is then used to see how the flow behaves around this point and this gives valuable information about the behavior of macroscopic quantities (like specific heat) at the critical point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the notion by itself is not terribly important if all you care about is integrating UV degrees of freedom. I don't think you need any &lt;em&gt;flow&lt;/em&gt; in classical theory. All you have to do is integrate some energetic interaction with an ambient field to obtain an effective mass (for a one concrete example). While renormalization group gives a useful framework for general understanding of scales of theories and their being effective, it's not really needed most of the time.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-01-07T14:12:49.710" CommentCount="6" />
  <row Id="2600" PostTypeId="1" AcceptedAnswerId="2611" CreationDate="2011-01-07T15:30:42.260" Score="4" ViewCount="24" Body="&lt;p&gt;I'm in the process of trying to understand X-Ray Absorption Fine Structure (XAFS) spectroscopy, but I think this question is more fundamental than merely as it pertains to XAFS.  I think I don't entirely understand the interplay between electronic, vibrational, and rotational excitations in molecules.  I understand that vibrational fine structure can be seen in electronic spectra.  But how do we know which vibrational state a transition will end up in?  Is it a Boltzmann distribution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to my main question. I realize that on the energy scale of XAFS, vibrational energy levels are not resolved; I just want to know if these transitions lead to the same type of vibrational state distribution that valence electronic transitions yield.  Is that the case?  Or if not, why not?&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2011-01-07T20:45:09.857" Title="Do XAFS excitations and subsequent relaxations lead to vibrationally hot molecules?" Tags="&lt;spectroscopy&gt;&lt;x-ray&gt;" AnswerCount="1" />
  <row Id="2601" PostTypeId="2" ParentId="2594" CreationDate="2011-01-07T15:31:33.370" Score="5" ViewCount="" Body="&lt;p&gt;The basic answer is indeed &quot;go read GSW or Polchinski&quot; but if your goal was to get from the&#xA;Polyakov action to a low-energy effective description as quickly as possible there are parts you could skip. In brief you want to&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Work out the spectrum of the first quantized string. The &quot;best&quot; method is modern, covariant quantization, but the quickest method is light-cone quantization. If you want to know how to do this for say type IIA or IIB superstring you need to read GSW Chap. 2 and 4.3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Deduce the existence of spacetime supersymmetry in the theory. This is the content of GSW Chap. 5.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is a unique low-energy action with the massless field content and spacetime supersymmetry of type IIA or IIB string theory, namely the actions for IIA and IIB Sugra, so you know on general grounds that the action for IIA or IIB supergravity must be the low-energy effective action for IIA or IIB string theory.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the general argument of 3. leaves you feeling uneasy you can remove this unease by computing a variety of 3 and 4 point tree amplitudes directly in string theory (now you need to read Chap. 7 of GSW), taking their low-energy limit, and comparing to the 3 and 4 point amplitudes you compute using IIA and IIB Sugra. You will find perfect agreement.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you want to generalize this to include additional matter fields you will need to extend your analysis to the heterotic string and/or learn how to include D-branes into your type II analysis. Now you need to start reading Polchinski or various reviews, but you'll have a good place to start if you have done 1,2 and part of 4.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T04:10:53.423" LastActivityDate="2011-03-02T04:10:53.423" CommentCount="1" />
  <row Id="2602" PostTypeId="1" CreationDate="2011-01-07T16:23:50.823" Score="4" ViewCount="209" Body="&lt;p&gt;If you take two ideal gases at different temperatures, and allow them to share energy through heat, they'll eventually reach a thermodynamic equilibrium state that has higher entropy than the original. The time evolution of that entropy increase is easy to predict, because we know the time evolution of $T$ and $U$ for each of the gases (assuming we have the necessary constants). This is OK.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take a system beyond the scope of thermodynamics. A single box containing a gas that is not in thermodynamic equilibrium (doesn't follow the Boltzman distribution). One would expect that gas to quickly thermalise, and reach said equilibrium. Entropy can still be defined using statistical mechanics, and the final state will have higher entropy than the initial state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for quantitative experimental evidence of this effect.&lt;br&gt;&#xA;&lt;strong&gt;Ideally, I'd like references measuring the time a gas takes to thermalise.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously this time depends on many factors and is not always doable. I wasn't more specific because I don't want to be picky, I'm looking for any experiments that verify it.&lt;/p&gt;&#xA;" OwnerUserId="724" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2011-01-08T23:44:43.473" LastActivityDate="2011-01-15T17:49:09.433" Title="Experiments that measure the time a gas takes to reach equilibrium" Tags="&lt;experimental-physics&gt;&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="2603" PostTypeId="2" ParentId="2596" CreationDate="2011-01-07T17:12:49.177" Score="5" ViewCount="" Body="&lt;p&gt;I'm not an expert in this topic too, but I'm trying wrap my head around it.&lt;br&gt;&#xA;Right now I'm trying to  make an adequate hierarchy of concepts related to renormalization.   Let me list them and tell how they are related:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Fields, Lagrangian (Hamiltonian) and coupling constants.   &lt;/li&gt;&#xA;&lt;li&gt;Perturbative calculations.&lt;/li&gt;&#xA;&lt;li&gt;Different scales.&lt;/li&gt;&#xA;&lt;li&gt;Self-similarity.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Quantum&lt;/strong&gt; fields.&lt;/li&gt;&#xA;&lt;li&gt;Ultraviolet divergences.&lt;/li&gt;&#xA;&lt;li&gt;Renormalization.&lt;/li&gt;&#xA;&lt;li&gt;Renormalization &lt;strong&gt;group&lt;/strong&gt; and running couplings.&lt;br&gt;&#xA;(Let me stress that the &quot;renormalization&quot; and &quot;renormalization group&quot; are different concepts.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Of course the concept of a field and the way to describe (1) it is a starting point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, It seems to me (while I can be wrong) that every time we talk about renormalization we always deal with some perturbative approach (2). There is always something that we want to neglect. And if there is a way to make calculations without any approximations then one needn't to use techniques related to renormalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of simplest examples is a hydrodynamics -- you don't want to &quot;get down&quot; to the level of molecules to describe a stream of water. You would like to work with some &quot;integral&quot; quantities, like viscosity. And the viscosity can be used to describe processes at many different scales (3): bloodstream, butterfly, submarine, internals of the star, e.t.c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydrodynamics works at different scales because of the self-similarity (4): by going several orders of magnitude larger you are still able to describe your system with the same Lagrangian, but, maybe, with some parameters changed. When one does the transition from one scale to another one always neglects some peculiarities (2), that occur at smaller scale. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the essence of &lt;a href=&quot;http://en.wikipedia.org/wiki/Renormalization_group&quot; rel=&quot;nofollow&quot;&gt;renormalization group&lt;/a&gt;(8) techniques. The changing parameters are also called the running couplings. I recommend you to read about Kadanoff transformation, to get more insight about it.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I never mentioned divergences by far. Because this is a slightly different topic. And one can use renormalization group even if there is no infinities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UV divergences appear due to our ignorance about smaller scales. When we talk about hydrodynamics we know that there is a &quot;fundamental scale&quot; -- the aforementioned molecules. But when we talk about &lt;strong&gt;quantum&lt;/strong&gt; fields (6) (like electromagnetic field or some fermion field) we don't know what is the &quot;fundamental&quot; scale for it. We don't even know if it exists at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different methods of dealing with the divergences are called the renormalization (7) methods. They are based on changes of the parameters of Larangian too, but now these changes are &quot;infinite&quot; because one have to &quot;compensate the infinities&quot; appearing from small scales. After cancelling the infinities this way one is still left with arbitrariness of choosing a finite values of the parameters. You can fix the parameters by getting them from experiment at certain scale(3) and use renormalization group (8) to go from one scale to another. &lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2011-01-11T15:25:35.667" LastActivityDate="2011-01-11T15:25:35.667" CommentCount="2" />
  <row Id="2604" PostTypeId="1" CreationDate="2011-01-07T17:19:41.587" Score="13" ViewCount="777" Body="&lt;p&gt;Background: I have written a pop-science book explaining quantum mechanics through &lt;a href=&quot;http://dogphysics.com/&quot; rel=&quot;nofollow&quot;&gt;imaginary conversations with my dog&lt;/a&gt;-- the dog serves as a sort of reader surrogate, popping in occasionally to ask questions that a non-scientist might ask-- and I am now working on a sequel. In the sequel, I find myself having to talk about particle physics a bit, which is not my field, and I've hit a dog-as-reader question that I don't have a good answer to, which is, basically, &quot;What purpose, if any, do higher-generation particles serve?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it in slightly more physics-y terms: The Standard Model contains twelve material particles: six leptons (the electron, muon, and tau, plus associated neutrinos) and six quarks (up-down, strange-charm, top-bottom). The observable universe only uses four, though: every material object we see is made up of electrons and up and down quarks, and electron neutrinos are generated in nuclear reactions that move between different arrangements of electrons and up and down quarks. The other eight turn up only in high-energy physics situation (whether in man-made accelerators, or natural occurances like cosmic ray collisions), and don't stick around for very long before they decay into the four common types. So, to the casual observer, there doesn't seem to be an obvious purpose to the more exotic particles. So why are they there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if there is some good reason why the universe as we know it has to have twelve particles rather than just four. Something like &quot;Without the second and third generations of quarks and leptons, it's impossible to generate enough CP violation to explain the matter-antimatter asymmetry we observe.&quot; Only probably not that exact thing, because as far as I know, there isn't any way to explain the matter-antimatter asymmetry we observe within the Standard Model. But something along those lines-- some fundamental feature of our universe that requires the existence of muons and strange quarks and all the rest, and would prevent a universe with only electrons and up and down quarks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is not &quot;why do we think there are there three generations rather than two or four?&quot; I've seen the answers to that here and elsewhere. Rather, I'm asking &quot;Why are there three generations rather than only one?&quot; Is there some important process in the universe that requires there to be muons, strange quarks, etc. for things to end up like they are? Is there some reason beyond &quot;we know they exist because they're there,&quot; something that would prevent us from making a universe like the one we observe at low energy using only electrons, up and down quarks, and electron neutrinos?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any pointers you can give to an example of some effect that depends on the presence of the higher Standard Model generations would be much appreciated. Having it already in terms that would be comprehensible to a non-scientist would be a bonus.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2011-02-21T17:42:00.530" Title="What Do We Get From Having Higher Generations of Particles?" Tags="&lt;particle-physics&gt;&lt;theory&gt;&lt;standard-model&gt;" AnswerCount="7" CommentCount="10" FavoriteCount="3" />
  <row Id="2605" PostTypeId="1" CreationDate="2011-01-07T18:34:10.150" Score="28" ViewCount="24798" Body="&lt;p&gt;When your body burns calories and you lose weight, obviously mass is leaving your body.  In what form does it leave?  In other words, what is the physical process by which the body loses weight when it burns its fuel?  Someone asked this question on the whiteboard at work.  Someone said it leaves the body in the form of heat but I knew this is wrong since heat is simply the internal kinetic energy of a lump of matter and doesn't have anything do with mass.  Obviously the chemical reactions going on in the body cause it to produce heat but this alone won't reduce its mass.&lt;/p&gt;&#xA;" OwnerUserId="1114" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-01-07T22:43:23.870" LastActivityDate="2011-06-03T18:55:43.990" Title="How Does Mass Leave the Body When you Lose Weight?" Tags="&lt;thermodynamics&gt;&lt;mass&gt;" AnswerCount="11" CommentCount="20" FavoriteCount="4" />
  <row Id="2606" PostTypeId="2" ParentId="2605" CreationDate="2011-01-07T19:02:32.693" Score="20" ViewCount="" Body="&lt;p&gt;Essentially, losing of weight occurs by means of burning fuels precisely like your car does when it burns petrol and emits exhaust gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only difference is that for humans that fuel is to be found in the form of sugars. The fat is what you want to get ultimately rid off, of course, but sugars are more easily processed and so this is what you are removing first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic aerobic cycle is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Citric_acid_cycle&quot; rel=&quot;nofollow&quot;&gt;Krebs cycle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/DfKoD.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to reach it first, glucose needs to be broken down first to pyruvate (by an anaerobic process of glycolysis) and then to acetyl-CoA. One can gain some energy from this reduction but not much. The real energy is hiding in the actual Krebs cycle but for it one needs (besides lots of other stuff) the mentioned acetyl-CoA and oxygen (this explains why you don't get enough energy when not breathing properly) producing the carbon dioxide and some energy that is stored in the $ATP$ (adenosine-triphosphate) and transported to wherever it is needed inside the cell (you are mostly interested in muscle contractions performed by muscle cells). So, you'll burn whatever amount of sugar you have ready in the body. You'll also lose carbon (initially stored in the glucose) by exhaling in the form of $CO_2$. There is also additional hydrogen produced and carried away in the form of $NADH$ and $FADH_2$. It's hard to estimate where it will end up though, as it is (similarly to $ATP$) used all over the organism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the body is not storing sugars in the form of glucose. Instead, they are stored as a glycogen (mainly in liver and muscles) which is a polysaccharide similar to starch. This is then quickly broken down to glucose as needed. But the body can keep only a small amount of glycogen (corresponding roughly to an hour of running, depending on one's fitness).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another form of storage of sugar. Body can convert it to fat. This is done when there is already enough glycogen in the organism. The body fat can then be &lt;a href=&quot;http://en.wikipedia.org/wiki/Fatty_acid_metabolism&quot; rel=&quot;nofollow&quot;&gt;reduced&lt;/a&gt; to acetyl-CoA (by lipolyses and then by beta oxidation), but this requires a lot more oxygen and so is not used when glycogen is at the disposal. But with regular exercise body can be trained to also burn greater proportion of fats than glycogen (this is of course necessary for long-range runners and cyclists because there is no way they would get enough energy just from glycogen).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a rough idea about the amount of mass you'll burn, read the calories content of some food. Sugar has something like 4kCal for 1g and fat 9kCal for 1g. One hour of running corresponds to something like 700kCal so if you are burning 50% sugars and 50% fat you'll be 100g lighter. All of these numbers are just very rough estimates depending on what kind of exercise you do and the general state of your body.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note that you'll also lose lots of water and minerals during the exercise. But I am not counting this to the mass balance as you need to replenish those in order to be healthy. Also, gradually some muscles will form, so this will actually &lt;em&gt;add&lt;/em&gt; weight.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-01-14T11:55:02.480" LastActivityDate="2011-01-14T11:55:02.480" CommentCount="19" />
  <row Id="2607" PostTypeId="2" ParentId="2605" CreationDate="2011-01-07T19:15:11.783" Score="19" ViewCount="" Body="&lt;p&gt;When you exercise, you &quot;burn&quot; more glucose, the simplified reaction for which (&lt;a href=&quot;http://en.wikipedia.org/wiki/Cellular_respiration&quot; rel=&quot;nofollow&quot;&gt;from Wikipedia&lt;/a&gt;) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\rm C_6H_{12}O_6 + 6~O_2 → 6~CO_2 + 6~H_2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when you exhale, the carbon in the carbon dioxide, and the hydrogen and the oxygen in the water vapor, came from the glucose being burned, thereby removing that mass from the body.&lt;/p&gt;&#xA;" OwnerUserId="292" LastActivityDate="2011-01-07T19:15:11.783" CommentCount="5" />
  <row Id="2609" PostTypeId="2" ParentId="2597" CreationDate="2011-01-07T19:46:42.917" Score="9" ViewCount="" Body="&lt;p&gt;There are a few different ways of answering this one.  For brevity, I'm going to be a bit hand-wavey.  There is actually still some research going on with this.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certain spacetimes will always have a conserved energy.  These are the spacetimes that have what is called a global timelike (or, if you're wanting to be super careful and pedantic, perhaps null) Killing vector.  Math-types will define this as a vector whose lowered form satisfies the Killing equation: $\nabla_{a}\xi_{b} + \nabla_{b} \xi_{a} = 0$.  Physicists will just say that $\xi^{a}$ is a vector that generates time (or null) translations of the spacetime, and that Killing's equation just tells us that these translations are &lt;b&gt;symmetries&lt;/b&gt; of the spacetime's geometry.  If this is true, it is pretty easy to show that all geodesics will have a conserved quantity associated with the time component of their translation, which we can interpret as the gravitational potential energy of the observer (though there are some new relativistic effects--for instance, in the case of objects orbiting a star, you see a coupling between the mass of the star and the orbiting objects angular momentum that does not show up classically).  The fact that you can define a conserved energy here is strongly associated with the fact that you can assign a conserved energy in any Hamiltonian system in which the time does not explicitly appear in the Hamiltonian--&gt; time translation being a symmetry of the Hamiltonian means that there is a conserved energy associated with that symmetry.  If time translation is a symmetry of the spacetime, you get a conserved energy in exactly the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, you can have a surface in the spacetime (but not necessarily the whole spacetime) that has a conserved killing tangent vector.  Then, the argument from above still follows, but that energy is a charge living on that surface.  Since integrals over a surface can be converted to integrals over a bulk by Gauss's theorem, we can, in analogy with Gauss's Law, interpret these energies as the energy of the mass and energy &lt;b&gt;inside&lt;/b&gt; the surface.  If the surface is conformal spacelike infinity of an asymptotically flat spacetime, this is the ADM Energy.  If it is conformal null infinity of an asymptotically flat spacetime, it is the Bondi energy.  You can associate similar charges with Isolated Horizons, as well, as they have null Killing vectors associated with them, and this is the basis of the quasi-local energies worked out by York and Brown amongst others.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can't have is a tensor quantity that is globally defined that one can easily associate with 'energy density' of the gravitational field, or define one of these energies for a general spacetime.  The reason for this is that one needs a time with which to associate a conserved quantity conjugate to time.  But if there is no unique way of specifying time, and especially no way to specify time in such a way that it generates some sort of symmetry, then there is no way to move forward with this procedure.  For this reason, a great many general spacetimes have quite pathological features.  Only a very small proprotion of known exact solutions to Einstein's Equation are believed to have much to do with physics.   &lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-07T21:05:46.310" LastActivityDate="2011-01-07T21:05:46.310" CommentCount="6" />
  <row Id="2610" PostTypeId="2" ParentId="2364" CreationDate="2011-01-07T19:54:51.720" Score="0" ViewCount="" Body="&lt;p&gt;Aris' use of polar is broader than the chemical/electromagnetic usage cited above, but his use may, strictly speaking, include that case as well.  The distinction is mainly one of scale.  In a polar fluid such as water, the polar nature comes from the molecules themselves, as opposed to, say, long polymer chains or a magnetizable suspended particles. (To cite another, very different example, a suspension of tiny pear-shaped particles, or even micro-organisms, in a gravitational field would also be polar.)  If the polar nature is confined to the molecular level, as for water, then the effects at the meso/macro-scopic scales are mainly confined to that substances chemical and physical properties (viscosity, thermal conductivity, melting/boiling point, and the nature of the liquid state itself) and are not apparent as non-symmetric stresses.  On the other hand, if the polar nature of the fluid is due to larger particles (or long chain molecules) then there can be non-symmetric stresses and, because the particles can transmit these stresses to the surrounding carrier fluid, the flows of this fluid will exhibit non-Newtonian flow characteristics.&#xA;So: I don't think Aris had (molecularly) polar fluids in mind, but I do think he had in mind certain suspensions (i.e. ferrofluids and MR fluids) and polymer fluids.  To me what is interesting is whether a polar suspension could be used effectively as a &quot;model system&quot; to examine the nature of molecular polar fluids, such as water.&lt;/p&gt;&#xA;" OwnerUserId="1124" LastActivityDate="2011-01-07T19:54:51.720" />
  <row Id="2611" PostTypeId="2" ParentId="2600" CreationDate="2011-01-07T20:45:09.857" Score="1" ViewCount="" Body="&lt;p&gt;You have different questions in the title and in the body of your post and I will answer both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electronic excitation happens very fast compared to nuclear motion so the molecule remains &quot;frozen&quot; when it is brought from an equilibrium to an excited state. If the equilibrium geometry of the excited state is very similar to the ground state then the molecule will remain still, but if this is not the case, the molecule will start to oscillate around the equilibrium point. This is known as &lt;strong&gt;Frank-Condon principle&lt;/strong&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Franck%E2%80%93Condon_principle&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; has nice figures to illustrate this point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exciting different electrons leads to different electronic states that have different equilibrium geometries so you cannot expect same vibrational distributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your title you also ask about &lt;em&gt;relaxation&lt;/em&gt; of excited molecule. Since the energy of the X-Ray photon is much higher than the ionization energy, the excited molecule will be ionized before it can relax vibrationally.&lt;/p&gt;&#xA;" OwnerUserId="701" LastActivityDate="2011-01-07T20:45:09.857" />
  <row Id="2612" PostTypeId="2" ParentId="193" CreationDate="2011-01-07T21:14:56.883" Score="0" ViewCount="" Body="&lt;p&gt;Are you asking for an intro level book or a more advanced book for someone who already has some background in those topics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an introductory level, I second the Schutz and Spivak recommended above. Penrose and Frankel are suitable only if you already have had an introductory course in those subjects, in my opinion. Frankel's introduction to manifolds is very condensed, and Penrose is really providing  a bird's eye view while skipping over many details beginners would need to build basic intuitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best introductory notes I've come across for manifolds as used in GR are David Malament's, which you can download &lt;a href=&quot;http://www.lps.uci.edu/malament/FndsofGR/GR.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-07T21:14:56.883" CommunityOwnedDate="2011-01-07T21:14:56.883" />
  <row Id="2613" PostTypeId="2" ParentId="2605" CreationDate="2011-01-07T23:03:39.860" Score="11" ViewCount="" Body="&lt;p&gt;As others have pointed out, in metabolism you breathe in oxygen and exhale carbon dioxide with the net reaction being&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$CH_2O + O_2 \to CO_2 + H_2O$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The carbon dioxide gets exhaled, and the water is lost through some combination of other things.  Let's think about whether this is a reasonable way to explain weight loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A breath is maybe one liter of air.  I breathe about ten times a minute, so that's 15,000 liters of air a day.  Air is about 20% oxygen, so 3,000 liters of oxygen.  Air weighs about 1kg/m^3, so about 3kg of oxygen.  The air you exhale still has some oxygen.  I don't know how good we are at extraction, but let's say we get a third of it, so 1kg of oxygen used by our bodies each day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon is lighter than oxygen and there's only one carbon exhaled per two oxygens, so that's about 300g of carbon lost per day.  Fat is basically a long carbon chain, so that says that if I don't eat, I'll lose about 300 kg of body mass a day (assuming I do drink water).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not too bad a rough estimate to say that you lose weight through your nose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember in middle school science class we learned about an experiment in which some guy a long time ago grew an entire tree in a pot, carefully covering the pot to keep stuff from getting into the soil.  The entire pot+tree system gained roughly the weight of the tree by the end.  So basically the tree got all its mass from the air.  You are pretty much an inverse tree.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-01-09T17:14:05.860" LastActivityDate="2011-01-09T17:14:05.860" CommentCount="2" />
  <row Id="2614" PostTypeId="2" ParentId="2605" CreationDate="2011-01-08T00:05:42.520" Score="32" ViewCount="" Body="&lt;p&gt;There's a lot of detail you &lt;em&gt;could&lt;/em&gt; go into with regard to this question, as is done in the other answers and comments, but I think the answer itself is pretty simple. Imagine a surface that just barely surrounds your body, as if you shrink-wrapped a body in plastic. By the law of conservation of mass (valid in non-relativistic physics), the only way your body can lose any amount of mass is for that amount of mass to pass out through the surface. So you just have to consider what bodily functions cause that to happen. I think they've all been identified in the comments:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Excretion&lt;/li&gt;&#xA;&lt;li&gt;Exhaling&lt;/li&gt;&#xA;&lt;li&gt;Sweating&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Actually, any dead skin cells, strands of hair, etc. that fall off you would also count, although my guess is that those represent a minor contribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a bonus, the &quot;shrink-wrap view&quot; also makes it easy to identify the ways in which you gain mass, by looking for all processes that cause matter to be drawn in through the invisible surface:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Eating&lt;/li&gt;&#xA;&lt;li&gt;Inhaling&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The thing is, when most people talk about losing weight, they're referring to a long-term average loss of mass, which means that the processes in the first list have to remove more mass over some extended period of time than the ones in the second list bring in. This clearly requires some of the preexisting mass in your body to be converted into the waste forms that you can dispose of through excretion, exhaling, and sweating. This preexisting mass generally tends to be body fat. The other answers do a pretty good job filling in the details of &lt;em&gt;how&lt;/em&gt; the fat gets converted to waste products.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-01-08T00:05:42.520" CommentCount="7" />
  <row Id="2615" PostTypeId="1" AcceptedAnswerId="2616" CreationDate="2011-01-08T00:13:23.350" Score="6" ViewCount="301" Body="&lt;p&gt;Having read the Wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Superfluid&quot; rel=&quot;nofollow&quot;&gt;superfluids&lt;/a&gt; I'm still not sure if stuff like liquid helium at the lambda point actually have surface tension or not. Is superfluidity the same thing? And are there any liquids with no surface tension at room temperature?&lt;/p&gt;&#xA;" OwnerUserId="961" LastActivityDate="2011-01-09T00:17:12.167" Title="Are there any liquids with zero surface tension?" Tags="&lt;fluid-dynamics&gt;&lt;surface-tension&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="2616" PostTypeId="2" ParentId="2615" CreationDate="2011-01-08T00:34:40.940" Score="4" ViewCount="" Body="&lt;p&gt;Surface tension is not a property of materials but of interfaces between two (or more) materials. It is implicit in its definition that the interface separates two kinds of materials that behave differently (otherwise the interface would be just some imaginary surface inside the one material with no physical meaning) and so there must always be some surface tension that sustains the physical interface.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-01-08T00:34:40.940" CommentCount="2" />
  <row Id="2617" PostTypeId="1" AcceptedAnswerId="2618" CreationDate="2011-01-08T01:45:43.433" Score="2" ViewCount="134" Body="&lt;p&gt;In elasticity theory, general equations of motion are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho \partial^2_t \overline{u} = \mu \nabla^2 \overline{u} + (\mu+\lambda) \nabla(\nabla \cdot \overline{u})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\overline u$ are displacements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a thin homogeneus rod with tractions and compressions in the x direction, stress tensor $S_{ij}$ is:&#xA;$$ \pmatrix{&#xD;&#xA;S_{xx} &amp;amp;0 &amp;amp;0 \\&#xD;&#xA;0 &amp;amp;0 &amp;amp;0 \\&#xD;&#xA;0 &amp;amp;0 &amp;amp;0 \\} $$&#xA;Young modulus is:&#xA;$$ E=\frac{S_{xx}}{\epsilon_{11}}=\frac{\lambda+\mu}{\mu(3\lambda+2\mu)} &#xD;&#xA;$$&#xA;where $\epsilon_{11}=\partial_x u_x$ is the deformation on x  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So equation of motion for x-displacements is:&#xA;$$&#xD;&#xA;\rho \partial^2_t u_x=\partial_x S_{xx}= E \partial^2_x u_x= \frac{\lambda+\mu}{\mu(3\lambda+2\mu)} \partial^2_x u_x&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this different to the result of general equation of motion with only dependence to x:&#xA;$$&#xD;&#xA;\rho \partial^2_t u_x=(2\mu+\lambda)\partial^2_x u_x&#xD;&#xA;$$&#xA;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or: why in a elastic rod has the sound smaller speed than in a generic homogeneus indefinite medium?&lt;/p&gt;&#xA;" OwnerUserId="558" LastEditorUserId="558" LastEditorDisplayName="" LastEditDate="2011-01-08T12:48:57.393" LastActivityDate="2011-01-08T12:48:57.393" Title="Young modulus and Sound Velocity in a continuus medium" Tags="&lt;elasticity&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="2618" PostTypeId="2" ParentId="2617" CreationDate="2011-01-08T02:06:27.587" Score="2" ViewCount="" Body="&lt;p&gt;Just check your math. Correct derivation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;{f \over \epsilon_{11}} &amp;amp; = &amp;amp; {S_{11} \over \epsilon_{11}}=(2\mu+\lambda)+\lambda {\epsilon_{22}+\epsilon_{33} \over \epsilon_{11}} = \\&#xD;&#xA;&amp;amp; = &amp;amp; (2\mu+\lambda)-{\lambda^2 \over \lambda + \mu} = {(2\mu + \lambda)(\lambda + \mu) - \lambda^2 \over \lambda + \mu}=\frac{\mu(3\lambda+2\mu)}{\lambda+\mu} &#xD;&#xA;\end{eqnarray}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is precisely what Young modulus is defined to be (see e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Elastic_modulus&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-01-08T02:13:26.883" LastActivityDate="2011-01-08T02:13:26.883" CommentCount="3" />
  <row Id="2619" PostTypeId="1" AcceptedAnswerId="2620" CreationDate="2011-01-08T02:35:48.323" Score="7" ViewCount="349" Body="&lt;p&gt;Let's suppose I have a Hilbert space $K = L^2(X)$ equipped with a Hamiltonian $H$ such that the Schrödinger equation with respect to $H$ on $K$ describes some boson I'm interested in, and I want to create and annihilate a bunch of these bosons.  So I construct the bosonic Fock space&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S(K) = \bigoplus_{i \ge 0} S^i(K)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S^i$ denotes the $i^{th}$ symmetric power.  (Is this &quot;second quantization&quot;?)  Feel free to assume that $H$ has discrete spectrum.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the new Hamiltonian on $S(K)$ (assuming that the bosons don't interact)?  How do observables on $K$ translate to $S(K)$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not entirely sure this is a meaningful question to ask, so feel free to tell me that it's not and that I have to postulate some mechanism by which creation and/or annihilation actually happens.  In that case, I would love to be enlightened about how to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, various sources (Wikipedia, the Feynman lectures) inform me that $S(K)$ is somehow closely related to the Hilbert space of states of a quantum harmonic oscillator.  That is, the creation and annihilation operators one defines in that context are somehow the same as the creation and annihilation operators one can define on $S(K)$, and maybe the Hamiltonians even look the same somehow.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is this?  What's going on here?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assume that I know a teensy bit of ordinary quantum mechanics but no quantum field theory.&lt;/p&gt;&#xA;" OwnerUserId="1046" LastEditorUserId="1046" LastEditorDisplayName="" LastEditDate="2011-01-08T02:45:58.607" LastActivityDate="2011-01-08T20:28:35.433" Title="What's the exact connection between bosonic Fock space and the quantum harmonic oscillator?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" />
  <row Id="2620" PostTypeId="2" ParentId="2619" CreationDate="2011-01-08T03:30:40.840" Score="3" ViewCount="" Body="&lt;p&gt;Let's discuss the harmonic oscillator first. It is actually a very special system (one and only of its kind in whole QM), itself being already second quantized in a sense (this point will be elucidated later).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, a general talk about HO (skip this paragraph if you already know them inside-out). It's possible to express its Hamiltonian as $H = \hbar \omega(N + 1/2)$ where $N = a^{\dagger} a$ and  $a$ is a linear combination of momentum and position operator). By using the commutation relations $[a, a^{\dagger}] = 1$ one obtains basis $\{ \left| n \right &amp;gt;$ | $n \in {\mathbb N} \}$ with $N \left | n \right &amp;gt; = n$. So we obtain a convenient interpretation that this basis is in fact the number of particles in the system, each carrying energy $\hbar \omega$ and that the vacuum $\left | 0 \right &amp;gt;$ has energy $\hbar \omega \over 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the above construction was actually the same as yours for $X = \{0\}$. Fock's construction (also known as second quantization) can be understood as introducing particles, $S^i$ corresponding to $i$ particles (so HO is a second quantization of a particle with one degree of freedom). In any case, we obtain position-dependent operators $a(x), a^{\dagger}(x), N(x)$ and $H(x)$ which are for every $x \in X$ isomorphic to HO operators discussed previously and also obtain base $\left | n(x) \right &amp;gt;$ (though I am actually not sure this is base in the strict sense of the word; these affairs are not discussed much in field theory by physicists). The total hamiltonian $H$ will then be an integral $H = \int H(x) dx$. The generic state in this system looks like a bunch of particles scattered all over and this is in fact particle description of a free bosonic field.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-01-08T03:37:04.800" LastActivityDate="2011-01-08T03:37:04.800" CommentCount="8" />
  <row Id="2621" PostTypeId="2" ParentId="2605" CreationDate="2011-01-08T03:45:40.233" Score="0" ViewCount="" Body="&lt;p&gt;I recently lost a good deal of weight and would say it seems the water mostly leaves through urination, not exhalation.  Basically, I could tell when my weight was going to be decreasing because I'd have a lot of pee in the middle of the night, without having much to drink before bed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the other answers say respiration byproducts are co2 Whig is exhaled and water, which, I think, makes it's way to the bladder at some point. &lt;/p&gt;&#xA;" LastActivityDate="2011-01-08T03:45:40.233" />
  <row Id="2622" PostTypeId="2" ParentId="2605" CreationDate="2011-01-08T04:07:43.973" Score="-2" ViewCount="" Body="&lt;p&gt;Scientifically, the answer can be expressed in a very deep level of detail.  But ultimately, the answer is simple: Heat.&lt;/p&gt;&#xA;" OwnerUserId="1139" LastActivityDate="2011-01-08T04:07:43.973" CommentCount="2" />
  <row Id="2623" PostTypeId="2" ParentId="2619" CreationDate="2011-01-08T05:55:45.063" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Reference:&lt;/strong&gt; Fetter and Walecka, &lt;em&gt;Quantum Theory of Many Particle Systems&lt;/em&gt;, Ch. 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hamiltonian for a SHO is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H = \sum_{i = 0}^{\infty}\hbar \omega ( a_i^{+} a_i + \frac{1}{2} ) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\{a^+_i, a_i\}$ are the creation and annihilation operators for the $i^\textrm{th}$ eigenstate (momentum mode). The Fock space $\mathbf{F}$ consists of states of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vert n_{a_0},n_{a_1}, ...,n_{a_N} \rangle  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which are obtained by repeatedly acting on the vacuum $\vert 0 \rangle $ by the ladder operators:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Psi = \vert n_{i_0},n_{i_1}, ...,n_{i_N} \rangle = (a_0^+)^{i_0} (a_1^+)^{i_1} \ldots (a_N^+)^{i_N} \vert 0 \rangle $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interpretation of $\Psi$ is as the state which contains $i_k$ quanta of the $k^\textrm{th}$ eigenstate created by application of $(a^+_k)^{i_k}$ on the vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above state is not normalized until multiplied by factor of the form $\prod_{k=0}^N \frac{1}{\sqrt{k+1}}$. If your excitations are bosonic you are done, because the commutator of the ladder operators $[a^+_i,a_j] = \delta_{ij}$ vanishes for $i\ne j$. However if the statistics of your particles are non-bosonic (fermionic or anyonic) then the &lt;strong&gt;order&lt;/strong&gt;, in which you act on the vacuum with the ladder operators, matters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, to construct a Fock space $\mathbf{F}$ you do not need to specify a Hamiltonian. Only the ladder operators with their commutation/anti-commutation relations are needed. In usual flat-space problems the ladder operators correspond to our usual fourier modes $ a^+_k \Rightarrow \exp ^{i k x} $. For curved spacetimes this can procedure can be generalized by defining our ladder operators to correspond to suitable positive (negative) frequency solutions of a laplacian on that space. For details, see Wald, &lt;em&gt;QFT in Curved Spacetimes&lt;/em&gt;.&#xA;Now, given &lt;strong&gt;any&lt;/strong&gt; Hamiltonian of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H = \sum_{k=1}^{N} T(x_k) + \frac{1}{2} \sum_{k \ne l = 1}^N V(x_k,x_l) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with a kinetic term $T$ for a particle at $x_k$ and a pairwise potential term $V(x_k,x_l)$, one can write down the quantum Hamiltonian in terms of matrix elements of these operators:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H = \sum_{ij} a^+_i \langle i \vert T \vert j \rangle a_i + \frac{1}{2}a^+_i a^+_j \langle ij \vert V \vert kl \rangle a_l a_k $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $|i\rangle$ is the state with a single excited quantum corresponding the action of $a^+_i$ on the vacuum. (For details, steps, see Fetter &amp;amp; Walecka, Ch. 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps resolves some of your doubts. Being as you are from math, there are bound to be semantic differences between my language and yours so if you have any questions at all please don't hesitate to ask.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-08T20:28:35.433" CommentCount="3" />
  <row Id="2624" PostTypeId="2" ParentId="2604" CreationDate="2011-01-08T06:16:54.570" Score="2" ViewCount="" Body="&lt;p&gt;At present, I'd have to agree with dmckee's &quot;Who ordered that?&quot; quote, in that the Standard Model must take the list of fundamental particles as an input, i.e., it provides no explanation (just as it does not explain color charge). I'd argue that CP violation isn't so much required by the theory (feel free to correct me here), as it is an observation of reality, like the particles themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some theories in development, such as String Theory, do provide a reason for precisely three families (as Jerry mentions). In the case of String Theory, it comes about because of allowable string oscillations, which themselves are dependent on the number of dimensions (compactified and extended). (The number of dimensions and the way they are compacted is more fundamental than the number of particle families, so I would argue that, while we may have gotten the number of dimensions in part to make the particle families work out, many other things, such as predicting the properties of the still theoretical graviton, also depend on the dimensional parameters. This leads me to make the claim that choosing the number of dimensions is more than a parameter dictated by the number of particle families, i.e. that dimensions predict three families, rather than three families being used to choose the number and form of the dimensions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, from a pop-sci standpoint, I'd have to say that currently accepted theory can't really explain why we have three families of particles, but theorists are hard at work on new theories, some of which can explain it as a consequence of something deeper (with an appropriate side note that the existing theories are fantastically good at explaining our world, but that we know they have very specific shortcomings in very special cases, and we won't be satisfied until we've cleared those up. I add this because I get tired of arguments of a religious nature that take the very small bit we don't understand and use that to claim we don't understand anything.)&lt;/p&gt;&#xA;" OwnerUserId="861" LastActivityDate="2011-01-08T06:16:54.570" CommentCount="3" />
  <row Id="2625" PostTypeId="1" AcceptedAnswerId="2626" CreationDate="2011-01-08T08:21:32.077" Score="8" ViewCount="490" Body="&lt;p&gt;I've just finished a class a few weeks ago which taught thermo and stat mech, and I still don't know the exact difference between the two. Can someone help clear this up for me? (Yeah it's sad, and even sadder that I got an A- in the course)&lt;/p&gt;&#xA;" OwnerUserId="202" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-01T07:36:16.157" LastActivityDate="2011-06-28T12:14:41.393" Title="Difference between thermodynamics and statistical mechanics" Tags="&lt;thermodynamics&gt;&lt;statistical-mechanics&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" />
  <row Id="2626" PostTypeId="2" ParentId="2625" CreationDate="2011-01-08T08:48:14.440" Score="10" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Statistical Mechanics&lt;/strong&gt; is the theory of the physical behaviour of macroscopic systems starting from a knowledge of the microscopic forces between the constituent particles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory of the relations between various macroscopic observables such as temperature, volume, pressure, magnetization and polarization of a system is called &lt;strong&gt;thermodynamics&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0750305401&quot; rel=&quot;nofollow&quot;&gt;first page from this good book&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2011-01-08T08:48:14.440" CommentCount="4" />
  <row Id="2627" PostTypeId="1" AcceptedAnswerId="2636" CreationDate="2011-01-08T09:48:36.543" Score="6" ViewCount="173" Body="&lt;p&gt;I got a question concerning the band strucutre of solids. &#xA;The reference I'm using is the book on solid state physics by Ashcroft/Mermin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is that I don't completely understand the reason why there exists a&#xA;band-index n. On the one hand Ashcroft gives a quite plausible explanation &#xA;saying that the general solution of the given eigenvalue problem H $\psi_k(r)$ = $E_{k}$ $\psi_k(r)$ can be decomposed into $\psi_k(r) = e^{i*k*r} u_{k}(r)$. Plugging this ansatz &#xA;into the Schroedinger-equation and applying H to the $e^{i*k*r}$ first we&#xA;obtain a new Hamiltonian $H_{k}$ that depends on the &quot;wavevector&quot; k and a &#xA;new eigenvalue-problem $H_{k} u_{k}(r) = E_{k} u_{k}(r)$. Now we fix k. The operator&#xA;$H_{k}$ acts on r and therefor produces a certain number of solution u_{n,k}(r).&#xA;These solution can be counted using the index n. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far so good. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Ashcroft states that the second proof of Bloch's theorem shows that &#xA;$u_{k}(r)=\sum_{G} c_{k-G} e^{i G r}$. The sum is taken over all reciprocal lattice vectors G. He says that from this way of writing the function $u_{k}$ it is obvious that multiple solution do exists. I don't really understand why that is the case. &#xA;First i thought that you may think of the n-th solution to be $u_{k,n}(r)=c_{k-G_{n}} e^{i G_{n} r}$ where $G_{n}$ denotes the n-th reciprocal lattice vector. &#xA;However this doesn't seem plausible to me, since you need to have the sum to &#xA;proof that $\psi_{k}$ = $\psi_{k+G}$ and therefor E(k)=E(k+G). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see I'm quite confused about all this. Actually I'm also a bit confused why&#xA;the reduction to the first brillpuin zone doesn't produce more than one Energy value for a given k. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I'd be more than happy if someone could help me. &#xA;Thanks in advance!! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See you.&lt;/p&gt;&#xA;" OwnerUserId="1156" LastActivityDate="2011-01-08T20:21:32.970" Title="Quantum numbers and the band structure of solids" Tags="&lt;solid-state-physics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="2628" PostTypeId="1" AcceptedAnswerId="2631" CreationDate="2011-01-08T10:50:41.230" Score="3" ViewCount="220" Body="&lt;p&gt;And expansion of space is equal to expansion of vacuum?&lt;/p&gt;&#xA;" OwnerUserId="238" LastActivityDate="2011-01-14T12:01:07.440" Title="&quot;Space&quot; in General Relativity and &quot;vacuum&quot; in Standard Model, is it the same thing?" Tags="&lt;general-relativity&gt;&lt;standard-model&gt;&lt;space&gt;&lt;vacuum&gt;" AnswerCount="2" />
  <row Id="2629" PostTypeId="2" ParentId="2508" CreationDate="2011-01-08T11:03:03.857" Score="1" ViewCount="" Body="&lt;p&gt;ftiaronsem is right, i think!   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The law of conservation of angular momentum for this system must be expressed as $I_m\omega_m+I_M\omega_M=0$ where $I_m$ and $I_M$ are the moments of inertia of the child and the pad about the axis which passes through the center of the mass of the whole system and $\omega_m$ and $\omega_M$ are corresponding angular velocities in the reference frame of the ground (or stationary water). More specifically, $\omega_M$ is angular velocity of the center of the pad.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mark Eichenlaub&lt;/strong&gt; argues that this is not sufficient and we need to include yet the angular momentum of the pad's translation as well as the angular momentum of the pad's spinning. I think this is a fundamental mistake. This is best explained by analogs in rotational and linear motion, i think.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider the case where the child moves along a diameter of the pad. This is a linear motion and the law of conservation of linear momentum for this system must be expressed as $mv_m+Mv_M=0$ where $m$ an $M$ are masses of the child and the pad and $v_m$ and $v_M$ are corresponding velocities in the reference frame of the ground (or stationary water). Let's  gather  the results together:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$mv_m+Mv_M=0$$  $$I_m\omega_m+I_M\omega_M=0$$  Moments of inertia $I_m,I_M$  and angular velocities $\omega_m,\omega_M$ are the rotational analogs of $m,M$ and $v_m,v_M$ and corresponding equations  must look like the same way.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ftiaronsem's final formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta = \frac{m\left(1 - \frac{m}{m+M}\right)^22\pi}{\left(\frac{M}{2} + M\left(\frac{m}{m+M}\right)^2 + m\left(1 - \frac{m}{m+M}\right)^2\right)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be simplified. For purposes of brevity let's introduce the ratio $x=\frac{m}{M}$. Then finally the angular displacement of the pad: &#xA;$$\theta=\frac{2x}{(1+x)(1+3x)}2\pi$$  An amazing thing of this result is that the angular displacement of the pad has a maximum absolute value. It reaches the maximum displacement at $x=\frac{m}{M}=\frac{1}{\sqrt{3}}$ where the maximum is:&lt;br&gt;&#xA;$$\theta_{max}=(2-\sqrt{3})2\pi=96.5^\circ$$ Another surprise is that &lt;strong&gt;if the child is infinitely heavy then the pad's rotation is zero!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="552" LastActivityDate="2011-01-08T11:03:03.857" CommentCount="39" />
  <row Id="2630" PostTypeId="1" AcceptedAnswerId="2640" CreationDate="2011-01-08T14:02:06.887" Score="7" ViewCount="341" Body="&lt;p&gt;So, I remember in college physics the prof using liquid nitrogen in a demonstration.  When he was done, he threw the container of LN at the front row, and of course it evaporated (or whatever) before it got to the students.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am watching a cooking show now and they are using LN -- if they touched that, what would happen?  Would it be possible for the chefs to pour the LN onto their skin accidentally, or would it evaporate before it reached their skin?&lt;/p&gt;&#xA;" OwnerUserId="1129" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-13T02:35:07.220" LastActivityDate="2011-08-13T02:35:07.220" Title="Dangers of using liquid nitrogen" Tags="&lt;freezing&gt;&lt;laboratory-safety&gt;" AnswerCount="4" CommentCount="9" />
  <row Id="2631" PostTypeId="2" ParentId="2628" CreationDate="2011-01-08T14:34:41.983" Score="1" ViewCount="" Body="&lt;p&gt;They are something completely different, but there is connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum Field Theory (which underlies the Standard Model) is dependent on background. It's possible to formulate it on the curved background but usually one works in the Minkowski space-time and special relativity. Vacuum is then a special state $\left | 0 \right &amp;gt;$ in the Hilbert space that is annihilated by all annihilation operators and has no a priori connection with the usual vacuum (i.e. empty space) you are thinking of. There can be multiple vacuum states (so called degeneracy of the vacuum corresponding to inequivalent sets of annihilation operators connected by non-unitary transformations) so then the interpretation is obviously muddled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also vacuum in Standard Model is not empty in the sense that it is able to produce particle-antiparticle pairs (or in different words, that Standard Model has non-trivial field content and interactions). But the most important point is that this vacuum is &lt;em&gt;static&lt;/em&gt; and has no effect on the curvature of the space-time whatsoever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that space in the General Relativity has none of the above properties and also that it is &lt;em&gt;dynamic&lt;/em&gt;. What you are after is a theory of quantum gravity (or at least some approximation) that would take the Standard Model vacuum interactions into account when solving Einstein's equations. Of course you can try to do this and obtain e.g. prediction for cosmological constant in terms of properties of Standard Model vacuum. Problem is that the predicted value is completely wrong and so this is one of the major problems on the road to quantum gravity.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-01-08T14:34:41.983" CommentCount="4" />
  <row Id="2632" PostTypeId="2" ParentId="2630" CreationDate="2011-01-08T14:51:56.843" Score="1" ViewCount="" Body="&lt;p&gt;Not recommended, however I recall doing just that as a student, it boils on contact with your skin, you barely feel it. If it hits a less conductive part, such as a callous, it might not boil but instead freeze it. I seem to remember things like zippers also getting quite cold.&#xA;   btw, it is threw not through.&lt;/p&gt;&#xA;" OwnerUserId="391" LastEditorUserId="391" LastEditorDisplayName="" LastEditDate="2011-01-08T17:46:27.400" LastActivityDate="2011-01-08T17:46:27.400" />
  <row Id="2633" PostTypeId="2" ParentId="2630" CreationDate="2011-01-08T14:59:39.710" Score="4" ViewCount="" Body="&lt;p&gt;Liquid nitrogen will not moisten human skin, so short contact with small amount of it should not be too harmful -- it would just float on a evaporated portion of itself like water over very hot pan; yet of course putting a hand into a container with it is not a good idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I have heard, the biggest problem is when you pour it on your shoes, because it moistures them creating a direct contact with skin and extending its length -- also it takes a significant amount of time to take shoes off.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2011-01-08T14:59:39.710" CommentCount="1" />
  <row Id="2634" PostTypeId="2" ParentId="2630" CreationDate="2011-01-08T14:59:59.273" Score="4" ViewCount="" Body="&lt;p&gt;You can get localized soft tissue damage rather like a burn from sustained contact with moderate amounts of cryogenic liquids, and large amounts can freeze flesh solid---which is really bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small amounts will dance on your skin because of the vapor barrier that develops as they vaporize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Treat cryogenic materials with respect. Think about what you're doing around them.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-01-08T14:59:59.273" />
  <row Id="2635" PostTypeId="2" ParentId="2605" CreationDate="2011-01-08T15:51:26.607" Score="0" ViewCount="" Body="&lt;p&gt;It is like an automobile engine.  Think if you stopped putting gas in your car and just left it running.  It would use up the existing fuel and the output would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Energy to propel the car released from breaking the bonds in the gas &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Heat as a by product of the combustion and friction on the internal systems &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Leftovers from combustion that aren't used by the car (H2O, CO, CO2, etc)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same principles at play in your body just a bit more biologically oriented.&lt;/p&gt;&#xA;" OwnerUserId="1160" LastActivityDate="2011-01-08T15:51:26.607" />
  <row Id="2636" PostTypeId="2" ParentId="2627" CreationDate="2011-01-08T16:21:49.930" Score="2" ViewCount="" Body="&lt;p&gt;Ashcroft and Mermin are occasionally not too careful about notation, and so the resolution is, as Marek mentioned in a comment, related to the fact that&#xA;$$ u_k = \sum_G c_{k-G}e^{iGr} $$&#xA;admits several solutions -- every set of coefficients $\{c\}$ corresponds to &lt;em&gt;some&lt;/em&gt; function $u_k$, but we're only interested in the $n$ solutions that solve the Schroedinger equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every delocalized band arises because of the hybridization of quantum states between unit cells. If there are $n$ states that the electron can occupy, their will be $n$ (possibly degenerate, depending on the symmetry of the $G$-space point) bands. I say this to clarify that there is no sense in which the band index corresponds to the set reciprocal lattice vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to answer your last question, you're absolutely right that reduction to the first Brillouin zone does give you multiple values of $E$ for a given $k \in IBZ$. If I remember correctly A&amp;amp;M introduce band theory by starting with the nearly free electron model and should show explicitly that this is the case. In the tight binding picture (sometimes called Linear Combination of Atomic Orbitals) the energies for $k$ outside the IBZ fall back onto their values in the IBZ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Continued:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, I wanted to point out that $E_n(k)=E(k+G_n)$ simply doesn't make any sense. Where did you read that? However the point is that the eigenvalues &lt;em&gt;that make up a particular band&lt;/em&gt; are periodic in $G$, so it is true that $E_n(k)=E_n(k+G)$. Now, what this means is that a description with any more than the first Brillouin zone is redundant. The fact that you have multiple bands does &lt;em&gt;not&lt;/em&gt; come from the reduction, though it can look that way at first, as A&amp;amp;M present it. I suggest skipping ahead to the chapter on tight binding to understand how multiple bands arise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Continued again:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, I didn't address this as clearly as I'd hoped. The reason a new quantum number appears at all is that we know from experience that $E(k)$ is not single-valued. The reason for this is that each atom has a set of discrete orbitals, and mathematically we can describe the motion of an electron in a lattice as tunnelings between these orbitals... However, these orbitals have very different symmetries, and we can not, for example, hop from an s-orbital on site-1 to a p-orbital on site-2. Thus each flavor of orbital symmetry will &lt;em&gt;basically&lt;/em&gt; correspond to a new band (with a lot of technical complications that will only muddy the point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, you're right about $E_n$ being bounded everywhere, and your reasoning is right as well. What was the question?&lt;/p&gt;&#xA;" OwnerUserId="791" LastEditorUserId="791" LastEditorDisplayName="" LastEditDate="2011-01-08T20:21:32.970" LastActivityDate="2011-01-08T20:21:32.970" CommentCount="5" />
  <row Id="2637" PostTypeId="2" ParentId="2627" CreationDate="2011-01-08T17:55:19.580" Score="1" ViewCount="" Body="&lt;p&gt;Thanks for your respose wsc. Let me maybe go a bit more into detail now. When we start solving the given eigenvalue problem of an single electron in a periodic potential we assume a solution of the form $\psi_{k}(r)=\sum_{k} C_{k} e^{ikr}$. Plugging this in leads to a set of algebraic equation. However one notices that for a fixed k only the coefficients $C_{k}$, $C_{k+G}$, ... are coupled. Hence we obtain a solution of the form  $\psi_{k}(r)=\sum_{G} C_{k-G} e^{i(k-G)r}$. So if we wanted the coefficient to depend on a quantum number n we should've known that beforehand and should've added this fact to our first ansatz. However I would say that at the beginning this is not so obvious. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly is there a specific formula for the eigenvalues $E_{n}(k)$. &#xA;I've read in a book that $E_{n}(k)=E(k+G_{n})$ where $G_{n}$ is the reciprocal lattice vector corresponding to the energy $E_{n}$. However this formula seems wrong to me and I also don't understand the reasoning here. Don't missunderstand me, I'm fine with 'existence'. I don't need an explicit formula. Still it would be kind of interesting to know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So thirdly concerning the reduction to the first Brillouin zone. &#xA;When I'm looking at a usual E-k-diagramm with $\pi / a &amp;lt; k &amp;lt; \pi / a$ as usual &#xA;and I'm fixing a certain k. Then I see that there is in general more than one energy corresponding to this k-vector. Now is this the result of the different energybands or is it the result of the reduction process? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me it seems as though that these additional energies corresponding to one k vector are the same anyway. For example: Fix k. Then the energies corresponding to this vector are&#xA;$E_{k}$, $E_{k+G}$, ... where G is a reciprocal lattice vector. However Ashcroft &#xA;derives the formule that $E_{k} = E_{k+G}$. So the energies are the same anyway. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your response in advance!&lt;/p&gt;&#xA;" OwnerUserId="1156" LastActivityDate="2011-01-08T17:55:19.580" />
  <row Id="2638" PostTypeId="2" ParentId="2615" CreationDate="2011-01-08T19:46:30.070" Score="6" ViewCount="" Body="&lt;p&gt;First of all, Marek is right that a surface tension exists only between two different materials (well, I would say between two different phases - for example water and ice).  So let's rephrase the question as &quot;Are there two phases with zero surface tension?&quot; and elaborate a little on the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface tension is the excess free energy (technically the 'grand potential') associated with the area of the interface between two phases. If the surface tension is positive (it always is), the system minimizes the free energy cost by minimizing the area of contact.  This leads, for example, to the spherical shape of a water droplet in coexistence with water vapor.  But if the surface tension were zero, we could deform the shape of the water droplet arbitrarily, with no free energy cost, so long as we didn't change its volume! We could even deform it to the point where, from the macroscopic point of view, the water and the water vapor seem to be perfectly mixed.  But this actually contradicts the fact that there was phase separation in the first place, because phase separation is an indication that a uniform (single-phase) system can lower its total free energy by splitting into two phases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So zero surface tension between two phases would be quite an unrealistic situation.  There are some situations in physics which are somewhat like having zero surface tension.  If you look at the surface tension between water and water vapor as the critical point is approached, the surface tension vanishes.  But at the critical point itself there is no distinction between the two phases (and therefore no interface), so we can't say that there is zero surface tension.  Another situation where it is tempting to say that surface tension has been made to vanish is when oil and water (which don't mix) are emulsified by adding a surfactant or emulsifier which lowers the surface tension.  Eventually, rather than two coexisting phases, we have a single phase with a lot of internal structure (little droplets of oil in water or vice versa).  But that is a single phase, not two coexisting phases!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I don't think that any two coexisting phases have zero surface tension.  In particular, liquid helium has a positive surface tension (with air I mean).  &lt;/p&gt;&#xA;" OwnerUserId="266" LastActivityDate="2011-01-08T19:46:30.070" CommentCount="8" />
  <row Id="2640" PostTypeId="2" ParentId="2630" CreationDate="2011-01-08T21:05:47.237" Score="10" ViewCount="" Body="&lt;p&gt;Liquid nitrogen boils when it comes in contact with skin, so small amounts of spatter are no danger at all-- the droplets just bounce off. I regularly pour a liter or so (a bit at a time) out on a lab table when I do liquid nitrogen demos, with no problems or safety gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest risk from the low temperature is getting it into fabric of some sort, which will hold it in closer proximity to skin for longer than the drops by themselves will. I have, on occasion spilled some on my pants, which is annoyingly cold, but not too bad unless you're wearing really tight clothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really, the biggest hazard from any of the nitrogen demos I do is not the temperature but the expansion. When it boils, it expands to something like 700 times the volume of the liquid, so if you put some in a sealed container, it can make a big bang. I know of a case where a grad student at MIT destroyed a bathroom with a 2-liter bottle of liquid nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use this expansion for a kind of cool demonstration if you take one of those little dropper bottles with the angled spouts that are common in chem labs, and seal a little nitrogen inside. Put it down on the floor, and it spins like a firework. The problem with that is, you almost always end up getting a little water vapor condensing in the spout, which plugs it up, and then the bottle will go bang. I had a student a few years ago who had one go off in his hand, and he said it stung pretty badly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen videos of people (Jearl Walker in particular) &quot;drinking&quot; liquid nitrogen by taking a small amount into their mouth, and holding it there for a second or so-- the instant boiling will keep it from giving you oral frostbite for a little while, and you can spit the liquid out after breathing out over it, which makes an enormous plume of steam. It's really cool to see, but kind of risky to do. I've never had the guts to try it myself.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2011-01-08T21:05:47.237" CommentCount="2" />
  <row Id="2641" PostTypeId="1" AcceptedAnswerId="2661" CreationDate="2011-01-08T23:37:17.487" Score="5" ViewCount="396" Body="&lt;p&gt;&lt;a href=&quot;http://green.blogs.nytimes.com/2011/01/07/hold-that-megawatt/&quot; rel=&quot;nofollow&quot;&gt;This story&lt;/a&gt; about the use of battery/freewheel based Frequency Regulators confused me about how the 60hz frequency of the North American power grid was set--saying that it was kept at that frequency by balancing load and supply. I used to think that it was only voltage which was affected by this balance, and that the frequency was determined by the speed of the rotors in the generator. EG see &lt;a href=&quot;http://en.wikipedia.org/wiki/Alternator_synchronization&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's page on Alternator Synchronization&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone help me understand the physics behind this?&lt;/p&gt;&#xA;" OwnerUserId="961" LastActivityDate="2011-01-10T01:33:06.000" Title="How does load affect frequency on the power grid?" Tags="&lt;electricity&gt;&lt;frequency&gt;" AnswerCount="1" />
  <row Id="2644" PostTypeId="1" AcceptedAnswerId="2899" CreationDate="2011-01-09T04:34:12.223" Score="11" ViewCount="1083" Body="&lt;p&gt;I've always assumed/been told that Newton's 2nd law is an empirical law — it must be discovered by experiment. If this is the case, what experiments did Newton do to discover this? Is it related to his studies of the motion of the moon and earth? Was he able to analyze this data to see that the masses were inversely related to the acceleration, if we assume that the force the moon on the earth is equal to the force the earth exerts on the moon? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia, the Principia reads:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lex II: Mutationem motus proportionalem esse vi motrici impressae, et fieri secundum lineam rectam qua vis illa imprimitur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Translated as: Law II: The alteration of motion is ever proportional to the motive force impress'd; and is made in the direction of the right line in which that force is impress'd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how did Newton come to this conclusion? I get that he knew from Galielo the idea of inertia, but this doesn't instantly tell us that the change in momentum must be proportional to the net force. Did Newton just assume this, or was there some experiment he performed to tell him this?&lt;/p&gt;&#xA;" OwnerUserId="807" LastEditorUserId="807" LastEditorDisplayName="" LastEditDate="2011-01-10T03:07:31.417" LastActivityDate="2011-08-25T16:26:33.633" Title="How did Newton discover his second law?" Tags="&lt;mechanics&gt;&lt;experimental-physics&gt;&lt;history&gt;" AnswerCount="8" CommentCount="17" FavoriteCount="1" />
  <row Id="2645" PostTypeId="1" AcceptedAnswerId="2650" CreationDate="2011-01-09T05:50:22.330" Score="3" ViewCount="275" Body="&lt;p&gt;I've heard it from a few people, and I've seen it popup here in the site a couple of times. There seems to be speculation (and studies?) towards this idea, and this is what I've picked up so far:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Living organisms are low entropy systems.&lt;/li&gt;&#xA;&lt;li&gt;The second law of thermodynamics is somehow strongly connected to the origin of life itself.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is a very interesting idea, unfortunately I've never seen anyone elaborate beyond that. I'd like to understand it a little better. Specifically:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why are the above statements true (neither is immediately obvious to me)?&lt;/li&gt;&#xA;&lt;li&gt;And what are the other arguments in favor of this theory?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'll accept both answers for and against this point of view, as long as it reasonably justifies &lt;em&gt;why&lt;/em&gt; the above statements are correct/false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: I believe the thermodynamic definition of entropy is the one commonly used with these arguments. And I'm not sure what definition they use for life. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Entropy_and_life&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; page is pretty vague on this.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="724" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2011-01-09T07:16:23.460" LastActivityDate="2011-01-09T14:38:20.077" Title="What are the arguments towards the Life-and-Entropy relation?" Tags="&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="4" CommentCount="11" FavoriteCount="2" />
  <row Id="2646" PostTypeId="2" ParentId="2645" CreationDate="2011-01-09T07:38:39.877" Score="0" ViewCount="" Body="&lt;p&gt;In the days since Boltzmann and Clausius our understanding of systems far from equilibrium has increased drastically. Developments in the past two decades have also shown that the 2nd law - as it was classically understood - was not the be all and end all when it came to understanding the emergence of complex systems such as those of life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good place to learn about these discoveries is with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fluctuation_theorem&quot; rel=&quot;nofollow&quot;&gt;Fluctuation Theorem&lt;/a&gt; which was first proposed in 1993 by Denis Evans, Cohen and Morriss. To quote from the abstract of the Wikipedia article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While the second law of thermodynamics predicts that the entropy of an isolated system should tend to increase until it reaches equilibrium, it became apparent after the discovery of statistical mechanics that the second law is only a statistical one, suggesting that &lt;strong&gt;there should always be some nonzero probability that the entropy of an isolated system might spontaneously decrease; the fluctuation theorem precisely quantifies this probability.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(emph. mine). Ultimately all the systems of life exploit this weakness of the 2nd law - that it is a statistical statement and not a microscopic law. What these Fluctuation theorems have demonstrated that given the right conditions an isolated system can exploit the possibility that its &quot;entropy might spontaneously decrease&quot;. In order to be able to exploit this possibility the system in question must be comparable in size to the thermal fluctuations in its environment. This is true for all biological motors such as those responsible for energy production via ATP, myosin motors which control muscle movement and artificially manufactured nano-machines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from these considerations, there is also the fact that life-bearing systems are &lt;em&gt;always&lt;/em&gt; far from equilibrium. When, for instance, &quot;Creationists&quot; try to use the 2nd law as an argument for, well, &quot;creation&quot; they forget that we live in the presence of a star which constantly supplies the earth with energy resulting in the planet being far from thermal equilibrium on a macroscopic scale so that no application of the 2nd law in its traditional form - i.e. without taking into account the Fluctuation Theorem - is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary:&lt;/strong&gt; If we want to even begin to speak sensibly of such questions we have to understand the statistical mechanics of non-equilibrium systems, which we have only just begun to do.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-09T07:38:39.877" CommentCount="12" />
  <row Id="2647" PostTypeId="2" ParentId="2425" CreationDate="2011-01-09T08:38:12.500" Score="0" ViewCount="" Body="&lt;p&gt;I don't think the equivalence principle is all that important in modern derivations of general relativity. Instead, a much more central role is given to diffeomorphism covariance. In fact, there are &lt;em&gt;modifications&lt;/em&gt; of general relativity &amp;mdash; like introducing dilaton couplings with dilaton gradients, or spinning particles moving in a spacetime with torsion &amp;mdash; which still satisfy diffeomorphism covariance, but violate the equivalence principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, the equivalence principle is more or less equivalent to the statement that freefalling particles only minimally couple to the metric tensor, and that there are no other geometrical fields, like a dilaton or a torsion tensor. So, if we make this assumption, and also assume the covariant dynamical equations for the metric tensor can only be at most second order in the derivatives, the Einstein field equations is just about the most general equation. And the source tensor that is being coupled to has to be symmetric and have a zero covariant divergence. According to the Coleman-Mandula theorem, just about the only field with these properties is the stress-energy tensor. So I guess you can kind of derive the Einstein field equations from the equivalence principle.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-09T14:55:32.867" LastActivityDate="2011-01-09T14:55:32.867" CommentCount="5" />
  <row Id="2648" PostTypeId="2" ParentId="2409" CreationDate="2011-01-09T09:01:06.403" Score="1" ViewCount="" Body="&lt;p&gt;We have a condensed matter analog of the Higgs mechanism. Colloquially, we say the gauge boson &amp;mdash; in this case, the photon &amp;mdash; &quot;eats up&quot; the Goldstone boson &amp;mdash; in this case, plasmons built up from a condensate of Cooper pairs &amp;mdash; giving rise to a new quasiparticle. Unlike photons, this quasiparticle has an energy gap in the dispersion relation. This is the condensed matter analog of a massive vector boson. Unlike a photon, this quasiparticle can have zero velocity, for instance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also true that if the energy of the photon before it reaches the superconductor is less than the energy gap, this photon will be reflected. If the initial energy is larger, we have a superposition of a transmitted and a reflected component for the wavefunction.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-01-09T09:01:06.403" CommentCount="3" />
  <row Id="2649" PostTypeId="1" CreationDate="2011-01-09T11:01:46.997" Score="5" ViewCount="247" Body="&lt;p&gt;In loop quantum gravity, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ashtekar_variables&quot; rel=&quot;nofollow&quot;&gt;canonical (Ashtekar) variables&lt;/a&gt; are chosen to be the densitized inverse triad $\mathbf{E}$ and some rotation connection field $\mathbf{A}$. To get the ordinary triad from $\mathbf{E}$, we need to take its &quot;square root&quot;. But we know as long as the volume factor, i.e. the square root of the determinant of $\mathbf{E}$ is nonzero, we always have two solutions for the triad. They are related by a branch cut around $|\mathbf{E}| = 0$. We also know that if this determinant is negative, the corresponding triad will be &lt;em&gt;imaginary&lt;/em&gt;! OK, you might now say there's no problem as long as we stick to solutions where $|\mathbf{E}|$ is always positive everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here is where another problem comes in. The state about which loop quantum gravity expansions are made is the state where $\mathbf{E}=0$ everywhere. Excitations over this state by &lt;a href=&quot;http://en.wikipedia.org/wiki/Wilson_loop&quot; rel=&quot;nofollow&quot;&gt;Wilson loops&lt;/a&gt; contain wavefunctional components with negative determinants as frequently as we have components with positive determinants. Does that mean loop quantum gravity predicts imaginary distances, negative areas and imaginary volumes? And besides, aren't we expanding about a highly irregular solution?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another related question is: People sometimes multiply the Hamiltonian constraint by the factor $\sqrt{| \mathbf{E} |}$, but this goes to zero whenever $| \mathbf{E} | = 0$, which actually &lt;em&gt;weakens&lt;/em&gt; the constraints by introducing additional solutions which were previously forbidden. Is such a procedure really justified?&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-28T17:16:04.577" LastActivityDate="2011-01-28T17:16:04.577" Title="Is taking the &quot;square root&quot; of the densitized inverse triad irregular in loop quantum gravity?" Tags="&lt;quantum-gravity&gt;&lt;loop-quantum-gravity&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="2650" PostTypeId="2" ParentId="2645" CreationDate="2011-01-09T11:48:44.767" Score="3" ViewCount="" Body="&lt;p&gt;The first point is simple to address. Gas that is located in the corner of the room has lower entropy than gas that is scattered all over the room because it has smaller number of accessible microstates that correspond to it. Similarly, you can consider one concrete organism that has very ordered structure. Most of the atoms just need to be where they need to be, otherwise you'd die; of course there is still variability explaining the diversity of the species, but that diversity doesn't extend all the way to your molecules flying around as a gas. So life is in a hypothetical small corner of the room representing every possible ordering of atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a notion that living organisms lower the entropy of their surroundings. This is indeed correct. Again, every ordered system must have lower entropy than unordered system (because it has lower number of accessible microstates) and so if you convert the heap of mud into a house, you'll lower entropy of your immediate surroundings. But in the process of ordering you'll exert great amount of work and heat and increase the total entropy of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the origin of life, I am not sure about the connection with second law. To me it's just a question of probability. If anything it seems to be an application of Poincaré recurrence theorem (which actually denies second law, or more precisely sets it out as only a statistical statement that is eventually bound to fail). Let's say proto-life is defined as certain small molecules that start to exhibit some signs of life (reproduction? self-correction? having certain metabolic processes? I'll leave that to experts on life). These are still complex ordered systems and so improbable to form, but the oceanic waters of the young Earth were probably big enough such that number of microscopic disordered systems was comparable with recurrence time associated to them. Poincaré recurrence theorem then says that in finite time those unordered systems will assemble themselves into ordered systems.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt; so why don't we see broken eggs assembling into original unbroken eggs? Answer: because it would take time $t \sim \exp({\Delta S \over k})$. So if you have enough time (or equivalently enough eggs), you'll see it assembled. Problem is these times are huge for macroscopic objects. But for proto-objects that started the evolution of life, that time should be reasonably small (like few hundreds of millions of years).&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-01-09T11:48:44.767" CommentCount="1" />
  <row Id="2651" PostTypeId="2" ParentId="2645" CreationDate="2011-01-09T12:31:14.013" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Living organisms are low entropy systems.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clearly the energy coming from the Sun is what powers life on Earth. While there are some microorganisms that use geothermal energy, this is true for the large majority of life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Earth%27s_energy_budget&quot; rel=&quot;nofollow&quot;&gt;Earth's energy budget&lt;/a&gt; is quite well understood, with the Sun pouring $174\times 10^{15} W$ worth of energy in it, and the Earth re-radiating all of it (assuming thermal equilibrium) through reflection (30%), radiation from the clouds (64%) and radiation from the ground (6%).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, since the energy coming from the Sun is also going out at the same rate, where does all energy that life needs comes from?&lt;br&gt;&#xA;It can be traced back through the food chain to &lt;a href=&quot;http://en.wikipedia.org/wiki/Photosynthesis&quot; rel=&quot;nofollow&quot;&gt;photosynthesis&lt;/a&gt;, which creates more complex and energetic molecules and generates waste heat (blackbody radiation). While the &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC224736/pdf/pnas00117-0088.pdf&quot; rel=&quot;nofollow&quot;&gt;overall entropy balance is positive&lt;/a&gt;, photosynthesis decreases the plant's entropy while increasing the environment entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your statement needs to be understood in this context: what powers life is not the energy of the Sun, but the fact that this energy has a low enough entropy that Earth can radiate the energy back via black body radiation while decreasing entropy locally.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The second law of thermodynamics is somehow strongly connected to the origin of life itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only thing I can think of around this statement is that some creationists have tried to argue against abiogenesis on the basis of the law of entropy - in the sense that complex, ordered systems like living beings have a lower entropy than a dead system with the same atoms without structure. Which is true - and the explanation of why abiogenesis does not necessarily violate the law of entropy is in my considerations above.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-01-09T12:31:14.013" />
  <row Id="2653" PostTypeId="2" ParentId="2645" CreationDate="2011-01-09T14:38:20.077" Score="1" ViewCount="" Body="&lt;p&gt;Entropy and the Second Law of Thermodynamics relate to life because life is made possible by the flows of energy implicit in the second law and the selective storage and manipulation of the biochemistry made possible by those flows.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Living organisms are low entropy&#xA;  systems.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In classical thermodynamics, entropy is a measure of the thermodynamic energy availability of a system. Thus a low entropy system has energy available in a form that may be used to perform work while a high entropy system has relatively less energy available. Alternately, entropy can be seen as a measure of disorder where low entropy means a relatively ordered system and high entropy means a disordered system. Finally, in statistical mechanics, entropy can also refer to information content where low entropy means relatively higher information content than high entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Living organisms are low entropy by all three related measures. The more complex the organism, the more true this statement is. Live organisms store energy and make it available for metabolic processes to move, eat, think, etc. Organisms are highly ordered systems, the most complex systems we have seen in the universe thus far. Organisms also have very high information content. Organic systems are controlled through a complex intermix of chemical, electrical, and genetic information that we do not even fully comprehend at this point, even for the most basic single celled organisms. They are the most highly ordered systems known.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The second law of thermodynamics is&#xA;  somehow strongly connected to the&#xA;  origin of life itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second law of thermodynamics states that in a closed system, entropy will tend to increase and that entropy will flow from the low entropy region to the high entropy region. This relates to life in two specific ways:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It makes life less likely, in general, since life must originate through a random fluctuation to a low entropy state. Since entropy normally increases, over time the entropy of the system will be higher and therefore less likely to originate life.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It makes life more likely on a planet like Earth that sits in the middle of the entropic gradient of the entropy flows from the very low entropy Sun outward to the relatively higher entropy solar system. The surface of the ocean is the edge of a very steep thermodynamic gradient where the energy from the sun mixes with the relatively cool water of the ocean. In these gradients, it is much more likely for random fluctuations to result in a very low entropy molecule like an amino acid or eventually DNA because there is a continuous energy flow from the Sun through the gradient. This energy makes possible various endothermic chemical reactions, increases the rate of chemical reaction, and offers a source of energy for continued conversion for metabolic purposes for any living organisms.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So both statements are true.&lt;/p&gt;&#xA;" OwnerUserId="948" LastActivityDate="2011-01-09T14:38:20.077" CommentCount="1" />
  <row Id="2654" PostTypeId="2" ParentId="1216" CreationDate="2011-01-09T15:11:17.637" Score="4" ViewCount="" Body="&lt;p&gt;Despite all I wrote in my other answer, there's a very interesting attempt by Xiao-Gang Wen to come up with emergent models of gravity starting from quantum lattice models with no gravity, and only nearest neighbor interactions. His work can be found at &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0606100&quot; rel=&quot;nofollow&quot;&gt;gr-qc/0606100&lt;/a&gt; and &lt;a href=&quot;http://arxiv.org/abs/0907.1203&quot; rel=&quot;nofollow&quot;&gt;arXiv:0907.1203&lt;/a&gt;. He managed to show that quasiparticles with no energy gap and a helicity of $\pm 2$ can emerge &lt;em&gt;without&lt;/em&gt; being accompanied by helicity $\pm 1$ or $0$ quasiparticles. Whether or not this model can be considered a model of gravity though is another matter.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-01-09T15:11:17.637" CommentCount="1" />
  <row Id="2656" PostTypeId="1" AcceptedAnswerId="13199" CreationDate="2011-01-09T17:56:06.307" Score="3" ViewCount="258" Body="&lt;p&gt;A &lt;a href=&quot;http://www.google.com/search?q=sun+jar&quot; rel=&quot;nofollow&quot;&gt;sun jar&lt;/a&gt; is an object that stores solar energy in a battery and then releases it during dark hours through a led.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a $65cm^2$ solar panel &lt;/li&gt;&#xA;&lt;li&gt;a 12h/12h light/dark cycle&lt;/li&gt;&#xA;&lt;li&gt;insolation of $2.61kWh/m^2/day$&lt;/li&gt;&#xA;&lt;li&gt;perfectly efficient components (i.e. without violating Entropy laws or other theoretical limits)&lt;/li&gt;&#xA;&lt;li&gt;light is emitted by an ideal &quot;white&quot; source&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How bright can we make the jar so that it shines for 12 hours?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Useful links:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Insolation&quot; rel=&quot;nofollow&quot;&gt;Insolation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Peukert%27s_law&quot; rel=&quot;nofollow&quot;&gt;Peukert's law&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_cell_efficiency&quot; rel=&quot;nofollow&quot;&gt;Solar cell efficiency&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Luminous_efficacy#Lighting_efficiency&quot; rel=&quot;nofollow&quot;&gt;Luminous efficacy&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-08-04T15:19:09.987" LastActivityDate="2011-08-04T20:36:41.150" Title="How bright can we make a sun jar?" Tags="&lt;optics&gt;&lt;entropy&gt;&lt;renewable-energy&gt;&lt;photoelectric-effect&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="2657" PostTypeId="2" ParentId="2656" CreationDate="2011-01-09T18:07:55.743" Score="0" ViewCount="" Body="&lt;p&gt;Well, basic answer is that it can be about twice &quot;brighter&quot; than sun (i.e. it will emit twice the amount of light absorbed), if LED emits only visible light, while battery absorbs everything from IR to UV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reality is of course much much worse.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-01-09T18:07:55.743" />
  <row Id="2658" PostTypeId="1" AcceptedAnswerId="2659" CreationDate="2011-01-09T20:23:07.120" Score="8" ViewCount="1450" Body="&lt;p&gt;I'm doing magnification and lens in class currently, and I really don't get why virtual and real images are called what they are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A virtual image occurs the object is less than the focal length of the lens from the lens, and a real image occurs when an object is further than focal length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By why virtual and real? What's the difference? You can't touch an image no matter what it's called, because it's just light.&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2011-01-10T17:17:11.967" LastActivityDate="2011-01-17T14:43:33.910" Title="Virtual vs Real image" Tags="&lt;optics&gt;&lt;geometric-optics&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="2659" PostTypeId="2" ParentId="2658" CreationDate="2011-01-09T21:20:57.613" Score="6" ViewCount="" Body="&lt;p&gt;You can project a real image onto a screen or wall, and everybody in the room can look at it. A virtual image can only be seen by looking into the optics and can not be projected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a concrete example, you can project a view of the other side of the room using a convex lens, and can not do so with a concave lens.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'll steal some image from Wikipedia to help here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First consider the line optics of real images (from &lt;a href=&quot;http://en.wikipedia.org/wiki/Real_image&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Real_image&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/geRzr.png&quot; alt=&quot;real images formed by a single convex lens of concave mirror&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the lines that converge to form the image point are all drawn solid. This means that there are actual rays, composed of photon originating at the source objects. If you put a screen in the focal plane, light reflected from the object will converge on the screen and you'll get a luminous image (as in a cinema or a overhead projector).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next examine the situation for virtual images (from &lt;a href=&quot;http://en.wikipedia.org/wiki/Virtual_image&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Virtual_image&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/E207U.png&quot; alt=&quot;virtual images formed by a single concave lens or convex mirror&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice here that the image is formed by a one or more dashed lines (possibly with some solid lines).  The dashed lines are  draw off the back of solid lines and represent the &lt;em&gt;apparent&lt;/em&gt; path of light rays from the image to the optical surface, but no light from the object ever moves along those paths. This light energy from the object is dispersed, not collected and can not be projected onto a screen. There is still a &quot;image&quot; there, because those dispersed rays all &lt;em&gt;appear to be coming&lt;/em&gt; from the image. Thus, a suitable detector (like your eye) can &quot;see&quot; the image, but it can not be projected onto a screen.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-01-17T14:43:33.910" LastActivityDate="2011-01-17T14:43:33.910" CommentCount="6" />
  <row Id="2660" PostTypeId="1" AcceptedAnswerId="2885" CreationDate="2011-01-09T21:27:35.897" Score="5" ViewCount="375" Body="&lt;p&gt;Initially Wheeler and Feynman postulated that, the electromagnetic field is just a set of bookkeeping variables required in a Hamiltonian description. This is very neat because makes the point of divergent vacuum energy a moot point (i.e: an example of asking the wrong question) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a few years later (1951), Feynman wrote to Wheeler that this approach would not be able to explain vacuum polarization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone knows what was the argument for saying so? I don't see how allowing both processes with entry and exit particles and processes that begin in pair-creation and end in pair-annihilation makes the existence of a field a requirement.&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-01-09T21:55:10.483" LastActivityDate="2011-02-01T22:51:32.377" Title="Wheeler-Feynman theory, QED without fields, vacuum polarization" Tags="&lt;quantum-field-theory&gt;&lt;polarization&gt;&lt;path-integral&gt;&lt;vacuum&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="2661" PostTypeId="2" ParentId="2641" CreationDate="2011-01-09T21:52:14.590" Score="5" ViewCount="" Body="&lt;p&gt;The physics is actually much easier than it seems at first glance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power generators are engines just like the everyday ones we see all around in our cars, lawnmowers, snowblowers, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except for new power sources like some wind and solar systems with electronic inverters, the vast majority of power is supplied by large rotating AC generators turning in synch with the frequency of the grid. The frequency of all these generators will be identical and is tied directly to the RPM of the generators themselves, generally 3600 RPM for gas turbines and 1800 RPM for nuclear plants. If there is sufficient power in the generators then the frequency can be maintained at the desired rate (i.e. 50Hz or 60Hz depending on the locale).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power from the individual generators will lead the grid in phase slightly by an amount roughly corresponding to the power they deliver to the grid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An increase in the power load is accompanied by a concurrent increase in the power supplied to the generators, generally by the governors automatically opening a steam or gas inlet valve to supply more power to the turbine. However, if there is not sufficient power, even for a brief period of time, then generator RPM and the frequency drops. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is much like what happens to a car on cruise control if you start going up a hill, if the hill is not too steep you can maintain speed, once you reach the limits of the torque supplied by the engine, the car and engine slow down. If the combined output of all the generators cannot supply enough power then the frequency will drop for the entire grid. All the generators slow down just like your car engine on a hill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For large grids the presence of many generators and a large distributed load makes frequency management easier because any given load is a much smaller percentage of the combined capacity. For smaller grids, there will be a much larger fluctuation in capacity as delays in matching power supplied are harder to manage when the loads represent a relatively larger percentage of the generated power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a battery systems like the one in the article is really designed to keep short-term fluctuations in power requirements from dropping the frequency because of lags in the governors and generators which require a finite time to adjust to the new power requirements. These &quot;frequency regulator&quot; power stations can supply very high power for short bursts to keep the power requirements even so that the other generators don't see too much load faster than they can respond due to mechanical limitations.&lt;/p&gt;&#xA;" OwnerUserId="948" LastEditorUserId="948" LastEditorDisplayName="" LastEditDate="2011-01-10T01:33:06.000" LastActivityDate="2011-01-10T01:33:06.000" CommentCount="2" />
  <row Id="2662" PostTypeId="2" ParentId="2658" CreationDate="2011-01-09T22:39:18.167" Score="4" ViewCount="" Body="&lt;p&gt;For a &lt;strong&gt;real image&lt;/strong&gt;, rays from a single source point converge to a single point on the other side of the lens. This means that a point on the image &lt;em&gt;remains well-defined&lt;/em&gt; after the optical transformation (refraction by lens).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a &lt;strong&gt;virtual image&lt;/strong&gt;, rays from a single source point &lt;em&gt;diverge&lt;/em&gt; after they pass through the lens. This means that the a point on the image &lt;em&gt;is no longer well-defined&lt;/em&gt; after the optical transformation. Realistically, what you'll see on a screen place at the other end of a diverging (concave) lens is a large haze of weak light (since the rays have diverged so greatly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that &lt;em&gt;convex&lt;/em&gt; (converging) single lenses always produce real images, while &lt;em&gt;concave&lt;/em&gt; (diverging) single lenses always produce virtual images. For &lt;em&gt;mirrors&lt;/em&gt; it's generally the other way round (there are exceptions though). Similarly for &lt;em&gt;double&lt;/em&gt; lenses, things are reversed (compared to single lenses).&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2011-01-10T18:16:21.490" LastActivityDate="2011-01-10T18:16:21.490" CommentCount="12" />
  <row Id="2663" PostTypeId="5" CreationDate="2011-01-09T22:56:49.410" Score="0" ViewCount="0" Body="" OwnerUserId="-1" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2011-01-09T22:56:49.410" LastActivityDate="2011-01-09T22:56:49.410" />
  <row Id="2664" PostTypeId="4" CreationDate="2011-01-09T22:56:49.410" Score="0" ViewCount="0" Body="Pertains to any question concerning the ideas that are formulated, wether mathematically or not, to explain or describe physical phenomena." OwnerUserId="-1" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2011-01-09T22:56:49.410" LastActivityDate="2011-01-09T22:56:49.410" />
  <row Id="2665" PostTypeId="5" CreationDate="2011-01-09T23:03:13.830" Score="0" ViewCount="0" Body="&lt;p&gt;Term introduced by Herman Minkowki to describe the ontological status of space and time in the theory of special relativity and subsequently also in general relativity and more modern theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence of the Lorentz transformations, time and space transform into each other when changing reference frame. This calls for a unified description: Minkowski spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term has outgrown its original context, and is now used in other theories.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2011-01-09T23:03:13.830" LastActivityDate="2011-01-09T23:03:13.830" />
  <row Id="2666" PostTypeId="4" CreationDate="2011-01-09T23:03:13.830" Score="0" ViewCount="0" Body="As a consequence of the Lorentz transformations, time and space transform into each other when changing reference frame. This calls for a unified description: Minkowski spacetime.&#xD;&#xA;" OwnerUserId="-1" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2011-01-09T23:03:13.830" LastActivityDate="2011-01-09T23:03:13.830" />
  <row Id="2667" PostTypeId="1" AcceptedAnswerId="2681" CreationDate="2011-01-09T23:39:23.850" Score="3" ViewCount="144" Body="&lt;p&gt;You've probably ridden on the fairground ride called the &quot;Tilt-A-Whirl&quot;, or--as Disneyland calls it--the &quot;Spinning Teacups&quot;, as well as other fairground rides that employ epicycles. You can really feel the centrifugal force strongly when your car spins complementary to the spin of the main rotor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, can that extra force be measured on Earth as our orbit around the Sun complements the Sun's orbit around the galactic hub? Has anyone done it?&lt;/p&gt;&#xA;" OwnerUserId="961" LastActivityDate="2011-01-10T15:45:16.780" Title="Can we measure the &quot;Tilt-A-Whirl&quot; effect of the Earth+Sun orbiting the galactic hub?" Tags="&lt;astrophysics&gt;&lt;centrifugal-force&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="2668" PostTypeId="2" ParentId="2644" CreationDate="2011-01-10T03:32:16.993" Score="0" ViewCount="" Body="&lt;p&gt;Ok, a bit more searching and I came across the &lt;a href=&quot;http://plato.stanford.edu/entries/newton-principia/&quot; rel=&quot;nofollow&quot;&gt;Stanford Encyclopedia of Philosophy&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In other words, the measure of the change in motion is the distance between the place where the body would have been after a given time had it not been acted on by the force and the place it is after that time. This is in keeping with the measure universally used at the time for the strength of the acceleration of surface gravity, namely the distance a body starting from rest falls vertically in the first second. The only special provision that Newton has to make is for non-uniform continuously acting forces, for which, in accord with Lemma 10, he takes the distance AB to vary “at the very beginning of the motion in the squared ratio of the times.”[21]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If this way of interpreting the second law seems perverse, keep in mind that the geometric mathematics Newton used in the Principia — and others were using before him — had no way of representing acceleration as a quantity in its own right. Newton, of course, could have conceptualized acceleration as the second derivative of distance with respect to time within the framework of the symbolic calculus. This indeed is the form in which Jacob Hermann presented the second law in his Phoronomia of 1716 (and Euler in the 1740s). But the geometric mathematics used in the Principia offered no way of representing second derivatives. (Newton employed curvature — that is, the circle “touching a curve” — in place of the second derivative with respect to distance throughout the Principia). Hence, it was natural for Newton to stay with the established tradition of using a length as the measure of the change of motion produced by a force, even independently of the advantage this measure had of allowing the law to cover both discrete and continuously acting forces (with the given time taken in the limit in the continuous case).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Under this interpretation, Newton's second law would not have seemed novel at the time. The consequences of impact were also being interpreted in terms of the distance between where the body would have been after a given time, had it not suffered the impact, and where it was after this time, following the impact, with the magnitude of this distance depending on the relative bulks of the impacting bodies. Moreover, Huygens's account of the centrifugal force (that is, the tension in the string) in uniform circular motion in his Horologium Oscillatorium used as the measure for the force the distance between where the body would have been had it continued in a straight line and its location on the circle in a limiting small increment of time; and he then added that the tension in the string would also be proportional to the weight of the body. So, construed in the indicated way, Newton's second law was novel only in its replacing bulk and weight with mass.[22]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I find this a bit hard to follow, but sounds like Newton is relying a lemma (assumption) that the distance an object falls varies to the squared ratio of the times, and making arguments about circular motion. So he essentially came to this conclusion on the basis of astronomical observations alone, is that correct? And how would you explain this to a high school student? &lt;/p&gt;&#xA;" OwnerUserId="807" LastEditorUserId="807" LastEditorDisplayName="" LastEditDate="2011-01-10T20:07:00.493" LastActivityDate="2011-01-10T20:07:00.493" CommentCount="2" />
  <row Id="2670" PostTypeId="1" AcceptedAnswerId="2676" CreationDate="2011-01-10T11:23:40.857" Score="9" ViewCount="665" Body="&lt;p&gt;Consider that we have a system described by a wavefunction psi(x). We then make an exact copy of the system, and anything associated with it, (including the inner cogs and gears of the elementary paticles, if any, aswell as the fabric of spacetime), but where all distances are multiplied by a number k, so psi(x) -&gt; psi(k*x), we consider the case k&gt;1 (if k=-1 this is just the parity operation, so for k&amp;lt;0 from the little i read about this we could express it as a product of P and &quot;k&quot; transformations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider then that all observables associated with the new system is identical to the original, ie we find that that the laws of the universe is invariant to a scale transformation x-&gt;kx.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Noether's theorem then, there will be an conserved quantity associated with this symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is; what would this conserved quantity be ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;An incomplete discussion regarding the existence of this symmetry is mentioned here:&#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/721/if-the-size-of-universe-doubled&quot;&gt;If the size of universe doubled&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit2:&#xA;I like the answers, but I am missing the answer for NRQM !&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T14:32:22.817" LastActivityDate="2011-07-06T13:21:14.570" Title="What is the conserved quantity of a scale-invariant universe?" Tags="&lt;quantum-mechanics&gt;&lt;general-relativity&gt;&lt;astrophysics&gt;&lt;conformal-field-theory&gt;" AnswerCount="5" CommentCount="15" FavoriteCount="1" />
  <row Id="2672" PostTypeId="1" AcceptedAnswerId="2869" CreationDate="2011-01-10T12:40:38.007" Score="7" ViewCount="298" Body="&lt;p&gt;Why, when we calculate the total cross section, we make the average other initial states and the sum over final states?&lt;/p&gt;&#xA;" OwnerUserId="592" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T15:22:00.017" LastActivityDate="2011-02-02T15:23:34.120" Title="Calculation of the cross section" Tags="&lt;quantum-mechanics&gt;&lt;scattering&gt;&lt;arrow-of-time&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="2673" PostTypeId="2" ParentId="2409" CreationDate="2011-01-10T12:40:46.287" Score="1" ViewCount="" Body="&lt;p&gt;In basic electrodynamics it is stated that light cannot &quot;get inside&quot; a metal. Therefore one just thinks of piece of metal as of boundary conditions for any light-related problems.  I don't see any difference with that approach when it comes to superconductors.  You can just think of superconductor as of ordinary metal with absolutiely the same conclusion about reflection of light off it's surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand it is obvious that some atoms or the metal must somehow interact with the fields. When we talk about electrodynamics of continuous media we deal with scales that are much larger than atomic scale. The statement about non-penetration of magnetic field inside of superconductor is valid for large scales as well, while it actually gets inside the media to the depth around $10^{-5}$ centimetres. In comparison to inter-atomic scales this is quite large. The same holds for &quot;light not getting inside metal&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to x-rays, I don't think that one can use classical electrodynamics at all, because wavelengths are starting to be comparable to the atomic sizes (1 nm for soft x-ray and 0.01 for hard x-ray against 0.05nm for Bohr radius). &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-01-10T12:40:46.287" CommentCount="1" />
  <row Id="2674" PostTypeId="2" ParentId="2672" CreationDate="2011-01-10T13:12:58.657" Score="5" ViewCount="" Body="&lt;p&gt;We do all the &quot;cross section business&quot; because we want to predict results of experiments.&lt;br&gt;&#xA;Let's take for example some particle with two polarizations states: &quot;+&quot; and &quot;-&quot;. You know that experimentalists will collide 1 000 000 pairs of particles, with polarisation of initial particles being unknown.  Best thing you can do is to hope that in experiment polarization would be evenly distributed. So that you will have around 250000 of &quot;++&quot; collisions, 250000 of &quot;+-&quot;, and the same for &quot;-+&quot; and &quot;--&quot;. And in order to predict the result you will need to do your calculations for each polarization configuration and average over them.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If in experiment the polarization is fixed, then you don't need to do the averaging. And if you know the polarization density matrix or some spectrum for the initial state, then you got to average according to them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next. You know that after collision they will detect some resulting particles. Let's say that your calculations predict 1000 of &quot;+&quot; polarized particles and 500 of &quot;-&quot; polarization. But it turns out that the detector do not detect polarization of final particles. Best thing you can do -- sum over final polarizations and predict 1500 for the final result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made an example with polarization, while it can be extended to any properties/degrees of freedom in the experiment. If you do not know some property in initial state -- you should average over them. If you do not know some property in final state -- sum over them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works also for total/differential cross section. If the detector can tell where does particle fly -- then you can work with differential cross section. But if the detector just tells you that &quot;there is a particle&quot; without any info on the direction -- then you will need to integrate the differential cross section (sum over final states) to get total cross section.  &lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2011-01-10T17:48:45.317" LastActivityDate="2011-01-10T17:48:45.317" CommentCount="5" />
  <row Id="2675" PostTypeId="2" ParentId="2528" CreationDate="2011-01-10T13:32:04.627" Score="2" ViewCount="" Body="&lt;p&gt;A new and more popular exposition directly from fields medalist &lt;strong&gt;Shing-Tung Yau: The shape of inner space&lt;/strong&gt;&lt;br&gt;&#xA;Find a review here: &lt;a href=&quot;http://plus.maths.org/content/node/5389&quot; rel=&quot;nofollow&quot;&gt;http://plus.maths.org/content/node/5389&lt;/a&gt;&lt;br&gt;&#xA;Find a very gentle and short intro here: &lt;a href=&quot;http://plus.maths.org/content/node/5388&quot; rel=&quot;nofollow&quot;&gt;http://plus.maths.org/content/node/5388&lt;/a&gt;&lt;br&gt;&#xA;Find a preview here: &lt;a href=&quot;http://books.google.de/books?id=M40Ytp8Os_gC&amp;amp;lpg=PP1&amp;amp;ots=3dHRt8v3KI&amp;amp;dq=The%20shape%20of%20inner%20space&amp;amp;hl=en&amp;amp;pg=PP1#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.de/books?id=M40Ytp8Os_gC&amp;amp;lpg=PP1&amp;amp;ots=3dHRt8v3KI&amp;amp;dq=The%20shape%20of%20inner%20space&amp;amp;hl=en&amp;amp;pg=PP1#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;&lt;br&gt;&#xA;Find the webpage here: &lt;a href=&quot;http://www.shapeofinnerspace.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.shapeofinnerspace.com/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2011-01-10T13:32:04.627" />
  <row Id="2676" PostTypeId="2" ParentId="2670" CreationDate="2011-01-10T14:24:05.150" Score="11" ViewCount="" Body="&lt;p&gt;The symmetry you are asking about is usually called a scale transformation or dilation and it, along with Poincare transformations and conformal transformations is part of the group of conformal isometries of Minkowski space. In a large class of theories one can construct an &quot;improved&quot; energy-momentum tensor $\theta^{\mu \nu}$ such that the Noether current corresponding to scale transformations is given by $s^\mu=x_\nu \theta^{\mu \nu}$. The spatial&#xA;integral of the time component of $s^\mu$ is the conserved charge. Clearly $\partial_\mu s^\mu = \theta^\mu_\mu$ so the conservation of $s^\mu$ is equivalent to the vanishing of the trace of the energy-momentum tensor. It should be noted that most quantum field theories are not invariant under scale and conformal transformations. Those that are are called conformal field theories and they have been studied in great detail in connection with phase transitions (where the theory becomes scale invariant at the transition point), string theory (the two-dimensional theory on the string world-sheet is a CFT) and some parts of mathematics (the study of Vertex Operator Algebras is the study of a particular kind of CFT).&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T04:10:33.157" LastActivityDate="2011-03-02T04:10:33.157" />
  <row Id="2677" PostTypeId="2" ParentId="2189" CreationDate="2011-01-10T14:42:01.493" Score="1" ViewCount="" Body="&lt;p&gt;The problem you are talking about is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Spherical_aberration&quot; rel=&quot;nofollow&quot;&gt;the spherical aberration&lt;/a&gt;. Spherical lenses are much easier to make, while from geometrical point of view the ideal focusing surface is a parabola. Since the light on optical instruments goes close to optical axis, one uses &lt;a href=&quot;http://en.wikipedia.org/wiki/Paraxial_approximation&quot; rel=&quot;nofollow&quot;&gt;the paraxial approximation&lt;/a&gt; where sphere and parabola are the same up to the quadratic term. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a large variety of &lt;a href=&quot;http://en.wikipedia.org/wiki/Aspheric_lens&quot; rel=&quot;nofollow&quot;&gt;aspheric lenses&lt;/a&gt;, with parabolic lenses among them. But it is usually simpler to use combinations of lenses to deal with spherical and other aberrations. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-01-10T14:42:01.493" />
  <row Id="2678" PostTypeId="2" ParentId="2670" CreationDate="2011-01-10T15:12:20.987" Score="7" ViewCount="" Body="&lt;p&gt;thank you for the nice question. It directly relates to the topics of &lt;strong&gt;conformal field theories&lt;/strong&gt;. I found a very nice &lt;a href=&quot;http://www.physicsforums.com/showthread.php?t=314188&quot; rel=&quot;nofollow&quot;&gt;thread in another forum&lt;/a&gt; where I guess your question has been answered.&lt;br&gt;&#xA;Nevertheless, I will try to summarize the main points here and maybe add some points.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Symmetries in General Relativity&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In general relativities, symmetries correspond to an &lt;strong&gt;isometry of the metric&lt;/strong&gt; $g=g_{ab}dx^a dx^b$, say $\varphi^\star g = g$. That means, if you move along the path of such a symmetry, it does not change. This can be expressed in terms of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lie_derivative&quot; rel=&quot;nofollow&quot;&gt;Lie-derivative&lt;/a&gt;.&lt;br&gt;&#xA;$$L_v g = 0$$&#xA;or&#xA;$$\nabla_{(a}v_{b)} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the parenthesis stands for &lt;a href=&quot;http://en.wikipedia.org/wiki/Symmetrization&quot; rel=&quot;nofollow&quot;&gt;symmetrization&lt;/a&gt; over indices and $v = \dot\varphi(t)$ is the vector field associated to $\varphi$. One can find very nice introductory calculations for this in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0226870332&quot; rel=&quot;nofollow&quot;&gt;Robert M. Wald: General Relativity&lt;/a&gt; and &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521010691&quot; rel=&quot;nofollow&quot;&gt;Hans Stephani's Introduction to Special and General Relativity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is a unit geodesic, further integration of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q=v_a n^a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;leads to &lt;strong&gt;conserved quantities&lt;/strong&gt; since&#xA;$$n^a \nabla_a \left(Q = n^b v_b\right) = n^a n^b\nabla_b v_a + v_b n^a \nabla_a n^b\equiv 0 $$&#xA;due to the Killing and geodesic equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Famous examples are mass $M$ (or energy) for a stationary spacetime or angular momentum $J$ for axial symmetry (yes, one can assign a spacetime an angular momentum, I found it puzzling in the first place),&#xA;$$M = 2\int_\Sigma \left( T_{ab}-\frac12 T^n_{\,n}g_{ab} \right)n^a \xi^b dV$$&#xA;$$J = -\int_\Sigma T_{ab}n^a \eta^b dV$$&#xA;where now $\xi$ is the stationary Killing vector, often $\xi = \partial_t$ and $\eta$, often $\eta=\partial_\varphi$ holds for the axial symmetry and $n$ is now vector perpendicular to a space-like hypersurface $\Sigma$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Conformal isometries&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now, the situation is a little bit different. A &lt;strong&gt;conformal Killing vector&lt;/strong&gt; $c$ now gives rise to a symmetry of the form&#xA;$$L_cg=\omega^2g$$&#xA;and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Conformal_Killing_equation&quot; rel=&quot;nofollow&quot;&gt;conformal Killing equation&lt;/a&gt;, implicitly defining $\omega$ now takes the form&#xA;$$\nabla_{(a}c_{b)} = \frac1n g_{ab}\nabla_d c^d$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, you force $\omega = 1$ but this is not of great importance as you will see next.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens to the &quot;conservation equation&quot;? We have&#xA;$$n^a \nabla_a \left( n^b c_b\right) = \frac1n \left( \nabla_d c^d \right) n^a n_a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is only zero if $n^a n_a = 0$, a null-geodesic. So, only for a very special class of movements, here light-particles, one will find a symmetry. But this was &lt;strong&gt;expected&lt;/strong&gt; since conformal transformations will not change angles thus light movement won't be affected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think that this is a conserved quantity in the sense of Emmy Noether.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS.: I apologize for any inconvenience concerning notation. I hope everything is clear from context.&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2011-01-11T15:03:35.567" LastActivityDate="2011-01-11T15:03:35.567" CommentCount="4" />
  <row Id="2679" PostTypeId="1" AcceptedAnswerId="2682" CreationDate="2011-01-10T15:26:00.860" Score="5" ViewCount="153" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_furnace&quot; rel=&quot;nofollow&quot;&gt;solar furnace&lt;/a&gt; is a device that concentrates the sun's light on a small point to heat it up to high temperature. One can imagine that in the limit of being completely surrounded by mirrors, your entire $4\pi$ solid angle will look like the surface of the sun, at about 6000K. The target will then heat up to 6000K and start to radiate as a blackbody, reaching thermal equilibrium with the sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: is there any way to surpass this temperature, perhaps by filtering the light to make it look like a BB spectrum at higher temp, then concentrating it back on the target?&lt;/p&gt;&#xA;" OwnerUserId="775" LastEditorUserId="869" LastEditorDisplayName="" LastEditDate="2011-01-10T20:55:24.100" LastActivityDate="2011-01-10T20:55:24.100" Title="What temperature can you attain with a solar furnace?" Tags="&lt;thermodynamics&gt;&lt;thermal-radiation&gt;&lt;radiation&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="2681" PostTypeId="2" ParentId="2667" CreationDate="2011-01-10T15:45:16.780" Score="3" ViewCount="" Body="&lt;p&gt;The force that you feel in a Tilt-A-Whirl is not possible in a galaxy and does not correspond to anything in our solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a Tilt-A-Whirl, you are subject to two additional constraints that are not present in the solar system galaxy interaction:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The base of the Tilt-A-Whirl platform is not flat so there are times when the Earth's gravity slows the spinning of the cars as you go uphill and times when the Earth's gravity speeds up the spinning of the cars as you go downhill as the overall platform rotates changing the plane of the individual cars as it goes around. There is no corresponding accelerating/decelerating force in the galaxy that affects the solar system.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is a physical constraint that keeps the individual cars in a Tilt-A-Whirl fixed to the larger rotating platform so the riders in the cars are kept at a fixed radius from the point of the car's rotation. In the galaxy, the Earth is free to move outward when the velocity increases so that it always follows an elliptical geodesic orbit which is a straight unaccelerated line through spacetime. Likewise, the variations in speed of the entire solar system as it revolves around the galaxy are not reflected in additional force because the system is not attached so it too follows an elliptical geodesic orbit rather than a circular one.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="948" LastActivityDate="2011-01-10T15:45:16.780" />
  <row Id="2682" PostTypeId="2" ParentId="2679" CreationDate="2011-01-10T18:07:49.490" Score="3" ViewCount="" Body="&lt;p&gt;Theoretically the answer is yes. That is because the sun is not a blackbody emitter, there is an excess of UV radiation. So if you were able to achieve radiative equilibrium with only UV light (which is maybe 1% of BB radiation at those temps), you could do it. Practically, I'd think it would be just about impossible, as your filter would have to have it's innermost surface at nearly 6000K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The solar UV primarily comes from the chromosphere and corona, which is heated (in some not too well understood way) by mechanical/magnetic energy derived from convective processes. The X-ray excess is even greater than the UV excess. Even the earth gives off detectable gamma rays, and that would be impossible thermally.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-01-10T18:07:49.490" />
  <row Id="2683" PostTypeId="2" ParentId="2189" CreationDate="2011-01-10T19:12:04.947" Score="4" ViewCount="" Body="&lt;p&gt;The last part of your question is the easiest to answer, so I'll get to that first. The best book on the fundamentals of optical design is &quot;Modern Optical Engineering&quot; by Warren J. Smith. It is not specific to aspheric optics, but does cover them in addition to the rest of geometrical optics and lens design. It is probably the single most common reference book among optical engineers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the rest of your question is a bit complicated, and needs a little bit of background, so bear with me for a moment. As has been mentioned, even an ideal lens will produce a focal spot of some minimum size, determined by the ratio of the lens focal length to its aperture (this quantity is called the &quot;f-number, or $f/\#$&quot;) and the wavelength of the light. This is what optical engineers call the diffraction limited spot size. For a circular aperture, the diameter of the diffraction limited spot size will be $$2.44 \times \lambda \times f/\#$$ where $\lambda$ is the wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as the $f/\#$ decreases (as the lens gets &quot;faster&quot;) the diffraction limited spot will become smaller. &lt;strong&gt;However&lt;/strong&gt;, any aberrations in the lens will also become more significant! This means that a very slow lens (one with a long focal length, relative to its aperture) can produce a diffraction limited spot even though it may have some aberration relative to an ideal lens, while a very fast lens will need to have a slightly aspheric shape to achieve diffraction limited performance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is important to understand because it means that, in some cases, a spherical lens can indeed focus light as close to a point as is physically possible, even though a sphere isn't the ideal shape.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is that ideal shape? Well again, it depends on a few things. For both lenses and mirrors, the ideal shape will change depending on the distance from the object plane to the lens, and from the lens to the image plane. In the case you've asked about, where the incoming light is collimated, optical engineers would say that the object plane is at infinity. In this case, as some other people have pointed out, the ideal shape for a mirror is indeed a parabola. However, for a lens this is not the case. &lt;strong&gt;As it turns out, the ideal shape for a lens to focus a collimated beam of light to a point is to have the first surface of the lens (the one the light hits first) be elliptical, and the back surface be hyperbolic.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lens designers usually specify the shape of a lens surface with the following equation: $$Z = \frac{C r^2}{1 + \sqrt{1-(1+\kappa) C^2 r^2}}$$ where $Z$ is the &quot;sag&quot; of the lens surface, or its departure from a plane tangent to the lens surface at the center of the lens, $r$ is the radial distance from the center of the lens, $c$ is the curvature of the lens (the reciprocal of its radius of curvature) and $\kappa$ is called the &quot;conic constant.&quot; It is the value of $\kappa$ which determines what sort of conic section describes the surface:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\kappa &amp;gt; 0$ Oblate ellipse&lt;/li&gt;&#xA;&lt;li&gt;$\kappa = 0$ Sphere&lt;/li&gt;&#xA;&lt;li&gt;$0 &amp;gt; \kappa &amp;gt; -1$ Prolate Ellipse&lt;/li&gt;&#xA;&lt;li&gt;$\kappa = -1$ Parabola&lt;/li&gt;&#xA;&lt;li&gt;$-1 &amp;gt; \kappa$ Hyperbola&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;On a related note, it is more than just the conic constant that can be adjusted to control aberrations. Even with purely spherical surfaces, the relative curvature of the front and back lens surface can be varied, while keeping the effective focal length constant. Adjusting this is more common than adding aspeheric surfaces to a lens, because aspheric surfaces are expensive to manufacture. Many optical supply companies even offer off-the-shelf optics with an ideal bending ratio for a given application. These are often sold as &quot;best form&quot; lenses.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-01-10T19:12:04.947" CommentCount="5" />
  <row Id="2684" PostTypeId="1" AcceptedAnswerId="2688" CreationDate="2011-01-10T19:44:50.060" Score="6" ViewCount="152" Body="&lt;p&gt;Assume a static metric with (known) components $g_{\mu\nu}$. I'd like to know what is the gravitational pull $g$ of a test particle placed on an arbitrary point $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational pull being defined as the acceleration the particle suffers as measured by an observer sitting in a reference frame fixed at the origin. &lt;strong&gt;What are (theoretically speaking) the steps one needs to take to find this acceleration?&lt;/strong&gt; &lt;em&gt;(no need to actually calculate for the general case, just list the general steps)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, &lt;strong&gt;what is the result for the simple case of the Schwarzschild metric?&lt;/strong&gt;&lt;br&gt;&#xA;In other words, what is the gravitational pull of a star or black hole? Will it matter if the test particle is moving or is static?&lt;/p&gt;&#xA;" OwnerUserId="724" LastActivityDate="2011-01-10T21:25:17.917" Title="How do I calculate the (aparent) gravitational pull with General Relativity?" Tags="&lt;general-relativity&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="2685" PostTypeId="1" CreationDate="2011-01-10T19:56:48.313" Score="13" ViewCount="373" Body="&lt;p&gt;We're all familiar with basic tenants such as &quot;information cannot be transmitted faster than light&quot; and ideas such as information conservation in scenarios like Hawking radiation (and in general, obviously). The Holographic Principle says, loosely, that information about a volume of space is encoded on its two-dimensional surface in Plank-sized bits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all these contexts, I can take &quot;information&quot; to mean predictive or postdictive capability, i.e. information is what enables us to state what the outcome of a measurement was or will be (locally). But what &lt;em&gt;is&lt;/em&gt; information, exactly? Do we have any kind of microscopic description of it? Is it just a concept and, if so, how can we talk about transmitting it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect this is probably as unanswerable as what constitutes an observer/measurement for wave function collapse, but I'd love to know if we have any formulation of what information is made of, so to speak. If I'm talking nonsense, as I suspect I may be, feel free to point this out.&lt;/p&gt;&#xA;" OwnerUserId="861" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2011-01-11T15:00:49.903" LastActivityDate="2011-01-19T02:24:38.647" Title="What is information?" Tags="&lt;theory&gt;&lt;information&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="6" />
  <row Id="2686" PostTypeId="2" ParentId="2644" CreationDate="2011-01-10T20:06:40.463" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/users/347/jerry-schirmer&quot;&gt;Jerry Schirmer&lt;/a&gt; and Tobais Kienzler offer the what seems to me to be a pretty good answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jerry says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It's kinetmatics to determine the acceleration of the moon. Geometry says that the acceleration of a circular orbit is v2r. You can measure the distance to the moon via parallax, and if you know the distance, you can infer the velocity from the length of the month. Newton's 2nd law is more a definition than a statement. Once you have the law of inertia, then you just presuppose when something deviates from constant motion, there must be some force, and the more deviation you get, the more force. It's circular unless you just define force this way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tobias Says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This may sound weird, but I never understood what's so special about it: There is momentum, and if it isn't constant there is a cause defined as force, and measurable by observing the change of momentum. The great thing however is the idea of generalization to obtain e.g. the law of gravity as something valid for all kinds of matter and not just the one observed in one experiment&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So is N2 really a way of defining force in terms of change in momentum? I had always heard it was a relationship that had to be proven by experiment, and that's certainly the only way I've seen it taught in school—via experiment. &lt;/p&gt;&#xA;" OwnerUserId="807" LastActivityDate="2011-01-10T20:06:40.463" CommentCount="1" />
  <row Id="2687" PostTypeId="1" CreationDate="2011-01-10T20:37:32.580" Score="2" ViewCount="257" Body="&lt;p&gt;I'm wondering what some standard, modern references might be for continuum mechanics.  I imagine most references are probably more used by mechanical engineers than physicists but it's still a classical mechanics question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This came up in a conversation with my father (who is a mechanical engineer).  I was curious to see the types of mathematics they use in stress analysis.  Complex analysis gets used to study 2-dimensional isotropic material here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N.I. Muskhelishvili. Some basic problems of the mathematical theory of elasticity. &#xA;3rd edition, Moscow-Leningrad. 1949. (Translated by J. R. M. Radok. Noordhoff. 1953.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but that's quite an old reference and my library does not have it.  Any favourite more modern and easy-to-find references? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm fine with mathematical sophistication (I'm a mathematician) but I'm not particularly seeking it out.  I'm looking for the kind of references that would be valued by physicists and engineers.  References &quot;for math types&quot; are fine too but that's not really what I'm after. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1185" LastActivityDate="2011-03-02T04:09:30.507" CommunityOwnedDate="2011-02-04T10:52:39.143" Title="Modern references for continuum mechanics" Tags="&lt;classical-mechanics&gt;" AnswerCount="7" />
  <row Id="2688" PostTypeId="2" ParentId="2684" CreationDate="2011-01-10T21:11:29.197" Score="5" ViewCount="" Body="&lt;p&gt;This problem is somewhat ill-formed.  It's difficult to even define what 'static at the origin' means--static relative to what?  What I will do is show the steps one needs to take to derive the motion of a particle, however.  The following will be more constructive than explanatory, as explaining all of this thoroughly would be close to a whole chapter in a relativity/differential geometry textbook.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the metric tensor $g_{ab}$, one can define the Christoffel symbols $\Gamma_{ab}{}^{c}$ by the equation$^{1}$ $\Gamma_{ab}{}^{c} = \frac{1}{2}g^{cd}\left(\partial_{a}g_{bd}+\partial_{b}g_{ad}-\partial_{d}g_{ab}\right)$.  While the Christoffel symbols serve an array of purposes in relativity, the simplest notion that one can tie to them is that they define a notion of parallel transport in the spacetime--a vector $v^{a}$ is parallel transported along a curve $x^{a}$ if $x^{a} \nabla_{a}v^{b} \equiv x^{a}\partial_{a}v^{b} + x^{a}v^{c}\Gamma_{ac}{}^{b} =0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can talk about paths whose tangent vectors are parallel translated relative &lt;b&gt;to themselves&lt;/b&gt;.  These paths are known as geodesics.  If we parameterize these curves with respect to the arc length subtended by the curve$^{2}$(which we interpret as the clock of the observer following the geodesic), then the curves satisfy the equation $\ddot x^{a} + \Gamma_{bc}{}^{a}\dot x^{b} \dot x^{c} = 0$.  If we choose the $x^{a}$ as a locally Cartesian orthonormal set of coordinates, then we can interpret $\ddot x^{a}$ as the acceleration of the $x^{a}$ position along the curve, and then interpret the other term as the acceleration of the curve.  Practically, this is a difficult game to interpret in general relativity, since time is one of the coordinates (making acceleration a subtle thing to define), and because our coordinates are completely arbitrary and need not even be (and in fact, usually are not) orthonormal.  Nonetheless, this interpretation can be useful for a few special cases--the Schwarzschild black hole metric being one of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, it's actually easiest to look at the conserved energy of a particle on the geodesic first, and then to define a force from this.  Here, in units where $G=c=1$, we have $E^{2} -1= \dot r ^{2} - \frac{2M}{r} + \left(\frac{L}{r}\right)^{2}  - \frac{2ML^{2}}{r^{3}}$, where E is the energy per unit mass of the test particle, $M$ is the mass of the central gravitating object, and $L$ is the angular momentum of the test particle.  Taking a time derivative of this, and dividing by the common factor $2\dot r$, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0=\ddot r + \frac{M}{r^{2}}-\frac{L^{2}}{r^{3}} + \frac{3ML^{2}}{r^{4}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first three terms should be familiar from Newtonian theory--they are precisely the Newtonian gravitational force term plus the centripetal inertia of the orbit.  Relative to the clock of an observer in orbit, so long as $\frac{3ML^{2}}{r^{4}}$ is small, Newtonian motion should be indistinguishable from Relativistic motion.  So, the only difference in acceleration comes from the rotation of the orbit.  It is precisely this term that created the anamolous acceleration of Mercury that was such a puzzle to the astronomers at the turn of the century.  When Einstein observed that this term was of the right sign and magnitude to explain this anamoly was when he truly believed that he had found the correct theory of gravity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^{1}$ where $\partial_{a}f$ is defined to be the derivative of $f$ with respect to the a$^{th}$ coordinate: $\partial_{a}f \equiv \frac{\partial f}{\partial x^{a}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^{2}$ this sounds complicated, but is in fact what we do when measuring angles on a circle in radians--how many radians I trace out tells me how much length along the circle I've traced out since I've left the origin.  Same basic idea here.&lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2011-01-10T21:25:17.917" LastActivityDate="2011-01-10T21:25:17.917" CommentCount="2" />
  <row Id="2689" PostTypeId="2" ParentId="2685" CreationDate="2011-01-10T21:31:40.093" Score="7" ViewCount="" Body="&lt;p&gt;Information is a purely mathematical concept, usually a a characteristic of uncertainty (of a probability distribution function), but can be interpreted in different ways. In the simplest form it is introduced in information theory as a difference between uncertainties of two distributions, with uncertainty being the logarithm of a number of possible equally-probable states of a discrete random variable. For continuous distribution, it can be introduced as a logarithm of an integral. Sometimes introduced proper information - a quantity which differs from negative entropy only by a constant independent of the distribution (this constant can be taken as zero).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus information is a difference of proper information (difference of negative entropy) of two states. The states are represented by probability distribution functions, thus information is a formal operator of two functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For continuous distributions (of which discrete case is a variant) proper information of distribution $w$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I[w]=-H(w)=-\int_{-\infty}^{+\infty}w(x)\log(w(x))dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and relative information of $w_2$ compared to $w_1$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I[w_2,w_1]=H(w_1)-H(w_2)=I(w_2)-I(w_1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I[w_2,w_1]=\int_{-\infty}^{+\infty}\log \left(\frac{w_1(x)^{w_1(x)}}{w_2(x)^{w_2(x)}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This operator is not much different from norm or angle in vector spaces. It is just one measure, attributed to members of the space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare this with the definition of norm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$||w||=\sqrt{\int_{-\infty}^{+\infty}w(x)^2dx}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;distance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D[w_1,w_2]=||w_1-w_2||=\sqrt{\int_{-\infty}^{+\infty}(w_1(x)-w_2(x))^2dx}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;angle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Phi[w_1,w_2]=\arccos \frac{\int_{-\infty}^{+\infty}w_1(x)w_2(x)dx}{\sqrt{\int_{-\infty}^{+\infty}w_1(x)^2dx}\sqrt{\int_{-\infty}^{+\infty}w_2(x)^2dx}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So think about information as of a mathematical quantity similar to angle.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastEditorUserId="1186" LastEditorDisplayName="" LastEditDate="2011-01-10T22:03:31.623" LastActivityDate="2011-01-10T22:03:31.623" CommentCount="9" />
  <row Id="2690" PostTypeId="1" AcceptedAnswerId="2692" CreationDate="2011-01-10T21:32:43.353" Score="9" ViewCount="754" Body="&lt;p&gt;According to Noether's theorem, all conservation laws originate from invariance of a system to shifts in a certain space. For example conservation of energy stems from invariance to time translation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of symmetry creates the conservation of mass?&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T15:21:25.383" LastActivityDate="2011-04-14T08:41:57.773" Title="What is the symmetry which is responsible for conservation of mass?" Tags="&lt;energy&gt;&lt;mass&gt;&lt;lagrangian-mechanics&gt;&lt;hamiltonian-mechanics&gt;&lt;symmetry&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="4" />
  <row Id="2691" PostTypeId="1" CreationDate="2011-01-10T21:42:54.223" Score="3" ViewCount="206" Body="&lt;p&gt;Here's the situation: a friend is working on a civil engineering project to make some equipment to decompress liquefied natural gas before pumping to households.  I don't know what the proper term for this bit of kit is calld but I will call it the &lt;strong&gt;decompresser&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently what happens in these stations is that the gas is decompressed and because the gas cools down during expansion, it has to be heated back to something like room temperature before going on its way through the gas pipes to homes.  They do this by burning kerosene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my friend says that they have designed a new decompresser that, in order to be more energy efficient, will generate electricity by decompressing the gas through a turbine.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a long time since I did any related physics but the simpleton-scientist-thought-process that I have going on in my head is something like the following: -&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if the pressurised gas is having to do work turning the turbine, then this energy has to come from somewhere &lt;/li&gt;&#xA;&lt;li&gt;thus the temperature of the resulting de-pressurized gas will be lower.  &lt;/li&gt;&#xA;&lt;li&gt;Because the temperature of the gas is lower, they would have to put more energy than they would have done to reheat the gas and they would not have any net energy gain.  They would have electricity but they would be burning an equivalent amount of kerosene. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now assuming that these civil engineers know a thing or two about thermodynamics, I readily admit that my thinking is wrong.  I would very much appreciate it if someone could put me right on whether or not they can generate electricity as well as have the gas come out at the same temperature and pressure and could they not just have used the extra kerosene to generate the same amount of electricity (ignoring the various differing efficiencies of energy conversion that are going on).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PS.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because I am clueless as to the actual answer, I will be relying heavily on the upvoting of + comments about the answers when it comes to marking an answer as the right 1!  &lt;/p&gt;&#xA;" OwnerUserId="1187" LastEditorUserId="1187" LastEditorDisplayName="" LastEditDate="2011-01-11T10:04:36.627" LastActivityDate="2011-02-22T17:09:19.187" Title="Decompressing gas whilst driving a turbine - where does the energy come from?" Tags="&lt;thermodynamics&gt;" AnswerCount="4" />
  <row Id="2692" PostTypeId="2" ParentId="2690" CreationDate="2011-01-10T21:50:49.473" Score="14" ViewCount="" Body="&lt;p&gt;Mass is only conserved in the low-energy limit of relativistic systems.  In relativistic systems, mass can be converted into energy, and you can have processes like massive electron-positron pairs annhillating to form massless photons.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What &lt;b&gt;is&lt;/b&gt; conserved (in theories obeying special relativity, at least) is mass energy--this conservation is enforced by the time and space translation invariance of the theory.  Since the amount of energy in the mass dominates the amount of energy in kinetic energy ($mc^{2}$ means a lot of energy is stored even in a small mass) for nonrelativistic motion, you get a very good approximation of mass conservation. out of the energy conservation.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-01-10T21:50:49.473" CommentCount="4" />
  <row Id="2693" PostTypeId="2" ParentId="2690" CreationDate="2011-01-10T22:24:53.140" Score="13" ViewCount="" Body="&lt;p&gt;Noether's theorem says that symmetries lead to conservation laws, not the converse. Conservation of mass doesn't follow from any of the obvious symmetries of nonrelativistic motion. Those symmetries are translations in space (leading to conservation of momentum), translations in time (conservation of energy), rotations (conservation of angular momentum), and boosts (i.e. changes to a frame moving at constant velocity with respect to the original frame, leading to conservation of center-of-mass motion). The algebra of these symmetries, known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Galilean_transformation&quot; rel=&quot;nofollow&quot;&gt;Galilean transformations&lt;/a&gt;, involves mass as a central charge. Since it commutes with everything in sight, I think it's fair to say that there's no nontrivial associated symmetry.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2011-01-10T22:24:53.140" CommentCount="2" />
  <row Id="2694" PostTypeId="2" ParentId="2691" CreationDate="2011-01-10T23:02:41.970" Score="0" ViewCount="" Body="&lt;p&gt;Ideal gas law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ PV = n R T $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the depressurization is an isothermal process then $T_1 = T_2 = T$ and $P_1 V_1 = P_2 V_2$, with $1$ and $2$ denoting the quantities before and after the depressurization respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work done is given by $ W = \int P dV$. For constant temperature $ P = nRT/V$, thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ W = nRT \int_{V_1}^{V_2} dV/V = nRT \, \ln(V_1/V_2) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as the system is well-insulated the work done by the gas during depressurization should drive the turbine without an appreciable drop in temperature. Of course, the devil lies in the details!&lt;/p&gt;&#xA;" LastActivityDate="2011-01-10T23:02:41.970" CommentCount="5" />
  <row Id="2695" PostTypeId="2" ParentId="2691" CreationDate="2011-01-10T23:03:37.890" Score="2" ViewCount="" Body="&lt;p&gt;I don't think you have a problem with thermo. The lower pressure gas will come out at a lower temperature. However heat is pretty common and cheap in the environment, so I suspect they simply plan to use free environmental heat to reheat the gas to nearly ambient temperature.  The energy in the first place came from the compression process (work = pressure times change in volume). The compressed gas was initially hot, but this heat was likely lost during storage by conduction through its container. The PV work done on the gas is stored as both kinetic energy (moving gas molecules), and internal energy within the molecules themselves. In any case, some of this work can be recovered during re-expansion, but whenever energy is transferred via non-adiabatic processes (heat flow) potential work is lost (i.e. system entropy increased).&lt;/p&gt;&#xA;" OwnerUserId="377" LastEditorUserId="377" LastEditorDisplayName="" LastEditDate="2011-02-22T17:09:19.187" LastActivityDate="2011-02-22T17:09:19.187" CommentCount="2" />
  <row Id="2696" PostTypeId="1" CreationDate="2011-01-11T00:17:23.180" Score="9" ViewCount="493" Body="&lt;p&gt;It is known that any accelerated observer is subject to a heat bath due to Unruh radiation. The principle of equivalence suggests that any stationary observer on the surface of a massive body should also experience heat bath. Indeed by substituting surface gravity g into the formula for Unruh radiation one can find the black body temperature for a hypothetical hyper-cool planet:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T = \frac{\hbar g}{2\pi c k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is $3.9766&amp;#215;10^{-20}\ K$ for Earth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one even can find the time which it will take for Earth to evaporate: $5.69&amp;#215;10^{50}$ years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the heat in the super-cold Earth cannot come out of nothing one should assume that it will come from decay of particles due to a certain mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes I heared an argument that an event horizon is needed for Hawking radiation to exist. But this can be countered by assumption of possibility of decay due to quantum virtual black holes (which inevitably should appear due to uncertainty principle, and the more massive and dense body is the greather concentration of virtual black holes inside it will be, eventually becoming similar to the concentration of bobbles inside a body of boiling water). Or just suggest that any massive body due to uncertainty principle can quantum tunnel into a black hole state so to emit Hawking radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the conclusion here? &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can we say that all massive bodies are surrounded by the atmosphere of heated vacuum?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is a weaker preposition: thermal state of surrounding vacuum does not mean energy transfer if the system is in thermodynamic equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Any body gradually evaporates, i.e. transfers its energy to the surrounding vacuum until completely vanishes?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is a stronger preposition and suggests emission of radiation al loss of mass.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastEditorUserId="1186" LastEditorDisplayName="" LastEditDate="2011-01-11T08:31:42.130" LastActivityDate="2011-08-15T04:48:08.520" Title="Do all massive bodies emit Hawking radiation?" Tags="&lt;thermodynamics&gt;&lt;black-holes&gt;&lt;vacuum&gt;&lt;hawking-radiation&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="3" />
  <row Id="2697" PostTypeId="2" ParentId="2696" CreationDate="2011-01-11T00:52:52.323" Score="3" ViewCount="" Body="&lt;p&gt;My previous answer is beside the point now that the question has been edited. There is a simpler example of a question of this type which has been analyzed in great detail in the literature, and that involves an electric charge which is stationary in a gravitational field. Since the power radiated is nonzero if a charge is accelerated one might, by the equivalence principle, expect such a charge to radiate. I believe the answer is that it does not, but that the analysis is subtle. You could start with the article D. Boulware, &quot;Radiation from a uniformly accelerated charge&quot;, Annals of Physics 124 (1980), 169-187 and work your way forward from there by looking at citations of that paper.&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T04:09:55.030" LastActivityDate="2011-03-02T04:09:55.030" CommentCount="9" />
  <row Id="2698" PostTypeId="2" ParentId="2687" CreationDate="2011-01-11T01:56:52.447" Score="3" ViewCount="" Body="&lt;p&gt;I don't know about very modern references, but it's hard to get better than Landau &amp;amp; Lifschitz: Theory of Elasticity, as far as I imagine most physicists are concerned.&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2011-01-11T01:56:52.447" CommunityOwnedDate="2011-02-04T10:52:39.143" CommentCount="1" />
  <row Id="2699" PostTypeId="2" ParentId="2696" CreationDate="2011-01-11T02:20:54.433" Score="4" ViewCount="" Body="&lt;p&gt;The answer is clearly no, but it is interesting to see what goes wrong with the argument. I think the problems lies in the distinction between Unruh’s effect and Hawking’s. In Unruh’s case one has to be careful what is and what is not implied: an accelerated detector in empty flat space will behave as if immersed in a heat bath, in that it detects particles distributed thermally. However, this does not imply there is radiation, in the sense of energy flow from one place to another, empty space is empty, even in Rindler coordinates. It is only for a real black hole that you have real radiation emanating from the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I vaguely remember a decent discussion of this in Birrell and Davies).&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-01-11T02:20:54.433" CommentCount="17" />
  <row Id="2700" PostTypeId="2" ParentId="2585" CreationDate="2011-01-11T02:30:19.077" Score="2" ViewCount="" Body="&lt;p&gt;Every state can be written in the way you mention, in terms of two states 1 and 2, for an appropriate choice of states 1 and 2. By itself it does not indicate any entanglement. What makes a state entangled is a specific property of the two states 1 and 2, namely that they are physical states belonging to two subsystems which do not interact with each other (for example systems spatially separated from each other). Only then it is interesting to talk about entanglement, which is roughly speaking the degree of correlation between the two states, which cannot be undone by operating on either one of the two subsystems separately.&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-01-11T02:30:19.077" CommentCount="2" />
  <row Id="2701" PostTypeId="1" AcceptedAnswerId="3026" CreationDate="2011-01-11T05:53:30.980" Score="2" ViewCount="223" Body="&lt;p&gt;What are the practical limits on generated wavelength in magnetron?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Too short question I guess :-D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, I don't have any more details :-) We know that magnetrons could be used efficiently for generating microwave for water heating, for radar applications... But what are the achivable wavelength limits?&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="930" LastEditorDisplayName="" LastEditDate="2011-01-11T08:42:37.913" LastActivityDate="2011-01-16T15:43:49.310" Title="Magnetron limits" Tags="&lt;untagged&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="2702" PostTypeId="1" AcceptedAnswerId="2822" CreationDate="2011-01-11T10:59:27.027" Score="2" ViewCount="330" Body="&lt;p&gt;Over and over I'm getting into the same trouble, so I'd like to ask for some help. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I need to solve some basic electrodynamics problem, involving magnetic fields, moving  charges or currents.&lt;/li&gt;&#xA;&lt;li&gt;But I forgot all this rules about &quot;where the magnetic field should go if the current flows up&quot;. I vaguely remember that it about hands or fingers or books or guns, and magnetic field should go somewhere along something, while current should flow along something else... But it doesn't help, because I don't remember the details.&lt;/li&gt;&#xA;&lt;li&gt;I do find some very nice rule and use it. And I think &quot;that one I will never forget&quot;.  &lt;/li&gt;&#xA;&lt;li&gt;...time passes...&lt;/li&gt;&#xA;&lt;li&gt;Go to step 1. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The problem is that you have to remember one choice from a dichotomy: &quot;this way or that way&quot;. And all the mnemonics I know have the same problem -- I got to remember &quot;left hand or right hand&quot;, &quot;this finger or that finger&quot;, &quot;inside the book or outside of the book&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe someone knows some mnemonics, that do not have such problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/SxrgK.gif&quot; alt=&quot;By Bruno Touschek&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-11T14:24:50.963" LastActivityDate="2011-01-13T23:43:41.017" Title="&quot;Magnetic mnemonics&quot;" Tags="&lt;electromagnetism&gt;&lt;mnemonic&gt;" AnswerCount="7" CommentCount="2" />
  <row Id="2703" PostTypeId="2" ParentId="2585" CreationDate="2011-01-11T12:14:36.963" Score="3" ViewCount="" Body="&lt;p&gt;Let's start from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_entanglement#Quantum_Mechanical_Framework&quot; rel=&quot;nofollow&quot;&gt;definition of entangled state&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Briefly -- if the state of your system can be described by separately defining the states of its components, then we call the state of this system a &lt;em&gt;separable state&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If such a description is impossible -- then the state is an &lt;em&gt;entangled state&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for both your examples it is impossible to factorize the states of individual particles in the description of the state of the total system. Therefore both of these states are entangled states. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-01-11T12:14:36.963" />
  <row Id="2704" PostTypeId="2" ParentId="2691" CreationDate="2011-01-11T12:58:34.237" Score="0" ViewCount="" Body="&lt;p&gt;If I undestood the situation correctly, the gas loses temperature during the decompression (hence expansion) accordingly to the law: &#xA;$$PV=nRT$$&#xA;it produces work hence losing energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then, apparently, you need to re-heat the gas at room temperature before pumping it in the gas pipes; i guess for technical reason (pumping a gas at a certaing pressure into pipes with a temperature too low can cause damages when the temperature get higher [due to insolation or stuff like that] thus incrementing the pressure inside the pipes).. in order to do that they burn kerosene.. &#xA;so energetically speaking their balance in negative since during the decompression they lose, in the surrounding, the heat lost by the expanding gas. Then they burn stuff to reheat the same gas.. for this reason i think they are using the heat lost by the decompressing gas to generate electricity (hence energy) that they can partly use to reheat at room temperature the gas that needs to be pumped into the pipes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite certainly (efficiency is a bitch..) their energy balance will be less than zero anyway, but less negative than with the less efficient decompressor..&lt;/p&gt;&#xA;" OwnerUserId="28" LastActivityDate="2011-01-11T12:58:34.237" />
  <row Id="2705" PostTypeId="2" ParentId="2702" CreationDate="2011-01-11T13:13:41.353" Score="2" ViewCount="" Body="&lt;p&gt;Three points I'll like to address:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Use cross product as much as possible. If I give you a and b make sure you know which direction a cross b is pointing in. This is essential for many things beyond EM. Then you can state a law of physics without referring to your left or right hand: for instance the magnetic force on a moving particle is q (v cross B).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course to define cross product you might have to use your hand... try to find a definition that involves your right hand (the reason is given later).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) To find the B field induced by a current point the thumb of your right hand towards the direction of the current. The fingers of your right hand are now curling in a certain direction (counterclockwise if the current is flowing into your face); this is the B field. This is non-negotiable; I'm sorry, the B field is defined such that this is true. And I am pretty certain there is no way of using your left hand to calculate this easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) To find E field induced by changing B field, read up on Lenz's law. If you're OK with the previous two this should be trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you should see why I recommended you use right hand to define cross product; you must use your right hand in step 2, so might as well use it in step 1 too. &lt;/p&gt;&#xA;" OwnerUserId="1193" LastActivityDate="2011-01-11T13:13:41.353" CommentCount="6" />
  <row Id="2706" PostTypeId="2" ParentId="2702" CreationDate="2011-01-11T14:01:52.010" Score="4" ViewCount="" Body="&lt;p&gt;To get the &lt;em&gt;right&lt;/em&gt; answer, you should use the &lt;em&gt;right&lt;/em&gt; hand, which happends to be the &lt;em&gt;right hand&lt;/em&gt;, it cannot possibly get simpler&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2011-01-11T14:08:12.767" LastActivityDate="2011-01-11T14:08:12.767" CommentCount="5" />
  <row Id="2707" PostTypeId="1" CreationDate="2011-01-11T15:01:24.183" Score="4" ViewCount="353" Body="&lt;p&gt;I was searching for the  eigensolutions of the two-dimensional &lt;a href=&quot;http://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation&quot; rel=&quot;nofollow&quot;&gt;Schrödinger equation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{i}\hbar \partial_t \mid  \psi \rangle = \frac{\mathbf{p}^2}{2m_e}\mid  \psi \rangle + V \mid  \psi \rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the potential is given by&#xA;$$V(\rho, \varphi)=\begin{cases} V_1 &amp;amp; \rho &amp;lt; R \\ -V_2 &amp;amp; \rho \geq R \end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using a space representation and &lt;a href=&quot;http://mathworld.wolfram.com/CylindricalCoordinates.html&quot; rel=&quot;nofollow&quot;&gt;cylindrical coordinates&lt;/a&gt;, $V_i \geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be happy if someone could point me to a reference or even give the solution here.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance&lt;br&gt;&#xA;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Request to close the question&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;As I can see in the comments, questions of this kind seem to be &lt;strong&gt;inappropriate&lt;/strong&gt;.&lt;br&gt;&#xA;The eigensolutions are given by something like&#xA;$$\psi_m(\mathbf{r},t)=e^{\mathrm{i}(m\varphi-\omega_m t)}\begin{cases} a_m J_m (k_{m,1} \rho) &amp;amp; \rho &amp;lt; R \\ b_m K_m (k_{m,2} \rho) &amp;amp; \rho \geq R \end{cases}$$&#xA;where the $a_m$ and $b_m$ can be calculated from the steadiness of $\psi$ and its spatial derivative in $R$. Furthermore, $k_{m,1/2} = \frac1\hbar \sqrt{\pm\,2m_e(\hbar\omega_m - V_{1/2})}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sorry for any inconvenience.&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2011-01-11T17:23:13.137" LastActivityDate="2011-01-11T17:23:13.137" ClosedDate="2011-01-19T00:18:50.237" Title="Solution for the Finite 2D Potential Well - Rotational Symmetry" Tags="&lt;quantum-mechanics&gt;" CommentCount="16" />
  <row Id="2708" PostTypeId="1" AcceptedAnswerId="2724" CreationDate="2011-01-11T15:27:25.957" Score="6" ViewCount="1029" Body="&lt;p&gt;What are the main practical applications that a Bose-Einstein condensate can have?&lt;/p&gt;&#xA;" OwnerUserId="592" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-11T15:41:23.563" LastActivityDate="2011-01-12T00:15:26.443" Title="Practical applications for a Bose-Einstein condensate" Tags="&lt;quantum-mechanics&gt;&lt;condensed-matter&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="2709" PostTypeId="1" AcceptedAnswerId="3166" CreationDate="2011-01-11T16:20:18.097" Score="3" ViewCount="254" Body="&lt;p&gt;In industry it's common to cut steel using 50-200W CO2 laser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, for making holes in PCB (35um of Cu) DPSS 355nm laser is usually used + secondary CO2 to drill through glass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why CO2 cut steel easily, but cannot be used to make holes in Copper?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPdate:&lt;/strong&gt; Will it help to add thin layer of black paint + preheat to 200C to get through damn 0.035mm of Cu?&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="930" LastEditorDisplayName="" LastEditDate="2011-01-17T16:36:46.773" LastActivityDate="2011-01-17T16:36:46.773" Title="Laser cutting/drilling of copper" Tags="&lt;laser&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="2710" PostTypeId="1" AcceptedAnswerId="2714" CreationDate="2011-01-11T16:28:41.427" Score="5" ViewCount="240" Body="&lt;p&gt;I've seen foliation used in the context of &quot;foliation of spacetime&quot; here and elsewhere in papers and such. Generally defined in reference to a &quot;sequence of spatial hypersurfaces.&quot; But I don't know what that means either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, I can imagine what these terms mean because of the English language meaning of the words. But what do these mean specifically in reference to the physics of spacetime?&lt;/p&gt;&#xA;" OwnerUserId="948" LastEditorUserId="948" LastEditorDisplayName="" LastEditDate="2011-01-11T17:39:42.523" LastActivityDate="2011-01-11T19:09:15.087" Title="What does &quot;foliation&quot; mean in the context of a &quot;foliation of spacetime?&quot;" Tags="&lt;general-relativity&gt;&lt;spacetime&gt;" AnswerCount="2" />
  <row Id="2711" PostTypeId="2" ParentId="2702" CreationDate="2011-01-11T16:30:04.337" Score="1" ViewCount="" Body="&lt;p&gt;I actually encounter the same problem all the time, but I just save myself by using my right hand. The mnemonic I have been taught for 2 is very, very simple. Just imagine grabbing a wire with your hand so that your thumb is along the direction of current. Now your fingers are curled around it and that's the direction of the magnetic field!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you need to find the direction of current just see the direction of the field and imagine wrapping something with it using your fingers. The direction your thumb points in is the direction of the current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh and can you remember the cases in which the left hand applies? If you do then just use the above rule with your left hand...&lt;/p&gt;&#xA;" OwnerUserId="1065" LastActivityDate="2011-01-11T16:30:04.337" />
  <row Id="2712" PostTypeId="2" ParentId="2644" CreationDate="2011-01-11T16:49:24.763" Score="0" ViewCount="" Body="&lt;p&gt;I don't if this is what Newton did, but I could prove it with an elaborate thought experiment. Think of an infinitely long smooth (frictionless) plane on which you could roll a ball. Imagine adding a ramp to this plane and roll a ball down that peak and onto the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, as Galileo had said it would keep on rolling towards infinity, but if you think about it what makes the ball start rolling in the first place? There has to be something pushing it or pulling it to make it do that. So, if you see the start then it has to be something to do with falling down. As, Galileo's experiments at Pisa had shown when 2 objects of different masses fall they hit the ground at the same moment and appear to fall at the same rate. So, the thing pulling them must adjust itself by how heavy that object is...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, this means that I don't have to care about how heavy my ball is. So, to just check if there is some relation over here. I could make a smooth ramp out of wood and just time a ball rolling down it from the same height again and again and again. I could also note where the ball is at different moments by putting some sort of scale next to that plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that I can correlate how fast it starts to go with what is pulling it. Now, this proves that there is something pushing it and that this ball seems to accelerate at the same rate, but what happens when they hit something? If we have balls of the same size and different mass and something that can give away down there then can we see how far the object gets moved over a rough surface? (how much work it does)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that mean somehow that these heavier balls have gained more inertia during the same time? So, that would mean acceleration and mass are correlated!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, essentially, I would go on like this looking at things, observing them and just testing my thoughts out until I could find something that could explain (a bunch of laws, perhaps?) why that damn ball behaves the way it behaves. :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;strong&gt;Note:&lt;/strong&gt; I am sorry for any spelling or grammatical errors. I am slightly dyslexic.]&lt;/p&gt;&#xA;" OwnerUserId="1065" LastActivityDate="2011-01-11T16:49:24.763" />
  <row Id="2713" PostTypeId="2" ParentId="2708" CreationDate="2011-01-11T16:58:59.733" Score="2" ViewCount="" Body="&lt;p&gt;Bose condensation is interesting, because it is a quantum mechanical effect, that often results in macroscopic objects with unusual properies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all Bose condensation is believed to be responsible for such phenomena as &lt;a href=&quot;http://en.wikipedia.org/wiki/Superfluid&quot; rel=&quot;nofollow&quot;&gt;superfluidity&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Superconductivity&quot; rel=&quot;nofollow&quot;&gt;superconductivity&lt;/a&gt;. I don't think that I should explain technological and, well, &quot;cognitive&quot; importance of those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I guess that you are mainly interested in &lt;a href=&quot;http://en.wikipedia.org/wiki/Bose%E2%80%93Einstein_condensate&quot; rel=&quot;nofollow&quot;&gt;the Bose condensate&lt;/a&gt; -- a cloud of atoms of ordinary (some gas) but very cold matter in come optical/magnetic trap. Physics of these is a rapidly developing area right now, so my information can be not the most up-to-date.   I know that people were thinking of making: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;some time/frequency standards with it&lt;/li&gt;&#xA;&lt;li&gt;some unusual chemical compounds/alloys with it (ultracold chemistry)&lt;/li&gt;&#xA;&lt;li&gt;use those clouds as high-resolution alternative for &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_beam_epitaxy&quot; rel=&quot;nofollow&quot;&gt;molecular beam epitaxy&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And, of course, the most bizarre application people consider is a quantum computing. I know that people managed to make those traps as small as 1cm. And that they made number of clouds in one trap, &quot;interacting&quot; with each other through spin entanglement: &lt;a href=&quot;http://www.itpro.co.uk/news/121086/trapped-atoms-could-advance-quantum-computing.html&quot; rel=&quot;nofollow&quot;&gt;see e.g. here&lt;/a&gt;.    &lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2011-01-11T21:29:46.870" LastActivityDate="2011-01-11T21:29:46.870" />
  <row Id="2714" PostTypeId="2" ParentId="2710" CreationDate="2011-01-11T17:14:39.590" Score="8" ViewCount="" Body="&lt;p&gt;If you want to solve equations of motion to describe the time evolution of a system, either classically or quantum mechanically, you need to impose initial condition at one point in time, and then under some conditions the entire evolution of the system (forward and backwards) is determined. This is the type of things physicists do all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, general relativity is a theory of spacetime, so it is not clear that any spacetime manifold will have well-defined evolution of the sort I described, where the conditions at a spatial slice at one point in time (called Cauchy surface) determines the system everywhere. For that to be true there has to be a way to separate what is the time direction at every point in spacetime. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this can be done you express the spacetime as a series of spatial slices which evolves in time (called foliation of spacetime), and you have now a problem which amounts to describing how those spatial slices evolve, which is a traditional initial value problem which physicists know and love. Manifolds for which this can be done are called globally hyperbolic, and those are the ones which are easier to discuss, though there are well-known examples of spacetimes which are not globally hyperbolic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you find one way to do it, one &quot;foliation&quot; of spacetime, usually there are many other ways, but the difficulty is usually in finding one way that works everywhere (it is always possible to do that separation only in some region of spacetime, but that exercise is not useful since you want to predict what happens everywhere, at any point in time).&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-01-11T19:09:15.087" LastActivityDate="2011-01-11T19:09:15.087" CommentCount="1" />
  <row Id="2715" PostTypeId="2" ParentId="2710" CreationDate="2011-01-11T17:17:31.613" Score="5" ViewCount="" Body="&lt;p&gt;The easiest answer is that it just means a surface of constant $t$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this isn't completely trivial like that, because, in general relativity, there are many possible choices of $t$ one can make.  For instance, take the Minkowski 2-plane, with coordinates $(t,x)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly, one foliation of this surface would be saying that you just pick every $t=$constant surface.  But, you have an infinity of more exotic choices, such as $t=\sin (x)+$cosntant, where each value for the constant picks a different surface.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do we do this?  The basic idea is to go and break up the notion of spacetime into a space with time evolution.  This is especially relevant if we're doing something like setting up two black holes as an initial condition, and then seeing what happens when we let them interact with each other.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-01-11T17:17:31.613" />
  <row Id="2716" PostTypeId="1" CreationDate="2011-01-11T18:29:42.143" Score="1" ViewCount="245" Body="&lt;p&gt;Suppose you have to specify the moment in time when a given event occurred, a &quot;zero time&quot;. The record must be accurate to the minute, and be obtainable even after thousands of years. All the measures of time we currently have are relative to a well defined zero, but the zero is not easy to backtrack exactly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possibility would be to take a sample of Carbon with a well defined, very accurate amount of 14C, and say: the event occurred when the 14C was x%. At any time, measuring the rate of decay, you would know when the event occurred. This however, requires a physical entity to measure, which may be lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way would be to give the time lapsed after a well defined series of solar eclipses. In order to define precisely the context, you would say a list of (say) five consecutive eclipses and the places on Earth where they were total, and then a time gap from the last of the set. At any time in the future, you can backtrack the specified conditions into a celestial mechanics program and find when the event occurred.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a standardized or well recognized method to do so?&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="217" LastEditorDisplayName="" LastEditDate="2011-01-11T20:41:48.960" LastActivityDate="2011-01-13T23:01:13.003" CommunityOwnedDate="2011-01-13T13:54:00.280" Title="How to describe a well defined &quot;zero moment&quot; in time" Tags="&lt;measurement&gt;&lt;time&gt;" AnswerCount="4" CommentCount="11" />
  <row Id="2717" PostTypeId="1" AcceptedAnswerId="2718" CreationDate="2011-01-11T19:14:39.907" Score="7" ViewCount="305" Body="&lt;p&gt;I have been taught that the $\pi^0$ particle contains either an up quark and an anti-up quark or a down and an anti-down. How can these exist without annihilating?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it is its own antiparticle, but it doesn't make sense that the up version and down version would annihilate when they meet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is what I've been taught a simplification - if so, in what state does this particle exist?&lt;/p&gt;&#xA;" OwnerUserId="954" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-11T19:46:31.133" LastActivityDate="2011-01-11T19:56:30.073" Title="How do $\pi^0$ particles exist?" Tags="&lt;particle-physics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="2718" PostTypeId="2" ParentId="2717" CreationDate="2011-01-11T19:51:16.167" Score="16" ViewCount="" Body="&lt;p&gt;Actually, the quark and antiquark &lt;em&gt;do&lt;/em&gt; annihilate with each other. It just takes some amount of time for them to do so. The actual time that it takes for any given pion is random, and follows an exponential distribution, but the &lt;em&gt;average&lt;/em&gt; time it takes is $8.4\times 10^{-17}\,\mathrm{s}$ according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Pion&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, which we call the lifetime of the neutral pion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you've learned is a simplification, in fact (it pretty much always is in physics). The actual state of a pion is a linear combination of the up state and the down state,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\sqrt{2}}(u\bar{u} - d\bar{d})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how it's able to be its own antiparticle: there aren't separate up and down versions of the neutral pion. Each one is a combination of both flavors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orthogonal linear combination,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\sqrt{2}}(u\bar{u} + d\bar{d})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;doesn't correspond to a real particle. (In a sense it &quot;contributes&quot; to the $\eta$ and $\eta&amp;#39;$ mesons, but I won't go into detail on that.)&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-11T19:56:30.073" LastActivityDate="2011-01-11T19:56:30.073" CommentCount="4" />
  <row Id="2719" PostTypeId="1" CreationDate="2011-01-11T20:05:21.670" Score="4" ViewCount="165" Body="&lt;p&gt;Given today's &lt;a href=&quot;http://www.physorg.com/news/2011-01-thunderstorms-hurling-antimatter-space-caught.html&quot; rel=&quot;nofollow&quot;&gt;revelation&lt;/a&gt; of the detection of terrestrial gammay-ray flashes (TGF) produced by thunderstorms and the associated pair production of an electron and a positron, how feasible is it to set up ground-based gamma ray detection station on the cheap using a &lt;a href=&quot;http://en.wikipedia.org/wiki/Geiger%E2%80%93M%C3%BCller_tube&quot; rel=&quot;nofollow&quot;&gt;Geiger-Muller tube&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electron/positron pair travels along the magnetic field of the earth and bounces back (additional question: why do they bounce back?) This is how Fermi could detect the positrons: they interacted with the satellite which happened to be in the way. The loss of energy due the interaction slow down the positrons which interact with electrons and produce two 511keV photons flying in opposite directions (&lt;a href=&quot;http://engphys.mcmaster.ca/undergraduate/outlines/3d03/3d3_lab1_gamma_intro.pdf&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: is it feasible at all to capture and detect positron annihilation as the pairs bounce back using a G-M tube? Given that there are 1500 thunderstorms a day I'm assuming each location on earth has a good chance of being a &quot;bounce site&quot; at least a few times a year.&lt;/p&gt;&#xA;" OwnerUserId="1121" LastActivityDate="2011-07-03T07:15:25.843" Title="Ground-based anti-matter detector on the cheap?" Tags="&lt;electromagnetism&gt;&lt;astrophysics&gt;&lt;electrodynamics&gt;" AnswerCount="1" />
  <row Id="2720" PostTypeId="2" ParentId="2708" CreationDate="2011-01-11T20:24:27.527" Score="1" ViewCount="" Body="&lt;p&gt;More recently, they have been &lt;a href=&quot;http://arstechnica.com/science/news/2011/01/pqe-2011-small-atoms-big-ideas-in-gravity-detection.ars&quot; rel=&quot;nofollow&quot;&gt;proposed&lt;/a&gt; as an array of inexpensive gravity wave detectors.    &lt;/p&gt;&#xA;" OwnerUserId="770" LastActivityDate="2011-01-11T20:24:27.527" />
  <row Id="2721" PostTypeId="1" AcceptedAnswerId="2722" CreationDate="2011-01-11T20:28:25.300" Score="5" ViewCount="366" Body="&lt;p&gt;Another Noether's theorem question, this time about electrical charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Noether's theorem, all conservation laws originate from invariance of a system to shifts in a certain space. For example conservation of energy stems from invariance to time translation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of symmetry creates the conservation of electrical charge?&lt;/p&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-01-28T15:04:01.577" Title="What is the symmetry which is responsible for preservation of electrical charges?" Tags="&lt;electromagnetism&gt;&lt;symmetry&gt;&lt;noethers-theorem&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="2722" PostTypeId="2" ParentId="2721" CreationDate="2011-01-11T20:32:33.520" Score="4" ViewCount="" Body="&lt;p&gt;Gauge invariance:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Charge_conservation#Connection_to_gauge_invariance&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Charge_conservation#Connection_to_gauge_invariance&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="266" LastActivityDate="2011-01-11T20:32:33.520" CommentCount="2" />
  <row Id="2723" PostTypeId="2" ParentId="2719" CreationDate="2011-01-11T20:53:39.847" Score="3" ViewCount="" Body="&lt;p&gt;First, positrons heading &lt;em&gt;down&lt;/em&gt; hit a lot of atmosphere, leading to a low probability of surviving to the ground because the positron will interact with an electron before it can get there. Up is not so bad; the tops of these storms are surprisingly high, and the atmosphere is pretty thin up there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If some &lt;em&gt;do&lt;/em&gt; make it (they won't) your big problem is going to be backgrounds. You face several distinct categories and they will almost certainly swamp your signal:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cosmic rays, including spallation products like Li-9, C-11, ...&lt;/li&gt;&#xA;&lt;li&gt;Environmental radiation from K-40, radon daughters, and long-lived products of atmospheric nuclear explosive tests...&lt;/li&gt;&#xA;&lt;li&gt;C-60 or U-238 in the material that make up your detector...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To even get started you'll need a energy sensitive detector (i.e. more sophisticated than a Geiger tube) and a cosmic veto of some kind. Low rate measurements near the Earth's surface are &lt;em&gt;hard&lt;/em&gt;. &lt;em&gt;Very&lt;/em&gt; hard.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Just to address the question of positron detection itself, this is standard tool if you control the source. You simply put two detectors back-to-back with the source in the middle ad trigger on a coincidence. A typical unit using NaI crystals might look like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+-----+-----+      +---------+      +-----+-----+&#xA;| PMT | NaI |      | source  |      | NaI | PMT |&#xA;+-----+-----+      +---------+      +-----+-----+&#xA;  \                                          /&#xA;   \   +------+     +-------+    +------+   /&#xA;    \--| disc |-----| coinc |----| disc |--/       &#xA;       +------+     +-------+    +------+&#xA;                        |&#xA;                   +---------+&#xA;                   | scaler  |&#xA;                   +---------+&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here &quot;disc&quot; represents a pulse discriminator, &quot;coinc&quot; a coincidence unit (i.e. and gate), and a scaler is a simple counting circuit.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-01-11T22:48:46.667" LastActivityDate="2011-01-11T22:48:46.667" CommentCount="3" />
  <row Id="2724" PostTypeId="2" ParentId="2708" CreationDate="2011-01-11T22:25:55.240" Score="8" ViewCount="" Body="&lt;p&gt;I assume you mean the relatively recent phenomenon of Bose-Einstein Condensation in dilute atomic vapors (first produced in 1995 in Colorado). The overall phenomenon of Bose-Einstein Condensation is closely related to superconductivity (in a very loose sense, you can think of the superconducting transition in a metal as the formation of a BEC of pairs of electrons), and that application would trump everything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary application of atomic BEC systems is in basic research areas at the moment, and will probably remain so for the foreseeable future. You sometimes hear people talk about BEC as a tool for lithography, or things like that, but that's not likely to be a real commercial application any time soon, because the throughput is just too low. Nobody has a method for generating BEC at the sort of rate you would need to make interesting devices in a reasonable amount of time. As a result, most BEC applications will be confined to the laboratory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the hottest areas in BEC at the moment is the use of Bose condensates (and the related phenomenon of degenerate Fermi gases) to simulate condensed matter systems. You can easily make an &quot;optical lattice&quot; from an interference pattern of multiple laser beams that looks to the atoms rather like a crystal lattic in a solid looks to electrons: a regular array of sites where the particles could be trapped, with all the sites interconnected by tunneling. The big advantage BEC/ optical lattice systems have over real condensed matter systems is that they are more easily tunable. You can easily vary the lattice spacing, the strength of the interaction between atoms, and the number density of atoms in the lattice, which allows you to explore a range of different parameters with essentially the same sample, which is very difficult to do with condensed matter systems where you need to grow all new samples for every new set of values you want to explore. As a result, there is a great deal of work in using BEC systems to explore condensed matter physics, essentially making cold atoms look like electrons. There's a good review article, a couple of years old now, by Immanuel Bloch, Jean Dalibard, and Wilhelm Zwerger (&lt;a href=&quot;http://rmp.aps.org/abstract/RMP/v80/i3/p885_1&quot; rel=&quot;nofollow&quot;&gt;RMP paper&lt;/a&gt;, &lt;a href=&quot;http://arxiv.org/abs/0704.3011&quot; rel=&quot;nofollow&quot;&gt;arxiv version&lt;/a&gt;) that covers a lot of this work. And people continue to expand the range of experiments-- there's a lot of work ongoing looking at the effect of adding disorder to these systems, for example, and people have begun to explore lattice structures beyond the really easy to make square lattices of the earliest work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a good deal of interest in BEC for possible applications in precision measurement. At the moment, some of the most sensitive detectors ever made for things like rotation, acceleration, and gravity gradients come from atom interferometry, using the wavelike properties of atoms to do interference experiments that measure small shifts induced by these effects. BEC systems may provide an improvement beyond what you can do with thermal beams of atoms in these sorts of systems. There are a number of issues to be worked out in this relating to interatomic interactions, but it's a promising area. Full Disclosure: My post-doc research was in this general area, though what I did was more a proof-of-principle demonstration than a real precision measurement. My old boss, Mark Kasevich, now at Stanford, does a lot of work in this area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other really hot area of BEC research is in looking for ways to use BEC systems for quantum information processing. If you want to build a quantum computer, you need a way to start with a bunch of qubits that are all in the same state, and a BEC could be a good way to get there, because it consists of a macroscopic number of atoms occupying the same quantum state. There are a bunch of groups working on ways to start with a BEC, and separate the atoms in some way, then manipulate them to do simple quantum computing operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a lot of overlap between these sub-sub-fields-- one of the best ways to separate the qubits for quantum information processing is to use an optical lattice, for example. But those are what I would call the biggest current applications of BEC research. None of these are likely to provide a commercial product in the immediate future, but they're all providing useful information about the behavior of matter on very small scales, which helps feed into other, more applied, lines of research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not by any stretch a comprehensive list of things people are doing with BEC, just some of the more popular areas over the last couple of years.&lt;/p&gt;&#xA;" OwnerUserId="166" LastEditorUserId="166" LastEditorDisplayName="" LastEditDate="2011-01-12T00:15:26.443" LastActivityDate="2011-01-12T00:15:26.443" />
  <row Id="2725" PostTypeId="1" AcceptedAnswerId="2730" CreationDate="2011-01-11T22:39:25.070" Score="7" ViewCount="311" Body="&lt;p&gt;I have read that the fine structure &lt;a href=&quot;http://www.physorg.com/news202921592.html&quot; rel=&quot;nofollow&quot;&gt;constant may well not be a constant&lt;/a&gt;. Now, if this were to be true, what would be the effect of a higher or lower value? (and why?)&lt;/p&gt;&#xA;" OwnerUserId="1123" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2011-01-12T12:07:10.990" LastActivityDate="2011-01-14T07:30:30.907" Title="Is the Fine Stucture constant constant?" Tags="&lt;astrophysics&gt;&lt;physical-constants&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="2726" PostTypeId="2" ParentId="2725" CreationDate="2011-01-11T23:00:15.460" Score="2" ViewCount="" Body="&lt;p&gt;We see the same spectral patterns arising from stars, galaxies, and dust clouds very far into space (which is to say very far into the past), and these features are dependent on the fine structure constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, yes. At least to a first approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been attempts (by reputable scientist) to form cosmologies to compete with the current big-bang-and-inflation scenario, and wondering about the constancy of various fundamental &quot;constants&quot; was one of the tools that people used in trying to form them. I don't know the exact state of any of these efforts right now, but I don't believe they have a lot of currency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that is not the final word...you &lt;em&gt;could&lt;/em&gt; assume a rapid change in the very early universe that approached the current value asymptotically such that $\alpha(t) \approx \alpha(\inf)$ for all time we can examine. This allows you to have funny things happening in the very early universe. However, in the absence of some failing of the conventional picture this seems pretty speculative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the observational evidence gets better and better, there is increasingly little room for such speculation, but only time (and a lot of effort, of course, but by people other than me...) will tell. &lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-01-11T23:25:00.487" LastActivityDate="2011-01-11T23:25:00.487" />
  <row Id="2727" PostTypeId="2" ParentId="1019" CreationDate="2011-01-11T23:02:44.553" Score="1" ViewCount="" Body="&lt;p&gt;I find people think that &quot;a bullet and a ball shot and dropped repectively from the same height will not hit the ground at the same time&quot;. &lt;/p&gt;&#xA;" OwnerUserId="1123" LastActivityDate="2011-01-11T23:02:44.553" CommunityOwnedDate="2011-01-11T23:02:44.553" CommentCount="1" />
  <row Id="2728" PostTypeId="2" ParentId="2716" CreationDate="2011-01-11T23:26:25.257" Score="0" ViewCount="" Body="&lt;p&gt;I think the standard method is &lt;a href=&quot;http://en.wikipedia.org/wiki/Coordinated_Universal_Time&quot; rel=&quot;nofollow&quot;&gt;UTC&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you are asking how to translate UTC into a form that people thousands of years from now will understand.  Here is a method off the top of my head: If you specify the relative positions of all of the planets and the moon, you should get uniquely defined precision to much less than a day over many millenia.  Then you can use local solar time with respect to some specified locations (e.g., summits of recognizable mountains) to get within a fraction of a second.&lt;/p&gt;&#xA;" OwnerUserId="761" LastActivityDate="2011-01-11T23:26:25.257" CommunityOwnedDate="2011-01-13T13:54:00.280" CommentCount="2" />
  <row Id="2729" PostTypeId="2" ParentId="1019" CreationDate="2011-01-11T23:31:01.733" Score="0" ViewCount="" Body="&lt;p&gt;A common misconception is that motion of massive objects with superluminal velocity is prohibited in General Relativity. This is not true as only local superluminal motion is prohibited.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastActivityDate="2011-01-11T23:31:01.733" CommunityOwnedDate="2011-01-11T23:31:01.733" />
  <row Id="2730" PostTypeId="2" ParentId="2725" CreationDate="2011-01-11T23:31:58.590" Score="6" ViewCount="" Body="&lt;p&gt;Another thing that would be changed by a varying fine structure constant would be that it would alter almost every electromagnetically mediated phenomenon.  All of the spectra of atoms would change.  What would also change would be the temperature at which atoms can no longer hold onto their electrons, since the strength of attraction between electrons and the nucleus would change.  This would then change the redshift at which the universe becomes transparent.  The end result would be that the cosmic background radiation would be coming from a different time in the universe's history than otherwise thought.  This would have consequences for the values of cosmological parameters.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you alter phenomena in this stage of the universe's history, though, you have to be quite careful to not disrupt the current predictions for how much hydrogen, helium, and heavy elements there are in the universe (while creating nuclei depends mainly on the strong interaction, electromagnetism does have something to do with determining the final energies of the nuclei, and so can't be completely neglected--changing the fine structure constant changes these cross-sections).  Current theory predicts these things with great accuracy, and changing things around, particularly particle physics parameters that govern the length of the neucleosyntheis era (which overlaps with, but is a subset of the time at which the universe is opaque) potentially make these observations not agree with theory.&lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2011-01-12T16:44:07.913" LastActivityDate="2011-01-12T16:44:07.913" CommentCount="7" />
  <row Id="2731" PostTypeId="1" AcceptedAnswerId="2734" CreationDate="2011-01-12T01:11:49.903" Score="4" ViewCount="148" Body="&lt;p&gt;The Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Deep_inelastic_scattering&quot; rel=&quot;nofollow&quot;&gt;deep inelastic scattering&lt;/a&gt; suggests that the experiment shows baryons have three point of deflections (corresponding to three quarks) and mesons have two points of deflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are the electrons fired in this experiment being detected, and how exactly do the two or three points of deflection appear in the data? Are they fired at a target consisting entirely of baryons, or are collisions with non-baryons somehow filtered from the data?&lt;/p&gt;&#xA;" OwnerUserId="786" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-01-12T05:03:29.413" LastActivityDate="2011-01-31T03:01:16.447" Title="How to count quarks using Deep Inelastic Scattering?" Tags="&lt;particle-physics&gt;&lt;experimental-physics&gt;&lt;quark-model&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="2732" PostTypeId="2" ParentId="2696" CreationDate="2011-01-12T01:35:56.453" Score="-2" ViewCount="" Body="&lt;p&gt;As the question is currently phrased, the answer is a straightforward one. Hawking radiation does not occur with any object except one with a horizon, i.e. a black hole. The arguments leading to Hawking radiation are subtle, but, conceptually, the idea (via Susskind in &lt;em&gt;The Black Hole War&lt;/em&gt;) is that quantum jitters become thermal jitters at the event horizon. (Susskind's explanation is much better, of course.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be a bit more complete and concrete, consider a standard quantum fluctuation of energy arising from the uncertainty relation between energy and time. Fluctuations on sufficiently small time scales will be large enough to result in particle-antiparticle pair creation. At the horizon of a black hole, the pair can be separated so that one particle falls into the black hole, while the other is shot out. This transforms a virtual particle (pair) into a real particle, which is then able to interact with other matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, one could say that this process is possible in any system with non-zero mass. However, the detailed and subtle mathematics of the process require an event horizon in order to occur. As a thought experiment, imagine that any mass could produce this effect. It would result in a wide-scale violation of the Uncertainty Principle because we would see changes in energy that exceeded those allowable over the associated time frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other objection is that black hole temperature, which is tied to Hawking radiation, increases as the mass of the black hole decreases. If &quot;normal&quot; mass were to behave in a similar way, elementary particles would have tremendous &quot;black-hole-like&quot; temperatures.&lt;/p&gt;&#xA;" OwnerUserId="1196" LastActivityDate="2011-01-12T01:35:56.453" CommentCount="16" />
  <row Id="2733" PostTypeId="2" ParentId="2702" CreationDate="2011-01-12T02:01:23.323" Score="1" ViewCount="" Body="&lt;p&gt;I'm just making these up as I go because we &quot;always&quot; use the right hand in the sense that our coordinate system (xyz) is always defined by the right hand rule, but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the cross-product, such as E = q(v cross B): ROD - right, out, down. If the first quantity (v) points to the right and the second (B) points out of the page, i.e. toward you, then the third (E) points down. I realize that &quot;out&quot; as out of the page may cause some problem, so you could substitute &quot;at,&quot; meaning the vector points &lt;em&gt;at&lt;/em&gt; you, and use RAD instead. I'd stick with ROD, though, so you don't think the A means &quot;away&quot; a year from now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For magnetic fields in wires: RAA (think of the music industry). If the current flows to the Right, the magnetic field Above the wire points Away from you. Since you know the magnetic field is a circle around the wire, this establishes the field direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all for positive charge/current, since that's the physics standard. In both these cases, use the mnemonic to establish the right hand rule and go from there, since that's probably easier than trying the rotate your problem to match the mnemonic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As one last note, if ROD (right, out, down) and RAA (right, above, away) slip, if you have a coordinate system drawn, x cross y = z will force you to remember it's your right hand to use.&lt;/p&gt;&#xA;" OwnerUserId="1196" LastActivityDate="2011-01-12T02:01:23.323" />
  <row Id="2734" PostTypeId="2" ParentId="2731" CreationDate="2011-01-12T04:42:57.363" Score="6" ViewCount="" Body="&lt;p&gt;There may be too many questions here. I'll try to hit some of the high points of the technologies, but be aware that you could write an entire dissertation on the matter (mine was on a closely related topic).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The incoming electron or proton beam is characterized by using &lt;em&gt;current&lt;/em&gt; monitors (inductive, resonant cavity, charge cups, etc) and by measuring its bending radius in known magnetic fields.&lt;/li&gt;&#xA;&lt;li&gt;Scattered electrons (after the collision) are detected by there ionization in interactions with matter, by Cerenkov radiation, and/or by transition radiation. Their momenta are again, characterized with magnetic fields.&lt;/li&gt;&#xA;&lt;li&gt;The detection of the reaction products use the same techniques as for the scattered electrons. &lt;/li&gt;&#xA;&lt;li&gt;Filtering the junk to get just the events you want is a &lt;strong&gt;big&lt;/strong&gt; topic. You put a lot of effort into designing the detector package, trigger, data acquisition, storage subsystem, and analytical programs to make it happen. This is what keeps grad students and post docs employed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, baryon targets are easy: put any matter made of protons and neutrons (i.e. everything) in front of the beam, or collide electrons on protons. Meson targets are hard: you can't just get a pile of pions because they decay. Fast. So you have to generate a meson beam (which is a bit of a trick in and of itself) and direct &lt;em&gt;it&lt;/em&gt; into either a fixed baryon target or another beam (electrons, say) in collider mode.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now some physics. You don't follow a single electron through multiple independent scatters in a single collision with a hadron (the time and distance scales are prohibitively small).{*} Instead, you chose a center of mass energy for the collision that will tend to suppress the effects of scattering from the whole composite object, and allow you to look as the electron-on-valence-parton cross-section. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That measurement tells you (after various corrects are factored in) the sum of the squares of the parton charges, and you already know the sum of the parton charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{*} Actually my dissertation concerned itself with detecting the effects of secondary scattering events in one very closely defined circumstance and how the rate of such scattering might{+} depend on a parameters of the collision called $Q^2$ (the squared four-momentum transfer). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;{+} The theorists had said we might see an interesting effect at 5--10 GeV$^2$. In the eight years it took to design the experiment (to get to about 8 GeV$^2$), get approval and beam time, and set it up they'd changed their minds. &quot;It's outside your experimental by a factor of two or three&quot; they said. It was a world class, high-precision null result.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-01-31T03:01:16.447" LastActivityDate="2011-01-31T03:01:16.447" CommentCount="4" />
  <row Id="2735" PostTypeId="2" ParentId="2702" CreationDate="2011-01-12T08:12:02.023" Score="1" ViewCount="" Body="&lt;p&gt;I see you don't like the other answers. I don't think anyone can help you and here's why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In EM there is &lt;em&gt;no&lt;/em&gt; way to get around arbitrariness of defining orientation (because the theory is P, as well as C and T symmetric). So you just have to remember the whole set orientation + hand trick. I assume you learned both your first name and the last name and there the arbitrariness is substantially larger than just choosing left or right. So just pick one and stick it into your brain. I hope you can learn a bit (pun intended).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, there might be a loop-hole. You can pick non-chiral theory (weak interactions) which tells you which orientation is &quot;correct&quot; (similarly like matter is more &quot;correct&quot; than antimatter in our universe). Quoting wikipedia on &lt;a href=&quot;http://en.wikipedia.org/wiki/Parity_%28physics%29#Parity_violation&quot; rel=&quot;nofollow&quot;&gt;parity violation&lt;/a&gt;: &lt;em&gt;&quot;The Standard Model incorporates parity violation by expressing the weak interaction as a chiral gauge interaction. Only the left-handed components of particles and right-handed components of antiparticles participate in weak interactions in the Standard Model.&quot;&lt;/em&gt; This is a useful piece of information of and by itself, so I hope you don't mind learning it.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-13T07:28:40.017" LastActivityDate="2011-01-13T07:28:40.017" CommentCount="6" />
  <row Id="2739" PostTypeId="1" CreationDate="2011-01-12T13:46:55.107" Score="2" ViewCount="322" Body="&lt;p&gt;Imagine an electromagnetic wave (a monochromatic one for example)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electric field amplitude, and its variations travel in the propagation direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if does really exist a propagation direction, what happen in other directions?&#xA;I mean, there is an oscilation in those other directions? &#xA;Does something (electric or magnetic field) move in those directions?&#xA;Does the speed of those direction depends on frequency?&#xA;or is this wave infinitely thin? It's no covering any space in other directions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen a lot of graph showing that the field variation is not in the direction of propagation, that &quot;means&quot;, based on polarization experiments that it's not a Longitudinal wave, it's a transverse wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I accept field is not variating in the propagation direction (it just go forward in that direction), then I could think it's variating in Other direction, so ok what direction? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps we can't say it's variating in any direction, it's not a spatial variable!!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't imagine how the field variation is distributed in space. (If it were distributed..)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any information welcome, thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I've already asked this at old site physic overflow, still searching an answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;some options&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1-Electromagnetic waves just move in one single spatial propagation direction.. then all spatial concepts that take account other directions are wrong, like &quot;polarization&quot; and &quot;transverse wave&quot; definition&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2-Electromagnetic waves moves in others directions too, then speed of propagation in those other directions could depend on frequency, and thats weird, but who knows!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3,4-- others..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2&lt;/strong&gt; (Add my comment as part of the question)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If there are infinite wave planes, then imagine a light that is turned OFF and then is turned ON, comparing the EM, first it is covering nothing, and then suddenly reach &quot;infinity&quot; (or at least very far..in fact any distance is enough), what is the speed of THAT propagation? because it is NOT a wave propagation, because is not in the direction of propagation, that infinite or very large field extends itself at infinite speed?&lt;/strong&gt;&lt;/p&gt;&#xA;" LastActivityDate="2011-01-17T12:46:02.467" Title="How does electromagnetic wave variation is distributed in space?" Tags="&lt;electromagnetism&gt;&lt;waves&gt;&lt;photons&gt;" AnswerCount="3" CommentCount="12" />
  <row Id="2740" PostTypeId="2" ParentId="2739" CreationDate="2011-01-12T13:50:41.750" Score="1" ViewCount="" Body="&lt;p&gt;I think you're trying to geometrise something that is not geometric. Your comment &quot;It's no covering any space in other directions&quot; is close to the truth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A field is defined as a (scalar or vector) quantity with a value at every point in space. In the case of classical electromagnetism, this simply means that at each point along the propagation of an electromagnetic wave has an electric and magnetic field vector associated with it. The wave is called &lt;em&gt;transverse&lt;/em&gt; because these vectors are both orthogonal to the direction of proportionate (phase) of the wave.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2011-01-12T13:50:41.750" CommentCount="3" />
  <row Id="2741" PostTypeId="2" ParentId="2739" CreationDate="2011-01-12T15:18:04.353" Score="2" ViewCount="" Body="&lt;p&gt;I'm not totally sure I'm parsing the question properly, but here's an attempt to answer part of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivation of electromagnetic waves from Maxwell's equations often presents &lt;em&gt;infinite plane waves&lt;/em&gt;, which have definite momentum but are extended throughout all of space. An example would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec{A}(x,y,z,t) = A_0 \hat{x} e^{i \omega(t - z/c)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This wave propagates in the $z$ direction with frequency $\omega$, as you can see from the dependence on $t - z/c$. From ${\vec E} = -\partial_t {\vec A}$ you see that the electric field points in the $x$ direction, and from ${\vec B} = \nabla \times {\vec A}$ you see that the magnetic field points in the $y$ direction. But the field exists &lt;em&gt;everywhere&lt;/em&gt;; no matter what point $(x,y,z)$ in space you look at, the field is oscillating there with amplitude $A_0$ and frequency $\omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as you know, in the real world electromagnetic waves don't look like this. When I turn on a lamp, EM waves come out of it, but they don't live everywhere in space; they propagate outward from the lamp and are absorbed or reflected by objects. So the infinite plane wave is an idealization. In the real world, a wave is a superposition of many plane waves. It might look like, for instance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_\mu(t,{\bf x}) = \int \frac{d^3{\bf k}}{(2\pi)^3} A({\bf k}) \epsilon_\mu(k) e^{ik \cdot x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I've switched to a relativistic notation here out of laziness; if it isn't familiar to you, you might want to look at the simpler example at the Wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Wave_packet&quot; rel=&quot;nofollow&quot;&gt;wave packets&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key points are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Real world waves aren't really exactly monochromatic, hence the integral over wave vectors ${\bf k}$.&lt;/li&gt;&#xA;&lt;li&gt;By adding up plane waves of different wave vectors and frequencies, with a frequency and wave vector-dependent amplitude $A({\bf k})$, you can construct solutions to Maxwell's equations that have different types of dependence on space and time.&lt;/li&gt;&#xA;&lt;li&gt;There exist (though I haven't demonstrated it here) solutions that look approximately like plane waves over some region of space, and so are very nearly propagating in a definite direction with definite frequency within that region, but which fall off to zero outside that region.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, the answer to one of your questions is that the infinite plane wave isn't &quot;infinitely thin&quot;; in fact, it's &quot;infinitely thick&quot; (spread out everywhere), but more realistic waves are localized over some finite region of space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give one more example: the beam that comes out of a laser is well-approximated as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_beam&quot; rel=&quot;nofollow&quot;&gt;Gaussian beam&lt;/a&gt;, with a field that decays like $e^{-kr^2/R(z)}$ as a function of transverse distance $r$. (Explaining what $R(z)$ means is beyond the scope of this answer, but you can read more at the link.)&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="19" LastEditorDisplayName="" LastEditDate="2011-01-12T15:28:09.817" LastActivityDate="2011-01-12T15:28:09.817" CommentCount="2" />
  <row Id="2742" PostTypeId="2" ParentId="2739" CreationDate="2011-01-12T15:38:17.977" Score="4" ViewCount="" Body="&lt;p&gt;I am still not sure I understand your question but from what I gathered you are confused about behavior of EM wave (let's say monochromatic) so I'll discuss properties of waves of increasing difficulty (eventually returning to EM).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EM waves are not easy to imagine because they are &lt;em&gt;vectorial&lt;/em&gt; in nature. That is, in every point of the space there is an arrow describing the magnitude and direction of the $E$ field. Let's simplify this a imagine that at every point there is just a single value, say pressure, so we are be talking about sound waves. It's useful to consider the notion of a wave front. This is a collection of points that have the same phase (think of the circles that form if you throw a stone into the water). For a monochromatic wave $$  p = p_0 \exp(\omega t - {\mathbf k \cdot r})$$&#xA;(characterized by the wave-vector $\mathbf k$ and frequency $\omega$), it will be just a plane. You can imagine that plane moving at the speed of propagation $c = {\omega \over |{\mathbf k}|}$ into the direction given by $\mathbf k$. Let's emphasize that all points of that plane will oscillate in the same manner; which simplifies the picture by letting us consider just one point from that plane. We are left with the following image&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/pDXK5.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that the direction of the propagation of the wave is the horizontal axis, the phase of the wave is carried out on the vertical axes and the plane is represented by any point of the curve (e.g. pick the top of the wave) that travels to the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's consider vector fields. These can be written as ${\mathbf E} = (E_x, E_y, E_z)$ and the previous discussion applies for every component of the field. So there are now three independent oscillations each with its own phase. Except that EM field is massless and so cannot oscillate in the direction of propagation (so called longitudinal polarization); this fact can be derived from Maxwell's equations. So this leaves us with two transversal polarizations $E_x$ and $E_y$ (taken as complex numbers because we also need to take account of their phases) by identifying the $\mathbf z$ axis with the direction of propagation $\mathbf k$. As in the previous case, for monochromatic wave the wavefront will be a plane. But now we have two independent polarization and instead of simple oscillation we will in general obtain an &lt;a href=&quot;http://en.wikipedia.org/wiki/Elliptical_polarization&quot; rel=&quot;nofollow&quot;&gt;elliptical polarization&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/BpBYC.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, single point on the curve represents a whole wavefront plane. Note that special cases include circular (when $E_y$ lags by $\pi /2$ behind $E_x$ but has equal magnitude) and linear polarizations (when $E_y$ and $E_x$ are in phase).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Notes&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;We haven't discussed the magnetic part of EM wave $\mathbf B$ because it's determined by just knowing the wave vector $\mathbf k$ and the $\mathbf E$ field. The important point being that it will be perpendicular to both. So the above discussion about the travel of wavefront applies also to $\mathbf B$ field.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The monochromatic waves are not realistic as you might imagine. They are infinite in extent (whole wavefront plane oscillates and it sweeps entire universe during its travel). In reality the waves travel as concentrated &lt;a href=&quot;http://en.wikipedia.org/wiki/Wave_packet&quot; rel=&quot;nofollow&quot;&gt;wave-packets&lt;/a&gt;. But the above discussion is still useful because we can use Fourier analysis to pass to the frequency image of the wave and decompose the packet into monochromatic waves.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If we are not in the vacuum then the picture is much more complex. The $\mathbf E$ and $\mathbf B$ fields need not be perpendicular to the $\mathbf k$ vector (because the photons gain effective mass in the medium thanks to refractive index), the energy travels in different direction than the wave itself and so on.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-01-12T15:38:17.977" CommentCount="1" />
  <row Id="2743" PostTypeId="1" AcceptedAnswerId="2760" CreationDate="2011-01-12T16:01:50.277" Score="18" ViewCount="894" Body="&lt;p&gt;This question appeared quite a time ago and was inspired, of course, by all the fuss around &quot;LHC will destroy the Earth&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a small black hole, that is somehow got inside the Earth. Under &quot;small&quot; I mean small enough to not to destroy Earth instantaneously, but large enough to not to evaporate due to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow&quot;&gt;Hawking radiation&lt;/a&gt;. I need this because I want the black hole to &quot;consume&quot; the Earth. I think reasonable values for the mass would be $10^{15} - 10^{20}$ kilograms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also let us suppose that the black hole is at rest relative to the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is:&lt;br&gt;&#xA;How can one estimate the speed at which the matter would be consumed by the black hole in this circumstances? &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-07-29T14:20:39.783" Title="How fast a (relatively) small black hole will consume the Earth?" Tags="&lt;astrophysics&gt;&lt;black-holes&gt;&lt;estimation&gt;" AnswerCount="6" CommentCount="7" FavoriteCount="7" />
  <row Id="2744" PostTypeId="1" CreationDate="2011-01-12T17:48:58.070" Score="1" ViewCount="469" Body="&lt;p&gt;It's been a couple years since I've taken a physics class, and I have been wondering about this basically since I bought my car:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I drive a 2010 Honda Fit, which has a street weight of &lt;a href=&quot;http://consumerguideauto.howstuffworks.com/2010-honda-fit-13.htm&quot; rel=&quot;nofollow&quot;&gt;approximately 2489 pounds&lt;/a&gt; (1129 kg).  One feature of the Fit is that the back seats can sit upright like a normal car, or they can completely fold down (&lt;a href=&quot;http://autocarmanual.com/automotive/fold-the-honda-fit-rear-magic-seat-down-to-reveal-more-space.jpg&quot; rel=&quot;nofollow&quot;&gt;this image&lt;/a&gt; from &lt;a href=&quot;http://autocarmanual.com/gallery-honda/2010-honda-fit-user-manual-overview/&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt; shows a folded seat next to an upright seat).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I rarely have more than one passenger in my car, I can leave the rear seats folded down almost all the time.  Having the seats folded down will result in a lower center of gravity which should increase my fuel efficiency (I think!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't been able to find an estimate on the weight of the rear seats, other than &quot;pretty substantial&quot; according to &lt;a href=&quot;http://www.fitfreak.net/forums/fit-interior-modifications/46083-back-seat-delete.html#post709667&quot; rel=&quot;nofollow&quot;&gt;this forum post&lt;/a&gt;.  But I'm not sure how much of a difference it will make overall, anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first part of my question is, assuming that raising or lowering the center of gravity affects acceleration and deceleration, how does one quantify that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, given that I may be able to move, say, 100 pounds of seat from the &quot;dead center&quot; of my vehicle to, say, 1-2 feet closer to the wheels, how can I estimate how that will affect my fuel economy?  Perhaps a more specific question would be, how can I estimate how that will affect my acceleration? (which is of course when I get the worst fuel economy, besides being stopped)&lt;/p&gt;&#xA;" OwnerUserId="1204" LastActivityDate="2011-01-13T23:38:02.783" Title="How does a lower center of gravity affect fuel economy?" Tags="&lt;classical-mechanics&gt;&lt;acceleration&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="2745" PostTypeId="1" AcceptedAnswerId="2749" CreationDate="2011-01-12T18:24:52.187" Score="9" ViewCount="242" Body="&lt;p&gt;I am sending a couple of questions which seem a bit more specific than others on this site, partially to probe if there is a point in doing so. Not sure what is the range of expertise here, and no way to find out without trying, so here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering what is known about QCD, or other field theories, in the regime of large density and low temperatures, specifically studied in the large N limit. I know of the qualitative picture at finite N, but lots of the instabilities (e.g. the superconducting ones) are suppressed at large N and replaced by other interesting phenomena. I am only aware at the moment of the DGR instability to form chiral density waves, and I am wondering what else exists in the vast and possibly quite old literature. Any pointers or entry points to that literature will be appreciated&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-12T21:24:58.197" LastActivityDate="2011-03-02T12:02:48.047" Title="QCD phase diagram in the large N limit" Tags="&lt;quantum-field-theory&gt;&lt;quantum-chromodynamics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="2746" PostTypeId="2" ParentId="2744" CreationDate="2011-01-12T18:27:10.083" Score="4" ViewCount="" Body="&lt;p&gt;The center of gravity should have no effect on economy, besides possibly some indirect effect via changing the load on front/rear wheel bearings or something like that. I would expect secondary effects like that to be insignificant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing the CG won't have an effect on acceleration either, unless you are accelerating at the limit of your tire's grip. A car shifts more of it's weight to the rear tires during acceleration, thus causing the front wheels (in a front wheel drive car) to slip at a lower acceleration than they otherwise would. Lowering the CG of your car will reduce this weight transfer to some degree. So, if you were in a road race it may be a good idea, but for normal driving I doubt that it matters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB: The same weight transfer during acceleration occurs in a rear wheel drive car, but because the wheels with more weight on them are now the driven wheels this actually improves acceleration. This is one reason why most sports cars are rear wheel drive -- they can accelerate much harder when tire friction is the limiting factor. ie., when accelerating from a standstill. &lt;/p&gt;&#xA;" OwnerUserId="869" LastEditorUserId="869" LastEditorDisplayName="" LastEditDate="2011-01-12T20:40:29.313" LastActivityDate="2011-01-12T20:40:29.313" />
  <row Id="2747" PostTypeId="1" AcceptedAnswerId="2823" CreationDate="2011-01-12T18:33:26.183" Score="9" ViewCount="282" Body="&lt;p&gt;I am sending a couple of questions which seem a bit more specific than others on this site, partially to probe if there is a point in doing so. Not sure what is the range of expertise here, and no way to find out without trying. This one is also not terribly focused, but nonetheless here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering if there are some well-known and well-studied examples of large N matrix models (in which the fields are adjoint rather than vectors) which are of use in describing some condensed matter phenomena. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of applications of matrix models in anything between nuclear physics to number theory, and there are well-known vector models which are useful in CM physics, but off the top of my head I cannot think about matrix models which are used to solve some condensed matter problems.  Quite possibly I am missing something obvious...References or brief descriptions will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-01-12T21:05:53.020" LastActivityDate="2011-01-13T23:50:15.600" Title="Matrix models and condensed matter physics" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="2748" PostTypeId="2" ParentId="2702" CreationDate="2011-01-12T18:33:44.420" Score="1" ViewCount="" Body="&lt;p&gt;As with all mnemonics, you just have to try lots until you find one that resonates with you ... Here's mine, I don't know if it will help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 2 &quot;hand&quot; rules for electromagnetism: the left-hand motor rule, and the right-hand current rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Left-hand motor rule&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left-hand motor rule shows what happens when a current-carrying wire moves in a magnetic field. Hold the thumb, first and second fingers mutually perpendicular. Then:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The thuMb gives Movement (or equivalently the THumb gives THrust)&lt;/li&gt;&#xA;&lt;li&gt;The First finger is the direction of the magnetic Field&lt;/li&gt;&#xA;&lt;li&gt;The seCond finger is the direction of the Current&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Together these give you the direction of force on a wire when current flows through it in a magnetic field (motor), or the direction of current induced in a wire when the wire moves in a magnetic field (inductor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Right-hand rule - direction of magnetic field around a wire&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Grab the wire with your right hand; the thumb points in the direction of the current, and the fingers curl around in the direction of the magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Left or right hand?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, I'm left-handed. So I associate the motor rule with my left hand - I do most stuff with my left, which is controlled by my motor neurones. Then the other rule uses the other hand. I accept that may not help if you're right-handed :-)&lt;/p&gt;&#xA;" OwnerUserId="474" LastActivityDate="2011-01-12T18:33:44.420" />
  <row Id="2749" PostTypeId="2" ParentId="2745" CreationDate="2011-01-12T19:26:12.287" Score="7" ViewCount="" Body="&lt;p&gt;If by large density you mean large baryon density, then I believe one of the fundamental large $N_c$ results is that at densities of order nuclear densities, but below the density where the baryons have dissolved into quarks, baryonic matter forms a crystalline structure. This has been analyzed in the Skyrme model. I think this paper by Klebanov was one of the first to point this out:  &quot;Nuclear Matter In The Skyrme Model,&quot;Igor R. Klebanov, Nucl.Phys.B262:133,1985.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AdS/QCD models can also be used to study QCD at large baryon chemical potential and low temperatures (and necessarily at large $N_c$). An instability to the condensation of vector mesons resulting from Chern-Simons couplings dictated by anomaly matching was found in a paper by me and S. Domokos, arXiv:0704.1604. I'm not aware of any evidence that this phenomenon happens in the real world, and even its existence at large $N_c$ should be regarded as conjectural. I don't know whether this instability is related to the DGR instability or not.&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2011-03-02T12:02:48.047" LastActivityDate="2011-03-02T12:02:48.047" CommentCount="1" />
  <row Id="2750" PostTypeId="1" AcceptedAnswerId="2754" CreationDate="2011-01-12T19:47:32.760" Score="12" ViewCount="516" Body="&lt;p&gt;What happens if you perform a double slit experiment near an event horizon, if one of the slits is outside, one is inside the event horizon?&lt;/p&gt;&#xA;" OwnerUserId="47" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-03-13T20:16:46.870" LastActivityDate="2011-03-13T20:16:46.870" Title="Double slit experiment near event horizon" Tags="&lt;black-holes&gt;&lt;quantum-mechanics&gt;&lt;double-slit-experiment&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="4" />
  <row Id="2751" PostTypeId="2" ParentId="2747" CreationDate="2011-01-12T19:49:08.457" Score="4" ViewCount="" Body="&lt;p&gt;Matrix models have been used to study RNA folding.  For example, see: &lt;a href=&quot;http://arxiv.org/pdf/q-bio/0411004v1&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/q-bio/0411004v1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="266" LastActivityDate="2011-01-12T19:49:08.457" CommentCount="1" />
  <row Id="2752" PostTypeId="2" ParentId="2744" CreationDate="2011-01-12T20:00:09.423" Score="1" ViewCount="" Body="&lt;p&gt;One of the sources of losses are the vibrations, i.e. the internal motions of the car, as it responds to the noneven road surface. Changing the CG would have an affect on the vehicles normal modes, and so probably has some effect. If the car had completely round tires, and the roadway was perfectly smooth, they would never be excited. It might not necessarily help however, the relative frequecies and modeshapes would be affected, but the devil is in the datails. In any case, some energy is lost to the excitation of these internal modes, and eventually to their damping via frictional forces. Car manufacturers pay a considerable amount of attention to these things, although they are more interested in the experience of the occupants, noise and harshness, but I doubt they optimize for rolling resitance.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-01-12T20:00:09.423" CommentCount="2" />
  <row Id="2753" PostTypeId="2" ParentId="2750" CreationDate="2011-01-12T20:01:08.700" Score="0" ViewCount="" Body="&lt;p&gt;I strongly suspect there would be no interference, as we would be able to determine which slit the particle went through.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-01-12T20:01:08.700" />
  <row Id="2754" PostTypeId="2" ParentId="2750" CreationDate="2011-01-12T20:13:47.350" Score="7" ViewCount="" Body="&lt;p&gt;This is just a little bit subtle, because at first it looks like nobody has causal access to both light rays, which is required for the existence of interference pattern. On the other hand the principle of equivalence seems to assure us that local experiments done by a freely falling observer would not be affected by the presence of the horizon. There is a sense in which this is true, as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an observer  freely falling into the horizon of a large black hole while conducting the double slit experiment, there will not be any change in the results of the experiment. Gravity is weak and nothing special happens when they cross the horizon, they will only discover they fell into a black hole sometime in the future when tidal forces will make life very uncomfortable. Any experiment localized within space and time will be the same up to tiny corrections due to weak gravity. This is true also if one of the slits (and therefore necessarily the light source) happens to be outside the horizon at the time of the experiment. Of course, for the freely falling observer that statement does not mean much, only the outside observer will be able to make meaningful distinction between inside and outside the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, for an observer staying outside the horizon, one of the slits is invisible, they don't have access to all the light rays, and they will not see any interference. In fact they will not even be aware that there is a double slit experiment going on. There is no problem there because they are not freely falling observers, and there is nothing that states that the result of their experiment is identical to the same experiment made in flat space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that infalling and outside observers describe the same phenomena so differently lead to many black hole paradoxes, where quantum mechanics and GR (or the principle of equivalence) just barely co-exist without contradiction. Lenny Susskind's popular book on black holes has a good discussion, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  If you jump into the black hole you can have access to the interference pattern after the fact. So, it looks like you can generate another paradox, in addition to the original one. Look at arxiv.org/abs/0808.2096 for the resolution of that one. Roughly speaking, after you jump into the black hole you have only finite time to make measurements before you encounter the singularity. This time is insufficient to get more information than you are entitled to by the rules of quantum mechanics. This statement requires a detailed calculation which is in that paper.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-01-21T19:49:47.387" LastActivityDate="2011-01-21T19:49:47.387" CommentCount="24" />
  <row Id="2755" PostTypeId="2" ParentId="2743" CreationDate="2011-01-12T20:20:52.957" Score="4" ViewCount="" Body="&lt;p&gt;If the black hole simply swalled matter, and didn't lose any energy, it probably isn't too hard a calculation, just assume the earth is unsupported mass that falls into the BH, which grows in mass as it adds more stuff. The problem, is we know this isn't how it would happen, and some significant fraction of swalled mass will be released as energy, maybe one to a few percent of mC**2. So the energy liberated from swallowing mass, is orders of magnitude greater per unit mass than an H bomb. Clearly most of the planets mass would be blown away, and only a small amount would end up incorporated into the BH. I'd bet this would happen extremely rapidly, and the shock wave that rips the planet apart would probably only take a few seconds. Note freefall time to the center of the earth is probably more like a half hour (order of magnitude), so most of the planet wouldn't even begin to fall before the released energy blasted it apart.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-01-12T20:20:52.957" />
  <row Id="2756" PostTypeId="2" ParentId="2743" CreationDate="2011-01-12T20:25:05.880" Score="0" ViewCount="" Body="&lt;p&gt;It would take a long time if we do a back of the envelope calculation.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the black hole would exert a force of 1g at around 20 km (assuming 10^20 kg of mass).&lt;/li&gt;&#xA;&lt;li&gt;if we can reasonably assume that the mass inside this sphere is going to be absorbed quickly, that would mean the black hole mass increases correspondingly.&lt;/li&gt;&#xA;&lt;li&gt;on the other hand this extra mass can be calculated to be around 10^20 kg too. So we can expect the 1g radius not to increase significantly.&lt;/li&gt;&#xA;&lt;li&gt;I believe that mass with less than 1g of pull will take a long tine to spiral inside the black hole, as its size (Shwartzchild radius) would be in the micrometer scale and the sizes involved in the kilometre scale.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-01-12T20:25:05.880" CommentCount="2" />
  <row Id="2757" PostTypeId="1" AcceptedAnswerId="2758" CreationDate="2011-01-12T20:35:07.623" Score="3" ViewCount="173" Body="&lt;p&gt;I'm trying to reproduce a result from a paper (T. Thatcher, &lt;a href=&quot;http://www.engr.uvic.ca/~struchtr/2008ICMMES06_G13BC.pdf&quot; rel=&quot;nofollow&quot;&gt;Boundary Conditions for Grad's 13 moment equations&lt;/a&gt;, equation (32), page 6), however, I haven't been able to do so. Hopefully someone can provide some help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general problem concerns the &lt;a href=&quot;http://en.wikipedia.org/wiki/Couette_flow&quot; rel=&quot;nofollow&quot;&gt;Couette flow&lt;/a&gt;: A flow passes through two infinite parallel plates at distance $L$ move with velocity $v_{w}^{0}$ and $v_{w}^{L}$ ($w$ refers to the wall) relative to each other in their respective planes, each with temperature $\theta_{w}^{0}$ and $\theta_{w}^{L}$. The velocity $v$ of the flow only depends on the $y$ coordinate, that is the coordinate perpendicular to the direction of the flow. After solving the Navier-Stokes-Fourier equations for this problem, you can find the following equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dv}{dy}=a=\text{constant}, \qquad \frac{d^{2}\theta}{dy^{2}}=-\frac{2\text{Pr}}{5}a^{2} \qquad \qquad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which are linearized. Using as boundary conditions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}&#xA;vv(L)-v_{w}^{L} = -\frac{2-\chi}{\chi}\alpha_{1}\sqrt{\frac{\pi}{2}}\text{Kn}\; a,  \qquad   \&#xA;\&#xA;v(0)-v_{w}^{0} = \frac{2-\chi}{\chi}\alpha_{1}\sqrt{\frac{\pi}{2}}\text{Kn}\; a  \qquad (2)&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}&#xA;\theta \theta(L)-\theta_{w}^{L} = -\frac{2-\chi}{\chi}\frac{5\beta_{1}}{4 Pr}\sqrt{\frac{\pi}{2}}\text{Kn}\frac{d\theta}{dy}, \quad  \&#xA;\&#xA;\theta(0)-\theta_{w}^{0} = \frac{2-\chi}{\chi}\frac{5\beta_{1}}{4 Pr}\sqrt{\frac{\pi}{2}}\text{Kn}\frac{d\theta}{dy} \qquad (3)&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the solution for the velocity should be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = \frac{v_{w}^{L}}{2}+a\left(\frac{y}{L}-\frac{1}{2}\right) \qquad \text{with} \quad a = \frac{v_{w}^{L}-v_{w}^{0}}{1+\frac{2-\chi}{\chi}\sqrt{2\pi} \alpha_{1}\text{Kn}} \qquad \qquad (4)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't be concerned about $\alpha_{1}$ or $\beta_{1}$; they are correction terms. $\text{Kn}$ and $\text{Pr}$ are, as usual, the Knudsen number and the Prandtl number, respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For now, I'm interested only in the solution for the velocity (equation $(4)$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; First and second attempts are wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First attempt:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use equation $(2)$ as $v-v_{w}^{L}=-\frac{2-\chi}{\chi}\alpha_{1}\sqrt{\frac{\pi}{2}}\text{Kn}\;a L$ and $v-v_{w}^{0}=\frac{2-\chi}{\chi}\alpha_{1}\sqrt{\frac{\pi}{2}}\text{Kn}\;a L$. Subtract the first from the second to obtain $a = \displaystyle \frac{v_{w}^{L}-v_{w}^{0}}{\frac{2-\chi}{\chi}\alpha_{1}\sqrt{2\pi}\text{Kn}L}$, and from equation $(1)$ for the velocity, $v = ay+\text{constant}$, evaluating at the plate localized in $L$ we have $v_{w}^{L}=aL+\text{constant}$, which gives us the constant, and &lt;em&gt;a&lt;/em&gt; solution but it's not the correct result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second attempt:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this can be useful: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take equation $(2)$ and solve for $a$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=\frac{-v_{w}^{L}+v_{w}^{0}}{-\frac{2-\chi}{\chi}\alpha_{1}\sqrt{2\pi}\text{Kn}L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we use the first part of equation $(1)$, so we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dv}{dy}=\frac{-v_{w}^{L}+v_{w}^{0}}{-\frac{2-\chi}{\chi}\alpha_{1}\sqrt{2\pi}\text{Kn}L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After integrating this equation, we evalute the velocity at the boundary $L$, which allows us to find the constant of integration. Substituting this constant, we obtain the following result&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v-v_{w}^{L}=\frac{v_{w}^{L}-v_{w}^{0}}{\frac{2-\chi}{\chi}\alpha_{1}\sqrt{2\pi}\text{Kn}}\left(\frac{y}{L}-1\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v-v_{w}^{L}=a\left(\frac{y}{L}-1\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not quite the result I'm looking for, but maybe this is the right path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Third attempt:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking $\displaystyle \frac{dv}{dy}=b$, it follows $v(L)=a L +b$ and $v(0)=b$. Using the boundary conditions to evalute $v(L)$ and $v(0)$ we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_{w}^{0}+\displaystyle\frac{2-\chi}{\chi}\alpha_{1}\sqrt{\frac{\pi}{2}}\text{Kn}a = b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_{w}^{L}-\displaystyle \frac{2-\chi}{\chi}\alpha_{1}\sqrt{\frac{\pi}{2}}\text{Kn}a = aL+b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After solving this system of equations: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a = \frac{v_{w}^{L}-v_{w}^{0}}{L+\text{Kn}\sqrt{2\pi}\alpha_{1}(\frac{2-\chi}{\chi})}$$&#xA;$$b = v_{w}^{0}-\frac{\left(v_{w}^{0}-v_{w}^{L}\right)\text{Kn}\sqrt{\frac{\pi}{2}}\alpha_{1}\left(\frac{2-\chi}{\chi}\right)}{L+\text{Kn}\sqrt{2\pi}\alpha_{1}\left(\frac{2-\chi}{\chi}\right)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting back in the original equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v(y)-v_{w}^{0}=a\left(y-\text{Kn}\sqrt{\frac{\pi}{2}}\alpha_{1}\left(\frac{2-\chi}{\chi}\right)\right)$$ &lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="18" LastEditorDisplayName="" LastEditDate="2011-01-13T19:03:10.870" LastActivityDate="2011-01-13T19:03:10.870" Title="Boundary conditions for Couette flow" Tags="&lt;fluid-dynamics&gt;&lt;mathematics&gt;&lt;couette-flow&gt;" AnswerCount="1" CommentCount="15" />
  <row Id="2758" PostTypeId="2" ParentId="2757" CreationDate="2011-01-12T23:15:07.043" Score="1" ViewCount="" Body="&lt;p&gt;Your mistake was to misinterpret the equation for the boundary conditions (2).  Since those two equations (one at each wall) are boundary conditions, they involve two different quantities v(y=0) and v(y=L). Unfortunately these were both written as v in the paper, which may have given you the impression that you could subtract the equations and thereby cancel v.  However, that is not the case.  The boundary conditions apply only at the boundaries, so they do not hold for v in general! The correct approach is to take the solution, which you know is v = ay + const. and to solve for the constants (a and const) by applying the boundary conditions to v(0) and v(L), that is to a(0) + const and to a(L) + const.  That will give you two linear equations and two unknowns (const and a). Presumably this gives the answer in the paper. But first you should fix the rest of your typos in equation (4), since it does not match that of the paper!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: There is a typo in the paper.  This is clear since the solution given in the paper does not obey $dv/dy = a$ but rather $dv/dy=a/L$.  I believe the correct solution has (y-L/2) rather than (y/L - 1/2).&lt;/p&gt;&#xA;" OwnerUserId="266" LastEditorUserId="266" LastEditorDisplayName="" LastEditDate="2011-01-12T23:37:39.090" LastActivityDate="2011-01-12T23:37:39.090" CommentCount="2" />
  <row Id="2759" PostTypeId="1" AcceptedAnswerId="9576" CreationDate="2011-01-12T23:32:37.070" Score="2" ViewCount="183" Body="&lt;p&gt;How would I figure out the Cartesian graph that describes a bar clamped flat for a length on one end with downward force being applied to the other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an idea that the bar will try to average out the stress over the free length.  Does it in fact evenly distribute stress over the whole free length?  If so, can I use that fact to calculate curvature at each point to derive a Cartesian graph?&lt;/p&gt;&#xA;" OwnerUserId="1213" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-11T20:18:59.433" LastActivityDate="2011-05-11T20:18:59.433" Title="What is the shape of a clamped bent bar?" Tags="&lt;statics&gt;&lt;stress-strain&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" />
  <row Id="2760" PostTypeId="2" ParentId="2743" CreationDate="2011-01-13T02:14:03.880" Score="13" ViewCount="" Body="&lt;p&gt;In the LHC, we are talking about mini black holes of mass around $10^{-24}kg$, so when you talk about $10^{15}-10^{20}kg$ you talk about something in the range from the mass of Deimos (the smallest moon of Mars) up to $1/100$ the mass of the Moon. So we are talking about something really big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Schwarzschild radius of such a black hole (using the $10^{20}$ value) would be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_s=\frac{2GM}{c^2}=1.46\times 10^{-7}m=0.146\mu m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can consider that radius to be a measure of the cross section that we can use to calculate the rate that the BH accretes mass. So, the accretion would be a type of Bondi accretion (spherical accretion) that would give an accretion rate &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{M}=\sigma\rho u=(4\pi R_s^2)\rho_{earth} u,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $u$ is a typical velocity, which in our case would be the speed of sound and $\rho_{earth}$ is the average density of the earth interior. &#xA;The speed of sound in the interior of the earth can be evaluated to be on average something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_s^2=\frac{GM_e}{3R_e}.$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the accretion rate is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{M}=\frac{4\pi}{\sqrt{3}}\frac{G^2M_{BH}^2}{c^4}\sqrt{\frac{GM_e}{R_e}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is an order of magnitude estimation that gives something like $\dot{M}=1.7\times10^{-6}kg/s$. If we take that at face value, it would take something like $10^{23}$ years for the BH to accrete $10^{24}kg$. If we factor in the change in radius of the BH, that time is probably much smaller, but even then it would be something much larger than the age of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that is not the whole picture. One should take also in to account the possibility of having a smaller accretion rate due to the Eddington limit. As the matter accretes to the BH it gets hotter since the gravitational potential energy is transformed to thermal energy (virial theorem). The matter then radiates with some characteristic luminosity. The radiation excerpts some back-force on the matter that is accreting lowering the accretion rate. In this case I don't thing that this particular effect plays any part in the evolution of the BH.&lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2011-01-13T02:14:03.880" CommentCount="6" />
  <row Id="2761" PostTypeId="2" ParentId="2687" CreationDate="2011-01-13T03:28:22.390" Score="2" ViewCount="" Body="&lt;p&gt;I understand there is a new book by Michael Spivak called &quot;Physics for Mathematicians&quot;&#xA;with Vol 1. being on Mechanics. Based on some notes for the first chapters, available &#xA;here &lt;a href=&quot;http://www.math.uga.edu/~shifrin/Spivak_physics.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uga.edu/~shifrin/Spivak_physics.pdf&lt;/a&gt; it looks pretty wonderful and you might particularly like it since it is aimed at teaching Mechanics to Mathematicians.&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T04:09:30.507" LastActivityDate="2011-03-02T04:09:30.507" CommunityOwnedDate="2011-02-04T10:52:39.143" CommentCount="1" />
  <row Id="2762" PostTypeId="2" ParentId="2687" CreationDate="2011-01-13T03:45:47.977" Score="1" ViewCount="" Body="&lt;p&gt;A good one is J. H. Heinbockel- &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1553691334&quot; rel=&quot;nofollow&quot;&gt;Introduction to Tensor Calculus and Continuum Mechanics&lt;/a&gt;. I guess that's the kind of mathematics an engineer would use, but I don't know for sure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can download it &lt;a href=&quot;http://www.math.odu.edu/~jhh/johnh.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2011-01-13T03:45:47.977" CommunityOwnedDate="2011-02-04T10:52:39.143" CommentCount="1" />
  <row Id="2763" PostTypeId="2" ParentId="2759" CreationDate="2011-01-13T04:42:18.707" Score="1" ViewCount="" Body="&lt;p&gt;I think if the boundary conditions were perfectly symmetric, you'd get a simple circular shape. In your case you have somewhat different conditions on both ends. The inside of the curve will be in compression, and the outside in tension. But I think your clamped end restricts some of the material degrees of freedom, so there is probably a transition region near that end where the curvature differs somewhat from the rest of the bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Real materials can be nonlinear. If the material weakens with greater stress -especially if it can plastically deform there can be a tendency for the strain to accumulate in a small region (like bending a coat hanger, and getting most of the bending in one small area).&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-01-13T04:42:18.707" />
  <row Id="2764" PostTypeId="2" ParentId="1626" CreationDate="2011-01-13T06:03:47.030" Score="0" ViewCount="" Body="&lt;p&gt;I have the exact same issue with my bathroom doors. It is generally cooler in there during the winter. Since its my bathroom, the doors suffer a heavy dose of water (its humid). But this problem ONLY occurs in the cooler months. Could it be the metal door hinges, handles, locks, nuts and bolts that contract?  If those mechanisms were to shrink in size, it would displace the door's positioning....thus preventing it from shutting.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-13T06:03:47.030" />
  <row Id="2765" PostTypeId="1" CreationDate="2011-01-13T07:25:22.173" Score="5" ViewCount="296" Body="&lt;p&gt;I've been looking with interest at a &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1012/1012.5166v1.pdf&quot; rel=&quot;nofollow&quot;&gt;recent biology paper&lt;/a&gt; claiming that DNA molecules give off electromagnetic signals which can cause the same types of molecules to be reconstructed at a remote location. The &lt;a href=&quot;http://science.slashdot.org/story/11/01/13/0017256/Nobel-Prize-Winner-Says-DNA-Performs-Quantum-Teleportation&quot; rel=&quot;nofollow&quot;&gt;Slashdot crowd&lt;/a&gt; seems to think the idea is pretty ridiculous, and I'm inclined to agree, but I still think it's worth digging into some of the physics they cite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A large part of the theoretical argument in the paper relies on what they call coherence domains. As far as I can tell, they're talking about a relatively large region of liquid (in this case water) in which all the molecules are in resonance with an oscillating electromagnetic field which is confined by the boundary of the region itself. I don't think I really &lt;a href=&quot;http://en.wikipedia.org/wiki/Grok&quot; rel=&quot;nofollow&quot;&gt;grok&lt;/a&gt; that description, and a Google search for further information turned up nothing relevant. I'd like to see some of the math that backs this up. So can anyone provide a more mathematical explanation of how these coherence domains are able to exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm assuming the whole idea is not totally bogus :-P)&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-08-23T07:45:33.640" Title="A quantitative explanation of coherence domains" Tags="&lt;quantum-field-theory&gt;&lt;fluid-dynamics&gt;&lt;many-body&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="2766" PostTypeId="2" ParentId="1268" CreationDate="2011-01-13T08:49:51.913" Score="5" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.fourmilab.ch/etexts/einstein/E_mc2/www/&quot; rel=&quot;nofollow&quot;&gt;Einstein's proof&lt;/a&gt; did not rely on having a rigid body. It relies on having a body with mass (obviously). To be more clear.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The paper only says &lt;em&gt;body&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;It does not rely on any rigid body property (like the size)&lt;/li&gt;&#xA;&lt;li&gt;It does not rely on any relativistic speed or condition on the body&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The proof merely involve how energy is measured by an observer stationary with the body and one stationary with the emitted waves.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let there be a stationary body in the system $(x, y, z)$, and let its energy--referred to the system $(x, y, z)$ be $E_0$. Let the energy of the body relative to the system $(\xi,\eta,\zeta)$ moving as above with the velocity $v$, be $H_0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact the paper is really easy and clear... :-)&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-13T10:35:38.270" LastActivityDate="2011-01-13T10:35:38.270" />
  <row Id="2767" PostTypeId="1" CreationDate="2011-01-13T09:48:26.100" Score="11" ViewCount="1039" Body="&lt;p&gt;Electromagnetic waves can be shielded by a perfect conductor. What about gravitational waves?&lt;/p&gt;&#xA;" OwnerUserId="1218" LastActivityDate="2011-02-07T00:25:55.153" Title="Is it theoretically possible to shield gravitational waves?" Tags="&lt;electromagnetism&gt;&lt;general-relativity&gt;&lt;gravity&gt;" AnswerCount="8" CommentCount="1" FavoriteCount="2" />
  <row Id="2768" PostTypeId="1" CreationDate="2011-01-13T10:33:30.800" Score="1" ViewCount="133" Body="&lt;p&gt;I am trying to derive a state space equation for human movement (i.e. walking, running, standing still) in some normal distribution sense. The following equations should not be so hard to follow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the state $\bar{x} = (x, \dot{x})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The state equation will be $\bar{x}(t) = A\bar{x}(t-1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the covariance matrix $C(t) = AC(t-1)A^{-1} + Q$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $A = \begin{bmatrix}&#xD;&#xA;&#xD;&#xA;1 &amp;amp; T \\&#xD;&#xA;&#xD;&#xA;0 &amp;amp; 1&#xD;&#xA;&#xD;&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $T$ is the time period, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and where $Q$ a &quot;process noise&quot; covariance matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q = \begin{pmatrix} \sigma_x^2 &amp;amp; \sigma_x \sigma_{\dot{x}} \\&#xD;&#xA;                     \sigma_x \sigma_{\dot{x}}  &amp;amp; \sigma_{\dot{x}}^2 \end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my view, the key of a correct model lies in getting correct values of $\sigma_x$ and $\sigma_{\dot{x}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to wiki, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Footspeed&quot; rel=&quot;nofollow&quot;&gt;fastest human&lt;/a&gt; speed recorded is 44.72 km/h.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you think?&lt;/p&gt;&#xA;" OwnerUserId="250" LastActivityDate="2011-01-16T14:57:23.310" Title="How to model the dynamics of human movement?" Tags="&lt;models&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="2770" PostTypeId="2" ParentId="2716" CreationDate="2011-01-13T11:20:22.350" Score="1" ViewCount="" Body="&lt;p&gt;You might be interested in this resource: &lt;a href=&quot;http://en.wikipedia.org/wiki/Clock_of_the_Long_Now&quot; rel=&quot;nofollow&quot;&gt;The Clock of the Long Now&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Clock of the Long Now, also called the 10,000-year clock, is a proposed mechanical clock designed to keep time for 10,000 years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In building it they discuss relevant topics such as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Longevity: The clock should be accurate even after 10,000 years, and must not contain valuable parts (such as jewels, expensive metals, or special alloys) that might be looted.&lt;/li&gt;&#xA;  &lt;li&gt;Maintainability: Future generations should be able to keep the clock working, if necessary, with nothing more advanced than Bronze Age tools and materials.&lt;/li&gt;&#xA;  &lt;li&gt;Transparency: The clock should be understandable without stopping or disassembling it; no functionality should be opaque.&lt;/li&gt;&#xA;  &lt;li&gt;Evolvability: It should be possible to improve the clock over time.&lt;/li&gt;&#xA;  &lt;li&gt;Scalability: To ensure that the final large clock will work properly, smaller prototypes must be built and tested.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution they are building is far from trivial.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-01-13T11:20:22.350" CommunityOwnedDate="2011-01-13T13:54:00.280" />
  <row Id="2771" PostTypeId="2" ParentId="2767" CreationDate="2011-01-13T12:13:31.410" Score="0" ViewCount="" Body="&lt;p&gt;Nature of gravitation is not understood completely yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I believe you can shield of sorts using black hole ;-)&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-01-13T12:13:31.410" CommentCount="4" />
  <row Id="2772" PostTypeId="1" CreationDate="2011-01-13T13:24:47.093" Score="4" ViewCount="268" Body="&lt;p&gt;Say there was some situation where you have a lot of subatomic particles interacting with each other and decided to draw (say, by joining Feynmann diagrams) those interactions- so that you got some sort of (?directed) graph... what kind of network would you obtain? Would it be scale-free? Would it be a complex network similar to a social network? What properties would it have?&lt;/p&gt;&#xA;" OwnerUserId="1221" LastEditorUserId="1221" LastEditorDisplayName="" LastEditDate="2011-01-13T15:33:34.837" LastActivityDate="2011-03-14T23:33:14.453" Title="Properties of graph of subatomic particle interactions" Tags="&lt;quantum-mechanics&gt;&lt;particle-physics&gt;&lt;mathematical-physics&gt;&lt;graph-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="2773" PostTypeId="2" ParentId="2767" CreationDate="2011-01-13T14:31:49.937" Score="5" ViewCount="" Body="&lt;p&gt;a possible realization for your &quot;spacetime engineering question&quot; could be &lt;a href=&quot;http://en.wikipedia.org/wiki/White_hole&quot; rel=&quot;nofollow&quot;&gt;white holes&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This hypothetical object can be described by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_metric&quot; rel=&quot;nofollow&quot;&gt;Schwarzschild metric&lt;/a&gt;&#xA;$$g = -c^2\left( 1-\frac{2Gm}{c^2 r}  \right)dt^2 + \left( 1-\frac{2Gm}{c^2 r}  \right)^{-1}dr^2 +r^2 d\Omega^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but in contrast to a black hole you &lt;strong&gt;interprete&lt;/strong&gt; the &lt;strong&gt;time direction&lt;/strong&gt; the other way round - &lt;strong&gt;nothing&lt;/strong&gt; will ever be able to &lt;strong&gt;enter&lt;/strong&gt; the object like nothing will ever leave a &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_hole&quot; rel=&quot;nofollow&quot;&gt;black hole&lt;/a&gt; (ok, maybe some Hawking radiation and virtual particles...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to construct another &quot;ordinary spacetime&quot; with &lt;a href=&quot;http://prd.aps.org/abstract/PRD/v68/i10/e104029&quot; rel=&quot;nofollow&quot;&gt;Maclaurin spheroids&lt;/a&gt; and counterrotating black holes on an axis of symmetry  (stationarity for a stable solution!) but my ability to visualize such a situation was not sufficient :)&lt;br&gt;&#xA;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Further notes&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;For those who are interested in &lt;strong&gt;experimental&lt;/strong&gt; realizations (in the sense of &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0505065&quot; rel=&quot;nofollow&quot;&gt;analogue gravity&lt;/a&gt;) of &lt;strong&gt;white-hole horizons&lt;/strong&gt; permitting light to enter a certain region via the nonlinear &lt;a href=&quot;http://en.wikipedia.org/wiki/Kerr_effect&quot; rel=&quot;nofollow&quot;&gt;Kerr-effect&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Prediction&lt;/strong&gt;: Philbin et al &lt;a href=&quot;http://www.sciencemag.org/content/319/5868/1367.abstract&quot; rel=&quot;nofollow&quot;&gt;Fiber-Optical Analog of the Event Horizon&lt;/a&gt;, Science 2008 (&lt;a href=&quot;http://arxiv.org/abs/0711.4796&quot; rel=&quot;nofollow&quot;&gt;arXiv-link&lt;/a&gt;)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Realizations&lt;/strong&gt;: &lt;a href=&quot;http://iopscience.iop.org/0295-5075/89/3/34004&quot; rel=&quot;nofollow&quot;&gt;Analogue gravity and ultrashort laser pulse filamentation&lt;/a&gt; (EPL) and &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v105/i20/e203901&quot; rel=&quot;nofollow&quot;&gt;Hawking Radiation from Ultrashort Laser Pulse Filaments&lt;/a&gt; (PRL).&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2011-01-19T09:28:59.883" LastActivityDate="2011-01-19T09:28:59.883" CommentCount="8" />
  <row Id="2774" PostTypeId="1" AcceptedAnswerId="2775" CreationDate="2011-01-13T14:34:00.773" Score="5" ViewCount="600" Body="&lt;p&gt;Imagine a bar &lt;/p&gt;&#xA;&#xA;&lt;p&gt;spinning like a helicopter propeller,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at $\omega$ rad/s because the extremes of the bar goes at speed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V = \omega * r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then we can reach near $c$ (speed of light)&#xA;applying some &lt;strong&gt;finite&lt;/strong&gt; amount of energy&#xA;just doing &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega = V / r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bar should be long, low density, strong to minimize the&#xA;amount of energy needed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example a $2000\,\mathrm{m}$ bar&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega = 300 000 \frac{\mathrm{rad}}{\mathrm{s}} = 2864789\,\mathrm{rpm}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a dental drill can commonly rotate at $400000\,\mathrm{rpm}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V$ (with dental drill) = 14% of speed of light&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I say this experiment can be really made&#xA;and bar extremes could approach $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you say? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our planet is orbiting at sun and it's orbiting milky way, and who knows what else, &#xA;then any earth point have a speed of 500 km/s or more agains CMB&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if we are orbiting something at that speed then there would be detectable relativist effect in different direction of measurements, simply extending a long bar or any directional mass in different galactic directions we should measure mass change due to relativity, simply because $V = \omega * r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what do you think?&lt;/p&gt;&#xA;" LastActivityDate="2011-04-28T19:33:39.470" Title="Rotate a long bar in space and get close to c speed" Tags="&lt;classical-mechanics&gt;&lt;mechanics&gt;&lt;special-relativity&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="1" />
  <row Id="2775" PostTypeId="2" ParentId="2774" CreationDate="2011-01-13T14:51:19.553" Score="11" ViewCount="" Body="&lt;p&gt;Imagine a rock on a rope. As you rotate the rope faster and faster, you need to pull stronger and stronger to provide centripetal force that keeps the stone on the orbit. The increasing tension in the rope would eventually break the it. The very same thing would happen with bar (just replace the rock with the bar's center of mass). And naturally, all of this would happen at speeds far below the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you imagined that there exists a material that could sustain the tension at relativistic speeds you'd need to take into account that signal can't travel faster than at the speed of light. This means that the bar can't be &lt;em&gt;rigid&lt;/em&gt;. It would bend and the far end would trail around. So it's hard to even talk about rotation at these speeds. One thing that is certain is that strange things would happen. But to describe this fully you'd need a relativistic model of solid matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People often propose arguments similar to yours to show Special Relativity fails. In reality what fails is our intuition about materials, which is completely classical.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-01-13T14:57:21.813" LastActivityDate="2011-01-13T14:57:21.813" CommentCount="5" />
  <row Id="2776" PostTypeId="2" ParentId="2774" CreationDate="2011-01-13T14:51:30.350" Score="4" ViewCount="" Body="&lt;p&gt;remember that in a three-dimensional description of &lt;a href=&quot;http://en.wikipedia.org/wiki/Special_relativity&quot; rel=&quot;nofollow&quot;&gt;special relativity&lt;/a&gt; the impulse of an object is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{p} = \gamma m \mathbf{v}$$&#xA;with the so-called Lorentz-factor&#xA;$$\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, do you think you can &lt;strong&gt;accelerate the masses&lt;/strong&gt; within the slab to a speed greater than light or do you think that something is wrong with your physical &lt;strong&gt;model of the system&lt;/strong&gt;?&lt;br&gt;&#xA;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2011-01-13T14:51:30.350" CommentCount="3" />
  <row Id="2777" PostTypeId="2" ParentId="2774" CreationDate="2011-01-13T14:59:30.917" Score="5" ViewCount="" Body="&lt;p&gt;I say no. Assuming all the practicalities work, you can get arbitrarily close to c. But not reach c. You can see this easily from the relativstic formula for kinetic energy:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_k = mc^2(\frac{1}{\sqrt{1-v^2/c^2}}-1)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $v$ approaches $c$, the energy you need to supply to a particle at the end of the bar tends to infinity.&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="125" LastEditorDisplayName="" LastEditDate="2011-04-28T19:33:39.470" LastActivityDate="2011-04-28T19:33:39.470" CommentCount="2" />
  <row Id="2779" PostTypeId="2" ParentId="2716" CreationDate="2011-01-13T15:43:21.003" Score="0" ViewCount="" Body="&lt;p&gt;I do not completely understand what you want, so I'll try to address three possibilities. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If you need the standard notation for a time measurement with (reasonably) large scales, then you are looking for &lt;a href=&quot;http://en.wikipedia.org/wiki/Julian_day&quot; rel=&quot;nofollow&quot;&gt;Julian Day&lt;/a&gt; (JD). It is standard for scientific/astronomical communities.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you are looking for &quot;one and only&quot; experimental procedure for precise measurement of large time intervals, then I don't think that there is one. There are a lot of methods with own good and band sides.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally. You are emphasising the question of &quot;zero time&quot;. And we know that the laws of physics are invariant under time translation -- there is no &quot;selected&quot; moment in time. Therefore any moment can be chosen as the reference point -- it's just a matter of convention. For the JD the reference point is the astronomical noon in Alexandria on the 1st of January 4713 year BC. This choice is mainly due to historical reasons. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-01-13T15:43:21.003" CommunityOwnedDate="2011-01-13T15:43:21.003" />
  <row Id="2780" PostTypeId="2" ParentId="2774" CreationDate="2011-01-13T16:07:32.533" Score="5" ViewCount="" Body="&lt;p&gt;In your calculations you assume that your propeller is a rigid body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot use that assumption, when your speeds are not much smaller than the speed of light. Because &quot;there are no rigid bodies in relativity&quot;. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-01-13T16:07:32.533" CommentCount="1" />
  <row Id="2782" PostTypeId="2" ParentId="2767" CreationDate="2011-01-13T16:54:48.123" Score="4" ViewCount="" Body="&lt;p&gt;One potential solution, related to a problem discussed by Kip Thorne, would be to construct an enormous array of resonant bar detectors that attenuate the amplitude of the GWs as they pass through. If it is big enough, and the frequency is correct, you could absorb all the energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more active device would be like noise-cancelling headphones: an array of quadrupole oscillators that can be tuned to exactly phase-cancel the incoming GWs, effectively reflecting them back to the source.&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2011-01-13T16:54:48.123" CommentCount="3" />
  <row Id="2783" PostTypeId="2" ParentId="2743" CreationDate="2011-01-13T17:25:35.120" Score="4" ViewCount="" Body="&lt;p&gt;Since I have much better answer from Vagelford -- I'll write my own version.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When matter falls on the black hole it gets fractioned and radiates. As far as I know (correct me if I'm wrong) one can estimate the radiated energy as $\simeq 0.05mc^2$. Where $m$ is the mass of the falling matter.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth's matter is pulled by the black hole gravitation and pushed away by the radiation.  Moreover, for the matter flow $J$ we have &quot;negative feedback&quot; system: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;bigger $J$  -&gt; more radiation -&gt; more matter is &quot;pushed away&quot;&lt;/li&gt;&#xA;&lt;li&gt;smaller $J$ -&gt; less radiation -&gt; more matter is &quot;pulled in&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The equilibrium between those forces corresponds to &#xA;already mentioned &lt;a href=&quot;http://en.wikipedia.org/wiki/Eddington_luminosity&quot; rel=&quot;nofollow&quot;&gt;Eddington luminosity&lt;/a&gt;:&lt;br&gt;&#xA;$L (J/s) = 1.3\cdot 10^{21} \frac{M}{M_{sun}}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equating $L=0.05Jc^2$ and going to $r_{sh}(m) = 3000 \frac{M}{M_{sun}}$, I obtain:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$J (kg/s) = 100 r_{sh}(m)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is remarkable, that the &quot;consumption speed&quot; for the $10^{20} kg$ black hole ($r_{sh} = 148.5\mu m$, &lt;a href=&quot;http://www.wolframalpha.com/input/?i=schwartzshilt+radius+for+10%5E20+kg&quot; rel=&quot;nofollow&quot;&gt;look here&lt;/a&gt;) will give you $1.48\cdot10^{-5}$ kg/s. Which is just order of magnitude larger than the estimate by Vagelford.   &lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2011-01-13T17:39:08.720" LastActivityDate="2011-01-13T17:39:08.720" CommentCount="3" />
  <row Id="2784" PostTypeId="2" ParentId="2767" CreationDate="2011-01-13T17:38:44.997" Score="2" ViewCount="" Body="&lt;p&gt;This is an interesting question. I don't know of any definite answer, but can point you to some (speculative) work done along these lines by Raymond Chiao. His arguments are quite fascinating and if they do end up working the implications would be breathtaking to say the least.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are links to two of his papers &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0601193&quot; rel=&quot;nofollow&quot;&gt;The Interface between Quantum Mechanics and General Relativity&lt;/a&gt; and &lt;a href=&quot;http://arxiv.org/abs/0903.3280&quot; rel=&quot;nofollow&quot;&gt;Laboratory-Scale Superconducting Mirrors for Gravitational Microwaves&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His basic argument is that the equivalence principle is a statement about &lt;em&gt;classical&lt;/em&gt; objects and must be modified when looking at the dynamics of quantum objects such as the cooper pairs in a BEC. He suggests, that the interaction between gravitational wave and a superconducting surface they are incident on can be amplified by a factor of $10^{42}$ due to this modification. Normally any such effects would be suppressed by the ratio of the strengths of the gravitational to the electromagnetic force ( $ \sim 10^{-42} $). However, due to the anomalous behavior of the cooper pairs of a superconductor this factor can in effect become of $\mathcal{O}(1)$. I leave it up to you to read through his papers and determine if his arguments are physically sound.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-13T17:38:44.997" CommentCount="1" />
  <row Id="2785" PostTypeId="1" CreationDate="2011-01-13T18:03:12.287" Score="5" ViewCount="98" Body="&lt;p&gt;Recently I started to play with some massive Gaussian models on a lattice. Motivation being that I work on massless models and want to understand the massive case because it seems easier to handle (e.g. cluster expansion makes sense thanks to contours carrying mass). Consider a Hamiltonian &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H(x_{\Lambda}) = \sum_{&amp;lt;i,j&amp;gt; \in E(\Lambda)} (x_i - x_j)^2 +\sum_{i \in \Lambda} U(x_i) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\Lambda$ a lattice (think mainly of ${\mathbb Z}^d$ with $d = 2$), $E(\Lambda)$ the set of its edges, $x_{\Lambda} \in (\Lambda \to {\mathbb R})$ and $U(y)$ a potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know what is known about this class of models in general. E.g. for a related class of models without mass but with arbitrary two-point function &lt;a href=&quot;http://www.springerlink.com/content/clxnha7e9r7mw8fq/&quot; rel=&quot;nofollow&quot;&gt;Funaki and Spohn&lt;/a&gt; showed that there is no phase transition if that function is convex.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1.&amp;nbsp;I wonder whether similar result is known for convex (replace with any other reasonable condition) $U$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Similarly, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2.&amp;nbsp;Is there a necessary condition on $U$ for there to be a phase transition? Try to give some examples.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;E.g. it seems natural that there will be a transition for a symmetric double-well model (with spontaneous symmetry breaking at low temperatures) and one could also investigate the case with one of the wells being more favorable. I think I should be able to prove these things with some work but I suppose it's already been done by someone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3.&amp;nbsp;Could you point me to a reference about some double-well models?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other kind of model I've been thinking about is (setting $\beta = 1$ and letting $p$ play the role of temperature)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\exp(-U(y)) = p \exp (-a y^2) + (1-p) \exp (-b y^2)$$&#xA;with $a$ suitably small and $b$ suitably large. This is a big well with a smaller well inside. Intuitively the system should sit in the small well at low temperatures ($p = 0$) and jump out and behave freely (with small mass $a$) at higher temperatures ($p = 1$), so this is a toy model of melting. Trouble is, I have no idea whether this really works and I can't decide whether there will or won't be a phase transition.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4.&amp;nbsp;Any ideas about this model? Pointing me to a reference would be wonderful but I am not sure this has been studied before.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-01-13T18:10:03.907" LastActivityDate="2011-01-13T18:10:03.907" Title="What is known about some massive Gaussian models on a lattice?" Tags="&lt;mathematical-physics&gt;&lt;statistical-mechanics&gt;&lt;condensed-matter&gt;&lt;lattice-model&gt;" />
  <row Id="2786" PostTypeId="1" AcceptedAnswerId="2814" CreationDate="2011-01-13T18:31:25.270" Score="4" ViewCount="562" Body="&lt;p&gt;Visualizations of hydrogen's wavefunctions / electron orbitals are abound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could not however locate a visualization of the wavefunction of a proton. The reason I was looking for one is to see whether the three quarks that make it up &quot;occupy&quot; disjoint regions of space (i.e. the maximum probability locations are separated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please point to one such visualization, or explain why they can't be created (or why my question is nonsensical, which is always a possibility...)&lt;/p&gt;&#xA;" OwnerUserId="786" LastActivityDate="2011-01-19T19:21:23.803" Title="Visualization of proton's wavefunction" Tags="&lt;quantum-mechanics&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" />
  <row Id="2787" PostTypeId="1" AcceptedAnswerId="2788" CreationDate="2011-01-13T18:34:52.717" Score="1" ViewCount="180" Body="&lt;p&gt;Supposed I have an position vector $$\vec{r}=\begin{pmatrix}&#xD;&#xA;10.0 &amp;amp; -30.0 &amp;amp; 25.0\end{pmatrix}$$ expressed in $\mathrm{millimeters}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the correct notation to display $\vec{r}$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\begin{pmatrix}&#xD;&#xA;10.0 &amp;amp; -30.0 &amp;amp; 25.0\end{pmatrix}\text{ mm}$&lt;/li&gt;&#xA;&lt;li&gt;$\begin{pmatrix}&#xD;&#xA;10.0\text{ mm} &amp;amp; -30.0\text{ mm} &amp;amp; 25.0\text{ mm}\end{pmatrix}$&lt;/li&gt;&#xA;&lt;li&gt;$\begin{pmatrix}&#xD;&#xA;10.0 &amp;amp; -30.0 &amp;amp; 25.0\end{pmatrix}\text{ in }\mathrm{mm}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If the answer is &lt;strong&gt;2.&lt;/strong&gt; then why add all those redundunt units when all elements of a vector have to be of the same unit. If you have a long list of values then usualy you present this a table with the units in the header (and not on each element). What if the units are complex (with powers and fraction), do we really have to write them out for each element?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you consicely write out a vector while describing the units those values are in also at the same time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. I did not post this in the &lt;code&gt;Math SE&lt;/code&gt; because they have never heard of units :-) and only physics deals with real situations.&lt;/p&gt;&#xA;" OwnerUserId="392" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2011-01-13T19:28:53.030" LastActivityDate="2011-01-13T21:49:24.260" Title="How to correctly show units with a vector" Tags="&lt;soft-question&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="2788" PostTypeId="2" ParentId="2787" CreationDate="2011-01-13T18:43:47.080" Score="5" ViewCount="" Body="&lt;p&gt;I would say 1. and 2. are correct. In the first you are multiplying by the scalar 1 mm.&lt;br&gt;&#xA;Also elements of vectors dont need to have the same units.&lt;br&gt;&#xA;Just consider the 4-vector (3 sec, 1 µm, 82 ly, 5.6 parsec)&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2011-01-13T18:54:53.510" LastActivityDate="2011-01-13T18:54:53.510" CommentCount="4" />
  <row Id="2789" PostTypeId="2" ParentId="2787" CreationDate="2011-01-13T19:11:04.357" Score="0" ViewCount="" Body="&lt;p&gt;I write it as 10i + 20j + 5k and set the scale somewhere before the calculations. Of course, you could add it after the vector, but that just doesn't look neat and tidy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could write the unit vector on the side and multiply the magnitude (with correct units) into it. That could also clarify things. I guess it just depends on how you grasp things and what's clearer to you. As long as you keep others in the loop then it should be okay.&lt;/p&gt;&#xA;" OwnerUserId="1065" LastActivityDate="2011-01-13T19:11:04.357" CommentCount="2" />
  <row Id="2790" PostTypeId="2" ParentId="2786" CreationDate="2011-01-13T19:21:49.643" Score="0" ViewCount="" Body="&lt;p&gt;At a very naive level your question amounts to asking what the bound state for three bodies should look like. I would suggest looking up &quot;Efimov trimers&quot;. These are many-body bound states predicted by Efimov some 40 years ago and recently experimentally observed. I'm not suggesting that their is a one-to-one correspondence between Efimov states and bound states of nucleons, but they should at the very least provide some sort of a picture. &lt;/p&gt;&#xA;" LastActivityDate="2011-01-13T19:21:49.643" />
  <row Id="2791" PostTypeId="1" CreationDate="2011-01-13T19:31:23.310" Score="4" ViewCount="276" Body="&lt;p&gt;Consider the following experiment: a double-slit set-up for firing electrons one at a time. Let's now add a second electron (orange), which is fired parallel to the first one, but in the opposite direction, and so as to pass closer to one slit than the other, and, importantly, above the plane in which interference occurs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/RCGXh.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The firing of the two electrons is timed so that the orange electron is &quot;closest&quot; to the purple electron somewhere around point P, i.e. after the latter has passed through the two slits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably the electrons will repel due to having charge. Presumably an interference pattern will still be produced by the purple electron if the experiment is repeated thousands of times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is, how exactly does the orange electron get deflected? If the experiment is repeated, would it always hit the same point on the opposite wall, or would it also start producing a sort of an interference pattern?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Thanks to user:Timwi for the diagram!&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="786" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-03-13T20:16:50.740" LastActivityDate="2011-03-13T20:16:50.740" Title="How do electrons interact if one of them had just exited the two slits of the double-slit experiment?" Tags="&lt;quantum-mechanics&gt;&lt;experimental-physics&gt;&lt;quantum-electrodynamics&gt;&lt;entanglement&gt;&lt;double-slit-experiment&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="2792" PostTypeId="1" CreationDate="2011-01-13T19:36:08.720" Score="2" ViewCount="213" Body="&lt;p&gt;I've been toying with the idea of making a 3D scanner that uses an &lt;a href=&quot;http://www.acroname.com/robotics/info/articles/sharp/sharp.html&quot; rel=&quot;nofollow&quot;&gt;IR distance sensor&lt;/a&gt; to find position vectors of an object in space and then translates that into a 3D computer model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the greatest challenges in this case is optical phenomena. If the surface has holes, or dimples that reflect the IR beam or cause diffraction, then getting an accurate reading would be nearly impossible, but that's an engineering problem. It would require several different passes from different angles to figure out what that real distance actually is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I would like to understand is there a way to use these &quot;glitches&quot; in the data to model the phenomena itself? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that this would make for a great experiment, but I don't know where to start and what to study. Essentially, I would like to derive from the data a theory of what interference was observed from the sensors and then model it to understand what it means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how should I try to;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Setting up the experiment&lt;/li&gt;&#xA;&lt;li&gt;Setting controls to verify the data&lt;/li&gt;&#xA;&lt;li&gt;Model the data&lt;/li&gt;&#xA;&lt;li&gt;Fabricate a theory to fit the observation&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; What I want to do is to create an experiment that uses these mistakes in data (weird readings upon hitting distorted surfaces) and create a model that shows what optical phenomena are at play over here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, you could say why not just see the surface and judge? That wouldn't be fun, now, would it? On a more serious note, what I am trying to do is just make my own experiment to actually measure whatever optical phenomena is at play. Explain it and perhaps derive an expression for it from the data itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I want to look at the data set correspond it with the physical model and see the variation in the distance reading vs. what it should have been and the surface and try to model why that happened with the 3D diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I am someone who just got out of high school. What do I learn and how do I go about actually doing this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I use optical phenomena as a sort of place holder for reflection, refraction, diffraction or whatever combination of these is at play. Sorry, for the confusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 2:&lt;/strong&gt; I guess the question title is misleading and so is the text. I changed / rephrased it to get more participation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, I know that it's hard to give a perfect answer to my question, but I would love to discuss this with someone and figure out what to learn and how. So, please go ahead take a shot. I would love to hear what you have on your mind.&lt;/p&gt;&#xA;" OwnerUserId="1065" LastEditorUserId="1065" LastEditorDisplayName="" LastEditDate="2011-01-14T18:17:01.277" LastActivityDate="2011-08-13T21:23:39.847" Title="How would you design an experiment to grok optical phenomena?" Tags="&lt;optics&gt;&lt;experimental-physics&gt;&lt;home-experiment&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="2793" PostTypeId="1" AcceptedAnswerId="2795" CreationDate="2011-01-13T19:36:13.223" Score="1" ViewCount="200" Body="&lt;p&gt;If you said someone had a velocity of -12mph and they were traveling north? Wouldn't it mean that they were traveling 12mph south?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a quote from &lt;a href=&quot;http://galileoandeinstein.physics.virginia.edu/lectures/momentum.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if something [object-x] moving to the right was taken to have positive momentum, then one should consider something [object-y] moving to the left to have negative momentum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But isn't Momentum a vector, so the direction should be specified separately to the number. What I mean is, object-y with -1200kg m/s and object-x with 1200kg m/s should both have momentum in the same direction. But this can't be because as the quote says they are moving in opposite direction.&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="4" LastEditorDisplayName="" LastEditDate="2011-01-14T01:29:10.717" LastActivityDate="2011-01-14T01:29:10.717" Title="negative vectors (eg velocity)" Tags="&lt;motion&gt;&lt;vectors&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="2794" PostTypeId="2" ParentId="2793" CreationDate="2011-01-13T19:46:39.473" Score="0" ViewCount="" Body="&lt;p&gt;In some sense you're right when you say &quot;the direction should be specified separately to the number&quot;. However you have to be careful about what you mean--if you say this then by 'number' you mean an unsigned quantity (think absolute value). When you start talking about the signs of numbers, either positive or negative, then that sign encodes information about the direction that something is moving. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note I have to define what 'positive' means and what 'negative' means, in the sense that positive doesn't always mean 'to the right'. It could mean 'to the left' as well. However once I make my choice I've set the meaning of positive and negative, and from then on 'positive' vectors point in that specific direction.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-13T19:46:39.473" CommentCount="1" />
  <row Id="2795" PostTypeId="2" ParentId="2793" CreationDate="2011-01-13T19:48:55.703" Score="3" ViewCount="" Body="&lt;p&gt;That quote is abit misleading, momentum is a vector, however a vector is neither negative nor positive, only its components can have this characteristic. The two objects you are describing does not have the same momentum, but they have the same &lt;em&gt;magnitdue&lt;/em&gt; of momentum (length of vector).&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-01-13T19:48:55.703" CommentCount="6" />
  <row Id="2796" PostTypeId="1" CreationDate="2011-01-13T19:55:34.037" Score="5" ViewCount="452" Body="&lt;p&gt;How can an electron distinguish between another electron and a positron?  They use photons as exchange particles and photons are neutral, so how does it know to repel or attract? &lt;/p&gt;&#xA;" LastActivityDate="2011-01-14T16:01:53.530" Title="How does charge work if photons are neutral?" Tags="&lt;electromagnetism&gt;&lt;quantum-electrodynamics&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="2797" PostTypeId="2" ParentId="2774" CreationDate="2011-01-13T20:00:44.470" Score="7" ViewCount="" Body="&lt;p&gt;There is a real object with relativistic speed of surface - &lt;a href=&quot;http://en.wikipedia.org/wiki/Millisecond_pulsar&quot; rel=&quot;nofollow&quot;&gt;millisecond pulsar&lt;/a&gt;. The swiftest spinning pulsar currently known, spinning 716 times a second. Surface speed of such pulsar with radius 16 km is about $7*10^7$ m/s or  24% speed of light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is calculated that pulsars would break apart if they spun at a rate of  more than 1500 rotations per second.&lt;/p&gt;&#xA;" OwnerUserId="238" LastActivityDate="2011-01-13T20:00:44.470" CommentCount="1" />
  <row Id="2798" PostTypeId="2" ParentId="1626" CreationDate="2011-01-13T20:04:13.580" Score="0" ViewCount="" Body="&lt;p&gt;I have a similar problem with a garage door (backdoor, not the car entrance). Here in California winters can be vert wet, while summers are bone dry. I've spent hours shaving wood off the edges, to keep it from getting too bad.&#xA;  Stresses and strains on other parts of the structure might also be to blame.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-01-13T20:04:13.580" />
  <row Id="2799" PostTypeId="2" ParentId="1626" CreationDate="2011-01-13T20:11:38.720" Score="0" ViewCount="" Body="&lt;p&gt;It's actually just that in winters or around that time water doesn't evaporate that fast. So, the cellulose in the wood expands due to the added water content and swells until it doesn't fit the door frame. (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's really interesting just how beautifully everything plays out in nature. If you think about it that property probably saved a lot of trees from dying. If wood expands in moisture that means the tree will be taller in just the right conditions. It contracts that means the tree will be shorter in just dry conditions. That, probably, reduces the surface area of evaporation and helps the tree to survive. I might be wrong, but it's mind boggling just how awesome nature is. She rocks. :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Cellulose is a polymer and I guess the expansion might have something to do with hydrogen bonding interacting with it's structure.&lt;/p&gt;&#xA;" OwnerUserId="1065" LastActivityDate="2011-01-13T20:11:38.720" />
  <row Id="2800" PostTypeId="2" ParentId="2796" CreationDate="2011-01-13T20:23:03.383" Score="3" ViewCount="" Body="&lt;p&gt;Well, one quick answer, if you want to answer this at the level of QED, is that there are two Feynman diagrams (one where an electron scatters off of a photon, which then scatters off of a positron; another where the electron and positron annihilate, and then the photon decays into an electron and positron) describing $e^{+} + e^{-} \rightarrow e^{+} + e^{-}$, while there is only one that describes $e^{-} + e^{-} \rightarrow e^{-} + e^{-}$ (electron scatters off of a photon, which scatters off of another electron).  The short of it is that you can't consider just QED vertices--you have to look at entire Feynman diagrams of processes, and at least know how many of them there are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, another answer is that, simply, the classical theory distinguishes these two processes--Maxwell's equations plus the Lorentz force law tell me that one process is attractive, and the other is repulsive.  Naïvely, one would expect that the quantum process, by land large, would mirror the quantum process.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-01-13T20:23:03.383" />
  <row Id="2801" PostTypeId="2" ParentId="2767" CreationDate="2011-01-13T20:33:00.557" Score="0" ViewCount="" Body="&lt;p&gt;You should read &lt;a href=&quot;http://en.wikipedia.org/wiki/The_First_Men_in_the_Moon&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1123" LastActivityDate="2011-01-13T20:33:00.557" />
  <row Id="2802" PostTypeId="2" ParentId="1906" CreationDate="2011-01-13T20:53:54.687" Score="2" ViewCount="" Body="&lt;p&gt;The first formula indeed follows from the second formula if we let $\omega\to0$. To see that, expand the fractions as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac1{\pm\hbar\omega + E^a - E^b} = \frac1{E^a-E^b}\left(1 \mp \frac{\hbar\omega}{E^a-E^b}\right) + \mathcal O(\omega^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to obtain $\sigma_{xy} = \sigma^1 + \sigma^2$ as the sum of a potentially divergent term&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sigma^1 = \frac{-ie^2}{V\omega} \sum_{a,b} f(E^a) \frac{\langle a|v_x|b \rangle \langle b|v_y|a \rangle + \langle a|v_y|b \rangle \langle b|v_x|a \rangle}{E^a - E^b} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and a term that looks like the first formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sigma^2 = \frac{-ie^2\hbar}{V} \sum_{a,b} f(E^a) \frac{- \langle a|v_x|b \rangle \langle b|v_y|a \rangle + \langle a|v_y|b \rangle \langle b|v_x|a \rangle}{(E^a - E^b)^2} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;To see that the &lt;strong&gt;first term&lt;/strong&gt; vanishes instead of diverging, we have to use the Heisenberg equation of motion $v_x = \frac{d}{dt}x = [H_0,x]$ which gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \langle a | v_x | b \rangle = \langle a | H_0 x - x H_0 | b \rangle = (E^a-E^b) \langle a | x | b \rangle $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \langle a|v_x|b \rangle \langle b|v_y|a \rangle + \langle a|v_y|b \rangle \langle b|v_x|a \rangle = (E^a-E^b) (\langle a|x|b \rangle \langle b|v_y|a \rangle - \langle a|v_y|b \rangle \langle b|x|a \rangle) .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The factors $(E^b-E^b)$ cancel and the remaining sum over $b$ becomes a sum over the identity $\sum_b |b\rangle\langle b| = 1$. Thus, we arrive at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sigma^1 = \frac{-ie^2}{V\omega} \sum_{a,b} f(E^a) \left(\langle a|xv_y - v_yx |a \rangle \right) = 0 .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since the commutator $[x,v_y]$ vanishes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;To see that the &lt;strong&gt;second term&lt;/strong&gt; is correct, we have to get the summation indices right. To do that, we have to rearrange the summation to obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sigma^2 = \frac{ie^2\hbar}{V} \sum_{a,b} (f(E^a)-f(E^b))\frac{\langle a|v_x|b \rangle \langle b|v_y|a \rangle}{(E^a - E^b)^2} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the limit $T\to0$, the difference of Fermi-Dirac distributions $f(E^a)-f(E^b)$ will be equal to&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$1$ if $E^a &amp;lt; E_F &amp;lt; E^b$&lt;/li&gt;&#xA;&lt;li&gt;$-1$ if $E^b &amp;lt; E_F &amp;lt; E^a$&lt;/li&gt;&#xA;&lt;li&gt;$0$ otherwise&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Using this and rearranging the summation again gives the Kubo formula in the first form.&lt;/p&gt;&#xA;" OwnerUserId="188" LastEditorUserId="188" LastEditorDisplayName="" LastEditDate="2011-01-13T21:02:52.430" LastActivityDate="2011-01-13T21:02:52.430" />
  <row Id="2803" PostTypeId="2" ParentId="2796" CreationDate="2011-01-13T20:58:19.443" Score="2" ViewCount="" Body="&lt;p&gt;It's a good question and it is more convenient to describe the process by the Feynman diagrams in the position representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one begins with the initial state that contains two charged particles whose wave functions are two wave packets separated by distance R, then the usual tree-level diagram involving the exchange of a photon quantifies the (small) change of the wave function by the (weak) electric field, in the position representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that the initial wave function is positive and real (which is possible for zero-velocity particles). The Feynman diagram (perturbatively) quantifies the change of the wave function. One may imagine, at some approximation, that it changes the two single-particle wave functions separately. Or you may imagine that one of the particles is superheavy and only the lighter particle's motion may be affected by the interaction (and the Feynman diagram).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The change of the lighter particle wave function, as evaluated from the Feynman diagram, vanishes at the center of the original wave packet. It has the opposite sign on the side closer to the other particle than on the opposite side. However, the overall sign of the Feynman diagram matters, too. In one case, the change of the wave function reduces the probability that the two particles are closer, and enhance the probability that they're distant - that's the repulsive interaction; for the opposite sign of the Feynman diagram, it's the other way around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, I cheated a little bit: the first change of the wave function has an &quot;i&quot; in it, and reduces the original zero-velocity wave function to a wave function of a particle with a small velocity; its sign will depend on the charges, just like in the previous paragraph. The previous paragraph is easier to talked about, so let me continue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Feynman diagram for 2 particles contains two electromagnetic vertices and each of them brings a multiplicative factor of &quot;Q&quot; - the electric charge of the corresponding particle. That's how the Feynman rules work. The Feynman diagram itself is a function of the positions (or momenta) that is therefore proportional to &quot;Q1.Q2&quot;. If that's positive, it means that the change of the wave function is positive - i.e. has the same sign as the initial wave function - for separations greater than the original one. That's how one gets the repulsive force because the final particles are further in average. If &quot;Q1.Q2&quot; is negative, one gets the attractive force for the same reason with the opposite sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, the Feynman diagram is a function proportional to &quot;Q1.Q2&quot; whose sign determines the interaction's being attractive of repulsive because it decides about the relative sign of the initial wave function and its change, and the change of the momentum or position of the initial particles is encoded in this relative sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To imagine that the intermediate photon is literally a &quot;ping pong ball&quot; that can only cause a repulsive interaction would be way too simplistic - especially because such an idea neglects the quantum mechanical fact that the positions of particles are encoded in wave functions. But in fact, one may explain the repulsive and attractive interactions in this picture, too. It's because assuming a positive energy, the momentum carried by the &quot;ping pong ball&quot; photon is &quot;negative&quot; for an attractive interaction.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-13T20:58:19.443" />
  <row Id="2805" PostTypeId="2" ParentId="2791" CreationDate="2011-01-13T21:13:57.220" Score="5" ViewCount="" Body="&lt;p&gt;This is a typical trade-off between the position properties of the purple electron and its wave properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the orange electron is sufficiently far so that it doesn't influence the purple electron much, everything will continue as before: the purple electron will produce an interference pattern while the orange one will draw one point on the photographic plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you may try to bring the orange electron closer. As you're bringing it closer, you increase the likelihood that it is repelled by the purple electron. The deflection is pretty much dictated by the vertical position of the purple electron on the picture, so the orange electron is de facto measuring this position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If most of the repulsive interaction occurs near the slits, the orange electron's motion is dominated by the which-slit information about the purple electron. The more accurately you measure it, by looking at the orange electron's path, the more you will disturb the interference pattern for the purple electron. The more you can read some &quot;position information&quot; from the orange electron's path, the less sharp will be the purple electron's interference pattern, and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to emphasize that all such problems can be exactly calculated - just use the proper quantum mechanics for two particles. The wave function - in the non-relativistic picture - is psi(x1,y1,z1,x2,y2,z2,t) where 1,2 are labels of the two electrons. This wave function is antisymmetric under the exchange of x,y,z for 1 and 2 and it evolves according to a Schrodinger equation for 2 particles and its squared absolute value knows about the probability that both electrons will be observed at any pair of places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One also has to be careful because the electrons are indistinguishable, so that if they get really close so that they could possibly get exchanged, one has to subtract (because of the Fermi statistics) the amplitudes in which they exchanged and one in which they have not, to keep their wave function antisymmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also want to emphasize that the electrostatic repulsion between two electrons in such experiments is tiny and should be more properly described as a &quot;quantized&quot; exchange of a single photon. Such a repulsive interaction always allows the possibility that additional photons are emitted (radiation from accelerating charges), and these extra photons may reduce the overall sharpness of the interference pattern further.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-13T21:13:57.220" CommentCount="3" />
  <row Id="2806" PostTypeId="2" ParentId="937" CreationDate="2011-01-13T21:27:06.323" Score="3" ViewCount="" Body="&lt;p&gt;I think it's helpful to think about the related question of how the &lt;em&gt;electric field&lt;/em&gt; gets out of a charged black hole.  That question came up in the (now-defunct) Q&amp;amp;A section of the &lt;em&gt;American Journal of Physics&lt;/em&gt; back in the 1990s.  Matt McIrvin and I wrote up an answer that was published in the journal.  You can see it at &lt;a href=&quot;https://facultystaff.richmond.edu/~ebunn/ajpans/ajpans.html&quot; rel=&quot;nofollow&quot;&gt;https://facultystaff.richmond.edu/~ebunn/ajpans/ajpans.html&lt;/a&gt; . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As others have pointed out, it's easier to think about the question in purely classical terms (avoiding any mention of photons or gravitons), although in the case of the electric field of a charged black hole the question is perfectly well-posed even in quantum terms: we don't have a theory of quantum gravity at the moment, but we do think we understand quantum electrodynamics in curved spacetime.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-01-13T21:27:06.323" />
  <row Id="2807" PostTypeId="2" ParentId="2787" CreationDate="2011-01-13T21:36:10.790" Score="1" ViewCount="" Body="&lt;p&gt;Both 1 and 2 are perfectly fine notation.  As someone who teaches introductory physics a lot, I would use either in class and accept either as correct from a student.  #3 expresses the meaning perfectly clearly but is not the usual form of expression.  I would certainly have no problem if a student wrote that in my class, for instance.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-01-13T21:36:10.790" />
  <row Id="2808" PostTypeId="1" AcceptedAnswerId="2812" CreationDate="2011-01-13T21:44:32.177" Score="1" ViewCount="121" Body="&lt;p&gt;I need the general expression for the lagrangian density of a linear elastic solid. I haven't been able to find this anywhere. Thanks.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-13T22:04:49.473" Title="Lagrangian density of linear elastic solid" Tags="&lt;mechanics&gt;&lt;lagrangian-mechanics&gt;&lt;elasticity&gt;" AnswerCount="1" />
  <row Id="2809" PostTypeId="2" ParentId="2767" CreationDate="2011-01-13T21:44:37.030" Score="16" ViewCount="" Body="&lt;p&gt;In a consistent theory of gravity, there can't exist any objects that can shield the gravitational field in the same way as conductors shield the electric field. It follows from the positive-energy theorems and/or energy conditions (roughly saying that the energy density cannot be negative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see why, just use the conductor to shield an ordinary electric field - which is what your problem reduces to temporarily for very low frequencies of the electromagnetic waves. The basic courses of electromagnetism allow one to calculate the electric field of a point-like charged source and a planar conductor: the electric field is identical to the original charge plus a &quot;mirror charge&quot; on the opposite side from the conductor's boundary. Importantly, the mirror charge has the opposite sign. In this way, one may guarantee that the electric field is transverse to the plane of the conductor. This fact makes the electromagnetic waves bounce off the mirror if you consider time-dependent fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to do an &quot;analogous&quot; thing for gravity, you would first have to decide what boundary conditions you want to be imposed for the gravitational field by the mysterious new object - what is your gravitational analogy of &quot;$\mathbf E$ is orthogonal to the conductor&quot;. The metric field has many more components. Most of them will require you to consider a negative &quot;mirror mass&quot; - but the mass in a region can't be negative, otherwise the vacuum would be unstable (one could produce regions of negative and positive energy in pairs out of vacuum, without violating any conservation laws).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Microscopically, one may also see why there can't be any counterpart of the conductor. The conductor allows to change the electric field discontinuously because it can support charges distributed over the boundary. The profile of the charge density goes like $\sigma\delta(z)$ if the conductor boundary sits at $z=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you would need a similar singular mass distribution to construct a gravitational counterpart. If the distribution failed to be positively definite, it would violate the positive-energy theorem or energy conditions, if you wish. If it were positively definite, it would create a huge gravitational field. Locally, the boundary of the gravitational &quot;conductor&quot; would have to look like an event horizon. But we know that the event horizons cannot shield the interior from the gravitational waves - the waves as well as everything else falls inside the black hole. Moreover, the price for such &quot;non-shielding&quot; is that you have to be killed by the singularity in a finite proper time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One may probably write some formal solutions that have some properties but they can't really work when all the behavior of gravity and the related consistency conditions are taken into account. One can't fundamentally construct a &quot;gravitational conductor&quot; because gravity means that the space itself remains dynamical and this fact can't be undone. In particular, in a consistent theory of gravity - in string/M-theory - you won't find any objects generalizing conductors to gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, there is one object that comes close to a &quot;gravitational conductor&quot; in M-theory - the Hořava-Witten domain wall, a possible boundary of the 11-dimensional spacetime of M-theory. It can be placed at $x_{10}=0$ and about $1/2$ of the components of the metric tensor become unphysical near this domain wall (boundary of the world that carries the $E_8$ gauge multiplet) because the domain wall also acts as an orientifold plane. But such an orientifold plane is not just some object (like a conductor) that is &quot;inserted&quot; into a pre-existing space that doesn't change. Quite on the contrary, the character of the underlying spacetime is changed qualitatively: the world behind the orientifold plane is &lt;em&gt;literally&lt;/em&gt; just a mirror copy of the world in front of the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is a lot of fundamentally incorrect thinking about gravity in all the answers that try to say &quot;Yes&quot;. Gravity is not another force that is inserted into a pre-existing geometrical space; gravity is the curvature and dynamics of the spacetime itself. Once we say that the space is dynamical, we can't find any objects that would &quot;undo&quot; this fundamental assumption of general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also some more detailed confusions in the other solutions. First, a white hole corresponds to a time-reversed black hole with the same (positive) mass but it is an unphysical time reversal because it violates the second law of thermodynamics: entropy has to increase which means that the (large entropy) black hole can be formed, but it cannot be &quot;unformed&quot;. But a white hole, which is forbidden thermodynamically (and microscopically, it corresponds to the same microstates as black hole microstates, they just never behave in any &quot;white hole&quot; way), is still something else than a negative-mass black hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A negative mass black hole isn't related to a positive-mass black hole by any &quot;reflection&quot;. It is a solution without horizons, with a naked singularity, and can't occur in a consistent theory of gravity because it would cause instability of the vacuum. (Also, naked singularities can't be &quot;produced&quot; by any generic evolution in 3+1 dimensions because of Penrose's Cosmic Censorship Conjecture.) So white holes and negative-mass black holes are forbidden for different reasons. However, even if you considered them, it would still fail to be enough to create a &quot;gravitational conductor&quot; which is nothing else than the denial of the fact that the geometry of the spacetime is dynamical, and one can't freely construct infinite, delta-function-like mass densities without completely changing the shape of spacetime.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-01-13T23:04:18.523" LastActivityDate="2011-01-13T23:04:18.523" CommentCount="5" />
  <row Id="2810" PostTypeId="2" ParentId="2787" CreationDate="2011-01-13T21:49:24.260" Score="1" ViewCount="" Body="&lt;p&gt;The correct notation is that of multiplying the whole thing by a unit (as a multiplication constant):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;(a,b,c)*mm&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2011-01-13T21:49:24.260" CommentCount="2" />
  <row Id="2811" PostTypeId="2" ParentId="1019" CreationDate="2011-01-13T21:52:41.073" Score="1" ViewCount="" Body="&lt;p&gt;How about this - that parity is one of the most fundamental symmetries (which is not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the experiment(s) showing parity violation. However, I guess what I still don't understand is the following, rotational symmetry is fundamental. And parity is just rotating 180 degrees. So if we say that all rotational symmetry is fundametal, why would a subset (rotating with a very specific angle - 180 degrees) not being fundamental? Isn't that self contradicting?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-13T21:52:41.073" CommunityOwnedDate="2011-01-13T21:52:41.073" CommentCount="1" />
  <row Id="2812" PostTypeId="2" ParentId="2808" CreationDate="2011-01-13T22:04:49.473" Score="1" ViewCount="" Body="&lt;p&gt;the Lagrangian density is $L=T-U$, the difference between the kinetic and potential energy density, as we're used to in all of mechanics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinetic energy density is $T = \rho v(x,y,z)^2/2$ where one has to calculate the density $\rho$ properly. The potential energy is more general and complicated,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U = \frac{1}{2} C_{ijkl}u_{ij}u_{kl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the tensor $C$ with four indices is called the elastic. The tensors $u$ are obtained from the displacement vectors as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ u_{ij} = \frac{1}{2} (\partial_i u_j + \partial_j u_i) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the density $\rho$ in the kinetic energy also depends on the displacement vectors, if you want to re-express it via the mass density at rest (without displacement).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For isotropic materials, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ C_{ijkl} = \lambda \delta_{ij} \delta_{kl} + \mu (\delta_{ik} \delta_{jl} + \delta_{il} \delta_{jk} ) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\lambda$ and $\mu$ are known as Lamé constants. For more general (but linear) crystals, however, $C$ is a general tensor symmetric under the $ij$ exchange, $kl$ exchange, and the exchange of $ij$ and $kl$ as pairs.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-13T22:04:49.473" CommentCount="1" />
  <row Id="2813" PostTypeId="2" ParentId="2264" CreationDate="2011-01-13T22:14:11.497" Score="1" ViewCount="" Body="&lt;p&gt;The usual derivation of the radiation multipoles involves writing the potentials as an integral over the source distribution, and then doing a Taylor expansion in the source coordinate.  In Griffiths's notation, we're evaluating &lt;strong&gt;A&lt;/strong&gt;( &lt;strong&gt;r&lt;/strong&gt; ) by performing an integral over source coordinates &lt;strong&gt;r&lt;/strong&gt;', and the Taylor expansion is in powers of the small quantity &lt;em&gt;r'&lt;/em&gt; / &lt;em&gt;r&lt;/em&gt;.  Electric dipole shows up at first order in this Taylor expansion, and both magnetic dipole and electric quadrupole show up at second order.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-01-13T22:14:11.497" />
  <row Id="2814" PostTypeId="2" ParentId="2786" CreationDate="2011-01-13T22:22:34.440" Score="3" ViewCount="" Body="&lt;p&gt;for the electron in the Hydrogen atom, the orbital motion doesn't interact with the electron's spin, so &quot;the wave function&quot; pretty much means just a complex $\psi(x,y,z)$. You may choose the electron's spin to be up or down independently of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you must realize that a wave function of three quarks has many more components. First of all, there are three particles in relative motion rather than two. It means that even if you decouple the center-of-mass coordinates, the relative wave function is a function of six coordinates, $x_1,y_1,z_1,x_2,y_2,z_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can't really visualize the full wave function in a simple way because it is a function of six variables rather than three and of course, it doesn't factorize into a product of functions of three variables (something that is strictly speaking true even for all atoms more complicated than the Hydrogen atom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, you must understand that the quarks have many discrete degrees of freedom that are not decoupled from the orbital motion in this case. Each of the three quarks has 1 of 3 colors and 1 of 2 possible values of the spin. The combinations of the color really force you to consider 27 combinations of the quark colors - 27 different wave functions (only 6 of them are nonzero and equal to each other, up to a sign) - and 8 combinations of the three quarks' spin. The spins are correlated as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the right wave function for 3 quarks is really a set of $27 \times 8$ complex functions of six variables. Of course, you may visualize various aspects of these functions by integrating it in space and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important that the strong force between the non-relativistic quarks is spin-dependent. That's why the protons' spin is not arbitrary - it equals $1/2$ instead. So one of the quarks' spin differs from the remaining two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you managed to invent ways how to visualize the $27\times 8$ wave functions of 6 variables describing the relative positions of the three quarks, with the best correlations between the colors, spins, and motion, it would still be a hugely oversimplified model of the proton. In reality, it is not true that the proton only contains 3 quarks. Those 3 quarks we normally talk about are analogous to &quot;valence electrons&quot; in an atom but there is also a large see of equally real gluons and quark-antiquark pairs - additional partons whose total color vanishes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To really describe the wave function of the proton, you need to talk about the right theory of the proton's structure - quantum chromodynamics (QCD) - which is an example of a quantum field theory - one discovered in the early 1970s. It has infinitely many degrees of freedom, like any field theory, so the right &quot;wave function&quot; is really a &quot;wave functional&quot; or a function of infinitely many variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must realize that many approximations valid in atomic physics break down. For example, the speed of electrons in the atoms is very small - effectively controlled by the fine-structure constant $1/137$ that tells you the speed in the units of the speed of light. That allows you to use non-relativistic quantum mechanics. However, the proton's &quot;strong&quot; fine-structure constant relevant for the &quot;three quarks only&quot; is close to one, so the speed of quarks in the proton is always comparable to the speed of light. Consequently, the extra relativistic energy/mass is comparable to the rest mass as well as the interaction energy, and the proton always has enough energy to create quark-antiquark pairs and so on. It's a mess where relativity is needed much like particle creation and annihilation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some sense, it's true that &quot;the 3 quarks&quot;, if you select them from the sea of the infinitely many partons, like to occupy three different regions in the ground state. But to be more accurate about what this statement actually means for the wave function(al) of the actual proton, you would have to go through a series of so many approximations and idealizations that it's not worth it. The goal of quantum field theory is not to visualize the structure of something; the goal is to calculate the results of the experiments and one can't really design good experiments that would probe the detailed shape of the wave function of the proton. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parton distribution functions are the closest observables to this information. However, their very assumption is that the proton has infinitely many rather than 3 partons (quarks, antiquarks, or gluons). To summarize, don't expect anything that is as valid yet as useful as the pictures for the atomic physics. Proton's analogy to an atom has severe limitations.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-13T22:22:34.440" CommentCount="4" />
  <row Id="2815" PostTypeId="2" ParentId="2793" CreationDate="2011-01-13T22:40:01.390" Score="0" ViewCount="" Body="&lt;p&gt;There is nothing wrong with the quote because it assumes that the only allowed (or considered) motion is to the left or to the right. So the text is explaining things in the context of mechanics with one spatial dimension. And one-dimensional vectors are isomorphic to ordinary numbers. Their first and only component may be positive or negative, so one may also talk about positive and negative vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is not possible for higher-dimensional vectors. For at least 2-dimensional vectors, one has to talk about components with respect to specific axes if he wants to discuss the &quot;signs of the momentum&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-13T22:40:01.390" />
  <row Id="2816" PostTypeId="2" ParentId="2772" CreationDate="2011-01-13T22:52:21.417" Score="1" ViewCount="" Body="&lt;p&gt;a difference between a social network and a Feynman diagram is that a Feynman diagram describes a possible history - something that occurs in the whole spacetime - while a social network describes a particular object that exists at one instant of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one can't talk about the &quot;Feynman diagram for particles that exist now&quot;. Instead, Feynman diagrams correspond to histories. And in fact, Feynman's main rule is to sum over all histories - all Feynman diagrams with the same external lines - before you get the result, and the result is interpreted as a probability amplitude when you're finished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, even if you tried to neglect the fact that the Feynman diagrams have to be summed up, and that Feynman diagrams are expressing histories in spacetime rather than objects in space, there would still be another serious problem with your question because the relationship in the graph - whether two particles interact i.e. whether they are &quot;Facebook friends&quot; of each other - isn't really well-defined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, all two particles may interact with each other except that the interaction may be weak and fail to influence any measurable quantities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one can't really construct Feynman diagrams analogous to social networks. But your question has many other aspects, including self-similarity. Yes, there exists a lot of scale-free, self-similar behavior in quantum field theories that describe elementary particles. So if one properly draws some &quot;other functions&quot; that describe the behavior of the system with massless particles only, she will get self-similar pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the &quot;typical&quot; trajectory of a particle in spacetime resembles the Brownian motion, or a random walk, at least in a certain parametrization. It is self-similar: if the timescale in a random walk is increased $k^2$ times, then the typical distances where the particle arrives get $k$ times longer, but the qualitative shape of the picture is unchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum field theory - in the scale-invariant massless limit - is full of similar chaotic, self-similar patterns of various dimensions. But they're much more like continuous functions of some real variables than combinatorial graphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address another intuition that is not quite adequate, quantum field theory implies that the &quot;number of elementary particles&quot; in a region isn't really a finite well-defined number. In some sense, the number of quarks in a proton, when properly counted, is infinite, and the more accurate &quot;resolution&quot; how to look inside the proton one chooses, the more quarks she sees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers,&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-13T22:52:21.417" />
  <row Id="2817" PostTypeId="2" ParentId="2716" CreationDate="2011-01-13T23:01:13.003" Score="1" ViewCount="" Body="&lt;p&gt;the special theory of relativity implies that there is no unique &quot;the timing system&quot; in the Universe. Each inertial system, depending on its velocity, has a different answer to the question whether two events occurred simultaneously, and if they didn't, what was the delay in between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to be able to quantify the delay between two events, you have to choose a reference frame. Most typically, you may associate your preferred reference frame with a particular pointlike object - such as the small golden seed that sits at the center of the Earth, or the tip of the Big Ben, or anything else. There is clearly no God-given preferred choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have such an object, you may measure the proper time along its world line in spacetime by some accurate clocks, e.g. by atomic clocks. You may also associate time with events away from the world line of your benchmark objects - e.g. as the time when the photons that got to the event were emitted from your benchmark object. Such a definition of time is nicely well-defined but is constant along &quot;light cones&quot; and for practical purposes, one may make subtractions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, there are many clocks and methods how to measure and coordinates how to describe events in spacetime but there is no privileged one. Quite on the contrary, relativity implies that the right labeling of &quot;time of an event&quot; depends not only on the preferred &quot;time zero&quot; - which is ambiguous due to the time-translational symmetry - but also because of the velocity that changes the notion of simultaneity and expands the duration of processes by the time dilation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers, LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-13T23:01:13.003" CommunityOwnedDate="2011-01-13T23:01:13.003" />
  <row Id="2818" PostTypeId="2" ParentId="2725" CreationDate="2011-01-13T23:09:39.723" Score="3" ViewCount="" Body="&lt;p&gt;Many very accurate measurements seem to imply that the fine-structure constant has been totally constant, within the error margin, since the Big Bang. And the few experiments that suggested otherwise have problems. Theoretically, it's likely that the fine-structure constant is completely constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you imagine that the fine-structure constant changes a bit, many things become different. For example, the stability and lifetimes of the neutron and nuclei will change abruptly. The modified fine-structure constant will also influence many other &quot;constants&quot; that define particle physics - such as the ratios of particle masses, fine-structure constants for other forces, mixing angles, and others - but you would have to describe what (dimensionless parameters) you want to keep fixed while you're changing the fine-structure constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From an &quot;anthropic&quot; viewpoint, it is important to notice that chemistry (minus nuclear physics, the care about isotopes etc.) would remain largely unaffected. It's because the atoms exist in the non-relativistic limit, and the fine-structure constant only determines the (order-of-magnitude estimate of the) speed of electrons in the atoms. Because the fine-structure constant is much smaller than one, the electrons in atoms may be approximated by non-relativistic mechanics. That would still be true if you e.g. doubled the fine-structure constant. So the ratios of the atomic frequencies etc. would remain pretty much unchanged, even if you doubled alpha, up to small corrections and splittings of the spectral lines (which are called the fine structure for a good reason).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is even plausible that as complicated molecules as the DNA could continue to work unaffected even if alpha were changed by many percent or dozens of percent, despite the fact that the functioning of the DNA depends on small detailed in energy differences. If the units were redefined properly, the world of biochemistry would be pretty much unchanged. However, that's because the quantum electrodynamics with electrons only depends on one dimensionful parameter only - the electron mass - and one dimensionless one - the fine-structure constant that moreover defines a different scaling for space and time in the non-relativistic limit. If you considered the behavior of high-energy physics which has many other elementary particles, a change of alpha would surely make a difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-13T23:09:39.723" />
  <row Id="2819" PostTypeId="2" ParentId="2743" CreationDate="2011-01-13T23:24:19.230" Score="1" ViewCount="" Body="&lt;p&gt;The black hole of mass $10^{20}$ kilograms is not as dangerous as it may look. It is $10^{28}$ Planck masses, so the radius is $10^{28}$ Planck lengths or $10^{-7}$ meters. The gravitational acceleration near its event horizon is $10^{-10}\times 10^{20} / 10^{-14}$ which is $10^{24}$ meters per second per second. Even meters from the black hole horizon, the acceleration reaches the Earth's acceleration on the surface. Centimeters from the black hole horizon, the acceleration would be enough to break the matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black hole would obviously try to find the minimum of the gravitational potential produced by the Earth, so it would sit and eventually stabilized at (and oscillated around) the Earth's center. If the Earth's center were imagined to be solid, I find it plausible that the black hole would break and eat some matter that is meters away from the micrometer-sized black hole. And the rest of the solid Earth could just sit around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is clearly not what would happen because the center of the Earth is liquid - because of the immense pressure. The liquid metals from meters around the black hole would simply flow to the black hole by reasonable speeds. The black hole would drink a lot of this fluid and its density would decrease in much of the Earth's core. It's questionable whether the solid layers of the planet could survive this reduced pressure. I can imagine that a black hole could actually be sitting inside the Earth and slowly drinking the liquid iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, we know for sure that this is not what generically happens inside the celestial bodies because it would also take place inside gas planets, and those could be fully swallowed in a relatively shorter time, producing lots of rotation along the way.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-13T23:24:19.230" />
  <row Id="2820" PostTypeId="2" ParentId="2744" CreationDate="2011-01-13T23:38:02.783" Score="0" ViewCount="" Body="&lt;p&gt;If you lower the center-of-mass of your car, it becomes more stable on the road. Consequently, you may afford to buy a new motor and bring your car closer to a formula one. Formulas have inverse wings to keep them stuck on the street at high speeds which is even better than a lowered center-of-mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you succeed to rebuild your car into a formula one, the impact on the fuel economy will be that the fuel consumption rate jumps by one order of magnitude. Instead of 7.5 liters per 100 kilometers that a normal car has, you will switch to 75 liters per 100 kilometers which is the average for formula one races.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you don't try to abuse your increased stability by replacing the motor or changing the speed, there will be no measurable effect on the fuel economy. In particular, if the speed of the car is constant, the friction forces etc. that slow the car down are independent of the height of the center of mass, so the fuel consumption is unchanged, too. Only when you accelerate or decelerate, there could be some secondary effects from misbalanced front and rear wheels but I don't expect them to be measurable.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-13T23:38:02.783" CommentCount="1" />
  <row Id="2821" PostTypeId="2" ParentId="1796" CreationDate="2011-01-13T23:43:23.327" Score="1" ViewCount="" Body="&lt;p&gt;The CADC's &lt;a href=&quot;http://www2.cadc-ccda.hia-iha.nrc-cnrc.gc.ca/cadcbin/meetings/getMeetings.pl?year=2010&quot; rel=&quot;nofollow&quot;&gt;International Astronomy Meetings List&lt;/a&gt; is probably my favourite aggregator of international astrophysics conferences.&lt;/p&gt;&#xA;" OwnerUserId="1252" LastActivityDate="2011-01-13T23:43:23.327" />
  <row Id="2822" PostTypeId="2" ParentId="2702" CreationDate="2011-01-13T23:43:41.017" Score="3" ViewCount="" Body="&lt;p&gt;Dear Kostya, if the electric field is a vector with an arrow, then the magnetic field is fundamentally not a vector: instead, it is an antisymmetric tensor with two indices, determining an &quot;oriented two-plane&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter carries the same information (3 different components) as a vector, and there is a convention given by the right-hand rule to switch from one to the other. A derived and related rule also determines the magnetic field of a solenouid and other things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, the convention to switch from the antisymmetric tensors to vectors could have been the other way around, too. So one has to remember something to know the convention; one can derive many things but not conventions. I agree that remembering the right hand operations is simple, especially because the word &quot;right&quot; also means &quot;correct&quot; and because the right-wing political opinions are the right ones while the left-wing political opinions are those that are left over.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-13T23:43:41.017" CommentCount="1" />
  <row Id="2823" PostTypeId="2" ParentId="2747" CreationDate="2011-01-13T23:50:15.600" Score="3" ViewCount="" Body="&lt;p&gt;Susskind and Polychronakos - and, independently, Hellerman and Van Raamsdonk - have constructed a matrix model - based on string-theoretical D0-branes, which is the right fundamental way to think about any adjoint matrix model - to describe the fractional quantum Hall effect. A review is here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://iopscience.iop.org/1751-8121/42/30/304006&quot; rel=&quot;nofollow&quot;&gt;http://iopscience.iop.org/1751-8121/42/30/304006&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also a Hellerman-Susskind realization of the quantum Hall effect in string theory&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0107200&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0107200&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, there has been a whole new industry started by the 1997 Maldacena's AdS/CFT correspondence. On the CFT side, one most typically has matrix-valued field theories, even though they're usually not just quantum-mechanical models but field theories in additional dimensions. Chern-Simons matrix models also have played a role.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AdS/something dualities have been successful to one extent or another in describing perfect fluids, Fermi liquids, non-Fermi liquids, superconductors, hydrodynamics, heavy ion physics, and other things. Phenomena are usually translated to the behavior of black holes in a higher-dimensional spacetime.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-13T23:50:15.600" CommentCount="1" />
  <row Id="2824" PostTypeId="1" AcceptedAnswerId="2877" CreationDate="2011-01-14T00:03:24.203" Score="3" ViewCount="825" Body="&lt;p&gt;When my wife uses her laptop, if I touch her skin, I can feel a buzz.  She doesn't feel the buzz, but she can hear it if I touch her ear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm guessing it's a faulty laptop, and she's conducting an electrical current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why would she not feel anything, and what would it be that she would be hearing when I touch her ear?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;More info:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect is only intermitent - it's pretty reliable in a single session on the laptop, but some sessions it won't happen and others it will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had the same sensation with a desk lamp that I had several years ago, with no moving parts (as far as I could tell)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect only occurs when I move my finger - if I'm stationary, I don't notice anything. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was playing with my son, and noticed the same buzz.  First I thought he was touching the laptop.  Then I realised he had skin-to-skin contact with my wife who was using the laptop.&lt;/p&gt;&#xA;" OwnerUserId="1254" LastEditorUserId="1254" LastEditorDisplayName="" LastEditDate="2011-01-15T09:16:52.013" LastActivityDate="2011-05-07T00:59:07.020" Title="Why does my wife's skin buzz when she's using her laptop?" Tags="&lt;electricity&gt;" AnswerCount="6" CommentCount="7" FavoriteCount="2" />
  <row Id="2825" PostTypeId="2" ParentId="2625" CreationDate="2011-01-14T00:07:08.057" Score="5" ViewCount="" Body="&lt;p&gt;Thermodynamics is an older theory that has studied work, energy, temperature, entropy and related concepts (including volume, pressure, polarization etc.) as continuous quantities describing continuous material objects. It was based on several principles - such as the conservation of energy and the growth of entropy - that prevent one from constructing perpetuum mobile gadgets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statistical mechanics is a newer theory that appreciates the atomic structure of the matter. Energy and entropy are distributed among the atoms and the temperature measures the average energy per one atom (or degree of freedom). Statistical mechanics allows one to derive the older laws and principles of thermodynamics by applying statistical methods on dynamics of a large number of atoms and molecules. For a particular model of a material, it even allows one to determine the material constants.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T00:07:08.057" />
  <row Id="2826" PostTypeId="2" ParentId="2594" CreationDate="2011-01-14T00:18:47.287" Score="2" ViewCount="" Body="&lt;p&gt;To answer your question including the full supergravity, explicitly enough, one needs an amount of space not much smaller than the usual textbooks. And supergravity usually appears in their second volumes only. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, one may divide the question to the proof of supersymmetry, and the proof of gravity in string theory. The proof that superstring theory obeys spacetime supersymmetry is a separate thing. A combination of gravity and supersymmetry immediately implies that the spacetime action has to be one of supergravity - this is pretty much by definition of supergravity, even though this verbal exercise doesn't really construct the action in components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the remaining question is why string theory predicts gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's because the world sheet action describes the proper area of a world sheet that propagates in a background spacetime. This spacetime may be infinitesimally changed by changing the background metric. Because the world sheet theory is scale-invariant, every infinitesimal deformation of the world sheet Lagrangian has to correspond to a marginal deformation, i.e. a dimension (1,1) tensor field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the state-operator correspondent, a local field of this dimension translates to an allowed, physical massless field in spacetime. Its spacetime spin is 2 because it carries the same indices as the background graviton. By adding a condensate of such closed strings in the gravitational vibrational pattern, one impacts the behavior of all other strings in the same way as if he deformed the background. This follows from the state-operator correspondence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the principle of equivalence is automatically satisfied: the graviton closed strings behave just like perturbations of the background, and they act on the motion of all objects and fields in the same way. The spacetime geometry itself follows Einstein's equations because they're the only diff-invariant locally Lorentz-invariant equations that they could satisfy - whose linear part is a wave equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More explicitly, one may show that one gets the Ricci tensor in spacetime because it's a consistency condition for the allowed backgrounds where strings may propagate. The string dynamics may only be defined by scale-invariant world sheet theories, and the change of the world sheet action with respect to the scale - also known as the beta-function (for each coupling constant) - has to vanish as a consequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the worldsheet, the metric tensor at each spacetime point is a coupling constant and the beta-function may be calculated from one-loop graphs in the world sheet theory, and it ends up being the Ricci tensor, roughly speaking. It follows that the background has to satisfy the right equations of general relativity at long distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The state-operator correspondence guarantees that the closed-string excitations also interact according to the same equations because the closed strings are equivalent to infinitesimal deformations of the background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supersymmetry is added as a lot of extra fermionic stuff and leads to supergravity in spacetime - which is still just gravity coupled to other matter fields in a way that preserves spacetime supersymmetry. See also&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2007/05/why-are-there-gravitons-in-string.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2007/05/why-are-there-gravitons-in-string.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T00:18:47.287" CommentCount="2" />
  <row Id="2827" PostTypeId="2" ParentId="2685" CreationDate="2011-01-14T00:28:00.217" Score="0" ViewCount="" Body="&lt;p&gt;the information is a dimensionless (unitless) - and, in this sense, &quot;purely mathematical&quot; - quantity measuring how much one has to learn to know something relatively to the point when he doesn't know it, expressed in particular units. Operationally speaking, it is the amount of RAM chips (or their parts) one needs to have so ehat they can remember some knowledge. Of course, by using the word &quot;knowledge&quot;, I am just avoiding the word &quot;information&quot;, and it is impossible to define any of these terms without any &quot;circular references&quot; because one has to know at least something to be able to define as elementary concepts as knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One bit of information is the knowledge needed to know whether a number that can be 0 or 1 with the same probability turned out (or will turn out) to be 0 or 1. In mathematics, a more natural unit than one bit is one &quot;e-bit&quot; which is such that 1 bit is ln(2) &quot;e-bits&quot;. The information needed to distinguish among &quot;N&quot; equally likely alternatives is ln(N) &quot;e-bits&quot;. The natural logarithm is always more natural than other logarithms - that's why it's called natural. For example, its derivative equals 1/x, without complicated constants. The formulae for the &quot;Shannon&quot; dimensionless information, assuming any probabilistic distribution, are given above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In physics, every physical system with some degrees of freedom may carry some information. In quantum information, the &quot;alternatives&quot; are usually associated with basis vector of the allowed Hilbert space of states. But in that context, one &quot;bit&quot; of information is usually referred to as &quot;qubit&quot; or &quot;quantum bit&quot; which means that in the real world, the alternatives may also be combined into arbitrary complex linear superpositions, like the postulates of quantum mechanics dictate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In discussions about causality, we mean that the spatially separated objects can't really influence each other. This is guaranteed by the Lorentz symmetry. In field theory, the condition is equivalent to the constraint that the space-like separated fields $\phi(x)$ and $\phi(y)$ commute with each other (or anticommute if both of them are fermionic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T00:28:00.217" />
  <row Id="2828" PostTypeId="2" ParentId="2690" CreationDate="2011-01-14T00:34:06.750" Score="1" ViewCount="" Body="&lt;p&gt;Noether's relationship implies that for every conserved quantity, there is a symmetry, and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the real world, mass may be converted to energy. For example, uranium power plants convert about 0.1% of the uranium mass to a huge energy, according to the $E=mc^2$ formula. So up to the conventional factor $c^2$, the total energy - including the latent one - and the total mass is the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its conservation is linked to the time-translational symmetry of the laws of physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the old world &quot;before Marie Curie&quot;, people didn't know any relativity or any other indications of relativistic physics such as radioactivity. So they believed that mass was conserved even if the energy is not included. In their understanding of the world, the total mass of the Universe was the sum of the rest masses of the electrons, protons, and other massive particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conservation law for the mass defined in this way didn't lead to any symmetries because the definition only depends on parameters - masses of all point masses are parameters - and not on dynamical, time-dependence quantities such as the positions or velocities. That's why Noether's theorem didn't apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not a real flaw of Noether's theorem because, as we know today, the total mass defined in the old way - and neglecting the mass increases from velocity (kinetic energy) and other forms of energy - is actually not conserved. This is no coincidence; a world compatible with general relativity doesn't allow any mass-like quantities to be non-dynamical. All quantities describing particular objects have to be dynamical i.e. changeable, and that's why Noether's theorem always holds in the world.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T00:34:06.750" CommentCount="1" />
  <row Id="2829" PostTypeId="2" ParentId="2670" CreationDate="2011-01-14T00:48:02.310" Score="3" ViewCount="" Body="&lt;p&gt;Jeff Harvey has of course provided you with the perfect, standardized answer: the scale invariance boils down to the tracelessness of the stress-energy tensor. But the tracelessness is not really a &quot;conserved quantity&quot; in the usual sense that you may have waited for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, one may transform the problem to something that is a conserved quantity in the usual sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, you may take your scale-invariant universe and insert a point-like object at a chosen point that I will call the origin. In quantum field theory, this is achieved by acting on the vacuum state with a local operator at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transformations proving scale invariance are just radial expansions that keep the origin untouched. The laws of physics are invariant under these transformations, by assumption, and this symmetry is equivalent to the conservation of the dimension of the operator from the previous paragraph. But its conservation not with respect to the normal evolution in time but evolution in the &quot;radial time&quot;, $\ln(r)$. Consequently, the dimensions of all operators are well-defined in scale-invariant theories. In scale-non-invariant theories, they would depend on the renormalization scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I added this verbal exercise in order to emphasize that the scale transformations in a scale-invariant theory are analogous - and in a very well-defined mathematical sense, equivalent - to ordinary translations in time. To be a bit specific, think about 2-dimensional Euclidean theories. The complex coordinate $z$ may be written as $\exp(a+ib)$. Here, $b$ is a periodic, angular variable with periodicity $2\pi$. However, $a$ is real and goes from $-\infty$ to $+\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scale transformations are nothing else than the ordinary translations in $a$ which are linked to a Hamiltonian. For example, you expand $z$ $e$-times by shifting $a$ by one. And indeed, scale invariance in 2 dimensions implies the full conformal invariance - under all transformations that preserve the angles - so instead of looking at the $z=x+iy$ plane, you may equally well look at the $a+ib$ plane where the original scale transformation looks like an ordinary translation in the $a$ direction. By conformal symmetry, the form of the action in the $z$ and $a+ib$ coordinates are identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In higher dimensions, it is not quite true that scale invariance (and Lorentz/rotational symmetry) implies the full conformal symmetry, but in the important cases, it is true, anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T00:48:02.310" />
  <row Id="2830" PostTypeId="2" ParentId="2658" CreationDate="2011-01-14T01:02:10.497" Score="0" ViewCount="" Body="&lt;p&gt;Operationally, one can distinguish real and virtual images by attaching a small but very powerful light bulb to the source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you insert your hand near the real image, the hand will get burned because the light rays from the source actually converge at the point of the real image. If you manage to insert your hand to the point of the virtual image, your hand won't burn because there are no light rays over there (and sometimes, there is a wall).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An ordinary flat mirror creates a virtual image because there are no light rays on the internal side of the mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The optics in the human eye creates a real image where the retina is located and sensitive cells actually read the information about the original source. The same mechanism works in digital cameras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Real images may be produced by converging lenses (eye, digital camera) and concave mirrors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Virtual images may be produced by diverging lenses and convex mirrors, but also by concave mirrors and converging lenses if you place the source within the focal length.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T01:02:10.497" />
  <row Id="2831" PostTypeId="2" ParentId="2743" CreationDate="2011-01-14T01:21:55.357" Score="0" ViewCount="" Body="&lt;p&gt;I just want to add that oscillations about the center of the Earth are dampened due to the momentum of the entering mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figures for volume of the mass continually eaten by the black hole differ by orders of magnitude going by previous posters.  But the consumed material as it falls will depend on the cross section of this volume times the radius of the Earth, or cross section times density for linear mass density of path of destruction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That mater has zero kinetic energy and a potential energy as a function of height.  Gravitational field is directly proportional to radius (due to continuous spherical distribution) so potential energy is $r^2$ function.  I write gravitational potential of BH as $C R^2 m$ where R is radius of Earth and m is mass (kg) of BH and C is some constant I'm not going to address.  Denote linear density of path of destruction as $l$ (kg/m), and integrate $C r^2$ potential to find $1/3 C R^3 l$ to center of Earth, or $2/3 C R^3 l$ to other side of Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume it eats material perfectly and there are no other interactions.  It begins with $C R^2 m$ energy (Newtonian!) and m mass.  It acquires $2 l R$ mass in one trip (assuming acquired mass is small relative to total and thus nearly touches surface again).  We find the deficit in specific potential energy at the end of its trip: (P.E._end/end_mass) / (P.E._start/start_mass)-1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{ \frac{ C R^2 m+\frac{2}{3} C R^3 l}{m+2 R l} }{ \frac{C R^2 m}{m} } - 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\frac{m}{m+2 R l}  \frac{R^2 m+\frac{2}{3} R^3 l}{R^2 m} - 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=\frac{1+\frac{2}{3} \frac{R l}{m} }{1+2 \frac{R l}{m}} - 1 = \frac{1+2 \alpha }{ 1+\frac{2}{3} \alpha} - 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\alpha = R l/m$ dimensionless parameter representing fraction of initial mass added by trip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We assume $\alpha \ll 1$ and Taylor expand at $\alpha=0$ to find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specific energy deficit after one trip $= -4/3 \alpha$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking more closely at alpha, write $\alpha = R A \rho/m$, where A is the cross sectional area I referred to and rho is the density of Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R = 6.4 \times 10^6 m$$&#xA;$$A = 1 cm^2$$&#xA;$$\rho = 4.0 g/cm^3$$&#xA;$$m = 10^{20} kg$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha = 2 R l/m = 2.56 \times 10^{-13}$ (fraction of BHs mass accumulated in half-trip, sounds good)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For change in height due to trip, use mgh approx and find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$( - \frac{4}{3} \alpha )  6,400,000 m = 2.18 x 10^{-6} m$$ Lower&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It falls 2.18 micro meters lower at the end of the trip.  Now, this scales directly with the area eaten, and thus with the square of the radius at which material is captured.  To get a factor of 1e6, that radius would need 10 meters versus 1 cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, dampening really IS SMALL, and the fate of the Earth would be dictated by how it eats matter while traveling at high speeds through the core.  I'm going to go tell people now that the reason the LHC is underground is so that a BH won't poke out the surface if an accident happens.  I love spreading disinformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: This was my first answer given on physics SE, so I've gone back and put the equations in the right format, although the organization of the answer probably reflects its bizarre history.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-07-29T14:20:39.783" LastActivityDate="2011-07-29T14:20:39.783" />
  <row Id="2832" PostTypeId="1" AcceptedAnswerId="2836" CreationDate="2011-01-14T01:40:46.527" Score="-2" ViewCount="439" Body="&lt;p&gt;There are 2 balls in a vacuum, next to each other but not touch. They are on the edge of a surface they will both leave the table at exactly the same time. One gets pushed harder than the other. The earth is completely flat. Apparently both balls will hit the floor at exactly the same time and even further be at the same vertical height from the floor at any point in time. How is this possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you push the one ball much much much harder than the first so that it is going at a considerable speed 100m/s compared to 1m/s say. The balls can't possible hit the floor at the same time, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the earth is round. You hit the one ball so hard it goes into orbit the other ball just falls of the table straight down like before. As the one ball has gone into orbit it will never reach the ground. However the slow ball will reach the ground much faster. How can a rule be different? If it doesn't work at the extreme then it shouldnt work at all. &lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2011-01-14T11:15:13.733" Title="2 balls falling hit the ground at the same time" Tags="&lt;speed&gt;&lt;motion&gt;&lt;time&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="2833" PostTypeId="1" CreationDate="2011-01-14T02:07:13.553" Score="1" ViewCount="113" Body="&lt;p&gt;If we observe light that originates X light years away, but it passes between black holes X/2 light years away, will it be normal or red shifted or blue shifted?    What if the black holes were X/4 or 3X/4 light years away?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have always wondered if the apparent red shift of distant galaxies must be due to relative velocity with us.&lt;/p&gt;&#xA;" OwnerUserId="1260" LastActivityDate="2011-01-14T14:04:08.920" Title="What is the color shift if light passes between two (or more) black holes?" Tags="&lt;black-holes&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="2834" PostTypeId="2" ParentId="2832" CreationDate="2011-01-14T02:14:02.677" Score="3" ViewCount="" Body="&lt;p&gt;The answer is that you are using a Discworld assumption (i.e. Earth is a plane) to derive conservation of momentum in the horizontal directions. Motion in this direction is then decoupled from the motion in vertical direction and so both balls have the same altitude at all times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you might imagine, Earth is round, so the above assumption fails at large distances. So you have to use full Newton's gravitational law and investigate possible trajectories. They are in general ellipses (which is the correct shape of the trajectory you thought was a parabola previously) and in particular circles. This is the trajectory that the orbiting ball will take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Usual disclaimers about ignoring all effects of air, uneven Earth's surface, spinning of the ball and everything else that hasn't been explicitly mentioned.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-01-14T02:14:02.677" />
  <row Id="2835" PostTypeId="2" ParentId="2832" CreationDate="2011-01-14T02:14:22.000" Score="-1" ViewCount="" Body="&lt;p&gt;Nowhere does it say that one ball will be faster than the other.  It only says that one is pushed harder than the other.  If the one pushed harder has more mass, then the amount of force on it might simply have been mecessary for the same acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, I read this as the ball is pushed horizontally -- perpendicular to the force of gravity.  That would have no affect on the fall to a hypothetical flat earth.&lt;/p&gt;&#xA;" OwnerUserId="1260" LastActivityDate="2011-01-14T02:14:22.000" />
  <row Id="2836" PostTypeId="2" ParentId="2832" CreationDate="2011-01-14T02:15:10.537" Score="5" ViewCount="" Body="&lt;p&gt;The confusion arises because there are two different versions of what Earth's surface looks like, and two different models of how gravity works between the case where the ball goes into orbit and the case where both balls hit the ground at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We often approximate gravity near the Earth's surface by saying that it is constant everywhere.  This approximation is pretty good if we are only interested in exploring distances much smaller than Earth's radius.  In this approximation, the two balls will always hit the ground at the same time, no matter what speed they leave the surface with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, in reality, the Earth is round, and if you push the ball so fast that it travels a distance which is comparable to the radius before it lands, you must take a more sophisticated picture of gravity and the Earth into account.  In this picture, it is possible that the fast moving ball will remain in orbit forever, or even fly away from the Earth never to return.&lt;/p&gt;&#xA;" OwnerUserId="1262" LastActivityDate="2011-01-14T02:15:10.537" />
  <row Id="2837" PostTypeId="2" ParentId="2833" CreationDate="2011-01-14T02:25:34.757" Score="3" ViewCount="" Body="&lt;p&gt;There are two ways that light frequencies can be shifted.  One is if the source of the light is moving toward or away from the observer.  The other is if the light has fallen into, or climbed out of a gravitational potential on its way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If light passes near a black hole, it will be blue-shifted since it is falling into a gravitational well.  However, if the light makes it back out of the gravitational well and makes it to Earth, it will have to climb out of the same well, getting red-shifted an equal amount.  So, on average there is no effect from having passed near the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy will be the same, but the direction may change since the black holes can act like a lens.&lt;/p&gt;&#xA;" OwnerUserId="1262" LastActivityDate="2011-01-14T02:25:34.757" CommentCount="1" />
  <row Id="2838" PostTypeId="1" AcceptedAnswerId="2856" CreationDate="2011-01-14T02:42:33.490" Score="5" ViewCount="638" Body="&lt;p&gt;In popular science books and articles, I keep running into the claim that the total energy of the Universe is zero, &quot;because the positive energy of matter is cancelled out by the negative energy of the gravitational field&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can't find anything concrete to substantiate this claim. As a first check, I did a calculation to compute the gravitational potential energy of a sphere of uniform density of radius R using Newton's Laws and threw in E=mc^2 for energy of the sphere, and it was by no means obvious that the answer is zero !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What is the basis for the claim - does one require General Relativity, or can one get it from Newtonian gravity ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What conditions do you require in the model, in order for this to work ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Could someone please refer me to a good paper about this ?&lt;/p&gt;&#xA;" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T14:36:59.957" LastActivityDate="2011-02-03T14:36:59.957" Title="Total energy of the Universe" Tags="&lt;general-relativity&gt;&lt;cosmology&gt;&lt;energy-conservation&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="2839" PostTypeId="1" AcceptedAnswerId="2855" CreationDate="2011-01-14T02:49:26.213" Score="5" ViewCount="309" Body="&lt;p&gt;Why is the number of molecules in a standard volume of gas at a standard temperature and pressure a constant, regardless of the molecule's composition or weight?&#xA;Let's say I have a closed box full of a heavy gas, one meter on a side.  It has a certain number of molecules inside.  I want to be able to add a lighter gas to the box without changing its internal pressure (or temperature), so I connect a cylinder to the side of the box, which holds a frictionless piston for expansion (the piston has a constant force applied to it, to maintain a constant pressure inside the box and allow the volume of the gas to grow as new gas is introduced into the box).&#xA;Now I add Helium to the box.  The piston moves back to maintain constant pressure, but why does the number of molecules in the box proper stay constant?&#xA;My mental image of this is that it would be like adding water to a bucket of marbles, and that, evidently, is wrong, but why is it wrong?&lt;/p&gt;&#xA;" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T14:37:38.770" LastActivityDate="2011-02-04T13:23:27.413" Title="Why is Avogadro's hypothesis true?" Tags="&lt;statistical-mechanics&gt;&lt;physical-chemistry&gt;&lt;pressure&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" />
  <row Id="2840" PostTypeId="2" ParentId="2685" CreationDate="2011-01-14T03:15:22.960" Score="3" ViewCount="" Body="&lt;p&gt;In short:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;information contained in a physical system = the number of yes/no questions you need to get answered to fully specify the system.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-01-14T03:15:22.960" CommentCount="2" />
  <row Id="2841" PostTypeId="2" ParentId="2838" CreationDate="2011-01-14T03:25:10.567" Score="0" ViewCount="" Body="&lt;p&gt;General relativity has difficulty in defining what is energy. In loose terms the difficulty boils down to the fact that gravitational energy can not be localized. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a speculative blog about these matters, see: &lt;a href=&quot;http://www.science20.com/hammock_physicist/square_root_universe&quot; rel=&quot;nofollow&quot;&gt;http://www.science20.com/hammock_physicist/square_root_universe&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-01-14T03:25:10.567" />
  <row Id="2843" PostTypeId="2" ParentId="2824" CreationDate="2011-01-14T03:38:06.060" Score="1" ViewCount="" Body="&lt;p&gt;I agree with jonathon that it is vibration, although it is far more likely to be the laptop fan, rather than the hard drive&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why she doesn't feel it? The human body has vibration receptors, and I'm sure she notices the laptop vibrating when she turns it on. However, like with all constant stimuli, but particularly touch, our brains are great at blocking them out. I suspect that the fact it happens to her whole body makes the effect even greater, as there is no sensory discrimination from different nerves&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You touch her, and two things are different. For you, it is a new sensation, and it is occurring in your hand alone. If you held her for a while, while focussing on something else, you would probably stop noticing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would argue that this is a physical effect, biophysical to be precise.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-14T03:38:06.060" />
  <row Id="2844" PostTypeId="2" ParentId="2838" CreationDate="2011-01-14T03:47:57.310" Score="3" ViewCount="" Body="&lt;p&gt;(Now I notice you're the same person who asked this at MathOverflow, where I've &lt;a href=&quot;http://mathoverflow.net/questions/38659/total-energy-of-the-universe/38704#38704&quot; rel=&quot;nofollow&quot;&gt;previously answered something similar&lt;/a&gt; -- if you didn't like the answer then, you won't like it now.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is really just expanding on Marek's comment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you compute the stress tensor in a field theory? You vary the action with respect to the metric and see what comes out: $T_{\mu\nu} = 1/\sqrt{-g} \frac{\delta S}{\delta g^{\mu\nu}}$. This makes sense in non-gravitational theories, and the $T^{00}$ component is the energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens if you do this in a gravitational theory? The metric is dynamical, and varying the whole action with respect to it gives you the equation of motion (i.e., Einstein's equation). So $T^{00}$, defined in this way, where you vary the &lt;em&gt;whole&lt;/em&gt; action (including the Einstein-Hilbert term), is just zero: it's the energy of the matter, $T^{00}$, plus the gravitational term, $-\frac{1}{8\pi G} G^{00}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what &quot;canceled out by the negative energy of the gravitational field&quot; means, but it's kind of a vacuous notion. I wouldn't waste time thinking too hard about the claims people make based on this idea. This isn't a physically useful notion of energy in a gravitational theory.&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="19" LastEditorDisplayName="" LastEditDate="2011-01-14T03:56:36.287" LastActivityDate="2011-01-14T03:56:36.287" CommentCount="1" />
  <row Id="2845" PostTypeId="2" ParentId="2838" CreationDate="2011-01-14T03:48:18.543" Score="1" ViewCount="" Body="&lt;p&gt;I'm a little out of my depths here, but I suspect you're asking about the density parameter, and will proceed on that assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the accepted big-bang-and-inflation scenario, and before for we had evidence for the existence of dark energy, it was possible to talk about the possible fate of the universe (open or closed) in terms of the initial expansion as balanced by the total mass only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in that model, for the universe to be as big, as dense, and as old as we see it, that balance must have been very nearly at the critical value between open and close (a geometry called &quot;flat&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This claim was allowed by measurement, and preferred on a philosophical basis by some theorists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try &lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;nofollow&quot;&gt;the wikipedia article on Friedmann equations&lt;/a&gt; for some more discussion. You're looking for $\Omega = \rho/\rho_c \approx 1$. Or there may be better links.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, however that the issues are changed rather a lot by the presence of dark energy in the universe. There is no chance of a closed geometry, and we are doomed to a cold and lonely ending in the far distance future as accelerating expansion rips the regions of low entropy ever further apart.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-01-14T14:54:34.640" LastActivityDate="2011-01-14T14:54:34.640" CommentCount="4" />
  <row Id="2846" PostTypeId="2" ParentId="2839" CreationDate="2011-01-14T04:43:11.280" Score="1" ViewCount="" Body="&lt;p&gt;&quot;My mental image of this is that it would be like adding water to a bucket of marbles, and that, evidently, is wrong, but why is it wrong?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the extent that a gas can be represented by the &quot;ideal gas&quot; model, the volume of the gas molecules is negligible compared to the volume of the container. Your intuition would apply in the limit of very high densities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the ideal gas case, you might think of the volume as being supported by pressure in a manner similar to how popping popcorn can support a volume in excess of the volume of the popcorn itself. It happens dynamically; the volume is almost entirely empty of molecules.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-14T04:43:11.280" />
  <row Id="2847" PostTypeId="2" ParentId="1901" CreationDate="2011-01-14T04:47:48.390" Score="-1" ViewCount="" Body="&lt;p&gt;Why would you need to compress the universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radius of the observable universe equals the Schwarzschild radius of the total mass of the observable universe. In terms of density the universe is dense enough to be a black hole. But it isn't a black hole, it behaves like a white hole. If you want to get a black hole, you'd rather change the direction of time so as to transform the Hubble expansion into a Hubble contraction.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastEditorUserId="1268" LastEditorDisplayName="" LastEditDate="2011-01-15T17:42:48.983" LastActivityDate="2011-01-15T17:42:48.983" />
  <row Id="2848" PostTypeId="2" ParentId="2644" CreationDate="2011-01-14T04:53:50.587" Score="0" ViewCount="" Body="&lt;p&gt;I'm tempted to assume that this doesn't have its origin in orbits. However, it can of course be used to describe why they occur in conjunction with a theory of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton would have been familiar with many textbook mechanics scenarios:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... the force pulling an object to the ground. He also calculated the centripetal force needed to hold a stone in a sling, and the relation between the length of a pendulum and the time of its swing. &lt;a href=&quot;http://www.newton.ac.uk/newtlife.html&quot; rel=&quot;nofollow&quot;&gt;http://www.newton.ac.uk/newtlife.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and hence may have been able to compare these forces - the weight of a large object, and the centripetal force of a stone - by their effects on known objects: snapping wires, uncoiling (or deforming) springs, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there you can start to quantify forces and compare them to changes in velocity and begin to postulate proportionality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's my take on it anyway - hope this helps!&lt;/p&gt;&#xA;" LastActivityDate="2011-01-14T04:53:50.587" />
  <row Id="2849" PostTypeId="2" ParentId="1019" CreationDate="2011-01-14T05:08:56.120" Score="1" ViewCount="" Body="&lt;p&gt;How about these three misconceptions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big bang theory tells us the universe started from a point&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Einstein we know all is relative&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed of light is a fundamental constant&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-01-14T05:08:56.120" CommunityOwnedDate="2011-01-14T05:08:56.120" />
  <row Id="2850" PostTypeId="2" ParentId="2656" CreationDate="2011-01-14T05:19:37.987" Score="0" ViewCount="" Body="&lt;p&gt;2.61kWh/m^2/yr * 65cm^2 = 0.002 watts; since the time to charge/uncharge is the same, this is both the power in and the power out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversion to light at 300 lumens/watt: 0.6 lumens &lt;a href=&quot;http://en.wikipedia.org/wiki/Light-emitting_diode&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Light-emitting_diode&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it shines in all directions, about 0.05 candela - so about as powerful as one-twentieth of a candle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So. Not very brightly, then.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-14T05:19:37.987" CommentCount="1" />
  <row Id="2851" PostTypeId="2" ParentId="75" CreationDate="2011-01-14T05:34:11.240" Score="2" ViewCount="" Body="&lt;p&gt;I always used to confuse stress and strain: most of my mnemonics involved making words out of initial letters.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;When you're stressed, you show the strain.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stress is what is applied to the material, strain is what it does in response - I always used to get these the wrong way around.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;E equals Fl/ea&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Young's Modulus = (force &amp;times; length) / (extension &amp;times; area)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck with the others: I suggest imagining a hard man (grizzled veteran) who is actually secretly limp wristed and camp (he bends unlike hard materials), a tough guy (showy, probably with twin pistols) who literally goes to pieces in difficult situations (imagine his brittle bones snapping)  and a strong man (lifting a dumbbell) who is crushed to a pancake by the weight.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-14T05:34:11.240" />
  <row Id="2852" PostTypeId="2" ParentId="1019" CreationDate="2011-01-14T06:41:21.603" Score="8" ViewCount="" Body="&lt;p&gt;The misconceptions about special relativity and quantum mechanics are quite well-known. A lot of the posts above discuss them in detail. So rather than doing that I'll list some misconceptions from general(say high school) physics:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;When a body rests on a surface the upward contact force acting on it is reaction to its weight.&lt;/strong&gt; This is obviously wrong as action and reaction act on different bodies.&lt;/li&gt;&#xA;&lt;li&gt;There's a lot of misconceptions about non-inertial(pseudo) force. My physics teacher once said that non-inertial forces arise only when the body is in &lt;em&gt;contact&lt;/em&gt; with an accelerating frame.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Nothing can move faster than light.&lt;/strong&gt; Of course it's false unless you add the phrase &quot;in vacuum&quot;. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cherenkov_radiation&quot; rel=&quot;nofollow&quot;&gt;Cherenkov radiation&lt;/a&gt; happens when some charged particle moves in a medium with speed greater than the speed of light in that medium. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Friction &lt;em&gt;always&lt;/em&gt; has to act in the opposite direction of overall motion.&lt;/strong&gt; Actually friction provides the necessary force for rolling without which no vehicle would ever run. The correct formulation is friction opposes the instantaneous motion of the point of contact.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Light always travel in straight lines.&lt;/strong&gt; Even without gravitational bending if we simply have a medium with a variable refractive index light will follow a curve through it. It's a nice application of Snell's law.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Newton's second law provides a definition of force.&lt;/strong&gt; It's a very widespread misconception unfortunately even among professional physics students. This strips Newton's second law of any physical content and forces(pun intended) it to become a tautology. Of course the actual content of the law is that the force is given by some other law(say gravitational or em) and it equals ma. For a persuasive discussion on this see the first volume of Feynman's Lectures on Physics. (I am very sorry that I forgot the chapter or page number).&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Newton's first law is derivable from second law.&lt;/strong&gt; The proof goes as follows : F=ma. If F=0 then a=0 since $m~{}\neq 0$ QED. The problem is without first law there is no notion of an inertial frame and the laws become pointless.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;In special relativity the hypothesis of the constancy of the speed of light in vacuum(c) with respect to all observers is redundant because it can be derived from the principle of relativity.&lt;/strong&gt; Of course c may vary without contradicting the principle of relativity. In fact, in Newtonian mechanics c is observer dependent it respects the principle of relativity. The constancy of c hypothesis gives the Lorentz transformation whereas in Newtonian mechanics we have the Gallilean transformation. If you are still not convinced then look at this formulation of special relativity without the second hypothesis. Google &lt;strong&gt;doubly special relativity&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1275" LastActivityDate="2011-01-14T06:41:21.603" CommunityOwnedDate="2011-01-14T06:41:21.603" CommentCount="4" />
  <row Id="2853" PostTypeId="2" ParentId="2725" CreationDate="2011-01-14T07:30:30.907" Score="0" ViewCount="" Body="&lt;p&gt;This is a purely speculative idea, but it seems to me that the fine-structure &quot;constant&quot; is actually a measure of the compressibility of the vacuum. In a theory of quantum gravity, the structure of the vacuum (on which quantum fields are defined) and that of the background geometry are necessarily interrelated, in the sense that the vacuum can be thought of as a condensate of quantum geometrical excitations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the large-scale geometry of the universe can indeed be given a description in terms of such a condensate then the material properties of the condensate would show fluctuations over long distances. One of these would be the &quot;vacuum&quot; compressibility or $\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Caveat Emptor:&lt;/strong&gt; There are a few key notions that must be elaborated on before the above argument can begin to stand on its own. These are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is a &quot;quantum geometric condensate&quot; and why should the vacuum of the standard model have such a description?&lt;/li&gt;&#xA;&lt;li&gt;A elucidation of the &lt;em&gt;precise sense&lt;/em&gt; in which one can think of $\alpha$ as a measure of the compressibility (or some other &quot;material&quot; characteristic) of the ground state of the resulting condensate.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The first proposition is one I firmly believe in, not least because of what what I have learned through my own research. The second is something I am hazy on at present, though I have a strong intuition that such a description exists and is even natural.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize, that in the absence of a clearer explanation of these two issues I am asking the reader to believe in my reasoning based more on a &quot;gut feeling&quot; than anything else. This might be too much to digest for some. &lt;strong&gt;However&lt;/strong&gt;, if for whatever reason, you are willing to give some credit to this line of reasoning then a cosmological variation in $\alpha$ is all but inevitable.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-14T07:30:30.907" />
  <row Id="2854" PostTypeId="2" ParentId="1796" CreationDate="2011-01-14T08:56:30.457" Score="1" ViewCount="" Body="&lt;p&gt;Adding two more places to look:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.slac.stanford.edu/spires/conferences/&quot; rel=&quot;nofollow&quot;&gt;SPIRES conference search&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://indico.cern.ch/&quot; rel=&quot;nofollow&quot;&gt;INDICO ar CERN&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-01-14T08:56:30.457" />
  <row Id="2855" PostTypeId="2" ParentId="2839" CreationDate="2011-01-14T09:13:29.157" Score="4" ViewCount="" Body="&lt;p&gt;Dear Wade, your good question is easily answered if you consider &quot;pressure&quot; to be a derived quantity, and let us derive it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An average molecule (or atom) of an ideal gas - and your proposition only holds for an ideal gas - has kinetic energy equal to &#xA;$$mv^2/2=3kT/2$$&#xA;It's because every degree of freedom carries $kT/2$ and there are three degrees of fredom in translations. Note that lighter molecules will move faster than the heavier ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we calculate the pressure? Well, put the molecule in a cubic box of volume $a^3$. It will hit the walls - the total surface of the cube is $6a^2$. We need to compute the average (over molecules) transfered momentum per unit time - this is called the force - and divide it by the area to compute the pressure of one molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force on the wall may be in $x,y,z$ directions. Let's consider the $x$ direction. The velocity of a particular molecule in the $x$-direction is $v_x$, so it takes $a/v_x$ of time to get from one side to the other side of the box in the $x$ direction. Once it gets to the other side, it bounces off the wall and changes the sign of the $x$-component of the velocity (and momentum). At this moment, the momentum $p_x$ clearly changes the sign - i.e. changes by $2p_x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the change of momentum $p_x$ per unit time is &#xA;$$F_x = 2p_x / (a/v_x) = 2p_x/(am/p_x) = 2p_x^2/am$$&#xA;That's equal to $4/a$ times the kinetic energy $K_x$ in the $x$-direction. The pressure is&#xA;$$p = (F_x+F_y+F_z) / 6a^2 = 3\times 4/a\times K_x / 6a^2 = 4K/a / 6a^2 = 2K/3a^3  $$&#xA;But as I have said, the average kinetic (motion) energy per molecule is $K=3kT/2$ where $T$ is the absolute temperature, so the pressure is&#xA;$$p = 2k/3a^3 = kT/a^3= kT/V$$&#xA;Note that we have just derived $pV=kT$ for one molecule or $pV=NkT$ for $N$ molecules - which is what we wanted. The mass of the molecule canceled: if the molecule is heavier, the average velocity at a given temperature is slower. But that doesn't matter - because the molecule has a greater momentum (because of the higher mass) which is compensated by the longer time it needs to get from one side to the other to transfer this momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for a fixed temperature, the pressure is independent of the molecule type.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T09:13:29.157" CommentCount="1" />
  <row Id="2856" PostTypeId="2" ParentId="2838" CreationDate="2011-01-14T09:18:46.173" Score="4" ViewCount="" Body="&lt;p&gt;A longer popular text why energy conservation becomes trivial (or violated) in general relativity is e.g. here, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2010/08/why-and-how-energy-is-not-conserved-in.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/08/why-and-how-energy-is-not-conserved-in.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize four of the points:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: In GR, spacetime is dynamical, so in general, it is not time-translational-invariant. Because it's not, one can't apply Noether's theorem to argue that there is a conserved energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: One can see this in details in cosmology: the energy carried by radiation decreases as the universe expands as every photon's wavelength is getting bigger; the cosmological constant has a constant energy density while the volume is increasing, so the total energy carried by the cosmological constant (dark energy), on the contrary, grows; the latter increase is the reason why the mass of the Universe is large - during inflation, the total energy grew exponentially for 60+ e-foldings, before it was converted to matter that gave rise to early galaxies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3: If one defines the stress-energy tensor as the variation of the Lagrangian with respect to the metric tensor, which is OK for non-gravitating field theories, on gets zero in GR because the metric tensor is dynamical and the variation - like all variations - has to vanish because this is what defines the equations of motion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4: In translationally-invariant spaces such as the Minkowski space, the total energy is conserved again because Noether's theorem may be revived; however, one can't &quot;canonically&quot; write this energy as the integral of energy density over the space; more precisely, any choice to distribute the total energy &quot;locally&quot; will depend on the chosen coordinate system&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-14T14:17:29.090" LastActivityDate="2011-01-14T14:17:29.090" CommentCount="6" />
  <row Id="2857" PostTypeId="2" ParentId="2832" CreationDate="2011-01-14T09:29:49.540" Score="0" ViewCount="" Body="&lt;p&gt;Dear Jonathan, the balls' height above the Earth is the same for both balls at each time because of the principle of relativity - that is true even in Newton's physics, without all the special effects that occur near the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principle says that the phenomena obey the same laws of physics from a viewpoint of a uniformly moving observer. So if you have two systems that are moving, without respect to one another, in the $x$ and/or $y$ direction, it doesn't change the fact that the &quot;depth&quot; at time $t$ after any ball leaves the desk goes like&#xA;$$-z = gt^2/2$$&#xA;In some sense, you may imagine that the forces and the laws of physics act on each component of the velocity separately. So the velocity of a ball in the horizontal direction (one of them) has no impact on its motion in the vertical direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to include an amusing correction: the Earth is not completely flat, after all haha.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because it is not flat now, the very speedy ball may escape to outer space. The principle of relativity doesn't hold anymore: the presence of the round Earth distinguishes objects that are at rest with respect to the surface from those that move horizontally. It's because if you move the whole physical system in a linear direction, the background looks different because the Earth changes its location. This wasn't the case for the flat Earth which was invariant under horizontal translations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, you may still introduce a principle of relativity, but with an extra correction. You may use two frames - one that is static with respect to the Earth and another one that is rotating around an axis through the Earth's center exactly in such a way that the speedy ball will look like a ball at rest. However, the laws in rotating frames are not quite identical to the laws in the inertial frames. Fictitious forces have to be added.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second, rotating frame, however, there is an additional force - the centrifugal force - that needs to be added to the laws of physics. Because of this centrifugal force, the speedy ball is able to escape from the gravitational field while the slow ball is not. More precisely, there's a centripetal force (into the center) that acts on the ball at rest - which is moving in the rotating frame - and this centripetal force keeps the (originally) slow ball in the Earth's gravitational field.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-14T10:24:15.737" LastActivityDate="2011-01-14T10:24:15.737" />
  <row Id="2858" PostTypeId="2" ParentId="2188" CreationDate="2011-01-14T09:52:44.193" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;No, electricity - or anything else in physics - can't be explained &quot;completely accurately&quot; in laymen's terms. Electricity is the simplest subgroup of electromagnetic phenomena which are caused by the electromagnetic field - an electric vector and magnetic vector that exists at each point of space and time - that is interacting with the matter. The electric charge is conserved and may be thought of as a kind of &quot;material&quot; but the analogy inevitably fails in many respects. In modern physics, electromagnetism arises from U(1) gauge symmetries that the laymen usually don't like to hear about.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The electric currents are all about moving electrons. The electrons' mass is about 1/2000 of the mass of the protons. But the density of electrons in a piece of material doesn't really depend on the current so the electric phenomena don't modify the mass. There is a small disclaimer: according to relativity, any form of energy corresponds to a mass according to the $E=mc^2$ formula by Einstein, so any energy - including electrostatic energy - increases the total mass of an object. But $c^2$ is about $10^{17}$ squared meters per squared seconds so the mass corresponding to a reasonable energy is just tiny.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Electrons are the only light charged objects inside the matter that may easily move. So if one wants to move the electric charge - what is known as the electric current - it has to be done through electrons. In principle, any other charged particle could do the same job but protons are heavy &quot;cores&quot; of the matter that is mechanically attached somewhere while other particles such as muons are unstable and they're not included in normal matter (at least not enough). Electrons are fundamental for electromagnetism exactly because they're the lightest electrically charged particles in the Universe (together with their antiparticles, positrons).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This is the same question. The electric current is by definition a transfer of the electric charge and the electrons are the only particles with charge that may be transferred through matter, so it is completely correct to say that in all regular materials, all electric phenomena boil down to motion - and interactions - of the electrons. Chemistry (and biology) boils down to the motion of electrons in electric fields, too. No disclaimers are necessary here: it is fully true for any situation that a lawyer could possibly encounter in his life.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T09:52:44.193" />
  <row Id="2859" PostTypeId="2" ParentId="2774" CreationDate="2011-01-14T10:02:07.307" Score="1" ViewCount="" Body="&lt;p&gt;Dear Hernan, as the distant parts of the bar are approaching the speed of light, they become heavier, so it becomes harder to accelerate them: you can never reach (or surpass) the speed of light. It doesn't matter whether you try to accelerate the &quot;final segments&quot; of the bar by jets or by their attachment to the rest of the bar that is being pushed in the middle: the speed of light can never be reached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to speak in terms of torques and moments of inertia (of the bar), the moment of inertia goes to infinity - much like the mass itself - when the velocity of some points on the bar approaches the speed of light. So much like you have to modify the formulae for masses of moving objects by relativistic effects, you need to modify the formulae for the moments of inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your statement that you need a finite energy to get to the speed of light is invalid. You would need an infinite energy. For speeds below the speed of light, the total energy that you need can simply be calculated as the sum of the kinetic energies of all the segments of the bar.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T10:02:07.307" />
  <row Id="2860" PostTypeId="1" CreationDate="2011-01-14T10:10:24.590" Score="2" ViewCount="295" Body="&lt;p&gt;i need to get a nice picture about how electron moves around nucleus ??? i find concept of probability and orbitals  quite difficult to understand ??&lt;/p&gt;&#xA;" OwnerUserId="1278" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-14T21:04:30.943" LastActivityDate="2011-01-14T21:04:30.943" Title="how does electron move around nucleus?" Tags="&lt;quantum-mechanics&gt;&lt;atoms&gt;&lt;electron&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="2861" PostTypeId="2" ParentId="2860" CreationDate="2011-01-14T10:20:42.697" Score="1" ViewCount="" Body="&lt;p&gt;Dear Ashok, it is an experimentally established fact that the motion of electrons inside atoms requires us to speak about the probabilistic distributions - and wave functions whose absolute value has to be squared to obtain the probability density. The equation for the wave functions is known and nicely enough, its solutions - the orbitals - can be drawn. This is much more &quot;user-friendly&quot; a situation than in many other portions of physics in which the objects can't be even drawn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before quantum mechanics was born, Niels Bohr had a simpler model of the Hydrogen atom that was similar to the Solar System: the electrons were just rotating along circles. He postulated that the circumference of the circle couldn't have been arbitrary but had to satisfy a &quot;discrete&quot; condition. This captured many properties of the Hydrogen atom but wasn't right in details: the details require quantum mechanics and the probabilistic language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're not the only one who has problems with it - many people, including famous physicists (not to speak about thousands of not so famous physicists today, and many physics fans), didn't want to accept the probabilistic description as fundamental. They thought there had to exist a more &quot;understandable&quot; description, similar to classical physics. However, these people - including Einstein - have been proved wrong.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T10:20:42.697" CommentCount="1" />
  <row Id="2862" PostTypeId="2" ParentId="2860" CreationDate="2011-01-14T10:24:01.310" Score="0" ViewCount="" Body="&lt;p&gt;I find Wikipedia explanation quite explicit: an atomic orbital is a mathematical function that describes the wave-like behavior of either one electron or a pair of electrons in an atom. This function can be used to calculate the probability of finding any electron of an atom in any specific region around the atom's nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, nobody &quot;knows&quot; what the electron does around the nucleus, but what we know is that the absolute value of the wave function is the probability density that the electron can be found in that point of space. Plot it and you get the nice pictures of the orbitals. Otherwise, the concept of &quot;movement&quot; is a classical one that you cannot extend directly to small scales.&lt;/p&gt;&#xA;" OwnerUserId="1279" LastActivityDate="2011-01-14T10:24:01.310" CommentCount="1" />
  <row Id="2863" PostTypeId="2" ParentId="2833" CreationDate="2011-01-14T10:36:49.647" Score="1" ViewCount="" Body="&lt;p&gt;If you assume that two equally heavy black holes are at rest relatively to one another (and you may need some arrangement to prevent them from attracting one another if they're too close), then the light emitted at frequency $f$ at a distance $R$ from the first black hole's center will be observed as $f$ at the same distance $R$ from the second black hole's center. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's because the light will be redshifted and then blueshifted by the same factor - assuming that the source and the observer are sitting at fixed distances from the black hole centers (and they may need jets to be saved from the attraction towards the black holes). Only the ratios of $g_{00}$ influences the frequency ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red shift is also caused by the Doppler effect, the relative motion of the source and the observer. The Doppler effect exists even in the flat space, in special relativity, i.e. in the absence of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We observe the distant galaxies to be &quot;redder&quot; than the nearby galaxies. It means that the same kinds of sources - the same stars - produce &quot;redder&quot; light if the stars sit in a distant galaxy. Even if you assumed that a gravitational red shift from the stars etc. influences the light coming from the other galaxies, the red shift (expressed as a ratio of frequencies) would still have to be equal for the nearby and distance galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not equal which is why the hypothesis that all the &quot;Hubble&quot; red shift is due to the gravity of the massive objects themselves is ruled out. In reality, we know that it comes from the relative motion. But it is true that in general relativity, one can't &quot;qualitatively&quot; distinguish these two causes of red shift. In particular, the relative motion of galaxies away from each other may also be interpreted as a gravitational field - as the increasing size of the component of the metric tensor $g_{xx}$ in the direction $x$ connecting them, using coordinates in which both galaxies sit at time-independent values of $x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When these &quot;unusual coordinates&quot; and interpretations are allowed, it is true that all red shift in general relativity may be interpreted as a (generalized) gravitational red shift.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T10:36:49.647" />
  <row Id="2864" PostTypeId="2" ParentId="2605" CreationDate="2011-01-14T10:45:04.760" Score="0" ViewCount="" Body="&lt;p&gt;When you lose weight, it is mostly fat that disappears from under your skin (or someone else's skin). How does the fat get out? Most of it doesn't directly evaporate. Instead, it is burned much like fuel in your car. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fat contains mainly carbon, hydrogen, and oxygen, much like typical organic compounds. When it's &quot;ideally&quot; burned, only water and carbon dioxide - $H_2 O$ and $CO_2$ - are left. The gas $CO_2$ gets out by breathing and farting (not sure about the relative significance) - in which you only inhale $O_2$, the oxygen - while $H_2 O$ is released from the body by urination and sweating. Well, maybe, some other material may also be added to the excrements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is too idealized because by burning fats, one also produces some &quot;less clean&quot; compounds that are added both to urine as well as excrements (you asked, I am just answering). Farts also contain other gases that carry a certain small fraction of the mass of the burned fat - including methane. Biology is complicated but the difference between animals and cars is not so dramatic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gaining weight is the nearly reverse process except that the carbon is not taken from the air or $CO_2$ - we would have to be plants able to do photosynthesis do achieve this goal. Instead, it is taken from the food.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T10:45:04.760" CommentCount="1" />
  <row Id="2865" PostTypeId="1" CreationDate="2011-01-14T10:49:44.870" Score="7" ViewCount="959" Body="&lt;p&gt;&lt;em&gt;EDIT: Further clarification in the context of answers/comments received to 20 Jan has been appended&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT: 21 Jan - Response to the Lubos Expansion appended [in progress, not yet complete]&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT: 23 Jan - Visser's calculations appended&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT: 26 Jan - Peter Shor's thought experiments rebutted&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Summary to Date (26 Jan)&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: are the Morris, Thorne, Yurtsever (MTY) and Visser mechanisms for converting a wormhole into a time machine valid? The objection to the former is that the &quot;motion&quot; of a wormhole mouth is treated in an inadmissable manner by the former, and that the &lt;em&gt;valid&lt;/em&gt; mathematical treatment of the latter is subsequently misapplied to a case in which a sufficient (and probably necessary) condition does not apply (existence of a temporal discontinuity). It is maintained that extant thought experiments lead to incorrect conclusions because in the former case correct treatment introduces factors that break inertial equivalence between an unaccelerated rocket and co-moving wormhole mouth, and in the latter case especially do not respect the distinction between &lt;em&gt;temporal coordinate values&lt;/em&gt; and spacetime &lt;em&gt;separations.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the detailed treatment of the Visser case is reproduced below, a valid argument in favour of a wormhole time machine must show how an interval $ds^2=0$ (the condition for a Closed Timelike Curve) obtains in the absence of a temporal discontinuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In considering the MTY (1988) paper, careful consideration should be given to whether the authors actually transport a wormhole mouth or just the coordinate frame that is convenient for describing a wormhole mouth if one happened to exist there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Issues concerning quantum effects, energy conditions, and whether any created time-machine could persist etc. are off topic; the question is solely about the validity of the reasoning and maths concerning time-machine creation from a wormhole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original postings in chronological order are below.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Over at Cosmic Variance Sean Carroll recommended this place for the quality of the contributions so I thought I would try my unanswered question here; it is most definitely for experts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is simple, and will be stated first, but I'll supplement the question with specific issues concerning standard explanations and why I am unable to reconcile them with what seem to be other important considerations. In other words, I'm not denying the conclusions I'm just saying &quot;I don't get it,&quot; and would someone please set me straight by e.g. showing me where the apparent counter-arguments/reanalyses break down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: &lt;strong&gt;How can a wormhole be converted into a time machine?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supplementary stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no problem with time-travel per se, especially in the context of GR. Godel universes, van Stockum machines, the possibilities of self-consistent histories, etc. etc. are all perfectly acceptable. The question relates specifically to the application of SR and GR to wormholes and the creation of time-differences between the mouths - leading to time-machines, as put forward in (A) the seminal Morris, Thorne, Yurtsever paper (Wormholes, Time Machines, and the Weak Energy Condition, 1987) and explained at some length in (B) Matt Visser's Book (Lorentzian Wormholes From Einstein To Hawking, Springer-Verlag, 1996).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A -- Context. MTY explore the case of an idealised wormhole in which one mouth undertakes a round trip journey (i.e. undergoes accelerated motion, per the standard &quot;Twins Paradox&quot; SR example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is unclear to me is how MTY's conclusions are justified given that the moving wormhole mouth is treated as moving against a Minkowskian background: specifically, &lt;strong&gt;can someone explain how the wormhole motion is valid as a diffeomorphism&lt;/strong&gt;, which my limited understanding suggests is the only permitted type of manifold transformation in relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elaborating... wormhole construction is generally described as taking an underlying manifold, excising two spherical regions and identifying the surfaces of those two regions. In the MTY case, if the background space is Minkowski space and remains undistorted, then at times t, and t' the wormhole mouth undergoing &quot;motion&quot; seems to identify different sets of points (i.e. different spheres have to be excised from the underlying manifold) and so there is no single manifold, no diffeomorphism. [Loose physical analogue: bend a piece of paper into a capital Omega shape and let the &quot;heels&quot; touch... while maintaining contact between them the paper can be slid and the point of contact &quot;moves&quot; but different sets of points are in contact]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm happy with everything else in the paper but this one point, which seems to me fundamental: moving one wormhole mouth around requires that the metric change so as to stretch/shrink the space between the ends of the wormhole, i.e. the inference of a time-machine is an artefact of the original approach in which the spacetime manifold is treated in two incompatible ways simultaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corollary: as a way of re-doing it consistently, consider placing the &quot;moving&quot; wormhole mouth in an Alcubierre style warp bubble (practicality irrelevant - it just provides a neat handle on metric changes), although in this case v is less than c (call it an un-warp bubble for subluminal transport, noting in passing that it is in fact mildly more practicable than a super-luminal transport system). As per the usual Alcubierre drive, there is no time dilation within the bubble, and the standard wormhole-mouth-in-a-rocket thought-experiment (Kip Thorne and many others) produces a null result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B -- Context. (s18.3 p239 onwards) Visser develops a calculation that begins with two separate universes in which time runs at different rates. These are bridged and joined at infinity to make a single wormhole universe with a temporal discontinuity. The assumption of such a discontinuity does indeed lead to the emergence of a time machine, but when a time-machine is to be manufactured within a single universe GR time dilation is invoked (one wormhole mouth is placed in a gravitational potential) to cause time to flow at different rates at the two ends of the wormhole (to recreate the effect described in the two universe case in which time just naturally flowed at different rates). However, in the case of a simple intra-universe wormhole there is no temporal discontinuity (nor can I see how one might be induced) and the application of the previously derived equations produces no net effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, as I read the explanations, neither SR nor GR effects create a time-machine out of a (intra-universe) wormhole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where have I gone wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Julian Moore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 1: ADDITIONAL COMMENTS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lubos' initial answer is informative, but - like several other comments (such as Lawrence B. Crowell's) - the answerer's focus is on the impossibility of an appropriate wormhole per se, and not the reasoning &amp;amp; maths used by Morris, Thorne, Yurtsever &amp;amp; Visser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree (to the extent that I understand them) with the QM issues, however the answer sought should assume that a wormhole can exist and then eliminate the difficulties I have noted with the creation of time differences between the wormhole mouths. Peter Shor's answer &lt;em&gt;assumes&lt;/em&gt; that SR effects will apply and merely offers a way to put a wormhole mouth in motion; the question is really, regardless of how motion might be created, does &amp;amp; how does SR (in this case) lead to the claimed effect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the GR case is simplest because the maths in Visser's book is straightforward, and if there is no temporal discontinuity, the equations (which I have no problem with) say no time machine is created by putting a wormhole mouth into a strong gravitational potential. The appropriate answer to the GR part of the question is therefore to show how a time machine arises in the absence of a temporal discontinuity or to show how a temporal &lt;em&gt;discontinuity&lt;/em&gt; could be created (break any energy condition you like, as far as I can see the absolute &lt;em&gt;discontinuity&lt;/em&gt; required can't be obtained... invoking QM wouldn't help as the boundary would be smeared by QM effects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lubos said that &quot;an asymmetry could gradually increase the time delay between the two spacetime points that are connected by the wormhole&quot;, I'm saying, &quot;My application of the expert's (Visser's) math says it doesn't - how is my application in error?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The SR case is much trickier conceptually. I am asserting that the wormhole motion of MTY is &lt;em&gt;in principle&lt;/em&gt; impossible because, not to put too fine a point on it, the wormhole mouth doesn't &quot;move&quot;. Consider a succession of snapshots showing the &quot;moving&quot; wormhole mouth at different times and then view them quickly; like a film one has an appearance of movement from still images, but in this case the problem is that the wormhole mouth in each frame is a &lt;em&gt;different&lt;/em&gt; mouth. If the background is fixed Minkowski space (ie remains undistorted) at times t and t' different regions of the underlying manifold have to be excised to create the wormhole at those times... so the wormhole manifolds are different manifolds. If the background is not fixed Minkowski space, then it can be distorted and a mouth can &quot;move&quot; but this is a global rather than a local effect, and just like the spacetime in an Alcubierre warp bubble nothing is happening locally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider two points A &amp;amp; B and first stretch and then shrink the space between them by suitable metric engineering: is there are time difference between them afterwards? A simple symmetry argument says there can't be, so if the wormhole mouths are treated as features of the manifold rather than objects in the manifold (as it seems MTY treat them) then the only way to change their separation is by metric changes between them and no time machine can arise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if a time-machine &lt;em&gt;could&lt;/em&gt; be created either way, it would indeed almost certainly destroy itself through feedback... but, to repeat, this is not the issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Robert Smith for putting the bounty on this question on my behalf, and thanks to all contributors so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2: Re The Lubos Expansion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lubos gives an example of a wormhole spacetime that appears to be a time-machine, and then offers four ways of getting rid of the prospective or resulting time machine for those who object in principle. Whilst I appreciate the difficulties with time-machines I am neither for nor against them per sunt, so I will concentrate on the creation issue. I have illustrated my interpretation of Lubos' description below.&#xA;&lt;img src=&quot;http://i.stack.imgur.com/D1wh7.jpg&quot; alt=&quot;Block Universe Wormhole&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, there is nothing in GR that in principle prevents one from having a manifold in which two otherwise spacelike surfaces are connected in such a way as to permit some sort of time travel. This is the situation shown in the upper part of the illustration. The question is how can the situation in the upper part of the illustration be obtained from the situation in the lower part?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the illustration below of a spacelike surface with a simple wormhole (which I think is a valid foliation of e.g. a toroidal universe). As time passes, the two mouths move apart thanks to expansion of space between them, and then close up again by the inverse process (as indicated by the changing separation of the dotted lines, which remain stationary)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/JT5eo.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 3: Visser's calculations reproduced for inspection&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Consider the result in the case where there is no temporal discontuity using the equations derived for the case where there is a discontinuity, given below&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Visser, section 18.3, p239&#xA;The general metric for a spherically symmetric static wormhole&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;ds^2~=~-e^{2\phi(l)}~dt^2~+~dl^2~+~r^2[d\theta^2~+~sin^2\theta~d\psi^2]~~~~~(18.35)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that &quot;there is no particular reason to demand that time run at the same rate on either side of the wormhole. More precisely, it is perfectly acceptable to have $ф(l=+\infty)~\neq~ф(l=-\infty)$&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reduce to (1+1) dimensions for simplicity and consider&#xA;$$&#xD;&#xA;ds^2~=~-e^{2\phi(l)}~dt^2~+~dl^2~~~~~~~~~~(18.36)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The range of l is (-L/2,+L/2) and l=-L/2 is to be identified with l=+L/2. &#xA;Define&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\phi_\pm\equiv\phi(l=\pm~L/2); ~\Delta\phi\equiv\phi_+~-~\phi_-~~~~~~~(18.37)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at the junction $l=\pm~L/2~~$ the metric has to be smooth, ie. $ds = \sqrt{g_{\mu\nu}{dx^\mu}{dx^\nu}}$ is smooth, implying&#xA;$$&#xD;&#xA;d\tau=e^{\phi_-}dt_-=e^{\phi_+}dt_+~~~~~~~~~(18.38)&#xD;&#xA;$$&#xA;Define the time coordinate origin by identifying the points&#xA;$$&#xD;&#xA;(0,-L/2)\equiv(0,+L/2)~~~~~~(18.39)&#xD;&#xA;$$&#xA;then the temporal discontinuity is&#xA;$$&#xD;&#xA;t_+=t_-e^{(\phi_-~-~\phi_+)}~=~t_-e^{(-\Delta\phi)}~~~~~~(18.40)&#xD;&#xA;$$&#xA;leading to the identification&#xA;$$&#xD;&#xA;(t_-,-L/2)\equiv(t_-e^{-\Delta\phi},+L/2)~~~~~~~~~(18.41)&#xD;&#xA;$$&#xA;which makes the metric smooth across the junction. Now consider a null geodesic, i.e. ds=0, which is&#xA;$$&#xD;&#xA;{dl\over{dt}}=\pm{e^{+\phi(l)}}~~~~~(18.42)&#xD;&#xA;$$&#xA;where the different signs correspond to right/left moving rays.&#xA;Integrate to evaluate for a right moving ray, with conventions $t_f$ is the final time and $t_i$ is the initial time&#xA;$$&#xD;&#xA;[t_f]_+=[t_i]_-+\int_{-L/2}^{+L/2}e^{-\phi(l)}dl~~~~~(18.43)&#xD;&#xA;$$&#xA;then apply the coordinate discontinuity matching condition to determine that the ray returns to the starting point at coordinate time&#xA;$$&#xD;&#xA;[t_f]_-=[t_f]_+e^{\Delta\phi} = [[t_i]_-+\oint{e^{-\phi(l)}}dl]e^{\Delta\phi}~~~~~(18.44)&#xD;&#xA;$$&#xA;A closed right moving null curve exists if $t[_f]_-=[t_i]_-$, i.e.&#xA;$$&#xD;&#xA;[t_i]^R_-={{\oint{e^{-\phi(l)}dl}}\over{e^{\Delta\phi}-1}}~~~~~(18.45)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 4: Peter Shor's thought experiments re-viewed&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peter Shor has acknowledged (at the level of &quot;I think I see what you mean...&quot;) both the arguments against wormhole time machine creation (the absence of the required temporal discontinuity in spacetime if GR effects are to be used, and that wormhole mouth motion requires metric evolution inconsistent with the Minkowksi space argument of MTY) but still believes that such a wormhole time-machine can be created by either of the standard methods offered a thought experiment. This is a counter to those thought experiments and whilst it does not constitute proof of the contrary (I don't think such thought experiments are rich enough to provide proof either way), I believe it casts serious doubt on their interpretation, thereby undermining the objections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The counter arguments rely on the key distinction between the &lt;em&gt;values&lt;/em&gt; of the time coordinate and the &lt;em&gt;separation&lt;/em&gt; of events ($ds^2$). Paragraphs are numbered for ease of reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Consider the classic Twins Paradox situation and the associated Minkowksi diagram. When the travelling twin returns she has the same t coordinate ($T_{return}$) as her stay-at-home brother (who said they had to be homozygous? :) ) As we all know, despite appearances to the contrary on such a diagram, the sister's journey is in fact shorter (thanks to the mixed signs in the metric), thus it has taken her &quot;less time&quot; to reach $T_{return}$ than it took her brother. Less time has elapsed, but she is not &quot;in the past&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Now consider the gravity dunking equivalent Twins scenario. This time he sits in a potential well for a while and then returns to his sister who has stayed in flat space. Again their t coordinate is the same, but again there is a difference in separations; this time his is shorter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Now for the travelling/dunking Twin substitute a wormhole mouth; then the wormhole mouths are brought together they do so at the same value of t. The moving mouths may have &quot;travelled&quot; shorter spacetime distances but they are not &quot;in the past&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) Suppose now we up the ante and give the travelling/dunking Twin a wormhole mouth to keep with them...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) According to the usual stories, Mr A can watch Ms A receding in her rocket - thereby observing her clock slow down - or he can communicate with her through the wormhole, through which he does &lt;em&gt;not&lt;/em&gt; see her clock slow down because there is no relative motion between the wormhole mouth and Ms A. Since this seems a perfectly coherent picture we are inevitably led to the conclusion that a time machine comes into existence in due course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(6) My objection to this is that there are reasons to doubt what is claimed to be seen through the wormhole, and if the absence of time dilation is not observed &lt;em&gt;through the wormhole&lt;/em&gt; we will not be led to the creation of a time machine. So, what &lt;em&gt;would&lt;/em&gt; one see through the wormhole, and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(7) I return to the question of the allowable transformations of the spacetime manifold. If a wormhole mouth is rushing &quot;through&quot; space, the space around it must be subject to distortion. Now, whilst there are reasons to doubt that one can arrange matter in such a way as to create the required distortion (the various energy condition objections to the original Alcurbierre proposal, for instance), we are less concerned about the &lt;em&gt;how&lt;/em&gt; and more concerned with the &lt;em&gt;what if&lt;/em&gt; (particularly since, if spacetime cannot &quot;move&quot; to permit the wormhole mouth to &quot;move&quot;, the whole question becomes redundant). The very fact that spacetime around the &quot;moving&quot; wormhole mouth is going to be distorted suggests at least the possibility that what is observed through the wormhole is consistent with what is observed the other way, or that effects beyond the scope of the equivalence principle demonstrate that observation through the wormhole is not equivalent to observing from an inertial frame. Unfortunately I don't have the math to peform the required calculations, but insofar as there is a principled objection to the creation of a time machine as commonly described, I would hope that someone would check it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(8) What then for the dunking Twin? In this case there is no &quot;motion&quot; of the wormhole mouth, so no compensating effects can be sought from motion. However, I believe that one can apply to the metric for help. Suppose that the wormhole mouth in the gravitational well is actually embedded in a little bit of flat space, then (assuming the wormhole itself is essentially flat) the curvature transition happens outside the mouth and looking through the wormhole should be like looking around it: Mr A seems very slowed down. If, instead, the wormhole mouth  is fully embedded in the strongly curved space that Mr A also occupies, then the wormhole cannot be uniformly flat and again looking through the wormhole we see exactly what we see around it (at least as far as the tick of Mr A's watch is concerned.) but the transition from flat to curved space (and hence the change in clock rates) occurs over the interior region of the wormhole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(9) Taken with the &quot;usual&quot; such thought experiments, we now have contradictory but equally plausible views of the same situations, and they can't both be right. I feel however that no qualitative refinements will resolve the issue, thus I prefer the math, which seems to make it quite plain that the usually supposed effects do not in fact occur. Similarly, if you disagree that the alternative view is plausible, since the &quot;usual&quot; result does not seem plausible to me,maths again provides the only common ground where the disagreement can be resolved. I urge others to calculate the round-trip separations using the equations provided from Visser's work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(10) I say the MTY paper is in error because it treats spacetime as flat, rigid Minkowskian and then treats the &quot;motion&quot; of a wormhole mouth in a way that is fundamentall incompatible with a flat rigid background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(11) I say Visser is in error in applying his (correct) inter-universe wormhole result to an intra-universe wormhole where the absence of the temporal discontinuity in the latter nullifies the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(12) These objections have been acknowledged but but no equally substantive arguments to undermine them (i.e. to support the extant results) has been forthcoming; they have not been tackled head on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(14) I am not comfortable with any of the qualitative arguments either for against wormhole-time machines; an unending series of thought experiments is conceivable, each more intricate and ultimately less convincing than the last. I don't want to go there; look at the maths and object rigorously if possible.&lt;/p&gt;&#xA;" OwnerUserId="1280" LastEditorUserId="1280" LastEditorDisplayName="" LastEditDate="2011-01-27T10:53:36.633" LastActivityDate="2011-01-27T10:53:36.633" CommunityOwnedDate="2011-01-26T16:20:07.400" Title="Wormholes &amp; Time Machines - for *experts* in GR/maths" Tags="&lt;general-relativity&gt;&lt;time-travel&gt;" AnswerCount="3" CommentCount="62" FavoriteCount="3" />
  <row Id="2866" PostTypeId="2" ParentId="1901" CreationDate="2011-01-14T11:02:34.143" Score="3" ViewCount="" Body="&lt;p&gt;About 73% of the energy of the Universe is stored in dark energy - the cosmological constant, most likely - which has a negative pressure numerically equal to the energy density. Because it is a cosmological &quot;constant&quot;, this portion of the mass of the Universe cannot be really compressed. So it is a problematic idea to include the dark energy into the &quot;mass that you want to compress&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only dark matter and visible matter - whose pressure is nearly zero - can really be &quot;compressed&quot;. So that would be about $1\times 10^{54}$ for the visible Universe. $2GM/c^2$ for this mass $M$ produces about 150 billion light years which is about 3 times larger than the radius of the observable Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We obtained the result that even if we only count the &quot;particulate&quot; component of the mass of the Universe, we find out that the Universe is actually smaller than the black hole. It is no contradiction because the Universe is not a static system. It would be impossible for the Universe to be kept this small. However, the galaxies are receding from each other which prevents the formation of a nearby event horizon. Instead, the closest horizon in the reality is the &quot;cosmic horizon&quot; - how far we can actually see because of the finite speed of light and finite age of the Universe. It depends on the observer but effectively makes everything beyond this horizon unphysical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our Universe, dominated by the dark energy, is already rather close to an empty de Sitter space which is, from many viewpoints, analogous to a black hole except that the interior of the visible part of the de Sitter space is analogous to the &lt;em&gt;exterior&lt;/em&gt; of a normal black hole, and the analogy of the &lt;em&gt;interior&lt;/em&gt; of a black hole is everything that is behind the cosmic horizon - where we don't see. It is misleading to create the analogy with the static black holes directly because our Universe is not static in the normal cosmological coordinates.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T11:02:34.143" />
  <row Id="2867" PostTypeId="2" ParentId="2832" CreationDate="2011-01-14T11:06:27.480" Score="0" ViewCount="" Body="&lt;p&gt;Your confusion arises because a lack of understanding general relativity.  Einstein tells us that gravity is a fictitious force.&#xA;//en.wikipedia.org/wiki/Fictitious_forces#Gravity_as_a_fictitious_force&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, Einstein tells us that all objects will follow timelike geodesics&#xA;//en.wikipedia.org/wiki/Geodesic_(general_relativity)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference in behaviour that you note are merely due to a change in initial conditions. In the examples you give the variables are position, momentum and direction, these determine the particular geodesic each ball will follow.  The difference in why one ball smacks into the earth and not the other is related to particles of the earth obstructing the path of the ball that was directed at its center.   If you could imagine that there was a precise hole through the earth along the path of the downward moving ball, you would see that it would in fact oscillate for eternity from one side of the earth to the other; just as the orbiting ball would oscillate around the earth for eternity (barring the influence of non-gravitational forces, or effects of cosmological constants, or extreme gravitional fields that create singularities).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking for a reasonable introduction to the subject of relativity, I would recommend Relativity DeMystified by David McMahon as a reasonable primer to Gravitation by Misner Thorne and Wheeler&lt;/p&gt;&#xA;" LastActivityDate="2011-01-14T11:15:13.733" />
  <row Id="2868" PostTypeId="2" ParentId="2264" CreationDate="2011-01-14T11:15:19.833" Score="1" ViewCount="" Body="&lt;p&gt;a good question. The discrepancy between the counting for electric and magnetic multipoles appears because in the potentials, the electric field is encoded in the potential $\phi$ while the magnetic field is encoded in the vector potential $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to find the electric field $E$, you take the gradient of $\phi$ combined with the time derivative of $A$ (both with a minus sign). There are no spatial derivatives in the second term so it is the dominant one, in a $1/r$ expansion. To obtain the magnetic field $B$, you need to take the curl of $A$ - which means its spatial derivatives. There is no way to get a term to $B$ from $\phi$ and $A$ that would contain no spatial derivatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a relativity viewpoint, spatial and temporal derivatives are of the same order. However, things are different for the analysis of static objects. It's only the spatial derivatives that produce an extra power of $1/r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the electric field $E$ is of the same order as the vector potential $A$ (the time derivative doesn't add any $1/r$) while the magnetic field $B$ has an extra $1/r$ relatively to the vector potential $A$. Quadrupoles have one extra $1/r$ relatively to the dipoles - the standard expansion in $1/r$. So to go from the electric dipoles to the next order, you either add the $1/r$ by switching from the electric to magnetic fields; or you get the extra $1/r$ by going from electric dipoles to electric quadrupoles. At any rate, the counting of powers is shifted by one between the electric and magnetic multipoles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T11:15:19.833" CommentCount="1" />
  <row Id="2869" PostTypeId="2" ParentId="2672" CreationDate="2011-01-14T11:42:01.160" Score="4" ViewCount="" Body="&lt;p&gt;The summing over final states and the averaging over initial states is a good observation that I always emphasize as the origin of the arrow of time. As soon as one considers mathematical logic, this asymmetry has to arise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are we summing over final states? Because &quot;we don't care&quot; about which of them occurs (and no one knows). We're calculating the probability - or cross section (which is the same thing, up to unproblematic coefficients) - of getting a final state $F_1$ or $F_2$. Both of them are OK so I use &quot;or&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the probability &#xA;$$P(F_1 \mbox{or} F_2) = P(F_1)+P(F_2)$$&#xA;is simply the sum if the two final states are mutually excluding - e.g. orthogonal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things are very different for the initial states. Now, it's not quite true that &quot;we don't care&quot; whether the state was $I_1$ or $I_2$. Instead, &quot;we don't know&quot; which of them it was (but we know that one of them was the right one). We have to make probabilistic assumptions about the initial state. The most &quot;balanced&quot; one is that each $I_1$ and $I_2$ appear with the 50 percent probability. There could also be an asymmetric choice of the &quot;priors&quot; - leading to a weighted average - but it's important that&#xA;$$P(I_1)+P(I_2)=1$$&#xA;The probabilities of the evolution include the calculate probabilities of transitions from $I_k$ to $F_l$ - given by the Feynman diagrams etc. - but for each $I_k$, we must also multiply this by the probability that $I_k$ occurred in the first place. This reduces the calculation by a factor of two. The total and averaged probability is&#xA;$$P(I\to F) = 1/N_{initial} \sum_{ij} P(I_i \to F_j)$$&#xA;The very asymmetry here - the extra factor $1/N_{initial}$ without its &quot;final&quot; counterpart - is the reason why low-entropy states are favored as initial states while the high-entropy ones are favored as final states. In particular, you may compute the ratio of the probabilities from I to F, and from F to I, and they will differ by the factor of&#xA;$$N_{final}/N_{initial} = exp(S_f-S_i)$$&#xA;or the exponential of the entropy difference between the final and initial state. This number's being very different from one is the reason why the transitions to lower-entropy states can't occur in practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many people - including professional physicists - remain extremely confused about these points when they suggest that the arrow of time remains mysterious. It doesn't. The second law of thermodynamics may be proved and it all boils down to your question and the right answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only assumptions I made are those about the addition of probabilities of assumptions and their effects - and these logical rules are fundamentally asymmetric when it comes to the role of the assumptions and their consequences. This logical arrow of time can't be removed from any reasoning about a world that depends on time - time only copies the logical relationship of implication. And this logical arrow of time is the source of the thermodynamic arrow of time as well.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T15:23:34.120" LastActivityDate="2011-02-02T15:23:34.120" CommentCount="2" />
  <row Id="2870" PostTypeId="2" ParentId="2649" CreationDate="2011-01-14T11:51:19.503" Score="2" ViewCount="" Body="&lt;p&gt;Yes, absolutely, the &quot;proper distances vanish&quot; is a singular point of the configuration space of general relativity and it is not permissible to &quot;expand around it&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you are also totally right about the &quot;imaginary roots&quot; problem. The map was chosen to be bilinear exactly because the loop quantum gravity physicists wanted the areas to be quantized in a simple way. So the triad $E$ really measures the proper areas and the proper distances must be taken as the square root of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is nothing physically legitimate about this operation - it is just a trick to pretend that the metric tensor is equivalent to a bulk gauge field. While the counting of components of the SU(2) gauge field could work, the physics doesn't work and the field redefinition is not one-to-one. The appearance of the imaginary roots is just one manifestation of the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other manifestations of the illegitimacy of the field redefinition from gravity to the &quot;new variables&quot;. In particular, the quantization of the areas themselves proves that the map is only valid locally on the configuration space, but not globally. The holonomies or Wilson loops constructed out of the gauge field are periodic variables; that's essentially why their canonical momenta are quantized - and the canonical momenta are the areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the proper gravity, with the continuous metric tensor, there is no quantization of the areas. In fact, such a quantization violates the local Lorentz symmetry because the &quot;spin network&quot; always picks a preferred reference frame, much like the luminiferous aether. This brutally breaks the Lorentz symmetry and destroys inertia: the entropy density is pretty much Planckian and an object moving through this material will instantly dissipate its energy to the &quot;spin network degrees of freedom&quot; and stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the field redefinition is only a game that is valid locally at the configuration space - because of the  right counting of the degrees of freedom - but is incompatible globally, because it is not one-to-one, and it incompatible with the required dynamics for the degrees of freedom. In the new variables, one can never get a theory that behaves as general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why the word &quot;gravity&quot; in &quot;loop quantum gravity&quot; is a misnomer. There can't be any gravity.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T11:51:19.503" CommentCount="1" />
  <row Id="2871" PostTypeId="2" ParentId="2628" CreationDate="2011-01-14T12:01:07.440" Score="2" ViewCount="" Body="&lt;p&gt;The space in GR and the vacuum of the SM are meant to represent the same physical object - the space out there (imagining me waving my hands and pointing to a random direction now).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the GR and the SM focus on different sets of phenomena that occur in that environment, so they say very different things about it. The Standard Model neglects gravity - which is the main thing that General Relativity wants to study. So in GR, the space can get curved and expanded; in the SM, it can't. In the SM, the vacuum has lots of activity and virtual particles; in GR, there is  nothing going on in the vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the real vacuum/space around us has the complicated properties from both GR and SM. It can get curved and expanded and there is a lot of microscopic virtual particle activity in it, too. The only known consistent theory that incorporates both GR-like and SM-like aspects of the vacuum - and anything else - is string/M-theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The word &quot;vacuum&quot; has different meanings in GR, SM, and string theory. While the meanings in GR and SM were sketched above, in string theory, we usually mean a particular &quot;empty space&quot; with the maximum (Poincare or de Sitter or anti de Sitter) symmetries. There are many solutions of string theory that fit into this category; their set is sometimes referred to as the &quot;landscape&quot;. So we often hear that we have $10^{500}$ semi-realistic vacua in string theory although the number is a guess rather than a result of a fully controllable calculation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of those vacua - we don't know which one - is identified with the vacuum/space we see around, and all its aspects that are included both in GR and SM can be extracted from string theory. Qualitatively, we know it to be the case - all of GR and SM effects can be extracted from the stringy vacua (or a subclass of them). We just don't know which one is exactly the right one so that we could calculate all properties of the real world, including the elementary particle masses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in the Standard Model or quantum field theory, the vacuum itself &quot;knows&quot; about the properties of all particles because they're constantly emerging and disappearing in the vacuum as virtual particles.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T12:01:07.440" />
  <row Id="2872" PostTypeId="1" AcceptedAnswerId="2893" CreationDate="2011-01-14T12:01:34.990" Score="1" ViewCount="169" Body="&lt;p&gt;i know that the claim is that nucleosynthesis models say that the observed abundance of deuterium is too high for dark matter to be baryonic, but couldn't deuterium be a waste product of civilizations?  you say that postulating civilizations is not the simplest explanation, but isn't it far more complex to postulate that we are unique (or an invisible form of matter with exotic physics for which we have no evidence)?  even from a single seed civilization, galactic colonization times are in the hundred million year range, so we should expect some evidence of environmental manipulation everywhere we look.  given the need for civilizations to hide from predators, isn't eliminating EM interactions/emissions the best strategy?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-14T15:13:58.100" Title="why can't the missing mass (evidence for dark matter) be matrioshka brains/dyson spheres?" Tags="&lt;dark-matter&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="2873" PostTypeId="1" AcceptedAnswerId="2879" CreationDate="2011-01-14T12:12:17.277" Score="6" ViewCount="461" Body="&lt;p&gt;This number is thrown around a lot so I'd like to understand its origin. I know that it counts compactifications of string theories on (not only) Calabi-Yau manifolds and probably also some other ingredients.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1.&amp;nbsp;What precisely are those ingredients that differ among the stringy vacua?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But as far as I know no one knows the precise number of Calabi-Yau 3-folds (it might still well be infinite). Is the estimate based on the number of presently found 3-folds?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2.&amp;nbsp;Where do the estimates come from?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="329" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T14:43:34.730" LastActivityDate="2011-02-04T07:18:25.227" Title="Where does the $10^{500}$ estimate for the number of stringy vacua come from?" Tags="&lt;string-theory&gt;&lt;quantum-gravity&gt;&lt;anthropic-principle&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="2874" PostTypeId="2" ParentId="2585" CreationDate="2011-01-14T12:16:19.523" Score="1" ViewCount="" Body="&lt;p&gt;Dear Sina, if you replace &quot;0&quot; and &quot;1&quot; by &quot;up&quot; and &quot;down&quot;, you get a similar state for two spins - which is referred to as the singlet. All these states are mathematically analogous except that the states &quot;0&quot; or &quot;1&quot;, or &quot;up&quot; and &quot;down&quot;, or &quot;plus&quot; and &quot;minus&quot; (as indices of your $\chi$) may mean physically different things - i.e. these states may influence the interactions of the system with other degrees of freedom differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the spin &quot;up&quot; and &quot;down&quot; likes to add some $-\mu.B$ energy in a magnetic field that depends on the direction of the spin. Other degrees of freedom interact differently - and must be prepared by different apparata, depending on the context. At the level of &quot;information&quot;, you always have two subsystems whose 1 qubit of information is correlated with the other in the same way; from the viewpoint of all physics, they can be very different things (just think about all the ways how qubits may be realized in quantum computers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the state of the form $|01-10\rangle$ is always entangled: the quantum numbers of the two fermions (or subsystems) $1,2$ in the state are nontrivially correlated. This doesn't prove any interaction - it just proves that they were prepared to have correlated properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that the state is entangled, regardless of the symbols, note that it cannot be written as a tensor product of a state for the fermion or subsystem 1, multiplied by another state of the subsystem or fermion 2. Equivalently, you may trace over the 2 degrees of freedom, to get a density matrix for the subsystem 1. And you will get $\mbox{diag}(0.5,0.5)$ which has a nonzero entropy $\rho \ln(\rho)$, proving that the state isn't poor. Because the induced 1-particle state isn't poor, it proves that the original state of the two particles was entangled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost all states in the multi-particle Hilbert space are entangled, of course. However, there are often reasons to assume that two systems are not entangled - because they didn't influence each other in the past (or at least not much).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T12:16:19.523" />
  <row Id="2875" PostTypeId="2" ParentId="2579" CreationDate="2011-01-14T12:27:02.217" Score="2" ViewCount="" Body="&lt;p&gt;these are good questions but already assume some fallacies in their very formulation. In particular, it is misleading to mix &quot;fuzzballs&quot; and &quot;Hawking evaporation&quot; in the same sentence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're talking about fuzzballs, then you are talking about very particular microstates of the black holes. There are many of them - $\exp(S)$ of them  where $S$ is the entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hawking evaporation - as a radiation of a temperature - was derived in the semiclassical approximation for an object that macroscopically behaves as a black hole with certain parameters. To obtain it from the accurate theory - string theory - you have to compute averages over the ensemble of all the microstates, and if you want to be 100% accurate, you have to tell us what your ensemble of states is and what are your weights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, this task can be done in full detail, at least in principle, for many black holes in particular descriptions of string theory such as Matrix theory or AdS/CFT. A result is that the low-energy approximation of course fully agrees with the conclusions derived by Hawking semiclassically. Moreover, it may be showed that the new physics of string theory guarantees that the information is preserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you talk about a particular fuzzball, you may also calculate its decay from string theory, at least in principle, but it will be a decay of a very generic and messy object and the decay shouldn't be called Hawking radiation. In particular, a fuzzball is a single microstate, so it should be said to have no entropy. By the relationship between entropy and area of event horizons, it also follows that fuzzballs don't have an event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, by averaging over the microstates or fuzzballs, one may derive the semiclassical results that look like temperature-T radiations. I won't write that these things can't be calculated in string theory because such a statement would be untrue. It is exactly a class of problems that string/M-theory has &quot;fully mastered&quot;, at least in principle - but also, in many case, in practice - with an accuracy that is often amazing. The open questions about &quot;quantum gravity&quot; effects in string theory are completely elsewhere than in the decay of black holes which is understood (including many of the stringy corrections to the Hawking approximation, and other things).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T12:27:02.217" CommentCount="2" />
  <row Id="2876" PostTypeId="2" ParentId="2563" CreationDate="2011-01-14T12:37:51.180" Score="0" ViewCount="" Body="&lt;p&gt;No, matter that has crossed the event horizon can never send any signals to the outer space anymore. In particular, it cannot become a part of a star that could shine (and be seen ouside the black hole). This would violate the very definition of a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, some hydrogen gas can get very hot while being swallowed to the black hole, but it will stay inside. Once it's inside, the life expectancy is not too high, anyway. A solar-mass black hole has a radius of about a few kilometers, so it takes a microsecond or so for all the matter to be destroyed at the singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even the galactic-center black hole whose mass is in millions of solar masses  only allows the matter inside to live for an extra second (or a few seconds) or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes are the final stage of the evolution of any bound system. They have the maximum entropy among all bound objects with the same mass (and charges). The only way how they can evolve is to evaporate by the Hawking radiation - which increases the entropy further. But it's only possible because the outgoing radiation is not bound and the total entropy may still grow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a back-transformation of a part of a black hole to a star would reduce the entropy - because a star has a vastly lower entropy than the same-mass black hole - which would also contradict the second law of thermodynamics.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T12:37:51.180" />
  <row Id="2877" PostTypeId="2" ParentId="2824" CreationDate="2011-01-14T12:48:45.857" Score="5" ViewCount="" Body="&lt;p&gt;This effect occurs very often when touching electronic devices that are connected to the power mains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can verify that it is the connect to the power mains: unplug the power adapter and all other connections to other devices connected to the power mains and try again.  The effect will be gone away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can best be felt also by the person touched if you touch with one finger only and move the finger (you can feel the difference on conducting surface areas of the device also).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are feeling is inducted power mains frequency so its probably 100 Hz (if you have a 50 Hz power mains or 120 Hz if you have 60 Hz power mains).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see this inducted voltage if you hold single-endedly the tip of a oscilloscope.&lt;/p&gt;&#xA;" OwnerUserId="1284" LastActivityDate="2011-01-14T12:48:45.857" CommentCount="2" />
  <row Id="2878" PostTypeId="2" ParentId="2497" CreationDate="2011-01-14T12:51:01.220" Score="7" ViewCount="" Body="&lt;p&gt;the conservation of energy is violated in cosmology - in all situations described by general relativity in which the time-translational symmetry of the &quot;background&quot; is broken. That's clearly the case of the Big Bang, too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again: By Noether's theorem, the conservation of energy is linked to the time-translational symmetry (the properties of the Universe don't depend on time) which is broken in an expanding Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one can see that the total energy/mass of the Universe is not conserved in time. In particular, the &quot;dust&quot; with zero pressure has a conserved total energy/mass. However, the energy carried by radiation - such as photons - decreases as $1/a$ - inverse linear size of the Universe - because the wavelength gets larger as well, which decreases the energy of each quantum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the contrary, the energy/mass carried by the dark energy is increasing with the volume because the energy density is constant - that's why dark energy is the normal realization is known as the cosmological constant. The density is constant but the volume of space is growing: the total energy is growing, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very likely that the huge mass of the Universe around us was created by inflation. During inflation, there was also a nonzero &quot;dark energy&quot; - energy density of the vacuum - which was constant while the volume of space was exponentially growing. This created lots of energy, and at the very end, a big part of the energy (kinetic energy of the inflaton which is proportional to the total energy etc.) was converted to ordinary particles that eventually gave rise to the galaxies etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the exponentially large mass of the Universe is a large, unnatural number, but this unnatural number is actually explained by inflation. As Alan Guth, the main father of inflation, said, the Universe is the ultimate free lunch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T12:51:01.220" CommentCount="2" />
  <row Id="2879" PostTypeId="2" ParentId="2873" CreationDate="2011-01-14T13:06:07.737" Score="10" ViewCount="" Body="&lt;p&gt;The most populous subclass of stringy vacua are type IIB flux vacua - studied in the KKLT paper. To find one such a vacuum solution of string theory, you need to specify the topology of a Calabi-Yau 3-fold, or 4-fold. There are just thousands of topologies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for each topology, you will find something like hundreds of cycles in the homology, and each cycle may carry a certain integer number of units of the generalized magnetic flux (the integral of the $p$-forms describing a magnetic flux from the Ramond-Ramond sector, besides the 3-form H-field from the NS-NS sector). There exists an inequality implying that these integers can't be arbitrarily high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have hundreds of cycles and each of them may have something like 10 different values of the flux, the number of discretely different possibilities will be like $10^{hundreds}$. For each choice of the integers, you may find a rather small number of solutions for all the shapes so that the potentials are minimized etc. So one gets an googol-like number of configurations of the RR fluxes; more precisely, this is a large number of supersymmetric AdS vacua (with a negative cosmological constant). It is believed that each of them may be dragged above zero to obtain a non-supersymmetric de Sitter background (which is needed) but there's no completely strict proof that the number of metastable de Sitter vacua is huge as well. The supersymmetric vacua are much more controllable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original &quot;modern&quot; vague claim that the number of vacua is a googol or its power was written down by Bousso and Polchinski in 2000:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0004134&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0004134&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You won't find the specific number $10^{500}$ over there which, I think, first appeared in a paper by Douglas. Douglas has made interesting fancy mathematical contributions to the counting of the vacua but I wouldn't count the &quot;fashionable&quot; value of the estimate to be among his major contributions. Search for 500 and references 8,9,10,11 or so e.g. in this paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0411173&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0411173&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will find a link to the KKLT paper, and others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One has to mention that there is nothing truly original about these &quot;large&quot; estimates. Wolfgang Lerche claimed that the number of certain vacuum solutions to string theory equations was something like $10^{1500}$ back in the 1980s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many other important phenomenological classes of string vacua are still thought to be vastly smaller. It is both good news and bad news. It is good news because you might think that they're much more unique and predictive. It is bad news because no one knows a convincing reason why these vacua should be able to produce the tiny observed value of the cosmological constant.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:18:25.227" LastActivityDate="2011-02-04T07:18:25.227" CommentCount="1" />
  <row Id="2880" PostTypeId="2" ParentId="1898" CreationDate="2011-01-14T13:18:01.910" Score="3" ViewCount="" Body="&lt;p&gt;Dear Dan, this is actually a very simple question. The phase velocity or group velocity of a photon may be smaller. But the energy of a single photon is always&#xA;$$E=hf$$&#xA;where $h$ is Planck's constant and $f$ is the frequency. This is true for quanta in any material - and not only for photons, in fact. It's also true for gravitons, electrons, muons, or any other particles. This relationship between energy and frequency of the wave associated with the particle is totally universal - and follows from the fact that energy (the Hamiltonian) generates the evolution in time, i.e. is given by the frequency for all periodic wave functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The frequency of a photon doesn't change anywhere - it must still make the same number of &quot;periods&quot; per second, wherever you look - imagine that you emit a packet that has 500 maxima and 500 minima of a wave, so the same number will be seen everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the energy of each photon remains constant as it moves through any environment. Of course, when it's absorbed, it gives its energy (or its part) to another particle.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T13:18:01.910" />
  <row Id="2881" PostTypeId="2" ParentId="822" CreationDate="2011-01-14T13:30:21.017" Score="2" ViewCount="" Body="&lt;p&gt;whether you call a similar concept &quot;fundamental&quot; is a matter of taste - and the proposition is just a meaningless emotional slogan. The angular momentum is surely an important quantity that is, in a very well-defined sense, as important as the normal momentum. Incidentally, both of them are conserved if the physical laws are symmetric with respect to translations and rotations, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the real question is why the spin in quantum mechanics can't be reduced to the orbital motion - i.e. to the &quot;linear motion&quot; and ordinary &quot;momentum&quot;. It's because the objects in quantum mechanics are described not just by their shape in space but by wave functions, and wave functions may be said to transform nontrivially (into something else) under rotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, if the wave function (or a field) is a vector or a tensor or, most typically, a spinor, then it means that in a different coordinate system, the values of the components of the wave function will be different. This is possible even in the case when the wave function (or field) is fully localized at one point, i.e. nothing is rotating &quot;orbitally&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angular momentum is defined by the change of the phase of the wave function under rotations, which may come from the dependence of the wave function on space, but also from the transformations of the components of the wave function among each other, which is possible even if everything is localized at a point. So even point-like objects may carry an angular momentum in quantum mechanics, the spin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the spin is a multiple of $\hbar/2$ and $\hbar$ is sent to zero in the classical limit, so in the classical limit, the spin as the internal angular momentum becomes zero and disappears, anyway. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another new feature of the spin is that unlike the angular momentum, it may be half-integer, not just a multiple of $\hbar$: also $\hbar/2$ is possible. That's because the wave functions (and fields) may transform as spinors that change the sign if they're rotated by 360 degrees. Only a rotation by 720 degrees is topologically indistinguishable from &quot;no rotation&quot;, so wave functions are obliged to return to their original values under a 720 degree rotation. But fermions change their signs under rotations by 360 degrees which corresponds to their half-integral spin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the word &quot;fundamental&quot; means that it can't be reduced to some other things such as a classical intuition about motion and rotation, then be sure that the spin is damn fundamental, much like the rest of quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T13:30:21.017" CommentCount="2" />
  <row Id="2882" PostTypeId="2" ParentId="2026" CreationDate="2011-01-14T13:32:36.013" Score="0" ViewCount="" Body="&lt;p&gt;It's simple to compute optical conductivity in 1/2 a lecture. It's not that exciting, but it is nice to see how you can use high powered techniques to derive something as simple as Ohm's law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to impress condensed matter types, computing the fermion spectral function and seeing marginal Fermi liquid behaviour is probably the best way to go.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-14T13:32:36.013" />
  <row Id="2883" PostTypeId="2" ParentId="2824" CreationDate="2011-01-14T13:36:35.730" Score="1" ViewCount="" Body="&lt;p&gt;1 -  Use a voltage detector pen in USB connector GND to see if there is a bad isolation between AC ground and DC ground (this is first option not because probability, but because it could be dangerous). It's normal a little fugue current between primary and secondary of transformer but the AC plug have 3 conectors, 2 for voltage and the third  is for GND (ground), perhaps your home installation haven't a good &lt;a href=&quot;http://en.wikipedia.org/wiki/Ground_%28electricity%29&quot; rel=&quot;nofollow&quot;&gt;grounding&lt;/a&gt;, check that! is very important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/HU9wq.gif&quot; alt=&quot;New Zeland gov picture&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 - Beware of what is on your wife's laptop screen when her skin buzz =P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.energysafety.govt.nz/templates/ContentTopicSummary____22978.aspx&quot; rel=&quot;nofollow&quot;&gt;picture source&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2011-02-01T12:06:36.723" CommentCount="1" />
  <row Id="2884" PostTypeId="2" ParentId="756" CreationDate="2011-01-14T13:38:51.447" Score="0" ViewCount="" Body="&lt;p&gt;when one talks about &quot;truly rigorous&quot; mathematical physics, there is really no good treatment of ordinary quantum field theory yet. So of course, there is no &quot;truly rigorous&quot; framework to discuss AdS/CFT whose one side is a quantum field theory and another side is something even more complicated - a description of a quantum gravity theory in terms of string/M-theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the physics level, nothing has changed since 1997 or early 1998: the claim was known in the right form from the very beginning. The string theory dynamics in the AdS space is exactly equivalent to the dynamics of the field theory on the boundary. While the field theory may be defined by the lattice - up to some issues with adjusting supersymmetry at long distances - the gravitating string theory side is only known from various limiting descriptions, including perturbative string theory, Matrix theory, and other dual descriptions, besides various terms calculable from SUSY etc. But there is no counterpart of the &quot;lattice&quot; that would allow us to define string theory &quot;completely exactly&quot; in any background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nathan Berkovits has made the longest steps to prove AdS/CFT in a Gopakumar-Vafa way, using his pure spinors etc: the world sheet of string theory directly degenerates into Feynman diagrams in the right limit. However, no physicist is too curious about such things because the map clearly works. I believe that the tests derived from the BMN-pp-wave duality - which show that even all strings and their interactions on the AdS side exactly match to the boundary CFT side - are among the most stringent tests of the AdS/CFT correspondence - especially for the canonical background AdS5 x S5 of type IIB related to N=4 SYM in d=4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This industry has transformed to the integrability business in recent years and increasingly began to overlap with the research of the people who study N=4 SYM scattering amplitudes via twistors although the role of the AdS gravity in the latter remains largely invisible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T13:38:51.447" CommentCount="1" />
  <row Id="2885" PostTypeId="2" ParentId="2660" CreationDate="2011-01-14T13:55:09.397" Score="5" ViewCount="" Body="&lt;p&gt;the original strategy of Feynman and wheeler was really about the desire to get rid of all self-interactions. In the modern language, it would eliminate most loop diagrams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, consider an electron propagator, in the modern language. One may attach a photon propagator on it. That modifies the electron's self-energy, and this is the kind of a term that the Wheeler-Feynman program wanted to eradicate completely. However, if you add another complexity to the photon propagator - namely an electron-positron loop in the middle - then it is a nontrivial contribution, especially because the vacuum polarization loop may be attached to different parts of the diagram as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their very idea would be that it is impossible for the same electron propagator to have two photon end points attached - that would be connected with one another. That would throw the baby out with the bath water. At any rate, no complete theory of their picture exists (or is mathematically possible) and their dreams and partial hints have only been a motivation for them to get the really important insights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T13:55:09.397" />
  <row Id="2886" PostTypeId="2" ParentId="2598" CreationDate="2011-01-14T14:01:21.953" Score="1" ViewCount="" Body="&lt;p&gt;Dear Marek, it has been showed that the paper by Rovelli was invalid for lots of reasons, including those related to yours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, as you hint, it is incorrect to treat the interior and exterior of the black hole asymmetrically because the location of the event horizon may only be determined a posteriori - after a star collapses. So there's no qualitative difference between the interior and the exterior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that in the &quot;real LQG&quot;, there would also be an entropy coming from the interior which would be volume-extensive. No one has ever showed that this term is absent; the absence is just a wishful thinking, so the proportionality law to the surface is just a result of an omission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, even if one removes the interior by hand, Rovelli's paper was showed incorrect. The numerical constant turned out to be incorrect, and newer calculations showed that even with the assumption that the black hole entropy comes from the horizon - which could make the area-law for the entropy tautological - the actual calculable entropy is actually not proportional to the area at all. The corrections to Rovelli's paper - showing that his neglecting of the higher spins etc. were invalid - appeared e.g. in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0407051&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/gr-qc/0407051&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0407052&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/gr-qc/0407052&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're looking for papers that show that it suddenly makes sense, you will be disappointed. Quite on the contrary, it has been showed that none of the early dreams that LQG could produce the right black hole entropy works.  This is also particular self-evident in the case of the quasinormal modes that were hypothesized to know about the &quot;right&quot; unnatural value of the Immirzi parameter - a multiplicative discrepancy in the Rovelli-like calculations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I showed that for the Schwarzschild, the result really contained $\ln(3)/\sqrt{2}$ and similar right things, but we also showed with Andy Neitzke - and with many other people who followed - that the number extracted for other black holes is totally different and excludes the heuristic conjecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So today, it's known that the relationship supported by the same Immirzi parameter on &quot;both sides&quot; was actually wrong on both sides, not just one. There is no calculation of an area-extensive entropy in LQG or any other discrete model of quantum gravity, for that matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-14T14:08:54.680" LastActivityDate="2011-01-14T14:08:54.680" CommentCount="9" />
  <row Id="2887" PostTypeId="2" ParentId="2833" CreationDate="2011-01-14T14:04:08.920" Score="1" ViewCount="" Body="&lt;p&gt;As the other posts say, there will be no net redshift. But there will be a net time delay, called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Shapiro_delay&quot; rel=&quot;nofollow&quot;&gt;Shapiro Delay&lt;/a&gt;, which should be observable if the light source is variable, and multiple images are resolved.&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2011-01-14T14:04:08.920" />
  <row Id="2888" PostTypeId="2" ParentId="41" CreationDate="2011-01-14T14:07:43.863" Score="5" ViewCount="" Body="&lt;p&gt;Hold on there now: don't be too hasteful in declaring TQFT's some artificial construct only benificial for abstract mathematics. This is simply not true, as there are examples of physical systems exhibiting behavior which is encaptured by the &lt;em&gt;phyics&lt;/em&gt; of a TQFT!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First things firsts: it's true that the model which physicists use for elementary particle physics does not use any topological field theories. This is the standard model, and it is based on a large construct of 4D QF Theories, such as QED, Yang-Mills theory, the Higgs boson, Dirac Field theory and so on. You can obtain properties of single particles, or the scattering amplitidues of a bunch of particles, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in condensed matter physics, such as metals, insulators, cold atom gases, lattice systems, you deal with not one or a few particles, but with many, many (&gt;10^10 --- 10^26) particles. Due to the impossibility of even attempting to write something that looks like a solution of the theory, starting from first principles, is simply impossible and highly ineffecient. Instead, one tries to come up with an effective model which encompasses the essence of the many-body problem. This is standard practice in condensed matter physics and frequently leads to an effective quantum field theory, which has little resemblence with the basic interactions of the original constituents (try to apply QED to an ionic lattice with electrons flowing around. Can you derive the electronic transport properties of such a system? The answer is: no, it's far too complex).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this elaboration have anything to do with topological quantum field theories? The reason is simple: some systems are &lt;em&gt;effectively&lt;/em&gt; described by a TQFT! Namely, the fractional quantum Hall effect. This effect arises as follows: two types of material are grown on top of each other. An interface between the two layers arises, and this interface &quot;confines&quot; the motion electrons. The electrons can only move in two out of three spatial dimensions. They are &lt;em&gt;effectively&lt;/em&gt; living in a two-dimensional world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without explaning what the quantum Hall effect actually is, I will instead tell you that the effective field theory associated with this effect is a Chern-Simons theory -- an example of a TQFT. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other examples exist as well. Topological insulators are likely to be connected to TQFT's as well (BF theories). I'm sure other examples follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The various links with CFT, quantum groups, knot theory, etc all have left footprints in this field. I'm sure there is plenty of &quot;translation&quot; of mathematical results to the physical implications left undone. The &quot;invariants&quot; you are talking about pop up as very physical observables. For instance, it's known since the 80's that the so-called conductivity of the fractional quantum Hall effect is nothing more but a certain Chern number. &lt;/p&gt;&#xA;" OwnerUserId="1287" LastActivityDate="2011-01-14T14:07:43.863" CommentCount="3" />
  <row Id="2889" PostTypeId="1" CreationDate="2011-01-14T14:23:07.403" Score="6" ViewCount="433" Body="&lt;p&gt;Consider an electromagnetic wave of frequency $\nu$ interacting with a stationary charge placed at point $x$. My question concerns the consistency of two equally valid quantum-mechanical descriptions of the electromagnetic wave. First I will describe the classical picture, then the two quantum mechanical descriptions, then I will ask experts for a conceptual unification of the two quantum mechanical descriptions. For parsimony I'll assume an extremely low frequency wave, but this is not strictly necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Classical description 1:&lt;/strong&gt;&#xA;At point $x$ the electromagnetic wave contributes a slowly varying electric field, and a slowly varying magnetic field. The charge at $x$ experiences a force due to the electric field, and begins to move. As the charge begins to move it experiences a force due to the magnetic field. Using the right-hand-rule it is easy to see that the net force on the charge is in the direction of motion of the electromagnetic wave. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Quantum mechanical description 1:&lt;/strong&gt;&#xA;In the quantum mechanical description of the same electromagnetic wave, real photons are traveling with momentum $\frac{h\nu}{c}$ (in the direction of motion of the above electromagnetic wave), and are absorbed by the charge at $x$, causing it to recoil in the above electromagnetic wave's direction of motion (due to conservation of momentum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Quantum mechanical description 2:&lt;/strong&gt;&#xA;(I will be assuming that in the quantum mechanical description of the electromagnetic field, the force due to the electric/magnetic fields between two moving charges is due to the exchange of virtual photons). At point $x$ the electromagnetic wave contributes a slowly varying electric field, and a slowly varying magnetic field. The charge at $x$ experiences a force due to the electric field, due to the exchange of virtual photons with the charge that produced the electric field. Similarly for the magnetic field. In other words, there are no real photons -- just the virtual photons mediating between the charge at $x$ and the charge whose movement created the electromagnetic wave in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, my question: how are descriptions 1 and 2 reconciled? In description 1, the origin of the electric and magnetic fields (a charge), and those fields' descriptions in terms of virtual photon exchange, is completely ignored. On the other hand in description 2 there are no real photons, and the virtual photons are longer-ranged (do they self-interact?). Are the two descriptions equivalent? If so, it must be that a real photon can be written in terms of &quot;virtual-photon&quot; basis states. What is such a decomposition called, and can someone point me to a discussion of it? &lt;/p&gt;&#xA;" OwnerUserId="1247" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-14T21:08:23.997" LastActivityDate="2011-01-14T21:08:23.997" Title="Properties of the photon: Electric and Magnetic field components" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;light&gt;&lt;photons&gt;&lt;quantum-electrodynamics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="2890" PostTypeId="2" ParentId="2872" CreationDate="2011-01-14T14:23:45.237" Score="0" ViewCount="" Body="&lt;p&gt;I might be talking non sense, but I'll try arguing with your logic. &lt;em&gt;(Is there any physics to be answered in there?)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're doing a lot more than just postulating civilization.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You're postulating they are on every single galaxy where dark matter effects have been found. (I'm no astrophysicist, but I believe dark matter effects have been verified on every single galaxy where we've looked.)&lt;/li&gt;&#xA;&lt;li&gt;You're postulating they reached a stage of galactic colonization, (again) on every single one of these cases. Note that if this stage really is on the range of 100 million years (which seems to be pure speculation on your part) than it is &lt;strong&gt;so&lt;/strong&gt; small compared to the age of galaxies. Considering the number of galaxies we've looked at, we should have found some that are not colonized &quot;yet&quot; (Statistically speaking).&lt;/li&gt;&#xA;&lt;li&gt;Not to mention, where did they get all the matter for that? Isn't dark matter density 3 times higher than ordinary matter density?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="724" LastActivityDate="2011-01-14T14:23:45.237" />
  <row Id="2891" PostTypeId="1" AcceptedAnswerId="4258" CreationDate="2011-01-14T14:42:04.507" Score="2" ViewCount="332" Body="&lt;p&gt;I've been searching and have not been able to find a formula for calculating the impedance of differential lines on inner layers where the dielectric above and below it is not symmetrical. I've seen plenty of examples of symmetrical stackups, but that would not help in my case. &#xA;Does anyone know what the formula would be for this?&#xA;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited for more information:&#xA;For those that don't know, a stackup consists of all the different layers of a PCB (Printed Circuit BOard). You can have, for example, a 4 layber PCB which has individual copper material &amp;amp; glass epoxy stacked on top of each other. This allows the board to have more layers for routing the nets of the circuits.&lt;/p&gt;&#xA;" OwnerUserId="1289" LastEditorUserId="1289" LastEditorDisplayName="" LastEditDate="2011-01-17T12:50:15.503" LastActivityDate="2011-01-31T01:44:56.277" Title="Impedance formula for an edge coupled differential stripline in an asymmetrical stackup?" Tags="&lt;electromagnetism&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="2892" PostTypeId="2" ParentId="2889" CreationDate="2011-01-14T14:59:03.453" Score="3" ViewCount="" Body="&lt;p&gt;first of all, the quantum mechanical description 1 in its current form is impossible. A charged particle can't simply absorb a real photon. This is most easily seen in the final particle's rest frame. There is no photon left so the total energy is just the rest mass of the particle, times $c^2$, but the initial state has a higher energy of the particle in this frame, because the particle was moving, and also an additional positive energy of the photon. So the energy couldn't have been conserved in this process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A charged particle may only absorb a real photon if it emits another one in another direction. So microscopically, it's always a process composed out of the Compton scattering subprocesses. If the number of real photons in the same state is large, they may be described as a classical wave. You may partially quantize the system, keeping the electromagnetic field classical and quantizing just the particle, or otherwise. The reason why this description agrees with the full quantum description in the classical limit is manifest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I don't quite follow the difference between your classical description and the quantum mechanical description 2. Concerning the usage of virtual photons, well, if you want to study the whole process using the tools of quantum field theory - including the pre-history when no electromagnetic wave existed - then of course the electromagnetic wave had to be created at some point, and some of the photons were absorbed by the charged particle. The charged particle became virtual for a little moment as well, before it emitted another photon which is needed for the energy conservation, as I have explained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the photon that was absorbed by the charged particle was virtual - it only existed for a finite amount of time. However, the electromagnetic wave was probably propagating for such a long time that even this photon may be called &quot;real&quot;. There is a simple relationship between virtual and real particles - real particles are the virtual ones that happen to sit exactly on the mass shell, so they satisfy $E^2-p^2=m^2$. This identity may be exactly checked only when $E,p$ are measured totally accurately - which means that the particles must exist indefinitely. If they don't exist indefinitely, then they're always &quot;virtual&quot; to some extent, but chances are that if they exist for a long time, you may also imagine that they're &quot;real&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;virtuality&quot; of a particle may be defined as the difference $E^2-p^2-m^2$ - the distance from the physical mass shell. If the virtuality is low, the virtual particle may exist for a long time and look &quot;real&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there are no &quot;virtual photon states&quot; in the Hilbert space. The Hilbert space only contains real particles. Virtual particles are an object that appears in the calculation of probability amplitudes for various processes - in the Feynman diagrams. Virtual particles are internal lines of Feynman diagrams, given by propagators that determine the 2-point function (correlator) of a quantum field. But they don't correspond to any physical states. There are no off-shell physical states in the Hilbert space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you have a history in which some particles exist for a finite amount of time, so that they're strictly speaking virtual from the Feynman-diagrammatic viewpoint, it is still true that at every moment, there must exist some real particles that are actually present. However, it is tough, misleading, ambiguous, and unnecessary to calculate the &quot;exact intermediate states&quot; in quantum field theory. Such objects - wave functionals - would also depend on the field redefinitions (of the quantum fields), renormalization schemes, and other things. It's actually very useful to avoid these things when they're not necessary and only talk about the things that can be measured - the cross sections that may be calculated from the scattering amplitudes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A problem with the &quot;wave functionals&quot; of the intermediate states is that they're only well-defined with respect to a reference frame - but virtually all regularizations we know to calculate the loop diagrams rely on the Lorentz symmetry. Because the Lorentz symmetry is obscure by the foliations of the spacetime, it becomes harder to &quot;regulate&quot; the exact wave functional at the loop level. Of course, at the classical or semiclassical level, one may describe very accurately what's going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your particular situation, there was no real problem because all the photons in the problem were really on-shell, and you may present them as real photons if you wish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-14T15:05:35.353" LastActivityDate="2011-01-14T15:05:35.353" CommentCount="5" />
  <row Id="2893" PostTypeId="2" ParentId="2872" CreationDate="2011-01-14T15:13:58.100" Score="1" ViewCount="" Body="&lt;p&gt;Physically, if you look at the low level of CMB anisotropy and the BAO power spectrum, you need some sort of mass that interacts gravitationally but is decoupled from the photons before recombination (i.e. - dark matter). Otherwise there would be no way to seed the galaxy formation that we see occuring at lower redshifts. That is to say, dark matter needs to be primordial, so that it can dodge Silk Damping and provide the graviational &quot;cores&quot; around which galaxies form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which would also imply that dark matter cannot be the construct of an alien civilization. &lt;/p&gt;&#xA;" OwnerUserId="1291" LastActivityDate="2011-01-14T15:13:58.100" />
  <row Id="2895" PostTypeId="2" ParentId="885" CreationDate="2011-01-14T15:47:57.837" Score="3" ViewCount="" Body="&lt;p&gt;While it is true that the function q-dot(t) is the derivative of the function q(t) w.r.t. time, it is not true that the value q-dot is at all related to the value q at a given point in time, since a value is just a number, not a function.  The action is a functional of q(t), and so it would make no sense to vary the action both w.r.t. q and q-dot.  But the&#xA;Lagrangian L(q,q-dot) is a function of the values q and q-dot, not a functional of the functions q(t) and q-dot(t).  We can promote L to a function of time if we plug in q(t) and q-dot(t) instead of just q and q-dot.  (Remember a functional turns a function into a number, e.g., S[q], whereas a function turns a value into a number, e.g., L(q,q-dot).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve for q(t) we extremize the action S, by demanding that it is extremal at every point, t.  This is equivalent to solving the Euler Lagrange equations at every point t.&#xA;Since at any point t the values q and q-dot are independent, they can be varied independently.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-14T15:47:57.837" />
  <row Id="2896" PostTypeId="2" ParentId="2796" CreationDate="2011-01-14T15:56:28.437" Score="1" ViewCount="" Body="&lt;p&gt;The wavefunction of a photon includes polarization as part of its description. The electric field is determined from the polarization, but in quantum electrodynamics, the electric field operator doesn't commute with the particle number operator. However, from the relative phases of diagram contributions with no photon exchange, one photon exchange, two photon exchange, etc., the direction of the electric field due to virtual photons depends upon the charge of the particle.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-14T16:01:53.530" LastActivityDate="2011-01-14T16:01:53.530" />
  <row Id="2897" PostTypeId="2" ParentId="160" CreationDate="2011-01-14T16:19:48.123" Score="0" ViewCount="" Body="&lt;p&gt;I whole heartedly suggest Lego. It's just amazing how many things about physics can be learnt from playing with the Lego Technic and Mindstorms series. When you're a child the most amazing thing in the world is making things and just watch them work. It's like magic, you know. So, you sit there furiously trying things, playing games in your mind and you start noticing things. You notice, while making a robotic arm, the farther away something heavy is from a joint the harder it is for the machine to lift stuff. Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's happening over there? Then you realize that how heavy something is depends on how far it is from you on a rod. Weird. So, then you play some more and figure out how to use this to your advantage. After a while, you start learning tings, noticing things and start using them in your creations. It's just something natural and before you know it. You're in love.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I still love physics to this day thanks to that one Lego set. It really is magic.&lt;/p&gt;&#xA;" OwnerUserId="1065" LastActivityDate="2011-01-14T16:19:48.123" CommunityOwnedDate="2011-01-14T16:19:48.123" />
  <row Id="2899" PostTypeId="2" ParentId="2644" CreationDate="2011-01-14T16:28:30.273" Score="2" ViewCount="" Body="&lt;p&gt;First of all, it would be preposterous to think that there was a simple recipe that Newton followed and that anyone else can use to deduce the laws of a similar caliber. Newton was a genius, and arguably the greatest genius in the history of science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second of all, Newton was inspired by the falling apple - or, more generally, by the gravity observed on the Earth. Kepler understood the elliptical orbits of the planets. One of Kepler's laws, deduced by a careful testing of simple hypotheses against the accurate data accumulated by Tycho Brahe, said that the area drawn in a unit time remains constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton realized that this is equivalent to the fact that the first derivative of the velocity i.e. the second derivative of the position - something that he already understood intuitively - has to be directed radially. In modern terms, the constant-area law is known as the conservation of the angular momentum. That's how he knew the direction of the acceleration. He also calculated the dependence on the distance - by seeing that the acceleration of the apple is 3,600 times bigger than that of the Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So he systematically thought about the second derivatives of the position - the acceleration - in various contexts he has encountered - both celestial and terrestrial bodies. And he was able to determine that the second derivative could have been computed from the coordinates of the objects. He surely conjectured very quickly that all Kepler's laws can be derived from the laws for the second derivatives - and because it was true, it was straightforward to prove him this conjecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, he had to discover the whole theory - both $F=ma$ (or, historically more accurately, $F=dp/dt$) as well as a detailed prescription for the force - e.g. $F=Gm_1m_2/r^2$ - at the same moment because a subset of these laws is useless without the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The appearance of the numerical constant in $F=ma$ or $p=mv$ is a trivial issue. The nontrivial part was of course to invent the mathematical notion of a derivative - especially because the most important one was the second derivative - and to see from the observations that the second derivative has the direction it has (from Kepler's law) and the dependence on the distance it has (from comparing the acceleration of the Moon and the apple falling from the tree).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It wasn't a straightforward task that could have been solved by anyone but it was manifestly simple enough to be solved by Newton. So he had to invent the differential calculus, $F=ma$, as well as the formula for the gravitational force at the same moment to really appreciate what any component is good for in physics.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T16:28:30.273" />
  <row Id="2900" PostTypeId="1" AcceptedAnswerId="2906" CreationDate="2011-01-14T16:30:47.210" Score="3" ViewCount="237" Body="&lt;p&gt;Why GPS/GLONASS/Galileo satellites are on low earth orbit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why geostationary orbit is so bad? Sattelites might be placed there 'statically' and more precise...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only problem I can see is navigation close to poles, but they have this problem anyway.&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-14T21:05:24.297" LastActivityDate="2011-02-28T13:25:39.767" Title="Why GPS is at LEO?" Tags="&lt;orbital-motion&gt;&lt;gps&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="2901" PostTypeId="1" CreationDate="2011-01-14T16:31:17.100" Score="3" ViewCount="470" Body="&lt;p&gt;Is it true that the whole galaxy is actually revolving, and powered by a black hole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has it been proven, and if it is true, how can our solar systems actually keep up the momentum to withstand the pull?&lt;/p&gt;&#xA;" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-05T10:24:29.233" LastActivityDate="2011-04-06T08:30:00.690" Title="Is there a black hole in the centre of the Milky Way?" Tags="&lt;astrophysics&gt;&lt;black-holes&gt;&lt;galaxy&gt;" AnswerCount="6" FavoriteCount="1" />
  <row Id="2902" PostTypeId="1" CreationDate="2011-01-14T16:35:29.823" Score="1" ViewCount="123" Body="&lt;p&gt;A Newtonian homogeneous density sphere has gravitational binding energy in Joules $U = -(3/5)(GM^2)/r$, G=Newton's constant, M=gravitational mass, r=radius, mks.  The fraction of binding energy to gravitational mass equivalent, $U/Mc^2$, is then (-885.975 meters)(Ms/r), Ms = solar masses of body, c=lightspeed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives ratios that are less than half that quoted for pulsars (neutron stars), presumably for density gradient surface to core and General Relativity effects (e.g., billion surface gees).  &lt;strong&gt;Please post a more accurate (brief?) formula acounting for the real world effects.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:  1.74 solar-mass 465.1 Hz pulsar PSR J1903+0327, nominal radius 11,340 meters (AP4 model), calculates as 13.6% and is reported as 27%.  A 2 sol neutron star calculates as 16.1% and is reported as 50%.  There is an obvious nonlinearity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-14T17:52:57.923" Title="Pulsar gravitational binding energy?" Tags="&lt;astrophysics&gt;&lt;relativity&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="2903" PostTypeId="2" ParentId="2901" CreationDate="2011-01-14T16:39:38.880" Score="8" ViewCount="" Body="&lt;p&gt;I was giving a talk about the galactic black hole at the center, Sagittarius A*, back in 1998. At that time, it was already clear to enlightened people that it had to be a black hole. An analysis of a two-temperature plasma helped to bring some new evidence that the object had a real event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black hole is huge but it is not &quot;galactically&quot; huge. Its mass is 4.2 million solar masses or so. This is of course large, in comparison with any star, but it is negligible if compared to - thousands of times smaller than - the mass of the Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it would be unreasonable to say that the black hole has a tremendous impact on the gravitational forces across the Milky Way. It is just a heavy single object but if one looks at the size of 5% of the Galaxy's diameter, the total amount of stars in such a region is already vastly larger than the mass of the black hole. Already in such small regions, the black hole is just a small droplet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes, just like any other heavy objects, are unable to &quot;power&quot; galaxies. Galaxies are composed of stars that move according to the laws of mechanics (or general relativity) - inertia modified by the gravitational force. (Today, we believe that most of the gravitational force is exerted by the dark matter that represents a majority of the galactic masses.) The dependence of the gravitational force on the distance from the center of the Galaxy determines the orbital velocity of the stars at every distance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every distribution of matter, we get some dependence of the gravitational force on the distance, and we can write down the velocities as a function of the distance for which the orbits remain circular. (And if the orbits are a bit elliptic, there is no problem with that, either.) Whatever the radial attractive force is, there always exists a velocity such that the gravitational attractive force exactly cancels against the centrifugal force. (More precisely, the gravitational force is the centripetal force.) So for any pull, there is a velocity such that one can withstand the pull, and it makes absolutely no difference whether a black hole contributes to the pull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while the object is interesting - and probably generic for most galaxies - it doesn't have any &quot;systemic&quot; importance for the functioning of the Galaxy. The radius of the object is millions of kilometers - something like 10 times the distance to the Moon. The matter around the black hole is being heated and &quot;cooked&quot; by the gravitational field and there's a high temperature. But if one managed to cross the event horizon, he could live for a few more seconds before he would be squeezed by the singularity at the black hole center.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T16:39:38.880" CommentCount="4" />
  <row Id="2904" PostTypeId="2" ParentId="2901" CreationDate="2011-01-14T16:41:57.973" Score="6" ViewCount="" Body="&lt;p&gt;There's a strong consensus among astrophysicists that there is a supermassive black hole at the center of our Galaxy (as there are in most large galaxies, apparently).  But as cool as that fact is, it's possible to make too much of it.  The black hole at the center of our Galaxy has a mass of a few million times that of the Sun, which is a tiny fraction of the mass of the whole Galaxy.  An object like the Sun, for instance, is attracted gravitationally toward the center of the Galaxy, but the black hole is responsible for only a tiny fraction of that attraction.  All of the other stuff is much more important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One common misconception out there about black holes is that they &quot;suck everything in.&quot;  A black hole of a certain mass is no better at pulling in a distant object than any other object of the same mass.  So if you weren't worried about the Solar System being sucked in toward the center of the Galaxy before you knew about the black hole, you shouldn't be any more worried about it afterwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit of self-promotion here: There's a bunch of stuff written about how to think about black holes, including Frequently Asked Questions about Black Holes, which I wrote back in the 1990s.  You can find it here: &lt;a href=&quot;http://cosmology.berkeley.edu/Education/BHfaq.html&quot; rel=&quot;nofollow&quot;&gt;http://cosmology.berkeley.edu/Education/BHfaq.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-01-14T16:41:57.973" CommentCount="4" />
  <row Id="2906" PostTypeId="2" ParentId="2900" CreationDate="2011-01-14T16:55:01.070" Score="5" ViewCount="" Body="&lt;p&gt;Why they're not on the geostationary orbit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's because the geostationary orbit - and indeed, there's just one such orbit, a one-dimensional curve in space - only exists above the equator while the GPS has to cover the whole planetary surface, including the points closer to the poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it's untrue that the GPS satellites are located at the LEO, either. Low Earth Orbit is defined as 160-2000 kilometers of altitude. However, the GPS satellite constellation is located roughly 20,200 km above the surface - over one half of the geostationary radius - in such a way that the position of each satellite returns to the same place twice per 24 hours. This is very convenient for synchronization and planning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Galileo satellites will be at altitude 23,222 km. It's also an intermediate circular orbit, much like for the 19,100 altitude of the GLONASS whose orbital period is about 11 hours.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-14T17:05:20.410" LastActivityDate="2011-01-14T17:05:20.410" CommentCount="2" />
  <row Id="2907" PostTypeId="1" AcceptedAnswerId="2910" CreationDate="2011-01-14T17:00:46.437" Score="3" ViewCount="335" Body="&lt;p&gt;this is my mental picture for how they travel without a medium, how (like water waves) some can't stay still, why they have wave and particle properties, energy/mass equivalence, conservation, etc.  it might capture uncertainty too -- i've heard that all waves have an uncertainty relation (say in their power spectrum), but i don't get why -- it seems like we can discuss waves with absolute precision.&lt;/p&gt;&#xA;" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-14T21:05:00.490" LastActivityDate="2011-01-15T19:31:04.790" Title="are particles &quot;knots&quot; or &quot;kinks&quot; of excitation in a field?" Tags="&lt;quantum-field-theory&gt;&lt;particles&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="2908" PostTypeId="2" ParentId="885" CreationDate="2011-01-14T17:03:39.950" Score="5" ViewCount="" Body="&lt;p&gt;The answer to your main question is already given -- you do not &lt;em&gt;vary&lt;/em&gt; coordiante and speed independently. But it seems that your main problem is about using coordinate and speed as independent variables.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me refer to &lt;a href=&quot;http://books.google.com/books?id=58SFj9x5RMEC&amp;amp;printsec=frontcover&amp;amp;hl=en#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this great book&lt;/a&gt;. The very first line of the book (where an author usually says to whom this book is devoted) is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/3DT26.png&quot; alt=&quot;William Burke&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that from time to time student do ask this question. But attempts to explain it &quot;top down&quot; are usually just lead to more and more questions. One really needs to make mathematical &quot;bottom up&quot; order in the topic. Well, as the name of the book suggest -- the mathematical discipline one needs is &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_geometry&quot; rel=&quot;nofollow&quot;&gt;differential geometry&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot retell all the details, but briefly it looks like this:   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You start with a configuration space $M$ of your system. $M$ is a (differentiable) &lt;a href=&quot;http://en.wikipedia.org/wiki/Differentiable_manifold&quot; rel=&quot;nofollow&quot;&gt;manifold&lt;/a&gt;, and $q$ are the coordinates on this manifold. &lt;/li&gt;&#xA;&lt;li&gt;Then there is a specific procedure, that allows you to add all the possible &quot;speeds&quot; at every given point of $M$. And you arrive at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tangent_bundle&quot; rel=&quot;nofollow&quot;&gt;tangent bundle&lt;/a&gt; $TM$, which is a manifold too, and ($q$,$\dot{q}$) are different coordinates on it. &lt;/li&gt;&#xA;&lt;li&gt;Lagrangian is a function on $TM$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="386" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-14T17:28:33.320" LastActivityDate="2011-01-14T17:28:33.320" CommentCount="2" />
  <row Id="2910" PostTypeId="2" ParentId="2907" CreationDate="2011-01-14T17:16:41.027" Score="3" ViewCount="" Body="&lt;p&gt;Actually that's not too far off the mark, although I'm not sure &quot;knot&quot; or &quot;kink&quot; is the best word. Quantum field theory, the best theory we currently have to describe particles, says that particles correspond to excitations of a field, which are kind of like waves in water; you could consider the surface of a pond &quot;excited&quot; whenever it's not flat. Just as with water waves, there's an infinite variety of &quot;shapes&quot; you can have for these excitations. For example, you could have a repeating wave, in which the surface of the water cycles up and down over a large area, or you could have just one wave front that just propagates across the water without spreading out very much. The former case is pretty typical for things like light waves, and the latter case is pretty typical for particles of matter, although technically any kind of field (whether it's the electromagnetic field for light, or a quark field in matter, or whatever) &lt;em&gt;can&lt;/em&gt; have any of the different types of excitations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, according to special (and general) relativity, even an object that is standing still is moving through time. So all of these excitations move through spacetime in one way or another. But only certain ones (the excitations in fields corresponding to massless particles) can move through space in such a way that they appear to us to be traveling at the speed of light.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-01-14T17:16:41.027" CommentCount="2" />
  <row Id="2911" PostTypeId="2" ParentId="2690" CreationDate="2011-01-14T17:23:18.023" Score="0" ViewCount="" Body="&lt;p&gt;For a superfluid, mass generates a change in the phase. That's because a superfluid is a superposition of states with different masses, i.e. a mass condensate. This also applies to supersolids. For other cases, states with different masses lie in different superselection sectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might object that for say, a helium-4 superfluid, a change in phase generates the number of helium-4 atoms, and not the mass per se, and that's true. But going the other way around, mass generates a change in the phase, scaled by the mass of a helium-4 atom. The reason for this asymmetry is the total number of other forms of matter decomposes the state space into superselection sectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This entire analysis presupposes Galilean relativity instead of special relativity. For the latter, mass isn't conserved.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-22T18:28:10.490" LastActivityDate="2011-01-22T18:28:10.490" />
  <row Id="2912" PostTypeId="2" ParentId="2901" CreationDate="2011-01-14T17:29:31.883" Score="1" ViewCount="" Body="&lt;p&gt;Clearly as all the above comments have stated, in the overall scheme of things its a tiny fraction of the overall mass of the galactic system. I wouldn't consider thats its main effect to be direct gravitation however. Supermassive black holes may roughly control the amount of gas in their surrounding galactic cores. As the gas density nearby gets higher, the BH feeds on some of it, and shines as a quasar. All that radiation tends to push the gas away. So there is a case to be made that supermassive black holes may regulate the amount of gas and young stars in the core regions of their galaxies. But they do this because of the vast energy radiating from their acretion disks when they actively feed. The issue is not so clearcut however, as regions of active star formation create massive stars whose stellar winds (and supernova explosions when they die) have much the same efect. Supermassive star clusters such as the Arches cluster which is close to the Milkyways blackhole probably formed there because of the graviational interaction of the gas and the BH.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-01-14T17:29:31.883" />
  <row Id="2913" PostTypeId="2" ParentId="2902" CreationDate="2011-01-14T17:39:03.093" Score="1" ViewCount="" Body="&lt;p&gt;Dear Uncle Al, your formula with the coefficient $3/5$ only holds in the Newtonian physics. But obviously, the Newtonian physics is not appropriate for the description of hugely heavy objects such as black holes or neutron stars that are not far from them. You need to use a more complete theory developed by the physicist named Albert Einstein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's called the special theory of relativity. Special relativity implies the $E=mc^2$ law - among other things you didn't mention in your question - while general relativity is needed to study very heavy objects that strongly curve the space such as the neutron stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if I am telling you something that you have already heard but it is far from clear according to the text of your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With proper definitions of the &quot;gravitational bound energy&quot; in general relativity, black holes may be said to store 100% of their energy in the gravitational binding energy. Again, the latter notion is ambiguous in general relativity. Neutron stars are typically very close from the radius-mass relationship that apply to black holes. So before they reach the tipping point where the collapse to a black hole is inevitable, the ratio will jump above your classical formula. That's where a nonlinearity comes from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, real neutron stars are really not uniform, either, so the formula at the top wouldn't apply even in the classical physics. The density closer to the center is larger. That effectively puts the matter closer together which increases the (absolute value of) the (negative) binding energy. The more massive neutron stars one considers, the more they tend to clump near the middle, so this is another source of the nonlinearity.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T17:39:03.093" />
  <row Id="2914" PostTypeId="1" AcceptedAnswerId="2923" CreationDate="2011-01-14T18:11:39.770" Score="2" ViewCount="140" Body="&lt;p&gt;I am working on a complex configuration of magnets and every time I make an experiment something unforseen happens. Now I believe I could speed up the development by sitting down and calculating the configuration in all scenarios. My question is however: Is there any software that can help me to calculate the forces between magnets. I am already working with AutoDesk as my CAD software.&lt;/p&gt;&#xA;" OwnerUserId="904" LastActivityDate="2011-01-14T20:09:13.630" Title="Software to calculate forces between magnets" Tags="&lt;magnets&gt;&lt;software&gt;" AnswerCount="1" />
  <row Id="2915" PostTypeId="2" ParentId="2907" CreationDate="2011-01-14T18:12:28.303" Score="4" ViewCount="" Body="&lt;p&gt;particles are ordinary quanta of the corresponding quantum fields - without any knots or other topologically nontrivial features. (You have to get used to the wave-particle duality, probabilities, and the uncertainty principle - they're fundamental features of the world around us.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is only true for &quot;weakly coupled particles&quot; that are directly described by a Lagrangian composed of &quot;free fields&quot; plus &quot;weak interactions&quot;. On the other hand, there can be many other objects that are not just quanta of waves but they have &quot;knots&quot; on them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that they are topological defects. Most typically, one can take the original fields and write down a classical solution of the equations of motion that is topologically nontrivial. The resulting configuration of fields has to exist in the quantum theory, too. It is preserved by the nontrivial topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1 spatial dimension, a topologically nontrivial configuration is one that drives a scalar field from 1 local minimum of the potential to another, as you move from the left to the right; it is known as the &quot;kink&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2 spatial dimensions, there are solutions known as &quot;vortices&quot; (or one &quot;vortex&quot;) in which the circular round trip around the center of the solution sees a field to go around a noncontractible curve in the configuration space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 3 spatial dimensions, we similarly find magnetic &quot;monopoles&quot; which are also some kinds of &quot;knots&quot; on the fields. More generally, the topological defects that correspond to real objects are known as &quot;solitons&quot;. This is contrasted with the &quot;instantons&quot; which are objects localized in the Euclidean spacetime; they don't describe static objects but rather special histories that contribute to various probability amplitudes for processes that appear at one &quot;instant&quot; of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another class of topologically nontrivial configurations are skyrmions, and if you want to see a representative that is really close to your &quot;knots&quot;, see a paper about knitted fivebranes in M-theory:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/9803108&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9803108&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the quantum theory, objects such as solitons - especially monopoles - may behave indistinguishably from the original quanta. There is sometimes a full symmetry between them - the most famous example is the S-duality of gauge theories in 4 dimensions. Electrically charged particles such as gauginos are exchanged with the magnetic monopoles - solitons - as the coupling $g$ is changed to $1/g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these things, when done properly, still satisfy all the postulates of quantum mechanics such as the nonzero commutators between observables (uncertainty principle). In string theory, topological defects include objects such as D-branes, NS5-branes, and in extra-dimensional theories in general, one also has special &quot;knitted&quot; configurations of the metric such as the Kaluza-Klein monopoles (magnetic monopoles extended in extra dimensions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-14T18:17:43.887" LastActivityDate="2011-01-14T18:17:43.887" />
  <row Id="2916" PostTypeId="1" AcceptedAnswerId="2918" CreationDate="2011-01-14T18:27:27.490" Score="6" ViewCount="588" Body="&lt;p&gt;What is the exact shape of the universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know of the balloon analogy, and the bread with raisins in it.&#xA;These clarify some points, like how the universe can have no centre, and how it can expand equally everywhere in all directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But they also raise some questions, like if you are on the surface of a balloon and travel in 1 direction, you will eventually return to your starting point, is it possible our universe have this feature? If it has, or had, would this be a symmetry of sorts (psi(x)=psi(x+R)), and as such have a conserved quantity associated with it (by Noether)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming &quot;small curled up dimesnions&quot; Wouldnt these dimensions have this type of symmetry, what are the associated conserved quantities?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it known exactly what the geomtrical shape of the universe is? (on a large scale)&#xA;(I am not talking about only the observable universe).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one define the &quot;size&quot; of a dimension, is this scale only applicable to curled up ones?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to describe to a layman the shape of the universe without resorting to inept analogies?&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-01-24T01:35:38.940" Title="Shape of the universe" Tags="&lt;general-relativity&gt;&lt;astrophysics&gt;&lt;cosmology&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="3" />
  <row Id="2917" PostTypeId="2" ParentId="2043" CreationDate="2011-01-14T18:34:21.443" Score="2" ViewCount="" Body="&lt;p&gt;your impression that the von Neumann algebras are not being talked about by the physicists is entirely valid. The operators on the Hilbert space may perhaps satisfy the definition of a von Neumann algebra, but it doesn't make the specific results from this portion of maths useful in physics. The von Neumann algebras are not linked to any &quot;specific piece&quot; of interesting knowledge or mechanisms that physicists have to learn. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An exception was algebraic or axiomatic quantum field theory which liked to talk about the von Neumann algebra but it has eventually become a fringe subdiscipline of theoretical physics. AQFT doesn't really work - it is not compatible with the most recent 40 years of fundamental physical insights about quantum field theory, such as the Renormalization Group. So the particular &quot;focus&quot; of the von Neumann algebra - in comparison with any algebra of operators on a linear space - are unlikely to be relevant for some important insights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from quantum field theory, the notion of von Neumann algebras is also sometimes mentioned by physicists who study statistical physics and other fields but I think it is correct to say that only physicists who have gone through some math education in the past may be seen to &quot;spontaneously&quot; start to use the concept of von Neumann algebras. The algebras have surely not become a standard topic of undergraduate or graduate courses directed at theoretical physicists and I think that even most top theoretical physicists don't exactly know what the algebras are and aren't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, John von Neumann who introduced them would think that they would become much more important in physics in those 80 years than they have become. Von Neumann may be counted as one of the founding fathers of quantum mechanics; among these fathers, he was clearly the most mathematical (abstract) one and many of his hints simply didn't become standard. That's also true for some other concepts he introduced to quantum mechanics, including quantum logic. Still, he was a smart guy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-14T18:34:21.443" />
  <row Id="2918" PostTypeId="2" ParentId="2916" CreationDate="2011-01-14T18:47:07.523" Score="7" ViewCount="" Body="&lt;p&gt;There are a bunch of questions here. Let me try to take them in order:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is it possible that our Universe has the feature that if you travel far enough you return to where you started?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Yes. The standard Big-Bang cosmological model is based on the idea that the Universe is homogeneous and isotropic. One sort of homogeneous spacetime has the geometry of a 3-sphere (like a regular sphere, but with one more dimension). In these cosmological models, if you travel far enough you get back to where you started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the best available data seem to indicate that the Universe is very nearly spatially flat. This means that, if we do live in a 3-sphere Universe, the radius of the sphere is very large, and the distance you'd have to travel is much larger than the size of the observable Universe. Even if that weren't true, the fact that the Universe is expanding would make it hard or impossible to circumnavigate the Universe in practice: no matter how fast you went (short of the speed of light), you might never make it all the way around.  Nonetheless, 3-sphere Universes, with the geometrical property you describe, are definitely viable cosmological models.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does this give rise to a symmetry by Noether's theorem?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Not really. Noether's theorem is generally applied to continuous symmetries (i.e., ones that can be applied infinitesimally), not discrete symmetries like this. The fact that space is homogeneous gives rise to a symmetry, namely momentum conservation, whether or not space has the 3-sphere geometry, but the symmetry you're talking about here doesn't give rise to anything extra.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Would small curled up dimensions have the same sort of symmetry?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'll leave this for someone else, I think. Not my thing.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is it known exactly what the geomtrical shape of the universe is?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;No, and don't let anyone tell you otherwise! Sometimes, especially in pop-science writing, people imply that we know a lot more about the global properties of the Universe than we do. We often assume things like homogeneity to make our lives simpler, but in fact we have precisely no idea what things are like outside of our horizon volume.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How to describe the &quot;size&quot; of a dimension?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If the Universe's geometry has enough symmetries, it makes sense to define an overall time coordinate everywhere. Then it makes sense to imagine a &quot;slice&quot; through spacetime that represents the Universe at an instant of time. If some of those slices have the geometrical property you're talking about, that traveling a distance R in a certain direction gets you back to your starting point, then it makes sense to call R the &quot;size&quot; of the corresponding dimension. If you can travel forever, then we say the size in that dimension is infinite.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is it possible to describe to a layman the shape of the universe without resorting to inept analogies?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All analogies are imperfect. I think the best you can do is use a bunch of them and try to convey the limitations of each.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-01-14T18:47:07.523" CommentCount="6" />
  <row Id="2919" PostTypeId="2" ParentId="2598" CreationDate="2011-01-14T18:50:25.837" Score="1" ViewCount="" Body="&lt;p&gt;[This was intended as a comment on Lubos' answer above, but grew too big to stay a comment.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(@Lubos) It is well understood that the horizon is, by definition, a trapping surface. Consequently external observers can gain no information about anything that happens in the interior once the trapping surface is formed. This is not an understanding peculiar to LQG. That is in fact what makes the results of LQG more robust in the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You state that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is no calculation of an area-extensive entropy in LQG or any other discrete model of quantum gravity, for that matter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An easy counterexample to that statement, for instance, is Srednicki's 1993 PRL &lt;a href=&quot;http://arxiv.org/abs/arXiv%3ahep-th/9303048&quot; rel=&quot;nofollow&quot;&gt;&quot;Entropy and Area&quot;&lt;/a&gt; (which has 359 citations so far). This paper shows that this entropy-area relation is a very universal aspect of plain old quantum field theory with no inputs whatsoever from loops or strings. Also, the papers you cite (by Domagala, Lewandowski and Meissner) - while these fix an error in Rovelli's work they are not intended to negate the basic procedure of counting states associated with quanta of area, but to reinforce it. So you may hate or love that specific paper by Rovelli, but that does not change the validity of the rest of the vast amount of work done on this topic in LQG. For a comprehensive bibliography I suggested looking up the references in Ashtekar and Lewandowski's 2005 &quot;LQG: Status Report&quot; paper and by doing arXiv searches for papers by Alejandro Corichi and collaborators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that Black Hole entropy should be determined &lt;em&gt;solely&lt;/em&gt; by counting the microscopic surface states of the horizon (and not those of the bulk interior) is something we know from Bekenstein and Hawking's work based on semiclassical QFT. Any microscopic theory, based on loops or strings or whatever, must ultimately yield the same results under coarse graining. LQG does this in a simple and natural way. The key lies in the notion of the &lt;strong&gt;area operator&lt;/strong&gt; - which by itself is a construction natural to and shared by &lt;em&gt;any&lt;/em&gt; theory of quantum geometry. Rovelli's paper is one the earliest (with Kirrill Krasnov, Baez and Ashtekar being among the other pioneers) which outlines the general notion. It is significant for these reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please allow me to stress that in no way am I trying to cast doubts on your (@Lubos') work with quasinormal modes and such. I have yet to properly understand that calculation and I also do not claim to have a universal understanding of all the work on black hole entropy from the loop perspective or otherwise. My hope is simply to refute the notion &quot;&lt;em&gt;that LQG actually doesn't work at all&lt;/em&gt;&quot;! This statement is unfounded and far more evidence than simply noting the error in Rovelli's paper is needed to back up such claims. Needless to say there are errors in the early papers on quantum mechanics, general relativity and string theory. Do those mistakes imply that either one of these frameworks &quot;doesn't work at all&quot;?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; There are some very recent papers which hopefully are big steps towards resolve the black hole entropy question in LQG, and should be of interest to some of the readers here - &lt;a href=&quot;http://arxiv.org/abs/1101.3660&quot; rel=&quot;nofollow&quot;&gt;Detailed black hole state counting in loop quantum gravity&lt;/a&gt; (published in PRD) and &lt;a href=&quot;http://arxiv.org/abs/1101.3662&quot; rel=&quot;nofollow&quot;&gt;Statistical description of the black hole degeneracy spectrum&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit (v2):&lt;/strong&gt; There are some persistent misunderstandings as reflected in the comments about the nature of the Ashtekar formulation. Let me restate, as I mentioned below, that &lt;strong&gt;Ashtekar's variables are nothing more than a canonical transformation&lt;/strong&gt; which lead to a simpler form of the ADM constraints. There are no assumptions about area quantization and such which go into the picture at this stage. Area and volume quantization is the outgrowth of natural considerations regarding quantum geometry. These were undertaken in the mid-90s, seven or eight years &lt;em&gt;after&lt;/em&gt; Ashtekar's original papers. Perhaps the single best and most comprehensive reference for the Ashtekar variables and more generally the complete framework of canonical quantum gravity is Thomas Thiemann's &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0110034&quot; rel=&quot;nofollow&quot;&gt;habilitation thesis&lt;/a&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-20T09:42:44.460" CommentCount="17" />
  <row Id="2920" PostTypeId="2" ParentId="2043" CreationDate="2011-01-14T19:09:12.770" Score="6" ViewCount="" Body="&lt;p&gt;[Once again reading @Lubos' answer sparked these memories in my mind. Thanks for the inspiration @Lubos :)]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@student - everything @Lubos says in this answer is valid. Given that von Neumann algebras are an exotic beast at present as far as their application in physics is concerned, I do know of three cases where they have had significant direct or indirect influence on theoretical physics.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The entire program of knot theory and manifold invariants etc - as represented in Witten's work on TQFT's (topological quantum field theories) - owes in large part to Vaugh Jones' discovery of a &lt;em&gt;knot invariant&lt;/em&gt; known as (obviously) the &lt;strong&gt;Jones Polynomial&lt;/strong&gt;. I know only the vague outline of how he was lead to this discovery but I do know that it happened in the course of his investigations on a particular class (type III?) of von Neumann algerbas.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Connes' non-commutative geometry program also has its roots in the study of von Nemann's algebras if I'm not mistaken. Non-commutative geometry is coming of age with a large number of applications ranging from methods of unifying the Standard Model particles to understanding the quantum hall effect. NCG also arises naturally in string inspired models of cosmology and inflation, &lt;a href=&quot;http://arxiv.org/abs/hep-th/0108190&quot; rel=&quot;nofollow&quot;&gt;[Reference]&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, &lt;a href=&quot;http://arxiv.org/abs/gr-qc/9406019&quot; rel=&quot;nofollow&quot;&gt;Connes and Rovelli&lt;/a&gt;  put forward the intriguing &quot;thermal time hypothesis&quot; in order to try to resolve some of the dilemmas regarding the notion of &quot;time&quot; evolution and dynamics which arise in theories of quantum gravity where the Hamiltonian is a pure constraint - as is the case in the &quot;Canonical Quantum Gravity&quot; program. Their construction rests on a certain property of von Neumann algebras. To quote from their abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... we propose ... that in a generally covariant quantum theory the physical time-flow is not a universal property of the mechanical theory, but rather it is determined by the thermodynamical state of the system (&quot;thermal time hypothesis&quot;). &lt;em&gt;We implement this hypothesis by using a key structural property of von Neumann algebras: the Tomita-Takesaki theorem, which allows to derive a time-flow, namely a one-parameter group of automorphisms of the observable algebra, from a generic thermal physical state&lt;/em&gt;. We study this time-flow, its classical limit, and we relate it to various characteristic theoretical facts, as the Unruh temperature and the Hawking radiation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Of course these are all rather specific and esoteric sounding applications so as @Lubos' noted, vNA's are far more being ubiquitous in theoretical physics.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-14T19:09:12.770" CommentCount="2" />
  <row Id="2921" PostTypeId="2" ParentId="2916" CreationDate="2011-01-14T19:23:56.727" Score="5" ViewCount="" Body="&lt;p&gt;the answer to the Magellan question for our Universe is actually &quot;No&quot;. If the Universe were a static 3-sphere, as Ted Bunn suggests, then you could &quot;swim&quot; around the Universe just like Megallan around the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But an important fact about our Universe is that its size is changing: the size can't stay constant just like the apple can't sit in the middle of a bedroom (in the air) - that's another thing that gravity according to general relativity guarantees. Already today, it resembles an empty de Sitter space (because the dark energy or cosmological constant is the majority of the energy in the Universe) that is exponentially expanding. The so-called Penrose causal diagram of the de Sitter space&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ej.iop.org/images/0264-9381/14/9/003/Full/img31.gif&quot; rel=&quot;nofollow&quot;&gt;http://ej.iop.org/images/0264-9381/14/9/003/Full/img31.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;shows that there are no time-like (mostly vertical) trajectories in it that would lead you to the same place, given by the $x$ coordinate. That's because the height of the diagram (square) is so limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it will never be possible to travel along a straight line, even if you approached the speed of light, and return to the same place where you started (and meet your civilization that didn't travel anywhere). The time-dependence of the size of the Universe is important for this conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spatial section of our Universe is nearly flat - that's because the total density (including dark energy and dark matter) is very close to the critical density, a fact guaranteed by the cosmic inflation. It may be slightly positively curved, like a 3-sphere (the surface of a 4-dimensional ball), or slightly negative curved (the surface of a hyperboloids resembling a saddle, but 3-dimensional one). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole spacetime geometry is close to the 4-dimensional de Sitter space that may be written as the following hyperboloid in 4+1 dimensions:&#xA;$$ -A^2+B^2+C^2+D^2+E^2=R^2 $$&#xA;where $R$ is the radius of the curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning another question you asked: if the Universe is compact, like a sphere, then the wave function may be simply thought of as a wave function of the sphere. Of course, if the sphere is parameterized by some coordinates such as the latitude and longitude, it is a periodic function at least of the latter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sphere has a full rotational symmetry, $SU(2)$. Extra dimensions in string/M-theory can't have any continuous symmetries. They're believed to be Calabi-Yau manifolds or very similar manifolds in which all the dimensions are knitted in a nontrivial way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a compact dimension of space and you write a wave function so that it satisfies $\psi(x) = \psi(x+2\pi R)$, then indeed, it is possible to  interpret this periodicity as a symmetry with respect to the (discrete) group of translations by multiples of $2\pi R$. While Noether's theorem is normally designed for continuous symmetries, you may say what the symmetry implies for such discrete symmetries as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The multiplicative generator of the symmetry is the translation by $2\pi R$ and it commutes with the Hamiltonian - because it is the indentity, after all. The generator may be written as $\exp(2\pi i R p)$ where $p$ is the momentum in the circular direction. And because it is equal to one, it means that the momentum is quantized in units of $1/R$ which is probably not too surprising. A problem here is that the translation by $2\pi R$ is not really a &quot;nontrivial symmetry&quot; in this case: it is an operation that is doing nothing at all. It keeps all objects invariant, so there is no nontrivial Noether's conserved quantity associated with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-14T19:29:39.453" LastActivityDate="2011-01-14T19:29:39.453" CommentCount="4" />
  <row Id="2922" PostTypeId="2" ParentId="2901" CreationDate="2011-01-14T19:37:28.357" Score="0" ViewCount="" Body="&lt;p&gt;I think the other responders adequately covered the &quot;revolving around&quot; part of the question and made the point that the mass of the black hole is insignificant compared to the mass of the galaxy as a whole.  I just wanted to point out that, despite that, it appears that the massive black holes at the centers of galaxies like our own do have some effect on their host galaxies. Astronomers have observed a strong (for astronomy) correlation between the mass of the central black holes and the mass of the host galaxy. Bigger galaxies have proportionally bigger black holes. We don't really know why, it may have to do with how radiation produced as the black hole consumes matter influences the flow of material into the galaxy.  &lt;/p&gt;&#xA;" OwnerUserId="1250" LastActivityDate="2011-01-14T19:37:28.357" CommentCount="2" />
  <row Id="2923" PostTypeId="2" ParentId="2914" CreationDate="2011-01-14T20:09:13.630" Score="4" ViewCount="" Body="&lt;p&gt;If you are interested in the force on a complex configuration of magnets, you most likely will need a 3D electromagnetics FEM solver, something like Ansoft Maxwell 3D, Opera 3D, or other comparable software packages. &lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2011-01-14T20:09:13.630" CommentCount="3" />
  <row Id="2925" PostTypeId="2" ParentId="2792" CreationDate="2011-01-14T21:45:04.910" Score="0" ViewCount="" Body="&lt;p&gt;Study of interference and the related application to 3D scanning has some aspects still to be resolved but it would be unwise to try to reinvent.&#xA;Much of the 3D imaging theorising and initial experimentation was completed during the last 3 decades and is now available thru academic and commercial contacts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I am always careful about recommending web references...&#xA; you really need to check out... &lt;a href=&quot;http://en.wikipedia.org/wiki/3D_scanner&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/3D_scanner&lt;/a&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-14T21:45:04.910" CommentCount="3" />
  <row Id="2926" PostTypeId="1" CreationDate="2011-01-14T21:49:39.813" Score="6" ViewCount="354" Body="&lt;p&gt;Some time ago I was talking to a professor in college about some of the fundamental aspects and origin of General Relativity. I was surprised to learn, in fact, that a pretty good approximation to GR can be achieved simply by using Special Relativity and applying a series of (infinitesimal?) Lorentz boosts to simulate acceleration/a gravitational field. In fact, I believe this is precisely the approach used by GPS system for accurate triangulation/location, as mathematically and computationally it is a somewhat simpler process than applying the Einstein field equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: has anyone else heard of this? An explanation of how exactly this approximation works, and where/why it fails in comparison to true GR (perhaps the strong gravitational field case?) would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2011-01-17T07:26:40.477" Title="Special Relativistic approximation to GR" Tags="&lt;general-relativity&gt;&lt;special-relativity&gt;&lt;relativity&gt;&lt;approximations&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="2927" PostTypeId="2" ParentId="2865" CreationDate="2011-01-14T22:01:36.043" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;this answer has been expanded at the end.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am convinced that macroscopic wormholes are impossible because they would violate the energy conditions etc. so it is not a top priority to improve the consistency of semi-consistent stories. At the same moment, I also think that any form of time travel is impossible as well, so it's not surprising that one may encounter some puzzles when two probably impossible concepts are combined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is a genuinely confusing topic. You may pick Leonard Susskind's 2005 papers about wormholes and time travel:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0503097&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/gr-qc/0503097&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0504039&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/gr-qc/0504039&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amusingly enough, for a top living theoretical physicist, the first paper has 3 citations now and the second one has 0 citations. The abstract of the second paper by Susskind says the following about the first paper by Susskind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In a recent paper on wormholes (gr-qc/0503097), the author of that paper demonstrated that he didn't know what he was talking about. In this paper I correct the author's naive erroneous misconceptions.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very funny. The first paper, later debunked, claims that the local energy conservation and uncertainty principle for time and energy are violated by time travel via wormholes. The second paper circumvents the contradictions from the first one by some initial states etc. The discussion about the violation of the local energy conservation law in Susskind's paper is relevant for your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that if you allowed any configurations of the stress-energy tensor - or Einstein's tensor, to express any curvature - it would also be possible for one throat of an initial wormhole to be time-dilated - a gravity field that is only on one side - and such an asymmetry could gradually increase the time delay between the two spacetime points that are connected by the wormhole. For example, you may also move one endpoint of the wormhole along a circle almost by the speed of light. The wormhole itself will probably measure proper time on both sides, but the proper time on the circulating endpoint side is shortened by time dilation, which will allow you to modify the time delay between the two endpoints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever you try to do, if you get a spacetime that can't be foliated, it de facto proves that the procedure is physically impossible, anyway. Sorry that I don't have a full answer - but that's because I fundamentally believe that the only correct answer is that one can't allow wormholes that would depend on negative energy density, and once one allows them, then he pretty much allows anything and there are many semi-consistent ways to escape from the contradictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Expansion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dear Julian,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am afraid that you are trying to answer more detailed questions by classical general relativity than what it can answer. It is clearly possible to construct smooth spacetime manifolds such that a wormhole is connecting places X, Y whose time delay is small at the beginning but very large - and possibly, larger than the separation over $c$ - at the end. Just think about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may cut two time-like-oriented solid cylinders from the Minkowski spacetime. Their disk-shaped bases in the past both occur at $t=0$ but their disked-shaped bases in the future appear at $t_1$ and $t_2$, respectively. I can easily take $c|t_1-t_2| &amp;gt; R$ where $R$ is the separation between the cylinders. Now, join the cylinders by a wormholes - a tube that goes in between them. In fact, I can make the wormhole's proper length decreasing as we go into the future.  It seems pretty manifest that one may join these cylinders bya  tube in such a way that the geometry will be locally smooth and Minkowski.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These manifolds are locally smooth and Minkowski, when it comes to their signature. You can calculate their Einstein's tensor - it will be a function of the manifold. If you allow any negative energy density etc. - and the very existence of wormholes more or less forces you to allow negative energy density - then you may simply postulate that there was an energy density and a stress-energy tensor that, when inserted to Einstein's equations, produced the particular geometry. So you can't possibly avoid the existence of spacetime geometries in which a wormhole produces a time machine sometime in the future just in classical general relativity without any constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only ways to avoid these - almost certainly pathological - configurations is to &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;postulate that the spacetime may be sliced in such a way that all separations on the slice are spacelike (or light-like at most) - this clearly rules &quot;time traveling&quot; configurations pretty much from the start&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;appreciate some kind of energy conditions that prohibits or the negative energy densities&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;impose other restrictions on the stress-energy tensor, e.g. that it comes from some matter that satisfies some equations of motion with extra properties&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;take some quantum mechanics - like Susskind - into account&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you don't do either, then wormholes will clearly be able to reconnect spacetime in any way they want. This statement boils down to the fact that the geometry where time-like links don't exist at the beginning but they do exist at the end may be constructed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the best&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-20T16:05:54.323" LastActivityDate="2011-01-20T16:05:54.323" CommunityOwnedDate="2011-01-29T11:42:54.297" CommentCount="5" />
  <row Id="2928" PostTypeId="1" AcceptedAnswerId="2929" CreationDate="2011-01-14T23:25:09.747" Score="2" ViewCount="221" Body="&lt;p&gt;So if you have a light bulb in a room, and you had a tool to measure the amount of light that's in the room, then let's assume the amount of light only caused by the bulb is &quot;1&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you place a mirror next to the bulb, does the amount of light in the room in crease to &quot;2&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you keep going so that the light in the room in creases to 3..etc with more mirrors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How far can you take this?  Also, is it actually adding light to the room, or is it simply causing the same photons to move faster (reflecting from one mirror to the next) making it seem like there is more light in the room, but there really isn't?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Made me think, so I thought I ask here :)&lt;/p&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-15T14:09:34.453" LastActivityDate="2011-01-15T15:55:32.627" Title="Light &amp; Mirrors" Tags="&lt;light&gt;&lt;photons&gt;&lt;energy-conservation&gt;" AnswerCount="3" />
  <row Id="2929" PostTypeId="2" ParentId="2928" CreationDate="2011-01-14T23:33:14.537" Score="7" ViewCount="" Body="&lt;p&gt;By &quot;light in the room&quot; I assume you mean Energy given off by the light bulb per second, this is commonly know as &quot;Power&quot; (measured in units of watts). You can also define how much power is hitting a certain area, for example the newspaper you may be reading. This new quantity is called Intensity (watts per area). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By putting mirrors in the room you never change the total &quot;Power&quot;, the energy coming out of the light bulb in a certain amount of time. Instead you change the intensity, i.e. how much light is hitting your newspaper, but it's always at the expense of a darker area of the room elsewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address the last part of your post, light always moves at the same speed. &lt;/p&gt;&#xA;" OwnerUserId="1311" LastActivityDate="2011-01-14T23:33:14.537" CommentCount="2" />
  <row Id="2930" PostTypeId="2" ParentId="2916" CreationDate="2011-01-15T00:49:54.127" Score="2" ViewCount="" Body="&lt;p&gt;Since only a fraction of the universe is actually observable, I am not sure if it possible to answer this question.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-01-15T00:49:54.127" CommentCount="1" />
  <row Id="2931" PostTypeId="1" CreationDate="2011-01-15T02:47:50.313" Score="4" ViewCount="179" Body="&lt;p&gt;I am currently reading journal articles about semiconductor physics in solar cells. What is injection level?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll try to start off with what I understand. Photons hitting the silicon cause its electrons to jump to a higher energy state. Some of these electrons jump high enough that they cross from the valence band into the conduction band, thus contributing to the flow of electric current. Now, how long do these electrons stay in the conduction band and keep flowing? I believe that is what is called carrier lifetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have just read that carrier lifetime depends on injection level. I think that injection level is the rate at which electrons jump into the conduction band. Am I on the right track?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is injection level?&lt;/li&gt;&#xA;&lt;li&gt;What is its unit of measure?&lt;/li&gt;&#xA;&lt;li&gt;What are some common ways of measuring it?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="293" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-15T03:31:19.557" LastActivityDate="2011-01-15T09:03:27.247" Title="What is injection level (semiconductor physics)?" Tags="&lt;solid-state-physics&gt;&lt;carrier-particles&gt;&lt;semiconductor-physics&gt;&lt;photovoltaics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="2932" PostTypeId="1" CreationDate="2011-01-15T03:17:55.043" Score="1" ViewCount="78" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/1008/why-isnt-dark-matter-just-matter&quot;&gt;Why isn&amp;#39;t dark matter just matter ?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;The evidence usually presented in support of dark matter is larger-than-expected masses in galaxy clusters: the mass required for the gravitational lensing is far larger than the mass detected in the galaxies themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why, then, is the assumption that this undetected extra mass is &lt;strong&gt;a previously-unknown kind of matter&lt;/strong&gt;? Why is that new concept required, instead of proposing that the estimates of total normal matter are wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of asking the same question could be: Why are we &lt;strong&gt;more sure of our estimate of the sum of normal matter&lt;/strong&gt; than of the evidence that there is more mass in galaxy clusters than estimated?&lt;/p&gt;&#xA;" OwnerUserId="1315" LastActivityDate="2011-01-15T03:17:55.043" ClosedDate="2011-01-15T03:27:58.653" Title="Why is “dark matter” a more parsimonious explanation than “more normal matter than we thought”?" Tags="&lt;cosmology&gt;&lt;dark-matter&gt;" CommentCount="1" />
  <row Id="2933" PostTypeId="1" CreationDate="2011-01-15T03:43:35.270" Score="2" ViewCount="382" Body="&lt;p&gt;I have had this question for some time now. Hopefully someone can answer it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Boltzmann_equation&quot; rel=&quot;nofollow&quot;&gt;Boltzmann equation&lt;/a&gt; is widely regarded as a cornerstone of statistical mechanics and many applications have been explored with a linearized version. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know that it's &lt;a href=&quot;http://www.physorg.com/news192967537.html&quot; rel=&quot;nofollow&quot;&gt;extremely hard&lt;/a&gt; to obtain exact solutions, which has started a considerable amount of investigation looking for an equally good (or acceptable) formalism to analyse systems that otherwise would be impossible or would take a great deal of computational resources to obtain a solution using the Boltzmann equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In spite of this, I never heard a precise description about the degree of accuracy (in comparison with experiments) that can be drawn from the Boltzmann equation. Obviously, I expect that accuracy depends on the system at hand, however, it would be great to hear about some specific examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recommended readings would also be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="18" LastEditorDisplayName="" LastEditDate="2011-01-15T05:42:18.883" LastActivityDate="2011-01-16T00:38:43.083" Title="Accuracy of the Boltzmann equation" Tags="&lt;statistical-mechanics&gt;&lt;kinetic-theory&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="0" />
  <row Id="2934" PostTypeId="1" AcceptedAnswerId="2938" CreationDate="2011-01-15T08:28:42.357" Score="5" ViewCount="465" Body="&lt;p&gt;We all experience things spinning, whether it's water down a drain, the earth on its axis, planets round the sun, or stars in a galaxy - even electrons round an atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why is spin so common in the universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there are some physical laws governing this, but perhaps my question is deeper than stating and explaining the law (conservation of angular momentum). &lt;/p&gt;&#xA;" OwnerUserId="1320" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-27T12:10:02.347" LastActivityDate="2011-01-27T12:10:02.347" Title="Why do galaxies and water going down a plug hole spin?" Tags="&lt;astrophysics&gt;&lt;angular-momentum&gt;&lt;spin&gt;&lt;hydrodynamics&gt;&lt;galaxy&gt;" AnswerCount="1" />
  <row Id="2936" PostTypeId="2" ParentId="2931" CreationDate="2011-01-15T09:03:27.247" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;the injection level is defined as $\delta n/p_0$ where $\delta n$ is the minority carrier (e.g. electrons') density excess at non-equilibrium while $p_0$ is the equilibrium density of the majority carriers (e.g. holes).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;From the formula which is a ratio of two very similar densities, it's clear that the injection level is dimensionless; &quot;high&quot; and &quot;low&quot; injection level usually means values that are much greater or much smaller than one&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The methods to determine the level depend on the context and cleverness of the physicist or engineer. In various contexts, it may be determined by the barrier formed at metal-dielectric interface; by the method described by Kobayashi and Kimura; from the absolute value of photoconductivity measured somewhere; from bias illumination; etc. I think it would be bad for me to pretend that I am actually an expert in any of these experimental things.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-15T09:03:27.247" CommentCount="1" />
  <row Id="2937" PostTypeId="2" ParentId="2928" CreationDate="2011-01-15T09:20:55.677" Score="3" ViewCount="" Body="&lt;p&gt;Dear Sev, first of all, a simple thing. Light in the vacuum always moves by the speed 299,792,458 meters per second: this fact is exactly true because of the modern definition of one meter. The speed of light in the air is just 0.03% smaller than the speed of light in the vacuum. Nothing ever travels faster than light in the vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, mirrors typically reflect less than 100% of coming light - something like 70%. But that's not the main problem here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, the energy is conserved, so you can't produce much more &quot;light&quot; by putting mirrors. You can't illuminate 100 households by having 1 light bulb and &quot;copying&quot; it by mirrors. Why not? While you may increase - and almost double - the &quot;amount of light&quot; that is hitting a particular area, this fact is (more than) balanced by the fact that the light that would be absorbed by the area occupied by the mirror itself is not absorbed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when it comes to the energy budget, ideal mirrors (that reflect 100% of light, just for the sake of simplicity) only rearrange the distribution of light - which areas finally absorb it and which areas just reflect it. The total amount of light that is absorbed is given by the total amount of light that is emitted on the light bulb - and it only depends on the light bulb (and its power).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider light as a &quot;practical thing allowing us something to to be seen&quot;, then you want the light to be reflected, e.g. by a book. But the counting for a book that reflects some light - so that we can read it - is similar as the counting for an object that absorbs the light. Mirrors may increase the amount of light reflected by a particular book, but they may not increase the amount of light reflected by the whole room, assuming it has a uniform albedo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at a light bulb and a nearby mirror, you may see &quot;two light bulbs&quot; and a doubled amount of light, so to say. But this is only true from certain directions. From other directions, the outcome is different and often opposite. For example, if you place your eyes behind the mirror, so that the light bulb is on the opposite side of the mirror than you, then you see no light bulb directly - and no unreflected light from a light bulb (and no light bulb light reflected only by mirrors).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This lesson is much more general. Mirrors - and any other gadgets - may move energy from one place to another, or transform it from one form to another. But they never change the total amount of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-15T09:20:55.677" CommentCount="2" />
  <row Id="2938" PostTypeId="2" ParentId="2934" CreationDate="2011-01-15T09:38:53.190" Score="5" ViewCount="" Body="&lt;p&gt;Dear Andrew, &#xA;an excellent question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let me begin with the electron because quantum mechanics is what I've been trained to do. If you find this portion inpenetrable, jump to the section Classical physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrons have an internal spin equal to 1/2. But when it comes to their rotation around the nucleus, electrons actually don't spin around the hydrogen nucleus or helium nucleus at all. The ground state has electrons in the &quot;1s&quot; state and the letter &quot;s&quot; means that the orbital angular momentum is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it's true that &quot;most states&quot; - like excited states of the hydrogen, and most states of other atoms - have spinning electrons. The reason is not hard to be explained mathematically: the angular momentum is a function of the other basic quantities (&quot;observables&quot;) that may change. And because the spin may be nonzero, it usually is nonzero. One needs to fine-tune things very accurately for the spin to be exactly zero and such things don't occur naturally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angular momentum in quantum mechanics is quantized and may be described by two quantum numbers, $l, m$, where $m$ goes between $-l$ and $+l$ with the spacing equal to one. Here, $m$ says what is the $z$-component of the angular momentum and $l$ determines its total magnitude. (The components relatively to other axes may be measured as well - $z$ is just a convention - but they cannot be measured simultaneously because the operators don't commute i.e. because of the uncertainty principle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The states with the angular momentum given by $l,m$ are described by wave functions that depend on the angular variables by the function $Y_{lm}(\theta,\phi)$, the spherical harmonics. Whenever $l,m$ is different from $0,0$, i.e. whenever the angular momentum is nonzero, the function $Y$ is non-constant, and vice versa. If you have anything that depends on the direction in quantum mechanics, it really implies that the angular momentum is nonzero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Classical physics&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angular momentum is conserved - both in classical physics and quantum physics - and the deeper reason for that is Noether's theorem. Emmy Noether has demonstrated that all systems whose laws are rotationally symmetric will exhibit a conserved quantity called the angular momentum, and vice versa. (She has done the same thing with many more - as well as the most general - pairs of symmetries and conservation laws, too.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am convinced that the rotational symmetry of the laws of physics is a deep thing - phenomena don't depend on the direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your universalistic summary of many different &quot;spinning systems&quot; surely sounds intriguing but I think that it is really not so deep from a physics viewpoint. The deep thing is that the angular momentum is conserved and why; but why a particular system has a high chance to end up spinning depends on the system and its details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, both the galaxies and the whirlpool are spinning for a very similar reason, indeed. It is the same reason why your spinning rate increases while you're skating and you shrink all your arms and hands closer to the axis of the rotation. Why? Well, the angular momentum is conserved. But it may be written as $J=I\omega$ where $J$ is the angular momentum, $I$ is the moment of inertia, and $\omega$ is the angular frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you shrink your arms and legs, you actually reduce $I$ because $I\approx MR^2$ where $M$ is the mass and $R$ is the rough distance of the mass from the axis. If you shrink $R$, by getting closer to the axis, you will also reduce $I$, and to keep $J$ constant which is a law of physics, you have to increase $\omega$: you will be spinning faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thing occurs with water in the bath tub. A liter of water that is approaching the sink was pretty much everywhere and had a reasonable angular momentum because of the chaotic motion of the water. However, when you suddenly compress all of the water in the sink - very near the axis of the sink - the liter of water that is just leaving will reduce its $R$, distance from the axis, and also $I$ as a consequence. So the angular frequency $\omega$ inevitably increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, for the galaxies that are being formed, lots of gas is moving in random directions, but the gravitational collapse shrinks the matter so that it is closer to the axis of the future galaxy. Because it is closer, the moment of inertia decreases, so the angular frequency has to increase just like before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So indeed, there is a similarity - and the same explanation - for many phenomena where things begin to spin more quickly than before. And these phenomena are cute and intriguing. But physics may explain them and the truly &quot;beautiful&quot; and &quot;deep&quot; explanations are much more abstract in character.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also a related question and answers about a ballerina who speeds up:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/3611/why-does-a-ballerina-speed-up-when-she-pulls-in-her-arms/3617&quot;&gt;Why does a ballerina speed up when she pulls in her arms?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Luboš&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-27T11:11:38.833" LastActivityDate="2011-01-27T11:11:38.833" CommentCount="2" />
  <row Id="2939" PostTypeId="2" ParentId="2567" CreationDate="2011-01-15T10:40:01.853" Score="2" ViewCount="" Body="&lt;p&gt;Dear Gway, LED lights are a nice discovery but if you want a superior light that really emits at all frequencies, light bulbs with wolfram are a choice. They're cheaper, emit at all frequencies, don't contain Hg or other serious poisons, and another advantage is that they can heat up your home. Unfortunately, light bulbs are getting banned at many &quot;progressive&quot; places such as the European Union.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black body radiation is nonzero at each frequency. It is an idealized radiation - which may be approximated by black objects (in the conventional sense). The most accurate natural source of black-body-like radiation is the cosmic microwave background but its temperature is 2.7 K. Black holes emit thermal Hawking radiation, too: but it hasn't been observed yet. However, the high frequencies carry a higher percentage of energy from a black body than the low frequencies. This can be circumvented by having a black body radiation from a whole array of sources that have different i.e. variable temperature. By taking a right mixture of the low-frequency ones, you may obtain a pretty much flat (or another sufficiently convex) spectrum over the visible (or another) interval as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If some light contains all frequencies, it is not coherent. Coherence requires the frequency - and usually the direction - to be universal for all the light. So the light containing all frequencies is pretty much by definition maximally incoherent. Coherence means that the information about the relative phases of the electromagnetic waves at two different points in spacetime is predictable and periodic. For light that contains all frequencies, it is unpredictable and aperiodic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-15T10:40:01.853" CommentCount="2" />
  <row Id="2940" PostTypeId="2" ParentId="2933" CreationDate="2011-01-15T10:57:07.867" Score="2" ViewCount="" Body="&lt;p&gt;the discrepancy between the Boltzmann equation and the experiment may always be blamed upon the experiments. If one does things right, the Boltzmann equation is theoretically completely exact (issues will be discussed below). However, it's pretty hard to measure the distribution functions &quot;directly&quot;, especially if you want the full dependence both on momenta and positions so one may say that the &quot;weak link&quot; is always on the experimental side (after errors are fixed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Boltzmann equation is mathematically accurate and a consequence of other laws of physics that have been tested pretty thoroughly. But there are a few subtleties here:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The Boltzmann equation in its general form contains a collision term which is a completely general function of all the variables - just like the distribution function itself. So unless one specifies what the collision term is, the Boltzmann equation is tautologically true for &lt;em&gt;some&lt;/em&gt; collision term and may be viewed as a definition of the collision term.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The collision term which is a function of all variables is often approximated by the molecular chaos Ansatz - which depends on the distribution functions of one particle in a bilinear way; and on a function describing the rate of collisions at a given place and given momenta. Of course, one has to carefully calculate - or indirectly measure - the latter function. Even when one does it, the Boltzmann equation with the molecular chaos condition included may start to fail when the molecules are not in real chaos. The closer one is to an &quot;ideal gas&quot;, the more accurately the molecular chaos assumption holds, but of course, the whole point of the equation is that it also works for non-ideal gases and liquids. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(It is a purely technical assumption to make further calculations more specific; obviously, the fact that Boltzmann or others made the molecular chaos assumption, to make their calculations specific, doesn't weaken the derivations of the H-theorem i.e. the second law of thermodynamics. It holds even without the molecular chaos assumption - the assumption or its equivalent is just needed to calculate the increase of entropy and other things quantitatively.)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Boltzmann equation is a classical beast that neglects quantum mechanics; however, one may still interpret the distribution functions as the appropriate expectation values - and these expectation values in quantum mechanics still have to satisfy an equation of the Boltzmann form. So the Boltzmann equation is valid even for the calculation of processes in the heavily quantum mechanical context. For example, it's been used to calculate the abundance of elements after nucleosynthesis - the period up to 3 minutes after the Big Bang when mostly light elements were produced in the hot &quot;nuclear liquid&quot; that was filling the Universe at the time. The abundances - mostly Hydrogen, some percentage of Helium and Lithium etc. - numerically agree with the observed percentages which is a nontrivial evidence for The Big Bang Theory (that will be aired on CBS at least until the 7th season, it was just decided). &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-15T11:02:25.010" LastActivityDate="2011-01-15T11:02:25.010" CommentCount="1" />
  <row Id="2941" PostTypeId="2" ParentId="1646" CreationDate="2011-01-15T12:52:34.147" Score="5" ViewCount="" Body="&lt;p&gt;a great topic. First, ten gigatesla is only the magnetic field near a magnetar - a special type of neutron star. They were discussed e.g. in this Scientific American article in 2003:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://solomon.as.utexas.edu/~duncan/sciam.pdf&quot; rel=&quot;nofollow&quot;&gt;http://solomon.as.utexas.edu/~duncan/sciam.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ordinary neutron stars have magnetic fields that are 1000 times weaker than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that in the magnetar stars, atoms are squeezed to cigars thinner than the Compton wavelength of the electron - which is in between the radius of the  electron (and also the radius of the nucleus) and the radius of the atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is such strong a field that many other things occur. For example, there is a box interaction between 4 photons, caused by a virtual electron loop. This is normally negligible - so we say that Maxwell's equations are linear in the electromagnetic fields. However, at such strong magnetic fields, the nonlinearity kicks in and one photon often splits into two, or vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there's a lot of new stuff going on in such fields. A magnetar that would be 1000 miles away would kill us due to diamagnetism of water in our cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Magnetars and fusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your idea to use magnetars to support fusion is creative, of course. But I think that to start fusion, you have to squeeze the nuclei closer than the Compton wavelength of the electron which is still $2.4 \times 10^{-12}$ meters, much longer than the nuclear radius. You would need to add two or three more orders of magnitude to the squeezing. A magnetar is not enough for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have such brutally deformed atoms, you can't neglect the nuclear reactions involving electrons - which are usually thought of as &quot;irrelevant distant small particles&quot; that don't influence the nuclear processes. However, if their wave functions are squeezed to radia that are substantially shorter than the Compton wavelength, their kinetic energy substantially increases. At the width of the wave function comparable to the Compton wavelength, the total energy/mass of the electron increases by O(100%) or so. This increase comes from the &quot;thin&quot; directions only but it is enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, note that the difference between the neutron mass and the proton mass is just 2.5 masses of the electron. So if you squeeze the electron so that its total energy increases more than 2.5 times, it becomes energetically favored for the protons inside your (not so) &quot;crystal&quot; to absorb the electron and turn into neutrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I believe that all the matter in a near proximity of the magnetar will actually turn into the same matter that the neutron star itself is made of. That will happen before the protons will have any chance to create new bound states such as iron nuclei (that you wanted to produce by fusion). You will end up with neutrons and almost no protons - the same state of matter that the star is built from itself. In some sense, I think that this shouldn't be surprising - if it is surprising for someone, he should have asked the question why there is no ordinary matter left on the neutron stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the timescale after which the electrons are absorbed to turn the protons into neutrons? Well, it's a process mediated by the weak nuclear interaction - like beta-decays. Recall that the lifetime of the neutron is 15 minutes but it is anomalously long a time because of some kinematical accidents. The normal objects of the same size - such as the extremely squeezed cigar-shaped atoms - would decay more quickly (into neutron and neutrinos, in this case). On the other hand, the electrons in the cigar-shaped atoms occupy a bigger region than the quarks in the neutron. But this can only add at most 4 orders of magnitude. To summarize, I think that within days or months, if not more quickly, the electrons would get swallowed to create neutrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the best&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-15T14:00:46.813" LastActivityDate="2011-01-15T14:00:46.813" />
  <row Id="2942" PostTypeId="2" ParentId="2926" CreationDate="2011-01-15T13:25:04.067" Score="3" ViewCount="" Body="&lt;p&gt;first, I don't know exactly whta &quot;sequence of infinitesimal Lorentz boosts&quot; you're referring to. A boost of the flat Minkowski space gives you a Minkowski space back, so you can't get a curved space by any sequence of boosts that act globally on the spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the adjective &quot;infinitesimal&quot; could be pretty much inconsequential. Transformations in physics are often represented as the product of infinitely many infinitesimal transformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the statement that the world - the Solar System - may be approximated by special relativity is obviously true. The curvature of spacetime is small and may be treated perturbatively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In pretty much any reference system you choose, and you may even choose the geocentric system, all the gravitational effects including the general relativistic effects may be incorporated as small corrections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, you may always define some operational coordinates in which the spacetime is nearly flat, so the metric is given by the flat Minkowski metric. The actual metric dictated by general relativity, whatever it is, may be expressed as a small correction to the Minkowski metric. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, such a correction may be calculated at arbitrary accuracy - not just the leading level - by perturbative expansions. This approach to general relativity is usually referred to as linearized gravity - even if one goes beyond the linearized level - and it is not only essential for the actual calculations of observable phenomena but also for a proper conceptual understanding of general relativity (and quantum gravity - the linearization is necessary to understand that there are gravitons, and what their mass, spin and physical polarizations are).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Einstein himself calculated the effects of a point mass - e.g. the Sun - on other bodies - such as the Mercury - in the perturbative scheme. It was before Schwarzschild gave an exact solution for Einstein's equations in the spherically symmetric case (when extrapolated everywhere in space, it is now referred to as the Schwarzschild black hole).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effects that the GPS has to take into account include, of course, the standard classical mechanics forces such as the centripetal force or the Coriolis force, as well as some special relativistic and general relativistic effects. But of course, when things are calculated in practice, people are de facto assuming that the whole world takes place on a Minkowski background and all the gravitational effects are small corrections that occur on the Minkowski arena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if Einstein hadn't discovered GR by methods of an ingenious physicists, engineers working for GPS would later find the right corrections in the process of adjusting their system and removing the discrepancies. They could experimentally measure the dependence on the daytime, latitude, and longitude of any effect that matters and that made their previous GPS plus software inaccurate. They wouldn't appreciate the beauty of GR right away, but of course, they could guess the right form of the correction terms from the experimental deviations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, to answer your last question, the perturbative approach to GR, expanding around an SR background, doesn't have to fail anywhere. It may be exact just like $\exp(x)$ may be exactly expressed by the Taylor expansion, $1+x+x^2/2+x^3/3!+\dots$. Of course, the perturbative approach becomes awkward, to say the least, for causally nontrivial setups such as black holes - including the interior; wormholes and other topologically nontrivial shapes of space; or some very global questions in cosmology. But when the spacetime is topologically trivial, perturbative expansions work. The flatter the space is, the more useful the perturbative expansions become.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-15T13:25:04.067" CommentCount="1" />
  <row Id="2943" PostTypeId="2" ParentId="2602" CreationDate="2011-01-15T13:51:59.797" Score="0" ViewCount="" Body="&lt;p&gt;I unfortunately don't have too much time to continue to research this question.  I can give direction to you for your own research.  I was somewhat intrigued by the question because the thrust of it is to imply that macroscopic physical laws are somehow invalid. To your point, classical laws do somewhat breakdown in the quantum regime do to phenomenon such as Bose-Einstein condensates:&#xA;//cua.mit.edu/ketterle_group/popular_papers/physics%20today%20v2.pdf&#xA;Which is obviously an area of active research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in general your question can be alternatively be understood as a test of diffusivity governed by the Arrhenius equation.  &lt;em&gt;The Science and Engineering of Materials Second Edition&lt;/em&gt; by Donald Askelund, chapter 5 on the Atom Movement of Materials, discuss Arrehenius equation in the context of movement of defects and imperfections in materials.  In the context provided though, one would assume that the vacuum surrounding an initial lump of gas as being a material (no I do not prescribe to aether, but this is a convenient assumption). So the question then is one of what the activation energy is to cause the atoms to move into the vacuum vs the potential energy available for such a reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick look for test of the Arrhenius equation will reveal some intersting results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you move further into this question you find discussion on reaction and diffusivity:&#xA;//www.scholarpedia.org/article/Reaction-diffusion_systems&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are multiple examples of diffusion theory underlying the processes used in several areas of technology.  So I'm not sure about the context and intent of your question.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-15T14:12:59.317" />
  <row Id="2944" PostTypeId="2" ParentId="2933" CreationDate="2011-01-15T14:33:01.367" Score="0" ViewCount="" Body="&lt;p&gt;I provided a response to question in the same context here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/2602/experiments-that-measure-the-time-a-gas-takes-to-reach-equilibrium&quot;&gt;Experiments that measure the time a gas takes to reach equilibrium.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, specifically to your question, this is a good quick overview:&#xA;&lt;a href=&quot;http://homepage.univie.ac.at/franz.vesely/sp_english/sp/node7.html&quot; rel=&quot;nofollow&quot;&gt;http://homepage.univie.ac.at/franz.vesely/sp_english/sp/node7.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where you find out that with certain simplifying assumptions, the Boltzmann distribution is a solution to the Boltzmann equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, it is interesting to understand how boltzmann distribution are related to the Arrhenius equation:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Arrhenius_equation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Arrhenius_equation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that these theories underpin most of 20th century technology I think is testament enough to accuracy within relevant regimes.&lt;/p&gt;&#xA;" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-15T21:05:15.853" LastActivityDate="2011-01-15T21:05:15.853" CommentCount="1" />
  <row Id="2945" PostTypeId="2" ParentId="2933" CreationDate="2011-01-15T14:34:42.910" Score="2" ViewCount="" Body="&lt;p&gt;The Boltzmann equation is an approximation. It is obtained by closing the so-called BBGKY hierarchy that links N-particle distribution functions to N+1 particle distribution functions. The closing is obtained by introducing an assumption, the 'molecular chaos' assumption. In short, molecular chaos asumes the absence of correlation in the incoming state for particles that oollide. &#xA;Detailed molecular dynamics simulations of many-particle model systems have revealed that the molecular chaos assumption indeed fails to describe quantitatively the transport coefficients of dense many-particle systems due to the presence of 'long-time tails'. In other words: correlations between particles don't die out exponentially. &#xA;Interestingly, the lower the number of dimensions, the stronger this effect. The Boltzmann equation is not a good approximation to 2D liquids.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-01-15T14:34:42.910" CommentCount="2" />
  <row Id="2947" PostTypeId="2" ParentId="2701" CreationDate="2011-01-15T15:04:36.907" Score="0" ViewCount="" Body="&lt;p&gt;I interpret the question as &quot;which frequencies/wavelengths can be generated by cavity magnetrons?&quot;&#xA;As far as the physics is concerned, I do not believe there are fundamental limits that constrain the frequencies generated. These frequencies are dictated by the cavity dimensions (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Cavity_magnetron&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Cavity_magnetron&lt;/a&gt;). Now there are many practical engineering constraints (which brings the question outside the realm of physics) that render cavity magnetrons practical devices mostly for generating cm wavelength EM radiation. And of course as far as the heating of food is concerned, a magnetron needs to emit radiation that is in resonance with polar molecules. And that restricts magnetron wavelengths to cm scales.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-01-15T15:04:36.907" CommentCount="3" />
  <row Id="2948" PostTypeId="2" ParentId="2567" CreationDate="2011-01-15T15:05:20.993" Score="2" ViewCount="" Body="&lt;p&gt;Sounds like you want a supercontinuum source:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.fianium.com/supercontinuum.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.fianium.com/supercontinuum.htm&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2011-01-15T15:05:20.993" />
  <row Id="2949" PostTypeId="1" CreationDate="2011-01-15T15:30:23.353" Score="5" ViewCount="247" Body="&lt;p&gt;Since the total mass-energy for the neutrino presumably does not change when a neutrino changes lepton flavor, though the mass is different, what compensates for the gain or loss of mass?  Does the propagation speed of the neutrino change?&lt;/p&gt;&#xA;" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-01-16T08:47:12.367" LastActivityDate="2011-01-17T04:28:48.457" Title="How can neutrinos oscillate though the lepton flavors have differing masses?" Tags="&lt;particle-physics&gt;&lt;energy-conservation&gt;&lt;neutrino&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="2950" PostTypeId="2" ParentId="1008" CreationDate="2011-01-15T15:41:03.723" Score="0" ViewCount="" Body="&lt;p&gt;The major missing link is the existence of a class of particles, which aside from gravity are sufficiently weakly interacting to fill the need. We already know of one kind of particle which has most of these characteristics, neutrinos. In this case physicists and cosmologists have theoretical reasons to believe the upper bound on the amount of mass in neutrinos is too small. But the mere existence of one class of weakly interacting particles should make the possibilty of another appear less novel. &lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-01-15T15:41:03.723" />
  <row Id="2951" PostTypeId="1" CreationDate="2011-01-15T15:48:40.747" Score="4" ViewCount="556" Body="&lt;p&gt;What are zero modes in quantum field theory, and what are they used for? Or, where can I read about them? I was never able to find a good introduction on the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am particularly interested in zero modes that appear in the context of Yang-Mills, QCD and QCD-like theories, at zero or finite temperature. My doubt consists on the nature of these zero modes in these theories: are they special field configurations, or quantum states, or something else? Basically, I'd like to know what's their physical significance.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-01T16:15:24.790" Title="What are zero modes?" Tags="&lt;quantum-field-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="2952" PostTypeId="1" AcceptedAnswerId="2965" CreationDate="2011-01-15T15:53:27.217" Score="5" ViewCount="478" Body="&lt;p&gt;For example: the role it might play in a theory of quantum gravity (ie causing space-time curvature)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that inertial mass can result from binding energy alone. Has the equivalence principle been tested on elementary particles (like the electron) whose mass would be entirely due to the Higgs coupling?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like an obvious question, but I have never heard it discussed before.&lt;/p&gt;&#xA;" OwnerUserId="1247" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-29T16:59:16.160" LastActivityDate="2011-03-29T16:59:16.160" Title="Does the equivalence between inertial and gravitational mass imply anything about the Higgs mechanism?" Tags="&lt;general-relativity&gt;&lt;high-energy-physics&gt;&lt;higgs-boson&gt;&lt;equivalence-principle&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="2953" PostTypeId="2" ParentId="2928" CreationDate="2011-01-15T15:55:32.627" Score="4" ViewCount="" Body="&lt;p&gt;I think that the other two answers miss an interesting point of your question. Lets put a light bulb inside 100% reflecting sphere. All emitted photons will stay inside the sphere and intensity of light will continuously increase to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Real mirrors are never 100% reflective but there are mirrors with 99.98 reflectivity (these are dichroic mirrors that work only in a narrow wavelength range). With such mirrors light can reflect 1000 times before escaping the sphere so light intensity inside the sphere will be 1000 times higher compared to a lamp without mirrors. The energy conservation is not violated because new photons are not created.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we just created is actually an &lt;em&gt;optical resonator&lt;/em&gt; although normally we have a one-dimensional case with two parallel mirrors and a laser pulse trapped between them (as in &lt;a href=&quot;http://en.wikipedia.org/wiki/Cavity_ring-down_spectroscopy&quot; rel=&quot;nofollow&quot;&gt;Cavity ring-down spectroscopy&lt;/a&gt;). Concave mirrors are used for better light trapping.&lt;/p&gt;&#xA;" OwnerUserId="701" LastActivityDate="2011-01-15T15:55:32.627" CommentCount="1" />
  <row Id="2954" PostTypeId="2" ParentId="2949" CreationDate="2011-01-15T16:00:52.840" Score="6" ViewCount="" Body="&lt;p&gt;There are a couple of misconceptions here.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The flavor states are not mass states. That is, the electron neutrino does not have a mass $m_{\nu_e}$ and the muon neutrino a mass of $m_{\nu_\mu}$. Rather, there are two different basis' in which to examine the neutrino. So a neutrino known to be $l$ flavored, is a mixture of mass states (numbered) like&#xA;$$ |l&amp;gt; = \sum_{i=1}^3 U_{li}^* |i&amp;gt; $$&#xA;where $|l&amp;gt;$ is a flavor state (for $l = e, \mu, \tau$); $|i&amp;gt;$ is a mass state; and $U$ is the unitary mixing matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutrinos interact in the flavor basis, but must propagate in the mass basis, so a neutrino mixing experiment probes the probability of detecting a neutrino in state $\beta$ and it was created in state $\alpha$&#xA;$$ P_{\alpha,\beta} = \left| &amp;lt;\beta|\alpha(t)&amp;gt; \right|^2 = \left| &amp;lt;\beta|Ue^{iEt}U^*|\alpha&amp;gt;\right|^2 $$&#xA;which is a pretty complicated expression in the full three-flavor analysis.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Nor is it a single mass state which propagates, all three do using the usual $e^{iEt}$ propagator, where $E = \sqrt{m_i^2 - p^2}$, which is where the mixing enters, because this reduces to oscillating trig functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expression above becomes&#xA;$$ P_{\alpha,\beta} = \left| \sum_i \sum_j U^*_{\alpha,i}U_{\beta,j} \sin \left(2X_{i,j} \right) &amp;lt;\beta|\alpha&amp;gt; \right|^2  $$&#xA;where $X_{i,j} = \frac{m_i^2 - m_j^2}{4E}L$, which can be reduced further because $&amp;lt;\beta|\alpha&amp;gt; = \delta_{\alpha\beta}$ and by using some trig identities, but all that is left as an exercise.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Finally, note that all the neutrinos we can interact with have energies measures in MeVs or GeVs, and all the mass states are understood to be less than 1 eV, so all neutrinos are ultra-relativistic: they move a the speed of light for nearly all practical purposes. (The exception here is the hope of comparing the arrival time of the neutrino and light wave-fronts from distant supernovae. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this were not the case, you would expect to the the probability distribution for a initially well defined neutrino pulse to differentiate by mass state as a function of time, with the leading edge being composed of the lightest state (i.e $m_1$ [$m_3$] if the normal [inverted] hierarchy obtains), and the trailing edge of the heaviest state ($m_3$ [$m_2$]). But those states would &lt;em&gt;still&lt;/em&gt; mix to all flavors, it just that the mixture would be time dependant.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-01-15T18:56:52.147" LastActivityDate="2011-01-15T18:56:52.147" />
  <row Id="2955" PostTypeId="2" ParentId="1008" CreationDate="2011-01-15T16:03:00.700" Score="6" ViewCount="" Body="&lt;p&gt;There is a very precise reason why dark planets made of 'ordinary matter' (baryons - particles made up of 3 quarks) cannot be the dark matter. It turns out that the amount of baryons can be measured in two different ways in cosmology:&#xA;- by measuring present-day abundances of some light elements (esp deuterium) which are very sensitive to the baryon amount, and&#xA;- by measuring the distribution of the hot and cold spots in the Cosmic Microwave background (CMB), radiation left over from the early universe that we observe today&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two methods agree spectacularly, and both indicate that baryons are 5% of the total stuff (energy/matter) in the universe.  Meanwhile, various measures of gravitational clustering (gravitational lensing, rotation of stars around galaxies, etc etc) all indicate that &lt;em&gt;total&lt;/em&gt; matter comprises 25% of the total. (The remaining 75% is in the infamous dark energy which is irrelevant for this particular question). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since 5% is much less than 25%, and since the errors on both of these measurements are rather small, we infer that most of the matter, about 4/5 ths (that is, 20% out of 25%) is 'dark' and NOT made up of baryons. &lt;/p&gt;&#xA;" OwnerUserId="1334" LastActivityDate="2011-01-15T16:03:00.700" />
  <row Id="2956" PostTypeId="2" ParentId="2926" CreationDate="2011-01-15T16:08:15.097" Score="1" ViewCount="" Body="&lt;p&gt;Special relativity is a good &lt;em&gt;local&lt;/em&gt; approximation to general relativity. Special relativity or any succession of Lorentz transformations will reveal nothing about the global aspects of gravitating systems.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-01-15T16:08:15.097" CommentCount="3" />
  <row Id="2957" PostTypeId="2" ParentId="2602" CreationDate="2011-01-15T17:49:09.433" Score="1" ViewCount="" Body="&lt;p&gt;&quot;The time a gas takes to reach equilibrium&quot; isn't really well-defined -- and even trying to interpret it in terms of an average sense can be tricky: in particular, there may very well be states which never reach equilibrium (consider very carefully placed and aimed gas moleculres in a box).  As far as physics is concerned though, the Boltzmann equation is what's believed to govern the approach to equilibrium, but one must put in some assumed &quot;collision term&quot; to actually put it to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of where your question may have been studied in current research, the issue of thermalization is obviously going to be a bigger issue in fields of physics when one is dealing with numbers of particles near the limits of validity of thermodynamics, so that's a sign that you might look for papers in the field of &quot;cold atoms&quot;.  See for example &lt;a href=&quot;http://link.aps.org/doi/10.1103/PhysRevA.64.034703&quot; rel=&quot;nofollow&quot;&gt;this 2001 paper by the group of Alain Aspect&lt;/a&gt; which carefully measures a &quot;thermalization time&quot; $\tau_{th}$ which they define via a Boltzmann equation-inspired model inspired by &lt;a href=&quot;http://dare.uva.nl/document/31522&quot; rel=&quot;nofollow&quot;&gt;a theory paper of Luiten et al&lt;/a&gt;.  These papers might be a good place to start to see how working physicists think about thermalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You frame your question in terms of &quot;ideal gases&quot; but as you'll notice from these papers, when you get down to numbers of particles and temperatures which are manageable directly in these experiments, quantum effects play a nontrivial role in the population dynamics.  (It's also worth giving some thought to the following: how can an ideal gas equilibrate if the particles are defined not to interact with each other?  Your answer will probably help you figure out better what exactly you're looking for.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're more interested in what happens to a large number of &quot;classical&quot; particles and how they thermalize purely through collisions, this is fairly easy to &lt;em&gt;simulate on the computer&lt;/em&gt; with molecular dynamics code, and then you can measure whatever you like with the computed particle trajectories.  Of course this isn't strictly a physical &quot;experiment&quot; but it is in my opinion a real test of the kinetic theory predictions all the same.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2011-01-15T17:49:09.433" CommentCount="1" />
  <row Id="2958" PostTypeId="2" ParentId="1479" CreationDate="2011-01-15T17:50:16.483" Score="5" ViewCount="" Body="&lt;p&gt;The correct answer is actually given by the person who asked the question (as well as Scott Wales), while most of the answers above give a misleading one. The shape is hourglass simply because Sloan Digital Sky Survey and 2 degree Field Sky Survey did not map out the whole sky, but only about 1/4 of it (for the SDSS). The other directions therefore appear dark in the visualization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is of course nothing special about the directions in which these surveys look. In fact, there exist other galaxy surveys that cover nearly the whole sky, though I guess they haven't produced fancy visualizations. (Moreover different surveys operate at different wavelengths of light and are complementary in this, and several other, ways). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that the Milky Way would obstruct some portion of the sky, but 1) this is a small fraction (20% or so of the sky), and 2) if you showed the sky in the Milky Way direction, it would actually appear bright, not black, for obvious reasons.&lt;/p&gt;&#xA;" OwnerUserId="1334" LastActivityDate="2011-01-15T17:50:16.483" />
  <row Id="2959" PostTypeId="2" ParentId="2792" CreationDate="2011-01-15T17:56:34.077" Score="0" ViewCount="" Body="&lt;p&gt;Perhaps an extreme case of your line of thought where you focus on the &quot;glitches&quot; in your detector and try to extract data out of them is the &quot;single pixel camera&quot;, where you take a series of random measurements from your very simple detector and reconstruct the image you're looking for.  Here's &lt;a href=&quot;http://terrytao.wordpress.com/2007/04/13/compressed-sensing-and-single-pixel-cameras/&quot; rel=&quot;nofollow&quot;&gt;a nice article by Terence Tao on the mathematics of these&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2011-01-15T17:56:34.077" />
  <row Id="2960" PostTypeId="2" ParentId="2951" CreationDate="2011-01-15T18:00:50.160" Score="2" ViewCount="" Body="&lt;p&gt;Zero modes are solution to an eigenvalue equation with zero eigenvalue. They appear many times in mathematics and physics, so it is hard to know where to point you to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly, bosonic zero modes indicate degeneracy in the system, for example such that results from symmetry. If you have a configuration of some classical field, think about it as some localized lump of energy, it can oscillate with different frequencies, called the normal modes of the system. They are classified by their frequency $\omega$ (or energy, $\hbar \omega$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of those normal modes are special in that they have zero frequency $\omega=0$, and then they are not oscillations at all, instead they are rigid translations of the whole system in some direction. This means there is  a family of solutions, for example localized in different positions, and the zero mode corresponds to translating the configuration to a new location. This is a zero mode because translation to a new direction does not cost you energy (unlike real oscillations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fermionic zero modes are even more interesting, via index theorems they are related to topology and anomalies, but you probably need more background in QFT (or topology) to appreciate those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Lubos has more quantitative discussion, you can read both to have a more complete picture).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference I'd recommend Rajaraman's &quot;Solitons and Instantons&quot;. The simple example of kink in two dimension, and its translational zero mode, should make the general pattern clear. I think there is also good discussion of fermionic zero modes, e.g. in the presence of an instanton, and their relation to topology via the Atiyah-Singer index theorem.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-02-01T16:15:24.790" LastActivityDate="2011-02-01T16:15:24.790" CommentCount="6" />
  <row Id="2961" PostTypeId="2" ParentId="123" CreationDate="2011-01-15T18:12:46.067" Score="5" ViewCount="" Body="&lt;p&gt;The best way of measuring the mass is in principle using gravitational lensing, as mentioned in the first posted answer to this question. However, applying this method has become feasible only relatively recently, partly because lensing is relatively rare and requires 'lucky' alignment of the source and the lens, and thus a large telescope survey to find enough of them and with an excellent spatial resolution (typically below an arcsecond) to resolve the source images. Moreover, in detail it is hard to apply. You don't know the 'density profile' (radial distribution of mass in the galaxy) &lt;em&gt;a priori&lt;/em&gt;, so you need to measure it. Therefore, ideally you need to have multiple sources of light behind a single galaxy at different angular locations, so that each source probes the galaxy mass M(R) enclosed at some radius. This can be done using the method called weak gravitational lensing (and it would take us beyond the scope of the question to talk about it). Anyway, this kind of measurement has started to happen only in recent years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traditionally in astrophysics, the mass of the galaxy is instead measured using various dynamical methods where velocities of stars are converted to the mass. But I disagree with Ernie that this is easy to do using Newton's Laws. The difficulties include the fact that you are only measuring the radial (and not tangential) velocity of the star(s) using spectroscopy; that the distribution of the mass of the galaxy may not be spherically symmetric, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traditionally people have used the virial theorem, most famously first used by Fritz Zwicky in 1933 to measure the mass of the Coma cluster of galaxies. In this approach, under some assumptions, one can relate the mass of the cluster to velocity dispersion of galaxies along the line of sight, and you can measure the latter as described in the earlier post to this question. (This brilliant logic, using existing measurements at the time, led Zwicky to announce that Coma must be dominated by invisible, dark, matter). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also other methods to get masses, but they are messy. Typically, one infers the mass-to-light ratio of the object (M/L), and then measures L to get M. These approaches have significant uncertainties. Therefore, as mentioned in the post above, gravitational lensing is in principle the cleanest method.  &lt;/p&gt;&#xA;" OwnerUserId="1334" LastActivityDate="2011-01-15T18:12:46.067" />
  <row Id="2962" PostTypeId="2" ParentId="2951" CreationDate="2011-01-15T18:17:41.773" Score="2" ViewCount="" Body="&lt;p&gt;Moshe is of course completely right, but let me be a bit less abstract for a while and list a few examples that cover most of the contexts where the term &quot;zero mode&quot; is used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one discusses the motion of an extended object, such as the string, the position of a point along the string - the point is parameterized by $\sigma$ - is $X^\mu(\sigma)$. The &quot;zero mode&quot; of $X$ is then nothing else than the average over the string - the center-of-mass coordinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dual quantity is the &quot;zero mode&quot; of the momentum - the integral of $P_\mu(\sigma)$: that's nothing else than the total momentum. These two zero modes, the center-of-mass position and the total momentum, are the quantities we would normally associate with a point-like particle. And indeed, in some approximation, the string may be considered to be a particle, and the zero modes are the usual properties of the particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one solves differential equations on compact manifolds, such as the Calabi-Yau manifolds, one may find zero modes, too. For example, we may have a Dirac equation&#xA;$$D^\mu \gamma_\mu \psi(x_1, \dots, x_6) = 0$$&#xA;The solutions $\psi$ to this massless Dirac equations are just zero modes. The equation above only included functions of the compact coordinates. But if you consider both 6 compact and the 3+1 non-compact coordinates, one may consider spinors $\psi$ whose dependence on the 6 compact coordinates is given by the &quot;zero mode&quot; - the solution of the equation above. Such a field will then behave as a massless particle in 4 dimensions - it will solve the 3+1-dimensional equation with a vanishing mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases above, there was an operator - either $-d^2/d\sigma^2$, or the Dirac operator $D^\mu \gamma_\mu$, that annihilated the solution and that's why the solution was ultimately called a &quot;zero mode&quot;. In this context, it's useful to mention what are the &quot;non-zero modes&quot; or &quot;normal modes&quot;. They are eigenvalues of the same operators with different eigenvalues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the operator $-d^2/d\sigma^2$ on the string may act on functions such as $\exp(in\sigma)$ and it will produce $n^2$ - the same with sines and cosines. These non-zero modes are eigenstates of an operator, so if you include all possible eigenvalues and the right degeneracy, you may reconstruct any function - by Fourier expansions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true for the Dirac example. All the spinors $\psi$ on the 6-dimensional manifold may be written as a combination of the &quot;modes&quot;, some of which are the zero modes but most of them are non-zero modes. When one discusses supersymmetry, the non-zero modes typically come in pairs and it can be proved, but the zero modes - massless particles in 3+1 dimensions, as I mentioned - may come non-paired. So the number of zero modes of the Dirac equation (more precisely the number of solutions minus the number of &quot;anti-solutions&quot; with some opposite charge or chirality etc.) is a special number known as the &quot;index&quot;. It is invariant under all continuous transformations. In heterotic string theory, it is interpreted as the number of generations (minus the number of antigenerations) of leptons and quarks we obtain in 3+1 dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may continue with solutions to Dirac and similar equations. But consider a different background, e.g. an instanton in gauge theory. And include the gauge field term to the Dirac operator. Then this total Dirac operator will again have some zero modes on $R^{3,1}$; the zero mode will be nonzero and nontrivial especially in the region where the instanton is located. Any fermionic particles in 3+1 dimensions that has a zero mode will actually appear as a factor in a complicated product of many fermionic fields - and this multi-product interaction term is actually induced by the instanton. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The zero mode is a $c$-number-valued solutions of the right differential equation with the right operator. But its real meaning is that the whole quantum field should be expanded into modes - zero modes and non-zero modes. Each term in the sum has a $c$-number valued function of the &quot;space&quot; or &quot;spacetime&quot; and it is multiplied by an operator corresponding to the mode. The operators that multiply the zero modes in the expansion play a special role.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of the Dirac operator, you may consider the differential operators for other fields, including bosonic fields such as the gauge field itself. The &quot;zero modes&quot; of the differential equation for the bosonic fields - essentially various forms of (covariantized) wave equations - describe deformations of the background. They're &quot;collective degrees of freedom&quot; of the instanton. It's called the 't Hooft interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the description &quot;collective degree of freedom&quot; applies to the string as well. The center-of-mass position and the total momentum are &quot;collective degrees of freedom&quot; for all the string &quot;bits&quot; along the string; they share these collective degrees of freedom. Some of the zero modes tell you how the whole object may move in space - like in the example of the string. But the fermionic zero modes can also tell you how the whole object may move in the superspace. And the zero modes of the instanton also include some modes whose variation determines changes of the size (and/or shape) of the instanton - or any other object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This text was only meant for the reader to have some more concrete ideas about the words by Moshe which are of course totally valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-15T18:25:24.813" LastActivityDate="2011-01-15T18:25:24.813" CommentCount="4" />
  <row Id="2963" PostTypeId="2" ParentId="1048" CreationDate="2011-01-15T18:37:48.847" Score="3" ViewCount="" Body="&lt;p&gt;I am not sure about the Lense-Thirring effect in this case, but it is definitely true that a global rotation would be described by one of the Bianchi homogeneous but anisotropic (not isotropic) models as mentioned by Moshe. For beautiful pictures as to how the Cosmic Microwave Background anisotropy sky would look in those models, see the paper by Andrew Pontzen for example (&lt;a href=&quot;http://arxiv.org/abs/0901.2122&quot; rel=&quot;nofollow&quot;&gt;Pontzen 2009&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the current CMB data, especially from the WMAP experiment, one can impose upper bounds on (and perhaps some day, detections of) those Bianchi models. This work goes at least as far back as Ted Bunn and collaborators' paper (&lt;a href=&quot;http://arxiv.org/abs/astro-ph/9605123&quot; rel=&quot;nofollow&quot;&gt;Bunn, Fereira &amp;amp; Silk 1996&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="1334" LastActivityDate="2011-01-15T18:37:48.847" CommentCount="1" />
  <row Id="2964" PostTypeId="1" AcceptedAnswerId="2968" CreationDate="2011-01-15T18:38:05.377" Score="5" ViewCount="374" Body="&lt;p&gt;There's something known as a &quot;resonating valence bond&quot; (RVB) state, which plays a role in at least some attempts to understand physics of high-$T_c$ superconductors. This, roughly, involves a state that's in a superposition (hence the &quot;resonating&quot; part of the name, if I understand correctly) of different ways to pair electrons into strongly-bonded spin singlets. My question is: what is a more precise definition of this sort of state? What's the underlying physics, when does it arise, and why is it interesting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Points an answer might address: is there a simple toy model for which this is the ground state, that sheds light on what sort of system it could arise in? Is there an interesting continuum limit, in which we can characterize this state in a more field-theoretic language? Are there particular kinds of instabilities such a state tends to be subject to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I know where I would start digging if I wanted to really understand this for myself, but mostly I'm asking it to probe the community and see what kind of expertise might be lurking here, since there haven't been so many condensed matter questions.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2011-01-15T19:35:46.287" Title="What is a resonating valence bond (RVB) state?" Tags="&lt;condensed-matter&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="2965" PostTypeId="2" ParentId="2952" CreationDate="2011-01-15T18:40:12.560" Score="3" ViewCount="" Body="&lt;p&gt;Dear user,&#xA;the equivalence between the inertial mass and gravitational mass tells us the following thing about the Higgs mechanism:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Any inertial mass produced or modified by the Higgs mechanism also has to produce or modify a source of gravity of the same magnitude. And vice versa, a gravitational mass produced by the Higgs mechanism also has to produce an inertial mass of the same magnitude.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why others - and myself - and confused about your question is that the equivalence principle tells us the very same thing not only about the Higgs mechanism but also about confinement, electrostatic attraction, spinning gyroscopes, or any other process, object, or mechanism that is taking place, has been taking place, or will take place. The equivalence principle is a totally universal law that holds for all objects and all processes in this Universe - and beyond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not true that the Higgs mechanism has a more special relationship to the equivalence principle than any other mechanism in non-gravitational physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning your more specific question about the experimental tests - of course that it has been tested. The available experimental tests of the equivalence principle show that all materials have the same ratio of inertial and gravitational mass up to the precision of $10^{-15}$ or so. Different materials have a different percentage of their mass coming from the electrons - the more neutrons a material has, the smaller the fraction is. So the mass stored in the electrons may go from 0.02% to 0.05%. While this is much smaller than 100%, it's surely enough to exclude the conjecture that the electron mass - as produced by the Higgs mechanism - doesn't obey the equivalence principle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The percentages above are just 3.5 orders of magnitude below 100%. So you still have 12 orders of magnitude left that prove that the mass produced by the electrons' interactions with the Higgs is exhibited both as inertial mass and gravitational mass - with the same ratio (one - in normal units) - as all other objects have. So once again, yes, all methods to obtain energy/mass i.e. $E=mc^2$ obey the equivalence principle and this fact has been tested with an amazing accuracy. The equivalence principle is true for masses produced by the Higgs mechanism, confinement, or anything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This fact is a problem for some &quot;cheap&quot; methods to solve the cosmological constant problem. The problem with the cosmological constant is that even things such as virtual objects in atomic physics, QCD, or anywhere create sources of vacuum energy density. It is not possible to throw them away because such a procedure would ultimately contradict the equivalence principle. So the mystery is why the cosmological constant is so tiny even though we may enumerate lots of possible sources that are much bigger and that appear at any conceivable scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-15T18:40:12.560" CommentCount="14" />
  <row Id="2966" PostTypeId="1" CreationDate="2011-01-15T19:25:09.873" Score="2" ViewCount="136" Body="&lt;p&gt;I got a question concerning the scattering of phonons and electrons. &#xA;I read an introductory explanation to this process that is somehow&#xA;not very satisfactory. It goes like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\psi_{k}$ and $\psi_{k&amp;#39;}$ be bloch-waves within a solid. &#xA;We denote the probability of transmission between these two states by $P_{k,k&amp;#39;}$. &#xA;This probability is according to quantum mechanical pertubation theory&#xA;proportional to $|&amp;lt;k&amp;#39;|H&amp;#39;|k&amp;gt;|^{2}$ where H' denotes the perturbed Hamiltonian &#xA;or the perturbed potential (caused by either phonons or impurities). &#xA;Now we assume that our two wave function are of the standard bloch-form. &#xA;Hence we obtain: &#xA;$|&amp;lt;k&amp;#39;|H&amp;#39;|k&amp;gt;| = \int{ d^{3}r u_{k&amp;#39;}(r)^{*} H&amp;#39;(r,t) u_{k}(r) e^{i(k-k&amp;#39;)r}}$  (Eq. 1)&#xA;where u_{k} and u_{k'} have lattice-periodicity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in an inelastic collision with of an electron and a phonon we have &#xA;by energy conservation:&#xA;E(k') - E(k) = +/- $h_{bar} \omega(q)$ &#xA;where E(k) and E(k') denote the electron energies and on the rights hand side&#xA;there is the energy of a phonon with a wavevector q. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now comes the tricky part of the analysis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now they say that the disturbed potential must include a dependency on $e^{iqr}$&#xA;Hence (Eq. 1) (which is the scattering probabilty) must include a matrix element&#xA;of the form $&amp;lt;k&amp;#39;|e^{iqr} |k&amp;gt;=\int{d^{3}r u_{k&amp;#39;}(r)^{*} u_{k}(r) e^{i(k-k&amp;#39;+q)r}}$ . (Eq. 2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is that? I don't see their point here. I want to remark here that I do have&#xA;heard an introductory quantum mechanics class and class in linear algebra. &#xA;However I neither understand this from a mathematical point of view nor from &#xA;a physical point of view. Can anyone give me a better explanation of this?&#xA;Please note that the course I'm taking is not a course in theoretical solid state physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the derivation goes on: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The say that since $u_{k}u_{k&amp;#39;}$ can be expanded into a fourier series of reciprocal lattice vectors. Fine - I agree that's legitimate. So assume: &#xA;$(u_{k}u_{k&amp;#39;})(r) = \sum_{G}f_{g} e^{-iGr}$. Plugging this into this equation &#xA;they claim the matrix element above  (Eq. 2) only does NOT vanish is k'-k+q=G &#xA;Well this doesn't seem plausible. I mean then $e^{i(k-k&amp;#39;+q-G)r}=1$. Integrating over all &#xA;space leads to infinity. Well I may agree that the integral over this term vanishes if the equality k'-k+q=G does not hold because of periodicity/symmetry. &#xA;Maybe one has to restrict the integral to the solid itself to keep it finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone give me a more detailed exlaination??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking forward to your responses. Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="1156" LastActivityDate="2011-01-16T17:49:52.663" Title="Scattering of phonons and electrons within solids" Tags="&lt;solid-state-physics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="2967" PostTypeId="2" ParentId="2907" CreationDate="2011-01-15T19:31:04.790" Score="3" ViewCount="" Body="&lt;p&gt;The short answer is no. Particles obey a very important property called &quot;cluster decomposition principle&quot;, while those type of lumps that you mentioned do not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What that means is that if you have a particle as an excitation of a theory, you might perfectly well have a two particle state as a solution when their separation goes to infinity. This is a essential ingredient to calculate S matrices and cross sections (that's almost everything that is useful in particle collision experiments). This is not true, in general, for these extended solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further details can be found in that excellent book Aspects of Symmetry, by Coleman. Check section 6.2.3: Lumps are like particles (almost).&lt;/p&gt;&#xA;" OwnerUserId="1337" LastActivityDate="2011-01-15T19:31:04.790" />
  <row Id="2968" PostTypeId="2" ParentId="2964" CreationDate="2011-01-15T19:35:46.287" Score="2" ViewCount="" Body="&lt;p&gt;RVB states were first coined in 1938 by Pauling in the context of organic materials and they were later extended to metals. Anderson revived the interest in this concept in 1973 when he claimed that they explained the Mott insulators. (Mott, not Matt and not Motl, which is a shame because I was born in 1973.) He wrote a new important paper in 1987 in which he described the copper oxide as an RVB state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one has a lattice of atoms etc. and there is qubit at each site - e.g. the spin of an electron - then the RVB state in the Hilbert space of many qubits is simply&#xA;$$|\psi\rangle = \sum |(a_1,b_1) (a_2,b_2) \dots (a_N,b_N)\rangle$$&#xA;which is a tensor product of &quot;directed dimer&quot; states of qubits which are simple singlets&#xA;$$|M,N\rangle = \frac{1}{\sqrt{2}} \left( |\uparrow_M\downarrow_N\rangle - |\downarrow_M\uparrow_N\rangle \right) $$&#xA;The sum defining the RVB state goes over all arrangements or methods how to divide the lattice into (vertical or horizontal or whatever dimensions you have) pairs of adjacent lattice sites. For each lattice site, one puts the corresponding two qubits (usually spins) into the singlet state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The singlet state above is $M,N$-antisymmetric, so one has to be careful about the signs. So all the tensor factors $(a_i,b_i)$ above are oriented and the orientation always goes in such a way that $a_i$ is a white site on a chessboard while $b_i$ is a black site on a chessboard, in the usual chessboard method to divide the lattice into two subsets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because all singlet states used in the RVB states are made out of nearest neighbors, it looks like a liquid, which is why the resulting material in this state is called the RVB liquid. (Imagine molecules in a liquid - they also like to interact with some neighbors only. If one doesn't rely on distant molecules to neutralize the spin, it is &quot;liquid-like&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea - related to the name - is that the information about the terms defining $|\psi\rangle$ is the information about which adjacent lattice sites are connected - these are the valence bonds (&quot;valence&quot; because the nearest neighbors are interacting through their valence electrons or degrees of freedom). However, a general term of this type could be claimed to evolve into another similar state where different links (valence bonds) are included to the creation of the singlets. If one tries to allow the valence bonds to jump anywhere - and switch from vertical to horizontal directions - one obtains a &quot;resonating&quot; system. This symmetrization (symmetric superposition) of all possibilities is a usual way to obtain a quantum eigenstate of the lowest energy, assuming that the different terms are able to change into each other by a transition amplitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The funny feature of this liquid state is that it is invariant under all the translations - those allowed by the lattice - and the rotations - allowed by the lattice, if any. This is very different from a particular chosen method how to divide the lattice sites (qubits) to pairs. By summing over all methods to divide into pairs, we reach a certain degree of &quot;democracy&quot; that gives the state very different and special properties - in comparison with some particle &quot;vertical crystals&quot; or other ways how you could orient the singlets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you may look at it from the other side. It is somewhat nontrivial to construct singlet states of the material, and the RVB state is the most democratic one. It is often useful to look at mathematical guesses that look special and the RVB state was no exception. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You seem to be interested in the high-$T_c$ superconductors. I believe that the critical paper in this direction was this 1987 paper&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://prb.aps.org/abstract/PRB/v35/i16/p8865_1&quot; rel=&quot;nofollow&quot;&gt;http://prb.aps.org/abstract/PRB/v35/i16/p8865_1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by Kivelson, Rokhsar, and Sethna. They asked a simple question - what are the excitations above the RVB state. A fascinating feature was that the excitations inherit only 1 of 2 key properties of the electron: there are spin-1/2 fermionic excitations - like the electron - but the shock is that they're electrically neutral; and there are charged excitations - like the electron - but they're spin-0 bosons (similar to solitons in polyacetylene).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a cool property that by choosing a pretty natural state, you may obtain totally unfamiliar excitations - of course, it's a common theme in condensed matter physics. I suppose that if they can talk about the mass of the excitations, and they're non-negative, they also have a Hamiltonian for which the state occurs, and they show that it is stable along the way. But you must read the full paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't mentioned the high-$T_c$ punch line yet. Of course, the bosonic charged excitations may produce a Bose gas and this Bose gas could exist at high temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course, one must be careful not to get carried away. The RVB state is not the only one that one can construct out of the spins. The experimental attempts to produce a full-fledged RVB liquids remained inconclusive, to put it lightly, and some previously believed applications of the RVB liquid are no longer believed to be true. For example, it was believed that the RVB state is a description of the disordering of anti-ferromagnets, but especially from a 1991 paper by Read and my ex-colleague Sachdev, it became much more likely that the spin-Peierls description is more likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting theoretical by-product of the RVB considerations were things related to cQED - strong coupling compact quantum electrodynamics - with a $\pi$-flux RVB state in the continuum limit. This bizarre theory also has the neutral spin-1/2 excitations; an infinite bare coupling; and has been studied nicely for $SU(N)$ and $Sp(2k)$ gauge groups. It must be assumed that the spin Peierls ordering doesn't develop in the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-15T19:35:46.287" CommentCount="4" />
  <row Id="2969" PostTypeId="2" ParentId="1361" CreationDate="2011-01-15T20:01:58.920" Score="5" ViewCount="" Body="&lt;p&gt;I guess you were asking about elastic photon-photon interaction. If that is the case, I remember reading some proposals (and argumentation of feasibility) of experiments using existing laser facilities in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PRL 96, 08362 (2006)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further references: hep-ph/0606136, hep-th/0512033&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea if the experiments were really made, but those references are quite cool :)&lt;/p&gt;&#xA;" OwnerUserId="1337" LastEditorUserId="1337" LastEditorDisplayName="" LastEditDate="2011-01-15T22:44:39.247" LastActivityDate="2011-01-15T22:44:39.247" />
  <row Id="2970" PostTypeId="2" ParentId="2239" CreationDate="2011-01-15T20:26:41.833" Score="2" ViewCount="" Body="&lt;p&gt;When the wing is brought near the ground, two things happen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The downwash velocity of the wing's trailing vortices is reduced (the ground literally &quot;blocks&quot; the downwash).  The effects are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1a) The effective angle of attack increases, so for a given geometric angle of attack the lift increases.  But in a gentle landing the lift is almost fixed (equal to the weight), so in this case the actual effect is to reduce the geometric angle of attack of the airplane so as to cancel the downwash reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1b) The induced drag decreases for a given lift.  Or equivalently, the span efficiency increases, sometimes above unity.  This has the same effect as increasing the thrust a bit, and it's what makes the airplane seem to want to &quot;float&quot; above the runway during the flare before finally settling down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Effects 1a) and 1b) depend on the dimensionless ratio  height_above_ground/wing_span. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The 2D-like flowfield near the wing's airfoil is modified so as to increase the pressure difference between its top and bottom, thus increasing the lift for a given angle of attack.  Again, if the lift is fixed, the actual effect will be to reduce the geometric angle of attack.  This effect simply adds to 1a) above.  It has no effect on the induced drag, so it doesn't influence 1b) above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This effect depends on the dimensionless ratio  height_above_ground/airfoil_chord, so it's only significant in low-wing airplanes.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-15T20:26:41.833" />
  <row Id="2971" PostTypeId="2" ParentId="2966" CreationDate="2011-01-15T20:38:50.453" Score="0" ViewCount="" Body="&lt;p&gt;everything you write is perfectly OK, except for the statement that &quot;it doesn't seem plausible&quot;. Yes, it is plausible and yes, the exponential is equal to one. That's the whole reason why the contribution for $G=k&amp;#39;-k+q$ to the interaction amplitude is nonzero. All the terms with different values of $G$ lead to oscillating functions in the complex plane that sum up to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, the integral of 1 over the whole space is infinite. And yes, you also know the way to make it finite: restrict the interval to the solid itself. It is not surprising - and it is physically correct - that the integral is infinite for an infinite solid. It's because the interaction Hamiltonian between the two plane wave states is proportional to the volume of the solid as well - the greater volume, the greater chance that the interaction will occur somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were calculating the things for a finite volume of the solid - e.g. a box - you would get manifestly finite numbers everywhere. However, some formulae would be unnecessary awkward and they would depend on the size and shape of the solid. That's why adult physicists learned from Paul Dirac how to use the distributions such as $\delta(G-G_0)$. For your problem, $G_0=k&amp;#39;-k+q$ is treated as a constant. The defining property of the $\delta$-function is that&#xA;$$\int_{-\infty}^\infty dG\,f(G) \delta(G-G_0) = f(G_0)$$&#xA;for any function $f(G)$. So the integral of $f(G)$ over $G$, weighted by the delta-function, only picks  the value at $G=G_0$. That's because $\delta(G-G_0)$ vanishes for all values  for which $G\neq G_0$. But for $G=G_0$, it is infinite and so large that the integral of $\delta(G)$ over $G$ is equal to one, and if you insert $f(G)$, it simply picks $f(G_0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The object $\delta(G-G_0)$ is not a function in the usual sense but it is extremely helpful and consistent in dealing with the integrals that appear in the Fourier transformations. The delta-function may be approximated by a function that is equal to $1/\epsilon$ for the argument being between $-\epsilon/2$ and $\epsilon/2$, and otherwise is equal to zero, in the limit where $\epsilon$ goes to zero. The delta-function can also be written as the Fourier transform of the function $1$ divided by $2\pi$:&#xA;$$\delta(G-G_0) = \frac{1}{2\pi}\int_{-\infty}^\infty dR\,\exp[i(G-G_0)R] $$&#xA;You may always imagine that the delta-function exercises are translated to a calculation at a finite volume of the space or solid; then the momenta such as $G$ become discrete - number of zeros of a standing wave, for example, times $1/R_{solid}$ - and the integral over $G$ is replaced by a summation. The function $\delta(G-G_0)$ is then replaced by a simple Kronecker delta symbol $\delta_{G,G_0}$ which is only nonzero - equal to one - if $G=G_0$. But you will pay the price that most formulae will contain powers of the volume and other things, and you need to remember the spacing of the momenta etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these extra things in the formulae will depend on the size and shape of the solid - or space(time) - that you picked. But at the end, you know very well that there exists a sensible physics in the infinite-volume limit that should be independent of the volume (because it was sent to infinity). For you, to learn how to deal with the Dirac distributions is important to do all these calculations effectively because the volume $V$ and all the problematic features about the spacing of the momentum evaporate from the formulae, and the fact that the formulae hold for any large piece of the solid (or space) becomes self-evident.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-15T20:43:56.753" LastActivityDate="2011-01-15T20:43:56.753" CommentCount="1" />
  <row Id="2972" PostTypeId="2" ParentId="1327" CreationDate="2011-01-15T22:20:08.830" Score="-1" ViewCount="" Body="&lt;p&gt;Contrary to some of the answers above, energy conservation does hold exactly in general relativity. It can be derived using Noether's Theorem and the time symmetry of the gravitational field equations when all fields including the gravitational metric are treated as dynamical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the Newtonian 1/r energy law is not applicable on a cosmological scale but it is true that the gravitational contribution to the energy equation is negative in general relativity as it is in Newtonian physics. Matter, radiation and dark energy contribute positive terms. Overall the energy in a given region of space which might expand over time changes only by an amount calculated as the flux of energy over the boundary of the region. This statement embodies the law of conservation of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many fallacies that are often repeated that make people think that energy conservation in general relativity is not exactly conserved. The most common is to think that Noether's theorem can only work in a static gravitational field. This is not true so long as the variable energy of the gravitational field itself is included in the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a fuller explanation the entries &lt;a href=&quot;http://blog.vixra.org/category/energy-conservation/&quot; rel=&quot;nofollow&quot;&gt;I posted on my blog&lt;/a&gt; can be consulted. Alternativly the Wikipedia entry about the &quot;Stress-energy-momentum pseudotensor&quot; is another valid analysis, although I prefer the covariant formalisms myself. &lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-15T22:20:08.830" CommentCount="7" />
  <row Id="2973" PostTypeId="2" ParentId="2838" CreationDate="2011-01-15T22:40:16.223" Score="-2" ViewCount="" Body="&lt;p&gt;The claim that the total energy in the universe is zero can be rigorously justified. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your specific questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) General Relaitivty is required. It does not apply for Newtonian gravity &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) It has to be assumed that classical general relativity with or without cosmological constant is correct and that the universe is spatially homogeneous on sufficiently large scales. If the universe is infinite the total energy is not really defined but it is still true that the total energy in an expanding volume of space is asymptotically zero when the region is large enough for the homogeneity of the universe to be a good enough approximation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Here is a &lt;a href=&quot;http://www.prespacetime.com/index.php/pst/article/view/81&quot; rel=&quot;nofollow&quot;&gt;link to a paper&lt;/a&gt; as requested.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-15T22:40:16.223" CommentCount="3" />
  <row Id="2974" PostTypeId="1" CreationDate="2011-01-15T23:16:10.170" Score="0" ViewCount="273" Body="&lt;p&gt;If the earth left the solar system for some reason. Assuming its moving at the same velocity it's currently exhibiting.  How long would it take for the atmosphere to freeze. Would we get methane clouds like Titan. Would there be an atmosphere or would it all be frozen on the ground? &lt;/p&gt;&#xA;" LastActivityDate="2011-01-16T09:48:38.793" Title="If the earth left the solar system for interstellar space. How long would it take for atmosphere to freeze" Tags="&lt;space&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="2975" PostTypeId="2" ParentId="2933" CreationDate="2011-01-15T23:45:13.483" Score="0" ViewCount="" Body="&lt;p&gt;@Robert Smith&#xA;Well you're asking a question about the accuracy of an equation.  The equation is created under certain assumptions about small objects sometimes called &quot;particles&quot; or &quot;atoms&quot;.  So if the equation accurately describes the system in question, then the behavior of the sytem should be described by solutions to the governing equations.  So questions as to accuracy of the equations, is really revealed by whether the systems behavior follows the solutions to the equations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the boltzmann's distribution is a solution to boltzmann's transport equation under certain conditions, it is the question of whether the system deviates from a boltzmann distribution under situations where it should hold that would indicate a situation where the boltzmann equation may no longer be valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you are still confused as how my answer relates to the question, it might be that you are not asking the right questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/astro-ph/pdf/9807/9807078v1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/astro-ph/pdf/9807/9807078v1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2011-01-15T23:45:13.483" CommentCount="2" />
  <row Id="2976" PostTypeId="2" ParentId="2604" CreationDate="2011-01-15T23:54:08.830" Score="-1" ViewCount="" Body="&lt;p&gt;The number of fermionic helicity states of the supersymetric standard model with massive neutrinos --if they are right neutrinos, as expected from seesaw and GUT-- and massive gauge bosons is 126. Of course the number of bosonic states is the same :-). You can add another two helicity states if the mass mechanism is the MSSM one. And of course, you can add another two helicity states if you put the gravitino in the bag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, with 3 generations and now that the neutrinos are massive, you win the possibility of fitting the game in a 128 fermion, which happens to be the dimension of a D=11 fermion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With any game of neutrinos, if you put them apart the extant fermions, they amount to 84 helicities. With three generations and a massive top, you can consider neutrinos as in the previous paragraphs and simply put apart the top quark and squark; again the extant &quot;light fermions&quot; amount to 84 helicities, and so they superpartners. Nothing useful has emerged of this, but it comes out from having three generations. Note that in D=11 SUGRA, an 84-component bosonic object is forced to exist, that complements the 44-component graviton.&lt;/p&gt;&#xA;" OwnerUserId="1335" LastEditorUserId="1335" LastEditorDisplayName="" LastEditDate="2011-02-21T17:42:00.530" LastActivityDate="2011-02-21T17:42:00.530" CommentCount="4" />
  <row Id="2977" PostTypeId="2" ParentId="2974" CreationDate="2011-01-15T23:58:48.137" Score="0" ViewCount="" Body="&lt;p&gt;Velocity relative to where? Not that crucial though =)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would take a matter of hours(maybe a few days) if you &lt;em&gt;instantly&lt;/em&gt; take it to an empty space. It would take a few weeks if you gradually move away from the Sun. Let me note that as Noldorin said the only way of heat transfer is via radiation, but since Earth is a relatively good radiator, it would not take very long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There would almost always be an atmosphere because there simply is enough gravity to attract gas, no matter what kind of gas it is. It will surely be dimmer&amp;amp;less dense though. There is not nearly enough methane on Earth to form a methane cycle similar to that of Titan. Titan possesses maybe hundreds of times more methane than Earth.&lt;/p&gt;&#xA;" OwnerUserId="504" LastActivityDate="2011-01-15T23:58:48.137" CommentCount="1" />
  <row Id="2978" PostTypeId="1" CreationDate="2011-01-16T00:26:13.137" Score="4" ViewCount="226" Body="&lt;p&gt;It is experimentally known that the equation of motion for a charge $e$ moving in a static electric field $\mathbf{E}$ is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}}{\mathrm{d}t} (\gamma m\mathbf{v}) = e\mathbf{E}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to show this using just Newton's law for the proper frame of $e$, symmetry arguments, the Lorentz transformations and other additional principles?&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-16T04:01:43.220" LastActivityDate="2011-01-16T04:01:43.220" Title="Derive $\frac{\mathrm{d}}{\mathrm{d}t}(\gamma m\mathbf{v}) = e\mathbf{E}$ from elementary principles?" Tags="&lt;special-relativity&gt;&lt;forces&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="2980" PostTypeId="1" CreationDate="2011-01-16T00:38:55.233" Score="1" ViewCount="300" Body="&lt;p&gt;A positron is odd under charge conjugation and parity reversal but nevertheless even with respect to time reversal. Is a theoretical positron which would be odd under all three symmetries (C, P, T) physical?&lt;/p&gt;&#xA;" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-16T01:54:48.563" LastActivityDate="2011-01-16T22:36:29.603" Title="Is &quot;real&quot; antimatter (odd under C, P, T) unphysical?" Tags="&lt;particle-physics&gt;&lt;antimatter&gt;&lt;symmetry&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="2981" PostTypeId="2" ParentId="2978" CreationDate="2011-01-16T00:40:33.197" Score="5" ViewCount="" Body="&lt;p&gt;I think this is a lot simpler than you suspect. It's really just Newton's 2nd law, and recognising the concept of momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dt} (\gamma mv) = \frac{dp}{dt} = F = eE$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Since classically the electric field $E$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Electric_field#Definition&quot; rel=&quot;nofollow&quot;&gt;defined&lt;/a&gt; as the force $F$ divided by the elementary charge $e$.)&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2011-01-16T00:40:33.197" CommentCount="4" />
  <row Id="2982" PostTypeId="1" CreationDate="2011-01-16T00:43:04.253" Score="1" ViewCount="163" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/1801/why-space-expansion-affects-matter&quot;&gt;Why space expansion affects matter?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Imagine two tiny spacecrafts that are moving with the Hubble flow and so are moving away from each other. Let's assume that they've been that way since the very early universe, never firing their engines, just drifting along their Hubble flow geodesics in a homogenous isotropic universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They then momentarily fire their engines so that they &quot;cancel&quot; the Hubble flow and have a fixed proper separation. Will they now start drifting apart again (presumably due to expansion)? Or will they stay at fixed proper separation, and maybe very slowly move towards each other due to their mutual gravitational attraction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess this is another way of asking whether expansion is kinematic and hence can be forgotten (so that &quot;Brooklyn isn't expanding&quot; because gravitational collapse and structure formation have erased memory of the expansion). Or maybe someone will help me refine this question and make me realize I just haven't thought things through completely?&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-01-16T01:08:17.080" ClosedDate="2011-01-16T01:42:47.427" Title="Why doesn't Brooklyn expand? (Or &quot;Is the expansion of the universe kinematic?&quot;)" Tags="&lt;gravity&gt;&lt;cosmology&gt;&lt;gravitational-collapse&gt;" AnswerCount="1" CommentCount="13" />
  <row Id="2983" PostTypeId="2" ParentId="2980" CreationDate="2011-01-16T00:45:02.053" Score="-1" ViewCount="" Body="&lt;p&gt;Actually, what you're describing - the transformation $(-c, -p, -t)$ - gives you exactly the same particle! (Not the antiparticle, not a &quot;real&quot; antiparticle - whatever that is...). In other words, a &quot;time-reversed&quot; positron is precisely an &lt;em&gt;electron&lt;/em&gt; (and vice-versa).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/CPT_symmetry&quot; rel=&quot;nofollow&quot;&gt;CPT symmetry&lt;/a&gt; &lt;em&gt;by definition&lt;/em&gt; implies that the transformation $(-c, -p, -t)$ is the identity transformation.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2011-01-16T00:50:11.267" LastActivityDate="2011-01-16T00:50:11.267" CommentCount="17" />
  <row Id="2984" PostTypeId="2" ParentId="2111" CreationDate="2011-01-16T00:57:18.337" Score="2" ViewCount="" Body="&lt;p&gt;I can not comment yet, so I will put it as an answer :-(   I like the energy approach, but why not to use Archimedes principle? First, substitute the air atmosphere by an extra ten meters of water around the straw, so that now the initial conditions are a vacuum straw inserted a lenght h in a fluid. The energy to produce such vacuum in the fluid you can see by Archimedes; and it is h/2 times g times the mass of the removed fluid. Let it move, and it can go up until filling a column of lenght h above the level, because the energy (now gravitational) of this column is, again, h/2 times its mass times g.  &lt;/p&gt;&#xA;" OwnerUserId="1335" LastEditorUserId="1335" LastEditorDisplayName="" LastEditDate="2011-01-16T03:09:53.210" LastActivityDate="2011-01-16T03:09:53.210" CommentCount="5" />
  <row Id="2985" PostTypeId="2" ParentId="2980" CreationDate="2011-01-16T01:00:03.110" Score="1" ViewCount="" Body="&lt;p&gt;T reversal simply reverses momenta. So as long as you can fire antiparticles in any direction you want, applying a T operation is trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying CP turns your left-handed particle into a right-handed antiparticle. Applying T to it then flips its momentum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the CPT theorem tells you then  is that the physics of any process involving particles and antiparticles of particular handedness is identical to the physics of the process in reverse with particles swapped for their antiparticles and all handedness reversed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the probability that an electron neutrino will oscillate into a tau neutrino after a certain distance (&lt;a href=&quot;http://en.wikipedia.org/wiki/Neutrino_oscillation&quot; rel=&quot;nofollow&quot;&gt;See Neutrino Oscillations&lt;/a&gt;) is equal to the probability that a tau antineutrino will oscillate into an electron antineutrino within the same distance. (Where the handedness of each has been reversed)&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-01-16T01:00:03.110" CommentCount="3" />
  <row Id="2986" PostTypeId="2" ParentId="2982" CreationDate="2011-01-16T01:08:17.080" Score="2" ViewCount="" Body="&lt;p&gt;This is a good question, but one that has a mundane explanation. Hubble expansion only holds over long distances at scales near the seperation of galaxies for systems that are not gravitionally bound.  So for us, our fate is likely one where we will eventually lose sight of other galaxies, but our galaxy will continue on alone for some time. There are some interesting questions about what happens when the temperature of the cmb drops below the temperatures of black holes in our galaxy (e.g. evaporation of black holes via hawking radiation).  Eventually, if black holes evaporate substantially, gravitational forces will relax and the kinetic energy of bodies may be sufficient to allow the galaxy to gradually expand.  At that point, some of the cold bodies of matter may have sufficient velocity to escape the galaxy's pull and become further flung.  Those may eventually be removed from view due to cosmological expansion.  In any case, the universe will eventually be composed of very far flung, cold objects.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-16T01:08:17.080" />
  <row Id="2987" PostTypeId="1" AcceptedAnswerId="2991" CreationDate="2011-01-16T01:17:56.987" Score="1" ViewCount="233" Body="&lt;p&gt;I thought light like singularity is where the geodesics end on a lightlike hypersurface and can't be extended anymore. I guess its different than light cone singularity. Lot's of places have mention of it, but I wanted to know its definition for sure.&lt;/p&gt;&#xA;" OwnerUserId="1349" LastActivityDate="2011-08-02T10:36:57.227" Title="What is light cone singularity? Has it any relation with light like singularity?" Tags="&lt;gravity&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="2988" PostTypeId="2" ParentId="2974" CreationDate="2011-01-16T01:19:55.747" Score="2" ViewCount="" Body="&lt;p&gt;If you could suddenly drop a big screen between the Earth and Sun, the initial rate of energy loss would equal the total solar energy input:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dE}{dt} = \alpha F A_e .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\alpha$ is the mean albedo, which Wolfram puts at $0.37$, the flux is $F = 1400\frac{\text{W}}{\text{m}^2}$, and the radius of the Earth is $R_e = 6.4 \times 10^6\text{ m}$, I get $7 \times 10^{16} W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the atmosphere masses roughly $5 \times 10^{18}\text{ kg}$ (Wolfram again), and the heat capacity of Nitrogen gas is $c_p = 30 \frac{\text{J}}{\text{mol K}} = 1070 \frac{\text{J}}{\text{kg K}}$. So the initial cooling will be around $1.3 \times 10^{-5}\text{ K/s}$, or about one Kelvin per day. That rate will drop by the fourth power of the temperature (use absolute units!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things left unconsidered:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I've neglected the oceans, and geothermal heat. Both effects will slow things down, the ocean considerable, geothermal, not so much. There is also a great deal of heat in the rock itself, but rock is a much better insulator than water, so it will have a smaller effect in the short term. Still, head into a deep mine or cave to stay warm, and watch out for the &lt;a href=&quot;http://mineshaftgap.net/&quot; rel=&quot;nofollow&quot;&gt;mineshaft gap&lt;/a&gt;!&lt;/li&gt;&#xA;&lt;li&gt;The heat of vaporization of water vapor and the heats of fusion of liquid water and carbon dioxide will slow things down while the phase transitions happen. There will be other phase transitions later on, but they won't matter to the frozen corpses...&lt;/li&gt;&#xA;&lt;li&gt;I've used a blackbody approximation which is surely not exact, but it should do for our purposes.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As Omega says in the comments all these factors are mitigating, and the real rate of temperature loss could be a factor of several (maybe even ten) lower than the above calculation. There are a lot of details that really matter, and things will change as the atmosphere changes (more clouds, then less, possible carbon dioxide clouds...) and the oceans freeze over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt;, the fact that the day/night temperature cycle is much larger than the daily change derived above suggests that the temperature near the Earth's surface is only loosely coupled to the mean atmospheric temperature (I guess that's not a big surprise as the surface is probably the warmest part of the atmosphere), so you can expect surface temperatures to drop much more rapidly than than the mean at first. How much faster? Well, like the day-to-night change initially, though I'd guess this will slow pretty quickly.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-01-16T02:09:10.530" LastActivityDate="2011-01-16T02:09:10.530" CommentCount="7" />
  <row Id="2991" PostTypeId="2" ParentId="2987" CreationDate="2011-01-16T02:05:49.447" Score="0" ViewCount="" Body="&lt;p&gt;You may have encountered it in a different context, but I recognize it from the topic of singularities of correlation functions in quantum field theory. In massless field theories such correlation fucntions becomes singular for points which are lightlike separated, and the structure of such singularities is determined by good physical principles such as locality and unitarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then again, I may be completely off the mark. In any event, except for the linguistic similarity, I don't think it has to do with null singularities in spacetime.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-01-16T02:17:18.200" LastActivityDate="2011-01-16T02:17:18.200" CommentCount="6" />
  <row Id="2992" PostTypeId="1" AcceptedAnswerId="3003" CreationDate="2011-01-16T02:11:39.130" Score="7" ViewCount="231" Body="&lt;p&gt;Sometimes in quantum cosmology, when we are thinking about 'wave functions of the universe' we have in mind some sort of formal path integral, where we include not just the variations in the dynamical fields (metric and so forth), but also possibly some sort of prescription for summing over all possible topologies.  At least, that seems to be some sort of heurestic guess often encountered in the literature (Hawking et al).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in String theory, admittedly a different context but there is a rather well defined notion of how this works (basically as a generalization of Feynman graphs over Riemann surfaces).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I don't understand exactly how this is supposed to work in quantum cosmology exactly (say in the Wheeler-De Witt context).  First of all there seems to be a massive amount of overcounting already at the metric level (where presumably one needs to mod out by all the diffeomorphisms), but how exactly does one deal with the topologies?  Do they only include connected topologies?  Is there even a mathematical formalism on how to even approach this problem?  Any good papers that deal with this in a comprehensive way?&lt;/p&gt;&#xA;" OwnerUserId="1350" LastActivityDate="2011-01-16T08:26:41.360" Title="Summing over topologies" Tags="&lt;cosmology&gt;&lt;quantum-gravity&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" />
  <row Id="2993" PostTypeId="2" ParentId="139" CreationDate="2011-01-16T02:18:49.807" Score="1" ViewCount="" Body="&lt;p&gt;The energy of a particle is gamma mc^2.  Neutrinos, say electron neutrinos, are produced in about 1-10 MeV units of energy.  We are not as certain about the absolue masses of neutrinos, but have measured estimates of the relative masses of neutrinos.  From the CKM and PMNS matrix and data input on neutrino oscillations the e-neutrino is estimated at about 1ev or so.  So that gamma is very large, gamma ~ 10^7.  Then compute from gamma^2 = 1/(1 - v^2) which gives a v (or v/c) =~ .9999998.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is the funny matter of the P_- = (1 - gamma^5)/2, which projects the helicity or spin in the opposite direction of the momentum --- left handedness and parity violation.  A massive neutrino means that in principle one could boost to a frame where the neutrino is right handed.  The small mass and large gamma is a sort of obstruction to that, though it is not absolute.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-16T02:18:49.807" />
  <row Id="2994" PostTypeId="2" ParentId="2978" CreationDate="2011-01-16T03:05:23.997" Score="7" ViewCount="" Body="&lt;p&gt;It pretty much is that simple. However, if for some reason you'd like an explicitly relativistic formulation, take a look at the Lorentz force law:&#xA;$$\frac{{\mathrm d}p^\mu}{{\mathrm d}\tau} = ev_\nu F^{\mu\nu}$$&#xA;For the derivative with respect to coordinate time that you want, we need to multiply through by $dτ/dt = 1/\gamma$. But for a constant electric field in Cartesian coordinates, the only nonzero components of $F^{\mu\nu}$ are $F^{0a} = -F^{a0}$ for a = 1,2,3, which are the electric field components. Thus, only the $v_0 = \gamma$ term can contribute, canceling factor brought in by time dilation factor. QED.&lt;/p&gt;&#xA;" OwnerUserId="1353" LastEditorUserId="1353" LastEditorDisplayName="" LastEditDate="2011-01-16T03:47:42.743" LastActivityDate="2011-01-16T03:47:42.743" CommentCount="3" />
  <row Id="2995" PostTypeId="1" CreationDate="2011-01-16T03:53:11.883" Score="0" ViewCount="73" Body="&lt;p&gt;Subscriber's phone line Resistance= 250 Oomi/km. &#xA;Voltage at phone station = 50V &#xA;Min. allowed  I = 20 mA at subscriber's phone.&#xA;Find distance of subscriber's phone line.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-16T03:53:11.883" ClosedDate="2011-01-16T03:56:20.433" Title="How to find distance of subscriber's phone line?" Tags="&lt;frequency&gt;" CommentCount="1" />
  <row Id="2996" PostTypeId="2" ParentId="634" CreationDate="2011-01-16T04:13:34.137" Score="2" ViewCount="" Body="&lt;p&gt;If you are interested in the origins, I think you must start the history with Sommerfeld. It is Sommerfeld who gets the fine structure levels, via an amazing cancellation of two mistakes (Bohr-Sommerfeld quantisation and lack of spin). So Dirac is sailing with a clear goal: his results must agree with Sommerfeld in some way. And of course Sommerfeld orbits are already the &lt;em&gt;relativistic&lt;/em&gt; correction to Bohr orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are more worried about the evolution towards field theory, I'd add Sakurai books to the Bjorken-Drell lot. Also Mandl.&lt;/p&gt;&#xA;" OwnerUserId="1335" LastEditorUserId="1335" LastEditorDisplayName="" LastEditDate="2011-01-16T09:42:50.557" LastActivityDate="2011-01-16T09:42:50.557" />
  <row Id="2997" PostTypeId="2" ParentId="1530" CreationDate="2011-01-16T04:45:40.523" Score="1" ViewCount="" Body="&lt;p&gt;Last time I did the sum, 201Hg to 197Au plus 4He did not need external energy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;200.970277 - (196.9665516 + 4.0026032) = 4.0037254 - 4.0026032 = 0.0011222. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Caveats: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I did not check the error term of atomic weigths, and&lt;/li&gt;&#xA;&lt;li&gt;The first decay, alpha to 197Pt, should be awesomely slow. In fact&lt;/li&gt;&#xA;&lt;li&gt;201Hg is considered stable in all the listings.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But the point is that the external energy would be recovered in this case.&lt;/p&gt;&#xA;" OwnerUserId="1335" LastActivityDate="2011-01-16T04:45:40.523" CommentCount="1" />
  <row Id="2998" PostTypeId="2" ParentId="1383" CreationDate="2011-01-16T05:30:58.923" Score="-1" ViewCount="" Body="&lt;p&gt;The speed of light is &lt;em&gt;not&lt;/em&gt; a fundamental constant. It is a conversion factor that creeps in due to the use of a clumsy system of units that uses different units for distance depending on direction in spacetime.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-01-16T05:30:58.923" />
  <row Id="3000" PostTypeId="1" CreationDate="2011-01-16T06:23:56.560" Score="1" ViewCount="82" Body="&lt;p&gt;I couldn't think of a better title because I'm not sure what's going on with this phenomenon--someone who has an answer, please edit to fit this better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I noticed that when I cook rice, at some point after most of the water has evaporated and the rice is settled at the bottom of the pot, there is a &quot;lattice&quot; of regularly-spaced gaps in how the grains at the top are arranged, with the holes extending down a grain or two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can try to explain this better if it doesn't make sense, but for now I can't do much better because it mystifies me. What's up here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Better tag suggestions are also welcome...)&lt;/p&gt;&#xA;" OwnerUserId="8" LastActivityDate="2011-01-16T09:31:26.053" Title="Settling of rice as it's cooked" Tags="&lt;everyday-life&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="3003" PostTypeId="2" ParentId="2992" CreationDate="2011-01-16T08:06:56.123" Score="5" ViewCount="" Body="&lt;p&gt;it's an important question but the available answers don't fully match this importance. However, there are several facts that &lt;em&gt;are&lt;/em&gt; known:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;First of all, you mentioned the redundancies coming from diffeomorphisms. Of course, any path integral calculation has to be divided by the diffeomorphism group, i.e. to guarantee that each class of geometries (or, more generally, configurations) that are diff-equivalent (or, more generally, gauge equivalent: diff symmetry is a kind of gauge symmetry) are only counted once. For gauge symmetries, the BRST quantization with the Faddeev-Popov ghosts (originally introduced by Feynman) is a poweful tool to do so. In supergravity, its extension with antighosts etc. has to be used and is known as the BV formalism which is mathematically heavier and &quot;more nonlinear&quot; than the BRST formalism, so only dozens of people in the world can really use the BV formalism.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the world sheet description of perturbative string theory, the summation over topologies is well-defined and produces Feynman diagrams of increasing order in the string coupling constant, as the number of &quot;handles&quot; increases or the Euler character of the manifold decreases. The reason why this expansion in terms of the metric works is that the metric in the world sheet theory is a fundamental, exact degree of freedom - so we're dealing with the correct theory. The metric is valid up to arbitrarily short distances on the world sheet because the world sheet theory is scale-invariant (at least at short distances) so if the metric is OK at one scale, it has to be right at any scale.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In quantum gravity in 3+1 dimensions or higher (and I will skip the special 2+1-dimensional case which has no physical polarizations of the graviton, which makes it look like things work &quot;automatically&quot; but they actually don't, even in this case, if one is accurate enough), the metric tensor is not a fundamental object that is perfectly well-defined at any length scale. Instead, near the Planck scale (and sometimes lower scale such as the string scale), it is supplemented by infinitely many massive fields that the metric can mix with and create non-trivial topological configurations that &quot;knit&quot; both the metric and other degrees of freedom etc. So the probable main reason why no known functioning description of quantum gravity in terms of a path integral exists for the metric in 3+1 dimensions or higher is that it is a wrong problem.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Quantum gravity is an extremely constrained problem and the &quot;quantization of the metric&quot; without any extra degrees of freedom and any extra powerful arrangements simply cannot produce a consistent theory, to all orders. At least, about 50 years of the efforts to &quot;map the landscape of possibilities&quot; how to define a theory of quantum gravity suggest that treating the metric tensor as the right and only degree of freedom in quantum gravity is a wrong way to proceed. So the quantization of the metric is only consistent as an approximation that only tells us how gravity works up a certain order in the expansion in the &quot;strength of the gravitational field&quot; - with the order measured by powers of $\hbar$. However, any topology change is inevitably a non-perturbative effect which is much more inaccessible than a 2-loop or 50-loop contribution. The quantization of general relativity can't even produce the right, unambiguous 2-loop contributions to the effective action and dynamics (because of well-known non-renormalizable divergences). So of course, it is also unable to produce different-topology-contributions to the effective action because they're equivalent to infinitely-many-loops contributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In supergravities, the non-renormalizable loop divergences may cancel at 2 loops and may appear at $k$ loops where $k$ is (much) greater than 2. In $N=8$ supergravity in $d=4$, all the perturbative divergences probably cancel. But the non-perturbative physics is still inconsistent - and the spontaneous topology change is always a non-perturbative effect. There are many ways to see that the supergravity Lagrangian cannot know about the nonperturbative effects. For example, its exact continuous exceptional noncompact symmetry is incorrect; it must be broken to its discrete group by quantum effects such as the Dirac quantization rule that guarantees that various electric and magnetic charges have to be quantized and belong to inverse lattices (with respect to one another). So even $N=8$ supergravity without anything else is unable to calculate the impact of topology-changing histories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People have speculated about many phenomena that quantum gravity, if it could be expressed as a path integral, could produce because of the summation over topologies. In particular, one could generically sum over spacetimes that always have some tiny handles or wormholes: this picture of the typical contribution to the path integral, envisioned by Wheeler, is referred to as the quantum form. Moreover, splitting and reconnecting &quot;baby universes&quot; were also described in the literature and they were hypothesized to be able to re-adjust values of coupling constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I won't mention any paper (with one exception below) because no paper of this kind that leads to any convincing calculation of anything exists. It has been pretty much established that a consistent theory of quantum gravity cannot be defined in the way you (and others) suggested. And many pictures that emerged from the only consistent theory of quantum gravity we have, string/M-theory, indicate that the picture of summing over many topologies is inadequate. In some vague sense, the new phenomena that have to exist in a functioning theory of quantum gravity make the short-distance spacetime more peaceful and de facto remove the contributions of nontrivial topologies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, it can't be a 100% accurate description and I will mention one paper by Iqbal, Nekrasov, Okounkov, and Vafa, anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0312022&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0312022&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In topological string theory, which may be thought of as a truncation of the spectrum of the full string theory where only special &quot;holomorphic&quot; states are allowed (which removes most of the dynamical states from the bulk), it is true that the whole path integral - or partition function - may be written down as a summation over all topologies of Calabi-Yau manifolds. So at least in this limited context, well-defined observables - partition sums of a system that may also be defined as a melting crystal - may be calculated by summing over infinitely many topologies of 6-dimensional geometries (which is higher than 2-dimensional ones).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, no corresponding calculation that would actually have an answer that has any relationship to the observations or other methods to justify that the answer is correct is known in any &quot;physical&quot; theories of quantum gravity with dynamical gravitons - in theories that actually contain the gravitational attraction (where the objects have a variable momentum). So at this moment, it surely looks like that the metric tensor is just an &quot;effective&quot; degree of freedom and this degree of freedom can be only used for problems at distances much longer than the Planck length. In particular, only calculations up to a fixed number of quantum loops - usually 0 or 1 (the classical limit and the semiclassical calculations) - can be done using this metric tensor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-trivial topologies that don't involve any curvature with curvature radius comparable to or smaller than the Planck scale become irrelevant in this picture because they're suppressed by something like $\exp(-A/A_{Planck})$ where $A$ are the typical $(D-2)$-dimensional areas of the &quot;handles&quot; that modify the topology. Needless to say, because $A_{Planck}$  is a tiny area, the exponential is zero for all practical (and most of the impractical) purposes. The exponentially suppressed term is much smaller than many other terms that are already incalculable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the answers above may get modified in the future and the absence of such a path-integral description with new topologies may sometimes be blamed on our current ignorance. It is plausible that a spacetime-based path-integral approach with different topologies exists; it remains puzzling how such a description can ever be consistent with the holographic principle (which says that the entropy in a finite region is finite and bounded by the surface). However, it seems almost guaranteed that such a hypothetical description would also include many other - perhaps dynamically generated - degrees of freedom besides the metric. The non-existence of a &quot;pure gravity&quot; in more than 3+1 dimensions seems to be a claim with an extensive amount of circumstantial evidence - a claim independent of any particular description - and is unlikely to be due to our lack of imagination. It seems to be a by-product of the research of quantum gravity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When research gets really profound, people sometimes find out that some of their basic assumptions - or &quot;dogmas&quot; - were really invalid. The possibility to define quantum gravity in terms of the &quot;metric tensor only&quot; seems to be an example of such a pre-scientific superstition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-16T08:26:41.360" LastActivityDate="2011-01-16T08:26:41.360" CommentCount="2" />
  <row Id="3004" PostTypeId="1" AcceptedAnswerId="3024" CreationDate="2011-01-16T08:17:21.107" Score="0" ViewCount="150" Body="&lt;p&gt;May be not neutrinos, but antineutrinos? Or both types?&#xA;In the last case, why they didn't annihilate and what is the ratio of relic neutrinos to relic antineutrinos? Is that ratio somehow related to the barion asymmetry? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference: &lt;a href=&quot;http://en.wikipedia.org/wiki/Relic_neutrinos&quot; rel=&quot;nofollow&quot;&gt;Relic neutrinos or cosmic neutrino background&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Like the cosmic microwave background radiation (CMB), the cosmic neutrino background is a relic of the big bang&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="238" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-01-16T10:11:34.627" LastActivityDate="2011-08-17T15:45:20.317" Title="How to calculate the density of relic neutrinos?" Tags="&lt;cosmology&gt;&lt;big-bang&gt;&lt;neutrino&gt;&lt;barion-asymmetry&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="3005" PostTypeId="1" AcceptedAnswerId="3006" CreationDate="2011-01-16T08:18:19.600" Score="1" ViewCount="626" Body="&lt;p&gt;I've started reading Peskin and Schroeder on my own time, and I'm a bit confused about how to obtain Maxwell's equations from the (source-free) lagrangian density $L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ (where $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ is the field tensor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting in for the definition of the field tensor yields $L = -\frac{1}{2}[(\partial_\mu A_\nu)(\partial^\mu A^\nu) - (\partial_\mu A_\nu)(\partial^\nu A^\mu)]$.  I know I should be using $A^\mu$ as the dynamical variable in the Euler-Lagrange equations, which become $\frac{\partial L}{\partial A_\mu} - \partial_\mu\frac{\partial L}{\partial(\partial_\mu A_\nu)} = - \partial_\mu\frac{\partial L}{\partial(\partial_\mu A_\nu)}$, but I'm confused about how to proceed from here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I should end up with $\partial_\mu F^{\mu\nu} = 0$, but I don't quite see why.  Since $\mu$ and $\nu$ are dummy indices, I should be able to change them: how do the indices in the lagrangian relate to the indices in the derivatives in the Euler-Lagrange equations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-16T09:09:37.690" Title="Derivation of Maxwell's equations from field tensor lagrangian" Tags="&lt;electromagnetism&gt;&lt;lagrangian-mechanics&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="3006" PostTypeId="2" ParentId="3005" CreationDate="2011-01-16T08:46:19.023" Score="1" ViewCount="" Body="&lt;p&gt;Well, you are almost there. Use the fact that&#xA;$$ {\partial (\partial_{\mu} A_{\nu}) \over \partial(\partial_{\rho} A_{\sigma})} = \delta_{\mu}^{\rho} \delta_{\nu}^{\sigma}$$&#xA;which is valid because $\partial_{\mu} A_{\nu}$ are $d^2$ independent components.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-01-16T08:46:19.023" />
  <row Id="3007" PostTypeId="2" ParentId="2980" CreationDate="2011-01-16T08:48:38.537" Score="2" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;first of all, &quot;even&quot; and &quot;odd&quot; are different words than &quot;positive&quot; and &quot;negative&quot; - although both pairs of adjectives form a $Z_2$ group. So a positron is not &quot;odd under charge conjugation&quot;. Instead, the charge conjugation of a positron is an electron. Because an electron is not &quot;minus positron&quot; (in the sense of vectors in the Hilbert space - they are two totally different vectors), it is not true that the positron is odd under C. A combination of the electron state and a positron state could be odd under C. However, if I remember well, the operator $C^2=-1$ for Dirac fermions, so the eigenvalues of $C$ are equal to $+i$ and $-i$. None of them is $-1$ so no fermionic state can really be &quot;odd&quot; under C.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Second, C is the only one among the three transformations C,P,T that changes the &quot;character&quot; of the particle: it turns electrons into positrons and vice versa. On the other hand, the transformations P,T, whenever they exist in the first place, are meant to to change the positions of particles in the space and time only. P is the parity, moving objects from $(x,y,z,t)$ to $(-x,-y,-z,t)$, while T is the time conjugation, changing the sign of $t$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The action of $T$ on a specific state of an instant of time is simple; effectively, one only changes the sign of all the velocities. The result is that with these velocities, if T is a symmetry, the particles will be moving in the opposite direction than previously. So the evolution into the future will be equivalent to the previous evolution in the past. Such a transformation T doesn't &quot;really&quot; commute with the Hamiltonian: it turns it into minus itself because the Hamiltonian is really the generator of translations in time - and time changes the sign under T. So T is not really a symmetry of objects in space: even when it is unbroken, it is really a symmetry of histories in the whole spacetime. You shouldn't think that T acts on objects as a symmetry: it acts on their whole histories in spacetime if you want to see that it is a symmetry. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;T is not a linear operator but an antilinear operator which is related; $T\lambda|\psi\rangle$ is equal to $\lambda^* T|\psi\rangle$, including the complex conjugation of the $\lambda$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Contrary to some claims in this thread, CPT - which is also an antilinear operator because it includes T - is not the identity. It is a nontrivial operation and a nontrivial operation is the same thing as &quot;not identity&quot;. CPT is a symmetry but a symmetry is something completely different than &quot;identity&quot;. A symmetry is an element of a symmetry group; the identity is just an element of a symmetry group that doesn't do anything and that differs from all the genuine symmetries. It's kind of interesting why so many people still can't get this simple point.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So if one uses the word &quot;odd&quot; in the proper physics sense, there doesn't exist any particle that is odd under P, and probably C, because these two operators square to minus one. Even if they or some of their products squared to plus one, the eigenstates of the operation would be a strange combination of the positron state and the electron state. We don't usually like to combine states with different electric charges because they belong to different superselection sectors: they can't possibly evolve from the other one so without a loss of generality, one may assume that the initial state is an element of one superselection sector, and it is guaranteed that no mixture of another superselection sector will ever be added by the evolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What your question may have actually wanted to ask was whether the actual states obtained by C,P,T,CP, and CPT transformations exist. For electrons and positrons, they always do. However, for neutrinos, the C-transformed and P-transformed states don't exist at all. That's because neutrinos are left-handed, antineutrinos are right-handed, but there are no right-handed neutrinos and left-handed antineutrinos which is what you would get from the first two by C (or P). The transformation of a state by T always exists - because the transformation by CP exists as well and CPT is a symmetry (whose transforms therefore exist as well). T only changes the velocities of particles - not the positions or charges (or other internal characteristics) of the particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if this is what you asked: The CPT transformation of a positron at $(x,y,z)$ moving in the direction $(v_x,v_y,v_z)$ is an electron at $(-x,-y,-z)$ moving in the direction $(v_x,v_y,v_z)$. Note that the signs of the velocities didn't change. Under CP, they would change the sign, but T changed the sign again. So if the positron is moving towards the origin, the CPT-transformed particle, the electron, will be moving away from the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-16T08:57:22.760" LastActivityDate="2011-01-16T08:57:22.760" CommentCount="1" />
  <row Id="3008" PostTypeId="2" ParentId="3005" CreationDate="2011-01-16T09:09:37.690" Score="1" ViewCount="" Body="&lt;p&gt;Dear amc,&#xA;first, write your Lagrangian density as&#xA;$$ L = -\frac{1}{4} F_{\mu\nu}F^{\mu\nu} = -\frac{1}{2} (\partial_\mu A_\nu) F^{\mu\nu} $$&#xA;Is that fine so far? The $F_{\mu\nu}$ contains two terms that make it antisymmetric in the two indices. However, it's multiplied by another $F^{\mu\nu}$ that is already antisymmetric, so I don't need to antisymmetrize it again. Instead, both terms give me the same thing, so the coefficient $-1/4$ simply changes to $-1/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the field equations force you to compute the derivatives of the Lagrangian with respect to $A_\mu$ and its derivatives. First of all, the derivative of the Lagrangian $L$ with respect to $A_\mu$ components themselves vanishes because the Lagrangian only depends on the partiial derivatives of $A_\mu$. Is that clear so far?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the equations of motion will be&#xA;$$0 = -\partial_\mu [\partial L / \partial(\partial_\mu A_\nu)] = \dots $$&#xA;Whoops, you already got to this point. But now, look at my form of the Lagrangian above. The derivative of the Lagrangian with respect to $\partial_\mu A_\nu$ is simply &#xA;$$-\frac{1}{2} F^{\mu\nu}$$&#xA;because $\partial_\mu A_\nu$ simply appears as a factor so the equations of motion will simply be&#xA;$$ 0 = +\frac{1}{2} \partial_\mu F^{\mu\nu} $$&#xA;However, I have deliberately made one mistake. I have only differentiated the Lagrangian with respect to $\partial_\mu A_\nu$ included in the first factor of $F_{\mu\nu}$, with the lower indices. However, $\partial_\mu A_\nu$ components also appear in $F^{\mu\nu}$, the second factor in the Lagrangian, one with the upper indices. If you add the corresponding terms from the Leibniz rule, the result is simply that the whole contribution will double. So the right equation of motion, including the natural coefficient, will be&#xA;$$ 0 = \partial_\mu F^{\mu\nu} $$&#xA;The overall normalization is important because this equation may get extra terms, like the current, whose coefficient is obvious, and you don't want to get a relative error of two between the derivative of $F$ and the current $j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-16T09:09:37.690" />
  <row Id="3009" PostTypeId="1" AcceptedAnswerId="3013" CreationDate="2011-01-16T09:27:21.130" Score="5" ViewCount="583" Body="&lt;p&gt;I understand that people explain (in layman's terms at least) that the presence of mass &quot;warps&quot; space-time geometry, and this causes gravity. I have also of course heard the analogy of a blanket or trampoline bending under an object, which causes other objects to come together, but I always thought this was a hopelessly circular explanation because the blanket only bends because of &quot;real&quot; gravity pulling the object down and then pulling the other objects down the sloped blanket.&#xA;In other words, to me, it seems that curved space wouldn't have any actual effect on objects unless there's already another force present. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how is curved space-time itself actually capable of exerting a force (without some source of a fourth-dimensional force)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I apologize for my ignorance in advance, and a purely mathematical explanation will probably go over my head, but if it's required I'll do my best to understand.&lt;/p&gt;&#xA;" OwnerUserId="1358" LastActivityDate="2011-02-06T17:14:55.893" Title="How exactly does curved space-time describe the force of gravity?" Tags="&lt;gravity&gt;&lt;forces&gt;&lt;spacetime&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" />
  <row Id="3010" PostTypeId="2" ParentId="2974" CreationDate="2011-01-16T09:28:01.000" Score="2" ViewCount="" Body="&lt;p&gt;the Earth is emitting thermal radiation corresponding to its temperature, and losing the energy (and decreasing temperature) in this way. Normally, this is cancelled - every day or every year - by the energy coming from the Sun. The second term would be absent in your thought experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So those 342 Watts per meter squared in average coming from the Sun (more precisely, only 250 or so because a part is reflected without any transformation of the energy form anyway, so it's like if it is not coming at all) would disappear from the budget. One needs to estimate the heat capacity of the atmosphere - and the relevant layers of the ocean that are capable to heat the atmosphere above it for a while. But you don't need to be that sophisticated. It's enough to look at the reality of weather - numbers that everyone knows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the continental climate, going from day to night lowers the temperature of the atmosphere by 1 Celsius degree in average (cloudiness makes the day-time difference smaller; sunny skies make it larger). Because what you say is just a permanent night, this rate of cooling would simply continue and continue. So the initial rate would be cooling by 1 degree per day or so; it would be slower above the ocean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the temperature of the Earth would be going down, the radiation would be dropping as the fourth power of the absolute temperature (in Kelvin degrees). The atmosphere would be getting cooler, partly condensed (if not frozen), and much thinner. So once the absolute temperature decreased by 20 percent (note that $1.2^4$ equals 2 or so) or about 50 Celsius degrees, the rate of cooling would drop to 1/2 of the original one. At any rate, you need a few months to get to minus 50 degrees or something like that. The ocean would still be oscillating for a long time and it would try to unmelt the thinner atmosphere right above it - just like the circulation of water beneath the ice in a pond. Note that deep ocean circulation takes 1,500 years or so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So of course, the ocean wouldn't freeze over for a very long time. But that's already a different question. Needless to say, heat almost doesn't get beneath the solid ground - kilometers below the rocks, they wouldn't notice that the Sun is gone for thousands of years or much longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-16T09:48:38.793" LastActivityDate="2011-01-16T09:48:38.793" />
  <row Id="3011" PostTypeId="2" ParentId="3000" CreationDate="2011-01-16T09:31:26.053" Score="2" ViewCount="" Body="&lt;p&gt;as rice cooks it releases starch. When the water level falls below the rice, the steam that is expelled creates the gaps that you noticed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These gaps are not, in fact &lt;em&gt;regularly&lt;/em&gt; distributed. They are in random positions and with random orientation. They do have a regular size corresponding to a grain of rice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is easy to grasp - the steam being released has a dropping pressure (the less water, the less steam being released). So what happens is that the gaps get smaller and smaller and finally they will be as small as possible, while being held in shape by the starch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also try to cover the whole surface, as steam is generated homogeneously and therefore has no preferred areas of the surface to generate these holes.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-01-16T09:31:26.053" CommentCount="2" />
  <row Id="3012" PostTypeId="2" ParentId="3009" CreationDate="2011-01-16T09:38:38.527" Score="3" ViewCount="" Body="&lt;p&gt;the trampoline analogy needs an extra source of gravity - because this is what the laymen, the recipients of the explanation, intuitively understand - but the real general relativity doesn't need any extra &quot;external&quot; gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, general relativity says that the space is getting curved by Einstein's equations,&#xA;$$G=T$$&#xA;where the left-hand side are numbers describing the curvature at a given point and the right hand side is the density of matter and momentum. I omit indices and constants haha. So general relativity says how the spacetime is curved under the influence of matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part of the story is that general relativity also says how matter moves in external geometry. It moves along &quot;geodesics&quot;, lines that are as straight as you can get.&#xA;$$\delta S_{action\,ie\,proper\,length} = 0$$&#xA;This actually means that the objects move along the predicted, seemingly curved trajectories. These trajectories are actually as straight in the curved spacetime as you can get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that there is a hemisphere  replacing a disk in the trampoline. So there exists a (nearly) straight line on the hemisphere - namely the equator near the junction with the rest of the trampoline. Note that the equator on the Earth is a maximum circle - so it is one of the straightest lines you can draw on the surface of Earth. The same is true for all actual trajectories that objects choose in spacetime of general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the hemisphere-above-trampoline example, particles can orbit around the equator of the attached hemisphere, just like planets, because it is the straightest and most natural line they can choose. I don't use any external gravity to explain the real gravity; instead, I use the principle that particles choose the most natural - the straightest - line they can find in the curved spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-16T09:46:26.300" LastActivityDate="2011-01-16T09:46:26.300" />
  <row Id="3013" PostTypeId="2" ParentId="3009" CreationDate="2011-01-16T10:10:32.340" Score="3" ViewCount="" Body="&lt;p&gt;Luboš's answer is of course perfectly correct. I'll try to give you some examples why the straightest line is physically motivated (besides being mathematically exceptional as an extremal curve).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image a 2-sphere (a surface of a ball). If an ant lives there and he just walks straight, it should be obvious that he'll come back where he came from with his trajectory being a circle. Imagine a second ant and suppose he'll start to walk from the same point as the first ant and at the same speed but into a different direction. He'll also produce circle and the two circles will cross at two points (you can imagine those circles as meridians and the crossing points as a north resp. south poles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, from the ants' perspective who aren't aware that they are living in a curved space, this will seem that there is a force between them because their distance will be changing in time non-linearly (think about those meridians again). This is one of the effects of the curved space-time on movement on the particles (these are actually tidal forces). You might imagine that if the surface wasn't a sphere but instead was curved differently, the straight lines would also look different. E.g. for a trampoline you'll get ellipses (well, almost, they do not close completely, leading e.g. to the precession of the perihelion of the Mercury).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So much for the explanation of how curved space-time (discussion above was just about space; if you introduce special relativity into the picture, you'll get also new effects of mixing of space and time as usual). But how does the space-time know it should be curved in the first place? Well, it's because it obeys Einstein's equations (why does it obey these equations is a separate question though). These equations describe precisely how matter affects space-time. They are of course compatible with Newtonian gravity in low-velocity, small-mass regime, so e.g. for a Sun you'll obtain that trampoline curvature and the planets (which will also produce little dents, catching moons, for example; but forget about those for a moment because they are not that important for the movement of the planet around the Sun) will follow straight lines, moving in ellipses (again, almost ellipses).&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-01-16T19:35:10.987" LastActivityDate="2011-01-16T19:35:10.987" CommentCount="9" />
  <row Id="3014" PostTypeId="1" AcceptedAnswerId="3017" CreationDate="2011-01-16T10:15:40.367" Score="14" ViewCount="1192" Body="&lt;p&gt;The simplistic undergrad explanation aside. I've never really understood what energy really is. I've been told that it's &lt;em&gt;something&lt;/em&gt; when converted from one kind of something to another kind does some &quot;work&quot;, as defined by us, but what is that something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, if the total amount of energy in the universe is finite and we cannot create energy. Then, where did it come from? I've learnt from thermodynamics where it &lt;em&gt;goes&lt;/em&gt;, but where does it come from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this sounds as something trivially simple, but there is so much going on over here and I just can't grasp what it is. Of course, I lack the mathematical understanding to grasp the subtle things the universe is doing, but what is it doing and how do I get to point of understanding what it's doing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;Note:&lt;/strong&gt; What prompted me to ask this was &lt;a href=&quot;http://physics.stackexchange.com/questions/296/is-energy-really-conserved/655#655&quot;&gt;this answer&lt;/a&gt;. I'm afraid that it just puzzled me further and I sat there staring at the screen for a good 10 minutes.)&lt;/p&gt;&#xA;" OwnerUserId="1065" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-27T11:17:06.410" LastActivityDate="2011-06-27T17:50:50.960" Title="What Is Energy? Where did it come from?" Tags="&lt;energy&gt;" AnswerCount="5" FavoriteCount="10" />
  <row Id="3015" PostTypeId="2" ParentId="3014" CreationDate="2011-01-16T10:42:06.380" Score="5" ViewCount="" Body="&lt;p&gt;I don't think the answer is trivially simple. I will try to give an explanation. In many problems of physics, what you are given is the initial and final states of the system. You don't know (or maybe no one does) what happens between these two states. Now there are quantities that you can measure before and after the system has undergone this change of state. The question is can you predict some of these quantities by knowing the others. Remember that we don't know the mechanism by which the system moves from these two states. But if you have something known as a conservation law, the problem becomes simple. (By saying that a quantity is conserved we mean that it doesn't change throughout some process). Suppose you have some magic function involving the quantities, which gives the same value no matter what the state of the system is, then you are done. The value of the function we call energy. And since its value doesn't change between these two states we say that its conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This excerpt is from Feynman Lectures:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a fact, or if you wish, a law, governing all natural phenomena that&#xA;  are known to date. There is no known exception to this law—it is exact so far as&#xA;  we know. The law is called the conservation of energy. It states that there is a&#xA;  certain quantity, which we call energy, that does not change in the manifold&#xA;  changes which nature undergoes. &lt;em&gt;That is a most abstract idea, because it is a&#xA;  mathematical principle; it says that there is a numerical quantity which does not&#xA;  change when something happens&lt;/em&gt;. It is not a description of a mechanism, or any-&#xA;  thing concrete; it is just a strange fact that we can calculate some number and when&#xA;  we finish watching nature go through her tricks and calculate the number again,&#xA;  it is the same. (Something like the bishop on a red square, and after a number of&#xA;  moves—details unknown—it is still on some red square. It is a law of this nature.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="966" LastActivityDate="2011-01-16T10:42:06.380" CommentCount="4" />
  <row Id="3016" PostTypeId="2" ParentId="3014" CreationDate="2011-01-16T10:43:08.407" Score="3" ViewCount="" Body="&lt;p&gt;To understand what energy is, it is necessary to understand the concept of work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work is defined as the action of a force over a path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ W=\vec{F}\cdot\vec{d}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this means? It describes how &quot;exerting&quot; or &quot;draining&quot; a particular action is. For example, imagine lifting a shopping bag of mass $10kg$ vertically by $1m$. This takes work, and exactly the following amount, given by the weight of the bag times the distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W= \vec{F}\cdot\vec{d} = Fd\cos{0}=mgd=10kg\times9.8ms^{-2}\times1m=98J$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy is classically defined as the capacity of a physical system to do work, or in other words: as you perform work, you exchange energy for some physical effect by doing work. Or in other terms again, by exerting a force over a distance you convert energy into work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our example, you need to use some form of energy to lift the shopping bag. The quantity you need is exactly the amount of work we calculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens to this work? It's converted to energy again - to gravitational potential energy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U_{final} = U_{initial} + W$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta U = U_{final} = U_{initial} = W = mgd$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the classical definition of gravitational potential energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in practice - we never see or measure energy directly. When energy changes form, it is called work, which we can measure. So work, in a way, is a &quot;transport&quot; concept for energy. Energy, on the other hand is like a &quot;reservoir&quot; of work in potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is energy a useful quantity? After all, work seems to be a more &quot;fundamental&quot; quantity from an experimental point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to this lies in the conservation law of energy. Work in itself describes a change in energy, so it's not a conserved quantity in itself unless you embed it in the more general concept of energy which is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, we can derive large swaths of classical mechanics using conservation of energy as a prime principle, together with the principle of least action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Caveats&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more advanced theories, conservation of energy is a much more complicated matter and does not apply as simply as in the classical sense. For example in SR, energy can be converted to apparent mass and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also very interesting mathematical properties of potential energy and its relation to forces and especially fields of forces. These explanations, though are way more abstract and mathematical - I assume you want an intuitive, instinctual explanation of what energy is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking for the former please see &lt;a href=&quot;http://physics.stackexchange.com/q/2144/66&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-16T11:08:21.403" LastActivityDate="2011-01-16T11:08:21.403" CommentCount="4" />
  <row Id="3017" PostTypeId="2" ParentId="3014" CreationDate="2011-01-16T11:20:27.207" Score="21" ViewCount="" Body="&lt;p&gt;Energy is any quantity - a number with the appropriate units (in the SI system, Joules) - that is conserved as the result of the fact that the laws of physics don't depend on the time when phenomena occur, i.e. as a consequence of the time-translational symmetry. This definition, linked to Emmy Noether's fundamental theorem, is the most universal among the accurate definitions of the concept of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the &quot;something&quot;? One can say that is a &lt;em&gt;number&lt;/em&gt; with units, a dimensionful quantity. I can't tell you that energy is a &lt;em&gt;potato&lt;/em&gt; or another material object because it is not (although, when stored in the gasoline or any &quot;fixed&quot; material, the amount of energy is proportional to the amount of the material). However, when I define something as a &lt;em&gt;number&lt;/em&gt;, it is actually a much more accurate and rigorous definition than any definition that would include potatoes. Numbers are much more well-defined and rigorous than potatoes which is why all of physics is based on mathematics and not on cooking of potatoes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Centuries ago, before people appreciated the fundamental role of maths in physics, they believed e.g. that the heat - a form of energy - was a material called the &lt;em&gt;phlogiston&lt;/em&gt;. But it is a long, long time ago when experiments were done to prove that such a picture was invalid. Einstein's $E=mc^2$ partly revived the idea - energy is equivalent to mass - but even the mass in this formula has to be viewed as a number rather than something that is made out of pieces that can be &quot;touched&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy has many forms - terms contributing to the total energy - that are more &quot;concrete&quot; than the concept of energy itself. But the very strength of the concept of energy is that it is universal and not concrete: one may convert energy from one form to another. This multiplicity of forms doesn't make the concept of energy ill-defined in any sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of energy's relationship with time above, the abstract definition of energy - the Hamiltonian - is a concept that knows all about the evolution of the physical system in time (any physical system). This fact is particularly obvious in the case of quantum mechanics where the Hamiltonian enters the Schrödinger or Heisenberg equations of motion, being put equal to a time-derivative of the state (or operators).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total energy is conserved but it is useful because despite the conservation of the total number, the energy can have many forms, depending on the context. Energy is useful and allows us to say something about the final state from the initial state even without solving the exact problem how the system looks like at any moment in between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work is just a process in which energy is transformed from one form (e.g. energy stored in sugars and fats in muscles) to another form (furniture's potential energy when it's being brought to the 8th floor on the staircase). That's when &quot;work&quot; is meant as a qualitative concept. When it's a quantitative concept, it's the amount of energy that was transformed from one form to another; in practical applications, we usually mean that it was transformed from muscles or electrical grid or battery or another &quot;storage&quot; to a form of energy that is &quot;useful&quot; - but of course, these labels of being &quot;useful&quot; are not a part of physics, they are a part of the engineering or applications (our subjective appraisals).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-08T12:16:23.553" LastActivityDate="2011-02-08T12:16:23.553" CommentCount="6" />
  <row Id="3018" PostTypeId="2" ParentId="3009" CreationDate="2011-01-16T11:28:06.140" Score="3" ViewCount="" Body="&lt;p&gt;There are actually two different parts of general relativity. They're often stated as&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Spacetime tells matter how to move&lt;/li&gt;&#xA;&lt;li&gt;Matter tells spacetime how to curve&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Point #1 is actually straightforward to explain: objects simply travel on the straightest possible paths through spacetime, called &lt;a href=&quot;http://en.wikipedia.org/wiki/Geodesic&quot; rel=&quot;nofollow&quot;&gt;geodesics&lt;/a&gt;. The paths only seem curved because of the warping of spacetime. If you're a physicist, then you would want to know that that fact can be derived from the principle of extremal action (with all the requisite mathematical details), but if you don't want to wade through the math, hopefully it at least makes sense that objects move on &quot;straight&quot; lines. There is no actual force involved when a massive (or even a massless) object's trajectory curves in response to gravity, because it doesn't take any force to keep something moving on a straight line. (I can definitely expand on this point if you want)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I mentioned that spacetime needs to be warped in order for objects' trajectories to appear curved to us despite them actually being &quot;straight.&quot; So the essence of point #2 is, why is spacetime warped in the first place? Physics doesn't have a good answer to that. Technically, we don't have an answer to point #1 either, but the &quot;straight line&quot; argument at least makes it seem plausible; unfortunately, there's no equivalent plausibility argument for why spacetime warps itself around matter. (Perhaps someday we will come up with one) All we can do right now is produce equations that describe &lt;em&gt;how&lt;/em&gt; spacetime behaves around matter, namely the &lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein_field_equations&quot; rel=&quot;nofollow&quot;&gt;Einstein equations&lt;/a&gt; which can be written $G_{\mu\nu} = 8\pi T_{\mu\nu}$ among other ways.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-01-16T11:28:06.140" />
  <row Id="3019" PostTypeId="2" ParentId="3009" CreationDate="2011-01-16T11:48:26.430" Score="0" ViewCount="" Body="&lt;p&gt;What Einstein's equation tell us, at a basic level, is that the curvature of space-time and stress-energy are the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for this law to be respected it is clear that the stress-energy of a test particle cannot be constant in a space-time with changing curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you can choose a coordinate set in which the stress energy tensor is represented by the mass-energy of the particle, then the practical effect you can observe is changing energy and momenta of the test particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you therefore observe the test particle, you will see it as having changing energy and momenta, and therefore derive a force driving these changes. This is what we call gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, general relativity gives a much deeper picture of gravity as a description of the curvature of space-time, so, in a way, gravity is an &lt;em&gt;observed effect&lt;/em&gt; of the curvature of space-time, or, if you like, an &lt;em&gt;observed effect&lt;/em&gt; of the distribution of mass and energy.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-01-16T11:48:26.430" />
  <row Id="3020" PostTypeId="1" AcceptedAnswerId="3030" CreationDate="2011-01-16T12:42:20.650" Score="7" ViewCount="319" Body="&lt;p&gt;Is the fact that weak eigenstates are not mass eigenstates completely arbitrary? Or is there a deeper reason for the existence of the PMNS and CKM matrices?&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-27T11:19:30.540" LastActivityDate="2011-02-04T07:19:33.370" Title="Why do leptons and quarks mix?" Tags="&lt;particle-physics&gt;&lt;standard-model&gt;&lt;fermions&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="3021" PostTypeId="2" ParentId="3014" CreationDate="2011-01-16T13:21:54.920" Score="2" ViewCount="" Body="&lt;p&gt;Energy is a convenient way to account for a system's ability to do useful work. Now there are certain modern qualifications we attach to energy, mostly that total energy of a closed system is always conserved (barring cosmological effects), which is now explained by the use of symmetry and Noether's theorem (as explained by other commenters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To try to get to  a more satisfying everyday notion of energy, it is best to resort back to the concept of useful work and accounting of it.  We understand that we exert effort by lifting an object from the ground to the top of a table.  For practical accounting, we need to understand how much effort was expended.  It was these sort of accounting problems that led early engineers to the concept of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy is useful to us only if it has the ability to change its current form into another form.  If there is a way for this to occur, we can think in terms of potential energy if it is stored form is static (like a gallon of gas) or we can think in terms of kinetic energy if the energy is stored in relative motion (like a moving car).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the energy is in a form that is useless to us, then we measure it in terms of entropy.  A closed system will have a maximum value of entropy associated with it.  If the entropy of a system is lower than its maximum entropy, then that system is called &quot;far from equilibrium&quot; and has a lot of usable energy internal to it, so there is a lot of work that the system can do internal to itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In everyday terms, we only think about energy it terms of the useful work that can be derived from it&lt;/strong&gt;.  So when we talk of selling energy in an energy market, what is being traded is a commodity that can be used to do work.  The way that they energy is stored is different, but in any case, when we by a certain amount of energy, we expect it to allow us to accomplish certain tasks in a predictable way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a somewhat simplified discussion, there is a lot more that can be added, and several clarifications that would be needed, but I don't know your level of understanding, so I want to see if this makes sense first.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-16T13:42:02.307" />
  <row Id="3023" PostTypeId="2" ParentId="2768" CreationDate="2011-01-16T14:57:23.310" Score="1" ViewCount="" Body="&lt;p&gt;When I look at the approach, my first inclination is to think you are stating that each possible configuration of position of human arms, legs, etc combined with the vector velocity of those components, represent a state of the human body.  In some sense then there should be some smooth mapping of one state to the next. It seems that you are saying that you should be able to predict within some level of certainty what the next state should be based on the last state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with the approach would be that you are suggesting that human motion is somehow not guided by a controller, so while there are errors introduced in the system that the controller (brain) can't compensate for, in general, the overall process is not probabilistic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the question you are asking may be more closely related to questions in psychometrics than system dynamics.  Where Bessel was the father of &quot;personal equations&quot; to correct for human induced error in astronomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=4YtB63kdK0AC&amp;amp;lpg=PA4&amp;amp;ots=1HGQEnpw2E&amp;amp;dq=bessel%20astronomy%20human%20error&amp;amp;pg=PA4#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=4YtB63kdK0AC&amp;amp;lpg=PA4&amp;amp;ots=1HGQEnpw2E&amp;amp;dq=bessel%20astronomy%20human%20error&amp;amp;pg=PA4#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can clarify the question it might help.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-16T14:57:23.310" />
  <row Id="3024" PostTypeId="2" ParentId="3004" CreationDate="2011-01-16T15:12:45.133" Score="3" ViewCount="" Body="&lt;p&gt;Back when I was in graduate school in the 1990s, the standard reference for this sort of thing was Kolb and Turner's book &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Special%3aBookSources/9780201626742#Libraries&quot; rel=&quot;nofollow&quot;&gt;The Early Universe&lt;/a&gt;&lt;/em&gt;. Even after all these years, that book's treatment of this subject is probably still a good place to look.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if there's no asymmetry-producing process for neutrinos (like baryogenesis), you still expect a relic neutrino background that's a thermal (Fermi-Dirac) distribution of both neutrinos and antineutrinos, with a temperature of about 2 K. The reason is that, at a certain time in the evolution of the Universe, the density dropped low enough that the neutrino number &quot;froze out&quot;: interactions that could change the number of neutrinos (such as primarily $e^- e^+ \leftrightarrow \nu_e\ \bar\nu_e$) became so rare that the time for any given particle to undergo such a reaction grew much longer than a Hubble time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's been a long time since I looked at &lt;a href=&quot;http://en.wikipedia.org/wiki/Baryogenesis&quot; rel=&quot;nofollow&quot;&gt;baryogenesis&lt;/a&gt; models with any care, but as I recall some models would be expected to produce an asymmetry in the neutrino sector as well. But in practice I don't think that would change the prediction much. The reason is that baryogenesis only has to produce a one part in $10^9$ asymmetry (a billion and one protons for every billion antiprotons). That produces very noticeable effects today, because there was essentially complete annihilation of the antiprotons. But neutrino freeze-out occurs much earlier, while neutrinos are still relativistic, so we don't think that that massive annihilation happened for neutrinos. So even if there is a neutrino-antineutrino asymmetry comparable to the asymmetry produced by baryogenesis, it should only result in a tiny difference in the number of neutrinos over antineutrinos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me put that another way. At early times, (temperature much greater than the proton mass), there were comparable numbers of photons, neutrinos, and protons. Baryogenesis resulted in an asymmetry of protons over antiprotons at that time. After that, nearly all of the protons and antiprotons annihilated, leaving the observed result that today there are a billion photons for every proton. But we expect the number of relic neutrinos to be of the same order as the number of photons, not protons, so a baryogenesis-level neutrino asymmetry won't be noticeable.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-08-17T15:45:20.317" LastActivityDate="2011-08-17T15:45:20.317" CommentCount="2" />
  <row Id="3025" PostTypeId="1" CreationDate="2011-01-16T15:13:29.413" Score="5" ViewCount="294" Body="&lt;p&gt;I have this relative simple-looking question that I haven't been able to solve for hours now, it's one of those questions that just drive you nuts if you don't know how to do it.&#xA;This is the scenario:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a &lt;a href=&quot;http://2.bp.blogspot.com/_P4EHIyposmM/TImv2-brnOI/AAAAAAAAAMw/nK-NO8JUSEU/s1600/1781_PotApp_FlexibleMetal_Spring_Big.jpg&quot; rel=&quot;nofollow&quot;&gt;spring&lt;/a&gt; that is on a flat surface, the springs details are like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;spring constant = 100N/m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;height = 0.1m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;mass = 0.5kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g = 10m/s^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there is nothing attached to the spring.&#xA;The initial force exerted on the surface is 5N.&#xA;I compress the spring halfway until the force exerted on the surface is double, now 10N and then let it go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (simple) oscillation starts, and at one point the force exerted on the surface will be 0N (weightless).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find out how much time has passed after letting it go, and reaching weightlessness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as in:&#xA;10(N)---time---&gt;0(N)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p.s. not homework, read comments.&lt;/p&gt;&#xA;" OwnerUserId="1369" LastEditorUserId="1369" LastEditorDisplayName="" LastEditDate="2011-01-17T18:17:04.050" LastActivityDate="2011-01-20T06:18:25.847" Title="Tricky spring on a surface question" Tags="&lt;forces&gt;&lt;momentum&gt;&lt;harmonic-oscillator&gt;&lt;spring&gt;" AnswerCount="2" CommentCount="26" />
  <row Id="3026" PostTypeId="2" ParentId="2701" CreationDate="2011-01-16T15:20:39.337" Score="2" ViewCount="" Body="&lt;p&gt;To produce higher frequencies by a magnetron, one needs &#xA;higher plate (anode) voltage and a stronger magnetic field, &#xA;to keep the electrons within the central cavity.&#xA;So, first limit is the magnet, if we exclude expensive &#xA;solutions like superconductors. &#xA;On the other hand, one could construct a wider central cavity, &#xA;but remember, the cathode surface is the inner wall of it! &#xA;This in turn would ask for a a lot of power to heat &#xA;this cathode surface to yellow heat. &#xA;Georg &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-01-16T15:43:49.310" LastActivityDate="2011-01-16T15:43:49.310" CommentCount="1" />
  <row Id="3028" PostTypeId="2" ParentId="3014" CreationDate="2011-01-16T16:23:46.043" Score="-1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;What is energy?&lt;/strong&gt;&lt;br&gt;&#xA;Energy is the capacity of a system to do work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Where does it come from?&lt;/strong&gt;&lt;br&gt;&#xA;It generally comes from another source of energy, as in &lt;em&gt;energy gets converted from one form to another&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Where does it &lt;em&gt;ultimately&lt;/em&gt; come from?&lt;/strong&gt;&lt;br&gt;&#xA;That my friend is a question for &lt;strong&gt;&lt;em&gt;MetaPhysics&lt;/em&gt;&lt;/strong&gt;.stackexchange.com, which sadly doesn't exist as of now. You might want to hop over to &lt;a href=&quot;http://area51.stackexchange.com&quot;&gt;Area51&lt;/a&gt; with a proposal.&lt;/p&gt;&#xA;" OwnerUserId="1371" LastActivityDate="2011-01-16T16:23:46.043" CommentCount="2" />
  <row Id="3029" PostTypeId="2" ParentId="129" CreationDate="2011-01-16T16:26:57.753" Score="5" ViewCount="" Body="&lt;p&gt;We need a mechanism to control the scattering amplitude of two $W^\pm$ particles. It violates a &quot;unitarity bound&quot; when going above the TeV scale. Arguments based in unitarity of scattering diagrams -with leptons- have worked in the past, to predict the W boson itself, and to predict the Z boson (as a way to cancel a lack of unitarity if only W particles were involved in some models). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yep, it should be in the LHC scale, or other alternative mechanism should be. Of course, such alternative mechanism could need other pieces at higher scale, even using a higgs mechanism, but then it would not be the &quot;SM Electroweak Higgs&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1335" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-17T11:03:36.903" LastActivityDate="2011-03-17T11:03:36.903" />
  <row Id="3030" PostTypeId="2" ParentId="3020" CreationDate="2011-01-16T16:28:41.397" Score="6" ViewCount="" Body="&lt;p&gt;Yes, there is a very good reason why leptons and quarks mix. It would be shocking if they didn't mix. Just to avoid confusions, we're not saying that leptons and quarks mix with each other: they don't because leptons are eigenstates of color or baryon charge with  different eigenvalues than quarks which are also eigenstates: this difference prevents mixing between leptons and quarks because the mass terms would violate the charge conservation (for color and/or baryon charge - the former would be an internal inconsistency; the latter would be consistent internally but it would imply that the proton would decay rapidly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, leptons mix with leptons and quarks mix with quarks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory - whether you mean quantum field theory, string theory, or anything else that may be found in the future - produces several generations of quarks. Let's talk about upper quarks, $u_1, u_2, u_3$ - normally called $u,c,t$. For each of them, the quantum field theory produces a 4-component Dirac spinor. It is actually more correct to decompose it into 2-component spinors, but let us use the 4-component formalism here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If those three Dirac spinors have some kinetic terms, they may always be diagonalized so that the kinetic terms have the form&#xA;i$$\sum_i \overline{u_i}\partial^\mu \gamma_\mu u_i$$&#xA;There could be some general matrix $I_{ij}$ multiplying $\overline{u_i}$ and $u_j$ but by a proper transformation of the $u$ fields, it can be set to the identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, once you bring the kinetic terms into this form - by choosing the right combinations - there is not much freedom left. Generally, the underlying quantum field theory or string theory or whatever will also produce arbitrarily bilinear interactions of the form &#xA;$$-\sum_{i,j} M_{ij}\overline{u_i} u_j$$&#xA;The matrix $M_{ij}$ may be chosen Hermitean because the antihermitean part would cancel, anyway, as the product of two $u$'s is Hermitian - it gets complex-conjugated under the exchange of $i,j$. But aside from the Hermiticity, $M_{ij}$ is a completely unknown and arbitrary matrix. There is no reason for it to be diagonal; even if it is a generic Hermitian matrix with off-diagonal terms, it satisfies all symmetries and consistency conditions we normally require. In particular, it conserves the electroweak SU(2) isospin, U(1) hypercharge, SU(3) color, the angular momentum (only spin in this case), and the rest of the Lorentz symmetry (and also the baryon number). It's not surprising because from the viewpoint of the values of all these charges, the three generations of the quarks are indistinguishable, so coupling $\overline{u_1}$ with $u_2$ is as good as coupling it with $u_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the kinetic term is still invariant under SU(3) rotations of the three quark flavors. You may use this surviving symmetry to diagonalize the matrix $M_{ij}$. In the right basis, it is diagonal. However, you won't be able to diagonalize the lower-type quark mass matrix at the same moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar mass matrix exists for the down-type quarks. However, the Standard Model has an SU(2) electroweak isospin symmetry which is very important, at any energy scale. For a chosen triplet of fields $u_1,u_2,u_3$ of the up-type quarks for which the mass matrix was diagonal, we have their SU(2) partners - obtained by applying an SU(2) generator - $d_1,d_2,d_3$. However, the mass matrix for these three down-type quarks, which is $M^d_{ij}$, another Hermitian matrix, can be and probably will be non-diagonal once again. Right now, we don't have any symmetry left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it means that the mass of the three down-type quarks is expressed by a $3\times 3$ Hermitian matrix that is not necessarily diagonal, and there is no useful residual symmetry left that would allow us to diagonalize the matrix. So we must accept it as a fact: the three mass eigenstates among the down-type quarks will be different from the SU(2) partners of the three mass eigenstates among the three up-type quarks. You can't do anything about it. The CKM matrix is what relates these two bases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case of the leptons is analogous if one includes the masses for the neutrinos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be shocking to find the matrix to be diagonal in the same basis because there is no real &quot;need&quot; - such as symmetry or renormalizability or anomaly cancellation - that would dictate that the off-diagonal elements of the mass matrix have to vanish. Because it doesn't have to happen, it probably won't happen - this principle is known as Gell-Mann's totalitarian (or anarchic) principle. All terms that are allowed by the symmetries and consistency will occur with nonzero coefficients - most likely with coefficients that are &quot;natural&quot; i.e. comparable to one (or the typical mass scale). In particular, this is also true for the off-diagonal elements of the mass matrices relating several generations of particles that are otherwise equivalent when it comes to their conserved charges. So the mixing generally occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nature seems to satisfy the anarchic principle, at least with some nonzero coefficients - not necessarily comparable to one. So it follows that it produces two pretty much random Hermitian matrices for the up-type quarks and down-type quarks, and the bases in which these two matrices can be diagonalized won't have the property that one basis may be obtained by SU(2)-rotating the other one. That's why the CKM matrix is not equal to the identity and one needs to discuss the mixing. Similarly for the PMNS matrix for the leptons. &quot;Guessing&quot; that the matrices should be diagonal is as unnatural as guessing that the wealth of Bill Gates is exactly 56,789,012,345 dollars. There's no reason why a &quot;nice&quot; guess like that should be true, so it will probably not be true. The same thing holds for the guess that the off-diagonal entries of the mass matrices vanish.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:19:33.370" LastActivityDate="2011-02-04T07:19:33.370" CommentCount="3" />
  <row Id="3031" PostTypeId="2" ParentId="357" CreationDate="2011-01-16T16:32:56.013" Score="-1" ViewCount="" Body="&lt;p&gt;Has anybody a pointer to the Kinect patent?? the device of the XBox measures distance using reflection of infrared light, so for sure it depends on a finite lightspeed to work.&lt;/p&gt;&#xA;" OwnerUserId="1335" LastActivityDate="2011-01-16T16:32:56.013" CommentCount="5" />
  <row Id="3032" PostTypeId="1" AcceptedAnswerId="3042" CreationDate="2011-01-16T16:50:25.113" Score="1" ViewCount="85" Body="&lt;p&gt;For example iron. A metal spoon heats up much quicker than a wooden/plastic one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-16T17:43:26.657" Title="Why do regular lattice elements heat up faster?" Tags="&lt;heat&gt;&lt;lattice-model&gt;" AnswerCount="1" />
  <row Id="3033" PostTypeId="2" ParentId="634" CreationDate="2011-01-16T16:51:28.540" Score="0" ViewCount="" Body="&lt;p&gt;In my opinion, a good introductory text for  Relativistic Quantum Mechanics is Quantum Field theory- by Itzykson and Zubor. Another text is Relativistic quantum mechanics of string fields by - Greiner and Muller &lt;/p&gt;&#xA;" LastActivityDate="2011-01-16T16:51:28.540" />
  <row Id="3035" PostTypeId="1" CreationDate="2011-01-16T16:53:27.137" Score="2" ViewCount="112" Body="&lt;p&gt;If a planet of radius R1 has a constant sub-surface temperature T0 at R0 &amp;lt; R1, what is the long-term equilibrium surface temperature T1?  Say we assume constant thermal diffusivity of the planet material, surface emissivity 1.0, no atmosphere, and no incoming radiation.  I figure the temperature profile is harmonic, i.e. T=a+b/R, for constants a and b, and that we can use (R0,T0) to eliminate one of these constants.  Is there enough information to obtain the other constant?&lt;/p&gt;&#xA;" LastActivityDate="2011-03-17T20:33:49.663" Title="What is planetary surface temperature given constant sub-surface temperature?" Tags="&lt;heat&gt;&lt;temperature&gt;" AnswerCount="3" />
  <row Id="3037" PostTypeId="1" CreationDate="2011-01-16T17:09:04.467" Score="5" ViewCount="212" Body="&lt;p&gt;The claim is often made that the discovery of the Higgs boson will give us information about the origin of mass. However, the bare masses of the up and down quarks are only around 5 MeV, quite a bit smaller than their &quot;constituent&quot; or &quot;dynamical&quot; mass of around 300 MeV. (Remember that a neutron, for example, is one up and two down quarks and has a total mass of 939 MeV.) What then, is the reasoning behind the claim that the Higgs will address the origin of mass when by far the majority of the mass of the neutron (and proton) is related instead to the dynamical breaking of chiral symmetry?&lt;/p&gt;&#xA;" OwnerUserId="1376" LastActivityDate="2011-07-10T09:34:18.033" Title="The contribution to mass from the dynamical breaking of chiral symmetry" Tags="&lt;mass&gt;&lt;high-energy-physics&gt;&lt;higgs-boson&gt;&lt;quantum-chromodynamics&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="3038" PostTypeId="1" AcceptedAnswerId="3047" CreationDate="2011-01-16T17:09:09.117" Score="3" ViewCount="250" Body="&lt;p&gt;Just assume that I understand that a field in quantum field theory is an operator-valued distribution. For simplicity, forget about the distribution and think about a function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\varphi:M \rightarrow L(H)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that assigns an operator to each point of spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain to me what (mathematically speaking) physicists mean by &quot;quanta&quot; of this field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: (follow up question): if one fixes an inertial system and a point $t_0$ in time (say 4:00pm), then there is &quot;space&quot; and a map &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^\dagger:\mathbb R^3 \rightarrow L(H) \rightarrow H$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(x,y,z)\mapsto a^\dagger(t_0,x,y,z) \mapsto a^\dagger(t_0,x,y,z) |0&amp;gt;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;, which creates &quot;quanta&quot;. Would it be correct to think about &quot;particles at the point (x,y,z)&quot; then one thinks about this map?&lt;/p&gt;&#xA;" OwnerUserId="1377" LastEditorUserId="1377" LastEditorDisplayName="" LastEditDate="2011-01-17T16:05:18.917" LastActivityDate="2011-01-17T16:05:18.917" Title="What are field quanta?" Tags="&lt;quantum-field-theory&gt;" AnswerCount="4" />
  <row Id="3039" PostTypeId="2" ParentId="3038" CreationDate="2011-01-16T17:21:28.523" Score="2" ViewCount="" Body="&lt;p&gt;The operators assigned to your field usually are creation operators. Quanta are the things these creation operators create. This can be photons, phonons, electron-hole-pairs etc. &lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-01-16T17:21:28.523" />
  <row Id="3040" PostTypeId="2" ParentId="2687" CreationDate="2011-01-16T17:27:25.140" Score="2" ViewCount="" Body="&lt;p&gt;I walked into my local Mech. E. department and had a chat with a continuum mechanist.  Two books he showed me that looked like pretty good answers were:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Marcelo Epstein and Marek Elzanowski.  Material inhomogeneities and their evolution: a geometric approach. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Marcelo Epstein.   The geometrical language of continuum mechanics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing I like about these references is they're fairly explicit about the kinds of mathematics they're using.  I haven't read them in much detail but a possible downside is there appears to be relatively few detailed families of applications. &lt;/p&gt;&#xA;" OwnerUserId="1185" LastEditorUserId="1185" LastEditorDisplayName="" LastEditDate="2011-01-20T15:51:23.457" LastActivityDate="2011-01-20T15:51:23.457" CommunityOwnedDate="2011-02-04T10:52:39.143" />
  <row Id="3041" PostTypeId="1" CreationDate="2011-01-16T17:36:23.897" Score="1" ViewCount="303" Body="&lt;p&gt;If two black holes collide and then evaporate, do they leave behind two naked sigularities ore? If there are two, can we know how they interact?&lt;/p&gt;&#xA;" OwnerUserId="1319" LastActivityDate="2011-05-25T16:46:14.377" Title="Black Hole Singularities" Tags="&lt;general-relativity&gt;&lt;mathematical-physics&gt;&lt;quantum-gravity&gt;&lt;black-holes&gt;&lt;hawking-radiation&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="3042" PostTypeId="2" ParentId="3032" CreationDate="2011-01-16T17:37:45.570" Score="2" ViewCount="" Body="&lt;p&gt;in physics jargon, metals are better conductors. They're better electrical conductors as well as heat conductors. It's because metals have lots of electrons that are shared by the whole material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each atom of the metal contributes some electrons that move in the potential of all atoms. The corresponding solution are Bloch waves that are completely delocalized. While respecting the Pauli exclusion principle, the electrons occupy states $\exp(ipx)$ for various values of the momentum $p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For electrical conductivity to exist, it must be true that some states of this kind are occupied, but there are still nearby states with slightly different values of $p$ that remain empty. A small voltage is enough to move electrons a little bit, without making &quot;too big amount of work&quot;, from the previously filled to the previously empty states - given by different values of $p$. For a voltage, the electrons will pick an asymmetric distribution in which &quot;positive&quot; values of $p$ are chosen by a higher percentage of electrons that the negative values of $p$ or vice versa (the sign is measured in the direction of the electric field). This asymmetry, which is favored because of the $Q\phi$ interactions with the voltage, will make the electric current flow in the direction correlated with the electric field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a related way, the metal is also a good heat conductor because the extra heat (energy) is also quickly transmitted to the chaotic motion of the shared electrons near the &quot;Fermi surface&quot; (the boundary between unoccupied and occupied shared states). Moreover, the very regular lattice is also more efficient in transferring the heat. The lattice of atoms is like a system of coupled harmonic oscillators - springs that excite the nearby springs, and so on. They're pretty strong springs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the wood and plastic materials are typically (electrical and heat) insulators because there are no shared electrons. The electrons are just sharply associated with the individual molecules and then there is a gap (an energy gap). There are no nearby free electron states which the electrons could occupy. You may still imagine that there is a &quot;band&quot; of shared electrons but this band is always completely filled with the electron, and the closest electron states that remain empty are separated by a nonzero, finite energy from the highest-energy occupied states - so you would need a huge amount of work to transfer an electron from one state to another one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, the electricity and heat can't get transferred by these electrons, so the heat ultimately propagates from one molecule of &quot;wood&quot; to the other by the interactions between the relevant nuclei (through their electrons, of course). However, the &quot;springs&quot; between the nearby molecules of &quot;wood&quot; or &quot;plastic&quot; - which are also typically much greater - are much weaker because the molecules are not arranged so nicely and tightly - because of the irregularity, there is a lot of lost space. Because the springs are weaker, it is harder for one molecule to move with the adjacent molecule, and the heat doesn't propagate too efficiently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-16T17:43:26.657" LastActivityDate="2011-01-16T17:43:26.657" />
  <row Id="3043" PostTypeId="1" CreationDate="2011-01-16T17:40:17.487" Score="3" ViewCount="181" Body="&lt;p&gt;How to calculate number of exchanged virtual pohotons per unit of time between two electromagnetically interacting objects?&lt;/p&gt;&#xA;" OwnerUserId="1380" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-06T20:09:21.377" LastActivityDate="2011-03-06T20:09:21.377" Title="Virtual photon counting" Tags="&lt;quantum-mechanics&gt;&lt;quantum-electrodynamics&gt;&lt;virtual-photons&gt;" AnswerCount="2" />
  <row Id="3044" PostTypeId="2" ParentId="3041" CreationDate="2011-01-16T17:43:43.053" Score="0" ViewCount="" Body="&lt;p&gt;The rules of general relativity give us a very specific set of rules for the interaction of the two black holes.  We can integrate thee equations to know what would happen if, for example, we had two black holes orbiting each other, or on a head-on collision course.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result that we get is that the two colliding black holes will not evaporate &lt;b&gt;or&lt;/b&gt; form a naked singularity.  Their horizons will coalesce into a single distorted horizon, which will then smooth itself out by emitting gravitational radiation.  At late times, you will have a single black hole with a mass less than the combined masses of the two initial black holes (the difference in mass will be the energy that is radiated away as gravitational radiation).  According to numerical simulations of collisions, no naked singularity should be formed for a generic collisions.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-01-16T17:43:43.053" CommentCount="4" />
  <row Id="3045" PostTypeId="2" ParentId="3038" CreationDate="2011-01-16T17:45:13.763" Score="4" ViewCount="" Body="&lt;p&gt;It goes more or less like this. You define what is known as a creation operator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a^{\dagger}(\vec{k}) = \int \varphi(x)i\overleftrightarrow{\partial}_0e^{-ikx}d\vec{x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and when this is applied on a distinguished vector of the Hilbert space called &quot;vacuum&quot; $|0\rangle$, which describes the ground state of the system, you create a quanta of the field:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|k\rangle = a^{\dagger}(\vec{k})|0\rangle $&lt;/p&gt;&#xA;" OwnerUserId="1337" LastActivityDate="2011-01-16T17:45:13.763" />
  <row Id="3046" PostTypeId="2" ParentId="2966" CreationDate="2011-01-16T17:49:52.663" Score="1" ViewCount="" Body="&lt;p&gt;Sorry I can't be more specific, but you will probably find the answer to the question in Fundamentals of Carrier Transport&#xA;By Mark Lundstrom (http://cobweb.ecn.purdue.edu/~lundstro/books.html)**. I also recommend that you listen to his lectures on phonon scattering on iTunesU or from the &lt;a href=&quot;http://NanoHub.org&quot; rel=&quot;nofollow&quot;&gt;NanoHub.org ECE 656 Lecture 23: Phonon Scattering 1&lt;/a&gt;. There are lots of tricks like this to help integrate equations. When you here him explain them it seems obvious! Good luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** Sorry I couldn't embed that link. The self importance of an administrator is currently limiting me to 1 link per answer! Furthermore, I actually intended this to be a comment to your question, but it seems they have crippled that too. &lt;/p&gt;&#xA;" OwnerUserId="1194" LastActivityDate="2011-01-16T17:49:52.663" />
  <row Id="3047" PostTypeId="2" ParentId="3038" CreationDate="2011-01-16T17:50:14.057" Score="5" ViewCount="" Body="&lt;p&gt;some authors of the questions expect a less mathematical answer than what they are given at the end. In a sense, you are the opposite example because you expect a more mathematical explanation than what the right explanation actually is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term &quot;quantum of a field&quot; is not representing a particular state or operator; it is meant to describe a particular &quot;object&quot; that an experimenter may measure. In the theoretical language, you could say that a quantum of a field is the difference between the states&#xA;$$|\psi\rangle\mbox{ and } a^\dagger|\psi\rangle$$&#xA;Well, the difference is multiplicative in a sense - it's what the creation operator $a^\dagger$ created. But note that the state $|\psi\rangle$ may be anything - a quantum of a field may be added to any configuration of other (or the same) particles. Most typically, we talk about quanta in momentum eigenstates, so $a^\dagger$ is usually expressed by&#xA;$$a^\dagger(p)  = C \cdot \int d^3x\,\exp(ipx) \phi_i(x)$$&#xA;where $p$ is the momentum of the quantum - which usually allows us to calculate the energy as well - and $\phi_x(x)$ is a component of the quantum field (which may be both bosonic or fermionic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the &quot;intuitive&quot; or &quot;experimental&quot; language, quanta of quantum fields are the actual particles (such as photons, Higgs bosons, or electrons). They're what the quantum fields are able to create (and annihilate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-16T17:50:14.057" CommentCount="1" />
  <row Id="3048" PostTypeId="2" ParentId="3041" CreationDate="2011-01-16T17:51:05.663" Score="2" ViewCount="" Body="&lt;p&gt;Since classical gravity is much stronger than quantum gravity, the collision will take place on much faster time scales than the eventual Hawking radiation. So, the two black hole will experience a fairly violent process, and eventually settle down to a static situation which is most likely a new black hole. That black hole will evaporate on much longer time scales, eventually either evaporating completely, or leaving behind some long-lived &quot;remnant&quot;. Nobody really knows what the black hole leaves behind after evaporating, but I don't think it will be a naked singularity.&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-01-16T17:51:05.663" />
  <row Id="3049" PostTypeId="2" ParentId="3035" CreationDate="2011-01-16T17:54:51.523" Score="0" ViewCount="" Body="&lt;p&gt;That is easy: an object at temperature T0 in a zero temperature environment (no incoming radiation) will radiate till its temperature is zero. So the long term equilibrium is zero.&#xA;The question as posed describes a hypothetical situation, there is no environment at zero temperature. The universe is filled with CMB radiation of finite temperature that drops in time due to the cosmic (accelerated) expansion. So the real situation is a wee bit more complicated...&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-01-16T17:54:51.523" CommentCount="2" />
  <row Id="3050" PostTypeId="2" ParentId="3037" CreationDate="2011-01-16T17:55:34.153" Score="1" ViewCount="" Body="&lt;p&gt;The extra mass of the proton and neutron is not due to chiral symmetry breaking. It is due to the energy in the electromagnetic and strong force fields. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If chiral symmetry were an exact symmetry of the Lagrangian then the pions and other mesons (not the baryons) would have zero mass due to spontaneous chiral symmetry breaking. The chiral symmetry is not exact due to the small bare masses of the quarks so the mesons are not exactly massless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this you can see that your question is a bit confused, but one part that is correct is that most of the mass in ordinary atomic matter is not due to the Higgs mechanism. When people say that the Higgs boson will give us information abut the origin of mass they mean the bare masses of non-composite particles such as electrons and quarks. &lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-16T17:55:34.153" CommentCount="2" />
  <row Id="3051" PostTypeId="2" ParentId="3037" CreationDate="2011-01-16T17:57:22.317" Score="2" ViewCount="" Body="&lt;p&gt;Your point is totally valid and it is a favorite point of David Gross who happened to co-discover Quantum Chromodynamics. Indeed, it is often said by the journalists that the Higgs boson is the God particle that gives everything its mass. Without this mass, everything would be massless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, most of the mass of the visible matter is stored in the mass of protons and neutrons. And approximately 98% of the mass of these nucleons comes from the complicated QCD interactions inside these particles which also create lots of virtual gluons and quark-antiquark pairs, aside from large kinetic energy of all these particles that makes their total mass substantially exceed the rest mass. Only 2% or so of the nucleons' mass boils down to the Higgs-generated rest mass of the up and down quarks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a bit subtle whether one should say that this 98% of the proton mass comes from the chiral symmetry breaking. It would be totally correct for the pions. Pions would be massless if the chiral symmetry were not broken; it would be a bit different with the nucleons because their mass would not be protected even if the chiral symmetry were exact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is a sense in which the God particle is necessary even for this dominant QCD-generated mass of the proton: you need nonzero quark masses to really break the chiral symmetry. The chiral symmetry breaking just amplifies this initial &quot;impulse&quot; - the nonzero rest masses - to break the chiral symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So of course, you and David Gross are right that the journalists exaggerate the importance of the God particle; most of the actual mass of the materials we know comes from the QCD mess. However, even the chiral symmetry breaking would be hard without the God particle.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:25:02.270" LastActivityDate="2011-02-04T07:25:02.270" CommentCount="2" />
  <row Id="3052" PostTypeId="2" ParentId="3043" CreationDate="2011-01-16T17:58:12.690" Score="3" ViewCount="" Body="&lt;p&gt;The short answer is that you just can't. Different ways of doing the accounting (e.g., different gauges) lead to different ways of counting up the virtual particles in any given situation.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-01-16T17:58:12.690" />
  <row Id="3053" PostTypeId="2" ParentId="3041" CreationDate="2011-01-16T17:59:36.073" Score="0" ViewCount="" Body="&lt;p&gt;The answer is neither two nor one. Black holes that collide and subsequently evaporate leave just a bunch of radiation. There is no remaining 'naked singularities'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the past there has been a lot of confusion about black hole evaporation. But questions like &quot;what happens with the information describing the micro state of the black hole&quot; have been answered. Key idea is the holographic principle and black hole complementarity. If you want to learn more about this, Lenny Susskind's book 'The Black Hole War' is a good starting point.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastEditorUserId="1268" LastEditorDisplayName="" LastEditDate="2011-01-16T18:06:47.250" LastActivityDate="2011-01-16T18:06:47.250" CommentCount="5" />
  <row Id="3054" PostTypeId="2" ParentId="3043" CreationDate="2011-01-16T18:14:38.893" Score="3" ViewCount="" Body="&lt;p&gt;Imagine an elastic collision of two protons - something that they also see at the LHC but it's not the most interesting kind of interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two protons will repel because of the electromagnetic interaction. Assume that the distance between them is never too small, relatively to the proton radius. But you may calculate the cross section of this process in quantum field theory in different ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may treat protons as elementary particles - which is OK if you don't break them to pieces and if you don't study their composition in any way. If that's so, it's directly a proton that emits the virtual photon, and it is absorbed by the other photon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at the leading order - which is pretty accurate - the process only involves a single virtual proton  - per the whole interaction. If you interpret the proton as a collection of three quarks, of course there will be many more interactions and each quark will interact with the quarks in the other proton - as well as its friends in the same proton - so there will be many more virtual particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the very fact that I could obtain the result &quot;one&quot; for the number of virtual protons that are exchanged should convince you that you may be imagining the &quot;exchange of the virtual photons&quot; too naively. A virtual photon is simply not a particle that objectively exists and that you can count. A virtual photon is a part of a mathematical expression - the Feynman diagram - that is used to evaluate the only thing that makes a physical sense here, namely the probability amplitude (producing the cross section) for a particular process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of virtual particles - the propagators - depends on the Feynman diagram. And you must actually sum all the Feynman diagrams (it's coming from the sum over different histories) - and each of them gives a different answer to the question how many virtual particles there have been - before you get the probability amplitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, even in &quot;individual diagrams&quot;, the number of virtual particles depends on the resolution you choose - whether you try to decompose particles into their internal structure or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, even if you collide two charged cellestial bodies, you may still say that they exchanged 1 virtual photon only. The virtual photons are just not real objects that you may count as apples. Only the final measurements are real - anything that happened in between is virtual and all histories are allowed and all histories contribute - and even the final measurements are probabilistic in character.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even if you were trying to count the number of &lt;em&gt;physical&lt;/em&gt;, real particles in a state, you could get misleading, divergent, and ambiguous numbers. The counting of the virtual particles is even much more ill-defined.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:26:42.520" LastActivityDate="2011-02-04T07:26:42.520" CommentCount="1" />
  <row Id="3055" PostTypeId="2" ParentId="3041" CreationDate="2011-01-16T18:23:56.547" Score="1" ViewCount="" Body="&lt;p&gt;you can't produce any naked singularity in any process in 3+1 dimensions. This statement was originally coined by Roger Penrose under the slogan &quot;Cosmic Censorship Conjecture&quot; (CCC). A naked singularity would be a singularity that could affect the world around it because it would have no horizon; I will explain these words later. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because dynamics seems problematic near a singularity - it's hard to calculate it by classical general relativity or even the descriptions of quantum gravity that we know - it looked like the Universe would be unpredictable if there could be &quot;naked singularities&quot;, i.e. singularities that are unprotected by event horizons. That was viewed as a problem by Penrose (and others), a potential inconsistency that the Universe should avoid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This thinking due to Penrose has been showed partly incorrect, at least in higher dimensions. A proper theory of quantum gravity simply can (and has to) produce predictions - at least in principle - what happens to an observer who can observe any singularity. And indeed, there have been counterexamples - thought experiments in higher dimensions - in which one can produce a naked singularity at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, it is still true that the CCC holds in 3+1 dimensions. One can't create naked singularities by any process - whether it involves 0 black holes, 1 black hole, or 2 black holes. Black holes usually have a singularity inside them but they are not naked exactly because they're black holes. A defining feature of a black hole is that they're holes that are black - which means that light can't escape from them. It can't escape from them because it is confined by the event horizon. The event horizon are the &quot;clothes&quot; in which the singularity is &quot;dressed&quot; - so it is not naked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So despite some popular misconceptions, the defining feature of the black holes is the event horizon, and not a singularity. A singularity is just a typical addition that comes with the black hole &quot;package&quot;. In Schwarzschild black holes, the singularity is spacelike. Nevertheless, once the black hole evaporates, nothing is left. At the very final stages of a black hole decay, the tiny black hole looks pretty much indistinguishable from a heavy particle species in ordinary particle physics (or string theory), and it decays to a few particles that we know from the colliders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two stabilized black holes, when they collide, emit some gravitational waves, and when they're sufficiently close (so that they don't just continue as two black holes), the rest of their mass=energy that hasn't been radiated away collapses into a single black hole that quickly gets stabilized as well. This black hole will proceed with the Hawking radiation and after a very long time, it disappears as well. Nothing is left. The research in the last 20 years or so has made it pretty much clear that there can be no remnants left etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the best&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-16T18:23:56.547" CommentCount="3" />
  <row Id="3056" PostTypeId="1" CreationDate="2011-01-16T18:29:34.057" Score="2" ViewCount="249" Body="&lt;p&gt;I've recently started studying the k.p method for describing electronic bandstructures near the centre of the Brillouin zone and I've been finding it hard to find any pedagogical references on the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In every paper I read there's simply a statement of the Hamiltonian and it is followed by the definition of the matrix elements. However, nowhere have I been able to find a source that derives (in full) the actual matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been reading the original 1955 paper by Luttinger and Kohn, as well as Physics of Photonic Devices by S.L. Chaung and the book on the k.p method by Van Voon, but in all cases, only a run up is given and the calculation of the actual matrix elements is not given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may not seem like such a big deal, but it is to a first timer when you have to try and work through the calculations yourself. With no guide, I have been able to derive some of the matrix elements, but that's using my own intuition as to which matrix elements of the momentum operator vanish and/or are equal to others and I'm not sure that I'm right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single pedagogical reference would be useful to confirm or dispel some of the things I believe to be true and would be very helpful for me to continue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" LastEditorUserId="1337" LastEditorDisplayName="" LastEditDate="2011-01-31T03:05:07.937" LastActivityDate="2011-01-31T03:17:20.197" Title="Modern and complete references for the k.p method?" Tags="&lt;quantum-mechanics&gt;&lt;group-theory&gt;&lt;semiconductor-physics&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" />
  <row Id="3057" PostTypeId="2" ParentId="3035" CreationDate="2011-01-16T18:36:06.393" Score="0" ViewCount="" Body="&lt;p&gt;Sorry for the confusion.  R0 is some given particular radius between 0 and R1, e.g. R0=5000, R1=6000 km.  R is the general radial variable, as in T=a+b/R.  T0 is the temperature at position R0, assumed constant throughout time.  &lt;/p&gt;&#xA;" LastActivityDate="2011-01-16T18:36:06.393" CommentCount="3" />
  <row Id="3058" PostTypeId="2" ParentId="3038" CreationDate="2011-01-16T18:46:55.150" Score="0" ViewCount="" Body="&lt;p&gt;I realize that my answers will never be to par as some of the other brains at this sight, but I want to throw my two cents in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we can imagine the situation in ordinary quantum mechanics, we understand that there are quantum states that a particle can be in that are described by a complete set of commutating observables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;//en.wikipedia.org/wiki/Complete_set_of_commuting_observables&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum field theory evolved to answer questions about the number of particles that may be in a closed system at a certain energy level.  Since relativity equates particle mass to energy, we run into the situation where the number and types of particles will change over time, just like the state of a system might change over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So QFT is largely concerned with the evolution of systems of particles, where the observables are the particles themselves.  So by analogy, the goal is to find a complete set of commutating observables (particles) that can be used to determine the exact quantum state of the closed system in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantization of a classical field into one described by particles is called the second quatization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;//en.wikipedia.org/wiki/One-electron_universe&lt;/p&gt;&#xA;" OwnerUserId="1383" LastActivityDate="2011-01-16T18:46:55.150" />
  <row Id="3059" PostTypeId="1" AcceptedAnswerId="3322" CreationDate="2011-01-16T18:49:58.407" Score="3" ViewCount="199" Body="&lt;p&gt;What is a good place to learn the details of symmetry breaking? What I am looking for is a more serious exposition than the wiki-article, which explains the details, especially the mathematical part, but at the same time less detail than the typical quantum field theory book. Preferably I would like to see a clean made up example with all the details, but not necessarily a real example, what one can see in most books, which naturally has a lot of details that are there because that's what the world is, and not to illustrate the idea. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope it is clear what I am looking for.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Just to clarify. I am looking for a reference, but if somebody is willing to write an example here, I would be more than happy.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-19T06:57:37.613" LastActivityDate="2011-01-19T11:32:53.040" Title="Symmetry breaking" Tags="&lt;symmetry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="3060" PostTypeId="2" ParentId="3041" CreationDate="2011-01-16T18:51:05.153" Score="0" ViewCount="" Body="&lt;p&gt;The answer is zero. When black holes evaporate due to Hawking radiation they will not leave any singularity. The singularities in black holes are inside the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when two black holes collide their singularities do merge to form one singularity inside the event-horizon of the merged black hole. Perhaps that is what you really wanted to know.  &lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-16T18:51:05.153" />
  <row Id="3061" PostTypeId="1" AcceptedAnswerId="3066" CreationDate="2011-01-16T19:16:16.033" Score="3" ViewCount="202" Body="&lt;p&gt;In modern cosmology, it is commonly stated to the public that as we look deeper and deeper into space, we are seeing further into the past.  CMB is the relic of the big bang after photon decoupling following recombination.  However, if we pushed further back in our conceptualization, eventually we get to the period of inflation, and then to a period before that where energy density of the universe was some large value (possibly infinite).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why would it be wrong to conceptualize the universe as being enclosed by an infinitely dense shell?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: For clarification, wikipedia has a good definition of observable universe, and answers in the affirmative on how every observer will see a different &lt;a href=&quot;http://en.wikipedia.org/wiki/Observable_universe&quot; rel=&quot;nofollow&quot;&gt;observable universe&lt;/a&gt;.  This question is whether there are valid approaches in theories to avoid issues of big bang singularites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update2: I scaled back the question and accepted an answer, I think I was being overly vague in the write up and plan on asking the question again.&lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="1383" LastEditorDisplayName="" LastEditDate="2011-01-16T22:03:00.197" LastActivityDate="2011-02-04T07:27:31.830" Title="Is the observable universe enclosed by an infinitely dense shell?" Tags="&lt;cosmology&gt;&lt;big-bang&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="3062" PostTypeId="1" CreationDate="2011-01-16T19:23:23.403" Score="7" ViewCount="311" Body="&lt;p&gt;Suppose we have two galaxies that are sufficiently far apart so that the distance between them increases due to Hubble's expansion.  If I were to connect these two galaxies with a rope, would there be tension in the rope?  Would the tension increase with time?  Is the origin of the tension some sort of drag between the expanding space and matter?&lt;/p&gt;&#xA;" OwnerUserId="1387" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:28:52.343" LastActivityDate="2011-02-19T14:58:25.700" Title="How Does Hubble's Expansion Affect Two Rope-Tied Galaxies?" Tags="&lt;general-relativity&gt;&lt;gravity&gt;&lt;cosmology&gt;&lt;space-expansion&gt;" AnswerCount="6" CommentCount="0" FavoriteCount="3" />
  <row Id="3063" PostTypeId="2" ParentId="3035" CreationDate="2011-01-16T19:29:25.830" Score="1" ViewCount="" Body="&lt;p&gt;You want to equate the conductive energy flux at the surface which is a constant times T1-T0 -given your geometry and conductivity you can determine the coupling factor, with the radiative heat flux sigma*T**4. You could also add in the CMB, which is simply the same sigma*T**4 using the CMB temperature. Then you simply have a nonlinear algebraic equation in T0 to solve. It should be solvable by iteration without much effort.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-01-16T19:29:25.830" CommentCount="2" />
  <row Id="3064" PostTypeId="2" ParentId="3062" CreationDate="2011-01-16T19:43:46.950" Score="3" ViewCount="" Body="&lt;p&gt;Yes, of course, there would be some tension in the rope. The rope would eventually break, and maybe it would be slowing the galaxies motion if it were a really tight rope (you can't get rope with the required rigidity to stop the motion of galaxies in Nature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one only considers a pair of galaxies only, the Hubble expansion doesn't really differ from the ordinary motion of two objects away from one another. They want to move along the natural trajectories - those we observe - so any rope trying to prevent them from doing so will be stretched by the force of inertia of these galaxies. If you prevent some objects to move in a natural way they like, you will always experience an inertial force. Whether you call this force (translated into a tension in the rope) as &quot;inertial&quot; or &quot;gravitational&quot; in the cosmological context is up to your taste: after all, the equivalence principle is what guarantees that the effects of gravity and acceleration are indistinguishable so both answers are &quot;equivalent&quot; from a GR viewpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the tension in the rope (well, I would say a spring) can be written as $k$ times the excess proper length, then the problem of its tension as a function of time is equivalent to the problem of the proper distance between the two galaxies as a function of time. This is nothing else than the $a(t)$ parameter used in cosmology. See some texts on the Friedmann equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Friedmann_equations&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that this $a(t)$ satisfies. As a result, $a(t)$ was given by various power laws as a function of time. As we're entering the era dominated by the cosmological constant, $a(t)$ becomes exponentially increasing in $t$. So already today, the tension in the rope is increasing kind of exponentially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, one has to be careful about the literal interpretation of these things. The signals about the tension in any real &quot;rope&quot; are propagating by the speed of sound which is usually much slower than the speed of light. So it would take a lot of time for the most of the internal part of the &quot;rope&quot; to learn that it is attached to any galaxies at the endpoints. So most likely, the rope would get torn apart at the very endpoints very quickly while the internal bulk of the rope would stay at rest. You would have to specify more precisely what kind of a rope you want to consider if you want to solve the &quot;engineering question&quot; rather than the conceptual question about the changing proper distances in a cosmology.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:28:16.557" LastActivityDate="2011-02-04T07:28:16.557" CommentCount="3" />
  <row Id="3065" PostTypeId="2" ParentId="3037" CreationDate="2011-01-16T20:05:48.480" Score="0" ViewCount="" Body="&lt;p&gt;EDIT of a previous -1 answer, to clarify...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could be, at all, that the putative Yukawian-Higgs mass is the same mechanism. You could have heard of Koide equation, a strange relationship between the masses of tau, mu and electron which acually was used to predict the mass of tau in the seventies. If you rewrite this equation as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ m_{e,\mu,\tau} = m_0 (1 + \sqrt 2 \phi_{e,\mu,tau})^2 $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angles are exactly separated 120 degrees, and the mass $m_0$ happens to be equal to the constituient mass of the quarks inside a hadron, 313 MeV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;===================== Old, obscured answer, follows ======================&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amazing thing is not that the  bare mass is different that the constituent mass, but that they can coincide for some flavour at all. The bare mass comes from Yukawa couplings to electroweak symmetry breaking, the constituent mass comes from QCD. The only argument I known for the bare mass of a quark to be in the GeV scale is that you expect the corrections to it to be of the same order of magnitude that the mass itself; setting the cut-off at planck scale and using SU(3) self interaction, it favours a mass about some GeVs (and using U(1), it favours the MeV scale).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the role of chiral symmetry breaking... From Goldberger-Treiman, you get a prediction for the mass of the nucleon of about 950-960 MeV. Is GT expected to be exact in the limit of chiral symmetry? if so, its breaking seens to be working to decrease the nucleon mass. (EDIT: Hmm, see the comments of Harvey in other answers. The question should be, &quot;is GT expected to be exact in the absence of &lt;em&gt;explicit&lt;/em&gt; chiral symmetry breaking?&quot;)&lt;/p&gt;&#xA;" OwnerUserId="1335" LastEditorUserId="1335" LastEditorDisplayName="" LastEditDate="2011-07-10T09:34:18.033" LastActivityDate="2011-07-10T09:34:18.033" CommunityOwnedDate="2011-07-10T09:34:18.033" />
  <row Id="3066" PostTypeId="2" ParentId="3061" CreationDate="2011-01-16T20:08:11.990" Score="4" ViewCount="" Body="&lt;p&gt;When we say &quot;the Universe&quot; without specifying &quot;when&quot;, we usually mean the slice of our spacetime at a constant value of $t$, the proper time since the Big Bang (at each point). This &quot;the Universe&quot; has no boundary or shells. Quite on the contrary, it is pretty much uniform at long enough distance scales - above 300 million light years or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The Universe&quot; that has a shell on the boundary is a different slice through the spacetime, namely the past light cone with the tip at the Earth now. Approximating the spacetime by the Minkowski space for a while, this past light cone is composed of events $(t,x,y,z)$ that satisfy&#xA;$$t=-\frac{1}{c}\sqrt{x^2+y^2+z^2}$$&#xA;where $t=0$ is identified with the present. Indeed, this past light cone is surrounded by a very thick and dense wall. But you can't say that this past light cone is &quot;the Universe&quot; now: it is a slice through spacetime that contains some places that are photographed now (those that are nearby) and some places that are photographed in a distant past (near the Big Bang, those that are very far).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what we see is bounded by a thick wall - and we see lots of quasars at a very high redshift (big distance, big excursion to the distant past) and we also see the cosmic microwave background - from the times when the Universe was 350,000 years or so old. We actually don't &quot;see&quot; before that point using photons because the Universe was not transparent at all when it was younger than 350,000 years - it was full of plasma that was absorbing, emitting, and changing the direction of photons all the time. Using other probes, we can perhaps see before the point 350,000 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it's still more accurate to say that the &quot;thick wall&quot; doesn't appear at distant places of the Universe at one moment; instead, it appeared everywhere in the Universe a long time ago.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:27:31.830" LastActivityDate="2011-02-04T07:27:31.830" CommentCount="0" />
  <row Id="3067" PostTypeId="2" ParentId="743" CreationDate="2011-01-16T20:10:26.060" Score="0" ViewCount="" Body="&lt;p&gt;And I have written another pedagogical article about renormalizations and IR divergences. I created a Google research group &quot;&lt;a href=&quot;http://groups.google.com/group/qed-reformulation&quot; rel=&quot;nofollow&quot;&gt;QED Reformulation&lt;/a&gt;&quot; and I run a &lt;a href=&quot;http://vladimirkalitvianski.wordpress.com/2009/05/22/problem-of-infinitely-big-corrections/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;blog&lt;/strong&gt;&lt;/a&gt; on this subject. It is an alternative view of the problem, and I think, it is much more physical than the mainstream one. It is always useful to see the problem from different points of view ;-).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Se also &lt;a href=&quot;http://www.science20.com/qed_reformulation_feasible/blog/ultimate_explanation_renormalizations-81791&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;this&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-08-20T10:05:15.270" LastActivityDate="2011-08-20T10:05:15.270" CommentCount="1" />
  <row Id="3068" PostTypeId="2" ParentId="2980" CreationDate="2011-01-16T20:11:09.453" Score="0" ViewCount="" Body="&lt;p&gt;Thank you all for your answers. I apologize if the question was a bit unclear. &quot;No matter, never mind&quot; if you misunderstood my original question which was “Real” antimatter i.e. -c -p -t , is it unphysical?&#xA;A positron is -c -p but nevertheless +t. Is a theoretical -c -p -t positron physical?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dear Lubos Motl,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was visualizing the end result geometrically (without mathematics) in 3D like you simply and clearly and with enough details show using mathematics which I do not however master: An electron moving forwards in time in the form of a rightwards expanding logarithmic spiral and a leftwards expanding logarithmic positron coming from the exactly opposite position and moving backwards in time. From one observer's point of view (at x,y,z) both particles expand/rotate to the right from the observer's at (−x,−y,−z)  point of view they both rotate to the left. (I would like to add that if this visualization is correct than I have myself understood it!). &#xA;Is it correct that a particle moving on a Moebius band surface undergoes these three transformations when it has moved one time around and restoring to the same particle when moving two times around as one sometimes can see in popular notes?&#xA;My original question was colored by the known fact that we do not detect the same amount of antimatter as matter in nature. It has to do with an asymmetry of &quot;histories in the whole spacetime&quot;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks for taking the time to answer!&lt;/p&gt;&#xA;" OwnerUserId="1391" LastEditorUserId="1391" LastEditorDisplayName="" LastEditDate="2011-01-16T22:36:29.603" LastActivityDate="2011-01-16T22:36:29.603" CommentCount="2" />
  <row Id="3069" PostTypeId="2" ParentId="3061" CreationDate="2011-01-16T20:15:29.750" Score="1" ViewCount="" Body="&lt;p&gt;What exactly do you mean by &quot;enclosed by an infinitely dense shell&quot;? Are you proposing to generalize the distant density in the past null-cone directions to all directions in 4D spacetime? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you intend to imply that 13.7 bln years in the future a big crunch is awaiting us, I can tell you this is definitely incompatible with observations (accelerated Hubble expansion, etc.)&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-01-16T20:15:29.750" CommentCount="2" />
  <row Id="3070" PostTypeId="1" CreationDate="2011-01-16T20:17:05.443" Score="2" ViewCount="162" Body="&lt;p&gt;In a typical photon experiment the photon is depicted as moving across the page, say from right to left.&#xA;Suppose we were actually able to witness such an experiment, from the side (to position of reader to a page). &#xA;If the photon is actually moving from left to right can I, standing at 90 degrees to the motion, see the photon? &lt;/p&gt;&#xA;" LastActivityDate="2011-01-18T23:40:20.717" Title="movement of photons" Tags="&lt;photons&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="3072" PostTypeId="2" ParentId="3070" CreationDate="2011-01-16T20:23:44.770" Score="1" ViewCount="" Body="&lt;p&gt;The simple answer is no, the eye can only detect photons by their direct interaction with the retina. In this case, the photon is not &quot;visible&quot;, since it is not itself incident on the eye, nor emits &quot;secondary&quot; photons that indicate its position.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2011-01-16T20:23:44.770" />
  <row Id="3073" PostTypeId="2" ParentId="3070" CreationDate="2011-01-16T20:25:42.347" Score="0" ViewCount="" Body="&lt;p&gt;Being very careful, we'll assume you mean a situation in which there exists only a single photon and two detectors placed at right angles to one another, all of which exists in a vacuum and effectively isolated from the rest of the universe: the answer would be 'no'. In practice of course, very few photons and detectors exist in such perfect isolation.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-16T20:25:42.347" />
  <row Id="3074" PostTypeId="2" ParentId="3070" CreationDate="2011-01-16T20:28:22.590" Score="0" ViewCount="" Body="&lt;p&gt;Your question is worded such that it presumes the photon to be localized within a size much smaller than the distance to the observer 'at the sideline'. In that case, no, this observer can't see the photon. You can't see a photon that does not hit your eye (in spacetime language the photon needs to travel along the past light cone of your pupil).&#xA;Now, of course a photon is not a classical particle, it has wavelike characteristics as well. So things are different (the photon can no longer be thought of as localized) when the wavelength of the photon becomes comparable to the distance to the observer. But then you can't say &quot;the photon moves from left to right across the page&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-01-16T20:28:22.590" />
  <row Id="3075" PostTypeId="1" AcceptedAnswerId="3091" CreationDate="2011-01-16T20:28:38.917" Score="6" ViewCount="343" Body="&lt;p&gt;What are the quantum mechanisms behind the emission and absorption of thermal radiation at and below room temperature? If the relevant quantum state transitions are molecular (stretching, flexing and spin changes) how come the thermal spectrum is continuous? What about substances (such as noble gases) which don't form molecules, how do they emit or absorb thermal radiation? Is there a semi-classical mechanism (with the EM field treated classically) and also a deeper explanation using the full apparatus of QFT?&lt;/p&gt;&#xA;" OwnerUserId="1394" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-01-16T21:37:59.853" LastActivityDate="2011-05-23T00:46:21.567" Title="Quantum mechanics of thermal radiation" Tags="&lt;quantum-mechanics&gt;&lt;thermal-radiation&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="3076" PostTypeId="1" AcceptedAnswerId="3078" CreationDate="2011-01-16T20:30:05.863" Score="4" ViewCount="312" Body="&lt;p&gt;&lt;em&gt;[insert obligatory statement of my lack of knowledge in physics]&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alright, so we have all seen the movies where someone gets blasted out of the airlock on their starship, or their suit decompresses while on a space walk. The poor schmoe usually either decompresses so violently that blood is oozing out of every orifice in their body, or they freeze instantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this I have two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Would the decompression really be that violent?&lt;/strong&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Clearly the drastic difference in pressure from a normal &quot;earth&quot; like environment to space would be bad, but would it be &lt;em&gt;that&lt;/em&gt; devastating.&lt;/li&gt;&#xA;&lt;li&gt;I vaguely remember that standard atmospheric pressure was something like 15 psi, which doesn't seem like enough to mess you up that bad.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Would you &lt;em&gt;actually&lt;/em&gt; freeze instantly in space?&lt;/strong&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Heat, or lack thereof is a measure of internal energy, but in a vacuum there wouldn't be anything to have internal energy, so does space even have a temperature?&lt;/li&gt;&#xA;&lt;li&gt;Wouldn't some form of matter have to be present in order to cool off? If there were no matter besides yourself and a few stray particles here and there, it seems like it would take a very long time to cool off.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1388" LastActivityDate="2011-01-27T23:29:30.293" Title="What would ACTUALLY happen to a person jettisoned into space?" Tags="&lt;heat&gt;&lt;pressure&gt;&lt;space&gt;&lt;atmosphere&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="3077" PostTypeId="2" ParentId="1756" CreationDate="2011-01-16T20:32:30.087" Score="1" ViewCount="" Body="&lt;p&gt;There may be three different &lt;strong&gt;r&lt;/strong&gt;-dependences of interaction: microscopic V(&lt;strong&gt;r&lt;/strong&gt;), effective elastic U(&lt;strong&gt;r&lt;/strong&gt;), and an effective inclusive W(&lt;strong&gt;r&lt;/strong&gt;) derived from an inclusive cross section; see &lt;a href=&quot;http://arxiv.org/abs/0806.2635&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;here&lt;/strong&gt;&lt;/a&gt; and in my weblog.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-01-16T21:07:25.150" LastActivityDate="2011-01-16T21:07:25.150" />
  <row Id="3078" PostTypeId="2" ParentId="3076" CreationDate="2011-01-16T20:36:27.713" Score="3" ViewCount="" Body="&lt;p&gt;What you see depicted in SF movies is nonsense:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/970603.html&quot; rel=&quot;nofollow&quot;&gt;http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/970603.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-01-16T20:36:27.713" CommentCount="6" />
  <row Id="3079" PostTypeId="2" ParentId="3070" CreationDate="2011-01-16T20:41:01.613" Score="1" ViewCount="" Body="&lt;p&gt;Dear Peter, the pictures are drawn to indicate that the photons are there. They are there even if you don't see them. There are many things that we can't see - or we can't see directly or at a given moment - but they still exist. And of course, you don't see the photons (with a wrong direction) by normal methods - you would have to collide them against something else that you could observe but that would change the propagation of the photons on the page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you want to be fancy, the answer is that you &lt;em&gt;can&lt;/em&gt; actually see photons by other photons. Because of quantum effects (with a virtual box-like loop of an electron, and four attached photons), the electromagnetic waves are slightly non-linear. So light can collide with other light so you could actually &quot;shine&quot; a very powerful beam of light to see another beam. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't do it in your kitchen but this phenomenon has actually been tested experimentally at SLAC, California. It works: see this thread:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/1361/scattering-of-light-by-light-experimental-status&quot;&gt;Scattering of light by light: experimental status&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-16T20:41:01.613" CommentCount="2" />
  <row Id="3081" PostTypeId="1" AcceptedAnswerId="3093" CreationDate="2011-01-16T20:49:41.353" Score="7" ViewCount="454" Body="&lt;p&gt;Updated:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;In order to fold anything in half, it must be &lt;code&gt;π&lt;/code&gt; times longer than its thickness, and that depending on how something is folded, the amount its length decreases with each fold differs.&quot;&lt;br/&gt; – &lt;a href=&quot;http://en.wikipedia.org/wiki/Britney_Gallivan&quot; rel=&quot;nofollow&quot;&gt;Britney Gallivan&lt;/a&gt;, the person who determined that the maximum number of times a paper or other finite thickness materials can be folded = 12.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematics_of_paper_folding&quot; rel=&quot;nofollow&quot;&gt;Mathematics of paper folding&lt;/a&gt; explains the mathematical aspect of this. &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know the physical explanation of this. Why is it not possible to fold a paper more than &quot;n&quot; (12) times?  &lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="1378" LastEditorDisplayName="" LastEditDate="2011-01-17T00:07:17.160" LastActivityDate="2011-03-10T12:20:40.653" Title="Why can't a piece of paper (of non-zero thickness) be folded more than &quot;n&quot; times?" Tags="&lt;mechanics&gt;" AnswerCount="2" CommentCount="13" FavoriteCount="3" />
  <row Id="3083" PostTypeId="2" ParentId="3076" CreationDate="2011-01-16T20:53:43.440" Score="1" ViewCount="" Body="&lt;p&gt;Most of what is shown in sci-fi on this is believed to be false. You do not freeze to death and your blood does not boil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Provided you do not try to hold your breath there are not likely to any ill effects for at least 30 seconds or so. If you do try to hold your breath as the de-pressurisation takes place you can suffer a &quot;burst lung&quot; i.e. an embolism, just as a diver would if he holds his breath during an ascent from depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some time in space the lack of oxygen will be the most damaging. A few minutes will lead to death.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this answer at &lt;a href=&quot;http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/970603.html&quot; rel=&quot;nofollow&quot;&gt;http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/970603.html&lt;/a&gt; where you can find a lot more details.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-16T20:53:43.440" />
  <row Id="3084" PostTypeId="1" AcceptedAnswerId="3089" CreationDate="2011-01-16T21:09:08.033" Score="3" ViewCount="214" Body="&lt;p&gt;On the &lt;a href=&quot;http://www.answers.com/topic/planck-length&quot; rel=&quot;nofollow&quot;&gt;Answers.com&lt;/a&gt; page on Planck length, I see two almost-same formulas for the Planck length that differ only by the use of h and hbar. However, the constants are the same, and my calculator gives the correct answer for hbar instead of h, so the first use of h was probably intended to mean hbar. Why doesn't the Oxford Dictionary (and my textbook!) use hbar instead?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: the equation (from Oxford dictionary?) I was talking about that uses h:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/2PaMT.jpg&quot; alt=&quot;planck stuff&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the equation from Wikipedia that uses hbar, but gives the same constant for the Planck length:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/S4oVI.png&quot; alt=&quot;wiki planck stuff&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="202" LastEditorUserId="202" LastEditorDisplayName="" LastEditDate="2011-01-16T21:34:50.533" LastActivityDate="2011-01-16T21:41:22.877" Title="Why do the formulas for planck's units use h instead of hbar?" Tags="&lt;planck&gt;&lt;formula&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="3085" PostTypeId="2" ParentId="3062" CreationDate="2011-01-16T21:09:36.767" Score="4" ViewCount="" Body="&lt;p&gt;The most interesting version of this question (I think) is this: Suppose that the rope has been attached long enough to bring the two galaxies to a stop relative to each other.  (d[physical distance]/d[cosmic time] = 0). For subsequent times, does the rope need to maintain any tension to keep the galaxies from starting to accelerate away from each other? If we cut the rope, does the distance to the galaxies start to increase, decrease, or stay the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer, it turns out, is that if the expansion of the Universe is decelerating (e.g., if there's no cosmological constant), then there need not be any tension in the rope after the galaxies have come to a stop, and if the rope is cut, the galaxies do not recede from each other. On the contrary, they fall toward each other. With a sufficiently large cosmological constant, on the other hand, they do recede after the rope is cut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Charley Lineweaver and Tamara Davis wrote some quasi-pedagogical articles on this, &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0104349&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/astro-ph/0104349&lt;/a&gt; being the main one. Personally, I think that this question is helpful in clarifying intuition about the &quot;meaning&quot; of expanding space. The tethered galaxy problem is part (but by no means all) of the argument we tried to make in &lt;a href=&quot;http://arxiv.org/abs/0808.1081&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0808.1081&lt;/a&gt; . (Sorry for the self-promotion, but I honestly do think it's a relevant citation, albeit not quite as relevant as the Davis-Lineweaver work.)&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-01-16T21:09:36.767" CommentCount="1" />
  <row Id="3086" PostTypeId="2" ParentId="3084" CreationDate="2011-01-16T21:16:03.577" Score="1" ViewCount="" Body="&lt;p&gt;That must be related to typesetting issues. Natural (Planck's) units have hbar=1, not h=1.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-01-16T21:16:03.577" />
  <row Id="3087" PostTypeId="2" ParentId="3062" CreationDate="2011-01-16T21:26:38.473" Score="0" ViewCount="" Body="&lt;p&gt;I think it might be clearest to first consider the case of a rope NOT connected to anything; a free, straightened rope in empty space. Each point along the rope feels the same effective force due to Hubble expansion; each point along the rope is slightly repelled by each of its neighboring points. Assuming the expansion is isotropic in space, there is no need to consider the speed of any forces propagating along the rope; the rope experiences a uniform tension that increases linearly with the rope's length. If the rope is long enough the stress on the rope will overcome its tensile strength, and it will snap.&lt;/p&gt;&#xA;" OwnerUserId="1247" LastActivityDate="2011-01-16T21:26:38.473" CommentCount="2" />
  <row Id="3088" PostTypeId="1" CreationDate="2011-01-16T21:28:35.713" Score="1" ViewCount="51" Body="&lt;p&gt;If U238 ions were circulating in a particle accelerator would their neutron absorption cross section spectra change?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-16T21:53:45.683" Title="If U238 ions were circulating in a particle accelerator would their neutron absorption cross section spectra change?" Tags="&lt;particle-physics&gt;&lt;accelerator-physics&gt;" AnswerCount="2" />
  <row Id="3089" PostTypeId="2" ParentId="3084" CreationDate="2011-01-16T21:28:56.713" Score="5" ViewCount="" Body="&lt;p&gt;the answers.com page you mentioned uses the following formula:&#xA;$$L_{Planck} = \sqrt{\frac{Gh}{2\pi c^3}}$$&#xA;Note that there is the $2\pi$ factor in the denominator - so $h/2\pi$ may be simplified as the usual $\hbar$. They probably weren't able to type this character, or wanted to avoid terminology and symbols that are only known to physicists. But there is no numerical error on the answers.com page. At any rate, the definition above is equivalent to&#xA;$$L_{Planck} = \sqrt{\frac{G\hbar}{c^3}}$$&#xA;which is the usual &quot;unreduced&quot; Planck length. See Wikipedia for the same formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Planck_length&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Planck_length&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numerically, it's $1.6 \times 10^{-35}$ meters. (Update: The Oxford Dictionary of English has a wrong formula - they omitted $2\pi$ and forgot to cross the $h$, too. But they clearly mean the same Planck length.) Sometimes, people also use the &quot;reduced&quot; Planck length which is more fancy and &quot;professional&quot; in a sense:&#xA;$$L_{Planck,reduced} = \sqrt{\frac{8\pi G\hbar}{c^3}}$$&#xA;Note that the $8\pi$ in the numerator may also be merged with the $\hbar$ to get $4h$ back - so the reduced Planck length is twice (because of the square root) the wrong Planck length that you would get by using $h$ instead of $\hbar$. But what is the real reason why $8\pi$ was added there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why $8\pi G$ appears instead of $G$ is because in some sense, $8\pi G$ is more natural a constant than $G$: this discussion is analogous to the treatment of $4\pi$ in electrodynamics. The constant $8\pi G$ is natural because the Einstein-Hilbert action is&#xA;$$S_{EH} = \int d^D x \frac{1}{16\pi G} R\sqrt{-g} $$&#xA;The most natural coefficient would be $1/2$ instead of $1/16\pi G$ which makes it natural to set $8\pi G=1$. The reduced Planck length is somewhat longer (five times or so) - less extremely tiny. Even more often, particle physicists talk about the Planck energy and the reduced Planck energy which are close to $10^{19}$ and $10^{18}$ GeV, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The convention for the constant $G$ was originally chosen by Newton who wanted to write the gravitational force as $GMm/r^2$. Well, it would be more natural to have the factor of $4\pi$ or $8\pi$ in the denominator, $\Gamma Mm/8\pi r^2$. You can see that $\Gamma$ is simply $\Gamma=8\pi G$, and it would be natural to set $\Gamma$ equal to one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that I don't have to explain why $\hbar$ is more natural than $h$ for adult physicists. The &quot;laymen&quot; versions of the formulae may be simpler with $h$ - but they deal with wavelength etc. Adult physicists know that the wavelength of the sine is proportional to $2\pi$. And the most fundamental equations, such as the Schrödinger's equation or the commutators of $[x,p]$, take a simpler form in terms of $\hbar$ than $h$, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to $G$: people had to choose the convention how to normalize $G$ in higher dimensions. The usual convention, as implicitly used above, is that the Einstein-Hilbert action always has the coefficient $1/16\pi G$. That implies that in $D$ spacetime dimensions, the force won't be $GMm/r^{D-2}$ but it will have some $D$-dependent numerical coefficients in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-16T21:41:22.877" LastActivityDate="2011-01-16T21:41:22.877" CommentCount="2" />
  <row Id="3090" PostTypeId="2" ParentId="2786" CreationDate="2011-01-16T21:31:48.277" Score="-1" ViewCount="" Body="&lt;p&gt;And I propose to look at my own pictures &lt;a href=&quot;http://vladimirkalitvianski.wordpress.com/2010/12/02/zoom-in-atom-or-unknown-physics-of-short-distances/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;here&lt;/strong&gt;&lt;/a&gt;. There is no proton wave function in atom but a wave function of the relative motion. It is a quasi-particle wave function. Proton is in a mixed state when in atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning always bound quarks, it is a non-linear problem with a strong coupling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And apart from the wave function with many arguments, one can think of the &quot;charge density&quot; $ \rho$(&lt;strong&gt;r&lt;/strong&gt;) determined with the charge form-factor(s) F(&lt;strong&gt;q&lt;/strong&gt;). This is more tangible.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-01-19T19:21:23.803" LastActivityDate="2011-01-19T19:21:23.803" CommentCount="4" />
  <row Id="3091" PostTypeId="2" ParentId="3075" CreationDate="2011-01-16T21:32:02.273" Score="4" ViewCount="" Body="&lt;p&gt;The thermal radiation associated with some object is typically described in terms of the &quot;black-body&quot; spectrum for a given temperature, given by the Planck formula. This formula is based on an idealization of an object that absorbs all frequencies of radiation equally, but it works fairly well provided that the object whose thermal spectrum you're interested in studying doesn't have any transitions with resonant frequencies in the range of interest. As the typical energy scale of atomic and molecular transitions is somewhere around an eV, while the characteristic energy scale for &quot;room temperature&quot; is in the neighborhood of 1/40 eV, this generally isn't all that bad an assumption-- if you look in the vicinity of the peak of the blackbody spectrum for an object at room temperature, you generally find that the spectrum looks very much like a black-body spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this arise from the interaction between light of whatever frequency and a gas of atoms or molecules having discrete internal states? The thing to remember is that internal states of atoms and molecules aren't the only degree of freedom available to the systems-- there's also the center-of-mass motion of the atoms themselves, or the collective motion of groups of atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The central idea involved with thermal radiation is that if you take a gas of atoms and confine it to a region of space containing some radiation field with some characteristic temperature, the atoms and the radiation will eventually come to some equilibrium in which the kinetic energy distribution of the atoms and the frequency spectrum of the radiation will have the same characteristic temperature. (The internal state distribution of the atoms will also have the same temperature, but if you're talking about room-temperature systems, there's too little thermal energy to make much difference in the thermal state distribution, so we'll ignore that.) This will come about through interactions between the atoms and the light, and most of these interactions will be non-resonant in nature. In terms of microscopic quantum processes, you would think of these as being &lt;a href=&quot;http://en.wikipedia.org/wiki/Raman_scattering&quot; rel=&quot;nofollow&quot;&gt;Raman scattering&lt;/a&gt; events, where some of the photon energy goes into changing the motional state of the atom-- if you have cold atoms and hot photons, you'll get more scattering events that increase the atom's kinetic energy than ones that decrease it, so the average atomic KE will increase, and the average photon energy will decrease. (Or, in more fully quantum terms, the population of atoms will be moved up to higher-energy quantum states within the box, while the population of higher-energy photon modes will decrease.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For thermal radiation in the room temperature regime, of course, the transitions in question are so far off-resonance that a Raman scattering for any individual atom with any particular photon will be phenomenally unlikely. Atoms are plentiful, though, and photons are even cheaper, so the total number of interactions for the sample &lt;em&gt;as a whole&lt;/em&gt; can be quite large, and can bring both the atomic gas and the thermal radiation bath to equilibrium in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've never seen a full QFT treatment of the subject, but that doesn't mean much. The basic idea of the equilibration of atoms with thermal radiation comes from Einstein in 1917, and there was a really good &lt;a href=&quot;http://www.phytem.ens-cachan.fr/telechargement/Optique_Quantique/Kleppner_Coef_Einstein.pdf&quot; rel=&quot;nofollow&quot;&gt;Physics Today article (PDF)&lt;/a&gt; by Dan Kleppner a few years back, talking about just how much is in those papers.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2011-01-16T21:32:02.273" CommentCount="6" />
  <row Id="3092" PostTypeId="2" ParentId="3075" CreationDate="2011-01-16T21:37:06.500" Score="1" ViewCount="" Body="&lt;p&gt;Addressing only the question of how a continuous spectrum arises from what appears to be a set of discrete spectra associated with individual modes of interaction with a solid pahse material...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get a hint one of the things that is going on here by considering the &lt;a href=&quot;http://en.wikipedia.org/wiki/M%C3%B6ssbauer_effect&quot; rel=&quot;nofollow&quot;&gt;Mössbauer effect&lt;/a&gt;. Individual photons are not required to interact with individual atoms, they &lt;em&gt;can&lt;/em&gt; (and therefore sometimes will) interact with larger units of matter. IN the case of the Mössbauer effect they exchange momentum with a large unit of crystalline lattice rather than just the nucleus they interact with. You use the adjective &quot;coherent&quot; to distinguish these interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is that a otherwise neat cluster of allowed energies is smeared out by the recoil associated with many different masses.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-01-16T21:37:06.500" />
  <row Id="3093" PostTypeId="2" ParentId="3081" CreationDate="2011-01-16T21:41:06.993" Score="15" ViewCount="" Body="&lt;p&gt;I remember that the question in your title was busted in &lt;a href=&quot;http://mythbustersresults.com/episode72&quot; rel=&quot;nofollow&quot;&gt;Mythbusters episode 72&lt;/a&gt;.&#xA;A simple google search also gives many other examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for single- vs alternate-direction folding, I'm guessing that the latter would allow for more folds. It is the thickness vs length along a fold that basically tells you if a fold is possible, since there is always going to be a curvature to the fold. Alternate-direction folding uses both flat directions of the paper, so you run out of length slightly slower. This would be a small effect since you have the linear decrease in length vs the exponential increase in thickness.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Thanks to &lt;a href=&quot;http://physics.stackexchange.com/users/518/gerry&quot;&gt;gerry&lt;/a&gt; for the key word (given in a comment above). I can now make my above guess more concrete.  The limit on the number of folds (for a given length) does follow from the necessary curvature on the fold. The type of image you see for this makes it clear what's going on&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/eOIWA.jpg&quot; alt=&quot;Fold diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a piece of paper with thickness $t$, the length $L$ needed to make $n$ folds is &lt;a href=&quot;http://oeis.org/A076024&quot; rel=&quot;nofollow&quot;&gt;(OEIS)&lt;/a&gt;&#xA;$$ L/t = \frac{\pi}6 (2^n+4)(2^n-1) \,.$$&#xA;This formula was originally derived by (the then Junior high school student) &lt;a href=&quot;http://pomonahistorical.org/12times.htm&quot; rel=&quot;nofollow&quot;&gt;Britney Gallivan in 2001&lt;/a&gt;. I find it amazing that it was not known before that time... (and full credit to Britney).&#xA;For alternate folding of a square piece of paper, the corresponding formula is&#xA;$$ L/t = \pi 2^{3(n-1)/2} \,.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both formulae give $L=t\,\pi$ as the minimum length required for a single fold. This is because, assuming the paper does not stretch and the inside of the fold is perfectly flat, a single fold uses up the length of a semicircle with outside diameter equal to the thickness of the paper. So if $L &amp;lt; t\,\pi$ then you don't have enough paper to go around the fold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's ignore a lot of the subtleties of the linear folding problem and say that each time you fold the paper you halve its length and double its thickness:&#xA;$ L_i = \tfrac12 L_{i-1} = 2^{-i}L_0 $ and $ t_i = 2 t_{i-1} = 2^{i} t_0 $,&#xA;where $L=L_0$ and $t=t_0$ are the original length and thickness respectively.&#xA;On the final fold (to make it n folds) you need &#xA;$L_{n-1} \leq \pi t_{n-1}$ which implies $L \leq \frac14\pi\,2^{2n} t$. &#xA;Qualitatively this reproduce the linear folding result given above.&#xA;The difference comes from the fact you lose slightly over half of the length on each fold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These formulae can be inverted and plotted to give the logarithmic graphs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/hY8bk.png&quot; alt=&quot;Number of folds given a length&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $L$ is measured in units of $t$. The linear folding is shown in red and the alternate direction folding is given in blue. The boxed area is shown in the inset graphic and details the point where alternate folding permanently gains an extra fold over linear folding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that there exist certain length ranges where you get more folds with alternate than linear folding. After $L/t = 64\pi \approx 201$ you always get one or more extra folds with alternate compared to linear. You can find similar numbers for two or more extra folds, etc...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Looking back on this answer, I really think that I should ditch my theorist tendencies and put some approximate numbers in here.  Let's assume that the 8 alternating fold limit for a &quot;normal&quot; piece of paper is correct. Normal office paper is &lt;a href=&quot;http://en.wikipedia.org/wiki/Paper#Types.2C_thickness_and_weight&quot; rel=&quot;nofollow&quot;&gt;approximately 0.1mm thick&lt;/a&gt;. This means that a normal piece of paper must be&#xA;$$ L \approx \pi\,(0.1\text{mm})  2^{3\times 7/2} \approx 0.3 \times 2^{10.5}\,\text{mm} &#xD;&#xA;  \approx .3 \times 1000 \, \text{mm} = 300 \text{mm} \,.&#xD;&#xA;$$&#xA;Luckily this matches the normal size of office paper, e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Paper_size#The_international_standard%3a_ISO_216&quot; rel=&quot;nofollow&quot;&gt;A4 is 210mm * 297mm&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last range where you get the same number of folds for linear and alternate folding is $L/t \in (50\pi,64\pi) \approx (157,201)$, &#xA;where both methods yield 4 folds. For a square piece of paper 0.1mm thick, this corresponds to 15cm and 20cm squares respectively. With less giving only three folds for linear and more giving five folds for alternating. Some simple experiments show that this is approximately correct.&lt;/p&gt;&#xA;" OwnerUserId="429" LastEditorUserId="429" LastEditorDisplayName="" LastEditDate="2011-01-18T07:23:09.723" LastActivityDate="2011-01-18T07:23:09.723" />
  <row Id="3094" PostTypeId="2" ParentId="3025" CreationDate="2011-01-16T21:42:14.013" Score="0" ViewCount="" Body="&lt;p&gt;Basically at a point on the spring where y is displacement from the equilibrium condition, you'll get a differential equation $d^2/dt^2 (Y \times density) = -d^2Y/dZ^2 \times k$ the spring constant. (sorry I can't use Latex)! If we postulate that solutions look like $e^{ikZ +i\omega t}$, $ 2\pi\omega$ will be the frequency. Plug in $\omega$ or $k$ and you can solve for the other one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then at $Z$=height of zero the boundary condition is displacement =0, this will imply that the spatial part of the solution looks like $\sin{(kx)}$. At the top of the spring $dY/dZ$ must be zero, else there would be an unbalanced force, that means that $kZ$ must be an odd multiple of $\pi/2$. The $\omega$ values that satisfy these conditions are your eigenvalues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next you need to discover the amplitudes for the infinately number of modes excited. i.E.&#xA;your initial displacement as a function of Z is proportional to Z, and you must find Ai such that&#xA;  $$Z=\sum_{i= odd N} Ai\cos{(ikZ)} (k*.1 = \pi/2)$$.&#xA;  You may need to lookup Fourier analysis to do this, but you should get a simple formula&#xA;for all the Ai. The solution at any future time will be the sum of these (note each term as its own frequency dependence). You should discover that when time is $\pi$ times the lowest eigenvalue, you've reversed the value of F everywhere, so that will be your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Undergrad physics should teach solving the wave equation, and show you how to apply these methods.&lt;/p&gt;&#xA;" OwnerUserId="391" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-16T21:55:17.693" LastActivityDate="2011-01-16T21:55:17.693" CommentCount="9" />
  <row Id="3095" PostTypeId="2" ParentId="3088" CreationDate="2011-01-16T21:47:59.233" Score="1" ViewCount="" Body="&lt;p&gt;The neutron cross-section is a function of energy &lt;em&gt;in the CoM frame for the pair&lt;/em&gt;:&#xA;$$ \sigma = \sigma(E) ,$$ &#xA;so unless the neutrons are boosted to match, they will be a different relative energy, and &lt;strong&gt;yes&lt;/strong&gt; the cross-section changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you haven't learned anything interesting by doing that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the question concerns neutrons that &lt;em&gt;have&lt;/em&gt; been boosted to move more-or-less in coincidence with the Uranium, then the principle of relativity tells us that &lt;strong&gt;no&lt;/strong&gt; the cross-section will not differ from that measured at rest in the lab frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, you haven't learned much.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-01-16T21:47:59.233" />
  <row Id="3096" PostTypeId="1" CreationDate="2011-01-16T21:51:39.273" Score="8" ViewCount="607" Body="&lt;p&gt;The theory of Divergent Series was developed by Hardy and other mathematicians in the first half of the past century, giving rigorous methods of summation to get unique and consistent results from divergent series. Or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In physics, it is said that the pertubative expansion for the calculation of QFT scattering amplitudes is a divergent series and that its summation is solved via the renormalisation group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some explicit connection between the mathematical theory and the physics formalism?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the question can have different answers for two different layers: the &quot;renormalized series&quot;, which can be still divergent, and the structure of counter-terms doing the summation at a given order. If so, please make clear what layer are you addressing in the answer. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1335" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:29:39.397" LastActivityDate="2011-07-10T22:43:55.607" Title="What is the relation between renormalization in physics and divergent series in mathematics?" Tags="&lt;mathematical-physics&gt;&lt;renormalization&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="5" />
  <row Id="3097" PostTypeId="2" ParentId="3088" CreationDate="2011-01-16T21:53:45.683" Score="0" ViewCount="" Body="&lt;p&gt;what clearly matters is the relative speed of the ions and the neutrons that may be absorbed; you won't change anything fundamental about the reaction if both particles are moving by the same speed, by the principle of relativity. For reasonably small energies, this is usually parameterized by the relative velocity $v$. See a graph how the cross section depends on $v$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.tpub.com/content/doe/h1019v1/css/h1019v1_113.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.tpub.com/content/doe/h1019v1/css/h1019v1_113.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite generally, the cross section decreases with $v$. For neutron kinetic energies below 1 eV (in the nucleus' rest frame), the cross section decreases as $1/v$ - it's the $1/v$ region. Then you have resonance peaks for neutron energy between $1$ and $100$ eV, before the cross section returns to a decrease that is getting slower and slower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph is only drawn up to neutron energies of order $1$ MeV. You don't really want to go to GeVs or even TeVs because the kinetic energy will then be large enough to create pions and the process itself will be dominated by the usual &quot;collider-like&quot; inelastic collisions rather than a simple absorption which is only an important process for relatively small relative velocities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course that if you accelerate neutrons against ions so that the center-of-mass energy will be in hundreds of GeVs or TeVs, then you will have a standard hadron collider and the results won't be too different from those at the LHC. After all, protons and neutrons don't differ too much. Only a tiny fraction of events will look like a smooth &quot;nuclear physics&quot; absorption. At very high energies and relativistic speeds, it is simply inappropriate to think about the processes in terms of nuclear physics - neutrons and ions - which is just an effective theory for small energies (and velocities). You should think in terms of subnuclear i.e. particle physics - it's a collision of high-energy quarks and gluons, just like at the LHC. That's why the science about quarks etc. belongs to something called the high-energy physics: you need it when the energies are high! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nuclear physics - such as neutron absorption - is pretty much a low-energy physics even though it is 1 million times higher an energy than the truly low-energy physics which contains atomic physics, chemistry, and biology. The low-energy physics concepts are only good enough approximations to describe the world when the energies are low enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-16T21:53:45.683" />
  <row Id="3098" PostTypeId="1" AcceptedAnswerId="3100" CreationDate="2011-01-16T21:56:27.677" Score="1" ViewCount="234" Body="&lt;p&gt;Decoherence times can be estimated and are inverse functions of mass.  Since there are no upper bounds on mass, can decoherence time be shorter than Planck time?&lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="1383" LastEditorDisplayName="" LastEditDate="2011-01-16T22:17:05.047" LastActivityDate="2011-01-16T22:22:35.227" Title="Can decoherence time be shorter than Planck time?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="3099" PostTypeId="2" ParentId="3096" CreationDate="2011-01-16T22:08:50.157" Score="0" ViewCount="" Body="&lt;p&gt;There are overlaps between QFT renormalization problems and mathematical approaches to divergent series summations (see for instance the section 'physics' in &lt;a href=&quot;http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_&lt;/a&gt;·&lt;em&gt;·&lt;/em&gt;· ).&#xA;However, most of the infinities encountered in modern physics are 'harder' than those that can be attacked by existing divergent series summation methods.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-01-16T22:08:50.157" />
  <row Id="3100" PostTypeId="2" ParentId="3098" CreationDate="2011-01-16T22:11:04.320" Score="1" ViewCount="" Body="&lt;p&gt;of course, the decoherence time may be - and quite often is - much shorter than the Planck time. There exists no law saying that &quot;all quantities with the units of seconds have to be longer than the Planck time&quot;. The only genuine limitation is that &quot;if you try to measure the timing of an event with the precision better than the Planck time, you will fail.&quot; This is an entirely different thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The decoherence time is the (approximate) time after which the off-diagonal elements of the density matrix between two macroscopically &quot;different&quot; states of a sufficiently macroscopic system decrease to a negligible fraction of its initial value. The processes that drive decoherence - that make the off-diagonal elements drop - are results of interactions of the macroscopic object with the environment. The more strongly the object interacts with the environment and the higher number of environmental degrees of freedom we are forgetting - i.e. tracing over - the more quickly the decoherence proceeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course that if you talk about Schrödinger's cat that is emitting lots of photons and other things, the decoherence time - for the &quot;dead&quot; and &quot;alive&quot; states - will be tiny. If it is not shorter than the Planck length, it's surely easy to consider just slightly bigger systems for which it will be shorter than the Planck time. The statement that it is shorter than the Planck time doesn't mean that you have been actually able to &quot;see&quot; events in spacetime with a trans-Planckian resolution, so it's no violation of the laws of quantum gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much more generally, there is a lot of confusion among the laymen about the statement that &quot;distances and times can't be shorter than the Planck scale&quot;. We must refine what distances and times we're talking about. Obviously, the statement may only hold for Lorentz-invariant quantities whose units are either seconds or meters. Not all &quot;lengths&quot; and &quot;times&quot; are equal or the same thing. It depends &quot;length of what&quot; or &quot;time of what&quot; you're talking about. Lengths of different things and times of different things have different properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only when the proper distances between some important points or events decrease beneath the Planck length, one may possibly face some contradictions with quantum gravity - or one produces a black hole that damages the experiment. But for example, the wavelength of an electromagnetic wave (which is the same thing as its periodicity in time, multiplied by $c$) can be much shorter than the Planck length. Arbitrarily short-wavelength photons exist; this is guaranteed by the theory of relativity. An observer may be accelerated to a speed that approaches the speed of light arbitrarily closely, and when it's done, the energy of the original photon may increase arbitrarily high - and the wavelength may become arbitrarily short by the Lorentz contraction. Nothing new is happening to such a photon: it's the same photon that had a low frequency in another inertial system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can't exist any restriction on the wavelength of waves (of photons etc.). But when proper distances inside an object (formally) drop beneath the Planck length, you may be sure that the laws of geometry have to be severely modified, to say the least; one should use something else than ordinary geometry to describe what is going on and where it is going on. Similarly for times. However, the density matrix linked to a macroscopic body is not a &quot;real object&quot;: it is just a set of numbers that can be used to predict probabilities of various observations of the macroscopic body. Such things can change arbitrarily quickly. And indeed, the more massive body you consider, the shorter the decoherence time will be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(When people misleadingly talk about the &quot;collapse of the wave function&quot;, this &quot;process&quot; may also occur even infinitely quickly. It's no violation of relativity because the wave function is not a &quot;real object&quot;: it is a set of numbers encoding the probabilities - and phases - of things we can predict. When one carefully checks whether some genuine information may be transmitted faster than light between two distant EPR entangled particles, the answer is a resounding No. This fact - causality - may be proved easily in quantum field theory. So the EPR particles won't allow you to speed up the communication with other galaxies.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-16T22:22:35.227" LastActivityDate="2011-01-16T22:22:35.227" />
  <row Id="3101" PostTypeId="1" AcceptedAnswerId="3105" CreationDate="2011-01-16T22:19:13.140" Score="0" ViewCount="107" Body="&lt;p&gt;During discussion on another question, the question of singularties in modern physical theories arose.  The big-bang is an obvious singularity in modern conceptions of the cosmos. Is it a goal of modern theories to find a way to remove the big bang singularity from theory, and how is that accomplished in theories being proposed today?&lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:31:05.510" LastActivityDate="2011-02-04T07:31:05.510" Title="Is it a goal of modern physical theory to avoid big bang singularities and how do they approach the problem?" Tags="&lt;cosmology&gt;&lt;big-bang&gt;" AnswerCount="1" />
  <row Id="3102" PostTypeId="2" ParentId="3098" CreationDate="2011-01-16T22:19:44.183" Score="0" ViewCount="" Body="&lt;p&gt;We should not claim that we understand Planck scale physics enough to answer this question. We can't even claim &quot;there are no upper bounds on mass&quot;. Sure, one can gather a lot of mass, but the mass (energy) associated with a degree of freedom (whatever that is in Planck scale physics) could very well be delimited.&#xA;In short: we can all speculate, but no-one can can answer this question.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-01-16T22:19:44.183" CommentCount="3" />
  <row Id="3103" PostTypeId="2" ParentId="3096" CreationDate="2011-01-16T22:35:33.917" Score="7" ViewCount="" Body="&lt;p&gt;You are conflating three conceptually different categories of &quot;regularizations&quot; of seemingly divergent series (and integrals).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The type of resummations that Hardy would talk about are similar to the zeta-function regularization - the example that is most familiar to the physicists. For example,&#xA;$$S=\sum_{n=1}^\infty n= -\frac{1}{12}$$&#xA;is the most famous sum. Note that this result is unique; it is a well-defined number. In particular, that allows one to calculate the critical dimension of bosonic string theory from $(D-2)S/2+1=0$ and the result is $D=26$. Fundamentally speaking, there is no real divergence in the sum. The &quot;divergent pieces&quot; may be subtracted &quot;completely&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the usual cases of renormalization - of a loop diagram - in quantum field theory, there are divergences. Renormalization removes the &quot;infinite part&quot; of these terms. A finite term is left but the magnitude of the term is &lt;em&gt;not&lt;/em&gt; uniquely determined, like it was in the case of the sum of positive integers. Instead, every type of a divergence in a loop diagram produces one parameter - analogous to the coupling constant - that has to be adjusted. Because the finite results can be &quot;anything&quot;, this is clearly something else than the zeta-regularization and, more generally, Hardy's procedures whose very goal was to produce unique, well-defined results for seemingly divergent expressions. Infinitesimally speaking, the Renormalization Group only mixes the lower-order contributions (by the number of loops) into a higher-order contribution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So these are two different things that one should distinguish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another category of problems that is different from both categories above: the summation of the perturbative expansions to all orders. It can be demonstrated that in almost all fields theories - and perturbative string theories as well - the perturbative expansions diverge. For a small coupling, one can sum them up to the smallest term, before the factorial-like coefficient begins to increase the terms again, despite the $g^{2L}$ suppression. The smallest term is of the same order as the leading non-perturbative contributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the very end, if the theory can be non-perturbatively well-defined - and both QCD-like theories and string theory can, at least in principle - the full function as a function of the coupling constant $g$ exists. But it just can't be fully obtained from the perturbative expansion. The Renormalization Group won't really help you because it only mixes the perturbative terms of another order to a perturbative diagram you want to calculate. If you don't know the non-perturbative physics, the equations of the Renormalization Group won't fill the gap because they will keep you in the perturbative realm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I have sketched three different things: in the Hardy/zeta problems, the answer to the divergent series was unique; in the particular $L$-loop diagrams in QFT, it wasn't unique but the infinite part was subtracted and the finite part was obtained by a comparison with the experiments; and in the perturbative expansion resummed to all orders, the sum actually didn't converge and indeed, it didn't know about all the information about the full result for a finite $g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last statement may have some subtleties; at least for some theories, the non-perturbative physics is fully determined by the perturbative physics. But I think it is not quite general and we have counterexamples - e.g. for AdS/CFT with orthogonal groups and different discrete values of $B$ etc. So it means that the perturbative expansion doesn't uniquely determine the theory non-perturbatively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the three examples differ at the level of &quot;what can be calculated&quot; and &quot;what cannot&quot;, they are different.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:29:21.300" LastActivityDate="2011-02-04T07:29:21.300" CommentCount="4" />
  <row Id="3104" PostTypeId="1" CreationDate="2011-01-16T22:49:47.977" Score="3" ViewCount="187" Body="&lt;p&gt;I'm having a hard time to find a good (and modern) classification scheme for phase transitions and related universality classes. Can someone recommend a paper/book/site? Detailed mathematical aspects are very much welcome.&lt;/p&gt;&#xA;" OwnerUserId="1397" LastActivityDate="2011-03-02T04:13:03.873" Title="Where can I find a good classification for phase transitions?" Tags="&lt;statistical-mechanics&gt;&lt;phase-transition&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="3105" PostTypeId="2" ParentId="3101" CreationDate="2011-01-16T22:57:41.670" Score="2" ViewCount="" Body="&lt;p&gt;I think it is not quite true that modern physics tries to &quot;avoid Big Bang singularities&quot;. Among many other things, modern physics tries to determine what is happening in the vicinity of the region that the Big Bang cosmology views as a singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By saying that the Big Bang singularity is completely avoided, you are already presupposing an answer. This possible answer may be a &quot;working hypothesis&quot; but it may also be incorrect. So it is not a &quot;goal of modern physics&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classical general relativity breaks down - and becomes unpredictive - in the presence of singularities. But that doesn't mean that the full theory that also knows everything about the short-distance physics has to avoid the singularities completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In string theory, the Big Bang singularity hasn't been understood yet - at least to the satisfaction of most string theorists. But many other singularities have been fully understood and it is not true that all of them disappeared. Some of them didn't disappear but the physics around them began well-defined, anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, time-like singularities - such as orbifold singularities; orientifold singularities; and conifold points - have been pretty much fully understood. There are new degrees of freedom and new phenomena that take place in their vicinity but in some sense, physics understands them as well as it understands the smooth space today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From some viewpoint - e.g. according to some &quot;probes&quot; (objects whose reactions we use to evaluate what's happening in the region) - the singularity may get regulated, replaced by a regular manifold with some typical length scales. Other probes see the singularity replaced by a non-commutative geometry that is also regulated. But some singularities according to some probes still look singular; the best geometrical description is still a geometry that has a full-fledged singularity at the original point. However, it is not necessarily a problem. Despite the singularity, physics at some (not quite) manifolds such as the conifold may be shown to be completely equivalent to physics at a completely smooth manifold (e.g. by mirror symmetry).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equivalence shows that physics at singular manifolds may be well-defined and predictive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The space-like singularities - like the singularity inside the Schwarzschild black hole and the Big Bang singularity - remain confusing. Some people even think that all questions involving the interior of a black hole or the very beginning of the Universe are inevitably ill-defined, at least to some extent, so there will never be a set of sharp observables that can be exactly calculated and discussed. I am personally not sure whether it is the case: it could be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course, the working hypotheses what's happening near the Big Bang - probably before inflation - also include some models with bounces; cyclic Universes; non-commutative geometry; a beginning of the Universe from &quot;zero&quot; that looks totally smooth after the Wick rotation to the Euclidean space (the Hartle-Hawking state); but also an abrupt tunneling into another Universe, and many other things. The collection of possibilities is pretty rich and some of them are more motivated than others. Of course, physics still doesn't know for sure which of the tools are truly relevant for the birth of the Universe.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:30:31.937" LastActivityDate="2011-02-04T07:30:31.937" />
  <row Id="3107" PostTypeId="1" AcceptedAnswerId="3112" CreationDate="2011-01-16T23:08:09.520" Score="5" ViewCount="188" Body="&lt;p&gt;Is it correct to visualize operators existing as matrices parameterized by spacetime coordinates in the context of QFT?&lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="1383" LastEditorDisplayName="" LastEditDate="2011-01-16T23:18:21.030" LastActivityDate="2011-02-04T07:31:45.570" Title="When one thinks of a field of operators in QFT, is it reasonable to think of a matrix being associated with each point in space time?" Tags="&lt;quantum-field-theory&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="3108" PostTypeId="2" ParentId="1847" CreationDate="2011-01-16T23:14:29.277" Score="5" ViewCount="" Body="&lt;p&gt;I think you are looking for things in the line of&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;R Haag &lt;em&gt;Local Quantum Physics: Fields, Particles, Algebras&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;R. F. Streater and A. S. Wightman, &lt;em&gt;PCT, Spin and Statistics, and All That&lt;/em&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;are you? Still, Zee's is a good idea, and some 1960 book to bridge the gap (EDIT: I was thinking some practical &quot;lets calculate&quot; book, as Bjorken Drell volumes, but from the point of view of fundamentals, also 1965 Jost &lt;em&gt;The general theory of quantized fields&lt;/em&gt; could be of interest for the OP). &lt;/p&gt;&#xA;" OwnerUserId="1335" LastEditorUserId="1335" LastEditorDisplayName="" LastEditDate="2011-01-17T01:54:59.153" LastActivityDate="2011-01-17T01:54:59.153" />
  <row Id="3109" PostTypeId="2" ParentId="3107" CreationDate="2011-01-16T23:21:39.043" Score="1" ViewCount="" Body="&lt;p&gt;Given that a quantum field is a field of operators and given that you can think of operators as (possibly infinite) matrices acting on a Hilbert (or, better, Fock) space, your assumption is correct. However, I wouldn't call it &quot;visualizing&quot; as I find operators more tangible than Matrices.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-01-16T23:21:39.043" />
  <row Id="3111" PostTypeId="2" ParentId="2596" CreationDate="2011-01-16T23:26:35.967" Score="1" ViewCount="" Body="&lt;p&gt;Marek wrote: &quot;First, the full notion of renormalization group, as studied in QFT, is definitely not needed in the classical theory....&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Marek, the renormalization of mass first appeared in Classical Electromagnetism, didn't it? Take the &quot;definition&quot;: $m_{physical} =  m_{bare} + \delta m$. This is one constraint for two addenda so there is a one-parametric &quot;invariance&quot; group even in the CED. As soon as the mass renormalization (discarding $\delta m$ to originally physical $m $) is done exactly, only once, it is not really interesting, to say the least. In QED this &quot;liberty&quot; is extended to the charge and is done perturbatively, but the main sense remains - the renorm-group is a &quot;liberty&quot; in choosing the two terms to satisfy one constraint: $e_{physical} = e_{bare}(\Lambda) + \delta e(\Lambda)$ where $(\Lambda)$ is a cutoff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we return to the original sense of renormalizations as to discarding unnecessary perturbative corrections to originally right, physical, fundamental constant values, no group appears, no stupid relationship between &quot;bare&quot; and &quot;physical&quot; constants are derived (no Landau pole), and everything is simple: one discards infinite or finite contributions of self-action. Self-action is an erroneous concept: it does not lead to any change (no action by definition), only to wrong terms to be discarded in the end. &lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-06-09T19:01:17.173" LastActivityDate="2011-06-09T19:01:17.173" CommentCount="4" />
  <row Id="3112" PostTypeId="2" ParentId="3107" CreationDate="2011-01-16T23:28:43.293" Score="6" ViewCount="" Body="&lt;p&gt;If you mean that QFT operators are matrices $M_{ij}$ whose indices $i,j$ are really points in space (or spacetime), so that the operator is really represented by the function $M(x,x&amp;#39;)$, then the answer is No. The actual matrices corresponding to QFT operators are much much larger than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An object expressed by $M(x,x&amp;#39;)$ is pretty much equivalent to operators in ordinary quantum mechanics of a single particle:&#xA;$$M(x,x&amp;#39;) = \langle x| \hat{M} |x&amp;#39;\rangle$$&#xA;The function $M(x,x&amp;#39;)$ knows everything about the operator $\hat{M}$ because we have evaluated all matrix elements of this operator with respect to a basis - in this case, the position eigenvector basis. (It is just space, not spacetime.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, quantum field theory has a much larger Hilbert space. Instead of the simple functions above, you may imagine that an operator is expressed by the following functional:&#xA;$$M[\phi(x,y,z),\phi&amp;#39;(x,y,z)] = \langle \phi(x,y,z)|\hat{M}| \phi&amp;#39;(x,y,z)\rangle$$&#xA;Note that the object on the left hand side is a functional - it is a function whose two arguments are functions of 3 variables themselves - field configurations of a Klein-Gordon field, in this case. Again, the left hand side knows everything about the operator $\hat{M}$ that acts on the Hilbert space of the Klein-Gordon quantum field theory. The formula above chose a specific basis - a &quot;truly&quot; continuous basis of field configurations of $\phi(x,y,z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this prescription may make the operators of QFT look more complicated than they really are. Any operator in any quantum mechanical theory may be specified by its matrix elements with respect to &lt;em&gt;any&lt;/em&gt; basis. And there are usually more intuitive bases. In particular, in free quantum field theories, a simple-to-imagine basis is the Fock space basis.&#xA;$$|0\rangle, a^\dagger_{i}|0\rangle, a^\dagger_{i} a^\dagger_{j}|0\rangle, \dots$$&#xA;It's the vacuum and all of its excitations by an arbitrary number of creation operators that add particles into any state. In this way, the Hilbert space is represented as an infinite-dimensional harmonic oscillator. The transformation &quot;matrix&quot; from the continuous basis of the field configurations to the Fock space basis may be found by copying the same transformation for a normal harmonic oscillator infinitely many times and taking the tensor product. (You don't have to mechanically repeat the same work infinitely many times, so this prescription may sound more intimidating than it is.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you meant that the quantum fields $\hat{F}(x,y,z)$ are operators and for each $x,y,z$, then the answer is almost Yes. They're operator distributions - that generalize operators much like the distributions such as the delta-function generalize the concept of a function. However, they're damn large matrices. They're matrices expressed with respect to a basis - for example one of the two bases above. But if this paragraph captures what you meant, then you were just asking whether operators are matrices. Indeed, at least morally, they are. But this fact is true in any quantum mechanical theory and isn't specific to QFT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An operator contains a lot of numbers and may always be represented as a matrix. However, the values of the matrix elements always depend on the basis you chose - sometimes in ways that can't be recognized quickly. However, there is something physical about each operator that doesn't depend on any basis: you may &quot;feel&quot; an operator. You may learn many of its properties. Some of them are more easily understood in one basis; other properties naturally lead to another basis. So it's useful to think about operators in a more general way than matrices in a particular fixed basis - because the latter viewpoint usually discourages one from gaining the insights that are more obvious in another basis. One shouldn't forget that the transition from one basis to another is just a &quot;trivial&quot; linear algebra that doesn't contain any special &quot;physics&quot; knowledge.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:31:45.570" LastActivityDate="2011-02-04T07:31:45.570" CommentCount="1" />
  <row Id="3113" PostTypeId="1" AcceptedAnswerId="3114" CreationDate="2011-01-16T23:42:27.563" Score="2" ViewCount="172" Body="&lt;p&gt;I am having this silly confusion! Suppose I have a system (a hamiltonian or an action say) and also suppose I have a perturbation parameter present (say only one in sight) in there, using which I can find out some kind of perturbative solution of that problem. Suppose now we also found out a non perturbative solution to that problem (by non-perturbative, I have in mind that the solution is not an expansion using the perturbation parameter) (well, origin of this question is that in string theory I have alpha^\prime and also g_s as perturbation parameter and then we also see D- branes as its non perturbative d.o.f. But I am asking here as a general scenario). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is it guranteed (I thought its not) that you have ``solved'' the problem once you know that particular perturbative solution and non-perturbative solution? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me rephrase: Unless I know the exact solution of a problem and the only way to solve that problem to us is such perturbative, non perturbative methods, how can we ever be sure that we've solved the problem and there are no more solutions? (probably it will vary from problem to problem, but can there be a problem with such ambiguity?) &lt;/p&gt;&#xA;" OwnerUserId="1349" LastActivityDate="2011-01-16T23:57:15.723" Title="perturbative solution+nonperturbative solution=full solution?" Tags="&lt;quantum-field-theory&gt;" AnswerCount="1" />
  <row Id="3114" PostTypeId="2" ParentId="3113" CreationDate="2011-01-16T23:52:05.730" Score="3" ViewCount="" Body="&lt;p&gt;I think that there may be some confusion about the term &quot;non-perturbative solution&quot; in the question. Perturbative methods are methods based on Taylor expansions in a small parameter, typically a coupling constant. These methods fail to capture some phenomena - e.g. instantons - which are non-perturbative phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-perturbative methods are any methods that don't rely on the perturbative expansions. They can usually produce insights or terms that cannot be extracted by perturbative methods themselves. The lattice QCD calculations or interactions induced by instantons are examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you say that you have a &quot;non-perturbative solution&quot;, the only thing that it can mean in my understanding of the words is that you simply have the exact solution. The adjective &quot;non-perturbative&quot; means that the solution is not just a &quot;perturbative approximation&quot;. That really means that it's exact, doesn't it? Assuming that it deserves to be called a solution at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you have a non-perturbative solution, then you have the exact solution, and you don't even need any additional perturbative solution; a perturbative solution may be obtained by Taylor-expanding the exact one, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I think that the question about the uniqueness is a completely different question. Some equations or conditions have a single solution; others may have many solutions. That's true perturbatively; and that's true exactly. Sometimes  people know all the solutions; sometimes they may be missing some of them. Two different perturbative solutions - unless they're equivalent by some transformation i.e. a change of the renormalization scheme - are approximations of two different exact solutions. The opposite is not quite true; two different exact solutions may accidentally reduce to the same perturbative approximation when Taylor-expanded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I may have misunderstood your question. Maybe by &quot;non-perturbative solutions&quot;, you meant some strong coupling expansions from the opposite side? This is not the same thing as &quot;non-perturbative&quot;. If one has two expansions from the opposite sides and they are consistent with each other, as in the S-dualities in field theories and string theory, it is a strong hint that the theory is OK everywhere in between but I would agree with you - if you wanted to say it - that it is not quite a proof. In supersymmetric theories, many important quantities may be calculated exactly anywhere in between and the function agrees with expansions from both sides. It's an even stronger circumstantial evidence that the theory is OK and solutions exist in the middle - however, it's not a full proof and it doesn't tell us what the solutions to any questions are &quot;in between&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers, LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-16T23:57:15.723" LastActivityDate="2011-01-16T23:57:15.723" CommentCount="1" />
  <row Id="3115" PostTypeId="2" ParentId="3107" CreationDate="2011-01-16T23:57:07.410" Score="1" ViewCount="" Body="&lt;p&gt;I'm not quite sure if you're asking the same question in the title of your post as you are in the body of your post. But the answer to the question in your title is yes, you can think of an operator field as having a matrix associated with each point in space. That's what a field is, after all: a mapping that associates some value (in this case, an operator, which can be represented by a matrix) with each point in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, keep in mind that there's a subtle difference between an operator and a matrix. An operator is an abstract entity that has some defined action on quantum states. There are various ways to describe the action of that operator, and giving the matrix representation in a certain basis is just one possible way. For example, in the basis $\{|1\rangle, |2\rangle, |3\rangle, \cdots\}$, the particle creation operator can be represented as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^\dagger = \begin{pmatrix}0 \\ 1 &amp;amp; 0 \\ &amp;amp; \sqrt{2} &amp;amp; 0 \\ &amp;amp; &amp;amp; \sqrt{3} &amp;amp; \ddots\end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but the same operator can also be represented by its action on a generic basis state,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;among other ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is, the operator itself is something more than any of these representations. So don't get stuck thinking that one operator corresponds to just one particular matrix. If you have an operator field, then sure, you can choose a particular basis and use that basis to create a matrix out of the operator at each spacetime point, and then you'll have a matrix at each point, like you were talking about. But you could choose a different basis, and have a different matrix representation at each point, but it'd still be the same field. You could even do certain things without picking a basis at all, and in that case you wouldn't have any matrices at all. You'd be working directly with the operators themselves.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-01-16T23:57:07.410" CommentCount="2" />
  <row Id="3117" PostTypeId="1" AcceptedAnswerId="3149" CreationDate="2011-01-17T00:01:56.513" Score="-2" ViewCount="291" Body="&lt;p&gt;(Note: I will let the question stand as is, it has generated a good answer, but the question is too &quot;wishy washy&quot; and needs to be tighter.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the uncertainty principle tells us that the product of the standard deviation of energy and the standard deviation in time must always be greater than h-bar. Since there is a unit element of action, is there a unit element of power as well?  In other words, if action is always present, then must power always be positive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarification: This is not intended to be a question related to people who think that there is &quot;vacuum energy/zero point energy/free energy&quot; that can be harvested.  This is more along the lines of whether the universe can self-perpetuate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarification 2: Self perpetuate I would define as the ability for the universe, through mechanisms of quantum fluctuations related to the uncertainty principle to produce another epoch similar to our own that would appear to be later in time, from our perspective, and after an apparent thermodynamic death of our current universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question restatement: Is quantizing action analagous to there being a source of power at the quantum level?&lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="1383" LastEditorDisplayName="" LastEditDate="2011-01-17T12:31:44.650" LastActivityDate="2011-02-14T15:00:09.843" Title="Can we view quantum fluctuations as a type of quantum power?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="3118" PostTypeId="2" ParentId="2644" CreationDate="2011-01-17T00:36:14.193" Score="0" ViewCount="" Body="&lt;p&gt;Keep reading the pricipia, say the first 20 or 30 pages... What theorems do Newton prove immediately after the ennuntiation of the laws? It is a good guess that the process of proving such theorems led him to depth thinking about the laws and axioms need for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An improvement respect to wikipedia/wikibooks is, as of today, the Newton Project, &lt;a href=&quot;http://www.newtonproject.sussex.ac.uk/&quot; rel=&quot;nofollow&quot;&gt;http://www.newtonproject.sussex.ac.uk/&lt;/a&gt; , where you can check the &quot;diplomatic&quot; versions, the pre-release versions of the texts, with corrections and variations by Newton himself.&lt;/p&gt;&#xA;" OwnerUserId="1335" LastActivityDate="2011-01-17T00:36:14.193" />
  <row Id="3119" PostTypeId="1" CreationDate="2011-01-17T00:37:10.617" Score="0" ViewCount="163" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/2774/rotate-a-long-bar-in-space-and-reach-c&quot;&gt;Rotate a long bar in space and reach c&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Sorry this is very naive, but it's bugging me. If you had a straight solid stick attached on one end and rotating around that attachment at a certain rpm, there would be a length at which the end of the stick would theoretically reach, with that rpm, the speed of light. Well, doesn't seem possible - what specifically would be the limitations that would prevent the end of the stick to reach the speed of light? What would happen?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-17T01:11:51.420" ClosedDate="2011-01-17T01:16:16.703" Title="Reaching speed of light" Tags="&lt;speed-of-light&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="3120" PostTypeId="2" ParentId="31" CreationDate="2011-01-17T00:45:37.377" Score="2" ViewCount="" Body="&lt;p&gt;Special relativity is based upon the idea of how the same events for different observers are located by each observer using their own rulers and clocks; their own measurement of space and time. It's also based upon the principle that the laws of physics don't change for observers travelling at different velocities to one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average lay person already has an intutive understanding of how they think events are seen to occur by observers travelling at different velocities to one another. It comes under the name of Galilean relativity after Galileo first introduced the idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a person driving a car will press the clutch at some time t1, and then press the break pedal a time t2 later at approximately the same location in his space; a zero space interval. On the other hand, someone on the road observng the driver will think these two events occured with a different space interval not equal to zero in his space, but separated by the same time intervale t2-t1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Special relativity instead proposes that the time inverval between events for different observers also changes in a way similar to that for space intervals. It also means that events being simultaneous with one another is relative, as Einstein emphasised in his 1905 paper.&lt;/p&gt;&#xA;" OwnerUserId="1345" LastActivityDate="2011-01-17T00:45:37.377" />
  <row Id="3121" PostTypeId="2" ParentId="3119" CreationDate="2011-01-17T01:02:41.770" Score="1" ViewCount="" Body="&lt;p&gt;In order for the bit of matter of mass $m$ at the very end of the stick to continue moving in a circular path of radius $R$ at a speed approaching the speed of light, it would need to be pulled toward the center with a force whose magnitude is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|F| = |p|\frac{|V|}{R} = \frac{1}{\sqrt{1-(v/c)^2}}\frac{mv^2}{R}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the centripetal force you learn about in introductory physics). That force becomes infinitely large as the speed v approaches the speed of light, very rapidly, and eventually exceeds the strength of any interatomic or intermolecular forces that might be trying to hold the object together.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2011-01-17T01:02:41.770" CommentCount="1" />
  <row Id="3122" PostTypeId="2" ParentId="3119" CreationDate="2011-01-17T01:11:51.420" Score="0" ViewCount="" Body="&lt;p&gt;Something else to consider is that if you begin with a rigid rod rotating about an axis penetrating its center of mass at a constant angular velocity, and then gradually extend the length of the rod by adding mass to the end(s) of the rod, by doing so you are increasing the rod's moment of inertia. The expression for rotational kinetic energy is 1/2*I*omega^2, where I is the moment of inertia of the rod, given a specific choice of axis. For one of the ends to reach the speed of light would require infinite mass to be added to the ends, and hence, an infinite amount of energy. This seems to be a physically unattainable situation. &lt;/p&gt;&#xA;" OwnerUserId="1401" LastActivityDate="2011-01-17T01:11:51.420" />
  <row Id="3123" PostTypeId="2" ParentId="2952" CreationDate="2011-01-17T02:02:44.663" Score="0" ViewCount="" Body="&lt;p&gt;The equivalence between inertial and gravitational masses (the weak equivalence principle) strictly holds only in metric theories of gravitation, such as Einstein's General Relativity. It may be violated e.g. in scalar-tensor theories of gravity (generalized Brans-Dicke theories) provided the scalar component non-universally couples to matter. The Higgs boson actually can play the role of such a scalar component mixed with the scalar graviton, since there is no reason to forbid a coupling of the Higgs field to the gravitational scalar curvature. Therefore, &lt;em&gt;in principle&lt;/em&gt;, the weak equivalence principle is broken. However, with the reasonable Higgs boson - curvature coupling, this violation is practically unobservable, because of smallness of Higgs-Yukawa couplings to the light generation matter and short-range nature of the Higgs exchange force. &lt;/p&gt;&#xA;" OwnerUserId="1402" LastActivityDate="2011-01-17T02:02:44.663" CommentCount="1" />
  <row Id="3124" PostTypeId="1" CreationDate="2011-01-17T02:21:40.513" Score="1" ViewCount="148" Body="&lt;p&gt;Is it possible to make an insulator have two poles like a magnet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, how to make it?&lt;/p&gt;&#xA;" OwnerUserId="652" LastActivityDate="2011-01-17T08:40:36.853" Title="Can an insulator have magnetic poles?" Tags="&lt;magnets&gt;" AnswerCount="2" />
  <row Id="3125" PostTypeId="1" AcceptedAnswerId="3283" CreationDate="2011-01-17T03:18:35.553" Score="7" ViewCount="411" Body="&lt;p&gt;For a spherical mirror, an object at the mirror's center has an image that is also at the center.  Its magnification is $-1$.  For a video showing this, see &lt;a href=&quot;http://www.youtube.com/watch?v=K_p5UoD6ljg&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you stand slightly behind the center of a large spherical mirror and hold out your right hand, as if to shake, you will see a reflection of yourself, upside down, holding out its left hand.  You can &quot;touch&quot; palm-to-palm with the reflection, or pass through it, but your hands won't be the in the &quot;shaking&quot; position.  Instead, the thumbs point opposite directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible, using any combination of mirrors and lenses in the geometric optics limit, to create an image at the same location as the source object, with magnification $+1$, so that you could appear to be shaking your own hand?  As you hold out your right hand, your image would need to appear right-side-up and also hold out its right hand.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a faked picture of what I'm thinking of.  To make it , I photographed my own hand, then used a computer to copy my hand, rotate it 180 degrees, line it up with the original, and &quot;ghost&quot; it out a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Shs89.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no requirement that the image hand track yours if you move around - a static handshake is enough.  I don't necessarily have to be able to view the image directly from where my head is, but it should be a real image so that if I make the air dusty, someone watching my hand from any angle would see the image.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-01-18T22:20:26.063" LastActivityDate="2011-07-15T15:56:54.307" Title="Can you shake your reflection's hand?" Tags="&lt;geometric-optics&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" />
  <row Id="3126" PostTypeId="2" ParentId="3020" CreationDate="2011-01-17T03:52:31.903" Score="-1" ViewCount="" Body="&lt;p&gt;Short speculative answer: I think that the leptons (electron, muon, and tau) mix because they are excitations of the same state. So nature can mix them the same way She can mix different excitations of a hydrogen atom.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-17T03:52:31.903" CommentCount="1" />
  <row Id="3127" PostTypeId="2" ParentId="2949" CreationDate="2011-01-17T04:18:44.740" Score="0" ViewCount="" Body="&lt;p&gt;The reason neutrino oscillations are confusing to those students who think carefully about them is partially because of the history of how the neutrinos were discovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Originally, it was thought that the neutrinos were massless and so the flavor eigenstates were the only states that existed. Then the neutrinos were named electron neutrino $\nu_e$, muon neutrino $\nu_\mu$ and tau neutrino $\nu_\tau$. But these were not the mass eigenstates. We usually call the mass eigenstates $\nu_1$, $\nu_2$, $\nu_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So rather than think of the situation as one involving the transmission of a single neutrino of a known (or unknown) mass, think of the situation as one involving three Feynman diagrams involving three different neutrinos $\nu_1$, $\nu_2$ and $\nu_3$. Each diagram contributes a complex number to the amplitude. By the rules of quantum mechanics, the three diagrams interfere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looked at this way, the mystery of neutrino oscillation becomes simply an interference that you're already familiar with. You'd have had the same type of interference if there were three possible energies of photons emitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a reference to this way of looking at things, see slide 18 and following in Smirnov's presentation:&#xA;&lt;a href=&quot;http://physics.ipm.ac.ir/conferences/lhp06/notes/smirnov1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://physics.ipm.ac.ir/conferences/lhp06/notes/smirnov1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-01-17T04:28:48.457" LastActivityDate="2011-01-17T04:28:48.457" />
  <row Id="3128" PostTypeId="2" ParentId="35" CreationDate="2011-01-17T04:37:47.697" Score="0" ViewCount="" Body="&lt;p&gt;I'll make some guesses about the motivation for the question. I think you're looking for a simple answer. You're choosing magnets because you feel that you understand how electric charges work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, if you bring two positive charges close to one another you will also feel a repulsion. The source of the energy for this repulsion is a place similar to, but a little less complicated than, the source of the energy for the repulsion of two magnets.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-17T04:37:47.697" />
  <row Id="3129" PostTypeId="2" ParentId="1530" CreationDate="2011-01-17T05:50:28.033" Score="3" ViewCount="" Body="&lt;p&gt;The following wikipedia link has a table of all the known isotopes of all the known elements. Gold has Z=79. According to the table, there's only one stable isotope of gold, $\;^{197}Au$ :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Table_of_nuclides_%28complete%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Table_of_nuclides_%28complete%29&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this you can figure out which isotopes decay into gold by considering the three possible decays (ignoring fission) beta decay, inverse beta decay, and alpha decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beta decay means a neutron decays into a proton, an electron and an anti neutrino. The electron and anti neutrino go away. So the result is that the number of neutrons (&quot;n&quot; in the above table) decreases by one while the number of protons (&quot;Z&quot; in the above table) increases by one. The result is that you move in the table one square diagonally towards the top right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inverse beta decay is the opposite of beta decay (more or less). The result is that you move one square diagonally towards the bottom left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alpha decay means the loss of a helium nucleus of 2 protons and 2 neutrons. So you move two squares diagonally to the top left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you want to look for chains that end with gold, do the following. Look at the isotopes 2 squares down and to the right (for alpha decay) or diagonally one square up right or down left (for beta decays). Now take a look at that isotope to see if it decays in the manner you've assumed. For example, $\;^{195}Hg$ is in a position to beta decay to gold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find out whether the isotope decays the way you've assumed, go to this website:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ie.lbl.gov/toi/perchart.htm&quot; rel=&quot;nofollow&quot;&gt;http://ie.lbl.gov/toi/perchart.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and click on the element, then the isotope. Clicking on Hg and then the isotope 195 shows that it does indeed decay to gold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this isotope can decay to gold in something like 46 ways. The different ways involve various excited nuclear states for either the mercury or the gold or both. Such excited states may also decay by emission of a photon (gamma ray).&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-17T05:50:28.033" CommentCount="1" />
  <row Id="3130" PostTypeId="2" ParentId="2926" CreationDate="2011-01-17T07:26:40.477" Score="4" ViewCount="" Body="&lt;p&gt;The modeling used in the GPS is based the static weak field metric&#xA;$$ds^2 = -(1+2\Phi)dt^2 + (1-2\Phi)dS^2,$$&#xA;where $dS^2$ is the Euclidean metric and $\Phi$ is the gravitational potential of the Earth, though only the monopole and quadrupole terms are used. The proper time of a clock is therefore determined by&#xA;$$d\tau = dt\sqrt{1+2\Phi-(1-2\Phi)\frac{dS^2}{dt^2}}\approx\left[1+\Phi-\frac{v^2}{2}\right]dt.$$&#xA;The above metric the Newtonian limit of linearized GTR for a stationary source, which in turn assumes the metric is a perturbation of the usual flat Minkowski one. (For the specific case of a spherically symmetric potential, the metric is also equiv/alent to the Schwarzschild geometry in isotropic coordinates with $Φ = -GM/r$ and the higher-order terms dropped.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of approximating gravity as acceleration found by infinitesimal Lorentz boosts in a flat background is not at all crazy by itself. A fairly canonical example is that of an observer undergoing a constant proper acceleration in the z direction in Minkowski spacetime:&#xA;$$ds^2 = -a^2z^2dt^2 + dS^2,$$&#xA;which can be thought of as performing the same infinitesimal Lorentz boost at each instant of proper time, producing a hyperbolic worldline. The Rindler chart looks a lot like Schwarzschild near the horizon at the equator and the zero azimuthal angle, under the substition $x = 2M(\theta-\pi/2)$, $y = 2M\phi$, $z = 4M(1-2M/r)$:&#xA;$$ds^2 = - \left(\frac{z}{4M}\right)^2dt^2 + \left[1 + {\mathscr{O}}\left(\frac{z}{4M},\frac{x}{2M}\right)^2\right]dS^2,$$&#xA;again after dropping the higher-order terms. But I don't see any obvious of doing so in the case of GPS, at least not without making the situation much more complicated than the very simple approximation given above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But at least that particular case is a bit more literal about &quot;special relativistic approximation to GTR&quot; than a perturbation of the Minkowski metric. How far can we get by assuming that the gravitational field is a field on Minkowski background? Some symmetric rank-2 tensor field $h_{\mu\nu}$ is a good candidate for the job, since it has the same number of local degrees as the metric in GTR, although we can do a scalar, vector, or what-have-you as well. It turns out that scalar and vector fields don't predict the gravitational deflection of light at all, and while $h_{\mu\nu}$ gets the right answer, it gets the precession of Mercury too high by a third. The symmetric rank-2 field also turns out to be formally identical to the usual perturbative method of linearized GTR that assumes $g_{\mu\nu} = \eta_{\mu\nu}+h_{\mu\nu}$, so in that case it's not anything particularly different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, this approach of a field on a Minkowski background is bound to fail eventually, not just because of the pitfalls of approximations, but of a more fundamental incompatibility. For suppose A and B are at different elevations in a static, non-uniform gravitational field, and A sends a monochromatic light pulse at B, of some set number of oscillations (or photons), taking some amount of A's proper time t to send and  B's proper time t' to receive. Due to gravitational redshift, the frequencies must be different, but the number of oscillations must remain the same: ft = f't', so $t\not=t&amp;#39;$. Therefore, gravitational redshift is incompatible with special relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If this is unclear, imagine following up with an identical pulse immediately thereafter, so that the signal trajectories are identical, forming a 'parallelogram' $A_{\text{send1}}B_{\text{receive1}}B_{\text{receive2}}A_{\text{send2}}$. If spacetime is flat, $t = A_{\text{send1}}A_{\text{send2}}$ should be equal to $t&amp;#39; = B_{\text{receive1}}B_{\text{receive2}}$, regardless of whether the signal trajectories are straight, so long as they're congruent. But gravitational redshift forces them to be different. This argument was originally made by A. Schild.)&lt;/p&gt;&#xA;" OwnerUserId="1353" LastActivityDate="2011-01-17T07:26:40.477" />
  <row Id="3132" PostTypeId="2" ParentId="35" CreationDate="2011-01-17T07:36:10.233" Score="1" ViewCount="" Body="&lt;p&gt;When Seth asks &quot;Where does this energy come from?&quot;, perhaps I could rephrase the question to &quot;Where does the force come from?&quot;. I think this is asking us to explain the nature of magnetism - why does it exist and do what it does - rather than describe how it manifests itself. (obviously my answer is not an answer, but a different question on the same topic). Indeed the same question can be asked of gravity ... but that would take us off topic!&lt;/p&gt;&#xA;" OwnerUserId="1320" LastActivityDate="2011-01-17T07:36:10.233" />
  <row Id="3133" PostTypeId="1" AcceptedAnswerId="3137" CreationDate="2011-01-17T07:46:58.593" Score="2" ViewCount="344" Body="&lt;p&gt;When water freezes in a pipe it can crack the pipe open. I assume this takes quite a lot of energy as when I try to crack a pipe it can be hard work!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think water freezing is a result of energy (heat) being lost from the water and out of the pipe into the freezing environment around it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what energy is cracking the pipe and how? When warm and not frozen there is more energy in the pipe than when frozen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My secondary question might be - is this a particular phenomenon of water or would other matter crack open a pipe when it freezes solid from liquid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Andrew&lt;/p&gt;&#xA;" OwnerUserId="1320" LastActivityDate="2011-01-17T14:48:42.333" Title="Why does frozen water burst a pipe?" Tags="&lt;energy&gt;&lt;water&gt;&lt;freezing&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="3134" PostTypeId="1" CreationDate="2011-01-17T07:56:56.050" Score="2" ViewCount="173" Body="&lt;p&gt;As I have understood it, the Standard Model includes particles that carry the different forces, e.g. the electromagnetic (EM) force, the gravitational (G) force. When talking about EM fields such as visible light or microwaves, the associated particle is said to be the photon. But what about a static EM field without any electricity, like a common household magnet? How does that magnet communicate its force? Via photons?&lt;/p&gt;&#xA;" OwnerUserId="1405" LastActivityDate="2011-01-17T08:38:45.937" Title="Is the force carrier of the magnetism in a common household magnet a photon?" Tags="&lt;electromagnetism&gt;&lt;magnets&gt;&lt;carrier-particles&gt;" AnswerCount="1" />
  <row Id="3136" PostTypeId="2" ParentId="3124" CreationDate="2011-01-17T08:18:36.010" Score="0" ViewCount="" Body="&lt;p&gt;you would have to explain what you mean by the poles. Insulators don't inherently create any inherent kind of an &quot;insulator field&quot; around them, so this non-existent field cannot have any poles, either. An insulator is a material that doesn't conduct - heat or electricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing is whether insulators can behave like dipole sources of the known fields such as electric and magnetic fields. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, permanent magnets I know of always have to display ferromagnetism or superconductivity. Ferromagnetic materials are similar to iron; superconducting materials are ... conducting as well. So permanent magnets have to be conductors, I believe. After all, if one is able to change the magnetization, it means that the angular momentum of the electrons or their spin is changed as well. If this is possible in an atom, the electrons may effectively jump to an adjacent atom, too, I think. But even though this proof is not complete, I do think that it's true that permanent magnets have to be conductors just like iron. As far as I know, there is no known counterexample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing the insulators could have are electric dipoles. Yes, insulators may act as electric dipoles, but only in an external electric field: the relative position of electrons in each atom moves a little bit relatively to the nuclei. This effect is known as polarization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They don't stay electric dipoles, however. They are not permanent dipoles. If a material is not a conductor (of electricity), it means that its electrons can't move much, so they're sitting in their molecules and in every molecule, there is an energetically optimal position where they want to sit. So I believe that (good) insulators can't be (good) permanent dipoles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-17T08:35:50.390" LastActivityDate="2011-01-17T08:35:50.390" />
  <row Id="3137" PostTypeId="2" ParentId="3133" CreationDate="2011-01-17T08:36:15.813" Score="6" ViewCount="" Body="&lt;p&gt;First of all, when you say that trying to crack a pipe is hard work, what you probably mean (in physics terms) is that it takes a large force. But that doesn't necessarily mean that it requires a lot of &lt;em&gt;energy&lt;/em&gt;. The energy used in a physical process like that is equal to the force times the distance over which the force is applied, and you don't have to push very far in order to crack a pipe. In fact, the pipe barely moves at all before it cracks, so even though the force required is quite large, it only acts over a tiny distance, and therefore it barely takes any actual energy. What little energy is required can come from the water itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain the &quot;how&quot; you have to consider molecular interactions. (Well you don't &lt;em&gt;have&lt;/em&gt; to, but I'm going to.) The energy of each pair of water molecules varies with the distance between them, in a manner shown (approximately) by the following graph (from &lt;a href=&quot;http://en.wikipedia.org/wiki/File:12-6-Lennard-Jones-Potential.png&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/5/5a/12-6-Lennard-Jones-Potential.png/800px-12-6-Lennard-Jones-Potential.png&quot; alt=&quot;Lennard-Jones potential&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll notice that there is a certain distance at which the energy is a minimum. This distance represents the &quot;natural&quot; equilibrium distance between molecules when there is no pressure. However, when the water is under pressure, the molecules get pushed together (because pressure is roughly akin to force), so their actual distance will be a little closer than the minimum of the graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water has the unique property that its &quot;natural&quot; density at a constant pressure reaches a maximum at a certain temperature, around 4 degrees Celsius, and that its frozen form (ice) is less dense than its liquid form. In other words, the equilibrium intermolecular distance (the minimum of the graph) is smallest at 4 degrees Celsius. If the water temperature is going to drop below 4 degrees Celsius, the minimum shifts a little to the right, which means one of two things has to happen either the water expands (if the intermolecular separation stays with the minimum of the graph), or its pressure rises (if the intermolecular separation creeps up the slope of the graph).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now think about the situation in a pipe. As long as the pipe stays intact, the water can't really expand at all. So the only option is for the pressure to increase. As the pressure increases, the force on the pipe increases as well, and you'll notice that because the slope of the curve is very steep, the force increases very rapidly. At some point, the force becomes large enough to overwhelm the bonds that hold the atoms/molecules in the pipe together, and at that point, the pipe cracks. Notice that in this theoretical model there's no need for any part of the pipe to have moved, which means the pipe could crack without any energy being used. (In practice, there are other things going on that do make it take a tiny bit of energy.)&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-01-17T08:36:15.813" CommentCount="14" />
  <row Id="3138" PostTypeId="2" ParentId="3134" CreationDate="2011-01-17T08:38:45.937" Score="1" ViewCount="" Body="&lt;p&gt;Hello and welcome to the cite.&lt;br&gt;&#xA;Yes -- the carrier for &lt;strong&gt;any&lt;/strong&gt; electromagnetic interaction is a photon. Including ordinary magnets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might ask: &quot;Why then can't I see those photons?&quot; Well, even from classical point view, the photons can have different wavelenghts. And with the magnet you deal with really large wavelenghts, while one cannot &quot;see&quot; radiowaves. While also I think that one cannot even use the concept of &quot;interacion carrier&quot; without working in a context of quantum field theory anyway. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-01-17T08:38:45.937" CommentCount="1" />
  <row Id="3139" PostTypeId="2" ParentId="3124" CreationDate="2011-01-17T08:40:36.853" Score="5" ViewCount="" Body="&lt;p&gt;In principle I can see no reason why an insulator cannot be magnetic, but in practice all pure materials that are ferromagentic seem to be conductors. This includes iron, nickel, cobalt and Gadolinium. Even the known plastic magnets are conductors or semi-conductors ( see &lt;a href=&quot;http://en.wikipedia.org/wiki/Plastic_magnet&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Plastic_magnet&lt;/a&gt; )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnetism can be produced by circulating electric currents round a coil. In this case the material clearly has to be a conductor. In ferrous magnets the magnetic field is produced by aligning the spins of the atoms, so the current that produces the magnetic field is really just the electrons orbiting the nucleus. It will only be an electrical conductor if the electrons are free to move between atoms. The properties of the materials that make it possible to move the atoms into alignment coincide with the properties that make electrical conductivity possible, but I don't think there is any general law of physics that says this has to be the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One solution would be to make a mixture of fine iron filings in a plastic insulator. The iron filings could be magnetised but the plastic would make the material an insulator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutrons have a small magnetic moment so in principle you can line them up in a magnetic field and switch off the field to leave a residual magnetism. Since they are neutral they cannot conduct a current. That may not count as an insulating material in the normal sense.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-17T08:40:36.853" CommentCount="3" />
  <row Id="3140" PostTypeId="2" ParentId="3125" CreationDate="2011-01-17T08:48:39.507" Score="3" ViewCount="" Body="&lt;p&gt;If you place twoflat  mirrors at right angles and look at them from the line that bisects the angle you will see yourself the right way up and with the right hand in the right place. If you stretch your hand out it will be as if you are about to shake hands with yourself. However the hands can not get to the point where they would touch before the mirror gets in the way.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-17T08:48:39.507" CommentCount="5" />
  <row Id="3141" PostTypeId="1" CreationDate="2011-01-17T09:38:49.113" Score="2" ViewCount="265" Body="&lt;p&gt;Can someone explain to me what the &lt;a href=&quot;http://arxiv.org/abs/hep-th/0301240&quot; rel=&quot;nofollow&quot;&gt;KKLT paper&lt;/a&gt; says, and what has and hasn't it achieved regarding the ability to construct solutions with a small positive or negative cosmological constant in string theory?&lt;/p&gt;&#xA;" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-01-17T10:34:34.557" LastActivityDate="2011-01-17T10:34:34.557" Title="Can/has string theory solved cosmological constant problem?" Tags="&lt;cosmology&gt;&lt;string-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="3142" PostTypeId="2" ParentId="3141" CreationDate="2011-01-17T10:09:33.253" Score="4" ViewCount="" Body="&lt;p&gt;Dear gob,&#xA;the KKLT 2003 paper&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0301240&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0301240&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is constructing a large number of stabilized vacua with a negative cosmological constant - anti de Sitter or AdS vacua - and, with a somewhat smaller certainty, many positive-cosmological-constant de Sitter or dS vacua derived from them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The precise paper is nontrivial because of the particular technical context how this is achieved; interpretation will be left to the end. It considers type IIB string theory on Calabi-Yau manifolds with a topology. This topology has typically lots of &quot;cycles&quot; - noncontractible submanifolds - and there can be an integer-valued generalized (NS-NS and/or R-R) magnetic flux through each &quot;cycle&quot;. By taking combinations, one may obtain googols (or powers of googols) of different stable points in the space of configurations of the &quot;string fields&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vacua they obtain at the beginning are AdS vacua. They're supersymmetric and the cosmological constant is negative. KKLT also show that there exist related vacua in which some &quot;antibranes&quot; are added. Those break supersymmetry but are metastable, with lifetimes that vastly exceed the current age of the Universe, so these vacua are potential candidates to correspond to our world. The KKLT computation of the de Sitter vacua's lifetime has come under some scrutiny and physicists differ in their opinion whether the existence of a large number of de Sitter vacua has been established; the situation is much clearer with the unbroken-supersymmetric AdS part of their construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Interpretation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These vacua are stable - that's a good thing because they don't include any exactly massless scalar fields whose value could spontaneously change, thus producing new (unobserved) long-range forces (and allowing the fundamental constants such as the fine-structure constant to change rapidly). The instabilities are nonperturbative - essentially quantum tunneling into other vacua. The precise choice which tunneling directions are favored has been updated by some newer papers, too. It remains a controversial technical question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems very likely that string theory's equations have many solutions - people often say $10^{500}$ although the number is not known at any accuracy - that qualitatively look like the Universe around us. By their sheer number, some of them will have a small value of the cosmological constant, comparable to the observed value. The smallness was only guaranteed by having many sufficiently (seemingly) &quot;random&quot; solutions to choose from. Some of them will have a small cosmological constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Anthropic selection&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A highly disputed question is whether we are living in a randomly chosen Universe or whether there exists a cosmological mechanism or an equation that picks a privileged vacuum (or at least a much smaller subset). As Weinberg argued decades ago, the existence of galaxies - which seems needed for the existence of intelligent beings - may act as a selection criterion because galaxies only arise if the cosmological constant is tiny, comparable to the observed one. According to the anthropic principle, we don't need to explain why the constants of Nature don't take values incompatible with life; if they did, there would be no one who could complain that the hopes for life were doomed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physicists defending the so-called &quot;anthropic principle&quot; usually say that there is no other selection criterion aside from the condition that galaxies and life may arise in the Universe. Moreover, they often assume that all the Universes with equal chances to produce life have the same &quot;prior probability&quot; - which means that we should live in a typical or random or average Universe among those that are compatible with life. As they usually admit, the actual probability distribution on the Universes is unknown and even in principle, we don't know an algorithm to calculate it. Even the most hardcore anthropic physicists know that they don't know whether the chances of a particular vacuum are increased by its longer lifetime, bigger volume, bigger expectation value of planets with life, and other factors. All these things are unknown and arguments favoring one answer over others are, so far, philosophical in character, not scientific.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the KKLT or related construction is right and we live in a rather random Universe selected from the string theory solutions, then string theory solves the cosmological constant problem. String theory surely admits solutions with nonzero values of the cosmological constant. And a sufficient number of candidates that produce a cosmological constant of a random size is predicted and the emergence of life is a sufficient criterion to choose a solution that agrees with the real world. In practice, it could be difficult to locate the right solution if there are googols (or powers of googol) of candidates and ours is pretty much random - and the other parameters of particle physics depend on the properties of the compactification and fluxes somewhat randomly, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other physicists, such as myself, are convinced that a deeper understanding of the structure of the space of vacua - the landscape - and the cosmological mechanisms that may related or comparing different places of the landscape will show that our vacuum is much more special and physics may ultimately become able to determine in which one we live. We're not there yet but it is fair to say that string theory has produced an internally consistent solution that may produce stable vacua with all the qualitative particle species and interactions we know as well as a reasonable value of the cosmological constant. Needless to say, it's the only framework known to modern physics that can offer such a realistic candidate description.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We won't know for some time whether this candidate description is valid. In particular, even if string theory is valid which is almost certainly the case, we won't know for a long time whether our Universe is close to one of the type IIB compactifications or whether it can only be described by another category of string models such as heterotic string theory; heterotic M-theory; type IIA braneworlds; M-theory on $G_2$ holonomy manifolds, or others. A much more detailed comparison of the vacua with the observed reality is needed to settle such questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-17T10:27:16.430" LastActivityDate="2011-01-17T10:27:16.430" />
  <row Id="3143" PostTypeId="2" ParentId="2014" CreationDate="2011-01-17T11:03:31.547" Score="1" ViewCount="" Body="&lt;p&gt;Wick's theorem works for an arbitrary number of states equipped with their creation and annihilation oscillators. In the same way, Wick's theorem for spins applies to a collection of an arbitrary number of spins - or qubits, which is the same thing if their spin is $1/2$. Wick's theorem is just a mathematical identity that works even for - and especially for - a finite number of qubits and/or number of creation and annihilation operators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it can't be the case that the source of the discrepancy hides in Wick's theorem. However, to uncover the reason of the discrepancy, you would probably have to be more specific about what you calculated (the expectation value in what state? Why isn't the original form of the expectation value already good as a &quot;result&quot;? In which form do you want the result?) for others to isolate the mistake or subtlety. Or others could even calculate the right result, avoiding mistakes - but it's not clear from the text what exactly you want them to calculate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Luboš&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-17T11:03:31.547" />
  <row Id="3144" PostTypeId="2" ParentId="3104" CreationDate="2011-01-17T11:12:59.710" Score="0" ViewCount="" Body="&lt;p&gt;I am afraid that your question is too ambitious. It's almost like asking &quot;where I can find a good resource of a theory of everything, including all emergent and complex phenomena in the world?&quot;. The question has many aspects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different physical systems have different numbers of different phases. In principle, there are infinitely many types of physical systems, so there are also infinitely many possible phase diagrams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is good to know a sufficient number of phase transitions - as well as their basic theories such as Landau's theory of second-order phase transitions - but it's hard to claim that one already knows &quot;all of them&quot; at a given moment - which is needed for claiming that they have been fully classified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, it's not always the case that the number of inequivalent phase transitions is increasing. Sometimes, it may be the other way around. For example, the AdS/CFT correspondence has shown that various phase transitions in gravitational theories - such as the Hawking-Page transition - are completely equivalent to some transitions in non-gravitational theories - such as the confinement/deconfinement transition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the insight of the last decade or so. Similar insights are being added even today. The science of phase transitions hasn't been closed. Phase transitions are pretty generic properties of most physical systems - so whenever one learns new things about some physical systems, it's likely that one also learns something about some of their phase transitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on what you really want, you may choose one of the books about phase transitions, e.g. from this list:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.amazon.com/s/ref=nb_ss_b?url=search-alias%3Dstripbooks&amp;amp;field-keywords=Phase+transitions&amp;amp;x=0&amp;amp;y=0&amp;amp;tag=lubosmotlsref-20&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/s/ref=nb_ss_b?url=search-alias%3Dstripbooks&amp;amp;field-keywords=Phase+transitions&amp;amp;x=0&amp;amp;y=0&amp;amp;tag=lubosmotlsref-20&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-17T11:12:59.710" CommentCount="1" />
  <row Id="3145" PostTypeId="1" CreationDate="2011-01-17T11:13:47.237" Score="4" ViewCount="112" Body="&lt;p&gt;In which color human eye is most sensitive ?&lt;/p&gt;&#xA;" OwnerUserId="1410" LastActivityDate="2011-01-18T03:26:21.063" Title="Sensitivity of eye" Tags="&lt;color&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="3146" PostTypeId="2" ParentId="3145" CreationDate="2011-01-17T11:18:24.670" Score="6" ViewCount="" Body="&lt;p&gt;Around 555nm wavelength - green color.&#xA;That's why green lasers are soo cool even at 10 mW :-D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/e/e9/Eyesensitivity.png&quot; rel=&quot;nofollow&quot;&gt;http://upload.wikimedia.org/wikipedia/commons/e/e9/Eyesensitivity.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/e/e9/Eyesensitivity.png&quot; rel=&quot;nofollow&quot;&gt;alt text&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="930" LastEditorDisplayName="" LastEditDate="2011-01-17T16:24:48.640" LastActivityDate="2011-01-17T16:24:48.640" CommentCount="2" />
  <row Id="3147" PostTypeId="2" ParentId="1530" CreationDate="2011-01-17T11:43:23.153" Score="1" ViewCount="" Body="&lt;p&gt;the alchemists have dreamed about the production of gold (Z=82) from some cheap material and lead (Z=79) was their favorite choice. They were just using a wrong science - namely primitive chemistry instead of nuclear physics. But otherwise their choice of lead was OK. And indeed, lead became the element that was transmuted into gold for the first time sometime in 1980 (and maybe even in 1972). See&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemistry.about.com/cs/generalchemistry/a/aa050601a.htm&quot; rel=&quot;nofollow&quot;&gt;http://chemistry.about.com/cs/generalchemistry/a/aa050601a.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One has to remove three protons which costs a lot of energy. Needless to say, the transmutation remains economically unacceptable. That's true for other strategies, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-17T11:43:23.153" CommentCount="3" />
  <row Id="3148" PostTypeId="2" ParentId="3133" CreationDate="2011-01-17T12:04:11.443" Score="1" ViewCount="" Body="&lt;p&gt;An answer to &#xA;Your second question on substances other than water: Yes, there are more examples of such behaviour,  antimony, gallium, Germanium, silicon, bismut and tin .  This happens, when the crystal structure is far from being a good fill of space (eg. diamond structure) and the chemical interactions who cause this, break down when melting occurs. Georg&#xA;PS &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Semantics: &#xA;A pipe will be ruptured by fr*&lt;em&gt;eezing&lt;/em&gt;* water. &#xA;Fr*&lt;em&gt;ozen&lt;/em&gt;* water is no harm.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-01-17T14:48:42.333" LastActivityDate="2011-01-17T14:48:42.333" CommentCount="2" />
  <row Id="3149" PostTypeId="2" ParentId="3117" CreationDate="2011-01-17T12:24:21.883" Score="0" ViewCount="" Body="&lt;p&gt;I don't fully understand what you're asking but let me try to answer, anyway.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The Universe is probably able to reproduce itself indefinitely. The most obvious physics mechanism that seems to coincide with your somewhat unusual terminology is known as &quot;eternal inflation&quot;. During eternal inflation, bubbles - fluctuations - in the spacetime of our Universe may suddenly become seeds of another Universe that may grow and contain billions of new galaxies after some time. In the daughter Universe, the same process may get repeated many times. The &quot;family tree&quot; of these Universes that &quot;self-perpetuate&quot; is known as the multiverse. It's conceivable that such a multiverse exists.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In general relativity, the energy conservation law becomes either invalid or vacuous, see e.g. &lt;a href=&quot;http://physics.stackexchange.com/questions/2838/total-energy-of-the-universe&quot;&gt;this question&lt;/a&gt;. GR allows spacetime to get curved and dynamical, so in general - and especially in cosmology - it doesn't have a time-translational symmetry. However, that symmetry is needed for Emmy Noether's theorem to be able to claim that a conserved energy exists.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Already today, the biggest portion of the energy in the Universe comes from the vacuum energy density, and this energy density is indeed receiving large (in fact, too large) contributions from the zero-point energy of quantum fields that exist - and are nonzero - exactly because of the vacuum fluctuations. These fluctuations contribute to the so-called cosmological constant - more generally known as &quot;dark energy&quot; - and this cosmological constant has a fixed energy density per unit volume. Because the volume can get (and is getting) larger in cosmology according to the general relativity and because the density is remaining constant, the total energy in the Universe is increasing, too.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At the beginning of the expansion of our Universe, when it was very small, there was probably a very similar period of accelerated expansion called the inflation that I already mentioned in point 1. The difference from the current expansion was that it was much faster and the energy density was greater than the current vacuum energy density by many dozens of orders of magnitude. In that era, the total energy of the Universe - carried by the &quot;vacuum energy density&quot; or, more precisely, the inflaton potential energy - increased exponentially, by a factor of $\exp(50)$ or more. When the era ended, this energy was largely converted (through the kinetic energy of the inflaton field) to mass that would become seeds of the galaxies - or &quot;structure&quot; - we inhabit today. Again, the cosmic inflation may also be partly blamed in the vacuum fluctuations which follow from quantum mechanics.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the first point, I mentioned the eternal inflation in which a point in space suddenly goes crazy and decides to produce a whole new Universe. The process of &quot;getting crazy&quot; depends on quantum mechanics as well. The small region of space is getting crazy because of another kind of a quantum fluctuation. So it's a random quantum process - essentially quantum tunneling (analogous to the ability of an object to spontaneously penetrate the wall, something that wouldn't be possible in non-quantum physics) - that is responsible for the emergence of the self-perpetuating multiverse, too.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I don't understand why you think that your question about the &quot;quantization of action&quot; that creates a new &quot;source of power&quot; is equivalent to the original question. What has the action have to do with it - except that it is a tool to define the laws of physics and we're talking about physics? Nevertheless, if by &quot;quantizing the action&quot;, you simply mean going to the quantum theory, it is surely true that the switching from classical physics to quantum physics - by quantizing the action, Hamiltonian, or anything we use to define physics - implies that the character of energy is changing. Energy of some systems may become discrete - that's why quantum mechanics was called quantum mechanics in the first place. Also, there are new terms contributing to energy that didn't exist in classical physics - such as the vacuum zero-point fluctuations. But the whole world is changing if one switches to a completely new - quantum - theory so it shouldn't be shocking that the allowed values and terms contributing to the energy change, too. Quantum mechanics is a completely new theory that changes our description of everything - and on the contrary, one has to use special arguments to show that something &quot;hasn't changed much&quot; (the classical limit of the quantum theory). Also, energy itself becomes an operator called the Hamiltonian whose value (the eigenvalues of the operator) can only be predicted probabilistically. The value measured in a particular copy of an experiment is random. A numerous repetition of the same measurement is needed to verify such predictions. Again, it's true for all predictions of quantum mechanics, not just energy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All the best, LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-17T12:40:48.120" LastActivityDate="2011-01-17T12:40:48.120" CommentCount="1" />
  <row Id="3150" PostTypeId="1" CreationDate="2011-01-17T12:54:24.983" Score="11" ViewCount="944" Body="&lt;p&gt;In late 2003, Edward Witten released a &lt;a href=&quot;http://arxiv.org/abs/hep-th/0312171&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; that revived the interest in Roger Penrose's twistors among particle physicists. The scattering amplitudes of gluons in $N=4$ gauge theory in four dimensions were expressed in a simple way using the twistor variables. Witten also proposed a particular model, the topological B-model on the $CP^{3|4}$ twistor space, to generate all these amplitudes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These methods began their own life but the topological B-model became largely silent, perhaps partly because the phenomenologists who fell in love with these things haven't been trained in string theory, especially not in the topological one. However, many twistor-related discoveries in the recent 3 years - which were made without Witten's constructive picture - lead me to ask whether Witten's theory actually knows about these matters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the &quot;dual superconformal symmetry&quot; was first noticed by &lt;a href=&quot;http://arxiv.org/abs/hep-th/0607160&quot; rel=&quot;nofollow&quot;&gt;Drummond et al.&lt;/a&gt; in 2006 and derived by stringy methods by &lt;a href=&quot;http://arxiv.org/abs/0710.1060&quot; rel=&quot;nofollow&quot;&gt;Alday &amp;amp; Maldacena&lt;/a&gt; in 2008 or so. The 3+1 dimensions on the CFT boundary may be T-dualized to produce another copy of the Yang-Mills theory that is superconformally invariant once again. Scattering amplitudes have been converted to the expectation values of piecewise linear Wilson loops in the dual theory - the segments have the directions and length of the light-like momenta of the scattering particles. My question is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Can you also &quot;T-dualize&quot; Witten's topological B-model to obtain another one in which the scattering amplitudes are computed in a different way?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you think that the answer is Yes, I would also like to know what is the &quot;dual prescription&quot; for the supersymmetric Yang-Mills amplitudes and whether the D1- and D5-branes in Witten's original models are replaced by other D1- and D5-branes or, for example, by D3-branes.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-08T12:29:26.297" Title="Is there a T-dual of Witten's twistor topological string theory?" Tags="&lt;mathematical-physics&gt;&lt;string-theory&gt;&lt;gauge-theory&gt;&lt;twistor&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="3151" PostTypeId="1" CreationDate="2011-01-17T13:18:55.673" Score="2" ViewCount="468" Body="&lt;p&gt;What &quot;breakthrough&quot; from a theoretical point of view is needed for solar energy to become feasible energy alternative?&lt;/p&gt;&#xA;" OwnerUserId="1413" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-13T10:03:21.530" LastActivityDate="2011-07-13T18:20:13.433" Title="What's the problem with solar energy?" Tags="&lt;energy&gt;&lt;electromagnetic-radiation&gt;&lt;renewable-energy&gt;" AnswerCount="6" CommentCount="9" />
  <row Id="3152" PostTypeId="2" ParentId="3151" CreationDate="2011-01-17T13:46:07.320" Score="5" ViewCount="" Body="&lt;p&gt;Although I am no big friend of the solar panels today - because they are subsidized - I think it is correct to say that from a theoretical point of view, we are already &quot;there&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solar panels are typically guaranteed to work for 10 years at their 90% capacity and/or for 25 years at their 80% capacity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The price of a piece of solar panel producing 1 Watt may be as small as USD 1.30 today, see&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ecobusinesslinks.com/solar_panels.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.ecobusinesslinks.com/solar_panels.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is a Watt when it's actually running and in average, it's only running at the maximum capacity for 2 hours a day in average. Clouds, nights, and bad angles are a problem. So a Watt panel only produces 2 Wh per day or 730 Wh per year or 10 kWh per 15 years or so which is the estimated lifetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So assuming no extra expenses when the panel is already installed, and no expenses for their liquidation, you pay the minimum of USD 1.30 per 10 kWh, or USD 0.13 per kWh. That's actually acceptably cheap - a few times cheaper than the electricity sold in many European countries and pretty much identical to the consumer electricity price in the U.S. No profit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, countries such as my homeland introduced lots of subsidies and lots of solar panels were installed that were not yet cheap enough; and they were installed in places with suboptimal inflow of solar radiation. The counting implied that the electricity was five times more expensive than what it was sold for - the difference had to be paid from the subsidies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think it's a purely technological question and we're on the edge of having solar panels that are profitable at the current market prices. Of course, you must also find the place where they're installed - on the arable land, you should also add the rent and/or opportunity costs for the food that could be grown up there. It's a barbarianism to use arable land for solar panels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there's a lot of useless space where the panels could work - e.g. Sahara - and in a couple of years or a decade, the solar panels may become OK economically. It's now a job for the engineers to improve the materials and other things. of course, the main problem for solar panels today remain the fossil fuel (especially coal) competitors that are simply cheaper, so offer the energy companies to guarantee a safe and healthy profit, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I want to say that if we ran out of the fossil fuels in some time, there would be no &quot;physics obstacle&quot; and almost no &quot;economic obstacle&quot; preventing us from switching to electricity that could be produced by solar panels. It's plausible that some totally new kind of a material will replace the materials used today - graphene (a layer of graphite, carbon) seems pretty promising at this moment (most existing solar panels contain silicon); it's conceivable that the evolution will be gradual and less dramatic. At any rate, there's no longer any reason to subsidize the solar panels. Quite on the contrary, the subsidies should be removed everywhere in the world to encourage the engineers to make the last steps needed for profitability.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:33:00.857" LastActivityDate="2011-02-04T07:33:00.857" CommentCount="1" />
  <row Id="3153" PostTypeId="1" AcceptedAnswerId="3154" CreationDate="2011-01-17T13:55:41.333" Score="1" ViewCount="106" Body="&lt;p&gt;Through unit analysis, one can identify the following relationship linking energy, action and power:&#xA;$energy^2 = action \times power$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, we rewrite this expression as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$power = \frac{energy^2}{action}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$action = \frac{energy^2}{power}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In light of this tight relationship, it seems odd that physicists prefer only to discuss energy and action when it comes to quantum field theory.  It would seem that the inverse relationship between action and power would lead to an alternative formulation where the goal is to find extremum of power instead of action.  So why is there very little discussion of power in physics?&lt;/p&gt;&#xA;" OwnerUserId="1383" LastActivityDate="2011-01-17T15:27:15.033" Title="Energy, power and action" Tags="&lt;energy&gt;&lt;effective-action&gt;" AnswerCount="1" />
  <row Id="3154" PostTypeId="2" ParentId="3153" CreationDate="2011-01-17T14:09:10.490" Score="4" ViewCount="" Body="&lt;p&gt;we say that your formulae are &quot;dimensionally correct&quot;, but that's it. They're dimensionally correct for a simple reason: energy is expressed in Joules which is the geometric average of Joule.seconds, the units of action, and Joules per second, the units of power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you can construct an equation that is dimensionally correct, it doesn't yet mean that it is a correct identity - and it is very far from being a demonstrably useful one. I can't imagine any sensible context in which if would be true that &quot;energy squared equals action times power&quot;. Moreover, even if such an identity existed, there could be a numerical coefficient, and it is pretty unlikely that it would be one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you're just playing with units - and yes, the left hand side has the same unit as the right hand side. That's a necessary condition but not a sufficient one for an equation to be OK. For such an equation to make sense, you must actually know &quot;power of what&quot;, &quot;energy of what&quot;, and &quot;action of what&quot; you are inserting into the equation. Otherwise you don't know what you're doing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no important law of &quot;extreme power&quot;. Why there should be one? Physical laws can't be discovered by randomly combining verbs and names of physical quantities. Science is not equivalent to the business of monkeys who randomly type letters and after some time, they inevitably write Hamlet just like Shakespeare. There is a law that systems that may get rid of some energy - or dissipate it into heat - will eventually do it. That means that they will minimize their energy. So energy minimization exists and is important for the description of the state of objects that become static.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there is the law of least action which can be used to derive the differential equations governing all of classical physics. But there's no law of &quot;extreme power&quot;. There's even no context that could be waiting for such a new law. So &quot;extremized power&quot; may be helpful for  consumers of electricity who try to save power, and/or for energy companies who try to maximize their power production. But both groups usually have some other considerations aside from the power itself, too. So your &quot;law&quot; never works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power itself is simply not as fundamental as energy or action, much like the permittivity change per unit viscosity and per unit entropy per day is not. The simple term &quot;power&quot; shouldn't confuse you - linguistic simplicity of some terms doesn't imply anything for physics. The &quot;power&quot; is still some &quot;energy per unit time&quot; which is a derived, rather than fundamental, concept. Energy and action are simply more fundamental and primary than power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: By the way, it's not difficult to produce as many dimensionally correct equations as many you want. In particular, every equation you write down is dimensionally OK in the Planck units because all quantities are dimensionless. Equivalently, one can always add the product of appropriate powers of universal constants - namely  $c,\hbar,G,k_{Boltzmann},\epsilon_0$ - to all terms in an equation to transform a dimensionally wrong equation into a correct one. So the nontriviality or the &quot;degree of shocking surprise&quot; of the fact that you managed to construct a dimensionally correct equation involving energy, action, and power is exactly zero.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-17T15:27:15.033" LastActivityDate="2011-01-17T15:27:15.033" CommentCount="2" />
  <row Id="3155" PostTypeId="2" ParentId="3104" CreationDate="2011-01-17T14:16:46.847" Score="2" ViewCount="" Body="&lt;p&gt;The basic classification of phase transitions is into first and second order transitions. The former is characterized by having a nonzero latent heat, e.g. the liquid-vapor transition of $H_2 O$. Often one can adjust a thermodynamic parameter to construct a line of first order transitions and this line terminates at a critical point where there is a second order transition. Second order transitions are characterized by an infinite correlation length (in field theory terms the mass parameter goes to zero). Critical points are often studied using the tools of Conformal Field Theory. The critical exponents which govern behavior near the critical point depend on the universality class which often depends on the symmetry of the system. The Wikipedia page &lt;a href=&quot;http://en.wikipedia.org/wiki/Phase_transition&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Phase_transition&lt;/a&gt; will give you a quick overview. I recall that the book &quot;Modern Theory of Critical Phenomenon&quot; by S. Ma has more detailed and pedagogical discussion of these matters.  &lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T04:13:03.873" LastActivityDate="2011-03-02T04:13:03.873" CommentCount="3" />
  <row Id="3156" PostTypeId="2" ParentId="3151" CreationDate="2011-01-17T14:33:35.593" Score="1" ViewCount="" Body="&lt;p&gt;I don't want this comment to be taken the wrong way, but I would argue that solar energy is already a successful energy alternative in many applications.  I think in large scale commercial markets in industrialized countries, I would agree that solar still has not reduced its costs to the point where it is a &quot;natural&quot; alternative.  However, solar definitely has viability in niche markets, starting with space and trickling all the way down to certain classes of consumer electronics (solar powered calculators and watches for instance).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long term cost effectiveness of solar is really a question centered around expected price of fuel in the future.  This problem gets tricky due to relationships in supply and demand and prices people are willing to pay for a commodity. This becomes very difficult to estimate, and puts industrial solar continuously on the threshold of profitability.  However, one economic model is to introduce solar to people who are in extreme environments, who can benefit from having access to electricity, but lack access to existing infrastructure.  Although that market is exceedingly small in industrialized countries, global markets are potentially much larger. &lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="1383" LastEditorDisplayName="" LastEditDate="2011-01-17T15:18:17.277" LastActivityDate="2011-01-17T15:18:17.277" CommentCount="1" />
  <row Id="3157" PostTypeId="1" AcceptedAnswerId="3165" CreationDate="2011-01-17T14:42:50.203" Score="2" ViewCount="164" Body="&lt;p&gt;So I'm pretty sure I'm approaching this problem in the wrong way and I need some guidance (my first hint is that I think I'm thinking about a quantum mechanical problem too classically)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there is an isolated molecule in the gas phase with an average cross-sectional area to be exposed to radiation of $A$.  (For my specific problem, the molecule is trapped in a superfluid Helium droplet, but I think the calculation should be roughly the same).  If the radiation source has a flux $f$ (in units of photons/second/square area/0.1% BW) at energy $E$, what is the probability of the molecule absorbing a photon within a given interaction time $t$ if the absorption probability at a given energy is $P(E)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is pretty easy to calculate if I treat the whole problem classically, i.e. like a ball and a target model.  For some reason, though, I get numbers that seem to be way too low if I do this.  I know it has to be more complicated than that, since light is a wave also.  What am I missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that transition probabilities are related to wavefunction overlap, etc.  Also, I should note that the radiation in my specific problem is in the hard x-ray region, though I don't think that should change the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for your help.&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2011-01-17T20:30:17.470" Title="Photon absorption probability for a given molecule in gas phase" Tags="&lt;quantum-mechanics&gt;&lt;spectroscopy&gt;" AnswerCount="3" />
  <row Id="3158" PostTypeId="1" AcceptedAnswerId="3163" CreationDate="2011-01-17T14:57:13.410" Score="11" ViewCount="1222" Body="&lt;p&gt;From everything I've read about quantum mechanics and quantum entanglement phenomena it's unobvious for me, why quantum entanglement is considered to be active link. I.e. it's stated every time that measurement of one particle &lt;em&gt;affects&lt;/em&gt; another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While in my head there is less magic explanation: the entangling measurement affect both particles in the way which makes their states identical, though unknown. In this case measuring one particle will reveal information about state of the other, but without magic &lt;em&gt;instant&lt;/em&gt; modification of remote entangled particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, I'm not the only one who had this idea. What are the problems associated with this view and why &lt;em&gt;magic&lt;/em&gt; view is preferred?&lt;/p&gt;&#xA;" OwnerUserId="1414" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:10:06.767" LastActivityDate="2011-02-05T17:33:32.533" Title="Why quantum entanglement is considered to be active link between particles?" Tags="&lt;quantum-information&gt;&lt;quantum-mechanics&gt;&lt;entanglement&gt;" AnswerCount="5" FavoriteCount="6" />
  <row Id="3160" PostTypeId="2" ParentId="3158" CreationDate="2011-01-17T15:21:50.470" Score="7" ViewCount="" Body="&lt;p&gt;In fact your view is quite official one; entanglement occurs just because both particles are described with one weave-function; the magic is in our classical habit of thinking that separate objects are described with separate &quot;coordinates&quot;.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2011-01-17T15:21:50.470" CommentCount="1" />
  <row Id="3161" PostTypeId="2" ParentId="3157" CreationDate="2011-01-17T15:23:18.833" Score="1" ViewCount="" Body="&lt;p&gt;Well my first guess that you shouldn't consider naive &quot;cross-sectional area&quot; of your molecule. It would be better If you calculated the cross section of absorption/scattering of photons on your molecule. It depends on different properties of your molecule and on the wavelength of the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But maybe your calculations are more or less correct. You said that you result &quot;too low&quot;, but hard x-ray photons are indeed very good at passing through matter. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-01-17T15:23:18.833" />
  <row Id="3162" PostTypeId="2" ParentId="3158" CreationDate="2011-01-17T15:24:26.767" Score="2" ViewCount="" Body="&lt;p&gt;Just a nice analogue a professor told me once: Imagine at home you put one glove in your coat without looking (and noticing it's only one of the two). After exiting the train you notice it's cold and you pull out that single glove. At this very instant you know it's either the left or the right glove, and you therefore know which one is left at home. However, no information was transmitted by your &quot;measurement&quot;. Of course in quantum mechanics this is more complicated because of the not entirely measurable wave function, but this is the basic idea.&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2011-01-17T15:24:26.767" CommentCount="4" />
  <row Id="3163" PostTypeId="2" ParentId="3158" CreationDate="2011-01-17T15:37:30.607" Score="30" ViewCount="" Body="&lt;p&gt;Entanglement is being presented as an &quot;active link&quot; only because most people - including authors of popular (and sometimes even unpopular, using the very words of Sidney Coleman) books and articles - don't understand quantum mechanics. And they don't understand quantum mechanics because they don't want to believe that it is fundamentally correct: they always want to imagine that there is some classical physics beneath all the observations. But there's none.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are absolutely correct that there is nothing active about the connection between the entangled particles. &lt;strong&gt;Entanglement is just a correlation&lt;/strong&gt; - one that can potentially affect all combinations of quantities (that are expressed as operators, so the room for the size and types of correlations is greater than in classical physics). In all cases in the real world, however, the correlation between the particles originated from their common origin - some proximity that existed in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People often say that there is something &quot;active&quot; because they imagine that there exists a real process known as the &quot;collapse of the wave function&quot;. The measurement of one particle in the pair &quot;causes&quot; the wave function to collapse, which &quot;actively&quot; influences the other particle, too. The first observer who measures the first particle manages to &quot;collapse&quot; the other particle, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This picture is, of course, flawed. &lt;strong&gt;The wave function is not a real wave.&lt;/strong&gt; It is just a collection of numbers whose only ability is to predict the probability of a phenomenon that may happen at some point in the future. The wave function remembers all the correlations - because for every combination of measurements of the entangled particles, quantum mechanics predicts some probability. But all these probabilities exist a moment before the measurement, too. When things are measured, one of the outcomes is just realized. To simplify our reasoning, we may forget about the possibilities that will no longer happen because we already know what happened with the first particle. But this step, in which the original overall probabilities for the second particle were replaced by the conditional probabilities that take the known outcome involving the first particle into account, is just a change of our knowledge - not a remote influence of one particle on the other. No information may ever be answered faster than light using entangled particles. Quantum field theory makes it easy to prove that the information cannot spread over spacelike separations - faster than light. An important fact in this reasoning is that the results of the correlated measurements are still random - we can't force the other particle to be measured &quot;up&quot; or &quot;down&quot; (and transmit information in this way) because &lt;strong&gt;we don't have this control even over our own particle&lt;/strong&gt; (not even in principle: there are no hidden variables, the outcome is genuinely random according to the QM-predicted probabilities).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend late Sidney Coleman's excellent lecture Quantum Mechanics In Your Face who discussed this and other conceptual issues of quantum mechanics and the question why people keep on saying silly things about it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2010/11/sidney-coleman-quantum-mechanics-in.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/11/sidney-coleman-quantum-mechanics-in.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-05T17:33:32.533" LastActivityDate="2011-02-05T17:33:32.533" />
  <row Id="3164" PostTypeId="1" CreationDate="2011-01-17T15:48:44.763" Score="8" ViewCount="799" Body="&lt;p&gt;When an atenna transmits radiowaves isn't it true that the electromagnetic pulse is radiated away from the accelerating electron as a spherical wave in all directions simultaneously, and if so how can the associated photon be &quot;everywhere&quot; on this rapidly expanding sphere?&lt;/p&gt;&#xA;" OwnerUserId="1413" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-01-17T16:21:29.770" LastActivityDate="2011-08-04T00:44:41.243" Title="How is a spherical electromagnetic wave emitted from an antenna described in terms of photons?" Tags="&lt;electrodynamics&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="2" />
  <row Id="3165" PostTypeId="2" ParentId="3157" CreationDate="2011-01-17T15:50:52.383" Score="1" ViewCount="" Body="&lt;p&gt;the set of quantities you offered us to calculate the result is strange, or at least unusual. In particular, there is nothing such as the (dimensionless) absorption probability $P(E)$ for a molecule to absorb a photon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absorption probability is given by the cross section which you called $A$. For every process, one has to determine the cross section again. Usually, the cross section is called $\sigma$ instead of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the best thing I could do with your numbers and functions would be to imagine that the cross section in your case was given by $\sigma(E) = P(E)\cdot A$. But it is really misleading to factorize the cross section in such ways. In particular, a universal &quot;cross section area $A$&quot; doesn't mean anything. Molecules are not hard balls that have a universal well-defined cross-sectional areas. They're nuclei surrounded by soft wave functions of electrons that reach arbitrarily far but are getting weaker with the distance - in fact, the (one) wave function for $N$ electrons lives in an $3N$-dimensional space rather than the ordinary $3$-dimensional space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of events, in your case absorption events, is simply the product of the integrated luminosity and the cross section $\sigma$. The integrated luminosity is the time integral of luminosity $L$. It is normally written as $L=\rho\times v$ where $\rho$ is the number density of the beam - in particles per unit volume - and $v$ is the velocity - for photons, it's the speed of light $c$. The quantity $L$ happens to be the same thing as your flux $f$. If there are some losses or extra percentages etc., you have to be careful about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantities such as the cross section are designed in such a way that you are allowed to imagine that the molecule is classical and literally has the cross-sectional area $\sigma$ - that's why they were designed - and you get the right probability. If you do think right, you will indeed find out that only a small percentage of the X-ray photons is absorbed. That's why X-rays are being used to see through human bodies at X-ray pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-17T15:50:52.383" CommentCount="2" />
  <row Id="3166" PostTypeId="2" ParentId="2709" CreationDate="2011-01-17T15:55:44.360" Score="5" ViewCount="" Body="&lt;p&gt;can You read German? here &#xA;&lt;a href=&quot;http://de.wikipedia.org/wiki/Laserschneiden&quot; rel=&quot;nofollow&quot;&gt;http://de.wikipedia.org/wiki/Laserschneiden&lt;/a&gt;&#xA;all Your questions are answered. The English pendant &#xA;is much shorter and does not deal Your problems.&#xA;In case You can't read that, some short excerpt:&#xA;Main problem with copper and Aluminium are high &#xA;reflectivity at 10 µm, the high thermal conductivity &#xA;and and no &quot;assistance&quot; from oxidation possible. &#xA;One special problem is &quot;Einstechen&quot; (initial penetration?). &#xA;I don't know an English  expression for that. It is the making of a &#xA;hole when starting a cut. This is a time consuming step &#xA;in laser cutting anyhow. &#xA;Now, making holes in a PCB is more or less &quot;Einstechen&quot; &#xA;all the time. &#xA;Georg&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-01-17T16:32:43.550" LastActivityDate="2011-01-17T16:32:43.550" CommentCount="6" />
  <row Id="3167" PostTypeId="2" ParentId="3164" CreationDate="2011-01-17T16:09:58.053" Score="11" ViewCount="" Body="&lt;p&gt;First of all, note that any realistic antenna emits a gigantic amount of photons at the radio frequencies. The energy of a single photon is $E=hf$ where $h=6.626\times 10^{-34}$ Js which is tiny, so if you have frequencies of order &quot;just a few Hertz&quot;, the energy of one photon will be a tiny fraction of one Joule. Antennas consume much more energy than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in reality, you emit trillions of photons that fly in all the directions more or less uniformly - well, vertical dipole antennas emit mostly in the horizontal directions etc. The number of photons is so huge that it makes no sense to talk about individual photons: classical electromagnetism described by Maxwell's equations is a totally satisfactory approximation for all practical purposes (and even many impractical ones).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you designed a similar experiment where you would only emit one photon, there would be one photon going away and its position i.e. direction would be undetermined. A photon is a particle that always respects the laws of quantum mechanics, including the uncertainty principle. If the frequency and angular momentum of the photon is known, its position - direction in which it propagates - is completely unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The photon is described by a probabilistic wave whose dependence on the space is pretty much the same as the dependence of $E+iB$ of a classical electromagnetic wave that you obtain if you emit lots of photons in the same state. The probability that the photon will be found at a point is proportional to the energy density $(E^2+B^2)/2$ of the classical electromagnetic wave that you emit by the same antenna if the number of photons is very large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for a single photon, you can't predict in what direction it will go. That's a basic feature of quantum mechanics that the evolution is not deterministic and outcomes of experiments can only be predicted probabilistically. If you know that you have only emitted one photon, the distant detectors will only detect one photon in one particular direction - but you can't be sure in which direction it will be. Again, for a dipole antenna, the nearly-horizontal directions will be preferred - the ratios of probabilities in different directions will follow the energy density of the corresponding classical electromagnetic wave.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:33:59.343" LastActivityDate="2011-02-04T07:33:59.343" CommentCount="4" />
  <row Id="3168" PostTypeId="1" AcceptedAnswerId="3172" CreationDate="2011-01-17T16:10:28.637" Score="7" ViewCount="419" Body="&lt;p&gt;While this concept is widely used in physics, it is really puzzling (at least for beginners) that you just have to multiply two functions (or the function by itself) at different values of the parameter and then average over the domain of the function keeping the difference between those parameters:&lt;br&gt;&#xA;$C(x)=\langle f(x&amp;#39;+x)g(x&amp;#39;)\rangle$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any relatively simple illustrative examples that gives one the intuition about correlation functions in physics? &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-02-04T07:34:49.247" Title="Why correlation functions?" Tags="&lt;correlation-function&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="3169" PostTypeId="2" ParentId="3168" CreationDate="2011-01-17T16:27:47.643" Score="3" ViewCount="" Body="&lt;p&gt;The correlation function you wrote is a completely general correlation of two quantities,&#xA;$$\langle f(X) g(Y)\rangle$$&#xA;You just use the symbol $x&amp;#39;$ for $Y$ and the symbol $x+x&amp;#39;$ for $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the environment - the vacuum or the material - is translationally invariant, it means that its properties don't depend on overall translations. So if you change $X$ and $Y$ by the same amount, e.g. by $z$, the correlation function will not change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, you may shift by $z=-Y=-x&amp;#39;$ which means that the new $Y$ will be zero. So&#xA;$$\langle f(X) g(Y)\rangle = \langle f(X-Y)g(0)\rangle = \langle f(x)g(0) \rangle$$&#xA;As you can see, for translationally symmetric systems, the correlation function only depends on the difference of the coordinates i.e. separation of the arguments of $f$ and $g$, which is equal to $x$ in your case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this should have explained the dependence on $x$ and $x&amp;#39;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what is a correlator? Classically, it is some average over the probabilistic distribution&#xA;$$\langle S \rangle = \int D\phi\,\rho(\phi) S(\phi)$$&#xA;This holds for $S$ being the product of several quantities, too. The integral goes over all possible configurations of the physical system and $\rho(\phi)$ is the probability density of the particular configuration $\phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum mechanics, the correlation function is the expectation value in the actual state of the system - usually the ground state and/or a thermal state. For a ground state which is pure, we have&#xA;$$\langle \hat{S} \rangle = \langle 0 | \hat{S} | 0 \rangle$$&#xA;where the 0-ket-vector is the ground state, while for a thermal state expressed by a density matrix $\rho$, the correlation function is defined as&#xA;$$\langle \hat{S} \rangle = \mbox{Tr}\, (\hat{S}\hat{\rho})$$&#xA;Well, correlation functions are functions that know about the correlation of the physical quantities $f$ and $g$ at two points. If the correlation is zero, it looks like the two quantities are independent of each other. If the correlation is positive, it looks like the two quantities are likely to have the same sign; the more positive it is, the more they're correlated. They're correlated with the opposite signs if the correlation function is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum field theory, correlation functions of two operators - just like you wrote - is known as the propagator and it is the mathematical expression that replaces all internal lines of Feynman diagrams. It tells you what is the probability amplitude that the corresponding particle propagates from the point $x+x&amp;#39;$ to the point $x&amp;#39;$. It is usually nonzero inside the light cone only and depends on the difference of the coordinates only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correlation functions involving an arbitrary positive number of operators are known as the Green's functions or $n$-point functions if a product of $n$ quantities is in between the brackets. In some sense, the $n$-point functions know everything about the calculable dynamical quantities describing the physical system. The fact that everything can be expanded into correlation functions is a generalization of the Taylor expansions to the case of infinitely many variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the scattering amplitude for $n$ external particles (the total number, including incoming and outgoing ones) may be calculated from the $n$-point functions. The Feynman diagrams mentioned previously are a method to do this calculation systematically: a complicated correlator may be rewritten into a function of the 2-point functions, the propagators, contracted with the interaction vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many words to physically describe a correlation function in various contexts - such as the response functions etc. The idea is that you insert an impurity or a signal into $x&amp;#39;$, that's your $g(x&amp;#39;)$, and you study how much the field $f(x+x&amp;#39;)$ at point $x+x&amp;#39;$  is affected by the impurity $g(x&amp;#39;)$.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:34:49.247" LastActivityDate="2011-02-04T07:34:49.247" />
  <row Id="3170" PostTypeId="1" AcceptedAnswerId="3171" CreationDate="2011-01-17T16:33:24.750" Score="3" ViewCount="170" Body="&lt;p&gt;Is there a way to know if a particle is acting as a wave or a particle? Alternatively, if an entangled particle was already measured?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A - Yes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So any experiment over an entangled particle that let you know if they are coherent or decoherent, so would let we know that the distant particle is same &quot;state&quot; (coherence or decoherence), so independiently of any quantum state measured, having detected that &quot;the wavicle has been measured&quot;, let you send information coded as decoherence time intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B - No&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, does still have sense to talk about a coherence/decoherence &quot;state&quot; if there is no way to detect it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wave has his meta-existence as a probability of a detectable particle interaction, but.. if there is no way to know about the system coherence or decoherence -before- it have &quot;expressed&quot; his behaviour (it's done always as a particle) then we could say that coherent state is never detectable?, I mean if you have measured a photon by a detector, you already know coherence is lost, but with engangled particles A, B, if you decoherence A, then what happend to B? only we could know that there is no &quot;random&quot;(umpredictable) value anymore.. then could we know if a particle is entangled with other? or explicitly could we know if a particle was measured or decoherenced from another entangled partner ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fantastic way to say it:&#xA;&quot;Suposse you are employed to discover if our entire world is being spied from other distant world via entangled particles and the random we see (and its subsecuent evolution) was already known via an engangled model of our world, so.. there is a way to know if someone had got access from a distant engangled particle?, or there is no way to know if any measured value was already determined or is the current experiment where it's first known and fixed (and last years as a lab note in a yellow paper..) or was the experiment just redundant, because the value was already known in an entangled distant experiment (and by a distant world, with beings laughing at our surprise) it doesn't matter if the value is random, anyway is a known value, and they saw it first&quot;&lt;/p&gt;&#xA;" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-18T00:47:43.053" LastActivityDate="2011-01-18T00:47:43.053" Title="Detecting coherence" Tags="&lt;quantum-mechanics&gt;&lt;faster-than-light&gt;&lt;entanglement&gt;" AnswerCount="1" />
  <row Id="3171" PostTypeId="2" ParentId="3170" CreationDate="2011-01-17T16:51:12.387" Score="2" ViewCount="" Body="&lt;p&gt;the answer is No, at least up to the moment after both particles in the entangled pair are measured. In fact, each particle in the world is always behaving both as a wave and as a particle. One of these two behavior may turn out to be more important - or exclusively important - for the description of a particular experiment. But which one it was can only be determined retroactively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is demonstrated by the &quot;delayed choice quantum eraser&quot; experiments, see some comments here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2010/11/delayed-choice-quantum-eraser.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/11/delayed-choice-quantum-eraser.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's an experiment in which some sloppy interpreters could think that the particle is already &quot;doomed&quot; and any possible interference pattern that the particle could have created has been damaged. However, the experiment may continue in such a way that the interference pattern is &quot;revived&quot; without any problem. The lesson is that one can never try to determine the &quot;results&quot; of anything before things are actually measured. All intermediate properties - and all histories, using Feynman's approach to quantum mechanics - can contribute and always do contribute to the quantity that matters, namely the probability amplitude whose only goal is to predict the probabilities of outcome at the very end of the experiment. When the measurement is completed, the outcome becomes an objective fact. But there are no objective facts about any system before it is measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like you can't say whether a particle is coming through slit 1 or slit 2 before you measure it - and if you don't interrupt the particle, both possible histories actually contribute to the interference pattern which is what you measure after many repetitions of the double slit experiment - you are also unable to say that a particle behaves &quot;only as a particle&quot; or &quot;only as a wave&quot; before anything is measured. Once again, both possible intermediate histories must be allowed to contribute to what's going on. The wave function has to be carefully evolved in time. And even though some portions of the wave function may later become irrelevant because they describe histories that were not realized, you can never make such a reduction prematurely. And you can't even imagine that in principle, one of the answers was already correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So be sure that you can't transmit any information by entangled particles faster than light. It's not possible by the measurements of their usual quantum numbers and it's not possible by your seemingly more sophisticated procedure which is actually not more sophisticated but it is the same thing with different types of measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you wrote that because I answered No, it doesn't make sense to talk about a coherent state. Well, a coherent state is a particular state that exists in a Hilbert space. (More precisely, a &quot;coherent state&quot; means something else than you say - it's a term denoting states of a shifted harmonic oscillator ground state - but you mean &quot;some state that exhibits coherence.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you make infinitely many repetitions of many experiments to test the character of a state, you may prove that it is a particular state, e.g. your &quot;coherent state&quot;, up to an overall phase. So the coherent state surely exists both as a mathematical object as well as an actual situation that leads to very particular predictions that differ from the predictions of any other state. So it's just incorrect to say that it makes no sense to talk about this state - or any other state. It makes a perfect sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What isn't right is to claim that a particular particle - or any other quantum object - is determined to have one property or another before it is measured. Indeed, whenever the wave function admits that both properties (up or down spin; or coherent or incoherent behavior) may be true, you must patiently admit that both possilibities are true until the very moment when all the measurements are completed. Then you may perhaps do some retroactive interpretations of what happened. But it's important that there can't be any interpretation before the measurements are over, and not even God knows whether two particles are coherent or incoherent with each other before all the manipulations with the particles are over and the particles are actually measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's how quantum mechanics works. States in the Hilbert space exist and maybe realized, but they can only be tested by statistically comparing their probabilistic predictions with repetitions of the same experiments. If you want to discuss a much more specific experiment and  notion of &quot;coherence&quot;, you will have to clarify your question a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-17T16:51:12.387" CommentCount="1" />
  <row Id="3172" PostTypeId="2" ParentId="3168" CreationDate="2011-01-17T16:57:09.423" Score="5" ViewCount="" Body="&lt;p&gt;a very &lt;a href=&quot;http://dx.doi.org/10.2971/jeos.2010.10035s&quot; rel=&quot;nofollow&quot;&gt;intuitive example&lt;/a&gt; for correlation functions can be seen in laser &lt;a href=&quot;http://en.wikipedia.org/wiki/Speckle&quot; rel=&quot;nofollow&quot;&gt;speckle&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Metrology&quot; rel=&quot;nofollow&quot;&gt;metrology&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you shine light on a surface which is rough compared to the wavelength, the resulting reflected signal will be somehow &lt;strong&gt;random&lt;/strong&gt;. This can also be stated as that you cannot say from one point of a signal how a neighbouring one looks like - they are &lt;strong&gt;uncorrelated&lt;/strong&gt;. Such a field is often referred to as &lt;strong&gt;speckle pattern&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This fact can be used. Suppose you take an image $A(x,y)$ of such a random scattered field, a movement of the image $$(x,y)\rightarrow (x+\delta_x, y+\delta_y)  = (x&amp;#39;,y&amp;#39;)$$ thus $$B(x,y) \approx A(x&amp;#39;,y&amp;#39;)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;will be clearly visible and since all information are statistical, one finds that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C(\Delta_x,\Delta_y) = \int B(x,y) A(x + \Delta_x, y + \Delta_y) dx dy $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;will only have a &quot;big&quot; contribution at $(\Delta_x,\Delta_y) \equiv (\delta_x, \delta_y)$ of some peaked form. The &lt;strong&gt;width of the peak&lt;/strong&gt; will be given by some physical properties of the illumination, roughness of the surface etc. - it directly corresponds to the &lt;strong&gt;local variation&lt;/strong&gt; of the field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we had now in the field some &lt;strong&gt;periodic variation&lt;/strong&gt; we could see that $C$ will have &lt;strong&gt;several peaks&lt;/strong&gt; corresponding to the image's (or field's) &lt;strong&gt;self-similarity&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, analyzing the correlation of a quantity will give you information on how fast it changes and if it is somehow self-similar.&lt;br&gt;&#xA;I hope you don't mind that I have chosen an application coming from a more &lt;strong&gt;practical&lt;/strong&gt; viewpoint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS.: More can be found in all the very rich works done by &lt;a href=&quot;http://en.wikipedia.org/wiki/Joseph_W._Goodman&quot; rel=&quot;nofollow&quot;&gt;Goodman&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2011-01-17T16:57:09.423" />
  <row Id="3173" PostTypeId="2" ParentId="401" CreationDate="2011-01-17T17:09:23.673" Score="1" ViewCount="" Body="&lt;p&gt;I use TKPAINT which still works very well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.netanya.ac.il/~samy/tkpaint.html&quot; rel=&quot;nofollow&quot;&gt;http://www.netanya.ac.il/~samy/tkpaint.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, one has to download ActiveTcl for Windows or its Tcl counterparts for Linux or whatever you use. It can draw filled or empty disks, ellipses, squares, rectangles, splines, rotate them, quickly copy them, move them, texts with many fonts, colors, grid, and it may be exported as EPS - encapsulated postscript as well - which is a standard way to embed similar diagrams in TeX papers on the arXiv and beyond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've used it in many papers when I was writing them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-17T17:09:23.673" CommunityOwnedDate="2011-01-17T17:09:23.673" />
  <row Id="3174" PostTypeId="2" ParentId="15" CreationDate="2011-01-17T17:15:29.080" Score="2" ViewCount="" Body="&lt;p&gt;To disprove the string theory I would say is not very likely; but there are two ways that would cause trouble in string theory: one theoretical and one by experiment. Both unlikely to happen soon, (an both unlikely to ever disprove it, because the string theory seems to be in the right way, at least until now by far.)&#xA;- If one would like to &quot;dissprove it&quot; at least in an initial approximation to cause some scepticism, should find a theory which solves at least as many problens as the string theory solves, and deals with at least some of the remaining problems of the string theory. (Even if this other theory exists, it will take several decades after its discovery to evolve anyway..)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Experimentally as everyone else said we need to achieve higher energies on earth, or some better way to  locate and detect these higher energies in universe possibly. To achieve or locate somewhere, the energies where string theory effects are detectable will take many many years from now-at least that is the common belief.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-The &quot;argument&quot; that sometimes I listen the string theory is not a theory or is wrong because it can not be verified or disproved is competely wrong. Simply because this is not an argument. &#xA;If we have to reach higher energies to see the string theory, then we just have to; is very possible that this is the physics. In any case even simpler physics stories like the one of neutrino which critised (with the critics of that time to be:'not even wrong' or the most optimistic &quot;there is no practical way of observing the neutrino&quot;); with Pauli and Fermi doing the theoretical prediction in ~1930 and the discovery of the Cowan and Reines in 1956 can teach us some things...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one sentence the answer to your last line is:&#xA;Until now there is no experiment or detector that can verify the string theory. In the future it is a belief that this will happen when we will be able to achieve or locate somewhere these higher energies that the stringy effects become detectable. &lt;/p&gt;&#xA;" OwnerUserId="1417" LastEditorUserId="1417" LastEditorDisplayName="" LastEditDate="2011-01-17T17:56:57.250" LastActivityDate="2011-01-17T17:56:57.250" />
  <row Id="3176" PostTypeId="2" ParentId="15" CreationDate="2011-01-17T18:24:24.963" Score="7" ViewCount="" Body="&lt;p&gt;The only way to 'test' string theory, is to actually figure out what it predicts first, which is ambigous at the moment.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contrary to most claims, String theory is actually a unique theory in that there are no adjustable free parameters.  However it has a large or possibly infinite amount of classical solutions or 'vacua'.  Most of these vacua look nothing like the real world, some (by some I mean it could be ~ 10^500) are very similar (in that they seem to correspond to low energy physics like the standard model), and at most one corresponds to the real world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have specified a vacua, you then have fixed the predictions identically -- for everything (thats what it means to be a theory of everything).  So for instance, you could compute the mass of the electron to however many decimal places and then test it in the lab and also the vacua might make an unambigous prediction for cosmological objects (eg the existence of cosmic strings or domain walls).  We don't know exactly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course absent a way to figure out which vacua corresponds to reality, we are left with the same problem that quantum field theory has.  Namely you actually have to go out and measure certain things first in order to make predictions (so for instance in the standard model we need to pin down the 26 adjustable free parameters, like the Yukawa couplings, masses of elementary particles and so forth).. But once you have done that, you can then predict infinitely many other things, like scattering cross sections. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in string theory, that would mean some way of wittling down 10^500 vacua to a more human number, like 50, 10 or better 1 or 0 (0 means the theory is falsified).  Which requires doing scattering experiments at the Planck scale, which means a particle accelerator roughly on the scale of the Milky Way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it may turn out that string theory makes unambigous falsifiable predictions, but that really requires a great deal of theoretical work b/c it implies knowing something about not just the vacua that corresponds to the real world, but also the 10^500 other ones if you get my meaning.  Still, we know that it does make some predictions.  For instance, the existence of the quanta of gravity is universal across all solutions.  Likewise, the fact that quantum mechanics and special relativity must hold is another robust prediction.&lt;/p&gt;&#xA;" OwnerUserId="1350" LastActivityDate="2011-01-17T18:24:24.963" CommentCount="3" />
  <row Id="3177" PostTypeId="2" ParentId="15" CreationDate="2011-01-17T18:25:42.720" Score="26" ViewCount="" Body="&lt;p&gt;you can disprove string theory by many observations that will almost certain not occur, for example:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;By detecting &lt;strong&gt;Lorentz violation&lt;/strong&gt; at high energies: string theory predicts that the Lorentz symmetry is exact at any energy scale; recent experiments by the Fermi satellite and others have showed that the Lorentz symmetry works even at the Planck scale with a precision much better than 100% and the accuracy may improve in the near future&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By detecting a &lt;strong&gt;violation of the equivalence principle&lt;/strong&gt;; it's been tested with the relative accuracy of $10^{-16}$ and it's unlikely that a violation will occur; string theory predicts that the law is exact&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By detecting a &lt;strong&gt;mathematical inconsistency&lt;/strong&gt; in our world, for example that $2+2$ can be equal both to $4$ as well as $5$; such an observation would make the existing alternatives of string theory conceivable alternatives because all of them are mathematically inconsistent as theories of gravity; clearly, nothing of the sort will occur; also, one could find out a previously unknown mathematical inconsistency of string theory - even this seems extremely unlikely after the neverending successful tests&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By experimentally proving that the &lt;strong&gt;information is lost&lt;/strong&gt; in the black holes, or anything else that contradicts general properties of quantum gravity as predicted by string theory, e.g. that the high center-of-mass-energy regime is dominated by black hole production and/or that the black holes have the &lt;strong&gt;right entropy&lt;/strong&gt;; string theory implies that the information is preserved in any processes in the asymptotical Minkowski space, including the Hawking radiation, and confirms the Hawking-Bekenstein claims as the right semiclassical approximation; obviously, you also disprove string theory by proving that &lt;strong&gt;gravitons don't exist&lt;/strong&gt;; if you could prove that gravity is an entropic force, it would therefore rule out string theory as well&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By experimentally proving that the world doesn't contain gravity, fermions, or isn't described by quantum field theories at low energies; or that the general &lt;strong&gt;postulates of quantum mechanics don't work&lt;/strong&gt;; string theory predicts that these approximations work and the postulates of quantum mechanics are exactly valid while the alternatives of string theory predict that nothing like the Standard Model etc. is possible&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By experimentally showing that the real world contradicts some of the general features predicted by all string vacua which are not satisfied by the &lt;strong&gt;&quot;Swampland&quot;&lt;/strong&gt; QFTs as explained by Cumrun Vafa; if we lived in the swampland, our world couldn't be described by anything inside the landscape of string theory; the generic predictions of string theory probably include the fact that gravity is the weakest force, moduli spaces have finite volume, and similar predictions that seem to be satisfied so far&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;By mapping the whole landscape&lt;/strong&gt;, calculating the accurate predictions of each vacuum for the particle physics (masses, couplings, mixings), and by showing that none of them is compatible with the experimentally measured parameters of particle physics within the known error margins; this route to disprove string theory is hard but possible in principle, too (although the full mathematical machinery to calculate the properties of any vacuum at any accuracy isn't quite available today, even in principle)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;By analyzing physics experimentally up to the Planck scale&lt;/strong&gt; and showing that our world contains neither &lt;strong&gt;supersymmetry&lt;/strong&gt; nor &lt;strong&gt;extra dimensions&lt;/strong&gt; at any scale. If you check that there is no SUSY up to a certain higher scale, you will increase the probability that string theory is not relevant for our Universe but it won't be a full proof&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A convincing observation of &lt;strong&gt;varying fundamental constants&lt;/strong&gt; such as the fine-structure constant would disprove string theory unless some other unlikely predictions of some string models that allow such a variability would be observed at the same time&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The reason why it's hard if not impossible to disprove string theory in practice is that string theory - as a qualitative framework that must replace quantum field theory if one wants to include both successes of QFT as well as GR - has already been established. There's nothing wrong with it; the fact that a theory is hard to exclude in practice is just another way of saying that it is already shown to be &quot;probably true&quot; according to the observations that have shaped our expectations of future observations. Science requires that hypotheses have to be disprovable in principle, and the list above surely shows that string theory is. The &quot;criticism&quot; is usually directed against string theory but not quantum field theory; but this is a reflection of a deep misunderstanding of what string theory predicts; or a deep misunderstanding of the processes of the scientific method; or both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In science, one can only exclude a theory that contradicts the observations. However, the landscape of string theory predicts the same set of possible observations at low energies as quantum field theories. At long distances, string theory and QFT as the frameworks are indistinguishable; they just have different methods to parameterize the detailed possibilities. In QFT, one chooses the particle content and determines the continuous values of the couplings and masses; in string theory, one only chooses some discrete information about the topology of the compact manifold and the discrete fluxes and branes. Although the number of discrete possibilities is large, all the continuous numbers follow from these discrete choices, at any accuracy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the validity of QFT and string theory is equivalent from the viewpoint of doable experiments at low energies. The difference is that QFT can't include consistent gravity, in a quantum framework, while string theory also automatically predicts a consistent quantum gravity. That's an advantage of string theory, not a disadvantage. There is no known disadvantage of string theory relatively to QFT. For this reason, it is &lt;em&gt;at least&lt;/em&gt; as established as QFT. It can't realistically go away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, it's been showed in the AdS/CFT correspondence that string theory is automatically the full framework describing the dynamics of theories such as gauge theories; it's equivalent to their behavior in the limit when the number of colors is large, and in related limits. This proof can't be &quot;unproved&quot; again: string theory has attached itself to the gauge theories as the more complete description. The latter, older theory - gauge theory - has been experimentally established, so string theory can never be removed from physics anymore. It's a part of physics to stay with us much like QCD or anything else in physics. The question is only what is the right vacuum or background to describe the world around us. Of course, this remains a question with a lot of unknowns. But that doesn't mean that everything, including the need for string theory, remains unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could happen - although it is extremely, extremely unlikely - is that a consistent, non-stringy competitor to string theory that is also able to predict the same features of the Universe as string theory can emerges in the future. (I am carefully watching all new ideas.) If this competitor began to look even more consistent with the observed details of the Universe, it could supersede or even replace string theory. It seems almost obvious that there exists no &quot;competing&quot; theory because the landscape of possible unifying theories has been pretty much mapped, it is very diverse, and whenever all consistency conditions are carefully imposed, one finds out that he returns back to the full-fledged string/M-theory in one of its diverse descriptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in the absence of string theory, it could hypothetically happen that new experiments will discover new phenomena that are impossible - at least unnatural - according to string theory. Obviously, people would have to find a proper description of these phenomena. For example, if there were preons inside electrons, they would need some explanation. They seem incompatible with the string model building as we know it today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even if such a new surprising observation were made, a significant fraction of the theorists would obviously try to find an explanation within the framework of string theory, and that's obviously the right strategy. Others could try to find an explanation elsewhere. But neverending attempts to &quot;get rid of string theory&quot; are almost as unreasonable as attempts to &quot;get rid of relativity&quot; or &quot;get rid of quantum mechanics&quot; or &quot;get rid of mathematics&quot; within physics. You simply can't do it because those things have already been showed to work at some level. Physics hasn't yet reached the very final end point - the complete understanding of everything - but that doesn't mean that it's plausible that physics may easily return to the pre-string, pre-quantum, pre-relativistic, or pre-mathematical era again. It almost certainly won't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-26T18:47:34.270" LastActivityDate="2011-01-26T18:47:34.270" CommentCount="12" />
  <row Id="3179" PostTypeId="1" CreationDate="2011-01-17T18:48:49.257" Score="3" ViewCount="254" Body="&lt;p&gt;Wick rotation makes sense for Dirac algebra - it has +--- and ++++ and ---- signatures. Just wondering if one can Wick rotate Pauli algebra from the standard +++ pauli matrices to +--.&lt;/p&gt;&#xA;" OwnerUserId="1423" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-18T00:48:19.887" LastActivityDate="2011-01-20T07:49:31.527" Title="Can Wick rotation be done on Pauli algebra to get from +++ to +--?" Tags="&lt;particle-physics&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="3180" PostTypeId="2" ParentId="15" CreationDate="2011-01-17T19:05:24.987" Score="7" ViewCount="" Body="&lt;p&gt;Since many people seem to have very odd ideas about this, let's address this from a much simpler point of view.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's suppose you have a friend who only knows math at the level of arithmetic of positive integers.  You try to tell him about the existence of negative numbers, and he tells you,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;That's stupid, there's obviously no such thing as &quot;negative&quot; numbers, how can I possibly measure something so stupid?  Can you have negative one apple?  No, you can't.  I can owe you positive one apple, but there's clearly no such thing as negative apples.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can you start to argue that there is such a thing as negative numbers?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very powerful first step is mathematical consistency.  You can list all of the abstract properties you believe to characterize everything about positive integer arithmetic:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For all a,b,c, a(b+c) = ab+ac&lt;/li&gt;&#xA;&lt;li&gt;For all a,b, a+b=b+a, ab=ba&lt;/li&gt;&#xA;&lt;li&gt;There exists a number, called 0, such that, for all a, a+0=0+a=a, a0=0a=0&lt;/li&gt;&#xA;&lt;li&gt;There exists a number, called 1, such that, for all a, 1a=a1=a&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(note that, in sharp contrast to the case of real numbers, the first property can be proved with induction, and need not be an axiom.  Similarly, other listed properties can be proved from other ones designated to be more basic if one wishes, which can not be done in the case of the reals.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, once you both agree that these axioms characterize the positive integers completely, you can show that these hypothetical negative numbers, based on their formal properties, are consistent with the above axioms.  What does this show?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The positive integers, with the addition of the negative integers, can do at least as much as the positive integers by themselves.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;STOP&lt;/strong&gt;  At this point, pause to realize how &lt;em&gt;powerful&lt;/em&gt; this constraint is!!  How many other ways could one generalize arithmetic, &lt;em&gt;at this level&lt;/em&gt;, to something else that is consistent with the properties you want?  Zero.  There is absolutely no other way to do it.  This is incredibly suggestive, and you should keep this in mind for the rest of the cartoon argument, and see how every argument that follows is secretly an aspect of this one!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend responds:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sure, you can write down toy models like that, and they may be consistent, but they don't correspond to reality.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, what else do you need to demonstrate to your friend to convince him of the validity of the negative numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You find something else they can do that you can't do with the positive numbers alone.  Simply, you can state that every positive-integer-valued algebraic equation does not have a solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x + 1 = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;does not have a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, it is a trivial fact that extending to the negative numbers allows you to solve such equations.  Then, all that's left to convince your friend of the validity of negative numbers is to show that this is equivalent to solving an (&quot;a priori&quot;) &lt;em&gt;different&lt;/em&gt; problem which only involved arithmetic of positive integers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x+1=0 &amp;lt;=&gt; y + 1 = 1 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, y = 0, and y=x+1 is equivalent to the other problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be complete, we also have to consider problems that are &quot;unique&quot; to the negatives, such as (-1)(-1) = 1, but in the realm of integers, these are trivial matters that are reducible to the above.  Even in the case of reals, given the other things we've shown, these consequences are almost &quot;guaranteed&quot; to work out intuitively obviously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, assuming your friend is a reasonable, logical, person, he must now believe in the validity of negative numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What have we shown?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Consistency, both with previous models and with itself&lt;/li&gt;&#xA;&lt;li&gt;The ability to solve new problems&lt;/li&gt;&#xA;&lt;li&gt;The reduction of &lt;em&gt;some&lt;/em&gt; problems in the new language to problems in the old language&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, to decide if this a good model for a particular system, you must look at the subset of problems that did not have a solution before, and see if the new properties characterize that system.  In this case, that's trivial, because the properties of negative numbers are so obvious.  In the case of applying more complicated things to describe the details of physical situations, it's less obvious, because the structure of the theory, and the experiments, is not so simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this apply to string theory?  What must we show to convince a reasonable person of its validity?  Following the above argument, I claim:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;String theory reproduces (by construction) general relativity&lt;/li&gt;&#xA;&lt;li&gt;String theory reproduces (by construction) quantum mechanics (and by the above, quantum field theory)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So string theory is at least as good as the rest of the foundations of physics.  Stop again to marvel at how powerful this statement is!  Realistically, how many ways are there to consistently and non-trivially write a theory that reduces to GR and QFT?  Maybe more than one, but surely not many!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question is--what new do we learn?  What additional constraints do we get out of string theory?  What problems in GR and QFT can be usefully written as &lt;em&gt;equivalent&lt;/em&gt; problems in string theory?  What problems can string theory solve that are totally outside of the realm of GR and QFT?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only the last of these is beyond the reach of current experiments.  The &quot;natural&quot; realm where string theory &lt;em&gt;dominates&lt;/em&gt; the behavior of an experiment is is at very high energies, or equivalently, very short distances.  Simple calculations show that these naive regions are well outside of direct detection by current experiments.  (Note that in the above example of negative numbers, the validity of the &quot;theory&quot; strictly in the corresponding realm didn't need to be directly addressed to make a very convincing argument; pause to reflect on why!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, theoretical &quot;problems&quot; with the previous theories, such as black hole information loss, can be solved with string theory.  Though these can't be experimentally verified, it's very suggestive that they admit the expected solution in addition to reproducing the right theories in the right limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two major successes of string theory that satisfy the other two requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AdS/CFT allows us to solve &lt;em&gt;purely&lt;/em&gt; field theory problems in terms of string theory.  In other words, we have solved a problem in the new language that we could already solve in the old language.  A bonus here is that it allows us to solve the problem precisely in a domain where the old language was difficult to deal with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;String theory also constrains, and specifies, the spectrum and properties of particles at low energies.  In principle (and in toy calculations), it tells us &lt;em&gt;all&lt;/em&gt; of the couplings, generations of particles, species of particles, etc.  We don't yet know a description in string theory that gives us &lt;em&gt;exactly&lt;/em&gt; the Standard Model, but the fact that it does constrain the low-energy phenomenology is a pretty powerful statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really, all that's left to consider to convince a very skeptical reader is that one of the following things is true:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is &lt;strong&gt;possible&lt;/strong&gt; for string theory to reproduce the Standard Model (e.g., it admits solutions with the correct gauge groups, chiral fermions, etc.)&lt;/li&gt;&#xA;&lt;li&gt;It is &lt;strong&gt;not possible&lt;/strong&gt; for string theory to reproduce the Standard Model (e.g., there is no way to write down chiral theories, it does not admit the correct gauge groups, etc.   This is the case in, e.g., Kaluza-Klein models.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I claim, and it is generally believed (for very good reasons), that the first of these is true.  There is no formal, complete, mathematical proof that this is the case, but there is absolutely no hint of anything going wrong, and we can get models very similar to the standard model.  Additionally, one can show that all of the basic features of the Standard Model, such as chiral fermions, the right number of generations, etc, are consistent with string theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also ask, what would it mean if string theory was wrong?  Really, this would signal that, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The theory was mathematically inconsistent (there is no reason to believe this)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At a fundamental level, either quantum mechanics or relativity failed in some fairly pathological way, such as a violation of Lorentz invariance, or unitarity.  This would indicate that a theory of everything would look radically different than anything written down so far; this is a very precarious claim--consider what would happen in the example of arithmetic in the above if there were something &quot;wrong&quot; with addition.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The theory is consistent, and a generalization of GR and QFT, but is somehow not a generalization in the right &quot;limit&quot; in some sense.  This happens in, e.g., Kaluza-Klein theory, where chiral fermions can't be properly written down.  In that case, a solution is also suggested by a sufficiently careful analysis (and is one potential way to get to string theory).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of these three possibilities, the first two are extremely unlikely.  The third is more likely, but given that it is known that all the basic features can show up, it would seem very strange if we could almost reproduce what we want, but not quite.  This would be like, in the arithmetic example, being able to reproduce all the properties we want, except for 1+ (-1) = 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're careful, you can phrase my argument in a more formal way, in terms of what it precisely means to have a consistent generalization, in the sense of formal symbolic logic, if you like, and see what must &quot;fail&quot; in order for the contrapositive of the argument to be true.  (That is, (stuff) =&gt; strings are true, so ~strings =&gt; ~(stuff), and then unpack the possibilities for what ~(stuff) could mean in terms of its components!)&lt;/p&gt;&#xA;" OwnerUserId="1403" LastActivityDate="2011-01-17T19:05:24.987" CommentCount="4" />
  <row Id="3181" PostTypeId="2" ParentId="822" CreationDate="2011-01-17T19:06:10.150" Score="-1" ViewCount="" Body="&lt;p&gt;Lubosh wrote: &quot;The angular momentum is defined by the change of the phase of the wave function under rotations, which may come from the dependence of the wave function on space, but also from the transformations of the components of the wave function among each other, which is possible even if everything is localized at a point. So even point-like objects may carry an angular momentum in quantum mechanics, the spin.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In QM it is impossible and is not necessary to impose &lt;strong&gt;R&lt;/strong&gt; = 0 (see my blog) to have a system at rest. On the contrary, one has to put &lt;strong&gt;P&lt;/strong&gt; = 0. It does not mean point-likeness but &lt;strong&gt;ubiquity&lt;/strong&gt; instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an article by R. Ohanian on &lt;a href=&quot;http://jayryablon.files.wordpress.com/2008/04/ohanian-what-is-spin.pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;spin&lt;/strong&gt;&lt;/a&gt;. But I am afraid it is finally a tautology or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the angular momentum is fundamental. I think that even in Classical Mechanics a description of anything with help of solely three coordinates &lt;strong&gt;R&lt;/strong&gt;(t) is too primitive. Generally everything is not point-like and rotates, roughly speaking. So the intrinsic angular momentum &lt;strong&gt;J&lt;/strong&gt; is as fundamental as the linear momentum &lt;strong&gt;P&lt;/strong&gt; (as well as color, charge, and flavor ;-).&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-01-17T23:02:57.040" LastActivityDate="2011-01-17T23:02:57.040" CommentCount="13" />
  <row Id="3182" PostTypeId="1" AcceptedAnswerId="3205" CreationDate="2011-01-17T19:10:53.707" Score="4" ViewCount="246" Body="&lt;p&gt;We learn that the quantum wave function $\Psi$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Wave_function_collapse&quot; rel=&quot;nofollow&quot;&gt;collapses&lt;/a&gt; when it interacts with a classic object (measurement). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question is:&lt;/strong&gt;&lt;br&gt; Why are there &lt;em&gt;classic&lt;/em&gt; objects after all, how did it all start? In a &lt;em&gt;quantum universe&lt;/em&gt; (and I understand we live in one) there should only be &lt;em&gt;superposition&lt;/em&gt;, how did these &lt;em&gt;classic&lt;/em&gt; beings came into existence (which &quot;decohere&quot; the others) in the first place? Is it a statistical necessity like &lt;em&gt;entropy&lt;/em&gt; - or doesn't, quite simply, anybody know the answer to that riddle? &lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-30T09:15:06.883" LastActivityDate="2011-01-30T09:15:06.883" Title="Why did decoherence start in the first place?" Tags="&lt;quantum-mechanics&gt;&lt;universe&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="3183" PostTypeId="2" ParentId="3179" CreationDate="2011-01-17T19:15:44.220" Score="3" ViewCount="" Body="&lt;p&gt;Wick rotation is a tool to simplify certain calculations, by defining certain functions by analytic continuation from Euclidean space where they can be evaluated more simply. Turns out that rotating the spinors (as well as the momenta) usually does not simplify calculations involving those spinors. The reason is that the structure of the spinor representations of the groups so(p,q) depends on the signature p-q in an interesting and intricate way. Since the Wick rotation changes the signature, it would also change the structure of spinors in a way that is not all that straightforward to keep track of. So, nothing is stopping you from rotating the spinors as well as all the positions and momenta, but usually it is not useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The actual rotation is fairly simple, just multiply the matrices you need to rotate by a factor of i).&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-01-17T22:07:51.527" LastActivityDate="2011-01-17T22:07:51.527" />
  <row Id="3184" PostTypeId="2" ParentId="3182" CreationDate="2011-01-17T19:26:40.803" Score="0" ViewCount="" Body="&lt;p&gt;In quantum mechanics you need the classical object for the formulation of the theory. For such questions you would need a more general theory, probably including gravity. Say the true theory of quantum gravity will give the answer.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2011-01-17T19:26:40.803" />
  <row Id="3185" PostTypeId="2" ParentId="401" CreationDate="2011-01-17T19:31:04.140" Score="1" ViewCount="" Body="&lt;p&gt;I've recently been introduced to &lt;a href=&quot;http://www.geogebra.org/cms/&quot; rel=&quot;nofollow&quot;&gt;GeoGebra&lt;/a&gt;, and while I haven't yet had the opportunity to use for any work, I love the interface. For geometrical diagrams it looks spectacular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also used XFig and gnuplot extensively. Particularly if you use $\LaTeX$, these tools serve their purpose very very well.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-01-17T19:31:04.140" CommunityOwnedDate="2011-01-17T19:31:04.140" />
  <row Id="3186" PostTypeId="1" AcceptedAnswerId="3188" CreationDate="2011-01-17T19:34:32.190" Score="1" ViewCount="108" Body="&lt;p&gt;There is a surprising number of papers seriously discussing the &quot;&lt;a href=&quot;http://focus.aps.org/story/v13/st3&quot; rel=&quot;nofollow&quot;&gt;feigel effect&lt;/a&gt;&quot;  This has been linked to the &lt;a href=&quot;http://peter.mapledesign.co.uk/writings/physics/2005_dissertation_The_Abraham-Minkowski_Controversy.pdf&quot; rel=&quot;nofollow&quot;&gt;Abraham-Minkowski controversy&lt;/a&gt; also &lt;a href=&quot;http://en.wikipedia.org/wiki/Abraham%E2%80%93Minkowski_controversy&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  Although there are good discussions picking apart the feigel effect calculations, there does not appear to be a conclusive refutation of the effect under certain experimental circumstances.  Much of this is linked to when the Poynting vector vanishes in the vacuum.  Is there a legitimate possibility for seeing the Feigel effect?  What would that mean about the nature of the poynting vector in physics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.esa.int/gsp/ACT/doc/ARI/ARI%20Study%20Report/ACT-RPT-PHY-ARI-041201-Grenoble_Feigel.pdf&quot; rel=&quot;nofollow&quot;&gt;The Feigel Process: Lorentz invariance, regularization, and experimental feasibility Final Report&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.esa.int/gsp/ACT/doc/ARI/ARI%20Study%20Report/ACT-RPT-PHY-ARI-041201-Koln_Feigel.pdf&quot; rel=&quot;nofollow&quot;&gt;Lorentz-invariant description of the Feigel Process for the Extraction of Momentum from a&#xA;vacuum Final Report&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1008/1008.3656v2.pdf&quot; rel=&quot;nofollow&quot;&gt;Alternative derivation of the Feigel effect and call for its experimental verification&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="1383" LastEditorDisplayName="" LastEditDate="2011-01-18T01:04:34.797" LastActivityDate="2011-01-18T01:04:34.797" Title="Poynting Vectors and Feigel Effects" Tags="&lt;quantum-field-theory&gt;&lt;experimental-physics&gt;" AnswerCount="1" />
  <row Id="3187" PostTypeId="2" ParentId="3182" CreationDate="2011-01-17T19:36:06.330" Score="4" ViewCount="" Body="&lt;p&gt;there are no God-given &quot;classical objects&quot; in our Universe. The whole Universe - and even multiverse, if one exists - obeys the laws of quantum mechanics. One of the postulates of quantum mechanics - that are universally valid - is that for any pair of states in the Hilbert space, their general complex linear superpositions are possible states as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classical physics is just an approximation and decoherence is a major process that clarifies the emergence of the approximate classical description. Decoherence explains why some objects suddenly start to behave in a pretty good agreement with the classical intuition. The fundamental truth is never classical, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Decoherence picks a privileged basis of the Hilbert space of a macroscopic object. But which basis it is depends on the Hamiltonian describing the evolution of the object - and its interactions with the Hamiltonian. The preferred basis is defined by the condition that in this basis, the density matrix becomes quickly diagonal if we trace over the environmental degrees of freedom. The diagonal entries of such a density matrix may be interpreted as ordinary &quot;classical&quot; probabilities. Quantum mechanics still predicts probabilities only, even in this limit: however, the coherence - the information about the relative phases of the states - disappears. That allows us to assume that one of the preferred states was already realized &quot;before&quot; someone looked at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which basis vectors are those preferred ones? For example, they include the Schrödinger's dead-cat-vector and the alive-cat-vector, in the popular thought experiment. Why are these vectors preferred? It's because each of them &quot;imprints&quot; the information about itself - and its shape in spacetime - into the environmental degrees of freedom in such a way that the states of the environment are orthogonal to the states of the environment produced by other preferred vectors. Because the Hamiltonians are usually local, the preferred states will differ by positions of macroscopic objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why when we trace over the environmental degrees of freedom, the off-diagonal elements quickly go to zero (because the corresponding entangled states of the environment are orthogonal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.karlin.mff.cuni.cz/~motl/entan-interpret.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.karlin.mff.cuni.cz/~motl/entan-interpret.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some pedagogical explanations of decoherence and its impact on the interpretational issues of quantum mechanics. Once again, there are no classical states or &quot;a priori classical states&quot; in any quantum mechanical system - or the real world. Which states will look like the seemingly preferred basis depends on the Hamiltonian; we say that this question is determined dynamically (because the Hamiltonian is what generates dynamics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The emergence of a legitimate classical description for large enough objects has nothing to do with gravity. Indeed, Newton's constant never appears in the description of decoherence and its speed. And the wave functions never &quot;collapse&quot;. In fact, decoherence is exactly the right description that replaces the popular misconceptions such as the &quot;collapse of a wave function&quot; as well as ad hoc separations of objects into &quot;microscopic quantum&quot; and &quot;macroscopic classical&quot; objects which is what the Copenhagen school had to do in the 1920s. That approach was effectively correct and is confirmed by decoherence - except that decoherence also explains where this boundary comes from and where (at what size or other quantities) this boundary (gradually) emerges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-17T19:41:30.593" LastActivityDate="2011-01-17T19:41:30.593" />
  <row Id="3188" PostTypeId="2" ParentId="3186" CreationDate="2011-01-17T20:05:28.137" Score="1" ViewCount="" Body="&lt;p&gt;Dear Humble,&#xA;good questions. I can't answer all your questions, but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the name of the author is Feigel, not Fiegel, and the paper is also a preprint here:&#xA;&lt;a href=&quot;http://arxiv.org/abs/physics/0304100&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/physics/0304100&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The controversies about the stress-energy tensor were inevitable. Only the total energy and momentum are conserved as a consequence of Noether's theorem, and how they're distributed in space may often be a matter of conventions. In particular, the Poynting vector may be defined in several different ways - and various automatically conserved pieces may be added, too. The integrals won't change. But for example, if there are crossing electric and magnetic fields in the vacuum, the usual Poynting vector $E\times B$ shows that the energy is flowing somewhere. This flow ends up &quot;circular&quot; if you look globally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feigel's paper seems to be just a generalization of the Casimir effect, with a slightly more complicated arrangement of things. Instead of the Casimir energy, he wants to play with the energy of the vacuum, and instead of the metallic boundaries, he plays with dielectric liquids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of these things changes that the total energy and the total momentum are exactly conserved, because of the symmetries. Whether the momentum is being extracted from the vacuum is a matter of interpretation. You may also say that it is extracted from a low-frequency electromagnetic wave that was emitted by another object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feigel suggests that there is a relevance of his setup for the Abraham-Minkowski arguments about the density of electromagnetic momentum in dielectric materials. But one must be careful. The energy carried by the Casimir effect is distributed nonlocally - it comes from modes that look like standing waves in between the metallic plates. So the Casimir energy is not &quot;naturally&quot; written as an integral of a density, I think. Please correct me if I am wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect this may be the case of any one-loop constructions of this sort, and Feigel's construction seems to be just a more complicated example. So I suspect the effect of the Feigel's phenomena are not just about changing some local densities either in the Abraham or in the Minkowski way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I would say that in principle, Quantum Electrodynamics or the Standard Model have standardized prescriptions for the stress-energy tensor that should be calculable in any context, including dielectric liquids in crossed electric and magnetic fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the best, LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-17T20:05:28.137" CommentCount="1" />
  <row Id="3189" PostTypeId="1" AcceptedAnswerId="3200" CreationDate="2011-01-17T20:09:28.820" Score="6" ViewCount="406" Body="&lt;p&gt;A &lt;a href=&quot;http://physics.princeton.edu/~mcdonald/examples/EM/barnett_prl_104_070401_10.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; was published claiming to resolve the Abraham-Minkowski controversy.  Is this paper viewed as definitive by physicists?&lt;/p&gt;&#xA;" OwnerUserId="1383" LastActivityDate="2011-06-22T01:17:30.730" Title="Is the Abraham-Minkowski controversy resolved?" Tags="&lt;electromagnetism&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="3190" PostTypeId="2" ParentId="3157" CreationDate="2011-01-17T20:30:17.470" Score="1" ViewCount="" Body="&lt;p&gt;It is not unusual in spectroscopy to define an absorption cross-section of a molecule in $cm^2$. Although it has no real physical meaning (it can even be larger that the molecule itself) and can be derived from Einstein coefficients, it is what you can directly measure by experiment $$cross-section\;\;\alpha= \frac{total\;energy\;absorbed\;/s}{total\;incident\;intensity\;(energy/s/area)}$$&#xA;The relation to the Einstein coefficient if simply $\alpha=\frac{\hbar\omega}{c}B_{21}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, transition probability is Einstein coefficient times photon density. You problem could be to correctly define a fraction of photons that are resonant with the transition that you study. This would greatly depend on the kind of light source that you use.&lt;/p&gt;&#xA;" OwnerUserId="701" LastActivityDate="2011-01-17T20:30:17.470" />
  <row Id="3191" PostTypeId="2" ParentId="15" CreationDate="2011-01-17T20:47:29.050" Score="3" ViewCount="" Body="&lt;p&gt;It is very hard to probe the scales of quantum gravity experimentally so there are very few possibilities for testing if string theory is incorrect as a unified theory encompassing quantum gravity. One of the few observations that has probed the Planck scale was the Fermi gamma ray telescope observation which showed that photons of different energy travel very close to the same speed over cosmic distances. If the result had shown a dispersion of speed it would have disproved string theory and encouraged physicist to look at other ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course any observation that disproves quantum theory or general relativity will also disprove string theory, but the real interest is in observations that relate directly to quantum gravity because the separate theories are already well established in their own regimes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although it is hard to get direct experimental input for any theory of quantum gravity the constraints on any theory from the logical requirement to combine quantum theory and gravity into one consistent theory for physics are already tremendously strong. In particular there should be some perturbative low energy limit that describes gravity in terms of gravitons which interact with matter. Despite much effort string theory is the only approach that can accomplish this and it is very hard to conceive of a second way. In fact it is very surprising that there is this one way because it requires almost miraculous cancellations of anomalies to make it work. This gives many people the confidence that string theory is the correct path to follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately there needs to be some definitive observation of a quantum gravitational effect that supports string theory. As I said, there are not many possibilities for such observations at present, but this would be a problem for any alternative theory of quantum gravity too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that we might get lucky and observe large extra dimensions at the LHC, but there is no reason to expect that. Another possibility that I think is a little more plausible is that supersymmetry would be observed and found to take a form that supports a supergravity origin. Still we have no moral right to expect the universe to hand us such an easy clue and string theory does not promise one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such difficulties do not mean that string theory is wrong as some opponents say. It just means that it is going to be difficult to explore quantum gravity empirically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There has still been constant progress in understanding string theory from the theoretical side and that is continuing. More work needs to be done on the non-perturbative side of string theory so that we can better understand its implications for cosmology. There is some hope that an observation of relic gravitational waves or even low frequency radio waves left over from the big bang may have a characteristic signature dependent on quantum gravitational effects. Again we have no moral rights to demand that such an observation will be forthcoming but it might if we are lucky.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-17T20:47:29.050" />
  <row Id="3192" PostTypeId="2" ParentId="3150" CreationDate="2011-01-17T20:48:10.550" Score="0" ViewCount="" Body="&lt;p&gt;The only thing which might be done is to cast the question in different forms.  The CY supermanifold $CP^{3|4}$ for the &quot;4&quot; corresponding to a spinor field and &quot;3&quot; coordinates might be cast into $ J^5(C) = R\oplus J^4\oplus C^4$, so the twistor components are contained in a $5\times5$ self adjoint matrix. By extension or analogue the question is whether this has some higher Jordan algebraic or a $J^3(O)$ realization.  The cubic form gives $OP^2 \sim OP^1$, which might (I stress might without any strong evidence) mean the $D1$ is dual to a $D2$ or $M2$.  The scalar part of this cubic form is the Chern-Simons form.  As for any duality with the $D5$ (or $NS5$ “black brane”) that would have to be determined.  The CS Lagrangian has a winding number transformation $L \rightarrow L + 2πNk$, which might then have a coordinate dual $x \rightarrow x + 2πiR$ winding or compactification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A chance to ponder per chance to solve.  This might be one way of trying to think about it.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-17T22:25:55.310" CommentCount="4" />
  <row Id="3193" PostTypeId="1" CreationDate="2011-01-17T20:53:29.373" Score="7" ViewCount="361" Body="&lt;p&gt;I understand that you can (in principle) measure whether &quot;free particles&quot; (no forces) experience accelerations in order to tell whether a frame is inertial. But fundamentally, what determines which frames are inertial (i.e. what principle selects in which frames free particles will not appear to accelerate)? I've been told that the cosmic microwaves determine the ultimate rest frame of the universe, but that doesn't make sense to me, since one can still ask why that frame is an inertial frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I understand that there are no real inertial frames in general relativity, but it seems like there certainly exists approximate inertial frames and we can ask why those frames are approximately inertial and not others. For example, in the frame of a person riding a merry go round, free particles appear to greatly accelerate; while in the frame of someone standing next to the merry go round there are no such great accelerations. Why does the guy (or gal) on the merry go round see free particles accelerating while the other guy doesn't. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you're gonna tell me that it's &quot;the rest of the stuff in the universe&quot; that determines whether the person on the merry go round sees free particles accelerate, I'll ask how you know that all that stuff is not spinning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this question sort of makes sense, it's been bothering me for a while and my study of relativity (most of special relativity and just the outline of general relativity) hasn't really clarified it for me much.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-04T07:35:32.117" Title="What determines which frames are inertial frames?" Tags="&lt;general-relativity&gt;&lt;special-relativity&gt;&lt;rotational-dynamics&gt;" AnswerCount="6" FavoriteCount="3" />
  <row Id="3194" PostTypeId="1" CreationDate="2011-01-17T20:58:42.383" Score="5" ViewCount="91" Body="&lt;p&gt;Imagine that I have a spherical particle of molecular weight $M$, volume $V$, and some experimentally observed center-of-mobility translational diffusion coefficient $D_{sphere}$ in water.  I take this sphere and compress it into a sheet with thickness $L$, keeping the $M$ and $V$ constant, and experimentally measure a new translational diffusion coefficient $D_{sheet}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I ab initio predict the ratio $\frac{D_{sheet}}{D_{sphere}}$?  Does the rigidity of the sheet matter?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-18T02:19:19.317" Title="Predicting the ratio of translational diffusion coefficients for a sphere and a sheet" Tags="&lt;experimental-physics&gt;&lt;biophysics&gt;" CommentCount="4" FavoriteCount="1" />
  <row Id="3195" PostTypeId="2" ParentId="3193" CreationDate="2011-01-17T21:10:54.220" Score="1" ViewCount="" Body="&lt;p&gt;Well, from the point of view of an inertial reference frame, angular momentum and linear momentum are conserved, and when you aren't in an inertial reference frame, they aren't conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, consider that you are on a spaceship accelerating away from a lone sun.  (ignoring an expanding universe for the moment).  In your frame, the sun is accelerating away from you, without anything to cause such an acceleration.  Momentum is not conserved from your frame of reference, and thus, you aren't in an inertial reference frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this helps.&lt;/p&gt;&#xA;" OwnerUserId="1256" LastActivityDate="2011-01-17T21:10:54.220" CommentCount="1" />
  <row Id="3196" PostTypeId="2" ParentId="3193" CreationDate="2011-01-17T21:16:56.053" Score="3" ViewCount="" Body="&lt;p&gt;In general relativity it is the gravitational field that determines which local frame of reference are local inertial frames. I talk about &quot;local&quot; frames because as you observed in your question, there are no global inertial reference frames in curved spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational field is represented by a metric tensor field. Given such a metric you can plot the timelike geodesics through any event. These are the trajectories that locally maximise the integrated proper-time calculated using the metric. Anything that moves along such a geodesic is in a local inertial frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cosmic microwave background does not define a preferred frame from the point of view of the equations of motion. It is just one of many reference frames that you can find in the universe by observation. In fact there are many different frames of reference that you could define based on just the cosmic background radiation, and they only coincide in a perfectly homogeneous universe, but the universe is not locally homogeneous. Furthermore these reference frames are in general mostly not inertial frames.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-17T21:16:56.053" CommentCount="7" />
  <row Id="3197" PostTypeId="2" ParentId="3193" CreationDate="2011-01-17T21:19:14.037" Score="5" ViewCount="" Body="&lt;p&gt;As you say, there's a perfectly sensible operational definition of an inertial frame: it's one in which free particles move with constant velocity. Even in general relativity, it makes sense to talk about inertial frames, but only locally. To be precise, an inertial frame is well-defined only in an infinitesimal neighborhood of a spacetime point, although in practice it's a sensible approximation to extend such a frame to a finite neighborhood, as long as the size is small compared to any length scales associated with spacetime curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that there are inertial frames is essentially an axiom of general relativity. The theory is based on the idea that spacetime has a certain geometric structure, which allows for the existence of geodesics, along which free particles travel. Within a sufficiently small neighborhood the geodesics near a given point &quot;look&quot; to a good approximation like what you'd get in an inertial frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there's not really a good answer to the question of why inertial frames exist: it's just part of the assumed framework of the theory. But that's not quite what you asked. You asked if there's a reason why a given frame S is inertial and a different frame S' isn't. It sort of depends on what you think would count as a reason. For a given spacetime geometry, the geodesics are well-specified (as solutions to a certain differential equation, or as curves that have certain geometrical properties). The inertial frames are the frames that make the geodesics look like straight lines. It's all terribly mathematically well-defined and self-consistent, but it may not have the intuitive feel of a &quot;reason why.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mention the possibility that the reason is &quot;all the other stuff in the universe.&quot; As you may know, this idea has a noble pedigree: it goes by the name of Mach's principle. Einstein was apparently quite enamored of Mach's principle when he was coming up with general relativity, and he would probably have been very happy if the theory had the property that the inertial frames were determined by all the other matter in the Universe. But general relativity's relationship with Mach's principle is complicated and problematic, to say the least. For instance, good old flat Minkowski spacetime is a perfectly valid solution to the equations of general relativity. That solution has well-defined inertial frames, even though there is no matter around to &quot;cause&quot; them.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-01-17T21:19:14.037" CommentCount="1" />
  <row Id="3198" PostTypeId="2" ParentId="2111" CreationDate="2011-01-17T21:23:08.193" Score="1" ViewCount="" Body="&lt;p&gt;Superman is going to manipulate the air pressure in the straw. To get the water to go up, he must provide a reduction in the pressure. It's clear that if this reduction is provided at a very slow rate, then he will not be able to significantly exceed 10 meters or so (as limited by atmospheric pressure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if he reduces the pressure quickly, it's at least possible that the water could reach a higher height. How high can the water go under this assumption? The idea is to use the momentum of the water to get the water higher, so the figure of merit will be the maximum speed of the water at surface level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reducing the pressure cannot move the water faster than the air and the air speed is limited by the speed of the gas molecules in the air or about 330 meters per second. By equating kinetic energy $0.5 m v^2$ with potential energy $mgh$, water with that initial speed can reach a height of $h = 0.5 v^2/g = $ 2775 meters. The height is small enough to justify the assumption that $g$ is a constant. Maybe you should add 10 meters for the usual vacuum effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hmmm. Ah, what the heck, I ought to just do the fracking calculation for how high the water goes in a wide straw when a vacuum is applied to it.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-01-19T04:51:18.313" LastActivityDate="2011-01-19T04:51:18.313" CommentCount="5" />
  <row Id="3200" PostTypeId="2" ParentId="3189" CreationDate="2011-01-17T21:41:27.597" Score="4" ViewCount="" Body="&lt;p&gt;good to ask about these historical controversies. It makes sense to review the controversy. In the medium, the density of momentum was (I set $c=\hbar=1$ everywhere)&#xA;$$E\times H$$&#xA;according to Max Abraham - that's equal to the Poynting vector which determines the flow of energy according to everyone - and&#xA;$$D\times B$$&#xA;according to Hermann Minkowski who argued that in the medium, the stress-energy tensor is asymmetric. In both cases, one can start with the product of the vacuum fields $E\times B$ and replace one of them by the material &quot;variation&quot; of the field. The two guys differed by whether they replaced the electric or the magnetic factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, a photon according to the first, Abraham form will have momentum&#xA;$$p=\omega/n$$&#xA;while the second, Minkowski definition will give&#xA;$$p=\omega n$$&#xA;Now, one may realize that the (complexified) electromagnetic wave depends on the position and the time as&#xA;$$\exp(-i\omega t+i k x)$$&#xA;where $k=\omega n$. The phase velocity is simply $\omega/k$ and it should be $1/n$, smaller than one, so it's clear that $k=\omega n$. So the wave function shows clearly that the conserved momentum of a single photon agrees with the Minkowski's template. By definition, the momentum is given by the $k$. Barnett agrees that the Minkowski momentum is what generates normal translations - of the vector potential and other fields - by conjugation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try to find out what are the arguments (and interpretations) that support the Abraham form, I think that all of them are incompatible with modern physics. As far as I can say, they're all based on the idea that the momentum of any particle should be equal to the &quot;mass times velocity&quot;. This is an extremely shaky assumption that is not really true in this case. In modern physics, the momentum has to be defined by a solid definition - and it is the &quot;quantity conserved, as showed by Noether's theorem, because of the spatial translational symmetry&quot;. One can show that this is $m_{total}v$ for ordinary particles in the vacuum but one cannot show this thing for a photon in a material - especially because it is not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very bizarre for Barnett to say that the &quot;canonical momentum is more subtle than the $mv$ momentun&quot;. The &quot;canonical momentum&quot; is the only momentum that is acceptable in modern physics and that may be generalized to new contexts. It's the only Noether-derived momentum. In this case, it's the Minkowski's momentum. The &quot;kinetic momentum&quot;, as he calls $mv$, is just an attempt to return to the 17th century as much as he can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For charged particles, there could be issues about $i\partial$ versus $i\partial-eA$, the velocity operator, but this doesn't occur for $A_i$ in the dielectric because the wave function of the photon cannot be transformed away by any gauge symmetries. Under the U(1) gauge symmetry with an oscillating parameter, $A$ may shift by a constant but it won't change the speed of its oscillations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other arguments supporting the Abraham form seem to be bogus, too. For example, the Abraham form is supported by the argument that the center-of-mass-energy continues in a uniform motion as the photon enters the material. However, this is an invalid assumption, too. The uniform motion of center-of-mass only holds - again according to Noether's theorem - in systems which respect the equivalence of inertial frames e.g. the Lorentz (or Galileo) symmetry. This symmetry is explicitly broken by the boundary between the vacuum and the dielectric material, so there is no reason why the corresponding conservation law - the conservation of the velocity of the center-of-mass motion - should hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, Minkowski was simply right while Abraham was wrong. But it's probably OK to &quot;redo&quot; the budget in such a way that a part of the momentum of the photon is attributed to the dielectric material when the photon enters it, and then it is returned back to the photon. In this way, one may justify the Abraham's form - and probably many other forms - but why should one really do it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stephen Barnett who wrote the 2010 paper above is a respectable optician but this whole thing was much ado about nothing. If you care about the sociology, I am pretty confident that almost no people in optics will oppose Barnett, and the people from other disciplines may say that this was really a trivial dispute. Minkowski was really right, and by the way, he was a much better physicist than Abraham from all other angles, too. For example, Abraham's model of the electron and the arguments used to support this model were truly painful. That's a very different league than Minkowski who was one of the first people to have understood special relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think that throughout the century, this problem was being studied by top physicists. To make things worse, the experiments are never testing these things &quot;directly&quot;, especially because one may always choose different interpretations where the momentum goes when the photon is changing the medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes,&#xA;Lubos&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: Let me mention one detail: Minkowski's stress-energy tensor is asymmetric, so if you define the density of the angular momentum purely in terms of this tensor, multiplied by $x$ in the usual way, the angular momentum won't be conserved. However, it's not a problem because this can be compensated by an extra contribution to the angular momentum that comes from the volume and surface density of the spins - including the internal angular momenta of the atoms of the materials. There are many ways how to attribute the conserved quantities (energy, momentum, angular momentum) to regions - only the total has to be conserved. The diverse interpretations will differ about all the functions, but they should ultimately predict the same predictions for the experiments when done right.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-17T21:50:53.200" LastActivityDate="2011-01-17T21:50:53.200" CommentCount="0" />
  <row Id="3201" PostTypeId="2" ParentId="2604" CreationDate="2011-01-17T21:54:30.930" Score="2" ViewCount="" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;The question:&lt;/em&gt;&lt;/strong&gt; &quot;I'm wondering if there is some good reason why the universe as we know it has to have twelve particles rather than just four.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;The short answer:&lt;/em&gt;&lt;/strong&gt; Our current standard description of the spin-1/2 property of the elementary particles is incomplete. A more complete theory would require that these particles arrive in 3 generations.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;The medium answer:&lt;/em&gt;&lt;/strong&gt; The spin-1/2 of the elementary fermions is an emergent property. The more fundamental spin property acts like position in that the Heisenberg uncertainty principle applies to consecutive measurements of the fundamental spin the same way the HUP applies to position measurements. This fundamental spin is invisible to us because it is renormalized away. What's left is three generations of the particle, each with the usual spin-1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a particle moves through positions it does so by way of an interaction between position and momentum. These are complementary variables. The equivalent concept for spin-1/2 is &quot;Mutually unbiased bases&quot; or MUBs. There are only (at most) three MUBs for spin-1/2. Letting a particle's spin move among them means that the number of degrees of freedom of the particle have tripled. So when you find the long time propagators over that Hopf algebra you end up with three times the usual number of particles. Hence there are three generations.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;The long answer:&lt;/em&gt;&lt;/strong&gt; The two (more or less classical) things we can theoretically measure for a spin-1/2 particle are its position and its spin. If we measure its spin, the spin is then forced into an eigenstate of spin so that measuring it again gives the same result. That is, a measurement of spin causes the spin to be determined. On the other hand, if we measure its position, then by the Heisenberg uncertainty principle, we will cause an unknown change to its momentum. The change in momentum makes it impossible for us to predict the result of a subsequent position measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As quantum physicists, we long ago grew accustomed to this bizarre behavior. But imagine that nature is parsimonious with her underlying machinery. If so, we'd expect the fundamental (i.e. before renormalization) measurements of a spin-1/2 particle's position and spin to be similar. For such a theory to work, one must show that after renormalization, one obtains the usual spin-1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A possible solution to this conundrum is given in the paper:&lt;br&gt;&#xA;Found.Phys.40:1681-1699,(2010), Carl Brannen, &lt;em&gt;Spin Path Integrals and Generations&lt;/em&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/1006.3114&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1006.3114&lt;/a&gt;&lt;br&gt;&#xA;The paper is a straightforward QFT resummation calculation. It assumes a strange (to us) spin-1/2 where measurements act like the not so strange position measurements. It resums the propagators for the theory and finds that the strange behavior disappears over long times. The long time propagators are equivalent to the usual spin-1/2. Furthermore, they appear in three generations. And it shows that the long time propagators have a form that matches the mysterious lepton mass formulas of Yoshio Koide.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Peer review:&lt;/em&gt;&lt;/strong&gt; The paper was peer-reviewed through an arduous process of three reviewers. As with any journal article it had a managing editor, and a chief editor. Complaints about the physics have already been made by competent physicists who took the trouble of carefully reading the paper. It's unlikely that someone making a quick read of the paper is going to find something that hasn't already been argued through. The paper was selected by the chief editor of Found. Phys. as suitable for publication in that journal and so published last year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chief editor of Found. Phys. is now Gerard 't Hooft. His attitude on publishing junk is quite clear, he writes&lt;br&gt;&#xA;&lt;strong&gt;How to become a bad theoretical physicist&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On your way towards becoming a bad&#xA;  theoretician, take your own immature&#xA;  theory, stop checking it for mistakes,&#xA;  don't listen to colleagues who do spot&#xA;  weaknesses, and start admiring your&#xA;  own infallible intelligence. Try to&#xA;  overshout all your critics, and have&#xA;  your work published anyway. If the&#xA;  well-established science media refuse&#xA;  to publish your work, start your own&#xA;  publishing company and edit your own&#xA;  books. &lt;strong&gt;&lt;em&gt;If you are really clever you&#xA;  can find yourself a formerly&#xA;  professional physics journal where the&#xA;  chief editor is asleep.&lt;/em&gt;&lt;/strong&gt;&#xA;  &lt;a href=&quot;http://www.phys.uu.nl/~thooft/theoristbad.html&quot; rel=&quot;nofollow&quot;&gt;http://www.phys.uu.nl/~thooft/theoristbad.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One hopes that 't Hooft wasn't asleep when he allowed this paper to be published.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Extensions:&lt;/em&gt;&lt;/strong&gt; My next paper on the subject extends the above calculation to obtain the weak hypercharge and weak isospin quantum numbers. It uses methods similar to the above, that is, the calculation of long time propagators, but uses a more sophisticated method of manipulating the Feynman diagrams called &quot;Hopf algebra&quot; or &quot;quantum algebra&quot;. I'm figuring on sending it in to the same journal. It's close to getting finished, I basically need to reread it over and over and add references:&lt;br&gt;&#xA;&lt;a href=&quot;http://brannenworks.com/E8/HopfWeakQNs.pdf&quot; rel=&quot;nofollow&quot;&gt;http://brannenworks.com/E8/HopfWeakQNs.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-02-18T21:31:48.273" LastActivityDate="2011-02-18T21:31:48.273" CommentCount="16" />
  <row Id="3202" PostTypeId="2" ParentId="2166" CreationDate="2011-01-17T21:55:06.550" Score="0" ViewCount="" Body="&lt;p&gt;IMO the answer is no. I share the Israel Pérez view  that time is not an entity 'de per se'. It is a mental construct percieved upon the existence of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See number 7 (pages 6/7) &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1012/1012.2423v1.pdf&quot; rel=&quot;nofollow&quot;&gt;A physicist’s view of the universe: a philosophical approach&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Understanding what time 'is', and what it is not, is fundamental to my answer.&#xA;Mathematically we can do whatever we like as long we dont contradict, but we need a  physical and philosophical insight when we try to understand the universe. This understanding narrows what can happen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole paper seems to me very interesting at several levels.&#xA;I think that he partially shares the view exposed by Espinosa in 'Ethics' about space/matter (available via the Guttenberg Project).&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-01-17T21:55:06.550" />
  <row Id="3203" PostTypeId="2" ParentId="3193" CreationDate="2011-01-17T22:11:04.533" Score="3" ViewCount="" Body="&lt;p&gt;In special relativity, the inertial frames are just postulated to exist. There is nothing wrong whatsoever with such an approach: they're frames in which objects will move by constant velocities if no forces act upon them. Newton needed pretty much the same thing to define the laws of mechanics. The important point about relativity (both Galilean and Einsteinian) is that if one frame is inertial, other frames moving uniformly with respect to this inertial frame are inertial frames, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is fully analogous to straight lines in the Euclidean geometry. (Inertial frames are just systems associated with observers whose world lines are straight lines in the spacetime - it's really the same thing in a different space.) Some lines on the paper simply are straight lines while others are not. One could also ask what is the principle that selects which lines are straight. Well, the principle is the set of axioms of the Euclidean geometry. One has to have a system that allows us to say things about the geometric objects - and being able to say whether a line is straight is among the &quot;tools&quot; we have to be given. If we describe the geometry in coordinates that we call Cartesian, then a straight line is given by $ax+by+c=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no confusion here unless someone deliberately tries to produce it. Asking who dared to make some systems inertial and others not is analogous to asking why mathematics discriminates against some numbers - because some of them are primes while others are not. Well, mathematics discriminates and it has the full right to do so. It's the very purpose of mathematics - and science - to discriminate all the time. Every time we ask a question, we want to hear the right answer and discriminate against all the other possible answers - the wrong ones. The right answer inevitably discriminates - it treats various objects or numbers asymmetrically. No maths or science could work if someone required the permanent democracy between everyone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general relativity, the spacetime is curved and a curved spacetime contains no reference frames - or systems of coordinates - in which the space would look flat. It's simply not flat. So there are no exact inertial frames in general relativity. In general relativity, one may only approximate the notion of an inertial system. One possible definition is that an inertial frame is a good approximation for local phenomena around freely falling objects. If an elevator is freely falling, you may call the frame associated with this elevator &quot;inertial&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, here on the Earth, it is not the usual choice. We usually say that the freely falling elevator is accelerating i.e. non-inertial. On the contrary, it is the non-falling elevator at rest that is inertial - even though it is not associated with geodesics in general relativity. The choice of the falling elevator has the advantage that you don't have to include the gravitational force among the forces that act on the objects inside the elevator. The only thing you have to include is the crash that will kill you: it's not the free fall but the collision with the ground that becomes your fate. :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you choose that the inertial system is linked to the elevator at rest, you have to add the attractive gravitational force to the equations for all objects on the Earth. Of course, that will make your description of high-velocity phenomena etc. a bit inaccurate. But it's simply the case that curved spacetimes - and nontrivial gravitational fields - can't be exactly described by special relativity (and its inertial frames) only. If this were possible, we wouldn't need general relativity. It's not possible and we need general relativity to describe gravity in the relativistic context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the broader cosmic context, away from the gravitational field of the Earth or the Sun, the approximate inertial frames can be defined by the freely moving objects. One of them will be the frame associated with the cosmic microwave background - the frame of spheres such that the total momentum hiding in the CMB photons that are crossing the sphere at each point of the sphere's surface is zero. The CMB frame determines not only what is the vanishing acceleration; it also determines what is the vanishing speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this benchmark in place, one may discuss the motion of the Sun (and the Solar System), our Galaxy, galaxy clusters and superclusters where we belong etc. relatively to the CMB cosmic frame. Those speeds are kind of known. But it's useful to recall that these speeds don't really mean that the Solar System frame is far from inertial. It's because uniform speeds don't spoil the inertial character of the reference frame. So even though the Sun is moving relatively to the CMB frame, by a rather high speed, the system associated with the Sun - and properly oriented relatively to some other galaxies etc. - is inertial with a huge accuracy.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:35:32.117" LastActivityDate="2011-02-04T07:35:32.117" CommentCount="4" />
  <row Id="3204" PostTypeId="2" ParentId="937" CreationDate="2011-01-17T22:17:16.040" Score="2" ViewCount="" Body="&lt;p&gt;While in many ways the question was already answered, I think it should be empasized that on the classical level, the question is in some sense backwards. The prior discussion of static and dynamic properties especially comes very close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's first examine a toy model of a spherically-symmetric thin shell of dust particles collapsing into a Schwarzschild black hole. The spacetime outside of the the shell will then &lt;em&gt;also&lt;/em&gt; be Schwarzschild, but with a larger mass parameter than the original black hole (if the shell starts at rest at infinity, then just the sum of the two). Intuitively, the situation is analogous to Newton's shell theorem, which a more limited analogue in GTR. At some point, it crosses the horizon and eventually gets crushed out of existence at the singularity, the black hole now gaining mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have the following picture: as the shell collapses, the external graviational field takes on some value, and as it crossed the horizon, the information about what it's doing can't get out the horizon. Therefore, he gravitational field &lt;em&gt;can't&lt;/em&gt; change in response to shell's further behavior, for this would send a signal across the horizon, e.g., a person riding along with the shell would be able to communicate across it by manipulating the shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, rather than gravity having a special property that enables it to cross the horizon, in a certain sense gravity can't cross the horizon, and it is that very property that forces gravity outside of it to remain the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the above answer assumed a black hole already, that doesn't matter at all, as for a spherically collapsing star the event horizon begins at the center and stretches out during the collapse (for the prior situation, it also expands to meet the shell). It also assumes that the situation has spherical symmetry, but this also turns out to not be conceptually important, although for far more complicated and unobvious reasons. Most notably, the theorems of Penrose and Hawking, as it was initially thought by some (or perhaps I should say &lt;em&gt;hoped&lt;/em&gt;) that any perturbation from spherical symmetry would prevent black hole formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also be wondering about a related question: if the Schwarzschild solution of GTR is a vacuum, does it make sense for a vacuum to bend spacetime? The situation is somewhat analogous to a simpler one from classical electromagnetism. Maxwell's equations dictate how the electric and magnetic fields change in response to the presence and motion of electric charges, but the charges alone do not determine the field, as you can always have a wave come in from infinity without any contradictions (or something more exotic, like an everywhere-constant magnetic field), and in practice these things are dictated by boundary conditions. The situation is similar in GTR, where the Einstein field equation that dictates how geometry are connected only fixes half of the twenty degrees of freedom of spacetime curvature.&lt;/p&gt;&#xA;" OwnerUserId="1353" LastActivityDate="2011-01-17T22:17:16.040" />
  <row Id="3205" PostTypeId="2" ParentId="3182" CreationDate="2011-01-17T22:42:19.280" Score="8" ViewCount="" Body="&lt;p&gt;&lt;em&gt;NOTE: this started out as a quick answer and kind of got out of hand... but everything below the second line is probably optional&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Actually, everything is quantum. The term &quot;classical object&quot; in that sense just refers to a system in which the classical approximation is valid, i.e. where quantum effects like uncertainty are negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about it like this: when you make a measurement, it changes the wavefunctions of both the measured particle and the detector. For example, suppose you're measuring the momentum of an elementary particle. When you make the measurement, the particle is going to lose some momentum and the detector gains the corresponding amount of momentum (or vice versa). However, the detector consists of a large number of particles, usually something in the $10^{20}$-ish range, so if you add or subtract one particle's worth of momentum, the change in the detector's wavefunction is pretty negligible. Because of this, you can use the classical approximation that the detector is unchanged by the measurement, and thus you say it's a classical object.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You can even do a quantitative analysis of this. But first, a quick aside on decoherence. Suppose your particle starts out in a state $\int\mathrm{d}p_p\ f_p(p_p)|p_p\rangle$ and the detector starts out in a state $\int \mathrm{d}p_d\ f_d(p_d)|p_d\rangle$, where $f_p$ and $f_d$ are the momentum-space wavefunctions of the particle and detector, respectively. The measurement process basically convolves the two states with a parameter that represents the momentum transferred, so the final state winds up being&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\mathrm{d}p_p\mathrm{d}p_d\mathrm{d}q\ f_p(p_p)f_d(p_d)f(q) |p_p - q\rangle |p_d - q\rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f(q)$ is some wavefunction describing the relative probabilities of various results of the measurement. In practice, we will read off a particular value of $q$, call it $q_0$, from the measuring device, which is mathematically represented by inserting a delta function $\delta(q - q_0)$ under the integral. That's what makes the states of the detector and particle appear to decohere,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\mathrm{d}p_p\mathrm{d}p_d\mathrm{d}q\ f_p(p_p)f_d(p_d)f(q)\delta(q - q_0) |p_p - q\rangle |p_d + q\rangle$$&#xA;$$\propto\int\mathrm{d}p_p\ f_p(p_p)|p_p - q_0\rangle\int\mathrm{d}p_d\ f_d(p_d)|p_d + q_0\rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose the detector contains 150 particles of the same type as the one being measured. Each of these 151 particles has a Gaussian momentum distribution $g(p)$, all with the same variance. For the particle wavefunction, $f_p(p_p) = g(p_p)$. But the detector wavefunction is a little more complicated, since you have to obtain it by &lt;a href=&quot;http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables&quot; rel=&quot;nofollow&quot;&gt;adding up 150 copies of the wavefunction&lt;/a&gt; $g(p)$ using the arithmetic of random variables; it winds up being a Gaussian with $\sqrt{150}$ times the width and $150$ times the mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, with the setup established, let's move on to the results. Before the measurement, the particle (green) and detector (red) wavefunctions look like this,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/kZRUp.png&quot; alt=&quot;Wavefunctions pre-measurement&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I've scaled the detector wavefunction down by 150 on the horizontal axis, so you're really looking at the &lt;em&gt;average&lt;/em&gt; momentum per particle in the detector. This way it's easier to get a sense of the &lt;em&gt;relative&lt;/em&gt; effect of any change in momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose I happen to measure $q_0 = 5$ in whatever units we're working with. Using the formula above, here's what the decohered wavefunctions look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/z0Gf5.png&quot; alt=&quot;Wavefunctions post-measurement&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the particle wavefunction has shifted quite a bit, but the (scaled) detector wavefunction has barely moved at all. So it's still probably a reasonable approximation to say that the detector has zero momentum, i.e. you can treat it like a classical object.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-01-17T22:42:19.280" />
  <row Id="3206" PostTypeId="1" AcceptedAnswerId="3207" CreationDate="2011-01-17T22:50:13.237" Score="4" ViewCount="161" Body="&lt;p&gt;Let $(u_x,u_y,u_z)$ be a unit vector. Then a normalized vector for spin-1/2 in this direction is given by normalizing a column from the pure density matrix: $\frac{1}{\sqrt{2+2u_z}}\left(\begin{array}{c}1+u_z\\u_x+iu_y\end{array}\right)$. This solution works except when $u_z=-1$. So this gives a normalization choice of spin vector that is continuous over the possible eigenstates of spin with the exception of spin down = $\left(\begin{array}{c}0\\1\end{array}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is impossible to choose normalized spin-1/2 vectors for all possible eigenstates of spin-1/2. This is a fact that is related to Berry-Pancharatnam phase. The best you can do is to leave a single point, as in the above, where there is an essential discontinuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'd like to know is what other j have spin-j vectors unable to be normalized. That is, given all possible eigenstates of spin-j, is it possible to choose one from each ray, such that the choice is continuous?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Partial Solution: Per Dr. Motl's answer below, I'll down the situation for a natural representation of spin-1. Choose&#xA;$S_x = i\left(\begin{array}{ccc}0&amp;amp;0&amp;amp;0\\0&amp;amp;0&amp;amp;+1\\0&amp;amp;-1&amp;amp;0\end{array}\right)$&#xA;$S_y = i\left(\begin{array}{ccc}0&amp;amp;0&amp;amp;-1\\0&amp;amp;0&amp;amp;0\\+1&amp;amp;0&amp;amp;0\end{array}\right)$&#xA;$S_z = i\left(\begin{array}{ccc}0&amp;amp;+1&amp;amp;0\\-1&amp;amp;0&amp;amp;0\\0&amp;amp;0&amp;amp;0\end{array}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to verify that these satisfy the $SU(2)$ rules $[S_x,S_y] = iS_z$, etc. Then the spin operator for spin in the $u = (u_x,u_y,u_z)$ direction is given by:&#xA;$S_u = i\left(\begin{array}{ccc}0&amp;amp;u_z&amp;amp;-u_y\\-u_z&amp;amp;0&amp;amp;u_x\\u_y&amp;amp;-u_x&amp;amp;0\end{array}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the solution with spin=0 in the $u$ direction is given by the vector $u$. And sure enough, it's impossible to choose the vectors for spin=1 as a continuous function of $u$.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-01-19T00:07:41.400" LastActivityDate="2011-02-04T07:36:29.263" Title="For which irreps can normalized spin vectors be chosen continuously, for all possible spins?" Tags="&lt;quantum-mechanics&gt;&lt;spin&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="3207" PostTypeId="2" ParentId="3206" CreationDate="2011-01-17T23:11:32.523" Score="2" ViewCount="" Body="&lt;p&gt;The fact that you can't find such a spinor-valued function on the sphere is analogous to the well-known theorem that &quot;you can't comb the sphere&quot; except that the bundle is a bit different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $j=1$, you can find a normalized continuous vector for each direction, namely the $u$ vector itself - properly transformed in the right basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I suppose you want the spin-$j$ vectors to be those of the maximum component $j_z=j$; strictly speaking, you haven't written it but it's the only interpretation I can imagine - that generalizes the $j=1/2$ case you described. That kills the previous paragraph because the vector $u$ itself as a function on the sphere is the solution for $j_z=0$. In particular, for $u=(0,0,1)$, the vector corresponding to the function $z$ is nothing else than the $m=0$ component of the triplet. Recall that $m=\pm 1$ correspond to the functions $x\pm iy$ on the sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were trying to solve the problem for $j=1$ and $j_z=\pm 1$, you would fail to find a solution again. In that case, the problem would become totally equivalent to the combing of the sphere - because you would be looking for tangential vectors to the sphere's surface - but it is impossible to comb the sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of these two no-go cases for nonzero $j_z$, I suppose that the problem can only be solved for $j$ being integer and $j_z=0$ in which case the solution is given by the symmetric tensor power of the solution for $j=1$, $j_z=0$, with the trace removed. I believe that all cases when you want a continuous $(2j+1)$-tuplet with $j_z$ along the variable axis being nonzero - which includes all half-integral $j$ cases as well as most integral $j$ cases, except those with $j_z=0$ - will suffer from the &quot;combing the sphere&quot; problem.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:36:29.263" LastActivityDate="2011-02-04T07:36:29.263" CommentCount="1" />
  <row Id="3208" PostTypeId="2" ParentId="3151" CreationDate="2011-01-18T00:01:15.360" Score="1" ViewCount="" Body="&lt;p&gt;Even in some less extreme environments, solar energy may make sense today. I have some panels on my roof in California, and the fact that peak solar production is highly correlated with peak demand, probably makes them beneficial for the grid as a whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, there are breakthroughs needed to make them a slamdunk. One area is efficiency, the best currently is around 40%, but this are very pricy multijunction cells, which only make economic sense for satelites, or for systems where sunlight is optically concentrated to circa a thousand times the unfocused intensity. I've seen some sciencedaily reports about projects to use transparent conductors to allow severalway junction cells, with perhaps 65% efficiency. If achievable this might make highly concentrated (optically focused) systems cost effective, as well as significantly reducing the landarea needed. We also have more standard (10% to 20%) silicon cells coming down in price -but this is heavily modulated by supply/demand, and currently demand is low. Nevertheless if medium efficiency panels (and balance of system components such as mounting and inverters) can continue to come down in price, the number of areas/applications where they make economic (as opposed to simple feelgood) sense should be on the rise. &lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-01-18T00:01:15.360" CommentCount="2" />
  <row Id="3209" PostTypeId="2" ParentId="1998" CreationDate="2011-01-18T00:23:49.257" Score="0" ViewCount="" Body="&lt;p&gt;The usual (and correct) method is to take the average speed over time. What you are defining is the average speed over distance. In computing average speeds, we have to take the average over time basically because speed has time in the denominator. That is, distance is given by the product of velocity and time so in computing distances correctly we have to average over time. In other words, what we seek is proportional to time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An area where the reverse definition would be appropriate is one where we are looking for an effect that is proportional to distance, but still depends on velocity. An example might be the wear and tear on roads, as a function of the speeds of the vehicles driving on it. Or a computation of the number of fatalities per mile. Average fuel consumption in gallons per mile might do the trick.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-18T00:23:49.257" />
  <row Id="3210" PostTypeId="2" ParentId="3193" CreationDate="2011-01-18T00:29:29.630" Score="0" ViewCount="" Body="&lt;p&gt;In ordinary mechanics, inertial frames are defined by constant-velocity observers. This is still true in GTR--experimentally, &quot;inertial motion&quot; means &quot;no proper acceleration&quot;, which can be measured by an accelerometer. The equivalence principle means that at least locally, mechanical systems act like there is no gravity, so on the conceptual level, GTR &quot;inherits&quot; inertial frames from ordinary mechanics, although in a more limited sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the formal perspective, it's exactly as axiomatic as the Euclidean geometry you referenced above. Say you have two curves intersecting at p on a differentiable manifold, and in some coordinates they have same derivative at p, so you say that they're in the same direction. The equivalence classes define the &lt;em&gt;tangent space&lt;/em&gt; at p (which can also be done in a variety of other ways), which is a vector space of &quot;directions from p&quot;. In particular, the velocity of a particle at some point is just the tangent vector of its curve in spacetime (wordline).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do we interpret inertial motion as &quot;having the same velocity&quot;? To do so, we need to be able to compare two vectors between the tangent spaces of &lt;em&gt;different points&lt;/em&gt;. Or, equivalent, to be able to transport vectors (and in general, tensors) from one point or another. The mathematical device that lets us do this is called the &lt;em&gt;connection&lt;/em&gt;, and like many things in mathematics, it is completely general, with infinitely many possible distict connections on the same manifold. &quot;Inertial motion&quot; or &quot;geodesic&quot; will then just be a curve whose tangent vector stays the same when transported between infinitesimally close points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where the GTR's postulate comes in. Assume that it is &quot;metric-compatible&quot;, which means that the inner product between two vectors likewise unchanged when transported along a curve. Also assume that the connection is &quot;torsion-free&quot;, which is a fancy way of saying that if you transport a vector around a small loop in spacetime back to the starting point, then to first order it is unchanged. It turns out that these conditions uniquely determine the connection. (Incidentally, in the second-order change in the loop transport operation defines the Riemann curvature.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, GTR postulates that the &quot;constant velocity&quot; curves are exactly the length-extremizing curves dictated by the metric tensor $g_{\mu\nu}$. And just like Euclidean postulates, it is quite possible to take a contradictory option. For example, Einstein-Cartan theory of gravity does not have the above assumptions, and in its torsion terms allows spin angular momentum and its &lt;em&gt;gravitational&lt;/em&gt; exchange with orbital angular momentum, unlike in GTR.&lt;/p&gt;&#xA;" OwnerUserId="1353" LastActivityDate="2011-01-18T00:29:29.630" />
  <row Id="3211" PostTypeId="2" ParentId="937" CreationDate="2011-01-18T00:30:30.407" Score="2" ViewCount="" Body="&lt;p&gt;The holographic principle gives a clue, as pointed out by David Zaslavsky.  The Schwarzschild metric element $g_{tt}~=~1 – r_0/r$, for $r_0~=~2GM/c^2$ gives a proper distance called the delay coordinate&#xA;$$&#xD;&#xA;r^*~=~r~+~r_0 ln[(r-r_00)/r_0]&#xD;&#xA;$$&#xA;which diverges $r^*~\rightarrow~-\infty$ as you approach the horizon.  What this means is that all the stuff which makes up the black hole is never seen to cross the horizon from the perspective of a distant outside observer.  The clock on anything falling into a black hole is observed to slow to a near stop and never cross the horizon.  This means nothing goes in or out of the black hole, at least classically.  So there really is not problem of gravity escaping from a black hole, for as observed from the exterior nothing actually ever went in.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-18T00:30:30.407" CommentCount="2" />
  <row Id="3212" PostTypeId="2" ParentId="3151" CreationDate="2011-01-18T01:20:58.927" Score="0" ViewCount="" Body="&lt;p&gt;Doped layers of graphene on a plastic substrate, or other thin film technologies, should permit one to buy solar cells in rolled sheets.  One would cut out sheets of the stuff, as if one were cutting pieces of wall paper, and then “plug &amp;amp; play” from there  These would be put on roof tops and other places.  Further, these would be lithographically produced, as opposed to using silicon melt or vapor deposition.  This would reduce the cost to maybe as low as \$10US per square meter.  This could be used to solarize lots of “dead” areas on top of buildings and other places.  About 5% of the Earth is covered by human structures, and even if 1% were solarized there could be a considerable solar energy input.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy future has to involve a matrixed approach with a diversity of sources.  Wind energy is playing a growing role.  I have thought that tethered submersible turbines in the ocean conveyer could generate a fair amount of energy.  Then nuclear power will have to be employed to fill in the down times on renewable sources.  By 2030 we will be well down the post peak on oil, and by 2050 the same with coal may hold true.  So we need to get our butts in gear over the energy issue.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-18T01:20:58.927" CommentCount="2" />
  <row Id="3213" PostTypeId="2" ParentId="3151" CreationDate="2011-01-18T01:32:06.560" Score="6" ViewCount="" Body="&lt;p&gt;The simple answer is &lt;em&gt;price&lt;/em&gt;. Solar panels are still highly cost inefficient (particularly in low-sunlight regions) compared to conventional methods of energy production.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be fair, there is really very little physics to be said about this. Efficiency, while far from 100%, is still pretty decent (typically from 10% to 30%). One interesting area just appearing is &lt;em&gt;organic solar panels&lt;/em&gt;, which are significantly cheaper to produce. If the issues of efficiency (already getting to decent levels I hear) and lifetime are improved, then this could be a real boost for solar panels.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-01-18T09:58:09.243" LastActivityDate="2011-01-18T09:58:09.243" CommentCount="1" />
  <row Id="3214" PostTypeId="1" CreationDate="2011-01-18T02:00:02.030" Score="2" ViewCount="143" Body="&lt;p&gt;Imagine that I have a polymer (approximated as a freely diffusing, freely jointed chain with some number of subunits 'N'), and I place this polymer into a sphere of some volume 'V'. Next, I proceed to add a series of infinitely thin, immobile chords of length 'L' to the inner walls of the sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because these chords are infinitely thin, they will not change the inside volume of the sphere, but should nevertheless place certain topological and geometric constraints on the behavior of the diffusing polymer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we quantify the change in entropy caused by the addition of these infinitely thin chords? For this calculation, what changes if we replaced the polymer with a monoatomic gas?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-18T11:53:25.350" Title="Entropy of a polymer contained in a sphere with infinitely thin chords" Tags="&lt;statistical-mechanics&gt;&lt;biophysics&gt;&lt;topological-entropy&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="3215" PostTypeId="2" ParentId="3193" CreationDate="2011-01-18T02:17:05.923" Score="1" ViewCount="" Body="&lt;p&gt;The previous answers have covered every philosophical and epistemological aspect, of the question of inertial reference frames (IRFs), that I could think of. What has not been mentioned is a &lt;em&gt;physical&lt;/em&gt; construction in terms of the kinematics of moving bodies. The reference for this answer is Michael Dickson's paper on &lt;a href=&quot;http://mdickson.net/pubs/Quantum_Reference_Frames.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Quantum Reference Frames&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This 106 year old construction is originally due to L. Lange (&quot;&lt;em&gt;Ueber das Beharrungsgesetz&lt;/em&gt;&quot;, Leipziger Berichte 37: 333-351, 1885). Lange's defines an inertial frame of reference as (quoting from Dickson):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a coordinate system in which each of some triple of force-free particles, moving from some common 'origin' in non-coplanar directions, moves in a straight line, the particles traveling mutually proportional distances in equal times. The law of inertia is then the claim that any other free particle will also move uniformly in such a coordinate system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once we have this working definition of an IRF, we can easily start to see where it can fail and thus shed light on the inherent physical limitations of the concept. What triples of particles existing in Nature can actually satisfy this construction? Three hard spheres moving inertially in a box with elastic boundary conditions (3D billiards), three celestial bodies - stars or galaxies, three elementary particles - each case defining an IRF on mesoscopic, macroscopic and microscopic scales respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that each such triple can serve as a basis for an IRF over some finite range of scales. So this discussion immediately suggests that in our non-scale-invariant Universe (i.e. with a hierarchy of scales - stars, billiards, atoms) there does not and cannot exist any single IRF valid on all scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more detailed discussion I strongly recommend reading Dickson's engrossing article.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-18T03:15:52.557" />
  <row Id="3216" PostTypeId="1" AcceptedAnswerId="3415" CreationDate="2011-01-18T02:47:52.610" Score="6" ViewCount="311" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Anthropic_principle&quot; rel=&quot;nofollow&quot;&gt;Anthropic Principle&lt;/a&gt; says &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;physical Universe must be compatible&#xA;  with the conscious life that observes&#xA;  it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I thought it is just philosophy mumbo-jumbo and it's not physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I could be wrong; I wonder whether is it possible to formulate it into a specific scientific theory that is falsifiable, at least in principle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument that &quot;the theory is true because we are here; our existence proves the theory and if we are not here, the theory is disproved&quot; is extremely unsatisfying, mainly because it can be formulated to support any kind of philosophical mumbo-jumbo.&lt;/p&gt;&#xA;" OwnerUserId="605" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-18T21:26:43.537" LastActivityDate="2011-01-20T14:16:23.883" Title="Can (or How) Anthropic Principle be Made into a Scientific Theory?" Tags="&lt;cosmology&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="3217" PostTypeId="2" ParentId="3179" CreationDate="2011-01-18T03:00:31.063" Score="-2" ViewCount="" Body="&lt;p&gt;This is meant to segue further from the answer above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $SL(2,~{\bf C})$ and $SU(1,~1)$ groups are defined&#xA;on four and three dimensional spacetime respectively.  The&#xA;generators of these groups exist on the principal bundles&#xA;$P(\Sigma,~{\cal G})$, where $\Sigma$ is a Cauchy surface (spatial&#xA;surface) of dimensions 3 and 2 for $SL(2,~{\bf C})$ and&#xA;$SU(1,~1)$.  The set of connections $A$ on the principle&#xA;bundle define Wilson loops $\oint Adx$ are maps&#xA;$\mu:[0,~1]~\rightarrow~\sigma$ for $\mu(0)~=~\mu(1)$.  An&#xA;element of $\cal G$ assigned to $\mu$ by the holonomy map&#xA;$H(\mu,~A)$ defines a function&#xA;$$&#xD;&#xA;F_\mu(A)~=~{1\over 2}Tr~H(\mu,~A)&#xD;&#xA;$$&#xA;that is invariant with respect to the group action of $\cal G$  Thus $F_\mu(A)$ is an element of ${\cal M}~=~A/{\cal G}$ or the moduli space.  The rotation $\sigma_3~\rightarrow~i\sigma_3$ will carry the separability condition to the noncompact case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group $SU(1,1)$ is related to $SU(2)$ by the signature change on the basis elements $\sigma_1,~\sigma_2,~\sigma_3$ of $SU(2)$.  For $\sigma_\pm~=~\sigma_1~\pm~\sigma_2$, the basis for $SU(1,~1)$ are then $\sigma_+,~\sigma_-,~\tau_3~=~i\sigma_3$.  Now consider a connection one-form&#xA;$$&#xD;&#xA;A~=~A^+\sigma_+~+~A^3\sigma_3&#xD;&#xA;$$&#xA;and a gauge transformation determined by the group action of&#xA;$g~\in~{\cal G}$, $g~=~e^{i\lambda\tau_3}$.  The gauge&#xA;transformed connection is then&#xA;$$&#xD;&#xA;A^\prime~=~g^{-1}Ag~+~g^{-1}dg~=~e^{-2\lambda}A^+\sigma_+~+~A^3\sigma_3,&#xD;&#xA;$$&#xA;where $d\lambda~=~A^3$.  Thus $\lambda$ is a parameterization of the gauge orbit for this connection.  This leads to the observation&#xA;$$&#xD;&#xA;\lim_{\lambda\rightarrow\infty}A(\lambda)~\rightarrow~A^3\sigma_3,&#xD;&#xA;$$&#xA;where $ A^+\sigma_+~+~A^3\sigma_3$ and $A^3\sigma_3$ have distinct holonomy groups and thus represent distinct points in the moduli space $\cal M$.  However by the last equation this gives&#xA;$$&#xD;&#xA;F_\mu(A^+\sigma_+~+~A^3\sigma_3)~=~ F_\mu(A^3\sigma_3),&#xD;&#xA;$$&#xA;which obtains similarly for any gauge invariant function.  Hence there exist two distinct points in the moduli space that define the same set of gauge invariant functions.  Hence there does not exist a measure over these two points that separates them and $\cal M$ is then nonHausdorff with a Zariski topology. &lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-01-18T03:43:24.270" LastActivityDate="2011-01-18T03:43:24.270" CommentCount="8" />
  <row Id="3218" PostTypeId="2" ParentId="3216" CreationDate="2011-01-18T03:09:49.383" Score="0" ViewCount="" Body="&lt;p&gt;The anthropic principle (at least in it's weak form) is neither philosophy nor science. It is simply a statement about the admissibility of evidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any precondition to an observation is not evidence about the whatever you are observing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Duh!&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-01-18T03:09:49.383" />
  <row Id="3219" PostTypeId="2" ParentId="3145" CreationDate="2011-01-18T03:26:21.063" Score="2" ViewCount="" Body="&lt;p&gt;Answer to the additional question: &quot;what is the quantum efficiency of the eye?&quot;&#xA;I remember old studies reporting that eight photons reaching the photoreceptors can be detected by a human observer as a flash of light in certain laboratory conditions. The more normal range would certainly be lower than 0.125 say 0.05 - 0.01. Cooling down laboratory animals thus reducing thermal noise in the retina enhances sensitivity but means longer integration times and lower spatial and temporal resolution.&lt;/p&gt;&#xA;" OwnerUserId="1391" LastActivityDate="2011-01-18T03:26:21.063" />
  <row Id="3220" PostTypeId="2" ParentId="2322" CreationDate="2011-01-18T03:31:10.097" Score="1" ViewCount="" Body="&lt;p&gt;is there a way to verify that the attributed quot was really uttered by Tesla?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything that can be verified as Tesla's work in this area exists?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the faster than light article by Hugo Gernback (that Sklivvz send a link to ), there is a patent number being mentioned 787412, filed May 16 1900. Is there a way to verify that?&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-01-18T06:21:00.857" LastActivityDate="2011-01-18T06:21:00.857" CommentCount="2" />
  <row Id="3222" PostTypeId="1" CreationDate="2011-01-18T04:04:32.387" Score="4" ViewCount="115" Body="&lt;p&gt;I have the second edition of Houghton's &quot;The Physics of Atmospheres&quot;.  In section 2.2 he says one can do simple radiative transfer calculations in a plane parallel atmosphere by assuming that there are two fluxes, one going vertically up and one vertically down.  In reality the radiation is going in all directions, but the two streams represent the amount integrated over the two respective hemispheres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in reality different rays will see a given layer of atmosphere as having different optical depths, depending on the angle a given ray makes with the normal.  Houghton says that detailed calculations show that one can take this into account by replacing the true thickness of a given layer dz with 5/3 dz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my question--  Where does that 5/3 factor come from?  When I do the calculation for an optically thin absorbing layer (as one would expect for something of infinitesmal thickness dz) I find that an initially isotropic stream (or rather, that half which is going either down or up) will have twice as much energy absorbed as would happen if all the radiation were travelling along a line normal to the layer.  So I would have guessed you'd replace dz with 2 dz rather than 5/3 dz, but Houghton is the expert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Houghton uses this 5/3 factor several times in the book.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2011-02-17T21:02:29.330" Title="Mean optical depth in plane parallel atmospheres" Tags="&lt;atmosphere&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="3223" PostTypeId="2" ParentId="3216" CreationDate="2011-01-18T04:18:32.230" Score="0" ViewCount="" Body="&lt;p&gt;I don't think the argument can be stated as &quot;The theory is true because we are here; our existence proves the theory and if we are not here, the theory is disproved&quot; because, as far as I know, the anthropic principle is not a statement about the truth of something. Instead, is a statement regarding the necessary conditions to produce intelligent life aware of certain features of its environment. The specifics are given depending on which anthropic principle  you are talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously the anthropic principle is not a theorem or a law or anything of that sort, but it adds what the common sense would add: In case you're dealing in physics with something that is restricted by a result you're expecting to obtain (for example, a process that permits the creation of life), then by avoiding solutions not allowing this final result, you're implicitly using the anthropic principle. However, I would find hard to argue that we actually need a fancy name for this kind of situation.&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2011-01-18T04:18:32.230" />
  <row Id="3224" PostTypeId="2" ParentId="3168" CreationDate="2011-01-18T05:09:13.883" Score="3" ViewCount="" Body="&lt;p&gt;Excellent question, Kostya. Lubos already gave a detailed answer using general arguments in the language of QFT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In astrophysics and cosmology, however, there is another, and very simple, reason why we use the correlation functions all the time. It turns out that the mean value of the function $f(\vec{x})$, denoted $\langle f(\vec{x})\rangle$, can often not be predicted by the theoretical model (e.g. hot Big Bang model with inflationary stage early on, cold dark matter at late times, etc... or whatever other model you wish to consider) -  while its correlation $\langle f(\vec{x})f(\vec{y})\rangle$ &lt;strong&gt;can&lt;/strong&gt; be predicted. Here $f$ can refer to any cosmological observable quantity, and $\vec{x}$ and $\vec{y}$ refer to spatial coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most common example would be to consider the excess density of dark matter, $f(\vec{x})\equiv \delta\rho(\vec{x})/\rho$, where $\rho$ is the mean density (units of which are kilograms per meter cubed for example) and $\delta\rho(\vec{x})$ is the excess over- or under-density at location $\vec{x}$, and over some region which I will not specify for simplicity of the argument. By definition, the mean of $f$ is zero, so we explicitly indicate that we are not interested in the mean (alternatively, we cannot easily get the mean density of the universe from first principles). But the correlation function, $\langle \delta\rho(\vec{x})\delta\rho(\vec{y})/\rho^2\rangle $ can be related to fundamental parameters of the universe, in particular details of the inflationary epoch, dark matter density, etc. Details of this are involved, and are taught in a graduate course in cosmology. Suffice it to say that theory predicts not the mean of the function (1-point correlation function), but rather its (co)variances (2-point correlation function). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, the two-point correlation function of $\delta\rho/\rho$ is related to the &quot;probability that, given an overdense region of dark matter at location $\vec{x}$, there is an overdense region at location $\vec{y}$&quot;, and this probability is determined by the good old law of gravity - and can be predicted from first principles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theory also in principle predicts the 3-point (e.g. $\langle f(\vec{x})f(\vec{y})f(\vec{z})\rangle$, and higher-point correlation functions, but those are both harder to calculate theoretically and measure observationally.  Nevertheless, there is a thriving subfield in particle physics and cosmology of predicting theoretically, and measuring observationally, these so-called higher-order correlation functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One final ingredient in all this is the role of measuring the correlation function. The angular averaging sign, $\langle\cdot\rangle$ implies that we should be averaging over different realizations of the system - that is, the universe - in the same &lt;em&gt;underlying cosmological model&lt;/em&gt;. This is clearly impossible, since we have only one universe to measure! Instead, we assume statistical homogeneity (which is the same as translational invariance from Lubos' post). Then, instead of averaging over the different universes, cosmologists average $f(\vec{x})f(\vec{y})$ over different locations ($\vec{x}, \vec{y}$) in our universe which have a fixed distance between the two points $|\vec{x}-\vec{y}|$. This way, using the statistical homogeneity assumption, we can get good measurements of the correlation function of any quantity we desire. &lt;/p&gt;&#xA;" OwnerUserId="1334" LastEditorUserId="1334" LastEditorDisplayName="" LastEditDate="2011-01-18T05:18:22.373" LastActivityDate="2011-01-18T05:18:22.373" CommentCount="1" />
  <row Id="3225" PostTypeId="2" ParentId="2322" CreationDate="2011-01-18T08:40:01.997" Score="7" ViewCount="" Body="&lt;p&gt;Tesla was an engineering giant but it is true that in most cases, he was just a crackpot when it came to theoretical physics. This &quot;theory of gravity&quot; is one of the major ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It wasn't really connected with gravity - the attraction of objects to the Earth etc. with a universal acceleration - by anything else than a wishful thinking. As expected for the practical guy, most of the support for his unusual claims came from experiments, and it was the electromagnetic experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tesla claimed that the vacuum was filled with a new kind of the aether, a rarified gas that he also called the Akasha. This name is relatively important in his theory so you can get some idea about the character of the theory. The Akasha was extremely elastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some spectacular experiments with some light emitted by either dielectric materials or conductors were made to support the claims, although they didn't have anything to do with the claims about the unifying theory. And at the end, it was found out that the experiments analyzed solidified air rather than any mysterious matter that could be filling the vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the theory was a complete nonsense. Tesla couldn't really distinguish the fundamental phenomena from the highly derived, environment-dependent ones.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T14:59:11.590" LastActivityDate="2011-02-02T14:59:11.590" />
  <row Id="3226" PostTypeId="2" ParentId="3025" CreationDate="2011-01-18T08:53:26.707" Score="1" ViewCount="" Body="&lt;p&gt;I assume for simplicity that the spring constant  has a quite a high value so that the settling down of the spring under its own weight is insignificant.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Designations:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x$-vertical displacement of the center of mass of the spring from its equilibrium position.&lt;br&gt;&#xA;$l$-vertical displacement of the top of the spring from its equilibrium position.&lt;br&gt;&#xA;$m$-the mass of the spring.&lt;br&gt;&#xA;$k$-the spring constant.&lt;br&gt;&#xA;$g$-gravitational acceleration.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all let's highlight the following relation: $$x=\frac{2}{3}l$$ Its derivation is elementary but too long to present here.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is write down the equation of the conservation of energy:&lt;br&gt;&#xA;$$m\frac{\dot{x}^2}{2}+\frac{3}{2}kx^2+mg(x_0-x)= \frac{3}{2}kx_0^2=const$$ $x_0=x(0)$ is an initial displacement of mass center of the spring from its equilibrium position. After differentiating with respect to $t$ we get the equation of the motion of the center of mass of the spring:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ddot{x}+\frac{3k}{m}x-g=0$$ According to initial conditions $x(0)=x_0= \frac{2}{3}l_0$ and $\dot{x}(0)=0$ the solution of this equation:&lt;br&gt;&#xA;$$x(t)=\frac{g}{\omega_0^2}+\left(x_0-\frac{g}{\omega_0^2}\right)cos(\omega_0t);\omega_0^2=\frac{3k}{m}$$ At the moment of the departure from the ground the following holds:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-mg=kl=\frac{3}{2}kx$$or $$x=-\frac{2g}{\omega_0^2}$$ Minus sign indicates that a vertical coordinate is above the equilibrium. Thus, the time we are looking for is:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t=\frac{1}{\omega_0}arccos\left(-\frac{3g}{-g+x_0\omega_0^2}\right)= \frac{1}{\omega_0}\left(\frac{\pi}{2}+arcsin\frac{3g}{x_0\omega_0^2-g}\right)$$ The formula has a meaning if&lt;br&gt;&#xA;$$x_0&amp;gt;\frac{4g}{\omega_0^2}$$ I would point out the assumption at the top of the post! For the given data this is probably not a good assumption. But as a first approximation maybe it fits.&lt;/p&gt;&#xA;" OwnerUserId="552" LastEditorUserId="552" LastEditorDisplayName="" LastEditDate="2011-01-20T06:18:25.847" LastActivityDate="2011-01-20T06:18:25.847" CommentCount="17" />
  <row Id="3227" PostTypeId="1" AcceptedAnswerId="3250" CreationDate="2011-01-18T10:06:12.197" Score="3" ViewCount="383" Body="&lt;p&gt;If one wanted to study string theory, what subjects should they learn first and what order should the study them?  I would imagine that there are opionions, but I suspect there is a natural order, or at least and heirarchy of subjects.&lt;/p&gt;&#xA;" OwnerUserId="1383" LastActivityDate="2011-01-18T16:20:13.527" Title="What would be a proper set of subjects, along with their order, for someone interested in string theory?" Tags="&lt;string-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="3228" PostTypeId="1" CreationDate="2011-01-18T10:09:52.837" Score="5" ViewCount="480" Body="&lt;p&gt;Say you have a vial of tritium and monitor their atomic decay with a geiger counter. How does an atom &quot;know&quot; when it's time to decay? It seems odd that all the tritium atoms are identical except with respect to their time of decay.&lt;/p&gt;&#xA;" OwnerUserId="1413" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-03T17:11:47.440" LastActivityDate="2011-06-04T01:37:25.390" Title="Is there any thing other than time that &quot;triggers&quot; a radioactive atom to decay?" Tags="&lt;quantum-field-theory&gt;&lt;nuclear-physics&gt;&lt;radioactivity&gt;&lt;radiation&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="2" />
  <row Id="3229" PostTypeId="1" AcceptedAnswerId="3231" CreationDate="2011-01-18T10:28:49.580" Score="3" ViewCount="213" Body="&lt;p&gt;The textbook approach to explaining T-dualities is to show that a type of T-duality transformation &quot;inverts the radius of the circle, that is, it maps $R\rightarrow\tilde{R} = \alpha&amp;#39;/R$ and it leaves the mass formula for the string invariant provided that the string winding number is exchanged with the Kaluza-Klein excitation number.&quot; &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521860695&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the broader classification of T-duality appears imply that theories are T-daul if there is any type of transformation that changes on theory into the other.  So the question is whether there are other approaches to showing T-duality, or is there a &quot;canonical&quot; approach for T-duality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521860695&quot; rel=&quot;nofollow&quot;&gt;String Theory and M-Theory: A Modern Introduction [Hardcover] Katrin Becker (Author), Melanie Becker (Author), John H. Schwarz (Author)&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-13T19:56:30.357" LastActivityDate="2011-02-13T19:56:30.357" Title="T-duality approaches" Tags="&lt;string-theory&gt;&lt;kaluza-klein&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="3230" PostTypeId="2" ParentId="3228" CreationDate="2011-01-18T10:29:55.287" Score="10" ViewCount="" Body="&lt;p&gt;Actually, all the atoms are identical. The time at which it is observed to decay is not an intrinsic property of a given atom, but rather an effect of quantum mechanics. For any given time bin, there is some finite amplitude for a transition to a decayed state, which in turn corresponds to a finite probability, since the particle(s) emitted will escape from the system once such a state is reached. This also means that the process is in irreversible, due to the open nature of the system. This works in the same way as atomic transitions when atoms emit photons (see the relevant &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_electron_transition&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each undecayed atom, in each time bin $T$ there is a probability of transitioning to the decayed state given by a fixed probability $p$ (which is independent of $T$, and depends only on the binning size). Thus between the time $t$ and $t+\Delta t$ there is a fixed probability $\Delta p = \lambda \Delta t$ of transitioning to the decayed state for any given atom. So if we have $N(t)$ undecayed nuclei at time $t$, then at time $t+\Delta t$ we should have $N(t+\Delta t) = (1-\lambda\Delta t)N(t)$. Rearranging thisa and taking the limit $\Delta t \to 0$ we obtain $dN/dt = -\lambda t$. Solving this equation yields the total number of nuclei left undecayed at time $t$ as $N(t) = N(0) e^{-\lambda t}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the point to take from all this is simply that the atoms are all identical and decay by a purely random process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: I forgot to mention that decay probability can be increased, for example via collision with another particle for the right energy, and this is exactly how fission based nuclear bombs work. Here though, again, there is nothing special about the particular atom decaying, and it is simply the particles involved in the collision that have the increased decay probability. (I must admit that I have pared this picture right down to the basics as otherwise it would need to be a far more technical discussion).&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditorDisplayName="" LastEditDate="2011-01-18T10:41:15.290" LastActivityDate="2011-01-18T10:41:15.290" CommentCount="8" />
  <row Id="3231" PostTypeId="2" ParentId="3229" CreationDate="2011-01-18T10:49:58.523" Score="3" ViewCount="" Body="&lt;p&gt;&quot;Any transformation that changes one theory into another&quot; (or the same) theory is not called T-duality. It is just a &quot;duality&quot;. A condition is that the two theories seemingly look different - otherwise the equivalence would be vacuous - but it must be true that their spectrum and the strength of interactions between their states must be totally isomorphic: physics has to be indistinguishable. A duality is therefore just a very fancy redefinition of coordinates that can't be quite done in the classical limit so you can't really write any &quot;explicit&quot; redefinition. But the impact is the same - the two theories behave in the same way. For each state/object on one side, you find a counterpart on the other side - whose geometrical interpretation may be very different but whose behavior is isomorphic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;T-duality in string theory&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T-duality is a duality that changes the geometrical properties of the target spacetime but that holds order-by-order in the expansion in terms of the string coupling in perturbative string theory. In particular, a T-duality must preserve the notion of the &quot;fundamental string&quot; (or &quot;world sheet&quot;) and its tension and it is a weak-weak duality when it comes to the interactions between strings. The world sheet of a given shape remains the world sheet of the same shape; however, the fields on it get transformed. That's T-duality in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite generally, T-dualities that we know may be interpreted as the duality you mentioned although it may also invert the radii of several circles (a torus) at the same moment, and the shape of these circles (and torus) may depend on other coordinates of spacetime (the so-called &quot;Buscher procedure&quot; is a standardized method to perform such a transformation). In particular, &quot;mirror symmetry&quot; relating two completely different Calabi-Yau shapes may be understood as a T-duality acting on a 3-dimensional toroidal fibers of the Calabi-Yau space (a special way to choose coordinates on the 6D shape so that 3 of them look like a torus whose shape depends on the remaining 3.) However, in recent years, people have also appreciated the importance of fermionic T-dualities that mostly act on fermionic fields on the world sheet but that are otherwise analogous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other dualities&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several other important kinds of dualities. S-dualities invert the weak and strong coupling. In particular, for a very large value of $g$ much greater than one, they find another theory with a weak coupling $g&amp;#39;=1/g$ or $1/g^2$ - which is much smaller than one - that is equivalent to the previous one. So S-dualities usually invert a small and large couplings, much like T-dualities exchange small and large radii of circles. A fundamental string doesn't remain a fundamental string under an S-duality; it usually gets transformed into another string, e.g. a D-string (also known as a D1-brane), that used to be heavy and unimportant when the coupling was weak but becomes the lightest and most important object when the coupling is strong. Again, an S-duality exchanges fundamental strings and D-strings, and similar pairs of objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, there are also U-dualities that combine the actions of S-dualities and T-dualities: they exchange a strong coupling with large radii with a weak coupling and small radii, among similar actions. In particular, M-theory (essentially 11-dimensional supergravity with the extra &quot;stringy&quot; stuff needed to make it consistent at all energies) compactified on a $k$-dimensional torus has a self-U-duality group which is something like $E_{k(k)}(Z)$, a discrete subgroup of a noncompact edition of the exceptional Lie groups (for $k\leq 8$: for $k\leq 5$, the exceptional group can be written in terms of non-exceptional groups). Such an exceptional U-duality group exchanges momenta in compact dimensions with wrapped membrane and fivebrane charges (and other discrete charges such as KK-monopole charges if there are at least 6 compact dimensions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several other equivalences between two seemingly different theories are also called &quot;dualities&quot; - and the AdS/CFT holographic correspondence itself is sometimes classified as a duality as well. It would triple this article if I wanted to cover all important kinds of dualities that appear in field theory and string theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why the existence of dualities was surprising&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, it's useful to stress that dualities - a key notion of theoretical physics of the last 20 years or so - didn't have to exist. More precisely, people thought that a theory compactified on an extremely tiny circle, or a theory extrapolated to an insanely strong coupling, is a completely new mysterious beast filled with sea dragons that behaves like nothing that the people had previously encountered. However, it turned out that the behavior of those &quot;mysterious limits&quot; is almost always governed by another &quot;mundane&quot; theory that the people have essentially known, and that the two are equivalent. It's like sailing to India and America and expecting totally new creatures over there - but finding the same animals and plants we're used to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever some radius is taken extremely tiny, coupling is made very strong, or even the number of colors in a gauge theory is sent to infinity, so that the perturbative techniques of the original theory totally break down, there always seems to be a new description that is totally weakly coupled. That's one of the lessons of physics since the mid 1990s - something that we didn't previously know but nowadays, we do know it. The fact that we seem to understand the &quot;opposite extreme&quot; limits of the previously known theories places an upper bound on the amount of &quot;new enigmas&quot; that may still be hiding in field theory and string/M-theory. If there are totally new enigmas, they have to hide at the intermediate values - like Atlantis in the middle of the Atlantic Ocean, if you wish. ;-)&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:37:26.983" LastActivityDate="2011-02-04T07:37:26.983" CommentCount="1" />
  <row Id="3232" PostTypeId="2" ParentId="3227" CreationDate="2011-01-18T11:17:19.597" Score="3" ViewCount="" Body="&lt;p&gt;Dear Humble, a great goal. One has to learn&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Arithmetics (I omit reading of digits, letters, and the Greek alphabet)&lt;/li&gt;&#xA;&lt;li&gt;Mechanics in terms of forces&lt;/li&gt;&#xA;&lt;li&gt;Advanced and abstract mechanics, involving Lagrangians and Hamiltonians&lt;/li&gt;&#xA;&lt;li&gt;Classical field theory - typically electromagnetism&lt;/li&gt;&#xA;&lt;li&gt;Quantum mechanics, and its non-relativistic models&lt;/li&gt;&#xA;&lt;li&gt;Special relativity and tensors (linear algebra is shared with the previous point)&lt;/li&gt;&#xA;&lt;li&gt;General relativity and some cosmology&lt;/li&gt;&#xA;&lt;li&gt;Basic quantum field theory (plus some basic background in experimental particle physics)&lt;/li&gt;&#xA;&lt;li&gt;Some advanced quantum field theory, including renormalization, anomalies etc.&lt;/li&gt;&#xA;&lt;li&gt;Some discrete group theory, but especially Lie groups, and their representations&lt;/li&gt;&#xA;&lt;li&gt;Geometry and topology of manifolds, and perhaps some algebraic geometry etc.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Some of the entries may be exchanged but it would be a bit awkward to classify exactly which of the $11!$ orderings are OK and which of them are not. ;-) Most people have not been totally strict about any ordering and they could still get to the point 12 which is string theory. ;-) The subjects are not quite isolated from each other, after all, and one simply learns facts as she needs them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I shouldn't have completely omitted thermodynamics and statistical physics, so please insert it between 5 and 6 and replace $11!$ by $12! = 479,001,600 $ in the previous paragraph. Well, because I have mentioned factorials, it may be good to say that some combinatorics is needed, too - although people usually get it in their leisure time via recreational mathematics. :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is fortunately not true that one has to master all those subjects at 100%. Only a certain fraction of insights about each of them is really needed to jump on string theory. On the other hand, the fraction that is needed is not negligible and kind of covers the whole extents of those simpler subjects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck, Luboš&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-18T11:24:46.653" LastActivityDate="2011-01-18T11:24:46.653" CommentCount="5" />
  <row Id="3233" PostTypeId="2" ParentId="3214" CreationDate="2011-01-18T11:53:25.350" Score="3" ViewCount="" Body="&lt;h2&gt;All states are accessible&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Basic assumption of statistical physics is that all &quot;allowed&quot; configurations are equally probable. And the entropy is related to the number of those configurations. Therefore entropy won't  change if all the configurations would be still accessible after introduction of the chords. Which is the case if length of joints of your polymer are much smaller than the characteristic length of/between the chords in the volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course even in this case the chords will affect the diffusion times. The configuration space will have some complex structure with interconnected &quot;islands&quot;, which must have some relation to &lt;a href=&quot;http://en.wikipedia.org/wiki/Glass&quot; rel=&quot;nofollow&quot;&gt;glasses&lt;/a&gt;. But the total number of states and, therefore, the entropy won't change. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Inaccessible states&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If length of joints becomes comparable to the size of the sphere and/or characteristic length of/between the chords, then the &quot;chordless&quot; configuration space will be broken in several mutually inaccessible subspaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think one can imagine an example when the number of those subspaces equal to the number of&lt;br&gt;&#xA;&quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stamp_folding&quot; rel=&quot;nofollow&quot;&gt;stamp foldings&lt;/a&gt;&quot;. And as far as I know this combinatorial problem is yet unsolved -- not only there is no closed expression, but not even guesses about the asymptotic behaviour (&lt;a href=&quot;http://www.chilton-computing.org.uk/acl/applications/number/p009.htm%5d&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-01-18T11:53:25.350" />
  <row Id="3234" PostTypeId="2" ParentId="1222" CreationDate="2011-01-18T12:00:46.573" Score="1" ViewCount="" Body="&lt;p&gt;I can't make sense of your quoted phrase &quot;The matter distribution cuts off the white hole solution.&quot;, but I'll try to answer the rest of the question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Oppenheimer-Synder model for black hole collapse is a solution of the equations of classical GR where a uniform speherically symmetrical dust cloud with no pressure or rotation collapses to form a black hole. The time reversal is also a valid solution of the field equations but it would represent a white hole and is therefore normally considered unphysical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside the sphere of dust the solution matches the well known cosmological Freidman, Robertson Walker solutions for the expanding universe. Outiside the collapsing sphere it looks like the Schwartzchild static black hole solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the cosmological solution, there are different cases where the universe can expand forever or recollapse. The same thing happens in the Oppenheimer-Synder solutuon. So there are cases where it starts with a white-hole and the dust expands outwards, but the dust does not have enough energy to escap so it recollapses to form a black hole. However, there are other cases where the dust does escape from the whitehole. The time reversal of this is also the formation of a blackhole without the whitehole in the past&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, these are all special cases of a more general class of spherically symmetric solutions called the Lamaitre-Tolman solutions.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-18T12:00:46.573" />
  <row Id="3235" PostTypeId="2" ParentId="3062" CreationDate="2011-01-18T12:47:53.150" Score="-2" ViewCount="" Body="&lt;p&gt;Against all the above opinions:&#xA;&lt;strong&gt;The rope will shrink&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;and maintain the normal tension between atomic structure&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A space expansion scenario is dual (almost) to a &lt;strong&gt;shrinking matter scenario&lt;/strong&gt;.&#xA;Unless we have a way to decide I will support my point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Already explained here:&#xA;&lt;a href=&quot;http://arxiv.org/abs/astro-ph/0208365&quot; rel=&quot;nofollow&quot;&gt;A relativistic time variation of matter/space fits both local and cosmic data&lt;/a&gt; &#xA;and here:&#xA;Cosmological Principle and Relativity - Part I  (arxiv astro-ph 0208365)&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-01-18T12:47:53.150" />
  <row Id="3236" PostTypeId="2" ParentId="3228" CreationDate="2011-01-18T12:54:58.933" Score="1" ViewCount="" Body="&lt;p&gt;I think if even there were a &quot;trigger variable&quot; for each atom, one would need to randomize it anyway to describe an ensemble of decaying atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, in case of atoms there is a stimulated emission - with help of photons coherent with the &quot;future&quot; photon. This shows that the &quot;environment&quot; is somewhat important. As soon as the environment is complicated and is hard to control, one can loosely think that the random character of decays is due to random character of the &quot;triggering QM environment&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-01-18T12:54:58.933" CommentCount="1" />
  <row Id="3237" PostTypeId="1" AcceptedAnswerId="3240" CreationDate="2011-01-18T13:16:21.550" Score="3" ViewCount="188" Body="&lt;p&gt;How to derive the selection rules $\Delta L= \pm 1$, $\Delta S=0$ for electron spectroscopy?&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2011-01-18T13:56:02.830" Title="Selection Rules in electron spectroscopy" Tags="&lt;quantum-mechanics&gt;&lt;spectroscopy&gt;" AnswerCount="1" />
  <row Id="3238" PostTypeId="2" ParentId="3189" CreationDate="2011-01-18T13:25:39.007" Score="3" ViewCount="" Body="&lt;p&gt;The sustained interest in the Abraham-Minkowski debate does not come from any theoretical concern-- theorists on both sides have always thought that they were unquestionably right, and the people on the other side were a bunch of incompetent hacks. The traditional arbiter of any such dispute between theories is experimental evidence, but that evidence has been ambiguous. Some experimental tests give results consistent with the Abraham formula, others with the Minkoski formula (the arxiv paper gives references to these).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recent paper is interesting not because it shows definitively which side is a bunch of incompetent hacks, but because it provides a framework for understanding how the seemingly contradictory experimental results can both be right. It will settle the question to the extent that this framework for reconciling the irreconcilable holds up under further investigation. Ideally, this would come through not just explaining past experiments, but through predicting some new result that can be verified experimentally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper in question is far too recent for the necessary experiments to have been done. It looks pretty good, but I imagine that a real partisan of one side or the other will be able to find some way to wriggle out, until the predictions have been tested experimentally.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2011-01-18T13:25:39.007" CommentCount="1" />
  <row Id="3239" PostTypeId="2" ParentId="197" CreationDate="2011-01-18T13:36:43.013" Score="2" ViewCount="" Body="&lt;p&gt;The notion of a point-like but charged particle is somewhat contradictory: on one hand we imagine it as localized in space (like a neutral particle); on the other hand, such a &quot;particle&quot; is &quot;long-handed&quot;, i.e., it is felt far away from it &quot;position&quot;. One may think that the charge is not point-like but is a &quot;part&quot; of a complicated, extended system. When you tear the system apart, you feel a resistance force. In case of a spring, this force increases with distance. In case of a shewing gum, it decreases ;-).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In QED there is a Coulomb gauge where the interaction potential is instant 1/r but acts between &quot;waves&quot;, not &quot;points&quot;. These waves are &quot;overlapping&quot; and we cannot really separate them.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-01-18T13:36:43.013" />
  <row Id="3240" PostTypeId="2" ParentId="3237" CreationDate="2011-01-18T13:56:02.830" Score="1" ViewCount="" Body="&lt;p&gt;these are selection rules for the electric dipole radiation. The transition from the excited state to a lower state of an atom is governed by the following matrix element:&#xA;$$C \cdot \langle f | \hat{p} | i\rangle$$&#xA;You could also write the sum of positions of electrons $\Sigma\hat{r}$ instead of their total momentum $\hat{p}$ in the middle. This simple form of the operator in the middle is because at low enough frequencies, i.e. long enough wavelength, the atom simply finds itself in a uniform field, anf the electric potential $\phi$ for a uniform field is linear in $\hat{r}$. Equivalently, the matrix element may be converted from $\hat{r}$ to $\hat{p}$ and vice versa by realizing that the commutator of the Hamiltonian with $\hat{x}$ is proportional to $\hat{p}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, the operator in the middle is a 3-dimensional vector that only acts on the positions or momenta of the electrons, not their spins. So it has to commute with the total spin operator $\hat{S}$ of the electrons, and $\Delta S=0$ as a consequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it is a vector, i.e. a $j=1$ object as far as the SO(3) transformations go, and by combining its $j=1$ angular momentum with the orbital angular momentum $L_i$ of the initial state, you may get the final $L_f$ being $L_i\pm 1$ or $L_i$ according to the basic rules of the addition of angular momentum. You may imagine that you're just adding two vectors of specified lengths, $L_i$ and $1$, and depending on their relative angle, the length of the sum may go from $L_i-1$ to $L_i+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your list didn't allow $L_f=L_i$ and I think you are right that it is typically forbidden as well because it violates parity. The angular momentum to parity goes like $(-1)^L$, I guess, but because the vector operator is parity-odd, the initial and final states have to differ by their parity as well, which means that only $\Delta L=\pm 1$ is allowed. Everyone, please correct me if I am saying something incorrectly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-18T13:56:02.830" CommentCount="3" />
  <row Id="3241" PostTypeId="2" ParentId="1442" CreationDate="2011-01-18T14:05:20.853" Score="1" ViewCount="" Body="&lt;p&gt;Recover the time standard?  No.  Time is the phase of an oscillator, i.e. keeping track of time is counting oscillations.  To recover the phase, you have to have another clock, because a clock is something that keeps track of the phase.  No memory = no knowledge of the time.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frédéric's reply concerns the &lt;em&gt;frequency&lt;/em&gt; of the oscillation, which is reproducible.  That will tell you the length of a second, but not which second (or fraction thereof) it is.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-18T14:05:20.853" CommentCount="1" />
  <row Id="3242" PostTypeId="2" ParentId="1222" CreationDate="2011-01-18T14:43:49.463" Score="1" ViewCount="" Body="&lt;p&gt;A white hole is in a sense a time reversed black hole.  If you take an Eddington-Finkelstein diagram and turn it upside down you have a white hole.  A physical sense of things suggests these do not exist.  The argument for “cutting off” is usually interpreted as how the Penrose conformal diagram with two square patches for time like regions and two triangular spacelike regions is reduced so the bottom spacelike region for the white hole is removed.  Pretty clearly there is an asymmetry between the two halves of the pure or “eternal” solution.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;White holes may have played a role in the early universe though.  The Hubble constant is ${\dot a}/a~=~H$, ${\dot a}~=~da/dt$, and $a$ the scale factor.  The FLRW differential equation of motion for a constant vacuum for the de Sitter spacetime is&#xA;$$&#xD;&#xA;({\dot a}/a)^2~=~8πG\Lambda/3,&#xD;&#xA;$$&#xA;which has an exponential solution.  So we have the scale factor evolving as&#xA;$$&#xD;&#xA;a~\simeq~\sqrt{3/8πG\Lambda}exp(\sqrt{8πG\Lambda/3}t).&#xD;&#xA;$$&#xA;This exponential expansion is what smoothed out anisotropies in the universe.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we consider this with some anisotropy which is rewound backwards in time.  So the small anisotropies rapidly clump back together during the inflationary period, which in a time reversed setting can mean the collapse of matter into black holes.  Now this is the time reversal of inflation, which means the inflationary period may have had white holes.  The white holes might have been some perturbation on the inflationary cosmology and their disappearance in the exponentially expanding space some aspect of inflation.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-18T14:43:49.463" />
  <row Id="3243" PostTypeId="1" CreationDate="2011-01-18T15:27:40.517" Score="6" ViewCount="146" Body="&lt;p&gt;I have a finite state ensemble with an energy functional (you can think of it as an ferromagnetic Ising model if you like), and I need very careful estimates of the partition function.  What methods are available to me to get reasonable estimates (in reasonable temperature regimes) of $Z(\beta)$?&lt;/p&gt;&#xA;" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-18T15:46:53.760" LastActivityDate="2011-02-27T21:24:21.277" Title="Estimating Partition functions" Tags="&lt;statistical-mechanics&gt;&lt;ising-model&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="3244" PostTypeId="1" CreationDate="2011-01-18T15:42:18.613" Score="7" ViewCount="238" Body="&lt;p&gt;At xmas, I had a cup of tea with some debris at the bottom from the leaves.  With less than an inch of tea left, I'd shake the cup to get a little vortex going, then stop shaking and watch it spin.  At first, the particles were dispersed fairly evenly throughout the liquid, but as time went on (and the vortex slowed, although I don't know if it's relevant) the particles would collect in the middle, until, by the time the liquid appeared to almost no longer be turning, all the little bits were collected in this nice neat pile in the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the physical explanation of the accumulation of particles in the middle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that it's something to do with a larger radius costing the particles more work through friction...&lt;/p&gt;&#xA;" OwnerUserId="1439" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-18T15:51:46.243" LastActivityDate="2011-01-20T00:57:17.353" Title="Vortex in liquid collects particles in center" Tags="&lt;fluid-dynamics&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="3245" PostTypeId="1" CreationDate="2011-01-18T15:49:13.907" Score="4" ViewCount="208" Body="&lt;p&gt;I'll be teaching a seminar for first-year undergraduates next year. The idea of my university's first-year seminar program is to expose students to exciting ideas and important texts in a somewhat interdisciplinary way. My course will focus on three epochs in the history of cosmology in which our ideas about the size of the Universe underwent radical expansions: the Copernican revolution, the early 20th century, and the present / recent past. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a lot of ideas for good undergraduate-friendly readings on the first two topics, but not so many for the last one. One idea I want to get at with them is that recent theories suggest that the observable Universe is a small and perhaps not even typical fraction of the entire Universe. I'd even love to get them arguing about the anthropic principle while I'm at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is this: Can you suggest good books, articles, etc. for me to consider in the syllabus for this course? Because of the interdisciplinary nature of the course, I'm happy to consider fiction, philosophy, and history as well as straight-up science. (For instance, Borges's story about the Library of Babel has a nice metaphorical connection to some of these ideas.) Just remember that these are kids fresh out of high school -- they're not ready for &lt;em&gt;Phys. Rev. D&lt;/em&gt;!&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T08:02:09.700" LastActivityDate="2011-02-03T08:02:09.700" Title="Undergraduate-friendly reading material on the multiverse?" Tags="&lt;cosmology&gt;&lt;education&gt;&lt;cosmological-inflation&gt;&lt;anthropic-principle&gt;" AnswerCount="1" />
  <row Id="3247" PostTypeId="1" CreationDate="2011-01-18T16:06:41.950" Score="13" ViewCount="381" Body="&lt;p&gt;If you drop a proton and a neutron in a gravitational field, they both fall, but the proton has a charge and accelerating charges radiate energy, so that leaves less kinetic energy for the proton and by this reasoning, it should fall more slowly than a charge-free object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue is discussed but not in the terms above in  Peierls's &quot;Surprises in Theoretical Physics&quot; in the chapter &quot;radiation in hyperbolic motion&quot;, but I didn't understand the chapter well enough (or at all) to apply it to my version of the question.  Peirls also refers to Pauli's Relativity book (section 32 gamma) but while Pauli claims there is no radiation from uniform hyperbolic motion, he does say there is radiation when two uniform rectilinear motions are connected by a portion of hyperbolic motion.   So I take it that would mean a proton at rest which falls for a second and then is somehow forced to maintain its newly acquired downward velocity from the fall without speeding up any further would have radiated.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2011-02-03T08:00:28.190" Title="Should a neutron fall faster than a proton?" Tags="&lt;electromagnetism&gt;&lt;general-relativity&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="3248" PostTypeId="2" ParentId="197" CreationDate="2011-01-18T16:10:38.653" Score="0" ViewCount="" Body="&lt;p&gt;A physical understanding of how forces are mediated is simply a quantum argument.  The momentum operator $p~=~-i\hbar\nabla$ is written in a gauge covariant form $p~\rightarrow~p~+~ieA$, for $A$ the vector potential.  This couples the particle to the electromagnetic field.  The momentum operator is usually expressed in the Dirac equation.  Now if you have two charged particles with momentum $p_1$ and $p_2$ they are coupled to the QED field, even if there are no real photons present.  The QED vacuum couples to the electrons, and this couples the electrons to virtual momenta of the QED field, or a $\delta p~=~A$.  A virtual photon (here thought of on a “tree level”) can then transfer a $\delta p$ of momentum from $p_1$ to $p_2$.  This means the two electrons scatter with momentum $p_1~-~\delta p$ and $p_2~+~\delta p$.  So the presence of uncertainty fluctuations is what transfers momentum from one particle to the other.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-18T16:10:38.653" />
  <row Id="3249" PostTypeId="2" ParentId="3247" CreationDate="2011-01-18T16:16:00.207" Score="3" ViewCount="" Body="&lt;p&gt;Yes, it should.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a lot of subtleties involved in some questions like this (see some articles by Stephen Parrott available on arxiv.org if you want to start digging into this), but fortunately in this particular case the answer is clear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest way to see it is just to apply energy conservation, as you say. There's no doubt that, long after the proton has fallen, radiation is escaping to infinity, carrying energy with it. That energy can only have come from the mechanical energy of the proton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not necessarily the case that the charged particle's acceleration is less than $g$ during its entire fall. The difference in acceleration can come only during the &quot;jerks&quot; near the beginning and end of the fall. In fact, that's what's predicted by the Lorentz-Dirac equation (the closest thing there is to a standard equation for the radiation reaction on an accelerated point charge). The proton takes a bit longer to get going, accelerates at $g$ for a while, and comes to a stop when it hits the floor. The difference in what happens to the charged and uncharged particles during the initial and final phases can account for the difference in energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lorentz-Dirac equation has all kinds of problems, but in this particular case its predictions are pretty much bound to be qualitatively right. The problems all come from treating the falling charge as a mathematical point, but if we treat it as a small sphere (which the proton is anyway!) of radius much less than any other length scale in the problem, the LD equation is a good approximation to the correct equation of motion.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-01-18T16:51:00.740" LastActivityDate="2011-01-18T16:51:00.740" CommentCount="11" />
  <row Id="3250" PostTypeId="2" ParentId="3227" CreationDate="2011-01-18T16:20:13.527" Score="3" ViewCount="" Body="&lt;p&gt;This is more of a comment than an answer, but I cannot leave comments, so here it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.phys.uu.nl/~thooft/theorist.html&quot; rel=&quot;nofollow&quot;&gt;http://www.phys.uu.nl/~thooft/theorist.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'t Hooft's list. It is not specific for string theory but it may be useful. &lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2011-01-18T16:20:13.527" CommentCount="1" />
  <row Id="3251" PostTypeId="2" ParentId="3244" CreationDate="2011-01-18T16:24:51.767" Score="3" ViewCount="" Body="&lt;p&gt;From the moment when You stop agitation,&#xA;the liquid will slow down at the bottom &#xA;and the wall of the cup, whereas the portion &#xA;at the surface maintains a higher rate of &#xA;rotation for some time. &#xA;This is caused by the friction at bottom &#xA;and wall.&#xA;This in turn causes a convection where the liquid &#xA;raises in the center and goes down along &#xA;the wall, because the centrifugal force is &#xA;higher at the surface. &#xA;At the bottom this convection goes from &#xA;the wall towards the center, thus sweeping &#xA;the tea leaves to the center.&#xA;Georg&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-01-18T16:24:51.767" CommentCount="5" />
  <row Id="3252" PostTypeId="2" ParentId="3245" CreationDate="2011-01-18T16:47:03.820" Score="4" ViewCount="" Body="&lt;p&gt;On January 25th, Brian Greene will release his 3rd popular book, The Hidden Reality,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2011/01/brian-greene-hidden-reality.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2011/01/brian-greene-hidden-reality.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is dedicated to the multiverse. A counterpart of the book by Lenny Susskind is The Cosmic Landscape,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0316013331&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Cosmic-Landscape-String-Illusion-Intelligent/dp/0316013331/ref=sr_1_1?s=books&amp;amp;ie=UTF8&amp;amp;qid=1295369061&amp;amp;sr=1-1&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may also select some review-like preprints on the arXiv - to choose randomly, Vilenkin and Garriga&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0711/0711.2559v3.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/arxiv/pdf/0711/0711.2559v3.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Try to find many papers e.g. in this way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://scholar.google.com/scholar?q=multiverse+anthropic&quot; rel=&quot;nofollow&quot;&gt;http://scholar.google.com/scholar?q=multiverse+anthropic&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A secret feature of all these papers is that they're not difficult at all and undergrads can get them. It's pretty much philosophy and most people feel very self-confident that they have something to say about these issues, and in some cases, they're even right.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T08:01:47.500" LastActivityDate="2011-02-03T08:01:47.500" CommentCount="4" />
  <row Id="3253" PostTypeId="2" ParentId="3247" CreationDate="2011-01-18T16:59:06.427" Score="1" ViewCount="" Body="&lt;p&gt;This is a very good question.  We think that from the equivalence principle that the proton and neutron should fall at the same rate.  This is a good approximation to the situation.  However, the equivalence principle holds for a local inertial frame in some small region of space.  Within that region the electric field of the falling charge is a nice radially symmetric field.  However, the field lines do not end there.  They extend throughout the space and thread through the curved space or spacetime.  As the charge approaches a radially symmetric gravity field the field lines will adjust to the curvature.  This means the charged particle will be associated with a changing electric and magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electric and magnetic field elements $F_\mu~=~F_{\mu\nu}U^\nu$.  The covariant constancy of the EM field $\nabla^\mu F_{\mu\nu}~=~0$ is used to derive $d^2F_\mu/ds^2$ with respect to some separation of the charge from another geodesic point --- which can be at $\infty$.  The result is then a calculation of $d^2U^\nu/ds^2$ with the geodesic deviation equation.  This is then equated to second derivatives of the fields $F_\mu$, which means radiation is emitted.  This is an attenuating process.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This argument is “approximate,” for the attenuation acts as an acceleration, which adjusts this from a pure geodesic deviation.  I worked this out in considerable detail a few years ago in response to a discussion on this matter. The analysis becomes rather involved.  Yet the result is that the EM field response to the infall of the proton will act as a sort of &quot;viscosity&quot; that slows it down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All one needs to do is to use the Larmor law $P~=~\mu e^2a^2/6\pi c$ and input the acceleration a = -GM/r^2 for Newtonian gravitation.  A Newtonian bit of thought illustrates how a gravity field can pull a charge and induce a radiation.  The radiation power $P~=~\int F\cdot dv$ $=~\int Pdt$ gives a radiation force (Abraham-Lorentz) law $F~=~\mu e^2{\dot a}/6\pi c$.  A relativistic result should recover a Newtonian result.  This radiation force is the resistance a charge experiences due to the emission of radiation.  The time derivative of the acceleration is clearly dependent on the changing radial distance of the charge from the gravitating body.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-01-18T18:14:27.007" LastActivityDate="2011-01-18T18:14:27.007" CommentCount="2" />
  <row Id="3254" PostTypeId="2" ParentId="3247" CreationDate="2011-01-18T17:03:42.660" Score="3" ViewCount="" Body="&lt;p&gt;This question is somewhat academic - and it has been controversial - but the more correct answer is No, it should not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is academic because the electrostatic force between two protons is about $10^{36}$ times stronger than the gravity between them. In reality, a proton will polarize any conductor on the ground and it will be attracted by a much bigger electrostatic force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you guarantee that such effects don't exist, then protons and neutrons - as well as everything else - fall exactly by the same acceleration. This follows from the equivalence principle. The principle is much more general than most people think. In a freely falling elevator, there is no way to find out whether the elevator is freely falling in a gravitational field, or in empty outer space without gravitational field. Different accelerations would surely allow us to distinguish the two cases and it would clash with the equivalence principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually we don't have to ask &quot;in what frame Maxwell's equations are valid&quot;. Are they more valid in a frame attached to the ground, or in a freely falling frame? The difference is tiny because the electromagnetism-induced accelerations are typically greater than the gravitational ones. Of course, the gravitational force always has to be added, too. But for example, for the protons in the LHC, gravity is negligible for the protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you're asking about a &quot;mixed effect&quot; that requires both electromagnetism and gravity. And a cleaner way to describe it is to use the freely falling frame. In that frame, the metric tensor is as flat as you can get, and the proton's and neutron's accelerations coincide. In fact, even in a very tall but thin cylinder observed over a very long time, one may always set the metric pretty much to constant, up to terms that go to zero when the cylinder is really thin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So classically, the field around the falling proton may be obtained in the freely falling frame, with a huge accuracy, and it will be just a field of a static proton - in this frame. Of course, if a field of a static proton is observed from another, e.g. relatively accelerating frame, it may look different. But  there won't be any real loss of energy that would allow the two accelerations to diverge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you tried to calculate &quot;how much energy&quot; the proton is supposed to radiate, the usual methods that are available in the flat space would fail. In the flat space, you usually &quot;attach&quot; the charged particle to the region at infinity, and solve Maxwell's equations with a source. However, all these equations are affected by the existence of the gravitational field - that is getting weaker as you go further from the Earth. Because of this weakening of gravity, and the terms by which gravity influences electromagnetism, you will find out that the radiation is zero in the freely falling frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum mechanically, one has to deal with the Unruh radiation etc. Frames that are relatively accelerating with respect to each other have different ideas what the ground state (vacuum) is. So an accelerating particle could interact with the Unruh quanta. This is another extra subtlety. I am confident that all actual measurements of the time needed to fall etc. has to coincide for protons and neutrons, even including $\hbar$ quantum corrections.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T08:00:28.190" LastActivityDate="2011-02-03T08:00:28.190" CommentCount="10" />
  <row Id="3255" PostTypeId="2" ParentId="3247" CreationDate="2011-01-18T17:13:58.617" Score="2" ViewCount="" Body="&lt;p&gt;The answer depends on the initial state of the electromagnetic field around the proton. It cannot be zero everywhere because the proton has charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electromagnetic field would of course affect the trajectory of the proton and determine whether it fell faster or slower than the neutron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you have not specified what that field is the question is incomplete and cannot be answered.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-18T17:13:58.617" CommentCount="2" />
  <row Id="3256" PostTypeId="2" ParentId="3244" CreationDate="2011-01-18T17:41:32.157" Score="-1" ViewCount="" Body="&lt;p&gt;The water molecules move more quickly along the periphery than in the center of the cup.&#xA;The resulting force on a macroscopic object is directed inward, as I try to show in diagram.&#xA;Oops, as I have no points to render images I will substitute for text art:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vertical lines are a representation of the velocity field&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                                          |&#xA;                                    |     |&#xA;                             |      |     |&#xA;         Center  _____|______|______|_____| Periphery&#xA;                 0    1      2      3     4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Any macroscopic object that extends, for instance, between radius 2 and 3 will suffer a force directed to the center. Tea leaves will move to the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE:&#xA;This answer is still incomplete/incorrect even after the clarifications. (Velez)&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-01-19T15:33:17.860" LastActivityDate="2011-01-19T15:33:17.860" CommentCount="11" />
  <row Id="3257" PostTypeId="1" CreationDate="2011-01-18T18:26:09.073" Score="3" ViewCount="323" Body="&lt;p&gt;I'm considering taking a hobby (taking math and physics classes, &lt;strike&gt;reading &lt;/strike&gt;attempting to read arXiv papers) and turning it into a career change. However, I have 3 kids and a wife who doesn't make a lot of money so I have no choice but to look at grim reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm willing to work hard, network with people, do research, etc. so let's leave effort aside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the statistical likelihood of getting a job as an actual theoretical physicist?&lt;/p&gt;&#xA;" OwnerUserId="1443" LastActivityDate="2011-01-19T02:21:52.893" ClosedDate="2011-01-19T18:48:45.663" Title="What is the statistical likelihood of getting a job as a theoretical physicist?" Tags="&lt;soft-question&gt;&lt;physics-careers&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="3258" PostTypeId="2" ParentId="3243" CreationDate="2011-01-18T18:57:36.980" Score="0" ViewCount="" Body="&lt;p&gt;Monte Carlo simulations would be ideal. You can also look at weak coupling and strong coupling expansions but don't extrapolate beyond any likely phase transitions. Mean field theory is another technique but the answer may be more qualitative than quantitative.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-18T18:57:36.980" CommentCount="2" />
  <row Id="3259" PostTypeId="2" ParentId="3257" CreationDate="2011-01-18T19:17:48.477" Score="5" ViewCount="" Body="&lt;p&gt;As an experimental physicist, I'm not perfectly placed to answer. However, the only job &lt;em&gt;as&lt;/em&gt; a theoretical physicist, that is, doing theoretical physics, at least as far as I'm aware, is in a research/invited/appointed post at a university, special institution (Institute for Advanced Study, etc.), or (maybe) a national lab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there are more Ph.D. students coming out of physics than there are university positions for them to occupy. For a short version of why it's unlikely to get a job as a theoretical physicist, take a look at the Many Applicants, Few Academic Posts on page two of this article: &lt;a href=&quot;http://www.scientificamerican.com/article.cfm?id=does-the-us-produce-too-m&amp;amp;page=2&quot; rel=&quot;nofollow&quot;&gt;http://www.scientificamerican.com/article.cfm?id=does-the-us-produce-too-m&amp;amp;page=2&lt;/a&gt; (the rest of the article is worth a read as well).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outcome of the situation is that a lot of physicists go into industry or other fields. But, industry doesn't use theoretical physicists as theoretical physicists. They use them in Wall Street for solving complex equations, for example, but not for physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd add to this that the competition in theoretical physics is rather significant, since it takes quite a special brain and a lot of training to make real headway on the deepest problems of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, unfortunately, I'd say that the likelihood of getting a job doing theoretical physics is rather slight. And, given your economic constraints, it's unlikely you could afford to go through the low-pay period of a Ph.D. (a requirement) and a post-doc (in practice, also a requirement) in order to get an appointment in the field. If, on the other hand, you want to learn theoretical physics, even get a degree in the field, and then use that knowledge elsewhere (Ph.D. only), you might be able to swing that.&lt;/p&gt;&#xA;" OwnerUserId="861" LastActivityDate="2011-01-18T19:17:48.477" />
  <row Id="3260" PostTypeId="2" ParentId="3257" CreationDate="2011-01-18T19:29:10.003" Score="3" ViewCount="" Body="&lt;p&gt;This is general information, since I really know rather little about you or your circumstances, but here it is anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Talking about probability in relation to getting a job is not really the right way to approach things, in my view. What you need is a lot of dedication, a certain amount of god-given talent, and a real passion. Perhaps that understates it, even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of practicality, you'll definitely need to start by getting a Bachelors or Masters in physics (4 years), after which you'll want to enter a PhD programme, write and defend your dissertation (~3 years), get a job as a post-doc for a few years, and finally become a researching theoretical physicist. The whole process can easily take up to 10 years, so it's nothing to be done lightly! And at the end of the day, there's nothing to guarantee you'll be successful; it's a highly competitive field. (They also say it's one for the young.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is, of course, the route you need to take these days if you want to enter academia/professional physics. There is nothing stopping you from pursuing all this yourself, using the relevant resources, and publishing papers yourself to gain reputation. IF the other way is harder, this is even harder. (Even the greatest minds in theoretical physics needed some amount of education in order to grow.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in theory, there is nothing stopping you, but it's a &lt;em&gt;very&lt;/em&gt; difficult path you're taking, especially if you did not begin it after high school. If you want any more specifics, I suggest you elaborate on your situation.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2011-01-18T19:34:14.320" LastActivityDate="2011-01-18T19:34:14.320" CommentCount="6" />
  <row Id="3261" PostTypeId="1" AcceptedAnswerId="3280" CreationDate="2011-01-18T19:35:29.227" Score="1" ViewCount="142" Body="&lt;p&gt;Suppose you had an isolated cloud of gas at a low temperature in a vacuum with no external sources of radiation (e.g. no CMB). The gas would clearly cool via the emission of low-energy photons. But where do these photons come from? The thermal energy of the gas is simply manifest as elastic collisions between atoms where no energy should be lost. Do we have to appeal to the vacuum state of the electromagnetic field?  But aren’t these virtual photons which could not be transformed into thermal radiation?&lt;/p&gt;&#xA;" OwnerUserId="1394" LastActivityDate="2011-01-21T17:30:28.973" Title="Cooling via the quantum vacuum?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;thermodynamics&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="3262" PostTypeId="1" AcceptedAnswerId="3305" CreationDate="2011-01-18T19:55:37.230" Score="7" ViewCount="306" Body="&lt;p&gt;String field theory (in which string theory undergoes &quot;second quantization&quot;) seems to reside in the backwaters of discussions of string theory.  What does second quantization mean in the context of a theory that doesn't have Feynman diagrams with point-like vertices?  What would a creation or annihilation operator do in the context of string field theory?  Does string field theory provide a context in which (a) certain quantities are more easily calculated or (b) certain concepts become less opaque?&lt;/p&gt;&#xA;" OwnerUserId="1445" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-01-19T08:02:11.683" LastActivityDate="2011-01-19T08:02:11.683" Title="What does second quantization mean in the context of string theory?" Tags="&lt;quantum-field-theory&gt;&lt;string-theory&gt;&lt;second-quantization&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="3263" PostTypeId="2" ParentId="3228" CreationDate="2011-01-18T20:22:05.003" Score="3" ViewCount="" Body="&lt;p&gt;You can trigger decay of certain nuclei with gamma rays, just like you can stimulate emission of photons from excited atoms with incoming radiation. You can even make a bomb if that is your kind of thing. &lt;a href=&quot;http://en.wikipedia.org/wiki/Induced_gamma_emission&quot; rel=&quot;nofollow&quot;&gt;Induced emission&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2011-01-18T20:22:05.003" />
  <row Id="3264" PostTypeId="1" AcceptedAnswerId="3269" CreationDate="2011-01-18T20:23:04.607" Score="1" ViewCount="194" Body="&lt;p&gt;Why is it that thermal radiation of a black body usually described by a spectral distribution function rather than an intensity vs frequency curve? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a vague explanation for this: Any measured intensity of radiation will contain a spectrum of frequencies which are impossible to separate (even with a very good spectrometer, one can only measure the intensity of an interval). So it is practically impossible to assign a particular intensity to a particular frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my reasoning correct?&lt;/p&gt;&#xA;" OwnerUserId="966" LastActivityDate="2011-01-19T10:09:47.187" Title="Thermal radiation spectrum of a blackbody" Tags="&lt;thermal-radiation&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="3265" PostTypeId="2" ParentId="3062" CreationDate="2011-01-18T20:30:55.280" Score="1" ViewCount="" Body="&lt;p&gt;A related question seems to be whether one can get useful work from the expansion of the universe.  The rope could be a long tether with inductive coils and magnets in a chain.  As the tether is pulled the magnets run through the solenoids and induce a current.  The tether could then transmit the power back to the “home galaxy.”  There would be loss of this energy due to $z~\simeq~Hd/c$ from more distant inductors along the tether, but you could in principle generate energy this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is curiously the dream of the ZPE energy guys who think they can hook a cable to the vacuum and get “free energy.” However the cosmological constant $\Lambda$ is very small so the amount of energy in any volume, or in this case length, of space is miniscule.  Tethering two galaxies together is a bit tough to accomplish I should think.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-18T20:30:55.280" />
  <row Id="3266" PostTypeId="2" ParentId="9" CreationDate="2011-01-18T20:35:58.100" Score="1" ViewCount="" Body="&lt;p&gt;Let us remember that the equations of motion with &lt;strong&gt;initial&lt;/strong&gt; conditions $q(0), (dq/dt)(0)$ were advanced first and the least action principle was formulated later, as a sequence. Although beautiful and elegant mathematically, the least action principle uses some future, &quot;boundary&quot; condition $q(t_2)$, which is unknown physically. There is no least action principle operating only with the initial conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, it is implied that the equations have physical solutions. This is so in the Classical Mechanics but is wrong in the Classical Electrodynamics. So, even derived from formally correct &quot;principle&quot;, the equations may be wrong on physical and mathematical level. In this respect, formulating the right physical equations is a more fundamental task for physicists than relying on some &quot;principle&quot; of obtaining equations &quot;automatically&quot;. It is we physicists who are responsible for correctly formulating equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In CED, QED, and QFT one has to &quot;repair on go&quot; the wrong solutions just because the physics was guessed and initially implemented incorrectly.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-01-18T22:17:43.910" LastActivityDate="2011-01-18T22:17:43.910" CommentCount="1" />
  <row Id="3267" PostTypeId="2" ParentId="3222" CreationDate="2011-01-18T20:39:09.827" Score="1" ViewCount="" Body="&lt;p&gt;I imagine you did some sort of integral like $\frac{\int \sin\theta \cos\theta \left(\frac{dz}{\cos\theta}\right) d\theta}{\int \sin\theta \cos\theta d\theta} = 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, even for optically thin atmospheres, the photons emitted at oblique angles ($\theta \approx \pi/2$), there is a chance for them to scatter into small angles and then escape, so the $dz/\cos\theta$ factor must be replaced with a higher-order approximation for the optical depth as a function of initial emission angle. I did this problem once, but remember it being rather messy, and I think I just wrote a mathematica program to get the answer, and it was close to 5/3.&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2011-01-18T20:39:09.827" CommentCount="2" />
  <row Id="3268" PostTypeId="1" CreationDate="2011-01-18T21:16:56.337" Score="0" ViewCount="58" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/3221/what-are-the-most-common-theories-and-their-categorization-motivation&quot;&gt;What are the most common theories and their categorization/motivation?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&#xA;Besides Field and Gauge theories what other types of theories are there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Found the answer at &lt;a href=&quot;http://en.wikipedia.org/wiki/Outline_of_physics&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Outline_of_physics&lt;/a&gt; &#xA;Just needed to google &quot;outline of Physics&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Major Theories : Standard Model, Quantum field theory, Quantum chromodynamics, Electroweak theory, Effective field theory, Lattice field theory, Lattice gauge theory, Gauge theory, Supersymmetry, Grand unification theory, Superstring theory, M-theory&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also if somebody can add the taxonomy tag to this &lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-01-18T22:16:46.350" LastActivityDate="2011-01-18T22:16:46.350" ClosedDate="2011-01-18T21:20:59.377" Title="Besides Field and Gauge theories what other types of theories are there? [taxonomy ]" Tags="&lt;theory&gt;" CommentCount="6" />
  <row Id="3269" PostTypeId="2" ParentId="3264" CreationDate="2011-01-18T21:19:24.683" Score="3" ViewCount="" Body="&lt;p&gt;It's true that you can't talk about intensity without referring to some specific frequency range, but the reason is the definition of intensity itself, not the inherent limits of our measuring devices. Intensity, also called &lt;a href=&quot;http://en.wikipedia.org/wiki/Irradiance&quot; rel=&quot;nofollow&quot;&gt;irradiance&lt;/a&gt;, is defined as power per unit area. Typically it takes into account radiation at all frequencies (or wavelengths), but you can also talk about the intensity within a particular band like $400\text{ nm}-700\text{ nm}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to create a graph of intensity versus frequency, you could certainly do so by dividing the frequency axis into bins and plotting one point for each bin representing the total intensity detected within that frequency range. Obviously, the smaller you can make the bins, the more detailed information you get about the spectrum, but the intensity values become correspondingly smaller (and harder to measure accurately) as the bins get smaller. Besides, if everyone did that, different plots from different experiments would be very difficult to compare unless they all decided on a standard bin size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to make data from different experiments comparable, you normalize the intensities by dividing each intensity measurement by the width of the bin. In the limit as the bin width goes to zero, this is just the spectral irradiance.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-01-18T21:19:24.683" CommentCount="2" />
  <row Id="3270" PostTypeId="2" ParentId="625" CreationDate="2011-01-18T21:48:10.400" Score="0" ViewCount="" Body="&lt;p&gt;In the context of quantum field theory Weinberg's advice to ignore the term &quot;second quantisation&quot; is good advice. However, to go beyond quantum field theory anything goes and some people have promoted the idea of multiple quantization as a speculative idea that could be fruitful. It's not a popular idea as you can see from the other answers, but the response to this question would not be complete without mentioning it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beware that the term &quot;third quantization&quot; is used in the context of quantum cosmology and does not really mean an extra quantization after second quantization. If you want to learn about the real thing try searching for terms like &quot;multiple quantization&quot;, &quot;iterated quantization&quot;, &quot;repeated quantization&quot;, &quot;fourth quantization&quot; or &quot;infinite quantization&quot; (and ignore anything about data compression.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will find that the results are speculative, varied and incomplete, but not always totally mad. I don't think people should get overexcited about the idea but it should not be blithely dismissed either. It's just something to keep in the back of your mind if you are trying to understand the structure of theories about quantum gravity for example.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-18T21:48:10.400" />
  <row Id="3271" PostTypeId="2" ParentId="3216" CreationDate="2011-01-18T22:15:04.197" Score="0" ViewCount="" Body="&lt;p&gt;From what I gathered from Wikipedia, the Anthropic principle has been used to explain the age of the universe and values of physical constants. The argument is that if the values of physical constants changed by a tiny amount, this would have huge consequence of life as we know it. Hence the perfect values for the constants. But in a &lt;a href=&quot;http://arxiv.org/abs/1101.2444&quot; rel=&quot;nofollow&quot;&gt;recent paper&lt;/a&gt; Don Page has argued that the value of cosmological constant is a bit larger. If it had been smaller and negative the universe would have had better conditions to support life (it would increase galaxy and star formation rates, thereby increasing the chance of life supporting environments). I think this flies directly in the face of the Anthropic Principle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, we have studied conscious life only on earth. There is no conclusive evidence to support that life elsewhere in the universe (if it existed) would have the same characteristics as that seen on earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also see the article on arXiv blog: &lt;a href=&quot;http://www.technologyreview.in/blog/arxiv/26276/&quot; rel=&quot;nofollow&quot;&gt;http://www.technologyreview.in/blog/arxiv/26276/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="966" LastActivityDate="2011-01-18T22:15:04.197" />
  <row Id="3272" PostTypeId="2" ParentId="625" CreationDate="2011-01-18T22:35:08.417" Score="0" ViewCount="" Body="&lt;p&gt;Second quantization is a way of recasting things.  Second quantization defines fields over the Fock space so formerly waves are now parameters of field amplitudes.  I have heard string theory called “third quantization,” but as I see it this is probably an abuse of language.  At one time when membranes were first considered the term fourth quantization was raised a few times, though I think more in jest.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end it is all just quantization, and Weinberg is probably right in ignoring  numberical ordering of quantization.  Writing nonrelativistic QM according to $a$ and $a^\dagger$ is called second quantization by some, but really nothing much has changed.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-18T22:35:08.417" />
  <row Id="3273" PostTypeId="2" ParentId="3216" CreationDate="2011-01-18T22:50:57.007" Score="2" ViewCount="" Body="&lt;p&gt;Anthropic reasoning is as good a scientific method as any others, in that it can produce experimentally testable predictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance it has predictions (For historical reasons these are mostly post-dictions I think) of the ranges of certain cosmological constants. Anthropic reasoning would say that these number must be within certain ranges because otherwise the universe would not be capable of supporting life, but we observe life so the constants must lie within ranges capable of supporting life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the basic argument goes like this &quot;We know A is impossible without B, but we know A is true, therefore B is true&quot;, which is perfectly valid reasoning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A=&gt;B and A) =&gt; B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Could downvoters please state what they dont understand?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the definition of a scientific method I use:&lt;br&gt;&#xA;-Take observation, in this case that life exist.&lt;br&gt;&#xA;-Use a formal system to derive testable predictions based on observation:&lt;br&gt;&#xA; Done as follows: Life is observed to exist, therefore physical constants must be within certain ranges.&lt;br&gt;&#xA;-Perform experimental test of prediction:&lt;br&gt;&#xA; Constants observed to lie within predicted values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observation fits prediction, the definition of a succesfull scientific theory.&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2011-01-19T17:47:40.287" LastActivityDate="2011-01-19T17:47:40.287" CommentCount="9" />
  <row Id="3274" PostTypeId="2" ParentId="1402" CreationDate="2011-01-18T22:51:40.997" Score="3" ViewCount="" Body="&lt;p&gt;Cutting a magnet in two pieces is like cutting a vector: you can change its length but you cannot change its direction (you cannot make it directionless). &lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-01-21T23:42:20.187" LastActivityDate="2011-01-21T23:42:20.187" />
  <row Id="3275" PostTypeId="2" ParentId="2501" CreationDate="2011-01-18T22:53:11.823" Score="4" ViewCount="" Body="&lt;p&gt;Just wanted to supplement the answers already posted with a few notes re: exoplanet eccentricity. In my understanding, the reason why exoplanets have a median eccentricity ~0.3 vs. almost circular orbits in the solar system is not quite satisfactorily explained just yet &#xA;(&lt;a href=&quot;http://adsabs.harvard.edu/abs/2008ApJ...686..603J&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; is still my favorite simulation that attempts to address the origin of eccentricities).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The relative importance of disk-planet and planet-planet interactions is not understood completely.  It is not clear whether Type I/Type II migration have an important role in shaping the final configuration at all. This is indicated by the number of planetary systems with large inclination wrt the star's equatorial plane discovered recently. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) There is probably some degree of bias in e-Np histograms, given that there is a degeneracy between a single-planet eccentric solutions and multiple-planets circular solutions to radial velocity observations. See, e.g., &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010ApJ...709..168A&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; and others:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;finding that (1) around 35% of the published eccentric one-planet solutions are &gt; statistically indistinguishable from planetary systems in 2:1 orbital resonance, (2) another 40% cannot be statistically distinguished from a circular orbital solution, and (3) planets with masses comparable to Earth could be hidden in known orbital solutions of eccentric super-Earths and Neptune mass planets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3) Planetesimals and resonance crossing have played an important role at some point in the history of the solar system re: the eccentricity evolution of Jupiter &amp;amp; Saturn (a &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010ApJ...716.1323B&quot; rel=&quot;nofollow&quot;&gt;recent paper about this&lt;/a&gt;). Why this led to low eccentricities in the solar system might reside in the relative configuration and mass ratio of J &amp;amp; S, the specific mass in planetesimals in the disk compared to other protoplanetary disks, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Large eccentricities, especially with small planets, tend to lead to instability and scattering, so invoking the anthropic principle to some degree is not totally unjustified...&lt;/p&gt;&#xA;" OwnerUserId="1252" LastActivityDate="2011-01-18T22:53:11.823" CommentCount="1" />
  <row Id="3276" PostTypeId="2" ParentId="822" CreationDate="2011-01-18T23:03:49.027" Score="0" ViewCount="" Body="&lt;p&gt;A hint of the special role of angular momentum happens when you look for its conjugate variable. It is angular position, which is &lt;em&gt;adimensional&lt;/em&gt;. And then you have that any product of a variable times its conjugate has units of action, which are the same units that angular momentum. So classical mechanics already tell us that something is going on.&#xA;(Caveat: that you can have the same units with scalar product and with cross product, and the physical meaning is different. If you have checked the pamphlets from German car and enginemakers, you could have noticed the unit &quot;Nm&quot;, newton times meter, and the unit &quot;Joule&quot;, used differently.)&lt;/p&gt;&#xA;" OwnerUserId="1335" LastActivityDate="2011-01-18T23:03:49.027" />
  <row Id="3277" PostTypeId="2" ParentId="822" CreationDate="2011-01-18T23:07:38.507" Score="1" ViewCount="" Body="&lt;p&gt;As for spin and extended particles, I'd say the contrary: it is not contrary to intuition that point particles have some intrinsic angular momentum, because a point looks as if it had some rotation invariance built-in. The surprising thing is that extended objects have this angular momentum, without a point for the rotational symmetry to pivot in.&lt;/p&gt;&#xA;" OwnerUserId="1335" LastActivityDate="2011-01-18T23:07:38.507" />
  <row Id="3278" PostTypeId="2" ParentId="3261" CreationDate="2011-01-18T23:27:50.223" Score="0" ViewCount="" Body="&lt;p&gt;You need to be a little clearer about your assumptions. You haven't said what holds the cloud of gas together. Is it contained in a zero temperature transparent box - e.g., a very deep non-rel. QM deep square well? What is the gas made of - atoms? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's atoms confined by a transparent box that only interacts with them by reflecting them, then the cooling would have to proceed by exciting internal electronic states of the atoms in occasional inelastic high-energy pair collisions, followed by energy loss via electromagnetic decay. No QFT required.&lt;/p&gt;&#xA;" OwnerUserId="1447" LastActivityDate="2011-01-18T23:27:50.223" CommentCount="4" />
  <row Id="3279" PostTypeId="2" ParentId="1916" CreationDate="2011-01-18T23:36:06.687" Score="-1" ViewCount="" Body="&lt;p&gt;I doubt that a BH is able to form.&#xA;I think that before the boundary condition is reached some mechanism acts, for example conversion of mass into radiation, which expels the excess weight out of the system.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-01-18T23:36:06.687" CommentCount="1" />
  <row Id="3280" PostTypeId="2" ParentId="3261" CreationDate="2011-01-18T23:38:24.890" Score="1" ViewCount="" Body="&lt;p&gt;The particles of the gas undergo a process called radiative collision. For example, two atoms collide, and you end up with the same two atoms plus one or more photons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Feynman diagram for this process would have two atoms (complicated bound states of nucleons and electrons) coming in, and the same two atoms plus some photons coming out. It should be easy to see that such Feynman diagrams are possible. (And just as for any other scattering process, the sum of the amplitudes of all possible diagrams is related to the cross section.)&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-01-18T23:38:24.890" CommentCount="2" />
  <row Id="3282" PostTypeId="1" AcceptedAnswerId="3378" CreationDate="2011-01-19T00:29:35.143" Score="12" ViewCount="427" Body="&lt;p&gt;There are a number of physical systems with phases described by topologically protected invariants (fractional quantum Hall, topological insulators) but what are the &lt;em&gt;simplest&lt;/em&gt; mathematical models that exhibit topological phases? Is the toric code as simple as we can go?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Just to be clear, I'm talking about phases meaning &lt;em&gt;states of matter,&lt;/em&gt; and not just the geometric phase that a wavefunction will pick up under parallel transport in a nontrivial configuration space. I'm looking for simple models where one can make a phase diagram, and as a function of the available couplings there is a change in some topological property of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For example, in the XY magnet, neglecting bound vortex-antivortex pair formation, there is an instability at finite temperature to creation a single vortex, which is topologically distinct from the vortex-free state.)&lt;/p&gt;&#xA;" OwnerUserId="791" LastEditorUserId="791" LastEditorDisplayName="" LastEditDate="2011-01-19T15:23:28.007" LastActivityDate="2011-03-09T14:51:14.917" Title="Simple models that exhibit topological phase transitions" Tags="&lt;mathematical-physics&gt;&lt;condensed-matter&gt;&lt;topological-insulators&gt;&lt;quantum-hall-effect&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="5" />
  <row Id="3283" PostTypeId="2" ParentId="3125" CreationDate="2011-01-19T00:34:44.107" Score="4" ViewCount="" Body="&lt;p&gt;Yes, it is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a mirror in the shape of a closed hemisphere - that is, half of a concave spherical mirror combined with a flat circular mirror passing through its center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are inside this hemispherical mirror, you will see a real image of yourself &lt;strong&gt;rotated&lt;/strong&gt; 180 degrees around the axis of the hemisphere. Here's one way to show this is true: The flat mirror creates a virtual image of you which is your reflection in the plane (duh), and the spherical mirror takes that as an object and creates a real image which is the inversion of that through the center of the sphere. The composition of the reflection and the inversion is a rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for practical reasons you don't want to be entirely enclosed by the mirror (it would be hard to get in and out, and you'd have to have your own light source...), so you could build just a part of the hemisphere. But the part you built would have to contain both some of the spherical surface and some of the flat surface for it to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have trouble understanding the orientation of everything, here is a concrete description: The flat mirror is a circle in the $z=0$ (horizontal) plane, $x^2+y^2\le1$. The spherical mirror is a hemisphere described by $z&amp;gt;0$, $x^2+y^2+z^2=1$. You're standing inside so your hand is at a point near the axis, for example, (0.01, 0, 0.2). What the mirrors do is this: For an object at a point $(x,y,z)$, the flat mirror creates an image at $(x,y,-z)$ (this image has opposite handedness because one coordinate is flipped). Then the spherical mirror takes that image at $(x,y,-z)$ and creates another image at $(-x,-y,z)$. This image has the same handedness as the original object. It is also &quot;right side up&quot; because the z coordinate is not flipped. The image of your right hand at (0.01, 0, 0.2) will appear as another right hand at (-0.01, 0, 0.2) which is in the appropriate orientation for you to shake hands.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditorDisplayName="" LastEditDate="2011-01-19T18:04:53.287" LastActivityDate="2011-01-19T18:04:53.287" CommentCount="7" />
  <row Id="3284" PostTypeId="2" ParentId="3261" CreationDate="2011-01-19T00:44:57.680" Score="0" ViewCount="" Body="&lt;p&gt;I have been pondering this question some.  I would answer yes, if I understand the question in one sense, but if I interpret the question another way I am tempted to say no.  The vacuum is a set of quantum occupation sites $\{|0_k\rangle\}$.  A hot body usually emits a blackbody spectrum of particles which fill these slots in a thermal distribution. So one can say the vacuum cools a body by permitting the emission of radiation   This is pretty standard fair.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one thing which I am pondering is whether this question is meant to ask whether the vacuum can absorb thermal energy in some way and remain a vacuum.  In the above example one has a vacuum plus a hot body, and the vacuum transforms into a set of boson states plus vacuum plus a cool body.  So in a box you end up with radiation and not a “pure vacuum.”  A vacuum can change its energy vev.  After all that is what all of this business of Higgs symmetry breaking and inflation and …, is all about.  So I might interpret the question as asking whether the thermal energy of a body can be absorbed by the vacuum so it changes its vev.  However, the vacuum is a zero temperature state.  So it is hard to see how one can cool an object with the vacuum without radiation emitted.  This also appears to be a violation of thermodynamics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There appears to be a loop hole, but I think it can be closed.  Suppose we have a hot body with a temperature $T$ and we accelerate it to $a$.  On the frame of the accelerated body there is a temperature $T’~=~2\pi a$ from the horizon, called Unruh radiation, which bounds its Rindler wedge.  On the accelerated frame the pure vacuum observed in an inertial frame becomes a vacuum plus a thermal distribution of boson states.  So it is tempting to think one could accelerate a body with $a$ so that for $T’~&amp;lt;~T$ and the radiation from the body crosses the Rindler horizon with a rate of energy flow that is faster than the energy it emits.  This state of affairs approximates the situation for a body accelerated near the event horizon of a black hole.  So for this hot body suspended above a black hole with a temperature $T~&amp;gt;~2\pi a$ the black hole will absorb more energy than it emits, as observed on that frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the problem?  For one the limiting case with $a~\rightarrow~0$ recovers the case above.  So the vacuum does not absorb this energy and change the vacuum vev.  The other problem is that an inertial observer does not detect Unruh radiation.  The body will absorb this radiation, but the inertial observer still detect nothing.  The reason is the time dilation slows down or redshift any radiation the body absorbs so as to cancel out any apparent heating up.  A body then will not be observed to actually heat up, or if it is hotter than the Unruh radiation it is not observed to be influenced by any exterior radiation source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What has bothered me is the near horizon approximation for a particle near a black hole. For the accelerated observer the life time of the exterior world races by in a flash.  For a stellar mass black hole it requires billions of g-forces to remain a few meters from the horizon, and to get within centimeters requires about a billion billion g's of acceleration.  If by some means an observer could do this the outside world would be racing by, say for a small proper time with $t~=~g^{-1} cosh(gs)$.  So the proper time element is $s~\simeq ~ g^{-1}ln(gt)$ for a time unit $t$ outside.  As a result for t the lifetime of the black hole $\sim~ 10^{67}$year, $g$ in units of distance $\sim~ 1cm ~ 10^{-10} sec~\sim~ 10^{-17}$year the proper time the observer on the accelerated frame observes the BH to evaporate is&#xA;$$&#xD;&#xA;s~\sim~ 5\times 10^{18} years.&#xD;&#xA;$$&#xA;This is much shorter than the BH life time measured by the exterior world. Assume you get that acceleration up to $10^{33}cm^{-1}$ or $10^{43}sec^{-1}$ or $10^{50}year^{-1}$, then you are hovering practically on the stretched horizon.  The BH evaporates in about $10^{-42}$ seconds, or close to the Planck unit of time!  Bang!; which means all that ingoing and outgoing radiation which interacts with the black hole hits you at once is a colossal thunderclap.  The event horizon appears for larger g close in to be more of a singularity, or a surface region of huge energy density that is radiating and absorbing energy at a ferocious rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think the result hold for a body suspended near a black hole as well.  The above argument means the same time dilation for the absorption and emission of radiation will hole in the BH case, just as with the accelerated frame case.  This question is interesting for this reason, and frankly this may still bother me for a while.  Yet I feel safe in saying that the vacuum will not cool a body by adjusting its vev in some way and without the emission of radiation.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-01-19T00:51:35.363" LastActivityDate="2011-01-19T00:51:35.363" CommentCount="2" />
  <row Id="3285" PostTypeId="1" AcceptedAnswerId="3290" CreationDate="2011-01-19T00:57:58.160" Score="0" ViewCount="212" Body="&lt;p&gt;One falls slower in a mine shaft than in free air. This is due to collisions with the walls. How should one model the terminal velocity in the presence of such collisions?&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1782" LastEditorDisplayName="" LastEditDate="2011-02-16T15:14:50.103" LastActivityDate="2011-02-16T15:14:50.103" Title="Terminal velocity for falling in a shaft" Tags="&lt;classical-mechanics&gt;&lt;mechanics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="3286" PostTypeId="2" ParentId="625" CreationDate="2011-01-19T01:02:19.700" Score="0" ViewCount="" Body="&lt;p&gt;There's a pretty interesting article where they use a trick that they call &quot;Third Quantization&quot; to study open fermi systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://iopscience.iop.org/1367-2630/10/4/043026&quot; rel=&quot;nofollow&quot;&gt;http://iopscience.iop.org/1367-2630/10/4/043026&lt;/a&gt; (open access no less!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not exactly what you have in mind, but as clearly illustrated by all of these other answers, &quot;third quantization&quot; is not really canon among physicists.&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2011-01-19T01:02:19.700" />
  <row Id="3287" PostTypeId="2" ParentId="3285" CreationDate="2011-01-19T01:10:09.367" Score="1" ViewCount="" Body="&lt;p&gt;Terminal velocity is terminal velocity, regardless of collisions (&lt;strong&gt;Edit:&lt;/strong&gt; I should clarify this... Terminal velocity is defined as the steady state velocity that you reach when a velocity dependent resistive force exactly cancels the force driving the motion [usually gravity]. Occasionally you'll hit a wall and velocity will be lost deforming the wall (or your body!) These collisions won't reduce the terminal velocity, but they will definitely reduce the average velocity and possibly prevent the falling object from ever even reaching terminal velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a first pass model, for fun, I'd suggest free fall in gravity, with the usual velocity dependent damping (air resistance) and a stochastic term that reduces the velocity according to a Poisson distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This of course won't change the &lt;em&gt;terminal velocity&lt;/em&gt; (which you'd get as the steady state in the collision-less limit), but it would give you a better estimate of the average velocity.&lt;/p&gt;&#xA;" OwnerUserId="791" LastEditorUserId="791" LastEditorDisplayName="" LastEditDate="2011-01-19T01:39:41.023" LastActivityDate="2011-01-19T01:39:41.023" CommentCount="2" />
  <row Id="3288" PostTypeId="2" ParentId="3282" CreationDate="2011-01-19T01:58:35.607" Score="1" ViewCount="" Body="&lt;p&gt;From what I understand, &quot;quantum phases&quot; or &quot;Berry-Pancharatnam phases&quot; are examples of &quot;topological phases&quot;. See Y Ben-Aryeh 2004 J. Opt. B: Quantum Semiclass. Opt. 6 R1, &quot;Berry and Pancharatnam Topological Phases of Atomic and Optical Systems&quot;, &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0402003&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/quant-ph/0402003&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under this definition of the terms, the simplest example of topological phase is the transition of states under unitary transformation in spin-1/2 for a single particle. This is the Berry or Pancharatnam phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a spin-1/2 particle that begins in a state of spin-up (+z). Its spin is then measured in the x and y directions and then in the z direction. If the results of these measurements are that its spin is +x, +y, and then +z, its return to the +z case will be with a quantum phase of $\pi/4$ as I now show:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\sigma_x$, $\sigma_y$, and $\sigma_z$ be the Pauli spin matrices. Then $(1+\sigma_x)/2$, $(1+\sigma_y)/2$, and $(1+\sigma_z)/2$ are the projection operators for spin in the +x, +y, and +z directions. The operator for a particle going from spin +z to +x to +y and back to +z in a series of measurements is the product of the projection operators which can be simplified as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1+\sigma_z)/2\;(1+\sigma_y)/2\;(1+\sigma_x)/2\;(1+\sigma_z)/2&#xD;&#xA;= (e^{i\pi/4}/\sqrt{8})\;(1+\sigma_z)/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we let $|+&amp;gt;$ be spin up, then we have that the amplitude for a particle going through this sequence of states is:&#xA;$&amp;lt;+|\; (1+\sigma_z)/2\;(1+\sigma_y)/2\;(1+\sigma_x)/2\;(1+\sigma_z)/2\; |+&amp;gt; = e^{i\pi/4}/\sqrt{8}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\pi/4$ is the topological phase. The $1/\sqrt{8}$ is the reduction in amplitude due to going through the measurements. I've typed this in from memory, it's not unlikely I got the sign wrong. :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, for spin-1/2 and spin-1, the topological phase is given by half the area (in steradians) carved out in the Bloch sphere by the sequence of states. For the above example, the area carved out is one octant. This has an area of $(4\pi) /8 = \pi/2$, so the topological phase is $\pi/4$. &lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-19T01:58:35.607" CommentCount="1" />
  <row Id="3289" PostTypeId="2" ParentId="2685" CreationDate="2011-01-19T01:59:30.190" Score="3" ViewCount="" Body="&lt;p&gt;Since there are already outstanding technical answers to this question, I think we should add some better philosophical underpinnings for you to explore that might help with gaining a better intuitive feel for what information is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Warren Weaver provided an excellent discussion on information theory in 1949 in his paper entitled &quot;&lt;a href=&quot;http://ada.evergreen.edu/~arunc/texts/cybernetics/weaver.pdf&quot; rel=&quot;nofollow&quot;&gt;Recent Contributions to The Mathematical Theory of Communication&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the paper he breaks down communications problems into three main categories: technical, semantic and effectiveness.  He further explains that the concept of information is purely derived to address the technical problem in communications theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple definition of information, provided by Weaver, is that &quot;information is a measure of one's freedom of choice when one selects a message&quot;; or more correctly, the logarithm of that freedom of choice.  Information is thus more clearly understood as a the number of combinations of component parts that are available to be chosen arbitrarily.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this sense, on can view it as a measure of randomness associated with a string of letters.  A great example is wheel of fortune.  When Pat Sajak shows you a the board with the white and green blocks, he as already provided you a lot of information by placing spaces between the white blocks, because he has drastically reduced the number of possible combinations that might be possible to fill in the white blocks.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum information (or entropy) of the board with 52 boxes or &quot;trilons&quot; and using 26 letters is $26^{52} = 3.8\times 10^{73}$ combinations or between $244$ and $245$ bits of information in binary. However, if there were only 11 boxes illuminated white, then the actual information of the board has suddenly dropped to $26^{11} = 3.6\times 10^{15}$ combinations giving an actual information content (or entropy) or $51$ to $52$ bits.  The relative information is $\dfrac{51}{244} = 0.21$ or 21%.  The redundancy is then given by $1 - 0.21 = 0.79$ or 79%.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Vanna flips boxes, she is decreasing the relative entropy and increasing the redundancy to a point where the probability of solving the puzzle becomes very high.  So in this sense, information, like entropy, is a measure of uncertainty about the system.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there are different types of uncertainty, one is the uncertainty associated with the freedom of choice of message, and the other is noise.  The uncertainty discussed in wheel of fortune example is due to the freedom of choice.  In a noiseless situation, we would expect the word or phrase that Vanna unveils to be exactly the one chosen before the show.  In a noisy environment, for instance, one where there is some probability of a crewmember mispelling the word  while setting up the blocks, then it is possible that the final word shown is not the one chosen before the show.  That uncertainty, or noise, is called equivocation, and is brought in by the environment itself.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distinction between a noisy and noiseless environment is very important.  William Tuller in 1949 published a paper &quot;&lt;a href=&quot;http://dspace.mit.edu/bitstream/handle/1721.1/4932/RLE-TR-114-14264176.pdf?sequence=1&quot; rel=&quot;nofollow&quot;&gt;THEORETICAL LIMITATIONS ON THE RATE&#xA;OF TRANSMISSION OF INFORMATION&lt;/a&gt;&quot; that proved that there was no limit in the amount of information that could be transmitted in a noiseless channel.  This was why Shannon's paper &quot;&lt;a href=&quot;http://www.stanford.edu/class/ee104/shannonpaper.pdf&quot; rel=&quot;nofollow&quot;&gt;Communication in the Presence of Noise&lt;/a&gt;&quot; was critical to communication theory in that it properly quantified what noise actually was, and how it affected communication and the transfer of information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, before finishing, it should be noted that Hartley in his 1928 paper &quot;&lt;a href=&quot;http://bbqshirt.org/wp-content/uploads/2010/07/transmission_of_information_1928b.pdf&quot; rel=&quot;nofollow&quot;&gt;Transmission of Information&lt;/a&gt;&quot; was the first to really give a modern definition of information and give it quantitative measure.  I would recommend reviewing that paper as a starting point.  Other significant contributions are made from other scientists, such as Wiener which is best captured in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/026273009X&quot; rel=&quot;nofollow&quot;&gt;Cybernetics&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a closing note, it is refreshing that the significance of quantum noise is beginning to be discussed, and I hope it continues in the future.&lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="1383" LastEditorDisplayName="" LastEditDate="2011-01-19T02:24:38.647" LastActivityDate="2011-01-19T02:24:38.647" CommentCount="1" />
  <row Id="3290" PostTypeId="2" ParentId="3285" CreationDate="2011-01-19T02:11:42.080" Score="2" ViewCount="" Body="&lt;p&gt;Have a look at: &lt;a href=&quot;http://iopscience.iop.org/0295-5075/60/2/220&quot; rel=&quot;nofollow&quot;&gt;http://iopscience.iop.org/0295-5075/60/2/220&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar situation is described in: &lt;a href=&quot;http://arxiv.org/abs/cond-mat/9904139&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/cond-mat/9904139&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1268" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-02-16T14:08:32.813" LastActivityDate="2011-02-16T14:08:32.813" CommentCount="1" />
  <row Id="3291" PostTypeId="2" ParentId="3257" CreationDate="2011-01-19T02:21:52.893" Score="0" ViewCount="" Body="&lt;p&gt;Stephen, &#xA;I am sure everything you are reading here is horribly depressing, but unfortuntely true.  Barring vast personal wealth, the normal path to practicing physics is largely for the young and single.  However, that said, there is nothing that would be better than to get a good education in a more accessible field, and bring a passion for math and physics to it.  I would recommend taking math and physics courses, even if they don't lead to a career of a physicist.  You will find that because people with a passion for math and physics is extremely rare, anyone who has skill and determination can bring a considerable amount of illumination to their field.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the question of what kind of math to learn on short notice is important, and my answer to that is statistics.  I would also spend time learning computer programming, nothing complex, but maybe some visual basic in order to create macros in excel.  If you understand those two things and can do it well, you have automatically positioned yourself ahead of 98% of the population.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1383" LastActivityDate="2011-01-19T02:21:52.893" />
  <row Id="3292" PostTypeId="2" ParentId="3179" CreationDate="2011-01-19T02:38:50.410" Score="0" ViewCount="" Body="&lt;p&gt;Since the original question asks about the Dirac algebra and Pauli matrices, I think the problem might best be answered by considering just these objects. This ignores the usual use of a Wick rotation in terms of changing the time variable by $t\to it$, but in the sense that the Dirac algebra and Pauli matrices define the Minkowski geometry I think this is a useful simplification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two common sorts of Clifford algebras, real and complex. In both cases one begins with a set of &quot;basis vectors&quot;. For the Dirac algebra the basis vectors are usually written as $\{\gamma^0,\gamma^1,\gamma^2,\gamma^3\}$ and for the Pauli algebra the basis vectors are $\{\sigma_x,\sigma_y,\sigma_z\}$. These basis vectors all anti-commute, and square to the various choices of signature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a real(complex) Clifford algebra one considers the real(complex) vector space defined with a basis given by all possible products of basis vectors. For the Dirac algebra, one ends up with sixteen basis elements. These are sometimes called the bilinear terms. There are $2^4 = 16$ and they come in five varieties, one scalar (S), one pseudoscalar (P), four vectors (V), four axial vectors (A), and six tensors (T).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Dirac algebra is a complex Clifford algebra, so its elements are 16-dimensional complex objects. Most people work with representations using 4x4 complex matrices. In this setting, a Wick rotation amounts to replacing a basis vector with $i$ times the basis vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are only three basis vectors in the Pauli algebra, there are only eight $= 2^3$ basis elements in the Clifford algebra. The eight basis elements are a single scalar 1, three vectors $\sigma_x,\sigma_y,\sigma_z$, three bivectors $\sigma_x\sigma_y$, $\sigma_x\sigma_z$, $\sigma_y\sigma_z$, and a single pseudoscalar $\sigma_x\sigma_y\sigma_z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the Pauli algebra pseudoscalar is somewhat strange. It squares to -1 and it commutes with all the other basis elements (and therefore with everything in the algebra). As such it acts just like an imaginary unit $i$. And so with the Pauli algebra, we use the real Clifford algebra instead of the complex one. Our usual representation has 8 real dimensions and these are kept in 2x2 complex matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the difference between the Pauli and Dirac algebras, when applying a Wick rotation, is that Pauli algebra is real while the Dirac algebra is complex. Hmmmm.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As a sort of aside on Wick rotations of Pauli spin matrices, the post here came to mind when I read Witten's latest paper (January 17, 2011). See equation (2.4), page 19:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_0 = \left(\begin{array}{cc}0&amp;amp;+1\\-1&amp;amp;0\end{array}\right)$,&#xA;$B_1 = \left(\begin{array}{cc}0&amp;amp;+1\\+1&amp;amp;0\end{array}\right)$,&#xA;$B_2 = \left(\begin{array}{cc}+1&amp;amp;0\\0&amp;amp;-1\end{array}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fivebranes and Knots&#xA;Edward Witten (2011)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1101.3216&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1101.3216&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is simply $B_0 = i\sigma_x, B_1=\sigma_y, B_2=\sigma_z$. These sorts of things happen constantly when one messes around with Clifford algebra as will be clear from the context.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-01-20T07:49:31.527" LastActivityDate="2011-01-20T07:49:31.527" CommentCount="2" />
  <row Id="3293" PostTypeId="1" AcceptedAnswerId="3310" CreationDate="2011-01-19T03:23:34.147" Score="2" ViewCount="85" Body="&lt;p&gt;Once again, an attempt to connect popular and real science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read that inflation implies our universe should look flat, or really close to flat, because our observable universe is only a really small part of a much larger &quot;inflationary bubble&quot;.&#xA;So, we are like people who can see only a very small patch of the Earth and think its flat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See, for example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.astro.ucla.edu/~wright/CMB-MN-03/inflating_bubble.html&quot; rel=&quot;nofollow&quot;&gt;http://www.astro.ucla.edu/~wright/CMB-MN-03/inflating_bubble.html&lt;/a&gt;   &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://nedwww.ipac.caltech.edu/level5/Guth/Guth5.html&quot; rel=&quot;nofollow&quot;&gt;http://nedwww.ipac.caltech.edu/level5/Guth/Guth5.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In that case, do we have any theoretical or empirical estimates of how large our inflationary bubble is compared to the observable universe ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read an article by Andrei Linde in Scientific American saying that the size of the bubble could be $10^{10^{12}}$ metres (1 followed by a trillion zeros). Is this generally accepted ?&lt;/p&gt;&#xA;" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-19T06:53:45.197" LastActivityDate="2011-01-19T09:12:57.333" Title="Size of our inflationary bubble" Tags="&lt;cosmology&gt;&lt;cosmological-inflation&gt;" AnswerCount="1" />
  <row Id="3294" PostTypeId="1" CreationDate="2011-01-19T03:33:28.400" Score="2" ViewCount="260" Body="&lt;p&gt;Wasn't the density of the universe at the moment after the Big Bang so great as to create a black hole? If the answer is that the universe/space-time can expand anyway what does it imply about what our universe looks like from the outside?&lt;/p&gt;&#xA;" OwnerUserId="1451" LastActivityDate="2011-08-19T18:49:58.767" Title="Big Bang snuffed by a black hole?" Tags="&lt;black-holes&gt;&lt;big-bang&gt;" AnswerCount="4" />
  <row Id="3295" PostTypeId="2" ParentId="3294" CreationDate="2011-01-19T04:48:15.447" Score="6" ViewCount="" Body="&lt;p&gt;I don't think that the question &quot;what does the universe look like from the outside?&quot; is very meaningful. Just because there is not outside for the universe. As for the black hole why should high density i.e. a lot of mass in little volume, cause the creation of a black hole? If you are thinking about the Schwarzschild solution (and radius), it describes a spherical object outside of which the space is empty, and as I said there is no outside for the universe. &lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2011-01-19T04:48:15.447" CommentCount="1" />
  <row Id="3296" PostTypeId="1" CreationDate="2011-01-19T05:09:38.080" Score="1" ViewCount="203" Body="&lt;p&gt;A little background: a Pinewood Derby car is a small wooden car that races down an inclined track, powered only by gravity. You are allowed to add weight to the car up to a certain limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.scoutorama.com/derby/derby_display.cfm?race_id=22&quot; rel=&quot;nofollow&quot;&gt;Here is a recommendation&lt;/a&gt; to add the weight to the upper back of the car, to maximize the potential energy. My own gut feel is to add it so that the weight is evenly distributed between all 4 wheels. I'd like some advice that is well grounded in science.&lt;/p&gt;&#xA;" OwnerUserId="1453" LastActivityDate="2011-01-19T20:25:57.307" Title="Where's the best place to add weight to a Pinewood Derby car?" Tags="&lt;classical-mechanics&gt;&lt;gravity&gt;" AnswerCount="3" CommentCount="11" />
  <row Id="3297" PostTypeId="1" AcceptedAnswerId="3300" CreationDate="2011-01-19T05:32:39.480" Score="-2" ViewCount="120" Body="&lt;p&gt;Since i was  a child i was told that current has  pushing or pulling effect. However, i never got a dc shock and i don't remember ac shock (i got it very long before).  The day before yesterday, an experienced electrician told me that dc has pulling effect ( most of the batteries supply only 12V it's not so much to worry but if supplied with very high voltage it is very deadly comparison to ac and) and ac has pushing effect. Logically i can't find any explanations. electricity is just flow of charge, and in our body, we have more electrons than protons (from C-14 decay) but still very less than enough to give shock or pushing or pulling effect.&lt;/p&gt;&#xA;" OwnerUserId="567" LastActivityDate="2011-01-19T06:24:25.033" ClosedDate="2011-01-19T06:52:43.810" Title="does current product push or pull effect" Tags="&lt;electricity&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="3298" PostTypeId="2" ParentId="3296" CreationDate="2011-01-19T05:33:31.163" Score="5" ViewCount="" Body="&lt;p&gt;UPDATE: Having totally misunderstood what a pinewood derby car was, my previous answer probably wasn't very helpful (I thought it was like a go-cart). Here is a proper answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, let me say the recommendation you read is still wrong, for reasons that will be seen in a minute. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Nuehp.jpg&quot; alt=&quot;Diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The car starts off inclined at an angle $\theta$, with the additional mass a distance $x$ behind the front axle, and a distance $h$ above the line passing through both axles. The change in height of the additional mass relative to the front axle is then given by $d_x + d_h$, where $d_x$ is the decrease in relative elevation of the point a distance $x$ behind the front axle on the line between the two axles, and $d_h$ is the decrease in relative elevation between the mass relative and this point. Clearly, $d_x = x \sin(\theta)$ and $d_h = h (\cos(\theta) - 1)$. Notice that $d_h$ is negative, since the change in relative elevation actually increases for positive $h$. Thus the total change in height (which is proportional to the additional potential energy) is given by $\Delta H = x\sin(\theta) - h(1-\cos(\theta))$. Thus, the best position for the additional mass is very low down (even below the line of the axles if possible), at the back.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditorDisplayName="" LastEditDate="2011-01-19T10:14:07.227" LastActivityDate="2011-01-19T10:14:07.227" CommentCount="10" />
  <row Id="3299" PostTypeId="2" ParentId="3294" CreationDate="2011-01-19T06:02:33.483" Score="2" ViewCount="" Body="&lt;p&gt;The standard ΛCDM model of the Big Bang fits obsersvations to the Friedmann-Robertson-Walker solutions of general relativity, which do not form black holes. Intuitively, the initial expansion is great enough to counteract the usual tendency of matter to gravitationally collapse. As far as we know, the universe looks about the same from every point on the large scale. It is a built-in assumption of the FRW family solutions, and sometimes called the &quot;Copernican principle.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't absolutely have to be right, of course, though in a sense it is the simplest possible empirically adequate model, and so is favored by Ockham's razor. There have been attempts to fit the astronomical observations to an isotropic and &lt;em&gt;inhomogeneous&lt;/em&gt; solution of GTR (meaning, we would be near &quot;the center&quot;), but to my knowledge they have been less than conclusive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an oversimplified model of spherical stellar collapse assumes that the star has uniform density and no pressure, the interior of which comes out to be equivalent to the k = +1 (positive curvature, closed) contracting FRW universe. The interior is smoothly patched to a Schwarzschild exterior. The k = 0 (flat) and k = -1 (open) cases can be thought of as the interior of such a star in the limit of infinite radius, collapsing from rest and with some finite velocity, respectively. They too can be smoothly patched to a Schwarzschild exterior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our the observed universe is expanding, but we still can say that's it's possible for the isotropic and homogeneous region we observe to have an &lt;em&gt;edge&lt;/em&gt;, or perhaps even be the the interior of a time-reversed black hole. But it should be emphasized that we have no empirical reason to believe that it's anything more exotic than a plain FRW universe. Though on a more serious alternatives, some models of cosmic inflation have our observed universe as one of many &quot;bubbles&quot; in an inflating background.&lt;/p&gt;&#xA;" OwnerUserId="1353" LastEditorUserId="1353" LastEditorDisplayName="" LastEditDate="2011-01-19T06:08:26.113" LastActivityDate="2011-01-19T06:08:26.113" CommentCount="1" />
  <row Id="3300" PostTypeId="2" ParentId="3297" CreationDate="2011-01-19T06:24:25.033" Score="1" ViewCount="" Body="&lt;p&gt;I suppose by &quot;pushing&quot; and &quot;pulling&quot; is meant to describe what happens with accidental electric shock.&#xA;What happens when one touches a live wire the body provides a conduit to ground and a current is set up. &lt;a href=&quot;http://en.wikipedia.org/wiki/Electric_shock&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; in Wikipedia explains the physiology:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;DC tends to cause continuous muscular contractions that make the victim hold on to a live conductor, thereby increasing the risk of deep tissue burns.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is a reason for the rumor.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-19T06:24:25.033" />
  <row Id="3302" PostTypeId="1" AcceptedAnswerId="3312" CreationDate="2011-01-19T06:37:09.170" Score="1" ViewCount="220" Body="&lt;p&gt;Quoting this &lt;a href=&quot;http://solar.calfinder.com/blog/solar-research/solar-even-more-affordable/&quot; rel=&quot;nofollow&quot;&gt;web site&lt;/a&gt; (which I haven't thoroughly verified for credibility yet):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The material is graphene, also known&#xA;  as graphite, a naturally-occurring&#xA;  mineral that forms a one-atom-thick&#xA;  sheet of honeycomb crystal lattice&#xA;  material (the ideal shape for solar&#xA;  cells) that has surprisingly robust&#xA;  electric characteristics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In what context is this ideal shape? In the nano or macro scale? Why?&lt;/p&gt;&#xA;" OwnerUserId="293" LastActivityDate="2011-02-04T07:37:47.777" Title="Why is honeycomb lattice the ideal shape for solar cells?" Tags="&lt;nanoscience&gt;&lt;photovoltaics&gt;" AnswerCount="2" />
  <row Id="3303" PostTypeId="2" ParentId="3262" CreationDate="2011-01-19T06:46:22.867" Score="3" ViewCount="" Body="&lt;p&gt;This is a matter of opinion, all I can state here is my opinion, and I’d fully expect other string theorists will have very different ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, second quantization in field theory is a way to proceed from one particle QM to free field theory whose quantization will in turn give many particles of the type you started with. The point is that this is a way to introduce free (or by generalization, weakly interacting) field theory. It is a process which is perturbative in nature and does not provide us with non-perturbaive information. Nevertheless it is useful, for example it is easier to discuss things like the vacuum structure, off-shell observables, and some other issues which are completely obscured in a first quantized formulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in string theory, the original formalism is first quantized, you discuss one string, or a fixed number of strings. It is natural to try to second quantize the theory, but in my view you'd be learning about perturbative string theory only in this process (as a sociological fact, all the non-perturbative knowledge we have about string theory does not connect very naturally to SFT). The formalism does have the advantage of having off-shell observables, like an ordinary quantum field theory, but it is not clear that in theory of gravity (which does not have local observables) this is necessarily a good thing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was all true for closed string field theory. There is a bit more motivation and results for open string field theory. Especially, there are some successes in exploring the space of (open string) vacua in this formalism (by discussing the so-called tachyon condensation and brane decay processes). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this context I can elaborate a bit on what it means to second quantize the theory. You have creation operators that create an open string (in a particular state). You can add interactions (turns out you only need cubic ones) and create Feynman diagrams whose propagators and vertices are extended. Turns out that the set of Feynman diagrams you get is precisely the set of world-sheets you can build in open string theory. When you build loop amplitude, you can see that you have closed strings propagating as well, in intermediate states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is hope and some indications that quantization of open SFT will lead in the future to non-perturbative formulation of string theory (including the closed strings), but the state of the art is not very close to this goal, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in my view the answer to both your questions is sadly negative, by and large, with some exceptions, but many really smart people continue working on the subject, and I may well be wrong in my expectations. I'd be happy if one of these smart people who has a more positive take will give their own answer here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I realize there may be obscure points in my reply, since I don't know your background. If you ask I'll try to explain things better).&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-01-19T06:52:30.630" LastActivityDate="2011-01-19T06:52:30.630" CommentCount="2" />
  <row Id="3305" PostTypeId="2" ParentId="3262" CreationDate="2011-01-19T07:58:02.467" Score="3" ViewCount="" Body="&lt;p&gt;Dear Andrew,&#xA;despite Moshe's expectations, I fully agree with him, but let me say it differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In QFT, we're talking about &quot;first quantization&quot; - this is not yet a quantum field theory but either a classical field theory or quantum mechanics for 1 particle. Those two have different interpretations - but a similar description. When it is &quot;second-quantized&quot;, we arrive to QFT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feynman diagrams in QFT may be derived from &quot;sums over histories&quot; of quantum fields in spacetime; for example, the vertices come from the interaction terms in the Lagrangian, and the propagators arise from Wick contractions of quantum fields. This is the &quot;second-quantized&quot; interpretation of the Feynman diagrams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a first quantized interpretation. You may literally think that the propagators are amplitudes for an individual particle to get from $x$ to $y$, and the vertices allow you to split or merge particles. You may think in terms of particles instead of fields. In QFT, this is an awkward approach because most particles have spins and it's confusing to write a 1-particle Schrödinger equation for a relativistic spin-one photon, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in string theory, spin is derived and the first-quantized interpretation is very natural. So the cylindrical world sheet describes the history of a closed string much like a world line describes the history of a particle. And it's enough to change the topology of the world sheet to get the interactions as well. So in string theory, one may produce the amplitudes &quot;directly&quot; from the first-quantized approach because the changed topology of the world sheet, which we sum over, knows all about multi-particle states and their interactions, too. We say that the interactions are already determined by the behavior of a single string.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Needless to say, like any Feynman diagrams, these sums over topologies are just perturbative in their reach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you may also write down string theory as a string field theory, in terms of quantized string fields in spacetime. Somewhat non-trivially, an appropriate interaction term - that &quot;knows&quot; about the merging and splitting of strings - may be constructed in terms of a &quot;star-product&quot; (a generalization of noncommutative geometry). In this way, string theory becomes formally equivalent to a quantum field theory with infinitely many fields in spacetime - for every possible internal vibration of the string, there is one string field in spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It used to be believed that this formalism would tell us much more than the perturbative expansions because, for example, lattice QCD in principle can be used to define the theory completely, beyond perturbative expansions. However, this belief has been showed largely untrue. At least so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's been shown that string field theory indeed offers an equivalent way to calculate all the amplitudes of perturbative string theory - especially for bosonic strings with external open strings (closed strings are possible, and surely appear as internal resonances, but they are awkward to include directly as external states; superstrings are probably possible but require a substantially heavier formalism).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, string field theory has been very useful to explicitly verify various conjectures about the tachyon potential in bosonic string theory (or, equivalently, about the fate of unstable D-branes which emerge as classical solutions in string field theory). These investigations, started by Ashoke Sen, led to some nice mathematical identities that had to work - because string theory works in all legitimate descriptions - but that were still surprising from a mathematical viewpoint. But all the physical insights confirmed by string field theory had already been known from more direct calculations in string theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So because string field theory is widely believed not to tell us anything really new about physics, only a dozen of string theorists in the world dedicate most of their time to string field theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moshe is surely no exception in thinking that it is not too important to work on SFT. Still, it is conceivable that sometime in the future, a more universal definition of string theory will be a refinement of string field theory we know today. However, it's also possible that this will never occur because it's not true: string field theory seems too tightly connected with a particular spacetime and with particular objects (strings) while we know that the true string theory finds it much easier to switch to another spacetime and other objects by dualities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-19T07:58:02.467" CommentCount="1" />
  <row Id="3306" PostTypeId="1" AcceptedAnswerId="3348" CreationDate="2011-01-19T08:06:31.723" Score="0" ViewCount="347" Body="&lt;p&gt;Consider two long wire with negligible resistance closed at one end of the resistance R (say a light bulb), and the other end connected to a battery (DC voltage). Cross-sectional radius of each wire is in  $x = 100$ times smaller than the distance between the axes of the wires.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At which value of resistance R (in ohms) the resultant force of the interaction of parallel conductors disappears?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Vxja0.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="552" LastEditorUserId="552" LastEditorDisplayName="" LastEditDate="2011-01-20T09:35:31.687" LastActivityDate="2011-02-16T08:42:57.023" Title="When is the force null between parallel conducting wires?" Tags="&lt;electromagnetism&gt;&lt;homework&gt;" AnswerCount="4" CommentCount="8" />
  <row Id="3307" PostTypeId="1" CreationDate="2011-01-19T08:09:53.167" Score="1" ViewCount="49" Body="&lt;p&gt;Consider a sensor which is located inside the solid substance. This sensor is capable of detecting the substance oscillations along each of the three axes (usually orthogonal, but generally, any non-degenerated (non-coplanar) basis). This sensor produces a recording of the detected oscillations, called a trace, containing the displacement sampled at high-frequency, capable to capture any oscillation frequency existing in the substance. There are three traces, one for each axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that at some period of time the sensor registers an interference of an event, consisting of the compression wave, and two shear waves. Traces now contain a recordings of the event, effectively, a projections of the oscillations on the sensor's axes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I now virtually re-orient the axes, that is, perform the linear transform of the traces, so that they will be oriented each along a corresponding wave displacement vector, and thus will contain only single wave event recording in each of the traces after the transformation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or should I ask this on &lt;code&gt;math&lt;/code&gt; ?&lt;/p&gt;&#xA;" OwnerUserId="1457" LastEditorUserId="1457" LastEditorDisplayName="" LastEditDate="2011-01-19T09:26:54.987" LastActivityDate="2011-01-19T09:26:54.987" Title="Reorienting a sensor axes according to particle displacement directions" Tags="&lt;waves&gt;&lt;plane-wave&gt;&lt;sensor&gt;" CommentCount="1" />
  <row Id="3308" PostTypeId="2" ParentId="3264" CreationDate="2011-01-19T08:12:15.477" Score="1" ViewCount="" Body="&lt;p&gt;Dear Finemann, &#xA;the &quot;spectral power distribution&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Spectral_power_distribution&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Spectral_power_distribution&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &quot;intensity vs frequency&quot; contain exactly the same information. They're so easily related that physicists wouldn't even consider them &quot;two different things&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spectral power distribution is the (infinitesimal) power per unit area and per unit wavelength:&#xA;$$ M_\lambda = \frac{\delta\Phi}{\delta A\delta \lambda} $$&#xA;You may simply write $\lambda$ as $c/f$ and use the right differentials of functions, $\delta\lambda=-(c/f^2)\times \delta f$ and you will get the analogous distribution in terms of the frequencies. The graph will be just flipped from the left to right (that's because of the irrelevant minus sign), and on one side, it will be stretched vertically and squeezed horizontally. There is no international ban on using the frequencies, and indeed, it's true that even top physicists usually prefer to parameterize radiation by its frequency rather than the wavelength. You may have seen &quot;popular&quot; sources that used frequencies instead of the wavelength but it is not because they were incorrect or misleading; they did copy conventions that common in research-level physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also think that the words &quot;intensity&quot; and &quot;power&quot; are different things. But &quot;intensity&quot; is just a popular term for &quot;irradiance&quot; which is just power per unit area, see&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Radiant_intensity&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Radiant_intensity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More properly, &quot;radiant intensity&quot; should be counted as power of a localized source per solid angle (note that if the radiation is radial, the power only depends on the solid angle and not the distance at which you absorb the light).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black body spectral power distribution is continuous: a black body is indeed able to emit at all frequencies. A usual disclaimer for continuous distributions should be added: the power emitted by a continuous power distribution at a particular exact frequency is always exactly zero. You need a whole interval of frequencies (or wavelengths) to get a nonzero answer. The (infinitesimal) power coming from this interval of frequencies will then be proportional to the (infinitesimal) width of the interval of frequencies. That's why we talk about the ratios of differentials etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-19T10:09:47.187" LastActivityDate="2011-01-19T10:09:47.187" />
  <row Id="3309" PostTypeId="1" CreationDate="2011-01-19T08:13:18.560" Score="7" ViewCount="414" Body="&lt;p&gt;How big is it by a truly quantum measurement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am thinking of comparing Science magazines &quot;Breakthrough of the Year&quot;  (BYOT) with the Zeilinger buckyball. &#xA;The BYOT is a piezoelectric mechanical oscillator (PO) the size of the cross section of a human hair.&#xA;It is placed in a superposition of its ground state and its first excited state.&#xA;The well known buckyball experiment is a two slit experiment using buckyballs.&#xA;(A third candidate might be a macroscopic Josephson junction oscillator conducting both ways at once.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have made some basic calculations.&#xA;For instance, the BYOT contains about 10^14 atoms compared to 60 or 72 atoms in the buckyball.&#xA;By this measure the BYOT is bigger by a factor of about 10^12.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the two slits are separated by 50 to  100 nanometers, or 10^-7 meters. &#xA;In its first excited state, the top of the BYOT PO moves about 10^-15 meters per cycle, according to my calculations.&#xA;By this measure the buckyball wins by a factor of about 10^8. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating energy of the moving parts, I find a much closer horserace, but the buckyball is about 100 times bigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, none of these calculations is at all quantum mechanical (QM).&#xA;ArXiv lists at least five papers proposing truly quantum mechanical measures of the size of a macroscopic Schrodinger cat.&#xA;The most recent is Lee and Jeong &lt;a href=&quot;http://arxiv.org/abs/1101.1209&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1101.1209&lt;/a&gt; which references the other four.&#xA;Can someone competent (or expert) in QM apply one or more of these quantum measures to the BYOT and the buckyball and tell me which is larger?&#xA;TIA. Jim Graber&lt;/p&gt;&#xA;" OwnerUserId="1172" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-19T10:14:08.787" LastActivityDate="2011-03-04T17:26:15.517" Title="What is the world's biggest Schrodinger cat?" Tags="&lt;quantum-mechanics&gt;&lt;experimental-physics&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="3310" PostTypeId="2" ParentId="3293" CreationDate="2011-01-19T09:09:17.270" Score="1" ViewCount="" Body="&lt;p&gt;The answer depends on how long inflation lasted, how rapidly it expanded, and how big the bubble was to start with. These parameters are obviously model dependent and there are many models none of which is entirely convincing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For arguments sake let's assume that inflation ends at the GUT scale with a doubling in size every Planck unit of time and with an initial diameter of the planck length. The GUT scale in time is 10000 planck units so the overall expansion factor is $2^{10000} \sim 10^{3000}$ . With this factor the initial size of $10^{-43}$ meters is almost irrelevant and we can say the final bubble size is around $10^{3000}$ meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously you could make different assumptions and get very different answers. I don't know what assumptions Linde was using but I guess he was taking the fastest inflation he thought was possible to get an upper limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is at least as interesting to consider the lower limit that is consistent with observation. This depends on how flat we observe the universe to be but it has to be bigger than the observed universe so not much smaller than $10^{25}$ meters&lt;/p&gt;&#xA;" OwnerUserId="1341" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-01-19T09:12:57.333" LastActivityDate="2011-01-19T09:12:57.333" CommentCount="2" />
  <row Id="3311" PostTypeId="1" CreationDate="2011-01-19T09:31:57.717" Score="1" ViewCount="181" Body="&lt;p&gt;Can the &lt;a href=&quot;http://www.nature.com/nphys/journal/v6/n4/full/nphys1629.html&quot; rel=&quot;nofollow&quot;&gt;gamma ray bursts connected to a new supernova&lt;/a&gt; &#xA;prevent the black hole from forming?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Quoting the abstract of the linked reference &quot;...has become clear that probably all long GRBs are associated with type-Ic supernovae resulting from massive stars &lt;strong&gt;that suffered substantial mass loss&lt;/strong&gt;...&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If enough mass is converted into radiated energy then the condition for BH formation can vanish?&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-01-19T15:26:08.257" LastActivityDate="2011-07-06T02:18:00.420" Title="Gamma-ray bursts: A supernova connection" Tags="&lt;black-holes&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="3312" PostTypeId="2" ParentId="3302" CreationDate="2011-01-19T09:48:57.837" Score="3" ViewCount="" Body="&lt;p&gt;The website is clearly supported by lots of money which doesn't guarantee that it reflects the most accurate scientific information. The very page you quoted says under the picture:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The material is graphene, also known as graphite...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, no. Graphene is not the same thing as graphite. Graphite is a 3-dimensional material used to produce pencils - and graphene is just one layer of graphite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can't expect everything to be &quot;quite right&quot;. The honeycomb lattice is useful to get a &quot;denser packing&quot; which, under some circumstances, could mean a higher density of the molecules that may transform light to electromagnetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is very subtle to compare &quot;different lattices&quot; of the &quot;same material&quot; because the same material usually can't form &quot;all lattices&quot;. So I suspect that the statement is just phenomenological in character - some of the most efficient solar cells that people have found had a honeycomb lattice but as far as I know, there's no &quot;universal proof&quot; that it has to be so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation would be much less ambiguous if one asked a well-defined geometric question - what is the best close packing of spheres, for example. However, the efficiency of solar cells also depends on more complicated physical properties of the material than just its lattice, so one shouldn't expect universal yet simple answers.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:37:47.777" LastActivityDate="2011-02-04T07:37:47.777" CommentCount="1" />
  <row Id="3313" PostTypeId="2" ParentId="3294" CreationDate="2011-01-19T10:00:54.903" Score="4" ViewCount="" Body="&lt;p&gt;A high enough energy density is a necessary condition but not a sufficient condition for black holes to form: one needs to have a center which will ultimately become the center of the black holes; one needs the matter that collapses to the black hole to have a low enough velocity so that gravity may squeeze it before the matter manages to fly away and dilute the density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter two conditions are usually almost trivially satisfied for ordinary chunks of matter peacefully sitting at some place of the Universe; but they're almost maximally violated by the matter density right after the Big Bang. This matter has no center - it is almost uniform throughout space - and has high enough velocity (away from itself) that the density eventually gets diluted. And indeed, we know that it did get diluted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, a collapse of matter (e.g. a star) into a black hole is an idealized calculation that makes certain assumptions about the initial state of the matter. These assumptions are clearly not satisfied by matter after the Big Bang. Instead of a collapse of a star, you should use another simplified version of Einstein's equations of general relativity - namely the Friedmann equations for cosmology. You will get the FRW metric as a solution. When it is uniform to start with, it will pretty much stay uniform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The visible Universe is, in some sense, analogous to a black hole. There exists a cosmic horizon and we can't see behind it. However, it is more correct to imagine that the interior of the visible space - that increasingly resembles de Sitter space because the cosmological constant increasingly dominates the energy density - should be viewed as an analogy to the exterior of a black hole. And it's the exterior of the visible de Sitter space that plays the role of the interior of a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship between (namely the ratio of) the mass and the radius for the visible Universe is not too far from the relationship between (or ratio of) the black hole mass and radius of the same size. However, it's not accurate, and it is not supposed to be accurate. The mass/radius ratio is only universal for static (and neutral) black holes localized in an external flat space and our Universe is clearly not one of them.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-19T17:03:19.617" LastActivityDate="2011-08-19T17:03:19.617" CommentCount="1" />
  <row Id="3314" PostTypeId="1" CreationDate="2011-01-19T10:21:30.057" Score="1" ViewCount="140" Body="&lt;p&gt;The question is: How to calculate the energy of electrostatic field of a newly created particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Electric_potential_energy#Energy_stored_in_an_electrostatic_field_distribution&quot; rel=&quot;nofollow&quot;&gt;usual formula&lt;/a&gt; involves integration &lt;em&gt;'all over space'&lt;/em&gt;.&#xA;According to GR I believe that &lt;em&gt;'all over space'&lt;/em&gt; means the interior volume with radius c*t_now, where t_now is the lifetime of the particle at the time of calculation and c is the light speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the energy stored in the field is a quantity dependent on time evolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/3246/conservation-of-energy-and-shrinking-matter-closed%3a&quot;&gt;this is a clarification of the question of PS: conservation of energy and shrinking matter&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-01-20T08:36:34.650" Title="energy field of a newly created particle" Tags="&lt;electrostatics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="3315" PostTypeId="1" AcceptedAnswerId="3369" CreationDate="2011-01-19T10:23:39.570" Score="6" ViewCount="285" Body="&lt;p&gt;This is intended to be a fun question.  Calorimetry used for calculating the heat generated from chemical changes has been around for centuries, however, I suspect the process for calculating food calories is much more complicated.  Can anybody provide a clear succinct, knowledgable answer?  Although I will accept answers with links, an independent answer would be preferred.&lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T14:44:49.267" LastActivityDate="2011-02-02T14:44:49.267" Title="How are the calories in food calculated?" Tags="&lt;energy&gt;&lt;physical-chemistry&gt;&lt;biophysics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="3316" PostTypeId="2" ParentId="2069" CreationDate="2011-01-19T10:32:27.063" Score="2" ViewCount="" Body="&lt;p&gt;We're not allowed to change the boundary, especially not in the direction of making a bigger piece of the world &quot;classical&quot; than what is allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the whole world - not just electrons but also the apparatus - obeys the laws of quantum physics. At some moment, however, the quantum phases, coherence, and interference become de facto impossible because they're lost in the noise of the environmental degrees of freedom. This process is called &quot;decoherence&quot; and once it happens, it is legitimate to consider the quantum-calculated probabilities to be ordinary classical probabilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Decoherence depends on the Hamiltonian and exactly determines where the boundary is. The important point is that if one assumed that the world is classical even at some &quot;more microscopic level&quot; than where it becomes classical - i.e. if one assumed that certain objects can't interfere and exhibit entanglement etc. even though they can - he would end up with incorrect results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See e.g. this introduction to decoherence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.karlin.mff.cuni.cz/~motl/entan-interpret.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.karlin.mff.cuni.cz/~motl/entan-interpret.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Niels Bohr was implicitly aware of decoherence but unfortunately, his explanations of his intuition weren't too coherent, either. That's why this insight had to wait for an explicit description until the 1980s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, the Copenhagen school adopted a pragmatic attitude. It didn't explain or calculate where the classical-quantum boundary was located. However, it placed it safely in the &quot;macroscopic realm&quot; so that no interference, entanglement, or other predictions of quantum mechanics were ever lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important to realize that one can use the full quantum description even with macroscopic objects - they satisfy quantum mechanics, too. I can treat all other observers and physicists as quantum systems who evolve into linear superpositions etc. and only calculate the probability that I see $A$ or $B$ at the very end - because I know that I won't &quot;perceive&quot; any bizarre linear superposition. It never hurts when you treat all things in the quantum mechanical framework. See Sidney Coleman's talk, &lt;em&gt;Quantum Mechanics In Your Face&lt;/em&gt;, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2010/11/sidney-coleman-quantum-mechanics-in.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/11/sidney-coleman-quantum-mechanics-in.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where this quantum treatment of the whole world is used in several pedagogical examples. On the other hand, when one deals with macroscopic enough objects, and traces over environmental degrees of freedom that are really impossible to keep track of, the density matrix becomes almost immediately and almost exactly diagonal in the &quot;preferred classical basis vectors&quot;. That allows us to treat the diagonal elements of the density matrix as classical probabilities. Quantum mechanics will still predict probabilities only and the results of individual events will be random. However, if the density matrix is kept diagonal, it allows us to assume - just like in classical physics - that the macroscopic degrees of freedom were in a particular state even before we measured (or saw) what they were. We won't reach any contradictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, once again, if we did make this assumption for any system that can still interfere and evolve quantum mechanically - i.e. if we imagined that any aspect of the wave function &quot;collapsed&quot; before we actually make the measurements - we would reach incorrect and contradictory predictions. So if you don't know where the boundary is, you should better assume that everything is quantum - because everything is quantum. Classical physics is just an approximation and one must appreciate that it is often or usually invalid.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-05T11:40:03.173" LastActivityDate="2011-02-05T11:40:03.173" CommentCount="2" />
  <row Id="3317" PostTypeId="2" ParentId="3059" CreationDate="2011-01-19T10:39:18.443" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0071543821&quot; rel=&quot;nofollow&quot;&gt;Quantum Field Theory Demystified&lt;/a&gt; by David McMahon chapter nine provides a good discussion on spontaneous symmetry breaking that is better than the wiki article.  It has a good level of technical calculation, provided at a senior college level, but written in a friendly style. If you have decent enough understanding of a Langrangian, then the description is good.  If not, no worries! The rest of the book will help.  &lt;/p&gt;&#xA;" OwnerUserId="1383" LastActivityDate="2011-01-19T10:39:18.443" />
  <row Id="3318" PostTypeId="1" CreationDate="2011-01-19T10:45:27.403" Score="3" ViewCount="47" Body="&lt;p&gt;In Peskin and Schroeder's chapter 12 about the renormalization group, it is stated that the parameter $\rho_m=m^2/M^2$, where $m$ is the mass and $M$ is the renormalization scale, is proportional to $T-T_c$. But this assertion is not derived in any detail, except for a handwaving one-line explanation &quot;$\rho_m$ is just the parameter that one adjust finely to bring the system to the critical temperature&quot;. Could someone elaborate this? In particular, why doesn't the parameter depend on the square or some other power of $T-T_c$?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-19T12:01:15.003" Title="Why is $\rho_m$ proportional to the deviation from critical temperature in critical phenomena?" Tags="&lt;quantum-field-theory&gt;&lt;statistical-mechanics&gt;&lt;renormalization&gt;" AnswerCount="1" />
  <row Id="3319" PostTypeId="2" ParentId="3315" CreationDate="2011-01-19T10:53:53.823" Score="4" ViewCount="" Body="&lt;p&gt;a great question. First, a detail: a person needs something like 2,000 kilocalories a day. At least that's how the unit is used today; in the past, the modern kilocalorie used to be called a calorie. Today, 1 kcal equals 4.182 kilojoules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, one could measure the kilocalories by direct calorimetry, i.e. in a bomb calorimeter. However, that would also burn some portions of the food that are not burned by the humans - as demonstrated by the fact that we sometimes need to use the restrooms to deposit some solids (with dietary fibers etc.) as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dietary value is only encoding the energy that can be extracted from the food by humans. So it is assumed that we can only get the energy from several basic &quot;elements of food&quot;, namely fat, ethanol, proteins, carbohydrates, organic acids, polyols (sweeteners, sugar alcohols), fibre, and erythritol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each food is chemically analyzed and the amount of these components is quantified. By multiplying the weight of each component by the &quot;energy density&quot; (for example, it's 9 kcal/g for fats), we obtain the kcal figure for the food. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose that these standardized figures - such as 9 kcal/g for fat - have been obtained by observing people's motion and heat for some time but I am actually not quite sure whether they were feeding someone with pure fat for some time haha. Someone may know an answer to this sub-question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-19T10:53:53.823" CommentCount="7" />
  <row Id="3320" PostTypeId="2" ParentId="3296" CreationDate="2011-01-19T11:04:32.063" Score="0" ViewCount="" Body="&lt;p&gt;Aside from position of weight, I would add that what you really want to reduce is the the rolling resistance of the car.  One of the first areas I would target is the contact surface area of the wheels with the track. I would also reduce the amount of weight at the edges of the wheel.  For instance, instead of using wide, fat wheels, I would use wheels that taper down into a v-shape.&#xA;This will minimize the energy being used to spin up the wheels, as well as reduce the energy being transferred through friction to the surface.  I would also do whatever it takes to reduce friction in whatever bearing mechanism there is between the wheels/shaft/body of the racer.  In addition, I would work to reduce as much weight in the wheels as possible.  Again, the energy absorbed by the wheels needs to be minimized as much as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So think along those lines...minimize rolling resistance, mimimize wheel weight, minimize wheel weight at the edge of the wheel. Maximize velocity at the bottom of the ramp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;addendum&#xA;I will add one little caveat, after looking at the track, there is an optimization problem here between energy lost due to friction, and the need to store energy to keep the vehicle moving.  A spin up of the wheels will store energy, which may be needed if your friction is high enough that you need that energy in the flatter portion of the track.  In any case, think in terms of transfer of energy from potenial to kinetic, what is stored on in the wheels, and what is lost due to friction, air resistance and rolling resistance.&lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="1383" LastEditorDisplayName="" LastEditDate="2011-01-19T11:18:32.327" LastActivityDate="2011-01-19T11:18:32.327" CommentCount="1" />
  <row Id="3321" PostTypeId="2" ParentId="3314" CreationDate="2011-01-19T11:14:58.447" Score="1" ViewCount="" Body="&lt;p&gt;classical physics doesn't have any well-defined framework to describe the &quot;creation of particles&quot;. So you have to talk about the real world - or, theoretical speaking, the world of quantum field theory (or something that contains it) - to be able to discuss the creation of particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in quantum field theory, you can't separate the energy coming from the field from the energy coming from the particles themselves in such a sharp way. In particular, even the lightest charged particle - the electron - has mass equal to 511 keV. That's big, and even if you compute the whole energy of the electrostatic field at the distance of the classical electron radius, $2.82 \times 10^{-15}$ meters, or shorter (not far from the radius of the proton, by the way), you will obtain a smaller energy than the rest energy of the electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to say that a (virtual) electron existed at least for a little while, you need to assume that the lifetime was at least the Compton wavelength of the electron (divided by $c$). That's about 100 times longer a distance. And even with this distance, the uncertainty of the energy is comparable to those 511 keV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This leads me to the key point which is the uncertainty principle for time and energy.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you determine the timing of the places where you measure the energy with a much better accuracy than the Compton wavelength of the electron (over $c$), then the uncertainty of the energy you measure will inevitably be much greater than the whole electron rest mass (times $c^2$). And if you determine the the timing with a worse accuracy, then you will inevitably include almost the whole electric field of the electron, and you will get the electron mass (times $c^2$) of 511 keV with a big accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, you won't be able to show any violation of the total energy which holds exactly in any system with a time-translational symmetry. You will be extremely far from finding such a contradiction: even if you could attribute parts of energy to fields and particles at every moment, as you suggested and which you cannot because of the uncertainty principle, you couldn't measure it with the same precision, and even if you could measure it, the variable mass of the electron at the &quot;center&quot; of the field could always compensate any violation of the conservation law, anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the really important message is that the total energy is conserved, it can be measured as long as we have a long enough time to measure it, and attributing energy to small pieces of space around a particle is not the right way to proceed - neither theoretically nor experimentally. Instead, one should ask what are the probabilities of various processes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;External and internal particles will always show that energy is conserved. However, energy can only be measured accurately if it is measured for a long enough time - longer than $\hbar/\Delta E$ where $\Delta E$ is the required precision. If you think about these matters, you will find out that you are trying to measure the amount and localization of energy - in space and/or in time - more accurately than the uncertainty principle allows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-19T11:14:58.447" CommentCount="1" />
  <row Id="3322" PostTypeId="2" ParentId="3059" CreationDate="2011-01-19T11:32:53.040" Score="3" ViewCount="" Body="&lt;p&gt;Since symmetry breaking has much wider applications, than just in QFT. It is investigated also by mathematcal physicists. I recommend the recent book called &quot;&lt;a href=&quot;http://books.google.com/books?id=tPOcCWjvUK8C&amp;amp;printsec=frontcover&amp;amp;hl=en&quot; rel=&quot;nofollow&quot;&gt;Symmetry breaking&lt;/a&gt;&quot; by Strocchi for mathematical treatment of the subject. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-01-19T11:32:53.040" />
  <row Id="3323" PostTypeId="2" ParentId="3318" CreationDate="2011-01-19T11:42:59.253" Score="2" ViewCount="" Body="&lt;p&gt;with some extra background knowledge, it is actually not a real handwaving but the full answer to your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have omitted the previous sentence which says &quot;In the statistical interpretation of scalar field theory, $\rho_m$ is just the parameter one must adjust finely to bring the system to the critical temperature.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this moment, the reader is probably supposed to know Landau's phenomenological theory of the second-order phase transitions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Landau_theory&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Landau_theory&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, there is also the fourth-order term (in the order parameter $\Psi$ which is interpreted as the scalar field $\phi$ in QFT) but the coefficient $r$ of the quadratic term - which gets interpreted as the mass term in QFT - is proportional to $(T-T_c)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it so? Because the action, or free energy, may be viewed as a result of a calculation at temperature $T$. The coefficients in the action - or free energy will be smooth around the critical temperature because nothing special happens at the level of the &quot;action&quot; at this point. So $dr/dT$ is finite. The phase transition - and all the complicated and sometimes fractional powers around the critical point - only arise once you try to find solutions to the system defined by the action - or free energy. The &quot;phase&quot; - whether the spins like to create domains etc. - is not &quot;directly&quot; encoded in the coefficients; the theory is always the same. Its solutions may be qualitatively different for different values of parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's what happens generically. Of course that you could try to imagine a system in which $dr/dT$ is either zero or infinite (different power laws) near the critical temperature $T_C$. But you would need some extra fine-tuning of the material or theory to obtain $dr/dT=0$, and even if you succeeded, such a behavior would be a higher-order phase transition. I don't think it's possible to have $dr/dT$ being infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-19T12:01:15.003" LastActivityDate="2011-01-19T12:01:15.003" />
  <row Id="3324" PostTypeId="1" CreationDate="2011-01-19T12:12:56.077" Score="2" ViewCount="333" Body="&lt;p&gt;Have you noticed that many questions and misconceptions in physics arise due to misuse of analogies, which were invented to &quot;explain in simple words&quot; some physical phenomena? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example there is a &quot;stretching rubber band&quot; analogy for Hubble expansion of the Universe,  which make people ask about the &quot;tension&quot; of this rubber band or speculating about &quot;cutting a hole&quot; in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure that there are a lot of such misuses. Do you have any remarkable examples of these (together with a proper treatment n of course)?&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-08-22T11:07:26.787" CommunityOwnedDate="2011-01-19T12:52:02.850" Title="Misused physics analogies" Tags="&lt;soft-question&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="3325" PostTypeId="1" CreationDate="2011-01-19T12:23:16.150" Score="2" ViewCount="189" Body="&lt;p&gt;When doing the classic double slit experiment is the time between emitting photons significant at all? Say, a single photon is emitted, the scientist waits T seconds, then emits another photon. Are there any values for T which do not cause interference patterns on the detector?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about if the slits are closed during the time between photons?&lt;/p&gt;&#xA;" OwnerUserId="526" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-03-13T20:16:47.820" LastActivityDate="2011-03-13T20:16:47.820" Title="Is Time Significant in the Double Slit Experiment" Tags="&lt;quantum-mechanics&gt;&lt;photons&gt;&lt;time&gt;&lt;double-slit-experiment&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="3326" PostTypeId="2" ParentId="3324" CreationDate="2011-01-19T12:40:34.247" Score="3" ViewCount="" Body="&lt;p&gt;The &quot;solar system&quot; analogy of the atomic nucleus and electrons is certainly a very misleading one, in many ways. Unfortunately, this is all too often the model taught until late high school or early undergraduate level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Bohr model, or more usually the fuller quantum mechanical description, is of course required to do any useful physics. The shell model can sometimes be sufficient for some chemistry, though it is often not justified properly.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2011-01-19T12:40:34.247" CommunityOwnedDate="2011-01-19T12:52:02.850" />
  <row Id="3327" PostTypeId="1" AcceptedAnswerId="3341" CreationDate="2011-01-19T13:06:27.220" Score="4" ViewCount="280" Body="&lt;p&gt;Very often when viewing pictures of the cosmos taken by telescopes, one can observe that larger/brighter stars do not appear precisely as points/circles on the image. Indeed, the brighter the light from star, the more we see this effect of four perpendicular rays &quot;shooting&quot; out from the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/LAA8M.jpg&quot; alt=&quot;Stars1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Taken from &lt;a href=&quot;http://hubblesite.org/gallery/album/pr2002010c&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt;.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: what are the optics responsible for this effect? I would suppose that both the hazy glow around the stars and the rays shooting outwards are optical effects created by the camera/imaging device, but really can't surmise any more than that. (The fact that the rays are all aligned supports this, for one.) A proper justification, in terms of (geometrical?) optics for both the glow and the rays would be much appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a few other examples of such images:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://grizzlyhugs.files.wordpress.com/2009/12/stars-sky-lg.jpg&quot; rel=&quot;nofollow&quot;&gt;http://grizzlyhugs.files.wordpress.com/2009/12/stars-sky-lg.jpg&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://spacespin.org/images/articles/hubble-large-small-stars-harmonious_3.jpg&quot; rel=&quot;nofollow&quot;&gt;http://spacespin.org/images/articles/hubble-large-small-stars-harmonious_3.jpg&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-30T09:18:29.983" LastActivityDate="2011-07-09T10:30:30.923" Title="Optical explanation of images of stars?" Tags="&lt;optics&gt;&lt;astronomy&gt;&lt;waves&gt;&lt;imaging&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="3328" PostTypeId="2" ParentId="3325" CreationDate="2011-01-19T13:07:17.033" Score="2" ViewCount="" Body="&lt;p&gt;No, the interaction between the photon source and the screen does not depend on anything in the future. There is no retrocausality in quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you only detect one &quot;position&quot; (in quotes because it depends on the resolution of your screen) at a time. But when you detect many events, even if they are ages apart, the interference pattern will build up.&lt;/p&gt;&#xA;" OwnerUserId="1337" LastActivityDate="2011-01-19T13:07:17.033" />
  <row Id="3330" PostTypeId="2" ParentId="3327" CreationDate="2011-01-19T13:14:34.740" Score="3" ViewCount="" Body="&lt;p&gt;a great question. These crosses around the stars are called &quot;diffraction spikes&quot;. They emerge because of wave properties of light - interference around rods that have to be inserted into the reflective telescope. See e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Diffraction_spike&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Diffraction_spike&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://apod.nasa.gov/apod/ap010415.html&quot; rel=&quot;nofollow&quot;&gt;http://apod.nasa.gov/apod/ap010415.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the best&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-19T13:14:34.740" CommentCount="2" />
  <row Id="3333" PostTypeId="1" AcceptedAnswerId="3336" CreationDate="2011-01-19T13:50:41.277" Score="3" ViewCount="109" Body="&lt;p&gt;We know that Spin is described with SU(2) and that SU(2) is a double cover of the rotation group SO(3). This suggests a simple thought experiment, to be described below. The question then is in three parts: (a) is this thought experiment theoretically sound?; (b) Can it be conducted experimentally?; (c) if so what has been the result?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experiment is to take a slab of material in which there are spin objects e.g. electrons all (or most) with spin UP. Then rotate that object 360 degrees (around an axis perpendicular to the spin direction), so that macroscopically we are back to where we started. Measure the electron spins. Do they point DOWN?&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:38:48.183" LastActivityDate="2011-02-04T07:38:48.183" Title="Basic Spin or Double Cover Experiment" Tags="&lt;experimental-physics&gt;&lt;spin&gt;&lt;fermions&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="3334" PostTypeId="1" AcceptedAnswerId="3340" CreationDate="2011-01-19T13:53:45.647" Score="4" ViewCount="301" Body="&lt;p&gt;I'm thinking equivalence principle, possibilities of unbounded space-time curvature, quantum gravity ...&lt;/p&gt;&#xA;" OwnerUserId="1394" LastEditorUserId="1394" LastEditorDisplayName="" LastEditDate="2011-01-19T18:41:01.543" LastActivityDate="2011-01-19T18:41:01.543" Title="Is there a maximum possible acceleration?" Tags="&lt;quantum-mechanics&gt;&lt;general-relativity&gt;&lt;quantum-gravity&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="3335" PostTypeId="2" ParentId="3306" CreationDate="2011-01-19T13:56:31.180" Score="1" ViewCount="" Body="&lt;p&gt;Since the wires are very long and very thin, we take them to be lines. The force between them is then proportional to the current, which is $$I = V/R.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like Georg mentioned, it is a trick question. The force will disappear only if there is no current, which means $R=\infty$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; On the other hand, Georg reminded me that the wires could have residual charges that would cause attraction and thus counteract the repulsion from a positive current. But in order to quantify that we would need more information.&lt;/p&gt;&#xA;" OwnerUserId="724" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2011-01-19T14:38:45.887" LastActivityDate="2011-01-19T14:38:45.887" CommentCount="9" />
  <row Id="3336" PostTypeId="2" ParentId="3333" CreationDate="2011-01-19T14:01:12.293" Score="4" ViewCount="" Body="&lt;p&gt;I think that you are confused. When you rotate something by 360 degrees, you won't change the direction in space of anything. You will only change the wave function to minus itself - if there is an odd number of fermions in the object (which is usually hard to count for large objects).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have electrons with spins pointing up and you rotate them around the vertical axis by any angle, whether it's 360 degrees or anything else, you will still get electrons with spin pointing up. This is about common sense - many spins with spin up give you a totally normal, &quot;classical&quot; angular momentum that can be seen and measured in many ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flip of the sign of the wave function can't be observed by itself because it is a change of phase and all observable probabilities only depend on the density matrix $\rho=|\psi\rangle \langle\psi|$ in which the phase (or minus sign) cancels. The phase - or minus sign - has nothing to do with directions in space. It is just a number. In particular, it is incorrect to imagine that complex numbers are &quot;vectors&quot;, especially if it leads you to think that they're related to directions in spacetime. They're not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would have to prepare an interference experiment of an object that hasn't rotated with the &quot;same&quot; object rotated by 360 degrees - and it's hard for macroscopic objects because the &quot;same&quot; object quickly decoheres and you must know whether it has rotated or not, so no superpositions can be produced. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, all detailed measurements of the spin with respect to any axis indirectly prove that the fermions transform as the fundamental representation of $SU(2)$. In particular, if you create a spin-up electron and measure whether it's spin is up with respect to another axis tilted by angle $\alpha$, the probability will be $\cos^2(\alpha/2)$. The only sensible way to obtain it from the amplitude is that the amplitude goes like $\cos(\alpha/2)$ and indeed, this function equals $-1$ for $\alpha$ equal to 360 degrees.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:38:20.633" LastActivityDate="2011-02-04T07:38:20.633" />
  <row Id="3337" PostTypeId="2" ParentId="1683" CreationDate="2011-01-19T14:05:01.477" Score="0" ViewCount="" Body="&lt;p&gt;With a vertical jet, Torricelli's law still holds because the displacement of the wagon is orthogonal to the acting forces, gravity plus (arguably, but orthogonal in any case) reaction force, so no work is used by the wagon, $\Delta W = {\bf F} \cdot {\Delta \bf x}=0$  and all the energy still goes to the water jet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we can calculate $m(t)$ as usual. Forget the drawing and use a square tank. The one in the drawing was calculated by Kepler, and it complicates the problem. Let the height of the water to be simply $h(t)={m(t)\over \rho S}$, ok? And $2 g h(t)$ is the square speed of the jet, the variation of mass follows $ m&amp;#39;(t)= - \rho s \sqrt { 2 g h(t)}$, and at the end we have &#xA;$$ m&amp;#39;(t) = - \sqrt {2 g \rho s^2 \over   S} \sqrt{ m(t)} $$&#xA;Which solves to $m(t)= m (1 - t \sqrt {g \rho s^2 \over 2 m S})^2 $ and tell us that the tank becomes empty at $t_f=\sqrt {2 m S \over g \rho s^2 }=\sqrt {2 h S^2  \over g s^2 }$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can plug this into &lt;a href=&quot;http://physics.stackexchange.com/questions/1683/mechanics-around-a-rail-tank-wagon/1689#1689&quot;&gt;Frederik &quot;&lt;strong&gt;wrong&lt;/strong&gt;&quot; solution&lt;/a&gt; $x(t)= l \ln {m+M \over m(t)+M}$ t o get the displacement &#xA;$$ x(t) = l \ln {m+M \over (1 - t/t_F)^2 m +M}$$&#xA;and the velocity&#xA;$$ \dot x(t)=  { 2 m l \over t_F} { (1-t/t_F) \over (1 - t/t_F)^2 m +M }&#xD;&#xA;= { 2 l (t_F-t) \over (t_F - t)^2 + {Mt_F^2 \over m} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in the limit of $M \ll m$, we get&#xA;$ \dot x(t)=  { 2 l \over (t_F - t)  } $ and thus $ \ddot x =  { 2 l \over (t_F - t)^2  }$, similar to other answers. Note that in this limit the speed at $t_F$ is infinite, but it is massless, so we can stop it anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another curious issue is that&#xA;$ \dot x(0) = { 2 l \over t_f (1 + M/m)} $&#xA;is not zero. It sounds strange, but consider that the initial speed of the jet is not zero neither.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Before to consider variants of Frederik solution, it is important to note that we have four blobs of mass playing some role.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the leaked water, $m-m(t)$&lt;/li&gt;&#xA;&lt;li&gt;the leaking water, $\Delta m= - m&amp;#39;(t) \Delta t$&lt;/li&gt;&#xA;&lt;li&gt;the cart mass, M&lt;/li&gt;&#xA;&lt;li&gt;the wagon water, m(t)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the leaking process, the leaked water is already inertial, will a horizontal momentum (in the railway direction) equal and opposite to the momentum of the other three masses, Or, for small $\Delta t$, equal to  $- (M + m(t)) V_{c+w}$. The questions to be fixed are: 1) which is the actual direction of the force by the water and the leaking water horizontal velocity: the velocity of the cart, the one of the CM of the water, or some other one? and 2) Does the acceleration of the cart changes enough the direction of &quot;gravity&quot; inside the cart (remember your last bus trip) to be considered a major perturbation of the problem? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Point 2 is most probably a red herring, at least in the approximation where $M \ll m(t)$, because in such case we don't have reasons to expect the accelerations of the cart and the [CM of the] water inside to be different. Remember that the &quot;horizontal gravity&quot; inside the wagon will be the difference of these accelerations. &lt;/p&gt;&#xA;" OwnerUserId="1335" LastEditorUserId="1335" LastEditorDisplayName="" LastEditDate="2011-01-24T12:15:45.640" LastActivityDate="2011-01-24T12:15:45.640" CommentCount="6" />
  <row Id="3338" PostTypeId="2" ParentId="3302" CreationDate="2011-01-19T14:06:36.610" Score="3" ViewCount="" Body="&lt;p&gt;&lt;em&gt;(the ideal shape for solar cells)&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, that is quite a jump, from graphene lattice to about some inches diameter &#xA;of a solar cell. In some German dialect, we would say: &#xA;&quot;von Kuchenbacken auf Arschbacken&quot;. This connection, as Lubos stated, is pure &#xA;nonsense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to solar cells: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;when such cells are produced from monocrystalline &#xA;silicon, the dies are circular disks. On the other hand, one needs cells &#xA;who fill the area on a roof. That is geometry, called &quot;tiles&quot;. You could use &#xA;triangles, squares (rectangles) and hexagons to &quot;pave&quot; a plane. Obviously the &#xA;hexagon is the figure to cut from the circular die with the least loss. &#xA;That is all, and to find some connection to the graphene structure You  need the brain of a &quot;artist&quot; in the ad department. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Georg&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-01-19T15:53:23.683" LastActivityDate="2011-01-19T15:53:23.683" />
  <row Id="3339" PostTypeId="2" ParentId="3333" CreationDate="2011-01-19T14:06:49.167" Score="3" ViewCount="" Body="&lt;p&gt;No that is not how it works. A 360 rotation multiplies the wave function by a factor -1 which by itself is not observable. It does not switch up and down spins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An experiment which demonstrated the effect would have to involve forming interference patterns between streams of electrons where one stream was being rotated through 360 degrees (e.g. using magnetic fields) I don't know if anything like it has been done.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-19T14:06:49.167" />
  <row Id="3340" PostTypeId="2" ParentId="3334" CreationDate="2011-01-19T14:19:32.693" Score="1" ViewCount="" Body="&lt;p&gt;The spit horizon in a Rindler wedge occurs at a distance $d~=~c^2/g$ for the acceleration $g$.  In spatial coordinates this particle horizon occurs at the distance $d$ behind the accelerated frame.  Clearly if $d~=~0$ the acceleration is infinite, or better put indefinite or divergent.  However, we can think of this as approximating the near horizon frame of an accelerated observer above a black hole.  The closest one can get without hitting the horizon is within a Planck unit of length.  So the acceleration required for $d~=~\ell_p$ $=~\sqrt{G\hbar/c^2}$ is $g~=~c^2/\ell_p$ which gives $g~=~5.6\times 10^{53}cm/s^2$.  That is absolutely enormous.  The general rule is that Unruh radiation has about $1K$ for each $10^{21}cm/s^2$ of acceleration.  So this accelerated frame would detect an Unruh radiation at $\sim~10^{31}K$.  This is about an order of magnitude larger than the Hagedorn temperature.  We should then use the string length instead of the Planck length $4\pi\sqrt{\alpha’}$ and the maximum acceleration will correspond to the Hagedorn temperature.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-19T14:19:32.693" CommentCount="6" />
  <row Id="3341" PostTypeId="2" ParentId="3327" CreationDate="2011-01-19T14:20:31.813" Score="8" ViewCount="" Body="&lt;p&gt;This is, as Lubos mentioned, an effect of the wave nature of light, and cannot be explained using geometrical optics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are seeing is called the Point Spread Function (PSF) of the imaging system. Because stars are so far away that they are effectively point sources of light (i.e. they are spatially coherent) their image will be the PSF of the imaging system. Up to a scale factor, the PSF is the Fourier transform of the pupil of the imaging system. For a lens system, the pupil is usually just a circle, so the PSF is the 2D Fourier transform of a circle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{J_1(2 \pi \rho)} {2 \pi \rho} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $J_1$ is the order 1 bessel function of the first kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, most modern telescopes are built with reflective optics, and there are various obscurations in the pupil due to the structures that support the secondary mirror. This more complicated pupil shape can produce a variety of artifacts in the PSF. The starburst pattern in your example images could be due to a simple &quot;plus&quot; shaped structure supporting the secondary mirror, but the effect is so strong that I suspect it was emphasized for creative effect. I'm not sure how the Hubble PSF looks, off the top of my head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, an image can be represented by the convolution of the ideal image $g(x,y)$ with the PSF, usually denoted $h(x,y)$. In the case of a point source (so $g(x,y)$ is a delta function, $\delta(x,y)$) it is trivial that the image is a copy of the PSF:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h(x,y)=\int_{-\infty}^{\infty} \delta(\xi,\eta) h(x-\xi, y-\eta) d\xi d\eta $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the case of a more complicated object, the convolution by the PSF acts to smooth or blur the image. This is why an out of focus camera produces blurry images. Although aberrations also degrade the image under a geometrical approximation, this is more accurate. The geometrical case and the wave optics (diffraction) result will become closer as the aberrations become large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes this effect is produced intentionally. You can actually buy filters for commercial cameras that have a fine grid of wires to produce this starburst effect for creative purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB: This answer ignores any discussion of phase effects in diffraction (because I'm short on time, I may update later). If you would like to learn about diffraction and the wave optics approach to imaging, the leading text on the planet is &quot;Introduction to Fourier Optics&quot; by J. Goodman. It is an absolutely spectacular book.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-01-19T14:20:31.813" CommentCount="8" />
  <row Id="3342" PostTypeId="1" CreationDate="2011-01-19T14:25:32.630" Score="4" ViewCount="208" Body="&lt;p&gt;It's known that gauge group for type-I superstring theory should be SO(32) for chiral anomalies to cancel. This is realized by presense of 16 space-time filling D9-branes, together with 16 space-time filling O9-planes with opposite charge: D9-branes bring U(16) gauge symmetry and due to orientifold planes the symmetry gets enhanced to SO(32). But what about other Dp-branes which can exist in type-I theory: wouldn't they bring anomaly - because open strings can end on those branes and create gauginos from gauge group different from required SO(32) - and will that anomaly cancel eventually (people make physical constrctions with other Dp-branes e.g. while studying BH in superstring theory)? Or in any picture we are always supposed to be able to perform enough T-duality transformations and get our 16 D9-branes and 16 O9-planes, but there will be additional branes. Probably the answer to my question is that all other possible branes in that picture will be part of 16 space-time filling branes, and therefore gauge YM group will be then SO(32) anyway and other gauge groups will be just subgroups of it - hence in the whole system anomalies will cancel, is it true?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-19T19:32:30.773" Title="Space-time filling D-branes in type-I superstring theory" Tags="&lt;string-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="3343" PostTypeId="1" AcceptedAnswerId="3347" CreationDate="2011-01-19T14:31:04.170" Score="11" ViewCount="677" Body="&lt;p&gt;One often hears the words &quot;string tension&quot; in string theory. But what does it really mean? In ordinary physics &quot;tension&quot; in an ordinary classical string arises from the fact that there are elasticity in the string material which is a consequence of the molecular interaction (which is  electromagnetic in nature). But string theory, being the most fundamental framework to ask questions about physics (as claimed by the string theorists) can not take such elasticity for granted from the start. So my question is, what does &quot;tension&quot; mean in the context of string theory? Perhaps this question is foolish but please don't ignore it.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-03T10:40:36.740" LastActivityDate="2011-02-08T07:23:50.323" Title="What is tension in string theory?" Tags="&lt;string-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="3344" PostTypeId="1" AcceptedAnswerId="4213" CreationDate="2011-01-19T14:31:27.210" Score="1" ViewCount="156" Body="&lt;p&gt;Eddy currents are induced in a metal plate when it experiences a changing magnetic flux. Is there a realistic visualization or simulation of eddy currents available? The only picture I found, on wikipedia, is a cartoon schematic. &lt;/p&gt;&#xA;" OwnerUserId="1462" LastActivityDate="2011-01-30T11:56:04.140" Title="Electrical eddy current visualization or simulation" Tags="&lt;electromagnetism&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="3345" PostTypeId="2" ParentId="3334" CreationDate="2011-01-19T14:45:28.073" Score="3" ViewCount="" Body="&lt;p&gt;a kind of important question. But according to quantum mechanics, you can't imagine that objects are moving along smooth, doubly differentiable trajectories - which you need to define acceleration. Instead, they're moving along all possible trajectories - I am using Feynman's path integral approach to quantum mechanics - and most of them are not differentiable even once. So the typical acceleration at a typical place of a trajectory in quantum mechanics is infinite. You could only study a &quot;finite bound&quot; on acceleration in classical physics and in classical (non-quantum) physics, there's no upper bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you may talk about the upper bounds on some &quot;correctly looking&quot; formulae for acceleration. For example, you may be able to &quot;derive&quot; that the maximum gravitational acceleration in quantum gravity is approximately equal to the Planck acceleration,&#xA;$$a_{Planck} = L_{Planck} / T_{Planck}^2 = \frac{\sqrt{\frac{\hbar G}{c^3}}}{\frac{\hbar G}{c^5}}=\sqrt{\frac{c^7}{\hbar G}} = 5.6 \times 10^{51}\,\mbox{m/s}^2$$&#xA;where the numerator and denominator depend on the Planck length and Planck time, respectively. Yes, it's huge. This upper bound holds because it's the acceleration on the surface (event horizon) of the smallest and most concentrated object. The most concentrated objects are black holes and the smallest black hole worth the name has a radius comparable to the Planck length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this bound only applies outside the black holes. Near the singularities inside the black hole, the accelerations may be formally larger. No one knows whether it makes sense to talk about the trans-Planckian accelerations. However, accelerations are not among the &quot;most fundamental quantities&quot; we use to describe physics according to its state-of-the-art theories, anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-19T14:45:28.073" CommentCount="4" />
  <row Id="3346" PostTypeId="1" AcceptedAnswerId="4002" CreationDate="2011-01-19T14:56:39.523" Score="3" ViewCount="127" Body="&lt;p&gt;So I know the answer to this question varies widely across mass spec techniques, masses, and, of course, budgets, but my question is about the &lt;em&gt;best&lt;/em&gt; case scenario for all of these variables (although I'm not really looking at proteins or anything of that size).  I realize that the concept of depletion spectroscopy relies heavily on very sensitive mass spectrometers, and I'm curious about the lower limits of signal detection for such methods.  Thus, I'm interested in the sensitivity, not resolution (&quot;mass resolving power&quot;), which seems to be a statistic that is difficult to find for most of the systems that I've looked at.  Some related follow-up questions are what mass spectrometry methods can and cannot be used with depletion spectroscopy and what methods/manufacturers are the gold standard in the field of mass spec?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for your answers.&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2011-01-27T15:08:07.757" Title="How small of a depletion signal can the best modern mass spectrometers detect?" Tags="&lt;spectroscopy&gt;&lt;mass-spectrometry&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="3347" PostTypeId="2" ParentId="3343" CreationDate="2011-01-19T15:03:23.480" Score="10" ViewCount="" Body="&lt;p&gt;A good question. The string tension actually &lt;em&gt;is&lt;/em&gt; a tension, so you may measure it in Newtons (SI units). Recall that 1 Newton is 1 Joule per meter, and indeed, the string tension is the energy per unit length of the string.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the string tension is not far from the Planck tension - one Planck energy per one Planck length or $10^{52}$ Newtons or so - it is enough to shrink the string almost immediately to the shortest possible distance whenever it is possible. Unlike the piano strings, strings in string theory have a variable proper length. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This minimum distance, as allowed by the uncertainty principle, is comparable to the Planck length or 100 times the Planck length which is still tiny (although models where it is much longer exist).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For such huge energies and velocities comparable to the speed of light, one needs to appreciate special relativity, including the $E=mc^2$ famous equation. This equation says that the string tension is also equal to the mass of a unit length of the string (times $c^2$). The string is amazingly heavy - something like $10^{35}$ kg per meter: I divided the previous figure $10^{52}$ by $10^{17}$ which is the squared speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Basic equations of perturbative string theory&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More abstractly, the string tension is the coefficient in the Nambu-Goto action for the string. What is it? Well, classical physics may be defined as Nature's effort to minimize the action $S$. For a particle in special relativity,&#xA;$$ S = -m\int d\tau_{proper} $$&#xA;i.e. the action is equal to (minus) the proper length of the world line in spacetime multiplied by the mass. Note that because Nature tries to minimize it, massive particles will move along geodesics (straightest lines) in general relativity. If you expand the action in the non-relativistic limit, you get $-m\Delta t+\int dt\, mv^2/2$, where the second term is the usual kinetic part of the action in mechanics. That's because the curved lines in the Minkowski space are shorter than the straight ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;String theory is analogously about the motion of 1-dimensional objects in the spacetime. They leave a history which looks like a 2-dimensional surface, the world sheet, which is analogous to the world line with an extra spatial dimension. The action is&#xA;$$ S_{NG} = -T\int d\tau d\sigma_{proper} $$&#xA;where the integral is supposed to represent the proper area of the world sheet in spacetime. The coefficient $T$ is the string tension. Note that it is like the previous mass (from the point-like particle case) per unit distance. It may also be interpreted as the action per unit area of the world sheet - it's the same as energy per unit length because energy is action per unit time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this moment, when you understand the Nambu-Goto action above, you may start to study textbooks of string theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Piano strings are made out of metallic atoms, unlike fundamental strings in string theory. But I would say that the most important difference is that the strings in string theory are allowed - and love - to change their proper length. However, in all the other features, piano strings and strings in string theory are much more analogous than the string theory beginners usually want to admit. In particular, the internal motion is described by equations that may be called the wave function, at least in some proper coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the strings in string theory are relativistic and on a large enough piece of world sheet, the internal SO(1,1) Lorentz symmetry is preserved. That's why a string carries not only an energy density $\rho$ but also a negative pressure $p=-\rho$ in the direction along the string.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-08T07:23:50.323" LastActivityDate="2011-02-08T07:23:50.323" CommentCount="4" />
  <row Id="3348" PostTypeId="2" ParentId="3306" CreationDate="2011-01-19T15:06:43.273" Score="1" ViewCount="" Body="&lt;p&gt;The magnetic force should be straightforward, you probably have a formula for it, or you can use the field strength from one wire, cross the current in the other to compute the force.&#xA;The charge is more difficult, there you have two oppositely charged cyllinders separated by a distance. You can need to figure out the voltage difference between them as a function of the charge denisty per unit length of the system. As a first approximation assume the charge is evenly distributed around the circumference of your wire, then you should be able to compute&#xA;the voltage difference between the two wires by integrating the electric field between them. I'm not going to do this (if it really is homework, you need to do it yourself). You should be able to use symmetry to reduce the amount of detailed math you need.   If you are really ambitious, you could calculate the effect of uneven charge distribution on each wire, i.e. there will be more charge on the surface closer to the other wire than on the outside -or at least show the approximate size of the correction factor. Should be a fun exercise.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-01-19T15:06:43.273" CommentCount="5" />
  <row Id="3349" PostTypeId="1" AcceptedAnswerId="3351" CreationDate="2011-01-19T15:23:25.153" Score="7" ViewCount="306" Body="&lt;p&gt;I know that there's a lot of evidence for extremely compact bodies. But is there any observation from which we can infer the existence of an actual &lt;em&gt;horizon&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if we are able to someday resolve directly the &quot;shadow&quot; of a possible black hole, isn't it still possible that it's just a highly compact body that's just gravitationally redshifting any radiation that it emits to beyond the detection capability of our telescopes? What would be indisputable evidence for a causal horizon?&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-03-14T06:36:29.757" Title="Evidence for black hole event horizons" Tags="&lt;astrophysics&gt;&lt;gravity&gt;&lt;black-holes&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="3350" PostTypeId="2" ParentId="3342" CreationDate="2011-01-19T15:41:55.693" Score="1" ViewCount="" Body="&lt;p&gt;Dear Mikhail,&#xA;good interest of yours. Some points that will hopefully store all the answers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Type I string theory is defined by having its orientifold O9-plane. It only has one such a plane; there is no sense in which there are 16 of them (you only look into one mirror; there are no open strings ending on orientifold planes, so no degeneracy may be introduced, either). Type I may be defined as type IIB with the spacetime-filling O9-plane - which makes all strings unorientable. And the 16 spacetime-filling D9-branes and their images have to be added to cancel the anomalies - either in spacetime, or in worldsheet.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In spacetime, the relevant anomaly for 9+1 dimensions is given by one-loop &quot;hexagon&quot; graph with gauginos and gravitinos and other chiral fermions running in the loop. That wouldn't cancel but in 1984, Green and Schwarz began the first superstring revolution when they showed that there is an extra tree-level graph, with a group of 2 and another group of 4 external lines attached to an internal propagator, and all the anomalies cancel for $SO(32)$ or $E_8 \times E_8$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The supersymmetric D-branes in type I string theory include D9-branes, D5-branes, and D1-branes. In type IIB string theory where we began, there were also D7-branes, D3-branes, and D(-1)-branes (D-instantons), but those three types were removed from the supersymmetric spectrum by the orientifold projection (or at least became non-supersymmetric and unstable). D1-branes become heterotic strings in the S-dual heterotic $SO(32)$ string theory; D5-branes become the heterotic fivebranes in the same strong-coupling limit.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The D1-branes of type I string theory - the heterotic strings - also have potential world volume anomalies and they do cancel because of the 1-9 strings (whose massless states are chiral on the D1-brane), open strings connecting the D1-branes with the 16 spacetime-filling D9-branes. This cancellation also works but isn't equivalent to the cancellation of the spacetime anomalies in type I - or the world sheet anomalies in type I, for that matter.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Similarly, the D5-branes become the heterotic fivebranes in the limit, so that's enough to see that they carry a $(1,0)$ supersymmetry in six dimensions. Because this asymmetric counting guarantees that the theory is left-right asymmetric, one must also check the anomalies and indeed, they still cancel. But this cancellation is again non-equivalent to the cancellations for D9-branes and D1-branes described above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Type I string theory may be T-dualized but what we get is not type I. In particular, type I is type IIB with an extra orientifold. So just like in any type IIB, the T-duality with respect to an odd number of circles gives you type IIA, and only if the number of circles is even, you get type IIB back. The T-duality with respect to $N$ circles transforms the O9-plane and D9-branes to O$(9-N)$-plane(s) and D$(9-N)$-branes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At the beginning, I said that there is &quot;one&quot; spacetime-filling O9-plane in type I. However, if you T-dualize $N$ dimensions as in the previous point, you will see that there are $2^N$ different loci with O$(9-N)$-planes. It's because if you have a periodic coordinate $\phi$ and the orientifold acts by flipping the sign of $\phi$, there are 2 fixed points on the circle of length $2\pi$, namely $\phi=0$ and $\phi=\pi$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If one T-dualizes D-branes, he has to deal with Wilson lines. Each $U(1)$ wrapped on a circle has a Wilson line around the circle which gets translated to the position of the dual (lower-dimensional) D-brane on the T-dual circle.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The T-dual theories to  type I are called type IA (also type I', read &quot;one prime&quot;) and perhaps type IAA etc., depending on the number of T-dualities. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The lower-dimensional D-branes arising by T-dualizing the original D9-branes may have different positions. The dilaton sources are only eliminated locally - and the dilaton stays constant across the spacetime - if we uniformly distribute the 16 D-branes and their images to the $2^N$ different loci of the orientifold O$(9-N)$-planes. For type IA theory, we deal with 2 loci of O8-planes and there are 8 (plus 8 images) D8-branes at each locus. So the gauge group is SO(16) at each of them. Similarly, for the doubly T-dualized type I, or IAA, one has 4 places with O7-planes and each of them carries an SO(8) group from the D7-branes. That division corresponds to the Wilson lines of the original type I theory whose $U(1)$ parts are maximally equally divided between trivial Wilson lines and $(-1)$ from the $U(1)$ groups.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are no anomalies localized on the D$(9-N)$ planes in the T-duals of type I. In particular, the D8-branes and other even-dimensional branes - in an orientifold of type IIA - are non-chiral. In particular, D-branes that are moved away from the orientifold planes become ordinary type II D-branes which are always non-chiral but carry no anomalies. Note that the bulk physics of type I differs from type IIB because it only has one gravitino, among other things; however, type IA, IAA, IAAA, IAAAA etc. are equivalent to type IIA, IIB, IIA, IIB... almost everywhere in the bulk and the gravitinos only miss some states exactly at the orientifold planes&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The T-duals for other choices of the Wilson lines - e.g. the $SO(32)$ point of type I on a circle itself - distribute the D9-branes asymmetrically and the dilaton is inevitably nonconstant in spacetime, increasing (or decreasing?) from the orientifold planes with a deficit of D-branes to those with an excess of D-branes (or to the D-branes themselves); this dilaton profile has to be taken into account when you talk about the anomalies; of course, localized D-branes in type II string theory have no anomalies, and neither have the O-planes, but the properly variable dilaton around all these objects - depending on the charges - plays the role in cancelling the anomaly; in some sense, the original anomaly in type I gets divided between the different places where the T-dual O-planes and D-branes are localized&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-19T15:49:45.433" LastActivityDate="2011-01-19T15:49:45.433" />
  <row Id="3351" PostTypeId="2" ParentId="3349" CreationDate="2011-01-19T15:55:36.847" Score="6" ViewCount="" Body="&lt;p&gt;At the galactic center, there is an object called Sagittarius A* which seems to be a black hole with 4 million solar masses. In 1998, a wise instructor at Rutgers made me make a presentation of this paper&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/astro-ph/9706112&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/astro-ph/9706112&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by Narayan et al. that presented a successful 2-temperature plasma model for the region surrounding the object. The paper has over 300 citations today. The convincing agreement of the model with the X-ray observations is a strong piece of evidence that Sgr A* is a black hole with an event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, even if you neglect the predictions for the X-rays, the object has an enormously low luminosity for its tremendously high accretion rate. The advecting energy is pretty &quot;visibly&quot; disappearing from sight. If the object had a surface, the surface would heat up and emit a thermal radiation - at a radiative efficiency of 10 percent or so which is pretty canonical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you may be dissatisfied by their observation of the event horizon as a &quot;deficit of something&quot;. You may prefer an &quot;excess&quot;. However, the very point of the black hole is that it eats a lot but gives up very little, so it's sensible to expect that the observations of black holes will be via deficits. ;-)&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-14T06:36:29.757" LastActivityDate="2011-03-14T06:36:29.757" CommentCount="3" />
  <row Id="3352" PostTypeId="1" AcceptedAnswerId="3353" CreationDate="2011-01-19T16:11:26.110" Score="7" ViewCount="330" Body="&lt;p&gt;Graphene has a honeycomb lattice (in the absence of defects and impurities). By considering the low-energy limit of the half-filled Hubbard model used to model the strongly interacting electron gas we find that the low-energy quasiparticles obey the dispersion relation for massless fermions. These details are all covered very nicely in a paper by Gonzalez, Guniea and Vozmediano &lt;a href=&quot;http://arxiv.org/abs/cond-mat/9208004&quot; rel=&quot;nofollow&quot;&gt;(reference)&lt;/a&gt; among others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might seem like I'm answering the question. I'm following this line of exposition because I don't want to assume that this is a topic something commonly known or understood outside the condensed matter community. Any answers which elaborated on these basics would be very useful as they would help make the discussion more broadly accessible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My &lt;em&gt;primary&lt;/em&gt; question is more about the implications this fact has for high-energy physics, in particular the question of emergent-matter in theories of quantum gravity. The case of graphene is a canonical example in that regard where one obtains &lt;em&gt;relativistic&lt;/em&gt;, &lt;em&gt;massless&lt;/em&gt; excitations in the low-energy corner of an otherwise non-relativistic system - the 2D electron gas (2DEG).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously I have my own beliefs in this regard and I will try to outline them in an answer. But I also want to solicit the communities views in this regard.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-20T09:19:43.053" Title="How do Dirac fermions arise in graphene, and, what significance (if any) does this have for high-energy physics?" Tags="&lt;condensed-matter&gt;&lt;graphene&gt;&lt;fermions&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="3353" PostTypeId="2" ParentId="3352" CreationDate="2011-01-19T16:21:45.710" Score="2" ViewCount="" Body="&lt;p&gt;The answer you'll get from most high-energy physicists is that there are no implications whatsoever. Lorentz invariance is extraordinarily well-tested: see, e.g., &lt;a href=&quot;http://arxiv.org/abs/0801.0287&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0801.0287&lt;/a&gt;. In particular, there are many relevant operators in the Standard Model that one would expect to be generated if physics at a high scale is not Lorentz-invariant. Even some irrelevant operators that one might naively expect to appear with order-one Planck-suppressed coefficients are constrained to have smaller coefficients. Adding in gravity only makes the problem worse. For instance, most attempts to generate emergent GR from nonrelativistic theories will have an extra scalar mode and run into massive phenomenological difficulties, because they aren't really gauging the full diffeomorphism group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be slightly more clear: there are cases (and the free relativistic fermion emerging in the long-distance limit of graphene is one of them) where lattice symmetries can forbid dangerous relevant operators. This shouldn't happen for the full Standard Model (I assume someone has written down a careful argument for this somewhere, but I don't know a reference offhand). Still, even for the graphene case there are irrelevant operators, and we have bounds on those too. Furthermore, once you start thinking about gravity you're more or less forced to give up the hope of an underlying highly symmetric lattice that forbids all the dangerous operators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more half-joking comment: this argument also tells you the correct answer to the &lt;a href=&quot;http://www.fqxi.org/community/essay&quot; rel=&quot;nofollow&quot;&gt;FQXi essay contest &quot;Is Reality Digital or Analog?,&quot;&lt;/a&gt; so if someone fleshes it out carefully they could possibly make up to $10k from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It is a good question, by the way; there's an obvious conventional wisdom from effective field theory that explains why you don't see high-energy theorists pursuing this sort of thing much, but from the outside it might not be so clear why such ideas don't generate much interest.)&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="19" LastEditorDisplayName="" LastEditDate="2011-01-19T18:48:48.707" LastActivityDate="2011-01-19T18:48:48.707" CommentCount="6" />
  <row Id="3354" PostTypeId="2" ParentId="3352" CreationDate="2011-01-19T16:28:21.457" Score="6" ViewCount="" Body="&lt;p&gt;The emergence of global symmetry at low energies is a familiar phenomena, for example Baryon number emerges in the context of the standard model as &quot;accidental&quot; symmetry. Meaning at low energies it is approximately valid, but at high energies it is not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this is the case is that it so happens that the lowest dimension operator you can write, with the matter content and symmetries of the standard model, is dimension 5. The effect is then suppressed by one power of some high energy scale - it is an irrelevant operator. This is a model independent way to characterize the possibility of the emergence of global symmetries at low energies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can then ask about Lorentz invariance - what are the possible violations of Lorentz invariance at low energies, and what is the dimensions of the corresponding operators. This depends on the matter content and symmetries - for the system describing graphene, there is such emergence. For anything containing the matter content of the standard model, there are lots and lots of relevant operators*, whose effect is enhanced at low energies - meaning that Lorentz violating effects, even small ones at high energies, get magnified as opposed to suppressed at observable energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, once we include gravity Lorentz invariance is now a gauge symmetry, which makes its violation not just phenomenologically unpleasant, but also theoretically unsound. It will lead to all the inconsistencies which necessitates the introduction of gauge freedom to start with, negative norm states and violations of unitarity etc. etc.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;At least 46, which were written down by Coleman and Glashow (Phys.Rev. D59, 116008). Relaxing their assumptions you can find even more. Each one of them would correspond to a new fine-tuning problem (like the cosmological constant problem, or the hierarchy problem).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-01-20T00:41:11.570" LastActivityDate="2011-01-20T00:41:11.570" />
  <row Id="3355" PostTypeId="1" CreationDate="2011-01-19T17:25:35.163" Score="4" ViewCount="440" Body="&lt;p&gt;One and the same electron in a p orbital and taking part in a common π (pi) bond has two lobes visualized as connecting through the nucleus. There is however zero probability of finding an electron at the plane through the nucleus at right angles to the lobes.&#xA;Where has the electron gone? Is the electron tunneling through the nucleus as a virtual photon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ed./ Without altering the meaning of the original question it is more clearly stated as: How does the electron cross the nucleus.&lt;/p&gt;&#xA;" OwnerUserId="1391" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-30T09:19:53.960" LastActivityDate="2011-01-31T01:03:34.673" Title="Zero probability of finding an electron in the nucleus" Tags="&lt;quantum-mechanics&gt;&lt;atoms&gt;&lt;schrodinger-equation&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="3356" PostTypeId="2" ParentId="3355" CreationDate="2011-01-19T17:45:40.657" Score="5" ViewCount="" Body="&lt;p&gt;the wave function describing an electron in a p-state ($L=1$) with $m=0$ indeed vanishes in the $z=0$ plane because the spherical harmonic $Y_{10}$ is proportional to $\cos(\theta)$ which vanishes in that plane - even at the very origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This wave function's squared absolute value describes the probability density that the electron is at a given point. This is true at any moment and it completely answers the question &quot;where the electron has gone&quot;. Because the state is stationary, it has gone to the very place where it has always been - in the p-state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could think that the electron is jumping in between the positive $z$ and negative $z$ half-clouds. And indeed, Feynman's path integral approach to quantum mechanics tells you that you must sum over all possible trajectories to get the transition amplitudes. They will always include trajectories that go in between the half-clouds - or anywhere else in the Universe. It's just true that most of them will nearly cancel. And it's true that the $l=1$, $m=0$ wave function vanishes at the $z=0$ plane. There is no contradiction here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the (wrong) Bohmian theory, the (fictitiously real point-like) electron is influenced by the &quot;quantum potential&quot; that repels it from the places where $\psi=0$. So in the p-state, the Bohmian electron would be repelled by the $z=0$ plane, too. If it began in the upper half-cloud, it would stay there, and the same thing is true for the lower half-cloud. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the Bohmian picture of physics is physically incorrect - I am just mentioning the point for the sake of completeness because you seem to be thinking about a &quot;real position of the electron&quot; at every moment - a fundamental misconception. According to quantum mechanics, the electron simply doesn't have any particular sharp position at every point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electron doesn't have to tunnel anywhere because you haven't showed that the sign of its $z$ has ever changed. But even if it changed, one wouldn't need any tunneling because there is no potential barrier that it would have to tunnel through. In particular, points where $\psi$ happens to vanish are not a &quot;wall&quot; in any sense. They're just zeros of the wave function. In the flawed Bohmian mechanics, points $x$ with $\psi(x)=0$ are singular (usually loci of bizarre solenoids), but in the real quantum mechanics which is totally linear in $\psi$, there is nothing special about points with $\psi(x)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even if the electron had to tunnel somewhere - which is never the case for a single atom - it could never become a virtual photon because such a strange transformation would violate the conservation of electric charge and spin, among many other things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies for debunking so many things - but your question wasn't a real question. It was closer to a sequence of approximately 37 independent fundamental misconceptions about physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-19T17:45:40.657" CommentCount="1" />
  <row Id="3357" PostTypeId="1" AcceptedAnswerId="3435" CreationDate="2011-01-19T17:53:01.297" Score="2" ViewCount="277" Body="&lt;p&gt;I read this &lt;a href=&quot;http://www.popsci.com/science/article/2011-01/if-quantum-teleportation-werent-spooky-enough-physicists-propose-time-teleportation&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; about time teleportation. &#xA;Can someone explain the concept better?&lt;/p&gt;&#xA;" OwnerUserId="425" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:39:35.763" LastActivityDate="2011-02-04T07:39:35.763" Title="What is time teleportation?" Tags="&lt;quantum-mechanics&gt;&lt;entanglement&gt;&lt;time-travel&gt;" AnswerCount="2" />
  <row Id="3359" PostTypeId="1" AcceptedAnswerId="3410" CreationDate="2011-01-19T18:25:08.023" Score="1" ViewCount="343" Body="&lt;p&gt;I did one module of physics for my GCSE one year ago which taught me about transverse EM waves &amp;amp; the EM spectrum, but since then, I do not understand how a wave would move in 3D space. Can someone show me some animation or something? I can understand it in 2D space (ie on a graph) but not 3D. I also read somewhere that they do not oscilate in space, but in electromagnetic field strength and direction? Is this true?&lt;/p&gt;&#xA;" OwnerUserId="1466" LastEditorUserId="4283" LastEditorDisplayName="" LastEditDate="2011-07-05T21:26:28.457" LastActivityDate="2011-07-05T21:26:28.457" Title="Visualizing Electromagnetic Waves in 3D Space" Tags="&lt;electromagnetism&gt;&lt;space&gt;&lt;visualization&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="2" />
  <row Id="3360" PostTypeId="1" CreationDate="2011-01-19T18:33:05.137" Score="0" ViewCount="328" Body="&lt;p&gt;Electromagnetism implies special relativity and then the universal constant &quot;c&quot;. And if we set c=1, the coupling constant has units of angular momentum (so in relativistic quantum mechanics we divide by $\hbar$ and we get the adimensional coupling $\alpha$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question, loose, is: In which explicit ways does this angular momentum appear in the classical formalism? Has it some obvious, useful meaning?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: some clarifications from the comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More explicitly, I have in mind the following. In classical no relativistic mechanics a circular orbit under, ahem, a central force equilibrates $F = K / r^2$ equal to $F= m v^2 / r$, and then we have $K = m r v^2 = L v.$ Thus when introducing relativity we can expect that the angular momentum for a particle orbiting in a central force will have a limit, the minimum possible value being $L_{min} = K/c$.  Note that this limit does not imply a minimum radius, we also have classically $K = L^2/ mr $, but m can be argued to be the relativistic mass, so when L goes towards its minimum, m increases and the radius of the orbit goes to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More edit: Given that it seems that my derivation of the Sommerfeld bound $L_{min} = K/c$ risks to be wrong, I feel I should point out that failure, it it is, is completely mine. The original derivation of this relativistic (not quantum!) bound appears in section II.1 of &lt;em&gt;Zur Quantentheorie der Spektrallinien&lt;/em&gt; (pages 45-47 &lt;a href=&quot;http://zs.thulb.uni-jena.de/servlets/MCRFileNodeServlet/jportal_derivate_00152120/19163561702_ftp.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) and also in Kap 5.2 of his &lt;a href=&quot;http://www.archive.org/stream/atombauundspekt00sommgoog#page/n9/mode/2up&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt;. The usual argument goes about generic ellipses and its stability properties.&lt;/p&gt;&#xA;" OwnerUserId="1335" LastEditorUserId="1335" LastEditorDisplayName="" LastEditDate="2011-01-27T10:41:40.243" LastActivityDate="2011-01-27T10:41:40.243" Title="Is there an explicit angular momentum in Maxwell equations?" Tags="&lt;electromagnetism&gt;&lt;special-relativity&gt;&lt;soft-question&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="3361" PostTypeId="2" ParentId="3325" CreationDate="2011-01-19T18:45:36.827" Score="5" ViewCount="" Body="&lt;p&gt;There is no time between photons for which the interference pattern goes away. Even if you send in one photon at a time, and wait until after it has been detected to send the next one, you will still see the detections build up to make an interference pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find lots of pages on the web describing experiments where the time between photons is large enough that they do not affect each other. I used the image from &lt;a href=&quot;http://ophelia.princeton.edu/~page/single_photon.html&quot; rel=&quot;nofollow&quot;&gt;this page at Princeton&lt;/a&gt; as illustrations in my book, but there are lots of others. You can also do the same thing with electrons, which again, has been done many times, including &lt;a href=&quot;http://www.hitachi.com/rd/research/em/doubleslit.html&quot; rel=&quot;nofollow&quot;&gt;this version from Hitachi&lt;/a&gt; which includes a nice movie of the pattern building up one electron at a time.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2011-01-19T18:45:36.827" />
  <row Id="3362" PostTypeId="1" AcceptedAnswerId="3377" CreationDate="2011-01-19T18:58:42.343" Score="3" ViewCount="195" Body="&lt;p&gt;A familiar trope within science-fiction is that of a large relativistic object hitting a planet such as the earth. This is normally an interstellar spacecraft or a kinetic weapon with a mass in the range 10&lt;sup&gt;3&lt;/sup&gt; - 10&lt;sup&gt;6&lt;/sup&gt; kg. But what would actually happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two scenarios seem to be: (a) the object creates a kind of conical tunnel through the earth with most of the material exiting on the far side; (b) the object dumps all of its kinetic energy within a few tens or hundreds of kilometres of the impact point and we have the equivalent of a conventional asteroid impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light transit time through an earth diameter of 12,800 km is just over 40 milliseconds. There’s not much time for lateral effects as the object barrels in. So what would happen if a 1 tonne object hit the earth, square on, at 0.99995c (&amp;gamma; = 100)?&lt;/p&gt;&#xA;" OwnerUserId="1394" LastActivityDate="2011-01-20T13:54:08.670" Title="Relativistic object impacts the earth" Tags="&lt;quantum-mechanics&gt;&lt;special-relativity&gt;" AnswerCount="3" />
  <row Id="3363" PostTypeId="2" ParentId="3309" CreationDate="2011-01-19T19:00:58.720" Score="6" ViewCount="" Body="&lt;p&gt;The largest genuine &quot;cat state&quot; sort of experiment I could find, measured by number of particles, was a couple of experiments on superconducting junctions which involved a few billion electrons. I have, of course, misplaced the references, but I believe &lt;a href=&quot;http://arxiv.org/abs/cond-mat/0004293&quot; rel=&quot;nofollow&quot;&gt;this arxiv paper&lt;/a&gt; is one of them. (Though looking around for the reference also turns up &lt;a href=&quot;http://iopscience.iop.org/0295-5075/89/3/30003/fulltext&quot; rel=&quot;nofollow&quot;&gt;this recent paper arguing the number is far smaller&lt;/a&gt;, at most a few thousand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that the &quot;Breakthrough of the Year&quot; paper (&lt;a href=&quot;http://www.nature.com/nature/journal/v464/n7289/full/nature08967.html&quot; rel=&quot;nofollow&quot;&gt;this Nature article&lt;/a&gt;) does not actually claim to have made a Schroedinger cat state. They have cooled a mechanical resonator to its ground state, and demonstrated some control over its state, but they haven't done all the work that would need to be done to demonstrate that what they have is a cat state (which would probably involve some sort of interference effects, that being the usual way to demonstrate something being in more than one state). Given what they &lt;em&gt;have&lt;/em&gt; done, it's not a big stretch to think that they will eventually do the cat state experiment, but they haven't published that yet.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2011-01-19T19:00:58.720" CommentCount="6" />
  <row Id="3364" PostTypeId="2" ParentId="3282" CreationDate="2011-01-19T19:22:45.683" Score="0" ViewCount="" Body="&lt;p&gt;The topological phase transitions you ask about I presume is the breakdown of a Landau electron fluid where quantum fluctuations are comparable to thermal fluctuations.  There is quite an extensive set of publications on this subject with heavy metals and quantum critical points.  Cubrovic has found parallels with AdS~CFT in such systems.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-19T19:22:45.683" CommentCount="1" />
  <row Id="3365" PostTypeId="2" ParentId="3342" CreationDate="2011-01-19T19:32:30.773" Score="0" ViewCount="" Body="&lt;p&gt;LM beat me to this question this morning.  There is not much I can add by writing.  Polchinski’s vol II of “String Theory” has from p 138 to 146, section titled T-duality of type I string, which covers much of this.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-19T19:32:30.773" />
  <row Id="3366" PostTypeId="2" ParentId="3309" CreationDate="2011-01-19T19:42:25.520" Score="-1" ViewCount="" Body="&lt;p&gt;The Dec 17 2011 Science has in its breakthrough of the year a discussion of “The first Quantum Machine.”  It is a vibrating device which has some superposition of vibrational states.  It is a maybe a sort of Schrodinger cat state.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-19T19:42:25.520" />
  <row Id="3367" PostTypeId="2" ParentId="3359" CreationDate="2011-01-19T19:48:41.970" Score="1" ViewCount="" Body="&lt;p&gt;You might enjoy &lt;a href=&quot;http://www.astronomynotes.com/light/s3.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.astronomynotes.com/light/s3.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1468" LastActivityDate="2011-01-19T19:48:41.970" CommentCount="1" />
  <row Id="3368" PostTypeId="2" ParentId="2767" CreationDate="2011-01-19T20:22:07.200" Score="1" ViewCount="" Body="&lt;p&gt;Suppose you were in a spacecraft in the neighborhood of two mutually orbiting black holes.  The system is decaying and producing gravity waves which you detect.  When the two BHs finally coalesce in the near future you know that lots of gravitational radiation will result and the quadrupole oscillation will turn you and your craft into pieces.  There is one way to shield your self from this, so to speak, which is to get out of Dodge fast!  If you accelerate to a high gamma velocity relative to this system you will Doppler shift the oncoming gravitational radiation in your frame.  If the radiation reaches you with a wavelength $\lambda~&amp;gt;&amp;gt;~L$, for $L$ the dimensions of your craft you will avoid being ripped apart.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electric field shields result from there being $\pm e$ charges, and an electric field can be stopped by a conducting shell or Faraday cage.  Similarly for magnetic shields, high $\mu$ metals have magnetic dipoles which conduct $B$ fields.  You don’t have this luxury with gravity.  With a gravity wave it should be possible to set up lots of bulk matter which oscillates and converts energy to heat.  Feynman used an argument with beads on a rod to show that energy from a gravity wave could be detected.  However, gravity waves couple to matter very weakly, so you need lots of material to attenuate a gravity wave.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible way to partially shield one’s self from a gravity wave would be to hide behind a black hole.  This might be about jumping out of the pan into the fire though.  However, the gravity wave will not pass through the black hole.  What will of course happen is that the black hole will absorb some $\delta mc^2$ of energy and adjust its horizon size to a larger radius.  This will then manifest itself in the reaction of spacetime near the black hole which will be in part in the form of gravity waves.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-19T20:22:07.200" CommentCount="1" />
  <row Id="3369" PostTypeId="2" ParentId="3315" CreationDate="2011-01-19T20:22:43.333" Score="4" ViewCount="" Body="&lt;p&gt;There's a few different ways of defining what you mean by 'Calories' in food.  There's the obvious physical/chemical definition of energy stored in the food, but I had an objection when I explained that &lt;a href=&quot;http://cooking.stackexchange.com/questions/66/how-to-calculate-the-calorie-content-of-cooked-food/500#500&quot;&gt;warmer food naturally has more Calories than cold food.&lt;/a&gt;  There's also the measure of absorbtion, or &lt;a href=&quot;http://www.fspublishers.org/ijab/past-issues/IJABVOL_5_NO_1/26.pdf&quot; rel=&quot;nofollow&quot;&gt;'digestibility', which can be determined by what nutrients pass through the body unabsorbed&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.fda.gov/Food/GuidanceComplianceRegulatoryInformation/GuidanceDocuments/FoodLabelingNutrition/FoodLabelingGuide/ucm064904.htm&quot; rel=&quot;nofollow&quot;&gt;US FDA has guidelines&lt;/a&gt; that basically suggest for their labeling requirements, it's based on estimates of calories in the component parts of the food:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;L12. Is it necessary to include a calorie conversion footnote which states that fat, carbohydrate, and protein furnish 9, 4, and 4 calories per gram, respectively?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Answer: No, the use of that footnote is optional. 21 CFR 101.9(d)(10)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But they also have rules on rounding:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;L31. What are insignificant amounts of nutrients?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Answer: These are the amounts that are permitted to be shown as zero on the Nutrition Facts label (e.g., less than 5 calories may be expressed as 0 calories) except that for total carbohydrate, dietary fiber, and protein, it is the amount that can be declared as “less than 1 g” on the Nutrition Facts label. 21 CFR 101.9(c)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems that less than 0.5g is required to be listed as '0g', even though they call for rounding at other times:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;L45. How should trans fatty acids be listed?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Answer: Trans fatty acids should be listed as “Trans fat” or “Trans” on a separate line under the listing of saturated fat in the Nutrition Facts label (see figure). The word “trans” may be italicized to indicate its Latin origin. Trans fat content must be expressed as grams per serving to the nearest 0.5-gram increment below 5 grams and to the nearest gram above 5 grams. If a serving contains less than 0.5 gram, the content, when declared, must be expressed as “0 g.” (21 CFR 101.9(c)(2)(ii)).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some items are allowed to take into account absorbtion (eg, Olestra, the fat that humans can't absorb), but there's also personal differences -- not all people absorb nutrients the same way, due to food intollerances.  There's also the question of how &lt;a href=&quot;http://cooking.stackexchange.com/questions/66/how-to-calculate-the-calorie-content-of-cooked-food&quot;&gt;cooking affects absorbtion&lt;/a&gt;, as some &lt;a href=&quot;http://www.beyondveg.com/tu-j-l/raw-cooked/raw-cooked-2a.shtml&quot; rel=&quot;nofollow&quot;&gt;studies have shown different digestibility between cooked and uncooked items.&lt;/a&gt;  Of course, nutrition labels don't have a way to explain that there's different effective Calories depending on how something's prepared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I go with the assumption that they're inaccurate estimates.  It's my understanding that the FDA doesn't even care unless the values are more than 20% off.  As such, there's no incentive for companies to test their product -- if they found out their values were wrong, they might be forced to correct them, and doing so could put them at a disadvantage to other groups that were 19.5% wrong in their values.  Also, as most food products are agricultural, there's going to be differences between tomatoes from one plant to another, and they just average these things out ... but we have no idea what the variance is for any of the values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;update&lt;/strong&gt; : Looking at what I wrote, I don't think I was really clear on the different perspectives of what 'Calories' in food are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;physics : total energy (chemical potential + thermal)&lt;/li&gt;&#xA;&lt;li&gt;chemistry : potential chemical energy only&lt;/li&gt;&#xA;&lt;li&gt;biology : chemical energy of the nutrients that were absorbed&lt;/li&gt;&#xA;&lt;li&gt;&quot;food science&quot; : an estimate of chemical energy that's likely to be absorbed by a human&lt;/li&gt;&#xA;&lt;li&gt;US FDA : really inacurate numbers vaguely related to the &quot;food science&quot; perspective&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="115" LastEditorUserId="115" LastEditorDisplayName="" LastEditDate="2011-01-20T17:07:13.627" LastActivityDate="2011-01-20T17:07:13.627" CommentCount="8" />
  <row Id="3370" PostTypeId="2" ParentId="3296" CreationDate="2011-01-19T20:25:57.307" Score="2" ViewCount="" Body="&lt;p&gt;Having been in the derby and lost as a kid, and as parent, it is true that the angle tapers back, so you could gain a small amount of potential energy by placing it back as far as possible. I think the biggest issue is keeping the wheel(s) from vibrating in the track and generating extra friction. The wheel is solid plastic with a hole drilled through it, and a nail goes through the axial hole, into a slit on the bottom of the wood block. I've seen people polish and/or oil the part of the nail that contacts the wheel, with little apparent results. I suspect it probably has more to do with how well aligned the axles are&#xA;to the direction of motion and how well balanced the wheels are.That may be largely luck of the draw. Perhaps one should time all possible combinations of wheels and nails (4 factorial squared), and select the fastest one. You might also try both combinations of wheel orientations, but that gives you 96*96 possible combinations, which is too many to try them all.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-01-19T20:25:57.307" CommentCount="2" />
  <row Id="3371" PostTypeId="2" ParentId="3247" CreationDate="2011-01-19T20:32:10.343" Score="2" ViewCount="" Body="&lt;p&gt;In addition to my original answer, I've scattered bits of an&#xA;argument through various comments in this thread, but I don't&#xA;think I've tied them all together very clearly. More importantly,&#xA;I think I made some mistakes. Let me try to say what I think is true,&#xA;and justify it more carefully than I have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the situation in which you initially hold a proton and&#xA;neutron at rest (relative to the Earth), drop them, and catch&#xA;them after they've fallen a certain distance.  I'm essentially&#xA;certain that the the proton falls slower than the neutron, in the&#xA;following specific sense: &lt;em&gt;the speed of the proton just before&#xA;you catch it will be less than the speed of the neutron&lt;/em&gt; (and&#xA;also the proton will arrive later than the neutron).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's why. Even if the proton doesn't radiate during most of its&#xA;fall, it does radiate for a brief period right when you drop&#xA;it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before trying to convince you that this last statement is true,&#xA;let me point out that there's certainly no equivalence-principle based&#xA;argument against it. At most, the equivalence principle says that during&#xA;the time the particle is in free fall, it shouldn't radiate. It doesn't&#xA;say anything about what happens during the transition from non-free-fall&#xA;to free fall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this, it seems clear to me that the burden of proof is on&#xA;anyone who says there's &lt;em&gt;no&lt;/em&gt; radiation during the transition&#xA;period. After all, we have a charge undergoing jerky motion. In&#xA;the absence of an equivalence-principle argument, the default&#xA;assumption would surely be that it radiates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's not a proof, of course. One thing that would count as a&#xA;proof would be to calculate the fields and determine the radiated&#xA;flux.  This would be hard to do in the full Schwarzschild&#xA;geometry, but a calculation in flat spacetime, replacing&#xA;observers at rest with respect to the Earth with&#xA;accelerated (Rindler) observers, wouldn't be hard. f I wanted to&#xA;do it, I'd probably start with the basic formalism set up in a&#xA;recent paper I just discovered by &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/andp.201000040/pdf&quot; rel=&quot;nofollow&quot;&gt;Maluf and&#xA;Ulhoa&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[One might question whether such a Schwarzschild-Rindler substitution would be justified. I'll just say that the difference between the two is simply tidal forces, and I see no reason that they're relevant in this situation. If you like, change the mass and radius of the Earth to make them much larger, while keeping $g$ constant. That weakens tidal forces still further, but it's very implausible, at least to me, that it bears on the radiation question.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even without doing such a calculation, I'm confident that there is&#xA;radiation during the jerk period. The reason is that the radiation&#xA;reaction force can be shown to be nonzero during this period.&#xA;The radiation reaction force is problematic for point charges, but&#xA;if we model the proton as a small sphere of charge (which, after all,&#xA;it really is!), with a radius much less than any other length&#xA;scale in the problem, then calculation of the radiation reaction force&#xA;is straightforward. You can look up how to do it in Jackson, and&#xA;there's a bunch of more recent literature. Specifically, Rohrlich&#xA;has many articles on the subject, but the monograph by &lt;a href=&quot;http://books.google.com/books?id=bZkaJZ5htiQC&amp;amp;lpg=PA133&amp;amp;ots=c8fZ_slGEJ&amp;amp;dq=yaghjian%20radiation&amp;amp;pg=PP1#v=onepage&amp;amp;q=yaghjian%20radiation&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Yaghjian&lt;/a&gt;&#xA;is the most complete reference. Anyway, the conclusion is completely&#xA;unambiguous: the radiation reaction force is nonzero during the jerk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me conclude with a &lt;em&gt;mea culpa&lt;/em&gt;: in some of my earlier comments, I &#xA;think I said that there was radiation even during a constant-acceleration&#xA;phase (although the radiation reaction force vanished then). I'm pretty&#xA;sure I was wrong about that. (My belief wasn't quite as stupid as it&#xA;might seem: that combination of radiation with no radiation reaction&#xA;is precisely what happens when a charge accelerates uniformly in&#xA;Minkowski space. It seems like a paradox, but it's not.)&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-01-19T20:32:10.343" />
  <row Id="3372" PostTypeId="1" AcceptedAnswerId="3374" CreationDate="2011-01-19T20:39:12.240" Score="-2" ViewCount="112" Body="&lt;p&gt;When an object is thrown upwards, when it eventually comes to rest and starts falling, for how long is it stationary? What about an particle in electric field having an initial velocity towards it's same charge? That too would come to rest and reverse velocity, the question is for how long is it at full stop with rest? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classical physics gives a time of 0, but is that correct? Is it really at rest for 0 seconds? That answer seems a bit counter intuitive.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-01-19T21:23:27.447" LastActivityDate="2011-01-19T21:23:27.447" ClosedDate="2011-01-19T21:30:35.100" Title="Non Classical mechanic answer to : length of time a thrown object spends in rest before falling down?" Tags="&lt;mechanics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="3374" PostTypeId="2" ParentId="3372" CreationDate="2011-01-19T20:50:17.573" Score="3" ViewCount="" Body="&lt;p&gt;Assuming you are talking about classical physics and point particles, the answer is zero seconds precisely. Why? Suppose the time were non-zero, so that particle stays at one place for at least time $T$. That means that it's velocity is zero for as much long and also acceleration would be zero and external forces would be zero. But that is a contradiction with the presence of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discussion of an extended body is similar because you can repeat everything that was said above for its center of mass. But the object might not rest at all because of its intrinsic rotation/vibration/etc.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-01-19T20:50:17.573" CommentCount="6" />
  <row Id="3375" PostTypeId="2" ParentId="3372" CreationDate="2011-01-19T21:03:31.330" Score="2" ViewCount="" Body="&lt;p&gt;Another result is &lt;a href=&quot;http://en.wikipedia.org/wiki/Earnshaw%27s_theorem&quot; rel=&quot;nofollow&quot; title=&quot;Earnshaw's_theorem&quot;&gt;Earnshaw's_theorem,&lt;/a&gt; which tells us that no particle can ever be held in a stable equilibrium in the presense of any other collection of interacting particles.&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2011-01-19T21:09:32.073" LastActivityDate="2011-01-19T21:09:32.073" CommentCount="1" />
  <row Id="3376" PostTypeId="2" ParentId="3324" CreationDate="2011-01-19T21:04:22.813" Score="2" ViewCount="" Body="&lt;p&gt;The problem in the rubber band analogy is not the analogy itself but the object that it uses becomes the focus. If instead of using a rubber band they used computer graphics with emphasis on galaxies further away accelerating faster then nobody would think of tension of the band.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The attention should be given only to the position of dots on the band and their relative motion from each other. But if the ether theory was being thought rubber band could have been considered as the ether!&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-01-19T21:04:22.813" CommunityOwnedDate="2011-01-19T21:04:22.813" />
  <row Id="3377" PostTypeId="2" ParentId="3362" CreationDate="2011-01-19T21:13:22.567" Score="6" ViewCount="" Body="&lt;p&gt;I think there would not be a hard collision of the object with the ground.  It would pass through a column of air of comparable mass, so I suspect the energy of the object would be consumed before it hits the ground.  This would be a sort of massive cosmic ray event.  The interaction energy of atoms would be comparable to Tevatron energy (1/10 of that or so) and what would reach the ground would be a huge pulse of secondary particles that generate an enormous thermal-mechanical shock wave.  At 10^3kg the energy would be $E~\simeq~\gamma mc^2$ or about $10^{19}J$ of energy.  One ton of explosives is 4.18 mega Joules of energy, so the energy released would be about the equivalent of $2\times 10^7$ megatons of explosives.  Clearly you would not want to be anywhere near this.  However, I suspect the damage done to the Earth’s surface would not be due to the body actually impacting the surface, but from this huge shock front.  So one would have to search out what is known about that sort of mechanics.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-19T21:13:22.567" CommentCount="2" />
  <row Id="3378" PostTypeId="2" ParentId="3282" CreationDate="2011-01-19T21:28:19.907" Score="7" ViewCount="" Body="&lt;p&gt;I think you need to define what you mean by a &quot;topological state of matter&quot;, since the term is used in several inequivalent ways. For example the toric code that you mention, is a very different kind of topological phase than topological insulators. Actually one might argue that all topological insulators (maybe except the Integer Quantum Hall, class A in the general classification) are only topological effects rather than true topological phases, since they are protected by discreet symmetries (time reversal, particle-hole or chiral). If these symmetries are explicitly or spontaneously broken then the system might turn into a trivial insulator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one of the simplest lattice models (much simpler that the toric code, but also not as rich) I know of is the following  two band model (written in k-space)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H(\mathbf k) = \mathbf d(\mathbf k)\cdot\mathbf{\sigma},$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\mathbf d(\mathbf k) = (\sin k_z, \sin k_y, m + \cos k_x + \cos k_y)$ and $\mathbf{\sigma} = (\sigma_x,\sigma_y,\sigma_z)$ are the Pauli matrices. This model belongs to the same topological class as the IQHE, meaning that it has no time-reversal, particle-hole or chiral symmetry. The spectrum is given by $E(\mathbf k) = \sqrt{\mathbf d(\mathbf k)\cdot\mathbf d(\mathbf k)}$ and the model is classified by the first Chern number&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_1 = \frac 1{4\pi}\int_{T^2}d\mathbf k\;\hat{\mathbf d}\cdot\frac{\partial \hat{\mathbf d}}{\partial k_x}\times\frac{\partial \hat{\mathbf d}}{\partial k_y},$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $T^2$ is the torus (which is the topology of the Brillouin zone) and $\hat{\mathbf d} = \frac{\mathbf d}{|\mathbf d|}$. By changing the parameter $m$ the system can go through a quantum critical point, but this can only happen if the bulk gap closes. So solving the equation $E(\mathbf k) = 0$ for $m$, one can see where there is phase transitions. One can then calculate the Chern number in the intervals between these critical points and find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_1 = 1$ for $0 &amp;lt; m &amp;lt; 2$, $C_1 = -1$ for $-2 &amp;lt; m &amp;lt; 0$ and $C_1 = 0$ otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus there are three different phases, one trivial and two non-trivial. In the non-trivial phases the system has quantized Hall response and protected chiral edge states (which can easily be seen by putting edges along one axes and diagonalizing the Hamiltonian on a computer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one takes the continuum limit, the model reduces to a 2+1 dimensional massive Dirac Hamiltonian and I think the same conclusions can be reached in this continuum limit but the topology enters as a parity anomaly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More information can be found here: &lt;a href=&quot;http://arxiv.org/abs/0802.3537&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0802.3537&lt;/a&gt; (the model is introduced in section IIB).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope you find this useful. &lt;/p&gt;&#xA;" OwnerUserId="1469" LastEditorUserId="1469" LastEditorDisplayName="" LastEditDate="2011-03-09T14:51:14.917" LastActivityDate="2011-03-09T14:51:14.917" CommentCount="4" />
  <row Id="3379" PostTypeId="2" ParentId="1453" CreationDate="2011-01-19T21:51:00.740" Score="-1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Non-perturbatively coupling a charge and its quantized electromagnetic field.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a better initial approximation for QED. It may prevent the theory from UV and IR infinities at one stroke and provide correct physical description even in the first Born approximation. The perturbative series will become quite different - with numerically small terms. If successful, this formulation of QED may serve as a model to other phenomena (interactions) in particle physics.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-01-19T22:22:31.050" LastActivityDate="2011-01-19T22:22:31.050" CommunityOwnedDate="2011-01-22T18:11:53.713" CommentCount="18" />
  <row Id="3380" PostTypeId="2" ParentId="2865" CreationDate="2011-01-19T21:58:09.147" Score="3" ViewCount="" Body="&lt;p&gt;If you have a charged black hole, and a strong enough magnet, you should be able to move the charged black hole without any theoretical difficulty. A wormhole mouth shouldn't be any different.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditorDisplayName="" LastEditDate="2011-01-19T22:08:54.193" LastActivityDate="2011-01-19T22:08:54.193" CommunityOwnedDate="2011-01-29T11:43:27.057" CommentCount="2" />
  <row Id="3381" PostTypeId="2" ParentId="1603" CreationDate="2011-01-19T22:06:12.463" Score="1" ViewCount="" Body="&lt;p&gt;Sean,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sorry I am responding to an old question but I have a very nice example of an application of advanced algebraic topology in physics (it's physics seen in experiments, not arbitrary speculative theories). It's the newly discovered &quot;topological insulators&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can topologically classify free Hamiltonians (hermitian matrices/operators) as a function of different symmetry classes and spatial dimension. It turns out that this can be done using topological K-theory (see periodic table in &lt;a href=&quot;http://arxiv.org/abs/1002.3895&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1002.3895&lt;/a&gt; ; table 1 on page 8).&#xA;There is a two-fold periodicity in symmetry-classes and dimension, originating from the Bott-periodicity of complex K-theory (classification of complex vector bundles up to stable equivalence). And there is a eight-fold periodicity in the other symmetry-classes originating from the Bott-periodicity of real K-theory. See more information here: &lt;a href=&quot;http://arxiv.org/abs/0901.2686&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0901.2686&lt;/a&gt; and &lt;a href=&quot;http://iopscience.iop.org/1367-2630/12/6/065010&quot; rel=&quot;nofollow&quot;&gt;http://iopscience.iop.org/1367-2630/12/6/065010&lt;/a&gt; (free access for both).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore I must mention that the 10-symmetry classes mathematically originate from Cartans classification of symmetric spaces and the labels in the above-mentioned table comes from this classification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I just saw that topological insulators were mentioned above, but not these aspects).&lt;/p&gt;&#xA;" OwnerUserId="1469" LastActivityDate="2011-01-19T22:06:12.463" CommentCount="1" />
  <row Id="3382" PostTypeId="2" ParentId="2865" CreationDate="2011-01-19T22:14:16.120" Score="0" ViewCount="" Body="&lt;p&gt;{\bf There is a more complete answer appended here}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with the conversion is at the point the two worm hole openings are separated by a lightlike interval.  This is a Cauchy horizon and there is this winding of paths, traversed by particles or photons, which pile up at the horizon.  This would include virtual particles or the vacuum as well as I see it.  The worm hole has a Lansczoc junction with a shell of negative mass-energy that permits the wormhole to exist.  The winding up of vacuum modes on the Cauchy horizon would introduce enormous fluctuation of positive energy which I think would destroy the wormhole.  I think this would destroy the solution, so that even if a wormhole can exist any attempt to convert it to a time machine would destroy it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other problem is that in QM we have that momentum is the generator of a position change.  Yet with multiply connected spacetimes there is a funny problem of an amgibuity.  A particle can travel from $x$ to $y$ by two types of momentum generators or transformation $e^{ipx}$.  So it is not entirely clear whether the wormhole is consistent with quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just a brief answer at this point.  I will try to work out a more complete version in a few days.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;addendum: This is not airtight, but it is worth a couple of evening’s work.  I think this gives a pretty good argument for why you can’t transform a wormhole into a time machine.  I will have to confess that I am a big enemy of these quirky spacetimes which give faster than light or time reversal results.  They are mathematical result, but quantum mechanics kills them in physics.  Further, the averaged weak energy condition is violated $T^{00}~&amp;lt;~0$, which means the quantum field which acts as this source has no lower bounded eigenvalue.  That is a complete disaster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A worm hole has a membrane or surface with a field that has an energy density.  This starts with a static wormhole as a necessary starting point.  Start with the Reissner-Nordstrom metric&#xA;$$&#xD;&#xA;ds^2~=~-F(r)dt^2~+~{1\over {F(r)}}dr^2~+~d\Omega^2,&#xD;&#xA;$$&#xA;for $F(r)~=~1~-~r_0/r$.  The event horizon with radius $r_0$ is replaced by a junction or thin shell at $r_0(t)~\rightarrow~r_0$ $+~\delta r(t)$.  This defines the openings of the wormhole in two regions.  The normal vectors to this sphere are&#xA;$$&#xD;&#xA;n^\mu~=~\pm\Big({{\dot r_0}\over{F(r_0)}},~\sqrt{F(r_0)~-~{\dot r_0}},~0,~0\Big),&#xD;&#xA; n_\mu~=~\pm\Big({\dot r_0},~{{\sqrt{F(r_0)~-~{\dot r_0}^2}\over {F(r_0)}}}, ~0, ~0\Big). &#xD;&#xA;$$&#xA;The sign is an indication of the direction of the normal on the sphere.  The extrinsic curvature is computed as $K_{\mu\nu}~=~{1\over 2}n^\sigma\partial_\sigma g_{\mu\nu}$ and the components are&#xA;$$&#xD;&#xA;K_{\theta\theta}~=~\pm {1\over r_0}\sqrt{F(r_0)~-~{\dot r_0}^2}&#xD;&#xA;$$&#xA;$$&#xD;&#xA;K_{tt}~=~\mp{1\over 2}{1\over{\sqrt{F(r_0)~-~{\dot r_0}^2}}}\Big({{\partial F(r_0)}\over{\partial r_0}}~-~{\ddot r_0}\Big). &#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy density $\rho~=~G^{00}$ $=~1/4\pi K_{\theta\theta}$ is integrated on a pillbox configuration to find the jump in the surface energy on the membrane, which is the jump in extrinsic curvature at $r~=~r_0(t)$.  For a static membrane the tension equals the energy density so there is an overall conservation of the stress-energy on the membrane.  The energy density is found from the ADM tensors&#xA;$$&#xD;&#xA;G^{00}~=~\rho~=~{1\over{2\pi r_0}}\sqrt{F(r_0)~-~{\dot r_0}^2}&#xD;&#xA;$$&#xA;and the tension is&#xA;$$&#xD;&#xA;\tau~=~-{1\over{4\pi}}(K_{\theta\theta}~-~K_{tt}).&#xD;&#xA;$$&#xA;The static membrane then infers the following equation&#xA;$$&#xD;&#xA;{{F(r_0)}\over{r_0^2}}~-~\Big( {{\dot r_0}\over {r_0}}\Big)^2~=~4\pi^2\tau^2.&#xD;&#xA;$$&#xA;The conservation of energy $\partial\rho /\partial t$ applied to this constraint equation results in the evolution equation&#xA;$$&#xD;&#xA;{\ddot r_0}~+~{1\over {r_0}}\Big(F(r_0)~-~{\dot r_0}^2\Big)~-~{{\partial F(r_0)}\over {\partial r_0}}~=~0. &#xD;&#xA;$$&#xA;This describes the dynamical evolution of the wormhole in a purely classical setting.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The evolution of the vacuum is governed by the discontinuity in the Einstein field at $r~=~r_0$.  This discontinuity is then given by&#xA;$$&#xD;&#xA;\lim_{\epsilon~\rightarrow~0}\int_{-\epsilon}^{+\epsilon} G^{00}dn~=~\delta\rho~=~-{1\over{2\pi r_0}}\sqrt{F(r_0)~-~(\dot r_0~+~U)^2},&#xD;&#xA;$$&#xA;where $U~=~\sqrt{U^\mu U_\mu}$ is the speed of the $r_1$ opening.  The value of this change in energy density is positive.  To induce the wormhole a field of some sort of field $\phi$ with a negative energy density $T^{00}(\phi)~&amp;lt;~0$.  This discontinuity is determined by this field, ie $\delta\rho~=~T^{00}(\phi)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With quantum mechanics we have a vacuum which winds through the wormhole.  The creation and annihilation operators $a^\dagger$ $a$ for this field are boosted relative to the moving opening.  This results in a Bogoliubov transformed operator for these operators with&#xA;$$&#xD;&#xA;b~=~a~cosh(g(s))~+~a^\dagger sinh(g(s)),~b^\dagger~=~ a^\dagger~cosh(g(s))~+~a sinh(g(s)).&#xD;&#xA;$$&#xA;The term $g(s)~=~gs$ for a constant acceleration.  We interpret this as a rapidity angle which varies with respect to the proper time $s$ and which diverges when the two openings have a lightlike separation.  The Hamiltonian is found to be&#xA;$$&#xD;&#xA;b^\dagger b~=~a^\dagger a~+~\big((a^\dagger)^2~+~a^2\big)cosh(g(s))sinh(g(s))&#xD;&#xA;$$&#xA;The expectation of the field remains the same for $E_n~=~\langle n|b^\dagger b|n\rangle$.  The off diagonal term is a form of the squeezed state operator, which removes the vacuum state of photons off quadrature.  The uncertainty in the momentum and positions are evaluated so that $\langle\Delta x\Delta p\rangle~=~(1/2)sinh(2gs)$ which is the evaluation of off diagonal term with completeness.  The entropy can be evaluated from &#xA;$$&#xD;&#xA;{k\over 2}ln\big(\langle\Delta x\rangle\langle\Delta p\rangle~-~\langle\Delta x\Delta p\rangle^2 \big)~=~k~ln(cosh(2g(s)).&#xD;&#xA;$$&#xA;This entropy is associated with the generation of photons from the vacuum, or as a form of Hawking radiation.  The energy of this radiation is positive. As the opening of the wormhole approach a lightlike separation this radiation will demolish the wormhole by overwhelming the negative energy at the junction.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-01-21T03:13:55.143" LastActivityDate="2011-01-21T03:13:55.143" CommunityOwnedDate="2011-01-29T11:43:08.600" CommentCount="6" />
  <row Id="3383" PostTypeId="2" ParentId="3243" CreationDate="2011-01-19T23:05:39.727" Score="1" ViewCount="" Body="&lt;p&gt;Good question! The reponse really depends on what you 'need': if you're sampling a small system and you're lucky enough to actually be able to count (or estimate...) the number of states $\mathcal{N}(E)$ for a given energy $E$, then you can explicitly work out $Z$. Such algorithms exist for the Ising model, but in the generic case you're not this lucky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is the possibility of 'naively sampling' the integral/sum you're doing: you can use any Monte Carlo algorithm (especially for Ising-like systems the literature is very rich) to generate a sequence of states and you use this representative sample to calculate $Z$ for any range of temperatures. (It's important though that you do not change $\beta$ during an MC run, since this tends to break detailed balance.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, like Greg P said, there are rather indirect methods, but they're sometimes rather subtle. You can go from histograms to path integrals (Wick rotated ones, that is!), but these methods all take time to understand. I'll just post my favorite reference on everything related to statistical physics/Monte Carlo, which is 'Statistical Mechanics: Algorithms and Computations' [Oxford Univ. Press] by Werner Krauth. Good luck!&lt;/p&gt;&#xA;" OwnerUserId="1471" LastActivityDate="2011-01-19T23:05:39.727" />
  <row Id="3384" PostTypeId="2" ParentId="3360" CreationDate="2011-01-19T23:18:19.147" Score="1" ViewCount="" Body="&lt;p&gt;I think R. &lt;a href=&quot;http://jayryablon.files.wordpress.com/2008/04/ohanian-what-is-spin.pdf&quot; rel=&quot;nofollow&quot;&gt;Ohanian&lt;/a&gt; gives the answer to the title question.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-01-19T23:18:19.147" CommentCount="8" />
  <row Id="3385" PostTypeId="1" AcceptedAnswerId="3387" CreationDate="2011-01-19T23:46:12.830" Score="1" ViewCount="264" Body="&lt;p&gt;A friend of mine is concerned about electromagnetic field negatively affecting their health and got a &quot;DMF meter&quot; to measure the field strength in their house in units of milliGauss. They are trying to compare the reading to a background level measured in mW/cm^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am skeptical if such a measurement makes sense, but this is &lt;strong&gt;NOT&lt;/strong&gt; meant to be a debate on whether cell phone towers, microwaves impact health: I just want to know if milliGauss and mW/cm^2 could possibly be referring to the same thing, and if so, what it might mean in terms of an electromagnetic field inside one's house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. The background rate they are using is 0.000000126mW/cm^2.&lt;/p&gt;&#xA;" OwnerUserId="228" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-20T00:12:39.950" LastActivityDate="2011-01-20T07:38:04.227" Title="gauss vs mW/cm^2: same thing?" Tags="&lt;electromagnetic-radiation&gt;&lt;si-units&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="3386" PostTypeId="2" ParentId="3244" CreationDate="2011-01-19T23:56:39.187" Score="0" ViewCount="" Body="&lt;p&gt;The word &lt;em&gt;viscosity&lt;/em&gt; hasn't been used above, and yet it's crucial to understand the problem. Since tea is viscous, it obeys a type of 'non-slip condition': it's completely at rest against the sides and bottom of the cup. The tea forms a small 'boundary layer' (of ~ 0.04 cm)on the bottom, where there is a large velocity gradient, since outside of this layer the tea behaves as if it were inviscid. This boundary layer is called an &lt;strong&gt;Ekman layer&lt;/strong&gt; - this should allow you to look up good references on the subject. Georg's model explained above is quite correct. I just wanted to add that this is a classical but very subtle exercise (you need to use Navier-Stokes in cylindrical coordinates in a rotating frame in two different regimes...), but in the end you can actually show that inside the boundary layer&#xA;$$v_r = -\frac{\Delta \Omega}{2} r e^{-\hat{z}}\sin{\hat{z}},$$&#xA;where $\Delta \Omega$ is the difference in rotational velocity between the bottom and the top of the cup, $\hat{z} = z \sqrt{\frac{\Omega}{\nu}}$ (thus giving a typical size for the boundary layer...), $\Omega$ is the typical rotational velocity and $\nu$ the tea's viscosity.&lt;/p&gt;&#xA;" OwnerUserId="1471" LastEditorUserId="1471" LastEditorDisplayName="" LastEditDate="2011-01-20T00:57:17.353" LastActivityDate="2011-01-20T00:57:17.353" />
  <row Id="3387" PostTypeId="2" ParentId="3385" CreationDate="2011-01-19T23:57:47.047" Score="4" ViewCount="" Body="&lt;p&gt;No, they do not. The short answer is that the Gauss (or milligauss) is a unit of &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_field&quot; rel=&quot;nofollow&quot;&gt;magnetic field&lt;/a&gt; strength, but milliwatts per centimeter squared is a measure of intensity (technically known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Irradiance&quot; rel=&quot;nofollow&quot;&gt;irradiance&lt;/a&gt;), the amount of energy delivered per unit time per unit surface area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For electromagnetic radiation, it is possible to convert between the two measurements, though. The time-averaged magnitude of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Poynting_vector&quot; rel=&quot;nofollow&quot;&gt;Poynting vector&lt;/a&gt;, denoted $\langle S\rangle$, gives you the irradiance of an electromagnetic wave when the wave impacts a surface perpendicular to its direction of travel. Assuming the wave has only a single frequency, you can calculate $\langle S\rangle$ from the electric or magnetic field as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle S\rangle = \frac{c}{2\mu_0}B_0^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $B_0$ is the amplitude of the magnetic field oscillations and $\mu_0$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_constant&quot; rel=&quot;nofollow&quot;&gt;magnetic constant&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm not sure if this device your friend has actually measures the magnetic field amplitude. But &lt;strong&gt;if&lt;/strong&gt; it does, you can plug that measurement in for $B_0$ and use the numerical value&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{c}{2\mu_0} = 0.119\frac{\text{mW}}{\text{cm}^2\,\text{mGs}^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(obtained from Google's calculator function) to calculate $\langle S\rangle$ in milliwatts per square centimeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Something I just thought of thanks to Stan's comments: the value you plug in for $B_0$ in the formula is the magnetic field amplitude of just the EM &lt;em&gt;wave&lt;/em&gt;, but if the device does measure magnetic field amplitude, it'll probably be picking up the Earth's magnetic field and any other contributions from, say, electronic devices in the area. So there might be a rather involved process to get from the reading on the device to the number that should be plugged into the formula. Depending on how the device works, it may not be as simple as subtracting off the value of a background reading.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-20T07:38:04.227" LastActivityDate="2011-01-20T07:38:04.227" CommentCount="6" />
  <row Id="3388" PostTypeId="2" ParentId="3385" CreationDate="2011-01-20T00:00:43.830" Score="1" ViewCount="" Body="&lt;p&gt;Are you sure they aren't webers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 mWb/cm$^2$ is 10 Tesla. Which is ridiculously huge. 1 milligauss, by comparison, is $10^{-7}$ Tesla.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using @David's conversion, we would have a flux density&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_0$ = $sqrt(\frac{\mu_0}{2c} * 1.26 * 10^{-7} mW/cm^2) \approx 5*10^{-11} T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is $5*10^{-4}$ millgauss... this doesn't seem right at all -- though the websites I've found all seem like they're trying to sell something and thus not worth linking to, this doesn't seem like the usual advertised background, which &lt;em&gt;seems&lt;/em&gt; to be on the order of milligauss. (Though I'd be glad to hear what these levels should be from someone who knows!)&lt;/p&gt;&#xA;" OwnerUserId="791" LastEditorUserId="791" LastEditorDisplayName="" LastEditDate="2011-01-20T00:37:46.950" LastActivityDate="2011-01-20T00:37:46.950" CommentCount="3" />
  <row Id="3389" PostTypeId="1" CreationDate="2011-01-20T01:13:00.913" Score="0" ViewCount="58" Body="&lt;p&gt;certainly if you calculate the maximum points in a parabolic motion for different values of angles, and join these points, the curve obtained is part of an elliptic curve. my question is: is there something else to hide the fact that the union of all these points as a result of this characteristic curve? or is it mere coincidence?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-20T01:13:00.913" ClosedDate="2011-01-20T01:30:42.583" Title="launching projectiles" Tags="&lt;astrophysics&gt;" CommentCount="8" />
  <row Id="3390" PostTypeId="1" AcceptedAnswerId="3400" CreationDate="2011-01-20T02:26:18.537" Score="6" ViewCount="246" Body="&lt;p&gt;Does anybody give a good textbook description of a quantum computer algorithm and how its different from an ordinary algorithm?&lt;/p&gt;&#xA;" OwnerUserId="1383" LastActivityDate="2011-01-20T09:29:20.530" Title="Can anybody provide a simple example of a quantum computer algorithm?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="3391" PostTypeId="2" ParentId="3359" CreationDate="2011-01-20T02:32:45.423" Score="0" ViewCount="" Body="&lt;p&gt;I think if you do a search you can find tons of examples, but here is another&#xA;&lt;a href=&quot;http://webphysics.davidson.edu/applets/EMWave/EMWave.html&quot; rel=&quot;nofollow&quot;&gt;http://webphysics.davidson.edu/applets/EMWave/EMWave.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1383" LastActivityDate="2011-01-20T02:32:45.423" />
  <row Id="3392" PostTypeId="2" ParentId="1453" CreationDate="2011-01-20T02:36:00.203" Score="6" ViewCount="" Body="&lt;p&gt;Greg P asked, so here it is. I think a proof of the AdS/CFT conjecture relating $N=4$ SYM with gauge group $SU(N_c)$ to IIB string theory on $AdS_5 \times S^5$ with $N_c$ units of five-form flux would be extremely important,&#xA;even if the proof was only a proof by physics standards. I say this not because I think the conjecture may be false, indeed there is overwhelming evidence that the conjecture is true. Rather I think it would be important because many of the most important applications of gauge/gravity duality require going beyond AdS/CFT to theories without supersymmetry or conformal symmetry (such as gravity duals of QCD) or to theories without Lorentz invariance (various condensed matter applications) or to theories on the gravity side which are not asymptotically AdS. For example there are proposals for a de Sitter correspondence and Verlinde has proposed &quot;emergent gravity&quot; bases on holographic ideas which as far as I can tell do not involve the input of data from the boundary of spacetime as one needs in AdS. A proof of the original conjecture might indicate more clearly whether these extensions of the idea are correct, or whether at some point things go wrong when you try to generalize. &lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T04:42:26.517" LastActivityDate="2011-03-02T04:42:26.517" CommunityOwnedDate="2011-01-22T18:11:53.713" />
  <row Id="3393" PostTypeId="2" ParentId="1603" CreationDate="2011-01-20T02:37:13.593" Score="0" ViewCount="" Body="&lt;p&gt;The fermion operators obey $b^2~=~{b^\dagger}^2$ $=~0$.  This is a form of the rule d^2 = 0.  Supersymmetry permit a cohomology of states $\psi~\in~ker(Q)/im(Q)$, which is a cohomology. The $Q$ obeys $Q^2~=~0$, physical states obey $Q\psi~=~0$, but where $\psi~\ne~Q\chi$.  This is the basis of BRST (Becchi, Rouet, Stora and Tyutin) quantization.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-20T02:37:13.593" CommentCount="2" />
  <row Id="3394" PostTypeId="1" AcceptedAnswerId="3396" CreationDate="2011-01-20T03:02:34.563" Score="6" ViewCount="178" Body="&lt;p&gt;The Standard Model gives non-zero mass to the electron via the coupling to the Higgs field. Issues of renormalizability aside, this is fundamentally unrelated to the fact that the electron couples to the EM field. However, if the Higgs mechanism did not operate - that is, if there were no vacuum symmetry breaking, the electron field would have no effective mass term. In QFT perturbation theory, this model offers no special difficulty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what is the classical limit of this theory, if it has one? Does the electron acquire a purely EM mass? If the low energy renormalized mass is set to zero (is there an obstacle to doing that?) what do the classical field configurations look like? My puzzlement is related to the classical description of the EM field of a relativistic charged particle - namely, that it becomes &quot;flattened&quot; in the direction of motion, just like a Lorentz-contracted sphere. That is, the field is weaker than for a motionless particle in the longitudinal direction, and stronger in the transverse directions. The limiting case is that the field becomes concentrated on a 2-d surface transverse to the particle location, where it has infinite strength - obviously an unphysical situation. So what actually happens?&lt;/p&gt;&#xA;" OwnerUserId="1447" LastActivityDate="2011-01-20T09:38:01.177" Title="What would the electromagnetic field of a massless electron look like" Tags="&lt;electromagnetism&gt;&lt;quantum-field-theory&gt;" AnswerCount="2" />
  <row Id="3395" PostTypeId="1" AcceptedAnswerId="3397" CreationDate="2011-01-20T03:04:50.440" Score="1" ViewCount="6438" Body="&lt;p&gt;Ok this is a really silly question, but I was really curious as to why this wouldnt work. I'm just starting my electricity and magnetism course, so I really dont know much about electricity. I know this is not possible, but why?&#xA;&lt;img src=&quot;http://i.stack.imgur.com/k2HD5.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T08:12:14.663" LastActivityDate="2011-02-03T08:12:14.663" ClosedDate="2011-01-20T09:42:53.550" Title="Laptop self charge cable?" Tags="&lt;electricity&gt;&lt;energy-conservation&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="3396" PostTypeId="2" ParentId="3394" CreationDate="2011-01-20T03:32:16.630" Score="3" ViewCount="" Body="&lt;p&gt;This has some bearing on the question in your title&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arXiv.org/abs/hep-th/9112020&quot; rel=&quot;nofollow&quot;&gt;http://arXiv.org/abs/hep-th/9112020&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an analogous problem in general relativity, the gravitational field generated by a massless particle is called the Aichelburg - Sexl metric. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Couple more comments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perturbation theory in the presence of massless charged fields is not unproblematic, there are infrared divergences one has to deal with. These are much more problematic than soft photon divergences. I think Weinberg's field theory book has a discussion of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Massless electrons would have an extra chiral symmetry, which should forbid the appearance of mass term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, I don't think these quantum field theory issues are directly related to your question, which is essentially one in classical electrodynamics.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-01-20T03:54:18.067" LastActivityDate="2011-01-20T03:54:18.067" CommentCount="8" />
  <row Id="3397" PostTypeId="2" ParentId="3395" CreationDate="2011-01-20T03:32:45.567" Score="5" ViewCount="" Body="&lt;p&gt;Sorry, but your self charging laptop can not work since it will violate the second law of thermodynamics. A computer processes information and during computation it generates heat. This heat is dissipated through the cooling system. You need continuous supply of usable energy from an external source to run the laptop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the second law of thermodynamics asserts that it is impossible to create a perpetual motion machine of the second kind. All real machines have efficiency less than 100% which means output power is less than input power. The difference is lost/ wasted as heat or other useless forms of energy. You must have noticed your laptop's temperature is high and making noisy sound during a heavy computation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you got the point.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-01-20T03:32:45.567" CommentCount="4" />
  <row Id="3398" PostTypeId="2" ParentId="3362" CreationDate="2011-01-20T03:36:42.600" Score="0" ViewCount="" Body="&lt;p&gt;The answer depends on whether it's heavy enough to reach the ground. If it's small enough, it will fall apart in air and the result will be an explosion. To compute the size of the explosion use the relativistic energy formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small meteors are slowed down by the air and reach the earth at their terminal velocity. These provide the samples you can buy on eBay. Meteors big enough to keep their natural speed of around 40,000 meters per second (i.e. 10 to 40 miles per second) vaporize on hitting the earth. This is what would happen to a relativistic meteor that was big enough to reach the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see why large meteors vaporize, note that the speed of sound in air is about 330 meters per second. This is the velocity for nitrogen molecules which have a molecular weight of 28 at say room temperature or around 300 Kelvin. Iron has an atomic weight of around twice that, so its velocity, at the same temperature, is around 0.7 as much (i.e. kinetic energy is given by $mv^2/2$ and the molecules in a gas at a given temperature all have the same average kinetic energy no matter how heavy they are, assuming they're a gas). So a meteor traveling at 40,000 meters per second has kinetic energy per atom of around 200 times the kinetic energy per atom of room temperature air. Thus the vapor produced is on the order of around 200 x 300K = 60,000 K. (Actual temperatures are a lot lower because the heat is spread around over more than just the meteor, the nearby rocks or whatever is also heated, but this gives the general idea.)&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-20T03:36:42.600" CommentCount="2" />
  <row Id="3399" PostTypeId="2" ParentId="1453" CreationDate="2011-01-20T04:30:00.223" Score="-1" ViewCount="" Body="&lt;p&gt;Does it make sense to talk about quantum gravity?  Is it possible that gravity is a classical field or a condensate which appears largely classical, but where the underlying quantum physics is entirely different.  Thirring fermions ~ sine Gordon solitons, where the SG solitons are a Euclideanize form of the hyperbolic dynamics on $AdS_2$.  The $AdS_2~\sim~ CFT_1$ tells us the isometry of the spacetime is equivalent to the group for conformal quantum mechanics on the boundary.  Might it then be the actual quantization is not with the spacetime, but with an underlying fermionic physics?  If so can this be generalized to $AdS_n$, for $n~&amp;gt;~2$?&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-20T04:30:00.223" CommunityOwnedDate="2011-01-22T18:11:53.713" CommentCount="7" />
  <row Id="3400" PostTypeId="2" ParentId="3390" CreationDate="2011-01-20T05:01:27.140" Score="11" ViewCount="" Body="&lt;p&gt;I'll avoid talking about Shor's algorithm and leave it for you to read on your own -- Shor's algorithm is &lt;em&gt;the&lt;/em&gt; quantum algorithm, and is the reason quantum computing has become such a hot topic, and is not only a must read, but Scott Aaronson has provided a better introduction than I could ever manage, &lt;a href=&quot;http://www.scottaaronson.com/blog/?p=208&quot; rel=&quot;nofollow&quot;&gt;http://www.scottaaronson.com/blog/?p=208&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead I'll guide you through Deutsch's algorithm, which solves a fantastically useless problem exponentially faster than any classical algorithm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine I have a function $f$ on a configuration of $n$ bits, $C=\{0,1\}^n$ that takes each configuration to a single bit. I promise you before hand that this function is either&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Constant - all configurations are mapped to the same value&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Balanced - exactly half of the configurations map to 1, the other half to 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So classically, in the &lt;em&gt;worst case&lt;/em&gt; you must evaluate the function for $2^{n-1}+1$ configurations to verify which category $f$ falls into.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now,&lt;/strong&gt; a quantum computer isn't just a parallel processor, where you can give it a superposition of the configurations and get back $f$ evaluated on all of them. At the end of the day, you have to make a measurement which destroys our carefully crafted superposition -- so we have to be clever! The fundamental feature of a quantum algorithm is that we use &lt;em&gt;unitary operations&lt;/em&gt; to transform our state and use &lt;em&gt;interference&lt;/em&gt; between states so that when we measure the state at the end, it tells us something unambiguous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So without further ado, the Deutsch algorithm for $n=2$ qubits (you can do this for many more qubits, of course, but you wanted a &lt;em&gt;simple&lt;/em&gt; example). The &quot;quantum version&quot; of our function is a unitary operation (a &quot;quantum gate&quot;) that takes me from $|a\rangle|b\rangle$ to $|a\rangle|b+f(a)\rangle$ (where the addition is modulo 2). Also at my disposal is a gate that takes any single bit and puts it in a particular superposition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|1\rangle\rightarrow(|0\rangle-|1\rangle)/\sqrt{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|0\rangle\rightarrow(|0\rangle+|1\rangle)/\sqrt{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step in the algorithm is to prepare my two qubits as $|0\rangle|1\rangle$ and then apply this transformation, giving me the state $(|0\rangle+|1\rangle)(|0\rangle-|1\rangle)/2$. Now I evaluate the function by applying the gate I described earlier, taking my state to&#xA;$$[|0\rangle(|0+f(0)\rangle-|1+f(0)\rangle)+|1\rangle(|0+f(1)\rangle-|1+f(1)\rangle)]/2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we stare at this carefully, and think about arithmetic mod 2, we see that if $f(0)=1$ the first term picks up a minus sign relative to its state before, and if $f(0)=0$ nothing happens. So we can rewrite the state as&#xA;$$[(-1)^{f(0)}|0\rangle(|0\rangle-|1\rangle)+(-1)^{f(1)}|1\rangle(|0\rangle-|1\rangle)]/2$$&#xA;or regrouping&#xA;$$(|0\rangle+(-1)^{f(0)+f(1)}|1\rangle)(|0\rangle-|1\rangle)/2$$&#xA;Now we shed the second qubit (we don't need it anymore, and it's not entangled with the first bit), and apply the second transformation I listed once more (and regrouping, the algebra is simple but tedious) -- at last, our final state is&#xA;$$[(1+(-1)^{f(0)+f(1)})|0\rangle+((1-(-1)^{f(0)+f(1)})|1\rangle]/2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And lastly, we measure! It should be clear from the state above that we've solved the problem... If $f$ is constant, then $f(0)=f(1)$ and we will always measure $|0\rangle$, and if $f$ is balanced then $f(0) \neq f(1)$ and we always measure $|1\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some final comments: hopefully this gives you some taste for the structure of a quantum algorithm, even if it seems like kind of a useless thing -- I strongly recommend going and reading the article I linked to at the beginning of this answer.&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2011-01-20T05:01:27.140" CommentCount="3" />
  <row Id="3401" PostTypeId="1" AcceptedAnswerId="3405" CreationDate="2011-01-20T05:17:42.020" Score="2" ViewCount="195" Body="&lt;p&gt;Quantum phase arises when a spin-j state is sent through a sequence of transitions that return it to its original position. For example with spin-1/2, a state picks up a complex phase of $\pi/4$ when it goes through three perpendicular (not orthogonal!) transitions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{&amp;lt;+z|+y&amp;gt;&amp;lt;+y|+x&amp;gt;&amp;lt;+x|+z&amp;gt;}{|&amp;lt;+z|+y&amp;gt;&amp;lt;+y|+x&amp;gt;&amp;lt;+x|+z&amp;gt;|} = e^{2i\pi/8}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $|+y&amp;gt;$ means the state with spin-1/2 in the +y direction. Since the above path covers an octant, or one eighth of the surface of a sphere, the total quantum phase for the whole sphere in the spin-1/2 case is $2i\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, for a spin state $|j,m,+z&amp;gt;$, we will have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{&amp;lt;j,m,+z|j,m,+y&amp;gt;&amp;lt;j,m,+y|j,m,+x&amp;gt;&amp;lt;j,m,+x|j,m,+z&amp;gt;}{|&amp;lt;j,m,+z|j,m,+y&amp;gt;&amp;lt;j,m,+y|j,m,+x&amp;gt;&amp;lt;j,m,+x|j,m,+z&amp;gt;|} = e^{2i n_{jm} \pi/8}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n_{jm}$ is an integer. The same phase is picked up for a spin-1 state with m=1, but a spin-1 state with m=0 picks up no phase at all. (In fact the m=0 states do not depend on the orientation.) At the moment, I believe that the j=1,m=-1 state will pick up n=-1 (in analogy with the spin-1/2 case), but I haven't bothered to actually run the computation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Yes, Dr. Motl found it. For a journal reference see &quot;Geometric Phases&quot; by Péter Lévay. The appropriate equation is (36) on page 19. Note that our $m$ is his $r-J$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/math-ph/0509064v1&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/math-ph/0509064v1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the method of looking at an infinitesimal rotation works because Berry-Pancharatnam phase doesn't depend on stuff like the arbitrary complex phases of spinors and of course there's no preferred region on the sphere of possible orientations for the spin axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, I was convinced that the j=1, m=+-1 case had the same coefficient as the j=1/2,m=+-1/2 case. This was because the geometric phase for light is the same as that for spin-1/2. But now it's clear that this is mixing apples and oranges.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-01-22T23:42:27.400" LastActivityDate="2011-01-22T23:51:27.673" Title="What is the quantum / Berry-Pancharatnam phase for a spin-j state with z-component m?" Tags="&lt;quantum-mechanics&gt;&lt;mathematical-physics&gt;&lt;spin&gt;&lt;berry-pancharatnam-phase&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="3402" PostTypeId="1" AcceptedAnswerId="4452" CreationDate="2011-01-20T05:42:00.493" Score="4" ViewCount="454" Body="&lt;p&gt;If the universe didn't have the relativity principle, would it be able to support life?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Life consists of very complicated organisms. The operation of these organisms depends on the laws of physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the laws of physics depended on absolute velocity, then it seems that life would have a more difficult task; organisms would have to adapt their biochemistry to the different absolute speeds of the planet as it moves with or against the motion of the sun around the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the laws of physics depended on the absolute gravitational potential, or on acceleration, then the biochemistry of life would have to adapt to the different accelerations / gravitational potential as life colonized higher altitudes. In addition, there would be a seasonal effect as the earth moves closer and farther away from the sun.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I think there's a way this question could be answered quantitatively. Begin with a modification of general relativity such as the post Newtonian parameters. See wikipedia article:&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Parameterized_post-Newtonian_formalism&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Parameterized_post-Newtonian_formalism&lt;/a&gt;&lt;br&gt;&#xA;Now analyze an important biological molecule whose shape is extremely important to life such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Hemoglobin&quot; rel=&quot;nofollow&quot;&gt;hemoglobin&lt;/a&gt;. Find out what range of post-Newtonian parameters are compatible with the operation of that molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I suspect that this is a research problem. If someone solves it, I presume they will publish it.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-01-31T01:18:58.757" LastActivityDate="2011-02-02T15:23:57.770" Title="Is relativity necessary for the existence of life?" Tags="&lt;general-relativity&gt;&lt;special-relativity&gt;&lt;universe&gt;&lt;biophysics&gt;" AnswerCount="8" CommentCount="7" FavoriteCount="4" />
  <row Id="3403" PostTypeId="2" ParentId="3314" CreationDate="2011-01-20T08:36:34.650" Score="0" ViewCount="" Body="&lt;p&gt;According to the usual laws of physics, charge cannot be created or destroyed. So it's impossible to create a particle with charge q without annihilating a particle with the same charge, or also creating a particle with charge -q, or some combination of these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you annihilated a particle with the same charge. Then your problem might better be described as one of &quot;which particle should you attribute the electrostatic energy to, the annihilated one or the created one?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if instead it was a case of creating two particles with charges +q and -q, then your problem could be described as one of having to separate the electrostatic energies of the two particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I don't see that there is any real problem with defining the energy of the electrostatic field in a particle. It's a matter of semantics. I think the universe has no opinion as to which particle we attribute the electrostatic energy.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-20T08:36:34.650" CommentCount="3" />
  <row Id="3404" PostTypeId="2" ParentId="1453" CreationDate="2011-01-20T09:11:28.903" Score="6" ViewCount="" Body="&lt;p&gt;Find a consistent and complete theory of quantum gravity combining quantum mechanics with general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, our universe is described by both quantum mechanics and general relativity. So, a more complete theory of the universe would have to incorporate quantum gravity somehow. However, combining the two has resisted decades of effort so far, and consistency and completeness turn out to be extremely stringent criteria in this case.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-21T16:57:37.063" LastActivityDate="2011-01-21T16:57:37.063" CommunityOwnedDate="2011-01-22T18:11:53.713" CommentCount="2" />
  <row Id="3405" PostTypeId="2" ParentId="3401" CreationDate="2011-01-20T09:15:10.903" Score="2" ViewCount="" Body="&lt;p&gt;I tend to think that you're right that the phase - but not the absolute value - may be decomposed into infinitesimal pieces of solid angle. And for each tiny solid angle, the phase you get is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not hard to see what happens with the $|jm\rangle$ eigenstate under a parallel transport around an infinitesimal solid angle $\Omega$. It simply gets transformed by the phase &#xA;$$\exp(i m \Omega).$$&#xA;In your notation, $n_{jm}=m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's because it reduces to exponentials of small rotations around $x,y,-x,-y$ axes, and the commutator boils down to $J_z$ for the usual reasons. For your example, $\Omega$ was $4\pi/8=\pi/2$ and $m$ was $1/2$. That's why you got $\exp(\pi i /4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, I realize that this answer seems to contradict the previous problem you posed because $\exp(4\pi i m)$ equals one for any conceivable value of $m$, so there should be no obstruction ever. So at this moment, I am not sure which answer is the right one. I would guess that my answer to your previous problem is completely wrong - there may exist bundles that can't be obtained simply as products of the non-existing low-spin bundles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there could also be an issue with the overall sign. Note that if the inner products between the two states remain real for any angle, you eventually hit the point where the inner product goes negative, but at this point, the absolute value changes the sign relatively to the inner product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-20T09:15:10.903" CommentCount="1" />
  <row Id="3406" PostTypeId="2" ParentId="1453" CreationDate="2011-01-20T09:15:59.883" Score="7" ViewCount="" Body="&lt;p&gt;Why is the cosmological constant $10^{-120}$ times smaller than the Planck scale? Why is the electroweak scale so many orders of magnitude smaller than the Planck scale? Why is the QCD scale comparable to the electroweak scale?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cosmological constant problem is especially acute because of zero-point energy corrections coming from quantum field theory. In a nonsupersymmetric theory, with a mismatch between the number of bosonic and fermionic fields, this would require an incredible fine-tuning. Even in a nonsupersymmetric theory where the number of bosonic and fermionic fields match, unless the masses also match, we still require an enormous fine-tuning. Even if we have supersymmetry, it has to be broken below the TeV scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dynamical mechanisms to solve this problem typically run into problems with the renormalization group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do the anthropic principle and the multiverse answer these questions?&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-21T16:53:59.847" LastActivityDate="2011-01-21T16:53:59.847" CommunityOwnedDate="2011-01-22T18:11:53.713" />
  <row Id="3407" PostTypeId="2" ParentId="3390" CreationDate="2011-01-20T09:29:20.530" Score="7" ViewCount="" Body="&lt;p&gt;Ok, since wsc gave you an explanation of Deutsch's algorithm, let me talk you through another, and more useful, algorithm: Grover's search algorithm. This is a search algorithm for an unsorted database, or for searching a black box for the input which results in a specific output (a very useful primitive in algorithms). For a set of $N$ entries, assuming there is only entry satisfying the search query, the algorithm takes only $O(\sqrt{N})$ queries, where as the best possible classical algorithm takes $O(N)$ queries of the database. Further, Grover's algorithm (with a small modification I won't go into here) is provably optimal. So, how does it work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume there is a database of $2^n$ entries. First we prepare a superposition of $2^n$ classical states with positive phase: $\mid S \rangle = 2^{-\frac{n}{2}} \sum_{x=1}^{2^n} | x \rangle$. This is can be done simply by applying a single quantum gate (in this case a Hadamard gate) to each of $n$ qubits initially prepared in the zero state. I'll represent the states encountered by a diagram, where the length of each line represents it's amplitude in the superposition, and whether it is direction from the center line to indicate its phase (in this case it will always be either positive or negative). The state we obtain after this initialisation step is then depicted as below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/nG7m9.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The algorithm itself consists of $r$ rounds which each consist of two steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The phase of a classical state in the superposition is flipped if (and only if) the entry in the database at that location matches the search query, as illustrated below. Here we assume entry $m$ is the entry matching the search query. Note that this is only a single query to the database, which is having different effects in each branch of the wavefunction. This is achieved by applying an operator $U_m = I - 2 | m \rangle \langle m |$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/psGuM.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The 'inversion about the mean' operator is applied. This is simply a unitary operator given by $U_S = 2 | S \rangle \langle S | - I$, where as before $| S \rangle = 2^{-\frac{n}{2}} \sum_{x=1}^{2^n} | x \rangle$. The effect of this operator is very simple: the amplitude $A_i$ of each entry $i$ is taken to $2\mu - A_i$, where $mu$ is the average amplitude. This is equivalent to inverting the distance above or below the mean. The first diagram below illustrates where the mean now lies, while the second shows the effect of applying this operator.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Mllrq.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/oeVIM.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly the effect of this operation is to amplify the amplitude of the entry matching the search query, and this will continue until the mean lies below the zero line (after $r$ rounds). This occurs after $O(\sqrt{N})$ queries. To see this, we first note that the state of the system after each round $i$ can always be written in the form $|\psi_i\rangle = \cos \theta_i |s\rangle + \sin \theta_i |m\mid$, where $|s\rangle = (2^n - 1)^{-\frac{1}{2}} \sum_{x\neq m} | x \rangle$. Clearly $|s\rangle$ and $|m\rangle$ are orthogonal, so we can represent any state encountered as a unit vector in the 2D plane spanned by $|s\rangle$ and $|m\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/9Z0XU.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider the effect of a single round. First $U_m$ is applied and then $U_S$ is applied. Note that $U_S = 2 |S\rangle\langle S| - I = (\frac{2(2^n - 1)}{2^n}-1)|s\rangle\langle s | + \frac{2\sqrt{2^n - 1}}{2^n}|s\rangle\langle m | + \frac{2\sqrt{2^n - 1}}{2^n}|m\rangle\langle s | + (\frac{2}{2^n} - 1)|m\rangle\langle m |$. When we multiply this by $U_m$ to get the total operation applied in a round $U_S U_m = (\frac{2(2^n - 1)}{2^n}-1)|s\rangle\langle s | - \frac{2\sqrt{2^n - 1}}{2^n}|s\rangle\langle m | + \frac{2\sqrt{2^n - 1}}{2^n}|m\rangle\langle s | + (\frac{2(2^n - 1)}{2^n})|m\rangle\langle m |$.&#xA;This is simply a rotation through a constant angle $\phi=\sin^{-1}(\frac{2\sqrt{2^n - 1}}{2^n}) \approx \frac{2}{\sqrt{2^n}}$ for large $n$. Since we initially start very close to the state $|s\rangle$, we thus require $r \approx \frac{2}{\sqrt{2^n}}$, meaning that we achieve the maximum amplitude for $|m\rangle$ in just ~$2\sqrt{N}$ queries of the database.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that this is impossible classically, since if you make $m$ queries to the database, there are still $N-m$ entries which have not been checked which may possibly match the search term, meaning that a linear number of entries need to be checked to insure even a constant probability of finding the correct entry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this is useful&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2011-01-20T09:29:20.530" CommentCount="1" />
  <row Id="3408" PostTypeId="2" ParentId="3394" CreationDate="2011-01-20T09:38:01.177" Score="0" ViewCount="" Body="&lt;p&gt;With a massless charged fermion, the electromagnetic coupling strength will be totally screened, with $1/e^2$ increasing logarithmically with the scale $R$. The electric field will be Coulomb multiplied by this logarithmic factor in $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, the massless charged fermion will be travelling at the speed of light, but this logarithmic screening will still happen.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-01-20T09:38:01.177" />
  <row Id="3409" PostTypeId="2" ParentId="2597" CreationDate="2011-01-20T10:38:34.207" Score="5" ViewCount="" Body="&lt;p&gt;Energy conservation does work perfectly in general relativity. The overall Lagrangian is invariant under time translations and Noether's Theorem can be used to derive a non-trivial and exact conserved current for energy. The only thing that makes general relativity a little different from electromagnetism is that the time translation symmetry is part of a larger gauge symmetry so time is not absolute and can be chosen in many ways. However there is no problem with the derivation of conserved energy with respect to any given choice of time translation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a long and interesting history to this problem. Einstein gave a valid formula for the energy in the gravitational field shortly after publishing general relativity. The mathematicians Hilbert and Klein did not like the coordinate dependence in Einstein's formulation and claimed it reduced to a trivial identity. They enlisted Noether to work out a general formalism for conservation laws and claimed that her work supported their view. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The debate continued for many years especially in the context of gravitational waves which some people claimed did not exist. They thought that the linearised solutions for gravitational waves were equivalent to flat space via co-ordinate transformations and that they carried no energy. At one point even Einstein doubted his own formalism, but later he returned to his original view that energy conservation holds up. The issue was finally resolved when exact non-linear gravitational wave solutions were found and it was shown that they do carry energy. Since then this has even been verified empirically to very high precision with the observation of the slowing down of binary pulsars in exact agreement with the predicted radiation of gravitational energy from the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula for energy in general relativity is usually given in terms of pseudo tensors such as those proposed by Laundau &amp;amp; Lifshitz, Dirac, Weinberg or Einstein himself. Wikipedia has a &lt;a href=&quot;http://en.wikipedia.org/wiki/Stress-energy-momentum_pseudotensor&quot; rel=&quot;nofollow&quot;&gt;good article&lt;/a&gt; on these and how they confirm energy conservation. Although pseudotensors are mathematically rigorous objects which can be understood as sections of jet bundles, some people don't like their apparent co-ordinate dependence. There are other covariant approaches such as the Komar Superpotential or a more &lt;a href=&quot;http://www.prespacetime.com/index.php/pst/article/view/81&quot; rel=&quot;nofollow&quot;&gt;general formula of mine&lt;/a&gt; which gives the energy current in terms of the time translation vector $k^{\mu}$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ J^{\mu}_G = \frac{1}{16\pi G} (k^{\mu}R - 2k^{\mu}\Lambda - 2{{k^{\alpha}}_{;\alpha}}^{\mu} + {{k^{\alpha}}_{;}}^{\mu}_{\alpha}+ {{k^{\mu}}_{;}}^{\alpha}_{\alpha})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite these general formulations of energy conservation in general relativity there are some cosmologists who still take the view that energy conservation is only approximate or that it only works in special cases or that it reduces to a trivial identity. In each case these claims can be refuted either by studying the formulations I have referenced or by comparing the arguments given by these cosmologists with analogous situations in other gauge theories where conservation laws are accepted and follow analogous rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One area of particular contention is energy conservation in a homogeneous cosmology with cosmic radiation and a cosmological constant. Despite all the contrary claims, a valid formula for energy conservation in this case can be derived from the general methods and is given by this equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ E = Mc^2 + \frac{\Gamma}{a} + \frac{\Lambda c^2}{\kappa}a^3 - \frac{3}{\kappa}\dot{a}^2a - Ka = 0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a(t)$ is the universal expansion factor as a funcrtion of time normalised to 1 at the current epoch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E$ is the total energy in an expanding region of volume $a(t)^3$. This always comes to zero in a perfectly homogeneous cosmology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M$ is the total mass of matter in the region&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c$ is the speed of light&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Gamma$ is the density of cosmic radiation normalised to the current epoch&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Lambda$ is the cosmological constant, thought to be positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\kappa$ is the gravitational coupling constant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K$ is a constant that is positive for spherical closed space, negative for hyperbolic space and zero for flat space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first two terms describe the energy in matter and radiation with the matter energy not changing and the radiation decreasing as the universe expands. Both are positive.&#xA;The third term is &quot;dark energy&quot; which is currently though to be positive and contributing about 75% of the non-gravitational energy, but this increases with time.&#xA;The final two terms represent the gravitational energy which is negative to balance the other terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation holds as a consequence of the well-known &lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;nofollow&quot;&gt;Friedmann cosmological equations&lt;/a&gt;, that come from the Einstein field equations, so it is in no sense trivial as some people have claimed it must be.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-20T10:38:34.207" CommentCount="5" />
  <row Id="3410" PostTypeId="2" ParentId="3359" CreationDate="2011-01-20T11:15:21.353" Score="1" ViewCount="" Body="&lt;p&gt;Ell, I sympathise. You look at the graphs showing an oscillating electromagnetic wave and misread the transverse axes as spatial. So you think an electromagnetic wave is like the wave you can flick along a skipping rope. No it isn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way to think about it is that at &lt;em&gt;each point in space&lt;/em&gt; where the wave is, there are two little arrows at right angles to each other and to the direction of motion of the wave. One arrow is the electric field, the other is the magnetic field. Suppose at some time the electric field arrow points to the right. As you watch, it gets shorter and shorter until it momentarily vanishes and then grows in the opposite direction to the left to the same length. Then it starts getting smaller again until it sloshes back to its original position. Meanwhile the other arrow, the magnetic field arrow at right angles, is sloshing upwards and downwards in a similar fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that &lt;strong&gt;you are to imagine the arrows as NOT extending in space around the point&lt;/strong&gt; - they really correspond to moment-by-moment meter-readings of the field strengths constituting the wave at exactly that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Note this corresponds to a plane-polarised wave - I'm trying not to make it too complicated].&lt;/p&gt;&#xA;" OwnerUserId="1394" LastEditorUserId="1394" LastEditorDisplayName="" LastEditDate="2011-01-20T11:21:12.420" LastActivityDate="2011-01-20T11:21:12.420" CommentCount="1" />
  <row Id="3411" PostTypeId="1" AcceptedAnswerId="3417" CreationDate="2011-01-20T11:16:16.200" Score="5" ViewCount="234" Body="&lt;p&gt;In 2010 there were press reports that CERN had identified unusual properties in particle behavour in collisions. One link &lt;a href=&quot;http://www.bbc.co.uk/news/science-environment-11391434&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a partial quote:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In some sense, it's like the particles talk to each other and they decide which way to go,&quot; explained CMS Spokesperson Guido Tonelli.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;There should be a dynamic mechanism there somehow giving them a preferred direction, which is not down to the trivial explanation of momentum and energy conservation. It is something which is so far not fully understood,&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been unable to find any scientific papers on this however. So has anyone read a scientific paper on this topic and able to enlighten?&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-30T09:03:04.527" LastActivityDate="2011-02-02T15:44:35.700" Title="Unusual particle effects at CERN" Tags="&lt;cern&gt;&lt;particles&gt;&lt;lhc&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="3412" PostTypeId="1" AcceptedAnswerId="3420" CreationDate="2011-01-20T12:26:06.370" Score="1" ViewCount="118" Body="&lt;p&gt;I had an idea how to design a &quot;quantum computer&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about designing interference-experiments where the design of the experiments itself represents algorithmical or mathematical problems that are coded in a way that the solution to that problem is interference = “true” and no interference = “false” as a result?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These results would be acquired instantaneously because the algorithm would not be processed step by step as with classical computers. If the design was changed on the run - thereby representing another problem - the solution would even be acquired faster than the quanta going through the apparatus as predicted by “delayed choice”-experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there already such ideas for “quantum computers” of this kind? They would not have the problem of decoherence but on the contrary decoherence vs. no-decoherence would be their solution set of the problems represented by the (flexible) experimental design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this make sense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Perhaps my question is not formulated well - perhaps I shouldn't call it &quot;quantum computer&quot;. My point is that any configuration with these double slit experiments - no matter how complicated - instantly shows an interference pattern when there is any chance of determining the way the photons took. My idea is to encode some kind of calculation or logical problem in this configuration and instantly find the answer by looking at the outcome: interference = “yes” and no interference = “no” – &lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2011-01-20T14:43:08.233" LastActivityDate="2011-01-20T17:28:36.077" Title="Using delayed choice interference experiments as a computing device" Tags="&lt;quantum-information&gt;&lt;quantum-computer&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="3413" PostTypeId="2" ParentId="3362" CreationDate="2011-01-20T13:54:08.670" Score="1" ViewCount="" Body="&lt;p&gt;I think that this question should be considered also as a materials physics question. There will certainly be a very high energy and a momentum hundreds of times greater than a traditional asteroid. The intention of the question is that the material not break up before collision. What happens at and after collision is probably vaporization. I say probably because a massless neutrino object could travel through the Earth. Otherwise the vaporization will happen quickly (nanoseconds), but not instantly. So can a material exist which will fail to vaporize until it has burrowed through the Earth some distance? There are discoveries of non-standard materials like metamaterials, but these as yet would be unsuitable. So it might be a research project to find such a material - or prove that none could exist (a proof which would require a broader theory of matter than available today as it would need to incude Dark Matter theory as well).&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-01-20T13:54:08.670" CommentCount="2" />
  <row Id="3414" PostTypeId="2" ParentId="1453" CreationDate="2011-01-20T13:56:09.970" Score="5" ViewCount="" Body="&lt;p&gt;Why was the initial state of the universe so special? Why was the initial entropy so low? See recent book by Sean Carroll. &lt;/p&gt;&#xA;" OwnerUserId="1458" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-22T23:00:05.477" LastActivityDate="2011-01-22T23:00:05.477" CommunityOwnedDate="2011-01-22T18:11:53.713" CommentCount="0" />
  <row Id="3415" PostTypeId="2" ParentId="3216" CreationDate="2011-01-20T14:16:23.883" Score="2" ViewCount="" Body="&lt;p&gt;The statement you quote is a truism, so not much science in it. There are however two ways you can extract science from the anthropic principle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is that it can give you non-trivial constraints on parameters in current theories. In most cases the existence of intelligent life itself is too much of a detail that enters the argument. It's more that you'd be saying we have observed that the universe is such that e.g. carbon exists in such and such an abundance, and this is only possible if parameter x is in some specific range, so you've constrained your parameter. That constraint might be more or less interesting, but nothing unscientific about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second one is that you might try to come up with a theory in which the &quot;evolution of life&quot; is well-defined in scientific terms, such that you can argue there's some optimization principle going on which results in some specific theory, that being the one we observe. However, there's some arguments that our standard model might not be the only one supporting the evolution of life (Google for: Weakless Universe). In any case, taking this road is going to involve a lot more than only physics because first you have to find a way to define &quot;life&quot; in mathematical terms.&lt;/p&gt;&#xA;" OwnerUserId="1458" LastActivityDate="2011-01-20T14:16:23.883" CommentCount="1" />
  <row Id="3416" PostTypeId="2" ParentId="3412" CreationDate="2011-01-20T14:16:31.183" Score="3" ViewCount="" Body="&lt;p&gt;no, you cannot represent a quantum bit in a quantum computer by &quot;coherence vs decoherence&quot; because any decoherence means that the computer ceased to be a quantum computer. So once the first &quot;zero&quot; would appear in your computer, you would start to get incorrect results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your answer to kakemonsteret that you may &quot;detect interference by seeing an interference pattern&quot; is a misunderstanding what should be seen and what is seen in quantum mechanics. In quantum mechanics, one only sees an interference pattern if the very same experiment is repeated many times. So this would be needed to determine the value of the qubit - which would also become a classical bit - for every operation, so the number of repetitions would grow exponentially with the number of operations needed to perform the quantum algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It just doesn't work. Decoherence may only occur at the end of the calculation in a quantum computer - when the result is being read. Any decoherence before that means that the quantum computer has broken down. Moreover, &quot;coherence vs decoherence&quot; is not really a quantum bit to start with - because you can't create linear superpositions of these two &quot;states&quot; with controllable phases - decoherence always has an undetermined phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-20T14:16:31.183" CommentCount="3" />
  <row Id="3417" PostTypeId="2" ParentId="3411" CreationDate="2011-01-20T14:21:49.977" Score="3" ViewCount="" Body="&lt;p&gt;The BBC report is so vague that one can't really determine what was observed. But here is a more coherent report:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2010/09/lhc-probably-sees-new-shocking-physics.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/09/lhc-probably-sees-new-shocking-physics.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The text above also contains the links to the scientific preprint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cms.web.cern.ch/cms/News/2010/QCD-10-002/QCD-10-002.pdf&quot; rel=&quot;nofollow&quot;&gt;http://cms.web.cern.ch/cms/News/2010/QCD-10-002/QCD-10-002.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is called&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Observation of Long-Range, Near-Side Angular Correlations in Proton-Proton Collisions at the LHC&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which kind of explains what was seen, too. But see my blog above for more details.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T15:44:35.700" LastActivityDate="2011-02-02T15:44:35.700" CommentCount="2" />
  <row Id="3418" PostTypeId="2" ParentId="2166" CreationDate="2011-01-20T14:50:05.237" Score="0" ViewCount="" Body="&lt;p&gt;One of the things that often get lost when talking about time travel is that actually the observer continues in his/her own time (!) but the rest of the universe goes back! So it is not &quot;you&quot; who is travelling back in time - on the contrary: if the was a shift in the time of the universe nobody would even realize!&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2011-01-20T14:50:05.237" />
  <row Id="3419" PostTypeId="2" ParentId="2597" CreationDate="2011-01-20T15:38:42.300" Score="1" ViewCount="" Body="&lt;p&gt;o carry on with Jerry Schirmer, the Killing vector defines isometries on a manifold.  If there exists a Killing vector $K_t~=~\partial/\partial t$ this means the momentum $K_t\cdot P~=~$ constant.  This then is a statement which can be interpreted as the constancy of an observable labeled energy.  As a rule of thumb, if a metric component involves time in an explicit way, and for example $K_t~\propto~\sqrt{g_{tt}(t)}$ is not proper or the action of this vector is not an isometry.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happens with the FLRW equation of cosmology.  In a de Sitter form we have &#xA;$$&#xD;&#xA;ds^2~=~dt^2~-~e^{\sqrt{\Lambda/3}t}(dr^2~+~r^2d\Omega^2),&#xD;&#xA;$$&#xA;which has a time dependency.  So we are not able to derive a conservation of energy from first principles.  The Ricci curvature is $R_{\mu\nu}~=~\Lambda g_{\mu\nu}$, and for $k~=~0$ the spatial curvature is zero.  The cosmological constant is dependent on the vacuum energy density, plus pressure terms.  With the equation of state $p~=~-\rho$, which approximates observational data pretty well, one can do some detailed balance work to show the universe is a net nothing and remains so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this connects with something deeper than just a &quot;detailed balance?  It might, and I suspect that Phillip's analysis connects with this.  The deSitter metric is a time dependent conformal theory of a flat metric.  For $g&amp;#39;~=~\Omega^2g$ the line element for $g&amp;#39;$ &#xA;$$&#xD;&#xA;ds^2~=~\Omega^2(du^2~-~d\sigma_{space}^2).&#xD;&#xA;$$&#xA;However for the time variable $du^2~=~\Omega^{-2}dt^2$ $\Omega$ is time dependent and&#xA;$$&#xD;&#xA;ds^2~=~dt^2~-~\Omega^2(t)d\sigma_{space}^2.&#xD;&#xA;$$&#xA;This recovers the de Sitter metric for $\Omega^2(t)~=~exp(\sqrt{\Lambda/3}t)$.  The de Sitter spacetime is then conformally equivalent to a flat spacetime which trivially has a $K_t~=~\partial/\partial t$.  So the spacetime we observe with the equation of state $p~=~-\rho$ is a class of spacetimes which are conformal to flat spacetime and which also preserve $E~=~constant$. I think that Phillip’s work on this matter projects out this special case of conformal spacetimes.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-01-20T15:55:25.783" LastActivityDate="2011-01-20T15:55:25.783" />
  <row Id="3420" PostTypeId="2" ParentId="3412" CreationDate="2011-01-20T15:48:46.350" Score="5" ViewCount="" Body="&lt;p&gt;Here is my understanding of what you are asking (and I believe it is a little different to Lubos's interpretation, so our answers will differ): Can you build a computer that uses the interference effects to perform computation, where you use the presence of light in a particular place to represent 1 and no light to represent 0?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is yes, though with certain caveats. First let me note that there is something already called a quantum computer which exploits quantum effects to outperform normal (classical) computers. Classical computation is a special case of quantum computation, so a quantum computer can do everything a classical computer can do, but the converse is not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have single photon detectors, you can use the interference effects of a network of beam splitters and phase plates together with the detectors to create a universal quantum computer. This is something called linear-optics quantum computing or LOQC. Perhaps the best known scheme is the &lt;a href=&quot;http://www.nature.com/nature/journal/v409/n6816/abs/409046a0.html&quot; rel=&quot;nofollow&quot;&gt;KLM proposal&lt;/a&gt;, due to Knill, Laflamme and Milburn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the caveats: you need to have a fixed finite number of photons, and you need to adapt the network based on earlier measurement results. This adaptive feed-forward is quite difficult to achieve in practice, though not impossible, and computation with such a setup has been demonstrated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A further interpretation of the question is whether it is sufficient to use such a linear network, but only make measurements at the end. This is an open question, though there is strong evidence that such a device is not efficiently simulable by a classical computer (see &lt;a href=&quot;http://www.scottaaronson.com/papers/optics.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; by Scott Aaronson). It is however not yet known whether you can implement universal classical or quantum computing on such a device.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditorDisplayName="" LastEditDate="2011-01-20T17:28:36.077" LastActivityDate="2011-01-20T17:28:36.077" CommentCount="1" />
  <row Id="3422" PostTypeId="2" ParentId="3402" CreationDate="2011-01-20T16:07:11.733" Score="0" ViewCount="" Body="&lt;p&gt;The main problem is that the sun wouldn't work - no mass-energy conversion. Recall all the unusual and baroque theories of the source of solar radiation before relativity and QM (a giant coal furnace, gravitational collapse, ...).&lt;/p&gt;&#xA;" OwnerUserId="1394" LastActivityDate="2011-01-20T16:07:11.733" CommentCount="3" />
  <row Id="3423" PostTypeId="2" ParentId="2166" CreationDate="2011-01-20T16:11:07.827" Score="3" ViewCount="" Body="&lt;p&gt;There are solutions to Einstein's field equations, which have closed timelike curves. For example Godel's solution. Would that constitute time travel, if you can reach the same point on your world line in finite time? One objection may be that such solution do not describe the universe, but examples as the Tipler's cylinder suggest that at least in theory we could modify parts of the universe to get closed timelike curves. Another objection may be that to talk about past and future you need linear order on time moments, which is not the case for closed curves.  &lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2011-01-20T16:11:07.827" />
  <row Id="3424" PostTypeId="2" ParentId="770" CreationDate="2011-01-20T16:11:30.353" Score="1" ViewCount="" Body="&lt;p&gt;Dear Splashhit,&#xA;the easiest way to destroy the Universe is to continue to drive cars for years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether you do it or not, the global mean temperature will increase from 15 °C to 16 °C or so in a century. When the temperature of the materials and organisms increases by as much as one Celsius degree, they will - according to the IPCC - transform into a deadly plasma that will kill all living life forms and evaporate all stars, ending with a complete cataclysm where the vacuum itself (scared by watching the horror on its most important planet) is eaten by nothingness. If you've ever experienced the staggering heat of 16 °C, you must know what terrible thing I am talking about. It's worse than hell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just kidding. There is obviously no threat whatsoever that would be linked to &quot;climate change&quot; as this meaningless tautology is often called. The IPCC is not destroying the Universe but it is surely destroying the ability of the people to think rationally about very basic things - and it is trying to destroy the world economy, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-20T16:11:30.353" CommunityOwnedDate="2011-01-20T16:11:30.353" />
  <row Id="3425" PostTypeId="2" ParentId="3402" CreationDate="2011-01-20T16:20:42.853" Score="-1" ViewCount="" Body="&lt;p&gt;Even with relativity organisms will meet different gravitational potentials at different altitudes. This reduces air density and pressure and causes breathing difficulties for nonadapted species.&#xA;Without General Relativity (or a similar theory) we might also have a problem putting together a Cosmological Theory. Thus the Universe might not get started at all....&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-01-20T16:20:42.853" CommentCount="5" />
  <row Id="3426" PostTypeId="2" ParentId="770" CreationDate="2011-01-20T16:25:07.233" Score="4" ViewCount="" Body="&lt;p&gt;You can't. Anything you do will only have an effect that effects an area that expands (at most) at the speed of light, so it only effects at most a sphere of radius $c t$ after time $t$. Anything else would violate the no-signalling principle which appears fundamental in physics. This means that the photon that just bounced off you into space will outrun the edge of any such an attempt to destroy the universe.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2011-01-20T16:25:07.233" CommunityOwnedDate="2011-01-20T16:25:07.233" />
  <row Id="3427" PostTypeId="2" ParentId="770" CreationDate="2011-01-20T16:39:41.887" Score="0" ViewCount="" Body="&lt;p&gt;Find a logical inconsistency and use it to prove that the universe does not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could kill all observers, that would make the existence of the universe an unprovable proposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately you can just wait for the Big Crunch.&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2011-01-20T16:52:44.717" LastActivityDate="2011-01-20T16:52:44.717" CommunityOwnedDate="2011-01-20T16:39:41.887" CommentCount="1" />
  <row Id="3428" PostTypeId="2" ParentId="3158" CreationDate="2011-01-20T16:49:06.533" Score="0" ViewCount="" Body="&lt;p&gt;I think that the best picture to understand this correlation is given by many-worlds interpretation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A singlet decomposes in a coupled pair of particles superposition |+⟩(A)|-⟩(B) + |-⟩(A)|+⟩(B)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So observer A sees a simple superposition of |+⟩ + |-⟩ (which is a partial trace of the global density matrix) and so do B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in many worlds interpretation; observer A will be split in a + and a - observer (and so will observer B). Now, where the correlation effect manifest itself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the 'coupling' effect is brought when observer A and observer B join together at subluminal speeds to compare notes of their measurements: (remember that according to many-worlds, we have two observers A and two observers B) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;observer A+ is disallowed by angular momentum conservation to interact with observer B+, (otherwise they will both agree that angular momentum was not conserved). Likewise, observer A- is disallowed to interact with observer B- by the same reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the remaining interactions between observers are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A+ interacts with B-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A- interacts with B+&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the final state is a superposition of |A+⟩|B-⟩ and |A-⟩|B+⟩, Which is interpreted as a 'correlation between remote observations'&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2011-01-27T15:12:06.260" LastActivityDate="2011-01-27T15:12:06.260" />
  <row Id="3429" PostTypeId="2" ParentId="770" CreationDate="2011-01-20T17:17:26.223" Score="0" ViewCount="" Body="&lt;p&gt;Ask God - not so sure he would trust you with those details, but at least you both get a laugh :)&lt;/p&gt;&#xA;" OwnerUserId="1478" LastActivityDate="2011-01-20T17:17:26.223" CommunityOwnedDate="2011-01-20T17:17:26.223" />
  <row Id="3430" PostTypeId="2" ParentId="1528" CreationDate="2011-01-20T17:34:32.777" Score="4" ViewCount="" Body="&lt;p&gt;when physicists played with the strong interactions, the linguistic creativity was at the top, partially due to the influence of people such as Murray Gell-Mann who is a keen linguist (and borrowed the word quark itself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flavor is something that distinguishes the quark species $u,c,t; d,s,b$ - especially in between the generations - and similarly for the charged leptons and neutrinos. They needed a comprehensible word that defines a certain &quot;property&quot; of the particles - but it's a completely new kind of property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the 3 possible QCD-charged versions of the quarks were named colors - red, green, blue are 3 independent colors we can see, so the analogy works so perfectly that sometimes people misunderstand that the 3 quark colors have nothing to do with the real colors haha.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the different generations were called flavors. While colors are kind of interchangeable - you can permute R,G,B to get a similarly meaningful picture, and then transform it back, and correspondingly, the $SU(3)$ symmetry is unbroken - flavors are qualitatively different. Bitter is not the same thing as sweet or sorrow or salty. The physicists clearly enumerated all senses we use to perceive &quot;properties of the world&quot; - vision, hearing, tasting etc. - and the vision (colors) was used for the confined QCD charges while the flavors (tasting) was used to distinguish the particles across the families.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up and down quarks were named in this way because they form the upper and lower component of the isospin doublet. Isospin $SU(2)$ is mathematically fully analogous to the $SU(2)$ group of spin that rotates the real space. A double - the fundamental representation of the spin $SU(2)$ - is conventionally described by a basis in which one state is spinning with spin up and the other has spin pointing down. So they used the same names - written as words - for the isospin as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That was the first two quarks. Hadrons containing the third lightest quark, the strange quark, were behaving a bit strangely, so Gell-Mann gave the strangely (partly) conserved quantum number the name strangeness, and the quark that carried it had to be strange as a consequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It became clear that it had to have an $SU(2)$ partner as well. They looked at the first three quarks and realized that the lower-charged quarks, bottom and strange, were named by some kind of inferior names. So they had to choose a random description for the new superior quark - they chose charm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While inventing names for the third family (or generation - none of those names has been eliminated yet), it was already a naming industry. So they chose truth for the superior species and beauty for the inferior species - because beauty is less important than the truth - apologies to all the hotties who are reading this text. However, it was too playful so the new up-like and down-like names which were more modest and had the same initial letter - top and bottom - were adopted instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many other properties one should use for the new internal quantum numbers. For example, I am convinced that R-parity in supersymmetry should be described as sex. Everything we observe is R-parity even and should be &quot;female&quot; - they have an even number of X-chromosomes. R-parity odd superpartners - to be seen - should be called &quot;male&quot; because they have an odd number of X chromosomes. ;-) If you collide two &quot;male&quot; particles, you may create a female one. But it is not a real sex (like when you emit an LSP): instead, it is a surgery when you pick the X-chromosomes from both fathers haha. Unfortunately, physicists are much less creative than in the 1960s so my terminology won't be adopted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-20T17:34:32.777" CommentCount="1" />
  <row Id="3431" PostTypeId="2" ParentId="3402" CreationDate="2011-01-20T19:06:13.053" Score="0" ViewCount="" Body="&lt;p&gt;There would be the problem of $E~=~mc^2$ and solar power.  We might imagine though some non-relativistic form of nuclear energy.  There is also the likely result that the spectra of elementary particles is determined by gravity.  QCD on the boundary of an AdS spacetime is equivalent to the symmetries of the graviton in the interior.  So without relativity the world might appear very differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is a bit of a “what if?” sort of question.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-20T19:06:13.053" CommentCount="1" />
  <row Id="3432" PostTypeId="1" CreationDate="2011-01-20T19:12:13.177" Score="20" ViewCount="817" Body="&lt;p&gt;What are the Physics/Astrophysics blogs you regularly read? I'm looking to beef up my RSS feeds to catch up with during my long commutes. I'd like to discover lesser-known gems (e.g. not well known blogs such as Cosmic Variance), possibly that are updated regularly and recently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is a list, give one entry per answer please. Feel free to post more than one answer.&lt;/p&gt;&#xA;" OwnerUserId="1252" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-02-10T07:18:21.767" LastActivityDate="2011-02-10T07:18:21.767" CommunityOwnedDate="2011-01-20T21:24:41.327" Title="Your favorite Physics/Astrophysics blogs?" Tags="&lt;soft-question&gt;&lt;big-list&gt;" AnswerCount="20" CommentCount="9" FavoriteCount="11" />
  <row Id="3433" PostTypeId="2" ParentId="3158" CreationDate="2011-01-20T19:18:54.790" Score="15" ViewCount="" Body="&lt;p&gt;I wish to complete @Luboš Motl's answer, to which I agree. My point is on why people continue to make this mistake of an active link. This mistake is connected with one of the most interesting properties of quantum mechanics, &lt;a href=&quot;http://en.wikipedia.org/wiki/Bell%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Bell's theorem&lt;/a&gt;. One can argue that any physical theory is an &lt;em&gt;hidden variable&lt;/em&gt; theory, the hidden variable being the description of the state of an object as written by the theoretician describing it. For quantum theory, the wavefunction of the object is the &lt;em&gt;hidden variable&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bell's theorem state that the prediction of quantum theory cannot be described by any &lt;em&gt;local&lt;/em&gt; hidden variable theory. More precisely, for any entangled state, you can find a set of measurement with statistics contradicting any local hidden variable theory. The two possible explanations are either :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Nature is not local : your physical description is a real physical object, and there is an active non-local link between the two entangled particle.&lt;/li&gt;&#xA;&lt;li&gt;Nature is not realist : your physical state is only an approximation and has no real meaning.&lt;/li&gt;&#xA;&lt;li&gt;Nature is not quantum.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(1) is much easier to explain and appears often in popular science, mainly because (2) is much more difficult to explain and accept. But I think most researcher working with entanglement prefer explanation (2). Einstein intuition was 3 (before Bell's theorem), because he could not accept (1) and (2). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, Einstein 1936 original paper on the &lt;a href=&quot;http://en.wikipedia.org/wiki/EPR_paradox&quot; rel=&quot;nofollow&quot;&gt;EPR paradox&lt;/a&gt; was on a case where you can easily find a local hidden variable theory. The state described it what is now called a two-mode squeezed state. It's Wigner function is positive and can therefore be interpreted as a classical probability distribution on the quadrature (position and momentum) measurements, the only one discussed in the EPR paper. Such classical analysis of entanglement can be theoretically very useful and help the intuition in some case without needing  any &lt;em&gt;spooky action at distance&lt;/em&gt;. However, as shown by Bell, such local hidden variable theory cannot be generic enough to encompass all quantum mechanics. &lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2011-01-20T19:18:54.790" CommentCount="6" />
  <row Id="3434" PostTypeId="2" ParentId="3402" CreationDate="2011-01-20T19:23:00.567" Score="0" ViewCount="" Body="&lt;p&gt;Special Relativistic dynamics is the reason that electrons have spin and behave as fermions. These properties of the electrons are the reason the various elements have different chemical properties or in other word the reason we have chemistry. Chemistry is the reason we have molecules. Without molecules there would be no life, thus without Special Relativity there would be no life.   &lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2011-01-20T19:23:00.567" CommentCount="3" />
  <row Id="3435" PostTypeId="2" ParentId="3357" CreationDate="2011-01-20T19:48:56.543" Score="2" ViewCount="" Body="&lt;p&gt;The preprint is here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1101.2565&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1101.2565&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Time teleportation, as described by the Australian ... physicists ... is a meaningless concept that is based on a misinterpretation of what entanglement is and what it is not. The true portion of their statements is well-known and trivial and they vaguely suggest things that are not true, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Entanglement is just a correlation between two or many objects - it is a correlation that may affect all properties of the two systems - so it allows two objects to be more completely &quot;identified&quot; than in classical physics. However, it is still a correlation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In normal situations, the correlation exists between two spatially separated objects. If one of them is measured and we know the result, we have to use the conditional probability - taking the known result of the measurement into account as a condition - to predict the probabilities of the other measurement. We may also run the logic in the opposite order and we obtain exactly the same results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course that the two objects that were measured may also be separated by time-like separations. The correlation will still exist and the predictions will be correlated just like before. Except that the mystery will evaporate. The entanglement is mysterious - at least for someone - exactly because it seems that one particle has to send a signal about its &quot;decision&quot; to the other particle (it sends no signals!) and such a signal would seemingly have to propagate faster than light which conflicts with relativity (no signal is needed for the correlation!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when the two measurements are time-like separated, there is no paradox, not even for the people who think that the entanglement needs &quot;signals&quot;. Even if a signal were needed, it could get from one particle to the other particle by a speed that is smaller than the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &quot;teleportation in time&quot; is just the same entanglement minus all the mystery that results from the spacelike separation, and such a &quot;teleportation in time&quot; may be viewed as nothing else than the later particle's behavior according to the correlated properties with the first particle that has decided about both. There is nothing mysterious about these things and it is not useful for anything because such an &quot;entanglement in time&quot; may be replaced by an ordinary storage of information. &lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:39:12.457" LastActivityDate="2011-02-04T07:39:12.457" CommentCount="2" />
  <row Id="3436" PostTypeId="1" AcceptedAnswerId="3437" CreationDate="2011-01-20T19:59:51.443" Score="5" ViewCount="647" Body="&lt;p&gt;I'm a big fan of the podcast Astronomy Cast and a while back I was listening to a Q&amp;amp;A episode they did. A listener sent in a question that I found fascinating and have been wondering about ever since. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the show &lt;a href=&quot;http://www.astronomycast.com/listeners/questions-shows/questions-show-decelerating-black-holes-earth-sun-tidal-lock-and-the-crushing-gravity-of-dark-matter/&quot; rel=&quot;nofollow&quot;&gt;transcript&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Arunus Gidgowdusk from Lithuania asks:&#xA;  “If you took a one kilogram mass and&#xA;  accelerated it close to the speed of&#xA;  light would it form into a black hole?&#xA;  Would it stay a black hole if you then&#xA;  decreased the speed?”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dr. Gay, an astrophysicist and one of the hosts, explained that she'd asked a number of her colleagues and that none of them could provide a satisfactory answer. I asked her more recently on Facebook if anyone had come forward with one and she said they had not.  So I thought maybe this would be a good place to ask.&lt;/p&gt;&#xA;" OwnerUserId="1479" LastActivityDate="2011-01-21T07:32:21.357" Title="If a 1kg mass was accelerated close to the speed of light would it turn into a black hole?" Tags="&lt;black-holes&gt;&lt;speed-of-light&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="3437" PostTypeId="2" ParentId="3436" CreationDate="2011-01-20T20:10:10.213" Score="14" ViewCount="" Body="&lt;p&gt;The answer is no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest proof is just the principle of relativity: the laws of physics are the same in all reference frames. So you can look at that 1-kg mass in a reference frame that's moving along with it. In that frame, it's just the same 1-kg mass it always was; it's not a black hole.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-01-20T20:10:10.213" CommentCount="3" />
  <row Id="3439" PostTypeId="2" ParentId="3436" CreationDate="2011-01-20T20:18:51.433" Score="5" ViewCount="" Body="&lt;p&gt;No, a 1kg mass would not turn into a black hole, even if it were zipping past you at very close to the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/31/lay-explanation-of-the-theory-of-relativity&quot;&gt;The principle of relativity&lt;/a&gt; is a fundamental idea in physics, and one consequence of it is that we can understand the physics of something that's moving by imagining we're moving alongside it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, you are watching people play pool on a train as it rushes past you.  You want to know whether a certain shot that's just been made will sink the 8-ball.  You figure it out by imagining you're inside the train and calculating everything you'd expect to happen from that simpler viewpoint where the pool table is stationary.  If the 8-ball goes into a certain pocket from that point of view, you can rest assured it will go into the same pocket if you analyze the situation again from your original vantage point on terra firma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the same principle to the 1kg mass, we see that moving along side it, it just looks like a normal mass, not a black hole.  Hence, from another point of view in which it moves close to the speed of light, it still looks like a normal mass, not a black hole.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-01-20T20:18:51.433" CommentCount="2" />
  <row Id="3441" PostTypeId="2" ParentId="3436" CreationDate="2011-01-20T20:55:15.243" Score="-1" ViewCount="" Body="&lt;p&gt;I am presuming the idea is the 1kg mass will length contract to below the Planck length.  It is either that or the relativistic energy (mass) $E~=~\gamma mc^2$ would be so large it would gravitationally implode.  The question though can be thought of according to what would happen to an observer on the mass.  The question could be turned around: Would the universe implode?  If a mass $M$ passes by a smaller mass $m~&amp;lt;&amp;lt;~M$ then one might think that $M$ could become a black hole and the small mass $m$ if close enough would become trapped in the black hole.  However, from the frame of the big mass $M$ the small mass is not a black hole.  This is a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ultra-relativistic mass will behave similar to a gravity wave as it passes another reference point. This Aichelburg-Sexl ultraboost has a plane wave pulse of spacetime.  The relativistic mass will result in a gravity wave pulse as detected by a stationary observer.  So there is a gravitational implication to such extreme relativistic boosts.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-20T20:55:15.243" />
  <row Id="3442" PostTypeId="1" AcceptedAnswerId="3446" CreationDate="2011-01-20T21:04:16.573" Score="0" ViewCount="219" Body="&lt;p&gt;Right now I am having this silly difficulty from the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;BTW, Conformal dimension/scaling dimension is -ve of mass dimension ..right?&lt;/li&gt;&#xA;&lt;li&gt;In p-63 of &lt;a href=&quot;http://arxiv.org/abs/hep-th/9905111&quot; rel=&quot;nofollow&quot;&gt;Magoo&lt;/a&gt;, after 3.15 eq, they said&#xA;a.) $\phi$ is dimensionless..why?&#xA;b.) They, after getting length dimension of boundary field as $\Delta d$, referred to 3.13 to comment that O operator has conformal dimension $\Delta$. I know its right..But as in the lhs of 3.13, the exponent must be dimension less, it seems O should have length or conformal dimension to be $-\Delta$..&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I know I am making a silly mistake here..Thanks for answering.&lt;/p&gt;&#xA;" OwnerUserId="1349" LastActivityDate="2011-02-04T07:39:55.543" Title="Question on dimensions of CFT operators (ref: MAGOO, hep-th/9905111)" Tags="&lt;string-theory&gt;&lt;conformal-field-theory&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="3443" PostTypeId="1" AcceptedAnswerId="3492" CreationDate="2011-01-20T21:34:49.703" Score="2" ViewCount="151" Body="&lt;p&gt;Consider two supermassive black holes of equal mass orbiting about their common centre of mass. Is it the case that a free-fall trajectory along the axis of rotation would be outside of either event horizon at all black hole separation distances &gt; 0 (based on the symmetry of the situation)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To rephrase this, you would be able to navigate a rocket along a path at right angles to the orbital plane and bisecting the line between the two black holes with no ill effects whatsoever even when the BHs were very close to each other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supplementary question:  what’s the shape of each of the event horizons prior to coalescence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I’m assuming that tidal forces would be small until the two singularities were extremely close. Note also that the point midway between the two BHs is the L1 Lagrangian point.&lt;/p&gt;&#xA;" OwnerUserId="1394" LastActivityDate="2011-01-21T13:37:47.170" Title="A flyby of orbiting supermassive black holes" Tags="&lt;general-relativity&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="3444" PostTypeId="1" AcceptedAnswerId="3451" CreationDate="2011-01-20T21:39:08.960" Score="2" ViewCount="190" Body="&lt;p&gt;Would Cherenkov radiation occur at the explosion of a nuclear bomb? Suppose it would not be occluded by smoke or anything else for that matter.&lt;/p&gt;&#xA;" OwnerUserId="1480" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-30T09:03:55.087" LastActivityDate="2011-02-02T15:43:51.520" Title="Cherenkov radiation in nuclear bomb" Tags="&lt;electrodynamics&gt;&lt;nuclear-physics&gt;&lt;cherenkov-radiation&gt;" AnswerCount="1" />
  <row Id="3445" PostTypeId="1" AcceptedAnswerId="3458" CreationDate="2011-01-20T21:54:02.600" Score="0" ViewCount="142" Body="&lt;p&gt;What would you consider the best online resources for learning the 3+1 ADM formalism and gauge invariant perturbation theory in cosmology? (Assuming intermediate level GR and QFT familiarity)&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-01-21T02:13:59.993" Title="References for cosmological perturbation theory" Tags="&lt;general-relativity&gt;&lt;cosmology&gt;" AnswerCount="3" FavoriteCount="0" />
  <row Id="3446" PostTypeId="2" ParentId="3442" CreationDate="2011-01-20T21:59:00.920" Score="2" ViewCount="" Body="&lt;p&gt;Just to be sure, MAGOO is the AdS/CFT Bible &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/9905111&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9905111&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whose authors are usually listed alphabetically. Concerning your questions,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Conformal dimension or scaling dimension or mass dimension are meant to be the same thing in the context of conformal field theories in more than 2 dimensions. In 2 dimensions, one may distinguish the left-moving (holomorphic) and right-moving (antiholomorphic) dimensions or their sum - but the separate chiral &quot;dimensions&quot; are usually called &quot;weights&quot;, anyway. The dimension pretty much counts the exponent of &quot;kilogram&quot; in the unit of the corresponding operator - except that in quantum mechanics, the exponent is often fractional. The dimensions are usually positive - they get mapped to the energy which is positive as well. Operators have positive dimensions of mass. If your cryptic &quot;-ve&quot; meant that there is a minus sign, then there is no minus sign. If you prefer to use units of length, then their powers are negative, but you should switch to masses as the base whose powers count the dimension.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a) The field $\phi$, as opposed to $\phi_0$, is dimensionless because it is an actual field in the AdS space (the bulk), unlike $\phi_0$ that is defined on the boundary. They mean that it is dimensionless under the conformal transformations of the boundary - and that's true because the conformal transformations are realized as simple isometries in the bulk, so they can't rescale the AdS field $\phi$ - at most, they move it to another point. Also, on that page, one deals with particular finite modified boundary conditions for $\phi$ at the boundary of the AdS space so it must be finite even at $z=0$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;b) In equation (3.15), the left hand side is dimensionless. The right hand side has a power of $\epsilon$ whose dimension is $length^{d-\Delta}$ because $\epsilon$ has units of length (on boundary), and $\phi_0$ whose dimension must be $length^{\Delta-d}$ as a consequence, to get a dimensionless product. In equation (3.13), the exponent on the left hand side has $d^4x$ whose dimension is $length^d$ - note that $d=4$ for $AdS_5$; $\phi_0$ whose dimension is $length^{\Delta-d}$ as I said in the previous sentence - note that the $d$ term in the exponent cancels; and $O$ whose dimension must therefore be $length^{-\Delta}$ which means $mass^{\Delta}$. As I said, the standard base for counting dimensions is mass, so we also say that $O$ has dimension $\Delta$. So I suspect that your sign error is simply caused by the point 1) - namely by your incorrect assumption that dimensions refer to the powers of length. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;When we say &quot;dimension&quot; without extra specifications, we mean the power of the mass, not length. It's a healthy convention because the operators end up having non-negative dimensions. You may want to remember the dimensions of basic operators in 4 dimensions. The identity operator is always dimensionless (and $x$-independent): the dimension is 0. Bosonic scalar fields $\phi_0$ and potentials $A_\mu$ have dimension 1, their field strength has dimension $2$, fermions like the Dirac $\psi$ have dimension 3/2, and all terms in the Lagrangian density have dimension $4$. These are classical dimensions; each derivative adds $1$ to the dimension. Products of these operators have dimensions that are sums of the dimensions of the factors - except that quantum mechanics also adds &quot;anomalous dimensions&quot; to this classical form of the dimension, so that the total dimensions may have fractional terms that are proportional to powers of $g$ etc.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:39:55.543" LastActivityDate="2011-02-04T07:39:55.543" CommentCount="3" />
  <row Id="3448" PostTypeId="1" CreationDate="2011-01-20T22:07:40.520" Score="0" ViewCount="136" Body="&lt;p&gt;Your replies to my question about being able to see a photon, from the side (answer, unanimous, “no”) have raised in me some additional questions. &#xA; Would it reasonable to think that in consequence, the photon, from any angle but directionally “in front”, is a black body? I ask because I am accustomed to think of the photon as a “light” particle, when perhaps it might be more correct to think of it as a “black” particle.&#xA;When an observer sees light from a photon, isn’t it true that in fact, what is actually being seen is the photon itself?&#xA;When the photon enters the eye of the observer, what happens (3.1) to the observer? (3.2)to the photon?&#xA;3.1 The photon has no mass, so the observer/photon combination remains at original  observer mass.&#xA; However, the photon has energy, what impact on the observer done this energy  input have?&lt;br&gt;&#xA;3.2 Upon entering the observer, does the photon disappear? If not what else can it do?&lt;/p&gt;&#xA;" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-21T10:25:55.493" LastActivityDate="2011-01-21T10:25:55.493" Title="Observing photons" Tags="&lt;speed-of-light&gt;&lt;photons&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="3450" PostTypeId="2" ParentId="3448" CreationDate="2011-01-20T22:25:40.373" Score="5" ViewCount="" Body="&lt;p&gt;Photons are not particles in the way that marbles or billiard balls are particles. You're never going to be able to &quot;see&quot; them by bouncing some other thing off them, because they will always move at exactly the speed of light. You could, in principle, reconstruct the path of a photon through some medium by looking at the perturbation of the medium caused by its passage, but in that case, it's debatable whether the photon that leaves the medium is really the &quot;same&quot; photon that entered, as the scattering process in the medium would involve repeated absorption and emission cycles. Which also points to one of the risks of talking about photons as discrete particles following a definite path, namely that they can be created and destroyed very easily through interactions with matter, and thus their identity is a little fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a photon strikes the eye of an observer, or the active surface in a detector, its energy goes into the medium. In an electronic detector, it usually knocks an electron loose in some manner, which then triggers some sort of cascade process that produces a measurable electronic pulse. In the eye, it causes some reconfiguration of light-sensitive molecules, which eventually produce a signal in the optic nerve that gets registered as a &quot;flash&quot; of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of an atom absorbing a photon, I suppose it is true that the apparent mass of the atom in its excited state would be infinitesimally higher than the ground-state atom's, by an amount $E_\gamma / c^2$, but that's such a trivial effect that nobody bothers to deal with it. To give you an idea of the scale, a typical atomic transition in the visible light range has an associated energy of a few eV, while the rest mass of a hydrogen atom is about 1,000,000,000 eV. You would never notice the effect (unless you're talking about high-energy gamma ray photons, in which case the result of absorbing a photon in the nucleus is usually to blast the nucleus apart after a very short time, often creating a bunch of new particles in the process).&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2011-01-20T22:25:40.373" CommentCount="2" />
  <row Id="3451" PostTypeId="2" ParentId="3444" CreationDate="2011-01-20T22:26:46.267" Score="4" ViewCount="" Body="&lt;p&gt;Of course, you could observe it in the water where the speed of light is much smaller than $c$ and electrons (beta-radiation) from the nuclear blast easily beat the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The air has index of refraction $1.0003$ so the speed of light is $0.9997c$. Note that $v^2/c^2$ is about $0.9994$ and when subtracted from 1, we get $0.0006$. The square root of it is about $0.025$ so charged particles get about 40 times heavier at this speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, this is about 40 GeV for protons or neutrons (and even more for nuclei), so you can't get this fast hadrons or nuclei in a nuclear blast. Even for electrons, 40 times the rest mass gives about 20 MeV which is, I believe, much more than the energy that an electron can ever get in any process of this kind. Moreover, fission really produces two smaller nuclei that carry the energy away, and the energy of a neutron that is created is just 2 MeV or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can't get Cherenkov radiation in the air by nuclear blast. Cherenkov radiation in the air is a strange animal, anyway: we usually talk about water or similar dense environments. How air usually enters these discussions is only via the supersonic analogy of the Cherenkov radiation - but when dealing with sound, it's not the real Cherenkov radiation. ;-)&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T15:43:51.520" LastActivityDate="2011-02-02T15:43:51.520" CommentCount="8" />
  <row Id="3452" PostTypeId="1" CreationDate="2011-01-20T23:10:40.680" Score="4" ViewCount="450" Body="&lt;p&gt;If one takes a bundle of wood up high to the mountains so its potential energy increases, would there be obtained more heat by burning it? &lt;/p&gt;&#xA;" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-21T22:26:16.647" LastActivityDate="2011-01-23T16:34:15.063" Title="Can gravitational potential energy be released in a fire?" Tags="&lt;thermodynamics&gt;&lt;energy&gt;&lt;newtonian-mechanics&gt;" AnswerCount="6" CommentCount="9" />
  <row Id="3453" PostTypeId="2" ParentId="3445" CreationDate="2011-01-21T01:20:22.597" Score="1" ViewCount="" Body="&lt;p&gt;Ma &amp;amp; Bertschinger (1995):&#xA;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1995ApJ...455....7M&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/1995ApJ...455....7M&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2011-01-21T01:20:22.597" />
  <row Id="3454" PostTypeId="2" ParentId="3443" CreationDate="2011-01-21T01:32:46.220" Score="0" ViewCount="" Body="&lt;p&gt;Even before the horizons merge to form a common horizon, the tidal forces may be very large, even for supermassive BHs. For a single black hole, particles originating from outside the BH cannot get closer than something like 1.5 times the horizon radius without getting captured. So a particle or photon that gets slightly closer to one of the two BHs would get captured. Thus, if you try to &quot;thread the needle&quot; you might get split in two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a picture of the shapes of the horizons, see &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0610122&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/gr-qc/0610122&lt;/a&gt; Fig 1.&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2011-01-21T01:32:46.220" CommentCount="1" />
  <row Id="3455" PostTypeId="2" ParentId="3432" CreationDate="2011-01-21T01:37:12.287" Score="7" ViewCount="" Body="&lt;h2&gt;&lt;a href=&quot;http://blogs.discovermagazine.com/cosmicvariance/&quot; rel=&quot;nofollow&quot;&gt;Cosmic Variance&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Science, Technology, and The Future&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; It would be great if someone who knows the blog well would write a few words about it. Just give a little more detail of what it's about. 3 sentences is more than enough. This is community wiki, so most people can edit it freely.&lt;/p&gt;&#xA;" OwnerUserId="775" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2011-01-21T22:07:47.793" LastActivityDate="2011-01-21T22:07:47.793" CommunityOwnedDate="2011-01-21T01:37:12.287" CommentCount="4" />
  <row Id="3456" PostTypeId="2" ParentId="743" CreationDate="2011-01-21T01:43:31.253" Score="2" ViewCount="" Body="&lt;p&gt;:-DDD I have written another pedagogical article about renormalization... Not, seriously, the article is blahblah and I had even forgotten I had written it. Probably it even contributed in the evaluation that expelled me from hep-th to hep-ph, who knows. But the list of references is useful and then it is an actual answer to the OP, so allow me to paste them here. It was at &lt;a href=&quot;http://arxiv.org/abs/hep-th/0208180&quot; rel=&quot;nofollow&quot;&gt;hep-th/0208180&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that some historical references (eg Borel 1928 and some comments in the body of the paper) are given only to suggest why people was not so afraid about divergences in 1930, it was even a hot topic in related areas.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;G.A Arteca, F.M. Fernandez, E.A. Castro, &lt;em&gt;Large Order Perturbation Theory and &#xA;Summation Methods in Quantum Mechanics&lt;/em&gt;, Lecture Notes In Chemistry, 53, Springer&lt;/li&gt;&#xA;&lt;li&gt;E. Borel, &lt;em&gt;Lessons sur las series divergentes&lt;/em&gt;, &#xA;ed. Gautier-Villars, 1928 (reprinted  editions Jacques Gabay)&lt;/li&gt;&#xA;&lt;li&gt;E. Brezin, J.C. LeGuillou, J. Zinn-Justin, &lt;em&gt;Perturbation theory at &#xA;large Order, I et II&lt;/em&gt;, Phys Rev D, v. 15, p. 1544 and Phys Rev D v. 15, p. 1558&lt;/li&gt;&#xA;&lt;li&gt;Ch. Brouder, &lt;em&gt;Runge-Kutta methods and renormalization&lt;/em&gt;, European &#xA;Physical Journal C v. 12, p. 521-534&lt;/li&gt;&#xA;&lt;li&gt;J.C. Butcher, &lt;em&gt;An Algebraic Theory of Integration Methods&lt;/em&gt; Math. Comp. v. 26, p. 79&lt;/li&gt;&#xA;&lt;li&gt;A. Connes and D. Kreimer hep-th/9912092, as well as D. Kreimer &#xA;             q-alg/9707029 and hep-th/0010059, and C-K hep-th/9904044&lt;/li&gt;&#xA;&lt;li&gt;F.J.Dyson, Phys Rev 85, p. 631 &lt;/li&gt;&#xA;&lt;li&gt;LY Chen, N. Goldenfeld, Y. Oono, Phys. Rev. E, v. 54 p. 376&lt;/li&gt;&#xA;&lt;li&gt;Feynman, &lt;em&gt;Space-Time Approach to NR Quantum Mechanics&lt;/em&gt; Rev Mod  Phys 20, p. 367&lt;/li&gt;&#xA;&lt;li&gt;M. Gell-Mann and F.E. Low  &lt;em&gt;Quantum Electrodinamics at Small Distances&lt;/em&gt;, &#xA;Phys. Rev. v. 95, p. 1300&lt;/li&gt;&#xA;&lt;li&gt;G t'Hooft, Nucl Phys B 35, p. 167; G. t'Hooft and M.Veltman, Nucl. Phys. B44 p. 189  &lt;/li&gt;&#xA;&lt;li&gt;D.J. Broadhurst and D. Kreimer,  &lt;em&gt;Renormalization tamed: 30 loop Pade Borel &#xA;Resumation&lt;/em&gt;, hep-th/9912093&lt;/li&gt;&#xA;&lt;li&gt;T. Kunihiro, for instance hep-th/9505166 and hep-th/9801196&lt;/li&gt;&#xA;&lt;li&gt;Polonyi, arxiv:hep-th/9409004, hep-th/9412042 and hep-th/9711061&lt;/li&gt;&#xA;&lt;li&gt;Tim R. Morris, hep-th/9802039&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1335" LastEditorUserId="1335" LastEditorDisplayName="" LastEditDate="2011-01-21T01:55:57.507" LastActivityDate="2011-01-21T01:55:57.507" />
  <row Id="3457" PostTypeId="2" ParentId="3402" CreationDate="2011-01-21T01:43:36.133" Score="-2" ViewCount="" Body="&lt;p&gt;another possibility is that, assuming life could evolve into highly intelligent and technically capable lifeforms, it could then propagate at exponentially fast rates and take over the universe.&lt;/p&gt;&#xA;" OwnerUserId="775" LastActivityDate="2011-01-21T01:43:36.133" CommentCount="1" />
  <row Id="3458" PostTypeId="2" ParentId="3445" CreationDate="2011-01-21T01:59:46.720" Score="1" ViewCount="" Body="&lt;p&gt;Hah, I just studied this a while ago with James Bardeen, so I would say he is the best resource for learning this!  Since you probably don't have access to the physical Bardeen, you can check out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physical Review D, Vol 22 no 8 (1980) &quot;Gauge-invariant cosmological perturbations&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physical Review D, Vol 40 no 6 (1989) &quot;Designing density fluctuation spectra in inflation&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a set of lecture notes I have sitting on my desk by him that claim they are &quot;to be published in Particle Physics and Cosmology&quot; which are dated 1988, so presumably they were published within the next year or so.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can find them, the talks are probably the easier of the three, and the first PrD article is the second easiest. The third paper is very nice, but more technically difficult.&lt;/p&gt;&#xA;" OwnerUserId="1403" LastActivityDate="2011-01-21T01:59:46.720" CommentCount="2" />
  <row Id="3459" PostTypeId="2" ParentId="3445" CreationDate="2011-01-21T02:13:59.993" Score="1" ViewCount="" Body="&lt;p&gt;The single best reference for learning about gauge-invariant cosmological perturbation theory is ch. 7 of Mukhanov's book &lt;em&gt;Physical foundations of Cosmology&lt;/em&gt;. Ok, not an online resource, but still the very best if that's what you're looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On doing a search for one of York's original papers I ran into &lt;a href=&quot;http://www.phy.olemiss.edu/~luca/Topics/gr/initial.html&quot; rel=&quot;nofollow&quot;&gt;this wonderful site&lt;/a&gt; by Luca Bombelli at Ole' Miss (University of Mississippi). This contains a very comprehensive bibliography on the initial-value problem. I would also recommend Robert Wald's GR book. It has very nice coverage of the IVP in chapters and in an appendix.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-21T02:13:59.993" CommentCount="1" />
  <row Id="3460" PostTypeId="2" ParentId="3432" CreationDate="2011-01-21T02:29:37.837" Score="2" ViewCount="" Body="&lt;h2&gt;&lt;a href=&quot;http://www.astrobetter.com/&quot; rel=&quot;nofollow&quot;&gt;AstroBetter&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Tips and Tricks for Professional Astronomers&lt;/p&gt;&#xA;" OwnerUserId="724" LastActivityDate="2011-01-21T02:29:37.837" CommunityOwnedDate="2011-01-21T02:29:37.837" />
  <row Id="3461" PostTypeId="2" ParentId="3432" CreationDate="2011-01-21T02:43:47.803" Score="7" ViewCount="" Body="&lt;h2&gt;&lt;a href=&quot;http://motls.blogspot.com/&quot; rel=&quot;nofollow&quot;&gt;The Reference Frame&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;it is the only one that is challenging.  &lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2011-01-21T02:54:52.157" LastActivityDate="2011-01-21T02:54:52.157" CommunityOwnedDate="2011-01-21T02:43:47.803" CommentCount="5" />
  <row Id="3462" PostTypeId="2" ParentId="3432" CreationDate="2011-01-21T03:03:11.587" Score="4" ViewCount="" Body="&lt;h2&gt;&lt;a href=&quot;http://www.wired.com/wiredscience/dotphysics/&quot; rel=&quot;nofollow&quot;&gt;Dot Physics&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This deals with popular topics from the perspective of an actual physicist (so, a lot of rants about instances when non-specialists get things wrong, and occasionally praise for when they get it right). It won't strain your brain, but I always find the posts to be fun to read, and pretty insightful. &lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-01-21T08:14:36.923" LastActivityDate="2011-01-21T08:14:36.923" CommunityOwnedDate="2011-01-21T03:03:11.587" CommentCount="6" />
  <row Id="3463" PostTypeId="2" ParentId="3357" CreationDate="2011-01-21T05:06:18.660" Score="3" ViewCount="" Body="&lt;p&gt;Some more details: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) In quantum physics, the behavior of objects is not deterministic. A &quot;quantum state&quot; gives you probabilities for the different possible behaviors. &quot;Entanglement&quot; means that the probabilities for two or more objects are correlated. This happens when objects interact with each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &quot;Teleportation&quot; is a way to transfer a quantum state from one object to another object, which uses entanglement with a third object and also ordinary signaling. You start with two objects that have interacted so they are maximally entangled with each other (maximal entanglement has a precise mathematical meaning). You let the original object (the one whose state you want to teleport) interact with one of these, you measure the results, and then you act on the second entangled object in a way determined by those results. This will recreate, in the second entangled object, the quantum state of the original object. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) If you have heard of Fourier analysis, you may know that a field can be expressed as a sum of &quot;modes&quot;, each with a specific wavelength and direction of movement. Classically you might just say how strong a field mode is (e.g. its Fourier coefficient, if it is a Fourier mode). But when you describe a field according to quantum mechanics, each field mode has to be described by a quantum state, in terms of probabilities. Because of the uncertainty principle, even in the lowest-energy state (the vacuum state, no particles present), each field mode has some probability of being active. Furthermore, the modes will interact (directly or indirectly), and so the probabilities of the different modes won't be independent - their quantum states will be entangled. And you can even imagine using this vacuum entanglement to perform quantum teleportation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) There has been some interest in experimentally detecting the entanglement of the vacuum field modes, but it would be difficult because your particle detector (that detects vacuum fluctuations in the field modes) has to undergo constant acceleration for the duration of the experiment. The Australian physicists thought of an alternative way to detect the mode correlations, where the detector remains at rest but is progressively retuned to mimic the effects of accelerated motion. So the behavior of the detector in the future is correlated with the behavior of the detector in the past, when it was responding to different modes. But they are still the same modes whose entanglement might have been detected in an equal-time experiment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) &quot;Teleportation in time&quot; is therefore just a form of the ordinary teleportation, in which there is a prominent time delay. In particular, what this paper describes is quantum teleportation via entangled field modes of the vacuum. Remember that for ordinary teleportation to work, you need information about the measurement results at the first step - they tell you what to do to the second entangled object, in order to recreate the original quantum state. So time teleportation also has to start with a measurement, stored somewhere in a conventional fashion (e.g. in a computer memory), and then the future will use that information to recreate the old quantum state by interacting with one of the vacuum field modes. &lt;/p&gt;&#xA;" OwnerUserId="1486" LastActivityDate="2011-01-21T05:06:18.660" CommentCount="1" />
  <row Id="3464" PostTypeId="2" ParentId="3432" CreationDate="2011-01-21T07:28:49.523" Score="5" ViewCount="" Body="&lt;p&gt;My personal blog is&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;a href=&quot;http://arcsecond.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Arcsecond&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;p&gt;It's not exclusively physics.  Since it's a personal blog, it's about whatever interests me at the moment.  When I do write about physics, the level is usually undergraduate (since that's my education level).  I usually update in irregular bursts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some example posts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Physics&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arcsecond.wordpress.com/2010/09/17/vietes-formula-and-spinning-pizza/&quot; rel=&quot;nofollow&quot;&gt;Viete's Formula and Spinning Pizza&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uses simple physics to derive Viete's Formula &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{2}{\pi} = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2+\sqrt{2}}}{2}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arcsecond.wordpress.com/2010/10/14/fucking-geomagnetism-how-does-that-work/&quot; rel=&quot;nofollow&quot;&gt;Fucking Geomagnetism, How Does That Work?&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Response to &lt;a href=&quot;http://blogs.discovermagazine.com/cosmicvariance/2010/10/13/insane-clown-posse-channels-walt-whitman/&quot; rel=&quot;nofollow&quot;&gt;Cosmic Variance's essay&lt;/a&gt; on Insane Clown Posse's &quot;Miracles&quot; video.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arcsecond.wordpress.com/2009/12/06/bounce-part-2/&quot; rel=&quot;nofollow&quot;&gt;Bounce, part 2&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why won't a ball bounce higher than it's dropped?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Math&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arcsecond.wordpress.com/2009/10/11/the-power-tower/&quot; rel=&quot;nofollow&quot;&gt;The Power Tower&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is $\sqrt{2}$ taken to the power of itself, over and over?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arcsecond.wordpress.com/2010/08/13/answer-lemming/&quot; rel=&quot;nofollow&quot;&gt;Answer, Lemming&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very cute puzzle from Ravi Vakil's &quot;A Mathematical Mosaic&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Memoir&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arcsecond.wordpress.com/2010/06/10/why-i-rode-a-bicycle-200-miles/&quot; rel=&quot;nofollow&quot;&gt;Why I Rode a Bicycle 200 Miles The Other Day&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arcsecond.wordpress.com/2009/02/14/pulse/&quot; rel=&quot;nofollow&quot;&gt;Pulse&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Essay&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arcsecond.wordpress.com/2010/11/07/simple-isnt-easy/&quot; rel=&quot;nofollow&quot;&gt;'Simple' Isn't 'Easy'&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Humor&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arcsecond.wordpress.com/2010/09/16/have-you-ever-noticed-a-panda-is-just-a-fat-slow-zebra/&quot; rel=&quot;nofollow&quot;&gt;Have You Ever Noticed A Panda Is Just a Fat, Slow Zebra?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arcsecond.wordpress.com/2009/09/10/why-i-dont-drink-soda/&quot; rel=&quot;nofollow&quot;&gt;Why I Don't Drink Soda&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arcsecond.wordpress.com/2010/01/09/a-review-of-reviews-of-racism-and-a-very-special-reimagining/&quot; rel=&quot;nofollow&quot;&gt;A Review of Reviews of Racism and a Very Sexy Reimagining&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Comic&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arcsecond.wordpress.com/2009/05/02/sine-error-comic-23/&quot; rel=&quot;nofollow&quot;&gt;Sine Error: Spare Me&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arcsecond.wordpress.com/2009/03/30/sine-error-comic-19/&quot; rel=&quot;nofollow&quot;&gt;Sine Error: Tricked You!&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-01-21T07:28:49.523" CommunityOwnedDate="2011-01-21T07:28:49.523" CommentCount="1" />
  <row Id="3465" PostTypeId="2" ParentId="3436" CreationDate="2011-01-21T07:32:21.357" Score="2" ViewCount="" Body="&lt;p&gt;While good, I think the other answers are currently missing one ingredient, so I'll post this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For particles traveling at constant velocity there is no event horizon, and so they act nothing like a black hole. Light from other regions of space will eventually reach it, unlike a black hole. Further, the forces between atoms in what ever matter constitutes the mass are co-moving and so there is no increased gravitational interaction between them. While the distances between them appear to change to an outside observer (as the mass is accelerated) once it reaches constant velocity they are fixed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What has not been mentioned in other answers is the effect of acceleration. When a particle is continuously accelerated there &lt;strong&gt;is&lt;/strong&gt; an apparent event horizon. See the relevant Wikipedia page &lt;a href=&quot;http://en.wikipedia.org/wiki/Event_horizon#Apparent_horizon_of_an_accelerated_particle&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. So this has some features that we associate with a black hole, however there are still major differences. An object undergoing constant acceleration does indeed behave like it is static in a constant gravitational field. However, in the case of such an object the direction of the equivalent field is constant (and in a constant direction) throughout the object. This is not true for the gravitational field of a black hole, which is spherically symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course once the particle stops accelerating the apparent horizon disappears.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2011-01-21T07:32:21.357" CommentCount="1" />
  <row Id="3466" PostTypeId="2" ParentId="3432" CreationDate="2011-01-21T07:33:11.093" Score="1" ViewCount="" Body="&lt;h2&gt;&lt;a href=&quot;http://skullsinthestars.com/&quot; rel=&quot;nofollow&quot;&gt;Skulls in the Stars&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Exceptional for its extremely clear, basic-level introductions to phenomena in optics I would otherwise never have heard of, and for interesting historical posts resulting from digging around in the archives of old journals and scientists' personal letters.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-01-21T07:40:01.053" LastActivityDate="2011-01-21T07:40:01.053" CommunityOwnedDate="2011-01-21T07:33:11.093" />
  <row Id="3467" PostTypeId="1" AcceptedAnswerId="3481" CreationDate="2011-01-21T07:48:22.917" Score="5" ViewCount="137" Body="&lt;p&gt;we often come across the phrase superselection sector. I have couple of questions from that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, I learned from some references about it (as a overview) and how is it related to superselection rules etc. There is of course this mathematical definition (from C* algebra, unfortunately I haven't gone through that), but I wanted to understand it physically. It seemed to me that basically if I construct two sectors (two hilbert spaces?) whose basis vectors don't cohere, then they belong to two different superselection sectors. My first question will be : Is that it or there  are more into it than meets the eye! Please.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, I encountered it in String theory. I will probably mention the place (I have an unrelated question here too!) where I encountered it in string theory. Its the argument using which we can guess the presence of D branes from SUGRA action. We see that in the low energy ST action, there can be a term (some p+1 form potential, properly coupled etc.) Now I saw this argument that &quot;this term has finite tension and hence its not finite energy excitation above the vacuum&quot; (why? and how did we guess about finite tension?- Is it because of the very fact that this term exists in the ST action?) and also its suggested that these objects (multi form potentials) are in different superselection sector.&#xA;  So again, please let me know if there is any general comment on superselection sector that you want to make in this context and also about the arguments.&lt;/p&gt;&#xA;" OwnerUserId="1349" LastActivityDate="2011-02-02T15:43:25.543" Title="Superselection sectors and $p+1$ form potentials in String Theory" Tags="&lt;quantum-mechanics&gt;&lt;string-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="3468" PostTypeId="1" AcceptedAnswerId="6413" CreationDate="2011-01-21T08:30:28.090" Score="8" ViewCount="674" Body="&lt;p&gt;Some speculation here:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.youtube.com/watch?v=_NMqPT6oKJ8&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=_NMqPT6oKJ8&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a possibility it would pass 'undetected' through your hand, or is it certain death?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you conclude it to be vital, or only loose your hand?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it simply make a small cylindrical hole through your hand, or is there some sort of explosion-effect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume your hand has a cross section of 50cm², and a thickness of 2cm, how much of the beam's energy would be transferred to your hand?&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2011-01-29T18:55:07.790" LastActivityDate="2011-03-06T01:56:30.417" Title="What would happen if you put your hand in front of the 7 TeV beam at LHC?" Tags="&lt;particle-physics&gt;&lt;high-energy-physics&gt;&lt;large-hadron-collider&gt;" AnswerCount="7" CommentCount="10" />
  <row Id="3469" PostTypeId="2" ParentId="3402" CreationDate="2011-01-21T08:49:10.617" Score="2" ViewCount="" Body="&lt;p&gt;one could tackle your question a little more formal if one looks at the limit $$c\rightarrow\infty\, .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can now ask what happens in the several theories and what implications there would be for life.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;A naive attempt&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;One starting point is indeed the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Physical_constant&quot; rel=&quot;nofollow&quot;&gt;physical constants&lt;/a&gt;. You can look for your favourite one depending on the speed of light and look on the corresponding site what it implies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, lets do this for one thing I just have chosen (almost) randomly: &lt;a href=&quot;http://en.wikipedia.org/wiki/Rydberg_constant&quot; rel=&quot;nofollow&quot;&gt;The Rydberg constant&lt;/a&gt; $R_\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is given by $$R_\infty = \frac{m_e e^4}{8\epsilon_0^2h^3 \mathbf{c}} \approx 1.097 \times 10^7 m^{-1}\, .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the article states, it is the &lt;strong&gt;most accurately measured&lt;/strong&gt; fundamental physical constant and its value can be &lt;strong&gt;derived&lt;/strong&gt; from first principles. Interesting to know, I always thought this was the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fine-structure_constant&quot; rel=&quot;nofollow&quot;&gt;fine-structure constant&lt;/a&gt; $\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We state that $$\lim_{c\rightarrow \infty}R_\infty = 0\, .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Rydberg constant has its interpretation as the lowest wavenumber $\lambda_{ion} = 1/R_\infty \rightarrow \infty$ that can &lt;strong&gt;ionize&lt;/strong&gt; the hydrogen atom. This is linked to some lowest energy $$E_{ion} = c \frac{h}{\lambda_{ion}} \rightarrow  \frac{m_e e^4}{8\epsilon_0^2h^2}\, .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it seems like we have &lt;strong&gt;won nothing at all&lt;/strong&gt;. The wavelength goes to infinity but the corresponding energy remains constant. Or do we run into further problems because we have to look at the permittivity of vacuum $\epsilon_0$ that is also linked to $c$ via $\mu_0\epsilon_0 = 1/c^2$. This is puzzling - we &lt;strong&gt;cannot answer&lt;/strong&gt; the question from this viewpoint, but earned a nice hint due to the fact that all we are discussing about corresponds to  wave propagation in electrodynamics.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Electromagnetic waves&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Wave propagation at a certain frequency $\omega$ through vacuum is described by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Helmholtz_equation&quot; rel=&quot;nofollow&quot;&gt;Helmholtz equation&lt;/a&gt;&#xA;$$\Delta A_{\mu} + \frac{\omega^2}{c^2} A_{\mu} = 0$$&#xA;which also holds for quantum electrodynamics as discussed in &lt;a href=&quot;http://physics.stackexchange.com/questions/560/does-anything-exist-in-the-intergalactic-space/1887#1887&quot;&gt;another thread&lt;/a&gt;. Here we can see clearly what happens if $c\rightarrow \infty$:&lt;br&gt;&#xA;The Helmholtz equation reduces to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Laplace_equation&quot; rel=&quot;nofollow&quot;&gt;Laplace equation&lt;/a&gt; &#xA;$$\Delta A_\mu = 0$$&#xA;which can be interpreted in a way that everything that happens, does it instantaniously - there is &lt;strong&gt;no retardation&lt;/strong&gt; anymore. This implies that &lt;strong&gt;everything happens at the same time&lt;/strong&gt;, at least in electrodynamics. In fact, this should also hold for all (effective) theories that can be described by interactions via light particles, or other &lt;a href=&quot;http://en.wikipedia.org/wiki/Massless_particle&quot; rel=&quot;nofollow&quot;&gt;massless particles&lt;/a&gt; since they also travel at $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to speak, the speed of light is something like a translation between space and time. If $c$ goes to infinity, maybe even the notion of time (and energy as the associated quantity) does not make sense.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what will happen, but one of the two cases seem to be plausible if $c\rightarrow\infty$:&lt;br&gt;&#xA;Either &lt;strong&gt;all will happen instantaniously&lt;/strong&gt;, or, maybe worse, everything will have to remain in a &lt;strong&gt;unchanged forever&lt;/strong&gt; (static) - I don't think that life as we think of it is possible under these circumstances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2011-01-21T08:49:10.617" />
  <row Id="3470" PostTypeId="2" ParentId="3467" CreationDate="2011-01-21T08:58:17.507" Score="0" ViewCount="" Body="&lt;p&gt;Glance through the first few pages of the book &quot;PCT, Spin and Statistics, and all that&quot;. You can read the section on superselection &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691070628&quot; rel=&quot;nofollow&quot;&gt;in the amazon preview.&lt;/a&gt; And at $20 used, you should have a copy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum mechanics is about interference. Stuff that is in different superselection sectors cannot be mixed. An example is a quark and an electron. As the PCT book states, you can write down a mixture using math, but you cannot produce it in the real world. So superselection sectors are a restriction on the Hilbert space.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-21T08:58:17.507" />
  <row Id="3471" PostTypeId="2" ParentId="3468" CreationDate="2011-01-21T09:05:47.857" Score="5" ViewCount="" Body="&lt;p&gt;Each bunch of protons circulating in the LHC at full power will have about 120 billion protons of 14TeV each, i.e. $1.6 \times 10^{24}$ eV which is about 250000 Joules. Compare that with a high powered rifle bullet carrying 1000 Joules or a small grenade that releases 600000 Joules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how much of that energy will be released in your hand? It is said that the protons at the LHC can penetrate 10m in copper so many will go through 2cm of flesh without losing much energy. This means the destructive energy released in the hand could be just 1/1000 of the total in the beam, so it might be like being hit with a small bullet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it will still be enough to do a lot of damage. It will also be in the form of high energy radiation which is going to spread out from the impact point. It's difficult to be sure how much energy will convert to different forms such as heat and blast, and in which directions, but whatever way it goes I think there will be a lot of initial damage and the radiation after effects will also have the potential to be fatal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One last edit: It might be worth mentioning that they can circulate up to 2800 of these bunches at once, so it might be more like being hit by 2800 small bullets in a fraction of a second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: question was changed from 14TeV to 7TeV for correct LHC beam energy, so use half the numbers above&lt;/p&gt;&#xA;" OwnerUserId="1341" LastEditorUserId="1341" LastEditorDisplayName="" LastEditDate="2011-01-21T13:12:15.263" LastActivityDate="2011-01-21T13:12:15.263" CommentCount="3" />
  <row Id="3472" PostTypeId="2" ParentId="3402" CreationDate="2011-01-21T09:09:27.463" Score="2" ViewCount="" Body="&lt;p&gt;First, the principle of relativity is the statement that the laws of physics are invariant in all frames of reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we have physical laws that are a function of the frame of reference, the answer to your question would ofcourse depend on exactly what laws you replace them with. If there were only some minute change in physical constants throughout space, we would not necessarily have a problem, I am proposing we wouldnt even have a theoretical problem, as many of these numbers seem arbitrary in the SM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Infact there are many more radical changes you can do to the physical laws that would not result in any (immediately) observable change at all, just think of string theory or most TOE ;)&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-01-21T09:09:27.463" />
  <row Id="3473" PostTypeId="2" ParentId="15" CreationDate="2011-01-21T09:34:13.513" Score="0" ViewCount="" Body="&lt;p&gt;String excitations, rspt lack thereof. Problem is however that unless you believe in a theory with a lowered Planck scale, you'll have to go to energies we'll never be able to reach in the lab to test this regime. But at least in principle it's falsifiable. &lt;/p&gt;&#xA;" OwnerUserId="1458" LastActivityDate="2011-01-21T09:34:13.513" />
  <row Id="3474" PostTypeId="2" ParentId="3468" CreationDate="2011-01-21T09:41:25.193" Score="2" ViewCount="" Body="&lt;p&gt;What Gibbs said (+1), except that because the beam is highly relativistic, the probability of radiation being sent in directions other than down the beam line is very low. You can see this by looking at the problem in the center of mass reference frame for a typical collision. Because special relativity increases the mass of the protons in the beam, the center of mass of the collision products will still be moving at nearly the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically it will drill a painful hole in your hand, something less than 1mm in diameter, depending on how its focused. The LHC beam dump takes a beam 0.3mm in diameter. For the effects of lower energy proton beams, see &lt;a href=&quot;http://www.aapm.org/meetings/09PRS/documents/Flanz_AAPM_Final.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.aapm.org/meetings/09PRS/documents/Flanz_AAPM_Final.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the reason those physicists in the youtube didn't have realistic answers is because they were thinking on their feet. I think this was an ambush question.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-21T09:41:25.193" CommentCount="2" />
  <row Id="3475" PostTypeId="2" ParentId="3468" CreationDate="2011-01-21T09:48:20.443" Score="7" ViewCount="" Body="&lt;p&gt;Well, there is the unbelievable story about a guy who actually put his head in a proton beam, the Russian scientist Anatoli Petrovich Bugorski. This happened at the U-70 synchrotron, near Moscow at the Institute for High Energy Physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the thing is, is that he actually didn't feel any pain. He did suffer from epyleptic attacks and damage to his skin and brain. Citing from:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.giftsandfreeadvice.com/free_advice/large-hadron-collider-anatoli-petrovich-bugorski-a-russian-scientist-survived-a-particle-accelerator-accident-and-didnt-develop-super-powers-or-die/&quot; rel=&quot;nofollow&quot;&gt;What would happen...?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bugorski was leaning over the piece of equipment when he stuck his head in the part through which the proton beam was running. Reportedly, he saw a flash “brighter than a thousand suns”, but did not feel any pain. The beam measured about 2000 gray when it entered Bugorski’s skull, and about 3000 gray when it exited after colliding with the inside of his head.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To be clear, a gray is the unit of absorbed energy, and 1 gray equals 1 Joule/kg. For comparison, CT scans work in the ~50 mGy range and radiation therapy in the ~50 Gy range. The bizarre thing is, if you're exposed to &gt;5 Gy it's usually lethal. But that applies to usual radiation,  gamma radiation. The effect of a proton beam is less known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The protons in the beam can be accelerated up to ~70 GeV, which is 1/2000 of the LHC range. There's also the beam luminosity, which is a measure for the number of particles flying through a unit area per unit time. The LHC can reach 10^34 1/(cm^2 s^2) and the U-70 about 10^32 1/(cm^2 s^2). So the most naive estimate would be an increase of 10^3 - 10^5 of the radiation dose -- but that's more like an upper bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It won't destroy your hand or blast it off. But the effects of radiation are quite severe, and you &lt;em&gt;will&lt;/em&gt; end up with a mutant hand.&lt;/p&gt;&#xA;" OwnerUserId="1287" LastEditorUserId="1287" LastEditorDisplayName="" LastEditDate="2011-01-21T12:54:56.663" LastActivityDate="2011-01-21T12:54:56.663" CommentCount="2" />
  <row Id="3476" PostTypeId="2" ParentId="721" CreationDate="2011-01-21T10:18:02.830" Score="0" ViewCount="" Body="&lt;p&gt;&quot;Suppose an exact copy of our universe is made, but where spatial distances and sizes are twice as large relative to ours. Would this universe evolve and function just as ours?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you ignore ideas like &lt;a href=&quot;http://en.wikipedia.org/wiki/Mach%27s_principle&quot; rel=&quot;nofollow&quot;&gt;those of Ernst Mach's&lt;/a&gt;, the answer is no. For example, if I were suddenly twice as tall, wide, and thick, but with the same electrons etc., I'd certainly be dead. To see this from the physical constants, note that there is a fundamental unit of distance, &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck_length&quot; rel=&quot;nofollow&quot;&gt;the Planck length,&lt;/a&gt; and this gives an absolute scale for the size of a human being (roughly $10^{35}$ Planck lengths).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if the characteristics of spacetime depend on the presence of other matter, as espoused by Mach, there might be a chance for the universe to evolve the same way, though I don't know of any such theory. And I should add that Mach didn't believe in atoms so he's a bit, eh, dated.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-21T10:18:02.830" />
  <row Id="3477" PostTypeId="1" AcceptedAnswerId="3482" CreationDate="2011-01-21T10:33:43.440" Score="2" ViewCount="181" Body="&lt;p&gt;The arrow of time is a concept that has been around for a while.  In modern cosmology, it has been argued that there may be different arrows of time in other universes.  Since these are also considered to be inaccessible, is it physically meaningful to talk of the arrow of time in other universes?&lt;/p&gt;&#xA;" OwnerUserId="1383" LastActivityDate="2011-03-10T10:05:07.413" Title="Is it physically meaningful to talk about the arrow of time in other universes?" Tags="&lt;cosmology&gt;" AnswerCount="4" />
  <row Id="3478" PostTypeId="2" ParentId="3432" CreationDate="2011-01-21T10:33:47.377" Score="5" ViewCount="" Body="&lt;h2&gt;&lt;a href=&quot;http://math.ucr.edu/home/baez/&quot; rel=&quot;nofollow&quot;&gt;John Baez's Stuff&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It is more mathematics, but a lot of physics/mathematical physics related &quot;stuff&quot; also. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-01-21T10:33:47.377" CommunityOwnedDate="2011-01-21T10:33:47.377" CommentCount="1" />
  <row Id="3480" PostTypeId="2" ParentId="3477" CreationDate="2011-01-21T10:52:27.130" Score="1" ViewCount="" Body="&lt;p&gt;If it &lt;em&gt;is&lt;/em&gt; physically meaningful to speak of &quot;other universes&quot;, then yes it is meaningful to speak of an arrow of time (and any other constructions in physics) in those universes. Taking this logic further, one could think of doing statistical mechanics over an ensemble of universes each with its own arrow of time - sort of like a spin-system - in the hopes that a &quot;common&quot; arrow of time could emerge. This is in spirit with how ferromagnetism emerges in spin-systems. Even though each spin considered individually is randomly oriented, on treating the many-body system we find collective behavior manifest itself in the form of a spontaneous, non-zero, macroscopic magnetization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feel free to criticize this answer for its speculative aspects. However, the question itself is in fairly speculative territory to begin with.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-21T10:52:27.130" CommentCount="2" />
  <row Id="3481" PostTypeId="2" ParentId="3467" CreationDate="2011-01-21T11:08:30.277" Score="3" ViewCount="" Body="&lt;p&gt;A superselection sector $H_S$ is one of the subspaces of the &quot;total Hilbert space&quot; $H$ such that one can prove that the transition amplitude from any state in $H_S$ into any state in another superselection sector $H_{S&amp;#39;}$ vanishes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in principle, one can divide the Hilbert space $H$ of an ordinary theory to superselection sectors with different eigenvalues of conserved quantities. We know that it's conserved so we can never get a state vector with a different values. However, that's not the dominant interpretation of the superselection sectors in particle physics because the sectors with a bigger energy or angular momentum can still contain regions with a smaller or different value of these quantities - plus another separated object that compensates it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particle physics, the superselection sectors are usually reserved for subspaces of the total Hilbert space whose states can't evolve into each other because it would require an infinite amount of energy than work. So typically, superselection sectors in quantum field theory or string theory refer to subspaces of the Hilbert space that correspond to different boundary conditions at infinity (in the space - inside the spacetime).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One would need to change an &quot;infinite portion&quot; of the space to switch from one sector to another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So any &quot;finite energy excitation&quot; of one vacuum - with some values of the low-energy parameters - automatically belongs to a different superselection sector than a &quot;finite energy excitation&quot; of another vacuum - with another value of the low-energy parameters (such as the cosmological constant, particle masses, and couplings). Even in quantum field theory, one may find several stationary points of a field, e.g. a scalar field. It most of the spacetime find this field near one place, it's automatically a different sector than if the place is different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding localized particles can't change the superselection sector in this sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you add (stable) infinite branes whose total mass is infinite, you surely do change the superselection sector. The waves can propagate along the branes but the branes are attached to &quot;infinity&quot; at some loci and a finite being can't ever change it. Infinite branes or fluxes that are changed otherwise across the infinite space is the only way how I can imagine that your points about the superselection sectors and form fields are related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum cosmology, the division into superselection really disappears because any realistic Universe, regardless of the size, may be produced out of &quot;almost nothing&quot; - e.g. in inflation. But in many contexts, we study asymptotically flat, anti de Sitter, or other infinite Universes and the superselection sectors are sharply separated.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T15:43:25.543" LastActivityDate="2011-02-02T15:43:25.543" />
  <row Id="3482" PostTypeId="2" ParentId="3477" CreationDate="2011-01-21T11:26:58.333" Score="0" ViewCount="" Body="&lt;p&gt;every universe that qualitatively resembles ours - which certainly includes any universe in a hypothetical &quot;multiverse&quot; where ours may belong - has to have a logical arrow of time. The Universe must know which slices are &quot;input&quot; and which of them are &quot;output&quot;. Proofs analogous to Boltzmann's H-theorem then guarantee that the universes also have a thermodynamic arrow of time and the two arrows coincide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All universes that are in a direct or indirect causal relationship with ours have to have the &quot;same&quot; arrow of time as ours - one that doesn't contradict our arrow of time. It must be possible to define the arrow &quot;uniformly&quot; across the whole spacetime and all of its regions. In particular, one can never mix both arrows of time within the same universe - or the same multiverse. That would be logically equivalent to having closed time-like curves - which are inevitably inconsistencies - because you could get to the future along one arrow and back to the past - which would still be to the future according to another arrow of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evolution by the standard laws, starting with any well-defined slice, automatically preserves the uniformity of the arrows of time even when one allows tunneling and eternal inflation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-21T13:36:33.383" LastActivityDate="2011-01-21T13:36:33.383" CommentCount="6" />
  <row Id="3483" PostTypeId="2" ParentId="3359" CreationDate="2011-01-21T11:53:25.103" Score="0" ViewCount="" Body="&lt;p&gt;There is a subtlety to be considered in these discussions on EM waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Classical Electrodynamics the &lt;strong&gt;E&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; fields (electric and magnetic) are vector functions of space (and time). So &lt;strong&gt;E&lt;/strong&gt; is &lt;strong&gt;E&lt;/strong&gt;( &lt;strong&gt;x&lt;/strong&gt;) for example, and can be measured throughout space. This is how the theory is introduced as well in elementary texts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maxwell's equations are equations for &lt;strong&gt;E&lt;/strong&gt;( &lt;strong&gt;x&lt;/strong&gt;) and &lt;strong&gt;B&lt;/strong&gt;( &lt;strong&gt;x&lt;/strong&gt;) and so the solutions (the waves) are also functions of &lt;strong&gt;x&lt;/strong&gt;. Thus the solutions describe waves in space and time. The diagrams and pictures linked in the other Answers well display these wave solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complication is that one also has quantum theory - which replaces all classical theories as the &quot;correct&quot; framework for physics. In the replacement of Electrodynamics we have Quantum Electrodynamics. The key difference is that it introduces abstract spaces (called Hilbert Spaces) and abstract mathematical objects in them. &quot;Abstract&quot; here means that these are not spacetime objects, but only indirectly get measured at spacetime points. So what happens is that the &lt;strong&gt;E&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; become abstract entities called &quot;operators&quot; which are measured at each spacetime point (or region). These &lt;strong&gt;E&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; operators obey algebraic laws - which turn out to be just Maxwell's equations in this format. So the wave solutions describe abstract objects (which can nevertheless be measured) associated with each point, rather than actual wave movements in the real x and y directions as the classical theory would have us believe.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-01-21T11:53:25.103" CommentCount="1" />
  <row Id="3484" PostTypeId="2" ParentId="3349" CreationDate="2011-01-21T11:57:58.863" Score="1" ViewCount="" Body="&lt;p&gt;This paper might be of interest:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Rates of type I X-ray Bursts from Transients Observed with RXTE: Evidence for Black Hole Event Horizons&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ronald A. Remillard, Dacheng Lin, Randall L.Cooper, Ramesh Narayan&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/astro-ph/0509758&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/astro-ph/0509758&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Abstract: We measure the rates of type I X-ray bursts, as a function of the bolometric luminosity, from a likely complete sample of 37 non-pulsing transients (1996-2004). Our goals are to test the burst model for neutron stars and to investigate whether black holes have event horizons. We find 135 type I bursts in 3.7 Ms of exposure for the neutron-star group, and the burst rate function is generally consistent with model predictions. However, for the black hole groups (18 sources), there are no confirmed type I bursts in 6.5 Ms of exposure, and the upper limits in the burst function are inconsistent with the model predictions for heavy compact objects with a solid surface. There are systematic spectral differences between the neutron-star and black-hole groups, supporting the presumption that physical differences underly the sample classifications. These results provide indirect evidence that black holes do have event horizons.&lt;/p&gt;&#xA;" OwnerUserId="1458" LastActivityDate="2011-01-21T11:57:58.863" CommentCount="1" />
  <row Id="3485" PostTypeId="2" ParentId="3477" CreationDate="2011-01-21T11:58:33.090" Score="1" ViewCount="" Body="&lt;p&gt;The possibilities depends on how diverse the alternative universe can be. Some alternatives may have two or more macroscopic time dimensions, or they may be just euclidean spaces with no time dimension. In that case the concept of arrow of time will not make any sense in those universes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other universes which have one time dimension like ours there may be a fixed arrow of time as we (appear to) have. It would make no sense to ask whether it pointed in the same or different direction to ours if there is no direct connection with our universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also possible that a universe with one time dimension has no discernible arrow of time because everything is perpetually in a high entropy state. One theory is that this would be the natural state in our universe except that for some as yet unexplained reason the big bang singularity had a low entropy that creates an arrow for time. I agree with this view but it is not proven and I don't think there is a consensus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may also be possible that a universe has different directions for the arrow of time at different times and places so long as it does not switch directly from forward to backward causing paradoxes. It can switch from backwards to forwards with a low entropy epoch at the changeover, or from forwards to backwards if there is a heat-death epoch between them with maximal entropy that prevents useful information passing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't know the laws of physics well enough to say which of these possibilities can be ruled out on theoretical grounds so they currently stand as just possibilities. There is also a valid philosophical question about whether such questions make sense if we can't observe these universes. Personally I think it is valid to consider them if we have theories that predict the existence of other univserses.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-21T11:58:33.090" />
  <row Id="3486" PostTypeId="1" AcceptedAnswerId="3553" CreationDate="2011-01-21T12:11:05.457" Score="0" ViewCount="208" Body="&lt;p&gt;I read occasionally popular science articles and from time to time encounter issues about quantum information teleportation. (this one for example &lt;a href=&quot;http://www.physorg.com/news193551675.html&quot; rel=&quot;nofollow&quot;&gt;http://www.physorg.com/news193551675.html&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I have the following understanding of basic principles:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;after measurement entangled particles have identical random state (i.e. we can not predict the result, but we know that it will be similar for both particles).&lt;/li&gt;&#xA;&lt;li&gt;&quot;decoding key&quot; should be sent by traditional channel&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As I understand the underlying process is the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Some amount of entangled particles is sent to A and B.&lt;/li&gt;&#xA;&lt;li&gt;Particles are measured at A and B and results R (equal in both places) are recorded. &lt;/li&gt;&#xA;&lt;li&gt;Knowing the results A computes a &quot;decoding key&quot; (some function F, that F(R) has some valuable meaning) which is sent by traditional means to B. &lt;/li&gt;&#xA;&lt;li&gt;B applies key to it's copy of R, and thus receives information.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The benefit is that amount of information that needs to be transmitted by traditional means is small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case everything stated before is correct, I have the question: how is this &quot;quantum information teleportation&quot; process differs from just sending two identical sets of random data to both destinations by traditional means?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could be wrong about understanding of principles of quantum information teleportation, in this case, please, correct me.&lt;/p&gt;&#xA;" OwnerUserId="1414" LastActivityDate="2011-01-21T22:14:07.520" Title="What are the benefits of quantum information &quot;teleportation&quot;?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-information&gt;&lt;entanglement&gt;&lt;teleportation&gt;" AnswerCount="1" />
  <row Id="3487" PostTypeId="2" ParentId="3443" CreationDate="2011-01-21T12:32:31.170" Score="1" ViewCount="" Body="&lt;p&gt;As I understand this question the two Black Holes initially have some distance D between their Event Horizons, a region of Space also containing L1. So yes I think that this journey outside the Event Horizons is theoretically possible assuming that they dont coalesce on you whilst on this journey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One practical problem, and hence risk, likely arises from the non-stability of L1 (at least in Newtonian 3 Body theory - I assume also true in General Relativity). This would mean that a journey which slightly deviated from it, could orbit the nearest BH (in Newtonian theory tangentially escaping the system is more likely). So determining the exact nature of this instability for GR for this setup is crucial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once orbiting a Black Hole - even although outside the Event Horizon - the next issue is whether your Spaceship has the energy to escape that orbit. Meanwhile with the two Black Holes converging around you there could still be problems staying out of one of the Event Horizons....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should point out that whilst researching this question, I have come across the &quot;Interplanetary Transport Network&quot; - &lt;a href=&quot;http://en.wikipedia.org/wiki/Interplanetary_Transport_Network&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Interplanetary_Transport_Network&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-01-21T13:06:17.197" LastActivityDate="2011-01-21T13:06:17.197" CommentCount="7" />
  <row Id="3488" PostTypeId="1" CreationDate="2011-01-21T12:34:26.090" Score="3" ViewCount="453" Body="&lt;p&gt;As you see things that are far smaller, a funny question about this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine there are many people in a row (all are same height)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    A            Other Guy B             C             D&#xA;    |                        &#xA;you |  distance     |        distance        distance &#xA;    | &amp;lt;--------&amp;gt;    |        &amp;lt;-------&amp;gt;   |  &amp;lt;-------&amp;gt;   &#xA;    |               |                    |             |&#xA;    |               |                    |             |&#xA;    ------------------------------------------------------  floor   &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From your point of View you see D much smaller than you, then..&#xA;Remember, if this is an &quot;effect&quot; then D is not smaller than you, it's just far away&#xA;(and this is not earth curvature, because D is smaller in all sense not only in height)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you have a laser pointer and want to point your partner's face to annoy him&#xA;so &#xA;how do you target it? &#xA;do you align the laser with the floor at the height of your own face believing in geometry? or do you point directly at the &quot;little person&quot; D ?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    A               B                    C             D&#xA;    |. . . . Laser path . . . . .  . &amp;gt;. . . . . . &amp;gt;. . . . . . &amp;gt;. . .  . &amp;gt;&#xA;    |  ^            |                           &#xA;you |  |parallel    |                    |&#xA;    |  |            |                    |             |     &#xA;    |  v            |                    |             |&#xA;    ------------------------------------------------------ floor   &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then you don't hit him, and the &quot;visual effect&quot; became so real that you start to think D is for Dwarf. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is something Wrong? Physics opinions are welcome!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer is that floor itself will seem not aligned eigther&#xA;so targeting the Dwarf you will be really aligned with floor&#xA;because floor itself will have perspective effect.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    A               B                    C             D&#xA;    |. . . . Laser path . . .   &#xA;    |               |        ''''''''''''......&#xA;you |               |                    |      ......&amp;gt;|&#xA;    |               |                    |      _______|___floor   &#xA;    |               |         ___________|------                        &#xA;    --------------------------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Anyway its weird&lt;/p&gt;&#xA;" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-22T23:48:12.217" LastActivityDate="2011-01-24T11:18:36.927" Title="Why do things that are far away seem smaller?" Tags="&lt;optics&gt;&lt;geometric-optics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" />
  <row Id="3489" PostTypeId="1" AcceptedAnswerId="3491" CreationDate="2011-01-21T12:34:32.103" Score="4" ViewCount="367" Body="&lt;p&gt;&lt;strong&gt;What is quantum entanglement?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please be pedagogical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I have updated my background under my profile.&lt;/p&gt;&#xA;" OwnerUserId="425" LastEditorUserId="425" LastEditorDisplayName="" LastEditDate="2011-01-21T13:03:11.570" LastActivityDate="2011-02-04T07:40:29.227" Title="What is quantum entanglement?" Tags="&lt;quantum-mechanics&gt;&lt;entanglement&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="3490" PostTypeId="2" ParentId="3488" CreationDate="2011-01-21T13:04:29.147" Score="5" ViewCount="" Body="&lt;p&gt;Dear Hernan,&#xA;I will kindly assume that your question isn't a hoax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objects that are further from the observing eyes look smaller because of geometric optics. If an object at distance $D$ from our eyes has size $S$, the rays from its endpoints will arrive to our eyes from the same angles as the rays from a smaller object of size $S/k$ whose distance is $D/k$, because of two simple similar triangles. The angle is essentially $S/D=(S/k)/(D/k)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single eye may only detect the direction from which a light ray is coming - the light ray will create a point on the retina that only depends on the angle from which it is coming (let's neglect focusing now - focusing either by lens in each eye or by the relative position of both eyes &lt;em&gt;is&lt;/em&gt; able to determine the distance of close enough objects), so it will think that the large object of size $S$ is as small as the smaller object of size $S/k$ just because the larger object is $k$ times further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A random picture sufficient to explain what I mean, see below:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/rmZPI.jpg&quot; alt=&quot;Similar triangles&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the picture above, &quot;A&quot; is the eye, the line interval &quot;BC&quot; is the closer and smaller object, and the &quot;DE&quot; interval is the larger but more distant object. The light rays from points &quot;C&quot; and &quot;E&quot; arrive to the eye &quot;A&quot; from the same direction, and similarly for the light rays from &quot;B&quot; and &quot;D&quot;, so the eye can't really distinguish the objects &quot;BC&quot; and &quot;DE&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2011-01-21T16:22:38.947" LastActivityDate="2011-01-21T16:22:38.947" CommentCount="6" />
  <row Id="3491" PostTypeId="2" ParentId="3489" CreationDate="2011-01-21T13:18:31.423" Score="5" ViewCount="" Body="&lt;p&gt;Entanglement is a quantum correlation between two (or many) objects - a correlation means that these two objects' properties are not independent of each other - which was created in the objects' common past when they were close to one another i.e. when they were two parts of the same physical system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum mechanics changes the character of possible &quot;properties&quot; that objects may have (the quantities describing objects's properties are usually called &quot;observables&quot; and they are represented by Hermitian operators on the Hilbert space), as well as the way how these properties are measured and predicted (just probabilistically), so it also changes the character and magnitude of correlations that the objects may exhibit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, quantum correlations may often be stronger - and affecting a large fraction of measurable properties of the objects - than what would be possible according to classical (i.e. non-quantum) physics. In classical physics, correlations have to satisfy e.g. the so-called Bell's inequalities in various situations but quantum mechanics - and the real world - can easily surpass these bounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically, objects and their properties in quantum mechanics are described by wave functions. To describe the state of two mostly independent objects, one has to take a wave function from the tensor product $H_1\otimes H_2$ of the Hilbert spaces describing the individual objects. The wave function in the tensor product implies probabilistic predictions for any pairs of properties of the first and second object; in general, they're not independent, and for each combination of the objects' properties, quantum mechanics (and the wave function) may remember an independent probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any vector in the tensor product that can't be written as a tensor product of vectors from &#xA;$H_1$ and $H_2$ (instead, it can only be written as a linear combination of such tensor products of vectors) is called entangled. In other words, it is non-entangled if it is a simple tensor product of two simpler vectors. If it is a simple tensor product, all probabilities of &quot;coupled properties&quot; of the pair of objects simply factorize to the probability of the first object, and probability of the second object, as you know from probabilities of independent phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest pedagogical example of an entangled state (as well as the entangled state that is most often found in literature) is&#xA;$$\frac{X_1\otimes Y_1 + X_2\otimes Y_2}{\sqrt{2}}$$&#xA;Because there are two terms with four different factors, you can't use the distributive law in any way that would allow you to rewrite it as a simple product. The letters $X,Y$ refer to the two objects and the labels $1,2$ refer to two different states of each of the two objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this state, if the property &quot;1 or 2&quot; is measured on $X$, one obtains the answers 1 or 2 with 50% probability for each: the coefficient of the wave function is $1/\sqrt{2}$ because these complex coefficients have to be squared to obtain the probability. However, because $X_1$ is &quot;coupled&quot; to $Y_1$ and $X_2$ is coupled to $Y_2$, the state and the machinery of quantum mechanics predict that the object $Y$ will be measured to have the same property: if $X$ is in 1, $Y$ is in 1, and the same for the state 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linear algebra - which is crucially important for quantum mechanics - allows one to reinterpret the state above as an &quot;identity operator&quot; so the correlation will exist regardless of the type of measurement that we perform both on $X$ and $Y$. For example, if the two states represent spins, the two particles will be correlated so that you will find out that they're polarized with respect to the same axis, if you measure both particles' polarizations with respect to the same, particular, but arbitrary axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be kind of impossible for two separated particles in classical physics that could only be perfectly correlated for one choice of the axis - but not another axis rotated by 45 degrees, for example - without a communication in between them. However, quantum mechanics predicts that such a 100% correlation &quot;regardless of the axis&quot; is not only possible but guaranteed by the state above and it requires no communication. Indeed, one can prove that relativistic theories in quantum mechanics - especially quantum field theory - don't allow one to transmit a single bit of information faster than light even though this would be needed in classical physics to guarantee the perfect correlation that quantum mechanics predicts for these experiments (and that the experiments confirm).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:40:29.227" LastActivityDate="2011-02-04T07:40:29.227" CommentCount="7" />
  <row Id="3492" PostTypeId="2" ParentId="3443" CreationDate="2011-01-21T13:37:47.170" Score="1" ViewCount="" Body="&lt;p&gt;You can't have a stable configuration of two orbiting black holes at all separations.  Even in the case of a test particle orbiting a single non-spinning black hole, the innermost stable orbit for the test particle is located at $r=6M$, with closer orbits plunging into the hole (circular orbits closer than $6M$ are unstable solutions analogous to balancing a rigid pendulum vertically.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have two black holes of appreciable mass, they will radiate away energy, gradually falling in closer and closer to each other, and then at some time before a common horizon is formed (also meaning before the &quot;center of mass&quot; of the system is inside the horizon), the two holes will hit their last stable orbit, and then will very rapidly combine, giving off a large burst of radiation, and an eventual end state of a single black hole.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a final aside, I should note that a spinning black hole's singularity is a ring, not a point, and that the center of this ring IS locally flat.  Since a binary black hole collision will definitely have net angular momentum (even if just from the orbital angular momenta of the holes), this means that the end state will almost certainly be a spinning hole, so it is not necessary that the &quot;center point&quot; ever not be locally flat.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-01-21T13:37:47.170" CommentCount="1" />
  <row Id="3493" PostTypeId="2" ParentId="3489" CreationDate="2011-01-21T13:40:44.750" Score="0" ViewCount="" Body="&lt;p&gt;A spin system has in the basis of the Pauli matrix $\sigma_z$ the states $|+\rangle$ and $|-\rangle$ for spin up and down.  The Pauli matrix acts on these states as&#xA;$$&#xD;&#xA;\sigma_z|\pm\rangle~=~\pm|\pm\rangle.&#xD;&#xA;$$ &#xA;Now these states are complex numbers, which means there are $2$ variables for each state and thus $4$ altogether.  However, there are constraints, such as the probability Born rule $1~=~P_+~+~P_-$, $P_\pm~=~|a_\pm|^2$ for a state $|\psi\rangle~=~a_+|+\rangle~+~a_-|-\rangle$, and irrelevance of a phase in real valued measurements.  So this reduces the number of variables from $4$ to $4~–~2~=~2$.  That is just what we would expect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let us consider two spin systems, say two electrons.  The use of electron spin state is not concrete, for these arguments hold just as well for polarization direction of photons.  So we have two sets of states and operators $\{\sigma_z, |\pm\rangle\}^1~ \{\sigma_z, |\pm\rangle\}^2$ denoted with an additional index $i~=~1,~2$ and we still have&#xA;$$&#xD;&#xA;\sigma ^i_z|\pm\rangle^i~=~\pm|\pm-\rangle^i.&#xD;&#xA;$$&#xA;We can form two independent states $|\psi\rangle^i~=~a^i_+|+\rangle^i~+~a^i_-|-\rangle^i$ for the two spin systems.  For each there are $4$ variables and $2$ constraints.  This gives $4$ degrees of freedom in total.  Yet we can compose these spin states in various ways.  One way of doing this is&#xA;$$&#xD;&#xA;|\psi\rangle~=~{1\over\sqrt{2}}(|+\rangle|-\rangle~+~e^{i\phi}|-\rangle |+\rangle),&#xD;&#xA;$$&#xA;where I have dropped the index $i$, and we just implicitly see the first and second $|\pm\rangle$ as $i~=~1$ and $2$.  This makes reading things clearer.  The $e^{i\phi}$ is a phase which for it equal $+$ and $–$ the state $|\psi\rangle$ is not an eigenstate of $\sigma^i$ and is an eigenstate of $\sigma ^i$ respectively.  So these are singlet and triplet state configurations.  This is an entangled state.  If you have access to $|\pm\rangle^1$ then you also have access to $|\pm\rangle^2$, and this holds no matter how far apart these states end up as.  You can entangle two electrons by overlapping their wave functions.  One that is done you can separate them arbitrarily far and they are still entangled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let us count the degrees of freedom for this state.  We have again $4$ variables for each $|\pm\rangle^i$ but now we have one constraint from Born rule and another from the entanglement state.  So you have $6$ independent variables with $4$ constraints giving $2$ in total.  This is the basic bipartite entanglement.  There are also n-partite entanglements, such as the W and GHZ states.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-21T13:40:44.750" CommentCount="1" />
  <row Id="3494" PostTypeId="2" ParentId="3468" CreationDate="2011-01-21T13:57:37.193" Score="2" ViewCount="" Body="&lt;p&gt;Back in the 1970s there were some researchers who put seeds in a beam path to see what would happen.  The plants which grew from the seeds were deformed.  One would have to look this up to get the details.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put your hand in the LHC beam one question is what would hurt the most; the damage due to high energy particle slamming into nuclei in your hand, or the high velocity shoe mark you would get on your butt.  A reasonable amount of the beam energy would be deposited in your hand, where primary damage would be from nuclei blasted to bits from the protons and secondary radiation from that.  You would have radiation damage.  However, I suspect it would be comparable to getting a radiation does from a synchrotron source.  Most damaged cells would die, those which are mildly damaged might have genome changes which could lead to transformed cells, maybe even runaway cell growth.  Runaway cell growth in somatic cell lines is what we call cancer.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-21T13:57:37.193" />
  <row Id="3495" PostTypeId="1" AcceptedAnswerId="3512" CreationDate="2011-01-21T14:05:33.403" Score="4" ViewCount="293" Body="&lt;p&gt;Consider standard quantum mechanics, but forget about the collapse of the wavefunction. Instead, use decoherence through interaction with the environment to bring the evolving quantum state into an eigenstate (rspt arbitrarily close by). Question: Can this theory be fundamentally deterministic? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one takes into account that the variables of the environment are not known, then the evolution is of course 'undetermined' in a probabilistic sense, but that isn't the question. Question is if fundamentally quantum mechanics with environmentally induced decoherence can be deterministic. Note that I'm not saying it has to be. I might be mistaken, but it seems to me decoherence could be followed by an actual non-deterministic process still, so the decoherence alone doesn't settle the question of determinism or non-determinism. Question is if one still &lt;em&gt;needs&lt;/em&gt; a non-deterministic ingredient? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: Please note that I asked whether the evolution can &lt;em&gt;fundamentally&lt;/em&gt; be deterministic, or whether it has to be non-deterministic. It is clear to me that for all practical purposes it will appear non-deterministic. Note also that my question does &lt;em&gt;not&lt;/em&gt; refer to the prepared state after tracing out the environmental degrees of freedom, but to the full evolution of system and environment. Does one need a non-deterministic ingredient to reproduce quantum mechanics, or can it with the help of decoherence be only apparently non-deterministic yet fundamentally deterministic?&lt;/p&gt;&#xA;" OwnerUserId="1458" LastEditorUserId="1458" LastEditorDisplayName="" LastEditDate="2011-01-23T10:48:17.093" LastActivityDate="2011-01-24T03:47:17.643" Title="Does decoherence need non-determinism?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="1" />
  <row Id="3496" PostTypeId="1" AcceptedAnswerId="3504" CreationDate="2011-01-21T14:26:53.303" Score="-4" ViewCount="384" Body="&lt;ul&gt;&#xA;&lt;li&gt;What are the main differences between these quantum theories?&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Quantum Mechanics &lt;/li&gt;&#xA;&lt;li&gt;Quantum Field theory&lt;/li&gt;&#xA;&lt;li&gt;Quantum Gravity&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask this question because when I asked a question before people talk about that I’m mixing different frameworks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have narrowed down my question.&lt;/p&gt;&#xA;" OwnerUserId="425" LastEditorUserId="425" LastEditorDisplayName="" LastEditDate="2011-01-24T09:47:38.777" LastActivityDate="2011-01-24T09:47:38.777" Title="What are the main differences between these quantum theories?" Tags="&lt;quantum-mechanics&gt;&lt;classical-mechanics&gt;&lt;quantum-field-theory&gt;&lt;quantum-gravity&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="3497" PostTypeId="1" CreationDate="2011-01-21T14:47:59.023" Score="2" ViewCount="180" Body="&lt;p&gt;&lt;a href=&quot;http://www.mbeckler.org/resistor_grid/&quot; rel=&quot;nofollow&quot;&gt;The numerical simulation&lt;/a&gt; of &lt;a href=&quot;http://physics.stackexchange.com/questions/2072/on-this-infinite-grid-of-resistors-whats-the-equivalent-resistance&quot;&gt;this nerdy question&lt;/a&gt; shows that the resistance decreases with the number of nodes along longest side, and converges to a finite value when the # of nodes approaches infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here comes my question: &lt;strong&gt;Can such grid or mesh design be utilized to reduce the resistance in wires/cables for long distance transport?&lt;/strong&gt; Let's say one unit-length segment of copper wire can be treated as one unit resistor, so a copper wire mesh is basically equivalent to a grid of resistors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there exist many different ways to reduce the resistance, such as using a thicker wire (increase the cross section area), lowering the temperature, or even changing the material. But here I specifically wonder what simple circuit or topological design can do the trick, and whether it is practical or not.&lt;/p&gt;&#xA;" OwnerUserId="749" LastEditorUserId="749" LastEditorDisplayName="" LastEditDate="2011-01-21T14:56:09.773" LastActivityDate="2011-01-21T16:45:28.440" Title="Special designs to reduce the electrical resistance of a wire" Tags="&lt;electromagnetism&gt;&lt;electricity&gt;" AnswerCount="2" />
  <row Id="3498" PostTypeId="1" CreationDate="2011-01-21T14:50:09.660" Score="2" ViewCount="169" Body="&lt;p&gt;This one should be a bit of a softball, but I can't find it explicitly stated anywhere on the internet, and my basic unit analysis doesn't seem to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a beam of synchrotron radiation with a brilliance $B$ at a given energy.  (For instance, the brilliance of the Advanced Light Source at Berkeley is given in a graph form &lt;a href=&quot;http://www-als.lbl.gov/index.php/beamlines/photon-source-parameters.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).  If I know the distance $d$ from the source and the area $A$ of the (for simplicity, unfocused) spot on my sample, how do I calculate the flux (units photons/sec) to my sample?  Another way to phrase the same question is given the distance from the source and the brilliance, how do I calculate the intensity (units photons/sec/area) of the photon source at my sample?  Am I missing some parameter?&lt;/p&gt;&#xA;" OwnerUserId="144" LastEditorUserId="144" LastEditorDisplayName="" LastEditDate="2011-01-22T15:58:02.903" LastActivityDate="2011-08-09T12:21:29.383" Title="Converting between brilliance, intensity, and flux" Tags="&lt;optics&gt;&lt;spectroscopy&gt;&lt;synchrotron-radiation&gt;&lt;unit-conversion&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="3499" PostTypeId="2" ParentId="3497" CreationDate="2011-01-21T14:54:18.103" Score="3" ViewCount="" Body="&lt;p&gt;No, using one wire between two points A and B will beat any other topology for the same R, in terms of (total mass of wire)/(distance A-to-B), thus also ofcourse less cost. Just imagine you got a 2xN grid, how long do you want to make the connections between the two wires, clearly as short as possible until you got 1 wire.&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2011-01-21T15:12:37.340" LastActivityDate="2011-01-21T15:12:37.340" />
  <row Id="3500" PostTypeId="1" AcceptedAnswerId="3501" CreationDate="2011-01-21T15:47:13.683" Score="12" ViewCount="312" Body="&lt;p&gt;It is an usual practice that any field theory starts with a suitable lagrangian density. It has been proved enormously successful. I understand, it automatically ensures valuable symmetries of physics to be preserved. But nevertheless the question about the generality of this approach keeps coming to my mind. My question is how one can be sure that this approach has to be always right and fruitful. Isn't it possible, at least from the mathematical point of view that a future theory of physics does not subscribe to this approach?&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-01-21T16:48:35.743" Title="How general is the Lagrangian approach in field theory?" Tags="&lt;quantum-field-theory&gt;&lt;lagrangian-mechanics&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="5" />
  <row Id="3501" PostTypeId="2" ParentId="3500" CreationDate="2011-01-21T16:08:46.237" Score="12" ViewCount="" Body="&lt;p&gt;That's an excellent question, which has a few aspects:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Can you quantize any given Lagrangian? The answer is no. There are classical Lagrangians which do not correspond to a valid field theory, for example those with anomalies.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do you have field theories with no Lagrangians? Yes, there are some field theories which have no Lagrangian description. You can calculate using other methods, like solving consistency conditions relating different observables.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Does the quantum theory fix the Lagrangian? No, there are examples of quantum theories which could result from quantization of two (or more) different Lagrangians, for example involving different degrees of freedom.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The way to think about it is that a Lagrangian is not a property of a given quantum theory, it also involves a specific classical limit of that theory. When the theory does not have a classical limit (it is inherently strongly coupled) it doesn't need to have a Lagrangian. When the theory has more than one classical limit, it can have more than one Lagrangian description.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prevalence of Lagrangians in studying quantum field theory comes because they are easier to manipulate than other methods, and because usually you approach a quantum theory by &quot;quantizing&quot; - meaning you start with a classical limit and include quantum corrections systematically. It is good to keep in mind though that this approach has its limitations.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-01-21T16:26:57.273" LastActivityDate="2011-01-21T16:26:57.273" CommentCount="2" />
  <row Id="3502" PostTypeId="2" ParentId="3432" CreationDate="2011-01-21T16:12:06.027" Score="4" ViewCount="" Body="&lt;h2&gt;&lt;a href=&quot;http://scienceblogs.com/startswithabang/&quot; rel=&quot;nofollow&quot;&gt;Starts With a Bang&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Similar in tone and style to Cosmic Variance with entertaining in-depth posts on various subjects in astrophysics and cosmology, including an excellent 8 part series on the history of the universe called &lt;a href=&quot;http://scienceblogs.com/startswithabang/2010/04/the_greatest_story_ever_told_-_7.php&quot; rel=&quot;nofollow&quot;&gt;&quot;The Greatest Story Ever Told&quot;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1479" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2011-01-21T22:12:02.607" LastActivityDate="2011-01-21T22:12:02.607" CommunityOwnedDate="2011-01-21T16:12:06.027" />
  <row Id="3503" PostTypeId="1" AcceptedAnswerId="3524" CreationDate="2011-01-21T16:12:17.473" Score="6" ViewCount="509" Body="&lt;p&gt;Consider a complex scalar field $\phi$ with the Lagrangian:&lt;br&gt;&#xA;$L = \partial_\mu\phi^\dagger\partial^\mu\phi - m^2 \phi^\dagger\phi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider also two real scalar fields $\phi_1$ and $\phi_2$ with the Lagrangian:&lt;br&gt;&#xA;$L = \frac12\partial_\mu\phi_1\partial^\mu\phi_1 - \frac12m^2 \phi_1^2&#xD;&#xA;    +\frac12\partial_\mu\phi_2\partial^\mu\phi_2 - \frac12m^2 \phi_2^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these two systems essentially the same? If not -- what is the difference?&lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-22T23:53:30.023" LastActivityDate="2011-08-13T08:07:58.483" Title="What is the difference between complex field and two scalar fields?" Tags="&lt;quantum-field-theory&gt;" AnswerCount="3" CommentCount="11" />
  <row Id="3504" PostTypeId="2" ParentId="3496" CreationDate="2011-01-21T16:20:42.897" Score="4" ViewCount="" Body="&lt;p&gt;Physics frameworks lead into each other as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mechanics leads to Statistical Mechanics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statistical Mechanics leads to Thermodynamics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is the quantum frameworks, which in their limiting case become the corresponding classical ones&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum Mechanics, limiting case Mechanics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum Statistical Mechanics, limiting case  Statistical Mechanics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is Quantum Field theory, which developed as a way to calculate crossections and is appropriate for particle physics, i.e. small dimensions, high energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mechanics leads to Neuton's Gravity Theory&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General Relativity has a limiting case Neuton's Gravity Theory&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no consistent quantum gravity outside of String Theory, which is the frontier being studied now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;String Theories have as limiting cases General Relativity and Quantum Field Theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All capitalized theories are &quot;physics  conceptual frameworks used for calculations&quot;&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-01-21T16:20:42.897" CommentCount="5" />
  <row Id="3505" PostTypeId="1" CreationDate="2011-01-21T16:21:34.737" Score="0" ViewCount="54" Body="&lt;p&gt;Ok, background - studying for the physics part of my radiology exams, and came across a question that went something like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An electron fired through a tungsten target loses energy by:&#xA;a) Bremsstrauhlung&#xA;b) collisions with bound target electrons&#xA;c) giving off characteristic radiation&#xA;d) all of the above&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, by the process of elimination d must be right, because both a and b are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question regards c - characteristic radiation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding of this process is the excitation of the electron to a higher energy state, usually a more peripheral valence shell, which then releases that extra energy when it drops back (or another drops in to fill the hole)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what I don't get is how a free electron (like one fired from the filament to the target) has energy states. By definition, it is unbound, and therefore there are no binding energies to create  quantised energy states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a property I am overlooking, is the question wrong, or is it a fudge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I apologise if this has been asked. I looked at the suggested answers, and did a google search of stack exchange, but my phone can't use the search bar here directly, so I could have missed something&lt;/p&gt;&#xA;" LastActivityDate="2011-01-21T17:25:55.147" Title="Free electrons and energy states" Tags="&lt;particle-physics&gt;" AnswerCount="1" />
  <row Id="3506" PostTypeId="2" ParentId="3503" CreationDate="2011-01-21T16:28:12.403" Score="6" ViewCount="" Body="&lt;p&gt;They're identical. Typically, we use complex fields if we have a $U(1)$ symmetry, or some more complicated gauge group with complex representations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, the same comment applies to whether we use Majorana spinors or Weyl spinors.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-21T16:39:23.170" LastActivityDate="2011-01-21T16:39:23.170" CommentCount="2" />
  <row Id="3507" PostTypeId="2" ParentId="3500" CreationDate="2011-01-21T16:33:04.427" Score="0" ViewCount="" Body="&lt;p&gt;As far as you're wondering about 'quantum' field theories, all bets are off - just take a look at the arXiv or on Google. However, most of those theories seem (to me) less well studied than regular QFT. They do have a lot of structure in common with normal field theory (you can still have a Hamiltonian, for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of classical fields, I might be giving some kind of trivial answer, but you could consider a (classical) field that obeys a differential equation that can not be derived from a Lagrangian through the Euler-Lagrange equations. Maybe someone with a background in PDE's can elaborate on this.&lt;/p&gt;&#xA;" OwnerUserId="1471" LastActivityDate="2011-01-21T16:33:04.427" />
  <row Id="3508" PostTypeId="2" ParentId="3497" CreationDate="2011-01-21T16:45:28.440" Score="4" ViewCount="" Body="&lt;p&gt;kakemonsteret's answer is perfectly correct, and complete as far as DC resistance goes, but people still make stranded wire such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Litz_wire&quot; rel=&quot;nofollow&quot;&gt;Litz wire&lt;/a&gt;. Why is that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have DC current flowing through a normal conductor, the current spreads out so that the current density is approximately constant throughout the cross section of the conductor. This is good for making low-resistance wires because you can make the wire thicker and the current actually flows through the whole cross-section, so the resistance decreases in proportion to the area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you have AC current flowing back and forth at a relatively high frequency (which depends on the relevant length dimensions), the current will not distribute itself evenly over the cross-section - instead it will become concentrated at the surface. This is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Skin_effect&quot; rel=&quot;nofollow&quot;&gt;skin effect&lt;/a&gt; and it's caused by the magnetic field of the current causing a Lorentz force on itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the skin effect, if you want to reduce the impedance (generalization of resistance) of a wire at some frequency, just making it thicker and thicker is not very effective. This is part of the reason why stranded and woven wire like Litz wire is used. If you look at the cable from a lightning rod that's designed to handle the lightning current, it will often have many copper strands woven together. Lightning current is DC, but the current ramps up from zero very quickly, which means the current signal has high-frequency components.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-01-21T16:45:28.440" CommentCount="1" />
  <row Id="3509" PostTypeId="2" ParentId="3500" CreationDate="2011-01-21T16:48:35.743" Score="0" ViewCount="" Body="&lt;p&gt;I was confused by this too.&lt;br&gt;&#xA;Maybe I'm wrong, but at some point I made my mind in a following way:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The most general way to describe 'quantum anything' is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Path_integral_formulation&quot; rel=&quot;nofollow&quot;&gt;Feynman path integral&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;In this picture for each 'path' we have to have an &lt;em&gt;action&lt;/em&gt; &lt;/li&gt;&#xA;&lt;li&gt;If we want causality and are agree that our world is Poincaré invariant, then the action must be &lt;em&gt;local&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;which means that action must be an integral of &quot;something&quot; over the space-time.&lt;/li&gt;&#xA;&lt;li&gt;This &quot;something&quot; is a Lagrangian&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Therefore questioning generality of the Lagrangian you are actually:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;either questioning the generality of Feynman path integral &lt;/li&gt;&#xA;&lt;li&gt;or questioning the validity of Poincaré invariance&lt;/li&gt;&#xA;&lt;li&gt;or the necessity of casuality&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Repeat -- I can be wrong. And I would be happy if someone points out my mistake. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-01-21T16:48:35.743" CommentCount="1" />
  <row Id="3510" PostTypeId="2" ParentId="1453" CreationDate="2011-01-21T17:03:43.123" Score="3" ViewCount="" Body="&lt;p&gt;Is information preserved in evaporating black holes in quantum gravity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there are already too many answers here relating to quantum gravity, but I think this question is important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to quantum field theory in curved spacetime, information crossing the event horizon will end up at the singularity where it will be destroyed. If we produce a pair of entangled particles outside the black hole, and throw one of them inside, it might appear as if we have converted a pure state into a mixed state, which is extremely problematic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of weird and undesirable things will happen if unitary time evolution is violated in quantum mechanics, which is why it's likely a complete theory of quantum gravity will lead to unitary time evolution. So, the question is now where is the information hidden?&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-01-21T17:03:43.123" CommunityOwnedDate="2011-01-22T18:11:53.713" />
  <row Id="3511" PostTypeId="2" ParentId="3505" CreationDate="2011-01-21T17:04:35.463" Score="1" ViewCount="" Body="&lt;p&gt;The (c) option is badly worded, in my opinion. The characteristic radiation is emitted by the &lt;em&gt;atom&lt;/em&gt; and this indirectly contributes to the original electron (say, A) losing energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how it happens, roughly: An electron A fired into the target can ionize one of the atoms which ejects an electron B. In doing so, the electron A loses energy. But the effect is seen when the ionized atom emits characteristic radiation when one of the higher shell electrons falls into the shell from which the atom ejected electron B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there's no question of free particles in bound states. Free particles by definition are not in bound states.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-01-21T17:25:55.147" LastActivityDate="2011-01-21T17:25:55.147" CommentCount="2" />
  <row Id="3512" PostTypeId="2" ParentId="3495" CreationDate="2011-01-21T17:13:18.023" Score="3" ViewCount="" Body="&lt;p&gt;Short answer to &quot;Can this theory be fundamentally deterministic?&quot;: No. Decoherence is the diagonalization of the density matrix in a preferred basis, with the off-diagonals vanishing at late times. Since you can get the same final diagonal matrix from several possible initial pure states of the system under consideration, there's a necessary loss of information and irreversibility. (I'm guessing this is what you meant by non-deterministic)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit more detail: Decoherence proceeds by the rapid establishment of entanglement-induced correlations between the system and the infinite degrees of freedom of the decohering environment. The second law prevents this process from being reversible (since S has to always increase, and S is zero for the pure state, while it is greater than zero for the decohered mixed state). If you take the second law to be fundamental, then the non-determinism here is fundamental too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: The updated question now refers to the full evolution of system + environment, in other words, the entire universe. Since there's nothing else for the universe to entangle with, it will remain in a pure state and evolve deterministically for ever if it always was in a pure state. I however don't know if the universe is in a pure state or a mixed state. Anyone?&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-01-24T03:47:17.643" LastActivityDate="2011-01-24T03:47:17.643" CommentCount="7" />
  <row Id="3513" PostTypeId="2" ParentId="3495" CreationDate="2011-01-21T17:17:21.610" Score="2" ViewCount="" Body="&lt;p&gt;UPDATE: It seems that we have not been answering the question WIMP intended. Here is an updated answer to deal with what I now understand to be the question: Given any unknown quantum state $|\psi\rangle$, can there be any deterministic process which will make it collapse onto a particular state $|\phi \rangle$, if $|\phi \rangle\langle \psi| \neq 0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to this question is no, because it violates the linearity of quantum mechanics, allowing us to distinguish between non-orthogonal states. This is trivial, because states orthogonal to $|\phi \rangle$ will have zero probability of collapsing onto it. This may not seem like a big deal, but it turns out that linearity is fundemantal to quantum mechanics on many levels. If we remove this constraint, then entanglement can be used to signal, and hence create problems with causality. No signalling seems one of the most fundamental features of physics, showing up in many independent theories (electromag, quantum mechanics, relativity, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how this can be done, consider an entangled state $\frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)$. This is the anti-symmetric state: for any basis $\sigma$ a measurement resulting in outcome $m$ will leave the other qubit in the opposite eigenstate of $\sigma$. Thus, if you could deterministically collapse onto the state $|0\rangle$ then you can be sure the your half of the EPR pair was not left in state $|1\rangle$ after the measurement on the other half. So, for Alice to communicate with Bob, she need only choose to measure in the $X$ or $Z$ basis. Measuring in $X$ will mean Bob receives the output $|0\rangle$ with probability 1, where as measuring in $Z$ will return result $|1\rangle$ with probability $\frac{1}{2}$. Although this is probabilistic, you can repeat the process arbitrarily many times to get exponentially close to perfect communication. This instantaneous communication breaks causality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you allow all states to collapse to the target state, then the only solution is a channel which swaps the state with another ancilla system. Systems which can perform such deterministic collapse can always be used to signal, as well as allowing all sorts of additional weirdness like efficient solutions to PSPACE-complete problems in computation and time travel. As a result, this is totally impossible within the current framework of physical theories, and there are very substantial reasons to believe that it is a feature of any physical theory that is valid in our world.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The answer is no, if by deterministic you mean possessing a local hidden variable interpretation. This follows directly from the observed violations of &lt;a href=&quot;http://en.wikipedia.org/wiki/Bell%27s_inequalities&quot; rel=&quot;nofollow&quot;&gt;Bell's inequality&lt;/a&gt;, which ever interpretation of quantum mechanics you choose (what you are referring to is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Everett_interpretation&quot; rel=&quot;nofollow&quot;&gt;Everett interpretation&lt;/a&gt; of quantum mechanics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bell's inequality works as follows: Given to possible local measurement operators ($A_i$ and $B_i$) at each of two localions $i \in \{1,2\}$, what is the maximum value of the expectation value of $\langle A_1 B_1 + A_1 B_2 + A_2 B_1 - A_2 B_2\rangle$. What Bell showed was that this can take on a value of at most 2 for any local hidden variables theory. However quantum mechanics allows it to take on values up to $2\sqrt{2}$, and many experiments have recorded violations of this inequality, showing values in the range $2 &amp;lt; v \leq 2\sqrt{2}$. This essentially rules out a local hidden variable model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, you mean can the unitary interaction of two particles give rise to decoherence, then the answer is yes, as follows: Imagine two particles initially in the state $1/\sqrt{2}(|0\rangle + |1\rangle)$. Now imagine they interact via an Ising interaction. After a certain time, they will be in the joint state $1/2(|00\rangle - |10\rangle - |01\rangle + |11\rangle)$. This is still a pure state, and so no decoherence has occurred. However, imagine one of these particles moves off far away (into the environment). If we only have access to one of these particles, then its reduced density matrix will be $1/2(|0\rangle \langle 0|+|1\rangle \langle 1|)$, which is simply a classical random distribution over the two orthogonal states, the same as would occur do to a collapse of the wavefunction.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditorDisplayName="" LastEditDate="2011-01-23T12:20:57.133" LastActivityDate="2011-01-23T12:20:57.133" CommentCount="6" />
  <row Id="3514" PostTypeId="2" ParentId="3503" CreationDate="2011-01-21T17:23:58.020" Score="2" ViewCount="" Body="&lt;p&gt;A complex scalar field represents a single charged particle whereas two real scalar fields &lt;strong&gt;may&lt;/strong&gt; represent two independent neutral particles. The difference is easy to note while imposing physical initial, boundary and/or normalization conditions which essentially depend on what you are describing - one charged or two different neutral particles. Two independent neutral scalars do not obey a superposition principle, one cannot mix them in one field.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-01-22T20:07:39.703" LastActivityDate="2011-01-22T20:07:39.703" CommentCount="7" />
  <row Id="3515" PostTypeId="2" ParentId="3261" CreationDate="2011-01-21T17:30:28.973" Score="0" ViewCount="" Body="&lt;p&gt;A basic part of the theory is to remember the Boltzmann distribution: N = exp (-E/kT) for each energy level E, where N is the occupation number and T is the Temperature (I am omitting subscripts). So for a given T (=/=0) not everything is at the ground state. These nonground state atoms can decay by spontaneous emission as stated in other Answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spontaneous Emission does have a modern (post Einstein) explanation in terms of the Quantum Vacuum however, which might also interest you.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-01-21T17:30:28.973" />
  <row Id="3516" PostTypeId="2" ParentId="3432" CreationDate="2011-01-21T17:34:35.707" Score="3" ViewCount="" Body="&lt;h2&gt;&lt;a href=&quot;http://www.science20.com/hammock_physicist&quot; rel=&quot;nofollow&quot;&gt;The Hammock Physicist&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Ok, time for some shameless self-promotion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like most science blogs, this blog focuses on an interested lay audience. I try to explain common misunderstandings, often by elucidating the physics with simple models and (animated) visuals. See for instance &lt;a href=&quot;http://www.science20.com/hammock_physicist/fibonacci_chaos_and_times_arrow&quot; rel=&quot;nofollow&quot;&gt;'Fibonacci Chaos'&lt;/a&gt;, &lt;a href=&quot;http://www.science20.com/hammock_physicist/limits_science_god_godel_gravity&quot; rel=&quot;nofollow&quot;&gt;'God, Godel, Gravity'&lt;/a&gt; and &lt;a href=&quot;http://www.science20.com/hammock_physicist/physical_reality_less_more&quot; rel=&quot;nofollow&quot;&gt;'Less Is More'&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastEditorUserId="1268" LastEditorDisplayName="" LastEditDate="2011-01-21T22:16:56.283" LastActivityDate="2011-01-21T22:16:56.283" CommunityOwnedDate="2011-01-21T17:34:35.707" />
  <row Id="3517" PostTypeId="2" ParentId="2558" CreationDate="2011-01-21T17:38:02.270" Score="1" ViewCount="" Body="&lt;p&gt;A black hole with an event horizon that is thermally equal to its environment will not be in equilibrium.  A black hole with the mass of about the moon and in free space would have the same temperature as the CMB. If it absorbs a unit of energy its horizon increases in size and it become cooler.  This means it will preferentially grow.  If it emits a quantum of Hawking radiation it will becomes smaller and preferentially radiates more radiation.  So there is no equilibrium.  This is because the effective heat capacity of an event horizon is negative.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no physically possible way to hold a black hole in an eternal state, except for one case.  This is a black hole in an anti de Sitter spacetime.  In this case the geodesic of the AdS are “repelling,” or equivalently any clock observed close to the boundary is seen to demark time at a faster rate.  So the BH will not crash into the boundary.  Also the Hawking radiation emitted by the BH will reach equilibrium with the AdS.  This is one reason researchers are so interested in AdS spacetimes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Around $10^{50}$ years from now there will be within regions bounded by cosmological horizons large galactic BHs.  The temperature of the horizon by Hawking-Gibbon radiation will be lower than the horizon temperature.  As a result these BHs will quantum radiate away so that around $10^{100}$ years into the future they evaporate away in a final flash of radiation.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-01-21T18:10:09.250" LastActivityDate="2011-01-21T18:10:09.250" />
  <row Id="3518" PostTypeId="1" CreationDate="2011-01-21T17:51:02.567" Score="1" ViewCount="180" Body="&lt;p&gt;If the space-time metric is expanding with the expansion of the universe, if I could travel back in time, would I be less dense than the matter in that previous era?  &lt;/p&gt;&#xA;" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-11T20:17:57.140" LastActivityDate="2011-05-12T06:37:03.297" Title="Expansion of the space-time metric" Tags="&lt;general-relativity&gt;&lt;cosmology&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="3520" PostTypeId="2" ParentId="3518" CreationDate="2011-01-21T18:00:00.870" Score="2" ViewCount="" Body="&lt;p&gt;The answer is 'no'. The Hubble expansion does not enlarge the distance between atoms in your body. It even does not enlarge distances between stars in our Milky Way. (Nor between the Milky Way and Andromeda for that matter.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question has been asked here more often. See for instance the answers in: &lt;a href=&quot;http://physics.stackexchange.com/questions/2110/why-does-space-expansion-not-expand-matter&quot;&gt;Why does space expansion not expand matter?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-01-21T18:00:00.870" />
  <row Id="3521" PostTypeId="1" AcceptedAnswerId="3522" CreationDate="2011-01-21T18:09:42.277" Score="7" ViewCount="269" Body="&lt;p&gt;This question was triggered by a comment of Peter Shor's (he is a skeptic, it seems.)  I thought that the holographic principle and ADs/CFT dealt with that, and was enough for Hawking to give John Preskill a baseball encyclopedia; but Kip Thorne is holding out, as are others.  Why?&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-21T18:14:58.667" LastActivityDate="2011-01-21T18:50:14.740" Title="Has the black hole information loss paradox been settled?" Tags="&lt;cosmology&gt;&lt;quantum-gravity&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="3" />
  <row Id="3522" PostTypeId="2" ParentId="3521" CreationDate="2011-01-21T18:25:10.403" Score="10" ViewCount="" Body="&lt;p&gt;It is a matter of opinion, largely dependent on what you mean by settled:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is information lost or is evolution unitary? There are arguments based on AdS/CFT which make it almost certain that the evolution of a system from pre-collapse matter to after collapse radiation is  unitary. This is because there is a dual description - an equivalent description of the system using different variables - in which unitary evolution is automatic. If this is true for black holes in AdS, it is hard to believe it is somehow different for black holes in flat space which are for all intent and purposes as close to their AdS cousins as we wish them to be. Based on that, most people I know are convinced what the right answer is.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How is the information preserved, what is the mechanism for the unitary evolution? In particular, what's wrong with Hawking's original arguments to support information loss? I think it is fair to say that we don't know the answer, at least we don't know how to phrase the answer in the original, gravitational, language. I think there is a lot to learn by making this answer, even if we don't dispute it, as explicit as possible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As a subset of that, were the arguments given by Hawking, based on Maldacena's superficially similar arguments, convincing? Lots of people are skeptical, including myself. There are some technical and conceptual assumptions that don't seem quite right in that proposed solution. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-01-21T18:50:14.740" LastActivityDate="2011-01-21T18:50:14.740" CommentCount="11" />
  <row Id="3524" PostTypeId="2" ParentId="3503" CreationDate="2011-01-21T18:46:54.547" Score="10" ViewCount="" Body="&lt;p&gt;There are some kind of silly answers here, except for QGR who correctly says they are identical. The two Lagrangians are isomorphic, the fields have just been relabeled. So anything you can do with one you can do with the other. The first has manifest $U(1)$ global symmetry, the second manifest $SO(2)$ but these two Lie algebras are isomorphic. If you want to gauge either global symmetry you can do it in the obvious way. You can use a complex scalar to represent a single charged field, but you could also use it to represent two real neutral fields. If you don't couple to some other fields in a way that allows you to measure the charge there is no difference. &lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T04:12:56.557" LastActivityDate="2011-03-02T04:12:56.557" CommentCount="2" />
  <row Id="3525" PostTypeId="2" ParentId="2750" CreationDate="2011-01-21T18:49:36.360" Score="1" ViewCount="" Body="&lt;p&gt;In order to perform a double slit experiment, you need to recombine the beams passing through both slits, and that is where the interference occurs. Since one slit is inside the event horizon, this interference location has to be inside the event horizon, in order to be reachable by photons/electrons/whatever passing through the &quot;inside slit&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An observer outside the black hole will therefore not be able to see whether the interference happened or not. And an observer inside the horizon can see the interference pattern, but it is not problematic because he can see both slit. However, he will not be able to describe us what he saw ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Clarify sentence 1 of §2, to answer @kakemonsteret's comment. &lt;/p&gt;&#xA;" OwnerUserId="373" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2011-01-22T16:44:44.000" LastActivityDate="2011-01-22T16:44:44.000" CommentCount="13" />
  <row Id="3526" PostTypeId="1" AcceptedAnswerId="3530" CreationDate="2011-01-21T19:16:33.093" Score="2" ViewCount="305" Body="&lt;p&gt;I'm looking for an article that sets up the framework described by Feynman in &lt;a href=&quot;http://web.ihep.su/dbserv/compas/src/feynman48c/eng.pdf&quot; rel=&quot;nofollow&quot;&gt;Space-Time Approach to Non-Relativistic Quantum Mechanics&lt;/a&gt;, but in Special Relativity.&lt;/p&gt;&#xA;" OwnerUserId="47" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T07:57:51.267" LastActivityDate="2011-02-03T07:57:51.267" Title="Special relativity version of Feynman's &quot;Space-Time Approach to Non-Relativistic Quantum Mechanics&quot;" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;special-relativity&gt;" AnswerCount="4" FavoriteCount="0" />
  <row Id="3527" PostTypeId="1" CreationDate="2011-01-21T19:31:31.573" Score="6" ViewCount="596" Body="&lt;p&gt;Why, when one releases 2 balls in Newton's cradle, two balls on the opposite side bounce out at approximately the same speed as the 1st pair, rather than one ball at higher speed, or 3 balls at lower speed?&lt;/p&gt;&#xA;" OwnerUserId="1498" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:41:06.557" LastActivityDate="2011-07-07T16:29:34.717" Title="Newton's cradle" Tags="&lt;energy-conservation&gt;&lt;newtonian-mechanics&gt;&lt;collision&gt;" AnswerCount="5" />
  <row Id="3528" PostTypeId="2" ParentId="3527" CreationDate="2011-01-21T19:42:03.763" Score="8" ViewCount="" Body="&lt;p&gt;It conserves both energy and momentum in the collision at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By design, when the balls collide the strings that hold them up are vertical (assuming balls are only swung from one side). This means there are no horizontal forces from the string on the balls so linear momentum in the direction of swing must be conserved in the collision. Energy is also nearly conserved provided not too much noise and heat are produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the two-ball case where each ball has mass $m$ and velocity $v$ at the time of collision. The kinetic energy will be $E = mv^2$ and the momentum $p = 2mv$. So suppose $n$ balls were to fly off the other side with velocity $u$. The energy would be $\frac{n}{2} mu^2$ and the momentum $nmu$. So by conservation we must have $mv^2 = \frac{n}{2} mu^2$ and $2mv = nmu$. It is not difficult to see that the only solution to these equation together is $n=2$ and $u=v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more complete analysis would rule out solutions with different balls leaving at different speeds, but I think this is enough to demonstrate the principle.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastEditorUserId="2445" LastEditorDisplayName="" LastEditDate="2011-06-28T22:34:04.277" LastActivityDate="2011-06-28T22:34:04.277" CommentCount="4" />
  <row Id="3529" PostTypeId="2" ParentId="3526" CreationDate="2011-01-21T19:43:30.957" Score="0" ViewCount="" Body="&lt;p&gt;The method introduced here by Feynman is the Path Integral approach to quantum mechanics. When trying to include special relativity, you also need to move from quantum mechanics to quantum field theory. The path integral approach is then also referred to as Functional Integration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many, many books on the subject since it's now considered a standard tool in quantum theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some references are:&#xA;A book by the man himself:&#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486477223&quot; rel=&quot;nofollow&quot;&gt;Feynman, Hibbs and Steiber&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tome on path integrals is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/9814273562&quot; rel=&quot;nofollow&quot;&gt;Kleinert&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some books treating Functional Integration and QFT are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1441919465&quot; rel=&quot;nofollow&quot;&gt;Nair&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201503972&quot; rel=&quot;nofollow&quot;&gt;Peskin and Schroeder&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...and pretty much every other modern day QFT book.&lt;/p&gt;&#xA;" OwnerUserId="1287" LastActivityDate="2011-01-21T19:43:30.957" />
  <row Id="3530" PostTypeId="2" ParentId="3526" CreationDate="2011-01-21T19:47:49.590" Score="3" ViewCount="" Body="&lt;p&gt;Dear mtrencseni,&#xA;the special relativistic counterpart of the 1948 Feynman paper you mentioned is the 1949 Feynman paper&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://web.ihep.su/dbserv/compas/src/feynman49b/eng.pdf&quot; rel=&quot;nofollow&quot;&gt;http://web.ihep.su/dbserv/compas/src/feynman49b/eng.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;called &quot;Spacetime Approach to Quantum Electrodynamics&quot;. Note that I used the same Russian server haha. Well, more precisely, Quantum Electrodynamics is one important example of a quantum field theory but the general methods in the Feynman paper were only updated in their &quot;technical details&quot; when people needed to get the similar description for any quantum field theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were expecting that the special relativistic version of the 1948 paper would still be essentially the same thing, with some $p^2/2m$ replaced by $mc^2/\sqrt{1-v^2/c^2}$, you must feel disappointed. But in fact, it is a reason for a huge happiness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact is that when special relativity is added to the quantum mechanical framework, one immediately encounters many effects that force us to use a fundamentally different classical starting point - field theory instead of quantum mechanics. Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, if you work with quantum mechanics - whether in an operator framework or in the path-integral approach - it doesn't respect the Lorentz symmetry. To respect the Lorentz symmetry, you need to switch from the non-relativistic one-particle Schrödinger equation to something like the Klein-Gordon equation or the Dirac equation. Both of them have a path-integral description as well although it is a bit subtle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the latter - the Dirac equation - because it's relevant for the same electron that used to be described by the non-relativistic Schrödinger equation. One can show that because of relativity, the equation inevitably predicts solutions with negative energy. While $E=p^2/2m$ only has non-negative values of energy, the condition $E^2-p^2 c^2-m^2 c^4$ has both positive and negative values of $E$ as solutions. You can't avoid it - the squaring of $E$ is a fundamental feature of special relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And indeed, the Dirac equation may be showed to have negative-energy solutions as well. If particles could have arbitrarily low energies, there would be an instability. You could get any energy from an electron, while the electron would be falling to arbitrarily low, negative energy levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nature avoids it because it tries to find the lowest-energy state, dissipate all the excess energy, and call the lowest-energy state &quot;the vacuum&quot; or &quot;the ground state&quot;. That's the only way to guarantee that it will be stable. So Nature actually fills all these negative-energy states - it has no choice. There is this &quot;Dirac sea&quot; of negative-energy electrons everywhere. By the Pauli exclusion principle, there can be at most one electron in each state. So the vacuum has 0 electrons in positive energy states and 1 electron in each negative energy state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a hole missing in this sea of negative-energy electron states, it will look like minus one electron with negative energy, i.e. it will be a positively charged particle, the positron, with a positive energy and positive charge. That's how Dirac predicted antimatter that was abruptly found. He got his Nobel prize for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, in relativity, you will find out that the pair creation of particles is inevitable. The number of particles can't be conserved. In some sense, it's because particles can move back and forth in spacetime - and the particle moving backwards in time is an antiparticle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all these reasons, you need to study quantum field theory if you want to combine special relativity with quantum mechanics. By quantizing a classical field, you get a system that looks like a system of particles once again. The number of particles coming from the quantum fields turns out to be integer (in the non-interacting limit) because of the same reason why the energy of a quantum harmonic oscillator is equally spaced. You will be able to recover the mechanics in the non-relativistic limit. But you won't be able to get rid of the phenomena that are implied by the fields - such as the pair creation and pair annihilation. They're real and important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relativistic non-field quantum mechanics is kind of inconsistent - or at least, it's not a viable approach to describe the real world. So you will only spend a very limited time with this concept - and you should eventually move to quantum field theory. This is true regardless of the formalism - Schrödinger's equation for the wave function, Heisenberg's equations for the operators, or Feynman's path integral approach. What I say about the fields is a physical insight and physical insights are independent of conventions and the choice of the mathematical machinery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-21T19:47:49.590" CommentCount="0" />
  <row Id="3531" PostTypeId="1" AcceptedAnswerId="3540" CreationDate="2011-01-21T19:53:21.197" Score="4" ViewCount="240" Body="&lt;p&gt;A distant star like the sun, thousands of light years away, could be so faint that only one photon might arrive per square meter every few hundred seconds. How can we think about such an arriving photon in wave packet terms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Years ago, in a popularisation entitled “Quantum Reality”, Nick Herbert suggested that the photon probability density function in such a case would be a macroscopic entity, something like a pancake with a diameter of metres in the direction transverse to motion, but very thin. (I know the wave packet is a mathematical construct, not a physical entity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have never understood how such a calculation could have been derived. After such a lengthy trip, tight lateral localisation suggests a broad transverse momentum spectrum. And since we know the photon’s velocity is c, the reason for any particular pancake “thickness” in the direction of motion seems rather obscure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Herbert then linked the wave packet width to the possibilities of stellar optical interferometry).&lt;/p&gt;&#xA;" OwnerUserId="1394" LastEditorUserId="1394" LastEditorDisplayName="" LastEditDate="2011-01-21T20:27:35.363" LastActivityDate="2011-01-22T12:27:15.900" Title="Photon wave packets from distant stars" Tags="&lt;quantum-mechanics&gt;&lt;quantum-optics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="3532" PostTypeId="2" ParentId="3526" CreationDate="2011-01-21T19:55:24.883" Score="0" ViewCount="" Body="&lt;p&gt;For the full framework, as MisterO explained, any modern textbook on QFT will do (I personally like Tony Zee's book (Quantum Field Theory In A Nutshell) most.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of the paper that you refer to, Feynman remarks &lt;em&gt;&quot;There are other ways of obtaining the Dirac equation which offer some promise of giving a clearer physical interpretation to that important and beautiful equation.&quot;&lt;/em&gt; I think he refers here to his checkerboard model that he never published (although years later he included it in the book he co-authored with Hibbs. So, as a stepping stone to the full framework, you might want to investigate this Feynman Checkerboard model (there is a lot of material on this toy model on the Internet). And of course the very accessible book QED by Feynman (which, however, is really directed to a lay audience and does not explore the relativistic aspects) can also be helpful to grok the basics of QFT.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastEditorUserId="1268" LastEditorDisplayName="" LastEditDate="2011-01-22T01:01:34.207" LastActivityDate="2011-01-22T01:01:34.207" />
  <row Id="3533" PostTypeId="2" ParentId="3527" CreationDate="2011-01-21T20:01:20.633" Score="1" ViewCount="" Body="&lt;p&gt;First, I have a free iPhone app for Newton's cradle - called Kinetic Balls - which has about 500 skins. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The events in the cradle are composed of many steps although they occur quickly after each other. In particular, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/qMBLf.jpg&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you have five balls, the first ball starts by colliding with the second ball, the second ball hits the third ball, the third ball hits  the fourth ball, the fourth ball hits the fifth ball. OK?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's study what happens when the first ball from the left hits the second ball from the left. The speed of the first ball is $v$ before it collides with the second one. Now, it simplifies things to look at the collision from the center-of-mass system of the first two balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Needless to say, the center of mass is moving by the speed which is the average of the speeds of the first two (equally heavy) balls. Because the first one is moving by $v$ to the right and the second one has the speed equal to $0$, the average is $v/2$, OK?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before the collision, the first ball is moving by the speed $v-v/2=v/2$ relatively to the center-of-mass frame, and the second ball is moving by the speed $0-v/2=-v/2$ relatively to the center-of-mass frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what happens after the collision? The balls can't penetrate through each other so their relative speed has to change the sign: the speeds get simply reversed. They change the sign. The situation is completely symmetric with respect to the exchange of the two balls combined with the reflection of &quot;left&quot; and &quot;right&quot; in the space. Consequently, the two final velocities (in the center-of-mass frame) must still be equal up to the opposite sign. The overall magnitude is guaranteed to be the same as before the collision, to conserve the kinetic energy $2\times mv^2/2$. The signs have to be opposite to one another - but also opposite to the initial ones because the balls can't penetrate one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the center-of-mass system, after the collision, the left ball will be moving by the speed $-v/2$ and the right ball will be moving by $+v/2$. Transforming back from the center-of-mass frame to the lab frame, we have to add $v/2$ again. So the first ball final speed will be zero, while the second ball's final speed will be $v/2+v/2=v$. OK?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the second ball is approaching the third ball at speed $v$ while the third ball is at rest. Take these two balls' rest frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, repeat the fairy-tale above four times - I could do it, it would be very exciting, but I want to save the server's hard disks - and you will end up with the situation in which the first four balls are at rest and the fifth one is moving by the speed $v$ to the right.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:41:32.970" LastActivityDate="2011-02-04T07:41:32.970" CommentCount="5" />
  <row Id="3534" PostTypeId="1" AcceptedAnswerId="3547" CreationDate="2011-01-21T20:10:15.837" Score="13" ViewCount="936" Body="&lt;p&gt;The common understanding is that, setting air resistance aside, all objects dropped to Earth fall at the same rate. This is often demonstrated through the thought experiment of cutting a large object in half, the halves of which clearly can't then fall more slowly just by being sliced in two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I believe the answer is that when two objects fall together, attached or not, they do &quot;fall&quot; faster than an object of less mass alone does. This is because not only does the Earth accelerate the objects toward itself but the objects also accelerate the Earth toward themselves. Considering the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_g = G m_1 m_2/d^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see that the force of gravity is dependent on BOTH the masses, not just that of the more massive object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course in everyday situations, we can for all practical purposes treat objects as falling at the same speed. But I'm hoping not for a discussion of practicality or what's measurable or observable, but what we think is actually happening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right or wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What really clinched this for me was considering dropping a small Moon-massed object close to the Earth and a small Earth-massed object close to the Earth. This made me realize that falling isn't one object moving toward some fixed frame of reference, but that the Earth is just another object and falling is multiple objects mutually attracting in space.&lt;/p&gt;&#xA;" OwnerUserId="1483" LastEditorUserId="1483" LastEditorDisplayName="" LastEditDate="2011-01-25T18:09:40.150" LastActivityDate="2011-02-18T22:50:41.370" CommunityOwnedDate="2011-01-25T01:11:38.203" Title="Don't heavier objects actually fall faster because they exert their own gravity?" Tags="&lt;gravity&gt;&lt;acceleration&gt;" AnswerCount="7" CommentCount="24" FavoriteCount="3" />
  <row Id="3536" PostTypeId="2" ParentId="3534" CreationDate="2011-01-21T20:33:25.967" Score="-2" ViewCount="" Body="&lt;p&gt;I always thought the reason a penny falls faster than a feather is air resistance.&#xA;There is real physics involved, but it's aerodynamics, not tides or gravity.&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-01-21T20:33:25.967" CommentCount="3" />
  <row Id="3537" PostTypeId="2" ParentId="3534" CreationDate="2011-01-21T20:36:54.857" Score="-2" ViewCount="" Body="&lt;p&gt;Drop a 5 lb iron barbell and a 25 lb iron barbell simultaneously.  The'll hit the ground simultaneously.  The only possible caveat is the recoil effect Ted Bunn brings up above.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-01-21T20:36:54.857" CommentCount="6" />
  <row Id="3538" PostTypeId="2" ParentId="3495" CreationDate="2011-01-21T20:39:48.603" Score="3" ViewCount="" Body="&lt;p&gt;I don't disagree with the other answers, but I want to try to use different words:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evolution of a quantum state is deterministic in the sense that it is given by the Hamiltonian. Quantum mechanics doesn't need anything beyond unitary evolution. So in that sense, the answer is deterministic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, decoherence means that eventually a quantum state may evolve into a superposition of very nearly orthogonal states, which for large enough systems will resemble to arbitrarily high precision the answer you might get from assuming that there is a nondeterministic, nonunitary process of &quot;wavefunction collapse.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all practical purposes, since we are large classical observers, we can observe only one such nearly-orthogonal combination, and the interference with other possible outcomes will be unmeasurably small. So, in this sense the outcome is &quot;nondeterministic.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this seems counterintuitive to you, consider something analogous about classical statistical mechanics and thermodynamics. If I start with a collection of gas molecules all bunched up in one corner of the room, it is a very atypical (low-entropy) state under any natural coarse-graining of phase space. Now, by entirely reversible interactions, it can become a typical (high-entropy) state with molecules scattered all over the room. This process appears to have lost information, in the sense that I would have to do many, many difficult measurements to ascertain that a short time before this was a very special state indeed. But really, the underlying physics is deterministic, so in principle the final state remembers where it came from, although for all practical purposes if I tried to evolve it backwards I would never discover the right answer. (To be clear, I'm not claiming a very sharp analogy here. But I'm saying that the notion that microscopically deterministic evolution can be consistent with &lt;em&gt;apparent&lt;/em&gt; or &quot;for all practical purposes&quot; loss of determinism in real observations is something that might be more intuitive in this context.)&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="19" LastEditorDisplayName="" LastEditDate="2011-01-21T21:14:38.637" LastActivityDate="2011-01-21T21:14:38.637" CommentCount="4" />
  <row Id="3539" PostTypeId="2" ParentId="3534" CreationDate="2011-01-21T20:57:27.070" Score="0" ViewCount="" Body="&lt;p&gt;Yes, a heavy object dropped from the same height will fall faster then a lighter one. This is true in the rest frame of either object. You can see this from $F=GmM/r^2=m*a=m*d^2r/dt^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fastest &quot;falling&quot; (since we are reredefining falling) object however is a photon, which has no mass.&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2011-01-25T18:19:45.793" LastActivityDate="2011-01-25T18:19:45.793" CommentCount="1" />
  <row Id="3540" PostTypeId="2" ParentId="3531" CreationDate="2011-01-21T20:57:56.427" Score="1" ViewCount="" Body="&lt;p&gt;first of all, the shape of the wave function of a photon that is emitted by an atom is independent of the number of photons because the photons are almost non-interacting and the atoms that emit them are pretty much independent of each other. So if an atom on the surface of a star spontaneously emits a photon, the photon is described by pretty much the same wave function as a single photon from a very dim, distant source. The wave function of many photons emitted by different atoms is pretty much the tensor product of many copies of the wave function for a single photon: they're almost independent, or unentangled, if you wish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The direction of motion of the photon is pretty much completely undetermined. It is just a complete nonsense that the wave function of a photon coming from distant galaxies will have the transverse size of several meters. The Gentleman clearly doesn't know what he is talking about. If the photon arrives from the distance of billions of light years, the size of the wave function in the angular directions will be counted in billions of light years, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's always the wrong &quot;classical intuition&quot; that prevents people from understanding that wave functions of particles that are not observed are almost completely delocalized. You would need a damn sharp LASER - one that we don't possess - to keep photons in a few-meter-wide region after a journey that took billions of years. Even when we shine our sharpest lasers to the Moon which is just a light second away, we get a one-meter-wide spot on the Moon. And yes, this size is what measures the size of the wave function. For many photons created in similar ways, the classical electromagnetic field pretty much copies the wave function of each photon when it comes to the spatial extent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the thickness of the wave packet. Well, you may just Fourier-transform the wave packet and determine the composition of individual frequencies. If the frequency i.e. the momentum of the photon were totally well-defined, the wave packet would have to be infinitely thick. In reality, the width in the frequency space is determined up to $\Gamma$ which is essentially equal to the inverse lifetime of the excited state. The Fourier transform back to the position space makes the width in the position space close to $c$ times the lifetime of the excited state or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not surprising: when the atom is decaying - emitting a photon - it is gradually transforming to a wave function in which the photon has already been emitted, aside from the original wave function in which it has not been emitted. (This gradually changing state is used in the Schrödinger cat thought experiment.) Tracing over the atom, we see that the photon that is being created has a wave function that is being produced over the lifetime of the excited state of the atom. So the packet created in this way travels $c$ times this lifetime - and this distance will be the approximate thickness of the packet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An excited state that lives for 1 millisecond in average will create a photon wave packet whose thickness will be about 300 kilometers. So the idea that the thickness is tiny is just preposterous. Of course, we ultimately detect the photon at a sharp place and at a sharp time but the wave function is distributed over a big portion of the spacetime and the rules of quantum mechanics guarantee that the wave function knows about the probabilistic distribution where or when the photon will be detected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thickness essentially doesn't change with time because massless fields or massless particles' wave functions propagate simply by moving uniformly by the speed $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-21T21:06:03.860" LastActivityDate="2011-01-21T21:06:03.860" CommentCount="2" />
  <row Id="3541" PostTypeId="1" CreationDate="2011-01-21T21:03:38.813" Score="4" ViewCount="1088" Body="&lt;p&gt;Wow--your answers to my original Q below clear up much of my confusion. I now have the daunting task of going over these nuggets and working up an equation-less (hopefully) explanation of the mass-less photon for non-physicist types.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, &lt;em&gt;from a layperson's view&lt;/em&gt;, it does seem remarkable that an existing piece of &lt;em&gt;matter&lt;/em&gt;--&#xA;which has to be made of &lt;em&gt;physical substance&lt;/em&gt;--could have zero mass at rest (though a photon is never at rest). It would be almost understandable if a piece of matter made of &lt;em&gt;nothing&lt;/em&gt; had zero mass, but that  seems to be an oxymoron, and &quot;nothing&quot; would equate to nonexistent, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you might find it interesting: I'm working on a writing project that posits we inhabit a universe that consists of matter (physical stuff) only, and that the NON-physical (aka supernatural) does not (and cannot) exist. For instance, if a purported supernatural phenomenon is found to actually exist, then by definition, its existence is proof that it is mundane/natural. All it would take to disprove this premise is reliable proof that ONE supernatural event has occurred. Despite thousands of such claims, that's never yet happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who else better than physicists to confirm my premise? However, I do wish the TV physicists would explain the terms they throw about, some of which mislead/confuse their lay viewers. Case in point: &quot;The universe is made up of matter and energy&quot; (without properly defining the term &quot;energy&quot; as a property of matter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is that laypersons are left with the impression that energy must therefore be something apart from or independent of matter (ie, nonphysical). Their use of the term &quot;pure energy&quot; without specifying exactly what that means adds to the confusion. (Thanks to your replies on this forum, I now understand that &quot;pure energy&quot; refers to photon particles.) However, &quot;psychics&quot; and other charlatans take advantage of such confusion by hijacking terms like energy (as in &quot;psychic energy&quot;), frequencies, vibrations, etc to give perceived scientific legitimacy to their claims that a supernatural spirit world, etc., exists. As you may realize, the majority of people in the US (per 2009 Harris Poll) and around the world believe in the existence of nonphysical/supernatural stuff such as ghosts and spirits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My purpose is to give laypersons the information they need to distinguish what's real from what's not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks so much for help...And, PLEASE, add any further comments you think might be helpful/insightful to better inform laypersons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;==============================&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read a number of the helpful Q&amp;amp;As on photons that mention the mass/mass-less issue. Do I understand correctly that the idea of mass-less (a rest mass of 0) may be just a convention to make the equations work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a layperson's view, it's difficult to understand how a particle of light (photon) can be mass-less. A physical object (everyday world-large or quantum-small) must have a mass. Yet, if my understanding is correct, the mass of a moving object/particle increases in proportion to its speed/velocity...so that at the speed of light, its mass would be infinite. A photon travels at the speed of light, but it obviously doesn't have infinite mass, right? Can someone formulate a practical explanation that can be understood by middle-school to high school kids? Much thanks for the help.&lt;/p&gt;&#xA;" OwnerUserId="1500" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-06T22:26:12.477" LastActivityDate="2011-07-12T09:17:58.677" Title="How can a photon have no mass and still travel at the speed of light?" Tags="&lt;special-relativity&gt;&lt;photons&gt;" AnswerCount="6" CommentCount="6" FavoriteCount="1" />
  <row Id="3542" PostTypeId="2" ParentId="3534" CreationDate="2011-01-21T21:13:03.067" Score="4" ViewCount="" Body="&lt;p&gt;The paradox appears because the &quot;rest frame&quot; of the Earth is not an inertial reference frame, it is accelerating. Keep yourself in the CM reference frame and, at least for two bodies, there is no paradox. Given an Earth of mass M, a body of mass $m_i$ will fall towards the center of mass $x_{CM}=(M x_M  + m_i x_i)/(M+m_i)$ with an acceleration $GM/(x_i-x_M)^2$. Note that $\ddot x_{CM}=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really we have only hidden the paradox, because of course $x_{CM}$ is different for each $m_i$. But this is a first step to formulate the problem in a decent inertial frame.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The paradox resurfaces again if you want to get rid of $(x_i-x_M)$. In most applications, now that you are in a non accelerating reference system, you want to consider distances related to it, ie $x_i-X_{CM}$. The solution is to redefine the mass. As &#xA;$x_i-x_{CM}= M (x_i - x_M) /(M+m_i)$, we can say that the object $i$ falls into the Mass Center with an acceleration $G{M^3 \over (M+m_ i)^2}{1 \over (x_i-x_{CM})^2}$ You could say that the actual mass of the &quot;earth at center of mass&quot; is this correction.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Once you are into the trick of changing the value of the mass, you can still stick to the reference frame of the earth. In this reference frame the quotient between force and acceleration is $Mm_i/M+m_i$ You can claim that this is the actual mass of the body during the calculation. This is called the &lt;em&gt;reduced mass&lt;/em&gt; $m_r$ of the system, and you can see that for small $m_i$, it is almost equal to $m_i$ itself. You can ever write some of the previous formulae using the reduced mass $m_r$ in combination with the original masses, for instance the above ${M^3 \over (M+m_ i)^2}= M {m_r^2\over m^2}$, but I am not sure of how useful it is.  In any case, you see that you were right about the &quot;heavier implies faster&quot; but that it is perfectly managed.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For three objects, m_1 and m_2 falling into M, the question is how to compare the case to m_1+m_2 falling into M. You separate the forces between internal, between 1 and 2, and external, against M. Look at the point $x_0= {m_1 x_1 + m_2 x_2 \over m_1+m_2}$ . This point it is not accelerated by the internal forces. And the external forces move them as&#xA;$$\ddot x_0={1 \over m_1+m_2} (m_1 {G M \over (x_1-x_M)^2} + m_2 {G M \over (x_2-x_M)^2})={F_1+F_2 \over m_1 + m_2}$$ &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is becoming long... ¡I can not put all the Principia in a single answer!. So you can forget all the previous stuff, consider it is just to a mean to fix notation and get some practice, an read the &lt;strong&gt;answer&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the two bodies are at the same distance $x$ of the &quot;external&quot; earth, they suffer the same external acceleration $g=GM/(x-x_M)^2$, and the same happens with $x_0$. If both bodies are in an approximation where $g$ can be considered constant, which was the case originally considered by Galileo (and the modern $g=9.8m/s^2$), then they have the same acceleration -and also the combined position $x_0$-. If they are not at the same distance nor in an approximation of constant -equal everywhere- field, then you can still save the movement of $x_0$ to work as if it were a gravitational force for some single mass $m_T$, but then the manipulation of the equations will produce in the relative positions of $x_1$ and $x_2$ some accelerations of the order of $1/(x_0-x_M)^3$. Such forces are the &quot;&lt;em&gt;tidal forces&lt;/em&gt;&quot;. &lt;/p&gt;&#xA;" OwnerUserId="1335" LastEditorUserId="1335" LastEditorDisplayName="" LastEditDate="2011-01-22T18:00:55.667" LastActivityDate="2011-01-22T18:00:55.667" CommentCount="6" />
  <row Id="3543" PostTypeId="1" AcceptedAnswerId="3548" CreationDate="2011-01-21T21:26:15.300" Score="4" ViewCount="184" Body="&lt;p&gt;This is a thing Iïve seen on many papers dealing with Warped Extra Dimensions, specifically on slices of AdS5. But the one where it appears more clearly is a lecture by Tony Gherghetta:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-ph/0601213&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-ph/0601213&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially what is done is that one builds a 5-dimensional theory in a slice of a Anti-De-Sitter space, the slice meaning that the fifth dimension has a small size and ends with a 4D brane at each extreme. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principle of least action applied to the 5D action leads us to two terms (eq. 7 in the paper above), one on the bulk (the vanishing of this one leads us to the bulk equation of motion) and one on the branes. The vanishing of this second term can be accomplished in two ways:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)  the variation of the field on the branes is zero&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) the term multiplying the variation is zero on the branes (let?s call it B)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, all authors affirm that, in absence of extra terms on the brane, this two conditions lead to Neumann or Dirichlet conditions for the field on the branes (in the paper above this is said just after equation 10). My question is: why is that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For scalar fields it can be shown that B equals the derivative of the field, so I can see the Neumann condition there (am I wrong?). But the 1st condition says the variation of the field is zero not the field itself. Clearly I am misunderstanding something here...&lt;/p&gt;&#xA;" OwnerUserId="1502" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-06T05:53:17.437" LastActivityDate="2011-02-06T05:53:17.437" Title="Understanding boundary conditions on slices of AdS5" Tags="&lt;quantum-field-theory&gt;&lt;variational-principle&gt;" AnswerCount="2" />
  <row Id="3545" PostTypeId="2" ParentId="3543" CreationDate="2011-01-21T21:37:26.160" Score="2" ViewCount="" Body="&lt;p&gt;&quot;Dirichlet&quot; in general means you specify the field value on the boundary (and don't allow it to fluctuate, setting the variation to zero), not necessarily that that boundary value is zero. But if it's nonzero, it corresponds in AdS/CFT to turning on a source for some operator, i.e. deforming the Lagrangian by a term like $J(x){\cal O}(x)$. So the &lt;em&gt;minimal&lt;/em&gt; thing to do, without changing the theory, is to set this source to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Fair warning: this is off-the-cuff, without thinking carefully about the $\epsilon \to 0$ limit that replaces a UV brane with conditions on the field behavior at the AdS conformal boundary. Saying that this is setting a source to zero might be a slightly wrong interpretation at finite but small $\epsilon$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interpretation of the case with an IR brane is more subtle, but you might imagine justifying any choice $\phi = \phi_0$ by adding a boundary-localized potential $V(\phi)$ with minimum at $\phi_0$ and taking the limit as it becomes steep.&lt;/p&gt;&#xA;" OwnerUserId="19" LastEditorUserId="19" LastEditorDisplayName="" LastEditDate="2011-01-21T21:52:54.457" LastActivityDate="2011-01-21T21:52:54.457" CommentCount="2" />
  <row Id="3546" PostTypeId="2" ParentId="3541" CreationDate="2011-01-21T21:39:05.770" Score="15" ViewCount="" Body="&lt;p&gt;There is absolutely nothing conventional about the mass of different particle species. For any particle moving in the vacuum, you may measure the total energy $E$ (including the latent energy) and the momentum $p$. It turns out experimentally - and Einstein's special theory of relativity guarantees - that the combination&#xA;$$E^2 - p^2 c^2$$&#xA;doesn't depend on the velocity but only on the type of the particle. It is a quantity describing the particle type and we call it&#xA;$$E^2 - p^2 c^2 = m_0^2 c^4$$&#xA;This determines the rest mass $m_0$ of the particle. The formula above works for any particle in the vacuum, any speed, and is always non-singular. Photons have $E=pc$ which implies that $m_0=0$. The rest mass of a photon is equal to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, that's also the reason why one can't really have a photon at rest, $v=0$. If a speed of something is $c$ in one reference frame, it will stay $c$ in any (non-singular) reference frame - that's another postulate of the special theory of relativity. So one can't ever make the speed of the photon zero by switching to another (non-singular) reference frame. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you want to see some values for all quantities, you may imagine that a photon at rest could exist and its total mass would be zero. At speed $v$, the mass is increased to&#xA;$$m_{total} = \frac{m_{0}}{\sqrt{1-v^2/c^2}}$$&#xA;For $m_0=0$ and $v=c$, the expression above is clearly a $0/0$ indeterminate form and its proper result may be anything. In particular, the correct value is any finite number. At the right speed, $v=c$, the massless photons can have any finite energy.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-06T19:18:46.763" LastActivityDate="2011-04-06T19:18:46.763" CommentCount="2" />
  <row Id="3547" PostTypeId="2" ParentId="3534" CreationDate="2011-01-21T21:47:29.453" Score="16" ViewCount="" Body="&lt;p&gt;Using your definition of &quot;falling,&quot; heavier objects do fall faster, and here's one way to justify it: consider the situation in the frame of reference of the center of mass of the two-body system (CM of the Earth and whatever you're dropping on it, for example). Each object exerts a force on the other of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = \frac{G m_1 m_2}{r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $r = x_2 - x_1$ (assuming $x_2 &gt; x_1$) is the separation distance. So for object 1, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{G m_1 m_2}{r^2} = m_1\ddot{x}_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and for object 2,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{G m_1 m_2}{r^2} = -m_2\ddot{x}_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since object 2 is to the right, it gets pulled to the left, in the negative direction. Canceling common factors and adding these up, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{G(m_1 + m_2)}{r^2} = -\ddot{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's clear that when the total mass is larger, the magnitude of the acceleration is larger, meaning that it will take less time for the objects to come together. If you want to see this mathematically, multiply both sides of the equation by $\dot{r}\mathrm{d}t$ to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{G(m_1 + m_2)}{r^2}\mathrm{d}r = -\dot{r}\mathrm{d}\dot{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and integrate,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G(m_1 + m_2)\left(\frac{1}{r} - \frac{1}{r_i}\right) = \frac{\dot{r}^2 - \dot{r}_i^2}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming $\dot{r}_i = 0$ (the objects start from relative rest), you can rearrange this to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{2G(m_1 + m_2)}\ \mathrm{d}t = -\sqrt{\frac{r_i r}{r_i - r}}\mathrm{d}r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where I've chosen the negative square root because $\dot{r} &amp;lt; 0$, and integrate it again to find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t = \frac{1}{\sqrt{2G(m_1 + m_2)}}\biggl(\sqrt{r_i r_f(r_i - r_f)} - r_i^{3/2}\cos^{-1}\sqrt{\frac{r_f}{r_i}}\biggr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $r_f$ is the final center-to-center separation distance. Notice that $t$ is inversely proportional to the total mass, so larger mass translates into a lower collision time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of something like the Earth and a bowling ball, one of the masses is much larger, $m_1 \gg m_2$. So you can approximate the mass dependence of $t$ using a Taylor series,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\sqrt{2G(m_1 + m_2)}} = \frac{1}{\sqrt{2Gm_1}}\biggl(1 - \frac{1}{2}\frac{m_2}{m_1} + \cdots\biggr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The leading term is completely independent of $m_2$ (mass of the bowling ball or whatever), and this is why we can say, to a leading order approximation, that all objects fall at the same rate on the Earth's surface. For typical objects that might be dropped, the first correction term has a magnitude of a few kilograms divided by the mass of the Earth, which works out to $10^{-24}$. So the inaccuracy introduced by ignoring the motion of the Earth is roughly one part in a trillion trillion, far beyond the sensitivity of any measuring device that exists (or can even be imagined) today.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-22T01:37:29.407" LastActivityDate="2011-01-22T01:37:29.407" CommentCount="4" />
  <row Id="3548" PostTypeId="2" ParentId="3543" CreationDate="2011-01-21T21:55:30.133" Score="2" ViewCount="" Body="&lt;p&gt;The answer is well-known from the analysis of open strings in perturbative string theory: at the end points, the coordinates $X(\sigma)$ must also satisfy either Neumann or Dirichlet boundary conditions for the very same reason that you mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, I think that you have almost answered your own question.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The variation of the field is zero. But the very point of the variational calculus - or the principle of the least action - is that the variations may change an allowed configuration to any other nearby allowed configuration. So if you want a constraint of the kind $\delta X(0)=0$ - or, in your AdS case, $\delta \phi|_{y*} = 0$, the only reason to guarantee it is to impose the condition $X(0)=const$ or $\phi|_{y*}=const$ for the allowed configurations. You may imagine that the constant is zero - it's just a convention. But because it's a constant, the variation - the difference of the value of $X(0)$ between two nearby allowed configuration - vanishes. Setting a scalar field at the boundary equal to a constant is called the Dirichlet boundary condition.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Alternatively, the boundary term in the variation - the second term in equation 7 - may vanish because the arbitrary nonzero $\delta \phi$ is multiplied by a vanishing coefficient. The coefficient is called $B_{curly}\phi$ in the equation but $B_{curly}$ is an operator proportional to $\partial_5$. So this method to make the boundary term vanish says that $\partial_5 \phi=0$ at the boundary, and this condition is called the Neumann boundary condition. Just to be sure, the $D_{curly}$ in the first term of equation 7 is also an operator - and it contains second derivatives. Note that the first term has second derivatives integrated over the 5D space, while the second term has first derivatives integrated over the 4D space, so the dimensions agree.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-06T05:52:23.833" LastActivityDate="2011-02-06T05:52:23.833" CommentCount="2" />
  <row Id="3549" PostTypeId="2" ParentId="3432" CreationDate="2011-01-21T22:12:08.177" Score="5" ViewCount="" Body="&lt;h2&gt;&lt;a href=&quot;http://blogs.uslhc.us/&quot; rel=&quot;nofollow&quot;&gt;US LHC Blogs&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A number of US scientists who work on the LHC explain various topics in physics. One particularly good series of posts is titled &lt;a href=&quot;http://blogs.uslhc.us/lets-draw-feynman-diagams&quot; rel=&quot;nofollow&quot;&gt;&quot;Let's Draw Feynman diagrams!&quot;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="724" LastActivityDate="2011-01-21T22:12:08.177" CommunityOwnedDate="2011-01-21T22:12:08.177" />
  <row Id="3550" PostTypeId="2" ParentId="3541" CreationDate="2011-01-21T22:12:25.060" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Do I understand correctly that the idea of mass-less (a rest mass of 0) may be just a convention to make the equations work?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, a photon really does have zero mass. You can think of it as a particle of &quot;pure energy&quot; if it helps you make sense of it, but the only sense in which that is valid is that a photon has energy but no mass.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yet, if my understanding is correct, the mass of a moving object/particle increases in proportion to its speed/velocity...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, actually mass is a relativistic invariant. Each object has its own particular intrinsic mass that is always the same no matter what speed the object moves at. What increases is the &lt;em&gt;energy&lt;/em&gt;. For massive objects, they are related by the formula Luboš cited,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = \frac{mc^2}{\sqrt{1 - v^2/c^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(note that my $m$ corresponds to his $m_0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a little historical motivation. In the early days of relativity, physicists quickly realized that as an object speeds up, it becomes harder and harder to increase its speed by a given amount. This is one way to express the reason why no object can exceed the speed of light. Now, we say that an object that is harder to move has more inertia, and the prevailing convention was that inertia should be related to mass, so it made sense that an object moving faster should have more mass, in some sense. Physicists coined the term &quot;relativistic mass&quot; to refer to this quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, they also recognized that something was special about the relativistic mass of an object at rest, since (for example) that was the minimum possible relativistic mass the object could ever have; it's the amount of mass that is intrinsic to the object. Appropriately enough, they labeled this the &quot;rest mass.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In later years, prompted in part by the discovery of general relativity, physicists realized that it makes more sense to say that inertia is related to &lt;em&gt;energy&lt;/em&gt;, not just mass. After all, GR tells you that there are other things besides mass that contribute to gravity, so it makes sense that there are other things besides mass that contribute to inertia. So the term &quot;relativistic mass&quot; fell out of favor (well, sort of; there are still a fair number of people who do use it) and physicists just started talking about energy instead. The two are simply related by the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = m_\text{rel}c^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the famous equation that people associate with Einstein, although they often forget the subscript &quot;rel&quot; and that leads to a lot of confusion ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that &quot;mass&quot; isn't being used to describe the total thing-that-is-related-to-inertia of a moving object, we can just use that word to describe the &quot;intrinsic mass&quot; or &quot;rest mass&quot; without any confusion. (Again, sort of; there are still a fair number of people who do talk about &quot;rest mass&quot; and it's possible to get confused when different groups with different conventions talk among each other.) So these days &quot;mass&quot; refers to what was previously known as &quot;rest mass.&quot; As I said, each object has its own particular mass, and for a photon, that happens to be zero. As Luboš said, only objects with zero mass can travel at the speed of light, otherwise the energy would be undefined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to make sense of the photon, remember Einstein's full equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E^2 = m^2 c^4 + p^2 c^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which says that energy comes from both mass and momentum. So even if a particle has no mass, it can still get its energy from momentum, and photons do indeed have both energy and momentum.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-06T18:02:12.617" LastActivityDate="2011-04-06T18:02:12.617" CommentCount="2" />
  <row Id="3551" PostTypeId="2" ParentId="3432" CreationDate="2011-01-21T22:12:55.913" Score="2" ViewCount="" Body="&lt;h2&gt;&lt;a href=&quot;http://www.symmetrymagazine.org/breaking/&quot; rel=&quot;nofollow&quot;&gt;Symmetry Breaking&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;an excellent general physics blog created by Fermilab and SLAC. &lt;/p&gt;&#xA;" OwnerUserId="724" LastActivityDate="2011-01-21T22:12:55.913" CommunityOwnedDate="2011-01-21T22:12:55.913" />
  <row Id="3552" PostTypeId="2" ParentId="3432" CreationDate="2011-01-21T22:13:15.693" Score="1" ViewCount="" Body="&lt;h2&gt;&lt;a href=&quot;http://www.astronomycast.com/&quot; rel=&quot;nofollow&quot;&gt;Astronomy Cast&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;this is actually a podcast, not a blog, but it is still fantastic. Two hosts, one a blogger, one an astrophysics go over various topics in astronomy, physics and cosmology. I particularly liked the &quot;&lt;a href=&quot;http://www.astronomycast.com/history/ep-183-history-of-astronomy-part-1-the-ancient-astronomers/&quot; rel=&quot;nofollow&quot;&gt;History of Astronomy&lt;/a&gt;&quot; series, but the posts on current discoveries are just as good. &lt;/p&gt;&#xA;" OwnerUserId="724" LastActivityDate="2011-01-21T22:13:15.693" CommunityOwnedDate="2011-01-21T22:13:15.693" />
  <row Id="3553" PostTypeId="2" ParentId="3486" CreationDate="2011-01-21T22:14:07.520" Score="3" ViewCount="" Body="&lt;p&gt;You're slightly mistaken about the mechanism for teleportation. The process does involve sharing an entangled pair of particles, but the individual states of those particles are never measured. Instead, a joint measurement is made of the state of particle whose state A wants to  teleport to B and one of the two entangled particles-- basically, asking &quot;are you in the same state, or different states?&quot; Based on the outcome of that measurement, which is sent to B via a classical channel, B does one of a small number of simple operations to the other entangled particle, which leaves it in exactly the state of the original particle that was being teleported.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantage of this process is that the state of A's original particle is never directly measured, which means that it arrives at B completely intact. This is the only sure way to transmit a single unknown quantum state without sending the particle in that state directly. You might think you could just make a copy of the state, and send the copy, but that is forbidden by the &quot;no-cloning&quot; theorem-- it is impossible to make an exact duplicate of a single quantum state unless you know in advance what that state is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key to the process is that the entangled pair of particles do not have well-defined states, but are correlated in a non-local fashion. You're not sending a set of random numbers with definite values from one place to another, you're sending an indeterminate state whose exact value will not be determined until it is measured, at which point its value will absolutely determine the state of a second particle a long distance away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process is of interest to people studying quantum information because it may be the surest way to move quantum information from one place to another. It's not going to allow you to build a Star Trek transporter any time soon, but it might be useful for building a quantum computer, or for connecting together quantum computers at distant locations.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2011-01-21T22:14:07.520" CommentCount="1" />
  <row Id="3554" PostTypeId="1" CreationDate="2011-01-21T22:15:30.357" Score="4" ViewCount="190" Body="&lt;p&gt;I have a mathematics background but am trying to educate myself a little about physics.  At the beginning of Feynman's QED book (not the popular one) is the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose all of the atoms in the universe are in a box.  Classically the box may be treated as having natural modes describable in terms of a distribution of harmonic oscillators with coupling between the oscillators and matter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess this is something that physicists learn, but I have never heard of it.  What is Feynman talking about and where can I learn more about it?  The Wikipedia article on harmonic oscillators gives no indication that physicists do this.&lt;/p&gt;&#xA;" OwnerUserId="1046" LastActivityDate="2011-01-22T14:42:21.813" Title="What's a good reference for this classical picture Feynman's talking about?" Tags="&lt;harmonic-oscillator&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="3555" PostTypeId="2" ParentId="3554" CreationDate="2011-01-21T22:24:09.970" Score="3" ViewCount="" Body="&lt;p&gt;This is a way of giving systematic meaning to the radiation continuum in the context of a set of discrete states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You assume some set of boundary conditions on the EM fields where they hit the box {1}, derive a set of allowed modes in terms of the geometry of the box {2}, then allow the box to expand without limit. Thus you arrive at a continuum of allowed modes.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;{1} Say $E = M = 0$ at the boundary as if the box were a very good conductor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{2} If the fields go to zero at the sides of the box then a half-integer number of wavelengths must fit, so only some wavelengths are allowed.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-01-21T22:24:09.970" CommentCount="5" />
  <row Id="3556" PostTypeId="2" ParentId="3534" CreationDate="2011-01-21T22:28:34.240" Score="3" ViewCount="" Body="&lt;p&gt;The new version of the question is much clearer than before, and I removed my down vote accordingly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is yes: in principle there is such an effect. When the mass of the dropped object is small compared to the mass of the planet, the effect is very small, of course, but in principle it's there.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-01-21T22:28:34.240" />
  <row Id="3557" PostTypeId="2" ParentId="3541" CreationDate="2011-01-21T22:48:43.147" Score="1" ViewCount="" Body="&lt;p&gt;Another point of view.  The smaller rest mass of the particle, the easier it is to be accelerated &lt;strong&gt;to the same speed&lt;/strong&gt;.&#xA;Light electron is easy to accelerate to the speed $0.9999c$ than heavy nucleus. Photon &lt;a href=&quot;http://pdglive.lbl.gov/Rsummary.brl?nodein=S000&amp;amp;exp=Y&quot; rel=&quot;nofollow&quot;&gt;with rest mass&lt;/a&gt; $&amp;lt; 10^{-18}eV$ may be considered as the lightest particle.&lt;/p&gt;&#xA;" OwnerUserId="238" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-01-22T09:28:45.507" LastActivityDate="2011-01-22T09:28:45.507" />
  <row Id="3558" PostTypeId="2" ParentId="3541" CreationDate="2011-01-21T22:59:22.063" Score="2" ViewCount="" Body="&lt;p&gt;You ask for a practical explanation that can be understood by interested middle/high-school kids. I think this comes close: &lt;a href=&quot;http://www.science20.com/hammock_physicist/whats_wrong_emc2&quot; rel=&quot;nofollow&quot;&gt;What's Wrong With $E=mc^2$?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass is the norm (=length) of the energy-momentum vector, and therefore invariant and conserved. The square norm of the energy-momentum vector for an object with energy $E$ and momentum $p$ is $E^2-p^2$. (see note below) The square root of this expression is mass ($m$). A photon has $E = p$, and hence zero mass. A massive particle at rest has $p = 0$ and hence $E = m$, Einstein's most famous equation (see footnote).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forget about 'relativistic mass'. This is a confusing term that does not add any understanding. Whenever you see the term 'relativistic mass', replace it with the term 'energy', as that is what it really is. A photon has energy (in proportion to it's frequency) but no mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you really think it is remarkable that there can be energy without mass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: In the above insert factors $c$ when &lt;em&gt;not&lt;/em&gt; working in natural units (i.e. replace $m$ with $m c^2$, and $p$ with $p c$)&lt;/p&gt;&#xA;" OwnerUserId="1268" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-02-26T14:32:24.270" LastActivityDate="2011-02-26T14:32:24.270" CommentCount="2" />
  <row Id="3559" PostTypeId="2" ParentId="3452" CreationDate="2011-01-22T02:31:08.297" Score="4" ViewCount="" Body="&lt;p&gt;Simply, no. Gravitational potential energy cannot be released by burning (combustion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitational potential energy is quite independent from chemical potential energy (which is ultimately electromagnetic in form). As of fundamental forces, they do not interact. Burning only converts &lt;em&gt;chemical potential energy&lt;/em&gt; (related to the binding energy of chemical compounds) to heat (thermal energy and EM radiation). Hence, gravity is basically irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saying this, there may be a small indirect effect of gravity. For a consideration of the possible effect of &lt;em&gt;air pressure&lt;/em&gt; (which varies with altitude), see Peter Shor's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Philip Gibbs also makes an interesting point relating to General Relativity. However, this relies on space-time curvature for a fixed observer, and is not as simple as considering gravitational potential.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2011-01-23T16:34:15.063" LastActivityDate="2011-01-23T16:34:15.063" CommentCount="21" />
  <row Id="3561" PostTypeId="2" ParentId="3452" CreationDate="2011-01-22T03:00:48.643" Score="-1" ViewCount="" Body="&lt;p&gt;The energy released in the fire is the difference between the energy of the fuel and the energy of the ash.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you carry wood up a mountain, it gains energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you burn wood on a mountain, the ashes also have more energy than the same ashes would have at sea level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wood has more energy on the mountain, but the ashes also have more energy. If you burn the wood and end up with the ashes, it doesn't matter whether you burn it up high or burn it down low, because the gravitational energy cancels out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy you put into the wood by carrying it up the mountain didn't go away, but the only way you can get it back is by letting the ashes fall down the mountain. (For example you could make the ashes pull a rope attached to a generator.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that in real life, the &quot;ash&quot; also includes the CO2 and other gases given off by the wood, so to do the experiment properly you'd have to burn it in some kind of large airtight container.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;APPENDIX: It seems like there are two other issues I need to address to satisfy some people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first issue could actually result in a measurable difference in the heat released from burning, and that's the difference in atmospheric pressure. What's going on is that when you completely burn the wood in O$_2$, the same amount of internal energy is always released, but what you actually measure is the enthalpy change $\Delta H = \Delta E + p \Delta V$. Let's look at the cases $\Delta V &amp;gt; 0$ and $\Delta V &amp;lt; 0$ separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the burning of the fuel releases more moles of gas than the moles of O$_2$ it consumes (which happens to be the case for burning real wood), then $\Delta V &amp;gt; 0$ so $\Delta H$ is a less negative number (smaller in magnitude) than $\Delta E$. What this means is that some of the energy from the fire went into pushing away the atmosphere rather than producing heat. The pressure $p$ is less on top of the mountain, so from this you would conclude that you get MORE heat after you carry it up the mountain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might GUESS that this extra heat energy is somehow related to the gravitational potential energy, but that would be WRONG as I'm about to show.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider a different fuel that consumes more moles of gas than it releases when burned. (For example, consider burning a strip of magnesium, which undergoes the simple reaction 2 Mg + O$_2$ $\rightarrow$ 2MgO (s).) In this case $\Delta V &amp;lt; 0$, so $\Delta H$ is more negative (greater in magnitude) than $\Delta E$, which means that you actually GAIN energy from the pushing of the atmosphere. Since the pressure is less on top of the mountain, this means that burning a strip of magnesium would provide LESS heat up there than it would at sea level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's going on here? The wood and the strip of magnesium both gain potential energy from being carried up the mountain, but one provides more heat up there, and the other provides less! This is because in both cases, the difference between the amounts of heat released at different altitudes is due to the difference in the energy of the atmosphere, NOT the gravitational energy of the solid fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;APPENDIX 2:&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditorDisplayName="" LastEditDate="2011-01-22T18:54:41.117" LastActivityDate="2011-01-22T18:54:41.117" CommentCount="5" />
  <row Id="3562" PostTypeId="2" ParentId="2750" CreationDate="2011-01-22T03:01:47.457" Score="0" ViewCount="" Body="&lt;p&gt;The question hinges upon how EPR pairs behave in the presence of an event horizon.  The operators become Bogoliubov transformed&#xA;$$&#xD;&#xA;b~=~a~cosh(g)~+~a^\dagger sinh(g),~b^\dagger~=~ a^\dagger~cosh(g)~+~a sinh(g).&#xD;&#xA;$$&#xA;These transformed operators mean that a Minkowski vacuum composed of regions inside and outside the horizon will involve terms ~ $sech(g)$.  A superposed or entangled state across the horizon will contain this factor, which for a large $g$ or acceleration near the horizon.  An EPR pair near the event horizon becomes entangled with the black hole, where this is similar to a measurement.  Another way of looking at this is the stationary observer near the horizon experiences a set of thermal states with a black body spectrum&#xA;$$&#xD;&#xA;cosh(g)~=~\prod_{\omega}\frac{1}{\sqrt{1~-~e^{2\pi\omega}}}&#xD;&#xA;$$&#xA;for $\omega$ the frequency of a Rindler particle.  The black hole acts as a thermal decoherence bath.  This loss of coherence of an EPR pair can occur outside the BH, for the path integral of the system contains paths with “probe” the black hole interior.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-22T03:01:47.457" CommentCount="2" />
  <row Id="3563" PostTypeId="1" CreationDate="2011-01-22T03:47:46.687" Score="1" ViewCount="200" Body="&lt;p&gt;Is it true that the visible light we see from the Sun comes from the electron-electron shell changes from the element &lt;strong&gt;iron&lt;/strong&gt; as they absorb and emit energy in the form of photons. This energy derived from the fusion reaction? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this process occurring deep in the sun so as to delay the final emmission of the photon out of the Sun by several hundred thousand years because of the density? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1510" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-23T00:30:48.840" LastActivityDate="2011-01-23T00:30:48.840" Title="The source of the light we see from the Sun" Tags="&lt;astrophysics&gt;" AnswerCount="2" />
  <row Id="3564" PostTypeId="2" ParentId="1601" CreationDate="2011-01-22T04:01:11.173" Score="2" ViewCount="" Body="&lt;p&gt;I recommend a recent book by Leon Takhtajan, &quot;Quantum mechanics for mathematicians&quot;. It starts with an introduction to classical mechanics aimed at mathematicians and explains the coordinate-free approach, among other things.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-22T04:01:11.173" />
  <row Id="3566" PostTypeId="2" ParentId="3563" CreationDate="2011-01-22T04:27:00.860" Score="4" ViewCount="" Body="&lt;p&gt;No.&#xA;   The basic energy comes from the conversion of four protons to a Helium nucleus in the sun's core. It is true that the opacity, which is the per unit mass absorbtivity is heavily influenced by the amount of &quot;metal&quot; present in the star. Astrophysicists refer to everything heavier than Lithium as Z (metal), so the abundance of these heavier elements contributes to the thermal structure. A lower abundance would mean lower opacity -easier for the heat to escape, which counterintuitively means the star would contract and burn even faster/hotter. But there are a lot more intermediate weight atoms (such as neon) in the sun than just iron. see:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements&lt;/a&gt;&#xA;so I really doubt the iron dominates the opacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case the photons we see, are emitted from the photosphere (the upper few thousand Km of the sun proper).&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-01-22T04:27:00.860" CommentCount="2" />
  <row Id="3567" PostTypeId="2" ParentId="3541" CreationDate="2011-01-22T05:39:44.420" Score="1" ViewCount="" Body="&lt;p&gt;I think you are confused.  Photons do not have mass, they have momentum.  &lt;/p&gt;&#xA;" OwnerUserId="1082" LastActivityDate="2011-01-22T05:39:44.420" />
  <row Id="3568" PostTypeId="1" AcceptedAnswerId="3685" CreationDate="2011-01-22T05:42:16.687" Score="3" ViewCount="209" Body="&lt;p&gt;Suppose I have a closed system with N molecules in it which are vibrating and all motion equations (rotation, translation and vibration) of the system are known along with any EM field equations in the region. Given all these information how do we &lt;strong&gt;calculate&lt;/strong&gt; (not measure) the &lt;strong&gt;temperature of this system&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1512" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-23T00:20:42.533" LastActivityDate="2011-01-23T21:20:56.493" Title="Temperature of a System of molecules" Tags="&lt;statistical-mechanics&gt;&lt;temperature&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="2" />
  <row Id="3569" PostTypeId="2" ParentId="3568" CreationDate="2011-01-22T06:09:55.457" Score="-2" ViewCount="" Body="&lt;p&gt;I got the answer from &lt;a href=&quot;http://www.physicsforums.com/showthread.php?t=300385&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperature is a purely statistical concept. It is defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1/(k T) = d[Log(Omega(E))]/dE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;em&gt;Omega(E) is the number energy eigenstates in a small interval around the energy E of the system.&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="1512" LastActivityDate="2011-01-22T06:09:55.457" CommentCount="2" />
  <row Id="3570" PostTypeId="2" ParentId="1402" CreationDate="2011-01-22T06:52:15.967" Score="2" ViewCount="" Body="&lt;p&gt;What you are talking about is the creation of magnetic monopoles (effectively magnetic charges). In classical electrodynamics, such objects are inconsistent with one of Maxwell's equations. Specifically the $\nabla \cdot B = 0$ equation specifically prohibits the existence of magnetic monopoles. In order for them to exist, you need to modify this equation so that it is proportional to the magnetic monopole density. While there are some unverified theories which could give rise to magnetic monopoles, all of our observations so far are consistent with the non-existence of monopoles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, the way magnets that you may have encountered work is via the magnetic field induced by moving charge, not by the presence of monopoles. Usually this field is caused by the angular momentum of charged electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this clarifies matters.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2011-01-22T06:52:15.967" CommentCount="1" />
  <row Id="3571" PostTypeId="2" ParentId="3568" CreationDate="2011-01-22T06:57:55.513" Score="1" ViewCount="" Body="&lt;p&gt;Dear AIB, the temperature can clearly be anything in the system you described. The temperature is not a universal constant calculable for each system: the temperature is a macroscopic quantity describing the state of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly speaking, the absolute temperature $T$ in Kelvins will be related to the kinetic energy of the average molecule by $E=5kT/2$ - assuming that two rotational degrees of freedom exist from the absolute zero to the temperature $T$ which they usually don't. The kinetic energy depends on the speed of the molecules - it is not calculable out of nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat capacity factor $5/2$ above should be replaced by a number that depends on the type of the molecule but for complicated enough stuff, the dependence won't be quite linear, anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-22T06:57:55.513" CommentCount="8" />
  <row Id="3572" PostTypeId="2" ParentId="2014" CreationDate="2011-01-22T07:20:43.647" Score="0" ViewCount="" Body="&lt;p&gt;As Lubos mentions Wick's Theorem is true for any number of particles. This is simply because it is giving a different way to express a two qubit operator, and hence has nothing to do with what system it is applied to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the magnetic field is 0 and the couplings are constant, then an XX spinchain can be diagonalized by a Fourier transform, so you may want to try the calculation via this approach. This will only work for a ring and not a linear chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that the problem you are having is indeed due to the finite length of the chain, but is hiding in a different place. Let me explain why: For a 2 qubit chain, the correlation operator commutes with the Hamiltonian, however this is not true for longer chains, and so we would expect to obtain different results for the $R=1$ correlation in the case of a two qubit chain and the case of a 3 qubit chain. Therefore, it seems clear that you are burying the chain length somewhere. If I had to guess, I would say it is probably an edge effect that has been skipped in doing the Jordan-Wigner transformation or in your explicit calculation, but obviously I can't be sure with out seeing both calculations explicitly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this is useful.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2011-01-22T07:20:43.647" />
  <row Id="3573" PostTypeId="2" ParentId="3432" CreationDate="2011-01-22T07:37:37.017" Score="2" ViewCount="" Body="&lt;h2&gt;&lt;a href=&quot;http://thebigblogtheory.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;The Big Blog Theory&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This is the blog of UCLA physicist &lt;a href=&quot;http://www.physics.ucla.edu/~saltzbrg/bio.html&quot; rel=&quot;nofollow&quot;&gt;David Saltzberg&lt;/a&gt;, who is the scientific advisor to the CBS sitcom The Big Bang Theory. The science (derived from the episodes) is simple. But you still can find some interesting stuff.&lt;/p&gt;&#xA;" OwnerUserId="966" LastActivityDate="2011-01-22T07:37:37.017" CommunityOwnedDate="2011-01-22T07:37:37.017" />
  <row Id="3574" PostTypeId="2" ParentId="3432" CreationDate="2011-01-22T08:02:44.617" Score="4" ViewCount="" Body="&lt;h2&gt;&lt;a href=&quot;http://www.science20.com/quantum_diaries_survivor&quot; rel=&quot;nofollow&quot;&gt;A Quantum Diaries Survivor&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The best place for detailed reports on new particle physics discoveries (and rumours)&lt;/p&gt;&#xA;" OwnerUserId="1341" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-25T17:03:17.390" LastActivityDate="2011-01-25T17:03:17.390" CommunityOwnedDate="2011-01-22T08:02:44.617" />
  <row Id="3575" PostTypeId="2" ParentId="3432" CreationDate="2011-01-22T08:04:36.563" Score="3" ViewCount="" Body="&lt;h2&gt;&lt;a href=&quot;http://resonaances.blogspot.com/&quot; rel=&quot;nofollow&quot;&gt;Resonances&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Another best place for detailed reports on new particle physics discoveries (and rumours)&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-22T08:04:36.563" CommunityOwnedDate="2011-01-22T08:04:36.563" />
  <row Id="3576" PostTypeId="1" AcceptedAnswerId="3578" CreationDate="2011-01-22T08:15:00.597" Score="4" ViewCount="143" Body="&lt;p&gt;How to mathematically formulate the Two Slit Experiment in a Lorentz invariant framework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a paper about this?&lt;/p&gt;&#xA;" OwnerUserId="47" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-07T18:03:17.517" LastActivityDate="2011-04-07T18:03:17.517" Title="How to mathematically formulate the Two Slit Experiment in Special Relativity?" Tags="&lt;quantum-mechanics&gt;&lt;special-relativity&gt;&lt;double-slit-experiment&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="3577" PostTypeId="2" ParentId="2558" CreationDate="2011-01-22T08:26:13.673" Score="0" ViewCount="" Body="&lt;p&gt;As you mentioned, it could evaporate away by Hawking radiation. We don't know how this ends for sure but it is most likely that it ends in a shower of elementary particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second possibility is that it keeps growing because matter or radiation keeps falling in. It could even reach a thermal equilibrium with its own radiation if held inside a heat proof box. Over very long time scales this scenario might not be viable because matter may decay and be dispersed by the universes expansion so that the black hole has to evaporate away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A third possibility is that it ends its life by falling into another black hole. The fate of that black hole then has the same possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There might be other ways a black hole could end that we don't really know about because we don't have all the theories. E.g. it might be possible for a black hole to be wiped out by a decay of the vacuum. It might also be possible for it to fall into a wormhole and pass into another universe. Perhaps in the other universe times arrow points the other way and it mutates into a white hole. I'm not saying these things can happen, we just don't know enough physics to say whether they can or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And don't forget that it can become a quantum supposition of different states that suffer different fates.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-22T08:26:13.673" />
  <row Id="3578" PostTypeId="2" ParentId="3576" CreationDate="2011-01-22T08:37:59.257" Score="4" ViewCount="" Body="&lt;p&gt;As several of us have argued beneath your other question here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/3526/special-relativity-version-of-feynmans-space-time-approach-to-non-relativistic&quot;&gt;Special relativity version of Feynman's &quot;Space-Time Approach to Non-Relativistic Quantum Mechanics&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;quantum mechanics combined with special relativity needs one to use quantum field theory (or a theory that is stronger than that).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one wants to study relativistic particle in the presence of slits, one may imitate it simply by putting various (absorbing) boundary conditions (for the fields whose excitations are the interfering particles) at the places where the boundaries of the material reside. The probability waves in space propagate according to the standard free equations, so interference measures the free particles' propagators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, the only relativistic particles whose interference may be observed in practice at present are photons. The mathematics of interference of individual photons is equivalent to the interference of classical electromagnetic waves - as they studied it since the early 19th century. The probability density is mapped to the energy density of an electromagnetic wave and the $E,B$ fields may be interpreted as the photon's wave function. The results for the interference patterns are, up to a normalization that can be determined from the total number of particles, identical to the case of the classical electromagnetic waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So papers for photons exist and the first ones were written in the early 19th century. Papers for other particle species don't exist because&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;they're simple sums of the well-known functions that govern the propagator of the corresponding probability waves in the space; for example, one only needs the simple, free one-particle Dirac equation to calculate the propagation of the electrons at any speeds &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the interference can only be measured for photons whose mathematical description is a very old story. For example, relativistic electrons must have a wavelength comparable to the Compton wavelength of the electron, something like $10^{-12}$ meters, or even shorter (ultrarelativistic electrons). The corresponding inteference pattern would probably be too tiny to be seen&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-22T08:43:40.017" LastActivityDate="2011-01-22T08:43:40.017" CommentCount="1" />
  <row Id="3579" PostTypeId="1" CreationDate="2011-01-22T09:52:55.483" Score="3" ViewCount="309" Body="&lt;p&gt;I came across a New Study at : &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1101/1101.2565v1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/arxiv/pdf/1101/1101.2565v1.pdf&lt;/a&gt; . Which talks about Time like quantum entanglement. What does that mean? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Comment added by L.Motl&lt;/strong&gt;: The same preprint has been discussed in &lt;a href=&quot;http://physics.stackexchange.com/questions/3357/what-is-time-teleportation&quot;&gt;this question&lt;/a&gt; and will be discussed in &lt;a href=&quot;http://physics.stackexchange.com/questions/6286/algorithm-definition-generalisation-time-entaglement-phenomennon&quot;&gt;this question&lt;/a&gt; in the future.&lt;/p&gt;&#xA;" OwnerUserId="1517" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-03T09:42:27.480" LastActivityDate="2011-07-03T05:15:21.333" Title="What is Time like Quantum Entanglement?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="3580" PostTypeId="1" AcceptedAnswerId="3583" CreationDate="2011-01-22T10:20:16.000" Score="5" ViewCount="376" Body="&lt;p&gt;I continue to find it amazing that something as “bulky” and macroscopic as a static magnetic or electric field is actually a manifestation of virtual photons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So putting on your QFT spectacles, look closely at the space near the pole of a powerful magnet – virtual photons! Now look between the plates of a charged capacitor – virtual photons again! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if it’s all virtual photons, how do we get the difference between a magnetic and electric field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[This question needs the tag &quot;magnetostatics&quot; as well].&lt;/p&gt;&#xA;" OwnerUserId="1394" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-03-06T20:08:52.033" LastActivityDate="2011-03-06T20:08:52.033" Title="Virtual photon description of B and E fields" Tags="&lt;quantum-field-theory&gt;&lt;quantum-optics&gt;&lt;quantum-electrodynamics&gt;&lt;electrostatics&gt;&lt;virtual-photons&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="3581" PostTypeId="2" ParentId="3580" CreationDate="2011-01-22T10:29:15.610" Score="1" ViewCount="" Body="&lt;p&gt;In QED, for &lt;em&gt;transverse&lt;/em&gt; photons, the electric field doesn't commute with the photon number operator. Neither does the magnetic field. In fact, the electric and magnetic fields don't even commute with each other. To get a state with a fixed electric field &amp;mdash; or at least with a small uncertainty in it in the quantum sense &amp;mdash; we need a superposition of states with different numbers of photons. This is called a squeezed state in quantum optics. The smaller the uncertainty in the electric field, the larger the corresponding uncertainty in the magnetic field and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photons come with a transverse polarization. The electric field is parallel to it, while the magnetic field is perpendicular to both the polarization and the wave vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about &lt;em&gt;longitudinal&lt;/em&gt; photons then? In the $\xi$-gauge, we have to impose additional constraints (Lorenz gauge and Gauss constraints), making the actual physical state a coherent state with a definite value for the Fourier transform of the longitudinal component of the electric field, which is equated with the Fourier transform of the charge density divided by the wave number. This applies to electrostatics. Longitudinal photons do not contribute to the magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about magnetostatics then? Only transverse photons can contribute to the magnetic field, and the analysis involving photon number operators applies. This time, we have a special case of a squeezed state, namely a coherent state.&#xA;We have to note though that the number of photons isn't conserved in time because of absorption and emission by the current source. This causes the coherent phase to kept stationary in time, instead of rotating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, unless a photon escapes into outer space and heads off to infinity without being absorbed by or interacting with any matter, then by the definition of a virtual photon as a photon propagator not contained in any of the external legs of a Feynman diagram, nearly all photons are virtual photons. And if a photon is detected in any way, it definitely has to be a virtual photon.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-02-14T15:11:52.353" LastActivityDate="2011-02-14T15:11:52.353" CommentCount="2" />
  <row Id="3582" PostTypeId="2" ParentId="3568" CreationDate="2011-01-22T10:48:03.213" Score="0" ViewCount="" Body="&lt;p&gt;You ofcourse understand that it's simply impossible to obtain all possible initial conditions (position, momentum and other internal degrees of freedom) for a large system. Furthermore, we actually have a good knowledge of the equations of motion that govern the dynamics of such a system. E.g. in classical physics you have some set of differential equations that follow from Newton's law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So suppose that we did know the inital conditions. Than there is still the other practical impossibility of actually determining the state of the system at some later time. We simply don't know how to solve these equations of motions analytically for large systems. Let alone the fact that we cannot keep track of each individual particle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why one turns to a &lt;strong&gt;&lt;em&gt;statistical&lt;/em&gt; description&lt;/strong&gt; of the system. In a statistical description you are not interested in what each individual particle is doing, but rather what the system is doing as a whole. So instead of making a statement like: particle 1 has momentum P and position X and particle 2 has some other properties, you would talk about the &lt;em&gt;probability&lt;/em&gt; of having a particle in some state and another in some other state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A key ingredient in this approach is the concept of &lt;strong&gt;time averaging&lt;/strong&gt;. The probabities mentioned before arise by averaging over a typical timescale. You assume that within that timescale the system explores all possible states it can end up in, as in, the system sits in all possible configurations it can during that timescale. It explores the &lt;em&gt;phase space&lt;/em&gt; of the system. The statistical state of the system is obtained through this time-averaging (which, mind you, is not done by actually starting with a state and then see what it does over a typical timescale. Rather, one assumes that the system will explore the phase space in some way, and you're not really interested in the way it does that).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this have to do with temperature? For that we need to introduce the concept of entropy first. The entropy is nothing more but the logarithm of the total number of states in the phase space that the system can evolve into. It is a measure of the size of the phase space that is available to the system. Again, this makes no reference to the particular state the system starts out in -- not the &lt;em&gt;microscopic&lt;/em&gt; details anyways. What it does require is the &lt;em&gt;total energy&lt;/em&gt; of the initial state. The reason should be clear: microscopically, the system can only evolve into other states with the same total energy. So when we talk about this time averaging, where the system explores the phase space, than the total energy serves as a constraint on the possible &quot;explorations&quot;. There are other constraints as well, such as the volume and number of particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy conservation places a severe constraint on the subspace of the phase space available to the system. At the same time the entropy is precisely determined by the size of this subspace. And this is where temperature comes in. By changing the total energy, the total entropy is changed -- the system now explores a different subspace of the phase space. Temperature is nothing more but the &lt;em&gt;rate of change&lt;/em&gt; of entropy. It is the derivative of the entropy with respect to the total energy, 1/T =  dS/dE. So it is in essence a measure on how the size of the phase space changes when you a little energy to the system (or extract some). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give a little more detail on this. You can also think of changing the entropy and see how the total energy of the system changes. You can also look at the change of total energy due to a change in the particle number, while keeping the entropy fixed. This leads to the chemical potential. Pressure and volume obey similar relations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So back to your question. How does one, in principle, calculate the temperature of the system? The only gedankenexperiment I can come up with is as follows: You let the system evolve. Then you perform a time averaging over this motion. You come up with some average state of the system, and, more importantly, the total entropy of the system. Now you add a little energy to the system and you the same thing. The change of entropy, divided by the change of energy is 1/T.&lt;/p&gt;&#xA;" OwnerUserId="1287" LastActivityDate="2011-01-22T10:48:03.213" CommentCount="2" />
  <row Id="3583" PostTypeId="2" ParentId="3580" CreationDate="2011-01-22T10:53:27.840" Score="3" ViewCount="" Body="&lt;p&gt;the wave function of a single photon has several components - much like the components of the Dirac field (or Dirac wave function) - and this wave function is pretty much isomorphic to the electromagnetic field, remembering the complexified values of $E$ and $B$ vectors at each point. The probability density that a photon is found at a particular point is proportional to the energy density $(E^2+B^2)/2$ at this point. But again, the interpretation of $B,E$ for a single photon has to be changed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So whether the field around an object is electric or magnetic or both is encoded in the &quot;polarization&quot; of the virtual photons.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may imagine that the photon has 6 possible polarizations or so, identified with the components of $E$ and $B$. Well, for a particular direction, it is really just the $E+iB$ combination that acts as the wave function, so there are only three polarizations for a given direction - and one of them (the longitudinal) is forbidden, too. ;-) But the qualitative point that there are many polarizations is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as emphasized repeatedly, you shouldn't imagine that a virtual photon is a real particle that can be counted. That's a reason why QGR's answer is pretty much irrelevant for your question because there is no operator counting virtual photons at all - so it makes no sense to ask whether it commutes with other operators. QGR may have thought about real photons but he hasn't answered your question, anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, static fields correspond to a vanishing frequency - because everything with a non-vanishing frequency will go like $\exp(i\omega t)$ or $\cos(\omega t)$. So if you want to describe the fields of electric sources and magnets as a collection of virtual photons, you must realize that the static nature of the field implies that the relevant fields will have the energy equal to zero. But the momentum is nonzero because the field depends on space - because of the sources. Such virtual photons are very far from being on-shell - they're very virtual, indeed. It is not too helpful to talk about virtual photons with particular frequencies and wavenumbers if there are electric sources in the middle of the region you want to describe. The Fourier analysis is only helpful for photons in a pretty much empty space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you &lt;em&gt;could&lt;/em&gt; calculate the probabilities of various outcomes for a charged particle in an external electric or magnetic field, produced e.g. by many spinning electrons, using Feynman diagrams - where the virtual photons are the internal lines. The Feynman diagrams would be able to calculate the force acting on the probe particle. Some terms in the force wouldn't depend on the velocity - the electric forces - while others would depend on the velocity - the magnetic ones. These different terms would always come from the &quot;same type&quot; of virtual photons but all these photons depend on the sources of the field, so you would of course get different results for electric and magnetic fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this stuff is confusing and really unnecessary. If you worry that quantum electrodynamics won't reproduce basic properties of electromagnetism - such as the difference between electricity and magnetism; or the difference between attractive and repulsive forces - then you shouldn't worry. It can be easily demonstrated that in the classical limit - e.g. for strong enough fields with a low enough frequency - the quantum electrodynamics (and the quantum field) directly reduces to the right classical limit, the classical electrodynamics (and the classical fields). Virtual photons are just a very helpful tool to study all kinds of processes similar to scattering. Their maths can be deduced from quantum fields - not the other way around - and these virtual photons don't happen to be useful to describe your kind of highly classical situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Luboš&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-22T10:53:27.840" CommentCount="1" />
  <row Id="3584" PostTypeId="1" AcceptedAnswerId="3586" CreationDate="2011-01-22T10:54:57.910" Score="6" ViewCount="370" Body="&lt;p&gt;This forum seems to agree that a billiard ball accellerated to ultra-relativistic speeds does not turn into a black hole.&#xA;(See recent question &quot;If a 1kg mass was accelerated close to the speed of light would it turn into a black hole?)&#xA;On the other hand, LHC scientists take seriously the formation of black holes by colliding protons.&#xA;Therefore I presume it will be agreed that two ultra-relativistic billiard balls colliding head on with the requisite energy will form a black hole.&#xA;The question is, suppose the two relativistic billiard balls are on antiparallel tracks that just miss each other, &#xA;but pass within the radius of the putative black hole formed by their combined mass-energy.&#xA;Will the black hole still form?  If not, why not?&lt;/p&gt;&#xA;" OwnerUserId="1172" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2011-01-22T13:17:11.190" LastActivityDate="2011-01-30T10:30:01.563" Title="If two ultra-relativistic billiard balls just miss, will they still form a black hole?" Tags="&lt;general-relativity&gt;&lt;gravity&gt;&lt;special-relativity&gt;&lt;black-holes&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="3585" PostTypeId="2" ParentId="3579" CreationDate="2011-01-22T11:07:56.030" Score="1" ViewCount="" Body="&lt;p&gt;In most quantum field theories, operators localized at regions which are timelike separated don't commute. Not only that, the state at the earlier time can affect the state at the later time, meaning they are not causally independent.&#xA;This means the relative states localized at timelike separated regions don't have a combined tensor product structure. Because of this, in most cases, it doesn't make any sense to speak of timelike entanglement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for the special case of 1+1D with only massless fields, the only modes are left-movers and right-movers travelling at the speed of light. As a result, timelike separated regions are causally independent, and operators localized therein do actually commute. So, we can have timelike entanglement in this case, but it doesn't generalize.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-22T11:18:00.773" LastActivityDate="2011-01-22T11:18:00.773" />
  <row Id="3586" PostTypeId="2" ParentId="3584" CreationDate="2011-01-22T11:08:33.953" Score="7" ViewCount="" Body="&lt;p&gt;first of all, while there are superficially plausible models in which the LHC may produce microscopic black holes, &lt;strong&gt;it is extremely unlikely that this will take place&lt;/strong&gt;. This possibility can only occur if there are &lt;strong&gt;large or warped extra dimensions&lt;/strong&gt; of space that are relevant already at the LHC energy scale - a TeV. However, even if those dimensions exist, the black holes will be among the &lt;strong&gt;last things that will be created&lt;/strong&gt;. The top researchers behind these models - such as Lisa Randall and Nima Arkani-Hamed (to choose the most famous people both from ADD and RS) - have written papers explaining the same fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if the black holes would be created, they would be &lt;strong&gt;indistinguishable&lt;/strong&gt; - at least to the laymen - &lt;strong&gt;from&lt;/strong&gt; other unstable elementary &lt;strong&gt;particles&lt;/strong&gt;. Of course, all worries about the fate of the Earth are based on confusing the properties of tiny black holes and the big ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, if you collide two billiard balls and at some moment, both objects fit into a volume that is &lt;strong&gt;sufficiently smaller than a number comparable to the Schwarzschild radius&lt;/strong&gt; of a black hole corresponding to the mass given by the center-of-mass energy of the two balls, then a black hole will form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The previous sentence contains an &lt;strong&gt;order-of-magnitude estimate&lt;/strong&gt;. It is very tough - and probably depending on the geometry of the balls and other things - to calculate the exact numerical coefficient. But if you're satisfied with the order-of-magnitude estimates, then it is true that if the balls even fit into the size of the black hole - from the center-of-mass inertial system's viewpoint - the black hole will form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formation of a black hole is a subtle thing. It will almost certainly not be the case that all the energy will be stored in the new black hole. An O(50%) part of the energy of the initial speedy balls will be &lt;strong&gt;radiated in the form of gravitational waves&lt;/strong&gt; before the black hole is formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But morally speaking, what you say is true. The balls don't have to directly collide. When they get close to each other, regardless of the fact that they will change the direction a bit, &lt;strong&gt;the space around them just curves&lt;/strong&gt; in the right way that leads to the emergence of the event horizon. After some short time, the initially anisostropic black hole radiates the energy (by &quot;ringing modes&quot;) and settles to the spherically symmetric shape we know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Luboš&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-22T11:08:33.953" CommentCount="1" />
  <row Id="3587" PostTypeId="2" ParentId="3452" CreationDate="2011-01-22T11:18:57.150" Score="1" ViewCount="" Body="&lt;p&gt;The question is about the effects of gravity on energy and can only be answered properly in the context of general relativity. The answer is that more energy is released when the wood is burnt on the top of the mountain, &lt;strong&gt;provided everything is measured from the point of view of a fixed observer&lt;/strong&gt; and we make some assumptions so that we can ignore uninteresting effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll assume that the strength of the gravitational pull at rest on the mountain top is the same as at the bottom, because otherwise we would have to consider other effects such as the tiny amount of energy in the wood due to its very slight compression under the gravitational field. This would change as the gravitational pull changed. I don't think that is what the question is meant to be about so I'll ignore it. I will just look at the direct effect of the potential energy as the question seems to suggest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy in relativity is a relative concept. It depends on the relative velocity of the various frames of reference. More significantly for this question, it depends on the position of an observer in a gravitational field. This effect can be measured by observing the frequency of radiation and how the measurement changes with altitude. In practice this has been done experimentally for even quite small vertical displacements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider first observers who are fixed relative to the bundle of wood and ask about the difference in released energy they can measure, if either the wood is burnt at the foot of the hill or at the top. By the equivalence principle there can be no difference unless they include the extra energy released when ashes or gasses released return to ground level. To avoid this effect let's assume that the wood is actually burnt within a sealed container so that only the radiated heat can escape. The conclusion then is that the observers must see exactly the same amount of heat released if they are fixed relative to the container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more interesting and correct way to look at the situation is from the point of view of observers fixed relative to the mountian. They could be high up in space collecting all the emitted radiation to measure its total energy or they could be doing the same thing from a lower altitude. The height will certainly make a difference to the total energy gathered because radiation is redshifted as it escapes from a gravitational mass. Someone higher up will measure less energy than someone lower down, but that was not the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is about how a fixed observer will measure the amount of radiated energy released from the combustion. (We can take this to be a fixed observer at large distance so that even people who do not agree that energy conservation makes sense in general relativity will agree that the ADM energy is well defined for this situation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer in this case has to be that when the wood is burnt on the mountain top, more energy will be released. This is because the radiation released from lower down will be redshifted more when it escapes to infinity so less energy will be measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to look at this is to consider what happens when someone uses energy to carry the wood to the top of the mountain where it is burnt, and then he returns the ashes and released gasses to ground level harvesting the energy released in doing so. Because the wood was burnt and it released energy, the ashes and gasses will actually weight slightly less afterwards according to mass energy equivalence, even when all the released ashes and gasses are collected. This means that less energy is released when the remains are returned to ground than were expended in lifting the original wood to the top. This would not have happened if the wood was burnt at ground level. By conservation of energy we have to conclude that more energy was released when the wood was burnt on the mountain top. Notice that I am not saying that more energy is released because the ashes return to ground level. The argument tells us that more energy must have been released during the combustion at altitude even before the ashes are returned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An exercise for the reader is to calculate the extra energy released using both the explanations I have given and show that the answer is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: To show this is not just hot air I'll give a formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $M$ is the mass of the before burning and energy $E$ is released in heat when burnt on the ground then the mass after burning is $M&amp;#39; = M - \frac{E}{c^2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy required to lift the wood up the hill a height $h$ is $V = Mgh$ and the enrgy released on taking the wood back down is $V&amp;#39; = M&amp;#39;gh = V - \frac{Egh}{c^2}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extra heat generated from burning at altitude is therefore $V-V&amp;#39; = \frac{Egh}{c^2}$ &lt;/p&gt;&#xA;" OwnerUserId="1341" LastEditorUserId="1341" LastEditorDisplayName="" LastEditDate="2011-01-22T20:38:50.323" LastActivityDate="2011-01-22T20:38:50.323" CommentCount="18" />
  <row Id="3588" PostTypeId="2" ParentId="3531" CreationDate="2011-01-22T12:27:15.900" Score="2" ViewCount="" Body="&lt;p&gt;Nigel,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to make another suggestion concerning this question. I have not read the 1985 book by Nick Herbert, but from online links I see that he was discussing the &quot;Hanbury Brown and Twiss&quot; mechanism for optical interferometry. Although this sounds like quite a mouthful and not very fundamental - in fact it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phenomenon has to do with photon ie boson quantum entanglement. The entanglement is between two photons from opposite sides of a star, allowing the star width to be measured. This technique was counter-intuitive apparently even to quantum physicists at the time. Here is a quote from Penrose 2004 (p598) on the matter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;When their method was first proposed, it met with great opposition from many (even distinguished) physicists, who argued that 'photons can only interfere with themselves, not with other photons'; but they had overlooked the fact that the 'other photons' were part of a boson entangled whole.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that NH is making some geometric claim about the shape of this photon entangled state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can follow the Wikipedia links on the HTB effect, or raise another Stack question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roy.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-01-22T12:27:15.900" CommentCount="1" />
  <row Id="3589" PostTypeId="2" ParentId="1402" CreationDate="2011-01-22T13:08:02.213" Score="1" ViewCount="" Body="&lt;p&gt;Magnetic monopoles appear in some Grand Unified Theories and SUSY models of particle physics, but are not possible in the standard model, so we don't know if they really can exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, even in stabdard electrodynamics you can create the illusion of magnetic monopoles by channelling the magnetic flux between two poles along a thin flux tube or &quot;needle&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently some condensed matter physicists were able to use this possibility within a spin ice system to make it seem like magnetic monopoles were moving in the material. This effect was described on some blogs at the time so I'm just going to link to the &lt;a href=&quot;http://backreaction.blogspot.com/2010/04/magnetic-monopoles-in-spin-ice.html&quot; rel=&quot;nofollow&quot;&gt;detailed one at Backreaction&lt;/a&gt; rather than try to repeat it myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, a truly isolated magnetic monopole that could be thrown into a black hole leaving its twin behind is only possible if some GUT theories are correct. People have tried to observe them by watching the current in a coil for the effect of a monopole that passes through it, but without (confirmed) success. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory, if they exist they should have been created in abundance in the very early universe at the time of the GUT scale energies. The theory of cosmological inflation was originally proposed to explain why we don't see them. They would be there but in such small numbers that we can't find one. Our only hope would be accelerate and collide particles at the GUT scale of $10^{12}$ TeV but that is currently beyond our wildest dreams.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-22T13:08:02.213" />
  <row Id="3590" PostTypeId="2" ParentId="3579" CreationDate="2011-01-22T13:11:49.050" Score="0" ViewCount="" Body="&lt;p&gt;The paper referenced in the question involves a transformation between Minkowski spacetime and Rindler spacetime.  This is related in some ways to the question on entanglement or EPR pairs near a black hole horizon.  An accelerated observer measure states in the region I they are causally connected to which are non-locally correlated across the event horizon into region II.  This is similar to the black hole near horizon situation with quantum EPR pairs and states correlated from the exterior timelike region to the interior spacelike region.  From the perspective of the Minkowski spacetime such spacelike correlations on the Rindler wedge are timelike.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-22T13:11:49.050" />
  <row Id="3593" PostTypeId="1" AcceptedAnswerId="4288" CreationDate="2011-01-22T13:45:47.843" Score="0" ViewCount="276" Body="&lt;p&gt;I asked a question recently where I wanted to know whether it was physically meaningful to talk about the arrow of time in other universes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although many people apparently have an intuitive notion of what is meant by &quot;physically meaningful&quot;, the only real context I have found is in terms of &quot;physically meaningful observables&quot; where in a quantum mechanical context we talk about the possible eigenvalues associated with eigenstates of an operator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H |\psi\rangle = \lambda|\psi\rangle$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The observables being possible outcomes of experiments or other operations on the state vector.  So in the context of the arrow of time, where there are claims of the possibility of a vector time, one would like to think of some possible operation that would produce set of possible values for that time.  So the first question one might ask is, &quot;what operation can I perform to observe the vector values of the arrow of time?&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would seem that there is some difference between defining such an operation and the possibility of it actually being executable.  So one might think of a thing being physically meaningful if one can define a set of executable operations (executabality being determined by some set of constraints).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I suspect there is a better definition, or a better way to conceptualize what is meant by &quot;physically meaningful&quot;, and I would like to read what those are. &lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2011-02-02T12:00:31.360" LastActivityDate="2011-02-02T12:00:31.360" Title="What is the definition of physically meaningful?" Tags="&lt;soft-question&gt;&lt;definition&gt;" AnswerCount="3" />
  <row Id="3596" PostTypeId="2" ParentId="3554" CreationDate="2011-01-22T14:35:37.140" Score="2" ViewCount="" Body="&lt;p&gt;I just purchased &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/9812563806&quot; rel=&quot;nofollow&quot;&gt;Feynman's Thesis&lt;/a&gt;, which provides some insight on how Feynman saw the world, and provides some context here.  One of the key issues Feynman was trying to reconcile in his Lagrangian approach was how to describe quantum mechanics without rely on a field defined by harmonic oscillators, from page 5:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In particular, the problem of the equivalence in quantum mechanics of direct interaction and interaction through the agency of an intermediate harmonic oscillator will be discussed in detail.  The solution of this problem is essential if one is going to be able to comapare a theory which considers field oscillators as real mechanical and quantized systems, with a theory which considers the field as just a mathematical construction of classical electrodynamics required to simplify the discussion of the interaction between particles.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have to understand that Feynman viewed matter as something different than the harmonic oscillator. Since mass is a parameter for a simple harmonic oscillator, we can see that in his discussion, Feynman didn't necessarily view matter as being the same thing as mass.  I suspect, matter would be viewed as the tangible reality that we are familar with, and quantum harmonic oscillators are the abstract entities that we use to describe behavior, so it is some way necessarily to map, or couple, the real to the astract.  &lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="1383" LastEditorDisplayName="" LastEditDate="2011-01-22T14:42:21.813" LastActivityDate="2011-01-22T14:42:21.813" CommentCount="2" />
  <row Id="3597" PostTypeId="1" CreationDate="2011-01-22T14:40:07.493" Score="1" ViewCount="91" Body="&lt;p&gt;Never mind whether the universe is &quot;fine tuned&quot; for anything in particular,just the idea that there is a nested hierarchy seems incredibly constraining on the outcome, and anything but accidental.&lt;/p&gt;&#xA;" OwnerUserId="1423" LastActivityDate="2011-01-22T17:16:41.450" ClosedDate="2011-01-23T00:16:35.767" Title="can a system with nested building blocks (atoms,cells) NOT be &quot;fine tuned&quot;?" Tags="&lt;universe&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="3598" PostTypeId="2" ParentId="3452" CreationDate="2011-01-22T14:47:59.007" Score="8" ViewCount="" Body="&lt;p&gt;Yes (at least in certain circumstances). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, let's assume that there's no ash left, and the combustion products are gasses that have the same density as the atmosphere. Now, the state of the system when you burn wood at the top of the mountain looks very much like the state of the system when you burn wood at the bottom of the mountain, so what happened to the energy you expended when you hauled the wood up the mountain? The answer is that the air pressure at the top of the mountain is lower than the air pressure at the bottom, so the fire at the top expends less work displacing air to account for the gasses released in combustion, and so produces more heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about in real life? You would also have to take into account the density of the gasses produced by combustion, and compare the potential energy of these gasses (taking into account the oxygen consumed) with the work done displacing the air. I'm not going to do the research to figure this out, but I think it very likely the answer is also &quot;Yes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADDED MATERIAL:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's been an objection to this answer, based on the fact that if you burn magnesium (say), which consumes oxygen without producing any gaseous combustion products, the fire at the top of the mountain produces less heat than the fire at the bottom. But consider what happens when you burn magnesium. You turn magnesium, Mg, into magnesium oxide, MgO. MgO is heavier than Mg. This means that if you haul magnesium up a mountain, burn it, and then bring the ashes down, you've actually gained energy in this process, since the ashes going down are heavier than the magnesium you took up. Where did this extra energy come from? It comes from the fact that the fire produces more heat on the bottom of the mountain.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditorDisplayName="" LastEditDate="2011-01-23T14:53:31.017" LastActivityDate="2011-01-23T14:53:31.017" CommentCount="12" />
  <row Id="3599" PostTypeId="2" ParentId="3593" CreationDate="2011-01-22T15:13:55.257" Score="3" ViewCount="" Body="&lt;p&gt;A conceptual description of any aspect of the world is &quot;physically meaningful&quot; if one can conceive of an empirical test to distinguish between the consequences of that description and its negation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, whilst certain tests might be theoretically satisfactory, they might not be very practical. One can test for the existence of magnetic monopoles, but the a priori probability of detecting one is negligible, thus it is difficult in practice to tell whether magnetic monopoles exist or do not, even though their existence is &quot;physically meaningful&quot; in the sense given above.&lt;/p&gt;&#xA;" OwnerUserId="1280" LastActivityDate="2011-01-22T15:13:55.257" CommentCount="1" />
  <row Id="3600" PostTypeId="2" ParentId="3526" CreationDate="2011-01-22T15:14:21.540" Score="1" ViewCount="" Body="&lt;p&gt;What is described above is something called the &quot;world-line approach&quot; to quantum field&#xA;theory, which is similar in spirit to the non-relativistic &quot;sum over particle paths&quot; picture of the Feynman/Hibbs book. See for instance&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Christian Schubert,&lt;br/&gt;&#xA;  &quot;Perturbative quantum field theory in the string inspired formalism&quot;,&lt;br/&gt;&#xA;  Phys.Rept.355 (2001) 73-234,&lt;br/&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/hep-th/0101036&quot; rel=&quot;nofollow&quot;&gt;hep-th/0101036&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As Luboš already pointed out, you do need to put in particle creation/annihilation more &#xA;or less by hand, but the method can nevertheless be a useful computational alternative to &#xA;standard quantum field theory methods (especially when dealing with non-trivial vacua, e.g. in the background of electromagnetic fields).&lt;/p&gt;&#xA;" OwnerUserId="26" LastActivityDate="2011-01-22T15:14:21.540" />
  <row Id="3601" PostTypeId="2" ParentId="2721" CreationDate="2011-01-22T15:15:10.850" Score="-1" ViewCount="" Body="&lt;p&gt;In CED written in terms of field strengths there is no a notion of gauge invariance. The charge value is a constant in time parameter by definition. There is also a continuity equation that governs charge flows. So it is a sequence of definitions and physical equations. Charge of a system is not a dynamical variable, nor a function of dynamical variables. The Noether theorem has nothing to do with its conservation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The masses, despite being constant, do not have a continuity equation in CED so they are not obliged to conserve ;-).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 1: I see this question is not so easy for many. OK, the charge value of one particle is constant by definition (like mass) so its conservation is a sequence of definition. Another matter - whether the &lt;strong&gt;system&lt;/strong&gt; charge is additive in particles? Does it evolve with time? Does it depend on interactions? To answer these questions, we have to employ the equations of motion. The charge continuity equation $\partial \rho /\partial t = div(\rho v)$ is valid for any &lt;strong&gt;v&lt;/strong&gt;, so the additivity is an exact sequence of this equation: $\rho$ is additive in particles and a single charge is constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the masses we can write such a continuity equations too but the system mass is generally not a sum of particle masses. The system mass is &lt;strong&gt;defined differently&lt;/strong&gt; as it depends also on interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2: The number of particles, charged or not, is also conserved in many theories. Do you really think it is a consequence of ambiguity in the potential definition?&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-01-28T15:04:01.577" LastActivityDate="2011-01-28T15:04:01.577" CommentCount="7" />
  <row Id="3602" PostTypeId="2" ParentId="3597" CreationDate="2011-01-22T15:18:16.540" Score="3" ViewCount="" Body="&lt;p&gt;You have to define exactly what you mean by 'nested hierarchy', but I suppose you deliberately pose your question in loose terms, and just wonder about systems that can be meaningfully described in markedly different ways at a hierarchy of levels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take a simple system as example: Conway's 'Game Of Life' (GoL). We all know this simple binary callular automaton and its local update rule. A hierarchy of structures is known to develop in GoL: gliders get produced in glider guns, these gliders move and interact, etc. Yet, the occurrence of such structures was not build into GoL. Someone just tried out some simple rules. It was only later when people started to simulate this system in more depth that a 'hierarchy of structures' was discovered. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might argue whether 'fine tuning' was applied by John Conway when he constructed this model, but in any case it appears that cellular automata with similar complex behaviors result for a wide range of cellular automata rules. So my answer is 'no'. It is entirely possible that a 'hierarchical system' is emergent from 'accidental rules'.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-01-22T15:18:16.540" />
  <row Id="3603" PostTypeId="2" ParentId="3432" CreationDate="2011-01-22T15:23:41.380" Score="4" ViewCount="" Body="&lt;h1&gt;&lt;a href=&quot;http://scienceblogs.com/principles/&quot; rel=&quot;nofollow&quot;&gt;Uncertain Principles&lt;/a&gt;&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Chad Orzel is awesome. Good stuff on AMO and sci-fi.&lt;/p&gt;&#xA;" OwnerUserId="412" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-01-23T11:49:46.097" LastActivityDate="2011-01-23T11:49:46.097" CommunityOwnedDate="2011-01-22T15:23:41.380" CommentCount="1" />
  <row Id="3604" PostTypeId="2" ParentId="3563" CreationDate="2011-01-22T15:30:19.960" Score="0" ViewCount="" Body="&lt;p&gt;I'd like to address the question of the delay between energy production by fusion and light emission at the visible surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start with, while the primary power source of every star is fusion, they also heat up by &lt;a href=&quot;http://en.wikipedia.org/wiki/Virial_theorem&quot; rel=&quot;nofollow&quot;&gt;Virial contraction&lt;/a&gt;. This is important because it means that proto-stars are very hot before fusion ever sets in (and indeed the heat is needed to trigger the fusion). Accordingly the following description applies to all stars, even at the very beginning of their life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interior of a star is composed of very hot, very dense, ionized gas (that is a plasma). This stuff is very opaque, meaning that the mean range of a photon is short (on the order of centimeters). So it is appropriately in some averaged point-of-view, to say that a photon travels a distance proportional to its range, interacts and is re-emitted in a random direction. Wash, rinse, repeat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can calculate how long is required for the random walk to move the photon up to the relatively transparent outer layer and allow emission into space. It turns out to be a long time: many thousands of year for a star like the sun. &lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-01-22T15:30:19.960" CommentCount="3" />
  <row Id="3605" PostTypeId="2" ParentId="2721" CreationDate="2011-01-22T15:55:00.693" Score="5" ViewCount="" Body="&lt;p&gt;Remember that voltage is always expressed as a &quot;potential difference.&quot; You can't measure the absolute value of voltage because everything is invariant when you add a constant voltage everywhere. That expresses a symmetry just like time translation invariance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you bring in the magnetic field this invariance or symmetry can be generalised to a bigger gauge invariance transforming the electromagnetic potential as a vector field. Charge particles are also described by fields such as Dirac spinors, which are multiplied by a phase factor under the action of this symmetry, making it a U(1) invariance. Electric charge is the conserved quantity that Noether's theorem gives for this symmetry.  &lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-22T15:55:00.693" />
  <row Id="3607" PostTypeId="1" AcceptedAnswerId="3610" CreationDate="2011-01-22T16:22:35.150" Score="2" ViewCount="120" Body="&lt;p&gt;Recently &lt;a href=&quot;http://arxiv.org/abs/1008.3907&quot; rel=&quot;nofollow&quot;&gt;J.Webb submitted paper&lt;/a&gt; which has extraordinary claim - that Fine Structure Constant is different in different directions in space! He (with others) measured $\alpha$ using quasar spectrum absorption method in different directions and found small variation. There is huge debate about this finding, but lets suppose it is true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you think spacial variation of Fine Structure Constant could have some influence to the Cosmic microwave background radiation? Because $\alpha$ influences the Electromagnetic interactions which is responsable for atom formation. And neutral atoms were the key for transparent universe in its early development sages. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could this be observed?&lt;/p&gt;&#xA;" OwnerUserId="1485" LastActivityDate="2011-01-22T17:05:33.897" Title="If Fine Structure Constant in not uniform in space what would that imply for cosmic background?" Tags="&lt;astrophysics&gt;&lt;cosmology&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="3608" PostTypeId="2" ParentId="1775" CreationDate="2011-01-22T16:33:18.040" Score="1" ViewCount="" Body="&lt;p&gt;If there is a maximum possible physical temperature it is well above anything we can reach experimentally and would require a complete theory of quantum gravity to understand it fully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutron stars are some of the hottest objects in the universe today with temperatures up to around 10 trillion degrees Kelvin ($10^{12} K$). Similar temperatures have been reached in heavy ion collisions recently at the Large Hadron Collider for very small volumes and times. At these temperatures even the protons and neutrons in nuclear matter are torn apart leaving just a plasma of quarks and glouns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But these temperatures are cool compared to the earliest moments of the big bang. According to our incomplete theories something really odd happens when you get to the Planck temperature which is around $10^{32} K$, so a good 20 orders of magnitude higher than anything we can produce.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At such temperature spacetime itself must be highly energised by gravitational interactions with hot matter. Some people think that spacetime passes through some kind of phase transition at this point, but if it does we have very little understanding of what kind of phase state lies beyond or whether temperatures can be raised further. Such understanding is in the realm of quantum gravity which is not yet fully developed. Such physics may describe the very earliest moments of the big bang and perhaps nowhere else in the universe.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-22T16:33:18.040" />
  <row Id="3610" PostTypeId="2" ParentId="3607" CreationDate="2011-01-22T17:05:33.897" Score="2" ViewCount="" Body="&lt;p&gt;if the fine-structure constant had different values at different places of space (which I don't believe for many reasons), then the size of the atoms e.g. in Planck units would depend on the location, too. Correspondingly, electrons and protons would &quot;sit&quot; into Hydrogen atoms at slightly different moments and different places would perform recombination - the birth of atoms when the Universe became nearly transparent and when the CMB was born - at slightly different moments. At different moments, the temperature was a bit different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a part of the observed anisotropy of the temperature of the CMB radiation could be blamed on the non-uniformity of the fine-structure constant in the times when the CMB was created. However, physicists would always think that the non-uniformity of alpha didn't arise &quot;just for fun&quot; but it was probably driven by something else. So they would ultimately reparameterize the influence of a variable $\alpha$ to the influence of something else, anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't see any alpha-effects in the detailed structure of the microwave background itself - except for the overall temperature discussed above. It's because the CMB radiation is a black body radiation which doesn't depend on the character of objects at the time when CMB emerged. Photons are massless and their properties - except for interactions with charged objects - don't depend on any alpha, either. In fact, the CMB radiation is the most accurate natural black body radiation we have ever observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Luboš&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-22T17:05:33.897" />
  <row Id="3611" PostTypeId="1" CreationDate="2011-01-22T17:08:32.037" Score="4" ViewCount="492" Body="&lt;p&gt;My friend thinks it's because she has less air resistance but I'm not sure.&lt;/p&gt;&#xA;" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-01-23T21:48:53.883" LastActivityDate="2011-01-23T21:48:53.883" Title="Why does a ballerina speed up when she pulls in her arms?" Tags="&lt;newtonian-mechanics&gt;&lt;angular-momentum&gt;&lt;rotational-dynamics&gt;" AnswerCount="4" />
  <row Id="3612" PostTypeId="1" AcceptedAnswerId="3647" CreationDate="2011-01-22T17:08:55.700" Score="9" ViewCount="295" Body="&lt;p&gt;This is rather a technical question for experts in General Relativity. An accessible link would be an accepable answer, although any additional discussion is welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GR has well known solutions relating to single Black Holes: Schwarzchild, Rotating &amp;amp; Rotating with Charge. These solutions demonstrate some non-trivial GR behaviour. However do there exist any (corresponding) binary star/black hole solutions? Because of the non-linearity of GR such a solution could well demonstrate additional properties to a &quot;solution&quot; that simply consisted of a pair of &quot;distant&quot; Schwarzchild solutions glued together.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-30T09:05:53.993" LastActivityDate="2011-02-02T15:42:05.670" Title="Binary Black Hole Solution of General Relativity?" Tags="&lt;general-relativity&gt;&lt;black-holes&gt;&lt;numerics&gt;&lt;analyticity&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="1" />
  <row Id="3613" PostTypeId="2" ParentId="3611" CreationDate="2011-01-22T17:16:33.977" Score="7" ViewCount="" Body="&lt;p&gt;No, it's caused by conservation of angular momentum. Reducing air resistance won't cause her (or anything else) to speed up without an external force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like linear momentum ($m v$), angular momentum ($r \times mv$) is a conserved quantity, where $r$ is the vector from the center of rotation. For a skater holding a static pose, for each particle making up her body, the contribution in magnitude to the total angular momentum is given by $m_i r_i v_i$. Thus bringing in her arms reduces $r_i$ for those particles. In order to conserve angular momentum, there is then an increase in the angular velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this answers your question.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2011-01-22T17:16:33.977" />
  <row Id="3614" PostTypeId="2" ParentId="3597" CreationDate="2011-01-22T17:16:41.450" Score="1" ViewCount="" Body="&lt;p&gt;your assumption is right at the naive level: the hierarchy seen in the objects' being composite at many levels implies that there have to be many different scales, and these many scales imply the natural existence of parameters of Nature that differ by many orders of magnitude. If all the parameters were random, you wouldn't expect a hierarchy - at least not one that has many layers - because most of the parameters would be of the same order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some sense, this is a more general statement than the hierarchy problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, just because some parameters are different by several orders of magnitude - e.g. the size of the atom and the size of its nucleus - it doesn't mean that there exists no non-fine-tuned theory that actually explains these hierarchies. So we know many parameters in Nature that are vastly different than their friends - and able to create hierarchies - but they don't require any unnatural fine-tuning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the emergence of molecules (and cells) as bound states of many atoms is inevitable. One can show that it is energetically favored for many atoms to create various larger molecules, so they will inevitably do so. The emergence of life is of course much more complex  and miraculous but there exists no evidence that life requires anything else than the laws of physics we know to be valid for atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to go inside atoms to see things that require new components of the physical explanation. You find nuclei and quarks inside them. And you may go to the very Planck scale - 19 orders of magnitude shorter than the size of the nuclei. That's where the &quot;matryoshka setup&quot; has to end because the geometry itself as we know it becomes inapplicable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However, various big ratios are understood.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the nucleus is much larger than the Planck scale, as I have just said. Although the difference is 19 orders of magnitude, one doesn't really need any excessive fine-tuning. It's because the strong interactions' &quot;fine-structure constant&quot; inevitably takes a value near the Planck scale which is of order one - something like $1/24$. However, it's running and getting stronger at longer distances - proportionally to the logarithm of the distance scale. Because the running is (naturally, calculably, and demonstrably) logarithmic, you need to go to exponentially longer distances than the Planck scale - namely the size of the proton - and that's where the force starts to confine the quarks. This is therefore the size of the protons that emerge out of these interacting quarks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other hierarchies - such as one between the Higgs mass and the Planck scale - are not understood in the same way at the present moment, or at least it hasn't quite been established which of the candidate explanations we possess is the right one (although all people &quot;in the know&quot; know that it's almost certainly supersymmetry haha). It's likely that a well-defined answer will be found in the future - for this hierarchy or any other hierarchy. However, even today, we already know enough to say that your general statement that &quot;any hierarchical architecture of matter requires fine-tuning&quot; is not correct, however attractive such an assertion could be for various emotional reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Luboš&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-22T17:16:41.450" CommentCount="1" />
  <row Id="3615" PostTypeId="1" CreationDate="2011-01-22T17:18:33.677" Score="1" ViewCount="471" Body="&lt;p&gt;I'm trying to simulate an RLC circuit using transfer function. Circuit is there: &lt;a href=&quot;http://i.stack.imgur.com/MC8ME.png&quot; rel=&quot;nofollow&quot;&gt;http://i.stack.imgur.com/MC8ME.png&lt;/a&gt; (I'm a new user therefore I cannot post images)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/W8N1q.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But I can, L.Motl...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Main current (I) is the output and V is the input. So far, I'm stuck at this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;TF = (Q1(s) + Q2(s)) / ((30 + s)(Q1(s) + Q2(s)) + 20 * Q1(s))&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I need to get rid of Q1 and Q2, but I cannot find a way.&lt;/p&gt;&#xA;" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-22T20:36:08.313" LastActivityDate="2011-01-26T00:41:44.973" Title="Transfer function of an RLC circuit" Tags="&lt;electricity&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="3616" PostTypeId="2" ParentId="3612" CreationDate="2011-01-22T17:22:22.953" Score="5" ViewCount="" Body="&lt;p&gt;There are no exact solutions, only approximations and &lt;a href=&quot;http://en.wikipedia.org/wiki/Numerical_relativity&quot; rel=&quot;nofollow&quot;&gt;numerical solutions&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't forget that orbiting black holes will radiate gravitational waves so any solution would have to include those and the corresponding decay of the orbit until the black holes coalesce.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-22T17:22:22.953" />
  <row Id="3617" PostTypeId="2" ParentId="3611" CreationDate="2011-01-22T17:31:46.327" Score="7" ViewCount="" Body="&lt;p&gt;Joe's answer is of course right and I gave it +1. However, let me say some slightly complementary things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever the laws of physics don't depend on the orientation in space, a number known as the angular momentum is conserved. For a rotating body - including the body of a lady - the angular momentum $J$ may be written as the product of the moment of inertia $I$ and the angular frequency $\omega$ (the number of revolutions per second, multiplied by $2\pi=6.28$):&#xA;$$ J = I \omega $$&#xA;The moment of inertia $I$ is approximately equal to&#xA;$$ I = MR^2 $$&#xA;where $M$ is the mass and $R$ is equal to the weighted average distance of the atoms (weighted by the mass) from the axis. (More precisely, I need to compute the average $R^2$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/fKVgH.gif&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It's up to you whether she is spinning clockwise or counter-clockwise.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the ballerina pulls in her arms, she becomes closer to the axis, and $R$ decreases. Her mass $M$ doesn't change but the moment of inertia $I$ decreases, too. Because $J=I\omega$ has to be conserved and $I$ decreased, $\omega$ inevitably increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also explain the increased angular frequency of the rotation in terms of forces and torques. If the arms move closer to the axis, they exert a torque on the ballerina that speeds her up. I would need some cross products here but I am afraid that wouldn't be fully appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These issues were also discussed here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/2934/why-do-galaxies-and-water-going-down-a-plug-hole-spin&quot;&gt;Why do galaxies and water going down a plug hole spin? &lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Why do galaxies and bathtub whirlpools spin?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-22T18:21:21.197" LastActivityDate="2011-01-22T18:21:21.197" CommentCount="14" />
  <row Id="3618" PostTypeId="1" CreationDate="2011-01-22T17:33:04.997" Score="12" ViewCount="1156" Body="&lt;p&gt;As an exercise I sat down and derived the magnetic field produced by moving charges for a few contrived situations. I started out with Coulomb's Law and Special Relativity. For example, I derived the magnetic field produced by a current &lt;em&gt;I&lt;/em&gt; in an infinite wire. It's a relativistic effect; in the frame of a test charge, the electron density increases or decreases relative to the proton density in the wire due to relativistic length contraction, depending on the test charge's movement. The net effect is a frame-dependent Coulomb field whose effect on a test charge is exactly equivalent to that of a magnetic field according to the Biot–Savart Law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question is:&lt;/strong&gt; Can Maxwell's equations be derived using only Coulomb's Law and Special Relativity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, and the &lt;em&gt;B&lt;/em&gt;-field is in all cases a purely relativistic effect, then Maxwell's equations can be re-written without reference to a &lt;em&gt;B&lt;/em&gt;-field. Does this still leave room for magnetic monopoles?&lt;/p&gt;&#xA;" OwnerUserId="1247" LastActivityDate="2011-07-16T20:10:24.587" Title="Can Maxwell's equations be derived from Coulomb's Law and Special Relativity?" Tags="&lt;electromagnetism&gt;&lt;special-relativity&gt;&lt;electricity&gt;&lt;electrodynamics&gt;&lt;magnetism&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="6" />
  <row Id="3619" PostTypeId="2" ParentId="3612" CreationDate="2011-01-22T17:35:49.053" Score="3" ViewCount="" Body="&lt;p&gt;According to general relativity, a pair of massive bodies that orbit each other emits gravitational waves - for analogous reasons to the reasons why accelerating charges in electrodynamics emit electromagnetic waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there can't be any static solutions resembling binary stars or binary black holes. The solutions have to be non-static and a complicated system of two orbiting bodies that emits gravitational waves - and eventually collapses into one object - clearly can't be solved analytically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These things are usually discussed numerically, see other answers. In particular, the 1993 Nobel physics prize was given for an observation of a pulsar whose frequency changes in time exactly in the right way to be explained as the loss of energy caused by the emission of gravitational waves as predicted and calculated by general relativity.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T15:42:05.670" LastActivityDate="2011-02-02T15:42:05.670" />
  <row Id="3620" PostTypeId="2" ParentId="3618" CreationDate="2011-01-22T17:47:33.407" Score="13" ViewCount="" Body="&lt;p&gt;Maxwell's equations do follow from the laws of electricity combined with the principles of special relativity. But this fact does &lt;em&gt;not&lt;/em&gt; imply that the magnetic field at a given point is less real than the electric field. Quite on the contrary, relativity implies that these two fields have to be equally real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the principles of special relativity are imposed, the electric field $\vec{E}$ has to be incorporated into an object that transforms in a well-defined way under the Lorentz transformations - i.e. when the velocity of the observer is changed. Because there exists no &quot;scalar electric force&quot;, and for other technical reasons I don't want to explain, $\vec{E}$ can't be a part of a 4-vector in the spacetime, $V_{\mu}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, it must be the components $F_{0i}$ of an antisymmetric tensor with two indices,&#xA;$$F_{\mu\nu}=-F_{\nu\mu}$$&#xA;Such objects, generally known as tensors, know how to behave under the Lorentz transformations - when the space and time are rotated into each other as relativity makes mandatory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The indices $\mu,\nu$ take values $0,1,2,3$ i.e. $t,x,y,z$. Because of the antisymmetry above, there are 6 inequivalent components of the tensor - the values of $\mu\nu$ can be&#xA;$$01,02,03;23,31,12.$$ &#xA;The first three combinations correspond to the three components of the electric field $\vec{E}$ while the last three combinations carry the information about the magnetic field $\vec{B}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I was 10, I also thought that the magnetic field could have been just some artifact of the electric field but it can't be so. Instead, the electric and magnetic fields at each point are completely independent of each other. Nevertheless, the Lorentz symmetry can transform them into each other and both of them are needed for their friend to be able to transform into something in a different inertial system, so that the symmetry under the change of the inertial system isn't lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only start with the $E_z$ electric field, the component $F_{03}$ is nonzero. However, when you boost the system in the $x$-direction, you mix the time coordinate $0$ with the spatial $x$-coordinate $1$. Consequently, a part of the $F_{03}$ field is transformed into the component $F_{13}$ which is interpreted as the magnetic field $B_y$, up to a sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, one may describe the electricity by the electric potential $\phi$. However, the energy density from the charge density $\rho=j_0$ has to be a tensor with two time-like indices, $T_{00}$, so $\phi$ itself must carry a time-like index, too. It must be that $\phi=A_0$ for some 4-vector $A$. This whole 4-vector must exist by relativity, including the spatial components $\vec{A}$, and a new field $\vec{B}$ may be calculated as the curl of $\vec{A}$ while $\vec{E}=-\nabla\phi-\partial \vec{A}/\partial t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You apparently wanted to &lt;em&gt;prove&lt;/em&gt; the absence of the magnetic monopoles by &lt;em&gt;proving&lt;/em&gt; the absence of the magnetic field itself. Well, apologies for having interrupted your research plan: it can't work. Magnets are damn real. And if you're interested, the existence of magnetic monopoles is inevitable in any consistent theory of quantum gravity. In particular, two poles of a dumbbell-shaped magnet may collapse into a pair of black holes which will inevitably possess the (opposite) magnetic monopole charges. The lightest possible (Planck mass) black holes with magnetic monopole charges will be &quot;proofs of concept&quot; heavy elementary particles with magnetic charges - however, lighter particles with the same charges may sometimes exist, too.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-08T05:47:06.593" LastActivityDate="2011-02-08T05:47:06.593" CommentCount="14" />
  <row Id="3622" PostTypeId="2" ParentId="1453" CreationDate="2011-01-22T19:03:09.543" Score="6" ViewCount="" Body="&lt;p&gt;What is the nature and parameters of the Higgs sector?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the Higgs particle exist? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, what is its mass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there Higgs multiplets e.g. as predicted by supersymmetry?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the last remaining question mark over the standard model and its resolution may take us to the next step beyond the standard model.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-22T19:03:09.543" CommunityOwnedDate="2011-01-22T19:03:09.543" CommentCount="1" />
  <row Id="3623" PostTypeId="2" ParentId="1453" CreationDate="2011-01-22T19:11:58.343" Score="14" ViewCount="" Body="&lt;p&gt;What makes up dark matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The existence of dark matter is well established cosmologically especially through recent gravitational lens observations of galaxies that collided. But we have no idea what it consists of. Ordinary baryonic matter is virtually rules out and galactic structure formation models suggest a weakly interacting massive particle is the best fit. So is that right and if so how do such particles fit into particle physics beyond the standard model?&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-22T19:11:58.343" CommunityOwnedDate="2011-01-22T19:11:58.343" />
  <row Id="3624" PostTypeId="2" ParentId="1453" CreationDate="2011-01-22T19:13:18.717" Score="4" ViewCount="" Body="&lt;p&gt;Shucks - i just repeat the question that M. Veltman asks about a dozen times in Facts and Mysteries - WHY ARE THERE THREE GENERATIONS OF FERMIONS ?  Halzen &amp;amp; Martin ask the same, as do a dozen others I could find.&lt;/p&gt;&#xA;" OwnerUserId="1423" LastActivityDate="2011-01-22T19:13:18.717" CommunityOwnedDate="2011-01-22T19:13:18.717" />
  <row Id="3625" PostTypeId="2" ParentId="1453" CreationDate="2011-01-22T19:37:18.163" Score="9" ViewCount="" Body="&lt;p&gt;What mechanism is responsible for the imbalance between matter and anti-matter in the universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did the universe start with such an imbalance already in place or is it due to some CP violating mechanism at high energy that was significant in the early universe? If so what interactions are involved?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently known CP violating mechanisms do not appear to be strong enough to account for the observed amount of matter in the universe, so the explanation is liekly to be someting new beyond the known standard model.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-22T19:37:18.163" CommunityOwnedDate="2011-01-22T19:37:18.163" CommentCount="1" />
  <row Id="3626" PostTypeId="2" ParentId="1453" CreationDate="2011-01-22T20:00:58.827" Score="11" ViewCount="" Body="&lt;p&gt;What is the explanation of sonoluminescence? Under the right circumstances, sound waves can cause a bubble in a liquid to emit light. But the mechanism by which this happens is not understood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the theories proposed to explain it are impressively exotic and far-fetched. Who knows what could come out of it?&lt;/p&gt;&#xA;" OwnerUserId="150" LastActivityDate="2011-01-22T20:00:58.827" CommunityOwnedDate="2011-01-22T20:00:58.827" CommentCount="1" />
  <row Id="3627" PostTypeId="2" ParentId="3452" CreationDate="2011-01-22T20:37:34.973" Score="2" ViewCount="" Body="&lt;p&gt;It seems to me that there is an ambiguity here that is causing some trouble in giving a definitive answer. The question does not specify how the material for the fire is burned, or what it consists of, and this can potentially give rise to different answers. Clearly if we elevate the material to 100km, some material won't burn, because of the lack of an oxygen rich atmosphere. On the other hand, some materials which are already oxygen rich can still be burned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To lift this ambiguity, and hopefully crystallize the question, lets consider a specific case: You have a sealed insulated contained in which a certain amount of fuel is burned with a fixed amount of oxygen. Further, let's assume there are no ashes left over. (Although it is relatively straight forward to expand this treatment to deal with that case, I think this gives a cleaner question). The container is then opened either at sea level, or at some elevation $H$. The question is now whether the latter process heats the atmosphere more than the first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to this question is clearly yes. In both cases exactly the same number and type of molecules are produced, and in each case these have the same kinetic energy. However the gas in the container that has been elevated also has some additional energy: the potential energy gained by each molecule due to it's relative elevation in the gravitational field. Once the containers are opened, in each case the gas will eventually come in to equilibrium with the atmosphere. Clearly in the latter case, the total energy of the atmosphere and gas as a whole is now higher than that of the first case by a factor of $m g H$, where $m$ is the mass of the gas released. As the average energy is increased, the temperature must also increase. This is because the possible energy levels for each molecule of the system are the same in both cases, the only way to achieve a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gibbs_state&quot; rel=&quot;nofollow&quot;&gt;Gibbs state&lt;/a&gt; with an increased energy expectation value is to increase the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this can finally resolve this discussion.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditorDisplayName="" LastEditDate="2011-01-22T21:00:27.463" LastActivityDate="2011-01-22T21:00:27.463" CommentCount="2" />
  <row Id="3628" PostTypeId="1" AcceptedAnswerId="3629" CreationDate="2011-01-22T20:42:18.417" Score="5" ViewCount="297" Body="&lt;p&gt;recently, I stumbled accross a concept which might be very helpful understanding quasiparticles and effective theories (and might shed light on an the question &lt;a href=&quot;http://physics.stackexchange.com/questions/2506/how-to-calculate-the-properties-of-photon-quasiparticles&quot;&gt;How to calculate the properties of Photon-Quasiparticles&lt;/a&gt;): the &lt;strong&gt;spectral function&lt;/strong&gt; $$A\left(\mathbf{k},\omega \right) \equiv -2\Im G\left(\mathbf{k},\omega \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as given e.g. in &lt;a href=&quot;http://prb.aps.org/abstract/PRB/v77/i8/e081412&quot; rel=&quot;nofollow&quot;&gt;Quasiparticle spectral function in doped graphene&lt;/a&gt; (&lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0708/0708.1133v2.pdf&quot; rel=&quot;nofollow&quot;&gt;on arXiv&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is widely used in &lt;strong&gt;many-body physics&lt;/strong&gt; of interacting systems and contains the information equivalent to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Green%27s_function&quot; rel=&quot;nofollow&quot;&gt;Greens function&lt;/a&gt; $G$. For free particles, $A$ has a $\delta$-peaked form and gets broader in the case of interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physical interesting thing is, as I read, quasiparticles of interacting systems can be found if $A$ is also somehow peaked in this case. I don't understand this relationship, hence my question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h3&gt;What is the relation of the spectral function's peak to the existence of quasiparticles in interacting systems?&lt;/h3&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance&lt;br&gt;&#xA;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-30T09:06:29.407" LastActivityDate="2011-01-30T09:06:29.407" Title="The Spectral Function in Many-Body Physics and its Relation to Quasiparticles" Tags="&lt;quantum-field-theory&gt;&lt;solid-state-physics&gt;&lt;quasiparticles&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="3629" PostTypeId="2" ParentId="3628" CreationDate="2011-01-22T20:48:37.207" Score="3" ViewCount="" Body="&lt;p&gt;Dear Robert, &#xA;the answer to your question is trivial and your statement holds pretty much by definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know, the Green's functions contain terms such as&#xA;$$G(\omega) = \frac{K}{\omega-\omega_0+i\epsilon}$$&#xA;where $\epsilon$ is an infinitesimal real positive number. The imaginary part of it is&#xA;$$-2\Im(G) = 2\pi \delta(\omega-\omega_0)$$&#xA;So it's the Dirac delta-function located at the same point $\omega$ which determines the frequency or energy of the particle species. At $\omega_0$, that's where the spectrum is localized in my case. If there are many possible objects, the $G$ and its imaginary part will be sums of many terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This delta-function was for a particle of a well-defined mass (or frequency - I omitted the momenta). If the particle is unstable, or otherwise quasi-, the sharp delta-function peak will become a smoother bump, but there's still a bump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because you didn't describe what you mean by &quot;peak&quot; more accurately, I can't do it, either. It's a qualitative question and I gave you a qualitative answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-22T20:48:37.207" CommentCount="3" />
  <row Id="3630" PostTypeId="2" ParentId="3615" CreationDate="2011-01-22T20:54:01.657" Score="2" ViewCount="" Body="&lt;p&gt;This circuit contains only passive components, and, by inspection, can be decomposed into series and parallel combinations.  You can solve it simply by writing down the complex impedance (as a function of frequency) for each of the components, and then combining those expressions using the usual rules for series and parallel combinations of impedances.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-01-22T20:54:01.657" />
  <row Id="3631" PostTypeId="2" ParentId="3452" CreationDate="2011-01-22T20:59:12.077" Score="-3" ViewCount="" Body="&lt;p&gt;I think the answer is very simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the combustion happens without smoke or gas rising for simplicity. All the mass stays at the same height, but it changes chemical form. There will be some slight variation due to binding energy, etc, but that behaviour is consistent at all heights and independent from the gravitational potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it is clear that whether you burn your fire at sea level or at 1000m, nothing changes from a gravitational potential standpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This said, there are differences, due to enthalpy, due to the concentration of oxygen in the atmosphere, and so on. Some of these have been considered in other answers and are correct - i think they are unrelated to the basic question though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the increased potential energy of the wood make a difference? No, it can't because the same mass ends up in the same relative place after the combustion, no matter at what height you perform the experiment.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-01-22T20:59:12.077" CommentCount="16" />
  <row Id="3632" PostTypeId="2" ParentId="3611" CreationDate="2011-01-22T21:06:02.783" Score="4" ViewCount="" Body="&lt;p&gt;A very simple explanation is the following: the arms of the ballerina are pulled outwards by the centripetal force she experiences by spinning. When she pulls her arms in, she is doing work by more than counteracting this force and this is what makes her spin faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is due to the fact that the spin velocity is related to the effort that the ballerina makes in pulling in her arms. The closer the arms, the more force she has to use to pull them in future or keep them in position, and the faster she spins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I was in high school, we did a more hardcore version of this experiment by sitting on a chair that could spin, holding two heavy weights outwards. Then someone would spin the poor test subject and ask him to pull in the weights... The results were quite scary... :-)&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-23T09:33:27.797" LastActivityDate="2011-01-23T09:33:27.797" CommentCount="17" />
  <row Id="3633" PostTypeId="2" ParentId="3612" CreationDate="2011-01-22T21:57:13.970" Score="3" ViewCount="" Body="&lt;p&gt;Franz Pretorius has worked on this and developed animations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://prl.aps.org/abstract/PRL/v95/i12/e121101&quot; rel=&quot;nofollow&quot;&gt;http://prl.aps.org/abstract/PRL/v95/i12/e121101&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The field is numerical relativity. Matthew Choptuik also, I believe, has done work on this.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-01-22T21:57:13.970" CommentCount="2" />
  <row Id="3634" PostTypeId="1" AcceptedAnswerId="3701" CreationDate="2011-01-22T22:02:49.157" Score="7" ViewCount="198" Body="&lt;p&gt;In Landau and Lipshitz's introductory book on Quanum Mechanics, &quot;Quantum Mechanics Non-Relativistic Theory, Third Edition: Volume 3&quot;, chapter XIV (page 433 in the edition on Amazon) is &quot;Addition of Angular Momenta&quot;. Interestingly, the first page is footnoted to the effect that the theory covered only applies in the limit of a binding force sufficiently weak that the angular momentum can be considered separately for the particles:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Strictly speaking, we shall always be&#xA;  considering (without explicitly&#xA;  mentioning the fact each time) a&#xA;  system whose parts interact so weakly&#xA;  that their angular momenta may be&#xA;  regarded as conserved in a first&#xA;  approximation.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;All the results given below apply, of&#xA;  course, not only to the addition of&#xA;  the total angular momenta of two&#xA;  particles (or systems) but also to the&#xA;  addition of the orbital angular&#xA;  momentum and spin of the same system,&#xA;  assuming that the spin-orbit coupling&#xA;  is sufficiently weak.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can someone provide a more complete explanation for this? I'm curious as to what fails as the binding force gets too strong. I'd like an intuitive explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the reasons for wanting to know more about this is that I suppose it has something to do with the &quot;proton spin crisis&quot;, that is, the question of how angular momentum is distributed among the components (quark, gluon &amp;amp;c.) of a proton. For example see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/0904.1735&quot; rel=&quot;nofollow&quot;&gt;A. W. Thomas, Int.J.Mod.Phys.E18:1116-1134,2009, &quot;Spin and orbital angular momentum of the proton&quot; 0904.1735&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it seems like an important part of understanding strongly bound states and it would be nice to better understand the limitations of the usual methods from an intuitive as well as through calculations.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-01-24T05:50:30.083" LastActivityDate="2011-04-08T15:12:30.433" Title="Addition of Angular Momementa in deeply bound situations, proton spin crisis" Tags="&lt;quantum-mechanics&gt;&lt;angular-momentum&gt;&lt;spin&gt;&lt;proton&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="3635" PostTypeId="1" AcceptedAnswerId="3652" CreationDate="2011-01-22T22:45:48.620" Score="3" ViewCount="273" Body="&lt;p&gt;This &lt;a href=&quot;http://physics.stackexchange.com/questions/3612/binary-black-hole-solution-of-general-relativity&quot;&gt;question&lt;/a&gt; on binary black hole solutions, led to me think about the similar question from the perspective of what we know about the Hydrogen atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prior to quantum mechanics, it was not understood what led to the stability of the Hydrogen atom against collapse of the electron orbits due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Bremsstrahlung&quot; rel=&quot;nofollow&quot;&gt;Bremsstrahlung&lt;/a&gt;, i.e. the emission of radiation due to the fact that it is in a non-inertial (accelerated) frame of reference. Bohr and Sommerfeld came up with a somewhat &lt;a href=&quot;http://en.wikipedia.org/wiki/Bohr%E2%80%93Sommerfeld_theory&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;ad-hoc&lt;/em&gt; procedure&lt;/a&gt; - the first ever instance of quantization - according to which in the quantum theory only those classical orbits are allowed the value of whose action is quantized in units of $\hbar$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_i p dq = 2 \pi n_i $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the integral is over the $i^{th}$ (closed) classical orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what I'm thinking of next has probably been thought of before but I haven't done a literature review to find out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classically, we expect the accelerating electron to radiate resulting in the catastrophic collapse of its orbit. However, in a &lt;em&gt;complete&lt;/em&gt; description we must also take the proton into consideration. It is also a charged object and as is well known from the two-body, inverse square law, central force problem (see eg. Goldstein) both the proton and electron orbit each other. Therefore the proton, being a charged object, must also radiate if we don't ignore its (accelerated) motion around the electron. An observer sitting at a distance $d \gg r$, where $r$ is the mean size of the two-body system, will measure radiation which is a superposition of that coming from both the electron &lt;em&gt;and&lt;/em&gt; the proton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is this: &lt;strong&gt;a).&lt;/strong&gt; What is the phase difference between the two contributions ($\mathbf{E}_e$ and $\mathbf{E}_p$) to the net electric field $\mathbf{E}$ as seen by this observer?, &lt;strong&gt;b).&lt;/strong&gt; What is the value of the total energy $E$ emitted by the electron-proton system, given by the integral of the Poynting vector across a closed surface enclosing the system, as seen by this observer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[ My motivation is to see if we can learn more about the Bohr-Sommerfeld quantization condition by considering the classical electrodynamics of the &lt;strong&gt;full&lt;/strong&gt; electron-proton system. The quantity $E$ will depend on the size and shape of the classical orbits of the two charged objects or more simply of their mean separation $E:=E(r)$. As we vary $r$ from $0$ to some value $r_{max} \ll d$ we would expect $E(r)$ to oscillate and have local minima for some classical orbits. If these classical minima occur for orbits which satisfy the Bohr-Sommerfeld condition then we would have established a connection between the full classical problem and its quantization. ]&lt;/p&gt;&#xA;" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-23T00:16:06.690" LastActivityDate="2011-03-21T11:00:17.750" Title="Radiation from a pair of charged objects orbiting each other" Tags="&lt;quantum-mechanics&gt;&lt;classical-mechanics&gt;&lt;electrodynamics&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="3636" PostTypeId="1" AcceptedAnswerId="3637" CreationDate="2011-01-22T22:53:04.887" Score="10" ViewCount="365" Body="&lt;p&gt;This &lt;a href=&quot;http://arxiv.org/abs/1009.1135&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; seems to show that $d=4, N=8$ supergravity is finite.  Yet the paper only has three citations in spires, and I certainly haven't heard talk of a new candidate theory of gravity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why isn't perturbative supergravity with some supersymmetry breaking principle, coupled with the standard model considered a possible theory of the universe?  Has someone checked the coupling to matter?  Is that the problem?&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-01-23T08:14:32.060" Title="Why is stringless supergravity not considered by many to be a candidate theory of quantum gravity?" Tags="&lt;quantum-field-theory&gt;&lt;particle-physics&gt;&lt;quantum-mechanics&gt;&lt;supergravity&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="3637" PostTypeId="2" ParentId="3636" CreationDate="2011-01-22T23:12:34.723" Score="10" ViewCount="" Body="&lt;p&gt;In order to not be entirely negative, I'll answer your question first and then provide a reason or two why research on the subject is interesting for other reasons.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The finiteness conjecture  has to do with perturbation theory. Even if true, it is still believed that in order for the theory to be non-perturbatively finite and otherwise consistent (e.g unitary), it has to include more degrees of freedom. The most plausible scenario is that the completion is the full string theory, for which N=8 SUGRA is very closely related.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The theory does not have enough structure to be a realistic description of nature by itself. For example it does not have chiral fermions, or a mechanism to break the extended SUSY spontaneously. All of these things become possible when you embed the theory within string theory.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you keep the theory as a QFT, but add some ingredients by hand to get more a realistic theory, the extended SUSY is broken and the magic is gone. Any finiteness conjecture is only valid for the pure (super)gravity case, and goes away as soon as you make the situation slightly less symmetric.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;On the other hand, purely as a theoretical laboratory it is fascinating that a quantum field theory which contains gravity is better behaved than expected, and the theory has a close relationship to other highly symmetric quantum field theories (for example N=4 SYM). There is great excitement and hope recently that by understanding precisely why it is finite (or at least well-behaved) we'd be able to understand the structure of quantum field theories better. It is in some sense the &quot;simplest quantum field theory&quot; (&lt;a href=&quot;http://arxiv.org/abs/0808.1446&quot; rel=&quot;nofollow&quot;&gt;arxiv.org/abs/0808.1446&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-01-23T05:55:20.307" LastActivityDate="2011-01-23T05:55:20.307" CommentCount="8" />
  <row Id="3638" PostTypeId="2" ParentId="1775" CreationDate="2011-01-22T23:23:46.357" Score="4" ViewCount="" Body="&lt;p&gt;There is an absolute maximum temperature, and it is $0^{-}$. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, that sounds silly, but look it up in L&amp;amp;L: Statistical Physics I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about an Ising paramagnet in an external field: At &quot;zero&quot; temperature (or actually $0^{+}$) the free energy of a system will be minimized by a unique minimum energy configuration. As we raise the temperature, the number of microstates with slightly higher energy grows rapidly, so we have a lower free energy in these entropically favorable configurations. Now we continue all the way to infinite temperature, at which point the system becomes completely disordered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But wait, what if we drive the system to even &lt;em&gt;higher&lt;/em&gt; energy? In that case there are fewer microstates and so the derivative that defines temperature goes negative, and the temperature that corresponds to these configurations is $-\infty$. This actually corresponds to the principle of &quot;population inversion&quot; in lasers. Anyway, higher and higher energy configurations (with their continually decreasing entropy) correspond to decreasing negative temperatures, until all of the spins point &lt;em&gt;against&lt;/em&gt; the external field at $T=0^-$.&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2011-01-22T23:23:46.357" CommentCount="4" />
  <row Id="3639" PostTypeId="2" ParentId="3401" CreationDate="2011-01-22T23:36:00.433" Score="0" ViewCount="" Body="&lt;p&gt;Just as a practice calculation, we can check the calculation for $j=1$.&lt;br&gt;&#xA;Let $L_x = \sqrt{1/2}\left(\begin{array}{ccc}0&amp;amp;1&amp;amp;0\\1&amp;amp;0&amp;amp;1\\0&amp;amp;1&amp;amp;0\end{array}\right)$,&#xA;$L_y = \sqrt{1/2}\left(\begin{array}{ccc}0&amp;amp;-i&amp;amp;0\\+i&amp;amp;0&amp;amp;-i\\0&amp;amp;+i&amp;amp;0\end{array}\right)$,&#xA;$L_z = \left(\begin{array}{ccc}1&amp;amp;0&amp;amp;0\\0&amp;amp;0&amp;amp;0\\0&amp;amp;0&amp;amp;-1\end{array}\right)$.&lt;br&gt;&#xA;Then the eigenvectors for spin $m=1$ in the +x, +y, and +z directions are:&lt;br&gt;&#xA;$&amp;lt;+x| = (1\;\;\;\;\;\;\sqrt{2}\;\;\;\;\;\;+1)/2$,&lt;br&gt;&#xA;$&amp;lt;+y| = (1\;\;\;\;\;\;i\sqrt{2}\;\;\;\;\;-1)/2$,&lt;br&gt;&#xA;$&amp;lt;+z| = (1\;\;\;\;\;\;\;\;0\;\;\;\;\;\;\;\;\;\;\;\;0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computing we find:&#xA;$\frac{&amp;lt;+x|+y&amp;gt;&amp;lt;+y|+z&amp;gt;&amp;lt;+z|+x&amp;gt;}{|&amp;lt;+x|+y&amp;gt;&amp;lt;+y|+z&amp;gt;&amp;lt;+z|+x&amp;gt;|} = \frac{(i/2)(1/2)(1/2)}{(1/2)^3} = e^{i\pi/2}$ as expected, just twice the angle of the spin-1/2 case (which gives $\pi/4$).&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-01-22T23:51:27.673" LastActivityDate="2011-01-22T23:51:27.673" CommentCount="1" />
  <row Id="3640" PostTypeId="2" ParentId="3618" CreationDate="2011-01-22T23:55:47.843" Score="7" ViewCount="" Body="&lt;p&gt;Lubos Motl's answer is very good, but I think it's worth saying one or two additional things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can regard magnetism as simply a byproduct of electricity, in the following sense: if you assume that Coulomb's Law is correct, and that special relativity is correct, and that charge is a Lorentz scalar (so that charge and current density form a 4-vector),  then you can derive all of Maxwell's equations. (Actually, you probably also need to assume the theory is linear as well, now that I think about it.) The undergraduate-level textbook by Purcell works this out very explicitly in a nice, pleasing way, and it's also in more advanced textbooks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some books gloss over the need to postulate that charge is a scalar. At least one textbook -- I don't remember which -- does emphasize it, and makes a convincing case that it's worth paying attention to. One way to see that it's not a trivial condition to impose is to consider the analogy with gravity -- that is, substitute mass for charge and gravity for electric field, and try to run the same argument. (Assume weak fields so that everything can be treated as linear if you like.) There are &quot;gravitomagnetic&quot; effects, but they're not related to regular gravity in the same way as the magnetic field is related to the electric field -- i.e., the gravitational analogues of Maxwell's equations look different from the regular Maxwell equations). One reason is the sign differences, of course -- like charges repel in one case and attract in the other. But a bigger reason is that the source of gravity is not a scalar: its density doesn't form part of a 4-vector, but rather of a rank-2 tensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But on a more philosophical (or perhaps semantic) level, I wouldn't jump from this fact to the conclusion that magnetism is &quot;merely&quot; a byproduct of electricity. At the very least, such language doesn't appear to be useful in understanding the theory or in using it! For instance, understanding how an electromagnetic wave can propagate from a distant galaxy to your eye is much easier and more natural if you look at it from the &quot;usual&quot; point of view.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-01-22T23:55:47.843" CommentCount="2" />
  <row Id="3641" PostTypeId="2" ParentId="3635" CreationDate="2011-01-23T00:01:37.260" Score="2" ViewCount="" Body="&lt;p&gt;I would guess that the contribution of the proton would be fairly negligible, because of the difference in mass. The classical radiated field amplitude will depend on the magnitude of the acceleration, and while the proton and electron both experience the same force, the proton is 1836 times heavier, so  the acceleration is reduced by a factor of 1836. Which means that the field amplitude will presumably be reduced by the same factor, in which case the phase relation isn't that important, because even if they were perfectly out of phase, the reduction of the total field due to the proton's contribution would be minimal.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2011-01-23T00:01:37.260" CommentCount="2" />
  <row Id="3642" PostTypeId="2" ParentId="3636" CreationDate="2011-01-23T00:34:48.413" Score="0" ViewCount="" Body="&lt;p&gt;There has been a recent spate of interest in computing high loop quantum gravity without strings.  In my opinion this is a huge laborious effort to capture what lies in string theory already.  $N = 8$ SUGRA with an $SU(8)$ symmetry plus an $E_{7(7)}$ symmetry which acts independently.  The $SU(8)$ acts upon the $133$ dimensions of the $E_(7(7)$, which we can think of for $E_{7(7)}({\bf R})$ as $133$ scalars.  The $133$ real parameters of $E_{7(7)}$ are ${\bf 133}~=~{\bf 28}~+~{\bf 35}~+~{\bf 35}~+~{\bf 35}$, where $\bf 28$ is an $SO(8)$ and combined with ${\bf 35}$  is an $SU(8)$.  The current is composed of these scalars as &#xA;$$&#xD;&#xA;J_\mu=~\sum_{a=1}^{133}J_\mu^a e_a&#xD;&#xA;$$&#xA;with the current conservation rule $\partial^\mu J_\mu~=~0$  If with fix the $SU(8)$ this is a coset rule $E_{7(7)}/SU(8)$ which subtracts out the ${\bf 28}~+~{\bf 35}~=~{\bf 63}$ of $SU(8)$ are leaves $70$ scalars. Thus we may think of $E_{7(7)}$ as&#xA;$$&#xD;&#xA;E_{7(7)}~=~SU(8)\times R^{70}.&#xD;&#xA;$$&#xA;The coset construction removes trouble some terms or provide counter terms.  It is then possible to compute currents, which are Noetherian conserved, in this coset construction which up to $7$ loops is $UV$ finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $E_{7(7)}({\bf R})$ is broken into a discrete group $ E_{7(7)}({\bf Z})$ by “quantization,” which in turn  contains the modular (or Mobius) group $SL(2,{\bf Z})$ of S duality and the T-duality group $SO(6,6,{\bf Z})$.  The Noether theorem operates for continuous symmetries, not for discrete ones.  However, a braid group or Yang-Baxter arguments may recover current conservation.  Further, a more general U duality description which meshes S an T dualities together may conserve current.  So there are open questions with the construction which have not been answered.  In an STU setting the lack of Noetherian currents should be replaced with Noetherian charges associated with qubits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $N~=~8$ and $d~=~4$ a paper by Green, Ooguri, Schwarz &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0704/0704.0777v1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/arxiv/pdf/0704/0704.0777v1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is not possible to decouple string theory from SUGRA.  There exists a set of states which make any decoupling inconsistent.  This calls into question the ability to compute the appropriate counter terms required to make a consistent SUGRA which is $UV$ finite. This paper attempts to circumvent this problem, but it must be realized this is a $7$ loop computation of considerable complexity which recovers results of string theory that are obtained rather readily.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These efforts are not without purpose though, for the construction of Noetherian currents, which I think will correspond to Noetherian charges associated with qubits and N-partite entanglements in STU theories.  However, as a replacement for string theory, reduction of gravity to a pure QFT, I doubt this will ever prove to be a complete approach.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-23T00:34:48.413" />
  <row Id="3643" PostTypeId="2" ParentId="3568" CreationDate="2011-01-23T01:13:55.290" Score="0" ViewCount="" Body="&lt;p&gt;For a gas of particles the microstates of the system have energy $E_i$, and the distribution of these states in a gas with temperature $T$ is given by&#xA;$$&#xD;&#xA;Z~=~\sum_i e^{-E_i\beta}&#xD;&#xA;$$&#xA;with $\beta~=~1/kT$.  It is not hard to see that&#xA;$$&#xD;&#xA;F~=~ln(Z)/\beta&#xD;&#xA;$$&#xA;which is the Helmholtz  free energy.  The internal energy is&#xA;$$&#xD;&#xA;U~=~-\Big(\frac{\partial Z}{\partial\beta}\Big).&#xD;&#xA;$$&#xA;The free energy $F~=~U~-~TS$ and $S~=~k(ln~Z~+~\beta U)$ can be used to write the partition function according to macroscopic varianbles $Z~=~e^{-F\beta}$.  $\beta$ may then be computed by taking a derivative with the Helmholtz free energy $F$&#xA;$$&#xD;&#xA;\beta~=~-\frac{\partial Z}{\partial\beta}.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-01-23T13:19:57.720" LastActivityDate="2011-01-23T13:19:57.720" CommentCount="4" />
  <row Id="3644" PostTypeId="1" CreationDate="2011-01-23T01:15:03.400" Score="5" ViewCount="517" Body="&lt;p&gt;Meaning, why is it the exact number that it is? Why not 2x10^8 /mps instead of 3? Does it have something to do with the mass, size or behavior of a photon? To be clear, I'm not asking &quot;how we determined the speed of light&quot;. I know there isn't a clear answer, I'm really looking for the prevailing theories.&lt;/p&gt;&#xA;" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-23T07:08:19.240" LastActivityDate="2011-08-15T13:15:09.963" Title="The origin of the value of speed of light" Tags="&lt;speed-of-light&gt;" AnswerCount="12" CommentCount="2" FavoriteCount="1" />
  <row Id="3645" PostTypeId="1" AcceptedAnswerId="3650" CreationDate="2011-01-23T01:23:35.683" Score="3" ViewCount="178" Body="&lt;p&gt;Coleman-Mandula is often cited as being the key theorem that leads us to consider Supersymmetry for unification.  An overview discussion with sufficient detail is missing in many popular texts.  So how does Coleman Mandula theory actually meet its no-go claim? &lt;/p&gt;&#xA;" OwnerUserId="1383" LastActivityDate="2011-01-23T07:31:17.997" Title="Can someone give a simple expose on Coleman Mandula theory and what Mandelstam variables are?" Tags="&lt;untagged&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="3647" PostTypeId="2" ParentId="3612" CreationDate="2011-01-23T01:28:50.163" Score="4" ViewCount="" Body="&lt;p&gt;For a very recent authoritative review of the numerical approach, see Centrella et. al.  &lt;a href=&quot;http://arxiv.org/abs/1010.5260&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1010.5260&lt;/a&gt;&lt;br&gt;&#xA;For the alternate parameterized post Newtonian approach, see Living Reviews of Relativity &lt;a href=&quot;http://relativity.livingreviews.org/Articles/subject.html&quot; rel=&quot;nofollow&quot;&gt;http://relativity.livingreviews.org/Articles/subject.html&lt;/a&gt;&lt;br&gt;&#xA;and look for articles number 2007-2, 2006-4 and 2003-6.&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-01-23T01:28:50.163" CommentCount="1" />
  <row Id="3648" PostTypeId="2" ParentId="3644" CreationDate="2011-01-23T01:45:18.467" Score="1" ViewCount="" Body="&lt;p&gt;c is just a fundamental property of our universe, the maximal speed of information propagation; yet it was defined to be exactly 299792458m/s to define a meter.  &lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2011-01-23T01:45:18.467" />
  <row Id="3649" PostTypeId="2" ParentId="3644" CreationDate="2011-01-23T01:46:38.917" Score="4" ViewCount="" Body="&lt;p&gt;Well, currently the speed of light is defined to be an exact number, with the second determined in terms of the electron transition times of cesium, and $c$ meters defined to be $c \times \left(1\, s\right)$.  So, the trite answer for this is that we defined it to be so.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think that the more careful answer would be that chemistry happens at very low energies compared to typical relativistic energies.  Since the energies are low, this means that the fundamental time scales of everyday life are much longer (in relativistic terms) than the fundamental length scales.  $c$ tells you how to convert from one to the other.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-01-23T01:46:38.917" />
  <row Id="3650" PostTypeId="2" ParentId="3645" CreationDate="2011-01-23T01:52:28.817" Score="2" ViewCount="" Body="&lt;p&gt;This  a bit of a sketch;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $S$-matrix acts on shift the state or momentum state of a particle.  A state with two particle states $|p, p’\rangle$ is acted upon by the $S$ matrix through the $T$ matrix&#xA;$$&#xD;&#xA;S~=~1~–~i(2\pi)^4 \delta^4(p~–~p’)T&#xD;&#xA;$$&#xA;So that $T|p, p’\rangle~\ne~= 0$.  For zero mass plane waves scatter at almost all energy.  The Hilbert space is then an infinite product of n-particle subspaces $H~=~\otimes_nH^n$.  As with all Hilbert spaces there exists a unitary operator $U$, often $U~=~exp(iHt)$, which transforms the states S acts upon.  $U$ transforms n-particle states into n-particle states as tensor products.  The unitary operator commutes with the $S$ matrix&#xA;$$&#xA;SUS^{-1}~=~[1 – i(2\pi)^4 \delta^4(p~–~p’)T]U[1~+~i(2π)^4 \delta^4(p~–~p’)T^\dagger]&#xA;$$$$&#xA;= U~+~i(2\pi)^4 \delta^4(p~–~p’)[TU~–~UT^\dagger] ~+~[(2\pi)^4 \delta^4(p~–~p’)]^2(TUT^\dagger).&#xA;$$&#xA;By Hermitian properties and unitarity it is not difficult to show the last two terms are zero and that the S-matrix commutes with the unitary matrix.  The Lorentz group then defines operator $p_\mu$ and $M_{\mu\nu}$ for momentum boosts and rotations.  The $S$-matrix defines changes in momentum eigenstates, while the unitary operator is generated by a internal symmetries $A_a$, where the index a is within some internal space (the circle in the complex plane for example, and we then have with some &#xA;$$&#xD;&#xA;[A_a,~p_\mu]~=~[A_a,~M_{\mu\nu}]~=~0.&#xD;&#xA;$$&#xA;This is a sketch of the infamous “no-go” theorem of Coleman and Mundula.  This is what prevents one from being able to place internal and external generators or symmetries on the same footing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way around this problem is supersymmetry.  The generators of the supergroup, or a graded Lie algebra, have 1/2 commutator group elements $[A_a,~A_b]~=~C_{ab}^cA_c$ ($C_{ab}^c$ = structure constant of some Lie algebra), plus another set of graded operators which obey&#xA;$$&#xD;&#xA;\{{\bar Q}_a, Q_b\}~=~\gamma^\mu_{ab}p_\mu,&#xD;&#xA;$$&#xA;which if one develops the SUSY algebra you find this is a loophole which allows for the intertwining of internal symmetries and spacetime generators.  One might think of the above anti-commutator as saying the momentum operator, as a boundary operator $p_\mu~= -i\hbar\partial_\mu$ which has a cohomology, where it results from the application of a Fermi-Dirac operator $Q_a$.  Fermi-Dirac states are such that only one particle can occupy a state, which has the topological content of $d^2~=~0$.  This cohomology is the basis for BRST quantization.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-23T01:52:28.817" CommentCount="2" />
  <row Id="3651" PostTypeId="2" ParentId="3635" CreationDate="2011-01-23T01:53:09.347" Score="2" ViewCount="" Body="&lt;p&gt;A better starting place for this study might be looking at the quantum mechanics of &lt;a href=&quot;http://physics.stackexchange.com/questions/236/how-does-positronium-exist&quot;&gt;positronium&lt;/a&gt;, which is a bound state of an electron and a positron.  In this case, both particles have equal masses, and therefore, would radiate equally.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, positronium isn't qualitatively very different from the hydrogen atom--the &lt;a href=&quot;http://en.wikipedia.org/wiki/Reduced_mass&quot; rel=&quot;nofollow&quot;&gt;reduced mass&lt;/a&gt; of the electron is obviously very different, and it is necessary to apply a correction for the fact that both charges are moving, but you get the same basic energy level structure that you do for the hydrogen atom.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-01-23T01:53:09.347" />
  <row Id="3652" PostTypeId="2" ParentId="3635" CreationDate="2011-01-23T02:17:42.237" Score="3" ViewCount="" Body="&lt;p&gt;Although a point charge does radiate when accelerated, this isn't necessarily true of all charge distritutions, where the radiation fields of all the charge elements can cancel in some cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This nonradiation condition has been known since 1910 when Paul Ehrenfest published a paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ungleichförmige Elektrizitätsbewegungen ohne Magnet- und Strahlungsfeld: Phys. Z. 11 (1910), 708-709)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information, have a look at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonradiation condition&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nonradiation_condition&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Nonradiation_condition&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Invisibility Physics: Acceleration without radiation, part I&#xA;&lt;a href=&quot;http://skullsinthestars.com/2008/04/19/invisibility-physics-acceleration-without-radiation-part-i/&quot; rel=&quot;nofollow&quot;&gt;http://skullsinthestars.com/2008/04/19/invisibility-physics-acceleration-without-radiation-part-i/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recent classic papers are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Goedecke, G. H. (1964). &quot;Classically Radiationless Motions and Possible Implications for Quantum Theory&quot;. Physical Review 135: B281–B288. doi:10.1103/PhysRev.135.B281&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Haus, H. A. (1986). &quot;On the radiation from point charges&quot;. American Journal of Physics 54: 1126&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="1345" LastEditorDisplayName="" LastEditDate="2011-01-23T02:37:59.190" LastActivityDate="2011-01-23T02:37:59.190" CommentCount="1" />
  <row Id="3653" PostTypeId="1" AcceptedAnswerId="3669" CreationDate="2011-01-23T03:16:03.160" Score="3" ViewCount="145" Body="&lt;p&gt;Is it possible to describe the Local F-theory models which seems to be relevant for particle physics phenomenology using Gauged Linear Sigma Models (GLSM), this includes describing the decoupling limit using the parameters of the GLSM? Actually, I am not sure if my question makes sense. Another related question is: Can elliptically fibered Calabi-Yau 4-folds be described as the IR limit of GLSM while also producing the global constraint of F-Theory compactifications, and also producing the Weierstrass form of the elliptic fibre. Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="1531" LastActivityDate="2011-02-22T08:02:47.820" Title="Linear sigma models for F-Theory" Tags="&lt;string-theory&gt;" AnswerCount="1" />
  <row Id="3654" PostTypeId="2" ParentId="3644" CreationDate="2011-01-23T03:54:55.217" Score="10" ViewCount="" Body="&lt;p&gt;Tom, would you have asked the question &lt;em&gt;&quot;why is the speed of light 1 ls/s&quot;&lt;/em&gt; if we happened to measure distance in lightseconds and time in seconds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The true answer to your question is: the speed of light is 1 if you measure distance and duration in compatible units, and it is whatever your system of units defines it to be if you adopt units that are more cumbersome. Another way of explaining is that speed - loosely speaking - corresponds to an angle in spacetime. And angles are dimensionless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know, this is not seen as a satisfactory answer. But that is because you ask the wrong question. The right question is &lt;em&gt;&quot;why is everything around us so slow? Why are the speeds we typically encounter for material objects around 10^-8 level?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-01-23T03:54:55.217" CommentCount="5" />
  <row Id="3655" PostTypeId="2" ParentId="3644" CreationDate="2011-01-23T04:09:39.410" Score="1" ViewCount="" Body="&lt;p&gt;A lot of the answers here seem to miss the thrust of the question, which is (I think), why do photons travel at the speed of light c, as opposed to some other speed. That is, given a definition of a meter as &quot;a stick that is this long&quot;, why does light take the particular amount of time to cross that distance that it does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that there is a set speed that any massless particle travels at, such as a photon, and that speed, c, is a fundamental property of our universe. Any time a physicist says something is fundamental, it means (s)he doesn't know why, it just is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be fair, you can explain why c is significant by appealing to relativity, the way we measure how time flows, the definition of the units we use when measuring it, etc. But, at the most basic level, c is a given, something we plug into equations, not something we get out of them. It is a property of light (and any massless particle), but it is one we have to observe the universe in order to find.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, the experiments used to determine the speed of light aren't especially unclear. There are several. Which one gave the first accurate answer is, perhaps, in some dispute, but the making of the measurement isn't a point of contention.&lt;/p&gt;&#xA;" OwnerUserId="1196" LastActivityDate="2011-01-23T04:09:39.410" />
  <row Id="3656" PostTypeId="1" CreationDate="2011-01-23T04:20:28.520" Score="7" ViewCount="360" Body="&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/1787/what-is-known-about-the-topological-structure-of-spacetime&quot;&gt;This question&lt;/a&gt; asks what constraints there are on the global topology of spacetime from the Einstein equations. It seems to me the quotient of any global solution can in turn be a global solution. In particular, there should be non-orientable solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But does quantum physics place any constaint? Because it seems to me that if space by itself is non-orientable then what happens to, say, neutral kaon interactions along two different paths that come back to the same spot with opposite orientations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So then, is there any reason why time cannot be non-orientable? For example my mental picture (two space dimensions suppressed) is of a disc. The big bang is the centre, time is the radial direction, space is the circumferential direction. A timelike geodesic that avoids black or white holes will start on the big bang, go out to the edge of the disc, continue on the opposite edge with time and orientation (and presumably matter/anti-matter) reversed, and return to the big bang (which is also therefore the same as the big crunch). The &quot;reflection time&quot; of the universe would be large enough that thermodynamic violations are not observed.&lt;/p&gt;&#xA;" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:17:34.653" LastActivityDate="2011-02-04T07:17:34.653" Title="Can spacetime be non-orientable?" Tags="&lt;general-relativity&gt;&lt;differential-geometry&gt;&lt;parity&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="3657" PostTypeId="2" ParentId="3644" CreationDate="2011-01-23T05:07:32.987" Score="6" ViewCount="" Body="&lt;p&gt;The particular value of $c$ depends on how long a meter is and how long one second is.  If meters were longer, for example, the speed of light would be a smaller number, even though light would still be as fast.  Viewed this way, physical measurements are ratios.  In this case, it's a ratio of the speed of light to a rather arbitrary speed - one meter per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One meter per second is roughly a walking speed.  So your question might be interpreted as, &quot;Why is the speed of light three hundred million times faster than a walking speed?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is very anthropocentric.  It is a question about how large we are (how many atoms are in our bodies), how much power our muscles can exert (the energy involved in chemical reactions), and how strong our bones and ligaments are (the strength of materials).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we would like to stick to physics, it will be more insightful to look at the speed of light as a ratio of something else.  We should look for some other speed set by nature, rather than a human-based speed, and compare the speed of light to that.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A typical candidate is to take Planck's constant $\hbar$ and the unit of electric charge $e$.  These can be combined to create a velocity $e^2/\hbar = 2.2*10^6 m/s$.  (In some systems of units, you need to include other &quot;constants&quot; like the permittivity of free space to convert the units.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is, roughly speaking, the speed of an electron in an atom.  An electron's energy is characterized by $E \approx e^2/r$, with $r$ the size of the orbit.  Its angular momentum comes in units of $\hbar$, so $L \approx \hbar \approx mvr$.  The virial theorem lets us write the energy as $E \approx mv^2$.  Using these facts, we can look for a way to estimate  the velocity.  $v = mv^2/mv \approx E/(L/r) \approx (e^2/r)/(L/r) = e^2/L = e^2/\hbar$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &quot;typical electron speed&quot; is about $\frac{1}{140} c$.  As a ratio, $e^2/\hbar c \approx \frac{1}{140}$.  This is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Fine-structure_constant&quot; rel=&quot;nofollow&quot;&gt;the fine structure constant&lt;/a&gt;.  It's very useful to know, because it's a number that describes the innate strength of the electromagnetic force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your original question becomes &quot;why is the fine structure constant $\frac{1}{140}$?&quot;, or &quot;Why is the speed of light $140$ when measured in fundamental units from quantum mechanics and electromagnetism?&quot;  Aside from a hokey invocation of the anthropic principle, I don't think there's an answer to this question, at least not yet.  A physical &quot;theory of everything&quot; might hope to derive the fine structure constant from some more basic idea, but this has not yet been achieved, and it is unknown whether it ever will be.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-01-23T05:07:32.987" CommentCount="3" />
  <row Id="3658" PostTypeId="2" ParentId="3593" CreationDate="2011-01-23T05:25:39.393" Score="1" ViewCount="" Body="&lt;p&gt;Physically meaningful means something that is, in principle, testable. That effectively means some measurement is possible that could answer a given question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the arrow of time, we might say that we could, in principle, measure how heat flows as a way of determining changes in entropy in order to make a comparison between another universe's arrow of time to ours. Of course, this presumes tons of other stuff that would be relevant to that specific question (equal numbers of time dimensions, entropy change as a measure of the arrow of time, laws governing entropy being similar between universes, the ability to measure this in more than one universe from a meaningful perspective, etc.), which is what makes that question interesting, but it underscores the basic idea of &quot;physically meaningful&quot; as testable or measurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much simpler examples that don't involve postulating other universes make the point more clearly. For example, we know it is not physically meaningful to talk about how close a radioactive atom is to decaying. By the above definition, this means that it is not possible, even in principle, to make measurements that allow you to answer the question &quot;How soon before this specific atom decays?&quot; It is only physically meaningful to talk about whether the atom has decayed or not, since this is something that can be measured, i.e. we can answer the question, &quot;Has this atom decayed yet?&quot;&lt;/p&gt;&#xA;" OwnerUserId="1196" LastEditorUserId="1196" LastEditorDisplayName="" LastEditDate="2011-01-23T05:47:31.923" LastActivityDate="2011-01-23T05:47:31.923" CommentCount="3" />
  <row Id="3659" PostTypeId="2" ParentId="3644" CreationDate="2011-01-23T05:33:40.170" Score="1" ViewCount="" Body="&lt;p&gt;The speed of light is not arbitrary. You can calculate the speed of propagation of small perturbations using Maxwell's equations, which gives $c=\frac{1}{\sqrt{\mu_0 \epsilon_0}}$. Thus when these to constants are fixed, so is the speed of light.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2011-01-23T05:33:40.170" CommentCount="7" />
  <row Id="3660" PostTypeId="1" CreationDate="2011-01-23T05:37:30.050" Score="0" ViewCount="90" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/2110/why-does-space-expansion-not-expand-matter&quot;&gt;Why does space expansion not expand matter?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Does cosmological expansion have any effects at the atomic level?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-23T06:13:38.710" ClosedDate="2011-01-23T06:41:03.797" Title="Effect of Cosmological Expansion at the Atomic Level" Tags="&lt;cosmology&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="3661" PostTypeId="2" ParentId="3660" CreationDate="2011-01-23T06:13:38.710" Score="0" ViewCount="" Body="&lt;p&gt;No. See: &lt;a href=&quot;http://physics.stackexchange.com/questions/3518/expansion-of-the-space-time-metric/3520#3520&quot;&gt;Expansion of the space-time metric&lt;/a&gt; and &lt;a href=&quot;http://physics.stackexchange.com/questions/2110/why-does-space-expansion-not-expand-matter&quot;&gt;Why does space expansion not expand matter?&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-01-23T06:13:38.710" />
  <row Id="3662" PostTypeId="1" AcceptedAnswerId="3960" CreationDate="2011-01-23T06:14:44.250" Score="7" ViewCount="467" Body="&lt;p&gt;In all the canonical approaches to the problem of quantum gravity, (eg. loop variable) spacetime is thought to have a discrete structure. One question immediately comes naively to an outsider of this approach is whether it picks a privileged frame of reference and thereby violating the key principle of the special relativity in ultra small scale. But if this violation is tolerated doesn't that imply some amount of viscosity within spacetime itself? Or am I writing complete nonsense here? Can anybody with some background in these approaches clear these issues?&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T07:53:27.620" LastActivityDate="2011-02-03T07:53:27.620" Title="Does the discreteness of spacetime in canonical approaches imply good bye to STR?" Tags="&lt;special-relativity&gt;&lt;loop-quantum-gravity&gt;" AnswerCount="7" CommentCount="5" />
  <row Id="3663" PostTypeId="2" ParentId="3656" CreationDate="2011-01-23T06:32:16.033" Score="5" ViewCount="" Body="&lt;p&gt;There is no mathematical difficulty of a non-time-orientable spacetime in GTR, and they can be generated by taking a quotient space rather easily, as you suggest. On the other hand, the same kind of action can be in some sense be &lt;em&gt;undone&lt;/em&gt; as well. For any spacetime $M$, let $M&amp;#39; = (p,o)$, where $p\in M$ and $o$ is a time orientation. Then this double covering spacetime, one for each orientation, will always be time-orientable (it could also be disconnected, in which case $M$ was time-orientable after all), so it's always possible to think of non-time-orientable spacetimes as incomplete pieces of time-orientable ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose for each symmetry C, P, T and all of their combinations, a process that violates them can be found. Then spacetime would have to be both spatially and temporally orientable, since the left/right and earlier/later distinction could be made experimentally. (Assume also that those physics are not just local.) Now, weak interactions uphold &lt;em&gt;only&lt;/em&gt; CPT, so that argument only works up to CPT-reversal. But it does suggest that spacetime is space-orientable if, and only if, it is time-orientable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that this only partly answers your question, but I don't know of any stronger conclusion. This was vaguely discussed in the monograph by Hawking &amp;amp; Ellis (Ch. 6 deals starts with this issue), and referenced Geroch's thesis. A similar more recent and accessible article by the same author on apparently similar topics is: &lt;a href=&quot;http://books.google.com/books?id=pxA4AAAAIAAJ&amp;amp;pg=PA212&quot; rel=&quot;nofollow&quot;&gt;Geroch, R., Horowitz, G. T., 1979. &lt;em&gt;Global Structure of Space-Times&lt;/em&gt;. In: &lt;em&gt;General Relativity, An Einstein Centenary Survey&lt;/em&gt;.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1353" LastActivityDate="2011-01-23T06:32:16.033" CommentCount="2" />
  <row Id="3665" PostTypeId="2" ParentId="3636" CreationDate="2011-01-23T06:45:27.047" Score="4" ViewCount="" Body="&lt;p&gt;Dear Jerry,&#xA;the $N=8$, $d=4$ &quot;non-stringy&quot; supergravity is&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;non-perturbatively inconsistent&lt;/li&gt;&#xA;&lt;li&gt;unacceptable phenomenologically&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Trying to fix either of these things leads one to string/M-theory. See&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Two roads from $N=8$ sugra to string theory&lt;/strong&gt;&#xA;  &lt;a href=&quot;http://motls.blogspot.com/2008/07/two-roads-from-n8-sugra-to-string.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2008/07/two-roads-from-n8-sugra-to-string.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The perturbative inconsistency may be seen in many ways: for example, the supergravity theory has $U(1)$ charges but produces no charged objects with respect to these $U(1)$'s. That's inconsistent because at least a newly formed black hole may confine these electric and magnetic fields and become charged. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electric and magnetic charges have to be quantized in inverse units, as seen by the Dirac quantization argument. It follows that the noncompact continuous exceptional $E_{7(7)}$ symmetry has to be broken to its discrete subgroup, the U-duality group. There are many ways to choose the lattice of allowed charges. These ways are related by the original continuous symmetry. In decompactification limits, the lightest of these charges (with smallest spacing) may be interpreted as Kaluza-Klein momenta with respect to new dimensions, and one discovers the 7 compactified dimensions of M-theory. It may also be showed that the other charges inevitably have the shape of string/M-theoretical membranes and fivebranes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no doubt today - and since the mid 1990s, in fact - that the supergravity theory is just a perturbative approximation to string/M-theory which is also why the supergravity community has been fully merged with the string/M-theory community. The people realize that they are working on the same theory and they are saying the same things. Ask Michael Duff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Phenomenology&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximal supergravity in four dimensions is left-right symmetric, and the high supersymmetry leads to too huge degenerate multiplets where spins differ by as much as $2$. The only acceptable supersymmetry is the minimal one where spins differ by $1/2$. The maximum supersymmetry implies that left-handed neutrinos couldn't exist and for each particle, there would have to be lots of very different superpartners. One couldn't get matter and gauge fields decoupled from gravity etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum supersymmetry cannot be broken down to a smaller one by field-theoretical mechanisms - except for an explicit breaking that just destroys all the finiteness virtues of the supergravity. However, it may be broken at the stringy level, by appreciating the extra 6-7 dimensions, and compactifying them differently. The resulting models are compactifications of string/M-theory. They preserve the perturbative finiteness by the added stringy species and they also lead to realistic phenomenology with all types of matter and interactions that we know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Joe Polchinski said, all roads lead to string theory. In the case of attempts to overcome limitations of supergravity, the previous sentence is not a slogan but rather an accurate description of the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-23T08:14:32.060" LastActivityDate="2011-01-23T08:14:32.060" />
  <row Id="3666" PostTypeId="2" ParentId="3644" CreationDate="2011-01-23T07:06:05.990" Score="2" ViewCount="" Body="&lt;p&gt;the speed of light was created by Nature to be one, the number whose multiplication influences nothing. But the primitive people who lived in spacetime and moved by speeds much smaller than $c=1$ - along small angles in the spacetime - were not able to see that their speeds were particular fractions of the maximum speed. The mankind remained that primitive until 1905 when Albert Einstein changed the story (with some marketing help by Hermann Minkowski in 1908).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even though space and time are fundamentally the same quantity measured in different directions, the people chose different units for length and duration. Some particular people chose $1/24/3600$ of the solar day because the powers of $60$ and $12$ etc. were quite popular - a lot of random messy history of mathematical conventions. They called the units one second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other people chose one meter as $1/40,000,000$ of the circumference of a meridian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In those randomly chosen units of distance and time - which were refined, to be more accurate, to the number of periods of various types of radiation - the speed of light $c=1$ could have been written as $299,792,458$ m/s. At least, the measurements became accurate enough so that the definition of one meter was changed in the 1980s to keep the speed of light in these units at least constant. So the speed I wrote is now actually exact, by definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adult physicists who work with relativistic theories use units where $c=1$. Similarly, adult quantum physicists use units with $\hbar=1$, $k_B=1$, and sometimes $G=1$ when they study general relativity (or quantum gravity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize: the numerical size of the universal constants has nothing to do with fundamental physics - it is all about human conventions (the units).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Luboš&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-23T07:06:05.990" CommentCount="2" />
  <row Id="3667" PostTypeId="2" ParentId="1453" CreationDate="2011-01-23T07:27:58.587" Score="2" ViewCount="" Body="&lt;p&gt;Why is the expansion of the universe accelerating?&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-01-23T07:27:58.587" CommunityOwnedDate="2011-01-23T07:27:58.587" CommentCount="1" />
  <row Id="3668" PostTypeId="2" ParentId="3645" CreationDate="2011-01-23T07:31:17.997" Score="2" ViewCount="" Body="&lt;p&gt;Mandelstam variables $s,t,u$ are quantities with units of squared momentum (or mass) that describe the Lorentz-invariant part of the information about the momenta and energy in $2\to 2$ scattering processes:&#xA;$$ s= (p_1+p_2)^2, \quad t=(p_1-p_3)^2,\quad u=(p_1-p_4)^2$$&#xA;If you Lorentz transform the incoming momenta $p_1,p_2$ as well as outgoing ones $p_3,p_4$, the quantities above don't change. So by relativity, all the nontrivial information about the collision is encoded in functions of $s,t,u$. Moreover, $s+t+u=4m^2$ if all four particles' masses are equal to $m$; this is generalized easily if they're different. The Mandelstam variables may also be generalized easily to the case of more than 4 external lines - there are more variables in that case. So the Mandelstam variables are a simple thing, given by the formulae above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Coleman-Mandula theorem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Coleman-Mandula theorem shows that theories with (bosonic) symmetry groups that mix the spatial (geometric) and internal parts, i.e. they don't have the form $G_{spacetime}\times G_{internal}$, the interactions essentially vanish, so such theories are not usable for any realistic modeling. So for example, if someone hypothesized that an $E_8$ symmetry may include both rotations in space as well as the Standard Model gauge group, his hypothesis would be instantly ruled out by the theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the proof, I think that they take the lightest scalar excitation in the given theory, two or several of them, and scatter them. The extra symmetries - besides energy and momentum etc. - will inevitably mean that some tensors have to be conserved in a collision, too. Because these tensors may only depend on the momenta of the particles we scatter and they have too many components that must remain unchanged, they can show that the momenta after the scattering have to be essentially equal to those before the scattering. This already means that the interactions kind of vanish universally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theorem assumes that the conserved quantities can't be spinors - with half-integral spin. When fermionic conserved quantities with half-integer spin - supersymmetries - are allowed, one finds out that it is possible to circumvent the original theorem, in theories with supersymmetry, because the conserved spinors are not too constraining. One gets supersymmetric theories but their possibilities are still restricted by a supersymmetric extension of the Coleman-Mandula theorem, the Haag-Lopuszanski-Sohnius theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-23T07:31:17.997" />
  <row Id="3669" PostTypeId="2" ParentId="3653" CreationDate="2011-01-23T07:35:44.253" Score="1" ViewCount="" Body="&lt;p&gt;concerning the first question, I don't think so. If I understand you well, you want to use GLSM's for a world sheet description of the vacuum. But a world sheet description of any string theory is only possible if the coupling constant in almost all of space is weak because the world sheet is the history of a (weakly coupled) string - so strings must remain the lightest objects etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the very point of F-theory - including the loci of the branes in the local models - is that the complexified coupling $\tau$ living in the fundamental domain may have very different values across the space, and usually undergoes nontrivial monodromies - which show that the strong coupling is identified with the weak coupling. So the world sheet of the fundamental string is exchanged with the world volume of the D-strings, and so on. That's why people have to use spacetime methods for F-theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know the answer to the elliptically fibered question you posed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-23T07:35:44.253" CommentCount="1" />
  <row Id="3670" PostTypeId="2" ParentId="3656" CreationDate="2011-01-23T07:40:50.810" Score="7" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Our spacetime cannot be unorientable&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's because the laws of physics describing our spacetime are not left-right symmetric. We say that they break the P-symmetry (parity) or that they are &quot;chiral&quot; (derived from a Greek word for the hand which is either left or right.) For example, a left-handed neutrino would turn into a right-handed neutrino if you made a round trip. But the right handed neutrinos can't exist, so the world would encounter a system error, trying to find out what happens with the neutrino after the orientation-changing round trip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could try to fix it by saying that it becomes a right-handed antineutrino which does exist. However, the CP-symmetry needed for this transformation is not an exact symmetry of our Universe, either. So it cannot work - just like before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In more general theories, you can have unorientable spacetimes.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, M-theory (and, similarly, string theory in ten dimensions) may be compactified on a Klein bottle or a Möbius strip. In fact, these theories have nice and well-defined properties. They preserve up to one half of the supersymmetries and we know various dual descriptions of them. In particular, the Möbius strip only has one boundary, so one gauge group per point in the large dimensions, and we know what the dual theory with one $E_8$ etc. is. Similarly, the Klein bottle has no boundaries. Again, we know what happens when M-theory gets compactified on a tiny Klein bottle. More complicated manifolds than the Klein bottle and the Möbius strip are also possible; however, they almost generally break all of supersymmetry - much like any other generic compactifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is not possible for compactifications to switch the arrow of time. Spacetimes must have the property that an arrow of time may be defined consistently across the whole spacetime. Unorientable spacetime manifolds, when it comes to the arrow of time, would have the same inconsistencies as spacetimes with closed time-like curve: you could assassinate your grandfather before he met your grandmother (because &quot;before&quot; could be reinterpreted as &quot;after&quot; by making an orientation-reversing round trip), and so on, leading to logical contradictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:17:11.407" LastActivityDate="2011-02-04T07:17:11.407" CommentCount="4" />
  <row Id="3671" PostTypeId="2" ParentId="3662" CreationDate="2011-01-23T07:45:30.123" Score="4" ViewCount="" Body="&lt;p&gt;Yes, sb1, all your statements are totally correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A discrete spacetime immediately implies lots of bulk degrees of freedom that remember the detailed arrangement of the discrete blocks - unless it is regular and unique. Consequently, the &quot;vacuum&quot; carries a gigantic entropy density - the Planck entropy density if the concept is applied at the Planck scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such an entropy density $(huge,0,0,0)$ immediately violates the Lorentz symmetry because in different reference frames, it will have nonzero spatial components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the privileged reference frame will mean that any moving object will instantly - within a Planck time or so - dissipate all of its kinetic energy to the extra degrees of freedom in the discrete structure. So the &quot;vacuum&quot; will actually behave as a superdense liquid that immediately stops any &quot;swimmer&quot;: inertia becomes totally impossible. One could continue with other flagrant contradictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these things dramatically violate basic observations of the real world. In fact, the Fermi satellite has verified that even at the Planck scale, the Lorentz symmetry works much better than up to O(100%) errors. Also, one can see that any non-stringy, non-QFT unification attempt for quantum gravity has considered the discreteness in the very sense we discuss, which is why it is instantly excluded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can use special dictionary and special proofs for each of them. For example, spin foam models define the proper area of surface $\Sigma$ more or less as the number of its intersections with the spin foam. However, it's clear that this number can't go to zero even if the shape of $\Sigma$ in spacetime is chosen to be light-like or near-light-like.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T07:53:05.763" LastActivityDate="2011-02-03T07:53:05.763" CommentCount="5" />
  <row Id="3672" PostTypeId="2" ParentId="3662" CreationDate="2011-01-23T08:48:33.977" Score="2" ViewCount="" Body="&lt;p&gt;Loop Quantum Gravity is an attempt to quantise four dimensional general relativity directly by canonical quantisation methods. If it succeeded it would have to be able to recover the original curved spacetime and gravitational dynamics as a classical limit, but in its current form it does not quite do that. In this context it does not even make sense to discuss whether STR is lost due to discreteness. (STR is taken to mean Special Theory of Relativity. More precisely in this context it means that we recover Lorentz symmetry for local reference frames) Some physicists such as Carlo Rovelli have not given up trying to fix the problems of LQG.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the mathematics of knot theory and spin networks that arises naturally in the LQG approach is itself very interesting and this may be telling us something about how a more successful theory of quantum gravity should work. There is too much ground to cover in discussing such possibilities so I'll mention just one feature of spin networks that gives a clue about how a discrete spacetime may not be in contradiction with STR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a preliminary point, canonical quantisation methods treat time and space differently which is why the question asks whether a preferred reference frame arises. The best answer to this is that canonical quantisation can be used in relativistic quantum field theory (for example in QED). It works and is equivalent to Lagrangian path integral methods that explicitly preserve Lorentz invariance, so no preferred reference frame arises in this case. However, since the symmetry is not explicitly preserved in canonical methods it has to be demonstrated that the final solution preserves it. As I've said already, this is the case in QFT but in LQG we don't get a final solution where this can even be tested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, The interesting thing to note is that you can use spin networks in three dimensional space to define a theory of 3D quantum gravity which is discrete but yet preserves diffeomorphism invariance (and therefore also local Lorentz invariance) due to algebraic relations that allow the spin networks to be modified without affecting the overall path integral sum. See e.g. &lt;a href=&quot;http://arxiv.org/abs/hep-th/9202074&quot; rel=&quot;nofollow&quot;&gt;arxiv:hep-th/9202074&lt;/a&gt; This result has inspired all the work on spin foams and group field theory which attempts to get a similar result for 4D quantum gravity. So far a fully working solution has not been found, but the message is that discrete spacetime does not necessarily have to violate Lorentz invariance and one day we may learn how to see that for a real theory of quantum gravity.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-23T08:48:33.977" CommentCount="7" />
  <row Id="3673" PostTypeId="2" ParentId="3662" CreationDate="2011-01-23T10:46:15.917" Score="2" ViewCount="" Body="&lt;p&gt;@sb1 you've opened up a can of worms here. This is a good question but it is subject to the standard misunderstandings about the nature of LQG and quantum geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is &lt;strong&gt;no&lt;/strong&gt; problem with the question of Lorentz Invariance (LI) in a discrete spacetime. &lt;em&gt;If your spacetime becomes discrete then your notion of LI must change accordingly.&lt;/em&gt; Keep in mind that LI is nothing more than a statement that physics should be causal &lt;a href=&quot;http://adsabs.harvard.edu/abs/1968PhRv..174.1606B&quot; rel=&quot;nofollow&quot;&gt;ref1&lt;/a&gt;. For a continuous $3+1$ manifold this requirement is expressed in terms of invariance of the space-time interval $ds^2 = -dt^2 + dx^2$. The corresponding symmetry group is the continuous Lie group $SO(3,1)$ or $SL(2,\mathbb{C})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a discrete spacetime $ds^2$ will have to be replaced by its discrete generalization $ \hat{d} s^2 = -\hat{d}t^2 + \hat{d}x^2 $ with $\hat{d}$ being the discrete (&quot;quantum&quot; or &quot;q&quot;) differential. The invariance group of this interval is $SL(2,\mathbb{Z})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a more physical route involving considerations of the transformation of punctures of black hole horizons w.r.t an external observer which leads us to see how $SL(2,\mathbb{C})$ must reduce to $SL(2,\mathbb{Z})$ in the event that the numbers of these punctures is small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also your intuition regarding the dissipative effects of discreteness, is correct:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But if this violation is tolerated doesn't that imply some amount of viscosity within space-time itself?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It absolutely does and this is a result that we have know from a completely different directions - primarily from AdS/CFT and the &lt;a href=&quot;http://arxiv.org/abs/1101.2451&quot; rel=&quot;nofollow&quot;&gt;fluid-gravity correspondence&lt;/a&gt; tells us that there is a &lt;strong&gt;lower bound&lt;/strong&gt; for the viscosity to entropy ratio of horizons - those of black holes or those experienced by accelerated observers in an otherwise flat spacetime. &lt;a href=&quot;http://arxiv.org/abs/0806.3165&quot; rel=&quot;nofollow&quot;&gt;Eling - Hydrodynamics of spacetime and vacuum viscosity&lt;/a&gt;, &lt;a href=&quot;http://arxiv.org/abs/0704.0240&quot; rel=&quot;nofollow&quot;&gt;Son and Starinets - Viscosity, Black Holes, and Quantum Field Theory&lt;/a&gt;. This leads to the question of information loss. Having a dissipative horizon seems to suggest that information is lost in quantum gravity. However, this is only as seen by a &quot;local&quot; observer - who only has access to a portion of the spacetime. For a &lt;em&gt;universal&lt;/em&gt; observer who has access to all the regions of the spacetime, this problem will not occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The debate around this topic is reminiscent of that surrounding Einstein's introduction of the notion that a Lorentzian manifold, rather than a Galilean one, was the right tapestry for events in spacetime. Then people struggled mightily to adapt all the physical inconsistencies of Newtonian's theory without having to give up the cherished Galilean property of absolute space and time. Now we see a similar reluctance to abandon the safe confines of continuous manifolds for a more general framework which can also describe discrete geometries. If one is willing to take this intellectual leap then there is no problem in becoming comfortable with a notion of LI which is adapted to the discrete setting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect a strong response from the usual suspect(s) ;)&lt;/p&gt;&#xA;" LastActivityDate="2011-01-23T10:46:15.917" CommentCount="17" />
  <row Id="3674" PostTypeId="2" ParentId="3618" CreationDate="2011-01-23T11:11:45.537" Score="1" ViewCount="" Body="&lt;p&gt;No, you can't. For several reasons. First, if you have E, to get the B field, you need additional assumptions about the structure of the theory, ie in more detail the field strength tensor, see above reply by Lubos. But in addition to this, even if you had the solution for a point charge, to get Maxwell's equations you need to know more than just having one solution. For example that they're linear, second order, and what the symmetry group is.  And if you've added that, you can derive the Maxwell equations from these assumptions anyway without even starting with the Coulomb field. &lt;/p&gt;&#xA;" OwnerUserId="1458" LastActivityDate="2011-01-23T11:11:45.537" CommentCount="1" />
  <row Id="3675" PostTypeId="2" ParentId="3611" CreationDate="2011-01-23T11:41:19.540" Score="3" ViewCount="" Body="&lt;p&gt;This is a fairly long answer, but I thought it would be fun to analyze the skater thinking solely about forces. Angular momentum comes in at the end, when it pops up inevitably.  I'll give a qualitative answer describing the forces in the system and how they cause spinning, then a quantitative answer to calculate the spin rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Qualitative Answer&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an ice skater pulls in her arms a legs, her arms and legs exert a &lt;a href=&quot;http://en.wikipedia.org/wiki/Torque&quot; rel=&quot;nofollow&quot;&gt;torque&lt;/a&gt; on her body, causing her to spin faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set a plate in front of you.  Try pushing on at in various places and in various directions.  You can imagine an arrow pointing in the direction of the force, starting from the point where you're pushing.  If this arrow points towards the center of the plate, the plate will not rotate.  Otherwise, it will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/GdFtZ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this picture, the red arrows indicate forces on the gray plate.  The force labeled &quot;Spin&quot; will induce some spinning in the plate (in addition to accelerating it as a whole) because the dotted line &quot;j&quot; of that force does not go through the center of the plate.  The force labeled &quot;NoSpin&quot; will only accelerate the plate and not cause any spinning because it lies on a line that passes through the center of the plate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is nothing special about the circular plate in this example.  Any other shape would work as well, but you would need to define the center point by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Center_of_mass&quot; rel=&quot;nofollow&quot;&gt;center of mass&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that the skater spins up as she pulls in her arms, we'll need to find the forces exerted on her body while she pulls her arms in.  Then we can see whether these forces point straight towards the center of her body or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll model an ice skater as a circle of mass $M$ with a massless stick pointing through it, and two more circles of mass $m$ on either end of the stick.  The skater can move the small circles inward or outward as she spins.  Here's a picture of the setup, along with the path her &quot;arms&quot; (the small circles) trace out if she does not pull them in at all while spinning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/T6Icw.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you imagine holding something heavy in your arms while spinning, your arms would feel as if they're being pulled out of their sockets.  In fact, they are.  You arms exert a force on the weights straight inwards towards your body, and the weights exert equal and opposite force straight outward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/xwhpG.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each force is color-coded according to which body it acts on.  $F_1$, for example, is the force exerted on the skater by the red arm.  (This force is really exerted on the stick, which is rigidly attached to the skater.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the blue forces lie on a line passing through the center of the skater, so the skater's spin rate doesn't change in this scenario.  As long as she leaves her arms out, she'll remain spinning the same speed (neglecting friction or other losses of energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we imagine the skater pulling her arms in.  If you watch the path the &quot;arms&quot; (small circles) trace out, you see a spiral shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/TA0L2.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, we show the two arms with both their past and future trajectories traced out.  The arms are spinning around the skater while simultaneously being pulled in.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will be more difficult to find the forces involved here.  The arms are no longer moving in simple circles.  However, at any given instant, there is a particular circle along which a given arm appears to move.  This is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Osculating_circle&quot; rel=&quot;nofollow&quot;&gt;osculating circle&lt;/a&gt;.  As before, there's a force on the arm pointing in towards the center of the osculating circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not the whole story, because we can no longer assume that the speed of the arms is constant.  Hence there may also be a force on the arms in the direction of their motion.  In the next picture, we'll draw only the forces on one arm, just to keep things from getting too cluttered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/yfQ4l.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This picture is zoomed in on one arm.  The blue circle is still the skater.  The green circle is the osculating circle.  $F_r$ is the centripetal force towards the center of the osculating circle that curves the arm's path, and $F_t$ is a (small) tangential force in the direction of motion that speeds the arm up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without knowing just how fast the arms are moving, how much they're accelerating, and the equation describing their trajectory, it's hard to tell precisely what the sizes of these forces will be.  However, we can see that the forces in general have no obligation to point towards the skater's center any more.  She can change her spin rate because the forces may not necessarily lie on lines that pass through her center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to see just what these forces are, we need to do an quantitative analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Quantitative Answer&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plan of this answer is to find the forces on the arms as they spiral in, then use Newton's third law to find the forces the arms exert on the stick.  Next, we'll relate these forces to the rate of change of the energy of the skater.  The energy of the skater can also be calculated from her motion directly, so we'll do that, take a time derivative, and compare to our previous expression.  This will reveal a conserved quantity, the angular momentum, which allows us to find the skater's rotation rate as a function of the initial conditions and the final distance of the arms from her center - the details of the shape of the spiral and how fast the arms are pulled in do not matter.  Finally, we'll see that the skater spins faster and faster as she pulls her arms in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll use &lt;a href=&quot;http://en.wikipedia.org/wiki/Polar_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;polar coordinates&lt;/a&gt; to describe the positions of the arms.  The radial coordinate $r$ of arm 1 is some function of its angular coordinate $\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r = f(\theta)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the definition $\omega = \dot{\theta}$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{r} = f&amp;#39;\omega$$&#xA;$$\ddot{r} = f&amp;#39;&amp;#39;\omega^2 + f&amp;#39;\dot{\omega}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Polar_coordinate_system#Vector_calculus&quot; rel=&quot;nofollow&quot;&gt;acceleration in polar coordinates&lt;/a&gt; is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{a} = \hat{r}(\ddot{r} - r \dot{\theta}^2) + \hat{\theta}(2\dot{r}\dot{\theta} ^2 + r\dot{\omega})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force on this arm is given by Newton's second law, $\vec{F} = m\vec{a}$.  The force the arm exerts on the stick is the negative of this, by Newton's third law.  This force of the arm on the stick is what we're interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the stick spins, the force of the arm on the stick does work on the stick, which goes into the kinetic energy of the skater (the stick itself has no mass).  The velocity of the point on the stick where the force is applied is $\vec{v} = f\omega\hat{\theta}$.  The power delivered by this force, doubled to include the work done by the other arm, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = \vec{F} \cdot \vec{v} = -2m\omega f(2f&amp;#39;\omega^2 + f\dot{\omega})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the rate of change of the kinetic energy of the skater.  That kinetic energy is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T = \frac{1}{2}M (R \omega)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $R$ the radius of the skater.  If we equate the power to the time derivative of the kinetic energy, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$MR^2\dot{\omega} = -2mf(2f&amp;#39;\omega^2 + f\dot{\omega})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you stare at this and scratch your head a moment, it's mathematically equivalent to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\textrm{d}}{\textrm{d}t}\left(\omega(MR^2 + 2mf^2)\right) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means we've discovered something that doesn't change in time - a conserved quantity.  It's called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Angular_momentum&quot; rel=&quot;nofollow&quot;&gt;angular momentum&lt;/a&gt;, and the part $MR^2 + 2mf^2$ is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Moment_of_inertia&quot; rel=&quot;nofollow&quot;&gt;moment of inertia&lt;/a&gt;.  We denote the angular momentum by $L$ and (switching back from $f$ to $r$) write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = \omega(MR^2 + 2mr^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $L$ is a constant, we can find $\omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega = \frac{L}{MR^2 + 2mr^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've got the angular frequency as a function of $r$ only -  the precise function form of $f$ didn't matter, and neither did how quickly we traversed the path.  As long as we know $L$ from the initial conditions, we've solved the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\textrm{d}\omega}{\textrm{d}r} = \frac{-4m L r}{(MR^2+2mr^2)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we see that, assuming $L$ is positive, as $r$ goes down, $\omega$ goes up, so the skater goes faster and faster as she pulls in her arms.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-01-23T11:41:19.540" CommentCount="3" />
  <row Id="3676" PostTypeId="1" CreationDate="2011-01-23T12:19:05.733" Score="4" ViewCount="192" Body="&lt;p&gt;What are some &quot;not so grand&quot; unsolved physics problems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg. problems that can very likely be solved (and papers published) by spending X months or years studying and working, it's just that nobody has done it yet?&lt;/p&gt;&#xA;" OwnerUserId="47" LastActivityDate="2011-01-23T12:19:05.733" CommunityOwnedDate="2011-01-23T12:20:13.190" ClosedDate="2011-01-24T00:39:49.660" Title="Not so grand unsolved physics problems" Tags="&lt;soft-question&gt;" CommentCount="5" FavoriteCount="0" />
  <row Id="3677" PostTypeId="2" ParentId="3495" CreationDate="2011-01-23T13:16:00.373" Score="0" ViewCount="" Body="&lt;p&gt;My answer is NO. There is a problem in many QM considerations that we have an isolated system obeying (deterministic!) Schrödinger equation that is a subject of mystical &quot;measurements&quot; that introduce non-deterministic behavior.  But one can't make a measurement and leave system isolated (indeed saying that system is isolated is always an approximation in QM, and much heavier than in CM). In fact measurement is an act of introducing interactions with measuring apparatus, measurer, coffee measurer drinks, ... -- so the measurement result can be in theory calculated, but would involve inaccessible and enormous amounts of information. This makes it practically non-deterministic, but fundamentally it is no better than classical chaos.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2011-01-23T13:16:00.373" CommentCount="2" />
  <row Id="3678" PostTypeId="2" ParentId="3495" CreationDate="2011-01-23T13:45:35.887" Score="1" ViewCount="" Body="&lt;p&gt;Not sure if you've ever heard of the De Broglie-Bohm pilot wave interpretation of QM, but it is a &lt;em&gt;fundamentally&lt;/em&gt; deterministic interpretation.&lt;/p&gt;&#xA;" OwnerUserId="1540" LastActivityDate="2011-01-23T13:45:35.887" CommentCount="1" />
  <row Id="3679" PostTypeId="2" ParentId="3662" CreationDate="2011-01-23T14:00:59.993" Score="-1" ViewCount="" Body="&lt;p&gt;As Lubos points out, LQG requires the imposition of a vast amount of data or degrees of freedom on spacetime.  The amount of data for separable states is larger than when those states are in entanglements.  The holographic principle also indicates how the degrees of freedom in an $N$ dimensional spacetime (say by counting vertices in a tessellation of the spacetime) may be reduced to that on an $N-1$ dimensional boundary or horizon.  This reduces the huge combinatorics the LQG work requires (which makes some of their papers nearly unreadable), and further this also yields some interesting results on entanglement entropy and correlations between fields in a space and its boundary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a cut off in the scale of spacetime given by the Planck length.  However, this does not mean spacetime is sliced and diced up, but rather indicates some limit on the information we can extract from spacetime.  This connects in some ways to entanglement entropy of a black hole, where the elemental unit of a black hole has a Planck mass.  This discrete system is then a limit on the information content we may observe about quantum gravity or spacetime, but it does not necessarily mean that spacetime has some sliced and diced characteristic that leads to actual violations of Lorentz symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the numbers of degrees of freedom may be reduced far further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Fermi spacecraft observed widely different wavelengths of photons from a very distant quasar.  These photons were emitted in a pulse.  The time of arrival was nearly simultaneous, which means the photons traversed a multi billion light year distance with no dispersion.  This dispersion is predicted by violations of the equivalence principle and Lorentz symmetry breaking by LQG was not observed.  These photons should couple to the quantum foam and exhibit a slight dispersion over this vast distance.  I think this puts the kibosh on a lot of LQG.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-23T14:00:59.993" CommentCount="4" />
  <row Id="3680" PostTypeId="2" ParentId="3656" CreationDate="2011-01-23T14:31:19.737" Score="-2" ViewCount="" Body="&lt;p&gt;The observable universe is not likely non-orientable.  It is interesting to question whether the brane of any spacetime cosmology can be some orientifold which corresponds to a non orientable space or spacetime.  The observable universe is likely some bubble nucleation or “pocket universe” on the $R^3$ of spacetime.  The whole thing is driven by a large cosmological constant, but where our observable universe is some local region where the vacuum energy was reduced to near zero.  This resulted in reheating or a latent heat of fusion we identify as the thermal big bang.  Now if that $R^3$ were replaced by a Klein bottle it is possible that our pocket universe is very small (a tiny bubble compared to the whole space) so we are not able to detect anything.  Clearly parity violations in weak interactions hold, so neutrinos do not make a round trip and come back right handed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not clear whether this question can be definitely answered.  There are some reported CMB signatures of possible “collisions” between bubbles.  The verdict is not in on that as I see it.  However, if this should turn out to be on the mark we might begin to ponder whether interactions between bubbles with different orientations are possible.  However, at this time I see no way this can be answered, and further to assume a non-orientable spacetime complicates things quite a bit.  So for the time I think it is safe to say spacetime is orientable.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-23T14:31:19.737" CommentCount="2" />
  <row Id="3682" PostTypeId="2" ParentId="3644" CreationDate="2011-01-23T14:46:55.873" Score="0" ViewCount="" Body="&lt;p&gt;Suppose that there is a SuperEarth with a city called SuperParis. In this city, the inhabitants keep an iron rod with two marks on it. This distance is called a supermeter, it's the basic unit of length on superearth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If by pure chance these people use the same unit of time as humans and if by pure chance 1 supermeter would be 299792458m here on earth - then what is the speed of light for the superpeople?&lt;/p&gt;&#xA;" OwnerUserId="1377" LastActivityDate="2011-01-23T14:46:55.873" />
  <row Id="3683" PostTypeId="2" ParentId="3644" CreationDate="2011-01-23T15:26:11.207" Score="0" ViewCount="" Body="&lt;p&gt;The speed of light is just a conversion factor from one coordinate direction to another.  The proper unit is “one,” or $1.0ly/y$ and so forth.  Now it has this funny set of units in cgs and so forth.  However, if you were to argue the speed of light could be different in these units then the Planck units, such as $\ell~=~\sqrt{G\hbar/c^3}$ would all rescale accordingly and as a result so would our rods and clocks.  This would make the rescaling completely unobservable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why $c~=~ 299,792,458m/s$ has to do with other constants of nature, such as the mass of the proton and so forth.  We measure the speed of light according to physical objects and it has this large value due to the physical dimensions of rods, which depends on the Bohr radius which in turn depends on mass of electrons and so forth.  The speed of light is so large, in part because gravity is very weak, and this really has to do with the fact elementary particles have little mass in comparison to the Planck mass.  If this huge disparity did not exist the natural unit for the speed of light is one Planck length per Planck time, which is just a unity.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-23T15:26:11.207" CommentCount="1" />
  <row Id="3685" PostTypeId="2" ParentId="3568" CreationDate="2011-01-23T17:22:29.160" Score="2" ViewCount="" Body="&lt;p&gt;AIB,&#xA;As you have read the temperature is a measure of the change of energy of a system to the change of the number of microstates, or:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T = \dfrac{\partial{E}}{\partial{S}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where: &#xA;$$S = \ln \Omega$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $\Omega$ is the number of microstates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is important to understand what a microstate is.  In general a microstate describes all of the exact positions and momentums of all the particles in a system (or particles in a box).  The microstate is often thought of as a point in a 6N dimension space, where N is the number of particles and the factor of 6 is related to there being 3 position components in a 3D space and 3 momentum components in 3D space (however, one could certainly consider other parameter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we could know all of the positions and momentums of all the particles than we would know the precise microstate of the system.  This precise microstate would have an exact energy associated with it.  However, we do not have an ability to know the precise microstate of a system, so we must appeal to a statistical concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at any one instant, there are a large number of microstates a system could be in.  Again, each microstate being a unique description of the exact position and momentum of all particles in the box.  What this also suggests, is that for any microstate, there are a large number of other microstates that would have the same energy associated with them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question asked was:&#xA;&quot;Given all these information how do we calculate (not measure) the temperature of this system?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to this question would be determined first by the range of all the free parameters and the values of those parameters that that would keep the energy constant.  The number of combinations possible within the energy constraint is the number of microstates of the systems, the log of which is the entropy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is tricky, because we would have to ask how the energy changes with a change in entropy.  This is where the partition function comes into play, assuming energy of each microstate is equivalent, if we look at the partion function we see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${Z} = \sum_i e^{-\frac{\partial{S}}{\partial{E}}{E_i}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z = Ne^{-\frac{1}{T}E_i}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$N = Ze^{\frac{1}{T}E_i}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can see that the temperature is a variable that controls the growth of states if we keep the value of Z constant. If I know an exact microstate from which to derive the energy, then I still won't necessarily know anything about the temperature until I can partition the relavant space in question.  In this sense, the temperture is a sort of hyperbolic phase factor which changes the range of states.&lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="1383" LastEditorDisplayName="" LastEditDate="2011-01-23T21:20:56.493" LastActivityDate="2011-01-23T21:20:56.493" CommunityOwnedDate="2011-01-23T19:55:17.817" />
  <row Id="3686" PostTypeId="2" ParentId="884" CreationDate="2011-01-23T17:45:58.937" Score="0" ViewCount="" Body="&lt;p&gt;Studies in Modern Algebra - edited by A.A. Albert.  (MAA)&lt;/p&gt;&#xA;" OwnerUserId="1423" LastActivityDate="2011-01-23T17:45:58.937" CommunityOwnedDate="2011-01-23T17:45:58.937" />
  <row Id="3687" PostTypeId="1" AcceptedAnswerId="3696" CreationDate="2011-01-23T17:50:49.797" Score="4" ViewCount="159" Body="&lt;p&gt;Under what conditions does a system with many degrees of freedom satisfy the equipartition theorem?&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-01-23T20:02:17.230" Title="For which systems is the equipartition theorem valid?" Tags="&lt;statistical-mechanics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="3688" PostTypeId="1" CreationDate="2011-01-23T18:05:27.220" Score="5" ViewCount="243" Body="&lt;p&gt;I'm asking for a qualitative explanation if there is one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My own answer doesn't work.   I would have guessed it's because when a gas has pressure the kinetic energy adds to the rest mass of a given quantity of the gas, so the pressure contribution would be equal to whatever energy density it contributes.   But that can't be right.   If one had an ideal monatomic gas where the atoms are randomly moving around at non-relativistic speeds, the kinetic energy per volume of the atoms is 1.5 times greater than the pressure, but in chapter 4 of Schutz's book &quot;A First Course in General Relativity&quot; (or any other GR text) he says that rho plus pressure (in units where c=1) plays the role of inertial mass density.  In my incorrect view the equation would be rho  plus 1.5 pressure &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is my answer wrong?   I'm guessing part of the problem is that the kinetic energy of the atoms is already part of the mass density term--that is, a hot gas of one mole of helium atoms would have a higher mass than a cold gas composed of helium.   Then the pressure is tacked onto that, which seems like counting it twice to me, but clearly I'm confused.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastActivityDate="2011-01-23T19:47:55.767" Title="Why does pressure act as a source for the gravitational field?" Tags="&lt;general-relativity&gt;&lt;fluid-dynamics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="3689" PostTypeId="2" ParentId="884" CreationDate="2011-01-23T18:10:36.400" Score="1" ViewCount="" Body="&lt;p&gt;Gerard t'Hooft has a website with a course in theoretical physics which is comprehensive.&#xA;It looks good, but I have only scanned it:&#xA;&lt;a href=&quot;http://www.phys.uu.nl/~thooft/theorist.html&quot; rel=&quot;nofollow&quot;&gt;http://www.phys.uu.nl/~thooft/theorist.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-01-23T18:10:36.400" CommunityOwnedDate="2011-01-23T18:10:36.400" />
  <row Id="3690" PostTypeId="2" ParentId="884" CreationDate="2011-01-23T18:48:17.617" Score="1" ViewCount="" Body="&lt;p&gt;A few books not mentioned come to mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feynman, Lectures on Gravitation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feynman, The Character of Physical Law&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geroch, Mathematical Physics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geroch Relativity A to B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Susskind, An Introduction to Black Holes, Information and the String Theory Revolution&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-23T18:48:17.617" CommunityOwnedDate="2011-01-23T18:48:17.617" />
  <row Id="3691" PostTypeId="2" ParentId="884" CreationDate="2011-01-23T19:02:54.083" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://books.google.ca/books?id=n8Mmbjtco78C&amp;amp;printsec=frontcover&amp;amp;dq=%22Quantum+field+Theory+in+a+Nutshell%22+by+A.+Zee&amp;amp;hl=en&amp;amp;ei=rXk8TebAGoG6sQOxy6DiAw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCkQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&quot;Quantum field Theory in a Nutshell&quot; by A. Zee&lt;/a&gt; is an excellent introduction to the Quantum Field Theory. Zee focuses on physics of main concepts behind QFT and omits trivial mathematical details. His approach is to consider the simplest example possible and then discuss the generalizations. I find it very useful and definitely wish I knew this book as an undergraduate. And the author's approach has the best traits of Feynman's books.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-01-23T19:02:54.083" CommunityOwnedDate="2011-01-23T19:02:54.083" />
  <row Id="3692" PostTypeId="2" ParentId="3688" CreationDate="2011-01-23T19:15:59.653" Score="-2" ViewCount="" Body="&lt;p&gt;I am going to give a largely physical argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The contribution of pressure to the curvature occurs if the motion of the molecules is relativistic.  If you have a gas with a certain pressure it means the molecules of the gas are moving rapidly.  The momentum of a molecule $p~=~\gamma mv$ will collide off of a wall and impart some momentum to the wall.  Given an area $A~=~x^2$ of a wall in a box, if it is massive enough the molecules bounce back with momentum $p’~=~-\gamma mv$.  For $N$ molecules impacting the wall in a time $\delta t$ this is a net momentum transfer $\delta p~=~2N\gamma mv$.  We assume the $\delta t$ is the time it takes each atom to travel between walls of the box $\delta t~=~x/v$.  In this way within that time interval nearly all the molecules impact the walls.  So there is a force&#xA;$$&#xD;&#xA;F~=~\frac{\delta F}{\delta t}~\simeq~2N\gamma mv^2/x&#xD;&#xA;$$&#xA;The pressure $P~=~F/A$ $=~2N\gamma mv^2/x^3$.  Now for the molecules highly relativistic $v~\simeq~c$ and this is an energy density.  So the pressure acts as a source of the gravity field --- it has mass-energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For cosmologies where galaxies are treated as particles the pressure is usually set to zero.  For the interior of a collapsing star the pressure becomes quite large and contributes to the source of the gravity field.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-23T19:15:59.653" CommentCount="2" />
  <row Id="3693" PostTypeId="2" ParentId="1018" CreationDate="2011-01-23T19:20:24.220" Score="0" ViewCount="" Body="&lt;p&gt;A particle excetuing simple harmonic motion should be, periodic its motion must be to n fro n the restoring force should always be directly propotional to displacement in th opposite direction i.e RF`directly propotional to negetive of displacemnt..&lt;/p&gt;&#xA;" LastActivityDate="2011-01-23T19:20:24.220" />
  <row Id="3694" PostTypeId="2" ParentId="3612" CreationDate="2011-01-23T19:38:43.360" Score="1" ViewCount="" Body="&lt;p&gt;A way of physically thinking about this is that a two body problem in general relativity does not generally have closed orbits.  If one of the bodies is very large and the other a small satellite the problem is integrable.  The periapsis (perihelion) advance of the small satellite is repeated with each orbit, which makes the problem integrable.  If the two bodies are of comparable mass the orbits of the two are perturbed in a manner which emulates a third orbiting body in Newtonian mechanics.  The three body problem is not integrable in general.  The periapsis advance of either mass adjusts to the changing position of the other mass, which “emulates” the presence of a third body.  Curiously, before Einstein people thought there was another planet near the sun which perturbed Mercury, what they called planet Vulcan.   If the two bodies are close enough and they are in an orbit with a quadrupole moment (Keplerian orbit, ellipses etc) there is the emission of gravity waves.  Gravity waves are mass-energy and contribute to the gravity field.  So a two body system in effect generates what might be thought of as a third body, or N-bodies.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-23T19:38:43.360" CommentCount="2" />
  <row Id="3695" PostTypeId="2" ParentId="3688" CreationDate="2011-01-23T19:47:55.767" Score="2" ViewCount="" Body="&lt;p&gt;Einstein's equations of general relativity say that&#xA;$$R_{\mu\nu}-\frac{1}{2}(R-2\Lambda) g_{\mu\nu} = \frac{8\pi G_N}{c^4}T_{\mu\nu}$$&#xA;Ignore the cosmological term proportional to $\Lambda$ for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left hand side is the Einstein tensor and the right hand side is proportional to the stress energy tensor. For low velocities, the dominant component of the equation is the $\mu\nu=00$ (time-time) component, and it effectively reduces to $\Delta \phi_{grav}=4\pi G\rho $, the Poisson equation for Newton's gravity, which implies all the inverse squared distance law, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in relativity, the energy is just 1 component of a 4-vector, the energy-momentum vector, and the density of anything is just one (the time-like) of 4 components of a vector which also includes the flux as the 3 spatial components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the mass or energy density $\rho$ becomes just the component $T_{00}$ of a whole symmetric tensor that has $4\times 3/2\times 1$ components (in four dimensions). Relativity implies that all of them are equally important because they can transform into each other by the Lorentz transformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the pressure appears as the doubly spatial components of the stress-energy tensor. Typically, $T_{xx}=T_{yy}=T_{zz}=p$, the pressure. For solids, this pressure is why the tensor contains the word &quot;stress&quot; - stress is a kind of pressure. For all materials, you may imagine that the pressure it the flux of the $p_x$ component of the momentum in the $x$-direction - that's why a gas or liquid will push a wall behind it. In relativity, all these components of the stress-energy tensor have to contribute to the corresponding components of the Einstein tensor (the curvature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I may return to the cosmological constant term. It is effectively the same thing as a stress energy tensor with $p=-\rho$, a negative pressure: you could put it on the right-hand side. Such a form of uniform matter density with a negative pressure deforms the Minkowski space into de Sitter space that is still &quot;maximally symmetric&quot;: dust with no pressure wouldn't be able to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am convinced that any valid - however qualitative - explanation why pressure curves the spacetime in general relativity has to boil down to Einstein's equations in one way or another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-23T19:47:55.767" />
  <row Id="3696" PostTypeId="2" ParentId="3687" CreationDate="2011-01-23T20:02:17.230" Score="3" ViewCount="" Body="&lt;p&gt;the system must be in a thermal equilibrium - a condition that allows one to assign a uniform temperature to the whole system. Strongly interacting systems reach it more quickly, weakly interacting systems may need a longer time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But once a system reaches the thermal equilibrium, it is guaranteed that each subsystem and/or each average degree of freedom of a certain kind carries the energy that is attributed to it by the equipartition theorem - e.g. $kT/2$ for a typical classical degree of freedom (translations or rotations). This is a statistical statement, so you need to look at many atoms or degrees of freedom of the same kind ($N$ of them) and take their average energy. When you do it right, the average energy will be given by the figure implied by the equipartition theorem plus minus an error - the relative error goes to zero as $1/\sqrt{N}$ for many atoms or degrees of freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, one must be careful what the actual prediction of the equipartition theorem for a &quot;degree of freedom&quot; is. The figure $kT/2$ only holds if the degree of freedom is de facto classical and more or less decoupled from others - at the given frequency. For very strongly coupled systems, the &quot;number of degrees of freedom&quot; is less clearly defined, and so is the energy predicted by the equipartition theorem for one of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-23T20:02:17.230" CommentCount="2" />
  <row Id="3697" PostTypeId="2" ParentId="2110" CreationDate="2011-01-23T20:05:12.230" Score="1" ViewCount="" Body="&lt;p&gt;This was written for a question that closed during my composition of this.  The question is how does the CC effect atomic physics, by Ashton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark energy has the mass-energy equivalent of a proton every 1-10 cubic meters.  That is a pretty diffuse energy.  An atom is on the scale of $10^{-8}$cm in length or has a volume of about $10^{-30}m^3$.  So about that proportion of a proton’s mass-energy worth of dark energy acts on an atom, or perturbs its atomic levels.  That is about $10^{-21}$ Gev or $10^{-12}$ev.  That is very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now your question is not entirely without merit.  Some very sensitive atomic measurements get atomic level splittings to within $10^{-6}$ev.  I will not say for certain, but these atomic-quantum optics people can be quite clever on the bench.  It is not entirely unimaginable that with squeezed states, entangled squeezed states of photons and electrons and so forth that this might be measured.  If there is an EM response due to a level splitting the wave would be around the sub Hertzian range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interaction Hamiltonian for the cosmological constant would be an inverted harmonic oscillator potential $H_{cc}~=~\Lambda r^2/3$.  Some analysis for avoided crossings of energy levels and states and the rest might not be an unreasonable thing to work on.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-23T20:05:12.230" />
  <row Id="3698" PostTypeId="1" CreationDate="2011-01-23T20:23:42.970" Score="4" ViewCount="404" Body="&lt;p&gt;The Schwarzschild metric reduces to the Minkowski metric in the limit of vanishing $M$, but the Hawking temperature which is proportional to $1/M$ diverges in the same limit. This would imply that flat spacetime has infinite rather than zero temperature. What am I missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This question is back up on the front page because I've posted an &lt;a href=&quot;http://physics.stackexchange.com/questions/3698/flat-space-limit-of-the-schwarzschild-metric-and-hawking-temperature-review/10510#10510&quot;&gt;answer of my own&lt;/a&gt; - let me know if I'm thinking about this right.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-05-28T16:37:20.740" LastActivityDate="2011-05-28T17:45:48.937" Title="Flat space limit of the Schwarzschild metric and Hawking temperature [REVIEW]" Tags="&lt;general-relativity&gt;&lt;black-holes&gt;" AnswerCount="8" FavoriteCount="1" />
  <row Id="3699" PostTypeId="2" ParentId="3698" CreationDate="2011-01-23T20:28:26.533" Score="4" ViewCount="" Body="&lt;p&gt;There's a topology change in the spacetime when you complete the $M\rightarrow 0$ limit--The Minkowski spacetime is non-singular, while the Schwarzschild spacetime contains a spacetime singularity for any (positive) value of $M$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also look at it in terms of the total energy available for a small mass BH is going to be miniscule, and you would expect the evaporation time to be rapid (especially since you're going to be triggering more decay modes at high temperature), with very hot evaporation remnants likely to scatter a large distance quickly, so its effect on the spacetime will be to produce a gas of high-temperature particles that quickly diffuse to an essentially zero temperature gas.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-01-23T20:28:26.533" CommentCount="11" />
  <row Id="3700" PostTypeId="2" ParentId="3698" CreationDate="2011-01-23T21:48:23.720" Score="3" ViewCount="" Body="&lt;p&gt;The lower the black hole mass, the higher the temperature of the Hawking radiation. This prediction is based on approximations that seem warranted as long as the black hole has mass $M &amp;gt;&amp;gt; M_\text{Planck}$. No-one knows what precise phenomena occur when the black hole mass approaches $M_\text{Planck}$. Drawing conclusions on the $M \to 0$ limit based on GR is certainly not justified.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastEditorUserId="1268" LastEditorDisplayName="" LastEditDate="2011-01-23T22:54:25.377" LastActivityDate="2011-01-23T22:54:25.377" CommentCount="1" />
  <row Id="3701" PostTypeId="2" ParentId="3634" CreationDate="2011-01-23T22:15:53.523" Score="4" ViewCount="" Body="&lt;p&gt;Consider two independent systems, two distant atoms, for example. Each atom has is own angular momentum $J_1$ and $J_2$. If these atoms are different and distant (non interacting), then the QM variables are separated, the total wave function becomes a product of the two atomic wave functions, and the total system angular momentum is a sum of the two. The system energy is determined with the sum of atomic energies and depends on $J_1$ and $J_2$ in a simple way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the atoms interact, one cannot introduce two independent atomic angular momenta since the variables &lt;strong&gt;are not separated&lt;/strong&gt;. In case of a weak interaction, one can use the approximation of non-interacting atoms as an initial approximation in the perturbation theory to calculate the interacting system properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of a strong interaction such an approximation may happen to be too poor, the perturbative corrections large, and the truncated perturbative solution far from the exact one. I think, in this case a quasi-particle approach is more helpful: a strongly coupled system may have low energy excitations like (nearly) independent quasi-particles with effective masses, charges, spins, etc. Then the approximate solution is built differently and there is a chance to guess it right.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-01-23T22:33:38.790" LastActivityDate="2011-01-23T22:33:38.790" />
  <row Id="3702" PostTypeId="2" ParentId="3698" CreationDate="2011-01-23T22:33:28.250" Score="1" ViewCount="" Body="&lt;p&gt;Jerry’s answer is pretty good, and I would say that topology, or $H^2({\cal M})$ is given by the evaluation of a curvature two-form $\Omega_{ab}~=~R_{abcd}\omega^c\wedge\omega^d$ on a two surface, and where $ R_{abcd}$ is the Riemann curvature tensor.  The result is a “charge,” which is a measure of the topology.  A very small black hole has a huge curvature near the horizon, which is different from the Minkowski spacetime of zero curvature.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-01-23T23:06:29.907" LastActivityDate="2011-01-23T23:06:29.907" />
  <row Id="3703" PostTypeId="1" AcceptedAnswerId="3704" CreationDate="2011-01-23T23:32:07.403" Score="9" ViewCount="428" Body="&lt;p&gt;Can a state be entangled without also being a superposition? (Please give an example)&lt;br&gt;&#xA;Can a state be a superposition without being entangled? (Again, an example please)&lt;br&gt;&#xA;And what about a cat state?&lt;br&gt;&#xA;I am royally confused, and a little bit of googling didn't help.&lt;br&gt;&#xA;This question is partly motivated by the recent news announcement (Nature, NYTimes)  that ten billion entangled pairs have been created.&lt;br&gt;&#xA;Are these spin pairs also in a superposition?&lt;br&gt;&#xA;Does this count as a cat state?&lt;/p&gt;&#xA;" OwnerUserId="1172" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-24T00:50:58.040" LastActivityDate="2011-01-24T14:26:34.493" Title="What's the difference between an entangled state, a superposed state and a cat state?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="3704" PostTypeId="2" ParentId="3703" CreationDate="2011-01-23T23:45:57.747" Score="13" ViewCount="" Body="&lt;p&gt;This goes on order from the more general to the more specific:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If A and B are two states, then you can form any linear combination to get superposed state, let's call it C. Of course, any state C can be expressed as superposition of two other states A and B, so calling a state &quot;superposed&quot; is not all that meaningful, unless there is something distinguishing the A,B states from all their superpositions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can only discuss entanglement if you first distinguish two (or more) sub-systems of the full system, for example if the distinguished sub-systems do not interact with each other, or if they are spatially separated. In this case states of system (1) can be entangled with states of system (2), if the total state cannot be written as a product $A=A_1 \times A_2$ where $A_1$ and $A_2$ are states of the separate subsystems. A more formal way to quantify that is the &quot;entanglement entropy&quot;, which tells you roughly how much information you are missing if you have access to one sub-system only.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The cat state is a particular example of an entangled state. Specifically in systems where two (or more) sub-systems consist of exactly one (qu)bit of information, which is labeled as 0 or 1, this is the state |00...0&gt; +|11...1&gt;. I am also told in comments there are generalizations beyond this system of n qubits, and more physical ways to characterize the state.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, if you decide that your product states as above are distinguished over any linear combination thereof, there is a meaningful comparison between what you call &quot;superposed&quot; state (which I have to say is not a standard  terminology) and the entangled states. An entangled state is one that cannot be expressed as product state, so it is necessarily a superposition of such states, say  $A_1 \times A_2$ and  $B_1 \times B_2$. The converse is not true: a superposition of product states may well be a product state itself.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-01-24T14:26:34.493" LastActivityDate="2011-01-24T14:26:34.493" CommentCount="6" />
  <row Id="3705" PostTypeId="1" CreationDate="2011-01-23T23:50:58.660" Score="5" ViewCount="326" Body="&lt;p&gt;Question: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If there are two conducting spherical shells and the inner shell is grounded, what will be &gt;the charge density in the inner shell if there is a charge Q placed on the outer shell?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, this is a HW problem, but I am not asking you guys to solve it for me... just show me the way :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a charge on the outer shell with radius $a$, the charge density will be $\frac{Q}{4\pi a^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That should induce, a charge density on the inner side of outer shell -- $\frac{Q}{4\pi (a-x)^2}$  where $x$ is thickness of the outer shell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the inner shell was never grounded, $\frac{Q}{4\pi b^2}$ ($b$ = radius of the inner shell) charge density would have been induced on the inner shell, now that it is grounded there would be no charge on it, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no need for a charge to be induced for the electric field to be 0 inside the inner shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to be my conclusion, but I fear it is too simple for the question. There might be something I am missing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: the hint of the problem says when a sphere is grounded, potential is infinity.&lt;/p&gt;&#xA;" OwnerUserId="1548" LastEditorUserId="1656" LastEditorDisplayName="" LastEditDate="2011-01-29T06:47:28.747" LastActivityDate="2011-01-29T06:47:28.747" Title="Charge density in concentric spheres" Tags="&lt;homework&gt;&lt;electrostatics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="3706" PostTypeId="1" CreationDate="2011-01-24T00:15:47.833" Score="5" ViewCount="360" Body="&lt;p&gt;I am not a professional physicist, so I may say something rubbish in here, but this question has always popped in my mind every time I read or hear anyone speak of particles hitting singularities and &quot;weird things happen&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to the question at hand, please follow my slow reasoning... As far as I've learned, to reach a black hole singularity, you must first cross an event horizon. The event horizon has this particular property of putting the external universe on an infinite speed to the falling observer. Now due to the Hawking radiation, and knowing that the cosmic background radiation is slowly dimming, sooner or later every black hole in this particular instance of inflation we are living in will evaporate, according to an external observer of said black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that every black hole has a finite timespan, as long as this universe survives that long. Now if we go back to the falling observer, we had already established that such an observer would see the outside universe &quot;speed up&quot; infinitely. This means that when the falling observer &quot;hits&quot; the event horizon, he will (or it if we speak about particles, which is clearer in this case), be immediately transported in time towards the final evaporation moment of the black hole. Either this or the particle gets some weird treatment. My point is, such a particle never gets to the singularity, because it has no time to get to it. The moment it crosses the event horizon, the black hole itself evaporates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where am I wrong here?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-24T14:40:06.030" Title="Does any particle ever reach any singularity inside the black hole?" Tags="&lt;black-holes&gt;" AnswerCount="5" CommentCount="5" />
  <row Id="3707" PostTypeId="2" ParentId="3703" CreationDate="2011-01-24T00:16:38.480" Score="0" ViewCount="" Body="&lt;p&gt;The difference between a superposition and entanglement is the following.  We consider a two slit experiment where a photon wave function interacts with a screen.  The wave vector is of the form&#xA;$$&#xD;&#xA;|\psi\rangle~=~e^{ikx}|1\rangle~+~e^{ik’x}|2\rangle&#xD;&#xA;$$&#xA;as a superposition of states for the slits labeled $1$ and $2$.  The normalization is assumed.  The state vector is normalized as&#xA;$$&#xD;&#xA;\langle\psi|\psi\rangle~=~1~=~\langle 1|1\rangle~+~\langle 2|2\rangle~+~ e^{i(k’~+~k)x}\langle 1|2\rangle~+~ e^{-i(k’~+~k)x}\langle 2|1\rangle&#xD;&#xA;$$&#xA;The overlaps $\langle 1|2\rangle$ and $\langle 2|1\rangle$ are multiplied by the oscillatory terms which are the interference probabilities one measures on the photoplate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now consider the classic situation where one tries to measure which slit the photon traverses.  We have a device with detects the photon at one of the slit openings.  We consider another superposed quantum state.  This is a spin space that is &#xA;$$&#xD;&#xA;|\phi\rangle~=~\frac{1}{\sqrt {2}}(|+\rangle~+~|-\rangle).&#xD;&#xA;$$&#xA;This photon quantum state becomes entangled with this spin state.  So we have&#xA;$$&#xD;&#xA;|\psi,\phi\rangle~=~e^{ikx}|1\rangle|+\rangle~+~e^{ik’x}|2\rangle|-\rangle&#xD;&#xA;$$&#xA;which means if the photon passes through slit number 1 the spin is + and if it passes through slit 2 the spin is in the – state.  Now consider the norm of this state vector&#xA;$$&#xD;&#xA;\langle|\psi,\phi|\psi,\phi\rangle~=~\langle 1|1\rangle\langle +|+\rangle~+~\langle 2|2\rangle\langle-|-\rangle~+~ e^{i(k’~+~k)x}\langle 1|2\rangle\langle +|-\rangle~+~ e^{-i(k’~+~k)x}\langle 2|1\rangle\langle-|+\rangle.&#xD;&#xA;$$&#xA;The spin states $|+\rangle$ and $|-\rangle$ are orthogonal and thus $\langle +|-\rangle $ and $\langle-|+\rangle$ are zero.  This means the overlap or interference terms are removed.  In effect the superposition has been replaced by an entanglement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This analysis does not tell us which state is actually measured, but it does tell us how the interference term is lost due to the entanglement of the system we measure with an instrument quantum state.  So one does not need to invoke an outright collapse to illustrate how a superposition is lost.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cat state is a form of the GHZ state where you have&#xA;$$&#xD;&#xA;|\psi\rangle~=~C(|0\rangle^{\otimes n}~+~|1\rangle^{\otimes n})&#xD;&#xA;$$&#xA;which are n-partite entanglements.  These types of states can violate Bell inequalities with one state, which illustrates quantum mechanics is not a purely statistical theory.  The statistics are derived from QM.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-24T00:16:38.480" />
  <row Id="3708" PostTypeId="2" ParentId="3703" CreationDate="2011-01-24T00:38:08.133" Score="6" ViewCount="" Body="&lt;p&gt;Any discussion of entanglement implies you are thinking of (at least) two separate systems. Superpositions on the other hand, can also apply to situations where you are only looking at a single system. Keeping this in mind:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Can a state be entangled without also&#xA;  being a superposition? (Please give an&#xA;  example)&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. An entangled state is by definition a state which cannot be written as a separable product in any basis. Whatever basis you choose, the state will be a superposition of products of the separate systems.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Can a state be a superposition&#xA;  without being entangled? (Again, an&#xA;  example please)&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course. A trivial example would be a single isolated system for which you can usually find some basis in which the state is in a superposition. And single isolated systems cannot be entangled with anything else because.. well.. they are single isolated systems - and like I said before, entanglement applies to a situation where you have divided your world into separate systems that can interact with each other.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;And what about a cat state?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A Schrodinger Cat state is usually used to refer to a &lt;em&gt;macroscopic&lt;/em&gt; system which is in a superposition of two or more states (in the eigenstate basis of some physical observable like energy or position). Real cats in geiger-counter-poison-rigged boxes probably won't be in such superpositions due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_decoherence&quot; rel=&quot;nofollow&quot;&gt;decoherence&lt;/a&gt; but there are some surprisingly large objects like buckyballs and macromolecules which have been shown to exist in superpositions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Are these spin pairs also in a&#xA;  superposition?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, in all bases, because they are entangled. Allegedly :)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Does this count as a cat state?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure because I haven't read the whole paper and also because the definition of a cat state depends on what you consider macroscopic.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-01-24T00:38:08.133" CommentCount="2" />
  <row Id="3709" PostTypeId="2" ParentId="3706" CreationDate="2011-01-24T01:16:44.557" Score="5" ViewCount="" Body="&lt;p&gt;Nothing is unusual to an observer falling into a black hole at the event horizon. He does not &quot;hit&quot; it. It is crossed with no fuss or bother.  As he falls farther and farther into the black hole, though, tidal gravitational forces &quot;spaghettifi&quot; him. I do not know what you mean by saying that the external universe is &quot;speeded up infinitely.&quot; The moment the particle crosses the event horizon, the black hole does not evaporate. The event horizon is not some sort of solid physical barrier.  The particle will approach the singularity, but GR breaks down at/near singularities.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-01-24T01:16:44.557" CommentCount="2" />
  <row Id="3710" PostTypeId="1" CreationDate="2011-01-24T01:24:13.663" Score="0" ViewCount="230" Body="&lt;p&gt;In math, fields are algebraic structure where basic arithmetic operations can be defined and certain axioms are satisfied.  Fields in physics assign values to each point in spacetime.  Fields in physics are also represented by continuous sets of complex numbers.  There is obviously some distinction yet some similarity between fields in math and fields in physics.  What are the similarities and distinctions between the two?&lt;/p&gt;&#xA;" OwnerUserId="1383" LastActivityDate="2011-01-24T01:24:13.663" ClosedDate="2011-01-24T01:29:14.533" Title="Field in physics vs. field in math" Tags="&lt;mathematical-physics&gt;" CommentCount="12" />
  <row Id="3711" PostTypeId="2" ParentId="2916" CreationDate="2011-01-24T01:35:38.940" Score="1" ViewCount="" Body="&lt;p&gt;From the NASA site: &quot;Recent measurements (c. 2001) by a number of ground-based and balloon-based experiments, including MAT/TOCO, Boomerang, Maxima, and DASI, have shown that the brightest spots are about 1 degree across. Thus the universe was known to be flat to within about 15% accuracy prior to the WMAP results. WMAP has confirmed this result with very high accuracy and precision. We now know that the universe is flat with only a 2% margin of error.&quot;&#xA;The WMAP results are given here :http://lambda.gsfc.nasa.gov/product/map/current/&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-01-24T01:35:38.940" CommentCount="1" />
  <row Id="3712" PostTypeId="1" AcceptedAnswerId="3717" CreationDate="2011-01-24T01:41:34.527" Score="0" ViewCount="113" Body="&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/hWGlJ.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem is giving me a lot of problems. So $E=k*q/d^2$. We'd want to find the distance from q1 to P, which is .1 meters (not cm) using pythagorean thereom. So we know k, which is just $9x10^9$ times q1 which is $-2.4u$ where $u=10^{-6}$ divided by $r^2$ which is just $.1^2$. Then I get $-216,000,000$. To get x component, I take that number multiplied by $cos45$ to come up with a final answer of $-108,000,000$. And for point q2, I dont think there is an x component for the electric field since its right below the point P. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What have I done wrong? &lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-24T04:18:04.013" LastActivityDate="2011-01-24T04:18:04.013" Title="Calculating electric field" Tags="&lt;electricity&gt;&lt;homework&gt;" AnswerCount="4" CommentCount="11" />
  <row Id="3713" PostTypeId="2" ParentId="3712" CreationDate="2011-01-24T01:49:30.747" Score="0" ViewCount="" Body="&lt;p&gt;First off, I don't want to answer the question for you, but your distance between q1 and point P seems to be incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should really be working with units as well - this will help you catch any mistakes you may be making.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your general reasoning seems to be on track though.&lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2011-01-24T01:49:30.747" CommentCount="2" />
  <row Id="3714" PostTypeId="1" CreationDate="2011-01-24T01:50:55.620" Score="5" ViewCount="443" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Method_of_image_charges&quot; rel=&quot;nofollow&quot;&gt;method of image charges&lt;/a&gt; is a well-known and very useful tool for solving problems in electrostatics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, when I was taught this method, it was presented simply as an algorithm. No real physical justification was given for its usage, and there was a complete absence of rigorous mathematics. The method has always been a little hazy to me since.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone could give a rigorous first-principles derivation of this method from Maxwell's equations (perhaps simply Coulomb's law?), it would be most appreciated.&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T07:52:21.397" LastActivityDate="2011-02-03T07:52:21.397" Title="Method of Images" Tags="&lt;electrostatics&gt;&lt;conformal-field-theory&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" />
  <row Id="3715" PostTypeId="2" ParentId="3714" CreationDate="2011-01-24T02:07:04.603" Score="2" ViewCount="" Body="&lt;p&gt;It takes advantage of the fact the solution is &lt;a href=&quot;http://en.wikipedia.org/wiki/Uniqueness_theorem_for_Poisson%27s_equation&quot; rel=&quot;nofollow&quot;&gt;unique&lt;/a&gt; and fixed by the boundary conditions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Any solutions is &lt;em&gt;the&lt;/em&gt; solution.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Any solution that gets the boundary conditions right is &lt;em&gt;a&lt;/em&gt; solution.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-01-24T02:07:04.603" CommentCount="11" />
  <row Id="3716" PostTypeId="2" ParentId="3712" CreationDate="2011-01-24T02:22:08.313" Score="0" ViewCount="" Body="&lt;p&gt;Note that the fields are vector quantities (that is they have direction as well as magnitude). The field from q1 points down and left, while the field from q2 points straight up.  (If you don't know why you should review your earlier studies until you do...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has implications for how they add. (This is another thing that you should already have studied.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By noting that the question asks only for the x-component of the resulting field you may be able to simplify your work. (Look again at the directions of the two fields...)&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-01-24T02:22:08.313" CommunityOwnedDate="2011-01-24T02:22:47.470" />
  <row Id="3717" PostTypeId="2" ParentId="3712" CreationDate="2011-01-24T02:29:39.533" Score="2" ViewCount="" Body="&lt;p&gt;The electric field is the vector sum&#xA;$$&#xD;&#xA;{\vec E}~=~\frac{1}{4\pi\epsilon_0}\Big(\frac{q_1{\bf n}_1}{2d^2}~+~\frac{q_2{\bf n}_2}{d^2}\Big)&#xD;&#xA;$$&#xA;so the components of the field are&#xA;$$&#xD;&#xA;E_x~=~\frac{1}{4\pi\epsilon_0}\frac{q_1}{2\sqrt{2}d^2}&#xD;&#xA;$$&#xA;$$&#xD;&#xA;E_y~=~\frac{1}{4\pi\epsilon_0}\frac{q_1}{d^2}\frac{1~+~2\sqrt{2}}{2\sqrt{2}}&#xD;&#xA;$$&#xA;The rest is plug and grind on numbers.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-01-24T02:35:43.540" LastActivityDate="2011-01-24T02:35:43.540" CommentCount="2" />
  <row Id="3718" PostTypeId="1" AcceptedAnswerId="3719" CreationDate="2011-01-24T02:38:54.050" Score="1" ViewCount="382" Body="&lt;p&gt;Below is attached for reference, but the question is simply about whether vectors used in physics in a vector space can be represented by complex numbers and whether they can be divided.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In abstract algebra, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Field_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;field&lt;/a&gt; is an algebraic structure with notions of addition, subtraction, multiplication, and division, satisfying certain axioms. The most commonly used fields are the field of real numbers, the field of complex numbers, and the field of rational numbers, but there are also finite fields, fields of functions, various algebraic number fields, p-adic fields, and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Any field may be used as the scalars for a vector space&lt;/strong&gt;, which is the standard general context for linear algebra. The theory of field extensions (including Galois theory) involves the roots of polynomials with coefficients in a field; among other results, this theory leads to impossibility proofs for the classical problems of angle trisection and squaring the circle with a compass and straightedge, as well as a proof of the Abel–Ruffini theorem on the algebraic insolubility of quintic equations. In modern mathematics, the theory of fields (or field theory) plays an essential role in number theory and algebraic geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In mathematics and physics, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Scalar_field&quot; rel=&quot;nofollow&quot;&gt;scalar field&lt;/a&gt; associates a scalar value to every point in a space. The scalar may either be a mathematical number, or a physical quantity. Scalar fields are required to be coordinate-independent, meaning that any two observers using the same units will agree on the value of the scalar field at the same point in space (or spacetime). Examples used in physics include the temperature distribution throughout space, the pressure distribution in a fluid, and spin-zero quantum fields, such as the Higgs field. These fields are the subject of scalar field theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In mathematics, an &lt;a href=&quot;http://en.wikipedia.org/wiki/Algebra_over_a_field&quot; rel=&quot;nofollow&quot;&gt;algebra over a field&lt;/a&gt; is a vector space equipped with a bilinear vector product. That is to say, it is an algebraic structure consisting of a vector space together with an operation, usually called multiplication, that combines any two vectors to form a third vector; to qualify as an algebra, this multiplication must satisfy certain compatibility axioms with the given vector space structure, such as distributivity. In other words, an algebra over a field is a set together with operations of multiplication, addition, and scalar multiplication by elements of the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_space&quot; rel=&quot;nofollow&quot;&gt;vector space&lt;/a&gt; is a mathematical structure formed by a collection of vectors: objects that may be added together and multiplied (&quot;scaled&quot;) by numbers, called scalars in this context. Scalars are often taken to be real numbers, but one may also consider vector spaces with scalar multiplication by complex numbers, rational numbers, or even more general fields instead. The operations of vector addition and scalar multiplication have to satisfy certain requirements, called axioms,... An example of a vector space is that of Euclidean vectors which are often used to represent physical quantities such as forces: any two forces (of the same type) can be added to yield a third, and the multiplication of a force vector by a real factor is another force vector. In the same vein, but in more geometric parlance, vectors representing displacements in the plane or in three-dimensional space also form vector spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In classical mechanics as in physics, the field is not real, but merely a model describing the effects of gravity. The field can be determined using Newton's law of universal gravitation. Determined in this way, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_field#In_classical_mechanics&quot; rel=&quot;nofollow&quot;&gt;gravitational field around a single particle is a vector field&lt;/a&gt; consisting at every point of a vector pointing directly towards the particle. The magnitude of the field at every point is calculated applying the universal law, and represents the force per unit mass on any object at that point in space. The field around multiple particles is merely the vector sum of the fields around each individual particle. An object in such a field will experience a force that equals the vector sum of the forces it would feel in these individual fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the force field is conservative, there is a scalar potential energy per unit mass at each point in space associated with the force fields, this is called gravitational potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gauss' law for gravity is mathematically equivalent to Newton's law of universal gravitation, but is stated directly as vector calculus properties of the gravitational field.&lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-16T18:48:08.343" LastActivityDate="2011-07-16T18:48:08.343" ClosedDate="2011-07-16T18:48:14.083" Title="Can vectors in physics be represented by complex numbers and can they be divided?" Tags="&lt;mathematics&gt;&lt;vectors&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="3719" PostTypeId="2" ParentId="3718" CreationDate="2011-01-24T02:59:03.457" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Can vectors in physics be represented by complex numbers?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Absolutely. There exists a direct isomorphism between the 2D Euclidean vector space and the Argand plane, for a start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, it is possible to talk of mathematical objects called &lt;a href=&quot;http://en.wikipedia.org/wiki/Quaternion&quot; rel=&quot;nofollow&quot;&gt;quaternions&lt;/a&gt; and use quaternion algebra analogously to vector algebra. Historically quaternions were used to represent geometrical operations and transformations in 3D space - in the days before vector algebra. (They still are used, especially in areas such as computer graphics where they offer one or two advantages over the simpler world of vectors.) In any case, the relationship to vector algebra is a very close one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can vectors in physics be divided?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, no, vector-vector division is not a well-defined operation. At least, not within the bounds of linear algebra. i.e. There exist none or multiple solutions to the equation $\vec{y} = \mathbb{A} \vec{x}$. (See the &lt;a href=&quot;http://mathworld.wolfram.com/VectorDivision.html&quot; rel=&quot;nofollow&quot;&gt;Wolfram page&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saying this, the concept of vector division has an interesting relationship with your first question. If we map vectors to complex numbers (or quaternions in &gt; two dimensions), we can use complex division or quaternion algebra respectively to define an analogous &quot;vector division&quot; operation. Note that quaternions can in fact be extended to higher dimensions, which allows for interesting possibilities. This study of this falls under the area of &lt;em&gt;Clifford algebras&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: vector-scalar division is of course well-defined, as is point-wise division of equal-length vectors, but I presume you are not referring to that.)&lt;/p&gt;&#xA;" OwnerUserId="13" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2011-01-24T17:33:19.237" LastActivityDate="2011-01-24T17:33:19.237" CommentCount="10" />
  <row Id="3720" PostTypeId="2" ParentId="3706" CreationDate="2011-01-24T03:00:40.427" Score="2" ViewCount="" Body="&lt;p&gt;For some videos simulating what you would see falling into a black hole look at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://jila.colorado.edu/~ajsh/insidebh/&quot; rel=&quot;nofollow&quot;&gt;http://jila.colorado.edu/~ajsh/insidebh/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outside universe does not speed up for an inertial observer falling into the black hole.  If an observer hovers over the event horizon then the exterior world does appear to speed up.  The observer who falls into a black hole will within a finite time period reach the singularity.  However, tidal forces grow enormously and the observer is pulled apart before actually reaching it.  In fact atoms and nuclei will be pulled apart before hand.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-24T03:00:40.427" />
  <row Id="3721" PostTypeId="2" ParentId="3698" CreationDate="2011-01-24T03:10:56.427" Score="1" ViewCount="" Body="&lt;p&gt;This problem is analogous to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ultraviolet_catastrophe&quot; rel=&quot;nofollow&quot;&gt;&quot;ultraviolet catastrophe&quot;&lt;/a&gt; in the pre-quantum understanding of black-body radiation. There, because one did not take into account the discrete character of radiation, the total energy emitted by a black-body was divergent. Planck came along and remedied the situation by noting that photons were emitted in discrete packets rather than the previous assumption that all frequencies occurred in the radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, as @Johannes noted in his answer, the calculations which lead to the prediction that a black hole radiates at a temperature $T \propto 1/M$ are valid only for macroscopic black holes. As and when, we understand how the discrete character of spacetime modifies this dependence, will we be able to answer this question. The simplest thing to do is to guess that such corrections will modify the temperature-mass dependence to be of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T \propto \frac{1}{M + M_O} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $M_O$ is the minimal size a black-hole can attain, leading to a finite maximum temperature that a system containing a decaying black hole can attain $T_{max} \sim 1/M_O$. The conjecture would be that a black-hole decays until we are left with a flat background populated by a gas of these black-hole &lt;em&gt;quanta&lt;/em&gt; of mass $M_O$ and temperature $T_{max}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is all very hand-wavy.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; A generalization of the above approach is to postulate that the exact form of the dependence of $T_{max}$ on $M$ is given by some function $f(M/M_O)$ which tends to $ 1/M $ in the limit $M/M_O \rightarrow \infty$.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-24T12:39:29.970" />
  <row Id="3722" PostTypeId="2" ParentId="3712" CreationDate="2011-01-24T03:49:08.377" Score="0" ViewCount="" Body="&lt;p&gt;Rather than answer the question, I assume you just started a physics course (my kids are in their first week this semester). This course should have had vector algebra, and probably other math as a prerequisite. If you don't have the prerequisites, then you have two choices, (1) drop the course and take it later after satisfying the prereqs, or (2) try to tough it out, which means you will have to rapidly pick up the missing prereqs. If you are a really good student, and the gaps aren't too great, I'd say go for it, otherwise option (1) would be your better choice. Note, option two may mean you have doubled or trippled the amount of work needed to pas the course, and you'll need to be an auto-didact (self learner). You can probably find free course notes on opencourseware, download the appropriate ones and see if you think you are up to the task.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-01-24T03:49:08.377" CommentCount="1" />
  <row Id="3723" PostTypeId="1" CreationDate="2011-01-24T04:06:39.447" Score="8" ViewCount="274" Body="&lt;p&gt;What substance will have the largest specific heat capacity integrated from T=0 to, say, room temperature?  In other words, given a finite amount of mass, what object or collection of objects has the largest number of degrees of freedom that can be excited as it absorbs energy starting from T=0?  Would it be a complicated molecular polymer that can be tangled in all sorts of ways, or some kind of gas of low-mass particles, or maybe a spin lattice of some sort?  Is there some kind of fundamental limit in the universe of the number of quantum degrees of freedom per mass or perhaps per volume that is allowed? &lt;/p&gt;&#xA;" LastActivityDate="2011-01-24T12:14:12.897" Title="Largest theoretically possible specific heat capacity?" Tags="&lt;statistical-mechanics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" />
  <row Id="3724" PostTypeId="2" ParentId="2558" CreationDate="2011-01-24T04:09:16.093" Score="0" ViewCount="" Body="&lt;p&gt;Hawking himself is now postulating that the universe may end in a &quot;big chill.&quot; Has anyone extrapolated how black holes will end under such a scenario?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-24T04:09:16.093" CommentCount="1" />
  <row Id="3725" PostTypeId="2" ParentId="3714" CreationDate="2011-01-24T04:35:42.377" Score="3" ViewCount="" Body="&lt;p&gt;In electrostatics, we want to solve &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson%27s_equation&quot; rel=&quot;nofollow&quot;&gt;Poisson's equation&lt;/a&gt; for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electric_potential&quot; rel=&quot;nofollow&quot;&gt;electric potential&lt;/a&gt; $\Phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla^2 \Phi  = 4 \pi \rho$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take, for example, free space, so $\rho = 0$.  The resulting equation is &lt;a href=&quot;http://en.wikipedia.org/wiki/Laplace%27s_equation&quot; rel=&quot;nofollow&quot;&gt;Laplace's equation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla^2 \Phi = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many solutions, for example $\Phi = 0$ or $\Phi = \Phi_0 e^{k(ix+y)}$ or others.  Further, if $\Phi_1$ and $\Phi_2$ are both solutions, then  $\alpha \Phi_1 + \beta \Phi_2$ is also a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Poisson's equation has the same difficulty.  If you have any solution to Poisson's equation, you can add any solution to Laplace's equation and get a new solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone said, &quot;I just threw a ball; predict its location as a function of time,&quot; we would be in a similar predicament.  We could determine a differential equation that describes the ball's motion, but there are many solutions, and we couldn't find the trajectory without knowing where the ball was thrown from and what its initial velocity was.  These are boundary conditions.  They take a differential equation with many solutions and narrow them down to one physically correct solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of Poisson's equation, one sufficient set of boundary conditions is the potential everywhere on the boundary of the area we're studying.  The &quot;boundary&quot; can be an infinity, if we want.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if there is a grounded conductor, we set $\Phi = 0$ everywhere on that conductor, then set $\Phi = 0$ at infinity.  If we can find any solution to Poisson's equation that also satisfies these two boundary conditions, it will be the only solution, and we will know the potential everywhere outside the conductor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove this uniqueness theorem, imagine there are two solutions, $\Phi_1$ and $\Phi_2$, to Poisson's equation.  Then $\Phi_3 = \Phi_1 - \Phi_2$ is a solution to Laplace's equation, and it has $\Phi_3 = 0$ on all the boundaries.  &lt;a href=&quot;http://mathworld.wolfram.com/LaplacesEquation.html&quot; rel=&quot;nofollow&quot;&gt;Any solution to Laplace's equation has no local minima or maxima&lt;/a&gt;, so the extrema must occur at the boudaries.  Since all boundaries are zero,  $\Phi_3 = 0$  and $\Phi_1 = \Phi_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the simplest application of the method of images.  There is an infinite grounded conducting sheet and a point charge $q$ sitting above it at distance $d$.  We need to find any solution to Poisson's equation such that $\Phi = 0$ at infinity and along the sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Discard the sheet momentarily, and consider another point charge $-q$ a distance $2d$ from the first one, so that it looks like its mirror image, with the plane where the conductor was serving as mirror (the original point charge is above the plane; the new one is below the plane an equal distance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a test charge sitting on the plane between the two mirror charges.  Because the mirror charges have opposite sign, the net force on the test charge has no component in the plane.  That means the plane is an equipotential surface, and the point charge can move out to infinity for free, where the potential is clearly zero.  Hence, the potential is zero at infinity and all along the plane.  The $\Phi$ you get in this scenario is easy to calculate because it's just the potential from two point charges, however, because it meets the boundary conditions, it is also the potential that arises from a point charge sitting above a conductor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method of images is not very general.  It will work whenever you can think of a clever way to set up an image charge so that the potential with the original charge and the image charge satisfies the same boundary conditions as your original problem.  You won't always be able to find a way to do that.  In a general case, you'll need to use Green's functions or numerical solution methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr /&gt;&#xA;&lt;b&gt;Reference:&lt;/b&gt; This is mostly from Chapter 3 of David Griffiths' &lt;i&gt;Introduction to Electrodynamics&lt;/i&gt;&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-01-24T04:35:42.377" CommentCount="1" />
  <row Id="3726" PostTypeId="2" ParentId="3723" CreationDate="2011-01-24T04:40:15.147" Score="4" ViewCount="" Body="&lt;p&gt;(1) If you allow a very long time for the substance to warm up, go with a kilogram of $\nu_1$, the lightest neutrino.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) If you require that the substance heat up using electromagnetism (i.e. photons), then go with a kilogram of electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) If you also require that the substance be electrically neutral and not decay, then a kilogram of hydrogen will fit the bill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) If you further require that the substance be chemically stable, then molecular hydrogen is almost as good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason these are the answers is because they have the lightest weights. So they have more degrees of freedom per kilogram.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-24T04:40:15.147" CommentCount="8" />
  <row Id="3727" PostTypeId="2" ParentId="3718" CreationDate="2011-01-24T05:13:41.937" Score="4" ViewCount="" Body="&lt;p&gt;In a certain sense, 3-d vectors can be divided in physics. To define the division, you begin by defining multiplication using the cross product. I.e. if $\vec{u} \times \vec{v} = \vec{w}$ then $\vec{w}/\vec{v} = \vec{u}$. This is a definition of division that is non commutative. It is also not well defined (as written) because you can always add multiples of $\vec{v}$ to $\vec{u}$ and still get a product of $\vec{w}$. That is, $\vec{u} \times \vec{v} = \vec{w}$ implies that $(\vec{u}+k\vec{v})\times\vec{v} = \vec{w}$. To make the division well defined you can require that $\vec{u}\cdot\vec{v}=0$ by subtracting off any component of $\vec{u}$ in the $\vec{v}$ direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before you have division, you might want to have inverses. That is, for a vector $\vec{u}$, we need to define a vector $\vec{v}$ such that $\vec{u}\vec{v} = 1$. In the usual approach to physics this is nonsense because &quot;1&quot; is not a vector. But it is possible to make sense out of this and a group of physicists have done just that. To get a &quot;1&quot; from a vector product you use the dot product. Then the inverse of a real vector is just the same vector, divided by its squared magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In actual practice, it's probably clear that you need both the dot product and the cross product to fully define vector multiplication. The discussion I've given here is intended only to whet the interest of the reader, an adequate understanding will require a lot of work. For example, in the first paragraph, w is always perpendicular to both u and v; how do you define division when w and u aren't perpendicular?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not very many physicists are interested in this sort of thing. It's studied by groups led by David Hestenes and makes up a part of what is called the &quot;Geometric Algebra&quot;. As you can see from the above, it requires that vectors be generalized. For example, since addition works with vectors, one must allow constructions such as the sum of a vector and a scalar such as $\vec{u} + 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you would expect, the geometric algebra also implies a geometric calculus (which includes all the usual vector calculus relations you're familiar with such as Stoke's theorem). To learn more about it will likely require a great deal of time. To start you can click around on his website:&lt;br&gt;&#xA;&lt;a href=&quot;http://geocalc.clas.asu.edu/html/Intro.html&quot; rel=&quot;nofollow&quot;&gt;http://geocalc.clas.asu.edu/html/Intro.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The geometric algebra is used in standard physics under a different name; the &quot;Dirac algebra&quot; or &quot;gamma matrices&quot; (so important to our modeling of the electron) are the geometric algebra in 3+1 dimensions. In this algebra, the usual spacetime or 4-vectors are generalized to include scalars, tensors, psuedo-vectors, and a pseudo-scalar.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-24T05:13:41.937" CommentCount="9" />
  <row Id="3728" PostTypeId="2" ParentId="3706" CreationDate="2011-01-24T05:24:36.183" Score="2" ViewCount="" Body="&lt;p&gt;What happens inside the event horizon of a black hole is at best a speculative question. Einstein's general relativity is the accepted gravitation theory. According to it, we cannot obtain information from an object dropped through the event horizon. So any answers you get will be sorely limited by the fact that no such experiment has ever been performed nor are we ever likely to perform one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current astronomical observations of the center of the galaxy suggest that Einstein's GR is working well fairly close to the event horizon. But GR blows up at the singularity so its predictions there are suspect at best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears that the only way of obtaining information about that singularity is to jump into a black hole. You will not be able to get information back to your friends, but you might find out yourself. On the other hand, tidal effects might kill you before you get close enough.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-24T05:24:36.183" CommentCount="1" />
  <row Id="3729" PostTypeId="2" ParentId="3705" CreationDate="2011-01-24T06:29:58.500" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Now, if the inner shell was never grounded, $\frac{Q}{4\pi b^2}$ ($b$ = radius of the inner shell) charge density would have been induced on the inner shell&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the inner shell were not grounded, it wouldn't be connected to anything. And if it wasn't connected to anything, how would it be able to have any charge density other than zero? Where would the extra charge come from?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;now that it is grounded there would be no charge on it, right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, as pointed out in the comments, the &lt;em&gt;potential&lt;/em&gt; is zero, but that doesn't mean the &lt;em&gt;charge&lt;/em&gt; is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you figure out the potential of the outer sphere? Once you know the potentials of both spheres, what else can you calculate using that information?&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-01-24T06:29:58.500" />
  <row Id="3730" PostTypeId="2" ParentId="3662" CreationDate="2011-01-24T07:13:05.847" Score="-1" ViewCount="" Body="&lt;p&gt;There are great answers above. One of the features of STR is the Klein-Gordon equation. I'd like to point out that the Klein-Gordon equation models the vibrations in an elastic solid. This is well known among engineers dealing with the subject. An elegant derivation is:&lt;br&gt;&#xA;R A Close, &lt;em&gt;Advances in Applied Cliord Algebras&lt;/em&gt;, &quot;Exact Description of Rotational Waves in an Elastic Solid&quot;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.springerlink.com/content/1873305m88k42145/&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/1873305m88k42145/&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/0908.3232&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0908.3232&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that despite the discrete substructure that might contribute to an elastic solid, the final result does satisfy the Klein Gordon equation quite well.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-24T07:13:05.847" CommentCount="1" />
  <row Id="3731" PostTypeId="1" AcceptedAnswerId="3733" CreationDate="2011-01-24T07:24:34.990" Score="7" ViewCount="180" Body="&lt;p&gt;The heterotic string is a combination of right-moving excitations from a D=10 superstring and left-moving excitations from a D=26 bosonic string, with the left-movers behaving as if the extra 16 dimensions are compactified. The heterotic string is also derived from D=11 M-theory, as an open 2-brane stretched between two &quot;end-of-the-world&quot; 9-branes (spatial boundaries; this is M-theory compactified on a line segment, and the 9-branes lie at the ends). So I am led to imagine a 27-dimensional theory, containing branes. We compactify 16 dimensions, and consider the worldvolume theory of two parallel 9-branes. When they are coincident, we get the heterotic string; when they are slightly separate, we get &quot;heterotic M-theory&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 27-dimensional fundamental theory has been discussed before (hep-th/9704158 section 4; hep-th/0012037; arXiV:0807.4899), but I don't see this particular line of thought discussed. &lt;/p&gt;&#xA;" OwnerUserId="1486" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-13T19:54:39.173" LastActivityDate="2011-02-13T19:54:39.173" Title="Heterotic string as worldvolume theory of two coincident 9-branes in 27 dimensions?" Tags="&lt;string-theory&gt;&lt;kaluza-klein&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="3732" PostTypeId="2" ParentId="3706" CreationDate="2011-01-24T07:44:00.257" Score="0" ViewCount="" Body="&lt;p&gt;This puzzle is an aspect of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_hole_information_paradox&quot; rel=&quot;nofollow&quot;&gt;black hole information loss paradox&lt;/a&gt; and a proposed solution is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Holographic_principle&quot; rel=&quot;nofollow&quot;&gt;holographic principle&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_hole_complementarity&quot; rel=&quot;nofollow&quot;&gt;black hole complementarity&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classical view of black holes is that any object which falls in ends its worldline on the singularity but an outside observer never sees this because the object appears to be frozen on the horizon. The paradox arises when this is considered in the light of quantum mechanics which tells us that the black hole can evapourate due to Hawking radiation. This means that the information about the object must be returned as part of the radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem can only be resolved with a theory of quantum gravity and although our theories of quantum gravity are incomplete some theorists have worked out some principles that govern how the solution might work. One part of the solution is the holographic principle that requires that the information is stored on the event horizon. The second part is black hole complementarity which says that the fate of the object is observer dependent. To an observer outside, the object stops at the horizon and gradually returns its energy and information to the surroundings as Hawking radiation. To an observer who falls into the black hole with the object its fate is very different. the object continues to pass the horizon and is destroyed when it hits the singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since two such observers can never meet up and compared notes there is no physical contradiction between these complementary views.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this is a speculative solution since it is far beyond anything we can currently test experimentally&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-24T07:44:00.257" CommentCount="1" />
  <row Id="3733" PostTypeId="2" ParentId="3731" CreationDate="2011-01-24T07:53:30.787" Score="4" ViewCount="" Body="&lt;p&gt;Dear Mitchell, this is a very nice research project - at least judging by the fact that I have made a similar proposal. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this very form, however, it can't be right because any hypothetical 27-dimensional theory fails to be supersymmetric and the supersymmetry breaking can't be quite undone. However, brave souls have played with the transmutation of string theories that are very different on the world sheet, see e.g. some of the papers by Simeon Hellerman and Ian Swanson:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.slac.stanford.edu/spires/find/hep/www?rawcmd=find+a+swanson+and+a+hellerman&amp;amp;FORMAT=WWW&amp;amp;SEQUENCE=&quot; rel=&quot;nofollow&quot;&gt;http://www.slac.stanford.edu/spires/find/hep/www?rawcmd=find+a+swanson+and+a+hellerman&amp;amp;FORMAT=WWW&amp;amp;SEQUENCE=&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fact, my specific version of the proposal had one mathematical piece of evidence that was much more specific than yours. You could imagine some $E_8$ group already in 27 dimensions. And a funny feature of the $E_8$ are its holonomy groups. The nontrivial one is the $\pi_3$ which is $Z$, and then the next nontrivial one is $\pi_{15}$. In normal M-theory, with the 3-form described following Diaconescu-Moore-Witten as the Chern-Simons form of an $E_8$ gauge field, $\pi_{3}$ is what allows fivebranes (codimension 5) to exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, $\pi_{15}$ of $E_8$ may create codimension 17 objects, and 27-17 = 10 which is the spacetime dimension of the Hořava-Witten domain wall. Very natural. So I would actually propose you modify the proposal so that $E_8$ already exists in the bulk of 27 dimensions and you create a variation of the DMW paper at the same moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, you will face a lot of trouble. The quantities are unstable, unprotected by supersymmetry, so even if the instabilities can be survived, you won't be able to match the precise numbers on both sides of a duality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, non-fermionic theories don't carry any gauginos and they have no anomalies, so you will be able to show no nontrivial anomaly cancellation that would be similar to the anomaly cancellation of heterotic M-theory, and so on. It is simply very hard to make a convincing story of a 27-dimensional origin of the heterotic string.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that even the ordinary bosonic M-theory remains highly inconclusive. So far, we have only presented some analogous construction for another string vacuum - an appendix to the papers you mentioned that are not terribly important (or famous) at this moment themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Lubos&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-24T07:53:30.787" />
  <row Id="3734" PostTypeId="1" AcceptedAnswerId="3736" CreationDate="2011-01-24T08:15:19.217" Score="1" ViewCount="115" Body="&lt;p&gt;Berry-Pancharatnam phase is the phase that quantum systems exhibit when they pass through a sequence of states and return to their original state. It's a complex phase and it is different from the usual complex phases in that it does not depend on the arbitrary complex phases present in quantum states. For an encyclopedia-level introduction to B-P phase for the purpose of this question, see&lt;br&gt;&#xA;Péter Lévay, Encyclopedia of Mathematical Physics, Elsevier 2006, &quot;Geometric Phases&quot;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/math-ph/0509064v1&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/math-ph/0509064v1&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a quark interacts with the weak force it typically changes flavor and emits or absorbs an electron and a neutrino (or anti-electron / anti-neutrino, as appropriate). This process is generally assumed to require a unitary matrix called the CKM matrix. The elements of this matrix can be measured in high dollar physics experiments but only in absolute value. That is, the complex phases are unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when the data is fit to a unitary CKM matrix, physicists have a choice in how to arrange the complex phases. There are four degrees of freedom in the data and five degrees of freedom in the arbitrary complex phases. The usual method is to use three of the degrees of freedom to (approximately) define the transition probabilities between generations. That is, one has $\{\theta_{12},\; \theta_{13},\; \theta_{23}\}$ for the &quot;mixing angles&quot; between the 1, 2, and 3 generations. (This is only approximate because the probability for going from the 1st to 3rd generation is not equal to the probability for going from the 3rd to the 1st generation.) The fourth degree of freedom is chosen to be $\delta$. If this fourth parameter is zero, then there can be no CP violation. The wikipedia article is a good introduction:&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cabibbo%E2%80%93Kobayashi%E2%80%93Maskawa_matrix&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Cabibbo%E2%80%93Kobayashi%E2%80%93Maskawa_matrix&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\delta$ angle enters the CKM matrix as a complex phase, that is, it appears as $e^{\pm i\delta}$. The other parameters $\theta_{12},\; \theta_{13},\; \theta_{23}$ are used in cosines and sines and are usually abbreviated for example as $\cos(\theta_{12}) = c_{12}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both these cases the central object, Berry-Pancharatnam phase and the $\delta$ angle, is a complex phase. And neither has anything to do with the arbitrary complex phases of quantum states. So my question is this: Can CP violation be defined in terms of a Berry-Pancharatnam phase?&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-03T02:11:03.453" Title="Is there a relationship between Berry-Pancharatnam phase and CP violation in quark mixing?" Tags="&lt;particle-physics&gt;&lt;standard-model&gt;&lt;berry-pancharatnam-phase&gt;&lt;cp-violation&gt;&lt;weak-force&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="3735" PostTypeId="1" AcceptedAnswerId="3738" CreationDate="2011-01-24T09:06:06.050" Score="2" ViewCount="138" Body="&lt;p&gt;I am puzzled by some data I see for a compressor. From what I read, the maximum discharge pressure reachable with a reciprocal compressor would vary depending on the nature of the gas used, and its molecular mass (maximum pressure lower with a heavier gas). This does not seem to fit with my quickly jotted notes from a perfect gas point of view, in which molecular mass shouldn't matter. Any explanation ?&lt;/p&gt;&#xA;" OwnerUserId="1559" LastActivityDate="2011-01-24T12:30:04.343" Title="Maximum pressure at a compressor outlet and molecular weight" Tags="&lt;thermodynamics&gt;" AnswerCount="1" />
  <row Id="3736" PostTypeId="2" ParentId="3734" CreationDate="2011-01-24T09:30:22.040" Score="1" ViewCount="" Body="&lt;p&gt;Dear Carl, first, one has to correct your statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The elements of this matrix can be measured in high dollar physics experiments but only in absolute value. That is, the complex phases are unknown.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not that the phases - beyond the usual 4 degrees of freedom of the CKM matrix - are &quot;unknown&quot;. &lt;strong&gt;Instead, these phases are unphysical.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They can be changed to whatever you like by a phase redefinition of the 3+3 = 6 mass eigenstates of the quarks - which doesn't change their being eigenstates. There are only 5 degrees of freedom in the CKM matrix that you may redefine by these 6 phases of the mass eigenstates: that's because the overall change of all 6 eigenstates by the same phase doesn't change the CKM matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the main question. &lt;strong&gt;Yes, of course, the CP-violating phase may be interpreted as a generalized Berry phase.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you must allow the normalization of the states to change during the monodromy. Now, if you get to the positive point, if you make a cyclic round trip around the three generations whose probability amplitude will be proportional to &#xA;$$V_{12}V_{23}V_{31},$$&#xA;the real matrix would mean that the complex phase of the product above would match the phase of a similar product&#xA;$$V_{13}V_{32}V_{21},$$&#xA;up to a sign. However, the phase of the product above becomes something else - a generic complex phase - for the CP-violating CKM matrix. Note that the ratio&#xA;$$\frac{V_{12}V_{23}V_{31}}{V_{13}V_{32}V_{21}} $$&#xA;is invariant under the phase redefinition of all six states, $d,s,b;d&amp;#39;,s&amp;#39;,b&amp;#39;$ (each of the 6 phases cancels among the numerator and denominator, I think, because each value of the index 1,2,3 appears once as the first index and once as the second index, in both numerator and denominator), so it may be viewed as an invariant definition of the CP-violating phase. I hope that the ratio is complex (unreal) for the actual CKM matrix; correct me if I am wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this sense, and only in this sense, you may interpret the CP-violating phase as a Berry phase. However, there's no &quot;parallel transport&quot; that would literally keep the character of the initial state physically unchanged. That's because there's no unbroken symmetry between the flavors in different generations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&#xA;LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-24T09:30:22.040" CommentCount="6" />
  <row Id="3737" PostTypeId="1" AcceptedAnswerId="3739" CreationDate="2011-01-24T09:36:11.390" Score="4" ViewCount="205" Body="&lt;p&gt;Have some trouble finding a useful reference to answer the following questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What's a typical electron-hole recombination time of semiconductors? And how does it depend on the temperature and the band gap? If there's no formula, examples will do.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What's the probability of an electron being kicked into the conduction band merely by thermal motion? Is it $\exp(-\Delta E/T)$, where $\Delta E$ is the band gap and $T$ the temperature?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That thermal excitation should lead to noise in the system, is that correct?&lt;/p&gt;&#xA;" OwnerUserId="1458" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-01-25T02:13:21.027" LastActivityDate="2011-02-04T07:24:21.103" Title="What's a typical electron-hole recombination time of semiconductors?" Tags="&lt;semiconductor-physics&gt;" AnswerCount="1" />
  <row Id="3738" PostTypeId="2" ParentId="3735" CreationDate="2011-01-24T09:43:55.720" Score="0" ViewCount="" Body="&lt;p&gt;centrifugal compressors actually achieve a higher pressure ratio with heavier gases - because their working depends on storing energy in the form of the kinetic energy of the gas, and it's easier if the gas has a higher molar mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship may go in the opposite direction for reciprocating compressors because the kinetic energy of the gases isn't helpful here. Instead, the gases have to be lifted and heavier gases probably consume more energy at some stage, or something like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, I think that the &quot;simple&quot; physics model you're thinking about - with pressure's being independent of the type of the (nearly ideal) gas - is correct and the reason for the dependence on the gas will be of an engineering character, rather than a fundamental physical character: it's probably due to a subtlety dependent on some detailed steps that make the device work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the best, LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-24T09:43:55.720" />
  <row Id="3739" PostTypeId="2" ParentId="3737" CreationDate="2011-01-24T10:02:50.003" Score="2" ViewCount="" Body="&lt;p&gt;Some observations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;in diodes, the recombination time is of order a nanosecond. It's as fast for similar materials with a high carrier density. The time will depend on the density but also other things and I am not aware of a simple universal formula. It's useful to know that the intra-band collisions that bring the bands into their well-known distributions are much faster and take as little as 0.1 picosecond. The inter-band processes' characteristic time may go to many microseconds or longer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;yes and no, the exponential determines the ratio of carriers that want to sit on the energetically disfavored side divided by those on the dominant side. However, if you talk about the probability that a particular process occurs, you have to specify &quot;in what time interval&quot; you want this process to occur (and how many electrons you watch etc.) - because the result will surely depend on $dt$. It's about the &quot;rate of the process&quot;. Moreover, in the equilibrium, recombination and generation are balanced.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;More generally, it is not true that the recombination appears primarily due to thermal noise that kicks the carrier through the whole band. Quite on the contrary, recombination - and the opposite process of generation - of the carriers in the real world is dominated by the impurities which locally produce new states for the carriers in the middle of the gap, making their jump easier (because it may be divided).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T07:24:21.103" LastActivityDate="2011-02-04T07:24:21.103" />
  <row Id="3743" PostTypeId="2" ParentId="2043" CreationDate="2011-01-24T10:34:17.363" Score="5" ViewCount="" Body="&lt;p&gt;as already mentioned, von Neumann algebras are at the heart of axiomatic approaches to quantum field theory and statistical mechanics, classical references to these topics are for the former (there are a lot more, of course)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hellmut Baumgärtel: &quot;Operatoralgebraic Methods in Quantum Field Theory&quot;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and for the latter&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ola Bratteli and Derek W. Robinson: &quot;Operator algebras and quantum statistical mechanics.&quot; (two volumes).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The basic idea is that the observables of a physical theory should have some algebraic structure, for example it should be possible to scale them, that is measure c*A instead of A. Even more, one should be able to measure any (measurable, no pun intended) function of any observable A, which is possible if A is a memeber of a von Neumann algebra by Borel functional calculus. The philosophy of axiomatic quantum field theory in the sense of Haag-Kastler is therefore that a specific QFT is specified by a net of von Neumann algebras fulfilling a specific set of axioms, and that everything else can be deduced from this net of algebras (for an example see the page on the nLab &lt;a href=&quot;http://nlab.mathforge.org/nlab/show/Haag-Kastler+vacuum+representation&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Lubos pointed out, this ansatz has been very succesful in proving a lot of model indepenent insights/theorems, like the PCT and spin/statistics theorem, but has not been successful in describing the standard model, as far as I know it is not possible to use this ansatz to calculate any number that could be compared to any experiment, which puts some criticism of string theory along these lines into perspective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it is possible to derive the Unruh effect and Hawking radiation using this framework in a much more rigorous fashion than it was done by the original authors, for more details see Robert M. Wald: &quot;Quantum field theory in curved spacetime and black hole thermodynamics.&quot; (Although somewhat outdated, this is still a good place to start.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two striking results where the deep connection between physical intuition and the (deep) mathematical theory of von Neumann algebras is visible involve the modular group of von Neumann algebras with a separating and cyclic vector:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the characterization of KMS states in statistical mechanics,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the Bisognano-Wichmann theorem connecting the automorphism of the modular group to the representation of the Lozentz group, for more ideas using modular theory see the paper &quot;Modular theory for the von Neumann algebras of Local Quantum Physics&quot; by Daniele Longo on the arXiv.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Bisognano-Wichmann theorem says that under specific conditions the modular group (of the von Neumann algebra associated with a wedge region in Minkowski space) coincides with the Lorentz boosts (that map the wegde onto itself), so here we have a very nontrivial connection of a mathematical object obtained from the structure theory of von Neumann algebras (modular theory) with an object coming from special relativity (a representation of the Lorentz group). &lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-01-24T10:34:17.363" />
  <row Id="3745" PostTypeId="2" ParentId="3714" CreationDate="2011-01-24T11:03:35.257" Score="3" ViewCount="" Body="&lt;p&gt;I think that Mark's answer is very good - and in fact, the Wikipedia article you quoted is not bad, either. But let me simplify it a bit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The task where the method is useful is to find the electric field created by charges and planar (or straight) or spherical (or circular) conducting planes - or more general geometries where the method still works but that are unlikely to be encountered &quot;naturally&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electric field has to satisfy the usual conditions, $\mbox{div}(\vec D)=\rho$ and $\mbox{curl}(E)=0$. At the surface of the conductor, $\vec{E}$ has to be perpendicular to the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the field induced by a configuration of charges and an infinite planar conducting plane? It's easy. You just put mirror charges on the other side from the surface of the planar conducting plane: they have the opposite signs of the charge and they're located just like the images in the mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a solution - and because of its uniqueness, the only solution - to the original problem because $\mbox{div}(\vec D)=\rho$ in the physical part of the space where the charges are located. The divergence of the electric field also contains terms that are nonzero in the unphysical part of the space, behind the mirror, but we don't trust our solution over there, anyway. We replace it by $\vec{E}=0$ over there because it's a conductor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most important check is that the electric field constructed as the sum from the real charges and their images is perpendicular to the surface of the conducting plane. That's easily seen because the field lines from the configuration of the charges are symmetric with respect to the plane of the conductor's surface, by construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one has the right divergence and the right curl of the electric field in the physical part of the space - and the right boundary conditions on its boundary, namely the perpendicularity of $\vec{E}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only other simple generalization of this method involves spherical (or circular, in 2D problems) surfaces of the conductor. The mirror charges are located somewhere again. Their charge has a different magnitude, the location has to be calculated, but when you do it right, it will still be true that the electric field will be perpendicular to the spherical or circular surface of the conductor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest way to understand why the mathematics works is to learn some conformal symmetry. By conformal transformations, the plane from the simplest &quot;flat mirror&quot; problem may be mapped to a spherical plane. The locations and magnitude of the charges and the mirror charges also transforms in a calculable way, allowing us to solve a broader set of problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, one may generalize the method to more complicated geometries of the surface of the conductor - more complicated configurations of charges may be needed and more complicated transformations (especially in 2D) may be helpful to find them; and the problem may also be generalized to problems with other boundary conditions on the surface - different components of the fields may be required to vanish. The set of all possible applications and generalizations can't be &quot;quite classified&quot;: it's important to actually understand why and how the method works and then you may apply it whenever you want.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T07:51:47.777" LastActivityDate="2011-02-03T07:51:47.777" CommentCount="1" />
  <row Id="3746" PostTypeId="2" ParentId="3488" CreationDate="2011-01-24T11:08:00.200" Score="2" ViewCount="" Body="&lt;p&gt;This is a good question, and what you are encountering is what is often called the inverse r-squared scaling law, or more simply, the &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/forces/isq.html&quot; rel=&quot;nofollow&quot;&gt;inverse square law&lt;/a&gt;.  In this case, we can write a crude relationship:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Size_{apparent} \propto Size_{actual}\dfrac{1}{R^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we think in terms of &lt;a href=&quot;http://en.wikipedia.org/wiki/Steradian&quot; rel=&quot;nofollow&quot;&gt;steradians&lt;/a&gt;, we can think in terms of the area of spheres we can take the actual two dimensional size of an object (say 1.5m high and 0.5m wide = 0.75 $m^2$); and if we say that object is 3m away, we would compare that area to the area of a sphere that has a radius of 3m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4\pi{R^2} = 4\pi{3^2} = 36\pi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{0.75}{36\pi} = \dfrac{1}{48\pi}steradians$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take that same object and move it out to 30m we find that its apparent size shrinks much faster than its linear distance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4\pi{R^2} = 4\pi{30^2} = 3600\pi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{0.75}{3600\pi} = \dfrac{1}{4800\pi}steradians$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the reason that an object appears very small as you move it further away from you.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you exclude curvature effects of the earth, the object that is further away from you because its apparent cross section is much smaller than if it were closer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In answer to your question, barring other physical effects (particularly curvature of the earth effects), if you wanted to hit an object of equal height that was further away with a laser, you would want to aim as straight and parrallel to the ground as possible, and above the ground at an appropriate height.&lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="1383" LastEditorDisplayName="" LastEditDate="2011-01-24T11:18:36.927" LastActivityDate="2011-01-24T11:18:36.927" />
  <row Id="3748" PostTypeId="2" ParentId="3662" CreationDate="2011-01-24T12:08:03.607" Score="3" ViewCount="" Body="&lt;p&gt;It's worth making an analogy here. LQG does not impose an fixed lattice structure. The only discreteness existing is that area and volume operators have a discrete spectrum, specifically meaning that there is a &quot;smallest&quot; area and volume possible before zero. This should be compared to the familiar spin algebra --- there is rotational invariance, but the states are not rotationally invariant (individually)! The situation is almost exactly analogous.&lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-01-24T12:08:03.607" CommentCount="1" />
  <row Id="3750" PostTypeId="1" AcceptedAnswerId="3757" CreationDate="2011-01-24T12:47:07.497" Score="9" ViewCount="457" Body="&lt;p&gt;This is a classic trick to do with a IR camera:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/Rh1fb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bu why is the plastic bag transparent, while the glasses aren't? I've also heard that water is not transparent in IR light. What causes this phenomena?&lt;/p&gt;&#xA;" OwnerUserId="250" LastEditorUserId="701" LastEditorDisplayName="" LastEditDate="2011-01-24T15:52:41.893" LastActivityDate="2011-09-02T08:25:22.353" Title="Why is a plastic bag transparent in infrared light?" Tags="&lt;optics&gt;&lt;spectroscopy&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="3751" PostTypeId="1" AcceptedAnswerId="3829" CreationDate="2011-01-24T12:47:33.133" Score="4" ViewCount="187" Body="&lt;p&gt;A previous &lt;a href=&quot;http://physics.stackexchange.com/questions/1628/is-the-curvature-of-space-time-a-smooth-function-everywhere-except-at-black-ho&quot;&gt;Stack question&lt;/a&gt; (before I joined) asking about continuity in GR received replies which suggested that Curvature would be discontinuous at say a planetary boundary (assume no atmosphere for simplicity). I will analyse some basics of this and then return to that question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that the Stress-Energy Tensor $T{_a}{_b}=0$ outside the body and is nonzero in the interior resulting in a discontinuity at the surface. This would imply that the Ricci Tensor $R{_a}{_b}$ is also discontinous at the boundary, and zero in the vacuum part as expected from the Einstein equations. However the Riemann Curvature Tensor $R{_a}{_b}{_c}{_d}$ (which generates the physically measurable accelerations) has contributions from the Weyl Curvature Tensor $C{_a}{_b}{_c}{_d}$ as well. In fact the Ricci Tensor &quot;hands over&quot; to the Weyl Tensor at the boundary: thus the Riemann Tensor stays non-zero there. However this &quot;hand over&quot; does not imply continuity, unless there is some GR Theorem which says that the Riemann Tensor stays continuous in this region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also in the Newtonian approximation the analogous role is played by the gravitational potential $\phi$ in the Poisson equation $\nabla^2 \phi = 4 \pi G\rho$. Clearly this shows a discontinuity too as the density $\rho$ suddenly drops off at the boundary. However the discontinuity is in the second derivative of the potential: the potential itself is continuous. This means that in exiting a planetary cave or mine one does not suddenly meet a change in Gravitational potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I do not know any theorem in GR which guarantees such continuity. The applicable in-the-large scenario might be the surface of a neutron star; there may be in-the-small particle models too.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-01-25T16:23:20.927" Title="The Matter-Vacuum Boundary in General Relativity" Tags="&lt;general-relativity&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="3752" PostTypeId="2" ParentId="3714" CreationDate="2011-01-24T12:57:10.957" Score="1" ViewCount="" Body="&lt;p&gt;The answers given so far are very good, but I want to add something that might help. The method of images begin by trying to &lt;em&gt;derive&lt;/em&gt; Coulomb's law. Coulomb's law is the Green's function of the Laplace equation. It's general form is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;G(\mathbf{x},\mathbf{x}^{\prime})=\frac{1}{R}+F(\mathbf{x},\mathbf{x}^{\prime})&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $F$ is a function that satisfies the Laplace equation in your domain. In the case that the domain extends to infinity and the potential fall-off fast, $F=0$ and you have the usual Coulomb law. However, in the case of a domain limited by conductors (well, any boundary condition for which the solution exists and is unique will do, I'm just using the usual case), where you have $G=0$ on the boundary, $F$ will have to be adjusted so that this condition is satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now comes the physical interpretation: since $F$ satisfies the Laplace equation inside your domain, but not outside, it can be interpreted as the potential due to a system of charges outside your boundaries. And those are exactly the image charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More details can be found in Jackson, &lt;em&gt;Classical Electrodynamics&lt;/em&gt;. Section 1.10: Formal solution of electrostatic boundary-value problems with Green Function.&lt;/p&gt;&#xA;" OwnerUserId="1337" LastActivityDate="2011-01-24T12:57:10.957" />
  <row Id="3753" PostTypeId="2" ParentId="3432" CreationDate="2011-01-24T13:15:36.803" Score="-3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://outrafisica.blogs.sapo.pt/299.html&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;OUTRA FISICA&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the end of the Dark Era of Cosmology. The author’s model is awesome and presents a challenge to our presumed knowledge.&#xA;Using ‘Plain Old physics’  ;-) he reveals a complete different History of the Universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space expansion? Big Bang? Dark Energy? Dark Matter? No thanks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read it in chronological order with google chrome, and reader, directly from Portuguese language (Chrome can translate to you).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To ease the entry I present here an ad-hoc translation of the first and the last posts (Creative Commons BY-NC Attribution-NonCommercial).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://outrafisica.blogs.sapo.pt/299.html&quot; rel=&quot;nofollow&quot;&gt;First post (Presentation):&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Another Physics&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Relativity and beyond. . ., unheard&#xA;  theories, fantastic ideas ,give birth&#xA;  to a new understanding of the&#xA;  Universe. Light rays of dawn will&#xA;  shine reflections of Future in the&#xA;  reborn eyes of four friends.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Presentation&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Mainly using the format of a&#xA;  conversation between four characters,&#xA;  I will try to present in language&#xA;  accessible to people without&#xA;  scientific training the essentials of&#xA;  the Theoryof Special Relativity and&#xA;  continue the line of thought of a&#xA;  specific group of scientists,including&#xA;  Poincaré, Einstein, Lorentz and&#xA;  Fitzgerald.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This path is different from that&#xA;  initiated by Minkowsky that led to the&#xA;  current Physics. The seemingly small&#xA;  initial differences lead to concepts&#xA;  as diverse as the universe of&#xA;  Copernicus was different from that of&#xA;  Ptolemy.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;There is also a profound difference in&#xA;  the methodological process. While the&#xA;  actual physics follows the simple&#xA;  methodology of Ptolemy (developing&#xA;  mathematical models on top of the&#xA;  observed data as obtained) here follow&#xA;  the methodology of Galileo (creating&#xA;  logically related models).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Let me introduce you the characters:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;George and Ann are two characters in&#xA;  mystery, because George has the&#xA;  knowledge of the Another Physics, Anna&#xA;  has a role that transcends this blog.&#xA;  Mario is a scientist with broad range&#xA;  of knowledge and open-minded to&#xA;  examine the ideas of George. Louise&#xA;  represents people whose education does&#xA;  not belong to science and technology.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://outrafisica.blogs.sapo.pt/&quot; rel=&quot;nofollow&quot;&gt;last post (&lt;em&gt;abstract&lt;/em&gt; of a foundational paper to be ;-)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Generalizing Relativity Principle to Comoving Coordinates allows the&#xA;  fitting of cosmic data with no dark&#xA;  matter or dark energy&lt;/strong&gt;   The standard&#xA;  space expansion model considers two&#xA;  kinds of coordinates: the atomic&#xA;  coordinates, where the validity of the&#xA;  Relativity Principle is established,&#xA;  having the length unit defined from&#xA;  atomic properties, and the comoving &#xA;  coordinates, where space expansion is&#xA;  null. In atomic coordinates Space&#xA;  expands, in comoving ones Matter&#xA;  decreases, evanesces.  In this work,&#xA;  as there is no observational evidence&#xA;  indicating that one of these&#xA;  coordinate systems should be any&#xA;  better at supporting physical laws, we&#xA;  postulate that Relativity Principle&#xA;  must hold in both systems. From this&#xA;  postulate we derive a model that fits&#xA;  both cosmic and local data with the&#xA;  same physical laws, needing no new&#xA;  parameters besides Hubble constant,&#xA;  namely no dark matter or dark energy.&#xA;  We also extend the generalization of&#xA;  the Relativity Principle to a more&#xA;  general, new kind of coordinates - the&#xA;  Field coordinates – of which the&#xA;  atomic and the comoving systems are&#xA;  just two particular cases.&#xA;         The generalization of the Relativity Principle presented here&#xA;  does not conflict with established&#xA;  theories, namely Special or General&#xA;  Relativity. It is presented a&#xA;  prediction that can be experimentally&#xA;  tested: an acceleration component in&#xA;  the rotation of isolated bodies. New&#xA;  results are obtained, such as the&#xA;  long-term evolution of Earth&#xA;  temperature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-01-24T13:15:36.803" CommunityOwnedDate="2011-01-24T13:15:36.803" />
  <row Id="3754" PostTypeId="1" AcceptedAnswerId="3755" CreationDate="2011-01-24T13:27:24.910" Score="4" ViewCount="344" Body="&lt;p&gt;In Physics class we were doing the two slit experiment with a helium-neon red laser. We used this to work out the wavelength of the laser light to a high degree of accuracy. On the piece of paper the light shined on there were patterns of interference, both constructive and destructive. My question is, when the part of the paper appeared dark, where did the energy in the light go?&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-03-13T20:16:40.410" LastActivityDate="2011-03-13T20:16:40.410" Title="Two slit experiment: Where does the energy go?" Tags="&lt;quantum-mechanics&gt;&lt;double-slit-experiment&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="3755" PostTypeId="2" ParentId="3754" CreationDate="2011-01-24T13:31:22.570" Score="7" ViewCount="" Body="&lt;p&gt;It goes to the brighter strips. In these regions there is a constructive interference, which is actually brighter than would be with just one source.&lt;/p&gt;&#xA;" OwnerUserId="1337" LastActivityDate="2011-01-24T13:31:22.570" CommentCount="2" />
  <row Id="3756" PostTypeId="2" ParentId="3751" CreationDate="2011-01-24T13:36:36.427" Score="1" ViewCount="" Body="&lt;p&gt;I don't know of a theorem of the kind you want, but the standard treatment of the matter-vacuum boundary in general relativity is given by the &lt;a href=&quot;http://www-dft.ts.infn.it/~ansoldi/Seminars/2004/FFP6/HTML/Level1.2.html&quot; rel=&quot;nofollow&quot;&gt;Israel junction conditions&lt;/a&gt;. Consider an infinitely thin shell of matter embedded in a manifold. The shell has two sides (assuming the shell is an orientable manifold) $S^+$ and $S^-$. The embedding of each of these surfaces in the background manifold is given by an extrinsic curvature $K^{\pm}_{ij}$. The difference in the extrinsic curvature between the two sides is related to the intrinsic curvature of the shell:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 8 \pi \left( R_{ij} - \frac{1}{2} g_{ij} R \right) = K^-_{ij} - K^+_{ij} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left hand side is also known as the Einstein tensor $G_{ij}$, i.e. the traceless part of the Ricci curvature $R_{ij}$. And this in turn is related to the energy-momentum tensor $T_{ij}$ of the matter constituting the shell by Einstein's equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ G_{ij} = 8\pi T_{ij} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For now this should be enough to get you going. I'll add more as I learn more. A good reference for this and other topics is the set of notes on advanced GR by Eric Poisson which you can find &lt;a href=&quot;http://www.physics.uoguelph.ca/poisson/research/notes.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2011-01-24T13:36:36.427" />
  <row Id="3757" PostTypeId="2" ParentId="3750" CreationDate="2011-01-24T13:48:00.097" Score="5" ViewCount="" Body="&lt;p&gt;the physical reason why this is happening is that &lt;a href=&quot;http://en.wikipedia.org/wiki/Absorption_%28electromagnetic_radiation%29&quot; rel=&quot;nofollow&quot;&gt;absorption&lt;/a&gt; of a medium is frequency dependent.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Mathematical description&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The most prominent example of a natural description might be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Beer%E2%80%93Lambert_law&quot; rel=&quot;nofollow&quot;&gt;Lambert-Beer law&lt;/a&gt; that states that the change of a quantity $q$, $dq/dx$ is related to its value at $x$ multiplied by a scalar factor $\lambda$, which might depend on some further parameters like the frequency $\nu$,&#xA;$$\frac{dq}{dx} = -\lambda(\nu) q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which has the solution&#xA;$$q(x) = q_0 e^{-\lambda(\nu) x}\, .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For our scenario, $x$ might be the distance a light ray already traveled into the plastic bag, $x\in[0,D_{bag}]$. In optics, one often uses the &lt;a href=&quot;http://en.wikipedia.org/wiki/Extinction_coefficient&quot; rel=&quot;nofollow&quot;&gt;extinction coefficient&lt;/a&gt; $n''(\nu)$, the imaginary part of the index of refraction or the absorbance $A_\lambda$ giving the same information for $\Delta x=\lambda$ (to clearify: $\lambda$ here is the wavelength, not the parameter above).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Electromagnetic case - penetration depth&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let us look directly at electrodynamics and how the Lambert-Beer law can be applied here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the absolute of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wave_vector&quot; rel=&quot;nofollow&quot;&gt;wave vector&lt;/a&gt; in a medium described by the (frequency dependent) refractive index $n(\omega) = n'(\omega) + \mathrm{i}n''(\omega)$ is given by $$k = \sqrt{n'(\omega) + \mathrm{i}n''(\omega)}\frac{\omega}{c}\, .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If now $n'  \gg n'' $ (which is true far away for resonances), we can expand it as&#xA;$$k = \sqrt{n'}\left(1 + \frac{\mathrm{i}}{2} \frac{n''}{n'} +\mathcal{O}\left(\frac{n''}{n'}\right)^2 \right)\frac{\omega}{c}\, .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, for a field $$\mathbf{E} \sim \mathbf{E_0} e^{\mathrm{i}{\mathbf{k\cdot r}-\omega t}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we find that $$\frac{|\mathbf{E}|}{|\mathbf{E_0}|} \leq \frac{1}{e}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;after a &lt;strong&gt;penetration depth&lt;/strong&gt; of &#xA;$$d_p \approx 2 \frac{\sqrt{n'}}{n''}\frac{c}{\omega}\, .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This quantity can be seen as a &lt;strong&gt;characteristic propagation length&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Light absorption of a plastic bag&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Since plastic bags are often made of &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyethylene&quot; rel=&quot;nofollow&quot;&gt;polyethylene&lt;/a&gt;, we have to search for the &lt;strong&gt;spectral properties&lt;/strong&gt; of it in the visible and infrared range to be able to &lt;strong&gt;answer the question&lt;/strong&gt;. A quick search revealed that in &lt;a href=&quot;http://www.jstor.org/stable/3575254?seq=4&quot; rel=&quot;nofollow&quot;&gt;Optical Properties of Polyethylene: Measurement and Applications&lt;/a&gt; on page four $n''(\nu)$ is given depending on the frequency in terms of &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronvolt&quot; rel=&quot;nofollow&quot;&gt;eV&lt;/a&gt; (visible light ranges roughly in 1.5 - 3 eV, IR is energetically lower). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With an $n''\leq 10^{-5}$ as may be interpreted from the figure given in the paper, we can calculate &#xA;$$d_p \approx 6\mathrm{cm}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if I did not miscalculate totally. So, we have to state that polyethylene cannot be accounted for the absorption alone as stated by &lt;a href=&quot;http://physics.stackexchange.com/users/701/gigacyan&quot;&gt;gigacyan&lt;/a&gt; and &lt;a href=&quot;http://physics.stackexchange.com/users/1370/georg&quot;&gt;Georg&lt;/a&gt; - &lt;strong&gt;other absorbing effects&lt;/strong&gt; need to be considered (&lt;em&gt;In the given case we can only speculate that it maybe was doping with carbon which causes the bag to appear black&lt;/em&gt;). These effects will result in a change of $n(\omega)$ including its imaginary part, hence resulting in a &lt;strong&gt;changed penetration depth&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this discussion is &lt;strong&gt;beyond the scope&lt;/strong&gt; of this answer - one would have to calculate the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronic_band_structure&quot; rel=&quot;nofollow&quot;&gt;electronic band structure&lt;/a&gt; of the material which is not an easy task of &lt;a href=&quot;http://en.wikipedia.org/wiki/Solid-state_physics&quot; rel=&quot;nofollow&quot;&gt;solid-state physics&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, I hope my explanations are sufficient to explain &lt;em&gt;makroskopic&lt;/em&gt; absorption phenomena including the one considered here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2011-01-24T18:09:32.390" LastActivityDate="2011-01-24T18:09:32.390" CommentCount="14" />
  <row Id="3758" PostTypeId="1" AcceptedAnswerId="3771" CreationDate="2011-01-24T13:53:38.653" Score="6" ViewCount="117" Body="&lt;p&gt;Topological insulators are materials known to have bulk insulator and metallic surface states. But, what is the origin of these metallic surface states? And how the topology of band could help the understanding of these states?&lt;/p&gt;&#xA;" OwnerUserId="1545" LastActivityDate="2011-01-24T16:26:27.060" Title="How metallic surfaces states can emerge in topological insulators?" Tags="&lt;solid-state-physics&gt;&lt;topological-insulators&gt;" AnswerCount="1" />
  <row Id="3759" PostTypeId="1" AcceptedAnswerId="3780" CreationDate="2011-01-24T14:14:26.067" Score="6" ViewCount="439" Body="&lt;p&gt;I know they have their adherents, but do more or less esoteric branches of mathematics such as Category Theory and/or Quantum Logic provide powerful tools for new theory development or are they just occasionally-useful presentational frameworks?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More particularly, is it worth investing the time to study these formalisms?&lt;/p&gt;&#xA;" OwnerUserId="1394" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-28T16:23:34.207" LastActivityDate="2011-02-28T16:23:34.207" Title="Do Category Theory and/or Quantum Logic add value in physics?" Tags="&lt;soft-question&gt;&lt;quantum-information&gt;&lt;category&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="3760" PostTypeId="2" ParentId="3731" CreationDate="2011-01-24T14:14:55.067" Score="1" ViewCount="" Body="&lt;p&gt;These researches are interesting.  This connects with the projective Cayley plane, which is important in the Jordan algebra.  Some of the structure discussed in the paper Hisham Sati&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0807/0807.4899v3.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/arxiv/pdf/0807/0807.4899v3.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;makes contact with the Jordan algebra. A constraint on the diagonal scalars of the Jordan matrix, which is similar to a light cone or infinite momentum condition reduces the system to $26$ dimensions.  Decomposed as $10~+~16$ appears to connect with a heterotic string in $10$ dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exceptional cubic matrix is an extension of the same spin structure by the isomorphism&#xA;$$&#xD;&#xA;J^3({\cal O})~\simeq~ R\oplus J^2({\cal O})\oplus{\cal O}.&#xD;&#xA;$$&#xA;The extension to the cubic matrix model with $V~\in~J^2({\cal O}$, involve $z_0~\in~R$, and $\theta~\in~{\cal O}^2$ for $V,~\theta$ vector and spinor elements in the $10~=~9~+~1$ dimensional spacetime.  The spinor representation of $so(9)$ when restricted to $so(8)$ splits as ${\bf 8}_1\oplus{\bf 8}_2$, so that $so(9)~\simeq~{\cal O}^2$.  The vector portion is given by the spin factor $J^2({\cal O})$ which splits according to $J^2({\cal O})~\simeq~({\bf 8}_0\oplus R)\oplus R$ in a representation of $10$ dimensional Lorentzian spacetime.  Hence the spatial coordinates $V~\in~{\bf 8}\oplus R$ are associated with their superpartners $\theta~\in~{\cal O}^2$ .  The three $\bf 8$ representations of $so(8)$ are mixed according to the triality operation in the cubic matrix&#xA;$$&#xD;&#xA;\left(\matrix{z_1 &amp;amp; {\cal O}_0 &amp;amp; {\bar{\cal O}}_2\cr&#xD;&#xA;{\bar{\cal O}}_0 &amp;amp; z_2 &amp;amp; {\cal O}_1\cr&#xD;&#xA;{\cal O}_2 &amp;amp; {\bar{\cal O}}_1 &amp;amp; z_0}\right)&#xD;&#xA;$$&#xA;The three $\cal O$s give $24$ degrees of freedom, which in addition to the 3 $z$'s gives a total of $27$ degrees of freedom.  The exceptional group $G_2$ is the automorphism on $\cal O$, or equivalently that $F_4\times G_2$ defines a centralizer on $E_8$.  The fibration $G_2~\rightarrow~S^7$ is completed with $SO(8)$, where the three ${\cal O}$'s satisfy the triality condition in $SO(8)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $so(9)$ is not the most general symmetry of $J^3(0)$.  There exists a permutation on the three scalars $z_0,~z_1,z_2$ and ${\cal O}^3$.  This means there is an additional automorphism $so(3)$.  The more general automorphism is then $F_4$.  The quotient between the $52$ dimensional $F_4$ and the $36$ dimensional $so(9)~\simeq~B_4$ defines the short exact sequence&#xA;$$&#xD;&#xA;F_4/B_4:1~\rightarrow~spin(9)~\rightarrow~F_{4\setminus 16}~\rightarrow~{\cal O}P^2~\rightarrow~1,&#xD;&#xA;$$&#xA;where $F_{4\setminus 16}$ means $F_4$ restricted to $36$ dimensions, which are the kernel of the map to the $16$ dimensional Moufang or Cayley plane ${\cal O}P^2$.  Geometrically the $F_4$ define the symmetry of the $24$cell, called the icositetrachoron or polyoctahedron, according to $24$ octahedral cells.  The $B_4$ also defines a more restricted symmetry on the $24$ cell according to $16$ tetrahedral cells and $8$ octahedral cells.  The $8$ octahedral cells define the ${\bf 8}_0$, or $so(8)$ in the $J^2({\cal O})$, while the $16$ tetrahedral cells are mapped to the ${\cal O}P^2$.  This means on the algebraic level $f_4~\simeq~ so(8)\oplus V\oplus \theta_1\oplus \theta_2$, which explicitly describes the triality condition the three octonions with the $so(8)$.  More generally according to octonions $f_4~\simeq~so({\cal O})\oplus {\cal O}^3$.  $f_4$ diagonalizes then the Jordan cubic matrix.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-24T14:14:55.067" />
  <row Id="3761" PostTypeId="2" ParentId="1869" CreationDate="2011-01-24T14:34:18.100" Score="-3" ViewCount="" Body="&lt;p&gt;I think that QFT and the string theory have the following flaw in common: they treat perturbatively coupling the things coupled permanently. This is a conceptual flaw that not always can be &quot;repaired&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-01-24T16:35:20.553" LastActivityDate="2011-01-24T16:35:20.553" CommentCount="5" />
  <row Id="3762" PostTypeId="2" ParentId="3706" CreationDate="2011-01-24T14:40:06.030" Score="1" ViewCount="" Body="&lt;p&gt;It would seem that the area beyond the event horizon is disconnected from the rest of the universe. If it is then we must consider larger questions such as: is it even governed by the same laws? Besides this though, the singularity is not necessarily a real object, it is simply the expression that GR can give no information about what happens to space-time in the center of a black hole. &#xA;We should further ask the question as to whether we can ever have a real theory of what goes on inside the black hole. We could model some equations, but what observer would be able to test them? Unless the area is not disconnected, in which case we need to reexamine GR. Changes to GR could change the very definition of what a black hole even is!!!!&#xA;So to rephrase what seems to be the consensus: no one knows + it seems we don't have even the foundations to solve this problem yet&lt;/p&gt;&#xA;" OwnerUserId="1567" LastActivityDate="2011-01-24T14:40:06.030" />
  <row Id="3763" PostTypeId="2" ParentId="3751" CreationDate="2011-01-24T14:51:26.407" Score="1" ViewCount="" Body="&lt;p&gt;The question touches on Birkhoff’s theorem.  A spherical distribution of matter has a gravity field entirely equivalent to a black hole of the same mass.  So if we were to ignore the matter in a star the interior configuration of spacetime must behave as if it were a Schwarzschild metric (or Reissnor-Norstrom, Kerr, etc).  Suppose we have a metric of with $g_{tt}~=~F(r)$ and $g_{rr}~=~1/g_{tt}$, and within a surface there is the $T^{ab}$ for material inside the body.  The Bianchi identity ${T^{ab}}_{;b}~=~0$ gives &#xA;$$&#xD;&#xA;p_{,r}~+~(m~+~p)\Big(\frac{F_{,rr}}{F_{,r}}~-~\frac{2F_{,r}}{F}\Big)~=~0&#xD;&#xA;$$&#xA;where now one matches the condition on the vacuum metric with this condition.  From there the metric $F(r)$ inside the body can be computed.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-01-24T15:11:49.760" LastActivityDate="2011-01-24T15:11:49.760" CommentCount="5" />
  <row Id="3764" PostTypeId="2" ParentId="1869" CreationDate="2011-01-24T15:09:11.027" Score="1" ViewCount="" Body="&lt;p&gt;There are two sides to this coin.  String/M theory is more than a theory, it is more in a way a framework of theories.  It is a vast territory, at least from a mathematical perspective.  A lot of it is not likely correct, or should we say does not manifest itself in anything physical.  On the other hand there appears to be universality to some stringy structure, where it is appearing in condensed matter physics.  Some stringy or AdS structure is making an appearance in heavy ion physics. So the other side of the coin is that stringy/M structures are too rich for the whole thing to be completely false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When looking at string physics it is probably best to hold more to basic physical ideas, such as how Susskind approaches things, and to remain close to mathematics which has some foundational aspect to it.  Highly rococo mathematics that deviates from “basics,” and theory which is increasingly complex might be held up to greater suspicion.  Where the dividing line lies here is a bit hard to define.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-24T15:09:11.027" />
  <row Id="3765" PostTypeId="2" ParentId="3698" CreationDate="2011-01-24T15:33:17.780" Score="0" ViewCount="" Body="&lt;p&gt;For asymptotically flat spaces, a black hole can never be in thermal equilibrium with the rest of spacetime. If it were, the rest of spacetime would be filled with radiation at the Hawking temperature, and the resulting nonzero stress-energy tensor everywhere is inconsistent with the assumption of asymptotic flatness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, assuming spacetime sufficiently far away from the black hole is vacuumlike (i.e. no cosmic microwave background radiation), we have a temperature gradient ranging from the Hawking temperature right outside the event horizon to zero very far away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But once you accept a temperature gradient, the temperature of the black hole no longer tells us anything about the temperature of spacetime far away from it. The Hawking temperature may be around the Planck temperature, but far away, the temperature is still approaches zero.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;AdS black holes are an entirely different matter. I know this question isn't about AdS black holes, but they're instructive as we can have thermal equilibrium &lt;em&gt;everywhere&lt;/em&gt; for AdS black holes. This is because we have a warp factor for clock rates which rescales the temperature as measured by a local observer compared to the temperature as measured by a distant observer. So, even at thermal equilibrium, the local temperature falls off exponentially beyond the AdS radius as we move away from the black hole. Any backreaction will remain finite because of this warp factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two cases to consider here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the black hole radius is smaller than the AdS radius&lt;/li&gt;&#xA;&lt;li&gt;the black hole radius is larger than the AdS radius&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the latter case, as the size of the black hole goes up, so does the temperature, and there's no upper limit to the temperature. This means the black hole has a positive heat capacity, and remains in thermal equilibrium with its environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the former case, the temperature goes down as the size of the black hole goes up, leading to a negative heat capacity. So, even if we start off with an &quot;equilibrium&quot; state, it is unstable. If the black hole expands, it becomes cooler than its environment, and so, it will absorb a net amount of radiation and expand even further beyond the AdS radius after which it will start to cool off, eventually settling down to the thermal state in the latter case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a U-shaped temperature relation as a function of black hole size, we can see it's not possible to have black hole temperatures below the AdS scale, $k$. Instead, we have a phase transition to a thermal state with no black holes.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-26T15:39:07.410" LastActivityDate="2011-01-26T15:39:07.410" CommentCount="2" />
  <row Id="3767" PostTypeId="1" AcceptedAnswerId="3788" CreationDate="2011-01-24T15:48:00.307" Score="9" ViewCount="405" Body="&lt;p&gt;John Cramer’s &lt;a href=&quot;http://en.wikipedia.org/wiki/Transactional_interpretation&quot; rel=&quot;nofollow&quot;&gt;transactional interpretation of quantum mechanics&lt;/a&gt; (TIQM) is billed as resolving the fuzzy agnosticism of the Copenhagen interpretation while avoiding the alleged ontological excesses of the Many Worlds Interpretation. Yet it has a low profile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this because no-one care anymore about ontology in physics, or is there something about TIQM which undermines belief in it?&lt;/p&gt;&#xA;" OwnerUserId="1394" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-28T21:24:40.547" LastActivityDate="2011-02-28T21:24:40.547" Title="The transactional interpretation of quantum mechanics" Tags="&lt;quantum-mechanics&gt;&lt;causality&gt;&lt;quantum-interpretations&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="3768" PostTypeId="2" ParentId="3750" CreationDate="2011-01-24T15:51:34.890" Score="1" ViewCount="" Body="&lt;p&gt;I am not a specialist in thermal vision so I will just answer your last question: &lt;strong&gt;what causes these phenomena?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electromagnetic radiation (e.g. visible light) propagates through matter by polarizing it. Oscillating electromagnetic field polarizes molecule that, in turn, begins to emit electromagnetic radiation, and this cycle continues until the light comes from the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except, often it doesn't. If the frequency of the field is close to a resonant frequency of the molecule, the photon can be absorbed and its energy will be converted to heat. These resonances can be of different nature: absorption in the visible range is caused by electrons whereas nuclear motion is resonant with lower frequencies in the IR range. These resonances are individual and can be used for identifying molecules by measuring light absorption versus wavelength (it's called &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectrophotometry&quot; rel=&quot;nofollow&quot;&gt;Spectrophotometry&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; In case of conductors like metals or, in this case, black carbon, free electrons are resonant to any frequency and these materials can observe light of any wavelength. However, as Georg pointed out, particles of carbon black are much smaller than the wavelength of thermal IR (10 $\mu$m) and this could be the reason of small absorption efficiency.&lt;/p&gt;&#xA;" OwnerUserId="701" LastEditorUserId="701" LastEditorDisplayName="" LastEditDate="2011-01-24T19:07:44.917" LastActivityDate="2011-01-24T19:07:44.917" />
  <row Id="3769" PostTypeId="1" AcceptedAnswerId="3782" CreationDate="2011-01-24T16:18:29.677" Score="10" ViewCount="400" Body="&lt;p&gt;Is there a cosmologist in the house?  I've got a basic understanding (with some degree of error) of some simple facts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Universe is a little over 13 billion years old.  Our galaxy is almost that old.&lt;/li&gt;&#xA;&lt;li&gt;Our solar system is roughly 4.6 billion years old.&lt;/li&gt;&#xA;&lt;li&gt;The heavier elements (Carbon, Oxygen, etc..) are only produced by stars.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When our solar system coalesced, the stuff that makes up the planets came from a previous generation of stars.  (Meanwhile our Sun is busy making its own slightly-heavier elements, but those aren't ours.  Those belong to the next &quot;generation&quot; after our star dies.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we're all sitting on a rock whose matter was spewed out of a star (or stars) that preceded this solar system.  Got it.  Don't need that lesson.  But what I want to know is...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many cycles of coalescing, fusing, doing the nucleosynthesis thing, exploding -- and then repeating the cycle -- have this atoms in this ball of rock I'm standing on been through?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you know this?  Can you tell by the make-up of the matter in the solar system?  Looking around the Orion arm to see remnants of our former home?  Wild ass guess?&lt;/p&gt;&#xA;" OwnerUserId="1570" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T07:49:59.820" LastActivityDate="2011-02-03T07:49:59.820" Title="How many times has the &quot;stuff&quot; in our solar system been recycled from previous stars?" Tags="&lt;cosmology&gt;&lt;nuclear-physics&gt;&lt;universe&gt;" AnswerCount="5" FavoriteCount="2" />
  <row Id="3770" PostTypeId="1" AcceptedAnswerId="3857" CreationDate="2011-01-24T16:24:37.343" Score="9" ViewCount="200" Body="&lt;p&gt;I've been very impressed to learn about kaluza-klein theory and compactification strategies. I would like to read more about this but in the meantime i'm curious about 2 different points. I have the feeling that there are no precise answers to these questions at the current moment but nonetheless i feel obligued to ask:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Does a compactification strategy has to impose a fixed thickness in the compactification or this arises as a dynamic consequence of the evolution equations? and if i start with a small thickness, does the dynamic in all cases keeps the thickness in a bounded value? are there results that addresses this? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; answer appears to be No, it is not assumed to be fixed. There is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Radion_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;effective scalar field&lt;/a&gt; that precisely describes this.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Given that the argument is that vibrations in the compactified dimensions determine the internal states, any significant gradient ratio in the compactification length would have to be probably measurable as either spontaneous decay or spontaneous stimulation events of particles. My question is, would experimental effects of any thickness gradients would be as difficult to measure as expected effects from a fixed but non-zero compactified dimension? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; a better rephrase of this question is that; &lt;em&gt;since energy of non-gravitational fields like electromagnetism, etc. is stored in vibrations along the compactified dimensions, it would seem that even a small gradient in the radion field (even 0.1%) should be measurable as a effective diffraction index (since a 0.1% variation in the length of the scale would affect each non-gravitational mode by that amount)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess even another way to frame the question is: is the low-energy phenomena we are able to see currently, invariant/non-dependent under small changes of scale (from point to point in space-time) of the compactification? isn't scale of compactification itself a vibrating degree of freedom?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So bottom line:&lt;/strong&gt; since we haven't see any low-energy consequences from small gradients in the radion field, can't we infer that the radion field, is, by whatever reasons (dynamics, broken symmetry, etc.) &lt;em&gt;effectively fixed&lt;/em&gt;? can't we currently estimate a bound in the radion gradient given the negative results mentioned above it would have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope i've made my questions clear and interesting enough. &lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-04-15T20:26:33.497" LastActivityDate="2011-04-15T20:26:33.497" Title="Measurement of kaluza-klein radion field gradient?" Tags="&lt;general-relativity&gt;&lt;gravity&gt;&lt;kaluza-klein&gt;&lt;compactification&gt;&lt;calabi-yau&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="3771" PostTypeId="2" ParentId="3758" CreationDate="2011-01-24T16:26:27.060" Score="3" ViewCount="" Body="&lt;p&gt;This won't completely answer your question, but maybe it will help. I remember when I was encountering the topic that I used to be completely bewildered by the discussion. I think two facts really helped me to understand things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;We are working in a non-interacting picture of electrons. This means that we only need to consider a single-particle Hamiltonian.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We are working in a lattice. This means that the single particle Hamiltonian is a direct sum of discrete Hamiltonians, indexed by a pseudo-momentum vector $k$ (i.e. point in the BZ): $H = \bigoplus_k H_k$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This should then frame the discussion. We are talking about insulators in the band sense, so there are no level-crossings, i.e. things are gapped. Let's assume that the Brilluoin zone is topologically a torus (i.e. we're going to work in 2D, rather than 3). The Hamiltonian $H_k$ should be a smooth function of $k$. Topological insulators are fundamentally to do with the possible topologies of $H_k$ as a function over the BZ; there is a theorem to the effect that there are topological invariants which are preserved as long as no level-crossing occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose that our material has a non-trivial topology in $H_k$ in the bulk. In vacuum, or generally against any other material with a trivial topology, we have to make the levels match up in the region of the boundary, which necessitates some level-crossing, i.e. closing the gap. This then generically demonstrates that there will be metallic states at the edges of topological insulators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this should be seen with caveats. I've swept a lot under the rug here (interactions, disorder, local vs bulk behaviour, etc.) I've also not treated things with the greatest possible generality (3D, other topologies of BZ's) or even rigour (I've simply ignored the crucial issue of where the Fermi level is --- mostly out of laziness). But maybe it helps. :-)&lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-01-24T16:26:27.060" />
  <row Id="3772" PostTypeId="2" ParentId="1869" CreationDate="2011-01-24T16:30:06.100" Score="1" ViewCount="" Body="&lt;p&gt;This question is a little confusing.  There are really no reasons to &quot;suspect it may be wrong&quot; (and there are surely many reasons to suspect it may be right!) in the sense you're probably thinking of.  It seems you're more likely asking the very different question: &quot;what things would we like string theory to do, but do not know how to do in it.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, of course, the list is very long because we would like to do everything with it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More concisely, the obvious point is that we'd like to be able to explicitly construct a model in string theory which reduces, at low energy, to the Standard Model plus classical gravity, and we do not know how to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could potentially wonder, as in the case of the old Kaluza-Klein theories, whether there is an obstruction to getting the correct structure.  In Kaluza-Klein, everything looks promising until you (or, Ed Witten) prove that while the theory does permit the correct gauge groups, it does not permit the correct representations (you don't get chiral theories).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in this case, we know that we can get both the right groups and representations out.  As far as I know, there are no good reasons to suspect that any particular property of the Standard Model is not allowed in string theory.  In fact, many models very similar to the Standard Model can be constructed, so in a sense, we can qualitatively get the right kinds of answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also have things like the AdS/CFT correspondence, which allows us to explicitly construct stringy duals of some ordinary field theories, so string theory is at least as right as those field theories are!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would say there are no theoretical reasons to expect string theory to be &quot;incorrect&quot; in any reasonable way, and there are only marginally good ones to suspect that it could be unable to give us the Standard Model.&lt;/p&gt;&#xA;" OwnerUserId="1403" LastActivityDate="2011-01-24T16:30:06.100" CommentCount="1" />
  <row Id="3773" PostTypeId="1" AcceptedAnswerId="3786" CreationDate="2011-01-24T16:43:51.553" Score="9" ViewCount="409" Body="&lt;p&gt;I've been looking at the, now very popular, graph of the SM Higgs decay branching ratios:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www-cdf.fnal.gov/physics/exotic/r2a/20050623.lmetbj_wh_tc/higgs_branch.jpg&quot; height=&quot;300&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see that the ZZ branching ratio has a funny dip around the $170\, GeV$, very different from the WW counterpart. It's true that graphs with Z will always have the $\gamma$ interference and, perhaps, this is the cause for this funny shape. However, the value of the mass for which this happens is &lt;strong&gt;not&lt;/strong&gt; something that I can intuitively explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that I have no idea of the right answer, let me ask: What causes this dip in the $H\rightarrow ZZ$ branching ratio around $M_H=170\, GeV$?&lt;/p&gt;&#xA;" OwnerUserId="1337" LastActivityDate="2011-03-14T06:38:59.597" Title="Shape of the Higgs branching ratio to ZZ" Tags="&lt;quantum-field-theory&gt;&lt;standard-model&gt;&lt;higgs-boson&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="3774" PostTypeId="2" ParentId="3759" CreationDate="2011-01-24T17:01:11.287" Score="3" ViewCount="" Body="&lt;p&gt;Category theory has some potential for physics.  Quantum logic I am less sanguine about.  It has always struck me as a way of expressing something we understand in set theoretic language.  It has always struck me as a formalistic study that brings little additional content.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Philip Goyal demonstrated how the above summation over intermediate points is equivalent in a discrete setting to the concatenation of measurements, where the summation is over all possible outcomes [1].  The complex amplitudes are products of complex numbers, and in a discrete setting this is a multiplication rule which requires complex numbers.  The result is that quantum mechanics is reduced to a simple system of associative and commutative mathematics of complex numbers with no reference to classical mechanics, or any notion of space or spacetime.  However, the one requirement is that the points intermediate to the initial and final points be intermediate in time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Goyal logic is a summation of Stern-Gerlach experiments.  The intermediate point corresponds to some intermediate measurement between the source of particles and the final Stern-Gerlach (SG) apparatus.  If the outcome of the intermediate SG apparatus is ignored, or no measurement is performed of their outcomes, the split beams recombine as a discrete summation.  So the intermediate SG apparatus represents a sum over elementary quantum events.  This summation in the complex algebra corresponding to this logic recovers the quantum superposition.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These summations over SG experiments occur in a sequence.  There is no ambiguity in the ordering of these events.  Further, the process appears well defined in a discrete setting.  The Goyal approach for discrete quantum mechanics, even if the number of elements is enormous, but not infinite, indicates some sort of quantization of time, and a discrete spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has elements of Zariski topology.   Consider the affine space $A_n$ as the n-dimensional space over a closed field $F$. The topology is constructed from closed sets defined by the polynomial set $S~\in~F$ by&#xA;$$&#xD;&#xA;V (S)~=~\{x~\in~A_n| f(x)~=~0;~\forall f~ \in~S\}&#xD;&#xA;$$&#xA;For two polynomials in the set S we have the following rules:&#xA;$$&#xD;&#xA;V (p_1)\cup V (p_2)~=~V (p_1\times p_2),~&#xD;&#xA;V (p_1)\cap V (p_2)~=~ V (p_1~+~p_2)&#xD;&#xA;$$&#xA;which serve as the representation map between the logic of outcomes and the algebra of quantum operations demonstrated by Goyal. This closed set topology defines the Zariski topology on the affine set $A_n$. So a connection to quantum mechanics exists within this system with respect to Zariski topology. This is the topology of {\’E}tale and Grothendieck, or topos theory. An overview of topos theory in physics is in [2] by Isham.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] P. Goyal, K. H. Knuth, J. Skilling, &quot;Origin of Complex Quantum Amplitudes and Feynman's Rules,&quot; {\it Phys. Rev.} {\bf A 81}, 022109 (2010) &lt;a href=&quot;http://arxiv.org/abs/0907.0909&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0907.0909&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] C. J. Isham, &quot;Topos Methods in the Foundations of Physics,&quot; {\it&quot;Deep Beauty,} ed. Hans Halvorson, Cambridge University Press (2010) &lt;a href=&quot;http://arxiv.org/abs/1004.3564&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1004.3564&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-24T17:01:11.287" CommentCount="2" />
  <row Id="3776" PostTypeId="2" ParentId="3773" CreationDate="2011-01-24T17:16:28.577" Score="2" ViewCount="" Body="&lt;p&gt;Nice question. I did notice this &quot;dip&quot; too, but didn't think too much about it. While  the explanation seems to be pretty simple:&lt;br&gt;&#xA;$M_W \simeq 80\,\mbox{GeV},\quad M_Z \simeq 91\,\mbox{GeV}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, around 170 GeV you are below the threshold for ZZ, but above the threshold for WW.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I also think that this dip tends to be overemphasised on such a graph. First because of the logarithmic scale, and also since the branching ratio is the &lt;em&gt;relative&lt;/em&gt; to the full width.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: In response to Luboš Motl's response I'll add two more points:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The reference to the  original of the plot: &lt;a href=&quot;http://arxiv.org/abs/hep-ph/9704448&quot; rel=&quot;nofollow&quot;&gt;hep-ph/9704448&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;In fact, below the threshold it is safe (with 1% accuracy) to assume that one boson is on shell.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="386" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2011-01-24T18:31:37.397" LastActivityDate="2011-01-24T18:31:37.397" />
  <row Id="3777" PostTypeId="1" AcceptedAnswerId="3791" CreationDate="2011-01-24T17:22:24.513" Score="8" ViewCount="292" Body="&lt;h3&gt;How much information is  contained in one qubit?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;A qubit is defined in Wikipedia as $a\left|0\right&gt; +b\left|1\right&gt;$, where a and b are complex numbers subject to $a^2 + b^2 = 1$.&lt;br&gt;&#xA;One complex number is equivalent to two real numbers, which suggests that a qubit is equivalent to four real numbers worth of information.&lt;br&gt;&#xA;Somehow I doubt this is true.&#xA;How much information is contained in one mathematical qubit, exactly transmitted?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;How much information is  contained in one physical photon?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Alice and Bob share an arbitrarily but not infinitely accurate clock.&#xA;They also have a shared cordinate frame (arbitrarily accurate relative location, no relative motion, no acceleration)&#xA;Alice and Bob can also have an arbitrarily complex shared coding system.&#xA;Alice is allowed to send one and only one photon to Bob between 12:00 and 1:00 oclock, and Bob receives it.&#xA;Alice sends the photon directly to Bob through an intervening vacuum.&#xA;If it makes a difference, assume the wavelength is exactly 500 nm.&lt;br&gt;&#xA;How much information can Alice send Bob?&lt;br&gt;&#xA;I assume it is not infinite, but only limited by the precision of the clocks,  the availability of bandwidth, Alices range of transverse motion, the accuracy of Bob's telescope, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume the photon is delivered by a single mode optical fiber, so Bob can gain no information by measuring the direction of the incoming photon.&#xA;Also assume that Eve puts in a variable delay of two to three hours, so Bob receives the photon between 2:00 and 4:00 and thus cannot gain any information from the time of receipt.&#xA;How much information can Alice now send Bob?&lt;br&gt;&#xA;Is it exactly one qubit, or something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We all know the plan where Alice transmits vertical for 0 and horizontal for 1,  Bob measures exactly vertical or exactly horizontal and one bit per photon is transmitted.&lt;br&gt;&#xA;Is there any improvement possible?&lt;/p&gt;&#xA;" OwnerUserId="1172" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-24T19:08:28.807" LastActivityDate="2011-02-02T15:40:43.707" Title="Informational capacity of qubits and photons" Tags="&lt;quantum-optics&gt;&lt;quantum-information&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="3778" PostTypeId="2" ParentId="3769" CreationDate="2011-01-24T17:30:49.360" Score="1" ViewCount="" Body="&lt;p&gt;First of all -- check &lt;a href=&quot;http://en.wikipedia.org/wiki/Metallicity&quot; rel=&quot;nofollow&quot;&gt;the metallicity&lt;/a&gt;, then &lt;a href=&quot;http://en.wikipedia.org/wiki/Nucleocosmochronology&quot; rel=&quot;nofollow&quot;&gt;the nucleocosmochronology&lt;/a&gt;. That gives the rough idea about the topic. And also that the topic is a currently developing area in cosmology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that the concise and experimentally confirmed theory of formation of first stars as well as theory of &lt;a href=&quot;http://en.wikipedia.org/wiki/Reionization&quot; rel=&quot;nofollow&quot;&gt;reionization&lt;/a&gt; are required to answer your question. And there no such theory (yet).&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-01-24T17:30:49.360" />
  <row Id="3779" PostTypeId="2" ParentId="3767" CreationDate="2011-01-24T17:43:54.013" Score="0" ViewCount="" Body="&lt;p&gt;I suspect a problem people have with TIQM is that Wheeler-Feynman absorber is woven into this.  The advanced potentials and the rest are a bit much for many to swallow.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-24T17:43:54.013" />
  <row Id="3780" PostTypeId="2" ParentId="3759" CreationDate="2011-01-24T17:48:47.940" Score="1" ViewCount="" Body="&lt;p&gt;My short answer is No, they're not too useful, but let me discuss some details, including positive ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Categories, especially derived categories, have been appropriate to describe D-brane charges - and not only charges - beyond the level accessible by homology and K-theory. See e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0104200&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0104200&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I feel it is correct to say that the string theorists who approached D-branes in this way did so because they first learned lots of category theory - in mathematics courses - and then they tried to apply their knowledge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure that a physicist would &quot;naturally&quot; discover the categories - or even formulated them in the very framework how they're usually defined and studied in mathematics. And on the contrary, I guess that the important qualitative as well as quantitative insights about the D-branes - including the complicated situations where category theory has been relevant - could have been obtained without any category theory, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course, people have different reactions to these issues and these reactions reflect their background. And I - a non-expert in category theory - could very well be missing something important that the category theory experts appreciate while others don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most famously, Joe Polchinski - the very father of the D-branes - reacted wittily to the notion that the D-branes should have been rephrased in terms of category theory. In a talk, he spoke about an analogy with a dog named Ginger. We tell Ginger not to do many things and do others, Ginger. What Ginger hears is &quot;blah blah blah blah Ginger blah blah blah&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a similar way, Polchinski reprinted &quot;what mathematicians say&quot;. It was a complicated paragraph about derived categories and their advanced methodology applied to D-branes. What Joe hears is &quot;blah blah blah blah D-branes blah blah blah blah T-duality blah blah D-branes blah.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some physicists also try to generalize gauge theory to some &quot;higher gauge theory&quot; using category theory but I don't think that there are any consistent and important theories of this kind. What they're doing is similar to the theories with $p$-forms and extended objects except that they don't do it right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As always, category theory may offer one a rigorous language to talk about analogies etc. - but I don't think that physicists need anything beyond the common-sense understanding how the method of analogies works. So if you learn category theory - which is pretty tough - I think you should have better reasons than a hope that the theory could be useful for physics.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-05T11:45:27.693" LastActivityDate="2011-02-05T11:45:27.693" CommentCount="2" />
  <row Id="3781" PostTypeId="2" ParentId="3767" CreationDate="2011-01-24T17:57:16.447" Score="6" ViewCount="" Body="&lt;p&gt;It's a combination of all these things and more. Most importantly, the TIQM interpretation is nonsense and all the positive words you hear about it are just unjustifiable hype promoted purely by John Cramer himself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ontology - or &quot;realism&quot;, as it is technically called in quantum mechanics - has been falsified in physics in the mid 1920s and it can never be &quot;unfalsified&quot; again. It's established that objects don't possess well-defined properties before they're measured. This insight continues to be hard to swallow for many people - however it doesn't mean that there is anything questionable about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The TIQM merges all the usual misconceptions about the &quot;real wave function&quot; with some very special inconsistencies such as retrocausality - the influence of the future on the past - that is obtained by a bizarre interpretation of the Feynman-Wheeler theory, a theory that turned out to be incorrect by itself (although it helped to stimulate Feynman and others to find the right rules of quantum field theory). Concerning the Feynman-Wheeler theory, its historical role, and some of its problems, see&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/2660/wheeler-feynman-theory-qed-without-fields-vacuum-polarization&quot;&gt;Wheeler-Feynman theory, QED without fields, vacuum polarization&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T07:50:43.177" LastActivityDate="2011-02-03T07:50:43.177" CommentCount="2" />
  <row Id="3782" PostTypeId="2" ParentId="3769" CreationDate="2011-01-24T18:01:11.917" Score="4" ViewCount="" Body="&lt;p&gt;A very qualitative way to look at it:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The Earth, and therefore &lt;em&gt;you&lt;/em&gt; are formed of the same material that contributes to the metallicity of our Sun&lt;/li&gt;&#xA;&lt;li&gt;Our Sun is a population I star, which means that it has a relatively high metallicity indicative of having formed after the heavy and short lived star of population II had already had their big blow offs.&lt;/li&gt;&#xA;&lt;li&gt;The population II stars divide into early and late groups, and all post-date the assumed population III stars.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;From this I conclude that a non-trivial number of the nucleons in your body have been part of &lt;em&gt;a few&lt;/em&gt; stars. Maybe as many as five. As Georg notes there has been time for the most prolific path to include many stars (dozens?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly all the carbon, nitrogen, oxygen and trace elements that make up your body have been part of &lt;em&gt;at least&lt;/em&gt; one star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of these facts shed much light on the average star-membership-history of the nucleons that make up your body.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-01-24T18:10:08.943" LastActivityDate="2011-01-24T18:10:08.943" CommentCount="1" />
  <row Id="3783" PostTypeId="2" ParentId="3769" CreationDate="2011-01-24T18:05:36.717" Score="3" ViewCount="" Body="&lt;p&gt;The &quot;generations&quot; of the stars are surely not as sharply separated as you suggest. It's true that the percentage of heavier elements - or &quot;metallicity&quot; - is increasing as the stars continue in their thermonuclear fusion. But most of the stuff that our Sun burns is still Hydrogen - and it's the same &quot;initial generation&quot; Hydrogen as any previous stars were using. Stars are being born continuously and the metallicity is usually a little bit higher than for the previous star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the era of &quot;nucleosynthesis&quot; that ended about 3 minutes after the Big Bang, mostly light elements were created in a thermal equilibrium at huge, &quot;nuclear&quot; temperatures that existed at those times. A good theory of nucleosynthesis predicts that most of the elements in the Universe should be hydrogen, with a smaller amount of helium and some trace amounts of lithium and other elements. The observations confirm that the predictions are pretty much accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heavier elements were created in previous stars. But it is not correct to suggest that the Sun is &quot;almost entirely&quot; built out of a recycled material. Quite on the contrary: it's more accurate to say that the Sun is mostly built from the hydrogen that was created in the first minutes after the Big Bang - and it is also &quot;contaminated&quot; by metals and other heavier elements from the previous stars. Those impurities are important for our life - and industry - but they're not important for the ability of the Sun to burn.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T07:49:30.337" LastActivityDate="2011-02-03T07:49:30.337" CommentCount="6" />
  <row Id="3784" PostTypeId="2" ParentId="3769" CreationDate="2011-01-24T18:16:58.063" Score="5" ViewCount="" Body="&lt;p&gt;Its likely a complicated mix, possibly including some virgin material, clouds of which occasionaly are absorbed into the galaxy, and some material that may have multiple recyclings. And not all the material that goes into a star is reacted before its blown off, so some (maybe much) of the hydrogen in the sun could have once been in the outer layers of another star. And even some first generation starts (pop III) that were low mass still haven't completed their lifecycles, so it is just a chaotic mess of gas and dust that gets pushed around by gravity and stellar winds for billions of years, plus some new gas that has recently fallen into the Milky Way all mixed up, but not so well mixed up as to be of uniform composition.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-01-24T18:16:58.063" />
  <row Id="3785" PostTypeId="2" ParentId="3718" CreationDate="2011-01-24T18:18:14.303" Score="0" ViewCount="" Body="&lt;p&gt;A complex number is an ordered pair of real numbers (a,b), that is also exactly the definiton of vectors in a two-dimensional real vector space. Most vectors in physics are not two dimensional, and in QM they need not even be real and can have infinite dimensions. Any vector can be divided by a scalar quantity. Vector division by vectors is not a well-defined concept, probably because it is useless. If you have an ordered set of complex numbers you can represent any vector you like as complex numbers, you just map one coordinate as (a,b), where b is 1 if a is to be taken as real, and 0 otherwise, this is a trivial concept and reflects the asinine nature of the question.&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-01-24T18:18:14.303" CommentCount="1" />
  <row Id="3786" PostTypeId="2" ParentId="3773" CreationDate="2011-01-24T18:20:18.820" Score="5" ViewCount="" Body="&lt;p&gt;The dip of the ZZ branching ratio - and all other branching ratios except for the WW branching ratio - near 170 GeV is caused by the increase of the total width (decay rate) at those masses. And the total width (decay rate) around 170 GeV increases exactly because the Higgs decays to the WW final states start to become possible. Because the total width goes up, the ratio of a (non-WW) partial width and the total width goes down - and this ratio is what we call the branching ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kostya wrote almost the same thing. But I want to emphasize a subtle point: note that in your graph, the branching ratio to WW is nonzero already from Higgs masses at 80 GeV or so. Similarly, the branching ratio to ZZ is nonzero from 90 GeV. How can a 90 GeV Higgs decay to two Z's, each of which has mass close to 90 GeV? Doesn't it violate energy conservation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that the graph shows the decays to off-shell particles, not the final states. A Higgs boson may decay to one virtual and one real Z-boson. The virtual particle continues in its decay. To check this hypothesis, note that all decay channels in the graph are composed out of two particles. The actual final states of the decay will often include (many) more than two particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the Higgs mass exceeds two times the mass of the W-bosons, the total width genuinely goes up because there's suddenly a lot of new &quot;phase space&quot; of the final states.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-14T06:38:59.597" LastActivityDate="2011-03-14T06:38:59.597" CommentCount="2" />
  <row Id="3787" PostTypeId="2" ParentId="1637" CreationDate="2011-01-24T18:21:56.430" Score="0" ViewCount="" Body="&lt;p&gt;&quot;Calculating&quot; is some word that can understood very differently.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;from &quot;scratch&quot; ie ab initio, only using physical values, nothing &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;from chemistry level. This was possible when I studied for the &lt;/p&gt;&#xA;&#xA;&lt;p&gt;molecule H2+ only. Today computers maybe are able to do such &lt;/p&gt;&#xA;&#xA;&lt;p&gt;calculations for molecules somewhat more complicated, but I do not &lt;strong&gt;know.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;from spectroscopic data, this is rather easy for 2-atomic molecules, &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;when spectroscopic data are simple. (statistical mechanics)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Method of Benson. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;this is an incremental method, the increments are derived from &lt;/p&gt;&#xA;&#xA;&lt;p&gt;experimental values for organic molecules where thermodynamic values are &lt;/p&gt;&#xA;&#xA;&lt;p&gt;known. The method is a kind of inter/extrapolation from known &lt;/p&gt;&#xA;&#xA;&lt;p&gt;to unknown. The increments are such for atoms, atom groups and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;types of bonds, sometimes some intramolecular non-bonding interactions &lt;/p&gt;&#xA;&#xA;&lt;p&gt;are put into calculation. One needs the the structural formula to start with, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;of course the Benson method is for organic molecules only. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The incentive is, that since the 30ties frequency of measurements of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;heats of combustion and specific heats has decreased, since the 50ties, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is almost zero. For the basic molecules, especially those conected &lt;/p&gt;&#xA;&#xA;&lt;p&gt;to mineral oil technology most thermodynamic values exist. But outside of this, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You look for data in vain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Benson_group_increment_theory&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Benson_group_increment_theory&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Analogon to Born-Haber cycle &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is a method used when You know all other &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thermodynamic values. Not often appicable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To convert values per mole in values per mass or volume &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is elementary, isn't it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Georg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I forgot another method: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So called &quot;force field calculation&quot;. Molecules are modelled &lt;/p&gt;&#xA;&#xA;&lt;p&gt;mechanically, between atoms (and atom groups) potentials &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for vibrations and internal rotations and nonbonding interactions &lt;/p&gt;&#xA;&#xA;&lt;p&gt;are set up and calculated for a set of  initial  distances and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;angles. This is repeated for different sets of distances etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some method for search of  overall minimum for total energy is applied, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;because the number of variables is several  dozen even for a &lt;/p&gt;&#xA;&#xA;&lt;p&gt;molecule of half a dozen atoms this is the crucial part. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I studied in the 70ties, the big machine of our university &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ran  some hours to calculate a molecule of 12 C atoms.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-01-24T20:33:52.323" LastActivityDate="2011-01-24T20:33:52.323" />
  <row Id="3788" PostTypeId="2" ParentId="3767" CreationDate="2011-01-24T18:27:25.997" Score="10" ViewCount="" Body="&lt;p&gt;Nobody has explained to me how Shor's quantum factorization algorithm works under the transactional interpretation, and I expect this is because the transactional interpretation cannot actually explain this algorithm. If it can't, then chances are the transactional interpretation doesn't actually work. (I have looked at some of the papers that purport to explain the transactional interpretation, and have found them exceedingly vague about the details of this interpretation, but assuming this interpretation is actually valid, maybe somebody else with more determination could figure these details out.)&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2011-01-24T18:27:25.997" CommentCount="12" />
  <row Id="3789" PostTypeId="2" ParentId="3777" CreationDate="2011-01-24T18:38:54.223" Score="-3" ViewCount="" Body="&lt;p&gt;If there is only one qubit with a probability $P(q)~=~1$ the Shannon information or entropy&#xA;$$&#xA;S~=~-\sum_ip_ilog(p_i) = -1~log(1)~=~0.&#xA;$$&#xA;If you include the vacuum with $P(0)$ and $P(1)$ the entropy is nonzero.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-24T18:38:54.223" CommentCount="4" />
  <row Id="3790" PostTypeId="2" ParentId="3777" CreationDate="2011-01-24T18:44:48.197" Score="5" ViewCount="" Body="&lt;p&gt;A qubit &lt;em&gt;is&lt;/em&gt; a unit of information, so the information included in one qubit is exactly one qubit. For most purposes, the information may be identified with the information of one bit. We call it a &quot;quantum bit&quot; because the two possibilities may be combined into arbitrary complex combinations such as $a|0\rangle + b|0\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the complex amplitudes $a,b$ cannot be measured - at least not by a single measurement. You can measure whether the system described by the qubit is found in the state $|0\rangle$ or $|1\rangle$ (and you may also do similar measurements with any orthogonal basis of the 2-dimensional Hilbert space). If you do so, you obtain either 0 or 1, nothing else. You obtain 0 with the probability $|a|^2$ and 1 with the probability $|b|^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you choose another basis and make the measurement, you get just one bit of information from the measurement. More precisely, you usually get less than one bit - namely $-p_a\ln(p_a)-p_b\ln(p_b)$ where $p_{a}=|a|^2$ and similarly for $b$. The expression is maximized for $p_a=p_b=1/2$ where its value is $\ln(2)$ of information (in &quot;nats&quot;) which is called one bit. For asymmetric choices of $p_a,p_b$, we get less than one bit of information from the measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you repeat the same experiment many times, or $K$ times, you may measure the amplitudes with the relative error of $1/\sqrt{K}$. However, it's the very point of quantum computation - the discipline where the notion of &quot;qubit&quot; actually becomes useful - that you only want to run the quantum algorithm once and get the result. If you needed to run the quantum algorithm many times, to measure the amplitudes, all the magic exponential speedup of quantum computers would be gone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, even if you repeatedly try to measure the amplitudes, you won't find the equivalent of four real numbers. First of all, the wave function of the qubit has to be normalized so that the total probability of both/all possibilities is equal to 100 percent. It means that &#xA;$$|a|^2+|b|^2=1.$$ &#xA;Moreover, the change of the overall phase i.e. the transformation&#xA;$$(a,b)\to (a\,\exp(i\phi),b\,\exp(i\phi))$$&#xA;is unphysical because the overall phase is unmeasurable by any tools, even in principle. So in fact, the number of measurable information - if you allow repeated measurements of the system in the same initial state - is not four real numbers but just two real numbers. Without a loss of generality, you may write&#xA;$$(a,b) = (A,e^{iB} \sqrt{1-A^2})$$&#xA;where $A,B$ are two real non-negative parameters. $A$ is between $0$ and $1$ while $B$ is between $0$ and $2\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I said that it's fallacious to imagine that the amplitudes $a,b$ are classical numbers. They're not classical numbers in any sense - it's always the same debate about the wave function that most laymen always want to imagine as a &quot;real classical wave&quot;. It's not a real classical wave. It's just a tool to predict probabilities. And one qubit is not the same thing as a &quot;large number of classical bits&quot;. Instead, it is one classical bit that has certain qualitatively new properties. As long as you will try to imagine that quantum mechanics is the same thing as classical physics, just with some bigger objects, you will misunderstand the essence of quantum physics. Quantum physics is qualitatively different than anything we know from the (seemingly) classical world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alice, Bob, and photon&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Alice and Bob send signals to one another and the only information they can adjust is the timing of one photon, then the information carried by this timing is simply $\ln(N)/\ln(2)$ bits where $N$ is the number of &quot;moments&quot; or &quot;intervals&quot; that they can distinguish. If you constrain them by what they're allowed to measure, then it's meaningless to talk about quantum bits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What they send to each other by the timing of the photon is some ordinary classical information. You would have to allow Alice and Bob to measure the interference of various possibilities - various timings - to turn them into a quantum computer, and then it could make some sense to talk about &quot;qubits&quot;. However, as you designed it, the information is classical and the unit should be called one bit.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T15:40:43.707" LastActivityDate="2011-02-02T15:40:43.707" />
  <row Id="3791" PostTypeId="2" ParentId="3777" CreationDate="2011-01-24T18:50:06.257" Score="10" ViewCount="" Body="&lt;p&gt;For (1), there is &lt;a href=&quot;http://en.wikipedia.org/wiki/Holevo%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;a theorem of Holevo&lt;/a&gt; that implies you cannot extract more than one bit of information from one qubit. You can indeed encode one bit of information, since the two inputs $| 0 \rangle$ and $| 1 \rangle$ (or any two orthogonal states) are distinguishable. If the sender and receiver share an entangled state, they can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Superdense_coding&quot; rel=&quot;nofollow&quot;&gt;superdense coding&lt;/a&gt; to send two bits using one qubit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (2), if the wavelength is exactly 500 nm (or at least as close as you can get in one hour), then by the uncertainty principle the photon must consist of an hour-long wave train which is a sine wave whose frequency gives 500 nm, and which (aside from its polarization) carries no information. You need non-zero bandwidth to transmit information. If you have a non-zero bandwidth, you can transmit information based on the frequency of the photon. The amount of information you can transmit depends on the bandwidth. The timescale of one hour limits the precision with which you can determine frequency, and this will give you the amount of information you can send. Note that if you use frequency to transmit the information, Alice's random delay doesn't hurt you. If you didn't have a random delay, then you could use the non-zero bandwidth to create wave packets that are localized in time, and use timing to transmit the same amount of information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's do some calculations. Let's suppose the frequency is 500 nm $\pm$ .5 nm; then the frequency is 6*10$^{14}$ $\pm$ 6*10$^{11}$ Hz, and the bandwidth is 1.2*10$^{12}$ Hz. Now, we have $TE \geq \hbar$, and $E = h \nu$, so $T\nu \geq \frac{1}{2\pi}$. If $T$ is 3600 sec, we can distinguish frequencies to accuracy 4.4*10$^{-5}$ Hz, so we get 2.7*10$^{16}$ different frequencies we can distinguish. Taking the log (base 2), this is around 55 bits. I suspect this is the most information that can be sent with one photon with the given bandwidth, but I don't know a proof of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all theoretical; doing this in practice would require a ridiculously accurate frequency filter. If you didn't have the random delay, using wave packets localized in time and a very accurate clock would work better. Even with the random delay, I wouldn't be surprised if there were clever, experimentally more feasible, ways of communicating with one photon.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditorDisplayName="" LastEditDate="2011-01-24T20:44:48.497" LastActivityDate="2011-01-24T20:44:48.497" CommentCount="2" />
  <row Id="3792" PostTypeId="1" AcceptedAnswerId="3860" CreationDate="2011-01-24T19:03:32.377" Score="4" ViewCount="436" Body="&lt;p&gt;What are the current most important &lt;em&gt;theoretical&lt;/em&gt; problems on quantum entanglement?&#xA;What is that we don't yet understand about how it works?&#xA;(Not considering interpretation etc problems)&lt;/p&gt;&#xA;" OwnerUserId="824" LastActivityDate="2011-01-25T17:13:18.387" CommunityOwnedDate="2011-01-24T19:37:04.507" Title="Unsolved entanglement problems" Tags="&lt;quantum-mechanics&gt;&lt;entanglement&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="3793" PostTypeId="2" ParentId="3792" CreationDate="2011-01-24T19:15:35.147" Score="0" ViewCount="" Body="&lt;p&gt;quantum entanglement is a basic property of multi-body quantum mechanics - or any quantum mechanics with at least two degrees of freedom - that has been understood from the mid 1920s. So for 85 years, there have been no open theoretical problems related to quantum entanglement and it has been understood &lt;em&gt;that&lt;/em&gt; it works as well as &lt;em&gt;how&lt;/em&gt; any doable experiment involving entanglement works even though the modern terminology related to quantum information has only been used for a few decades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely&#xA;Luboš&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-24T19:15:35.147" CommunityOwnedDate="2011-01-24T19:37:04.507" CommentCount="13" />
  <row Id="3794" PostTypeId="1" AcceptedAnswerId="3849" CreationDate="2011-01-24T20:24:25.847" Score="8" ViewCount="278" Body="&lt;p&gt;I was reading this interesting recent review on arxiv about particle identification:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/1101.3276&quot; rel=&quot;nofollow&quot;&gt;Particle Identification&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In figure 2, there is an interesting comparison between the CMS and ATLAS calorimeter performances. I hope the paper author won't mind if I reproduce the image here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img809.imageshack.us/img809/1180/calo.png&quot; height=&quot;250&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see that the CMS EM calorimeter has much better resolution than the ATLAS one, and I'm certain that this is due to the almost unfair comparison between an homogenous and a sampling calorimeter. No surprise here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the ATLAS hadronic calorimeter has much better resolution than CMS's. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Why is that? Is this just a by-product of the reduced size of the CMS hadronic calorimeter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Are those resolutions final or can the collaborations make them better when they have more data? How?&lt;/p&gt;&#xA;" OwnerUserId="1337" LastEditorUserId="1337" LastEditorDisplayName="" LastEditDate="2011-01-24T20:29:04.237" LastActivityDate="2011-01-26T13:20:38.920" Title="ATLAS and CMS calorimeters" Tags="&lt;experimental-physics&gt;&lt;high-energy-physics&gt;&lt;large-hadron-collider&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="3795" PostTypeId="1" CreationDate="2011-01-24T20:41:15.273" Score="6" ViewCount="256" Body="&lt;p&gt;The title says it all. Why are snowflakes &lt;a href=&quot;http://en.wikipedia.org/wiki/Snowflake#Gallery&quot; rel=&quot;nofollow&quot;&gt;symmetrical in shape&lt;/a&gt; and not a mush of ice? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a property of water freezing or what? Does anyone care to explain it to me? I'm intrigued by this and couldn't find an explanation.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-25T06:13:16.557" Title="Why are snowflakes symmetrical?" Tags="&lt;water&gt;&lt;freezing&gt;&lt;ice&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" />
  <row Id="3796" PostTypeId="2" ParentId="3795" CreationDate="2011-01-24T20:54:22.860" Score="1" ViewCount="" Body="&lt;p&gt;You have an appropriate nickname, snowflake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When water freezes, you get ice. Ice, like many solid materials, forms a crystalline structure. In the case of water, the crystalline structure may be attributed to the hydrogen bond, a special kind of an attractive interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a big chunk of ice will have a crystalline structure - preferred directions, translational symmetry, and some rotational symmetries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/uBzD8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about a snowflake?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A snowflake differs from a big chunk of ice by its being small. Even more importantly, it is in the process of growing. You should think about the process in which the snowflake was created.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the beginning, it was small. A few atoms formed a small piece of crystal. Such a small piece of crystal almost always has some hexagonal (or different) symmetry. What happens if you wait for a little while and it continues to freeze?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, the water molecules are added to the crystal because it's energetically favored: vapor turns into ice - and you need vapor to create snowflakes because liquid water freezes differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nature adds one water molecule at a time. The molecules always try to choose the most energetically favored position on the frozen body. Because these laws of creation of a snowflake are symmetric with respect to the rotational symmetries, it follows that any symmetry that exists at the beginning - a hexagonal symmetry of a small number of molecules in the initial crystal - will be preserved. It's pretty much inevitable that all the arms are growing approximately equally, so the initial symmetry group is preserved and becomes a symmetry of a macroscopic object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more difficult question is actually why the snowflakes are so diverse and beautiful. Most likely, it is not an accident that they achieve one shape or another. Changing pressure, temperature, or humidity as a function of time changes the conditions that determine the optimal places where the new vapor molecule should be added. But even if pressure, temperature, and humidity depends on time, the hexagonal symmetry is still maintained.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-24T21:12:37.423" LastActivityDate="2011-01-24T21:12:37.423" CommentCount="4" />
  <row Id="3797" PostTypeId="1" AcceptedAnswerId="3809" CreationDate="2011-01-24T20:55:37.350" Score="1" ViewCount="128" Body="&lt;p&gt;Whereas I can calculate the &lt;strong&gt;Chern number&lt;/strong&gt; of a quantum state (or band) from the integration of the Berry curvature in all space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I infer the topology of the quantum state from this result? What is the physical meaning of a quantum state with non-zero Chern number?&lt;/p&gt;&#xA;" OwnerUserId="1545" LastEditorUserId="1545" LastEditorDisplayName="" LastEditDate="2011-01-24T21:20:56.540" LastActivityDate="2011-01-25T02:05:45.710" Title="How can I infer the topology of a quantum state (or band) from its Chern number?" Tags="&lt;quantum-mechanics&gt;&lt;berry-pancharatnam-phase&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="3798" PostTypeId="1" AcceptedAnswerId="3801" CreationDate="2011-01-24T21:25:42.300" Score="3" ViewCount="332" Body="&lt;p&gt;Watching quantum mechanics lectures and it was mentioned that it is pointless/meaningless to try to talk/question things that can not be tested/measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a principle? And if so what is it's name?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also does this apply to questions other than Quantum mechanics? E.g. does it make sense to ask if earth was the only object in universe if gravity would still exist? Although it seems intuitive to answer yes, yet it is something that can never be tested/verified. It seems almost as meaningless as to ask if the universe was only made of a unicorn would it have gravity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an analogy consider glottogony, at some point it was banned as it seemed to be an unanswerable problem. &lt;a href=&quot;http://en.wikipedia.org/wiki/Origin_of_language&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Origin_of_language&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2011-08-30T08:17:12.677" LastActivityDate="2011-08-30T08:17:12.677" Title="What is the name of the principle saying it is meaningless to talk/ask questions that can not be measured/tested?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="3799" PostTypeId="1" AcceptedAnswerId="3822" CreationDate="2011-01-24T21:33:19.330" Score="14" ViewCount="1172" Body="&lt;p&gt;Cold fusion is being mentioned a lot lately because of some new setup that apparently works. This is an unverified claim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See for example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://hardware.slashdot.org/story/11/01/24/1550205/Italian-Scientists-Demonstrate-Cold-Fusion&quot; rel=&quot;nofollow&quot;&gt;http://hardware.slashdot.org/story/11/01/24/1550205/Italian-Scientists-Demonstrate-Cold-Fusion&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.physorg.com/news/2011-01-italian-scientists-cold-fusion-video.html&quot; rel=&quot;nofollow&quot;&gt;http://www.physorg.com/news/2011-01-italian-scientists-cold-fusion-video.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.wipo.int/pctdb/en/wo.jsp?WO=2009125444&amp;amp;IA=IT2008000532&amp;amp;DISPLAY=DOCS&quot; rel=&quot;nofollow&quot;&gt;http://www.wipo.int/pctdb/en/wo.jsp?WO=2009125444&amp;amp;IA=IT2008000532&amp;amp;DISPLAY=DOCS&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.journal-of-nuclear-physics.com/files/Rossi-Focardi_paper.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.journal-of-nuclear-physics.com/files/Rossi-Focardi_paper.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;While we should give the scientific community time to evaluate the set up and eventually replicate the results, there is undoubtedly some skepticism that cold fusion would work at all, because the claim is quite extraordinary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the past, after Fleischmann and Pons announced their cold fusion results, in perfectly good faith, they were proven wrong by subsequent experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the experimental realities that make Fleischmann and Pons style cold fusions experiments easy to get wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the same risks apply to this new set up?&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-08-18T20:40:30.613" Title="Why is cold fusion considered bogus?" Tags="&lt;experimental-physics&gt;&lt;experimental-evidence&gt;&lt;cold-fusion&gt;" AnswerCount="13" CommentCount="9" FavoriteCount="3" />
  <row Id="3800" PostTypeId="2" ParentId="3754" CreationDate="2011-01-24T21:48:04.777" Score="3" ViewCount="" Body="&lt;p&gt;Try integrating the power in the resulting interference pattern; you will find that energy is indeed conserved.  In the case where the slit is illuminated by a plane wave and observed in the far-field, this is very easy:  the interference pattern is simply the Fourier transform of the pattern of slits, and conservation of energy is then given by Parseval's theorem!&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-01-24T21:48:04.777" />
  <row Id="3801" PostTypeId="2" ParentId="3798" CreationDate="2011-01-24T21:56:52.813" Score="9" ViewCount="" Body="&lt;p&gt;It's not really a single principle - it's a philosophy and in the context of philosophical discussions about science, it is usually known as positivism. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Positivism&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Positivism&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As any philosophy, it cripples the penetrating power of science if it is extended too far - and every philosophy ultimately fails. The thought experiment about the Earth in the Universe is just one among millions of examples. Positivism could have a problem with the whole concept of thought experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it was very useful and important for the development of quantum mechanics to realize that science doesn't &lt;em&gt;have&lt;/em&gt; to talk about things that can't be measured, i.e. that theories that deny the existence of things that can't be observed are just fine, it still remains true that science also &lt;em&gt;can&lt;/em&gt; talk about concepts that can't be observed, such as quarks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's up to the scientific method to decide whether an auxiliary concept or a theory that isn't accessible to observations has an explanatory power that justifies its validity - and the answer may be different in each individual situation.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-24T21:56:52.813" CommentCount="7" />
  <row Id="3802" PostTypeId="2" ParentId="3799" CreationDate="2011-01-24T22:02:47.387" Score="0" ViewCount="" Body="&lt;p&gt;Update : I guess it is the same reason that multi universe theory is considered bogus by some : &quot;fantastical speculation, disconnected from the reality that we can access empirically&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.scientificamerican.com/blog/post.cfm?id=is-speculation-in-multiverses-as-im-2011-01-28&quot; rel=&quot;nofollow&quot;&gt;http://www.scientificamerican.com/blog/post.cfm?id=is-speculation-in-multiverses-as-im-2011-01-28&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Old content down here :&#xA;I wanted to send this as a comment to your question but I don't have enough rep to do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the book &quot;Forbidden Science&quot;, Cold fusion is discussed, and the attitude of scientific community mentioned. The book &quot;The trouble with physics&quot; also gives a glimpse into the physics community.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe instead of wondering why it is considered bogus, should change it to considered bogus by some. I haven't been able to find a verification to the story that before the planes were invented there were &quot;mathematical&quot; proofs floating around that why it was impossible for machines made of metal to fly. ( any one has references for this? )&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-01-31T21:08:48.367" LastActivityDate="2011-01-31T21:08:48.367" CommentCount="9" />
  <row Id="3803" PostTypeId="2" ParentId="3798" CreationDate="2011-01-24T22:06:53.043" Score="1" ViewCount="" Body="&lt;p&gt;I'm not an expert on philosophical terminology, but I'd go for &lt;a href=&quot;http://plato.stanford.edu/entries/operationalism/&quot; rel=&quot;nofollow&quot;&gt;operationalism&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-01-24T22:06:53.043" />
  <row Id="3805" PostTypeId="2" ParentId="3799" CreationDate="2011-01-24T22:44:19.410" Score="5" ViewCount="" Body="&lt;p&gt;The Fleischmann and Pons device relied on calorimetry (measuring the energy balance in terms of heat) maintained over multiple day time spans to ascertain that something unexpected was happening in the cell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is experimentally tricky, as it requires high precision temperature measurements to be maintained against a consistent reference, and relies on an understanding of how the calorimeter may or may not be losing heat though unmonitored channels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short a large heat loss is being subtracted from a large heat input, and both measurements have some uncertainty. That's a warning sign in any experiment, but not a death knell &lt;em&gt;if&lt;/em&gt; the uncertainty can be quantified with sufficient accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation is complicated by the time dependent heat loss that Fleischmann and Pons reported. If real, that would indicate some unexpected process at work, though we have no way of know &lt;em&gt;a priori&lt;/em&gt; if that is fusion or some energy storage-and-release mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seem to have been lack of consistency in reports of energy gain or neutrons that swung opinion against any fusion actually happening in the F&amp;amp;P cell.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;BTW--People who kept at it for a few years after the consensus turned against were engaged in good science. There was always some small possibility that the process was dependent on some un-measured and un-controlled factor. To find out, one would need to accumulate a non-trivial set of working cells and then investigate how they were different from non-working cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as the man said &lt;em&gt;&quot;If at first you don't succeed, try, try again. Then quit. There's no use being a damn fool about it.&quot;&lt;/em&gt; Sooner or later you just have to give up.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-01-24T22:59:43.410" LastActivityDate="2011-01-24T22:59:43.410" CommentCount="4" />
  <row Id="3806" PostTypeId="2" ParentId="3799" CreationDate="2011-01-24T22:44:54.370" Score="5" ViewCount="" Body="&lt;p&gt;There are a few reasons.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) There was never any clear reason why electrifying palladium should create pressures sufficient to ignite a fusion reaction.  Without a mechanism, this seems to be the most ridiculously radical and sensational conclusion possible, even if the calorimetry says that electrified palladium creates net energy somehow.  Why not start with simpler explanations than saying that it's fusion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) the only evidence anyone has ever offered was calorimetry of a small-scale experiment.  But calorimetry is a complicated thing--you have to model the insulation of the system just right, and you have to measure the heat inputs for the system correctly, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) No fusion products have ever been observed--fusion would be plausible if they were able to show that the reactions produced some helium-3 or lithium or whatever.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) The setup they describe in these experiments seems quite simple--just submerge some palladium underwater, and run electricity through it.  If this really produces a fusion reaction, why not just scale it up and build a reactor?  I'm sure if they could power a building with a palladium fusion reactor (or even just a light bulb), every critic would instantly shut up and get behind them.  And they never seem to explain why they don't just do this.  I'm never going to believe in your revolutionary new form of cheap energy unless you just start producing cheap energy with it (or at least explain why you can't).  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-01-24T22:44:54.370" CommentCount="5" />
  <row Id="3807" PostTypeId="2" ParentId="3799" CreationDate="2011-01-25T00:20:26.830" Score="1" ViewCount="" Body="&lt;p&gt;If there had been a fusion process there would have also been neutron production.  There were some claims of neutrons, but they were not verified.  There is another problem with neutrons, if there was fusion Pons and Fleischmann would have been irradiated by neutrons.  They might in fact have gotten sick or died.  There was something amiss with the whole thing from the start.  If they were intending to get cold fusion, why did they not shield themselves?  If they did not intend to get cold fusion, but later suspected they had it, why did they not then shield against neutrons?  They played around with this set up for some time, and if they really thought they were getting fusion, why did they not set the apparatus behind lead bricks?  Either they were stupid or they were scamming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea that interionic lattice potential could force nuclei together is ridiculous.  There is a prospect for a bosonization or condensate physics.  Here the D or T enter into the same quantum state and might transition into H^4.  This is implausible though.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-25T00:20:26.830" CommentCount="4" />
  <row Id="3808" PostTypeId="2" ParentId="3798" CreationDate="2011-01-25T01:24:29.477" Score="0" ViewCount="" Body="&lt;p&gt;There is a difference between to measure and to test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If something can't be measured (by devices or the senses) it is not a scientific concept according to old positivism. Pure positivism in use was/is for instance operationalism which thus meant that a concept is meaningful only if we have some way to measure it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way a theory can or cannot be tested has more to do with the scientific method. According to Popper a theory is not scientific if it can't be falsified in contrast to positivists considering a theory scientific if it (among other things) can be proved true by confirmation.&lt;/p&gt;&#xA;" OwnerUserId="1391" LastActivityDate="2011-01-25T01:24:29.477" />
  <row Id="3809" PostTypeId="2" ParentId="3797" CreationDate="2011-01-25T01:25:09.473" Score="1" ViewCount="" Body="&lt;p&gt;I am in part trying to understand this myself.  The Berry phase is computed from differential forms, such as the one-forms $\omega$ constructed from states&#xA;$$&#xD;&#xA;\omega~=~\langle\psi|d\psi\rangle&#xD;&#xA;$$&#xA;and with the covariant differential $D~=~d~+~\omega$ the two-forms&#xA;$$&#xD;&#xA;\Omega~=~D\omega~=~d\omega~+~\omega\wedge\omega&#xD;&#xA;$$&#xA;The tensor components of the 2-form $F$ are elements of a self-adjoint principal bundle $P$.  The determinant of these elements&#xA;$$&#xD;&#xA;det\Big|1~+~\frac{ixF}{2\pi}\Big|~=~\sum_nc_jx^n&#xD;&#xA;$$&#xA;which is a characteristic polynomial which represents the Chern class.  Each $c_n(P)$ is an element of $H^{2n}(M)$.  So the curvature form for the Berry phase, or the Fubini-Study metric $\Omega=~dz\wedge d{\bar z}/(1~+~|z|^2)^2$ is evaluated $\int\Omega~=~2\pi i$ and gives $c_1~=~1$  So there is a nontrivial cocycle on the “2-level.  For this projective geometry there are alternating Betti numbers $1,~0$ for even and odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had some product of states $\prod_n |\psi_n\rangle$, say in an entangled state etc, you could apply the differential $d$ up to $n$ times and form and $n$-form.  For instance the product $|\psi_1,~\psi_2\rangle$ $=~|\psi_1\rangle|\psi_2\rangle$ defines the one-form&#xA;$$&#xD;&#xA;\omega~=~d|\psi_1,~\psi_2\rangle~=~d|\psi_1\rangle|\psi_2\rangle~+~|\psi_1\rangle d|\psi_2\rangle&#xD;&#xA;$$&#xA;and one could then build up a system of differential forms on various chains.  The analogue of the projective geometry for this is a $G_2(V)$ Grassmannian and this continues up for n-product spaces.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-01-25T02:05:45.710" LastActivityDate="2011-01-25T02:05:45.710" CommentCount="1" />
  <row Id="3810" PostTypeId="1" AcceptedAnswerId="3817" CreationDate="2011-01-25T02:07:13.343" Score="0" ViewCount="100" Body="&lt;p&gt;Proton cuircuits and proton motive force are part of standard discussion in biology and processes involving &lt;a href=&quot;http://www.pnas.org/content/102/27/9709.long&quot; rel=&quot;nofollow&quot;&gt;photosynthesis&lt;/a&gt;.  The sort of proton currents discussed in biology are obviously slightly different than the magnetically confined proton currents in an acclerator like LHC, however, although we are very familar with electronic circuitry, I was wondering if there are example of protonic circuitry outside of biological processess.&lt;/p&gt;&#xA;" OwnerUserId="1383" LastActivityDate="2011-01-25T08:55:24.513" Title="Examples of circuitry using proton currents" Tags="&lt;biophysics&gt;" AnswerCount="1" />
  <row Id="3811" PostTypeId="2" ParentId="1121" CreationDate="2011-01-25T02:28:56.933" Score="3" ViewCount="" Body="&lt;p&gt;NASA actually experimented with producing current in &lt;a href=&quot;http://www-istp.gsfc.nasa.gov/Education/wtether.html&quot; rel=&quot;nofollow&quot;&gt;tethers suspended from the space shuttle&lt;/a&gt;.  The current generated was sufficient to melt the tether.&lt;/p&gt;&#xA;" OwnerUserId="1383" LastActivityDate="2011-01-25T02:28:56.933" />
  <row Id="3812" PostTypeId="2" ParentId="3799" CreationDate="2011-01-25T02:45:27.500" Score="2" ViewCount="" Body="&lt;p&gt;Very simple! They can never repeat the results in a scientific way to demonstrate to others that it works. What's the point of science if we simply ignore the scientific method? If they did truly come up with something then they wouldn't have a need to be secretive and not show exactly what they did. If somehow they got it through accident then that is not science. Once they are able to reproduce the results and show others and others can reproduce them then it becomes useful and becomes science. (even if cold fusion is possible it is useless if we can't ever reproduce it)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why cold fusion is considered bogus is mainly because many people have tried and all have failed... and those that have claimed to succeed never proved they did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No one knows if cold fusion is possible but given all the schemes and cons that have happened in the past it's easier to be skeptical than not. If someone does discover some way then it will be VERY easy to prove and if they are true scientists they won't have any issues doing so.&lt;/p&gt;&#xA;" OwnerUserId="1551" LastActivityDate="2011-01-25T02:45:27.500" CommentCount="3" />
  <row Id="3813" PostTypeId="2" ParentId="3795" CreationDate="2011-01-25T02:54:58.040" Score="1" ViewCount="" Body="&lt;p&gt;Not all snowflakes are symmetrical. One can disrupt the symmetry quite easily by introducing impurities or some mechanical artifact. In nature, snowflakes have plenty of time to form and it is more natural for them to form symmetric shapes because of the molecular structure of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, when there is more time for the molecules to move about and position themselves they will do so in a way that is in accordance with some crystalline structure that the molecule can exhibit.(you can use other things besides water to get &quot;crystals&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nature of the shape depends on many many factors but to see that such diversity can come from something simple you just have to look at IFS's. Basically you take some very simple rules and generate a huge number of variations by making small changes in the rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Iterated_function_system&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Iterated_function_system&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1551" LastActivityDate="2011-01-25T02:54:58.040" CommentCount="1" />
  <row Id="3814" PostTypeId="2" ParentId="527" CreationDate="2011-01-25T02:59:56.117" Score="0" ViewCount="" Body="&lt;p&gt;The design frequency for microwave ovens is 2450 MHz.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-25T02:59:56.117" CommentCount="1" />
  <row Id="3815" PostTypeId="1" CreationDate="2011-01-25T03:17:18.293" Score="2" ViewCount="159" Body="&lt;p&gt;I need a solution to the heat equation that shows temp increase in an object, e.g. a cube or sphere, in sunlight.  The object is assumed to be exposed on all sides except one.  It is a solid object with a certain surface emissivity and heat capacity.  In other words, I don't care what is inside the object.  It absorbs some light from the sun and heats up, and heat is radiated and convected in air so that it reaches an equilibrium temperature.  The solution presumably uses the parabolic PDE and the Neumann boundary condition.  A Matlab formulation would be great!&lt;/p&gt;&#xA;" LastActivityDate="2011-04-25T22:41:35.823" Title="Solar heating of an object in air" Tags="&lt;thermodynamics&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" />
  <row Id="3816" PostTypeId="2" ParentId="3798" CreationDate="2011-01-25T03:32:19.913" Score="2" ViewCount="" Body="&lt;p&gt;Old answer: the &quot;verification principle&quot;.  The context was positivism, operationalism, or instrumentalism.  The idea was that if a statement cannot be empirically verified it cannot be said to have meaning.  This &quot;principle&quot; fell by the wayside when philosophers such as Bertrand Russell realized that the principle itself could not be empirically verified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recent: Please see the excellent discussion of this question in the context of string theory by Gordon Kane in the Nov. 2010 issue of Physics Today.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-25T03:32:19.913" />
  <row Id="3817" PostTypeId="2" ParentId="3810" CreationDate="2011-01-25T04:15:30.757" Score="1" ViewCount="" Body="&lt;p&gt;Current in any acid electrolyte (eg a lead accumulator) is carried &#xA;predominantely by  hydronium ions (real protons do not exist in water). &#xA;Due to the about tenfold mobility of hydronium ions compared to all other &#xA;cations, in acid solution charge transport is almost by hydronium alone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Georg &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-25T08:55:24.513" LastActivityDate="2011-01-25T08:55:24.513" CommentCount="7" />
  <row Id="3818" PostTypeId="1" AcceptedAnswerId="3821" CreationDate="2011-01-25T04:30:55.037" Score="0" ViewCount="132" Body="&lt;p&gt;Assuming I have a body travelling in space at a rate of 1000m/sec. Let's also assume my maximum deceleration speed is 10m/sec*sec. How can I calculate the minimum stopping distance of the body?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the formulas I can find seem to require either time or distance, but not one or the other.&lt;/p&gt;&#xA;" OwnerUserId="1581" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-25T04:40:37.833" LastActivityDate="2011-01-25T04:44:14.630" Title="Stopping Distance (frictionless)" Tags="&lt;kinematics&gt;&lt;acceleration&gt;" AnswerCount="2" />
  <row Id="3819" PostTypeId="2" ParentId="3815" CreationDate="2011-01-25T04:34:33.830" Score="1" ViewCount="" Body="&lt;p&gt;To the extent that convection is important, LOL. Temperature differences drive the fluid flows, which drive the heat loss. It would be a lot easier with a sphere than a cube, but the real problem is relating surface temperature with heat loss. If you put it in a vacuum, so you only had radiative losses, it would be easily tractable.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-01-25T04:34:33.830" />
  <row Id="3820" PostTypeId="2" ParentId="3818" CreationDate="2011-01-25T04:44:11.933" Score="2" ViewCount="" Body="&lt;p&gt;The formula you want is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_f^2 = v_i^2 + 2a(x_f - x_i)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's one of the basic kinematic formulas taught in high school (or even middle school) physics classes.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-01-25T04:44:11.933" CommentCount="6" />
  <row Id="3821" PostTypeId="2" ParentId="3818" CreationDate="2011-01-25T04:44:14.630" Score="1" ViewCount="" Body="&lt;p&gt;If the speed is $1000 m/s$ and the deceleration is $10 m/s$, it will take $100s$ to stop.  The average speed in that time is $500 m/s$, so the distance traveled is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$500m/s*100s = 5*10^4m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working through the same logic with an initial speed $v$ and a deceleration $a$, the final distance $d$ traveled before stopping is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d = v_{avg}*t = (v/2)*(v/a) = \frac{v^2}{2a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula becomes more interesting when you learn a bit more physics because it's simple example of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Work_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;work-energy theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-01-25T04:44:14.630" CommentCount="1" />
  <row Id="3822" PostTypeId="2" ParentId="3799" CreationDate="2011-01-25T04:53:26.740" Score="19" ViewCount="" Body="&lt;p&gt;This was beautifully answered theoretically right away at the 1989 APS session in NY, I think by Koonin.    Theoretically, for any sort of fusion one needs to overcome the Coulomb repulsion of the relevant nuclei, on the order of MeV in order to allow the nuclei to get close enough for their wave functions to overlap and fuse.  Because of the phenomenom of quantum mechanical tunnelling, this can be reduced to tens to hundreds of kev.    So temperatures of &gt;&gt; 10^5 K, or cold muons (which outweigh electrons by 200x) are required to reduce the internuclear distance (as in muon catalyzed cold fusion, a real phenomenon), or some other special mechanism is required to allow this close approach. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for any sort of chemically catalyzed fusion, i.e. via the valence electrons, to take place, the binding energy of the two H atoms to the catalyst would have to be so high, that the particular configuration of the low energy valence electrons, etc. would necessarily be entirely irrelevant to the problem, i.e. whatever their arrangement they could not possibly catalyze the fusionable nuclei to approach close enough to fuse.   So no clever packing arrangement, quasiparticles, special adsorbtion, special crystal lattice structures, etc. could ever alter this conclusion.   Whatever was happening at such low energy scales would appear as a kind of irrelevant fluff compared to the energy scale of the internuclear distance necessary for fusion.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore valence electron catalyzed cold fusion would violate the fundamental laws of quantum mechanics, nuclear physics, etc.   &lt;a href=&quot;http://www.nature.com/nature/journal/v340/n6228/abs/340045a0.html&quot; rel=&quot;nofollow&quot;&gt;Leggett and Baym&lt;/a&gt; also published an argument like this around the same time (summarized for free &lt;a href=&quot;http://books.google.com/books?id=wIp85LMKp-8C&amp;amp;pg=PA29&amp;amp;lpg=PA29&amp;amp;dq=fusion+barrier+leggett+baym&amp;amp;source=bl&amp;amp;ots=g8EWSOOoRR&amp;amp;sig=BbGxF0vg9bIsb0iBs3DzGv6iZXY&amp;amp;hl=en&amp;amp;ei=wpk-TYKUIIWglAeU4dDmAg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;sqi=2&amp;amp;ved=0CBcQ6AEwAA#v=onepage&amp;amp;q=fusion%20barrier%20leggett%20baym&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).  Koonin and Nauenberg published an accurate calculation &lt;a href=&quot;http://www.nature.com/nature/journal/v339/n6227/abs/339690a0.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, showing that if the mass of the electron were 5-10 times larger than it really is, chemically calalyzed fusion could work.   Note however, that the reaction rate depends on the electron mass very, very strongly, so that this remains impossible in our universe. &lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2011-02-26T04:07:13.527" LastActivityDate="2011-02-26T04:07:13.527" CommentCount="10" />
  <row Id="3823" PostTypeId="2" ParentId="3795" CreationDate="2011-01-25T06:08:54.850" Score="3" ViewCount="" Body="&lt;p&gt;K Libbrecht has a nice paper that answers your question in considerable detail and has some nice pictures-- his homepage:&#xA;&lt;a href=&quot;http://www.its.caltech.edu/~atomic/publist/kglpub.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.its.caltech.edu/~atomic/publist/kglpub.htm&lt;/a&gt;&#xA;Scroll down to the article in American Scientist in his publications list &#xA;&quot;The Formation of Snow Crystals,&quot; K. G. Libbrecht, American Scientist 95, 52-59 (2007). View pdf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the pdf is &lt;a href=&quot;http://www.its.caltech.edu/~atomic/publist/AmSci2007.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-01-25T06:13:16.557" LastActivityDate="2011-01-25T06:13:16.557" />
  <row Id="3824" PostTypeId="1" AcceptedAnswerId="3827" CreationDate="2011-01-25T06:25:32.313" Score="4" ViewCount="584" Body="&lt;p&gt;Given that the electrical force is so much stronger than gravitational force at atomic levels, why is it that it's the gravitational force between you and the earth that keeps you on the ground rather than the electrical force between you and the earth?&lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="41" LastEditorDisplayName="" LastEditDate="2011-01-25T21:24:56.447" LastActivityDate="2011-04-22T20:33:14.920" Title="Electrical force vs gravitational force" Tags="&lt;gravity&gt;&lt;electricity&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="3825" PostTypeId="2" ParentId="3824" CreationDate="2011-01-25T06:32:29.630" Score="1" ViewCount="" Body="&lt;p&gt;I assume we are not supposed to give direct answers to homework questions. If I'm wrong I'll be happy to edit this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mohabitar, to answer this you should consider what properties of an object cause electrical forces to be exerted on it. Do you posses these properties under normal circumstances? What about the same question as applied to gravitational force?&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-01-25T06:32:29.630" CommentCount="6" />
  <row Id="3826" PostTypeId="2" ParentId="1557" CreationDate="2011-01-25T07:01:18.587" Score="1" ViewCount="" Body="&lt;p&gt;Special relativity does not rule out tachyons which travel faster than the speed of light and whose speed increases with decreasing energy. Also, the Alcubiere drive (and metric) allows a warp bubble to travel (expand) and supraluminal speeds (providing one ignores theoretical problems with building one:)&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-01-25T07:01:18.587" CommentCount="5" />
  <row Id="3827" PostTypeId="2" ParentId="3824" CreationDate="2011-01-25T07:53:50.313" Score="8" ViewCount="" Body="&lt;p&gt;Because there's no negative gravitational mass. Electric charge can be neutralized, gravitational attraction not.&lt;/p&gt;&#xA;" OwnerUserId="1458" LastActivityDate="2011-01-25T07:53:50.313" />
  <row Id="3828" PostTypeId="2" ParentId="1971" CreationDate="2011-01-25T09:32:06.730" Score="1" ViewCount="" Body="&lt;p&gt;There's no simple and surely no analytic function describing the daily cycles of the temperature. It's not hard to see why: while you could add some higher harmonics etc., the actual process clearly contains many unsmooth points. During the day, when the Sun is visible, the temperature typically increases, and the rate of increase is maximum around the noon.  The maximum temperature during the day is reached around 2 p.m. in average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, after the sunset, the Sun's warming effect goes pretty much strictly to zero. So the second derivative of the temperature is discontinuous after the sunset. Moreover, the time of the sunrise and sunset depend on the seasons and latitude. You can't get any universal answer for what you're looking for, and even a function that would be OK would have to depend on several parameters and contain unsmooth functions such as the absolute value.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-25T09:32:06.730" />
  <row Id="3829" PostTypeId="2" ParentId="3751" CreationDate="2011-01-25T09:42:45.140" Score="4" ViewCount="" Body="&lt;p&gt;Roy, your wishful thinking is manifestly impossible. If the tensor $T_{\mu\nu}$ is discontinuous, and it surely is on the surface of a solid, then Einstein's equations guarantee that the Einstein tensor $G_{ab}$ is discontinuous as well - up to a normalization, it's the same tensor. It follows that the Ricci tensor and Riemann tensor, $R_{\mu\nu}$ and $R_{\kappa\lambda\mu\nu}$, must also be discontinuous because the Einstein tensor $G_{\mu\nu}$ can be easily calculated both from the Ricci tensor as well as from the Riemann tensor, so if the Ricci or Riemann tensor were continuous, the Einstein tensor would have to be continuous, which is an obvious contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just proved the opposite theorem that the Riemann tensor is discontinuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should realize that the Riemann tensor has a higher number of components than the Ricci (or Einstein) tensor, so its continuity - which means the continuity of all of its components - is an even stronger condition than the continuity of the Ricci (or Einstein) tensor. The argument above proves that none of these tensors is continuous in the presence of solids - which is why there can't be any theorem saying the opposite thing (it would be wrong). Another question is whether the Weyl tensor is continuous near such boundaries. I don't know the answer. The answer could be easily calculated from the very formula for the Weyl tensor. &lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-25T09:42:45.140" CommentCount="5" />
  <row Id="3830" PostTypeId="2" ParentId="1453" CreationDate="2011-01-25T10:17:02.273" Score="3" ViewCount="" Body="&lt;p&gt;Are there extra spatial dimensions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since many problems become exceptionally easier to solve with an assumption of extra spatial dimensions, I think this is one of the greatest questions of all time.&lt;/p&gt;&#xA;" OwnerUserId="1383" LastActivityDate="2011-01-25T10:17:02.273" CommunityOwnedDate="2011-01-25T10:17:02.273" />
  <row Id="3832" PostTypeId="2" ParentId="1453" CreationDate="2011-01-25T10:56:55.667" Score="7" ViewCount="" Body="&lt;p&gt;What is the explanation for all the seemingly arbitrary dimensionless parameters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is so special about &lt;img src=&quot;http://i.stack.imgur.com/gHelW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the biggest mystery of science today I think&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2011-01-25T11:07:58.683" LastActivityDate="2011-01-25T11:07:58.683" CommunityOwnedDate="2011-01-25T10:56:55.667" CommentCount="5" />
  <row Id="3833" PostTypeId="1" AcceptedAnswerId="3837" CreationDate="2011-01-25T11:00:13.717" Score="4" ViewCount="190" Body="&lt;p&gt;One of the great unheralded advances made in the history of science was the ability to determine the age of earth based on the decay of isotopic uranium.  Based on the apparent abundance of uranium in the early earth, what conclusions can be drawn about the star that proceded the sun?&lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T15:40:14.643" LastActivityDate="2011-02-02T15:40:14.643" Title="Age of the earth and the star that proceded the sun" Tags="&lt;astrophysics&gt;&lt;nuclear-physics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="3834" PostTypeId="1" AcceptedAnswerId="5356" CreationDate="2011-01-25T11:00:35.070" Score="2" ViewCount="196" Body="&lt;p&gt;Could you please help me understand how one can measure output frequency of free electron laser &lt;strong&gt;(provided that we know size of magnetic domains and electron energy)&lt;/strong&gt;? This should be a function of magnetic domains size &amp;amp; electron velocity? But electron velocity should drop due to photon emission - does that means that FEL have a &quot;wideband&quot; spectrum? Why military users focus on Röntgen radiation while UV in air transparancy window might have less air absorption?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bold is still unanswered :-)&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-09-04T21:59:16.657" LastActivityDate="2011-09-04T21:59:16.657" Title="Generated wavelength of free electron laser" Tags="&lt;laser&gt;&lt;wavelength&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="3836" PostTypeId="2" ParentId="2644" CreationDate="2011-01-25T11:47:13.907" Score="0" ViewCount="" Body="&lt;p&gt;It could just well be that Newton was given the idea of the inverse square law from a contemporary genius, Robert Hooke. See here for more details:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Newton&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Newton&lt;/a&gt;'s_law_of_universal_gravitation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Even if Hooke originated the idea (how?) it was Newton's genius that propelled the inverse square law to the far reaches of time and space!&lt;/p&gt;&#xA;" OwnerUserId="1413" LastActivityDate="2011-01-25T11:47:13.907" />
  <row Id="3837" PostTypeId="2" ParentId="3833" CreationDate="2011-01-25T11:49:06.153" Score="3" ViewCount="" Body="&lt;p&gt;Dear Humble, the cosmic nucleosynthesis (first three minutes) only produced hydrogen, helium, lithium, and beryllium. All heavier elements came from slow fusion inside living stars and, especially the heaviest ones, originate from dying stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/1phxn.gif&quot; alt=&quot;The periodic table&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regular fusion-related processes inside stars (see the list at the bottom of the answer) only produce elements up to $Z=70$ or so: I included the periodic table for your convenience. The even heavier elements, especially gold and uranium, were produced by three extra processes&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;s-process&lt;/li&gt;&#xA;&lt;li&gt;r-process&lt;/li&gt;&#xA;&lt;li&gt;rp-process&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The s-process depends on the existence of elements in the iron group. An extra neutron may be absorbed (probably coming from reactions inside red giants), increasing $A$ by one, and if an unstable element is produced in this way, a neutron in the nucleus beta-decays by emitting an electron. Additional neutrons may be absorbed and the process may continue. Nuclei in the &quot;valley of beta stability&quot; can be produced in this way. &quot;S&quot; stands for &quot;slow&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the contrary, the r-process is &quot;rapid&quot;. The neutrons are absorbed in a similar way but in the cores of supernovae. The seed nucleus is usually Ni-56. The rp-process, which may occur in the neutron stars and elsewhere, is also rapid, &quot;r&quot;, but the particle is that absorbed is a proton, therefore &quot;p&quot; in &quot;rp&quot;. Logically, unlike the previous two, it produces nuclei on the &quot;proton-rich side&quot; of the stable valley.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The uranium we observe on the Earth probably comes from all these processes - and from many stars - there is arguably no &quot;the star&quot; that preceded the Sun. In particular, our Earth hasn't orbited any other star before the Sun because it is as old as the Sun, at least this is what is believed. The hydrogen used by our Sun couldn't have been &quot;recycled&quot; and it began to burn soon after the sufficient collapse - it couldn't have been recycled from elsewhere. The heavier elements were recycled from many places. There was probably no &quot;permanent region&quot; that was inheriting the brand &quot;Solar System&quot;.  These issues were discussed yesterday:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;How many times has the stuff of the Sun been recycled?&lt;/strong&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/3769/how-many-times-has-the-stuff-in-our-solar-system-been-recycled-from-previous-st&quot;&gt;How many times has the &quot;stuff&quot; in our solar system been recycled from previous stars?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let me mention that it's not a problem for the heavy material to spread across large distances of the Cosmos. For example, an exploding supernova shoots most or all the matter by the speed 1% of the speed of light. So in 400 years, the material from the Sun - if it went supernova (it won't) - reaches Proxima Centauri and in less than 100 million years, it may reach almost any point in the Milky Way. Even the Solar System is moving at speed of 0.1% of the speed of light which is enough to move matter by light years in thousands of years. It's silly to imagine that the material had to wait on the same place from an &quot;ancestor star&quot;, being saved for some humans on some Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be useful to list all processes of stellar nucleosynethesis, not only those linked to uranium:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;pp-chain / CNO cycle / α process / Triple-α / Carbon burning / Ne burning / O burning / Si burning / R-process / S-process / P-process / Rp-process&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-25T12:45:48.687" LastActivityDate="2011-01-25T12:45:48.687" CommentCount="2" />
  <row Id="3838" PostTypeId="2" ParentId="3799" CreationDate="2011-01-25T12:19:02.823" Score="4" ViewCount="" Body="&lt;p&gt;Some &#xA;of the (partly rhetorical) questions in this thread &#xA;are not about physics, but physicists (or chemists).&#xA;To understand this &quot;how could he/she/they?&quot; I recommend &#xA;some reading on &quot;N-Rays&quot;, &quot;Piltdown Man(skull)&quot;, &#xA;or &quot;Polywater&quot;. Especially the last topic is very &#xA;interesting, because there were (at least) two groups &#xA;involved, not just one or two persons. &#xA;Error, envy, &quot;Erfolgsdruck&quot;, self-betrayal, vanity,&#xA;what did I forget? &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-08-02T09:43:35.347" LastActivityDate="2011-08-02T09:43:35.347" CommentCount="6" />
  <row Id="3839" PostTypeId="2" ParentId="1557" CreationDate="2011-01-25T12:34:03.707" Score="2" ViewCount="" Body="&lt;p&gt;As for why you can't go just a bit higher. It's not a problem of having the energy, the problem is transferring it the particle you want to accelerate. Those particles are accelerated using electromagnetic fields generated in superconductor devices. There is a limit on how large these fields can be made, since when the magnetic field is too big, the superconductive state is lost and all hell break loose (the temperature is not the only thermodynamic variable in superconductors, you can increase the Gibbs function by increasing the magnetic field as well). You also have other less &quot;thermodynamically fundamental&quot; issues, but let's forget about them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if you want to accelerate a bit more, you have to either make the acceleration path even longer or make the particles go in circles and pass the accelerating region many times. The first case is not doable, since the size would be longer than any laboratory we already have. The second case has limitations too. You have to keep the particles in a stable beam for a long time, the particles lose some energy while going around the circular path, and so on...&lt;/p&gt;&#xA;" OwnerUserId="1337" LastActivityDate="2011-01-25T12:34:03.707" />
  <row Id="3841" PostTypeId="1" AcceptedAnswerId="3842" CreationDate="2011-01-25T13:13:08.353" Score="5" ViewCount="96" Body="&lt;p&gt;Imagine opening a water tap in order to have a smooth and cylindrical outflow and then slowly decrease the flow by adjusting the knob. At a certain moment, the side profile of the flow will become uneven, and even if you decrease the scale of the faucet, the water shall start to drip. It is possible to determine, based on the thermodynamic conditions of the system (pressure, temperature, and surface tension of water) the dropping condition?&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2011-01-26T22:01:48.473" Title="Dropping condition" Tags="&lt;thermodynamics&gt;&lt;surface-tension&gt;" AnswerCount="1" />
  <row Id="3842" PostTypeId="2" ParentId="3841" CreationDate="2011-01-25T13:49:01.417" Score="3" ViewCount="" Body="&lt;p&gt;The answer is no. Thermodynamical variables like pressure, temperature etc. are not sufficient to determine the flow completely. The reason is, the flow of water is too much chaotic and too much sensitive to its initial conditions. Even a slight change in the initial conditions will make the water flow differently. As each drop leaves the faucet it creates oscillations in the residue of the nozzle which in turn sets the initial condition of the next drop. Since it is almost impossible to know all the details of the system the flow can not be completely predicted. However there are laws by which you can calculate some patterns with much more precision. The whole process comes under the subject of nonlinear dynamics. Follow these links &lt;a href=&quot;http://www.seas.harvard.edu/softmat/downloads/pre2000-02.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.seas.harvard.edu/softmat/downloads/pre2000-02.pdf&lt;/a&gt;&#xA;http://www.pas.rochester.edu/~AdvLab/10-DrFaucet/DrippFaucet.pdf&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-01-25T14:48:15.960" LastActivityDate="2011-01-25T14:48:15.960" />
  <row Id="3845" PostTypeId="1" CreationDate="2011-01-25T14:36:29.107" Score="6" ViewCount="194" Body="&lt;p&gt;Some weeks ago, there was lots of talk about this CDF paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/1101.0034v1&quot; rel=&quot;nofollow&quot;&gt;Evidence for a Mass Dependent Forward-Backward Asymmetry in Top Quark Pair Production&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where they measured a much higher asymmetry than the one predicted in the SM. In trying to understand the paper, I had some question which I thought would fit in a discussion here.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Experimental question&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;When all $t\bar{t}$ events are considered, the asymmetry is merely $1.4\, \sigma$ above the SM prediction. Well, no one would be excited with such small mismatch. However, in the paper above, they separate the data in 2 bins of $t\bar{t}$ mass, above and below $450\, GeV$ and claim a $3.6\, \sigma$ difference for the high mass bin. Oh! 3.6 is &lt;strong&gt;way&lt;/strong&gt; more exciting (most of the people said).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my question is: where did this $450\, GeV$ come from? In the paper they claim:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We find that the significance of the asymmetry at high mass is maximized when the bin division is at $M_{t\bar{t}} = 450 GeV/c^2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can this be justified statistically? I mean, can you simply ignore the rest of your data and focus on the subset which gives the largest statistical evidence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that I'm not necessarily questioning the CDF paper, which is well written, I'm just asking whether people should have been so excited about the significance of this &quot;evidence&quot;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Theoretical question&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I was trying to understand where this asymmetry comes from. There is this paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/1003.5827v3&quot; rel=&quot;nofollow&quot;&gt;Renormalization-Group Improved Predictions for Top-Quark Pair Production at Hadron Colliders&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which discusses, among many other things, the reason for the asymmetry. They say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These &lt;em&gt;[the $t\bar{t}$ asymmetry]&lt;/em&gt; arise if, in the interference of one-loop and tree-level diagrams, the top-quark fermionic line and the light-quark fermionic line are connected by three gluons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is that so obvious? I really don't understand the reasoning. Would someone care to explain why this is the reason in more details?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They even have a Feynman diagram which, they claim, depicts the asymmetry (I thought it would be useful to have it here in case someone wants to refer to it while answering).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img834.imageshack.us/img834/92/ttbar.png&quot; height=&quot;150&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1337" LastEditorUserId="1337" LastEditorDisplayName="" LastEditDate="2011-01-25T14:57:48.417" LastActivityDate="2011-01-25T19:26:17.937" Title="$t\bar{t}$ asymmetry" Tags="&lt;quantum-field-theory&gt;&lt;high-energy-physics&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="3846" PostTypeId="1" CreationDate="2011-01-25T14:56:07.687" Score="10" ViewCount="561" Body="&lt;p&gt;I'm looking for simple problems in theoretical mechanics that are impossible or unreasonably difficult to solve by means of &quot;brute-force&quot; numerical integration of &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler-Lagrange_equation&quot; rel=&quot;nofollow&quot;&gt;Newton&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler-Lagrange_equation&quot; rel=&quot;nofollow&quot;&gt;Euler-lagrange&lt;/a&gt; equations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in these beacuse I noticed that kind of &quot;computer-&lt;a href=&quot;http://en.wikipedia.org/wiki/Nihilism&quot; rel=&quot;nofollow&quot;&gt;nihilism&lt;/a&gt;&quot; point of view is getting popular (at least among some students):&#xA;A person says, that &quot;&lt;em&gt;in the end we anyway doing real stuff by computer simulation&lt;/em&gt;. &lt;em&gt;And for the numerical values of parameters we are usually able to numerically obtain the result with a given precision. So we just need to know how to write down the equations&lt;/em&gt;&quot;.&lt;br&gt;&#xA;And, therefore, &quot;&lt;em&gt;there is no need to learn all that complicated stuff in theoretical mechanics&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from obvious counter-arguments for this, I'd like to show that there are basic problems you are unable to solve without &quot;the complicated stuff&quot;.&lt;br&gt;&#xA;Let me give an example of such a problem:&lt;br&gt;&#xA;&lt;strong&gt;Given:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A center, that creates some strange field with the potential $U(r)=-\frac{\alpha}{r^3}$. (Mysterious planet)&lt;/li&gt;&#xA;&lt;li&gt;A body with mass $m$ scattering off this center. (Our space ship.)&lt;/li&gt;&#xA;&lt;li&gt;A radius R, at which we want to stay as long as possible.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Find:&lt;/strong&gt; &#xA;the impact parameter $\rho$ and the energy $E_0$ for our body, so it will stay in the &quot;ring&quot; $R&amp;lt;r&amp;lt;2R$ for as long as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is easily formulated. And it is easy to solve even for &quot;newbies&quot; in theoretical mechanics.&#xA;The specific feature of the problem  -- there is no reasonable way of solving it by doing straightforward computer simulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you propose other examples of problems with these properties?  &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-06-24T07:19:48.620" CommunityOwnedDate="2011-01-25T18:01:46.793" Title="Basic mechanics problems, unsolvable by brute-force numerical integration" Tags="&lt;classical-mechanics&gt;&lt;numerics&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="3" />
  <row Id="3847" PostTypeId="2" ParentId="3846" CreationDate="2011-01-25T15:35:43.097" Score="2" ViewCount="" Body="&lt;p&gt;I think this is just a question of defining what you mean by &quot;solve&quot;. All physical problems, most definitely so in classical mechanics, can be posed as differential equations for which solutions (i.e. trajectories of the dependent variables) can be found &lt;em&gt;at least&lt;/em&gt; through numerical integration. In this sense, as far as I know, no one has identified any non-computable phenomena in physics. &lt;em&gt;*&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does not however mean that all quantities of interest (which aren't explicit dependent variables in your differential equations) can simply pop-out of a computer program. Computers can only give you answers for questions that you define perfectly and precisely - and the best such example is a PDE for which you want numerical solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A computer can give you precise values for the impact parameter and energy that you have asked for in your question, but in order for it to do that, you will have to define to the computer what impact parameter and energy are, and in trying to do so you will have solved your mechanics problem :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, it is easy to ask a computer for trajectories, since it can calculate these from the relevant (P)DEs. But to get other quantities of interest such as energy or impact parameter, you will have to extract them from the physics of the problem and translate it into a form the computer will understand, which essentially involves doing some more work on paper.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;*&lt;/em&gt; (Is there a better way to leave a footnote?)&#xA;Of course, you probably already know that most problems are not analytically solvable (classic example is the full 3-body problem) and also that the most interesting numerical problems are simply intractable - the prominent example being the quantum simulation of many-body systems in condensed matter physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if it turns out that some new physics that is discovered is indeed non-computable, there are those who claim that it might be relevant to the problem of consciousness (See Penrose's &quot;Shadows of the Mind&quot; - and I'll leave you to be the judge of that)&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-01-25T16:46:55.033" LastActivityDate="2011-01-25T16:46:55.033" CommunityOwnedDate="2011-01-25T18:01:46.793" CommentCount="6" />
  <row Id="3848" PostTypeId="2" ParentId="1869" CreationDate="2011-01-25T15:38:55.950" Score="2" ViewCount="" Body="&lt;p&gt;I don't know any reason to suspect string theory may be  incorrect, but the closest analogy to renormalisation infinities is the fact that we only know string theory as a perturbation series from a fixed background, plus a collection of non-perturbative dualities that relate different versions. By rights there should be a better formulation from which all the perturbative string theories can be derived directly. M-theory is a name sometimes used in this connection but we still don't know what it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if string theory were not a theory of physics there would be good reason from the mathematics to think that such a formulation exists because otherwise all the relationships between the different string theories would be just creepy coincidences. There really has to be some underlying framework that explains them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is comparable in some ways with the original view of renormalisation which made people question the original perturbative formulations of quantum field theories. Once non-perturbative formulations such as lattice gauge theory emerged they were able to understand renormalisation as a scaling behaviour related to critical points in the theory and things were better understood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;String theory has other problems such as the need to understand how a vacuum can be selected that explains low energy particle physics and the cosmological constant. We also need to understand better what happens to spacetime in string theory at Planckian energies. These problems are often highlighted because they are the reason why string theory cannot yet make any testable predictions, but I don't think they can be resolved properly until the underlying principles of the theory are known. So that is the key issue to resolve first.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-25T15:38:55.950" CommentCount="3" />
  <row Id="3849" PostTypeId="2" ParentId="3794" CreationDate="2011-01-25T15:54:35.297" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;1) Why is that? Is this just a by-product of the reduced size of the CMS hadronic calorimeter&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My investigations got me into this book: &quot;&lt;a href=&quot;http://www.worldscibooks.com/physics/7349.html&quot; rel=&quot;nofollow&quot;&gt;AT THE LEADING EDGE. The ATLAS and CMS LHC Experiments&lt;/a&gt;&quot;. I think that most of the general questions about the design of those detectors are answered in the book. While seems that it is very difficult to point out a particular reason for the design decisions. In &lt;a href=&quot;http://lss.fnal.gov/archive/2010/pub/fermilab-pub-10-210-ppd.pdf&quot; rel=&quot;nofollow&quot;&gt;chapter 10&lt;/a&gt; the ideas beyond the design of the HCAL for CMS are explained. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the CMS the tracker and ECAL are emphasised, so the decision was to put the HCAL inside the solenoid for better magnet design. Another requirement is a easy maintenance. Since you was critical to that, I'm citing:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A &lt;em&gt;unique&lt;/em&gt; feature of CMS is its moving-ring-based structure, allowing for very&#xA;  good access and maintenance of the detector elements. This design feature had the&#xA;  tradeoﬀ for the HCAL in that the readout system (front-end electronics system)&#xA;  had to be placed inside the magnetic volume. Any alternative location for the&#xA;  photodetectors was so far away that there would have been prohibitively large light&#xA;  loss in the long clear optical ﬁbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the HCAL has to be small while incorporating the photodetectors. And everything has to operate in 4T magnetic field -- the point is that photomultiplier tubes cease to work there. I'm citing:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Unfortunately, phototubes lose their gain quickly in a magnetic ﬁeld (due to inability to focus the electrons). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So It seems that answer to the question (1) are: reduced size, magnetic field and modularity. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2) Are those resolutions final or can the collaborations make them better when they have more data? How?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It depends on what do you mean under &quot;final&quot;. These resolutions were measured in beam-tests at SPS. The description of those tests are given in references [10] and [11] or the paper in question. Part of the modules were put under the various beams at SPS. Citing ref. about ALICE [10]:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;5.7.2 Hadronic end-cap performance.&lt;/strong&gt;&lt;br&gt;&#xA;  About 25% of the series production modules were exposed to beams of muons, electrons and pions&#xA;  with energies up to 200 GeV at the CERN SPS [150]&lt;br&gt;&#xA;  ...&lt;br&gt;&#xA;  &lt;strong&gt;5.7.4 Tile-calorimeter performance&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;5.7.4.1 Stand-alone performance&lt;/strong&gt;&lt;br&gt;&#xA;  Approximately 12% of all production modules of the tile calorimeter have been measured extensively in dedicated test-beam periods at the CERN SPS.&lt;br&gt;&#xA;  ...&lt;br&gt;&#xA;  &lt;strong&gt;5.7.4.2 Combined LAr and tile calorimeter test-beam measurements&lt;/strong&gt;&lt;br&gt;&#xA;  The combined performance of the barrel LAr electromagnetic and tile calorimeters was measured&#xA;  in 1996 in the H8 beam at the CERN SPS. The setup used prototype modules of the two calorimeters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In that sense -- those resolutions are final.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But also I asked some experimentalist and he told me that there is a procedure which is called &quot;the understanding&quot; of a detector. You have to measure the performance of the detector as a whole, accounting for everything: geometry, triggers, e.t.c. As far as I understood both detectors are not completely understood yet.&lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2011-01-26T13:20:38.920" LastActivityDate="2011-01-26T13:20:38.920" CommentCount="4" />
  <row Id="3852" PostTypeId="1" AcceptedAnswerId="3885" CreationDate="2011-01-25T16:23:03.097" Score="4" ViewCount="519" Body="&lt;p&gt;Some explanations of the device base it on a simple echo of light: &quot;The camera transmits invisible near-infrared light and measures its “time of flight” after it reflects off the objects. Time-of-flight works like sonar: If you know how long the light takes to return, you know how far away an object is.&quot; (&lt;a href=&quot;http://www.wired.com/gadgetlab/2010/11/tonights-release-xbox-kinect-how-does-it-work/&quot; rel=&quot;nofollow&quot;&gt;wired.com&lt;/a&gt;). Is it so easy? On other hand, the developers, Primesense, speak about &quot;a sophisticated parallel computational algorithm to decipher the received light coding&quot;. And they claim to avoid noise from ambiance light and to be able to obtain a resolution of 1cm at 2 meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A motivation to ask this here in physics is that some other questions were asking for measurement of light speed with household devices.&lt;/p&gt;&#xA;" OwnerUserId="1335" LastActivityDate="2011-01-25T22:31:18.393" Title="How does the Kinect device work?" Tags="&lt;optics&gt;&lt;applied-physics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="3853" PostTypeId="2" ParentId="3751" CreationDate="2011-01-25T16:23:20.927" Score="1" ViewCount="" Body="&lt;p&gt;If there is a discontinuity in the Ricci tensor at the boundary, this would show up as a Dirac delta contribution to the covariant derivative of the Ricci tensor there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the Bianchi identity $R_{\mu\nu\rho\sigma;\tau}+R_{\mu\nu\sigma\tau;\rho}+R_{\mu\nu\tau\rho;\sigma}=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Decompose the Riemann tensor as the sum of the Weyl tensor plus Ricci contributions. After this decomposition, group the terms in the Bianchi identity into Weyl terms, and Ricci terms. Generically, the Ricci terms have a Dirac delta contribution. So, in order to sum up to zero, so must some terms of the covariant derivative of the Weyl tensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, we generically have a discontinuity in the Weyl tensor.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-01-25T16:23:20.927" CommentCount="5" />
  <row Id="3855" PostTypeId="2" ParentId="3824" CreationDate="2011-01-25T16:38:48.693" Score="6" ViewCount="" Body="&lt;p&gt;Actually, if you were at rest on the ground, the van der Waals repulsion between the atoms of the soles of your feet and the atoms of the ground &amp;mdash; which is electrical in origin &amp;mdash; exactly cancels the gravitational force between your &lt;em&gt;entire&lt;/em&gt; body and the &lt;em&gt;entire&lt;/em&gt; mass of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the van der Waals force isn't even a direct Coulomb force between two simple charges, but a much weaker indirect electrical repulsion between neutral atoms due to a deformation of the electron clouds. I know you might argue the repulsive van der Waals force is actually due to the Pauli exclusion principle, and that is also true. After fixing the positions of the nuclei, the physical ground state &amp;mdash; which is given by a totally antisymmetric electron wavefunction &amp;mdash; has a higher energy than some unphysical electron wavefunctions which aren't totally antisymmetric. But if we restrict ourselves to the physical wavefunction, the immediate repulsive force can still be traced back to electrostatics.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-27T16:38:31.483" LastActivityDate="2011-01-27T16:38:31.483" />
  <row Id="3856" PostTypeId="2" ParentId="3845" CreationDate="2011-01-25T16:40:24.213" Score="1" ViewCount="" Body="&lt;p&gt;yes, the threshold was chosen at 450 GeV because this turned out to maximize the deviation as the number of standard deviations, given the data they have acquired. If the effect is real, then 450 GeV is the actual boundary where the &quot;new physics&quot; responsible for this effect begins to occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, the choice of the boundary at 450 GeV is cherry-picking or a legitimate measurement depending on the question whether the effect is real or not. Only in the future, when this question is settled, people will know. But of course, experimenters are always trying to look for real - i.e. most promising - deviations from the theory. That's the reason why they're doing the experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course, the fact that they had to make a choice somewhat diminishes the p-value - the strength - of their experimental evidence. Imagine that they could choose the boundary anywhere in an interval whose length is 500 GeV, and they had to choose it with the precision 20 GeV to get an effect that is at least 3.5 sigma (I don't claim that the numbers 500 GeV and 20 GeV are exact - you should substitute the right ones instead). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that they had about 25 places to cherry-pick from. The 3.6-sigma evidence is equivalent to the probability of noise being approximately &quot;1 in 2,000&quot;, but because the cherry-picking of the threshold increases the probability of a &quot;fake 3.6-sigma signal&quot; by the factor of 25, as argued at the beginning of the inerval, the 1-in-2,000 really decreases to 1-in-2,000/25 = 1-in-80.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the cherry-picking is reflected in the p-value, then, given my numbers, the signal decreases to something like 2.5 sigma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Theoretical question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, the statement is not obvious. This is why the paper contains not only the sentences you quoted but also additional sentences. If you continued to read the paper, the following page would answer your question. The formulae (120) and (121) exactly tell you the color-dependent coefficients of the main and crossed terms and quantify the difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the tree-level diagrams produce no asymmetry, it's clear that one has to go to non-tree diagrams i.e. loop diagrams (plus nonperturbative effects). Among the non-tree diagrams, one loop diagrams are the most important ones, usually, followed by the two-loop diagrams. Whenever internal loops appear in the diagrams, one wants to take them to be heavy fermions with the right numbers because the corresponding Yukawa couplings are maximized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The loop diagrams lead to more complex mathematical expressions, so it's not surprising that the main amplitude and the crossed ones are no longer equal. To get the probability of a specific process with specific initial and final states - which is what is being calculated here - one first has to calculate the probability amplitude. The probability amplitude is the sum over all Feynman diagrams (or histories) with the same initial and final states; they may constructively and destructively interfere. And because the loop diagram is asymmetric with respect to the exchange of the top quark and its antiparticle, the interference will lead to an asymmetric result, too.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-25T17:29:23.840" LastActivityDate="2011-01-25T17:29:23.840" />
  <row Id="3857" PostTypeId="2" ParentId="3770" CreationDate="2011-01-25T16:51:49.410" Score="2" ViewCount="" Body="&lt;p&gt;This is the problem of radion stabilization for KK theories. The radion is the effective 4D field which measures the size of the compactified dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Calabi-Yau compactifications with unbroken SUSY, the radion turns out to be a moduli, which is experimentally unsatisfactory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many different radion stabilization mechanisms out there, e.g. fluxes, Goldberger-Wise, etc. . In general relativistic theories, the mechanism has to be dynamical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After quantization, the radion would show up as a massive particle. The radion also doubles as a dilaton in the effective 4D theory as $M_{4D}^2$ (4D Planck mass squared) is proportional to the compactification volume.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-01-25T16:51:49.410" CommentCount="2" />
  <row Id="3858" PostTypeId="2" ParentId="3846" CreationDate="2011-01-25T16:56:04.523" Score="1" ViewCount="" Body="&lt;p&gt;I think you can construct such an example by considering the large scale properties or evolution of some chaotic system, like a gas, you would have a problem (time-wise) if you were to track each particle in CM, but thermodynamic formulas gives instant answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;This one:&#xA;Given two point-particles, will they ever impact?&#xA;GL simulating and checking r(t)=0.&#xA;Whereas angular momentum consideration gives instant answer.&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2011-01-25T17:44:13.033" LastActivityDate="2011-01-25T17:44:13.033" CommunityOwnedDate="2011-01-25T18:01:46.793" CommentCount="7" />
  <row Id="3859" PostTypeId="1" AcceptedAnswerId="3866" CreationDate="2011-01-25T17:05:04.860" Score="3" ViewCount="167" Body="&lt;p&gt;When I dip a paper towel in a cup of water the water gets drawn up due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Capillary_action&quot; rel=&quot;nofollow&quot;&gt;capillary action&lt;/a&gt;. How is this reconciled with conservation of energy, as it would seem on the surface that the potential energy of the system has gone up with no work being done?&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T07:48:51.633" LastActivityDate="2011-02-03T07:48:51.633" Title="Capillary action and conservation of energy" Tags="&lt;energy-conservation&gt;&lt;capillary-action&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="3860" PostTypeId="2" ParentId="3792" CreationDate="2011-01-25T17:13:18.387" Score="3" ViewCount="" Body="&lt;p&gt;I see two current main axis of research around entanglement. I have not really followed the last years of research, so I don't know exactly which questions are open and which have been solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I sadly have no time to elaborate this week, but since its a community wiki, others can complete my answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Entanglement characterization&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is on how to characterize entanglement as a resource and how to interconvert various entanglement form into others. While the problem has essentially been solved for two-party-entanglement, many open question subsists in n-party-entanglement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a recent technical review, you can read the Horodecki family paper in Review of Modern Physics &lt;a href=&quot;http://rmp.aps.org/abstract/RMP/v81/i2/p865_1&quot; rel=&quot;nofollow&quot;&gt;(restricted access)&lt;/a&gt; &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0702225&quot; rel=&quot;nofollow&quot;&gt;(arXiv version)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Finding quantum physics from information theoretic consideration&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well known that quantum entanglement does not allow any faster than light communication. This is called the &quot;no-signalling&quot; principle. However, the no-signalling principle does not forbids theories with stronger correlations than quantum mechanical entanglement (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Nonlocality#Generalising_nonlocality&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.) It has been shown that if one could build such a &quot;super-entangled&quot; machine, one could perform various unintuitive communication-complexity related tasks (see &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0501159v1&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://arxiv.org/abs/0905.2292&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for technical papers about this, and &lt;a href=&quot;http://scienceblogs.com/pontiff/2009/05/information_causality.php&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://complementaryslackness.wordpress.com/2009/06/01/information-causality/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for blog posts about the second paper). The interesting open question is the following: can we find all the quantum-entanglement correlations (and the quantum mechanics formalism) from reasonable information-theoretic constraints similar to the no-signaling principle.&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2011-01-25T17:13:18.387" CommunityOwnedDate="2011-01-25T17:13:18.387" />
  <row Id="3861" PostTypeId="1" CreationDate="2011-01-25T17:26:13.423" Score="6" ViewCount="172" Body="&lt;p&gt;Dirac came up with a general theory of constraints, including second-class constraints. To quantize such systems, he first computed the Dirac bracket &lt;em&gt;classically&lt;/em&gt;, and only then &quot;promoted&quot; the classical Dirac bracket into a commutator. However, this leads to operator ordering ambiguities over and above what already exists for the Poisson bracket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any more direct way of coming up with a quantum Dirac bracket and Hamiltonian operator which resolves such ambiguities? If the system is symmetrical, symmetry considerations can often single out the &quot;correct&quot; quantization, but what about asymmetrical systems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, the correct quantum prescription ought to match what we get from using path integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if the second-class constraint is linear, the quantum commutator and Hamiltonian is also straightforward. However, even relatively simple models can have subtleties. Consider the toy model&#xA;$$L=\frac{\dot{q}^2}{2} -f[q]\frac{F^2}{2}$$&#xA;where F is an auxiliary variable and $f$ is a function of $q$. The classical second-class constraints are given by $p_F \approx 0$ and $fF\approx 0$. Using path integrals as a guide,&#xA;$$\int \mathcal{D}q\, \mathcal{D}F\, \exp\left( \frac{i\dot{q}^2}{2} - \frac{ifF^2}{2} \right) \propto \int f^{-1/2}\mathcal{D}q\, \exp\left( \frac{i\dot{q}^2}{2} \right),$$ which corresponds to the following operator ordering for the Hamiltonian:&#xA;$$\widehat{H} = \frac{1}{2} \widehat{f}^{1/4}\widehat{p}\widehat{f}^{-1/2}\widehat{p}\widehat{f}^{1/4}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about the generic nonlinear case? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;On another note, is there a formalism where we can impose the Dirac bracket &lt;em&gt;after&lt;/em&gt; quantization, rather than before?&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-27T17:28:11.577" LastActivityDate="2011-02-02T05:15:09.080" Title="How do we resolve operator ordering ambiguities when quantizing generic nonlinear second-class constraints?" Tags="&lt;mathematical-physics&gt;&lt;hamiltonian-mechanics&gt;&lt;quantization&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="3862" PostTypeId="1" CreationDate="2011-01-25T17:29:42.060" Score="6" ViewCount="399" Body="&lt;p&gt;The so called consistent histories interpretation is claimed to be a correction of the Copenhagen Interpretation. One of its aim, as much as I can see is to show that observers don't have any special role and a history is represented as a sequence of projection&#xA;operators in the Heisenberg picture. It has been convincingly shown that due to environmental decoherence the density matrix will be almost diagonalized within a short time (expressed in terms of a certain basis of interest). It will smell just like a &quot;collapse&quot; of the wave function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we want to see the whole picture, we see that superposition still exists in the entangled whole. My question is whether this interpretation is actually a &quot;many worlds interpretation&quot; in disguise? If not, then why not?&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-28T21:25:00.640" LastActivityDate="2011-05-27T18:37:11.103" Title="Is the &quot;consistent histories&quot; interpretation of QM a &quot;many worlds interpretation&quot; in disguise?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-interpretations&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" />
  <row Id="3863" PostTypeId="1" AcceptedAnswerId="3864" CreationDate="2011-01-25T17:44:17.243" Score="1" ViewCount="342" Body="&lt;p&gt;I feel this is somehow a stupid question, but I don't know the true answer. What happens to an astronaut who's floating in a spaceship in space when it begins to move? Will the astronaut not move until he smashes onto a wall in the spaceship? Or will he move with the spaceship due to gravity? Or does it depend on the speed/acceleration  of the spaceship; if it's too fast, the astronaut will smash onto a wall and if it's enough slow, he'll move with the spaceship?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for helping me out!&lt;/p&gt;&#xA;" OwnerUserId="977" LastActivityDate="2011-01-25T18:19:08.817" Title="What happens to an astronaut who's floating in a spaceship (in space) when it begins to move?" Tags="&lt;gravity&gt;" AnswerCount="3" />
  <row Id="3864" PostTypeId="2" ParentId="3863" CreationDate="2011-01-25T17:58:27.933" Score="6" ViewCount="" Body="&lt;p&gt;He will start to move with the spaceship due to the air pressure inside (the ship pushes the air, and the air pushes him) and the gravity of the ship pulling him, but unless it is accelerating &lt;em&gt;really&lt;/em&gt; slowly he will smash into something, it is not so different from beeing in an accelerating car, infact its exactly the same if you consider only the horizontal direction.&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-01-25T17:58:27.933" CommentCount="3" />
  <row Id="3865" PostTypeId="2" ParentId="3863" CreationDate="2011-01-25T17:59:12.477" Score="3" ViewCount="" Body="&lt;p&gt;The gravitational force of the spaceship on the astronaut is tiny, effectively zero. It would take a really huge object (e.g. a small moon) to have any appreciable effect on an astronaut or anything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an astronaut is in a spaceship and the spaceship accelerates, for all practical purposes, the astronaut will not move along with the spaceship unless he's attached or connected to it in some way, until one of the walls hits him.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-01-25T17:59:12.477" />
  <row Id="3866" PostTypeId="2" ParentId="3859" CreationDate="2011-01-25T18:07:44.540" Score="2" ViewCount="" Body="&lt;p&gt;as the Wikipedia article you linked explains, capillary action is due to the &quot;surface tension&quot; (cohesion within the liquid) as well as &quot;adhesion&quot; (attraction between the water and the paper).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, surface tension&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Surface_tension&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Surface_tension&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is the energy per unit area of the surface of the liquid. The molecules of water inside the liquid are nicely sitting in a potential &quot;hole&quot; induced by the adjacent molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the molecules of water that are on the surface lack the attractive force to the now non-existent neighbors on the side where the liquid no longer exists. Because attractive forces are related to a negative potential energy, the lack of them leads to a positive potential energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, water will contain some extra positive energy called the &quot;surface tension&quot;. It's a form of potential energy but it can only be interpreted in this way at the molecular level. As hinted in the previous paragraph, the surface tension is a contribution to the energy that is proportional to the surface. That's why the surface tension causes water to take the shape of spherical droplets, among many other things, because the sphere minimizes the surface among shapes of the same volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, adhesion - between the paper and water in this case - adds a negative energy proportional to the surface of contact. Again, it may be explained by the (now added, not lacking) negative potential energy between the molecules of the water and the molecules of the paper that attract. To reduce the energy, water tries to touch the paper as much as it can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only the total energy is conserved. The energy needed to lift the water is obtained from the surface tension and adhesion. Microscopically, you may literally imagine that the molecules of water at the top were attracted by the nearby (and so far higher) molecules of paper.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-25T18:07:44.540" CommentCount="1" />
  <row Id="3867" PostTypeId="2" ParentId="3863" CreationDate="2011-01-25T18:19:08.817" Score="2" ViewCount="" Body="&lt;p&gt;Dear huy, when a spaceship is flying to the Moon or when the International Space Station is orbiting the Earth, both the spaceship and the astronaut are moving by the same speed, so the relativity velocity is zero. Moreover, gravity determines the acceleration of all of them. The principle of equivalence implies that when the previous sentence holds, all effects will proceed exactly as in the absence of gravity and acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, rockets have to be accelerated to get them to the speed. When they're accelerated, astronauts are pushed to their seats and their faces get deformed by the inertial force that is pushing them from one direction and the force from the seat or wall that is pushing them in the opposite directions. Astronauts should be able to withstand the acceleration of several $g$ - multiples of the Earth's gravitational field. They're trained on the Earth - in centrifugal gadgets, the vomit comet (even Stephen Hawking tried it), and otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a spaceship accelerates by a big acceleration, it's a good idea to fasten your belt because indeed, the astronaut body floating inside the spaceship has absolutely no reason to accelerate at the same moment. It will continue to float by the same speed, so if the spaceship accelerates, the relative position of the astronaut and the spaceship will accelerate, too. The astronaut will hit the wall much like when he falls - by a constant acceleration - from a wall on Earth. The gravitational force of the spaceship or the astronauts are unmeasurably tiny; they make no effect. Even the gravity of Mr Everest is hard to measure or perceive by &quot;ordinary tools&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-25T18:19:08.817" />
  <row Id="3868" PostTypeId="2" ParentId="3862" CreationDate="2011-01-25T18:38:01.803" Score="2" ViewCount="" Body="&lt;p&gt;Dear sb1, the top experts behind the Consistent Histories often credit Hugh Everett - the father of the many-worlds interpretation - with making the foundational quantum physics community focus on the &quot;histories&quot; as the basic entity whose probability quantum mechanics predicts. However, this is where the overlap ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People behind Consistent Histories usually admit that their interpretation - my favorite one - is just a refinement of the probabilistic Copenhagen interpretation. Nothing essential has changed; the predictions are still fundamentally probabilistic. Consistent Histories is the framework that incorporated the explanations of decoherence - the key process that calculates the boundary of the classical and quantum world - as the first one (and maybe still only one). Many-worlds interpretation is just a semi-popular psychological framework to think about quantum mechanics - and it hasn't been useful to do any actual, new calculations. One doesn't really know how to extract the numerical values of the probabilities from the many worlds, at least not in a way that would tell us more than any other interpretations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Niels Bohr is said to have understood the essence of decoherence but he was unable to transform his understanding into comprehensible explanations or equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Decoherence is not quite the collapse of a wave function&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you say that there is a similarity between decoherence and the collapse of the wave function. Indeed, decoherence is what replaces the flawed ideas about the &quot;collapse&quot; and it is the key process that creates &quot;classical objects&quot; for which our non-quantum intuition is OK. However, all the details are completely different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, decoherence can be fully derived from the standard equations of quantum mechanics - from evolution combined with tracing over the environmental degrees of freedom. It doesn't try to distort the laws of quantum mechanics in any way: it just shows a generic and previously unappreciated consequence of these equations that is responsible for explaining lots of old puzzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the contrary, the idea of a &quot;collapse of the wave function&quot; always tries to incorrectly treat the wave function as a real classical wave. For it to &quot;physically&quot; collapse, there must be new mechanisms and new terms added to the equations that make it collapse at a certain point. Needless to say, all these mechanisms are fictitious and all the mechanisms responsible for such a huge effect can be easily excluded. Quantum mechanics doesn't imply anything of the sort and the tests of quantum mechanics are not compatible with anything of the sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important technical difference between decoherence and the collapse of the wave function is that decoherence actually doesn't decide which result will be measured. It just diagonalizes the density matrix to $\mbox{diag}(p_1,p_2,\dots,p_N)$ where $p_i$ then play the role of the probabilities that the individual preferred basis vectors will be detected. Because the information about the relative phases is getting quickly lost, the off-diagonal elements rapidly converge to zero. But decoherence never transforms the density matrix with the many $p$'s to something like $(0,0,1,0,\dots,0)$. Never, ever. Quantum mechanics remains the only player here and its predictions remain and will always stay probabilistic. One can never and one will never restore determinism and decoherence doesn't try to do anything of the sort.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-25T18:38:01.803" CommentCount="2" />
  <row Id="3869" PostTypeId="2" ParentId="3846" CreationDate="2011-01-25T19:21:08.533" Score="0" ViewCount="" Body="&lt;p&gt;I remember during my PhD I had a discussion with my colleague about mechanics problems that were still unsolved. He mentioned that it was not proven mathematically that a coin falling on the floor will, while settling down, speed up and produce that characteristic &quot;ringing sound&quot;. I don't know what the current status of that problem is and how to look for it, since I don't know the name of the problem, if it has one. But it sounds like something interesting. It seems the difficulty in solving this problem is to see how far one can simplify it without throwing out what is essential for explaining the phenomenon.&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2011-01-25T19:21:08.533" CommunityOwnedDate="2011-01-25T19:21:08.533" />
  <row Id="3870" PostTypeId="2" ParentId="3845" CreationDate="2011-01-25T19:26:17.937" Score="-1" ViewCount="" Body="&lt;p&gt;I am pondering what role the triangle diagram plays here.  There appear to be anomaly cancellations involved with this $t$$\bar t$ asymmetry.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-25T19:26:17.937" />
  <row Id="3871" PostTypeId="1" CreationDate="2011-01-25T19:37:03.933" Score="7" ViewCount="601" Body="&lt;p&gt;I have never seen a “causality operator” in physics. When people invoke the informal concept of causality aren’t they really talking about &lt;em&gt;consistency&lt;/em&gt; (perhaps in a temporal context)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you allow material object velocities &gt; c in SR you will be able to prove that at a definite space-time location the physical state of an object is undefined (for example, a light might be shown to be both on and off).  This merely shows that SR is formally inconsistent if the v &amp;lt;= c boundary condition is violated, doesn’t it; despite there being a narrative saying FTL travel violates causality? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: this is a spinoff from the question:  &lt;a href=&quot;http://physics.stackexchange.com/questions/3767/the-transactional-interpretation-of-quantum-mechanics&quot;&gt;The transactional interpretation of quantum mechanics&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1394" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T07:46:33.663" LastActivityDate="2011-02-21T06:26:59.423" Title="Is causality a formalised concept in physics?" Tags="&lt;mathematical-physics&gt;&lt;special-relativity&gt;&lt;causality&gt;" AnswerCount="9" CommentCount="11" FavoriteCount="4" />
  <row Id="3872" PostTypeId="2" ParentId="3759" CreationDate="2011-01-25T19:46:50.360" Score="3" ViewCount="" Body="&lt;p&gt;Nigel,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few comments. Firstly I think we should separate the questions for Category Theory and Quantum Logic here, as I think they are rather different in a sense I will explain below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Category Theory: My view is that most physicists look for insights and theories from a Geometric viewpoint. This somewhat unites Relativity Theorists and String Theorists. Also the Fibre Bundle model of particle physics (Yang-Mills) is largely a geometric structure too. So mainstream physicists are likely to continue to look to geometric models for answers and insights. This doesnt have to be &quot;narrowly geometric&quot; however, as there is some interesting work on Topological insights into physics: solitons, etc. However the most interesting material to me has been the topological discoveries which have a geometric form. Even Statistical Mechanics has some topological theorems of this sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said this there is a &quot;school&quot; of fundamental &quot;pre-geometry&quot; physicists (which can be found at another site that some of the Stack physics group use called the &quot;Institute for Fundamental Questions&quot;). This link is to an author there I have briefly interacted with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.fqxi.org/community/forum/topic/482&quot; rel=&quot;nofollow&quot;&gt;http://www.fqxi.org/community/forum/topic/482&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They may not use Category Theory itself however, but some are looking for a way to derive spacetime geometry and quantum theory (thereby providing a framework for Quantum Gravity) from something else: but what can that &quot;something else&quot; mathematically be? Chris Isham quoted in another Answer is motivated by this too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Futhermore although Category Theory is billed as a &quot;unifying&quot; theory, physicists tend to get excited by more specific unifications between specific models: the String Theorists have shown us that too, although such unifications also (seem to) happen outside of both String Theory and Category Theory too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: There is an online Category Theory, Philosophy and Physics &lt;a href=&quot;http://ncatlab.org/nlab/show/HomePage&quot; rel=&quot;nofollow&quot;&gt;Wiki&lt;/a&gt;: a bit technical in detail but some sections outline why the researchers believe that Category Theory is useful in their area e.g in Path Integral Modelling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unifying attraction of Category Theory in a less Geometric world, like Abstract Algebra or Software Engineering Logics is higher - it might have value there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum Logic: Unlike mathematical theories or Software Logics this arose as a result of a theorem from Von Neumann about what Quantum Mechanics was supposed to be telling us about physical reality. However this theorem has been disputed, see &lt;a href=&quot;http://write4science.com/Assets/pdfs/Global%20logic.pdf&quot; rel=&quot;nofollow&quot;&gt;http://write4science.com/Assets/pdfs/Global%20logic.pdf&lt;/a&gt; (around page 61 of 74), and although controversy still exists quantum logic per se seems to dead. It has left us with related lattice structures though, and again these might prove useful to those interested in a novel &quot;Quantum Gravity Foundation&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extra Note: Chris Isham has also managed to keep QL going with his Temporal Quantum Logic &lt;a href=&quot;http://en.wikipedia.org/wiki/HPO_formalism&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/HPO_formalism&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-01-31T15:17:53.913" LastActivityDate="2011-01-31T15:17:53.913" CommentCount="2" />
  <row Id="3873" PostTypeId="1" CreationDate="2011-01-25T20:10:29.943" Score="3" ViewCount="318" Body="&lt;p&gt;I am trying to calculate sqrt(-g) in terms of a background metric and metric perturbations, to second order in the perturbations. I know how to expand tensors that depend on the metric, but I don't know how to expand the metric determinant. Does anyone know how to to this or (even better) know of a source that does second order perturbation theory in GR?&lt;/p&gt;&#xA;" OwnerUserId="1589" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T14:46:30.670" LastActivityDate="2011-02-02T14:46:30.670" Title="How do I calculate the perturbations to the metric determinant?" Tags="&lt;general-relativity&gt;&lt;differential-geometry&gt;" AnswerCount="5" />
  <row Id="3874" PostTypeId="2" ParentId="3871" CreationDate="2011-01-25T20:20:39.197" Score="3" ViewCount="" Body="&lt;p&gt;In the axiomatic approach to quantum field theory, sometimes also called local or algebraic quantum field theory, pioneered by Araki, Haag, Kastler, Bogoljiobov et. alt., causality is formalized as an axiom, most often called the &quot;locality&quot; axiom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is this: To every bounded open subset of Minkowski spacetime we associate an operator algebra, all selfadjoint elements of this algebra represent all observables of this region, that is everything that is measurable in this region. Then algebras associated to two spacelike separated regions are assumed to commute, this is the locality or causality axiom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When two observables aka selfadjoint operators commute, this means of course that measuring aka observing the first will have no effect on measuring aka observing the second and vice versa, therefore there cannot be any causal relationship of the events of measuring them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW: The Reeh-Schlieder theorem seems - intuitively - to violate causality/locality, so it is interesting to note that it is possible to prove this theorem &lt;strong&gt;without&lt;/strong&gt; invoking the locality axiom. The reason for this is that the Reeh-Schlieder theorem is about entanglement effects which don't violate locality in the sense of SR. &lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-01-25T20:20:39.197" CommentCount="3" />
  <row Id="3875" PostTypeId="2" ParentId="3873" CreationDate="2011-01-25T20:24:38.747" Score="1" ViewCount="" Body="&lt;p&gt;I guess you basically answered the question yourself. You might have to treat &#xA;$$g = \det g_{\mu\nu}$$&#xA;and the negative squareroot of it as a function of the metric as you would do for a scalar (rank zero tensor).&lt;br&gt;&#xA;The definition of the $\det$ can be found e.g. on &lt;a href=&quot;http://en.wikipedia.org/wiki/Determinant&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;.&lt;br&gt;&#xA;Sincerely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2011-01-25T20:24:38.747" CommentCount="0" />
  <row Id="3876" PostTypeId="2" ParentId="3873" CreationDate="2011-01-25T20:35:32.480" Score="3" ViewCount="" Body="&lt;p&gt;I confess that I haven't looked at this sort of calculation in ages, but I have done other calculations involving perturbative expansions of determinants, and in those contexts the identity&#xA;$$&#xD;&#xA;\ln(\det M) = {\rm Tr}(\ln M)&#xD;&#xA;$$&#xA;has been really useful. If $g=g^{(0)}+\epsilon g^{(1)}+\epsilon^2g^{(2)}\ldots$ for some small parameter $\epsilon$, you can write out the power series of $\ln g$ to any desired order, take the trace, and exponentiate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be some reason I'm not thinking of why this doesn't work in this context, in which case never mind.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-01-25T20:35:32.480" CommentCount="9" />
  <row Id="3877" PostTypeId="1" CreationDate="2011-01-25T20:37:11.143" Score="0" ViewCount="58" Body="&lt;p&gt;if a spinning magnet kept coming in contact with a like pole will it keep spinning&lt;/p&gt;&#xA;" LastActivityDate="2011-01-25T20:37:11.143" ClosedDate="2011-01-25T20:39:26.750" Title="If like poles keep meeting" Tags="&lt;magnets&gt;" CommentCount="6" />
  <row Id="3878" PostTypeId="2" ParentId="3871" CreationDate="2011-01-25T20:44:30.133" Score="4" ViewCount="" Body="&lt;p&gt;Dear Nigel, causality is not an observable (quantity) with a value and a unit; so it is not identified with any operators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Causality is a principle. In a broader scientific and colloquial context, causality is any property of the relationship between the cause and its effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in physics, we mean something more particular by causality. In classical physics, we mean the following proposition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If a cause takes place at time $t_1$ and its effect at time $t_2$, then $t_1&amp;lt;t_2$ must hold.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, the cause precedes its effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's obvious that if the principle above would be violated, the world would become logically inconsistent. Events at time $t_1$ could cause some events at time $t_2$ which would generally cause different events at time $t_1$, producing contradictory answers to the question what happened at time $t_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Looking at causality from a relativistic viewpoint&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the special theory of relativity, the statement above must still hold for the history of spacetime to be free of logical contradictions. However, special relativity is based on the principle of relativity that says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Laws of physics take the same form in all inertial frames - those that are in uniform motion relatively to one chosen inertial system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This must be true for all laws, including the principle of causality itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this principle of relativity is combined with the principle of locality above, we may actually derive a stronger statement. In relativity, the delay between two events depends on the inertial system: simultaneity of events is relative, we say. So two events may be chronologically ordered in the opposite way if you switch into a different inertial system. However, spacelike separated events remain spacelike separated events; and an event in the future (or past) light cone of another event stays in the same cone from the viewpoint of all inertial frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the principle of relativity to the principle of causality, we may derive a stronger, relativistic principle of causality:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If a cause takes place at point $P_1$ in spacetime and if its effect takes place at point $P_2$ in spacetime, then $P_2$ must belong to the future light cone of $P_1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a stronger statement than the original one (about the ordering of $t_1$ and $t_2$): the relativistic causality implies the ordinary causality, but something more (it implies the non-relativistic condition from the viewpoint of all relativistic inertial frames). A cause is not only unable to affect its past, like in the non-relativistic causality; it is unable to affect the spacelike-separated points in the spacetime, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any violation of the relativistic causality - which means that causes may only influence their future light cones - would lead to the same logical contradictions that I explained in the non-relativistic context. In particular, you wrote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This merely shows that SR is formally inconsistent if the $v \leq c$ boundary condition is violated, doesn’t it; despite there being a narrative saying FTL travel violates causality?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, indeed. However, special relativity is demonstrably a valid theory of our spacetime (at least locally). So the &quot;mere&quot; inconsistency of special relativity that you mentioned, in a somewhat incomprehensibly dismissive tone, would automatically mean an inconsistency of the whole Universe which is a pretty serious problem. There's no doubt that there can't be any signals that move faster than light. Logical consistency is an omnipresent and unquestionable assumption in all of physics (and maths), so one is always allowed - and encouraged - to assume it. When we assume it, we may easily show that faster-than-light motion violates causality. In fact, relativistic causality is &lt;em&gt;exactly&lt;/em&gt; what bans faster-than-light motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am convinced that this text explains - and fully unmasks - all deeper and more foundational facts and arguments behind the notion of causality in physics.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-25T22:08:36.270" LastActivityDate="2011-01-25T22:08:36.270" CommentCount="4" />
  <row Id="3879" PostTypeId="1" CreationDate="2011-01-25T21:17:13.993" Score="1" ViewCount="248" Body="&lt;p&gt;I was just cooking some noodles and staring at the pot waiting for them to cook made me wonder... Will my food cook quicker when the water is boiling and bubbling, or when it is at a temperature where it is about to start boiling but there are no bubbles in the water yet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My instinct would be that food will cook quicker just before the water starts boiling because steam conducts heat slower than water and so when the food is in contact with the steam bubbles it receives less thermal energy. I'm basing this assumption off the fact that a steam bath/Turkish bath which is AFAIK at 70 degrees C and close to 100% relative humidity feels a lot colder than sticking my finger in a cup of water at 70 C. Although this begs another question, does the food even come in contact with the steam or does the &lt;em&gt;desire&lt;/em&gt; of the bubble to stay round (minimum energy configuration) actually prevent it from even coming in contact with the actual food?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what is it that intrinsically makes one substance a better thermal conductor than another? I mean what is heat anyways--I was always taught that it was the &quot;kinetic energy of the particles&quot;, so are we talking $0.5mv^2 + 0.5I\omega^2$ where $m$ is the mass of the individual atom? If so, what is the relationship that equates this to degrees Kelvin? And even then, that makes sense for a gas, but what about a liquid or a solid where particles are not free to move around like they can in a gas, is it potential elastic energy of the particles displacing from the optimum minimum energy lattice configuration?&lt;/p&gt;&#xA;" OwnerUserId="734" LastActivityDate="2011-01-25T22:29:40.330" Title="Do my noodles cook quicker when the water is boiling or when it is just about to boil?" Tags="&lt;thermodynamics&gt;&lt;water&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="3880" PostTypeId="2" ParentId="3873" CreationDate="2011-01-25T21:18:29.233" Score="4" ViewCount="" Body="&lt;p&gt;Dear David, as Ted already wrote, but didn't optimize the answer for your problem, there is an identity&#xA;$$\det g = \exp(\mbox{Tr}(\ln g))$$&#xA;You may easily take the square root of it; let us assume that $g$ is positive definite for a while; the resulting formulae will obviously work for all matrices.&#xA;$$\sqrt{\det g} = \exp\left(\frac{1}{2}\mbox{Tr}(\ln g)\right)$$&#xA;However, as he also realized, the formula is not terribly useful for your purpose, anyway, because you would need to differentiate a logarithm of a matrix which is even harder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is better to differentiate the determinant directly with respect to the individual components of the metric tensor explicitly. The determinant of a $d \times d$ matrix is the sum of the $d!$ products over all permutations covering all columns and rows, with the sign counting the parity of the permutation. What is the following derivative?&#xA;$$ \frac{\partial (\det g)}{\partial g_{\mu\nu}}$$&#xA;Well, it's not hard to calculate. Among the $d!$ terms in the determinant, you only pick those that contain $g_{\mu\nu}$. I will return to the evaluation after a paragraph about a technicality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to be sure, $g_{\mu\nu}$ and $g_{\nu\mu}$ are treated as independent variables and the calculation is formally done for general asymmetric matrices. At the very end, we may realize that the metric is a symmetric tensor which simplifies some formulae and allows us to be more sloppy about the ordering. But at no point, we will actually be forced to distinguish the derivatives with respect to diagonal and off-diagonal elements of the metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, so the derivative of the determinant with respect to a particular matrix element is the corresponding subdeterminant of the $(d-1) \times (d-1)$ matrix in which the $\mu$-th row and $\nu$-th column were omitted. You know that the inverse matrix is defined using a similar subdeterminant - however, it must be divided by the whole determinant. It leads us to the identity:&#xA;$$ \frac{\partial (\det g)}{\partial g_{\mu\nu}} = \pm \mbox{subdet} (g_{omitted\,\mu,\nu}) =  (\det g) g^{\mu\nu}$$&#xA;The $\pm$ symbol indicates that I want to stay sloppy about the overall sign. However, the sign is fixed in the following step and regardless of $\mu\nu$, it is a plus sign. This is very clear if you choose e.g. $(\mu,\nu)=(1,1)$ - and the formula must have a nice form for all values of the indices. The determinant had to be reinserted to the right hand side because the inverse matrix elements are defined as ratios with the determinant in the denominator, so I had to multiply it back. If you wrote the determinant as the exponential my answer started with, the exponential - the determinant - would be recycled back to the expression because the derivative of an exponential is the exponential times the derivative of the argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the $+---$ or $-+++$ signature, the determinant is negative. The right sign is still the same&#xA;$$ \frac{\partial (\det g)}{\partial g_{\mu\nu}} = \pm \mbox{subdet} (g_{omitted\,\mu,\nu}) =  (\det g) g^{\mu\nu}$$&#xA;because the previous formula really holds for any matrix, regardless of the signature. Now, write &#xA;$$g_{\mu\nu} = \eta_{\mu\nu}+h_{\mu\nu}$$&#xA;where $\eta$ doesn't necessarily have to have the simple Minkowski form. You can Taylor-expand the determinant in $h$:&#xA;$$\det(g_{\mu\nu}) = \det(\eta_{\mu\nu}) + h_{\mu\nu} \frac{\partial (\det g)}{\partial g_{\mu\nu}}|_{g=\eta} + \frac{1}{2!} h_{\kappa\lambda}h_{\mu\nu} \frac{\partial^2 (\det g)}{\partial g_{\kappa\lambda}\partial g_{\mu\nu}}|_{g=\eta} + \dots$$&#xA;I am convinced that now you can solve it. The second derivatives with respect to the metric are not too much harder; they're the first derivatives of the first derivative. However, you know how to differentiate metric components themselves with respect to $g_{\mu\nu}$; and you also know how to differentiate the determinant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, it's helpful to tell you how to differentiate the inverse metric components with respect to the original metric components, too:&#xA;$$\frac{\partial g_{\kappa\lambda}}{\partial g_{\mu\nu}} = \delta^\mu_\kappa \delta^\nu_\lambda$$&#xA;That was the simple Kronecker delta differentiation. Now the more complex one, with the inverse metric&#xA;$$\frac{\partial g^{\kappa\lambda}}{\partial g_{\mu\nu}} = -g^{\kappa\mu}g^{\lambda\nu}$$&#xA;This formula may be determined from the fact that the derivative of $\delta^\pi_\rho$, which may be obtained as the product of $g^{\pi\alpha}g_{\alpha\rho}$, with respect to any $g_{\delta\epsilon}$ vanishes. The Leibniz rules produces two terms for it, and that's why one of them has to have a minus sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that you know how to Taylor-expand the square root as well, and how to insert Taylor expansions into each other - derivatives of composite functions. All of it follows standard formulae. You may forget that you're dealing with matrices; you may simply imagine that all the expressions are functions of $d^2$ independent variables $g_{\mu\nu}$.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-25T21:23:39.897" LastActivityDate="2011-01-25T21:23:39.897" CommentCount="3" />
  <row Id="3881" PostTypeId="1" AcceptedAnswerId="3883" CreationDate="2011-01-25T21:25:49.837" Score="4" ViewCount="153" Body="&lt;p&gt;When a quantum system have a double degenerescence at one point, the Hamiltonian should be proportional to Pauli matrices near this point (also known as &lt;em&gt;diabolic point&lt;/em&gt;) [&lt;a href=&quot;http://ajp.aapt.org/resource/1/ajpias/v78/i7/p661_s1?isAuthorized=no&quot; rel=&quot;nofollow&quot;&gt;Ref.&lt;/a&gt;]. But, why the Hamiltonian have this form?&lt;/p&gt;&#xA;" OwnerUserId="1545" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-27T11:20:51.460" LastActivityDate="2011-01-27T11:20:51.460" Title="Why the Hamiltonian near degeneracies should be proportional to Pauli matrices?" Tags="&lt;quantum-mechanics&gt;&lt;berry-pancharatnam-phase&gt;&lt;lie-algebra&gt;" AnswerCount="1" />
  <row Id="3882" PostTypeId="2" ParentId="3759" CreationDate="2011-01-25T21:35:01.003" Score="0" ViewCount="" Body="&lt;p&gt;Look at this and see if it helps&#xA;&lt;a href=&quot;http://arxiv.org/pdf/0905.3010v2&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/0905.3010v2&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="304" LastActivityDate="2011-01-25T21:35:01.003" CommentCount="1" />
  <row Id="3883" PostTypeId="2" ParentId="3881" CreationDate="2011-01-25T21:39:11.563" Score="5" ViewCount="" Body="&lt;p&gt;Dear Leandro, it's because the Pauli matrices, together with the $2\times 2$ identity matrix, form a full real basis of all the Hermitian $2\times 2$ matrices (note that $2\times 2$ Hermitian matrices depend on four real parameters), and the identity matrix is irrelevant in a Hamiltonian because it's just a conventional energy shift that acts on all vectors equally and doesn't create any subtleties such as line crossing so one may omit it in any discussion of interesting physical effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;em&gt;any&lt;/em&gt; $2\times 2$ Hermitian matrix that &quot;matters&quot; - and yes, Hamiltonians have to be Hermitian operators - is a real linear combination of the three Pauli matrices. They don't have to be interpreted as a spin of any kind. But they're still a more natural basis than any other basis of the Hermitean matrices because the product of any pair of the matrices generates a multiple of another matrix in the basis, thus simplifying all the calculations. (Analogy with the conventional $SO(3)$ generators is a more transparent way to see the power of this basis.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even if you chose a less clever basis than the Pauli matrices, you could derive the same results. The equations could just become a bit more cumbersome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Line crossing etc. is first observed at degeneracy 2. However, one may also study $3\times 3$ or $N\times N$ matrices. Conventional bases of the Hermitian matrices are called the Gell-Mann matrices in the case of $SU(3)$ - because Gell-Mann generalized the Pauli matrices when he studied the strong force where $SU(3)$ enters in two ways. The $N\times N$ Hermitian matrices depend on $N^2$ real parameters. One usually takes the identity matrix to be one of the basis vectors; the remaining $N^2-1$ vectors are &quot;generators of $SU(N)$&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-25T21:39:11.563" CommentCount="1" />
  <row Id="3884" PostTypeId="2" ParentId="3879" CreationDate="2011-01-25T21:44:01.433" Score="8" ViewCount="" Body="&lt;p&gt;&lt;em&gt;My instinct would be that food will cook quicker just before the water starts boiling because steam conducts heat slower than water and so when the food is in contact with the steam bubbles it receives less thermal energy.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is absolutely the opposite of reality! &#xA;A steam bubble in contact with a body with a temperature less than &#xA;100°C will condense on this body, transferring about 540 cal/g!.&#xA;In fact heating a wall by condensing steam is one of the most &#xA;efficient heat transfer methods. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now Your primary question: Those noodeles &quot;cook&quot; faster &#xA;where they experience the higher temperature. In a pot on a fire, &#xA;the surface of the water will be as low as maybe 80 or 90 degrees, &#xA;when bubbles raise at the bottom, but do not reach the surface yet. &#xA;(Again here: heat transfer by bubble evaporation is extremely &#xA;fast, same reasons as for condensation) &#xA;10 or 20 °C less are a lot when kooking noodles (wheat starch) &#xA;this may lower the reaction velocity to half the value at 100 °C,&#xA;or even lower (rule of thumb for such reaction)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermal conductivity: most amorph solids/liquids have similar, low &#xA;heat conduction. Crystalline solids are medium, metals are much higher, &#xA;the thermal conductivity being related to the electrical conductivity. &#xA;Google for:&#xA;&quot;Wiedemann Franz Lorenz&quot; &#xA;Thermal conductivity of gases can be  calculated by kinetic gas theory, &#xA;this was &quot;triumph&quot; for Clausius Maxwell and Bolzmann. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Georg&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-01-25T22:29:40.330" LastActivityDate="2011-01-25T22:29:40.330" CommentCount="6" />
  <row Id="3885" PostTypeId="2" ParentId="3852" CreationDate="2011-01-25T21:46:40.790" Score="5" ViewCount="" Body="&lt;p&gt;No. It does not measure time of flight. Kinect is, deep down, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Structured_Light_3D_Scanner&quot; rel=&quot;nofollow&quot;&gt;structured light scanner&lt;/a&gt;, meaning that it projects an infrared pattern (so invisible for us). According to the underlying technology firm PrimeSense, &lt;a href=&quot;http://www.primesense.com/?p=535&quot; rel=&quot;nofollow&quot;&gt;the structured light code is drawn with an infrared laser&lt;/a&gt;. This pattern is then read by an infrared camera and the 3D information is reconstructed from the distortions of the pattern. This results in a depth channel which is made available through USB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to see the pattern, you may have some luck by turning off all the lights in the room, turn on the kinect, and try to use your cellphone camera. Generally, these camera sensors are sensitive to IR, which appears as green. You may verify if this is the case by trying the same with a TV remote and pressing the buttons. The LED should turn green.&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="217" LastEditorDisplayName="" LastEditDate="2011-01-25T21:52:52.373" LastActivityDate="2011-01-25T21:52:52.373" CommentCount="6" />
  <row Id="3886" PostTypeId="2" ParentId="3871" CreationDate="2011-01-25T21:48:57.323" Score="0" ViewCount="" Body="&lt;p&gt;The Causality structure of a Spacetime might be best modelled by a partial order between events. An event is a primitive concept in spacetime theory, though one is welcome to question that primitiveness in a yet more fundamental theory. Penrose's Twistor Theory is another theory (this time geometric) in which spacetime points (events) are derived, not fundamental.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However given that we assume events, we can introduce a partial order between them. Introduce some conditions - or derive a null cone structure from the partial order. So one can ask for conditions on the PO to derive a null cone from it. Then from a Global perspective one can introduce further conditions to ensure that this PO does not have x &amp;lt; y and y &amp;lt; x. Of course this is the Closed Timelike Curve condition in General Relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in GR/SR a &quot;cause&quot; of x is just anything in the past lightcone of x. In Quantum Theory this gets more interesting, particularly in the context of the TQM: here something needs to be &quot;emitted&quot; and &quot;received&quot; - much like a telephone message. TQM might not be correct, but this telecom analogy shows that one could go deeper into what a &quot;cause&quot; really is. I suspect that the place for such a theory is in the &quot;Fundamental Framework for Quantum Gravity&quot; - should such a revolutionary Framework come about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: It has become clear that this question was motivated by a criticism of retrocausality in a related post on Transactional Quantum Mechanics (TQM). Rather than continue with a general discussion of Causality in Physics I shall note that Cramer addresses this question in his papers (Transactional Interpretation of Quantum Mechanics, 1986 and cites therein). He introduces two notions of Causality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Strong Causality&lt;/strong&gt; Cause always precedes effect in any reference frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Weak Causality&lt;/strong&gt; Ditto, but only for macroscopic observations and observer-to-observer communication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cramer's cancellation argument implies that there is no Weak Causality Violation in his theory/interpretation. If we accept this, then the question is whether it is acceptable to have strong causality violation in a theory of physics. I can envisage further debate on this.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-01-27T12:51:36.367" LastActivityDate="2011-01-27T12:51:36.367" />
  <row Id="3887" PostTypeId="2" ParentId="3871" CreationDate="2011-01-25T21:54:53.320" Score="0" ViewCount="" Body="&lt;p&gt;From a statistical perspective, whenever you can establish that two type of events (described as states (e1,e2) of a pair systems S1xS2 ) have high correlation (correlations basically stands for the probability of system S1 be in state e1 and S2 be in state e2 simultaneously), you can say that both events are 'causally related'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being said that, there is no formal way to establish (at least from a purely statistical point of view) which event causes which one, other than your willingness to accept a particular convention of time order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In QFT, There is a very formal notion of causality established by the conmutator of field operators as functions of space-time, for instance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left[ \phi(x_\mu),\phi(x&amp;#39;_\mu)\right]=\theta (x_\mu x&amp;#39;_\mu) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when this conmutator is zero it means there is no possible correlation between physical events. the quantity in the right side is different from zero when the two space-time points are time-like to each other (inside the light cone of influence)&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-02-03T20:45:03.827" LastActivityDate="2011-02-03T20:45:03.827" CommentCount="3" />
  <row Id="3888" PostTypeId="2" ParentId="3873" CreationDate="2011-01-25T22:35:46.550" Score="3" ViewCount="" Body="&lt;p&gt;Just for posterity, I thought I'd put up the answer I calculated using the helpful answers given above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $$g_{\mu \nu}=b_{\mu \nu}+h_{\mu \nu}$$ where $b$ is the background metric and $h$ is the perturbation; and raise/lower indeces on $h$ using $b$. Then to second order in $h$ we have: $$\sqrt{-g}=\sqrt{-b}\left(1+\frac{1}{2}{h^{\mu}}_{\mu}+\frac{1}{8}{h^{\mu}}_{\mu}{h^{\nu}}_{\nu}-\frac{1}{4}{h^{\mu}}_{\nu}{h^{\nu}}_{\mu}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure this is correct as I checked it on Mathematica for a general metric.&lt;/p&gt;&#xA;" OwnerUserId="1589" LastEditorUserId="1589" LastEditorDisplayName="" LastEditDate="2011-01-25T22:52:44.087" LastActivityDate="2011-01-25T22:52:44.087" CommentCount="1" />
  <row Id="3889" PostTypeId="1" AcceptedAnswerId="3954" CreationDate="2011-01-25T22:48:30.587" Score="4" ViewCount="195" Body="&lt;p&gt;$:\!\!\hat\phi(x)^2\!\!:$, for example, constructed from the real Klein-Gordon quantum field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a Wightman field, the Wightman function $\left&amp;lt;0\right|\hat\phi(x)\hat\phi(y)\left|0\right&amp;gt;$ is a distribution, which is certainly the case for the real Klein-Gordon quantum field --- call it $C(x-y)$ in this case. In contrast, the expected value $\left&amp;lt;0\right|:\!\!\hat\phi(x)^2\!\!:\,:\!\!\hat\phi(y)^2\!\!:\left|0\right&amp;gt;$ is a product of distributions, in fact for the real Klein-Gordon quantum field it's $2C(x-y)^2$, which seems to make this normal-ordered product of quantum fields not a Wightman field. $C(x-y)^2$ is well-enough behaved off the light-cone, but on the light-cone it has a $[\delta((x-y)^2)]^2$ component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $:\!\!\hat\phi(x)^2\!\!:$ is not a Wightman field, then is it nonetheless in the Borchers' equivalence class of the free field? If so, why so? A (mathematically clear) citation would be nice!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if $:\!\!\hat\phi(x)^2\!\!:$ is not in the Borchers' equivalence class of the free field, because it isn't a distribution, is it nonetheless empirically equivalent to the free field at the level of S-matrix observables, as is proved to be the case for Borchers' equivalence classes (Haag, &lt;em&gt;Local Quantum Physics&lt;/em&gt;, $\S$ II.5.5), even though it is manifestly not empirically equivalent to the free field at the level of Wightman function observables?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reading of the Wightman fields literature of the late 1950s and 1960s is far from complete, which may be why I haven't so far found clear answers for these questions.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-01-27T00:38:53.057" LastActivityDate="2011-01-27T11:17:24.147" Title="Is a normal-ordered product of free fields at a point a Wightman field?" Tags="&lt;quantum-field-theory&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="3890" PostTypeId="2" ParentId="3846" CreationDate="2011-01-25T23:00:20.757" Score="1" ViewCount="" Body="&lt;p&gt;Not sure if this has been mentioned but the classic N-Body problem is computationally intractable. Hence all the statistical methods when N is somewhat large. This is because the problem is effectively O(N!) since every body must interact with every other body. &lt;/p&gt;&#xA;" OwnerUserId="1551" LastActivityDate="2011-01-25T23:00:20.757" CommunityOwnedDate="2011-01-25T23:00:20.757" />
  <row Id="3891" PostTypeId="2" ParentId="3873" CreationDate="2011-01-25T23:03:58.897" Score="1" ViewCount="" Body="&lt;p&gt;The rest of these answers are fine, but another way of doing this computation is to write the metric determinant as $g =\frac{1}{4!} \epsilon^{abcd}\epsilon^{efgh}g_{ae}g_{bf}g_{cg}g_{df}$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where you define $\epsilon^{abcd}$ by the conditions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\epsilon^{txyz}$ is equal to 1, $&lt;/li&gt;&#xA;&lt;li&gt;$\epsilon^{abcd}=-\epsilon^{bacd}=-\epsilon^{acdb}=-\epsilon^{abdc}$&lt;/li&gt;&#xA;&lt;li&gt;$\epsilon^{abcd}=0$ if any two indices take the same value&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then, along with some algebra, you can see that $g^{ab} = \frac{1}{3!g} \epsilon^{aecd}\epsilon^{bfgh}g_{ef}g_{cg}g_{df}$, and then doing the variation is easy.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-01-25T23:03:58.897" />
  <row Id="3892" PostTypeId="1" AcceptedAnswerId="3909" CreationDate="2011-01-25T23:50:51.890" Score="3" ViewCount="282" Body="&lt;p&gt;I was thinking of, for example a Schwarzchild metric at r=0, i.e. the gravitational singularity, a point of infinite density. I realise that there are different types of singularities--timelike, spacelike, co-ordinate singularities etc. In a short discussion with Lubos, I was a bit surprised when I assumed they are idealized and I believe he feels they exist. I am not a string theorist, so am not familiar with how singularities are dealt with in it.  In GR, I know the Penrose-Hawking singularity theorems, but I also know that Hawking has introduced his no-boundary, imaginary time model for the Big Bang, eliminating the need for that singularity. Are cosmic strings and other topological defects singularities or approximations of them (if they exist).  In what sense does a singularity exist in our universe? --as a real entity, as a mathematical or asymptotic idealization, as a pathology in equations to be renormalized or otherwise ignored, as not real as in LQG, or as real in Max Tegmark's over-the-top &quot;all mathematical structures are real&quot;? &lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-02-02T21:37:29.420" Title="Do singularities have a &quot;real&quot; as opposed to mathematical or idealized existence?" Tags="&lt;general-relativity&gt;&lt;cosmology&gt;&lt;black-holes&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="3894" PostTypeId="2" ParentId="3833" CreationDate="2011-01-26T00:07:09.233" Score="1" ViewCount="" Body="&lt;p&gt;Obviously we have a mix from multiple sources, dying stars dumping heavy elements into the interstellar medium, which over billions of years is turbulently partially mixed. I did see a paper a few years back that argued that even small bodies (asteroids) of only a few KM diameter melted. The implication being that there must have been enough Aluminum-27 (the principal radioactive decay product of supernova) to supply enough internal heating to melt them, and that implied these bodies had formed with a few hundred thousand years of the SN explosion. I don't know if that paper has stood the test of time, but there might be some properties of objects in the solar system that we think have been unchanged since the beginning, that will give us better information about the protosolar nebula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earth is more problematic, as it has been melted/differentiated (Iron Nickel sank to the core etc.), and blasted by numerous collisions with other bodies, so interesting information about shorter lived isotopes has probably been lost. The supposed collision with a Mars sized object forming the moon, would clearly have substantially rearranged stuff.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-01-26T00:07:09.233" CommentCount="1" />
  <row Id="3895" PostTypeId="2" ParentId="3615" CreationDate="2011-01-26T00:29:26.197" Score="1" ViewCount="" Body="&lt;p&gt;You must use Kirckhoff laws for alternate circuits. So, the tranfer function that you'll find can be used for any input signal, with Fourier Transform (because transfer function is the output of the system for a delta signal input in frequency domain).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\{\begin{aligned}&#xD;&#xA;&amp;amp;V_I e^{i\omega t}= R (I_1(t)+I_2(t)) + j \omega L (I_1(t)+I_2(t))+ R I_1(t) \\&#xD;&#xA;&amp;amp;V_I e^{i\omega t}= R (I_1(t)+I_2(t)) + j \omega L (I_1(t)+I_2(t))+  \frac{1}{j \omega C} I_2(t) \\&#xD;&#xA;\end{aligned}\right. $$&#xA; or, if you aren't interested to know $I_1$ and $I_2$ in separated rings you can consider the parallel of R2 and C:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_I e^{i\omega t}= RI(t) + j \omega L I(t) + \frac{1}{\frac{1}{j\omega L}+ j\omega C}I(t) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you haven't sayed where is Vout...but with these approach, you can easily calculate current and tension in every point of circuit.&lt;/p&gt;&#xA;" OwnerUserId="558" LastEditorUserId="558" LastEditorDisplayName="" LastEditDate="2011-01-26T00:41:44.973" LastActivityDate="2011-01-26T00:41:44.973" />
  <row Id="3896" PostTypeId="2" ParentId="3892" CreationDate="2011-01-26T00:32:16.560" Score="0" ViewCount="" Body="&lt;p&gt;The singularity in a Schwarzschild metric is a three dimensional space where the Weyl curvature diverges.  There are reasons why we might think this is “quantized,” for otherwise the quantum evaporation would present the outside world with a singularity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An observer that follows a string into a black hole will observe the tidal forces on the string increase.  It is extended along the radial direction and its energy increases.  The extension is along the $X^+$ direction, and the gauge for the system is set on the $X^-$ direction.  This is opposite the gauge condition the exterior observer employs in observing the string fill up the stretched horizon of the black hole.  The analysis for the string on the horizon can be found in Susskind &amp;amp; Lindsey “Black Holes, Information and the String Revolution”  If it is desired I can work out the case of the string which falls into the black hole.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The string falling towards the singularity becomes highly excited and reaches the singularity in some state.  It probably ends as an open string on a brane dual to the NS5-brane.  The string ends in this state of affairs, or it reaches its upper temperature limit and becomes something we are not well acquainted with.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density of states for a string with respect to modes n is&#xA;$$&#xD;&#xA;\eta(n)~\sim~ exp(4\pi n \sqrt{\alpha’})&#xD;&#xA;$$&#xA;that defines a partition function $Z~=~ \int \eta(n)exp(-n/T)dn$.  The temperature is computed by $1/T~=~\partial Z/\partial n$ and the path integral diverges for a temperature greater than &#xA;$$&#xD;&#xA;T_H~=~4\pi \sqrt{\alpha’}&#xD;&#xA;$$&#xA;which is the Hagedorn temperature.  This is proportional to the reciprocal of the string length. The entropy of the system is the logarithm of the density of states the $S~\sim~ 1/nT_H$, which in the large n limit is zero.  The modes number is given by $n~=~1/(\sqrt{d}M_s)$, for d the number of degrees of freedom and $M_s$ the string mass.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These physical conditions occur before the Planck energy is reached.  Consequently the Schwarzschild singularity is valenced by quantum mechanics.  The singularity is valenced by, or really replaced with, a D-brane with a gas of II strings or a D-brane with a gas of D0-brane solitons or … .&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-26T00:32:16.560" />
  <row Id="3897" PostTypeId="2" ParentId="3734" CreationDate="2011-01-26T00:45:43.800" Score="0" ViewCount="" Body="&lt;p&gt;Dr. Motl's answer is fairly complete. I'll add a few details, and introduce the Jarlksog invariant $J_{CP}$, show that it gives a measure of CP violation, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any product of pure density matrices that begins and ends with the same pure density matrix $\hat{x}$ is some complex number $k$ times that pure state. We write:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hat{x}\hat{y}\hat{z}...\hat{x} = k_{xyz...x}\;\hat{x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\hat{x},\hat{y},\hat{z}...$ are pure density matrices and $k_{xyz...x}$ is a number. If the left side happens to be zero, we define $k$ to be zero as well. The $k$ are observables. For example, the transition probability between $\hat{x}$ and $\hat{y}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leaving off the neutrinos, an up, charm, or top quark $\{u,c,t\}$ can emit a $W^+$ and become a down, strange, or bottom quark $\{d,s,b\}$:&lt;br&gt;&#xA;$\{u,c,t\} \to W^+\;\;+\;\;\{d,s,b\}.$&lt;br&gt;&#xA;Similarly the $\{d,s,b\}$ can emit a $W^-$:&lt;br&gt;&#xA;$\{d,s,b\} \to W^-\;\;+\;\;\{u,c,t\}.$&lt;br&gt;&#xA;The two types of quarks $\{d,s,b\}$ and $\{u,c,t\}$ define two bases for the 3-dimensional Hilbert space. The transition amplitudes define a unitary matrix known as the CKM matrix:&lt;br&gt;&#xA;$V_{CKM} = \left(\begin{array}{ccc}&#xD;&#xA;\langle u|d\rangle &amp;amp;\langle u|s\rangle &amp;amp;\langle u|b\rangle \\&#xD;&#xA;\langle c|d\rangle &amp;amp;\langle c|s\rangle &amp;amp;\langle c|b\rangle \\&#xD;&#xA;\langle t|d\rangle &amp;amp;\langle t|s\rangle &amp;amp;\langle t|b\rangle \end{array}\right)$&lt;br&gt;&#xA;In elementary particles literature the CKM matrix is defined with the weak force boson interaction included so $(u,c,t)^t = V_{CKM}\gamma^0(1-\gamma^5)/2(d,s,b)^t$. See, for example, Byron P. Roe. Particle Physics at the New Millenium. Springer-Verlag,&#xA;1996. Our abbreviation is the usual quantum information theory liberty of ignoring force bosons; either way one obtains the same $V_{CKM}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find Berry-Pancharatnam phases in $V_{CKM}$ we must consider transitions between pairs of states such as $\{d,s\}$ and $\{u,c\}$. To work this out, let's define the projection operators using hats, that is, define $\hat{s} = |s\rangle \langle s|$, etc. Then the observable for the transition sequence $d\to c \to s \to u \to d$ is a complex number $k_{duscd}$ defined by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k_{duscd}\hat{d} = \hat{d}\hat{u}\hat{s}\hat{c}\hat{d}\;\;$ or&lt;br&gt;&#xA;$k_{duscd} = \langle d|u\rangle \langle u|s\rangle \langle s|c\rangle \langle c|d\rangle  = V_{du}\;V_{cu}^*\;V_{sc}\;V_{dc}^*$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $V_{jk}$ are the entries in the CKM mixin matrix. The $k_{duscd}$ is a Jarlskog invariant. See Cecilia Jarlskog, &quot;Commutator of the quark mass matrices in the standard&#xA;electroweak model and a measure of maximal CP nonconservation&quot;, Phys. Rev. Lett., 55:1039–1042, 1985. Note that $k_{duscd}^* = k_{dcsud}$; complex conjugation reverses the ordering. For a measure of CP or T violation, this is exactly what we want, that is, CP violation will be the difference between the forward going process and the backwards going one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\{d,s,b\}$ form a complete basis we have:&lt;br&gt;&#xA;$\hat{s} = 1 - \hat{d}-\hat{b}.$&lt;br&gt;&#xA;Substituting the above we find:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k_{duscd}-k_{dcsud} = (k_{du1cd}-k_{dudcd}-k_{dubcd}) - (k_{dc1ud}-k_{dcdud}-k_{dcbud}),$&lt;br&gt;&#xA;$=0-k_{dudcd}-k_{dubcd}-0+k_{dcdud}+k_{dcbud},$&lt;br&gt;&#xA;$=k_{dcbud}-k_{dubcd}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $J_{CP}$ for transitions between $\{d,b\}$ and $\{u,c\}$ is equal to the $J_{CP}$ for transitions between $\{d,s\}$ and $\{u,c\}$. More generally, $J_{CP}$ is an invariant of the $3\times 3$ CKM matrix, that is, it does not depend (except for sign) on the choice of pairs of states considered. And since we've written it in terms of pure density matrices, there is no dependence on the arbitrary complex phases of the rows and columns of the $V_{CKM}$ matrix. All CP violations in the quarks are proportional to $J_{CP}$.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-02-03T02:11:03.453" LastActivityDate="2011-02-03T02:11:03.453" CommentCount="2" />
  <row Id="3898" PostTypeId="2" ParentId="3871" CreationDate="2011-01-26T01:02:57.377" Score="0" ViewCount="" Body="&lt;p&gt;Perhaps the question is too broad. The fundamental principle underlying causality is locality. We can observe causality indirectly by observing locality. By this I mean that the very existence of a coordinate system is the very manifestation of locality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of it this way; the motion of an observer 'causes' lengths to appear smaller. In other words, the 'effect' is the shortening of measured length. You can imagine that the relative motion is a cause and that length contraction is an effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to measure this effect we postulate a coordinate system. When we do this, it turns out, that we predict a finite maximal speed. Which happens to be the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Quantum Mechanics the issue of locality is completely defined by the Heisenberg Uncertainty Principle (HUP) which is equivalent to the treatment of locality in General Relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, the HUP allows you to measure precisely the position of a particle/field. In other words, the HUP allows you to define a coordinate system. This is sufficient to define locality and hence sufficent to embody causality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of it this way; the existence of an observer causes a coordinate system to appear. In QM, the observer is the 'cause' and the coordinate system is an 'effect'.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-26T01:02:57.377" />
  <row Id="3899" PostTypeId="2" ParentId="3871" CreationDate="2011-01-26T02:01:49.803" Score="4" ViewCount="" Body="&lt;p&gt;Causality becomes much more subtle whenever theories are statistical or probabilistic. When we see a correlation, it may be that one event caused the other, but it may be that there is a common cause or that there is just a chance correlation that would disappear if we do more of the same data gathering. For the notion of &quot;common cause&quot;, the standard old-time reference is Reichenbach. Try &lt;a href=&quot;http://plato.stanford.edu/entries/physics-Rpcc/&quot; rel=&quot;nofollow&quot;&gt;the &quot;Reichenbach's common cause principle&quot; entry&lt;/a&gt; in &lt;a href=&quot;http://plato.stanford.edu/&quot; rel=&quot;nofollow&quot;&gt;the Stanford Encyclopedia of Philosophy&lt;/a&gt;, and other entries on causality therein. Often quite a good reference for Philosophy of Science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mathematics of causality in modern statistical physics is very simple, but it has subtle consequences. Quantum field theory distinguishes between the brute random fluctuations of the vacuum state (which are caused, if they are caused by anything, by the random fluctuations that were there in the past, which were caused, if they were caused ...) and causal relationships between measurements. If two measurements are associated with regions of space-time that are at space-like separation from each other, quantum field theories predict that correlations will be observed in the recorded data, but we say that &lt;em&gt;ideal&lt;/em&gt; measurement devices do not cause such correlations. If we don't have measurement devices that are close enough to this ideal, we may have to make allowances for the non-ideal details of the real devices to make the theory match the data. Somewhat non-standardly, I would say that ideal quantum measurements that are at time-like separation from each other do cause some component of the correlations we observe in the recorded data from such measurements. That's different from ideal classical measurement devices, which record data while changing neither the physical state nor the data recorded by other ideal classical measurement devices. But this is a research project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apropos of the Transactional Interpretation starting point, &lt;a href=&quot;http://plato.stanford.edu/entries/qm-action-distance/&quot; rel=&quot;nofollow&quot;&gt;the SEP entry &quot;Action at a Distance in Quantum Mechanics&quot;&lt;/a&gt; might be interesting. &lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-01-26T02:01:49.803" CommentCount="1" />
  <row Id="3900" PostTypeId="2" ParentId="3846" CreationDate="2011-01-26T02:05:37.827" Score="4" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://www-m3.ma.tum.de/m3old/bornemann/challengebook/index.html&quot; rel=&quot;nofollow&quot;&gt;SIAM 100-Digit Challenge&lt;/a&gt; springs to mind. Problem 10 gives the flavor of the type of problems put forward in this challenge:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A particle at the center of a 10 x 1 rectangle diffuses until it hits the boundary. What is the probability that it hits at one of the short ends (rather than at one of the long sides)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer needs to be accurate to at least 10 digits. And no, straightforward Monte Carlo is not going to help you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And... there is an exact solution...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a strict mechanical problem, but it brings across the message that certain physics problems are best solved analytically. If you insist on a mechanical problem, you should look at problem 2 which is basically a billiard problem.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastEditorUserId="1268" LastEditorDisplayName="" LastEditDate="2011-01-26T02:14:29.923" LastActivityDate="2011-01-26T02:14:29.923" CommunityOwnedDate="2011-01-26T02:05:37.827" />
  <row Id="3901" PostTypeId="1" AcceptedAnswerId="3904" CreationDate="2011-01-26T03:04:12.747" Score="9" ViewCount="374" Body="&lt;p&gt;By now everybody knows that gravity is non-renormalizable, what is often lacking is a simplified mathematical description of what that means.  Can anybody provide such a description?&lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="1383" LastEditorDisplayName="" LastEditDate="2011-01-26T10:51:09.267" LastActivityDate="2011-01-26T18:08:59.277" Title="What is a good mathematical description of the Non-renormalizability of gravity?" Tags="&lt;general-relativity&gt;&lt;gravity&gt;&lt;renormalization&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="3" />
  <row Id="3902" PostTypeId="2" ParentId="3901" CreationDate="2011-01-26T03:18:02.637" Score="2" ViewCount="" Body="&lt;p&gt;The simplified way to understand gravity's non-renormalizability is to realize that one needs to make an infinite number of infinite scaling which is not possible. In general, if the number of counterterms in the Lagrangian, required to cancel the divergences is infinite then the process of renormalization fails.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-01-26T03:45:01.183" LastActivityDate="2011-01-26T03:45:01.183" />
  <row Id="3903" PostTypeId="1" AcceptedAnswerId="3910" CreationDate="2011-01-26T03:47:02.227" Score="6" ViewCount="400" Body="&lt;p&gt;The textbooks I have available explain that due to the infinite degrees of freedom of a field, the relevant object in QFT is the lagrangian density. A lagrangian is then obtained for the field by integrating over space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find the justification for this procedure unclear. In classical mechanics, the lagrangians of two particles may be added only if the particules do not interract. Does it mean that the lagrangian density concept is only valid for a free field? What happens in the case of interracting particles?&lt;/p&gt;&#xA;" OwnerUserId="1559" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T15:39:13.117" LastActivityDate="2011-02-02T15:39:13.117" Title="Why is lagrangian density correct?" Tags="&lt;quantum-field-theory&gt;&lt;lagrangian-mechanics&gt;&lt;locality&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="3904" PostTypeId="2" ParentId="3901" CreationDate="2011-01-26T04:06:25.197" Score="5" ViewCount="" Body="&lt;p&gt;How about a description in words instead? I think the mathematics is not the point so much. Other answers lay out the mechanics in detail. The short answer is that the theory has infinitely many divergences which cannot be absorbed by fixing finitely many couplings to their observed values. This means it is not predictive, or in more fancy words does not provide algorithmic compression: the amount of input it requires equals the amount of output it produces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the talk about divergences masks the physics of the situation, which I'll try to describe:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any theory has couplings which describe the possible interactions of the system. When you probe the system at different length scales, or energy scales, those couplings change in a calculable way. This is the process of renormalization, which fundamentally has nothing to do with infinities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Renormalizable theory is one that can be continued to high energies, or short distances, without encountering any difficulties. This means all couplings remain small, and all calculations are reliable. In non-renormalizable theories, the higher energies you probe the system, the stronger the couplings become, at some stage they become infinite. This means that you cannot do reliable calculations, which is normally an indication that there are some physical effects that you are missing. There are many examples where such physical effects are well-understood, for example new degrees of freedom (which are not visible at low energies) become important at those energy scales. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In gravity, when you calculate in perturbation theory, for small perturbations around flat space, it looks like the coupling becomes strong and something is missing. The mechanics of seeing this is straightforward and not that illuminating. Perhaps one intuitive way to understand this is that in gravity the interaction coupling is governed by the energy, so almost by definition it grows with energy...Most likely scenario is that we need new degrees of freedom to complete the theory at short distances. One suggestion for &quot;UV completion&quot; is string theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another scenario which may apply to gravity. Once the couplings become strong, you cannot any more reliably calculate anything, and that includes their scale dependence. So it is a logical possibility that they will stabilize at some finite value, and don't go all the way to infinity. This scenario is called asymptotic safety. In my mind this is very unlikely, and there is no indication* this scenario applies to gravity, but it remains a logical possibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*I know about the various claims, this is a statement of my personal opinion.)&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-01-26T18:08:59.277" LastActivityDate="2011-01-26T18:08:59.277" CommentCount="1" />
  <row Id="3906" PostTypeId="1" AcceptedAnswerId="3913" CreationDate="2011-01-26T04:26:19.397" Score="6" ViewCount="329" Body="&lt;p&gt;Does anyone have a good explanation of the physics and vectors of force involved in the skateboarding trick the ollie  (where the skater jumps and causes the skateboard to rise off the ground with him)?&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="1160" LastEditorDisplayName="" LastEditDate="2011-01-26T17:17:25.320" LastActivityDate="2011-03-10T16:13:56.157" Title="Physics of a skateboard ollie" Tags="&lt;forces&gt;&lt;vectors&gt;&lt;sports&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="3907" PostTypeId="2" ParentId="3903" CreationDate="2011-01-26T04:44:48.587" Score="10" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In classical mechanics, the lagrangians of two particles may be added only if the particules do not interract.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wouldn't say that. You can always write a Lagrangian $L$ for a system of two particles. In general, it takes the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = L_1 + L_2 + L_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $L_i$ is an interaction term that depends on the coordinates and/or velocities of both particles. If and only if the particles &lt;em&gt;don't&lt;/em&gt; interact, $L_i = 0$, and only in that case can you write the Lagrangian as the sum of individual particle Lagrangians $L_1$ and $L_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar idea applies in quantum field theory. Remember that QFT Lagrangian densities take forms like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{L}(\phi, \partial\phi) \sim (\partial\phi)^2 - m^2\phi^2 - \sum_n g_n\phi^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there are many different kinds, but in general there is always a kinetic term which involves the derivatives of the fields, and other terms which represent either the mass of the field or interactions between the field and itself or other fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in a sense, a derivative is a way of coupling the values of some object at different spacetime points. So it should make sense that the kinetic term of the actual Lagrangian&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L_\text{kin} \sim \int\mathrm{d}^3\mathbf{x}\ (\partial\phi)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;couples the values of the field $\phi$ at different points in spacetime. This is analogous to the term $L_i$ in the classical Lagrangian which involves the coordinates of multiple particles, except here, coordinates are replaced by fields and particles are replaced by locations. So you have a term that couples the fields at different spacetime points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice, though, that in the rest of the Lagrangian, there are no derivatives. This means that outside of the kinetic term, there is no connection between what happens at different points in spacetime. Specifically, the interaction terms&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int\mathrm{d}^3\mathbf{x}\ \sum_n g_n\phi^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are local, which means that all field interactions occur at a &lt;em&gt;single&lt;/em&gt; spacetime point. This is a simple way to ensure that interactions don't proceed differently when viewed from different reference frames. So there's no problem with integrating the interaction terms over all of space.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-01-26T04:44:48.587" />
  <row Id="3908" PostTypeId="2" ParentId="3901" CreationDate="2011-01-26T06:29:50.690" Score="4" ViewCount="" Body="&lt;p&gt;One of the simplest ways I know of to see the non-renormalizability of gravity is to note that the coupling constant in this case (Newton's constant $G$) is dimension-full with dimensions of $L^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The action for GR is of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S_{EH} = \frac{1}{8\pi G} \int d^4 x \sqrt{-g} R $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we try to expand this action as a power series in $1/G$ we see that each additional term has an extra power of $1/L^2 \sim k^2$, rendering the integration over momenta divergent in each term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is in contrast to QED where the coupling constant is the fine structure constant $\alpha = e^2/\hbar c \sim 1/137$ which is a dimensionless number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; For a more comprehensive and informed account see @Lawrence's answer below.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-26T18:00:24.193" />
  <row Id="3909" PostTypeId="2" ParentId="3892" CreationDate="2011-01-26T07:05:04.443" Score="2" ViewCount="" Body="&lt;p&gt;Dear Gordon, I hope that other QG people will write their answers, but let me write mine, anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, you need to distinguish the types of singularities because their character and fate is very different, depending on the type. You rightfully mentioned timelike, spacelike, and coordinate singularities. I will divide the text accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Coordinate singularities&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coordinate singularities depend on the choice of coordinates and they go away if one uses more well-behaved coordinates. So for example, there seems to be a singularity on the event horizon in the Schwarzschild coordinates - because $g_{00}$ goes to zero, and so on. However, this singularity is fake. It's just the artifact of using coordinates that differ from the &quot;natural ones&quot; - where the solution is smooth - by a singular coordinate transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as the diffeomorphism symmetry is preserved, one is always allowed to perform &lt;em&gt;any&lt;/em&gt; coordinate transformation. For a singular one, any configuration may start to &lt;em&gt;look&lt;/em&gt; singular. This was case in classical general relativity and it is the case for any theory that respects the symmetry structure of general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion is that &lt;em&gt;coordinate singularities can never go away&lt;/em&gt;. One is always free to choose or end up with coordinate systems where these fake singularities appear. And some of these coordinate systems are useful - and will remain useful: for example, the Schwarzschild coordinates are great because they make it manifest that the black hole solution is static. Physics will never stop using such singularities. What about the other types of the singularities?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Spacelike singularities&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most famously, these include the singularity inside the Schwarzschild black hole and the initial Big Bang singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite lots of efforts by quantum cosmologists (meaning string theorists working on cosmology), especially since 1999 or so, the spacelike singularities remain badly understood. It's mainly because they inevitably break all supersymmetry. The existence of supersymmetry implies the existence of time-translational symmetry - generated by a Hamiltonian, the anticommutator of two supercharges. However, this symmetry is brutally broken by a spacelike singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So physics as of 2011 doesn't really know what's happening near the very singular center of the Schwarzschild black hole; and near the initial Big Bang singularity. We don't even know whether these questions may be sharply defined - and many people guess that the answer is No. The latter problem - the initial Big Bang singularity - is almost certainly linked to the important topics of the vacuum selection. The eternal inflation answers that nothing special is happening near the initial point. A new Universe may emerge out of its parent; one should quickly skip the initial point because nothing interesting is going on at this singular place, and try to evolve the Universe. The inflationary era will make the initial conditions &quot;largely&quot; irrelevant, anyway. However, no well-defined framework to calculate in what state (the probabilities...) the new Universe is created is available at this moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mentioned the no-boundary initial conditions. I am a big fan of it but it is &lt;em&gt;not&lt;/em&gt; a part of the mainstream description of the initial singularity as of 2011 - which is eternal inflation. In eternal inflation, the initial point is indeed as singular as it can get - surely the curvatures can get Planckian and maybe arbitrarily higher - however, it's believed by the eternal inflationary cosmologists that the Universe cannot really start at this point, so they think it's incorrect to imagine that the boundary conditions are smooth near this point in any sense, especially in the Hartle-Hawking sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Schwarzschild singularity is different - because it is the &quot;final&quot; spacelike singularity, not an initial condition - and it's why no one has been talking about smooth boundary conditions over there. Well, there's a paper about the &quot;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0310281&quot; rel=&quot;nofollow&quot;&gt;black hole final state&lt;/a&gt;&quot; but even this paper has to assume that the final state is extremely convoluted, otherwise one would macroscopically violate the predictions of general relativity and the arrow of time near the singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the spacelike singularities remain badly understood, there exists no solid evidence that they are completely avoided in Nature. What quantum gravity really has to do is to preserve the consistency and predictivity of the physical theory. But it is &lt;em&gt;not&lt;/em&gt; true that a &quot;visible&quot; suppression of the singularities is the only possible way to do so - even though this is what people used to believe in the naive times (and people unfamiliar with theoretical physics of the last 20 years still believe so).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Timelike singularities&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The timelike singularities are the best understood ones because they may be viewed as &quot;classical static objects&quot; and many of them are compatible with supersymmetry which allowed the physicists to study them very accurately, using the protection that supersymmetry offers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And again, it's true that most of them, at least in the limit of unbroken supersymmetry and from the viewpoint of various probes, remained very real. The most accurate description of their geometry is singular - the spacetime fails to be a manifold, i.e. diffeomorphic to an open set near these singularities. However, this fact doesn't lead to any loss of predictivity or any inconsistency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest examples are orbifold singularities. Locally, the space looks like $R^d/\Gamma$ where $\Gamma$ is a discrete group. It's clear by now that such loci in spacetime are not only allowed in string theory but they're omnipresent and very important in the scheme of things. The very &quot;vacuum configuration&quot; typically makes spacetime literally equal to the $R^d/\Gamma$ (locally) and there are no corrections to the shape, not even close to the orbifold point. Again, this fact leads to no physical problems, divergences, or inconsistencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the string vacua compactified on spaces with orbifold singularities are equivalent - dual - to other string/M-theory vacua on smooth manifolds. For example, type IIA string theory or M-theory on a singular K3 manifold is equivalent to heterotic strings on tori with Wilson lines added. The latter is non-singular throughout the moduli space - and this fact proves that the K3 compactifications are also non-singular from a physics viewpoint - they're equivalent to another well-defined theory - even at places of the moduli spaces where the spacetime becomes geometrically singular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same discussion applies to the conifold singularities; in fact, orbifold points are a simple special example of cones. Conifolds are singular manifolds that include points whose vicinity is geometrically a cone, usually something like a cone whose base is $S^2\times S^3$. Many components of the Riemann curvature tensor diverge. Nevertheless, physics near this point on the moduli space that exhibits a singular spacetime manifold - and physics near the singularity on the &quot;manifold&quot; itself - remains totally well-defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This fact is most strikingly seen using mirror symmetry. Mirror symmetry transforms one Calabi-Yau manifold into another. Type IIA string theory on the first is equivalent to type IIB string theory on the second. One of them may have a conifold singularity but the other one is smooth. The two vacua are totally equivalent, proving that there is absolutely nothing physically wrong about the geometrically singular compactification. We may be living on one. The equivalence of the singular compactifications and non-singular compactifications may be interpreted as a generalized type of a &quot;coordinate singularity&quot; except that we have to use new coordinates on the whole &quot;configuration space&quot; of the physical theory (those related by the duality) and not just new spacetime coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very clear by now that some singularities will certainly stay with us and that the old notion that all singularities have to be &quot;disappeared&quot; from physics was just naive and wrong. Singularities as a concept will survive and singular points at various moduli spaces of possibilities will remain there and will remain important. Physics has many ways to keep itself consistent than to ban all points that look singular. That's surely one of the lessons physics has learned in the duality revolution started in the mid 1990s. Whenever physics near/of a singularity is understood, we may interpret the singularity type as a generalization of the coordinate singularities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, one should discuss lots of exciting physics that was found near singularities - especially new massless particles and extended objects (that help to make singularities innocent while preserving their singular geometry) or world sheet instantons wrapped on singularities (that usually modify them and make them smooth). All these insights - that are cute and very important - contradict the belief that there's no &quot;valid physics near singularities because singularities don't exist&quot;. Spacetime manifolds with singularities do exist in the configuration space of quantum gravity, they are important, and they lead to new, interesting, and internally consistent phenomena and alternative dual descriptions of other compactifications that may be geometrically non-singular.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-26T08:39:45.543" LastActivityDate="2011-01-26T08:39:45.543" CommentCount="4" />
  <row Id="3910" PostTypeId="2" ParentId="3903" CreationDate="2011-01-26T07:29:29.077" Score="7" ViewCount="" Body="&lt;p&gt;You are confusing the concepts of &quot;interactions&quot; and &quot;nonlocality&quot;. In realistic field theories, including all theories we ever used to study phenomena in the world around us, the interactions exist but they keep the physics local.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As David mentioned, the Lagrangian density takes the form&#xA;$${\mathcal L} = \sum_i \left[ (\partial_\mu \phi_i)^2 + m^2 \phi_i^2 \right] + O(\phi^{3+n}) $$&#xA;The sum over $i$ of the terms bilinear in $\phi$ or its first derivatives produces the free particles. But the higher-order terms - that I only wrote under the $O$ symbol - which are cubic, quartic, or of even higher orders - are responsible for all the interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the electromagnetic interaction between two charged objects boils down to their common interaction with the electromagnetic field via the interaction&#xA;$${\mathcal L_{em}} = j_\mu A^\mu $$&#xA;where $j_\mu$ is the 4-vector including the charge density and the flux. Because of this local term at one point, the electromagnetic field is perturbed by the first charge. The electromagnetic field $A^\mu$ continues to propagate to another charge particle, much like if it were a free field, and then the local interaction term of the type above &quot;clicks&quot; again and makes the second particle accelerate according to the first particle's position and charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This description is particularly optimized for quantum field theory where the photon is called the virtual particle - or a messenger of the interaction. However, even in classical field theory, one may use a similar language. Even the Lagrangian of a classical field theory that describes the electromagnetic interactions involving charge matter has a local form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understand, what you were proposing was that there would be bilocal or otherwise non-local terms in the Lagrangian&#xA;$$ L = \int d^{d-1} x\,{\mathcal L}_{local} + \int d^{d-1} x\, d^{d-1} y\, F(x) G(y) $$&#xA;which would directly attract or repel some densities $F,G$ that exist at any pair of points $x,y$, right? This never works in field theory at the fundamental level. The Lagrangian above is not local - it is not an integral of a density - so fields at point $x$ would be immediately influenced by fields at any other point $y$. This would violate locality - an action at a distance - and it would contradict relativity because when combined with the principle of relativity, a violation of locality also means a violation of causality (the rule that causes must precede their effects).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the bilocal Lagrangian above may be &quot;approximately derived&quot; by &quot;integrating out the electromagnetic field&quot;. I can't explain exactly what it means, especially because this term is only standard in quantum mechanics (in classical physics, it corresponds to &quot;solving $A_\mu$ away from the equations&quot;). However, let me just say the conclusion. The bilocal interaction terms between two charges you are thinking about may be &quot;approximately&quot; derived from the fully local Lagrangian I began with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because special relativity is a well-established fact about the reality and we have never observed any &quot;action at a distance&quot; - or interactions between two separated bodies occur because of a &quot;messenger&quot; that has to move along a path connecting the two objects - all Lagrangians of field theories we ever study are written as integrals of a Lagrangian density. This feature is called &quot;locality&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T15:38:27.470" LastActivityDate="2011-02-02T15:38:27.470" />
  <row Id="3911" PostTypeId="1" AcceptedAnswerId="3918" CreationDate="2011-01-26T08:14:09.573" Score="1" ViewCount="138" Body="&lt;p&gt;Trying to secure a wall hanging using magnets; me and a coworker came up with an interesting question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When the hanging is hung using 1 magnet, the weight of it causes it to quickly drag the magnet down and the hanging drops. Using &lt;em&gt;n&lt;/em&gt; magnets retards this process; causing it to fall more slowly, but does there exist a number of magnets &lt;em&gt;m&lt;/em&gt; such that their combined strength will prevent the hanging from slipping, entirely and permanently?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because this doesn't make for a very good question; we worked at it and arrived at a similar one; but slightly more idealized:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A weight is suspended, perfectly still, from a wire in a frictionless vacuum. If the mass of the weight is too great; it will gradually distend the cable, causing it to snap and release the weight; but will a light enough weight hang there indefinitely, or will the mass of the weight (and indeed the cable) cause the cable to snap sooner or later?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1598" LastActivityDate="2011-01-26T11:36:22.860" Title="Will tensile strength keep a cable from snapping indefinitely?" Tags="&lt;mechanics&gt;&lt;material-science&gt;" AnswerCount="3" FavoriteCount="0" />
  <row Id="3912" PostTypeId="1" CreationDate="2011-01-26T08:17:25.317" Score="2" ViewCount="202" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/9/hamiltons-principle&quot;&gt;Hamilton&amp;#39;s Principle&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;The Lagrangian formulation of Classical Mechanics seem to suggest strongly that &quot;action&quot; is more than a mathematical trick. I suspect strongly that it is closely related to some kind of &quot;laziness principle&quot; in nature - Fement's principle of &quot;least time&quot;, for example, seems a dangerously close concept - but I cannot figure out these two principles are just analogous, or if there's something deeper going on. Am I missing something obvious? Why is action&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A = \int T-V \, dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and what interpretation does the $T-V$ term in it have?&lt;/p&gt;&#xA;" OwnerUserId="214" LastActivityDate="2011-01-26T15:02:07.560" ClosedDate="2011-01-26T15:23:34.510" Title="Does Action in Classical Mechanics have a Interpretation?" Tags="&lt;classical-mechanics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="3913" PostTypeId="2" ParentId="3906" CreationDate="2011-01-26T08:27:02.200" Score="8" ViewCount="" Body="&lt;p&gt;A &lt;a href=&quot;http://www.youtube.com/watch?v=8ahDQQDp9es&quot; rel=&quot;nofollow&quot;&gt;slow-motion video&lt;/a&gt; of an ollie makes the physics clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main idea is that the skateboarder pushes down hard on one side out past the wheels, torquing the board up into a hop, then pushes down with the other foot to level the board out and make it appear to stick to the feet.  The skateboarder never lifts the board - the only necessary upward force in the whole process is the normal force from the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We begin by pushing down hard on the right-hand side of the skateboard, to the right of the wheels.  This creates a large torque about the right wheel and makes the skateboard start to rotate up off the ground.  The upward momentum is provided by a large normal force from the ground.  Here is a picture at the first moment, when the skateboard has zero velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Pokl9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red arrows show the forces.  The blue arrow shows the acceleration of the center of mass.  The torque about the right wheel is into your monitor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The skateboard gains a significant amount of momentum upwards in this way.  If we left it alone, it would fly up in the air, flipping, and moving to the right.  Instead, we push down on it on the left hand side.  This induces a counter torque and reduce's the skateboard's upwards momentum.  Here is a picture before the skateboard leaves the ground, right when the force on the left wheel is first applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/u9lf2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red arrows are forces and the blue arrow is the acceleration again.  The green arrow is the velocity of the center of mass.  The torque is now in the opposite direction - out of the page.  In the next moment, the skateboard's center of mass will continue to rise and move a little to the right, but also the skateboard will begin losing angular momentum.  It will stop rotating, then rotate back to the flat position.  In the mean time, its inertia will raise it off the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now both feet push on the board as we fall.  Here is a picture of the board at its highest point, with the velocity again zero, and the feet pushing equally down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/AwWna.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of these descriptions are not perfectly precise.  For example, the skateboard does not need to be perfectly flat or have equal force on both sides right when it attains the greatest height in its trajectory.  I omitted gravity (which is not really essential, except for the skateboarder).  There may be some other small forces involved, etc.  &lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-01-26T08:27:02.200" CommentCount="4" />
  <row Id="3914" PostTypeId="2" ParentId="3911" CreationDate="2011-01-26T08:58:08.970" Score="0" ViewCount="" Body="&lt;p&gt;Ultimately, the weight will fall.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the weight falls, it will collide with the floor and convert its gravitational potential energy into heat, resulting in an increase in entropy, so the process of the weight eventually falling is thermodynamically favored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible (and somewhat ludicrous) mechanism is &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_tunnelling&quot; rel=&quot;nofollow&quot;&gt;quantum tunneling&lt;/a&gt;.  A bizarre result of this mechanism is that the time we expect to wait before the weight tunnels increases exponentially with the square root of the weight's mass.  That  means a light object will fall to the ground before a heavy object, by a long shot!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a somewhat silly mechanism for a macroscopic weight to fall, but it's a theoretical limit.  My guess is that realistically the weight will fall because small statistical fluctuations in the wire and weight due to their thermal energies will eventually cause molecular-scale defects in crystal structure of the wire, evaporation of atoms from the wire, etc.  When enough of these defects have accumulated the wire strength will be low enough that the weight falls.  If you had a very light weight and you kept the box very cold and extremely-well mechanically isolated from all surroundings, it would take an incredibly long time for the weight to fall.  Nonetheless, the second law of thermodynamics says fall it must, eventually.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-01-26T08:58:08.970" />
  <row Id="3915" PostTypeId="2" ParentId="3911" CreationDate="2011-01-26T10:40:04.337" Score="0" ViewCount="" Body="&lt;p&gt;Metals normally have crystalline, ordered structure. An atom moving from one appropriate site of the crystal lattice to the next one has to overcome a high energy barrier. If the tension is too small to distort the crystal structure, atoms will stay in their places and your weight will hang forever&lt;sup&gt;*&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;*&lt;/sup&gt; By forever I mean long time for any practical purpose, like thousands of years. Sometimes atoms do randomly jump inside the crystal. By putting a tension we slightly favor one direction for these jumps so eventually the cable will stretch but it will rust much faster.&lt;/p&gt;&#xA;" OwnerUserId="701" LastActivityDate="2011-01-26T10:40:04.337" />
  <row Id="3916" PostTypeId="2" ParentId="3618" CreationDate="2011-01-26T11:05:51.297" Score="1" ViewCount="" Body="&lt;p&gt;by Hans de Vries (*): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://physics-quest.org/Magnetism_from_ElectroStatics_and_SR.pdf&quot; rel=&quot;nofollow&quot;&gt;The simplest, and the full derivation of Magnetism as a Relativistic side efect of ElectroStatics&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He uses only Electrostactic field and the non-simultaneity to obtain the Magnetic Field. He does explain it better than Purcell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnetic field is a side effect of movement in the electric field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) Hans de Vries has a very interesting online book (not yet finished) in his site, and he offers another pearl, not related to this post, but I feel compelled to share: The Lorentz contraction is a real effect and not only 'a referential effect' as we are tempted to believe.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-01-26T11:05:51.297" CommentCount="4" />
  <row Id="3917" PostTypeId="1" AcceptedAnswerId="3919" CreationDate="2011-01-26T11:07:25.383" Score="8" ViewCount="103" Body="&lt;p&gt;In &lt;a href=&quot;http://arxiv.org/abs/cond-mat/9907002v1&quot; rel=&quot;nofollow&quot;&gt;this review&lt;/a&gt; of the QHE, Steve Girvin makes the following statement (bottom of pg. 6, beginning of Sec. 1.1.1):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As one learns in the study of scaling in the localization transition, resistivity (which is what theorists calculate) and resistance (which is what experimental- ists measure) for classical systems (in the shape of a [d-dimensional] hypercube) of size L are related by [12,13]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$ R = \rho L^{(2-d)} $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where $R$ is the &lt;em&gt;resistance&lt;/em&gt; of the hypercube and $\rho$ its &lt;em&gt;resistivity&lt;/em&gt;. For $d=2$ it is clear that the resistance and the resistivity of the sample are the same, and as Girvin goes on to say, this fact is crucial for the universality of the quantization of the resistivity plateaus in the QHE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;strong&gt;Is there a simple physical explanation of the above relationship?&lt;/strong&gt; Girvin cites two papers. Both are RMP: &lt;a href=&quot;http://rmp.aps.org/abstract/RMP/v57/i2/p287_1&quot; rel=&quot;nofollow&quot;&gt;&quot;Disordered Electronic Systems&quot;&lt;/a&gt; by Lee and Ramakrishnan and &lt;a href=&quot;http://arxiv.org/abs/cond-mat/9609279&quot; rel=&quot;nofollow&quot;&gt;&quot;Continuous Quantum Phase Transitions&quot;&lt;/a&gt; by Sondhi et al. I am hoping that someone can spare me the effort of going through these two papers by providing an answer.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-26T12:53:44.233" Title="Resistance of a two-dimensional sample" Tags="&lt;condensed-matter&gt;&lt;quantum-hall-effect&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="3918" PostTypeId="2" ParentId="3911" CreationDate="2011-01-26T11:36:22.860" Score="2" ViewCount="" Body="&lt;p&gt;Your two questions are not really related, in my thoughts. &#xA;The first one is about friction of some magnets clutched to a ferromagnetic wall. &#xA;The second is about failiure of some &quot;wire&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both are strange and unnessecary mixtures of idealized classical &#xA;mechanics and some real world problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, first Question is really:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;does friction (at rest) last forever?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and second: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;does a &quot;wires&quot; stability against rupture last forever? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And answer(s): Yes in a surrounding of appropriate idealization, no in real world.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Georg&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-01-26T11:36:22.860" />
  <row Id="3919" PostTypeId="2" ParentId="3917" CreationDate="2011-01-26T12:13:40.430" Score="7" ViewCount="" Body="&lt;p&gt;It's all about scale/units. Resistivity is, by definition, the resistance you would measure if you had a &quot;wire&quot; of unit length and unit cross-sectional area: $$R = \rho * \textrm{length} / \textrm{cross-sectional area}.$$ For your hypercube of lengths $L$, this would correspond to a length of $L$ and cross-sectional area of $L^{d-1}$: $$R = \rho * L / L^{d-1} = \rho * L^{-(d-2)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the key is that resistivity has units which change! In 2D it happens to be exactly Ohms, but that actually hides a bit of &quot;pure geometry&quot;. It's really something like &quot;Ohms per square&quot; --- but the same rectangle has a different resistance depending on which way it is...&lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-01-26T12:13:40.430" CommentCount="2" />
  <row Id="3920" PostTypeId="2" ParentId="3794" CreationDate="2011-01-26T13:06:00.540" Score="2" ViewCount="" Body="&lt;p&gt;Compact Muon Solenoid is the extended name of CMS and points to the decision to have good muon identification. Good electromagnetic calorimetry ensures that photons and electrons  are measured well. &#xA;The choice of a strong field 4Tesla, to Atlas 2Tesla,  ensures better track momentum measurements. Having a limited volume for hadronic calorimetry is the price payed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rational is that muons electrons and photons give a better handle to new physics than jets en gross. The hadronic calorimeter is useful to trigger on missing Et for picking physics channels, but it will be the accuracy of muon and electron/photon measurements that will allow discrimination of various channels for the new physics expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it was Feynman who said: &quot;if you want to see the insides of watches you do not bang them against each other, you take a screw driver&quot;.  It is the reason why neutrino experiments and e+e- experiments were so successful in mapping the Standard Model: Photons electrons and muons because of the electroweak  interaction are the screw drivers (time reversed in the hadron collider case).&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-01-26T13:16:52.630" LastActivityDate="2011-01-26T13:16:52.630" CommentCount="1" />
  <row Id="3921" PostTypeId="2" ParentId="3360" CreationDate="2011-01-26T13:44:34.110" Score="2" ViewCount="" Body="&lt;p&gt;In SR with a central force (not gravity) you get from &lt;a href=&quot;http://en.wikipedia.org/wiki/Born_coordinates&quot; rel=&quot;nofollow&quot;&gt;Born coordinates&lt;/a&gt;, or substitution of classicla experssion   $v-&amp;gt;gamma*v$&lt;br&gt;&#xA;$F=k/r^2=m*(gamma*v)^2/r=L*v*gamma^2/r^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With c=1 this gives, $L=k/(v*gamma^2)=k*(1-v^2)/v$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the limit of that as v-&gt;1 gives L=0 as expected.&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-01-26T13:44:34.110" CommentCount="7" />
  <row Id="3922" PostTypeId="1" AcceptedAnswerId="3923" CreationDate="2011-01-26T13:49:46.880" Score="2" ViewCount="323" Body="&lt;p&gt;I have read in a couple of places that Ψ(p) and Ψ(q) are Fourier transforms of one another (e.g. Penrose).  But isn't a Fourier transform simply a decomposition of a function into a sum or integral of other functions?  Whereas the position and momentum wavefunctions are essentially different but related.  They must preserve expectation values like the relationship of classical mechanics, p=m(dq/dt) (where p and q are now expectation values).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a momentum wave packet that has a positive expectation value constant over time implies a position wave packet that moves over time in some direction. Simply saying there is Fourier transform seems to obscure this important relation.&lt;/p&gt;&#xA;" OwnerUserId="1602" LastActivityDate="2011-01-26T15:42:47.887" Title="What is the relation between position and momentum wavefunctions in quantum physics?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="3923" PostTypeId="2" ParentId="3922" CreationDate="2011-01-26T14:14:23.567" Score="6" ViewCount="" Body="&lt;p&gt;Dear user1602, yes, $\psi(x)$ and $\tilde\psi(p)$ are Fourier transforms of one another. This answers the only real question you have asked. So if one knows the exact wave function as a function of position, one also knows the wave function as a function of momentum, and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, there is no &quot;wave function&quot; that would depend both on $x$ and $p$. Indeed, such a &quot;wave function&quot; would violate a basic principle of quantum mechanics, the uncertainty principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wave function - that only depends on $x$ or that only depends on $p$ - remembers everything that a particle can and needs to remember about its position and momentum. For example, a good wave function describing a particle localized around $x_0$ and moving with momentum around $p_0$ is given by&#xA;$$\psi_{x_0,p_0}(x) = C \exp\left(-K(x-x_0)^2 + ip_0 x/\hbar\right)$$&#xA;The constant $K$ determines the width but you see that because of the quadratic term, the wave function is only non-vanishing near $x_0$. On the other hand, the $ipx$ term guarantees that the particle is moving to the right with the right momentum. It's all encoded in the changing phase of the wave function. The more quickly the phase of $\psi(x)$ changes with $x$, the higher is the momentum of the particle. If the phase rotates clockwise or counter-clockwise, the particle is moving to the right or to the left, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Fourier transform of the wave function above is something like&#xA;$$\tilde \psi_{x_0,p_0}(p)  = C&amp;#39; \exp\left(-(p-p_0)^2/K&amp;#39; - ip x_0/\hbar\right)$$&#xA;Just try it. Schrödinger's equation will guarantee that the wave packet is moving in the right direction - and by the right speed - encoded in $p_0$, and the center-of-mass position of the packet will change accordingly, too. The normalization constants $C,C&amp;#39;$ are physically irrelevant but may be chosen to normalize the state vectors to unity. The parameters $K,K&amp;#39;$ specifying the width are equal, up to a multiplication by a numerical constant and a power of $\hbar$: but it's true that the width in the $x$ representation is inverse to the width in the $p$ representation. That's implied by the uncertainty principle, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not true that one needs &quot;wave functions&quot; that would depend both on position and momentum. It's the whole point of the uncertainty principle that you may only specify the amplitudes with respect to one of these quantities - the other one doesn't commute with it. If one chooses $\psi(x)$, the position operator is a multiplication by $x$ and the momentum $p$ is simply the operator $-i\hbar\partial/\partial x$. Similarly, for $\tilde\psi(p)$, the momentum operator is the multiplication by $p$ and the position operator $x$ equals $+i\hbar \partial/\partial p$. It's pretty much symmetric with respect to $x,p$.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-26T14:14:23.567" CommentCount="2" />
  <row Id="3924" PostTypeId="2" ParentId="3912" CreationDate="2011-01-26T15:02:07.560" Score="1" ViewCount="" Body="&lt;p&gt;Before variations, T - V = L (Lagrangian) is a function of unknown independent variables. They are not the equation solutions yet, one cannot use them as solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the solutions are found, they can be used to calculate quantitatively some useful conserved quantities like the system energy, momentum, angular momentum expressed via L and its derivatives (see Noether theorems).&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-01-26T15:02:07.560" />
  <row Id="3925" PostTypeId="2" ParentId="3922" CreationDate="2011-01-26T15:42:47.887" Score="5" ViewCount="" Body="&lt;p&gt;If you have a state $|\psi\rangle $, the position wave function is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\langle \vec{x}\mid \psi\rangle&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the momentum wave function is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\langle \vec{p}\mid \psi\rangle = \int\;d\vec{x}\; \langle\vec{p}\mid\vec{x}\rangle\langle \vec{x}\mid\psi\rangle&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why the two expressions are &quot;different but related&quot; is the same why they preserve the expectation values. It is called &lt;strong&gt;completeness relation&lt;/strong&gt; $\int\; d\vec{x}\;\mid\vec{x}\rangle\langle\vec{x}\mid = 1$, $\int\; d\vec{p}\;\mid\vec{p}\rangle\langle\vec{p}\mid = 1$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\int\;d\vec{x}\; d\vec{x}^{\prime} \langle \psi\mid\vec{x}\rangle\langle\vec{x}\mid A\mid\vec{x}^{\prime}\rangle\langle\vec{x}^{\prime}\mid\psi\rangle&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;=\int\;d\vec{x}\; d\vec{x}^{\prime}\; d\vec{p}\; d\vec{p}^{\prime} \langle \psi\mid\vec{p}\rangle\langle \vec{p}\mid\vec{x}\rangle\langle\vec{x}\mid A\mid\vec{x}^{\prime}\rangle\langle \vec{x}^{\prime}\mid\vec{p}^{\prime}\rangle\langle\vec{p}^{\prime}\mid\psi\rangle&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;=\int\; d\vec{p}\; d\vec{p}^{\prime} \langle \psi\mid\vec{p}\rangle\langle\vec{p}\mid A\mid\vec{p}^{\prime}\rangle\langle\vec{p}^{\prime}\mid\psi\rangle&#xA;\end{equation}&lt;/p&gt;&#xA;" OwnerUserId="1337" LastActivityDate="2011-01-26T15:42:47.887" />
  <row Id="3926" PostTypeId="2" ParentId="3901" CreationDate="2011-01-26T16:20:46.030" Score="0" ViewCount="" Body="&lt;p&gt;Often infinite perturbative corrections are a sign of a &quot;bad start&quot; in the perturbation theory (providing the exact equations are correct). In Physics one often starts from incorrect equations too and &quot;repairs&quot; the wrong solutions &quot;on go&quot;, for example, by discarding infinite corrections to the fundamental constants. It &quot;works&quot; only in case of very simple &quot;perturbation&quot; Hamiltonians. In GR it is not the case (as in many other non-renormalizable QFT).&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-01-26T16:20:46.030" CommentCount="1" />
  <row Id="3927" PostTypeId="2" ParentId="1135" CreationDate="2011-01-26T16:30:57.720" Score="-2" ViewCount="" Body="&lt;p&gt;I think in general, one might interpret a distinction between kinematical and dynamical models as loosely referring to phenomenological descriptions (kinematical) versus. descriptions obtained from first principles (dynamical)&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-01-26T16:30:57.720" CommentCount="1" />
  <row Id="3928" PostTypeId="1" AcceptedAnswerId="3938" CreationDate="2011-01-26T16:43:32.697" Score="1" ViewCount="121" Body="&lt;p&gt;Previous posts such as &lt;a href=&quot;http://physics.stackexchange.com/questions/907/type-of-stationary-point-in-hamiltons-principle&quot;&gt;this&lt;/a&gt; ask about types of stationary point in Hamilton's Principle. There is, however, another aspect to discuss: the question as to whether the extremal path is unique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One geometric way to envisage this is to assume that multiple paths are simultaneously extremal. I believe that this is an explanation for lenses, but I have not seen lenses explained as multiple classical solutions to Hamilton's Principle. (The multiple paths being the 360 degrees of rays between source and focus, etc also demonstrable through Fermat's principle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can generalise lenses, but also consider a simpler case. Let the surface of a sphere be the action (phase space) surface which is minimized in classical paths. Thus (ignore antipodals here) between two points A and B the geodesic is the unique classical path. In quantum form the WKB approximation would no doubt have constructive maxima on this path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if the sphere has a disk (containing that geodesic) cut out, the shortest path now has exactly two choices: around one or the other rim from A to B. Presumably WKB would maximize the quantum paths on these two (although I havent proved this). If so then classically we have a quantum-like phenomenon: a particle has a choice in going from A to B. Experimentalists might see this and wonder whether the particle went from A to B via the LHS, the RHS or both....&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-26T21:51:39.560" LastActivityDate="2011-01-26T21:51:39.560" Title="Multiple classical paths from Hamilton's principle" Tags="&lt;hamiltonian-mechanics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="3929" PostTypeId="2" ParentId="3928" CreationDate="2011-01-26T17:00:01.380" Score="1" ViewCount="" Body="&lt;p&gt;Dear Roy, for a chosen initial configuration $x_i(t_i)$ and final configuration $x_f(t_f)$, there can exist more than one local extrema of the action. However, this point is irrelevant in classical physics. At every moment, including the initial moment $t_i$, the particle also has a well-defined velocity $\dot x_i(t_i)$, and the principle of least action is just a way to derive the differential equations that determine the motion. For a given $x(t_i)$ and $\dot x(t_i)$, the evolution will be inevitably unique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, the initial velocity either says that the particle will avoid the removed disk from the left side, or from the right side, or it will hit the missing disk (and perhaps gets reflected from it) - unless you include some potential that repels the particle from the disk, there is no reason why the particle should be obliged to avoid it. So no issue of the type you mention exists in classical physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Quantum mechanics&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation is different in quantum mechanics. All trajectories contribute and as the double-slit experiment shows, there may be interference between many classical histories. The interference pattern in the double-slit experiment may be obtained by adding the neighborhood of &quot;two classical trajectories&quot; only - these trajectories are piecewise linear and go through the two slits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum tunneling may also be phrased in terms of a contribution of complexified trajectories in the complexified time - that are also local extrema of the action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A much more interesting situation arises in quantum field theories. The vacuum - that lasts eternally - is a global minimum of the action. However, in the Euclidean spacetime, they may also exist other local minima. Because they're local minima, they also solve the classical equations of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum field theory, such solutions are called &lt;em&gt;instantons&lt;/em&gt; because they are localized both in space and in the Euclidean time (near one instant of time). They contribute to probabilities of various processes because one must sum over all histories, including those that get mapped to the instanton if one uses the Euclidean formalism (via the Wick rotation). In particular, they produce the 't Hooft interaction in gauge theories - a product of all fermions in the theory (a fact that is derived from the fermionic zero modes on the background of the instanton).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instantons have to be stable (minima of the Euclidean action, not maxima) and they're typically protected by a topological charge by which they differ from the vacuum configuration (some &quot;winding number&quot; or &quot;homotopy&quot;). There can also exist unstable solutions - the saddle points that are minima with respect to most directions but they are maxima with respect to a finite number of directions in the configuration space. Such saddle-like mixed extrema are called &lt;em&gt;sphalerons&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-26T17:00:01.380" CommentCount="2" />
  <row Id="3930" PostTypeId="2" ParentId="3901" CreationDate="2011-01-26T17:03:05.023" Score="3" ViewCount="" Body="&lt;p&gt;The number of divergences is not finite.  In QED and other YM theories the divergences are finite and may be absorbed into the renormalization of mass or gauge charge.  The rules for anomaly cancellations may be established in a finite description as well.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With quantum gravity in a QFT setting has a quadratic momentum dependence on the graviton vertex $V[k]~\sim~k^2$.  This is due to the form the action takes in an expansion of the metric density ${\tilde g}_{\mu\nu}~=~g^{1/2}g_{\mu\nu}$ with&#xA;$$&#xD;&#xA;{\tilde g}_{\mu\nu}~=~\eta_{\mu\nu}~+~\kappa^2{\phi^\alpha}_\mu\phi_{\alpha\nu}.&#xD;&#xA;$$&#xA;The action $L~\sim~\kappa^{-2}\phi^{\alpha\mu;\alpha}{\phi_{\alpha\mu}}^{;\alpha}$ gives a three point vertex function which is quadratic in the momentum.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A general Feynman diagram will also have internal lines $I[k]~\sim~1/k^2$ and loops with $L[k]~\sim~\int^kd^4p$.  So the internal portion of a Feynman diagram will have internal lines, vertices and loops.  The Euler characteristic for a graph &#xA;$$&#xD;&#xA;1~=~V[k]~-~I[k]~+~L[k]&#xD;&#xA;$$&#xA;is used in conjunction with the degree of divergence of these parts of the graph $D_V~=~2$ $D_I~=~-2$ $D_L~=~4$ with a total divergence $D~=~4L[k]~-~2I[k]~+~2V[k]$, so that&#xA;$$&#xD;&#xA;D~=~2(L[k]~+~1).&#xD;&#xA;$$&#xA;Consequently the divergence has an unbounded growth with the order of each Feynman diagram.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed out above the gravitational constant has units of $[G]~=~Area$, which differs from the fine structure constant $\alpha~=~e^2/\hbar c$  and other gauge couplings which are unitless.  The dimensional content of the gravitational constant is related to the problem of quadratic vertices.  The interest in holography and AdS/CFT correspondence is with how gravity in an AdS space with negative Gaussian curvature may be replaced by quantum field on the boundary.  So the divergence in a naïve QFT theory of gravity may be substituted with a stringy theory on the boundary of a space where these divergences do not exist.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-01-26T17:10:03.410" LastActivityDate="2011-01-26T17:10:03.410" CommentCount="1" />
  <row Id="3931" PostTypeId="2" ParentId="1135" CreationDate="2011-01-26T17:10:48.123" Score="0" ViewCount="" Body="&lt;p&gt;Another thing we can say is that kynematics problems are a class of mathematical problems. For example, all treatement of kynematics in Classical Mechanics can be done only with Differential Geometry of Curves, without using Newton' Second Law. Even Special Relativity is a problem of Differential Geometry, but in Minkowski Spacetime.&lt;/p&gt;&#xA;" OwnerUserId="1585" LastActivityDate="2011-01-26T17:10:48.123" />
  <row Id="3932" PostTypeId="1" CreationDate="2011-01-26T17:25:00.813" Score="0" ViewCount="75" Body="&lt;p&gt;Don't they pour down just the same amount of water at the end of the day?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean in the tropical countries.&lt;/p&gt;&#xA;" OwnerUserId="1603" LastEditorUserId="1603" LastEditorDisplayName="" LastEditDate="2011-01-26T17:45:03.317" LastActivityDate="2011-01-26T17:45:03.317" ClosedDate="2011-01-26T18:36:49.883" Title="Why is sometime it raining heavily but in a short time but in other time not so heavy but long?" Tags="&lt;water&gt;&lt;geophysics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="3933" PostTypeId="1" AcceptedAnswerId="3939" CreationDate="2011-01-26T17:32:46.147" Score="8" ViewCount="639" Body="&lt;p&gt;In our Physics lab we have a 1 milliwatt (0.001W) helium neon laser. Despite the low power, we were cautioned not to even look at reflections of the beam as it could cause permanent eye damage - why is such a low power level hazardous? I have a 3W LED which I can look directly into which is more than 3,000x more powerful.&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-27T11:18:39.513" LastActivityDate="2011-01-27T11:18:39.513" Title="Why is a 1mW laser dangerous?" Tags="&lt;optics&gt;&lt;home-experiment&gt;&lt;laser&gt;&lt;power&gt;&lt;led&gt;" AnswerCount="7" CommentCount="7" FavoriteCount="3" />
  <row Id="3934" PostTypeId="2" ParentId="3932" CreationDate="2011-01-26T17:45:00.587" Score="0" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;light rains that last for a short time&lt;/li&gt;&#xA;&lt;li&gt;light rains that last for a long time&lt;/li&gt;&#xA;&lt;li&gt;heavy rains that last for a short time&lt;/li&gt;&#xA;&lt;li&gt;heavy rains that last for a long time&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And of course, anything in between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of precipitation that you get varies; it is not true that you always get the same amount of water and it is meaningless cherry-picking to compare different rains just because they produce the same total amount of water. Rains are usually started in regions where enough moisture and upward motion is present. However, depending on the situation, the humidity may differ, and so may the pressure gradient and temperature gradient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a certain pressure and/or temperature, droplets start to form, grow, and fall. The greater the pressure gradient (pressure difference between two points separated by a fixed distance) is, the bigger chance that you get speedy, heavy rains. So heavy rains often come with hurricanes - when the pressure gradient is large, and so is usually the wind speed. Fast rains also become more likely when the temperature gradient is high - the temperature strongly depends on the location. However, there are also many other factors that influence the character and strength of rains. For example, very heavy and fast &quot;torrential rains&quot; have been linked to the solar activity although the link remains somewhat inconclusive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may want to read the Wikipedia article about rain:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rain&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Rain&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It also tells you how the rain is usually classified and what are the typical geometries at different places that lead to rain - things depend on the altitude, mountains around, latitude, and other things. The atmosphere is a complex system and the character of rains is determined by pretty much all of its complexities.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-26T17:45:00.587" />
  <row Id="3935" PostTypeId="2" ParentId="3933" CreationDate="2011-01-26T17:47:34.830" Score="5" ViewCount="" Body="&lt;p&gt;Because the light from the laser is collimated. This means that you have all this energy focused on a small spot. The larger power from a non collimated light source is spatially spread and the effective amount of energy that reaches your retina per time is much smaller.&lt;/p&gt;&#xA;" OwnerUserId="1337" LastActivityDate="2011-01-26T17:47:34.830" />
  <row Id="3936" PostTypeId="2" ParentId="3933" CreationDate="2011-01-26T17:48:00.617" Score="1" ViewCount="" Body="&lt;p&gt;If you have ever looked at a laser beam expanded and projected on a screen you may have noticed it has a speckled appearance.  A red HeNe beam spot after it passes through a divergent lens is not a uniform illumination.  Those spots are little regions with much higher irradiance.  If that beam goes directly into the eye, where the lens focuses the light, some retinal cells can be zapped.  If you get even low power laser beam directly into the eye with considerable repetition you can accumulate retinal damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course now this is compounded by the green lasers of considerably higher power that knuckleheads use to play as Starwars light sabers or worse to flash at aircraft.  These are only to point at the sky, where it reflects off the sodium layer and lets one point out stars etc&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-26T17:48:00.617" CommentCount="5" />
  <row Id="3937" PostTypeId="2" ParentId="3933" CreationDate="2011-01-26T17:49:18.197" Score="5" ViewCount="" Body="&lt;p&gt;As far as I know you have to worry about how much energy being deposited per surface area. And the area of the &quot;hotspot&quot; of the laser can be very small. So the deposited energy it enough to kill cells on your retina.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually there is a whole &lt;a href=&quot;http://en.wikipedia.org/wiki/Laser_safety&quot; rel=&quot;nofollow&quot;&gt;article on Wikipedia&lt;/a&gt; about it.&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-01-26T17:49:18.197" CommentCount="1" />
  <row Id="3938" PostTypeId="2" ParentId="3928" CreationDate="2011-01-26T17:54:39.410" Score="1" ViewCount="" Body="&lt;p&gt;[Another comment to answer transplant]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like you're asking about a classical analog to the superselection sectors of quantum mechanics. One situation where this occurs in classical mechanics is when considering particle motion of a manifold with non-trivial topology - i.e. with holes and handles. In such cases there can be more than one extremal path from A to B. An example is an arcade pin-ball machine, if you're familiar with those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, when you talk about optics it is important to keep in mind that there are two different regimes, those of wave optics and geometric optics. In the second case one has well-defined &quot;trajectories&quot; and you can find extremal trajectories. Not so in the first case.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-26T17:54:39.410" />
  <row Id="3939" PostTypeId="2" ParentId="3933" CreationDate="2011-01-26T17:57:31.833" Score="12" ViewCount="" Body="&lt;p&gt;Dear Thomas, the diameter of the beams of these HeNe lasers is between 0.5 and 1 millimeter, so the power 1 mW is coming to $10^{-6}$ squared meters or so. The ratio of power and area is $10^{-3}/10^{-6} = 10^{+3}$ Watts per squared meter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, when a 3W LED is watched from the distance 0.1 meters, the power of 3 W is divided to $4\pi R^2 = 0.13$ squared meters, so the power per units area is 3/0.13 = 23 Watts per squared meter. In both cases, the eye will refocus the beam so the power per unit area of the retina will actually be much higher in both cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this estimate, the laser has about 40 times higher power per unit area than the LED light bulb. Of course, it's small, so it will only burn one small point in your retina, but that's a bad enough problem. ;-)&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-27T11:09:59.087" LastActivityDate="2011-01-27T11:09:59.087" CommentCount="6" />
  <row Id="3940" PostTypeId="2" ParentId="3906" CreationDate="2011-01-26T17:58:04.290" Score="2" ViewCount="" Body="&lt;p&gt;An important part I think is that the board acts as a spring.  When the skater pushed down to leap up this stores energy in the board which is released to pop the board up right under the skater's feet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My son does all of these things with great ease, and does fancy stuff on half pipes and the like.  If I tried these things I suspect a month later I would get some hospital bills.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-26T17:58:04.290" />
  <row Id="3941" PostTypeId="1" CreationDate="2011-01-26T18:12:34.843" Score="5" ViewCount="212" Body="&lt;p&gt;One of the key elements of any quantum mechanical system is the spectrum of the Hamiltonian.  But what about in quantum field theory?  It seems as if nobody ever discusses the spectrum of a system at all -- or have I missed something?  Just what role does the spectrum of any given potential play in field theory?  To put it another way:  does knowledge of a spectrum in QM help to understand the analogous problem in QFT?&lt;/p&gt;&#xA;" OwnerUserId="1605" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-26T18:40:27.477" LastActivityDate="2011-01-26T21:54:06.207" Title="The energy spectrum in quantum field theory" Tags="&lt;quantum-field-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="3942" PostTypeId="2" ParentId="3941" CreationDate="2011-01-26T18:25:04.780" Score="4" ViewCount="" Body="&lt;p&gt;Because of Lorentz Invariance, the spectrum of the Hamiltonian is always continuous. This does not mean bound state problems are not interesting, but the relevant discrete spectrum to discuss is the spectrum of (Lorentz invariant) masses. For an example think about bound state of electron and positron (positronium), in which they orbit each other. There is a discrete spectrum of masses of this composite object, to do with what is the angular momentum of their orbit around the center of mass. But of course, if you boost the system as a whole the energy of that moving object (which unlike the rest mass is not Lorentz invariance) will change and can take continuous values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other reason you hear less about the spectrum in QFT is because the problem of finding the spectrum in bound state problems is a lot harder (to define and to calculate). If you want an entry point to relevant discussions you can look for references for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bethe%E2%80%93Salpeter_equation&quot; rel=&quot;nofollow&quot;&gt;Bethe-Salpeter equation&lt;/a&gt;, and to &lt;a href=&quot;http://en.wikipedia.org/wiki/Lattice_QCD&quot; rel=&quot;nofollow&quot;&gt;lattice QCD&lt;/a&gt;, two of the many approaches that discuss spectrum in QFT.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-26T18:37:13.267" LastActivityDate="2011-01-26T18:37:13.267" CommentCount="7" />
  <row Id="3943" PostTypeId="2" ParentId="3933" CreationDate="2011-01-26T18:44:47.580" Score="4" ViewCount="" Body="&lt;p&gt;1 mW is in fact not extremely dangerous. &#xA;Such lasers are sold to anybody as pointing devices. &#xA;But the &quot;philosophy&quot; of security is a lot about &quot;habits&quot;. &#xA;Handling of a laser should be trained like handling &#xA;a gun: never aim at a person!&#xA;BTW those 3 or more Watt LEDs can be dangerous as well!&#xA;Georg&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-01-26T18:44:47.580" CommentCount="4" />
  <row Id="3944" PostTypeId="1" AcceptedAnswerId="3981" CreationDate="2011-01-26T18:48:25.390" Score="4" ViewCount="111" Body="&lt;p&gt;I read that if we have a system with two co-existing phases with  chemical potentials respectively $\mu_1$ and $\mu_2$ then, at the equilibrium, the concentrations $X_1$ and $X_2$ are related by the relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X_1=X_2e^{[\frac{-(\mu_1-\mu_2)}{kT}]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that true? How can I derivate this expression?&lt;/p&gt;&#xA;" OwnerUserId="592" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2011-01-27T11:08:08.140" LastActivityDate="2011-01-27T11:13:38.427" Title="Boltzman distribution for chemical potentials" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="3945" PostTypeId="1" AcceptedAnswerId="3947" CreationDate="2011-01-26T19:18:49.430" Score="5" ViewCount="211" Body="&lt;p&gt;In somes references of Relativistic Quantum Mechanics, the one-particle states are given by representation theory of Poincaré algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I mimics this for the non-relativistic case? States in non-relativistic quantum mechanics could be seen as representation of Galilei algebra (or Galilei group)?&lt;/p&gt;&#xA;" OwnerUserId="1545" LastActivityDate="2011-01-26T20:04:25.997" Title="Could I see the quantum states as representations of the Galilei algebra (or Galilei group)?" Tags="&lt;quantum-mechanics&gt;&lt;lie-algebra&gt;&lt;group-representations&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="3947" PostTypeId="2" ParentId="3945" CreationDate="2011-01-26T19:57:48.220" Score="3" ViewCount="" Body="&lt;p&gt;Dear Leandro, in the relativistic case, the full Hilbert space is a unitary representation of the full Poincaré group. However, one-particle states are representations of the translations simply because they carry momenta $p$ and under translations by $\Delta x^\mu$, they get multiplied by the phase $\exp(ip\cdot\Delta x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The translation generators of the Poincaré groups commute with each other, so you may diagonalize the Hilbert space and divide it into subspaces or &quot;sectors&quot; with fixed values of $d$-momenta - imagine $d=4$ in the most well-known dimensionality. These subspaces have well-defined transformation laws under the translations and under the boosts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you look at the subspace corresponding to the momentum $k^\mu$, it contains states that are eigenstates of translations. They map to completely different states - in different sectors - under the boosts. However, we want to classify particles with momentum $k^\mu$ according to the rotations. Why is it so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's because the little group is defined as the subgroup of the Lorentz group that preserves $k^\mu$ - so it's exactly the group that maps the sector with a given value of $k^\mu$ onto itself. For massive particles, the little group is $SO(d-1)$; for massless particles, it is a semidirect product of $SO(d-2)$ with some $d-2$ semi-light-like mutually commuting generators of the Lorentz group, $J_{i-}$. For tachyons, if they could exist, the little group would be $SO(d-2,1)$ - a signature flip from the massive particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the particles are then classified according to representations of the little group. For the massless particles, only the compact part $SO(d-2)$ is really important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the Galilean group is a contraction (infinite $c$ limit) of the Poincaré group; it has the same number of generators. In fact, the search for the representations would follow the very same prescription. You may also define energy and momenta - except that in non-relativistic theories, these two types of a quantity are not related by any speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you could also look at a given sector with a fixed energy and fixed 3-momentum, and find out that the sector is only preserved by the little group. Again, the boosts - in the non-relativistic case, they're simply $\vec x&amp;#39;=\vec x+\vec v t$ - would map the sector onto other sectors. The little group would always be $SO(d-1)$, like for massive relativistic particles, because $c$ is sent to infinity so no particles can move by the speed of light or faster. All of them resemble the relativistic, slowly moving massive particles. Well, you could have &quot;instantaneous particles&quot; (a new type of representation) that move by the infinite speed but they would cause lots of problems to the theory - and you could integrate their fast effects out immediately, anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, you will find out that the non-relativistic particles simply carry some energy, some momentum, and transform under the little group - the usual $(2j+1)$-tuplet we know from non-relativistic quantum mechanics in the case $d=4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a somewhat different question how one could construct and classify off-shell fields and field theories (and their Lagrangians if they have one). Obviously, the time coordinate may be treated as a scalar and off-shell fields only have to form representations under $SO(d-1)$. The non-relativistic theories are much less constrained because one may break a spacetime tensor from relativity into time-like, space-like, and perhaps various mixed components and treat them separately, erase some of them while keeping others, give them different interactions etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theories may be nonlocal - because there is no speed-of-light limit on the velocity - which makes the theories even less constrained. The invariance under the non-relativistic boosts is also easier to satisfy.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-26T20:04:25.997" LastActivityDate="2011-01-26T20:04:25.997" CommentCount="1" />
  <row Id="3948" PostTypeId="2" ParentId="3889" CreationDate="2011-01-26T19:59:20.513" Score="0" ViewCount="" Body="&lt;p&gt;the real free bosonic quantum field with mass m &gt; 0 is a Wightman field, because one can define it as an operator valued distribution that satisfies the Klein-Gordon equation in the sense that $ \phi(\Box f + m^2 f) = 0 $ for all test functions f. The product $\phi(x) \phi(x)$ is then simply not defined for well known reasons (you don't have to look at the two point functions, already the field itself is not well defined). I assume that you know all this and your question was if physicists know a trick around this problem: As far as I know there is no such trick within the Wightman framework. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to make sense of this &quot;field&quot; I think we would need to leave the Wightman axioms and allow more severe singularities of two point functions on the diagonal, which are usually described via operator product expansions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum: When we have a representation of a distribution as a boundary value of an analytic function, it is not true that we can consistently define e.g. the square of this distribution as the boundary value of the square of the analytic function (if this would work, we would have found a solution to a problem where our math friends have already proven that there is no solution, namely to define $\delta^2(x)$). The reason for this is that we are actually talking about an equivalence class of analytic functions instead of one unique analytic function (equivalence means equal up to an analytic function that has an analytic extension to the boundary). Multiplication is not well defined on the equivalence classes. For more details see the buzzword &quot;hyperfunction&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditorDisplayName="" LastEditDate="2011-01-26T23:46:43.147" LastActivityDate="2011-01-26T23:46:43.147" CommentCount="4" />
  <row Id="3949" PostTypeId="2" ParentId="437" CreationDate="2011-01-26T20:34:06.810" Score="2" ViewCount="" Body="&lt;p&gt;There is a slight confusion in this question. In quantum field theory, the Dirac equation and the Schrödinger equation have very different roles. The Dirac equation is an equation for the field, which is not a particle. The time evolution of a particle, ie, a quantum state, is always given by the Schrödinger equation. The hamiltonian for this time evolution is written in terms of fields which obey a certain equation themselves. So, the proper answer is: Schrödinger equation with a hamiltonian given in terms of a massless vector field whose equation is nothing else but Maxwell's equation.&lt;/p&gt;&#xA;" OwnerUserId="1337" LastActivityDate="2011-01-26T20:34:06.810" />
  <row Id="3950" PostTypeId="1" CreationDate="2011-01-26T20:35:56.120" Score="0" ViewCount="133" Body="&lt;p&gt;What would the size of the universe be if it were physically possible to remove all of the empty space, leaving only matter?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-26T21:52:26.307" ClosedDate="2011-01-26T21:48:08.147" Title="Volume of Matter in the Universe" Tags="&lt;cosmology&gt;&lt;condensed-matter&gt;&lt;vacuum&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="3951" PostTypeId="2" ParentId="3950" CreationDate="2011-01-26T21:16:47.880" Score="1" ViewCount="" Body="&lt;p&gt;Yes, well making some assumptions, you would simply get a black hole with a radius $2GM/c^2$, so plug in values.&#xA;You would have to decide if the radius is to the horizon of the observable universe.&#xA;Wiki entry &quot;observable universe&quot; estimates the combined mass of normal and dark matter to be around $1.5 \times 10^{53}$ kg. The size of the &quot;entire&quot; universe cannot be estimated. And &quot;empty space&quot; is hard to characterize--as Frank Wilczek (via Sidney Coleman) says, &quot;Nothing is unstable.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-01-26T21:52:26.307" LastActivityDate="2011-01-26T21:52:26.307" CommentCount="1" />
  <row Id="3952" PostTypeId="2" ParentId="3933" CreationDate="2011-01-26T21:47:52.237" Score="1" ViewCount="" Body="&lt;p&gt;A laser is coherent, collimate and intense&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be a small amount of power but all that comes down a precise beam, in the same phase, at the same time. Your eye is a lens and focuses it into an even smaller spot, burning the retina.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try staring at a 200w bulb for 1 minute, note the temporary black spots when you look away. That 200w is radiating in all directions, is incoherent, and wastes a huge amount as heat. The 0.1mw laser delivers more in a shorter time period to your retina than the bulb will&lt;/p&gt;&#xA;" OwnerUserId="1612" LastActivityDate="2011-01-26T21:47:52.237" />
  <row Id="3953" PostTypeId="2" ParentId="3941" CreationDate="2011-01-26T21:54:06.207" Score="0" ViewCount="" Body="&lt;p&gt;I just want to note that in the first chapters of your favourite texbook, whe you quantize the basic scalar field, you are actually finding a spectrum of your hamilonian, don't you?&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-01-26T21:54:06.207" />
  <row Id="3954" PostTypeId="2" ParentId="3889" CreationDate="2011-01-26T21:55:34.807" Score="2" ViewCount="" Body="&lt;p&gt;If I am not wrong the product is well defined, because the 2 point funtion is boundary value of an analytic function in some tube (see Streater/Wightman) with certain bounds.&#xA;More genral there is 1-1 correspondence between translation invariant tempered distributions which satisfy the spectrum condition and analytic functions in this mentioned tube and &#xA;an easy way to define the product of the distribution is the product of the &#xA;analytic functions which lies in the same class and therefore gives a well defined distribution on the boundary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really easy is the massless case in let's say 4D. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W(x-y) \sim \frac1{(x-y+i\epsilon(x_0-y_0))^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W^2(x-y) \sim \frac1{(x-y+i\epsilon(x_0-y_0))^4}$&#xA;where the two-point function is just the boundary value of the analytic function $1/z^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1 dimension, that means a chiral field on a light ray, it is even more drastic. &#xA;There the Wick (=normal ordered) square of the free fermion is a free boson.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The answer to the question &quot;Is a normal-ordered product of free fields at a point a Wightman field?&quot; is: YES!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For convinience I give you a reference, which is &quot;General Principles of Quantum Field Theory&quot; by Bogolubov, Logunov, Oksak, Todorov (1990) &#xA;p. 344&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ex. 8.16 Prove that the Wick Monomials satisfy the Wightman Axioms&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt; Tim: I can not comment on your yet. I did not say you can multiply general distribution which are boundary values by multiplying the function, just for a certain class but maybe I was sloppy because I thought this was standard. &#xA;The $\delta$ is not in this class, because it does not fullfill a &quot;spectrum contion&quot;, i.e. the fourier transform is supported in some convex cone with non-empty dual cone. The fourier transform of $\delta(x)$ is constant so does not fall into this class. &#xA;But distributions which have this property on their fourier transformation you can multiply:&#xA;see the standard textbook Reed Simon Volume 2 - page 92 Ex 4. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope my references clear the confusion.&lt;/p&gt;&#xA;" OwnerUserId="1611" LastEditorUserId="1611" LastEditorDisplayName="" LastEditDate="2011-01-27T11:17:24.147" LastActivityDate="2011-01-27T11:17:24.147" CommentCount="8" />
  <row Id="3955" PostTypeId="1" CreationDate="2011-01-26T22:36:29.073" Score="9" ViewCount="442" Body="&lt;p&gt;This question arose in a seminar today about the solar wind...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my vagueish understanding of the problem - please correct if you see errors!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 'classical' picture of atmospheric electricity is that the Earth as a whole is neutral, but that thunderstorms maintain a voltage of around +300kV at the electrosphere with respect to the Earth's surface, with a current of around 1 kA slowly discharging around 500 kC of total charge separation. The solar wind is supposed to neutralise any net charge that might be there between the Earth as a whole and the solar wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, positive and negative charges in the solar wind are differently trapped in the van Allen belts, from which they can then descend to the Earth's atmosphere, which implies that a net charge can be developed due to this differential leakage. This begs the question of whether there are any estimates of the total net charge. I've hunted in the literature but have found little useful material other than Dolezalek's 1988 paper: &lt;a href=&quot;http://www.springerlink.com/content/u057683112l148x5/&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/u057683112l148x5/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone offer an explanation, or point me to some more relevant papers?&lt;/p&gt;&#xA;" OwnerUserId="1610" LastActivityDate="2011-02-08T11:27:47.923" Title="What is the net charge of the Earth?" Tags="&lt;electrostatics&gt;&lt;geophysics&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="3956" PostTypeId="2" ParentId="3955" CreationDate="2011-01-26T22:59:42.283" Score="0" ViewCount="" Body="&lt;p&gt;My understanding for this is that electrons weigh a lot less than anything else so they have higher velocities (at the same temperature, because of thermodynamics). For matter to escape the earth requires a minimum velocity. So the result is that the earth loses more electrons than positive ions (or negative ions). The result is that the earth is positively charged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, in the earth's atmosphere, the transfer of charge is due to rain. Charged ions and electrons are attracted to raindrops for the same reason that charges are attracted to a conducting surface, that is, the image charges. But electrons are lighter than ions and so electrons get to the raindrops before ions and so raindrops are net negatively charged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since raindrops fall to earth, the earth ends up negatively charged while the atmosphere above it is positively charged. This is the source of the &quot;sky voltage&quot;:&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sky_voltage&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sky_voltage&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-02-08T04:49:48.297" LastActivityDate="2011-02-08T04:49:48.297" />
  <row Id="3957" PostTypeId="1" AcceptedAnswerId="4732" CreationDate="2011-01-26T23:32:42.650" Score="9" ViewCount="174" Body="&lt;p&gt;The heliopause is now estimated to be something around 100 AU (1 AU = Astronomical unit = about the earth sun distance). See the wikipedia article:&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Heliosphere&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Heliosphere&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From reading a book on NASA's Voyager mission, I learned that before the launch of these spacecraft, the expectation was that the heliopause was at around Jupiter or Saturn's orbit which is about 5 to 10 AU. The book says that as the spacecraft continued to move away from the sun, the space physicists kept increasing their estimate of the position of the heliopause. If it's about an energy or flux balance, then the estimate would be defined by a certain area of the sphere where the heliopause occurs, and since areas are proportional to the square of the radius they got the number wrong by as much as $(100/5)^2 = 400$. That's a huge underestimate of the sun's output or a huge overestimate of what goes on in interstellar space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book doesn't explain why it is that early estimates were wrong and I didn't see an explanation. Perhaps someone knows and will give a nice intuitive explanation for the estimates.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-07T01:22:35.537" Title="Why were space physicists wrong about the location of the heliopause?" Tags="&lt;space&gt;&lt;solar-system&gt;&lt;plasma-physics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="3958" PostTypeId="2" ParentId="2598" CreationDate="2011-01-26T23:34:38.077" Score="3" ViewCount="" Body="&lt;p&gt;Dear Marek.   The distinction between distinguishable and indistinguishable microstates is the following. For an observer outside the BH, two microstates are distinguishable if they can affect the future evolution of the observer differently. Two microstates with a different geometry of the horizon are distinguishable. Instead, if the geometry differs only inside the horizon, there is no way the outside observer can be affected by the difference.   Why is this relevant for the entropy?  Because the entropy is a quantity that characterizes the heat exchanges with a system. These exchanges are determined by the number of different distinguishable microstates the system can be in, and not by the total number of states. If a system has a part which is completely isolated, including thermally, then its states are irrelevant for the thermodynamical behavior of the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that the entropy depends on which observer sees it? Yes of course, but this is well known. The entropy depends a lot on the observer; for instance it depends on the macroscopic quantities chosen to describe the system. A system has an entropy only after you specify how you are looking at it, namely which are the macroscopic quantities that you use to describe it. Then the entropy is determined by the number of states at those macroscopic parameters fixed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, the story of BH entropy in Loop Gravity has much evolved since that paper of mine, and many more things have been understood.  I think that the BH counting in LQG is a success, but I also think that the problem is not resolved, and the situation is still perplexing. I am not convinced by the idea that the solution is just fixing a parameter to make it come out right.   If anybody is interested in what I think today about the black hole entropy calculations in LQG, the place to look is my very recent review &lt;a href=&quot;http://fr.arxiv.org/abs/1012.4707&quot; rel=&quot;nofollow&quot;&gt;http://fr.arxiv.org/abs/1012.4707&lt;/a&gt;, which is written for a large audience, and where I try to asses the state of the field, including the BH entropy problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Lubos Motl often writes about my work and about LQG, I think I should at least say what I think about his criticisms.  What Lubos says about loop quantum gravity is complete nonsense. He does not know the field at all, and just keeps repeating that everything in LQG is wrong, presenting technical arguments that sound informed, but are always empty.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best to everybody,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carlo Rovelli &lt;/p&gt;&#xA;" OwnerUserId="1613" LastEditorUserId="1613" LastEditorDisplayName="" LastEditDate="2011-01-28T14:54:49.270" LastActivityDate="2011-01-28T14:54:49.270" CommentCount="4" />
  <row Id="3959" PostTypeId="2" ParentId="3957" CreationDate="2011-01-26T23:51:37.447" Score="3" ViewCount="" Body="&lt;p&gt;There was a considerable stretch when getting answers to within an order of magnitude was pretty good on topics like stellar structure and cosmology. Things improved---first slowly but later on quite markedly---so that now we are in the age of &quot;precision cosmology&quot; (a loss in terms of jokes about putting the error bar on the exponents, alas). But individual topics might have languished if they didn't attract much attention.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-01-26T23:51:37.447" CommentCount="2" />
  <row Id="3960" PostTypeId="2" ParentId="3662" CreationDate="2011-01-26T23:51:39.983" Score="5" ViewCount="" Body="&lt;p&gt;Here is as I see the problem.  The best way to understand it is to think what happens with rotations and quantum theory. Suppose that a certain vector quantity V=(V1,V2,V3) is such that its components are always multiples of a fixed quantity d. Then one is tempted to say that obviously rotational invariance is broken because if I take the vector V=(d,0,0) and rotate it a bit, I get V=(cos(phi) d, sin(phi) d,0), and cos(phi) d is smaller than d.  Therefore, either rotational invariance is broken, or the vector components can be smaller. Right?  No, wrong. Why? Because of quantum theory. Suppose now that the quantity V is the angular momentum of an atom. Then, since the atom is quantum mechanical, you cannot measure all the 3 components together. If you measure one, you can get say eithe 0, or hbar, or 2hbar ..., that it, precisely multiples of a fixed quantity. Now supose you have measured that the x component of the angular momentum was hbar. Rotate slwly the atom. Do you measure them something a bit smaller that hbar? No!  You measure again either zero, or hbar ... what changes continuously is not the eigenvalue, namely the quantity that you measure, but rather the probabilities of measuring one or the other of those eigenvalues. &#xA;Same with the Planck area in LQG.  If you measure an area, (and if LQG is correct, which all to be seen, of course!) you get a certain discrete number. If you boost the system, you do not meqsure the Lorentz contracted eigenvalues of the area: you measure one or the other of the same eigenvalues, with continuously changing probabilities.&#xA;And, by the way, of course areas are observables. For instance any CERN experiment that measures a total scattering amplitude amplitude is measuring an area. Scattering amplitudes are in cm2, that is are areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;carlo rovelli &lt;/p&gt;&#xA;" OwnerUserId="1613" LastActivityDate="2011-01-26T23:51:39.983" CommentCount="2" />
  <row Id="3961" PostTypeId="2" ParentId="2649" CreationDate="2011-01-27T00:02:10.310" Score="1" ViewCount="" Body="&lt;p&gt;No, LQG does not predict imaginary distances, negative areas and imaginary volumes. It is sufficient to remember the geometrical interpretation of the triad. A triad with a negative&#xA;determinant is simply a left handed rather than a right handed triad. Therefore it does not describes funny spaces with imaginary volumes, but just the same good old space with where the basic triad is left handed rather than right handed.  This simply guides you in giving the proper definitions of area and volume, putting the right sign and the right absolute values where it belongs. There may be papers where those absolute values have been placed wrong, but when things are done properly, no imaginary volumes !&#xA;carlo rovelli &lt;/p&gt;&#xA;" OwnerUserId="1613" LastActivityDate="2011-01-27T00:02:10.310" CommentCount="6" />
  <row Id="3962" PostTypeId="2" ParentId="2407" CreationDate="2011-01-27T00:20:18.937" Score="12" ViewCount="" Body="&lt;p&gt;Here is the way I would try to explain Loop Quantum Gravity to my grand mother. Loop Quantum Gravity is a quantum theory. It has a Hilbert space, observables and transition amplitudes. All these are well defined.  Like all quantum theories, it has a classical limit. The conjecture (not proven, but for which there are many elements of evidence), is that the classical limit is standard General Relativity. Therefore the &quot;low energy effective action&quot; is just that of General Relativity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main idea of the theory is to build the quantum theory, namely the Hilbert space, operators and transition amplitudes, without expanding the fields around a reference metric (Minkowski or else), but keeping the operator associated to the metric itself.  The concrete steps to write the theory are just writing the Hilbert space, the operators and the expression for the transition amplitudes. This takes only a page of math. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result of the theory are of three kind. First, the operators that describe geometry are well defined and their spectrum can be computed. As always in quantum theory, this can be used to predict the &quot;quantization&quot;, namely the discreteness, of certain quantities. The calculation can be done, and area and volume are discrete. therefore the theory predicts a granular space. This is just a straightforward consequence of quantum theory and the kinematics of GR. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, it is easy to see that in the transition amplitudes there are never ultraviolet divergences, and this is pretty good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there are more &quot;concrete&quot; results. Two main ones: the application to cosmology, that &quot;predicts&quot; that there was big bang, but only a bounce: And the Black Hole entropy calculation, which is nice, but not entirely satisfactory yet. &#xA;Does this describe nature?  We do not know...   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;carlo rovelli &lt;/p&gt;&#xA;" OwnerUserId="1613" LastActivityDate="2011-01-27T00:20:18.937" CommentCount="5" />
  <row Id="3963" PostTypeId="2" ParentId="2435" CreationDate="2011-01-27T00:25:25.467" Score="0" ViewCount="" Body="&lt;p&gt;In LQG there is an Hilbert space, which can be built along the lines you mention.  Then we know that in quantum mechanics there are in general several useful basis in the Hilbert space of a system.  For instance, for an harmonic oscillator useful bases are the one that diagonalizes the energy, or that which diagonalizes the position.&#xA;Well, the spin network state are just the state sin a particularly useful basis in the Hilbert space of the theory. &#xA;It is useful because it diagonalizes certain operators. Area and volume. Therefor it has a nice physical interpretation.  Like the discrete basis that diagonalizes the energy of the harmonic oscillator describes states of N quanta of energy, similarly the spin network basis describes &quot;quanta of space&quot;, connected to one another...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;carlo rovelli&lt;/p&gt;&#xA;" OwnerUserId="1613" LastActivityDate="2011-01-27T00:25:25.467" CommentCount="1" />
  <row Id="3964" PostTypeId="2" ParentId="3933" CreationDate="2011-01-27T00:30:09.997" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;The laser is most likely not dangerous at all when used normally&lt;/strong&gt;, because it is very low power and within the visible portion of the spectrum.   The idea is that it is sufficiently low power that your &lt;strong&gt;blink reflex&lt;/strong&gt; will protect your eye before any damage occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The danger of a given laser source depends on its power, power density, and wavelength, as these determine what part of the eye will be damaged, how quickly, and whether you will blink.  The Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Laser_safety&quot; rel=&quot;nofollow&quot;&gt;Laser Safety&lt;/a&gt; gives a fairly comprehensive summary of the matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As others have written, the main reason why lasers are dangerous in general is because they can concentrate a large amount of power into a small area.  The school most likely wants to instill a sense of respect for lasers and inspire good operating habits (both admirable goals) even though in this case they are probably over-stating the danger.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-01-27T00:35:57.910" LastActivityDate="2011-01-27T00:35:57.910" CommentCount="2" />
  <row Id="3966" PostTypeId="2" ParentId="2435" CreationDate="2011-01-27T03:31:01.780" Score="0" ViewCount="" Body="&lt;p&gt;The spin along each each edgelink is also from my understanding a spin-connection.  A discrete parallel translation around a lattice gives a curvature, which is a two-form evaluated through the area of the translation.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-27T03:31:01.780" />
  <row Id="3967" PostTypeId="1" CreationDate="2011-01-27T03:46:14.920" Score="4" ViewCount="597" Body="&lt;p&gt;The one difficulty I see with LQG is that it requires an enormous number of degrees of freedom, &lt;em&gt;e.g.&lt;/em&gt; these spin variables in the net.  This is in contrast to stringy holographic theory where the fields in a space are equivalent to fields on a boundary or a horizon of one dimension lower.  In this setting entropy of a black hole is the entanglement entropy of states interior and exterior to the black hole.  This reduces the amount of data, and thus entropy, required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there suggestions, conjectures or maybe serious theory which attempts to describe the spin variables of LQG according to such entanglements in string-brane theory?&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2011-01-27T11:21:29.383" LastActivityDate="2011-01-27T11:21:29.383" Title="Can Loop Quantum Gravity connect in any way with string theory?" Tags="&lt;string-theory&gt;&lt;quantum-gravity&gt;&lt;loop-quantum-gravity&gt;" AnswerCount="4" />
  <row Id="3969" PostTypeId="2" ParentId="2300" CreationDate="2011-01-27T05:08:02.267" Score="0" ViewCount="" Body="&lt;p&gt;Yes. I think we are missing something here. It might help if we knew what he means by power. If it is power consumption for heating and cooling a city, we can make some gueses as to plausible parameterizations. I'm assuming his hours are hours of the year (8760 hours/year), but then humidity doesn't seem right to me, it should rise at night and be at a minimum during the afternoon (but maybe these are a 24hour averages?). If is is heating and cooling, then t is a proxy for the amount of sunlight available, but it would be much better to have that as an independent variable as it is probably quite important (and quite variable depending upon cloud cover). &lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-01-27T05:08:02.267" />
  <row Id="3970" PostTypeId="2" ParentId="3967" CreationDate="2011-01-27T06:03:00.373" Score="-1" ViewCount="" Body="&lt;p&gt;At a fundamental level there seems to be an incompatibility.  String theory requires an infinite amount of other fields in addition to the simple metric tensor.  LQG meanwhile pretends to merely be a humble theory of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But string theory would be inconsistent if it was just about gravity, so at a basic level there seems to be a clash.  I believe the only way around that would be if LQG were to add extra structure on top of their theory, and I believe the state of the art is far from being able to manage much if any matter degrees of freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure others will bring up other salient incompatibilities, but I am a little skeptical about those simply b/c it seems that LQG is still in a very raw and preliminary stage of development and the literature seem to change the tone and their predictions every year or so.  For instance, not long ago we were led to believe that Lorentz breaking was an element of LQG, but no apparently that is not a correct statement.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose the correct answer would be to say that even posing the question is premature. &lt;/p&gt;&#xA;" OwnerUserId="1350" LastActivityDate="2011-01-27T06:03:00.373" />
  <row Id="3971" PostTypeId="2" ParentId="3967" CreationDate="2011-01-27T07:25:14.590" Score="4" ViewCount="" Body="&lt;p&gt;I would say that the it is the other way around. The number of degrees of freedom in LQG is not &quot;enormous&quot;, and it is much &lt;em&gt;less&lt;/em&gt;, and not much &lt;em&gt;more&lt;/em&gt; that in strings.  The spin variables on the net are just the same variables as in classical general relativity, namely the metric, except that there is a cut off at small scale. So, there are infinitely many &lt;em&gt;less&lt;/em&gt; variables than in classical general relativity.  because the degrees of freedom with wavelenght smaller than the Planck length exist in classical GR but not in LQG. It is true that in strings one might expect an holographic principle that reduces the degrees of freedom to the boundary theory, but this is not a boundary a theory in our physical spacetime. It is a boundary theory in a spacetime of higher dimensions. So, the boundary theory has still more dimensions than our space, and therefore the number of degrees of freedom is still much bigger that in LQG.&lt;br&gt;&#xA;About the stage of development, certainly both loops and strings are very preliminary, and not fully understood, but I would definitely say that LQG is far better understood (it is a simpler theory than strings). In LQG we know the fundamental degrees of freedom and we can write the basic equations of the full theory in a compact form in a few equations.  See for instance my recent review paper &lt;a href=&quot;http://fr.arxiv.org/abs/1012.4707&quot; rel=&quot;nofollow&quot;&gt;http://fr.arxiv.org/abs/1012.4707&lt;/a&gt;.  In strings we do not yet know the fundamental degrees of freedom, and we only know certain &quot;corners&quot; of the theory, with many indications that these different corners fit into a single scheme. But the actual single scheme we do not know yet.  So, the basic theoretical situation in LQG is simple and clear, not so in strings.&#xA;Finally, no the predictions in LQG have not been changing. In fact, I wish there were solid predictions. There is none for the moment, like for strings. What exists in both cases is suggestion of possibilities. For instance strings suggested that perhaps the gravitational force could change at a measurable distance because of the extra dimensions, that supersymmetric particles be seen at lower energy, that black holes be formed at CERN. Nothing of this has been true so far, but this does not invalidate strings, because the theory is perfectly compatible with these effects existing but also compatible with these effects not existing. In very much the same way, it had been suggested that LQG might be combatible with violations of Lorentz invariance, and these seem to have been quite ruled out now by observations. But LQG is perfectly compatible with these Lorentz violation not being there. In fact, personally I have always been very skeptical at he suggestions that LQG might lead to Lorentz violation. If you read my old papers, I have always insisted that the theory is perfectly compatible with Lorentz invariance, and I could not see a source of Lorentz violation. These papers, for instance arXiv:gr-qc/0205108, which we wrote almost 10 years ago, indicate that LQG is locally Lorentz invariant, and were written long before the recent indications against Lorentz violation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the compatibility between Loops and Strings, I really do not know.  It is true that all the incompatibilities that Columbia indicates are there.  But on the other hand we do not yet know the fundamental degrees of freedom of string theory.  If there is a fundamental description of strings, this should be background independent; perhaps could it resemble somehow LQG?  And in LQG there is no hint at the solution of the unification problem. Would it lead towards something more resembling strings?  If I had to bet, I would say no, the two paths are really different, but I would not rule out the possibility a priori.  I agree that the question is a bit premature, but I would say that the reason is not hat LQG is in a raw stage. LQG is a well and clearly defined theory, what is hard is to compute out of it. The reason is more because strings are in a raw stage, since we do not know a fundamental formulation of the theory, its basic degrees of freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;carlo rovelli &lt;/p&gt;&#xA;" OwnerUserId="1613" LastActivityDate="2011-01-27T07:25:14.590" CommentCount="1" />
  <row Id="3973" PostTypeId="2" ParentId="3967" CreationDate="2011-01-27T08:04:07.867" Score="7" ViewCount="" Body="&lt;p&gt;Dear Lawrence, an equivalence between LQG and string theory - or an LQG-like description of string theory physics - has surely been an attractive idea for many physicists (myself included) but it is impossible because of fundamental differences in virtually all general features and predictions of both frameworks.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;As you correctly mentioned, the counting of the degrees of freedom disagrees. &lt;strong&gt;String theory respects the holographic principle.&lt;/strong&gt; It also means that the entropy within a volume is always bounded by the surface in Planck units. On the other hand, &lt;strong&gt;LQG admits an arbitrary, volume-extensive, Planckian entropy density&lt;/strong&gt; - in fact, it predicts a Planckian entropy density of the vacuum (the information about the details of the spin network). For a related class of examples, LQG always predicts a volume-extensive term in the black hole entropy, too. It can only be &quot;cured away&quot; by erasing it (together with the whole black hole interior) by hand and pretending it was never there. It's important to mention that the infinite multiplicity of &quot;string fields&quot; is just an artifact of a formalism - string field theory. One can't add an arbitrary number of excitations of all these kinds into a finite volume (because they would gravitationally collapse). The only truly &quot;physically invariant&quot; measure of the number of degrees of freedom boils down to entropy and ST - as a holographic theory - predicts a much smaller entropy than non-holographic theories such as LQG. In particular, string theory vacua are unique and carry no entropy density.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;LQG breaks the local Lorentz symmetry&lt;/strong&gt; while string theory exactly preserves is. Because the Fermi satellite has showed that there is no Lorentz violation at the Planck scale, LQG was falsified. (It was falsified in many other ways, too.) String theory remains compatible with the observations. The preservation of Lorentz symmetry in string theory may be seen e.g. perturbatively, by considering strings propagating on a target spacetime. The $SO(d-1,1)$ symmetry of the spacetime directly arises from the $SO(d-1,1)$ global symmetry rotating fields (representing spacetime coordinates) on the world sheet. The violation of Lorentz symmetry in LQG may be seen from the fact that the hypothetical solution - a spin network - picks a privileged reference frame, analogous to the luminiferous aether. In this frame, the entropy density is huge, essentially Planckian, and in all other reference frames, there would be a huge entropy flow in a direction, which would break the rotational symmetry. In the preferred frame, the motion of all objects instantly stops as their kinetic energy is dissipated to the thermal energy of the spin network which is a de facto infinite heat bath.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;String theory implies that the space is smooth and almost flat at long distances.&lt;/strong&gt; Although there is no fully-general &quot;no-go theorem&quot;, all partial models and circumstantial evidence suggest that LQG in any form can never predict a smooth space at long distances. It gets crumbled. For this reason, it doesn't even make sense to ask whether LQG reproduces Einstein's equations at long distances - there are no long distances in LQG.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;LQG implies that there can't be any forces and elementary particles aside from gravity.&lt;/strong&gt; String theory predicts that gravity has to exist, much like non-gravitational forces and particle species. The absence of other forces in LQG isn't a cosmetic problem that can be fixed. The strength of other forces doesn't go to zero, not even at the Planck scale. In fact, there exist general arguments that gravity has to be the weakest force - just like it is in the real world - so a valid microscopic description can never start by neglecting the non-gravitational forces because it is really gravity that is a correction, not the other way around. String theory predicts the right &quot;draft&quot; of the world with spin-1/2 fermions, spin-1 gauge bosons, potential for gauge anomalies and a nontrivial anomaly cancellation, chiral fermions and chiral interactions, Higgs bosons, Higgs mechanism, confinement of non-Abelian gauge fields, running couplings and other phenomena related to the renormalization group, and so on, while &lt;strong&gt;LQG has nothing whatsoever to do with particle physics&lt;/strong&gt; and is incompatible pretty much with all the basic concepts of particle physics I enumerated. The contrast becomes even stronger if we realize that string theory has also led to (or at least inspired) some of the most novel, explanatory, and important models of beyond-the-standard-model phenomenology such as supersymmetry, models with extra dimensions, deconstructions, and others that are currently studied by a big portion of phenomenologists, even those who don't consider themselves string theorists in any sense.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;String theory includes dualities.&lt;/strong&gt; They're transformations that totally rearrange the degrees of freedom and change their interpretation. Quantum mechanics is totally crucial for those S-dualities, T-dualities, U-dualities, holographic dualities, and other dualities to work. On the other hand, LQG doesn't imply any dualities. More generally, it doesn't employ quantum mechanics in any deep way. It is just a variation of the ancient Greek models of atoms whose properties are promoted to operators - but this promotion never leads to anything interesting.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;LQG doesn't admit supersymmetry, wants to avoid extra dimensions, strings, extended objects, etc.&lt;/strong&gt; So LQG is unlikely to be a dual description of any aspect of string theory. It's been established in string theory that supersymmetry is an omnipresent, fundamental symmetry that has to appear in all semi-realistic models at some scale. Extra dimensions are needed for consistency. On the other hand, LQG starts by assuming that none of these things exist, and even though the appearance of extended objects etc. is generic in consistent field theories and vacua of string theory, the LQG research is based on the assumption that they must be avoided. This leads me to a much general point.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;String theory is a natural theory based on objectively important mathematical structures and relationships.&lt;/strong&gt; Physicists are discovering these features, much like Columbus was discovering America. They are learning new things - and they are identifying the previous errors in their reasoning. On the other hand, LQG is a man-made theory. It is being invented in a similar way as Edison was inventing the light bulb. Preconceptions are what ultimately decides about the shape of the theory. LQG is being constructed step by step. That's why one can never make any solid statements about anything - and one can make no statements that he wouldn't believe at the beginning. This strikingly differs from string theory that implies unique answers to many fundamental questions. For example, it implies that the equivalence principle, local Lorentz symmetry, and constancy of the universal constants have to hold in general. All these issues are permanently open in LQG because someone may always modify the theory in a different way tomorrow. One may learn new conceptual insights about physics - and mathematics - from string theory. That's different from LQG that is designed to depend on no nontrivial mathematics that would be difficult for average undergraduate students. Consequently, one can never learn anything about physics, space, time, or mathematics from LQG. The whole enterprise is meant to find justifications for a predetermined opinion that quantum gravity can be approached in this simple-minded way - a strategy that is not unsimilar to scientists proving the Intelligent Design or geocentrism. So far, however, no justifications have been found.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The multiplicities of possibilities how the vacuum may look like according to string theory boil down to &lt;strong&gt;solutions of objective equations that we pretty much understand&lt;/strong&gt;: the rules of the game are constant. That's very different from LQG where new models are created at any point by arbitrarily changing the rules of the game. That's related to the previous point that string theory makes some general predictions, even when the right &quot;vacuum&quot; is unknown. LQG can never make any predictions of this kind.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Information is lost in LQG.&lt;/strong&gt; Indeed, it is a local theory of a very naive type so even if space and black hole were possible in LQG, one could show that the assumptions of Hawking's original argument are satisfied which implies that the information cannot get out of the black hole for causal reasons even if the black hole could evaporate (which is established in string theory but surely not in LQG). On the other hand, string theory implies that there exist subtle nonlocalities in the bulk spacetime that imply that the information gets out. This answer is known to be valid because there often exist dual descriptions of the stringy physics where unitarity is manifest.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;LQG tries to use ill-defined observables and ignore the well-defined ones.&lt;/strong&gt; In particular, the area of a surface isn't well-defined at the Planckian accuracy in a theory where measurements cannot measure distances shorter than the Planck scale. This is why all statements of LQG about the &quot;quantization of areas&quot; cannot be operationally or otherwise defined. On the other hand, string theory implies that areas of small surfaces are not well-defined observables and automatically leads us to the physically meaningful observables such as the scattering amplitudes for gravitons - which can't be calculated in LQG. Scattering amplitudes may be measured experimentally and they satisfy important theoretical constraints such as unitarity, too: they're the right way to parameterize &quot;all predictions&quot; of a meaningful relativistic quantum theory. Quite generally, string theory automatically addresses quantities that are important in high-energy physics while LQG is disconnected from all the 20th century physics and its key concepts.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To summarize, the differences between the technical properties of the two frameworks as well as very philosophy what it means to do good science are completely insurmountable.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-27T10:15:53.397" LastActivityDate="2011-01-27T10:15:53.397" CommentCount="14" />
  <row Id="3975" PostTypeId="2" ParentId="3076" CreationDate="2011-01-27T10:10:53.143" Score="1" ViewCount="" Body="&lt;p&gt;Your circulatory system puts enough extra pressure on your blood to keep it from boiling in a vacuum, and space itself doesn't conduct heat, so all the heating/cooling would be due to radiation or evaporation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think your exposed skin would swell, and fluids would evaporate and dry out your tongue, throat, and lungs, and your eardrums might burst, which would be painful. If you tried to move around you might risk overheating, since there's no way to cool yourself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, there's a chance your sweat-glands might still work, since they cool with evaporation.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-27T10:10:53.143" />
  <row Id="3976" PostTypeId="2" ParentId="3955" CreationDate="2011-01-27T10:12:33.207" Score="1" ViewCount="" Body="&lt;p&gt;The solid/liquid &quot;earth&quot; is negative, upper atmosphere is &#xA;positively charged.&lt;br&gt;&#xA;(close to thunderstorms polarity can be reverse) &#xA;German wikipedia says a bit more on this than the &#xA;English pendant, but it is not as clear as I liked with &#xA;respect to overall load of earth relative to &quot;space&quot;. &#xA;Georg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS &#xA;&lt;strong&gt;positive ions in air have a mobility of 1.37 exp-4 (m/s)/(V/m),&#xA;negative Ions have 1.89 exp-4 (....). Thermal Electrons as such do not &#xA;exist in air, they attach to molecule aggregates. For that any reasoning &#xA;about &quot;weight&quot; of electrons doing something faster in air, is wrong.&lt;/strong&gt;  &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-02-08T11:27:47.923" LastActivityDate="2011-02-08T11:27:47.923" CommentCount="1" />
  <row Id="3977" PostTypeId="2" ParentId="2111" CreationDate="2011-01-27T10:24:12.993" Score="2" ViewCount="" Body="&lt;p&gt;Trick question, he'd use his super strength to bend the straw into an Archimedes' screw, then hold it at an angle to the surface of the water and rotate it about the axis.&#xA;This lets him draw it up to any height, and then he can drain the world's oceans to prove a point or do whatever other superdickery he's trying to do.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-27T10:24:12.993" />
  <row Id="3978" PostTypeId="2" ParentId="3967" CreationDate="2011-01-27T10:42:00.137" Score="1" ViewCount="" Body="&lt;p&gt;It's a very good question and I can't match the expert knowledge of those who have already answered, but I also can't resist making a few points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In LQG it is hard to calculate anything because it does not give a classical limit that looks like smooth spacetime. If such a limit existed we would expect to be able to look at small deviations from flat spacetime and it is hard to see how that could lead anywhere except to a perturbative theory of gravitons. From the particle physics world we know that the supergravity/superstring route is probably the only way to do that. Some people might say that there could be another unknown way or that LQG would somehow avoid such a perturbative limit but let's assume otherwise until there is some good explanation of how that would work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case LQG could only work if it included matter as string theory does. Matter might be emergent but that would mean that LQG has to work as it is with a classical limit and that does not seem to be the case, so probably LQG needs matter put in as some additional degrees of freedom. I think Lee Smolin tried to generalise LQG to look more like string theory in the early days before he gave up and became more dismissive of string theory. For example, he and others looked for higher dimensional version and supersymmetric version of LQG but there was nothing very promising. I think it would be wrong for a younger generation to assume that no progress can be made with such a connection just because others could not find it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the string theory side the fundamental issue is that it's degrees of freedom and underlying principles are not known or fully understood. LQG has spins and knots. Spin half entities are qubits which also arise in string theory. This does not mean there is a connection because such entities arise everywhere as representations of symmetries. However, LQG and string theory share similar origins out of gauge theories and they share some mathematical structures. The areas where they are least well understood are also the areas where we might expect to see connections if there are any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally I think that theorists have to take Dirac's advise to look for elegant mathematical structures and be driven by them. When they find them in relation to one approach it suggests that the approach has some promise. Both string theory and LQG passed this test when trying to solve the same problem but failed so far to make contact with experiment. I think you have to keep a broad view and look for connections between the mathematics of these and other approaches that look interesting. Unfortunately sociological issues driven by the way funding is allocated seems to discourage people from looking at the bigger picture.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course some experimental input would also help.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-01-27T10:42:00.137" CommentCount="4" />
  <row Id="3981" PostTypeId="2" ParentId="3944" CreationDate="2011-01-27T11:13:38.427" Score="1" ViewCount="" Body="&lt;p&gt;It is just the &lt;a href=&quot;http://en.wikipedia.org/wiki/Boltzmann_distribution&quot; rel=&quot;nofollow&quot;&gt;Boltzmann distribution&lt;/a&gt;, with the energy being given by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_potential&quot; rel=&quot;nofollow&quot;&gt;chemical potential&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2011-01-27T11:13:38.427" />
  <row Id="3982" PostTypeId="2" ParentId="1019" CreationDate="2011-01-27T11:34:54.680" Score="2" ViewCount="" Body="&lt;p&gt;Very low friction at skating due to &quot;pressure melting&quot; of water ice. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-01-27T11:34:54.680" CommunityOwnedDate="2011-01-27T11:34:54.680" CommentCount="3" />
  <row Id="3983" PostTypeId="1" CreationDate="2011-01-27T11:36:24.693" Score="12" ViewCount="597" Body="&lt;p&gt;There exists this famous Haag's theorem which basically states that the interaction picture in QFT cannot exist. Yet, everyone uses it to calculate almost everything in QFT and it works beautifully. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Why? More specifically to particle physics: In which limit does the LSZ formula work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Can someone give me an example of a QFT calculation (of something measurable in current experiments, something really practical!) in which the interaction picture fails miserably due to Haag's theorem?&lt;/p&gt;&#xA;" OwnerUserId="1337" LastEditorUserId="1337" LastEditorDisplayName="" LastEditDate="2011-01-27T11:57:03.693" LastActivityDate="2011-02-13T23:37:19.350" Title="Haag's theorem and practical QFT computations" Tags="&lt;quantum-field-theory&gt;&lt;mathematical-physics&gt;" AnswerCount="4" FavoriteCount="3" />
  <row Id="3984" PostTypeId="2" ParentId="3871" CreationDate="2011-01-27T11:39:18.857" Score="0" ViewCount="" Body="&lt;p&gt;As the originator of this question I have reviewed and learned something from all the answers posted so far. I would like to summarise my own views here.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Where did this query come from? From a question about the &lt;a href=&quot;http://physics.stackexchange.com/questions/3767/the-transactional-interpretation-of-quantum-mechanics&quot;&gt;Transactional Interpretation of Quantum mechanics&lt;/a&gt;  (TIQM), where said  theory’s reliance upon “retrocausality” (‘causality’ backwards in time) was held to be a fatal defect. This kind of causality-argument is common in physics: we say that faster-than-light travel in SR is ruled out because it would violate causality.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Such causality arguments are conducted in what you might call the “metalanguage of physics”: technical English which supports and explains formal results. However, the arbiter in the end is the maths, so how do we interpret the notion of causality within the formalism?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Physical theories are defined by mathematical relationships between entities (observed and unobserved) usually expressed by equations (think Schrodinger, Dirac, the Lorentz transformation).  If we say that event E1 “causes” event E2, several answers here suggest that the interpretation of causality in the formal theory is that:  (i) &lt;em&gt;if E1 is postulated to occur then the theory logically implies that E2 must occur as well&lt;/em&gt;; (ii) &lt;em&gt;E2 is within or on the future light-cone of E1&lt;/em&gt; (we say “cause precedes effect”).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;However, it’s possible that condition (ii) is too stringent.  While logical entailment is obviously an essential part of any formalised theory, our smuggling in of the word “future” is already an extra assumption. Our fundamental theories do not impose a specific past-future direction on the time dimension. This means that if you reverse the film, the events you see are still consistent with our fundamental theories.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Sometimes people use causality-like words in the physics metalanguage without conventional time-ordering condition (ii). For example, a possible Feynman diagram for &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aElectron-positron-scattering.png&quot; rel=&quot;nofollow&quot;&gt;electron-positron scattering&lt;/a&gt; has a narrative that an electron travelling backwards in time from the future encounters a (normal) electron, they exchange a virtual photon and continue on their way scattered. The 'cause' of the scattering event was the arrival of the future electron. Many textbooks mention this way of thinking but we don’t mind because the underlying theory gives consistent results which accord with observation. Perhaps TIQM is like this despite its narrative of retrocausation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;So my conclusion is that we have to be careful about arguments concerning a theory’s validity relying upon causation arguments couched in physics’ metalanguage. It’s not a slam-dunk. Sometimes if a theory violates conventional “cause precedes effect” causation it indicates a breakdown in the underlying mathematics, normally inconsistency. At other times a 'causation' argument is just a way of talking about the entailment of the theory in an innovative or whimsical way, and the theory is actually OK. Go look at the maths.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;NOTE: there is a whole separate discussion about why, in natural language, we think so naturally in terms of cause and effect. It links to discussions about the arrow of time and why we do seem to be unhappy about running the film backwards as a valid picture of reality. That is a whole separate issue but still, I suspect, part of physics judging by the number of recent books on the subject.&lt;/p&gt;&#xA;" OwnerUserId="1394" LastActivityDate="2011-01-27T11:39:18.857" CommentCount="1" />
  <row Id="3985" PostTypeId="2" ParentId="3983" CreationDate="2011-01-27T11:57:35.450" Score="4" ViewCount="" Body="&lt;p&gt;Dear Rafael, every theorem is only as powerful as its assumptions (and propositions). The answers are clearly that&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The LSZ formula always works for the field theories where it's used.&lt;/li&gt;&#xA;&lt;li&gt;No actual calculation relevant to physics fails because of Haag's theorem. Haag's theorem is just a philosophy.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Haag's theorem is morally wrong because it studies the question whether the operators in the interacting theory are &quot;strictly&quot; unitarily equivalent to those in the free theory:&#xA;$$O_{interacting} = U O_{free} U^{-1}$$&#xA;Not surprisingly, Haag finds out that such a unitary equivalence doesn't exist. This is not surprising because, as we know, operators acquire anomalous dimensions from the interactions (and quantum effects), among other deviations from the classical intuition, and the naive algebra that is valid in the free theory simply no longer applies to the interacting theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the addition of the interactions also modifies the commutation relations between the fields that &quot;directly&quot; create and annihilate the particles - at least the low-energy effective fields. For example, the quantum effects produce effective Lagrangians that contain higher-derivative terms, including new terms with time-derivatives, and the latter modify the canonical momenta and/or the canonical commutation relations. When one is rigorous, many things change when the interactions are added. Haag only assumed that &quot;some&quot; things change, so his results are inconsequential for physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, this 1955 theorem is obsolete - much like most of the former discipline that used to be known as algebraic quantum field theory or axiomatic field theory - and be sure that interacting quantum field theories exist - lattice QCD is an example of a specific way how to define them - and it is equally true that the perturbative approximation of all the physical amplitudes may be calculated by the usual perturbative methods, with the extra philosophy and rigorous refinements given e.g. by the LSZ formalism you mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Haag's theorem was invented as an attempt to show that there was something wrong with one of the first loop diagrams people understood - the vacuum polarization graph. Mr Haag didn't like them. However, there is nothing wrong with the loop diagram - or any other loop diagrams that became the bulk of knowledge about particle physics in the subsequent decades. The developments in renormalization showed that the calculations, including the loops, are totally valid. The renormalization group made some further progress - it explained why the theories are universal and why the subtraction of infinities work. Haag's theorem became misleading and obsolete in the 1970s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the LSZ formalism uses the &quot;adiabatic hypothesis&quot;, the assumption that one may neglect the interactions between the particles in the asymptotically distant past. By slowly and continuously turning on the coupling constant, we may map the states of free particles to the states describing particles in the interacting Hilbert space. This is possible as long as all distances between the particles are large. However, this procedure wouldn't work for general configurations of nearby particles - so one can't promote this trick into a full-fledged &quot;canonical&quot; unitary equivalence between the free and interacting Hilbert spaces. There is clearly no such a &quot;natural&quot; or &quot;unique&quot; or &quot;canonical&quot; isomorphism because the free and interacting theories are physically inequivalent. When understood rationally, Haag's theorem is not saying anything else than this self-evident proposition. However, such an isomorphism is not needed to calculate physically meaningful quantities such as the scattering amplitudes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least from the viewpoint of physics as an empirical science, it should be clear that the actual calculations in QFT are good science - one that has made predictions and has passed tests comparing the predictions with experiments - while Haag's theorem is not because it hasn't predicted anything that has passed empirical tests. Haag's theorem tries to find problems with the fact that quantum field theory contains new effects such as renormalization that don't appear in quantum mechanics with a finite number of degrees of freedom. However, these extra effects of quantum field theory are real and essential and they lead to no inconsistencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Haag's theorem is not a tool to do particle physics; it's an excuse for someone who doesn't want to study particle physics. As every theorem, it says &quot;A implies B&quot;. Because we know that B is incorrect - perturbative QFT clearly works - it follows that the assumptions A aren't right.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-27T13:27:12.983" LastActivityDate="2011-01-27T13:27:12.983" CommentCount="7" />
  <row Id="3986" PostTypeId="1" AcceptedAnswerId="3988" CreationDate="2011-01-27T12:08:43.370" Score="7" ViewCount="363" Body="&lt;p&gt;This thought experiment is originally due to Sir Isaac Newton. We have a sphere of water floating freely in an opaque box in intergalactic space, held together by surface tension and not rotating with respect to the distant galaxies. Now we set the box and water to rotate about some axis and we notice that the sphere flattens into an oblate spheroid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the water know it’s spinning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: Newton thought this proved the concept of absolute rotation with respect to a preferred spatial frame of reference. Perhaps these days we can do better, or different?&lt;/p&gt;&#xA;" OwnerUserId="1394" LastEditorUserId="1394" LastEditorDisplayName="" LastEditDate="2011-01-27T14:33:02.793" LastActivityDate="2011-01-27T16:32:33.273" Title="Newton's Bucket" Tags="&lt;general-relativity&gt;&lt;classical-mechanics&gt;&lt;relativity&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="3987" PostTypeId="1" AcceptedAnswerId="3989" CreationDate="2011-01-27T12:09:58.783" Score="3" ViewCount="201" Body="&lt;p&gt;Why, in somes cases, such as spin 1/2 systems like in the Stern-Gerlach experiment, the representation of some kets has a term with an complex exponential? For example:&#xA;$ \vert S_{x}; + \rangle = \dfrac{1}{\sqrt{2}} \vert + \rangle + \dfrac{1}{\sqrt{2}} \exp [i \delta] \vert - \rangle $&lt;/p&gt;&#xA;" OwnerUserId="1618" LastActivityDate="2011-01-28T18:21:48.587" Title="Phase on Hilbert Spaces" Tags="&lt;quantum-mechanics&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="3988" PostTypeId="2" ParentId="3986" CreationDate="2011-01-27T12:24:31.660" Score="4" ViewCount="" Body="&lt;p&gt;Dear Nigel, Newton had to postulate an absolute space. In fact, he used his physics insights to support the idea of a &quot;spirit&quot; that is filling the space - a paradigm this greatest scientist and a devoted Christian was as passionate about as about physics itself. The absolute space determined geometry everywhere except that it didn't know about any preferred velocity; it only knew about preferred accelerations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Inertial systems in classical physics&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton's laws of physics were valid in inertial frames only. If the laws have the usual forms in one frame, one can show that they also have the same form in all frames that are moving by a constant speed in the same direction. But one can also show that the form of the laws changes if we switch to a different system that is accelerating or spinning because this system is not inertial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between inertial and non-inertial frames is surely a basic postulate of classical mechanics and it is one that is extremely well established by the experiments, too. Newton's bucket is one of the simple ways to show that rotating frames and non-rotating frames simply differ, so the hypothesis (assumed in between lines of your question) that there is a &quot;complete democracy&quot; between all frames, regardless of their rotation, is instantly falsified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Special relativity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar &quot;absolute structures&quot; filling space and time survived in relativity as well, despite Einstein's original fascination with the so-called Mach's principle that de facto wanted to deny that the rotating bucket behaves differently than the non-rotating one. General relativity ultimately rejected Mach's principle even though one may see some individual effects - memories - predicted by general relativity that are similar to those discussed by Mach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In special relativity, there exists a &quot;metric tensor&quot; in the whole spacetime that tells all the buckets - and all other objects - whether they're rotating (and accelerating) or not. If they're not rotating, the metric will be given by &#xA;$$\eta(x,y,z,t)=\mbox{diag}(-1,+1,+1,+1)$$&#xA;I chose the sign convention randomly. However, if one transforms this metric to a frame that is inertial - it is spinning or accelerating - the metric tensor will be transformed into a different one, namely a set of 10 non-constant functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;General relativity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The very same thing is true in general relativity where the metric tensor becomes dynamical and may be curved by the presence of heavy objects. It is still true that the metric in non-rotating frames will be given by &#xA;$$ds^2 =-c^2dt^2+dx^2+dy^2+dz^2$$&#xA;which is just a different way of writing the metric $\eta$ a few lines above. However, if one transforms this metric tensor to a spinning frame, one gets a different metric tensor. The deviation from the flat space metric may be interpreted as a &quot;gravitational field&quot;. The equivalence principle guarantees that the effect of gravitational fields is indistinguishable from the effect of inertial forces resulting from spin or acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the extra corrections in the metric tensor will know all about the centrifugal, centripetal, and Coriolis forces that are responsible for the modified shape of the water surface, among many other effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, the bucket - and all other objects - know how to behave and whether they're spinning because they interact with the metric tensor that fills the whole spacetime and that allows one to distinguish straight lines (or world lines) from the curved lines (or world lines) at any point. It's important to realize that the metric tensor, while it allows to distinguish accelerating (curved) lines from the non-accelerating (straight) lines, can't distinguish &quot;moving objects&quot; from &quot;objects at rest&quot;. This is the principle of relativity underlying both Einstein's famous theories but in this general form, it was true already in Newton's mechanics - and realized by Galileo himself.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-27T12:24:31.660" CommentCount="6" />
  <row Id="3989" PostTypeId="2" ParentId="3987" CreationDate="2011-01-27T12:30:50.050" Score="2" ViewCount="" Body="&lt;p&gt;Your notation is misleading. Certainly, the eigenvector of the $S_x$ operator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;S_x = \frac{\hbar}{2}\begin{pmatrix}0 &amp;amp; 1 \cr 1&amp;amp; 0\end{pmatrix}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\mid S_x,+\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\cr 1\end{pmatrix}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With an arbitrary choice of global phase. Ok, so this is what I guess you meant by $\mid S_x,+\rangle$. Now, the most general state for this system is given by two numbers (the two original complex numbers need four real numbers, but one you remove by the normalization requirement and the second by the freedom of choosing a global phase). These are usually represented by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bloch_sphere&quot; rel=&quot;nofollow&quot;&gt;Bloch sphere&lt;/a&gt;. Your state is just a set of states in a circle on this sphere, but not necessarily an eigenvector of $S_x$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit (01/27/2010)&lt;/strong&gt;: In view of the claim that I'm not answering the question, let's try to fully understand the origin of this phase. Let's say you want to measure the spin in a direction given by the unit vector:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\vec{n} = (\sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a common exercise in QM to show that the eigenvalues of $\vec{S}\cdot\vec{n}$ are&#xA;$\pm 1$ and that the eigenvector associated to the eigenvalue $+1$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\mid\vec{S}\cdot\vec{n},+\rangle = \cos(\theta /2)\mid +\rangle + \sin(\theta /2)e^{i\varphi}\mid -\rangle&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the basis are eigenvectors of $S_z$ and with the state written up to an arbitrary global phase, as I explained before. If you have have trouble finding this answer, look it up in Sakurai's book, chapter 3. He derives this result over and over again. Therefore, by looking at the state in the question, we immediately identify $\theta=\pi/2$ and $\varphi$ as the phase you left undetermined $\varphi = \delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, this state is a spin 1/2 system which, if you measure $\cos\varphi\; S_x + \sin\varphi\; S_y$ (meaning, if you align your Stern-Gerlach device in the direction $\vec{n}=(\cos\varphi,\sin\varphi,0)$), you will get always the result $+1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's the origin of your phase. But your notation is misleading, indeed.&lt;/p&gt;&#xA;" OwnerUserId="1337" LastEditorUserId="1337" LastEditorDisplayName="" LastEditDate="2011-01-27T13:44:47.853" LastActivityDate="2011-01-27T13:44:47.853" CommentCount="4" />
  <row Id="3990" PostTypeId="2" ParentId="3986" CreationDate="2011-01-27T12:33:41.913" Score="3" ViewCount="" Body="&lt;p&gt;A synopsis of what Lubos wrote: It is possible to tell what an inertial reference frame is, locally (in an infinitesimal neighborhood of any point of spacetime), with respect to the local gravitational field (it's the one that is &quot;freely falling&quot;). The bucket &quot;knows&quot; that it is rotating because it rotates with respect to the local inertial frame, that is because it &quot;rotates relative to the local gravitational field&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-01-27T12:33:41.913" CommentCount="5" />
  <row Id="3991" PostTypeId="2" ParentId="3987" CreationDate="2011-01-27T12:36:05.293" Score="3" ViewCount="" Body="&lt;p&gt;Dear Rodrigo, all amplitudes - the coefficients in front of the states such as $|+\rangle$ and $|-\rangle$ you mentioned - are complex numbers in quantum mechanics. See e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2010/08/why-complex-numbers-are-fundamental-in.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/08/why-complex-numbers-are-fundamental-in.html&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Why complex numbers are fundamental in physics&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In particular, coefficients $1$ and $\exp(i\delta)$ for any real $\delta$ are physically equivalent if one places them in front of the whole wave function. It's because all predicted probabilities only depend on $\psi(x)\psi^*(y)$ where the phase cancels: note that $\exp(i\delta)^* = \exp(-i\delta)$ which is also the inverse to $\exp(i\delta)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The imaginary unit is necessary because the commutator of $[x,p]$ has to be $i\hbar$, to guarantee the uncertainty principle - one may show that the commutator of two Hermitian (&quot;real&quot;) operators is antihermitian (&quot;pure imaginary&quot;). Also, Schrödinger's equation, Heisenberg equations, and Feynman path integrals have to contain $i$ in their basic form, too. That's because wave functions have to oscillate in the complex plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relative phases between two basis vectors are very important and physical, however. They may be measured. It's because we may measure the probability that a given state vector has another property. That's given by the square of the absolute value of the inner product of this state vector with a particular eigenvector of the property. All phases influence the result because they influence the inner product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relative phases between component waves coming from the two slits in a double-slit experiment decide about the location of the interference maxima and minima. Phases decide about many other important things. Wave functions with energy $E$ depend as $\exp(i\delta)$ on time where $\delta = Et/i\hbar$; the dependence on space is $\exp(i\delta)$ where $\delta = px/\hbar$ where $p$ is the momentum. And I could go on and on and on (angular momentum $j_z$ and many other things).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The particular vector you wrote is the eigenstate of a component of the spin; it is the &quot;up&quot; eigenstate with respect to an axis in the $xy$ plane whose relative angle from the axis $x$ is $\delta$. It can be seen by writing the vector $(1,1)/\sqrt{2}$ which is the &quot;up&quot; eigenstate with respect to the $x$-axis. Then you may rotate this vector around the axis $z$ by angle $\delta$. This changes the two components differently because the $+$ and $-$ states have the opposite $j_z$ angular momentum. You get $(\exp(-i\delta/2),\exp(+i\delta/2))$ where the factor $1/2$ comes from the fact that we deal with a spin-1/2 system. After you multiply this vector by an overall phase $\exp(i\delta/2)$ that is physically irrelevant, as explained at the beginning, we obtain the vector you wrote. You should have perhaps used $x&amp;#39;$ in the description of the vector because it's not the normal $x$-axis but an axis whose direction depends on $\delta$.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-27T12:43:07.063" LastActivityDate="2011-01-27T12:43:07.063" />
  <row Id="3992" PostTypeId="2" ParentId="3983" CreationDate="2011-01-27T12:42:02.157" Score="4" ViewCount="" Body="&lt;p&gt;A comment on what Lubos wrote (it would seem that I need more reputation to post comments :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are still people who are living and are doing research in AQFT, so maybe AQFT is obsolete for some of us, but it is not dead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We still don't have a full mathematically rigorous understanding of QFT. But everybody is allowed to ignore this, of course, and use the computational tools of QFT that have proven their value. Haag's theorem tells us that we simply don't know why the computational tools of perturbative QFT work so well, but as far as concrete computations go, as long as you get the numbers right, this does not need to concern you. Therefore you won't find any calculations that did not work out because of Haag's theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any rigorous construction of an interacting 4D QFT will have to avoid Haag's theorem in one way or another, however. But if the endeavor to do research into this direction is worth the effort is of course a matter of discretion.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-01-27T12:42:02.157" CommentCount="6" />
  <row Id="3993" PostTypeId="1" CreationDate="2011-01-27T12:49:42.063" Score="1" ViewCount="171" Body="&lt;p&gt;In a couple of letters dated 1923, Pauli writes to Sommerfeld and Lande about Zeeman effect, and he describes (a piece of) quantisation in a peculiar way: the substitution of the &quot;&lt;em&gt;Differentialquotient&lt;/em&gt;&quot; $$ {d \over dj} {1 \over j} $$ ($j$ is called a &lt;em&gt;impulsquantenzahlen&lt;/em&gt;? is it an angular momentum, isn't it?) by the &quot;&lt;em&gt;Differenzen quotient&lt;/em&gt;&quot; $$\frac 1j - \frac 1{j-1}$$ (see page 3 of &lt;a href=&quot;http://cdsweb.cern.ch/record/83732/files/sommerfeld_0463-2.pdf&quot; rel=&quot;nofollow&quot;&gt;letter to Sommerfeld&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The later being argued as plausible from the integration of $\int {dj \over j^2}$ (footnote in page 4 of &lt;a href=&quot;http://cdsweb.cern.ch/record/83027/files/lande_0343.pdf?version=1&quot; rel=&quot;nofollow&quot;&gt;letter to Landé&lt;/a&gt;). I wonder, did the concept of &quot;&lt;em&gt;Differenzen quotient&lt;/em&gt;&quot; survive in the literature, as a quantisation method or even as some fundational concept of quantum mechanics?&lt;/p&gt;&#xA;" OwnerUserId="1335" LastEditorUserId="1335" LastEditorDisplayName="" LastEditDate="2011-01-27T17:02:31.547" LastActivityDate="2011-01-27T17:02:31.547" Title="Interpretation of QM as discrete derivative" Tags="&lt;quantum-mechanics&gt;&lt;history&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="3994" PostTypeId="2" ParentId="3644" CreationDate="2011-01-27T13:02:00.470" Score="0" ViewCount="" Body="&lt;p&gt;We measure relations between objects that belong to our local environment. &#xA;Our rulers are based on atomic properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;c&lt;/em&gt;&lt;/strong&gt; is &lt;strong&gt;a property of space&lt;/strong&gt; or, as said above, is a property of the universe, not of the light, because I share the viewpoint of &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1012/1012.2423v1.pdf&quot; rel=&quot;nofollow&quot;&gt;Israel Perez&lt;/a&gt; that matter and space are different expressions of the same entity, following Spinoza (Ethics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;c&lt;/em&gt;&lt;/strong&gt; is a relation, a ratio, between length and time intervals. If you check the definitions on time and length units it becomes clear that they are not independently defined, one is defined with help of the other. Then one must conclude that &lt;strong&gt;&lt;em&gt;c&lt;/em&gt;&lt;/strong&gt; is really a fundamental property that express how perturbations can evolve within space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar behaviour is watched in the waves in a lake. &#xA;It is a property of the medium (vacuum, field, aether,space,...many names)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This explains why light propagates always in the same way regardless of the motion of the source or the receiver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the value of the measure of c: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that we can only measure the two-way speed of light, not the one-way (as Poincaré). We only measure the ratio (L/T) and this is an absolute constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that in this world (or in the distant past) the atoms have the double of the size of our local atoms, then the relation L/T has to be the same, and physical laws keep invariant, (the electrons will take a longer time, compared to our local time, to evolve around the proton  and the emmited light will be redened :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do not make any 'absolute' measure on lenghts or time durations. Fundamentally we can not do it in another way.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-01-27T13:02:00.470" />
  <row Id="3995" PostTypeId="2" ParentId="3993" CreationDate="2011-01-27T13:46:26.877" Score="2" ViewCount="" Body="&lt;p&gt;the &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Difference_quotient&quot; rel=&quot;nofollow&quot;&gt;difference quotient&lt;/a&gt;&quot; (click) which is the straightforward translation of &quot;Differenzen quotient&quot; is simply the ratio of differences whose limit defines the derivative:&#xA;$$\frac{\Delta F}{\Delta X} = \frac{F(X+\Delta X)-F(X)}{\Delta X}$$&#xA;This simple ratio, as all simple mathematical expressions, obviously wasn't invented by Wolfgang Pauli and it appears at many places in physics (although fewer places than derivatives - because fundamental equations of physics depend on continuous variables). In your example, $X\equiv j$ and $\Delta j\equiv 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, a speculative partially hand-written letter from 1923, years before quantum mechanics was understood, is no longer relevant to understand quantum mechanics. Bohr's old model of the hydrogen atom has been superseded by quantum mechanics and this letter is even more vague, imprecise, and ephemeral than Bohr's old model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course that if $j$ is discrete, one cannot differentiate with respect to $j$ - at least not without an analytical continuation. The difference quotient would be the closest thing to differentiation one could find. On the other hand, there is usually no good reason to differentiate with respect to $j$, so the previous two sentences are largely inconsequential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Luboš&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-27T14:04:52.330" LastActivityDate="2011-01-27T14:04:52.330" CommentCount="1" />
  <row Id="3996" PostTypeId="2" ParentId="2014" CreationDate="2011-01-27T13:59:52.853" Score="0" ViewCount="" Body="&lt;p&gt;XX (and more generally XY) spin models became very popular lately, especially in connection to quantum-information theoretic problems. The XY model (without magnetic field though) has been introduced and &lt;em&gt;properly&lt;/em&gt; solved in 1961 by Lieb, Schultz and Mattis (Ann. Phys. 16, 407, (1961)). By the way this is a great reference to start with, if you didn't read this paper already. Nonetheless, in many recent articles, this model is solved incorrectly, and this might be the origin of the discrepancy that you observed. &#xA;Let me address your point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I imagine that the average that you mean is the ground state average: i.e. you digonalize the Hamiltonian, take the eigenvector related to the lowest eigenvalue, and compute statistical averages with this vector. &#xA;This said, to specify completely the problem you must assign some boundary conditions (BC). For many reasons (which I won't enter here), the most appropriate boundary condition is the periodic one (PBC) for the spins. With this I mean $\sigma^\alpha_{L+1}=\sigma^\alpha_1$, $L$ being the system size. When you map the spin problem to a Fermionic one through the Jordan-Wigner (JW) mapping, the Hamiltonian becomes that of free Fermions plus a peculiar border term. This is because the JW string accumulates in the bond which couples spins 1 and L. The result is that an XY (or XX) spin model with PBC is equivalent to a free-fermion model with a parity dependent boundary conditions (the parity operator is $P\sim \prod \sigma^z_i \sim (-1)^N$, and $N$ is the total number operator in the Fermi picture, and the $\sim$ symbol ignores potential factors of $i$). In the parity 1 sector (which comprises the ground state) BC are &lt;em&gt;antiperiodic&lt;/em&gt; (i.e. $c_{L+1}=-c_1$ for the Fermi operators $c_i$). This boils down to &lt;em&gt;antiperiodic&lt;/em&gt; quasimomenta when you do the Fourier transform: $k_n = \pi (2n+1)/L$ with $n=0,1,\ldots,L-1$. &#xA;If you instead choose open boundary conditions, there is no funky border term in the fermionic problem either, but the diagonalization of an open chain is slightly more cumbersome. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could be one source of errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another potential source of error is degeneracy (but this shouldn't be your case). When you diagonalize the Hamiltonian you should check that the ground state is unique. In a quasi-free systems you have degenerate ground states when the one-particle dispersion becomes exactly zero for some quasimomentum. This for instance, is known to happen in a pure free fermionic chain with nearest neighbour interaction and PBC,  when the size $L$ is a multiple of 4. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said, a great reference is Lieb, Schultz and Mattis article, where the &lt;em&gt;precise&lt;/em&gt;  diagonalization procedure is presented. Boundary conditions are also discussed in a recent paper of mine (PRA 81, 060101 (2010)) or in the book of M. Henkel, &lt;em&gt;Conformal invariance and critical phenomena&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best,&#xA;Lorenzo&lt;/p&gt;&#xA;" OwnerUserId="1619" LastActivityDate="2011-01-27T13:59:52.853" CommentCount="0" />
  <row Id="3997" PostTypeId="2" ParentId="3986" CreationDate="2011-01-27T14:11:05.317" Score="1" ViewCount="" Body="&lt;p&gt;If this is really about General Relativity then this is a description of a rotating body in General Relativity. As this question impinges on some issues in my General relativity Stack questions I shall make a few remarks about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General Relativity provides a solution (which we can think of for now as a generalised metric) given some conditions: usually matter conditions. Your basic scenario is of a bounded perfect fluid which is understood in GR. The other aspect of your condition is that it is rotating. A google search shows that for bounded matter a full rotational GR solution may not yet be known. The mechanism by which it is studied is that of a perturbation of a non-rotating solution based around the Schwarzchild solution. This is the model for a rotating star and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matter in the solution will follow the metric, which has curved spacetime inside and around the fluid. The surface tension and other things like that are meant to be incorporated in the Stress-Energy Tensor: if those features are present the fluid isnt perfect, and so another layer of approximation is used in practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that this rotating scenario also generates (weak) gravitational waves!&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-01-27T14:11:05.317" CommentCount="1" />
  <row Id="3998" PostTypeId="1" AcceptedAnswerId="4000" CreationDate="2011-01-27T14:14:11.907" Score="3" ViewCount="196" Body="&lt;p&gt;Can someone explain (as rigorously as possible) what is involved in analytically continuing, say, the Schwarzschild solution to the Kruskal manifold? I understand the two metrics separately but I'm not sure how analytic continuation is used, since I can't really see how the process of extending the domain of a complex function has anything to do with extending a manifold through a coordinate change.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T14:20:59.117" LastActivityDate="2011-02-03T14:20:59.117" Title="What is a maximal analytic extension?" Tags="&lt;general-relativity&gt;&lt;mathematical-physics&gt;&lt;cosmology&gt;&lt;black-holes&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="3999" PostTypeId="1" CreationDate="2011-01-27T14:29:15.263" Score="4" ViewCount="190" Body="&lt;p&gt;Last month, &lt;a href=&quot;http://physics.stackexchange.com/questions/1683/mechanics-around-a-rail-tank-wagon&quot;&gt;a puzzle here in physics.stackexchange&lt;/a&gt; asked to calculate the dynamics of the outflow in a loose (over rails, no friction) tank using Torricelli's law as a simplification. The simplification turned out to be problematic; one would prefer some solution where the height $h(0)=h_0$ starts decreasing with the usual condition $h&amp;#39;(0)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A first try to generalize is to use the potential energy stored in the tank to provide kinetic energy both to the CoM of the fluid and to the outflow. Consider for instance two horizontal equal nozzles, so that the CoM only goes down vertically. While this situation drives to a manageable second order differential equation, I am not sure if it is good enough. Particularly it seems that when we close the nozzle(s), so that the CM stops, this kinetic energy should change to be stored as internal energy, does it? We could imagine some configuration where we open the nozzle an interval $\Delta t$, close it, open it at the next interval, &amp;amp;c, so that we force this conversion to internal, not CoM, energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is there some generic condition where we can grant that all the energy goes either to the outflow or to the fluid CoM? If not, there is some usual parameter measuring how much of the energy goes to CoM and how much goes to internal (vorticity or so)?&lt;/p&gt;&#xA;" OwnerUserId="1335" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2011-01-27T15:15:03.727" LastActivityDate="2011-02-01T10:07:32.987" Title="Common approximations for fluid outflow beyond Torricelli's law" Tags="&lt;fluid-dynamics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4000" PostTypeId="2" ParentId="3998" CreationDate="2011-01-27T14:44:48.583" Score="3" ViewCount="" Body="&lt;p&gt;Dear dbrane, some basic terminology refinement could be useful. The words &quot;maximal&quot; and &quot;analytic&quot; are two more or less independent adjectives of the &quot;extension&quot;. The word &quot;maximal&quot; means that &quot;it cannot be extended further&quot;. On the other hand, the word &quot;analytic&quot; refers to the standard &quot;analytic functions&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analytic functions are infinitely differentiable functions such that if you write the Taylor expansion around any point, it converges to the exact original function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Real vs complex variables in analyticity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite Tim's warning, you are very right that the word &quot;analytic&quot; in the context of general relativity is linked to &quot;analytic&quot; in the context of functions of complex variable. The only difference is that in general relativity, we typically substitute real values for the spacetime coordinates only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the definition of an analytic function of a complex variable and an analytic function of a real variable is totally analogous. For complex functions of complex variables, it's still true that analytic functions are infinitely differentiable so that the Taylor expansion converges to the full function at each point of the &quot;domain&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, analytic functions of the complex variable are much more constrained than &quot;analytic functions of two real variables&quot;, namely the real and imaginary part: that's because the analytic functions of complex variables must be holomorphic - independent of the complex conjugate variable. In fact, &quot;analytic&quot; and &quot;holomorphic&quot; are exactly equivalent adjectives when it comes to functions of complex variables. So the right analogy is between holomorphic functions and real analytic functions of one (rather than two) real variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Extending solutions in general relativity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let's return to general relativity. In that case, the spacetime coordinates are real. A solution we start with - e.g. the Schwarzschild solution - is usually not maximally extended to start with: it has coordinate singularities and one can't get beyond them by reading the solution, even though geodesics continue through those points in the real space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An extension of this solution can be obtained if we cleverly redefine the spacetime coordinates so that the space around the coordinate singularity - in this case, the Schwarzschild event horizon - becomes regular. By this step, we get rid of the coordinate singularity and the metric tensor becomes nondegenerate even in the locus of the previous coordinate singularity (event horizon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we do so, it becomes clear that the locus of the horizon appears at a finite locus in spacetime, and because the metric is smooth on one side, the metric tensor may be continued as a collection of analytic functions of the new spacetime coordinates. This continuation is totally analogous to the continuation in the complex case: we may write the Taylor expansion around a given point near the previous boundary and simply extrapolate it as far as we can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may continue it as far as we can and if the Taylor expansion diverges somewhere, we may try to continue from another point to get even further. Again, even in the new coordinates, we may encounter coordinate singularities as we are extending the spacetime. In order to extend the solution further, we may choose even better coordinates, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process ultimately stops because the spacetime is surrounded either by asymptotic infinity - infinite volume where trajectories may be extended to an infinite proper length - or by genuine (curvature) singularities that cannot be extended by any coordinates. Geodesics physically terminate at those real singularities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My description above is pretty much a mechanical recipe how to proceed. However, in practice, one always needs to be clever at each point, to know which coordinates should be chosen to get as far as you can, and so on. He may also find out that there are several maximal extensions although I am not sure and I cannot mention any well-known examples now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A useful example are the kruskal-Szekeres coordinates for the Schwarzschild solution&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Kruskal-Szekeres_coordinates&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Kruskal-Szekeres_coordinates&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which are the maximal analytic continuation of the neutral black hole geometry.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-27T14:50:53.270" LastActivityDate="2011-01-27T14:50:53.270" CommentCount="2" />
  <row Id="4001" PostTypeId="2" ParentId="3955" CreationDate="2011-01-27T15:00:26.987" Score="3" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;However, positive and negative charges&#xA;  in the solar wind are differently&#xA;  trapped in the van Allen belts, from&#xA;  which they can then descend to the&#xA;  Earth's atmosphere, which implies that&#xA;  a net charge can be developed due to&#xA;  this differential leakage. This begs&#xA;  the question of whether there are any&#xA;  estimates of the total net charge.&#xA;  I've hunted in the literature but have&#xA;  found little useful material other&#xA;  than Dolezalek's 1988 paper:&#xA;  &lt;a href=&quot;http://www.springerlink.com/content/u057683112l148x5/&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/u057683112l148x5/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Right I see an important misunderstanding here!&#xA;Positive and negative charges from the solar wind are NOT differently trapped.  When we talk about the van Allen belts we are talking about high energy particles (MeV); low energy particles are still there, though in the majority of the outer belt the density is very low.  Its fair to say that the electrons and protons go through different acceleration processes (its also worth noting that there are two inner belts, a positive and a negative that overlap as well as sitting on the region of the magnetosphere called the plasmasphere - a cold dense plasma that extends to about 4 Earth radii)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the same waves that scatter relativistic waves into the loss cone and cause them to precipitate (electromagnetic ion-cyclotron waves) also scatter lower energy protons into the atmosphere.  Over very short time scales there will be a charge difference, but on average quasi-neutrality will be maintained. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When precipitation occurs you have to remember that you are dealing with the ionosphere which is itself a plasma and therefore quasi-neutral. Current systems do form due to the collisions of ions with the neutral atmosphere. In the D-region, where MeV electrons deposit when they precipitate (in fact they cause extra ionisation and extend the D-layer downwards) there is a whole mixed bag of high collisions, attachment processes and recombination processes involving all sorts of complex ion-chemistry, the less energetic protons will deposit at higher altitudes so there would be a small charge gains in different regions but due to high conductivity in the E layer (120 km) and possible current systems that would not be the case for long. *&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the atmosphere stuff is not my bag but you might like to go and look at the work of Dr. Martin Fullerkrug of the university of Bath who does some very cool work on sprites and whatnot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*I am not saying that exactly equal numbers of protons and electrons precipitate at a given moment but looking at the system over even relatively small time-scales the net charge would be damn close to zero.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-27T15:00:26.987" CommentCount="3" />
  <row Id="4002" PostTypeId="2" ParentId="3346" CreationDate="2011-01-27T15:08:07.757" Score="1" ViewCount="" Body="&lt;p&gt;Since there are no specialists in depletion mass spectrometry, I will try to answer in a more general way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With depletion spectroscopy you look at a small variation of a large signal so what you need is not high sensitivity but signal stability and high dynamic range of the detector. I assume that you have a continuous stream of ions, otherwise pulse to pulse fluctuations of ion concentrations will be the major limiting factor. Probably, you won't need superb mass resolution so you would want to buy a quadrupole mass spectrometer - these are cheap, compact and there are plenty of companies that make them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dynamic range depends on the ion detector and since you would be looking for a wide dynamic range and stability with time, the very best choice is a simple Faraday cup. You can be sure that, whatever your experiment is, the sensitivity will be limited not by the detector but by fluctuations of your signal - most likely, by how stable your &lt;em&gt;depleting factor&lt;/em&gt; is.&lt;/p&gt;&#xA;" OwnerUserId="701" LastActivityDate="2011-01-27T15:08:07.757" CommentCount="2" />
  <row Id="4003" PostTypeId="2" ParentId="815" CreationDate="2011-01-27T16:29:52.540" Score="1" ViewCount="" Body="&lt;p&gt;I have been attacking this question in a series of papers over many years, My last 4-5 papers on gr-qc are relevant. Pardon me for not repeating them here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Louis Crane&lt;/p&gt;&#xA;" LastActivityDate="2011-01-27T16:29:52.540" CommentCount="2" />
  <row Id="4004" PostTypeId="2" ParentId="3986" CreationDate="2011-01-27T16:32:33.273" Score="4" ViewCount="" Body="&lt;p&gt;The answers that have already been posted are correct, but @kakemonsteret raises a followup question in the comments that's worth addressing:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Lets say you are spinning somewhere in&#xA;  outer space, can you know you are&#xA;  spinning, ie can you rule out that the&#xA;  forces you feel are not caused by a&#xA;  mass distribution somewhere ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This question is getting at some ideas about &lt;a href=&quot;http://en.wikipedia.org/wiki/Mach%27s_principle&quot; rel=&quot;nofollow&quot;&gt;Mach's principle&lt;/a&gt; and its relation to general relativity, which is a somewhat complex subject. But there is a well-known effect in general relativity that bears directly on this question: the &lt;a href=&quot;http://en.wikipedia.org/wiki/Frame-dragging&quot; rel=&quot;nofollow&quot;&gt;Lense-Thirring effect&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a large spinning spherical massive shell. The local inertial reference frames inside the shell will be &quot;dragged&quot; around by the rotating mass, so that they rotate with respect to the &quot;fixed stars&quot; (i.e., the inertial reference frames far outside the shell). So if you lived inside this shell, and felt like you weren't rotating, you would &quot;really&quot; be rotating relative to the fixed stars. If you then started turning the opposite way at just the right rate, you could make it so that you weren't &quot;really&quot; rotating relative to the fixed stars, but you felt like you were.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I put scare quotes aroung the word &quot;really&quot; there for a reason: in general relativity, the most natural meaning to ascribe to the phrase &quot;really rotating&quot; is &quot;rotating with respect to your local inertial frame&quot; -- that is, if you feel like you're rotating (or if your Newton's bucket indicates you're rotating), then you are. But if you define &quot;really rotating&quot; to mean rotating with respect to very distant inertial objects, then yes, you can feel like you're rotating, even when you're not &quot;really rotating,&quot; due to being surrounded by lots of spinning mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Needless to say (I presume), this is all very much in-principle stuff: the frame-dragging effect is very small in practice.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-01-27T16:32:33.273" />
  <row Id="4005" PostTypeId="1" AcceptedAnswerId="4338" CreationDate="2011-01-27T16:43:46.173" Score="4" ViewCount="285" Body="&lt;p&gt;Recently I saw some physical problems that can be modeled by equations with fractional derivatives, and got a doubt: is it possible to write an action that results in an equation with fractional derivatives? For example, treat a hypothetical physical system with the principle of least action and find a &quot;wave equation&quot; with the derivative in time $3/2$? Or does such a question not make sense?&lt;/p&gt;&#xA;" OwnerUserId="1579" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2011-01-28T10:34:36.507" LastActivityDate="2011-02-01T00:40:26.280" Title="The derivation of fractional equations" Tags="&lt;mathematical-physics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="4006" PostTypeId="1" CreationDate="2011-01-27T16:50:06.920" Score="4" ViewCount="174" Body="&lt;p&gt;FYI: I imagine the physics of thunderstorms involves wind stripping electrons from condensed water cells (raindrops or snowflakes). Like a Van De Graaff generator, sort of; I'm also using Wilson's cloud chamber concept of an ion at the core of a droplet. Not sure if global warming promises higher winds, but I think so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acceptable answers: will address the seemingly polynomial temperature dependence of the thunderstorm probability. Extra thanks if the answer says something about other contributions of expected climate changes (eg the faster, more severe changes in the weather, but not the rise in average global temperature).&lt;/p&gt;&#xA;" OwnerUserId="1520" LastEditorUserId="1520" LastEditorDisplayName="" LastEditDate="2011-01-27T23:59:37.570" LastActivityDate="2011-01-27T23:59:37.570" Title="Why are cold thunderstorms rare?" Tags="&lt;geophysics&gt;&lt;lightning&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="4007" PostTypeId="2" ParentId="3698" CreationDate="2011-01-27T16:50:18.177" Score="0" ViewCount="" Body="&lt;p&gt;Assuming you're not interested in thermal equilibrium everywhere, then as the black hole mass approaches the Planck mass, the thermodynamic limit for black hole microstates breaks down, and we can no longer speak of its temperature meaningfully.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-01-27T16:50:18.177" />
  <row Id="4008" PostTypeId="2" ParentId="4005" CreationDate="2011-01-27T16:52:46.380" Score="1" ViewCount="" Body="&lt;p&gt;Fractional derivatives are nonlocal, but actions are usually assumed to be local.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-01-27T16:52:46.380" CommentCount="5" />
  <row Id="4009" PostTypeId="2" ParentId="4006" CreationDate="2011-01-27T17:15:00.237" Score="3" ViewCount="" Body="&lt;p&gt;Thunderstorms &#xA;get their energy from condensation of humidity in rising air. &#xA;Enthalpy of condensation of water is very high compared to &#xA;other liquids. &#xA;So in summertime, very humid air is lifted by a cold front &#xA;inrush or some mechanism which brings wet warm air below colder air&#xA;(very important for those tornadoes in great plains) thunderstorms &#xA;are common. &#xA;In winter the air in a &quot;warm sector&quot; of  a cyclone is much colder,&#xA;the humidity which is carried by this air is low: &#xA;Not enough &quot;gas/petrol&quot; to run a thunderstorm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Georg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS &#xA;In case that &quot;global warming&quot; will produce winters as warm/humid as &#xA;summers today, yes, thunderstorms will be more common, obviously.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-01-27T17:15:00.237" />
  <row Id="4010" PostTypeId="1" AcceptedAnswerId="4014" CreationDate="2011-01-27T18:43:56.897" Score="3" ViewCount="461" Body="&lt;p&gt;Yesterday I looked underwater with my eyes open (and no goggles) and I realized I can't see anything clearly. Everything looks very, very blurry. My guess is that the eye needs direct contact with air in order to work properly. With water, the refraction index is different, and the eye lens are not able to compensate for correct focalization on the retina.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right ? If so, what lenses should one wear in order to see clearly while under water ?&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="217" LastEditorDisplayName="" LastEditDate="2011-01-27T20:16:08.760" LastActivityDate="2011-01-28T16:48:27.560" Title="Eyes open under water" Tags="&lt;optics&gt;&lt;water&gt;&lt;refraction&gt;&lt;eye&gt;&lt;vision&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="4011" PostTypeId="1" CreationDate="2011-01-27T18:53:21.343" Score="0" ViewCount="129" Body="&lt;p&gt;The role of &quot;time&quot; in fundamental physics has changed a lot in recent decades. Some physcists believe time is an emergent property rather than a most fundamental parameter of nature laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when reviewing the relativity chapter of my undergrad textbook, I found it really difficult (at least for me) to remove the temporal dimension from its conceptual framework. For example, can we formulate Lorentz spacetime interval without time parameters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I am not asking how to formulate the spacetime interval with only three independent parameters. What I wonder is, given a physical phenomenon that between two events there exists a quantity, which is an invariance of all inertial reference frames, very similar to the distance between two points in a Euclidean space, can we write down an equation of this quantity without a parameter of time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantity is not necessarily the same thing as we know it in relativity. Any alternative would be fine as long as it tells the same story.&lt;/p&gt;&#xA;" OwnerUserId="749" LastEditorUserId="749" LastEditorDisplayName="" LastEditDate="2011-01-27T19:45:01.453" LastActivityDate="2011-01-27T19:45:01.453" ClosedDate="2011-01-27T19:13:30.937" Title="How to formulate Lorentz invariance without the temporal dimension?" Tags="&lt;spacetime&gt;&lt;relativity&gt;" CommentCount="4" />
  <row Id="4012" PostTypeId="2" ParentId="4010" CreationDate="2011-01-27T19:24:31.817" Score="3" ViewCount="" Body="&lt;p&gt;Snell’s law works with a change in the index of refraction across two media.  The difference is greater for air on cornea than it is for water on cornea.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-27T19:24:31.817" />
  <row Id="4013" PostTypeId="2" ParentId="4010" CreationDate="2011-01-27T19:38:46.597" Score="4" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Am I right ? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If so, what lenses should one wear in order to see clearly while under water ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You don't need extra lens you have one in your eyes, just use goggles that makes a layer of air between the water and your eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you decide to put a convergent lens in front of your eyes it won't work because your eye will still not be able to focus different depths. So, you will only be able to focus rays coming from a fixed distance, what is not very useful.&lt;/p&gt;&#xA;" OwnerUserId="1337" LastActivityDate="2011-01-27T19:38:46.597" />
  <row Id="4014" PostTypeId="2" ParentId="4010" CreationDate="2011-01-27T19:41:59.927" Score="8" ViewCount="" Body="&lt;p&gt;You can't see clearly underwater for a couple of reasons.  One is the thickness of your lens, but the main one is the index of refraction of your cornea. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, here's the Wikipedia picture of a human eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/5tAHv.png&quot; alt=&quot;http://en.wikipedia.org/wiki/File:Schematic_diagram_of_the_human_eye_en.svg&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cornea&quot; rel=&quot;nofollow&quot;&gt;According to Wikipedia&lt;/a&gt;, two-thirds of the refractive power of your eye is in your cornea, and the cornea's refractive index is about 1.376.  The refractive index of water (&lt;a href=&quot;http://www.google.com/search?sourceid=chrome&amp;amp;ie=UTF-8&amp;amp;q=refractive+index+of+water&quot; rel=&quot;nofollow&quot;&gt;according to Google&lt;/a&gt;) is 1.33.  In water, your cornea bends light as much as a lens in air whose refractive index is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1.376-1.33}{1.33} + 1 = 1.034$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means you're losing about 90% of your cornea's refractive power, or 60% of your total refractive power, when you enter the water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question becomes whether your lens can compensate for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't find a direct quote on how much you can change the focal distance of your lens, but we can estimate that your cornea is doing essentially nothing, and ask whether your lens ought to be able to do all the focusing itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a spherical lens with index of refraction $n$ sitting in a medium with index of refraction $n_0$, the &lt;a href=&quot;http://www.edmundoptics.com/technical-support/optics/understanding-ball-lenses/&quot; rel=&quot;nofollow&quot;&gt;effective focal length is&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f = \frac{nD}{4(n-n_0)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The refractive index of your vitreous humor is about 1.33 (like water), and the refractive index of your lens, &lt;a href=&quot;http://en.wikipedia.org/wiki/Lens_%28anatomy%29&quot; rel=&quot;nofollow&quot;&gt;according to Wikipedia&lt;/a&gt;, varies between 1.386 and 1.406.  Let's take 1.40 as an average.  Then, plugging in the numbers, the effective focal distance of a spherical eye lens would be five times its diameter.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia picture of a human eye makes this look reasonable - a spherical lens might be able to do all the focusing a human eye needs, even without the cornea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that your eye's lens isn't spherical.  From the same Wikipedia article&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;In many aquatic vertebrates, the lens is considerably thicker, almost spherical, to increase the refraction of light. This difference compensates for the smaller angle of refraction between the eye's cornea and the watery medium, as they have similar refractive indices. [2] Even among terrestrial animals, however, the lens of primates such as humans is unusually flat.[3]&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, the reason you can't see well underwater is that your eye lens is too flat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wear goggles, the light is refracted much more as it enters the cornea - the same amount as normal.  If you want to wear some sort of corrective lenses directly on your eye like contact lenses, they should have a refractive index as low as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Googling for &quot;underwater contact lens&quot;, I found &lt;a href=&quot;http://sportsillustrated.cnn.com/vault/article/magazine/MAG1075502/index.htm&quot; rel=&quot;nofollow&quot;&gt;an article about contact lenses made with a layer of air&lt;/a&gt;, allowing divers to see sharply underwater.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-01-27T19:41:59.927" CommentCount="2" />
  <row Id="4015" PostTypeId="1" AcceptedAnswerId="4018" CreationDate="2011-01-27T19:42:47.247" Score="2" ViewCount="244" Body="&lt;p&gt;My question is the following. Are there solutions to the Einstein field equations, which have the property that there is a hypersurface of constant time and to the past of that surface space is empty (Minkowski space-time) and to the future it is not (non-vanishing stress-energy tensor)? At first thought it seems strange to have nothing and suddenly something. On the other hand define $g_{\mu\nu}(x)$ to be the Minkowski metric for the past of the surface and any functions (sufficiently smooth) that match it on the surface and are different than the Minkowski for the future. There are many ways (infinitely many) to do this. Then one can define the stress-energy tensor using Einstein's equations. This would seem to work, but there are certain conditions on the tensor, for example positivity, or other restrictions for physical reasons that I am unaware of, so it may be there are no such functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the questions is: does anyone know of an explicit example or a more convincing argument than &quot;there are so many functions, there has got to be some that work&quot;? Of course a reason why it doesn't work, if that is the case, would be good too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.    &lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-01-27T21:41:43.560" LastActivityDate="2011-02-19T14:51:51.117" Title="Empty universe in the past, non-empty in the future" Tags="&lt;general-relativity&gt;" AnswerCount="5" />
  <row Id="4016" PostTypeId="2" ParentId="1453" CreationDate="2011-01-27T19:59:08.907" Score="2" ViewCount="" Body="&lt;p&gt;The Fundamental Properties of Space and Time at an Event.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some specifics are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is Time a logical derivable construct from some non-time based fundamental theory?&lt;/li&gt;&#xA;&lt;li&gt;Is Time discrete or continuous?&lt;/li&gt;&#xA;&lt;li&gt;Is Space(-Time) discrete - and if so at the Planck level?&lt;/li&gt;&#xA;&lt;li&gt;Directionality (and perhaps dimensionality) of Time&lt;/li&gt;&#xA;&lt;li&gt;Dimensionality of Space (the String Theory issue)&lt;/li&gt;&#xA;&lt;li&gt;Substructure (and any physics) below any Planck scale &quot;minimum distance&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These kinds of issues are of course at the root of many conceptual and calculational issues in the theories around today.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-01-27T19:59:08.907" CommunityOwnedDate="2011-01-27T19:59:08.907" />
  <row Id="4017" PostTypeId="2" ParentId="4015" CreationDate="2011-01-27T20:08:44.683" Score="1" ViewCount="" Body="&lt;p&gt;Dear MBN, if you allow the future metric to be sourced by an arbitrary $T_{\mu\nu}$ that doesn't have to be derived from any actual type of matter or fields, then your spliced spacetime may be constructed. However, it's simply because Einstein's equations become tautologies if you allow $T_{\mu\nu}$ to be anything. For any geometry (metric), you may find a $T_{\mu\nu}$ profile such that Einstein's equations are satisfied - just calculate the &quot;right&quot; stress-energy tensor from Einstein's equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just take the Minkowski space for $t&amp;lt;0$ and extend the metric tensor components to arbitrary functions of $t$ that are infinitely differentiable, even at $t=0$, but that are non-constant for positive $t$. For example, you may take such functions to be &#xA;$$g_{\mu\nu}(t) = g_{\mu\nu}(t&amp;lt;0) + C_{\mu\nu} \exp(-1/t^2)$$&#xA;for positive $t$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, calculate the curvature and Einstein's tensor out of this arbitrary metric, and you will know what $T_{\mu\nu}$ should be declared to be the source of this gravitational field. However, you will never derive such a &quot;suddenly turning on&quot; matter source out of any well-defined equations. It's because you would face a similar problem e.g. for the electromagnetic field that could source the strange spliced gravitational field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the electromagnetic source also can't be turned on &quot;suddenly&quot;, unless it's sourced by an electric charge distribution $j_\mu$ that also has to be turned on suddenly. So something has to be &quot;externally inserted&quot; to any physical system to achieve the change of the behavior in $t=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's more important is that you can't get your spliced spacetime out of the vacuum Einstein's equations, i.e. those with $T_{\mu\nu}=0$. Einstein's equations are second-order partial differential equations for the metric components $g_{\mu\nu}(t,x,y,z)$. Throw away all the indices and useless dimensions and solve a morally similar problem, the equation&#xA;$$d^2 x/dt^2 = 0$$&#xA;Clearly, this equation has a unique solution for given initial conditions. Well, in this particular form, the solutions have to be linear functions, but more generally, the solution is uniquely determined by the initial conditions. The same can be proved for the partial differential equations from a broad class - including Einstein's equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, you can say that the change of the behavior can only be achieved if some external sources of the fields are manually added at $t=0$. Of course, if you gradually turn them on, the response will be gradually increasing, too. But there is nothing mysterious about it. If you suddenly &quot;buy&quot; some matter from another spacetime and insert it to Einstein's equations starting from $t=0$, e.g. by an infinitely smooth function, the metric starts to get curved in a similar way after $t=0$, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the stress-energy tensor has to have a vanishing covariant divergence (because the Einstein's tensor has the same property identically), it will constrain the kinds of stress-energy tensors that may be turned in this way. It's likely that the required stress-energy tensor - e.g. one calculate from the metric I wrote in a displayed equation above - always violates some (or all) energy conditions right after $t=0$. In particular, the components of the required $T_{\mu\nu}$ will be &quot;mostly spacelike&quot; which is forbidden by the null or dominant energy conditions.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-27T20:14:31.977" LastActivityDate="2011-01-27T20:14:31.977" CommentCount="1" />
  <row Id="4018" PostTypeId="2" ParentId="4015" CreationDate="2011-01-27T20:10:24.107" Score="6" ViewCount="" Body="&lt;p&gt;I believe that no such spacetime exists, if the matter is assumed to satisfy an inequality known as the &lt;em&gt;dominant energy condition&lt;/em&gt;. The dominant energy condition says that, if $\xi$ is a future-directed timelike vector, then $-{T^a}_b\xi^b$ is a future-directed timelike or null vector (sign conventions, etc., following Wald's book &lt;em&gt;General Relativity&lt;/em&gt;). Heuristically, this condition means that an observer at any location will always measure an energy-momentum 4-current in his vicinity that is flowing at less than or equal to the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this condition, one can show that, if space is empty at one time (i.e., if there is a spacelike Cauchy hypersurface along which $T=0$), then it vanishes at all times (Wald, p. 219). If in addition spacetime is Minkowski along some Cauchy surface, then the initial value theorems say that it's Minkowski at all times (Wald, Chapter 10).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heuristically, the above argument says that, if there's no matter at one time, but there is at a later time, then matter must have popped out of nowhere. The dominant energy condition doesn't allow that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know whether the result holds if you assume something weaker than the dominant energy condition (e.g., the aptly-named weak energy condition). Certainly you need &lt;em&gt;some&lt;/em&gt; sort of restriction on the allowed properties of the matter, as Lubos Motl says: if any $T_{ab}$ is allowed, then any smooth metric is allowed.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-01-27T20:16:17.733" LastActivityDate="2011-01-27T20:16:17.733" CommentCount="1" />
  <row Id="4019" PostTypeId="2" ParentId="1453" CreationDate="2011-01-27T21:46:47.990" Score="0" ViewCount="" Body="&lt;p&gt;What is mass/Inertia? What is gravity?&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-01-27T21:46:47.990" CommunityOwnedDate="2011-01-27T21:46:47.990" />
  <row Id="4020" PostTypeId="2" ParentId="4006" CreationDate="2011-01-27T22:01:30.963" Score="1" ViewCount="" Body="&lt;p&gt;Thunderstorms, are convective storms, driven by convective heat transfer from lower layers of the atmosphere to higher up. During winter in temperate lattitudes the surface radiation balance (solar shortwave absorbed, minus net infrared radiation upwards) is negative or weakly positive, so very little convective activity ocurrs. While in the summer, the surface radiation balance is strongly positive, so the excess heat is carried upwards by convection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thunderstorms do not require a cold front, (although it helps). Thunderstorms are nearly a daily occurrence in some parts of the tropics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if the frequency of thunderstorms will increase or decrease due to global warming. Convection is driven by temperature constrasts with respect to height, not to absolute temperature. There should be fewer cold fronts, because the poles will warm more than the tropics, this would tend to reduce the numbers. But, there will also be more moisture in the air which would tend to increase the numbers. The high end storms seem to be becoming more frequent, and this is thought to be becuse at higher temperatures the relative amount of latent energy of water vapor is higher, which tends to allow the strongest storms to sustain higher inflows thus drawing moisture in even faster.&lt;/p&gt;&#xA;" OwnerUserId="377" LastEditorUserId="377" LastEditorDisplayName="" LastEditDate="2011-01-27T23:32:10.107" LastActivityDate="2011-01-27T23:32:10.107" CommentCount="1" />
  <row Id="4021" PostTypeId="2" ParentId="3987" CreationDate="2011-01-27T22:18:00.233" Score="2" ViewCount="" Body="&lt;p&gt;You'll know the Pauli matrices $\sigma_1=\left(\begin{matrix}0 &amp;amp; 1\\1 &amp;amp; 0\end{matrix}\right)$, $\sigma_2=\left(\begin{matrix}0 &amp;amp; -i\\i &amp;amp; 0\end{matrix}\right)$, and $\sigma_3=\left(\begin{matrix}1 &amp;amp; 0\\0 &amp;amp; -1\end{matrix}\right)$. Take a spinor such as your $v=\frac{1}{\sqrt{2}}\left(\matrix{1 \\e^{i\delta}}\right)$, taking your basis vectors $\left|+\right&amp;gt;,\left|-\right&amp;gt;$ to be $\left(\matrix{1 \\0}\right)$ and $\left(\matrix{0 \\1}\right)$, and construct the 3-vector $(v^\dagger\!\sigma_1\! v, v^\dagger\!\sigma_2\! v,v^\dagger\!\sigma_3\! v)$, taking both the transpose of $v$ and the complex conjugate, which is called the Hermitian conjugate; you'll find that it's $(\cos(\delta),\sin(\delta),0)$. This is only saying what Rafael and Lubos already said.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Update: There's a significant&#xA;addition to this, pushing a little&#xA;beyond the level of your question,&#xA;which is that as well as the 3-vector&#xA;$(v^\dagger\!\sigma_1\! v,&#xD;&#xA;   v^\dagger\!\sigma_2\!&#xD;&#xA;   v,v^\dagger\!\sigma_3\! v)$, we can&#xA;also construct the &lt;em&gt;null&lt;/em&gt; 4-vector&#xA;$(v^\dagger v, v^\dagger\!\sigma_1\!&#xD;&#xA;   v, v^\dagger\!\sigma_2\!&#xD;&#xA;   v,v^\dagger\!\sigma_3\! v)$, where&#xA;$(v^\dagger v)^2$ is the same as the&#xA;sum of the squares of the other three&#xA;components. This is important when we&#xA;come to relativistic systems. The&#xA;relative phase you asked about still&#xA;corresponds to a rotation of the&#xA;3-vector part of this null 4-vector,&#xA;but now there's also a possibility of&#xA;Lorentz boosts introducing a relative&#xA;phase, as well as rotations.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you change the complex phase $\delta$, the new spinor corresponds to a different 3-vector, rotated through the difference between the angles. If you consider a spinor such as $\left(\matrix{\cos(\alpha) \\ \sin(\alpha)e^{i\delta}}\right)$ you get a vector somewhere between the $x-y$ plane and the $z$ axis. This turns out to be very useful and significantly different from other ways of constructing vectors. On the other hand, if you just multiply the spinor by $e^{i\alpha}$, you'll get the same 3-vector. In Physics we just throw away the overall complex phase of a spinor, but in Mathematics we can consider something like the real and complex parts of the complex 3-vector $(v^T\sigma_2\sigma_i v)$, which turn out to be two more 3-vectors that are always orthogonal to $(v^\dagger\sigma_i v)$ and the same length. Nice! The mapping $v^\dagger\mapsto v^T\sigma_2$ is known as the &quot;charge conjugate&quot; in Physics, $v\mapsto v^c$; it looks a little screwy in this notation, but it can be constructed more naturally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interpretation of spinors is a minefield, however. A spinor can as well represent rotations, or a triad of vectors, as a single vector, and for Dirac spinors there has at times been a whole industry devoted to trying to figure out ways to think about them, largely because there is not as simple a relationship between Dirac spinors and 4-vectors and rotations in Minkowski space. Perhaps best to think of them as just happening to be the right number of degrees of freedom, and being a &lt;em&gt;representation&lt;/em&gt; of the Lorentz group in a useful way, when we're describing experimental results. &lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-01-28T18:21:48.587" LastActivityDate="2011-01-28T18:21:48.587" />
  <row Id="4022" PostTypeId="1" CreationDate="2011-01-27T22:46:56.227" Score="2" ViewCount="188" Body="&lt;p&gt;Measuring a qubit and ending up with a bit feels a little like tossing out infinities in renormalization. Does neglecting the part of the wave function with a vanishing Hilbert space norm amount to renormalizing of Hilbert space?&lt;/p&gt;&#xA;" LastActivityDate="2011-01-28T02:20:23.410" Title="Renormalization and Infinites" Tags="&lt;mathematical-physics&gt;&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="4023" PostTypeId="2" ParentId="4022" CreationDate="2011-01-27T22:53:20.590" Score="1" ViewCount="" Body="&lt;p&gt;I am maybe a bit uncertain what you are asking, but from what I understand the answer would be no.  Renormalization is a procedure for absorbing infinities in an interacting field theory.  A quantum bit is really just a state, but referred to in information theoretic terms.  The two physics are not directly related as such.  In a measurement if one considers it as a collapse there is a new normalization (renormalization?) of the system state, which is just the state vector which pertains to the measurement outcome.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-27T22:53:20.590" />
  <row Id="4024" PostTypeId="2" ParentId="3076" CreationDate="2011-01-27T23:06:10.953" Score="2" ViewCount="" Body="&lt;p&gt;I saw a video last year of a person testing a pressure suit in a vacuum chamber.  This occurred back in 1966.  The video has been removed unfortunately.  What happened is a pressure hose detached and the air pressure ran out of his suit.  He actually stood up conscious for about 30 seconds, but then passed out.  He fell off of a stage, which frankly looked painful.  The crew outside repressurized the chamber and within a minute the guy was alert and stood up.  According to this man he could feel the saliva in his mouth start to boil as if he were drinking a soft drink.  Otherwise there were no ill effects.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-27T23:06:10.953" />
  <row Id="4025" PostTypeId="2" ParentId="3076" CreationDate="2011-01-27T23:29:30.293" Score="0" ViewCount="" Body="&lt;p&gt;If you remember the infamous teenage date malady, the hicky. I think you'd see something like that as the blood would begin to ooze through your skin. But, I think this effect takes several minutes, so you'd already have died from lack of Oxygen.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-01-27T23:29:30.293" CommentCount="1" />
  <row Id="4026" PostTypeId="2" ParentId="4022" CreationDate="2011-01-28T00:40:24.027" Score="5" ViewCount="" Body="&lt;p&gt;No, those are two very different processes (as far as I understand).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Renormalization: When you are calculating vacuum expectation values, for instance $\langle \Omega\mid T(\phi(\mathbf{p})\phi(0))\mid \Omega\rangle$, you discover that these values are infinite. However, you can interpret this infinity, in a consistent manner, as the value of this correlation function at other momentum $\mathbf{p}^{\prime}$ and a finite part that relates the correlation function at the two different momenta. Nothing is really lost in the renormalization procedure, it is just a matter of how to introduce a measured quantity (the correlation function at this other momentum) into the theory.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Measurement: The measurement concerns a certain state $\mid \psi\rangle$ coupled to the a measurement device. Originally, before being coupled, the pure state has entropy equal to zero. Later, by the time evolution of the coupled system, the system being measured has, after tracing over the measurement device states, entropy larger than zero. The difference is the information lost by the system in the process. So, something is lost in the measurement process, contrary to renormalization.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1337" LastEditorUserId="1337" LastEditorDisplayName="" LastEditDate="2011-01-28T02:20:23.410" LastActivityDate="2011-01-28T02:20:23.410" />
  <row Id="4027" PostTypeId="1" AcceptedAnswerId="4032" CreationDate="2011-01-28T02:11:40.070" Score="7" ViewCount="219" Body="&lt;p&gt;LQG formulations have a minimum length/area. Since say, a Planck area can always&#xA;be boosted, any minimum area in space can be shrunk. Do LQG proponents worry about local Lorentz invariance violation, and if not, why not?  In LQG, does considering length to be a quantum operator really get rid of the boost problem?&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-01-29T08:08:12.030" Title="Are there any versions of LQG that claim to not violate Lorentz symmetry?" Tags="&lt;quantum-gravity&gt;&lt;loop-quantum-gravity&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="2" />
  <row Id="4028" PostTypeId="2" ParentId="1453" CreationDate="2011-01-28T03:08:17.203" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;What is the exact form of Exchange-correlation (xc) functional in Density Functional Theory (DFT)?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The entire community of Condensed Matter Physics and Quantum Chesmitry want to know that answer. (:&lt;/p&gt;&#xA;" OwnerUserId="1545" LastActivityDate="2011-01-28T03:08:17.203" CommunityOwnedDate="2011-01-28T03:08:17.203" CommentCount="1" />
  <row Id="4029" PostTypeId="1" CreationDate="2011-01-28T03:27:52.350" Score="0" ViewCount="125" Body="&lt;p&gt;Pitch sequences are classificed disregarding the effects of five musical transformations: octave shift, permutation, transposition, inversion, and cardinality change. These may be modelled according to 32 families of quotient spaces for octave , transposition, triad permutations etc.  Theses structures are moduli spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general question asked is then what commonality do you think there is between music  physics and the foundations of the universe?&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-28T03:27:52.350" ClosedDate="2011-01-28T03:58:19.793" Title="What is the relationship between the structures of music and the universe?" Tags="&lt;group-theory&gt;" CommentCount="3" />
  <row Id="4030" PostTypeId="2" ParentId="4015" CreationDate="2011-01-28T05:11:45.427" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there solutions to the Einstein field equations, which have the property that there is a hypersurface of constant time and to the past of that surface space is empty (Minkowski space-time) and to the future it is not (non-vanishing stress-energy tensor)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is essentially what happens in inflationary theory. Though, not quite in the strong mathematical sense that you are seeking. There one starts out with a scalar field in an expanding background metric, usually deSitter. What happens is that the exponential expansion of spacetime amplifies quantum fluctuations in the scalar so that the end of the inflationary phase we are left with a scalar with a scale-invariant spectrum of perturbations. It is these perturbations that go on to see structure formation in the early Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Einstein's equations are satisfied by the initial scalar field + deSitter system. Of course these considerations involve QFT on a curved background so one has to go beyond the framework of classical GR in calculating the spectrum of perturbations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read more about this in Chapter 8 of Mukhanov's excellent book &quot;Physical Foundations of Cosmology&quot;.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-28T05:11:45.427" CommentCount="3" />
  <row Id="4031" PostTypeId="1" AcceptedAnswerId="4035" CreationDate="2011-01-28T08:57:48.417" Score="7" ViewCount="227" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Can anyone tell me where can I read about the notion of &quot;short&quot; and &quot;long&quot; representations? Like what they are etc. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;From where can I learn the arguments which show that the bosonic subalgebra of $\cal{N}=2$ and $\cal{N}=3$ Lie superalgebra are given by $SO(3,2)\times SO(2)$ and $SO(3,2)\times  SO(3)$ respectively? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I have often seen the following statement being made about the above which I do not understand, that primary states of these algebras are labeled by $(\Delta,j,h)$ where $\Delta$ is the scaling dimension of the primaries, $j$ is its spin and $h$ is its $R-$charge (or R is its charge highest weight)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Can someone explain the above labeling scheme. I don't understand much of the above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somehow the above doesn't seem to fit with my elementary understanding of what are primary operators in 1+1 CFT which are defined using particular forms of the OPE with stress-energy tensor.   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Then one argues that for $j\neq 0$ unitarity forces $\Delta \geq \vert h\vert + j+1$ and for $j=0$ unitary representations occur when $\Delta = h$ or $\Delta \geq \vert h \vert + 1$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I haven't been able to trace the above argument in any reference book or exposition. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It is said that the isolated representations which saturate the above unitarity bound for $j=0$ are all short. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Apparently the Witten Index in this context is defined as $Tr(-1)^Fx^{\Delta +j}$ and it vanishes on all long representations but is nonzero on all short representations. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Further to understand the state content of all unitary representations in superconformal algebras people defined two different kinds of Witten indexes, $\cal{I}^+$ and $\cal{I}^-$ as,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$${\cal I}^+ = Tr(-1)^Fx^{\Delta+j}e^{-\beta(\Delta-j-h)}$$&#xA;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\cal I}^+ = Tr(-1)^Fx^{\Delta+j}e^{-\beta(\Delta-j+h)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above are apparently independent of $\beta$. I would like to know why. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently the first index above receives contributions only from states with $\Delta = j+h$ and all such states probably are annihilated in some sense and have something to do with the &lt;code&gt;cohomology of the supercharge with charges $(\frac{1}{2},-\frac{1}{2},1)$ Similarly somehow the second index above receives contributions only from states with $\Delta = j-h$ and all such states probably are also annihilated in some sense and have something to do with the&lt;/code&gt;cohomology of the supercharge with charges $(\frac{1}{2},\frac{1}{2},1)$ &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Eventually I see these indices to be writable as products (sometimes infinite) of rational polynomials in $x$. I don't know how it happens.  This way of writing has something to do with &quot;single/multitrace operators/primaries&quot; (another concept which I don't understand!) &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would be grateful to hear of explanations of the above arguments and constructions and also if detailed references/expositions exist for the above. &lt;/p&gt;&#xA;" OwnerUserId="1637" LastEditorUserId="1637" LastEditorDisplayName="" LastEditDate="2011-01-28T09:25:29.263" LastActivityDate="2011-01-28T17:06:34.480" Title="Superconformal theories" Tags="&lt;quantum-field-theory&gt;&lt;conformal-field-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="4032" PostTypeId="2" ParentId="4027" CreationDate="2011-01-28T09:45:16.033" Score="2" ViewCount="" Body="&lt;p&gt;This has been asked and answered before: see &lt;a href=&quot;http://physics.stackexchange.com/questions/3662/does-the-discreteness-of-spacetime-in-canonical-approaches-imply-good-bye-to-str&quot;&gt;Does the discreteness of spacetime in canonical approaches imply good bye to STR?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, this question has popped up many times on other sites such as physicsforums: &lt;a href=&quot;http://www.physicsforums.com/showthread.php?t=281951&quot; rel=&quot;nofollow&quot;&gt;http://www.physicsforums.com/showthread.php?t=281951&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is roughly that LQG does not in fact violate Lorentz invariance. The discretisation of area and volume operators does not imply a broken symmetry, any more than discretisation of angular momentum states imply breaking of rotational symmetry --- symmetries in quantum theories are equations of the operator algebra, not of the states!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also: &lt;a href=&quot;http://arxiv.org/abs/1012.1739&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1012.1739&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-01-28T09:45:16.033" CommentCount="6" />
  <row Id="4033" PostTypeId="2" ParentId="4031" CreationDate="2011-01-28T10:12:08.780" Score="2" ViewCount="" Body="&lt;p&gt;The functional space of all possible field configurations of a primary field is always a representation of the conformal group.&#xA;For most choices of tensors and conformal weights, this representation is irreducible, and the primary field together with its associated tower of secondary fields constructed from taking derivatives of the primary field forms a covariant representation of it. This is called a &lt;strong&gt;long representation&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for some special cases, we may write down covariant constraints which drastically reduce the representation to massless modes only. Such representations are &lt;strong&gt;short representations&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, given a primary scalar field $\phi$ with a conformal weight of $1$, the constraint $\square \phi = 0$ is conformally covariant and gives us a short representation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or we might have a primary vector field $\mathbf{A}$ with a conformal weight of $1$ and the constraint $\partial^\nu \left( \partial_\mu A_\nu - \partial_\nu A_\mu \right) = 0$ quotiented by the gauge transformation $\mathbf{A} \to \mathbf{A} + d\lambda$ for scalar primary fields $\lambda$ with zero conformal weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or a primary 2-form $\mathbf{F}$ with a weight of $2$, and constraints $d\mathbf{F} = 0$ and $\partial^\nu F_{\mu\nu} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The conformal group of d+1 dimensional conformal spacetime is $SO(d+1,2)$. Conformal superalgebras can be either real, complex or pseudoreal depending upon the number of dimensions. For the real case, we can have the optional $R$-symmetry $SO(\mathcal{N})$. For the complex case, it's $U(\mathcal{N})$ or sometimes just $SU(\mathcal{N})$. For the pseudoreal case, it's $\text{Sp}(\mathcal{N})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have left and right-handed Weyl spinors which are either Majorana-Weyl or pseudoreal, we can have an $(\mathcal{M}, \mathcal{N})$ superconformal algebra with left- and right-handed $R$-symmetries.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In the covariant representation of a superconformal group, the stabilizer of a point is generated by the Lorentz group, a dilatation, some proper conformal generators $\mathbf{K}$, fermionic conformal generators $S^{\alpha i}$, $R$-symmetry generators, and possibly, some central extensions. $\Delta$ is the weight under the dilatation. The spin is the representation under the Lorentz group, and we have to specify the representation of the $R$-symmetry. $\mathbf{K}$ and $S^\alpha$ generate a tower of components with different dilatation weights. For physical representations, we have to assume there is a lower bound to this tower. The primary field corresponds to the eigenspace of the lowest eigenvalue.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;1+1D is a special case entirely, and we have two copies of the Virasoro algebra, complete with central extensions.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-28T17:06:34.480" LastActivityDate="2011-01-28T17:06:34.480" CommentCount="1" />
  <row Id="4034" PostTypeId="1" AcceptedAnswerId="4041" CreationDate="2011-01-28T10:14:30.750" Score="3" ViewCount="224" Body="&lt;p&gt;I find myself greatly fascinated by physics and astronomy, though I really do not understand most of it. For example, I avidly watch programs and read books that discuss our current search for a Unified Theory which will unite the major forces (Strong Interactive, Weak Interactive, Electromagnetism on the one side, with Gravitation on the other). I also try to understand a little bit about our work with subatomic particles, the search for the Higgs particle, and our use of tools like the Large Hadron Collider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite this enthusiasm, I find myself asking one major question: What might come of definitively finding a Unified Theory (Theory of Everything)? Most of the material I absorb seem to imply that there are major technological advances from such an accomplishment, but rarely provide examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously we cannot predict accurately all of the potential for the outcome of this scientific progression, but I get the impression that physicists and astronomers have a great deal of pending problems that await the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a layperson, what kinds of advancements and breakthroughs might one realistically expect?&lt;/p&gt;&#xA;" OwnerUserId="1638" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-01-28T12:22:23.780" LastActivityDate="2011-01-28T15:42:28.963" Title="Help understanding the potential outcome of solving the Unified Theory in modern physics" Tags="&lt;theory-of-everything&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="4035" PostTypeId="2" ParentId="4031" CreationDate="2011-01-28T10:16:52.660" Score="5" ViewCount="" Body="&lt;p&gt;Dear Anirbit, great questions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Long and short representations&lt;/strong&gt; are not difficult to be defined and most introductory texts to supersymmetry explain them. However, some of them don't use this particular terminology, so let me tell you: long multiplets are representations that transform nontrivially under all supercharges. So you can't find a &lt;strong&gt;supercharge that annihilates the whole representation&lt;/strong&gt;. Consequently, if there are $N$ real supercharges, the dimension of the representation has to be $2^{N/2}$. On the other hand, short multiplets - also known as &lt;strong&gt;BPS multiplets&lt;/strong&gt; - are annihilated by a subset of the supercharges, some holomorphic or chiral subset of the spinor(s), so the irreducible short multiplets have a lower dimension than $2^{N/2}$. Because some supercharges annihilate the whole multiplet, their anticommutator also annihilates it, and these anticommutators are given by energy minus some central charges (bosonic generators that appear on the right hand side of the SUSY algebra, together with the energy-momentum). That's why the short multiplets also minimize the energy among the charged states - they saturate the BPS bound and the short multiplets are also called &lt;strong&gt;BPS states&lt;/strong&gt;. Massless particles' supermultiplets are often BPS but there can also be massive short multiplets. &quot;Most&quot; of the multiplets in the spectrum of a theory are long. The subsets of supercharges that annihilate a representation may be &quot;really large&quot; in which case the representation is even smaller, &quot;hypershort&quot;. This terminology is often used but depends on the context (the group).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The groups you write are relevant only for superconformal theories &lt;strong&gt;in certain dimensions&lt;/strong&gt;, namely $d=2+1$ in your case. &lt;strong&gt;In 3 dimensions&lt;/strong&gt;, the Lorentz group is $SO(2,1)$ and the conformal group - the first factor in the bosonic subgroup of the superconformal group - is obtained by adding 1 spatial and 1 temporal dimension (this is a universal rule in any dimension, a rule that becomes self-evident in AdS/CFT where you add $1+1$ dimensions because the larger group is the isometry of an anti de Sitter space which is a hyperboloid inserted to a space with an extra time coordinate and the extra holographic spatial coordinate), so it is $SO(3,2)$ in both cases. The spinors in $d=2+1$ dimensions are real (Majorana), so if you have $N=2$ spinors or $N=3$ spinors, the rotation group that rotates the internal index labeling the different spinors of the supercharges is inevitably $SO(2)$ or $SO(3)$ which is the other factor you mentioned. The bosonic subgroup of the superconformal groups is simply the product of the &quot;spacetime part&quot; and the &quot;internal part&quot; (R-symmetry): they have to be separated in this way by the Coleman-Mandula theorem.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;The primary fields or operators&lt;/strong&gt; or highest-vector states etc. are operators or states such that you may obtain all other elements of the representation by the raising and lowering operators. For each algebra, there is a different set of raising and lowering operators - and different number of them. The more &quot;factors&quot; your algebra has, the more labels you need. That's why your actual labels you learned from the $2$-dimensional conformal algebra are obviously not applicable for different algebras in different dimensions. The dimension of an operator is a label of a conformal primary field in any spacetime dimension (or it is a sum of other labels). However, in $d=2+1$ dimensions, you also need to know how the field behaves under the rotation of spatial dimensions in $SO(2,1)$, that's given by the spin, and how it behaves under the $SO(2)$ or $SO(3)$ R-symmetry group, which is given by the R-charge.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You're asking a lot. This answered the first three points. OK, but let me continue with your other questions and complaints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The inequality (and all similar inequalities) may be derived by writing $\Delta\pm h-j-1$ as a &lt;strong&gt;square a real supercharge, or a sum of such squares, so it cannot be negative&lt;/strong&gt;. &lt;strong&gt;This is a general trick in all of supersymmetric theories.&lt;/strong&gt; For $j=0$, $|h|+j+1$ reduces to $|h|+1$ which explains that there are unitary representations for such values of $\Delta$ if the values are allowed by the algebra (integer or half-integer spin etc.). I don't understand your representations that are supposed to exist for $\Delta=|h|$ which violates the bound. However, there may exist some exotic or irregular representations of similar groups with &quot;smaller dimensions&quot;, too. The representation theory of $SL(2,R)$ itself - which is isomorphic to $SO(2,1)$ - is very subtle and has many kinds of representations and even many types of unitary reps.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are three more points then:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Representations that saturate the BPS bound are usually &lt;strong&gt;short because&lt;/strong&gt; the bound's total quantity (which is non-negative, the quantity of the $\Delta-h$ type) may be written as the &lt;strong&gt;squared real supercharges, or their sum&lt;/strong&gt; (such as $QQ^\dagger$ for a non-hermitean $Q$), and if the quantity vanishes, it follows that the supercharges have to vanish across the representation as well - of course, assuming a positive-definite Hilbert space and (which is related) unitary representation. The invariance of the whole rep under a subset of supercharges is how we defined &quot;shortness&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Witten's index vanishes for all long representations&lt;/strong&gt; because one may always pair the basis vectors of the long multiplet to objects of the form $|\psi\rangle$, $Q|\psi\rangle$, for a well-chosen $Q$ and a basis of $\psi$'s, and the contribution of the two vectors in the pair to the Witten's index is opposite up to the sign (which is flipped by the addition of the $Q$), so the pair's contribution cancels. On the other hand, when you construct the index properly, &lt;strong&gt;short representations will contribute nonzero&lt;/strong&gt; because the sign of the Witten index contribution is constructed so that it is positive for the whole short representation or at least there's no cancellation. The &quot;would-be&quot; partners in the pairs are zero because $Q|\psi\rangle=0$ across the short multiplet for the same $Q$ on the short representation, because it's short.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;You need two indices&lt;/strong&gt;, with the opposite signs in front of $h$, because, as explained above, you want to show that $\Delta$ is bigger than $|h|+j+1$ which is equivalent to showing it's bigger than $h+j+1$ as well as $-h+j+1$. In general, indices are &lt;strong&gt;independent&lt;/strong&gt; of $\beta$ or any other &lt;strong&gt;continuous parameter&lt;/strong&gt; because they're indices. If you change $\beta$ or any other continuous parameter of the theory, the long multiplets may shift their mass but their contribution to the index will still vanish. The short multiplets either keep their mass, or they may shift it - away from the BPS bound - but that's possible only if they combine with the &quot;mirror&quot; short multiplet(s) into a long multiplet and those two (or many) short multiplets' contribution to the Witten's index was zero before the deformation, anyway. So because the (short) multiplets that contribute nonzero may only &quot;go long&quot; and disappear in pairs (or, on the contrary, they may appear by decomposing a long multiplet that previously existed and went to zero mass where it can split), the Witten's index is invariant under all continuous deformations of the theory, including the changes of $\beta$. That's its main property making it a powerful calculational tool.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You wrote exactly what bosonic generators the two types of short multiplets are annihilated by. For those generators of the $j\pm h$ type, you also find &lt;em&gt;different&lt;/em&gt; sets of supercharges that square to $j+h$ or $j-h$, respectively. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I didn't understand what you meant by $x$ but it's plausible that the reason is that this was your proposal and it was based on a misunderstanding. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At any rate, I think that I have answered all your questions.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-28T10:44:05.183" LastActivityDate="2011-01-28T10:44:05.183" CommentCount="4" />
  <row Id="4036" PostTypeId="2" ParentId="4034" CreationDate="2011-01-28T10:59:40.480" Score="4" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The supreme task of the physicist is to arrive at those universal elementary laws from which the cosmos can be built up by pure deduction.&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;em&gt;A. Einstein.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no (and wasn't) any &quot;practical applications&quot; in the physicist's supreme goal.  &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-01-28T10:59:40.480" />
  <row Id="4037" PostTypeId="1" CreationDate="2011-01-28T11:26:55.630" Score="4" ViewCount="266" Body="&lt;p&gt;Now that an equivalence of &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1101/1101.2451v2.pdf&quot; rel=&quot;nofollow&quot;&gt;Navier Stokes and Einstein&lt;/a&gt; equations has been established, and it is known solutions to &lt;a href=&quot;http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;amp;id=pdf_1&amp;amp;handle=euclid.cmp/1103859246&quot; rel=&quot;nofollow&quot;&gt;Einstein-Maxwell-Boltzmann&lt;/a&gt; exist and are unique, and it is known that &lt;a href=&quot;http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;amp;id=pdf_1&amp;amp;handle=euclid.cmp/1103859246&quot; rel=&quot;nofollow&quot;&gt;Einstein equations in a wave guage with electromagnetic source with a lorentz gauge&lt;/a&gt; also have unique solutions that exist, what are the next steps?  What are the good references on research for existence and uniqueness for einsten solutions without electromagnetic sources?  Also, what does it mean physically that there are unique solutions to Einstein Field equations when electromagnetic sources are included?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those struggling with the concept of existance and uniqueness, I think &lt;a href=&quot;http://www.physicsforums.com/showthread.php?t=85862&quot; rel=&quot;nofollow&quot;&gt;this post&lt;/a&gt; provides some good insight.&lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-01-28T21:10:17.047" LastActivityDate="2011-02-08T17:27:33.293" Title="Existence and uniqueness of solutions for Einstein equations" Tags="&lt;general-relativity&gt;&lt;mathematics&gt;&lt;theory&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="4038" PostTypeId="1" CreationDate="2011-01-28T11:29:57.017" Score="2" ViewCount="82" Body="&lt;p&gt;A barrel/drum with a diameter of 60cm is rotating at 20RPM to get a good mixing of the fluid contained (type thick oil). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At what RPM should a barrel of 30cm rotate to get the same mixing efficiency with the same fluid? To get the same peripheral velocity, the RPM should be 40RPM, but will this give the same mixing of the fluid? I guess gravitational and centripetal forces play a role her...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;More information to my otherwise incomplete question :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The barrel is rotating around the length axis and the degree of fill is 90%.&lt;/p&gt;&#xA;" OwnerUserId="1639" LastEditorUserId="1639" LastEditorDisplayName="" LastEditDate="2011-01-28T11:50:13.183" LastActivityDate="2011-01-28T17:50:38.977" Title="Mixing of fluid in a rotating barrel" Tags="&lt;fluid-dynamics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="4040" PostTypeId="1" CreationDate="2011-01-28T12:00:49.053" Score="2" ViewCount="481" Body="&lt;p&gt;How the notion of &lt;em&gt;weak measurement&lt;/em&gt; resolves &lt;a href=&quot;http://en.wikipedia.org/wiki/Hardy%27s_paradox&quot; rel=&quot;nofollow&quot;&gt;Hardy's paradox&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="824" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-28T16:10:20.197" LastActivityDate="2011-02-23T15:23:38.930" Title="Weak measurement and Hardy's paradox" Tags="&lt;quantum-mechanics&gt;&lt;entanglement&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="4041" PostTypeId="2" ParentId="4034" CreationDate="2011-01-28T12:06:26.300" Score="9" ViewCount="" Body="&lt;p&gt;It will all depend on what the Theory of Everything will be, no?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can only look back at the technological progress that appeared once we mastered electromagnetism,  and then quantum mechanics and nuclear physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maxwell unified electricity and magnetism in his theory, in the middle of the nineteenth century. His work in producing a unified model of electromagnetism is one of the greatest advances in &lt;a href=&quot;http://en.wikipedia.org/wiki/James_Clerk_Maxwell&quot; rel=&quot;nofollow&quot;&gt;physics&lt;/a&gt;.It took decades for applications to appear and they are still appearing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same can be said about the theoretical formulation of Quantum Mechanics which carried the unification further and gave us from transistors to superconductivity decades after wards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ditto for nuclear physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can only extrapolate, and not predict, that given a rigorous theory of everythinq, unexpected applications will be inevitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true  that physicists and other researchers are in for the intellectual excitement of frontier research, not for finding applications. Applications are the realm of engineering, once the theory is known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the cost of the LHC, it probably costs less than an airplane carrier, which at a time of peace  is wasted money too. Considering that the world wide web came out of the last accelerator, LEP, as a spin off, and not only this, the money is well invested even so. Already the GRID, a technology of large data transfers and manipulations, is pushing the frontier and will surely find applications in industry in the future.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-01-28T12:39:15.463" LastActivityDate="2011-01-28T12:39:15.463" CommentCount="1" />
  <row Id="4042" PostTypeId="2" ParentId="4038" CreationDate="2011-01-28T12:29:42.737" Score="2" ViewCount="" Body="&lt;p&gt;Knowing a bit about process engineers' attitude to science I would say you just need to have same Reynolds number. $U\approx\Omega R$, $L\approx R$, then $Re\approx\frac{\Omega R^2}{\nu}$ so 80RPM.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2011-01-28T12:29:42.737" CommentCount="1" />
  <row Id="4043" PostTypeId="2" ParentId="4040" CreationDate="2011-01-28T12:57:56.640" Score="2" ViewCount="" Body="&lt;p&gt;Dear Eelvex, first of all, the term &quot;paradox&quot; is misleading, much like for all &quot;paradoxes&quot; in physics. Quantum mechanics, when applied properly, doesn't lead to any paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right, quantum mechanical solution to the &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v68/i20/p2981_1&quot; rel=&quot;nofollow&quot;&gt;Hardy's 1992 thought experiment&lt;/a&gt; concludes that there is a nonzero probability that the electron and positron will be detected in two particular detectors $d^+, d^-$. As I will explain in the second part of the answer, where the whole &quot;paradox&quot; is clarified, the real paradox has nothing to do with any specific dynamics of antiparticles or annihilation. It's all about quantum information and the usual differences between classical physics and quantum mechanics - the same conceptual issues that arise in Bell's inequalities or the GHZ state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Where is the paradox?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only paradox arises if you try to imagine that the particle and the antiparticle objectively possessed some &quot;well-defined properties&quot; - especially the location - in the past. If you do so, the situations in which the particles end up detected in $d^+,d^-$ correspond to &quot;histories&quot; in which the particles had to meet and annihilate - but they didn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hardy's method is just a particularly striking way to show why the realistic reasoning is inapplicable in the world of quantum phenomena. The paradox only arises because of &quot;retrodictions&quot; using a classical intuition. If you only apply the proper probabilistic quantum rules and never imagine that the system was represented by any &quot;hidden variables&quot; at each moment, and you just evolve the wave function according to the right equations and calculate the probability at the end, you will make the right prediction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation has been measured experimentally. And indeed, the prediction of quantum mechanics was confirmed: the electron and the positron may be detected at &quot;classically contradictory&quot; pair of detectors in a certain fraction of cases. Weak measurements have been used to detect the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people have continued to assume that there could be a way how to &quot;retrodict&quot; what was happening according to the final results, and they found the weak measurement  - producing weak values - to be useful. See e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yakir Aharonov et al.: &lt;strong&gt;Revisiting Hardy's Paradox: Counterfactual Statements, Real Measurements, Entanglement and Weak Values&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0104062&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/quant-ph/0104062&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As far as I can see, they just deduce one more &quot;classically paradoxical&quot; outcome for each original Hardy-like outcome. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many wrong things have been said about Hardy's paradox during the two decades. For example, some people such as Henry Stapp have argued that even in ordinary quantum mechanics, one needs acausality to explain the observation. This assertion was showed incorrect by David Mermin:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;N. David Mermin: &lt;strong&gt;Nonlocal character of quantum theory?&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/abs/quant-ph/9711052&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/quant-ph/9711052&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The nonlocality and acausality doesn't exist in these experiments - or any other phenomena in the Universe - and it doesn't help to explain them. Instead, the lack of &quot;realism&quot; (i.e. the non-existence of any sharp properties of physical systems prior to the measurement) is the right explanation why these things work as they work. By focusing on the &quot;weak measurements&quot;, you seem to be on a wrong track. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A comprehensible explanation of Hardy's setup also appears on page 35 of this review:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Frank Laloe: &lt;strong&gt;Do we really understand quantum mechanics?&lt;/strong&gt; &lt;br&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0209123&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/quant-ph/0209123&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In particular, I like the fact that this review makes it very transparent that the thought experiment has nothing to do with any particular dynamics of particles, antiparticles, and their annihilation. It's all about the basic logic and correlations that work differently in quantum physics than in any classical or &quot;realistic&quot; model of physics. The electron-positron setup is just a particular realization of the &quot;paradox&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My summary and solution of the &quot;paradox&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me reformat Laloe's presentation. Imagine we have a system of two particles which are associated with letters &quot;a&quot; and &quot;b&quot;, respectively. On each particle, we may perform two kinds of measurements, either &lt;em&gt;A&lt;/em&gt; or &lt;em&gt;A'&lt;/em&gt; for the article &quot;a&quot; and either &lt;em&gt;B&lt;/em&gt; or &lt;em&gt;B'&lt;/em&gt; for the particle &quot;b&quot;. Imagine that the un-primed measurements are spin measurements with respect to a vertical axis and the primed measurements are with respect to a different, general axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of these measurements may produce the result $-1$ or $+1$. There are three qualitatively different types of measurements: either both &quot;a&quot; and &quot;b&quot; are tested on their &quot;unprimed&quot; polarization, or one of them is measured by the &quot;primed&quot; gadget and the other by the &quot;unprimed&quot; gadget, or both of them have their &quot;primed&quot; quantities measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as we can easily show, it is possible to design the apparatus so that the following three conditions are guaranteed. Each condition constrains the results with respect to one of the three kinds of a measurement - either the &quot;two unprimed measurements&quot; or &quot;mixed&quot; or &quot;two primed measurements&quot;. The conditions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;After the doubly unprimed measurements, $A=+1$ and $B=+1$ sometimes occurs&lt;/li&gt;&#xA;&lt;li&gt;After the mixed measurements, $A=+1$ and $B&amp;#39;=+1$ never occurs, and $A&amp;#39;=+1$ and $B=+1$ never occurs, either&lt;/li&gt;&#xA;&lt;li&gt;After the doubly primed measurements, $A&amp;#39;=-1$ and $B&amp;#39;=1$ never occurs.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can you prepare the particles &quot;a&quot; and &quot;b&quot; so that regardless of the two experimenters' choice of &quot;primed&quot; vs &quot;unprimed&quot;, all conditions above will be universally satisfied? The answer of any realistic theory, whether it's local or not, is &quot;no&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first condition guarantees that it may happen that the particles are sometimes prepared to be measured as $A=B=1$. However, for these cases, the second condition guarantees that $B&amp;#39;=1$ - because $B&amp;#39;=+1$ is impossible because we already have $A=1$. And similarly, $A&amp;#39;=-1$ - because $A&amp;#39;=+1$ is impossible because we already have $B=+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We just derived that in the situations in which the particles are prepared to be observed with $A=B=+1$, we have $A&amp;#39;=B&amp;#39;=-1$ which violates the third condition. It sounds like impenetrable logic. Whenever you assume that the particles already objectively have some properties prepared for any experiments that the experimenters may choose, it follows that the three rules above are incompatible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in quantum mechanics, they're totally compatible, and the compatibility has been demonstrated experimentally, too. With respect to the primed bases, the right two-particle state that satisfies all the three conditions may be written as&#xA;$$|\Psi\rangle = -\cos\theta \left(  |A&amp;#39;=+1,B&amp;#39;=-1\rangle  + |A&amp;#39;=-1,B&amp;#39;=+1\rangle  \right) + \sin\theta  |A&amp;#39;=+1,B&amp;#39;=+1\rangle$$&#xA;where the angle between the &quot;primed&quot; and &quot;unprimed&quot; axes was chosen to be $2\delta$ and we consider the spin equal to $1/2$. Note that one of the four basis vectors, the vector with $A&amp;#39;=B&amp;#39;=-1$, is omitted, so we automatically satisfy the third condition. By a proper transformation to the rotated bases, we may also check that the second condition is satisfied. That's why the coefficients were chosen to be cosines and sines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the first condition is also satisfied because the coefficient of the $A=B=+1$ can be seen to be nonzero - after the simple rotation of both bases. So quantum mechanics implies that two particles may be arranged in a state such that all three conditions - for all three types of paired measurements that the two experimenters may decide to perform - will be satisfied. This rules out all &quot;realistic&quot; interpretations of reality, whether they're local or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is different about the quantum logic that it doesn't allow us to derive the classical conclusion? Well, the second condition says that the state $|\Psi\rangle$ is annihilated by two projection operators,&#xA;$$  P(A=1) P(B&amp;#39;=1)|\Psi\rangle = 0\,\mbox{ and }\,P(A&amp;#39;=1) P(B=1)|\Psi\rangle = 0 $$&#xA;It's also annihilated by the following operator, as dictated by the third condition,&#xA;$$ P(A&amp;#39;=-1) P(B&amp;#39;=-1) |\Psi\rangle = 0. $$&#xA;Classically, we could derive that the state must also be annihilated by&#xA;$$ P(A=1) P(B=1) |\Psi\rangle = 0 $$&#xA;which is a projection operator relevant in the first condition. The classical logic was showed above. However, quantum mechanically, the state doesn't have to be annihilated by the most recent projection operator because the projection operators $P(A)$ and $P(A&amp;#39;)$ don't commute with one another, and similarly for the particle &quot;b&quot;. This characteristic fact of quantum mechanics - that observables don't commute with each other, not even the projection operators describing Yes/No properties of systems - guarantees that we can't derive the third displayed equation above from the previous two lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The very Yes/No properties of a particle don't commute with each other, so we can't ever imagine that a particle is ready to react to different kinds of measurements at the same moment. In some sense, it's just another form of the uncertainty principle, optimized for projection operators and binary properties in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See a blog version of this answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2011/01/hardys-paradox-kills-all-realistic.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2011/01/hardys-paradox-kills-all-realistic.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-23T15:23:38.930" LastActivityDate="2011-02-23T15:23:38.930" />
  <row Id="4044" PostTypeId="2" ParentId="4034" CreationDate="2011-01-28T13:10:07.847" Score="1" ViewCount="" Body="&lt;p&gt;Sean Carroll at the blog Cosmic Variance had an interesting post on this: &lt;a href=&quot;http://blogs.discovermagazine.com/cosmicvariance/2010/09/23/the-laws-underlying-the-physics-of-everyday-life-are-completely-understood/&quot; rel=&quot;nofollow&quot;&gt;The laws underlying the physics of everyday life are completely understood&lt;/a&gt;. This suggests there are &lt;em&gt;no technologies&lt;/em&gt; within the range of energies we can handle on earth which would be transformed by a future 'theory of everything' (super HEP experiments aside). It's a shame because I was looking forwards to those &lt;a href=&quot;http://keelynet.com/gravity/nasaag.htm&quot; rel=&quot;nofollow&quot;&gt;strange rotating devices&lt;/a&gt; which would locally nullify gravity!&lt;/p&gt;&#xA;" OwnerUserId="1394" LastEditorUserId="1394" LastEditorDisplayName="" LastEditDate="2011-01-28T15:42:28.963" LastActivityDate="2011-01-28T15:42:28.963" CommentCount="2" />
  <row Id="4045" PostTypeId="2" ParentId="822" CreationDate="2011-01-28T13:15:04.190" Score="2" ViewCount="" Body="&lt;p&gt;Existence of a spin of a particle is of course an indication that the particle is in fact composed of space-separated parts. This does not mean though that the particle is composed of other particles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, at least a part of the electron's spin is currently known to be in fact the orbital momentum of the quantum vacuum fluctuations which are involved by electron's core into rotation. This part is known as anomalous angular momentum of electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is photon where the spin can be explained as an order in which the energy contained in electric and magnetic fields rotates around the axis laid along the direction of the photon's propagation.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastActivityDate="2011-01-28T13:15:04.190" />
  <row Id="4046" PostTypeId="2" ParentId="4034" CreationDate="2011-01-28T13:18:05.197" Score="3" ViewCount="" Body="&lt;p&gt;Here is a broad answer as one cannot predict what new discoveries will entail unless they happen. I think then that technological applications will come from any &quot;emergent structures&quot; associated with this theory. So a few cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;No emergent structures&lt;/strong&gt;. This would be the case if such a unified theory was simply &quot;tidying up&quot; the mathematics of existing theories. This might happen if the unified theory just gave a new (mathematical) perspective on existing known results which managed to (just) unify those theories and no more. This would also imply that existing known physics was very close to the entire physical truth already; but perhaps not expressed in the right way mathematically as yet. So there may be no extra technological consequences.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Mathematical Improvements.&lt;/strong&gt; This is similar to the previous case except that the mathematics is calculationally superior in the Unified Theory. This doesnt always happen, but it might then allow us to solve equations that are too complex right now. This probably would translate into more efficient technologies - at least where the maths allowed exploration of physical solutions that are too hard to work with right now. A lay person might not notice these improvements over computational improvements generally, however. On a good day we might get something like High Temperature SuperConductors, maybe as a later spinoff.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;New Fields&lt;/strong&gt;. These might be at such a scale that any technological advantage would be expensive to undertake. Nuclear Fusion research might be a prototype for this: theoretically available but practically still decades away despite (presumably) understanding the theory. If there were accessible new fields however, there would be technological consequences. History's classic example is Electromagnetism. The definition of &quot;accessible&quot; here would be critical - maybe such fields would only be measurable near large mass objects like Jupiter - eg Higgs field?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;New Spaces&lt;/strong&gt;. If physics finds new spaces, then they will get explored. In a sense  the Hilbert space of quantum mechanics is a new space which gets explored by examining  unusual shapes (like squeezed light, Bose-Einstein Condensates) that can arise. These have  technological consequences, in fibre optics and elsewhere.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Everything else is Science Fiction speculation, which you can read and wonder...&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-01-28T13:18:05.197" CommentCount="2" />
  <row Id="4047" PostTypeId="1" AcceptedAnswerId="4093" CreationDate="2011-01-28T14:31:28.057" Score="5" ViewCount="261" Body="&lt;p&gt;Consider a particle in a potential well. Let’s assume it’s a simple harmonic oscillator potential and the particle is in its ground state with energy E&lt;sub&gt;0&lt;/sub&gt; = (1/2) &amp;#8463;&amp;omega;&lt;sub&gt;0&lt;/sub&gt;. We &lt;strong&gt;measure&lt;/strong&gt; its position (measurement-1) with a high degree of accuracy which localises the particle, corresponding to a superposition of momentum (and therefore energy) states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we &lt;strong&gt;measure&lt;/strong&gt; the particle’s energy (measurement-2) and happen to find that it’s E&lt;sub&gt;10&lt;/sub&gt; = (21/2) &amp;#8463;&amp;omega;&lt;sub&gt;0&lt;/sub&gt;. Where did the extra energy come from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the textbooks it’s claimed that the extra energy comes from the act of observation but I wonder how that could work. Measurement-1 which probed the position of the particle can’t have delivered to it a precise amount of energy, while measurement-2 might just have been passive. No doubt there is entanglement here between the particle state and the measuring device but where, and which measurement?&lt;/p&gt;&#xA;" OwnerUserId="1394" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T14:22:46.887" LastActivityDate="2011-02-03T14:22:46.887" Title="Energy conservation and quantum measurement" Tags="&lt;quantum-mechanics&gt;&lt;energy-conservation&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="4048" PostTypeId="1" AcceptedAnswerId="4161" CreationDate="2011-01-28T14:49:11.567" Score="6" ViewCount="230" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Feynman_checkerboard&quot; rel=&quot;nofollow&quot;&gt;Feynman Checkerboard Wikipedia article&lt;/a&gt; states: &quot;There has been no consensus on an optimal extension of the Chessboard model to a fully four-dimensional space-time.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it hard to extend it to more than 1+1 dimensions?&lt;/p&gt;&#xA;" OwnerUserId="47" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-29T18:46:03.273" LastActivityDate="2011-01-30T01:33:59.943" Title="Why is it hard to extend the Feynman Checkerboard to more than 1+1 dimensions?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="2" />
  <row Id="4049" PostTypeId="1" AcceptedAnswerId="4055" CreationDate="2011-01-28T14:49:35.967" Score="8" ViewCount="430" Body="&lt;p&gt;What is the meaning, mathematical or physical, of the anti-commutator term?&#xA;$\langle ( \Delta A )^{2} \rangle \langle ( \Delta B )^{2} \rangle \geq \dfrac{1}{4} \vert \langle [ A,B ] \rangle \vert^{2} + \dfrac{1}{4} \vert \langle \{ \Delta A, \Delta B \} \rangle \vert^{2}$,&#xA;where $\Delta A, \Delta B, A$ and $ B$ are operators.&#xA;The inequality is still true, and the anti-commutator term &quot;strengthens&quot; the inequality, but why does it appear?&lt;/p&gt;&#xA;" OwnerUserId="1618" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2011-01-28T15:52:24.243" LastActivityDate="2011-01-28T18:09:33.253" Title="Meaning of the anti-commutator term in the uncertainty principle" Tags="&lt;quantum-mechanics&gt;&lt;uncertainty-principle&gt;" AnswerCount="2" />
  <row Id="4050" PostTypeId="2" ParentId="4047" CreationDate="2011-01-28T14:51:25.460" Score="4" ViewCount="" Body="&lt;p&gt;Dear Nigel, if you measure the particle's position and find it in a small region, you also change its state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you correctly wrote, a localized wave packet (I don't talk about a delta-function state vector whose average kinetic energy would be infinite) can be rewritten as a linear superposition of energy eigenstates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It means that before you measured the energy, there was a nonzero probability for the energy to be $(21/2)\hbar\omega$, and this particular outcome was ultimately realized. There is no violation of the energy conservation law here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you may be annoyed by is that the final energy of the electron usually isn't equal to the expectation value of energy in the state before the measurement. It's surely true. But there is no reason why it should be. The expectation value isn't any &quot;objective value&quot; of energy. It's just a statistical average of many possibilities, and only some of them will be realized, as dictated by the probabilities predicted by quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish, the process of measurement violates the &quot;conservation of the expectation value of energy&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Needless to say, this &quot;violation&quot; can't be used to obtain any sharp contradiction with the conservation law for the &quot;actual&quot; energy. You may only interpret the measured energy in a &quot;classical way&quot; after decoherence, which means after the measuring apparatus has interacted with the environmental degrees of freedom (that are needed for decoherence). If you prepare your harmonic oscillator plus the apparatus in a state whose total energy is sufficiently sharply well-defined to reveal the &quot;violation of the conservation law for the expectation value&quot;, the environmental degrees of freedom will inevitably spoil this accuracy.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-28T14:51:25.460" CommentCount="4" />
  <row Id="4051" PostTypeId="2" ParentId="1453" CreationDate="2011-01-28T14:55:34.960" Score="1" ViewCount="" Body="&lt;p&gt;What the heck happens when we do a measurement in quantum theory? That is why should the system jump to an Eigen state? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why should nice unitary evolution suddenly do something weird just because someone did a measurement? What exactly is a measurement anyway?&lt;/p&gt;&#xA;" OwnerUserId="1643" LastActivityDate="2011-01-28T14:55:34.960" CommunityOwnedDate="2011-01-28T14:55:34.960" />
  <row Id="4052" PostTypeId="2" ParentId="4048" CreationDate="2011-01-28T15:02:08.170" Score="3" ViewCount="" Body="&lt;p&gt;Dear mtrencseni, the very page you quoted has an answer in the very following sentences:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two distinct classes of extensions exist, those working with a fixed underlying lattice (19,20) and those that embed the two dimensional case in higher dimension (21,22). The advantage of the former is that the sum-over-paths is closer to the non-relativistic case, however the simple picture of a single directionally independent speed of light is lost. In the latter extensions the fixed speed property is maintained at the expense of variable directions at each step.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is clearly no reason why similar mathematical curiosities should exist in higher dimensions. Nevertheless people tried to find a higher-dimensional counter along two different competing lines. Both of them remain highly inconclusive, to put it very generously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most obvious reason why the Feynman checkerboard is hard in higher dimensions is that the Dirac equation has many components and the individual spinor components cannot associated with any simple discrete properties of a lattice, such as the direction where the particle moves in the first step. The fact that the geometry of direction on a line is so simple is the reason why the Feynman checkerboard in 1+1 dimensions, and only in 1+1 dimensions, may be used as a bookkeeping device for the helicity components etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the higher-dimensional lattices don't include lines that move &quot;exactly by the speed of light&quot; (45 degrees) in generic directions. That's also different from 1+1 dimensions where the Southwest and Southeast directions are the only two light-like directions. This simple fact makes it hard to obtain a relativistic theory from a particle moving on a lattice.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-28T15:02:08.170" />
  <row Id="4053" PostTypeId="2" ParentId="1019" CreationDate="2011-01-28T15:05:24.957" Score="2" ViewCount="" Body="&lt;p&gt;Notion of simultaneity. Because of speed of light is so big, it looks true in our day to day affairs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it really is a non existent thing [due to special relativity]. &#xA;2 people in 2 different places can't say &quot;at the same instant&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1643" LastActivityDate="2011-01-28T15:05:24.957" CommunityOwnedDate="2011-01-28T15:05:24.957" />
  <row Id="4054" PostTypeId="2" ParentId="4047" CreationDate="2011-01-28T15:18:39.480" Score="-1" ViewCount="" Body="&lt;p&gt;A QM wave function describes an ensemble of measurements, not the single measurement output. The corresponding potential does not describe the perturbation due to measurement. If you only measure the system energy, it will be $E_0$, of course, because you measure the eigenstate of the corresponding operator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the coordinate operator does not commute with the Hamiltonian, the ground state is not its eigenstate so you will get dispersion in coordinate measurements. While measuring coordinate you may introduce a perturbation that changes the initial state. Now you may have a superposition of eigenstates of Hamiltonian. No wonder you may then find energies different from the ground state.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-01-28T15:47:40.993" LastActivityDate="2011-01-28T15:47:40.993" CommentCount="3" />
  <row Id="4055" PostTypeId="2" ParentId="4049" CreationDate="2011-01-28T15:24:01.270" Score="4" ViewCount="" Body="&lt;p&gt;Dear Rodrigo, it's an interesting stronger version of the uncertainty principle for general operators $A,B$ that I've never seen before but I just verified it holds. Just to be sure, the anticommutator is simply&#xA;$$\{A,B\}\equiv AB+BA.$$&#xA;I like when the braces are only used for pairs of Grassmannian objects but people use it as a bookkeeping device to simplify $AB+BA$ in all situations. Nothing difficult about the notation. Note that the commutator and anticommutator appear totally symmetrically in the inequality, a fact we will derive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see why the stronger inequality holds, open Wikipedia here&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Uncertainty_principle#Mathematical_derivations&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Uncertainty_principle#Mathematical_derivations&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where only the simpler version of the inequality (without the squared anticommutator) is proved by combining two inequalities. The first one,&#xA;$$ ||A\psi||^2 ||B\psi||^2 \geq |\langle A\psi|B\psi\rangle|^2 $$&#xA;remains unchanged. However, the second inequality from the Wikipedia article may be strengthened to a full-fledged equality&#xA;$$  |\langle A\psi|B\psi\rangle|^2  = \left| \frac{1}{2i} \langle \psi | AB-BA | \psi \rangle \right|^2 + \left| \frac{1}{2} \langle \psi | AB+BA | \psi \rangle \right|^2 $$&#xA;This identity simply says that the squared absolute value of a complex number is the sum of the squared real part and the squared imaginary part (which was omitted on Wikipedia). Combining the previous two inequalities, one gets your &quot;stronger&quot; uncertainty principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, the equation derived above is uselessly weak unless the expectation values of $A,B$ vanish themselves. It can be strengthened into yours by repeating the same prodedure for $\Delta A = A-\langle A\rangle$ and similarly $\Delta B = B-\langle B\rangle$ instead of $A,B$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote what the anticommutator means mathematically and why the inequality is true. Now, what does the anticommutator term mean physically? I don't know what this question mean. It's a term in an equation that I can read and explain for you again. The precise answers in physics &lt;em&gt;are&lt;/em&gt; given by mathematics. So I guess that the answer you want to hear is that it means nothing physically, it's just pure mathematics. This fact doesn't mean that it can't be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, in normal cases, the stronger version is not &quot;terribly&quot; useful because the anticommutator term is only nonzero if there is a &quot;correlation&quot; in the distributions of $A,B$ - i.e. if the distribution is &quot;tilted&quot; in the $A,B$ plane rather than similar to a vertical-horizontal ellipse which is usually the case in simple wave packets etc. Maybe this is what you wanted to hear as the physical explanation of the anticommutator term - because $AB+BA$ is just twice the Hermitean part of $AB$, it measures the correlation of $A,B$ in the distribution given by the wave function - although the precise meaning of these words has to be determined by the formula.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-28T16:23:48.043" LastActivityDate="2011-01-28T16:23:48.043" CommentCount="6" />
  <row Id="4057" PostTypeId="2" ParentId="1019" CreationDate="2011-01-28T15:46:02.427" Score="4" ViewCount="" Body="&lt;p&gt;As an instructor, I have great difficulty teaching Newton's 3rd law: &quot;For every action there's an opposite and equal reaction.&quot; This is very basic and very old physics but it's hard to teach. A typical example of the false reaction force answer is the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I hold an apple in my hand. The earth&#xA;  pulls down on the apple with a&#xA;  gravitational force. What is the&#xA;  reaction force to this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer most of them will give is that the reaction force is &quot;my hand pushing up on the apple.&quot; Arggghhhh! Of course the reaction force is &quot;the apple pulling up on the earth.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Students fail to realize that the opposite and equal reactions have to be between the same pair of objects. That is, forces arise as pairs. I wish they'd just rename the law so that it makes it more clear that the reaction force has to operate between the same pair of objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I demonstrate the law by holding a long spring in my hands and telling them that forces are like this spring. When it applies a force on one end it applies a force on the other (assume massless spring). Next quarter I'm going to try some more extreme measures on this, clearly I'm failing.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-28T15:46:02.427" CommunityOwnedDate="2011-01-28T15:46:02.427" />
  <row Id="4058" PostTypeId="1" AcceptedAnswerId="4062" CreationDate="2011-01-28T16:47:41.767" Score="3" ViewCount="173" Body="&lt;p&gt;It's possible apply the variational method to the brownian motion ? I mean, one of requisites on $y(t)$ is that it must be continuous and $\partial_t{y(t)}$ too, and in this case, $\partial_t{y(t)}$ is not.&lt;/p&gt;&#xA;" OwnerUserId="1579" LastActivityDate="2011-01-28T21:08:55.953" Title="Variational method applied to brownian motion" Tags="&lt;mathematical-physics&gt;&lt;lagrangian-mechanics&gt;" AnswerCount="2" />
  <row Id="4059" PostTypeId="2" ParentId="4010" CreationDate="2011-01-28T16:48:27.560" Score="0" ViewCount="" Body="&lt;p&gt;The reason goggles work is because the lenses are flat. That way the interface between air and water doesn't have any focal length properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of saying the same thing: You can see fish perfectly clearly in an aquarium with flat sides. The only effect is that the fish seem a little closer than they actually are. Goggles do the same thing for the same reason.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-28T16:48:27.560" />
  <row Id="4060" PostTypeId="2" ParentId="4047" CreationDate="2011-01-28T16:55:45.007" Score="1" ViewCount="" Body="&lt;p&gt;What is interesting about this question is that the answers may display its author's Interpretation of QM. Let us first review the order of events here:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Particle is prepared in Ground State $\psi_0$ (an Energy Eigenstate)&lt;/li&gt;&#xA;&lt;li&gt;Particle is position measured with observable X (measurement-1)&lt;/li&gt;&#xA;&lt;li&gt;Particle is now in a Position Eigenstate&lt;/li&gt;&#xA;&lt;li&gt;Particle is now Energy measured (measurement-2)&lt;/li&gt;&#xA;&lt;li&gt;Particle is in state $\psi_{21/2}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In this example energy has been given to the particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly this is similar to an atom's electron becoming excited. If this has happened via a collision or photon absorption during either measurement, then the particle will only absorb an integral quantums worth of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However since the Hamiltonian in this case is $H=1/2(p^2 + x^2)$, then $[X,H]=\hbar p\neq0$.&#xA;So the quantum formalist can simply take the Copenhagen view that since the observables dont commute no guarantees can be given about their eigenvalues staying the same after such measurement history. And no further explanation as to &quot;why&quot; need be given beyond &quot;quantum fluctuation&quot; (or &quot;quantum randomness&quot; if you prefer). Thus there is no &quot;microtheory&quot; explaining where the energy (in this case - maybe extra momentum in another case) came from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further considerations: (1) if one entertains a &quot;hidden variable&quot; philosophy then this experiment needs to be explained by that means. (2) This experiment has similarities to the quantum vacuum explanations for spontaneous emission. If such a model could apply here then we could say that the particle has had a random payoff from the quantum vacuum, perhaps mediated by one of the measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also any entanglement between measurement instrument and particle is supposed to have ended shortly after measurement as a result of decoherence between a classical object and a quantum one.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-01-29T10:51:00.460" LastActivityDate="2011-01-29T10:51:00.460" CommentCount="3" />
  <row Id="4061" PostTypeId="1" AcceptedAnswerId="4066" CreationDate="2011-01-28T16:57:21.450" Score="1" ViewCount="86" Body="&lt;p&gt;Question goes: &quot;An anvil hanging vertically from a long rope in a barn is pulled to the side and raised like a pendulum 1.6 m above its equilibrium position. It then swings to its lowermost point where the rope is cut by a sharp blade. The anvil then has a horizontal velocity with which it sails across the barn and hits the floor, 10.0 m below. How far horizontally along the floor will the anvil land?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I've been working a while on it and what i got was&#xA;PE at side = 1.6 * 10 (gravitational constant) * m (mass of anvil).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;KE at lowermost point = 1/2 m * v^2 = PE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;16m = 1/2 m * v^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;16 = 1/2 v^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;32 = v^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v = sqrt(32)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to the free-fall problem, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;10 = 1/2 g * t^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;20 = g * t^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 = t ^ 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;t = sqrt(2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the distance traveled horizontally = sqrt(2) * sqrt(32) = 1.414 meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I checked the model-answers and the answer was 8 meters, would someone please explain this? (The model-answer might be wrong though)&lt;/p&gt;&#xA;" OwnerUserId="1375" LastActivityDate="2011-01-28T17:23:12.543" Title="Mechanical energy problem" Tags="&lt;mechanics&gt;&lt;gravity&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="4062" PostTypeId="2" ParentId="4058" CreationDate="2011-01-28T17:01:27.823" Score="1" ViewCount="" Body="&lt;p&gt;Dear gsAllan, in the real world, the Brownian motion is the result of collisions of the moving object we study - a pollen particle - with small (water) molecules that have some positions and that may be described by the variational method - either in classical mechanics or in quantum field theory (where I mean the Feynman path integral by the variational method), at least in principle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you do so with all the real-world degrees of freedom of all the water molecules that are unmeasurable in practice, you will find out that the system is described by continuous functions and the pollen particle moves continuously all the time. There will exist a typical &quot;mean path&quot; and &quot;mean time between collisions&quot; and during this time, the velocity of the pollen particle will be finite (and piecewise linear, with the turning points that may be either sharp or smoothened into a finite acceleration, depending on the description of the collisions: whether you repel them by an infinite potential wall or a finite repulsive potential).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you're apparently talking about, however, is a simplified or idealized description of the Brownian motion where the force is &quot;random&quot;, i.e. when one doesn't claim to be able to keep track of the positions of all the water molecules and where one only wants to compute the statistical averages (e.g. of the distance how far the pollen particle gets). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And moreover, the randomness of the force is extrapolated to arbitrarily short distance scales. In that case, the velocities are not continuous and it would make no sense to describe the motion by the variational principle because the action itself would have to contain some &quot;random&quot; parameters as well. Because the energy of the pollen particle may be always transferred from/to the water molecules, the energy conservation is inconsequential, too. Systems where &quot;nothing useful is conserved&quot; are usually not usefully described by the actions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The very purpose of the variational principle is to find the single one unique canonical correct solution for the trajectory. But if there are random forces, there is no single correct trajectory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Quantum mechanics&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is an interesting relationship of the Brownian motion and the quantum version of the variational principle. In quantum mechanics, the action may also be declared to be the &quot;primary object&quot; defining all of dynamics. The approach is called the Feynman path integral definition of quantum mechanics. One sums the phase $\exp(iS/\hbar)$ over all trajectories, where $S$ is the action of the trajectory, $i$ is the imaginary unit, and $\hbar$ is the reduced Planck's constant, and interprets the sum as the probability amplitude for a transition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly enough, the typical trajectories that contribute to the Feynman path integral look much like the trajectories of the Brownian motion! So the Brownian motion paths have a profound importance for physics formulated in terms of the action - but only at the quantum level.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-28T17:01:27.823" />
  <row Id="4063" PostTypeId="1" AcceptedAnswerId="4070" CreationDate="2011-01-28T17:05:36.227" Score="2" ViewCount="160" Body="&lt;p&gt;I am an engineer and have been away from Physics since High school (not due to lack of interest).....so plz excuse me for if my question appears a bit naive or trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I always use to wonder how the experimental physicists discover new particles every now and then whose dimensions/properties/mass/charge several order of magnitudes below that of anything that is visible/perceptible. So as engineers do, I guess they also set up a extremely complicated equipment and do some pretty complicated experiments and measure some physical quantity, say (pardon me if my example is poor) an voltage or a magnetic field intensity or could be anything depending on the experimental setup. Now they compare this measured voltage variation with that of what is expected theoretically and then go on to prove the hypothesis. ( This is the hardest thing I could imagine).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my request is where can i find a set of data (preferable a continuous variation of a physical parameter with respect to another...may be sampled at sufficient sampling frequency) along with the context of experiment (as minimal as possible but sufficient) so that i can carry out some processing of data in my own way so that i can verify the hypothesis or any such a thing. Simply put i need some really cool real world data ( in the form of signal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please suggest me if any such thing is available on the internet or where could I find one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIt : Please suggest me something which involves signal processing.excuse me if my question is vague....I really need something which involves signal processing whether is astronomical signal or something related to particle physics.&lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="540" LastEditorDisplayName="" LastEditDate="2011-01-28T17:22:36.020" LastActivityDate="2011-01-29T22:07:33.857" CommunityOwnedDate="2011-01-29T21:34:39.710" Title="A reference request for real world experimental data" Tags="&lt;particle-physics&gt;&lt;astrophysics&gt;&lt;experimental-physics&gt;&lt;astronomy&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="4064" PostTypeId="2" ParentId="4063" CreationDate="2011-01-28T17:12:54.470" Score="1" ViewCount="" Body="&lt;p&gt;Dear Rajesh, you want the data announced by experimenters that were used to discover new particles. That's simple. Write the URL&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-ex/xxxxxxx&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-ex/xxxxxxx&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where the seven crosses are random digits - the first one should better be 9 or 0 because the first two are a year. The third and fourth digit is the month, and the last three digits shouldn't be too high. You can play with it and find tens of thousands of papers of the kind that you described. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, one of them will be&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-ex/9503003&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-ex/9503003&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which happens to be the paper on the observation of the top quark, the most recent elementary particle that was discovered in 1995.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could also give you tens of thousands of similar papers and solutions to your problem in the field of condensed matter physics, chemistry, biology, geology, or dozens of other disciplines of science. You see the plethora of solutions so perhaps, your question was a little bit too vague? ;-)&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-28T17:12:54.470" CommunityOwnedDate="2011-01-29T21:34:39.710" CommentCount="1" />
  <row Id="4066" PostTypeId="2" ParentId="4061" CreationDate="2011-01-28T17:23:12.543" Score="0" ViewCount="" Body="&lt;p&gt;For any nonzero gravitational acceleration we know:&#xA;(1) Horizontal kinetic energy at impact = .16 time vertical KE at impact  1.6/10&#xA;(2) Thus horizontal velocity at impact is .4 times vertical velocity at impact.&#xA;(3) The time averaged vertical velocity is half of the end velocity.&#xA;  So the average horizontal vel is .8 times the average vertical velocity, it should travel .8 times as far horizontally as vertically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also sqrt(2)*sqrt(32) ==sqrt(64) , so you needn't use your calculator. It is a good idea to have a reasonable guess as to what the result is, before trusting in calculator results (its so easy to type it in wrong).&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-01-28T17:23:12.543" CommentCount="1" />
  <row Id="4067" PostTypeId="1" AcceptedAnswerId="4101" CreationDate="2011-01-28T17:33:11.077" Score="-1" ViewCount="182" Body="&lt;p&gt;Is there any theory in which every particle can be further subdivided into any number of particles and the total number of particles any where in the space time are infinity in theory and only due to practical constraints that we are capable of observing the known particles in the current World.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also could you please explain the nature of the two widely accepted theories which explain the small scale and large scale universe which are QP and GR repectively in the current context of the question. &lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-04T22:17:26.273" LastActivityDate="2011-02-04T22:17:26.273" Title="Question on particles" Tags="&lt;mathematical-physics&gt;" AnswerCount="2" />
  <row Id="4068" PostTypeId="1" AcceptedAnswerId="4083" CreationDate="2011-01-28T17:34:35.270" Score="17" ViewCount="665" Body="&lt;p&gt;I'm wondering about current efforts to provide mathematical foundations and more solid definition for quantum field theories. I am aware of such efforts in the context of the simpler topological or conformal field theories, and of older approaches such as algebraic QFT, and the classic works of Wightman, Streater, etc. etc . I am more interested in more current approaches, in particular such approaches that incorporate the modern understanding of the subject, based on the renormalization group. I know such approaches exists and have had occasions to hear interesting things about them, I'd be interested in a brief overview of what's out there, and perhaps some references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Thanks for all the references and the answers, lots of food for thought! As followup: it seems to me that much of that is concerned with formalizing perturbative QFT, which inherits its structure from the free theory, and looking at various interesting patterns and structures which appear in perturbation theory. All of which is interesting, but in addition I am wondering about attempts to define QFT non-perturbatively, by formalizing the way physicists think about QFT (in which the RNG is the basic object, rather than a technical tool). I appreciate this is a vague question, thanks everyone for the help.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-01-28T20:30:31.777" LastActivityDate="2011-03-25T23:35:18.863" Title="Formalizing Quantum Field Theory" Tags="&lt;quantum-field-theory&gt;&lt;mathematical-physics&gt;" AnswerCount="5" CommentCount="11" FavoriteCount="9" />
  <row Id="4069" PostTypeId="2" ParentId="1453" CreationDate="2011-01-28T17:38:16.790" Score="-4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;The atom is invariant thru time?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NO - The atom is invariant thru time:&lt;/strong&gt;&#xA;This answer is naturally the expected one. It is 'common sense'. &lt;strong&gt;&lt;em&gt;The physics building is rooted upon this answer&lt;/em&gt;&lt;/strong&gt;, but this question has not being asked, nor answered, nor discussed. And Physics continues as usual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;YES - The atom is NOT invariant thru time:&lt;/strong&gt;&#xA;It is NOT the expected answer. If it happens to be the valid answer then Physics continues as usual, after a revolution. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How sure we are that the atom is 'invariant'?&lt;/strong&gt;&lt;br&gt;&#xA;Because our local lab is our reference and we can not measure directly any variability it follows that &lt;strong&gt;we are not sure&lt;/strong&gt;.&#xA;This fact justifies our natural answer, but we can not know the rightness of this answer until we explore.&#xA;One possible way to explore it is to look back the distant universe in search of clues.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How to decide that the space expansion model is better than a shrinking or  'evanescent matter'  model ?&lt;/strong&gt;&#xA;Without further reasoning one can say that one model appears to be the dual of the other, like an image in a mirror.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If we can not assert that the atom is invariable then the &lt;strong&gt;consequences&lt;/strong&gt; can be dramatic:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;No space expansion&lt;/li&gt;&#xA;&lt;li&gt;No Big Bang&lt;/li&gt;&#xA;&lt;li&gt;No Inflation Era&lt;/li&gt;&#xA;&lt;li&gt;No Dark Energy&lt;/li&gt;&#xA;&lt;li&gt;No Dark Matter&lt;/li&gt;&#xA;&lt;li&gt;A better understanding of the nature of TIME&lt;/li&gt;&#xA;&lt;li&gt;A model of the Universe with only one parameter: The Hubble Constant&lt;/li&gt;&#xA;&lt;li&gt;A better understanding of the past and the future history of the Earth and of Life&lt;/li&gt;&#xA;&lt;li&gt;A different and interesting Large Scale Structure evolution of the Universe is attainable.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We must be conscient that a long future awaits the Humanity and the Physics must evolve. Under a closer scrutiny several fundamental notions, our present believes, can peril.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I tried to persuade our community, whith the above lines, that there exists merit in this quest and it is of the utmost importance, and surely it is an unsettled question.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This subject of a possible 'atom evanescence' was already studied in the arxiv paper linked bellow,  and it was never criticized nor discussed. In there is presented a model that conforms to the Universe evolution and the experimental barionic data, with only one parameter - the Hubble Constant. The equations of the model can be written in the back of an envelope, somewhat ironically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As any model it was constructed upon one hypothesis, as above described, and applying it upon data. The paper includes a comparison between the SM and the new model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/astro-ph/pdf/0208/0208365v1.pdf&quot; rel=&quot;nofollow&quot;&gt;A relativistic time variation of matter/space fits both local and cosmic data (arxiv astro-ph 0208365)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-01-28T17:38:16.790" CommunityOwnedDate="2011-01-28T17:38:16.790" CommentCount="1" />
  <row Id="4070" PostTypeId="2" ParentId="4063" CreationDate="2011-01-28T17:44:06.100" Score="1" ViewCount="" Body="&lt;p&gt;The raw data from modern particle physics experiments are many terabytes (even petabytes) in size, and quite complicated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For collider experiments the detectors are compound, layered devices with three or more different technologies used by five or more distinct subsystems, plus ancillary monitoring of detector performance, temperature and humidity conditions in the experimental hall, data provided by the accelerator operating crew on the state of the beam, and on and on and on. There are ten of thousands of individual detector channels and hundreds of &quot;slow&quot; devices (like the thermoments, magnet currents, beam current monitors, etc...). All of this has been pre-filterd by the trigger hardware (and exactly what filtering was applied changes over time). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For neutrino experiments the data are detailed information about the charge detected from photo-tubes (some combination of total charge in a window, peak voltage, peak time, onset time, and/or digitized wave forms) for hundreds or thousands of PMTs. Plus environmental monitoring like that done by the collider people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases there scads of calibration data, changes in operating conditions throughout the data taking period, and sometimes replacement of re-tuning of sub-systems part way through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is typically many tens of thousands of lines of custom computer code for opening and processing the data files. Code written by physicists. Now, particle physicists are a little more professional about coding then some of their peers, but that does &lt;em&gt;not&lt;/em&gt; mean state of the art process and beautiful code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It generally takes many thousands of grad-student and post-doc hours to reduce this to something physics can be extracted from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a reason we call this &quot;Big Science&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;That said, &lt;strong&gt;you generally &lt;em&gt;can&lt;/em&gt; get the data&lt;/strong&gt;. Eventually. (Each collaboration will hold theirs for a while to insure they get to publish first.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you get it? &lt;strong&gt;Just ask.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you'll have to provide your own storage (and possibly copying hardware); come to where the data are kept; understand that the documentation will be scattered over hundreds or thousands of internal (to the collaboration) documents written as they went along by diverse authors some of whom have English as a second or third language (and may evidence some idiosyncrasies); and that help interpreting all this will be terse as these people have moved on and have other projects keeping them busy. And you may have to convince the people with the data that you have the capacity to manage it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The availability of partly processed data sets is not something I am as sure about, but you could try asking for that too. The worst that can happen is you get told &quot;No&quot;. But even if you can get this, don't image that it is easy to work with.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If I haven't dissuaded you, let me suggest a practical method for getting started. Go to the nearest university that has a nuclear or particle physics group, and ask to help out. Really. There is always a need for lab monkeys, and you &lt;em&gt;will&lt;/em&gt; learn as you go along because you can't do the work if they don't teach you stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the process you'll&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Learn how some of the sub-systems work. Get a feel for what kind of raw data they return and how it is processed into less raw data. If you ask people will tell you how the less raw data can be transformed into still more physics-like information and eventually reconstructed into particles.&lt;/li&gt;&#xA;&lt;li&gt;Make some contacts in the business. Begin able to say &lt;em&gt;&quot;I work with Prof. Smith and Podunk U.&quot;&lt;/em&gt; is much better than &lt;em&gt;&quot;I'm interested.&quot;&lt;/em&gt; when it comes to getting access to data.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-01-28T18:02:22.257" LastActivityDate="2011-01-28T18:02:22.257" CommunityOwnedDate="2011-01-29T21:34:39.710" />
  <row Id="4071" PostTypeId="1" AcceptedAnswerId="4090" CreationDate="2011-01-28T17:45:17.337" Score="8" ViewCount="314" Body="&lt;p&gt;Here are some depictions of electromagnetic wave, similar to the depictions in other places:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://idol.union.edu/malekis/Phys%20&amp;amp;%20Pol%202010/Images/e&amp;amp;mWave.gif&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;http://micro.magnet.fsu.edu/primer/java/wavebasics/basicwavesjavafigure1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;http://www.who.int/entity/mediacentre/factsheets/images/fs205.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't there an error? It is logical to presume that the electric field should have maximum when magnetic field is at zero and vise versa, so that there is no moment when the both vectors are zero at the same time. Otherwise one comes to a conclusion that the total energy of the system becomes zero, then grows to maximum, then becomes zero again which contradicts the conservation law.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-18T21:02:17.057" LastActivityDate="2011-07-18T21:02:17.057" Title="Popular depictions of electromagnetic wave: is there an error?" Tags="&lt;electromagnetism&gt;&lt;waves&gt;&lt;energy-conservation&gt;&lt;plane-wave&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="4072" PostTypeId="2" ParentId="4063" CreationDate="2011-01-28T17:45:33.023" Score="0" ViewCount="" Body="&lt;p&gt;At least in particle physics, you will not have access to this low-level information that you want unless you are part of the experimental group which is actually making the measurement. There are many reasons for that, some political other more physical. Let's comment quickly on both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, it is a lot of work to measure these particles and the collaborations want to use the data themselves before allowing anyone else to look at it. So, they won't show you the raw data before they extract everything they can. &quot;Ok, show me data from old experiments&quot;, you may say. Now come another problem, the amount of data is simply overwhelming. It is not something you can record in a DVD and carry home or download from megaupload.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the more physical problem. You are specifically asking for very low level, raw data. To transform voltages into particles, there are lots of other assumptions involved. For instance, how the interaction between the particles and the detectors happens, how the signal on the detectors are processed and so on. Every experiment spends years, literally, understanding these interaction to develop softwares which allow the &lt;em&gt;reconstruction&lt;/em&gt; of the signal and testing them to know whether the accuracy of this reconstruction is enough to measure whatever you want to measure properly. Well, even if you are given access to the raw data, who can guarantee that you will reconstruct it properly? The people who measured the data don't want any other person drawing wrong conclusions with their work. It's a kind of scientific responsibility. The only information you will really have access today is the high-level, already analysed data that can be found in any paper (see Lubos' answer above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are interesting proposals to make more information available publicly. One that you may want to look at is the &lt;a href=&quot;http://arxiv.org/abs/1010.2506v1&quot; rel=&quot;nofollow&quot;&gt;Recast&lt;/a&gt; system. But something similar for all experiments is far from being reality.&lt;/p&gt;&#xA;" OwnerUserId="1337" LastActivityDate="2011-01-28T17:45:33.023" CommunityOwnedDate="2011-01-29T21:34:39.710" CommentCount="5" />
  <row Id="4073" PostTypeId="2" ParentId="4071" CreationDate="2011-01-28T17:49:55.000" Score="6" ViewCount="" Body="&lt;p&gt;There is no contradiction with conservation law, you are just applying it wrongly. To do it right, you have to consider a small closed region and check whether the variation of energy inside this domain is the same as (minus) the energy flux through its boundaries. In the case of a electromagnetic wave, when the energy density decreases inside the region, it just means that the energy has left the region you are considering.&lt;/p&gt;&#xA;" OwnerUserId="1337" LastActivityDate="2011-01-28T17:49:55.000" CommentCount="10" />
  <row Id="4074" PostTypeId="2" ParentId="4038" CreationDate="2011-01-28T17:50:38.977" Score="0" ViewCount="" Body="&lt;p&gt;If I take thick oil to mean what I think it does, this should be a very viscous fluid, so Reynolds number, which relates inertial forces to viscous forces should be so small as to not be applicable. I think in the purely linear viscous regime, that good solutions exist. My first guess would be that you need the same number of rotations to achieve the same degree of mixing, so 20RPM should probably achieve the same rate of mixing per unit of time. Of course for the smaller barrel, you may be able to run it faster and have the process take less time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a way to measure the quality of mixing, you might be able to experiment with it. Perhaps the density difference between two samples of the oil. My assumption of number of rotations comes from assuming the geometry is what matters, i.e. the free surface is roughly horizontal, and the flow pattern is the same as you change size/rotation speed. Obviously at some rotation speed, the flow will be too slow so the the free surface (air bubble) may simply be carries around the barrel, so I'm guessing there is a max rotation rate, and I'm not sure how that scales with size/viscosity.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-01-28T17:50:38.977" />
  <row Id="4076" PostTypeId="1" AcceptedAnswerId="4947" CreationDate="2011-01-28T18:00:47.797" Score="3" ViewCount="283" Body="&lt;p&gt;About 1905 Einstein published a work about diffusion of hard spheres &#xA;and brownian motion. One effluence of that is the so called &#xA;&quot;Viscosity Equation&quot; which was/is very important for dertermining &#xA;moleculat weight of macromolecules by viscosity of solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ias.ac.in/initiat/sci_ed/resources/chemistry/Viscosity.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ias.ac.in/initiat/sci_ed/resources/chemistry/Viscosity.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Intrinsic_viscosity&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Intrinsic_viscosity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I like to know: Was there any relation to the other &lt;/p&gt;&#xA;&#xA;&lt;p&gt;topics Einstein worked on? Is something known on Einsteins reason &lt;/p&gt;&#xA;&#xA;&lt;p&gt;work on Viscosity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Georg&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-02-10T16:37:51.967" Title="What was Einsteins reasons for the work on Viscosity/Brownian Motion" Tags="&lt;history&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="4077" PostTypeId="2" ParentId="4049" CreationDate="2011-01-28T18:09:33.253" Score="4" ViewCount="" Body="&lt;p&gt;Dear Rodrigo, I am not aware of any direct physical meaning of the anti-commutator term, but it is useful when you want to pin down the states that saturate the inequality in the Heisenberg principle. Obviously, two conditions must be met if an equality in the usual uncertainty principle should occur: the anti-commutator term must vanish and the Cauchy-Schwarz inequality (see the comment by Luboš Motl) must be saturated. The latter happens if and only if the vectors $\Delta A|\psi\rangle$ and $\Delta B|\psi\rangle$ are collinear, let's say $\Delta A|\psi\rangle=\lambda\Delta B|\psi\rangle$. This is equivalent to $(A-\lambda B)|\psi\rangle=(\langle A\rangle-\lambda\langle B\rangle)|\psi\rangle$, that is, $\psi$ is an eigenvector of $A-\lambda B$. But then the expectation value of the anti-commutator becomes $\langle\{\Delta A,\Delta B\}\rangle=(\lambda+\lambda^*)\langle(\Delta B)^2\rangle$, which vanishes only if $\lambda$ is purely imaginary unless, of course, $|\psi\rangle$ is an eigenvector of $B$ in which case the whole inequality is trivial. So in the end, the state $|\psi\rangle$ saturates the inequality in the (usual formulation of the) uncertainty principle if and only if it is an eigenstate of $A-\lambda B$ for some purely imaginary $\lambda$. This happens for example for the coherent states of the harmonic oscillator.&lt;/p&gt;&#xA;" OwnerUserId="1641" LastActivityDate="2011-01-28T18:09:33.253" />
  <row Id="4078" PostTypeId="2" ParentId="4071" CreationDate="2011-01-28T18:11:17.427" Score="3" ViewCount="" Body="&lt;p&gt;Shorter answer: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Close to the antenna (less than some wavelengths) the phase shift &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is 90 Degrees indeed. This is the situation in classical pictures &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for a dipole transmitting &quot;Hertzian Waves&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further down, the phase shift between electric and magnetic field &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is zero degrees, this is the thing displayed in the pictures You linked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Georg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the English expressions for &quot;Nahfeld&quot; and &quot;Fernfeld&quot;? &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-01-28T18:30:01.490" LastActivityDate="2011-01-28T18:30:01.490" CommentCount="2" />
  <row Id="4079" PostTypeId="1" AcceptedAnswerId="4084" CreationDate="2011-01-28T18:25:39.140" Score="5" ViewCount="513" Body="&lt;p&gt;As human beings, we observe the world in which we live in three dimensions. However, it is certainly theoretically possible that more dimensions exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any direct or indirect evidence supporting a fourth dimension?&lt;/p&gt;&#xA;" OwnerUserId="1646" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-31T10:03:07.667" LastActivityDate="2011-05-31T10:03:07.667" Title="Experimental evidence of a fourth spatial dimension?" Tags="&lt;spacetime&gt;&lt;experimental-evidence&gt;&lt;kaluza-klein&gt;&lt;geometry&gt;&lt;dimensionality&gt;" AnswerCount="6" CommentCount="5" />
  <row Id="4080" PostTypeId="2" ParentId="4079" CreationDate="2011-01-28T18:32:39.380" Score="2" ViewCount="" Body="&lt;p&gt;Since you were here yesterday, you're certainly here at the moment and we hope you're here tomorrow, there's some pretty compelling evidence for an additional dimension!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be less flippant, here is a more vivid example. Suppose an infinitely thin disc (like a coin with no discernible thickness) came into existence on your desk Tuesday midday and remained there until Thursday midday when it vanished. We can think of this object as a &lt;em&gt;cylinder&lt;/em&gt; with cross section the disc and &quot;length&quot; two days. Time is being treated here as analogous to the third spatial direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To describe a world which extends in time as well as in 3D space we need that extra time coordinate which constitutes a fourth dimension. As other answers indicate, in relativity the time dimension is distinguished from spatial directions by a sign change in the definition of space-time interval.&lt;/p&gt;&#xA;" OwnerUserId="1394" LastEditorUserId="1394" LastEditorDisplayName="" LastEditDate="2011-01-28T20:42:50.670" LastActivityDate="2011-01-28T20:42:50.670" CommentCount="4" />
  <row Id="4081" PostTypeId="2" ParentId="4037" CreationDate="2011-01-28T18:33:11.500" Score="1" ViewCount="" Body="&lt;p&gt;Classically, there are initial configurations in general relativity without closed timelike curves which can be shown to evolve into future states with closed timelike curves. The boundary between both regions is the Cauchy horizon, and evolution across the Cauchy horizon is nondeterministic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For anti de Sitter space without any conformal boundary conditions, we also have nondeterministic evolution, even far away from conformal infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, classically, anything can come out of a white hole singularity.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Probably the best bet for dealing with questions about existence and uniqueness is to work with the ADM time-slicing Hamiltonian approach.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-02-08T17:27:33.293" LastActivityDate="2011-02-08T17:27:33.293" />
  <row Id="4082" PostTypeId="2" ParentId="4079" CreationDate="2011-01-28T18:50:15.233" Score="3" ViewCount="" Body="&lt;p&gt;Assuming you mean spatial dimensions, Nima Arkani-Hamed has some ideas and papers on the subject. Here is a popular article:&#xA;&lt;a href=&quot;http://www.nature.com/nature/journal/v433/n7021/full/433010a.html&quot; rel=&quot;nofollow&quot;&gt;http://www.nature.com/nature/journal/v433/n7021/full/433010a.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of a fourth dimension was popularized by Charles Hinton in the late 1800s: &lt;a href=&quot;http://en.wikipedia.org/wiki/Charles_Howard_Hinton&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Charles_Howard_Hinton&lt;/a&gt;&#xA;and, of course, by Edwin Abbott (Flatland).&#xA;Then, there are the modern takes on it, like The Higher Dimensions Family Church :)&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Carlton_Pearson&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Carlton_Pearson&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-01-28T18:50:15.233" CommentCount="1" />
  <row Id="4083" PostTypeId="2" ParentId="4068" CreationDate="2011-01-28T18:54:39.307" Score="6" ViewCount="" Body="&lt;p&gt;There are a number of high level mathematicians who are working on giving a more mathematically precise description of perturbative QFT and the renormalization procedure.&#xA;For example there is a recent paper by Borcherds &lt;a href=&quot;http://arxiv.org/pdf/1008.0129&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/1008.0129&lt;/a&gt;, work&#xA;of Connes and Kreimer on Hopf algebras and the work of Bloch and Kreimer on mixed Hodge structures and renormalization &lt;a href=&quot;http://www.math.uchicago.edu/~bloch/monodromy.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uchicago.edu/~bloch/monodromy.pdf&lt;/a&gt; just to name a few. To be honest, I am not mathematically sophisticated enough to judge what has been accomplished in these papers, but I think there are some problems in QFT which will probably involve some rather high-powered mathematics of the type being developed in these papers. For example, the current attempt to reformulate N=4 SYM in terms of Grassmannians apparently has some connection to rather deep mathematical objects called Motives. Results on the degree of transcendentality which show up in perturbative N=4 SYM amplitudes also seem beyond what physicists really understand and&#xA;I believe the presence of transcendental objects (like $\zeta(3)$) in QFT amplitudes provides some of the motivation for the work of Bloch and Kreimer. I'm not an expert on this stuff, so perhaps someone else will chime in with a more complete explanation and additional references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: One more reference which is closer to the spirit of the original question is a book in progress by Costello on perturbative quantum field theory treated from the Wilsonian, effective field theory point of view. Notes are available online at &lt;a href=&quot;http://www.math.northwestern.edu/~costello/renormalization&quot; rel=&quot;nofollow&quot;&gt;http://www.math.northwestern.edu/~costello/renormalization&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T04:03:29.200" LastActivityDate="2011-03-02T04:03:29.200" CommentCount="6" />
  <row Id="4084" PostTypeId="2" ParentId="4079" CreationDate="2011-01-28T18:55:39.797" Score="6" ViewCount="" Body="&lt;p&gt;No. Unless you consider time as the fourth dimension. Also current research suggests that spacetime becomes two dimensional at high energies and very small distances. So, yes, dimensionality of spacetime is a mutable concept in modern physics. But no experimental evidence, indirect or direct, has been found for a fourth spacelike or a second timelike direction.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Someone is bound to question my statement that &quot;spacetime becomes two dimensional at high energies and very small distances&quot;. My backup reference for this is Carlip's recent paper &lt;a href=&quot;http://arxiv.org/abs/1009.1136v1&quot; rel=&quot;nofollow&quot;&gt;The Small Scale Structure of Spacetime&lt;/a&gt;, where he considers inputs from five or six different theoretical lines of research. Of course, there is bound to be a spectrum of opinion on this issue.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-28T19:07:22.173" CommentCount="1" />
  <row Id="4085" PostTypeId="2" ParentId="4079" CreationDate="2011-01-28T18:58:51.600" Score="1" ViewCount="" Body="&lt;p&gt;In Newtonian physics, there are 3 dimensions of space and there is time. Everything happens in these 3 dimensions of space and time is only a parameter which measures the evolution of things (for example the change in the positions of a particle). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Special Relativity, it was postulated that the 3 dimension of space and time should be considered under equal footing and that time should also be seen as a dimension. The fact that that was the case was shown by proving the validity of relativistic transformations between various reference frames which mixed time with space. But because a reference frame in Special Relativity can be considered as a universal frame for that particular observer, it introduced time in an absolute sense and thus one could consider again time to be only a parameter for that particular observer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General Relativity established the full equivalence of time and space as parts of spacetime, in the sense that it fully geometrised gravity and the place where everything happens is the 4-dimensional spacetime with 3 dimensions of space and 1 of time. The evidence supporting the geometric view of spacetime are those supporting the Einstein equivalence principle which states that every free falling frame doesn't experience any gravitational effects or that an accelerated observer can't distinguish with local experiments his accelerated frame from a gravitational field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to quote some experiments that verify the Einstein equivalence principle, we have:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The redshift of photons climbing the gravitational potential,&lt;/li&gt;&#xA;&lt;li&gt;The time delay of clocks in different gravitational fields,&lt;/li&gt;&#xA;&lt;li&gt;The Eotvos-type experiments that show that all objects regardless their composition fall with the same acceleration,&lt;/li&gt;&#xA;&lt;li&gt;The constancy of physical constants.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="383" LastEditorUserId="383" LastEditorDisplayName="" LastEditDate="2011-01-28T19:12:44.760" LastActivityDate="2011-01-28T19:12:44.760" />
  <row Id="4086" PostTypeId="2" ParentId="4079" CreationDate="2011-01-28T19:00:46.833" Score="0" ViewCount="" Body="&lt;p&gt;Not data, but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The string theorists and other TOE people would say that a single theory exhibiting all the required symmetries only has representations at higher dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then they deal &lt;em&gt;&quot;So why can't we see the other ones?&quot;&lt;/em&gt; by giving them periodic boundary conditions on length scales smaller than we have been able to probe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be &quot;extra&quot; spacial dimensions, but would not have any interesting effects on day to day life.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-01-28T19:00:46.833" CommentCount="2" />
  <row Id="4088" PostTypeId="2" ParentId="4068" CreationDate="2011-01-28T19:25:15.947" Score="5" ViewCount="" Body="&lt;p&gt;you'll find a lot of information on the nLab, the open online Wiki of a bunch of people working on n-categories. You should really click around and see what is there, here is the page about the &quot;functorial&quot; POV on QFT, the formalization of the Schrödinger picture of QFT, including TQFTs:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ncatlab.org/nlab/show/FQFT&quot; rel=&quot;nofollow&quot;&gt;Functorial quantum field theory&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;there is also a page about the Heisenberg picture aka axiomatic quantum field theory:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ncatlab.org/nlab/show/AQFT&quot; rel=&quot;nofollow&quot;&gt;AQFT&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The wiki software written by Jaques Distler has a nice search function, use it! You'll find there is a lot about (formalizing) string theory, too, also about the work of Jacob Lurie et alt. on TQFTs, a long page about CQFT, and references about recent work on the picture of perturbative QFT and renormalization groups from the AQFT point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, since you asked specifically for the latter, here is the direct link (but this is also on the nLab, together with a whole plethora of other resources):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Romeo Brunetti, Michael Duetsch, Klaus Fredenhagen: &lt;a href=&quot;http://arxiv.org/abs/0901.2038&quot; rel=&quot;nofollow&quot;&gt;Perturbative Algebraic Quantum Field Theory and the Renormalization Groups&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There is also information about Connes' work on formalizing the standard model and unifying it with gravity by using noncommutative spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides, if anyone does not find anything that should be there, go to the nForum and tell the folks over there about it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Explanation of &quot;noncommutative spaces&quot;: When you take a real smooth manifold like a spacetime, for example, this manifold is completely described by the algebra of charts. In fact, the very definition of manifold can be done this way. Every property of the manifold corresponds to a property of the algebra of charts. This algebra is commutative, of course. Connes' great idea of &quot;noncommutative geometry&quot; is that we could replace the commutative algebra of charts with a noncommutative operator algebra and see what geometric concepts we could transfer from the commutative to the noncommutative setting. Operator algebras ($C^*$-algebras, to be more precise) are then considered as a noncommutative analog of charts of a &quot;noncommutative&quot; space.  Connes's did a lot of work on the standard model and perturbative QFT using this idea, but unfortunately it is mathematically quite sophisticated. For a good introduction for physicists, see for example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;J. Madore: &quot;An Introduction to Noncommutative Differential Geometry and its Physical Applications&quot;, 2nd edition, Cambridge University Press&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This book also explains ideas of extending classical spacetime with noncommutative aspects.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditorDisplayName="" LastEditDate="2011-01-29T10:06:03.100" LastActivityDate="2011-01-29T10:06:03.100" CommentCount="2" />
  <row Id="4089" PostTypeId="2" ParentId="3644" CreationDate="2011-01-28T19:29:08.697" Score="0" ViewCount="" Body="&lt;p&gt;The speed of light is a property of spacetime, which has a Minkowski metric, which means it has a Null Space, which marries space and time into Velocity, which is invariant. One might have a look at David Hestenes &quot;Space Time Algebra&quot; and various of his papers, or visit a site on Geometric Algebra (ie, Clifford Algebra ).  The geometry of spacetime is impervious to units of measurement - whatever floats your boat. One prof used furlongs per fortnight.&lt;/p&gt;&#xA;" OwnerUserId="1423" LastActivityDate="2011-01-28T19:29:08.697" />
  <row Id="4090" PostTypeId="2" ParentId="4071" CreationDate="2011-01-28T19:29:15.933" Score="9" ViewCount="" Body="&lt;p&gt;The depictions you're seeing are correct, the electric and magnetic fields both reach their amplitudes and zeroes in the same locations. Rafael's answer and certain comments on it are completely correct; energy conservation does &lt;em&gt;not&lt;/em&gt; require that the energy density be the same at every point on the electromagnetic wave. The points where there is no field do not carry any energy. But there is never a time when the fields go to zero &lt;em&gt;everywhere&lt;/em&gt;. In fact, the wave always maintains the same shape of peaks and valleys (for an ideal single-frequency wave in a perfect classical vacuum), so the same amount of energy is always there. It just moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To add to Rafael's excellent answer, here's an explicit example.&#xA;Consider a sinusoidal electromagnetic wave propagating in the $z$ direction. It will have an electric field given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{E}(\mathbf{r},t) = E_0\hat{\mathbf{x}}\sin(kz - \omega t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the curl of this and you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla\times\mathbf{E}(\mathbf{r},t) = \left(\hat{\mathbf{z}}\frac{\partial}{\partial y} - \hat{\mathbf{y}}\frac{\partial}{\partial z}\right)E_0\sin(kz - \omega t) = -E_0 k\hat{\mathbf{y}}\cos(kz - \omega t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using one of Maxwell's equations, $\nabla\times\mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{\partial\mathbf{B}(\mathbf{r},t)}{\partial t} = -E_0 k\hat{\mathbf{y}}\cos(kz - \omega t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrate this with respect to time to find the magnetic field,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{B}(\mathbf{r},t) = -\frac{E_0 k}{\omega}\hat{\mathbf{y}}\sin(kz - \omega t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing this with the expression for $\mathbf{E}(\mathbf{r},t)$, you find that $\mathbf{B}$ is directly proportional to $\mathbf{E}$. When and where one is zero, the other will also be zero; when and where one reaches its maximum/minimum, so does the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an electromagnetic wave in free space, conservation of energy is expressed by Poynting's theorem,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial u}{\partial t} = -\nabla\cdot\mathbf{S}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left side of this gives you the rate of change of energy density in time, where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u = \frac{1}{2}\left(\epsilon_0 E^2 + \frac{1}{\mu_0}B^2\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the right side tells you the electromagnetic energy flux density, in terms of the Poynting vector,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{S} = \frac{1}{\mu_0}\mathbf{E}\times\mathbf{B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Poynting's theorem just says that the rate at which the energy density at a point changes is the opposite of the rate at which energy density flows away from that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you plug in the explicit expressions for the wave in my example, after a bit of algebra you find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial u}{\partial t} = -\omega E_0^2\left(\epsilon_0 + \frac{k^2}{\mu_0\omega^2}\right)\sin(kz - \omega t)\cos(kz - \omega t) = -\epsilon_0\omega E_0^2 \sin\bigl(2(kz - \omega t)\bigr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(using $c = \omega/k$) and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla\cdot\mathbf{S} = \frac{2}{\mu_0}\frac{k^2}{\omega}E^2 \sin(kz - \omega t)\cos(kz - \omega t) = \epsilon_0 \omega E_0^2 \sin\bigl(2(kz - \omega t)\bigr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus confirming that the equality in Poynting's theorem holds, and therefore that EM energy is conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the expressions for both sides of the equation include the factor $\sin\bigl(2(kz - \omega t)\bigr)$ - they're not constant. This mathematically shows you the structure of the energy in an EM wave. It's not just a uniform &quot;column of energy;&quot; the amount of energy contained in the wave varies sinusoidally from point to point ($S$ tells you that), and as the wave passes a particular point in space, the amount of energy it has at that point varies sinusoidally in time ($u$ tells you that). But those changes in energy with respect to space and time don't just come out of nowhere. They're precisely synchronized in the manner specified by Poynting's theorem, so that the changes in energy at a point are accounted for by the flux to and from neighboring points.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-28T21:27:42.340" LastActivityDate="2011-01-28T21:27:42.340" CommentCount="2" />
  <row Id="4091" PostTypeId="2" ParentId="4067" CreationDate="2011-01-28T19:54:19.793" Score="0" ViewCount="" Body="&lt;p&gt;In answer to your question, experimental data up to now have given us a number of particles that can be incorporated in the general framework of String Theory, which is a theory that also incorporates the quantization of Gravity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;String Theory does not allow an infinite number of particles if you think of them as being nested like russian dolls, more and more composite. It allows for a mathematical description of the Standard Model which incorporates particles we have seen, also predicts new particles but not in the nested form you are suggesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many types of infinities: The energy levels of a potential well are infinite, but at some point the distinction of the levels becomes trivial energetically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment we believe that quarks and leptons are the real &quot;atoms&quot; , i.e. non composite, components of nature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If these experimental facts are contradicted by new data, it will be a big revolution in theory, including string theory which is aiming at being the theory of everything.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-01-28T19:54:19.793" />
  <row Id="4092" PostTypeId="2" ParentId="4079" CreationDate="2011-01-28T19:56:01.973" Score="3" ViewCount="" Body="&lt;p&gt;If we consider the heat capacity of a mono-atomic gas, it will answer the question how many independent degrees of freedom exist in space. Experiment gives three space dimensions.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-01-28T20:31:00.060" LastActivityDate="2011-01-28T20:31:00.060" />
  <row Id="4093" PostTypeId="2" ParentId="4047" CreationDate="2011-01-28T20:16:06.140" Score="2" ViewCount="" Body="&lt;p&gt;For a high degree of accuracy you would have to probe the particle with a high energy (short wavelength) photon so there is plenty of energy that can go into vibrational excitation. After such hard hit the particle will be smeared across a wide range of states $$\Psi=a_0*\Psi_0+a_1*\Psi_1+...$$This is not an entanglement but a simple superposition of eigenstates. The expectation value of the particle's energy $\bar{E}=\sum_{i}a_i^2E_i$ should not be equal to the energy of any particular state and the second measurement will yield $E_i$ with $a_i^2$ probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the extra energy comes from an interaction with a probe particle and it doesn't have to be precisely equal to the energy of a certain vibrational state.&lt;/p&gt;&#xA;" OwnerUserId="701" LastActivityDate="2011-01-28T20:16:06.140" CommentCount="1" />
  <row Id="4094" PostTypeId="1" AcceptedAnswerId="4115" CreationDate="2011-01-28T20:18:30.683" Score="7" ViewCount="459" Body="&lt;p&gt;The planck length is defined as $l_P = \sqrt{\frac{\hbar G}{c^3}}$. So it is a combination of the constants $c, h, G$ which I believe are all Lorentz invariants. So I think the Planck length should also be a Lorentz invariant! &#xA;But there seem to be some confusion about that, see e.g. the following paper&#xA;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0112090&quot; rel=&quot;nofollow&quot;&gt;Magueijo 2001: Lorentz invariance with an invariant energy scale&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The combination of gravity $G$, the&#xA;  quantum $h$ and relativity $c$ gives rise&#xA;  to the Planck length, $l_p$ or its&#xA;  inverse, the Planck energy $E_p$ . These&#xA;  scales mark thresholds beyond which&#xA;  the old description of spacetime&#xA;  breaks down and qualitatively new&#xA;  phenomena are expected to appear. ...&#xA;  This gives rise immediately to a&#xA;  simple question: in whose reference&#xA;  frame are $l_P$ and $E_P$ the thresholds for&#xA;  new phenomena?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if $l_P$ is a Lorentz invariant their is no question about that. $l_P$ is the same in all reference frames! Another confusing issue is that the Planck mass (from which the Planck length is derived) is often derived by setting equal the Compton length $\lambda_C = \frac{h}{m_0 c}$ ( a Lorentz invariant 4-length) and the Schwarzschild length $r_{s} = \frac{2Gm}{c^2}$ (which I believe is not a Lorentz invariant, since in the derivation of the Schwarzschild metric it is assumed to be a 3-length, measuring a space distance). But since Compton wavelength and Schwarzschild radius are not lengths of the same kind I think such a derivation is not valid. &#xA;So my question is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the Planck length a Lorentz invariant and if so, how to derive it then without using the Compton wave length and the Schwarzschild radius ?&lt;/p&gt;&#xA;" OwnerUserId="1648" LastActivityDate="2011-07-20T23:29:29.410" Title="Is the Planck length Lorentz invariant?" Tags="&lt;general-relativity&gt;&lt;special-relativity&gt;&lt;quantum-gravity&gt;&lt;planck&gt;" AnswerCount="5" CommentCount="10" FavoriteCount="3" />
  <row Id="4095" PostTypeId="2" ParentId="4037" CreationDate="2011-01-28T20:25:36.007" Score="-2" ViewCount="" Body="&lt;p&gt;The formal classification of solutions to Einstein field equations is the Penrose-Petrov-Pirani scheme.  This classifies solutions according to eigen-Killing vectors of the Weyl tensor.  These solutions range from the type D solutions for black holes to the type N solutions for gravity waves.  In between there are type I II and III solutions for Robinson-Trautman spacetimes.  These turn out to have a relationship to each other, where type D solutions are near source terms and type N solutions are far field.  The other solutions are in between.  This is a gravitational analogue of the near and far field solutions for Mazwell’s equations.  A source on these solutions is in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H. Stephani, D. Kramer, M. MacCallum, C. Hoenselaers, E. Herlt, Exact Solutions of Einstein’s Field&#xA;Equations, Cambridge: Cambridge University Press. (2003)&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-28T20:25:36.007" CommentCount="1" />
  <row Id="4096" PostTypeId="2" ParentId="4094" CreationDate="2011-01-28T20:44:12.710" Score="1" ViewCount="" Body="&lt;p&gt;When it is transverse, it is Lorentz invariant. When it is longitudinal - it is not ;-).&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-01-28T20:44:12.710" />
  <row Id="4097" PostTypeId="2" ParentId="4076" CreationDate="2011-01-28T20:50:21.703" Score="4" ViewCount="" Body="&lt;p&gt;Although the notion of viscosity in gases and liquids was known well before 1906, as well as its derivation from statistical mechanics, the atomistic structure of matter was still under experimentally discovering, so it was a hot topic at that time.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-01-28T20:50:21.703" />
  <row Id="4098" PostTypeId="2" ParentId="4058" CreationDate="2011-01-28T21:02:18.373" Score="3" ViewCount="" Body="&lt;p&gt;expanding a bit on Lubos' answer: Brownian motion as a stochastic process in the sense of mathematics does not describe a trajectory (or many trajectories or a probability measure on trajectories) of point particles in the sense of classical mechanics. When this mathematical tool is used to model physical systems some - severe - assumptions about these systems are usually made. For more details have a look at the references listed here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Azimuth project: &lt;a href=&quot;http://www.azimuthproject.org/azimuth/show/Stochastic+differential+equation&quot; rel=&quot;nofollow&quot;&gt;stochastic differential equations&lt;/a&gt;,&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;especially the reference&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;William T. Coffey, Yuri P. Kalmykov, and John T. Waldron: The Langevin equation. With applications to stochastic problems in physics, chemistry and electrical engineering.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Brownian motion reappears in quantum mechanics as Lubos has explained, a classic reference is&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Barry Simon: Functional Integration and Quantum Physics&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That's the same Simon that wrote the 4 part classic book about analysis and operator theory with Michael Reed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stochastic processes of continuous time live in function spaces and have therefore an infinite amount of degress of freedom. A calculus of variations has therefore to be a calculus of variations on infinite dimensional spaces. Such a calculus exists, it is usually called Malliavin or white noise calculus.&#xA;There has been limited success to provide a rigorous construction of the Feynman path integral using white noise calculus (the buzzword is &quot;Feynman path integral as a Hida distribution&quot;), see for example&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Zhi-yuan Huang and Jia-an Yan: &quot;Introduction to Infinite Dimensional Stochastic Analysis&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and especially chapter V paragraph 3.3, &quot;Application to Feynman integrals&quot;, and paragraph 4, &quot;applications to quantum physics&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: It is usually considered helpful if one explains what one already knows about the topic one is asking about and what the reasons for asking are. This is helpful for people in order to taylor the answer to the needs of the one who asked :-)&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditorDisplayName="" LastEditDate="2011-01-28T21:08:55.953" LastActivityDate="2011-01-28T21:08:55.953" />
  <row Id="4099" PostTypeId="2" ParentId="4094" CreationDate="2011-01-28T21:07:50.820" Score="1" ViewCount="" Body="&lt;p&gt;This is a funny question, and Marek does point out that a constant is simply a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Planck length is a length because it is the square root of the gravitational constant, G times some units that are units in naturalized units.  The speed of light is unity and the Planck constant is momentum-length, which are reciprocals of each other.  The gravitational constant in naturalized units is [Area], which connects up with the entropy per unit area of an event horizon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Planck area squared $\ell_p^2~=~G\hbar/c^3$ is a unit of a black hole event horizon, and for $n$ of these unit areas $A~=~n\ell$ this is associated with units of action,&#xA;$$&#xD;&#xA;A~=~n\hbar(G/c^3).&#xD;&#xA;$$&#xA;We may also write $n$ = $M/m_p$ for $m_p~=~\sqrt{\hbar c/G}$ the Planck mass.  The area is also proportional to the entropy $S~=~kA/4\ell^2$ and so the event horizon is also a measure of the number of degrees of freedom unavailable for observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These units are all expressed according to black hole horizons.  The Planck area is computed from the area of a de Broglie wave that is equal to its horizon area.  So these are expressed according to invariant (horizons), and are thus invarant.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-28T21:07:50.820" CommentCount="1" />
  <row Id="4100" PostTypeId="2" ParentId="4068" CreationDate="2011-01-28T21:27:19.927" Score="-2" ViewCount="" Body="&lt;p&gt;I think they have already given you the necessary references before me. I return to your phrase: &quot;... in addition I am wondering about attempts to define QFT non-perturbatively ...&quot;. I think the maximum that we may hope for is a partially taking into account some interactions. In this sense this part is non-perturbative. This may give us a better initial approximation (in/out states); the rest of interactions can be taken into account perturbatively. I started such an activity, see &lt;a href=&quot;http://vladimirkalitvianski.wordpress.com/2009/05/22/problem-of-infinitely-big-corrections/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;here&lt;/strong&gt;&lt;/a&gt;, but no renormalization group appears in my approach, sorry. Besides, it is not formalizing but reformulating QFT basing on physical phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please, don't kill me for that!&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-01-28T22:21:53.900" LastActivityDate="2011-01-28T22:21:53.900" CommentCount="0" />
  <row Id="4101" PostTypeId="2" ParentId="4067" CreationDate="2011-01-28T21:34:55.830" Score="2" ViewCount="" Body="&lt;p&gt;You may consider every excited state of an atom $\Psi_n$ as another particle. Then their number may be infinite even in this simple example. The higher energy of the system, the higher $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In QFT each particle has occupation numbers. During reactions some occupation numbers decrease, some other increase. Factually QFT equations are the balance equations governing the occupation numbers while interactions.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-01-28T22:19:32.513" LastActivityDate="2011-01-28T22:19:32.513" />
  <row Id="4102" PostTypeId="1" AcceptedAnswerId="4132" CreationDate="2011-01-28T21:44:12.993" Score="12" ViewCount="574" Body="&lt;p&gt;This question might be trivial or it might not even have a deep answer, but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the Lagrangian a function of the position and velocity (possibly also of time) and why are dependences on higher order derivatives (acceleration, ...) excluded?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a good reason for this or is it simply &quot;because it works&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sam&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2011-01-29T10:27:49.273" Title="Why are only derivatives to the first order relevant?" Tags="&lt;classical-mechanics&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="2" />
  <row Id="4105" PostTypeId="2" ParentId="4102" CreationDate="2011-01-28T22:06:37.117" Score="2" ViewCount="" Body="&lt;p&gt;Well, the usual physics in classical mechanics is formulated in terms of second-order differential equations. If you are familiar with the process of deriving Euler-Lagrange equations from the Lagrangian then it should be natural that the kinetic term must be proportional to $(\partial_t x)^2$ to reproduce that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'd considered more general Lagrangians (which you are certainly free to) you would obtain arbitrarily complicated equations of motions but these wouldn't correspond to anything physical. Nevertheless, some of those equations might describe some mathematical objects (because Lagrangian formalism and calculus of variations isn't inherent only to physics but also to lots of other mathematical disciplines).&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-01-28T22:06:37.117" CommentCount="2" />
  <row Id="4106" PostTypeId="2" ParentId="4102" CreationDate="2011-01-28T22:07:44.303" Score="-1" ViewCount="" Body="&lt;p&gt;If we assume, say, a second derivative in the Lagrangian, the Euler-Lagrange equations which minimize action &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A[q] = \int_{x_1}^{x_2}  L(x,q,q&amp;#39;,q&amp;#39;&amp;#39;)\, dx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial q&amp;#39;} + \frac{d^2}{dt^2}\frac{\partial L}{\partial q&amp;#39;&amp;#39;} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a fourth order differential equation. However, this can't be the case as we already know that $q&amp;#39;&amp;#39;=F/m$, i.e acceleration is determined by Force, which is &quot;outside&quot; the initial conditions. In a gravitational force field, for example, you know, a piori, the forces at every point in the system, and hence the acceleration at every point in the system is already known. A fourth order DE would lead to an internal inconsistency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deeper question to ask, I suppose, is why $F=mq&amp;#39;&amp;#39;$, not $F=mq&amp;#39;&amp;#39;&amp;#39;$ or $F=mq&amp;#39;&amp;#39;&amp;#39;&amp;#39;$. I won't pretend to know the answer to this, but I suspect there might be one.&lt;/p&gt;&#xA;" OwnerUserId="214" LastEditorUserId="214" LastEditorDisplayName="" LastEditDate="2011-01-28T22:12:55.583" LastActivityDate="2011-01-28T22:12:55.583" />
  <row Id="4107" PostTypeId="2" ParentId="4102" CreationDate="2011-01-28T22:21:33.703" Score="2" ViewCount="" Body="&lt;p&gt;There are formulations involving higher order derivatives, however, you made a fair characterization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think a rule of thumb would be to start looking for the simplest Lagrangian you can think of. In the general case, a good Lagragian should obey homogeneity of space, time and isotropy of space which means that it can't contain explicitly the position, time and velocity $\vec{v}$, respectively. Then, the simplest allowed possibility is to have a Lagrangian with a velocity squared. Since we don't need to look for more conditions to be fulfilled, there is no need to add terms involving higher derivatives or combinations of other terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see this procedure at work (quite a few times, actually) in Landau &amp;amp; Lifshitz, The Classical Theory of Fields.&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2011-01-28T22:21:33.703" CommentCount="4" />
  <row Id="4108" PostTypeId="1" CreationDate="2011-01-28T22:23:45.467" Score="6" ViewCount="406" Body="&lt;p&gt;I am aware that momentum is the thing which is conserved due to symmetries in space (rotational symmetry, translaitonal symmetry, etc). I am aware that in some systems, the generalized momentum, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_j = {\partial \mathcal{L} \over \partial \dot{q}_j}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is not simply $mv$. One example of this might be a charged particle in an electric field, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_j = mv_i + {q \over c} A_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone provide other examples where this is the case (simple, preferabily) and an intuition perhaps of why the &quot;old&quot; definition of momentum breaks down?&lt;/p&gt;&#xA;" OwnerUserId="214" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T14:24:13.530" LastActivityDate="2011-06-22T14:17:48.233" CommunityOwnedDate="2011-01-28T22:32:59.833" Title="Examples where momentum is not equal to $mv$?" Tags="&lt;classical-mechanics&gt;&lt;momentum&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="4109" PostTypeId="2" ParentId="4102" CreationDate="2011-01-28T22:29:06.983" Score="1" ViewCount="" Body="&lt;p&gt;Excellent question, and one that I've never really found a &lt;em&gt;completely&lt;/em&gt; satisfactory answer for. But consider this: in elementary classical mechanics, one of the fundamental laws is Newton's second law, $\mathbf{F} = m\mathbf{a}$, which relates the force on an object to the object's acceleration. Now, most forces are exerted &lt;em&gt;by&lt;/em&gt; one particular object &lt;em&gt;on&lt;/em&gt; another particular object, and the value of the force depends only on the positions of the source and &quot;target&quot; objects. In conjunction with Newton's second law, this means that, in a classical system with $N$ objects, each one obeys an equation of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ddot{\mathbf{x}}_i = \mathbf{f}(\{\mathbf{x}_j|j\in 1,\ldots,N\})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathbf{f}$ is some vector-valued function. The point of this equation is that, if you have the &lt;em&gt;positions&lt;/em&gt; of all the objects, you can compute the &lt;em&gt;accelerations&lt;/em&gt; of all the objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By taking the derivative of that equation, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\dddot{\mathbf{x}}}_i = \mathbf{f&amp;#39;}(\{\mathbf{x}_j\})\{\dot{\mathbf{x}}_j\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm getting quite loose with the notation here ;p) This allows you to compute the jerk (third derivative) using the positions and velocities. And you can repeat this procedure to get a formula (at least in some abstract sense) for any higher derivative. To put it in simple terms, since Newton's second law relates functions which are two orders of derivative apart, you only need the 0th and 1st derivatives, position and velocity, to &quot;bootstrap&quot; the process, after which you can compute any higher derivative you want, and from that any physical quantity. This is analogous to (and in fact closely related to) the fact that to solve a second-order differential equation, you only need two initial conditions, one for the value of the function and one for its derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The story gets more complicated in other branches of physics, but still, if you look at most of them you will find that the fundamental evolution equation relates the value of some function to its first and second derivatives, but no higher. For example, in quantum mechanics you have the Schrodinger equation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$i\hbar\frac{\partial\Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + U(x)\Psi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or in quantum field theory, the Klein-Gordon equation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{\partial^2\phi}{\partial t^2} + \frac{\partial^2\phi}{\partial x^2} - m^2\phi = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and others, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Maxwell%27s_equations&quot; rel=&quot;nofollow&quot;&gt;Maxwell's equations&lt;/a&gt; (equivalently, the wave equation that can be derived from them) in classical electromagnetism. In each case, you can use a similar argument to at least motivate the fact that only position or its equivalent field and its first derivative are enough to specify the entire state of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you might still wonder &lt;em&gt;why&lt;/em&gt; the equations that describe the universe relate functions that are only two derivatives apart, rather than three or four. That part is a mystery, but one that falls in the realm of philosophy rather than physics.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-01-28T22:29:06.983" CommentCount="1" />
  <row Id="4111" PostTypeId="2" ParentId="4108" CreationDate="2011-01-28T22:36:03.600" Score="1" ViewCount="" Body="&lt;p&gt;I think if you use non Cartesian coordinates in the expression &lt;strong&gt;p&lt;/strong&gt; = m &lt;strong&gt;v&lt;/strong&gt;, you will obtain something more generalized.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-01-28T22:36:03.600" CommunityOwnedDate="2011-01-28T22:36:03.600" CommentCount="3" />
  <row Id="4112" PostTypeId="2" ParentId="4102" CreationDate="2011-01-28T22:50:22.053" Score="2" ViewCount="" Body="&lt;p&gt;There are implications for causality when a equation of motion contains higher than second derivatives of the fields, EM radiation from charged bodies goes over the derivative of the acceleration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i don't know the details of WHY but this book should give more details:&#xA;(Causality and Dispersion Relations)&#xA;&lt;a href=&quot;http://books.google.com/books?id=QDzHqxE4anEC&amp;amp;lpg=PP1&amp;amp;dq=causality%20dispersion%20relations&amp;amp;pg=PP1#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=QDzHqxE4anEC&amp;amp;lpg=PP1&amp;amp;dq=causality%20dispersion%20relations&amp;amp;pg=PP1#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-01-28T22:50:22.053" />
  <row Id="4113" PostTypeId="2" ParentId="1557" CreationDate="2011-01-28T22:59:35.353" Score="0" ViewCount="" Body="&lt;p&gt;From SR, the speed of light is always $c$ in every intertial frame. Accelerating a particle to $c$ would mean the velocity of light wasn't $c$ in the frame of the particle. The Lorentz transformations ensure you can't do this where you can show that the relationship between the acceleration of the particle measured in the lab, $a$, and the particle's frame, $a&amp;#39;$, is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a&amp;#39; = \gamma^3 a$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the velocity of the particle approaches $c$, $\gamma$ approaches infinity and $a$ approaches zero for a finite $a&amp;#39;$.&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="1345" LastEditorDisplayName="" LastEditDate="2011-01-28T23:35:29.807" LastActivityDate="2011-01-28T23:35:29.807" />
  <row Id="4114" PostTypeId="2" ParentId="4108" CreationDate="2011-01-28T23:13:02.837" Score="1" ViewCount="" Body="&lt;p&gt;Momentum defined as $mv$ isn't conserved in collisions approaching $c$, whereas relativistic momentum, $\gamma mv$, is. Hence the redefinition of momentum.&lt;/p&gt;&#xA;" OwnerUserId="1345" LastActivityDate="2011-01-28T23:13:02.837" CommunityOwnedDate="2011-01-28T23:13:02.837" />
  <row Id="4115" PostTypeId="2" ParentId="4094" CreationDate="2011-01-28T23:29:06.083" Score="4" ViewCount="" Body="&lt;p&gt;A possible answer to the last part of the question: the article &lt;a href=&quot;http://ajp.aapt.org/resource/1/ajpias/v78/i9/p925_s1&quot; rel=&quot;nofollow&quot;&gt;Six Easy Roads to the Planck Scale&lt;/a&gt;, Adler, Am. J. Phys., 78, 925 (2010) contains multiple &quot;derivations&quot; that you might (or might not) find more satisfactory than the one you mention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the rest of the question is concerned, others have made the most relevant points. I think a fair summary of what Magueijo is getting at is something like the following: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One frequently hears that &quot;interesting new physics&quot; happens when some length $l$ is less than the Planck length. The Planck length is manifestly Lorentz invariant. The other length $l$, if it is the physical length of some object, is manifestly not Lorentz invariant. What meaning, then, can one assign to such statements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that reasonable people can differ over whether this is an interesting question. I don't find it manifestly insane, myself.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-01-28T23:29:06.083" CommentCount="6" />
  <row Id="4116" PostTypeId="1" AcceptedAnswerId="4117" CreationDate="2011-01-28T23:45:44.077" Score="7" ViewCount="310" Body="&lt;p&gt;I know this is an astronomy question, but no such stackexchange site exists. So here I am, asking about the physics of the solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the rotation period of the moon equals its revolution period. It's just so astonishing that these 2 values have such a small difference. I mean, what is the probability of these 2 values to be practically the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't believe this to be a coincidence. It's just too much for a coincidence. What could have caused this?&lt;/p&gt;&#xA;" OwnerUserId="1654" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-29T00:22:09.450" LastActivityDate="2011-03-10T15:29:28.790" Title="Why does the moon face earth with the same side?" Tags="&lt;astrophysics&gt;&lt;astronomy&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="4117" PostTypeId="2" ParentId="4116" CreationDate="2011-01-28T23:50:24.223" Score="7" ViewCount="" Body="&lt;p&gt;This is a gravitational phenomenon known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Tidal_lock&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;tidal lock&lt;/strong&gt;&lt;/a&gt;. It is closely related to the phenomenon of &lt;em&gt;tides&lt;/em&gt; on Earth, hence the name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tidal locking is an effect caused by the gravitational gradient from the near side to the far side of the moon. (That is, the continuous variation of the gravitational field strength across the Moon.) The end result is that the Moon rotates around its own axis with the same period as which it rotates around the Earth, causing the face of one hemisphere always to point towards the Earth.&lt;/p&gt;&#xA;" OwnerUserId="13" LastActivityDate="2011-01-28T23:50:24.223" CommentCount="5" />
  <row Id="4118" PostTypeId="1" AcceptedAnswerId="4127" CreationDate="2011-01-28T23:52:49.213" Score="5" ViewCount="326" Body="&lt;p&gt;Is the number of states in the Universe countable? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What framework could be used to answer the question in the title?&lt;/p&gt;&#xA;" OwnerUserId="405" LastActivityDate="2011-06-15T20:37:48.157" Title="How many bytes can the observable universe store?" Tags="&lt;cosmology&gt;&lt;quantum-information&gt;&lt;information&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="4120" PostTypeId="2" ParentId="4116" CreationDate="2011-01-29T00:19:40.480" Score="4" ViewCount="" Body="&lt;p&gt;Begin by imagining that the moon isn't quite a perfect sphere. One side is just a little bigger than the other. As the moon rotates, the heavier face will swing around towards the earth a little faster, and it will swing away from the earth a little slower, since it feels a stronger gravitational attraction via its larger mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since gravity is a conservative force, you might think that this continues forever - but it doesn't! The moon isn't totally rigid; rocks can slide around both on the surface and even inside the moon. The heavier lump actually slides &lt;em&gt;through&lt;/em&gt; the moon to try to stay facing the earth, which causes friction inside the moon. That friction heats up the rocks, but the heat is slowly lost into space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have a conservation of energy problem: the rotational energy of the moon is being converted to thermal energy in the rocks, and that thermal energy is being slowly leaked out of the system. The only resolution is that the rotation slowly stops over many millions of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, let's return to our assumption of a small mass imbalance. Is this true? Almost certainly - all we need is the fact that the moon isn't a truly &lt;em&gt;perfect&lt;/em&gt; sphere.&lt;/p&gt;&#xA;" OwnerUserId="1655" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-01-29T18:53:26.840" LastActivityDate="2011-01-29T18:53:26.840" CommentCount="5" />
  <row Id="4121" PostTypeId="1" CreationDate="2011-01-29T00:21:38.640" Score="3" ViewCount="235" Body="&lt;p&gt;In Weinberg: Gravitation and Cosmology chapter 2.10 (Relativistic Hydrodynamics) the speed of sound is derived as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_s^2 = \left(\frac{\partial p}{\partial \rho}\right)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the equation of state is given as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p=(\gamma - 1) (\rho - nm)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\gamma = 4/3$ for the case of a relativistic gas and $\gamma=5/3$ for the case of a nonrelativistic monoatomic gas ($nm$ is the rest mass density and $\rho$ the energy density). &#xA;Then Weinberg computes the sound speed for the relativistic case as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_s= \sqrt{\frac{1}{3}}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;in natural units and states that this is still safely less than unity (the speed of light). However what he doesn't show is the same calculation for the nonrelativistic case. This would yield &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_s^2 = \frac{\partial ((5/3 - 1) (\rho - nm))}{\partial \rho} = 2/3$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so the speed of sound in the non-relativistic case would actually be $v_s =\sqrt{\frac{2}{3}}$. So the speed of sound for a non-relativistic gas is actually higher than the speed of sound for an relativistic gas!? However one usually computes the non-relativistic speed of sound with another equation of state &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p=K\rho^{\gamma}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which yields &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_s^2=\frac{p}{\rho} = K \gamma \rho^{\gamma-1} = \gamma \frac{p}{\rho}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a completely different result compared to the one we get from the equation of state given by Weinberg. So is the equation of state given by Weinberg wrong? If so, what is the correct equation of state for a relativistic gas and what is the real maximum speed of sound for a relativistic gas? If not, what is wrong with my calculation of the non-relativistic sound speed based on the equation of state given by Weinberg? &lt;/p&gt;&#xA;" OwnerUserId="1648" LastActivityDate="2011-01-29T06:59:30.717" Title="How to compute the speed of sound in relativistic hydrodynamic?" Tags="&lt;special-relativity&gt;&lt;sound&gt;&lt;speed&gt;&lt;hydrodynamics&gt;" AnswerCount="1" />
  <row Id="4122" PostTypeId="2" ParentId="4118" CreationDate="2011-01-29T00:34:11.597" Score="0" ViewCount="" Body="&lt;p&gt;The de Sitter vacuum has a cosmological horizon at $r~=~\sqrt{3/\Lambda}$ for $\Lambda$ the cosmological constant.  This is about $10^{-54}cm^{-2}$  This puts the cosmological horizon at about $10^{10}ly$ or about $10^{18}cm$.  Now apply the Bekenstein bound $S~=~kA/4L_p^2$ for the horizon area $A~=~4\pi r^2$ or about $10^{37}cm^2$.  The entropy is $S~=~klog(N)$ and we divide by the Boltzmann constant to get the entropy according to bits $S_{bit}~\sim~10^{80}$.  Now consider that logarithm and take the exponential of this.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-29T00:34:11.597" CommentCount="3" />
  <row Id="4123" PostTypeId="1" CreationDate="2011-01-29T02:42:41.967" Score="0" ViewCount="153" Body="&lt;p&gt;We've got a hot star in the middle of a gas cloud. We point a spectrometer at the star, calculating the following attributes of our line of sight at the star through the cloud:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Total number of various elements and molecules (H, H2, C, O, N, CO, lots of ions of all of the above) along the line of sight (units are number/cm^2)&lt;/li&gt;&#xA;&lt;li&gt;Average temperature of the atoms in our line of sight&lt;/li&gt;&#xA;&lt;li&gt;Average radial velocity of the atoms in our line of sight&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All the radial velocities are roughly the same, so we infer that the line of sight is more or less a single coherent blob of gas. Good!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the star itself which illuminated the line of sight is moving about 40 km/s towards us, while the gas cloud chunk we observed is moving at around 20 km/s away. Also, from comparing the luminosity of the star to the temperature of the gas, we conclude that the gas cloud is a light year or two away from the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given all this information, here is your task: If we take another spectrum of the &lt;em&gt;same line of sight&lt;/em&gt; ten years from now, will it be the same blob of gas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will tell you right away that the radial velocity will only make the gas 0.002LY closer to the star, which is way less than what we need. So, in other words, given all the above, &lt;strong&gt;can you conclude anything about the transverse velocity of the cloud?&lt;/strong&gt; Can you tell anything about the bulk flow and determine whether it is probable that the gas in our line of sight has moved with respect to the star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My physical intuition says that you'd expect yes, since the radial velocities are so different, but I'm not totally sure - the radial velocity doesn't necessarily imply anything about the transverse velocity. Or does it?&lt;/p&gt;&#xA;" OwnerUserId="1655" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-01-31T17:09:17.957" LastActivityDate="2011-01-31T18:01:16.133" Title="Given temperature, composition, column density, and radial velocity, can I find the bulk flow of a gas cloud?" Tags="&lt;astrophysics&gt;&lt;fluid-dynamics&gt;&lt;statistical-mechanics&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="4125" PostTypeId="2" ParentId="3705" CreationDate="2011-01-29T05:08:25.760" Score="0" ViewCount="" Body="&lt;p&gt;As the problem is spherically symmetric, you know that the potential must go as $1/r$ in free space. Let $Q$ be the charge at the outer sphere and $Q&amp;#39;$ the charge at the inner sphere; then the potential must be expressed as $\frac{A\,(Q+Q&amp;#39;)}{r}$ for the space outside the outer shell ($r&gt;a$) and as $\frac{A\,Q&amp;#39;}{r}+K$ at the space between the inner shell and the outer shell ($a&gt;r&gt;b$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get the value of $Q&amp;#39;$ and $K$ by matching both expressions for the potential at $r=a$ (the potential doesn't jump at surface charge distributions) and by forcing a zero potential at $r=b$ (you have two equations with two unknowns).&lt;/p&gt;&#xA;" OwnerUserId="1656" LastActivityDate="2011-01-29T05:08:25.760" CommentCount="4" />
  <row Id="4126" PostTypeId="2" ParentId="4108" CreationDate="2011-01-29T06:19:19.647" Score="5" ViewCount="" Body="&lt;p&gt;the momentum is the quantity that is conserved as a result of the symmetry with respect to translations. This is the so-called Emmy Noether's theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only reason why we had $p=mv$ in classical mechanics was that the Lagrangian only depended on $\dot q$ by the kinetic energy $m\dot q^2/2$ term. By the differentiation, you may see that $p=m\dot q$. However, whenever the Lagrangian is different, the relationship between momentum and velocity gets modified. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You quoted one example. Another example is any theory in relativity where&#xA;$$ p = \frac{m_0 v}{\sqrt{1-v^2/c^2}} $$&#xA;which may also be derived from the Lagrangian. Note that in the relativistic case, the Lagrangian is (minus) the proper time of the world line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases, the momentum gets modified but people have disagreed what it is. A recently discussed example here was the Abraham-Minkowski controversy about the momentum of a photon in a dielectric material:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/3189/is-the-abraham-minkowski-controversy-resolved&quot;&gt;Is the Abraham-Minkowski controversy resolved?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For photons, the momentum is $p=E/c$ in the vacuum - this itself is different from massive slow particles - but this relationship can get multiplied or divided by the index of refraction $n$, and arguments arose from this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In field theory, one has totally different formulae for the momentum. There are no particles i.e. no velocities of a few objects. Instead, the momentum is distributed in the field - e.g. electromagnetic field - and every region of space carries a momentum density. For example, the momentum density of the electromagnetic field is $\vec E\times \vec H$ or $\vec D\times \vec B$ - which of them is &quot;true&quot; was a famous controversy, equivalent to the controversy about the momentum of a single photon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other fields obviously have other formulae for the momentum density.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="2151" LastEditorDisplayName="" LastEditDate="2011-06-21T20:51:16.823" LastActivityDate="2011-06-21T20:51:16.823" CommunityOwnedDate="2011-01-29T06:19:19.647" CommentCount="2" />
  <row Id="4127" PostTypeId="2" ParentId="4118" CreationDate="2011-01-29T06:32:25.540" Score="3" ViewCount="" Body="&lt;p&gt;Dear J.F. Sebastian, for a few decades, physicists thought that the cosmic microwave background - photons created 350,000 years after the Big Bang that fill the space and currently correspond to thermal radiation at 2.7 Kelvin degrees - carried most of the entropy of the Universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Entropy, denoted $S$, is the physical quantity representing the information that can be carried by the &quot;arrangement of atoms&quot; or other microscopic building blocks. It's the &quot;useless information&quot; that we can't measure in isolation and that is responsible for thermodynamic phenomena. Most of the information cannot be decoded - the question what part of this entropy may be employed as memory in functioning memory chips is an engineering question but it is a small part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The entropy's unit is &quot;Joule per Kelvin&quot;. However, if you divide it by $k$, the Boltzmann's constant, you get a dimensionless number that measures the information in &quot;nats&quot;. One bit is equal to $\ln(2)\approx 0.69$ nats, so the dimensionless entropy is approximately the same thing as the number of (useless) bits that the atoms carry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Microwave background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the entropy of the cosmic microwave background?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The approximate &quot;radius&quot; of the visible Universe is $10^{60}$ Planck lengths, so the volume is $10^{180}$ Planck lengths. However, the CMB temperature is just $10^{-32}$ Planck temperatures or so, which means that the volume has to be counted in units of volume that are $10^{96}$ times larger. We get about $10^{180-96}\approx 10^{85}$ photons in CMB and each of them carries one bit or so. &lt;strong&gt;So the CMB entropy is about $10^{85}$ bits.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Large black holes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it was found that most galaxies store a huge black hole in their center and black holes actually maximize the entropy that can be squeezed into a fixed volume, or carried by a fixed amount of bound mass. At our galactic center, the Sgr A* black hole has mass about 4 million solar masses or $10^{37}$ kg which is about $10^{45}$ Planck masses, so the radius is also about $10^{45}$ Planck lengths and the area is $10^{90}$ Planck areas, producing $10^{90}$ bits of entropy just from the single black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because there are approximately $10^{11}$ galaxies in the Universe, &lt;strong&gt;we get $10^{101}$ bits of entropy carried by the galactic black holes&lt;/strong&gt; which is much higher than the CMB entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cosmic holographic bound on entropy&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be fundamental to mention that the ultimate entropy of the Universe is bound by the area of the de Sitter horizon in Planck units. The radius of the horizon is about $10^{60}$ Planck lengths so the area is $10^{120}$ Planck areas. &lt;strong&gt;The largest entropy that our Universe may carry is therefore about $10^{120}$ bits.&lt;/strong&gt; In some sense, we may say that this huge entropy is already &quot;out there&quot; at the cosmic horizon today - but we may attribute it to &quot;everything that is behind the horizon&quot; and we don't see. However, there could be lots of matter inside the horizon and its entropy could approach those $10^{120}$ bits as well. In particular, a single black hole that would grow enough to almost touch the cosmic horizon (which would shrink along the way) would carry almost $10^{120}$ bits, too. However, such a black hole will never exist, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that Lawrence, who calculated the final bound, ended up with a figure is wrong by 40 orders of magnitude. His numbers have errors at each step. First of all, the radius of the cosmic horizon is $10^{28}$ centimeters rather than $10^{18}$ centimeters he wrote; he apparently forgot to add the speed of light or to distinguish seconds and years. This gave him the first 20 orders of magnitude of mistake. He made two more errors that produced the remaining 20 orders of magnitude.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-29T07:28:59.357" LastActivityDate="2011-01-29T07:28:59.357" CommentCount="0" />
  <row Id="4128" PostTypeId="2" ParentId="4121" CreationDate="2011-01-29T06:59:30.717" Score="3" ViewCount="" Body="&lt;p&gt;Dear asmaier, sometimes it's useful to look into the real world to avoid some simple mistakes. The actual speed of sound in the air is 340 meters per second which is about one millionth of the speed of light. The squared speed of sound is one trillionth of the squared speed of light, so your claim that the non-relativistic speed of sound is comparable to the speed of light is surely incorrect, isn't it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first formula by Weinberg that you quoted is universally valid but you apply it incorrectly. Well, Weinberg doesn't make these errors so that his first $\sqrt{1/3}$ result for the relativistic gas is correct, too. However, in particular, $(\rho-nm)$ is supposed to measure the energy above the latent energy of $E=mc^2$, because only this &quot;purely kinetic energy&quot; contributes to the pressure; it must still be multiplied by $(\gamma-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, $(\rho-nm)$ is negligibly small in the non-relativistic limit. So your $c\sqrt{2/3}$ for the nonrelativistic speed of sound is clearly invalid. In fact, $(\rho-nm)$ measures the kinetic energy of the molecules, $mv^2/2$; the factor of $m$ cancels. Because it will get differentiated and multiplied by $(\gamma-1)$, and we take the square root at the end, you may see that the speed of sound comes out approximately equal to the speed of molecules in the air, just slightly smaller than that (mainly because of the factor of $1/2$ in the kinetic energy $mv^2/2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that the (root mean square) speed of air molecules is 500 meters per second and the speed of sound is 340 meters per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no contradiction between the relativistic formulae and the non-relativistic ones, which are the limits of the former, and the non-relativistic materials' speed of sound is vastly smaller than the speed of light.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-29T06:59:30.717" />
  <row Id="4129" PostTypeId="1" CreationDate="2011-01-29T07:37:33.810" Score="3" ViewCount="511" Body="&lt;p&gt;A quick wikipedia search will reveal the many different approaches to the measurement problem in quantum theory.&#xA;What I find strange is that they even make up all these weird hypothesis like true randomness, many worlds, bohm and other interpretations when there isn't any ToE they're working it out from.&#xA;I'm just a layman so maybe my assumption is wrong, but wont a fundamental theory inevitably change quantum theory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is really a realistic model where superpositions are nothing but lack of information on our part not likely to come from the ToE?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my layman mind it seems premature to even begin to try to answer these questions if a theory of everything isn't in place.&#xA;Or is there actually a reason that people are trying to solve this now?&#xA;Such as: no matter what the ToE is, quantum theory wont change, superpositon will always exist etc.?&lt;/p&gt;&#xA;" OwnerUserId="1658" LastActivityDate="2011-02-01T10:11:23.000" Title="Could the measurement problem be solved by string theory / other ToE?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="5" FavoriteCount="2" />
  <row Id="4130" PostTypeId="2" ParentId="4129" CreationDate="2011-01-29T07:56:41.117" Score="2" ViewCount="" Body="&lt;p&gt;Dear Qurious, for all practical and experimentally testable purposes, the measurement theory was understood in the late mid 1920s, soon after quantum mechanics was discovered, and the &quot;measurement problem&quot; was solved at the level of phenomenology. The old interpretation of quantum mechanics left some questions open but from an empirical viewpoint, they were &quot;academic questions&quot; only. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later progress, especially the derivations of decoherence in the 1980s, has confirmed the Copenhagen picture and derived the classical-quantum boundary, a missing piece in the old Copenhagen approach, from the rules of many-body quantum mechanics itself. All problems related to the measurement theory have been answered in principle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;People may prefer different philosophies to interpret quantum mechanics - consistent histories; many worlds; the old positivist Copenhagen interpretation - but they ultimately agree about the predictions for all experiments so their philosophies are physically equivalent. What's important is that the predictions may be made for every situation now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So the measurement problem can't be solved by quantum field theory, string theory, or anything else, because it has been solved for a long time before these newer layers of physical knowledge started to be studied.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum field theory, string theory, theories describing states of condensed matter physics, nuclear physics, atomic and molecular physics, optics, and other subdisciplines of physics that depend on the quantum phenomena take the universal postulates of quantum mechanics as exactly valid principles that cannot be modified in any way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern developments in the research of entanglement, original started by Einstein and his collaborators (especially &lt;a href=&quot;http://cid-9cd81cfa06ff7718.office.live.com/self.aspx/.Public/entan-interpret.pdf&quot; rel=&quot;nofollow&quot;&gt;Bell's inequalities&lt;/a&gt;, &lt;a href=&quot;http://motls.blogspot.com/2010/11/sidney-coleman-quantum-mechanics-in.html#ghzm&quot; rel=&quot;nofollow&quot;&gt;GHZM state&lt;/a&gt;, and &lt;a href=&quot;http://physics.stackexchange.com/questions/4040/weak-measurement-and-hardys-paradox/4043#4043&quot;&gt;Hardy's paradox&lt;/a&gt;), have eliminated all doubts that the new quantum description of the reality has to be taken seriously and all &quot;classical&quot; models of the quantum phenomena may be falsified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So string theory - and even quantum field theory - don't change anything whatsoever about the shared framework of quantum mechanics and its postulates. The question whether our Universe follows the known laws of quantum mechanics has been answered, de facto for 85 years, and most likely, nothing will ever change about these basic matters again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no real problem over here and the postulates of the quantum dynamical framework below will stay with us. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The world is associated with a Hilbert space&lt;/li&gt;&#xA;&lt;li&gt;Every complex linear superposition of two allowed vectors is allowed&lt;/li&gt;&#xA;&lt;li&gt;Every observable is associated with a linear Hermitean operator&lt;/li&gt;&#xA;&lt;li&gt;The squared absolute values of probability amplitudes - complex inner products with eigenvectors of the observables - determine the probabilities that one or another result will be observed&lt;/li&gt;&#xA;&lt;li&gt;Nothing except for the probabilities may be predicted; it is incorrect to imagine that any physical system that could be affected by quantum mechanics has any well-defined, determined properties (variable quantities) prior to the measurement&lt;/li&gt;&#xA;&lt;li&gt;Evolution is encoded in the Hamiltonian or the action, via the Schrödinger's equation, Heisenberg equations of motion for the operators, Feynman's path integral, or another equivalent mathematical formalism&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What quantum field theory and string theory are doing is to find and investigate the more correct Hilbert spaces and the &quot;Hamiltonians&quot; (or equivalent structures) acting on them that define the dynamics. There are also people who study theories based on the basic assumption of &quot;denial of quantum mechanics&quot; but they have nothing to do with research in particle physics or string theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it is hypothetically conceivable that sometime in the future, one (or several) of the postulates will be falsified experimentally. If that occurs, quantum field theories, string theory, and many other theories in nuclear, condensed matter, and other branches of physics will be showed invalid or incomplete, too. But the &quot;fear&quot; that this could happen doesn't mean that science shouldn't build on the assumptions that seem to be valid according to the current state of the art. If the &quot;fear&quot; were enough to stop research, science would become impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;There is nothing &quot;premature&quot; about taking the principles of quantum mechanics as established facts, especially because they have been known for 85 years and every single experiment that has been done in those 85 years has confirmed that they are correct.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In his excellent 1994 talk &lt;a href=&quot;http://media.physics.harvard.edu/video/index.php?id=SidneyColeman_QMIYF.flv&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Quantum Mechanics In Your Face&lt;/em&gt;&lt;/a&gt;, eminent physicist Sidney Coleman compared the resistance to quantum mechanics to the resistance to the heliocentric theory. People used to say that the geocentric beliefs (much like the belief that the world has to be classical) existed because it &quot;looks like&quot; the world is geocentric (or classical). How would it look like if the world were heliocentric (or quantum)? Well, it would look like the real world. Welcome home. By the way, Coleman's talk was given 17 years ago but most people still haven't accepted quantum mechanics as a fact.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-29T09:00:58.663" LastActivityDate="2011-01-29T09:00:58.663" CommentCount="28" />
  <row Id="4131" PostTypeId="1" CreationDate="2011-01-29T10:20:29.983" Score="6" ViewCount="392" Body="&lt;p&gt;Here is a scan from an old Soviet textbook for school children: &lt;img src=&quot;http://i.stack.imgur.com/HCdJh.png&quot; alt=&quot;http://storage5.static.itmages.ru/i/11/0129/h_1296295703_dda36c65eb.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It shows the table of quarks and antiquarks of different generations, colours, spins.&#xA;The book also includes similar tables of gluons and their interactions with quarks and between themselves.&#xA;They all are coded in this fancy cake-shape code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard model table includes Z and W-bosons, gluons as well as graviton and gravitino.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you think about such manner of teaching physics and in general, what do you think about appropriateness of teaching Standard Model in junior school?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some other images from the book (quark-gluon interactions, hadron decay):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/1NZlv.png&quot; alt=&quot;quarks-gluon interaction&quot;&gt;&lt;img src=&quot;http://i.stack.imgur.com/1OuOp.png&quot; alt=&quot;decay of hadrons&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Illustration in a chapter about quantum parity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/mXrKX.png&quot; alt=&quot;quantum parity&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tha book has chapters about quantum chromodynamics, spontaneous symmetry breaking, quantum parity and so on.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastEditorUserId="1664" LastEditorDisplayName="" LastEditDate="2011-01-29T18:39:59.473" LastActivityDate="2011-01-29T18:39:59.473" ClosedDate="2011-01-29T21:32:28.573" Title="What do you think about teaching Standard Model in school?" Tags="&lt;particle-physics&gt;&lt;education&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="1" />
  <row Id="4132" PostTypeId="2" ParentId="4102" CreationDate="2011-01-29T10:27:49.273" Score="12" ViewCount="" Body="&lt;p&gt;I reproduce a blog post I wrote some time ago:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We tend to not use higher derivative theories. It turns out that there &#xA;  is a very good reason for this, but that reason is rarely discussed in &#xA;  textbooks. We will take, for concreteness, $L\left(q,\dot q, \ddot &#xD;&#xA;  q\right)$, a Lagrangian which depends on the 2nd derivative in an &#xA;  essential manner. Inessential dependences are terms such as $q\ddot q$ &#xA;  which may be partially integrated to give ${\dot q}^2$. Mathematically, &#xA;  this is expressed through the necessity of being able to invert the &#xA;  expression $$P_2 = \frac{\partial L\left(q,\dot q, \ddot &#xD;&#xA;  q\right)}{\partial \ddot q},$$ and get a closed form for $\ddot q &#xD;&#xA;  \left(q, \dot q, P_2 \right)$. Note that usually we also require a &#xA;  similar statement for $\dot q \left(q, p\right)$, and failure in this &#xA;  respect is a sign of having a constrained system, possibly with gauge &#xA;  degrees of freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the non-degeneracy leads to the Euler-Lagrange equations in &#xA;  the usual manner: $$\frac{\partial L}{\partial q} - &#xD;&#xA;  \frac{d}{dt}\frac{\partial L}{\partial \dot q} + &#xD;&#xA;  \frac{d^2}{dt^2}\frac{\partial L}{\partial \ddot q} = 0.$$ This is then &#xA;  fourth order in $t$, and so require four initial conditions, such as &#xA;  $q$, $\dot q$, $\ddot q$, $q^{(3)}$. This is twice as many as usual, and &#xA;  so we can get a new pair of conjugate variables when we move into a &#xA;  Hamiltonian formalism. We follow the steps of Ostrogradski, and choose &#xA;  our canonical variables as $Q_1 = q$, $Q_2 = \dot q$, which leads to &#xA;  \begin{align} P_1 &amp;amp;= \frac{\partial L}{\partial \dot q} - &#xA;  \frac{d}{dt}\frac{\partial L}{\partial \ddot q}, \ P_2 &amp;amp;= &#xA;  \frac{\partial L}{\partial \ddot q}. \end{align} Note that the &#xA;  non-degeneracy allows $\ddot q$ to be expressed in terms of $Q_1$, $Q_2$ &#xA;  and $P_2$ through the second equation, and the first one is only &#xA;  necessary to define $q^{(3)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can then proceed in the usual fashion, and find the Hamiltonian &#xA;  through a Legendre transform: \begin{align} H &amp;amp;= \sum_i P_i \dot{Q}_i - &#xA;  L \ &amp;amp;= P_1 Q_2 + P_2 \ddot{q}\left(Q_1, Q_2, P_2\right) - L\left(Q_1, &#xA;  Q_2,\ddot{q}\right). \end{align} Again, as usual, we can take time &#xA;  derivative of the Hamiltonian to find that it is time independent if the &#xA;  Lagrangian does not depend on time explicitly, and thus can be &#xA;  identified as the energy of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we now have a problem: $H$ has only a linear dependence on &#xA;  $P_1$, and so can be arbitrarily negative. In an interacting system this &#xA;  means that we can excite positive energy modes by transferring energy &#xA;  from the negative energy modes, and in doing so we would increase the &#xA;  entropy &amp;mdash; there would simply be more particles, and so a need to &#xA;  put them somewhere. Thus such a system could never reach equilibrium, &#xA;  exploding instantly in an orgy of particle creation. This problem is in &#xA;  fact completely general, and applies to even higher derivatives in a &#xA;  similar fashion.&lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-01-29T10:27:49.273" />
  <row Id="4134" PostTypeId="2" ParentId="4129" CreationDate="2011-01-29T10:52:44.660" Score="-3" ViewCount="" Body="&lt;p&gt;&quot;I believe in determinism ...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What a determinism can be while dealing with a very complicated thing, with many degrees of freedom? Determinism arises when you make average of some reality.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-01-29T13:09:09.647" LastActivityDate="2011-01-29T13:09:09.647" CommentCount="7" />
  <row Id="4135" PostTypeId="2" ParentId="4131" CreationDate="2011-01-29T10:59:48.563" Score="4" ViewCount="" Body="&lt;p&gt;Of course, we don't know the age of the soviet schoolchildren here but I think we can be confident that they wouldn't have the background to understand the quark model. So it's just sterile dogma - empty facts for memorisation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crucial point which distinguishes physics, and science in general is that it's experimental. The theory is not taken on trust or authority, in the end it has passed experimental tests. This is the critical insight that school students need to absorb, through doing experiments and seeing how the theory they're taught is anchored in practical experience. Even future theorists need to understand that lesson.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a way to introduce 20th century ideas such as QM and RT in such a way at school level I applaud it, but based on the experience of TV popular science I'm sadly skeptical.&lt;/p&gt;&#xA;" OwnerUserId="1394" LastActivityDate="2011-01-29T10:59:48.563" CommentCount="22" />
  <row Id="4136" PostTypeId="2" ParentId="4131" CreationDate="2011-01-29T11:04:25.263" Score="0" ViewCount="" Body="&lt;p&gt;where do You live, which countries schools do You have in mind? &#xA;There are countries, where it was fine if children had &lt;strong&gt;some&lt;/strong&gt; school. &#xA;There is a big country in western hemisphere, where more and more &#xA;religious nonsense is thought. &#xA;And to Your questions core: &#xA;99.9 % of physics is outside the capacity of &quot;junior&quot; schools. &#xA;Of course printers of schoolbooks woud appreciate to include &#xA;Your topic in new (sic!) books, I think it was more important &#xA;if Juniors get a &lt;strong&gt;solid base&lt;/strong&gt;.&#xA;Georg&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-01-29T11:04:25.263" />
  <row Id="4137" PostTypeId="1" AcceptedAnswerId="4142" CreationDate="2011-01-29T11:35:00.997" Score="3" ViewCount="400" Body="&lt;p&gt;I have read about the existence of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Nonlinear_Schr%C3%B6dinger_equation&quot; rel=&quot;nofollow&quot;&gt;non-linear scrhödinger equation&lt;/a&gt;. What is its utility and application? And how can it be derived? Is it in a relativistic or non-relativistic context?&lt;/p&gt;&#xA;" OwnerUserId="558" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-29T18:44:22.603" LastActivityDate="2011-01-29T18:44:22.603" Title="Non-linear Schrödinger equation" Tags="&lt;quantum-mechanics&gt;&lt;mathematical-physics&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="4138" PostTypeId="2" ParentId="4129" CreationDate="2011-01-29T11:47:17.410" Score="1" ViewCount="" Body="&lt;p&gt;The &quot;measurement problem&quot; connected with QM indeed gains much of its longevity because of the randomness that QM introduced into a fundamental physics that had previously been successful within its domain and (seemingly) deterministic. Physicists had worked with Statistical theories (where the statistics &lt;em&gt;did&lt;/em&gt; come from ignorance of details) before, however, and have learned to adapt as none of this Interpretational debate affects the basic equations (like Schrodinger's).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A key part of the question is &quot;wont a TOE/fundamental theory change quantum theory?&quot;. A simple answer is that we dont know: it might, but it might not touch quantum theory - it will have other work to do. A more subtle aspect of this is that in physics when a replacement theory comes along it has to respect the successes (gained over many decades or centuries) of its predecessor. As quantum theory is quite mathematical such a new theory also has to &quot;respect&quot; many of the theorems now associated with quantum theory as well as the experimental results. It is this factor which makes discussion on this topic lengthy: as some theorems (but not all such theorems) occasionally get finessed. Example theorems include: Bell's Theorem, Kochen-Specker Theorem - now Kochen-Specker-Conway. Conway has recently (controversially) introduced a concept like &quot;free will&quot; to explain a theorem!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These type of theorems all incorporate the sort of properties and issues of superposition that the question mentions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally one (perhaps even quite likely) possibility is that the TOE will be just a set of (experimentally validated) equations. Then interpretational questions may still remain. No doubt modified by the knowledge of exactly what equations work, but still subject to (what will then be described as philosophical) debate.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-01-29T11:47:17.410" CommentCount="2" />
  <row Id="4139" PostTypeId="2" ParentId="4131" CreationDate="2011-01-29T11:59:45.390" Score="3" ViewCount="" Body="&lt;p&gt;First of all I should say that I find this particular way of presenting quarks rather unfortunate. Not only it delivers very little physics information, but it's even quite misleading (I mean those fancy shapes). On the other hand, I do believe that teaching some modern physics at high school is extremely important. Of course the students do not have the background to understand how quantum mechanics works - maybe apart from the semiclassical approaches such as Bohr quantization which should be avoided at any cost since it only leads to confusion. Of course, as Nigel Seel pointed out, the students should absorb the theory-experiment relation and understand their role. However, an equally (and sometimes even more) important part of the education of possible future scientists is a motivation for choosing a research career. Thus, the students should get in touch with modern physics as soon as possible so that they are not discouraged by the fact that they have to go through many years of study of theoretical background that was developed more than 100 years ago. In fact, one can get their hands on and have a lot of fun with physics without really understanding deeply the underlying theory. I expect objections that classical mechanics and all the other standard high school material can also be made interesting, yet I think that for the sake of motivation, nothing can beat showing the students a (though tiny) part of the actual current research.&lt;/p&gt;&#xA;" OwnerUserId="1641" LastActivityDate="2011-01-29T11:59:45.390" CommentCount="6" />
  <row Id="4140" PostTypeId="2" ParentId="4137" CreationDate="2011-01-29T12:16:45.817" Score="2" ViewCount="" Body="&lt;p&gt;You can get some more information about the derivation from the related Wikipedia article about the Gross-Pitaevskii equation which is used in the context of Bose-Einstein condensation. There, one starts with a quantum field theory whose Hamiltonian is nonlinear in the fields, and replaces the field operator by a classical function that describes the condensate. So the reason why the same equation appears in so many different systems is that they are all described by an effective field theory with a nonlinear (fourth-order) interaction term. (The form of this effective theory is dictated by symmetry and the relevant degrees of freedom at low energy.) In the semiclassical approximation, this effective field theory reduces to the nonlinear &quot;Schroedinger&quot; equation.&lt;/p&gt;&#xA;" OwnerUserId="1641" LastActivityDate="2011-01-29T12:16:45.817" CommentCount="6" />
  <row Id="4141" PostTypeId="1" AcceptedAnswerId="4144" CreationDate="2011-01-29T12:23:02.107" Score="5" ViewCount="599" Body="&lt;p&gt;Dirac once said that he was mainly guided by mathematical beauty more than anything else in his discovery of the famous Dirac equation. Most of the deepest equations of physics are also the most beautiful ones e.g. Maxwell's equations of classical electrodynamics, Einstein's equations of general relativity. Beauty is always considered as an important guide in physics. My question is, can/should anyone trust mathematical aesthetics so much that even without experimental verification, one can be fairly confident of its validity? (Like Einstein once believed to have said - when asked what could have been his reaction if experiments showed GR was wrong - Then I would have felt sorry for the dear Lord)&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T15:36:11.217" LastActivityDate="2011-02-02T15:36:11.217" Title="Why beauty is a good guide in physics?" Tags="&lt;soft-question&gt;" AnswerCount="7" CommentCount="10" FavoriteCount="2" />
  <row Id="4142" PostTypeId="2" ParentId="4137" CreationDate="2011-01-29T12:35:41.010" Score="6" ViewCount="" Body="&lt;p&gt;Broadly speaking nonlinear Schrödinger equations are used to describe wave propagation in nonlinear media. There are many examples, like deep-water waves (where the linear shallow water approximation to the Navier-Stokes equations is not valid), or nonlinear optics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is for example an effect called &quot;Kerr effect&quot; in optical fibers where the refractive&#xA;index depends on the intensity of the optical pulse. You can find a detailed derivation of a nonlinear Schrödinger equation describing electromagnetic waves in such a medium in the book&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;M. J. Ablowitz, B. Prinari, A. D. Trubatch: &quot;Discrete and Continuous Nonlinear Schrödinger Systems&quot;, Cambridge University Press, 2004&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Edit: Fluids/Hydrodynamics: It is not easy to find a reference for the use of Schrödinger equations in Hydrodynamics, maybe because most authors write for a target audience that is not fluent in quantum mechanics, I don't know. But here is one:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Andrew Majda and Andrea Bertozzi: &quot;Vorticity and Incompressible Flow&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;chapter 7.1, &quot;Simplified Asymptotic Equations for Slender Vortex Filaments: The Self-Induced Approximation, Hasimoto's Transform and the Nonlinear Schrödinger Equation&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditorDisplayName="" LastEditDate="2011-01-29T15:37:19.787" LastActivityDate="2011-01-29T15:37:19.787" CommentCount="3" />
  <row Id="4143" PostTypeId="2" ParentId="3468" CreationDate="2011-01-29T12:36:41.633" Score="0" ViewCount="" Body="&lt;p&gt;It will look much like having placed am arm in a laser beam. I once gave blood for analysis and the medic used a laser penetrator to get the blood rather than a needle. It was quick and without any pain, the skin cells evaporated in a moment.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastActivityDate="2011-01-29T12:36:41.633" CommentCount="1" />
  <row Id="4144" PostTypeId="2" ParentId="4141" CreationDate="2011-01-29T12:49:27.660" Score="4" ViewCount="" Body="&lt;p&gt;Dear sb1, a good question. Well, beauty is a good guide in physics research but only for those whose sense of beauty is aligned with Nature's sense of beauty. ;-) Dirac was among them, at least when he was writing down his beautiful equation, but many others have a different sense of beauty that can easily lead them off the track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right sense of beauty is linked to the equations' rigidity and uniqueness. If a woman is beautiful, you may think that not a single thing could be improved about her. Every correction would damage this beauty. The same thing holds for the beautiful theories and equations in physics that simply &quot;fit together&quot;. An important characteristic that can make a theory more rigid and constrained is symmetry - but it is not the only characteristic that can do so. For example, the nontrivial cancellation of various a priori conceivable theoretical problems - such as anomalies - also constrains theories and makes them &quot;prettier&quot; relatively to theories that haven't had to pass any similar theoretical tests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are the equations and theories that &quot;fit together&quot; more likely to be the right description of Nature? Well, unless they're already falsified, they have many fewer parameters waiting to be adjusted than the competing - so far unfalsified - theories that are not so beautiful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;posterior&quot; (after the comparison with the reality) probability that the &quot;not so beautiful&quot; candidate theory is valid is, by the Bayesian logic, multiplied by the probability $P(g=g_0)$ that the parameters $g$ take the right values to agree with the reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the overall prior probability for the &quot;beautiful&quot; and &quot;ugly&quot; classes of theories are chosen to be equal, then the &quot;ugly&quot; theory is punished by the extra factor of $P(g=g_0)$, so it becomes less likely that it is the right theory that describes the observations. A more constrained point in the space of theories (constrained by symmetries and special consistency advantages) gets a &quot;higher weight&quot; because it's qualitatively different from the more &quot;generic&quot; or &quot;uglier&quot; points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nature has apparently chosen some repeatable laws that apply everything in the Universe (and maybe beyond) and that predict millions of phenomena from a very small amount of information about the laws that has to be known in advance. So it makes sense to extrapolate this observation and assume that the laws of physics are as constrained as possible, and in this sense, they must &quot;fit together&quot; and be &quot;beautiful&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But again, one has to be very careful about this method to look for theories that becomes very unscientific unless the &quot;beauty&quot; of the mathematical structures may be justified by some technical arguments.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-29T12:56:16.887" LastActivityDate="2011-01-29T12:56:16.887" CommentCount="1" />
  <row Id="4145" PostTypeId="1" CreationDate="2011-01-29T12:55:59.167" Score="-1" ViewCount="158" Body="&lt;p&gt;Particular solutions, or solutions in general, can be thought of as a combination of waves within a finite period of time.  The creation of the waves are generally constrained by limits of physics of oscillators.  We also typically associate certain types of solutions with number of particles and certain types of oscillators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the definable finiteness, it would seem that one could create a mathematical space that was sufficiently large such that each conceivable solution produced by oscillators in the universe would be defined as a point in that space.  So in some wild way, it is in fact the determination and mapping of the space of solutions that that creates the solutions, and it is merely up to a group of scientists to later discover the points of interest in that space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not being a scientist myself, what sort of physical parameters and variables would I consider to define the space of solutions in physics?&lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="1383" LastEditorDisplayName="" LastEditDate="2011-02-01T11:07:02.413" LastActivityDate="2011-02-01T11:07:02.413" ClosedDate="2011-01-29T15:58:33.977" Title="&quot;How can you map or define the space of solutions?&quot;" Tags="&lt;applied-physics&gt;&lt;acoustics&gt;" AnswerCount="2" CommentCount="10" />
  <row Id="4146" PostTypeId="1" CreationDate="2011-01-29T13:11:34.603" Score="3" ViewCount="177" Body="&lt;p&gt;I haven't seen any reference which explains these things and I am not sure of all the steps of the argument or the equations. I am trying to reproduce here a sequence of arguments that I have mostly picked up from discussions and I would like to know of references for the background details and explanations. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It seems that in general the Witten Index (${\cal I}$) can be written as $ Tr (-1)^F \prod_i x_{i}^{C_i}$ where $x_is$ are like the fugacities of the conserved quantity $C_is$ and $C_is$ form a complete set of simultaneously measurable operators.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know the motivation for the above and especially about calling the above an ``index&quot; even if it depends on the fugacities which doesn't seem to be specified by the Lagrangian of the theory or by the underlying manifold? Even with this dependence does it reproduce some intrinsic property of either the underlying manifold or the theory? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Say $\phi$ and $\psi$ are the bosonic and fermionic component fields of a superfield and ${\cal D}$ be the superderivative. Now when the above tracing is done over operators of the kind $Tr({\cal D}^n\phi)$ and $Tr({\cal D}^n\psi)$ it is probably called a &quot;single trace letter partition function (STLP)&quot; and when it also includes things like products of the above kind of stuff it is called a ``multi-trace letter partition function (MTLP)&quot; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would like to know about the exact/general definitions/references for the above terminologies and  the motivations behind them.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Probably for adjoint fields, if $f(x)$ is a STLP then apparently in some limit (&quot;large N&quot;?) one has, MTLP = $\prod _{n=1} ^{\infty} \frac{1}{1-f(x^n)}$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;One also defines something called the ``full STLP (FSTLP)&quot; where one probably includes terms also of the kind $Tr({\cal D}^n \phi ^m {\cal D}^p \psi ^q)$ and then the MTLP can be gotten from that as MTLP = $exp [ \sum _{m=1} ^ {\infty} \frac{1}{m} FSTLP]$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The upshot all this is probably to show that MTLP is the same as Witten Index (in some limit?). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I don't understand most of the above argument and I would be happy to know of explanations and expository references which explain the above concepts (hopefully beginner friendly!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I see these most often come up in the context of superconformal theories and hence references along that might be helpful especially about the representations of the superconformal group.)  &lt;/p&gt;&#xA;" OwnerUserId="1637" LastEditorUserId="1637" LastEditorDisplayName="" LastEditDate="2011-01-30T10:00:30.223" LastActivityDate="2011-01-30T14:58:56.220" Title="Witten Index and letter partition function" Tags="&lt;quantum-field-theory&gt;&lt;conformal-field-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="4147" PostTypeId="1" CreationDate="2011-01-29T13:22:30.250" Score="11" ViewCount="234" Body="&lt;p&gt;Let us consider the following &lt;a href=&quot;http://en.wikipedia.org/wiki/Thought_experiment&quot; rel=&quot;nofollow&quot;&gt;Gedankenexperiment&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cylinder &lt;strong&gt;rotates&lt;/strong&gt; symmetric around the $z$ axis with &lt;a href=&quot;http://en.wikipedia.org/wiki/Angular_velocity&quot; rel=&quot;nofollow&quot;&gt;angular velocity&lt;/a&gt; $\Omega$ and a plane wave with $\mathbf{E}\text{, }\mathbf{B} \propto e^{\mathrm{i}\left(kx - \omega t \right)} $ gets scattered by it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We assume to know the &lt;em&gt;isotropic&lt;/em&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Permittivity&quot; rel=&quot;nofollow&quot;&gt;permittivity&lt;/a&gt; $\epsilon(\omega)$ and &lt;a href=&quot;http://en.wikipedia.org/wiki/Permeability_%28electromagnetism%29&quot; rel=&quot;nofollow&quot;&gt;permeability&lt;/a&gt; $\mu(\omega)$ of the cylinder's material &lt;em&gt;at rest&lt;/em&gt;. Furthermore, the cylinder is infinitely long in $z$-direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The static problem ($\Omega = 0$) can be treated in terms of &lt;a href=&quot;http://en.wikipedia.org/wiki/Mie_theory&quot; rel=&quot;nofollow&quot;&gt;Mie Theory&lt;/a&gt; - here, however, one will need a &lt;strong&gt;covariant&lt;/strong&gt; description of the system for very fast rotations (which are assumed to be possible) causing nontrivial transformations of $\epsilon$ and $\mu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence my question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h3&gt;What is the scattering response to a plane wave on a fastly rotating cylinder?&lt;/h3&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.personal.uni-jena.de/~p3firo/PhysicsSE/RotatingDisc.png&quot; alt=&quot;RotatingDisc&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance  &lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-31T02:32:14.743" LastActivityDate="2011-03-02T12:33:06.307" Title="Covariant Description of Light Scattering at a fastly rotating Cylinder" Tags="&lt;electromagnetism&gt;&lt;special-relativity&gt;&lt;electrodynamics&gt;&lt;scattering&gt;" CommentCount="11" FavoriteCount="4" />
  <row Id="4148" PostTypeId="2" ParentId="4146" CreationDate="2011-01-29T13:25:51.530" Score="3" ViewCount="" Body="&lt;p&gt;Dear Anirbit, a good question. To figure out the answer, it's a good idea to realize why people talk about indices in the first place. What is a particular formula called the index good for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, it's good because it doesn't change under continuous deformations. Consequently, there may be many methods to calculate the same number - and, which is related, many ways to interpret this number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in your previous question about short multiplets,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/4031/superconformal-theories&quot;&gt;Superconformal theories&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;all of us discussed what the index is and why it is invariant. All answers to your new question are already written there. More or less. The purpose of the index is that it is invariant under continuous changes of parameters. That includes the changes of your variables $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The terms $\ln(x_i) C_i$ - which exponentiate to $x_i^{C_i}$ - may be absent in your Lagrangian or Hamiltonian but you may always add them to the Lagrangian or Hamiltonian. So they &lt;em&gt;are&lt;/em&gt; examples of the continuous parameters that don't change the value of the index. As long as it is a full-fledged index, the result will be independent of all quantities $x_i$. The more quantities $x_i$ you add, the more ways to calculate the index may be found. So you may try to calculate them for $x=1$, $x=0$, $x=\infty$, or another value (or combination of values, to be more general).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's useful to mention that the index is sometimes generalized to similar concepts such as &quot;elliptic genus&quot; which are only invariant under the deformation of some parameters but they explicitly depend on others, usually holomorphically or in a similar way. They're not just &quot;numbers&quot; describing the theory but whole &quot;functions&quot;. Such expressions are useful in theories where the short multiplets may still possess some quantities that may change continuously. So the elliptic genus doesn't just &quot;count&quot; the short multiplets which is what the full-fledged index essentially does (with some weights); instead, it counts some function of the quantities that the short multiplets possess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, in each case, one has to prove that the quantity is invariant under the transformations of some kinds and it depends on others.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-29T13:25:51.530" CommentCount="2" />
  <row Id="4149" PostTypeId="2" ParentId="4131" CreationDate="2011-01-29T13:31:48.267" Score="1" ViewCount="" Body="&lt;p&gt;It's chronologically impossible to write about M-theory in Soviet textbook. The good thing is when bright kids have access to information and can ask knowledgeable people the questions. &lt;/p&gt;&#xA;" LastActivityDate="2011-01-29T13:31:48.267" CommentCount="1" />
  <row Id="4150" PostTypeId="2" ParentId="4145" CreationDate="2011-01-29T13:40:37.893" Score="-1" ViewCount="" Body="&lt;p&gt;I premise that your question is a bit different to normal question about physics here, For a composer (I'm a composer for example... ;-) ), characters of sound for our hearing are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-intensity: that is connected to intensity of the wave (amplitude, frequency, density of air...and you must include the fact that the response of our ears is logarithmic (Weber-Fechner law)...);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-lenght;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-height: that is connected to frequency, but here the human ears have logarithmic &#xA;response...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-tone-colour: a violin has a different sound to a piano...this depend from fourier spectrum of the waves produced from instruments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These 4 charachters are specified by composer with musical notation. So, because they are associated with physical and measurable magnitudes, you can associate them to numbers. &#xA;If we consider only  intensity, lenght and height, we associate every wound at a point with values of these parametres as coordinates in a space (I think it will have strange topologycal proprieties, that sure won't vectorial space...). A melody is rapresented so with a set of points in this space.&#xA;When you have many voices (many instruments, or a piano or a organ that are polyphonic instrument) you must consider cartesian product of many of that spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tone colour complicate the problem, because the number of its fourier armonycs is infinite;-) But I think that with a laborious work you can define it in this situation too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question is so connected with another question: &quot;maths is invented or discovered?&quot; I think the first option... ;-) &lt;/p&gt;&#xA;" OwnerUserId="558" LastEditorUserId="558" LastEditorDisplayName="" LastEditDate="2011-01-29T14:31:14.100" LastActivityDate="2011-01-29T14:31:14.100" />
  <row Id="4151" PostTypeId="2" ParentId="4137" CreationDate="2011-01-29T14:03:25.110" Score="1" ViewCount="" Body="&lt;p&gt;The nonlinear Schrödinger equation (NLSE) in one dimension is&#xA;$$&#xD;&#xA;i\frac{\partial\psi}{\partial t}~+~\frac{\partial^2\psi}{\partial x^2}~+~f(|\psi|)\psi~=~0.&#xD;&#xA;$$&#xA;For $f(|\psi|)~=~|psi|^2$ this is the cubic Schrödinger equation.  This is related to a couple of relativistic equations, such as the quartic equation for the Higgs field.  The Dirac equation $i\gamma^\mu\partial_\mu\psi~+~({\bar\psi}\gamma^\mu \psi)\gamma_\mu\psi$ is the Thirring model of fermion condensates, related to BCS theory. This is a related nonlinear wave equation.  In general the NLSE describes the motion of quantum waves through nonlinear media or in some process where the locality of standard Lagrangians (quadratic etc) is replaced with some nonlocal field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A separable solution to the NLSE $\psi(x,t)~=~u(x)v(t)$ may be found with $v(t)~=~exp(-i\omega t)$ and the spatial function is given by a DE $u_{xx}~+~\omega u$ $f(|u|)u~=~0$defined as&#xA;$$&#xD;&#xA;\int\frac{du}{\sqrt{Au^2~-~2F(u)~+~B}}~=~C~\pm~x,~F(u)~=~\int uf(|u|)du&#xD;&#xA;$$&#xA;which is a traveling wave solution.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="724" LastEditorDisplayName="" LastEditDate="2011-01-29T16:38:29.403" LastActivityDate="2011-01-29T16:38:29.403" CommentCount="2" />
  <row Id="4152" PostTypeId="2" ParentId="4129" CreationDate="2011-01-29T14:14:29.140" Score="9" ViewCount="" Body="&lt;p&gt;We know from the successful experiments demonstrating Bell's inequality and entanglement that the universe is not classical, but quantum mechanical. This means that it behaves in very non-intuitive ways. The &quot;measurement problem&quot; arises because we can't formulate a good intuitive picture of what's going on. One way that we try to explain it, the &quot;Copenhagen interpretation,&quot; makes calculations relatively easy, but has a mysterious step, &quot;measurement,&quot; which eludes our intuition. Explaining this measurement step is known as the &quot;measurement problem,&quot; and since it is fundamentally a problem with our intuition, there may be no solution (note that adherents of the many-worlds interpretation and the de Broglie-Bohm pilot-wave interpretation claim to have solved the measurement problem, but in each case they have replaced it with something equally abhorrent to our intuitions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum mechanics isn't just a theory of the universe, but really a framework used for constructing all sorts of theories (the Standard model, string theory, non-relativistic quantum mechanics). &#xA;For a theory of everything, there are two possibilities. The first is that it's still fundamentally quantum mechanical, in which case the situation will remain much the same and the measurement problem will not go away. The second is that it is fundamentally something other than quantum mechanical. Because it still has to approximate quantum mechanics, it's not going to behave according to our classical intuition. I think this means that it would very likely be even more non-intuitive than quantum mechanics, and the measurement problem would be replaced by something even more difficult to understand.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditorDisplayName="" LastEditDate="2011-01-29T14:20:21.960" LastActivityDate="2011-01-29T14:20:21.960" CommentCount="13" />
  <row Id="4153" PostTypeId="2" ParentId="4145" CreationDate="2011-01-29T14:22:28.370" Score="-2" ViewCount="" Body="&lt;p&gt;This is a philosophical question on the lines of: &quot;how many monkeys would it take and for how long to reproduce the works of Shakespeare&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also borders on the question of whether consciousness can be described with mathematical tools or it is a meta level to the mathematical descriptions. Example of meta-level: the alphabet and &quot;Romeo and Juliet&quot;. The latter is a meta level to the level of alphabet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that sense the songs are meta levels to sounds and words, and the rules of meta levels are different from the rules of the base. Example from physics: statistical mechanics was found to be the base level of Thermodynamics, each framework having independent axiomatic and mathematical descriptions. So even though the particles participating in an ensemble   define a mathematical space on which thermodynamics is based, the thermodynamic description has a mathematical  order with independent solutions appropriate to the thermodynamic view. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same holds for songs from your hypothetical space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Discovery&quot; in the everyday sense has no meaning because one would have to explore zillions of ensembles to hit on one that has a good solution for a song. Probability 0 .&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-01-29T14:22:28.370" CommentCount="2" />
  <row Id="4154" PostTypeId="2" ParentId="4068" CreationDate="2011-01-29T15:14:41.470" Score="2" ViewCount="" Body="&lt;p&gt;The question here is how to organize a response. Names can be given, but I hope I can give some conceptual order. Hopefully other answers will find other ways to organize their response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wightman axioms &lt;em&gt;are&lt;/em&gt; classic. I take the approach here of organizing how other approaches sit with the Wightman axioms, even though they may not be axiomatic. A useful critique of the Wightman axioms can be found in Streater R F, Rep. Prog. Phys. 1975 38 771-846. More recent is the assessment of Fredenhagen, Rehren, and Seiler, “Quantum Field Theory: Where We Are” in &lt;a href=&quot;http://arxiv.org/abs/hep-th/0603155&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0603155&lt;/a&gt;, which I recommend. In my scheme here, however, the approach you particularly ask about, formalizations of the renormalization group, do not figure, because, as you say, they have a completely different starting point. I’d say that the starting point is perhaps the concept of Feynman integrals rather than the renormalization group in itself, but I’d also say that that’s a quibble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a large question of what we hope to achieve by axiomatizing. (1) We can loosen the axioms, so that we have more models, some of which might be useful in Physics, but we have to figure out which on a case-by-case basis. This makes Engineering somewhat quixotic. (2) We can tighten the axioms, with the ambition that all the models are useful in Physics, but some Physically useful models might be ruled out. Mathematicians are often happy to work with axioms that a Physicist would consider too tight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the Wightman axioms, more-or-less in Haag’s presentation in “Local Quantum Physics”:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;The state space&lt;/strong&gt;&#xA;(a) is a separable Hilbert space. There are people trying to use non-associative algebras, amongst other things. &#xA;(b) which supports a representation of the Poincaré group. There are people doing QG, QFT on CST, and many ways of breaking Lorentz symmetry at small scales.&#xA;(c) There is a unique Poincaré invariant state. Thermal sectors don’t satisfy this. Non-unique vacuums are an old favorite, but the vacuum state is pervasive in Particle Physics.&#xA;(d) The spectrum of the generator of translations is confined to the closed forward light-cone. This is an elephant, IMO. The underlying reason for this is “stability”, which has no axiomatic formulation. The belief that the spectrum condition is necessary for stability may rely on classical thinking, particularly on the primacy of the Hamiltonian or Lagrangian. Feynman integrals for loops introduce negative frequencies, however, so there’s something of a case against it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;The observables&lt;/strong&gt;&#xA;(which, implicitly, correspond in some way to statistics of experimental data)&#xA;(a) Are operator-valued distributions.&#xA;People have introduced other Generalized function spaces. Haag-Kastler tightens this, to bounded operators, but the mapping from space-time regions to operators is looser. In Particle Physics, the S-matrix, which discusses transitions between free field states on time-like hyperplanes at t=+/-infinity, has been the supreme observable for decades: trying to reconcile this with the Lorentz invariant operator-valued distributions of the Wightman axioms pretty much killed the latter. Condensed matter Physics, optics, etc., take correlation functions quite seriously, which seems to me to be at the heart of the split between Particle and other Physicists. Another elephant.&#xA;(b) Are Hermitian.&#xA;There’s a complex structure. People have also introduced quaternions in various ways.&#xA;(c) The fields transform under the Poincaré group. This goes with 1b.&#xA;(d) The observables are jointly measurable at space-like separation, but in general are not jointly measurable at time-like separation. Stepping away from the Poincaré group almost always results in violation of this axiom. Random fields, which are always jointly measurable operator-valued distributions, and the differences between them and QFT, are something I have published on.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The Haag-Kastler approach to some extent brings the states and observables into the single structure of von Neumann algebras, but essentially the distinction of linear operators and their duals remains. Refusing to split the world into states and observables, which we might call “holism”, makes Physics almost impossible. There’s always the question of exactly where the Heisenberg cut should be put, but pragmatically we just put it &lt;em&gt;somewhere&lt;/em&gt;. Bell tries to square that circle while still doing Physics in his ‘Against “Measurement” ’, and Bohm as much as left Physics behind. There are people trying to do that kind of thing, but I find very little of it useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Returning back to earth, there’s also a question of how we deform a system that we have managed to construct so that it’s significantly different in interesting ways. This isn’t in the axioms, but the standard has been to deform the Hamiltonian or Lagrangian. Both methods, however, require a choice of one or two space-like hypersurfaces, which goes against the spirit of the Poincaré group. Algebraic deformations, the other known alternative (others?), have hardly left the ground because the constraints of positive energy, microcausality, and the primacy of the S-matrix have hitherto ruled them out (I have also published on this, based on Lie fields from the 1960s). If we deform the algebra of observables instead of the dynamics, the question arises of what &quot;stability&quot; might be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is of course a question whether one ought to start from the Wightman axioms at all, but one has to choose somewhere. Then, with Lee Smolin, one has to set off into the valleys, hoping to find a bigger hill.&#xA;Best wishes.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-01-29T15:14:41.470" />
  <row Id="4155" PostTypeId="2" ParentId="4141" CreationDate="2011-01-29T15:27:30.990" Score="2" ViewCount="" Body="&lt;p&gt;Steven Weinberg has an interesting comment on this subject:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Aesthetically motivated simplicity&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Einstein made what from the&#xA;  perspective of today's theoretical&#xA;  physics is a deeper mistake in his&#xA;  dislike of the cosmological constant.&#xA;  In developing general relativity, he&#xA;  had relied not only on a simple&#xA;  physical principle--the principle of&#xA;  the equivalence of gravitation and&#xA;  inertia that he had developed from&#xA;  1907 to 1911--but also on a sort of&#xA;  Occam's razor, that the equations of&#xA;  the theory should be not only&#xA;  consistent with this principle but&#xA;  also as simple as possible. In itself,&#xA;  the principle of equivalence would&#xA;  allow field equations of almost&#xA;  unlimited complexity. Einstein could&#xA;  have included terms in the equations&#xA;  involving four spacetime derivatives,&#xA;  or six spacetime derivatives, or any&#xA;  even number of spacetime derivatives,&#xA;  but he limited himself to second-order&#xA;  differential equations.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This could have been defended on&#xA;  practical grounds. Dimensional&#xA;  analysis shows that the terms in the&#xA;  field equations involving more than&#xA;  two spacetime derivatives would have&#xA;  to be accompanied by constant factors&#xA;  proportional to positive powers of&#xA;  some length. If this length was&#xA;  anything like the lengths encountered&#xA;  in elementary-particle physics, or&#xA;  even atomic physics, then the effects&#xA;  of these higher derivative terms would&#xA;  be quite negligible at the much larger&#xA;  scales at which all observations of&#xA;  gravitation are made. There is just&#xA;  one modification of Einstein's&#xA;  equations that could have observable&#xA;  effects: the introduction of a term&#xA;  involving no spacetime derivatives at&#xA;  all--that is, a cosmological constant.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But Einstein did not exclude terms&#xA;  with higher derivatives for this or&#xA;  for any other practical reason, but&#xA;  for an aesthetic reason: They were not&#xA;  needed, so why include them? And it&#xA;  was just this aesthetic judgment that&#xA;  led him to regret that he had ever&#xA;  introduced the cosmological constant.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Since Einstein's time, we have learned&#xA;  to distrust this sort of aesthetic&#xA;  criterion. Our experience in&#xA;  elementary-particle physics has taught&#xA;  us that any term in the field&#xA;  equations of physics that is allowed&#xA;  by fundamental principles is likely to&#xA;  be there in the equations. It is like&#xA;  the ant world in T. H. White's &lt;em&gt;The&#xA;  Once and Future King&lt;/em&gt;: Everything that&#xA;  is not forbidden is compulsory.&#xA;  Indeed, as far as we have been able to&#xA;  do the calculations, quantum&#xA;  fluctuations by themselves would&#xA;  produce an infinite effective&#xA;  cosmological constant, so that to&#xA;  cancel the infinity there would have&#xA;  to be an infinite &quot;bare&quot; cosmological&#xA;  constant of the opposite sign in the&#xA;  field equations themselves. Occam's&#xA;  razor is a fine tool, but it should be&#xA;  applied to principles, not equations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Physics Today&lt;/em&gt;, Nov. 2005, p.32)&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2011-01-29T15:27:30.990" CommentCount="3" />
  <row Id="4156" PostTypeId="1" AcceptedAnswerId="4159" CreationDate="2011-01-29T15:38:26.913" Score="7" ViewCount="446" Body="&lt;p&gt;Recently I talked about QFT with another physicist and mentioned that the Quantum Field Theory of a fermion is a quantisation of its one-particle quantum mechanical theory. He denied this and responded that he rather sees the single particle QM as the non-relativistic limit of a QFT. He elaborated that the energies encountered are all much smaller than the particles mass, so we can ignore all multi-particle excitations in Fock Space and get an effective Hilbert space consisting of all single-particle excitations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In turn, I asked what the corresponding limit of QED quantum mechanics of the massless photon is, and he of course responded that there can't be a nonrelativistic limit of QED exactly because of the masslessness. But there is classical ED, the classical limit of QED.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is taking the classical or the nonrelativistic limit the same, or does one include the other, or is there some deep difference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question &lt;a href=&quot;http://physics.stackexchange.com/questions/850/what-does-a-field-theory-mean&quot;&gt;What does a Field Theory mean?&lt;/a&gt; has something to do with it, but does not fully answer my question.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2011-07-27T23:20:57.797" Title="Are Classical Field Theory and Quantum Mechanics of a single particle (nonrelativistic or &quot;classical&quot;) limits of Quantum Field Theory?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;special-relativity&gt;&lt;electrodynamics&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="4157" PostTypeId="2" ParentId="4141" CreationDate="2011-01-29T15:57:00.357" Score="3" ViewCount="" Body="&lt;p&gt;Although I agree, beauty can be seductive and may be related to our evolutionary development of &quot;pattern forming&quot; in the brain--we seem inclined to find symmetry beautiful.  Effective theories are often ugly as sin---take a look at the Standard Model Lagrangian and tell me it is beautiful :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://nuclear.ucdavis.edu/~tgutierr/files/stmL1.html&quot; rel=&quot;nofollow&quot;&gt;http://nuclear.ucdavis.edu/~tgutierr/files/stmL1.html&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Download the plain version of the Standard Model Lagrangian Density: [ps][pdf][tex][txt]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Or download a &quot;fun yet soul-crushing exam question&quot; based on it: [ps][pdf][tex][txt]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was tempted to ask the exam question here to see what response it evoked :)&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-01-30T22:57:11.600" LastActivityDate="2011-01-30T22:57:11.600" CommentCount="5" />
  <row Id="4158" PostTypeId="2" ParentId="4156" CreationDate="2011-01-29T15:58:33.637" Score="4" ViewCount="" Body="&lt;p&gt;Some quick remarks (I hope that these are helpful to formulate more precise versions of your question):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;Recently I talked about QFT with another physicist and mentioned that the Quantum Field Theory of a fermion is a quantisation of its one-particle quantum mechanical theory.&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think you meant to say that the Hilbert space of the QFT is constructed as the fermionic Fock space from the one particle Hilbert space, which is usually called &lt;strong&gt;second&lt;/strong&gt; quantization, right?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;He elaborated that the energies encountered are all much smaller than the particles mass, so we can ignore all multi-particle excitations in Fock Space and get an effective Hilbert space consisting of all single-particle excitations.&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Particle creation is only one aspect of special relativity that is melded into quantum mechanics by quantum field theory. You also have to handle the non-zero probability of a particle described by nonrelativistic quantum mechanics to propagate with a velocity greater than c (the vacuum velocity of light). Take a free massive spin-zero boson in one dimension localized at a point $x$ at a time $t=0$, the time evolution is described by a Schrödinger equation. At an arbitrary small time $t \gt 0$ the wavefunction will be a Gaussian, which means there is a non-zero probability to find the particle at an arbitrary distance from $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore any limit $\lim_{c \to \infty}$ has to look at the dynamics as well, not only at the state space. &#xA;In what sense the Schrödinger equation describes the dynamics of elementary particles with nonzero mass in the limit $\lim_{c \to \infty}$ is explained in every textbook of QFT that I know of.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;...he of course responded that there can't be a nonrelativistic limit of QED exactly because of the masslessness.&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Yes, that's just true, because the classical theory (Maxwell's equations) is already relativistic.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;So is taking the classical or the nonrelativistic limit the same, or does one include the other, or is there some deep difference?&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm sorry but I don't understand this question. From a pure formal point of view, the classical limit is $\lim_{h \to 0}$, and the nonrelativistic limit is $\lim_{c \to \infty}$, and the examples that you cite yourself make already clear that this is not the same.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-01-29T15:58:33.637" />
  <row Id="4159" PostTypeId="2" ParentId="4156" CreationDate="2011-01-29T16:01:16.360" Score="9" ViewCount="" Body="&lt;p&gt;Dear Turion, the Dirac quantum field may be formally obtained by quantizing the Dirac equation which is a &lt;em&gt;relativistic&lt;/em&gt; but single-particle quantum mechanical equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The non-relativistic limit of the single-particle Dirac equation is the Pauli equation which is essentially the non-relativistic Schrödinger equation for a wave function with an extra 2-fold degeneracy describing the spin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get from the non-relativistic Schrödinger equation for an electron to a quantum field theory with a quantized Dirac field, you therefore&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Have to add the spin and going to the Pauli equation - easy&lt;/li&gt;&#xA;&lt;li&gt;Guess the right relativistic generalization of the Pauli equation - it's the single-particle Dirac equation that also has negative-energy solutions&lt;/li&gt;&#xA;&lt;li&gt;Realize that the negative-energy solutions are inconsistent in the single-particle framework, so you have to second-quantize the wave function and obtain the Dirac quantum field&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This sequence of steps is formal. One can't really &quot;deduce&quot; things in this order, at least not in a straightforward way. After all, the step 1 needed a creative genius of Pauli's caliber, the step 2 needed a creative genius of Dirac's caliber, and the step 3 needed a collaboration of dozens of top physicists who developed quantum field theory. Quite on the contrary, as you were correctly told, the meaningful well-defined operations go exactly in the opposite order - but it doesn't follow the steps above. You begin with the quantum field theory, including the Dirac field, which is the right full theory, and you may take various limits of it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The non-relativistic limit is of course something totally different than the classical limit. The nonrelativistic limit is still a quantum theory, with probabilities etc. - but it doesn't respect the special role of the speed of light. On the other hand, the classical limit is something totally different - a classical deterministic theory that respects the Lorentz symmetry, and so on. Let us discuss the limits of quantum electrodynamics separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Classical limits&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classical, $\hbar\to 0$, limit of QED acts differently on fermions and bosons. The bosons like to occupy the same state. However, to &quot;actually&quot; send $\hbar$ to zero, you need quantities with the same units that are much larger than $\hbar$: $\hbar$ goes to zero relatively to them. What quantities may you find? Well, the electromagnetic field may carry a lot of energy in strong fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you get a classical limit by having many photons in the same state. They combine into classical electromagnetic fields that are governed by classical Maxwell's equations; note that classical Maxwell's equations are &quot;already&quot; relativistic although people before Einstein hadn't fully appreciated this symmetry (although Lorentz wrote the &quot;redefinition&quot; down without realizing its relationship with the different inertial frames or symmetry groups, for that matter). You just erase hats from the similar Heisenberg equations for the electromagnetic fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, for extremely high frequencies, the number of photons won't be large because they carry a huge energy. So for high frequencies, you may also derive another classical limit - based on &quot;pointlike particles&quot;, the photons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fermions, e.g. electrons described by the Dirac equation, obey the exclusion principle. So you can't have many of them. There is at most one particle per state. In the quantum mechanical theories, it has an approximate position and momentum that don't commute. The classical limit is where they do commute. So the classical limit must inevitably produce mechanics for the fermions - with positions and momenta of individual particles. As I mentioned, this picture may be relevant for high-energy bosons, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Nonrelativistic limit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The non-relativistic, $c\to\infty$, limit of QED is something completely different. It is still a quantum theory. Because the photons propagate by the speed of light and the speed is sent to infinity, the electromagnetic waves propagate infinitely quickly in the non-relativistic limit. That means that the charged (and spinning or moving charged) objects instantly influence each other by electric (and magnetic) fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to fermions, you undo one of the steps at the beginning: you reduce the speed of the electrons. Assuming there are no positrons for a while, the non-relativistic limit where the velocities are small will prevent you from the creation of fermion-antifermion pairs. So the number of particles will be conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it makes sense to decompose the Hilbert space into sectors with $N$ particles for various values of $N$ and you're back in multi-body quantum mechanics. They will also have the spin, as in the Pauli equation, and they will interact via instantaneous interactions - the Coulomb interaction and its magnetic counterparts (combine the Ampére and Biot-Savart laws for $B$ induced by currents with the usual magnetic forces acting on moving charges and spins). You will get the usual non-relativistic quantum mechanical Hamiltonian used for atomic physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There will be no waves because they move by the infinite speed. You won't be able to see them. But they won't destroy the conservation of energy etc. because in the non-relativistic limit, the power emitted by accelerating charges goes to zero because it contains $1/c^3$ or another negative power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the non-relativistic limit, the photons just disappear from the picture, and their only trace will be the instantaneous interactions of the Coulomb type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Classical non-relativistic limit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you may apply both limiting procedures at the same moment. Then you get non-relativistic point-like classical electrons interacting via the Coulomb and similar instantaneous interactions.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-29T16:15:59.027" LastActivityDate="2011-01-29T16:15:59.027" CommentCount="3" />
  <row Id="4160" PostTypeId="1" CreationDate="2011-01-29T16:17:25.097" Score="0" ViewCount="173" Body="&lt;p&gt;The pedestrian way I have asked this masks its intent. It is &quot;fun stuff&quot;. Ambrose Bierce, a contemporary of Mark Twain, wrote &quot;The Devil's Dictionary&quot; published in 1911 under that title.&#xA;&lt;a href=&quot;http://www.alcyone.com/max/lit/devils/&quot; rel=&quot;nofollow&quot;&gt;http://www.alcyone.com/max/lit/devils/&lt;/a&gt;&#xA;An example from it is- &#xA;&quot;Conservative --(noun) A statesman who is enamoured of existing evils, as distinguished from the Liberal, who wishes to replace them with others.&quot; &#xA;I was wondering if folks would like to submit examples from physics of words designed to confuse the lay public and provide fodder for journalists and New Age authors---an example would be, say, &quot;charm&quot;. Or examples of misuse of physics words or concepts--eg Deepak Chopra &quot;Quantum Healing&quot;. I hope you find this strange question about colorful language appealing and not without charm; if not, we will need a curmudgeon label :)&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-01-29T16:36:34.160" LastActivityDate="2011-01-29T16:36:34.160" ClosedDate="2011-01-29T17:16:47.000" Title="What are some examples of nonstandard word use in physics?" Tags="&lt;fun-stuff&gt;" CommentCount="10" />
  <row Id="4161" PostTypeId="2" ParentId="4048" CreationDate="2011-01-29T16:32:35.557" Score="1" ViewCount="" Body="&lt;p&gt;There's an interesting paper on the subject, written at a lower level than some, on my website here:&lt;br&gt;&#xA;&lt;a href=&quot;http://brannenworks.com/plavchan_feynmancheckerboard.pdf&quot; rel=&quot;nofollow&quot;&gt;http://brannenworks.com/plavchan_feynmancheckerboard.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It notes that there are issues with the particle in 3 dimensions apparently being superluminal with a speed of at least $\sqrt{3}\;c$.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-29T16:32:35.557" />
  <row Id="4162" PostTypeId="2" ParentId="4048" CreationDate="2011-01-29T16:33:35.603" Score="2" ViewCount="" Body="&lt;p&gt;In 1+1 dimensional spacetime you can split the operator $d^{2}/dt^{2} - d^{2}/dx^{2}$ in a very simple way into a product: $(d/dt - d/dx)(d/dt + d/dx)$. This is not possible in higher dimensions.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-01-30T01:33:59.943" LastActivityDate="2011-01-30T01:33:59.943" CommentCount="3" />
  <row Id="4163" PostTypeId="2" ParentId="4131" CreationDate="2011-01-29T16:35:45.877" Score="3" ViewCount="" Body="&lt;p&gt;While I agree with @Sigel arguments, that the substructure of matter should be delayed until the appropiate age for understanding its process of discovery, I think that it should exist an appropiate way to answer the question &lt;em&gt;if the kids ask it&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can not just answer &quot;&lt;em&gt;turtles until the end&lt;/em&gt;&quot; not to appeal to mistery or legend. And while we can explain that the technical details of the argument will be addressed at later ages, we must still stress that they can be addressed by the use of reason and experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that the teacher could need some support material in order to focus on how to answer this question -substructure of matter- if it is asked. The booklet above contains some good ideas, but it should never be a main teaching material, it should be support material.&lt;/p&gt;&#xA;" OwnerUserId="1335" LastEditorUserId="1335" LastEditorDisplayName="" LastEditDate="2011-01-29T16:52:30.657" LastActivityDate="2011-01-29T16:52:30.657" />
  <row Id="4164" PostTypeId="2" ParentId="3468" CreationDate="2011-01-29T17:02:36.987" Score="1" ViewCount="" Body="&lt;p&gt;In 2003, there was an accident in the Tevatron similar to what is being asked here. One of the CDF Roman Pots went inadvertently down into the beam. It actually just touched the halo because, I guess, everything shut off automatically after that. I remember seeing some pictures back then... you may want to google it.&lt;/p&gt;&#xA;" OwnerUserId="1337" LastActivityDate="2011-01-29T17:02:36.987" CommentCount="1" />
  <row Id="4165" PostTypeId="2" ParentId="2249" CreationDate="2011-01-29T17:57:00.553" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Using the resistors analogy seen in the image above, applying a CIP is equivalent to setting the resistors in parallel for each of the spin orientations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's only true if the spacing between FM layers is large. In order for the current in plane configuration to work, the spacing must be less than the mean free path so that electrons traversing the device are scattered by adjacent FM layers. If the adjacent FM layers are anti-parallel, both spin up and spin down will spend some time in a high scattering region. If they are parallel, one spin direction will always see high scattering in the magnetic layers, while the other will always see low scattering. The latter is the low resistance channel that gives rise to the change in resistance of the device. &lt;/p&gt;&#xA;" OwnerUserId="443" LastActivityDate="2011-01-29T17:57:00.553" CommentCount="1" />
  <row Id="4166" PostTypeId="2" ParentId="3584" CreationDate="2011-01-29T19:30:51.537" Score="-2" ViewCount="" Body="&lt;p&gt;I believe a simpler and logically equivalent question would be: Is is possible to accelerate a massive object to the point that it will become a black hole? I examine the simplest definition of a black hole: an object from which light cannot escape or it's escape velocity is greater than the speed of light. Since no massive object can be accelerated to c- light will always be able to escape (or be reflected from such an object). This as you said was agreed upon. If we have two such &quot;ultra-relativistic&quot; objects we can easily determine the physics from the rest frame of one of the objects. That is- one billiard ball (chosen at rest) will see the other billiard ball moving by at the appropriate Lorentz adjusted velocity, no instant black hole. You're question specifically says they pass by and do not collide. From this I am assuming that you are trying to allow the combined kinetic energies of the objects to allow it's transition to a black hole. This brings us back to the beginning of my argument. Now- if a collision were to occur- that could be a different story. For example, collisions in the LHC will allow the energy of motion to change into mass and vice a versa.&lt;br&gt;&#xA;And yes- the gravitational field does remain the same regardless of velocity. Einstein's Field Equations are Lorentz invariant as it appears the rest of physical law is as well.  &lt;/p&gt;&#xA;" OwnerUserId="1567" LastActivityDate="2011-01-29T19:30:51.537" CommentCount="4" />
  <row Id="4167" PostTypeId="2" ParentId="2285" CreationDate="2011-01-29T20:15:58.780" Score="1" ViewCount="" Body="&lt;p&gt;That Heisenberg did not know about Matrices ... If he read Weyl  then he would have known.&lt;/p&gt;&#xA;" OwnerUserId="1423" LastActivityDate="2011-01-29T20:15:58.780" CommunityOwnedDate="2011-01-29T20:15:58.780" CommentCount="2" />
  <row Id="4168" PostTypeId="1" CreationDate="2011-01-29T20:37:53.300" Score="3" ViewCount="164" Body="&lt;p&gt;I have a pot of vigorously boiling water on a gas stove. There's some steam, but not alot. When I turn off the gas, the boiling immediately subsides, and a huge waft of steam comes out. This is followed by a steady output of steam that's greater than the amount of steam it was producing while it was actually boiling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is there more steam after boiling than during boiling? Also, what's with the burst of steam when it stops boiling?&lt;/p&gt;&#xA;" OwnerUserId="1666" LastActivityDate="2011-01-29T20:59:52.707" Title="Why is there more steam after a pot of water *stops* boiling?" Tags="&lt;thermodynamics&gt;&lt;water&gt;&lt;phase-transition&gt;&lt;everyday-life&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="4169" PostTypeId="2" ParentId="4168" CreationDate="2011-01-29T20:59:52.707" Score="4" ViewCount="" Body="&lt;p&gt;I have read that true steam is clear (transparent) water vapor.  According to this theory, the white &quot;steam&quot; you see is really a small cloud of condensed water vapor droplets, a fine mist in effect.  So what you are seeing is not more steam, but more condensation and more mist.  The speed with which the steam/vapor/mist rises and disperses may also change.&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-01-29T20:59:52.707" CommentCount="5" />
  <row Id="4170" PostTypeId="1" CreationDate="2011-01-29T21:04:54.177" Score="2" ViewCount="65" Body="&lt;p&gt;just wondering if it is a distinction without a difference - it seeming a bit weird that one thing can decay into different things.&lt;/p&gt;&#xA;" OwnerUserId="1423" LastActivityDate="2011-01-30T00:55:40.837" Title="is the Z boson one entity or are there as many entities as decay pairs, but they are equivalent and lumped together" Tags="&lt;particles&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="4171" PostTypeId="2" ParentId="4170" CreationDate="2011-01-29T21:22:04.573" Score="2" ViewCount="" Body="&lt;p&gt;No more weird than that a single particle can take different paths through different slits with different amplitudes.    In this case it's mutually exclusive different decay channels.  And yes, on occasion there can be interference between them.    An excited atom can also decay through different channels to different final states in the same way.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2011-01-29T21:22:04.573" />
  <row Id="4172" PostTypeId="1" AcceptedAnswerId="4183" CreationDate="2011-01-29T21:31:41.637" Score="10" ViewCount="380" Body="&lt;p&gt;The way it is sometimes stated is that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if we have a certain amount of &quot;free will&quot;, then, subject to certain assumptions, so must some elementary particles.&quot;(&lt;a href=&quot;http://en.wikipedia.org/wiki/Free_will_theorem&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is confusing to me, but it seems to be an amazing theorem. It has been interpreted as ruling out hidden variable theories, but there is still some dissent. Lubos has a &lt;a href=&quot;http://motls.blogspot.com/2007/12/john-conway-70th-birthday.html&quot; rel=&quot;nofollow&quot;&gt;good discussion&lt;/a&gt; of it on his blog in the birthday blog for John Horton Conway.   I assume it means that the outcomes of microscopic measurements are not deterministic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does the theorem assume and what does it prove?&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2011-01-29T23:46:20.520" LastActivityDate="2011-01-30T16:46:44.400" Title="What does John Conway and Simon Kochen's &quot;Free Will&quot; Theorem mean?" Tags="&lt;quantum-mechanics&gt;&lt;mathematical-physics&gt;" AnswerCount="2" CommentCount="17" FavoriteCount="3" />
  <row Id="4173" PostTypeId="2" ParentId="4170" CreationDate="2011-01-29T21:42:26.143" Score="2" ViewCount="" Body="&lt;p&gt;I can offer a different way to think about it. Many different particles are electric charged, right? Do you think this is strange?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, the same happens with the weak nuclear force. Many different particles are charged under this force and, therefore, interact via a Z boson.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing is that the Z, different from the photon, has a non-zero mass. So you start thinking about it more particle-like than the photon, because when it is mediating an interaction it can come as close to the mass shell as one want. But that's not always a very good way to interpret it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow... the idea is that many things are charged under the weak force. And there is nothing odd about it.&lt;/p&gt;&#xA;" OwnerUserId="1337" LastEditorUserId="1337" LastEditorDisplayName="" LastEditDate="2011-01-29T21:53:25.347" LastActivityDate="2011-01-29T21:53:25.347" />
  <row Id="4174" PostTypeId="1" AcceptedAnswerId="4228" CreationDate="2011-01-29T21:55:16.857" Score="1" ViewCount="138" Body="&lt;p&gt;Suppose a large neutron star were to be spun-up by a particular pattern of mass-accretion. The increased centrifugal force could presumably mitigate the increased gravity thus delaying gravitational collapse. How far could this process be taken?&lt;/p&gt;&#xA;" OwnerUserId="1394" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-01-30T19:27:56.050" LastActivityDate="2011-01-30T23:01:29.043" Title="High-spin neutron star" Tags="&lt;general-relativity&gt;&lt;black-holes&gt;&lt;neutron-stars&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="4175" PostTypeId="2" ParentId="4174" CreationDate="2011-01-29T22:03:04.603" Score="0" ViewCount="" Body="&lt;p&gt;The problem with this idea is that you get a fairly modest reduction of the pressure at the center of the star because the poles of rotation don't feel the effect. The detailed calculation would have to be relativistic &lt;em&gt;and&lt;/em&gt; assume an equation of state for neutronium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last I heard theory and observation could put only fairly broad limits on the real equation of state.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-01-29T22:34:47.060" LastActivityDate="2011-01-29T22:34:47.060" CommentCount="2" />
  <row Id="4176" PostTypeId="2" ParentId="4063" CreationDate="2011-01-29T22:07:33.857" Score="0" ViewCount="" Body="&lt;p&gt;You can download raw (or processed) data from various observational cosmology projects (e.g. COBE and WMAP) here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://lambda.gsfc.nasa.gov/product/map/current/&quot; rel=&quot;nofollow&quot;&gt;http://lambda.gsfc.nasa.gov/product/map/current/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a project, you could use this to re-compute the famous angular power spectrum of the cosmic microwave background radiation.  &lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-01-29T22:07:33.857" CommunityOwnedDate="2011-01-29T22:07:33.857" />
  <row Id="4177" PostTypeId="2" ParentId="4141" CreationDate="2011-01-29T22:09:38.247" Score="5" ViewCount="" Body="&lt;p&gt;There's nothing particularly special about physics as far as beauty goes. And there are plenty of examples of ugly realities in physics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) The values of the CKM mixing matrix are close to unity but not quite. It would be more beautiful if there was some pattern to it. Instead, all we have is the experimental values. Same applies to the MNS mixing matrix, but not such a large degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) It would be more symmetric if elementary particles did not depend on handedness. In fact, before the discovery of the handedness of the weak interaction it seems to have been widely believe (on the basis of beauty and simplicity) that there was no such dependency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) The masses of the various particles seem quite contrived. No one knows where they come from and their values are all over the map. This is not beautiful. Beauty would have been all the leptons having masses differing by factors of 2, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) The renormalization needed in QFT is quite ugly. Nature is beautiful partly because She avoids having to cancel one infinity against another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) While general relativity has a simple and beautiful set of assumptions, the results are ugly even for the simplest case of a non rotating mass, that is, a black hole. Such an object has a singularity at the center. Singularities are certainly not beautiful.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-29T22:09:38.247" CommentCount="5" />
  <row Id="4178" PostTypeId="2" ParentId="4141" CreationDate="2011-01-29T22:17:35.490" Score="5" ViewCount="" Body="&lt;p&gt;Take a preference for &quot;beautiful&quot; theory as a meta-rule like Occam's Razor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to work a lot of the time, but like the Razor, it must give way to data.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-01-29T22:17:35.490" />
  <row Id="4179" PostTypeId="2" ParentId="4174" CreationDate="2011-01-29T22:20:21.730" Score="2" ViewCount="" Body="&lt;p&gt;Here is a link to a paper where rotationally delayed collapse is seriously considered:&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/0804.0594&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0804.0594&lt;/a&gt;&lt;br&gt;&#xA;It is found likely to occur in some cases, but the delays are of order 10 milliseconds,&lt;br&gt;&#xA;(which is significant for gravitational wave detection.)&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-01-29T22:20:21.730" CommentCount="1" />
  <row Id="4180" PostTypeId="1" CreationDate="2011-01-29T22:51:11.837" Score="4" ViewCount="386" Body="&lt;p&gt;sometimes I get &quot;charged&quot; and the next thing I touch something that conducts electricity such as a person, a car, a motal door, etc I get shocked...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to avoid this so if I suspect being &quot;charged&quot; I try to touch something that does not conduct electricity (such as a wooden table) as soon as possible, in the belief that this will &quot;uncharge me&quot;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is it true that touching wood will uncharge you?&lt;/li&gt;&#xA;&lt;li&gt;How and when do I get charged? I noticed that it happens only in parts of the years, and after I get out of the car...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-03-02T12:01:51.060" Title="How to avoid getting shocked" Tags="&lt;electricity&gt;" AnswerCount="5" />
  <row Id="4181" PostTypeId="2" ParentId="4180" CreationDate="2011-01-29T22:57:07.067" Score="1" ViewCount="" Body="&lt;p&gt;That's incredibly annoying, I agree. Can I make a joke with physics content here? I always thought a great idea would be to carry a capacitor with you. Then, from time to time, you discharge yourself with it (probably with a resistor, so that you actually avoid getting shocked), charging the capacitor. When you pass by that person you don't like very much, you discharge the capacitor at him. See?, two problems solved at once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;:)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that the joke is told, let me try to answer the questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is it true that touching wood will uncharge you?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;No, you have to touch a conductor. Wood won't work.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How and when do I get charged? I noticed that it happens only in parts of the years, and after I get out of the car...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It happens when the air is very dry. I used to live in a city where the humidity in the air is huge and that never happened. In the winter the air gets very dry and these shocks are more common. So, you can move to a rain forest and your problem will also be solved.&lt;/p&gt;&#xA;" OwnerUserId="1337" LastEditorUserId="1337" LastEditorDisplayName="" LastEditDate="2011-01-29T23:02:36.060" LastActivityDate="2011-01-29T23:02:36.060" />
  <row Id="4182" PostTypeId="2" ParentId="4180" CreationDate="2011-01-29T23:03:13.607" Score="3" ViewCount="" Body="&lt;p&gt;Free or nearly-free electrons on stuff like wool are getting rubbed off onto you so that your body holds some sort of net total electric charge. When you touch a metal door (or any piece of metal) then the electrons want to spread out to balance themselves between you and the metal. Since the metal conducts electricity very well, they fly off you very quickly which heats up the air in between you and the metal, giving you the painful &quot;shock&quot; feeling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wood won't uncharge you very well, since it doesn't conduct electricity very well. The only ways to prevent getting shocked are either not building up charge in the first place, or constantly touching metal so that the charges get released way before they can build up - in effect spreading out your shocks to many smaller shocks you can't feel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many ways to build up a static charge, but it is generally much, much easier when the air is dry. Since cold air is drier, this means you probably build up a charge more quickly in the winter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for how you build up charge in the first place, it is usually by rubbing certain materials together. Plastic or rubber rubbing against wool carpeting or clothes (or any sort of hair) will do it, which is the most common cause for people in their day-to-day lives. If you want to stop it, you could consider using dryer sheets, which use a substance which happens to be conductive to soften your clothes - this makes you constantly discharge as you touch your own clothes, so you achieve the &quot;many tiny shocks&quot; method alluded to above. Or, you could use a humidifier in your home, which adds water to the air, making static much harder to build up.&lt;/p&gt;&#xA;" OwnerUserId="1655" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-01-29T23:27:56.370" LastActivityDate="2011-01-29T23:27:56.370" />
  <row Id="4183" PostTypeId="2" ParentId="4172" CreationDate="2011-01-29T23:14:46.340" Score="3" ViewCount="" Body="&lt;p&gt;I shall attempt here to give an explanation of the meaning of the theorem with a limited background. Issues such as the validity of the proof I shall leave aside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Free Will Theorem (assuming SPIN, TWIN, and FIN)].&#xA;&quot;If the choice of directions in which to perform spin 1 experiments is not a&#xA;function of the information accessible to the experimenters, then the responses&#xA;of the particles are equally not functions of the information accessible to them.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This theorem is a combination of ingredients which explain the hypotheses: The EPR setup of two particles (the SPIN  and TWIN axioms); special relativity in a limited form (the FIN (later MIN) axiom); the Kochen-Specker &quot;non-existence of a function&quot; theorem/paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So loosely we can imagine the typical EPR setup with two spacelike separated Observers making independent &quot;choices&quot; as to which of several axes to measure spin in two correlated particles: A and B, say. Let A's measurement be at time $t_A$. Then the conclusion is that no function of past properties (past light cone to $t_A$) predicts the outcome for A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the key phrase is &quot;no function&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This theorem was primarily motivated to further exclude a fully deterministic interpretation of QM (obviously with hidden variables). Such hidden variables would give rise to a function of them - which here doesnt exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A secondary issue - where the &quot;free will&quot; comes from - is whether this just reconfirms an essential randomness in QM. Well the argument here, I believe, is that they interpret &quot;random&quot; to mean &quot;a random function of&quot; - but since no function exists even a random function doesnt exist (of any earlier properties). The responses of the particle A is determined at time $t_A$ based on no prior information (or information on B) - just the &quot;free&quot; choices of the experimenters as to what to measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A link to the Strong Free Will Paper: &lt;a href=&quot;http://www.ams.org/notices/200902/rtx090200226p.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.org/notices/200902/rtx090200226p.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-01-29T23:14:46.340" CommentCount="1" />
  <row Id="4184" PostTypeId="1" AcceptedAnswerId="4186" CreationDate="2011-01-29T23:20:26.343" Score="12" ViewCount="661" Body="&lt;p&gt;Effective theories like &lt;em&gt;Little Higgs models&lt;/em&gt; or &lt;em&gt;Nambu-Jona-Lasinio model&lt;/em&gt; are non-renormalizable and there is no problem with it, since an effective theory does not need to be renormalizable. These theories are valid until a scale $\Lambda$ (ultraviolet cutoff), after this scale an effective theory need a UV completion, a more &quot;general&quot; theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Standard Model, as an effective theory of particle physics, needs a more general theory that addresses the phenomena beyond the Standard Model (an UV completion?). &lt;strong&gt;So, why should the Standard Model be renormalizable?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1545" LastEditorUserId="13" LastEditorDisplayName="" LastEditDate="2011-01-30T21:18:27.820" LastActivityDate="2011-02-12T16:49:43.673" Title="Why should the Standard Model be renormalizable?" Tags="&lt;quantum-field-theory&gt;&lt;particle-physics&gt;&lt;standard-model&gt;&lt;renormalization&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="4185" PostTypeId="2" ParentId="4184" CreationDate="2011-01-29T23:32:30.360" Score="3" ViewCount="" Body="&lt;p&gt;Because we happen to be working at the right energy scale. In general, if there are renormalizable interactions around, they dominate over nonrenormalizable ones, by simple scaling arguments and dimensional analysis. Before the electroweak theory was developed, the Fermi theory of the weak interactions was nonrenormalizable because the leading interactions people saw were beta decay, which is a dimension-6 operator, and the structure of the W and Z bosons hadn't been uncovered yet. Now we're at high enough energy that we've seen the underlying renormalizable interactions responsible for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are other heavy particles we don't know about, there certainly will be higher dimension operators around that we should add to the Standard Model. The fact that we haven't seen their effects yet is something that worries those of us who are hoping for discoveries soon....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I should add that, since we know neutrinos have mass, the Standard Model isn't really a renormalizable theory anymore. Not that this is relevant for most of particle physics most of the time.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2011-01-29T23:32:30.360" CommentCount="1" />
  <row Id="4186" PostTypeId="2" ParentId="4184" CreationDate="2011-01-29T23:34:47.500" Score="9" ViewCount="" Body="&lt;p&gt;The short answer is that it doesn't have to be, and it probably isn't. The modern way to understanding any quantum field theory is as an effective field theory. The theory includes all renormalizable (relevant and marginal) operators, which give the largest contribution to any low energy process. When you are interested in either high precision or high energy processes, you have to systematically include non-renormalizable terms as well, which come from some more complete theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back in the days when the standard model was constructed, people did not have a good appreciation of effective field theories, and thus renormalizability was imposed as a deep and not completely understood principle. This is one of the difficulties in studying QFT, it has a long history including ideas that were superseded (plenty of other examples: relativistic wave equations, second quantization, and a whole bunch of misconceptions about the meaning of renormalization). But now we know that any QFT, including the standard model, is expected to have these higher dimensional operators. By measuring their effects you get some clue what is the high energy scale in which the standard model breaks down. So far, it looks like a really high scale.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-02-02T04:32:12.777" LastActivityDate="2011-02-02T04:32:12.777" CommentCount="3" />
  <row Id="4187" PostTypeId="2" ParentId="4174" CreationDate="2011-01-29T23:39:19.740" Score="2" ViewCount="" Body="&lt;p&gt;The maximum rotation rate of a neutron star is: $\Omega=(M/R^3)^{1/2}$ where M is the non-rotating mass and R the Radius. I admit that I havent worked this out, but I found it in a thesis I have been studying in connection with another Stack question &quot;Geometrical and Physical Aspects of Rotating Neutron Star Models&quot;. Around equation (4.5), there is further discussion of conditions and variants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.icg.port.ac.uk/theses/white.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.icg.port.ac.uk/theses/white.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that White Dwarfs are similar objects which are held together by electron degeneracy pressure - and may obey similar equations.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-01-29T23:47:34.910" LastActivityDate="2011-01-29T23:47:34.910" CommentCount="5" />
  <row Id="4188" PostTypeId="1" AcceptedAnswerId="4206" CreationDate="2011-01-30T00:01:05.780" Score="3" ViewCount="232" Body="&lt;p&gt;One often comes across actions written with an extra auxiliary field, with respect to which if you vary the action, you get the equation of motion of the auxiliary field, which when plugged into the original action lets you retrieve a more familiar looking action without the auxiliary field. An example is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{equation*} S=\int d\tau e^{-1}(\tau)\eta_{\mu\nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau} \end{equation*}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which when solved for the einbein $e$ gives you the familiar square root form of the action. Why would you want this extra auxiliary field apart from having an action which is not in an annoying square root form? Why, for example, does GSW mention on Page 18 of &quot;Superstring Theory&quot; that &quot;the role of $e(\tau)$ is to ensure the action is invariant under reparametrizations of $\tau$&quot; when you can have reparametrization invariance even in the square root form of the action?&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T15:35:46.257" LastActivityDate="2011-02-02T15:35:46.257" Title="What's the point of having an einbein in your action?" Tags="&lt;mathematical-physics&gt;&lt;gauge-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="4189" PostTypeId="1" AcceptedAnswerId="4190" CreationDate="2011-01-30T00:05:10.233" Score="0" ViewCount="130" Body="&lt;p&gt;So I think my algebra is wrong somewhere.&#xA;Lets say you have an object that goes under SHM with some initial conditions (amplitude is 5m. The piston is at 5m at t = 0 and period is 20 seconds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay so $x(t) = A sin(\omega_0 + \phi)$. $\phi$ for us is 0. The frequency = $0.05 Hz$, angular frequency = $\frac{\pi}{10}$ and period of the piston = $20s$. &#xA;The maximum velocity is is when the first derivative $x&amp;#39;(t) = \omega_0 A cos(\omega_0 t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum velocity is when the cosine function is 1 at x = 0. So the maximum velocity is $v(t) = A\omega_0$ so that means $cos(\omega_0 t)= 1$ but $\omega_0 = \frac{\pi}{10}$ so the missing value is t. But thats trivial to find, since cosine function is 1. So $t = 20s$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here is my issue. The maximum velocity can only be at x = 0 (when kinetic energy is max, and potential energy is zero). But the value t = 20s represents the object at x = Amplitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what am I doing wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: I just realized that if  $x(t) = A sin(\omega_0 + \phi)$ then the initial conditions are not satisfied. When t = 0, sin function goes to 0 and hence everything is 0 but that dosnt satisfy initial conditions. When $t = 0$, displacement should be $5m$ However, if i replace $sin$ with $cos$ then everything works. Is that okay?&lt;/p&gt;&#xA;" OwnerUserId="551" LastEditorUserId="551" LastEditorDisplayName="" LastEditDate="2011-01-30T00:58:16.013" LastActivityDate="2011-01-30T01:57:32.043" Title="max velocity of SHM" Tags="&lt;classical-mechanics&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="4190" PostTypeId="2" ParentId="4189" CreationDate="2011-01-30T00:21:04.680" Score="0" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But the value t = 20s represents the object at x = Amplitude.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Check this calculation again. Remember your formula is  $x(t) = A sin(\omega_0 t)$, and you should get full credit on your homework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(or at least mostly full credit. There are many values of $t$ that satisfy $cos(\omega_0 t)=1$, right? you've found only one, which isn't the most general answer you could give.)&lt;/p&gt;&#xA;" OwnerUserId="1655" LastActivityDate="2011-01-30T00:21:04.680" CommentCount="11" />
  <row Id="4192" PostTypeId="2" ParentId="4180" CreationDate="2011-01-30T00:33:24.893" Score="5" ViewCount="" Body="&lt;p&gt;Carry some metal in your pocket. When you suspect you are carrying an electric charge, take the metal (a coin?) out of your pocket and touch it to something grounded.&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2011-01-30T00:33:24.893" />
  <row Id="4193" PostTypeId="2" ParentId="4170" CreationDate="2011-01-30T00:55:40.837" Score="1" ViewCount="" Body="&lt;p&gt;A closer similar weirdness is that a particle can have different charges, eg colour and electric, and then interact via the exchange of different particles. Worse, a charge for a non abelian force (such as colour, or also electroweak theory itself) immediately has different particles for the same charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is not problem here. It is perfectly possible for an elementary Z to decay into different pairs, and in fact it does with the probabilities predicted in the standard model. &lt;/p&gt;&#xA;" OwnerUserId="1335" LastActivityDate="2011-01-30T00:55:40.837" CommentCount="1" />
  <row Id="4194" PostTypeId="2" ParentId="4188" CreationDate="2011-01-30T01:15:53.093" Score="-1" ViewCount="" Body="&lt;p&gt;You can choose a coordinate gauge where $e~\rightarrow~e(d\tau’/d\tau)$ so it is gauge invariant under reparameterizations.  This means the value of $e$ is arbitrary and may be chosen in a gauge such that it is unity.  So this einbein is a gauge imposed quantity.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-30T01:15:53.093" />
  <row Id="4195" PostTypeId="2" ParentId="1453" CreationDate="2011-01-30T01:44:14.563" Score="3" ViewCount="" Body="&lt;p&gt;A mathematical solution of the plasma physics that causes the reversal of the sun's magnetic field every 11.5 years.&lt;/p&gt;&#xA;" LastActivityDate="2011-01-30T01:44:14.563" CommunityOwnedDate="2011-01-30T01:44:14.563" />
  <row Id="4196" PostTypeId="2" ParentId="4189" CreationDate="2011-01-30T01:57:32.043" Score="0" ViewCount="" Body="&lt;p&gt;What is the solution of $\sin(\omega t)=0$ ? It is $t=n\,\frac{\pi}{\omega}$ where $n$ is 0, 1, 2, 3, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your case the answer is $t=n\,10$ or $t=0,10,20\ldots$.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-01-30T01:57:32.043" />
  <row Id="4197" PostTypeId="2" ParentId="887" CreationDate="2011-01-30T03:13:52.943" Score="1" ViewCount="" Body="&lt;p&gt;Search term &quot;relativistic lattice boltzmann&quot; gives over 9000 results on Google Scholar. For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;S. Succi, &lt;a href=&quot;http://www.jstor.org/pss/3066322&quot; rel=&quot;nofollow&quot;&gt;Lattice Boltzmann equation for relativistic quantum mechanics&lt;/a&gt;, Philosophical Transactions: Mathematical, Physical and Engineering Sciences&#xA;Vol. 360, No. 1792, Discrete Modelling and Simulation of Fluid Dynamics (Mar. 15, 2002), pp. 429-436 &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1670" LastActivityDate="2011-01-30T03:13:52.943" />
  <row Id="4198" PostTypeId="2" ParentId="887" CreationDate="2011-01-30T03:34:08.527" Score="1" ViewCount="" Body="&lt;p&gt;Lattice Boltzmann take on 1+1 dimensional quantum field theory:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;S. Succi, &lt;a href=&quot;http://iopscience.iop.org/1751-8121/40/26/F07&quot; rel=&quot;nofollow&quot;&gt;Lattice Boltzmann method for quantum field theory&lt;/a&gt;, 2007 J. Phys. A: Math. Theor. 40 F559&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1670" LastActivityDate="2011-01-30T03:34:08.527" />
  <row Id="4199" PostTypeId="1" AcceptedAnswerId="4222" CreationDate="2011-01-30T04:19:59.110" Score="5" ViewCount="202" Body="&lt;p&gt;A continuous charge distribution flowing as a constant current in a closed loop doesn't radiate. Is it therefore true that as you increase the number of proton bunches in the LHC, while keeping the total charge constant, the synchrotron radiation decreases?&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="1345" LastEditorDisplayName="" LastEditDate="2011-01-30T18:47:24.610" LastActivityDate="2011-01-30T18:47:24.610" Title="Does bunching reduce synchrotron radiation?" Tags="&lt;electrodynamics&gt;&lt;synchrotron-radiation&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" />
  <row Id="4200" PostTypeId="1" AcceptedAnswerId="4203" CreationDate="2011-01-30T04:40:04.583" Score="9" ViewCount="278" Body="&lt;p&gt;Stephen Wolfram's &quot;A New Kind of Science&quot; hit the bookstores in 2002 with maximum hype.&#xA;His thesis is that the laws of physics can be generated by various cellular automata--simple programs producing complexity. Occasionally (meaning rarely)  I look at the NKS blog and look for any new applications. I see nothing I consider meaningful. Is anyone aware of any advances in any physics theory resulting from NKS?&#xA;While CA are both interesting and fun (John Conway, Game of Life), as a theory of everything, I see problems. The generator rules are deterministic, and they are local in that each cell state depends on its immediate neighbors. So NKS is a local deterministic model of reality. Bell has shown that this cannot be. Can anyone conversant with CA comment?&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-01-30T06:19:11.613" Title="Is Stephen Wolfram's NKS, an attempt to explain the universe with cellular automata, in conflict with Bell's Theorem?" Tags="&lt;quantum-mechanics&gt;&lt;theory-of-everything&gt;&lt;bells-inequality&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="4201" PostTypeId="1" AcceptedAnswerId="4208" CreationDate="2011-01-30T05:16:32.417" Score="6" ViewCount="337" Body="&lt;p&gt;I have read Sean Carrol's book. I have listened to Roger Penrose talk on &quot;Before the Big Bang&quot;. Both are offering to explain the mystery of low entropy, highly ordered state, at the Big Bang.&#xA;Since the second law of thermodynamics is considered a fundamental law of nature, and since it states that in a closed system entropy either must stay the same or increase, the entropy at the time of the Big Bang must have been much lower than it is now. Also, the thermodynamic arrow was explained by Boltzmann in 1896 embodied in S=klnW which was inscribed on his tombstone. W is the number of distinct microstates of the system. It seems to me that trivially this number will be less in the past than in the future since entropy obeys the 2nd law. This determines the thermodynamic arrow of time. Why do we need more explanation of a &quot;fundamental law&quot;?&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-01-30T08:58:32.940" Title="Why does the low entropy at the big bang require an explanation? (cosmological arrow of time)" Tags="&lt;cosmology&gt;&lt;thermodynamics&gt;&lt;entropy&gt;&lt;arrow-of-time&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="4202" PostTypeId="2" ParentId="4200" CreationDate="2011-01-30T05:24:01.033" Score="4" ViewCount="" Body="&lt;p&gt;Most of these automata models are deterministic in the same sense as pseudorandom number generators are. For example in the lattice gas models the deterministic rules end up generating noise and large scale fluctuations in accord to the Navier-Stokes equations (including turbulence, although this is computationally impractical because of the large lattice dimensions required for reducing the lattice viscosity). The lattice gas game turned in the late eighties from noisy discrete particle automata to smooth distribution based lattice Boltzmann mesoscopic scale continuous-value automata (see Guy R. McNamara and Gianluigi Zanetti, &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v61/i20/p2332_1&quot; rel=&quot;nofollow&quot;&gt;Use of the Boltzmann Equation to Simulate Lattice-Gas Automata&lt;/a&gt;, Phys. Rev. Lett. 61, 2332–2335 (1988) ), so that's where you find most relevant advances these days.&lt;/p&gt;&#xA;" OwnerUserId="1670" LastEditorUserId="1670" LastEditorDisplayName="" LastEditDate="2011-01-30T06:19:11.613" LastActivityDate="2011-01-30T06:19:11.613" />
  <row Id="4203" PostTypeId="2" ParentId="4200" CreationDate="2011-01-30T05:35:00.340" Score="8" ViewCount="" Body="&lt;p&gt;Wolfram's early work on cellular automata (CAs) has been useful in some didactical ways. The 1D CAs defined by Wolfram can be seen as minimalistic models For systems with many degrees of freedom and a thermodynamic limit. Insofar these CAs are based on a mixing discrete local dynamics, deterministic chaos results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from these didactical achievements, Wolfram's work on CAs has not resulted in anything tangible. This statement can be extended to a much broader group of CAs, and even holds for lattice gas automata (LGAs), dedicated CAs for hydrodynamic simulations. LGAs have never delivered on their initial promise of providing a method to simulate turbulence. A derivative system (Lattice Boltzmann - not a CA - has some applications in flow simulation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is against this background that NKS was released with much fanfare. Not surprisingly, reception by the scientific community has been negative. The book contains no new results (the result that the 'rule 110 CA' is Turing complete was proven years earlier by Wolfram's research assistant Matthew Cook), and has had zero impact on other fields of physics. I recently saw a pile of NKS copies for sale for less than $ 10 in my local Half Price Books store. &lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-01-30T05:35:00.340" CommentCount="1" />
  <row Id="4204" PostTypeId="2" ParentId="4201" CreationDate="2011-01-30T05:45:03.727" Score="0" ViewCount="" Body="&lt;p&gt;The key issue is: To allow galaxy and star formation, the universe had to start up in an extremely (but not perfectly) homogeneous state. The fact that the initial inhomogeneities were of size 1 part in a gadzillion (as opposed to zero) is what is puzzling.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-01-30T05:45:03.727" CommentCount="7" />
  <row Id="4205" PostTypeId="2" ParentId="4201" CreationDate="2011-01-30T05:52:08.237" Score="0" ViewCount="" Body="&lt;p&gt;Because this violates the assumption that every state ought to be equally likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The low entropy of the big bang leads to the second law of thermodynamics, not the other way around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, as Boltzmann had shown, for almost all states constrained to have a given macroscopic description at a given time, entropy will increase both toward the past as well as the future. For the entropy to have been continually decreasing as we move back to the past for billions of years requires an incredible amount of fine-tuning of the correlations between almost all the particles in the present.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-01-30T05:52:08.237" CommentCount="5" />
  <row Id="4206" PostTypeId="2" ParentId="4188" CreationDate="2011-01-30T06:51:52.490" Score="4" ViewCount="" Body="&lt;p&gt;first, you and Moshe may have somewhat misunderstood the comment that &quot;the purpose of introducing $e$ is to guarantee the reparameterization symmetry of the action&quot;. This wasn't meant to express the circular argument that we need to add a redundant field in order to have a redundancy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, I think that the authors simply wanted to write that &quot;in order for this action - which we already chose to study - to be equivalent to the einbein-frei form, the new field $e$ has to be unphysical, so the reparameterization has to be insured, so the factor of $e$ has to be included.&quot; If the $e$ were simply omitted in this form of the action, the action wouldn't be reparameterization-invariant, so it wouldn't be equivalent to the &quot;proper length&quot; of the world line we want to describe: the proper length of a world line &lt;em&gt;is&lt;/em&gt; independent of the way how we parameterize the world line to calculate the integral (the proper length), so any equivalent description has to have the same property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation for gauge redundancies&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this particular case, the main technical reason is that by introducing the new auxiliary variable, the einbein, you get rid of the square root. Note that without this variable, the action would depend on the standard Lorentz gamma factor, $1/\sqrt{1-v^2/c^2}$. From the Lagrangian, it would imprint itself into the Hamiltonian as well. Hamiltonians that depend on square roots of functions of observables are hard - relatively to what you get with $e$. In the world sheet generalization, the Hamiltonian would even include an integral of a square root which is really bad - especially if you want to Fourier-transform things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the underlying physics actually doesn't need the messiness as can be seen by introducing clever redundancies. There is no square root in the action including the einbein - but it's still equivalent. By introducing this fake degree of freedom, or a redundancy, we actually convert the equations of motion for $x$ to a simple wave equation, $\nabla_\mu\nabla^\mu x^\alpha=0$. Well, it's a wave equation in at least $1+1$ dimensions. In $0+1$ dimensions, it's just an equation for a uniform motion of $x^\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case of other gauge symmetries&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reparametrization of $\tau$ is a redundancy of the system, but so is any gauge symmetry, for example the colorful $SU(3)$ in QCD or the diffeomorphism group in general relativity (which is nothing else a higher-dimensional generalization of this einbein example: $e(\tau)\equiv \sqrt{g_{00}}$ is the redundant field, and gravity has no physical components in $0+1$ dimensions). You could also work without it - and the twistor description of $N=4$ $d=4$ $SU(N)$ gauge theory actually has no locality so it has no local gauge symmetry in spacetime. Other descriptions that don't have any gauge redundancy include the unitary gauge in spontaneously broken gauge theories; AdS/CFT that replaces the redundancy by a totally different one in the dual description; the S-dual description that replaces the gauge group with the magnetic monopoles' dual gauge group; and many others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of having local symmetries - or gauge redundancies - is always either to simplify the equations or to make some of their symmetries manifest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, if you want elementary particles with spin equal to one or higher, and if you want a manifest Lorentz symmetry in spacetime and manifest locality, you need local vector- or tensor-valued fields such as $A_\mu$ and $g_{\mu\nu}$ that also have time-like components. In a quantum theory, the time-like components would produce negative-norm states: the theory could predict negative probabilities if these polarizations would stay in the spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, gauge invariance is the only consistent way to remove them. Because gauge symmetry is a &quot;symmetry&quot; in the sense that it commutes with the Hamiltonian, the gauge invariance of the initial state guarantees the gauge invariance of the final state, so unphysical, gauge-dependent states are not produced by the evolution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, the choice of a redundant equivalent description is always made in order to make some or several of the properties below manifest:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Unitarity combined with manifest Lorentz invariance: gauge symmetry is needed to remove the ghosts&lt;/li&gt;&#xA;&lt;li&gt;Locality - one needs local tensor fields&lt;/li&gt;&#xA;&lt;li&gt;Simple, harmonic-oscillator-like underlying character of the dynamics that would be masked by square roots in the einbein-frei formulation (typical in the case of diffeomorphisms of the world volumes)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Cheers, LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-30T06:51:52.490" />
  <row Id="4207" PostTypeId="2" ParentId="4184" CreationDate="2011-01-30T07:09:43.903" Score="6" ViewCount="" Body="&lt;p&gt;the Standard Model just happens to be perturbatively renormalizable which is an advantage, as I will discuss later; non-perturbatively, one would find out that the Higgs self-interaction and/or the hypercharge $U(1)$ interaction would be getting stronger at higher energies and they would run into inconsistencies such as the Landau poles at extremely high, trans-Planckian energy scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the models where the Higgs scalar is replaced by a more convoluted mechanism are not renormalizable. That's not a lethal problem because the theory may still be used as a valid effective theory. And effective theories can be non-renormalizable - they have no reason not to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why physicists prefer renormalizable field theories is that they are more predictive. A renormalizable field theory's predictions only depend on a finite number of low-energy parameters that may be determined by a comparison with the experiments. Because with a fixed value of the low-energy parameters such as the couplings and masses, a renormalizable theory may be uniquely extrapolated to arbitrarily high scales (and it remains predictive at arbitrarily high scales), it also means that if we postulate that the new physics only occurs at some extremely high cutoff scale $\Lambda$, all effects of the new physics are suppressed by positive powers of $1/\Lambda$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This assumption makes the life controllable and it's been true in the case of QED. However, nothing guarantees that the we &quot;immediately&quot; get the right description that is valid to an arbitrarily high energy scale. By studying particle physics at ever higher energy scales, we may equally well unmask just another layer of the onion that would break down at slightly higher energies and needs to be fixed by another layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My personal guess is that it is more likely than not that any important extra fields or couplings we identify at low energies are inherently described by a renormalizable field theory, indeed. That's because of the following reason: if we find a valid effective description at energy scale $E_1$ that happens to be non-renormalizable, it breaks down at a slightly higher energy scale $E_2$ where new physics completes it and fixes the problem. However, this scenario implies that $E_1$ and $E_2$ have to be pretty close to one another. On the other hand, they must be &quot;far&quot; because we only managed to uncover physics at the lower, $E_1$ energy scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The little Higgs models serve as a good example how this argument is avoided. They adjust things - by using several gauge groups etc. - to separate the scales $E_1$ and $E_2$ so that they only describe what's happening at $E_1$ but they may ignore what's happening at $E_2$ which fixes the problems at $E_1$. I find this trick as a form of tuning that is exactly as undesirable as the &quot;little hierarchy problem&quot; that was an important motivation of these models in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The history has a mixed record: QED remained essentially renormalizable. The electroweak theory may be completed, step-by-step, to a renormalizable theory (e.g. by the tree unitarity arguments). The QCD is renormalizable, too. However, it's important to mention that the weak interactions used to be described by the Fermi-Gell-Mann-Feynman four-fermion interactions which was non-renormalizable. The separation of scales $E_1$ and $E_2$ in my argument above occurs because particles such as neutrons - which beta-decay - are still much lighter than the W-bosons that were later found to underlie the four-fermion interactions. This separation guaranteed that the W-bosons were found decades after the four-fermion interaction. And this separation ultimately depends on the up- and down-quark Yukawa couplings' being much smaller than one. If the world were &quot;really natural&quot;, such hierarchies of the couplings would become almost impossible. My argument would hold and almost all valid theories that people would uncover by raising the energy scale would be renormalizable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General relativity is a big example on the non-renormalizable side and it will remain so because the right theory describing quantum gravity is not and cannot be a local quantum field theory according to the old definitions. As one approaches the Planck scale, the importance of non-renormalizable effective field theories clearly increases because there is no reason why they should be valid to too much higher energy scales - at the Planck scale, they're superseded by the non-field quantum theory of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the best, LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-30T07:09:43.903" />
  <row Id="4208" PostTypeId="2" ParentId="4201" CreationDate="2011-01-30T07:13:36.013" Score="1" ViewCount="" Body="&lt;p&gt;This is somewhat controversial issue. But let me present the reasons, as far as I understood, why people like Sir Penrose thinks so. Their arguments are roughly as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The basic microscopic laws of physics are perfectly time symmetric. They are not biased in any time direction past or future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Second law follows from the fact that that given an initial condition of a system which is not in the most probable state will tend to go towards the most probable state by the same microscopic laws. Since number of disordered states are much higher the system will become more and more disordered with time. Accordingly its entropy will increase until a maximum value when the system comes to the thermal equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Since the microscopic laws are time symmetric the same argument can be made towards the past time direction as well. Given an an initial condition of a system which is not in the most probable state should go towards more disordered (high entropy) states towards past as well. That's what the mathematics of the laws tells us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) This is against our experience. Either all the parts of the universe we are observing (including our memories of past) has just undergone a HUGE fluctuation &lt;strong&gt;right now&lt;/strong&gt; to give the impression that there was a more ordered past (which is crazy) OR the system was &lt;strong&gt;already&lt;/strong&gt; even more ordered (low entropy) and more special in the past. But that means &lt;strong&gt;even more&lt;/strong&gt; huge a fluctuation. This reasoning will lead us to conclude that at the moment of big bang the universe was extra ordinarily ordered and most special. It should be so special that it requires explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Critics often point out that prediction and retrodiction is not the same thing forgetting that when one talks about the very &quot;arrow of time&quot; no one can say with justification which is prediction and which is retrodiction. Other than that it is also questionable whether second law can be applied this way to the whole universe or not.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-01-30T08:58:32.940" LastActivityDate="2011-01-30T08:58:32.940" CommentCount="5" />
  <row Id="4209" PostTypeId="2" ParentId="4199" CreationDate="2011-01-30T08:40:21.717" Score="4" ViewCount="" Body="&lt;p&gt;Dear John, a good question. You may want to read a relevant paper about the closely related question for the late SSC collider:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://mafurman.lbl.gov/SSC-N-143.pdf&quot; rel=&quot;nofollow&quot;&gt;http://mafurman.lbl.gov/SSC-N-143.pdf&lt;/a&gt;&lt;br&gt;&#xA;  &lt;strong&gt;Bunch-Length Dependence of Power Loss for the SSC&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The beam has $M$ bunches in the orbit. Each of them carries $Ne$ of electric charge. All of the particles orbit by frequency $f_0$ (revolutions per second, in Hertz). We define the product, the bunch current, to be $I_b=N e f_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In equation 12, you will see the result:&#xA;$$Power = 1.101 Z_0 M I_b^2 \sigma_{\phi}^{-4/3} $$&#xA;Here, $Z_0$ is just the impedance of the vacuum, $4\pi/c = 377 \Omega$; they use some Gaussian units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More importantly, $\sigma_\phi$ is just $\sigma_z/\rho$, the angular &lt;em&gt;root mean square&lt;/em&gt; size of the bunch. You see that it's the only quantity whose increase makes the power decrease. If you spread the bunches around the ring, you're getting closer to your &quot;closed loop current&quot; that doesn't radiate, indeed. In practice, you don't want to spread the bunches completely because you wouldn't know the timing of the collisions. In real applications, $\sigma_\phi$ is much smaller than one, giving you a significant increase to the synchrotron radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula is simply proportional to the number of bunches. If they're separated, each of them loses the same energy per revolution. Without a loss of generality, you may imagine that we only consider one bunch, $M=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, for a fixed $f_0$ - which is given by the size of the tunnel and the speed of light, assuming that the particles are near the speed of light - the power radiated by the bunch is actually proportional to $N^2$. If you double the number of charged particles in the bunch, the synchrotron radiation quadruples! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's because the energy density (and flux) is proportional to the squared electric (and magnetic) fields, and those - derived from the Liénard-Wiechert potentials - are linear in the charges (and currents) that produce the electromagnetic fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So once again, the power that is radiated is not proportional to the &quot;density&quot; of protons in the bunch but to its square! In this sense, bunching makes the synchrotron radiation worse, not better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you shouldn't think that it is a catastrophe. In the designed conditions for the LHC, one proton only loses something like 6.7 keV of energy per revolution which is a billionth of those 7 TeV they ultimately want to get (in 2011, they decided to continue at 3.5 TeV). Why is it so small for the hadron colliders? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, for the lepton colliders, you lose a lot because the synchrotron radiation is proportional to $\gamma^6$ and the Lorentz factor $\gamma$ has to be 2,000 times higher for electrons than for protons to achieve the same energy; see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Synchrotron_radiation#Derivation&quot; rel=&quot;nofollow&quot;&gt;derivation&lt;/a&gt;. Take the sixth power of that to see the impact of the light particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the hadron colliders, the main limitation is of course the magnetic field you need to keep the protons on their circular orbit. That's why you need to have all the superconducting magnets. For colliders with light particles that need a huge $\gamma$, the synchrotron radiation is very important. That's also why linear accelerators are often preferred for the leptons. Well, you won't get rid of the full synchrotron radiation because you still need to accelerate the leptons to have some fun - so there will still be a component of the acceleration in the direction of the velocity even though the straight tunnel may liberate you from the &quot;centripetal&quot; acceleration transverse to the velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To return to the closed loop, yes, I do think that you would turn the synchrotron radiation from the circular motion off completely if you distributed the bunches uniformly - even for leptons. It would be just like a wire with a current. However, there would still be a synchrotron radiation from the acceleration in the forward direction that you need to accelerate the particles to high speeds in the first place.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-01-30T08:45:52.113" LastActivityDate="2011-01-30T08:45:52.113" CommentCount="9" />
  <row Id="4210" PostTypeId="2" ParentId="3527" CreationDate="2011-01-30T10:32:18.503" Score="1" ViewCount="" Body="&lt;p&gt;Let's start with an observation from billiards.  Say the red ball is stationary and you hit it dead-on with the cue ball at speed $v$.  The cue ball will stop and the red ball will continue on at speed $v$.  Lubos gave a nice, simple description of this in &lt;a href=&quot;http://physics.stackexchange.com/questions/3527/newtons-cradle/3533#3533&quot;&gt;his answer&lt;/a&gt;.  You can see it happening at the beginning of &lt;a href=&quot;http://www.youtube.com/watch?v=QF7ffrkHM1Y&amp;amp;feature=related&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine a row of balls (red-orange-yellow-green-blue-violet, say) lined up perfectly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/aoi5F.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You hit the red one dead-on with the cue, and the cue stops and the red ball goes ahead.&#xA;After just a moment, the red ball hits the orange ball.  The red one stops and the orange on goes on ahead.  Then the orange one hits the yellow one and stops, etc.  Ultimately the violet ball comes out of the line, and all the other balls in the line have just moved down the table a little way. One ball in, one ball out. You can see something pretty much like this happen 50 seconds into &lt;a href=&quot;http://www.youtube.com/watch?v=QF7ffrkHM1Y&quot; rel=&quot;nofollow&quot;&gt;the same video&lt;/a&gt; (it's not quite perfect, though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we do it again.  Line up all the balls, but this time roll the cue ball towards the red ball, and roll the 8-ball right after it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/lxcRq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cue ball strikes the red ball and stops; the red ball goes forward, as before.  The red ball hits the orange ball.  This time, that's not the only thing happening.  Simultaneously with that collision, the 8-ball runs into the stopped cue ball.  After these collisions there are two balls in motion - the orange ball in front, and the cue ball, which has now had two collisions, one in front and one from behind.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process repeats, going down the line, always with two balls in motion with one stationary ball in between. Near the end, the two moving balls are the violet (last) ball and green (third-from-last) ball.  The violet ball is free, but the green ball impacts the blue one, and that's the last collision.  The result is that the blue and violet balls come away from the line up.  Everything else is stopped.  Two balls went in, and two come out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here it is not hard to see that we could send in three, four, or even 100 balls (in theory) and get the same number out again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty with this explanation is that in Newton's cradle, the balls physically touch one another, and so the stop-start argument does not apply in the same way.  This simple analysis should make Newton's cradle plausible, but a full argument relies on studying the continuum mechanics associated with the collisions between solid bodies. &lt;a href=&quot;http://www.maths.tcd.ie/~garyd/Publications/Delaney_2004_AmJPhys_Rocking_Newtons_Cradle.pdf&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt;, mentioned by Georg in the comments, provides such an analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:  I didn't watch the YouTube video all the way through and don't necessarily vouch for everything it says about physics.  I just wanted examples of those shots.  Also, the balls in billiards are rolling, which can sometimes make a difference compared to Newton's cradle, but here we will assume it does not.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-06-28T20:58:39.923" LastActivityDate="2011-06-28T20:58:39.923" CommentCount="9" />
  <row Id="4211" PostTypeId="2" ParentId="4184" CreationDate="2011-01-30T10:53:04.780" Score="0" ViewCount="" Body="&lt;p&gt;The standard model need not be renormalizable. It's completion should be.&lt;/p&gt;&#xA;" OwnerUserId="1458" LastActivityDate="2011-01-30T10:53:04.780" />
  <row Id="4212" PostTypeId="1" CreationDate="2011-01-30T10:58:11.810" Score="0" ViewCount="323" Body="&lt;p&gt;Suppose I have the following double split experiment set up:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a monochromatic electron source of low intensity, which we can model as emitting a single electron at a time with energy $T$.&lt;/li&gt;&#xA;&lt;li&gt;a diffraction screen at a distance $A$ with two macroscopic slits equidistant from the source with width $w$ and distance from the center of the screen $\frac{d}{2}$.&lt;/li&gt;&#xA;&lt;li&gt;a detector screen sitting parallel to the diffraction screen at a distance $B$ which consists essentially of a lattice of ammeters each connected with a square surface of side $r$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;http://i.stack.imgur.com/s6SyA.jpg&quot; alt=&quot;Double slit experiment&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classical quantum mechanics, and experiments, tell us that the following behaviour is experienced:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Only one random ammeter is disturbed each time an electron is fired&lt;/li&gt;&#xA;&lt;li&gt;The probability distribution of the random ammeters being fired is as given by QM.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The interaction of the electron with the detector is not modeled, the electron is essentially modeled as a wave with interference and which ammeter is fired, essentially an effect of particle behaviour, cannot be discussed within this picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the measurement or decoherence of the electron wave function corresponds to this switch between wave-like and particle-like models of the electron - which is clearly unsatisfying, intellectually, because QM is vague about this - it merely associates mathematical operations with the procedure, without giving any particular physical or mathematical justification of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The questions: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do further refinements of QM (QED or QFT or ST) give any better explanation of what justifies the measurement &quot;recipe&quot; of classical QM by correctly modelling the electron/detector interaction?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;How, in layman terms, is the change of picture from wave to particle modeled (if it is at all)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="66" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-03-13T20:17:00.223" LastActivityDate="2011-03-13T20:17:00.223" Title="How is the double slit experiment modeled in contemporary physical theories?" Tags="&lt;quantum-field-theory&gt;&lt;string-theory&gt;&lt;quantum-electrodynamics&gt;&lt;double-slit-experiment&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="0" />
  <row Id="4213" PostTypeId="2" ParentId="3344" CreationDate="2011-01-30T11:27:45.063" Score="0" ViewCount="" Body="&lt;p&gt;You can see accurate visualizations of (simulated) eddy currents in multiple papers. One example is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.asc.tuwien.ac.at/~schoeberl/wiki/publications/multiharmonic2.pdf&quot; rel=&quot;nofollow&quot;&gt;Efficient Solvers for Nonlinear Time-Periodic Eddy Current Problems&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1656" LastEditorUserId="1656" LastEditorDisplayName="" LastEditDate="2011-01-30T11:56:04.140" LastActivityDate="2011-01-30T11:56:04.140" CommentCount="1" />
  <row Id="4214" PostTypeId="2" ParentId="4212" CreationDate="2011-01-30T11:32:02.423" Score="1" ViewCount="" Body="&lt;p&gt;Dear sklivvz, the very same question was asked a few days ago. Quantum field theory, string theory, or any other viable theory that may supersede quantum mechanics directly reduces to non-relativistic quantum mechanics in the non-relativistic limit and changes nothing about the basic postulates of quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It means that you may find the corresponding low-energy, low-speed (multi-)particle states in the Hilbert space of QFT or string theory or anything else - essentially creation operators acting on the relevant vacuum - and you may prove that the QFT or stringy Hamiltonian acts on these states exactly as the non-relativistic Hamiltonian does, plus corrections that go like positive powers of $1/c$ and may be neglected in the non-relativistic limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So nothing changes about quantum mechanics and the double-slit experiment in QFT or string theory and it's likely that those things will never change. Cheers, LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-30T11:32:02.423" CommentCount="3" />
  <row Id="4215" PostTypeId="2" ParentId="4141" CreationDate="2011-01-30T12:24:02.257" Score="1" ViewCount="" Body="&lt;p&gt;&quot;Doing physics is a lot like making love to a beautiful woman...&quot;, Swiss Toni knows best.&lt;/p&gt;&#xA;" OwnerUserId="1670" LastActivityDate="2011-01-30T12:24:02.257" />
  <row Id="4218" PostTypeId="2" ParentId="4212" CreationDate="2011-01-30T13:40:19.040" Score="0" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So, the measurement or decoherence of&#xA;  the electron wave function corresponds&#xA;  to this switch between wave-like and&#xA;  particle-like models of the electron -&#xA;  which is clearly unsatisfying,&#xA;  intellectually, because QM is vague&#xA;  about this - it merely associates&#xA;  mathematical operations with the&#xA;  procedure, without giving any&#xA;  particular physical or mathematical&#xA;  justification of it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Quantum mechanics doesn't have two models of the electron. For intuitive purposes it can be useful to visualize very broad wave packets (that are almost a momentum eigenstate) as &quot;waves&quot; and very localized wave packets (that are almost a position eigenstate) as &quot;particles&quot;, but the QM description is unique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Copenhagen interpretation assumes that a measurement of position &quot;collapses&quot; the (position) wavefunction to a very localized wave packet with a width in the order of the measurement uncertainty. Decoherence explains the apparent collapse of the wavefunction by an entanglement between the system and environmental degrees of freedom due to the measurement process.&lt;/p&gt;&#xA;" OwnerUserId="1656" LastActivityDate="2011-01-30T13:40:19.040" CommentCount="10" />
  <row Id="4219" PostTypeId="2" ParentId="4212" CreationDate="2011-01-30T13:41:01.443" Score="0" ViewCount="" Body="&lt;p&gt;I find it strange that even after 85 years since its inception, quantum theory has not been fully appreciated! I don't know what you meant by the term &quot;classical quantum mechanics&quot;. I guess, it reflects your desire to &quot;explain&quot; quantum theory in terms of classical intuition. But what one should do is really the opposite. As far as your question is concerned, quantum theory in  its current formulation has no problem. No agonizing scrutiny of the basic postulates are at all required. &quot;Measurement&quot; in QT has always confused some people and doubtless will continue to do so. But once decoherence is understood, it is not such a big deal.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-01-30T13:41:01.443" CommentCount="6" />
  <row Id="4220" PostTypeId="2" ParentId="4180" CreationDate="2011-01-30T13:55:17.660" Score="2" ViewCount="" Body="&lt;p&gt;Here is a list how to slowly discharge before getting out of the car: &lt;a href=&quot;http://www.wikihow.com/Get-out-of-a-Car-Without-Getting-Shocked-by-Static-Electricity&quot; rel=&quot;nofollow&quot;&gt;http://www.wikihow.com/Get-out-of-a-Car-Without-Getting-Shocked-by-Static-Electricity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember trucks used to have a trailing chain to discharge any accumulation from the friction of wheels. These days the design must take care of that because one no longer sees this.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-01-30T13:55:17.660" CommentCount="1" />
  <row Id="4221" PostTypeId="2" ParentId="3327" CreationDate="2011-01-30T14:11:49.550" Score="1" ViewCount="" Body="&lt;p&gt;These optical effects of course have nothing to do with the size of stars.  Until recently the size of a star could not be imaged on a photoplate or CCD.  An image can have diffraction effects from the aperture and other optical elements.  However, these are not directly related to the size of a star and our estimates on stellar diameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first estimate of a star size is with the bolometric luminosity.  The emission of energy is proportional to the fourth power of the effective absolute temperature $T$, and the area is proportional to the square of the diameter of the star, we have the diameter $D$ proportional to the square root of the luminosity $L$ and the square of the effective temperature $T$. If $D$, $L$ and $T~=~5800K$ for the Sun, then for another star with temperature $T’$ and luminosity $L’$ D/D' = √(L/L')(5800/T)^2.&#xA;$$&#xD;&#xA;D’~=~D\sqrt{L’/L}(T’/T)^2.&#xD;&#xA;$$&#xA;This is a black body or Stephan law approach to calculating the diameter of a star&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more exact size of a star can be deduced optically for a telescope that his at it resolution limit.  The physical effect is the Hanbury Brown and Twiss effect.  An electromagnetic wave with phase $e^{i\omega t}$ will reach two detectors with a relative phase $e^{i\phi}$.  The irradiances on the two detectors will then be &#xA;$$&#xD;&#xA;I_1~=~E^2 e^{i\omega t},~ I_2~=~E^2 e^{i\phi}e^{i\omega t}&#xD;&#xA;$$&#xA;The correlation function between them $\langle I_1I_2\rangle$ over a measurement time $T$ is $T^{-1}\int_0^T I_1I_2dt$, which for sufficiently long time is given by &#xA;$$&#xD;&#xA;\langle I_1I_2\rangle ~=~\lim_{T\rightarrow\infty}T^{-1}\int_0^T I_1I_2dt~=~\frac{E^4}{4}\Big(1~+~\frac{1}{2}cos(2\phi)\Big)&#xD;&#xA;$$&#xA;The phase difference can exist for a large telescope.  The Airy pattern is $I(r)~=~I(0)(2J1(z)/z)^2, $z~=~2π[r/(2\lambda f/d)]$, where $d$ is the diameter of the aperture.  If this is made small enough the HBT phase or correlation may be detected and used to more directly measure the diameter of a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I included this to illustrate something related to this issue, and to iclear a common misconception about how images of stars are often thought to convey direct information about their sizes.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-30T14:11:49.550" />
  <row Id="4222" PostTypeId="2" ParentId="4199" CreationDate="2011-01-30T14:34:50.033" Score="7" ViewCount="" Body="&lt;p&gt;Synchrotron radiation can be coherent and incoherent. Coherent SR arises when electrons are grouped into short bunches so that the entire bunch emits SR as a whole. Quantum mechanically, in coherent SR the photon emission from different electrons in a bunch sum up at the amplitudes level and constructively interfere. In the incoherent SR they sum up at the level of intensity, and there is no interference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incoherent SR does not care how electrons are distributed along the ring, while the coherent SR is obviously boosted up in the presence of strong bunching. So, the more homogeneously you distribute the electrons, the less the effect of coherent SR will be and the less overall SR you'll have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's look at the incoherent SR. Theoretically, you are right: if we managed to create the &lt;strong&gt;absolute homogeneous&lt;/strong&gt; charge distribution along the ring, we would (classically) have no SR at all because charge distribution does not change in time. The point is that this is not feasible experimentally, at least for the accelerators and the beams we have. That would require putting electrons in a well-defined quantum state of the radial motion and a well-defined angular quantum number m for the azimuthal dependence, and the accelerator technology is very far from that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is another thing which mimics that situation closely. People have managed recently to put freely propagating electrons in states with well-defined orbital angular momentum (m as high as 75), see &lt;a href=&quot;http://www.sciencemag.org/content/331/6014/192.abstract&quot; rel=&quot;nofollow&quot;&gt;this paper in Science&lt;/a&gt; for details, and they really see the annular distribution for the electron density. For such a state there exists a reference frame where the electron does not move along the z axis but just rotated as a whole in the transverse plane (with some radial distribution) around the symmetry axis. This rotation is not driven by any force, it's just the peculiar superposition of plane waves that creates this steady pattern. So in this case you can say that the electron indeed circulates but does not emit any SR.&lt;/p&gt;&#xA;" OwnerUserId="59" LastActivityDate="2011-01-30T14:34:50.033" CommentCount="3" />
  <row Id="4223" PostTypeId="2" ParentId="4146" CreationDate="2011-01-30T14:58:56.220" Score="2" ViewCount="" Body="&lt;p&gt;This appears to be related to twisted or broken supersymmetry.  For unbroken supersymmetry the total Hamiltonian ${\cal H}~=~N{\cal H}_b~+~N{\cal H}_f$ for $N$ strings with the bosonic operators $a,~a^\dagger$ and the fermionic operators $b,~b^\dagger$ for N strings is:&#xA;$$&#xD;&#xA;{\cal H}~=~{N\over 2}\sum_k (\omega(k)a_k^\dagger a_k~+~{N\over 2}\sum_k \omega(k)b_k^\dagger b_k.&#xD;&#xA;$$&#xA;There exists an operator $\lambda$ so that $\omega(k)/2~=~\omega log(\lambda_k)$.  Thus by supersymmetry we may write the Hamiltonian as&#xA;$$&#xD;&#xA;{\cal H}~=~N\sum_k (log(\lambda_k)a_k^\dagger a_k~+~N\sum_k log(\lambda_k)b_k^\dagger b_k.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Wick rotate the integrand in the path integral $e^{-it H_b}$ under the substitution $it~\rightarrow~\beta$ and find that&#xA;$$&#xD;&#xA;Tre^{-NH_b\beta}~=~\sum_{i=1}^\infty\lambda_i^{-N\omega\beta}~=~\zeta(N\omega\beta).&#xD;&#xA;$$&#xA;This is a generalized zeta function, and for both bosons and fermions,&#xA;$$&#xD;&#xA;Tr\big((-1)^{N_b}e^{-N{\cal H}_b\beta}e^{-N{\cal H}_f\beta}\big)~=~e^{-N({N}_b~-~{N}_f)\omega\beta}~=~1&#xD;&#xA;$$&#xA;since ${N}_b~=~{N}_f$ in unbroken supersymmetry.  As such we then have that&#xA;$$&#xD;&#xA;Tr\big((-1)^{N_F}e^{-N{\cal H}_f\beta}\big)~=~{1\over{\zeta(N\omega\beta)}}.&#xD;&#xA;$$&#xA;and the fermion partition function may be written with the Mobius function $\mu$ as&#xA;$$&#xD;&#xA;Tr\big((-1)^{N_f}e^{-NH_f\beta}\big)~=~\sum_i{{\mu(\lambda_i)}\over{\lambda_i}^{N\omega\beta}}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The supersymmetry is partially broken, only half the stringy bosons have superpairs then the bosonic Hamiltonian takes the form&#xA;$$&#xD;&#xA;{\cal H}_B~=~2N\sum_klog(\lambda_k)c^\dagger_k c_k~+~N\sum_k log(\lambda_k)f_k^\dagger f_k.&#xD;&#xA;$$&#xA;where $c_k~=~a_ka_k$ and $c_k^\dagger~=~a_k^\dagger a_k^\dagger$.  The operators $f_k$ and $f^\dagger_k$ are fermionic operators \lq\lq recovered\rq\rq$~$from the bosonic Hamiltonian.  Now if we compute $Z_F(\beta)$ we find that&#xA;$$&#xD;&#xA;Z_F(\beta)~=~ Tr\big((-1)^{N_F}e^{-N\sum_k log(\lambda_k)f^\dagger_k f_k\beta} \big)\times 1&#xD;&#xA;$$&#xA;$$&#xD;&#xA;~=~ Tr\big((-1)^{N_F}e^{-N\sum_k log(\lambda_k)f^\dagger_k f_k\beta} e^{-2N\sum_k log(\lambda_k)c^\dagger_k c_k\beta}e^{-2N\sum_k log(\lambda_k)b^\dagger_k b_k\beta}\big)&#xD;&#xA;$$&#xA;$$&#xD;&#xA;~=~Tr\big((-1)^{N_f}e^{-\big(2N{\cal H}_F~+~N{\cal H}_B\big)\beta}\big).&#xD;&#xA;$$&#xA;where the unity involves the old fermion and boson operators for unbroken supersymmetry.  Now $Z_F(\beta)$ is&#xA;$$&#xD;&#xA;Z_F(\beta)~=~ \sum_{n=1}^\infty{{\mu(n)}\over{n^{2N\omega\beta}}}~=~ {{\zeta(N\omega\beta)}\over{\zeta(2N\omega\beta)}}.&#xD;&#xA;$$&#xA;Here the $c,~c^\dagger$ boson fields have superpartners and the $f,~f^\dagger$ fields do not.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-30T14:58:56.220" CommentCount="2" />
  <row Id="4224" PostTypeId="2" ParentId="4174" CreationDate="2011-01-30T15:29:18.643" Score="0" ViewCount="" Body="&lt;p&gt;Neutron stars are stable end points of stellar evolution.  So long as their mass is less than whatever the modern, updated version of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chandrasekhar_limit&quot; rel=&quot;nofollow&quot;&gt;Chandrasekhar mass&lt;/a&gt; is (something like 1.5-2.2 times the mass of the sun), they won't collapse to a black hole.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they're over the Chandrasekhar mass, then the core will  be the most compact part of the NS, and centrifugal forces will be less important, because the mass will be close to the center of rotation, and the core will collapse first, and you'll get a NS supernova.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't get the premise of this question.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-01-30T15:29:18.643" CommentCount="1" />
  <row Id="4225" PostTypeId="2" ParentId="4199" CreationDate="2011-01-30T16:37:17.677" Score="-2" ViewCount="" Body="&lt;p&gt;I am afraid that the &lt;strong&gt;radiation&lt;/strong&gt; cannot cancel everywhere so it is better to say the radiation does not occur in case of a constant current. This is so because, according to Maxwell equations, it is not acceleration of a single charge that creates the radiation but the current time-dependence at a given point. In other words, different sources radiate differently and it is not reduced to the sum of radiations. The total filed is determined differently: superposition of fields is not a sum of radiations! The same is valid in the opposite case of short bunches where the radiative losses are proportional to the charge squared ( = source-dependent phenomenon).&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-01-30T16:49:57.843" LastActivityDate="2011-01-30T16:49:57.843" CommentCount="10" />
  <row Id="4226" PostTypeId="1" AcceptedAnswerId="5305" CreationDate="2011-01-30T16:44:51.330" Score="6" ViewCount="220" Body="&lt;p&gt;The latest I read is 3 but that was in Oct. With Lene Hau of Harvard's &quot;frozen light&quot; and with quantum donuts, newer strategies for stabilization are appearing, but the problem of keeping the qubit in superposition for long enough to sample it seems boggling.&#xA;&lt;a href=&quot;http://www.sciencedaily.com/releases/2009/03/090309105026.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencedaily.com/releases/2009/03/090309105026.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.seas.harvard.edu/haulab/publications/HauPublications_All.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.seas.harvard.edu/haulab/publications/HauPublications_All.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-02-16T21:07:55.483" Title="How many stabilised qubits have been achieved in Quantum Computing?" Tags="&lt;quantum-information&gt;&lt;condensed-matter&gt;&lt;quantum-optics&gt;" AnswerCount="1" />
  <row Id="4227" PostTypeId="2" ParentId="4172" CreationDate="2011-01-30T16:46:44.400" Score="1" ViewCount="" Body="&lt;p&gt;The &lt;em&gt;meaning&lt;/em&gt; is that either one shows that one or more of the axioms or some hidden premise is not acceptable for some reason, or one accepts the conclusion; or one can live in a state of &quot;there's something wrong with that, I just can't find it&quot; for as long as forever takes. If that &lt;em&gt;means&lt;/em&gt; sleepless nights, so be it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FWIW, a long time ago I chose to &lt;em&gt;stop talking about particles&lt;/em&gt;. I haven't followed the Conway-Specker literature, since I argued some years ago that Bell inequalities cause almost no problem for random fields and no-one has so far contradicted &lt;em&gt;that&lt;/em&gt; argument [go at it now, if you like, &quot;Bell inequalities for random fields&quot;, &lt;a href=&quot;http://dx.doi.org/10.1088/0305-4470/39/23/018&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;J. Phys. A: Math. Gen.&lt;/em&gt; &lt;strong&gt;39&lt;/strong&gt; (2006) 7441-7455&lt;/a&gt;, &lt;a href=&quot;http://arxiv.org/abs/cond-mat/?0403692&quot; rel=&quot;nofollow&quot;&gt;cond-mat/0403692&lt;/a&gt;, if you haven't previously decided it's wrong and decided not to publish a rebuttal]. Conway and Specker introduce assumptions that &lt;em&gt;are&lt;/em&gt; fairly reasonable for classical particles but it's easy to exhibit that they do not hold for random fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The MIN axiom, in particular, is generally not true for quantum fields, because there &lt;em&gt;are&lt;/em&gt; correlations at space-like separation in QFTs, without, because of microcausality, any need for there to be transmission of causal effects between space-like separated regions.&#xA;The MIN axiom is also generally not true for random fields. There are correlations, but there is no causality, because there is microcausality at time-like and light-like separation as well as at space-like separation. The correlations, to your taste, can either be pre-existing (the Conspiracy Loophole, if you like, but the conspiracy is probabilistic, not deterministic) or, instrumentally, they're there just because they're there (that is, we observe them, so they're there). The FIN axiom is harder, because then one has to be convinced that negative frequency components of the random field do not cause trouble, but of course classical electromagnetism lives happily with negative frequencies without problems of causality outside the light-cone, negative frequencies occur in QFT loop diagrams without difficulty, and also frequency is not linearly related to energy for classical fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For random fields, since they're not well known, you can try my approach in &quot;Equivalence of the Klein-Gordon random field and the complex Klein-Gordon quantum field&quot;, &lt;a href=&quot;http://dx.doi.org/10.1209/0295-5075/87/31002&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;EPL&lt;/em&gt; &lt;strong&gt;87&lt;/strong&gt; (2009) 31002&lt;/a&gt;, &lt;a href=&quot;http://arxiv.org/abs/0905.1263&quot; rel=&quot;nofollow&quot;&gt;arXiv:0905.1263 quant-ph&lt;/a&gt;, where a few references to other people's work can also be found. It's a niche, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Move on&lt;/em&gt;, is what this theorem means; in my case on to random fields, but it's best if we all make different choices. You might, alternatively, be happy to drop locality and/or classical states and observables and/or something else, or to lose sleep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I think you could do much worse than to accept the answer on Luboš Motl's blog, which I found very interesting, so I voted up his comment on Roy Simpson's Answer as proxy.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-01-30T16:46:44.400" CommentCount="5" />
  <row Id="4228" PostTypeId="2" ParentId="4174" CreationDate="2011-01-30T16:53:17.307" Score="1" ViewCount="" Body="&lt;p&gt;Excerpt from a report I once wrote on this topic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://bytepawn.com/pdf/Exciting.pdf&quot; rel=&quot;nofollow&quot;&gt;http://bytepawn.com/pdf/Exciting.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitationally bound compact objects --- such as &lt;strong&gt;neutron stars&lt;/strong&gt; --- cannot have pulsar periods less than 0.3msec. The pulsar period is of course radius and mass dependent, but&#xA;is bound from below by a neutron star just on the verge of forming a black hole&#xA;singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Strange quark stars&lt;/strong&gt; are hypothetical objects qualitatively diﬀerent from other&#xA;compact stars such as neutron stars. First, they are composed of strange quark&#xA;matter, a hypothetical form of matter consisting of equal part u, d and s quarks.&#xA;This phase must be of lower energy than hadronic matter for a strange star to be&#xA;stable to quark fusion. Second, unlike neutrons stars which are held together by&#xA;gravity, strange stars are bound by the strong interaction. As strange stars are not bound by gravity, they are free to occupy the sub-millisecond pulsar region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limiting angular velocity for a gravitationally bound star goes as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(M/R^3)^{1/2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This comes from Newtonian arguments (centrifugal vs. gravitation force). The GR version turns out to have the same form, just a different multiplicative constant (&quot;accurate to better than 10%&quot; according to Glendenning). So, to maximise the angular velocity, you maximise M and minimize R. But, if you do that, you make the start more compact, and eventually it forms a black hole! Quantitatively, there's your limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But note that the limit is not one number it's actually a function mass M. To get rid of R, use the fact that according to Einstein's equations of stellar structure (Oppenheimer-Volkoff equations for hydrostatic equilibrium)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M/R &amp;lt; 4/9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end the formula for the rotational period is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P &amp;gt; 0.167 M / M_{sun} [msec]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more please check the book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387989773&quot; rel=&quot;nofollow&quot;&gt;Compact Stars&lt;/a&gt; by N. K. Glendenning, it's very good.&lt;/p&gt;&#xA;" OwnerUserId="47" LastEditorUserId="47" LastEditorDisplayName="" LastEditDate="2011-01-30T23:01:29.043" LastActivityDate="2011-01-30T23:01:29.043" CommentCount="2" />
  <row Id="4229" PostTypeId="1" AcceptedAnswerId="4246" CreationDate="2011-01-30T17:42:50.803" Score="2" ViewCount="84" Body="&lt;p&gt;Can one construct a filter out of nothing but pure metal that will pass only right circular polarized radiation and reflect left circularly polarized radiation?&lt;br&gt;&#xA;What would it look like?  A helical spring in a tube?&lt;br&gt;&#xA;What about a twisted metalic ribbon inside a circular conducting tube?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An axial mode helical antenna transmits circularly polarized radiation.&lt;br&gt;&#xA;If I put an axial mode antenna (perhaps with a resistive load?) in a waveguide, can I create a filter that passes only say left circularly polarized radiation?&lt;br&gt;&#xA;I would ideally like to do it with only conductive (i.e. not resistive) material.&lt;br&gt;&#xA;I would also like it to reflect (not absorb) radiation of the opposite circular polarization.&lt;br&gt;&#xA;Can someone tell me how to do this?&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-01-31T00:42:29.893" Title="Can one make an axial mode circularly polarized radio wave filter with purely conducting materials?" Tags="&lt;electromagnetic-radiation&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4230" PostTypeId="2" ParentId="4184" CreationDate="2011-01-30T18:09:27.760" Score="-6" ViewCount="" Body="&lt;p&gt;And here we are! We do not need renormalizability anymore! Congratulations, physicists! Everything is effective and does not work in some limits. How natural it is!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But wait, even an effective theory should predict phenomena occurring with the &lt;strong&gt;probability 1&lt;/strong&gt;. Soft radiation, for example. Does QED predict soft radiation if the first Born approximation? No! The same statement is valid for those &quot;effective&quot; theories that treat the essential coupling perturbatively. Therefore we have to reformulate our effective theories so that at least the most probable events are obtained automatically in the first Born approximation. Otherwise those &quot;effective&quot; theories are &lt;strong&gt;completely ineffective&lt;/strong&gt;. I think we have a very huge problem in QFT - we miss everything as we have to justify our &quot;on go&quot; fixes and reparations of our initially dead theory. And when it does not work, they say &quot;it's OK anyway -  the theory is &quot;effective&quot;&quot;. What a shame! It is presenting a failure as an achievement!&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-01-31T15:02:13.753" LastActivityDate="2011-01-31T15:02:13.753" CommentCount="6" />
  <row Id="4231" PostTypeId="2" ParentId="4141" CreationDate="2011-01-30T18:33:02.090" Score="-1" ViewCount="" Body="&lt;p&gt;Beauty is harmful to theoretical physics if it is canonized. Then you factually deal with a beautiful mathematical physics but not with physics. Remember P. Dirac. He wrote his famous equation to satisfy the unitarity requirement or so. Later on his equation was adopted for a quite different use - QFT with antiparticles and pair creations rather than for description of one electron. Moreover, up to his death P. Dirac insisted that QED he'd created had a wrong interaction (or the total Hamiltonian). He invited researchers to write better Hamiltonians because despite &quot;beauty&quot; of his equation, the physics in QED was implemented in a wrong way. Many remember the UV and IR divergences in this respect. I will tell you: the ugliness (wrongness) of the current QED is easily seen even in the first Born approximation where one obtains the famous elastic cross section, i.e., even before encountering divergences. The first Born approximation does not predict the soft radiation - the process occurring with unity probability. Than means a very bad initial approximation for interacting particles: the field and &quot;particle&quot; subsystems are decoupled in it. No wonder one obtains an &quot;explosion&quot; of the perturbative corrections in the next approximation.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-01-30T22:20:01.367" LastActivityDate="2011-01-30T22:20:01.367" CommentCount="7" />
  <row Id="4232" PostTypeId="2" ParentId="4180" CreationDate="2011-01-30T22:01:46.363" Score="3" ViewCount="" Body="&lt;p&gt;My brother, an electrical engineer, used to carry around a 1 megaohm resistor&#xA;during the dry winter months when you easily get a shock after walking across a carpet and touching a light switch or another person. If you hold one lead of the resistor in your hand and touch the light switch or whatever else you are touching with the other the discharge turns from a nasty shock into a very mild and kind of amusing fizzle. It's kind of fun and I'm sure you can find a resistor with large enough resistance lying around in your local physics lab, so give it a try! Of course I don't know anybody geeky enough to actually use this as a practical solution.&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2011-03-02T12:01:51.060" LastActivityDate="2011-03-02T12:01:51.060" />
  <row Id="4233" PostTypeId="1" AcceptedAnswerId="4235" CreationDate="2011-01-30T22:47:01.743" Score="3" ViewCount="139" Body="&lt;p&gt;I think that, by now, it's understood that the gluon propagator in QCD has a dynamically generated mass. Ok, so my question is the following: where does the extra polarization degree of freedom come from? Or, asking in another way: suppose you try to define an S matrix for QCD, apart from the usual problems for doing so, would it be unitary? How?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a Higgs mechanism, it is clear that the extra degree of freedom comes from &quot;eating&quot; the Higgs, as they say. But and where does is come from in theories where the mass is dynamically generated?&lt;/p&gt;&#xA;" OwnerUserId="1337" LastActivityDate="2011-03-02T12:01:21.823" Title="Polarization of the gluon" Tags="&lt;quantum-field-theory&gt;&lt;quantum-chromodynamics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="4234" PostTypeId="1" AcceptedAnswerId="4249" CreationDate="2011-01-30T23:00:08.250" Score="1" ViewCount="302" Body="&lt;p&gt;What are the factors that affect the density of an atom?&lt;/p&gt;&#xA;" OwnerUserId="158" LastActivityDate="2011-01-31T01:05:26.653" Title="Why is the relationship between atomic number and density not linear?" Tags="&lt;atoms&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="4235" PostTypeId="2" ParentId="4233" CreationDate="2011-01-30T23:02:49.823" Score="5" ViewCount="" Body="&lt;p&gt;The asymptotic states of QCD are gauge invariant. They can include mesons which are quark-anti quark bound states and glueballs (which are roughly speaking bound states of gluons) but not gluons themselves. It doesn't really make any sense to say that the gluon propagator has a dynamically generated mass as this is a very gauge dependent statement and gluons are not asymptotic states with a well defined mass. Glueballs are massive and can have various spins, but there no puzzle regarding counting degrees of freedom because one is not starting with a perturbative state with fewer degrees of freedom and adding an additional degree of freedom as in the Higgs mechanism. This is not to say that the mass generation for glueballs is obvious. As a matter of fact, proving that pure glue QCD has a mass gap will win you a million dollars from the Clay Institute.&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2011-03-02T12:01:21.823" LastActivityDate="2011-03-02T12:01:21.823" CommentCount="1" />
  <row Id="4236" PostTypeId="1" CreationDate="2011-01-30T23:12:23.907" Score="2" ViewCount="253" Body="&lt;p&gt;I have searched the web for good answers to why refraction occurs when light moves from one medium to another with different density. I have limited background in physics and want to know if there is an easy way to understand this without having to go back to school for years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most explanations come in some variant of the &quot;Marching line of soldiers&quot; analogy, where a straight line of soldiers change pace when they hit an boundary. I have no problem understanding the analogy, my problem is understanding how this is even relevant to light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The analogy does not work for light as a particle. The particle does not &quot;know&quot; of particles around it, and should follow a straight line. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) How does a single ray of light (wave), as part of a &quot;wave front&quot; know what other rays are doing? &lt;/p&gt;&#xA;" OwnerUserId="1675" LastEditorUserId="1675" LastEditorDisplayName="" LastEditDate="2011-01-31T21:32:43.027" LastActivityDate="2011-08-20T04:48:02.530" Title="Refraction question" Tags="&lt;light&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="4237" PostTypeId="2" ParentId="4236" CreationDate="2011-01-30T23:36:27.107" Score="1" ViewCount="" Body="&lt;p&gt;This is an interesting question in that it is one that involves mostly the understanding of the questioner. A similar question asked by someone who used different terminology would deserve a different answer. The one I'm providing is, I hope, compatible with the understanding of the questioner. It is not a complete explanation of the situation as would be suitable for a physics grad student. With that caveat...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The particle does not &quot;know&quot; of&#xA;  particles around it, and should follow&#xA;  a straight line.&#xA;  In contemporary physics, light is considered as both a particle (the photon) and also a wave (not in the sense of the waves of electricity and magnetism, but instead in the sense of a quantum wave). A quantum wave does know about more than just the classical path.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;How does a single ray of light (wave),&#xA;  as part of a &quot;wave front&quot; know what&#xA;  other rays are doing?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the way this question is written, I'm assuming that the writer sees light as a sort of classical point particle (whose trajectories would follow &quot;rays&quot;), but with the added feature of being a wave (and therefore having an existence along the ray, as is necessary for a wave to have a definite energy or momentum). The question is sophisticated in that the writer understands that light has wave properties and I'll try to move that understanding a little ways further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light (even light that is &quot;classical&quot; that is, that follows Maxwell's laws of electricity and magnetism) does not follow rays. A ray is an idealization that applies only when the width of the ray is large compared to the wavelength of the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could try to create a ray of light more narrow than a wavelength of the light by shining the light on a sheet of copper (or other thin opaque material) with a single hole drilled in it. If we make the hole smaller than the wavelength of light than it stands to reason that the ray that comes out of it will have a diameter smaller than the wavelength of light. But in fact, this is not what happens. Instead, light is scattered by the hole and diverges in all directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect is similar to what would happen if you had an obstruction to ocean waves but with a single hole in it. Assuming the hole is small compared to the wavelength of the ocean waves, the waves that come through it will not appear as a ray of waves, but instead will spread away from the hole in all directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than &quot;refraction&quot; this feature of light is called &quot;diffraction.&quot; The wikipedia article on diffraction ( &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffraction&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Diffraction&lt;/a&gt; ) has a useful photo of water waves diffracting. Light operates the same way. I've added red lines to show the barrier and green arrows to show the direction that the waves are moving. The wave begins at the bottom left and moves towards the top right:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Gc7lI.jpg&quot; alt=&quot;Diffraction Demonstration&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-30T23:36:27.107" CommentCount="2" />
  <row Id="4238" PostTypeId="1" AcceptedAnswerId="4282" CreationDate="2011-01-30T23:48:03.783" Score="22" ViewCount="680" Body="&lt;p&gt;Why do we have an elementary charge $e$ in physics but no elementary mass? Is an elementary mass ruled out by experiment or is an elementary mass forbidden by some theoretical reason? &lt;/p&gt;&#xA;" OwnerUserId="1648" LastActivityDate="2011-08-26T05:41:36.047" Title="Why do we have an elementary charge but no elementary mass?" Tags="&lt;mass&gt;" AnswerCount="7" FavoriteCount="2" />
  <row Id="4239" PostTypeId="1" CreationDate="2011-01-30T23:57:39.150" Score="3" ViewCount="93" Body="&lt;p&gt;I heard the term the other day, but it seems strange to me. My understanding is that neutron stars are made up of neutrons; and neutrons (having no charge) shouldn't be magnetised. &lt;/p&gt;&#xA;" OwnerUserId="158" LastActivityDate="2011-03-02T03:56:34.463" Title="What is a magnetised neutron star?" Tags="&lt;astrophysics&gt;&lt;neutron-stars&gt;&lt;neutron&gt;" AnswerCount="2" />
  <row Id="4240" PostTypeId="2" ParentId="4233" CreationDate="2011-01-30T23:59:49.650" Score="2" ViewCount="" Body="&lt;p&gt;I'm not sure this is useful, but I suppose that the problem with an S-matrix for gluons is that the gluons are not free. That is, an S-matrix deals with free particles that interact, perhaps exchange bodily fluids (i.e. charge or whatever), and then escape to infinity. But the real questions with gluons have to do with bound states. So the problem is the absence of free particles at the beginning and end of the interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I want to believe that an S-matrix &quot;formalism&quot; exists even with bound particles. Instead of letting the participants escape to infinity, you assume that they eventually return to their initial state. Then perhaps you can derive some restrictions based on the requirement that the initial state repeating. This is, sort of like a dual to the usual S-matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the wikipedia article on S-matrix theory: &lt;a href=&quot;http://en.wikipedia.org/wiki/S-matrix&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/S-matrix&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-30T23:59:49.650" CommentCount="1" />
  <row Id="4241" PostTypeId="2" ParentId="4238" CreationDate="2011-01-31T00:12:08.573" Score="1" ViewCount="" Body="&lt;p&gt;I think it's beause do not have a fundamental understanding of mass. If we did, maybe that fundamental unit would have some relationship (i.e. tiny fraction) with the Planck mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current effort in that direction probably begins with understanding the Higgs. There are several competing theories of the Higgs. They don't even agree on the number of such particles. So in that sense, the ball is in the experimentalist court.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-31T00:12:08.573" />
  <row Id="4242" PostTypeId="2" ParentId="4234" CreationDate="2011-01-31T00:20:41.200" Score="-1" ViewCount="" Body="&lt;p&gt;I think your argument is assuming that the density of a material is due to the density of the atoms and I'll try to answer it as such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is largely because of chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the noble gasses all have low density because they're gasses. They're gasses because they don't combine chemically. They don't combine chemically because all their atomic orbitals are already filled. And their atomic orbitals are filled because they have just the &quot;right&quot; number of electrons. On the other hand, the alkali metals all have a single extra electron. They all form metals and so are considerably denser than the noble gasses right next to them in atomic number.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-31T00:20:41.200" CommentCount="3" />
  <row Id="4243" PostTypeId="1" AcceptedAnswerId="4255" CreationDate="2011-01-31T00:23:36.883" Score="8" ViewCount="457" Body="&lt;p&gt;I can understand that on small scales (within an atom/molecule), the other forces are much stronger, but on larger scales, it seems that gravity is a far stronger force; e.g. planets are held to the sun by gravity. So what does it mean to say that &quot;gravity is the weakest of the forces&quot; when in some cases, it seems far stronger?&lt;/p&gt;&#xA;" OwnerUserId="158" LastActivityDate="2011-08-03T08:58:33.190" Title="What does it mean to say &quot;Gravity is the weakest of the forces&quot;?" Tags="&lt;gravity&gt;&lt;forces&gt;" AnswerCount="7" />
  <row Id="4245" PostTypeId="1" CreationDate="2011-01-31T00:34:07.940" Score="0" ViewCount="145" Body="&lt;p&gt;The alkali earth metals form the two left-most column of the periodic table of the elements, other than hydrogen. See wikipedia articles:&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Alkali_metal&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Alkali_metal&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Alkaline_earth_metal&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Alkaline_earth_metal&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now &quot;alkali&quot; means that these are in a sense the opposite of &quot;acid&quot;. One of the ancient uses of alkali materials is in the conversion of corn into &lt;a href=&quot;http://en.wikipedia.org/wiki/Hominy&quot; rel=&quot;nofollow&quot;&gt;hominy&lt;/a&gt; (or, for those in the southern US, &lt;a href=&quot;http://en.wikipedia.org/wiki/Grits&quot; rel=&quot;nofollow&quot;&gt;grits&lt;/a&gt; similar to posole, polenta or farina ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conversion of corn to hominy is of ancient origin. From a nutrition point of view it is useful because it increases the availability of the lysine and tryptophan proteins in the corn. See &lt;a href=&quot;http://www.practicallyedible.com/edible.nsf/pages/hominy&quot; rel=&quot;nofollow&quot;&gt;http://www.practicallyedible.com/edible.nsf/pages/hominy&lt;/a&gt; This helps prevent the nutritional disease pellagra &lt;a href=&quot;http://en.wikipedia.org/wiki/Pellagra&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Pellagra&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In ancient times, hominy was manufactured by soaking corn in a mixture of water and ashes. This is equivalent to the modern method, which is to soak the corn in lye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is this: why is it that ashes are alkali?&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-03T23:49:09.370" ClosedDate="2011-02-03T23:56:30.167" Title="Why are the alkali and alkaline earth metals present in the ashes of burnt wood?" Tags="&lt;physical-chemistry&gt;&lt;chemical-compounds&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="4246" PostTypeId="2" ParentId="4229" CreationDate="2011-01-31T00:42:29.893" Score="3" ViewCount="" Body="&lt;p&gt;From the way your question is written, I believe that you already know how to make a wave guide that will split horizontal from vertical waves. I'm assuming that you also know how to combine and split radio waves. Furthermore, I'm going to assume that you can build a &quot;splitter&quot;, that is a device where radio waves come in and are split into two outputs, one where the horizontal waves go the other where the vertical waves go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So build a horizontal and a vertical wave splitter. Combine their outputs and inputs so that they are sufficiently out of phase and the combined apparatus will be a circular splitter. This is a matter of applying the law of superposition.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-31T00:42:29.893" />
  <row Id="4247" PostTypeId="1" AcceptedAnswerId="4259" CreationDate="2011-01-31T00:47:50.263" Score="1" ViewCount="252" Body="&lt;p&gt;This is more of a math question and one, furthermore, that I know the final answer to.  What I am asking is more of a &quot;how do I get there&quot; question as this question was generated during a self study situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for a flat plate with a meniscus on it at some contact angle the form is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a y''$/(1+(y')^2)^{3/2}$ + b g y = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where a is surface tension, b is density, and g is the usual 9.8 m$/s^2$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I know that when you &quot;A first integration, together with the boundary condition dy/dx=y=0 as x goes to infinity yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1$/(1+y'^2)^{1/2}$=1-(bg/2a)$y^2$    &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--From de Gennes on Menisci&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is with the integration.  How do I integrate something like the 1st equation?  I find myself running in circles.  Worse, I know that this type of integration is something I've run across numerous times, but I can't seem to find the methodology in the current stack of books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I beg, can someone either tell me where to look to find the appropriate methodology to attack this problem or take me through the process?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be grateful for as long as it would be useful for you.&#xA;Thanks,&#xA;Sam&lt;/p&gt;&#xA;" OwnerUserId="1037" LastActivityDate="2011-01-31T02:36:21.090" Title="Solving a Young Laplace equation for a meniscus against a flat plate" Tags="&lt;mathematics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="4248" PostTypeId="2" ParentId="4234" CreationDate="2011-01-31T00:51:55.203" Score="2" ViewCount="" Body="&lt;p&gt;Examine the densities of di-atomic gaseous elements ($\text{H}_2$, $\text{N}_2$, $\text{O}_2$) at STP. Compare to their respective atomic weights. Also look at the densities of the noble &lt;em&gt;gasses&lt;/em&gt; at STP. Again, note how they scale roughly with the atomic weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By choosing materials that are more or less ideal gases I've isolated the system from the effects of almost all the messy chemistry that cause other states of matter to not scale (the exception is the question of how many atoms go into each molecule which is why I separated my examples into two groups).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the condensed states of mater (liquid and solids) the chemistry involves itself more strongly in terms of inter-molecular forces. This is most evident in crystalline solids as the geometry of the crystal sets the number density of atoms in terms of both the length of inter-atomic bonds and the symmetries of the crystalline structure.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-01-31T01:02:11.273" LastActivityDate="2011-01-31T01:02:11.273" />
  <row Id="4249" PostTypeId="2" ParentId="4234" CreationDate="2011-01-31T00:59:35.583" Score="4" ViewCount="" Body="&lt;p&gt;At an atomic scale, there are two things that go into the density, which is $\rho=\frac{\mbox{mass}}{\mbox{volume}}$. First, we have the number density, which is  $n=\frac{\mbox{number of atoms}}{\mbox{volume}}$. Then we also have the mass per atom, $\mu=\frac{\mbox{mass}}{\mbox{atom}}$. It is easy to see by combining these equations that the density is then $\rho=n\mu$ - it depends on both number density and atomic weights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the mass per atom, $\mu$, goes up linearly (very roughly speaking) as you go up in atomic number, the number density does &lt;em&gt;not&lt;/em&gt; go up linearly. This is because interactions between the actual atoms cause them to cluster closer together, so you get a higher value of $n$ in the equation above for some atoms, but a much lower one for others. For example, in that graph you linked to above:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/6iOi9.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those spikes in density are caused by much higher number density, which is in turn caused by metallic attractions between the atoms. The deepest valleys are the noble gases which have virtually no mutual attractions whatsoever.&lt;/p&gt;&#xA;" OwnerUserId="1655" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-01-31T01:05:26.653" LastActivityDate="2011-01-31T01:05:26.653" />
  <row Id="4250" PostTypeId="2" ParentId="3355" CreationDate="2011-01-31T01:03:34.673" Score="-1" ViewCount="" Body="&lt;p&gt;The nucleus is not at a single point in space. It's a quantum object and so does not have a determinate position. Consequently, in a real physical atom there is no single point that the electron has to go through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The apparent problem arises only from our approximations. To get a problem where the nucleus didn't move we'd have to have an infinite mass nucleus and this would lead to a gravitational singularity. Even requiring the charge of the nucleus originate from a single point would require singularities. These are not physical problems, they are math problems.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-31T01:03:34.673" CommentCount="2" />
  <row Id="4251" PostTypeId="2" ParentId="4245" CreationDate="2011-01-31T01:07:59.153" Score="1" ViewCount="" Body="&lt;p&gt;Where I live we have a problem with a bushy plant called Tamarix.  This is a nasty invasive plant that takes over everything.  It is also practically indestructible.  It is called salt cedar because it salinates soils.  These voracious bushes have long tap roots which reach the water table and pull up water with alkaline salts.  The plant then secretes most of this into the soil, further damaging ecosystems.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an extreme example, but trees usually tap root into the water table where there are dissolved salts and alkaline metal compounds.  These then accumulate in the xylem.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-31T01:07:59.153" CommentCount="1" />
  <row Id="4252" PostTypeId="2" ParentId="4243" CreationDate="2011-01-31T01:11:49.063" Score="4" ViewCount="" Body="&lt;p&gt;Gravity seems stronger because it's always attractive. Of the other 3 interactions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Electromagnetism has positive and negative charges, so it only manifests macroscopically when there is a charge imbalance.&lt;/li&gt;&#xA;&lt;li&gt;The weak and strong interactions are intrinsically short-ranged.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1656" LastActivityDate="2011-01-31T01:11:49.063" CommentCount="1" />
  <row Id="4253" PostTypeId="2" ParentId="4243" CreationDate="2011-01-31T01:13:31.713" Score="0" ViewCount="" Body="&lt;p&gt;This is indeed something one has to be careful about, because, after all, gravity scales with the mass of the particles in question, whereas the other forces scale with the electric charge or the magnetic moment. It appears that one compares apples with pears. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I believe the declaration that gravity is the &quot;weakest&quot; of the forces stems indeed solely from its irrelevancy on the scale of particle physics.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-01-31T01:13:31.713" CommentCount="1" />
  <row Id="4254" PostTypeId="2" ParentId="4239" CreationDate="2011-01-31T01:16:09.100" Score="2" ViewCount="" Body="&lt;p&gt;In spite of the name, neutron stars also contain protons and electrons. This is required for equilibrium with respect to weak-interaction processes which can convert neutrons into protons and electrons. Since neutron stars also contain protons and electrons, they can contain electric currents which generate magnetic fields. It is thought that the protons actually form a superconductor and the magnetic field in most of the interior of a neutron star is carried my quantized magnetic vortices, much as in a type II superconductor.&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T03:56:34.463" LastActivityDate="2011-03-02T03:56:34.463" />
  <row Id="4255" PostTypeId="2" ParentId="4243" CreationDate="2011-01-31T01:19:09.553" Score="13" ViewCount="" Body="&lt;p&gt;When we ask &quot;how strong is this force?&quot; what we mean in this context is &quot;How much stuff do I need to get a significant amount of force?&quot; Richard Feynman summarized this the best in comparing the strength of gravity - which is generated by &lt;em&gt;the entire mass of the Earth&lt;/em&gt; - versus a relatively tiny amount of electric charge:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And all matter is a mixture of&#xA;  positive protons and negative&#xA;  electrons which are attracting and&#xA;  repelling with this great force. So&#xA;  perfect is the balance however, that&#xA;  when you stand near someone else you&#xA;  don't feel any force at all. If there&#xA;  were even a little bit of unbalance&#xA;  you would know it. If you were&#xA;  standing at arm's length from someone&#xA;  and each of you had one percent more&#xA;  electrons than protons, the repelling&#xA;  force would be incredible. How great?&#xA;  Enough to lift the Empire State&#xA;  building? No! To lift Mount Everest?&#xA;  No! The repulsion would be enough to&#xA;  lift a &quot;weight&quot; equal to that of the&#xA;  entire earth!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another way to think about it is this: a proton has both charge and mass. If I hold another proton a centimeter away, how strong is the gravitational attraction? It's about $10^{-57}$ newtons. How strong is the electric repulsion? It's about $10^{-24}$ newtons. How much stronger is the electric force than the gravitational? We find that it's $10^{33}$ times stronger, as in 1,000,000,000,000,000,000,000,000,000,000,000 times more powerful!&lt;/p&gt;&#xA;" OwnerUserId="1655" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-01-31T01:40:16.890" LastActivityDate="2011-01-31T01:40:16.890" CommentCount="1" />
  <row Id="4256" PostTypeId="1" AcceptedAnswerId="4388" CreationDate="2011-01-31T01:19:11.753" Score="3" ViewCount="267" Body="&lt;p&gt;I am wondering: The noble 'gases' are inert because they have closed shells and don't want to give that up. But the noble metals, such as Copper, Silver, Rhodium, Gold, don't seem to have this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, Rhodium has electron configuration $4d^{8}\, 5s^1$. To me, this looks very similar to the alkali metals, which we all know are 'very' reactive. So why doesn't Rhodium happily give up its s-electron? Same with Silver, $4d^{10}\, 5s^1$. Why not give up the s electron?&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-02-01T17:24:37.270" Title="Why are the noble metals inert?" Tags="&lt;physical-chemistry&gt;" AnswerCount="3" />
  <row Id="4257" PostTypeId="2" ParentId="4243" CreationDate="2011-01-31T01:31:04.983" Score="8" ViewCount="" Body="&lt;p&gt;When we say that gravity is much weaker then the other forces we mean that its &lt;strong&gt;coupling constant&lt;/strong&gt; is much smaller than the coupling constants of other forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about a coupling constant as a parameter that says how much energy there will be in per &quot;unit of interacting stuff&quot;. This is a very rough definition but it will serve our purpose. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you determine the coupling constants of all different forces, you discover that, in decreasing order, strong, eletromagnetic and weak forces are much, much stronger than gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need around $10^{32}$ (that is 100,000,000,000,000,000,000,000,000,000,000) times more &quot;stuff interacting&quot; to get around the same energy scale with gravity if you compare it with the weak force. Moreover, the difference between strong, weak and electromagnetic forces among themselves isn't nearly as extreme as the difference between gravity and the other forces.&lt;/p&gt;&#xA;" OwnerUserId="673" LastEditorUserId="673" LastEditorDisplayName="" LastEditDate="2011-01-31T03:59:47.803" LastActivityDate="2011-01-31T03:59:47.803" CommentCount="7" />
  <row Id="4258" PostTypeId="2" ParentId="2891" CreationDate="2011-01-31T01:44:56.277" Score="0" ViewCount="" Body="&lt;p&gt;Of course it doesn't matter what voltage the voltage planes are held at so assume both are at ground. Now, take advantage of symmetry: assume that the voltages on the two strips are opposite. Use the same logic as with the presence of a charge near a conducting plane implies the presence (for calculation purposes) of an image charge. Therefore the problem reduces to one of determining the impedance of a single strip in a trough. I'll draw something up in paint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/MusfF.jpg&quot; alt=&quot;stackup&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be able to look up the impedance of the above single asymmetric stripline. If not, you can calculate it. A reference I quickly found that seems useful is:&lt;br&gt;&#xA;&lt;a href=&quot;http://lss.fnal.gov/archive/tm/TM-1270.pdf&quot; rel=&quot;nofollow&quot;&gt;http://lss.fnal.gov/archive/tm/TM-1270.pdf&lt;/a&gt;&lt;br&gt;&#xA;but I bet you can find better in your electrical engineering library.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-31T01:44:56.277" CommentCount="4" />
  <row Id="4259" PostTypeId="2" ParentId="4247" CreationDate="2011-01-31T01:52:42.210" Score="3" ViewCount="" Body="&lt;p&gt;I think it's a simple substitution: recognizing the derivative of some part of the expression in another part of the expression... in a useful way. :-D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivative of $1 + (y&amp;#39;)^2$ almost appear in the numerator, but the $y&amp;#39;$ is lacking. Let's try multiplying the equation by $y&amp;#39;$, as the second term looks easier:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a y&amp;#39;&amp;#39; y&amp;#39;}{(1+(y&amp;#39;)^2)^\frac{3}{2}} + b g yy&amp;#39; = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can integrate the first term using $u = 1+(y&amp;#39;)^2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int dx \frac{a y&amp;#39;&amp;#39; y&amp;#39;}{(1+(y&amp;#39;)^2)^\frac{3}{2}} = \int du \frac{a}{2}u^{-\frac{3}{2}} = -au^{-\frac{1}{2}} = \frac{a}{(1+(y&amp;#39;)^2)^\frac{1}{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going for the second term (here using just $dy = y'\,dx$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int dx\,b g y y&amp;#39; = b g \int dy\,y = \frac{b g}{2}y^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting it all together with the constant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a}{(1+(y&amp;#39;)^2)^\frac{1}{2}} + \frac{b g}{2}y^2 = C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only the boundary conditions remain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{x \rightarrow +\infty} y = lim_{x \rightarrow +\infty} y&amp;#39; = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a}{(1+(0)^2)^\frac{1}{2}} + \frac{b g}{2}0^2 = C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a = C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replacing this value for $C$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a}{(1+(y&amp;#39;)^2)^\frac{1}{2}} + \frac{b g}{2}y^2 = a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying by $1/a$ and reordering terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{(1+(y&amp;#39;)^2)^\frac{1}{2}} + \frac{b g}{2a}y^2 = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{(1+(y&amp;#39;)^2)^\frac{1}{2}} = 1 - \frac{b g}{2 a}y^2$$&lt;/p&gt;&#xA;" OwnerUserId="1656" LastEditorUserId="1656" LastEditorDisplayName="" LastEditDate="2011-01-31T02:11:55.977" LastActivityDate="2011-01-31T02:11:55.977" CommentCount="1" />
  <row Id="4261" PostTypeId="2" ParentId="4243" CreationDate="2011-01-31T01:59:11.760" Score="1" ViewCount="" Body="&lt;p&gt;Gravity is weak because the masses of elementary particles are so small.  Gravity has a natural mass unit, $m_p~=~\sqrt{\hbar c/G}$, the Planck mass, which is about $10^{-5}$ g.  The proton is $22$ orders of magnitude less massive.  So the stuff which makes up the world is elementary particle “styrofoam stuff” which gravity couples to.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be seen as well with IIA strings and their S-dual heterotic strings.  Those heterotic strings just do not like to stay on our brane, which they have no end points to form Chan-Paton factors or Dirichlet boundary conditions on the brane with.  They slip through our brane as if nothing is there.  Their S-dual strings are open strings on the brane, but with puny masses --- far less than the Planck mass or the mass corresponding to the string tension.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-01-31T02:58:22.427" LastActivityDate="2011-01-31T02:58:22.427" CommentCount="12" />
  <row Id="4263" PostTypeId="2" ParentId="3056" CreationDate="2011-01-31T02:07:32.720" Score="4" ViewCount="" Body="&lt;p&gt;Any good book in Semiconductor Physics will have a description of the k.p method. Try Fundamentals of Semiconductor Physics by Peter Yu and Manuel Cardona. Another reference for Kane Model and EFA are chapters 2 and 3 of &quot;Wave Mechanics Applied to Semiconductor Heterostructures&quot; by Gerald Bastard. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a more mathematically/group theory oriented discussion, try the first chapters on the extended Kane model in &quot;Spin-Orbit Coupling Effects in Two-Dimensional Electron and Hole Systems&quot; by Roland Winkler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the rationale is very simple and it's a simple consequence of Bloch's theorem, that states that the eigenstates of an electron sitting in a periodic potential can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\psi_{k}(\vec{x}) = e^{i\vec{k}\cdot\vec{x}} u_{\vec{k}}(\vec{x})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, this is very important, $u(\vec{x})$ have the same periodicity of the potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you can just plug this identity in Schrödinger's equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(\frac{p^2}{2m} + V(\vec{x})\right)\psi_{\vec{k}}(\vec{x})  = E(\vec{k}) \psi_{\vec{k}}(\vec{x})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you will eventually find a differential equation for the $u_{\vec{k}}(\vec{x})$ functions(*):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(\frac{p^2}{2m} + \frac{\hbar}{m} \vec{k}\cdot\vec{p} + V(\vec{x})\right)u_{\vec{k}}(\vec{x})  = E(\vec{k}) u_{\vec{k}}(\vec{x})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that this looks like an eigenvalue problem for a $\vec{k}$-dependent hamiltonian. For each value of $\vec{k}$, you can solve this problem and find the eigen-energies. This will give rise to the energy bands. Notice the term $\vec{k}\cdot\vec{p}$ in this hamiltonian, that gives the method its name. It looks like this k-dependent hamiltonian is just the old hamiltonian plus a perturbation term proportional to $\vec{k}\cdot\vec{p}$. (**) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you can construct a matrix representation of this eigenvalue problem by projecting it on a basis. For a convenient choice of basis you must specify the symmetry of the specific material you are dealing with. You are going to study the irreducible representations of the symmetry group of the crystal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick is that the function $u_{\vec{k}}(\vec{x})$ must be invariant under the same symmetry group as the original hamiltonian &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H = \frac{p^2}{2m} + V(\vec{x})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the case, then you can easily choose a basis in which this hamiltoniean is diagonal! You just have to choose vectors that transform according to the irreducible representations of this group. Now you only have to worry about the k.p term. The matrix hamiltonian wil be a diagonal part plus an off-diagonal perturbation. But this off-diagonal term is not arbitrary. If you study how the operator $\vec{k}\cdot\vec{p}$ transforms relatively to the various irreps of the group, you'll have lots of zeros. You can determine all of the terms in the hamiltonian that will be non-zero with this procedure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you just have a matrix with a bunch of constants in each entry where symmetry tells you that the hamiltonian shouldn't in principle be zero. Now you have to find this constants. You have two possibilities: you get data from experiments, or from ab initio simulations. This is as far as you can go by just using symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that until this point this is all exact. No approximations were made. The next step from the Kane model is to include the possibility of slightly broken translational symmetry. This takes us to the Envelope Function Approximation (EFA). But this is already getting too long and confuse! :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please take a look in the Yu-Cardona book. There's a really nice, pedagogical discussion about this there. Bastard's book have a good derivation of the Kane matrix and the EFA matrix. Wrinkler's book have a more mathematically rigorous derivation from group theory arguments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need experimentally measured constants for your Kane matrix there's a very comprehensive review by Vurgaftman et al for III-V semiconductors: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Band parameters for III-V compound semiconductors and their alloys. &lt;em&gt;J. Appl. Phys.&lt;/em&gt;, v. 89, n. 11, p. 5815–5875, 2001.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) if you have spin-orbit interactions or other relativistic effects this will complicate a bit the derivation but it will remain essentially valid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(**) There's a crucial difference: this new eigenvalue problem have different boundary conditions! You can solve it just inside a unit cell, cause $u(\vec{x})$ is periodic. And the boundary condition is given by this periodicity requirement.&lt;/p&gt;&#xA;" OwnerUserId="673" LastEditorUserId="673" LastEditorDisplayName="" LastEditDate="2011-01-31T03:17:20.197" LastActivityDate="2011-01-31T03:17:20.197" CommentCount="1" />
  <row Id="4264" PostTypeId="2" ParentId="4256" CreationDate="2011-01-31T02:19:58.743" Score="3" ViewCount="" Body="&lt;p&gt;Gold, as a metal, is, like an noble gas, not very reactive. But if you take a single gold atom it is very reactive. It combines with other gold atoms to form gold metal, and it reacts chemically with other atoms. For example, the Colorado, US town of Telluride is named after a gold-tellurium mineral found there, see Calaverite:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.galleries.com/minerals/sulfides/calaveri/calaveri.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.galleries.com/minerals/sulfides/calaveri/calaveri.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So these metals are not noble as compared to noble gasses. They are reactive. But I think you're asking the question &quot;why are these metals noble while the other ones aren't?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The noble metals are the least &quot;reactive&quot;. See&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Reactivity_series&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Reactivity_series&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As shown in the table in the above wikipedia article, the noble metals are those where it is very difficult to remove electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take a look at silver, $4d^{10}5s^1$. If we are to remove one electron it's going to be the one which has the highest energy. Now normally you'd think that would be the $5s^1$ electron. But instead, on all of the heavier metals, the S levels have LOWER energies than you'd expect. And so when you singly ionize silver the result would be the loss of one of the 4d electrons. See&lt;br&gt;&#xA;&lt;a href=&quot;http://www.chemguide.co.uk/atoms/properties/atomorbs.html&quot; rel=&quot;nofollow&quot;&gt;http://www.chemguide.co.uk/atoms/properties/atomorbs.html&lt;/a&gt;&lt;br&gt;&#xA;for a chart showing the energies of the first few electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spreading out of the orbitals like this is due to the shielding of the nucleus by the lower energy electrons. That is, the S orbitals penetrate the closest to the nucleus and so they see a nucleus which, in effect, has a higher charge than the P, D, and F orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall effect is that the chemistry of heavier metals is more a function of the outermost (or most energetic) F orbital, if present. If it's not present, then the most energetic D orbital, then most energetic P orbital. The S orbital is never the most energetic.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-01-31T02:19:58.743" CommentCount="1" />
  <row Id="4265" PostTypeId="2" ParentId="4238" CreationDate="2011-01-31T02:21:55.390" Score="0" ViewCount="" Body="&lt;p&gt;The coupling constant for gauge theories is dimensionless, such as the fine structure constant $\alpha~=~e^2/(4\pi\epsilon_0\hbar c)$ $\simeq~1/137$.  Mass has naturalized units of reciprocal length.  This makes the establishment of a charge more reasonable, and a unitless number is something which is benchmarked as being an absolute constant.  In other words, if $\alpha$ changed it would be a pure numerical variation.  Occasionally there are claims of this.  A quantity which has an actual dimension in units is so in relationship to other quantities.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a question related to the problem of quantum gravity.  The Planck mass $m_p~=~\sqrt{\hbar c/G}$ can be thought of as the fundamental unit of reciprocal length, and the gravitational constant $G$ has units of area.  This area corresponds to the unit area of a black hole event horizon.  For a Yang-Mill field theory the coupling constant functions in a field which is unitary.  By contrast units of mass are related to this reciprocal length, which in turn is not just a unit involving gravitational modes, but also the degeneracy of modes which have an entropy --- or entanglement entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So mass does not quantize in quite the elementary fashion we might expects with charge and other coupling parameters for interactions.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-31T02:21:55.390" CommentCount="3" />
  <row Id="4266" PostTypeId="2" ParentId="4243" CreationDate="2011-01-31T02:31:02.877" Score="-1" ViewCount="" Body="&lt;p&gt;The Randall-Sundrum model explains it. The other forces are confined to the brane which  we consider to be our universe. The brane is embedded in higher dimensional space where some of the dimensions may be compactified, but others could be larger or even infinite (a 5 dimensional anti-de Sitter space in which a (3+1)dim brane is embedded.All particles except the graviton are bound to the brane.) Higher dimensional space is called the bulk. If gravity is not confined to our brane and can penetrate into the bulk, that would explain its weakness. The problem with the extreme difference in strengths of the forces is termed the hierarchy problem (weak force=$10^{32}$grav force). There are other explanations involving supersymmetry.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-01-31T02:57:52.303" LastActivityDate="2011-01-31T02:57:52.303" CommentCount="5" />
  <row Id="4267" PostTypeId="2" ParentId="4247" CreationDate="2011-01-31T02:36:21.090" Score="3" ViewCount="" Body="&lt;p&gt;Another solution (hopefully, the answer given by mmc is alright, but I followed another procedure):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this change of variable $p=\displaystyle \frac{dy}{dx}$, we have $\displaystyle p\dot p = \frac{d^{2}y}{dx^{2}}$ (keep in mind that $y$ depends on $x$ and $p$ depends on $y$). Substituting in the original equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a p \dot p}{(1+p^{2})^{\frac{3}{2}}}=-bgy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We notice that the left-hand side can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dy}\left(-\frac{a}{(1+p^{2})^{\frac{1}{2}}}\right)=-bgy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After integration we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{a}{(1+y&amp;#39;^{2})^{\frac{1}{2}}}=-\frac{bgy^{2}}{2}+C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using your boundary conditions, we find that $C=-a$. Therefore, we obtain the equation you were looking for&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{(1+y&amp;#39;^{2})^\frac{1}{2}}=1-\frac{bgy^{2}}{2a}$$&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2011-01-31T02:36:21.090" />
  <row Id="4268" PostTypeId="1" AcceptedAnswerId="4270" CreationDate="2011-01-31T02:44:44.787" Score="0" ViewCount="97" Body="&lt;p&gt;I'm in the process of working on a physics related game.  I'm looking to find the maximum velocity of an object given it's mass and the force acting on it when it is traveling horizontally.  I believe there must be a method of calculating this but I'm unaware of it.  Is there a formula for this scenario?  I'm unaware if this is the same calculation as terminal velocity but I don't believe it is.  Thanks for any help!&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="1677" LastEditorDisplayName="" LastEditDate="2011-01-31T02:59:14.017" LastActivityDate="2011-01-31T02:59:14.017" Title="Determining Maximum Velocity of an object traveling horizontally" Tags="&lt;velocity&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="4269" PostTypeId="1" AcceptedAnswerId="4279" CreationDate="2011-01-31T02:46:32.453" Score="1" ViewCount="149" Body="&lt;p&gt;Suppose I know that Selenium has electron configuration $[Ar] 4s^2 3d^{10} 4p^4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I use this information to then come to the conclusion that its possible oxidation states are 6, 4, 2, 1 and -2?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the 6: Give up all s electrons to reach $Ar$ configuration&#xA;I also understand the -2: Obtian two electrons to complete the 4th shell.&#xA;I can guess why 4 is plausible: Give up all your 4p electrons&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But 2 and 1? A guess for 2 would be to give up the 4s electrons, but why should it? And for the 1 I have no idea. Maybe give up one electron in the 4th shell and have it replaced by one of the 3d electrons?&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-01-31T02:53:35.893" LastActivityDate="2011-01-31T16:14:44.453" Title="Obtain oxidation numbers from looking at electron configuration?" Tags="&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="4270" PostTypeId="2" ParentId="4268" CreationDate="2011-01-31T02:51:18.877" Score="2" ViewCount="" Body="&lt;p&gt;Suitable for game physics and BoTE calculations:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Assume a functional form for the frictional resistances (all the apply from rolling, sliding and fluid (wind or water)), and solve for the total resistance equal to the driving force.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If that does not limit the speed to a reasonable velocity and this even takes place on a planet, compute the orbital velocity at that elevation. Above that speed a downward force will have to be applied to maintain local horizontalness for the motion...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Toward (1):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;rolling and sliding friction tend to be independent of speed, except that heat dissipation may cause secondary effects. &lt;/li&gt;&#xA;&lt;li&gt;Wind resistance is generally proportional to $v^2$ except in the neighborhood of the speed of sound. &lt;/li&gt;&#xA;&lt;li&gt;Water resistance for surface vessels is complicated with several parts proportional to square (viscous) and cubic (wave making) terms at least.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-01-31T02:58:33.283" LastActivityDate="2011-01-31T02:58:33.283" />
  <row Id="4273" PostTypeId="2" ParentId="3527" CreationDate="2011-01-31T04:50:04.080" Score="9" ViewCount="" Body="&lt;p&gt;Contrary to what is stated in many textbooks, energy-momentum conservation alone cannot explain the behavior of Newton’s cradle. For N balls we have two equations and N final velocities to calculate. Hence, conservations laws can do the job only for N=2. This means that if we want to give an explanation of the cradle behavior based on conservation laws, we have to split the N-ball collision into a sequence of two-ball collisions, as done in some of the answers given to the question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with this approach is that it assumes that initially the balls are not touching one another, which is not the case in most cradles. One might argue that it doesn’t really matter that the balls are initially in contact, only that the collision time (the time during which momentum is transferred from one ball to the other) is much smaller than the time it takes for the mechanical perturbation to cross one ball. However, the typical collision time is of the order of 0.1 ms, much bigger than the propagation time of about 0.01 ms. Under such conditions one should expect wave propagation along the ball chain to play a significant role in the cradle’s dynamics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a nice paper ( &lt;a href=&quot;http://dx.doi.org/10.1119/1.12407&quot; rel=&quot;nofollow&quot;&gt;http://dx.doi.org/10.1119/1.12407&lt;/a&gt; ) showing that wave propagation along the ball line must be dispersion-free if we are to find equal numbers of incoming and outgoing balls. The authors also provide an interesting picture of how the system decides how many balls it is going to send away. Basically, the first impact produces two wave pulses moving in opposite directions along the ball line. The pulses are reflected at the ends of the line and meet again at the point where the ball chain breaks up. It is easy to see that this break-up point is symmetrical (with respect to the middle of the line) to the impact point, which explains why the number of outgoing balls is equal to that of the incoming ones.   &lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2011-01-31T04:50:04.080" CommentCount="2" />
  <row Id="4274" PostTypeId="2" ParentId="4239" CreationDate="2011-01-31T05:40:38.797" Score="4" ViewCount="" Body="&lt;p&gt;The neutron has no net charge, but it does have a net magnetic moment. As an aside, this simple fact provides strong evidence that the neutron is a composite particle (made of smaller things like quarks and gluons), because if it were a neutral elementary particle we would not expect it to have any magnetic moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we know that the neutron is composite, and it definitely has a magnetic moment (it can be measured). Therefore if you have a macroscopic object made of neutrons whose spins are polarized (their angular momentum vectors tend to point one direction rather than another), it will be magnetized. In fact, it will be very strongly magnetized compared to ordinary matter simply because nuclear matter is so much more dense.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-01-31T05:40:38.797" CommentCount="5" />
  <row Id="4275" PostTypeId="2" ParentId="4238" CreationDate="2011-01-31T05:45:06.507" Score="4" ViewCount="" Body="&lt;p&gt;Charge comes from discrete symmetries and is  countable and additive.&#xA;Mass comes from &lt;em&gt;continuous&lt;/em&gt; 4d space,  is exchangeable with energy and, in quantum mechanical dimensions not linearly additive, thus not countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have an elementary quantum of mass, m_q. In the world we know two such quanta would not end up as 2*m_q. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One would add the four vectors and take the measure in 4space, and square root it, to get the invariant mass of two of them, etc for higher numbers at will. Given a mass, you could never know/count of how many m_q it is composed. It is a continuum. Whereas charge is simply additive and countable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way an elementary particle rest masses could be a linear sum of m_qs is for there to be no binding energy, and experiments tell us the elementary particles are bound, if stable.  If there were no binding energy then the composites would crumble into the constituent m_q with the slightest scattering.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-01-31T10:15:37.237" LastActivityDate="2011-01-31T10:15:37.237" CommentCount="3" />
  <row Id="4276" PostTypeId="2" ParentId="4238" CreationDate="2011-01-31T06:55:57.670" Score="5" ViewCount="" Body="&lt;p&gt;Dear asmailer, the reason is simple and completely understood: the electric charge is the generator of a $U(1)$ symmetry which is compact and may be parameterized by an angle, $\phi$. So wave functions may only depend on the angle $\phi$ in a periodic way, $\exp(iQ\phi)$ where $Q$ is integer (or an integer multiple of $e/3$, if I look at the elementary $U(1)$ rescaled by a factor of three that also allows quarks).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the mass is nothing else than the energy measured in the rest frame. The energy generates translations in time - and time is noncompact. So the corresponding phase $\exp(Et/i\hbar)$ isn't constrained by any condition of periodicity. So the energy is continuous even in the rest frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the other frames, the continuous character of the energy is even more obvious because the &quot;already continuous&quot; rest mass is multiplied by the Lorentz factor $1/\sqrt{1-v^2/c^2}$ which changes - and has to change - continuously as we vary the velocity; the latter is required by the principle of relativity. So the mass and energy are continuous, have to be continuous, and will always remain continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could continue to ask &quot;why&quot; and in fact, you could get even deeper answers. You could ask why time is not periodic - which was used for the continuity of energy in a particular frame. Well, time has to be &quot;aperiodic&quot; because a periodic time would cause the grandfather paradox and other bad things - closed time-like curves. Time is also unbounded in the future because we live in a space with the positive cosmological constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, groups such as $U(1)$ have to be compact and are compact in any quantum theory of gravity. This was argued e.g. by Cumrun Vafa in his Swampland program. For $U(1)$, the situation is simpler: the electric charge has to be quantized because of the Dirac quantization rule and because of the existence of the magnetic monopoles which is also guaranteed in a consistent theory of quantum gravity as was explained in another question on this server.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-31T06:55:57.670" CommentCount="7" />
  <row Id="4277" PostTypeId="2" ParentId="4238" CreationDate="2011-01-31T07:12:47.933" Score="1" ViewCount="" Body="&lt;p&gt;Mass is determined by how a particle interacts with the Higgs boson(s). Mass is also&#xA;determined by the relativistic mass-energy equation $E^2=m^2c^4+p^2c^2$ or more simply $m=\sqrt{(E^2-(pc)^2)}/c^2$  The energy values are a continuum, so there is no discrete elementary mass unit. In General Relativity, things are more complicated. In Stationary spacetimes, for example, the gravitational potentials (metrics) are not functions of time and the ST has time translational symmetry, so energy is conserved-- but while the stress-energy tensor is Lorentz covariant, in a non-isolated system, the system exchanges energy-momentum with its environment and its &quot;mass&quot; is not invariant--once again, no elementary or fundamental mass.&#xA;I hope this is what you meant, but maybe you just meant the Planck mass...&#xA;Frank Wilczek spends quite awhile in his popular book &quot;The Lightness of Being&quot; trying to say what mass is and isn't. He does quite a good job in non-technical terms-&#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B004HEXSXG&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Lightness-Being-Ether-Unification-Forces/dp/B004HEXSXG/ref=sr_1_1?s=books&amp;amp;ie=UTF8&amp;amp;qid=1296458301&amp;amp;sr=1-1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-01-31T07:24:32.453" LastActivityDate="2011-01-31T07:24:32.453" CommentCount="2" />
  <row Id="4278" PostTypeId="1" AcceptedAnswerId="4280" CreationDate="2011-01-31T07:29:45.520" Score="1" ViewCount="168" Body="&lt;p&gt;Under the Lorentz transformations, quantities are classed as four-vectors, Lorentz scalars etc depending upon how their measurement in one coordinate system transforms as a measurement in another coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proper length and proper time measured in one coordinate system will be a calculated, but not measured, invariant for all other coordinate systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what kind of invariants are proper time and proper length?&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T14:29:07.143" LastActivityDate="2011-02-03T14:29:07.143" Title="What kind of invariants are proper time and proper length?" Tags="&lt;special-relativity&gt;&lt;tensor-calculus&gt;" AnswerCount="2" />
  <row Id="4279" PostTypeId="2" ParentId="4269" CreationDate="2011-01-31T07:55:16.133" Score="1" ViewCount="" Body="&lt;p&gt;Dear Lagerbaer, first, this is an inorganic chemistry question. Chemistry is a subset of physics but it probably has special servers with questions and answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, one may get a rough idea about the oxidation numbers by your method but it's not universally valid. The oxidation number of an atom depends on the molecule in which the atom is incorporated. The &quot;shape&quot; of the atom (and its electron shells) in a molecule depends on and is also influenced by the other atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, I think that both parts of this sentence&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I understand the 6: Give up all s electrons to reach Ar configuration I also understand the -2: Obtian two electrons to complete the 4th shell.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;had bugs. I guess you meant that the oxidation number 6 is achieved by giving up all 4s and 4p electrons (rather than &quot;all s electrons&quot;: 4p is not s) while -2 is obtained by getting two electrons to complete the 4p6 shell - which will not yet complete 4d and other 4th shells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, in principle, you may get any oxidation number between the minimum and the maximum value. And the minimum and maximum values usually have an explanation of the kind you offered for +6 and -2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more electronegative the partner atoms will be, the higher (or more positive or less negative) the oxidation number of an atom will be. So there's no reason why it shouldn't happen that the partner atoms in the molecule will only remove or add a subset (a priori, any subset) of the removed or added electrons that you discussed in the case of +6 and -2. One has to check individual molecules to figure out whether a particular oxidation number is realized somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Selenium's oxidation numbers&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, selenium belongs to Group 16 - Oxygen, Sulphur, Selenium, etc. Their oxidation number is -2 unless they're combined with more electronegative elements - namely Oxygen or halogens. If they are, they may possess some positive oxidation numbers. To see that your list is the right one, one would have to go through all possible compounds - or at least all possible &quot;neighborhoods of a Selenium atom&quot; - and list all possibilities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, such things may be calculated from many-body quantum mechanics (assuming that we have a good enough definition of the oxidation state - it depends on some conventions). However, chemists don't do such things - instead, they learn many cases and acquire some experience.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-31T07:55:16.133" />
  <row Id="4280" PostTypeId="2" ParentId="4278" CreationDate="2011-01-31T08:07:18.677" Score="3" ViewCount="" Body="&lt;p&gt;Dear John, it doesn't matter whether a quantity is measured or calculated. In some sense, you may say that the proper time or proper length are measured in the other frames, too. Calculation is sometimes a necessary part of the measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Because the values aren't affected by the transformations, proper lengths and proper times are formally scalars&lt;/strong&gt;, much like the rest mass which is the proper length of the energy-momentum vector, if you wish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we usually don't use the word &quot;scalar&quot; for such quantities because they're just some &quot;somewhat unnatural&quot; functions of vectors. For example, the proper length is $\sqrt{dx^\mu dx_\mu}$ assuming that $dx^\mu$ (and the convention for the metric) is spacelike. So fundamentally, the proper length is some information about a vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the proper length is just some &quot;partial information&quot; about a vector, you can't build a &quot;field theory&quot; out of proper lengths themselves. In relativistic field theory, we talk about &quot;scalar fields&quot; and &quot;vector fields&quot; and other &quot;tensor fields&quot;, aside from &quot;spinor fields&quot; etc. However, a fundamental theory could have no elementary field associated with the &quot;proper length&quot; even if such constructions made sense: it would have the whole &quot;vector field&quot; while the length of this vector would be a &quot;derived quantity&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-31T08:07:18.677" CommentCount="2" />
  <row Id="4282" PostTypeId="2" ParentId="4238" CreationDate="2011-01-31T09:29:21.440" Score="12" ViewCount="" Body="&lt;p&gt;Let me add two references to points already mentioned in this discussion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today, there is no reason known why the electric charge has to be quantized. It is true that the quantization follows from the existence of magnetic monopoles and the consistency of the quantized electromagnetic field, which was shown first by Dirac, you'll find a very nice exposition of this in&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gregory L. Naber: &quot;Topology, geometry and gauge fields.&quot; (2 books, of the top off my head I don't know if the relevant part is in the first or the second one).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;AFAIK there is no reason to believe that magnetic monopoles do exist, there is no experimental evidence and there is no compelling theoretical argument using a well established framework like QFT. There are of course more speculative ideas (Lubos mentioned those).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AFAIK there is no reason why mass should or should not be quantized (in QFT models this is an assumption/axiom that is put in by hand, even the positivity of the energy-momentum operator is an axiom in AQFT), but a mass gap is considered to be an essential feature of a full fledged rigorous theory of QCD, for reasons that are explained in the problem description of the Millenium Problem of the Clay Institute that you can find here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.claymath.org/millennium/Yang-Mills_Theory/&quot; rel=&quot;nofollow&quot;&gt;Yang-Mills and Mass Gap&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-01-31T09:29:21.440" CommentCount="4" />
  <row Id="4283" PostTypeId="2" ParentId="4278" CreationDate="2011-01-31T09:49:47.273" Score="1" ViewCount="" Body="&lt;p&gt;allow me a somewhat informal discussion: Pour yourself a cup of coffee, you see that your cup does not move, your can does not move, and that the coffee flowing from the can ends up in your cup.&#xA;Now, every observer from every other reference frame will agree with you that the coffee ends up in the cup, from his point of view. This is what we call a Lorentz scalar: Every observer agrees upon it, from the point of view of his reference frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proper time and proper length are not Lorentz scalars in this sense. In this case, we first single out a special reference frame R, it is the reference frame where a special object is at rest. Next we have to tell every observer how his reference frame R' is related to R. Every other observer now has to measure time and length in his reference frame R' and is then able to calculate, using the additional information of the relation of R and R', what the corresponding measurements would have told him if he lived in R. If every observer does this calculation correctly, they'll agree upon proper time and proper length :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ergo: Proper time and proper length are simply observables defined with respect to a special reference frame. There is no buzzword for this comparable to Lorentz scalar or Lorentz vector, nor is there the need for one, IMHO.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-01-31T09:49:47.273" CommentCount="1" />
  <row Id="4284" PostTypeId="1" AcceptedAnswerId="4308" CreationDate="2011-01-31T09:56:53.383" Score="3" ViewCount="395" Body="&lt;p&gt;Is there a tutorial explanation as to how decoherence transforms a wavefunction (with a superposition of possible observable values) into a set of well-defined specific &quot;classical&quot; observable values &lt;em&gt;without&lt;/em&gt; the concept of the wavefunction undergoing &quot;collapse&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean an explanation which is less technical than that decoherence is &quot;the decay or rapid vanishing of the off-diagonal elements of the partial trace of the joint system's density matrix, i.e. the trace, with respect to any environmental basis, of the density matrix of the combined system and its environment..&quot; (&lt;a href=&quot;http://en.wikipedia.org/wiki/Decoherence#Density_matrix_approach&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edited title to label the question more clearly + textual clarification.]&lt;/p&gt;&#xA;" OwnerUserId="1394" LastEditorUserId="1394" LastEditorDisplayName="" LastEditDate="2011-01-31T16:10:20.337" LastActivityDate="2011-01-31T16:28:14.270" Title="A tutorial explanation of decoherence?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="4285" PostTypeId="2" ParentId="4284" CreationDate="2011-01-31T10:09:40.910" Score="3" ViewCount="" Body="&lt;p&gt;Dear Nigel, the Wikipedia explanation is not too technical, but let me try to offer an even less technical one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In ordinary quantum mechanics, the wave function remembers lots of complex numbers - the so-called probability amplitudes. These numbers may be added and combined in many ways and the absolute values of the squares of these sums are interpreted as probabilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is different from classical (non-quantum) physics which contains no complex numbers &quot;beneath&quot; the probabilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Probabilistic statistical physics&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there exists a statistical formulation of classical physics that does deal with probabilities, but ordinary &quot;classical ones&quot;. There are no complex numbers beneath them. There simply exists a set of possibilities - e.g. that you throw dice and you get 1,2,3,4,5, or 6 - and classical physics with an uncertain, probabilistically given initial state, may calculate that each of these final outcomes has probability of 1/6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this probabilistic classical physics, one doesn't claim to know the positions and momenta $p,x$ of all particles. Instead, one works with the probabilistic distribution function $\rho(x_i,p_j)$ that is a function of all the possible positions and momenta of all the particles. The function describes the probability density that the particles are located in a small volume around a given point $(x_i,p_i)$ of the phase space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the latter form of classical physics that quantum mechanics reduces to after decoherence. Decoherence produces a preferred set of possibilities that may be measured and it calculates the probabilities of each of them, much like you would in the probabilistic version of classical physics. But of course, the underlying theory is still quantum theory and it can only make probabilistic predictions. So decoherence will never actually find a way to decide which of the outcomes is realized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, there is no &quot;collapse&quot; into a single outcome, just like there is no collapse in the probabilistic classical physics. The density matrix, obtained by tracing the wave function, is not an &quot;objective state of reality&quot;: there exists no &quot;objective state of reality&quot;. Much like the classical distribution function, it's just a probability distribution describing our incomplete state of knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The role of decoherence is to choose a preferred set of outcomes that can be measured - e.g. &quot;alive cat&quot; and &quot;dead cat&quot; - and explain why all the other linear superpositions of the preferred outcomes are illegitimate. This result &quot;bans&quot; further interference and other typically quantum phenomena. That allows us to think about the evolution using the classical intuition. But it's still true that the evolution is indeterministic - and it will always be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A review of decoherence&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For others, a more technical explanation of decoherence is e.g. here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.karlin.mff.cuni.cz/~motl/entan-interpret.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.karlin.mff.cuni.cz/~motl/entan-interpret.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-31T10:09:40.910" CommentCount="5" />
  <row Id="4286" PostTypeId="1" CreationDate="2011-01-31T10:39:47.503" Score="0" ViewCount="66" Body="&lt;p&gt;I have built two time machines (finally) based on two different technologies. ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to race them and see which one takes me from 12/12/1212 to 11/11/1111 (dd/mm/yyyy) first. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I do this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not speed then what else can tell that one time machine is better than the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: Please hurry up! my clients will be here any minute for test ride ;-) &lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2011-01-31T10:44:53.640" LastActivityDate="2011-01-31T10:44:53.640" ClosedDate="2011-01-31T11:18:11.987" Title="How are we going to measure the speed of time machine?" Tags="&lt;general-relativity&gt;&lt;time-travel&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="4287" PostTypeId="1" CreationDate="2011-01-31T10:47:38.800" Score="0" ViewCount="118" Body="&lt;p&gt;Ive been watching a video about dark mater and Joe al lot of the mass is missing in our universe. And astronomers got to this by measuring the speed that stars orbit he Center of the galaxy and when they did this I didn't match the amount of gravity/mass in the galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I thought that the mass of the black hole in the Center of the galaxy was measured by the stars orbiting it? To me it sounds like a catch 22? But I dot understand it very well. So could someone explain it further?&lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2011-01-31T14:09:07.870" Title="How is the mass of black hole at the centre of our galaxy measured?" Tags="&lt;astrophysics&gt;" AnswerCount="2" />
  <row Id="4288" PostTypeId="2" ParentId="3593" CreationDate="2011-01-31T11:05:21.060" Score="0" ViewCount="" Body="&lt;p&gt;This &lt;a href=&quot;http://streamer.perimeterinstitute.ca/mediasite/viewer/NoPopupRedirector.aspx?peid=ff8f97a4-2848-4b93-98d9-1218de2070af&amp;amp;shouldResize=False#&quot; rel=&quot;nofollow&quot;&gt;talk&lt;/a&gt; begins to cover some of this question in detail at around 6:30.  I haven't completed viewing it but it is moving further into the rabbit hole. Found also &lt;a href=&quot;http://motls.blogspot.com/2011/01/twistor-minirevolution-goes-on.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just the first of many notes.  I find the analogy of N=4 SYM to the harmonic oscillator very fascinating.  Also the implication that physical reality is about the structure of scattering amplitudes is fantastic.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This entry was given down check, however, I was intending to work iteratively.I will continue to update periodically as I move forward. 2/1/2011&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Update one&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer this question we must define three key components first:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;a href=&quot;http://en.wikipedia.org/wiki/Physical_system&quot; rel=&quot;nofollow&quot;&gt;Physical System&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;a href=&quot;http://en.wikipedia.org/wiki/Data&quot; rel=&quot;nofollow&quot;&gt;Data&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) &lt;a href=&quot;http://en.wiktionary.org/wiki/interconnection&quot; rel=&quot;nofollow&quot;&gt;Interconnection&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A physical system is a type of &lt;a href=&quot;http://en.wikipedia.org/wiki/System&quot; rel=&quot;nofollow&quot;&gt;system&lt;/a&gt; that deals with analyzing the physical universe.  &lt;a href=&quot;http://en.wiktionary.org/wiki/physical&quot; rel=&quot;nofollow&quot;&gt;Physical&lt;/a&gt; is having to do with &lt;a href=&quot;http://en.wikipedia.org/wiki/Physics&quot; rel=&quot;nofollow&quot;&gt;physics&lt;/a&gt; which has to do with analyzing &lt;a href=&quot;http://en.wikipedia.org/wiki/Nature&quot; rel=&quot;nofollow&quot;&gt;nature&lt;/a&gt; which has to do with &lt;a href=&quot;http://en.wikipedia.org/wiki/Phenomenon&quot; rel=&quot;nofollow&quot;&gt;phenomena&lt;/a&gt; which has to do with &quot;any event that is observable, however commonplace it might be, even if it requires the use of instrumentation to observe, record, or compile data concerning it.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Data is the &quot;lowest level of abstraction from which information and then knowledge are derived&quot;; suggesting that data is the most fundamental element associated with an observation.  Starting with a set of data, nothing about a physical system can be said using that data unless the data and physical system have some interconnection.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With those three components: physical system, data and interconnection; we might be able to explore a definition of physically meaningful.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To be continued, despite the trolls...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="1383" LastEditorDisplayName="" LastEditDate="2011-02-02T11:08:12.237" LastActivityDate="2011-02-02T11:08:12.237" CommentCount="3" />
  <row Id="4289" PostTypeId="1" AcceptedAnswerId="4850" CreationDate="2011-01-31T11:21:11.397" Score="10" ViewCount="886" Body="&lt;p&gt;Recently, there was a rapid communication published in Phys.Rev.D (&lt;a href=&quot;http://prd.aps.org/abstract/PRD/v83/i2/e021502&quot; rel=&quot;nofollow&quot;&gt;PRD 83, 021502&lt;/a&gt;), titled &lt;em&gt;&quot;Gravity is not an entropic force&quot;&lt;/em&gt;, that claimed that &lt;a href=&quot;http://prd.aps.org/abstract/PRD/v67/i10/e102002&quot; rel=&quot;nofollow&quot;&gt;an experiment performed in 2002 with ultra cold neutrons in a gravitational field&lt;/a&gt;, disprove Verlinde's entropic approach to Gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The neutrons experiment gave results consistent with the predictions of Newtonian gravity for the lowest energy state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, the author claims that the fact that Verlinde's entropic force comes from a thermodynamic process that is irreversible (or approximately reversible), leads to non-unitarity in the evolution of quantum systems. The non-unitarity then exponentially suppresses the eigenfunctions, predicting results very much different than the Newtonian. Thus, that experiment is in contradictions to what is expected if Verlinde's approach is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are, &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;First of all, is there anything else essential that I am missing?&lt;/li&gt;&#xA;&lt;li&gt;Is there any response to that argument? &lt;/li&gt;&#xA;&lt;li&gt;Is that a fatal problem with Verlinde's entropic approach? &lt;/li&gt;&#xA;&lt;li&gt;Is that a fatal problem for any entropic approach?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;updates on the discussion: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There is also this recent comment &lt;a href=&quot;http://arxiv.org/abs/1104.4650&quot; rel=&quot;nofollow&quot;&gt;arxiv.org/abs/1104.4650&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Once more: gravity is not an entropic force &lt;a href=&quot;http://arxiv.org/abs/1108.4161&quot; rel=&quot;nofollow&quot;&gt;arxiv.org/abs/1108.4161&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="383" LastEditorUserId="383" LastEditorDisplayName="" LastEditDate="2011-09-01T07:57:15.223" LastActivityDate="2011-09-01T07:57:15.223" Title="Is Gravity an entropic force after all?" Tags="&lt;quantum-mechanics&gt;&lt;gravity&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="1" />
  <row Id="4290" PostTypeId="2" ParentId="4287" CreationDate="2011-01-31T12:13:50.190" Score="2" ViewCount="" Body="&lt;p&gt;You mix up two different observations-&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Black hole evidence/mass at the very center comes from orbits of stars very &#xA;close to galactic center around some dozen of light years. This was done by some &#xA;infrared camera, which is able to &quot;see&quot; through the dust around galactic &#xA;center&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Missing mass in entire galaxy relates to the orbits outside of the &#xA;central bulge, &quot;up&quot; to those at the rim of galaxy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Georg&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-01-31T12:13:50.190" CommentCount="2" />
  <row Id="4291" PostTypeId="2" ParentId="3769" CreationDate="2011-01-31T12:20:26.823" Score="1" ViewCount="" Body="&lt;p&gt;There are three general populations of stars, based on their &lt;a href=&quot;http://en.wikipedia.org/wiki/Metallicity&quot; rel=&quot;nofollow&quot;&gt;metallicity&lt;/a&gt;, which is the abundance of heavy elements on their atmosphere. That abundance is an indicator of the abundance of heavy elements of the initial cloud that gave birth to the star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have Population I stars that are metal-rich, Population II that are metal-poor and Population III that are metal-free. Population III stars are the first stars that formed in our universe and they must have been very massive and sort lived. Population II are the second generation of stars that created the heavier elements. Finally, Population I stars are the most recently created stars that also have heavier elements that were created in the Population II stars. &#xA;Our Sun is a Population I star, so in a sense the Sun is a third generation Star. Taking into account the age of the universe and the average life-time of the more massive stars that form the heavier elements in supernovas, then the assumption that the Sun has used the material of at least two previous stars is reasonable.    &lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2011-01-31T12:20:26.823" />
  <row Id="4293" PostTypeId="2" ParentId="4284" CreationDate="2011-01-31T12:40:51.863" Score="2" ViewCount="" Body="&lt;p&gt;The Wikipedia article is rather a lengthy description of decoherence which undoubtedly also prompts this question. Part of the reason is that decoherence is not, in a certain sense, a truly fundamental mechanism amending quantum theory, but an applied mechanism introduced to &quot;explain&quot; collapse using ideas from Statistical Mechanics (those density matrices and concepts of &quot;bath&quot;/environment) supplementing those of quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Borrowing a key equation from the Tutorial mentioned by Lubos, one writes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H = H_c + H_e + H_i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the Hamiltonian $H$ understood from a decoherence perspective: its ingredients are $H_c$ which is the simpler component (= system under study) Hamiltonian discussed in elementary textbooks; $H_e$ is the environment and $H_i$ the Interaction (including the experiment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is saying &quot;dont just look at the system alone; look at the entire environment+system&quot; from a conceptual perspective. Abstractly there is no collapse here; although thermodynamic-like arguments are then used to give the (local) appearance of collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To follow this up a Tutorial can do two things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Explain the thermodynamic aspect a bit more;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Explain what happens to &quot;collapse&quot; at the $H$ level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this helps a little.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-01-31T16:18:02.407" LastActivityDate="2011-01-31T16:18:02.407" />
  <row Id="4296" PostTypeId="2" ParentId="4287" CreationDate="2011-01-31T14:09:07.870" Score="2" ViewCount="" Body="&lt;p&gt;For the most part the mass of the super massive black hole (SMBH) at the galaxy center is estimated in much the same way the solar mass is estimated by Newtonian mechanics.  In fact general relativistic elements are not used much, where GR contributes heavily for orbits with a few number of Schwarzschild radii, or semi-major axis length.  The motion of stars in the galactic center have been observed over a number of years and the velocities and accelerations known.  It is then straight forwards to use Newtonian mechanics or $v~=~\sqrt{2GM/r}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good source for this is the UCLA Galactic Center web site:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astro.ucla.edu/~ghezgroup/gc/&quot; rel=&quot;nofollow&quot;&gt;http://www.astro.ucla.edu/~ghezgroup/gc/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/1nfiw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data animation of stellar trajectories:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.astro.ucla.edu/~ghezgroup/gc/pictures/orbitsOverImage10.shtml&quot; rel=&quot;nofollow&quot;&gt;http://www.astro.ucla.edu/~ghezgroup/gc/pictures/orbitsOverImage10.shtml&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is interesting, just click on the animation hyperlink below.  A couple of stars are on highly eccentric orbits which take then somewhat close to the SMBH.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-31T14:09:07.870" />
  <row Id="4297" PostTypeId="2" ParentId="4256" CreationDate="2011-01-31T14:17:07.440" Score="2" ViewCount="" Body="&lt;p&gt;There are some misconceptions here: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Any comparison to noble gases is unhistoric, noble metals &#xA;were called in this way since long times, first maybe thousands &#xA;of years ago for the fact that they did not corrode. &#xA;This includes gold and silver, copper is not thought to be &#xA;&quot;noble&quot; traditionally, because copper and bronze get some patina. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The chemical reason for that is electrochemical potential, &#xA;this makes it hard to oxidize gold and silver, because the have large &#xA;positive potentials. Copper is borderline. Thus Platinum metals &#xA;(known from about 1800 only) are named &quot;noble&quot; too. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;reactivity&quot; a more kinetic phenomenon, is not a good reason. &#xA;Nobility is defined on equilibrium values, ie. chemical potential. &#xA;And in this case chemical potential is easily accessible! &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Beware of arguments to &quot;remove&quot; electrons from Atoms and the like! &#xA;Chemistry happens not between atoms in gas phase as far as &#xA;metals are concerned! Gold is dissolved in aqua regia, or in a cyanidic &#xA;lye, or You have it as an anode with a certain positive potential. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Georg&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-01-31T16:08:20.853" LastActivityDate="2011-01-31T16:08:20.853" />
  <row Id="4298" PostTypeId="2" ParentId="4289" CreationDate="2011-01-31T14:36:09.187" Score="1" ViewCount="" Body="&lt;p&gt;I think there has been some confusion over this matter.  Of course if makes little sense to think a trajectory around a black hole will exhibit an entropy increase.  Verlinde proposed an entropy force of gravity from which Newton's law of gravity may be derived.  This is a thermodynamic principle for the entropy variation of a holographic screen &#xA;$$&#xD;&#xA;\Delta S~=~2\pi k_B(mc/\hbar)\Delta x&#xD;&#xA;$$&#xA;where $\Delta x$ is the distance between the holographic screen and a test particle of mass $m$.  The entropy is then in increments of $2\pi k_B$ according to displacements of the screen equal to the Compton wavelength $\lambda~=~\hbar/mc$.  The standard entropy formula $S~=~k_BA/4L_p^2$ indicates a proportionality with respect to area.  For the radius of the screen adjusted $S_0~\rightarrow~S~=$ $S_0~+~\Delta S$ by the radial change in the screen $r~=~r_0~+~\Delta r$ then&#xA;$$&#xD;&#xA;ΔS~=~k_B/4L_p^2(A~–~A_0)~=~(2\pi k_B/L_p^2)r\Delta r~=~(2\pi k_Bc^3/G\hbar)r\Delta r,&#xD;&#xA;$$&#xA;which is linear in the radial displacement.  By equating $\hbar/mc~=~G\hbar/rc^3$ gives a radius $r~=~Gm/c^2$, which is appropriate for the Newtonian result, but is half the Schwarzschild result.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The entropy for an orbit of a test mass is constant, and this entropy is a measure of the holographic screen.  So if you place a Gaussian surface around a gravitating radially symmetric body that cloaks the configuration of the body, the entropy of the screen is the maximum entropy of the system.  For a particle orbiting the body the entropy is constant, or $\Delta S~=~0$, for the screen remains constant. &lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-01-31T14:36:09.187" CommentCount="2" />
  <row Id="4299" PostTypeId="1" AcceptedAnswerId="4304" CreationDate="2011-01-31T14:38:09.360" Score="8" ViewCount="465" Body="&lt;p&gt;See image. This is from an experiment we did in physics class. We shone a sodium light at a convex lens on top of a sheet of glass - and this image was captured by a USB microscope. I know what causes the main rings - but what about the surrounding ones, and why are they different colours?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/BAaQx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="869" LastEditorDisplayName="" LastEditDate="2011-01-31T15:04:55.653" LastActivityDate="2011-02-03T17:09:26.163" Title="Newton's rings: What causes the other rings?" Tags="&lt;optics&gt;&lt;light&gt;&lt;newton&gt;&lt;interference&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="4300" PostTypeId="1" AcceptedAnswerId="4307" CreationDate="2011-01-31T14:42:44.157" Score="4" ViewCount="94" Body="&lt;p&gt;I don't know hydrodynamics, but I wonder how one would compute resonance modes of a cubic box of water which we shake. I believe the waves would directly depend on the height of water and the width and length of the box. Is it a bit like a waveguide for electodynamics?&lt;/p&gt;&#xA;" OwnerUserId="545" LastActivityDate="2011-01-31T15:48:44.193" Title="Resonance modes of a cubic box of water which we shake" Tags="&lt;waves&gt;&lt;water&gt;&lt;simulation&gt;&lt;hydrodynamics&gt;" AnswerCount="1" />
  <row Id="4301" PostTypeId="2" ParentId="4299" CreationDate="2011-01-31T14:50:25.633" Score="1" ViewCount="" Body="&lt;p&gt;My first thought was &quot;some grains of dirt&quot; :=)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But than I realized that the most prominent (8) &quot;ghost patterns&quot; &#xA;are arranged in a square. As fas as I can see, those ghosts &#xA;are identical in diameters to the main fringes, from that &#xA;I conclude that this is some artifact from the microscope &#xA;or the (USB?) camera. &#xA;Is there a chance for some caleidoscope effect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Georg&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-01-31T14:50:25.633" />
  <row Id="4302" PostTypeId="2" ParentId="4299" CreationDate="2011-01-31T14:52:40.693" Score="1" ViewCount="" Body="&lt;p&gt;The extra patterns are not part of the 'ideal' Newton ring experiment. The grid symmetry of the spacings of the rings suggests it is not due to the actual lens (assuming it is convex), which should be radiallyy symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am guesssing that the extra rings are due to the light source being a cluster of LEDs, or maybe the natures of the microscope lens, or possibly an artifact the CCD ...&lt;/p&gt;&#xA;" OwnerUserId="1072" LastActivityDate="2011-01-31T14:52:40.693" />
  <row Id="4303" PostTypeId="2" ParentId="3754" CreationDate="2011-01-31T14:56:08.043" Score="1" ViewCount="" Body="&lt;p&gt;Energy is indeed conserved. The link below gives a nice explanation...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://skullsinthestars.com/2010/04/07/wave-interference-where-does-the-energy-go/&quot; rel=&quot;nofollow&quot;&gt;http://skullsinthestars.com/2010/04/07/wave-interference-where-does-the-energy-go/&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2011-01-31T14:56:08.043" />
  <row Id="4304" PostTypeId="2" ParentId="4299" CreationDate="2011-01-31T14:57:50.213" Score="10" ViewCount="" Body="&lt;p&gt;You are observing a &lt;s&gt;Moiré&lt;/s&gt; Aliasing pattern (thanks to &lt;a href=&quot;http://physics.stackexchange.com/users/869/colin-k&quot;&gt;Colin K&lt;/a&gt; for the correction). This means that the optical resolution of your USB microscope is good and that the image sensor has no anti-aliasing filter. When image details are smaller than the pixel size, such patterns will appear. And they will be colored because in color sensors each pixel is only sensitive to one color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I took &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aNewton-rings.jpg&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia image&lt;/a&gt; of Newton's rings and downsized it by nearest neighbor method to get a picture that is very similar to yours:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/fmTZ6.jpg&quot; alt=&quot;Newton's rings: moire&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="701" LastEditorUserId="701" LastEditorDisplayName="" LastEditDate="2011-01-31T15:53:14.113" LastActivityDate="2011-01-31T15:53:14.113" CommentCount="3" />
  <row Id="4305" PostTypeId="2" ParentId="4299" CreationDate="2011-01-31T14:58:25.193" Score="11" ViewCount="" Body="&lt;p&gt;This is a manifestation of &quot;Frequency Aliasing&quot; which shows up in lots of situations where you have data which is sampled at discrete intervals. For a more in-depth explanation of aliasing in general, you should look for references on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem&quot; rel=&quot;nofollow&quot;&gt;Shannon-Nyquist Sampling Theorem&lt;/a&gt;. In brief however, it is sufficient to explain that when data is sampled at some given resolution, called the sampling period $T_s$, then any component of the signal with a frequency above $\frac{1}{2T_s}$ (this is called the Nyquist frequency) will be &quot;aliased&quot;, meaning they will appear to have a frequency of $\frac{1}{T_s}-f$ where $f$ is the frequency of the signal component in question. In this case the sampling period $T_s$ is the spacing of the pixels in the camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see from the un-aliased central rings, the frequency of the rings increases radially. This is a spatial frequency, as opposed to temporal frequency which people are more used to, but the mathematics is the same. As the ring frequency increases, it crosses the Nyquist frequency of the camera, and you see the pattern repeated as the frequency components that compose it are aliased below the Nyquist frequency of the camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, it is pretty straightforward to calculate where the secondary rings will show up (if you allow me to heavily gloss over some of the details). The spherical surface of the lens can be approximated as a parabolic surface $$y \propto r^2 $$ where $y$ is the height of the surface from the reference plate, and $r$ is the radial distance from the center of the primary rings. The frequency of the rings is related to the slope of the surface, which in this case is clearly linear in $r$, so without bothering to compute any constant factors, we can define $$f =Ar$$ where $f$ is the ring frequency and $A$ is a constant with units of $\frac{1}{\mbox{length}^2}$ or $\frac{\mbox{frequency}}{\mbox{length}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we could do some Fourier transforms and derive the exact form of the aliased ring image, but that is a bit more $\LaTeX$ than I feel like typing, and I suspect it is a bit beyond the scope of the question. Instead, I'll just calculate the location of the centers of the aliased ring patterns. So where are the centers of the ring patterns? They are the locations where the apparent ring frequency is zero. This is trivial for the central pattern, of course. Now where will the first set of aliased rings lie? We want to set the aliased frequency to zero, and solve for $r$; so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{lcr}&#xD;&#xA;\frac{1}{T_s}-f = 0 &amp;amp;\mbox{and} &amp;amp;f=Ar\\&#xD;&#xA;\frac{1}{T_s}-Ar = 0&amp;amp;\\&#xD;&#xA;r=\frac{1}{AT_s}&amp;amp; \mbox{or} &amp;amp; r=\frac{2f_n}{A}&#xD;&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computing the value of $A$ is a bit unnecessary for this explanation, so I'll simply say that it will depend on the wavelength of the light and the radius of curvature of the lens you are using. The resulting expression gives a simple description of where the rings will show up, and could serve as a guide for which lenses your microscope could measure without aliasing (i.e. when the aliasing won't appear until $r$ is larger than your field of view). Clearly, as the pixel spacing of your camera grows, aliasing becomes more of a problem, as it occurs closer and closer to the central rings; and as the radius of curvature of the lens increases (as the lens becomes less curved) aliasing becomes less problematic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also interesting is that the radial position of the rings in the diagonal direction is larger. This would seem counter-intuitive given the expression derived above, but keep in mind that the explanation I've given is vastly simplified. If you endeavor to derive a more general result using the Fourier transform, the diagonal rings will be correctly described.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The color effects are due to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bayer_filter&quot; rel=&quot;nofollow&quot;&gt;Bayer filter&lt;/a&gt; on the camera. This filter is what allows the camera to capture color images, but it also has the effect of giving a slightly different sampling rate for each of the different colors. This means that the aliased pattern will have some color effects, but this is not &quot;real&quot;, in the sense that it is only an effect of the camera itself. Bayer color effects can be seen in other situations as well, including situations that are more commonly described as Moiré effects than aliasing. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/6a2iO.jpg&quot; alt=&quot;Shirt with aliasing&quot;&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moiré and aliasing effects are, of course, closely related, and are in some sense the same thing; but frequency aliasing is a very general concept that is applicable to any sort of sampled data, while the Moiré effect is generally the name given to the patters resulting from overlapping periodic patterns, rather than sampling specifically.&lt;/p&gt;&#xA;" OwnerUserId="869" LastEditorUserId="869" LastEditorDisplayName="" LastEditDate="2011-02-03T17:09:26.163" LastActivityDate="2011-02-03T17:09:26.163" />
  <row Id="4307" PostTypeId="2" ParentId="4300" CreationDate="2011-01-31T15:38:30.647" Score="4" ViewCount="" Body="&lt;p&gt;Yes it is a bit like waveguide for electrodynamics. You've got &lt;a href=&quot;http://en.wikipedia.org/wiki/Airy_wave_theory#Mathematical_formulation_of_the_wave_motion&quot; rel=&quot;nofollow&quot;&gt;absolutely the same equation&lt;/a&gt;. The difference is in boundary conditions -- instead of &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirichlet_boundary_condition&quot; rel=&quot;nofollow&quot;&gt;Dirichlet&lt;/a&gt; you've got to use &lt;a href=&quot;http://en.wikipedia.org/wiki/Neumann_boundary_condition&quot; rel=&quot;nofollow&quot;&gt;Neumann boundary conditions&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ZqoUF.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While of course it works only if you are able no neglect all the non-linear effects -- your waves are low enough.&lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2011-01-31T15:48:44.193" LastActivityDate="2011-01-31T15:48:44.193" CommentCount="4" />
  <row Id="4308" PostTypeId="2" ParentId="4284" CreationDate="2011-01-31T15:58:38.243" Score="7" ViewCount="" Body="&lt;p&gt;The Copenhagen interpretation consists of two parts, unitary evolution (in which no information is lost) and measurement (in which information is lost). Decoherence gives an explanation of why information appears to be lost when it in actuality is not. &quot;The decay of the off-diagonal elements of the wave function&quot; is the process of turning a superposition:&#xA;$$\sqrt{{1}/{3}} |{\uparrow}\rangle + \sqrt{{2}/{3}} |{\downarrow}\rangle$$ into a probabilistic mixture of the state $|\uparrow\rangle$ with probability $1/3$, and the state $|\downarrow\rangle$ with probability $2/3$. You get the probabilistic mixture when the off-diagonal elements go to 0; if they just go partway towards 0, you get a mixed quantum state which is best represented as a density matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Decoherence is basis dependent. That is, you need to write the density matrix in some basis, and then decoherence is the process of reducing the off-diagonal elements in that basis. How do you decide which basis? What you have to do is look at the interaction of the system with its environment. Quite often (not always), this interaction has a preferred basis, and the effect of the interaction on the system can be represented by multiplying the off-diagonal elements in the preferred basis by some constant. The information contained in the off-diagonal elements does not actually go away (as it would in the Copenhagen interpretation) but gets taken into the environment, where it is experimentally difficult or impossible to recover.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditorDisplayName="" LastEditDate="2011-01-31T16:28:14.270" LastActivityDate="2011-01-31T16:28:14.270" CommentCount="3" />
  <row Id="4309" PostTypeId="2" ParentId="4289" CreationDate="2011-01-31T16:41:30.177" Score="3" ViewCount="" Body="&lt;p&gt;Dear Vagelford,&#xA;you're totally right. Gravity cannot be an entropic force because&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;its phenomena would be irreversible&lt;/li&gt;&#xA;&lt;li&gt;the degeneracy of the states coming from the entropy would destroy the interference patterns that have been measured e.g. by neutron interferometry.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;More than a year ago, this was also explained on my weblog&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2010/01/erik-verlinde-comments-about-entropic.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/01/erik-verlinde-comments-about-entropic.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and Erik Verlinde much like some of his junior Dutch colleagues tried to react but as far as I can say, none of their reactions has ever made any sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The neutron interferometry experiments are pretty impressive. They not only show that the interference pattern survives the action by the force of gravity. But it is exactly as shifted as the equivalence principle implies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in fact, the changes of the phases have been measured so accurately that the experimenters may deduce not only the zeroth order gravitational acceleration but also the higher-order corrections to it such as the tidal forces. All of these effects preserve the interference pattern - which wouldn't be possible if there were many states representing a macroscopic configuration - and this pattern exactly moves and behaves according to general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For these two and other reasons, gravity cannot be entropic. We also know it from the explicit models in the AdS/CFT correspondence and elsewhere: only event horizons may produce a large entropy of this order. A cold binary star doesn't carry any entropy associated with the gravitational attraction, certainly not an entropy comparable to the black hole entropy which is what Erik Verlinde claims.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a multi-million euro grant has already paid by some European politicians to endorse this &quot;research&quot; so it may be unreasonable to expect that too many people aside from the two of us will say these obvious things too comprehensibly and loudly. After all, many people can be bought very easily and inexpensively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One additional disclaimer: If you originally encountered the proofs that gravity can't be entropic on my blog, you shouldn't treat this answer as an independent confirmation of my previous claims. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the best, LM&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-01-31T16:41:30.177" CommentCount="7" />
  <row Id="4310" PostTypeId="1" AcceptedAnswerId="4311" CreationDate="2011-01-31T16:48:40.073" Score="3" ViewCount="120" Body="&lt;p&gt;In an interacting theory I expect there to be caustics, resonances, and other situations in which &lt;em&gt;some&lt;/em&gt; observables would give an infinite experimental result. Of course, these are idealized states and observables -- if a real device's measurement results are quite accurately modeled by such an observable, and we created a state in which its expected value is large enough, the device would be destroyed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An idealized world in which models never predict infinite results seems somehow different from the world we live in. Even though experiments never return the measurement result &quot;infinity&quot;, they do return the measurement result &quot;I'm so sorry, I'm breaking now, it's bigger than you thought it could be&quot;. The Wightman axioms do not require operators to be bounded, and to me seem much better for it. QFT as it's used in practice isn't constructed within the Wightman axioms, but it's much less constructed within the Haag-Kastler axioms, and, it seems to me, particularly for this reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I take boundedness to mean that all the eigenvalues of an operator in its action on a Hilbert space of idealized Physical states are required to be finite. This is much stronger than requiring the expected values of an operator to be finite for a dense subset of the Hilbert space (or other, relatively weaker, requirements).&#xA;It's certainly mathematically more convenient to use bounded operators (because we don't have to keep track of for which states we get a finite result for a given measurement, because they all do), but is that enough? At least, is this acceptable as part of a major attempt to axiomatize theoretical Physics? Is it obvious enough to be an axiom?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm prompted to ask this question in this way by a comment in Doplicher's &quot;The principle of locality: Effectiveness, fate, and challenges&quot; &lt;a href=&quot;http://jmp.aip.org/resource/1/jmapaq/v51/i1/p015218_s1&quot; rel=&quot;nofollow&quot;&gt;J.Math.Phys. 51, 015218 (2010)&lt;/a&gt;, &lt;a href=&quot;http://arxiv.org/abs/0911.5136&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0911.5136&lt;/a&gt;, which I'm reading this morning, where he sets out on the first page (2nd page in the arXiv) that &quot;In quantum mechanics the observables are given as bounded operators on a fixed Hilbert space&quot;, which seems a specially sanitized version of QM, insofar as it rules out position, momentum, and energy observables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, this question is asking, as always, have I got something (very) wrong?&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-01-31T19:58:20.733" Title="Does boundedness of observables in the Haag-Kastler axiomatization rule out interactions?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="4311" PostTypeId="2" ParentId="4310" CreationDate="2011-01-31T17:09:32.457" Score="3" ViewCount="" Body="&lt;p&gt;If you have an algebra of unbounded observables, the idea is to take the algebra&#xA;of all bounded measurable functions of these observables.&lt;br&gt;&#xA;In quantum mechanics this leads to the Weyl form: e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Stone%E2%80%93von_Neumann_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Stone%E2%80%93von_Neumann_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt;&#xA;A short comment on the title. In two-dimensions certain models were constructed in the algebraic framework which interact, models with so called factorizing S-matrix. Lechner showed in his thesis for a class of S-matrix that there is an algebra of local observables and that the theory is asymptotic complete etc.&#xA;The existence in the Wightman setting is harder to establish, it goes under the name form factor programm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt;&#xA;Sorry still not able to comment: Can one measure something infinite in finite time. I don't think so?!&lt;/p&gt;&#xA;" OwnerUserId="1611" LastEditorUserId="1611" LastEditorDisplayName="" LastEditDate="2011-01-31T19:58:20.733" LastActivityDate="2011-01-31T19:58:20.733" CommentCount="3" />
  <row Id="4312" PostTypeId="2" ParentId="4310" CreationDate="2011-01-31T17:10:30.380" Score="4" ViewCount="" Body="&lt;p&gt;The Haag-Kastler axioms and the Wightman axioms are &lt;em&gt;almost equivalent&lt;/em&gt;, AFAIK the equivalence has been proven with some additional technical assumptions that are believed to hold true for all &quot;relevant&quot; physical systems. (It's a matter of fact that all experts of axiomatic QFT work with some form of the Haag-Kastler axioms only, not with the Wightman axioms.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See for example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Borchers, Yngvason: &quot;From Quantum Fields to Local von Neumann Algebras&quot;, Rev.Math.Phys. Special issue, 1992, p.15-47.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Edit: The problem is that taking the algebra of all bounded functions of Wightman fields may result in a net of algebras that does not fulfill the locality axiom, for more details the the Borchers/Yngvason paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a quick and dirty pro/contra list:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Pro: Some observables do not have an upper bound, e.g. you can always increase the impulse of an electron, no matter how big that already is (in theory).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Contra: A detector has an upper bound for every observable it can detect, and that is what is actually measured in an experiment. The situation of the pro-bullet won’t arise because we measure observables in bounded regions of spacetime, and the energy contained in a bounded region is always finite (we are excluding artifacts like black holes from our consideration, of course).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Contra to even asking the question: Theories of bounded and unbounded operators/observables are physically equivalent modulo mathematical subtleties (see paper above), meaning they produce the same numbers that can be compared to experiments.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As for the construction of models with interactions: During the past two decades there has been some progress in the task of constructing QFTs with interactions within the Haag-Kastler axioms, see for example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Romeo Brunetti, Michael Duetsch, Klaus Fredenhagen: &quot;Perturbative Algebraic Quantum Field Theory and the Renormalization Groups&quot; (&lt;a href=&quot;http://arxiv.org/abs/0901.2038&quot; rel=&quot;nofollow&quot;&gt;arXiv&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The problems in constructing such a theory is not related to the fact that the Wightman axioms are about unbounded operators and the Haag-Kastler axioms are about bounded ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: It is true that AQFT is vastly more complicated than QM and has some surprising and counterintuitive features. Example: The Reeh-Schlieder theorem implies that local algebras cannot contain a particle count operator, see the nLab page for more details: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ncatlab.org/nlab/show/Reeh-Schlieder+theorem&quot; rel=&quot;nofollow&quot;&gt;Reeh-Schlieder theorem&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Edit: The philosophy of the Haag-Kastler approach is very different from the Wightman axioms however, in the Wightman axioms the fields are put in by hand. The philosophy of the Haag-Kastler axioms is that the theory is fully specified by specifying the observables, that is by prescribing a local net of von Neumann algebras. Everything else should be constructed from this net, including &quot;fields&quot;. This can indeed be done in certain situations, the relevant result is known as the &quot;Doplicher-Roberts reconstruction theorem&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditorDisplayName="" LastEditDate="2011-01-31T19:32:06.983" LastActivityDate="2011-01-31T19:32:06.983" CommentCount="2" />
  <row Id="4313" PostTypeId="1" AcceptedAnswerId="4321" CreationDate="2011-01-31T17:15:44.523" Score="9" ViewCount="416" Body="&lt;p&gt;I was thinking of the noted 1980 paper by Sidney Coleman and Frank de Luccia--&quot;Gravitational effects of and on vacuum decay&quot;-- about metastable vacuum states that could tunnel to a lower energy &quot;true vacuum&quot; with catastrophic results. I suspect the answer is that if our vacuum is a false one, we wouldn't &quot;see&quot; the true one til it hit us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://prd.aps.org/abstract/PRD/v21/i12/p3305_1&quot; rel=&quot;nofollow&quot;&gt;http://prd.aps.org/abstract/PRD/v21/i12/p3305_1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-01-31T18:52:24.770" LastActivityDate="2011-01-31T20:15:22.670" Title="Are we living in a false vacuum? Is there any way to tell?" Tags="&lt;quantum-mechanics&gt;&lt;astrophysics&gt;&lt;soft-question&gt;&lt;vacuum&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="4314" PostTypeId="1" AcceptedAnswerId="4323" CreationDate="2011-01-31T17:25:45.083" Score="2" ViewCount="228" Body="&lt;p&gt;Suppose an advanced undergraduate student has reached a moderate level of understanding on electrodynamics.&#xA;Where should he &lt;strong&gt;focus&lt;/strong&gt; on, to sharpen his &lt;strong&gt;problem-solving skills&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Practicing integrals and/or other &lt;strong&gt;mathematical tools&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Studying &lt;strong&gt;theoretical results&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Working on the &lt;strong&gt;physical meaning&lt;/strong&gt; and applications of ED equations.&lt;/li&gt;&#xA;&lt;li&gt;Something else.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I understand this is a somewhat subjective question but please try to be as objective as possible: state what really &lt;strong&gt;works&lt;/strong&gt; based on your experience as a teacher/TA for example&lt;/p&gt;&#xA;" OwnerUserId="824" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-01T01:48:43.163" LastActivityDate="2011-02-01T01:48:43.163" Title="Studying electrodynamics problems" Tags="&lt;soft-question&gt;&lt;electrodynamics&gt;&lt;education&gt;" AnswerCount="2" CommentCount="10" />
  <row Id="4316" PostTypeId="2" ParentId="4313" CreationDate="2011-01-31T17:51:37.553" Score="1" ViewCount="" Body="&lt;p&gt;Let me simplify it to this level: you've got a ball moving in 1D. And you observe that the ball is in a stable equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you assume that the potential for it has a form of a polynomial of 4th degree in x, with $x=0$ at the position of the ball:&lt;br&gt;&#xA;$V(x) = Ax^4+Bx^3+Cx^2$&lt;br&gt;&#xA;(No additive constant and no linear term because we are in the equilibrium.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find the coefficients of this polynomial by testing small oscillations around the equilibrium. For this you need to find the dependence of frequency of oscillations on amplitude. And suppose that you have found the following values:&lt;br&gt;&#xA;$A=1,B=-16/3,C=6$&lt;br&gt;&#xA;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=extrema+x%5E4-16x%5E3/3%2B6x%5E2&quot; rel=&quot;nofollow&quot;&gt;By investigating this function&lt;/a&gt; you realise that it has another minimum, which is lower than the current one. So you are in &quot;false vacuum&quot; and you realised it without moving your system into it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, of course this conclution is based on the assumption that the potential has this particular polynomial form and that whole your model works. But, in my opinion, this is the only context, it which the question makes sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can always invent a model where true vacuum is created at some particular energy or even at some particular process. And in order to verify this we must get to the energy either destroying everything, either disproving the theory. Personally, I don't think that this is a sound scientific theory. But if allow yourself this &quot;model freedom&quot;, then the answer to your question is: &quot;yes the only way to check if we are in the false vacuum is to create the true one&quot;. &lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2011-01-31T20:15:22.670" LastActivityDate="2011-01-31T20:15:22.670" CommentCount="8" />
  <row Id="4317" PostTypeId="2" ParentId="4123" CreationDate="2011-01-31T18:01:16.133" Score="1" ViewCount="" Body="&lt;p&gt;The radial velocity can not give us any information about transverse velocity directly. But it provides a clue that the star and gas bulk are not belonging to the same dynamical system. So they are unlikely to keep still to each other on the projected 2-D plane.&#xA;If the comparative transverse movement is at the same level of radial velocity, 60km/s for example, after 10 years the gas bulk will move $60 / 3E+5 * 10 = 0.018 ly$, which is 20 times larger than the Pluto's orbit (0.00085 ly at most). It is a small scale for a gas bulk. But it can be big enough for elements' column density to change (due to elements distribution and projection effect). You could observe slight shape changes of some absorption lines. But as mentioned by &lt;code&gt;gigacyan&lt;/code&gt;, with only one star it is really hard to get the direction and velocity of transverse movement. If you make some assumption, like taking the bulk as a sphere, then you can explore something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's more, both star and gas bulk are not isolated in the space, there are velocity limitations due to their position in the Galaxy, on the disk, in the bulge or halo etc. You can give some reasonable estimates base on these.&lt;/p&gt;&#xA;" OwnerUserId="518" LastActivityDate="2011-01-31T18:01:16.133" CommentCount="3" />
  <row Id="4318" PostTypeId="2" ParentId="3754" CreationDate="2011-01-31T18:05:52.647" Score="0" ViewCount="" Body="&lt;p&gt;It goes where the photons go. Best explanation I've seen is in Feynman &quot;QED: the strange theory of light and matter&quot;. One does not have to unlearn anything. It is even more interesting after reading his more technical works.&lt;/p&gt;&#xA;" OwnerUserId="1423" LastActivityDate="2011-01-31T18:05:52.647" />
  <row Id="4319" PostTypeId="1" AcceptedAnswerId="4337" CreationDate="2011-01-31T18:52:03.820" Score="0" ViewCount="261" Body="&lt;p&gt;I have tried to answer the problem correctly, and I have found similar (and even an identical) problem in my textbook, yet I still can't seem to yield the correct answer with this specific problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example problem in my textbook: A crate is hauled 8m up a ramp under a constant force of 200 N applied at an angle of 25 degrees to the ramp. Find the work done. W = F * D cos(25) = (200)(8)cos(25) = 1450J.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone give me any advice on how to solve it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A woman exerts a horizontal force of 4 pounds on a box as she pushes it up a ramp that is 5 feet long and inclined at an angle of 30 degrees above the horizontal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the work done on the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think that the work would simply be (4)(5)cos(30), but that is not the correct answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I've figured out how to accept answers now, so I'll accept the most helpful answer.)&lt;/p&gt;&#xA;" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-11T20:13:20.237" LastActivityDate="2011-05-11T20:13:20.237" Title="Homework Help - Physics / Calculus III / Force-Work Problem" Tags="&lt;homework&gt;&lt;mechanics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="4320" PostTypeId="1" AcceptedAnswerId="4324" CreationDate="2011-01-31T18:58:08.753" Score="5" ViewCount="200" Body="&lt;ul&gt;&#xA;&lt;li&gt;I came across an argument which seems to say that the &quot;constraint of Gauss's law&quot; enforces gauge theory on compact spaces to be such that physical states over which the partition function sums over be gauge invariant. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would like to hear of explanations of the above argument.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Also the above seems to lead (quite non-obviously to me) to the conclusion that these physical states correspond to traces of products of operators acting on the Fock space vacuum. It is not clear to me as to how this trace is defined such that even after tracing it remains an operator. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;{Very often one seems to want these operators to be in the &quot;adjoint of&quot; the gauge group. The meaning and motivation of this demand is not clear to me. (I am familiar with the notion of adjoint representation of Lie groups)}&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Related to the above is another claim I see which seems say that massless modes will be absent for any gauge group Yang-Mill's theory and any matter content if the theory is on a compact space. Is the above correct? Why (whether yes or no)? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In such scenarios is the terminology of &quot;basic excitations&quot; of a theory the same thing as single particle states? How are these single particle states in general related to the physical states constructed above? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which of these is what is referred to when people talk of &quot;modes&quot; of a QFT? &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If there is a gauge symmetry by definition it will commute with the Hamiltonian and hence the states of the theory at every energy level will form a representation of the gauge group. Can something be said about its reducibility or not? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The claim seems to be that if the there are say $n_E$ quanta at the energy level $E$ (transforming under say representation $R_E$ of the gauge group) then when counting its contribution to the partition function the boltzman factor has to further weighted by the number of $1$ dimensional representations (&quot;singlets&quot; ?) in the $n$-fold symmetric (for bosons) or the anti-symmetric (for fermions) tensor power of $R_E$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be glad to know of explanations of the above. &lt;/p&gt;&#xA;" OwnerUserId="1637" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T15:30:33.563" LastActivityDate="2011-02-02T15:30:33.563" Title="Modes of a QFT and irreducible representation of the gauge group" Tags="&lt;quantum-field-theory&gt;&lt;gauge-theory&gt;&lt;group-representations&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="4321" PostTypeId="2" ParentId="4313" CreationDate="2011-01-31T19:08:04.927" Score="4" ViewCount="" Body="&lt;p&gt;Inflation is a rapid stretching which result in cosmic smoothness and uniformity on large scales; as such, inflation is a key component of almost all fundamental cosmological scenarios. Not only does inflation explain the overall uniformity of the universe, but quantum fluctuations during inflation plant the seeds that grow into galaxies and clusters of galaxies that exist today.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The potential for the inflationary potential early in the universe is a de Sitter form.  The FLRW equations are&#xA;$$&#xD;&#xA;\Big(\frac{\dot a}{a}\Big)^2~=~\frac{8\pi G\Lambda}{3}~-~\frac{k}{a^2},&#xD;&#xA;$$&#xA;where we assume $k~=~0$ for the generally flat space we appear to observe.  The early inflationary universe was driven by a scalar field which generated this vacuum energy where $V(\phi)~=~-a\times\phi$, $a$ a constant.  This set the early cosmological constant for the de Sitter expansion with a vacuum energy about 13 orders of magnitude smaller than the Planck energy.  The universe had more vacuum energy density than quark-gluon field density in a hadron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lagrangian for a scalar field is $L~=~(1/2)\partial^a\phi\partial_a\phi~–~V(\phi)$ and in QFT we work with the Lagrangian density ${\cal L}~=~L/vol$ so the action $S~=~\int d^3xdt{\cal L}(\phi, \partial\phi)$. We run this into the Euler-Lagrange equation $\partial_a(\partial{\cal L}/\partial(\partial_a\phi))~-~\partial{\cal L}/\partial\phi~=~0$, and keep in mind $vol~\sim~x^3$. This gives a dynamical equation&#xA;$$&#xD;&#xA;\partial^2\phi ~-~ (3/vol^{4/3})\partial_a\phi~–~ \frac{\partial V(\phi)}{\partial\phi}~=~ 0.&#xD;&#xA;$$&#xA;If we assume the inflaton field is more or less constant on the space for a given time on the Hubble frame this DE may be simplified to&#xA;$$&#xD;&#xA;{\ddot\phi}~–~(3/vol^{4/3}){\dot\phi}~–~\frac{\partial V(\phi)}{\partial\phi}~=~0&#xD;&#xA;$$&#xA;That middle term is interesting for it is a sort of friction. It indicates the inflaton field, the thing which drives the inflationary expansion, is running down or becoming diffused in the space. The potential function here is complicated and not entirely known, but it is approximately constant, or some small decrease with the value of $\phi$.   What then happens, which is not entirely understood, is that the field experiences a phase transition, the potential becomes $V(\phi)~\sim~\phi^2$ with a minimum about 110 orders of magnitude smaller than it was in the unbroken phase. The phase transition has a latent heat of fusion that is released and this is the reheating.  If the vacuum is a false vacuum then the $V(\phi)~\sim~\phi^4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means the accelerated expansion of the universe should be driven by either of these field and the a force which drives the field:&#xA;$$&#xD;&#xA;F~=~-\frac{\partial V}{\partial\phi}&#xD;&#xA;$$&#xA;which is larger for the steep potential, or the quartic.  During this period a quantum fluctuation in the field is typically $\delta\phi~=~\pm\sqrt{V(\phi)}$.  For the inflationary period the variation in the field due to the force is $\delta\phi_F~=~F/V$ $~sim~\phi^{-1}$ and the quantum fluctuation in the scalar field $\delta\phi_q~=~\pm const\sqrt{\phi}$  The quantum fluctuations can become larger than that classical variation in the field when&#xA;$$&#xD;&#xA;\delta\phi_F~=~\delta\phi_q~\rightarrow~\phi~\simeq~a^{1/3}&#xD;&#xA;$$&#xA;For the reheating potential $V(\phi)~=~b\phi^n $, $n~=~2,~4$ the condition for the fluctuation equal the classical field variation is&#xA;$$&#xD;&#xA;\phi~\simeq~(n^2/a)^{1/{(n+2}}&#xD;&#xA;$$&#xA;For $n~=~4$ the field may vary far less for the quantum fluctuation to equal the classical variation.  If this happens for $n~=~4$ we would expect the universe to tunnel into a lower energy vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/E1YP0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now turn to some data H. V. Peiris and R. Easther, JCAP 0807, 024 (2008) arXiv:0805.2154 astro-ph.  This figure illustrates joint 68% (inner) and 95% (outer) bounds on two variables which characterize the primordial perturbations, derived from a combination of WMAP and SuperNova Legacy Survey data. The predictions for our two inflationary models are superimposed. The numbers refer to the logarithm of the size of universe during the inflationary era. Cosmological perturbations are generated when this quantity is around $60$, so $\phi^4$ inflation is not consistent with the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we are probably out of the danger zone for having one of Coleman-Luccia’s vacuum transitions which destroys everything.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-01-31T19:13:16.187" LastActivityDate="2011-01-31T19:13:16.187" CommentCount="4" />
  <row Id="4322" PostTypeId="1" AcceptedAnswerId="4329" CreationDate="2011-01-31T19:21:16.803" Score="0" ViewCount="169" Body="&lt;p&gt;Is there a &lt;em&gt;time + two spatial dimension&lt;/em&gt; representation of a Minkowski-space surface which could be constructed within our own (assumed Euclidean) 3D space such that geometric movement within the surface would intuitively demonstrate the “strange&quot; effects of the Lorentz transformation (length contraction, time dilation)? Perhaps by making manifest the idea of a hyperbolic rotation (the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rapidity&quot; rel=&quot;nofollow&quot;&gt;rapidity&lt;/a&gt;)?&lt;/p&gt;&#xA;" OwnerUserId="1394" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-02-01T10:28:03.100" LastActivityDate="2011-02-01T10:28:03.100" Title="Is there an intuitive geometric view of the effects of Lorentz transformations?" Tags="&lt;special-relativity&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="4323" PostTypeId="2" ParentId="4314" CreationDate="2011-01-31T19:28:02.887" Score="3" ViewCount="" Body="&lt;p&gt;In my humble experience, solving Griffiths problems gets you good at solving Griffiths problems, but not much more. Typically, they've already done/seen the math required, so I'd only work on the calculus part if they're really struggling. Studying theoretical results doesn't seem a great idea either; they'll pick that up along the way or in class. What I prefer is to study more 'extended' problems, i.e. a real life example for the (classical) literature, guide them through the development and finish with a comparison to experimental results. This way, they plough through the maths but also need to do a minimum of interpreting. However, if your goal is just to get them great at working Griffiths-like problems, this might be overkill.&lt;/p&gt;&#xA;" OwnerUserId="1471" LastActivityDate="2011-01-31T19:28:02.887" />
  <row Id="4324" PostTypeId="2" ParentId="4320" CreationDate="2011-01-31T19:42:16.923" Score="5" ViewCount="" Body="&lt;p&gt;It's a lot of questions but they have pretty easy answers, so here they are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Gauss's law is just $\mbox{div }\vec D=\rho$ in electrodynamics. Note that it contains no time derivatives so it's not really an equation describing evolution: it's an equation restricting the allowed initial conditions. More generally, it's the equation of motion that you get by varying the Lagrangian with respect to $A_0$, the time component of the gauge field, so the corresponding equation of motion counts the divergence of the electric field minus the electric sources (charge density) that must be equal to it. This difference is nothing than the generator of the overall $U(1)$ group, or any other group, if you consider more general theories, so the classical equation above is promoted to the quantum equation in which $(\mbox{div }\vec D-\rho)|\psi\rangle=0$ which just means that the state $|\psi\rangle$ is gauge-invariant.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The traces you are encountering here - in non-Abelian theories - are just traces over fundamental (or, less frequently, adjoint) indices of the Yang-Mills group. They're different than traces over the Hilbert space. You must distinguish different kinds of indices. Tracing over some color indices doesn't change the fact that you still have operators.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Adjoint of a gauge group&quot; is clearly the same thing as &quot;Adjoint representation of the Lie group that is used for the gauge group.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It's not true that all compact spaces eliminate all massless fields - for example, the Wilson line of a gauge field remains a perfectly massless scalar field on toroidal compactifications in supersymmetric theories - but in most other, generic cases, it's true that compactification destroys the masslessness of all fields. All the Fourier (or non-zero normal) components of the fields that nontrivially depend on the extra dimensions - the Kaluza-Klein modes - become massive because of the extra momentum in the extra dimensions. But even the &quot;zero modes&quot; become massive in general theories because of the Casimir-like potentials resulting from the compactification.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Basic excitations are not &quot;the same thing&quot; as one-particle states. In fact, we want to use the word &quot;excitation&quot; exactly in the context when the goal is to describe arbitrary multi-particle states. But the basic excitations are just the creation operators (and the corresponding annihilation operators) constructed by Fourier-transforming the fields that appear in the Lagrangian, or that are elementary in any similar way.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You haven't constructed any &quot;specific states above&quot; so I can't tell you how some other states you haven't described are related. In this respect, your question remained vague. All of them are some states with particles on the Hilbert space - but pretty much all states may be classified in this way.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A mode of a quantum field is the term in some kind of Fourier decomposition, or - for more general compactifications and backgrounds - another term (such as the spherical harmonic) that is an eigenstate of energy i.e. that evolves as $\exp(E_n t/i\hbar)$ with time. So for example, a field $X(\sigma)$ for a periodic $\sigma$ may be written as the sum below. The individual terms for a fixed $n$ - or the factor $X_n$ or the function that multiplies it (this terminology depends on the context a bit) - are called the modes. &#xA;$$\sum_{n\in Z} X_n  \,\exp(in\sigma - i|n|\tau)$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Gauge symmetry has to commute with the Hamiltonian because we want to ban the gauge-non-invariant states, and by banning them in the initial state, they have to be absent in the final state, too. So it has to be a symmetry. On the other hand, the representation theory is trivial because, as I said at the very beginning, we require physical states to be gauge-invariant; that was the comment about Gauss's law: states have to be annihilated by all operators of the type $\mbox{div }\vec D-\rho$ which are just generators of the gauge group at various points. In other words, all of the physical states have to be singlets under the gauge group. That's why the term &quot;symmetry&quot; is somewhat misleading: some people prefer to call it &quot;gauge redundancy&quot;. So the Hilbert space is surely completely reducible - to an arbitrary number of singlets. You may reduce it to the smallest pieces that exist in linear algebra - one-dimensional spaces.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I just explained you again why all the physical states are singlets under the gauge group. The fact that $n$-particle states with identical particles are completely symmetric or completely antisymmetric reduces to the basic insight that in quantum field theory, particles are identical and their wave function has to be symmetric or antisymmetric (for bosons and fermions) because the corresponding creation operators commute (or anticommute) with each other.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T15:26:09.170" LastActivityDate="2011-02-02T15:26:09.170" CommentCount="9" />
  <row Id="4325" PostTypeId="1" CreationDate="2011-01-31T20:20:06.223" Score="2" ViewCount="142" Body="&lt;p&gt;My guess at a list of them would be: spin foams, casual sets, non-commutative geometry, Machian theories, twistor theory or strings and membranes existing in some higher-dimensional geometry...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, what have I missed? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if there is a string theorist or similar on here that could briefly shed some light on what's involved with these approaches I'd be most grateful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Editted/expanded from here on]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK so I'm not having much luck getting a response to this so I suppose I've not been clear enough. I'll try to expand a little:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In early 2010 I attended this inagural lecture by string theorist- Prof. Mavromatos entitled 'MAGIC strings'. In it he proposes that some string theory models may violate Lorentz symmetry at the Planck scale resulting in a kind-of foamy spacetime that could be observed by differing arrival times of photons of different energies reaching us from distant astronomical sources. See &lt;a href=&quot;http://www.kcl.ac.uk/news/events_details.php?year=2010&amp;amp;event_id=2178&quot; rel=&quot;nofollow&quot;&gt;http://www.kcl.ac.uk/news/events_details.php?year=2010&amp;amp;event_id=2178&lt;/a&gt; or here for one of the papers: &lt;a href=&quot;http://iopscience.iop.org/1742-6596/174/1/012016&quot; rel=&quot;nofollow&quot;&gt;http://iopscience.iop.org/1742-6596/174/1/012016&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, as I mentioned in the comment below, Prof. Sir Roger Penrose mentions that Wheeler and others have strongely argued that if we could examine spacetime at the Planck scale we would see a tubulent chaotic situation (from quantum-mechanical vacuum fluctuation I suppose) or perhaps a discrete granular one. Penrose goes on to list some other approaches that may suggest how this discete stucture may manifest itself. Hence my list above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have studied much QM, introductory QFT and the Standard Model as well as some basic GR but I have no formal experience of string theory. I therefore wondered if anyone here could jot down a sentace or two to explain what's involved with each of the above approaches and add to the list if I've overlooked some alternative popular(ish) approaches.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="1696" LastEditorDisplayName="" LastEditDate="2011-02-01T17:10:12.500" LastActivityDate="2011-02-01T17:10:12.500" Title="What are the current (popular(ish)) approaches to modelling the quantum nature of spacetime at the Planck scale?" Tags="&lt;string-theory&gt;&lt;quantum-gravity&gt;&lt;spacetime&gt;&lt;quantum-mechanics&gt;&lt;vacuum&gt;" CommentCount="5" FavoriteCount="1" />
  <row Id="4328" PostTypeId="2" ParentId="4236" CreationDate="2011-01-31T21:38:51.820" Score="2" ViewCount="" Body="&lt;p&gt;Thanks to Marek, for pointing out the lectures given by Richard Feynman in these two videos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://vega.org.uk/video/programme/45&quot; rel=&quot;nofollow&quot;&gt;http://vega.org.uk/video/programme/45&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://vega.org.uk/video/programme/46&quot; rel=&quot;nofollow&quot;&gt;http://vega.org.uk/video/programme/46&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About half way into the second video he explains the concept of refraction of photons. The lectures can be understood without deep knowledge of physics. You should set aside about 120 minutes. &lt;/p&gt;&#xA;" OwnerUserId="1675" LastActivityDate="2011-01-31T21:38:51.820" CommentCount="1" />
  <row Id="4329" PostTypeId="2" ParentId="4322" CreationDate="2011-01-31T21:39:46.153" Score="2" ViewCount="" Body="&lt;p&gt;I think what you are looking for is given in the first chapter of the epic text &quot;Spinors and Spacetime&quot; by Wolfgang Rindler and Roger Penrose. Or at least it is &lt;strong&gt;if what you're asking for is a simple and clear geometric construction that illustrates the effects of Lorentz transformations on the bulk (3+1) geometry.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ortxj.jpg&quot; alt=&quot;The celestial sphere - from Penrose and Rindler&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Fingers crossed .... &lt;strong&gt;IMHO&lt;/strong&gt; the use of this image and the ones below is covered under &quot;fair use&quot;. If it is deleted blame the copyright regime.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To cut a long story short, you identify points on the celestial sphere $\mathcal{S}^-$ with a light ray as seen by an observer at the center of the sphere. A stereographic projection allows to map points on $\mathcal{S}^-$ to the complex plane $\mathbb{C}$. The action of Lorentz boosts on the observer translates into the action of an $SL(2,\mathbb{C})$ element on the points of $\mathcal{S}^-$. The result is shown in the figure below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/45Nis.jpg&quot; alt=&quot;Effect of Lorentz transformations on $\mathcal{S}^{\pm}$&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know of any other constructions which so vividly illustrates the geometrical effects of Lorentz transformations. I have left out many details for which I once again recommend the amazing text by Penrose and Rindler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All hail the copyright gods.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; In response to the comments, I answered the question the best I could understand it. I've emphasized the relevant sentence in the first para.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-01T09:12:28.167" CommentCount="6" />
  <row Id="4331" PostTypeId="1" AcceptedAnswerId="4357" CreationDate="2011-01-31T22:33:23.693" Score="8" ViewCount="436" Body="&lt;p&gt;If the S-Matrix is the only observable, that rules out both generalized free fields and Wick-ordered polynomials of generalized free fields as interesting Physical models, because both result in a unit S-matrix. Neither possibility has been developed since the 1960s when these results were proved, and when the S-Matrix was ascendant as the only observable in Particle Physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the S-matrix is not the only quantum field observable, which it certainly seems not to be in Condensed Matter Physics and in Quantum Optics, to name just two fields in which Wightman or correlation functions play a large part in modeling, does that encourage us to construct Wick-ordered polynomials of generalized free fields as algebraic, non-dynamical deformations of free fields? In such a construction, the point is to avoid deforming a Hamiltonian or Lagrangian evolution, instead working with deformations of the observables, requiring for example that all observables and all states will be constructed as Wick-ordered polynomials in a Wick-ordered operator-valued distribution such as  $\hat\Phi(x)=:\!\!\hat\phi(x)+\lambda\hat\phi^\dagger\hat\phi(x)^2\!\!:$. The Hamiltonian is taken to be a derived quantity in this approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The S-Matrix seems remarkable, in that it requires us to set up a hyperplane at an initial time $t_I$ and another at a final time $t_F$, and another two hyperplanes between which the interaction will be nontrivial, at times $t_I&amp;lt;t_A&amp;lt;t_B&amp;lt;t_F$, all of which is not Lorentz-invariant, then take the limits $t_I\rightarrow-\infty,t_A\rightarrow-\infty,t_B\rightarrow+\infty,t_F\rightarrow+\infty$. It seems that in no other branch of Physics would we construct such an idealization and say it's the only possible way to report the experimental data. In other words, I wonder, going further, whether the S-matrix even is an observable? Particle Physics seems to be the only field in Physics that seems to think it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;post-Answer addition.&lt;/strong&gt; It's Greenberg who proves in JMathPhys 3, 31(1962) that Wick-ordered finite polynomials in the field have a trivial S-matrix, and I think, without looking it up, that it's also Greenberg who introduced and ruled out generalized free fields. I've resisted doing something else because Wick-ordered polynomials in a generalized free field seem to give such a lot of freedom to create models that surely they can be useful, which I know is not a good reason, but I had to feel how bad it is before I could move on. I suppose I was almost ready to give up this way of doing things before I asked this question. While Tim van Beeks' response is definitely of interest, and Matt's Answer is clearly the &quot;right&quot; answer, but doesn't go into the periphery of the question in the right way to help me, it was Marcel's response that particularly pushed me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My comment to Marcel indicates the way I'm now going to take this, to functions of the field such as $\hat\Phi(x)=\tanh{(\hat\phi(x))}$. Insofar as we can say that the measured value of $\hat\phi(x)$ is almost always $\pm\infty$ in the vacuum state, because the expected value $\left&amp;lt;0\right|\hat\phi(x)\left|0\right&amp;gt;$ is zero and we can say, loosely, that the variance is $\infty$ in the vacuum state, presumably $\left&amp;lt;0\right|\delta(\hat\Phi(x)-\lambda)\left|0\right&amp;gt;$ is non-zero only for $\lambda=\pm 1$ (though with some worries about this construction). It's not clear that we even have to introduce normal-ordering. $\tanh{(\hat\phi(x))}$ is of course a bounded operator of the unbounded Wightman free field, whereas no nontrivial polynomial in the field $\hat\phi(x)$ can be a bounded operator. The particular choice of $\hat\Phi(x)=\tanh{(\hat\phi(x))}$ is clearly a particular coordinatization; if we change the coordinatization by taking a function of $\hat\Phi(x)$, we in general get a measurement operator that results in two discrete values of the field, mapping $\pm 1$ to $a,b$ respectively. If we take something different from the real Klein-Gordon field, my aim as I envisage it four hours after first imagining it is to map something like an $SU(3)$ invariant Wightman field, say, to a finite number of discrete values. If such an $SU(3)$ symmetry is unbroken, the relationships between each of the discrete values will all be the same, but if the $SU(3)$ symmetry is broken, there presumably has to be a coordinate-free way in which the relationships between the different discrete values are different. There will, I now suppose, have to be enough distinction between raising and lowering operators between different sectors of the theory and measurement operators to allow there to be a concept of an S-matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may look crazy, but I'll also put on the table here why the structures of the Feynman graphs formalism encourage me. We introduce connected Feynman graphs at different orders to calculate $n$-point connected Wightman functions. Although we typically expand the series in terms of the number of loops, we can alternatively expand the series in terms of the number of points we have introduced between the $n$ points at which we measure; the extra points ensure that there are infinite numbers of different paths between the $n$ points. With something like $\hat\Phi(x)$, we introduce an infinite number of paths directly between the $n$ points, without introducing any extra points, so we need, in effect, a transformation of the superposition of an infinite number of Feynman path integrals into a superposition of an infinite number of weighted &lt;em&gt;direct&lt;/em&gt; paths between the $n$ points. Getting the weights on the direct paths right is of course rather important, and I also imagine the analytic structure has to be rather carefully done, particularly if I don't use normal-ordering. I'll do any amount of work to avoid renormalization, even in its modern gussied up form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me significant that the quantum field $\hat\Phi(x)=\tanh{(\hat\phi(x))}$ is not reducible. A number of proofs concerning Wightman fields rely on this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone else understands this (or reads this far) I'll be surprised. In any case I expect it will look very different a few years down the road if I ever manage to get it into a journal. Although I've worked in and around the Wightman axioms for the last few years, I find it interesting that I can now feel some pull towards something like the Haag-Kastler axioms. &lt;em&gt;Lots&lt;/em&gt; of work to do! Thank you all! Good luck with your own crazy schemes!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;That's the question (and the to me unexpected state of play a day later). Completely separately, as an example&lt;/strong&gt;, &lt;em&gt;to show the way I'm going with this, hopefully&lt;/em&gt; (which, a day later, looks as if it will be only a background concern for the next little while, but I think not likely to be completely forgotten by me), the real-space commutator of the creation and annihilation operator-valued distributions of the free field of mass $m$ is, in terms of Bessel functions, $$C_m(x)=\frac{m\theta(x^2)}{8\pi\sqrt{x^2}}\left[Y_1(m\sqrt{x^2})+i\varepsilon(x_0)J_1(m\sqrt{x^2})\right]$$&#xA; $$\qquad\qquad+\frac{m\theta(-x^2)}{4\pi^2\sqrt{-x^2}}K_1(m\sqrt{-x^2})-\frac{i}{4\pi}\varepsilon(x_0)\delta(x^2).$$&#xA;If we take a weighted average of this object with the normalized weight function&#xA;$w_{\alpha,R}(m)=\theta(m)\frac{R(Rm)^{\alpha-1} {\rm e}^{-Rm}}{\Gamma(\alpha)}, 0&amp;lt;\alpha\in \mathbb{R},\ 0&amp;lt;R$, $\int w_{\alpha,R}(m)C_m(x)dm$, we obtain the commutator of a particular generalized free field, which can be computed exactly in terms of Hypergeometric functions and which at space-like separation $\mathsf{r}=\sqrt{-x^\mu x_\mu}$ is asymptotically&#xA;$\frac{\Gamma(\alpha+1)R^{\alpha}}{\pi\Gamma(\frac{\alpha+1}{2})^2\bigl(2\mathsf{r})^{\alpha+2}}$, and at time-like separation $\mathsf{t}=\sqrt{x^\mu x_\mu}$ is asymptotically&#xA;$$\frac{\cos{(\frac{\pi\alpha}{2})}\Gamma(\frac{\alpha}{2}+1)R^\alpha}&#xD;&#xA;                        {4\sqrt{\pi^3}\Gamma(\frac{\alpha+1}{2})\mathsf{t}^{\alpha+2}}&#xD;&#xA;-i\frac{R^\alpha}{4\Gamma\left(-\frac{\alpha}{2}\right)\Gamma\left(\frac{\alpha+1}{2}\right)\mathsf{t}^{\alpha+2}},$$&#xA;except for $\alpha$ an even integer. At small space-like or time-like separation, the real part of this generalized free field is $-\frac{1}{4\pi^2x^\mu x_\mu}$, identical to that of the massless or massive free particle, independent of mass, but we can tune the 2-point function at large distances to be any power of the separation smaller than an inverse square law. On the light-cone itself, the delta-function component is again identical to that of the massless or massive free particle, independent of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is of course an infinity of possible normalized weight functions, a half-dozen of which I have worked out exactly and asymptotically, and somewhat obsessively, by use of MAPLE and Gradshteyn &amp;amp; Ryzhik, though I've managed to stop myself at the moment.&#xA;In a subsequent edit, I can't resist adding what we obtain if we use the weight function $w_{\mathsf{sm}[R]}(m)=\frac{\theta(m)R\exp{\left(-\frac{1}{mR}\right)}}{2(mR)^4},\ 0&amp;lt;R$ [using 6.591.1-3 from Gradshteyn &amp;amp; Ryzhik]. This function is smooth at $m=0$, and results at time-like separation in&#xA;$$\left\{Y_2\!\left(\sqrt{\frac{2\mathsf{t}}{R}}\right)+iJ_2\!\left(\sqrt{\frac{2\mathsf{t}}{R}}\right)\right\}&#xD;&#xA;  \frac{K_2\!\left(\sqrt{\frac{2\mathsf{t}}{R}}\right)}{8\pi R^2}&#xD;&#xA;   \asymp \frac{\exp{\left(-i\sqrt{\frac{2\mathsf{t}}{R}}+\frac{\pi}{4}\right)}&#xD;&#xA;                \exp{\left(-\sqrt{\frac{2\mathsf{t}}{R}}\right)}}&#xD;&#xA;               {16\pi\sqrt{R^3\mathsf{t}}}$$&#xA;and at space-like separation in&#xA;$$\frac{K_2\!\left((1+i)\sqrt{\frac{\mathsf{r}}{R}}\right)&#xD;&#xA;        K_2\!\left((1-i)\sqrt{\frac{\mathsf{r}}{R}}\right)}{4\pi^2R^2}&#xD;&#xA;   \asymp \frac{\exp{\left(-2\sqrt{\frac{\mathsf{r}}{R}}\right)}}&#xD;&#xA;                {16\pi\sqrt{R^3\mathsf{r}}}.$$&#xA;I wish I could do this integral for more general parameters, but hey! This weight function has observable effects only at space-like and time-like separation $\mathsf{r}&amp;lt;R$ and $\mathsf{t}&amp;lt;R$, so this is essentially unobservable if $R$ is small enough. The kicker is that this decreases faster than polynomially in both space-like &lt;em&gt;and&lt;/em&gt; time-like directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generalized free field construction always results in a trivial $n$-point function for $n&amp;gt;2$, however by introducing also Wick-ordered polynomials of these generalized free fields, we can also tune the $n$-point connected correlation functions, which are generally non-trivial and to my knowledge finite for all $n$. All this is far too constructive, of course, to prove much. I think I propose this more as a way to report the $n$-point correlation functions in a manner comparable to the Kallen-Lehmann representation of the 2-point correlation function than as something truly fundamental, because I think it does not generalize well to curved space-time.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-02-02T16:57:18.240" LastActivityDate="2011-02-03T00:33:57.693" Title="Is the S-Matrix the only quantum field observable?" Tags="&lt;quantum-field-theory&gt;" AnswerCount="5" FavoriteCount="2" />
  <row Id="4332" PostTypeId="2" ParentId="4322" CreationDate="2011-01-31T22:49:24.617" Score="0" ViewCount="" Body="&lt;p&gt;Since this question is about learning Special Relativity with an alternative to the Minkowski Diagram to aid understanding (and dropping one - even two - dimensions shouldnt cause much harm for that purpose), might I recommend consideration of the Bondi K-Calculus?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the &quot;K&quot; is introduced into the geometry, which represents the relativistic Doppler term: it is additive and cancels out much of the hyperbolic weirdness in basic Minkowski accounts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick G-search found this simple Tutorial: &lt;a href=&quot;http://www.math.ku.edu/~lerner/m291/SR_Lecture2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.ku.edu/~lerner/m291/SR_Lecture2.pdf&lt;/a&gt; with drawings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia links to another Tutorial (without diagrams as far as I can tell). &lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-01-31T22:49:24.617" />
  <row Id="4333" PostTypeId="2" ParentId="4331" CreationDate="2011-01-31T23:22:23.513" Score="1" ViewCount="" Body="&lt;p&gt;I'm not sure I understand what your question really is, so I'll try to answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Is the S-matrix the only observable in QFT?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, other observables are for example the total values of charges like the electric charge, that is all numbers that specify a superselection sector correspond to non-trivial observables. The S-matrix may be one of the most &lt;em&gt;interesting&lt;/em&gt; observables because most experiments in high energy physics are about scattering processes, however...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Is the S-matrix an observable in QFT?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, in traditional Lagrangian QFT one simply assumes that the S-matrix exists and has all the nice properties that one wishes for, so in this framework the question is not very interesting. It is very hard and non-trivial in axiomatic QFT, on the other hand, since you'll find that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b.1) defining the concept of a &quot;particle&quot; is not trivial and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b.2) the S-matrix may not be defined because the theory is not asymptotically complete: In QM this question is completely answered, in relativistic axiomatic QFT it seems to be an open problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A recent review paper is this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Detlev Buchholz, Stephen J. Summers: &quot;Scattering in Relativistic Quantum Field Theory: Fundamental Concepts and Tools&quot; (&lt;a href=&quot;http://arxiv.org/abs/math-ph/0509047&quot; rel=&quot;nofollow&quot;&gt;arxiv&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And again all physicists are free to choose if the failure of incorporating &quot;simple and obvious concepts&quot; is a catastrophic failure of AQFT, or if this difficulty hints at some deeper truth that we still don't understand in QFT. Your choice.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-01-31T23:22:23.513" CommentCount="5" />
  <row Id="4334" PostTypeId="1" AcceptedAnswerId="4424" CreationDate="2011-02-01T00:06:41.737" Score="0" ViewCount="341" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Could life on earth survive a large pole shift caused by an asteroid collision?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I became aware that there are people who believe that the earth's pole suddenly shifts. That is, its rotational axis changes rather than its magnetic axis. This is listed in wikipedia as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cataclysmic pole shift hypothesis&lt;/strong&gt;&#xA;The cataclysmic pole shift hypothesis is the conjecture that there have been rapid shifts in the relative positions of the modern-day geographic locations of the poles and the axis of rotation of a planet. For the Earth, such a dynamic change could create calamities such as floods and tectonic events.&lt;br&gt;&#xA;...&lt;br&gt;&#xA;The field has attracted pseudoscientific authors offering a variety of evidence, including psychic readings.&lt;br&gt;&#xA;...&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cataclysmic_pole_shift_hypothesis&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Cataclysmic_pole_shift_hypothesis&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While watching a chess game at the local club a fellow member told me that he'd found the solution for where oil came from. It was long and involved. One of its features required that the earth's axis be changed by 90 degrees due to the collision of an asteroid a few million years ago. The idea was that the earth's axis used to be in the plane of the earth's orbit, but then it was knocked into its current rotation axis tilt of around 23 degrees. This changed the climate and buried a lot of stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axis shift seemed to be an obvious weak link with the theory in that it involves the simplest physics. I told him that any asteroid large enough to change the earth's axis an appreciable amount (i.e. 70 degrees!) would cause so much damage as to destroy all life on the surface of the planet including its oceans. I'd have forgotten about it, but the other night, while listening to the radio, a guest appeared who had an idea that was similar. So this seems like a common target for pseudoscience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so my question is this: Would an asteroid collision sufficient to immediately change the earth's rotation axis wipe out all life on the planet?&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-02-07T21:48:05.543" LastActivityDate="2011-02-07T21:51:50.783" Title="Could life survive a pole shift caused by an asteroid collision?" Tags="&lt;classical-mechanics&gt;&lt;thermodynamics&gt;&lt;angular-momentum&gt;&lt;rotational-dynamics&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="2" />
  <row Id="4335" PostTypeId="2" ParentId="4314" CreationDate="2011-02-01T00:23:06.373" Score="0" ViewCount="" Body="&lt;p&gt;If you're going to stay in physics, you're going to have to take the physics GRE. So why not study the E&amp;amp;M questions on the sample GRE tests? They're nice simple questions and test your understanding of physics, not your understanding of mathematics. You can get the sample questions here:&lt;br&gt;&#xA;&lt;a href=&quot;http://grephysics.net/ans/&quot; rel=&quot;nofollow&quot;&gt;http://grephysics.net/ans/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-01T00:23:06.373" CommentCount="1" />
  <row Id="4337" PostTypeId="2" ParentId="4319" CreationDate="2011-02-01T00:31:27.927" Score="0" ViewCount="" Body="&lt;p&gt;Your answer &quot;(4)(5) cos(30)&quot; with units of &quot;foot-lbs&quot; is correct for the problem as you've stated it, however there are other forces on the box. You might try including gravity.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-01T00:31:27.927" />
  <row Id="4338" PostTypeId="2" ParentId="4005" CreationDate="2011-02-01T00:40:26.280" Score="1" ViewCount="" Body="&lt;p&gt;When I've seen fractional derivatives I've assumed that one place where they would naturally arise is in physical situations where there's a fractional dependency on time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, random walks typically result in movement proportional to $\sqrt{t}$. Googling for &quot;fractional+derivative+random+walk&quot; on arxiv.org finds some papers that explore this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.google.com/search?hl=en&amp;amp;client=firefox-a&amp;amp;hs=RPk&amp;amp;rls=org.mozilla%3Aen-US%3Aofficial&amp;amp;q=fractional+derivative+random+walk+site%3Aarxiv.org&amp;amp;aq=f&amp;amp;aqi=&amp;amp;aql=&amp;amp;oq=&quot; rel=&quot;nofollow&quot;&gt;http://www.google.com/search?hl=en&amp;amp;client=firefox-a&amp;amp;hs=RPk&amp;amp;rls=org.mozilla%3Aen-US%3Aofficial&amp;amp;q=fractional+derivative+random+walk+site%3Aarxiv.org&amp;amp;aq=f&amp;amp;aqi=&amp;amp;aql=&amp;amp;oq=&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm wondering if there's a way of relating some of the diffusion versions of QM with fractional derivatives.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-01T00:40:26.280" />
  <row Id="4339" PostTypeId="2" ParentId="4319" CreationDate="2011-02-01T01:20:03.497" Score="1" ViewCount="" Body="&lt;p&gt;I asked my professor. (4)(5)cos(30) is correct when the calculation uses degrees, but it was incorrect because it uses radians. The answer is technically (4)(5)cos(pi/6). Thank you all for the help.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-01T01:20:03.497" />
  <row Id="4340" PostTypeId="1" CreationDate="2011-02-01T02:07:33.217" Score="3" ViewCount="290" Body="&lt;p&gt;I have a question. The only finite mathematical framework that incorporates both the standard model of particle physics and gravity under one umbrella that I am aware of is string theory. I would like to know whether there are any other mathematical possibilities exist which do not depend on supersymmetry and still consistent with the standard model and gravity and produce finite answers. In a nutshell my question is: can there be any alternative to string theory? (Remember, I am not talking about only gravity. I am talking about gravity as well as other phenomena).&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-01T02:44:03.450" LastActivityDate="2011-07-31T04:35:00.453" Title="Question on other possible theories which are generalizations of the standard model with incorporation of gravity" Tags="&lt;gravity&gt;&lt;standard-model&gt;&lt;theory-of-everything&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="4341" PostTypeId="2" ParentId="4334" CreationDate="2011-02-01T02:28:10.790" Score="5" ViewCount="" Body="&lt;p&gt;Quite clearly this man’s theory is balderdash.  A massive collision could change the tilt of the Earth’s axis, but that would be one hell of a torque.  The collision would not be with an asteroid either, but with another planet that might be as big as Mars.  Clearly nothing like this happened in recent geological history, though the Earth did suffer a collision of this magnitude a few 100 million years after gravitational accretion.  A chunk from the collision spalled off to form the moon.  This other planet probably was in some resonance condition around a Lagrange point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be a devastating collision, which would doubtless change this planet into something else.  The solar system has not experienced such collsions since its early evolution.  The solar system is now staid and respectable.  Velikovskian things just do not happen.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-01T02:28:10.790" CommentCount="1" />
  <row Id="4342" PostTypeId="2" ParentId="4340" CreationDate="2011-02-01T02:36:25.417" Score="-1" ViewCount="" Body="&lt;p&gt;The Loop Quantum Gravity folks have not been able to get elementary particle theory into their picture very well.  They have been working with braids which have twists, which start to sound a bit like string to me.  There seems to be a trend where all roads lead to string theory.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-01T02:36:25.417" CommentCount="2" />
  <row Id="4343" PostTypeId="1" AcceptedAnswerId="4553" CreationDate="2011-02-01T02:50:16.033" Score="1" ViewCount="387" Body="&lt;p&gt;I have a series of measurements of temperature and relative humidity, together with mosquito capture data. Because mosquitoes are sensitive to desiccation, it's reasonable that RH may be useful in predicting their activity. But I recently heard of a value called saturation deficit or vapor pressure deficit which has been well correlated with evapotranspiration in plant leaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it accurate to say saturation deficit is the &quot;perceived dryness&quot; or &quot;drying power&quot; of the air?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can saturation deficit be calculated directly from temperature and RH? Is it dependent on elevation, air pressure or anything else (all measurements are &amp;lt;600m above sea level).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we calculate it? I have found two methods online which give wildly different results. &lt;/p&gt;&#xA;" OwnerUserId="1680" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-01T04:19:22.627" LastActivityDate="2011-02-03T18:27:16.497" Title="how can I calculate vapor pressure deficit from temperature and RH?" Tags="&lt;temperature&gt;&lt;atmospheric-science&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="4344" PostTypeId="2" ParentId="4334" CreationDate="2011-02-01T02:55:23.030" Score="2" ViewCount="" Body="&lt;p&gt;Even more than the collision, you'd be hard pressed to explain why the Earth's axis would lie in the orbital plane in the first place. The &quot;default&quot; inclination for the major planets in the solar system is close to perpendicular to the orbital plane (well understood since the Laplace theory of planet formation, with some caveats), and inclinations as large as that of Uranus and Venus required &lt;em&gt;very&lt;/em&gt; major collisions.&lt;/p&gt;&#xA;" OwnerUserId="1252" LastActivityDate="2011-02-01T02:55:23.030" />
  <row Id="4345" PostTypeId="1" AcceptedAnswerId="4348" CreationDate="2011-02-01T03:42:15.420" Score="4" ViewCount="215" Body="&lt;p&gt;So we have two particles (A and B) that are entangled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand, entanglement isn't destroyed, it is only obscured by subsequent interactions with the environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particle A goes zooming off into outer space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10 years later, Particle B becomes incorporated into my brain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The day after that, an alien scientist measures the entangled property on Particle A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will have some immediate non-local effect on Particle B won't it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And since B's state has been altered (in some sense), and it is part of my brain, then my brain state has been altered as well, hasn't it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe only a tiny amount, obscured by the many environmental interactions that the two particles have been subjected to since the initial entanglement, but in a way that is real and at least conceivably significant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if that is true, then to the extent that mental states supervene on brain states, my mental state would also have been altered by non-local effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is that wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1700" LastActivityDate="2011-03-08T16:13:53.270" Title="Entangled particles" Tags="&lt;quantum-mechanics&gt;&lt;entanglement&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="4346" PostTypeId="2" ParentId="3799" CreationDate="2011-02-01T04:02:36.833" Score="0" ViewCount="" Body="&lt;p&gt;Actually to my knowledge scientists have been able reproduce some of the experiments successfully, they just don't fully understand the process as of yet.  Research is being done to come up with a working LENR theory (what it's known as in the physics community) to accurately model what is happening (unexplained excess heat generation).  As some have said yes it's a very complex process and it's still possible that the energy is liberated from some other means (non-nuclear) but I would hardly say the it's completely been proven on way or another.  A good website for learning more is &lt;a href=&quot;http://lenr-canr.org/&quot; rel=&quot;nofollow&quot;&gt;http://lenr-canr.org/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1701" LastActivityDate="2011-02-01T04:02:36.833" CommentCount="3" />
  <row Id="4348" PostTypeId="2" ParentId="4345" CreationDate="2011-02-01T04:18:18.057" Score="9" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;From what I understand, entanglement isn't destroyed, it is only obscured by subsequent interactions with the environment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Depends on how you view it. There is an explanation of quantum measurement (called decoherence) in which this is true. I will not be using that explanation in this post because it's unnecessarily complicated.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This will have some immediate non-local effect on Particle B won't it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nope. When an alien scientist measures particle A, it does not have an &lt;em&gt;effect&lt;/em&gt; on particle B. Specifically, it does not induce any sort of change in particle B that can be detected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, we would say that the quantum state of particle B (technically: the combined quantum state of the pair AB) has changed, but changes of this nature in the quantum state can't be detected. For example, suppose the entangled state, written in a basis of energy levels, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\psi_1\rangle =  \frac{1}{\sqrt{2}}\left(|1\rangle|2\rangle + |2\rangle|1\rangle\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that if particle A has an energy of 1 (in unspecified units), particle B will have an energy of 2, and vice versa. Suppose the alien scientist measures the energy of particle A to be 1. The state of the pair AB becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\psi_2\rangle = |1\rangle|2\rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the next day, when your brain measures the energy of particle B, it will get 2 as the result. But as far as your brain is concerned, the energy of particle B could have just been 2 all along. Or it could have had some arbitrary probability distribution with 2 as one of the possible results. The fact that the state changed from $|\psi_1\rangle$ to $|\psi_2\rangle$ has no effect, and could not have been detected by your brain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And if that is true, then to the extent that mental states supervene on brain states,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If by that you mean that your thoughts are determined by quantum states, that's definitely false.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-02-01T04:18:18.057" CommentCount="5" />
  <row Id="4349" PostTypeId="1" AcceptedAnswerId="4421" CreationDate="2011-02-01T04:35:43.390" Score="7" ViewCount="325" Body="&lt;p&gt;W and Z bosons are observed/discovered. But as force carrying bosons they should be virtual particles, unobservable? And also they require to have mass, but if they are virtual they may be off-shell, so are they virtual or not. &lt;/p&gt;&#xA;" OwnerUserId="1702" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-02-01T22:59:00.907" LastActivityDate="2011-02-02T05:28:53.393" Title="Are W &amp; Z bosons virtual or not?" Tags="&lt;standard-model&gt;&lt;carrier-particles&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="4350" PostTypeId="2" ParentId="4343" CreationDate="2011-02-01T05:00:51.610" Score="1" ViewCount="" Body="&lt;p&gt;By the name I think it is simply the difference between vapor pressure at saturation and the actual vapor pressure. The later should simply be the relative humidity times the vapor pressure at saturation. You should be able to find some tables (or formulas for vapor pressure as a function of temperature).&#xA;   You can look up vapor pressure in Wikipedia. They give a formula they call the Antoine equation. Also more theoretically you could look at the Clausius-Clapeyron equation, but I think it would be easier just to plug in the Antoine equation instead.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-02-01T05:00:51.610" CommentCount="3" />
  <row Id="4351" PostTypeId="2" ParentId="4349" CreationDate="2011-02-01T05:09:04.283" Score="6" ViewCount="" Body="&lt;p&gt;Photons are force carrying bosons and come in both virtual and real varieties. There is nothing wrong with that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Virtual means off-shell, and real means on-shell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even on-shell weak bosons decay very quickly, however, because there are plenty of modes with the right quantum numbers and much lower total mass (and thus lots of phase space).&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-02-01T05:09:04.283" CommentCount="12" />
  <row Id="4352" PostTypeId="2" ParentId="4322" CreationDate="2011-02-01T05:28:31.250" Score="0" ViewCount="" Body="&lt;p&gt;I think &lt;a href=&quot;http://arxiv.org/abs/0708.2483&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; is what you are looking for.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-02-01T05:28:31.250" CommentCount="1" />
  <row Id="4353" PostTypeId="1" AcceptedAnswerId="4355" CreationDate="2011-02-01T05:38:47.977" Score="2" ViewCount="193" Body="&lt;p&gt;This just started to bother me after reading yet another entangled particle question, so I hate to ask one myself, but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have two entangled particles and take a measurement of one, we know, with certainty, what the outcome of same measurement of the other particle will be when we measure it. But, as with all fundamental QM, we also say the second particle does not have a definite value for the measurement in advance of making it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the particle pairs are always described by an appropriate wave function, and measurement of half the pair changes the probabilities associated with each term. But, and this is the part that bothers me, with knowledge of the first measurement, we know the outcome of the second measurement. There's nothing strange or forbidden or anything about this, no non-local affects, no FTL communication, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But does it still make sense to talk about the second particle &lt;em&gt;not&lt;/em&gt; being in a definite state prior to the measurement? Or can we consider a local measurement of one half of an entangled pair to be a measurement of the other half? The latter option feels distasteful because it implies communication between the particles when communication is possible, which is obviously wrong, but I find it hard to pretend the second particle in the pair, whose state I know by correlation, is somehow not in that state before I confirm it by measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should clarify that my uneasiness comes from a comparison with the usual &quot;large separation&quot; questions of entangled particles, where our ignorance of the first measurement leads us to write an &quot;incorrect&quot; wavefunction that includes a superposition of states. Although this doesn't affect the measurement outcome, it seems there is a definite state the second particle is in, regardless of our ignorance. But in this more commonly used scenario, it seems more fair to say the second particle really isn't in a definite state prior to measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, I realize there's no contradiction in two different observers having different descriptions of the same thing from different reference frames, so I could accept that a local measurement of half the pair does count as a measurement of the other half, while a non-local measurement doesn't, but the mathematics of the situation clearly point to the existence of a definite state for the second particle, whereas we always imply it doesn't have a well-defined state (at least in the non-local case) before measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, my own thinking is that the second particle is in a definite state, even if that state could, in the non-local separation case, be unknown to the observer. But that answer feels too much like a hidden variables theory, which we know is wrong. Any thoughts?&lt;/p&gt;&#xA;" OwnerUserId="1196" LastEditorUserId="1196" LastEditorDisplayName="" LastEditDate="2011-02-01T06:04:45.727" LastActivityDate="2011-02-01T18:42:30.097" Title="Local measurement of entangled particle pairs and interpretation of state" Tags="&lt;quantum-mechanics&gt;&lt;entanglement&gt;" AnswerCount="2" />
  <row Id="4354" PostTypeId="2" ParentId="4331" CreationDate="2011-02-01T06:24:05.670" Score="8" ViewCount="" Body="&lt;p&gt;Correlation functions of local operators are the other observables that field theorists talk about most of the time: things like $\left&amp;lt;{\cal O}_1(x_1) {\cal O}_2(x_2) \ldots\right&amp;gt;$. But there's no shortage of observables in quantum field theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation where you'll hear people say that the S-matrix is the only observable is quantum gravity in asymptotically flat space; this is because local operators don't give gauge-invariant physical observables in theories with gravity.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2011-02-01T06:24:05.670" CommentCount="3" />
  <row Id="4355" PostTypeId="2" ParentId="4353" CreationDate="2011-02-01T06:51:15.793" Score="1" ViewCount="" Body="&lt;p&gt;According to quantum mechanics, the particles - whether it is the first particle or the second particle or any other particle in the Universe - refuse to have any well-defined state or property prior to the measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only &quot;kind of&quot; exception is the case - relevant for a maximally entangled scenario - in which we are interested in a property of the second particle - or any other particle - that is predicted to take a particular value with probability equal to 100 percent. Of course, if the probability is 100 percent, then you can be sure what the measured property will be, and you may assume that this value of the property exists even before the measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for the very same particle that has some value of the quantity equal to something at 100 percent, there still inevitably exist other observables that are not known. (Just design a Hermitian observable with random off-diagonal matrix elements.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the basis of eigenstates of those other properties, the probability amplitudes are generic, and some of the options have probabilities that differ from 0 percent as well as 100 percent. For those quantities - and it is a majority of observables - the usual prescriptions of quantum mechanics hold: the value is not determined prior to the measurement. It is not just unknown to the physicists: it is unknown to Nature. A picture in which those observables are determined leads to wrong predictions and contradictions.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-01T18:42:30.097" LastActivityDate="2011-02-01T18:42:30.097" CommentCount="4" />
  <row Id="4356" PostTypeId="1" AcceptedAnswerId="4370" CreationDate="2011-02-01T07:06:41.093" Score="0" ViewCount="217" Body="&lt;p&gt;My question is the title of a 1991 paper by Richard Gott and Li-Xin Li:&#xA;&lt;a href=&quot;http://arxiv.org/abs/astro-ph/9712344&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/astro-ph/9712344&lt;/a&gt;&#xA;and is also a subject of his popular book, &quot;Time Travel in Einstein's Universe&quot;&#xA;Ultimately with cosmology, the chicken and egg question reduces to &quot;How can something come from nothing?&quot; Gott and Li use the concept of a &quot;jinn&quot; (or djinn-genie)---that is something that loops back in time in a closed timelike curve--In the movie, &quot;Somewhere in Time&quot;, an elderly Jane Seymour places a pocket watch in young Christopher Reeve's jacket and tells him to come back to her Later, he travels back in time and gives her the watch. The watch is a djinn. Where did it come from? Gott argues that in say the first Planck time, the universe could have pulled a closed timeline curve and created itself. Other than the immense fun value of this thought, is using the concept of CLC in this way valid if it is done in less than the Planck time without violating causality?&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-02-01T15:51:58.650" LastActivityDate="2011-02-01T18:05:37.170" Title="Can the Universe create itself?--Is Gott's Use of CTC in Planck time or less valid?" Tags="&lt;general-relativity&gt;&lt;cosmology&gt;&lt;big-bang&gt;&lt;time-travel&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" />
  <row Id="4357" PostTypeId="2" ParentId="4331" CreationDate="2011-02-01T07:47:34.087" Score="1" ViewCount="" Body="&lt;p&gt;Do I understand your question correct? You ask: if the S-matrix is the only quantity which really can distinguish interesting models from free models?&#xA;Then I would say no. But there are no interesting models in 4D, fullfilling let's say the Wightman axioms, are there? &#xA;In lower dimension for example conformal field theories are almost like free ones, but they have a very interesting representation theory, with anyionic sectors braiding etc, which make them also mathematically interesting. But there scattering is trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your example taking quasi free fields (with a certail Kallen-Lehmann density) doesn't give something new, neither does taking Wick polynomials of these fields, because it gives you essential the same model (same Borchers class).&lt;/p&gt;&#xA;" OwnerUserId="1611" LastActivityDate="2011-02-01T07:47:34.087" CommentCount="6" />
  <row Id="4358" PostTypeId="2" ParentId="3799" CreationDate="2011-02-01T07:53:11.957" Score="2" ViewCount="" Body="&lt;p&gt;This new &quot;cold fusion&quot; &lt;a href=&quot;http://www.journal-of-nuclear-physics.com/&quot; rel=&quot;nofollow&quot;&gt;reported&lt;/a&gt; in what is really a blog is a commercial enterprise to all intents and purposes. Their claims are so large, that either their  constructs will be successful or they will eat their hat. We do not have long to wait.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they are successful, the theory will be found. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One note about crystals ( they are using Ni crystals) and large energies: Along the crystal axis high energy muon beams go through intact, without interacting with the coulomb barrier. Hard to find references, I know Tom Ypsilantis was working on this in the '80s. Here is a &lt;a href=&quot;http://www.sciencedirect.com/science?_ob=ArticleURL&amp;amp;_udi=B6TJN-3VXHMCP-1N&amp;amp;_user=10&amp;amp;_coverDate=10%2F02%2F1996&amp;amp;_rdoc=1&amp;amp;_fmt=high&amp;amp;_orig=search&amp;amp;_origin=search&amp;amp;_sort=d&amp;amp;_docanchor=&amp;amp;view=c&amp;amp;_searchStrId=1626404885&amp;amp;_rerunOrigin=google&amp;amp;_acct=C000050221&amp;amp;_version=1&amp;amp;_urlVersion=0&amp;amp;_userid=10&amp;amp;md5=7e3cdf35964c197191f73167a6b64131&amp;amp;searchtype=a&quot; rel=&quot;nofollow&quot;&gt;proposal for a muon collider&lt;/a&gt; that uses the concept.&#xA;So if the announcement is not snake water, an appropriate solid state/nuclear theory will emerge.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-02-01T07:53:11.957" CommentCount="1" />
  <row Id="4359" PostTypeId="1" CreationDate="2011-02-01T09:09:33.113" Score="6" ViewCount="276" Body="&lt;p&gt;In a gauge theory like QED a gauge transformation transforms one mathematical representation of a physical system to another mathematical representation of the same system, where the two mathematical representations don't differ at all with respect to observables. Gauge transformations are therefore manifestations of a true redundandcy of the mathematical descriptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In GR the role of diffeomorphisms is different, a diffeomorphism represents a change of the reference frame. Different observers in different reference frames will of course have different results when they measure the same observable/event. In this sense I agree with Raymond Streater (see &lt;a href=&quot;http://www.mth.kcl.ac.uk/~streater/lostcauses.html#XXII&quot; rel=&quot;nofollow&quot;&gt;Diff(M)&lt;/a&gt;) that it is misleading to say that Diff(M) is a gauge group (if you disagree with me please explain).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In AQFT one associates observables (selfadjoint operators) with bounded open subsets of a spacetime, these observables represent what is observable in the given domain of space and time. A detector that is operated for two hours in a laboratory would be, for example, represented by such an observable (this is only approximately true, because the Reeh-Schlieder theorem says that it is not possible to use a truly localized observable, but one has to use an &quot;approximately local observable&quot; instead).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that this line of reasoning will stay true even if one day there is a theory of quantum gravitation. But from time to time I read statements like &quot;local observables are not gauge invariant in a theory with (quantum) gravity and therefore cannot exist/ are not valid observables&quot;. (If my phrasing of the statement is wrong, please explain and correct it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have never read about an explanation of this statement and would like to hear about one. Isn't a detector, for example, a (approximatly) local observable and won't detectors exist within a theoretical framework of (quantum)  gravity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: A little of explanation of &quot;observables&quot; in GR: I am aware that in GR only &quot;events&quot; make sense as an observable, but of course not, for example, the spacetime coordinates of a point of spacetime, see the discussion of Einstein's hole argument on the nLab:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ncatlab.org/nlab/show/spacetime&quot; rel=&quot;nofollow&quot;&gt;spacetime&lt;/a&gt;, see paragraph &quot;Einstein's hole argument&quot;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When a detector detects a particle, I assume that this is an event that is observable because it is defined by the proximity of a localized field excitation and the detector, and the fact that the detecter makes &quot;ping&quot; is a fact that all observers in all reference frames agree upon.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditorDisplayName="" LastEditDate="2011-02-01T09:43:58.983" LastActivityDate="2011-02-01T18:38:06.810" Title="Diff(M) as a gauge group and local observables in theories with gravity" Tags="&lt;quantum-field-theory&gt;&lt;quantum-gravity&gt;&lt;gauge-theory&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" />
  <row Id="4360" PostTypeId="2" ParentId="4359" CreationDate="2011-02-01T09:26:59.843" Score="3" ViewCount="" Body="&lt;p&gt;At the level of representation theory, the diffeomorphism group in general relativity and its extensions plays the very same role as Yang-Mills symmetries in gauge theories. In both cases, one may define the action of the symmetry on the operators. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases, only the gauge-invariant states - singlets - are allowed in the physical spectrum. That's why the representation theory of the gauge group plays no role at the level of the Hilbert space - only the singlets are relevant. That's why the &quot;gauge symmetries&quot; always just reduce the number of independent degrees of freedom and many people prefer to call them &quot;gauge redundancies&quot; rather than &quot;gauge symmetries&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases, the latter condition of gauge invariance (&quot;physical states are singlets&quot;) must totally hold for transformations that converge to the identity at infinity. In both cases, there are subtleties for transformations that change the asymptotic region (fields at spatial infinity). In both cases, the gauge invariance of the physical states arises as the quantum version of the Gauss's law - the subset of Maxwell's or Einstein's equations that don't contain any time derivatives and may therefore be considered constraints on the initial state rather than evolution equations: it's the $\mbox{div} D=\rho$ equation in the case of electromagnetism. The corresponding operator $(\mbox{div} D-\rho)$ has to annihilate the physical states $|\psi \rangle$ in the quantum theory (which is non-trivial if the quantum theory is formulated in terms of the redundant fields, the gauge potentials). A total analogy exists for Einstein's equations (extrinsic curvature on the slice enters the constraint).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The difference between the two gauge groups is just in the &quot;details&quot; - how they act on the spacetime.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Yang-Mills transformations are local, so $\phi(x,y,z,t)$ only changes by other fields at the same point $(x,y,z,t)$: imagine a phase transformation of a charged field $\phi$, for example. In the case of diffeomorphism, the change is nonlocal: the field at one point depends on the fields at another point $(x&amp;#39;,y&amp;#39;,z&amp;#39;,t&amp;#39;)$ before the transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This technical difference changes the character of gauge-invariant observables - and only gauge-invariant observables may correspond to numbers that make a physical sense and may be measured. Because the gauge transformations in the gauge theory are local, one may construct gauge-invariant local operators such as $\mbox{Tr}(F_{\mu\nu} (x,y,z,t)F^{\mu\nu}(x,y,z,t))$, to choose a random example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of general relativity, such quantities are not gauge-invariant because e.g. the Ricci scalar $R(x,y,z,t)$ is transformed to the Ricci scalar at another point $R(x&amp;#39;,y&amp;#39;,z&amp;#39;,t&amp;#39;)$, so even the Ricci scalar at a given point is not gauge-invariant. To construct gauge-invariant observables in general relativity, one has to be sensitive to - e.g. integrate over - the whole space (or spacetime). For example, the ADM energy is gauge-invariant in asymptotically flat backgrounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A physical apparatus that exists in a gravitational theory is not represented by any local observable - in the technical sense of &quot;local&quot; - because its location is not a gauge-invariant quantity. If you have a small device in the vicinity of $(x,y,z,t)=(0,0,0,0)$, its measured results may be expressed as a functional of the fields in the vicinity of $(0,0,0,0)$. However, that's only true in one coordinate system. In other words, it is only true before you make a general gauge transformation. After the gauge transformation, the form of the observable corresponding to the quantity measured by the apparatus is expressed by a different formula involving the physical fields - the new formula depends on fields at different values of $(x,y,z,t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may &quot;know&quot; that $(0,0,0,0)$ is &quot;physically&quot; the same point as $(7,2,-3,5)$ in some new coordinates but the mathematics doesn't know it: the form of the expression is changed so the quantity is not gauge-invariant. In the same way, you could claim that you know that a &quot;red quark field&quot; before the $SU(3)$ transformation is &quot;physically&quot; the same thing as a &quot;green quark field&quot; after the transformation - because you also know the transformation. But exactly because the &lt;em&gt;form&lt;/em&gt; of the field that corresponds to the &quot;same physical thing&quot; depends on the transformation, we say that colorful fields in QCD - and local operators in GR - are &lt;em&gt;not&lt;/em&gt; gauge-invariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may pinpoint the location of the gadget by defining its proper distances from $d-1=3$ points A,B,C at &quot;infinity&quot; or &quot;far enough&quot; where you already require the legitimate gauge transformations (coordinate redefinitions) to be trivial. But the calculation of the point - and the fields at this point - will depend on the metric tensor between the apparatus and the points A,B,C. So the definition of the observables that represent the values measured by the apparatus is manifestly and inevitably non-local. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, there are no gauge-invariant observables in a theory with a coordinate reparametrization symmetry. I claim that the text above makes it totally clear but if it is not clear, please write down something that you think is a gauge-invariant local quantity - as a function of the basic degrees of freedom - in a theory with the general covariance and I will show you why it is not a gauge-invariant local quantity. There aren't any.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-01T18:38:06.810" LastActivityDate="2011-02-01T18:38:06.810" CommentCount="7" />
  <row Id="4361" PostTypeId="2" ParentId="3999" CreationDate="2011-02-01T10:07:32.987" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://pdfcast.org/pdf/dfgdsg&quot; rel=&quot;nofollow&quot;&gt;&quot;An exact solution to the draining reservoir problem of the incompressible and non-viscous liquid. Seok-In Hong&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to satisfy h'(0)=0 and energy conservation.&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-02-01T10:07:32.987" CommentCount="2" />
  <row Id="4362" PostTypeId="2" ParentId="4129" CreationDate="2011-02-01T10:11:23.000" Score="2" ViewCount="" Body="&lt;p&gt;I am optimistic that string theory in its final form will explain quantum mechanics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest way for this to occur would be for the final theory to be intrinsically quantum; not a &quot;quantization&quot; of a classical theory, though it must still have a classical limit, an approximation which behaves classically. This would mean that quantum mechanics and the final theory were a package deal, mathematically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I mean a genuine explanation. The truly problematic feature of quantum mechanics is not nondeterminism, it's antirealism. If quantum mechanics were a classically random theory, we would have a picture of what goes on between measurements, e.g. a &quot;Brownian-motion-like&quot; evolution of physically localized properties. But quantum probabilities arise from complex numbers. This allows destructive interference to occur: the contributions of histories to the path integral can cancel. This cannot happen in classical probability theory, where probabilities are always nonnegative and thus always add in a nondecreasing way, and it is the main obstruction to a realist account of quantum mechanics. A classically probabilistic mechanics that reproduces quantum mechanics - like Edward Nelson's stochastic mechanics - violates the spirit of special relativity, because it involves a random walk in all directions of configuration space. A nonlocally deterministic mechanics like Bohmian mechanics is also conceptually nonrelativistic. So instead we hear that things don't have properties until they are measured. That's &quot;antirealism&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of a third realist approach to quantum mechanics, and how it might be realized by string theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a similarity between quantum field theory in imaginary time and classical statistical mechanics at finite temperature that has been noted many times. This might suggest an explanation of quantum probability in terms of classical probabilistic dependencies which are local in space and time, but which give rise to spacelike correlations via a zigzag in time - forward along one lightcone, backward along another. There are a number of proposed formulations of quantum mechanics which have this aspect of bidirectional causality, but none of them provides a rigorous derivation of the quantum framework from ordinary probability theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way that &lt;em&gt;might&lt;/em&gt; happen is if you were calculating probabilities on a space-time manifold that was not time-orientable. If you had an initial condition, a final condition, and a classical probability distribution over the number of closed timelike curves in between, maybe you'd end up with something like the transition probabilities of quantum mechanics, even with a locally deterministic law of field propagation on the manifold. But planck-scale time loops involve very high curvatures, and so in a string theory framework we'd be talking about the heavy degrees of freedom of the string. Anyway, the idea is that the fundamental theory is &quot;classical&quot; in the sense of offering an objective description of nature, and that the peculiarities of quantum mechanics arise simply because you're doing probability theory on a manifold whose topology is not globally determined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, if I look at what's actually happening in string theory, there are formal developments which &lt;em&gt;could&lt;/em&gt; be the beginning of movement beyond quantum mechanics. In promoting the recent work on twistors in gauge theory, Arkani-Hamed says it offers not just emergent space-time but also emergent quantum mechanics, because unitarity is not assumed. And Witten had a paper last year deriving nonrelativistic quantum mechanics from topological string theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, Witten was still using path integrals, and it could be that the work of Arkani-Hamed et al is also just quantum mechanics from another angle, rather than a glimpse of something deeper. And what I wrote about quantum mechanics arising from classical probabilities that are bidirectional in time could also be wrong. But it helps to have a few such possibilities in view, when opposing antirealism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if some form of Bohmian mechanics were the only viable realist interpretation of quantum mechanics, I think we would be rationally obliged to favor it over antirealism. What I actually expect, however, is that in the process of formulating the final physical theory, the incremental development of new quantum formalisms (see the end of point 6 in the list by Lubos), carried out for technical rather than philosophical reasons, will ultimately produce a version of the quantum framework where you don't need to talk about measurements and observables at the fundamental level. There will simply be relationships among things that exist according to the theory. And at that point antirealism will no longer be necessary. &lt;/p&gt;&#xA;" OwnerUserId="1486" LastActivityDate="2011-02-01T10:11:23.000" CommentCount="2" />
  <row Id="4363" PostTypeId="2" ParentId="3767" CreationDate="2011-02-01T11:01:12.390" Score="4" ViewCount="" Body="&lt;p&gt;I have come across a very simple question or critique of the Transactional Interpretation in an article &lt;a href=&quot;http://www-physique.u-strasbg.fr/cours/l3/divers/meca_q_hervieux/Articles/Nine_form.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Nine formulations of Quantum Mechanics&quot;&lt;/a&gt;. The multiple authors of this 2002 AMJ paper are QM physics teachers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They ask how &quot;two particle&quot; transactional handshakes work: are there &quot;two handshakes across spacetime&quot; or &quot;one handshake across Configuration Space&quot;? Without an answer to this question they are unable to &quot;report on how the TI differentiates between bosons and fermions&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-01T11:01:12.390" />
  <row Id="4364" PostTypeId="1" AcceptedAnswerId="4368" CreationDate="2011-02-01T11:21:36.293" Score="1" ViewCount="96" Body="&lt;p&gt;The positive mass conjecture was proved by &lt;a href=&quot;http://www.doctoryau.com/papers/PositiveMassConjecture.pdf&quot; rel=&quot;nofollow&quot;&gt;Schoen and Yau&lt;/a&gt; and later reproved by &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.cmp/1103919981&quot; rel=&quot;nofollow&quot;&gt;Witten&lt;/a&gt;.  Total mass in a gravitating system must be positive except in the case of flat minkowski space, where energy is zero.  Since QG is intended to be a theory of interaction with force particles called gravitons, one may begin to wonder if the interactions are in fact the important defining features of the space in question.  So does a theory with interactions also require that space be curved?&lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="1383" LastEditorDisplayName="" LastEditDate="2011-02-02T11:00:15.057" LastActivityDate="2011-02-02T11:00:15.057" Title="Does the positive mass conjecture indicate a necessity of interactions in our universe?" Tags="&lt;theory&gt;" AnswerCount="1" />
  <row Id="4365" PostTypeId="2" ParentId="4331" CreationDate="2011-02-01T11:36:13.310" Score="1" ViewCount="" Body="&lt;p&gt;The exact S-matrix elements with finite number of the initial and final photons in QED are equal to zero so such processes are unobservable (never happen). For observation you need an infinite number of photons (=inclusive picture). This makes me think that the theory should be reformulated so that this fact (dominating inelastic processes) would be taken into account automatically, like in &lt;a href=&quot;http://arxiv.org/abs/0806.2635&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;atom-atomic&lt;/strong&gt;&lt;/a&gt; potential scattering (consider the case $n &amp;gt;&amp;gt; 1$, when the threshold of excitations is negligible).&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-01T12:26:55.177" LastActivityDate="2011-02-01T12:26:55.177" CommentCount="2" />
  <row Id="4366" PostTypeId="1" AcceptedAnswerId="4382" CreationDate="2011-02-01T12:01:23.593" Score="1" ViewCount="193" Body="&lt;p&gt;In discussions around a recent &lt;a href=&quot;http://physics.stackexchange.com/questions/4172/what-does-john-conway-and-simon-kochens-free-will-theorem-mean/4227#4227&quot;&gt; Stack answer&lt;/a&gt; which covered axiomatic quantum mechanics, Bell's Theorem and Random Fields (related to axiomatic QFT) the phrase &quot;Heisenberg Cut&quot; was used. The basic question is what does this mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having reviewed &lt;a href=&quot;http://www.phys.tue.nl/ktn/Wim/qm1.htm#Heis_cut&quot; rel=&quot;nofollow&quot;&gt;this discussion&lt;/a&gt; I can provide some background to set a context for the question. In (Von Neumann) QM a distinction is made between :  OBSERVER [Cut] QUANTUM OBJECT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This simplistic distinction has been critized later in that article and elsewhere such that the term &quot;Heisenberg Cut&quot; has fallen into disuse. If one wishes to make distinctions in a QM experiment then here is a larger proposal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ENVIRONMENT [Cut1] QUANTUM OBJECT [Cut2] CLASSICAL INSTRUMENT [Cut3] OBSERVER [Cut4] O-O&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here O-O is any hypothetical observer-observer as discussed in the &quot;Wigner's Friend&quot; concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So has the concept of &quot;Heisenberg Cut&quot; been re-introduced in modern axiomatic QM/QFT and if so what is its formalisation (ie how can one make experimental or Bell Theorem-like deductions from it?)&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-01T18:35:57.703" Title="Formalization (and meaning) of Heisenberg Cut" Tags="&lt;quantum-mechanics&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="4367" PostTypeId="2" ParentId="4366" CreationDate="2011-02-01T12:33:08.437" Score="0" ViewCount="" Body="&lt;p&gt;The term &quot;Heisenberg cut&quot; has only been used in the philosophical discussions about the character of quantum mechanics. It's the classical-quantum boundary that should be placed somewhere in between the observed quantum objects and observer's perceptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's never a mistake to place the cut closer to the observer - to treat a larger set of phenomena using the machinery of quantum mechanics because at the end, quantum mechanics applies everywhere. On the other hand, it could be a mistake to treat some systems or their properties classically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;minimal&quot; location of the Heisenberg cut - one that treats a maximum fraction of the world classically - may be calculated by decoherence. In modern quantum mechanics, decoherence is what defines the classical-quantum boundary - but the term &quot;Heisenberg cut&quot; is rarely used for this boundary in modern physics.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-01T18:35:57.703" LastActivityDate="2011-02-01T18:35:57.703" CommentCount="2" />
  <row Id="4368" PostTypeId="2" ParentId="4364" CreationDate="2011-02-01T12:58:28.373" Score="1" ViewCount="" Body="&lt;p&gt;Dear Humble, because non-gravitational yet interacting field theories such as QCD or the Standard Model exist and they don't predict a curved space, the answer to your question is clearly No, interactions don't imply that the spacetime has to be curved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the curved spacetime follows from many other assumptions - or combinations of assumptions - for example from the requirement that the gravitational force (respecting the equivalence principle) simultaneously exists with the relativistic Lorentz invariance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theorems you mentioned clearly &lt;em&gt;assumed&lt;/em&gt; that the spacetime is allowed to become curved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, I assumed that you agree that the existence of interactions is a property of a theory, not a property of a configuration. But what about the possibility that you meant the &quot;existence of interactions&quot; to be a property of a state, or a configuration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the positive-energy theorem implies that the energy is strictly positive with the single exception of an empty Minkowski space, it follows that if you also agree that the Minkowski space &quot;has no interactions in it&quot;, then every state that has interactions &quot;in it&quot; has nonzero energy and consequently has to lead to a curved spacetime.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-01T12:58:28.373" CommentCount="1" />
  <row Id="4369" PostTypeId="2" ParentId="4366" CreationDate="2011-02-01T14:05:29.177" Score="0" ViewCount="" Body="&lt;p&gt;in addition to what Lubos said about QM, the philosophy of axiomatic QFT is to construct/describe QFTs without any reference to concepts of classical physics. As far as I know there is not even a concept for a &quot;classical limit&quot; in axiomatic QFT. Especially &quot;macroscopic classical&quot; devices like detectors are modeled in AQFT via an observable, that is a selfadjoint operator, and that is a pure quantum concept. Therefore I don't think that you'll find a formalization of something like the &quot;Heisenberg cut&quot; in axiomatic QFT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the interpretation of the measurement process etc., this is usually left to the philosophical interpretation of QM, from a pure philosophical point of view there is no conceptual difference of the interpretation of QM and that of QFT, which is the reason why most people working on this topic concentrate on the technically much simpler QM.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-02-01T14:05:29.177" CommentCount="6" />
  <row Id="4370" PostTypeId="2" ParentId="4356" CreationDate="2011-02-01T14:55:59.337" Score="0" ViewCount="" Body="&lt;p&gt;I hope I understand this question properly.  The text appears to focus on time loops, or closed timelike curves (CTCs), while the title of the question concerns the creation of a universe from nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The holographic principle of black holes tells us the field theoretic information of strings on the event horizon is completely equivalent to field theoretic information in the spacetime one dimension larger outside.  This physics is observed on a frame stationary with respect to the black hole.  The question naturally arises; what physics is accessed by the observer falling through the event horizon on an inertial frame?  This question is important for the black hole small enough to exhibit fluctuations comparable to its scale.  A sufficiently small quantum black hole will be composed of strings in a superposition of interior and exterior configurations or states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motion of a string onto a black hole approximates the dynamics of a string in a Rindler wedge.  The Rindler wedge is defined by the frame of an accelerated observer, which is equivalent to the frame of a stationary observer near a black hole event horizon.  The observer witnesses the final emission of radiation by the string just above the event horizon, where upon the string becomes frozen eternally on the particle horizon.  Of course in the Rindler wedge case the string proceeds onwards on its geodesic or string world sheet with no apparent change due to this observed state of affairs.  This is approximated as well with the black hole, where the string passes through the event horizon unaffected so long as the radius of curvature is much smaller than the string length.  This picture persists until the string approaches the center or singularity of the black hole.  At this point the Rindler wedge model departs from reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interior perspective or the physics of a string as measured by an observer falling with the string, is outside the domain of the holographic principle.  Once the string passes through the event horizon it evolves on a domain of causal support not included in the data set accessible to an observer on an accelerated frame stationary with respect to the black hole.  The observer that falls through the event horizon observes the further evolution of the string beyond the frozen state the stationary observer finds as its final state.  Further, the string evolves into a different state as it approaches the singularity.  There the string will begin to experience a rapidly growing Weyl curvature.  The stationary observer measures transverse modes of the string on the black hole horizon, while longitudinal coordinates are compressed to near the Planck length.  The observer falling in with the string will witness the string distended by the growing Weyl curvature in the direction of motion.  Consequently, this observer witnesses the extension of longitudinal extension of the string.  The frozen state of the string measured by the exterior observer is cancelled by Hawking radiation which escapes later.  The string is entangled with the black hole, and there is a superposition of configuration spaces for the string; the exterior and interior configuration variables.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may put a new twist on the holographic result that events in spacetime do not have the a realism as understood classically.  Quantum mechanics removes a measure of reality with the existence of incompatible measurements of observables which exist in incommensurate commuting sets of operators.  The invariant interval, or event, is left as something which has an ontology or “realism.”  However, this indicates that a realism to any event is not supportable on fundamental grounds.  Further, superposition of exterior and interior states of a black hole prevents a sharp distinction between pre-selected and post-selected states.  This means that cosmic censorship, chronology protection are aspects of “classical reality,” where underneath the ontology of ordered events or their invariant meaning is lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The creation of the universe from nothing might involves something called the biverse.  The de Sitter spacetime is a solution of the scale factor $a(t)~=~\sqrt{\frac{3}{\Lambda}}cosh(t\sqrt{\frac{\Lambda}{3}})$, which has this strange backwards part of the hyperboloid.  Hawking and others have considered the idea of a universe connected at this minimal scale factor as two universes which are time reversed.  It is possible that this is an elementary model for how a “blob” of vacuum energy in a spacetime cosmology quantum tunnels across a potential barrier to form a nascent spacetime cosmology.  The blog of vacuum energy is annihilated at one side of the potential by its “opposite” (other half of the biverse), where the “opposite” is annihilated at the other side of the potential by the occurrence of the vacuum blob.  This is similar to the tunneling of an electron across a barrier, where we can think of it as annihilated by a positron traveling backwards in time, which in turn is connected to an electron on the other side of the potential barrier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of the day one might question what are the role of the Hawking-Penrose energy conditions, such as the averaged weak energy condition $T^{00}~\ge~0$.  It makes sense that the gravitational states are the Hartle-Hawking states which are degenerate and zero.  So ultimately the physical states of the universe globally are zero, or in other words the net total of everything is zero. &lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-02-01T18:05:37.170" LastActivityDate="2011-02-01T18:05:37.170" CommentCount="2" />
  <row Id="4371" PostTypeId="2" ParentId="893" CreationDate="2011-02-01T15:33:24.030" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/The_Feynman_Lectures_on_Physics&quot; rel=&quot;nofollow&quot;&gt;The Feynman Lectures on Physics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;University level, but very readable, with a focus on concepts. When I was in high school, these three volumes got me interested in becoming a physicist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set is pricey, but you can probably borrow a copy from your library. There are abridged versions too:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Six Easy Pieces&lt;/li&gt;&#xA;&lt;li&gt;Six Not-So-Easy Pieces&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One of the more recent editions of the Feynman Lectures includes &quot;Feynman's tips on Physics&quot;, a real gem in which Feynman gives tips on how to approach solving physics problems.&lt;/p&gt;&#xA;" OwnerUserId="1462" LastEditorUserId="1462" LastEditorDisplayName="" LastEditDate="2011-02-01T15:38:59.793" LastActivityDate="2011-02-01T15:38:59.793" CommunityOwnedDate="2011-02-01T15:33:24.030" />
  <row Id="4372" PostTypeId="1" CreationDate="2011-02-01T15:38:07.240" Score="2" ViewCount="198" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;First law: Every body remains in a state of rest or uniform motion (constant velocity) unless it is acted upon by an external unbalanced force. This means that in the absence of a non-zero net force, the center of mass of a body either remains at rest, or moves at a constant speed in a straight line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Newton%27s_laws_of_motion&amp;amp;oldid=410908453&quot; rel=&quot;nofollow&quot;&gt;Wikipedia &amp;mdash; Newton's laws of motion&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Doesn't entropy affect all objects though, since they are all in the closed system of the universe at large, and therefore they are all subject to slowing down, regardless of the containing medium, given enough time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess what I'm curious is, can there ever be a body that will remain at uniform motion or uniform rest given that entropy must exist?&lt;/p&gt;&#xA;" OwnerUserId="1690" LastActivityDate="2011-02-01T22:16:40.450" Title="Does entropy apply to Newton's First Law or does &quot;acted upon&quot; always require an external factor?" Tags="&lt;entropy&gt;&lt;motion&gt;" AnswerCount="4" />
  <row Id="4375" PostTypeId="2" ParentId="4372" CreationDate="2011-02-01T15:54:28.960" Score="1" ViewCount="" Body="&lt;p&gt;Yes, it will (in the classical picture) continue forever, even with its entropy increasing, the entropy increase just means that some of the potential energy within the body will turn into heat (=kinetic energy), however the center of mass motion is unaffected.&#xA;Entropy increase says nothing about slowing down, rather the opposite.&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-02-01T15:54:28.960" CommentCount="20" />
  <row Id="4376" PostTypeId="2" ParentId="2660" CreationDate="2011-02-01T15:56:17.443" Score="0" ViewCount="" Body="&lt;p&gt;In CED one can obtain an exact solution of equations and exclude the electromagnetic field from the &quot;mechanical&quot; equations. In QED it was impossible and the field remained in some approximation in the perturbative calculations. Perturbative modifications of this field were called &quot;vacuum polarization&quot; effects. They thought to be &quot;real physical effects&quot;; that is why R. Feynman was of that opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would draw the following analogy: let us consider an atom-atomic scattering in the first Born approximations. It is described with unperturbed atomic wave functions and the Coulomb potentials acting between charges. The exact solutions differs from such a picture: atoms are polarized while scattering, for example, and the cross section is slightly different. It is like using &quot;modified&quot; Coulomb potentials or even more complicated description if one refers to the first Born approximation picture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the Coulomb forces remain themselves in the exact equation and in the exact solution. It is simply the exact solution differs from the approximate one and nothing else. No true &quot;vacuum polarization&quot; takes place in our atomic problem. Similarly, I think, it is in QED: the exact solution is different from any perturbative one and some corrections are called &quot;vacuum polarization&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-01T22:51:32.377" LastActivityDate="2011-02-01T22:51:32.377" />
  <row Id="4377" PostTypeId="2" ParentId="4372" CreationDate="2011-02-01T15:56:31.580" Score="0" ViewCount="" Body="&lt;p&gt;If you ignore the microscopic explanation of entropy, entropy is just an internal state of a system, on par with the system's volume, or the number of particles in the system.  If you have a gas with a fixed entropy (let's store it in a vacuum tube so it doesn't escape, and let's give the tube infinite insulation so it doesn't leech out any heat) out in deep space, and you throw it, it will just happily trail off in a straight line forever with a constant entropy, volume, and number of particles.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You get changes in entropy and whatnot only when the system in question interacts either with another system or its environment.  But it really is best thinking of these things, at least on a macroscopic scale, as internal degrees of freedom of the system.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-02-01T15:56:31.580" CommentCount="4" />
  <row Id="4378" PostTypeId="2" ParentId="4353" CreationDate="2011-02-01T16:01:01.567" Score="0" ViewCount="" Body="&lt;p&gt;Before measurement, we have a bipartite entanglement between the two particles. After measurement, we have a multipartite entanglement between the two particles, the measuring apparatus, and the environment, which includes the observers. It's only with respect to an Everett branch of the wavefunction that we have definite values.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-01T16:01:01.567" />
  <row Id="4379" PostTypeId="2" ParentId="4345" CreationDate="2011-02-01T16:07:01.403" Score="0" ViewCount="" Body="&lt;p&gt;Even with quantum entanglement, quantum information still can't be transmitted faster than light. This is the no-signaling theorem.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-01T16:07:01.403" />
  <row Id="4380" PostTypeId="2" ParentId="4372" CreationDate="2011-02-01T16:09:12.380" Score="-1" ViewCount="" Body="&lt;p&gt;Unfortunately, you have no idea about the concept of entropy. First of all, the concept of entropy has no meaning for a single rigid particle. Secondly energy and entropy are not same. You can define energy of a single particle but not entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Newtonian mechanics energy of any closed system of particles remains constant. If no net force is applied then the center of mass will move with a constant velocity. The individual particles will move and collide with each other. Their velocities will change in each collision. But the sum of their kinetic energies will remain constant. The total energy of the system will be energy due to the motion of the center of mass of the system plus the total internal kinetic energy of the colliding particles. The total internal kinetic energy of the particles are the internal energy of the system which we call heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Entropy on the other hand is quite different from energy. It is a measure of unavailability of workable energy of a system. You may also think it as disorder of a system. While energy remains constant for a closed system entropy almost always increases with time. The more entropy a system has the more difficult it is to extract energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a nutshell, Energy of a closed system will always remain constant but entropy for any real process will increase.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-01T16:15:24.540" LastActivityDate="2011-02-01T16:15:24.540" CommentCount="7" />
  <row Id="4381" PostTypeId="2" ParentId="4372" CreationDate="2011-02-01T16:09:27.110" Score="1" ViewCount="" Body="&lt;p&gt;No, it does not &quot;apply&quot;, just by definition. &#xA;Newtons mechanics deals about idealized masses moving forever, &#xA;if there is no external influence. This kind of mechanics does &#xA;not know about thermodynamics!&#xA;But even for a real mass moving somewhere (lets, just for fun, &#xA;assume we talk about a probe, which is about to leave &#xA;our sun system) entropy is not an issue. Lets assume, there is &#xA;a explosive load in the probe, and it is ignited. This of &#xA;course will increase the entropy of the probe, because it is transformed &#xA;into a cloud, nevertheless the center of gravity of that cloud will march on! &#xA;Georg&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-02-01T22:16:40.450" LastActivityDate="2011-02-01T22:16:40.450" CommentCount="2" />
  <row Id="4382" PostTypeId="2" ParentId="4366" CreationDate="2011-02-01T16:16:42.417" Score="1" ViewCount="" Body="&lt;p&gt;I take the heart of the Heisenberg cut to be the way we calculate probabilities and expected values in QM. For elementary QM, for some measurement described by an operator $\hat M$, the expected value in a state described by a density matrix $\hat\rho$ is given by the trace $\left&amp;lt;\!M\right&amp;gt;=\mathsf{Tr}\left(\!\hat M\hat\rho\!\right)$. What we put in the $\hat\rho$ is what is in our model universe. The measurement operator describes our measurement apparatus, which is &lt;em&gt;not&lt;/em&gt; in the model universe, but instead describes how our measurement apparatus gets information out of the model universe. There's an almost-symmetry between the ways $\hat M$ and $\hat\rho$ appear; it's almost as if there's a measurement apparatus universe as well as the model universe. Different measurement apparatuses can affect each other in the measurement apparatus universe without changing the model universe, which is called measurement incompatibility (did I just create an interpretation of QM? Do I know this one? I guess it's too glib, sadly.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: We can extend the mathematics in many different ways, but one deserves mention because it's of great practical value. We can introduce transformations $\hat T_i$ that operate between the preparation apparatus and the measurement apparatus, in which case we have $\left&amp;lt;\!M\right&amp;gt;=\mathsf{Tr}\left(\!\hat M\hat T_n\cdots\hat T_2\hat T_1\hat\rho\!\right)$. We can at any time say that $\hat M\hat T_n\cdots\hat T_5$, say, or any other part of this list (without changing the order), is our measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Anyway&lt;/em&gt;, to some extent we can move stuff from the model universe into the measurement apparatus universe and vice versa, although we may have to get into technical stuff like POVMs to do it. Once we go to quantum field theory, there's a tight relationship between the measurement apparatus universe and the model universe, because we use the same Lego blocks to build measurements and states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The separation into states and measurements is absolutely fundamental in QM. It's how the relationship between Hilbert spaces and experimental results works, which causes trouble when people want to do cosmology, with &lt;em&gt;everything&lt;/em&gt; in the model universe. Although I hadn't previously thought about &lt;em&gt;calculating&lt;/em&gt; where one should put the separation, I can see that if one chooses a particular accuracy that one wants one's model of an experimental apparatus to achieve relative to one's real apparatus, that might put a limit on where one can put the Heisenberg cut. I'm not sure, however, that one can't always improve the sophistication of one's description of a measurement, particularly if one is willing to go to POVMs. I suppose, however, that putting people inside your model universe is always going to be in the realm of toy models. The separation into states and measurement famously comes under the microscope in Bell's article `Against &quot;Measurement&quot; '.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, I see you went to Willem de Muynck, who is perhaps a little idiosyncratic, but I've often found his a good counterpoint of view.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-02-01T16:39:40.927" LastActivityDate="2011-02-01T16:39:40.927" CommentCount="1" />
  <row Id="4384" PostTypeId="1" AcceptedAnswerId="4387" CreationDate="2011-02-01T16:41:30.093" Score="7" ViewCount="515" Body="&lt;p&gt;I would like to know the physical meaning of the Legendre transformation, if there is any? I've used it in thermodynamics and mechanics and it seemed only a change of coordinates?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1579" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-02T10:38:59.237" LastActivityDate="2011-05-02T02:15:19.023" Title="Physical meaning of Legendre transformation" Tags="&lt;classical-mechanics&gt;&lt;thermodynamics&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="4385" PostTypeId="2" ParentId="2167" CreationDate="2011-02-01T16:45:45.647" Score="0" ViewCount="" Body="&lt;p&gt;Not without some sort of external power source. Extremely low power neodymium magnet pulse motors are a reality though, and can be powered from the ambient energy in the environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: has to be said though, that extremely low power magnet motors are more like electromechanical oscillators than motors.&lt;/p&gt;&#xA;" OwnerUserId="1670" LastEditorUserId="1670" LastEditorDisplayName="" LastEditDate="2011-02-01T17:27:03.767" LastActivityDate="2011-02-01T17:27:03.767" CommentCount="5" />
  <row Id="4386" PostTypeId="2" ParentId="4384" CreationDate="2011-02-01T16:46:23.773" Score="6" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://arxiv.org/abs/0806.1147&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0806.1147&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-01T17:36:08.760" LastActivityDate="2011-02-01T17:36:08.760" CommentCount="1" />
  <row Id="4387" PostTypeId="2" ParentId="4384" CreationDate="2011-02-01T17:23:21.880" Score="5" ViewCount="" Body="&lt;p&gt;See &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Legendre_transformation#Applications&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Legendre_transformation#Applications&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In theoretical physics, the basic or defining mathematical properties of the Legendre transformation are used to switch between one form of the energy - or &quot;potential&quot;, as the generalized energies are called in thermodynamics - to another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is important to switch between the Lagrangian in abstract mechanics that depends on $x,v$ (positions and velocities) to the Hamiltonian, the true energy that depends on $x,p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In thermodynamics, the number of applications and &quot;types of switches&quot; is even higher. You may go from energy to enthalpy or Helmholtz free energy or Gibbs free energy by Legendre-transforming with respect to various variables. The transform goes back and forth. As the Wikipedia example explains, there are other useful variables that you may Legendre-transform with respect to, including the charge and voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may consider the Legendre transformation to be a &quot;mere&quot; redefinition of variables - but that's why it's so important in practice. In reality, the different ways to describe the system that differ by a Legendre transformation are &quot;equally fundamental&quot; or &quot;equally natural&quot; so it's often useful to be familiar with all of them and to know what is the relationship between them. The relationship is given by the Legendre transformation.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-01T18:08:37.907" LastActivityDate="2011-02-01T18:08:37.907" CommentCount="5" />
  <row Id="4388" PostTypeId="2" ParentId="4256" CreationDate="2011-02-01T17:24:37.270" Score="3" ViewCount="" Body="&lt;p&gt;Some more misconceptions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ChemGuide website quoted above might be a useful reference for &quot;UK-based exam purposes&quot; as stated there, but it certainly does not help in solving the question. The arguments given above that followed the comments on ChemGuide are inaccurate. A simple quantum chemistry calculation of gold in its ground state will give you that the electron in the s orbital (A1G) is the most energetic in this atom. Hence, ionization will most easily be accomplished by removal of this electron, and not of d electrons, and this is easily proved by another computation for ionized gold, which will show you that the 5d orbitals will remain filled while the 6s orbital is no longer occupied. Actually, it is known that if the most external d shell is filled, the energies of these orbitals will be effectively lowered, and there is a very high probability that the ionized electron will not come from it, but from more energetic s or p orbitals. (I have just done a few of these calculations in order to make sure this point is right)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to analyze why some metals are more inert than others, various effects come into play. Relativistic effects, such as the contraction of s orbitals, for example, are a major factor in making gold less reactive than silver, and in lowering the oxidation potential of gold. So, in addition to looking at chemical potentials when discussing the inertness of metals in different environments, it is better not to reduce the arguments to simple electron configuration trends which usually work quite well for main group elements, since, though they might generate insights for the understanding of the behavior of metals, these insights might be either right or wrong.&lt;/p&gt;&#xA;" OwnerUserId="1695" LastActivityDate="2011-02-01T17:24:37.270" CommentCount="2" />
  <row Id="4389" PostTypeId="1" AcceptedAnswerId="4391" CreationDate="2011-02-01T18:00:03.510" Score="1" ViewCount="176" Body="&lt;p&gt;Why is the speed of sound given so much importance in Astrophysics? For example in gas outflow (and accretion) problems, we often calculate the sonic point (the point at which the outflow speed becomes equal to the speed of sound). Is it simply because speed of sound is a convenient number that can be used to scale down the variables, or is something deep involved? &lt;/p&gt;&#xA;" OwnerUserId="966" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-02T10:38:29.767" LastActivityDate="2011-02-02T10:38:29.767" Title="Speed of sound in astrophysics" Tags="&lt;astrophysics&gt;&lt;fluid-dynamics&gt;&lt;sound&gt;" AnswerCount="2" />
  <row Id="4390" PostTypeId="2" ParentId="4389" CreationDate="2011-02-01T18:10:22.597" Score="2" ViewCount="" Body="&lt;p&gt;One thing I can think up immediately is that the speed of sound sets the point at which shocks are produced, and shocks produced by an outflow are an important source of heating of the interstellar or intercluster media.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-02-01T18:10:22.597" />
  <row Id="4391" PostTypeId="2" ParentId="4389" CreationDate="2011-02-01T18:15:56.087" Score="3" ViewCount="" Body="&lt;p&gt;For one thing the speed of sound determines the rate at which disturbances can propagate in the medium or the &quot;cone of causality&quot;. And this characteristic scale in turn determines at what point the system becomes unstable and heads towards gravitational collapse. In cosmology this is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jeans_length&quot; rel=&quot;nofollow&quot;&gt;Jeans instability&lt;/a&gt;. The Jeans length - i.e. the characteristic scale which demarcates perturbations which will induce gravitational collapse from stable oscillations of the medium is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lambda_J = c_s \left( \frac{\pi}{G\rho} \right)^{1/2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c_s$ is the speed of sound in the gas and $\rho$ is the mean density.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-01T18:15:56.087" CommentCount="2" />
  <row Id="4392" PostTypeId="2" ParentId="3799" CreationDate="2011-02-01T19:24:50.047" Score="-1" ViewCount="" Body="&lt;p&gt;I think politics is the strongest reason, what ever the truth may be.&lt;/p&gt;&#xA;" OwnerUserId="1670" LastActivityDate="2011-02-01T19:24:50.047" CommentCount="8" />
  <row Id="4393" PostTypeId="1" AcceptedAnswerId="4396" CreationDate="2011-02-01T20:47:42.670" Score="3" ViewCount="493" Body="&lt;p&gt;Apologies in advance if this is too basic a question for physics.stackexchange. I don't want to dumb down this venerable institution. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My wife and I just watched &lt;a href=&quot;http://www.youtube.com/watch?v=XSz4Q6rOqoM&quot; rel=&quot;nofollow&quot;&gt;this TV ad for Kit-Kat&lt;/a&gt; where a crew of crane operators line their wrecking balls up to form an enormous Newton's Cradle. It left us wondering if the basic premise is feasible. If the wrecking balls were perfectly aligned would it work? I imagine the lack of lateral stability inherent in having only one anchor point per &quot;ball&quot; would be a problem, but what else (if anything) stops this from working for real?&lt;/p&gt;&#xA;" OwnerUserId="1712" LastActivityDate="2011-02-02T03:27:34.120" Title="Is the giant Newton's cradle in the Kit-Kat ad feasible?" Tags="&lt;energy&gt;&lt;momentum&gt;" AnswerCount="4" />
  <row Id="4394" PostTypeId="1" AcceptedAnswerId="4397" CreationDate="2011-02-01T20:55:25.060" Score="7" ViewCount="297" Body="&lt;p&gt;Hi I have a doubt (I'm not very expert in statistical mechanics, so sorry for this question). We consider a gas of hydrogen atoms with no interactions between them.&#xA;The partition function is:&#xA;$$&#xD;&#xA;Z=\frac{Z_s^N}{N!}&#xD;&#xA;$$&#xA;where $Z_s$ is the partition function of one atom. So we write&#xA;$$&#xD;&#xA;Z_s =Tr\{e^{\beta \hat H}\}&#xD;&#xA;$$&#xA;and we must consider sum to discrete and continous spectrum. So I'd write for the contribute of the discrete spectrum:&#xA;$$&#xD;&#xA;Z_{s_{disc}}=\sum_{n=1}^{+\infty}n^2 e^{\beta \frac{E_0}{n^2}}&#xD;&#xA;$$&#xA;but this serie doesn't converge. &#xA;For continuum spectrum, I'm not be able to write the contribute to the sum, because I have infinite degeneration, so where's my mistake?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have thought that the spectrum tha I considered for energy values is for free atom in the space, and partition function maybe is defined for systems with finite volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, this doubt however would me ask a problem. Can whe prove that operator $e^{-\beta \hat H}$ always has got finite trace for halll phsical systems? &lt;/p&gt;&#xA;" OwnerUserId="558" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-02T10:36:54.717" LastActivityDate="2011-02-02T10:49:33.073" Title="Partition function of a hydrogen gas" Tags="&lt;quantum-mechanics&gt;&lt;statistical-mechanics&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="0" />
  <row Id="4395" PostTypeId="1" AcceptedAnswerId="4402" CreationDate="2011-02-01T21:14:13.410" Score="3" ViewCount="117" Body="&lt;p&gt;There is not-so-rough evidence that &lt;a href=&quot;http://arxiv.org/abs/1101.5476&quot; rel=&quot;nofollow&quot;&gt;at very large scale the universe is flat&lt;/a&gt;. However we&#xA;see everywhere that there are local lumps of matter with positive curvature. So i have several questions regarding this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Does the fact that a manifold with a) asymptotic (space) curvature zero and b) local inhomogeneities with positive (space) curvature imply that there will be regions with negative (space) curvature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) a Region of negative (space) curvature implies dark energy in that region? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) assuming answer to both 1) and 2) are true: does this represent an independent confirmation of dark energy? or there is somehow an geometric relationship relating asymptotic flatness to accelerated expansion (the traditional reason to introduce dark energy in the first place)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDITED: to reflect distinction between space and space-time curvatures. &lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-02-01T22:06:20.317" LastActivityDate="2011-02-01T22:06:20.317" Title="asymptotic curvature of the universe and correlation with local curvature" Tags="&lt;general-relativity&gt;&lt;cosmology&gt;&lt;dark-energy&gt;&lt;curvature&gt;" AnswerCount="2" />
  <row Id="4396" PostTypeId="2" ParentId="4393" CreationDate="2011-02-01T21:15:26.447" Score="6" ViewCount="" Body="&lt;p&gt;Theoretically it would be feasible, as the law of conservation of energy is a universal principle, regardless of how large the system is. I think is would be difficult to realize a very large model like shown in the ad. First of all, the balls aren't exactly rigid. The larger the balls get, the larger the area of contact becomes, thereby increasing dissipation of energy. Coming to the ad, it is clearly a product of modern day computer graphics. If we assume a length of 50 m for the hanging chain, the time period of the pendulum, $2\pi\sqrt{\frac{L}{g}}$ ~ 14 s. Clearly the time period of the pendulum in the video is much much smaller. Further in reality, the balls would initially break-up due the elastic response of the balls (the balls are a little 'springy' and obey &lt;a href=&quot;http://en.wikipedia.org/wiki/Hooke%27s_law&quot; rel=&quot;nofollow&quot;&gt;Hooke's Law&lt;/a&gt;). Then each ball would oscillate about a mean position (yes all the balls move! We don't notice this as the time period of the individual oscillations is quite small compared to the time period of the main balls). And finally due to dissipation of energy, the balls would settle down. A very realistic simulation has been done on Newton's Cradle by &lt;a href=&quot;http://www.maths.tcd.ie/~garyd/Publications/Delaney_2004_AmJPhys_Rocking_Newtons_Cradle.pdf&quot; rel=&quot;nofollow&quot;&gt;Stefan Hutzler et al&lt;/a&gt;. A video of the simulation can be found &lt;a href=&quot;http://www.maths.tcd.ie/~garyd/MOVIES/CRADLE/GW_Delaney_Cradle_disp.mpg&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="966" LastEditorUserId="966" LastEditorDisplayName="" LastEditDate="2011-02-01T21:31:35.373" LastActivityDate="2011-02-01T21:31:35.373" CommentCount="1" />
  <row Id="4397" PostTypeId="2" ParentId="4394" CreationDate="2011-02-01T21:17:15.527" Score="8" ViewCount="" Body="&lt;p&gt;This is quite a subtle problem:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Notice that you have the same problem if you have only a single hydrogen atom.&lt;/li&gt;&#xA;&lt;li&gt;You've ignored all the unbound states of the hydrogen atom, which make thing even worse (they are higher in energy, and have even larger degeneracies).&lt;/li&gt;&#xA;&lt;li&gt;Divergent partition function is actually not really a problem --- all physical properties will depend on derivatives of the logarithm of it.&lt;/li&gt;&#xA;&lt;li&gt;However, you will find that at any finite temperature things like entropy (which are physical) does indeed diverge.&lt;/li&gt;&#xA;&lt;li&gt;This is again not so bad if you realise that you've effectively got an infinite system --- the entropy per volume should still be finite.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Concretely, you should regularise things by considering a finite box of volume V, with N particles. This will modify the relevant states (and their energies) in correct (but incredibly hard to compute) ways to render all physical properties finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively the effect is due to the n'th bound state having a radius $O(n^2)$. Thus you can approximate the effects of a finite boundary by simply removing all states above a certain size --- the divergences are so slow that in practice physical quantities will only depend very weakly on the cut-off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://scienceblogs.com/builtonfacts/2011/01/the_hydrogen_partition_functio.php&quot; rel=&quot;nofollow&quot;&gt;http://scienceblogs.com/builtonfacts/2011/01/the_hydrogen_partition_functio.php&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed043p364&quot; rel=&quot;nofollow&quot;&gt;http://pubs.acs.org/doi/abs/10.1021/ed043p364&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-02-01T21:17:15.527" />
  <row Id="4398" PostTypeId="2" ParentId="4393" CreationDate="2011-02-01T21:19:22.790" Score="2" ViewCount="" Body="&lt;p&gt;The &quot;art&quot; of alignment would make this very difficult, maybe impossible. &#xA;And, this would work for the first passage only, on &quot;return&quot; the alignment &#xA;will be no more good enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem were those &quot;wrecking balls&quot;. Such balls will not be made &#xA;from the super-hard steel one needs for a good working cradle. &#xA;Cradles are made from balls produced for ball bearings, but a wrecking ball &#xA;has to be less hardened, to avoid the risk that it might split in work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Georg&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-02-01T21:19:22.790" />
  <row Id="4399" PostTypeId="2" ParentId="3861" CreationDate="2011-02-01T21:23:34.173" Score="1" ViewCount="" Body="&lt;p&gt;If you think that there are simple and unambiguous ways to construct theories, then you were misled by someone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every achievement was reached by trying different possibilities. Only more-or-less successful constructions were left in the end (different in different particular cases).&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-02-01T21:23:34.173" />
  <row Id="4400" PostTypeId="2" ParentId="4393" CreationDate="2011-02-01T21:23:50.993" Score="2" ViewCount="" Body="&lt;p&gt;This was funny enough for me to break away from work.  In principle this would work, but the crade has each ball suspended by two lines at an angle.  That keeps the balls elastically colliding in a line.  The wreaking balls are suspected by a single cable.  Unfortunately I think the set up might only work for a couple of periods before the balls start to get out of allignment.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-01T21:23:50.993" />
  <row Id="4401" PostTypeId="2" ParentId="4395" CreationDate="2011-02-01T21:35:20.783" Score="0" ViewCount="" Body="&lt;p&gt;The universe is flat spatially, but the space is being stretched with time on the Hubble frame.  This means how the space is embedded in spacetime is such that there is curvature, such as a “time-time” Ricci curvature $R^{tt}$.  This solution to the Einstein field equations is such that the pressure is equal to the negative of the energy density of the vacuum.  So dark energy, which is associated with this pressure is due to a positive energy density.  The Hamiltonian for this is $H~=~\Lambda x^2/6$, which is similar to the spring potential.  However, the force acts in the same direction as the displacement.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-01T21:35:20.783" CommentCount="5" />
  <row Id="4402" PostTypeId="2" ParentId="4395" CreationDate="2011-02-01T21:51:47.013" Score="5" ViewCount="" Body="&lt;p&gt;You have to be careful to distinguish between curvature of &lt;em&gt;space&lt;/em&gt; and curvature of &lt;em&gt;spacetime&lt;/em&gt;. When we say that the Universe is flat on large scales, we're talking about space -- that is, about a slice through spacetime at constant cosmic time. With respect to spatial curvature, statement 1 is correct: we do have zero curvature on average, and positive curvature in some regions, which implies negative curvature in other regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But statement 2 doesn't follow from statement 1, because in this case we want to talk about &lt;em&gt;spacetime&lt;/em&gt; curvature. To be specific, ordinary matter produces positive spacetime curvature (i.e. a positive Ricci scalar), and dark energy produces negative spacetime curvature. But spatial curvature and spacetime curvature are different things.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-01T21:51:47.013" CommentCount="4" />
  <row Id="4404" PostTypeId="1" AcceptedAnswerId="4405" CreationDate="2011-02-01T22:03:33.687" Score="3" ViewCount="158" Body="&lt;p&gt;I'm having trouble getting my head around the time dilation paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observer A and B are at the same &quot;depth&quot; in a gravity well. Observer B then descends into the well. A will observe B's time as going slower than their own. B will observe A's time as going faster than their own. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens if B were to ascend the well back to A's depth, would B's local time speed back up to the same rate as A's, but B would be younger (relative to A)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the paradox caused by relative motion (ignoring gravity)? If A is moving relative to B, A and B will both observe the other's time as going slower. If A and B were together initially, then B moves away and returns, do their clocks agree? they can't both be younger than each other :s (i get thats the paradox, but what explanation resolves it?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1713" LastActivityDate="2011-02-02T07:05:30.157" Title="Time Dilation - what happens when you bring the observers back together?" Tags="&lt;relativity&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="4405" PostTypeId="2" ParentId="4404" CreationDate="2011-02-01T22:12:22.673" Score="3" ViewCount="" Body="&lt;p&gt;Calculations show that younger will be the observer who suffered accelerations/decelerations.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-02-01T22:12:22.673" CommentCount="6" />
  <row Id="4406" PostTypeId="2" ParentId="4404" CreationDate="2011-02-01T22:29:19.603" Score="4" ViewCount="" Body="&lt;p&gt;Answer to the first question: yes, B will be younger than A.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-01T22:29:19.603" />
  <row Id="4407" PostTypeId="1" AcceptedAnswerId="4410" CreationDate="2011-02-01T23:29:23.863" Score="2" ViewCount="189" Body="&lt;p&gt;Of course there is pretty hot stuff too. For example Suns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But isn't the vacuum of space the perfect containment for heat? And shouldn't the rare collision with particles even heat up an object that floats around in space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why aren't they hot from time to time? Is it because they loose their heat by radiant heat?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if so does radiant heat emission always occur until the object is the same temperature as the surrounding or does it depend on other factors? &lt;/p&gt;&#xA;" OwnerUserId="1714" LastActivityDate="2011-02-02T02:57:49.947" Title="Why is everything floating in space so cold?" Tags="&lt;heat&gt;&lt;space&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="4408" PostTypeId="2" ParentId="4407" CreationDate="2011-02-01T23:57:39.210" Score="1" ViewCount="" Body="&lt;p&gt;When a gas expands, it cools.  It turns out that the universe has expanded very significantly since the Big Bang, meaning that it is now very cool.  If any gas is to be placed in the vacuum of the universe, then, absent something like gravity to hold it together, it will expand and thereby cool to the temperature of the medium around it.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-02-01T23:57:39.210" />
  <row Id="4409" PostTypeId="2" ParentId="4407" CreationDate="2011-02-02T00:06:22.083" Score="1" ViewCount="" Body="&lt;p&gt;What is everything? If it is a cosmic ray or the solar wind, it is very hot. The coldest&#xA;it can get is 2.72 K because that is the temperature of the cosmic background radiation, and if there were no other sources of radiation, the object would come into equilibrium with this. In practice, there is always some form of other energy streaming through space. So the answer depends on which neighborhood you are hanging in.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-02-02T00:06:22.083" />
  <row Id="4410" PostTypeId="2" ParentId="4407" CreationDate="2011-02-02T00:08:18.357" Score="4" ViewCount="" Body="&lt;p&gt;Well, the basic law of thermodynamics tells you that the heat flows from the hotter object to the colder one. But what this means in a given situation will depend on the precise mechanism of that heat flow, i.e. elementary interaction that makes it possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's discuss few of them and see where it takes us.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First, there's the usual &lt;em&gt;conductive&lt;/em&gt; heat transfer. This is mediated by the collisions of the touching materials (and even more macroscopically by an electromagnetic force. Obviously, this effect is irrelevant in the vacuum.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Second, there's &lt;em&gt;evaporation&lt;/em&gt;. Even if you have a perfect crystal that is very stable, there's a possibility that a large fluctuation will occur and one of its molecules will be &quot;kicked out&quot; of it and set flying into the empty space. Because the molecule needs to have a (lot) higher energy than the rest of the molecules around it to be able to get free from the crystal, the average remaining energy and therefore also the temperature will decrease. But this effect is more relevant for non-crystals (because the molecules are much less tightly bound). E.g. if water were put into the free space it would freeze by evaporation very quickly.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Last, but not least, &lt;em&gt;radiation&lt;/em&gt;. Any object at non-zero temperature will have lots of its constituents in excited energy states and there will be a non-zero probability for returning to a lower energy state and emission of a photon. In more simple terms, kinetic energy is converted to an electromagnetic energy. This effect is most relevant for black objects which absorb all the radiation (an ideal form of which is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_body&quot; rel=&quot;nofollow&quot;&gt;black body&lt;/a&gt;) and least relevant for white objects which reflect everything (note that the terms black and white are to be considered here in generalized sense, not just pertaining to visible light).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Well, these effects are most important for the usual matter. But there is also the &lt;em&gt;dark matter&lt;/em&gt; which doesn't interact electromagnetically (therefore the name), so it wouldn't cool by photonic radiation (but might cool by emission of other particles with which it interacts). In conclusion, the heat transfer depends very much on the precise microscopic interaction that the given object is able to undergo.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-02-02T00:08:18.357" />
  <row Id="4411" PostTypeId="1" AcceptedAnswerId="4439" CreationDate="2011-02-02T00:12:04.080" Score="1" ViewCount="203" Body="&lt;p&gt;I know about these different energy conditions in GR, namely strong, weak and null, but never really understood the full physical significance of them or for example how to 'derive' them or how compelling is a certain condition (for that particular one to be applied). I know that they exist because you need some constraints for $T_{\mu\nu}$ to satisfy as otherwise its completely unconstrained. Can someone please elaborate?&lt;/p&gt;&#xA;" OwnerUserId="1349" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-02T10:25:49.447" LastActivityDate="2011-02-06T12:34:22.723" Title="Significances of energy conditions" Tags="&lt;general-relativity&gt;&lt;gravity&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="4412" PostTypeId="2" ParentId="4407" CreationDate="2011-02-02T00:24:07.970" Score="2" ViewCount="" Body="&lt;p&gt;Anything in space is at the same temperature as the blackbody temperature of radiation it is immersed in, or an integration of the blackbody radiation per sterradian angle.  If that is a close orbit around the sun the temperature is hot.  The temperature in intergalactic space is 2.7K.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-02T00:24:07.970" CommentCount="1" />
  <row Id="4413" PostTypeId="2" ParentId="2378" CreationDate="2011-02-02T00:48:05.820" Score="0" ViewCount="" Body="&lt;p&gt;Our location is around 13.7 billion light years away from it in time.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-02-02T00:48:05.820" CommentCount="1" />
  <row Id="4414" PostTypeId="2" ParentId="4411" CreationDate="2011-02-02T00:51:57.680" Score="1" ViewCount="" Body="&lt;p&gt;The energy momentum tensor at a point $p~\in~M$ in the weak energy condition obeys the condition $T_{ab}U^aU^b~\ge~0$, for $U^a$ a tangent vector in $T_pM$.  The tensor then has components $T^{00}~=~\rho$ and $T^{ii}~=~p_i$ as energy density and pressure along the geodesic through the point $p$.  The dominant energy condition is that for $T_{ab}U^a$ a non-spacelike vector $T_{ab}U^aU^b~\ge~0$.  The strong energy condition is $T_{ab}U^aU^b~\ge~(1/2){\it tr}T U^aU_a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These conditions are not derived, but imposed.  They do give conditions on causality and Cuachy conditions.  Violations of them remove these conditions, which permit the existence of closed time-like curves.  The averaged weak energy condition is $T^{00}~\ge~0$.  This condition exists for some quantum field, where we have $\langle T^{00}\rangle$.  Assume this momentum-energy tensor is defined by a scalar field.  We then have&#xA;$$&#xD;&#xA;T_{ab}~=~\phi_{;a}\phi_{;b}~-~\frac{1}{2}(\phi_{;c}\phi_{;d}g^{cd}~+~m^2\phi^2)&#xD;&#xA;$$&#xA;If we assume it is massless and we just consider $T^{00}~=~(1/2)\phi_{;t}\phi_{;t}$.  The field is expanded into operators $a,~a^\dagger$, and if we ignore for the sake of argument the covariant derivative, energy is $\sim~a^\dagger a$.  We all know that quantum systems have a vacuum, some bounded minimal energy value.  Yet if the averaged weak energy condition is violated then this operator always returns a negative value, unbounded below.  That is a disaster, for it means the quantum system can cascade down an endless ladder of states and emit a vast amount of radiation.  For this reason it is generally thought the weak energy condition holds.  However, this has not been proven to be a consequence of more fundamental physics.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-02T00:51:57.680" />
  <row Id="4415" PostTypeId="2" ParentId="4411" CreationDate="2011-02-02T00:52:27.927" Score="1" ViewCount="" Body="&lt;p&gt;An energy condition is merely a restriction on the stress-energy tensor conjectured to hold for all physically reasonable matter.  In other words, it’s just a guess at the precise definition of “physical reasonableness”.   As such, energy conditions are not derived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, every energy condition yet proposed has failed.  This means that for each known condition an example exists of matter considered physically reasonable that violates it.   These examples usually stem from quantum field theory, although there are classical examples based on the existence of (an as yet undetected) fundamental scalar field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Averaged energy condition are also known to be violated by examples considered physically reasonable.  See for example &lt;a href=&quot;http://arxiv.org/abs/0910.5925&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0910.5925&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1716" LastEditorUserId="1726" LastEditorDisplayName="" LastEditDate="2011-02-02T18:18:26.750" LastActivityDate="2011-02-02T18:18:26.750" CommentCount="1" />
  <row Id="4417" PostTypeId="1" CreationDate="2011-02-02T01:22:45.483" Score="1" ViewCount="141" Body="&lt;p&gt;Not long ago I submitted an article to PRE. One referee was in favor and two against (this not more than 3 months ago). Rather than pressing, I decided to submit it elsewhere. A single referee report was negative on the grounds that the article was unclear, though the details helped me revise the article. Is it all right to resubmit a substantially revised article and new supporting material to PRE under the original article ID number, or am I obligated to make a new submission? In the event of the former, should (must?) I mention the intervening submission?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-02T02:50:46.273" ClosedDate="2011-02-02T10:20:36.677" Title="Submitting an article to another journal" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="4418" PostTypeId="1" CreationDate="2011-02-02T01:26:36.027" Score="3" ViewCount="122" Body="&lt;p&gt;In string theory I came across the fact that there are difficulties in describing the coupling of R-R fields with world sheets in RNS formalism and it can be done in GS formalism only. Can someone explain the reason(s) behind that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I think if you have a space time where you can quantize strings and you have non zero R-R fields, then you need to use GS formalism to quantize strings there. It seems to me that these two are related and again my question is- why is that? Do we always need to use GS formalism there?&lt;/p&gt;&#xA;" OwnerUserId="1349" LastActivityDate="2011-02-02T18:19:27.383" Title="R-R fields in RNS formalism" Tags="&lt;string-theory&gt;" AnswerCount="2" />
  <row Id="4419" PostTypeId="2" ParentId="4417" CreationDate="2011-02-02T01:32:49.703" Score="5" ViewCount="" Body="&lt;p&gt;It seems like your questions are answered here: &lt;a href=&quot;http://pre.aps.org/info/polproce.html&quot; rel=&quot;nofollow&quot; title=&quot;PRE Editorial Policies&quot;&gt;http://pre.aps.org/info/polproce.html&lt;/a&gt;, in particular:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Authors should state whether the paper they submit has been previously considered for publication in any of the APS journals (Physical Review Letters, other Physical Review journals, or Reviews of Modern Physics) and supply the code number assigned by that journal.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1252" LastActivityDate="2011-02-02T01:32:49.703" />
  <row Id="4420" PostTypeId="2" ParentId="4418" CreationDate="2011-02-02T01:44:16.457" Score="3" ViewCount="" Body="&lt;p&gt;In the RNS formalism the vertex operators for Ramond fields are not polynomials in the worldsheet fields. If you want to include them as part of the background, you'd have to deal with non-polynomial actions on the worldsheet. This is technically difficult. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to deal with this difficulty is to use a different method to quantize the worldsheet of the string, like the GS method or the Berkovits method. These are all equivalent, but some things that are difficult in one language are easier in another (and vice versa). In the quantization methods where spacetime SUSY is manifest on the worldsheet, RR fields are simpler, but seems to me many other things are more difficult, but maybe I am biased by my own education here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example how such quantization works, take a look at this paper by &lt;a href=&quot;http://arxiv.org/abs/hep-th/9902098&quot; rel=&quot;nofollow&quot;&gt;Berkovits-Vafa-Witten&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-02-02T01:44:16.457" CommentCount="3" />
  <row Id="4421" PostTypeId="2" ParentId="4349" CreationDate="2011-02-02T02:00:57.497" Score="6" ViewCount="" Body="&lt;p&gt;Seems to me there is a confusion between various concepts, let me try to clear it up:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Virtual particle is one that doesn't live forever, at some stage it gets converted to something else. As Jeff points out, none of us lives long enough to tell the difference, so the distinction between virtual and non-virtual is a matter of degree. Particles that live for a long time are declared &quot;real&quot;, and particles that decay quickly are called &quot;virtual&quot;. These are just names, there is no implication that &quot;virtual&quot; particles don't really exist, like white unicorns and other mythical creatures, those are all real measurable effects you can see with your own eyes...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Any particle can be either real or virtual, whether or not it is massive, whether or not it is bosonic force carrier, or fermionic matter. There is a sense in which massive particles tend to live shorter life (because they have more opportunities to decay), but this is just a rule of thumb.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Off-shell can be taken here to be synonymous with &quot;virtual&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-02-02T05:28:53.393" LastActivityDate="2011-02-02T05:28:53.393" CommentCount="1" />
  <row Id="4422" PostTypeId="2" ParentId="4418" CreationDate="2011-02-02T02:19:37.980" Score="0" ViewCount="" Body="&lt;p&gt;The easiest way to write the curved background action is to covariantize the vertex operator. But in the RNS version of the superstring, to write the RR vertex you need to break worldsheet supersymmetry and mix ghost and matter fields. You can look at that famous paper by Friedan, Martinec and Shenker if you don't know how it is done:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://newton.kias.re.kr/~brane/2004/Y.%20Kim/fms.pdf&quot; rel=&quot;nofollow&quot;&gt;Conformal Invariance, Supersymmetry and String Theory&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without supersymmetry as a guiding principle, is very difficult to write the right action and no one really knows how to do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Green-Schwarz or pure spinor version, there is not much problem to do so. See, for instance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/hep-th/0112160&quot; rel=&quot;nofollow&quot;&gt;Ten-Dimensional Supergravity Constraints from the Pure Spinor Formalism for the Superstring&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The GS superstring has problems to be quantized and this is usually made in the light-cone gauge, but it is not always that this gauge exists. So, I guess that the the real known way to treat a superstring in a RR background as a quantum CFT is with the pure spinor action.&lt;/p&gt;&#xA;" OwnerUserId="1337" LastEditorUserId="1337" LastEditorDisplayName="" LastEditDate="2011-02-02T18:19:27.383" LastActivityDate="2011-02-02T18:19:27.383" />
  <row Id="4423" PostTypeId="2" ParentId="4349" CreationDate="2011-02-02T02:33:54.463" Score="2" ViewCount="" Body="&lt;p&gt;They can be real, no problem with that. However, all Z and W observed are virtual. And yes, they are off-shell, whatever that means to you. We actually measure their width, for instance, &lt;a href=&quot;http://arxiv.org/pdf/0909.4814&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/0909.4814&lt;/a&gt; (it's more or less 2 GeV around a mass of more or less 80-90 GeV).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is observed is a pole in the S matrix for some final states in particle collisions. So, really, what is measured is a certain differential cross section, to which is fit a distribution with this pole structure. From the values, both the mass and the width are read out.&lt;/p&gt;&#xA;" OwnerUserId="1337" LastEditorUserId="1337" LastEditorDisplayName="" LastEditDate="2011-02-02T05:11:56.837" LastActivityDate="2011-02-02T05:11:56.837" />
  <row Id="4424" PostTypeId="2" ParentId="4334" CreationDate="2011-02-02T02:45:32.563" Score="3" ViewCount="" Body="&lt;p&gt;Let's begin by choosing coordinates. Let the orbit of the earth define the x-y plane. Assume that just before the collision the earth's orbital axis is pointing in the x-direction, that is $(1,0,0)$. I'll assume that the earth absorbs the asteroid, and that after the collision the axis has been changed to $(\sin(27),0,\cos(27))$ where 27 is the degree measure of the earth's axis currently. So the relative change in angular momentum for the earth is&lt;br&gt;&#xA;$(1,0,0) - (\sin(27),0,\cos(27)) = (0.55,0,-.89)$&lt;br&gt;&#xA;which is a vector of length 1.05 and so the angular momentum of the asteroid has to be 1.05 times the angular momentum of the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make the approximation that the earth is a sphere of constant density. Then its angular momentum is given by $0.4 M_e\omega\;R_e^2$ where $M_e,R_e$ are the mass and radius of the earth and $\omega$ is its angular rate of rotation. We have&lt;br&gt;&#xA;$M_e = 6\times 10^{24}\;kg\;\;\;R_e = 6.4\times 10^6\;m,\;\;\;\omega = 2\pi/(24\;hours) = 2\pi/ (86400\;sec)$.&lt;br&gt;&#xA;I'm using sloppy approximations here; the earth is not constant density, I'm not using the sidereal day, etc. This gives $7\times 10^{33}\;kg\;m^2/s$ as the approximate angular momentum contributed by the asteroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula for angular momentum is mass x velocity x radius. This is maximum when the radius is maximum; for an asteroid hitting the earth this happens when the radius is equal to the earth's radius, i.e. $ 6.4\times 10^6\;m$. Dividing the asteroid's angular momentum by this give its linear momentum:&lt;br&gt;&#xA;$P_a = 7\times 10^{33} / 6.4\times 10^6 = 10^{27}\;kg\;m/s$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason a relatively small asteroid can wipe out all life on earth is due to the kinetic energy it carries. Upon collision, the kinetic energy is converted to heat. If the asteroid is large and fast enough, the heat will increase the temperature of the atmosphere enough to boil off the oceans and even vaporize the salt deposits left over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since kinetic energy is proportional to the square of velocity while momentum is only proportional to velocity, we will assume that our asteroid is moving as slowly as possible. The escape velocity of the earth is 11 km per second; to achieve a slower velocity an asteroid would have to be very lucky. So in calculating the kinetic energy of the asteroid, we'll assume a very conservative speed of 5 km/sec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting $10^{27} = 5000 m$ we find the mass of the asteroid as $2\times 10^{23}$ kilograms. Assuming a specific density of 5, or 5000 kg per cubic meter, this gives a radius for the asteroid of 2000 kilometers or a diameter of 5000 km. This is far more than enough to destroy all life on the planet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The asteroid's kinetic energy is&lt;br&gt;&#xA;$0.5\times 2\times 10^{23}\times (5000)^2 = 2.5\times 10^{30}$ Joules&lt;br&gt;&#xA;$= 6\times 10^{14}$ megatons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the collision would result in the release of kinetic energy equal to 600 million million hydrogen bombs going off, each with an equivalent energy of one million tons of high explosives. The earth's surface area is 500 million square kilometers or 500 million million square meters. So the energy release is equivalent to having a 1 megaton bomb going off on each square meter of the earth's surface. And since the vaporized rock from the blast is lighter than rock, this vapor will condense on the surface and apply its heat to the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a video of a 500 km asteroid hitting the earth (1/1000 of the volume necessary to change the earth's axis) see:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.youtube.com/watch?v=vZiZU42sn6w&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=vZiZU42sn6w&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/eK5b0.jpg&quot; alt=&quot;Video frame&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-02-07T21:51:50.783" LastActivityDate="2011-02-07T21:51:50.783" CommentCount="1" />
  <row Id="4425" PostTypeId="2" ParentId="4417" CreationDate="2011-02-02T02:50:46.273" Score="1" ViewCount="" Body="&lt;p&gt;If it were me, I'd send an email to the editor who reviewed the first tine you sent it. Ask him if the changes make it worth resubmitting. The fact that they agreed to review it the first time, and that it got a 3rd reviewer, are good points in its favor.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-02T02:50:46.273" />
  <row Id="4426" PostTypeId="2" ParentId="4407" CreationDate="2011-02-02T02:57:49.947" Score="0" ViewCount="" Body="&lt;p&gt;Actually space is the perfect environment for making things cold. The night sky is cold enough that people in the desert can leave trays of water out and find them frozen in the morning, even though the night-time air temperature never dropped to freezing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happened is that the trays radiated their heat off into space. Of course space radiated heat back to the tray, but the stars are so far between and far away that the amount of heat that gets here is very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't apply during the daytime. In short, things in space get hot if they're in sunlight, and they cool down otherwise.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-02T02:57:49.947" />
  <row Id="4427" PostTypeId="2" ParentId="4393" CreationDate="2011-02-02T03:27:34.120" Score="1" ViewCount="" Body="&lt;p&gt;One way of looking at the question is to rephrase it as follows: &quot;A small steel ball, raised to a height of a few times its diameter, bounces well. Does this behavior scale to a large steel ball?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a question for engineers. To bounce well means that the collision is &quot;elastic&quot;. A sufficiently large ball will collide inelastically and this will ruin the equations of  Newton's cradle. (That is, while momentum is still conserved in an inelastic collision, the momentum will not be transferred through the intermediate balls to the one on the end because this requires conservation of energy.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there's an engineering limit to how large you can make Newton's cradle. It's due to the requirement that the collisions be elastic. To find that limit you have to know the strength of the materials involved. I might add some calculations and try and figure this out, it's a cool question...&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-02T03:27:34.120" />
  <row Id="4428" PostTypeId="1" AcceptedAnswerId="4440" CreationDate="2011-02-02T04:45:29.097" Score="3" ViewCount="262" Body="&lt;p&gt;I'm trying to explain in simple terms what the weak interaction does, but I'm having trouble since it doesn't resemble other forces he's familiar with and I haven't been able to come up (or find on the web) with a good, simple visualization for it. &lt;/p&gt;&#xA;" OwnerUserId="1252" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-02-18T16:30:24.940" LastActivityDate="2011-02-18T16:30:24.940" Title="How to explain the weak force to a layman?" Tags="&lt;particle-physics&gt;&lt;weak-force&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="4429" PostTypeId="2" ParentId="2407" CreationDate="2011-02-02T04:48:01.943" Score="1" ViewCount="" Body="&lt;p&gt;The answers above already cover the necessary ground, but they did not mention one 'angle' of it that i find particularly illuminating: &lt;em&gt;holonomy-flux algebras&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Ashtekar variables describing the metric ($SU(2)$-valued decomposition of the metric) can be understood in terms of its holonomies — and, much like it's done in gauge theories, one can deal with these holonomies' algebras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find this interesting because it seems to link to Gauge Theories in a very particular way, à la Wilson Loops. And, if you squint a bit, ;-) , you can also think of 'flux compactification' (string theory; cf. Green-Schwarz-Witten, Vol 2, chapter 14, or &lt;a href=&quot;http://arxiv.org/abs/hep-ph/0107039&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-ph/0107039&lt;/a&gt;) as a much broader 'object' in Physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just my 2¢.&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2011-02-02T04:48:01.943" CommentCount="3" />
  <row Id="4430" PostTypeId="2" ParentId="3861" CreationDate="2011-02-02T05:15:09.080" Score="0" ViewCount="" Body="&lt;p&gt;There's a nice little book that deals with this question of yours in a mathematically rigorous way: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/038798318X&quot; rel=&quot;nofollow&quot;&gt;Mathematical Topics between Classical and Quantum Mechanics (Springer Monographs in Mathematics)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end of the day, it boils down to studying &lt;a href=&quot;http://en.wikipedia.org/wiki/Symplectic_geometry&quot; rel=&quot;nofollow&quot;&gt;Symplectic geometry&lt;/a&gt; and defining more appropriately what is meant by 'quantization'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Weyl_quantization&quot; rel=&quot;nofollow&quot;&gt;Deformation (Weyl) quantization&lt;/a&gt; and &lt;a href=&quot;http://arxiv.org/abs/math-ph/0208008&quot; rel=&quot;nofollow&quot;&gt;Geometric Quantization&lt;/a&gt;, they attack your questions heads-on.&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2011-02-02T05:15:09.080" />
  <row Id="4431" PostTypeId="2" ParentId="4394" CreationDate="2011-02-02T05:33:25.883" Score="2" ViewCount="" Body="&lt;p&gt;The contribution of the continuum states to the partition function is not really a problem. Actually it solves a problem, canceling the divergence on the sum over the discrete spectrum and leaving as result the so called Planck-Larkin partition function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z_{discrete}=\sum_{n=1}^\infty n^2 [\exp (-\beta E_n)-1+\beta E_n]\; , $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $E_n=R(1-1/n^2)$ (the ground state is at zero energy) and $R$ is the ionization energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good paper on this subject is F.J.Rogers, The Astrophysical Journal 310, p. 723. &lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2011-02-02T05:33:25.883" CommentCount="2" />
  <row Id="4432" PostTypeId="1" AcceptedAnswerId="4441" CreationDate="2011-02-02T05:34:07.613" Score="3" ViewCount="190" Body="&lt;p&gt;The qubit is a big topic of quantum information theory. A qubit is a single quantum bit. Physical examples of qubits include the spin-1/2 of an electron, for example, see page 39 of Preskill:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.theory.caltech.edu/people/preskill/ph229/notes/chap5.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.theory.caltech.edu/people/preskill/ph229/notes/chap5.pdf&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum mechanics, two variables are called complementary if knowledge of one implies no knowledge whatsoever of the other. The usual example is position and momentum. If one knows the position exactly, then the momentum cannot be known at all. And to the extent that a situation can exist where we know something about both, there is a restriction, Heisenberg's uncertainty principle, that relates the accuracy of our knowledge:&lt;br&gt;&#xA;$\sigma_x\sigma_p \ge \hbar/2$&lt;br&gt;&#xA;where $\sigma_x$ and $\sigma_p$ are the RMS errors in the position and momentum, and $\hbar$ is Planck's constant $h$ divided by $2\pi$. The same relationship obtains for other pairs of complementary variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The units of $\hbar$ are that of angular momentum. Since spin-1/2 has units of angular momentum, it's natural that its complementary variable has no units. This is typically taken to be angle. That is, the usual assumption of quantum mechanics is that the complementary variable to spin is angle. For example, see Physics Letters A Volume 217, Issues 4-5, 15 July 1996, Pages 215-218, &quot;Complementarity and phase distributions for angular momentum systems&quot; by G. S. Agarwal and R. P. Singh, &lt;a href=&quot;http://arxiv.org/abs/quant-ph/9606015&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/quant-ph/9606015&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But at the same time, in quantum information theory, the concept of &quot;mutually unbiased bases&quot; has to do with complementary variables in a finite Hilbert space. The usual example of this is that spin-1/2 in the x or y direction is complementary to spin in the z direction. In other words, &lt;strong&gt;&lt;em&gt;in quantum information theory, the usual complementary variable to spin is not taken to be angle, but instead is taken to be spin itself.&lt;/em&gt;&lt;/strong&gt; For example, see J. Phys. A: Math. Theor. 43 265303, &quot;Mutually Unbiased Bases and Complementary Spin 1 Observables&quot; by     Paweł Kurzyński, Wawrzyniec Kaszub and Mikołaj Czechlewski, &lt;a href=&quot;http://arxiv.org/abs/0905.1723&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0905.1723&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But according to the Heisenberg uncertainty principle, spin can only be its own complementary variable if we have $\hbar=1$. Of course it's possible to choose coordinates with $\hbar=1$ -- this is common in elementary particles, but what I'm asking about is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a compatible way to interpret the two different choices for the complementary variable to spin angular momentum? For example, can we also interpret spin as an angle?&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-07-27T23:53:21.287" Title="The complementary variable to the qubit and spin-1/2" Tags="&lt;quantum-information&gt;&lt;spin&gt;&lt;uncertainty-principle&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="4433" PostTypeId="1" CreationDate="2011-02-02T05:34:09.937" Score="1" ViewCount="89" Body="&lt;p&gt;With the impending landfall of Cyclone Yasi, we were pondering how much water would be dropped by storm after hearing them say there would be about a metre of rain falling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that this is the amount of rain falling on one square metre, and is constant across the main storm, and that the storm is a regular circle of 500 kilometres diameter, we arrived at:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3.1415 * 250 000 * 250 000 = 196 343 750 000     cubic meters&#xA;                           = 196 343 750 000 000 litres&#xA;                           = 196 344             gigalitres&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;By comparison - we have a reservoir which holds about 20 gigalitres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if we assumed a constant falloff from the centre to the edge - we would still get 49 086 gigalitres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then figured out how much water was dropped in QLD in December (209 mm average over the month) which caused the severe floods. Queensland covers 1 852 642 square kilometres:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0.209 * 1 852 642 000 000 = 387 202 178 000     cubic meters&#xA;                          = 387 202 178 000 000 litres&#xA;                          = 387 202             gigalitres&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which means that this storm will dump a significant amount of water on a very small area...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything wrong with our assumptions - are these calculations right?&lt;/p&gt;&#xA;" OwnerUserId="1719" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-02T10:14:57.497" LastActivityDate="2011-02-02T10:14:57.497" Title="How to calculate the volume of water from the measured rainfall?" Tags="&lt;atmospheric-science&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="4434" PostTypeId="2" ParentId="4428" CreationDate="2011-02-02T05:45:24.090" Score="1" ViewCount="" Body="&lt;p&gt;I like the history oriented approach, when explaining something to a layman. In this case you could start by briefly explaining the fundamental forces and how we needed a new force to model beta decay.&lt;/p&gt;&#xA;" OwnerUserId="301" LastActivityDate="2011-02-02T05:45:24.090" />
  <row Id="4435" PostTypeId="2" ParentId="4433" CreationDate="2011-02-02T06:01:51.683" Score="2" ViewCount="" Body="&lt;p&gt;I don't think there is anything wrong, but depending on the application you should consider studying a more complex dynamics. Consider how much water enters your system (per unit of time) and how much is drained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see that 1 meter of water in 1 square meter sounds a lot! but if it's in a period of one week then the water will probably drain or evaporate at almost the same rate. No flood in this case.&lt;/p&gt;&#xA;" OwnerUserId="301" LastActivityDate="2011-02-02T06:01:51.683" CommentCount="2" />
  <row Id="4436" PostTypeId="1" AcceptedAnswerId="4446" CreationDate="2011-02-02T06:14:35.770" Score="2" ViewCount="134" Body="&lt;p&gt;It is known from experiments that the dielectric constant of a solvent might decrease in regions where there is a strong electric field, for example, near a highly charged ion in an infinitely dilute solution. In such a case the dielectric constant of the liquid might be described as function of the distance from the ion (considering a monoatomic ion, so that we can approximate the dielectric function by a spherically symmetric one), and this description is supposed to capture all the effects of the strong electric field on the dielectric medium. What I am trying to understand here and cannot actually find a good theoretical discussion of is why the dielectric constant would be reduced in regions where the electric field is strong, i.e., what is the molecular mechanism of dielectric saturation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnetic saturation usually happens when the external magnetic field is strong enough so that all the magnets are oriented in a manner which will maximize the interactions between the field and the particles. One cannot help to imagine that in dielectric saturation the dielectric medium will polarize itself in a &quot;maximal&quot; way, in order to maximize the interactions with the strong electric field. It is also not hard to imagine that at some saturation point (a certain electric field), the relaxation/reorientation/etc of the solvent in response to the electric field would increase its energy instead of decreasing it, and hence, it becomes unfavorable for the liquid to polarize itself. The most favorable configuration of this system (strong electric field at a small region of a dielectric liquid) would then be attainable by some other kind of reorganization of the solvent which would reduce its effective polarity, thus effectively decreasing its dielectric constant function. If the argument here given is true, then I would be interested in this molecular reorganization that follows the dielectric saturation, although I can imagine it is probably specific to the actual molecules that make up this dielectric medium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any insights on this issue will be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to clarify the origin of the term:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dielectric saturation is a popular term in solvation chemistry and materials science. It is known to arise at regions in a dielectric medium where there is a strong field, and the liquid/material responds nonlinearly. Effectively, i.e., what experiments find, is that a  reduction in the dielectric constant of the system arises at the region where the field is strong. You can find this term easily in continuum solvent models literature, such as in the 1st chapter (&quot;Modern Theories of Continuum Models&quot;) of the recent book &quot;Continuum Solvation Models in Chemical Physics&quot;, by Mennucci and Cammi, which is available here &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/9780470515235.ch1/summary&quot; rel=&quot;nofollow&quot;&gt;http://onlinelibrary.wiley.com/doi/10.1002/9780470515235.ch1/summary&lt;/a&gt; provided one has access to it.&lt;/p&gt;&#xA;" OwnerUserId="1695" LastEditorUserId="1695" LastEditorDisplayName="" LastEditDate="2011-02-02T16:22:43.710" LastActivityDate="2011-02-02T23:29:27.920" Title="What is the mechanism of dielectric saturation?" Tags="&lt;condensed-matter&gt;&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="4437" PostTypeId="2" ParentId="3432" CreationDate="2011-02-02T06:16:47.497" Score="1" ViewCount="" Body="&lt;p&gt;&#xA;&lt;a href=&quot;http://www.google.ca/search?q=cocktail+party+physics&amp;amp;ie=utf-8&amp;amp;oe=utf-8&amp;amp;aq=t&amp;amp;rls=org.mozilla%3aen-US%3aofficial&amp;amp;client=firefox-a&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Cocktail Party Physics&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jennifer Ouellette, spousal unit of Sean Carroll, organized this and has some of her female friends writing guest blogs on rotation.&#xA;The topics are mostly physics with a discursive and historical slant, and are at popular level. Jennifer is an intelligent English major who writes science well. I threw this in both because the writing is good, but also hopefully to attract more females to physics stack. Now that I have disclosed my motivation, my real favorite blog has to be &lt;a href=&quot;http://motls.blogspot.com/&quot; rel=&quot;nofollow&quot;&gt;The Reference Frame&lt;/a&gt; for its virtuoso physics &#xA;improvisations by Lubos.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-02-02T06:57:55.137" LastActivityDate="2011-02-02T06:57:55.137" CommunityOwnedDate="2011-02-02T06:16:47.497" />
  <row Id="4438" PostTypeId="2" ParentId="4404" CreationDate="2011-02-02T07:05:30.157" Score="1" ViewCount="" Body="&lt;p&gt;As I've been brought to understand this - the person who has descended down the gravity and then returned has not been accelerating at a constant rate throughout the journey (the change in direction requires a change in acceleration), and this causes the differential aging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the same thing for the original twin paradox where one of them travels at near light speed and then returns. This may be oversimplifying things - but you seem to be after a lay explanation.&lt;/p&gt;&#xA;" OwnerUserId="1719" LastActivityDate="2011-02-02T07:05:30.157" />
  <row Id="4439" PostTypeId="2" ParentId="4411" CreationDate="2011-02-02T07:26:09.083" Score="2" ViewCount="" Body="&lt;p&gt;The energy conditions are just various possible generalizations of the statement that &quot;energy density cannot be negative&quot; to the whole stress-energy tensor. The reason why energy density cannot be allowed to be negative - at least in some sense - is that if arbitrary positive- and negative-energy regions were allowed, the empty vacuum would become unstable: it could spontaneously change to regions with positive energy and regions with negative energy, the the &quot;energy density gap&quot; would keep on growing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that a condition of this kind has to hold and prevent the Universe from developing regions of a large negative energy density but we are somewhat uncertain what the precise condition satisfied by the Universe is. As I mentioned, there are many ways how to generalize the condition $T_{00}\geq 0$ to the whole tensor - one that also includes the momentum density and the fluxes of energy and momentum (the latter is the very &quot;stress&quot;). Do we require the energy density to be greater than zero or even the absolute value of the pressure? And so on...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most important conditions are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;null energy condition&lt;/li&gt;&#xA;&lt;li&gt;dominant energy condition&lt;/li&gt;&#xA;&lt;li&gt;weak energy condition &lt;/li&gt;&#xA;&lt;li&gt;strong energy condition&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;See their mathematical conditions at:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Energy_condition#Mathematical_statement&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Energy_condition#Mathematical_statement&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's somewhat likely that the null energy condition is indeed satisfied in the whole Universe - any macroscopic region of it - and the same may be true for the weak and dominant one (dominant one is strictly stronger than the weak one). However, one may write down models of somewhat exotic matter where the strong energy condition is violated. Although lots of &quot;data&quot; are known, it remains somewhat uncertain what is the precise condition that follows from the fundamental theory that knows all about the allowed types of matter - string theory. I forgot to say that all these conditions may be modified by a special treatment of the vacuum energy density coming from the cosmological constant.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-02T07:26:09.083" CommentCount="1" />
  <row Id="4440" PostTypeId="2" ParentId="4428" CreationDate="2011-02-02T07:34:44.453" Score="2" ViewCount="" Body="&lt;p&gt;The weak force &quot;looks&quot; different because in the first (and still most important) reincarnation we have encountered it - namely beta-decay (including the decay of the neutron) - the force seems to be a contact interaction: it has an extremely short range, essentially zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, any phenomenon that differs from the indefinite existence of an object that moves in the same direction by the same speed forever requires a force to be explained. The force required for the beta-decay is the weak nuclear force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the decay seems to &quot;directly&quot; transform a neutron into a proton, electron, and antineutrion, a closer investigation of the force that began in the 1960s has demonstrated that this force is actually analogous to other forces, including electromagnetism, because its range is finite (nonzero). It's only limited because it's mediated by the W and Z bosons which are, unlike photons, massive. So the force doesn't get &quot;too far&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in our modern description of the forces, electromagnetism and the weak force have to be described by a unified &quot;electroweak&quot; theory and they mix with one another. At distances much shorter than the range of the W/Z bosons, the electromagnetic and weak forces become equally strong and, in some proper sense, indistinguishable.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-02T07:34:44.453" />
  <row Id="4441" PostTypeId="2" ParentId="4432" CreationDate="2011-02-02T07:45:19.323" Score="3" ViewCount="" Body="&lt;p&gt;&quot;Composite&quot; observables such as the angular momentum have no &quot;unique&quot; complementary observables. In fact, the three components of the angular momentum are not really &quot;independent&quot; in the sense of spanning a proper configuration space. Only two functions of the three angular momentum components - conventionally $j^2$ and $j_z$ - may be selected into a basis of mutually commuting observables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have this basis, you may talk about other observables that don't commute with them. There are many. Of course, they include the other components, e.g. $j_x, j_y$, of the angular momentum. If you want a treatment that is analogous to the treatment of the momenta and positions, of course that the angles $\theta,\phi$ are the natural dual variables to $j,m$. You may either use the basis of spherical harmonics, $Y_{lm}$, or you may choose the continuous basis of delta-functions located at particular values of $(\theta,\phi)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, for internal spin - especially the half-integer-valued spin - there is no orbital rotation so there is no $(\theta,\phi)$ basis of the Hilbert space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why there's no unique answer is that the spin is &quot;composite&quot; and the full Lagrangian can't be written as a function of the angular momentum only - and even if it could, there are many ways to do so. In particular, a radial motion away from the origin carries $\vec j=0$ but it is still nontrivial. For such motion, the parameterization via the angular momentum would go singular. You would need both the angular momentum and the ordinary one - but then the variables would be redundant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For higher-dimensional space, it becomes even more clear that the angular momentum cannot describe a proper basis of the configuration or phase space because the angular momentum has many components - $d(d-1)/2$ of them - which becomes (much) higher than the actual number of components describing the motion of a point-like particle, $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, the basic assumption that there exists &quot;the&quot; dual variable for an arbitrary observable you choose in any theory, is incorrect. Nevertheless, if you want the most accurate analogy of the relationship of the momentum- and position-basis, it's the basis of spherical harmonics and the delta-functions on the sphere and it only works for orbital angular momentum.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T07:51:19.003" LastActivityDate="2011-02-02T07:51:19.003" CommentCount="1" />
  <row Id="4442" PostTypeId="2" ParentId="4428" CreationDate="2011-02-02T09:13:37.077" Score="2" ViewCount="" Body="&lt;p&gt;I would actually emphasize the &lt;em&gt;difference&lt;/em&gt; between the forces, rather than the similarity. Although we (as theorists) like to bundle the whole shebang into a &quot;neat&quot; $U(1)\times SU(2) \times SU(3)$ gauge structure (and possible some gauge version of gravity), it doesn't mean that reality has to be that neat (e.g. chirality of electroweak, neutrino masses, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Electromagnetism is long ranged, and drops off in strength with distance.&lt;/li&gt;&#xA;&lt;li&gt;Strong force is actually also long-ranged, but gets stronger with distance! This causes the side effect that trying to separate a pair of opposite charges causes pair creation, and so we always see neutral composite particles.&lt;/li&gt;&#xA;&lt;li&gt;Weak force is intrinsically short ranged (order $1/M_W$), and primarily it does not transmit a force --- but transmutates particles. Electrons go to neutrinos, quarks mix, etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And for gravity, say whatever your favourite quantum gravity picture say it is :-)&lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-02-02T09:13:37.077" />
  <row Id="4443" PostTypeId="2" ParentId="4394" CreationDate="2011-02-02T10:02:27.467" Score="2" ViewCount="" Body="&lt;p&gt;For a &lt;em&gt;canonical&lt;/em&gt; ensemble consisting of a single hydrogen atom in infinite space, the electron will ionize sooner or later. This reduces the free energy. The actual energy will go up, but the infinite increase in entropy available from infinite space will more than compensate for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when we go over to a gas of hydrogen atoms, even a very dilute gas, things change drastically. The atoms are no longer independent of each other; if one atom ionizes, the probability the electron of another atom becomes bound to the original atom goes up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly enough, if we start off with a nonzero temperature less than the Bohr temperature, but adjust the density so that the average volume per atom is sufficiently large exponentially, the free energy is actually minimized for a &lt;em&gt;plasma&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-02-02T10:49:33.073" LastActivityDate="2011-02-02T10:49:33.073" />
  <row Id="4444" PostTypeId="1" AcceptedAnswerId="4458" CreationDate="2011-02-02T10:11:59.547" Score="1" ViewCount="130" Body="&lt;p&gt;I would be glad if someone can help me understand the argument in appendix B.1 and B.2 (page 76 to 80) of &lt;a href=&quot;http://arxiv.org/abs/hep-th/0310285&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument in B.1 supposedly helps understand how the authors in that paper got from equation 3.6 to 3.8 on page 18 of the same paper. These 3 equations form the crux of the calculation done in this paper and unfortunately I am unable to see this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the appendix B.2 they calculate certain rational functions which are completely mysterious to me. Like I can't understand what it means to say that $\frac{1}{1-x}$ is the partition function of the operator $\partial$. Similarly one can see such functions in the equations B.7, B.8 and B.10 of that paper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Curiously these polynomials had also appeared on page 6 and 7 in &lt;a href=&quot;http://arxiv.org/abs/hep-th/9908001&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; long before the above paper. I am completely unable to understand how the series of polynomials between equation 15 to 20 of this paper were gotten and what they mean. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't seen any book ever discuss the methods being used here. I would be glad to hear of some pedagogic and expository references on the background of all this. &lt;/p&gt;&#xA;" OwnerUserId="1637" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-02T10:19:24.027" LastActivityDate="2011-02-02T13:47:20.633" Title="Single trace partition function" Tags="&lt;quantum-field-theory&gt;&lt;gauge-theory&gt;" AnswerCount="1" />
  <row Id="4445" PostTypeId="1" AcceptedAnswerId="4447" CreationDate="2011-02-02T10:40:48.597" Score="8" ViewCount="282" Body="&lt;p&gt;A jar is filled with two types of balls, red and green. Red balls have radius $r_1$ and mass $m_1$, green balls have radius $r_2$ and mass $m_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If initially the balls are randomly placed throughout the jar, and we are to shake this jar, generally one type of balls will tend to gather at the top. Why does this happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are necessary conditions on $r_1$, $r_2$, $m_1$, $m_2$ such that green balls are more likely to occupy the upper part of the jar after shaking?&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-02-03T09:18:02.787" LastActivityDate="2011-02-03T09:18:02.787" Title="Shaking a jar of balls" Tags="&lt;classical-mechanics&gt;&lt;mechanics&gt;&lt;statistical-mechanics&gt;&lt;energy&gt;&lt;intuition&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="4446" PostTypeId="2" ParentId="4436" CreationDate="2011-02-02T11:02:40.023" Score="1" ViewCount="" Body="&lt;p&gt;I never heard about such an effect, but it is now 40 years ago &#xA;since I studied such things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what You tell, I can &quot;tinker&quot; the following thoughts: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;All reasoning is on the timescale of water molecule reorientation &#xA;and proton hopping (tunneling?) in water structure.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The dielectric is a very polar, protic solvent, presumably water. &#xA;(few other solvents dissolve ions, polar aprotic almost never, exept &#xA;ion pairs, but this is a different story) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The dielectric constant (DC) of protic solvents is &quot;produced&quot;  by &#xA;orientation polarisation, this is reason for the DCs being very high &#xA;compared to nonpolar substances. EG benzene about 2.23 , whereas water &#xA;has DC of 81 ! &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ions in water have a first shell of water molecules around, which &#xA;is immobilized, depending on charge density of the ion a second and &#xA;third layer are attached, with increasing orientational freedom. &#xA;Exchange of this layers to the bulk of the liquid can be very slow. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;These water shells will behave not like water, DC will be as high &#xA;as we know it from aprotic or apolar solvents. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This would explain &quot;regions&quot; of lower DC, as You stated, but &#xA;I would not call that &quot;saturation&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is some nonlinearity in a lot of dielectrics (used in nonlinear optics)&#xA;but I hesitate to call such small nonlinearities a &quot;saturation&quot;. &#xA;At best those may be the onset of saturation. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I hope there is something that might help You in Your search. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Georg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somebody doing measurement of DC in aqueous solutions over a big range &#xA;of frquencies (up to &quot;Microwave&quot;) and does not know about water structure &#xA;or the chemistry of solvatation, might interpret his results naively &#xA;as dielectric &quot;saturation&quot; of the solvate shell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW where did You read/hear about this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What You write about the origin of the term, fits well to my answer. &#xA;I am not happy with the word &quot;saturation&quot; still, but that &#xA;is of course a matter of convention. As a chemist, I look upon &#xA;such a stongly solvated ion as an chemical compound, being in &#xA;equilibrium with the bulk water, not as a &quot;saturation&quot;. &#xA;Manfred Eigen might shiver :=(&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-02-02T23:29:27.920" LastActivityDate="2011-02-02T23:29:27.920" CommentCount="2" />
  <row Id="4447" PostTypeId="2" ParentId="4445" CreationDate="2011-02-02T11:17:18.100" Score="5" ViewCount="" Body="&lt;p&gt;I know about several attemps to formulate a theory for this problem (it would seem that this is commonly called the &quot;Brazil Nut Problem&quot; in the English literature, I'm familiar with the &quot;Paranuss-Problem&quot; from the German literature), see&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Daniel C. Hong, Paul V. Quinn, Stefan Luding: &quot;The Reverse Brazil Nut Problem: Competition between Percolation and Condensation&quot; (&lt;a href=&quot;http://xxx.lanl.gov/abs/cond-mat/0010459&quot; rel=&quot;nofollow&quot;&gt;arXiv&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Troy Shinbrot und Fernando J. Muzzio: &quot;Reverse Buoyancy in Shaken Granular Beds&quot;, Phys. Rev. Lett. 81, 4365–4368 (1998)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-02-02T11:17:18.100" />
  <row Id="4448" PostTypeId="1" CreationDate="2011-02-02T11:21:28.353" Score="-7" ViewCount="91" Body="&lt;p&gt;&lt;em&gt;UPDATED: AT LEAST 5 PEOPLE SAVED FROM EXISTENTIAL CRISIS&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In classical thermodynamics, an isolated system is one where internal energy sums to zero, since energy is zero in a space with a minkowsi metric, are minkowki spaces considered isolated?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am surprised about the how difficult it is for people to understand a simple question.  As it is well known, an isolated system &lt;a href=&quot;http://en.wikipedia.org/wiki/Isolated_system&quot; rel=&quot;nofollow&quot;&gt;is a physical system that does not interact with its surroundings&lt;/a&gt;. It is also one where the internal energy of an isolated system is constant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, taken together, by definition, there is no way for an isolated system to provide any information to its surroundings, so there is no way that the isolated system could contribute to the total energy of whatever surroundings it is in, for that would require some sort interaction with the surroundings.  So it should be intuitive that in the context of its surroundings, the energy of an isolated system is zero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there can not be a change in internal energy, and the internal energy of a system is the sum of all the kinetic and potential energies, then it is a perfectly true statement to say that all the internal energy sums to zero.  A fact that is true by definition (whether people abuse the term isolated system is an altogether different issue).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the first sense that the question makes sense.  In the context of GR, if mass-energy creates curvature, and an isolated system has no mass-energy contribution to its surroundings, then it seems natural to understand that system as not creating curvature and in fact being defined as flat (e.g. having a Minkowski Metric).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This concept should not be new to anyone, and in fact has been discussed by individuals as famous as Hawking within the context of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Zero-energy_universe&quot; rel=&quot;nofollow&quot;&gt;zero energy universe&lt;/a&gt;.  Since the universe is proposed as perhaps being the only hypothetical example of a physically isolated system, this represent the second sense that the above question makes sense.  In fact, we find out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Due to quantum uncertainty energy fluctuations such as electron and its anti-particle a positron can arise spontaneously out of nothing but must disappear rapidly. The lower the energy of the bubble, the longer it can exist. &lt;strong&gt;A gravitational field has negative energy. Matter has positive energy. The two values cancel out provided the universe is completely flat.&lt;/strong&gt; In that case the universe has zero energy and can theoretically last forever.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also find this whole issue being relevant in any theory that proposes that matter and energy is in fact trapped at the event horizon of a black hole.  In the context of &lt;a href=&quot;http://scienceworld.wolfram.com/physics/BirkhoffsTheorem.html&quot; rel=&quot;nofollow&quot;&gt;Birkhoff's theorem&lt;/a&gt;, there is a corallary that &quot;the metric inside a spherical cavity inside a spherical mass distribution is the Minkowski metric&quot;.  Since the Minkowski metric is a metric of spacetime, it is related to an earlier question I asked: &lt;a href=&quot;http://physics.stackexchange.com/questions/3061/is-the-observable-universe-enclosed-by-an-infinitely-dense-shell&quot;&gt;&quot;Is the observable universe enclosed by an infinitely dense shell?&quot;&lt;/a&gt; from which it is tempting to conclude that any observer in a universe with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble_radius&quot; rel=&quot;nofollow&quot;&gt;Hubble radius&lt;/a&gt; will always see a flat space because infinite mass in a thin, nearly spherical shell must create a pretty flat space inside the observed cavity.  This is the third sense in which the question is relavent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we go back an &lt;a href=&quot;http://physics.stackexchange.com/questions/4364/does-the-positive-mass-conjecture-indicate-a-necessity-of-interactions-in-our-uni&quot;&gt;earlier question on the positive mass conjecture&lt;/a&gt; and read the paper by Witten, we find the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Although there is no satisfactory way to define the local energy density when&#xA;gravity is present, one can define the total energy of a gravitating system. The&#xA;total energy (and momentum and angular momentum) of a gravitating system can&#xA;be defined in terms of the asymptotic behavior, at large distances, of the&#xA;gravitational field. However, it is far from obvious that the total energy so defined&#xA;is always positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is an old conjecture that this total energy is in fact always strictly positive,&#xA;except for flat Minkowski space, which has zero energy. This matter has been&#xA;studied by a variety of means.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also know from Beckenstein and Hawking that black holes (objects that occur in &lt;a href=&quot;http://en.wikipedia.org/wiki/General_relativity&quot; rel=&quot;nofollow&quot;&gt;GR&lt;/a&gt;, Georg, but not really in chemistry ;-) ), are &lt;a href=&quot;http://www.scholarpedia.org/article/Bekenstein-Hawking_entropy&quot; rel=&quot;nofollow&quot;&gt;in fact thermodynamic systems&lt;/a&gt;.  So the fact that they do have a temperature, and are observable is at least of some interest to people who think about metrics and spacetime, which is the fourth sense that this question is relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I guess in summary, I don't really understand why this question is hard for anyone to understand...although if it turns out that in some very unlikely situation that somehow I rocked the world of physics so bad that I shook the house of cards to the ground, just let me know.&lt;/p&gt;&#xA;" OwnerUserId="1383" LastEditorUserId="1383" LastEditorDisplayName="" LastEditDate="2011-02-03T02:14:52.123" LastActivityDate="2011-02-03T02:14:52.123" ClosedDate="2011-02-03T08:45:23.843" Title="If energy in a flat space is zero, are we saying that a flat space is isolated?" Tags="&lt;general-physics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="4449" PostTypeId="1" AcceptedAnswerId="4455" CreationDate="2011-02-02T11:28:31.560" Score="4" ViewCount="262" Body="&lt;p&gt;I have a particle at rest. At $t = 0$ a periodic force like $F_0 \sin\omega t$ starts acting on my particle. Can such a force transport my particle to infinity when $t  \to \infty$? Please answer this question without solving the mechanical problem, just by intuition.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-02-03T00:05:28.883" LastActivityDate="2011-02-03T01:43:49.993" Title="Is a periodic force capable of transporting a particle to large distances?" Tags="&lt;classical-mechanics&gt;&lt;intuition&gt;" AnswerCount="4" CommentCount="9" FavoriteCount="0" />
  <row Id="4450" PostTypeId="2" ParentId="4445" CreationDate="2011-02-02T11:32:39.607" Score="5" ViewCount="" Body="&lt;p&gt;It turns out that larger balls tend to move up &lt;em&gt;even&lt;/em&gt; if their density is larger than the density of smaller balls. Intuitively it happens because for the small balls on the bottom of a large ball it much more difficult to move than for the ones on the top.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is no rigorous answer to your question. It is a whole area of &lt;a href=&quot;http://www.springer.com/materials/journal/10035&quot; rel=&quot;nofollow&quot;&gt;Granular Matter&lt;/a&gt; physics for one to research. Models, different simulations, experiments, e.t.c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will also note that the separation between different balls strongly depends on many different properties -- not only sizes and masses. Friction, air flows, boundaries, way of shaking can strongly influence the phenomenon, while there is no  understanding why this is happening.  &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-02-02T11:32:39.607" CommentCount="7" />
  <row Id="4451" PostTypeId="2" ParentId="4449" CreationDate="2011-02-02T11:46:12.887" Score="4" ViewCount="" Body="&lt;p&gt;Here is what my intuition says:&lt;br&gt;&#xA;$m\ddot{x} = F_0\sin\omega t, x(0)=0, \dot{x}(0)=0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{x}(t)=\frac{F_0}{m\omega}-\frac{F_0}{m\omega}\cos\omega t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x(t)=\frac{F_0}{m\omega^2}\left(\omega t - \sin\omega t\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hmmmm... Yes it can!&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-02-02T11:46:12.887" CommentCount="0" />
  <row Id="4452" PostTypeId="2" ParentId="3402" CreationDate="2011-02-02T12:41:13.027" Score="1" ViewCount="" Body="&lt;p&gt;This question has been modified into a more specific form concerning two structures: Haemoglobin and the PPN Formalism for Gravitation. So I shall make some general comments about how to consider this: a linked Supplementary question might then be the best means to progress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Haemoglobin&lt;/strong&gt; is a large (class of) molecules, which have high complexity. Approx molecular weight 68000 (where Hydrogen=1). They are in many terrestrial life forms and they are a key component of blood. There are undoubtedly many outstanding scientific questions concerning the origin and dynamics of such a large molecule. One aspect in particular is the whole issue of &quot;protein folding&quot; which would help form them efficiently and successfully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of physics there are two aspects: Fundamental physics aspects and local environment aspects. The latter are the most important in practice with e.g Temperature undoubtedly being important; although pressure (ie blood pressure) is likely a function of local gravitation too. So there is the whole question of the biomechanics of a body in differing gravitational fields (surface of Earth, surface of Moon, in Vacuum, on Jupiter, etc) to consider. As the body is the &quot;manufacturing unit&quot; for this molecule its well being is important too - and Vacuum conditions and Jupiter conditions are generally considered hostile to life as we understand it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of the fundamental physics and laws, the most important here are those of Chemistry which come from Quantum Mechanics. In a hypothetical different universe with a different electric charge for example undoubtedly different molecules would form; perhaps no molecules can form, only atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of our universe (whatever gravitational laws really apply: Newton, Einstein, etc) the local strength of the gravitational field will be an important parameter in the body formation. In terms of biochemistry it probably only has an indirect effect from its contribution to environment (air,sea) pressure and other thermodynamic variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these issues from the persective of life history and presence on other planets and environments is important research in the field of &lt;a href=&quot;http://en.wikipedia.org/wiki/Astrobiology&quot; rel=&quot;nofollow&quot;&gt;astrobiology&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;PPN formalism&lt;/strong&gt; was invented to compare Einstein's General Relativity with competing theories via delicate astronomical measurements: but GR remains the best theory in terms of these tests. The best way for a gravity theory parameter to affect quantum parameters would be if, in a theory of Quantum Gravity, it was found that electric charge,say, was a function of the Gravitation constant (which then might not be constant). So that theory might have additional solutions with unusual parameters which might or might not allow the formation of complex molecules.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-02-02T15:23:57.770" LastActivityDate="2011-02-02T15:23:57.770" CommentCount="3" />
  <row Id="4453" PostTypeId="1" AcceptedAnswerId="4461" CreationDate="2011-02-02T12:44:06.363" Score="2" ViewCount="221" Body="&lt;p&gt;I didn't have much luck getting a response to this question before so I have tried to reword and expand it a little:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In early 2010 I attended this inaugural lecture by string theorist- Prof. Mavromatos entitled 'MAGIC strings'. In it he proposes that some string theory models may violate Lorentz symmetry at the Planck scale resulting in a kind-of foamy spacetime that could be observed by differing arrival times of photons of different energies reaching us from distant astronomical sources. See &lt;a href=&quot;http://www.kcl.ac.uk/news/events_details.php?year=2010&amp;amp;event_id=2178&quot; rel=&quot;nofollow&quot;&gt;http://www.kcl.ac.uk/news/events_details.php?year=2010&amp;amp;event_id=2178&lt;/a&gt; or here for one of the papers: &lt;a href=&quot;http://iopscience.iop.org/1742-6596/174/1/012016&quot; rel=&quot;nofollow&quot;&gt;http://iopscience.iop.org/1742-6596/174/1/012016&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, in 'Cycle's of Time' that I read recently, Prof. Sir Roger Penrose mentions (page 203) that Wheeler and others have strongly argued that if we could examine spacetime at the Planck scale we would see a turbulent chaotic situation (from vacuum fluctuations of the quantum fields I suppose) or perhaps a discrete granular one. Penrose goes on to list some other approaches that may suggest how this discrete structure may manifest itself. Loosely transcribed these are: &lt;strong&gt;spin foams, casual sets, non-commutative geometry, Machian theories, twistor theory, [EDIT] loop quantum gravity, or strings and membranes&lt;/strong&gt; existing in some higher-dimensional geometry...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have studied some QM, introductory QFT and the Standard Model as well as some basic GR but I have no formal experience of string theory.  My questions are therefore:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What's involved with each of the above approaches? I.e. in what way does the spacetime become discretised? (Particularly in string theory)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are there any other popular(ish) approaches that should be added to the list?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Supplementary query, with GR being a background-independent theory, I fail to see how one can end up with discretised spacetime without it being a pre-defined background onto which a theory of the dynamics would have to be 'bolted-on'??&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Please forgive my ignorance if what I have said is misinformed, all comments and elucidations would be most welcome.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-02-02T19:22:30.310" LastActivityDate="2011-02-02T19:22:30.310" Title="How could spacetime become discretised at the Planck scale?" Tags="&lt;general-relativity&gt;&lt;string-theory&gt;&lt;quantum-gravity&gt;&lt;spacetime&gt;&lt;vacuum&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="4454" PostTypeId="2" ParentId="4448" CreationDate="2011-02-02T12:53:31.250" Score="0" ViewCount="" Body="&lt;p&gt;I think that the question will need re-wording in light of the comment. Firstly if this is a question about general relativity then even without considering thermodynamics Minkowski spaces are only &quot;toy&quot; solutions: they dont describe anything. A more sophisticated version of this question might ask about a GR solution (like Schwarzchild) which actually described something, and then ask how this solution would relate to Minkowski Space. I see that there are related Stack questions which discuss similar issues: if they dont meet your needs try again...&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-02T12:53:31.250" />
  <row Id="4455" PostTypeId="2" ParentId="4449" CreationDate="2011-02-02T13:10:27.977" Score="6" ViewCount="" Body="&lt;p&gt;Kostya has provided an excellent mathematical proof that it can. I will try to give a more intuitive explanation. Suppose the particle started with zero initial speed. Since the force is given as $F_0\sin(\omega t)$, the time for which the particle accelerates and decelerates would be the same (think about the graph of sin). Therefore for some time, the particle will gain speed. Soon the particle will decelerate and lose speed and instantaneously come to rest, before being accelerated again. Note that the in the whole cycle of acceleration and deceleration the particle doesn't change its direction. Thus the particle can be transported to any distance you want provided you wait long enough.&lt;/p&gt;&#xA;" OwnerUserId="966" LastEditorUserId="966" LastEditorDisplayName="" LastEditDate="2011-02-02T13:16:21.887" LastActivityDate="2011-02-02T13:16:21.887" CommentCount="1" />
  <row Id="4457" PostTypeId="2" ParentId="4453" CreationDate="2011-02-02T13:38:38.117" Score="0" ViewCount="" Body="&lt;p&gt;Qftme, I think that part of the problem you have with this question is the lack of a Uniform approach to these extra theories. You may have try the papers on their own; although there might be recent conference proceedings on the subject of Quantum Gravity, with a paper on each approach.  You will see that the Penrose book does have references for these theories too. For example a paper on &quot;Causal Sets as discrete spacetime&quot; is arXiv:gr-qc/0309009.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Penrose's own approach (Twistors) itself derives from an older (non-relativistic) model of his called &quot;Spin Networks&quot;. In the Penrose cases it is not (just) that spacetime is discrete but that spacetime is derived from another structure. In the &quot;Spin Network&quot; model this was a combinatoric construction, hence discrete. In Twistors it is a space called, appropriately enough &quot;Twistor Space&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure whether Loop Quantum Gravity is on your list explicitly. There have been recent Stack discussions about this with emphasis on whether it is Lorentz invariant. Without wanting to go into the details of that discussion, just to say that a spacetime &quot;box&quot; of a certain size say $\hbar$ by $\hbar$ by $\hbar$ by $\hbar$ - a discretization of existing spacetime - will have a different size in a different lorentz frame. So is this simple approach consistent with Special Relativity? Different answers to that yield different approaches too.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-02T13:38:38.117" CommentCount="1" />
  <row Id="4458" PostTypeId="2" ParentId="4444" CreationDate="2011-02-02T13:47:20.633" Score="2" ViewCount="" Body="&lt;p&gt;Their convention for the partition function is explained in the second paragraph before the equation (B.7) of the paper by Shiraz et al.  It is&#xA;$$z(x) = \sum_{operators} x^{\Delta_{operator}}$$&#xA;Note that this is just a different way of writing the usual $\mbox{Tr }\exp(-\beta H)$ if you identify $\exp(-\beta)\equiv x$ and $\Delta\equiv H$. Yes, the dimension is the same thing as the Hamiltonian (of the radial quantization) and it is often helpful to avoid exponentials and write powers of $x$ only, so therefore the exponential redefinition of $\beta$ vs $x$. The trace is the summation over the basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They're calculating the partition function of a whole theory, not the $\partial$ operator itself. So the partition function is the sum over operators, as described above. In this simple case, the operators are $\partial_i \partial_j \dots \phi$, i.e. arbitrary derivatives of $\phi$ by $d$ different partial derivative symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivatives with respect to different directions commute with each other and are completely independent. So imagine $d=1$ for a while, only one direction. In that case, you have operators &#xA;$$\phi, \partial \phi, \partial^2 \phi, \dots$$&#xA;and their dimensions are&#xA;$$\Delta=0,1,2,\dots$$&#xA;plus the dimension of $\phi$ if it were nonzero. The partition sum is the sum of $x^\Delta$ over these operators which means&#xA;$$1+x+x^2+\dots = \frac{1}{1-x}.$$&#xA;The sum is obtained as geometric series. Note that the coefficients of the Taylor expansion are simply equal to one: there is no source where you could have gotten something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the operators in the $d$-dimensional space may be obtained by acting with some derivatives in the 1st direction; some in 2nd, and so forth, on $\phi$. So the space of operators is a tensor product of spaces from each of the $d$ directions, and the partition sum is therefore the product of the partition sums from the individual directions, i.e. the $d$-th power of $1/(1-x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other factors multiplying the total partition function but you haven't asked about it, and I can't explain every detail in a 50-page paper you haven't asked about. But yes, the other paper you mentioned almost certainly uses the same basic insight about the geometric series.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-02T13:47:20.633" CommentCount="2" />
  <row Id="4459" PostTypeId="1" AcceptedAnswerId="4468" CreationDate="2011-02-02T13:52:29.470" Score="2" ViewCount="278" Body="&lt;p&gt;Given a collection of point-particles, interacting through an attractive force $\sim \frac{1}{r^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing only $m_1a=\sum_i \frac{Gm_1m_i}{r^2}$ and initial conditions we can deduce the motion of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently we can observe that three quantities remains constant&#xA;A) center of mass of the system&#xA;B) total energy&#xA;C) angular momentum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we derive these 3 facts directly from $m_1a=\frac{Gm_1m_2}{r^2}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these quantities conserved for any attractive force $\sim\frac{1}{r^n}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given any monotonically decreasing force for $r$ in $(0,\infty)$, which are the conserved quantities?&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="673" LastEditorDisplayName="" LastEditDate="2011-02-02T23:01:08.170" LastActivityDate="2011-02-27T22:45:51.360" Title="Conserved quantities in generalized n-body problem" Tags="&lt;classical-mechanics&gt;&lt;gravity&gt;&lt;homework&gt;&lt;newtonian-mechanics&gt;&lt;many-body&gt;" AnswerCount="6" CommentCount="4" FavoriteCount="1" />
  <row Id="4460" PostTypeId="1" CreationDate="2011-02-02T13:59:55.860" Score="2" ViewCount="163" Body="&lt;p&gt;Knowing &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;* earth spin and translation movement around sun,&#xA;* sun have a big magnetic field,&#xA;* a variable flux can generate a induced voltage (and current) into the circuit&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Would it be posible to receive in a coil that energy? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps it's just enough to detecting the earth movement relative to magnetic field,&#xA;or if it were strong enough even for use the energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't seem to be so strong because if it where perhaps every coil in earth without a current generated by this flux variation.&lt;/p&gt;&#xA;" LastEditorUserId="1370" LastEditorDisplayName="Hernan Eche" LastEditDate="2011-08-09T09:11:18.243" LastActivityDate="2011-08-09T09:11:18.243" Title="Using energy from sun magnetic field" Tags="&lt;electromagnetism&gt;&lt;energy&gt;&lt;sun&gt;" AnswerCount="1" />
  <row Id="4461" PostTypeId="2" ParentId="4453" CreationDate="2011-02-02T14:02:30.007" Score="1" ViewCount="" Body="&lt;p&gt;Let me try to address your questions, even though just the first one seems quite heavy in itself.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Spacetime Discreteness: let me give you links to references that are relevant to your questions, than i'll make some general comments. &lt;a href=&quot;http://arxiv.org/abs/gr-qc/9905087&quot; rel=&quot;nofollow&quot;&gt;An Introduction to Spin Foam Models of Quantum Gravity and BF Theory&lt;/a&gt;; &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0410049&quot; rel=&quot;nofollow&quot;&gt;Spacetime in String Theory&lt;/a&gt;; &lt;a href=&quot;http://arxiv.org/abs/hep-th/0303037&quot; rel=&quot;nofollow&quot;&gt;The quantum structure of spacetime at the Planck scale and quantum fields&lt;/a&gt;; &lt;a href=&quot;http://arxiv.org/abs/hep-th/0006166&quot; rel=&quot;nofollow&quot;&gt;Meaning of Noncommutative Geometry and the Planck-Scale Quantum Group&lt;/a&gt;; &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0309009&quot; rel=&quot;nofollow&quot;&gt;Causal Sets: Discrete Gravity (Notes for the Valdivia Summer School)&lt;/a&gt;; &lt;a href=&quot;http://users.ox.ac.uk/~tweb/00001/&quot; rel=&quot;nofollow&quot;&gt;On the Origins of Twistor Theory&lt;/a&gt; — this should get you going. As for string theory and spacetime discreteness, let me say that, in a crude way, the $\alpha$ that appear in the Action in this link &lt;a href=&quot;http://en.wikipedia.org/wiki/Superstring#The_mathematics&quot; rel=&quot;nofollow&quot;&gt;Superstring theory&lt;/a&gt;, called 'string tension', is basically what 'measures' this.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Other approaches not listed: i think your list is fairly complete. But, you didn't list &lt;a href=&quot;http://en.wikipedia.org/wiki/Loop_quantum_gravity&quot; rel=&quot;nofollow&quot;&gt;Loop quantum gravity&lt;/a&gt; — maybe you were thinking of it, or maybe it fits in one of your named categories: i just thought i'd make it explicit.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;GR discreteness: to me, this is a subtler question, in the sense that once you have discretized spacetime (for one reason or another), you should not expect that the other [geometric] structures remain 'continuous' — in fact, there's a whole branch of research dealing with 'quantum groups' and 'discretized' (or 'latticized': think computer simulations) theories. The point being that if you discretized all of your ingredientes, you still maintain a certain relation among them (e.g., discrete gauge symmetry, or $q$-gauge symmetry). The bottom line is that you can perfectly define a theory where all ingredients are properly 'discretized', and so it maintains its relevant features (recovering the continuum theory in some limit). As a side note, it's worth seeing that it's possible to discretize theories at the level of differential forms, à la &lt;a href=&quot;http://ccom.ucsd.edu/~astern/pdfs/cambridge20081121.pdf&quot; rel=&quot;nofollow&quot;&gt;Discrete Differential Forms, Gauge Theories, and Regge Calculus (PDF)&lt;/a&gt; (and similar constructions by several other folks). In this sense, many of the relevant properties are kept even after discretization (quite robust method).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I hope this can get this discussion started.&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2011-02-02T14:02:30.007" CommentCount="2" />
  <row Id="4462" PostTypeId="2" ParentId="4459" CreationDate="2011-02-02T14:08:47.047" Score="1" ViewCount="" Body="&lt;p&gt;*How can we derive these 3 facts directly from m1*a=G*m1*m2/r^2 ?*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of them!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First two are conserved in any motion in classic mechanic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One does not need any version of a centric potential. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third fact needs some &quot;circular&quot; motion of, course, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but as any top or a mass circling some other by means of a rope show,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the square law is not necessary (or unique) at least. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Georg&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-02-02T14:08:47.047" CommentCount="10" />
  <row Id="4463" PostTypeId="2" ParentId="4459" CreationDate="2011-02-02T14:15:37.683" Score="3" ViewCount="" Body="&lt;p&gt;These quantities will be conserved by any system whose laws respect translational and rotational symmetry and are static in time. Respecting translational symmetry means that interactions don't depend on the positions of individual particles, just on differences of those positions. Respecting rotational invariance in turn means that interactions depend only on the magnitude of those differences. Being static means that the interactions don't depend explicitly on time or velocities. So this means that the two-point interaction must look like $V({\mathbf x}, {\mathbf y}) = f(|{\mathbf x - \mathbf y}|)$ (with $f$ suitably smooth). So yes, in particular your $r^{-n}$ potential will work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in general you could also consider more general interactions e.g. three- or more-point interactions. But to conserve the said quantities the interactions must still be singlet states of the respective groups.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-02-02T14:15:37.683" />
  <row Id="4464" PostTypeId="2" ParentId="4094" CreationDate="2011-02-02T14:17:44.950" Score="1" ViewCount="" Body="&lt;p&gt;To answer this question one should first define the Planck scale operatively, i.e. to define how different Lorentz observers would measure it in (perhaps thought) experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually one defines &lt;em&gt;coupling&lt;/em&gt; constants in terms of local, (quasi) static experiments that each observer can perform in its rest frame. Thus G, for example, is measured in such experiment and therefore is, by definition, Lorentz invariant, in spite of being a dimensionful quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c on the other hand clearly cannot be measured that way, but it is not a coupling constant, but a dimensionful observer-independent scale used to define what Lorentz transformations are (Lorentz transformations are such linear spacetime transformations that allow for the presence of an invariant velocity scale c).&lt;/p&gt;&#xA;" OwnerUserId="1721" LastActivityDate="2011-02-02T14:17:44.950" CommentCount="1" />
  <row Id="4465" PostTypeId="2" ParentId="4460" CreationDate="2011-02-02T14:22:00.943" Score="1" ViewCount="" Body="&lt;p&gt;In case of a so called &quot;magnetic storm&quot; you could collect some energy. Every time such a storm hits the earth, the magnetic field of earth is a bit compressed and then released. On such occasions in electrical mains systems in Canada (the big high voltage network loops) voltages are induced, fuses blow and so on (Canada is rather close to the magnetic pole). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might happen in Alaska too, but are there such mains systems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Georg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a wiki on the topic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Geomagnetic_storm&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Geomagnetic_storm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Omega for the &quot;reminder&quot;&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="869" LastEditorDisplayName="" LastEditDate="2011-03-04T20:19:35.020" LastActivityDate="2011-03-04T20:19:35.020" CommentCount="2" />
  <row Id="4466" PostTypeId="2" ParentId="3432" CreationDate="2011-02-02T14:45:08.567" Score="4" ViewCount="" Body="&lt;h1&gt;&lt;a href=&quot;http://arsphysica.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Ars Physica&lt;/a&gt;&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Although mostly in brazilian portuguese (pt_BR — even though in the days of &lt;a href=&quot;http://translate.google.com/&quot; rel=&quot;nofollow&quot;&gt;Google Translate&lt;/a&gt; i don't know how big a problem this actually represents ;-), it's a collective effort including high energy theorists and experimentalists, condensed matter theorists, and neuroscientists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the topic selection is quite varied, and so is the level of presentation: from popularization pieces all the way to full fledged research.&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2011-02-02T14:45:08.567" CommunityOwnedDate="2011-02-02T14:45:08.567" />
  <row Id="4467" PostTypeId="1" CreationDate="2011-02-02T15:11:37.873" Score="0" ViewCount="237" Body="&lt;p&gt;I think I am over thinking this question: A point charge of -2 microC is located in the center of a hollow sphere. The internal and external radius of the sphere is given by 6.5 and 9.5 cm, the charge density is 7.35 E(-4). Plot electric field vs. distance from the center of the sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thoughts were the region before r1 it would just be density*area, for r1 to r2 the value would be zero, and for a radius greater than r2 out of the sphere it would have an inverse 1/r^2 curve. Am I horribly off? Any help would be greatly appreciated, thanks&lt;/p&gt;&#xA;" LastEditorUserId="701" LastEditorDisplayName="" LastEditDate="2011-02-02T15:19:35.857" LastActivityDate="2011-02-02T15:19:35.857" Title="Electric Field for a hollow sphere with point charge" Tags="&lt;homework&gt;&lt;electrostatics&gt;" CommentCount="4" />
  <row Id="4468" PostTypeId="2" ParentId="4459" CreationDate="2011-02-02T15:17:46.260" Score="7" ViewCount="" Body="&lt;p&gt;Actually, this is the basic stuff every mechanics textbook should have.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Center of mass is the most basic it needs just Newton laws:&lt;br&gt;&#xA;Second: $m_i\ddot{\vec{r}_i}= \vec{F}_i$&lt;br&gt;&#xA;And third: $\sum_i\vec{F}_i = 0$&lt;br&gt;&#xA;Summing over i one obtains:&lt;br&gt;&#xA;$\frac{d^2}{dt^2}\left(\sum_i m_ir_i\right) = \sum_i\vec{F}_i = 0$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For the total energy you need those forces to be potential and independent on time:&lt;br&gt;&#xA;$F_i = -\frac{d}{d\vec{r}_i}U(\vec{r}_1,\vec{r}_2,...)$&lt;br&gt;&#xA;Then just take the total energy :&#xA;$E = \sum_i \frac{m|\vec{r}_i|^2}{2}+U$&lt;br&gt;&#xA;And differentiate with respect to time:&lt;br&gt;&#xA;$\frac{dE}{dt} = \sum_i m_i(\ddot{\vec{r}}_i\dot{\vec{r}}_i)+\sum_{i} \frac{dU}{d\vec{ r}_i}\dot{\vec{r}}_i = 0$    &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, for the angular momentum $U$ must be rotational invariant:&lt;br&gt;&#xA;$U(R\vec{r}_1,R\vec{r}_2,R\vec{r}_3,...) = U(\vec{r}_1,\vec{r}_2,\vec{r}_3,...)$&lt;br&gt;&#xA;where R is the rotation matrix. Consider now very small (infinitesimal) rotation:&lt;br&gt;&#xA;$R\vec{r}_i = \vec{r}_i+\left[\delta\vec\phi\times\dot{\vec{r}}_i\right]$&lt;br&gt;&#xA;Substituting and expanding, one can get:&lt;br&gt;&#xA;$U+\delta\vec\phi\sum_i\left[\frac{dU}{d\vec{r}_i}\times\vec{r}_i\right]=U$&lt;br&gt;&#xA;Which works for every angle $\delta\phi$, so the following must hold&#xA;$\sum_i\left[\frac{dU}{d\vec{r}_i}\times\vec{r}_i\right]=0$ &#xA;&amp;nbsp;&lt;br&gt;&#xA;Finally take the angular momentum:&lt;br&gt;&#xA;$\vec{M} = \sum_i m_i \left[\vec{r}_i\times\dot{\vec{r}}_i\right]$&lt;br&gt;&#xA;And differentiate with respect to time:&lt;br&gt;&#xA;$\frac{d\vec{M}}{dt} = \sum_i  \left[\vec{r}_i\times m_i\ddot{\vec{r}}_i\right]+&#xD;&#xA;\sum_i  m_i \left[\dot{\vec{r}}_i\times \dot{\vec{r}}_i\right] = \sum_i\left[\frac{dU}{d\vec{r}_i}\times\vec{r}_i\right]=0$  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Phew... You can see that I never used the &quot;pairwise&quot; interactions. And I only needed the basic assumptions about the potential energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning &quot;extra&quot; things that you cans say for your systems -- check &lt;a href=&quot;http://en.wikipedia.org/wiki/Virial_theorem&quot; rel=&quot;nofollow&quot;&gt;the Virial theorem&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-02-02T15:17:46.260" />
  <row Id="4469" PostTypeId="2" ParentId="4459" CreationDate="2011-02-02T15:37:16.293" Score="-3" ViewCount="" Body="&lt;p&gt;There are many more conserved quantities in a potential N-body problem but not all of them are &lt;strong&gt;additive&lt;/strong&gt; on particles. This fact is not related to the form of potentials at all.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-27T22:45:51.360" LastActivityDate="2011-02-27T22:45:51.360" CommentCount="4" />
  <row Id="4470" PostTypeId="1" AcceptedAnswerId="4472" CreationDate="2011-02-02T15:48:06.693" Score="4" ViewCount="226" Body="&lt;p&gt;Could someone explain the correspondence between lines in twistor space and minkowski space-time points? a basic derivation would suffice&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-02-07T15:41:11.590" Title="twistor-spacetime correspondence" Tags="&lt;twistor&gt;&lt;minkowski-space&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="4471" PostTypeId="1" AcceptedAnswerId="4488" CreationDate="2011-02-02T15:53:29.570" Score="4" ViewCount="616" Body="&lt;p&gt;What would happen if instead of $F=m*d^2x/dt^2$, we had $F=m*d^3x/dt^3$ or higher? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, I have always seen a justification for ~1/r^2 forces as the &quot;forces beeing divided equally over the area of a sphere of radius r&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why the 2 in $F=m*d^nx/dt^n$ ?&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-02T16:52:49.730" LastActivityDate="2011-02-03T23:08:59.787" Title="Why F=ma and not F=ma'?" Tags="&lt;classical-mechanics&gt;&lt;intuition&gt;" AnswerCount="6" CommentCount="3" />
  <row Id="4472" PostTypeId="2" ParentId="4470" CreationDate="2011-02-02T16:04:56.360" Score="6" ViewCount="" Body="&lt;p&gt;The ordinary twistor space is parameterized by $(\lambda^\alpha,\mu_{\dot\alpha})$. Here, the $\alpha$ is a 2-valued $SL(2,C)$ spinor index of one chirality and the dotted index is its complex conjugate, the index of the opposite chirality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the level of spinors, vectors are equivalent to &quot;spintensors&quot; with one undotted and one dotted index.&#xA;$$ V_\mu = \sigma_\mu^{\alpha \dot\alpha} V_{\alpha \dot \alpha}.$$&#xA;This is a basic fact about the Lie algebras. $SO(3,1)$ is locally isomorphic to $SL(2,C)$ - they're the same 6-dimensional Lie groups - and the 4-vector is the tensor product of ${\bf 2}$ and ${\bf \overline{2}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're unfamiliar with this equivalence of vectors of &quot;spintensors&quot; with two indices, notice that the components of a 4-vector may be organized as&#xA;$$v^{\alpha\dot\alpha} = \left( \begin{array}{cc} v^0+v^3 &amp;amp; v^1-i v^2\\ v^1+i v^2 &amp;amp; v^0-v^3 \end{array} \right) $$&#xA;Note that the determinant of this matrix - a natural function of the matrix elements - is simply $v^\mu v_\mu$. The whole matrix may be understood as the appropriate combination of the three Pauli matrices - plus the &quot;time-like Pauli&quot; (identity) matrix multiplied by the time component of the vector. Sorry if my signs deviate from the prevailing convention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, it has only been a story about the vectors or spinors. What about the twistors? Well, it's a simple one-line formula. If I have a point $x^\mu$ which is equivalent to the $x^{\alpha \dot\alpha}$ matrix - as any 4-vector - I may simply write an equation&#xA;$$ \lambda^\alpha = x^{\alpha \dot\alpha} \mu_{\dot\alpha} $$&#xA;Note that it is a set of two complex linear equations - for $\alpha=0,1$ - so it defines a linear object. For a different value of $x^\mu$, I get different equations. Moreover, the equations link the $\lambda$ and $\mu$ objects which are coordinates on the twistor space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I have to explain why the equations above define a line. First, the $\lambda$ and $\mu$ objects are pairs of complex numbers, so in total, we have four complex coordinates. However, the twistor space is a projective space. Note that if the equations above are satisfied for some $\lambda$ and $\mu$ - four complex numbers - they will also be satisfied if you multiply both $\lambda$ and $\mu$ by an arbitrary complex number (the same one for both). This projectivity holds universally: $\mu$ is naturally scaled in the same way as $\lambda$ because it may be understood as a dimensionally inverse object to $\lambda^{\dot\alpha}$ (an object that appeared for the first time in this answer) that scales in the inverse way relatively to $\lambda^\alpha$ in order to keep e.g. the vector $\lambda^\alpha \lambda^{\dot\alpha}$ constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the twistor space is really a complex projective space, $CP^3$, and the equations above are two complex conditions, so we're left with a one(-complex)-dimensional object, a complex line. Spacetime points are in one-to-one correspondence to complex lines in the twistor space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many more things may be translated. For example, if two spacetime points are separated by a null interval, the corresponding two lines in the twistor space intersect. The intersection  - a point in the twistor space - may be identified with a null line in the Minkowski space, and I could derive many other things of this kind.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-02T16:04:56.360" CommentCount="2" />
  <row Id="4473" PostTypeId="2" ParentId="4471" CreationDate="2011-02-02T16:15:52.700" Score="3" ViewCount="" Body="&lt;p&gt;It is because the evolution of mechanical system is fully determined by initial coordinates and speeds. Therefore, your equation must be of second order, otherwise setting initial accelerations and &quot;speeds of increase of accelerations&quot; would be necessary.  &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-02-02T16:15:52.700" CommentCount="3" />
  <row Id="4474" PostTypeId="1" CreationDate="2011-02-02T16:19:17.113" Score="3" ViewCount="272" Body="&lt;p&gt;In Lorentzian AdS space there are both normalizable and non normalizable solutions and we also know (at least for scalar fields in bulk) what do they correspond to in the boundary. But I saw the calculation only for scalar fields. Can someone please give me a reference where people have calculated these modes for a gauge fields, say for a graviton field? McGreevy's lecture note says the relation $\Delta(\Delta-D)=m^2L^2$ gets modified to $(\Delta+j)(\Delta+j-D)=m^2L^2$ for form $j$ fields. Does this mean for other fields too the normalizable and non normalizable behavior remains the same: namely $Z_0^{\Delta_+}$ and $Z_0^{\Delta_-}$, as $z_0\rightarrow 0$ ($\Delta_{\pm}$ are two solutions of course)? How can that be?&lt;/p&gt;&#xA;" OwnerUserId="1349" LastActivityDate="2011-03-02T04:09:06.077" Title="Normalizable and non normalizable modes of gauge fields in AdS/CFT" Tags="&lt;string-theory&gt;" AnswerCount="2" />
  <row Id="4475" PostTypeId="2" ParentId="4471" CreationDate="2011-02-02T16:32:17.423" Score="6" ViewCount="" Body="&lt;p&gt;Because the second derivative is fully determined by the external force, it is an experimental fact. The force can be directly measured with a dynamo-meter. It is not an abstraction. So the acceleration is known as soon as the force is known and vice versa. The trajectory depends also on the initial conditions which are independent from the force but reference frame-dependent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In CED there is an equation (Lorentz-Abraham one) with a third derivative in time. It has non physical (runaway) solutions.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-03T15:30:20.103" LastActivityDate="2011-02-03T15:30:20.103" CommentCount="5" />
  <row Id="4476" PostTypeId="2" ParentId="4474" CreationDate="2011-02-02T16:32:21.423" Score="3" ViewCount="" Body="&lt;p&gt;At the end of Sec 3.3.1 of the MAGOO review (hep-th/9905111) you will find a useful list of the relationship between conformal dimension $\Delta$ and masses $m$ for scalars, spinors, vectors, p-forms, first order (d/2)-forms, spin $3/2$ and massless spin $2$ fields along with a list of references to the literature where these various cases were analyzed. &lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T04:09:06.077" LastActivityDate="2011-03-02T04:09:06.077" CommentCount="2" />
  <row Id="4477" PostTypeId="2" ParentId="4449" CreationDate="2011-02-02T16:39:10.870" Score="1" ViewCount="" Body="&lt;p&gt;Can and must. The momentum at any given time is&#xA;$$&#xD;&#xA;p(t)=\int_0^t F(t&amp;#39;)dt&amp;#39;={F_0\over\omega}(1-\cos\omega t),&#xD;&#xA;$$&#xA;which oscillates back and forth between zero and $F_0/\omega$ and averages to a positive value over any whole number of periods. &lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-02T16:39:10.870" CommentCount="2" />
  <row Id="4478" PostTypeId="2" ParentId="4470" CreationDate="2011-02-02T16:48:07.603" Score="2" ViewCount="" Body="&lt;p&gt;Given Luboš answer, a good place to learn about this stuff is straight from the horse's mouth:&lt;a href=&quot;http://books.google.com/books?id=BOR6cgAACAAJ&quot; rel=&quot;nofollow&quot;&gt;Spinors and Space-Time: Volume 1, Two-Spinor Calculus and Relativistic Fields&lt;/a&gt; and &lt;a href=&quot;http://books.google.com/books?id=f0mgGmtx0GEC&quot; rel=&quot;nofollow&quot;&gt;Spinors and space-time: Spinor and twistor methods in space-time geometry&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2011-02-02T16:48:07.603" CommentCount="2" />
  <row Id="4479" PostTypeId="1" AcceptedAnswerId="4480" CreationDate="2011-02-02T17:15:30.433" Score="3" ViewCount="217" Body="&lt;p&gt;Given: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Wave_plate&quot; rel=&quot;nofollow&quot;&gt;half wave plate&lt;/a&gt; freely floating in space. &lt;/li&gt;&#xA;&lt;li&gt;Circularly polarized light, falling perpendicularly to it.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The plate changes polarisation of the beam to the opposite one. Therefore it receives angular momentum and starts to rotate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does energy comes from?&lt;br&gt;&#xA;What would happen with a single photon passing through the plate?  &lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T17:53:13.370" LastActivityDate="2011-02-06T23:11:30.203" Title="Half wave plate and angular momentum" Tags="&lt;optics&gt;&lt;angular-momentum&gt;&lt;energy-conservation&gt;&lt;polarization&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="4480" PostTypeId="2" ParentId="4479" CreationDate="2011-02-02T17:33:33.380" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Summary: the energy change is negligible and if it is not, the energy difference comes from a frequency change of the photon.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One must realize that unless the plate was already quickly rotating before the experiment, the energy stored in the rotation of the plate at the end is negligible relatively to the energy of the photon for the same reason why the electron carries most of the kinetic energy in the hydrogen atom even though both electron and proton are orbiting around the shared center-of-mass. In the latter case, the energy of a particle goes like $p^2/2m$, so for a fixed value of $p^2$ - and indeed, the proton's and electron's $p$ only differ by a sign - the lighter particle carries much more kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is totally true for the rotational motion as well. The kinetic energy of rotation is $J^2/2I$ where $J$ is the angular momentum and $I$ is the moment of inertia. For a single photon, the angular momentum relative to the direction of the photon's motion, $J_1=-\hbar$, is changed to $J_2=+\hbar$. The overall change is $J=+2\hbar$ which becomes the angular momentum of the plate. When you square it, you clearly get a negligible number - that is divided by an $O(1)$ value of the moment of inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the energy obtained in the form of the increased rotation, caused by a single photon's change of the polarization, is tiny. If you study where this small amount comes from, you will find out that the photon's frequency is decreased by a tiny fraction so that its reduced energy exactly compensates the increase of the kinetic energy of the plate. But we can only see this change as nonzero if we allow the plate to have an arbitrary angular frequency before the experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not too difficult to explicitly check that this statement works: if $J$ jumps by $2\hbar$, the energy $J^2/2I$ jumps by $J\times \Delta J/I=2J\hbar/I$. Because $E=\hbar\omega_\gamma$ for photons, the energy conservation law requires that the frequency of the photon decreases by $\delta \omega = -2J/I$ where $J$ is the plate's angular momentum before it changed the photon's polarization. But indeed, $-2J/I=-2\omega$ where $\omega$ is the angular frequency of the plate before the experiment. That's exactly the frequency change that the photon experiences. Why? Because the periodicity of the photon is conserved if measured relatively to the rotating plate - so its angular frequency changes from $|-\omega_\gamma|$ to $|+\omega_\gamma|$. But relatively to the inertial system, it's changed from $|-\omega_\gamma-\omega|$ to $|+\omega_\gamma-\omega|$ i.e. it drops by $2\omega$. It's just like comparing sidereal days and solar days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would happen with a single photon? I was working with a single-photon case from the beginning because it's a reasonable approach. A large electromagnetic wave may always be represented as a coherent state of many photons, so one just multiplies the results for a single photon. Of course, for a single photon, the &quot;electromagnetic wave&quot; must actually be represented as a wave function determining the probability amplitudes. At any rate, it is true that such wave plates may guarantee that every single photon that enters with a particular polarization leaves with another polarization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you measure the &quot;exact&quot; polarizations of the final photon, you will get the right one with probability of 100 percent. If you measure a different one, you may get odds between 0 and 100 percent. Every photon will be ultimately detected at a single point; the wave function knows about the odds.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-02T19:42:40.660" LastActivityDate="2011-02-02T19:42:40.660" />
  <row Id="4481" PostTypeId="1" AcceptedAnswerId="4482" CreationDate="2011-02-02T17:39:21.797" Score="3" ViewCount="207" Body="&lt;p&gt;Rb83 is unstable and &lt;a href=&quot;http://atom.kaeri.re.kr/cgi-bin/nuclide?nuc=Rb-83&quot; rel=&quot;nofollow&quot;&gt;decays to Kr-83&lt;/a&gt;. Mode of  the decay is &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_capture&quot; rel=&quot;nofollow&quot;&gt;electron capture&lt;/a&gt;. Rb85 is &lt;a href=&quot;http://atom.kaeri.re.kr/cgi-bin/nuclide?nuc=Rb85&quot; rel=&quot;nofollow&quot;&gt;stable&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nuclei Rb83 and Rb85 have the same charge. Rb85 is heavier than Rb85, but gravitation is too weak to take into account. So why orbital electron falls into the nucleus of Rb83? What is the reason? &lt;/p&gt;&#xA;" OwnerUserId="238" LastActivityDate="2011-02-02T19:48:45.800" Title="Why doesn't orbital electron fall into the nucleus of Rb85, but falls into the nucleus of Rb83?" Tags="&lt;nuclear-physics&gt;&lt;electron&gt;&lt;radiation&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="4482" PostTypeId="2" ParentId="4481" CreationDate="2011-02-02T17:44:56.423" Score="3" ViewCount="" Body="&lt;p&gt;It is not a matter of &quot;falling in&quot;: all s orbitals have non-trivial probability densities at the center. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &lt;em&gt;is&lt;/em&gt; about energy balance in the nucleus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kr-83 is a lower energy configuration than Rb-83 by enough to make up for the neutrino and the gamma(s). Evidently Kr-85 is not a sufficiently lower energy state than Rb-85.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-02-02T19:48:45.800" LastActivityDate="2011-02-02T19:48:45.800" CommentCount="2" />
  <row Id="4483" PostTypeId="2" ParentId="4411" CreationDate="2011-02-02T18:03:47.727" Score="0" ViewCount="" Body="&lt;p&gt;It might also help to have a sense of the physical meaning of each energy condition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Trace EC&lt;/strong&gt;:  Density must at least equal the sum of the principal pressures&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Strong EC&lt;/strong&gt;: Matter must gravitate toward matter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dominant EC&lt;/strong&gt;:  Energy must not flow faster than light&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Weak EC&lt;/strong&gt;: Local mass-energy density must not be negative&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Null EC&lt;/strong&gt;: Stress-energy experienced by a light ray must not be negative&lt;/p&gt;&#xA;" OwnerUserId="1726" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-02-06T12:34:22.723" LastActivityDate="2011-02-06T12:34:22.723" />
  <row Id="4484" PostTypeId="2" ParentId="4471" CreationDate="2011-02-02T18:09:01.057" Score="6" ViewCount="" Body="&lt;p&gt;Rephrasing in a Lagrangian manner, this is equivalent to asking about higher-derivative theories. See: &lt;a href=&quot;http://physics.stackexchange.com/questions/4102/why-are-only-derivatives-to-the-first-order-relevant&quot;&gt;Why are only derivatives to the first order relevant?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-02-02T18:09:01.057" />
  <row Id="4485" PostTypeId="2" ParentId="4479" CreationDate="2011-02-02T18:43:32.917" Score="3" ViewCount="" Body="&lt;p&gt;Lubos is right: the (tiny) energy comes from the frequency shift of the final photon. That's one of the forms of the &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v78/i13/p2539_1&quot; rel=&quot;nofollow&quot;&gt;rotational Doppler effect&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="59" LastActivityDate="2011-02-02T18:43:32.917" />
  <row Id="4486" PostTypeId="2" ParentId="4471" CreationDate="2011-02-02T18:43:57.393" Score="3" ViewCount="" Body="&lt;p&gt;Another reason is that if n did not equal 2, some of the symmetries of the Newton equation would be lost. For example, classically, physics on a microscopic scale is time reversal invariant. We can see this from the Newton equation because if x(t) is sent to x(-t), the 2nd time derivative ensures that the negatives cancel. If n was an odd number, we would not observe this symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If n was smaller than 2, then Galilean transformations would not leave the equation invariant. If we send x(t) to x(t) + vt, then the second time derivative kills the vt term on the LHS. If n=1, then this would not be possible, and relative velocity would be meaningless (even non-relativistically). If n was some number larger than 2, then transformations that send x(t) to x(t) + b(t^m), where m is less than n, would be a symmetry of the Newton equation. But relative accelerations, jerks, etc. produce observable discrepancies, so that should not be possible either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, this may not really answer &quot;why&quot; in the sense that you are asking, but the fact that the equation matches observations is enough in science to justify it. The Newton equation is basically experimental fact, like Vladimir says. &lt;/p&gt;&#xA;" OwnerUserId="1635" LastActivityDate="2011-02-02T18:43:57.393" />
  <row Id="4487" PostTypeId="2" ParentId="4459" CreationDate="2011-02-02T19:20:07.800" Score="3" ViewCount="" Body="&lt;p&gt;The total momentum is also conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noether's theorem relates conserved quantities to continuous symmetries of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the inverse square law, there's also a less known symmetry&#xA;$x \to e^\lambda x$, $t \to e^{2\lambda} t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we shouldn't expect any more because of chaos; this isn't an integrable system.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-02T19:20:07.800" />
  <row Id="4488" PostTypeId="2" ParentId="4471" CreationDate="2011-02-02T19:23:12.587" Score="4" ViewCount="" Body="&lt;p&gt;In statics, you can still have a force without acceleration so $F$ is independent of $a$. $F$ is the cause of the change in the position of an object initially at rest in some frame. To give it physical meaning, you have to define how it's to be measured and one way would be to define 1 unit of F causing one unit of compression in some standard spring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $F$ causes a body at rest to change its position, then over a time dt the postion has changed by dx. Your job as a physicist is to construct an equation relating F to the change in velocity of the body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with all this in mind, what would happen if $F=m*d^3x/dt^3$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would mean that even though $F$ is the cause behind the change in velocity of a body, there are some changes in the velocity possible where $F = 0$ such as for $a = const$. You would end up with particles accelerating in arbitary directions for $F = 0$.&lt;/p&gt;&#xA;" OwnerUserId="1345" LastActivityDate="2011-02-02T19:23:12.587" CommentCount="2" />
  <row Id="4489" PostTypeId="2" ParentId="4047" CreationDate="2011-02-02T19:31:52.610" Score="0" ViewCount="" Body="&lt;p&gt;Thinking about your question has led me to the following conclusion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there is a a conserved quantity $Y$, and an isolated system $S$. Let's also suppose the isolation is temporarily lifted to measure an observable $\widehat{X}$ of $S$. This measurement is performed by a measuring apparatus $A$. For simplicity, let's lump together the rest of the universe outside the system including the apparatus and the environment into $A$. Let's further assume we have a tensor product structure between $S$ and $A$ for the Hilbert space. Let $\widehat{Y}_S$ and $\widehat{Y}_A$ be the restriction of $\widehat{Y}$ to $S$ and $A$ respectively.  Then, the sum $\widehat{Y}_S + \widehat{Y}_A$ has to be conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's suppose $\widehat{Y}_S$ and $\widehat{X}$ don't commute. Let's see what happens if we further make an assumption very common in measurement theory that for systems starting out in an eigenstate of $X$, a perfect measurement will leave it in an eigenstate of $X$ with the same eigenvalue. The Hilbert space of $A$ decomposes into eigenspaces of a pointer operator $\widehat{P}$ such that after a measurement of an eigenstate of $X$, $A$ ends up in an eigenstate of $\widehat{P}$ with an eigenvalue equal to the eigenvalue of $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $\widehat{Y}_S$ and $\widehat{X}$ don't commute, there exists an eigenvalue of $X$ such that all nonzero eigenstates of it aren't also $\widehat{Y}_S$ eigenstates. With all these assumptions in place, an initial state of $\sum_n c_n \left| x_n \right\rangle$ will evolve into&#xA;$$\sum_n c_n \left| x_n \right\rangle \otimes \left| \chi_n \right\rangle$$&#xA;where $\left| \chi_m \right\rangle$ and $\left| \chi_n \right\rangle$ are orthogonal if $m \neq n$ because they have different pointer values. Let's further assume $A$ is initially in an eigenstate of $\widehat{Y}_A$. Assume that the system is initially in an eigenstate of $\widehat{Y}_S$ with eigenvalue $y_m$ which isn't also an eigenstate of $X$. Then, this state can be expressed as $\sum_n U_{mn} \left| x_n \right\rangle$ with $U_{mn}$ being nonzero for at least two different values of $n$. The final state is&#xA;$$\sum_{np} U_{mn} U_{pn}^* \left| y_p \right\rangle \otimes \left| \chi_n \right\rangle$$&#xA;If we look at the contribution to the sum for one of the values of $n$ for which $U_{mn}$ is nonzero, we find the combined $S+A$ system can't be in an eigenstate of $\widehat{Y}$. Precisely because the $\chi$'s are orthogonal, the contributions from different $n$'s can't cancel, and this remains true for the wavefunction as a whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course, $\widehat{Y}$ generates a symmetry, and $\widehat{X}$ is not invariant under this symmetry. So, to measure $X$, the initial state of the apparatus can't possibly be invariant under this symmetry either. So, we shouldn't expect $A$ to start in an eigenstate of $\widehat{Y}_A$. The measurement process itself is expected to change $\widehat{Y}_S$, but we have a compensating change in $\widehat{Y}_A$.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-02T19:31:52.610" />
  <row Id="4490" PostTypeId="2" ParentId="4474" CreationDate="2011-02-02T19:46:24.610" Score="2" ViewCount="" Body="&lt;p&gt;As Jeff Harvey points out, you can find all of the original calculations for the various form fields in the references of the MAGOO article. But you are better off working the result out for yourself, and it's not a difficult calculation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are only concerned with the asymptotics of the fields, so you can make some simplifications. Start with the Poincaré patch of AdS (where the metric is especially easy to work with)&#xA;$$ ds^2 = \frac{\ell^2}{z^2}\,(dz^2 + \eta_{ab} dx^a dx^b) ~.$$ &#xA;Now focus on the $z$ dependence of the field, ignoring any dependence on the boundary coordinates. Assume the field scales like $z^{\Delta}$ as $z \to 0$. Evaluate the equation of motion with this ansatz, and the condition for $\Delta$ follows. For a scalar field this calculation takes just a few lines. Working it out for massive p-forms is a bit more involved: you will need to expand the covariant derivative of a (p+1)-form field strength and then work out the z dependence of the terms involving Christoffel symbols. But the simple form of the metric in the Poincaré patch keeps this from getting too complicated, and eventually you'll get the result you quoted for $\Delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that this calculation works equally well for Euclidean and Lorentzian AdS. For a discussion of the differences between the two, see hep-th/9805171.&lt;/p&gt;&#xA;" OwnerUserId="1727" LastActivityDate="2011-02-02T19:46:24.610" CommentCount="5" />
  <row Id="4491" PostTypeId="1" AcceptedAnswerId="10774" CreationDate="2011-02-02T20:17:04.107" Score="1" ViewCount="109" Body="&lt;p&gt;Kostyas question for angular momentum and half wave plates &#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/4479/half-wave-plate-and-angular-momentum&quot;&gt;Half wave plate and angular momentum&lt;/a&gt;&#xA;made me think a little bit. It took me some seconds to &quot;swallow&quot; the answer. :=) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I started to think further, whether such a plate, rotating in the &#xA;right direction, could lead to a slightly shorter wavelength? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And further, what happens in an optical active substance, which &#xA;&quot;rotates&quot; the plane of polarisation of a light beam? &#xA;Of course no momentum is changed, there is only a extremely small &#xA;difference of interaction of the two circular polarized waves &#xA;(representing the plane polarized beam) with the substance. &#xA;Is there any &quot;mechanic&quot; reaction of the &quot;sample&quot;? &#xA;Might a small torque be excerted along the axis parallel to the &#xA;direction of the beam? &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-02-03T12:50:47.853" LastActivityDate="2011-06-04T20:24:41.190" Title="Interaction of polarized beam to optical active substance" Tags="&lt;optics&gt;&lt;angular-momentum&gt;&lt;polarisation&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4492" PostTypeId="1" AcceptedAnswerId="4500" CreationDate="2011-02-02T20:41:25.970" Score="2" ViewCount="351" Body="&lt;p&gt;We have learned that Torque is equal to a force that is perpendicular to a radius (displacement);  however, I just cannot grasp one of the study questions we received:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A hammer thrower accelerates the&#xA;  hammer (mass = 7.30 kg) from rest&#xA;  within four full turns (revolutions)&#xA;  and releases it at a speed of 30.0&#xA;  m/s. Assuming a uniform rate of&#xA;  increase in angular velocity and a&#xA;  radius of 1.50 m, calculate the&#xA;  following answers.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(a)the angular acceleration:  7.96 rad/s2&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(b) the (linear) tangential acceleration: 11.94 m/s2&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(c) the centripetal acceleration just before release: 600 m/s2&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(d) the net force being exerted on the hammer by the athlete just before release: 4380 N&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(e) the angle of this force with&#xA;  respect to the radius of the circular&#xA;  motion is___&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So basically, I have answered the angular acceleration, tangential acceleration, centripetal acceleration and the net force, but I cannot determine the angle.  Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1729" LastEditorUserId="1729" LastEditorDisplayName="" LastEditDate="2011-02-02T21:44:14.503" LastActivityDate="2011-02-09T02:55:11.727" Title="Torque homework" Tags="&lt;classical-mechanics&gt;&lt;homework&gt;&lt;torque&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="0" />
  <row Id="4493" PostTypeId="1" CreationDate="2011-02-02T21:04:32.600" Score="1" ViewCount="280" Body="&lt;p&gt;As the galaxy is moving and the solar system orbiting the galaxy and the Earth orbiting the sun. So how fast is each object moving and what is the fastest we move at?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we even know how fast the galaxy is moving that is not relative to another galaxy (although I guess velocity has to be measured relative to something)&lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2011-02-03T00:55:10.540" Title="How fast is earth moving through the universe" Tags="&lt;astrophysics&gt;&lt;velocity&gt;" AnswerCount="4" />
  <row Id="4494" PostTypeId="2" ParentId="4493" CreationDate="2011-02-02T21:17:10.900" Score="5" ViewCount="" Body="&lt;p&gt;Velocity does indeed have to be measured relative to something. We can measure our radial velocity relative to any other astronomical object we care to, by measuring Doppler shifts. But if you want to know our velocity &quot;relative to the Universe as a whole&quot; rather than relative to any one object, we have to be a bit careful to define our terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the Universe appears to be approximately homogeneous and isotropic, it makes sense to define a &quot;rest frame&quot; at any given point. (The rest frames at different points are moving with respect to each other -- that's what it means to say that the Universe is expanding.) This &quot;rest frame&quot; is essentially the frame in which the stuff surrounding that point appears to be moving isotropically (the same in all directions). In practice, the best way to define that rest frame is to find the frame in which the cosmic microwave background appears the same in all directions (has no dipole moment, to be precise). Relative to this frame, the local group of galaxies is moving at about 600 km/s (&lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_microwave_background_radiation&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; gives precise numbers and probably citations that I'm too lazy to look up).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People sometimes worry about whether the existence of a preferred &quot;rest frame&quot; of this sort is in conflict with the principle of relativity. The answer is that it isn't. There are a couple of ways to see why. One is to note that the principle of relativity says that the &lt;em&gt;laws of physics&lt;/em&gt; have no preferred frames, but &lt;em&gt;particular solutions&lt;/em&gt; to the laws can have preferred frames. Another way of putting it, which I prefer, is that the &quot;rest frame&quot; we use in cosmology is simply the center-of-momentum frame of a bunch of particles (namely the CMB photons in our neighborhood). In other contexts, we're not surprised or worried by the fact that a bunch of particles have a rest frame, so why should we worry about it here?&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-02T21:17:10.900" />
  <row Id="4495" PostTypeId="2" ParentId="4493" CreationDate="2011-02-02T21:22:20.613" Score="2" ViewCount="" Body="&lt;p&gt;The Earth is moving by 30 km/s around the Sun and relatively to the Sun. The Sun is orbiting the center of our Galaxy, the Milky Way, by the speed of about 200-250 km/s. Our Galaxy is moving relatively to the Local Group where it orbits and the Local Group falls toward the Virgo Cluster of Galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the latter two velocities are small relatively to the Local Group's velocity relatively to the cosmic microwave background, the closest approximation of &quot;the right frame&quot; that cosmology may offer you: the Local Group is moving by 600 km/s relatively to the cosmic microwave background. That's 0.2% of the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, it's important to appreciate that all velocities are relative, all observers who are moving by constant velocities relatively to each other may use the same laws of physics (the principle of relativity), and the cosmic microwave background is so weak that it doesn't break the democracy between various reference frames in a significant way.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-02T21:22:20.613" />
  <row Id="4496" PostTypeId="2" ParentId="4493" CreationDate="2011-02-02T21:26:41.433" Score="1" ViewCount="" Body="&lt;p&gt;Solar system is moving at $368\pm 2$ km/sec relative to the microwave background in the direction towards the constellation Leo.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastEditorUserId="1186" LastEditorDisplayName="" LastEditDate="2011-02-03T00:55:10.540" LastActivityDate="2011-02-03T00:55:10.540" CommentCount="2" />
  <row Id="4497" PostTypeId="2" ParentId="4493" CreationDate="2011-02-02T21:36:32.837" Score="0" ViewCount="" Body="&lt;p&gt;First you have to think about what you mean by space. The concept of how fast you are moving through the universe sounds like you are assuming an absolute space as envisioned by Newton rather than a relational space as promoted by Leibniz. So, as worded, it seems that you need an absolute background spacetime, which is not the case with special relativity. The space described by inertial reference frames in areas of small curvature is usually used as &quot;background space&quot; and, as Ted says, descibes an isotropic and homogenous space. Motion is only relative to other inertial frames, and if you choose a rest frame, you have to define one yourself.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-02-02T21:36:32.837" CommentCount="1" />
  <row Id="4498" PostTypeId="1" CreationDate="2011-02-02T21:37:05.027" Score="1" ViewCount="155" Body="&lt;ol&gt;&#xA;&lt;li&gt;First, what determines the strength of an RF signal?&lt;/li&gt;&#xA;&lt;li&gt;Why is it said that a cellphones radiates more energy when the reception from signal tower  is lower in strength.&lt;/li&gt;&#xA;&lt;li&gt;Why is it said that RF gets transmitted along wires such as in wired headsets making them act as antennae?&lt;/li&gt;&#xA;&lt;li&gt;Is it then safer to use bluetooth headsets over wired headsets or direct usage?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: Do excuse if this is not purely in the realm of physics.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="619" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-02T22:05:47.143" LastActivityDate="2011-03-04T23:04:18.967" Title="Cellphone RF radiation" Tags="&lt;electromagnetic-radiation&gt;&lt;radio&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4499" PostTypeId="2" ParentId="3892" CreationDate="2011-02-02T21:37:29.420" Score="0" ViewCount="" Body="&lt;p&gt;Of course, not. Singularities are prohibited by the Cosmic Censorship principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even more: no matter can ever move under the so-called &quot;event horizon&quot; because as one approaches the horizon or any other potential well, the time slows, and on the horizon it slows infinitely. Thus if the horizon existed, nobody could reach it in finite time, and the black hole existence time is finite, so any black hole would evaporate before the falling observer reaches the event horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The singularity and event horizon terminology is useful only in theory which does not account for quantum effects and thermodynamics.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastActivityDate="2011-02-02T21:37:29.420" />
  <row Id="4500" PostTypeId="2" ParentId="4492" CreationDate="2011-02-02T21:58:32.547" Score="1" ViewCount="" Body="&lt;p&gt;You know the tangential and centripetal acceleration so it's good to use these directions as base. And you know that the direction of the force is the same as the acceleration. Meaning, you have the direction of the force in the plane the hammer is rotating $\vec{v} = (-600, 11.94)$. In this same plane you know the direction of the radius: it's centripetal $\vec{u} = (1,0)$. Now you just calculate the angle between $\vec{v}$ and $\vec{u}$.&lt;/p&gt;&#xA;" OwnerUserId="1337" LastActivityDate="2011-02-02T21:58:32.547" CommentCount="1" />
  <row Id="4501" PostTypeId="2" ParentId="4498" CreationDate="2011-02-02T22:03:24.650" Score="1" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;Phone's radio stack is programmed to both conserve battery and maintain the connection with base station as long as possible. So, it will increase its sending power when it receives weak signal and reduce it when the received signal is strong to save battery power.&lt;/li&gt;&#xA;&lt;li&gt;It is just that changing EM field induces current in all conductors, so the signal from phone will induce some current in headphone cables (and vice-versa, current induced in cable by base station signal may be delivered to phone). It is not effective at all, but may help in very bad reception conditions or when phone's antenna is dumped or shielded by some material around it.&lt;/li&gt;&#xA;&lt;li&gt;Health is a difficult issue since human body is just a too complex and too inseparable system to predict its behavior with good confidence. Yet the picture emerging from current studies suggest that there is no significant influence.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="68" LastActivityDate="2011-02-02T22:03:24.650" CommentCount="6" />
  <row Id="4502" PostTypeId="1" CreationDate="2011-02-02T22:23:05.313" Score="8" ViewCount="482" Body="&lt;p&gt;Recently I got the book &quot;On the Shoulders of Giant&quot; from Stephen Hawkings. It consists of more than 1000 pages of classical publications in physics. However 900 pages are given to the work of Copernicus, Galilei and Newton and 100 pages are given to the papers of Einstein. Therefore I think this books gives a wrong impression about what is really important for our knowledge in physics today. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could fill a book with the most important papers relevant for our knowledge in physics today, which ones would you select?    &lt;/p&gt;&#xA;" OwnerUserId="1648" LastEditorUserId="1648" LastEditorDisplayName="" LastEditDate="2011-02-02T22:42:47.573" LastActivityDate="2011-02-03T20:30:41.797" CommunityOwnedDate="2011-02-02T23:14:13.237" Title="What are the most important papers in physics?" Tags="&lt;soft-question&gt;" AnswerCount="8" CommentCount="5" FavoriteCount="3" />
  <row Id="4503" PostTypeId="1" CreationDate="2011-02-02T22:32:17.003" Score="1" ViewCount="152" Body="&lt;p&gt;Where/what is emitting cosmic background radiation, and when did it come into existance, was immediately after the big bang?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the universe isn't 3D in the traditional sense, and I don't pretend to understand it. However if the big bang originated from a single point, and the cosmic radiation was emitted from that point at the very start, then unless he explosion/expansion/inflation (whatever happened) wasn't faster the speed of light, then the CMB would have reached the edge of the universe and gotten out of it.&lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2011-02-03T22:28:29.000" Title="Cosmic background radiation" Tags="&lt;astrophysics&gt;&lt;cosmology&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="4504" PostTypeId="2" ParentId="4503" CreationDate="2011-02-02T22:46:05.580" Score="5" ViewCount="" Body="&lt;p&gt;When the Universe was very young, the matter in it was very hot. That matter emitted thermal radiation, in the same way that other hot things do. For quite a while, the radiation was in equilibrium with the matter: photons of radiation were constantly being both emitted and absorbed by the matter (mostly the electrons that were zipping around).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a certain point, the matter &quot;decoupled&quot; from the radiation, meaning that interactions between photons and other particles became very rare. The main reason for this is that most of the matter formed neutral atoms, rather than consisting of free charged particles (electrons and nuclei). Neutral atoms don't interact with radiation nearly as strongly as charged particles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that time on, most of the photons that were in existence simply flew through space, not interacting with anything. Those are the CMB photons that we see today. &lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-02T22:46:05.580" />
  <row Id="4505" PostTypeId="2" ParentId="4503" CreationDate="2011-02-02T22:52:05.963" Score="1" ViewCount="" Body="&lt;p&gt;The CMB was decoupled from the &quot;hot bath&quot; approximately 380000 years after the Big Bang.&#xA;The moment was the end of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Photon_epoch&quot; rel=&quot;nofollow&quot;&gt;Photon epoch&lt;/a&gt; and called &lt;a href=&quot;http://en.wikipedia.org/wiki/Recombination_%28cosmology%29&quot; rel=&quot;nofollow&quot;&gt;&quot;the recombination&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question about the speed of light is ok, while it deals with different topic. There is nothing wrong about two objects moving with speeds higher than the speed of light avay from each other. It is just forbidden to transmit signals with the higher than light speed. Just get to the usual analogy  with &quot;ants on a ballon&quot;: two ants running away from each other with maximal speed 1cm/s and the baloon expansion adds some extra speed.&lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2011-02-03T21:19:56.273" LastActivityDate="2011-02-03T21:19:56.273" />
  <row Id="4506" PostTypeId="1" CreationDate="2011-02-02T22:52:56.557" Score="1" ViewCount="175" Body="&lt;p&gt;I have a top of unknown mass that has a moment of inertia $I=4\times 10^{-7} kg \cdot m^2$. A string is wrapped around the top and pulls it so that its tension is kept at 5.57 N for a distance of .8 m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could somebody help me derive some equations to help with this? Or to get me in the right direction? I have been trying to derive some sort of equations from $E=\frac{I \cdot \omega ^{2}}{2}$ but I cant get anywhere without ending up at radius = radius or mass = mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need the final angular velocity.&lt;/p&gt;&#xA;" OwnerUserId="1732" LastEditorUserId="701" LastEditorDisplayName="" LastEditDate="2011-02-03T08:08:48.333" LastActivityDate="2011-04-04T10:27:33.600" Title="Homework about spinning top" Tags="&lt;homework&gt;&lt;forces&gt;&lt;angular-momentum&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="4507" PostTypeId="2" ParentId="4502" CreationDate="2011-02-02T22:53:24.187" Score="2" ViewCount="" Body="&lt;p&gt;The number of citations a paper receives is a good indicator of its importance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Top Cited Articles of All Time (2010 edition) can be found at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.slac.stanford.edu/spires/topcites/2010/alltime.shtml&quot; rel=&quot;nofollow&quot;&gt;http://www.slac.stanford.edu/spires/topcites/2010/alltime.shtml&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.37791&#xA;Review of Particle Physics&#xA;By Particle Data Group (Claude Amsler et al.). &#xA;Citations are counted for all versions of the RPP, most recent version is: &#xA;Published in:Phys.Lett.B667:1-1340,2008 &#xA;[3679 Total citations in HEP] &#xA;[37799 Total Citations to all copies of RPP in HEP]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.7328&#xA;A Model of Leptons&#xA;By Steven Weinberg (MIT, LNS). &#xA;Published in:Phys.Rev.Lett.19:1264-1266,1967 &#xA;[7328 Total citations in HEP] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.7135&#xA;The Large N limit of superconformal field theories and supergravity&#xA;By Juan Martin Maldacena (Harvard U.). &#xA;Published in:Adv.Theor.Math.Phys.2:231-252,1998, Int.J.Theor.Phys.38:1113-1133,1999 (arXiv: hep-th/9711200) &#xA;[7192 Total citations in HEP]&lt;/p&gt;&#xA;" OwnerUserId="1345" LastActivityDate="2011-02-02T22:53:24.187" CommunityOwnedDate="2011-02-02T23:14:13.237" CommentCount="2" />
  <row Id="4508" PostTypeId="2" ParentId="4502" CreationDate="2011-02-02T22:55:03.403" Score="4" ViewCount="" Body="&lt;p&gt;The birth of computational physics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N. Metropolis, A.W. Rosenbluth, M.N. Rosenbluth, A.H. Teller, and E. Teller (1953). &quot;Equation of State Calculations by Fast Computing Machines&quot;. Journal of Chemical Physics 21 (6): 1087–1092. doi:10.1063/1.1699114.&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2011-02-02T22:55:03.403" CommunityOwnedDate="2011-02-02T23:14:13.237" CommentCount="1" />
  <row Id="4509" PostTypeId="2" ParentId="4502" CreationDate="2011-02-02T23:03:52.097" Score="2" ViewCount="" Body="&lt;p&gt;These days, the most important papers are generally the ones that get the most citations.&#xA;Papers' citations are listed at Spires (or Inspires) at &lt;a href=&quot;http://inspirebeta.net/&quot; rel=&quot;nofollow&quot;&gt;Inspires&lt;/a&gt;.&#xA;Simply look up an author then click on his/her name and go to citations. They rank the papers by means of citations from &quot;unknown papers&quot; to &quot;renowned papers&quot;. It is hard to rank the older guard of physicists this way.&#xA;Stephen Hawking collected what he regards as important original papers of the greats(I note that you have that) in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/076241698X&quot; rel=&quot;nofollow&quot;&gt;&quot;On the Shoulders of Giants&quot;&lt;/a&gt; and a similar book of original math papers &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0762430044&quot; rel=&quot;nofollow&quot;&gt;And God Created the Integers&lt;/a&gt;&#xA;As an aside, Sidney Coleman said that if he had seen farther than others it was because he was standing behind the shoulders of dwarfs....:)&#xA;Oh, I see you have On the Shoulders,...well others are George Johnson's , The Ten Most Beautiful Experiments, or The Great Equations, by Robert Crease.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-02-02T23:12:16.917" LastActivityDate="2011-02-02T23:12:16.917" CommunityOwnedDate="2011-02-02T23:14:13.237" />
  <row Id="4510" PostTypeId="2" ParentId="4502" CreationDate="2011-02-02T23:07:34.757" Score="4" ViewCount="" Body="&lt;p&gt;The two papers of fundamentals of &lt;strong&gt;Density Functional Theory&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) P. Hohenberg and W. Kohn. &lt;a href=&quot;http://prola.aps.org/abstract/PR/v136/i3B/pB864_1&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Phys. Rev.&lt;/em&gt; 136, B864–B871 (1964)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) W. Kohn and L. J. Sham. &lt;a href=&quot;http://prola.aps.org/abstract/PR/v140/i4A/pA1133_1&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Phys. Rev.&lt;/em&gt; 140, A1133–A1138 (1965)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Kohn-Sham paper have 10,575 citations and the Hohenberg-Kohn have 8,714 citations. These papers are importants for Condensed Matter Physics, Atomic and Molecular Physics, Quantum Chemistry and Nanoscience in general. It is a prime example of interdisciplinarity.&lt;/p&gt;&#xA;" OwnerUserId="1545" LastEditorUserId="1545" LastEditorDisplayName="" LastEditDate="2011-02-02T23:28:06.887" LastActivityDate="2011-02-02T23:28:06.887" CommunityOwnedDate="2011-02-02T23:14:13.237" />
  <row Id="4511" PostTypeId="2" ParentId="4449" CreationDate="2011-02-02T23:35:57.707" Score="0" ViewCount="" Body="&lt;p&gt;A particle is transported by the velocity, $ds = vdt$, rather the acceleration, $ds = 1/2adt^2$ and hence force, because of the dt rather than $dt^2$ dependence.&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="1345" LastEditorDisplayName="" LastEditDate="2011-02-03T01:43:49.993" LastActivityDate="2011-02-03T01:43:49.993" />
  <row Id="4512" PostTypeId="2" ParentId="4502" CreationDate="2011-02-02T23:46:35.520" Score="1" ViewCount="" Body="&lt;p&gt;In the field of statistical physics there is nothing more famous than Onsager's solution for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Square-lattice_Ising_model&quot; rel=&quot;nofollow&quot;&gt;Ising ferromagnet on a square lattice&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://prola.aps.org/abstract/PR/v65/i3-4/p117_1&quot; rel=&quot;nofollow&quot;&gt;Crystal Statistics. I. A Two-Dimensional Model with an Order-Disorder Transition&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper itself isn't that important nowadays because there are both better solutions and better pedagogical approaches to the problem. Nevertheless, historically it was the first full solution of a non-trivial physical system which exhibits phase transitions.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-02-02T23:46:35.520" CommunityOwnedDate="2011-02-02T23:46:35.520" CommentCount="1" />
  <row Id="4513" PostTypeId="2" ParentId="4502" CreationDate="2011-02-03T00:07:25.067" Score="3" ViewCount="" Body="&lt;p&gt;I know it's been mentioned already here, but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://prl.aps.org/abstract/PRL/v19/i21/p1264_1&quot; rel=&quot;nofollow&quot;&gt;A Model of Leptons&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is worth every bit of its (almost) 4 pages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently, the one that changed everything in theoretical physics is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/9711200&quot; rel=&quot;nofollow&quot;&gt;The Large N limit of superconformal field theories and supergravity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;perhaps it already deserves to be in this listing.&lt;/p&gt;&#xA;" OwnerUserId="1337" LastActivityDate="2011-02-03T00:07:25.067" CommunityOwnedDate="2011-02-03T00:07:25.067" />
  <row Id="4514" PostTypeId="1" AcceptedAnswerId="4524" CreationDate="2011-02-03T00:26:50.267" Score="5" ViewCount="371" Body="&lt;p&gt;This is a nice question when you find it out, and I am really looking for a proper answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take &lt;a href=&quot;http://en.wikipedia.org/wiki/Mercury_%28element%29&quot; rel=&quot;nofollow&quot;&gt;quicksilver&lt;/a&gt; (Hg) in the periodic table. It has one proton more than Gold (melting point 1337.33 K), and one less than Thallium (melting point 577 K). It belongs to the same group as Zinc (692.68 K) and Cadmium (594.22 K). All not very high melting points, but still dramatically higher than quicksilver (234.32 K). When his neighbors melt, quicksilver vaporizes (at 629.88 K).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the reason for this exceptional behavior of quicksilver ?&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-03T01:22:48.723" LastActivityDate="2011-02-04T11:12:54.947" Title="Why is quicksilver (mercury) liquid at room temperature?" Tags="&lt;physical-chemistry&gt;&lt;metals&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="4515" PostTypeId="2" ParentId="4331" CreationDate="2011-02-03T00:33:57.693" Score="0" ViewCount="" Body="&lt;p&gt;By definition (and Wigner's Theorem) the S matrix is a unitary operator such that $|0&amp;gt;_{out}=S|0&amp;gt;_{in}$. In quantum mechanics an Observable is a Hermitian operator (Penrose has suggested changing this to a Normal Operator with complex eigenvalues). Either way an S Matrix will not be an observable (at least not in a QFT that obeys the rules of QM anyway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was the question of the title.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-03T00:33:57.693" CommentCount="2" />
  <row Id="4516" PostTypeId="2" ParentId="2110" CreationDate="2011-02-03T00:40:43.280" Score="0" ViewCount="" Body="&lt;p&gt;The right answer is 3. Space expansion manifests itself like a repelling force.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastActivityDate="2011-02-03T00:40:43.280" />
  <row Id="4517" PostTypeId="2" ParentId="4514" CreationDate="2011-02-03T00:48:42.313" Score="2" ViewCount="" Body="&lt;p&gt;Gold has its 6s valence shell unfilled, so is more reactive than Hg. thallium has a valence electron in the 6p shell which again is unfilled. The valence shell of mercury, 6s2 is filled, and is drawn closer to the nucleus because the proximity causes the electron to move at relativistic speeds thereby causing its rest mass to increase, which draws it closer yet to the nucleus. Thus it is even less reactive with other mercury atoms and thus has a lower melting point.&#xA;See &lt;a href=&quot;http://www.madsci.org/posts/archives/1997-05/862179191.Ch.r.html&quot; rel=&quot;nofollow&quot;&gt;Mercury&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-02-03T00:56:44.690" LastActivityDate="2011-02-03T00:56:44.690" CommentCount="1" />
  <row Id="4518" PostTypeId="2" ParentId="4503" CreationDate="2011-02-03T01:07:51.837" Score="0" ViewCount="" Body="&lt;p&gt;To understand this we can start with spacetime cosmology with matter and radiation.  The expansion of the universe is based on pretty simple logic.  Without laying down arguments I just state the Friedman-Lemaitre-Robertson-Walker FLRW energy (so called energy) equation for the evolution of a scale parameter of spatial distance a,&#xA;$$&#xD;&#xA;\Big(\frac{\dot a}{a}\Big)^2~=~\frac{8\pi G}{3}\rho~-~\frac{k}{a^2}&#xD;&#xA;$$&#xA;for $\rho$ the energy density.  The Hubble parameter or constant with space at each time is $H~=~{\dot a}/a$. We set $k~=~0$ for a flat space ${\mathbb R}^3$ to match observations.  Now the reasoning is that energy density for photons scales inversely with the length of the box.  The box is thought of as a resonance cavity that is equivalent to a situation where the number of photons that leave is approximately equal to the number of photons that enter.  During the radiation dominated period things were in a near equilibrium, so this is not out of line with some physical reasoning.  In a stat-mech course an elementary problem of N-photons in a box uses the same logic, the energy of the photons scales inversely with the size of the box.  So the energy of photons $E~=~hc/\lambda$, and the wave length scales with the scale factor a.  So the density scales as $\rho~=~hc/a^4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with this et up let us propose a time dependency on the scale factor a with time $a~\sim~t^n$.  Put this into the &quot;energy equation&quot; and turn the crank and you find that $n~=~1/2$.  The scale factor grows as the square root of time.  This is an energy equation, and the balance tells us that the loss of energy in photons is equal to the gain in gravitational potential energy. This connects well with Newtonian analysis and the Pound-Rebka experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may continue further, for the photons in a box exert a pressure on the sides of the box $p~=~F/a^2$, and the force induces an increment of change in the size of the box $dE~=~Fdx$.  The force is distributed on 3 different directions and so $p~=~\rho/3$.  This may then be used in the equation $pV~=~NkT$ to find that for $p~\sim~a^{-4}$ and $V~\sim~ a^3$ with the above $E~\sim~1/\lambda$ that $\lambda~\sim~ 1/T$, which is Wein's law for the wavelength as the peak of the BB curve.  The proportionality of the energy density with scale factor and temperature also gives $E~\sim~T^4$.  So this physics is remarkably in line with laboratory understanding of the basic thermodynamics of radiation.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-03T01:07:51.837" />
  <row Id="4519" PostTypeId="1" CreationDate="2011-02-03T02:06:49.567" Score="4" ViewCount="332" Body="&lt;p&gt;I've come across this quote attributed to Misner and Thorne from their book, Gravitation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'Mathematics was not sufficiently refined in 1917 to cleave apart the demands for &quot;no prior geometry&quot; and for a geometric, coordinate-independent formulation of physics. Einstein described both demands by a single phrase, &quot;general covariance.&quot; The &quot;no prior geometry&quot; demand actually fathered general relativity, but by doing so anonymously, disguised as &quot;general covariance&quot;, it also fathered half a century of confusion.'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was the confusion?&lt;/p&gt;&#xA;" OwnerUserId="1345" LastActivityDate="2011-02-03T16:10:10.617" Title="How did &quot;no prior geometry&quot; father 50 years of confusion?" Tags="&lt;general-relativity&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="4520" PostTypeId="2" ParentId="3983" CreationDate="2011-02-03T02:58:14.003" Score="3" ViewCount="" Body="&lt;p&gt;Much has been said… but i think there's a crack that i can poke.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;References&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;First off, let me list a few of the relevant references for this topic: this way we can set the basis of the discussion and agree on a certain 'common theme of knowledge' (i understand there may be dissent and disagreement on the choice given below, but i believe that the historical relevance of the works is self-evident — this list is by no means exhaustive, and i mean no disrespect for works i may have forgotten to list). There it goes,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691070628&quot; rel=&quot;nofollow&quot;&gt;PCT, Spin and Statistics, and All That&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540610499&quot; rel=&quot;nofollow&quot;&gt;Local Quantum Physics: Fields, Particles, Algebras (Theoretical and Mathematical Physics)&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0198517734&quot; rel=&quot;nofollow&quot;&gt;Mathematical Theory of Quantum Fields (International Series of Monographs on Physics)&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540510583&quot; rel=&quot;nofollow&quot;&gt;Finite Quantum Electrodynamics&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3642086357&quot; rel=&quot;nofollow&quot;&gt;Perturbative Quantum Electrodynamics and Axiomatic Field Theory (Theoretical and Mathematical Physics)&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540347623&quot; rel=&quot;nofollow&quot;&gt;Quantum Field Theory I: Basics in Mathematics and Physics: A Bridge between Mathematicians and Physicists (v. 1)&lt;/a&gt;, and &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540853766&quot; rel=&quot;nofollow&quot;&gt;Quantum Field Theory II: Quantum Electrodynamics: A Bridge between Mathematicians and Physicists&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821847058&quot; rel=&quot;nofollow&quot;&gt;Quantum Field Theory (Mathematical Surveys and Monographs)&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521115779&quot; rel=&quot;nofollow&quot;&gt;Mathematical Aspects of Quantum Field Theory (Cambridge Studies in Advanced Mathematics)&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521060257&quot; rel=&quot;nofollow&quot;&gt;Quantum Field Theory for Mathematicians (Encyclopedia of Mathematics and its Applications)&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540769544&quot; rel=&quot;nofollow&quot;&gt;Mathematical Theory of Feynman Path Integrals: An Introduction (Lecture Notes in Mathematics)&lt;/a&gt;, and &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540579850&quot; rel=&quot;nofollow&quot;&gt;White Noise Calculus and Fock Space (Lecture Notes in Mathematics)&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h3&gt;Comments&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;With that out of the way, let me make some comments. With a bit of luck i won't digress too much to the point of loosing the original path…&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Historically, there where two movements with separate names: &lt;em&gt;Axiomatic&lt;/em&gt; (or &lt;em&gt;Algebraic&lt;/em&gt;) Quantum Field Theory, and &lt;em&gt;Constructive&lt;/em&gt; Quantum Field Theory. While the former is marked by the reference (1) above, the latter is summarized in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387964770&quot; rel=&quot;nofollow&quot;&gt;Quantum Physics: A Functional Integral Point of View&lt;/a&gt;. Both of them faced similar obstacles during their times: the first incarnation of Wightman's Axioms did not allow for [spontaneous] symmetry breaking (a fact later corrected by Ray Streater); while Glimm &amp;amp; Jaffe's book still insists that a [Feynman] Path Integral is associated to an unique quantum theory (something we know not to be true based on the Path Integral's dependence on its parameters (aka, coupling constants): different sets of parameters yield distinct QFTs). In fact, this is the very reason why i did not mix these topics (Algebraic and Constructive QFTs) in the list of refs above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Haag's Theorem (and also Haag-Kastler's theorem), historically, belongs to the field of &lt;a href=&quot;http://www.math.ufl.edu/~sjs/aqft.html&quot; rel=&quot;nofollow&quot;&gt;Algebraic/Axiomatic QFT&lt;/a&gt;. But, later, with the further developments lead by &lt;a href=&quot;http://www.lqp.uni-goettingen.de/&quot; rel=&quot;nofollow&quot;&gt;Haag and his followers&lt;/a&gt;, this field naturally evolved into what is presently known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Local_quantum_field_theory&quot; rel=&quot;nofollow&quot;&gt;Local Quantum Physics&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are ways to rigorously define Euclidean QFTs in the lattice and then take its continuum (thermodynamic) limit. But, of course, i can't remember (or retrieve from my archives) the reference. IIRC, it was a group from Boston University… but, more importantly, i know who will be able to remember this: Pedro (lqpman ;-). In any case, B. Simon has a bunch of work done in this area (namely $P(\phi)_2$ Euclidean QFT and its relation to Statistical Mechanics), and it's not difficult to convince oneself that rigorous QFTs can be defined appropriately (in different ways, using different techniques).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are several ways to think of a QFT (or Gauge Theory, as you wish) and, as such, to formulate it in a variety of rigorous ways. Let me present one which i think is somewhat intuitive and more straightforward. In every single 'real life scenario' you have a maximum available energy, ie, an UV-cutoff (call it $\Lambda$). With this simple realization you can start to construct your ingredients more rigorously: cast your QFT in a lattice where the spacing is given by $a = 1/\Lambda$. What i'm saying is that, in 'real life', we're always dealing with an Effective Field Theory of a form or another — and this is not altogether bad, for we can use its UV-scale to build a lattice where we're going to define our theory. This is one way to do it… another way is to use this realization of a UV-cutoff but rather than use it to define a latticized QFT, we can use it to define a &lt;a href=&quot;http://en.wikipedia.org/wiki/Vertex_operator_algebra&quot; rel=&quot;nofollow&quot;&gt;Vertex Operator Algebra&lt;/a&gt;. The bottom line, in both cases, is that we're dealing with the problem of multiplying distributions (aka generalized functions) by defining some form of &lt;a href=&quot;http://www.scholarpedia.org/article/Operator_product_expansion&quot; rel=&quot;nofollow&quot;&gt;OPE&lt;/a&gt;: what lattices and VOAs do is to define the particular &quot;point splitting&quot; that your theory &quot;likes&quot; (ie, that makes your theory &quot;well behaved&quot;). So, what you are effectively doing (pardon the pun) is defining your QFT via a particular choice of &quot;point splitting&quot; (OPE), be it with the help of a lattice, be it via VOAs. In the end of the day, however, what matters is that you, somehow, &quot;discretized&quot; your theory… and, as such, you can deal with it and circumvent Haag's theorem (which is only valid in the thermodynamic/continuum limit). This is what's behind the curtains.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In some sense, this same discussion could be done regarding Statistical Mechanical models, e.g., the Potts Model, which is completely determined by its Transfer Matrix (analogous to the $S$ Matrix in QFT). However, when you try and take this continuum/thermodynamic limit… things get complicated really quickly. Of course this is a 'technical' point, a 'mathematical detail', and so on and so forth… but, i thought this was the whole point of this question… if i'm wrong, by all means, disregard these 'nitty picky'-points of mine. In any event, it is because of this limit to infinite degrees-of-freedom that QFT is more than just &quot;S Matrix theory&quot; or naïve extensions of Quantum Mechanics: as the saying goes, &lt;em&gt;more is different&lt;/em&gt;. ;-)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Anyway, this is getting long and its getting late (ie, i'm getting hungry ;-)… and my point wasn't to give a scholastic answer, but to frame this discussion into its appropriate path. I hope this helps a bit.&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2011-02-03T02:58:14.003" />
  <row Id="4521" PostTypeId="2" ParentId="4492" CreationDate="2011-02-03T03:29:49.017" Score="0" ViewCount="" Body="&lt;p&gt;For linear acceleration you know you find distance by&lt;br&gt;&#xA;Distance = 0.5 x acceleration x time^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would solve your problem for the linear case. Try looking around for something similar in the rotational case. There are a bunch of ways of doing the problem even starting with my hint.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-03T03:29:49.017" />
  <row Id="4522" PostTypeId="1" AcceptedAnswerId="4555" CreationDate="2011-02-03T03:56:03.160" Score="1" ViewCount="2122" Body="&lt;p&gt;One of the curses of 2-liter bottles is that when carbonated beverages are stored in them for long periods (i.e. days between drinks) at least the last liter ends up being flat. However, if after every pour the bottle itself is crushed such that the remaining volume of air above the liquid is as small as possible, and such that the structure of the bottle will prevent it from deforming back to its original shape (easily done by crushing the dome first and then working your way down), then the soda will stay nice and fizzy down to pretty much the bitter end, when it becomes impossible to crush the bottle any more but a significant air volume remains. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the answer has to do with vapor pressures and equilibrium, but when I try and articulate my thoughts to others their eyes glaze over. Can anyone come up with a broadly accessible, fairly pithy, and yet technically correct explanation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize this is perhaps slightly more on the chemistry side of things than physics, but (1) there's no chemistry Stack Exchange site, and (2) &lt;a href=&quot;http://physics.stackexchange.com/questions/1302/why-cant-co2-mix-back-with-the-liquid-after-a-soda-bottle-has-been-shaken&quot;&gt;this post&lt;/a&gt; seemed reasonable precedence.&lt;/p&gt;&#xA;" OwnerUserId="1737" LastActivityDate="2011-03-18T22:32:54.693" Title="Why will crushing a partially empty 2-liter bottle keep the soda more fizzy?" Tags="&lt;pressure&gt;" AnswerCount="3" CommentCount="9" />
  <row Id="4523" PostTypeId="2" ParentId="4506" CreationDate="2011-02-03T04:00:20.317" Score="1" ViewCount="" Body="&lt;p&gt;You should be able to calculate the work done by pulling the string.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should also be able to write down an equation for the amount of work necessary to accelerate an object with a given MOI to some arbitrary angular velocity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That should be a good start.&lt;/p&gt;&#xA;" OwnerUserId="1737" LastActivityDate="2011-02-03T04:00:20.317" />
  <row Id="4524" PostTypeId="2" ParentId="4514" CreationDate="2011-02-03T04:03:03.620" Score="4" ViewCount="" Body="&lt;p&gt;Although the question has been partially answered, there is a superb reference on this topic which will certainly give you some of the deep, and not so deep insights needed to understand the answer to this question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed068p110&quot; rel=&quot;nofollow&quot;&gt;http://pubs.acs.org/doi/abs/10.1021/ed068p110&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, both the contraction of the s(1/2) orbitals predicted by the Dirac equation, and the filled valence shell of Hg are the major causes for the odd physical properties of Hg. However, there are other effects that should be considered and the paper above is pretty clear on those. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me reformulate in order to make things clearer and more specific:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for liquid Hg can be stated simply: the outer electrons of Hg (6s2) that participate in metallic bonds are &quot;less available&quot; to bonding (which might be observed for example by looking at binding energies of clusters of Hg, dimers, etc) then other common metals, and hence the interaction between Hg atoms is much less strong compared to other metal-metal bonds. The explanation for the &quot;less availability&quot; of the 6s electrons is the contraction of the 6s orbitals, caused by the high speeds achieved by those electrons. This effect is promptly predicted by the Dirac equation. Now, one would ask: what about Au? It has a 6s electron and it is a very stable solid metal, which means it forms strong metallic bonds with other gold atoms. The question then turns to be, what makes metallic bonds strong in gold and weak in Hg? And, here comes the indispensable valence shell argument: gold has only one electron in its 6s orbital, while Hg has two electrons in the 6s orbital. It turns out that this creates a large difference in bonding: we can use a simple qualitative molecular orbital argument to understand why. Let's imagine a simple picture of the bonding between two Au atoms, such that only the 6s electron of each atom contributes to the bonding. That being true, 2 MOs are formed in the process of bonding, and the 6s electron from each atom occupies the bonding orbital, while the antibonding MO is unoccupied. The bonding MO has a lower energy than the 6s orbitals, which stabilizes the dimer, while the antibonding has a higher energy than the 6s orbitals, making it unfavorable for a bond to form. For Au, as we just saw the 6s electrons would be in the bonding orbital and the metallic bond would be favorable. Now, this scheme is easily extendable for Hg, and the surprise here is that since the 6s of Hg has 2 electrons, in order for Hg to dimerize, 2 electrons from our simplified model would go to the bonding molecular orbital, while the other two would be in the antibonding molecular orbital. Therefore, the stabilizing effect generated by electrons in the bonding orbitals is now circumvented by the destabilization provoked by the two electrons in antibonding molecular orbitals. In these circumstances, the metallic bond would be unfavorable and the interaction between the Hg atoms will be weak, i.e., dispersion will probably dominate. By scaling these arguments to a much larger number of Hg atoms we can see that at room temperature solid Hg would not be stable, just like a bunch of other systems in which the constituents interact mainly via van der Waals forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A question that might arise from the analysis above is why cadmium or zinc which have 5s and 4s orbitals filled are solids at room temperature, and the reason is that in this case the s orbitals do not suffer appreciable relativistic contraction, and therefore it is not unfavorable for these atoms to generate solids in which those electrons participate in metallic bonds. We can note that Hg(2+)-Hg(2+) is a quite stable ion in which the Hg(2+) also have filled shells (though the 6s orbitals are unoccupied now). The reason for this last case is that the most external valence subshell is the 5d, with 10 electrons. This subshell is much more diffuse than the 6s shell and the relativistic effects are opposite, i.e., the d orbitals are extended compare to the non-relativistic case, and hence the electrons would be more easily engaged in interionic bonding, even if they are held by dispersion forces only, since the polarizability of Hg2+ is much larger compared to Hg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arguments above were derived from a simplified model, but others more complex based on those might be easily derived. I know that there are weak spots, but as noted in my comments, experimental facts and calculations converge with those explanations, and the paper referenced above shows part of this in more depth, and even explains other interesting relativistic effects such as the &quot;inert pair effect&quot; that general chemistry professors love to talk about, but in most of the times, do not know what it really is.&lt;/p&gt;&#xA;" OwnerUserId="1695" LastEditorUserId="1695" LastEditorDisplayName="" LastEditDate="2011-02-04T00:06:39.847" LastActivityDate="2011-02-04T00:06:39.847" CommentCount="6" />
  <row Id="4525" PostTypeId="2" ParentId="4492" CreationDate="2011-02-03T05:28:53.033" Score="0" ViewCount="" Body="&lt;p&gt;The force acted by the thrower on the hammer has two components: centripetal and tangential. What you actually computed in point (d) was only the centripetal component, related to the centripetal acceleration. If you also compute the tangential component of the force, you'll be able to calculate:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the correct net force, using Pythagoras' theorem&lt;/li&gt;&#xA;&lt;li&gt;the angle at which this force is acting, using some basic trigonometry (a sketch of the situation seen from above will help you determine the correct relation to use)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1725" LastActivityDate="2011-02-03T05:28:53.033" />
  <row Id="4526" PostTypeId="2" ParentId="4522" CreationDate="2011-02-03T05:28:57.387" Score="-1" ViewCount="" Body="&lt;p&gt;If you left your cola out in the open it would get very flat because all the CO2 would leave it and blow away. Putting it in a bottle with no air it stays fizzy. Having it in a bottle with some air is between these two extremes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't much CO2 when its shipped in a filled-up bottle because there are two activities going on. CO2 leaving the drink for the air, and CO2 going back into the drink from the air. Over a long time (probably a matter of hours) these processes have to balance, and from then on you'll not lose any more CO2. When they are in balance it will amount to a certain percentage of CO2 in the air and a (different) percentage of CO2 in the water. For example, there might be 1g of CO2 in each cubic centimeter of air (I don't know, but could look the number up), while there's 2g of CO2 in each cubic centimeter of drink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two effects. First, some of the CO2 gets out of the drink and goes into the air in the bottle. The more air the more CO2 you lose this way. Second, each time you open the bottle the same thing happens again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It takes a while for your drink to become flat because it starts with quite a lot of CO2. To get an idea of how much CO2 is in your bottle, look for the youtube videos showing what happens when you drop certain candies into it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/zH9vW.jpg&quot; alt=&quot;CO2 erruption from wikipedia&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Soda_and_candy_eruption&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Soda_and_candy_eruption&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a lot of CO2, but eventually, if you open up that bottle enough, most of it will get out.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-02-03T05:43:17.137" LastActivityDate="2011-02-03T05:43:17.137" />
  <row Id="4527" PostTypeId="2" ParentId="4519" CreationDate="2011-02-03T07:41:31.523" Score="3" ViewCount="" Body="&lt;p&gt;I don't know specifically what the confusion they're thinking of was, having not been around during that half-century, but I can easily make some educated guesses.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;prior geometry,&quot; I believe what MTW mean is any aspect of the geometry that is externally fixed or non-dynamical.  So any additional constraints on the mathematical structure other than $G = 8 \pi T$ or the underlying structure (metric, connections, etc) really fall under this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first obvious case is &lt;a href=&quot;http://en.wikipedia.org/wiki/Mach%27s_principle&quot; rel=&quot;nofollow&quot;&gt;Mach's principle&lt;/a&gt;, which contains speculation about the correct notion of physically meaningful reference frames, and physically meaningful notions of motion.  This was thought to be important, and lead to Einstein's ideas behind the equivalence principle.  After GR was formulated, people (including Einstein) asked if Mach's principle was consistent with, or derivable from, GR / the equivalence principle, but it turns out that it is not.  (There are, however, some remnants of the idea left in GR.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before people realized this, it did lead to a lot of confusion, even among some of the big-name people, and some of that confusion seems to have lasted a lot longer than it should have.  But trying to impose Mach's principle on top of GR would constitute a &quot;prior geometry&quot; in a sense, and would either render GR inconsistent with observation, or make it mathematically inconsistent, depending on how one does it, so this could be one of the things they had in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second obvious case that jumps to mind is a little more subtle (but was still certainly understood by MTW's time), and is the issue that, if you want to break up the metric as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g_{\mu\nu} = \eta_{\mu\nu} + \epsilon h_{\mu\nu}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or more generally, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g_{\mu\nu} = \overline{g}_{\mu\nu} + \epsilon h_{\mu\nu}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for an appropriately defined small $\epsilon$, what are the correct interpretations of $g$, $\overline{g}$, and $h$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can surely, without making any additional assumptions, write the metric as the sum of two parts, and write Einstein's equations in terms of these two parts without changing anything.  But you want to &quot;decouple&quot; the &quot;background&quot; and &quot;perturbed&quot; parts of the metric in order to allow you to solve some kinds of problems, such as gravitational waves.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems, though, once you have done this, you may have done something strange.  You now have two independent metrics, which describe two seemingly independent, unrelated (&quot;decoupled&quot;) spacetimes, and have two independent sets of symmetries.  So it now seems you may have, in addition to the prior geometry of $\overline{g}$, an ambiguity in matching coordinates from $g$ to ones in the components $\overline{g}$ and $h$ due to their independent diffeomorphism symmetries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This ambiguity lead to some confusion over if this kind of method is applicable (it is), if you have to be careful in using it (mostly, you don't), and if it introduces any fundamental conceptual difficulties (it doesn't).  (I don't remember how well MTW explains this, but you can find some detailed descriptions in some of the newer GR books, IIRC, so I won't describe it here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, this confusion started in considering gravitational waves, and was later picked up by people trying to study quantum gravity as a canonically quantized version of GR in this way.  The trouble in the quantization case was, through misunderstandings related to this prior geometry issue, people thought it okay to introduce what amounted to &#xA;actual &quot;physical&quot; prior geometry, and thought that this problem that had been understood by the gravitational wave people was telling them something important and fundamental about how quantum gravity worked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This prior geometry, in the quantum gravity case, I believe amounts to problems like selecting a preferred reference frame that would typically be quite obvious macroscopically, or breaking Lorentz invariance to (e.g.) a discrete subset, or causing the theory to become inconsistent.  Unfortunately, it seems some of the people working on various &quot;alternative&quot; approaches to quantum gravity still haven't learned this lesson, and continue to try to write down theories like this.  But to the rest of the world, this issue has been well-understood for a very long time.&lt;/p&gt;&#xA;" OwnerUserId="1403" LastActivityDate="2011-02-03T07:41:31.523" />
  <row Id="4529" PostTypeId="2" ParentId="4519" CreationDate="2011-02-03T08:11:45.373" Score="1" ViewCount="" Body="&lt;p&gt;&quot;Prior geometry&quot; in the sense of MTW is any kind of geometry that is not influenced by the distribution of gravitational sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unique selling point of general relativity is that the geometry of spacetime is determined dynamically by the &quot;self consistency&quot; Einstein field equation. Especially there is no a priori meaning of a point in spacetime, space and time have no meaning until the gravitational field is specified. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me add some aspects from this viewpoint to Jeremy's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) a coordinate-free description of geometric objects was not well developed (vector spaces, manifolds), although I suspect that e.g. Gauss already thought this way. The theory of differential forms for example was not developed (Edit: that's not quite right, see the comments) and not well known for several decades after the publication of GR. (In fact, Feynman did not understand differential forms, which motivated his approach to GR as a (quantum) field theory in his &quot;gravitation lectures&quot; where he re-discovered the graviton.) For a long time the distinction of coordinate singularities and true singularities was not clear to many physicists, which you can see for yourself when you study papers about gravitational waves from the 1950ties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Einstein himself had severe problems with the concept of general covariance that manifested in his struggle with the &quot;hole problem&quot;. For a detailed discussion from a modern viewpoint please have a look at &lt;a href=&quot;http://ncatlab.org/nlab/show/spacetime&quot; rel=&quot;nofollow&quot;&gt;spacetime&lt;/a&gt; on the nLab. This is closely related with the Mach principle(s), of course, you'll find a very interesting discussion of this in the book&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Carlo Rovelli: &quot;Quantum Gravity&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditorDisplayName="" LastEditDate="2011-02-03T09:24:13.890" LastActivityDate="2011-02-03T09:24:13.890" CommentCount="5" />
  <row Id="4532" PostTypeId="1" AcceptedAnswerId="4533" CreationDate="2011-02-03T12:35:41.870" Score="2" ViewCount="99" Body="&lt;p&gt;This question is inspired by the following comment:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the strings in string theory are relativistic and on a large enough piece of world sheet, the internal SO(1,1) Lorentz symmetry is preserved. That's why a string carries not only an energy density ρ but also a negative pressure p=−ρ in the direction along the string.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;by Lubos at the end of his answer to the question &lt;a href=&quot;http://physics.stackexchange.com/questions/3343/what-is-tension-in-string-theory&quot;&gt;&quot;What is tension in String Theory?&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see how having a $SO(1,1)$ symmetry for the worldsheet leads to a negative pressure. I have the following questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Why does a string carry negative pressure?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can a gas of strings then be treated as a substance which has a negative equation of state:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ w = \frac{p}{\rho} = -1 $$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If this is possible then the next natural question is: what are the implications for the cosmological constant problem?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background material to give some context for questions 2 and 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One part of the cosmological constant problem is an explanation of what form of matter could seed cosmological expansion. A cosmological constant term $\Lambda$ in the action for GR is equivalent to having a medium which satisfies the negative equation of state (relation between density and pressure). One simple example of matter with a negative equation of state is a homogenous, isotropic scalar field [for details see any book on cosmology with a chapter on inflation]. If strings carry negative pressure and if a string gas can be treated as matter with $w\sim-1$ then one would have a far more natural alternative to a scalar field.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-03T14:11:33.060" Title="Strings with negative pressure" Tags="&lt;quantum-field-theory&gt;&lt;string-theory&gt;&lt;cosmological-constant&gt;" AnswerCount="1" />
  <row Id="4533" PostTypeId="2" ParentId="4532" CreationDate="2011-02-03T12:48:51.183" Score="3" ViewCount="" Body="&lt;p&gt;Consider a patch of a world sheet and choose the coordinates so that it is extended in the $t$ and $z$ directions, setting $x=y=0$ - and similarly for the other transverse dimensions I omitted here. The spacetime dimensions is $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement that the strings have tension equal to $T$ means that the energy density has to be&#xA;$$T_{tt} = T \delta^{(D-2)} s(x,y,\dots).$$&#xA;However, the world sheet is required to preserve the $SO(1,1)$ symmetry rotating the $t,z$ axes, so it must be true that a part of the stress energy tensor, the components &#xA;$$T_{tt}, T_{tz}, T_{zt}, T_{zz}$$&#xA;have to be proportional to the 1+1-dimensional metric tensor because multiples of the metric tensor are the only tensors that are invariant under the Lorentz transformations, $SO(1,1)$ in this case. It follows that the $T_{tz}$ and $T_{zt}$ components have to vanish while&#xA;$$T_{zz} = -T \delta^{(D-2)} (x,y,\dots).$$&#xA;The doubly spatial components of the stress-energy tensor represent the pressure but don't forget that the doubly transverse components such as $T_{xx}$ continue to vanish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one has a gas of randomly oriented strings and he averages over all directions of the string, the average $T_{ii}$ will be $-T_{tt}/(D-1)$ where $(D-1)=3$ in our 4-dimensional spacetime so that $p=-\rho/3$. That's the standard pressure from cosmic strings - domain walls would have $p=-2\rho/3$ for very similar reasons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The appearance of $1/3$ is not hard to understand: all the rotated strings will have the same trace over the spatial part of $T_{ii}$. The $z$-oriented string has this trace equal to $-\rho/3$ so after the averaging over directions i.e. over $SO(d-1)=SO(3)$, when the spatial part becomes proportional to $\delta_{ij}$, the coefficient of this Kronecker delta inevitably has to become $-\rho/3$ to preserve the spatial trace.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The negative pressure of strings is not too important if the strings remain small and compact but it is important e.g. in string gas cosmology&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://scholar.google.com/scholar?hl=en&amp;amp;q=string-gas-cosmology+negative-pressure&amp;amp;btnG=Search&amp;amp;as_sdt=0,5&amp;amp;as_ylo=&amp;amp;as_vis=0&quot; rel=&quot;nofollow&quot;&gt;http://scholar.google.com/scholar?hl=en&amp;amp;q=string-gas-cosmology+negative-pressure&amp;amp;btnG=Search&amp;amp;as_sdt=0,5&amp;amp;as_ylo=&amp;amp;as_vis=0&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So your number $p=-\rho$ is incorrect; it should be $p=-\rho/(D-1)$ because the pressure only exists in the direction along the string, so the pressure in all directions get diluted by the averaging over directions. That's why the string gas has a different equation of state than the cosmological constant. Like any gas, string gas picks a preferred reference frame, unlike the cosmological constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The precision measurement are enough to exclude the possibility that dark energy boils down to a network of cosmic strings or domain walls because their $w$ is just too far from the &quot;approximately observed&quot; $w=-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, $w$ is defined as $p/\rho$ rather than $\rho/p$.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T14:11:33.060" LastActivityDate="2011-02-03T14:11:33.060" CommentCount="2" />
  <row Id="4534" PostTypeId="2" ParentId="4502" CreationDate="2011-02-03T14:02:09.233" Score="1" ViewCount="" Body="&lt;p&gt;For this question, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;all citation indexes are nonsense. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Papers are cited for some time, until the topics are &#xA;down to textbook level for some time. If Copernicus, &#xA;Newton and Galileis papers were cited by the same &#xA;standards as Einsteins papers are cited, citation &#xA;attachments would double in many papers. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The question reflects the hype with Einstein, &#xA;especially very vigorous in USA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other Gods beside Einstein, and Hawking is their prophet!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Georg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And: last but not least, the old papers are in latin. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-02-03T15:55:40.827" LastActivityDate="2011-02-03T15:55:40.827" CommunityOwnedDate="2011-02-03T14:02:09.233" CommentCount="3" />
  <row Id="4535" PostTypeId="2" ParentId="78" CreationDate="2011-02-03T14:27:48.520" Score="1" ViewCount="" Body="&lt;p&gt;They repel because of the same reason why, say, atoms or molecules repel: because of Pauli principle which is the basis of elastiсity and solidity of matter. Why you cannot move through the walls? Why the dish holds food? Because all these objects are composed of fermions, the particles of substance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is opposed to say, light which can go through another light ray without any interaction. Because light is composed of bosons - the quants of radiation.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastActivityDate="2011-02-03T14:27:48.520" />
  <row Id="4536" PostTypeId="2" ParentId="4502" CreationDate="2011-02-03T14:29:57.473" Score="4" ViewCount="" Body="&lt;p&gt;E. Wigner, &quot;On Unitary Representations of the Inhomogeneous Lorentz Group&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the apocryphal story of how it got published.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: It was originally rejected by one of the leading physics journals as not being of general interest. Then Wigner mentioned it to von Neumann (his classmate from school back in Hungary). J. von Neumann told Wigner to send it to the Annals of Math., where von Neumann was the chief editor. It was immediately accepted there! Years later, Wigner received a leter from the AMS, saying that after a survey his paper had been found to be one of the most influential papers.  &lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2011-02-03T20:30:41.797" LastActivityDate="2011-02-03T20:30:41.797" CommunityOwnedDate="2011-02-03T14:29:57.473" CommentCount="1" />
  <row Id="4537" PostTypeId="1" AcceptedAnswerId="4543" CreationDate="2011-02-03T15:29:11.553" Score="5" ViewCount="429" Body="&lt;p&gt;recently I have had &lt;a href=&quot;http://physics.stackexchange.com/questions/4372/does-entropy-apply-to-newtons-first-law-or-does-acted-upon-always-require-an-e/4380#4380&quot;&gt;some exchanges with @Marek&lt;/a&gt; regarding entropy of a single classical particle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I always believed that to define entropy one must have some distribution. In Quantum theory, a single particle can have entropy and I can easily understand that. But I never knew that entropy of a single rigid classical particle is a well defined concept as Marek claimed. I still fail to understand that. One can say that in the classical limit, the entropy of a particle in QT can be defined and that corresponds the entropy of a single classical particle. But I have difficulty accepting that that gives entropy of a single Newtonian particle. In my understanding, If a system has entropy then it also should have some temperature. I don't understand how one would assign any temperature to a single classical particle. I came across &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1005/1005.2751v3.pdf&quot; rel=&quot;nofollow&quot;&gt;a paper where there is a notion of &quot;microscopic entropy&quot;&lt;/a&gt;. By no means, in my limited understanding, it corresponded to the normal concept of entropy. I am curious to know, what is the right answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is, is it possible to define entropy of a single classical particle?&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="673" LastEditorDisplayName="" LastEditDate="2011-02-03T20:31:17.457" LastActivityDate="2011-02-03T20:31:17.457" Title="Can a single classical particle have any entropy?" Tags="&lt;thermodynamics&gt;&lt;statistical-mechanics&gt;&lt;entropy&gt;" AnswerCount="4" FavoriteCount="3" />
  <row Id="4538" PostTypeId="2" ParentId="3871" CreationDate="2011-02-03T15:32:25.810" Score="-2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Causality as a consequence of finite energy:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 - in the universe there is more than one object with finite energy content each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;Reductio ad absurdum'&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - If there is no limit on the speed of transmission of energy (light, electric force, gravitational ,...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 - then at the same time a moving object (with finite energy content) is here and there. You need twice the energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 - then at the same time an object is here and in any other place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 - then that object must have an infinite amount of energy distributed throughout the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 - then there is no room or energy for more than that object in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5  contradicts 0, then 1 is false.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-02-03T17:29:43.167" LastActivityDate="2011-02-03T17:29:43.167" />
  <row Id="4539" PostTypeId="2" ParentId="4537" CreationDate="2011-02-03T15:37:35.417" Score="0" ViewCount="" Body="&lt;p&gt;Given a probability distribution,&#xA;$$S=-k\sum_i p_i \ln p_i.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless the probability distribution is of the form $e^{-\beta \left( H - \mu N \right)}/Z$, or reasonably close to it, we can't really define the temperature of the particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should also add that if the probability distribution is a Dirac delta function, the entropy goes to zero (at least if we introduce an ultraviolet regulator...).&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-02-03T17:57:42.847" LastActivityDate="2011-02-03T17:57:42.847" CommentCount="3" />
  <row Id="4540" PostTypeId="2" ParentId="4519" CreationDate="2011-02-03T15:52:25.967" Score="3" ViewCount="" Body="&lt;p&gt;You ask about the history.  Perhaps the easiest answer is to refer you to the long reference in Wikipedia, which will give you more than fifty pages of explanation:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.pitt.edu/~jdnorton/papers/decades_re-set.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.pitt.edu/~jdnorton/papers/decades_re-set.pdf&lt;/a&gt;&lt;br&gt;&#xA;But you probably already knew about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One short answer is that “general covariance” and “no prior geometry” are two separate things, but that was not realized at first.  At least general covariance has a very clear mathematical definition.  But it was not what Einstein wanted or needed.  After it was realized that something else, or something additional was needed, it took a while for that something else to be clearly defined.  Of course, the something else turned out to be what MTW call “no prior geometry”.  Nowadays it is more often called “background free” or “background independence”.  It is still controversial, particularly as to how strong it should be and as to how it applies or does not apply to string theory and loop quantum gravity.&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-02-03T15:52:25.967" CommentCount="1" />
  <row Id="4541" PostTypeId="2" ParentId="4519" CreationDate="2011-02-03T16:10:10.617" Score="2" ViewCount="" Body="&lt;p&gt;This is an interesting question, in part about the history of Einstein's formulations of General Relativity and about its foundations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To deal with &quot;prior geometry&quot; first here is an example metric that was current in Einstein's time (and mentioned in MTW) as a possible form of a geometric gravity theory: $g=\phi^2\eta$ where $\eta$ is the Minkowski metric. So $g$ is the spacetime metric, but there is really a prior metric around too $\eta$. All the variation in $g$ just comes from the $\phi$ function defined on the prior geometry. So Einstein wanted a &quot;prior geometry free geometry&quot; in effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;confusion&quot; referred to by MTW and by this Question is the physical status of the principle known as &quot;General Covariance&quot;. Einstein claimed much for this principle but almost immediately Kretschmann (1917) described it as &quot;physically meaningless&quot;. A very closely related question was whether &quot;General Relativity&quot; was correctly named - had there actually been a physical discovery here which generalised the relativity principle of Special Relativity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kretschmann made the prescient claim in his argument that any (worthwhile) physical theory could be made &quot;generally covariant&quot; - even Newtonian Gravitation. MTW chapter 12 describes the modern differential formulation of the Newton-Cartan theory which seems to vindicate Kretschmann.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The confusion lasted sixty odd years because physicists somewhat lost interest in GR until the 1960s: the textbooks prior to that time did not include the modern language of differential geometry, which somewhat explains what Einstein was getting at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However even after MTW debate continued into aspects of the foundations of GR. Einstein had initially said it was based on three principles:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;General Covariance&lt;/li&gt;&#xA;&lt;li&gt;Principle of Equivalence&lt;/li&gt;&#xA;&lt;li&gt;Mach's Principle&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The three form a somewhat inter-related package, and all have been challenged or rewritten in various ways post Einstein. What often happens in modern mathematical texts on GR is that (some or all) of these are ignored and the theory is presented as a mathematical theory with Tensors on Manifolds. In an undergraduate GR text sitting on my desk right now, the role of &quot;fundamental principles&quot; above has actually been replaced by the need for equivalence to the Newtonian theory in the weak field limit: they are not mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A modern student of GR probably should learn it that way and then learn what later authors have done to these principles. It turns out to be an interesting story for those interested in fundamentals: &quot;weak&quot; versus &quot;strong&quot; covariance, mathematical restrictions on Tensor forms, etc. The contemporary significance of this work would be whether it has significance in &quot;Quantum Gravity&quot; - there is a &quot;Quantum Equivalence Principle&quot; for example in the literature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the history of this problem in GR I have found &lt;a href=&quot;http://www.pitt.edu/~jdnorton/papers/decades_re-set.pdf&quot; rel=&quot;nofollow&quot;&gt; this reference.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-03T16:10:10.617" />
  <row Id="4542" PostTypeId="2" ParentId="4459" CreationDate="2011-02-03T16:57:46.457" Score="-2" ViewCount="" Body="&lt;p&gt;All codes of the N-body simulation make use of instantaneous propagation of the speed of gravity.&#xA;The other answers follow the same criterion that is good enough for local simulations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;As a consequence of the finite speed of propagation of gravity and the motion of the masses there is a dissipative component.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the bodies are never static, the masses continuously lose energy to space because there is no known mechanism that reverses the direction of flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(An equivalent effect applies to the radiation of charged particles by virtue of its continued accelerated motion.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from &lt;strong&gt;Jorgen Kalckar and Ole Ulfbeck experiment&lt;/strong&gt; (a) (energy of two masses and a spring):&#xA;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The difference between these two&#xA;  energies is lost by each mass: it is&#xA;  taken away by space-time, in other&#xA;  words, it is radiated away as&#xA;  gravitational radiation&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;gravity can behave like a wave:&#xA;  gravity can radiate&#xA;  ...&#xA;  All this follows from the expression of universal gravity when applied to moving observers, with the requirement that neither observers&#xA;  nor energy can move faster than c.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(a) &lt;strong&gt;&lt;em&gt;motionmountain&lt;/em&gt;&lt;/strong&gt;  free online book, ch 18, pag 526/527: Motion in General Relativity, Gravitational Waves.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-02-03T16:57:46.457" CommentCount="1" />
  <row Id="4543" PostTypeId="2" ParentId="4537" CreationDate="2011-02-03T16:59:59.960" Score="5" ViewCount="" Body="&lt;p&gt;Entropy is a concept in thermodynamics and statistical physics but its value only becomes indisputable if one can talk in terms of thermodynamics, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do so in statistical physics, one needs to be in the thermodynamic limit i.e. the number of degrees of freedom must be much greater than one. In fact, we can say that the thermodynamic limit requires the entropy to be much greater than one (times $k_B$, if you insist on SI units).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the thermodynamic limit, the concept of entropy becomes independent of the chosen ensembles - microcanonical vs canonical etc. - up to corrections that are negligible relatively to the overall entropy (either of them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single particle, much like any system, may be assigned the entropy of $\ln(N)$ where $N$ is the number of physically distinct but de facto indistinguishable states in which the particle may be. So if the particle is located in a box and its wave function may be written as a combination of $N$ small wave packets occupying appropriately large volumes, the entropy will be $\ln(N)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the concept of entropy is simply not a high-precision concept for systems away from the thermodynamic limit. Entropy is not a strict function of the &quot;pure state&quot; of the system; if you want to be precise about the value, it also depends on the exact ensemble of the other microstates that you consider indistinguishable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider larger systems with $N$ particles, the entropy usually scales like $N$, so each particle contributes something comparable to 1 bit to the entropy - if you equally divide the entropy. However, to calculate the actual coefficients, all the conceivable interactions between the particles etc. matter.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-03T16:59:59.960" CommentCount="16" />
  <row Id="4544" PostTypeId="1" AcceptedAnswerId="4548" CreationDate="2011-02-03T17:03:46.787" Score="2" ViewCount="81" Body="&lt;p&gt;As we know, fermions are subject to exchange interactions that limit the densities they can achieve. However bosons (simple or composite) are not constrained by this, which implies physical phenomena like the Bose-Einstein condensate, where the main component is Helium-4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: Are there physical limits to the density that can be achieved with Bose-Einstein condensates made up from composite bosons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my suspicion is that there is a critical density above which the fermion character of the fermion components will play up, but i couldn't find any mention about this anywhere&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-02-03T18:56:21.063" Title="existing bounds on maximum density achieved by a Bose condensate" Tags="&lt;condensed-matter&gt;&lt;bose-einstein-condensate&gt;&lt;exchange-interaction&gt;&lt;spin-statistics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="4546" PostTypeId="1" AcceptedAnswerId="4550" CreationDate="2011-02-03T17:24:44.593" Score="5" ViewCount="236" Body="&lt;p&gt;On John Baez's website, &lt;a href=&quot;http://math.ucr.edu/home/baez/entropy.html&quot; rel=&quot;nofollow&quot;&gt;http://math.ucr.edu/home/baez/entropy.html&lt;/a&gt;, he discusses the problem of how entropy increases when a cloud of ideal gas collapses gravitationally (no black holes - keeping it simple).  But he doesn't give away the answer - instead he implies that, I presume, either gravitons or photons are released during collapse and these increase the overall system entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please answer which it is, gravitons or photons?  If gravitons (or gravity waves), then does this mean that pre-20th century physicists, with no notion of gravity waves, would have concluded that gravitational attraction violates the 2nd law?  And if photons, then what about if the gas particles were totally neutral - wouldn't they not emit any EM radiation in that case?  Or am I wrong that gravitons or photons are the answer? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1743" LastActivityDate="2011-02-03T22:30:43.017" Title="Please clarify how entropy increases when matter gravitationally coalesces" Tags="&lt;gravity&gt;&lt;statistical-mechanics&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="4547" PostTypeId="2" ParentId="4544" CreationDate="2011-02-03T17:34:13.443" Score="1" ViewCount="" Body="&lt;p&gt;There's a repulsive van der Waals force between the atoms. This shows up in the second quantized formalism as&#xA;$$H = \int d^3x \left[ \frac{\hbar^2}{2m} \nabla \Psi^\dagger \cdot \nabla \Psi - \mu\Psi^\dagger \Psi  \right] + \int d^3x\, d^3y\, \frac{1}{2}V\left( \vec{y} - \vec{x} \right) \Psi^\dagger\left( \vec{x} \right) \Psi^\dagger\left( \vec{y} \right) \Psi\left( \vec{x} \right) \Psi\left(\vec{y}\right)$$&#xA;The density is given by the expectation value $\left\langle \Psi^\dagger\Psi \right\rangle$, but you can see the repulsive potential acts as a brake requiring far greater and greater pressure to achieve higher and higher densities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond a certain density, the van der Waals approximation treating two atoms close to each other as independent subsystems breaks down.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-03T17:34:13.443" />
  <row Id="4548" PostTypeId="2" ParentId="4544" CreationDate="2011-02-03T17:38:31.813" Score="1" ViewCount="" Body="&lt;p&gt;It's a good question. The answer is that the bound on the density is given by the requirement that the interactions between the bosons have to remain weak for the Bose-Einstein condensate to exist. In practice, the helium-4 atoms have to be further away from each other than their radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why it is so? Well, if you're talking about the bosons occupying the &quot;same state&quot;, it really means that you are constructing a multi-particle state in the multi-particle theory. If you want the energy of this state to be simply given by the sum of the energies of the individual bosons - i.e. $N$ times the energy of the one-particle state - you must guarantee that you have the right Hamiltonian which is essentially the Hamiltonian for the bosons in an external potential, without any significant interaction term in between the bosons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sufficiently strongly interacting Hamiltonian for the bosons couldn't be solved that easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try to push the composite bosons really close to each other, i.e. by lowering the temperature extremely close to the absolute zero, the interactions between them will start to matter which will prevent you from approximating the Hamiltonian by a sum of many one-particle terms. Consequently, the right description is in terms of the component particles - which are often fermions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's believed by many condensed matter physicists that the ultimate state of any bound matter very near the absolute zero is a superconductor or a Fermi liquid - and I don't know. Consider helium-3 as an example. I am actually not sure what one gets at superextremely low temperatures.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-03T18:56:21.063" LastActivityDate="2011-02-03T18:56:21.063" CommentCount="3" />
  <row Id="4549" PostTypeId="2" ParentId="4546" CreationDate="2011-02-03T17:50:18.190" Score="0" ViewCount="" Body="&lt;p&gt;Thermodynamically, only materials which are totally transparent to electromagnetic radiation will fail to emit photons when heated. The stellar clouds are hardly transparent, especially when dense enough.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-03T17:50:18.190" />
  <row Id="4550" PostTypeId="2" ParentId="4546" CreationDate="2011-02-03T18:02:09.087" Score="4" ViewCount="" Body="&lt;p&gt;The answer would be actually much simpler if the collapse produced a black hole. It can be easily shown to have entropy vastly exceeding the entropy of any gas of the same mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning your main question, the answer is, of course, that &lt;em&gt;any&lt;/em&gt; system with many degrees of freedom - both in classical physics as well as quantum physics - always satisfies the second law of thermodynamics. The second law may always be proved - quite generally. The proofs are the proofs of the H-theorem or its generalization for any physical system you consider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just think about the &quot;balls&quot; in the phase space - any phase space - how it gets deformed via the time evolution. The &quot;smoothened&quot; version of this evolved &quot;spaghetti&quot; has a higher volume whose logarithm represents the entropy increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you didn't allow the molecules to emit photons when they collide, they wouldn't ever shrink spontaneously by obeying the laws of gravity. The probability that a molecule slows down (or gets closer) under the gravitational influence of the other molecules would be equal to the probability that it speeds up (or gets further) - in average. If you introduce some objects and terms in the Hamiltonian that allow inelastic collisions, these inelastic collisions will selectively slow down the molecules that happened to be closer to each other, which is the mechanism that will be reducing the average distance between the molecules (the actual rate will depend on the gravitational attraction, too).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote photons because, obviously, the probability of the emission of a photon is much higher for real-world gases because most of their interactions are electromagnetic interactions. Because a photon carries as much entropy as a graviton would, but you produce many more photons by random collisions, the entropy increase is stored in the photons. The entropy carried by gravitons is smaller by dozens of orders of magnitude.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-03T18:02:09.087" CommentCount="5" />
  <row Id="4551" PostTypeId="2" ParentId="4546" CreationDate="2011-02-03T18:17:52.297" Score="7" ViewCount="" Body="&lt;p&gt;The situation Baez describes is of a virialized ball of gas shrinking gravitationally. As he points out (and further emphasizes in his &lt;a href=&quot;http://math.ucr.edu/home/baez/entropy2.html&quot; rel=&quot;nofollow&quot;&gt;hint&lt;/a&gt;), this process doesn't conserve energy. Once you know where the energy is going, you've got the answer to what's happening with the entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your guesses about what's carrying off the energy (photons and gravitons) are both good ones, and in some contexts &quot;photons&quot; is the answer. (I know of no astrophysical context in which gravitons are responsible for a significant fraction of the energy loss.) In other astrophysical contexts the answer is actually something else: it's just gas particles &quot;evaporating off.&quot; The way a cloud of gas often shrinks is that some particles acquire speeds greater than escape velocity and fly off, leaving behind less-energetic particles. A small fraction of the particles may escape, but they carry off enough energy to cause the collapse, and enough entropy to satisfy the second law.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-03T18:17:52.297" CommentCount="5" />
  <row Id="4552" PostTypeId="2" ParentId="4537" CreationDate="2011-02-03T18:18:25.073" Score="3" ViewCount="" Body="&lt;p&gt;(The question has already been answered by Lubos but perhaps I can elaborate on the quantum aspect.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a single classical particle the answer, in general, is 'no' with some exceptions as discussed near the end. And when the number of particles $n \gt 1$ the concept of entropy becomes clear only in the limit that $ n \gg 1$ or the thermodynamic limit (as Lubos mentioned in his answer).  With a quantum 'particle' however, the situation is somewhat different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a qubit, a single spin $1/2$ object, whose state at any given moment is specified by a vector $\vert \psi \rangle$ in a Hilbert space $H$. This is the simplest quantum analog of a classical particle one can think of. In contrast to the classical situation a many-body quantum system can exist either in &lt;em&gt;pure&lt;/em&gt; or &lt;em&gt;mixed&lt;/em&gt; states. The above specification of the qubit as a state vector is only possible if the state is pure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;em&gt;many-body&lt;/em&gt; state will not, in general, be describable by a state in a Hilbert space. One must use the more general notion of a &lt;em&gt;density matrix&lt;/em&gt; for the correct description. A general density matrix $\rho$ is written in terms of the state-vectors $\vert \psi_s \rangle$ as :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \rho = \sum_s p_s \vert \psi_s \rangle \langle \psi_s \vert $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p_s$ is the &quot;fraction of the ensemble&quot; [&lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_state#Mixed_states&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt;] in each pure state $\vert \psi_s \rangle $. For instance an ensemble of spin-$1/2$ particles would be described by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \rho = p_{\uparrow} \vert \psi_{\uparrow} \rangle \langle \psi_{\uparrow} \vert + p_{\downarrow} \vert \psi_{\downarrow} \rangle \langle \psi_{\downarrow} \vert $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p_{\uparrow}$ ($p_{\downarrow}$) is the fraction of particles with spin up (down). Now, on the face of it, it seems one should be able to use this definition for a &lt;em&gt;single&lt;/em&gt; particle. It is important to keep in mind that though one can &lt;em&gt;a priori&lt;/em&gt; assign probabilities to the two states of a single particle, experimentally these probability amplitudes can be confirmed only by repeating a measurement of the particle's spin many times over. The result of a single measurement is a single number signifying which eigenstate the particle collapsed into due to the measurement process. To obtain a probability one must take the average over many such numbers, which requires more than one measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the entropy loses meaning as the measure of the lack of information external observers have about a system. The &quot;system&quot; is no longer a qubit with a given state or density matrix. Instead it is a qubit which has undergone &lt;em&gt;repeated&lt;/em&gt; cycles of the measurement process consisting of state preparation, time evolution and finally collapse to an eigenstate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question boils down to the fact that probability amplitudes can only be obtained experimentally by generating an ensemble of observations. &lt;strong&gt;A single system cannot be assigned an entropy&lt;/strong&gt;. A system consisting of an &lt;em&gt;observational ensemble&lt;/em&gt; for qubits - either with a single or few qubit and many measurements or many qubits and few measurements - can be assigned an entropy. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Extra material&lt;/strong&gt; The classical notion of Gibbs-Shannon entropy can be extended to the quantum version called the von-Neumann entropy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S_{cl} = -\sum_k p_k \ln(p_k) \Rightarrow S_{qm} = - Tr\left[\rho \ln(\rho)\right] $$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First Edit&lt;/strong&gt; @Marek makes some pertinent observations in the comments, which I feel are important enough to be included in this edit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;In contrast to the classical situation a many-body quantum system can exist either in pure or mixed states.&quot; what contrast? In classical physics it is also either pure (a single point in the phase space, which is the analogue of the Hilbert space in the classical physics) or a mixture of a pure states which can be described by a measure on the phase space. Everything is completely the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The key difference lies in the different effect of measurement on a classical and quantum process. In the absence of a unified description of measurement spanning all domains, we are left with two notions of measurement. In the quantum mechanical case an irreducible error is present in &lt;strong&gt;any&lt;/strong&gt; measurement. This is not the case for a classical mechanical system where measurements can be done to an arbitrary accuracy. Of course, in a unified description of measurement, a classical system would utimately be understood in terms of many-body quantum systems subject to the effects of decoherence.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;A single system cannot be assigned an entropy&quot; -- of course it can be assigned an entropy. Just because you can't measure everything at once doesn't mean it doesn't make sense to define it :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not talking about &quot;everything&quot; but only of the very definite concept of &quot;probability&quot;. Probability is a statistical notion. You cannot do statistics with one data point.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because if you'd said so, you might as well throw whole quantum theory out of the window, which ultimately doesn't allow you to talk about anything else than probabilities and for those you need multiple measurement as well... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well yes, of course, you need multiple measurements for quantum probabilities. That is the essence of my argument.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-03T19:27:38.403" CommentCount="9" />
  <row Id="4553" PostTypeId="2" ParentId="4343" CreationDate="2011-02-03T18:27:16.497" Score="1" ViewCount="" Body="&lt;p&gt;From &lt;em&gt;[The ASCE Standardized Reference Evapotranspiration Equation]&lt;/em&gt;&lt;a href=&quot;http://books.google.com/books?id=Sq4FEqD0Jc4C&amp;amp;lpg=PP1&amp;amp;dq=asce%20standardized%20reference%20evapotranspiration&amp;amp;pg=PP1#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given T is temperature in degrees Celsius, and RH is relative humidity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturation Vapor Pressure (es) =&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0.6108 * exp(17.27 * T / (T + 237.3))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Actual Vapor Pressure (ea) =&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;RH / 100 * es &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Vapor Pressure Deficit =&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ea - es&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Why this is a meaningful measurement: &quot;The strain under which an organism is placed in maintaining a water balance during temperature changes is much more clearly shown by noting the vapor pressure deficit than by recording the relative humidity.&quot; Anderson, D. B. 1936. Relative humidity or vapor pressure deficit. Ecology 17, no. 2: 277–282.&lt;/p&gt;&#xA;" OwnerUserId="1680" LastActivityDate="2011-02-03T18:27:16.497" />
  <row Id="4554" PostTypeId="2" ParentId="4537" CreationDate="2011-02-03T19:02:08.033" Score="8" ViewCount="" Body="&lt;p&gt;First of all we must distinguish between two things that are called entropies. There's a  microscopic entropy, also called Shannon Entropy, that is a functional over the possible probability distributions you can assign for a given system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle H[p] = -\sum_{x \in \mathcal{X}}\; p(x) \log(p(x))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathcal{X}$ is the set where your variable x takes values. And there's a &quot;macroscopic entropy&quot;, that is merely the value of the functional above calculated for a specific family of distributions parametrized by some variable $\theta$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(\theta)=-\sum_{x \in \mathcal{X}}\; p(x|\theta) \log(p(x|\theta))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what happens in thermodynamics and equilibrium statistical physics is that you have a specific family of distributions to substitute in the first expression: the Gibbs equilibrium distribution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(x | V, T, N) = \frac{1}{Z}e^{-\frac{E(x)}{T}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, as an example, we have as parameters the volume, temperature and number of particles, and E(x) is the energy of the specific configuration x. If you substitute this specific family of distributions on $H[p]$, what you'll have is the thermodynamic equilibrium entropy, and this is what physicists usually call entropy: a state function depending on parameters of the Gibbs distribution (as opposed to a functional that associate a real value for each possible choice of distributions). Now, to find what is the apropriate physical equilibrium for this system when those parameters are allowed to vary, you must maximize this entropy (1). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here it's common to make the following distinction: x is a microscopic variable that specify the detailed configuration of the system, and V, T and N are macroscopic parameters. It doesn't need to be so. In the specific case of statistical physics the origin of the distribution function is the fact that there are so many degrees of freedom that it's impossible (and even undesirable) to follow them all, so we are satisfied with a statistical description. Under this assumptions it's natural to expect that the distribution would be over microscopic variables with macroscopic parameters. But this is not the only reason why one would use a distribution function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could have other sources of ignorance. As an example, you could have the following problem: suppose we recently discovered a new planet on a solar system where there' 2 more planets. It's position $\vec{x}$ and velocity $\vec{v}$ at a given instant $t = 0$ have been measured within some precision $\sigma_x$ and $\sigma_v$. Let's assume that the sources of possible errors in the measures are additive. Then it's reasonable to assume that we have a gaussian probability distribution for the position of the planet:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle p(\vec{x}(0), \vec{v}(0) | \sigma_x, \sigma_v) =\frac{1}{Z} \exp\left(-\frac{x(0)^2}{2\sigma_x} -\frac{v(0)^2}{2\sigma_v} \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where Z is some normalization constant. Now suppose we want to predict this planets position in the future given the current positions of the other planets and their uncertainties. We would have a distribution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle p(\vec{x}(t), \vec{v}(t) | \vec{x}_i(0), \vec{v}_i(0), \sigma_{x,i},\sigma_{v,i})= \displaystyle p(\vec{x}(0), \vec{v}(0) | \sigma_x, \sigma_v)\prod_{i=1}^{2}\displaystyle p(\vec{x}_i(0), \vec{v}_i(0) |  \sigma_{x,i}\sigma_{v,i}) \times$ &#xA;$\times p(\vec{x}(t), \vec{v}(t) | \vec{x}(0), \vec{v}(0),\vec{x}_1(0), \vec{v}_1(0), \vec{x}_2(0), \vec{v}_2(0))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p(\vec{x}(t), \vec{v}(t) | \vec{x}(0), \vec{v}(0),\vec{x}_1(0), \vec{v}_1(0), \vec{x}_2(0), \vec{v}_2(0))$ would take Newton's equations of motion into account. Note that there's a small number of particles here: just 3. And the only source of &quot;randomness&quot; is the fact that I don't know the positions and velocities precisely (for a technological reason, not a fundamental one: I have limited telescopes, for example). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can substitute this distribution in the definition of entropy and calculate an &quot;macroscopic entropy&quot; that depends on the positions, velocities and measurement precisions of the other planets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(x_i, v_i,\sigma_{x,i},\sigma_{v,i})  = - \int d\vec{x} d\vec{v} p(\vec{x}, \vec{v} | t, \vec{x}_i, \vec{v}_i, \sigma_{x,i},\sigma_{v,i}) \log \left[p(\vec{x}, \vec{v} |\vec{x}_i, \vec{v}_i, \sigma_{x,i},\sigma_{v,i})\right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this entropy means? Something quite close to what thermodynamic entropy means!!! Is the logarithm of the average configuration space volume where I expect to find the given planet in instant t (2)!!! And it's the entropy of a 'single particle'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no problem with that. I can even have situations where I must maximize this entropy! Suppose I don't know the position planet 2, but I do know all three planets have coplanar orbits. There are well defined procedures in information and inference theory that say to me that one way of dealing with this is to find the value of $\vec{x}_2$ that maximizes the entropy, subject to the constraint that all orbits are in the same plane, and then substitute this value in the original distribution. This is often called &quot;principle of maximum ignorance&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are interpretations of thermodynamics and statistical physics as an instance of this type of inference problem ( please refer to the works of E. T. Jaynes, I'll give a list of references below). In this interpretation there's nothing special on the fact that you have many degrees of freedom besides the fact that this is what makes you ignorant about the details of the system. This ignorance is what brings probabilities, entropies and maximum entropy principles to the table. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refrasing it a bit, probabilities and entropies are a part of your description when ignorance are built in your model. This ignorace could be a fundamental one - you can't know something about your system; could be a technical one - you could know if you had better instruments; and even, as in the case of statistical physics, a deliberate one - you can know, at least in principle, but you choose to leave detail out cause it isn't relevant in the scale you're interested in. But the details about &lt;strong&gt;how&lt;/strong&gt; you use probabilities, entropies and maximum entropy principles are completely agnostic to what the sources of your ignorance are. They are a tool for dealing with ignorance, no matter the reasons why you are ignorant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) For information-theoretic arguments why we have to maximize entropy in thermodynamics please refer to E. T. Jaynes' famous book &quot;Probability Theory: The Logic of Science&quot; (3) and this series of articles:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jaynes, E. T., 1957, &lt;a href=&quot;http://bayes.wustl.edu/etj/articles/theory.1.pdf&quot; rel=&quot;nofollow&quot;&gt;Information Theory and Statistical Mechanics&lt;/a&gt;, Phys. Rev., 106, 620&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jaynes, E. T., 1957, &lt;a href=&quot;http://bayes.wustl.edu/etj/articles/theory.2.pdf&quot; rel=&quot;nofollow&quot;&gt;Information Theory and Statistical Mechanics II&lt;/a&gt;, Phys. Rev., 108, 171.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another interesting source:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Caticha, A., 2008, &lt;a href=&quot;http://arxiv.org/pdf/0808.0012&quot; rel=&quot;nofollow&quot;&gt;Lectures on Probability, Entropy and Statistical Physics&lt;/a&gt;, arxiv:0808.0012&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) This can be given a rigorous meaning within information theory. For any distribution p(x) let the set $A_\epsilon$ be defined as the smallest set of points with probability greater than $1 - \epsilon$. Then the size of this set must be of order:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$log |A_\epsilon| = S + O(\epsilon)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For another form of this result see the book &quot;Information Theory&quot; by Cover and Thomas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Some of Jaynes's rants about quantum theory in this book may appear odd today, but let's excuse him. He committed some errors too. Just focus on the probability theory, information theory and statistical physics stuff which is quite amazing. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) It seems that dealing with this kinds of problems from Celestial Mechanics was actually one of the first problems that made Laplace interested in probabilities, and apparently he used it in calculations on Celestial Mechanics. The other problem that took his attention towards probability theory was... gambling! Hahaha...&lt;/p&gt;&#xA;" OwnerUserId="673" LastEditorUserId="673" LastEditorDisplayName="" LastEditDate="2011-02-03T19:56:27.270" LastActivityDate="2011-02-03T19:56:27.270" CommentCount="4" />
  <row Id="4555" PostTypeId="2" ParentId="4522" CreationDate="2011-02-03T19:33:24.900" Score="4" ViewCount="" Body="&lt;p&gt;I started writing a catchy but long explanation, but since you understand it and want something short for others, let's try this...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put too much salt in a glass of water, you saturate the water and end up with salt sitting at the bottom of the glass. If the temperature changes, the amount of salt that can dissolve changes (more for higher temperature, but you can leave that out). (For extra pithy-ness, leave this paragraph out entirely.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For gases, besides liquid temperature, gas pressure matters. More pressure means more dissolves. When you open the soda and lose the factory-provided pressure, the gas pressure above the soda is suddenly lower, so carbon dioxide starts leaving the soda. It keeps doing this until &quot;enough&quot; CO2 is in the space above the soda. More space means you need more CO2 to fill it up. So, if you crush the bottle to leave less space, less CO2 escapes from the soda, and it stays fizzy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this glosses over a lot of usefully clarifying stuff, such as the concepts you mentioned in your post, but it keeps it short. If you can hold their attention long enough, I would throw in a comment about how only the CO2 pressure matters, not the general gas pressure, just so they don't buy those worthless &quot;pump air into your soda bottle&quot; devices.&lt;/p&gt;&#xA;" OwnerUserId="1196" LastEditorUserId="1196" LastEditorDisplayName="" LastEditDate="2011-02-03T20:55:44.813" LastActivityDate="2011-02-03T20:55:44.813" CommentCount="1" />
  <row Id="4556" PostTypeId="2" ParentId="4546" CreationDate="2011-02-03T21:29:49.857" Score="0" ViewCount="" Body="&lt;p&gt;The result is generic to matter which is under greater pressure due to gravitational implosion.  The hydrostatic equilibrium of a star is the condition where the internal pressure of the material keeps the star in a static configuration against the inward push of gravity&#xA;$$&#xD;&#xA;\frac{dP}{dr}~=~-\frac{GM\rho}{r^2},&#xD;&#xA;$$&#xA;for $P$ the pressure and $\rho$ the density of matter.  We may think of an adiabatic situation where a small variation in the stellar radius occurs, where the star shrinks inwards from $R$ to $R~-~\Delta R$.  The density of a unit of material is then $\rho~=~3m/(4\pi r^3)$, which we sum up to $M$.  We consider the equation of state as the natural gas law $PV~=~NkT$.  So the change in the pressure is&#xA;$$&#xD;&#xA;\Delta P~=~\int_P^{P+\Delta P}dP~=~\int_R^{R~-~\Delta R}\frac{3GM^2}{4\pi r^5}dr&#xD;&#xA;$$&#xA;$$&#xD;&#xA;=~-\frac{15GM^2}{4\pi r^4}\Big|_R^{R~-~\Delta R}~=~-\frac{15GM^2}{4\pi}\Big(\frac{1}{(R~-~\Delta R)^4}~-~\frac{1}{R^4}\Big)~\simeq~\frac{15GM^2}{\pi R^4}\frac{\Delta R}{R}&#xD;&#xA;$$&#xA;This means the change in the pressure is positive, as we might expect, and the change in the temperature within a unit volume in the star is similarly positive $\Delta T~=~\frac{V}{Nk}\Delta P$.  Then the entropy is $\Delta S~=~c\Delta T/T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we expect the star to increase its luminosity, and the hydrostatic equation of Eddington was derived to compute that.  This applies up to the point the implosion increases the gravity field enough to red shift light.  The delay coordinate&#xA;$$&#xD;&#xA;r^*~=~r~+~2GM log\Big(\frac{r~-~2GM}{2GM}\Big)&#xD;&#xA;$$&#xA;maybe used to compute the red shift.  General relativity then predicts a dimming of the star as it collapses.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-02-03T22:30:43.017" LastActivityDate="2011-02-03T22:30:43.017" CommentCount="1" />
  <row Id="4557" PostTypeId="2" ParentId="4503" CreationDate="2011-02-03T22:28:29.000" Score="1" ViewCount="" Body="&lt;p&gt;Briefly: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big bang did not happen at a single point and explode into some existing space.  Instead, the big bang happened &quot;everywhere&quot;; since then, space itself has expanded from the initial singularity into the present size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first the universe was incredibly hot and dense, so hot and dense that elementary particles could not yet combine to form neutral atoms, and particles, including photons, could only travel a short distance before hitting and interacting with other particles.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the universe expanded, it cooled and and became less dense.  Eventually the density crossed beneath a threshold allowing photons to travel essentially forever without running into something; we say that the universe became transparent to photons at this time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nearly all of the photons that were in the cosmic soup at that time are still here.  Because the expansion of the universe is an expansion of all space, the expansion carried the photons with it, so they are everywhere.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These photons, with their wavelength stretched by the intervening expansion of space, form the cosmic microwave background radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good book on the early universe is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0465024378&quot; rel=&quot;nofollow&quot;&gt;The First Three Minutes&lt;/a&gt;  by Steven Weinberg.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-02-03T22:28:29.000" />
  <row Id="4558" PostTypeId="1" AcceptedAnswerId="4567" CreationDate="2011-02-03T22:54:54.917" Score="8" ViewCount="347" Body="&lt;p&gt;One of the main sources of subtlety in the AdS/CFT correspondence is the role played by boundary terms in the action. For example, for a scalar field in AdS there is range of masses just above the Breitenlohner-Freedman bound where there are two possible quantizations and which one you get depends on what boundary terms you add to the action. Boundary terms are also essential in the treatment of first-order Lagrangians for fermions and self-dual tensor fields. These all involve the &quot;UV&quot; boundary as $z \rightarrow 0$ in Poincare coordinates. Then there are dual models of QCD like the hard-wall model where one imposes an IR cutoff and imposes boundary conditions at the IR boundary and/or adds IR boundary terms to the action. My question is a bit vague, but basically I would like references to reviews, books or papers that give a good general treatment of the variational principle when one has to be careful about boundary terms. It would help if they clearly distinguish the requirements that follow from mathematical consistency from those that are imposed because of a desire to model the physics in a certain way.&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="838" LastEditorDisplayName="" LastEditDate="2011-02-04T18:09:03.960" LastActivityDate="2011-02-04T18:09:03.960" Title="Treatment of boundary terms when applying the variational principle" Tags="&lt;variational-principle&gt;&lt;ads-cft&gt;&lt;reference-request&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="4559" PostTypeId="2" ParentId="4471" CreationDate="2011-02-03T23:08:59.787" Score="3" ViewCount="" Body="&lt;p&gt;There is a deeper reason for $F~=~\frac{d^2x}{dt^2}$  Within the Galilean group it is an invariant with respect to all changes of frame $x’~=~x~+~vt$.  The acceleration of a body is not something which can be made to vanish by boosting to another Galilean frame as&#xA;$$&#xD;&#xA;F’~=~\frac{d^2x’}{dt^2}~=~\frac{d^2x}{dt^2}~+~\frac{d^2vt}{dt^2}&#xD;&#xA;$$&#xA;where for constant $v$ the second term is clearly zero.  The next higher derivative $dF/dt~=~mda/dt$, called a jerk” is also invariant, as are all $d^nx/dt^n$, but the acceleration contained in $T^2_p$ is the lowest element on the jet $T^n_p$ $n~\ge~2$ which is invariant.  Further, odd powers of $n$ would not be time reverse invariant under $t~\rightarrow~-t$&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-03T23:08:59.787" />
  <row Id="4560" PostTypeId="2" ParentId="4245" CreationDate="2011-02-03T23:49:09.370" Score="1" ViewCount="" Body="&lt;p&gt;Partial answer: A fire increases the temperature. This separates things which are volatile from things which are non volatile. Examining the periodic table of the elements we see that the base forming oxides are on the far left (i.e. the alkali and alkali earths), while the acid forming oxides are on the far right (i.e. the halogens). So what is happening is that the acid forming atoms are more volatile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, the problem reduces to the question, &quot;why is it that the oxides of the alkali and alkali earths metals are non volatile, while the halogens and their oxides are volatile?&quot; This is a question that can be answered completely in physics; it's due to their electron configurations.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-03T23:49:09.370" />
  <row Id="4561" PostTypeId="1" AcceptedAnswerId="4562" CreationDate="2011-02-03T23:57:08.377" Score="2" ViewCount="114" Body="&lt;p&gt;In the periodic table of the elements, only a few of the elements are gases at standard temperature and pressure. Those elements include the noble gases some of the halogens, and a few of the elements next to them, i.e. oxygen and nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why are these elements gases?&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-02-04T23:06:12.143" LastActivityDate="2011-02-04T23:06:12.143" Title="Why are the lighter halogens gases?" Tags="&lt;thermodynamics&gt;&lt;physical-chemistry&gt;&lt;chemical-compounds&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="4562" PostTypeId="2" ParentId="4561" CreationDate="2011-02-04T00:09:12.217" Score="1" ViewCount="" Body="&lt;p&gt;The [EDIT] lighter [/EDIT] halogens are quite eager to form covalent bounds and become diatomic molecules. These are strongly bound. Because charge is distributed symmetrically among these bounds, the resulting molecules are not strongly polarized, so the interaction with other molecules is relatively weak. Therefore, you neither have solids nor liquids with them.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditorDisplayName="" LastEditDate="2011-02-04T15:58:54.817" LastActivityDate="2011-02-04T15:58:54.817" CommentCount="5" />
  <row Id="4563" PostTypeId="2" ParentId="2687" CreationDate="2011-02-04T00:19:08.070" Score="2" ViewCount="" Body="&lt;p&gt;Mathematical foundations of elasticity by Marsden and Hughes. Published 1983 and then republished by Dover 1994, so while not exactly recent it's very modern compared to the one by Muskhelishvili. It uses differential forms, lie groups, and bifurcation theory and all that. &lt;/p&gt;&#xA;" LastActivityDate="2011-02-04T00:19:08.070" CommunityOwnedDate="2011-02-04T10:52:39.143" CommentCount="1" />
  <row Id="4564" PostTypeId="2" ParentId="4561" CreationDate="2011-02-04T00:31:35.957" Score="-1" ViewCount="" Body="&lt;p&gt;They are gases because God wanted the Germans to use Chlorine gas in WWI :-).  It is basically for the reason Lagerbaer states.  They have one open p-orbital available and bind in diatomic molecules.  The perturbation on the other closed p-orbitals is small so there is little dipolar interaction.  You have to make then very cold to get a liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don’t know what liquid Cl would be like, but something tells me it is a nasty fluid.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-04T00:31:35.957" CommentCount="2" />
  <row Id="4565" PostTypeId="2" ParentId="4558" CreationDate="2011-02-04T00:49:27.157" Score="1" ViewCount="" Body="&lt;p&gt;Here is a general treatment, &lt;a href=&quot;http://arxiv.org/abs/hep-th/0402184&quot; rel=&quot;nofollow&quot;&gt;Ishibashi and Wald&lt;/a&gt;, in a slightly more formal language than normal in this type of discussion. Not sure if this is what you are looking for though.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-02-04T00:49:27.157" />
  <row Id="4566" PostTypeId="2" ParentId="2687" CreationDate="2011-02-04T04:11:50.203" Score="1" ViewCount="" Body="&lt;p&gt;My favorite is: &quot;Mechanical Vibrations&quot;. I find it interesting because it gives the tensor notation for small vibrations. It's very practical oriented and it's cheap (Dover published):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486647854&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Mechanical-Vibrations-Dover-Books-Engineering/dp/0486647854&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-04T04:11:50.203" CommunityOwnedDate="2011-02-04T10:52:39.143" />
  <row Id="4567" PostTypeId="2" ParentId="4558" CreationDate="2011-02-04T04:48:44.050" Score="6" ViewCount="" Body="&lt;p&gt;I'm not sure how far back you want to go, but one of the earliest &quot;careful&quot; treatments (in the Hamiltonian formalism) is in the paper by Regge and Teitelboim:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;T. Regge and C. Teitelboim, “Role Of Surface Integrals In The Hamiltonian Formulation Of General Relativity,” Annals Phys. 88, 286 (1974)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Analogous work in the Lagrangian formalism didn't happen until much more recently. See the paper by Mann and Marolf, or the treatment in this relevant but shameless self-promotion:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mann, Marolf, McNees, and Virmani, &quot;On the Stress Tensor for Asymptotically Flat Gravity,&quot;&#xA;(http://arxiv.org/abs/0804.2079)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As far as calculations specific to AAdS spacetimes there is the paper by Skenderis and Papadimitriou&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Papadimitriou and Skenderis, &quot;Thermodynamics of Asymptotically Locally AdS Spacetimes,&quot;&#xA;(http://arxiv.org/abs/hep-th/0505190)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;which attempts to formulate things in a language similar to the Wald paper referenced by Moshe. All of these papers are concerned with the consistent treatment of the variational problem on spacetimes with a spatial infinity. The references of the last two will turn up additional relevant works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should immediately discount any paper which claims that the inclusion of the Gibbons-Hawking-York surface term gives a good variational problem for a gravitational theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This topic -- the proper variational formulation of gravitational theories -- is one of my main interests. I'm trying to take the Red Line from Loyola down to U of C for seminars this term (as teaching allows), so perhaps we can talk about it some time. &lt;/p&gt;&#xA;" OwnerUserId="1727" LastActivityDate="2011-02-04T04:48:44.050" CommentCount="6" />
  <row Id="4568" PostTypeId="2" ParentId="2455" CreationDate="2011-02-04T04:54:54.337" Score="0" ViewCount="" Body="&lt;p&gt;In humans (and vertebrates) the retinal structure evolved basically inverted in that the photons have to pass through layers of neurons and blood vessels before they hit the rods and cones. Wolves and some other vertebrates have a reflective membrane(the tapetum lucidum)  behind the photoreceptors that reflect the photons back through the rods and cones and provide a double pass amplification (causing &quot;eyeshine&quot;). Humans lack this.&#xA;The retina has ten layers which process signals from the photoreceptors, whose surface membranes contain retinol, which is isomerised by the photon energy, which affects ion channels, and may result in an action potential (necessarily leaving out all sorts of steps) which is preprocessed by retinal ganglion cells and information compressed and conducted along the optic nerve axons to the visual processing areas of the brain.&#xA;Vision has been extensively studied and is immensely complicated. Much of the energy takes part in chemical reactions and in multiple action potentials (electrochemical) in the preprocessing in the 10 layer retina.  Francis Crick (physicist, of DNA fame) and Christof Koch have written much about it--Koch's book, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0974707708&quot; rel=&quot;nofollow&quot;&gt;The Quest for Consciousness&lt;/a&gt; is fairly technical neuroscience focussing on the visual system.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-02-04T06:07:46.480" LastActivityDate="2011-02-04T06:07:46.480" CommentCount="1" />
  <row Id="4569" PostTypeId="1" AcceptedAnswerId="4596" CreationDate="2011-02-04T08:58:13.320" Score="2" ViewCount="262" Body="&lt;p&gt;I've heard the claim that some aspects of string theory are used to improve Monte-Carlo simulations of lattice QCD, for example by people working at the LHC.&#xA;I know a bit about Monte-Carlo methods in lattice QCD, but have no idea of how string theory ideas are applied to QCD at all, so I'd like to ask for &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;a short explanation of how and why string theory ideas are applied to QCD in general and&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;how they can be used in Monte-Carlo simulations of lattice QCD in particular.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;References are welcome, too, of course. Maybe I should have looked for them myself, but I'm sure I'll get some useful recommendations much faster this way :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Let's narrow the question down to: How can ideas in string theory be used to improve Monte-Carlo simulations of lattice QCD? (Lattice QCD is to be understood in the sense of strict SM without any ideas beyond the SM like supersymmetry, Ads/Cft etc.)&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditorDisplayName="" LastEditDate="2011-02-04T09:18:51.290" LastActivityDate="2011-03-02T04:42:12.507" Title="Lattice QCD and string theory" Tags="&lt;string-theory&gt;&lt;gauge-theory&gt;&lt;lattice-model&gt;&lt;qcd&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="4571" PostTypeId="2" ParentId="2839" CreationDate="2011-02-04T11:11:17.633" Score="1" ViewCount="" Body="&lt;p&gt;I think the question has been anwered. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for the sake of brevity, two more:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is true due to equipartition theorem. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(theorem, not law!)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is true by experimental evidence. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-02-04T11:11:17.633" />
  <row Id="4574" PostTypeId="2" ParentId="3164" CreationDate="2011-02-04T12:10:57.863" Score="-1" ViewCount="" Body="&lt;p&gt;One addition to Lubos answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radio waves, emitted by antennas are never spherical! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;AFAIK Hertz himself undertook the painstaking work to calculate &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the picture of the wave around a dipole and published that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;in one of the first papers on electromagnetic waves.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this talk about &quot;spherical&quot; waves arises &lt;/p&gt;&#xA;&#xA;&lt;p&gt;from sound wave analogies in popular books and brains. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Georg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think its not a question of &quot;radio&quot; or dipole antennas, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but transversal waves, only longitudinal waves can be &quot;spherical&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-02-05T11:45:46.327" LastActivityDate="2011-02-05T11:45:46.327" CommentCount="3" />
  <row Id="4575" PostTypeId="1" AcceptedAnswerId="4578" CreationDate="2011-02-04T12:52:27.107" Score="1" ViewCount="141" Body="&lt;p&gt;Here :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.newscientist.com/article/dn20084-neutron-star-seen-forming-exotic-new-state-of-matter.html&quot; rel=&quot;nofollow&quot;&gt;http://www.newscientist.com/article/dn20084-neutron-star-seen-forming-exotic-new-state-of-matter.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are news on superfluidity in a neutron star. The necessary bosons they say are pairs of neutrons. So far, so good. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then they postulate the production of neutrinos in the formation of those neutron pairs. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When neutrons pair up to form a superfluid they release neutrinos which should pass easily through the star,...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What kind of reaction is that? n + n =&gt;  neutrino + &quot;n-n&quot; + &quot;XYZ&quot; ? &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-02-18T16:28:49.433" LastActivityDate="2011-02-18T16:28:49.433" Title="neutrinos by formation of &quot;neutron pairs&quot;" Tags="&lt;astrophysics&gt;&lt;neutron-stars&gt;&lt;neutrino&gt;" AnswerCount="1" />
  <row Id="4577" PostTypeId="2" ParentId="2839" CreationDate="2011-02-04T13:23:27.413" Score="3" ViewCount="" Body="&lt;p&gt;Avogadro's hypothesis holds only for ideal gases. For example, if we take a van der Waals gas with equation of state&#xA;$$ (P + n^2 a / V)(V - n b) = nRT $$&#xA;it is easy to see that, for fixed $P$, $V$, $T$, the number of moles $n$ depends on the parameters $a$ and $b$, which in general will change from one type of gas to the other. &lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2011-02-04T13:23:27.413" />
  <row Id="4578" PostTypeId="2" ParentId="4575" CreationDate="2011-02-04T13:33:04.693" Score="1" ViewCount="" Body="&lt;p&gt;The phenomenon was first predicted in this paper:&lt;br&gt;&#xA;&quot;&lt;em&gt;Neutrino pair emission from finite-temperature neutron superfluid and the cooling of young neutron stars&lt;/em&gt;&quot;&lt;br&gt;&#xA;&lt;em&gt;Flowers E. G., Ruderman M., Sutherland P. G., 1976, ApJ, 205,541&lt;/em&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.adsabs.harvard.edu/abs/1976ApJ...205..541F&quot; rel=&quot;nofollow&quot;&gt;PDF here&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process they are describing is actually $n^*n^*\to\nu\overline\nu$, where $n^*$ is a quasiparticle excitation above the superfluid condensate. It is the annihilation of those quasipartilces that produces those neutrinos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the &quot;real&quot; neutrons do not annihilate, but the description of the process in terms of the &quot;real&quot; neutrons is way too complex, because there is a whole medium made of constantly interacting neutrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is once again shows the usefulness of a concept of a quasiparticle.&lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-02-04T15:07:04.597" LastActivityDate="2011-02-04T15:07:04.597" CommentCount="3" />
  <row Id="4579" PostTypeId="2" ParentId="3229" CreationDate="2011-02-04T14:16:57.633" Score="0" ViewCount="" Body="&lt;p&gt;T-duality involves an interchange $x_9~\rightarrow~x_9~+~2\pi R$ with the compactification of a dimension. A particle running around this circle has a momentum $\sim~1/R$, and this does connect in part with the idea of a reciprocal lattice. The mass of the particle running around the circle is $M~=~n/R$. A closed string can also wrap around a circle. The winding of a string $w$ contributes energy $E~=~2\pi wRT$, $T~=~1/(2\pi\alpha’)$ the string. The mass is related to the winding number by &#xA;$$&#xD;&#xA;M^2~=~\Big(\frac{n}{R}\Big)^2~+~\Big(\frac{wR}{\alpha’}\Big)^2~+~\frac{1}{\alpha’}(2N~+~w~-~2)&#xD;&#xA;$$&#xA;This has the reciprocal structure as well. The connection between the lattice and compactification and string/brane wrappings will have to be brought in tighter with a connection to compactified spaces.  T-duality relates type IIA superstring theory to type IIB superstring theory, and it relates heterotic $SO(32)$ superstring theory to heterotic $E_8\times E_8$ superstring theory.  There is an interchange between mode and winding numbers with $R~\rightarrow~\alpha’/R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S-duality is a relationship between two similar field theories with reciprocal coupling strengths.  This is based on the Montenen-Olive relationship which is a form of the Bohr-Sommerfeld quantization rule.  Given charges $q,~g$ for a similar gauge field then&#xA;$$&#xD;&#xA;qg~=~n\hbar.&#xD;&#xA;$$&#xA;A low energy limit of a type IIB SUGRA with a complex field $\phi$ decomposed into an axion and diliton field as the real and imaginary parts $\phi~=~\psi~+~iexp(X)$ has an $SL(2,{\mathbb R})$ structure, where the discrete subset $SL(2,{\mathbb Z})$ has a modular (or M{\”o}bius) structure, where the field has a linear fractional transformation&#xA;$$&#xD;&#xA;\phi~\rightarrow~\frac{a\phi~+~b}{c\phi~+~d},~ad~-~bc~=~1&#xD;&#xA;$$&#xA;where for $b~=~c~=~1$ and $a~=~d~=~0$ we have another reciprocating relationship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The T and S dualities may be composed in various ways to define this less well understood U-duality.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-02-04T14:28:05.753" LastActivityDate="2011-02-04T14:28:05.753" />
  <row Id="4580" PostTypeId="1" CreationDate="2011-02-04T14:32:39.767" Score="0" ViewCount="247" Body="&lt;p&gt;Why is the amount of energy transferred to a driven mechanical oscillator largest when the drive frequency is the same as the natural frequency of the oscillator?  Why are they exactly out of phase by π/2?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarifying example experimental setup: &lt;a href=&quot;http://av.ph.tum.de/Experiment/1000/Grafik/b1605.gif&quot; rel=&quot;nofollow&quot;&gt;http://av.ph.tum.de/Experiment/1000/Grafik/b1605.gif&lt;/a&gt;&lt;/p&gt;&#xA;" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-02-04T19:21:45.517" LastActivityDate="2011-02-04T19:21:45.517" Title="Energy transfer to a driven oscillator" Tags="&lt;energy&gt;&lt;resonance&gt;" AnswerCount="5" CommentCount="7" />
  <row Id="4581" PostTypeId="1" AcceptedAnswerId="4584" CreationDate="2011-02-04T15:11:54.247" Score="2" ViewCount="248" Body="&lt;p&gt;Taking off from the Rutherford-Bohr model,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/R8c0f.png&quot; alt=&quot;Rutherford-Bohr model&quot;&gt;&lt;br&gt;&#xA;&lt;strong&gt;Figure 1.&lt;/strong&gt; &lt;em&gt;Rutherford-Bohr model.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when an electron absorbs energy (from light, heat, or otherwise) it jumps to a higher energy level, or shell. Then when the electron loses that energy, it reverts back to a lower-numbered shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that the atomic orbital model is now more widely accepted, how does the above concept relate to it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/88Kwj.jpg&quot; alt=&quot;atomic orbitals&quot;&gt;&lt;br&gt;&#xA;&lt;strong&gt;Figure 2.&lt;/strong&gt; &lt;em&gt;Atomic orbitals 1s, 2s, 2px, 2py, and 2pz.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the orbital model is based on the premise that we can never pinpoint the exact location of an electron at any time, hence the &lt;em&gt;electron cloud&lt;/em&gt;, or &lt;em&gt;probability density&lt;/em&gt; representation. Let's take a hydrogen atom for example. Its lone electron will be likely to be found in the volume defined by 1s. If that is excited to a higher energy level, will it be more likely to be found in the volume defined by 2s?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about for other elements with more electrons? When excited, will they follow the electron placement rules as shown below? (I wonder why I couldn't find 2px, 2py, 2pz in the illustration below, but that's probably for another question)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/qPYcN.png&quot; alt=&quot;Electron placement&quot;&gt;&lt;br&gt;&#xA;&lt;strong&gt;Figure 3.&lt;/strong&gt; &lt;em&gt;Electron placement.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may have some terms and concepts mixed up, so corrections will be much appreciated. Also, please keep the jargon to a minimum, if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;All pictures are from Wikipedia.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="293" LastActivityDate="2011-02-04T20:22:52.020" Title="How does electron excitation relate to atomic orbitals?" Tags="&lt;quantum-mechanics&gt;&lt;energy&gt;&lt;atoms&gt;&lt;orbital-motion&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="4582" PostTypeId="2" ParentId="4580" CreationDate="2011-02-04T15:33:04.247" Score="0" ViewCount="" Body="&lt;p&gt;I think what you are raising a question about is a special case of the double pendulum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Double_pendulum&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Double_pendulum&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-04T15:33:04.247" CommentCount="2" />
  <row Id="4583" PostTypeId="2" ParentId="4580" CreationDate="2011-02-04T15:42:07.317" Score="2" ViewCount="" Body="&lt;p&gt;If you have two decoupled oscillators, they satisfy differential equations&#xA;$$-\frac{d^2}{dt^2}x_i=\omega^2_{i} x_i$$&#xA;where $i=1,2$. The solutions are clearly multiples of $\cos(\omega_i t+\phi_i)$. Now, consider two interacting oscillators. Each oscillator must know about the phase of the other, so the simplest dependence is to add a multiple of $x_2$ (a linear multiple) to the equation for $x_1$ and vice versa:&#xA;$$-\frac{d^2}{dt^2}\vec x=\Omega \cdot \vec x$$&#xA;Here, $\Omega$ (capital omega) is a $2\times 2$ matrix, and if you don't know how to multiply matrices, you should learn it. The matrix $\Omega$ has diagonal elements $\omega_1^2$ and $\omega_2^2$ but we have just added real nonzero off-diagonal elements $\Omega_{12}$ and $\Omega_{21}$, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a system of two linear differential equations may be easily solved by diagonalizing the matrix $\Omega$. In other words, there exist two linear combinations of the two equations such that the second time-derivative of the combination of $x_1,x_2$ will only depend on the multiple of the same combination of $x_1,x_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diagonalizing the matrix means to find the eigenvalues. The eigenvalues of the $2\times 2$ matrix $\Omega$ are solutions to the charcteristic equation&#xA;$$0 = \det(\Omega-\lambda\cdot{\bf 1}) = (\omega_1^2-\lambda)(\omega_2^2-\lambda)-\Omega_{12}\Omega_{21}$$&#xA;That's a quadratic equation for the eigenvalue $\lambda$ that has two solutions&#xA;$$\lambda_\pm = \frac{\omega_1^2+\omega_2^2\pm\sqrt{(\omega_1^2-\omega_2^2)^2+4\Omega_{12}\Omega_{21}}}{2} $$&#xA;Note that under the square root, there is a difference of the squared frequencies. It's because the $+2\omega_1^2\omega_2^2$ term got overcompensated by $-4\omega_1^2\omega_2^2$ and switched the sign. For a fixed coupling between the two degrees of freedom, i.e. for a fixed $\Omega_{12}\Omega_{21}$, the square root - representing the difference between the two eigenfrequencies - is minimized for $\omega_1^2$ close to $\omega_2^2$. This proximity is what you want for a resonance, an effective transfer of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the energy is being transferred from one oscillator to the other by the (small) off-diagonal $\Omega_{12}$ and $\Omega_{21}$ elements, the kinetic energy stored in $x_1$ (plus the corresponding potential energy) will be slowly moved to the kinetic energy stored in $x_2$ (plus the corresponding potential energy). Energy conservation implies that the energies of the two oscillators have to go like&#xA;$$E_1 = E \cos^2(\omega_d t), \quad E_2 = E \sin^2(\omega_d t)$$&#xA;because they sum up to constant (coefficients omitted). That means that $x_1^{max}$ itself has to go like $\cos(\omega_d t)$ and similarly for $x_2^{max}$ and $\sin$. The phase difference between $x_1$ and $x_2$ is therefore $\pm \pi/2$.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-02-04T16:39:25.203" LastActivityDate="2011-02-04T16:39:25.203" CommentCount="4" />
  <row Id="4584" PostTypeId="2" ParentId="4581" CreationDate="2011-02-04T15:49:13.177" Score="5" ViewCount="" Body="&lt;p&gt;The Rutherford model of the atom didn't respect any quantization: it was a classical planetary model. The Bohr-Sommerfeld model had the quantization of the allowed orbit from your first picture; however, you conflated these two models and spoke about &quot;Rutherford-Bohr&quot; model which has never existed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third thing that you conflated is the actual quantum mechanical equation that describes the atom correctly - in the non-relativistic limit - while neither the Rutherford model nor the Bohr model are correct in details. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The states $1s,2s,2p,3s,3p,3d,\dots$ that you refer to on your second and third picture only exist in the correct quantum mechanical model that predicts three quantum numbers for the electron, $n,l,m$ (if we ignore the spin). The Bohr model only predicts (incorrectly) one quantum number $n$, so it would only have states $n=1,2,3,4$ and no extra $s,p,d$ labels that distinguish different values of $l$. In some sense, the Bohr model has angular momentum $l=n$ and it doesn't allow any values $l&amp;lt;n$ while the correct quantum mechanical models only allows $l&amp;lt;n$ but all of them - it predicts $l=0,1,2,\dots n-1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you should recognize the different models. The Rutherford model is classical and hopeless - and only included the insight that the nuclei are smaller than the atoms. It didn't know anything correct about the motion of the electron. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Bohr-Sommerfeld model knew something &quot;qualitative&quot; about the motion of the electron, namely that there was something quantized about it, but it was still too classical and it was the wrong model that only happened to &quot;predict&quot; the right energies after some adjustments but this agreement for the Hydrogen atom was completely coincidental and related to the fact that the Hydrogen atom may be solved exactly (and the answer for the allowed energies is very simple).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer how you can see $1s,2s,2p,\dots$ states in the Bohr (or even Rutherford) model is obviously that you can't see them because the Bohr and Rutherford models are invalid models of these detailed features of the atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you decided to learn quantum mechanics and abandoned the naive ideas such as the Rutherford and Bohr-Sommerfeld models, you could also discuss other properties of the electron states in the Hydrogen atom. For example, the states $2px, 2py, 2pz$ from your first picture are particular complex linear combinations of the usual basis of states $2p_{m=-1},2p_{m=+1},2p_{m=0}$. In fact, $2pz=2p_{m=0}$ while $2px\pm i\cdot 2py =  2p_{m=\pm 1}$, up to normalization factors.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T15:56:58.063" LastActivityDate="2011-02-04T15:56:58.063" />
  <row Id="4585" PostTypeId="2" ParentId="4580" CreationDate="2011-02-04T16:19:42.047" Score="0" ViewCount="" Body="&lt;p&gt;If the frequencies are different, the averaged influence of one oscillator on the other cancels out over many cycles. Only if the frequencies are close together do we get the possibility of a resonance with influences building up constructively cycle after cycle.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-04T16:19:42.047" />
  <row Id="4586" PostTypeId="1" AcceptedAnswerId="4588" CreationDate="2011-02-04T16:23:17.703" Score="2" ViewCount="405" Body="&lt;p&gt;I have been reading Hawking's 'A Brief History of Time' and it has gotten me thinking about Einstein's theory of relativity, in that it assumes that an object must have infinite mass if it is to be traveling at the speed of light (please correct if I'm wrong in my beginner's knowledge of physics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, do light waves have any sort of measurable mass?  Or in that same vein, do sound waves?&#xA;Is it somehow possible?&lt;/p&gt;&#xA;" OwnerUserId="1752" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-05T00:26:46.783" LastActivityDate="2011-02-05T00:26:46.783" Title="Do light and sound waves have mass" Tags="&lt;special-relativity&gt;&lt;waves&gt;&lt;mass&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="4587" PostTypeId="2" ParentId="4580" CreationDate="2011-02-04T16:31:01.517" Score="0" ViewCount="" Body="&lt;p&gt;I guess your question is related to the phenomenon called &quot;resonance&quot;. I shall give you a descriptive answer since you couldn't follow Lubos's rather mathematical answer. Hannes, whenever an oscillator is given some forced vibration it is forced to vibrate with the frequency of that 2nd oscillator and not with its own natural frequency. What does this mean? It means the two oscillators are vibrating without synchronization. They are not vibrating with &quot;in phase&quot;. Most of the time they are &quot;out of phase&quot;. The result is loss of energy in the form of heat and sound. On the other hand if the frequencies of the two oscillators match then they vibrate in perfect synchronism and they are always in phase. The loss becomes minimum. Almost all the energy of the two oscillators are manifested in the kinetic energy of the two oscillator system. Their amplitude becomes maximum. This is an example of resonance between two mechanical oscillators but the same principle holds in more general cases e.g. electrical circuit with a register, inductor and capacitor. When you tune your radio to a particular channel what you actually do is change the frequency of the circuit to match the signal frequency of that channel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope, you have understood the phenomenon of resonance.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-04T16:57:43.340" LastActivityDate="2011-02-04T16:57:43.340" CommentCount="0" />
  <row Id="4588" PostTypeId="2" ParentId="4586" CreationDate="2011-02-04T16:45:25.710" Score="7" ViewCount="" Body="&lt;p&gt;The particles of light waves - the photons - have the rest mass $m_0$ equal to zero. However, at the speed of light, $v=c$, the total mass &#xA;$$ m= \frac{m_0}{\sqrt{1-v^2/c^2}} $$&#xA;is increased to an indeterminate form, $0/0$, which should be evaluated as a finite number. The photons - and everything else - carry the total mass that is proportional to the total energy via the famous $E=mc^2$ relation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, this mass may be measured. For example, uranium nuclear power plants burn the uranium and reduce its mass by 0.1 percent or so because the waste products (the nuclei) are actually a little bit lighter. This energy may be completely transformed to the radiation coming from light bulbs - and the light from these light bulbs carry 0.1 percent of the uranium mass away. This mass is a source of gravitational field and adds inertia to boxes with this light etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sound is different. The speed of sound is much smaller than the speed of light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While &quot;phonons&quot; in low-temperature condensed matter physics - particles of sound - are analogous to photons in many respects, and $E=mc^2$ still applies, the same is not true for sound waves in the air etc. Because the temperature of the air is nonzero, the &quot;ground state&quot; - the lowest-energy state at fixed conditions, with the minimum number of &quot;sound quanta&quot; or &quot;phonons&quot; - is not really unique. Instead, there are many states of the air &quot;without any sound&quot; which correspond to chaotic configurations of the air molecules. So one can't consistently divide the energy of the air to the energy of its ground state and the energy of the phonons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course, if you produce some loud sounds, they will carry lots of energy in the air and the mass of the air will inevitably increase by $m=E/c^2$ which is, well, not too high because $c^2$ is a large number.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T16:50:36.173" LastActivityDate="2011-02-04T16:50:36.173" CommentCount="3" />
  <row Id="4590" PostTypeId="2" ParentId="4580" CreationDate="2011-02-04T18:04:25.413" Score="0" ViewCount="" Body="&lt;p&gt;The test setup pertains to &lt;strong&gt;forced&lt;/strong&gt; oscillations, and not coupling of two oscillators. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_oscillator&quot; rel=&quot;nofollow&quot;&gt;Driven Harmonic Motion&lt;/a&gt; for more information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The response amplitude depends on the ratio of (driven frequency) to (natural frequency) and is infinite when the two are equal and there is no damping. Damping, dampens then amplitude a but, and also changes the phase of the response. The reason is that the coupling force acts in the direction of motion, adding power (and hence energy) to the system.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-02-04T18:04:25.413" />
  <row Id="4591" PostTypeId="1" AcceptedAnswerId="4592" CreationDate="2011-02-04T18:48:36.700" Score="5" ViewCount="401" Body="&lt;p&gt;[Question suggested by @Lawrence B. Crowell's answer to the question on &lt;a href=&quot;http://physics.stackexchange.com/questions/3229/t-duality-approaches&quot;&gt;T-duality&lt;/a&gt;]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three parts to my question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A.  What is the action for $N=4$ SUSY Yang-Mills and what is the physics of the various terms in the action?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. Give a simple explanation for the origin of &lt;a href=&quot;http://en.wikipedia.org/wiki/Montonen%E2%80%93Olive_duality&quot; rel=&quot;nofollow&quot;&gt;Montonen-Olive duality&lt;/a&gt; in this theory. This duality maps the physics at gauge coupling $g$ to the physics at $1/g$, i.e. this is a duality between the strong and weak coupling regimes of the theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. &lt;strong&gt;Auxiliary Question&lt;/strong&gt;: Explain how one can decompose a SUSY gauge field into its bosonic and fermionic parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;strong&gt;Background&lt;/strong&gt;: I am familiar with non-abelian gauge theory and have had a class on SUSY some years back. So I pick up on the jargon but am missing a fuller picture.]&lt;/p&gt;&#xA;" LastActivityDate="2011-02-04T20:15:09.327" Title="$N=4$ supersymmetric yang-mills theory and S-duality" Tags="&lt;quantum-field-theory&gt;&lt;duality&gt;&lt;supersymmetry&gt;&lt;yang-mills&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="4592" PostTypeId="2" ParentId="4591" CreationDate="2011-02-04T20:09:32.983" Score="3" ViewCount="" Body="&lt;p&gt;A. The action of $N=4$ SYM (Super Yang-Mills theory) in $d=4$ is the simple dimensional reduction of the 9+1-dimensional SYM, the maximal dimensional SYM that exists. The latter is&#xA;$$S = \int d^{10} x\mbox{ Tr } \left( -\frac{1}{4} F_{\mu\nu}F^{\mu \nu} + \overline{\psi}D_\mu \gamma^\mu \psi \right)$$&#xA;where $D$ is the covariant derivative and $\psi$ is a real chiral spinor in 9+1 dimensions which has 16 real components, leading to 8 fermionic on-shell degrees of freedom. The dimensional reduction reduces $d^{10}x$ to $d^4 x$ but it also renames 6 &quot;compactified&quot; spatial components $A_\mu$ as six scalars $\Phi_I$ in $d=4$. The derivatives in the corresponding 6 directions are set to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one looks what fields and interactions we get in $d=4$ - it's straightforward to write the action - it's one gauge field; four Weyl fermions; six real scalars. All those fields transform as adjoint of the gauge group - the most popular ones are $SU(N)$. They have the standard kinetic terms; standard cubic couplings of the gauge field to all the other fields; the usual quartic Yang-Mills self-interaction of the gauge field; quartic interaction of the gauge field and the scalars; Yukawa cubic couplings for the 6 scalars that arise from the gauge interactions of the fermions in 9+1 dimensions; quartic potential for the scalars which is equal to the squared commutator. Everything has to be traced over the gauge group. All these interactions are related and determined by supersymmetry - by 16 real supercharges. The individual vertices of Feynman diagrams are self-evident but the true physics behind all of them is related by symmetries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. The simplest explanation of the S-duality is to represent the gauge theory as the low-energy limit of the dynamics of a stack of D3-branes in type IIB string theory. The S-duality group - actually it's an $SL(2,Z)$ group because it may also act on the $\theta$-angle (RR-axion) - is directly inherited from the same S-duality group of type IIB string theory. In particular, the $g\to 1/g$ may be interpreted in the F-theory description of type IIB as the exchange of the 11th and 12th (infinitesimal) dimension of F-theory. One may also get the gauge theory as the compactification of the $d=6$ (2,0) superconformal field theory on a tiny two-torus, and $SL(2,Z)$ acts in the obvious way - again, the exchange of the two radii is the $g\to 1/g$ map. It's a non-perturbative duality so there's no simple perturbative &quot;field redefinition&quot; that would prove it at the classical level. However, one may make many consistency checks that the duality seems viable - e.g. one may found the magnetic monopole solutions that turn to light elementary excitations at the strong coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. The $N=4$ $d=4$ vector multiplet contains all the physical fields in the theory and I have already written what they are: a vector field with 2 physical polarizations, 6 real scalars, and 8 fermionic degrees of freedom from 4 Weyl fermions, carrying the $SU(4) \approx SO(6)$ R-symmetry group. It's not too helpful to use superspace for $N=4$ theories, unless one wants to break it to $N=1$ or $N=2$. Too big supersymmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree with Jeff that those things can be found in first (SUSY) chapters of any modern introductory textbook or other literature on advanced quantum field theory or string theory so in this sense, this question is a theft of the time of other users of this server.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, I also want to mention that the $N=4$ theory is arguably the &quot;most important&quot; or &quot;simplest&quot; $d=4$ non-gravitational theory, by the modern criteria, and the action above is far from the only one - and maybe even from the most elementary - way to describe this theory. This theory, by the AdS/CFT correspondence, is also dual i.e. exactly equivalent to type IIB string theory on $AdS_5\times S^5$. The $N=4$ SYM also has the &quot;dual superconformal symmetry&quot; that, together with the original superconformal symmetry, generates an infinite-dimensional &quot;Yangian symmetry&quot;. Twistor techniques are particularly useful for the computation of scattering amplitudes in this $N=4$ SYM and many of the twistor researchers think that the twistor formulae are more fundamental and elementary ways to describe physics of the SYM than the perturbative action above.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-04T20:15:09.327" LastActivityDate="2011-02-04T20:15:09.327" CommentCount="8" />
  <row Id="4593" PostTypeId="2" ParentId="4581" CreationDate="2011-02-04T20:22:52.020" Score="0" ViewCount="" Body="&lt;h3&gt;Using orbitals to describe multi-electron atoms:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Hydrogen (and hydrogen-line atoms like $He^+$) is a two-body system and for it we can solve the Schrödinger equation &lt;em&gt;analytically&lt;/em&gt; and thus obtain wavefunctions for electron orbitals that are known as 1s, 2s etc. It is a natural idea to use these orbitals to &lt;em&gt;approximate&lt;/em&gt; electronic structure of other atoms and it works not so bad for smaller ones. However, every electron added to the atom will influence distribution of other electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: Helium has 1s2 configuration which means there are two electrons of equal energy. However, it the energy 24.6 eV is needed to remove the first electron and 54.4 eV for the second.&lt;/p&gt;&#xA;" OwnerUserId="701" LastActivityDate="2011-02-04T20:22:52.020" />
  <row Id="4594" PostTypeId="1" AcceptedAnswerId="4595" CreationDate="2011-02-04T22:55:09.750" Score="1" ViewCount="177" Body="&lt;p&gt;Should they be taken out of the wrapping and stored as single units, or should I leave them in the package? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT to add I mean, the package is 24 bottles bundled together. Should I just leave them in the crate, or disperse them around the fridge?&lt;/p&gt;&#xA;" OwnerUserId="1755" LastEditorUserId="1755" LastEditorDisplayName="" LastEditDate="2011-02-06T04:18:31.777" LastActivityDate="2011-02-06T13:43:15.893" Title="How should 500ml water bottles be stored in the fridge so they will be coldest?" Tags="&lt;temperature&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="4595" PostTypeId="2" ParentId="4594" CreationDate="2011-02-04T23:05:50.097" Score="1" ViewCount="" Body="&lt;p&gt;That depends, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;if the water in the bottles is warm and You want to cool it down &#xA;as fast as possible, then unwrap them. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If there is enough time to cool or the water is cold already, &#xA;then leave the package intact, it might be useful later to &#xA;protect the water from heating up fast, when the bottle is removed &#xA;from fridge for use. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Georg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT to Todds edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is not much &quot;wrapping&quot; in this case, but the outermost &lt;/p&gt;&#xA;&#xA;&lt;p&gt;bottles are a shield for the inner ones. So, the answers stay the same. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-02-06T13:43:15.893" LastActivityDate="2011-02-06T13:43:15.893" />
  <row Id="4596" PostTypeId="2" ParentId="4569" CreationDate="2011-02-05T00:09:55.567" Score="3" ViewCount="" Body="&lt;p&gt;I don't know that much about lattice QCD, but I do know a few things about string theory and the gauge/string duality approach to QCD. I'm pretty sure the answer to your question is that ideas in string theory have not been used to improve Monte-Carlo simulations of QCD. Unified versions of string theory can't since they involve physics much above the QCD scale. Gauge/string duality models are still too crude, typically giving answers that are sometimes accurate to 15% but not in a well controlled approximation. The only way that I can see there being a smidgen of truth to the claim about using string ideas to improve lattice QCD is through the renormalization group improvement of lattice actions for QCD, an idea which I believe goes back to Symanzik. If you google &quot;lattice QCD renormalization group improvement&quot; you will find a number of reviews. Some ideas from string theory have improved our ability to do perturbative QCD calculations, and the RG improvement techniques are based on perturbative calculations, so it is just possible that this is the link that was mentioned to you. However I am not aware of this actually being the case, that is of a calculation of an RG improved lattice action that required new techniques from string theory in order to do the perturbative QCD calculations.&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T04:42:12.507" LastActivityDate="2011-03-02T04:42:12.507" CommentCount="1" />
  <row Id="4597" PostTypeId="1" AcceptedAnswerId="4611" CreationDate="2011-02-05T00:14:02.670" Score="3" ViewCount="415" Body="&lt;p&gt;How the Green's functions and the Quantum Mechanics are related? Do they can be used to solve the Schrödinger equation of an particle subjet to some potential that is not a Dirac's delta? And the proprieties of some Green's functions that are symmetrical, i.e. $ G(x|\xi) = G(\xi|x)^{\ast} $, has some relation with the propriety of the inner product $ \langle \alpha \vert \beta \rangle = \langle \beta \vert \alpha \rangle^{\ast} $?&lt;/p&gt;&#xA;" OwnerUserId="1618" LastActivityDate="2011-02-05T09:31:49.263" Title="Green functions in Quantum Mechanics" Tags="&lt;quantum-mechanics&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="4598" PostTypeId="2" ParentId="4586" CreationDate="2011-02-05T00:26:27.703" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;an object must have infinite mass if it is to be traveling at the speed of light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, that's not true at all. An object's mass is a fixed property that doesn't change, regardless of what speed it travels at. But its &lt;em&gt;energy&lt;/em&gt; does change. The energy increases with increasing speed, according to the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = \frac{mc^2}{\sqrt{1 - v^2/c^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $m$ is the mass, an inherent property of the object, and $E$ is the energy. What you're probably thinking about is the fact that an object that has a nonzero mass ($m &gt; 0$) can never travel at the speed of light, because it would require an infinite amount of energy for it to do so. But zero-mass particles like the photon (the quantum of light) can travel at the speed of light without requiring infinite energy. (In fact, they can't travel at any slower speed.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people use the term &quot;mass&quot; to mean the quantity that I'm calling energy (in different units), and the term &quot;rest mass&quot; when they want to refer to what I call the mass. In that case, one would say that a material object traveling at the speed of light would have infinite mass. I refer you to &lt;a href=&quot;http://physics.stackexchange.com/questions/3541/follow-up-how-can-a-photon-have-no-mass-and-still-travel-at-the-speed-of-light/3550#3550&quot;&gt;another answer&lt;/a&gt; I've written for more information on the historical context of the terms.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But, do light waves have any sort of measurable mass? Or in that same vein, do sound waves?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, but they do have energy. For light waves, the energy is related to the frequency $\omega$ of the wave,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = n\hbar\omega$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($n$ is the number of photons), and for sound waves, the energy is related to the amplitude (particle displacement) $\xi$ and the frequency $\omega$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = A\rho \xi^2\omega^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ is the cross-sectional area of the sound wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could calculate an &quot;equivalent mass&quot; as $m_\text{eq} = E/c^2$, which would tell you the amount of mass it would take to have the same energy (at rest) as a given light or sound wave. If it were possible to convert the energy of the wave into mass directly, $m_\text{eq}$ would be the amount of mass you'd get. But that's the only sense in which a wave has mass.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-02-05T00:26:27.703" />
  <row Id="4599" PostTypeId="1" CreationDate="2011-02-05T01:07:01.723" Score="0" ViewCount="157" Body="&lt;p&gt;I have an $n$-dimensional simple Gaussian function,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x) = \exp\bigl(-a|x - k|^2\bigr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a$ is a scalar, $x$ is a vector, and $k$ is a vector. $k$ is the center of the Gaussian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume I have two such Gaussian functions with different values of $a$ and $k$.  Assume I apply the &lt;a href=&quot;http://en.wikipedia.org/wiki/Antisymmetrizer&quot; rel=&quot;nofollow&quot;&gt;antisymmetrizer operator&lt;/a&gt; to each Gaussian and then try to compute the inner product between these two functions.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the expression for the inner product of 2 anti-symmetrized Gaussian functions?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How does the value of the inner product depend on the dimension of the vector space of the domain of the gaussian functions?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;How would you approach this problem?&lt;/p&gt;&#xA;" LastEditorUserId="391" LastEditorDisplayName="" LastEditDate="2011-02-05T03:22:52.927" LastActivityDate="2011-02-05T03:22:52.927" Title="How do I compute the inner product between two antisymmetized Gaussian basis functions?" Tags="&lt;quantum-mechanics&gt;&lt;homework&gt;&lt;mathematics&gt;" CommentCount="2" />
  <row Id="4600" PostTypeId="1" AcceptedAnswerId="4602" CreationDate="2011-02-05T01:11:11.203" Score="2" ViewCount="194" Body="&lt;p&gt;Today I saw the derivation of the wave equation in class, and I did not understand the following step. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are modeling a uniform-density string as being made up of tiny masses spaced a small amount $h$ apart connected by springs obeying Hooke's law. Let $y_i(t)$ be the vertical position of the $i$'th such particle. The mass of this particle is $(h/L) M$ where $L$ is the total length, $M$ is the total mass. On the other hand, the displacement of the string between the $i$'th and the $i+1$st particle can be shown to be approximately proportional to $y_{i+1}(t) - y_i(t)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that this leads to the equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{hM}{L} y_i&amp;#39;&amp;#39;(t) = k \left( [y_{i+1}(t) - y_i(t)] + [y_{i-1}(t) - y_i(t)] \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but that is, in fact, wrong; the derivation by instructor gives sticks a $1/h$ on the right hand side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain to me (i) why the right hand side needs a $1/h$ (ii) intuitively, what is happening? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first guess is that the spring constant $k$ should be proportional to the length of the spring, keeping everything else constant? But that seems weird to me; it seems to me like  if you have a spring with spring constant $k$, then cutting the same spring into two halves should result in two springs each with half the length but the same spring constant. Can someone unconfuse me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="1758" LastActivityDate="2011-02-05T02:47:06.357" Title="A basic question on the derivation of the wave equation" Tags="&lt;waves&gt;" AnswerCount="2" />
  <row Id="4601" PostTypeId="2" ParentId="4597" CreationDate="2011-02-05T01:22:19.950" Score="-4" ViewCount="" Body="&lt;p&gt;An interesting system to study to understand this is the Poincare half disk or half plane. ${\cal H}^2$. It also illustrates the role of Laplacian operators, Green’s functions and resolvents.  The Laplace-Beltrami operator on a Riemannian manifold with a Gaussian metric is&#xA;$$&#xD;&#xA;\Delta~=~\sum_{ij}{1\over\sqrt{g}}{\partial\over{\partial x^i}}\Big(\sqrt{g}g^{ij}{\partial\over{\partial x^j}}\Big),&#xD;&#xA;$$&#xA;for $g~=~|det(g_{ij}|$.  The Laplacians for the Poincare half plane and disk are&#xA;$$&#xD;&#xA;\Delta_{{\cal H}^2}~=~y^2\Big({{\partial^2}\over{\partial x^2}}~+~{{\partial^2}\over{\partial y^2}}\Big),~\Delta_{{\cal D}}~=~(\alpha^2~-~x^2~-~y^2)^2\Big({{\partial^2}\over{\partial x^2}}~+~{{\partial^2}\over{\partial y^2}}\Big).&#xD;&#xA;$$&#xA;The Laplacian commutes with all group elements $g~\in~Iso({\cal H}^2)$ $\Delta T_g~=~T_g\Delta$, so the metric is invariant under these isometries. The Laplacian satisfies the the differential equation $(\Delta~+~\lambda)f(z)~=~0$ defined as the kernel of the resolvent $(\Delta~+~\lambda)^{-1}$ by the equation&#xA;$$&#xD;&#xA;(\Delta~+~\lambda)^{-1}f(z)~=~\int (z,~z^\prime,~\lambda)f(z^\prime)d\mu(z^\prime),&#xD;&#xA;$$&#xA;with the harmonic condition $(\Delta~+~\lambda)G(z,~z^\prime,~\lambda)~=~\delta(z,~z^\prime)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This space is such that the Laplaican $\Delta$ has eigenvalue $-2$, or a negative Gaussian curvature.  This space is a model for the $AdS_2$ spacetime.  I hope that with this example you can see answers to you questions, such as the symmetry under interchange.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-02-05T01:27:26.950" LastActivityDate="2011-02-05T01:27:26.950" CommentCount="8" />
  <row Id="4602" PostTypeId="2" ParentId="4600" CreationDate="2011-02-05T01:24:16.747" Score="1" ViewCount="" Body="&lt;h2&gt;Just addressing the question of the spring constant of the cut down spring pieces:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For a typical material spring you will find that cutting it in half result in two springs with &lt;em&gt;twice&lt;/em&gt; the spring constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand why imagine that you tie the two half back together again. At their natural length the system look like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;O~~~x~~~O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where &lt;code&gt;O&lt;/code&gt; is an end, &lt;code&gt;x&lt;/code&gt; the place you tied them together and &lt;code&gt;~&lt;/code&gt; a piece of spring.&#xA;Now you pull them apart by a distance $2s$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;O~~~~x~~~~O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;because the two halves are identical they each stretch by $s$. You also know that each spring exerts the &lt;em&gt;same&lt;/em&gt; force $F$ on &lt;code&gt;x&lt;/code&gt; (because &lt;code&gt;x&lt;/code&gt; is not accelerating and neither is either segment of spring).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for the combined spring&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F = -k_c (2s) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and for each half-spring&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F = -k_h s $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus $k_h = k_c*2$, and you can generalize this to a fragment of length $h$ taken from a total length of $L$ to find that you need a factor of $L/h$. &lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-02-05T02:03:31.517" LastActivityDate="2011-02-05T02:03:31.517" CommentCount="1" />
  <row Id="4603" PostTypeId="2" ParentId="4600" CreationDate="2011-02-05T02:20:55.183" Score="1" ViewCount="" Body="&lt;p&gt;Probably you got into trouble because of the detailed quantity $\frac{hM}{L}$, which is simply the mass of small weights. However, how many of them are there? Let's suppose you string is made up of $N$ small weights, each weight separated by a small space of length $h$. Therefore, we can see that the total length is $L=N h$. Likewise, the total mass should be $M=Nm$, right?. Having this in mind, we can find that the expression $\frac{hM}{L}$ is actually the mass $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice also that your $k$ is actually the stiffness of each spring. However, the total stiffness is given by $K = \frac{k}{N}$. Substituting this value in your equation, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{hM}{L} y_i&amp;#39;&amp;#39;(t) = KN \left( [y_{i+1}(t) - y_i(t)] + [y_{i-1}(t) - y_i(t)] \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we know can put $N$ in terms of $h$, that is, $N = \frac{L}{h}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ y_i&amp;#39;&amp;#39;(t) = \frac{K L^{2}}{M} \left( \frac{y_{i+1}(t) - y_i(t) + y_{i-1}(t) - y_i(t)}{h^{2}} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There you have the $h$ you were looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, take a look at the difference inside the parentheses. They are very interesting because those terms provide a nice way to understand the Laplacian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: I realized I didn't answer your specific question &quot;Why does the right hand needs an (extra) h?&quot;. Well, you're equating two forces, therefore, the right hand needs to have something involving second order derivatives. You need that $h$ to obtain such derivative (that is, the whole thing inside the parentheses) when the separation between weights tends to $0$. However, in my opinion, it's easier to understand the derivation from first principles. Also, you could make this clearer if you write explicitly the dependence of $y$ on $h$. &lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="18" LastEditorDisplayName="" LastEditDate="2011-02-05T02:47:06.357" LastActivityDate="2011-02-05T02:47:06.357" CommentCount="1" />
  <row Id="4604" PostTypeId="2" ParentId="893" CreationDate="2011-02-05T03:24:12.173" Score="1" ViewCount="" Body="&lt;p&gt;For those who don't already know a lot of physics, &lt;a href=&quot;http://www.amazon.com/s/ref=nb_sb_noss?url=search-alias%3Daps&amp;amp;field-keywords=george+gamow&amp;amp;x=0&amp;amp;y=0&quot; rel=&quot;nofollow&quot;&gt;George Gamow's books&lt;/a&gt; are both informative and playful. &lt;a href=&quot;http://www.amazon.com/gp/search/ref=sr_tc_2_0?rh=i%3Astripbooks%2Ck%3AMartin+Gardner&amp;amp;keywords=Martin+Gardner&amp;amp;ie=UTF8&amp;amp;qid=1296876058&amp;amp;sr=1-2-ent&amp;amp;field-contributor_id=B000AP8X8G&quot; rel=&quot;nofollow&quot;&gt;Martin Gardner's books&lt;/a&gt; of his Sci Am Mathematical Games stimulate critical thought and are fun. Maybe also something by Steven Weinberg like The First Three Minutes.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-02-05T03:24:12.173" CommunityOwnedDate="2011-02-05T03:24:12.173" />
  <row Id="4605" PostTypeId="1" CreationDate="2011-02-05T03:53:58.890" Score="0" ViewCount="269" Body="&lt;p&gt;Is it correct to say that kinetic energy is a scalar?&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="1345" LastEditorDisplayName="" LastEditDate="2011-02-05T13:57:36.620" LastActivityDate="2011-02-05T13:57:36.620" ClosedDate="2011-02-05T13:41:33.147" Title="Is kinetic energy a scalar?" Tags="&lt;kinetic-theory&gt;" AnswerCount="3" CommentCount="10" />
  <row Id="4607" PostTypeId="2" ParentId="4605" CreationDate="2011-02-05T04:35:51.043" Score="0" ViewCount="" Body="&lt;p&gt;I would say it depends on the context. In an Euclidean 3 space it is a scaler (provided its total energy is kinetic) for non relativistic case. In 4D relativistic case, it is a component of a 4 vector. &lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-05T09:43:03.267" LastActivityDate="2011-02-05T09:43:03.267" CommentCount="12" />
  <row Id="4608" PostTypeId="2" ParentId="4605" CreationDate="2011-02-05T04:35:59.907" Score="1" ViewCount="" Body="&lt;p&gt;A scalar is something that doesn't transform under coordinate transformations.  A vector is something that transforms &quot;like a vector,&quot; in other words, its coordinate transforms are realized as a local multiplication by a linear operator (e.g., a matrix).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Newtonian mechanics, it's simple to see that kinetic energy, being proportional to the square of a vector (a length) doesn't change under the allowable coordinate transformations (rotations and translations) in Newtonian mechanics, so it is a scalar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In relativity, we mix up space and time coordinates with coordinate transformations, so there, it transforms as the time component of a 4-vector.  This is because energy comes from symmetries associated to time translation.  Symmetries associated with space translations are associated with conservation of momentum.  Because coordinate transforms in relativity mix up space and time coordinates, we have a single 4-vector whose components are (energy, momentum in x direction, momentum in y direction, momentum in z direction) instead of separate notions of energy and momentum.&lt;/p&gt;&#xA;" OwnerUserId="1403" LastActivityDate="2011-02-05T04:35:59.907" CommentCount="9" />
  <row Id="4609" PostTypeId="2" ParentId="4605" CreationDate="2011-02-05T08:55:55.380" Score="2" ViewCount="" Body="&lt;p&gt;Just to add some explicitness to the above answers: take an isolated particle at rest; it's KE is zero. Now switch to a reference frame with relative velocity $\beta$ wrt the particle. In this frame, it's KE is $$KE&amp;#39; = E - mc^2 = (\gamma - 1)mc^2 = \frac{1}{2}mv^2 + O(\beta^4).$$ We see that $KE&amp;#39; \neq KE,$ thus it's not a scalar.&lt;/p&gt;&#xA;" OwnerUserId="1471" LastActivityDate="2011-02-05T08:55:55.380" />
  <row Id="4610" PostTypeId="2" ParentId="4597" CreationDate="2011-02-05T09:15:33.617" Score="0" ViewCount="" Body="&lt;p&gt;In more 'down-to-earth' QM, you use Green's functions to find the density of states. I'm deprived of my books so at a loss for giving a good reference, but the idea is to calculate $$G(x,x&amp;#39;;E) = \langle x, (E - H)^{-1} x&amp;#39; \rangle,$$ where $H$ is the system's Hamiltonian. You can then define a spectral function $F(x,x&amp;#39;,E) = -\frac{1}{\pi} lim_{\epsilon \rightarrow 0} \text{Im } G(x,x&amp;#39;,E+i\epsilon),$ whose trace is the density of states: $$\mathcal{N}(E) = \int F(x,x,E) dx.$$ &#xA;Finally, you can also use this formalism to calculate other expectation values, with formulas like (modulo an incorrect prefactor) $\langle A \rangle = -\frac{1}{\pi} \text{Im Tr}(AG).$ So yes, they are symmetrical, but they can not really be used to 'solve' a Schrödinger's equation, only on a formal level. That's why they're useful though: they're used all the time in many-body QM/solid state physics, where you'll never 'solve' the problem but can learn lots of interesting stuff by indirect approaches, as the one used above.&lt;/p&gt;&#xA;" OwnerUserId="1471" LastActivityDate="2011-02-05T09:15:33.617" CommentCount="2" />
  <row Id="4611" PostTypeId="2" ParentId="4597" CreationDate="2011-02-05T09:31:49.263" Score="1" ViewCount="" Body="&lt;p&gt;Schrödinger equation is a linear partial differential equation, so sure, you can use the usual formalism of Green's functions to solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First let's recall how the stuff works.&#xA;Suppose $L$ is the linear operator and $D$ are the boundary conditions and we want to solve equations $Lu = f$ and $Du = 0$ for $u$. Using the identity property of the convolution $g*\delta = g$ one is motivated to solve the simpler equation $LG = \delta$ and then one finds $u = G*f$ because $$L(G*f) = (LG)*f = \delta*f = f$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for the time-independent Schrödinger equation the following should be useful. If the operator (understood also with the given boundary conditions) also has a complete basis of eigenvectors $\left\{\left|\phi_n\right&amp;gt;\right\}$ corresponding to eigenvalues $\left\{\lambda_n\right\}$ then the Green's function can easily be seen to be $$G(x, x&amp;#39;) = \sum_n {\phi_n(x)^* \phi_n(x&amp;#39;) \over \lambda_n}$$ (just apply the operator $L$ to it and use that $L \left|\phi_n\right&amp;gt; = \lambda_n \left|\phi_n\right&amp;gt;$. So again we can see that $G$ is in a sense an inverse of $L$ (and indeed it is often written simply as $L^{-1}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it turns out there is a deeper connection between Green's functions and quantum mechanics via Feynman's path integral if we pass to the time dependent Schrödinger equation. I am not going to derive all the stuff here but suffice it to say that Green's function takes on the meaning of a propagator of the particle. Namely, the probability amplitude that the particle gets from the event (t, x) to the event (t', x') is a Green's function of the time-dependent Schrödinger equation $G(x,t;x&amp;#39;,t&amp;#39;) = \left&amp;lt;x\right| U(t,t&amp;#39;) \left|x&amp;#39;\right&amp;gt;$. So yes, the fact that the Green's function is symmetric is precisely because it can be interpreted as an inner product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This stuff generalizes further to quantum field theory and Green's functions are among the basic objects of study there.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-02-05T09:31:49.263" CommentCount="1" />
  <row Id="4613" PostTypeId="2" ParentId="3164" CreationDate="2011-02-05T11:06:33.937" Score="6" ViewCount="" Body="&lt;p&gt;What follows is not a proper answer to your question, but the statement of a fact worth knowing.  There is no way to create a (perfectly) spherical wave:  &quot;A spherically symmetric vacuum solution to Maxwell's equations is always static.&quot;  (Pappas, Am. J. Phys., 52 (255), 1984.)  Also:  H.F. Mathis:  &quot;A short proof that an isotropic antenna is impossible&quot;, Proc. IRE, 59 (979), 1951.  It's an amusing application of Brouwer's &quot;hairy ball&quot; theorem.&lt;/p&gt;&#xA;" OwnerUserId="1761" LastActivityDate="2011-02-05T11:06:33.937" />
  <row Id="4616" PostTypeId="2" ParentId="32" CreationDate="2011-02-05T13:55:52.537" Score="1" ViewCount="" Body="&lt;p&gt;Before &lt;/p&gt;&#xA;&#xA;&lt;p&gt;one calculates Coriolis force on a Rankine whirl, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;one should show, that some kind of whirl is &quot;necessary&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or ineviltable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, the two directions were equivalent solutions, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and one might calculate the chances of coriolis &lt;/p&gt;&#xA;&#xA;&lt;p&gt;force to be  a significant contribution in the decision &lt;/p&gt;&#xA;&#xA;&lt;p&gt;of direction. In a theoretical calculation without some &lt;/p&gt;&#xA;&#xA;&lt;p&gt;residual rotation or any asymmetry, a whirl will never develop.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's just an example of unstable equilibrium&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-02-05T14:05:02.777" LastActivityDate="2011-02-05T14:05:02.777" />
  <row Id="4617" PostTypeId="1" AcceptedAnswerId="4624" CreationDate="2011-02-05T14:08:13.440" Score="8" ViewCount="255" Body="&lt;p&gt;It is always told as a fact without explaining the reason. WHY two objects get charged by rubbing? Why one object get negative charge and other get positive charge?&lt;/p&gt;&#xA;" OwnerUserId="847" LastActivityDate="2011-09-04T23:49:19.350" Title="Why two objects get charged by rubbing" Tags="&lt;electrostatics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="4618" PostTypeId="2" ParentId="4617" CreationDate="2011-02-05T14:17:19.460" Score="3" ViewCount="" Body="&lt;p&gt;&lt;em&gt;It is always told as a fact without explaining the reason.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is true for beginners textbooks maybe, not all. To understand the phenomenon, You must know that different materials have different ability to attract electrons or lose electrons when in mechanical &lt;strong&gt;contact&lt;/strong&gt;. Rubbing is just a means to get a good contact, it is not necessary per se. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experiments to prove this are made by dipping some wax cylinder (fastened on a insulating handle) into a liquid. When removed from the liquid, both are mutually &quot;loaded&quot;. If You are Interested, google for &quot;contact potential&quot; and &quot;electrophorus&quot; (or electrophor?)&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-09-04T23:49:19.350" LastActivityDate="2011-09-04T23:49:19.350" CommentCount="1" />
  <row Id="4619" PostTypeId="1" CreationDate="2011-02-05T14:40:22.110" Score="7" ViewCount="476" Body="&lt;p&gt;The increasing water pressure as you go deeper is generally explained in terms of the weight of the water column above the observation point pressing down. The question, then, is what would happen if you had a big blob of water in free fall, say 100m in diameter-- not big enough to produce large gravitational forces on its own-- and swam to the center of it. I'm imagining some sort of contained environment, here-- a giant space station, or the planet-sized envelope of air in Karl Schroeder's Virga novels, that sort of thing-- so you don't worry about the water boiling off into vacuum or freezing solid, or whatever. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would there be any pressure differential between the surface of the blob and the center of the blob? At first glance, it would seem not, or at least not beyond whatever fairly trivial difference you would get from the self-gravity of the water. But it's conceivable that there might be some other fluid dynamics thing going on that would give you a difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that matter, what difference would you expect between the pressure inside the water and outside the water? We know from shots of astronauts goofing around that water in free fall tends to stay together in discrete blobs. This is presumably some sort of surface tension effect. Does that lead to a higher pressure inside the water than out? How much of a difference would that be? Or would you not particularly notice a change from sticking your head into a blob of free-falling water? (Other than, you know, being wet...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is just an idle question, brought on by thinking about the equivalence principle, and thus free-falling frames. It occurred to me that somebody here might know something about this kind of scenario, so why not post it?)&lt;/p&gt;&#xA;" OwnerUserId="166" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-05T15:24:50.183" LastActivityDate="2011-02-05T16:03:16.657" Title="Water pressure in free fall" Tags="&lt;fluid-dynamics&gt;&lt;pressure&gt;&lt;equivalence-principle&gt;&lt;free-fall&gt;" AnswerCount="4" />
  <row Id="4621" PostTypeId="2" ParentId="4619" CreationDate="2011-02-05T14:46:19.077" Score="2" ViewCount="" Body="&lt;p&gt;The same pressure throughout.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the same argument used to show that horizontal bands under a gravitational field must be at equal pressure: if they weren't there would be a net flow.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-02-05T14:46:19.077" CommentCount="2" />
  <row Id="4622" PostTypeId="2" ParentId="208" CreationDate="2011-02-05T14:52:49.010" Score="1" ViewCount="" Body="&lt;p&gt;This is an old question, but I thought it might be worth chewing on a bit. The loss of mass due to fusion in the sun is piffle.  The Earth’s orbital radius will change more likely due to interactions with the other planets.  The first order perturbations in the orbital elements of the Earth are its eccentricity and right ascension.  The change in the orbital radius or the sem-major axis distance is higher order.  However, that can occur and there is an over all orbital drift in planetary orbits which is chaotic in nature.  The Earth is in a near 1/12 orbital resonance with Jupiter.  The Earth may over the next billion years shift away from this and enter into a near 1/11 orbital resonance with Jupiter, where our orbital radius is about 1.06AU.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This early Earth may have been at .83AU relative to today’s orbit very early on.  This is an orbital resonance of about 16 with Jupiter.  The sun had a power output of 70% of current power.  If you factor these together you get a solar irradiance on the Earth comparable to today.  If the Earth had the same orbital radius as today, even factoring in a $CO_2$ atmosphere temperatures would be $30C$ cooler than today.  Curiously if Earth does drift outwards this delays the solar death of the Earth.  If Earth remains at the current radius temperatures will become intolerant in 500 million years for complex life.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some numerical analyses of this I have run.  The interaction with Jupiter results in a periodic oscillation, and a computation over a longer period of time result in a drift which pushes the Earth outwards on average by about $4.2km/sec$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/KqZtN.jpg&quot; alt=&quot;Perburbation of Earth orbit by Jupiter&quot;&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/icw0T.jpg&quot; alt=&quot;Long term drift of Earth orbit&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bf[addendum]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is in part due to alpha Centuri’s commets.  One big uncertainty is with understanding the early Earth.  I did some homework on this and at 1AU about the warmest the Earth could have been is about -25C with various estimates.  Of course this is my interpretation of geo-modelling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbital dynamics is based on computer modeling.  This is a general plot of 45,000 years.  I should have posted this image.  This illustrates the “signal” in these long runs, where the low frequency stuff has the largest amplitude.  This is the main signal for an outwards drift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/QuptJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does extend the future for life on Earth.  If this planet stays at 1AU the prognosis becomes grim about 500 million years from now.  The planet will start to reach temperatures 30C higher than today and complex life will begin to die out, and further in a billion years oceans will start to boil.  That will really foul things up.  However, with the outwards drift these time frames are almost doubled.  The luminosity increase in the sun will accelerate faster in time and over take this.  The outwards range on this is 2.5 billion years before the oceans start boiling.  Once the oceans start boiling this planet will transform into a 400C version of Venus.  So I figure complex life on this planet, life which emerged with the Cambrian revolution 550 million years ago, might have a good 750 to maybe 1000 million years ahead of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I first read about the future time frame of life on Earth my mind instantly questioned what happened going back in time.  It implies a very cold early Earth; one where it seems the development of life would have been far more difficult.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-02-06T00:52:14.247" LastActivityDate="2011-02-06T00:52:14.247" CommentCount="1" />
  <row Id="4623" PostTypeId="2" ParentId="4619" CreationDate="2011-02-05T14:55:23.530" Score="8" ViewCount="" Body="&lt;p&gt;indeed there would be a (very small) and homogenous pressure &#xA;within the blob, coming from surface tension. &#xA;This pressure is calculated by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kelvin_equation&quot; rel=&quot;nofollow&quot;&gt;Kelvin Equation&lt;/a&gt;&#xA;and is significant in small droplets (reason for small &#xA;droplets to have  a higher vapour pressure than bulk liquid) &#xA;In Your 100 m blob, this extra pressure is negligible of course. &#xA;There is another thing in liquids the so called internal &#xA;pressure, caused by the cohesion forces. (more theoretical)&#xA;But this You cannot sense in Your blob, because &#xA;Your body per se always has this internal pressure. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-05T15:24:28.640" LastActivityDate="2011-02-05T15:24:28.640" CommentCount="2" />
  <row Id="4624" PostTypeId="2" ParentId="4617" CreationDate="2011-02-05T15:05:53.320" Score="7" ViewCount="" Body="&lt;p&gt;The effect in which two objects get charged by rubbing and remain charged is called the &lt;strong&gt;triboelectric effect&lt;/strong&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Triboelectric_effect&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Triboelectric_effect&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where the root &quot;tribo&quot; means friction in Greek. Friction is actually unnecessary: contact is enough in principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This effect shouldn't be confused with the (Volta or Galvani) &quot;contact potential&quot; between metals which only exists as long as the two metals remain in contact, and especially not with &quot;contact electrification&quot; which was a name of a scientifically incorrect theory of electricity at the end of the 18th century that attempted to overgeneralize the interpretation of the triboelectric effect. &quot;Electrophorus&quot; was a gadget, first produced by Volta, that used the triboelectric effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cause of the triboelectric effect is adhesion - the atoms on the surface literally form chemical bonds. Materials such as fur are ready to lose electrons and become positively charged while the materials such as ebonite or glass gain electrons and become negatively neutral. To get some idea about which atoms are likely to lose or gain electrons, it's useful to know their electronegativity:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity#Electronegativities_of_the_elements&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Electronegativity#Electronegativities_of_the_elements&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The redder atom, the higher electronegativity, and the more likely it is for the atom to gain electrons and become negatively charged. That's especially true for light halogens (fluorine, chlorine) and oxygen. That's partly why glass - with lots of $SiO_2$ - likes to get negatively charged in the triboelectric effect. Even sulfur (40% of ebonite) has a higher electronegativity than e.g. carbon and hydrogen that are abundant in the fur which is why fur loses electrons and becomes positively charged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the actual arrangement of the atoms in the molecules matters, too. So this overview of the periodic table was just an analogy, not a reliable way to find out the results of the triboelectric effect.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-05T16:17:12.150" LastActivityDate="2011-02-05T16:17:12.150" CommentCount="3" />
  <row Id="4625" PostTypeId="2" ParentId="4619" CreationDate="2011-02-05T15:25:12.833" Score="1" ViewCount="" Body="&lt;p&gt;I know less about this than you do, and I think you answered your own question correctly;&lt;br&gt;&#xA;The only significant force is the self gravity of the water.&lt;br&gt;&#xA;Just for fun, I calculated the acceleration at the surface and got approximately 10^-6 g.&lt;br&gt;&#xA;Therefore the pressure at the center should be approximately 10^-5 atmospheres.&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-02-05T15:25:12.833" CommentCount="3" />
  <row Id="4626" PostTypeId="1" CreationDate="2011-02-05T16:00:27.907" Score="0" ViewCount="208" Body="&lt;p&gt;If I have a vector A=4i+3j and B=5i-2j, how can I find the vector product AxB? I know that given the angle, its C=AB sin theta, but how can I solve this without the angle?&lt;/p&gt;&#xA;" LastEditorUserId="391" LastEditorDisplayName="" LastEditDate="2011-02-05T19:36:45.133" LastActivityDate="2011-02-05T19:36:45.133" ClosedDate="2011-02-05T22:00:57.500" Title="Vector product in 2 dimensions" Tags="&lt;homework&gt;&lt;vectors&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="4627" PostTypeId="2" ParentId="4619" CreationDate="2011-02-05T16:03:16.657" Score="0" ViewCount="" Body="&lt;p&gt;I'm not sure how it would behave if the internal pressure is less than the vapor pressure of water at that temperature. If the absolute pressure is lower than that thermodynamics would tend to favor the formation/growth of steam bubbles. You bubble would be (nonlinearly) unstable to boiling. I used the term nonlinear, because surface tension increases the internal pressure in bubbles, and only bubbles that exceed a certain critical size would grow. So your liquid very much resembles soda water, provide some surfaces or bubbles on which the vapor phase can grow and it woul start fizzing.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-02-05T16:03:16.657" CommentCount="3" />
  <row Id="4628" PostTypeId="2" ParentId="4626" CreationDate="2011-02-05T16:13:17.130" Score="0" ViewCount="" Body="&lt;p&gt;By the vector product of 2-dimensional vectors $(a_1,a_2)$ and $(b_1,b_2)$, you probably mean the third component of the vector product of $(a_1,a_2,0)$ and $(b_1,b_2,0)$ which is clearly $a_1 b_2 - a_2 b_1$ (the first two components of the product vanish), as indicated by your comment that the magnitude of the vector product is $\sin(\theta)$ just like for 3-dimensional vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the vector product of $4i+3j$ and $5i-2j$ in your notation is $4\times (-2) -3\times 5=-23$. If you expressed the vectors as complex numbers $z_1=4+3i$ and $z_2=5-2i$, you could get the vector product as $\mbox{Re}(i z_1 z_2^*)$.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-05T16:13:17.130" />
  <row Id="4629" PostTypeId="2" ParentId="4626" CreationDate="2011-02-05T16:13:19.017" Score="1" ViewCount="" Body="&lt;p&gt;Vector Product of 2 vectors $ai + bj + ck$ and $Ai + Bj + Ck$ is given by the determinant $  \left| \begin{array}{ccc}&#xD;&#xA;\  i &amp;amp; j &amp;amp; k \\&#xD;&#xA;a &amp;amp; b &amp;amp; c \\&#xD;&#xA;A &amp;amp; B &amp;amp; C \end{array} \right|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example it is $  \left| \begin{array}{ccc}&#xD;&#xA;\  i &amp;amp; j &amp;amp; k \\&#xD;&#xA;4 &amp;amp; 3 &amp;amp; 0 \\&#xD;&#xA;5 &amp;amp; -2 &amp;amp; 0 \end{array} \right|$ $=$ $-23k$ &lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-05T16:54:27.297" LastActivityDate="2011-02-05T16:54:27.297" />
  <row Id="4630" PostTypeId="2" ParentId="4626" CreationDate="2011-02-05T16:39:33.377" Score="0" ViewCount="" Body="&lt;p&gt;Luboš's answer is correct (and efficient), but, for someone new to vector algebra, perhaps confusing. The easy way to explain this is to do the cross product for each component of the vectors, rather than the whole vector at once. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for example, using your definition for the cross product, take only the x-component of the vectors and plug them in. You'll get (4)(5)sin(0)=0. A little thought would tell you that you can skip this for vector components with the same unit vector, since sine of the angle between them will always give zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking one more example, grab the x-component of A and the y-component of B. You get (4)(-2)sin(90)=-8. Remember when doing this that you must use the counterclockwise angle between the vectors, starting from the first vector, or you won't get the minus sign you should for the jXi component (unless you use the right-hand rule to determine whether it's in the positive or negative k direction, in which case you can use the smaller angle between the vectors).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second way you could approach this is to convert your A and B vectors into magnitude and angle expressions. |A| = sqrt(4^2 + 3^2)=5, and angle is given by the inverse tangent of 3/4. This, by the way, is a terrible way to do this sort of problem, as it creates a lot of unnecessary work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, the matrix method that Luboš explains is by far the best method, as it's efficient and is the easiest to accomplish, especially with 3D vectors. If you don't know how that method works, you can grab the following Powerpoint presentation for a good explanation: &lt;a href=&quot;http://www.whrhs.org/21331011516827430/lib/21331011516827430/Cross_Product.ppt&quot; rel=&quot;nofollow&quot;&gt;http://www.whrhs.org/21331011516827430/lib/21331011516827430/Cross_Product.ppt&lt;/a&gt;. If you don't have a program that can open that file, feel free to google &quot;matrix method for cross product,&quot; although many explanations, while correct, are less easy to follow than the one linked.&lt;/p&gt;&#xA;" OwnerUserId="1196" LastActivityDate="2011-02-05T16:39:33.377" CommentCount="2" />
  <row Id="4631" PostTypeId="1" AcceptedAnswerId="4633" CreationDate="2011-02-05T16:56:31.010" Score="3" ViewCount="168" Body="&lt;p&gt;The usual way to model a vibrating membrane is by using the wave equation. Is it possible to do that from &quot;within&quot;? Probably the answer is yes, but where can I see it done explicitly. What I mean is can we model the membrane as a two dimensional Riemannian manifold without any specific embedding in $\mathbb R^3$ plus equations involving the metric (or form, tensors etc.) such that at any time it is isometric to the surface given by the graph of the function satisfying the wave equation at that time. Higher dimensional generalizations are also interesting.   &lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2011-02-05T17:17:43.157" Title="A question about a vibrating membrane" Tags="&lt;mathematical-physics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="4632" PostTypeId="2" ParentId="4594" CreationDate="2011-02-05T17:05:50.547" Score="0" ViewCount="" Body="&lt;p&gt;Left long enough, no matter how you store anything at a fixed temperature, it will equilibrate with its environment and reach the same temperature. So store them however you want, they'll be just as cold in the refrigerator in any case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though, I would point out that a) some parts of the fridge are colder than others, so that'll make a difference, and b) as Georg points out, the speed that they cool off in the fridge (and warm when taken out) will depend on packaging.&lt;/p&gt;&#xA;" OwnerUserId="1196" LastActivityDate="2011-02-05T17:05:50.547" />
  <row Id="4633" PostTypeId="2" ParentId="4631" CreationDate="2011-02-05T17:17:43.157" Score="1" ViewCount="" Body="&lt;p&gt;A part of your question sounds like the covariant treatment of membranes (and higher-dimensional branes) in string/M-theory. Of course, to do actual quantitative calculation, one has to choose a particular embedding and world volume coordinates along the membrane, and impose the coordinate redefinition symmetry (the same thing is done for strings).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the embedding in some actual spacetime still exists. If it is guaranteed or required to exist, it makes no sense to pretend that it doesn't exist: the system will effectively be all about wave equations for the transverse coordinates (to the brane).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to encode the curvature of the brane in its metric tensor only, assuming that the metric tensor behaves just like an induced metric from a higher-dimensional space, you would get different equations of motion. The simplest equations for the metric are Einstein-like equations which are second-order in the metric; however, the induced metric is proportional to the derivatives of spatial coordinates, so the Einstein-like equations would be third-order in the spacetime coordinates, and moreover nonlinear to contract the odd number of indices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I probably don't understand what exactly you have in mind - and apologies, it's probably because what you have in mind is impossible mathematically.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-05T17:17:43.157" CommentCount="5" />
  <row Id="4634" PostTypeId="2" ParentId="4594" CreationDate="2011-02-05T17:29:03.693" Score="1" ViewCount="" Body="&lt;p&gt;In the bottom, the temperature will be lowest there. Also you should take the bottle out of the wrapping, and the water out of the bottle for fastest cooling, and the air out of the freezer, otherwise you use power to cool the wrapping and air etc.&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-02-05T17:29:03.693" />
  <row Id="4635" PostTypeId="1" CreationDate="2011-02-05T19:01:31.077" Score="0" ViewCount="173" Body="&lt;p&gt;I'm trying to solve a problem where I have an object resting on an inclined plane, with the angle of the plan being alpha, and the weight being w. I'm having trouble figuring out how I can calculate the component of the weight parallel to the plane. I also want to find out the weight component perpendicular to the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't want an outright answer, more of an explanation to help me understand.&#xA;Thanks!&lt;/p&gt;&#xA;" LastEditorUserId="701" LastEditorDisplayName="" LastEditDate="2011-02-05T19:45:52.970" LastActivityDate="2011-03-08T04:04:22.903" Title="Understanding weight on an inclined plane" Tags="&lt;homework&gt;&lt;vectors&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="4636" PostTypeId="1" AcceptedAnswerId="4639" CreationDate="2011-02-05T19:12:56.603" Score="4" ViewCount="243" Body="&lt;p&gt;I premise that I haven't yet studied General Relativity, but in Relativistic Electrodymaics I have knowed impulse-energy tensor of Electromagnetic Field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know in Einstein's equations there is impulse-energy tensor $T_{\mu\nu}$ too:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;R_{\mu \nu} - {1 \over 2} g_{\mu \nu} R + \Lambda g_{\mu \nu} = \frac{8  \pi G}{c^4} T_{\mu \nu}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose that it's connected with matter in the space, being that equation about gravitation theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: has got any sense put in the Einstein's field equation the impulse-energy tensor of EM field? If yes, what is the physical meaning? What are the consequences?&lt;/p&gt;&#xA;" OwnerUserId="558" LastActivityDate="2011-02-05T20:24:46.150" Title="Einstein's Field equations and impulse-energy tensor" Tags="&lt;general-relativity&gt;&lt;classical-electrodynamics&gt;&lt;field-theory&gt;" AnswerCount="1" />
  <row Id="4637" PostTypeId="1" AcceptedAnswerId="4640" CreationDate="2011-02-05T20:10:48.953" Score="3" ViewCount="372" Body="&lt;p&gt;I'm sorry if this is too newbie.&lt;br&gt;&#xA;As I understand, light is what is more generally called &quot;electromagnetic radiation&quot;, right?&lt;br&gt;&#xA;The energy radiated by a star, by an antenna, by a light bulb, by your cell phone, etc.. are all the same king of energy: electromagnetic energy, i.e. photons traveling through space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, so good? (if not please clarify)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, there are these things called &quot;electromagnetic fields&quot;, for example earth's magnetic field, or the magnetic field around magnets, the electric field around a charge or those fields that coils and capacitors produces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here is my question:&lt;br&gt;&#xA;Are these two things (electromagnetic fields and electromagnetic radiation) two forms of the same thing? or they are two completely different phenomena?&lt;br&gt;&#xA;If they are different things, What does light (radiation) have to do with electromagtetism?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I'm not asking a complex theoretical answer, just an intuitive explanation. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1768" LastActivityDate="2011-02-06T06:17:47.997" Title="Electromagnetic fields vs electromagnetic radiation" Tags="&lt;electromagnetism&gt;&lt;light&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="5" />
  <row Id="4638" PostTypeId="2" ParentId="4637" CreationDate="2011-02-05T20:22:42.910" Score="1" ViewCount="" Body="&lt;p&gt;Electromagnetic waves are a special case of electromagnetic fields characterized by a synchronous time and space dependence of perpendicular electric and magnetic field components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whew! Too many words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, it is worth studying isolated electric fields, isolated magnetic fields, the combined effects of static fields taken together &lt;em&gt;and&lt;/em&gt; the wave behavior because all these cases come up in real problems.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-02-05T20:22:42.910" />
  <row Id="4639" PostTypeId="2" ParentId="4636" CreationDate="2011-02-05T20:24:46.150" Score="3" ViewCount="" Body="&lt;p&gt;The stress-energy tensor $T_{\mu\nu}$ on the right side of the Einstein equation characterizes all of the various forms of &quot;stuff&quot; in the spacetime. If there are electromagnetic fields in the spacetime, then the stress-energy tensor of the electromagnetic field is part of that $T_{\mu\nu}$, along with contributions from other forms of energy, mass, etc.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-05T20:24:46.150" CommentCount="3" />
  <row Id="4640" PostTypeId="2" ParentId="4637" CreationDate="2011-02-05T20:29:42.637" Score="5" ViewCount="" Body="&lt;p&gt;Electromagnetic radiation consists of waves of electric and magnetic fields, but not all configurations of electric and magnetic fields are described as &quot;radiation.&quot; Certainly static fields, like the Earth's magnetic field and the other fields you describe, are not called &quot;radiation.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a standard technical definition of electromagnetic radiation, but roughly speaking, we think of a configuration of electromagnetic fields as constituting radiation when it has &quot;detached&quot; from its source and propagates on its own through space. One of Maxwell's equations says, in effect, that a changing magnetic field produces an electric field. Another says that a changing electric field produces a magnetic field. An electromagnetic wave results from these two processes producing a steady flow of radiated energy that persists far from the source.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-05T20:29:42.637" CommentCount="2" />
  <row Id="4641" PostTypeId="1" AcceptedAnswerId="4643" CreationDate="2011-02-05T21:17:14.673" Score="3" ViewCount="182" Body="&lt;p&gt;According to this picture&lt;br&gt;&#xA;&lt;a href=&quot;http://mynasadata.larc.nasa.gov/images/EM_Spectrum3-new.jpg&quot; rel=&quot;nofollow&quot;&gt;http://mynasadata.larc.nasa.gov/images/EM_Spectrum3-new.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the light emitted by an object depends on its temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That makes perfect sense when we heat a metal. As its temperature raises we see it red at first, then orange, then yellow... and then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we heated it even more should we see the metal turning green? and then blue?&lt;/p&gt;&#xA;" OwnerUserId="1768" LastEditorUserId="1768" LastEditorDisplayName="" LastEditDate="2011-02-05T22:57:38.107" LastActivityDate="2011-02-09T01:19:36.017" Title="Light emitted by an object according to its temperature" Tags="&lt;light&gt;&lt;electromagnetic-radiation&gt;&lt;temperature&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="4642" PostTypeId="2" ParentId="4641" CreationDate="2011-02-05T21:34:19.510" Score="1" ViewCount="" Body="&lt;p&gt;The phenomenon you're referring to is &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_body_radiation&quot; rel=&quot;nofollow&quot;&gt;black body radiation&lt;/a&gt;. When a material gets hot, its electrons are in excited states. Occasionally, an electron will drop down to a lower state, emitting a photon. The higher the temperature, the higher the frequency of emitted light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why the heating element in your oven turns red. It's also how incandescent light bulbs work. Most of the light we see from stars is due to black body radiation, and we can infer the temperature of a star by its color because of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation that gives energy emitted as a function of frequency and temperature is &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck%27s_law&quot; rel=&quot;nofollow&quot;&gt;Plank's Law&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/uhA55.png&quot; alt=&quot;Plank's law&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1770" LastActivityDate="2011-02-05T21:34:19.510" CommentCount="2" />
  <row Id="4643" PostTypeId="2" ParentId="4641" CreationDate="2011-02-05T21:36:43.907" Score="9" ViewCount="" Body="&lt;p&gt;You would be unlikely to see green. The problem is that in order to see green, you would need the spectrum of emitted light to peak at green and have relatively little contribution from other frequencies. This, for example, is the reason you do not see green stars (There is a cute &lt;a href=&quot;http://books.google.ca/books?id=7papZR4oVssC&amp;amp;pg=PA293&amp;amp;lpg=PA293&amp;amp;dq=feynman+green+star&amp;amp;source=bl&amp;amp;ots=erW-gdlK21&amp;amp;sig=Yz-t5yKHVraZteRcdKp-f7Ktt7A&amp;amp;hl=en&amp;amp;ei=_8NNTdLtHYK6sQOltLmmCg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBUQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Feynman story about this&lt;/a&gt;). An explanation of &lt;a href=&quot;http://www.google.ca/search?q=color+temperature&amp;amp;ie=utf-8&amp;amp;oe=utf-8&amp;amp;aq=t&amp;amp;rls=org.mozilla%3aen-US%3aofficial&amp;amp;client=firefox-a&quot; rel=&quot;nofollow&quot;&gt;color temperature&lt;/a&gt; is given on wikipedia. Here is the path a black body takes with temperature increase--a &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aPlanckianLocus.png&quot; rel=&quot;nofollow&quot;&gt;Planckian locus diagram&lt;/a&gt;&#xA;.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-02-05T21:42:37.547" LastActivityDate="2011-02-05T21:42:37.547" CommentCount="1" />
  <row Id="4644" PostTypeId="2" ParentId="1094" CreationDate="2011-02-05T21:47:56.790" Score="1" ViewCount="" Body="&lt;p&gt;In short, just build a &lt;a href=&quot;http://en.wikipedia.org/wiki/Faraday_cage&quot; rel=&quot;nofollow&quot;&gt;Faraday Cage&lt;/a&gt; out of conducting material with a mesh size much smaller than the wavelength of the electromagnetic wave you want to block.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-02-05T21:47:56.790" />
  <row Id="4645" PostTypeId="1" CreationDate="2011-02-05T22:28:58.880" Score="5" ViewCount="198" Body="&lt;p&gt;Many equations of motion can be derived from a variational principle. To take a simple example, the wave equation $h^{ij} \partial_i \partial_j u = 0$ (where $h^{ij}$ is the Minkowski metric $\mathrm{diag}[-1,+1,+1,+1]$) can be derived from the Lagrangian density&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L = \frac{1}{2} h^{ij} (\partial_i u) (\partial_j u)$ :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is $\frac{\delta L}{\delta \partial_i u} = h^{ij} \partial_j u$, from which follows the equation of motion via the usual ansatz $\frac{\delta L}{\delta u} - \partial_i \frac{\delta L}{\delta \partial_i u} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This calculation neglects boundary conditions, which it should not since the ansatz implies an integration by parts which has a boundary term. Assuming e.g. I want to solve the wave equation in a box $0 \le x \le 1$, how would I e.g. impose Dirichlet or von Neumann boundary conditions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can think of two approaches:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Assume that $u$ lives in a function space that already satisfies the boundary condition. However, this seems complicated in general -- is this workable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Impose a constraint (via a Lagrange multiplier) that corresponds to the boundary condition. This seems weird because the multiplier would live only on the boundary of the domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do these approaches work? Is there another way?&lt;/p&gt;&#xA;" OwnerUserId="1547" LastActivityDate="2011-03-02T04:12:46.510" Title="How does one formulate boundary conditions in a variational approach?" Tags="&lt;variational-principle&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="4646" PostTypeId="2" ParentId="4637" CreationDate="2011-02-05T22:31:07.690" Score="5" ViewCount="" Body="&lt;p&gt;The first thing to know, if you don't already, is that physicists define a &quot;field&quot; to be a a value associated with each point in spacetime. The electromagnetic field is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Tensor_field&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;tensor field&lt;/em&gt;&lt;/a&gt;, meaning that at each point in spacetime, it has a value given by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_tensor&quot; rel=&quot;nofollow&quot;&gt;electromagnetic tensor&lt;/a&gt;, which is essentially a set of 6 numbers. These numbers can't just take on any old values, though; the ways that the numbers change as you move through space and time are constrained by &lt;a href=&quot;http://en.wikipedia.org/wiki/Maxwell%27s_equations&quot; rel=&quot;nofollow&quot;&gt;Maxwell's equations&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you can take two of Maxwell's equations (in a vacuum) and combine them to get the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_wave_equation&quot; rel=&quot;nofollow&quot;&gt;wave equation&lt;/a&gt;, which tells you that when there is no other matter around, any disturbance in the electromagnetic field will propagate in the form of a wave. A propagating disturbance in the EM field is what we call electromagnetic radiation.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-02-05T22:31:07.690" CommentCount="2" />
  <row Id="4647" PostTypeId="2" ParentId="4641" CreationDate="2011-02-05T22:31:30.610" Score="0" ViewCount="" Body="&lt;p&gt;To see the green color, you have to use a green glass (a filter). The higher temperature, the brighter the green light emission.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-02-05T22:31:30.610" />
  <row Id="4648" PostTypeId="2" ParentId="728" CreationDate="2011-02-06T00:18:45.337" Score="3" ViewCount="" Body="&lt;p&gt;Lagrangian solution: Use Cartesian coordinates (x,y) and let the height of the hill define the potential. Each point (x,y) gives a unique position of the ball which therefore uniquely determines its potential (which is not in general equal to the height since the ball can lean over at various angles but this is a detail that does not alter the math here). Thus the potential part of the Lagrangian can be written as V(x,y).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following the useful and obvious comment by Mark Eichenlaub that $\omega = Rv$, and noting that the kinetic energy depends only on the squares of velocity and angular momentum, and calculating that $\vec{v} = (\dot{x},\dot{y},h_x\dot{x}+h_y\dot{y})$ where $h_x,h_y$ are the slopes of the hill in the x and y directions, the kinetic energy can be written in the form&lt;br&gt;&#xA;$T(x,y,\dot{x},\dot{y}) = \alpha(\dot{x}^2A+\dot{y}^2B+\dot{x}\dot{y}C)$&lt;br&gt;&#xA;where $A,B,C$ depend only on the slope at a given position, and $\alpha$ depends on the moment of inertia and is different between the rolling and sliding ball. (I.e. the sliding ball has the same energy as a rolling ball with zero moment of inertia.) Note, per Qmechainc that $A,B,C$ do depend on position through the slope. Of course the potential energy depends only on the position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations of motion are:&lt;br&gt;&#xA;$\frac{d}{dt}\frac{\partial T}{\partial \dot{x}} = \frac{\partial (T+V)}{\partial x}$&lt;br&gt;&#xA;and similarly for y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $t\to t/\sqrt{\alpha}$. This makes $dt\to dt/\sqrt{\alpha}$ and $\partial \dot{x} \to \sqrt{\alpha} \partial \dot{x}$. So the operator acting on T is unaffected by this transformation:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{dt}\frac{\partial }{\partial\dot{x}} \to \frac{d}{dt}\frac{\partial }{&#xD;&#xA;\partial\dot{x}}$,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since T is only quadratic in the velocities it transforms as the square of the velocities, i.e. $\dot{x}^2 \to \dot{x}^2/\alpha$:&lt;br&gt;&#xA;$T\to T/\alpha$&lt;br&gt;&#xA;This eliminates $\alpha$ from the equations of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, a similar argument shows that the answer to the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;Does the path a ball slides down a hill depend on the magnitude of the force of gravity?&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is also no, which is a more intuitive way of understanding the answer.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-04-25T00:15:09.290" LastActivityDate="2011-04-25T00:15:09.290" CommentCount="7" />
  <row Id="4649" PostTypeId="1" AcceptedAnswerId="4681" CreationDate="2011-02-06T01:15:57.300" Score="2" ViewCount="124" Body="&lt;p&gt;A local FM radio station transmitting at 89.3 MHz recently &lt;a href=&quot;http://twitter.com/#!/WRKF/status/32914964863778816&quot; rel=&quot;nofollow&quot;&gt;announced&lt;/a&gt; that it would be running at 50% power due to freezing weather and a forecast of ice accumulation, as &quot;&lt;a href=&quot;http://twitter.com/#!/WRKF/status/33543424346947584&quot; rel=&quot;nofollow&quot;&gt;when ice is forecast ... it can damage our antenna if we run at full power&lt;/a&gt;.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the effect of ice on (the electrical properties of) an antenna?&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-02-06T20:03:33.257" LastActivityDate="2011-02-06T20:03:33.257" Title="What is the effect of ice on an antenna?" Tags="&lt;electrodynamics&gt;&lt;electromagnetic-radiation&gt;&lt;antennas&gt;" AnswerCount="2" />
  <row Id="4652" PostTypeId="1" CreationDate="2011-02-06T01:37:04.943" Score="5" ViewCount="169" Body="&lt;p&gt;How do we know that dark matter is dark. In that it doesnt give out any light or absorb any. It is impossible for humans to be watching every single wavelengths. For example what about wavelengths that are too big to detect on Earth?&lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2011-02-06T02:33:19.107" Title="How do we know that dark matter is dark?" Tags="&lt;dark-matter&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="4653" PostTypeId="2" ParentId="4645" CreationDate="2011-02-06T01:39:24.213" Score="0" ViewCount="" Body="&lt;p&gt;Normally when one does a variational calculation, the action at the boundary is held fixed. So the variation at the boundary is zero by definition. Note that the goal of the variational technique is to derive a differential equation, that is, a relationship between various rates of change. As such, boundary conditions don't come into play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do need to impose boundary conditions, you do that when you solve the differential equation.&lt;/p&gt;&#xA;" OwnerUserId="150" LastActivityDate="2011-02-06T01:39:24.213" />
  <row Id="4654" PostTypeId="1" CreationDate="2011-02-06T01:41:05.573" Score="2" ViewCount="97" Body="&lt;p&gt;Am I correct it remembering that unless a pulsars beams plane faces earth we can not detect them. And that similarly inbetween the pulses we can't see them either?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so how does this differ from dark matter? Isn't it possible that there are a lot more pulsars that we just can't see them because they are pointing the wrong way. &lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2011-04-07T02:14:22.663" Title="Ability to detect pulsars" Tags="&lt;astrophysics&gt;&lt;dark-matter&gt;&lt;pulsars&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="4655" PostTypeId="2" ParentId="4652" CreationDate="2011-02-06T01:48:52.560" Score="-1" ViewCount="" Body="&lt;p&gt;Nobody is saying that dark matter &lt;em&gt;necessarily&lt;/em&gt; doesn't emit or reflect any EM radiation at all. The reason we call it dark matter is that it doesn't emit radiation in the visible spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original discovery was simply that the &lt;em&gt;stars&lt;/em&gt; in galaxies did not have enough mass to account for the measured mass of the galaxy. Astronomers therefore realized that there had to be some stuff that wasn't emitting visible light, hence they called it &quot;dark.&quot; Originally, it was thought that brown dwarfs (very small stars without sustained fusion) could be the dark matter; those wouldn't emit visible light, but they would certainly emit infrared radiation. However, later experiments ruled that out, along with other forms of &quot;normal&quot; (baryonic) matter known to exist. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point in time, I believe we know from experimental observations that dark matter is not only dark in the visible spectrum, it appears to emit/reflect negligible amounts of radiation in all other frequency bands that we've checked. But that's just an observation. I don't know of any requirement that dark matter &lt;em&gt;can't&lt;/em&gt; interact with EM radiation at all. Most theories about the nature of dark matter do predict that it's totally (or almost totally) decoupled from the EM field. But there is still the possibility that it isn't.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-02-06T01:48:52.560" CommentCount="2" />
  <row Id="4656" PostTypeId="2" ParentId="4652" CreationDate="2011-02-06T02:24:22.810" Score="2" ViewCount="" Body="&lt;p&gt;There is indeed very good reason to believe dark matter is dark - apart from all the evidence from &quot;missing mass&quot; in luminosity counts and gravitational lensing studies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This comes from theories of large-scale structure formation:&#xA;That there has always had to be some sort of matter that doesn't interact electromagnetically at all is crucial to most scenarios of large-scale structure formation. The density fluctuations in the present universe would be too large than what would be predicted if there were only ordinary baryonic matter that interacted only electromagnetically. With dark matter, you can have something that gravitates yet decouples from radiation much before baryonic matter does. This allows the dark matter to form gravitational wells (under collapse) which have a much longer time to expand with the universe. By the time ordinary matter decouples from radiation and joins the rest of the expansion flow, the ordinary matter will quickly fall into these large gravitational wells of the dark matter that have had far more time to grow. This, in a way, amplifies density perturbations in the early universe and allows large-scale structure to form. (to the extent that we see it today in the form of clusters and galaxies)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The required amount of dark matter calculated, in this way, in order to observe the present scale of density fluctuations matches very well with the amount of dark matter required to explain galactic rotation curves, gravitational lensing, etc. So there's excellent agreement that all of these are due to the same thing - some sort of matter which doesn't interact electromagnetically at all, viz. dark matter.  &lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-02-06T02:32:24.910" LastActivityDate="2011-02-06T02:32:24.910" />
  <row Id="4657" PostTypeId="2" ParentId="4652" CreationDate="2011-02-06T02:32:48.140" Score="4" ViewCount="" Body="&lt;p&gt;If dark matter emitted very long wave lengths of electromagnetic radiation it would mean it is composed of charged particles.  There is no escape from that conclusion.  Somebody might propose that dark matter is some strange configuration of charged particles which acts as a very long wavelength antenna.  That might be a good model, but there is a hitch with it.  If it emits longer wavelength radiation is must be colder.  The Wien displacement law is that $\lambda~=~c/T$, for $c$ a constant, which may be calculated, but is not relevant here.  For $T~=~2.7$K the wavelength of radiation is $.1cm$.  Let me assume that there does exist some very long wavelength radiation, say $100m$ as the lower bound on this.  The ratio of temperatures gives that&#xA;$$&#xD;&#xA;T_{100m}~=~\frac{10^{-3}}{10^2}2.7K~=~2.7\times 10^{-5}K &#xD;&#xA;$$&#xA;So all this has to be is a very cold gas, which is the hitch.  This gas is much colder than the background radiation and not in equilibrium.  So from some physical grounds this is not likely, and dark matter is most likely not some ordinary form of matter that interacts by EM.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-06T02:32:48.140" CommentCount="1" />
  <row Id="4658" PostTypeId="2" ParentId="4652" CreationDate="2011-02-06T02:33:19.107" Score="0" ViewCount="" Body="&lt;p&gt;Aside from the very good theoretical reasons dbrane gives, I'd say that we don't have observational evidence that DM is truly dark. Some of the largely discredited MACHO candidates, like large numbers of red dwarfs, brown dwarfs, planetissimals etc. have been rejected for theoretical -not observational reasons. DM, that is lightly emitting is not currently detectable.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-02-06T02:33:19.107" />
  <row Id="4659" PostTypeId="2" ParentId="4654" CreationDate="2011-02-06T02:40:42.907" Score="1" ViewCount="" Body="&lt;p&gt;There are lots of things such as isolated neutron stars that are not pulsars, isolated stellar mass black holes, and perhaps yes your example of pulsars &quot;pointing the wrong way&quot; that we cannot see.&lt;br&gt;&#xA;But there are arguments from microlensing experiments such as OGLE and MACHO that there can not be enough of them to constitute the dark matter.&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-02-06T02:40:42.907" CommentCount="4" />
  <row Id="4660" PostTypeId="2" ParentId="728" CreationDate="2011-02-06T03:00:15.337" Score="2" ViewCount="" Body="&lt;p&gt;I posed the problem to my physics club, and someone pointed out that the calculation is easier if we do it around the point of contact with the hill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the ball have moment of inertia $I$ about a point on its edge, and make $\vec{\omega}$ the angular velocity around the point where the ball contacts the hill.  We'll try to show that the path doesn't change when you change the moment of inertia with fixed radius.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ball sliding down the hill is equivalent to a ball rolling down the hill with all its mass located at the center of the ball, so if we show that all rolling balls (with different moments of inertia but the same radius) follow the same path, the sliding ball follows that path, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Introduce a parameter $\lambda$ measuring the length along the spatial path of the ball, so that when we take a derivative with respect to $\lambda$, we are looking at how values change in space rather than in time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The direction of the path is described by a unit tangent vector $\vec{v}/v$.  The path curves in space, and this is described by the derivative of the tangent vector with respect to lambda.  We want to show that this is independent of $I$.  The quantity we're examining is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\textrm{d}}{\textrm{d}\lambda} \frac{\vec{v}}{v} = \frac{\textrm{d}\vec{v}/\textrm{d}\lambda}{v} - \frac{\vec{v}}{v^2} \frac{\textrm{d}v}{\textrm{d}\lambda}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are going to describe the physics in terms of torque and angles, so let's define $\vec{R}$ as a vector pointing from the point of contact with the hill to the center of the ball, and write &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{v} = \vec{\omega}\times\vec{R}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This lets us do some calculus and algebra to re-express the change in the tangent vector, which we previously wrote in terms of $\vec{v}$, in terms of $\vec{\omega}$.  I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\omega R}\left( \left( \frac{\textrm{d}\vec{\omega}}{\textrm{d}\lambda} - \frac{\vec{\omega}}{\omega} \frac{\textrm{d}\omega}{\textrm{d}\lambda}\right) \times \vec{R} + \vec{\omega} \times \frac{\textrm{d}\vec{R}}{\textrm{d}\lambda}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final term on the right is clearly independent of $I$ because it depends the radius, which is assumed to be the same.  We'll drop that term in future discussion because we're only interested in whether this change in the tangent is independent of $I$.  For the same reason, we don't care that the stuff in parentheses is being crossed with $\vec{R}$, or divided by $R$, so we'll simplify to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\omega}\left( \frac{\textrm{d}\vec{\omega}}{\textrm{d}\lambda} - \frac{\vec{\omega}}{\omega} \frac{\textrm{d}\omega}{\textrm{d}\lambda} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need to look at the physics.  The goal is to find expressions for $\textrm{d}\vec{\omega}/\textrm{d}t$ and $\textrm{d}\omega/\textrm{d}t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torque $\vec{\tau}$ is the same for all symmetric balls of equal mass and radius, and we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I \frac{\textrm{d}\vec{\omega}}{\textrm{d}t} = \vec{\tau}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the chain rule,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\textrm{d}\vec{\omega}}{\textrm{d}\lambda} = \frac{\textrm{d}\vec{\omega}}{\textrm{d}t}\frac{\textrm{d}t}{\textrm{d}\lambda} = \frac{\textrm{d}\vec{\omega}}{\textrm{d}t}\frac{1}{v} =  \frac{\textrm{d}\vec{\omega}}{\textrm{d}t}\frac{1}{\omega R}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\textrm{d}\omega}{\textrm{d}\lambda} = \frac{\textrm{d}\omega}{\textrm{d}t}\frac{1}{\omega R}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega = (\vec{\omega}\cdot\vec{\omega})^{1/2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\textrm{d}\omega}{\textrm{d}t} = \frac{\textrm{d}\vec{\omega}/\textrm{d}t\cdot\vec{\omega}}{\omega}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This lets us plug things back in to the expression we were examining.  Simplifying the algebra,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\omega}\left( \frac{\textrm{d}\vec{\omega}}{\textrm{d}\lambda} - \frac{\vec{\omega}}{\omega} \frac{\textrm{d}\omega}{\textrm{d}\lambda} \right) = \frac{1}{I\omega^2 R}\left(\vec{\tau} - \frac{\vec{\omega}(\vec{\tau}\cdot\vec{\omega})}{\omega^2} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I \omega^2$ is the twice the kinetic energy, and therefore a function only of the position.  It doesn't change as we change $I$.  $\vec{\tau}$ is also independent of $I$.  Finally, $\frac{\vec{\omega}(\vec{\tau}\cdot\vec{\omega})}{\omega^2}$ is also independent of $I$, because although changing $I$ changes the size of $\omega$, in this expression all the units of $\omega$ cancel out.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We conclude that we can take two balls with different spherically-symmetric mass distributions and the same radius and roll them down the same path over the hill.  The sliding ball is identical to the limit where $I \to m R^2$.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-03-15T22:09:56.850" LastActivityDate="2011-03-15T22:09:56.850" CommentCount="15" />
  <row Id="4661" PostTypeId="2" ParentId="4654" CreationDate="2011-02-06T03:05:58.890" Score="0" ViewCount="" Body="&lt;p&gt;The following paper is about some putative microlensing of black holes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/astro-ph/pdf/0203/0203257v4.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/astro-ph/pdf/0203/0203257v4.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar considerations can involve neutron stars.  Cold and dark bodies in interstellar space form some part of the DM, but a small percentage.  Microlensing has found that MACHO contribution to dark matter is small.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-06T03:05:58.890" />
  <row Id="4662" PostTypeId="1" CreationDate="2011-02-06T03:21:11.293" Score="1" ViewCount="150" Body="&lt;p&gt;So I am studying for my exam on monday for analytical mechanics. I am kind of stuck on a question, so wondering if someone can give me a pointer. I am actually looking for some pointers and some techniques (hints and tips) to get me through this. the topic is just simple velocity as a function of time, velocity as a function of displacement, quadratic air resistance, oscillations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here is the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A block of wood is projected up an&#xA;  inclined plane with initial speed&#xA;  $v_0$. If the inclination of the plane&#xA;  is $30^\circ$ and the coefficient of&#xA;  sliding friction is $\mu_k = 0.1$,&#xA;  find the total time for the block to&#xA;  return to the point of projection.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Heres my work:&#xA;$$F_{\text{net}} = -F_f = -\mu_k \cdot m \cdot g \cdot sin(30) = m \cdot \frac{dv_x}{dt}$$ &#xA;so the m's cancel and solving the differential equation you get the following:&#xA;$$ \int_{v_0}^v dv = \int_0^t -\mu_k \cdot g \cdot sin(30) dt$$&#xA;$$v-v_0 = -\mu_k \cdot sin(30) \cdot g\cdot t$$&#xA;so &#xA;$$ v(t) = v_0 - 0.98\cdot sin(30) \cdot t $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so if i set that to 0 i know the time when it reaches the top and starts to slide backwards. Now how would I know when it reaches the point of projection? Would it just be double the time?&lt;/p&gt;&#xA;" OwnerUserId="551" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-02-06T03:32:28.617" LastActivityDate="2011-02-06T23:22:34.507" Title="How long does it take a block to slide up and down a plane with friction?" Tags="&lt;classical-mechanics&gt;&lt;homework&gt;" AnswerCount="3" />
  <row Id="4663" PostTypeId="2" ParentId="4635" CreationDate="2011-02-06T03:21:19.593" Score="0" ViewCount="" Body="&lt;p&gt;Imagine the inclined plane, and imagine an arrow pointing straight down, representing your weight (the force of gravity). Then imagine rotating that picture so that the plane is horizontal, with the arrow rotated sideways by alpha. Imagine now a horizontal x axis, and a vertical y axis, like they always draw in the textbooks. The x and y axes should cross (intersect) at the tail of the weight arrow. Now imagine a horizontal line that starts at the tip of the arrow, and ends at the (vertical) y axis. That line is the component of the weight that is parallel to the inclined plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then use what you know about sines and cosines (trigonometry) to calculate the amount of weight in the directions parallel and perpendicular to the incline. &lt;/p&gt;&#xA;" OwnerUserId="1462" LastEditorUserId="1462" LastEditorDisplayName="" LastEditDate="2011-02-06T03:28:52.543" LastActivityDate="2011-02-06T03:28:52.543" />
  <row Id="4664" PostTypeId="2" ParentId="4635" CreationDate="2011-02-06T03:27:25.313" Score="2" ViewCount="" Body="&lt;p&gt;Here's a quick sketch:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/DhhST.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity is the vector $u$.  Its components in the plane and against the plane are $v$ and $w$ respectively.  You want to find $v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle between the plane and horizontal is the same as the angle between $w$ and $u$, which allows you to find a simple trig relation to solve the problem.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-02-06T03:27:25.313" />
  <row Id="4665" PostTypeId="2" ParentId="4662" CreationDate="2011-02-06T03:31:55.653" Score="2" ViewCount="" Body="&lt;p&gt;You've forgotten to include the force of gravity.   When you do, you'll find that the time to come to a rest going up the plane is shorter than the time to go back down.  That's because as the block goes up the plane, both gravity and friction are pushing it back down, but as the block comes down the plane, gravity is pulling it down while friction is trying to push it back up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the same reason, if you throw a ball up in the air, it takes longer to come down than to go up.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-02-06T03:31:55.653" />
  <row Id="4666" PostTypeId="2" ParentId="4645" CreationDate="2011-02-06T05:24:46.320" Score="5" ViewCount="" Body="&lt;p&gt;The boundary term when you vary your action is $\int \delta u n^i \partial_i u$ where $n^i$&#xA;is a unit normal vector to the boundary and the integral is over the boundary. You want this term to be stationary under the variation. You can achieve this by taking $n^i \partial_i u=0$, that is Neumann boundary conditions, or you can demand that $u$ has a fixed value at the boundary which requires that one only consider variations $\delta u$ that vanish on the boundary, that is Dirichlet boundary conditions. Both lead to a consistent variational principle.&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T04:12:46.510" LastActivityDate="2011-03-02T04:12:46.510" CommentCount="3" />
  <row Id="4667" PostTypeId="2" ParentId="4637" CreationDate="2011-02-06T05:27:26.140" Score="0" ViewCount="" Body="&lt;p&gt;Electromagnetic radiations are electromagnetic waves in the electromagnetic field. It is disturbance of the electromagnetic field propagated with the velocity of light. Electromagnetic field itself is a seat of energy. $u = 1/2 (\epsilon_0E^2 + B^2/\mu_0)$&#xA; Where $u$ is the energy density of the em field. Other symbols carry their usual meanings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During propagation of electromagnetic waves, energy is carried from one place to another in accordance with &lt;a href=&quot;http://en.wikipedia.org/wiki/Poynting%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Poynting's theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-06T05:32:58.440" LastActivityDate="2011-02-06T05:32:58.440" />
  <row Id="4668" PostTypeId="2" ParentId="4637" CreationDate="2011-02-06T06:17:47.997" Score="4" ViewCount="" Body="&lt;p&gt;&lt;b&gt;Electric Field&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a simple way to build a device to detect an electric field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a normal, air-filled balloon and tie a string to it.  Hold it by the string.  It should hang straight down, due to the gravitational force on it.  However, by tapping the balloon you see that it only takes a little bit of a push to move the balloon around.  If it experiences a constant force, for example due to a steady, light wind, the balloon string will point at an angle.  The angle of the balloon string is essentially a force detector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rub the balloon against your hair (or borrow someone else's hair if you don't have enough).  The balloon now has some charge on it.  If you hold it by the string, it still hangs straight down most of the time.  However, if there is an electric field present, the balloon will move somewhat in the direction of the electric field.  The direction the string points indicates the direction of the electric field, and the deflection of the string from vertical indicates the strength of the electric field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you hold the balloon near a wall, or near your sweater, it will likely start to deflect.  This indicates that the wall or your sweater is creating an electric field.  (This happens by &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrostatic_induction&quot; rel=&quot;nofollow&quot;&gt;electrostatic induction&lt;/a&gt;.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you walk around to different places, you find the direction and strength of the field is different everywhere.  Even if you stay in one place, you may find that the direction and strength of the field is changing in time.  By making a whole array of balloons all over a giant hall and watching all their deflections, you can map out the entire electric field.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can visualize it as a bunch of arrows in space, the same way you might visualize the velocity of the air, which moves at different speeds in different directions everywhere.  However, the arrows do not indicate anything is moving; they just indicate the deflection a balloon would have if it were there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also visualize the electric field by imagining the arrows everywhere grow into each other, forming lines.  For example, here's the wikipedia picture of the electric field lines for a dipole (one positive and one negative charge sitting nearby each other).  Nothing is moving in this picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ncbGv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Magnetic Field&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnetic fields are very much like electric fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically, your balloon could detect a magnetic field by moving it around and observing the forces on it, but that is not practical.  A simple magnetic field detector is a compass.  A compass points in the direction of the magnetic field.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also get an idea for how strong the magnetic field is by twisting the compass around in a circle.  This will set the needle swinging back and forth.  The faster the oscillations, the stronger the magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can visualize magnetic fields directly because little slivers of iron can act as tiny compasses.  By spreading a bunch of them out around a magnet, we can see the outlines of the magnetic field lines.  Here's the Wikipedia picture for this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/EeHNi.png&quot; alt=&quot;enter image description here&quot;&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a magnetic dipole, and as you can see it bears a strong resemblance to the electric dipole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Relationship between Electric and Magnetic Fields&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that electric and magnetic fields are related to each other.  Charged particles create electric fields.  However, if those same charges start moving, they create magnetic fields.  If you try to use a compass near a wire carrying DC current, you'll see the needle deflected by the magnetic field created by the moving charges in the wire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, electric fields and magnetic fields can create each other according to precise mathematical rules called &lt;a href=&quot;http://en.wikipedia.org/wiki/Maxwell%27s_equations&quot; rel=&quot;nofollow&quot;&gt;Maxwell's equations&lt;/a&gt;.  Any time an electric field changes in time, it creates a magnetic field that &quot;curls&quot; around it (loosely speaking - you have to learn vector calculus for the precise statement).  Similarly, a changing magnetic field creates an electric field that curls around it in the same way. This is called &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_induction&quot; rel=&quot;nofollow&quot;&gt;electromagnetic induction&lt;/a&gt;&quot; (and is a different use of the word &quot;induction&quot; than when the balloon induced an electric field in the wall).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Electromagnetic Radiation&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rules for the relationship between electric and magnetic fields work out so that you can get propagating waves of electric and magnetic fields traveling through space.  Very roughly speaking, the changing electric field creates a changing magnetic field, which creates a changing electric field, etc, and the whole thing propagates forward at the speed of light.  To truly see how this works, you'll have to learn the math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make an electromagnetic wave, just take something with charge and shake it.  If you take that balloon you rubbed against your hair and start shaking it back and forth, you're creating electromagnetic waves (their wavelength is hundreds of thousands of kilometers, though).  If you could shake the balloon back and forth about a quadrillion times per second, you would actually see light emitted from the balloon.  At slightly lower frequencies you could emit microwaves from it to cook your food or, lower still, listen to it on your radio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for what an electromagnetic wave is, it is just a changing electric and magnetic field.  If an electromagnetic wave came past you, you could detect it with your balloon by watching the balloon vibrate back and forth, or with your compass in the same way.  However, most electromagnetic waves have frequencies too high to notice with an instrument as coarse as a balloon or a compass.  Instead, we detect electromagnetic waves with things like film, CCD's and antennas.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-02-06T06:17:47.997" CommentCount="3" />
  <row Id="4669" PostTypeId="1" CreationDate="2011-02-06T06:20:02.000" Score="6" ViewCount="917" Body="&lt;p&gt;Can you provide a list of the most important discoveries/breakthroughs in physics recently?&#xA;By recent, I mean the past decade or so. All branches of physics are welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I am interested in major physics breakthroughs/discoveries which haven't become well-known yet outside their narrow specialties. Most breakthroughs in the 90s like string dualities and the accelerating universe have already become common knowledge. &lt;/p&gt;&#xA;" LastActivityDate="2011-02-16T01:54:44.867" CommunityOwnedDate="2011-02-06T07:31:38.553" Title="What are the most important discoveries/breakthroughs in physics recently?" Tags="&lt;soft-question&gt;" AnswerCount="11" CommentCount="1" FavoriteCount="6" />
  <row Id="4670" PostTypeId="2" ParentId="4669" CreationDate="2011-02-06T06:52:34.243" Score="3" ViewCount="" Body="&lt;p&gt;In theoretical side, I think Witten's discovery that all five string theories are different limits of one unique theory and in experimental side it has to be the surprising discovery that the universe is expanding in an accelerating manner.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-02-06T06:52:34.243" CommunityOwnedDate="2011-02-06T07:31:38.553" CommentCount="6" />
  <row Id="4671" PostTypeId="2" ParentId="4669" CreationDate="2011-02-06T07:03:17.493" Score="1" ViewCount="" Body="&lt;p&gt;1) Black hole radiation (Hawking radiation)&#xA;2) Discovery of the dark energy&#xA;3) Evaluation of the BH entropy from string theory.&lt;/p&gt;&#xA;" OwnerUserId="1774" LastActivityDate="2011-02-06T07:03:17.493" CommunityOwnedDate="2011-02-06T07:31:38.553" CommentCount="10" />
  <row Id="4672" PostTypeId="1" CreationDate="2011-02-06T07:09:49.407" Score="6" ViewCount="324" Body="&lt;p&gt;There has been some heated debate as to whether the laws of physics allow for traversable wormholes. Some physicists claim we require exotic matter to construct wormholes, but then others counter the Casimir effect with ordinary matter is sufficient. But these same physicists seldom come up with an explicit solution or state of ordinary matter keeping the throat of a wormhole open via the Casimir effect. Yet others claim with extra dimensions, a Gauss-Bonnet interaction is sufficient to keep the wormhole throat open, but opposing physicists claim such a term can't arise from string theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is, do traversable wormholes exist as solutions to string theory? &lt;/p&gt;&#xA;" OwnerUserId="1776" LastActivityDate="2011-02-07T14:24:03.500" Title="Do traversable wormholes exist as solutions to string theory?" Tags="&lt;general-relativity&gt;&lt;quantum-field-theory&gt;&lt;string-theory&gt;&lt;quantum-gravity&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="4673" PostTypeId="2" ParentId="4669" CreationDate="2011-02-06T08:11:41.000" Score="7" ViewCount="" Body="&lt;p&gt;One of the biggest recent developments in high-energy physics is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutrinos#Direct_detection_of_flavor_oscillation_in_solar_neutrinos&quot; rel=&quot;nofollow&quot;&gt;confirmation of neutrino oscillations&lt;/a&gt; and the consequent realization that neutrinos must have some non-zero mass.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-02-06T22:23:39.447" LastActivityDate="2011-02-06T22:23:39.447" CommunityOwnedDate="2011-02-06T08:11:41.000" CommentCount="4" />
  <row Id="4674" PostTypeId="1" CreationDate="2011-02-06T08:22:08.343" Score="5" ViewCount="254" Body="&lt;p&gt;Perturbation theory presumes we have a valid family of models over some continuous (infinitely differentiable, in fact) range for some parameters, i.e. coupling constants. We have some special values for the coupling constants characterizing the unperturbed model, which presumably, is relatively easy to solve. We also assume the family of models transform smoothly under the coupling constants. Then, we perform a Taylor series analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if the landscape of valid quantum gravity models is discrete? Even though superstring theory admits a dilaton modulus over 10 uncompactified dimensions, what about the landscape of models we get after compactifying 6 spatial dimensions with nonzero fluxes and some branes, and maybe some orbifolding? We still have moduli when supersymmetry remains unbroken, but what about the metastable states where SUSY is broken? What is the Taylor series of a Dirac delta function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about perturbation theory from the perspective of path integrals? With path integrals, the Wheeler-DeWitt constraint shows up in a different guise as a projection operator. We start off with some wavefunctionals, and then take the functional integral over some finite time interval T. In the limit as T goes to infinity, we are left with a projection operator singling out WDW solutions. But what happens when we interchange the order in which we take the limit of the coupling constant going to zero, and T going to infinity? If the spectrum of the Hamiltonian constraint is discrete, and varies with the coupling constant, such an interchange won't be valid! This is a fancy way of saying for most choices of coupling constants, the projection operator is zero.&lt;/p&gt;&#xA;" OwnerUserId="1776" LastActivityDate="2011-02-06T14:47:11.083" Title="Does perturbation theory break down for quantum gravity?" Tags="&lt;quantum-gravity&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" />
  <row Id="4675" PostTypeId="2" ParentId="2435" CreationDate="2011-02-06T08:53:45.473" Score="0" ViewCount="" Body="&lt;p&gt;Let me answer to the part of the question regarding the relation between spin networks and functionals of the connection.  Each spin network defines precisely one functional of the connection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, take the simplest spin network formed by just one loop.  The corresponding functional of the connection is the &quot;trace of the holonomy&quot; of the connection. That is the trace of the group element obtained integrating the connection along the loop. Physically, this is a number that says how much a vector comes back rotated if you parallel transport it along the loop.  This of course depends on the connection, so it is a functional of the connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A functional of the connection is a quantum state, like functions of x are quantum states psi(x) is usual quantum theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It then turns out that these functionals have the property that they diagonalize the area and volume. This like the state psi(x)=exp{ipx} being an eigenstate of the momentum. So, spin networks are functionals of the connection that are eigenstates of the spatial geometry (this is the result of a calculation).&lt;/p&gt;&#xA;" OwnerUserId="1613" LastActivityDate="2011-02-06T08:53:45.473" CommentCount="2" />
  <row Id="4676" PostTypeId="2" ParentId="4674" CreationDate="2011-02-06T09:03:38.560" Score="2" ViewCount="" Body="&lt;p&gt;The perturbative expansion of general relativity breaks at the two-loop level: one produces a term in the effective action which is cubic in the Weyl tensor, as showed by Goroff and Sagnotti, which has a UV-divergent coefficient, and which has to be canceled by a counterterm. This introduces a new unknown finite coupling to the &quot;no longer Einstein-Hilbert action&quot;, too. Because of that, one ultimately gets an infinite number of unknown couplings and loses predictivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that general relativity is perturbatively non-renormalizable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that there has to be new physics that determines all the unknown parameters of the low-energy physics (and, typically, adds qualitatively new phenomena at high energies, too). The spectrum of possibilities is given by the &quot;landscape&quot; of string/M-theory. All versions of the Minkowski or anti de Sitter or de Sitter space span a complicated set with many components. Some of these components are discrete; we call them the stabilized vacua. Some of them have residual parameters - the moduli - and those are very interesting mathematically (and they are usually calculable, and often have some unbroken supersymmetry) but they are unacceptable phenomenologically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only the stabilized vacua are realistic candidates for a theory of the real world. The unstabilized ones violate the equivalence principle, allow the fine-structure constant and similar constants to vary, and lead to new, unobserved long-range forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, around each vacuum, it's still true that the amplitudes may be Taylor-expanded in any coupling constant that happens to be weak. The fact that only one value of a coupling constant - or a scalar field, if we're in the gravitational context - is the right one is seen as the existence of a potential for this scalar field. If we find ourselves away from the minimum (or extremum) of this potential, there will exist non-zero one-point functions for this scalar field that drive the Universe back to the minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the computation is done properly, the scattering amplitudes are analytic functions of the energy-momentum vectors &quot;almost everywhere&quot;. This fact is guaranteed by the locality (or even approximately locality) of the physical phenomena in spacetime. So there can't be any delta-functions here - except for those that impose the conservation laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is indeed true that the Minkowski-like or anti-de-Sitter-like &quot;empty space&quot; solutions only exist for the right values of the couplings that minimize the potential; the physical spectrum of &quot;empty space&quot; states is strictly vanishing away from the right stabilized value of the moduli. But this fact shouldn't be viewed as a discontinuity in the underlying maths. Instead, you should imagine that for wrong values of the coupling constants, there exist &quot;analogous&quot; solutions which are not &quot;empty space&quot;. Instead, in these solutions, the scalar fields oscillate around their preferred value for which the potential is minimized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The states &quot;don't disappear&quot; as you move to wrong values of the coupling constant. Instead, they just fail to be translationally symmetric in time. In this sense, there's no discontinuity and the calculations of physical amplitudes and other observables never involve delta-functions of the scalar fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much more generally, however, it is of course conceivable that the perturbative expansions break down for many known and unknown reasons. It's been known for a long time that the perturbative expansions ultimately diverge; and they don't capture all the physical phenomena, anyway. If we resum the terms up to the minimal one (before they start to blow up again) in a divergent expansion in a coupling constant, the minimum term - an uncertainty of the sum - is of the same order as the first non-perturbative contributions to the amplitudes (various instantons). But attempts to &quot;resum the perturbative expansions more properly&quot; are not the only methods to approach the non-perturbative physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if a perturbative expansion were &quot;qualitatively&quot; failing even for a very small value of the coupling constant, it would probably prove that the theory is inconsistent. This shouldn't happen. Even if it happened in a hypothetical theory, you would have to think how this theory allows to calculate something, at least in principle - without that, you shouldn't talk about a theory at all.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-06T09:22:39.557" LastActivityDate="2011-02-06T09:22:39.557" />
  <row Id="4677" PostTypeId="1" AcceptedAnswerId="4694" CreationDate="2011-02-06T09:28:24.890" Score="1" ViewCount="102" Body="&lt;p&gt;Inspired by &lt;a href=&quot;http://physics.stackexchange.com/questions/4652/how-do-we-know-that-dark-matter-is-dark&quot;&gt;How do we know that dark matter is dark?&lt;/a&gt; and &lt;a href=&quot;http://physics.stackexchange.com/questions/519/what-is-the-temperature-of-the-surface-and-core-of-a-neutron-star-formed-12-billi&quot;&gt;What is the temperature of the surface and core of a neutron star formed 12 billion years ago now equal to?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="238" LastActivityDate="2011-02-06T15:25:24.413" Title="What about the neutron stars with CMB temperature as dark matter?" Tags="&lt;astrophysics&gt;&lt;neutron-stars&gt;&lt;dark-matter&gt;" AnswerCount="3" />
  <row Id="4678" PostTypeId="2" ParentId="4677" CreationDate="2011-02-06T09:41:14.440" Score="3" ViewCount="" Body="&lt;p&gt;The inner temperature of a newly formed neutron star is around $10^{11}-10^{12}$ Kelvin degrees. It emits lots of neutrinos and cools to $10^{6}$ Kelvin within years. That's a high enough temperature behind the fact that neutron stars emit thermal X-rays. The temperature may decrease a little bit but getting to $2.7$ Kelvin is, well, unrealistic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But more generally, it's a good idea that the dark matter is pretending to be cosmic microwave radiation. You could replace neutron stars by small black holes that emit at the CMB temperature, too. The temperature would have to be fine-tuned to the right temperature (otherwise we would see it as extra anisotropy in WMAP pictures) - and unlikely to keep it as the Universe expands and the CMB temperature cools down. ;-) If this were true, the Universe would have to fine-tuned to confuse observers who happen to live 13,730,002,011 years after the Big Bang. :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, no, I am actually wrong. It's not excessively fine-tuned because objects near the CMB temperature don't cool down anymore. But it's still difficult for matter in any reasonable state to cool down that much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, even if you don't consider neutron stars but more general large objects as an explanation of dark matter - they're referred to MACHO, RAMBO etc. - they seem to be disfavored as a solution to the dark matter. It seems that most of the dark matter has to be cold dark matter, and WIMP (Weakly Interacting Massive Particle) arguably represents a majority of the mass of the dark matter.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-06T11:48:42.277" LastActivityDate="2011-02-06T11:48:42.277" CommentCount="7" />
  <row Id="4679" PostTypeId="1" CreationDate="2011-02-06T10:41:33.700" Score="0" ViewCount="69" Body="&lt;p&gt;Could you explain me everything about treating particle as wave and particle in one ?&lt;/p&gt;&#xA;" OwnerUserId="58" LastActivityDate="2011-02-06T10:41:33.700" ClosedDate="2011-02-06T10:49:56.367" Title="How electron as a wave looks like" Tags="&lt;particle-physics&gt;&lt;waves&gt;" CommentCount="2" />
  <row Id="4680" PostTypeId="1" AcceptedAnswerId="4682" CreationDate="2011-02-06T11:04:48.243" Score="2" ViewCount="164" Body="&lt;p&gt;Where exactly does CMB come from. I've seen it in documentaries as a huge sphere with Earth in the middle. But if all this radiation was ejected from the start of the universe some time after the big bang; why can we see it? Surely the radiation should be travelling away from us? Just like every galaxy is?&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-07T20:49:00.250" LastActivityDate="2011-02-07T20:49:00.250" Title="Where does CMB come/emit from" Tags="&lt;astrophysics&gt;&lt;cosmology&gt;" AnswerCount="1" />
  <row Id="4681" PostTypeId="2" ParentId="4649" CreationDate="2011-02-06T11:18:33.657" Score="2" ViewCount="" Body="&lt;p&gt;Ice, like water, has a much higher dielectric constant than air. This can detune the resonant frequency of the antenna so there is now an impedance mismatch with the feedline (transmission line feeding the antenna) at the frequency you're using. This is manifested as an increase in VSWR (voltage &lt;a href=&quot;http://en.wikipedia.org/wiki/Standing_wave_ratio&quot; rel=&quot;nofollow&quot;&gt;standing wave ratio&lt;/a&gt;). Basically this means that, since more power is being reflected from the antenna back into the feedline, you need to have a higher power in the feedline to get the same radiated power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the second question here: &lt;a href=&quot;http://www.michwave.com/bbnetwork/faq/troubleshooting.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.michwave.com/bbnetwork/faq/troubleshooting.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-02-06T11:18:33.657" CommentCount="1" />
  <row Id="4682" PostTypeId="2" ParentId="4680" CreationDate="2011-02-06T11:24:21.177" Score="5" ViewCount="" Body="&lt;p&gt;The radiation was &quot;produced&quot; about 380,000 years after the Big Bang, and it was produced at every point of the Universe. From the very beginning, it's been (almost) uniform (the same at all places) and isotropic (the same in all directions). Since that time, the radiation was moving in all directions, essentially without any interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cosmic microwave radiation &quot;decoupled&quot; - separated - from the rest of the matter in the Universe in this era we call &quot;decoupling&quot;. Before the decoupling, the temperature of the Universe was so high that electrons and protons were largely separated in a plasma filling the Universe. Plasma carries a lot of random electric charge that severely interacts with photons all the time - so the plasma was opaque for the radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, after the &quot;decoupling&quot;, the atoms were formed for the first time - mostly Hydrogen atoms. Hydrogen atoms are neutral and their interactions with the photons are much weaker so the Universe became essentially transparent. The photons - and everything else in the Universe - at the moment of decoupling had a certain high temperature (thermal equilibrium, about 3000 Kelvin) which means that their spectrum was Planck's black-body thermal spectrum corresponding to the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that moment, photons were moving without any interactions and their wavelength was increasing proportionally to the size of the Universe. That also means that the energy of each photon was decreasing by the same factor; the temperature of the black body radiation did the same thing. That's why the current CMB temperature is just 2.7 Kelvin. You may see that the Universe's linear dimensions expanded about 1,000 times from the decoupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that 13.73 billion years over 380,000 years is substantially more than 1,000. That's because in the early stages, the expansion of the Universe was &quot;decelerating&quot; as a function of time. Only in recent few billions of years, the expansion got actually accelerating because of dark energy that gradually became important.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the WMAP probe detects a photon of the cosmic microwave background, this collision with the telescope is the first interaction of this photon since the moment when the Universe was 380,000 years old. This fact allows you to to deduce how far is the point when the photon was born - or when it last interacted with another object. The birth place is clearly a point in the direction where the photon is coming from. The distance is always the same so all the photons we see here today had to be produced at a particular spherical shell in spacetime. The center of the shell is &quot;our place in the past&quot; and the radius is such that the photons from the shell, when travelling inwards, exactly needed those 13.7 billion years of the cosmic time to get here.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-06T11:30:39.517" LastActivityDate="2011-02-06T11:30:39.517" />
  <row Id="4683" PostTypeId="2" ParentId="4677" CreationDate="2011-02-06T12:28:30.123" Score="1" ViewCount="" Body="&lt;p&gt;After &#xA;formation of the neutron star, some  superfluid &#xA;phase developes, the energy produced by bonding that &#xA;neutron pairs is radiated as neutino/antineutrino  pairs:&#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/4575/neutrinos-by-formation-of-neutron-pairs&quot;&gt;neutrinos by formation of &quot;neutron pairs&quot;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-02-06T12:28:30.123" />
  <row Id="4684" PostTypeId="2" ParentId="4649" CreationDate="2011-02-06T12:58:30.050" Score="2" ViewCount="" Body="&lt;p&gt;What Keenan &lt;/p&gt;&#xA;&#xA;&lt;p&gt;surmises, is simply fact. &#xA;Ice has rel. dieelectric constant of about 2 *&#xA;at 100 MHz. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(That is what nibot forgot to &#xA;write in his question. When an American speaks &#xA;of &quot;Radio&quot; this is usually medium wave still!) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such FM antennas are well tuned, and as in any &#xA;transmitter the &quot;standing wave ratio&quot; (SWR) is &#xA;monitored or controlled if possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This SWR controls the amount of power reflected &#xA;back to the output stage. In case of detuning, &#xA;this will destroy the output stage of the transmitter,&#xA;if it is running at borderline of thermal dissipation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Georg&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;this is a very old figure, but more recent ones only can be &#xA;higher, no solid with a &quot;normal&quot; density can be less than DC  = 2 . &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-02-06T18:44:37.597" LastActivityDate="2011-02-06T18:44:37.597" />
  <row Id="4686" PostTypeId="2" ParentId="4674" CreationDate="2011-02-06T13:42:38.100" Score="-4" ViewCount="" Body="&lt;p&gt;The perturbation theory for QG breaks down as well as for any other QFT: one obtains infinite perturbative corrections. The only difference is that the UV infinities in QG cannot be discarded in a &quot;consistent&quot; way. They say the QG is &quot;non renormalizable&quot;. That means it is not normal initially and is not normal after &quot;reparations&quot; (renormalizations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a simple criterion to check if a theory physically and mathematically reasonable. This criterion is not very popular but it reveals a fallacious theory in the first Born approximation, i. e., even before encountering infinities. This criterion is the following: if the first Born approximation cannot catch the processes whose probability is unity (soft radiation), then you will obtain an explosion of perturbative corrections. If one misses the processes that happen always, one is doing something rather wrong, and no wonder the perturbative corrections will try to &quot;violently correct&quot; that bad start: the perturbation theory initial approximation is too far from the exact solution and needs infinite corrections and a their non linear summation to obtain a finite result. This is a simple physical and mathematical explanation of origin of those &quot;difficulties&quot; that are usually hided later on under rug.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Downvoters, if you disagree, give your disagreement statements, please.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-06T14:47:11.083" LastActivityDate="2011-02-06T14:47:11.083" CommentCount="3" />
  <row Id="4687" PostTypeId="2" ParentId="4662" CreationDate="2011-02-06T13:52:15.107" Score="1" ViewCount="" Body="&lt;p&gt;In case that in such a problem the figure of friction &#xA;at rest is given, You should check, whether the block&#xA;might stick in the uppermost position.&#xA;Such kind of traps are common, depending on &quot;niceness&quot; of Prof.  :=(&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-02-06T13:52:15.107" CommentCount="2" />
  <row Id="4688" PostTypeId="2" ParentId="4669" CreationDate="2011-02-06T14:00:27.097" Score="3" ViewCount="" Body="&lt;p&gt;I know this was more than a decade ago, but the discovery and formulation of the AdS/CFT correspondence stating that a conformal field theory in d dimensions without gravity is equivalent to a theory of quantum gravity over an anti de Sitter background one dimension higher was a major breakthrough.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-06T14:00:27.097" CommunityOwnedDate="2011-02-06T14:00:27.097" CommentCount="1" />
  <row Id="4689" PostTypeId="2" ParentId="4669" CreationDate="2011-02-06T14:10:11.783" Score="3" ViewCount="" Body="&lt;p&gt;This also happened more than a decade ago, but the discovery of anisotropies in the cosmic microwave background of the order of $10^{-5}$ by the COBE satellite was also groundbreaking. Subsequent measurements by WMAP have shown the power spectrum is approximately scale-invariant and Gaussian.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-06T14:10:11.783" CommunityOwnedDate="2011-02-06T14:10:11.783" />
  <row Id="4690" PostTypeId="1" CreationDate="2011-02-06T14:17:37.450" Score="4" ViewCount="168" Body="&lt;p&gt;Most string theory compactifications analyzed so far have as backgrounds a conformal field theory corresponding to a nonlinear sigma model with a Calabi-Yau target space, or some relatively classical background, possibly with orbifolding. But as long as the central charge of the Virasoro algebra is right, and worldsheet superconformal symmetry is respected, any other superconformal field theory will do just as well. Why haven't string theorists looked at such models more often?&lt;/p&gt;&#xA;" OwnerUserId="1776" LastActivityDate="2011-03-02T06:40:41.937" Title="Why aren't string compactifications by generic conformal field theories considered more often?" Tags="&lt;string-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="4691" PostTypeId="2" ParentId="4669" CreationDate="2011-02-06T14:29:34.153" Score="2" ViewCount="" Body="&lt;p&gt;The discovery of a landscape of compactifications in string theory with exponentially many KKLT compactifications is another recent groundbreaking discovery. The landscape made the strong anthropic principle respectable for the first time, and solved fine-tuning problems like the cosmological constant problem.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-06T14:29:34.153" CommunityOwnedDate="2011-02-06T14:29:34.153" CommentCount="1" />
  <row Id="4692" PostTypeId="2" ParentId="4669" CreationDate="2011-02-06T14:34:10.203" Score="5" ViewCount="" Body="&lt;p&gt;The isolation of a graphene --- a single layer of graphite --- was another breakthrough. 2D graphene has many unusual properties, all related to a highly delocalized and strongly entangled band of electrons.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-06T14:34:10.203" CommunityOwnedDate="2011-02-06T14:34:10.203" />
  <row Id="4693" PostTypeId="2" ParentId="4669" CreationDate="2011-02-06T14:55:02.833" Score="-7" ViewCount="" Body="&lt;p&gt;I think my idea of &quot;electronium&quot; as a compound system including the electron and &quot;its&quot; quantized electromagnetic field deserves some attention. It is a non perturbative solution of some coupled QED equations advanced as a physical and mathematical ansatz. It may help construct divergence-less theories.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-06T15:11:53.510" LastActivityDate="2011-02-06T15:11:53.510" CommunityOwnedDate="2011-02-06T14:55:02.833" CommentCount="6" />
  <row Id="4694" PostTypeId="2" ParentId="4677" CreationDate="2011-02-06T15:25:24.413" Score="1" ViewCount="" Body="&lt;p&gt;There were extensive searches in the early and mid 1990's for astronomical micro-lensing events, and these concluded that either the Milky Way is anomalous or there just aren't enough MACHOs to be a big part of the missing mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MACHO?&lt;/strong&gt; That's MAssive Compact Halo Object. Anything small, heavy, cool and gravitationally bound to a galaxy. Cool neutron stars would count.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Micro-lensing?&lt;/strong&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_microlensing&quot; rel=&quot;nofollow&quot;&gt;If a MACHO passes &lt;em&gt;very&lt;/em&gt; precisely between a telescope and a distant star there is a detectable change in the intensity of the light from the star due to gravitational lensing&lt;/a&gt;, and that change follows a predictable curve depending on the intervening mass and the impact parameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, with a sensitive (you need to be able to image down to 20th magnitude or better to get much data), computerized telescope you watch a nearby galaxy (say the Magellanic clouds) &lt;em&gt;every night&lt;/em&gt; and count. With enough data you can make a pretty good estimate of the total mass of these things in the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside: When I was an undergrad I worked on a small, cooled CCD telescope for a physics prof. And we'd try to image micro-lensing events reported by IAU telegrams. We were struggling with the tracking software, so it was a struggle to get imaging below 17th magnitude, but we did have a good run where we got a light curve that matched the big boys pretty well.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-02-06T15:25:24.413" />
  <row Id="4695" PostTypeId="2" ParentId="4672" CreationDate="2011-02-06T16:07:38.323" Score="-1" ViewCount="" Body="&lt;p&gt;It might be best to assume that a string solution for wormholes exist and see what the consequences are.  Wormholes are similar to black holes, but where the even horizon is replaced by a membrane of some type of quantum field which causes geodesics to diverge.  The Hawking-Penrose energy condition, in particular the weak energy condition, gives geodesics which focus inwards in a spacetime diagram.  In order to get a wormhole it requires that inward focusing geodesics be “defocused” near or at the region where the event horizon of a black hole would otherwise exist.  This means the geodesics are defocused into some other region of spacetime.  The wormhole is then two 3-balls cut out of spacetime and where the boundaries of the two balls have points identified with each other.  The special field which violates the Hawking-Penrose energy conditions defines a junction in spacetime where the curvature exhibits an abrupt change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In string theory the action defines the world sheet area&#xA;$$&#xD;&#xA;S~=~-\frac{T}{2}\int d^2\sigma\sqrt{h}h^{ab}(\sigma)g_{\mu\nu}\partial_a X^\mu \partial_b X^\nu&#xD;&#xA;$$&#xA;where $T$ is the string tension, $h_{ab}$ is the metric for the string world sheet with coordinates $\sigma~=~(\tau,~\sigma^1)$, $g_{\mu\nu}$ is the spacetime metric and $X^\mu$ is the coordinates of the string parameterized in its modes $\alpha_n$ ${\tilde\alpha}_n$.  We then want to expand this string into terms which correspond to spacetime curvatures and then to look at the momentum-energy curvature.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The string coordinates are functions of the spacetime coordinates $X^\mu~=~X^\mu$ where we consider the string coordinates expanded as $X^\mu ~\rightarrow~X^\mu~+~\delta X^\mu$.  The variations in the string $\delta X^\mu~=~Y^\mu(\sigma)$ are small oscillations of the string which occur around the string path $\sigma^1$.  Now consider $\partial_aX^\mu$ in the action expanded by these variation.  The linear term in $Y^\mu$ will be the covariant derivative $Y^\nu\nabla_\nu(\partial_aX^\mu)$.  Plugging this in and using the geodesic equation the action for the string in first order with these oscillations&#xA;$$&#xD;&#xA;S~\simeq~-\frac{T}{2}\int d^2\sigma\sqrt{h}h^{ab}\big(g_{\mu\nu}~+~R_{\mu\alpha\nu\beta}Y^\alpha Y^\beta\big)\partial_a X^\mu \partial_bX^\nu&#xD;&#xA;$$&#xA;The term $ R_{\mu\alpha\nu\beta}Y^\alpha Y^\beta~\sim~R_{\mu\nu}Y^2$ is negative, which defocuses geodesics.  This Lagrangian determines the momentum energy tensor &#xA;$$&#xD;&#xA;T_{ab}~=~-\frac{2}{T\sqrt{h}}\frac{\delta S}{\delta h^{ab}}&#xD;&#xA;$$&#xA;which is traceless and the field equation have $\frac{\delta S}{\delta h^{ab}}~=~0$.  This then leads to a form of the action as a formula for the area of the world sheet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then have a funny situation here.  This negative Ricci curvature defines a momentum energy tensor in spacetime indices where $T^{00}~&amp;lt;~0$.  Further, the state space which constructs this $\langle~|T^{00}|~\rangle$ is not bounded below.  This means the negative curvature may become arbitrarily large.  This suggests a serious contradiction, for this can imply the world sheet area of a string can be negative, and arbitrarily negative.  It is unclear exactly what is meant by a negative area for a string world sheet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black hole has an event horizon with an area $A$, where entropy is $S~=~k~A/4L_p^2$.  Each unit of area contributes a unit of entropy, and is further identified according to units of $G$ with naturalized units $area$.  The event horizon in the holographic setting if covered by strings, and the modes of the string define the degenerate set of states of the black hole.  Hence we may think of the horizon area as a summation over the string world sheet areas, which are positive and are identified with a positive entropy.  The wormhole from a stringy perspective has then a funny appearance, where there are negative areas and negative entropies.  If wormholes exist it is not hard to see that one could connect up with the interior of a black hole and reduce the entropy of the system by accessing states in the interior.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-06T17:36:51.933" CommentCount="0" />
  <row Id="4696" PostTypeId="1" AcceptedAnswerId="4699" CreationDate="2011-02-06T16:13:59.493" Score="0" ViewCount="176" Body="&lt;p&gt;For &lt;strong&gt;S&lt;/strong&gt; and &lt;strong&gt;S'&lt;/strong&gt; in standard configuration, the Galilean transformations are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;x' = x - vt, y' = y, z' = z, t' = t&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the Lorentz transformations for v &amp;lt;&amp;lt; c:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;x' = x - vt, y' = y, z' = z, t' = t - vx/c^2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it looks as if the Galilean transformations become increasingly accurate  for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;vx -&gt; 0, v &amp;lt;&amp;lt; c&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And exact for v = 0 for all x. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, all text books I've come across state that the Galilean transformatons become more accurate for the condition &lt;strong&gt;v &amp;lt;&amp;lt; c&lt;/strong&gt; only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what are the conditions under which the Galilean transformations become more accurate and why?&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-07T10:53:55.663" LastActivityDate="2011-02-09T22:16:36.873" Title="Under what conditions do the Galilean transformatons become more accurate?" Tags="&lt;classical-mechanics&gt;&lt;special-relativity&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="4697" PostTypeId="2" ParentId="4690" CreationDate="2011-02-06T16:28:26.043" Score="4" ViewCount="" Body="&lt;p&gt;Such models have been looked at in a fair amount of detail. For example, &quot;Gepner&quot; models&#xA;are constructed by tensoring together superconformal minimal models. Asymmetric orbifolds which have the correct superconformal worldsheet symmetry have also been constructed. The problem is that if you break spacetime susy when you construct these theories directly as SCFTs you typically have tachyons and no good way to figure out the ground state of the system. If you preserve spacetime supersymmetry then many (all?) of the Gepner models turn out to describe special points in the moduli space of Calabi-Yau compactifications. So in this case at least, sigma-models with CY target end up being the more general construction. Some of the asymmetric orbifold constructions are not equivalent to CY as far as I know, but they are quite specialized and I don't know of any that have very realistic phenomenology. So basically the answer is that people tend to focus on the CY case because it is richer and you can do more with it. Of course there are probably some interesting constructions based on the SCFT approach you are asking about that have not been found yet. For example, I don't know of any Gepner model constructions of $G_2$ holonomy compactifications (I'd be happy to be corrected if these do in fact exist).&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T06:40:41.937" LastActivityDate="2011-03-02T06:40:41.937" />
  <row Id="4698" PostTypeId="2" ParentId="4696" CreationDate="2011-02-06T16:54:44.507" Score="0" ViewCount="" Body="&lt;p&gt;I have not quite understood your problem. For $v &amp;lt;&amp;lt; c$, $vx/c^2$ is negligible unless $x$ is huge. Hence $t\prime = t$ We get the Galilean transformation back. I guess, you want to get back the Galilean transformation &quot;exactly&quot;. No John, that never happens. All we can say is Galilean transformation is &lt;em&gt;sufficiently accurate&lt;/em&gt; for description of transformations from one inertial frame to another for the $v &amp;lt;&amp;lt; c$ case. If the separation between two events is huge then yes, even for ordinary velocities you can't use Galilean transformations, you have to use Lorentz transformations.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-07T02:39:22.387" LastActivityDate="2011-02-07T02:39:22.387" CommentCount="1" />
  <row Id="4699" PostTypeId="2" ParentId="4696" CreationDate="2011-02-06T17:12:34.960" Score="3" ViewCount="" Body="&lt;p&gt;If I understand correctly, the question is about whether, in deriving the Galilean transformation as an approximate limiting case of the Lorentz transformation, it is necessary to impose the requirement that $vx$ is small, &lt;em&gt;in addition&lt;/em&gt; to the &quot;obvious&quot; requirement that $v$ is small. The answer is yes, it is necessary. Of course, since $vx$ is not a dimensionless quantity, we have to specify what we mean when requiring that it is small. The specific requirement is that $vx/c^2\ll \Delta t$, where $\Delta t$ is the precision with which we wish to calculate time intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that, in the inertial reference frame in which you are at rest, two stars explode simultaneously, one here and one in the Andromeda galaxy. Consider the same two events in the reference frame of someone walking at a liesurely pace. The Lorentz transformation indicates that they will be separated by a time interval of about 2 days. If you care about levels of precision less than that, you can't use the Galilean transformation, even though $v/c$ is small.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-06T17:12:34.960" />
  <row Id="4700" PostTypeId="1" CreationDate="2011-02-06T17:15:38.723" Score="2" ViewCount="411" Body="&lt;p&gt;Why can't photons have a mass? Could you explain this to me in a short and mathematical way?&lt;/p&gt;&#xA;" OwnerUserId="58" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-02-10T10:06:26.037" LastActivityDate="2011-02-10T10:06:26.037" Title="Why can't photons have a mass" Tags="&lt;mass&gt;&lt;photons&gt;&lt;quantum-electrodynamics&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="4701" PostTypeId="2" ParentId="4700" CreationDate="2011-02-06T17:27:08.893" Score="1" ViewCount="" Body="&lt;p&gt;In the context of special relativity, anything that travels at the speed of light can't have a nonzero rest mass. One way to see this is that the kinetic energy of an object of mass $m$ moving at speed $v$ is&#xA;$$&#xD;&#xA;mc^2\left({1\over\sqrt{1-v^2/c^2}}-1\right),&#xD;&#xA;$$&#xA;which tends to infinity as $v\to c$. Physically, this means that it would cost an infinite amount of energy to raise a massive particle up to speed $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as special relativity is concerned, it's logically possible that photons do have mass and travel at speeds (slightly) less than $c$. (This would mean that the quantity $c$ that occurs in special relativity should not be called &quot;the speed of light.&quot;) The experimental limits on this possibility are extremely severe, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may have misguessed the level of your question and the sort of answer you're looking for. For instance, there are separate reasons for believing the photon to be strictly massless based on gauge invariance. &lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-06T17:27:08.893" CommentCount="0" />
  <row Id="4702" PostTypeId="2" ParentId="4700" CreationDate="2011-02-06T17:35:49.980" Score="0" ViewCount="" Body="&lt;p&gt;According to the special theory of relativity, any particle with a finite mass requires an infinite amount of energy to reach the speed of light. Therefore no particles with any intrinsic mass can travel with the speed of light.&#xA;The energy required to attain a speed $v$ is given by $E$ = $\frac{mc^2}{\surd1-v^2/c2}$ - $mc^2$&#xA;As $v$ approaches $c$, $E$ approaches $\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only massless particles are allowed to travel at the speed of light. Photon is massless, hence it can travel with the speed of light. The energy of a photon is given by $E = pc$ where p is the momentum of the photon.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-06T17:54:59.260" LastActivityDate="2011-02-06T17:54:59.260" CommentCount="10" />
  <row Id="4704" PostTypeId="1" CreationDate="2011-02-06T18:19:54.500" Score="2" ViewCount="267" Body="&lt;p&gt;This question is directed mostly at people giving lectures on black holes, but input by other physicists or students is very much appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Do you know a good (home)-experiment with a black hole analog (such as water in a bathtub) that allows to discuss most of the pertinent concepts and features of a black hole?&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify what I mean by 'good' let me provide what I consider as a good experiment with a white hole analog (see exercise 2.2 in &lt;a href=&quot;http://quark.itp.tuwien.ac.at/~grumil/pdf/ex2.pdf&quot; rel=&quot;nofollow&quot;&gt;this exercise sheet&lt;/a&gt;), namely a hydraulic jump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This experiment allows you to discuss in very simple terms what a white hole is (the white hole region is visibly distinct from the exterior because of their different water depths). It is also quite easy to show experimentally that you cannot send information into the white hole (in the form of shallow water wave excitations aka 'ripplons'). Moreover, the white hole exhibits some interesting features in the near horizon region (see the concentric rings in the picture on the back of the exercise sheet). Most importantly, this experiment can be done basically by anyone - there is no need for access to lasers, waveguides, Laval nozzles, supersonic flow etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of the hydraulic jump experiment is to provide a pedagogic introduction to white holes. I would like to have a comparable (non-Gedanken-)experiment available to introduce black holes. &lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-02-11T00:27:17.903" Title="Black hole analog experiment?" Tags="&lt;black-holes&gt;&lt;home-experiment&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="4705" PostTypeId="2" ParentId="4704" CreationDate="2011-02-06T18:29:12.880" Score="1" ViewCount="" Body="&lt;p&gt;How deep do you want the analogy to go?  If you just want some region of a fluid where information cannot escape, all you need to do is pump liquid out of the volume quickly enough that you get fluid flows speeds that exceed the speed of sound in the fluid.  I don't know how easy it is to do this and not make the flow completely turbulent, and it would only be an analogy, as a &quot;swimmer&quot; with sufficient strength could overcome the flow, but the basic idea of water signals not being able to escape would be present.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-02-06T18:29:12.880" CommentCount="1" />
  <row Id="4706" PostTypeId="2" ParentId="4700" CreationDate="2011-02-06T19:05:25.037" Score="5" ViewCount="" Body="&lt;p&gt;The other answers explain that there's no paradox but they don't explain why the particular particle called photon is massless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's massless because it is the messenger particle responsible for electromagnetism which is a long-range force. Its range is infinite so the mass has to be zero. One may view the Coulomb potential as the zero-mass limit ($m\to 0$) of the Yukawa potential&#xA;$$V(r) = \frac{\exp(-mr)}{r} $$&#xA;OK, so why is it massless and why the range is infinite? It's because of the unbroken $U(1)$ gauge invariance for the electromagnetic field that acts on the electromagnetic gauge potential as&#xA;$$A_\mu\to A_\mu+\partial_\mu \lambda$$&#xA;The mass term (in the Lagrangian) for a gauge field would have the form $m^2 A_\mu A^\mu/2$ and it is not invariant under the gauge invariance above. The gauge invariance is needed to make the time-like mode $A_0$ unphysical - otherwise it would produce quanta with a negative norm (because of the opposite sign in the signature for the timelike direction) which would lead to negative probabilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, gauge fields may consistently become massive via the Higgs mechanism - like the W-bosons and Z-bosons. Then they lead to short-range forces. Beta-decay is mediated by the W-bosons.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-06T19:05:25.037" />
  <row Id="4707" PostTypeId="2" ParentId="2230" CreationDate="2011-02-06T19:08:42.080" Score="2" ViewCount="" Body="&lt;p&gt;John Moffat and Moffat and Albrecht and Magueijo have variable speed of light theories where the speed varied in the early universe and is not a constant. Majueijo has a poplular book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B000EXYZLI&quot; rel=&quot;nofollow&quot;&gt;Faster Than The Speed of Light&lt;/a&gt; outlining his theories. IMO the book is quite outrageous and insults various people.  I mention this answer for completeness only as I believe the speed of light in a vacuum is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space can expand faster than the speed of light, but no information can be transmitted.&#xA;See the &lt;a href=&quot;http://en.wikipedia.org/wiki/Alcubierre_drive&quot; rel=&quot;nofollow&quot;&gt;Alcubierre warp drive&lt;/a&gt; for some fun.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-02-06T19:49:20.953" LastActivityDate="2011-02-06T19:49:20.953" />
  <row Id="4708" PostTypeId="2" ParentId="4704" CreationDate="2011-02-06T19:28:34.140" Score="4" ViewCount="" Body="&lt;p&gt;This is a water flow analogy with a black hole.  Take a fish tank filled with guppies and put a large drain or siphon in the tank.  This opening has to be big enough to set up a decent water flow.  When the water starts flowing the fish will generally swim against the current.  Assume all the guppies are the same size and generally swim at the same speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/3XjLw.jpg&quot; alt=&quot;model configuration&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the rate they flow is a vector sum of the water flow and the swimming velocity in the absence of water flow.  This will provide a visual model for how photons move in spacetime.  The flow of water is a model for the flow of spacetime points.  The guppies represent photons (modulo the fact they change direction at will), and the guppies within some approximate radius will tend to be taken up by siphon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/fzufz.gif&quot; alt=&quot;flow of spacetime point in a Schwarzschild geometry&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-06T19:28:34.140" CommentCount="2" />
  <row Id="4709" PostTypeId="1" AcceptedAnswerId="4717" CreationDate="2011-02-06T19:43:28.560" Score="2" ViewCount="150" Body="&lt;p&gt;If I poured water into my tea, would I see more or less of the bottom of the tea-cup?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, there would be as many particles blocking as many photons, and so I'd see the bottom just as clearly as before.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2011-02-06T19:49:25.353" LastActivityDate="2011-02-06T23:10:37.830" Title="Will watering tea down make it clearer?" Tags="&lt;optics&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="4710" PostTypeId="2" ParentId="4709" CreationDate="2011-02-06T20:06:41.743" Score="-1" ViewCount="" Body="&lt;p&gt;Index of refraction is related to density, as your intuition predicted. The question is how would tea and water look like at a molecular level. Without a microscope I can only guess, of course, but the tea as being big solid chunks floating around in the water might explain.&lt;/p&gt;&#xA;" OwnerUserId="301" LastActivityDate="2011-02-06T20:06:41.743" CommentCount="1" />
  <row Id="4711" PostTypeId="2" ParentId="2230" CreationDate="2011-02-06T20:19:01.607" Score="0" ViewCount="" Body="&lt;p&gt;That the speed of light is invariant is a property of Minkowski spacetime, and there should be plenty on that in Wiki - or search for 'geometric algebra' or Clifford Algebra.&lt;/p&gt;&#xA;" OwnerUserId="1423" LastActivityDate="2011-02-06T20:19:01.607" />
  <row Id="4712" PostTypeId="2" ParentId="4672" CreationDate="2011-02-06T20:47:49.737" Score="4" ViewCount="" Body="&lt;p&gt;It is known that Lorentzian wormholes can exist only if they respond to matter sources that violate some energy conditions. This is a general phenomena: a lot of what makes gravity special is that it is universally attractive. If you violate energy conditions then gravity can become repulsive, and then it is less distinguished from other forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it is believed that negative energy matter does not exist, mostly because this will cause instability (producing more of it will lower the energy indefinitely, so no reason for such production to stop at any point). It has never been observed, which is another reason to doubt it exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some attempts to involve quantum mechanics in order to circumvent this classical result (for example the Casimir effect can result in negative energies). However, if quantum effects are to violate a classical result, there is no way they are small. Then, you have to understand the full quantum mechanics, you cannot reach any conclusion based on the first quantum corrections. Same comment applies to any other attempt to add effects that are normally small (like higher derivative terms). Either you understand the full picture or you cannot come to any reliable conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also an independent problem, which is that wormhole spacetimes are unstable - small fluctuations tend to be magnified in parts of the geometry and therefore change it drastically. So even if you can write down a metric, it is not clear it is a good approximation to a real physical situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having explained some of the issues, it should be clear that you have to understand quite a bit of quantum gravity to address these questions. I don't think that string theory is currently well-understood enough to resolve the question, or even to change anybody's mind either way. For what it is worth, my feeling is that what we know so far indicates wormholes don't exist.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-02-07T14:24:03.500" LastActivityDate="2011-02-07T14:24:03.500" />
  <row Id="4716" PostTypeId="2" ParentId="4709" CreationDate="2011-02-06T21:12:30.090" Score="0" ViewCount="" Body="&lt;p&gt;You are right -- there is an approximated formula for this, called &lt;a href=&quot;http://en.wikipedia.org/wiki/Beer%E2%80%93Lambert_law&quot; rel=&quot;nofollow&quot;&gt;Lamber-Beer law&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A=\epsilon l c$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ is absorbance (log of quotient of light intensities before and after the solution), $\epsilon$ is substance dependent coefficient (constant), $l$ is the size of layer of solution and $c$ is the concentration. Let's now say that the glass cross-section is $S$, and initial height of fluid inside is $h$; adding water up to height $\alpha h$ will decrease concentration $c$ $\alpha$ times, but increase $l$ $\alpha$ times, so absorbance will stay the same.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2011-02-06T21:12:30.090" CommentCount="2" />
  <row Id="4717" PostTypeId="2" ParentId="4709" CreationDate="2011-02-06T21:17:45.100" Score="4" ViewCount="" Body="&lt;p&gt;There is some misconception in Tims question already: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;as many particles blocking as many photons, and so I'd see the bottom just as clearly as before&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Blocking of photons will not impair clarity at all!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One has to distinguish absorption and scattering! Tea is a solution of &#xA;some dyestuffs in water (optically) this dyestuff absorbs light of &#xA;certain wavelengths (making the tea looking yellow/brown) but the tea &#xA;is not made turbid (&quot;Unclear&quot;) by that absorption. (Think of some color filter &#xA;for cameras!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absorption is governed by &lt;a href=&quot;http://en.wikipedia.org/wiki/Beer%E2%80%93Lambert_law&quot; rel=&quot;nofollow&quot;&gt;Lambert/Beer law&lt;/a&gt;. &#xA;The law shows, that doubling the length of the liquid column is &#xA;&quot;neutralized&quot; by reducing concentration to half of original value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question of turbidity, In practice a tea may contain &#xA;some particles from lime in the water, often some agglomerates from &#xA;lime and some tanning agents in the tea. On top there is some &#xA;fat/oil which drift as droplets or collect on the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any particle floating in the tea, which is about 50 nm up to &#xA;some µmeters and having a index of refraction different enough&#xA;from the index of water, will cause some scattering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This scattering will blur the picture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the scattering in such suspensions like tea or milk &#xA;are governed by &lt;a href=&quot;http://en.wikipedia.org/wiki/Mie_theory&quot; rel=&quot;nofollow&quot;&gt;Mie scattering&lt;/a&gt;. Scattering/concentrations/length &#xA;is not as simple as Lambert-Beer, but at low concentrations &#xA;one can assume an analog dependence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some practical note: when having a good (low lime) water and &#xA;and the leaves are in a bag with good filter action, and You &#xA;wait for a moment until the oil accumulates at the surface, &#xA;tea will be a nearly perfectly clear liquid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Georg &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-06T23:10:37.830" LastActivityDate="2011-02-06T23:10:37.830" CommentCount="4" />
  <row Id="4718" PostTypeId="1" AcceptedAnswerId="4720" CreationDate="2011-02-06T21:50:12.067" Score="2" ViewCount="139" Body="&lt;p&gt;Im a writing a school report regarding the electric energy and potential of a capacitor.&#xA;In short:&#xA;An experiment was carried out where we charged a capacitor, and then decharged it with an electric engine, lifting up a predefined weight. We then measured how high the weight was lifted with different voltages.&#xA;i then found the effiency of the system by comparing the mechanical energy with the electrical energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second half of the report is then to find the electric potential of the capacitor, but I'm not to strong at electrics, and can't find anything about this in my physics book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is now; how can i find the electric potential of the capacitor?&lt;/p&gt;&#xA;" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-06T23:22:42.370" LastActivityDate="2011-02-06T23:22:42.370" Title="Electric potential in a capacitor" Tags="&lt;electricity&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="4719" PostTypeId="2" ParentId="4700" CreationDate="2011-02-06T22:04:07.580" Score="1" ViewCount="" Body="&lt;p&gt;I think the central issue is the invariant interval and invariant mass.  A particle moving in spacetime has the interval $ds^2~=~dt^2~-~dx^idx_i$.  There is the corresponding invariant mass $m^2~=~E^2~-~p^2$, which is the momentum spacetime interval. So let us consider the plane wave $\psi~=~exp(-i{\vec k}\cdot{\vec x}~+~i\omega t)$ $=~exp(-ik^\mu x_\mu)$.  The Laplacian operator $\Delta~=~\nabla^2~-~\partial^2/\partial t^2$ applied to $\psi$ is&#xA;$$&#xD;&#xA;(\nabla^2~-~\partial/\partial t)~=~(\omega^2~-~k^2)\psi~=~\hbar^{-2}(E^2~-~p^2)\psi.&#xD;&#xA;$$&#xA;This is an eigenvalued problem with $\Delta\psi~=~\lambda\psi$.  If the particle has mass this eigenvalue is the mass squared.  This means there is dispersion as $|k|~=~\sqrt{\omega^2~-~m^2} $ and for $\omega~=~2\pi/\lambda$ we then have that&#xA;$$&#xD;&#xA;|k|~=~c\sqrt{2\pi/\lambda^2~-~m^2}&#xD;&#xA;$$&#xA;The velocity of a wave is then only exactly $c~=~1$, or $|k|~=~2\pi/\lambda$, with of course $kc~=~\omega$, for $m~=~0$, and otherwise we contradict our assumption of the wave existing on a null interval $ds^2~=~0$.  If the invariant interval in the spacetime is zero, then the corresponding invariant mass in the momentum-spacetime must be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above Laplacian in the case of a photon is predicted as the wave equation operator by Maxwell's equations.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-06T22:04:07.580" />
  <row Id="4720" PostTypeId="2" ParentId="4718" CreationDate="2011-02-06T23:00:49.607" Score="1" ViewCount="" Body="&lt;p&gt;From the context, I'm going to assume that by &quot;electric potential&quot; you mean the energy in the capacitor, rather than the usual meaning of &quot;electric potential&quot; which is simply the voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula for electric potential energy (in Joules) is $0.5 C V^2$, where C is the capacitance in Farads, and V is the voltage in volts. Here is a reference which has the formula. (Ignore the calculus for now, you'll get there eventually.)&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Capacitor#Energy_storage&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Capacitor#Energy_storage&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-06T23:00:49.607" CommentCount="1" />
  <row Id="4721" PostTypeId="2" ParentId="1345" CreationDate="2011-02-06T23:10:45.510" Score="0" ViewCount="" Body="&lt;p&gt;The speed of light is a vector quantity and vector summation works perfectly well for it (at least in Special Relativity). You just cannot change the frame of reference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if you have one object moving at c in one direction and another object moving at 1/2c in the opposite direction, then the middle between the two will move at c/2 in the same direction as the first. This is from point of view of a stationary observer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between the two objects grows as 3/2c. This is in the stationary reference frame of course, the objects themselves will see each other moving at speed of light.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastActivityDate="2011-02-06T23:10:45.510" />
  <row Id="4722" PostTypeId="2" ParentId="4718" CreationDate="2011-02-06T23:11:26.820" Score="1" ViewCount="" Body="&lt;p&gt;I think what you are asking is the energy stored by a capacitor and the amount of work it can do.  Voltage is charge divided by capacitance, $V~=~q/C$.  See below for the reason for this.  We know that the power $P$ (wattage) in a circuit is given by $P~=~VI$.  $I$, the current, is the time rate of change in the charge, or $I~=~dq/dt$.  Power is the time rate of change in energy $P~=~dE/dt$  and so we have $dE~=~Vdq$  So put all of this together and you get the work $W~=~\int dE$ done by a capacitor is&#xA;$$&#xD;&#xA;W~=~\int \frac{qdq}{C}~=~\frac{q^2}{2C}~=~\frac{qV}{2}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The capacitor is defined as a charge per unit area.  If you have a surface with a constant charge distribution $\sigma$ so that $q~=~\int\sigma da$ $\simeq~\sigma A$ the electric field is a constant vector field normal to the surface.  The inside of a capacitor is two such plates with a charge across two plates.  Electric field has units of $volt/m$ and so the voltage $V~=~\int Edx$.  Capacititance is $C~=~q/A$, which is the charge distribution on the flat plate.  This all leads to $V~=~q/C$&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-06T23:11:26.820" CommentCount="1" />
  <row Id="4723" PostTypeId="1" AcceptedAnswerId="4727" CreationDate="2011-02-06T23:19:30.003" Score="1" ViewCount="269" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A damped harmonic oscillator with&#xA;  $m = 10$kg, $k = 250$N/m, and&#xA;      $c = 60$kg/s is subject to a driving force given by $F_0\cos(\omega t)$,&#xA;  where&#xA;      $F_0 = 48$N.&#xA;      (a) What value of $\omega$ results in steady-state oscillations with maximum&#xA;      amplitude?&#xA;      (b) What is the maximum amplitude? What is the phase shift at the&#xA;      resonance&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Studying for my exam tomorrow, I am kinda of confused to how to even begin this problem. What does it mean by steady-state solution? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know maximum amplitude occurs when velocity is zero. &#xA;The diff equation looks as follows:&lt;br&gt;&#xA;$$a + \frac{cv}{m} + \frac{kx}{m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thats all I know. Please do not give me the answer, but please tell me how I can approach this.&lt;/p&gt;&#xA;" OwnerUserId="551" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-06T23:24:48.563" LastActivityDate="2011-02-07T00:06:02.393" Title="Finding the steady state solution of a damped oscillator" Tags="&lt;classical-mechanics&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="4724" PostTypeId="2" ParentId="4662" CreationDate="2011-02-06T23:22:34.507" Score="0" ViewCount="" Body="&lt;p&gt;This is an algebra physics problem, no need for calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity force (in the direction of sliding) is $mg\sin(30)$. Friction force is therefore $0.1mg\cos(30)$ and opposes velocity. From these you get the accelerations (different for going up and down). Now move the block up to its turning point using the going up acceleration (which is the one with the largest magnitude), and from its position, figure out the time to get back down using the coming down acceleration.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-06T23:22:34.507" />
  <row Id="4725" PostTypeId="1" CreationDate="2011-02-06T23:46:03.687" Score="1" ViewCount="141" Body="&lt;p&gt;Hey guys,&#xA;while learning thermodynamics i wondered how a the principle of a heat pump would look at a microscopic level, not on a quantum mechanical level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learned that when a hot and a less hot body are in direct contact, the heat is transferred by the kinetic energy of the atoms, so to speak the vibrating atoms of the hot body kick the ones of the cooler ones and so transfer heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a heat pump, to maintain a increase in total entropy one needs to add mechanical work to get a cool body transfer heat to a body with a higher temperature&#xA;(I hope this is correct).&#xA;But how do the atoms of the cooler body interact with those of the warmer? If they would collide (along with their smaller kinetic energy) with those of the warmer body (along with their greater kinetic energy) I would intuitively say that nothing would happen...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how does that work? :)&lt;/p&gt;&#xA;" OwnerUserId="1789" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-07T10:53:38.360" LastActivityDate="2011-02-07T10:55:06.243" Title="How to vizualize a heat pump microscopically?" Tags="&lt;thermodynamics&gt;&lt;heat&gt;&lt;entropy&gt;" AnswerCount="2" />
  <row Id="4726" PostTypeId="2" ParentId="603" CreationDate="2011-02-06T23:55:22.097" Score="0" ViewCount="" Body="&lt;p&gt;You already answered yourself: this is because the principle of casualty does not prohibit such motion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tangential motion and moving away from another object with superluminal velocity is not prohibited because in such processes the casualty cannot brake. What is prohibited is approaching another object with superluminal velocity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only reason why we do not see objects moving away from each other at superluminal speeds is because moving from some object at v&gt;c usually means approaching another object with v&gt;c, and this is prohibited. But in the case of the expansion of Universe it is possible to move away from one object without approaching another, and that's why superluminal growth of distances between two galaxies separated by big distances is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even more, due to expansion of universe any light rays going from us (or any other atationary observer) have velocity somewhat greater than c and any light rays going toward us have velocity smaller than c. This of course means also that any two stationary observers move away from each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is only one case where one can imagine superluminal approaching of two objects: when both are under the horizon of a black hole. When an object approaches a black hole, he reaches superluminal tangential velocity at ergosphere (which is outside the black hole horizon so the object can return) and the radial velocity reaches c exactly at the horizon. Mathematically this means that inside the horizon the object should have radial velocity greater than c. But this is not the case: once object reached c (or near c) at the horizon, the time stops for him and he remains there until the black hole explodes (finally evaporates). &lt;/p&gt;&#xA;" OwnerUserId="1186" LastActivityDate="2011-02-06T23:55:22.097" />
  <row Id="4727" PostTypeId="2" ParentId="4723" CreationDate="2011-02-07T00:06:02.393" Score="1" ViewCount="" Body="&lt;p&gt;As Mark Eichenlaub said in his comment, start by writing down $F=ma$, and including all the forces (damping force, spring, driving force) on the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is a differential equation for the unknown function $x(t)$. You're looking for a &quot;steady state&quot; solution, which just means a solution in which $x(t)$ is a sinusoidal oscillation: $x(t)=A\cos(\omega t+\phi)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are such steady solutions for any $\omega$ you care to name. Find the value of $\omega$  that leads to a solution with the biggest amplitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, sorry if this sounds rude, but you should probably also look up &quot;resonance&quot; and &quot;driven harmonic oscillator&quot; in your textbook. There's bound to be a bunch of useful stuff there.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-07T00:06:02.393" />
  <row Id="4728" PostTypeId="2" ParentId="1134" CreationDate="2011-02-07T00:07:18.957" Score="2" ViewCount="" Body="&lt;p&gt;The form of a black hole horizon behaves as a viscous liquid. It flattens at poles when BH is rotating, there can be waves on the surface of a black hole when it eats something and the energy of waves then radiated as gravitational waves so the surface becomes smooth again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As something falls on a BH, its diameter increases, but not simultaneously everywhere (the BH can be so large that even light takes some time to reach the other side, so are the gravitational waves). When two BHs are approaching each other they also distend towards each other as two drops of attracting liquids, and then they fuse. The energy of the surface waves is again radiated as gravitational rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Closer to your question, being between two black holes would not help to escape, just the opposite, the surfaces of the both BHs expand in the direction of each other so you would be in even greater trouble. To escape from between two BHs to the infinity you'd have to fight the gravitation of both.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastActivityDate="2011-02-07T00:07:18.957" />
  <row Id="4729" PostTypeId="2" ParentId="2767" CreationDate="2011-02-07T00:25:55.153" Score="0" ViewCount="" Body="&lt;p&gt;Yes, gravitational waves can be shielded. As gravitons are the particles which can interact with matter, the matter can absorb gravitons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would depend on the wavelength however. For uniform substance the rate of absorption would generally increase with the wavelength. The rate of absorption also can be increased by constructing mechanical oscillators with frequency in resonance with that of gravitational waves. Such oscillators would transform gravitational waves into mechanical energy.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastActivityDate="2011-02-07T00:25:55.153" />
  <row Id="4730" PostTypeId="2" ParentId="2175" CreationDate="2011-02-07T01:03:31.457" Score="9" ViewCount="" Body="&lt;p&gt;The information about the pushes will be received on the other end with the speed of sound in the substance of the pole. For any real mmaterial it is much slower than the speed of light (for a steel rod it would be about 1000 m/s). &lt;/p&gt;&#xA;" OwnerUserId="1186" LastActivityDate="2011-02-07T01:03:31.457" CommentCount="1" />
  <row Id="4731" PostTypeId="1" AcceptedAnswerId="4757" CreationDate="2011-02-07T01:19:33.293" Score="8" ViewCount="326" Body="&lt;p&gt;1.In classical mechanics, using Newton's laws, the ellipticity of orbits is derived. It is also said that the center of mass is at one of the foci. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.Each body will orbit the center of the mass of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is : Are the assumptions in 1 and 2 correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Follow up question : Assuming the distance from the centre of the mass to each body remains the same, do we have two bodies orbiting the centre of the mass of the system in an elliptical or circular orbit? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally : With elliptical orbits, if the heavier mass is supposed to be in one of the foci, if there is any significance to second focus, what is it? Is it a Lagrange point by any chance or does it have some other mathematical property?&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-02-07T09:00:39.913" LastActivityDate="2011-02-07T16:15:45.453" Title="Significance of the second focus in elliptical orbits" Tags="&lt;classical-mechanics&gt;&lt;gravity&gt;&lt;newtonian-gravity&gt;" AnswerCount="3" />
  <row Id="4732" PostTypeId="2" ParentId="3957" CreationDate="2011-02-07T01:22:35.537" Score="11" ViewCount="" Body="&lt;p&gt;A brief history of the misapplication of magnetohydrodynamics to the analysis of the solar wind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1959: Soviet satellite Luna 1 directly observed the solar wind for the first time and measured its strength.&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Luna_1&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Luna_1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as of 1959, by direct experimental observation, it was known that the heliopause was at least the radius of the earth or R⊙.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pneuman and Kopp 1971 Model: According&#xA;  to a more complex but still simplified&#xA;  MHD [MagnetoHydroDynamics] model of the coronal structure&#xA;  (ISP p. 114-117 etc., the model of&#xA;  Pneuman and Kopp 1971), the dipolar&#xA;  magnetic field lines form closed loops&#xA;  if they originate at solar latitudes&#xA;  of less than about 45° (above or below&#xA;  the solar equator). However, those&#xA;  arising greater than about 45° are&#xA;  open field lines that may curve around&#xA;  the closed region to some extent but&#xA;  eventually extend far into space in&#xA;  all directions, at least beyond a&#xA;  heliocentric distance of about 2 R⊙.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mcgoodwin.net/pages/spacephysics_ess471.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.mcgoodwin.net/pages/spacephysics_ess471.pdf&lt;/a&gt;&#xA;(page 36)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the only paper I've been able to find that approximately reads on the magnetopause question. This is a highly cited paper and it's early enough to influence the expectations at the Voyager launches (1977). So I believe that the Pneuman and Kopp paper gave the expectation that the heliopaue would be at around 2 R⊙ based on MHD calculations. Since this was a huge error, I've not been able to find any better detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man who developed MHD was Hannes Alfvén. He got the 1970 Nobel prize in physics for this. His Nobel prize lecture was partially dedicated to the task of claiming that his theory was being abused. In particular, he noted that the space physics situation was out of control. From his lecture, I've italicized the parts having to do with space physics predictions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Plasma physics, space research and the origin of the solar system&lt;/strong&gt;&lt;br&gt;&#xA;[ Nobel Prize Lecture, 1970, by Hannes Alfvén ]&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... The cosmical plasma physics of&#xA;  today is far less advanced than the&#xA;  thermonuclear research physics. &lt;strong&gt;&lt;em&gt;It is&#xA;  to some extent the playground of&#xA;  theoreticians who have never seen a&#xA;  plasma in a laboratory. Many of them&#xA;  still believe in formulae which we&#xA;  know from laboratory experiments to be&#xA;  wrong.&lt;/em&gt;&lt;/strong&gt; The astrophysical&#xA;  correspondence to the thermonuclear&#xA;  crisis has not yet come. The reason&#xA;  for this is that &lt;strong&gt;&lt;em&gt;several of the basic&#xA;  concepts on which the theories are&#xA;  founded, are not applicable to the&#xA;  condition prevailing in cosmos&lt;/em&gt;&lt;/strong&gt;. They&#xA;  are &quot;generally accepted&quot; by most&#xA;  theoreticians, they are developed with&#xA;  the most sophisticated mathematical&#xA;  methods and it is only the plasma&#xA;  itself which does not &quot;understand&quot;,&#xA;  how beautiful the theories are and&#xA;  absolutely refuses to obey them. It is&#xA;  now obvious that we have to start a&#xA;  second approach from widely different&#xA;  starting points.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If you ask where the border goes&#xA;  between the first approach and the&#xA;  second approach today, an approximate&#xA;  answer is that it is given by the&#xA;  reach of spacecrafts. This means that&#xA;  &lt;strong&gt;&lt;em&gt;in every region where it is possible&#xA;  to explore the state of the plasma by&#xA;  magnetometers, electric field probes&#xA;  and particle analyzers, we find that&#xA;  in spite of all their elegance, the&#xA;  first approach theories have very&#xA;  little to do with reality&lt;/em&gt;&lt;/strong&gt;. It seems&#xA;  that the change from the first&#xA;  approach to the second approach is the&#xA;  astrophysical correspondence to the&#xA;  thermonuclear crisis. ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://nobelprize.org/nobel_prizes/physics/laureates/1970/alfven-lecture.pdf&quot; rel=&quot;nofollow&quot;&gt;http://nobelprize.org/nobel_prizes/physics/laureates/1970/alfven-lecture.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above lecture includes a table with a detailed comparison between the &quot;first approach&quot; and &quot;second approach&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem in estimating the heliopause was mostly due to theoreticians overestimating their understanding of the limitations of MHD. In particular, the MHD equations fail when electric currents are strong enough to overcome the magnetic field. This breaks the MHD assumption that ions and electrons remain pinned to magnetic field lines.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-07T01:22:35.537" CommentCount="3" />
  <row Id="4733" PostTypeId="2" ParentId="4725" CreationDate="2011-02-07T01:40:55.303" Score="1" ViewCount="" Body="&lt;p&gt;I'll bite. You have a small chamber where you keep a gas. To heat it up you apply mechanical force to it so as to decrease its volume. To cool it you let it do work on your mechanical force so that its volume increases. This pumps heat in and out so move the two heat baths appropriately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely there are some more elegant methods that could be constructed physically. I would suggest spin interactions as a way of making a heat pump. Another place to look is in the electrically powered heat pumps now used to cool some integrated circuits. See:&lt;br&gt;&#xA;&quot;Thermoelectric Effect&quot;&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Thermoelectric_effect&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Thermoelectric_effect&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-07T01:40:55.303" />
  <row Id="4734" PostTypeId="2" ParentId="970" CreationDate="2011-02-07T01:55:15.233" Score="1" ViewCount="" Body="&lt;p&gt;There is an ongoing research regarding your question and some solutions have been proposed. I recommend &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0609024&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/gr-qc/0609024&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0701096&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/gr-qc/0701096&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/1011.2219&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1011.2219&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In short, it is proposed that the internal energy of infalling observer is fully transformed into kinetic energy and then into radiation. This radiation is called &quot;pre-Hawking&quot; radiation. Although there are some counter-arguments.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastActivityDate="2011-02-07T01:55:15.233" />
  <row Id="4735" PostTypeId="1" AcceptedAnswerId="4737" CreationDate="2011-02-07T05:55:11.653" Score="1" ViewCount="145" Body="&lt;p&gt;As you may have seen, a &lt;a href=&quot;http://blogs.discovermagazine.com/badastronomy/2011/02/04/breaking-sofa-sized-asteroid-gives-us-a-close-shave/&quot; rel=&quot;nofollow&quot;&gt;small asteroid had a near miss with the earth a few days ago&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a physics teacher teaching momentum, I think this could be an excellent problem for my students, however, I don't know how to determine the mass or velocity of this object. I could get a decent estimate of the mass simply by assuming its a spherical rock, but I still don't see how to determine the velocity. I can get the &lt;a href=&quot;http://www.minorplanetcenter.net/mpec/K11/K11C12.html&quot; rel=&quot;nofollow&quot;&gt;orbital elements&lt;/a&gt;:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/m2jch.png&quot; alt=&quot;Orbital Elements of 2011 CQ1&quot;&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the 'V' term the magnitude of the velocity, and if so, what are those units? km/s? How could I get a heading for the asteroid? &lt;/p&gt;&#xA;" OwnerUserId="807" LastActivityDate="2011-02-07T20:30:48.080" Title="How to determine velocity and mass of asteroid 2011 CQ1" Tags="&lt;astrophysics&gt;&lt;mechanics&gt;" AnswerCount="1" />
  <row Id="4736" PostTypeId="2" ParentId="4731" CreationDate="2011-02-07T06:41:15.547" Score="4" ViewCount="" Body="&lt;p&gt;Your assumptions are correct. The classical solution to the two-body problem is that each mass goes in an elliptical orbit, and one focus of each ellipse is at the center of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the special case that the distance from the center of mass to each body remains constant, that means both orbits are circular, a circle being a special case of an ellipse. Note that the radii of the two circles will not be equal unless the objects have the same mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I can think of no physical relevance to the second focus, even in the limiting case that one of the masses is much much larger than the other (and therefore stays at rest at the first focus). In particular, it is not a Lagrange point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had a brief thought that if you put another large, fixed mass at the second focus, you could keep the same elliptical orbit, but this is in fact incorrect because the time-dependent orbit is not symmetrical between the two foci (the small mass spends a longer time away from the large mass), so you can't superimpose them to get another simple elliptical orbit. The solution to the circular restricted three-body problem (Euler's three-body problem) is &lt;strong&gt;not&lt;/strong&gt; an ellipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically the second focus is just a curiosity and not physically relevant.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-02-07T06:41:15.547" CommentCount="3" />
  <row Id="4737" PostTypeId="2" ParentId="4735" CreationDate="2011-02-07T07:16:32.807" Score="0" ViewCount="" Body="&lt;p&gt;I guess it's better to read articles optimized for human readers, e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.pagef30.com/2011/02/asteroid-2011-cq1-to-pass-by-earth-at.html&quot; rel=&quot;nofollow&quot;&gt;http://www.pagef30.com/2011/02/asteroid-2011-cq1-to-pass-by-earth-at.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The relative velocity in the Earth's frame was 10 km/s when it was closest, 5471 km. Your webpage says that MOID (closest distance) was 0.0001 AU but it turned out to be 3 times closer at the end. A picture of the direction is included on the page I linked, too. The radius 1 meter wouldn't pose a problem if it hit us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the table, you have the ephemeris:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ephemeris&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Ephemeris&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, YYYY MM DD.## is the date (fourth of February 2011) followed by numbers 00-06 which are probably just labels (in the table I see above). It's separated by a space from the HH MM.MM time, changing from 8:20 to 8:21 or so (decimal point in minutes in between - about a minute was listed by you). What follows is R.A. or right ascension&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Right_ascension&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Right_ascension&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(the entries are 14) which is the latitude on the Earth above which it occurs (where we see it), with the rotation of the Earth subtracted. Note that it is not moving much because the &quot;horizontal&quot; orbits of both bodies around the Sun are relatively synchronized. Then you get the declination&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Declination&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Declination&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is the ordinary Earth's latitude above which the object is located. It's changing between 37 and 43 (plus means North Hemisphere): the planes of orbits differ. From the changes of these two coordinates, you may deduce the angular velocity. The declination changed from 37.5 to 42.5 i.e. by 5 degrees = $pi/36$ radian. At distance of 5471 km I announced above, it's 477 kilometers. It took some time - the difference between 8:21.60 and 8:20.73 which is 0.87 minute i.e. 52 seconds, so you can see that the speed was about 10 km/s in your table&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The table also includes Delta which I am not sure about now; distance from the Sun in AU called $r$ (which is obviously not too far from one), and elongation&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Elongation_%28astronomy%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Elongation_%28astronomy%29&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is the angle between the object and the Sun, as seen from the Earth. The distance from the Earth is not directly listed because it's not directly measured. One only knows the angle and the redshift. $V$ is not the velocity you want - which I calculated above.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="292" LastEditorDisplayName="" LastEditDate="2011-02-07T20:30:48.080" LastActivityDate="2011-02-07T20:30:48.080" CommentCount="1" />
  <row Id="4738" PostTypeId="1" CreationDate="2011-02-07T07:26:51.137" Score="5" ViewCount="277" Body="&lt;p&gt;Last month I asked about a &lt;a href=&quot;http://physics.stackexchange.com/questions/3731/heterotic-string-as-worldvolume-theory-of-two-coincident-9-branes-in-27-dimension&quot;&gt;27-dimensional origin of the heterotic string&lt;/a&gt;. Now I'm looking at Witten's &lt;a href=&quot;http://arxiv.org/abs/0706.3359&quot; rel=&quot;nofollow&quot;&gt;&quot;Three-dimensional gravity revisited&quot;&lt;/a&gt;, where he proposes that pure gravity on AdS&lt;sub&gt;3&lt;/sub&gt; is dual to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Monstrous_moonshine&quot; rel=&quot;nofollow&quot;&gt;&quot;monstrous moonshine&quot;&lt;/a&gt; CFT, which is bosonic string theory on a &lt;a href=&quot;http://en.wikipedia.org/wiki/Leech_lattice&quot; rel=&quot;nofollow&quot;&gt;&quot;Leech lattice&quot;&lt;/a&gt; orbifold. Wouldn't that mean we were also talking about bosonic M-theory (the 27-dimensional theory) on AdS&lt;sub&gt;3&lt;/sub&gt; x R&lt;sup&gt;24&lt;/sup&gt;/Λ&lt;sub&gt;24&lt;/sub&gt;? But that would mean that on this occasion, the emergence of the &quot;M&quot; dimension was the same thing as the emergence of the radial AdS dimension. Does that even make sense? &lt;/p&gt;&#xA;" OwnerUserId="1486" LastActivityDate="2011-02-07T15:16:03.500" Title="Bosonic M-theory and moonshine gravity" Tags="&lt;string-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="4739" PostTypeId="2" ParentId="4738" CreationDate="2011-02-07T08:34:41.420" Score="2" ViewCount="" Body="&lt;p&gt;Mitchell, if you search for &quot;Josh Grey&quot; at&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.physics.harvard.edu/QFT/SidneyfestBlog.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.physics.harvard.edu/QFT/SidneyfestBlog.htm&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;you will see a comment that we had a related idea about the AdS compactifications of the bosonic string/M-theory in 2000. The advantage of having the bosonic string itself is that the one-loop beta function of $AdS_5\times S^{21}$ actually cancels just like for the $N=4$ supersymmetric background: the $11/3$ coefficient from the beta-function knows about the $2\times 11=22$ dimensions in which the $S^{21}$ sphere resides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of those backgrounds of the bosonic string/M-theory may be justified by any solid arguments, and they probably don't exist. In particular, there are no RR fluxes that could support the curvature of the sphere. That's true for your $AdS_3$ case of bosonic M-theory, too. Witten's CFT used in the monstrous moonshine has the Leech lattice instead of a sphere but the lattice cannot be decompactified so it's somewhat physically meaningless to link it to the decompactified bosonic string.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Witten's CFT itself, despite the name of its author, has its problem and probably can't be generalized to a larger curvature radius of the $AdS_3$, as shown by Davide Gaiotto using group-theoretical arguments. And bosonic M-theory itself is the least justified among all those building blocks. The lack of supersymmetry guarantees that all numerical comparisons between the two sides of the dualities fail. Moreover, there are no chiral fermions and no anomalies, so there's no nontrivial counterpart of the $E_8$ anomaly cancelation on the supersymmetric M-theory domain walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the analogies are suggestive but supersymmetry prevents you from seeing that the backgrounds are stable, dual, and well-behaved, and if you ask me, I think that it's more likely that this is not just an illusion: the lack of supersymmetry simply makes these backgrounds genuinely pathological (unstable and other things). Supersymmetry plays a crucial role for consistent string/M-theory solutions and all these bosonic models should be viewed as mere toy models to learn the physics that really makes sense, and it's the physics of (either spontaneously broken or unbroken) supersymmetric vacua.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-07T08:34:41.420" />
  <row Id="4740" PostTypeId="1" CreationDate="2011-02-07T10:03:11.350" Score="5" ViewCount="308" Body="&lt;p&gt;Why massless particle can't exceed speed of light? &lt;/p&gt;&#xA;" OwnerUserId="58" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-02-07T14:05:59.333" LastActivityDate="2011-02-07T14:05:59.333" Title="Why can't massless particle exceed speed of light?" Tags="&lt;speed-of-light&gt;&lt;particles&gt;" AnswerCount="1" />
  <row Id="4741" PostTypeId="2" ParentId="4740" CreationDate="2011-02-07T10:21:27.470" Score="4" ViewCount="" Body="&lt;p&gt;It's just what it means for a particle to be massless. For a general motion of a particle in the vacuum, the rest mass $m_0$ satisfies &#xA;$$m_0^2 c^4 = E^2-p^2 c^2$$&#xA;If the left hand side is zero, it follows that&#xA;$$E=|p|c$$&#xA;which means that the energy-momentum vector is null (light-like). In general, the speed may be calculated from the direction of the energy-momentum vector as $v=pc^2/E$ and it equals $v=c$ if the formula above is satisfied. Massless particles are not only unable to move faster than light; they're unable to move slower than light, too. The massless particles must move exactly by the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general absence of any signals that move faster than light is required by the special theory of relativity; a signal or particle that would be moving faster than light would be equivalent (from a viewpoint of a different inertial system) to a signal or particle that moves backwards in time, which would lead to logical contradictions (killing grandfather before he met your grandmother etc.).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-07T10:34:07.383" LastActivityDate="2011-02-07T10:34:07.383" />
  <row Id="4742" PostTypeId="2" ParentId="4725" CreationDate="2011-02-07T10:55:06.243" Score="1" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;microscopic&quot; has a definite meaning in &#xA;thermodynamics. This is the level of smallest &#xA;particles one deals with, usually molecules or atoms. &#xA;A heat pump on that scale is purely theoretical. &#xA;(Laplac's demon could be named a heat pump, when &#xA;somebody would control the valve in an appropriate way) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;the heat is transferred by the kinetic energy of the atoms. so to speak the vibrating atoms&lt;/em&gt; &#xA;What is tranferred? Kinetic energy or vibrational quanta? &#xA;That is not the same! &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In general, when You learn thermodynamics, &#xA;forget all &quot;atoms&quot;! Thermodynamics does not know of atoms, &#xA;This theory works on &quot;matter&quot; without any thought of the structure. &#xA;Next step then will be statistical thermodynamics, where &#xA;You will learn about a bridge to the microscopic world of atoms.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-02-07T10:55:06.243" CommentCount="4" />
  <row Id="4743" PostTypeId="1" CreationDate="2011-02-07T11:09:52.327" Score="3" ViewCount="222" Body="&lt;p&gt;I had earlier also asked a question about super conformal theories and I am continuing with that, now with more specific examples. I am quite puzzled with it given that I see no book explaining even the basics of this. I am merely picking this up from stray notes and papers and mostly discussions.  So for ${\cal N} = 2$ superconformal algebra in $2+1$ dimensions the symmetry group was $SO(3,2)\times SO(2)$ and possibly the primary states of this algebra are labelled by the tuple $(\Delta, j,h)$ where $\Delta$ is the scaling dimension and $j$ is the spin and $h$ is its $R$ charge (or whatever it means to call it the $R$ charge highest weight) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess that in this context by “primary state” one would mean those which are annihilated by the special conformal operators (K) as well as the $S$ operators of the algebra.  Though confusingly enough the terminology seems to be used for operators as well. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I would like to know where can I find the complete superconformal algebra to be listed which can enable me to find the $(\Delta, j,h)$ of the states which are gotten by acting other operators n the primaries. I guess its only the momentum operators and the supersymmetry that is left to create the tower of the “descendants” &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess that once one knows the above algebra I can understand these mysterious tables that I see listing the “states” corresponding to the supersymmetry operators. If for the identity operator the tuple is $(0,0,0)$ then I would like to understand how for the $Q$ the possible labels are apparently $(0.5,\pm 0.5,\pm 1)$ and for $Q^2$ the labels are $(1,0,\pm 2)$, $(1,0,0)$ and $(1,1,0)$. For $Q^3$ the labels are $(1.5,0.5,\pm 1)$ and for $Q^4$ the labels are $(2,0,0)$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would like to know how the above labels for the operators are derived. It gets harder for me to read this since for the operators the R symmetry as well as the spinorial index are both being suppressed.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Now consider a primary labelled by $(\Delta, j,h)$ such that it is in the long representation and hence $\Delta &amp;gt;j+\vert h\vert +1$. Then I see people listing something called the “conformal content” of this representation labelled by the above state. For the above case the conformal content apparently consists of the following states, $(\Delta, j,h)$, $(\Delta+0.5, j\pm 0.5,h\pm 1)$, $(\Delta + 1 , j,h \pm 2)$, $(\Delta +1 , j+1,h)$, twice $(\Delta + 1, j,h)$, $(\Delta + 1, j-1,h)$, $(\Delta + 1.5, j\pm 0.5,h \pm 1)$ and $(\Delta + 2, j,h)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would like to know what exactly is the definition of “conformal content” and how are lists like the above computed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar lists can be constructed for various kinds of short representations like those labelled by $(j+h+1,j,h)$ ($j, h \neq 0$), by $(j+1,j,0)$, by $(h,0,h)$, by $(0.5,0,\pm 0.5)$, by $(h+1,0,h)$ and $(1,0,0)$.  Its not completely clear to me a priori as to why some of these states had to be taken out separately from the general case, but  I guess if I am explained the above two queries I would be able to understand the complete construction. &lt;/p&gt;&#xA;" OwnerUserId="1637" LastActivityDate="2011-02-07T16:29:05.887" Title="Superconformal algebra" Tags="&lt;quantum-field-theory&gt;&lt;conformal-field-theory&gt;&lt;supersymmetry&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="4744" PostTypeId="2" ParentId="4743" CreationDate="2011-02-07T11:26:44.543" Score="1" ViewCount="" Body="&lt;p&gt;Anirbit, please check e.g. the AdS/CFT review &quot;MAGOO&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/9905111&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9905111&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where many questions are just answered. Search for conformal primary, superconformal primary, algebra, etc. Most of your questions are basic textbook stuff and you shouldn't be asking 3 paragraphs - which require 5 paragraphs to be answered - about every single simple step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is nothing confusing about using the term &quot;primary field&quot; for the operators because conformal field theories imply the state-operator correspondence, a one-to-one map between the operators and the states of the Hilbert space. This map can be simply obtained in the radial quantization and it is one of the first things you should have learned about conformal field theories. The transformation of the operators under the generators of a group is given by the commutator - but the result is nothing else than the operator corresponding to the transformed state (by the simple rule).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question: You can't determine the dimensions and charges of all the primary fields &quot;just from the algebra&quot;. The spectrum obviously depends on the theory - and there are many conformal or superconformal theories that share the same algebra. It's not true that all physics is encoded in the symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $(\Delta,j,h)$ charges of various states - primary fields in this case - are simply the eigenvalues. For operators, as I said, they're also eigenvalues, but the symmetry generator $G$ has to act on the operator $O$ as the commutator, $[G,O]$. For example, the identity's commutators universally vanish so all the dimensions and charges of the identity (constant) operator are equal to zero. For supercharges, they're equal to something else and some of the signs differ for different components of the supercharge. However, individual supercharges usually square to zero, so only some products of the form $Q^2$ or $Q^3$ are nonzero which is why the charges and dimensions of the higher powers of $Q$ are less ambiguous than if all the products of components of $Q$ were nontrivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you know the definitions and some maths, you must be able to answer all your questions above because they're simple. If you won't be able to jump into the rhythm where you can answer most of similar questions yourself, or by finding the right literature by yourself, you should give CFTs up because there's no point in learning a math-loaded set of questions by the method of listening/memorizing the answer to every individual question that you hear from someone else. The only conceivable meaningful purpose of learning similar things - such as portions of maths and physics - is to become able to answer subsequent questions without the help of others. If it can't work for you, please just give it up.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-07T11:32:27.737" LastActivityDate="2011-02-07T11:32:27.737" CommentCount="2" />
  <row Id="4745" PostTypeId="2" ParentId="4743" CreationDate="2011-02-07T12:04:27.830" Score="2" ViewCount="" Body="&lt;p&gt;There are usually two books that I recommend that provide some background knowledge with a bias to the more mathematically inclined reader:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Martin Schottenloher: &quot;A Mathematical Introduction to Conformal Field Theory&quot; (no superconformal field theory, but explains the connection to the Wightman axioms)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Blumenhagen, Plauschinn: &quot;Introduction to Conformal Field Theory, With Applications to String Theory&quot;. (With a section on supersymmetric CFT and boundary CFT).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is true that most papers on CFT don't explain and don't define the most basic concepts, so that there is a significant gap if you go from an introduction to QFT to the CFT literature, but the books above should bridge that gap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sidenote: The state-operator correspondence is sometimes called the Reeh-Schlieder property, which characterizes QFTs where the Reeh-Schlieder theorem is true. In these QFTs there is a unique vacuum state |0&gt; that is separating and cyclic for all local operator algebras so that there is a 1:1 correspondence of operators A with the state that the vacuum gets mapped to, A |0&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditorDisplayName="" LastEditDate="2011-02-07T16:29:05.887" LastActivityDate="2011-02-07T16:29:05.887" CommentCount="1" />
  <row Id="4746" PostTypeId="1" CreationDate="2011-02-07T12:34:24.440" Score="3" ViewCount="274" Body="&lt;p&gt;I read this report and summarise here but my question is - if quantum mechanics will make glass melt at temperatures near absolute zero and it is near absolute zero then wouldn't this be a huge issue for satellites and space shuttles? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencedaily.com/releases/2011/02/110202102748.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencedaily.com/releases/2011/02/110202102748.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scientists Use Quantum Mechanics to Show That Glass Will Melt Near Absolute Zero&#xA;ScienceDaily (Feb. 4, 2011)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prof. Eran Rabani of Tel Aviv University's School of Chemistry and his colleagues at Columbia University have discovered a new quantum mechanical effect with glass-forming liquids. They've determined that it's possible to melt glass -- not by heating it, but by cooling it to a temperature near absolute zero.&#xA;This new basic science research, to be published in Nature Physics, has limited practical application so far, says Prof. Rabani. But knowing why materials behave as they do paves the way for breakthroughs of the future. &quot;The interesting story here,&quot; says Prof. Rabani, &quot;is that by quantum effect, we can melt glass by cooling it. Normally, we melt glasses with heat.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1320" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-02-10T11:56:32.043" LastActivityDate="2011-02-10T13:16:40.893" Title="Glass melting at near absolute zero?" Tags="&lt;quantum-mechanics&gt;&lt;temperature&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="4748" PostTypeId="1" CreationDate="2011-02-07T13:55:50.080" Score="6" ViewCount="238" Body="&lt;p&gt;Why is there a deep mysterious relation between string theory and number theory (Langlands program), elliptic curves, modular functions, the exceptional group $E_8$, and the Monster group as in Monstrous Moonshine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely it's not just a coincidence in the Platonic world of mathematics.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-07T13:55:50.080" ClosedDate="2011-02-07T15:25:45.083" Title="Why is there a deep mysterious relation between string theory and number theory, elliptic curves, $E_8$ and the Monster group?" Tags="&lt;mathematical-physics&gt;&lt;string-theory&gt;" CommentCount="7" FavoriteCount="2" />
  <row Id="4749" PostTypeId="1" AcceptedAnswerId="4776" CreationDate="2011-02-07T14:07:20.393" Score="3" ViewCount="358" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/3500/how-general-is-the-lagrangian-approach-in-field-theory&quot;&gt;How general is the Lagrangian approach in field theory?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;A lot of physics, such as classical mechanics, General Relativity, Quantum Mechanics etc can be expressed in terms of Lagrangian Mechanics and Hamiltonian Principles. But sometimes I just can't help wonder whether is it ever &lt;em&gt;possible&lt;/em&gt; (in the future, maybe) to discover a physical law that can't be expressed in terms of Lagrangian Equations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or to put it in other words, can we list down for all the physical laws that can be expressed in terms of Lagrangian equations, what are the mathematical characteristics of them( such as it must not contain derivatives higher than 2, all the solutions must be linear etc)?&lt;/p&gt;&#xA;" OwnerUserId="605" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-07T18:02:15.257" LastActivityDate="2011-02-07T18:05:49.003" ClosedDate="2011-02-08T10:33:52.493" Title="Is there any physics that cannot be expressed in terms of Lagrange equations?" Tags="&lt;lagrangian-mechanics&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="2" />
  <row Id="4751" PostTypeId="2" ParentId="4749" CreationDate="2011-02-07T14:12:07.800" Score="0" ViewCount="" Body="&lt;p&gt;If you have a classical theory specified by some partial differential equations, you can automatically come up with a Lagrangian by introducing a Lagrange multiplier for each PDE.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-07T14:12:07.800" CommentCount="2" />
  <row Id="4752" PostTypeId="2" ParentId="4746" CreationDate="2011-02-07T14:31:33.237" Score="4" ViewCount="" Body="&lt;p&gt;This is part of what I found in &quot;ScienceDaily&quot;:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Until now, structural quantum glasses had never been explored -- that is, what happens when you mix the unique properties in glass and add quantum effects. Prof. Rabani was challenged to ask: if we looked at the quantum level, would we still see the hallmarks of a classical glass?&#xA;What the researchers unearthed is a new and unique hallmark, showing that quantum glasses have a unique signature. Many materials he says can form a glass if they're cooled fast enough. Even though their theory is not practical for daily use: few individuals own freezers that dip down nearly 500 degrees below zero.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the feeling, that they deal about &quot;quantum glass&quot;, as oposed to &#xA;classical glass! Some mix up of imcompatible topics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thanks to Carls link it is now clear: This is about &quot;quantum liquids/glasses&quot; only!&#xA;Ability to read is of course a prerequisite to realize that. For amateurs: a typical silicate glass (that what Your windows are made of) is a network of polyanionic SiO2-units  with big voids, where the  alkali- and earth-alkali ions are located. The covalent (and strictly localized!) Si-O bonds, which make up this network have a bonding energy of around 440 kJ/mol, any idea of melting at some Kelvins is plain nonsense. Melting of glasses occurs when some of those bonds break, this is the only possibility &#xA;to get some mobility in this 3-dimendional network&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-02-10T11:55:59.200" LastActivityDate="2011-02-10T11:55:59.200" />
  <row Id="4753" PostTypeId="2" ParentId="4738" CreationDate="2011-02-07T14:31:42.633" Score="-2" ViewCount="" Body="&lt;p&gt;It might be that the ultimate structure of the universe is monstrous-moonshine.  The potential role is with the interior of that $\hbar,~i$ planar region with the various string theories connected by a region which is not well known. If the $27$ dimensional system is the Jordan matrix algebra $J^3({\cal O})$, there are three scalar elements on the diagonal and the $E_8$, or octonions, ${\cal O}_i,~i~-~1,~2,~3$.  The infinite momentum constraint on the diagonals, with metric signature $[-,+,+]$ reduces this to $26$ dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Borcherds symmetries in M-theory&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dedicated to Pr. S. Hawking on his 60th birthday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pierre Henry-Labord`ere, Bernard Julia and Louis Paulot&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/hep-th/pdf/0203/0203070v2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/hep-th/pdf/0203/0203070v2.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;should be of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The connections with $n~=~2,~3,~4$ n-qubit systems is with the automorphism groups of the groups, or in $24$ dimensions $Aut(M_{24}),~{\mathbb Z}_n~=~\Lambda_{24}$, which is the Leech lattice.  The bipartite, tri-partite (qutrite) and 4-qubit entanglements are associated with the particular sublattice systems, $D_4$, $E_8$ and $\Lambda_{16}$, where the last is the Barnes-Wall lattice.  The above infinite momentum constraint on the $3$ scalars reduces this to symmetries on a $2$ dimensional hyperbolic space with the two dimensional group, $SL(2,R)$, which are the isometries of the $AdS_2$.  Without the constraint this is the isometries of $AdS_3$.  This then does segue into Witten’s three dimensional gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What Lobos indicates with this being a bosonic theory is worthy of note.  However, with string theory we start with the bosonic string in $26$ dimensions, which is the dimension where the Kac-Moody anomaly cancellation occurs.  The structures above have strong connections with $26$ dimensional string.  The dimensions of $24$, $26$ and $27$ are particularly interesting.  $24$ dimensions is the dimension of the Leech lattice, $26$ is half of the $F_4$ group, $dim(F_4)~=~52$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discrete group $PSL(2,~7)$ is the automorphism group of the octonionic Fano plane.  For complex-octonions the automorphism is $F_4$ of 52 dimensions. $F_4$ is equivalent to adding 16 short roots vectors to the four roots of the group $SO(9)$.  This is also the symmetry of the 24-cell.  The short roots define the quotient $F_4/SO(9)$ which defines the sequence:&#xA;$$&#xD;&#xA;F_4/SO(9):~1~\rightarrow~spin(9)~\rightarrow~F_{4\setminus 52}~\rightarrow~{\cal O}P^2,&#xD;&#xA;$$&#xA;where $4\setminus 52$ means the group is restricted to $36$ of the $52$ dimensions.  The ${\cal O}P^2$ defines the Cayley plane. $F_4$ is the automorphism group of the Jordan algebra with the symmetric product $X\odot Y~=~(1/2)(AY~+~YX)$.  The group $F_4$ acts transitively on ${\cal O}P^2$ with isotropy subgroup $spin(9)$.  For the octo-octonions the group is the exceptional $E_8\times E_8$ with heterotic structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does ultimately connect to supergravity.  It does so in much the same way that the bosonic string connects with supersymmetric string theory.  The $F_4/B_4$ gadget fits into this.  Suppose we have that $SO(16)~\rightarrow~~SO(14,2)$.  Now let us restrict this to $SO(14,1)$ the Lorentz spacetime.  We then have the further reduction due to the lightcone gauge or the “infinite momentum frame.  This reduces dimensions further, yet we may not want to do that here.  We have the $SO(16)\times SO(10)$ in the type II theory corresponding to the 26-dimensional bosonic string.  This embeds into the larger $SO(26)$.  So let us do our reductions on $SO(24,2)$, to $SO(24,1)$, then the infinite momentum or light cone gauge reduction is to SO(23,1).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $SO(24)$ group (here Euclideanized for simplicity) is the string theory of quantum gravity.  The $E_8\times E_8$ corresponds to the left and right moving modes on the string.  Consider a heterotic string with $26$ dimension for left moving modes, which correspond to $10$ right moving modes.  The k-metric is then $k^2~=~k_l^2~–~k_r^2$, which for $10$ of the $k_l$ modes $k_l~=~k_r$ component by component $1$ to $10$, then the k-metric is $16$ dimensional.  This is the origin of the $E_8\times E_8$ or equivalently $SO(2*dim(k))~=~SO(32)$.  The bosonic portion in $26$ dimensions contains two dimensions which are tachyonic vacuum modes and the remaining an $SO(24)$ field with tensor $\Omega_{ab}$.  The string operators $\alpha^a_{-1}$ projects the vacuum $|0\rangle$ to $|\Omega_{ab}\rangle$ by &#xA;$$&#xD;&#xA;\alpha^a_{-1}\alpha^b_{-1}|0\rangle~=~|\Omega^{ab}\rangle&#xD;&#xA;$$&#xA;This field has a traceless portion which is the spin-2 graviton, the trace term a dilaton operator and the anti-commuting portion define a gauge field theory in the adjoint representation.  This is a decomposition into $SO(8)$ and $SO(16)$, where the $SO(8)$ portion is conformal gravity and the $SO(16)$ gives the roots of $E_8$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $AdS_n$, $n~=~2,~3$ gravity then has a correlation with the standard $SO(24)$ bosonic string gravity in Green, Schwarz &amp;amp; Witten.  I think there is a great prospect for the future of this, which should lead to a monstrous-moonshine realization of the universe.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-02-07T15:16:03.500" LastActivityDate="2011-02-07T15:16:03.500" CommentCount="1" />
  <row Id="4754" PostTypeId="1" CreationDate="2011-02-07T14:42:53.140" Score="1" ViewCount="237" Body="&lt;p&gt;What happens in string theory when the temperature exceeds the Hagedorn temperature? Is that even possible? If yes, what is the nature of the phase transition and the phase beyond that? What happens to spacetime in the new phase?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-07T16:11:00.613" Title="What happens in string theory beyond the Hagedorn temperature?" Tags="&lt;string-theory&gt;&lt;statistical-mechanics&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="4755" PostTypeId="2" ParentId="4731" CreationDate="2011-02-07T15:23:44.617" Score="1" ViewCount="" Body="&lt;p&gt;It is possible to calculate the position of the second focus given only the radius vector and the momentum of the body. For this purpose consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Laplace%E2%80%93Runge%E2%80%93Lenz_vector&quot; rel=&quot;nofollow&quot;&gt;laplace-Runge-Lenz&lt;/a&gt; vector $\vec{A}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are at the &quot;main&quot; focus (the one with the attraction center), then $-\vec{A}$ points into the direction of the second focus. The length of this vector is $|\vec{A}| = mke$. Therefore the position of the second focus is &lt;a href=&quot;http://en.wikipedia.org/wiki/Latus_rectum#Introduction&quot; rel=&quot;nofollow&quot;&gt;given by&lt;/a&gt;:&lt;br&gt;&#xA;$\vec{R}_{F2}= -2a\vec{e}=&#xD;&#xA;-\frac{2\vec{e}}{1-e^2}\frac{L^2}{mk}=-\frac{2L^2\vec{A}}{m^2k^2-|A|^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use the notations from Wikipedia:&lt;br&gt;&#xA;$e$ -- is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Eccentricity_%28orbit%29&quot; rel=&quot;nofollow&quot;&gt;eccentricity&lt;/a&gt; and $\vec{e}$ -- is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Eccentricity_vector&quot; rel=&quot;nofollow&quot;&gt;eccentricity vector&lt;/a&gt;&lt;br&gt;&#xA;$L = \vec{r}\times\vec{p}$ -- angular momentum&lt;br&gt;&#xA;$k$ -- is a parameter that describes strength of the central force&#xA;$a$ -- &lt;a href=&quot;http://en.wikipedia.org/wiki/Latus_rectum#Introduction&quot; rel=&quot;nofollow&quot;&gt;major semiaxis&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, substituting $A^2=m^2k^2+2mEL^2$ ($E$ is negative):&lt;br&gt;&#xA;$\vec{R}_{F2}=\frac{\vec{A}}{mE}$&lt;br&gt;&#xA;or&lt;br&gt;&#xA;$\vec{R}_{F2}=\frac{\vec{p}\times\vec{L}}{mE}-\frac{k\vec{r}}{Er}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is remarkably simple -- It is LRL vector divided by the mass*Energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though I cannot invent an intuitive interpretation for it.    &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-02-07T15:23:44.617" />
  <row Id="4756" PostTypeId="2" ParentId="4754" CreationDate="2011-02-07T15:39:35.820" Score="-1" ViewCount="" Body="&lt;p&gt;The density of states for a string with respect to modes n is&#xA;$$&#xD;&#xA;\eta(n)~\sim~ exp(4\pi n \sqrt{\alpha’})&#xD;&#xA;$$&#xA;that defines a partition function&#xA;$$&#xD;&#xA; Z~\simeq~\int\eta(n)exp(-n/T)dn.  &#xD;&#xA;$$&#xA;The temperature is computed by $1/T~=~\partial Z/\partial n$ and the path integral diverges for a temperature greater than &#xA;$$&#xD;&#xA;T_H~=~4\pi\sqrt{\alpha’}&#xD;&#xA;$$&#xA;which is the Hagedorn temperature.  This is proportional to the reciprocal of the string length. The entropy of the system is the logarithm of the density of states the $S~\sim~ 1/nT_H$, which in the large $n$ limit is zero.  The modes number is given by $n~=~1/(\sqrt{d}M_s)$, for $d$ the number of degrees of freedom and $M_s$ the string mass&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point there is new physics.  Hagedorn temperature occurs when physics scales below the string length, and pertains to a domain between the string length and the Planck length.  This the UV limit of string theory, which has correspondences with the IR limit by holography.  A gas of strings percolate into a single system (one large string), which has a correspondence with the Lorentz contraction and “spreading”of strings on the event horizon of a black hole at the IR limit.  At or near the Hagedorn temperature there is likely to be quantum fluctuations which exchange the various string types.  This in M-theory corresponds to the wrapping-mode T-duality, modular S-duality and such structures which exchange string types.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would conjecture that just as with the meson string, two quarks connected by a chromodynamic flux tube, strings are similar, and above the Hagedorn temperature become “gas” of D0-branes or particles.  The endpoints of strings (Chan-Paton factors or Dirichlet conditions on Dp-branes) become a plasma of D0-partons.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-07T15:39:35.820" />
  <row Id="4757" PostTypeId="2" ParentId="4731" CreationDate="2011-02-07T15:40:33.093" Score="9" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/6j9DX.gif&quot; alt=&quot;Tides &amp;amp; second focus&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second (empty) focus is relevant in the theory of tides. In an elliptical orbit, the line joining the planet and the empty focus rotates at the same frequency as the mean motion of the planet; therefore, if spin rotation period is equal to the orbital period (the planet is locked in a synchronous rotation), the planet rotates with one face pointing to the empty focus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Importantly, a tidal bulge will try to point to the massive object (the occupied focus) while the planet itself will be pointing to the empty focus, causing a &quot;librational tide&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1252" LastActivityDate="2011-02-07T15:40:33.093" CommentCount="1" />
  <row Id="4758" PostTypeId="1" AcceptedAnswerId="4783" CreationDate="2011-02-07T15:40:41.390" Score="6" ViewCount="271" Body="&lt;p&gt;I would like to know how important the &quot;cosmic censorship conjecture&quot; is? Should a quantum theory of gravity must obey this? It was never rigorously proved in classical GR too. What would be the consequences if it turns out that CCC (weak) does not hold?&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-02-07T20:32:02.620" LastActivityDate="2011-03-02T03:55:08.880" Title="How important is the cosmic censorship conjecture?" Tags="&lt;general-relativity&gt;&lt;quantum-gravity&gt;" AnswerCount="4" />
  <row Id="4759" PostTypeId="2" ParentId="4470" CreationDate="2011-02-07T15:41:11.590" Score="3" ViewCount="" Body="&lt;p&gt;I would like to add some further points to the answers above on the Twistor Space &amp;lt;--&gt; Spacetime correspondence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Twistor space &lt;strong&gt;T&lt;/strong&gt; is a four complex dimensional space with elements described by $(Z^0,Z^1,Z^2,Z^3)$ or $Z^{\alpha}=(\omega^A,\pi_{A&amp;#39;})$ in spinor terms. The incidence relation between Minkowski points and Twistors is given (in spinor form) by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega^A=ix^{AA&amp;#39;}\pi_{A&amp;#39;}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In addition to a different representation of $x^{AA&amp;#39;}$ to that of Lubos it should also be noted that there is a 4-fold ambiguity in the representation of a spacetime point by a Twistor $(Z^{\alpha},-Z^{\alpha},iZ^{\alpha},-iZ^{\alpha})$ are all the same point as there are two sets of double cover involved.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Twistor Space contains a norm &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$||Z||=Z^AZ^c_{A}=\omega^A\pi^c_{A} + \omega^{cA&amp;#39;}\pi_{A&amp;#39;}$ where $^c$ denotes spinor/twistor complex conjugation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two sets of reductions to Twistor space to obtain the Minkowski space correspondence:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;||Z||=0 these are the &lt;em&gt;null&lt;/em&gt; Twistors - the space sometimes denoted &lt;strong&gt;N&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;T&lt;/strong&gt; modulo  $\lambda Z^{\alpha}$ - the space denoted &lt;strong&gt;PT&lt;/strong&gt; -projective Twistor Space.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Putting these two together we get &lt;strong&gt;PN&lt;/strong&gt; - projective Null Twistor Space. It is &lt;strong&gt;PN&lt;/strong&gt; which is mapped to Minkowski space via the incidence relation ie projective null twistors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This raises the question as to what to more general cases of Twistor mappings correspond to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general &lt;strong&gt;PT&lt;/strong&gt; maps to Complexified Minkowski space. This space has the $(x^0,x^1,x^2,x^3)$ become complex numbers and the metric becomes a complex analytic extension of the Minkowski metric. Non-null Twistors become complex planes in that space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mapping to Complexified Minkowski space had several attractions in the original Twistor theory including the idea that in a curved space context maybe the disappearance of a point as a Singularity could be seen as just its disappearance from real Minkowski space, but not the more basic Twistor space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally in the Incidence relation we are assuming that $\pi_{A&amp;#39;}$ is non-zero. In that case the Incidence relation could be geometrically considered to be a mapping to a Compactified completion of Complex Minkowski Space which includes the null cone at Infinity and thereby the other constructs of Compactified Minkowski space used in Penrose diagrams.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-07T15:41:11.590" CommentCount="4" />
  <row Id="4760" PostTypeId="2" ParentId="4754" CreationDate="2011-02-07T15:59:33.590" Score="4" ViewCount="" Body="&lt;p&gt;This depends a bit what you mean by string theory. I will take this to mean perturbative string theories. Also, my opinion below does not reflect a complete consensus, but I think it is the majority view. There is also lots of work on this, quite a bit of it before my time, I might be missing something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in free string theories, there is exponential density of states at high energies. This means there is a “Hagedorn” temperature, a temperature beyond which the partition function $Tr(e^-{\beta H})$ does not exist.  The same appears to be happening in QCD, where there is an exponential density of almost stable resonances in some limit (the large N limit) - this is the context in which Hagedorn was discussing this temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is an interesting question what happens as you heat up the system and approach this temperature. One has to distinguish two cases: the string coupling is precisely zero, or the string coupling is small but finite (which of course is the more sensible case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At zero coupling, as you heat the system, you have competition between two situations. One is the conventional picture: gas of strings with different energy levels which are thermally distributed. On the other hand you can have a single excited string, which as a single object does not have a thermodynamic description. The reason you can have competition between such situations is because the single string has many “internal” degrees of freedom, so it can individually have large density of states. What happens near the Hagedorn temperature is that as you put more energy into the system, you excite a single string rather than change the thermal distribution of many strings. You can see the breakdown of the thermodynamic description because the specific heat of the system, which is a measure of the strength of the thermal fluctuations, diverges near the Hagedorn temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the coupling is finite, no matter how small, we have to consider interactions between the strings. The results of the classic paper by &lt;a href=&quot;http://www.sciencedirect.com/science?_ob=ArticleURL&amp;amp;_udi=B6TVC-471XMJK-RK&amp;amp;_user=10&amp;amp;_coverDate=12%2F05%2F1988&amp;amp;_rdoc=1&amp;amp;_fmt=high&amp;amp;_orig=search&amp;amp;_origin=search&amp;amp;_sort=d&amp;amp;_docanchor=&amp;amp;view=c&amp;amp;_acct=C000050221&amp;amp;_version=1&amp;amp;_urlVersion=0&amp;amp;_userid=10&amp;amp;md5=febcbcb84d908079a0896ef4bd3a5ef1&amp;amp;searchtype=a&quot; rel=&quot;nofollow&quot;&gt;Atick and Witten&lt;/a&gt; suggest that well before you reach the Hagedorn temperature, the system will go through a first order phase transition and never really reach the Hagedorn temperature. The nature of the new phase was not clarified in that paper, and it probably depends on which string theory it is precisely. One universal expectation is that since strings attract gravitationally, the phase transition involves gravitational collapse of the system.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-02-07T16:11:00.613" LastActivityDate="2011-02-07T16:11:00.613" />
  <row Id="4761" PostTypeId="1" AcceptedAnswerId="4766" CreationDate="2011-02-07T16:03:01.120" Score="1" ViewCount="189" Body="&lt;p&gt;The hotter something is glowing the more white/blue it appears. A dying medium sized star expands, cools and becomes a red giant for a while, but eventually it is going to gravitationally collapse (once enough Iron (Fe) is accumulated in the core). Then it blows the outer layers away and what is left collapses into a white dwarf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes the dwarf shine? and why is it white? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the luminosity decreases as the object cools down, or is there some other reaction that keeps it glowing for a long time? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can a white dwarf turn brown or black never to be seen again?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do all white dwarfs turn into Neutron stars eventually?&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-02-07T20:46:43.787" Title="Why do white dwarfs shine white?" Tags="&lt;neutron-stars&gt;&lt;white-dwarfs&gt;&lt;stellar-evolution&gt;" AnswerCount="3" />
  <row Id="4762" PostTypeId="2" ParentId="4758" CreationDate="2011-02-07T16:04:05.300" Score="1" ViewCount="" Body="&lt;p&gt;At this stage we really do not know.  There are plenty of reasons to think the CCC holds.  Classical gravity does not predict the CCC, and we must impose it.  In the thread &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/4672/do-traversable-wormholes-exist-as-solutions-to-string-theory&quot;&gt;Do traversable wormholes exist as solutions to string theory? &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I argue this leads to some funny predictions with elementary string theory.  So it might be that string theory rules out solutions to the Einstein field equations which violate the CCC.  If this does not happen it could then be quantum gravity, in what ever form it might end up as, prevents violations of the CCC in the classical limit.  In that setting there might be quantum paths which traverse spacetime in the opposite time direction, but where these paths destructively interfere at the WKB or classical limits. Another possibility is that violations of the CCC are permitted, but where the physical solutions are simply very unstable.  So a wormhole might exist, but where the smallest perturbation send the solution into another form.  So on the solution space,or moduli, these solutions are isolated points --- give it a tweek and it moves off that solution type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So these are not answers, but possible ways to think about the problem.  At this time we have no definative answer.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-07T16:04:05.300" />
  <row Id="4763" PostTypeId="2" ParentId="4758" CreationDate="2011-02-07T16:04:53.303" Score="1" ViewCount="" Body="&lt;p&gt;It's actually the case that we expect cosmic censorship to be false for a quantum theory.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why is that one consequence of classical relativity, any of the positive energy conditions, and cosmic censorship is that the area of a black hole must always increase (the so-called Area Increase Theorem).  But, if there is Hawking radiation coming from the black hole, then the area of the black hole must decrease.  If this is the case, the apparent horizon of the black hole fails to trap matter inside the black hole.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, you haven't contradicted cosmic censorship--we can just interpret the Hawking radiation as a negative-energy matter distribution.  If your starting spacetime &lt;a href=&quot;http://www.google.com/imgres?imgurl=http://www.astro.umd.edu/~chris/Teaching/ASTR340_Spring_2002/HW4_files/image004.jpg&amp;amp;imgrefurl=http://www.astro.umd.edu/~chris/Teaching/ASTR340_Spring_2002/HW4.htm&amp;amp;usg=__R-Nh8na9FnODdP03hjgtFVXI7A4=&amp;amp;h=399&amp;amp;w=624&amp;amp;sz=19&amp;amp;hl=en&amp;amp;start=0&amp;amp;sig2=tEvpqhDFrLmNf47yD6jdkg&amp;amp;zoom=1&amp;amp;tbnid=SZ5b5sgoPqrl-M:&amp;amp;tbnh=128&amp;amp;tbnw=205&amp;amp;ei=gBZQTd3zM4n1gAeC0ZnsDw&amp;amp;prev=/images%3Fq%3Dnordstrom%2Bspacetime%26um%3D1%26hl%3Den%26sa%3DN%26biw%3D1158%26bih%3D673%26tbs%3Disch:1&amp;amp;um=1&amp;amp;itbs=1&amp;amp;iact=hc&amp;amp;vpx=369&amp;amp;vpy=83&amp;amp;dur=380&amp;amp;hovh=179&amp;amp;hovw=281&amp;amp;tx=196&amp;amp;ty=102&amp;amp;oei=gBZQTd3zM4n1gAeC0ZnsDw&amp;amp;esq=1&amp;amp;page=1&amp;amp;ndsp=25&amp;amp;ved=1t:429,r:2,s:0&quot; rel=&quot;nofollow&quot;&gt;had a timelike singularity, however, as is the case for a charged or spinning black hole&lt;/a&gt;${}^{1}$, it will then be possible for a null ray originating from the singularity to leave the black hole interior and reach null infinity--light from the singularity will escape the hole.  This is very definitely a violation of cosmic censorship.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{1}$The picture on the right in the link is the penrose-carter diagram for the Nordstrom spacetime, which represents a black hole with a net charge.  The upward direction represents motion toward the future, and 45${}^{\circ}$ angles represent the paths of light rays.  The diamond directly below the pictured x-axis reprensents the portion of the spacetime containing a black hole (and not a white hole, which would be the diamonds above and below this one).  You can see the CC-violating path by letting the pictured horizon move upward--in this case, the 45${}^{\circ}$ rays from the bottom of the singularity will be able to leave the BH interior and reach the &quot;ordinary space&quot; region.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-02-07T16:04:53.303" CommentCount="2" />
  <row Id="4764" PostTypeId="2" ParentId="4749" CreationDate="2011-02-07T16:05:43.863" Score="-1" ViewCount="" Body="&lt;p&gt;You cannot model friction very well with Lagrangian Mechanics.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-02-07T16:05:43.863" CommentCount="7" />
  <row Id="4765" PostTypeId="1" CreationDate="2011-02-07T16:07:41.880" Score="0" ViewCount="153" Body="&lt;p&gt;We distinguish between the states of matter: gas, liquid and solid. Possibly we could add the plasma state and/or the superconductive state as new states of matter. Phase transistions at certain temperature perhaps with some other conditions should have to exist. What do you think, does it make sense?&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2011-02-07T22:12:29.067" ClosedDate="2011-02-07T17:58:32.513" Title="Matter in superconductive state" Tags="&lt;condensed-matter&gt;&lt;plasma-physics&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="4766" PostTypeId="2" ParentId="4761" CreationDate="2011-02-07T16:08:49.120" Score="3" ViewCount="" Body="&lt;p&gt;White dwarves used to be the interior of a star, which was the hottest part of the star.  They shine white because they are still very hot from this past part of their history.  As they age, they will cool, and as they cool, they will lose temperature, and their &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_body&quot; rel=&quot;nofollow&quot;&gt;blackbody profile&lt;/a&gt; will shift to redder and redder colors, and eventually into the infared and radio ranges where they won't seem to shine at all to the naked eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, a white dwarf state is a stable final state of a star, so long as it does not interact further with any matter.  If that happens, it is possible to have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Type_Ia_supernova&quot; rel=&quot;nofollow&quot;&gt;White Dwarf supernova&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-02-07T16:08:49.120" CommentCount="5" />
  <row Id="4767" PostTypeId="2" ParentId="4749" CreationDate="2011-02-07T16:11:14.747" Score="0" ViewCount="" Body="&lt;p&gt;This question is rather generic in several different respects. For example it is not clear that even Schrodinger's equation is an equation that can be &quot;expressed in Hamiltonian principles&quot;. Yes we have $\delta/\delta t \Psi=H \Psi$ but is this an expression in &quot;Hamiltonian Principles&quot;? How different must the equation become to not meet this requirement - assuming it does now, perhaps adding a non-linear term?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore &quot;Hamilton's Principle&quot; does not itself apply in the Quantum context, although the action paths that it introduces are used in Feynman Path Integrals. Hamilton's Principle being a classical concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another generality is in the range of Physics. The whole area of Thermodynamics comes to mind. Now there are phase space explanations, but is that &quot;Hamiltonian Principles?&quot;&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-07T16:11:14.747" />
  <row Id="4768" PostTypeId="2" ParentId="4761" CreationDate="2011-02-07T16:14:12.890" Score="2" ViewCount="" Body="&lt;p&gt;Last week I wrote an answer to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/4546/please-clarify-how-entropy-increases-when-matter-gravitationally-coalesces/4556#4556&quot;&gt;Please clarify how entropy increases when matter gravitationally coalesces&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I illustrate how a collapsing body will increase its entropy.  This also corresponds to an increased temperature.  The imploded white dwarf has its internal energy compressed into a very small volume.  Hence they initially shine at a very high temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[further]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interior of a white dwarf has Fermi-Dirac statistics, but the gas on the outer layers is probably more ordinary.  Further, that is what we detect.  Consider a stellar core imploding from about $1.0R_{sol}$, probably more like less than this, which we will write at $5\times 10^{5}$, to about $5\times 10^3$.  A rough estimate would be the natural gas law, in fact in the form Boyle wrote it in&#xA;$$&#xD;&#xA;T~=~\frac{V}{V_0}T_0~\simeq~10^6\times T_0&#xD;&#xA;$$&#xA;The initial temperature of the core is about $10^7$K, which gives a pretty outstanding temperature of about $10^{13}$K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not realistic, for as the white dwarf forms it radiates most of its energy to eject off the outer layers into a nebular cloud.  So the outer layer of the core collapses inwards more at a temperature comparable to a stellar surface.  So that knocks this estimate down three or four orders of magnitude.  In fact a newly minted white dwarf as a temperature of about $10^7$ K  So direct simple calculations are not very accurate.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-02-07T18:30:47.827" LastActivityDate="2011-02-07T18:30:47.827" CommentCount="2" />
  <row Id="4770" PostTypeId="2" ParentId="4761" CreationDate="2011-02-07T16:19:53.663" Score="3" ViewCount="" Body="&lt;p&gt;To amplify a bit bit on Jerry's answer.&#xA;  Because of its small surface area, and large thermal mass (typically about a half the mass of the sun) the cooling time of white dwarves is billions of years. As he says they do cool, however the universe isn't old enough to have created condensed red dwarves. The stars currently called red dwarfs, are main sequence (hydrogen burning) low mass stars with lifetimes on the order of a trillion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess, we've all simply forgotten to answer his other question &quot;do all white dwarves become neutron stars/&quot;. The answer is no. The white dwarfs pressure is maintained by electron degeneracy pressure, they do not contract appreciably as they cool down, and dense as they are they are orders of magnitude less dense than nuclear matter. A white dwarf has to exceed the Chadreshekar mass for core collapse in order to become a neutron star. And if enough mass is added to a white dwarf to exceed that limit, you get a thermonuclear runaway reaction leading to a type 1a supernova instead.&lt;/p&gt;&#xA;" OwnerUserId="377" LastEditorUserId="377" LastEditorDisplayName="" LastEditDate="2011-02-07T20:46:43.787" LastActivityDate="2011-02-07T20:46:43.787" />
  <row Id="4771" PostTypeId="2" ParentId="4765" CreationDate="2011-02-07T16:31:28.753" Score="1" ViewCount="" Body="&lt;p&gt;Yes, there are other phases of matter besides solid, liquid, gas, and phase transitions between these other phases work in a similar way to transitions between the more familiar phases. As another example, helium undergoes a phase transition between a &quot;normal&quot; fluid and a superfluid at a temperature of about 2 K.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-07T16:31:28.753" />
  <row Id="4772" PostTypeId="2" ParentId="4765" CreationDate="2011-02-07T16:35:15.560" Score="1" ViewCount="" Body="&lt;p&gt;Naming plasma the fourth state of matter is quite reasonable and is done since long. &#xA;For superconductors this is nonsense, because it is nor universal, (some metals only) &#xA;and it is something limited to a small part of the electrons in such a metal. &#xA;Matter is all the matter, not just some electrons. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-02-07T16:35:15.560" />
  <row Id="4773" PostTypeId="2" ParentId="4758" CreationDate="2011-02-07T16:48:49.113" Score="0" ViewCount="" Body="&lt;p&gt;If the cosmic censorship conjecture does not hold, there are problems with causality in that the universe would cease to be deterministic. One could say that QM is probabilistic, but that is a misunderstanding. The Schrodinger wave equation is deterministic. Quantum mechanics is deterministic about the probabilities. General relativity is deterministic.&#xA;Situations where the CCC seem to be violated, as in the Kerr-Newman metric for spinning black holes, may not occur because of other limiting factors arising -&quot;-Solving the above effective potential equation for the maximum under the given conditions results in a maximum potential of exactly e^2-1/2! Testing other values shows that no particle with enough angular momentum to violate the censorship conjecture would be able to enter the black hole, because they have too much angular momentum to fall in.&quot;--from Hartle's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0805386629&quot; rel=&quot;nofollow&quot;&gt;Gravity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue is far from settled, although, gut reaction is that it holds. (Eventhough, when asked for his gut reaction about some other thing, Carl Sagan replied &quot;I prefer not to think with my gut.&quot; :)&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-02-07T16:48:49.113" CommentCount="2" />
  <row Id="4774" PostTypeId="2" ParentId="4765" CreationDate="2011-02-07T17:20:32.190" Score="1" ViewCount="" Body="&lt;p&gt;A phase transition is a Thermodynamic (therefore bulk matter) concept which pertains to abrupt changes in the physical properties of a system which occur on a boundary of a Phase Diagram. In the classic case this Phase Diagram has axes for Temperature and Pressure. Water has the three classic phases, plus a &quot;supercritical&quot; region above the critical point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of Superconductivity the physical change occurs on the temperature conductivity graph, but as remarked in another Answer this is not a phase transition universal to all matter.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-02-07T22:12:29.067" LastActivityDate="2011-02-07T22:12:29.067" CommentCount="2" />
  <row Id="4776" PostTypeId="2" ParentId="4749" CreationDate="2011-02-07T18:05:49.003" Score="1" ViewCount="" Body="&lt;p&gt;Hamilton's dynamics occurs on a phase space with an equal number of configuration and momentum variables $\{q_i,~p_i\}$, for $i~=~1,\dots n.$  The dynamics according to the symplectic two form ${\underline{\Omega}}~=~\Omega_{ab}dq^a\wedge dp^b$ is a Hamiltonian vector field&#xA;$$&#xD;&#xA;\frac{d\chi_a}{dt}~=~\Omega_{ab}\partial_b H,&#xD;&#xA;$$&#xA;with in the configuration and momentum variables $\chi_a~=~\{q_a,~p_a\}$ gives &#xA;$$&#xD;&#xA;{\dot q}_a~=~\frac{\partial H}{\partial p_a},~{\dot p}_a~=~-\frac{\partial H}{\partial q_a}&#xD;&#xA;$$&#xA;and the vector $\chi_a$ follows a unique trajectory in phase space, where that trajectory is often called a Hamiltonian flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a system the bare action is $pdq$ ignoring sums.  The Hamiltonian is found with imposition of Lagrangians as functions over configuration variables.  This is defined then on half of the phase space, called configuration space.  It is also a constraint, essentially a Lagrange multiplier.  The cotangent bundle $T^*M$ on the configuration space $M$ defines the phase space.  Once this is constructed a symplectic manifold is defined.  Therefore Lagrangian dynamics on configuration space, or equivalently the cotangent bundle defines a symplectic manifold.  This does not mean a symplectic manifold defines a cotangent bundle.  The reason is that symplectic or canonical transformations mix the distinction between configuration and momentum variables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result there are people who study bracket structures which have non-Lagrangian content.  The RR sector on type IIB string is non-Lagrangian.  The differential structure is tied to the Calabi-Yau three-fold, which defines a different dynamics.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-07T18:05:49.003" />
  <row Id="4777" PostTypeId="1" CreationDate="2011-02-07T19:36:05.770" Score="3" ViewCount="159" Body="&lt;p&gt;a fight from east to west, (e.g. Australia melbourne to perth) takes longer than travel backwards. (west to east).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the reason? &lt;/p&gt;&#xA;" OwnerUserId="1799" LastActivityDate="2011-02-07T20:03:38.240" Title="time difference with airplane travel direction" Tags="&lt;time&gt;&lt;airplanes&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="4778" PostTypeId="1" AcceptedAnswerId="4781" CreationDate="2011-02-07T19:49:05.350" Score="3" ViewCount="249" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Deuterium#Approximated_wavefunction_of_the_deuteron&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article on deuterium&lt;/a&gt; says this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The deuteron wavefunction must be&#xA;  antisymmetric if the isospin&#xA;  representation is used (since a proton&#xA;  and a neutron are not identical&#xA;  particles, the wavefunction need not&#xA;  be antisymmetric in general).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wonder why does the wave function need to be antisymmetric when isospin representation is used. I assume that if two somehow different particles are exchanged the total wavefunction changes sign. Is it so? Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1800" LastEditorUserId="1800" LastEditorDisplayName="" LastEditDate="2011-02-07T19:55:31.897" LastActivityDate="2011-02-07T20:09:50.917" Title="Why must the deuteron wavefunction be antisymmetric?" Tags="&lt;nuclear-physics&gt;&lt;quantum-mechanics&gt;&lt;symmetry&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="4780" PostTypeId="2" ParentId="4777" CreationDate="2011-02-07T20:03:38.240" Score="5" ViewCount="" Body="&lt;p&gt;The speed of an aircraft during the flight is ideally kept constant relatively to the air surrounding the aircraft. However, in the moderate zone - between 30 and 60 degrees of latitude on both hemispheres - the &lt;strong&gt;dominant winds are called westerlies&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Westerlies&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Westerlies&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;for a good reason. They blow from the West to the East. So it's faster to get from Perth to Melbourne because it's in the direction of the wind; but it takes a longer time to fly from Melbourne to Perth because it's against the wind and, again, the natural speed of the aircraft is measured relatively to the air (i.e. wind).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why do the westerlies have this direction?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's because of the &lt;strong&gt;Coriolis force&lt;/strong&gt;. The Coriolis force - a &quot;fictitious&quot; force in a rotating frame that adds to the centrifugal one - steers winds (and rivers) in the clockwise direction on the Northern Hemisphere and counter-clockwise direction on the Southern hefmisphere.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Coriolis_effect&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Coriolis_effect&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why does the Coriolis force create westerlies - which are particularly strong on the Souther Hemisphere e.g. in Australia? It's because the normal air that heats up near the equator expands and it would travel to one of the poles. However, this air going towards the poles gets steered by the Coriolis force, so before it reaches e.g. the Southern pole, it's already moving in the East direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The westerlies are weaker on the Northern Hemisphere as well as on the summer-having hemisphere because when it's so, the wind from the equator is strong enough (amplified by the land on the Northern Hemisphere, or by the summer) so that it usually manages to reach the pole (without drifting to the East), anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very, very far but my uncle lives in Melbourne while a generous IT expert who lives in Perth sent me an iPod Touch for a relatively modest and doable help with a project half a year ago. ;-)&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-07T20:03:38.240" />
  <row Id="4781" PostTypeId="2" ParentId="4778" CreationDate="2011-02-07T20:09:50.917" Score="1" ViewCount="" Body="&lt;p&gt;The neutron and proton may be viewed as the same particle - sometimes referred to as the nucleon. A proton is a &quot;nucleon with the isospin up&quot; and the neutron is a &quot;nucleon with the isospin down&quot;. With this qualification, it's still true that nucleons are identical fermions, so their total wave function has to be antisymmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the simplest Ansatz, the total wave function is the tensor product of the isospin wave function; spin wave function; and orbital (spatial) wave function. The odd number of those has to be antisymmetric for the result to be antisymmetric. The (anti)symmetry of each factor is governed by the total isospin (0 antisymmetric or 1 symmetric); total spin (0 antisymmetric or 1 symmetric); orbital angular momentum (even means symmetric and odd means antisymmetric).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-07T20:09:50.917" CommentCount="6" />
  <row Id="4782" PostTypeId="1" CreationDate="2011-02-07T23:54:56.263" Score="5" ViewCount="202" Body="&lt;p&gt;I'm working on an AI for a pinball-related video game.  The ultimate goal for the system is that the AI will be able to fire a flipper at the appropriate time to aim a pinball at a particular point on the table.  We are using an engine that will handle the underlying physics interactions when the ball is hit (Unreal).  I'm trying to quasi-realistically model the physics for the AI's &quot;thinking&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a reasonable understanding of Newtonian physics but I'm going to need some help making sure I'm correct about everything I may need to account for.  For the purposes of simplification, I'm considering the table to be vertical on an XY axis with gravity just being a small fraction of actual gravity accelerating down on the Y axis (yes, this means I'm ignoring friction between the ball and table, for now).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By my understanding, I've got the work with the following when a ball is rolling along a flipper.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The ball will have velocity along both X and Y axes as it rolls to the flipper.&lt;/li&gt;&#xA;&lt;li&gt;The flipper will have a range of motion (minimum and maximum angles) and a Torque value.&lt;/li&gt;&#xA;&lt;li&gt;When the flipper fires, it will apply force to the ball based on the flipper's Torque, the distance the ball is from the flipper's rotation point, and the length of the arc the contacted point on the flipper needs travel.&lt;/li&gt;&#xA;&lt;li&gt;The applied force will accelerate the ball back up the table following a new trajectory.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My question(s) are: Am I overlooking an important interactions in this system?  Or am I misinterpreting anything?  Does it seem like this is a reasonable modeling for video game?  Et cetera?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1802" LastActivityDate="2011-02-08T08:19:45.713" Title="Trying to model pinball physics for game AI" Tags="&lt;newtonian-mechanics&gt;&lt;simulation&gt;&lt;software&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="4783" PostTypeId="2" ParentId="4758" CreationDate="2011-02-08T00:49:32.113" Score="3" ViewCount="" Body="&lt;p&gt;Lehner and Pretorius have recently given some persuasive numerical evidence that there are generic violations of cosmic censorship which arise in the time evolution of black strings in 5d gravity, aka the Gregory-LaFlamme instability &lt;a href=&quot;http://arxiv.org/pdf/1006.5960&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/1006.5960&lt;/a&gt; . The failure of  cosmic censorship certainly does not imply the downfall of causality. Classically it means we can see a singularity without it being hidden behind a horizon. This seems to me like a very good thing. We all believe that there is some UV completion of gravity, although different camps have different views on what this is. This theory will cut off the singularity through some combination of classical modifications to Einstein gravity and quantum effects. Being able to see such a direct effect of the UV completion would be a wonderful thing, so I think we should all hope that there are also generic violations of cosmic censorship in four dimensions, although as far as I know the jury is still out on this.&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T03:55:08.880" LastActivityDate="2011-03-02T03:55:08.880" />
  <row Id="4784" PostTypeId="1" AcceptedAnswerId="4786" CreationDate="2011-02-08T01:05:56.490" Score="5" ViewCount="511" Body="&lt;p&gt;One thing I've heard stated many times is that &quot;most&quot; or &quot;many&quot; physicists believe that, despite the fact that they have not been observed, there are such things as magnetic monopoles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I've never really heard a good argument for why this should be the case. The explanations I've heard tend to limit themselves to saying that &quot;it would create a beautiful symmetry within Maxwell's equations&quot; or similar. But I don't see how the fact that &quot;it would be beautiful&quot; is any kind of reason for why they should exist! That doesn't sound like good science to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So obviously there must be at least another reason apart from this to believe in the existence of natural magnetic monopoles. (I'm aware that recent research has shown that metamaterials (or something similar) can emulate magnetic monopole behaviour, but that's surely not the same thing.) Can anyone provide me with some insight?&lt;/p&gt;&#xA;" OwnerUserId="1803" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-08T05:57:23.527" LastActivityDate="2011-03-02T06:40:59.517" Title="Why do physicists believe that there exist magnetic monopoles?" Tags="&lt;electromagnetism&gt;&lt;soft-question&gt;&lt;magnetic-monopoles&gt;" AnswerCount="6" FavoriteCount="2" />
  <row Id="4785" PostTypeId="2" ParentId="4784" CreationDate="2011-02-08T01:18:37.880" Score="3" ViewCount="" Body="&lt;p&gt;I'd be interested to know who told you such a thing. I gather that the existence of magnetic monopoles is a natural consequence of many (most? all?) grand unified theories, so that's probably what whoever it is is talking about. But there are quite a few physicists out there who are more swayed by experimental evidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you combine the GUTs that predict the existence of monopoles with non-inflationary hot big bang theory, you predict that monopoles should have been produced in huge numbers in the early Universe, leading to the puzzle of why we don't see them around today. Inflation was originally devised at least in part to solve this problem: if the monopoles were produced before inflation, then they would have been &quot;inflated away&quot; to unobservably low density today (i.e., less than one in the observable Universe).&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-08T01:18:37.880" />
  <row Id="4786" PostTypeId="2" ParentId="4784" CreationDate="2011-02-08T01:41:27.277" Score="5" ViewCount="" Body="&lt;p&gt;Aqwis, it would help in the future if you mentioned something about your background because it helps to know what level to aim at in the answer. I'll assume you know E&amp;amp;M at an undergraduate level. If you don't then some of this explanation probably won't make much sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part one goes back to Dirac. In E&amp;amp;M we need to specify a vector potential $A_\mu$. Classically the electric and magnetic fields suffice, but when quantum mechanics is included you need $A_\mu$. The vector potential is only defined up to gauge transformations $A_\mu \rightarrow g(x)(A_\mu + \frac{i}{e} \partial_\mu ) g^{-1}(x)$ where $g(x)=\exp(i \alpha(x))$. The group involved in these gauge transformations is the real line (that is the space of possible values of $\alpha$) if electric charge is not quantized, but if charge is quantized, as all evidence points to experimentally, then the group is compact, that is it is topologically a circle, $S^1$. So to specify a gauge field we specify an element of $S^1$ at every point in spacetime. Now suppose we don't know for sure what goes on inside&#xA;some region (because we don't know physics at short distances). Surround this region with a sphere. We can define our gauge transformation at every point outside this region, but now we have to specify it on two-spheres which cannot be contracted to a point. At a fixed radial distance the total space of angles plus the gauge transformation can be a simple product, $S^2 \times S^1$ but it turns out there are other possibilities. In particular you can make what is called a principal fibre bundle where the $S^1$ twists in a certain way as you move around the $S^2$. These are characterized by an integer $n$, and a short calculation which you can find various places in the literature shows that the integer $n$ is nothing but the magnetic monopole charge of the configuration you have defined. So charge quantization leads to the ability to define configurations which are magnetic monopoles. So far there is no guarantee that there are finite energy objects which correspond to these fields. To figure out if they are finite energy we need to know what goes on all the way down to the origin inside our region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part two is that in essentially all models that try to unify the Standard Model you find that there are in fact magnetic monopoles of finite energy. In grand unified theories this goes back to work of 't Hooft and Polyakov. It also turns out to be true in Kaluza-Klein theory and in string theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there are three compelling reasons to expect that magnetic monopoles exist. The first is the beauty of a deep symmetry of Maxwell's equations called electric-magnetic duality, the second is that electric charge appears to be quantized experimentally and this allows you to define configurations with quantized magnetic monopole charge, and the third is that when you look into the interior of these objects in essentially all unified theories you find that the monopoles have finite energy.&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T06:40:59.517" LastActivityDate="2011-03-02T06:40:59.517" />
  <row Id="4787" PostTypeId="2" ParentId="728" CreationDate="2011-02-08T02:38:36.917" Score="1" ViewCount="" Body="&lt;p&gt;Although my intuition was wrong (like Peter Shor's) and while I admire both Mark Eichenlaub's (self) and Carl Brannen's detailed and exact answers, what does it to me is the intuitive response (with correction based on comments) that Jerry Schirmer gave, although I would take the opposite conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simple fact that the reaction force needed to create the lossless friction must be exactly opposed to the angular acceleration (which can only be caused by the effects of gravity) means that the accelerations are colinear. That's it really, no need to say more, so they follow the same path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, kudos for a very well defined problem. As the minute the ball flies or touches in two points, the friction thing will go mad with infinite acceleration, or break the rule, and the ball's spin will dramatically alter its course (not counting on atmospheric effects either ;-)).&lt;/p&gt;&#xA;" OwnerUserId="1804" LastActivityDate="2011-02-08T02:38:36.917" />
  <row Id="4788" PostTypeId="1" CreationDate="2011-02-08T02:51:47.667" Score="2" ViewCount="666" Body="&lt;p&gt;What causes gravity? Why is there attraction between masses? Is it due to time or space distortion?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-08T07:59:37.380" Title="What causes gravity?" Tags="&lt;gravity&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="4789" PostTypeId="1" AcceptedAnswerId="4811" CreationDate="2011-02-08T03:24:48.797" Score="14" ViewCount="421" Body="&lt;p&gt;I am doing some self-study in between undergrad and grad school and I came across the beastly Wigner-Eckart theorem in Sakurai's quantum mechanics.  I was wondering if someone could tell me why it is useful and perhaps just help me understand a bit more about it.  I have had two years of undergrad mechanics and I think I have a reasonably firm grasp of the earlier material out of Sakurai, so don't be afraid to get a little technical.  Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1807" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-02-08T11:39:33.917" LastActivityDate="2011-02-09T01:03:50.083" Title="What is the usefulness of the Wigner-Eckart theorem?" Tags="&lt;quantum-mechanics&gt;&lt;symmetry&gt;&lt;group-theory&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="10" />
  <row Id="4790" PostTypeId="2" ParentId="4788" CreationDate="2011-02-08T03:42:04.907" Score="1" ViewCount="" Body="&lt;p&gt;I would say, it is rather vague to ask this &quot;why&quot; questions in physics when you are talking about fundamental things. We may say that gravity is a consequence of a deep symmetry of nature, called general covariance, which says that every reference frame should be equally suitable for the description of the laws of physics and it follows that the geometry of the spacetime is affected by the distribution of energy which is described by the Einstein's equations in simplest way. Particles follow straightest line called geodesic in this curved geometry. However this is actually &quot;how&quot; gravity works. What causes gravity may still be a vague question since gravity, as per current wisdom, is a fundamental interaction in nature. You just can not &quot;explain&quot; it from other simpler facts. Some recent attempts try to project it as less fundamental (like gravity is an entropic force). But that is very much controversial.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-08T05:12:30.813" LastActivityDate="2011-02-08T05:12:30.813" CommentCount="6" />
  <row Id="4791" PostTypeId="2" ParentId="4782" CreationDate="2011-02-08T05:41:58.923" Score="2" ViewCount="" Body="&lt;p&gt;I would need to see - and understand - your code but it is not clear from the text whether you also remember and take care of the rotation of the ball - its angular frequency (and the internal angular momentum, which is proportional). I am convinced that this is totally necessary to get some realistic dynamics and it's where most of the pinball fun is all about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In typical situations, the ball is rolling on the table without any &quot;creeping&quot; but this condition still allows the spin around the axis transverse to the table. This ball obtains this spin when it's hit (or when it collides) from the side, and when it hits another object while it's (the ball is) spinning, it will reflect in a direction that is influenced by the spin. Tennis players etc. are using the spin to confuse the opponent all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, I am not sure whether the flipper gives the ball a fixed torque - as opposed to a fixed &quot;power&quot; in the normal direction (transverse to the flipper's tangent near the contact point with the ball) or some other function or a fixed &quot;velocity&quot; in the normal direction (doesn't it just achieve that the ball's normal velocity away from the flipper ends up being equal to the flipper's high velocity?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, I think that the degrees of freedom in your approximation are really simple. If the ball moves along the plane and never jumps, then there are 2 components of the velocity and 3 components of the angular momentum. Two components of the angular momentum are, however, determined by the condition that the ball is rolling not creeping. The y velocity is accelerating as you said and the spin is conserved unless the ball is in the contact in which case you have to calculate the right force and torque.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-08T05:41:58.923" />
  <row Id="4792" PostTypeId="2" ParentId="4784" CreationDate="2011-02-08T05:56:39.613" Score="6" ViewCount="" Body="&lt;p&gt;The symmetry between the exchange of the electric and magnetic fields would become more complete. Indeed, this is not a proof that the magnetic monopoles exist. Moreover, the symmetry is broken, for example because the mass and charges of the two types of charged particles substantially differ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the actual state-of-the-art reasons why magnetic monopoles have to exist were explained e.g. in our discussion with Carlo Rovelli below my answer to this question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/3618/can-maxwells-equations-be-derived-from-coulombs-law-and-special-relativity/3620&quot;&gt;Can Maxwell's equations be derived from Coulomb's Law and Special Relativity?&lt;/a&gt;&lt;br&gt;&#xA;  &lt;strong&gt;Can Maxwell's equations be derived from Coulomb's law and special relativity?&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In any consistent theory of quantum gravity, one may pair-create at least black holes that have a nonzero magnetic flux through the event horizon; the two black holes have the opposite value, of course. There exists an instanton solution that guarantees that such a process has a nonzero probability. This solution can't be &quot;banned&quot; or &quot;denied&quot; without a rather severe violation of the laws of locality. It implies that there exist magnetically charged monopole states - light black holes - that are as heavy as the Planck scale (the lightest black holes worth the name). But they can be lighter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In grand unified theories (GUT), one may write down completely regular, smooth solutions that put the old singular solution by Dirac for the &quot;Dirac monopole&quot; explained in another answer on a firmer ground. Such Dirac monopoles have to be as heavy as the GUT scale which is lighter than the Planck scale. All the intuition in this and previous paragraphs is field-theoretical in character but it is confirmed by our only theory whose reach goes beyond field theory, namely by string theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would bet that these arguments may still sound too mathematical to you but they are &lt;em&gt;extremely good science&lt;/em&gt;. The true logic behind physics &lt;em&gt;is&lt;/em&gt; based on mathematics and the mathematics &lt;em&gt;is&lt;/em&gt; beautiful for those who constantly adopt Nature's sense of beauty as more knowledge is being collected.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-08T05:56:39.613" />
  <row Id="4793" PostTypeId="2" ParentId="4789" CreationDate="2011-02-08T06:08:35.140" Score="2" ViewCount="" Body="&lt;p&gt;The Wigner-Eckart Theorem&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Wigner_Eckart_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Wigner_Eckart_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is a formula that tells us about all &quot;simple constraints&quot; that group theory - the mathematical incarnation of the wisdom about symmetries, especially in the $SO(3)\approx SU(2)$ case (rotations in a three-dimensional space) - implies about matrix elements of tensor operators - those that transform as some representation of the same symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dependence on the indices labeling basis vectors of the representations - $m_1, m_2, m_3$ in the $SU(2)$ case - is totally determined. Instead of thinking that some matrix elements depend on many variables, physicists may realize that the symmetry guarantees that the matrix elements only depend on a few labels labeling the whole &quot;multiplets&quot; of the states and operators rather than on all the labels identifying the individual components. It's always critically important to know how much freedom or how much uncertainty there is about some observables - for example, experimenters don't want to repeat their experiment $(2J+1)^3$ times without a good reason - and we would get a totally wrong idea without this theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that the theorem is used all the time, check e.g. these 5700 papers&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://scholar.google.com/scholar?q=wigner-eckart+theorem&amp;amp;hl=en&amp;amp;lr=&amp;amp;btnG=Search&quot; rel=&quot;nofollow&quot;&gt;http://scholar.google.com/scholar?q=wigner-eckart+theorem&amp;amp;hl=en&amp;amp;lr=&amp;amp;btnG=Search&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;many of which are highly cited ones. The topics of the papers include optics, nanotubes, X-rays, spectroscopy, condensed matter physics, mathematical physics involving integrable systems, quantum chemistry, nuclear physics, and virtually all other branches of physics that depend on quantum mechanics. In many other cases, the theorem is being used without mentioning its name, and its generalizations are being used all the time in advanced theoretical physics, e.g. in the contexts with groups that are much more complicated than $SO(3)\approx SU(2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's interesting to note that in some sense, quantum mechanics allows the symmetry to impose as many constraints as classical physics. In classical physics, we could start with an initial state labeled by numbers $I_i$, apply some operations depending on parameters $O_j$, and we would obtain a final state described by parameters $F_k$. Classical physics would tell us Yes/No - whether we can get from $I_i$ via $O_j$ to $F_k$: that's the counterpart of the quantum probability amplitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rotational $SO(3)$ symmetry - which has 3 parameters (3 independent rotations - or the latitude; longitude of the axis, and the angle) would only tell us that if we rotate all objects $I_i,O_j,F_k$ by the same rotation, we obtain a valid proposition again (Yes goes to Yes, No goes to No). So the dependence on 3 parameters - corresponding to the 3 rotations - is eliminated. In quantum mechanics, we also eliminate the dependence on 3 parameters - in this case $m_1,m_2,m_3$, the projections $j_z$ for the two state vectors and for the operator sandwiched in between them. In some proper counting, this is true for any $d$-dimensional group of symmetries, I think.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-08T08:59:20.543" LastActivityDate="2011-02-08T08:59:20.543" />
  <row Id="4794" PostTypeId="2" ParentId="4788" CreationDate="2011-02-08T07:38:17.263" Score="-4" ViewCount="" Body="&lt;p&gt;&quot;gravity, as per current wisdom, is a fundamental interaction in nature&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Science is not about &quot;current wisdom&quot; :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Science is about Hypothesis (read Henri Poincare about Science and Hypothesis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity may be an illusion like people think that heat was like a fluid. It is an illusion because we think in term of &quot;action at distance&quot; even in term of time/space bent whereas the paradigm may be totally different. This guy has an interesting theory &lt;a href=&quot;http://www.youtube.com/watch?v=SbTuFFr61Qk&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=SbTuFFr61Qk&lt;/a&gt; he's not a physicist but a computer software engineer, but his intuition may be right in the end. He wrote a book where he relates Gravity to his Hyperdimensions.&lt;/p&gt;&#xA;" OwnerUserId="1811" LastActivityDate="2011-02-08T07:38:17.263" />
  <row Id="4795" PostTypeId="2" ParentId="4784" CreationDate="2011-02-08T07:41:08.963" Score="0" ViewCount="" Body="&lt;p&gt;I'll bite. Why did the charge monopoles survive inflation in this scheme?&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-02-08T07:41:08.963" CommentCount="2" />
  <row Id="4796" PostTypeId="2" ParentId="4788" CreationDate="2011-02-08T07:59:37.380" Score="0" ViewCount="" Body="&lt;p&gt;Hypothetically, Gravitons could also be attributable to why gravity exists. An explanation of gravitons would be too extensive for this answer box. You can read more on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Graviton&quot; rel=&quot;nofollow&quot;&gt;Gravitons wiki page&lt;/a&gt; or google for more reference on the topic.&lt;/p&gt;&#xA;" OwnerUserId="1812" LastActivityDate="2011-02-08T07:59:37.380" CommentCount="2" />
  <row Id="4797" PostTypeId="2" ParentId="3618" CreationDate="2011-02-08T08:03:21.347" Score="1" ViewCount="" Body="&lt;p&gt;With Coulomb's law and special relativity you can derive Ampere's law, which gives you magnetostatics. What's missing for electrodynamics is the displacement current ($\frac{1}{c^2} \frac{\partial E}{\partial t}$), which is a source of magnetic field arising from time-varying electric field, and not a result of the motion of electric charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relativity has only two postulates:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The laws of physics are the same in all inertial reference frames&lt;/li&gt;&#xA;&lt;li&gt;All inertial observers measure the same speed for light in vacuum.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Relativity, by itself, does not mandate that electric fields (or electric potential for that matter) must travel at the speed of light. To derive the Maxwell equations, you need an additional postulate, and that is provided by the wave equation (for electric potential) in Section 4 of the reference in Helder's answer. Without this additional postulate (that changes in electric potential propagate at the speed of light), you cannot derive the displacement current from Coulomb's law and relativity alone.&lt;/p&gt;&#xA;" OwnerUserId="1462" LastEditorUserId="1462" LastEditorDisplayName="" LastEditDate="2011-02-08T08:11:14.890" LastActivityDate="2011-02-08T08:11:14.890" />
  <row Id="4798" PostTypeId="2" ParentId="4782" CreationDate="2011-02-08T08:19:45.713" Score="1" ViewCount="" Body="&lt;p&gt;When the ball is hit with the flipper, two velocity components should be considered: parallel to the flipper $v_{\parallel}$ and perpendicular to it $v_{\perp}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_{\parallel}$ will change if the ball is spinning - as Luboš said, this definitely should be included in your model. You can start with a crude approximation $\Delta v_{\parallel}\propto \omega$ for $\omega&amp;lt;\omega_0$ and $\Delta v_{\parallel}=const$ otherwise. $\omega_0$ would be the ball angular speed at which it starts to slip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_{\perp}$ changes its sign (elastic collision) and gets additional kick from the flipper: $v_{\perp}=-{\gamma}v_{\perp}+v_{flip}$, where $\gamma$ is elasticity coefficient close to 1 and $v_{flip}=\omega_{flip}r$, $r$ is the distance from the flipper axis to the point where the ball hits it.&lt;/p&gt;&#xA;" OwnerUserId="701" LastActivityDate="2011-02-08T08:19:45.713" />
  <row Id="4799" PostTypeId="1" AcceptedAnswerId="4801" CreationDate="2011-02-08T09:12:39.737" Score="5" ViewCount="260" Body="&lt;p&gt;How are photons counted?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the experimental setup used to count photons from a laser or even a lamp? Of course, in the case of the lamp, I would be able to count only the photons that pass through an area sensor in a particular observation point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If at all possible, a setup that can be done at home is preferred--one that avoids expensive instruments as much as possible (I may be crazy to even think this is possible). The key point is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We can't count photons like we count sheep. So how do we infer from the effects of single photons and count from there?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can start with $E = \frac{hc}{\lambda}$, and all other manifestations of energy, like heat or motion.&lt;/p&gt;&#xA;" OwnerUserId="293" LastEditorUserId="293" LastEditorDisplayName="" LastEditDate="2011-02-08T09:19:29.210" LastActivityDate="2011-02-08T20:46:36.780" Title="How to count photons" Tags="&lt;photons&gt;&lt;quantum-optics&gt;&lt;home-experiment&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="4800" PostTypeId="2" ParentId="4789" CreationDate="2011-02-08T09:14:39.343" Score="8" ViewCount="" Body="&lt;p&gt;First, W-E theorem is just a simple (just bear with me, I know that the theorem can appear formidable if not explained properly) statement about the decomposition of the tensor product of representations into its irreducible components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have a group $G$ and a tensor operator $A_r$ transforming under a(n irreducible) representation $\Gamma^1$ of it ($r$ counting the number of components of the operator, e.g. 3 for the usual angular momentum in 3 dimensions). Also suppose that you have additional (irreducible) representation $\Gamma^2$ with basis $\left\{ \left| \psi_n \right&amp;gt;\right\}$. It is easy to check that then vectors $\Psi_{rn} \equiv A_r \left|\psi_n\right&amp;gt;$ transform as a tensor product $\Gamma^1 \otimes \Gamma^2$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In the following we use that the group $G$ is compact in order to be able to decompose its representations into irreducible components. This can be weakened but in general representations of non-compact groups don't have to be reducible to their irreducible components; there's pretty subtle mathematics behind representation theory of non-compact groups; even seemingly simple ones as $SL(2, \mathbb{R})$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, we'll decompose the representation as $$\Gamma^1 \otimes \Gamma^2 = \bigoplus_{\alpha} \Gamma^{\alpha}$$ where $\alpha$ runs over irreducible representations of $G$ (possibly repeated). This amounts to finding a more suitable basis for vectors $\Psi_{rn}$. We will write $\Phi_{\alpha m}$ for that basis (with $m$ indexing the components of representation $\Gamma^{\alpha}$). Then we can write $$\Psi_{rn} = \sum_{\alpha, m} U^{\alpha m}_{rn} \Phi_{\alpha m}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, back to the problem: we are interested in computing some element such as $\left&amp;lt;\omega_k \right| A_r \left | \psi_n\right&amp;gt;$ with $\omega_k$ transforming in some $\Gamma^3$ representation. Thanks to &lt;a href=&quot;http://en.wikipedia.org/wiki/Schur_orthogonality_relations&quot; rel=&quot;nofollow&quot;&gt;Schur orthogonality relations&lt;/a&gt; and the fact that we decomposed $A_r \left | \psi_n\right&amp;gt;$ into its irreducible components we can see that for this element to be non-zero, there has to $\Gamma^3$ representation in the $\Gamma^{\alpha}$ decomposition. That already saves us a lot of computation. If there is no $\Gamma^3$ present there then we're finished, all those elements will be zero. And if it is present, we only have to carry out calculations corresponding to just the $\Gamma^3$ part of the decomposition (which will usually be a small part of the total).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, I think the above might have been a little confusing, so let's try an example. The canonical one is with $SO(3)$ of course. Suppose we want to compute something like&#xA;$\left&amp;lt;j m\right| \mathbf X \left|j&amp;#39; m&amp;#39;\right&amp;gt;$ with $\mathbf X$ being the position operator (which transforms under $SO(3)$ vector irrep). So we are interesting of doing a tensor product ${\mathbf 3} \otimes (\mathbf{2j +1})$ which can be shown to be equal to $(\mathbf{2j-1}) \oplus (\mathbf {2j+1}) \oplus (\mathbf {2j+3})$  (supposing $j$ is high enough for simplicity). So $j&amp;#39;$ has to be equal to one $j-1, j, j+1$ and moreover $m = m_X + m&amp;#39;$ (if $\mathbf X$ is written in a diagonal basis of the vectorial representations one can decompose it into operators having eigenvalues of m_X = $-1, 0, 1$) will need to hold (this is again a consequence of orthogonality relations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the most important point (at least for me) is about the &lt;strong&gt;decomposition of the tensor product into irreducible components&lt;/strong&gt;. Supposing you can carry out this decomposition (which can be pretty hard at times), the calculation of matrix elements will simplify greatly and you can immediately decide stuff such as whether some molecule with a given symmetry will radiate in IR (which amounts to computing matrix elements of a dipole operator between eigenstates of that molecule). I am sure you can imagine lots of other similar stuff you may compute yourself. Applications of this theorem are pretty much limitless.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-02-08T09:21:41.507" LastActivityDate="2011-02-08T09:21:41.507" CommentCount="4" />
  <row Id="4801" PostTypeId="2" ParentId="4799" CreationDate="2011-02-08T09:42:46.857" Score="6" ViewCount="" Body="&lt;p&gt;A single photon can easily be detected by a &lt;a href=&quot;http://en.wikipedia.org/wiki/Photomultiplier&quot; rel=&quot;nofollow&quot;&gt;photomultiplier&lt;/a&gt;.  The basic idea is that a photon hitting a metal plate in the tube ejects an electron from the metal plate by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Photoelectric_effect&quot; rel=&quot;nofollow&quot;&gt;photoelectric effect&lt;/a&gt;.  An electric field inside the photomultiplier then accelerates the electron until it slams into another metal plate, releasing a bunch of electrons.  These are then accelerated to a third metal plate, etc.  The end result is a sizable current we can measure.  The Wikipedia article is quite good and has more detail.  You might be able to build a crude one at home with a lot of dedication, but it's a delicate device requiring a vacuum and quality electronics.  These devices work for IR to UV light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also measure individual photons with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Scintillation_counter&quot; rel=&quot;nofollow&quot;&gt;scintillation counter&lt;/a&gt;.  I used these in a couple of undergraduate labs to detect x-ray radiation from nuclear processes.  They work by detecting when a photon (usually high-energy) ionizes an atom in some particular substrate, so they're tuned to detect photons at certain ranges of frequencies.  The scintillator does not directly detect these photons, but converts them to several lower-energy photons that we can detect with other means to infer the high-energy photons' presence, so you'll need some more electronics to go with it.  Still, we were able to watch single-photon events get counted in lab.  (Thanks dmckee for clarification in comments).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;rod&quot; photoreceptors in your eye &lt;a href=&quot;http://en.wikipedia.org/wiki/Photoreceptor_cell#Advantages&quot; rel=&quot;nofollow&quot;&gt;may be able to detect single photons&lt;/a&gt;.  So you can detect single photons at home without any equipment, under the right circumstances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One device you can fairly easily build at home is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cloud_chamber&quot; rel=&quot;nofollow&quot;&gt;cloud chamber&lt;/a&gt;, but it will detect mostly $\alpha$ and $\beta$ radiation rather than photons.  However, you might see trails from high-energy photons (gamma radiation).  There should be lots of sets of instructions on the web for how to build one.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-02-08T20:46:36.780" LastActivityDate="2011-02-08T20:46:36.780" CommentCount="5" />
  <row Id="4802" PostTypeId="2" ParentId="4799" CreationDate="2011-02-08T09:47:16.980" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;We can't count photons like we count sheep.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With photomultipliers counting of photons is possible,&#xA;(maybe there are semiconducting devices today as well?)&#xA;Of course those detectors will count the photons passing &#xA;through their sensitive area at best. There is a quantum &#xA;yield of course, a certain detector eg might count 90 % of &#xA;the photons falling onto it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several problems of course, one is attanuation. &#xA;Such detectors have a dead time during/after a count. &#xA;So, the photon stream has to be small enough, to avoid &#xA;more than one photon falling in within that period. &#xA;When this is the case, one can insert some attanuating filter &#xA;between light source and detector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second is the problem of most lamps radiating unevenly in &#xA;many/all directions of space. For this problem one will &#xA;use an averaging/spatial integrating device, called &#xA;&quot;&lt;a href=&quot;http://www.ise.fraunhofer.de/areas-of-business-and-market-areas/applied-optics-and-functional-surfaces/areas-of-competence/characterization-methods/angle-dependent-reflection-and-transmission-measurements-with-a-large-ulbricht-sphere&quot; rel=&quot;nofollow&quot;&gt;Ulbrichts sphere&lt;/a&gt;&quot;. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-02-08T11:18:44.243" LastActivityDate="2011-02-08T11:18:44.243" />
  <row Id="4803" PostTypeId="2" ParentId="4799" CreationDate="2011-02-08T10:25:42.080" Score="4" ViewCount="" Body="&lt;p&gt;Here is a reference for counting photons in a cavity with very sophisticated equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hal.archives-ouvertes.fr/docs/00/16/56/29/PDF/Gerlin_et_al_2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://hal.archives-ouvertes.fr/docs/00/16/56/29/PDF/Gerlin_et_al_2.pdf&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good CCD camera basically works at the single photon level.&lt;br&gt;&#xA;Amateur astronomers who photograph faint stars by stacking images are working at this level.&#xA;You can get more info from Sky &amp;amp; Telescope or Astronomy magazines.&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-02-08T10:25:42.080" CommentCount="3" />
  <row Id="4804" PostTypeId="2" ParentId="3150" CreationDate="2011-02-08T12:29:26.297" Score="2" ViewCount="" Body="&lt;p&gt;Luboš would know this already (he's acknowledged in this paper), but &lt;a href=&quot;http://arxiv.org/abs/hep-th/0402128&quot; rel=&quot;nofollow&quot;&gt;Neitzke and Vafa&lt;/a&gt; conjectured in 2004 that the mirror manifold of $CP^{3|4}$ is a quadric surface $Q$ in $CP^{3|3}$ x $CP^{3|3}$, and mirror symmetry is a type of T-duality. There were a few follow-ups, including a paper by &lt;a href=&quot;http://arxiv.org/abs/hep-th/0410014&quot; rel=&quot;nofollow&quot;&gt;Sinkovics and Verlinde&lt;/a&gt; which studies classical $N=4$ super-Yang-Mills on the quadric, which in the very last paragraph asks whether the quantum scattering amplitudes can also be recovered from $Q$. After that, I can find nothing. But at least it's a place to begin! &lt;/p&gt;&#xA;" OwnerUserId="1486" LastActivityDate="2011-02-08T12:29:26.297" CommentCount="1" />
  <row Id="4805" PostTypeId="1" AcceptedAnswerId="4809" CreationDate="2011-02-08T12:32:55.293" Score="6" ViewCount="241" Body="&lt;p&gt;I was reading a post about space mining, specially lunar mining. I was thinking about what would change in Earth's orbit if we start bringing tons of rocks to it? I mean, in a huge scale.&#xA;So, would space mining change Earth's orbit in a way to produce any dangerous effects?&#xA;If the system in question is not a planet/satellite like Earth/Moon, would that change anything? Would bringing tons of rocks from Mars be any different than Moon?&lt;/p&gt;&#xA;" OwnerUserId="1815" LastActivityDate="2011-02-08T13:44:12.880" Title="Effects of space mining on Earth's orbit" Tags="&lt;gravity&gt;" AnswerCount="4" />
  <row Id="4807" PostTypeId="2" ParentId="4805" CreationDate="2011-02-08T13:03:00.757" Score="1" ViewCount="" Body="&lt;p&gt;The earth accumulates cosmic dust at about 40 tons per day (according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_dust#Detection_methods&quot; rel=&quot;nofollow&quot;&gt;Wikipedia: Cosmic Dust&lt;/a&gt;). This seems to have no effect on gravity and the orbit of the earth around the sun. For space mining to have an effect, the mined amount will have to be significantly larger.&lt;/p&gt;&#xA;" OwnerUserId="1816" LastActivityDate="2011-02-08T13:03:00.757" CommentCount="1" />
  <row Id="4808" PostTypeId="2" ParentId="4805" CreationDate="2011-02-08T13:25:34.657" Score="1" ViewCount="" Body="&lt;p&gt;The redistribution of mass due to space activities is very small.  To the extent there is any space mining which happens it is that we mine out materials from the Earth and over the last several decades sent somewhere around 10 thousand tons of material into space.  Given that the mass of the Earth is about $6\times 10^{21}$ metric tons the mass of stuff we have sent into space is a very small percentage.  The perturbation is tiny,  Any mining from space is also going to be a tiny perturbation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commercial activities in space have been with massless forms, in particular with information.  The next possible step might be with solar power satellites.  Again this is a massless form.  Sending spacecraft into space to mine materials is not likely to work any time in the near future.  The Saturn Apollo rocket weighed in at 3000 tons of highly processed materials, which brought back 50kg of lunar rocks. From an economic or ratio of materials and energy in verses materials and energy back this clearly is a loss.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-08T13:25:34.657" CommentCount="2" />
  <row Id="4809" PostTypeId="2" ParentId="4805" CreationDate="2011-02-08T13:30:00.713" Score="3" ViewCount="" Body="&lt;p&gt;No, there would be no detectable - and surely no dangerous - changes to the Earth's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for the sake of an argument, imagine that we double our coal reserves by bringing coal from another place - and even some of the precious metals are too expensive to be brought by spaceships at this moment. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth's coal reserves are something like 1 trillion tons which is $10^{15}$ kg. Let's bring this amount from another celestial body - it's about 10 orders of magnitude less than what we can do now but let's imagine we double our coal reserves in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth's mass is $6\times 10^{24}$ kg, so the coal reserves are approximately $10^{-10}$ of the Earth's mass. Now, if all the extraterrestrial coal landed by a slow speed relatively to the Earth, the Earth's velocity wouldn't change at all; only the mass would increase and the heavier Earth would continue along exactly the same trajectory as before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now, imagine that the coal lands at some speed, e.g. the safe speed that the space shuttles used to take. It's about $350$ km/h. If you don't get approximately this low, there's a risk that your coal will burn in the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $10^{-10}$ of the Earth's mass has a relative velocity that is $350$ km/h, and let's imagine that all the momentum will go in the same direction - we could make the coal space shuttles land at different places if we wanted - then the velocity of the Earth will change by $350\times 10^{-10}$ km/h which is $3.5\times 10^{-5}$ m/h or $10^{-8}$ m/s. The speed of Earth around the Sun is approximately $3\times 10^{4}$ m/s, so we only change it by a trillionth. Correspondingly, the eccentricity of the Earth's orbit could change at most by one trillionth. We would have a hard time to detect this change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, you would need to increase the amount of resources you bring roughly by 10 orders of magnitude (which means by 20 orders of magnitude relatively to what we can achieve today) to produce any threat for the Earth. But even if you brought the whole Moon here to Earth, $7\times 10^{22}$ kg (eight orders of magnitude heavier than the Earth's coal reserves), there would be no significant change of the orbit because the speed of the Moon is approximately the same as the speed of the Earth - they're bound together. Well, the shape of the Earth could change a bit if we tried to incorporate the Moon too quickly. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would have to bring a big fraction of Mars to the Earth (Mars is both heavier and has a substantially different speed) to change the eccentricity of the Earth's orbit by a significant amount and I assure you that this will remain in the realm of science fiction for many, many centuries if not forever. If you brought 1/3 of Mars to the Earth, you would also have to build mountains that are 1000 kilometers high - more than 100 times Mount Everest, around the whole Earth. And it would still not be too dangerous as far as the orbital characteristics go. Of course, there could be a danger for the people who suddenly have 1000 kilometers of rock above their heads. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question clearly seems to be an artifact of the unscientific doomsday scenarios that have been presented as science in recent years - e.g. the doomsday caused by CO2 in the atmosphere. Even relatively to the atmosphere, our additions of CO2 are  negligible - we're changing the number of molecules in the atmosphere by 2 parts per million (0.0002%) per year. But the atmosphere is just one millionth of the Earth's mass, so our annual CO2 emissions only redistribute something like 2 parts per trillion of the Earth's mass every year. Clearly, all those changes are irrelevant from a &quot;mechanical&quot; viewpoint and they're arguably irrelevant from a climatic viewpoint, too.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-08T13:44:06.613" LastActivityDate="2011-02-08T13:44:06.613" CommentCount="1" />
  <row Id="4810" PostTypeId="2" ParentId="4805" CreationDate="2011-02-08T13:33:14.773" Score="1" ViewCount="" Body="&lt;p&gt;The answers so far miss the key &lt;strong&gt;physical&lt;/strong&gt; aspect: planetary orbit does not depend on the mass of the planet (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion&quot; rel=&quot;nofollow&quot;&gt;Kepler's laws&lt;/a&gt;) because planet's mass is much smaller then the mass of the Sun. This means that even if you &lt;em&gt;double&lt;/em&gt; Earth's mass by mining Mars completely, it will stay on the same orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: this is assuming that Earth's velocity stays the same, but then, if people will be able to move these enormous masses, they could change Earth's orbit at will. &lt;/p&gt;&#xA;" OwnerUserId="701" LastEditorUserId="701" LastEditorDisplayName="" LastEditDate="2011-02-08T13:44:12.880" LastActivityDate="2011-02-08T13:44:12.880" CommentCount="2" />
  <row Id="4811" PostTypeId="2" ParentId="4789" CreationDate="2011-02-08T13:53:11.747" Score="6" ViewCount="" Body="&lt;p&gt;I will not get into theoretical details -- Luboš ad Marek did that better than I'm able to.&lt;br&gt;&#xA;Let me give an example instead: suppose that we need to calculate this integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int d\Omega (Y_{3m_1})^*Y_{2m_2}Y_{1m_3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $Y_{lm}$ -- are &lt;a href=&quot;http://en.wikipedia.org/wiki/Spherical_harmonics&quot; rel=&quot;nofollow&quot;&gt;spherical harmonics&lt;/a&gt; and we integrate over the sphere $d\Omega=\sin\theta d\theta d\phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This kind of integrals appear over and over in, say, spectroscopy problems. Let us calculate it for $m_1=m_2=m_3=0$:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int d\Omega (Y_{30})^*Y_{20}Y_{10} = \frac{\sqrt{105}}{32\sqrt{\pi^3}}\int d\Omega \cos\theta\,(1-3\cos^2\theta)(3\cos\theta-5\cos^3\theta)=$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ = \frac{\sqrt{105}}{32\sqrt{\pi^3}}\cdot 2\pi \int d\theta\,\left(3\cos^2\theta\sin\theta-14\cos^4\theta\sin\theta+15\cos^6\theta\sin\theta\right)=\frac{3}{2}\sqrt{\frac{3}{35\pi}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hard work, huh? The problem is that we usually need to evaluate this for all values of $m_i$. That is 7*5*3 = 105 integrals. So instead of doing all of them we got to exploit their symmetry. And that's exactly where the Wigner-Eckart theorem is useful:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int d\Omega (Y_{3m_1})^*Y_{2m_2}Y_{1m_3} = \langle l=3,m_1| Y_{2m_2} | l=1,m_3\rangle = C_{m_1m_2m_3}^{3\,2\,1}(3||Y_2||1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{m_1m_2m_3}^{j_1j_2j_3}$ -- are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Clebsch%E2%80%93Gordan_coefficients&quot; rel=&quot;nofollow&quot;&gt;Clebsch-Gordan coefficients&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(3||Y_2||1)$ -- is the reduced matrix element which we can derive from our expression for $m_1=m_2=m_3=0$:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{3}{2}\sqrt{\frac{3}{35\pi}} = C_{0\,0\,0}^{3\,2\,1}(3||Y_2||1)\quad \Rightarrow \quad (3||Y_2||1)=\frac{1}{2}\sqrt{\frac{3}{\pi}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the final answer for our integral is:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int d\Omega(Y_{3m_1})^*Y_{2m_2}Y_{1m_3}=\sqrt{\frac{3}{4\pi}}C_{m_1m_2m_3}^{3\,2\,1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is reduced to calculation of the Clebsch-Gordan coefficient and there are a lot of, &lt;a href=&quot;http://en.wikipedia.org/wiki/Table_of_Clebsch-Gordan_coefficients&quot; rel=&quot;nofollow&quot;&gt;tables&lt;/a&gt;, programs, reduction and summation formulae to work with them. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-02-08T13:53:11.747" CommentCount="3" />
  <row Id="4812" PostTypeId="1" AcceptedAnswerId="4818" CreationDate="2011-02-08T16:36:02.760" Score="1" ViewCount="82" Body="&lt;p&gt;A useful way of thinking (not only) oceanic waves is to consider them as a superimposition of linear modes: the elevation η of the sea surface is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: $\eta({\bf x}, t) = \int_{-\infty}^{\infty} \hat{\eta}({\bf k}) e^{i({\bf k \cdot x} - \omega t)} d{\bf k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\omega$ and ${\bf k}$ are related by a dispersion relation, and $\hat{\eta}({\bf k})$ is a sort of time independent Fourier transform. $\hat{\eta}$ is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: $\hat{\eta}({\bf k}) = const \int_{-\infty}^{\infty} e^{-i{\bf k \cdot x}}(\eta({\bf x}, 0) + \frac{i}{\omega({\bf k})} \frac{\partial \eta({\bf x}, 0)}{\partial t}) d{\bf x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;meaning that if we know $\eta({\bf x}, 0)$ and its time derivative we can determine the amplitude of all the normal modes, and neglecting non linear interactions between the modes we know the evolution of the wave. Clear and smart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the simples way to obtain relation 2 from relation 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="312" LastActivityDate="2011-02-08T17:30:09.430" Title="Time evolution of wave spectrum" Tags="&lt;fluid-dynamics&gt;&lt;waves&gt;&lt;mathematics&gt;" AnswerCount="1" />
  <row Id="4813" PostTypeId="2" ParentId="4784" CreationDate="2011-02-08T16:47:43.500" Score="0" ViewCount="" Body="&lt;p&gt;Magnetic Field is not a fundamental entity as we can see in the document of Hans de Vries (*):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://physics-quest.org/Magnetism_from_ElectroStatics_and_SR.pdf&quot; rel=&quot;nofollow&quot;&gt;The simplest, and the full derivation of Magnetism as a Relativistic side efect of ElectroStatics&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He uses Electrostatic field and the non-simultaneity to obtain the Magnetic Field. &#xA;Thus it is a side effect of movement in the Electrostatic field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beautiful, in my opinion, is Nature simplicity that forces us to reduce to fewer laws, as time goes by. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) Hans de Vries has a very interesting online book 'Understanding &#xA;Relativistic Quantum Field Theory' (not yet finished) in his site&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-02-08T16:47:43.500" />
  <row Id="4814" PostTypeId="2" ParentId="4784" CreationDate="2011-02-08T17:01:30.493" Score="0" ViewCount="" Body="&lt;p&gt;If the $U(1)$ electromagnetism (more accurately, hypercharge) gauge group is compact (as opposed to $\mathbb{R}$), and there's an ultraviolet regularization of the gauge theory in space, then for purely topological reasons, magnetic monopoles have to exist.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-08T17:01:30.493" />
  <row Id="4815" PostTypeId="1" CreationDate="2011-02-08T17:13:50.567" Score="5" ViewCount="229" Body="&lt;p&gt;Quantum electrodynamics based upon Euler-Heisenberg or Born-Infeld Lagrangians predict photons&#xA;to move according to an effective metric which is dependent on the background electromagnetic&#xA;field. In other words, photon trajectories are curved in presence of electromagnetic fields,&#xA;meaning that an effective gravity is acting upon. If part of fermion masses is allegedly of&#xA;electromagnetic origin, the question why their trajectories are not affected by this&#xA;effective gravity naturally comes to mind.&lt;/p&gt;&#xA;" OwnerUserId="1823" LastEditorUserId="1823" LastEditorDisplayName="" LastEditDate="2011-02-10T21:34:34.537" LastActivityDate="2011-03-12T22:32:46.187" Title="Why can't fermions be affected by effective gravity in non-linear quantum electrodynamics?" Tags="&lt;quantum-gravity&gt;&lt;quantum-optics&gt;&lt;quantum-electrodynamics&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="4816" PostTypeId="1" AcceptedAnswerId="4819" CreationDate="2011-02-08T17:16:00.620" Score="5" ViewCount="226" Body="&lt;p&gt;I recall from my undergraduate statistical mechanics and QM classes that Planck's Law may be derived fairly straight-forwardly by considering the density of states of EM radiation in thermal equilibrium with a particle inside a cubic box. Indeed the derivation on &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck%27s_law#Derivation&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; wiki page is just the same and appears (IMO) to be fairly concise and accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wondered therefore, that if we invoke all the magic of QFT (such as the fields being promoted to operators and higher (than tree-level) order particle interactions taken into account,) can a similar derivation still be carried out? (Or indeed a dissimilar one?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Associated to which, would such a derivation (if it is even possible) require consideration of QED alone or would the effects of the broken electro-weak symmetry have a significant effect on recreating the correct radiation spectrum?&lt;/p&gt;&#xA;" OwnerUserId="1696" LastActivityDate="2011-02-09T20:29:18.937" Title="Is there a fully quantum-field theoretic treatise of Planck's Law for black-body radiation?" Tags="&lt;quantum-field-theory&gt;&lt;statistical-mechanics&gt;&lt;quantum-mechanics&gt;&lt;planck&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="4817" PostTypeId="1" CreationDate="2011-02-08T17:27:34.223" Score="0" ViewCount="67" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/4788/what-causes-gravity&quot;&gt;What causes gravity?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;If gravity is mainly the warp of space around a large object, that is curves in space time; how does that give weight to an object held in your hand? There should be a reason it falls. I hope i worded this right because it's driving me crazy. I have read some Richard Feynman and Brian Green books but I have little math. &lt;/p&gt;&#xA;" LastActivityDate="2011-02-08T17:27:34.223" ClosedDate="2011-02-08T17:48:24.517" Title="Why does an object seem to have weight" Tags="&lt;general-relativity&gt;" CommentCount="0" />
  <row Id="4818" PostTypeId="2" ParentId="4812" CreationDate="2011-02-08T17:30:09.430" Score="2" ViewCount="" Body="&lt;p&gt;It's the inverse Fourier transform which is used for all waves, not just ocean waves. The simplest way to obtain (2) from (1) is to guess (2) and prove that it's right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to prove (2) given (1). Just calculate $\partial\eta(\vec x, t)/\partial t$. You will get the same formula as (1) except that there will be an extra coefficient of $-i\omega=\omega/i$ inserted in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you're really ready to prove (2). Substitute (1) instead of $\eta(\vec x,t)$, but use $\vec k&amp;#39;$ for the integration variable in (1), instead of $\vec k$, so that it doesn't get confused with $\vec k$ in (2). The dependence of the integral on $\vec x$ will be&#xA;$$\exp[-i(\vec k-\vec k&amp;#39;)x]$$&#xA;and if you integrate this over $d^3 x$, you will get the famous delta-function:&#xA;$$(2\pi)^3 \delta^{(3)}(\vec k-\vec k&amp;#39;)$$&#xA;Because we're integrating over $\vec k&amp;#39;$ and the integral contains a delta-function, we may replace $\vec k&amp;#39;$ by $\vec k$ everywhere. So (2) reduces to &#xA;$$\hat \eta(\vec k) = const \cdot (2\pi)^3\cdot \hat\eta(\vec k)\cdot e^{-i\omega t} (1+1)$$&#xA;(in the second term, $\omega/\omega$ and the $i$ factors canceled so we got one) which is indeed correct for &#xA;$$const =\frac{e^{i\omega t}}{2(2\pi)^3}$$&#xA;As you can see, it would be enough to use one of the terms and avoid the time derivative. However, this is just an artifact of your oversimplification of equation (1). The equation (1) as you wrote it only contains terms with &quot;positive values of $\omega$&quot; so $\eta(\vec x)$ clearly can't be real. If you wrote (1) as the real part of what you did, you would get some extra terms with the opposite $\omega$ but the same $k$ which wouldn't cancel, and you would have to get rid of them by hand. Adding the time-derivative with the right coefficient, so that they contribute equally $(1+1)$ to the &quot;right term&quot;, is exactly the right way to cancel the wrong term. But I invite you to check it yourself.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-08T17:30:09.430" CommentCount="1" />
  <row Id="4819" PostTypeId="2" ParentId="4816" CreationDate="2011-02-08T18:07:39.733" Score="3" ViewCount="" Body="&lt;p&gt;The derivation of the Planck distribution takes into account the quantization of the electromagnetic field: any particular wave-vector of the field can have quantized energies, which are interpreted as the number of photons times the energy of each one of them. At the time Planck derived his formula, this is all he needed. He had no formulation of a physical theory in which the electromagnetic field was quantized, but he showed that when you manage to get one, that quantization alone will be sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum field theory, this is built-in as the starting point: you promote each field to an operator, and when you &quot;second quantize&quot;* you satisfy the requirement for the deriving the Planck spectrumm - the energy in each mode of the electromagnetic field is quantized. I believe second quantization of the Maxwell field (no need to add electrons to get QED) can be found in every text on the subject. After you do that, the derivation of the Planck spectrum is not all that different from Planck's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is only the starting point though. Once you have QFT you can add interactions and calculate corrections to the spectrum and any other thermodynamic quantity - due to QED, electroweak symmetry breaking, and what have you. Those will be normally small corrections**. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the spectrum, the corrections comes effectively from adding interactions between the photons. The single photon stays massless (because of gauge invariance) and therefore its spectrum does not get modified. But the energy of two photons is no longer the sum of their individual energies. This should correct the measured spectrum in a small way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As a side comment: those calculations in thermal field theory are not that simple, as compared to usual loop integrals you get at zero temperature.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Just for the record, I strongly dislike this term for other reasons, but here is seems appropriate.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;** It is interesting whether or not such corrections have been measured - thermal effects of QED. I don't know the answer to that.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-02-09T20:29:18.937" LastActivityDate="2011-02-09T20:29:18.937" CommentCount="8" />
  <row Id="4820" PostTypeId="1" AcceptedAnswerId="4825" CreationDate="2011-02-08T18:48:50.150" Score="2" ViewCount="93" Body="&lt;p&gt;Is 89MHZ station emitting photons of 89MHZ frequency? (I mean $\nu$ in $E=h\nu$).&lt;/p&gt;&#xA;" OwnerUserId="58" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-08T19:18:16.483" LastActivityDate="2011-02-08T19:18:16.483" Title="Radio waves and frequency of photon" Tags="&lt;photons&gt;&lt;frequency&gt;&lt;radio&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="4821" PostTypeId="1" AcceptedAnswerId="4826" CreationDate="2011-02-08T18:53:15.240" Score="9" ViewCount="223" Body="&lt;p&gt;A photon emitted from a receding source (Doppler redshift) has less energy when detected at an observer's location. Please explain the energy loss from the perspective of energy conservation.&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-02-08T19:27:25.407" Title="Redshifted Photon Energy" Tags="&lt;energy-conservation&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="4822" PostTypeId="2" ParentId="4821" CreationDate="2011-02-08T18:59:27.207" Score="2" ViewCount="" Body="&lt;p&gt;If it is a gravitational redshift, to a first, non-rigorous order, the energy loss is due to the fact that it is moving in a gravitational field, and thus is gaining potential energy while losing kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is a redshift due to the actual motion of the object, then the energy lost in the redshift is imparted to the object doing the emitting since energy and momentum are conserved in the emission process--it is an energy transfer due to recoil.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-02-08T18:59:27.207" CommentCount="6" />
  <row Id="4823" PostTypeId="2" ParentId="4820" CreationDate="2011-02-08T19:01:33.593" Score="1" ViewCount="" Body="&lt;p&gt;Well, of course! But what is that (h(v)&amp;lt;-  ??&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-02-08T19:01:33.593" CommentCount="1" />
  <row Id="4825" PostTypeId="2" ParentId="4820" CreationDate="2011-02-08T19:16:18.717" Score="2" ViewCount="" Body="&lt;p&gt;Oneat, yes, and I understood what you meant &quot;I mean this one&quot;. ;-) You meant that it's the frequency $\nu$ (nu) that appears in the formula $E=h\nu$ for the photon energy, as opposed to $\omega$ that appears in $E=\hbar\omega$ and that differs by a factor of $2\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever you hear &quot;$\mbox{Hz}$&quot; or any multiple of it, it means that the frequency is defined in the same convention as the frequency in $E=h\nu=hf$, without the slashed $\hbar$ (hbar). If someone wanted to express the angular frequency $\omega=2\pi\nu$, he would have to write the unit as $\mbox{s}^{-1}$ instead of $\mbox{Hz}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dimensionally, $\mbox{Hz}$ is the same thing as $\mbox{s}^{-1}$. However, people deliberately use both units so that the inverse second is reserved for the angular frequency $\omega$ and the Hertz is reserved for the old-fashioned frequency $\nu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nu=89 \mbox{MHz}$ is the same thing as $\omega=5.6\times 10^{8} \mbox{s}^{-1}$.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-08T19:16:18.717" />
  <row Id="4826" PostTypeId="2" ParentId="4821" CreationDate="2011-02-08T19:27:25.407" Score="8" ViewCount="" Body="&lt;p&gt;Consider the following scenario: I am on a train moving away from you. I throw a ball to you. The speed of the ball as measured by you when you catch it, is less than the speed of the ball as measured by me when I threw it. Where did the energy go?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This situation is precisely the same as the Doppler shift situation you describe. In both cases, there's no problem with energy conservation, because the energies in question are measured in two different reference frames. Energy conservation says that, in any given reference frame, the amount of energy doesn't change. It says nothing about how the energy in one frame is related to the energy in another frame.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-08T19:27:25.407" CommentCount="9" />
  <row Id="4827" PostTypeId="1" CreationDate="2011-02-08T20:13:34.883" Score="-2" ViewCount="382" Body="&lt;p&gt;I have a ball attached to a spring and the spring is attached to a wall. There is no gravity for simplicity. In the rest RF the oscillating ball energy is conserved: T + U = const. In a moving RF it is not conserved. I would like to see the shortest answer to the question &quot;Why?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;|           &amp;lt;--  --&gt;&lt;br&gt;&#xA;|/\/\/\/\/\/\/&lt;strong&gt;O&lt;/strong&gt;&lt;br&gt;&#xA;|&lt;/p&gt;&#xA;&#xA;&lt;p&gt;=====&gt;  $V_{RF}--&amp;gt;$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; For those who has doubt - I choose the moving RF velocity equal to the maximum ball velocity, for example. Then the ball with its maximum velocity in the rest RF looks as still and does not have any potential energy in the moving RF (the maximum velocity is attained at the equilibrium position where no force acts). So T = 0 and U = 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the farthest right position the ball looks as moving with $-V_{RF}$ and having a potential energy $kx^2/2$, both energy terms are positive ($v^2 &gt; 0, x^2 &gt; 0$). So now the total energy of the ball is positive and thus is not conserved. It oscillates from zero to some maximum value. Why?&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-30T18:33:05.743" LastActivityDate="2011-08-30T18:33:05.743" ClosedDate="2011-02-08T22:33:14.670" Title="Is the total energy conserved in a moving reference frame?" Tags="&lt;classical-mechanics&gt;" AnswerCount="3" CommentCount="13" />
  <row Id="4828" PostTypeId="2" ParentId="4827" CreationDate="2011-02-08T21:01:35.027" Score="3" ViewCount="" Body="&lt;p&gt;Energy is &lt;strong&gt;not&lt;/strong&gt; a Lorentz scalar (or Galilean scalar). In different reference frames, the values of energies wil also be different, but this does not mean that there is no energy conservation. The energy is still conserved in &lt;strong&gt;each&lt;/strong&gt; reference frame.&lt;/p&gt;&#xA;" OwnerUserId="1545" LastActivityDate="2011-02-08T21:01:35.027" CommentCount="2" />
  <row Id="4829" PostTypeId="2" ParentId="4827" CreationDate="2011-02-08T21:14:43.707" Score="1" ViewCount="" Body="&lt;p&gt;Suppose that the wall has some large but finite mass. Then the actual motion of the system involves oscillations of the wall as well as oscillations of the mass. If you work out the total energy of the entire system, you'll find that it's conserved.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-08T21:14:43.707" CommentCount="12" />
  <row Id="4830" PostTypeId="1" AcceptedAnswerId="4835" CreationDate="2011-02-08T21:30:39.963" Score="5" ViewCount="174" Body="&lt;p&gt;First, I should point out that this question was raised by a particle physics Professor whose lessons I attended last year. I don't recall exactly how the question was phrased so if anyone would like to suggest an edit feel free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, here are some statements that to the best of my knowledge are correct:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In QM and the Standard Model of particle physics all fundamental particles of the same species are completely equivalent and indistinguishable.&lt;/li&gt;&#xA;&lt;li&gt;The wavefunction for every real particle exists throughout the entire Universe. (I'm not sure how to phrase this statement in terms of QFT but I'm pretty sure there is still a non-zero probability of finding any given particle popping up at the edge of the universe.)&lt;/li&gt;&#xA;&lt;li&gt;The ground state of the electron in a hydrogen atom has a unique, universal energy eigenvalue.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My question is therefore if (despite their distinct positions) all electrons in hydrogen atoms have the same energy, and their wavefunctions extend across the entire Universe, how does this not violate the Pauli Exclusion Principle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember my Professor saying the resolution to this was that the electrons must therefore have minutely differing energies; either agreed between them from the first instances of the big bang or resulting from the interfering 'tales' of their wavefunction. Clearly the first option doesn't explain hydrogen atoms formed more recently so the second would seem more promising. I should note that by &quot;minute difference&quot; I think he said something like the order of 10^-20 eV so well out of the realms of potential measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like I said I may not have remembered and phrased this question correctly. Please don't crucify me for any 'school-boy' mistakes but constructive criticism in the comments would be appreciated to help me phrase it better.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastActivityDate="2011-06-24T03:04:41.230" Title="Does particle indistinguishability and quantised enery levels (in bound states) violate the Pauli Exclusion Principle?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="4831" PostTypeId="1" AcceptedAnswerId="4834" CreationDate="2011-02-08T21:47:51.763" Score="5" ViewCount="315" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Give me a lever long enough and a&#xA;  fulcrum on which to place it, and I&#xA;  shall move the world.&#xA;                &lt;br&gt;-Archimedes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How long would that lever have to be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is to say, how long a lever would be needed on Earth, to lift a sphere with the mass of Earth if a human of average size were to sit on the other side of the lever?&lt;/p&gt;&#xA;" OwnerUserId="1829" LastEditorUserId="1829" LastEditorDisplayName="" LastEditDate="2011-02-08T22:01:00.503" LastActivityDate="2011-08-15T05:19:42.720" Title="How long would a lever have to be to move the world?" Tags="&lt;mechanics&gt;" AnswerCount="3" CommentCount="11" />
  <row Id="4832" PostTypeId="2" ParentId="4827" CreationDate="2011-02-08T21:57:51.070" Score="3" ViewCount="" Body="&lt;p&gt;We must consider the kinetic energy of the wall to see energy conservation.  Taking the limit where the wall's mass goes to infinity does not alleviate this responsibility because although the wall's oscillations become infinitely small in that case, the oscillations in its energy do not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this roughly, imagine that the wall is very massive and moving sinusoidally.  In the center of mass frame, the maximum speed of the wall is $s$.  Then the fluctuations in its kinetic energy are proportional to $M s^2$.  Thus, in the rest frame, we can make the wall's energy fluctuations arbitrarily small by increasing $M$, because $s^2$ will decrease more rapidly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now go into a moving frame where the average speed of the wall is $v$.  The fluctuations in its energy are now proportional to $M(v+s)^2 - Mv^2 \approx 2Mvs$ in the limit $v&amp;gt;&amp;gt;s$.  For fixed $v$, we can make $s$ smaller and smaller by increasing $M$, but as we do so the fluctuations in the energy of the wall don't change size because $s \propto 1/M$.'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next note that the fluctuations in the energy of the ball in this frame are proportional to $mvs&amp;#39;$, with $s&amp;#39;$ the speed of the ball in this frame.  By conservation of momentum, $Ms = ms&amp;#39;$, so the fluctuations in energy of the ball and wall are about the same size in the moving reference frame.  Thus, we must account for both of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll give a fuller answer in one dimension for simplicity.  I'll first find the motion in the center of mass frame, then transform it and show that energy is conserved in the tranformed frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the ball have mass $m$ and the wall have mass $M$.  The spring constant is $k$.  Let the ball's position from equilibrium be $x$ and the wall's position be $X$.  The Lagrangian is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{L} = \frac{1}{2}\left(M\dot{X}^2 + m\dot{x}^2 - k(X-x)^2\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations of motion are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ -k(X-x) = M\ddot{X}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k(X-x) = m\ddot{x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding these gives the conservation of momentum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\textrm{d}}{\textrm{d}t}(M\dot{X} + m\dot{x}) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the center of mass frame this momentum is zero, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{x} = \frac{M\dot{X}}{m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the initial condition $x(0) = X(0) = 0$ we can integrate this to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X = -\frac{m}{M}x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation of motion for the small mass becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k(\frac{1}{M} + \frac{1}{m}) x = \ddot{x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with the solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = A \sin(\omega t + \phi)$$&#xA;$$X = -\frac{m}{M}A\sin(\omega t + \phi)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\omega^2 = k(\frac{1}{M} + \frac{1}{m})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now boost to a frame moving at speed $v$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = A \sin(\omega t + \phi) - vt$$&#xA;$$X = -\frac{m}{M}A\sin(\omega t + \phi) - vt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The potential energy is in the spring&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U = \frac{1}{2}k(X-x)^2 = \frac{1}{2}k \left(A (1+\frac{m}{M})(\sin(\omega t + \phi) \right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinetic energy is in the motion of the ball and wall&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T = \frac{1}{2} \left(A \omega \cos(\omega t + \phi)\right)^2 \left(m + M(\frac{m}{M})^2\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add these and applying the identity $\sin^2x+\cos^2x = 1$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = \frac{1}{2}k\left(A(1+\frac{m}{M})\right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is constant in time.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-02-08T22:10:55.457" LastActivityDate="2011-02-08T22:10:55.457" CommentCount="8" />
  <row Id="4833" PostTypeId="2" ParentId="4830" CreationDate="2011-02-08T21:59:51.037" Score="2" ViewCount="" Body="&lt;p&gt;The Pauli exclusion principle is a direct &lt;strong&gt;consequence&lt;/strong&gt; of indistinguishability, so there is no way to indistinguishable particle violate this rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have two or more indistinguishable particles, the total wavefunction is an antisymmetric combination of single-particles wavefunctions. The overlap of these wavefunctions will decrease the total energy in an amount known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Exchange_interaction&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;exchange energy&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exchange energy is important for the electronic structure calculations of atoms, molecules and solids, but is not so important when there are a large distance between these electrons.&lt;/p&gt;&#xA;" OwnerUserId="1545" LastActivityDate="2011-02-08T21:59:51.037" />
  <row Id="4834" PostTypeId="2" ParentId="4831" CreationDate="2011-02-08T22:14:37.430" Score="4" ViewCount="" Body="&lt;p&gt;The mass of Earth is $6\times10^{24}kg$. If Archimedes can lift 60 kg, he would need a lever with an arm ratio of $10^{23}:1$. So if the short arm is one meter long, the lever length will be $10^{23}$ meters plus one. Also, note that he would have to push the lever for $10^{20}$ meters to shift the Earth just by one millimeter.&lt;/p&gt;&#xA;" OwnerUserId="701" LastEditorUserId="701" LastEditorDisplayName="" LastEditDate="2011-02-08T22:31:07.213" LastActivityDate="2011-02-08T22:31:07.213" CommentCount="6" />
  <row Id="4835" PostTypeId="2" ParentId="4830" CreationDate="2011-02-08T22:46:56.560" Score="2" ViewCount="" Body="&lt;p&gt;Electrons are indistinguishable and the Pauli Exclusion Principle applies to them. Theoretical solution for the 1s wavefunction of an electron in hydrogen atom $\Psi_{1s}=Ne^{-r/a}$ is calculated neglecting all particles in the world except one electron and one proton. If another electron exists in the same quantum state, both would avoid each other and the wavefunction would be distorted leading to a change in energy &lt;strong&gt;of a total system&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I emphasize it because - remember - electrons are indistinguishable and interaction of two hydrogen atoms cannot lead to electron A having lower energy than electron B. This does not contradict PEP &amp;mdash; the principle applies to quantum states, not potential energies. Both hydrogen atoms will exhibit the same shift of electron energy levels. It may help to visualize the system as almost dissociated hydrogen molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the wavefunction decays much faster than the dipole electrostatic potential of an atom, so for distances larger than $a$ (Bohr radius, $0.53 A $) the effect of wavefunction overlap is negligible compared to electrostatic interactions with other atoms.&lt;/p&gt;&#xA;" OwnerUserId="701" LastEditorUserId="701" LastEditorDisplayName="" LastEditDate="2011-06-24T03:04:41.230" LastActivityDate="2011-06-24T03:04:41.230" CommentCount="7" />
  <row Id="4836" PostTypeId="2" ParentId="4789" CreationDate="2011-02-09T01:03:50.083" Score="2" ViewCount="" Body="&lt;p&gt;And if you are interested in &quot;where&quot; it might come useful: Some of the selection rules for optical transitions can be obtained from it, and I faintly recall that it helps rewriting the Hamiltonian for Spin-Orbit coupling into a much more convenient form.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-02-09T01:03:50.083" />
  <row Id="4837" PostTypeId="1" CreationDate="2011-02-09T01:05:50.933" Score="2" ViewCount="244" Body="&lt;p&gt;I was wondering, within a static train, for a fly which is currently floating in the air and not moving at all, what will happen if:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The train starts to accelerate and move forward. Will the floating fly experience inertia and being pushed backward?&lt;/li&gt;&#xA;&lt;li&gt;If the train is completely evacuated (without air), will the floating fly experience inertia too?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1833" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-10T10:09:50.233" LastActivityDate="2011-02-10T10:09:50.233" Title="What happens to a fly floating in the air when the train starts moving" Tags="&lt;forces&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="4838" PostTypeId="2" ParentId="4837" CreationDate="2011-02-09T01:12:57.933" Score="3" ViewCount="" Body="&lt;p&gt;The fly will experience only whatever forces are acting on it, which is &lt;em&gt;none&lt;/em&gt; in a vacuum. The same way a person on ice skates would not experience any forces on him as the train moved forward and he remained stationary. That is because there is no horizontal component of the contact force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW: A fly cannot fly in a vaccum.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-02-09T01:12:57.933" CommentCount="2" />
  <row Id="4839" PostTypeId="2" ParentId="4641" CreationDate="2011-02-09T01:19:36.017" Score="1" ViewCount="" Body="&lt;p&gt;I'll show you an interactive diagram which lets you vary the temperature of a blackbody and see its effect on the radiated wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But before you go there, recall that the visible spectrum is within the range of about $0.4$ - $0.8$ $\mu m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the &lt;a href=&quot;http://www.pveducation.org/pvcdrom/properties-of-sunlight/blackbody-radiation&quot; rel=&quot;nofollow&quot;&gt;page with the diagram&lt;/a&gt;. As you increase the temperature, you'll find that the emitted radiation peaks over the range of wavelengths of visible light. So you're correct that the blackbody emits virtually the entire rainbow at very high temperatures.&lt;/p&gt;&#xA;" OwnerUserId="293" LastActivityDate="2011-02-09T01:19:36.017" />
  <row Id="4841" PostTypeId="1" AcceptedAnswerId="12762" CreationDate="2011-02-09T03:25:13.127" Score="4" ViewCount="458" Body="&lt;p&gt;I am aware about different interpretations of quantum mechanics out there but would mostly like to see an answer from the perspective of Copenhagen interpretation (or relative quantum mechanics if you wish).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let an observer being a man with brain consisting of molecules and atoms. According the basic principles of quantum mechanics each of these particles has a wave function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: is there a combined wave function of all those particles which constitute the observer? Can such wavefunction be (in theory) determined by the observer himself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the observer cannot isolate himself from his own brain, this would mean that the wave function, at least the part which determines his thoughts is permanently collapsed (i.e. the measurement happens instantly once the state changes). Does this &quot;permanently collapsed&quot; wave function imply special physical properties of the observer's own brain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does knowing his own thoughts constitute a measurement? Which moment should be counted as the moment of the collapse of wave function when making measurements on own brain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretend an observer tries to measure the wave function of his own brain by a means of an X-ray apparatus or other machinery and read his own thoughts. Would not his own knowledge of that measurement or its results invalidate the results thus making the whole measurement impossible? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the behavior of particles which constitute the observer's brain differ statistically (acoording his measurements) from the behavior of particles which constitute the brains of other people?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a connection with quantum immortality here?&lt;/p&gt;&#xA;" OwnerUserId="1186" LastEditorUserId="1186" LastEditorDisplayName="" LastEditDate="2011-02-09T03:32:55.470" LastActivityDate="2011-07-28T05:27:43.230" Title="What is the wavefunction of the observer himself?" Tags="&lt;quantum-mechanics&gt;&lt;measurement&gt;" AnswerCount="6" CommentCount="4" FavoriteCount="2" />
  <row Id="4842" PostTypeId="2" ParentId="4841" CreationDate="2011-02-09T04:05:17.877" Score="0" ViewCount="" Body="&lt;p&gt;A measurement is a type of entanglement.  If you measure a photon at the opening of a slit in a double slit experiment then you can think of there being a spin state at this opening.  If the photon passes through it the spin changes direction.  So the wave function for the photon in the double slit experiment is&#xA;$$&#xD;&#xA;\psi(x)\rangle~=~C(e^{ikx}|1\rangle~+~e^{ik(x’)}|2\rangle)&#xD;&#xA;$$&#xA;where the x and x’ denote the different paths through the two slits. If you compute the modulus squared you get&#xA;$$&#xD;&#xA;\langle\psi|\psi\rangle~=~|C|^2(2~+~e^{ik(x’-x)}\langle 1|2\rangle~+~ e^{-ik(x&amp;#39;-x)}\langle 2|1\rangle)&#xD;&#xA;$$&#xA;The exponential terms give the interference between the superposed states $|1\rangle$ $|2\rangle$ which have a nonzero overlap $\langle 1|2\rangle~\ne~0$.  We now consider the coupling of a spin to this&#xA;$$&#xD;&#xA;\psi(x)\rangle~\rightarrow~C(e^{ikx}|1\rangle|+\rangle~+~e^{ik(x’)}|2\rangle|-\rangle)&#xD;&#xA;$$&#xA;which serves as the detector. Since $\langle +|-\rangle~=~0$ if you compute $\langle\psi|\psi\rangle$ the interference term is gone.  This gives a reason for why one can’t measure which slit the particle travels through.  We have replaced a superposition of states type of nonlocality with an entanglement.&#xA;Now in doing this we have not addressed the question of how one actually observes the spin.  One might presume we entangle some other states and do so up some “chain.”  This of course leads to the Schrodinger cat problem.  A cat or a human being, our brains and the like are not described well as single quantum systems.  In fact they are messy thermal systems with high entropy.  This is one problem with the whole idea of quantum consciousness, which never got out of the starting gates as a serious physical problem.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-09T04:05:17.877" CommentCount="3" />
  <row Id="4843" PostTypeId="2" ParentId="4830" CreationDate="2011-02-09T04:10:14.740" Score="2" ViewCount="" Body="&lt;p&gt;First, the wave states for the electrons of atom A are different from the wave states for the electrons of atom B; the two waves have their maxima at two different regions. So there's no violation of the HUP for them (actually in combining them into a single multi-particle wave function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what you're really supposed to do is to compute an overall wave function for all the electrons. If you have 8 electrons, this means having 8 copies of the universe (x,y,z,t) coordinates. And the wave function is antisymmetric under the exchange of two electrons.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-09T04:10:14.740" />
  <row Id="4844" PostTypeId="1" AcceptedAnswerId="4871" CreationDate="2011-02-09T05:10:58.543" Score="2" ViewCount="251" Body="&lt;p&gt;We've previously discussed why it is that spinning tops do not fall over, see: &lt;a href=&quot;http://physics.stackexchange.com/questions/271/why-dont-spinning-tops-fall-over&quot;&gt;Why don't spinning tops fall over?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as the highest rated answer notes, the angular momentum of the spinning top is &quot;quite high&quot;. On the other hand, I know that if the angular momentum of the top is zero it will, in fact, fall over. This suggests that increasing the angular momentum cannot stop the top from falling over, but instead can only increase the time (but this turns out not to be the case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So suppose a top with mass concentrated in a spherical body M of radius S with center situated a distance R to a point a horizontal table at an angle $\theta$ while rotating at angular speed $\omega$. How long does it take for the top to fall over?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some simplifying assumptions: (1) Assume the point of the top is fixed on the table. (2) Assume that the time at which the top falls over is when $\theta = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Ch8dP.jpg&quot; alt=&quot;Drawing&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-02-09T22:41:04.597" LastActivityDate="2011-02-10T11:36:46.893" Title="How long for a frictionless top to fall over?" Tags="&lt;classical-mechanics&gt;&lt;angular-momentum&gt;&lt;rotational-dynamics&gt;" AnswerCount="3" CommentCount="11" />
  <row Id="4846" PostTypeId="2" ParentId="4815" CreationDate="2011-02-09T07:29:58.913" Score="1" ViewCount="" Body="&lt;p&gt;In the presence of a background electromagnetic field, electromagnetic fields travel along a deformed light cone which is smaller than the &quot;relativistic light cone&quot;. However, charged fermions can still travel faster than electromagnetic waves as long as they are still slower than the &quot;relativistic speed of light&quot;. They emit Cherenkov radiation while doing so. &lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-09T07:29:58.913" CommentCount="3" />
  <row Id="4847" PostTypeId="1" AcceptedAnswerId="4848" CreationDate="2011-02-09T09:15:05.233" Score="4" ViewCount="370" Body="&lt;p&gt;As I read in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0679454438&quot; rel=&quot;nofollow&quot;&gt;The Road to Reality&lt;/a&gt; by &lt;a href=&quot;http://en.wikipedia.org/wiki/Roger_Penrose&quot; rel=&quot;nofollow&quot;&gt;Roger Penrose&lt;/a&gt;, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Joukowsky_transform&quot; rel=&quot;nofollow&quot;&gt;Joukowsky transform&lt;/a&gt; $$w(z) = \frac12\left( z + \frac1z \right)$$&#xA;after &lt;a href=&quot;http://en.wikipedia.org/wiki/Nikolai_Zhukovsky&quot; rel=&quot;nofollow&quot;&gt;Nikolai Zhukovsky&lt;/a&gt; (transcribed in several versions from &lt;a href=&quot;http://ru.wikipedia.org/wiki/%D0%96%D1%83%D0%BA%D0%BE%D0%B2%D1%81%D0%BA%D0%B8%D0%B9,_%D0%9D%D0%B8%D0%BA%D0%BE%D0%BB%D0%B0%D0%B9_%D0%95%D0%B3%D0%BE%D1%80%D0%BE%D0%B2%D0%B8%D1%87&quot; rel=&quot;nofollow&quot;&gt;Никола́й Его́рович Жуко́вский&lt;/a&gt;) can be used to &lt;strong&gt;calculate the flow&lt;/strong&gt; of a non-viscious, incompressible and irrotational flow &lt;strong&gt;around an airfoil&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be done since the solution of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Potential_flow_around_a_circular_cylinder&quot; rel=&quot;nofollow&quot;&gt;potential flow around a cylinder&lt;/a&gt; is known in full analyticity and the given transform &lt;a href=&quot;http://en.wikipedia.org/wiki/Conformal_map&quot; rel=&quot;nofollow&quot;&gt;conformally maps&lt;/a&gt; a circle on an airfoil-like geometry. I don't understand this argumentation, so:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h3&gt;How is the Joukowsky Transform used to calculate the Flow of an Airfoil?&lt;/h3&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An &lt;strong&gt;example&lt;/strong&gt; of such a transformation is given in the mentioned Wikipedia article:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/d/d5/Jouktrans.png&quot; alt=&quot;RotatingDisc&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="1501" LastEditorDisplayName="" LastEditDate="2011-02-09T09:26:29.483" LastActivityDate="2011-02-09T13:00:52.980" Title="How is the Joukowsky Transform used to calculate the Flow of an Airfoil?" Tags="&lt;fluid-dynamics&gt;&lt;hydrodynamics&gt;&lt;aerodynamics&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="4848" PostTypeId="2" ParentId="4847" CreationDate="2011-02-09T09:33:55.577" Score="4" ViewCount="" Body="&lt;p&gt;The crux of the argument is that we can treat complex analytic (holomorphic) functions as functions in 2D, and their real and imaginary parts (separately) are solutions of Laplace's equation ($\nabla^2 \psi = 0$), due to the Cauchy-Riemann condition. Conformal maps such as the one you cite map analytic functions to analytic functions, i.e generate new solutions from old ones. Thus, by knowing a trivial solution (such as around the cylinder), we can generate the flow around a new object by finding a conformal map to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relevant details:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Laplace's equation solves potential flow problems: incompressible, inviscid, curl-free flow (though we are allowed rotational flow around finite objects --- the resulting singularity is technically outside of the domain).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We can somewhat relax the need for fully holomorphic functions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Such mappings tend to mess up your boundary conditions at infinity --- so it may be quite hard in general to find such mappings.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-02-09T09:33:55.577" CommentCount="4" />
  <row Id="4849" PostTypeId="1" CreationDate="2011-02-09T10:03:37.063" Score="5" ViewCount="783" Body="&lt;p&gt;I am in a debate with a friend about the value of string theory in physics. He is concerned that we are wasting valuable intellectual and financial resources on a path that is fanciful and can't ever hope to be verified by experiment and evidence (11 - 20 dimensions etc). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My point is not to agree with string theory but to argue that maths is powerful and capable of produce new ideas that can be verified. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;His question to me is&#xA;&quot;Can math can provide new and viable ideas without an original idea based on the natural world or through observation&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't think of any good examples where the study of maths has resulted in new theory about physics. Can anyone?&lt;/p&gt;&#xA;" OwnerUserId="1320" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-10T00:40:07.053" LastActivityDate="2011-03-02T04:46:26.323" Title="Can pure maths create new theories in physics or does the &quot;idea&quot; ALWAYS come before the math?" Tags="&lt;soft-question&gt;&lt;mathematics&gt;" AnswerCount="13" CommentCount="11" FavoriteCount="1" />
  <row Id="4850" PostTypeId="2" ParentId="4289" CreationDate="2011-02-09T10:06:55.253" Score="4" ViewCount="" Body="&lt;p&gt;This question has introduced me to the whole &quot;entropic force&quot; area which has several papers during 2010. I see that there are references to &quot;entropic force&quot; explanations for Coulomb's law and other areas too. Here is a &lt;a href=&quot;http://www.science20.com/hammock_physicist/it_bit_entropic_gravity_pedestrians&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; to a simple introduction to these ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Verlinde paper and others however are deriving Newtons Law, Einstein's GR etc as &lt;em&gt;classical&lt;/em&gt; theories. The underlying formulation of course being a stochastic behaviour of unknown microstates. Despite the presence of $\hbar$ and the motivation from the Black Hole area formulae the Verlinde paper does not introduce an explicit link with quantum mechanics. Thus there is no derivation of Schrodinger's equation and no introduction of $\Psi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Kobakhidze paper says &quot;One starts with a &quot;holographic screen” S which contains macroscopically large number of microscopic states which we denote as $|i(z)&amp;gt;i, i(z) = 1, 2, ...,N(E(z), z).$ The screen is then described by the mixed state&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho(z)=\Sigma p_{i(z)}|i(z)&amp;gt;&amp;lt;i(z)|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However Verlinde does not explicitly introduce microstates as quantum states, with density matrices etc, although this is a tempting extension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it might be that this is the only sensible quantum development of the stochastic basis of the &quot;entropic idea&quot;, but Verlinde has not taken it. So what is disproved is a theory that Verlinde has not written down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said this, there is a resemblance between &quot;entropy&quot; and the idea of introducing &quot;stochastics&quot; into quantum theory. One such attempt is known as &quot;Stochastic Quantum Electrodynamics&quot; described in this Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Stochastic_electrodynamics&quot; rel=&quot;nofollow&quot;&gt;overview&lt;/a&gt;. As you will see from the summary this has had successes with e.g. the Unruh effect, but problems modelling genuine quantum phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I dont know whether anyone has considered combining the two areas directly.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-09T10:06:55.253" CommentCount="1" />
  <row Id="4851" PostTypeId="2" ParentId="4844" CreationDate="2011-02-09T10:10:10.367" Score="1" ViewCount="" Body="&lt;p&gt;I think, once launched, the top will never fall over. Such are kinematic restrictions in this problem. For example, $\theta = const$ if it exceeds the minimum angle necessary for the top to be detached from the table. Of course, I imply that the initial conditions on the angular momentum are such that $\theta &amp;gt; \theta_{min}$.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-02-09T10:10:10.367" CommentCount="2" />
  <row Id="4852" PostTypeId="2" ParentId="4837" CreationDate="2011-02-09T10:13:58.600" Score="1" ViewCount="" Body="&lt;p&gt;The fly &#xA;will feel a wind blow when the train accelerates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Re: Vacuum: A &quot;neighbor&quot; next table in &quot;Quantitative Lab&quot; &#xA;(2nd semester Chemistry, 1968) did some experiment, &#xA;wich I remember vividly. He placed a fly in a large &#xA;(about 5 ltrs) desiccator and started to evacuate. &#xA;Results: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;at about 100 to 50 mmHg the fly fell down, but revived on &#xA;aeration again. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;at about 20 to 50  mmHg the fly was damaged irreversibly &#xA;(so called death) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-02-09T12:28:53.680" LastActivityDate="2011-02-09T12:28:53.680" CommentCount="4" />
  <row Id="4853" PostTypeId="2" ParentId="4849" CreationDate="2011-02-09T10:17:57.393" Score="7" ViewCount="" Body="&lt;p&gt;Mathematical develoments can lead to new physics, this is how anti-particles were predicted by Dirac. See also &lt;a href=&quot;http://physics.stackexchange.com/questions/4141/why-beauty-is-a-good-guide-in-physics&quot;&gt;Why beauty is a good guide in physics?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-02-09T10:17:57.393" CommentCount="3" />
  <row Id="4854" PostTypeId="2" ParentId="4849" CreationDate="2011-02-09T10:18:56.817" Score="-2" ViewCount="" Body="&lt;p&gt;Mathematical ideas serve to do mathematics. Often some mathematical ideas are related to those one uses in physics; then one deals with the &quot;mathematical physics&quot;. The true theory that describes some experimental facts should be based on them. In other words, any physical theory should be phenomenological first of all. Otherwise it is a mathematical physics branch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is rather practical. Those who think they are capable of inventing a Theory of Our Everything (TOE) are involved in a mad rush for fame and Nobel prize and they make too many groundless promises. It became difficult to raise a voice of reason. Even clear failures are represented as &quot;achievements&quot; or &quot;insights&quot; now.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-09T16:25:24.107" LastActivityDate="2011-02-09T16:25:24.107" />
  <row Id="4855" PostTypeId="2" ParentId="4849" CreationDate="2011-02-09T10:21:01.880" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Physics without math is blind and math without physics is lame.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mostly mathematics and physics goes hand in hand. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Group theory &lt;/li&gt;&#xA;&lt;li&gt;Tensor calculus&lt;/li&gt;&#xA;&lt;li&gt;Stokes theorem in Vector calculus &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These mathematical theories came first and then utilized later on by physicists. &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2011-02-09T10:21:01.880" CommentCount="8" />
  <row Id="4856" PostTypeId="2" ParentId="4847" CreationDate="2011-02-09T10:31:15.653" Score="1" ViewCount="" Body="&lt;p&gt;A classic textbook on this topic from the perspective of Complex Analysis is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Churchill: Complex Variables and Applications. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chapter 8 of that book discusses Fluid Flow (amongst other applications). The aerofoil transform itself is introduced in some Exercises, although the earlier theory describes the assumptions. Of course the basic assumption is that the fluid plane x-y describes all the relevant properties, since it is a 2-D model.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-09T10:31:15.653" CommentCount="1" />
  <row Id="4857" PostTypeId="2" ParentId="4849" CreationDate="2011-02-09T10:40:20.060" Score="1" ViewCount="" Body="&lt;p&gt;Speaking of Dirac, he also came up with the Dirac equation, creation and annihilation operators, magnetic monopoles, and the theory of constraints, in particular second class constraints and the Dirac bracket.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-09T10:40:20.060" CommentCount="4" />
  <row Id="4858" PostTypeId="2" ParentId="4841" CreationDate="2011-02-09T10:52:17.177" Score="-3" ViewCount="" Body="&lt;p&gt;Look in the mirror and you will see your wave functions squared ;-).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, I am joking. In the mirror you see an inclusive picture, not elastic one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The observer himself consist of so many degrees of freedom with so tiny differences between energetic levels that it is impossible to keep it in a pure state. There are permanent transitions leading to decoherence.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-09T11:05:05.293" LastActivityDate="2011-02-09T11:05:05.293" CommentCount="12" />
  <row Id="4859" PostTypeId="2" ParentId="4849" CreationDate="2011-02-09T11:22:05.653" Score="3" ViewCount="" Body="&lt;p&gt;The question is where you draw the line separating the &quot;original idea about the natural world&quot; from ensuing mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think a particularly good example is non-euclidean geometry and Gauss: After Gauss understood that the axiom of paralles is indeed an axiom that can be replaced by other axioms, and does not follow from the other axioms, he tried to measure the angles of large triangles during is work on land surveying. He wanted to find out if physical space is Euclidean or if it is not, motivated by the pure mathematical insight that other geometries are possible.&#xA;From our viewpoint the question was incomplete, of course, Riemann and Gauss should have thought about the geometry of spacetime instead of that of space only, but I think it is safe to say that these purely mathematically motivated line of reasoning paved the path for Einstein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is Maxwell's prediction of electromagnetic waves and their velocity, which he got from analyzing his equations, but in this case you could of course argue that his equations were extracted from observational research done by Faraday and others.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-02-09T11:22:05.653" />
  <row Id="4860" PostTypeId="2" ParentId="4849" CreationDate="2011-02-09T12:28:38.830" Score="5" ViewCount="" Body="&lt;p&gt;This question is the analogous of : Is Romeo and Juliet the result of grammar and syntax or is there an external input necessary? In my opinion mathematics as far as physics goes, is a tool. A beautiful tool, tools are very important for creating stuff, but still physics is a meta level to mathematics. Mathematics is necessary for rigorous physics but not sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is of course the school of Pythagoras, &quot;everything is the music of the spheres&quot; can be replaced by &quot; everything can be described with mathematics&quot;,  and in this point of view mathematics comes first. In this view everything exists in potentia as a mathematical concept waiting to be born.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Theory of Everything is found, maybe the latter will be true. Until then, I vote that physics uses mathematics as a necessary tool.  I suspect that Godel's theorem ( can there be a TOE?) in some form or another will somehow still set mathematics in a tool position, necessary but not sufficient. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-02-09T12:28:38.830" CommentCount="4" />
  <row Id="4861" PostTypeId="2" ParentId="4847" CreationDate="2011-02-09T13:00:52.980" Score="2" ViewCount="" Body="&lt;p&gt;Check this reference:&#xA;&lt;a href=&quot;http://www.grc.nasa.gov/WWW/K-12/airplane/map.html&quot; rel=&quot;nofollow&quot;&gt;http://www.grc.nasa.gov/WWW/K-12/airplane/map.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be sure that your browser supports java -- worth it. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-02-09T13:00:52.980" CommentCount="1" />
  <row Id="4862" PostTypeId="1" CreationDate="2011-02-09T14:02:23.020" Score="1" ViewCount="436" Body="&lt;p&gt;Does polarized light interfere?&lt;/p&gt;&#xA;" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-09T17:18:44.503" LastActivityDate="2011-03-07T08:59:37.230" Title="Interference of polarized light" Tags="&lt;optics&gt;&lt;light&gt;&lt;polarization&gt;&lt;interference&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="4863" PostTypeId="2" ParentId="4862" CreationDate="2011-02-09T14:15:39.107" Score="-1" ViewCount="" Body="&lt;p&gt;Yes, it does and this property is independent from a particular polarization. So non polarized light gives the same interference pattern.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-09T16:22:03.800" LastActivityDate="2011-02-09T16:22:03.800" CommentCount="3" />
  <row Id="4864" PostTypeId="2" ParentId="4849" CreationDate="2011-02-09T14:44:02.157" Score="5" ViewCount="" Body="&lt;p&gt;Perhaps the most drastic mathematical concept that was developed without any reference to physics is the idea of imaginary numbers, which came to their full glory in physics only through quantum mechanics. This is one of the reasons why Feynman called the Euler identity $e^{i\pi}+1=0$ &quot;the most remarkable formula in mathematics&quot;. &#xA;See also Feynman's popular book &quot;QED: The Strange Theory of Light and Matter&quot;, which places emphasis on complex numbers and their physical importance in quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more recent example would be the Atiyah-Singer index theorem, which started out as pure math and now is a valuable tool in physics (see also the question &lt;a href=&quot;http://physics.stackexchange.com/questions/1858/where-is-the-atiyah-singer-index-theorem-used-in-physics&quot;&gt;Where is the Atiyah-Singer index theorem used in physics?&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-02-09T14:44:02.157" />
  <row Id="4865" PostTypeId="2" ParentId="4844" CreationDate="2011-02-09T14:55:32.960" Score="3" ViewCount="" Body="&lt;p&gt;Given a nonzero moment of inertia $I$, the transverse extension of the top from the rod has to be at least $\sqrt{I/M}$ because we can't have negative mass distributions. Because of this finite transverse extension, a spin-orbit cross coupling between the orbital angular momentum $L$ and the spin $S$ will have to exist. But this by itself isn't sufficient to transfer spin into orbital angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crucial question to ask is, is the distribution of mass of the rod rotationally symmetric about the rod axis? If yes, the kinetic and gravitational potential energy doesn't depend upon the orientation angle $\alpha$ of the top about the rod axis. However, if there are slight rotational asymmetries, we will have an $\alpha$ dependence leading to wobbles. These wobbles transfer spin into orbital angular momentum, and if sufficient angular momentum is transfered, the top will topple.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-09T14:55:32.960" CommentCount="3" />
  <row Id="4866" PostTypeId="2" ParentId="4862" CreationDate="2011-02-09T15:03:16.697" Score="6" ViewCount="" Body="&lt;p&gt;Yes. In fact, light will only interfere with light of the same polarization. If you take a &lt;a href=&quot;http://en.wikipedia.org/wiki/Mach%E2%80%93Zehnder_interferometer&quot; rel=&quot;nofollow&quot;&gt;Mach–Zehnder interferometer&lt;/a&gt;, for example, and put a polarization rotating optic (a waveplate) in one of the arms, the interference pattern will lose contrast. If the polarization is rotated 90 degrees, the pattern will vanish completely.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-02-09T15:03:16.697" CommentCount="2" />
  <row Id="4867" PostTypeId="2" ParentId="4849" CreationDate="2011-02-09T15:20:39.253" Score="11" ViewCount="" Body="&lt;p&gt;It is implicit in the wording of the question that string theory is an example of math coming first, but this is false. String theory grew out of Regge theory which was and is a phenomenological theory of the strong interactions which applies to scattering processes at high energies but small momentum transfer. This in turn is connected to the Regge trajectories observed in the spectrum of QCD, that is mesons and baryons tend to lie on straight lines in a plot of angular momentum $J$ vs mass squared $M^2$. This is the same as the spectrum of a relativistic string. This is where string theory came from, a phenomenological model of the strong interactions, not from math.&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T04:46:26.323" LastActivityDate="2011-03-02T04:46:26.323" CommentCount="2" />
  <row Id="4868" PostTypeId="1" AcceptedAnswerId="4873" CreationDate="2011-02-09T15:20:54.260" Score="4" ViewCount="102" Body="&lt;p&gt;I'm looking at Jeans (thermal) escape of hydrogen from the early Earth's atmosphere. I understand how to calculate the rate in (g cm^-2 s^-1) using the number density, average particle velocity, and integrating the tail of the boltzmann distribution, but those units are frustrating me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I convert to mass per second? Multiply by the surface area of the atmosphere? This seems to give me a mass-loss rate that is too low for the Earth (that is, it would seem that the Earth should still have hydrogen).&lt;/p&gt;&#xA;" OwnerUserId="1291" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-09T17:18:22.007" LastActivityDate="2011-02-09T18:17:52.410" Title="Estimating hydrogen loss by Jeans escape" Tags="&lt;thermodynamics&gt;&lt;atmosphere&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4869" PostTypeId="2" ParentId="4849" CreationDate="2011-02-09T15:41:43.650" Score="3" ViewCount="" Body="&lt;p&gt;I think this question is too philosophical and I am not sure whether it is within the scope of this site or not. However, I think the question you have asked is very important and deserve serious discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traditionally mathematics always played the role of a tool. Since nature is written in the language of mathematics any physical idea found its natural expression in it. It has also realized that no pure idea (pure mathematical ideas in case of physics), however beautiful, by itself can lead to any real physical truth. One needs to have empirical knowledge about the world. Science should start with observation, go with it and ultimately suggest/lead us to new realm of experiences. Since a scientific theory is essentially a guess, experimental verification should always be the ultimate judge of a scientific theory. That's how science has progressed always.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, one can not help observing a key feature about the relationship between mathematics and physics. The more physics has advanced the possible scope of mathematical manipulation became ever more constrained. Theories became evermore rigid and unique. You just try to tinker with a minor element of a theory and the whole structure immediately collapses. This has brought a new insight. The insight that in an advanced stage of science one can rely with far more confidence on the purely formal nature of inquiry than when it was primitive. Mathematical consistency may itself become a powerful guide in the search of the laws of nature. Experimental verification of ideas are still the ultimate judge but mathematics may play the role of a very strict judge on initial validation checks of physical ideas.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-02-09T15:41:43.650" CommentCount="0" />
  <row Id="4870" PostTypeId="2" ParentId="4700" CreationDate="2011-02-09T16:33:57.607" Score="1" ViewCount="" Body="&lt;p&gt;There is nothing special about the photon having zero mass. Although zero is the smallest mass any particle can have, it is as good as any other value. In this sense, there is no &lt;strong&gt;mathematical proof&lt;/strong&gt; that the photon &lt;strong&gt;has to have&lt;/strong&gt; zero mass, this is a purely experimental fact. And, to our best knowledge, the photon mass is consistent to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to describe a theory with a zero mass vector in a manifestly relativistic way, you have to have gauge invariance. This is a mathematical fact. As is the fact that if you force this symmetry to be quantum mechanically exact, the mass will not receive quantum corrections (perturbatively, at least). Gauge theories can be shown to have all sorts of other nice features (like IR finiteness, if you sum enough virtual and real diagrams) and that makes us believe that at low energies they are the right theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one would be inverting the logical order within physics if one says that the mass of the photon is zero because EM is described by a gauge theory. EM is described by a gauge theory because the photon has zero mass. There would be no problem with special relativity either. The fact that the maximal velocity is the same as the velocity of light in the vacuum is, again, an experimental fact (equivalent to the one we are discussing here) but by no means necessary by any mathematical theorem.&lt;/p&gt;&#xA;" OwnerUserId="1337" LastEditorUserId="1337" LastEditorDisplayName="" LastEditDate="2011-02-09T17:28:12.660" LastActivityDate="2011-02-09T17:28:12.660" CommentCount="2" />
  <row Id="4871" PostTypeId="2" ParentId="4844" CreationDate="2011-02-09T16:52:21.973" Score="1" ViewCount="" Body="&lt;p&gt;Well if you are really interested in Gyroscopes what about the equations of motion to get us started? Here I have copied some from &lt;a href=&quot;http://www.gyroscopes.org/math2.asp&quot; rel=&quot;nofollow&quot;&gt;http://www.gyroscopes.org/math2.asp&lt;/a&gt; which are in Euler-like coordinates: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;q = nutation angle - nutation rate q'= n&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f = precession angle - precession rate f'= p&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y = spin angle - spin rate y'= s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and primes denote differentiation wrt time. We are assuming that s is constant - this is because there is no friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$SM_x=I_xn&amp;#39;-I_yp^2cosqsinq+I_zpsinq(pcosq+s)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$SM_y=I_y(pncosq)-I_zn(pcosq+s)+I_xpncosq$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$SM_z=I_z(-pnsinq)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expression for $SM_z$ simplifies because of the symmetry in the moments of inertia which in our case with the sphere on a rod R are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I_x=mR^2+2/5mr^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I_y=mR^2+2/5mr^2 = I_x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I_z=2/5mr^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this is an external torque gyroscope - as we are on a table on some gravitational environment. The external torque applies in the x direction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_x=mgRsinq$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that all of these equations are written in terms of the nutation angle q. In these coordinates q=0 when it is upright and $q=90^0$ when it is fallen over (although this is an approximation since it falls over at $q=90-arctan(r/R)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intuition behind your question is that when there is no spin s=0, there is no precession p=0. In this case the &quot;fall&quot; corresponds to a nutation from a small angle $q_0$ down to $q=90^0$ - and this nutation is the only motion. As s&gt;0 these equations will generate a non-zero precession p&gt;0, also the nutation will now reach a maximum $q_{max}$. The requirement is that s is sufficiently large that $q_0-q_{max}&amp;lt;90^0$. If this condition is not met then the top will still hit the surface after some rotation (precession angle f in these coordinates).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one wished to take this further analytically then we could approximate for small initial angle $q_0$. So $sin q_0=q_0$ and $cos q_0=1$. Then we consider the x axis terms for the external torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_x=M_x$ - which holds at all times&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$mgRq_0=I_xn&amp;#39;-I_yp^2q_0+I_zpq_0(p+s)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This quadratic in p will determine it at t=0 in terms of other parameters (but n' might not be a constant - from graphical solutions I expect it to be sinusoidal). Likewise one can determine the &quot;falling over&quot; equation when $q=90^0$. If p=p(s) does not satisfy the &quot;falling over&quot; equation then the gyroscope will continue without falling. If it does we can integrate to determine q and hence q(t).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT: Additional point on Gyroscope Intuition&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comments suggest that the motion described by the equations is not directly in accord with intuition. This is likely to be because a real Gyroscope has some elements of Friction. This friction has the effect of a damping term in the above equations. The friction will be basically a function of s, but n=n(s) through the equations, and the only acceleration term here is n'. Thus the friction, for a realistic Gyroscope, will damp the Nutation oscillations that otherwise occur. This would make the nutation an approximate constant of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Treating nutation as a constant, simplifies the equations by setting n'=n=0 and so q, cos q and sin q become constants. However from an intuition perspective it reduces the number of (Euler) degrees of freedom from 3 to 2. So in a real Gyroscope we might see only the two degrees of freedom: spin and precession. Trying to intuit from this the behaviour in the remaining nutation dimension could be misleading as it ordinarily does not display behaviour in that dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A related physical system which does display nutation is the &quot;spinning plate on a pole&quot; scenario. As the spin of the plates reduces through friction they start to wobble. If a wobble is too great the plate will fall over, even although it is still spinning a little.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-02-10T11:36:46.893" LastActivityDate="2011-02-10T11:36:46.893" CommentCount="1" />
  <row Id="4872" PostTypeId="2" ParentId="4862" CreationDate="2011-02-09T16:53:22.363" Score="7" ViewCount="" Body="&lt;p&gt;Let's do some math in order not to be unsubstantiated.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;1. Perpendicular polarizations.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;First wave $E_{1x} = E_0\,\cos\omega t$,second wave $E_{2y} = E_0\,\cos(\omega t+\Delta)$.&lt;br&gt;&#xA;Where $\Delta$ -- is a phase difference between waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total field: $\vec{E} = E_0\left(\vec{i}\cos\omega t+\vec{j}\cos(\omega t+\Delta)\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intensity: $I_\perp\sim \langle|\vec{E}|^2\rangle = E_0^2\langle\cos^2\omega t+\cos^2(\omega t+\Delta)\rangle$&lt;br&gt;&#xA;where average means: $\langle f(t)\rangle = \frac{1}{T}\int_0^{T}\,dt\,f(t)$,  so that $\langle cos^2(\omega t+\Delta)\rangle = \frac{1}{2}$ for any $\Delta$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally we got $I_\perp\sim E_0^2$, which is independent on the phase difference between the waves.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;2. Parallel polarizations.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;First wave $E_{1x} = E_0\,\cos\omega t$,second wave $E_{2x} = E_0\,\cos(\omega t+\Delta)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total field: $\vec{E} = \vec{i}E_0\left(\cos\omega t+\cos(\omega t+\Delta)\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intensity: $I_\parallel\sim E_0^2\langle\cos^2\omega t+2\cos\omega t\cos(\omega t+\Delta)+\cos^2(\omega t+\Delta)\rangle=E_0^2(1+\cos\Delta)$, which nicely depends on the phase shift between the waves.  &lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2011-03-07T08:59:37.230" LastActivityDate="2011-03-07T08:59:37.230" CommentCount="12" />
  <row Id="4873" PostTypeId="2" ParentId="4868" CreationDate="2011-02-09T17:06:35.180" Score="3" ViewCount="" Body="&lt;p&gt;I think your calculation might be correct, because the loss of hydrogen isn't accurately modeled by Jeans escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember in my statistical mechanics class we had the same problem as homework, and a lot of people got the answer &quot;it should all be gone by now&quot;, but that was because they erroneously used an atomic mass of 1 (instead of 2 for $H_2$ molecules). At least one person used the atomic mass of 2 and got a quite different answer like &quot;most of it should still be here&quot;. Then the professor actually emailed the author of the problem and explained, and he replied back something like &quot;Well I'll be darned!&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you calculate the Jeans escape rate for $H_2$ molecules and assume nothing else is going on, you get that most of them should still be here. However, it's easy to think of other effects that could make it disappear faster. For one thing, UV light (and to some extent solar wind / cosmic rays) ionizes $H_2$ and dissociates it to $H$ in the upper atmosphere, which justifies using an atomic mass of 1. I think just as important, however, is the fact that the outermost wisps of the atmosphere are constantly being &quot;blown away&quot; by the solar wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, it's also possible that you just made a mistake, but if you get an answer that's sort of in-the-ballpark in order of magnitude, but still not nearly enough to explain the current lack of $H_2$ in the atmosphere, this is probably what's going on.)&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditorDisplayName="" LastEditDate="2011-02-09T18:17:52.410" LastActivityDate="2011-02-09T18:17:52.410" />
  <row Id="4874" PostTypeId="2" ParentId="937" CreationDate="2011-02-09T18:02:05.027" Score="2" ViewCount="" Body="&lt;p&gt;In my opinion this is an excellent question, which manages to puzzle also some accomplished physicists. So I do not hesitate to provide another, a bit more detailed, answer, even though several good answers exist already.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that at least part of this question is based upon an incomplete understanding on what it means to mediate a static force from a particle physics point of view. As others have mentioned in their answers already, you encounter a similar issue in the Coulomb problem in electrodynamics. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let me answer your question from a field theory point of view, since I believe this concurs best with your intuition about particles being exchanged (as apparent from the way you phrased the question). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, no gravitational waves can escape from inside the black hole, as you hinted already in your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, no gravitational waves have to escape from inside the black hole (or from the horizon) in order to mediate a static gravitational force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity waves do not mediate the static gravitational force, but only quadrupole or higher moments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to think about forces in terms of particles being exchanged you can view the static gravitational force (the monopole moment, if you wish) as being mediated by &quot;Coulomb-gravitons&quot; (see below for the analogy with electrodynamics). Coulomb-gravitons are gauge degrees of freedom (so one may hesitate to call them &quot;particles&quot;), and thus no information is mediated by their &quot;escape&quot; from the black hole. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is quite analog to what happens in electrodynamics: photon exchange is responsible for the electromagnetic force, but photon waves are not responsible for the Coulomb force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photon waves do not mediate the static electromagnetic force, but only dipole or higher moments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can view the static electromagnetic force (the monopole moment, if you wish) as being mediated by Coulomb-photons. Coulomb-photons are gauge degrees of freedom (so one may hesitate to call them &quot;particles&quot;), and thus no information is mediated by their &quot;instantaneous&quot; transmission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, this is precisely how you deal with the Coulomb force in the QFT context. In so-called Bethe-Salpeter perturbation theory you sum all ladder graphs with Coulomb-photon exchanges and obtain in this way the 1/r potential to leading order and various quantum corrections (Lamb shift etc.) to sub-leading order in the electromagnetic fine structure constant.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In summary, it is possible to think about the Schwarzschild and Coulomb force in terms of some (virtual) particles (Coulomb-gravitons or -photons) being exchanged, but as these &quot;particles&quot; are actually gauge degrees of freedom no conflict arises with their &quot;escape&quot; from the black hole or their instantaneous transmission in electrodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An elegant (but perhaps less intuitive) way to arrive at the same answer is to observe that (given some conditions) the ADM mass - for stationary black hole space-times this is what you would call the &quot;black hole mass&quot; - is conserved. Thus, this information is provided by boundary conditions &quot;from the very beginning&quot;, i.e., even before a black hole is formed. Therefore, this information never has to &quot;escape&quot; from the black hole. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;On a side-note, in one of his lectures Roberto Emparan posed your question (phrased a bit differently) as an exercise for his students, and we discussed it for at least an hour before everyone was satisfied with the answer - or gave up ;-) &lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-02-09T18:02:05.027" />
  <row Id="4875" PostTypeId="2" ParentId="4841" CreationDate="2011-02-09T18:13:47.067" Score="1" ViewCount="" Body="&lt;p&gt;This question was phrased in terms of the Copenhagen Interpretation, although it is tempting to discuss how other Interpretations would deal with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Copenhagen Interpretation (according to the Wikipedia article) does have several formulations and paradoxes connected with it. This example could also be said to stretch the Copenhagen Interpretation. The most similar problem amongst the classics would be the Wigner's Friend argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copenhagen is saying that $\Psi$ is determined only after an experiment, by means of classical instruments by an Observer. So this leaves quite a gap when trying to understand &quot;Brain Function&quot; and &quot;Consciousness&quot;. The first question is whether &quot;consciousness&quot; arises from the quantum behavior of the atoms in the brain. Copenhagen does not address this, neither does Quantum Mechanics more generally. Of course Neuroscientists such as Hameroff are interested in this idea. The problem, from a Copenhagen and classical philosophic perspective are similar: &quot;who or what is observing the brain and collapsing the wave function?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Copenhagen does not really have an answer to this, Hameroff is more interested in a different type of QM Interpretation. In this interpretation the collapse happens for physical reasons very quickly and frequently. There is no observer to do the collapsing, as Copenhagen would require.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Decoherence approach also collapses $\Psi$ in a thermodynamic way, so it doesnt have such a direct connection with Observers either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A variant of this question which is less Neuroscience oriented, is the question of the $\Psi$ for the Universe. If this obeys a Schrodinger-like equation (called Wheeler-deWitt equation) and we assume Copenhagen, then again who does the Observing? Here any Observers are part of the Universe and so there would seem to be a self-referential problem if we take an Observer-centric interpretation like Copenhagen.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-09T18:13:47.067" CommentCount="2" />
  <row Id="4876" PostTypeId="2" ParentId="4831" CreationDate="2011-02-09T18:18:49.027" Score="4" ViewCount="" Body="&lt;p&gt;In addition to @gigacyan answer: length of the lever, as he said, $10^{23}$ meters or 10 million light years.&#xA;For comparison - Andromeda Galaxy is  the nearest spiral galaxy to the Milky Way approximately 2.5 million light-years away.&#xA;It takes 10 thousand years to push the lever for  $10^{20}$ meters at the speed of light. &lt;/p&gt;&#xA;" OwnerUserId="238" LastActivityDate="2011-02-09T18:18:49.027" />
  <row Id="4877" PostTypeId="2" ParentId="4849" CreationDate="2011-02-09T18:32:32.980" Score="4" ViewCount="" Body="&lt;p&gt;The successful historical relationship between Mathematics and Physics seems to be when there is some pre-existing Mathematical Theory M (e.g. Group Theory, Topology, Non-Euclidean Geometry - or special cases of these) and an evolving Physical theory P. P might be expressed in a different kind of mathematics, if at all. Then a mapping is found which maps P onto M, except that M has more equations or components than this mapping includes. So the question becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Do the missing components of M map onto some (so far) unobserved properties of P?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When this is successful (as in many of the other examples cited) then historians say that mathematics has proven valuable for Physics (yet again).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example of this phenomenon might be the SU(3) classification of some particles which had a gap in the representation when mapped onto known particles; the gap mapped onto the $\Omega$ particle. Didnt Gell-Mann get a Nobel prize for that one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;String Theory seems a little bit different from this classic scenario (but it is not the only such) where there is a deliberate attempt to develop Mathematics to model known (and perhaps unknown) Physics. This might be seen as a form of anticipation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some mathematicians take a stronger view than the account given here, in that they believe that the Physical Universe is fundamentally mathematical in character. Often, like Penrose, they may have specific types of mathematics in mind with that claim. So from this perspective the development of that mathematics is valuable over and above any current experimental data. A similar belief seems to underlay the String Theory efforts.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-02-09T18:52:17.657" LastActivityDate="2011-02-09T18:52:17.657" />
  <row Id="4878" PostTypeId="2" ParentId="4837" CreationDate="2011-02-09T19:12:34.240" Score="0" ViewCount="" Body="&lt;p&gt;We needn't think about a train to develop a mental picture of the physics involved - actually, that may over-complicate it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you ever caught a fly in a bottle? If so, you'll notice that the fly's motion is relatively independent of the bottle's motion - until the fly hits a wall of the bottle, of course. Moving the bottle does (practically) nothing to the fly's motion because the bottle is exerting very (very!!) little gravitational pull on the floating fly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the fly's motion will remain (practically) unaffected by moving the bottle, and a large train is in effect just a much larger container than the bottle, with more room for the fly to move until it hits the wall of a train. &lt;/p&gt;&#xA;" LastActivityDate="2011-02-09T19:12:34.240" CommentCount="3" />
  <row Id="4879" PostTypeId="2" ParentId="4849" CreationDate="2011-02-09T19:27:05.277" Score="2" ViewCount="" Body="&lt;p&gt;Well, it often leads both ways. With Ed Witten, physics has led to new math and a Fields medal for him. And the other way, Galois and group theory has led to all sorts of goodies in physics with gauge theory etc. I had voted to close, but am now apologizing :) Recently, &lt;a href=&quot;http://www.twistordiagrams.org.uk/&quot; rel=&quot;nofollow&quot;&gt;Andrew Hodges&lt;/a&gt;, an Oxford mathematician and writer of an excellent biography of Alan Turing, called &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0099116413&quot; rel=&quot;nofollow&quot;&gt;Enigma&lt;/a&gt; wrote a paper with Nima Arkani-Hamed--&lt;a href=&quot;http://www.twistordiagrams.org.uk/papers/index.html&quot; rel=&quot;nofollow&quot;&gt;Twistors&lt;/a&gt;, and Alain Connes, another Fields medal mathematician, has been using &lt;a href=&quot;http://www.twistordiagrams.org.uk/papers/index.html&quot; rel=&quot;nofollow&quot;&gt;non-commutative geometry&lt;/a&gt; for interesting speculations on physics (even if he has admitted problems with the theory---great man.)&#xA;Then, there is Mad &lt;a href=&quot;http://space.mit.edu/home/tegmark/crazy.html&quot; rel=&quot;nofollow&quot;&gt;Max Tegmark and his theory that all mathematical stuctures have a physical reality&lt;/a&gt; which is the ultimate in Platonism :)&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-02-09T19:27:05.277" />
  <row Id="4880" PostTypeId="2" ParentId="4849" CreationDate="2011-02-09T19:31:04.333" Score="6" ViewCount="" Body="&lt;p&gt;Well, it often leads both ways. With Ed Witten, physics has led to new math and a Fields medal for him. And the other way, Galois and group theory has led to all sorts of goodies in physics with gauge theory etc. I had voted to close, but am now apologizing :) Recently, &lt;a href=&quot;http://www.twistordiagrams.org.uk/&quot; rel=&quot;nofollow&quot;&gt;Andrew Hodges&lt;/a&gt;, an Oxford mathematician and writer of an excellent biography of Alan Turing, called &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0099116413&quot; rel=&quot;nofollow&quot;&gt;Enigma&lt;/a&gt; wrote a paper with Nima Arkani-Hamed--&lt;a href=&quot;http://www.twistordiagrams.org.uk/papers/index.html&quot; rel=&quot;nofollow&quot;&gt;Twistors&lt;/a&gt;, and Alain Connes, another Fields medal mathematician, has been using &lt;a href=&quot;http://www.alainconnes.org/docs/einsymp.pdf&quot; rel=&quot;nofollow&quot;&gt;non-commutative geometry&lt;/a&gt; for interesting speculations on physics (even if he has admitted problems with the theory---great man.)&#xA;Then, there is Mad &lt;a href=&quot;http://arxiv.org/abs/0704.0646&quot; rel=&quot;nofollow&quot;&gt;Max Tegmark and his theory that all mathematical stuctures have a physical reality&lt;/a&gt; which is the ultimate in Platonism :)&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-02-09T21:37:01.587" LastActivityDate="2011-02-09T21:37:01.587" />
  <row Id="4881" PostTypeId="1" AcceptedAnswerId="4883" CreationDate="2011-02-09T21:10:00.087" Score="5" ViewCount="191" Body="&lt;p&gt;Assume two uncharged non-rotating black holes traveling straight at each other with no outside forces acting on the system. What is thought to happen to the kinetic energy of these two masses when they collide?  Is the excess energy lost through gravitational radiation?  What would the effect of these gravity waves be on matter or energy that they encounter?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-09T22:39:18.930" Title="Black holes in a head-on collision" Tags="&lt;astrophysics&gt;&lt;gravity&gt;&lt;black-holes&gt;" AnswerCount="3" />
  <row Id="4882" PostTypeId="1" AcceptedAnswerId="4903" CreationDate="2011-02-09T21:16:10.670" Score="4" ViewCount="197" Body="&lt;p&gt;I am becoming familiar with the Geometric Optics approximation in General Relativity which (to summarise) says that EM waves follow null geodesics under the geometric optics approximation. In the context of General Relativity the Geometric Optics approximation amounts to the assumption that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda &amp;lt;&amp;lt; R$ where $\lambda$ is the wavelength and R is some measure of curvature (~ 100 meters near Black Holes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So under this approximation EM waves follow null geodesics. Additionally the &quot;photon&quot; model can make sense too, and we can say &quot;photons follow null geodesics&quot;. But what about the &quot;general case&quot;? In the general case the order of curvature is the same as (or less) than that of the wavelength. So the usual statements do not obviously apply. Where is the full case discussed and what implications are there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give some extreme examples for those who have not thought about this before:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In a Schwarzchild model at R=2M photons can get into circular null orbits. But clearly this provides an IR cut-off as they would be in a Bohr-like orbit with a max wavelength $\lambda_{max}=4\pi M$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The usual &quot;equivalence principle&quot; may fail as the local trivialisation cannot extend far enough into the curvature region (and $\lambda$ is larger still). So the equivalence laboratory can detect such waves, but cannot measure their behaviour and prove that they follow geodesics (local straight lines).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I am struggling to find any coherent account of this, just odd facts and claims.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; In large part this question is answered by the comment from Genneth, and the links related to this Wikipedia article: &lt;a href=&quot;http://en.wikipedia.org/wiki/Maxwell%27s_equations_in_curved_spacetime&quot; rel=&quot;nofollow&quot;&gt;Maxwell's equations in curved spacetime&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-22T03:03:30.373" LastActivityDate="2011-05-22T03:03:30.373" Title="Low frequency waves in General Relativity" Tags="&lt;general-relativity&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4883" PostTypeId="2" ParentId="4881" CreationDate="2011-02-09T21:17:49.357" Score="4" ViewCount="" Body="&lt;p&gt;Some energy is lost to gravitational radiation. Some probably ends up in the final black hole (i.e., $m_{\rm final}$ could be greater than the sum of the two initial masses). Figuring out the proportions of these two is not trivial, I would imagine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational waves striking matter would not have any terribly dramatic effect. Once you're a decent way away from the black holes, the strain amplitude is low. They'd shake things around a bit.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-02-09T22:39:18.930" LastActivityDate="2011-02-09T22:39:18.930" />
  <row Id="4884" PostTypeId="2" ParentId="4881" CreationDate="2011-02-09T21:55:18.420" Score="3" ViewCount="" Body="&lt;p&gt;When two black hole collide the horizon area of the resulting black hole must exceed the sum of the areas of the two initial black holes.  This places an upper limit on the amount of gravitational radiation which is produced.  If the two initial black holes have equal masses $M$ then the finial black hole must obey&#xA;$$&#xD;&#xA;4\pi(2M_f)^2~=~16\pi(M_f)^2\ge~4\pi (2M) ^2~+~4\pi (2M)^2~=~32\pi M^2&#xD;&#xA;$$&#xA;and so $M_f~\ge~\sqrt{2}M$.  The amount of gravitational radiation emitted is then $E~\le~(2~-~\sqrt{2})M$  which is $.585M$&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-09T21:55:18.420" />
  <row Id="4885" PostTypeId="2" ParentId="4881" CreationDate="2011-02-09T21:59:01.357" Score="4" ViewCount="" Body="&lt;p&gt;During the collision of two black holes gravitational radiation is emitted. After the collision process the gravitational radiation eventually dies off and a single black hole remains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To treat this 2-body system quantitatively is a major task for numerical relativity, which has seen some big breakthrough after the work by &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0507014&quot; rel=&quot;nofollow&quot;&gt;Frans Pretorius&lt;/a&gt;. It is quite impractical, if not impossible, to deal with the general relativistic 2-body problem analytically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational radiation emitted in such a process - from the inspiralling phase through the merger to the ringdown of the final state black hole - has a characteristic profile that may facilitate the discovery of corresponding gravity waves with an experiment like (advanced) LIGO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to give a simple upper bound on the maximum of energy of the gravitational radiation emitted in such a process using Hawking's area theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose for simplicity that the initial state consists approximately of two Schwarzschild black holes with equal masses $m$ and the final state approximately of a single black hole with mass $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy $E$ of the gravitational radiation emitted during the collision is given by $E=2m-M$. So you get a maximal amount of gravity waves if $M$ is as small as possible, $M=M_{\rm min}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hawking's area theorem implies that the area of the final black hole is bigger or equal to the sum of the areas of the two initial black holes. Using area $\propto$ mass$^2$ leads to the inequality $M^2 \geq 2m^2$. The final black hole mass becomes minimal if this inequality is saturated, $M_{\rm min}=\sqrt{2}m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, the maximal amount of energy radiated away in the collision of two equal mass Schwarzschild black holes is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_{\rm max} = 2m-\sqrt{2}m = (2-\sqrt{2}) m \approx 0.29 \times 2m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the &quot;best case&quot; scenario you can extract at most 29% of the initial energy in the form of gravitational radiation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can play the same game with Schwarzschild black holes of different masses or Kerr-Newman black holes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your second question, the gravitational waves emitted in such a process are not different (apart from their characteristic pattern alluded to above) from gravitational waves emitted by other sources. You may check the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_wave&quot; rel=&quot;nofollow&quot;&gt;Wikipage&lt;/a&gt; for an animation of how the two polarizations of a gravitational wave distort a ring of particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While writing this answer Lawrence B. Crowell posted his answer, which also contains the upper bound on gravitational radiation derived above.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-02-09T21:59:01.357" />
  <row Id="4888" PostTypeId="1" CreationDate="2011-02-10T00:18:10.270" Score="3" ViewCount="226" Body="&lt;p&gt;Lets say you're an undergrad physics student with a lot of &quot;sense&quot; for technology and limited theoretical abilities. Now you need some kind of a career advice about the specific field of physics to specialize in. So you ask your senior colleagues: what is the current area of research that is promising most technological advances in the near future (~ 10-15 years)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p.s. hope this question is valid, though predicting future isn't (yet) accurately answerable&lt;/p&gt;&#xA;" OwnerUserId="1800" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-10T00:41:13.993" LastActivityDate="2011-02-10T17:23:14.143" Title="Most techologically applicable physics research area" Tags="&lt;soft-question&gt;&lt;physics-careers&gt;&lt;general-physics&gt;" AnswerCount="7" CommentCount="1" />
  <row Id="4889" PostTypeId="2" ParentId="4888" CreationDate="2011-02-10T01:05:00.830" Score="4" ViewCount="" Body="&lt;p&gt;Broadly speaking and somewhat obviously - condensed matter physics. More specifically, and this is a very biased answer - quantum cryptography and given an extra 10-20 years - quantum computation.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-02-10T01:05:00.830" CommentCount="1" />
  <row Id="4890" PostTypeId="1" CreationDate="2011-02-10T01:24:51.553" Score="4" ViewCount="258" Body="&lt;p&gt;I want to calculate stress matrix in a cube with two faces parallel to x axis and perpendicular to z axis (sorry I don't know how can I put a picture in this post).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two force uniform distributions (that we'll indicate with p) over this two surfaces: the superior is in the x direction, the inferior in the -x direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'll have only shear stress, and a shear-modulus $\mu$ dipendence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We suppose uniform stress tensor in the cube, because every infinitesimal dV of medium is in statical equilibrium with +pdS force with x direction by the superior infinitesimal dV and -pdS force by the inferior infintesimal dV for 3rd Newton's law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remembering that any stress over a surface is $t_{ij}n{j}$, where $n_j$ is the normal versor to the surface, we must write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_{ij}n_1=0$ because we haven't any force on the surfaces perpendicular to the x-axis; so the first column is composed from three 0;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_{ij}n_2=0$ because we haven't any force on the surfaces perpendicular to the y-axis;so the second column is composed from three 0 too;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_{ij}n_3=p n_1$, because we have the force-distribution p over the surfaces that are perpendicular to z axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_1$ is $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$,  $n_2$ is $\begin{pmatrix} 0 \\ 1\\ 0 \end{pmatrix}$, $n_3$ is $\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$. So the matrix $T_{ij}$ becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix}&#xD;&#xA;0 &amp;amp; 0 &amp;amp; p\\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA;\end{pmatrix}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this haven't any sense, because stress-tensor must be simmetric for conservation of angular momentum. Where is the mistake?&lt;/p&gt;&#xA;" OwnerUserId="558" LastEditorUserId="558" LastEditorDisplayName="" LastEditDate="2011-02-10T11:46:17.933" LastActivityDate="2011-02-10T11:46:17.933" Title="Stress tensor in a cube with shear forces" Tags="&lt;material-science&gt;&lt;elasticity&gt;&lt;continuum-mechanics&gt;" AnswerCount="2" />
  <row Id="4891" PostTypeId="2" ParentId="4849" CreationDate="2011-02-10T02:13:14.227" Score="-2" ViewCount="" Body="&lt;p&gt;Math is infinite in size so given any finite number of data points that each is taken to a finite accuracy, there has to be an infinite number of theories that are consistent with the data. Another way of saying the same thing is that there are a lot more beautiful, but incorrect, theories of physics than correct physics.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-10T02:13:14.227" />
  <row Id="4892" PostTypeId="2" ParentId="4862" CreationDate="2011-02-10T02:29:11.280" Score="2" ViewCount="" Body="&lt;p&gt;As others have noted, you will not get any intensity modulation from the interference of two linearly polarized light beams with orthogonal polarizations. It's worth noting, though, that this does not mean that beams with perpendicular polarizations don't affect each other. In fact, a counter-propagating pair of beams with orthogonal linear polarizations-- the so-called &quot;lin-perp-lin&quot; configuration-- is the best system for understanding the Sisyphus cooling effect, the explanation of which was a big part of the &lt;a href=&quot;http://nobelprize.org/nobel_prizes/physics/laureates/1997/illpres/doppler.html&quot; rel=&quot;nofollow&quot;&gt;1997 Nobel Prize in Physics&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The superposition of two counter-propagating linearly polarized beams with orthogonal polarizations doesn't give you any modulation of intensity, but does create a polarization gradient. For the lin-perp-lin configuration, you get alternating regions of left- and right-circular polarization, and combined with optical pumping this lets you set up a scenario where you can cool atomic vapors to extremely low temepratures. This makes laser cooling vastly more useful than it would be otherwise, and allows all sorts of cool technologies like atomic fountain clocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not interference in the sense that is usually meant, but it is a cool phenomenon that results from overlapping beams with different polarizations. So you shouldn't think that just because it doesn't produce a pattern of bright and dark spots it's not interesting.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2011-02-10T02:29:11.280" />
  <row Id="4893" PostTypeId="2" ParentId="4888" CreationDate="2011-02-10T02:38:50.237" Score="5" ViewCount="" Body="&lt;p&gt;Depends on what you mean by &quot;sense&quot; for technology. Condensed matter physics is the obvious answer, but that's partly because there are more condensed matter physicists than any other sort (particle physics gets all the press, but the condensed matter division of the APS is the largest). A great many of those people are employed in industry, so it's an area with good career prospects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for specific areas within condensed matter, it's a little difficult to say. Materials science is the most directly applied area, I think, so if you're talking about a student who's good at making things and conscientious about experimental design and so forth, that's probably the direction to go. Semiconductors and superconductors have bigger upside, though it's not clear that anybody is ever going to push high-Tc superconductors to a comercially useful regime, and that area probably requires a little more theoretical ability to understand what's going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum information and quantum computing is extremely cool, and a lot of the experimental processes are relatively concrete and don't require much mathematical sophistication to understand. They're not really going to be pumping out commercial products any time soon, though, so they're likely to remain somewhat more academic in nature. Which may or may not be a negative feature, depending on what the student in question is really after.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2011-02-10T02:38:50.237" />
  <row Id="4894" PostTypeId="2" ParentId="4888" CreationDate="2011-02-10T02:47:40.473" Score="3" ViewCount="" Body="&lt;p&gt;Optics and/or optical engineering has been good for me. It's interesting and there are plenty of jobs that are highly applied, but are still tied very closely to active areas of research.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-02-10T02:47:40.473" />
  <row Id="4895" PostTypeId="2" ParentId="4888" CreationDate="2011-02-10T04:39:57.943" Score="1" ViewCount="" Body="&lt;p&gt;Go with biophysics. The next big revolution in industry is going to biology related; genomes and all that. There will be lots of high dollar applications in just about everything. Medicine, agriculture, silviculture, animal husbandry (or whatever fancy thing they call it nowadays), fishing, construction, biofuels, eventually even stuff like consumer goods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will be the &quot;plastics&quot; of the 2020s.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-10T04:39:57.943" />
  <row Id="4896" PostTypeId="2" ParentId="4888" CreationDate="2011-02-10T04:42:29.880" Score="1" ViewCount="" Body="&lt;p&gt;Medical biophysics would be good. For example, at TRIUMF UBC, they are working on designing liquid xenon PET scanner. There are DNA chips etc. Combining the above answers, optical computing is also sounds interesting--see Lene Hau's work &quot;freezing&quot; light, or work on quantum donuts.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-02-10T04:42:29.880" />
  <row Id="4897" PostTypeId="1" AcceptedAnswerId="4900" CreationDate="2011-02-10T04:57:09.737" Score="1" ViewCount="95" Body="&lt;p&gt;I'm looking for the most readable &lt;strong&gt;&lt;em&gt;quantitative&lt;/em&gt;&lt;/strong&gt; explanation, with the least amount of difficult mathematics. Rather than an answer with just a number, I'd like to see the derivation from the CKM matrix.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-10T06:03:17.073" Title="How much CP violation does the weak force cause?" Tags="&lt;particle-physics&gt;&lt;cp-violation&gt;&lt;weak-force&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="4898" PostTypeId="2" ParentId="1019" CreationDate="2011-02-10T05:42:41.763" Score="3" ViewCount="" Body="&lt;p&gt;The concept that quantum mechanics undermines determinism. The Schrodinger wave equation evolution is completely deterministic. The results of measurements are probabilistic, but this does not mean that the various superposed states do not have causes. This is not the same thing as a hidden variable theory. The probabilities are deterministic.&#xA;&lt;a href=&quot;http://arxiv.org/abs/quant-ph/0212095&quot; rel=&quot;nofollow&quot;&gt;T'Hooft&lt;/a&gt; has some interesting ideas on a determinism underlying QM (not the same thing as saying the wave equation is deterministic).  I am not arguing that qm is in all senses deterministic, but it isn't completely non-deterministic either.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-02-10T05:42:41.763" CommunityOwnedDate="2011-02-10T05:42:41.763" />
  <row Id="4899" PostTypeId="1" CreationDate="2011-02-10T05:49:19.660" Score="2" ViewCount="148" Body="&lt;p&gt;what are the conditions to check point  type singularity in a bianchi type model ?&#xA;bianchi type model are of Type I,II,III,IX,IV or u can say we use different Bianchi type models  having some specific metric or line element as i know .&lt;/p&gt;&#xA;" OwnerUserId="1843" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2011-02-15T15:59:15.307" LastActivityDate="2011-02-15T15:59:15.307" Title="Singularities in Bianchi models in general relativity ( physical science)" Tags="&lt;general-relativity&gt;&lt;mathematical-physics&gt;&lt;string-theory&gt;&lt;cosmology&gt;" CommentCount="2" />
  <row Id="4900" PostTypeId="2" ParentId="4897" CreationDate="2011-02-10T06:03:17.073" Score="4" ViewCount="" Body="&lt;p&gt;A clear recent review of flavor physics, including CP violation, is in the &lt;a href=&quot;http://arxiv.org/abs/1005.3106&quot; rel=&quot;nofollow&quot;&gt;TASI lectures by Gedalia &amp;amp; Perez&lt;/a&gt;. The parametrization-independent measure of how much CP violation is present in the Standard Model is called the &quot;Jarlskog invariant&quot;; it's explained in those lectures, but might be a useful keyword if you're searching for other resources. If you want to understand why the observed CP violation is insufficient to explain the baryon asymmetry we observe in the universe around us, you might try &lt;a href=&quot;http://arxiv.org/abs/hep-ph/0609145&quot; rel=&quot;nofollow&quot;&gt;these lectures on baryogenesis by Jim Cline&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2011-02-10T06:03:17.073" CommentCount="3" />
  <row Id="4901" PostTypeId="2" ParentId="4890" CreationDate="2011-02-10T06:46:56.743" Score="0" ViewCount="" Body="&lt;p&gt;The stress tensor is only symmetric if either the spin density describing the density of &lt;em&gt;internal&lt;/em&gt; angular momentum is zero. For most materials, this is the case, because we can't align most of the spins of the constituent atoms or molecules in the same direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine the cube you mentioned embedded as a subsystem of a much larger material system. With an asymmetric stress-tensor, the forces will cause the cube to spin relative to the material it is embedded in, which for most materials, will be prevented by molecular binding forces between the molecules or atoms on both sides of the boundary. In other words, the cube can't spin relative to the material just outside the boundary. This is due to come counterforce due to molecular binding and shows up as a counterforce on the surfaces normal to the $x$-direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also have to realize this stress analysis only applies in the limit as the cube size becomes infinitesimal. Otherwise, we can have unbalanced forces on the different surfaces, which only means the shape of the cube becomes deformed in time.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-10T06:46:56.743" CommentCount="4" />
  <row Id="4902" PostTypeId="2" ParentId="3432" CreationDate="2011-02-10T07:11:02.260" Score="4" ViewCount="" Body="&lt;h2&gt;&lt;a href=&quot;http://www.ellipsix.net&quot; rel=&quot;nofollow&quot;&gt;Ellipsix Informatics&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Since a few other people have posted their own blogs, I thought it wouldn't be inappropriate for me to do the same. This is my personal website, not a dedicated physics blog, but the posts I make about physics and/or information technology are the current &quot;main attraction.&quot; The posts are based on whatever I happen to be thinking about when I have time to write something. Mostly I like to focus on applications of physics to everyday life (i.e. situations that would arise outside a lab).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the best posts are the ones about Mythbusters episodes, e.g.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ellipsix.net/blog/2010/12/death-rays-and-thermal-radiation.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ellipsix.net/blog/2010/12/death-rays-and-thermal-radiation.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ellipsix.net/blog/2010/04/calculating-terminal-speed.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ellipsix.net/blog/2010/04/calculating-terminal-speed.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ellipsix.net/blog/2010/01/bouncing-bullets.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ellipsix.net/blog/2010/01/bouncing-bullets.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-02-10T07:11:02.260" CommunityOwnedDate="2011-02-10T07:11:02.260" />
  <row Id="4903" PostTypeId="2" ParentId="4882" CreationDate="2011-02-10T07:20:41.313" Score="1" ViewCount="" Body="&lt;p&gt;Roy, indeed, geometric optics with trajectories of light rays is only applicable when $\lambda\ll R_{curvature\,radius}$. The beginning of your question suggests that you are aware of this condition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the rest of your question is all about your refusal of the condition. So it's hard to understand whether you actually understand the condition or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, your conclusion 1 seems to be an artifact of the negligence of the condition because it clearly makes no sense to talk about &quot;orbits&quot; of photons whose wavelength is longer than the black hole radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I don't know why you would say 2. The equivalence principle always applies locally - it's how it's defined - and this principle is 100% valid in the real world. You can't disprove the equivalence principle by violating its assumptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right way to treat light and electromagnetic waves outside the regime of validity of geometric optics is to treat them as waves, i.e. using field theory. This is what you may be asking although I haven't really understood your question. So one has to learn Maxwell's equations. If the term &quot;Maxwell's equations&quot; sounds new to you, then it could have been a bit counterproductive and illogical to study general relativity (before Maxwell's equations) but don't panic, I did the same mistake 25 years ago. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classical electromagnetic waves in general relativity are described by the Einstein-Maxwell system; quantum radiation - including the concept of photons - requires you to study quantum electrodynamics on a curved background. Quantum field theories on curved backgrounds bring some novelties such as particle production - different Hamiltonians and different ground states of these Hamiltonians.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-10T08:30:44.577" LastActivityDate="2011-02-10T08:30:44.577" CommentCount="1" />
  <row Id="4904" PostTypeId="2" ParentId="4862" CreationDate="2011-02-10T07:23:16.357" Score="2" ViewCount="" Body="&lt;p&gt;Your question is quite vague, but in short, the answer is: &lt;strong&gt;yes&lt;/strong&gt;, look it up on &lt;a href=&quot;http://en.wikipedia.org/wiki/Interference_%28wave_propagation%29&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;. But let's be more precise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as the light intensity is low enough to not obtain &lt;a href=&quot;http://en.wikipedia.org/wiki/Nonlinear_optics&quot; rel=&quot;nofollow&quot;&gt;nonlinear effects&lt;/a&gt;, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Superposition_principle&quot; rel=&quot;nofollow&quot;&gt;superposition principle&lt;/a&gt; of &lt;strong&gt;linear optics&lt;/strong&gt; is valid. That means, the amplitudes of two electromagnetic (EM) fields sum up, thus yielding interference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since the amplitudes are vectors (while the intensity, being related to the absolute squares of the amplitudes, is a scalar), the &lt;strong&gt;interference depends on the relative polarization&lt;/strong&gt;, the total intensity for two &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_polarization&quot; rel=&quot;nofollow&quot;&gt;linearly polarized&lt;/a&gt; EM waves is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = I_1 + I_2 + 2\sqrt{I_1I_2}\cos(\Delta\varphi)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Delta\varphi$ denotes the angle between the two polarizations. You see that for perpendicular polarization the cosine term vanishes, the intensities just add up and you obtain no interference, while for antiparallel polarization ($\Delta\varphi = 180°$) you obtain destructive interference since the cosine becomes -1. In case you wonder about energy conservation (being proportional to the intensity) keep in mind that only the global energy is conserved while local fluctations are ok.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One final note: The whole thing &lt;strong&gt;only works for a well-defined phase relation&lt;/strong&gt; between two EM waves. That is, only &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;spectral components&lt;/a&gt; of the same wavelength can interfere, and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Coherence_length&quot; rel=&quot;nofollow&quot;&gt;coherence length&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Coherence_time&quot; rel=&quot;nofollow&quot;&gt;time&lt;/a&gt; need to be large enough - you won't obtain perfect interference if your light source flickers due to heat for example.&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-02-10T07:38:00.390" LastActivityDate="2011-02-10T07:38:00.390" />
  <row Id="4905" PostTypeId="2" ParentId="4890" CreationDate="2011-02-10T08:03:44.443" Score="2" ViewCount="" Body="&lt;p&gt;Boy, the right stress tensor for similar static situations is symmetric, indeed. It's not hard to see why: the stress tensor knows about the density of forces and an asymmetry would destroy the equilibrium. See&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stress_(mechanics)#Equilibrium_equations_and_symmetry_of_the_stress_tensor&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Stress_(mechanics)#Equilibrium_equations_and_symmetry_of_the_stress_tensor&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last paragraph of the section above contains a proof why the stress tensor is symmetric. You may say that the antisymmetric part would act as a torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example, both $T_{xz}$ and $T_{zx}$ are equal to the shear stress you denoted $p$. In particular, it is not true that $T\cdot n_x=0$, as you wrote in the first step (with confusing extra indices), because the stress tensor measures the internal forces and not just forces that you actively add by your hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The internal forces respect the symmetry between the $x$ and $z$ axes. The only asymmetric tensor was one you wanted to &quot;prescribe&quot; to the system. But you can't prescribe arbitrary properties and behavior to the physical system: the objects' properties and behavior obey the laws of physics rather than your expectations. In particular, laws of physics guarantee that the shear stress will treat any pair of axes symmetrically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should imagine that the squares in an atomic grid in the $xz$ plane are deformed to rhombi. But the rhombic curvature exists relatively to both axes $x,z$. If you insert a probe into the solid that will measure the internal tension, be sure that you will get totally identical results when you push the $z=0$ plane in the $x$ direction as if you push the $x=0$ plane in the $z$ direction. The result of both things is to change the angle between the (former) $x=0$, $z=0$ planes in the solid. There's no difference.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-10T08:03:44.443" />
  <row Id="4906" PostTypeId="1" AcceptedAnswerId="4917" CreationDate="2011-02-10T08:35:24.807" Score="4" ViewCount="122" Body="&lt;p&gt;I know that it must have something to do with (gauge?) theories with N=2 supersymmetry, BPS states and even black holes, but most papers on the subject are too technical for me. So what is wall crossing?&lt;/p&gt;&#xA;" OwnerUserId="1844" LastActivityDate="2011-02-10T19:12:52.850" Title="What is wall crossing?" Tags="&lt;quantum-field-theory&gt;&lt;string-theory&gt;" AnswerCount="1" />
  <row Id="4907" PostTypeId="1" CreationDate="2011-02-10T08:48:47.633" Score="5" ViewCount="173" Body="&lt;p&gt;In the early days of quantum electrodynamics, the most popular gauge chosen was the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gupta%E2%80%93Bleuler_formalism&quot; rel=&quot;nofollow&quot;&gt;Gupta-Bleuler gauge&lt;/a&gt; stating that for physical states,&#xA;$$\langle \chi | \partial^\mu A_\mu | \psi \rangle = 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this gauge is practically never used now. Why? Is there anything wrong or inappropriate with the Gupta-Bleuler gauge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is the Gupta-Bleuler gauge related to the $R_\xi$ gauge?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-10T13:53:23.123" Title="Why is the Gupta-Bleuler gauge unfashionable?" Tags="&lt;mathematical-physics&gt;&lt;quantum-electrodynamics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="4908" PostTypeId="1" AcceptedAnswerId="4914" CreationDate="2011-02-10T08:56:15.833" Score="4" ViewCount="212" Body="&lt;p&gt;In several papers (including a recent one by Banks and Seiberg) people mention a &quot;folk-theorem&quot; about the impossibility to have global symmetries in a consistent theory of quantum gravity. I remember having heard one particular argument that seemed quite reasonable (and almost obvious), but I can't remember it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found other arguments in the literature, including (forgive my sloppiness):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In string theory global symmetries on the world-sheet become gauge symmetries in the target space, so there is no (known) way to have global symmetries&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;in AdS/CFT global symmetries on the boundary correspond to gauge symmetries in the bulk so there again there is no way to have global symmetries in the bulk&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The argument in the Banks-Seiberg paper about the formation of a black hole charged under the global symmetry&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I find none of these completely satisfactory. Does anybody know of better arguments?&lt;/p&gt;&#xA;" OwnerUserId="1844" LastActivityDate="2011-02-10T14:17:17.130" Title="Global symmetries in quantum gravity" Tags="&lt;string-theory&gt;&lt;quantum-gravity&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="4909" PostTypeId="1" AcceptedAnswerId="4910" CreationDate="2011-02-10T09:44:47.070" Score="1" ViewCount="170" Body="&lt;p&gt;I've been wondering about this question for a while. If you have alpha and beta particles released from a radioactive core, how do they ionise surrounding particles?&lt;/p&gt;&#xA;" OwnerUserId="1846" LastActivityDate="2011-02-10T10:10:03.993" Title="How do alpha and beta particles ionise surrounding particles?" Tags="&lt;particle-physics&gt;&lt;nuclear-physics&gt;&lt;particles&gt;" AnswerCount="2" />
  <row Id="4910" PostTypeId="2" ParentId="4909" CreationDate="2011-02-10T10:05:38.940" Score="1" ViewCount="" Body="&lt;p&gt;When alpha particles (helium nuclei) or beta particles (electrons) are released in a radioactive decay they carry significant kinetic energy. As they go through the surrounding material they bump here and there occasionally kicking off some electrons from the surrounding atoms which are then ionised, until they finally stop.&lt;/p&gt;&#xA;" OwnerUserId="1844" LastActivityDate="2011-02-10T10:05:38.940" CommentCount="4" />
  <row Id="4911" PostTypeId="2" ParentId="4909" CreationDate="2011-02-10T10:10:03.993" Score="0" ViewCount="" Body="&lt;p&gt;Alpha particles are the cores of $He$, i.e. lacking the two electrons they carry a charge of $+2e$, while beta particles are either electrons or positrons and thus carrying a charge of $\mp 1e$. Positrons meeting other atoms' electrons annihilate thus leaving a positive ionization. The other particles may just carry enough momentum to transfer the energy required for the ionization of shell electrons.&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2011-02-10T10:10:03.993" CommentCount="0" />
  <row Id="4912" PostTypeId="2" ParentId="4907" CreationDate="2011-02-10T10:11:30.060" Score="4" ViewCount="" Body="&lt;p&gt;See: &lt;a href=&quot;http://en.wikipedia.org/wiki/BRST_formalism#Pre-BRST_approaches_to_gauge_fixing&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/BRST_formalism#Pre-BRST_approaches_to_gauge_fixing&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short summary version:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gupta-Bleuler gauge (Lorentz), like many others (Coulomb), works fine for QED, but is technically challenging/impossible to push to the non-Abelian case (of Yang-Mills theory). The modern view point is via something called the BRST transform, which is capable of dealing with the full general case.&lt;/p&gt;&#xA;" OwnerUserId="1501" LastEditorUserId="1501" LastEditorDisplayName="" LastEditDate="2011-02-10T10:17:23.657" LastActivityDate="2011-02-10T10:17:23.657" CommentCount="1" />
  <row Id="4913" PostTypeId="2" ParentId="4746" CreationDate="2011-02-10T10:22:25.017" Score="0" ViewCount="" Body="&lt;p&gt;A preprint has appeared for this:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accepted for publication in Nature Physics., Thomas E. Markland, Joseph A. Morrone, B. J. Berne, Kunimasa Miyazaki, Eran Rabani, David R. Reichman, &lt;em&gt;Quantum fluctuations can promote or inhibit glass formation&lt;/em&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/1011.0015&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1011.0015&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the article does apply to glasses in general, it is not a danger to our spacecraft because their temperatures are nowhere near absolute zero. In fact, even intergalactic space is quite warm, compared to absolute zero, because of the cosmic microwave background radiation (which has a temperature of about 2.7 degrees Kelvin).&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-10T10:22:25.017" />
  <row Id="4914" PostTypeId="2" ParentId="4908" CreationDate="2011-02-10T10:24:59.313" Score="3" ViewCount="" Body="&lt;p&gt;Perhaps this is just rephrasing your last explanation, so I am not sure if you consider this as a &quot;better argument&quot;, but I'll give you a good reference for further reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum gravity may break global symmetries because the global charge can be eaten by virtual black holes or wormholes, see &lt;a href=&quot;http://arxiv.org/abs/hep-th/9502069&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-02-10T10:24:59.313" CommentCount="4" />
  <row Id="4915" PostTypeId="2" ParentId="2791" CreationDate="2011-02-10T11:04:40.943" Score="1" ViewCount="" Body="&lt;p&gt;I think that Luboš Motl has covered this rather well, but I get the idea that what you're looking for is a discussion on how to calculate the electron density for the combination of a single &quot;free&quot; electron and another electron that's gone through the interference. I'll give you the flavor of the calculation without actually doing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\psi_p(x)$ be the wave function for the purple electron that went through the slit where $x$ means a 3-d point and I'm leaving off the time variable by assuming that there's no dependence on it. This is a complex function and when you evaluate it on the screen you'll find that its squared magnitude $|\psi_p(x)|^2$ gives you the interference pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And let $\psi_o(x)$ be the wave function for the orange electron. Then the combined wave function is antisymmetrized:&lt;br&gt;&#xA;$\psi(x_1,x_2) = \psi_o(x_1)\psi_p(x_2) - \psi_p(x_1)\psi_o(x_2)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the probability that you get one electron at the point $x_1$ and the other at the point $x_2$ is proportional to $|\psi(x_1,x_2)|^2$. This calculation assumes that the electrons do not interact.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As Dr. Motl notes, so long as the two electrons don't get extremely close this will be quite accurate and the result will be the hole drilled by the orange electron and the scattering pattern from the other. To add the interaction you will have to use quantum field theory. In that theory you begin with the above calculation for $\psi(x_1,x_2)$ and add corrections to it. The first correction will be the addition of a single photon that is emitted by one electron and absorbed by the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The photon calculation is an example of a Feynman diagram. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Feynman_diagram&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Feynman_diagram&lt;/a&gt;&lt;br&gt;&#xA;The pretty drawing is:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/LtNoM.png&quot; alt=&quot;Feynman drawing&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the mathematics is kind of messy and should be the subject of a separate question.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-10T11:04:40.943" CommentCount="2" />
  <row Id="4916" PostTypeId="2" ParentId="147" CreationDate="2011-02-10T11:16:18.500" Score="2" ViewCount="" Body="&lt;p&gt;It's possible to do correct quantum mechanics without believing that particles get altered by 360 degree rotations. Use the &quot;density matrix&quot; form instead of &quot;wave function&quot;.&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Density_matrix&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Density_matrix&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To convert a quantum wave state $\psi(x)$ or $|a\rangle$ to a density matrix, multiply the ket by the bra as in:&lt;br&gt;&#xA;$\psi(x) \to \rho(x,x&amp;#39;) = \psi(x)\psi^*(x&amp;#39;)$&lt;br&gt;&#xA;$|a\rangle \to |a\rangle \langle a|$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the bras and kets take complex phases, i.e.&lt;br&gt;&#xA;$e^{+i\alpha}|a\rangle \equiv e^{-i\alpha}\langle a|$&lt;br&gt;&#xA;the complex phases cancel. This is more general than the -1 you get by a 360 degree rotation, but a factor of -1 is also a phase and so it's also canceled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short while the state vectors or wave functions take a -1 on 360 degree rotation, the (pure) density matrices are left unchanged.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-10T11:16:18.500" />
  <row Id="4917" PostTypeId="2" ParentId="4906" CreationDate="2011-02-10T11:31:40.113" Score="2" ViewCount="" Body="&lt;p&gt;Wall crossing is any discontinuous change of an integer (or at least rational) quantity - or, more generally, any qualitative change of the spectrum etc. - that occurs when one moves to the opposite side from a &quot;wall&quot; in a moduli space or parameter space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A would-be index may suddenly change discontinuously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wall - a codimension one locus in the parameter space - is then referred to as the &quot;wall of marginal stability&quot;. One one side, an object may be stable while it is unstable on the other side. It is typically unstable because a decay suddenly becomes plausible because the hypothetical decay product get light enough, if you wish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objects are typically BPS objects on the stable side and they can also be BPS black holes or any other BPS objects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see objects that may be exactly stable, BPS, but that are also sufficiently diverse, $N=2$ supersymmetry or eight supercharges is an ideal number of supercharges. That's why those considerations played an important role in the Seiberg-Witten insights about $N=2$ gauge theories, among related situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paul Aspinwall liked to say that 8 supercharges is the optimum number for nice physics. Only Nature didn't manage to choose the number 8 for supercharges, but that's not Aspinwall's fault.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-10T19:12:52.850" LastActivityDate="2011-02-10T19:12:52.850" />
  <row Id="4918" PostTypeId="1" AcceptedAnswerId="4920" CreationDate="2011-02-10T12:38:10.643" Score="-1" ViewCount="172" Body="&lt;p&gt;Okay, so I just read through huge sections of &lt;a href=&quot;http://gordonwatts.wordpress.com&quot; rel=&quot;nofollow&quot;&gt;http://gordonwatts.wordpress.com&lt;/a&gt;, which I found quite interesting. There are some questions that really came to my mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do a lot of particle physicists use windows computers? (relative to other physicists?) And why? In most other academic departments, the professors seem to gravitate towards Linux/MACs. But Watts writes a lot of his programs for Windows only, which shows that there's definitely a lot of people who use it there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does particle physics research always stall until we get the sexiest new supercollider? (like, is it pretty much stalled until CERN starts putting out substantial output?) We haven't heard of anything from particle physics for quite some time (of course, we will soon, from CERN). But from the likes of particle physics blogs, I'm just getting the impression that everyone is talking about what's possible, rather than what results came out. It seems that condensed matter physics, biophysics, and astrophysics are the fields that are getting the most press.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're an experimental particle physicist, does that mean that you just use computational techniques to reduce/analyze the data now? In other words, could you be an experimental physicist without even knowing how to use laboratory apparatus? (just like how you can be an astrophysicist without knowing how to use a telescope these days).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1576" LastActivityDate="2011-02-10T13:22:29.690" ClosedDate="2011-02-10T14:09:39.547" Title="How does particle physics compare to other fields of physics?" Tags="&lt;particle-physics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="4919" PostTypeId="2" ParentId="4746" CreationDate="2011-02-10T13:16:40.893" Score="3" ViewCount="" Body="&lt;p&gt;The wider question here is about Glass formation and its properties which have been described as one of the main open questions in condensed matter physics. There is even a Paradox connected with current theories to explain glass formation temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One point to note is the definition of &quot;glass&quot; in all this. Our familiar example is the $SiO_2$ glass, but in the condensed physics context it means anything which enters the &quot;glass regime&quot; in the Temperature phase diagrams. The example which seems to be closer to this paper is &quot;Polymer Glass&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Papers identified by Carl Brannen, there was a mathematical model and computer simulation of a Quantum Theory which extended an existing classical theory of Glass formation. This showed that at the low Temperature quantum level there would indeed be a second &quot;liquid&quot; state forming. The classical model just had particles becoming more mobile at higher T, but at the quantum level the rise in wavelength allowed a second route to mobility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of practical applications it depends how general this model is, and what substances are &quot;glassy&quot; - rather than solid - at these low temperatures to begin with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of spacecraft, NASA has had a problem with the glass-solid boundary (in a polymer glass) at low temperatures before: this was the O-ring problem which caused the Challenger disaster.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-10T13:16:40.893" CommentCount="1" />
  <row Id="4920" PostTypeId="2" ParentId="4918" CreationDate="2011-02-10T13:22:29.690" Score="0" ViewCount="" Body="&lt;p&gt;Physics departments in the U.S. - and elsewhere - use lots of Linux computers and other Unix flavors, including Solaris, and this gets inherited by the particle physics community, too. It has various image and ideological causes but they're not the topic to be discussed on this server. Obviously, many people use Windows and some people use Macs as well. That's true in any community.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physicists who are genuinely compatible and deal with lots of data clearly can't have any problems to communicate with someone who uses a widespread operating system such as Windows. If I take my estimated statistics of the talks I have organized in a few years, most of them use Windows laptops, 1/4 uses Linux, and 1/4 uses Macs. That's my estimate. A significant fraction likes to have several operating systems on their computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, experimental particle physics depends on large experiments, and they take a lot of time to be built and started. So there are slow waves of the activity. The physicists may only take new exciting data when the data are available: I would believe that this tautological statement doesn't have to be explained even though this is what you seem to be asking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, experimenters are doing a lot of work when they prepare and plan new experiments or evaluate the data that were collected years ago. So there's always some work. Moreover, theorists are largely independent of the collider &quot;cycle&quot;. The more theoretical or formal a particle physicist is, the more his or her research is independent of the particular phase of the collider &quot;cycle&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most particle physicists are linked to experiments such as those at CERN and Fermilab, so they obviously don't make their particle physics experiment in their garage or living room. The colliders need extremely powerful computers to evaluate the data. At the top capacity, the LHC (1) will produce 15 petabytes (15 million gigabytes) of data per year. This has to be transmitted and evaluated by the &quot;grid&quot; etc. Obviously, questions about the user interface are not the primary questions here. It's the very immense amount of information that is the main problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, I think that one can be an experimental particle physicist without any real experience with &quot;tabletop experiments&quot;, but in that case, he or she can only do certain jobs - which are still needed but they don't cover anything. There have to be lots of experimental physicists who know how to create and run small apparata which become parts of detectors etc. However, teams such as one at CERN include something like 10,000 physicists so be sure that the roles are divided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think that condensed matter physics, biophysics, and astrophysics receive &quot;most press&quot;. Well, there are many new cosmic objects someone may see, so they of course produce a larger number of &quot;topics&quot; but each discovery generates a much smaller feedback than the corresponding advances in particle physics. On the contrary, I think that all those low-energy fields realize that particle physics is the top discipline of physics which is why some of them try to sell their research by (sometimes bogus or exaggerated) relationships with particle physics. So you may sometimes read articles such as &quot;graphene may uncover the secrets of the Higgs&quot;, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physicsworld.com/cws/article/news/44994&quot; rel=&quot;nofollow&quot;&gt;http://physicsworld.com/cws/article/news/44994&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In practice, the opposite articles don't exist because everyone knows that the low-energy physics questions are the more mundane ones.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-10T13:22:29.690" CommentCount="1" />
  <row Id="4921" PostTypeId="1" AcceptedAnswerId="4924" CreationDate="2011-02-10T13:27:31.803" Score="4" ViewCount="259" Body="&lt;p&gt;If the universe is &lt;em&gt;spatially infinite&lt;/em&gt; (and assuming, if it makes a difference, that we don't have eternal inflation), what actually happened 13.7 billion years ago? Was the energy density infinite (or &quot;very large&quot;) at each point in &lt;strong&gt;R&lt;/strong&gt;&lt;sup&gt;3&lt;/sup&gt;? Or did &lt;strong&gt;R&lt;/strong&gt;&lt;sup&gt;3&lt;/sup&gt; itself collapse to some other structure?  &lt;/p&gt;&#xA;" OwnerUserId="1394" LastActivityDate="2011-02-12T16:59:20.143" Title="The Big Bang in an infinite universe" Tags="&lt;general-relativity&gt;&lt;cosmology&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="4922" PostTypeId="2" ParentId="147" CreationDate="2011-02-10T13:30:48.493" Score="2" ViewCount="" Body="&lt;p&gt;About this:&lt;br&gt;&#xA;&lt;strong&gt;How are some useful ways to imagine a particle with spin 1/2 to make a 360° turn without returning to it's original position (the wave function transforms as: Ψ→−Ψ).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a nice example of such an objects -- &quot;the Dirac scissors&quot;:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/0wy8i.png&quot; alt=&quot;Dirac scissors&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture is from the book by Penrose and Rindler &quot;Spinors and space-time.&quot; I suggest to read it.&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-02-10T13:30:48.493" />
  <row Id="4923" PostTypeId="2" ParentId="629" CreationDate="2011-02-10T13:44:11.800" Score="7" ViewCount="" Body="&lt;p&gt;I drew up this picture below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/3Srpx.jpg&quot; alt=&quot;delivery of braking force between the ground and vehicle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where upon returning to this page I see this question is 3 months old.  However, I will answer this anyways, because nobody got the answer right in my opinion.  The picture illustrates the springs or struts in the under carriage of the vehicle.  In decelerating these communicate the deceleration from the road or rails to the vehicle.  The deceleration is due to a static friction (holonomic or no slipping) between the tire and the ground, $F_b~=~-kx$.  Here the braking friction is equal to the spring force the struts deliver.  The braking force has a dependency on the velocity, where $F_b(v)~=~ F\theta(v)$, which is a heaviside function that turns off when the velocity equals zero.  This means the spring has a slight distension when the vehicle comes to a rest and this then gives that little force forwards we experience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is often that people who do theoretical physics have never done things like rebuild a car engine.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-10T13:44:11.800" CommentCount="2" />
  <row Id="4924" PostTypeId="2" ParentId="4921" CreationDate="2011-02-10T13:54:01.427" Score="2" ViewCount="" Body="&lt;p&gt;The laws of physics - field theory and/or general relativity - are local (and in string theory, they're approximately local). So it doesn't make much difference whether a slice of the Universe is compact or noncompact. Near the Big Bang, the proper distances were shrunk almost to zero. But in cosmology, it still makes sense to use the coordinates in which the geometry of the slice is&#xA;$$ds^2 = a(t)^2 (dx^2+dy^2+dz^2)$$&#xA;where $a(t)$ is an overall rescaling factor that goes to zero, $a(t)\to 0$, for $t\to 0$. The parameter $t$ may also be reparameterized so that the $t=|r|$ lines are null which is useful to clarify the causal structure of the spacetime. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, near $t=0$, physics works &quot;locally&quot; in the coordinates $(x,y,z)$ above, regardless of the fact that $a(t)$ goes to zero. Objects moving along lines with variable $t$ but fixed $(x,y,z)$ correspond to static objects. You shouldn't imagine that objects are getting further in the $(x,y,z)$ coordinates near $t=0$ just in order to keep a proper distance fixed. No real objects are trying to keep their proper distance fixed as the Universe expands; the expansion of the proper distances is very real and you shouldn't try to deny it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any nonzero $t$, the $R^3$ stays the very same $R^3$ (and $S^3$ or $H^3$ would stay the same manifolds as well), and the laws of physics can't be blindly extrapolated to the strict $t=0$ point, anyway. So it makes no physical sense to ask what was &quot;before&quot; the Universe was an $R^3$. There was just a singularity at $t=0$; at every moment that physics can discuss, the geometry was $R^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eternal inflation has some new things to say - which would be relevant if eternal inflation were right - about these matters but you explicitly asked this topics to be avoided.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-10T13:59:21.443" LastActivityDate="2011-02-10T13:59:21.443" CommentCount="4" />
  <row Id="4925" PostTypeId="1" CreationDate="2011-02-10T13:59:51.307" Score="3" ViewCount="175" Body="&lt;p&gt;Why do Faddeev-Popov ghosts decouple in BRST? What is the physical reason behind it? Not just the mathematical reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If BRST quantization is specifically engineered to make the ghosts decouple, how does this engineering work?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-12T23:37:01.660" Title="Why do Faddeev-Popov ghosts decouple in BRST?" Tags="&lt;quantum-field-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="4926" PostTypeId="2" ParentId="629" CreationDate="2011-02-10T14:05:10.643" Score="1" ViewCount="" Body="&lt;p&gt;The effect comes from ones own &quot;springs&quot; (muscles keeping back) which last longer than the train's braking for a fraction of a second. This was explained several times already here, but there is one technical phenomenon not yet mentioned:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moment of &quot;switch&quot; from friction by braking, and the transition to friction at rest. This is a kind of &quot;snap in&quot; followed by an oscillation of the suspension system (Which is much shorter and smaller in trains compared to cars). This &quot;snap in&quot; when the wheels stop to turn, is a short extra&quot;impulse&quot; in deceleration. It may contribute to that feeling being &quot;pushed&quot; back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another effect is the train being compressed between the wagons (buffers). When the train stops and the brakes are released immediately, the train relaxes, and due to the big mass of the locomotive the wagons at the rear end will be pushed back. But that will feel like an extra braking for someone in one of the rearmost wagons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last is some physiology: all our senses, including the feel of strain in our tissue, do get &quot;used to&quot; when the same condition carries on. You know maybe of the red afterimage, when you look at a green spot for a minute, and then you turn away? Similarly, we get used to the feel of &quot;deceleration&quot; and when the braking stops, we react as if we felt some acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another effect is by &quot;preparedness&quot;: a driver in a car is prepared to whatever he &quot;does&quot;. He senses those phenomena much less than a passenger in a car or train.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know whether all these effects are important, but an experiment might be to have a kind of safety belt in the train. This would be tightened and you would try not to &quot;do&quot; something. Then you tell your feelings here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum:&lt;/strong&gt;&#xA;For a &quot;rigorous&quot; theory of the effect, one has to have &#xA;a model of the passenger first. Because the reactions &#xA;are well below (too fast) conscious doing, a closed loop &#xA;control circuit trying to hold the body upright is the &#xA;appropriate description. &#xA;The train can be described as being braked in a way that &#xA;a constant force  decelerates the train. &#xA;When the train comes to rest, the force drops to zero.&#xA;This is not the exact truth, but lacking a more sophisticated &#xA;&quot;brake function&quot; this one will do. &#xA;To see what happens principally to the passenger needs no calculation. &#xA;It's just the well-known reaction of a closed loop control (PID)&#xA;circuit to a step function. There will be time lag (&quot;dead time&quot;), proportional reaction with some overshoot. Integral part  will &#xA;regulate that &quot;getting used to&quot; phenomenon. The closed loop is comprised &#xA;of the passenger's labyrinth organ (accelerometer) and eyes, &#xA;and nerves and muscles. &#xA;To solve this, one needs the values of this control circuit, maybe &#xA;well known to &quot;sports physics&quot;?&#xA;The resulting movement, especially acceleration of the passenger's head &#xA;will tell the &quot;feeling&quot; of the passenger. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Another addendum:&lt;/strong&gt; &#xA;The people programming/constructing flight simulators will &#xA;have the figures for the control loop, especially the &#xA;accelerometer/equilibrium organ in our ear.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="185" LastEditorDisplayName="" LastEditDate="2011-02-17T09:30:50.347" LastActivityDate="2011-02-17T09:30:50.347" CommentCount="3" />
  <row Id="4927" PostTypeId="2" ParentId="4908" CreationDate="2011-02-10T14:17:17.130" Score="-4" ViewCount="" Body="&lt;p&gt;If one builds a QG in a flat space-time, a la Logunov's RTG, then one may have global symmetries. But it is forbidden to say and is punished, beware.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-02-10T14:17:17.130" />
  <row Id="4928" PostTypeId="2" ParentId="629" CreationDate="2011-02-10T14:26:00.280" Score="1" ViewCount="" Body="&lt;p&gt;Bit late to the game, but if we put a glass of water on the train in these circumstances you will see the water pushed to the &quot;front&quot; side of the glass as the train slows and then, when the train comes to a stop,  swish to the &quot;back&quot; side of the glass, repeating in broadly diminishing cycles until coming to a rest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The human body is certainly more rigid than a water in a glass, but your organs (and brain which is pretty heavy) will follow the waters pattern of motion and kick back. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now try the same thought experiment with a glass of frozen solid water. No kickback. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the answer is fluid dynamics combined with deceleration. &lt;/p&gt;&#xA;" OwnerUserId="1320" LastActivityDate="2011-02-10T14:26:00.280" CommentCount="3" />
  <row Id="4929" PostTypeId="2" ParentId="4921" CreationDate="2011-02-10T14:43:22.343" Score="0" ViewCount="" Body="&lt;p&gt;Your question is very good.  I find the idea of eternal inflation to be problematic.  The vacuum which induces this frantic inflation is a scalar field which gets stretched beyond the horizon length.  In this inflationary situation the vacuum energy density is about 14 orders of magnitude lower than the Planck energy density.  This means any region has a Hubble volume at subatomic scales and the inflaton field is stretched across regions where it is not causally complete.  It gets frozen, in a sense.  It also becomes attenuated by its own inflationary action.  So the eternal inflation concept has always struck me as questionable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The potential for the inflationary potential early in the universe is a de Sitter form.  The FLRW equations are&#xA;$$&#xD;&#xA;\Big(\frac{\dot a}{a}\Big)^2~=~\frac{8\pi G\Lambda}{3}~-~\frac{k}{a^2},&#xD;&#xA;$$&#xA;where we assume $k~=~0$ for the generally flat space we appear to observe.  The early inflationary universe was driven by a scalar field which generated this vacuum energy where $V(\phi)~=~-a\times\phi$, $a$ a constant.  This set the early cosmological constant for the de Sitter expansion with a vacuum energy about 13 orders of magnitude smaller than the Planck energy.  The universe had more vacuum energy density than quark-gluon field density in a hadron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lagrangian for a scalar field is $L~=~(1/2)\partial^a\phi\partial_a\phi~–~V(\phi)$ and in QFT we work with the Lagrangian density ${\cal L}~=~L/vol$ so the action $S~=~\int d^3xdt{\cal L}(\phi, \partial\phi)$. We run this into the Euler-Lagrange equation $\partial_a(\partial{\cal L}/\partial(\partial_a\phi))~-~\partial{\cal L}/\partial\phi~=~0$, and keep in mind $vol~\sim~x^3$. This gives a dynamical equation&#xA;$$&#xD;&#xA;\partial^2\phi ~-~ (3/vol^{4/3})\partial_a\phi~–~ \frac{\partial V(\phi)}{\partial\phi}~=~ 0.&#xD;&#xA;$$&#xA;If we assume the inflaton field is more or less constant on the space for a given time on the Hubble frame this DE may be simplified to&#xA;$$&#xD;&#xA;{\ddot\phi}~–~(3/vol^{4/3}){\dot\phi}~–~\frac{\partial V(\phi)}{\partial\phi}~=~0&#xD;&#xA;$$&#xA;That middle term is interesting for it is a sort of friction. It indicates the inflaton field, the thing which drives the inflationary expansion, is running down or becoming diffused in the space. The potential function here is complicated and not entirely known, but it is approximately constant, or some small decrease with the value of $\phi$.   What then happens, which is not entirely understood, is that the field experiences a phase transition, the potential becomes $V(\phi)~\sim~\phi^2$ with a minimum about 110 orders of magnitude smaller than it was in the unbroken phase. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So nucleation bubbles, such as the observable universe with a small cosmological constant, form so long as the inflaton field is large.  However, as it rolls downwards and becomes attenuated bubble nucleation may decrease.  It is like soda water which loses its fizziness.  So the whole $R^3$ may by now be “flat,” and our observable universe is due to one bubble out of a vast number produced spread across this $R^3$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what started this whole shebang?  It may be due to the $Dp$-brane collisions, or due to the tunneling of some vacuum region near a black hole singularity in another spacetime, or … .  There are a number of possible ways this may have been generated.  So we might imagine some vacuum “blob” near a black hole singularity which tunnels out of that spacetime and generates a nascent deSitter spacetime.  For the Reissnor-Nordstrom metric with $g_{tt}~=~1~-~r_0/r~-~\Lambda r^2/3$ then near the signularity the Wheeler-deWitt (WDW) euqation is approximately&#xA;$$&#xD;&#xA;\Big[\frac{\partial^2}{\partial a^2}~-~\frac{9\pi^2}{4G^2}\Big(a^2~-~\frac{\lambda}{3}a^4\Big)\Big]\Psi[a]~=~0,&#xD;&#xA;$$&#xA;where there is this pottential $U(a)~=~Ca^2~-~Da^4$ the quantum field tunnels through.  This is then a barrier for the production of a cosmology.  If $k~=~1$ for this blob then this is a tiny $R^3$ region which tunnels into a new spacetime.  So somehow if the observable universe is flat with $k~=~0$ there is a topology change.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The WDW equation defines Hartle-Hawking states.  A recent paper by Ashoke Sen &lt;a href=&quot;http://arxiv.org/abs/1101.4254&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1101.4254&lt;/a&gt; illustrates how the $AdS_2$ boundary is a $CFT_1$ which corresponds to Hartle-Hawking states.  This means there may be some “braney” correspondence between the above model with the WDW and $Dp$-brane physics.  So the blob which tunnels from the neighborhood of a black hole singularity might amount to the quantum “deposition” of IIB strings onto a brane.  This then kicks off the inflationary spacetime, or think of dropping the Mentos into the coke bottle, which sets off lots of bubble nucleation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is admittedly rather speculative.  I am suggesting this as a possible way of thinking about this, and way which might get around this little issue you raise with the apparent problem of there being an infinite amount of energy on an $R^3$.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-02-10T14:51:14.157" LastActivityDate="2011-02-10T14:51:14.157" CommentCount="5" />
  <row Id="4930" PostTypeId="1" AcceptedAnswerId="4934" CreationDate="2011-02-10T14:54:03.527" Score="4" ViewCount="220" Body="&lt;p&gt;I've read some papers recently that talk about gapped Hamiltonians or gapless systems, but what does it mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Is an XX spin chain in a magnetic field gapped? Why or why not?&lt;/p&gt;&#xA;" OwnerUserId="873" LastEditorUserId="873" LastEditorDisplayName="" LastEditDate="2011-02-10T16:55:59.287" LastActivityDate="2011-02-10T19:42:32.120" Title="What does it mean for a Hamiltonian to be gapped?" Tags="&lt;quantum-mechanics&gt;&lt;condensed-matter&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="4931" PostTypeId="2" ParentId="4888" CreationDate="2011-02-10T14:54:40.667" Score="2" ViewCount="" Body="&lt;p&gt;It's getting to the point where building nanoscale devices is commercially practical, so maybe nanophotonics, nanomaterials, nanofluid dynamics, nanoengineering, nanomechanics (?), etc. This area may be very important for practical quantum computing, but there are many much nearer applications. And some of these are being studied in engineering departments rather than physics departments, but you should be able to find related research topics in physics.&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2011-02-10T14:54:40.667" />
  <row Id="4932" PostTypeId="1" CreationDate="2011-02-10T14:55:46.407" Score="3" ViewCount="193" Body="&lt;p&gt;Does Wick rotation work for quantum gravity? The Euclidean Einstein-Hilbert action isn't bounded from below.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-10T19:09:25.147" Title="Does Wick rotation work for quantum gravity?" Tags="&lt;quantum-gravity&gt;" AnswerCount="2" FavoriteCount="4" />
  <row Id="4933" PostTypeId="2" ParentId="4930" CreationDate="2011-02-10T15:18:13.830" Score="2" ViewCount="" Body="&lt;p&gt;Gapped and gapless are usually attributes for many-body Hamiltonians. A gapped Hamiltonian is simply one for which there is a non-zero gap between the ground state and the first excited state.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-02-10T15:18:13.830" CommentCount="3" />
  <row Id="4934" PostTypeId="2" ParentId="4930" CreationDate="2011-02-10T15:18:58.980" Score="3" ViewCount="" Body="&lt;p&gt;This is the distinction between continuous and discrete spectrum, but only considering the low energy excitations. For an Hamiltonian with gapped spectrum, the lowest eigenvalue is separated by a gap from $E=0$. For example dispersion relation of the form $E=|k|$ is an example of gapless spectrum, and $E=|k+m|$ is an example of gapped one, where $k$ is the wave vector (which can be any real number) and $m$ is the mass (= gap).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This distinction leads to qualitative difference in the physics - for example the difference between a material being a conductor or an insulator. Many times also, the gap is generated by interesting physics (like the mass gap in Yang-Mills theory, or the gap in BCS superconductivity).&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-02-10T17:02:10.247" LastActivityDate="2011-02-10T17:02:10.247" CommentCount="5" />
  <row Id="4935" PostTypeId="1" AcceptedAnswerId="4945" CreationDate="2011-02-10T15:22:22.250" Score="8" ViewCount="292" Body="&lt;p&gt;Does a relativistic version of quantum thermodynamics exist? I.e. in a non-inertial frame of reference, can I, an external observer, calculate quantities like magnetisation within the non-inertial frame? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be interested to know if there's a difference between how to treat thermodynamics in a uniformly accelerated reference frame and in a non-uniformly accelerated reference frame. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="873" LastActivityDate="2011-02-11T04:10:23.220" Title="Is there a relativistic (quantum) thermodynamics?" Tags="&lt;quantum-mechanics&gt;&lt;thermodynamics&gt;&lt;relativity&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="3" />
  <row Id="4937" PostTypeId="1" AcceptedAnswerId="4951" CreationDate="2011-02-10T15:43:03.060" Score="10" ViewCount="418" Body="&lt;p&gt;The Higgs mechanism allows massless fields to acquire mass through their coupling to a scalar field. But if the masses cannot be predicted because the couplings have to be fixed, what really is the utility of the Higgs mechanism? Instead of saying &quot;Here are &lt;em&gt;a priori&lt;/em&gt; couplings; the Higgs mechanism generates mass.&quot;, I could just as well say &quot;Here are &lt;em&gt;a priori&lt;/em&gt; masses. Period.&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the Higgs mechanism is crucial to electroweak unification, but I have the same question there. Why does electromagnetism and the weak force have to be unified? Even if the couplings of the photon, Z and W bosons became related on unification, this is still at the cost of introducing new parameters - so it's not really clear to me that something has been explained or tidied up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do either the Higgs mechanism or electroweak unification tell us something new? Do either make any predictions that don't come at the cost of extra parameters? (I'm not really challenging anything here; I'm sure the answer to both questions is 'yes' - I just want to fill the gaps in my understanding as I study the Standard Model)&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-02-11T06:12:47.857" Title="What is the need for the Higgs mechanism and electroweak unification?" Tags="&lt;quantum-field-theory&gt;&lt;particle-physics&gt;&lt;standard-model&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="4938" PostTypeId="2" ParentId="2490" CreationDate="2011-02-10T15:50:27.897" Score="0" ViewCount="" Body="&lt;p&gt;Bells have evolved into the &quot;bell&quot; shape because a vibrating bell produces different partial frequencies depending on metallic thickness and shape. My &lt;a href=&quot;http://www.allsaintswokinghambells.org.uk/AbBells/AbBells.html&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt; says that &quot;The shape of the bell has evolved so that the main partials are in roughly the right relationship. Removing small amounts of metal from different parts of the inside of the bell adjusts the different frequencies. A modern bell is cast deliberately thick, and then tuned on a vertical boring machine (a giant lathe). Before this, metal was chipped away with a tuning hammer, a practice that persisted into the early 20th century, despite the introduction of tuning machines from the late 18th century.&quot;&lt;br&gt;&#xA;In some bells, the tuning is done by stripping inner annular rings from the bell. Obviously this is much easier than trying to remove the proper amt of metal from other shapes.&#xA;Thanks for an interesting question :)&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-02-10T15:50:27.897" CommentCount="3" />
  <row Id="4939" PostTypeId="2" ParentId="4937" CreationDate="2011-02-10T16:09:31.960" Score="0" ViewCount="" Body="&lt;p&gt;Electroweak unification means that there is a symmetry between electromagnetic and weak interactions -- you can use them interchangeably. In reality this is not the case -- $W$ and $Z$ bosons have mass, while photon haven't. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Higgs mechanism provides a way for a &lt;em&gt;spontaneous symmetry breaking&lt;/em&gt; between those interactions: Lagrangian of standard model is electroweak symmetric, while the vacuum is not due to non-zero vacuum expectation value for the Higgs field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same applies for fermions -- you cannot introduce mass terms for quarks and leptons in the Lagrangian, while preserving the electroweak symmetry. But it is possible to introduce electroweak symmetric Yukawa terms, coupling Higgs field to the fermions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;br&gt;&#xA;I don't think that Higgs mechanism can &quot;tell us something new&quot;. It is just a simplest way to ensure spontaneous symmetry breaking. While electroweak unification means that those interactions are gauge interactions and establishes the gauge symmetry itself. The classification of fermions into three generations is also done from the &quot;electroweak point of view&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you can argue that this systematization or classification is not &quot;something new&quot;. But from such a point of view one can criticise almost every theoretical construction that attempts to predict results of future experiments.&lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2011-02-10T16:33:25.263" LastActivityDate="2011-02-10T16:33:25.263" CommentCount="3" />
  <row Id="4940" PostTypeId="2" ParentId="4937" CreationDate="2011-02-10T16:10:23.573" Score="-5" ViewCount="" Body="&lt;p&gt;In a phenomenological approach one introduces masses without problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is in the &quot;local gauge invariance&quot; approaches where one needs &lt;strong&gt;a fix&lt;/strong&gt; because the gauge approach does not give masses, unfortunately. Technically it is is implemented as &quot;coupling&quot; to Higgs in some way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I never considered the &quot;local gauge invariance principle&quot; as reasonable or physical. It is one of many blind ways to &quot;construct&quot; interacting theories; it does not guarantee anything physical and does not save us from infinities in calculations. It's been done by analogy with QED which itself has problems. &lt;strong&gt;This way needs fixes.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say the Higgs is a price for choosing a &quot;mathematical&quot; rather than physical approach to constructing theories. It is a wrong guess, in my opinion. There are  more physical ideas on this subject but they remain unknown and thus unexploited.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-10T17:34:49.460" LastActivityDate="2011-02-10T17:34:49.460" CommentCount="2" />
  <row Id="4941" PostTypeId="1" CreationDate="2011-02-10T16:12:13.900" Score="3" ViewCount="216" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Van_Eck_phreaking&quot; rel=&quot;nofollow&quot;&gt;Van Eck phreaking&lt;/a&gt;, the ability to reconstruct distally the text on a CRT or LCD screen using the leaking em from the target computer, was in the news about five to ten years ago. It is talked about as if it is easily done, but I see little hard confirmation. The NSA has something called project &lt;a href=&quot;http://en.wikipedia.org/wiki/TEMPEST&quot; rel=&quot;nofollow&quot;&gt;TEMPEST&lt;/a&gt; to study this, but I think most of it is classified.  Does anyone know how good van Eck phreaking is, or is it mostly hot air?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question also leads to attempts by neuroscientists to read brain states to control movement etc (ie flying a plane by thought, using brainwave em to control machines (paraplegics etc)...some progress has been made.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-02-10T18:25:38.707" LastActivityDate="2011-02-11T00:00:07.193" Title="Does van Eck phreaking really work, or is it an urban myth?" Tags="&lt;electromagnetism&gt;&lt;electronics&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="4943" PostTypeId="1" CreationDate="2011-02-10T16:23:18.170" Score="4" ViewCount="182" Body="&lt;p&gt;The work done during a process between two equilibrium states can be described by thermodynamics. Even when process itself is out of equilibrium, the thermodynamic laws can still be used, though calculating the work is much more difficult. But if the initial or final states, or both, are not in equilibrium, can the work done in driving the system from one to the other be calculated? How?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; @Roy @genneth @Marek I do mean far out of equilibrium. There wasn't a nonequilibrium tag, and I don't have enough reputation points to create one! I can't seem to post comments today, so I'll write this here instead: I'm familiar with Jarzynski's equality and while it is very useful, it is still only valid between equilibrium initial and final states (though the final state being in equilibrium can be relaxed). I'd like to know if there's any way work (or heat) can be defined or calculated when the initial and final states are out of equilibrium, possibly very out of equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit2:&lt;/strong&gt; @Roy I'd like to know if it's possible in a system with an initial state which is not in equilibrium which is then driven to a final value which is still not in equilibrium. I don't want to make any other assumptions if possible. So I don't necessarily expect local equilibrium to hold, although I'd still be interested to know if work between the 2 nonequilibrium states can be found in that case.&lt;/p&gt;&#xA;" OwnerUserId="873" LastEditorUserId="873" LastEditorDisplayName="" LastEditDate="2011-02-12T10:07:13.467" LastActivityDate="2011-02-12T10:07:13.467" Title="Can the work done between two non-equilibrium states be calculated?" Tags="&lt;thermodynamics&gt;&lt;non-equilibrium&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="4944" PostTypeId="2" ParentId="4937" CreationDate="2011-02-10T16:31:21.973" Score="0" ViewCount="" Body="&lt;p&gt;It is true that recognizing that  the data have an SU2xU1 symmetry one has a number of parameters and one may say that problem number one has been reduced to problem number two which has the same number of unknowns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me give the often quoted example of the epicycles in the geocentric system and the ellipses in the heliocentric one. The number of parameters are the same, and if you go to a planetarium program and go to the geocentric frame, the epicycles will appear in all their glory. Nevertheless there is nobody now who would ask, &quot;what is the use of ordering the data in the heliocentric system&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asking &quot;why should electromagnetism be unified with weak theory&quot; is a bit like asking &quot;why have a heliocentric system&quot;.The answer is that in both cases, the data fall into an ordered form effortlessly. And then we were led to higher symmetries (SU3xSU2xU1) and more inclusive theories. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should have added that the clarity introduced by  symmetries/order once manifested lead to calculable theories which can describe the data and make predictions for new observations. The Higgs is one such prediction coming out from the theories describing the quark model symmetries. If it is not found, a rethink will be in order, (though there are papers which claim that a Higgs mechanism can be a composite manifestation, not necessarily one particle).&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-02-11T06:12:47.857" LastActivityDate="2011-02-11T06:12:47.857" CommentCount="18" />
  <row Id="4945" PostTypeId="2" ParentId="4935" CreationDate="2011-02-10T16:34:54.953" Score="5" ViewCount="" Body="&lt;p&gt;There is a classic treatise on &quot;Relativity, Thermodynamics and Cosmology&quot; from R. Tolmann from the 1930s - it is still referenced in papers today. This generalises Thermodynamics to Special Relativity and then General Relativity. As a simple example the transformation law for Temperature is stated as: $T=\sqrt(1-v^2/c^2)T_0$ when changing to a Lorentz moving frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is that &quot;entropy density&quot; $\phi$ is introduced, which is also subject to a Lorentz transformation. Finally this becomes a scalar with an associated &quot;entropy 4-vector&quot; in GR. The Second Law is expressed using these constructs by Tolmann.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is some discussion in Misner, Thorne and Wheeler too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course both these texts also include lots of regular General Relativity Theory which you may not need.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-10T16:34:54.953" CommentCount="1" />
  <row Id="4946" PostTypeId="2" ParentId="4941" CreationDate="2011-02-10T16:36:41.910" Score="2" ViewCount="" Body="&lt;p&gt;I saw a &quot;computer&quot; show on TV on this topic in those days, &#xA;they had the equipment and it worked, at least for some &#xA;dozen meters (Not in HD-quality:=) &#xA;I doubt, that this would have worked out of an business building, &#xA;where several CRT would transmit at the same time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But all this was CRTs! There was a voltage at the CRTs &#xA;control grids (Wehnelt) which was modulated with the frequencies &#xA;of brightness changes for the CR running along a line. &#xA;The frequency spectrum is up to some dozen MHz, something &#xA;easily radiated by the circuit, if not shieleded properly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today, the flatscreens do radiate much less/maybe not enough &#xA;for spys even when close. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the nature of the signal, things like letters or &#xA;vector graphic symbols were easily to recognize, but &#xA;half-tone pictures were not. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-02-10T16:52:27.623" LastActivityDate="2011-02-10T16:52:27.623" />
  <row Id="4947" PostTypeId="2" ParentId="4076" CreationDate="2011-02-10T16:37:51.967" Score="4" ViewCount="" Body="&lt;p&gt;From my studies of Einsteins work I am under the impression that the most important motive for his works stems from his interest in understanding physics from the point of view that matter exists of atoms and molecules that follow paths in a space-time continuum.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the sense that he was strongly influenced by the idea of a deterministic universe - the Kant-Laplace universe. This point of view easily explains why he couldn't accept quantum physics where the concept of probablities - without underlying mechanism - supersedes determinism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the comment of space_cadet is important: &quot;[...] the atomic hypothesis was not universally accepted before 1905 [...] a direct link between the atomic hypothesis and the observed macroscopic properties of matter&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my view his work on the SRT has the same motive: understanding space-time relations from the point of view of light rays imagined as moving more or less like macroscopic objects.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2011-02-10T16:37:51.967" CommentCount="1" />
  <row Id="4948" PostTypeId="2" ParentId="4925" CreationDate="2011-02-10T16:43:59.937" Score="3" ViewCount="" Body="&lt;p&gt;Dear Student, as Moshe says, the reason why the Faddeev Popov ghosts &quot;decouple&quot; is that they're designed to decouple. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, they - and all the formulae that depend on them - are designed so that the excitations of these ghosts, as well as unphysical excitations of more ordinary physical fields - such as the time-like and longitudinal components of gauge fields - are cleanly separated from the physical polarizations - such as transverse polarizations of the photons and other gauge bosons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The addition of the extra BRST ghosts and the BRST charge makes things particularly natural.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rough reason why the right physical states survive - as BRST cohomologies - while the wrong states don't survive is easily seen in the case of commuting gauge symmetries, generated by generators $L_i$. In that case, the BRST charge is&#xA;$$ Q = \sum_i c^i L_i$$&#xA;The ghosts $c^i$ anticommute with each other - and square to zero. You see that the single condition&#xA;$$ Q|\psi \rangle = 0 $$&#xA;automatically implies that &#xA;$$L_i |\psi\rangle = 0,$$&#xA;among other things, so that the states $|\psi\rangle$ are automatically gauge-invariant. Also, states that are gauge transformations of each other become equivalent if we define $Q$ variations of all states to be unphysical. In particular, if &#xA;$$|\chi\rangle = \sum_i \lambda^i b_i |\phi\rangle$$&#xA;then &#xA;$$Q|\chi\rangle = \sum_i \lambda^i L_i |\phi\rangle$$&#xA;is an arbitrary gauge variation of the $|\phi\rangle$ vector (which was assumed to be annihilated by the $c$'s in this case), which is identified with the vector $0$ in the physical Hilbert space by the BRST cohomology business.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Excitations of $b,c$ themselves are unphysical&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The states that contain nothing else than unwanted excitations of $b_i$ fail to be BRST-closed (they are not annihilated by $Q$, much like the components of the gauge field that violate the Gauss's law) while the states that contain just unwanted excitations of $c_i$ are BRST-trivial (much like the pure gauge configurations of the gauge field). So even the excitations of $b,c$ become unphysical, together with two non-transverse excitations of the gauge field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole BRST machinery is only useful for non-Abelian (non-commuting) groups but the machinery above may be easily generalized. A $cbc\cdot f/2$ term ($f$ are the structure constants) has to be added to the BRST charge to preserve $Q^2=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The presentation above is mathematical because the BRST machinery &lt;em&gt;is&lt;/em&gt; just a mathematical trick. There is no real physics. Indeed, by the very definition, the whole toolkit added by the BRST formalism is by definition unphysical. It is by definition a mathematical trick. Everyone who is telling you that he or she is discovering new physics by playing with the BRST formalism itself is lying to you; it is not possible. There's no physics in the BRST formalism; it's just a very useful modern tool to do physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let me add one more comment why the unphysical states - the states that are not BRST-closed; and the states that are BRST-exact - decouple. If the initial state at $t=0$ is physical, it remains physics at later moments - because $Q$ commutes with the Hamiltonian $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At $t=0$, it is legitimate to identify physical; states that differ by $Q|\phi\rangle$ for any $|\phi\rangle$ because all these differences of the form $Q|\phi\rangle$ are orthogonal to all physical states $\langle \kappa|$ because&#xA;$$\langle\kappa|Q|\phi\rangle =0.$$&#xA;The latter is true because you may act with $Q$ on the left and the result is zero because $\langle \kappa|$ was BRST-closed by assumption.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-10T16:50:11.270" LastActivityDate="2011-02-10T16:50:11.270" />
  <row Id="4949" PostTypeId="2" ParentId="4888" CreationDate="2011-02-10T17:00:30.810" Score="2" ViewCount="" Body="&lt;p&gt;I think Shor is in part right about nano-scaled technology.  This was something Feynman wrote about in 1960!  This is a fascinating talk he have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.zyvex.com/nanotech/feynman.html&quot; rel=&quot;nofollow&quot;&gt;http://www.zyvex.com/nanotech/feynman.html&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The future of technology probably follows three rules.  The first rule I call the stone-tool rule.  The second is historical continuity, and the third is what I call the anthro-entropy principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stone-tool rule is that we are wrong in saying people gave up stone tools, in particular blades, in favor of bronze and later iron ones because metals were better.  This is clearly wrong.  If you have the time do some look at Folsom tips and other late Paleolithic stone tools.  A well flaked blade has a concoidal fracture which sharpens to a single atom.  It is almost a form of nano-tech.  Well made stone blades in the hands of a skilled worker are far superior to metal blades.  We shifted to metal blades because they can be made in greater numbers much more quickly, not because they are superior.  We substituted metal parts with plastic, because plastic is cheaper and easier to form objects with, but is usually inferior in quality.  Future technology will follow to some degree the same rule, we will have smaller cheaper devices in far greater numbers which do not last as long, become obsolete faster, but which can be accessed easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thomas Hobbes commented on how an ounce of gold bought one a week’s lodging, food and other expenses.  Though gold prices have been in a bit of a peak of late, the rule curiously holds true today on average.  This is an example of historical continuity.  Another example is with our forms of entertainment, which is a large sector of economic activity.  The Roman constructed large coliseums for the purpose of gladiatorial games and “violencia.”  Lots of engineering, hard work and investment capital went into this activity.  We really do much the same, but in a much more sophisticated form.  We have used our basic knowledge of physics to make plasma screens with video productions of very violent content.  The average American watches in a lifetime about 50,000 killings in this vicarious or virtual setting.  Historical continuity again, and any technology in the future with enhances the stimulus of the senses and heighten the endorphin levels of people through such violent content will doubtless be a winner.  Other forms of spectacles hold as well, sports, rock concerts and their ancient comparison with chariot races and other circuses.  The other constant is war-craft, which since the ancient world has consistently taken up 20-25% percent of economic activity.  While it is nice to think the peacemakers will inherit the Earth, the safe bet is on the defense contractors.  The next big form of activity is with food production, and agriculture based technologies will continue to move forwards.  This of course has clear implications for bio-tech and genetic engineering.  Finally the last is medical, where the just as the shaman of Pleistocene cultures held high position, so too do doctors and their ability to permit us to “beat the reaper,” at least for another day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final is the anthro-entropy principle.  Our species has an amazing capacity to exploit its environment.  It is what intelligence gives you the ability to do, and so far any environmental limit encountered can be circumvented.  So we are a species of life which, unlike all others, has no permanent biological constraints upon it by other life forms.  This permits our numbers to grow and for us to exert greater control and exploitation of the world.  The net result is we use up stores of natural energy and materials and replace them with entropy and garbage.  The trend is also exponential, where in more recent times we appear to be little more than a brainy ground ape on an exponential rampage.  Any future technology which facilitates this progression will win the day as well.  This includes energy efficiency technologies, where by increasing efficiency and reducing costs this ends up permitting far greater numbers of people to use the technology and ends up facilitating more consumption and ultimately more entropy production.  Further, even though our population growth is starting to slow, it is being done by permitting vast numbers of people to consume far more.  In 1960 the world population was 3 billion, but based on per capita use of energy and materials our current population would be over 20 billion, based on a 1960 consumption level.  So we are replacing one form of growth with another.  Look up Jevon’s paradox for examples of how this works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above little matter will not be solved by economics or politics.  It will end in much the same way locust swarms come to an end: They eat everything up and die out.  The one thing we will not be able to overcome is when we use up the natural world itself and have nothing left to work with.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-02-10T17:23:14.143" LastActivityDate="2011-02-10T17:23:14.143" />
  <row Id="4950" PostTypeId="2" ParentId="4935" CreationDate="2011-02-10T17:09:49.553" Score="3" ViewCount="" Body="&lt;p&gt;Jane, you had to mean quantum statistical mechanics. Thermodynamics only emerges in the thermodynamic limit of statistical mechanics, which requires a large number of degrees of freedom, and when you have a large number of degrees of freedom, the relevant quantities automatically behave classically, too. In this sense, thermodynamics is always classical (non-quantum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum statistical physics is easily promoted to special relativity. But one must understand that a thermal ensemble is given by &#xA;$$\rho\approx \exp(-\beta H)$$&#xA;where $H$ is the energy, the generator of translations in time. You may extend it in a relativistically invariant way,&#xA;$$\rho\approx \exp(-\beta H - \vec\beta_p \cdot \vec p)$$&#xA;where $\vec\beta_p$ are spatial components of inverse temperature. However, the density matrix above is equivalent - by a Lorentz transformation - to the previous one. Only the norm of the 4-vector $(\beta,\beta_p)$ has an invariant physical meaning, and you may always go to the frame where the spatial components vanish. In this frame, you get an ordinary thermal ensemble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, another question is whether one knows how to calculate the thermal properties of objects in a particular relativistic theory. That involves new mathematical challenges. But the definition of the objects - e.g. the thermal density matrix - is totally unaffected by having a relativistic theory.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the formula for the thermal state is actually identical for relativistic quantum statistical mechanics - and obviously, it's also identical in the classical limit of it, the relativistic classical statistical mechanics. The 4-vector of the inverse temperature had to be time-like, otherwise the density matrix would be divergent (the momentum-dominated &quot;energy&quot; wouldn't be bounded from below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general relativity, you may define the temperature (and its preferred reference frame) locally, and locally, the physics reduces to that of special relativity which was explained above. However, there's no real ensemble with a fixed global temperature globally, except for geometries that are static - that have a time-like Killing vector field. That's because to define the thermal ensemble, you need a periodic Euclidean time (the energy generating time translations enters the exponential in the density matrix), and if the original solution nontrivially depended on time, you couldn't make the time periodic in the imaginary direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many fancy thermal effects in curved spaces - such as particle production; Unruh effect; and Hawking radiation. The most exciting portion is black hole thermodynamics. However, to get there, one must first understand why the extension of statistical mechanics to the special relativistic context isn't really conceptually new. It may be useful if you asked a more advanced question about the references when you figure out what is actually out there so that your question becomes somewhat more well-defined.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-10T17:09:49.553" CommentCount="2" />
  <row Id="4951" PostTypeId="2" ParentId="4937" CreationDate="2011-02-10T17:23:49.823" Score="8" ViewCount="" Body="&lt;p&gt;Dbrane, aside from &quot;beauty&quot;, the electroweak unification is actually needed for a finite theory of weak interactions. The need for all the fields found in the electroweak theory may be explained step by step, requiring the &quot;tree unitarity&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is explained e.g. in this book by Jiří Hořejší:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/9810218575&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/dp/9810218575/&lt;/a&gt;&lt;br&gt;&#xA;  &lt;strong&gt;Google books:&lt;/strong&gt;&lt;br&gt; &lt;a href=&quot;http://books.google.com/books?id=MnNaGd7OtlIC&amp;amp;printsec=frontcover&amp;amp;hl=cs#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=MnNaGd7OtlIC&amp;amp;printsec=frontcover&amp;amp;hl=cs#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sketch of the algorithm is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beta-decay changes the neutron to a proton, electron, and an antineutrino; or a down-quark to an up-quark, electron, and an anti-neutrino. This requires a direct four-fermion interaction, originally sketched by Fermi in the 1930s, and improved - including the right vector indices and gamma matrices - by Gell-Mann and Feynman in the 1960s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this 4-fermion interaction is immediately in trouble. It's non-renormalizable. You may see the problem by noticing that the tree-level probability instantly exceeds 100% when the energies of the four interacting fermions go above hundreds of GeV or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to fix it is to regulate the theory at higher energies, and the only consistent way to regulate a contact interaction is to explain it as an exchange of another particle. The only right particle that can be exchanged to match basic experimental tests is a vector boson. Well, they could also exchange a massive scalar but that's not what Nature chose for the weak interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there has to be a massive gauge boson, the W boson.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One finds out inconsistency in other processes, and has to include the Z-bosons as well. One also has to add the partner quarks and leptons - to complete the doublets - otherwise there are problems with other processes (probabilities of interactions, calculated at the tree level, exceed 100 percent). It goes on and on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end, one studies the scattering of two longitudinally polarized W-bosons at high energies, and again, it surpasses 100 percent. The only way to subtract the unwanted term is to add new diagrams where the W-bosons exchange a Higgs boson. That's how one completes the Standard Model, including the Higgs sector. Of course, the final result is physically equivalent to one that assumes the &quot;beautiful&quot; electroweak gauge symmetry to start with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a matter of taste which approach is more fundamental and more logical. But it's certainly true that the form of the Standard Model isn't justified just by aesthetic criteria; it can be justified by the need for it to be consistent, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, 3 generations of quarks are needed for CP-violation - if this were needed. There's not much other explanation why there are 3 generations. However, the form of the generations is tightly constrained, too - by anomalies. For example, a Standard Model with quarks and no leptons, or vice versa, would also be inconsistent (it would suffer from gauge anomalies).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-10T19:41:45.790" LastActivityDate="2011-02-10T19:41:45.790" CommentCount="10" />
  <row Id="4952" PostTypeId="2" ParentId="4932" CreationDate="2011-02-10T18:02:33.953" Score="3" ViewCount="" Body="&lt;p&gt;The Wick rotation is surely a natural method to apply even in the gravitational case. And by the way, it's very likely that the Euclidean calculation is always more fundamental and well-behaved than the Minkowski one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you correctly say, the Einstein-Hilbert action isn't bounded from below (and above) in the Euclidean space. But it is of course unbounded in the Minkowski spacetime. So by switching to the Euclidean spacetime, you surely don't make things worse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even more importantly, the unboundedness isn't a sign of a problem with a particular method: it is an actual physical insight. The action can go negative because of the &quot;conformal mode&quot; of the metric tensor. If you visualize the metric tensor as a collection of scalars, the &quot;conformal mode&quot; - one linked to the determinant of the metric - is the single scalar field with the wrong sign of the kinetic term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This difference of the sign arguably has cosmological consequences. It's also the reason why moduli spaces of gravitational theories are covered by &quot;well-understood&quot; patches in 2+1 dimensions or higher but they produce new, unknown regions in 1+1 dimensions or lower:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/9811194&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9811194&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Euclidean spacetime representation is arguably even more important in quantum gravity than it is in non-gravitational quantum field theories. In particular, the Euclidean continuation of the black hole solutions make the right value of the temperature transparent: the temperature is such that the asymptotic periodicity of the solution at infinity is $\beta$ (in the Euclidean spacetime), the interior of the black hole is totally removed from the solution, and the event horizon is a smooth places of the solution without any deficit (or excess) angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because general relativity allows many topologies and many ways to choose the time coordinate globally or locally, it allows many ways to Wick-rotate the spacetime. This diversity recently began to be used by brave physicists. I mean e.g. this paper by Ashoke Sen:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1101.4254&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1101.4254&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just to be sure, the Hartle-Hawking wave function is another key paradigm in the Euclideanized quantized general relativity. It postulates that the initial conditions near the Big Bang are such that the Euclidean spacetime is smooth around the origin.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-10T18:02:33.953" />
  <row Id="4953" PostTypeId="2" ParentId="4930" CreationDate="2011-02-10T18:49:45.763" Score="1" ViewCount="" Body="&lt;p&gt;I would just like to add a little to these answers in light of the Edit to the question which introduces &quot;XX Spin Chains&quot; as a context for this question. I have found a Tutorial on Spin Chains &lt;a href=&quot;http://www.physics.miami.edu/nepomechie/primer.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Basically they are N spins on a line. Here is the Hamiltonian from that paper where N=2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_{12}=J/4(\sigma_1^x\sigma_2^x+\sigma_1^y\sigma_2^y+\sigma_1^z\sigma_2^z - I \times I)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the sign of J this has either 3 degenerate ground solutions, plus one excited solution or one ground solution. This is a basic model of ferromagnetic/antiferromagnetic states. In this case the solutions have a gap. They will still have a gap for general N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However many developments of this largely integrable model have happened in recent papers, with an applied continuous magnetic field for example. In some of these cases the model may be gapless. There is also the question of what the model implies in the Thermodynamic limit $N -&amp;gt; 00$.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-10T18:49:45.763" />
  <row Id="4954" PostTypeId="2" ParentId="4932" CreationDate="2011-02-10T18:54:21.280" Score="3" ViewCount="" Body="&lt;p&gt;Good answer from Lubos, as always. My take is slightly different, perhaps less optimistic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ordinary quantum field theory in flat spacetime, Wick rotation is a great calculational tool for many things you might be interested in: vacuum correlation functions of time ordered products of local operators, or static quantities in thermal equilibrium. One does Wick rotation, and then you can think about the path integral literally as an integral, albeit infinite dimensional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, even in this simpler context, try to calculate more complicated quantities and you'd find that Euclidean prescription either doesn't exist or is not all that useful. Examples are: time dependent quantities in thermal equilibrium, observables in non-equilibrium situations, retarded or advanced correlation functions  etc. etc. For many of these calculations there is a prescription of &lt;em&gt;complexifying&lt;/em&gt; time, and specifying an appropriate contour in complex time (or frequency) plane. This is subtle business, and even when this works it is clearly not just &quot;rotation&quot; to Euclidean time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seems to me the strongest argument to thinking about Euclidean formulation as being more fundamental, even though it doesn't cover all possible situations that Lorentzian physics presents us with, is the desire to understand the path integral literally as an integral, or at least an analytic continuation of something which can be defined as a convergent integral. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps this is somewhat well-motivated in QFT, but it is probably not the case in quantum gravity, away from the semi-classical limit. There is no sense in which you integrate over metrics to define quantum gravity. The Einstein-Hilbert action is not renormalizable so you cannot think about the path integral over metrics as giving you any information beyond leading semi-classical (saddle point) approximation. I think for these, and many more elaborate reasons, many people I know are suspicious of any claims made using Euclidean  quantum gravity beyond that leading order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the specific question about the conformal factor: it has the wrong sign, so the Euclidean path integral does not converge and you have to define it by further analytic continuation of the conformal factor. As always with analytic continuation, this can have some subtlties, and when we are presented with choices of prescription we are less able to be guided by physics to understand the meaning of the various possibilities.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-02-10T19:09:25.147" LastActivityDate="2011-02-10T19:09:25.147" CommentCount="1" />
  <row Id="4955" PostTypeId="1" CreationDate="2011-02-10T19:03:37.723" Score="4" ViewCount="140" Body="&lt;p&gt;Hello this is my first question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an open string you can pick different boundary conditions for the endpoints along different directions of space time. For example, you can choose Dirichlet conditions in some directions and Neumann along others. This gives us D-branes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, why can't you choose Neumann b.c. along some directions, and closed string b.c. along some others? Is there some easy way to see this is nonsense?&lt;/p&gt;&#xA;" OwnerUserId="1856" LastActivityDate="2011-02-19T17:18:24.310" Title="Open/Closed Branes" Tags="&lt;string-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="4956" PostTypeId="2" ParentId="1626" CreationDate="2011-02-10T19:07:14.617" Score="0" ViewCount="" Body="&lt;p&gt;I have the same problem with my doors in Concord California.  The reason seems to be the cememt slab.  During the Dry weather the ground is dry and the house slumps.  In the winter when the soil is wet the house rises up and everything fits differently.  I could water my slab (plant close to the wall some water loving plants) or live with it.  Water costs and hassle keep me from doing that so it keeps settling.  You could plane off the door so it's a little smaller to make it fit in the winter and be a little small in summer.  Just observe where it sticks and with your wood plane shave a little off till you like it.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-10T19:07:14.617" />
  <row Id="4957" PostTypeId="2" ParentId="4955" CreationDate="2011-02-10T19:26:54.320" Score="3" ViewCount="" Body="&lt;p&gt;A good idea. ;-) Let's look at it. You have, for example, boundary conditions&#xA;$$X^1(\sigma+\pi)=X^1(\sigma), \quad X^{\prime 2}(\sigma=0)=X^{\prime 2}(\sigma=\pi)=0$$&#xA;The boundary terms in the variation of the action cancel separately for the different coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see that it's not quite periodic, so the string still has preferred points, &#xA;$$\sigma=0, \quad \sigma=\pi$$&#xA;We may still call them the &quot;end points&quot; of the hybrid string. So the coordinate $X^2$ behaves like for an open string. But $X^1$ has to be periodic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you quantize the string, you get left-movers and right-movers for $X^1$, but only one set of standing waves for $X^2$ etc. So this will also produce some bizarre zero-point energy. But I think that at the level of the free strings, the theory could be well-defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I think you won't be able to define any consistent interactions although I don't have a strict proof. Your hybrid string is neither open nor periodic. So the world sheet has boundaries that are &quot;partly identified&quot; with other points of the boundary. It's like a closed world sheet with a preferred line drawn on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you try to define the interacting theory, you have to decide whether the number of these &quot;bizarre lines&quot; on the world sheet - or bizarre, partly identified pairs of end-points, is conserved or not. If it is conserved, you couldn't pair-create such hybrid strings, so they wouldn't be too interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that a sensible theory would have to allow these strange pairs of end-points to be created and destroyed. But if it were so, then the world sheet would include singular points - the points in which the cut disappears and your hybrid string transmutes into a normal closed string (or vice versa). My guess is that this process would create a singular points on the world sheet that could return all the ultraviolet divergences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I am not sure whether you could preserve the conformal symmetry. In particular, I don't understand how the state-operator correspondence would work. Closed string states are dual to operators in the bulk of the world sheet; open string states are equivalent to operators on the boundary of the world sheet. But what about your hybrid strings? Maybe, they are dual to the operators at the points where the cut ends. ;-) This could return some unacceptable UV problems to the world sheet, or not. I am not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should try to think about it. If you could find such a new bizarre object that may actually exist in string theory, you could become as important as Joe Polchinski who found D-branes. It's unlikely but you should try. ;-)&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-10T19:26:54.320" CommentCount="1" />
  <row Id="4958" PostTypeId="2" ParentId="4930" CreationDate="2011-02-10T19:42:32.120" Score="0" ViewCount="" Body="&lt;p&gt;A short remark for the &quot;edited&quot; part of your question (whether there is a gap in&#xA;the XX chain or not). The XX spin chain in a magnetic field, i.e., the model defined by the Hamiltonian&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H = \sum_i (\sigma^{x}_i \sigma^{x}_{i+1} + \sigma^{y}_i \sigma^{y}_{i+1} + h \sigma^{z}_i) $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is gapped when $|h| &amp;gt; 1$. This is not a very difficult results,  it comes out immediately if you do the usual Jordan-Wigner and a Fourier transformation a la the famous paper of Lieb, Schultz and Mattis (Ann. Phys. 16, 407, (1961)) (although there the $\sigma^{z}_i$ terms are missing, but they are not hard to incorporate).  &lt;/p&gt;&#xA;" OwnerUserId="1855" LastActivityDate="2011-02-10T19:42:32.120" />
  <row Id="4959" PostTypeId="1" AcceptedAnswerId="4965" CreationDate="2011-02-10T19:44:59.917" Score="5" ViewCount="605" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Noether%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Noether's theorem&lt;/a&gt; is one of those surprisingly clear results of mathematical calculations, for which I am inclined to think that some kind of intuitive understanding should or must be possible. However I don't know of any, do you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Independence of time &amp;lt;=&gt; energy conservation.&lt;br&gt;&#xA;Independence of position &amp;lt;=&gt; momentum conservation.&lt;br&gt;&#xA;Independence of direction &amp;lt;=&gt; angular momentum conservation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the mathematics leads in the direction of Lie-algebra and such but I would like to discuss whether this theorem can be understood from a non-mathematical point of view also.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2011-02-11T22:25:13.213" Title="Can Noether's theorem be understood intuitively?" Tags="&lt;noethers-theorem&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="3" />
  <row Id="4960" PostTypeId="2" ParentId="4959" CreationDate="2011-02-10T19:59:54.377" Score="-2" ViewCount="" Body="&lt;p&gt;I can only tell that those conserved quantities you have listed above are additive in particles: $P = \sum p_i(t)$, for example. But there are those that are not additive! They do not have special names. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For N differential equations there are as many integrals of motion as the initial conditions or so. Some of them can be casted sometimes in the additive form but generally (when there are no symmetries) the total number of integrals of motion remains the same. They all are simply non additive (more messy, if you like). So I would answer that symmetries help combine some integrals of motion as conserved quantities additive in all particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 1: Maybe Noether's theorem shows explicitly what the conserved quantities are whereas from equations it may be not so evident to derive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 2: I got -4. Is my reasoning really that bad?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 3: a page from Landau:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/jbecX.jpg&quot; alt=&quot;Conservation Laws&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 4: An example of integrals of motion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Z1n7z.jpg&quot; alt=&quot;1D Integrals of motion&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-11T22:25:13.213" LastActivityDate="2011-02-11T22:25:13.213" CommentCount="36" />
  <row Id="4961" PostTypeId="1" CreationDate="2011-02-10T20:22:57.187" Score="4" ViewCount="269" Body="&lt;p&gt;One statement I've heard many times is that QFT is &quot;defined&quot; by the lattice, or that the &quot;only&quot; definition of QFT is on the lattice (when such definition exists, e.g in pure Yang-Mills theory). I've heard that from many people I respect, but I have my doubts. Specifically - the lattice is an algorithmic definition of those quantities that can be calculated in Euclidean spacetime. As I said in my answer to &lt;a href=&quot;http://physics.stackexchange.com/questions/4932/does-wick-rotation-work-for-quantum-gravity&quot;&gt;this question&lt;/a&gt;, this is a proper subset of all physical quantities you may be interested in calculating in general, and many physical situations do not have Euclidean formulation. Is there some response to this objection I am missing, or am I being naive somehow?&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-02-11T18:00:11.763" LastActivityDate="2011-02-11T18:00:11.763" Title="Is quantum field theory defined by its lattice regularization?" Tags="&lt;quantum-field-theory&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="4962" PostTypeId="2" ParentId="4943" CreationDate="2011-02-10T20:29:51.110" Score="2" ViewCount="" Body="&lt;p&gt;This is a topic in Non-Equilibrium Thermodynamics. There is a standard concept in Thermodynamics of &quot;Thermodynamic Force&quot;, &quot;Thermodynamic Flux&quot; and so on. In the Physical Chemistry context you might be familiar with &quot;Affinity&quot; and &quot;Chemical Potential&quot;. These are the mechanisms used to explain chemical reaction directions, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to summarise this large area with an example two substances S1 (with Temperature T1) and S2 (with Temperature $T2 &amp;lt; T1$) are in thermal contact, the combined system is not in equilibrium. The Thermodynamic Force here is: $F=(1/T2 - 1/T1)$. This is derivable from the formula for change in Entropy in this situation:$dS = -dS1/T1 + dS2/T2$. In general a &quot;thermodynamic force&quot; causes a change in Entropy - a situation that can only arise in non-Equilibrium situations. The thermodynamic force will become zero when T1=T2 and the system S1+S2 is in equilibrium. So the idea is that Thermodynamic force models Entropy change (in a mechanics-like manner). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Associated with this Force is the Flux $J_Q$: the time dependent construct&lt;br&gt;&#xA;$dQ/dt = J_Q = \alpha(T1-T2)$ where $\alpha$ is the Fourier coefficient of heat conductivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general the First Law of Thermodynamics holds the key to the constructions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dU=TdS - PdV + \Sigma \mu_i dN_i $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(here the $\mu_i$ are the chemical potentials of the ith particle species) - useful if they are created or destroyed (as in a chemical or nuclear reaction.)  This equation could be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dU=TdS - \Sigma X_i dx_i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the $X_i$ are the generalised forces and $x_i$ are the generalised conjugate variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the extra variables include electrical potentials we can have thermo-electric equations, etc. These can describe the flows in electric-chemical batteries and the like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One point of debate is whether and to what extent thermodynamic variables are really &quot;local&quot; as would be required in a true continuum based theory. Thus how valid is &quot;Temperature at a point&quot; etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These topics are covered in Thermodynamics texts e.g. Callen: &quot;Introduction to Thermodynamics and Thermostatics&quot; or Prigogine &quot;Modern Thermodynamics&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-02-10T20:35:35.583" LastActivityDate="2011-02-10T20:35:35.583" CommentCount="8" />
  <row Id="4963" PostTypeId="2" ParentId="4955" CreationDate="2011-02-10T20:34:05.323" Score="1" ViewCount="" Body="&lt;p&gt;It looks like open string with kinda &quot;Dirichlet&quot; boundary condition in some direction - in the one where string is closed - but when D-brane which enables this &quot;Dirichlet&quot; b.c. is allowed to move. This movement would be center-of-mass movement plus rippling - when we consider free string.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-10T21:40:42.113" />
  <row Id="4964" PostTypeId="2" ParentId="4961" CreationDate="2011-02-10T20:37:28.697" Score="-4" ViewCount="" Body="&lt;p&gt;As far as I understand, a lattice formulation makes it possible to employ &quot;naturally&quot; the famous Wilson's reasoning about scales and &quot;effective&quot; theories. Factually there is nothing physical in this QFT &quot;definition&quot; since the lattice step is arbitrary unlike in the Wilson approach to the critical phenomena. I would call it a &quot;fog&quot; to hide real physical problems in.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-02-10T20:37:28.697" />
  <row Id="4965" PostTypeId="2" ParentId="4959" CreationDate="2011-02-10T20:42:33.210" Score="2" ViewCount="" Body="&lt;p&gt;It's intuitively clear that the energy most accurately describes how much the state of the system is changing with time. So if the laws of physics don't depend on time, then the amount how much the state of the system changes with time has to be conserved because it's still changing in the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way, and perhaps even more intuitively, if the laws don't depend on position, you may hit the objects, and hit them a little bit more, and so on. The momentum measures how much the objects depend on space, so if the laws themselves don't depend on the position on space, the momentum has to be conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angular momentum with respect to an axis is determining how much the state changes if you rotate it around the axis - how much it depends on the angle (therefore &quot;angular&quot; in the name). So the symmetry is linked to the conservation law once again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your intuition doesn't find the comments intuitive enough, maybe you should train your intuition because your current intuition apparently misses the most important properties of time, space, angles, energy, momentum, and angular momentum. ;-)&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-10T20:42:33.210" CommentCount="10" />
  <row Id="4966" PostTypeId="2" ParentId="4961" CreationDate="2011-02-10T20:50:37.920" Score="1" ViewCount="" Body="&lt;p&gt;I can only guess that &quot;QFT is defined on the/a lattice&quot; means that certain infinities in Lagrangian QFT come from defining QFT on a smooth (Lorentzian or Euclidiean) manifold, while many physicists suspect that the &quot;smoothness&quot; is an approximation that breaks down on a certain length scale (like the Planck length), and incorporating the true - quantum - nature of spacetime will cure the QFT-infinities.&#xA;In this sense, defining QFT on a lattice is a better approximation than defining QFT on a smooth manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would object, however, that there is no single, unifying framework for doing QFT, Lagrangian QFT is only one approach. There is of course algebraic/local/axiomatic QFT that does not have any problems to define QFTs on globally hyperbolic Lorentzian manifolds at all - it does have a problem with the construction of interesting models, but in the last two decades there has been some progress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There has been one specific approach that elevates operator product expansions to a defining property, an axiom, of QFT, it started somewhen around this paper:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;S. Hollands, R. M. Wald: &lt;a href=&quot;http://arxiv.org/abs/0805.3419&quot; rel=&quot;nofollow&quot;&gt;Quantum field theory in curved spacetime, the operator product expansion, and dark energy&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;An introduction and a toy example can be found in this thesis:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Jan Holland: &lt;a href=&quot;http://arxiv.org/abs/0906.5468&quot; rel=&quot;nofollow&quot;&gt;Construction of operator product expansion coefficients via consistency conditions&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is a very new approach, which I cite mainly to illustrate that there is a lot of room left for the development of an unifying, axiomatic approach to QFT which, when found, may change our thinking about QFT significantly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm also aware of the fact that many people in quantum gravity think that everything about QFT is said and done: well, shoot me.)&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-02-10T20:50:37.920" CommentCount="1" />
  <row Id="4967" PostTypeId="2" ParentId="4961" CreationDate="2011-02-10T20:58:25.333" Score="2" ViewCount="" Body="&lt;p&gt;Moshe, I hope others will give their answer and I would also like to read it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I agree with you that the Euclidean calculations are &quot;only&quot; good for some rather important quantities that are easy to be calculated in the Euclidean setup - and they should be continued to the Minkowski spacetime. Some time-dependent questions surely make it more natural to try to solve things directly in the Minkowski spacetime. However, the full set of on-shell scattering amplitudes or off-shell Green's functions is a very important set of observables and, in some sense, contains &quot;all the dynamical knowledge&quot; about the theory. And this dynamical knowledge is much more naturally calculated in the Euclidean setup which is why people like me wouldn't be afraid to say that the Euclidean calculations are more fundamental.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somewhat independently of that, there's the question of lattice QFT. Lattice QCD is usually meant to denote the Monte Carlo calculations etc. and they're calculated in the Euclidean spacetime. However, you may also discretize the spatial coordinates only, do the Hamiltonian lattice QCD, and keep the Minkowski space. I would separate the Euclidean spacetime from the latticization as two different and largely independent procedures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Discretizing spacetime is a straightforward procedure to approximate a real theory if we want to calculate it on a digital computer. Many people think &quot;discretely&quot; and they have a problem to distinguish the reality from the computer models. That's why they think that the discrete models are the &quot;real thing&quot;, and lattice QCD is their definition of choice. Of course, there is nothing scientific about this attitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nature has no trouble to work with continuous quantities - and fields that depend on continuous variables (and even infinitely many variables, if necessary) - and even the physicists, who are much more limited than Nature, have learned many mathematical ways that allow them to directly calculate the results in the continuous spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Needless to say, some properties of the system change when the spacetime is Wick-rotated to the Euclidean one (or back); and when the spacetime is discretized. The first transformation changes the reality conditions for the spinors and produces various problems with the fermions. The second transformation - the discretization - also produces its related problems with the fermions, the fermion doubling problem etc. Moreover, it breaks the continuous translational symmetry, and because the supersymmetry anticommutes to translations, lattices inevitably break (most of) the supercharges, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if one views supersymmetry as a key property of a theory, its loss is a huge problem and mostly invalidates the idea that the lattice definition is &quot;more real&quot; or &quot;more fundamental&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, the latticization is just one particular way - one of the conceptually simplest or most straightforward ways - to define a theory that flows to the desired quantum field theory at long distances. And some people want to confine their reasoning into the lattice because they believe that their duty as scientists is to behave as similarly to a mindless digital computer as possible. (Even computers may deal with non-lattice approaches, but that requires them and the programmers to learn some serious maths and not just mechanical operations on a lattice.) As far as I know, there is no scientific justification of the idea that the &quot;right&quot; definition of a field theory should ultimately boil down to a lattice.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-10T21:04:26.607" LastActivityDate="2011-02-10T21:04:26.607" />
  <row Id="4968" PostTypeId="2" ParentId="2625" CreationDate="2011-02-10T21:03:48.997" Score="1" ViewCount="" Body="&lt;p&gt;The point of contact between the two disciplines is taken to be the function for Entropy S which is a function of the thermodynamic variables $S=S(U, V, N_i)$ etc, with U being internal energy, V volume. Sometimes this is inverted as: $U=U(S,V,N)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a &lt;strong&gt;Thermodynamics&lt;/strong&gt; perspective equations for temperature and other relations can be established by taking derivatives of S or U wrt the variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Statistical Mechanics&lt;/strong&gt; aims to derive S explicitly as a function $S = k \Sigma f_i ln f_i$ where $f_i$ is a possible microstate. The totality of these microstates, and their associated energy, is determined from a partition function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both subjects can diverge from here too. Thermodynamic equations met in courses generally refer to equilibrium conditions. Many systems are in non-equilibrium and some thermodynamic theories try to explain this condition and how it changes. Some might argue that the regular thermodynamic concepts like Entropy are not even well defined far from equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statistical Mechanics would differentiate between &quot;classical&quot; systems and various types of &quot;quantum systems&quot; - which have different partition functions and energy behaviour. Thermodynamics is thus often viewed as a limiting classical theory (like Newtonian mechanics), but some Thermodynamic concepts (like Chemical Potential and Temperature) find themselves in quantum partition functions of Statistical Mechanics. In addition Entropy and Temperature are associated with Black Holes for reasons that dont currently translate into Statistical Mechanics.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-10T21:03:48.997" />
  <row Id="4969" PostTypeId="2" ParentId="4959" CreationDate="2011-02-10T21:05:25.363" Score="3" ViewCount="" Body="&lt;p&gt;Here are my two cents. Read the proof it will help you understand and build intuition because it is constructive. It explicitly shows you what the conserved quantity is, given the group of symmetries. If it is too hard to follow and you can't see the forest because of the trees, try a few examples it should help. Also here is a link that may help a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.ucr.edu/home/baez/noether.html&quot; rel=&quot;nofollow&quot;&gt;http://math.ucr.edu/home/baez/noether.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2011-02-10T21:05:25.363" />
  <row Id="4970" PostTypeId="2" ParentId="4961" CreationDate="2011-02-10T21:19:16.630" Score="0" ViewCount="" Body="&lt;p&gt;Lattice QCD is define on a grid with spacing $a$, and where on the edgelinks there are gauge potentials.  The reciprocal lattice spacings are momentum cut-offs, and must be set so that the heaviest quark mass is $m_q~&amp;gt;&amp;gt;~1/a$.  So the lattice spacing is not entirely arbitrary.  A discrete difference in gauge potentials defines the fields &#xA;$$&#xD;&#xA;F_{ij}~=~(A_j(x’)~-~A_i(x”))~-~(A_i(x”’)~-~A_j(x”’))~-~g(A_j(x’)A_i(x”)~-~A_i(x“’)A_j(x“’))&#xD;&#xA;$$&#xA;where the coordinates are illustrated in the figure.  The internal gauge indices are implied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Abaj0.jpg&quot; alt=&quot;Plaquette in a quantum lattice&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This evaluates the gauge potential around a square or plaquette, which is in a large grid.  The field is then evaluated as one evaluates a magnetic field as $B~=~-\nabla\times A$ as evaluated in the area of the plaquette.  This means the Lagrangian for the quantum system is periodic, and is similar in some ways to a Bloch potential.  The limit that one takes the lattice spacing small enough the funny periodic nature of the Lagrangian due to the lattice plays an insignificant role.  This is one reason the inverse lattice spacings have to be smaller than the quark masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the lattice is a finite element approach to QFT, which is set to the cut off scale of the theory.  Lattice QCD is worked because the nonlinear nature of the theory makes explicit calaculations exceedingly difficult, particularly due to inverse screening.  This make the computation of the hadron spectrum tough where one does not has asymptotic freedom to work with.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-02-10T21:30:54.600" LastActivityDate="2011-02-10T21:30:54.600" CommentCount="12" />
  <row Id="4971" PostTypeId="2" ParentId="4935" CreationDate="2011-02-10T21:34:37.343" Score="3" ViewCount="" Body="&lt;p&gt;There's always a relativistic version of any theory in a weak sense that we can construct a system that is Lorentz *co*variant. The question is then what Lorentz non-invariant entities have to be added. As Luboš implies, the additional structure in the case of quantum statistical fields is a time-like 4-vector. In particular, the quantum vacuum state is invariant under the action of the Poincare group, whereas a thermal state is invariant only under the subgroup of the Poincare group that leaves a particular time-like 4-vector invariant. If we look at the $n$-point functions of the Klein-Gordon free field in the vacuum state and in a thermal state, taking $T^\mu$ to be invariant, both states are Gaussian, so the 2-point function gives full information, but the Fourier transforms of the 2-point functions are given by $2\pi\delta(k^2-m^2)\theta(k_0)$ and by $\coth{\!\left(\!\frac{\hbar \sqrt{|\mathbf{k}|^2+m^2}}{2\mathsf{kT}}\!\right)} 2\pi\delta(k^2-m^2)\theta(k_0)$, respectively. The $|\mathbf{k}|^2$ is the length of the 3-vector component of the wave-number relative to the time-like 4-vector $T^\mu$. This deforms the measure on the mass-shell, but it is still on the mass-shell. You can find a derivation of this, more-or-less, in my arXiv:quant-ph/0411156, which is also published in &lt;em&gt;Phys. Lett.&lt;/em&gt; &lt;strong&gt;A&lt;/strong&gt; 338, 8-12(2005).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we have a quantum statistical &lt;em&gt;free&lt;/em&gt; field theory. Interacting QFT only exists &lt;em&gt;perturbatively&lt;/em&gt; in 1+3 dimensions, and hence questionably, and the situation for quantum statistical field theory is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: A small change of $\tanh{}$ to $\coth{}$! The wrong way up for the way I stated it here. The $\coth{\!\left(\!\frac{\hbar \sqrt{|\mathbf{k}|^2+m^2}}{2\mathsf{kT}}\!\right)}$ factor behaves more-or-less as $1/\sqrt{|\mathbf{k}|^2+m^2}$ when $|\mathbf{k}|$ is small, low frequency, and $m\hbar$ is small relative to $\mathsf{kT}$, making the measure on the mass-shell effectively the same as the classical thermal state in that case, but behaves as a constant, $1$, when  $|\mathbf{k}|$ is large, high frequency, making the measure on the mass-shell effectively the same as the quantum vacuum state in &lt;em&gt;that&lt;/em&gt; case.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-02-11T04:10:23.220" LastActivityDate="2011-02-11T04:10:23.220" />
  <row Id="4972" PostTypeId="1" AcceptedAnswerId="4974" CreationDate="2011-02-10T22:38:46.877" Score="3" ViewCount="205" Body="&lt;p&gt;I was hoping for an answer in general terms avoiding things like holonomy, Chern classes, Kahler manifolds, fibre bundles and terms of similar ilk. Simply, what are the compelling reasons for restricting the landscape to admittedly bizarre Calabi-Yau manifolds? I have &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0465020232&quot; rel=&quot;nofollow&quot;&gt;Yau's semi-popular book&lt;/a&gt; but haven't read it yet, nor, obviously, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0071498702&quot; rel=&quot;nofollow&quot;&gt;String Theory Demystified&lt;/a&gt; :)&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-13T19:56:10.343" LastActivityDate="2011-02-13T19:56:10.343" Title="Why (in relatively non-technical terms) are Calabi-Yau manifolds favored for compactified dimensions in string theory?" Tags="&lt;string-theory&gt;&lt;kaluza-klein&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="4973" PostTypeId="1" CreationDate="2011-02-10T22:42:46.137" Score="1" ViewCount="409" Body="&lt;p&gt;Let us consider a classical mechanical system of N particles in a constant external field. We have 3N coordinates and 3N velocities, so totally 6N unknown variables. We have 6N ordinary differential equations of the first order for them or 3N equations of the second order. It is known that such a system has 6N-1 conserved quantities, see Landau-Lifshits, V. 1, Classical Mechanics, Chapter about Conservation Laws. In absence of symmetries these conserved quantities are rather complicated analytical expressions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of some symmetries one can construct conserved quantities &lt;strong&gt;additive on particles&lt;/strong&gt;. Everybody knows about total energy, momentum, and angular momentum conservation laws. But the number of conserved quantities remains to be 6N-1. It means the symmetries may help us construct &quot;less messy&quot; integrals of motion than in general case. We obtain simplification of analytical expressions but not &lt;strong&gt;new&lt;/strong&gt; integrals of motion in case of symmetries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question I wrote in title has, in my opinion, the following answer: &lt;strong&gt;Symmetries do not lead to &lt;em&gt;additional&lt;/em&gt; conservation laws but to &lt;em&gt;simplification&lt;/em&gt; of existent ones.&lt;/strong&gt;&#xA;To a great extent it is due to simplification of equation system in case of symmetries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of integrals of motion in 1D case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Nw5Su.jpg&quot; alt=&quot;1D Integrals of motion&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-04-18T17:27:45.450" LastActivityDate="2011-04-18T17:27:45.450" ClosedDate="2011-02-11T00:18:36.553" Title="Do symmetries increase the number of conserved quantities?" Tags="&lt;classical-mechanics&gt;&lt;noethers-theorem&gt;&lt;conservation-laws&gt;&lt;integrals-of-motion&gt;" CommentCount="25" />
  <row Id="4974" PostTypeId="2" ParentId="4972" CreationDate="2011-02-10T22:46:02.607" Score="3" ViewCount="" Body="&lt;p&gt;There is one simple reason: in such scenario the physics at the string scale has supersymmetry. Supersymmetry (more technically $N=1$ supersymmetry) has some nice phenomenological features that make it an attractive bridge between low energy physics and string theory. The existence of this symmetry translates directly to the requirement that the compactification manifold is Calabi-Yau.&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-02-10T22:46:02.607" CommentCount="2" />
  <row Id="4975" PostTypeId="2" ParentId="4972" CreationDate="2011-02-10T22:46:37.017" Score="2" ViewCount="" Body="&lt;p&gt;Since the word &quot;supersymmetry&quot; did not appear in your list of forbidden words let me give you this answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because Calabi-Yau manifolds leave unbroken some part of the original supersymmetry, which is advantageous for model building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are alternatives to Calabi-Yaus, like flux compactififcations or large extra dimensions.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-02-10T22:46:37.017" CommentCount="2" />
  <row Id="4976" PostTypeId="1" AcceptedAnswerId="4984" CreationDate="2011-02-10T23:03:45.310" Score="6" ViewCount="184" Body="&lt;p&gt;I've read that for a Bose-Einstein gas in 1D there's no condensation. Why this happenes? How can I prove that?&lt;/p&gt;&#xA;" OwnerUserId="558" LastActivityDate="2011-02-11T00:46:38.757" Title="Bose-Einstein condensate in 1D" Tags="&lt;condensed-matter&gt;&lt;bose-einstein-condensate&gt;" AnswerCount="3" />
  <row Id="4977" PostTypeId="2" ParentId="4941" CreationDate="2011-02-10T23:22:51.690" Score="4" ViewCount="" Body="&lt;p&gt;I spent many years working in the video graphics design industry. One of our problems is the opposite problem, that is, we have limits on how much electromagnetic radiation our products can produce. Before we can ship any new design, we have to test it to make sure it meets the limits. (The tests are done on samples, not on every item shipped.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limits have to be low enough that the equipment doesn't cause interference for radio and TV. You would think that would eliminate the problem, but in fact, all equipment puts out some unwanted radiation. So I have no doubt at all that the radiation is there to be analyzed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Display systems have a clock that moves the data. On some modern systems the clock is &quot;spread spectrum&quot; but it should be possible to figure out the spread algorithm. It's not like they design these things to be impervious to decryption, instead the purpose with spread spectrum is to spread the noise produced by your equipment over a wider frequency range. This lowers the spikes in your noise spectrum and keeps them under the limits. In other words it allows you to radiate more while still staying in compliance with government limits. See the wikipedia article:&lt;br&gt;&#xA;&lt;strong&gt;Spread Spectrum clock signal generation&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Spread_spectrum#Spread-spectrum_clock_signal_generation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Spread_spectrum#Spread-spectrum_clock_signal_generation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone wanted me to do this, I would try to design an antenna that (a) Is very very directional. That way you can point it at the device you want to receive from. (b) I would work hard on understanding the interfering signals and subtract them from the signal, either digitally or in analog. (c) Understand the spread spectrum of the target device and design a system to cancel it out. And just typing this, I've already got some ideas that I'm sure would work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more general problem, where you don't know what make of equipment you're trying to receive from, is going to be a lot tougher, but it should be doable by the guys in black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I have no doubt at all that, given a paycheck and some time, I could extract those pictures.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-10T23:22:51.690" />
  <row Id="4978" PostTypeId="2" ParentId="4976" CreationDate="2011-02-10T23:30:38.023" Score="-1" ViewCount="" Body="&lt;p&gt;Look at the derivation of the critical temperature of a Bose gas. In there, you should get nonsensical results for one dimension. This is because the density of states even of non-interacting particles depends on the dimension.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-02-10T23:30:38.023" CommentCount="1" />
  <row Id="4979" PostTypeId="2" ParentId="4959" CreationDate="2011-02-10T23:30:50.527" Score="3" ViewCount="" Body="&lt;p&gt;Well, I don't know about any intuitive explanation besides intuition gained by understanding the underlying math (mainly differential geometry, Hamiltonian mechanics and group theory). So with the risk of not giving you quite what you want, I'll try to approach the problem mathematically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know Hamiltonian mechanics then the statement of the theorem is exceedingly simple. Assume we have a Hamiltonian $H$. To this there is associated a unique Hamiltonian flow (i.e. a one-parameter family of symplectomorphisms -- which is just a fancy name for diffeomorphisms preserving the symplectic structure) $\Phi_H(t)$ on the manifold. From the point of view of Lie theory, the flow is a group action and there exists its generator (which is a vector field) $V_H$ (this can also be obtained from $\omega(\cdot, V_H) = dH$ with $\omega$ being the symplectic form). Now, the completely same stuff can be written for some other function $A$, with generator $V_A$ and flow $\Phi_A(s)$. Think of this $A$ as some conserved quantity and of $\Phi_A(s)$ as a continuous family of symmetries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, starting from Hamiltonian equation ${{\rm d} A \over {\rm d} t} = \left\{A,H\right\}$ we see that if $A$ Poisson-commutes with $H$ it is conserved. Now, this is not the end of the story. From the second paragraph it should be clear that $A$ and $H$ don't differ that much. Actually, what if we swapped them? Then we'd get ${{\rm d} H \over {\rm d} s} = \left\{H,A\right\}$. So we see that $A$ is constant along Hamiltonian flow (i.e. conserved) if and only if $H$ is constant along the symmetry flow (i.e. the physical laws are symmetric).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So much for why the stuff works. Now, how do we get from symmetries to conserved quantities? This actually isn't hard at all but requires some knowledge of differential geometry. Let's start with most simple example.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Translation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This is a symmetry such that $x \to x^&amp;#39; = x + a$. You can imagine that we move our coordinates along the $x$ direction. With $a$ being a parameter, this is a symmetry flow. If we differentiate with respect to this parameter, we'll get a vector field. Here it'll be $\partial_x$ (i.e. constant vector field aiming in the direction $x$). Now, what function on the symplectic manifold does it correspond to? Easy, it must be $p$ because by differentiating this we'll get a constant 1-form field $dp$ and then we have to use $\omega$ to get a vector field $\partial_x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other way to see that it must be $p$: suppose you have a wave $\exp(ipx)$. Then $\partial_x \exp(ipx) = ip \exp(ipx)$ so momentum and partial derivatives are morally the same thing. Here we're of course exploiting the similarity between Fourier transform (which connects $x$ and $p$ images) and symplectic structure (which combines $x$ and $p$).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Rotation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Now onto something a bit harder. Suppose we have a flow &#xA;$$\pmatrix{x \cr y} \to \pmatrix{x&amp;#39; \cr y&amp;#39;}=  \pmatrix{\cos(\phi) &amp;amp; \sin(\phi) \cr - \sin(\phi) &amp;amp; \cos(\phi)} \pmatrix {x \cr y} $$&#xA; This is of course a rotational flow. Here we'll get a field $y {\rm d}x - x {\rm d} y$ and the conserved quantity of the form $y p_x - x p_y$ which can in three dimensions be thought of as a third component of angular momentum $L_z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the above was done mainly for illustrative purposes as we could have worked in polar coordinates and then it would be actually the same problem as the first one because we'd get the field $\partial_{\phi}$ and conserved quantity $p_{\phi}$ (which &lt;em&gt;is&lt;/em&gt; angular momentum).&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-02-10T23:30:50.527" CommentCount="13" />
  <row Id="4980" PostTypeId="2" ParentId="4921" CreationDate="2011-02-10T23:39:46.200" Score="0" ViewCount="" Body="&lt;p&gt;I will treat this question as primarily about General Relativity, and its Cosmological metric solution as discussed by Lubos. So I would just like to add a few points to that answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It needs to be understood that the Big Bang was the creation of Space-Time in General Relativity: thus in particular Space was created at the Big Bang. The use of &lt;strong&gt;R&lt;/strong&gt;$^3$ in the question I take to be a synonym for &quot;Space&quot;. Looking at the wording in the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;the universe is spatially infinite&quot; - this might be interpreted as a FLRW solution (a Cosmological solution like Lubos's) which has negative or zero curvature. The other FLRW case is positive curvature which would not be considered &quot;spatially infinite&quot;. EDIT I shall assume that the question refers to the zero curvature case, in which case the FLRW metric simplifies to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ds^2 = dt^2 - a(t)^2(dx^2+dy^2+dz^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like any GR solution there is a question as to what this means, especially Cosmologically. There is clearly a Euclidean &lt;strong&gt;R&lt;/strong&gt;^3 metric here and it is associated with constant times in this metric, giving a family of &lt;strong&gt;R&lt;/strong&gt;^3 spaces. In this sense there is an &lt;strong&gt;R&lt;/strong&gt;^3 model in GR. However from the perspective of an individual observer O it is worth noting that O's particle horizon is finite ie the Observer doesnt &quot;see&quot; &lt;strong&gt;R&lt;/strong&gt;^3 at a given time t, just elements of their past light cone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is of interest to note that a transformation of the t coordinates renders this equation as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ds^2 = a(T)^2(dT^2 - dx^2+ dy^2+ dz^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is conformally equivalent to Minkowski space - as a model for the Universe!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is about t=T=0 however. In Penrose Conformal Space-Time diagrams and elsewhere the traditional notion that the Singularity is a &quot;point&quot; is replaced by it being a &quot;3-surface&quot; (drawn as a wavy line in these diagrams). From this perspective we could view the T=0 singularity as a 3-surface too ie perhaps &lt;strong&gt;R&lt;/strong&gt;^3. So the notion that the Big Bang is a zero dimensional point could be misleading here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In either case it becomes a question as to whether there was a pre-Big Bang state - perhaps one which generated the 3-surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Penrose has in fact pursued this idea in the Conformal Cyclic Cosmology model. Although I shall not give the full details, in this approach a mathematical technique called &quot;Conformal rescaling&quot; is applied at or near the Big Bang. What this does is expand (or &quot;inflate&quot;) the metric from a point to an entire 3-surface (not necessarily &lt;strong&gt;R&lt;/strong&gt;$^3$ ). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted that &quot;Conformal rescaling&quot; maintains the light cone and causality structure of a metric but expands (or contracts) infinite spaces into finite regions (as in the Conformal diagrams alluded to earlier). Thus in a sense it is more &quot;inflationary&quot; than Inflation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This 3-surface is taken to be the end surface after a previous aeon. So the Big Bang is not a point, nor is it the beginning of the Universe, just the beginning of an &quot;aeon&quot;. Penrose reckons that, more or less, he can deduce this from GR without much new Quantum Theory. This theory claims to be able to explain any large scale structures in CMB that even &quot;Inflation&quot; might not be able to explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CCC idea is tentative in several respects and does not as yet include Quantum effects which it will surely need to become a full theory of pre-Big Bang.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-02-12T16:59:20.143" LastActivityDate="2011-02-12T16:59:20.143" CommentCount="2" />
  <row Id="4981" PostTypeId="2" ParentId="4976" CreationDate="2011-02-10T23:46:37.690" Score="1" ViewCount="" Body="&lt;p&gt;This is obviously pure geometry; precisely because the density of states of zero energy is not approaching 0 in $d&amp;lt;3$ (it behaves like $E^{\frac{d-2}{2}}$); (probably) the simplest proof can be done by showing that this explodes the number of particles.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2011-02-10T23:46:37.690" />
  <row Id="4982" PostTypeId="2" ParentId="4941" CreationDate="2011-02-11T00:00:07.193" Score="1" ViewCount="" Body="&lt;p&gt;My general reaction is to be pretty skeptical. Essentially, you are trying to take an imperfect signal (signal plus noise), and trying to reconstruct the original perfect signal. Thats sounds like an inverse operator. Inverse operators often have high condition numbers, i.e. they amplify the effects of signal noise. If the resultant noise is too large, the output signal will be seriously corrupted. So my guess, is it might work in some idealized cases, but for most realworld applications system noise effectively destroys information.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-02-11T00:00:07.193" CommentCount="1" />
  <row Id="4984" PostTypeId="2" ParentId="4976" CreationDate="2011-02-11T00:46:38.757" Score="3" ViewCount="" Body="&lt;p&gt;The claim is often that there is no condensation in $d&amp;lt;3$. The other answers are correct, but let's be clear, there are actually &lt;em&gt;two&lt;/em&gt; assumptions present in the claim:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Assume you have $N$ noninteracting bosons in $d$-dimensions in a hypervolume $L^d$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Assume that these bosons have an energy-momentum relationship of $E(p) = Ap^s$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, the way we calculate the critical temperature ($1/\beta_c$) for BEC requires satisfying the equation&#xA;$$\int_0^\infty \frac{\rho(E)dE}{e^{\beta_c E}-1}=N$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\rho(E)$ is the density of states. Whether this integral is convergent or not depends on the values of &lt;strong&gt;both&lt;/strong&gt; $s$ and $d$. The details of the proof are up to you though. :)&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2011-02-11T00:46:38.757" CommentCount="3" />
  <row Id="4985" PostTypeId="2" ParentId="4959" CreationDate="2011-02-11T01:40:19.497" Score="1" ViewCount="" Body="&lt;p&gt;Since it is a mathematical theorem whose physical content you know already, it is difficult to discuss it without mathematics. But still I will try to present it in a simple way. It may help if we understand how it is derived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally we look for an invariance of the action under a symmetry transformation with a time independent parameter. This is then a trivial mathematical identity. Now it is observed that if the dynamical variables obey the equations of motions then action becomes stationary even if the parameter is time dependent. We observe that the variation of the action - which must be zero since the action is stationary - can only depend on the integration of the time derivatives of the parameter.  Now integrate by parts to take all the time derivatives off it and keep the rest in the integrand. Since the parameter is arbitrary, its co efficient in the integral must be zero. Now this coefficient is time derivative of something whose time derivative is zero. Therefore this &quot;something&quot; is constant or conserved in time.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-11T01:46:34.043" LastActivityDate="2011-02-11T01:46:34.043" />
  <row Id="4986" PostTypeId="1" AcceptedAnswerId="4992" CreationDate="2011-02-11T03:24:40.623" Score="1" ViewCount="111" Body="&lt;p&gt;With regard to the recent arXiv article:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J. D. Shelton, &lt;em&gt;Eddy Current Model of Ball Lightening&lt;/em&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/1102.1224&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1102.1224&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if this is a reasonable explanation of ball lightening, or if there is such an explanation. The paper is somewhat technical and E&amp;amp;M is one of my worst subjects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please feel free to edit this question to one better suited, or if you don't have the rep, add a comment suggesting changes.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-11T08:03:28.530" Title="Could ball lightening be a form of plasma?" Tags="&lt;electromagnetism&gt;&lt;plasma-physics&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="4987" PostTypeId="2" ParentId="4986" CreationDate="2011-02-11T04:04:11.213" Score="-1" ViewCount="" Body="&lt;p&gt;I think that &quot;magnetophosphenes&quot; like those produced by transcranial magnetic stimulation,&#xA;are more likely--ie a strong fluctuating magnetic field produced by nearby lightning strikes causes visual hallucination in the occipital cortex (visual region). These are reproducible using TMS but other mechanisms like plasmas are possible.&#xA;&lt;a href=&quot;http://arxiv.org/abs/1005.1153&quot; rel=&quot;nofollow&quot;&gt;Transcranial stimulability of phosphenes by long lightning electromagnetic pulses&#xA;Authors: J. Peer, A. Kendl&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-02-11T04:04:11.213" CommentCount="6" />
  <row Id="4988" PostTypeId="1" AcceptedAnswerId="4989" CreationDate="2011-02-11T05:01:04.583" Score="7" ViewCount="276" Body="&lt;p&gt;I know that in string theory, D-branes are objects on which open strings are attached with Dirichlet boundary conditions. But what exactly is a brane? Are they equally fundamental objects like string? If so then do they also vibrate? If the visible universe itself is not a brane then what is the dynamics of these branes within the universe? Do individual D-Branes interact, collide? Can an open string tear itself off from the D-brane? If so what are the results?&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-02-11T07:21:40.497" Title="What is a D-brane?" Tags="&lt;string-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="4989" PostTypeId="2" ParentId="4988" CreationDate="2011-02-11T06:36:45.917" Score="3" ViewCount="" Body="&lt;p&gt;Branes are (usually) extended objects; $p$-branes are objects with $p$ spatial dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D-branes are a special and important subset of branes defined by the condition that fundamental strings can end on the D-branes. This is literally the technical definition of D-branes and it turns out that this simple fact determines all of the properties of D-branes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perturbatively, fundamental strings are more fundamental than branes or any other objects. In that old-fashioned description, D-branes are &quot;solitons&quot; - configurations of classical fields that arise from the closed strings. They are analogous to magnetic monopoles - which may also be written as classical configurations of the &quot;more fundamental fields&quot; in field theory. In a similar way, D-branes' masses diverge for $g\to 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-perturbatively, D-branes and other branes are equally fundamental as strings. In fact, when $g$ is sent to infinity, some D-branes may become the lightest objects - usually strings of a dual (S-dual) theory. When we include very strongly coupled regimes (high values of the string coupling constant $g$), there is a brane democracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to the perturbative realm. The condition that open strings can end on D-branes - and nowhere else - means that there exists a particular spectrum of open strings stretched between such D-branes. By quantizing these open strings, we obtain all the fields that propagate along (and in between) such D-branes. The usual methods (world sheets of all topologies, now allowing boundaries) allow us to calculate all the interactions of these modes, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, D-branes also vibrate. But because their tension goes to infinity for $g\to 0$, you need even more energy to excite these vibrations than for strings. The quanta of these vibrations are particles identified with open strings - that move along these D-branes but are stuck on them. The insight that the D-branes are dynamical and may vibrate, and the insight that they carry Ramond-Ramond charges (generalizations of the electromagnetic field one obtains from superstrings whose all RNS fermionic fields are periodic on the world sheet) were the main insights of Joe Polchinski in 1995 that made D-branes essential players and helped to drive the second superstring revolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other branes typically have qualitatively similar properties as D-branes but one must use different methods to determine these properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we quantize a D-brane, we find open string states which are scalars corresponding to the transverse positions. It follows that D-branes may be embedded into the spacetime - in any way. The shape oscillates according to a generalized wave equation again. Also, all D-branes carry electromagnetic fields $F_{\mu\nu}$ in them. These fields are excited by the endpoints of the open strings that behave as quarks (or antiquarks). For a stack of $N$ coincident branes, the gauge group gets promoted to $U(N)$. The electric flux inside the D-branes may be viewed as a &quot;fuzzy&quot; continuation of the open strings that completes them to &quot;de facto closed strings&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those fields have superpartners in the case of the supersymmetric D-branes which are stable and the most important ones, of course. D-branes may collide and interact much like all other objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most appropriate interaction that allows the open strings to &quot;disconnect&quot; from D-branes is the event in which two end points (of the opposite type, if the open strings are oriented) collide. Much like a quark and antiquark, these two endpoints may annihilate. In this process, an open string may become a closed string - which may escape away from the D-brane. The same local process of &quot;annihilation of the endpoints&quot; may also merge two open strings into one. Such interactions are the elementary explanations of all the interactions between the fields produced by the open strings - for example between the transverse scalars and the electromagnetic fields within the D-brane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from that, some branes may also be open branes, and end on another kind of branes. The latter brane always includes some generalized electromagnetic fields that are sourced by the endpoints or end curves or whatever is the $(p-1)$-dimensional geometry of the boundary of the former brane.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-11T07:21:40.497" LastActivityDate="2011-02-11T07:21:40.497" CommentCount="3" />
  <row Id="4990" PostTypeId="1" AcceptedAnswerId="4991" CreationDate="2011-02-11T06:40:51.303" Score="3" ViewCount="400" Body="&lt;p&gt;My professor in class went a little over chaos theory, and basically said that Newtonian determinism no longer applies, since as time goes to infinity, no matter how close together two initial points are, the distance between them will increase greatly. But why isn't this merely a matter of the imprecision of our measuring instruments? If we can somehow know our initial conditions &lt;em&gt;exactly&lt;/em&gt;, wouldn't we still be able to calculate what the system will be like at some time &lt;em&gt;t&lt;/em&gt; in the future?&lt;/p&gt;&#xA;" OwnerUserId="202" LastActivityDate="2011-02-11T13:09:56.270" Title="Chaos theory and determinism" Tags="&lt;determinism&gt;&lt;chaos-theory&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="4991" PostTypeId="2" ParentId="4990" CreationDate="2011-02-11T07:19:09.200" Score="6" ViewCount="" Body="&lt;p&gt;Well, yes. In a purely mathematical world where you can specify initial conditions exactly, chaotic systems are fully deterministic. It's not like a quantum system with wavefunction collapse, whose evolution can never be specified exactly by the initial conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in practice, we can never specify (or know) the initial conditions exactly. So there will always be some uncertainty in the initial conditions, and it makes sense to characterize the behavior of a system &lt;a href=&quot;http://en.wikipedia.org/wiki/Lyapunov_exponent&quot; rel=&quot;nofollow&quot;&gt;in terms of its response to this uncertainty&lt;/a&gt;. Basically, a chaotic system is one in which any uncertainty in the state at time $t=0$ leads to exponentially larger uncertainties in the state as time goes on, and a non-chaotic system is one in which any initial uncertainty in the state decays away or at least stays steady with time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the former (chaotic) case, given that we can't know the initial conditions to infinite precision, there will always be some time after which predictions of the behavior of the system become essentially meaningless - the uncertainty becomes so large that it fills up most of the state space. This is &lt;em&gt;effectively&lt;/em&gt; similar to the behavior of a truly non-deterministic (e.g. quantum) system, in that our ability to make predictions about it is limited, so some people call chaotic systems non-deterministic.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-02-11T07:19:09.200" CommentCount="8" />
  <row Id="4992" PostTypeId="2" ParentId="4986" CreationDate="2011-02-11T08:03:28.530" Score="4" ViewCount="" Body="&lt;p&gt;Ball lightning could definitely be some atmospheric pressure plasma phenomenon. You can make a pretty impressive ball plasma by discharging a kilojoule-scale capacitor bank into a bucket of salt water. Check out &lt;a href=&quot;http://www.pppl.gov/colloquia_pres/WC17SEP08_GWurden.pdf&quot; rel=&quot;nofollow&quot;&gt;Free-Floating Atmospheric Pressure Ball Plasma&lt;/a&gt;. In most of those pictures they're using a copper sulfate solution, but that's not essential (sodium chloride also works). These ones only last a (significant) fraction of a second, but I'm sure if you made a larger one (e.g. by a lightning strike), they could last longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, this was the subject of a killer science fair project: &lt;a href=&quot;http://www.youtube.com/watch?v=SE6sbaNsKoc&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=SE6sbaNsKoc&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-02-11T08:03:28.530" />
  <row Id="4993" PostTypeId="2" ParentId="4990" CreationDate="2011-02-11T09:17:23.183" Score="6" ViewCount="" Body="&lt;p&gt;I think the following from the wikipedia entry clears up well the terminology:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Chaos theory is a field of study in applied mathematics, with applications in several disciplines including physics, economics, biology, and philosophy. Chaos theory studies the behavior of dynamical systems that are highly sensitive to initial conditions; an effect which is popularly referred to as the butterfly effect. &lt;strong&gt;Small differences in initial conditions (such as those due to rounding errors in numerical computation) yield widely diverging outcomes for chaotic systems, rendering long-term prediction impossible in general.&lt;a href=&quot;http://www.uwm.edu/~aatsonis/&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;&lt;/strong&gt; This happens even though these systems are deterministic, meaning that their future behavior is fully determined by their initial conditions, with no random elements involved.[2] In other words, &lt;strong&gt;the deterministic nature of these systems does not make them predictable.&lt;/strong&gt;[3] This behavior is known as deterministic chaos, or simply chaos.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bold mine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that chaos results in completely deterministic systems when  small errors in initial conditions yield highly divergent solutions. It is the &quot;exactly&quot; in your question that is unattainable, that will be a tiny bit off in chaotic situations,(highly nonlinear response to input parameters) even in computer solutions, because one cannot be more accurate than the computer bits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that this does not mean that there are no mathematical methods to study the behavior of such systems. There are, and can be predictive in bulk. I would give as an example the study of &lt;a href=&quot;http://www.uwm.edu/~aatsonis/&quot; rel=&quot;nofollow&quot;&gt;Tsonis et al&lt;/a&gt; who have studied climate with a neural net chaotic model using as inputs bulk behavior of atmospheric and ocean currents.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-02-11T09:52:22.993" LastActivityDate="2011-02-11T09:52:22.993" CommentCount="1" />
  <row Id="4994" PostTypeId="1" AcceptedAnswerId="4996" CreationDate="2011-02-11T09:49:42.087" Score="3" ViewCount="169" Body="&lt;p&gt;Once you understand the formalism, I think it's clearest to say the critical dimension of the space-time arises because we need to cancel the central charge of the (super)conformal ghosts on the worldsheet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But suppose I need to explain this to a person who only knows very basic QFT. How would you explain the concept of the critical dimension?&lt;/p&gt;&#xA;" OwnerUserId="1864" LastActivityDate="2011-02-11T10:42:11.640" Title="Pedagogical explanations of critical dimensions of string theories" Tags="&lt;string-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="4995" PostTypeId="1" CreationDate="2011-02-11T10:29:34.003" Score="4" ViewCount="367" Body="&lt;p&gt;I’m trying to better my understanding of the thermodynamics and momentum balance of pipe flows. The following situation, however, is making me scratch my head and I’ve found no help in my books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a pipe of constant section in which an ideal gas is flowing with negligible friction. Somewhere along this pipe there is a cooler, and Tin &gt; Tout. Assume the cooler is frictionless as well and of constant section equal to that of the pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will there be a pressure drop in the pipe due to the cooling?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I have done some more research on the topic and it seems that what I’m considering is a case of a so-called Rayleigh flow. I tried to solve it this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p = \rho r T$ (equation of state)&#xA;$p + \rho v^2 = Cst$ (conservation of momentum)&#xA;$\rho v = Cst$ (conservation of mass)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 3 unknowns: p, $\rho$ and v. I assume the temperature profile in the flow is known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this for p gives me a second degree polynomial with two solutions for p. I am not sure yet how to discriminate between the two.&lt;/p&gt;&#xA;" OwnerUserId="1559" LastEditorUserId="1559" LastEditorDisplayName="" LastEditDate="2011-02-14T10:33:11.683" LastActivityDate="2011-08-13T19:23:39.060" Title="Pressure drop in a pipe due to cooling" Tags="&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="4996" PostTypeId="2" ParentId="4994" CreationDate="2011-02-11T10:42:11.640" Score="3" ViewCount="" Body="&lt;p&gt;There are about 10 &quot;significantly different&quot; ways of calculating the critical dimension but all of them require some maths.&#xA;n&#xA;For the case of bosonic string theory, one may show that the $bc$ system carries the central charge $-3k^2+1$ for $k=3$ which is $-26$, so it has to be canceled by 26 bosons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, one may define the light-cone gauge string. Only $D-2$ transverse dimensions produce oscillators. There are oscillators&#xA;$$\alpha_n^i$$&#xA;where $i$ has $D-2$ different values and $n$ is an integer. The squared mass (in spacetime) of the string i.e. particle is given by the total excitation of all the harmonic oscillators&#xA;$$\sum_{n&amp;gt;0,i} \alpha_{-n}^i \alpha^i_n$$&#xA;The oscillators $\alpha_n$ are normalized as $\sqrt{n}\hat a$ for a normal annihilation operator of the harmonic oscillator, so the excitations are weighed by $n$. Consequently, the zero-point energy is&#xA;$$\frac{D-2}{2}\sum_{n=1}^\infty n$$&#xA;The sum of positive integers is equal to $-1/12$ - by the zeta-function regularization or any equally good one (the correct final result of the sum has been known for a century or more). The first excited state - excited by $\alpha_{-1}^i$ - adds one to the energy and it is a spacetime vector. However, it only has $D-2$ components rather than $D-1$ components, so it must correspond to a massless particle. Consequently,&#xA;$$\frac{D-2}{2} \left(-\frac{1}{12}\right)+1 = 0$$&#xA;has to hold. It follows that $D-2=24$, $D=26$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculations of the superstring critical dimension, $D=10$, have to account for the world sheet fermions as well (which can be added via several equivalent machineries, either as spacetime vectors, RNS, or spacetime spinors, GS). Alternatively, one may adopt a spacetime perspective and explain why 10 and 11 are the right dimensions in which maximally supersymmetric string theory and M-theory may live.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some sense, a basic knowledge of QFT is enough to understand those calculations - but less than the basic knowledge of QFT is just not enough. As far as I know, there is no honest &quot;high school level&quot; derivation of the critical dimension - even though the final result may be understood by the high school kids.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-11T10:42:11.640" CommentCount="3" />
  <row Id="4997" PostTypeId="1" CreationDate="2011-02-11T11:47:26.500" Score="2" ViewCount="122" Body="&lt;p&gt;Consider gWZW action $S_{gWZW}(g,A)=S_{WZW}(g)+S_{gauge}(g,A)$, where $S_{WZW}$ is usual WZW action being sum of sigma model and WZ terms for field $g$ taking values in group $G$ and $S_{gauge}=\int d^2\sigma\text{Tr}(A_+\partial _-gg^{-1}-A_-g^{-1}\partial _+g-g^{-1}A_+gA_-+A_+A_-)$ is gauge action which depends on field $g$ and gauge fields $A_\pm$, the later take values in subalgebra of group $G$ algebra which we want to gauge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to equations of motion following from gWZW action field strength&#xA;$F_{+-}=\partial _+A_--\partial _-A_+ +[A_+,A_-]$ vanishes on-shell. To show it one shall project equation of motion for field $g$ on gauge subalgebra and substitute constraints for gauge fields $A_\pm$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;strong&gt;are we allowed&lt;/strong&gt; to use $F_{+-}=0$ equation of motion while studying symmetries of action - that is are we allowed to substitute $F_{+-}=0$ in variation of action justifying it by claim that fields $A_\pm$ are non-dynamical?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have natural doubts that this is allowed because as described above equation $F_{+-}=0$ is obtained not just with the help of equations following from variation of action by non-dynamical $A_\pm$ fields but also using $g$ field equations. This is topical issue e.g. when one deals with sort of &lt;em&gt;supersymmetric extension of gWZW model&lt;/em&gt; (some non-ordinary or potentially-deformed kinds of them - because there's no necessity to set $F_{+-}=0$ in usual supersymmetric gWZW model). If so to obtain $F_{+-}=0$ equation one also needs to use fermionic fields equations of motion.&lt;/p&gt;&#xA;" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-11T19:53:21.890" LastActivityDate="2011-03-13T20:32:53.397" Title="Vanishing of field strength in gauged WZW model" Tags="&lt;variational-principle&gt;&lt;supersymmetry&gt;" AnswerCount="1" />
  <row Id="4999" PostTypeId="2" ParentId="4997" CreationDate="2011-02-11T12:22:30.743" Score="1" ViewCount="" Body="&lt;p&gt;Your intuition is right: you are not allowed to assume the equations of motion if you're determining the symmetries of the action. Quantum mechanically, the theory may be defined as a path integral over all configurations - including (and especially) those that don't satisfy the classical equations of motion - and all of them contribute to the transition amplitudes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the symmetry didn't exist for the configurations that don't obey the equations of motion, the calculated amplitudes wouldn't obey the symmetry, either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While you're not allowed to assume the equations of motion, you do allow variations of the Lagrangian that are total derivatives - so that the action which is the integral is still preserved assuming a good behavior at infinity. In some contexts, this &quot;symmetry up to total derivatives&quot; plays a similar role to using the equations of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was there some particular symmetry you wondered about here? The essence of your question is general and has no special relationship with the WZNW models.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-11T12:22:30.743" />
  <row Id="5000" PostTypeId="2" ParentId="185" CreationDate="2011-02-11T12:26:05.830" Score="2" ViewCount="" Body="&lt;p&gt;One answer that is missing in the good summaries above  is the real reason why there are two experimental setups at the LHC. Independent verification from experiments with different systematic detector and computing/methodology  errors. A fairly recent example was the 3sigma announcement of the &lt;a href=&quot;http://cdsweb.cern.ch/record/476830/files/ep-2000-138.pdf&quot; rel=&quot;nofollow&quot;&gt;Higgs at 114GeV by ALEPH&lt;/a&gt;, which was reduced in sigma to undetectable when the other three experiments did not see the resonance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Detector errors should be evident, different accuracies and methods of gathering information might introduce an unexpected &quot;signal&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computing and methodology errors are more insidious and are mostly grounded on the sociological observation that large groups of people, even if they are physicists, can take off in the wrong direction given enough enthusiasm (herd/pack mentality). History has recorded the alternating neutral currents, for example, &quot;now you see them, now you don't,&quot; because such heavy names were behind the &quot;discoveries&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at least two independent confirmations are absolutely necessary.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-02-11T12:26:05.830" />
  <row Id="5001" PostTypeId="2" ParentId="629" CreationDate="2011-02-11T12:43:03.083" Score="1" ViewCount="" Body="&lt;p&gt;Another way of observing this effect is to have a car go up a gentle slope at some speed, put it in neutral gear, wait for it to stop and then hitting the breaks. Most people expect that using the breaks when the car is at rest (even if just momentarily) will have no perceptible effect, and they are surprised to feel a 'jerk' very similar to the one mentioned in the question.     &lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2011-02-11T12:43:03.083" />
  <row Id="5002" PostTypeId="2" ParentId="4990" CreationDate="2011-02-11T13:09:56.270" Score="5" ViewCount="" Body="&lt;p&gt;Classical mechanics is perfectly integrable for two bodies as a closed or isolated system.  However, early on it was found that problems existed, where Newton found he could not find a solution for the motion of the planets in a complete form.  He made his famous statement that God had to readjust the solar system now and them.  Poincare solved the Sweden prize for a solution to the stability of the solar system by demonstrating no such solution existed.  This is what opened the door for chaos theory, where Poincare developed methods of separatrices and perturbations on them.  For general systems it turns out that Newtonian mechanics is not integrable,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classical mechanics for systems with three or more bodies can\rq t be solved in closed form.  For any $N$ body problem there are 3 equations for the center of mass, $3$ for the momentum, $3$ for the angular momentum and one for the energy.  These are $10$ constraints on the problem.  An N-body problem has $6N$ degrees of freedom.  For $N~=~2$ this means the solution is given by a first integral with degree $2.$  For a three body problem this first integral has degree $8$.  This runs into the problem that Galois illustrated which is that any root system with degree $5$ or greater generally have no algebraic roots.  First integrals for differential equations are functions which remain constant along a solution to that differential equation.  So for $8$ solutions there is some eight order polynomial $p_8(x)~=~\prod_{n=1}^8(x~-~\lambda_n)$, with $8$ distinct roots$\lambda_n$ that are constant along the $8$ solutions.  Since $p_8(x)~=~p_5(x)p_3(x)$, a branch of algebra called Galois theory tells us that fifth order polynomials have no general algebraic system for finding its roots, or a set of solutions that are algebraic.  This means that any system of degree higher than four are not in general algebraic.  At the root of the $N$-body problem Galois theory tells us there is no algebraic solution for $N~\ge~3$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A problem in classical mechanics is the vanishing denominator problem for three bodies. This has a Hamiltonian  $H( J,~\theta)~=~H_0(J,~\theta)~+~\epsilon H_1(J,~\theta)$ for $J~=~(J_1,~J_2)$ and $\theta~=~(\theta_1,~\theta_2)$.   Here a generating function written according to the variable $J^\prime$&lt;br&gt;&#xA;$$&#xD;&#xA;S(J^\prime,~\theta)~=~\theta\cdot J^\prime~+~i\epsilon\sum_{n_1,n_2}{{H_{1,n_1,n_2}}\over{n_1\omega_1(J^\prime)~+~n_2\omega_2(J^\prime)}}e^{ n_1\omega_1(J^\prime)~+~n_2\omega_2(J^\prime)}&#xD;&#xA;$$&#xA;will be divergent for the resonant condition $n_1\omega_1(J^\prime)~+~n_2\omega_2(J^\prime)~=~0$.  This resonance condition has lead many to presume that the solar system can not be stable with resonance conditions.  Yet the solar system is replete with near resonance conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of resonance conditions that exist on the real line are dense.  Within any $\epsilon$ neighborhood there will exist a countably infinite number of possible resonance conditions that correspond to rational numbers.  As the orbit of a planet drifts it will pass through these resonance conditions and be chaotically perturbed.  It is to be expected that for simple rational numbers, such as $1/12$ for Earth and Jupiter, rather than $1003/12000$ strong resonances occur.  For more complex rational numbers it might be expected that the instability will be weaker.  In other words if the ratio of frequencies are \lq\lq sufficiently irrational\rq\rq$~$so that&#xA;$$&#xD;&#xA;\Big|{{\omega_1}\over{\omega_2}}~-~{m\over s}\Big|~&amp;gt;~{{k(\epsilon)}\over{s^{2.5}}},~\lim_{\epsilon~\rightarrow~0}k(\epsilon)~\rightarrow~0&#xD;&#xA;$$&#xA;the orbit is more stable.  So an orbit that is removed from a “strong resonance”” condition near a simple rational number will be more stable than an orbit that is near an orbit with a simple rational ratio of frequencies.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the basis for Greenberg’s Hamiltonian approach to chaos theory.  This is called deterministic for the differential equations are time reverse invariant so the motion of a particle is absolutely determined.  However. if you have a slight variation in the initial conditions of that particle is may in general end up arbitrarily far away from its starting point.  The tiny variation $\delta z~=~(\delta q,~\delta p)$ becomes amplified by an exponential map $\delta z~\rightarrow~exp(\lambda t)\delta z$, for $\lambda$ the Lyapunov exponent.  Any error in the specification of the initial conditions of a body results in the amplification of this error.  From an algorithmic perspective a truncation results in numerical overflow errors which grow.  So the dynamics of a particle can’t be integrated by computer to arbitrary accuracy into the future, even though nature actually does determine its dynamics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;W. Zurek took this a bit further and considered how quantum fluctuations, where $\delta z$ is set by the Heisenberg uncertainty principle.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-11T13:09:56.270" CommentCount="1" />
  <row Id="5003" PostTypeId="2" ParentId="4995" CreationDate="2011-02-11T15:16:38.940" Score="-1" ViewCount="" Body="&lt;p&gt;If the gas does not accumulate or condense in the pipe, during a given time frame $\Delta t$ there should be the same amount of substance entering the pipe and exiting the pipe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the gas is &lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_gas&quot; rel=&quot;nofollow&quot;&gt;ideal&lt;/a&gt;, we then have $\dfrac{P_{in}V_{in}}{T_{in}}=\dfrac{P_{out}V_{out}}{T_{out}}$, so either pressure drops, or outgoing volume of gas per unit of time is less than incoming volume of gas per unit of time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the impression that the answer to this question depends on the boundary conditions: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;if the gas is in contact with a fixed-pressure environment, e.g. it exits into an atmosphere at the same pressure as the incoming gas, then volume will decrease, so the effect of the cooling will be to increase the gas density.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if the pipe is long enough before being in contact with another environment, while keeping a fixed section, then conservation of momentum does indicate that the incoming amount of substance will keep the same velocity (cooling will decrease the overall kinetic energy of the gas molecules, but not their overall momentum). With constant section $S$ and constant velocity $v=\dfrac{\Delta L}{\Delta t}$, the same volume $\Delta V=S\Delta L$ will be coming in and going out of the cooling zone in the time $\Delta t$, so it's the pressure that must decrease.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;depending on the length of the pipe after the cooler and before contact with a fixed pressure (and non-constant section) environment, there may be a combination of both effets through a gradient of pressure.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1725" LastActivityDate="2011-02-11T15:16:38.940" CommentCount="4" />
  <row Id="5004" PostTypeId="2" ParentId="1626" CreationDate="2011-02-11T15:34:47.547" Score="0" ViewCount="" Body="&lt;p&gt;I have doors like that too.  Probably everyone in a climate where it gets reasonably cool does too.  Other interesting effects include cracks that open/close based on temperature.  Appears that the house shifts slightly during seasonal changes.  Possibly due to temperature differences that the house generates in the ground resulting in asymmetrical movement?  Humidity appears to be a near non-issue (it does have some effect, but never enough to make the difference between working and not working).&lt;/p&gt;&#xA;" OwnerUserId="919" LastActivityDate="2011-02-11T15:34:47.547" />
  <row Id="5005" PostTypeId="1" AcceptedAnswerId="5011" CreationDate="2011-02-11T16:32:50.910" Score="7" ViewCount="374" Body="&lt;p&gt;In my opinion, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Grassmann_number&quot; rel=&quot;nofollow&quot;&gt;Grassmann number&lt;/a&gt; &quot;apparatus&quot; is one of the least intuitive things in modern physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember that it took a lot of effort when I was studying this. The problem was not in the algebraic manipulations themselves -- it was rather psychological: &lt;em&gt;&quot;Why would one want to consider such a crazy stuff!?&quot;&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later one just gets used to it and have no more such strong feelings for the anticommuting  numbers. But this psychological barrier still exists for newbies.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to explain or motivate one to learn the Grassmann numbers? Maybe there is a simple &lt;strong&gt;yet practical&lt;/strong&gt; problem that demonstrates the utility of those? It would be great if this problem would provide the connection to some &quot;not-so-advanced&quot; areas of physics, like non-relativistic QM and/or statistical physics. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-02-11T18:38:19.120" Title="&quot;Velvet way&quot; to Grassmann numbers" Tags="&lt;quantum-field-theory&gt;&lt;mathematical-physics&gt;&lt;grassmann-numbers&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="7" />
  <row Id="5006" PostTypeId="2" ParentId="5005" CreationDate="2011-02-11T16:42:35.667" Score="3" ViewCount="" Body="&lt;p&gt;Nope. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can't be any role for the Grassmann numbers outside quantum physics. The reason is simple: they can't take any &quot;values&quot; from a particular &quot;set&quot;. Instead, they only become analogous to the regular commuting numbers once we use them as intermediate steps and we integrate or differentiate over them or set them equal to zero. Indeed, the integrals are algebraically analogous to the integral over commuting variables. However, the overall sign in the presence of the Grassmann numbers is always a bit undetermined - and one must square the result to get rid of the ambiguity. That's why the Berezin (Grassmann) integrals must always be interpreted as probability amplitudes and quantum mechanics is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One may say that a commuting number may be a product of two (or another even number of) Grassmann numbers - so a commuting number has an even number of previously unknown building blocks. But the measurable number - the number that Nature offers as a part of Her &quot;user interface&quot; to the consumers i.e. observers - is always commuting. However, at the fundamental level, in the core of Her inner workings, the Grassmann numbers are as natural as the commuting numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Grassmann numbers are the $c$-numbers for fermionic creation or annihilation operators in the same way as the ordinary commuting $c$-numbers are the numbers for the ordinary bosonic creation or annihilation operators (and many other operators). That's the closest analogy. It's natural for two operators to &quot;almost commute&quot; with one another or &quot;almost anticommute&quot; with one another. It must be one of the two options because if one repeats the interchange twice, we must get back to the same object, so the factor we pick by interchanging them must satisfy $S^2=+1$, leaving $S=\pm 1$ as the only possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A person must first understand why the Fermi-Dirac statistics is as natural as the Bose-Einstein statistics; then he can understand that at a deeper level, the Grassmann numbers are exactly as natural as the commuting numbers. However, everyone must also be ready that he doesn't have any &quot;material experience&quot; with the Grassmann numbers because they can't play any role in the classical limit - they can't appear in the &quot;user interface&quot; of Nature.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-11T16:49:12.510" LastActivityDate="2011-02-11T16:49:12.510" CommentCount="5" />
  <row Id="5007" PostTypeId="2" ParentId="5005" CreationDate="2011-02-11T16:53:38.383" Score="0" ViewCount="" Body="&lt;p&gt;You may also utilize Grassmann-odd numbers while considering generalization of some simple classical models to supersymmetric version of theirs. So for example in the paper&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/0705.2249&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0705.2249&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there're considered various supersymmetrizations of Landau model - that is the model describing charged particle moving in the plane orthogonal to uniform magnetic field. The supersymmetric version roughly speaking replaces commuting coordinates of this plane by anticommuting Grassmann-odd coordinates. Subsequent quantization gives quantum model with Landau levels being solutions of stationary Schroedinger equation.&lt;/p&gt;&#xA;" OwnerUserId="1866" LastActivityDate="2011-02-11T16:53:38.383" CommentCount="2" />
  <row Id="5008" PostTypeId="1" AcceptedAnswerId="5019" CreationDate="2011-02-11T16:58:39.053" Score="8" ViewCount="334" Body="&lt;p&gt;I am a computer scientist interested in network theory. I have come across the Bose-Einstein Condensate (BEC) because of its connections to complex networks. What I know about condensation is the state changes in water; liquid-gas. Reading the wikipedia articles on the subject is difficult, not because of the maths, but the concept doesn't seem to be outlined simply. Other source share this approach going straight into the topic without a gentle introduction to set the scene for the reader. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate a few paragraphs describing the problem at hand with BEC (dealing with gas particles right? which kind, any kind? only one kind? mixed types of particles? studying what exactly, their state changes?), what effects can occur (the particles can form bonds between them? which kind of bonds? covalent? ionic?), what do we observe in the BEC systems (some particles form many bonds to particles containing few bonds? The spatial configurations are not symmetric? etc), and what degrees of freedom exist to experiment with (temperature? types of particles? number of particles?) in these systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best,&lt;/p&gt;&#xA;" OwnerUserId="813" LastActivityDate="2011-06-17T21:29:55.240" Title="trying to understand Bose-Einstein Condensate (BEC)" Tags="&lt;quantum-mechanics&gt;&lt;particle-physics&gt;&lt;statistical-mechanics&gt;&lt;condensed-matter&gt;&lt;bose-einstein-condensate&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="5010" PostTypeId="2" ParentId="5005" CreationDate="2011-02-11T17:18:57.003" Score="3" ViewCount="" Body="&lt;p&gt;One of your questions was whether there exists a use of the Grassmann number apparatus in statistical physics. The answer is yes. This is not surprising since there are many deep connections between QFT and statistical physics (even classical one). For example, the Grassman technique can be used as a tool for calculating the partition function of the dimer model - this model originates from the following question: if diatomic molecules are adsorbed on a surface, forming a single layer, how many ways can they be arranged, what is the entropy of the system? For a work on this, see: &quot;Grassmann Variables and Exact Solutions for Two-Dimensional Dimer Models&quot; ( &lt;a href=&quot;http://arxiv.org/abs/cond-mat/9711156&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/cond-mat/9711156&lt;/a&gt; ). One can also use Grassmann techniques in the field of classical spin systems, a review of this can be found in &quot;Grassmann techniques applied to classical spin systems &quot; ( &lt;a href=&quot;http://arxiv.org/abs/0905.1104&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0905.1104&lt;/a&gt; ).&lt;/p&gt;&#xA;" OwnerUserId="1855" LastEditorUserId="1855" LastEditorDisplayName="" LastEditDate="2011-02-11T18:38:19.120" LastActivityDate="2011-02-11T18:38:19.120" CommentCount="8" />
  <row Id="5011" PostTypeId="2" ParentId="5005" CreationDate="2011-02-11T17:33:10.750" Score="7" ViewCount="" Body="&lt;p&gt;I don't have an answer to the question &quot;why would one want to consider such crazy stuff in &lt;em&gt;physics&lt;/em&gt;?&quot; since I don't know much physics, but as a mathematics student I do have an answer to the question &quot;why would one want to consider such crazy stuff in &lt;em&gt;mathematics&lt;/em&gt;?&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What physicists call Grassmann numbers are what mathematicians call elements of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Exterior_algebra&quot; rel=&quot;nofollow&quot;&gt;exterior algebra&lt;/a&gt; $\Lambda(V)$ over a vector space $V$. The exterior algebra naturally arises as the solution to the following geometric problem. Say that $V$ has dimension $n$ and let $v_1, ... v_n$ be a basis of it. We would like a nice natural definition of the $n$-dimensional volume of the paralleletope defined by the vectors $\epsilon_1 v_1 + ... + \epsilon_n v_n, e_i \in \{ 0, 1 \}$. When $n = 2$ this is the standard parallelogram defined by two linearly independent vectors, and when $n = 3$ this is the standard paralellepiped defined by three linearly independent vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing about the &lt;em&gt;naive&lt;/em&gt; definition of volume is that it is very close to having really nice mathematical properties: it is almost multilinear. That is, if we denote the volume we're looking at by $\text{Vol}(v_1, ... v_n)$, then it is almost true that $\text{Vol}(v_1, ... v_i + cw, ... v_n) = \text{Vol}(v_1, ... v_n) + c \text{Vol}(v_1, ... v_{i-1}, w, v_{i+1}, ... v_n)$. You can draw nice diagrams to see this readily. However, it isn't actually completely multilinear: depending on how you vary $w$ you will find that sometimes the volume shrinks to zero and then goes back up in a non-smooth way when really it ought to keep getting more negative. (You can see this even in two dimensions, by varying one of the vectors until it goes past the other.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To fix that, we need to look instead at &lt;em&gt;oriented&lt;/em&gt; volume, which can be negative, but which has the enormous advantage of being completely multilinear and smooth. The other major property it satisfies is that if any of the two vectors $v_i$ agree (that is, the vectors are linearly dependent) then the oriented volume is zero, which makes sense. It turns out (and this is a nice exercise) that this is equivalent to oriented volume coming from a &quot;product&quot; operation, the exterior product, which is anticommutative. Formally, these two conditions define an element of the top exterior power $\Lambda^n(V)$ defined by the exterior product $v_1 \wedge v_2 ... \wedge v_n$, and choosing an element of this top exterior power (a volume form) allows us to associate an actual number to an $n$-tuple of vectors which we can call its oriented volume in the more naive sense. If $V$ is equipped with an inner product, then there are two distinguished elements of $\Lambda^n(V)$ given by a wedge product of an orthonormal basis in some order, and it's natural to pick one of these as a volume form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alright, so what about the rest of the exterior powers $\Lambda^p(V)$ that make up the exterior algebra? The point of these is that if $v_1, ... v_p, p &amp;lt; n$ is a tuple of vectors in $V$, we can consider the subspace they span and talk about the $p$-dimensional oriented volume of the paralleletope given by the $v_i$ in this subspace. But the result of this computation shouldn't just be a number: we need a way to do this that keeps track of &lt;em&gt;what subspace&lt;/em&gt; we're in. It turns out that mathematically the most natural way to do this is to keep in mind the requirements we really want out of this computation (multilinearity and the fact that if the $v_i$ are not linearly independent then the answer should be zero), and then just define the result of the computation to be the &lt;em&gt;universal thing&lt;/em&gt; that we get by imposing these requirements and nothing else, and this is nothing more than the exterior power $\Lambda^p(V)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This discussion hopefully motivated for you why the exterior algebra is a natural object from the perspective of geometry. Since Einstein, physicists have been aware that geometry has a lot to say about physics, so hopefully the concept makes a little more sense now.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let me also say something about how modern mathematicians think about &quot;space&quot; in the abstract sense. The inspiration for the modern point of view actually derives at least partially from physics: the only thing you can really know about a space are observables defined on it. In classical physics, observables form a commutative ring, so one might say roughly speaking that the study of commutative rings is the study of &quot;classical spaces.&quot; In mathematics this study, in the abstract, is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Algebraic_geometry&quot; rel=&quot;nofollow&quot;&gt;algebraic geometry&lt;/a&gt;. It is a very sophisticated theory that encompasses classical algebraic geometry, arithmetic geometry, and much more, and it is in large part because of the success of this theory and related commutative ring approaches to geometry (topological spaces, manifolds, measure spaces) that mathematicians have gotten used to the slogan that &quot;commutative rings are rings of observables on some space.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, quantum mechanics tells us that the actual universe around us doesn't work this way. The observables we care about don't commute, and this is a big issue. So mathematically what is needed is a way to think about noncommutative rings as &quot;quantum spaces&quot; in some sense. This subject is very broad, but roughly it goes by the name of &lt;a href=&quot;http://en.wikipedia.org/wiki/Noncommutative_geometry&quot; rel=&quot;nofollow&quot;&gt;noncommutative geometry&lt;/a&gt;. The idea is simple: if we want to take quantum mechanics completely seriously, our spaces shouldn't have &quot;points&quot; at all because points are classical phenomena that implicitly require a commutative ring of observables, which we &lt;em&gt;know&lt;/em&gt; is not what we actually have. So our spaces should be more complicated things coming from noncommutative rings in some way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Grassmann numbers satisfy one of the most tractable forms of noncommutativity (actually they &lt;em&gt;are&lt;/em&gt; commutative if one alters the definition of &quot;commutative&quot; very slightly, but never mind that...), and even better it is a form of noncommutativity that is clearly related to something physicists care about (the properties of fermions), so anticommuting observables are a natural step up from commuting observables in order to get our mathematics to align more closely with reality while still being able to think in an approximately classical way. &lt;/p&gt;&#xA;" OwnerUserId="1046" LastActivityDate="2011-02-11T17:33:10.750" CommentCount="10" />
  <row Id="5012" PostTypeId="2" ParentId="4972" CreationDate="2011-02-11T17:43:54.397" Score="1" ViewCount="" Body="&lt;p&gt;We can have compactifications over 7D manifolds with a $G_2$ holonomy, or an 8D manifold with an $SO(7)$ holonomy. We can have orbifolds, or flux compactifications. We can have warped compactifications like $AdS_5 \times S^5$.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-11T17:43:54.397" />
  <row Id="5013" PostTypeId="2" ParentId="4961" CreationDate="2011-02-11T17:48:21.613" Score="1" ViewCount="" Body="&lt;p&gt;Unless someone can show how to come up with chiral fermions and avoid the fermion doubling problem over a lattice, it's clear lattice regularizations can't cover everything. That person would have to bypass the Nielsen-Ninomiya theorem.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-11T17:48:21.613" CommentCount="1" />
  <row Id="5014" PostTypeId="1" CreationDate="2011-02-11T17:55:28.167" Score="2" ViewCount="423" Body="&lt;p&gt;What are some good sources to learn the mathematical background of Quantum Mechanics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am talking functional analysis, operator theory etc etc...&lt;/p&gt;&#xA;" LastActivityDate="2011-02-12T20:32:09.700" Title="Mathematical background for Quantum Mechanics" Tags="&lt;quantum-mechanics&gt;&lt;mathematics&gt;&lt;education&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="5015" PostTypeId="2" ParentId="5014" CreationDate="2011-02-11T18:16:18.587" Score="3" ViewCount="" Body="&lt;p&gt;You will have more than enough math for the first two semesters of quantum mechanics if you are taking functional analysis in a mathematics course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The immense majority of quantum mechanics books will have the requisite math in an appendix.  That is true whether or not you want the more sophisticated mathematical treatments (for instance Von Neumman's classic text), or the less rigorous nuts and bolts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while you may have to look elsewhere for the theory of distributions (eg the proper way to handle the Dirac delta function) for the most part you don't really need to know it in order to do the physics, or rather it just clutters notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This more or less also holds true for Quantum field theory (where eg you need to know some group theory) and again most texts will discuss the necessary material, although of course depending on how far you get it will lead into active research areas where knowing some esoteric maths sometimes does come in handy.&lt;/p&gt;&#xA;" OwnerUserId="1350" LastActivityDate="2011-02-11T18:16:18.587" CommentCount="1" />
  <row Id="5017" PostTypeId="2" ParentId="5014" CreationDate="2011-02-11T18:31:53.350" Score="2" ViewCount="" Body="&lt;p&gt;Dover books are good and cheap. For example, here is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486453081&quot; rel=&quot;nofollow&quot;&gt;Mathematics for Quantum Mechanics&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-02-11T18:31:53.350" />
  <row Id="5018" PostTypeId="2" ParentId="5014" CreationDate="2011-02-11T18:36:00.583" Score="1" ViewCount="" Body="&lt;p&gt;The most important background is the extension of linear algebra to infinite-dimensional vectorial spaces. So you introduce Banach and Hilbert spaces, $L^p$ and note that only $L^2$ (that's the space of quantum waves functions) is a Hilbert space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must study linear operator on $L^2$, and $l^2$: many attentions must be given to adjoint operators, hermitian and antihermitian operators, unitary operators, proiectors...After that, definition of the norm of a vector and a operator and limited and nonlimited operator (a limited operator is a continuous operator), and the Riesz theorem.&#xA;If you study Lebesgue's measure theory is better.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of tensor product of many hilbert space (only a finite tensor product of hilbert spaces is a hilbert space again) is important too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last but not least, Fourier analisys, the notion of a complete and orthonormal basis, scalar products and generalized Fourier series; Green functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these definitions and arguments can be found in this book for example: Reed M , Simon B , Methods Of Modern Mathematical Physics, that is mathematically very rigorous and accurate.&lt;/p&gt;&#xA;" OwnerUserId="558" LastEditorUserId="558" LastEditorDisplayName="" LastEditDate="2011-02-11T18:41:44.230" LastActivityDate="2011-02-11T18:41:44.230" />
  <row Id="5019" PostTypeId="2" ParentId="5008" CreationDate="2011-02-11T19:15:11.190" Score="7" ViewCount="" Body="&lt;p&gt;First and foremost, the BEC systems studied in detail today do not involve the formation of any bonds between atoms. Bose-Einstein Condensation is a quantum statistical phenomenon, and would happen even with noninteracting particles (though as a technical matter, that's impossible to arrange, but you can make a condensate and then manipulate the interactions so they are effectively non-interacting, and the particles remain a condensate). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;high school physics&quot; version of what happens at the BEC transition is this: particles with integer intrinsic spin angular momentum are &quot;bosons,&quot; and many of them can occupy the same energy state. This is in contrast to particles with half-integer spin, such as electrons, termed &quot;fermions,&quot; which are unable to be in exactly the same quantum state (this feature of electrons accounts for all of chemistry, so it's a Good Thing). When we talk about a confined gas of atoms, quantum mechanics tells us that we must describe it in terms of discrete energy states, spaced by a characteristic energy depending on the details of the confinement. Because of this, the two classes of particles have very different behaviors in large numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lowest-energy state for a gas of fermions is determined by the number of particles in the gas-- each additional particle fills up whatever energy state it ends up in, so the last particle added goes in at a much higher energy than the first particle added. For this reason, the electrons inside a piece of metal have energies comparable to the hot gas in the Sun, because there are so many of them that the last electron in ends up moving very rapidly indeed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lowest-energy state for a gas of bosons, on the other hand, is just the lowest-energy state available to them in whatever system is confining them. All of the bosons in the gas can happily pile into a single quantum state, leaving you with a very low energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that, as you cool a gas of bosons, you will eventually reach a point where the gas suddenly &quot;condenses&quot; into a state with nearly all of the particles occupying a single state, generally the lowest-energy available state. This happens with material particles because the wave-like character of the bosons becomes more and more pronounced as you lower the temperature. The wavelength associated with them, which at room temperature is many times smaller than the radius of the electron orbits eventually becomes comparable to the spacing between particles in the gas. When this happens, the waves associated with the different particles start to overlap, and at some point, the system &quot;realizes&quot; that the lowest-energy state would be for all the particles to occupy a single energy level, triggering the abrupt transition to a BEC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This transition is a purely quantum effect, though, and has nothing to do with chemical bonding. In fact, strictly speaking, the dilute alkali metal vapors that are the workhorse system for most BEC experiments are actually a metastable state-- at the temperatures of these vapors, a denser gas would be a solid. They form a BEC, though, because the density of these gases is something like a million times less than the density of air. The atoms are too dilute to solidify, but dense enough to sense each others' presence and move into the same energy state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The underlying physics is described in detail in most statistical mechanics texts, though it's often dealt with very briefly and in an abstract way. There are decent and readable descriptions of the underlying physics in &lt;em&gt;The New Physics for the Twenty-first Century&lt;/em&gt; edited by Gordon Fraser, particularly the pieces by Bill Phillips and Chris Foot, and Subir Sachdev.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2011-02-11T19:15:11.190" CommentCount="1" />
  <row Id="5020" PostTypeId="1" AcceptedAnswerId="5021" CreationDate="2011-02-11T19:39:05.603" Score="5" ViewCount="193" Body="&lt;p&gt;Moshe's recent questions on &lt;a href=&quot;http://physics.stackexchange.com/questions/4068/formalizing-quantum-field-theory&quot;&gt;formalizing quantum field theory&lt;/a&gt; and &lt;a href=&quot;http://physics.stackexchange.com/questions/4961/is-quantum-field-theory-defined-by-its-lattice-regularization&quot;&gt;lattices as a definition of field theory&lt;/a&gt; remind me of something I occasionally idly wonder about, and maybe this site can tell me the answer. Are there any mathematicians working on defining quantum field theory by beginning with a rigorous definition of CFTs, and working from there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I ask is that I think this is how most of us in physics think about quantum field theory (that is, in a Wilsonian way): to define a QFT, you start with a UV fixed point, and deform it with some relevant operator. So if you had a general theory of CFTs, you would know how to understand how CFTs respond to external sources for operators, and getting a more general QFT would &quot;just&quot; mean turning on a spatially homogeneous source for some operator and seeing how it responds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other object we study is &quot;effective field theory,&quot; which you might imagine in this language is a CFT that serves as the IR fixed point, together with some notion of equivalence class of irrelevant operators deforming &quot;up&quot; away from that point (being agnostic about whether you ever reach a UV fixed point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very (extremely) naively, I would suspect mathematicians might have better luck studying the space of CFTs rather than trying to begin with all QFTs. And you might imagine this approach would be well-suited for questions physicists might care about (like, say, whether there is an &quot;&lt;em&gt;a&lt;/em&gt;-theorem&quot; or something similar, analogous to the &lt;em&gt;c&lt;/em&gt;-theorem in 2 dimensions, characterizing RG flows as irreversible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axiomatic/algebraic/constructive field theory seems to worry about &lt;em&gt;all&lt;/em&gt; kinds of field theory at once, and other mathematicians seem to be trying to dig up interesting structure in perturbation theory, which I'm not sure will ever lead to progress in nonperturbatively understanding field theory. I know there are some mathematicians who work on CFTs. (I found &lt;a href=&quot;http://mathoverflow.net/questions/37063/what-do-mathematicians-currently-do-in-conformal-field-theory-or-more-general-fi&quot; rel=&quot;nofollow&quot;&gt;this MathOverflow question&lt;/a&gt; that has a lot of links to work by mathematicians on CFTs, for instance.) But I wonder if any have tried to work on CFTs as a &lt;em&gt;route&lt;/em&gt; to understanding QFT more generally.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2011-02-11T20:10:26.060" Title="CFTs and formalizing quantum field theory" Tags="&lt;quantum-field-theory&gt;&lt;mathematical-physics&gt;&lt;conformal-field-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="5021" PostTypeId="2" ParentId="5020" CreationDate="2011-02-11T20:10:26.060" Score="2" ViewCount="" Body="&lt;p&gt;CFT consists for most mathematicians - who are interested in this topic - currently of the study of vertex operator algebras, see this question on math overflow:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://mathoverflow.net/questions/53988/what-is-the-motivation-for-a-vertex-algebra/&quot; rel=&quot;nofollow&quot;&gt;motivation for vertex algebra&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You can find a little bit more about the topic and the work of several mathematicians here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ncatlab.org/nlab/show/vertex+operator+algebra&quot; rel=&quot;nofollow&quot;&gt;vertex algebras on the nLab&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As you can see from the answers on mathoverflow, vertex algebras were not invented for the study of CFT, and that they form an axiomatic abstraction of operator algebra products was noted only later. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A personal and very subjective note: One should not underestimate the amount of theoretical physics that is necessary to understand what a QFT is to physicists. Most mathematicians that encounter physics for the first time since highschool through some QFT framework seem to be quite taken aback by the high intrance fee they'd have to pay to understand this. This is my own, personal explanation for the observtion that most mathematicians study the formal machinery only, in order to use it to prove some new mathematical theorems, but only very rarely in order to better understand what physicists do. Although you'll find quite a lot of work by quite a lot of rather famous mathematicians if you follow the links above, AFAIK there is none doing the work you describe.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-02-11T20:10:26.060" CommentCount="3" />
  <row Id="5022" PostTypeId="1" AcceptedAnswerId="5023" CreationDate="2011-02-11T21:15:53.303" Score="4" ViewCount="163" Body="&lt;p&gt;I'm trying to learn how to do a many-body path integral for &lt;strong&gt;both fermions and bosons&lt;/strong&gt;, and I'm stuck. I'm following Altland and Simons - Condensed Matter Field Theory, chapter 4. On page 167, equation 4.27 is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;Z = \int \prod_{n=1}^N d(\bar{\psi}^n,\psi^n) e^{-\delta \sum_{n=0}^{N-1}[\delta^{-1}(\bar{\psi}^n - \bar{\psi}^{n+1}).\psi^n + H(\bar{\psi}^{n+1},\psi^n)]}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I've set $\mu=0$ from the equation in the book). The limit $N \rightarrow \infty$ is then taken which involves various things, but the part I don't understand is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\lim_{N \rightarrow \infty} \delta^{-1}(\bar{\psi}^n - \bar{\psi}^{n+1})) \rightarrow -\partial_\tau \bar{\psi}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is fine, but the next is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;Z = \int D(\bar{\psi},\psi) e^{-S[\bar{\psi},\psi]}, \hspace{4mm} S[\bar{\psi},\psi] = \int_0^\beta [\bar{\psi} \partial_\tau \psi + H(\bar{\psi},\psi)]&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how do you get from $-\partial_\tau \bar{\psi}$, which is $-\partial_\tau \bar{\psi} \psi$ in $Z$, to $+\bar{\psi} \partial_\tau \psi$? If this was only for fermions, I would guess that the Grassmann variable $\psi$ and the derivative $\partial_\tau$ anticommute which is where the minus sign comes from. But the book says it's valid for bosons as well as fermions; for bosons, the $\psi$ is a complex number, and so I wouldn't expect the minus sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="873" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-11T21:29:03.273" LastActivityDate="2011-02-11T21:53:58.213" Title="Do derivatives anticommute with Grassmann variables and complex numbers in a many-body path integral?" Tags="&lt;condensed-matter&gt;&lt;path-integral&gt;&lt;fermions&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="5023" PostTypeId="2" ParentId="5022" CreationDate="2011-02-11T21:39:33.250" Score="6" ViewCount="" Body="&lt;p&gt;Jane, $\partial_\tau$ is clearly a derivative with respect to a bosonic time $\tau$, so it &lt;em&gt;commutes&lt;/em&gt; with everything else (except for functions of $\tau$ itself, with which it has a nonzero commutator), rather than anticommutes. Only if both objects have a fermionic character (if both of them are Grassmann-odd), they anticommute with one another (or they have an anticommutator that can be evaluated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no sign error in the formulae, however. You asked a good question: how do you get from &#xA;$$-\partial_\tau \bar\psi \psi$$&#xA;to &#xA;$$+\bar\psi\partial_\tau \psi$$&#xA;One needs a bit of patience to answer this question. Note that in the two expressions, a different variable is differentiated. In the first one, it's $\bar\psi$ that is differentiated; in the second one, it's $\psi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't just move derivatives around. Even for bosonic functions, $uv&amp;#39;$ is something else than $u&amp;#39;v$, isn't it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the two expressions are not &quot;obviously equal&quot;, not even up to a sign, and to convert one to the other, you must carefully integrate by parts. Note that&#xA;$$\partial_\tau (\bar\psi \psi) = \partial_\tau\bar\psi \psi + \bar\psi\partial_\tau\psi. $$&#xA;This &quot;Leibniz rule&quot; proceeded just like for the derivative of products of bosonic factors because I had to bubble $\partial_\tau$ through the $\psi$'s, and $\partial_\tau$ is a bosonic object. If I were writing down a Leibniz rule for a Grassmannian derivative, I would have to change the sign everytime the derivative would bubble through a Grassmann-odd factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here we deal with bosonic $\tau$-derivatives so the Leibniz rule is just like it has always been. So it implies that up to a total derivative - namely the left-hand side  $\partial_\tau (\bar\psi \psi)$ that integrates to zero over the periodic Euclidean time - the two terms on the right hand side are opposite to one another. That's where the minus sign came from.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-11T21:53:58.213" LastActivityDate="2011-02-11T21:53:58.213" CommentCount="6" />
  <row Id="5024" PostTypeId="2" ParentId="5008" CreationDate="2011-02-11T22:01:22.747" Score="1" ViewCount="" Body="&lt;p&gt;Each particle can be described entirely by its quantum mechanical state (1), which are a set of properties which distinguish one particle from another. More precisely, a quantum mechanical state is a particular combination of values for these properties. (i.e. if for two particles, these properties all match, they are in the same state). For fundamental particles (i.e. electrons), the state is the only way to distinguish the particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Associated with each quantum mechanical state is an energy which can be calculated if we know the state. The state in the system that has the lowest energy is known as the &lt;em&gt;ground state&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a significant number of particles co-exist in the ground state we have a &lt;strong&gt;Bose-Einstein condensate&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Low temperatures are generally required for the ground state to exist since temperature imparts energy to particles, and so 'kicks' them out of the ground state into a state with a higher associated energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Briefly looking at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bose%E2%80%93Einstein_condensation%3a_a_network_theory_approach&quot; rel=&quot;nofollow&quot;&gt;wikipedia page on this topic&lt;/a&gt;, it seems that the network theory analogy is concerned with the 'particles' of a system going from having a large range of properties to some number sharing the same properties, i.e. condensing into a single 'state'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example given is a traffic jam. Before the cars hit the jam, they would have had a range of speeds, once the cars hit the jam, they have zero speed.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(1) When we talk about Bose-Einstein Condensates, the word &lt;em&gt;state&lt;/em&gt; means the &lt;em&gt;quantum mechanical state&lt;/em&gt; and generally not &lt;em&gt;solid, liquid, gas&lt;/em&gt; etc.&lt;/p&gt;&#xA;" OwnerUserId="1072" LastEditorUserId="1072" LastEditorDisplayName="" LastEditDate="2011-06-17T21:29:55.240" LastActivityDate="2011-06-17T21:29:55.240" CommentCount="3" />
  <row Id="5025" PostTypeId="1" CreationDate="2011-02-12T00:04:47.413" Score="1" ViewCount="170" Body="&lt;p&gt;I was watching the documentary Carl Sagan did about gravity (I believe it's quite old though) and wondered about space being &quot;flat&quot; and that mass creates dents in this plane as shown at about 3 minutes in this clip (&lt;a href=&quot;http://www.youtube.com/watch?v=Y-db4iC0aHw&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=Y-db4iC0aHw&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this simply a metaphor or is it something more than that? Does gravity follow this just instead of being flat the dent is &quot;rotated&quot; in all directions?&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-02-12T06:28:07.573" LastActivityDate="2011-02-12T16:20:08.710" Title="Space as &quot;flat&quot; plane" Tags="&lt;astrophysics&gt;&lt;gravity&gt;" AnswerCount="3" />
  <row Id="5026" PostTypeId="1" CreationDate="2011-02-12T00:08:33.090" Score="1" ViewCount="194" Body="&lt;p&gt;With the form y(x,t)=Asin(kx-wt+O)&#xA;there are two variables, how do I find the velocity? I don't know I can apply derivative with two variables....&lt;/p&gt;&#xA;" LastActivityDate="2011-02-12T23:20:20.053" Title="How do you find the velocity function of a mechanical wave?" Tags="&lt;waves&gt;" AnswerCount="3" />
  <row Id="5027" PostTypeId="1" AcceptedAnswerId="5044" CreationDate="2011-02-12T00:11:20.997" Score="6" ViewCount="164" Body="&lt;p&gt;Imagine a linear decrease in concentration from left to right. Using Fick's first law,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$J = -D \frac{d \psi}{d x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all x, from left to right, we have the same flux amount because the decrease is linear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $J(x) = m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Fick's second law,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d \psi}{d t} = -D \frac{d^2 \psi}{d x^2} = \frac{d J}{d x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\frac{d J(x)}{d x} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so dJ/dx is just 0 since the 2nd derivative of a line is 0. Yet this seems unintuitive. I would expect as long as there is a concentration gradient, there should be a change in concentration at each point until the concentration is completely uniform. There must be an error in my math or reasoning, where is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify the boundary conditions, imagine a closed box with no out flow or in flow at the edges.&lt;/p&gt;&#xA;" OwnerUserId="1873" LastEditorUserId="1873" LastEditorDisplayName="" LastEditDate="2011-02-12T05:29:52.373" LastActivityDate="2011-02-12T18:10:28.013" Title="Why is the answer to this diffusion example unintuitive?" Tags="&lt;fluid-dynamics&gt;&lt;diffusion&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5028" PostTypeId="2" ParentId="5026" CreationDate="2011-02-12T00:33:18.107" Score="2" ViewCount="" Body="&lt;p&gt;You can (and, in fact, have to) apply the derivative to find the velocity, but it takes a bit of careful reasoning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, think about this: what exactly is the speed of a wave? It's the speed at which a particular point on the wave's structure moves. Points on the wave's structure are identified by their &lt;em&gt;phase&lt;/em&gt;, which is the argument of the $\sin$ function. For instance, a peak is identified by phase $\phi = \frac{n\pi}{2}$, where $n$ is an odd integer. So you're looking for the speed of a point of &lt;em&gt;constant phase&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know that, you can just implicitly differentiate the expression for phase,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi = kx - \omega t + \phi_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;keeping in mind that $\phi$ is constant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}}{\mathrm{d}t}\phi = \frac{\mathrm{d}}{\mathrm{d}t}[kx - \omega t + \phi_0]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;giving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 = k\frac{\mathrm{d}x}{\mathrm{d}t} - \omega$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\omega}{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the expression for the speed of a sinusoidal wave.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-02-12T00:33:18.107" CommentCount="3" />
  <row Id="5029" PostTypeId="1" AcceptedAnswerId="5034" CreationDate="2011-02-12T00:33:54.877" Score="2" ViewCount="131" Body="&lt;p&gt;Recently it has been studied non-abelian anyons in some solid-state systems. These states are being studied for the creation and manipulation of qubits in quantum computing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But, how these non-abelian anyons can arise if all particles involved are fermions?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, how electronic states can have different statistics than the fermionic if all electrons are fermions?&lt;/p&gt;&#xA;" OwnerUserId="1545" LastActivityDate="2011-02-12T03:29:45.053" Title="How Non-abelian anyons arise in solid-state systems?" Tags="&lt;statistical-mechanics&gt;&lt;solid-state-physics&gt;&lt;anyons&gt;" AnswerCount="2" />
  <row Id="5030" PostTypeId="2" ParentId="5027" CreationDate="2011-02-12T00:49:02.883" Score="4" ViewCount="" Body="&lt;p&gt;The error is in your intuition. Your calculation is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that might help your intuition is to think about what happens at the edges of the region under consideration. There must be a steady inflow from one end and a steady outflow from the other end. Fluid is continually flowing &quot;downhill&quot; (from high concentration to low), but the source at the &quot;top&quot; constantly replenishes things, so the concentration gradient remains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if that helps, but it's the best I can do.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-12T00:49:02.883" CommentCount="4" />
  <row Id="5031" PostTypeId="1" AcceptedAnswerId="5040" CreationDate="2011-02-12T00:54:05.797" Score="6" ViewCount="375" Body="&lt;p&gt;One thing I know about black holes is that an object gets closer to the event horizon, gravitation time dilation make it move more slower from an outside perspective, so that it looks like it take an infinite amount of time for the object to reach the event horizon. It seems like a similar process should slow the formation of the black hole itself: As the star collapses, its gravitational time dilation make itself collapse more slowly. This make me wonder, are what astronomers claim to be black holes really black holes, or are they stars that progressively make themselves more similar to one without actually reaching the stage of having an event horizon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Contemplating one answer, I realize the question is ambiguous. What does finite time mean in general relativity. Here is a less ambiguous question: Is there a connected solution of 3+1 dimensional general relativity with one space-like slice not have a singularity, and another space-like slice having one.&lt;/p&gt;&#xA;" OwnerUserId="1859" LastEditorUserId="1859" LastEditorDisplayName="" LastEditDate="2011-02-12T03:02:30.183" LastActivityDate="2011-04-12T07:28:00.490" Title="Can black holes form in a finite amount of time?" Tags="&lt;general-relativity&gt;&lt;black-holes&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="5032" PostTypeId="1" AcceptedAnswerId="5033" CreationDate="2011-02-12T01:33:24.267" Score="3" ViewCount="215" Body="&lt;p&gt;Absorption spectra are a result of light of a certain wavelength exciting an atom from a lower energy level to a higher one and at the same time being absorbed. However, the atom should eventually go back down to its lower energy state, and at the same time emit a photon of same frequency it absorbed earlier. Overall, no changes to the star's spectrum should occur.&lt;/p&gt;&#xA;" OwnerUserId="1859" LastEditorUserId="185" LastEditorDisplayName="" LastEditDate="2011-02-12T12:11:48.333" LastActivityDate="2011-02-12T14:02:02.993" Title="Why do stars have absorption spectra?" Tags="&lt;astronomy&gt;&lt;spectroscopy&gt;" AnswerCount="3" />
  <row Id="5033" PostTypeId="2" ParentId="5032" CreationDate="2011-02-12T01:44:27.357" Score="3" ViewCount="" Body="&lt;p&gt;When the energy level drops and the photon is re-emitted, it is scattered or re-emitted in a random direction, so less of that frequency reaches your telescope/spectroscope. It is sort of analogous to the &quot;why is the sky blue&quot; question. What you are looking at is direct line photons(line of sight), so photons passing through the star's gaseous layers are depleted in this line of sight by the random scattering.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-02-12T01:51:40.297" LastActivityDate="2011-02-12T01:51:40.297" CommentCount="3" />
  <row Id="5034" PostTypeId="2" ParentId="5029" CreationDate="2011-02-12T01:52:04.913" Score="4" ViewCount="" Body="&lt;p&gt;The point is that the anyons are &lt;em&gt;not&lt;/em&gt; electronic states at all. As you've rightly noted, the electrons are fermions, and there's nothing that's going to make them forget that, but very rarely are condensed matter systems so simple that electrons are appropriate degrees of freedom to work with. Instead, the fractional quantum Hall states conjectured to give rise to non-abelian anyonic degrees of freedom are complex many-body states, and the anyons are &lt;em&gt;emergent,&lt;/em&gt; rather than fundamental degrees of freedom, arising from the collective behavior of the electrons. For a more transparent example of this phenomena, look up spin-charge separation in 1D electronic systems.&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2011-02-12T01:52:04.913" />
  <row Id="5035" PostTypeId="2" ParentId="5031" CreationDate="2011-02-12T01:56:36.793" Score="4" ViewCount="" Body="&lt;p&gt;You are simply looking at it from an observer's viewpoint. Yes, looking from outside, matter tends to asymptotically approach but never reach the event horizon. If you were part of that matter spiraling into a black hole, there would be no problem reaching the horizon, crossing it, and going right down to the singularity.  The event horizon is not a physical barrier.&#xA;You could be free falling, and your time would not be infinitely dilated. So the answer is yes they can form easily in a finite time.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-02-12T02:03:31.307" LastActivityDate="2011-02-12T02:03:31.307" CommentCount="1" />
  <row Id="5036" PostTypeId="2" ParentId="1628" CreationDate="2011-02-12T02:13:38.293" Score="0" ViewCount="" Body="&lt;p&gt;Well other than black hole singularities, there are other cosmological topological defects like cosmic strings, domains, and textures (and magnetic monopoles) that would not be considered as smooth. These are regions where there is or has been a phase transition.&#xA;Of course, these things may not exist, but string theory predicts monopoles, so they had better ;)&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-02-12T02:13:38.293" />
  <row Id="5037" PostTypeId="2" ParentId="5025" CreationDate="2011-02-12T03:05:44.980" Score="0" ViewCount="" Body="&lt;p&gt;Lets look at Einstein's field equations:&#xA;$G{\mu\nu}=8\pi.T{\mu\nu}$  where $G{\mu\nu}=R{\mu\nu}-1/2g{\mu\nu}R$&#xA;The left side is the curvature of space (the metric). The right side is the stress-energy-momentum tensor, which is the totality of what is producing the curvature. $g{\mu\nu}$ is the metric tensor , R, the scalar curvature, and $R{\mu\nu}$ the Ricci curvature tensor, but the terminology doesn't matter here. Simply the equation is saying the the curvature of space (the metric) is produced by what is in the space (pressure, energy etc)&lt;br&gt;&#xA;John Wheeler, who always used colorful language said, &quot;Matter tells space how to curve.&#xA;Space tells matter how to move.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-02-12T03:29:32.040" LastActivityDate="2011-02-12T03:29:32.040" />
  <row Id="5038" PostTypeId="2" ParentId="5029" CreationDate="2011-02-12T03:06:07.177" Score="3" ViewCount="" Body="&lt;p&gt;The dichotomy of bosonic and fermionic behaviour essentially arises because of the nature of the rotation group in dimensions greater than 2+1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exchange of 2 particles (which determines the statistics) introduces the same phase that you get when you rotate a particle by 2pi - this really comes from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Spin-statistics_theorem&quot; rel=&quot;nofollow&quot;&gt;spin-statistics theorem&lt;/a&gt; that tells you that exchanges (that determine statistics) and rotations (that generate spin) are in some sense identical. So if we can understand how rotations differ in 2D space and 3D space, we can see qualitatively how the statistics in 2D and 3D should differ too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will not go into the details but you can see in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521558859&quot; rel=&quot;nofollow&quot;&gt;Sternberg's excellent book&lt;/a&gt; the topological differences between the rotation groups SO(2) and SO(3) in 2D and 3D space respectively that lead to the restriction of phase changes in 3D to +1 or -1, but produce no such restriction in 2D space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So planar systems can produce &lt;a href=&quot;http://en.wikipedia.org/wiki/Fractional_statistics&quot; rel=&quot;nofollow&quot;&gt;&quot;anyonic&quot; statistics&lt;/a&gt; (that interpolate between fermionic and bosonic). The answer to your question of how fermions can display this behaviour is (as pointed out in another answer) that the anyonic states are &lt;em&gt;collective&lt;/em&gt; excitations of a large number of fermions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in a planar &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_Hall_effect&quot; rel=&quot;nofollow&quot;&gt;Quantum Hall&lt;/a&gt; system, the wavefunctions of all the electrons in the system rearrange and overlap and conspire to produce an overall collective wavefunction (best approximated by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Laughlin_wavefunction&quot; rel=&quot;nofollow&quot;&gt;Laughlin wavefunction&lt;/a&gt;) for quasiparticles and quasiholes. These wavefunctions have a form that produces an &lt;a href=&quot;http://en.wikipedia.org/wiki/Aharanov-Bohn_effect&quot; rel=&quot;nofollow&quot;&gt;Aharanov-Bohm phase&lt;/a&gt; change on exchanging two quasiparticles/holes (taking one around another &quot;half times&quot;) that differs from +1 or -1. So these are systems whose quasiparticle excitations (which are collective &quot;emergent&quot; excitations of the underlying fermionic electrons) produce abelian anyonic statistics. Similar Quantum Hall systems with degenerate ground states produce non-abelian anyonic statistics because you no longer have just one phase but a whole matrix of phases transforming your degenerate subspace as it picks up statistical phases on exchange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I suppose similar things happen in other condensed matter systems were anyons are expected, but I wouldn't know the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize: you need &#xA;1) a planar (2D) system to exploit the topological subtleties of the SO(2) rotation group&#xA;2) a large number of fermions who conspire to have collective/emergent anyonic wavefunctions, like in QHall systems&#xA;3) and a degenerate ground state if you want non-abelian and not just abelian statistics&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-02-12T03:29:45.053" LastActivityDate="2011-02-12T03:29:45.053" />
  <row Id="5039" PostTypeId="2" ParentId="5032" CreationDate="2011-02-12T04:01:17.183" Score="2" ViewCount="" Body="&lt;p&gt;A similar explanation, is that the intensity at any given frequency is the optical depth weighted average of the plank (black body) function. As the light ray goes deeper (maybe comes from deeper down is more intuitive) for each delta in optical depth the local temperature determines the ratio of emission to absorption. In the center of an absorption line, the opacity is higher, and thus the depth in the photosphere for a given optical depth is less than it is outside of the line. Since in the photosphere, temperature decreases with height, the spectral lines reflect a lower emission temperature. A more intuitive, but less exact way to think of it, is to consider the beam of light to be formed at an optical depth of unity. In the center of an absortion line, that depth is much less, so the strength of the plank function is much lower (reflecting the lower temperature higher up in the stellar atmosphere).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For very very strong lines, you can get the opposite effect -emission lines, as the temperature in the stars chromosphere and corona increase with height, and the closer to the line center you are the higher the effective height of formation.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-02-12T04:01:17.183" />
  <row Id="5040" PostTypeId="2" ParentId="5031" CreationDate="2011-02-12T04:48:21.207" Score="1" ViewCount="" Body="&lt;p&gt;(this answer addresses the new question)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence of the singularity theorems, it is not only possible but (arguably) inevitable for singularities to form in a finite amount of &quot;time&quot; in a physically reasonable spacetime. The word &quot;time&quot; in this context means &quot;proper time along a specific timelike geodesic&quot;. For example, if there is a trapped surface* in spacetime, then a singularity will appear within a finite amount of proper time (along a timelike geodesic) in the future of that surface; so, an observer sitting in a collapsing star will reach the singularity in finite time. Thus, the collapse of matter is one possible way to create a singularity &quot;out of nothing&quot;. If your spacetime is globally hyperbolic and you foliate it by Cauchy surfaces you can say in a much more &quot;universal&quot; way that the singularity didn't exist at time [;t_0;] and came to exist at time [;t_1;]. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should point out that the singularities are a generic feature of physically reasonable spacetimes; take a look at the Hawking-Penrose theorem - it applies in very general situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as the original question was about black holes and not singularities, I should advise you to make a clear distinction between the two concepts. Trapped surfaces form due to the condensation of matter (this is the famous Schoen-Yau theorem), and under a certain extra hypothesis, these surfaces will be hidden inside black holes. This extra hypothesis is the well-known (weak) Cosmic Censorship Conjecture (CCC). If it does not hold, gravitational collapse can create naked singularities, that is, singularities not &quot;causally hidden&quot; by the event horizon of a black hole. Much of what is known in general about black holes depend crucially on the CCC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*A trapped surface is a two-dimensional spacelike compact surface such that the null geodesics departing from it are accelerating towards each other - mathematically, we say that the expansion of the congruence of future-directed null geodesics orthogonal to the surface is negative.&lt;/p&gt;&#xA;" OwnerUserId="565" LastEditorUserId="565" LastEditorDisplayName="" LastEditDate="2011-02-12T04:56:53.300" LastActivityDate="2011-02-12T04:56:53.300" CommentCount="4" />
  <row Id="5041" PostTypeId="2" ParentId="5025" CreationDate="2011-02-12T05:19:31.963" Score="0" ViewCount="" Body="&lt;p&gt;In the context of general relativity theory the following things happen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the absence of matter, spacetime remains flat. The relevant space is a 4 dimensional Minkowski space. It has some similarities with Euclidean flat geometry, which you have learnt in the school. There are also important difference. In an Euclidean space the symmetry group is Euclidean group whereas in Minkowki space it is called a Poincare group. The former space has only space like dimensions, the latter has 3 space and 1 time like dimensions. However the 4 dimensional Minkowski space is also like a table top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if there is some mass energy the geometry of the surrounding spacetime no longer remains Minkowskian, It changes to a more general geometry (although locally within an infinitesimal region it remains Minkowskian). The geometry of Minkowski space is analogous/corresponds to geometry of a plane surface and the latter more general geometry is analogous/corresponds to a geometry of a curved surface. In this curved geometry the sum of three angles of a triangle may be more than or less than $\pi$ depending on the curvature. This curvature does not move itself in static case. Only a particle follow the straightest possible path. Since the spacetime is itself curved the particle appears to be moving in a curved path as if by a force.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-12T08:26:43.343" LastActivityDate="2011-02-12T08:26:43.343" CommentCount="4" />
  <row Id="5042" PostTypeId="2" ParentId="414" CreationDate="2011-02-12T05:45:04.067" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.ams.org/samplings/math-history/prime-chaos.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ams.org/samplings/math-history/prime-chaos.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-02-12T05:45:04.067" />
  <row Id="5043" PostTypeId="2" ParentId="5031" CreationDate="2011-02-12T05:57:33.770" Score="1" ViewCount="" Body="&lt;p&gt;To begin with, there is a connected solution of 3+1 GR in which particles fall to the singularity in finite time. In particular, Gullstrand-Painleve coordinates do this. The big difference with Schwarzschild coordinates is that the speed of light depends on direction: light moves into a black hole faster than it moves out.  See:&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the formation of a black hole in these coordinates, see:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phys.Rev.D79:101503,2009, J. Ziprick, G. Kunstatter,  &lt;em&gt;Spherically Symmetric Black Hole Formation in Painlevé-Gullstrand Coordinates&lt;/em&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/0812.0993&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0812.0993&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the generalization of Gullstrand-Painleve coordinates to the rotating black hole, see the very readable paper that gives an intuitive explanation for what is going on, see:&lt;br&gt;&#xA;Am.J.Phys.76:519-532,2008, Andrew J. S. Hamilton, Jason P. Lisle, &lt;em&gt;The river model of black holes&lt;/em&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0411060&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/gr-qc/0411060&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: that the above paper is peer reviewed and shows that yes indeed, particles falling past the event horizon travel with speeds greater than the 1 (in GP coordinates). In GR, the speeds of objects depend on the choice of coordinates. Consequently, this exceeding of the speed 1 is not equivalent to exceeding the speed of light. In GP coordinates, a light beam moving towards the singularity inside the event horizon also moves at speed greater than 1. Consequently, there is no violation of special relativity.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-04-10T21:35:35.950" LastActivityDate="2011-04-10T21:35:35.950" CommentCount="5" />
  <row Id="5044" PostTypeId="2" ParentId="5027" CreationDate="2011-02-12T06:10:17.847" Score="11" ViewCount="" Body="&lt;p&gt;As Ted Bunn said, the linear concentration profile is only a steady state if there is a steady inflow at one end and a steady outflow at the other. This net flow is what preserves the concentration gradient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the &quot;closed box&quot; boundary condition instead, there is indeed an error in your reasoning because the linear profile is no longer a steady state. So, to make things explicit, you should have instead:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left.J(x)\right|_{t=0} = m$$&#xA;$$\left.\frac{\partial\psi}{\partial t}\right|_{t=0} = 0$$&#xA;for all x in the &lt;strong&gt;interior&lt;/strong&gt; of the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, these results do not imply that $\psi(x)$ is always constant. At time $t=0$, there is a constant flow from left to right, but because the box is closed this means that the concentration at the left edge of the box is decreasing and the concentration at the right edge is increasing (even though it hasn't yet begun to change anywhere in the interior - If you want, you can say that $\partial\psi/\partial t(t=0)$ has the form of two Dirac delta functions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way I know to get the full solution is expanding in a Fourier series. For concreteness, say the box extends from $x=-1/2$ to $x=1/2$. The correct basis of eigenfunctions to use for this boundary condition contains functions whose derivative is zero at the edges of the box, namely $\sin(n \pi x)$ for odd n and $\cos(n \pi x)$ for even n. Since the initial condition is an odd function, the cosines don't appear. Also, for convenience, set the initial slope equal to $\pi^2/4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi(x,t=0) = \frac{\pi^2}{8} - \frac{\pi^2}{4} x = \frac{\pi^2}{8} - \sum_{n\text{ odd}} \frac{\sin(n\pi x)}{n^2}$$&#xA;(where the last equality is from the well-known Fourier series of a triangle wave)&#xA;$$\psi(x,t) = \frac{\pi^2}{8} - \sum_{n\text{ odd}} \frac{\sin(n\pi x)}{n^2} e^{-nt/\tau}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\tau$ is a time scale that depends on the diffusion constant and dimensions (if you want, I can work out what it actually is, but it's irrelevant for the discussion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you plot this function at different $t$ values increasing from zero, you can clearly see that the concentration is becoming smoothed out and tending toward a uniform concentration of the average value, $\pi^2/8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, even though at $t = 0$ it seems like the concentration isn't changing anywhere ($\partial\psi/\partial t$ = 0), it immediately &lt;strong&gt;begins&lt;/strong&gt; changing, and diffusion does eventually lead to a uniform concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some plots I made, using terms through $n=15$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/3wdE2.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/u3Ny4.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/olzlF.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/r2Vh5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditorDisplayName="" LastEditDate="2011-02-12T18:10:28.013" LastActivityDate="2011-02-12T18:10:28.013" CommentCount="2" />
  <row Id="5045" PostTypeId="2" ParentId="5014" CreationDate="2011-02-12T07:58:12.573" Score="0" ViewCount="" Body="&lt;p&gt;A good place to look for book recommendations for mathematical physicists is this page of John Baez:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://math.ucr.edu/home/baez/books.html&quot; rel=&quot;nofollow&quot;&gt;books&lt;/a&gt;, how to learn math and physics&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I have to agree with the others that the best way to learn the mathematical background of QM is to learn QM, you'll see yourself what kind of mathematical tools you'll have to study further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, here are a few tips: If you decide to attend a math course on functional analysis, see that it is about linear operators on Hilbert space leading to the spectral theorem. That's the part of functional analysis you'll need first in QM. &quot;Functional Analysis&quot; is a broad topic, and many math faculties start with more abstract stuff you'll need only later (like Banach algebras or topological vector spaces).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of a book that is taylored for the special needs of physicists learning QM:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nino Boccara: &quot;Functional Analysis. And Introduction for Phycisists&quot;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is short, it explains everything in detail and covers the essential topics for QM:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Measure and Integration&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lebesgue Spaces&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Hilbert Spaces&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Distributions, Fourier and Laplace Transforms&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Linear Operators, Bounded and Unbounded, Spectral Theory&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-02-12T07:58:12.573" />
  <row Id="5046" PostTypeId="1" CreationDate="2011-02-12T09:20:58.580" Score="12" ViewCount="623" Body="&lt;p&gt;Can I use normal metal antenna to light for example red one ?&lt;/p&gt;&#xA;" OwnerUserId="58" LastActivityDate="2011-02-12T10:03:52.300" Title="Can I use antenna as light source?" Tags="&lt;light&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="4" />
  <row Id="5047" PostTypeId="2" ParentId="5046" CreationDate="2011-02-12T10:01:22.897" Score="6" ViewCount="" Body="&lt;p&gt;Yes, if you heat it :) But assuming you want to use it in the usual way the answer is no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The color of light depends on its wavelength (assuming it's monochromatic; otherwise it will be some mixture of its components). Normal antennas have length scale on the order of centimeters to meters which is somewhat longer than microwave radiation which is still very much outside the visible spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get visible light, you need wavelengths on the order of 100 nm (and specifically for red light around 600 nm). Where do you get such short lengths? Well, light carries energy inversely proportional to its wavelength. It turns out that these energies corresponding to visible light are often present in the absorption/emission energies of usual atoms (which is ultimately why some objects look red). Of course, this argument should in fact run backwards: the fact that our eye responds to visible light is because that's the kind of radiation to be found around us. But I digress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as alluded to in the beginning, you can also get red light by heating stuff. This is because at certain temperature the thermal energy will be just right to produce (mostly) red light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, to produce red light (in particular coherent, as in lasers) you need to use the above observation about absorption and emission and base your device on semiconductors, crystals, etc.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-02-12T10:01:22.897" CommentCount="3" />
  <row Id="5048" PostTypeId="2" ParentId="5046" CreationDate="2011-02-12T10:03:52.300" Score="26" ViewCount="" Body="&lt;p&gt;with your question you tackle one of the most rapidly evolving fields at the moment: &lt;a href=&quot;http://en.wikipedia.org/wiki/Nanophotonics&quot; rel=&quot;nofollow&quot;&gt;Nanophotonics&lt;/a&gt;. Let me explain in the following why this is indeed the case.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Dipole antennas&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It was &lt;a href=&quot;http://en.wikipedia.org/wiki/Heinrich_Hertz&quot; rel=&quot;nofollow&quot;&gt;Heinrich Hertz&lt;/a&gt;, student of &lt;a href=&quot;http://en.wikipedia.org/wiki/Gustav_R._Kirchhoff&quot; rel=&quot;nofollow&quot;&gt;Kirchhoff&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Hermann_von_Helmholtz&quot; rel=&quot;nofollow&quot;&gt;Helmholtz&lt;/a&gt; who tried to verify &lt;a href=&quot;http://en.wikipedia.org/wiki/Maxwell%27s_equations&quot; rel=&quot;nofollow&quot;&gt;Maxwells equations&lt;/a&gt; by generating electromagnetic radiation building a &lt;a href=&quot;http://en.wikipedia.org/wiki/Dipole_antenna&quot; rel=&quot;nofollow&quot;&gt;dipole antenna&lt;/a&gt; out of a somehow transformed &lt;a href=&quot;http://en.wikipedia.org/wiki/RLC_circuit&quot; rel=&quot;nofollow&quot;&gt;RLC-circuit&lt;/a&gt;:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/UWwqi.gif&quot; alt=&quot;emerging of a dipole&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the physical ground on which we have to investigate your question to get an idea if optical antennas are possible and if you could use some of your ones to create red light. So let's look at the &lt;strong&gt;characteristic value&lt;/strong&gt; of such an antenna: its length.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Length Scales&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Roughly speaking, to be able to generate radiation with the mentioned form of antennas, one needs to impose a &lt;a href=&quot;http://en.wikipedia.org/wiki/Standing_wave&quot; rel=&quot;nofollow&quot;&gt;standing wave&lt;/a&gt; on the antenna. Assuming that the reflection at the ends of the antenna is somehow perfect, we find that an antenna will have a length of approximately $$L \approx \frac\lambda2\,.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_spectrum&quot; rel=&quot;nofollow&quot;&gt;electromagnetic spectrum&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/f/f1/EM_spectrum.svg/1000px-EM_spectrum.svg.png&quot; alt=&quot;EM-Spec&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;red light has a wavelength of about 600 to 700nm, your antenna will have $$L\approx 300nm$$ which is incredibly small. This is the reason, antennas in optical frequencies are subject to &lt;strong&gt;current research&lt;/strong&gt;, see e.g. &lt;a href=&quot;http://www.sciencemag.org/content/308/5728/1607&quot; rel=&quot;nofollow&quot;&gt;Resonant Optical Antennas&lt;/a&gt; by Mühlschlegel et al.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, from this point of view it seems &lt;strong&gt;very unlikely&lt;/strong&gt; that a Layman will be able to produce an antenna that can emit light at optical frequencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2011-02-12T10:03:52.300" CommentCount="1" />
  <row Id="5049" PostTypeId="2" ParentId="5032" CreationDate="2011-02-12T11:08:00.757" Score="1" ViewCount="" Body="&lt;p&gt;Stars have plenty of &quot;cold&quot; gas (atoms in their ground states). If you put some gas on the way from a light source to you, some frequencies will be less represented (dark absorption lines) than non resonance frequencies. Atoms for such frequencies are like a fog on the light way - they capture and diffuse the resonance frequencies to the whole solid angle $4\pi$. So you observe them as less represented than the neighboring non resonance frequencies in the continuous spectrum.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-12T14:02:02.993" LastActivityDate="2011-02-12T14:02:02.993" />
  <row Id="5050" PostTypeId="2" ParentId="5026" CreationDate="2011-02-12T11:18:25.317" Score="1" ViewCount="" Body="&lt;p&gt;David and Mark explained how one can estimate the velocity $v_x$ of the form &lt;strong&gt;propagation&lt;/strong&gt; along the propagation direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another velocity, say, the vertical velocity $v_y$ at a given place which is quite different and is determined with the wave amplitude, frequency, and time: $v_y = A\omega cos(\omega t - kx - \phi_0)$. It is variable.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-02-12T11:18:25.317" />
  <row Id="5051" PostTypeId="1" AcceptedAnswerId="5053" CreationDate="2011-02-12T11:23:59.797" Score="0" ViewCount="159" Body="&lt;p&gt;I have heard before that the 4th dimension is time, however, another theory makes a lot more sense to me. This is that the 4th dimension is the third dimension stacked on top of each other in a similar in which 3d objects are just many 2d planes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen many articles related to the 4thdimension being time, but how do we know this?&lt;/p&gt;&#xA;" OwnerUserId="1878" LastActivityDate="2011-02-12T11:43:30.907" ClosedDate="2011-02-12T13:35:42.530" Title="What is the 4th dimension?" Tags="&lt;spacetime&gt;&lt;dimensionality&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="5052" PostTypeId="2" ParentId="629" CreationDate="2011-02-12T11:30:58.270" Score="1" ViewCount="" Body="&lt;p&gt;A simple experiment:&lt;br&gt;&#xA;Take a pendulum which is suspended using some support like a wooden frame.&lt;br&gt;&#xA;Tape what happens when you stop.&lt;br&gt;&#xA;Compare with video of you on the same train during the same stop.&lt;br&gt;&#xA;Scientific enough and simple.&lt;br&gt;&#xA;If the pendulum's motion differs from yours, your explanation is correct.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry don't have numerical or theoretical explanation. &lt;/p&gt;&#xA;" OwnerUserId="1847" LastActivityDate="2011-02-12T11:30:58.270" CommentCount="5" />
  <row Id="5053" PostTypeId="2" ParentId="5051" CreationDate="2011-02-12T11:43:30.907" Score="-2" ViewCount="" Body="&lt;p&gt;It is all in what you want to describe mathematically. You can have an N dimensional space and yes, you could &quot;visualize&quot; the analogue of two dimensions going into three. These are Euclidean spaces , i.e. the metric is ds*&lt;em&gt;2=dx&lt;/em&gt;*2+dy*&lt;em&gt;2+dz&lt;/em&gt;*2+.... up to N terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time is the fourth dimension in current physics because we are attempting to describe and predict motions and interactions of matter and light in a mathematical manner, and the equations are such that they simplify when time is assumed to be the fourth dimension in what is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Pseudoeuclidean_space&quot; rel=&quot;nofollow&quot;&gt;pseudo Euclidean space.&lt;/a&gt;, and in our case dt**2 has a negative sign.&#xA;It is what the physics comes out with that makes time the fourth dimension.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-02-12T11:43:30.907" CommentCount="7" />
  <row Id="5054" PostTypeId="1" AcceptedAnswerId="5056" CreationDate="2011-02-12T12:19:53.253" Score="0" ViewCount="125" Body="&lt;p&gt;The mass of a nucleus if less than the mass of the protons and nucleus. The difference is knows as binding energy of the nucleus. This nuclear binding energy is derived from the strong nuclear force.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is that does the strong nuclear force provide this energy? How does the mass decrease if the force provides the energy?&lt;/p&gt;&#xA;" OwnerUserId="1847" LastEditorUserId="1847" LastEditorDisplayName="" LastEditDate="2011-02-12T12:30:16.847" LastActivityDate="2011-02-12T15:28:46.617" Title="Mass into energy" Tags="&lt;energy&gt;&lt;binding-energy&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="5055" PostTypeId="2" ParentId="4669" CreationDate="2011-02-12T13:36:43.657" Score="7" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Iron-based high-Tc superconductors&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cuprate high-Tc superconductors completely revolutionised the field of superconductivity in the late 80's but proved very difficult to understand and over 20 years later the problem still stands. A new class of high-Tc materials, the iron-pnictides, were discovered in 2008 which have provided a whole new set of system to study.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron-pnictides are generally much simpler in structure than the cuprates meaning the crystal growers can generally make larger, better crystals and the theorists have an easier time of it as well.&lt;/p&gt;&#xA;" OwnerUserId="1072" LastActivityDate="2011-02-12T13:36:43.657" CommunityOwnedDate="2011-02-12T13:36:43.657" />
  <row Id="5056" PostTypeId="2" ParentId="5054" CreationDate="2011-02-12T13:37:53.687" Score="1" ViewCount="" Body="&lt;p&gt;The question is clear enough to answer.  The question does involve mass defect.  The best place to start is with ordinary problems.  We can consider a nonrelativistic problem of a particle in a potential well, $V(r)~=~-r^n$ for some power of $n$.  For mathematical reasons $n~=~\pm 2$ have nice properties with closed form solutions.  This is input into a Schrodinger equation&#xA;$$&#xD;&#xA;i\hbar\frac{\partial}{\partial t}\psi(r,t)~=~-\frac{\hbar^2\nabla^2}{2m}\psi(r,t)~+~V\psi(r,t)&#xD;&#xA;$$&#xA;So if we consider a stationary phase $\psi(r,t)~=~e^{-i\omega t}\psi(r)$  and the frequency determines energy $E~=~\hbar\omega$ &#xA;$$&#xD;&#xA;\hbar\omega~=~\frac{p^2}{2m}~+~V&#xD;&#xA;$$&#xA;where the potential energy is negative.  If the potential energy were “turned off” so the particles are free then energy is larger.  The energy of system is then lower, and if this energy $E~=~mc^2$ is some appreciable fraction of the mass-energy of the system $E’~=~Mc^2$, say $m/M~~\simeq~.01$ to $.1$ the system is not highly relativistic but the mass equivalence is measurable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For atomic physics the energy levels of electrons are on the order of electron volts, while the mass of electrons are $.51$Mev.  So the mass defect is pretty small.  For nuclear physics of nucleons and mesons the energy levels are on the order of $10Mev$ while the collection of nucleons has mass-energy in multiples of $1Gev$.  The energy levels are determined by $\pi^0,~\pi^\pm$ mesons, which are the intermediary gauge bosons between the nucleons ${P,~N}$.  This in fact forms a doublet which has energy level splitting due to the electric charge difference.  The above model may be made more exact if the mesons are considered to be similar to a photon, with a gauge potential &#xA;$$&#xD;&#xA;{\vec A}(k)~=~{\vec n}(a_ke^{-ikr}~+~a^\dagger_ke^{ikr})&#xD;&#xA;$$&#xA;which interacts with a dipole formed from the nucleon doublet ${\vec{\cal P}}~=~p{vec\sigma}$ in an interaction Hamiltonian&#xA;$$&#xD;&#xA;H_{int}~=~-{\vec{\cal P}}\cdot{\vec A}(k).&#xD;&#xA;$$&#xA;We are only considering interactions with one momentum or wave number.  Now expand that out and keep terms $a(k)\sigma^+$ and $a^\dagger\sigma^-$, which is the rotating wave approximation in atomic interactions with photons.  This makes the above Schrodinger equation and potential more exact.  This interaction Hamiltonian will then reproduce the mass-defect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may be further improved of course.  The nucleon doublet is SU(2), and the meson potential may be extended from this naïve U(1) approximation to SU(2) as well.  The momentum operator may be made covariant with respect of the gauge potential and the theory refined further.  In fact the Yang-Mills theory was derived to understand this isopin theory of nuclear physics as understood in the 1950s.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-02-12T15:28:46.617" LastActivityDate="2011-02-12T15:28:46.617" CommentCount="4" />
  <row Id="5057" PostTypeId="1" CreationDate="2011-02-12T14:58:54.003" Score="0" ViewCount="270" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/15/what-experiment-would-disprove-string-theory&quot;&gt;What experiment would disprove string theory?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;We carefully observe things, observe patterns and then build theories that predict. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;String theory is frequently criticized for not providing quantitative experimental prediction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the problems that prevent this theory from producing quantitative experimental predictions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there no experiment suggested by string theorists to verify validity of their theory? &#xA;Is the problem mathematical? &lt;em&gt;(or it just requires many dimensional equipments...just kidding).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not criticizing the theory because to do that I should understand it first, but I haven't studied it. I just want to know. &lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-02-12T15:50:26.410" LastActivityDate="2011-02-13T02:12:16.547" ClosedDate="2011-02-12T23:09:58.043" Title="Does string theory provide quantitative experimental predictions?" Tags="&lt;string-theory&gt;&lt;experimental-physics&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="2" />
  <row Id="5058" PostTypeId="2" ParentId="4669" CreationDate="2011-02-12T15:01:17.417" Score="2" ViewCount="" Body="&lt;p&gt;There are many fantastic discoveries slightly outside your time-horizon, like the discovery of a cosmological constant or the AdS/CFT correspondence, but I'll respect your time-horizon:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;LHC - the main discovery so far was that LHC is working ;-), which is highly non-trivial; the real discoveries will come this decade, like Higgs or, if it exists at the TeV scale, new physics (like SUSY)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Carbon - nanotubes and graphene will keep physicists and engineers busy for at least a decade (engineers probably much longer)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;WMAP - the WMAP data are very accurate and allow to put relevant bounds on the cosmological parameters, thereby already ruling out some speculative modesl of our Universe&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Quark Gluon Plasma - LHC converts lead into quark gluon plasma; RHIC has been converting gold into quark gluon plasma; this has tremendously enhanced our understanding about the QCD phase diagram and also allowed applications of the AdS/CFT correspondence, like the famous shear viscosity over entropy density calculation&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-02-12T15:01:17.417" CommunityOwnedDate="2011-02-12T15:01:17.417" />
  <row Id="5059" PostTypeId="2" ParentId="5057" CreationDate="2011-02-12T15:14:43.590" Score="9" ViewCount="" Body="&lt;p&gt;Let me make an analogy that is due to Wati Taylor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take Einstein and, say, Ketterle, lock them in a room and give them two stones, an ordinary watch, paper and pencil. Then give them the task to verify general relativity (GR) experimentally. They will fail, inevitably.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not because GR is &quot;wrong&quot; or, as some might put it, &quot;not-even-wrong&quot;, but because the experimental equipment is not adequate for a stringent test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately for GR, there were observations about perihelion shifts and light bending that eventually convinced the skeptics (or, more probably, the skeptics died out).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;String theory is a quantum theory of gravity that makes a bunch of predictions on the Planck scale. However, we have no current experiment that is good enough to be sensitive to Planck scale physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, the same problem basically applies to ANY quantum theory of gravity - they are usually safely away from experiment...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, string theory is more ambitious and makes not only quantum gravitational predictions, but also restricts the low energy physics. It is not clear, however, if these restrictions are tight enough to be of observational relevance. I am sure that others will have to say more about the landscape of string vacua, so I will stop here.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-02-12T15:14:43.590" CommentCount="19" />
  <row Id="5060" PostTypeId="2" ParentId="5025" CreationDate="2011-02-12T16:20:08.710" Score="1" ViewCount="" Body="&lt;p&gt;The picture by Sagan is somewhat of a simplification of the true geometry - so in a sense that picture is a metaphor. The interpretation given there of Einstein's equations is indeed that of curvature of 4 dimensional Space-Time (not just 3 dimensional space).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in these higher dimensional geometries there is more than one notion of curvature: Scalar Curvature &lt;strong&gt;R&lt;/strong&gt;; &lt;strong&gt;Ricci&lt;/strong&gt; curvature; &lt;strong&gt;Riemann&lt;/strong&gt; Curvature and &lt;strong&gt;Weyl&lt;/strong&gt; Curvature. The Scalar curvature exists in all dimensions and is simply a function giving the curvature at each point. For an embedded (2-dim) sphere it is &lt;strong&gt;R&lt;/strong&gt;$=2/r^2$ (twice the Gaussian curvature). So for a sphere it is non-zero everywhere on its surface. Higher dimensional spaces use these other Curvature objects (built like generalised matrices and called Tensors) to represent their curvature more precisely than a single number. In higher dimensions the Scalar represents a kind of &quot;average curvature&quot; (at each point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corresponding to these different notions of curvature, there are different notions of &quot;flatness&quot; as we will see below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the Einstein equations directly equate the &lt;strong&gt;Ricci&lt;/strong&gt; curvature to something; and in the example shown in the Sagan excerpt it was equated to zero. &lt;strong&gt;Ricci&lt;/strong&gt;$ = 0$ is the Einstein Vacuum equation alluded to in other answers, and is appropriate because outside a star there is a vacuum. This has an immediate mathematical consequence that &lt;strong&gt;R&lt;/strong&gt;$=0$ ie the scalar is zero as well! So in this sense the vacuum is flat (called Ricci-flat).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there is still curvature around in that space, so it is not Minkowski (ie Euclidean) flat. The experimental demonstration of this was the bending of light rays near the Sun (assuming as one does in Einstein's theory that light rays measure the &quot;straight lines&quot;). So where does the curvature come from if it is Scalar and Ricci flat? The answer is that it is not Riemann flat: the tensor &lt;strong&gt;Riemann&lt;/strong&gt;$\neq 0$. However this does not quite explain the origin of curvature here. Expressed very loosely we have the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Weyl&lt;/strong&gt; = &lt;strong&gt;Riemann&lt;/strong&gt; - &lt;strong&gt;Ricci&lt;/strong&gt; - &lt;strong&gt;R&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the real source of curvature in the Sagan excerpt is the &lt;strong&gt;Weyl&lt;/strong&gt; component of the Riemann tensor: everything else is zero. Now we come to the representation problem that Sagan had: the Weyl tensor is always zero in two and three dimensions. In other words the kind of curvature it represents does not exist in two and three dimensions: only four and above dimensions have this kind of curvature. So it cannot be directly represented on a 2 or 3 dimensional picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead what Sagan appears to have represented here is the gravitational potential (like in Newton's theory) but expressed as space curvature. It is not completely wrong perhaps, but it is not quite correct and so is just a metaphor.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-12T16:20:08.710" CommentCount="1" />
  <row Id="5061" PostTypeId="1" CreationDate="2011-02-12T16:24:19.653" Score="0" ViewCount="256" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/3922/what-is-the-relation-between-position-and-momentum-wavefunctions-in-quantum-physi&quot;&gt;What is the relation between position and momentum wavefunctions in quantum physics?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;What is the relation between the position-space wavefunction $\psi(x)$ and the momentum-space wavefunction $\phi(p)$? I know that you can go back and forth between them using the Fourier transform but...why? Thanks. &lt;/p&gt;&#xA;" OwnerUserId="1885" LastActivityDate="2011-02-12T16:24:19.653" ClosedDate="2011-02-12T16:53:42.143" Title="Position-space and Momentum-space wave functions" Tags="&lt;quantum-mechanics&gt;&lt;schrodinger-equation&gt;" CommentCount="1" />
  <row Id="5062" PostTypeId="1" AcceptedAnswerId="5067" CreationDate="2011-02-12T16:32:42.837" Score="2" ViewCount="449" Body="&lt;p&gt;In continuation of &lt;a href=&quot;http://physics.stackexchange.com/questions/4959/can-noethers-theorem-be-understood-intuitively&quot;&gt;another question&lt;/a&gt; about Noether's theorem I wonder whether there exists some kind of relationship between this theorem and the Heisenberg uncertainty principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because both the principle and the theorem relate energy with time, momentum with space, direction with angular momentum. When this is a general fact then e.g. electrical charge and electrostatic potential(*) should be partners in an uncertainty relationship too. Are they?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel that these results look so basic and general that I hope that a pure physical reasoning (without math or only with a minimal amout of math) exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also compare &lt;a href=&quot;http://physics.stackexchange.com/questions/3922/what-is-the-relation-between-position-and-momentum-wavefunctions-in-quantum-physi&quot;&gt;this question&lt;/a&gt; where again momentum and space are connected, this time through a Fourier transform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) i.e. electric potential and magnetic vector potential combined.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2011-04-02T19:56:45.000" LastActivityDate="2011-09-03T05:40:28.840" Title="Noether's theorem vs. Heisenberg uncertainty principle" Tags="&lt;uncertainty-principle&gt;&lt;noethers-theorem&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" />
  <row Id="5064" PostTypeId="1" CreationDate="2011-02-12T17:08:11.180" Score="4" ViewCount="416" Body="&lt;p&gt;Inspired by &lt;a href=&quot;http://physics.stackexchange.com/questions/2407/give-a-description-of-loop-quantum-gravity-your-grandmother-could-understand&quot;&gt;this question&lt;/a&gt;, let me ask a similar question. Is it possible to do the same (give a description of M-theory your grandmother could understand)for M theory? While I know even experts don't understand it properly, I still hope some basic ideas may be expressed. Is it just a hypothetical non-perturbative formulation of quantume gravity whose different limits for varisous coupling constants are five string theories and 11d supergravity or theorists know about it in more detail?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;As far as my grandmother is concerned let me inform that she has a Phd (HEP her specialization) and quite a smart woman. She has also a good knowledge of GR. So she won't mind technical staff. What she finds frustrating is that there is no rigorous definition of M-theory available to her. In her age (80) she doesn't hope to understand every details of M-theory but she certainly can find some definition of M-theory helpful.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-13T01:44:26.040" LastActivityDate="2011-08-07T17:50:09.540" Title="Give a description of M-theory your grandmother can understand" Tags="&lt;string-theory&gt;&lt;quantum-gravity&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="2" />
  <row Id="5065" PostTypeId="2" ParentId="5064" CreationDate="2011-02-12T17:49:46.520" Score="6" ViewCount="" Body="&lt;p&gt;Here is something, which may be aiming a little low...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main two ways we describe our universe, quantum mechanics and general relativity, contradict each other when applied simultaneously. This seems to point out that the quantum nature of spacetime itself needs to be understood better. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to resolve theoretical problems with our current understanding of spacetime is to embed it in a larger theoretical structure, which has powerful underlying symmetries. Those symmetries cannot be too restrictive: they should be enough not only to make the model well-behaved, but also to be consistent with the not-quite-so-symmetric world we see around us. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This way of thinking has led to the theoretical structure of string theory and then M-theory. To study the structure of the theory, it is useful to first concentrate on the most symmetric situations, even though these are the most removed from our world. At first, this led to study of supersymmetric string theories in 10 dimensions (higher dimensional theories are more symmetric - their Lorentz invariance is larger and more restrictive). Later it turned out that those are all secretly related to an even larger  and more symmetric structure, dubbed M-theory, which describes all the previously known string theories as well as 11-dimensional supergravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The story is not finished, we only have bits and pieces of the underlying structure that is M-theory. But, we do have many indications we are on the right track. As always with deep structures we found side benefits in the form of unexpected applications in mathematics and physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the unexpected discoveries is that quantum gravity is not all that different from other parts of physics, and sometimes conventional physics can be reformulated in different variables to make it equivalent to a quantum gravitational theory. Using classical and semi-classical gravity calculations then helps us explore conventional physics in regimes otherwise inaccessible. This is the whole subject of holography and its applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what we seem to have found is, instead of a specific model to describe our universe at short distances (or high energies), a whole new language in which we describe and discuss physics - and not just high energy physics. Where precisely this is going to lead is anyone’s guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if your grandmother feels patronized, has more knowledge of physics and would like to ask more specific questions, I can try to add more details. It is a very large subject...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(see also the answers to &lt;a href=&quot;http://physics.stackexchange.com/questions/2532/what-evidence-exists-for-string-theory-viability#2539&quot;&gt;this similar question&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="4521" LastEditorDisplayName="" LastEditDate="2011-08-07T17:50:09.540" LastActivityDate="2011-08-07T17:50:09.540" CommentCount="9" />
  <row Id="5066" PostTypeId="1" AcceptedAnswerId="5070" CreationDate="2011-02-12T18:12:01.187" Score="5" ViewCount="315" Body="&lt;p&gt;Liquid and solid chemical fuels in rockets are very expensive and inefficient. I have heard of solar sails but what are the most realistic space travel fuels that will be used in the future to get close to the speed of light?&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditorDisplayName="" LastEditDate="2011-02-12T18:34:28.827" LastActivityDate="2011-03-24T18:26:19.537" Title="What are the most realistic ways of high speed space propulsion?" Tags="&lt;speed-of-light&gt;&lt;universe&gt;&lt;space&gt;&lt;rocket-science&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="1" />
  <row Id="5067" PostTypeId="2" ParentId="5062" CreationDate="2011-02-12T18:24:00.503" Score="5" ViewCount="" Body="&lt;p&gt;Noether theorem is as valid in CM(*) as in QM(**). It deals with conservation laws and symmetries. In CM the variables are certain, in QM they may be uncertain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HUP belongs to QM and gives a limitation on canonically conjugated variable uncertainties in a given state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If some variable in QM is uncertain, it does not mean its expectation value is not conserved. A superposition of free motions states $e^{ipr}$ is also a free motion state although the momentum, for example, may be uncertain. The dynamics of the momentum expectation value is determined with an external force, like in CM (see the Ehrenfest's equations). No external force, no variation of the expectation value &amp;lt;&lt;strong&gt;p&lt;/strong&gt;(t)&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I do not see any relationship between HUP and Noether.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) Classical mechanics&#xA;(**) Quantum mechanics&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="139" LastEditorDisplayName="" LastEditDate="2011-02-13T08:28:30.703" LastActivityDate="2011-02-13T08:28:30.703" />
  <row Id="5068" PostTypeId="2" ParentId="5014" CreationDate="2011-02-12T20:32:09.700" Score="0" ViewCount="" Body="&lt;p&gt;So I already have had two courses on QM and two more courses that depend on QM. Yet, I still don't understand what a Hilbert Space really is.. And why it is used and not any other space in QM.. And a bunch of other questions about the formulation that are not answered in any of the 'problem solving' QM books I have gone through till now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for all the recommendations.. I will go through them and see how they go.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2011-02-12T20:32:09.700" />
  <row Id="5070" PostTypeId="2" ParentId="5066" CreationDate="2011-02-12T20:55:17.927" Score="5" ViewCount="" Body="&lt;p&gt;In a lot of ways this is a technology---rather than physics---question, but lets look at some limits imposed by physics.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For rockets there are two numbers that matter: the velocity relative the spacecraft with which the fuel can be expelled (called the specific impulse) and the fraction of the original mass that is fuel. For very high mass fractions the spacecraft can have delta V equal to several times the specific impulse. That means that if you plan to stop at the other end your maximum velocity perhaps twice is the specific impulse. If you need to come home without refueling it is around the specific impulse.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That's the killer for fast burn rockets. We still use them for launch from the surface because they can develop the necessary thrust (our high efficiency engines (ion drives and the like) are all low thrust at the present time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That suggests one of two strategies:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Leave your engine behind. This is the mechanism used by laser sails, star wisps, etc.&lt;/li&gt;&#xA;&lt;li&gt;Pick up your fuel as you go. The strategy of ram-scoops.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Sail technology is--at least in principle--within reach of our current competence, but building and running one capable of efficient interstellar travel is beyond the current Gross Planetary Product. Star wisps are a subset of sails, but require considerable expertise in nanotechnology before they are useful even for exploration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second disadvantage of sails for traveling to other start in person, is that you're counting on politicians and policy makers you left behind to continue funding your trip...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ram-scoops are a big unknown, as we don't know how to build them as yet. A BOTE calculation for a minimal, naive hydrogen fusion ram-scoop limits their top speed to about 14% of c (under the maximum drag assumption that the fuel must be brought to rest relative the craft). &lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-03-05T01:56:24.437" LastActivityDate="2011-03-05T01:56:24.437" CommentCount="4" />
  <row Id="5071" PostTypeId="2" ParentId="629" CreationDate="2011-02-12T21:35:03.017" Score="1" ViewCount="" Body="&lt;p&gt;The explanation comes from standard mechanics where a body is in equilibrium when the forces acting on, and torques about the centre of mass sum to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the train is decelerating, there are torques acting on your body where it makes contact with the train via the seat or floor, and you automatically alter your body postition to create counter-torques so that your body remains in equilibrium. When the train suddenly stops, this net counter torque is no longer balanced, and so you rotate about your centre of mass in a direction that pulls you back.&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="1345" LastEditorDisplayName="" LastEditDate="2011-02-12T21:42:46.550" LastActivityDate="2011-02-12T21:42:46.550" />
  <row Id="5072" PostTypeId="1" AcceptedAnswerId="5102" CreationDate="2011-02-12T22:03:38.417" Score="4" ViewCount="306" Body="&lt;p&gt;Here's a question on something I've been wondering about for quite some time. (I am not a physicist.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand correctly, according to Einstein's General Theory of Relativity, mass results in a distortion in space-time. In turn, the motion of the mass is affected by the distortion. A result of the interplay between mass and space-time is that the 'force' of gravity may be explained away. Masses are not subject to a force, but are merely following a 4-dimensional space-time geodesic; gravity is just geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yet physicists are searching for exchange particles for the force of gravity, and are trying to unify quantum mechanics with relativity, or to unify the weak/strong/electromagnetic forces with that of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What have I missed? Are these different communities of physicists? Does relativity explain only part of the story of masses acting under gravity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is gravity a force or not? Is it only an apparent force or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can such an apparent force 'generate' exchange particles? Are the exchange particle and geometric models both different views of the same underlying truth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A side question might be: why can't the other forces be explained away similarly? Or is that what is happening with all this talk of small extra dimensions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd appreciate any illumination on this matter, or suggested reading (preferably at the 'popular science' or undergraduate level).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Thanks for all the replies and responses. Much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="1888" LastEditorDisplayName="" LastEditDate="2011-02-15T23:55:37.357" LastActivityDate="2011-02-15T23:55:37.357" Title="Why is there a search for an exchange particle for gravity?" Tags="&lt;quantum-mechanics&gt;&lt;general-relativity&gt;&lt;gravity&gt;&lt;geometry&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="5073" PostTypeId="1" AcceptedAnswerId="5079" CreationDate="2011-02-12T22:04:40.130" Score="8" ViewCount="722" Body="&lt;p&gt;Can someone summarize, with references if possible, all of the alternatives to the simplest model (that requires only a single scalar Higgs field with the Mexican Hat potential) of spontaneous electroweak symmetry breaking?&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-02-13T11:42:06.807" CommunityOwnedDate="2011-02-15T01:40:48.167" Title="What are the alternatives to the Higgs mechanism?" Tags="&lt;quantum-field-theory&gt;&lt;particle-physics&gt;&lt;standard-model&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="3" />
  <row Id="5074" PostTypeId="1" AcceptedAnswerId="5106" CreationDate="2011-02-12T22:13:06.353" Score="0" ViewCount="187" Body="&lt;p&gt;What impulse should be applied to an object of mass m, having a known coefficient of friction u to get to a distance d ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update 14.02.2011: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I still wasn't able to find an answer to this. Could you please share the solution? I need an &quot;Idiot's solution&quot; for this. I'm not experienced enough to find a formula by myself starting from your clues. I've searched the forums but they take into account other parameters. Thank you !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update 13.02.2011: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a representation of what I need: &lt;img src=&quot;http://i.stack.imgur.com/Pc7qa.png&quot; alt=&quot;http://img412.imageshack.us/img412/2192/20110213112259.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's been a long time since I studied the principles of physics, sorry if I did not make myself clear in the first post. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried formulas like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)m * a = u * m * g  =&gt; a = u * g&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)a = v / t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)d = v * t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from these three equations, distance = u * g * t^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The impulse p = m*v , so I get p= m * u * g * t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the time is not known and it is not important how long it takes for the object to get to target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this formula (p= m * u * g * t) seem right? The mass, u, g are known, but what about time?&lt;/p&gt;&#xA;" OwnerUserId="1890" LastEditorUserId="1890" LastEditorDisplayName="" LastEditDate="2011-02-14T20:47:01.433" LastActivityDate="2011-02-14T20:47:01.433" Title="Impulse - distance question" Tags="&lt;mechanics&gt;&lt;homework&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="5075" PostTypeId="1" AcceptedAnswerId="5098" CreationDate="2011-02-12T22:29:19.363" Score="5" ViewCount="285" Body="&lt;p&gt;Stephen Wolfram in his book &quot;A New Kind of Science&quot; touches on a model of space itself based on automata theory. That it, he makes some suggestions about modelling not only the behaviour of matter through space, but the space itself in terms of state machines (a notion from computing). Here, the topology of space arises from a small-scale connection lattice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wondered whether any theoretical work is being undertaken along these lines within the physics community.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for my interest in this regards one of the mysteries of quantum mechanics, that of quantum entanglement and action at distance. I wondered whether, if space is imagined as having a topology that arises from a notion of neighbourhood at a fine level, then quantum entanglement might be a result of a 'short circuit' in the connection lattice. That is two points at a distance through 'normal' space might also still be neighbours at a fundamental level; there might be a short strand of connectivity in addition the all the long strands relating the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I think Richard Feynman also alluded to this sort of model with his take on quantum electro-dynamics.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Thanks to all for the replies and comments.&lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="1888" LastEditorDisplayName="" LastEditDate="2011-02-15T23:57:18.603" LastActivityDate="2011-02-15T23:57:18.603" Title="Is anyone studying how the topology of space arises from more fundamental notions?" Tags="&lt;quantum-mechanics&gt;&lt;spacetime&gt;&lt;models&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="5076" PostTypeId="2" ParentId="5072" CreationDate="2011-02-12T22:46:05.580" Score="2" ViewCount="" Body="&lt;p&gt;General relativity, currently our best model of gravity, not only predicts the gravitational force as a geometric effect, it also predicts the existence of gravitons, the &quot;exchange particle for gravity&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A theory without gravitons or gravitational waves (I'll use them interchangably) could exist, but as we have already indirect evidence for gravitational waves from the &lt;a href=&quot;http://en.wikipedia.org/wiki/PSR_B1913%2B16&quot; rel=&quot;nofollow&quot;&gt;Hulse-Taylor pulsar&lt;/a&gt; it seems unlikely that gravitons do not exist in Nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We shall see in the coming decade, since &lt;a href=&quot;http://www.advancedligo.mit.edu/&quot; rel=&quot;nofollow&quot;&gt;advanced LIGO&lt;/a&gt; is bound to find gravitational waves if they exist.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-02-12T22:46:05.580" CommentCount="2" />
  <row Id="5077" PostTypeId="2" ParentId="5057" CreationDate="2011-02-12T22:53:37.893" Score="2" ViewCount="" Body="&lt;p&gt;The most obvious stringy signature would be the observation of Regge resonances at energies close to the string scale. If the extra dimensions are very large, i.e. string scale is very low (I'm extremely skeptical about this possibility), such signatures could even be observed by the LHC. See this paper for the detailed computations of the scattering amplitudes and the crossections: &lt;a href=&quot;http://arxiv.org/abs/0807.3333&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0807.3333&lt;/a&gt;. Read the summary at the end of the paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more generic/less specific, prediction is the existence of superpartners at some scale below the string scale. In the phenomenologically interesting scenarios, e.g. Calabi Yau compactifications of the Heterotic string, one obtains some type of N=1 D=4 supergravity. Unfortunately, there is no unique prediction for the details of the sparticle spectrum because there exist different mechanisms of supersymmetry breaking and the spectrum depends on that. However, in such compactifications, one generically expects some type of gravity mediation possibly mixed with high scale gauge mediation. Furthermore, one also has to specify in which corner of M-theory one is working, e.g. Heterotic, Type IIB, Type IIA, M-theory on G2, F-theory etc, which results in certain restrictions on the form of the superpotential and the Kahler potential. For example, in the G2 corner without fluxes the superpotential is purely non-perturbative because all the compactification moduli enjoy the PQ symmetry inherited from the gauge symmetry of the 11D supergravity 3-form. Thus, one can make a generic statement that the Yukawa couplings will have exponential hierarchies and the scale of SUSY breaking may be exponentially suppressed relative to the Planck scale. Furthermore, in this sector SUSY breaking is naturally gravity mediated because in 7 dimensions the 3-cycles supporting visible and hidden sectors generically do not intersect, etc. Once the mechanism of SUSY breaking and the M-theory patch are specified, one may be able to compute the sparticle spectrum at the string/GUT scale and put strong constrains on it from the top down and combine those with the bottom up requirements. In this way one can get several testable scenarios parameterized by very few phenomenological dials. The whole exponentially large landscape issue may be effectively decoupled when one is interested in these types of questions (SUSY breaking and the sparticle spectrum). To be more specific about the last point, in Type IIB flux vacua, the contribution of the fluxes to the superpotential can take on an exponentially large number of values, however, the corresponding F-terms for the complex structure moduli are still zero and the only phenomenologically relevant parameter will be the value of the flux superpotential, which is just one input parameter, whose detailed microscopic dependence on the fluxes is irrelevant for the computation of the sparticle spectrum!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another generic prediction of string compactifications comes in the imprint of the non-trivial topology on the particle spectrum in 4D. In particular, string compactifications typically imply the existence of a number of particles with similar properties. The multiplicity of SM generations is one such example and while it's still not clear why there are only three generations, it's clear that having multiple generations is generically expected. Of particular interest are so-called axions, which are ultra-light pseudoscalar particles - the partners of some (or all in the G2 case) of the geometric moduli. One of these axions can naturally provide a dynamical solution to the strong CP problem and the PQ symmetry that makes it so light can be directly traced back to the gauge symmetry of the corresponding RR-type field in 10D. Depending on the topology, there may actually be hundreds of such particles whose existence would be a complete mystery from the 4D effective field theory point of view. The experimental implications of such an &quot;Axiverse&quot; is described in detail here: &lt;a href=&quot;http://arxiv.org/abs/0905.4720&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0905.4720&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a related note, in a generic compactification one also expects a large number of moduli fields - scalars in 4D EFT. These fields gain large masses (generically at or well above the gravitino mass scale) and interact with the visible sector via Planck suppressed operators. Their presence can have a major impact on cosmology because some of the geometric moduli end up as light as the gravitino and can be quite long-lived if SUSY breaking scale is low. They may therefore come to dominate the energy density of the universe after inflation and the standard thermal cosmological history must be revised. This is a very active area of research and there are many good papers on the topic.&lt;/p&gt;&#xA;" OwnerUserId="1892" LastEditorUserId="1892" LastEditorDisplayName="" LastEditDate="2011-02-13T02:12:16.547" LastActivityDate="2011-02-13T02:12:16.547" />
  <row Id="5078" PostTypeId="2" ParentId="5074" CreationDate="2011-02-12T23:05:08.723" Score="0" ViewCount="" Body="&lt;p&gt;What property of an object's motion is related to impulse? That is, if you know the impulse, and you know that the object started from rest, what can you conclude about its state of motion after the impulse?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coefficient of friction has one primary purpose in life: to allow you to calculate the force of friction. What is the relationship between those two? Once you know the force, and the information that you know by answering the questions in the previous paragraph, try to chart a path from there to the answer you're looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't speak for others here, but personally I'd be much more inclined to try to give you some more specific help if you show me something about the efforts you've already made.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-12T23:05:08.723" />
  <row Id="5079" PostTypeId="2" ParentId="5073" CreationDate="2011-02-12T23:16:27.477" Score="7" ViewCount="" Body="&lt;p&gt;I'm pretty sure that even the brief summarization of &lt;em&gt;all&lt;/em&gt; the alternatives will take a book or two. I will try to give a review of basic things from my perspective. Let me from the beginning note that the following classification is not accurate -- different classes may and do overlap. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;More scalar doublets (multiplets)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;First of all one can introduce more scalar multiplets. Two Higgs doublet model (2HDM) is the most favored, because it is also naturally arises from MSSM. NHDMs are also considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doublets are usually considered, because there is a basic constraint on the quantum numbers of the fields, coming from &quot;rho parameter&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho = \frac{M_W^2}{M_Z^2\cos^2\theta_w}=1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which can be satisfied only if $(2T+1)^2-3Y^2=1$ with the most natural solution $Y=1,T=1/2$. Of course there are other solutions, leading to bigger values, but I've never seen anyone seriously considering those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is still a lot of freedom to impose some extra discrete symmetries, continuous symmetries, the way these scalars interact with fermions, e.t.c., which leads to many subclasses of such models.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Composite Higgses&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The central example is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Little_Higgs&quot; rel=&quot;nofollow&quot;&gt;Little Higgs&lt;/a&gt; model where Higgs arrives as a (pseudo-)goldstone boson from some higher global symmetries. Changing the underlying symmetry one obtains the whole class of such models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extra gauge symmetries are also considered -- they are usually &lt;a href=&quot;http://en.wikipedia.org/wiki/Dynamical_symmetry_breaking&quot; rel=&quot;nofollow&quot;&gt;broken dynamically&lt;/a&gt;. &lt;a href=&quot;http://en.wikipedia.org/wiki/Technicolor_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;Technicolor&lt;/a&gt; was the most popular one -- now it not so favored, while I don't think that it was refuted completely. &lt;a href=&quot;http://en.wikipedia.org/wiki/Top_quark_condensate&quot; rel=&quot;nofollow&quot;&gt;Top condensate&lt;/a&gt; is another dynamical model. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Originated from extra dimensions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Lots of geometries, compactifications, boundary conditions -- I feel completely lost with those. Most popular are &lt;a href=&quot;http://en.wikipedia.org/wiki/Higgsless_model&quot; rel=&quot;nofollow&quot;&gt;higgsless models&lt;/a&gt; -- attempts to get rid of Higgses completely. They are usually based on some specific boundary conditions. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Of course the list in incomplete. There are a lot of different &quot;mixtures&quot; between those models, usually with some new funny names. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://iopscience.iop.org/0034-4885/74/2/026201&quot; rel=&quot;nofollow&quot;&gt;nice recent reference&lt;/a&gt; that reviews some of the mentioned models going more deeply into some tehnical details.   &lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2011-02-13T11:42:06.807" LastActivityDate="2011-02-13T11:42:06.807" CommunityOwnedDate="2011-02-15T01:40:48.167" CommentCount="1" />
  <row Id="5080" PostTypeId="2" ParentId="5026" CreationDate="2011-02-12T23:20:20.053" Score="0" ViewCount="" Body="&lt;p&gt;What do you think is meant by the velocity of a wave?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, it's the velocity of any point on the wave, so pick one where y(x1,t1) = C say. Over an additional time t2, the point will have moved an additional distance x2 and since we're looking at the same point, this means y(x1+x2, t1+t2) = C also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're told that y(x,t)=Asin(kx-wt+O), so Asin(kx1-wt1+O) = C, so kx1-wt1+O = D. At an additional time t2 and distance x2, k(x1+x2) - w(t1+t2) + O = D. Subtracting these two expressions from one another gives,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k(x1+x2)-K(x1) - w(t1+t2) + w(t1) = 0,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;((x1+ x2)-(x1))/((t1+t2)-(t1)) = w/k&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v = w/k&lt;/p&gt;&#xA;" OwnerUserId="1345" LastActivityDate="2011-02-12T23:20:20.053" />
  <row Id="5081" PostTypeId="2" ParentId="4925" CreationDate="2011-02-12T23:37:01.660" Score="2" ViewCount="" Body="&lt;p&gt;This is an interesting question. And I know that Lubos has already written a nice and complete answer but I think I can add some of my own words here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll concentrate here on the non-abelian YM case, but the results here are quite general. The idea is that the BRST &lt;em&gt;quantum action&lt;/em&gt; can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;S(qu) = S(cl) + \int\; d^4x\; s\Psi&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Psi$ is related to a certain surface over which you rotate the antifields of the action, or, as is more commonly stated, to the gauge fixing function. For non-abelian YM, the usual way to write is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\Psi = b_a(F^a+\frac{1}{2}\xi d^a)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This terms appears because we insist on describing a massless vector boson as a four component field while there are only two physical degrees of freedom. For external states there is no problem since only the two degrees of freedom appears, and that's ultimately the physical reason why the BRST ghosts are engineered to decouple. For internal lines, ie, propagators, the four components contribute and the extra two non-physical degrees of freedom spoil unitarity. This term above fixes it. However, you don't want your theory to depend on the unphysical choice of $\Psi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the natural way is to ask that the physical states are in the cohomology of the charge $Q$ associated to the operator $s$, ie, $sF = [F,Q]$ and this implements mathematically the decoupling (the rest follows from what Lubos already said, that one ghost excitation is not BRST-closed and the other's is BRST-exact).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although everyone calls this action &lt;em&gt;the quantum action&lt;/em&gt;, is does not include any quantum term. The argument above will continue to hold quantum mechanically if not only the action but also the measure is BRST invariant. Well, it turns out that the YM measure is not BRST invariant (this is a regulator-dependent statement!!), but one can show that the variation is always $s\chi$, ie, BRST-exact and everything continues to work as in the classical case.&lt;/p&gt;&#xA;" OwnerUserId="1337" LastActivityDate="2011-02-12T23:37:01.660" />
  <row Id="5083" PostTypeId="1" AcceptedAnswerId="5096" CreationDate="2011-02-13T01:36:32.677" Score="3" ViewCount="103" Body="&lt;p&gt;I'm contemplating particle-hole symmetry, and as an example I am looking at either an electron moving along a hypothetical lattice of hydrogen ions, or a hole moving along a hypothetical lattice of helium atoms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to some lecture notes I found, the hopping integral I get when I treat this in a tight-binding ansatz is positive for holes and negative for electrons, and now I want to see why this is so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For electrons, I can understand the result: The atomic wavefunctions are just the 1s hydrogen wavefunctions, and they are positive everywhere, so the matrix element with the kinetic energy operator gives something negative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for holes I am not sure how I would have to proceed. Can I just say that the relevant Hamiltonian for holes is the negative of the Hamiltonian for electrons because, after all, a hole with energy $E$ would be an electron with energy $-E$ missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of, if that's wrong, how would I come to the conclusion that the hopping integral for electrons is negative and for holes it's positive?&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-02-13T07:34:57.740" Title="Electron-hole symmetry in H and He" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" />
  <row Id="5084" PostTypeId="2" ParentId="5074" CreationDate="2011-02-13T01:46:09.940" Score="1" ViewCount="" Body="&lt;p&gt;We assume the friction force is constant.  Then the question is equivalent to, &quot;What is the momentum of an object with mass $m$ as it strikes the ground if you drop it from height $d$, on a planet where gravity is $g/\mu$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy of the object scales linearly with $m$, $d$, and gravity.  The energy times twice the mass is the square of the momentum, so the momentum should scale as the square root of the distance dropped and square root of gravity, and should vary linearly in $m$.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-02-13T01:46:09.940" />
  <row Id="5085" PostTypeId="2" ParentId="5072" CreationDate="2011-02-13T02:03:46.680" Score="3" ViewCount="" Body="&lt;p&gt;General relativity is a purely classical theory means it does not take the uncertainty principle of quantum mechanics into account. Although it is a very good description of gravity in large scale it should not be a good description for very small scale since in those very small scale quantum effects can not be ignored. There are at least two instances, the big bang and black hole where GR predicts mathematical singularities. That means in those situations GR fails to be a good description of the gravitational interaction. According to quantum mechanics every classical field have a quantum analogue. The classical Field of Maxwell, called electromagnetic field, has a quantum analogue. We picture the electromagnetic interaction by messenger particles called virtual photons which has spin 1. This virtual photons are the particles of electromagnetic field. The disturbances in this field called electromagnetic waves, which according to quantum field theory are emission of real photons. There is no contradiction here. This is a basic feature of quantum theory. We call it wave particle duality. Similarly for gravitation the quantum analogue of the relevant classical field can be thought of as consisting of virtual gravitons of spin 2. Gravitational interactions are mediated by these virtual gravitons. When a real gravitons are emitted we call it gravitational waves. Gravitons are messenger particles of gravity.  Again there is no contradiction here.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-13T06:30:14.317" LastActivityDate="2011-02-13T06:30:14.317" />
  <row Id="5086" PostTypeId="1" AcceptedAnswerId="5103" CreationDate="2011-02-13T02:14:12.040" Score="3" ViewCount="189" Body="&lt;p&gt;Could you please suggest the software, where I can load my 3D model and see how it behave on various conditions (speed - preferably including supersonic, temperature, pressure)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both free &amp;amp; commercial variants are interesting.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-02-17T13:13:44.910" Title="Software for simulating supersonic aerodynamics" Tags="&lt;simulation&gt;&lt;software&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="5087" PostTypeId="2" ParentId="5066" CreationDate="2011-02-13T02:17:03.787" Score="2" ViewCount="" Body="&lt;p&gt;The most realistic way is to get rid of humans, and let robots travel. They don't eat and can spend 1'000'000 years in flight without being bored too much. This way is achievable even with current tech. ;-)&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-02-13T02:17:03.787" CommentCount="2" />
  <row Id="5088" PostTypeId="2" ParentId="5064" CreationDate="2011-02-13T02:18:24.233" Score="2" ViewCount="" Body="&lt;p&gt;I tend to explain this with analogues with instruments.  I say imagine a string, like a guitar string or a piano string.  However, this string has some interesting properties.  The tension in the string is due to the quantum uncertainty principle.  Because quantum mechanics says we are not able to identify the position of the string to complete precision while also knowing its momentum, this means there is a constant wavy motion which occurs.  This waviness happens whether we are pulling on the string or not.  So the string has its own tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the open string there are these flapping open ends, a bit like bullwhips cracking in space.  This means that waves are reflected off the ends and waves going up and down are the same.  This is different from a closed string, which is a loop that have waves going clockwise and counter-clockwise (left &amp;amp; right) which are independent.  Now let us suppose we take that open string and put in on a sound board. The end points are now held fixed and a wave which approaches the endpoint is reflected off by this fixed point.  However, the sound board, which we will now call a membrane, or D-brane, may adjust in response.  So that reflected wave on the string may be changed in a way so that it has an independent part from the incoming wave.  This means the wave dynamics on that open string tied to D-brane becomes similar to the wave dynamics on the closed string.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way the behavior of quantum strings with respect to these D-branes, which are similar to Fermi surfaces or “defects” in space (with a follow on discussion on Fermi surfaces etc) can be transformed from one type to the other.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean really, if you are talking to “grandmas” discussions about symmetries and the like will be little more than soporific.  Gotta keep it really simple, even if maybe slightly wrong. In the case of sb1's gramma this could be a sort of spring board for going into deeper discussions.  However, this gives at least a physical sense of M-theory.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-13T02:18:24.233" CommentCount="6" />
  <row Id="5089" PostTypeId="2" ParentId="414" CreationDate="2011-02-13T02:48:12.737" Score="2" ViewCount="" Body="&lt;p&gt;I am on thin ice here, but I know that people in number theory study modular forms, and this is connected to partition functions for example of conformal field theory.&lt;/p&gt;&#xA;" OwnerUserId="1611" LastActivityDate="2011-02-13T02:48:12.737" />
  <row Id="5091" PostTypeId="2" ParentId="414" CreationDate="2011-02-13T03:53:55.737" Score="0" ViewCount="" Body="&lt;p&gt;Here is a nice article from a news channel:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sciencemag.org/content/274/5295/2014.full&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencemag.org/content/274/5295/2014.full&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="159" LastActivityDate="2011-02-13T03:53:55.737" />
  <row Id="5092" PostTypeId="2" ParentId="5066" CreationDate="2011-02-13T04:03:07.800" Score="5" ViewCount="" Body="&lt;p&gt;I wrote a semi-popular book on this subject a couple of years ago:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/9812706178&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Can-Star-Systems-Be-Explored/dp/9812706178/ref=sr_1_3?ie=UTF8&amp;amp;s=books&amp;amp;qid=1297567209&amp;amp;sr=8-3&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of this was worked out.  I illustrate the relativistic rocket and the photon sail.  The photon sail is clearly the most reasonable of these two.  I also suggest using electromagnetically accelerated nano-probes, where millions of tiny probes are sent towards another star, with the prospect that a few might set up shop there and build up a robotic colony of sorts.  The solar sail is capable of reaching a $\gamma~=~1.4$, or .$69c$, though a more realistic figure is $\gamma~=~1.15$ or about $v~=~.5c$.  This analysis involve the redshifting of light on the frame of the sail craft and so forth.  To accomplish this requires placing large Fresnel lenses in space which collimate light towards the sail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is not much point in sending probes towards extrasolar systems faster than $\gamma~=~2$, or $v~=~.86c$.  The time saved becomes smaller and the energy costs enormous.  A $\gamma~=~2$ probe has as much kinetic energy as its rest mass, which would be a considerable feat to achieve.  At $.5c,~.7c$, and $.86c$ probes could be send within the interstellar neighborhood and signals returned in reasonable time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ramjet concept is useful IMO from one main reason.  Suppose various instruments were packed onto a spacecraft with an acceleration of one gravity, or larger.  These instruments look at the CMB and maybe particle detectors.  The machine reaches a high $\gamma$ in order to probe the universe as it appeared earlier on.  The CMB would be blue shifted into the IR, and maybe other detectors which can detect weakly interacting particles which pass through the early radiation dominated period of the universe.  This is pretty speculative, but maybe not impossible in principle.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-13T04:03:07.800" />
  <row Id="5093" PostTypeId="2" ParentId="5064" CreationDate="2011-02-13T06:32:06.453" Score="3" ViewCount="" Body="&lt;p&gt;M-theory is a conjectured 11-dimensional theory with the property that, compactified on a circle, it becomes Type IIA string theory, and, compactified on a line segment, it becomes &lt;em&gt;E&lt;sub&gt;8&lt;/sub&gt; x E&lt;sub&gt;8&lt;/sub&gt;&lt;/em&gt; heterotic string theory. At some level of description, it should consist of 2- and 5-dimensional supermembranes coupled to the fields of 11-dimensional supergravity. An exact formulation is not known, but &lt;a href=&quot;http://arxiv.org/abs/1012.0459&quot; rel=&quot;nofollow&quot;&gt;arXiV:1012.0459&lt;/a&gt; reviews recent progress. &lt;/p&gt;&#xA;" OwnerUserId="1486" LastActivityDate="2011-02-13T06:32:06.453" />
  <row Id="5094" PostTypeId="1" AcceptedAnswerId="5101" CreationDate="2011-02-13T06:43:54.603" Score="1" ViewCount="152" Body="&lt;p&gt;Witten (and earlier, Bodner) hypothesized that strange matter (up, down, strange quarks) should be more stable than &quot;regular&quot; nuclear matter(The strange matter hypothesis). That is that the typical hadrons in matter are metastable, and will eventually decay into strange matter because it can exist at lower energy configuations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was some publicity about strangelets as a disaster scenario for the RHIC and the LHC, and a very interesting paper was commissioned by the &lt;a href=&quot;http://www.bnl.gov/rhic/disaster.htm&quot; rel=&quot;nofollow&quot;&gt;Brookhaven National Laboratory&lt;/a&gt; to evaluate risks of various disasters possible in the RHIC: The authors are &lt;a href=&quot;http://www.bnl.gov/rhic/docs/rhicreport.pdf&quot; rel=&quot;nofollow&quot;&gt;Busza, Jaffe, Sandweiss, and Wilczek&lt;/a&gt;. A conclusion is that positively charged and neutral strangelets would be no problem, and that negatively charged ones, which would be a huge problem because they would convert all the nuclear matter to strange matter, would be very unlikely based on cosmic ray evidence (ie lack of them) and other considerations. There are theories that strange matter occurs in the cores of neutron stars and may convert them completely resulting in denser, smaller stars than predicted for nuclear matter, but I am not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we expect the LHC to produce negatively charged strangelets? I think the answer to the question is NO, none will be found or produced, but the reasoning from cosmic rays is not definitive imo( I guess I should have said, &quot;but asymptotically close to it&quot;--G). The Jaffe paper above has merit I believe not just as a physics document, but also as public relations, and has historical value, so I wanted to draw attention to it.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-02-13T16:52:18.477" LastActivityDate="2011-02-13T16:52:18.477" Title="Will negatively charged strangelets be produced by the LHC?" Tags="&lt;particle-physics&gt;&lt;accelerator-physics&gt;&lt;lhc&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="5095" PostTypeId="2" ParentId="5075" CreationDate="2011-02-13T07:29:25.707" Score="3" ViewCount="" Body="&lt;p&gt;This is essentially the philosophy which emerges from the spin-network picture of spacetime. A nice example of this is &lt;a href=&quot;http://arxiv.org/abs/0903.5303&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; by Smolin and Prescod-Weinstein which argues that the differing notions of locality arising due to the graph like nature of spacetime, might be relevant for understanding dark energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the notion of treating spacetime itself as a cellular automaton goes as far back as Konrad Zuse who designed and built some of the first digital computers. He has a nice monograph on this called &quot;Rechrender Zaum&quot; or &quot;Rendering Space&quot; which you can find along with other useful information on &lt;a href=&quot;http://www.idsia.ch/~juergen/digitalphysics.html&quot; rel=&quot;nofollow&quot;&gt;Jurgen Schmidhuber's&lt;/a&gt; page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many people who have recently done work along these lines, such as Seth Lloyd, Livine, Terno, Girelli, and others (&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0603008&quot; rel=&quot;nofollow&quot;&gt;ref1&lt;/a&gt;, &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0501135&quot; rel=&quot;nofollow&quot;&gt;ref2&lt;/a&gt;). What precise implications such CA models have for topology is not clear. What is certain is that such models will naturally incorporate non-trivial topologies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model of Feynman that you refer to is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Feynman_checkerboard&quot; rel=&quot;nofollow&quot;&gt;checkerboard model&lt;/a&gt;. This was his attempt to deduce the dynamics of an electron in a simplified setting from a stochastic model in which &lt;em&gt;a priori&lt;/em&gt; the electron does not obey any predetermined equations of motion. One finds that the Dirac equation for the electron (in $1+1$ dimensions only) &lt;em&gt;emerges&lt;/em&gt; from a path-integral calculation in this model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sidebar:&lt;/strong&gt; Wolfram's book has received a great deal of criticism, some of it no doubt well-deserved. However it takes a great deal of courage to put forward a hypothesis such as that of the &quot;computational universe&quot; (my terminology) which has the potential to overturn 500 years of conventional wisdom.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-13T07:29:25.707" />
  <row Id="5096" PostTypeId="2" ParentId="5083" CreationDate="2011-02-13T07:34:57.740" Score="2" ViewCount="" Body="&lt;p&gt;Yes, a hole with energy $E$ is the same as an electron with a negative energy $E$ missing - that's why it's called a hole and that's how Paul Dirac first encountered it in the relativistic context (in the form of positrons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A positively-charged positron may look more &quot;particle-like&quot; but one may describe it as the very same holes in the otherwise filled sea of negative-energy electron states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases - semiconductors and positrons - you may assume that the negatively-charged electrons are the only &quot;real&quot; particles. However, you will always derive the existence of positively-charged holes that behave &quot;just like electrons&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find states such that the energy $E$ is an increasing function of the momentum, the system will first try to fill the low-momentum, low-energy states, and you may add additional higher-energy, higher-momentum particles (electrons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But some part of the spectrum may have the property that the energy $E$ is a decreasing function of the momentum $k$. In that case, the electron states with a higher value of momentum are filled first, and you're adding them &quot;inwards&quot;. This is counterintuitive, so it's more logical to exchange the convention for what we mean by a &quot;filled state&quot; and what we mean by an &quot;empty state&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we do so, we also change the charge and energy of the particle in each state. Consequently, we will deal with positive-charge holes whose energy $E$ behaves just like $-E$ of the original electrons, and increases with $k$ just like for ordinary electron states. The only difference will be in the charge.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-13T07:34:57.740" CommentCount="1" />
  <row Id="5097" PostTypeId="1" CreationDate="2011-02-13T07:43:32.247" Score="3" ViewCount="173" Body="&lt;p&gt;What is the current status of the swampland? Have people expanded the list of swampland criteria?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If let's say I wish to figure out if the Standard Model belongs to the swampland, are there any concrete steps I can use to go about it, at least in principle? If let's say technical difficulties aren't an objection, and we have access to a supercomputer with top notch programmers and mathematics code, is there any algorithmic procedure to show whether or not the Standard Model belongs to the swampland?&lt;/p&gt;&#xA;" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-13T13:17:22.603" LastActivityDate="2011-02-13T13:17:22.603" Title="What is the current status of the swampland?" Tags="&lt;string-theory&gt;&lt;swampland&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="5098" PostTypeId="2" ParentId="5075" CreationDate="2011-02-13T07:49:21.990" Score="3" ViewCount="" Body="&lt;p&gt;Space_cadet mentioned already work about deriving spacetime as a smooth Lorentzian manifold from more &quot;fundamental&quot; concepts, there are a lot of others -like &lt;a href=&quot;http://en.wikipedia.org/wiki/Causal_sets&quot; rel=&quot;nofollow&quot;&gt;causal sets&lt;/a&gt;, but the motivation for the question was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The reason for my interest in this regards one of the mysteries of quantum mechanics, that of quantum entanglement and action at distance. I wondered whether, if space is imagined as having a topology that arises from a notion of neighbourhood at a fine level, then quantum entanglement might be a result of a 'short circuit' in the connection lattice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not convinced that such an explanation is possible or warranted, the reason for this is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Reeh%E2%80%93Schlieder_theorem&quot; rel=&quot;nofollow&quot;&gt;Reeh-Schlieder theorem&lt;/a&gt; from quantum field theory (I write &quot;not convinced&quot; because there is some subjectivity allowed, because the following paragraph describes an aspect of axiomatic quantum field theory which may become obsolete in the future with the development of a more complete theory): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It describes &quot;action at a distance&quot; in a mathematically precise way. According to the Reeh-Schlieder theorem there are correlations in the vacuum state between measurements at an arbitrary distance. The point is: The proof of the Reeh-Schlieder theorem is independent of any axiom describing causality, showing that quantum entanglement effects do not violate Einstein causality, and don't depend on the precise notion of causality. Therefore a change in spacetime topology in order to explain quantum entanglement effects won't work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Discussions of the notion of quantum entanglement often conflate the notion of entanglement as &quot;an action at a distance&quot; and Einstein causality - these are two different things, and the first does not violate the second.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-02-13T07:49:21.990" CommentCount="2" />
  <row Id="5099" PostTypeId="2" ParentId="5075" CreationDate="2011-02-13T07:49:23.937" Score="3" ViewCount="" Body="&lt;p&gt;In general, if you construct a manifold out of a combinatorial graph, or out of patches, then you're not finding anything &quot;more fundamental&quot;. You're just describing the most straightforward and most superficial definition of the concept of topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that this very modest sentence contradicts the whole philosophy of Stephen Wolfram's book, and the reason behind this contradiction is that Stephen Wolfram's book is misguided in this very basic respect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to find more fundamental laws of physics that produce topology, you always deal with phases of a physical system that have to be found by solving some equations; and/or with a geometric interpretation of individual terms in an expansion of a physical observable that doesn't have a topological interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see the latter example, check e.g. the quantum foam paper&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0309208&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0309208&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;by Okounkov (a Fields medalist), Reshetikhin, and (most importantly) Vafa. A partition sum of a topological string theory may be obtained as a sum over manifolds of many different topologies - but it may also be viewed as an expansion of a function that describes the propagation of (topological) strings in a flat background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both ways of writing the partition sum are also equivalent to the partition sum of a melting crystal. Their paper is a particular, quantitative realization of John Wheeler's concept of a &quot;quantum foam&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other contexts of physics, different topologies are simply allowed and they have to be summed over. That's the case of perturbative string theory where the scattering amplitudes are calculated as sums over world sheets of all topologies (Riemann surfaces).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sum may be obtained in various other formalisms where it looks &quot;derived&quot;. In the non-perturbative light cone gauge description of perturbative string theory, the so-called matrix string theory,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/9701025&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9701025&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/abs/hep-th/9703030&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9703030&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the very states containing $K$ strings are obtained from the same Hilbert space, by identifying various holonomies of the gauge field around a circular dimension of a Yang-Mills theory. The possible holonomies for which low-energy states exist may be labeled by permutations of $N$ eigenvalues, and the cycles from which these permutations are composed may literally be interpreted as strings of various lengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, 1 string is topologically different from $K$ strings for $K\neq 1$ but in matrix string theory, all these states are configurations of a single field theory in different limits. Analogously, the sum over histories will contain world sheets of all topologies which are generated as histories of switching in between the phases of the Yang-Mills theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In string theory, you get generically dual descriptions of a compactification in which the topology of the space is completely different. For example, mirror symmetry relates two very different topologies of a six-dimensional Calabi-Yau manifold. M-theory or string theory compactified on a four-dimensional K3 manifold is equivalent to heterotic strings on tori; the individual excitations of the heterotic string are mapped to nontrivial submanifolds of the K3 manifold, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the summary is that there are many exciting ways in which topology of space may turn out to be &quot;emergent&quot;, or at least &quot;as fundamental as other, non-topological descriptions&quot;, but none of them is similar to your &quot;discretized&quot; template.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-13T07:49:23.937" />
  <row Id="5100" PostTypeId="2" ParentId="5097" CreationDate="2011-02-13T07:57:25.847" Score="3" ViewCount="" Body="&lt;p&gt;Yoshiko, the Standard Model belongs to the swampland for some easy reasons - you don't need a supercomputer for that. First of all, you should add gravity to the Standard Model - in some approximate way that people would normally do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When you add gravity, the Standard Model still belongs to the swampland.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, it predicts that there are no magnetic monopoles. That's forbidden in the &quot;landscape&quot; of consistent theories of quantum gravity because one may demonstrate that pairs of black holes that carry the magnetic monopole charge may be pair-produced because of an instanton that enables this process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are probably many other pathological properties of the Standard Model that guarantee it is in the swampland but the map of all the conditions remains poorly understood. Let me just mention another example: the Standard Model perturbatively conserves the lepton and baryon charges. In the allowed theories of quantum gravity, there can't exist exact global symmetries (which don't produce any new local fields) and the baryon and lepton charges would violate this rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you generalized the term &quot;Standard Model&quot; and included all theories that look like the Standard Model but may be modified at the GUT scale or another high energy scale (e.g. GUT theories themselves where magnetic monopoles can be easily constructed; and electroweak instantons allow one to violate at least $B$ or $L$ separately), then you would have to conclude that the generalized Standard Model almost certainly belongs to the &quot;landscape&quot; and is allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all, quantum gravity has to describe the real world. However, there are no rigorous ways to mathematically prove such a modest proposition at this moment. What is available at this moment are various constraints that may show that various theories are bad and belong to the swampland. One of those that were studied in recent years is the condition that &quot;gravity has to be the weakest force&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0601001&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0601001&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course, the known conditions are useful to eliminate various pathological &quot;cousins&quot; of the Standard Model. The Standard Model itself is close to the actual laws of Nature so it doesn't suffer from &quot;unfixable&quot; pathologies - only from pathologies that arise from the incompleteness of the Standard Model.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-13T08:04:19.600" LastActivityDate="2011-02-13T08:04:19.600" />
  <row Id="5101" PostTypeId="2" ParentId="5094" CreationDate="2011-02-13T08:20:37.107" Score="4" ViewCount="" Body="&lt;p&gt;Dear Gordon, apologies but I think that the paper by Busza, Jaffe, Sandweiss, and Wilczek is very robust. In particular, it is a physics paper, not a public relations paper, and one could argue that your comment claiming otherwise may be offending to the authors. It is a physics paper about a question that physicists usually consider funny so most physicists don't really study this topic seriously but these four authors &lt;em&gt;did&lt;/em&gt; study it seriously. The paper is &quot;less technical&quot; and looks &quot;softer&quot; than many other papers by these and similar authors but that's not because they did a bad job; it's because the very topic of the paper is an amusing schoolkid's conspiracy theory, not a cutting-edge question about state-of-the-art physics. So the relevant arguments may be accessible to the schoolkids, too. But that doesn't mean that they're wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your question whether the LHC will produce negatively charged strangelets is obviously No, at a 99.9999999999+ percent level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what exactly you find unsatisfactory about the paper - and, sorry to say, I agree with David that it is not a good approach to pose a &quot;question&quot; in which you just declare that you don't trust a paper, without offering a glimpse of a rational reason - but the paper actually mentions many other reasons why the answer is No - not just the cosmic rays argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cosmic rays are arguably the most waterproof reason why this hypothetical catastrophe - and pretty much any catastrophe of this type - has to be wrong. It is waterproof exactly because it is phenomenological in character and only depends on the observation of a long-lived Universe with long-lived stars etc. It doesn't strongly depend on our theoretical assumptions. So even if we're making some incorrect assumptions, it's still true that planets can't be destroyed in this way because this would have happened many times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you allow theorists to actually trust their theories, which is what they're allowed to do in all other papers (papers in scientific disciplines that have found something can't start and don't start from scratch), there are multiple other reasons why it can't happen (and they're described in the paper), namely&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;QCD overwhelmingly indicates that the metastable stranglet matter of the relevant size would be positively charged; see e.g. &lt;a href=&quot;http://arxiv.org/abs/hep-ph/0008217&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-ph/0008217&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The matter would be unstable for small sizes - because of destabilizing surface effects&lt;/li&gt;&#xA;&lt;li&gt;Strangelets are not easily produced by hadron-hadron collisions&lt;/li&gt;&#xA;&lt;li&gt;Strangelets have never been seen in the outer space at all; the neutron stars collapsed into the stable strangelet matter (as you mentioned) would be called &quot;quark stars&quot; but they haven't been seen yet&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Some research also indicates that strangelets can only be locally stable but they're not ground states similar to the nuclei, see e.g. &lt;a href=&quot;http://arxiv.org/abs/hep-ph/0604093&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-ph/0604093&lt;/a&gt; ... Another paper leads to mixed results about this question: &lt;a href=&quot;http://arxiv.org/abs/hep-ph/0209184&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-ph/0209184&lt;/a&gt; ... Also, strangelets failed to appear in various collision experiments, see e.g. STAR Collaboration 2005 (gold-gold) &lt;a href=&quot;http://arxiv.org/abs/nucl-ex/0511047&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/nucl-ex/0511047&lt;/a&gt; ... Quark-gluon plasma was hypothesized to be linked to the Centauro events, see e.g. &lt;a href=&quot;http://arxiv.org/abs/hep-ex/0209008&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-ex/0209008&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretty much every point above would be enough to conclude that those stable strangelets can't be produced, and if you really believe that they will be produced, I can offer you 1,000:1 odds to get &quot;insured&quot;. In recent years, it has become popular to indefinitely spread myths about various hypothetical catastrophes caused by &quot;small events&quot;, regardless of the actual physics evidence that such catastrophes can't occur, but I don't think this server should adopt this approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning your question whether the &quot;Bodner-Witten&quot; (I would also add Farhi and Jaffe) strange quark matter (SQM) hypothesis is correct, it depends what you exactly mean by this hypothesis. If you just mean that SQM as a form of quark-gluon plasma (QGP) has a lower energy density than the nuclear matter, then it looks relatively likely (but surely not guaranteed), and supported by lattice QCD etc. If you mean that it can be found somewhere in the Universe, the answer is we don't know, but if it's somewhere, it's in artifacts of supernova explosions and gamma-ray bursts (which provide the particles with a sufficiently extreme environment). If you mean the hypothesis that SQM is negatively charged and a part of ordinary stars and solar systems such as ours, the answer as seen by the contemporary science is almost certainly No. It's very important to distinguish those different propositions and questions, exactly because their conflation leads to irrational fears etc. (The same disclaimer holds for the the global warming threats or any other catastrophe scenario. One must be very clear about what he means that e.g. &quot;global warming is real&quot; because different questions have different answers.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper by Busza et al. offers 20+ pages of evidence that such things can't happen, and the cosmic-rays argument is just the most waterproof one among them. And of course, it's a whole paper because it's meant to be much more detailed and careful an analysis of the question than one answer or several answers on a Physics Stack Exchange. It's just not quite kosher to de facto deny that those 20+ pages exist and to offer one paragraph without any technical content that tries to suggest that the paper is not robust.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-13T09:40:43.967" LastActivityDate="2011-02-13T09:40:43.967" CommentCount="8" />
  <row Id="5102" PostTypeId="2" ParentId="5072" CreationDate="2011-02-13T08:26:06.433" Score="2" ViewCount="" Body="&lt;p&gt;An addendum to the answers of Daniel Grumiller and sb1: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The major difference of the gravitational field and other fields is that according to general relativity the gravitational field &lt;em&gt;defines&lt;/em&gt; space and time and therefore &lt;em&gt;defines&lt;/em&gt; the relation of events. It is true that it is possible to do an &quot;arbitrary&quot; split of a certain linear approximation of the gravitational field into a &quot;flat background&quot; and &quot;waves&quot; propagating on this background. In principle this kind of reasoning is a violation of the very idea that the gravitational field defines the background of spacetime in a holistic way, and it was the subject of a lot of discussions if this approximation is of any use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is considered to be settled by the observational evidence that bistar systems loose energy in exact the way that the &quot;graviational wave approximation&quot; predicts, as cited by Daniel Grumiller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The existence of &lt;em&gt;gravitons&lt;/em&gt; is a &lt;em&gt;conjecture&lt;/em&gt; based on the assumption that gravitational waves exhibit the same quantum nature as classical waves, e.g. waves in classical electromagnetism. At the basis of this conjecture is the idea that it should be possible to split the gravitational field in a part defining the background, and then having gravitational waves propagating on this background and exhibiting the same wave-particle duality as other waves. It would then be possible to treat quantum gravitational effects in a semi-classical approximation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there is no observational evidence for this, this conjecture is still the subject of controversy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are these different communities of physicists?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some have a strong believe in the existence of gravitons, some think that quantum gravity needs a bigger conceptual step than only gravitions, and some do both, so, yes, there are different communities believing different things.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does relativity explain only part of the story of masses acting under gravity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It explains everything in a classical setting with not too strong forces alright, but does not exlpain quantum effects or what happens when forces get so strong that singularities occur.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is gravity a force or not? Is it only an apparent force or not?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is a force, it is an apparent force in the sense that classical GR says that you feel it because you live in an accelerating reference frame. Both statements are valid in the classical setting and are independent of the quantum nature of gravity, and in particular of the existence of gravitons.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can such an apparent force 'generate' exchange particles? Are the exchange particle and geometric models both different views of the same underlying truth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, see above (geometric model = classical setting, exchange particle = semi-classical approximation). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;why can't the other forces be explained away similarly? Or is that what is happening with all this talk of small extra dimensions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The gravitational field is fundemantally different from other fields (see above), and this has no connection to extra dimensions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'd appreciate any illumination on this matter, or suggested reading (preferably at the 'popular science' or undergraduate level).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The problem is that if you are able to ask this question, you're already beyond the popular science level. I'd really like to recommend to you an introducory class on QFT and one on GR, there you'd get the best answer to your question :-)&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-02-13T08:26:06.433" CommentCount="6" />
  <row Id="5103" PostTypeId="2" ParentId="5086" CreationDate="2011-02-13T08:46:53.450" Score="2" ViewCount="" Body="&lt;p&gt;If you want something that looks working and don't care much about the details, the standard solution is Fluent.&lt;br&gt;&#xA;The nearest Open Source option is &lt;a href=&quot;http://www.openfoam.com/&quot; rel=&quot;nofollow&quot;&gt;OpenFoam&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2011-02-13T08:46:53.450" />
  <row Id="5104" PostTypeId="2" ParentId="5094" CreationDate="2011-02-13T08:49:00.140" Score="1" ViewCount="" Body="&lt;p&gt;What Lubos said. &#xA;There was also a big brouhaha about black holes produced by the LHC a while ago in the news. I read the report, and it was written before the large string dimension black holes that people hope to see at the LHC. The cosmic rays summary in the paper would cover this disaster also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course it is a type of anthropic principle: since the universe has reached 14+billion years and we exist then such catastrophes are very very rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember back in the beginning 1980's when proton decay was the fashion. At the same time lasers and induced de excitations of pumped crystals were still new. I remember some  worry, after realizing that we are sitting on a pumped up ball of putative decay products of proton decay; until putting down the numbers and seeing that we are saved by the very small value of hbar.( another anthropic input).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We already have a quark gluon plasma at the LHC :). I do not think that strangelets are in the program. There is &lt;a href=&quot;http://lsag.web.cern.ch/lsag/LSAG-Report_add.pdf&quot; rel=&quot;nofollow&quot;&gt;a report on this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer is no, there is no danger . &lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-02-13T08:49:00.140" CommentCount="2" />
  <row Id="5106" PostTypeId="2" ParentId="5074" CreationDate="2011-02-13T10:18:34.283" Score="1" ViewCount="" Body="&lt;p&gt;In terms of the equations that you have presented equn 1) is correct. The problem is with 2) and especially 3). The reason is that this object does not have a constant velocity during motion. So you will need to either remember or calculate the formulae for an object with an initial velocity and moving under acceleration. It is true that these formulae will also contain time t terms, but there are ways to eliminate the t term, which might occur once you see the corrected formula for motion under acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to begin with the velocity of the object at time t is the initial velocity plus a term from the (constant) acceleration applied for time t.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I am not sure how much Calculus you remember. Equation 2) would be more accurate as $a=dv/dt$. With some basic calculus manipulations you can derive the equations. If you dont quite remember, then take a guess and try to confirm with some calculus you do remember or directly.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-13T10:18:34.283" CommentCount="8" />
  <row Id="5107" PostTypeId="1" CreationDate="2011-02-13T10:50:58.540" Score="2" ViewCount="116" Body="&lt;p&gt;If we try to work out the details of string theory over a worldsheet with genus 1, does that mean we have periodic time on the worldsheet? But if we have periodic time, and in some portions of the worldsheet, the target space time $X^0$ increases with $\tau$, won't it have to decrease with $\tau$ somewhere else on the worldsheet? But wouldn't such a decrease mean we have an anti-string mode?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-13T20:05:46.870" Title="String worldsheets with periodic time?" Tags="&lt;string-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5108" PostTypeId="2" ParentId="5107" CreationDate="2011-02-13T11:33:30.250" Score="2" ViewCount="" Body="&lt;p&gt;Dvořák, yes, unless the spacetime coordinate $X^0$ itself is a periodic variable, and it shouldn't be to avoid closed time-like curves, the $X^0$ has to decrease with $\tau$ as much as it increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And indeed, you may say that there are &quot;antistrings&quot;: after all, one-loop diagrams such as torus correspond to the creation of particle-antiparticle pairs - which may be called &quot;string-antistring&quot; pairs. Of course, if we only talk about whole string and not their particular excited states, an antistring is nothing else than the original string - it's just oppositely oriented in space (it may have the opposite winding numbers, if there are any, too). Note that the reversed orientation of the time on the world sheet may be fully compensated by the reversed orientation of the spatial coordinate along the world sheet - only the overall $\epsilon_{\mu\nu}$ antisymmetric tensor on the world sheet knows about the orientation of the whole world sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it's usually considered that the world sheet is Euclidean (signature) and such a world sheet should naturally be embedded into the Euclidean spacetime, too. In the Euclidean spacetime, there is no sharp difference between future-directed and past-directed vectors; after all, all vectors in the Euclidean spacetime are spacelike. So the sharp difference in the future/past sign disappears. The coordinate $\tau$ is just an auxiliary variable and one shouldn't be shocked that functions are not monotonic functions of $\tau$.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-13T20:05:46.870" LastActivityDate="2011-02-13T20:05:46.870" />
  <row Id="5109" PostTypeId="1" CreationDate="2011-02-13T12:57:22.037" Score="2" ViewCount="129" Body="&lt;p&gt;When radiation from a colder source arrives at a warmer surface there is some debate about what happens next. To make the question more concrete lets say that the colder source is at temperature 288K. The warmer surface is at 888K and has emissivity of 1.&#xA;3 possibilities&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;We ignore such radiation because it cannot happen.&lt;/li&gt;&#xA;&lt;li&gt;The radiation is subtracted from the much larger radiation of every wavelength leaving the hotter surface.&lt;/li&gt;&#xA;&lt;li&gt;The radiation is fully absorbed and its effect is to be re radiated at characteristic temperature of 888K (plus infinitesimally small T increase due to radiation absorption).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I would have thought that 2. and 3 are more plausible than 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both 2 and 3 satisfy the Stephan Boltzmann equation.&#xA;3 however seems to imply that the radiation from colder object is transformed into much higher quality radiation and a possible second law of thermodynamics infringement.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-13T14:14:53.243" Title="Colder surface radiates to warmer surface" Tags="&lt;thermodynamics&gt;" AnswerCount="4" />
  <row Id="5110" PostTypeId="2" ParentId="5109" CreationDate="2011-02-13T13:14:10.687" Score="3" ViewCount="" Body="&lt;p&gt;The option three is valid -- the outgoing radiation is only dependent on the radiating object. Second law of thermodynamics is not violated since it must be applied to the whole system (and so to the &lt;strong&gt;net&lt;/strong&gt; radiative exchange). &lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2011-02-13T13:14:10.687" />
  <row Id="5111" PostTypeId="2" ParentId="5109" CreationDate="2011-02-13T13:21:15.483" Score="5" ViewCount="" Body="&lt;p&gt;Obviously, the option 3 is the right option. There is no violation of the 2nd law, since the emission rate is higher than the absorption rate for the hotter body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a closed system of two objects. One is hotter and the other is colder. What will happen? Both will radiate energy and both will absorb energy from the incident radiation. The hotter object will radiate energy at a higher rate than it will absorb energy. Opposite is for the colder object. As a result after sufficient time both will come to same temperature and thermal equilibrium will be established. In this condition both object will radiate and absorb energy at the same rate.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-13T13:39:14.397" LastActivityDate="2011-02-13T13:39:14.397" />
  <row Id="5112" PostTypeId="2" ParentId="5109" CreationDate="2011-02-13T14:04:49.817" Score="3" ViewCount="" Body="&lt;p&gt;The hotter object absorbs external radiation as well as it absorbs its own radiation inside the body before it reaches its surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat loss of a hot object (radiated power) is determined with the object temperature but the &lt;strong&gt;rate of cooling down&lt;/strong&gt; (if there is cooling down $dT/dt &amp;lt; 0$) is smaller in case of additional external source of energy in form of radiation coming from exterior.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-13T14:14:53.243" LastActivityDate="2011-02-13T14:14:53.243" />
  <row Id="5113" PostTypeId="2" ParentId="5109" CreationDate="2011-02-13T14:09:59.187" Score="1" ViewCount="" Body="&lt;p&gt;Edit: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS Wladimir has given an equivalent answer slightly ahead of me. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-02-13T14:09:59.187" />
  <row Id="5114" PostTypeId="1" AcceptedAnswerId="5117" CreationDate="2011-02-13T15:02:19.910" Score="3" ViewCount="194" Body="&lt;p&gt;When people talk about the first superstring revolution they often mention the miraculous cancellation of anomalies via the Green Schwarz mechanism. My question is whether such a string-theoretic mechanism is also at work when the 4D gravitational and gauge-gravitational anomalies are tackled? In this context, would it be fair to say that a possible discovery of superpartners at the LHC, which automatically implies some version of N=1 D=4 supergravity, imply that stringy couplings (higher order in alpha') must be present in the corresponding lagrangian to cancel the anomalies? What type of coupling are those?&lt;/p&gt;&#xA;" OwnerUserId="1892" LastActivityDate="2011-03-02T04:46:04.080" Title="Gravitational and gauge-gravitational anomalies in N=1 D=4 supergravity coupled to a SUSY gauge theory with chiral matter" Tags="&lt;string-theory&gt;&lt;quantum-gravity&gt;&lt;supergravity&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="5115" PostTypeId="2" ParentId="5072" CreationDate="2011-02-13T15:10:47.843" Score="1" ViewCount="" Body="&lt;p&gt;Firstly I would like to question whether anyone is really experimentally &lt;em&gt;searching&lt;/em&gt; for a graviton. The effect would be too weak to detect with current technology. There is the closely related concept of &quot;gravitational wave&quot; which is a &quot;curvature wave&quot; in General Relativity and that is being searched for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are probably two main reasons for expecting the graviton idea to have some merit.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The Gravitational Wave is a classical wave, and all other classical waves (like with Electromagnetism) have found a particle form in Quantum Theory and Quantum Field Theory. The familiar concept here is &quot;wave particle duality&quot;. So the expectation is that under a theory of &quot;Quantum Gravity&quot; there would be a graviton. This point does not in itself prejudice the fundamental form of that Quantum Gravity theory: it might still be geometric in its foundations.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The equations which describe quantum particles and fields when expressed in a notation called &quot;2-spinor form&quot; have a basic pattern and form. So an electron is described via Dirac equations as $\phi_A$. This relates to spin 1/2 which is all correct for the electron. Spin 1 particles/fields (e.g. the Photon) are described by $\phi_{AB}$ in this form of Maxwell's equations. And so the pattern continues with N labels A,B, ..., K describing a spin N/2 particle. Well Einstein's General Relativity equations when written in this spinor form have $\phi_{ABCD}$. So this suggests a spin 2 - even without any linearisation or quantisation, etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The big difference between Einstein's equations and these others of lower spin is that the Einstein equations are non-linear (representing space-time curvature) whereas these other equations are linear (essentially representing a linear underlying space). So one photon plus one photon equals two photons, but one graviton plus one graviton would not quite equal two gravitons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arena in which the debate continues is that of Quantum Gravity in which different camps take different views as to the nature of any underlying space. These debates are reflected in some of the other Stack questions.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-13T15:10:47.843" />
  <row Id="5116" PostTypeId="2" ParentId="5114" CreationDate="2011-02-13T15:11:36.283" Score="0" ViewCount="" Body="&lt;p&gt;In 4D, we can have an axion mechanism. We have the axion-gauge coupling $\int d^4x\, d^2\theta \, \Phi W^\alpha W_\alpha$. But there are no gauge or gravitational anomalies in MSSM!&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-13T15:11:36.283" CommentCount="1" />
  <row Id="5117" PostTypeId="2" ParentId="5114" CreationDate="2011-02-13T15:59:53.753" Score="6" ViewCount="" Body="&lt;p&gt;There are no purely gravitational anomalies in $D=4$. The one source of gauge-gravitational anomalies is a triangle diagram with one gauge vertex and two graviton vertices. This vanishes provided that ${\rm Tr}_L Q=0$ where the trace runs over all left-handed fermions and $Q$ is the gauge generator with the potential anomaly. In the SM the only potential nonzero contribution arises from taking $Q=Y$, where $Y$ is the generator of the&#xA;$U(1)$ part of $SU(3) \times SU(2) \times U(1)$. In the SM with standard fermion assignments this trace vanishes. In string compactifications one often gets additional $U(1)$ symmetries, and sometimes one finds these are anomalous by the above criterion. In such situations one finds a version of the Green-Schwarz mechanism involving a coupling of an axion-like mode&#xA;to the gauge field which ends up giving a mass to the $U(1)$ gauge field. The axion-like field arises in string theory as a two-form field $B$, but in $D=4$ one can dualize to a scalar via $H=dB=*da$.&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T04:46:04.080" LastActivityDate="2011-03-02T04:46:04.080" CommentCount="15" />
  <row Id="5118" PostTypeId="1" AcceptedAnswerId="5119" CreationDate="2011-02-13T16:12:40.850" Score="3" ViewCount="259" Body="&lt;p&gt;Is this just a historical artifact - that the particle physics community decided at some point to call all of the pre-oscillation physics by the name the &quot;Standard Model&quot;? The reason I ask is because I often see articles and books say something to the effect &quot;the strongest hint of physics beyond the SM are the non-zero neutrino masses&quot; as if this is something significant and mysterious - whereas from what I gathered from the answer to a &lt;a href=&quot;http://physics.stackexchange.com/questions/3020/why-do-leptons-and-quarks-mix&quot;&gt;question I asked previously&lt;/a&gt; , lepton mixing is something natural and unsurprising. So why aren't neutrino oscillations considered part of the SM? I am not asking out of any sociological interest but because I want to make sure I haven't underestimated the significance of the discovery of neutrino oscillations.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-02-18T16:23:38.067" LastActivityDate="2011-03-02T04:45:39.820" Title="Why are neutrino oscillations considered to be &quot;beyond the Standard Model&quot;?" Tags="&lt;particle-physics&gt;&lt;standard-model&gt;&lt;neutrino&gt;" AnswerCount="4" />
  <row Id="5119" PostTypeId="2" ParentId="5118" CreationDate="2011-02-13T16:26:31.143" Score="9" ViewCount="" Body="&lt;p&gt;The historical formulation of the SM involved one Higgs doublet and only renormalizable couplings, the latter being due to the focus at the time on achieving a renormalizable formulation of the weak interactions. With these restrictions neutrinos are massless and do not oscillate. To get neutrino masses you need to extend this framework either by adding&#xA;non-renormalizable dimension 5 operators, which one would naturally expect to be there in the framework of effective field theory, or you have to add renormalizable couplings involving new fields, typically including SM singlet Weyl fermions (i.e. right-handed neutrinos) and a SM singlet Higgs field. How much of an extension of the SM this really involves is subjective. There were many theoretical papers speculating on such extensions before the actual discovery of neutrino oscillations. &lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T04:45:39.820" LastActivityDate="2011-03-02T04:45:39.820" />
  <row Id="5120" PostTypeId="2" ParentId="5118" CreationDate="2011-02-13T17:43:28.790" Score="4" ViewCount="" Body="&lt;p&gt;Because neutrinos were still widely considered massless at the time &quot;The Standard Model&quot; was formulated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could argue that we're on StandardModel v2.3ish at this point and that the up-to-date release includes massive, mixing neutrinos, but that just leads to a confusion of terminology.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-02-13T17:43:28.790" CommentCount="1" />
  <row Id="5121" PostTypeId="2" ParentId="5118" CreationDate="2011-02-13T18:22:18.553" Score="2" ViewCount="" Body="&lt;p&gt;I'll add reference to the &quot;&lt;a href=&quot;http://arxiv.org/abs/0706.0684v1&quot; rel=&quot;nofollow&quot;&gt;Ten Lectures on the ElectroWeak Interactions&lt;/a&gt;&quot; by Barbiery.&#xA;In my opinion -- the best reading on the electroweak physics.&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-02-13T18:22:18.553" />
  <row Id="5122" PostTypeId="2" ParentId="2690" CreationDate="2011-02-13T18:47:27.390" Score="-3" ViewCount="" Body="&lt;p&gt;The rest mass of a particle in Classical Electrodynamics (as well as the particle charge) is a phenomenological parameter (number) in differential equations of motion. It is defined to be constant. No dynamics can change it, just by definition. It is not expressed via dynamical variables and its conservation is not due to some symmetry. $dm/dt = 0, de/dt = 0$ are experimental facts, if you like. Of course, physical models and their equations should be &lt;strong&gt;compatible&lt;/strong&gt; with such definitions. Some &quot;theoretical&quot; hypothesis are incompatible with it, for example, self-acting particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one obtains &quot;corrections&quot; to the particle mass (or/and  charge) in course of perturbative calculations, then the theory formulation is wrong. In some theories they discard such &quot;corrections&quot; and call it &quot;renormalizations&quot;. P. Dirac was unhappy about that &quot;renormalization prescription&quot; and invited researchers to change the original equations, not the results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Classical Electrodynamics the sum of particle masses $\sum m_i$ is also conserved but it does not define the system mass (the latter is just defined differently).&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-14T15:50:53.187" LastActivityDate="2011-02-14T15:50:53.187" />
  <row Id="5123" PostTypeId="2" ParentId="419" CreationDate="2011-02-13T19:17:27.427" Score="0" ViewCount="" Body="&lt;p&gt;For a single particle motion $div(v) = [div(dr/dt)] = d/dt [div(r)]$. $div(r) = 3 = const.$ But the time-derivative of a constant is zero, so $div(v) = 0$. It is not an automatic assumption, it is a calculation result ;-).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: somebody voted down. OK, &lt;strong&gt;v&lt;/strong&gt; = &lt;strong&gt;v&lt;/strong&gt;(t) so any $\partial v_i/\partial x_k = 0$ by definition of &lt;strong&gt;v(t)&lt;/strong&gt;. What else do you need as explanation?&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-13T21:19:48.203" LastActivityDate="2011-02-13T21:19:48.203" />
  <row Id="5124" PostTypeId="1" CreationDate="2011-02-13T19:23:03.320" Score="8" ViewCount="204" Body="&lt;p&gt;Let $X$ be a manifold with $G_2$ holonomy and $\Phi$ be the fundamental associative 3-form on $X$. Let $*\Phi$ be the dual co-associative 4-form on $X$. Now consider a particular associative 3-cycle $Q\in X$, which is Poincare dual to the co-associative 4-form $*\Phi$, i.e. $*\Phi=\lambda\times PD(Q)$, where $\lambda\in R_{+}$. Is there a standard mathematical notation to describe such an associative 3-cycle? Are there any topological restrictions on the properties of such a 3-cycle, e.g. can it be rigid ($S^3$ or $S^3/Z_N$)?&lt;/p&gt;&#xA;" OwnerUserId="1892" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-13T19:27:03.923" LastActivityDate="2011-03-04T01:28:00.997" Title="Question about associative 3-cycles on G2 manifolds" Tags="&lt;mathematical-physics&gt;&lt;string-theory&gt;&lt;kaluza-klein&gt;&lt;geometry&gt;" AnswerCount="3" />
  <row Id="5125" PostTypeId="2" ParentId="5124" CreationDate="2011-02-13T19:34:45.557" Score="5" ViewCount="" Body="&lt;p&gt;A very sensible (and advanced) question. Yes, it is typical for 3-cycles to have the topology you mentioned. For example, this paper&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0011190&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0011190&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;studies only associative 3-cycles that are either $S^3$ or orbifolds of it, or the hyperbolic 3-space $H^3$ and orbifolds of that. And in some cases, the spherical cycle is rigid. The text implicitly says that even with this topology, it doesn't have to be rigid, however. Also, other topologies than those listed above are almost certainly possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think that there's a standardized notation to describe associative cycles of this kind. After all, the set of such cycles is not a linear space or a group or any meaningful algebraic structure like that. It's just a set. So you could only write that something is an element of that, or some union or intersection of this set, or another set is a subset, or vice versa, and these statements are simple enough to be expressed in words. But I don't know the terminology of everyone (and not even most people in that field), however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mathematical introduction to associative 3-cycles (aside from basic, the text studies the cycles in $R^7$), using &quot;lemmas&quot; and similar words unfamiliar to a physicist, is e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.math.hmc.edu/seniorthesis/archives/2001/iweiner/iweiner-2001-thesis.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.hmc.edu/seniorthesis/archives/2001/iweiner/iweiner-2001-thesis.pdf&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://www.springerlink.com/content/8451j84w08j28432/&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/8451j84w08j28432/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first PDF is free, the second - more general review of all calibrated cycles - is not.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-13T19:53:34.480" LastActivityDate="2011-02-13T19:53:34.480" CommentCount="5" />
  <row Id="5126" PostTypeId="2" ParentId="3117" CreationDate="2011-02-13T20:02:58.013" Score="-3" ViewCount="" Body="&lt;p&gt;Let us consider an atom in its ground state with $E_0$. The energy is perfectly defined and conserved so there are &lt;strong&gt;no energy fluctuations&lt;/strong&gt; in this state. In some books one can find statements like &quot;there are coordinate and momentum fluctuations in such a state&quot;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\partial p /\partial t = [H,p]$  or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may produce an illusion of possibility to harvest the fluctuation energy. But there is nothing to harvest because the energy is not fluctuating! One can say that in the energy definition those uncertainties in the particle coordinate and momentum add up so that the fluctuating parts &quot;cancel&quot; (the sum is not fluctuating). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The meaning of &quot;fluctuations&quot; is uncertainty of the dynamical variable in QM: apart from an expectation value, there are deviations, etc., and generally there is a probability distribution. There are, however, states with certain values (eigenstates) of dynamical variables. This is the case of energy for the vacuum state, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can acquire/loose energy in course of interaction with something else, as usual, not due to fluctuations of non-commuting with the Hamiltonian variables.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-14T15:00:09.843" LastActivityDate="2011-02-14T15:00:09.843" />
  <row Id="5127" PostTypeId="2" ParentId="3983" CreationDate="2011-02-13T20:49:00.593" Score="-3" ViewCount="" Body="&lt;p&gt;Lubosh wrote: &quot;... perturbative QFT clearly works...&quot;. No, it miserably fails in the initial approximation (&quot;bare&quot; particles, no soft radiation predicted) and in course of search of the solutions by iterations (infinite corrections to the initial approximation). That is why there are so many questions to it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is comparable with experimental data is a renormalized and IR summed up result (a &quot;repaired solution&quot;) which is quite different from the original solution. And even after that there are conceptual and mathematical difficulties in the theory. Besides, there are non-renormalizable theories where attempts to &quot;repair solutions on go&quot; fail hopelessly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QFT, as a human invention, suffers from severe problems. It is very far from a desired state and needs repairing. Some times renormalizations &quot;work&quot; but not always, and we are far from the statement &quot;&lt;a href=&quot;http://motls.blogspot.com/2010/11/quantum-field-theory-has-no-problems.html&quot; rel=&quot;nofollow&quot;&gt;QFT has no problem&lt;/a&gt;&quot;. We should try other constructions. I disagree with the Lubosh's statement &quot;this is not possible&quot;, especially if with help of renormalizations and IR contribution summation we go away from initially wrong approximation and obtain reasonable results. I believe we may start from a better initial approximation, eliminate those problems, and arrive at the final results directly. Denying such opportunities is not wise, to say the least.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-13T23:37:19.350" LastActivityDate="2011-02-13T23:37:19.350" />
  <row Id="5128" PostTypeId="1" CreationDate="2011-02-13T21:28:08.917" Score="0" ViewCount="258" Body="&lt;p&gt;Can we describe our universe without it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example; Explaining a muon able to hit Earth, by its time dilation only?&#xA;Or a spinning disc contracting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it exist, and if it does, does it do so from all frames of existence, that means, if you observe a Lorentz contraction, will the person in that frame, 'experience' it too?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By that I also mean that arguing that no 'frames of reference' can be 'simultaneous' considering a 'event', won't cut to the cheese/chase whatever, here :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume a shared universe, there should be a 'real' contraction, or not be a 'real' contraction. To have it both ways would definitely hurt my head :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-14T07:03:14.397" LastActivityDate="2011-02-14T07:03:14.397" Title="Lorentz contraction, what is it?" Tags="&lt;special-relativity&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="5129" PostTypeId="2" ParentId="5128" CreationDate="2011-02-13T21:42:41.040" Score="4" ViewCount="" Body="&lt;p&gt;The Lorentz contraction is the reduction of the length of a moving object that follows from its motion - one of the main effects implied by Einstein's special theory of relativity. If the length of the object in the direction of motion is $L_0$, if measured at rest, then the other observer who sees the object as moving at speed $v$ will see the length reduced to &#xA;$$L = L_0 / \gamma$$&#xA;where the Lorentz factor $\gamma=1/\sqrt{1-v^2/c^2}$ is greater than one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Muons' average lifetime is something like 2 microseconds. By the speed of light, they would travel 600 meters in average. However, their Lorentz factor is of order $\gamma\approx 10+$. This means that a very large fraction of the muons produced at the top of the atmosphere are easily able to get through the atmosphere, something like 8 kilometers, before they decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &quot;increased ability&quot; to get very far is interpreted in the following ways:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In the Earth's frame, the muons' life processes are slowed down by time dilation, by the factor of $\gamma$. This effectively increases their lifetime by an order of magnitude, so they're able to fly not for 2 microseconds times $c$ but for 20+ microseconds times $c$ which is equal to 8 kilometers or so.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the muon's inertial frame, where the muon is moving down, the vertical thickness of the atmosphere is Lorentz-contracted by the factor of $\gamma$, so even though the muon only lives for 2 microseconds in this frame, it will also be able to get through the 8 kilometer atmosphere because the thickness seems to be less than a kilometer for the muon.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As you can see, the effect exists, is real, can't be made unreal, can't be forgotten, can't be denied, and so on. You have asked the same question many times, and I hope that I answered your question many times, too. The Lorentz contraction is a basic consequence of the special theory of relativity and there's no doubt that this prediction is valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But no, the object itself doesn't see itself contracted - just like it doesn't see itself moving. In its inertial frame, the rest frame, it can't experience the Lorentz contraction because it's at rest. The speed is $v=0$ which means $\gamma=1$. The whole point of the Lorentz contraction is that it only appears because of the relative motion of the object and the inertial frame that keeps track of the object. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I wanted to avoid the verb &quot;observe&quot; to emphasize that this is not an optical effect: the Lorentz contraction is about how long the object actually is, not how long it looks like when the finite speed of light emitted or reflected by this object is taken into account.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using your terminology, the very point of the special theory relativity - the point that gave it its name - is that the Universe is &lt;em&gt;not&lt;/em&gt; shared. The standard terminology is that the space and time are &lt;em&gt;not&lt;/em&gt; absolute. Many quantities - such as the length of the objects or the duration of a process - depend on the observer, especially his velocity. That's why we say that those quantities - including space and time - are relative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You may say that &quot;everything in the Universe&quot; is shared, but this &quot;everything&quot; has to be defined as the whole &lt;em&gt;spacetime&lt;/em&gt;. A particular slicing of the spacetime to space and time, which is needed to measure the length at a given moment, or the duration of a process, is not shared.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why this fundamental insight due to Einstein, one that represents one of the two main pillars of modern physics, is called &lt;em&gt;relativity&lt;/em&gt;, even though this theory also depends on some quantities' being invariant - and Einstein himself used to prefer the name &quot;theory of invariants&quot; for relativity, at least in the early days. However, he also adopted the name &quot;relativity&quot; later and this name has a good justification, one that was described above. Apologies for the headaches that this fact may cause to you - but it's Nature's &quot;fault&quot;, not mine and not even Einstein's.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-13T21:47:54.717" LastActivityDate="2011-02-13T21:47:54.717" CommentCount="7" />
  <row Id="5130" PostTypeId="2" ParentId="5128" CreationDate="2011-02-13T22:21:26.997" Score="1" ViewCount="" Body="&lt;p&gt;If there were a spaceship that is 200m long when you view it at rest, and then the ship accelerated to about 87% of the speed of light, you would view the spaceship's length to be 100m.  That is Lorentz contraction, but it is not a mechanical effect.  It is purely kinematic - it's an artifact of the way we make measurements.  The ship is not being crushed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The passengers inside the spaceship would still believe that the ship's length was 200m because the body of the ship is not the only thing getting shorter.  Everything inside the ship gets shorter by the same amount.  For example, a ruler would get shorter by a factor of 2.  Therefore, if people inside lined up a bunch of rulers to measure the ship, they'd get the value 200m, because from your point of view they're measuring something 100m long using rulers that are half their normal length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We might wonder why the ship doesn't suffer any ill-effects from being squashed like this.  If I flatten a soda can by a factor of two by putting it in a vice and squeezing, it bursts and crumples.  Why don't the soda cans in the spaceship burst?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I put a soda can in a vice and squeeze, the front and back of the soda can get closer together.  If I had a ruler next to the can, the distance between the front and back, measured by the ruler, would decrease.  But that is not so in the spaceship.  The soda can shrinks, and so does the ruler, and so as far as the ruler and soda can are concerned, there's no contraction going on at all.  You might be clever and say, &quot;What if I put the ruler in the vice along with the soda can?&quot;, but this would not make everything shorter - just those two.  For example, the vice has plates that are getting closer to each other when measured against the size of the vice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The source of length contraction, and the intuitive problems it causes, can be traced back to our intuition about the relativity of simultaneity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider how you would make a measurement of the spaceship's length.  What you would do is lay down a giant measuring tape, and wait until the spaceship comes up to it.  Then, right when it gets up to your tape, you would measure where the front of the spaceship is, where the back is, and subtract them to get the length of 100m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The people inside the spaceship would object to this procedure.  You have to be very careful about taking your readings of the positions of the front and back of the ship at the same time.  The problem is that in relativity, &quot;the same time&quot; is dependent on your reference frame.  The people inside the ship would say that your clocks at the front and back of your tape measure are not synchronized, and that you took your readings at different times, and that's why you erroneously measured their 200m ship as being 100m long.  (In fact, they think your tape measure is marked wrong because from their point of view it is the length-contracted thing.  The people in the space ship think that if your clocks were synchronized correctly, you'd have read 400m off your length-contracted tape measure.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get around this simultaneity disagreement, maybe you and the passengers agree on a different procedure.  You'll set up a stopwatch, and measure how much time it takes for the ship to pass you.  Then you multiply by the speed of the ship to get the length.  Doing this, you still get 100m.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time, the passengers still call foul, but for a different reason.  They think your stop watch is running slow due to time dilation.  In fact it's running slow by a factor of two, so you recorded half as much time as you should have, and incorrectly measured their 200m ship as 100m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So length contraction is unavoidable and it's real, but it's not a matter of things &quot;really getter shorter&quot; or not.  It's a matter of disagreement on how to measure lengths and times when you're moving in different reference frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What everyone will agree on is the proper time $\tau$ between two events, defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau = \sqrt{\Delta t^2 -  \left(\frac{\Delta x}{c}\right)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta t$ is the time separation between two events (you start your stopwatch, you stop it), and $\Delta x$ is the space separation between them.  Different observers will have different time or space separations, and therefore different lengths of the ship, but this proper time (or for spacelike-separate events the spacetime interval $s^2 = -(c\tau)^2$) will be agreed on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a nice parable at the beginning of Taylor and Wheeler's &lt;i&gt;Spacetime Physics&lt;/i&gt;.  Imagine two surveyors.  One of defines the cardinal directions based on geographic north, and the other based on magnetic north.  They would tell you different things when you asked how far north New York is of Los Angeles.  However, if you asked them the total distance, you'd get the same answer.  The surveyors have different coordinate systems to measure the same thing, and so the components of the separation are different.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In relativity &quot;distance&quot; becomes just one component of the spacetime separation, just like in space &quot;north/south distance&quot; is just one component of the space separation.  Observers in different reference frames measure different distance components because they are using different coordinates for the same underlying spacetime events.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-02-13T22:21:26.997" CommentCount="14" />
  <row Id="5131" PostTypeId="2" ParentId="5066" CreationDate="2011-02-13T22:21:47.717" Score="0" ViewCount="" Body="&lt;p&gt;I suppose that one could develop some sort of slingshot-like approach along side of some other approach using gravitaional pull of stars. My best guess is to look at technolgies that has yet to be discovered or developed. The current discussion seems limited to existing theory. Consider if similar discussion occurred 100 years ago about placing all of my phonograph records into a single device that I could carry with me and listen at whim to the different selection therein. We could very easily be discussing mini records, stacked in a tube with multiple needles to each disc. The techology of what would become an MP3 player would never occur to any in the discussion. This is why I enjoy - good - science fiction that tries to propose theory that is far beyond current and conventional understanding. I'm not sure but I would imagine that a better understanding of light itself may lead to a propulsion approach that far superior that what most consider.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-13T22:21:47.717" />
  <row Id="5132" PostTypeId="1" AcceptedAnswerId="5169" CreationDate="2011-02-13T23:09:39.433" Score="4" ViewCount="159" Body="&lt;p&gt;I am currently taking a graduate level class on lasers. The primary focus of the class is on the design and engineering aspects of lasers, e.g. resonator design. However the first portion of the class is an overview of the quantum mechanics of laser processes in semi-classical terms: transition probabilities for a 2-level dipole atom in an external (classical) field, perturbation theory, and matrix QM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an undergraduate degree in physics, so the basics of QM are familiar to me, but it has been years since I seriously applied any of that knowledge. In addition, the professor is moving through this material very rapidly, as it is not the primary focus of the class. I can follow well enough that I'm not worried about my grade, but I'm feeling a bit cheated out of a more thorough understanding of this topic, which I would really like to have. Not only is this subject matter an appropriate level of advancement over my existing knowledge, it is also the basis for something (lasers) which I will likely spend a career dealing with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody recommend a text which covers this topic, and would be appropriate for somebody with my background? I would value a clearly written description over absolute rigor, but I am by no means looking for a laypersons description of QM. For example, I am quite comfortable already with the notation and concepts in the textbook &quot;Quantum Mechanics of atoms, molecules, solids, nuclei, and particles&quot; By R. Eisberg and R. Resnick. The ideal reference would be at a similar level of sophistication, but provide a more in-depth look at laser processes in particular.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-02-15T12:11:14.333" Title="What is a good reference for the quantum mechanical description of lasers?" Tags="&lt;quantum-mechanics&gt;&lt;laser&gt;&lt;reference-request&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="5133" PostTypeId="2" ParentId="5128" CreationDate="2011-02-13T23:26:43.737" Score="0" ViewCount="" Body="&lt;p&gt;You're making an issue of something that can be seen in everyday life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone in a car driving past you will say they waved to you at (x1',y1', z1', t1') and crashed at (x1',y1', z1', t2') using their space and time coordinates, whereas you claim  these events occured at (x1, y1, z1, t1 = t1') and (x2, y2, z2, t2 = t2') in your frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you see the difference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a difference in the space interval you both measure between these two events, and so you could view it as a space contraction of some sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Special relativity takes it a stage further with the introduction of a time contraction between events, between different frames.&lt;/p&gt;&#xA;" OwnerUserId="1345" LastActivityDate="2011-02-13T23:26:43.737" CommentCount="2" />
  <row Id="5134" PostTypeId="1" AcceptedAnswerId="5135" CreationDate="2011-02-14T00:40:04.693" Score="1" ViewCount="214" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hopf_algebra&quot; rel=&quot;nofollow&quot;&gt;Hopf algebra&lt;/a&gt; appears in recent papers that systematize renormalization of quantum field theory (QFT). For example see &lt;a href=&quot;http://www.google.com/search?q=connes+hopf+site%3Aarxiv.org&amp;amp;ie=utf-8&amp;amp;oe=utf-8&amp;amp;aq=t&amp;amp;rls=org.mozilla:en-US:official&amp;amp;client=firefox-a&quot; rel=&quot;nofollow&quot;&gt;Connes' work and citing papers&lt;/a&gt; or a paper referenced &lt;a href=&quot;http://physics.stackexchange.com/questions/4068/formalizing-quantum-field-theory&quot;&gt;here on PSE&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R. E. Borcherds, &quot;Renormalization and quantum field theory&quot;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/1008.0129&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1008.0129&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to be an effort that is separate from string theory, but I wonder if they are compatible. Since &quot;compatible&quot; is perhaps too loose of a term, I'm asking:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do string theory and Hopf algebra renormalization seem to have no intersection?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the reader feels the question could be improved by rewriting, please so assist.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-02-14T01:23:59.097" LastActivityDate="2011-03-02T03:54:46.387" Title="Why do string theory and Hopf algebra renormalization seem to have no intersection?" Tags="&lt;quantum-field-theory&gt;&lt;string-theory&gt;&lt;renormalization&gt;&lt;hopf-algebra&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5135" PostTypeId="2" ParentId="5134" CreationDate="2011-02-14T01:00:23.327" Score="3" ViewCount="" Body="&lt;p&gt;Why do you want to renormalize a theory that gives UV finite answers?&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T03:54:46.387" LastActivityDate="2011-03-02T03:54:46.387" CommentCount="1" />
  <row Id="5137" PostTypeId="2" ParentId="5124" CreationDate="2011-02-14T01:30:11.730" Score="3" ViewCount="" Body="&lt;p&gt;I'm not aware of a standard notation. Associative 3-folds can be rigid, but need not be.&#xA;Although it doesn't address your question in any kind of generality, associative 3-cycles can be understood quite easily for a special class of G2 manifolds of the form $(X \times S^1)/Z_2$ where $X$ is a Calabi-Yau 3-fold and the involution acts as $(\sigma_X,-1)$ where $\sigma_X$ is an anti-holomorphic involutive isometry of $X$. There are then &quot;holomorphic&quot; 3-cycles of the form $(\Sigma_2^- \times S^1)/Z_2$ where $\Sigma_2^-$ is a holomorphic curve mapped to $- \Sigma_2^-$ by $\Sigma_X$. This is rigid if $\Sigma_2^-$ is rigid.&#xA;There are also &quot;Lagrangian 3-cycles&quot; of the form $\Sigma^+/Z_2$ where $\Sigma^+$ is a cycle mapped to plus itself by $\sigma_X$. An example of a rigid &quot;Lagrangian 3-cycle&quot; in a G2 manifold is given in Appendix C of hep-th/9907026.&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T03:54:23.437" LastActivityDate="2011-03-02T03:54:23.437" CommentCount="1" />
  <row Id="5139" PostTypeId="2" ParentId="419" CreationDate="2011-02-14T01:50:38.963" Score="2" ViewCount="" Body="&lt;p&gt;Despite your notation, I'm guessing that you're asking about partial derivatives. To tex a partial derivative use &quot;\partial x&quot; as in &quot;$\partial x$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sort of confusion arises in Lagrangian calculations and I'll bet that's where you came upon it. The effect arises from our mathematics. It appears with only a single dimension x so I'll illustrate the effect that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we left everything in terms of position (with velocity defined as $dx/dt$, then the gradient of the velocity wouldn't be zero. That's because the velocity does, in fact, depend on position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, when we do classical mechanics with Lagrange's equation, we think of the Lagrangian as being a function of position, velocity, and perhaps time. So we write it as $L(x,\dot{x},t)$. The effect of this way of looking at the problem is that we double the number of variables (from $x$ to $x,\dot{x}$), but we eliminate the need for second derivatives. This makes the problem actually easier to solve, but to be consistent, we have to make our partial derivatives (i.e. partials with respect to position or velocity) apply only to the position or velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can quickly verify that the Lagrangian&lt;br&gt;&#xA;$L(x,y,\dot{x},\dot{y}) = 0.5m\dot{x}^2 +0.5m\dot{y}^2 - mgy$&lt;br&gt;&#xA;leads to the equations of motion that you expect for a body of mass m moving in a gravitational field:&lt;br&gt;&#xA;$m\ddot{x} = 0,\;\;m\ddot{y} = -g$,&lt;br&gt;&#xA;and that this happens only if you follow the rules you've been given for calculating partial derivatives. Maybe this will ease your mind. Also see:&lt;br&gt;&#xA;&lt;strong&gt;Why does Calculus of Variations work?&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/885/why-does-calculus-of-variations-work&quot;&gt;Why does calculus of variations work?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the place where partial derivatives used to bother me the most was in thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in short, we don't automatically assume it. It happens when we use math in certain ways.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-14T01:50:38.963" CommentCount="2" />
  <row Id="5141" PostTypeId="1" AcceptedAnswerId="5151" CreationDate="2011-02-14T02:56:46.110" Score="3" ViewCount="181" Body="&lt;p&gt;I was thinking the other day about what sort of exotic materials one would find in the depths of a planet. I have heard theories about how an enormous diamond might be found in the centre of gas giants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current theories have a ball of iron at Earth's core - would that be like the iron up here on the surface or does the pressure change its structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's the question: Aside from a carbon becoming diamond, what materials change structure under intense pressure? &lt;/p&gt;&#xA;" OwnerUserId="1910" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-02-14T13:12:57.333" LastActivityDate="2011-02-15T19:08:41.257" Title="Aside from carbon, what materials change structure under intense pressure?" Tags="&lt;pressure&gt;" AnswerCount="5" />
  <row Id="5142" PostTypeId="2" ParentId="5141" CreationDate="2011-02-14T03:21:00.730" Score="3" ViewCount="" Body="&lt;p&gt;A pretty recent example is given by the perovskite-to-&lt;a href=&quot;http://en.wikipedia.org/wiki/Post-perovskite&quot; rel=&quot;nofollow&quot;&gt;&quot;post perovskite&quot;&lt;/a&gt; structural change of magnesium silicate at sufficiently high temperature and pressure. It doesn't seem particularly exotic, but perovskite materials are exceptionally stable and ubiquitous, and discovering a new high-pressure phase was a major coup for deep earth geophysics.&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2011-02-14T03:21:00.730" CommentCount="1" />
  <row Id="5143" PostTypeId="2" ParentId="5141" CreationDate="2011-02-14T04:57:09.537" Score="6" ViewCount="" Body="&lt;p&gt;Water has a &lt;a href=&quot;http://www.lsbu.ac.uk/water/phase.html&quot; rel=&quot;nofollow&quot;&gt;very complicated phase diagram&lt;/a&gt; with several high pressure states.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-02-14T04:57:09.537" />
  <row Id="5144" PostTypeId="1" AcceptedAnswerId="5146" CreationDate="2011-02-14T05:25:10.213" Score="4" ViewCount="76" Body="&lt;p&gt;It is known that in F-theory compactifications on CY 4-folds one can get gauge groups with very large ranks. The largest single factor* gauge group for compact CY 4-folds I found in the literature is the mind-blowing SO(7232) in this paper: &lt;a href=&quot;http://arxiv.org/abs/hep-th/9706226&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9706226&lt;/a&gt; . One can imagine that there exist 4D compactifications of F-theory, which don't have Heterotic duals, with gauge groups that are even bigger in rank. Now, my question is whether there exists an upper bound on the ranks of single factors* in product gauge groups in either F-theory compactifications on CY 4-folds or in G2 holonomy compactifications of M-theory? An order of magnitude answer, if known, would be good enough :) . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) Thanks to @Luboš Motl for correcting my question!&lt;/p&gt;&#xA;" OwnerUserId="1892" LastEditorUserId="1892" LastEditorDisplayName="" LastEditDate="2011-02-14T14:24:00.923" LastActivityDate="2011-02-14T14:24:00.923" Title="Is there an upper bound on the gauge group rank in F-theory compactifications on CY 4-folds?" Tags="&lt;mathematical-physics&gt;&lt;string-theory&gt;&lt;geometry&gt;" AnswerCount="1" />
  <row Id="5145" PostTypeId="2" ParentId="5141" CreationDate="2011-02-14T07:58:51.940" Score="2" ViewCount="" Body="&lt;p&gt;This is a common feature of many (if not most) crystalline materials. Iron in the inner core of Earth presumably has $\epsilon$ structure. To see the difference between $\alpha$ and $\epsilon$ structures of iron, see pictures at the bottom of &lt;a href=&quot;http://cep.xor.aps.anl.gov/newsletter/martinHighlight/index.html&quot; rel=&quot;nofollow&quot;&gt;this webpage&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="701" LastActivityDate="2011-02-14T07:58:51.940" CommentCount="4" />
  <row Id="5146" PostTypeId="2" ParentId="5144" CreationDate="2011-02-14T08:01:34.033" Score="3" ViewCount="" Body="&lt;p&gt;Stringpheno, you are slightly confused about - and you underestimate - the gauge group of the model you linked to. Its rank is not 7232/2, as it would be if the gauge group were SO(7232). The full gauge group of the model actually has rank 302,896 - about 100 times higher than what you say - and SO(7232) is just one of the 251 factors of the gauge group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't remember the exact number but I think that this gauge group is still far from being the world record when it comes to the rank of gauge groups in F-theory &lt;em&gt;geometric engineering&lt;/em&gt;, the term you may want to search for if you're interested in these fun constructions. I believe that there are 6-dimensional compactifications with 8 supercharges that admit even simple groups (or at least products of a smaller number of very large factors) as gauge groups which are larger than the product above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can clearly exist &quot;infinite&quot; gauge groups but if there is a limit for finite ranks is unknown to me. I believe that these very high gauge groups only appear in F-theory, not in M-theory on G2 manifolds, and it's arguably not just due to the limited creativity of the geometric engineers.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-14T09:41:35.167" LastActivityDate="2011-02-14T09:41:35.167" CommentCount="3" />
  <row Id="5147" PostTypeId="1" CreationDate="2011-02-14T08:52:46.357" Score="0" ViewCount="181" Body="&lt;p&gt;I'm a mathematician studying Arnold's Mathematical Methods of Classical Mechanics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On p. 83 the following definition is given.&#xA;Let $M$ be a differentiable manifold, $TM$ its tangent bundle, and $L:TM \to \mathbf{R}$ a differentiable function. A map $\gamma: \mathbf{R} \to M$ is called a motion in the lagrangian system with configuration manifold $M$ and lagrangian function $L$ if $\gamma$ is an extremal of the functional $\Phi(\gamma)=\int_{t_0}^{t_1}L(\gamma^&amp;#39;)dt$,&#xA;where $\gamma&amp;#39;$ is the velocity vector $\gamma&amp;#39;(t) \in TM_{\gamma(t)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Is it correct then that geodesics in $M$ (if $M$ is Riemannian) are just a special case of this construction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Homogeneous spaces, i.e. manifolds of the form $G/K$, where $G$ is a Lie group and $K$ a closed subgroup, provide interesting examples of manifolds. &#xA;What would a physical system look like which has $G/K$ as a configuration space? I remember hearing something along the lines &quot;global $G$-invariance, local $K$-invariance&quot; but I'm not sure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what interpretation do the geodesics then have in this model?&lt;/p&gt;&#xA;" OwnerUserId="1912" LastActivityDate="2011-02-14T12:02:09.747" Title="Geodesics and trajectories" Tags="&lt;lagrangian-mechanics&gt;" AnswerCount="2" />
  <row Id="5148" PostTypeId="2" ParentId="5147" CreationDate="2011-02-14T09:35:41.160" Score="3" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes, the geodesics are obtained in this way if $L(\gamma&amp;#39;)$ is the proper length of the vector. However, the Arnold construction is not meant to describe geodesics only - but any motion of particles with any configuration space (with velocities added). The Lagrangian doesn't have to be identified with a proper length in general, and for most systems in mechanics, it is not.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;These quotients are omnipresent in supergravity theories, among others. For example, 11-dimensional supergravity compactified on a 7-torus and dimensionally reduced is a 4-dimensional theory known as $N=8$, $d=4$ supergravity with 133-63 = 70 scalars spanning the $E_{7(7)}/SU(8)$ coset. The global and local interpretation of the groups is exactly as you suggest: the $E_{7(7)}$ exceptional group in the numerator is a global symmetry while the denominator, $SU(8)$, is a local symmetry. It's because you may define an action of $E_{7(7)}$ on the whole $E_{7(7)}/SU(8)$ quotient (it's a symmetry of the rest of the theory, too). However, most of the generators are spontaneously broken. Only the generators of $SU(8)$ are unbroken: the vacuum state is invariant under them which is why you don't get new vacua by $SU(8)$ transformations, so there are no new scalars that would label these non-existent new vacua. That's why it's quotient.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The remaining generators of $E_{7(7)}(R)$ change the vacuum in supergravity. They change it to a &quot;different vacuum&quot; whose physics is, however, completely equivalent: it's a global symmetry. I added the label $R$ to emphasize it is a continuous group. In quantum gravity, however, electric and magnetic charges have to be quantized because of the Dirac quantization rule. They have to belong to a lattice and the lattice is changed under almost any $E_{7(7)}(R)$ transformation: so these transformations map a vacuum to a physically distinct vacuum and $E_{7(7)}$ is no longer a global symmetry. Only its discrete subgroup of elements that preserve the lattice, the $E_{7(7)}(Z)$ group, remains an exact symmetry of M-theory on $T^7$. It's also known as the U-duality group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, maybe, I should have avoided supergravity. It would be enough to mention the Goldstone bosons. Scalars span a $G/K$ space is $G$ is the symmetry group of the action while $K$ is the unbroken subgroup. For each broken generator, you obtain one boson - a coordinate parameterizing the quotient. The supergravity story was an unnecessarily advanced variation of the same topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me mention a simple nontrivial example. Imagine that you describe the orientation of an object by an element of $SO(3)$. Now, imagine that the object is an axially symmetric gyroscope that preserves the $SO(2)$ symmetry. The actual &quot;shape&quot; of the gyroscope in space won't be affected by this $SO(2)$ symmetry, so the orientation will only be given by an element of $SO(3)/SO(2)$ which is isomorphic to a two-sphere, $S^2$.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-14T09:35:41.160" />
  <row Id="5149" PostTypeId="1" CreationDate="2011-02-14T09:45:10.530" Score="5" ViewCount="195" Body="&lt;p&gt;This is something I don't ever seem to hear about, except regarding QCD (&quot;lattice QCD&quot;). What about QED? Is numerical integration always inferior to hand-calculating Feynman diagrams in perturbation theory? What about numerical simulation of the full time-evolution of a quantum field (rather than just calculating correlation functions)? Here are some specific questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Numerical solutions to the Schrodinger equation. Has the time evolution of a complex quantum system ever been numerically simulated in order to shed light on the measurement problem? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Has the time evolution of a quantum field ever been numerically simulated? What about QED? If not, how much more computing power is needed? Could it be useful? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Ignoring lattice QCD, is numerical integration ever useful for calculating physical quantities in the SM? For example analytic computation of the terms in perturbation theory yields a value of the anomalous magnetic moment of the electron to something like 10 decimal places. How many decimal places would straight-up brute-force numerical integration get you? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) What is the current state of the field? Are there any numerical-analysis-aided-achievements in theoretical particle physics outside lattice QCD worth noting?&lt;/p&gt;&#xA;" OwnerUserId="1247" LastActivityDate="2011-02-14T10:01:46.083" Title="What is the status of applying numerical analysis to QM/QFT problems" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;computational-physics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="5150" PostTypeId="1" CreationDate="2011-02-14T09:45:16.460" Score="2" ViewCount="222" Body="&lt;p&gt;Did spacetime started with the Big bnag? I mean, was there any presence of this space-time we are experiencing now before big bang? And could there be presence/existence os any other space-time before the big bang?&lt;/p&gt;&#xA;" OwnerUserId="1913" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-02-14T10:23:27.277" LastActivityDate="2011-02-15T10:46:11.740" Title="Did spacetime start with the Big bang?" Tags="&lt;cosmology&gt;&lt;spacetime&gt;&lt;big-bang&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="5151" PostTypeId="2" ParentId="5141" CreationDate="2011-02-14T09:47:56.457" Score="3" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;theories about how an enormous diamond&#xA;  might be found in the centre of gas&#xA;  giants.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hmmm, because Diamond is still somewhat away from &#xA;a closest package, the giant planets may contain a &#xA;sort of carbon denser than diamond, metallic &#xA;in properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, all substances (not only crystalline) &#xA;will transform into something with a densest package &#xA;of atomic cores and a delocalized electron gas: &#xA;metals, if only pressure is high enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For hydrogen, some experimental research was/is &#xA;done to find this &quot;metallic&quot; form. I remember an article &#xA;in SCIAM some years ago. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course all this is valid only if temperature is low enough. &#xA;(For that reason we speak of planets cores!)&#xA;But at the end of pressure scales in neutron stars all materia &#xA;transforms to neutrons irrespective of enormous temperatures,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;would that be like the iron up here on&#xA;  the surface or does the pressure&#xA;  change its structure?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One difference to &quot;surface iron&quot; is obvious: the innermost core is solid, in spite &#xA;of the temperature. And one has to be aware that the iron in the core is not &#xA;pure, there will be some other metals and carbon in the melt of outer core. &#xA;At the border of outer core and mantle sometimes sulfides are assumed. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-02-14T13:23:41.943" LastActivityDate="2011-02-14T13:23:41.943" />
  <row Id="5152" PostTypeId="2" ParentId="5149" CreationDate="2011-02-14T10:01:46.083" Score="4" ViewCount="" Body="&lt;p&gt;You can't define QED as a strict continuum limit of a &quot;lattice QED&quot; simply because pure QED is inconsistent at extremely short distance scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fine-structure constant &quot;runs&quot; and at energies of the form $\exp(137 C)$ times mass of the electron, where $C$ is a number of order one and $137$ stands for the inverse fine-structure constant, the coupling diverges and prevents one from defining the theory at even higher energies (or shorter distances).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Landau problem is completely invisible to all orders of the perturbative expansion which is why QED is perturbatively well-defined. In some sense, the Landau pole - the scale where QED becomes inconsistent - is the closest analogy of the &quot;characteristic mass scale&quot; for QED which plays a similar role as the QCD scale for QCD. It's the scale where the coupling becomes of order one and wants to run to infinity (even though QED and QCD make it run in the opposite directions). That's too bad because physics becomes inconsistent at the QED Landau pole scale - a very different situation from asymptotically free QCD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lattice QCD itself is pretty demanding - even though one only deals with a rather limit number of sites. The size of the box as well as the size of the lattice spacing are not &quot;spectacularly different&quot; from the QCD scale, or the radius of the proton. However, QED doesn't have any privileged scale of this kind (except for the bad Landau pole scale), so there's no interesting regime or scale at which the lattice QED would be useful, anyway. Note that lattice QCD really becomes powerful to study the effects for which the strong coupling constant surpasses one or so (hadrons and their collisions, for example). That never happens for the QED fine-structure constant. The latter is not confining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one used an approximation that circumvents the Landau-pole problem (this approximation would inevitably have to remain silent about all nonperturbative effects because they can't be extracted from QED itself), and if he fine-tuned the UV couplings properly, he could &quot;numerically integrate&quot; QED, at least in principle. All the local Green's functions in QED are perturbatively well-defined which means that one could in principle integrate QED over finite periods of time, too. But because of technical limitations, nothing like that can be done in practice, at least not in any helpful way. The only thing that one could realistically get by this method would be to derive the RG flows and changes of the coupling constant over an order of magnitude or two (at the energy scale): you would substitute the QED Lagrangian at your lattice spacing scale, and derive the effective QED laws at a longer scale - which would be pretty much the same laws but with modified couplings and other parameters. And these things can be derived theoretically, without any help from computers, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that it's fair to say that only QCD and its equally confining cousins can be - and has been - usefully studied by these numerical techniques. And even in that case, one still deals with the arguably eternal problems of QCD - problems with fermion doubling and other fermion problems; problems to naturally impose all of supersymmetry, and so on.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-14T10:01:46.083" />
  <row Id="5153" PostTypeId="2" ParentId="5147" CreationDate="2011-02-14T10:56:20.000" Score="3" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;(1) Is it correct then that geodesics in M (if M is Riemannian) are just a special case of this construction?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, it is correct to say this. Give a Riemannian manifold $M$ with a metric $g_{ab}$ &lt;em&gt;and&lt;/em&gt; an affine connection $\Gamma^a_{bc}$, one can construct various geometric invariants on $M$ such as the Riemann and Ricci tensors. Now for a point particle propagating on this &lt;strong&gt;static&lt;/strong&gt; manifold, i.e without taking General Relativity into account, one could write various Lagrangians of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ L = \sqrt{|g|} ( K_m - V_m) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $|g|$ is the absolute value of the determinant of $g_{ab}$ and $K_m$ and $V_m$ are kinetic and potential terms respectively which depending on the position and velocity of the particle or more generically is energy-momentum n-vector $j^a$. A generic example of the kinetic term in this case is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ K_m := \frac{1}{2} \left( \partial^a j_a \right)^2 + \frac{1}{2}(\partial^a j^b \partial_a j_b) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $V_m$ is some polynomial in $j^a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we wish to incorporate the dynamics of the manifold itself (i.e. move into the domain of GR) then our Lagrangians must contain &quot;kinetic&quot; and &quot;potential&quot; terms for the various geometric invariants. These quantities however lack a simple expression for a general manifold and are subsumed into the Ricci scalar:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ L = \sqrt{|g|}(K_m - V_m + \mathcal{R}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the Ricci scalar $\mathcal{R} = g^{ab}\mathcal{R}_{ab} = g^{ab}g^{cd} \mathcal{R}_{acdb} $ is the trace of the Ricci tensor which in turn is a partial trace of the Riemann tensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, geodesics on a Riemannian manifold $M$ with a metric $g_{ab}$ &lt;em&gt;and&lt;/em&gt; an affine connection $\Gamma^a_{bc}$ are given by solutions (integral curves) of the of the geodesic equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v^a\nabla_a v^b = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $v := v^a \partial/\partial_a $ is a vector field on $M$, the covariant derivative is given in terms of the affine connection $\nabla_a v_b = \partial_a v_b + \Gamma^c_{ab}v_c $ and indices are raised and lowered with the metric $ v^a = g^{ab} v_b $. Your question amount to asking if there exists a Lagrangian $L(v^a)$ whose equation of motion is this geodesic equation. The answer is 'yes'. The following Lagrangian&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ L = \sqrt{g^{ab} v_a v_b} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is nothing more than the proper distance along the particle's wordline, yields the geodesic equation on variation w.r.t $\delta v_a$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(2) Homogeneous spaces, i.e. manifolds of the form G/K, where G is a Lie group and K a closed subgroup, provide interesting examples of manifolds. What would a physical system look like which has G/K as a configuration space? I remember hearing something along the lines &quot;global G-invariance, local K-invariance&quot; but I'm not sure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An example of a physical system with this configuration space is MacDowell-Mansouri gravity. A beautifully written reference for this is the review paper &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0611154&quot; rel=&quot;nofollow&quot;&gt;MacDowell-Mansouri gravity and Cartan geometry&lt;/a&gt; by Derek Wise, a former student of John Baez.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There one starts with a five-dimensional spacetime with symmetry group $G$ which can be deSitter, Anti-deSitter or Minkowski with a lie-algebra valued antisymmetric tensor $B$ and the curvature of the gauge connection $F$. The action is that of a topological BF theory given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S_{BF} = \int tr B \wedge F  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can identify $B:= B_{\mu\nu}^{IJ}$ with the wedge product of two fermion fields: $ B^{IJ} = \psi^I \wedge \psi^J $ which transform in the fundamental representation of the gauge group. Furthermore, as in the case of the BCS mechanism, let us assume that the dynamics of the system contains a four-fermion term which acquires a vev (vacuum expectation value) due to the formation of a condensate of these fermions. Such a term results in the action:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S&amp;#39;_{BF} = \int tr B \wedge F - \frac{G\Lambda}{6} B \wedge \star B $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\star B$ is the Hodge dual of $B$. $G$ and $\Lambda$ are Newton's constant and the cosmological constant respectively. The formation of the condensate can be physically described by writing the five-dimensional gauge connection in the following form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ {}^5 A = &#xD;&#xA;\left( \begin{array}{cc}&#xD;&#xA;{}^4 A &amp;amp;&amp;amp; \frac{1}{l}\{e^0,e^{i} \} \\&#xD;&#xA;\frac{1}{l} \{ e^0, \epsilon e^i \} &amp;amp;&amp;amp; 0&#xD;&#xA;\end{array} \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where ${}^4 A$ is a four-dimensional connection and $\{e^0,e^{i}\}$ (where $ i \in {1,2,3}$) is a vier-bien (tetrad). $\epsilon = \{-1,0,1\}$ for $G = \{ SO(4,1)$ (deSitter), $ISO(3,1)$ (Minkowski), $SO(3,2)$ (Anti-deSitter) $\}$ respectively. The group $H$ in each case is $SO(3,1)$ and the resulting theory has gauge group $\{SO(4,1)/SO(3,1), ISO(3,1)/SO(3,1),SO(3,2)/SO(3,1)\}$ respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting theory describes general relativity in &lt;em&gt;four&lt;/em&gt; dimensions with the addition of topological terms such as the Nieh-Yan and Pontyargin terms in the action. For more details see Wise's excellent paper reference above and also the seminal paper (&lt;a href=&quot;http://arxiv.org/abs/hep-th/0501191&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;) by Friedel and Starodubstev who first proposed this formulation of the MacDowell-Mansouri mechanism.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-14T12:02:09.747" CommentCount="2" />
  <row Id="5154" PostTypeId="2" ParentId="5150" CreationDate="2011-02-14T11:22:11.780" Score="2" ViewCount="" Body="&lt;p&gt;Gulshan,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main theory which describes Space-Time and from which the prediction of the Big Bang comes is called General Relativity, from Einstein. This theory has several mathematical solutions and cosmologists worked to determine the most accurate. There are a class of alternatives but they all have the property that the equations which describe this solution have a singularity at T=0. Furthermore when this situation is examined physically it seems that there is a high density of all the Universe's matter there and then. So it is called the Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Singularity means that some terms become infinite and others unhelpfully become zero. So General Relativity has not been able to predict (or retrodict) what happens before, or how this process really began. The general assumption has been that it was some kind of giant Quantum Event. This assumption, when explained using a more complete theory of Quantum Gravity, may yet be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in the last few years, several mathematical cosmologists have taken seriously the idea that there was a Pre-Big Bang. Part of the reason for this may be because of the Cosmic Background Radiation data from satellites like WMAP. This data shows larger scale structure in the early universe than the older theories would have predicted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular Roger Penrose has developed a view that the period since the Big Bang should be called an aeon, and that there were earlier aeons each infinitely long. This makes the Big Bang a kind of transition period between two aeons. The theory is speculative in several respects, but it is based on some mathematical constructions in General Relativity. This theory is called Conformal Cyclic Cosmology (CCC for short).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A recent short paper &lt;a href=&quot;http://arxiv.org/ftp/arxiv/papers/1011/1011.3706.pdf&quot; rel=&quot;nofollow&quot;&gt;Concentric circles in WMAP data may provide evidence of violent pre-Big-Bang activity&lt;/a&gt; gives the general idea. Although it is technical in places it demonstrates the kind of evidence that is motivating this theory. There are references in that paper to a book and other papers which describe that theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other theories around too, which suggest a pre-Big Bang model, perhaps other answers will mention those.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-02-14T11:27:46.013" LastActivityDate="2011-02-14T11:27:46.013" CommentCount="3" />
  <row Id="5155" PostTypeId="2" ParentId="5150" CreationDate="2011-02-14T13:55:33.083" Score="-1" ViewCount="" Body="&lt;p&gt;Bear in mind that the real solution for the de Sitter spacetime is a scale factor&#xA;$$&#xD;&#xA;a(t)~=~\sqrt{3/\Lambda}cosh(t\sqrt{\Lambda/3})&#xD;&#xA;$$&#xA;for $\Lambda$ the cosmological constant.  $\Lambda$ was very large in the early universe, and it is not unreasonable to think that the universe was connected by the “throat” to the other half of the hyperboloid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is similar to the problem of the white hole, which is the other half of the Schwarzschild metric.  We generally ignore that, and so too most often the de Sitter spacetime is physically considered to be an exponentially expanding space where $cosh(x)~\simeq~e^x$ for large $x$.  So we start with the FLRW energy equation&#xA;$$&#xD;&#xA;\Big(\frac{\dot a}{a}\Big))^2~=~{8\pi G\rho}{3}~–~\frac{k}{a^2}&#xD;&#xA; $$&#xA;Here &quot;dot&quot; means time derivative.  This equation can be derived using Newton's laws, or the energy of a projectile moving in a gravity field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other half of the hyperboloid might physically involve an instanton state, or tunneling state. The Schrodinger equation for a particle moving in one dimension with some potential $V$ is&#xA;$$&#xD;&#xA;i\hbar\frac{\partial\phi}{\partial t}~=~-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}~-~V(x)\psi&#xD;&#xA;$$&#xA;If we consider a stationary case with a phase $\psi(x,t)~=~\psi(x)e^{-iEt/\hbar}$ the left had term just becomes $E\psi$, where $E$ is the energy of the particle.  Now let us rearrange things so that &#xA;$$&#xD;&#xA;\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}~=~(E~-~V(x))\psi&#xD;&#xA;$$&#xA;For a particle moving in space we set $\psi(x)~\sim~exp(ikx)$, do the two derivatives and cancel out the ψ(x).&#xA;$$&#xD;&#xA;k^2~=~(2m/\hbar^2)(E~–~V).&#xD;&#xA;$$&#xA;The funny thing is that for $V~&amp;gt;~E$ we have an imaginary $k$.  This means that the kinetic energy is in a funny sense negative, which is not something you expect in classical mechanics.  For a system of this sort it is in a classically forbidden region, and in more general systems there may be some dispersion $\omega~=~\omega(k)~=~vk~+~\dots$, which leads to an imaginary frequency.  The phase for the system is $exp(i\phi)~=~exp(i\omega t)$.  The imaginary quantity associated with the angular frequency ω may be reassigned to the time $t$, that is a mathematical triviality.  So in some of these problems it is useful to use this and work with imaginary time, or what is sometimes called Euclideanized time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the other half of the hyperboloid might be physically modeled to be a tunneling of a cosmology across a potential boundary.  This might then be an instanton due to a “blob” of vacuum energy which quantum escapes from another spacetime.  So from that setting if may be seen that the universe has some sort of precursor. or is quantum mechanically tied to another spacetime.  This might then be seen in the setting of a multi-cosmological universe or within what is called the multiverse.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-14T13:55:33.083" CommentCount="11" />
  <row Id="5156" PostTypeId="1" AcceptedAnswerId="5175" CreationDate="2011-02-14T15:13:07.820" Score="3" ViewCount="224" Body="&lt;p&gt;I have several questions regarding the &quot;big bounce&quot; theory. It appears to be popular among LQG researchers. My questions are as as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) How one reconciles it with the fact that it is now experimentally established that the universe is expanding in an accelerating manner?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The entropy of the universe should have been very very low (maybe zero) at the time of the big bang/ big bounce. What was the entropy &quot;before&quot; that event? Was it even lower? If it was lower then doesn't that mean entropy decreases indefinitely? On the other hand if we accept the entropy was higher then doesn't that violate the second law of thermodynamics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) What observational evidence can confirm pre-bounce physics? According to GR there are some points near the bang/bounce where one simply can't easily define time in a meaningful way. Even for a quantum gravity theory, is it possible for any theory to extrapolate itself &quot;before&quot; the &quot;bounce&quot;? If there are no observational consequences after the bounce shouldn't one apply Occam's razor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/HeKrp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-02-14T23:30:21.417" Title="Question regarding the validity of the big bounce" Tags="&lt;cosmology&gt;&lt;quantum-gravity&gt;&lt;big-bang&gt;&lt;loop-quantum-gravity&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="5157" PostTypeId="2" ParentId="5156" CreationDate="2011-02-14T15:55:08.383" Score="2" ViewCount="" Body="&lt;p&gt;1) Isn't really a problem--aside from inflation, you'd expect matter to be dominating the cosmological constant in the very early universe--we have only recently entered the era where the CC dominates matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) is an open research question right now--no one really knows the answer to this.  I guess the hope would be that the bounce event somehow sorts out microstates or something.  But, to my knowledge, no one, in any field, has really satisfactorily solved the entropy problem in cosmology.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) people have talked about there maybe being some sort of fingerprint on the CMB coming from a bounce--I heard talk that there would be rings in the CMB from pressure waves generated by the bounce.  There was a small amount of hubhub when people supposedly saw these rings, which were later dismissed as a confirmation bias problem.  That said, I don't think the phenomenology of this has been taken too far--there may well be observational fingerprints to this that no one has thought of.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you take the claims of people waiting a big bounce too seriously--saying that a model predicts that this can happen is not the same thing as saying that this definitely &lt;b&gt;did&lt;/b&gt; happen, and that the model is correct.  And it's not uncommon to have a big picture prediction to precede observational evidence for that big picture prediction.  Even if it turned out that the big bounce was unobservable, this model might still tell you something about the behaviour of plasma heated to Planck temperatures, or something.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-02-14T15:55:08.383" CommentCount="3" />
  <row Id="5158" PostTypeId="1" AcceptedAnswerId="5260" CreationDate="2011-02-14T16:56:23.580" Score="5" ViewCount="212" Body="&lt;p&gt;My sister is in 10th grade. She doesn't seem to understand the concept of time varying functions (current, light, sound wave forms etc). I explained her in easiest possible terms. She got it but not very comfortable with the concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, instead of explaining. I simply wanted her to take a stopwatch &amp;amp; measure &lt;em&gt;&quot;something&quot;&lt;/em&gt; and tabulate all the values. Then plot them on a graph paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm thinking of that &quot;something&quot; that she can measure. I want something that has following properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It must vary every second or two. (because it would be more intuitive than something that varies after a while).&lt;/li&gt;&#xA;&lt;li&gt;Its variation should be function of some known function like. &lt;strong&gt;Periodic function like Sine or Cosine would be great&lt;/strong&gt; but any known function would do.&lt;/li&gt;&#xA;&lt;li&gt;Please see the tag &quot;home-experiment&quot;. I mean that :)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I cannot think of any such thing. I thought may be you could help.  &lt;/p&gt;&#xA;" OwnerUserId="447" LastActivityDate="2011-02-16T09:27:21.773" Title="Example of a time varying function which can be easily measured" Tags="&lt;experimental-physics&gt;&lt;soft-question&gt;&lt;home-experiment&gt;" AnswerCount="9" CommentCount="2" FavoriteCount="1" />
  <row Id="5159" PostTypeId="2" ParentId="5158" CreationDate="2011-02-14T17:11:10.993" Score="0" ViewCount="" Body="&lt;p&gt;Something like the raise in temperature of water that you use for cooking. Very home-experiment-like.&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2011-02-14T17:11:10.993" CommentCount="3" />
  <row Id="5160" PostTypeId="2" ParentId="5158" CreationDate="2011-02-14T17:32:44.970" Score="1" ViewCount="" Body="&lt;p&gt;Set up a simple pendulum with a string and a heavy object--a barbell will do.  You can hang it from a tree branch if you need space to do it.  The period varies as $T=2\pi\sqrt{\frac{L}{g}}$, where $L$ is the length of the pendulum, and $g$ is the local acceleration due to gravity, roughly $32 {\rm \frac{ft}{s}}$.  It shouldn't be too hard to adjust the length of your pendulum so that the period is something sensible and easily measurable--a ten foot pendulum should have a period of about 4 s, so it should be possible to measure position versus time for a variety of positions along the weight's path.&lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2011-02-14T17:48:10.533" LastActivityDate="2011-02-14T17:48:10.533" CommentCount="3" />
  <row Id="5161" PostTypeId="2" ParentId="5158" CreationDate="2011-02-14T17:38:12.330" Score="7" ViewCount="" Body="&lt;p&gt;Height of water accumulated in a bucket from a constant supply source. If the bucket is cylindrical height will be a linear function of time. If it is conical it will be a parabolic function etc.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-02-14T17:38:12.330" CommentCount="2" />
  <row Id="5162" PostTypeId="2" ParentId="5158" CreationDate="2011-02-14T18:24:37.763" Score="1" ViewCount="" Body="&lt;p&gt;Note outside temperature every day at fixed times, &#xA;if You have a minimum/maximum thermometer, note this values as well. &#xA;This will make a fine weather plot. This might become a science project.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-02-14T18:24:37.763" />
  <row Id="5163" PostTypeId="2" ParentId="5156" CreationDate="2011-02-14T18:31:54.373" Score="0" ViewCount="" Body="&lt;p&gt;On the entropic behavior (issue 2): the entropy would behave pretty much as in a big bang scenario &lt;em&gt;at both sides of the bounce&lt;/em&gt;. There is no issue with the 2nd law of thermodynamics. The entropy would be minimum at the bounce itself, and conscious beings at either side of the bounce would experience time to increase in the direction the entropy increases (away from the bounce).&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-02-14T18:31:54.373" />
  <row Id="5164" PostTypeId="2" ParentId="2792" CreationDate="2011-02-14T18:33:40.237" Score="0" ViewCount="" Body="&lt;p&gt;@Anna what you are asking about is the problem of &lt;em&gt;recording&lt;/em&gt; holographic images of 3D objects, the much harder counterpart of the problem of &lt;em&gt;displaying&lt;/em&gt; a recorded hologram as a 3D image. I've thought about this a bit and it seems there is no simple solution if you want to obtain an image with a 360 degree view, with only one optical device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, someone has managed to rig Microsoft's controller to act as a 3D camera. Perhaps replicating this feat might make a nice project for you! Of course, there is a finite field of view and the example is far from a working &quot;holographic recorder&quot; of some sort. The &lt;a href=&quot;http://www.youtube.com/watch?v=7QrnwoO1-8A&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt; of the demonstration is pretty mind-blowing though!&lt;/p&gt;&#xA;" LastActivityDate="2011-02-14T18:33:40.237" CommentCount="2" />
  <row Id="5165" PostTypeId="2" ParentId="5158" CreationDate="2011-02-14T18:42:30.073" Score="0" ViewCount="" Body="&lt;p&gt;I wouldn't use abstract concepts or things that are too related to physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try with f(t) = &quot;which program is on air on a given TV channel&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is time-varying, it is easily measurable AND it is something she will likely be acquainted with. And it has some periodicity inside (like news, it is always at the same hour).&lt;/p&gt;&#xA;" OwnerUserId="1917" LastActivityDate="2011-02-14T18:42:30.073" CommentCount="2" />
  <row Id="5166" PostTypeId="1" CreationDate="2011-02-14T18:49:30.520" Score="2" ViewCount="213" Body="&lt;p&gt;Supposing it's possible to see some distant galaxies with an earth telescope, then, at the tip of the telescope lens there are photons comming from the distant galaxy...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if I extend my hand in a dark night without visible moon, and far from city lights, photons from distant galaxies are in my hand skin... if it is true... is kind of amazing...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then at that moment if photons of EVERY visible galaxy (from an earth telescope) are in my hand, then why my hand didn't bright? knowing (from photoelectric effect) that photons act as a whole or don't act, then why my hand didn't bright?&lt;/p&gt;&#xA;" OwnerUserId="1916" LastEditorUserId="1916" LastEditorDisplayName="" LastEditDate="2011-02-15T11:41:38.290" LastActivityDate="2011-02-15T11:41:38.290" Title="What happend with the light from all the galaxies visibles from an earth telescope?" Tags="&lt;astronomy&gt;&lt;photons&gt;&lt;universe&gt;" AnswerCount="1" />
  <row Id="5167" PostTypeId="2" ParentId="5166" CreationDate="2011-02-14T19:07:13.730" Score="9" ViewCount="" Body="&lt;p&gt;I'm not sure, but I think this question is about what is generally known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Olbers%27_paradox&quot; rel=&quot;nofollow&quot;&gt;Olbers's paradox&lt;/a&gt;: If, at any given moment, light from all light sources in the Universe are illuminating any given point, one might expect every point to be extremely bright.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be precise, consider all of the light sources in a spherical shell of radius $R$ and thickness $dR$. If the Universe is uniform, the number of such sources is proportional to $R^2\,dR$. The apparent brightness of each source is proportional to $1/R^2$, so the total light striking your hand from all such sources is simply proportional to $dR$. If the Univers is infinite, then the total that results from integrating all such sources is infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, distant sources will be blocked by nearby sources, so the power striking your hand won't be infinite, but it will be as large as f the entire sky were filled with sources of the same temperature as a typical star. (To see this, note that every line of sight ends on the surface of a star. The surface brightness (watts per steradian) due to a source doesn't depend on distance: the brightness goes down like $1/R^2$, but so does the solid angle. So the entire sky has the surface brightness of a star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the paradox. What's the resolution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main thing is simply that we shouldn't integrate out to $R=\infty$. The Universe has a finite age, which means we can see out only to a finite distance. If you consider only the sources within our observable volume, the integrated brightness of all those galaxies still turns out to be very small. Galaxies are luminous, but they're far away. (There's also the fact that distant sources are receding from us, so the light from them is highly redshifted.)&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-14T19:07:13.730" CommentCount="12" />
  <row Id="5169" PostTypeId="2" ParentId="5132" CreationDate="2011-02-14T19:38:09.823" Score="4" ViewCount="" Body="&lt;p&gt;One of the standard texts for this kind of thing (the quantum mechanics of lasers, without all the technical details you'd need to know to design a real one) is Loudon's Quantum Theory of Light.  I've got the 2nd edition, I think he's up to the 3rd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my edition he does the &quot;pre-Quantum&quot; explanation of lasers (i.e. in terms of Einstein A &amp;amp; B coefficients) in about 40 pages.  Over the next 200 pages or so he does the fully quantum treatment of light (quantization of the field, creation and annihilation operators, etc) and then revisits the laser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first 40 pages should cover the &quot;clearly written description&quot; criteria, and - as far as it goes - it's relatively rigorous as well.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-02-14T19:38:09.823" />
  <row Id="5170" PostTypeId="1" AcceptedAnswerId="5173" CreationDate="2011-02-14T20:16:40.700" Score="1" ViewCount="255" Body="&lt;p&gt;Specifically, does the focal length change? How can this be rationalized?&lt;/p&gt;&#xA;" OwnerUserId="733" LastActivityDate="2011-02-14T21:03:24.123" Title="What happens when you cut a biconvex lens in half?" Tags="&lt;optics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="5171" PostTypeId="2" ParentId="5158" CreationDate="2011-02-14T20:35:10.887" Score="1" ViewCount="" Body="&lt;p&gt;Width of chest while breathing.   Height while growing.   loudness of heart sounds in a stethescope.   Distance while walking (try walking in circles as well).    Physiology makes it personal.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2011-02-14T20:35:10.887" />
  <row Id="5172" PostTypeId="2" ParentId="5170" CreationDate="2011-02-14T20:45:33.157" Score="1" ViewCount="" Body="&lt;p&gt;This entry about the Lens' maker equation in Wikipedia may help you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lens_(optics)#Lensmaker.27s_equation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lens_(optics)#Lensmaker.27s_equation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can calculate the ratio of focal lenghts before and after cutting by doing $R_2\rightarrow\infty$ and $d\rightarrow d/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The focal power of the second surface that now is plane will be smaller (actually 0) and, unless this is compensated by a huge decrease in thickness, the lens will have longer focal distance. That's what intuitively seems to happens with the dimensions used in real-life lenses.&lt;/p&gt;&#xA;" OwnerUserId="1337" LastEditorUserId="1337" LastEditorDisplayName="" LastEditDate="2011-02-14T20:53:38.950" LastActivityDate="2011-02-14T20:53:38.950" CommentCount="1" />
  <row Id="5173" PostTypeId="2" ParentId="5170" CreationDate="2011-02-14T21:03:24.123" Score="4" ViewCount="" Body="&lt;p&gt;It might help to think about the symmetry of a sliced biconvex lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/hNaJP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the biconvex lens can focus light from point F1 to F2, both distance $f$ from the lens, then when you cut the lens in half, each half will have a focal length equal to $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The focal length of the biconvex lens is $f/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general lensmaker's equation, &lt;a href=&quot;http://en.wikipedia.org/wiki/Lens_%28optics%29&quot; rel=&quot;nofollow&quot;&gt;from Wikipedia&lt;/a&gt;, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/m2XpD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-02-14T21:03:24.123" />
  <row Id="5175" PostTypeId="2" ParentId="5156" CreationDate="2011-02-14T22:36:02.327" Score="2" ViewCount="" Body="&lt;p&gt;This connects in some ways to what I wrote in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/5150/did-spacetime-start-with-the-big-bang&quot;&gt;Did spacetime start with the Big bang?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other side of the de Sitter “throat” is a cosmology which implodes.  Now it is important that everyone not take the solutions to Einstein’s field equations literally.  Johannes is right in fact, the entropy increases from the minimal region in the throat with a time arrow heading in opposite directions.  Then according to our time arrow entropy decreases on the “other side.”  This is not different from the white hole part of the Schwarzschild solution.  The white hole horizon rapidly decreases and the entropy declines as well.  However, we do not have white holes in the current state of the universe, so they are not taken as physically the same as a black hole.  They may have existed in the inflationary universe, where the rewind of anisotropy in the reverse time direction might lead to black holes --- or white holes in the forwards time direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other side may correspond to some quantum “anti-universe.”  An electron may be thought to tunnel through a potential barrier by being annihilated by a positron, which in turn is in a pair $e,~e^+$ and the electron on the other side of the barrier escapes free.  This other half of the universe, or “biverse” might be compared to such tunneling.  So this other universe should not be taken in some concrete sense, but more as a QFT or mathematical gadget.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data does suggest our universe will expand endlessly and will no recollapse.  So the cyclic model appears to be ruled out.  Our observable universe is not likely to recollapse unless there is some unexpected potential function in the mini-superspace which reverses the current accelerated expansion.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-14T22:36:02.327" CommentCount="2" />
  <row Id="5176" PostTypeId="1" CreationDate="2011-02-14T22:41:02.373" Score="0" ViewCount="148" Body="&lt;p&gt;I've always wanted to learn physics and never really got around to it. What would you as a community recommend for an introduction to physics book? I figured it would be a good idea to ask here as there are a lot of experts and I'd like to get you opinions. &lt;/p&gt;&#xA;" OwnerUserId="1920" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-15T09:30:12.913" LastActivityDate="2011-02-15T09:30:12.913" CommunityOwnedDate="2011-02-15T01:41:13.407" Title="Basic intro to physics" Tags="&lt;soft-question&gt;&lt;education&gt;&lt;books&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" />
  <row Id="5177" PostTypeId="2" ParentId="5156" CreationDate="2011-02-14T23:30:21.417" Score="0" ViewCount="" Body="&lt;p&gt;This question can be related to the linked question as remarked by Lawrence. In that question I mentioned the Penrose CCC theory and it has to deal with exactly the issues in this question. In short the answers are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The Penrose CCC mathematics is independent of the overall K curvature, but it is generally working to the current evidence that the Universe is expanding. Not only that, but surprisingly, the previous universe (aeon) was expanding too! How this can be fitted together is contained in the details of the CCC idea, but a core part is the use of Conformal transformations. Note that the previous aeon lasted for infinite local time (useful in the next point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Penrose has suggested that the information loss connected with Black Holes is responsible for reducing Entropy again. Basically the loss of this matter (in the previous aeon's black holes) reduced the number of degrees of freedom available. So he is postulating that Entropy will balance out come the next Bang. Of course something has to give, and Penrose is comfortable with one consequence that Quantum Unitary does not hold in this Black Hole context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) As mentioned in the linked Answer, there is a recent Arxiv paper on analysis of circles in WMAP data, which Penrose claims provides evidence of gravitational radiation from huge black hole collisions in the earlier aeon.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-14T23:30:21.417" CommentCount="2" />
  <row Id="5178" PostTypeId="2" ParentId="2219" CreationDate="2011-02-15T01:21:11.067" Score="2" ViewCount="" Body="&lt;p&gt;For freshman level physics, the canonical text is Halliday, Resnick, and Walker's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471758019&quot; rel=&quot;nofollow&quot;&gt;Fundamentals of Physics&lt;/a&gt; (at least in the US). It will require calculus. Don't worry about the hefty price tag - you can get an old edition and be fine. It covers everything you'd learn in the first year as a physics student.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Halliday, Resnick, and Walker is good for teaching you how to do physics problems and great at the fundamentals, but it isn't really very fun. If you want something that's more fun to read, you should definitely go for &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Feynman_Lectures_on_Physics&quot; rel=&quot;nofollow&quot;&gt;the Feynman Lectures&lt;/a&gt;. They're amazing, pitched at a basic level but full of deep detail. You won't learn how to solve actual real-world problems especially well, but they are full of insight on every page and really well-written.&lt;/p&gt;&#xA;" OwnerUserId="1655" LastActivityDate="2011-02-15T01:21:11.067" CommunityOwnedDate="2011-02-15T01:41:13.407" CommentCount="1" />
  <row Id="5179" PostTypeId="2" ParentId="5158" CreationDate="2011-02-15T01:37:55.783" Score="2" ViewCount="" Body="&lt;p&gt;For a slower function, try the angle of the sun over the course of a day. To measure, just put a stick in the ground and look at the shadow. Also a great way to introduce some trig as well as the effect of amplitude vs season.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some of the faster suggestions here like pendulums, try recording it with a video camera if you have one. The recording function on most hand held cameras should be fine, and then you can step through the video a few frames at a time to measure your function.&lt;/p&gt;&#xA;" OwnerUserId="1863" LastActivityDate="2011-02-15T01:37:55.783" />
  <row Id="5180" PostTypeId="1" AcceptedAnswerId="5187" CreationDate="2011-02-15T02:17:02.333" Score="2" ViewCount="103" Body="&lt;p&gt;Venus (0.815 Earth mass) supports a CO2 atmosphere of 93 bars at its surface. Earth's N2/O2 atmosphere has a surface pressure of one bar.&#xA;What are the factors that limit the mass of atmospheric gases on a planet?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-02-15T17:36:39.123" Title="Planetary Atmospheres" Tags="&lt;astronomy&gt;" AnswerCount="3" />
  <row Id="5181" PostTypeId="1" AcceptedAnswerId="5198" CreationDate="2011-02-15T02:24:16.833" Score="3" ViewCount="176" Body="&lt;p&gt;I'm studying Richardson Model in second quantization. There are many initial points that I don't understand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-we supposed an actractive force between 2 electrons, due to electron-phonon interaction: an electron generates a phonon that is absorbed by the other electron, and we prove that the resultant potential is actractive. But this is really stronger that Coulomb repulsive potential? How can estimate this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Why it neglects actractive force between electrons with opposite spin? (+-) They have an opposite magnetic momentum too...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-why we consider anyhow only the actrations between electrons with  opposite spin?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Why we suppose ${k_fR} &amp;lt;&amp;lt;1$ , where $K_f$ is $\hbar\cdot$Fermi momentum and R the interaction distance? &lt;/p&gt;&#xA;" OwnerUserId="558" LastEditorUserId="558" LastEditorDisplayName="" LastEditDate="2011-02-15T13:06:10.960" LastActivityDate="2011-02-15T13:06:10.960" Title="BCS theory, Richardson model and Superconductivity" Tags="&lt;material-science&gt;&lt;fermions&gt;&lt;bose-einstein-condensate&gt;&lt;superconductivity&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="5183" PostTypeId="2" ParentId="5180" CreationDate="2011-02-15T03:01:55.837" Score="1" ViewCount="" Body="&lt;p&gt;Large planets tend to be gas giants, where Jupiter is gas all the way to a core made of metallic hydrogen and some rock.  The pressures inside there are very extreme.  A gas giant up to some 50 times the mass of Jupiter is a brown dwarf, and more massive than that and you start getting fusion.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-15T03:01:55.837" CommentCount="5" />
  <row Id="5185" PostTypeId="1" AcceptedAnswerId="5234" CreationDate="2011-02-15T03:57:45.110" Score="5" ViewCount="267" Body="&lt;p&gt;The famous AdS/CFT calculation of the shear viscosity/entropy ratio for strongly coupled $N=4$ SYM relates the shear viscosity to the absorption cross section for fluctuations of the metric onto a black hole in AdS which in turn can be related to the absorption cross section for a massless scalar field. On the AdS side it seems clear that dissipation and relaxation back to an equilibrium configuration after a disturbance means that the disturbance has been absorbed by the black hole, and this would seem to imply that the disturbance propagates from small $z$ to large $z$ in Poincare coordinates where $z \rightarrow 0$ is the UV boundary. That is, the energy of the disturbance propagates from the UV to the IR where it is dissipated. First question: is this the right physical picture? Second question: this seems to imply on the CFT side that the dissipation of energy due to shear viscosity also involves a transfer of energy from the UV to the IR. Is this consistent with what is known in fluid mechanics for viscous flow? If so, can you provide a reference?&lt;/p&gt;&#xA;" OwnerUserId="838" LastActivityDate="2011-02-15T23:14:56.423" Title="Scale dependence of energy dissipation in viscous flow via AdS/CFT" Tags="&lt;fluid-dynamics&gt;&lt;ads-cft&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="5186" PostTypeId="2" ParentId="2501" CreationDate="2011-02-15T05:37:15.113" Score="2" ViewCount="" Body="&lt;p&gt;So far, this is an unknown question that is the subject of current research. For example:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sean N. Raymond, David P. O'Brien, Alessandro Morbidelli, Nathan A. Kaib, &quot;Building the Terrestrial Planets: Constrained Accretion in the Inner Solar System&quot;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To date, no accretion model has&#xA;  succeeded in reproducing all observed&#xA;  constraints in the inner Solar System.&#xA;  These constraints include 1) the&#xA;  orbits, in particular the small&#xA;  eccentricities, and 2) the masses of&#xA;  the terrestrial planets -- Mars'&#xA;  relatively small mass in particular&#xA;  has not been adequately reproduced in&#xA;  previous simulations; 3) the formation&#xA;  timescales of Earth and Mars, as&#xA;  interpreted from Hf/W isotopes; 4) the&#xA;  bulk structure of the asteroid belt,&#xA;  in particular the lack of an imprint&#xA;  of planetary embryo-sized objects; and&#xA;  5) Earth's relatively large water&#xA;  content, assuming that it was&#xA;  delivered in the form of water-rich&#xA;  primitive asteroidal material.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/0905.3750&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0905.3750&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-02-15T05:58:41.077" LastActivityDate="2011-02-15T05:58:41.077" />
  <row Id="5187" PostTypeId="2" ParentId="5180" CreationDate="2011-02-15T06:13:56.500" Score="2" ViewCount="" Body="&lt;p&gt;One would  start by the molecular composition of what constituted finally a planet. 1 bar is approximately 1 kilogram per square centimeter. The heavier the gases found on a planet the larger the pressure on the surface given the same size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the reason why Venus has heavier and/or more gases and earth less so, is another story.&#xA;Pick in the planetary lottery :).Planets do not have the same molecular composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extra heat from the sun with respect to earth and closer approach should play a role in light gases escaping from the atmosphere. In the wikipedia article on Venus there are more  explanations offered.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-02-15T06:13:56.500" />
  <row Id="5188" PostTypeId="1" AcceptedAnswerId="6854" CreationDate="2011-02-15T06:38:16.430" Score="-2" ViewCount="187" Body="&lt;p&gt;I know the basics of solar system like how earth moves around sun, and that we have so many planets, elliptical orbit of earth, and how far is sun from earth etc etc. I want to take a step back and understand how mankind has been able to construct this knowledge of solar system piece by piece.Questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- Sun is something we see on daily basis, so it would not be strange if humankind got curious to know about it, and made theories,observations,maths to prove it. How did Copernicus and others came up with the theory that earth moves around sun in elliptical orbit. How did they prove it? Has there been any actual way by which we have actually observed sun being in center and earth moving around in elliptical orbit. or we made such theory and result of this theory have been proved by other observations like day,night,change in weather at a point and at different points on earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- How did human have an inkling that there are other stuff on space apart from sun and moon and some stars, like mars,jupiter and other planets, because these planets are not visible from earth. Did human accidently bump into these planets while moving around in spacecraft like they bumped into America.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- When was the first man made satellite made? When did the first spacecraft go into space. Weren't people afraid to step their foot outside their spacecraft in an utter uncertainty now knowing what waits them on an alien planet, living condition on it etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- How has the distance between sun and earth been calculated? and speed of light? and size of sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- What are the most used tools used to acquiring such knowledge apart from telescope? What is the range of most powerful telescopes that man has made? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- And why are humans so curious to know more and more about Mars in particular? Is it because it is relatively easier to study Mars for some scientific reason? how we do scientist intend to put their Mars knowledge to practical use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- What is the difference between spacecraft and rocket? What is the application of each?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="1093" LastEditorUserId="1093" LastEditorDisplayName="" LastEditDate="2011-02-15T07:28:04.657" LastActivityDate="2011-03-13T22:08:04.633" ClosedDate="2011-03-13T22:13:04.317" Title="History of Man Acquired knowledge of solar system" Tags="&lt;astrophysics&gt;&lt;gravity&gt;&lt;cosmology&gt;&lt;astronomy&gt;&lt;space&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="5189" PostTypeId="2" ParentId="2219" CreationDate="2011-02-15T07:03:49.837" Score="0" ViewCount="" Body="&lt;p&gt;the Feynman Lectures are the thing if you want to learn physics. He was the best.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2011-02-15T07:03:49.837" CommunityOwnedDate="2011-02-15T07:03:49.837" />
  <row Id="5190" PostTypeId="1" AcceptedAnswerId="5196" CreationDate="2011-02-15T07:10:35.653" Score="2" ViewCount="307" Body="&lt;p&gt;I know the basics of solar system like how earth moves around sun, and that we have so many planets, elliptical orbit of earth, and how far is sun from earth etc etc. I want to take a step back and understand how mankind has been able to construct this knowledge of solar system piece by piece.Questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- Sun is something we see on daily basis, so it would not be strange if humankind got curious to know about it, and made theories,observations,maths to prove it. How did Copernicus and others came up with the theory that earth moves around sun in elliptical orbit. How did they prove it? Has there been any actual way by which we have actually observed sun being in center and earth moving around in elliptical orbit. or we made such theory and result of this theory have been proved by other observations like day,night,change in weather at a point and at different points on earth.&lt;/p&gt;&#xA;" OwnerUserId="1093" LastEditorUserId="1093" LastEditorDisplayName="" LastEditDate="2011-02-27T06:47:25.747" LastActivityDate="2011-02-27T06:47:25.747" Title="How do we know that earth moves around sun and that too in elliptical orbit" Tags="&lt;gravity&gt;&lt;astrophysics&gt;&lt;cosmology&gt;&lt;astronomy&gt;&lt;space&gt;" AnswerCount="4" />
  <row Id="5191" PostTypeId="2" ParentId="5190" CreationDate="2011-02-15T07:23:12.313" Score="0" ViewCount="" Body="&lt;p&gt;This is a very general question best answered by reading up on material found in &lt;a href=&quot;http://en.wikipedia.org/wiki/Heliocentric_system&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short answer is that forces like the gravitational force have orbits for circling masses that are elliptical in nature, where the sun sits at one focus of the ellipse. Circles are an ellipse where the two foci overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you read the wiki article, you will see that a lot of data were gathered over the centuries on the motion of the planets on the celestial sphere, the geocentric system. Copernicus realized that a heliocentric system would fit the data simply and beautifully instead of the epicycle system that had dominated astronomy till then. The orbits were ellipses.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-02-15T07:23:12.313" CommentCount="3" />
  <row Id="5192" PostTypeId="1" AcceptedAnswerId="5203" CreationDate="2011-02-15T07:35:11.640" Score="4" ViewCount="211" Body="&lt;p&gt;How has the distance between sun and earth been calculated by scientists?&#xA;and size of sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="1093" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-02-15T08:10:52.343" LastActivityDate="2011-02-16T15:57:44.170" Title="How is distance between sun and earth calculated?" Tags="&lt;astronomy&gt;&lt;newtonian-mechanics&gt;&lt;astrometrics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="5193" PostTypeId="2" ParentId="5190" CreationDate="2011-02-15T07:38:55.897" Score="3" ViewCount="" Body="&lt;p&gt;The direction in which the celestial bodies are located is seen directly in the telescopes (two angular coordinates); the distance from the Earth is the hard part. For planets, we may measure the distance by waiting for a radar echo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aristarchus has invented a useful method to get a third piece of information: when the Moon is exactly half-full, Sun-Moon-Earth form a right angle with the Moon angle being 90 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance from the Earth to the Sun has also been measured by the transit-of-Venus method - to measure the time it takes to cross the Sun. Another method is parallax - measuring and comparing the apparent locations of the celestial objects at the same time, as seen from two places on the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Tycho Brahe&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, the angular information - 2/3 of the coordinates - is the most accurate one. Tycho Brahe, a royal astronomer in Prague, measured the positions of planets extraordinarily accurately. The fact that he believed the geocentric system couldn't affect his precision. After all, you may describe everything from the Earth's reference frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His young assistant Johannes Kepler - who realized that Copernicus' heliocentric model was right - used those accurate data by Brahe to derive his Kepler's laws. Note that even if you don't know how the distance from the Earth is changing as a function of time, Kepler's laws make many very constraining predictions for each moment of time - where the planets and the Sun should be seen. Kepler has simply made the right guess and verified the predictions of this guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Prague, Brahe was afraid that Kepler could try to get rid of him, so that Kepler would become the main royal astronomer himself. At the end, Brahe died because of not-quite-understood problems with the bladder. I don't claim that Kepler poisoned him. ;-)&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-15T07:38:55.897" CommentCount="2" />
  <row Id="5194" PostTypeId="1" AcceptedAnswerId="5197" CreationDate="2011-02-15T08:12:16.677" Score="0" ViewCount="216" Body="&lt;p&gt;How is the speed of light calculated? My knowledge of physics is limited to how much I studied till high school, but one way that comes to my mind is if we throw light from one point to other (of known distance) and measure time taken, we could know the speed of light? but do we have such fine time measuring tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="1093" LastEditorUserId="1093" LastEditorDisplayName="" LastEditDate="2011-02-15T08:22:29.820" LastActivityDate="2011-02-16T05:10:42.640" Title="How is the speed of light calculated?" Tags="&lt;electromagnetism&gt;&lt;light&gt;&lt;astronomy&gt;&lt;waves&gt;&lt;speed-of-light&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="5195" PostTypeId="1" AcceptedAnswerId="5229" CreationDate="2011-02-15T08:14:43.763" Score="0" ViewCount="374" Body="&lt;p&gt;Suppose we have a cylindrical resistor, with resistance given by $R=\rho\cdot l/(\pi r^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $d$ be the distance between two points in the interior of the resistor and let $r\gg d\gg l$. &#xA;Ie. it is approximately a 2D-surface (a rather thin disk).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the resistance between these two points?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $r,l\gg d$, (ie a 3D volume), is the resistance $0$ ?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Clarification:&#xA;A voltage difference is applied between two points a distance $d$ apart, inside a material with resistivity $\rho$, and the current is measured, the proportionality constant $V/I$ is called $R$. The material is a cylinder of height $l$ and radius $r$, and the two points are situated close to the center, we can write $R$ as a function of $l$, $r$ and $d$, $R(l,r,d)$, for small $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The questions are then:&lt;br&gt;&#xA;What is $$ \lim_{r \rightarrow \infty} \lim_{l \rightarrow \infty}  R(l,r,d) $$&lt;br&gt;&#xA;What is $$ \lim_{r \rightarrow \infty} \lim_{l \rightarrow 0} R(l,r,d) $$&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2011-02-16T09:53:59.497" LastActivityDate="2011-02-17T14:33:30.620" Title="Resistance between two points on a conducting surface" Tags="&lt;electromagnetism&gt;&lt;electricity&gt;&lt;electrostatics&gt;&lt;resistors&gt;" AnswerCount="4" CommentCount="13" />
  <row Id="5196" PostTypeId="2" ParentId="5190" CreationDate="2011-02-15T08:35:20.197" Score="7" ViewCount="" Body="&lt;p&gt;In his lecture &lt;a href=&quot;http://www.youtube.com/watch?v=7ne0GArfeMs&quot; rel=&quot;nofollow&quot;&gt;The Cosmic Distance Ladder&lt;/a&gt; (video), mathematician Terence Tao describes the history of how mankind has successfully mapped the solar system and beyond. In particular, he describes why Copernicus put the sun in the center (reason: he discovered that the sun is dozens of times bigger than the earth) and how Kepler found his laws of planetary motion (He needed a reference point that was fixed in place, but all planets move. His brilliant idea was too choose Mars as the reference point, but only looking at it after a full orbit, when its position returns to the original one).&lt;/p&gt;&#xA;" OwnerUserId="188" LastActivityDate="2011-02-15T08:35:20.197" />
  <row Id="5197" PostTypeId="2" ParentId="5194" CreationDate="2011-02-15T09:02:27.467" Score="5" ViewCount="" Body="&lt;p&gt;From Wikipedia:&lt;br&gt;&#xA;Presently, the speed of light in a vacuum is defined to be exactly 299,792,458 m/s (approximately 186,282 miles per second). The fixed value of the speed of light in SI units results from the fact that the metre is now defined in terms of the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different physicists have attempted to measure the speed of light throughout history. Galileo attempted to measure the speed of light in the seventeenth century. An early experiment to measure the speed of light was conducted by Ole Rømer, a Danish physicist, in 1676. Using a telescope, Ole observed the motions of Jupiter and one of its moons, Io. Noting discrepancies in the apparent period of Io's orbit, Rømer calculated that light takes about 22 minutes to traverse the diameter of Earth's orbit.[4] Unfortunately, its size was not known at that time. If Ole had known the diameter of the Earth's orbit, he would have calculated a speed of 227,000,000 m/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another, more accurate, measurement of the speed of light was performed in Europe by Hippolyte Fizeau in 1849. Fizeau directed a beam of light at a mirror several kilometers away. A rotating cog wheel was placed in the path of the light beam as it traveled from the source, to the mirror and then returned to its origin. Fizeau found that at a certain rate of rotation, the beam would pass through one gap in the wheel on the way out and the next gap on the way back. Knowing the distance to the mirror, the number of teeth on the wheel, and the rate of rotation, Fizeau was able to calculate the speed of light as 313,000,000 m/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Léon Foucault used an experiment which used rotating mirrors to obtain a value of 298,000,000 m/s in 1862. Albert A. Michelson conducted experiments on the speed of light from 1877 until his death in 1931. He refined Foucault's methods in 1926 using improved rotating mirrors to measure the time it took light to make a round trip from Mt. Wilson to Mt. San Antonio in California. The precise measurements yielded a speed of 299,796,000 m/s.&lt;/p&gt;&#xA;" OwnerUserId="1847" LastActivityDate="2011-02-15T09:02:27.467" CommentCount="2" />
  <row Id="5198" PostTypeId="2" ParentId="5181" CreationDate="2011-02-15T09:59:13.243" Score="2" ViewCount="" Body="&lt;p&gt;I will just answer the first part of question: is phonon attraction stronger than Coulomb?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short answer: No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longer answer: Nothing (in condensed matter) is ever stronger than the Coulomb force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longest answer: There are two aspects to consider. First is the self-screening of the electrons, which will add a mass term to the photons, giving a Yukawa-esque potential ($e^{-mr}/r$). Second, although we always work in momentum space, it's actually unhelpfully abstract in this case. You need to imagine what the force looks like in real space and time. The Coulomb force is a 1/r^2 instaneous repulsion, but it decays very fast (instaneous) in time. The screened Yukawa/Thomas-Fermi force decays less slowly (exponentially with scale $m$) in time. The phonon attraction, however, is a &lt;em&gt;retarded&lt;/em&gt; force, meaning it only acts some finite time after the electron has left. Thus if you imagine a plot of the potential as as function of spacetime, there is a singularity at the origin, but there is a small dip after it! In terms of dynamics this is even easier to understand: electrons cause small lattice deformations as they move around, but the lattice deformations need time to be set up and relax again, and this time-scale is larger than the time it takes an electron to get across the distortion; thus another electron sees a small, attractive hole into which it could fall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the Fermi liquid is so unstable (it has too many low energy degrees of freedom) &lt;em&gt;any&lt;/em&gt; attractive interaction in any form will push it towards a broken symmetry state, such as superconductivity (or charge density waves, etc.). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So although the phonon attraction isn't &quot;stronger&quot;, it can beat Coulomb repulsion. It just has to be clever about it.&lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-02-15T09:59:13.243" CommentCount="4" />
  <row Id="5199" PostTypeId="2" ParentId="3306" CreationDate="2011-02-15T10:19:04.027" Score="1" ViewCount="" Body="&lt;p&gt;I wrestled with this problem more than a week and got quite an unexpected result, i think. I am using SI system.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A starting point is that there are excessive surface charges at each of the wires, flows a current through them or not. Therefore, besides the magnetic force $F_m$ it is necessary to take into account also the electrical force $F_e$ between wires. Let the unit length of the wires have a charge equal to $\lambda (\frac{C}{m})$. Then the electric force acting on unit length of wire from other wire:&lt;br&gt;&#xA;$$F_e=\lambda E=\frac{2\lambda^2}{4\pi\epsilon_0d}$$ where $d$ is distance between wires.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnetic force acting on the unit length of wire:&lt;br&gt;&#xA;$$F_m=\frac{\mu_0}{4\pi}\frac{2I^2}{d}$$ where I is a direct current through wires. It follows from $F_m=F_e$ that  $$\left(\frac{\lambda}{I}\right)^2=\epsilon_0\mu_0$$ Next we notice that $$\lambda=CU=CIR_0$$ or $$\frac{\lambda}{I}=CR_0$$ and thus $$CR_0=\sqrt{\epsilon_0\mu_0}$$ where C is a capacitance per unit length for a pair of parallel wires, U is a voltage between wires (due to the battery), $R_0$ is the resistance, $\epsilon_0=8.85*10^{-12}\frac{F}{m}$ and $\mu_0=4\pi*10^{-7}\frac{H}{m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Capacitance per unit length for a pair of parallel wires is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C=\frac{\pi \epsilon_0}{ln\left(\frac{d}{r}\right)}= \frac{\pi \epsilon_0}{ln(x)}$$ Here $r&amp;lt;&amp;lt;d$  is assumed. $r$ is a radius of wires and $d$ is a distance between wires.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, required resistance:&lt;br&gt;&#xA;$$R_0=\sqrt{\frac{\mu_0}{\epsilon_0}}\frac{ln(x)}{\pi}=552.4\Omega$$ An unexpected result is that this resistance does not depend on the applied voltage.&lt;/p&gt;&#xA;" OwnerUserId="552" LastEditorUserId="552" LastEditorDisplayName="" LastEditDate="2011-02-16T08:42:57.023" LastActivityDate="2011-02-16T08:42:57.023" CommentCount="2" />
  <row Id="5200" PostTypeId="2" ParentId="5150" CreationDate="2011-02-15T10:46:11.740" Score="0" ViewCount="" Body="&lt;p&gt;I seriously doubt that.&#xA;There are quite a few theories trying to discuss &quot;the beginning of the universe&quot;, and not all of them agree with having a &quot;beginning&quot; to the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have concrete facts to support what I'm about to say so consider it as &quot;just another idea.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Our whole universe was in a hot dense place, then fourteen billion years ago expansion started.&quot;&#xA;So there was a big explosion, and everything tried to get away from every other thing. This process has been going on for about fourteen billion years, and it will eventually stop. Because a big portion of the attracting/repelling forces (electromagnetic force which &lt;em&gt;can&lt;/em&gt; attract and repel) will have been neutralized (like a stone which &lt;em&gt;does&lt;/em&gt; have electrons and protons but doesn't attract or repel), and the attracting forces (gravity) will take over. The universe starts to get smaller and denser, and collapse into a small hot dense place, and in &quot;hot places&quot;, the repelling forces will eventually take over, and so another bang happens, and it will create a universe much like of our own.&lt;/p&gt;&#xA;" OwnerUserId="1927" LastActivityDate="2011-02-15T10:46:11.740" />
  <row Id="5201" PostTypeId="2" ParentId="5132" CreationDate="2011-02-15T12:11:14.333" Score="3" ViewCount="" Body="&lt;p&gt;It sounds like what you're asking about is basically atomic physics, rather than laser physics. In which case, I recall &lt;em&gt;Optical Resonance and Two-Level Atoms&lt;/em&gt; by Allen and Eberly being useful when I took that back in grad school. There's a good deal of it &lt;a href=&quot;http://books.google.com/books?id=1q0ae-XNmWwC&amp;amp;printsec=frontcover&amp;amp;dq=eberly+allen+atoms&amp;amp;source=bl&amp;amp;ots=mOW8Yf8Xuu&amp;amp;sig=LHDjp9NX-TH3qfP9XMsjrlnAY1g&amp;amp;hl=en&amp;amp;ei=fGtaTbm7O4KUtwfzzYzxCw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBcQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;available on Google Books&lt;/a&gt; if you want to check it out. It's short, hits most of the phenomena involving the interaction of light and atoms that you might want to know about, and it's a Dover paperback, so it's cheap for a physics text. It might assume more knowledge than you have, though. Anyway, it's probably worth a look.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxford University Press has a series of &quot;Masters&quot; books pitched for advanced undergrads and beginning grad students that might also have something worthwhile. I've used the &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B000WEPHBQ&quot; rel=&quot;nofollow&quot;&gt;Quantum Optics&lt;/a&gt; volume by Fox, and the &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0198506961&quot; rel=&quot;nofollow&quot;&gt;Atomic Physics&lt;/a&gt; one by foot, and between them they probably cover everything you're after.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2011-02-15T12:11:14.333" CommentCount="2" />
  <row Id="5202" PostTypeId="1" AcceptedAnswerId="5204" CreationDate="2011-02-15T12:40:11.967" Score="3" ViewCount="159" Body="&lt;p&gt;I often encountered the term &quot;localization&quot; in the context of instantons, as for example in the work of Nekrasov on extensions of Seiberg-Witten theory to N=1 gauge theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone give a &quot;intuitive&quot; explanation of the concept of localization as well as a &quot;simple&quot; concrete realisation of it?&lt;/p&gt;&#xA;" OwnerUserId="1844" LastActivityDate="2011-02-15T13:28:29.367" Title="What is &quot;localisation&quot; of instantons?" Tags="&lt;quantum-field-theory&gt;&lt;gauge-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="5203" PostTypeId="2" ParentId="5192" CreationDate="2011-02-15T12:50:51.333" Score="4" ViewCount="" Body="&lt;p&gt;The most precise measures of this distance are from radars in the 1960s. However, the distance has been known, though roughly, since the Ancient Times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aristarchus of Samos (310BC - 230BC) used the angle between the Earth-Moon axis and the Earth-Sun when the Moon is in First Quarter (elongation of the Moon, E) and then, with simple trigonometry, could deduce the distances: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;cos E = distance (Earth-Moon) / distance (Earth-Sun)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since he had already computed the Earth-Moon distance from the duration of lunar eclipses, he could conclude on the Earth-Sun distance. His results were false, because of too loose measure of the angle, but his method was very accurate. &#xA;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Aristarchus_On_the_Sizes_and_Distances&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; for more details. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method was explored in 1672 by Cassini and Richer: they measured the parallax (i.e. the variation in angle when seen from different places) under which Mars was seen in Cayenne and Paris, at the moment of opposition. From this, they deduced the distance Earth-Mars. &#xA;Then, using the Kepler law&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a^3 / p^2 = constant&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(where a is the distance between the planet and the Sun, and p the sideral time) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;they could figure out what was the distance to the Sun. &lt;/p&gt;&#xA;" OwnerUserId="1928" LastActivityDate="2011-02-15T12:50:51.333" />
  <row Id="5204" PostTypeId="2" ParentId="5202" CreationDate="2011-02-15T13:22:50.243" Score="4" ViewCount="" Body="&lt;p&gt;Dear inovaovao, in general, localization is the fact that certain integrals over some variables $\mu$ may be proved to be equal to lower-dimensional integrals or even the sum over contributions from a discrete - e.g. finite - set of points in the $\mu$-space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple conceptual example that doesn't require any advanced Seiberg-Witten-like machinery - but that is the &quot;master setup&quot; for everything to be explained below - is the classical limit of a path integral: the otherwise infinite-dimensional path integral de facto localizes on the classical trajectories. Although the path integral integrates over histories that are not classical solutions, the integral localizes on histories that are solutions. Here, let me emphasize, we're not changing the rules of the game of Feynman's approach to quantum mechanics: we're showing that the resulting formulae are fully equivalent to others where the integral gets simpler or lower-dimensional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the points where the path integral gets localized are just local (rather than global) extrema of the action - the instantons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Seiberg-Witten case, one has to calculate contributions of instantons - topologically nontrivial solutions of the classical equations of motion (stationary points of the action which are local minima but not the global ones). The instanton solutions themselves are spanned by many parameters: for $k$-instanton solutions of $SU(N)$ gauge theory, the number of parameters, as seen via the ADHM construction, is essentially $4kn$. It remembers the embedding of the instantons in $SU(N)$, their size, and their positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the instanton contributions are still integrals over the $4kN$-dimensional moduli space of the instantons. That's progress relatively to the infinite-dimensional path integral we could have started with. But the simplification may go on. Even this integral over the moduli space of instantons can be reduced to a lower-dimensional integral, and often a sum, over individual points of the instanton moduli space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The typical instantons that contribute are usually &quot;point-like&quot; and singular in some sense (in other contexts, the relevant places where the integrals localize may be labeled by unusually complex values of some parameters) and a special treatment has to be made to regulate these potentially ill-defined expressions, but it can be done. It's been pursued in many papers co-authored by Nikita Nekrasov, such as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/9711108&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9711108&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/abs/hep-th/9801061&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9801061&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/abs/hep-th/9803265&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9803265&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/abs/hep-th/9712241&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9712241&lt;/a&gt;&lt;br&gt;&#xA;  ...&lt;br&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/abs/hep-th/0206161&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0206161&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and many others. The required formulae that allow you to prove that the integrals localize are somewhat nontrivial mathematically. After all, Nikita Nekrasov and his collaborators are pretty powerful mathematicians. But in some sense, these formulae are always generalizations of the fact that integrals in the complex plane can be rewritten in terms of residues of the poles.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-15T13:28:29.367" LastActivityDate="2011-02-15T13:28:29.367" CommentCount="3" />
  <row Id="5205" PostTypeId="2" ParentId="3306" CreationDate="2011-02-15T13:39:05.507" Score="1" ViewCount="" Body="&lt;p&gt;Hmmmm...,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the same formula is for the impedance of a Lecher line!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z = 1/pi sqrt(µ0/eps0) ln(d/r) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes the formula is given with arcosh(d/2r) instead of ln(d/r), &#xA;I assume that is a variant more precise for small d/r = x. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Funny isn't it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;+1 for posting this solution!&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-02-15T21:46:17.000" LastActivityDate="2011-02-15T21:46:17.000" CommentCount="3" />
  <row Id="5206" PostTypeId="2" ParentId="2687" CreationDate="2011-02-15T14:45:46.663" Score="1" ViewCount="" Body="&lt;p&gt;The classic reference is Truedell and Noll's &quot;Non-linear field theories of mechanics&quot;, published in 1965 and republished (in several languages) 4 or 5 times since.  Other books worth using:  Mort Gurtin (1981, 2010), R. Bowen (available on his Texas A&amp;amp;M website), L. Malvern (1969 or so, old but good).&lt;/p&gt;&#xA;" LastActivityDate="2011-02-15T14:45:46.663" CommunityOwnedDate="2011-02-15T14:45:46.663" />
  <row Id="5207" PostTypeId="1" AcceptedAnswerId="5208" CreationDate="2011-02-15T14:55:06.720" Score="6" ViewCount="538" Body="&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/4748/why-is-there-a-deep-mysterious-relation-between-string-theory-and-number-theory&quot;&gt;This question&lt;/a&gt; has led me to ask somewhat a more specific question. I have read somewhere about a coincidence. Numbers of the form $8k + 2$ appears to be relevant for string theory. For k = 0 one gets 2 dimensional string world sheet, For k = 1 one gets 10 spacetime dimensions and for k = 3 one gets 26 dimensions of bosonic string theory. For k = 2 we get 18. I don't know whether it has any relevance or not in ST. Also the number 24, which can be thought of as number of dimensions perpendicular to 2 dimensional string world sheet in bosonic ST, is the largest number for which the sum of squares up to 24 is itself a square. $(1^2 + 2^2 + ..+24^2 = 70^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is it a mere coincidence or something deeper than that?&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="838" LastEditorDisplayName="" LastEditDate="2011-02-15T15:04:21.397" LastActivityDate="2011-03-02T04:45:06.857" Title="Number of dimensions in string theory and possible link with number theory" Tags="&lt;string-theory&gt;&lt;mathematics&gt;" AnswerCount="4" CommentCount="13" FavoriteCount="1" />
  <row Id="5208" PostTypeId="2" ParentId="5207" CreationDate="2011-02-15T15:19:23.877" Score="4" ViewCount="" Body="&lt;p&gt;There is definitely something deep going on, but there is not yet a deep understanding of what it is. In math the topology of the orthogonal group has a mod 8 periodicity called Bott periodicity. I think this is related to the dimensions in which one can have Majorana-Weyl spinors with Lorentzian signature which is indeed $8k+2$. So this is part of the connection and allows both the world-sheet and the spacetime for $d=2,10$ to have M-W spinors. The $26$ you get for $k=3$ doesn't have any obvious connection with spinors and supersymmetry, but there are some indirect connections related to the construction of a Vertex Operator Algebra with the Monster as its symmetry group. This involves a $Z_2$ orbifold of the bosonic string on the torus $R^{24}/\Lambda$ where&#xA;$\Lambda$ is the Leech lattice. A $Z_2$ orbifold of this theory involves a twist field of dimension $24/16=3/2$ which is the dimension needed for a superconformal generator. So the fact that there are $24$ transverse dimensions does get related to world-sheet superconformal invariance.&#xA;Finally, the fact you mentioned involving the sum of squares up to $24^2$ has been exploited in the math literature to give a very elegant construction of the Leech lattice starting from the Lorentzian lattice $\Pi^{25,1}$ by projecting along a null vector $(1,2, \cdots 24;70)$ which is null by the identity you quoted. I can't think of anything off the top of my head related to $k=2$ in string theory, but I'm sure there must be something.&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T04:45:06.857" LastActivityDate="2011-03-02T04:45:06.857" CommentCount="3" />
  <row Id="5209" PostTypeId="2" ParentId="5207" CreationDate="2011-02-15T16:01:23.340" Score="-1" ViewCount="" Body="&lt;p&gt;I don't know about the particular example that you mention, but there are certainly some interconnections with special numbers in mathematics and in string theory/supersymmetry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One worked out example is the connection of possible dimensions of supersymmetry in dimensions 3, 4, 6 or 10 which is connected to the existence of normed division algebras in dimensions 1, 2, 4 and 8. For more details see&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;John C. Baez, John Huerta: &lt;em&gt;Division Algebras and Supersymmetry I&lt;/em&gt;  (&lt;a href=&quot;http://arxiv.org/abs/0909.0551&quot; rel=&quot;nofollow&quot;&gt;arXiv&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and related work about higher gauge theory.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-02-15T16:01:23.340" CommentCount="2" />
  <row Id="5210" PostTypeId="2" ParentId="5195" CreationDate="2011-02-15T16:02:09.877" Score="-3" ViewCount="" Body="&lt;p&gt;&quot;Resistance between two points&quot; is not a well-defined concept.  If  c  is a curve connnecting  point  A  to point  B, the electromotive force  V  along  c  is a well-defined thing (it's the circulation of the electric vector field along  c), and resistance would be the factor  R  such that  V = RI  when  I  is the intensity flowing through a thin tube borne by  c.  But it's clear that intensity  depends on the cross section of this virtual tube, a non-defined entity here.   So we have no proper definition of &quot;resistance  R  between  A  and  B&quot; that would allow us to compute it.&lt;/p&gt;&#xA;" OwnerUserId="1761" LastActivityDate="2011-02-15T16:02:09.877" CommentCount="2" />
  <row Id="5212" PostTypeId="2" ParentId="5195" CreationDate="2011-02-15T16:48:51.987" Score="1" ViewCount="" Body="&lt;p&gt;I think the answer is infinite because of the singular nature of a point. If we assume a steady state situation then the divergence of the potential is zero, by symmetry in a full 2D model the current density J will scale like 1/r. This implies the voltage scales like log(r), which diverges as r goes to zero. In the 3D case its even worse, as J scales as 1/r**2, and thus voltage scales as 1/r, which diverges even faster as r goes to zero. Note that we can compute restance, by fixing the current, and computing the voltage difference. The problem is that the voltage difference doesn't converge in the vicinity of a point current source/sink.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-02-15T16:48:51.987" CommentCount="2" />
  <row Id="5213" PostTypeId="2" ParentId="2508" CreationDate="2011-02-15T16:54:28.163" Score="0" ViewCount="" Body="&lt;p&gt;This answer is just for the purpose of discussion. It will be edited sometimes and It may contain wrong conclusions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Martin: I am trying to image the motion you are describing. Unfortunatelly I have some severe difficulties, since I always run into some kind of contradiction. Therefore I have drawn two motions, in which the pad is not rotating. Please tell me which one you think is appropriate for this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motion one:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/ivLui.png&quot; alt=&quot;Assuming counterclockwise baby motion&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motion two: &#xA;&lt;img src=&quot;http://i.stack.imgur.com/jZtGK.png&quot; alt=&quot;Assuming counterclockwise baby motion&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for one time drawing a clockwise and one time drawing a counterclockwise baby-motion. As I noticed, it was already too late to change. But this should make no difference for the sake of argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please also tell, if you think of a different motion of the pad and try describing / or drawing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ftiaronsem&lt;/p&gt;&#xA;" OwnerUserId="393" LastActivityDate="2011-02-15T16:54:28.163" CommentCount="6" />
  <row Id="5214" PostTypeId="2" ParentId="5180" CreationDate="2011-02-15T17:09:05.727" Score="0" ViewCount="" Body="&lt;p&gt;It mainly has to do the the planets distribution of volatile (gaseous at ambient temperature) substances, and the distribution of them. In the earth most of the CO2 is bound up in carbonate minerals. High temperatures favor silicates over carbonates, and drive off the CO2 as gas. The reverse reaction, CO2 plus silicates is said require liquid water (I don't know the reason), and hence isn't available on dry Venus (I think the hydrogen was lost because Venus lacks a magnetic field to shield it from the solar wind), and the CO2 ends up in the atmosphere. On mars, CO2 is a major component of the atmosphere, but the low temperatures mean it condenses out as frost during the polar winter, so the atmosphere can't contain very much. Also the history of the planet can also have an effect on the amounts of volatiles. If the planet was very hot during its some phase ot its history the volatiles may have been lost at that time. The earth probably lost most of its volatiles during the massive moon forming collision, and the current inventory of volatiles probably came from asteroid and comet impacts since that time.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-02-15T17:09:05.727" />
  <row Id="5215" PostTypeId="1" CreationDate="2011-02-15T17:29:43.933" Score="4" ViewCount="78" Body="&lt;p&gt;In quantum gravity, ADM wavefunctional solutions have to satisfy the Wheeler-DeWitt equation. This leads to timelessness. What happens if we have a time periodic solution? In classical general relativity, a time periodic solution just means that and no more. But when combined with a timeless wavefunctional, if the same configuration occurs twice or more often, the coefficient of the configuration component in the wavefunctional has to be exactly the same both times around, both in magnitude and in phase. That is because we have no disambiguating clock external to the timeless wavefunctional, which is not multivalued. But this is exactly the description of a closed timelike curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume space is compact so we do not have to worry about asymptotic infinity.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-15T17:29:43.933" Title="When periodic solutions are combined with timelessness, do we get closed timelike curves?" Tags="&lt;quantum-gravity&gt;" FavoriteCount="2" />
  <row Id="5217" PostTypeId="1" AcceptedAnswerId="5218" CreationDate="2011-02-15T18:20:45.760" Score="1" ViewCount="220" Body="&lt;p&gt;First, discrete examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a computer screen I can specify any &quot;2D&quot; point with just one single number &#xA;(pixel order starting count from first at upper left, and going on, left2right and up2down like reading till last one at right bottom corner) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I don't need two numbers to specify position, just one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same can extend to other &quot;dimensions&quot; I can use same trick to describe a point inside a discrete cube, once have covered a screen like &quot;area&quot; descend a plane level&lt;/p&gt;&#xA;&#xA;&lt;p&gt;plane&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 .....5&#xA;6 .....10&#xA;11.....15&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;then a cube (made of above planes)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0  ____ 15&#xA;16 ____ 31&#xA;32 ____ 47&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;so a single number can map any point in a 3d discrete cube&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what about continuous one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;same thing using differential dx instead of discrete points or planes, and that's all&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(of course as differential is not well defined as discrete this won't be so easy)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway my question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;what is a good definition of dimention that avoid these tricks? there is one?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally another way to take 3d to 1D, is a reversible transformation like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3d point = (x,y,z)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;W = x + y * 2^20000 + z * 2^40000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the only &quot;restriction&quot; for this method is that equivalent 3d coords should be &#xA;lesser than 2^20000, a very easy thing, at least in the known universe even in Plank's units&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(20000 and 40000 could of course be changed for lesser numbers, &#xA;but I like concrete examples)&lt;/p&gt;&#xA;" OwnerUserId="1916" LastEditorUserId="1916" LastEditorDisplayName="" LastEditDate="2011-02-15T18:26:37.813" LastActivityDate="2011-02-15T20:27:49.130" Title="what are dimensions?" Tags="&lt;space&gt;&lt;dimensionality&gt;" AnswerCount="5" />
  <row Id="5218" PostTypeId="2" ParentId="5217" CreationDate="2011-02-15T18:31:01.863" Score="5" ViewCount="" Body="&lt;p&gt;It's true that if you just have a set of points, with no additional mathematical structure, the notion of &quot;dimension&quot; is problematic as you say. But the spaces we deal with in physics usually have extra structures that make the notion well-defined. Often, the definition works by making precise a notion of different &quot;directions&quot; at a given point, and then finding a way of counting how many of those directions are independent of each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, we often work with vector spaces (in which it makes sense to talk about adding vectors, etc.). The dimension of a vector space is well-defined: it's the maximum number of independent vectors you can find (such that no linear combination of them add to zero). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, we often talk about geometrical spaces such as differentiable manifolds, in which there's a notion of &quot;smoothness&quot; of some sort. Once again, these manifolds have a well-defined notion of dimension, and once again it's essentially the number of independent directions you can identify at any given point. In fact, for a smooth manifold one way to define the dimension is to note that you can smoothly map a subsection of the manifold onto a vector space, and then figure out the dimension of the vector space as before.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-15T18:31:01.863" />
  <row Id="5219" PostTypeId="2" ParentId="5217" CreationDate="2011-02-15T18:53:35.217" Score="1" ViewCount="" Body="&lt;p&gt;HDE,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ted Bunn has outlined the mathematical answer, but you have come at this question as a programmer with a programming example. Well in that case consider the programming construct of an unbounded 2 dimensional Array. Here there are two independent values for each pixel, and there is no limit to how large each array dimension value can become. In your model there is always a limit as it &quot;wraps around&quot;. With such an Array (admittedly a slight idealisation for most real languages) we have two independent values for each pixel: hence two dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason dimensions are used is partly that there is no assumed limit for each direction/dimension in Physics. If we &lt;em&gt;knew&lt;/em&gt; that the Universe had exactly N particles and M states (as some occasionally postulate), then the mathematics used to model it might be able to borrow from the simpler 1 dimensional computer models.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-15T18:53:35.217" CommentCount="2" />
  <row Id="5220" PostTypeId="2" ParentId="5217" CreationDate="2011-02-15T18:58:45.743" Score="1" ViewCount="" Body="&lt;p&gt;You can interpret the number of dimensions as the minimum amount of information required to uniquely specify an object in the &quot;space&quot; you are interested in. What you are doing, essentially, is providing an enumeration system with a specified counting scheme (&quot;turning about the edges&quot;). However when you subtract one piece of information, you are introducing another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is qualitative, and I do not know how to make it rigorous, but here is an example of what I mean to say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us say I wish to describe the position of points lying on the circumference of a circle (2D plane). Then you can either &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt; construct one of the coordinate systems. A point in such a system is described by a pair of numbers that is unique for every point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, another way would be to &#xA;&lt;strong&gt;2.&lt;/strong&gt;  fix a point on the circumference and denote it your origin, and then say that every other point is labeled by just &lt;em&gt;one&lt;/em&gt; _number_ which is the distance the point would have moved &lt;strong&gt;along&lt;/strong&gt; that circle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in #&lt;strong&gt;2&lt;/strong&gt; you have introduced a constraint (additional information).  The reduction of the number of degrees of freedom by the introduction of holonomic constraints in classical mechanics might serve as an analogy. &lt;/p&gt;&#xA;" OwnerUserId="1671" LastEditorUserId="1671" LastEditorDisplayName="" LastEditDate="2011-02-15T20:27:49.130" LastActivityDate="2011-02-15T20:27:49.130" CommentCount="3" />
  <row Id="5221" PostTypeId="2" ParentId="5217" CreationDate="2011-02-15T19:07:24.093" Score="2" ViewCount="" Body="&lt;p&gt;I'll just add to the others' answers by saying that I'm not sure the passage from discrete to continuous is so easy: even a finite continuous square could not be &quot;numbered&quot; (i.e.: a function (x,y) -&gt; z defines such that each z is unique for a point) in the way you mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;trick&quot; in discrete space is that you can use the modulo to split the number in two distinct scalar quantities, while the same is not true for real (continuous) numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Thanks to Ted Bunn, who has illuminated me to the existence of &lt;a href=&quot;http://en.wikipedia.org/wiki/Space_filling_curve&quot; rel=&quot;nofollow&quot;&gt;space-filling curves&lt;/a&gt;. The article talks about the existence of a bijective relation between a closed surface and a segment, which is all that is needed to extend the ideas in the question.&lt;/p&gt;&#xA;" OwnerUserId="1933" LastEditorUserId="1933" LastEditorDisplayName="" LastEditDate="2011-02-15T19:27:27.493" LastActivityDate="2011-02-15T19:27:27.493" CommentCount="0" />
  <row Id="5222" PostTypeId="2" ParentId="5141" CreationDate="2011-02-15T19:08:41.257" Score="1" ViewCount="" Body="&lt;p&gt;Lots of materials change under pressure. Many of the products you use everyday wouldn't exist without high pressure processes. Practically any material has a denser form when subjected to high intensity pressure. It's also likely to change the chemical bonding in the process of applying the high pressure as bond energy is inter-reliant on pressure and temperature. Increasing the internal pressure and seeing the results of this would be a good experiment to carry out on a material, as temperature is constant (ideally) when pressure increases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best results are when you combine temperature and pressure to create denser materials. Sintering (not pressure-less sintering mind you) is a great example of this, although generally if you do it with metals you'll want to do it in an inert gaseous atmosphere. Lots of nuclear materials rely on sintering, ytteria oxide powder is a good example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron at the core, depending on whether it has impurities and is at a given phase will certainly be denser at the core. I'm going to assume given the nature of solid solutions and other minerals down there that it'll probably be quite dense.    &lt;/p&gt;&#xA;" OwnerUserId="1936" LastActivityDate="2011-02-15T19:08:41.257" CommentCount="1" />
  <row Id="5223" PostTypeId="2" ParentId="5217" CreationDate="2011-02-15T19:11:38.600" Score="3" ViewCount="" Body="&lt;p&gt;Let me try to answer differently, in some sense more mathematically, and more directly to the point. Intuitively, the dimensionality is the number of independent numbers you need to identify a point. However, the redefinitions of the coordinates need to be continuous which prohibits your &quot;labeling of the pixels&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, one can easily define one-to-one maps from $R^d$ to $R$. For example, the point&#xA;$$(0.147346,0.295002,0.139523)$$&#xA;may be identified with a single number&#xA;$$0.121493759305402623.$$&#xA;I have simply taken the digits from all the three coordinates - from 1st; 2nd; 3rd; 1st; 2nd; 3rd, and so on. However, this map from $R^3$ to $R$ is not continuous so one is not allowed to do it when he computes the dimension. In the example above, only the 18 digits behind the decimal point(s) were rearranged but the same thing can clearly be done with the fully precise real numbers, too. In &quot;set theory&quot;, an abstract branch of mathematics, they would interpret this construction as a proof that the sets $R$ and $R^3$ have the same cardinality (a generalized number of elements): the set theorists call the number of elements in either set simply &quot;continuum&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In mathematics, the dimensionality - the number of dimensions - is defined by concepts such as Hausdorff dimension:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hausdorff_dimension&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hausdorff_dimension&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You try to cover your manifold by a minimum number $N$ of balls of small radius $r$; a notion of a distance has to be available. A ball is a set of point with the distance from the center smaller or equal to $r$. How many balls you need? Well, the volume scales like $r^d$, so you will need $V/r^d$ balls or so. If you take the logarithm of the number of balls, it will be $\ln(V)-d\ln(r)$, and by sending $r\to 0$ i.e. $\ln(r)\to-\infty$, you may extract the coefficient $d$ by a limiting procedure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amusingly, this definition also works for fractals that can have fractional dimensions. The simplest ones have dimensionalities that are ratios of logarithms of simple numbers (e.g. integers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In physics, we use the space to define theories and we differentiate the coordinates and fields with respect to time (and space, in the case of field theory). That's why the condition of &quot;continuity&quot; or &quot;smoothness&quot; is automatically required in physics. In particular, local quantum field theories must live in a spacetime with a well-defined number of spacetime dimensions. Relabeling the coordinates in your &quot;pathological way&quot; doesn't preserve the &quot;differentiable structure&quot; on the spacetime manifold, and because physics depends on derivatives (differentiation), such redefinitions would make physical laws meaningless. Also, quantum field theories are defined on a spacetime that has a well-defined metric tensor; so I can use the Hausdorff definition to count the dimension, too. (One should avoid the &quot;balls&quot; for the Minkowski case which has an indefinite metric.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these matters become ill-defined in quantum gravity, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Hausdorff dimension definition, I needed to consider arbitrarily small balls. But there are no objects smaller than the Planck length - the characteristic (tiny) distance scale of quantum gravity. Consequently, the limiting procedure cannot be performed in quantum gravity. It follows that the number of dimensions of spacetime isn't quite well-defined. Only the dimensions that are &quot;large&quot; - where the balls may still be much smaller than the size of these dimensions - have a physical meaning. However, the number of dimensions that are as small as the Planck length or so (or as curved as the Planck length curvature radius) can't be defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And indeed, there are often equivalent descriptions of a theory that actually disagree about the number of dimensions. For example, M-theory on a K3 manifold - which has 11 dimensions in total - is equivalent to heterotic strings on a 3-torus - which only have 10 spacetime dimensions. Also, the AdS/CFT correspondence shows that theories in spacetimes of different dimensions (by one) are equivalent: the radial, &quot;holographic&quot; dimension is invisible in the Lagrangian of CFT which is linked to its significant curvature.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-15T19:19:42.087" LastActivityDate="2011-02-15T19:19:42.087" CommentCount="1" />
  <row Id="5224" PostTypeId="1" AcceptedAnswerId="5225" CreationDate="2011-02-15T19:45:54.237" Score="2" ViewCount="302" Body="&lt;p&gt;The Schrodinger equation governs the possible time evolution of a wave function, expressed as a partial differential equation.  Isn't this equivalent to the simpler equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega = \hbar k^2/2m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. any wave function that satisfies this dispersion relation will also satisfy Schrodinger's equation, and the equation above is a lot easier to understand?  &lt;/p&gt;&#xA;" OwnerUserId="1602" LastEditorUserId="1602" LastEditorDisplayName="" LastEditDate="2011-02-15T21:51:23.597" LastActivityDate="2011-02-15T22:12:01.983" Title="Why is it necessary to represent Schrodinger's equation as a partial differential equation?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" />
  <row Id="5225" PostTypeId="2" ParentId="5224" CreationDate="2011-02-15T19:52:28.440" Score="11" ViewCount="" Body="&lt;p&gt;Your equation is the right solution to Schrödinger's equation in the momentum-energy representation. However, it's only that simple for Schrödinger's equation with no potential, $V(x,y,z)=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's zero, the solution (or, similarly, the reformulation of the equation) is as easy as the algebraic relationship you wrote - but it's also uninteresting for the same reason. The interesting cases have e.g. the Coulomb potential $k/r$ or the harmonic oscillator potential $kx^2/2$ and they can't be &quot;solved&quot; in the simple way you sketched. For a nonzero potential, the problem is genuinely equivalent to a partial differential equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, that doesn't mean that it's the only way in which the problem may be formulated or solved. Both the harmonic oscillator and the Hydrogen atom may be solved (i.e. their spectrum may be found) algebraically, by the creation and annihilation operators in the harmonic oscillator case, or by a hidden $SO(4)$ symmetry in the Hydrogen atom case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A general Schrödinger's equation in quantum mechanics is really an ordinary differential equation for the state vector; the &quot;spatial derivatives&quot; only appear as the action of particular operators on the Hilbert space. Some of these operators - e.g. the momenta in the position representations - are conveniently represented as partial derivatives with respect to spatial coordinates but that's only the case if we use a &quot;continuous basis&quot; for the Hilbert space.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-15T19:52:28.440" CommentCount="2" />
  <row Id="5226" PostTypeId="1" AcceptedAnswerId="5511" CreationDate="2011-02-15T19:52:53.617" Score="9" ViewCount="400" Body="&lt;p&gt;A truly reversible thermodynamic process needs to be infinitesimally displaced from equilibrium at all times and therefore takes infinite time to complete.  However, if I execute the process slowly, I should be able to get close to reversibility.  My question is, &quot;What determines when something is slow?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For definiteness, let's take an insulating cylindrical piston with cross-sectional area $A$ and original length $L_0$.  There is an ideal gas inside with $n$ molecules of gas with mass-per-molecule $m$.  The temperature is $T_0$, and the adiabatic index is $\gamma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plan to expand the piston adiabatically to length $L_1$, taking a time $t$ to do so.  If I take $t$ to be long enough, the process will be nearly-reversible.  However, $t$ being long does not mean &quot;one minute&quot; or &quot;one year&quot;.  It means $t &amp;gt;&amp;gt; \tau$ for some &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau = f(A, L_0, L, n, m, T, k_B, \gamma)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is $\tau$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From purely dimensional considerations, I guess the relationship is something like &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau = \sqrt{\frac{mLL_0}{k_bT}}f(n, L/L_0, A/L^2, \gamma)$$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I don't have a strong physical explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit&lt;/b&gt;&#xA;A meaningful answer should let me do the following:  I take a certain example piston and try expanding it a few times, putting it in a box so I can measure the heat released to the environment.  I calculate the entropy change in the universe for the expansions.  After doing several expansions, each slower than the last, I finally get $\Delta S$ for the universe down to a number that I think is sufficiently small.  Next, I plan to repeat the experiment, but with a new piston that has different dimensions, different initial temperature, etc.  Based on my results for the previous piston, how can I figure out how long I should take to expand the new one to achieve a similar degree of reversibility on the first try?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, the pressure is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = \frac{nk_BT}{V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the speed of sound is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = \sqrt{\frac{\gamma k_b T}{m}}$$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I'm happy to have an answer in terms of these or other derived quantities.  Formulas for entropy and thermodynamic potentials can be found in &lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_gas&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-02-16T18:22:27.733" LastActivityDate="2011-02-22T09:32:30.470" Title="How slow is a reversible adiabatic expansion of an ideal gas?" Tags="&lt;thermodynamics&gt;&lt;reversibility&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="5227" PostTypeId="2" ParentId="5224" CreationDate="2011-02-15T20:12:36.847" Score="4" ViewCount="" Body="&lt;p&gt;Adrianos, similar but relativistic dispersion relation was written by Louis de Broglie for a free electron. The corresponding solutions was a plane wave $e^{i(pr-Et)/\hbar}$. It's a wave in a 3D space. E. Schroedinger wrote the Klein-Gordon equation for such a wave (partial derivatives needed). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then he tried to include an external potential and arrived at his non-relativistic equation with an external potential where there is no dispersion relation (there are discrete energy levels). Read history of QM.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-15T22:12:01.983" LastActivityDate="2011-02-15T22:12:01.983" CommentCount="1" />
  <row Id="5228" PostTypeId="1" CreationDate="2011-02-15T20:41:32.830" Score="2" ViewCount="140" Body="&lt;p&gt;On a principal bundle, at each point you have a tangent vector space. At a given point, the vectors tangent to the fiber form the vertical vector space. Then the vector space at that point is a direct sum of the vertical vector space and what's called the horizontal vector space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all standard stuff in gauge theory. What I don't intuitively understand is why is the horizontal vector space not unique?&lt;/p&gt;&#xA;" OwnerUserId="1937" LastActivityDate="2011-02-15T23:02:36.543" Title="On a principal bundle, why is the horizontal vector space not unique?" Tags="&lt;mathematical-physics&gt;" AnswerCount="2" />
  <row Id="5229" PostTypeId="2" ParentId="5195" CreationDate="2011-02-15T21:33:07.357" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Potential for 2D problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start with a 2D disk and try to solve the general problem for infinitesimally flat disk.  I will change notations a bit -- the surface resistance will be $\sigma$ and the radius of the disk will be $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with basic electrodynamics:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec{j} = -\sigma\frac{\partial u}{\partial \vec{r}},\, div\vec{j}=0\,\Rightarrow\,\Delta u = 0$ with the boundary condition: $\vec{n}\cdot\vec{j} = 0 \Rightarrow \vec{n}\frac{\partial u}{\partial \vec{r}} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's first consider the current $I$ flowing into the surface in the centre and uniformly flowing away from the edges. solution for potential is well known:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U(r,\phi) = -\frac{I}{2\pi\sigma}\, \ln r$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use the conformal map $z\to a\frac{z-s}{a^2-s*z}$ to &quot;shift the centre&quot; into the point $s=x_{source}+iy_{source}$. The potential is then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U(r,\phi) = -\frac{I}{2\pi\sigma}\, \ln\left|a\frac{re^{i\phi}-s}{a^2-s^*re^{i\phi}}\right|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I substract the similar potential, with different parameter $d=x_{drain}+iy_{drain}$ to compensate the outgoing flow. Obtaining:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U(r,\phi) = -\frac{I}{2\pi\sigma}\, \ln\left|\frac{re^{i\phi}-s}{re^{i\phi}-d}&#xD;&#xA;\cdot\frac{a^2-d^*re^{i\phi}}{a^2-s^*re^{i\phi}}\right|$ or $U(z) = -\frac{I}{2\pi\sigma}\, \ln\left|\frac{z-s}{z-d}&#xD;&#xA;\cdot\frac{a^2-d^*z}{a^2-s^*z}\right|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the harmonic function, satisfying the boundary conditions. &lt;a href=&quot;http://www.wolframalpha.com/input/?i=Log%28Abs%28%28x%2Biy-2%29%289%2Bx%2Biy%29%2F%28%28x%2Biy%2B1%29%289-2%28x%2Biy%29%29%29%29%29&quot; rel=&quot;nofollow&quot;&gt;You can play here with it.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Interpretation of the solution&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The potential is divergent in points $s$ and $d$. This happens because the resistance is strongly dependend on the microscopic details of the problem. Indeed -- as you get closer to the source -- all your current have to pass through smaller and smaller amount of conductor. And in the limit of infinitely small source you get infinite resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Formulation issue&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I admit that while solving I first fixed the current and then found the potential, while you formulated the problem differently -- &lt;em&gt;&quot;set the potential here and there and find the current&quot;&lt;/em&gt;. But let us use logic:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Nonzero current leads to infinite voltage: $I\neq0\,\Rightarrow\,\Delta U \to \infty$.&lt;/li&gt;&#xA;&lt;li&gt;If $A\Rightarrow B$, then $!B\Rightarrow !A$.&lt;/li&gt;&#xA;&lt;li&gt;$\Delta U\mbox{-finite}\,\Rightarrow\,I=0$ &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;At finite voltage you'll get zero current or, equivalently, infinite resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What happens in 3D case?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same thing. Just consider single pointlike source -- and the potential $U\sim\frac{1}{r}$ is divergent. Don't need to go into further details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;Cutoffs&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to move on I introduce the &quot;cutoffs&quot; -- new small (real) quantities $\epsilon_{s,d}$ which denoting &quot;sizes&quot; of the source and the drain. Using them I obtain the voltage:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U(d+\epsilon_d)-U(s+\epsilon_s)=\frac{I}{2\pi\sigma}\left[\ln\frac{\epsilon_s}{|s-d|}+\ln\frac{\epsilon_d}{|s-d|} +\ln\left|\frac{a^2-s^*d}{a^2-|s|^2}\cdot\frac{a^2-sd^*}{a^2-|d|^2}\right|\right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scales&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting together everything above. One can say that in the problem there are  &lt;em&gt;four&lt;/em&gt; (or, even five) scales:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Radius of the disk.&lt;/li&gt;&#xA;&lt;li&gt;Thickness of the disk. &lt;/li&gt;&#xA;&lt;li&gt;Distance between contacts $|s-d|$&lt;/li&gt;&#xA;&lt;li&gt;Sizes of those contacts $\epsilon_{s,d}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Since you are talking about &quot;points&quot; -- then first we have to take $\epsilon_{s,d}\to0$, right? But if $\epsilon_{s,d}$ is much smaller that any other scale then they introduce divergent contribution into the resistance. And any other detail of the problem becomes irrelevant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the answer to your question is: &lt;strong&gt;The resistance between two points is infinite, whatever the geometry of the problem is.&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2011-02-17T14:33:30.620" LastActivityDate="2011-02-17T14:33:30.620" />
  <row Id="5230" PostTypeId="2" ParentId="5228" CreationDate="2011-02-15T22:05:09.963" Score="1" ViewCount="" Body="&lt;p&gt;It's not unique simply because its definition - it's another vector space in the direct sum that must be equal to the total vector space - doesn't determine it uniquely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the vertical vector space is uniquely determined - you may perform a simple test whether a particular vector from the total vector space belongs to the vertical vector space or not - a similar test doesn't exist for the horizontal vector space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's like writing the two-dimensional plane as a direct sum of two vector spaces. The vertical vector space may be determined - for example, it may contain all vertical vectors of the form $(0,y)$ for a real $y$. However, the horizontal vector space may be chosen to be the space of all vectors of the form $x(1,a)$ where $a$ is any fixed number (labeling different choices of the horizontal vector space) and $x$ takes any real value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could think that the vectors $(x,0)$ are preferred - i.e. the choice $a=0$ is privileged. But that's only true if you want to construct an orthogonal basis - or if you had a reason to take the adjective &quot;horizontal&quot; literally. In general, the two vector spaces don't have to be orthogonal, and in some cases, an inner product isn't even available. And there's no God-given definition of &quot;horizontality&quot; (unlike my Cartesian example). The two-dimensional plane may be generated from pretty much any basis with 2 vectors, so even if you choose one of them, the other vector may be anything (except for multiples of the first one) to get the right direct sum. And all these solutions are equally good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example above easily generalizes to the case when both spaces are multi-dimensional.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-15T22:05:09.963" CommentCount="1" />
  <row Id="5232" PostTypeId="1" AcceptedAnswerId="5248" CreationDate="2011-02-15T23:01:40.963" Score="1" ViewCount="236" Body="&lt;p&gt;The quarks and leptons chiral states have hypercharges of $0,\pm 1/3,\pm 2/3, \pm 3/3, \pm 4/3, \pm 6/3$. The value of $\pm 5/3$ is prominently missing. Is there a theoretical principle which picks out these values?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In looking for papers on arXiv with hypercharge 5/3, I've found only a handful of papers. One is a preprint from December 2010 (See page 16 &quot;KK (Kaluza-Klein) towers&quot;) : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Giuliano Panico, Mahmoud Safari, Marco Serone, (2010) &lt;em&gt;Simple and Realistic Composite Higgs Models in Flat Extra Dimensions&lt;/em&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/1012.2875&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1012.2875&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then they come up as a vector diquark here and in a few similar papers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Journal of High Energy Physics 2002, 024, (2002), Engin Arik, Serkant A. Çetin, Orhan Çakir, and Saleh Sultansoy, &lt;em&gt;A Search for Vector Diquarks at the CERN LHC&lt;/em&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/hep-ph/0109011v3&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-ph/0109011v3&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is there a theoretical principle that picks out which weak hypercharge values might be observed in nature?&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-02-15T23:15:59.200" LastActivityDate="2011-03-02T04:44:18.060" Title="What restricts the value of weak hypercharge from being 5/3?" Tags="&lt;particle-physics&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="5233" PostTypeId="2" ParentId="5228" CreationDate="2011-02-15T23:02:36.543" Score="1" ViewCount="" Body="&lt;p&gt;Giving a horizontal bundle is equivalent as giving a connection. It is one of the definitions actually. Connections are not unique. They are extra structure on the bundle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ehresmann_connection&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Ehresmann_connection&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2011-02-15T23:02:36.543" />
  <row Id="5234" PostTypeId="2" ParentId="5185" CreationDate="2011-02-15T23:14:56.423" Score="1" ViewCount="" Body="&lt;p&gt;Perhaps I am just displaying my ignorance on the subject, but I will stick out my neck and try to answer - I am grateful for comments, especially if my picture is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ad first question: I would have pictured the dissipation as follows. By switching on some perturbation you bring the black hole into an excited state that can be treated perturbatively as long as it is a small enough excitation. This then leads to quasi-normal ringing of the black hole. The perturbed black hole state falls back to the unperturbed black hole state by emission of gravitational waves towards the AdS boundary. Thus, the dissipation process on the gravity side rather seems to be energy transported from the UV to the IR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ad second question: By the UV/IR connection the quasi-normal ringing on the gravity side should on the fluid side correspond to energy in the IR being dissipated into the UV. I would have thought that this is precisely what happens in a viscous fluid: you have some long wavelength motion that gets decelerated due to friction, which is basically small wavelength excitations that then thermalize with the heat bath.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References on quasi-normal modes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In AdS: hep-th/9909056 &#xA;In general relativity: gr-qc/9909058&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-02-15T23:14:56.423" CommentCount="3" />
  <row Id="5235" PostTypeId="1" AcceptedAnswerId="5240" CreationDate="2011-02-15T23:15:11.663" Score="4" ViewCount="148" Body="&lt;p&gt;If someone (or some robot) on the moon were to point a laser at the Earth, how many watts would the laser need, to be easily seen with the un-aided eye from Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously this depends on a number of factors. Assume the beam is made divergent enough to cover latitudes 60S to 60N, wavelength 532nm, and the beam originates from a fully lit area of the moon, and the observing is done in full darkness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can work out watts/meter-square from the laser, but don't know how to work out the watts/meter-square from the moon, and don't have a clue about how the eye would perceive the contrast.&lt;/p&gt;&#xA;" OwnerUserId="717" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-16T10:42:47.713" LastActivityDate="2011-02-16T10:42:47.713" Title="Visibility of moon-based laser from Earth" Tags="&lt;optics&gt;&lt;radiometry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5236" PostTypeId="1" AcceptedAnswerId="5263" CreationDate="2011-02-16T00:51:18.643" Score="2" ViewCount="272" Body="&lt;p&gt;We know that an operator A in quantum mechanics has time evolution given by Heisenberg equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\frac{i}{\hbar}[H,A]+\frac{\partial A}{\partial t}=\frac{d A}{d t}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we derive from this that &#xA;$$&#xD;&#xA;A(t)=e^{\frac{i}{\hbar}Ht}A(0)e^{-\frac{i}{\hbar}Ht} \qquad ?&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;L.M.: I added $i/\hbar$ in front of $[H,A]$.&lt;/p&gt;&#xA;" OwnerUserId="558" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-03-22T13:15:25.093" LastActivityDate="2011-03-22T13:15:25.093" Title="Time evolution in quantum mechanics" Tags="&lt;quantum-mechanics&gt;&lt;mathematical-physics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="5238" PostTypeId="2" ParentId="5190" CreationDate="2011-02-16T01:14:48.613" Score="0" ViewCount="" Body="&lt;p&gt;I really liked Ethan Siegel's explanation of this over at ScienceBlogs:&#xA;&lt;a href=&quot;http://scienceblogs.com/startswithabang/2010/09/geocentrism_was_galileo_wrong.php&quot; rel=&quot;nofollow&quot;&gt;http://scienceblogs.com/startswithabang/2010/09/geocentrism_was_galileo_wrong.php&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-02-16T01:14:48.613" />
  <row Id="5239" PostTypeId="2" ParentId="5235" CreationDate="2011-02-16T01:32:29.093" Score="6" ViewCount="" Body="&lt;p&gt;I afraid if you going to light entire earth with a &quot;laser&quot; it will not be visible from earth.&#xA;With the same success you can take 10kW xenon lamp which is easily focusable to the required beam for way lower price, weight and better efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rough estimation: You are lighting 4000km x 4000km area, human eye has surface 7x7mm, so you will 'see' 0.32x10^-18 of the power you emit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 photon (555nm) has energy 1.54x10-12 J, so if you will have 1000W laser, your eye will get 1 photon each 5000 seconds. Which is obviously very very far away from being visible.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-02-16T01:32:29.093" CommentCount="2" />
  <row Id="5240" PostTypeId="2" ParentId="5235" CreationDate="2011-02-16T01:51:23.857" Score="6" ViewCount="" Body="&lt;p&gt;There are two things that make it difficult to see the laser: 1) the large area illuminated (as BarsMonster pointed out) and 2) the resolving power of the eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eye has an angular resolution of roughly 10^-4 radians (from the diffraction limit), so your eye will &quot;blur&quot; the light from the laser with all the light from the moon's surface within an angle of 10^-4.  Using the earth-moon distance of 4e8 m, that means you're competing with the light from a 10^9 m^2 area.  Since you say it's a fully lit area of the moon, I'll take an intensity of 1 kW/m^2 for the solar intensity and assume the moon is white (a bad assumption) for a total power of 10^12 W.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The light from the moon's surface is scattered into something like 2-pi of solid angle (as opposed to the laser).  So the fraction of light that will be competing with the laser (assuming the laser diameter at the earth is 4000 km) is roughly ((4000 km)/(400000 km))^2= 10^-4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we want the intensity from the laser to equal the intensity of the unresolved scattered light from the moon (which, as a ballpark figure, should make it noticeably green and thus &quot;easily seen&quot;), we need a laser that's 10^8 Watts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck building that.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-02-16T01:51:23.857" CommentCount="4" />
  <row Id="5242" PostTypeId="2" ParentId="4669" CreationDate="2011-02-16T01:54:44.867" Score="-5" ViewCount="" Body="&lt;p&gt;Three ideas with disruptive potential remain undiscussed and are public since 2002/3. All Darkness can be enlightened now. They do not colide with SR/GR nor QM and do not need any speculative entity. The authors provide physical models, and math, that fits the observed data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - &lt;strong&gt;Is it spacetime or, is it space + time?&lt;/strong&gt; How sure we are?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 - &lt;strong&gt;Is it space expansion or a matter shrinking scenario, or a combination of both?&lt;/strong&gt; How sure we are?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 - &lt;strong&gt;a particle model by Douglas Pinnow 'The Resonant Universe'&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questioning our rooted beliefs (or myths?):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - We used to assume that the atom sizes are absolute and we have no physical basis to assert that. This presumption is an obfuscation to the understanding of the properties of the universe and time. This paper analyses the effects that arise from the perspective  of an &lt;strong&gt;'instant observer'&lt;/strong&gt;, leading to the natural split of spacetime into space + time.  A &lt;strong&gt;scale-invariant, and self-similar&lt;/strong&gt;, model of the Universe is presented there. All we know is that the ratio 'space/matter' is getting greater. Because we use 'atomic properties' to take measures we can explore 'more space', 'less matter' or both, as time goes by.&#xA;An independent derivation of the Einstein Special Relativity is also presented. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/physics/pdf/0205/0205033v1.pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Cosmological Principle and Relativity - Part I&lt;/strong&gt;  (arxiv physics&#xA;0205033)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;2 - This paper extends the above one and is easier to follow. The author presents a model of our Universe that is scale-invariant (self-similar) where Dark Energy is not invited and has only one parameter, the Hubble constant. He explores the consequences of it in the past history, and future, of the Solar System and the Earth and the primordial ambient of Life's Origin. The comparison is made with the values of SM. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/astro-ph/pdf/0208/0208365v1.pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;A relativistic time variation of matter/space fits both local and&#xA;cosmic data&lt;/strong&gt; (arxiv astro-ph&#xA;0208365)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;3 - Pinnow presents a monograph with a new model of particles, based only on EM (Electromagnetism), and with only one parameter - the electron mass. He derives the main particles properties to within 1% of their values, and the barionic masses well bellow 1%, that does not suffer the barionic 'spin crisis'. His predictions are different from the SM expectations, no Higgs particle ;) and can be tested with the data obtained at LHC. He follows the steps of Goedecke, Haus and others, on the non-radiating condition of the EM field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;resuming:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 1) and 2) - a ONE parameter model - there is no space expansion, no BB, no Inflation era, no DE, no DM, no cosmological constant, and make me understand better some issues like the GRBs properties, the 'isotope stability valley' and understand the why of the measured variability of the fine structure constant  (by Webb et al), and the Pioneer 10 and 11 anomaly, and supermassive BH, etc,etc ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 3) - a ONE parameter model - All particle mass, the rest mass, can be converted to energy avoiding the formation of BH. Thanks, because I feel unconfortable with Black Holes, and this is not possible under SM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a follower, not the author. As the mammals were born in the time of the dinosaurs also revolutionary ideas are born on the fringes of mainstream thinking. &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-02-16T01:54:44.867" CommunityOwnedDate="2011-02-16T01:54:44.867" />
  <row Id="5243" PostTypeId="1" CreationDate="2011-02-16T02:08:32.160" Score="2" ViewCount="524" Body="&lt;p&gt;There's a conventional wisdom that the best way to minimize the force impact of a punch to the head is to lean into it, rather than away from it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true? If so, why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Hard to search for where I got this CW, but heres &lt;a href=&quot;http://www.xomba.com/how_to_take_a_blow_to_the_head&quot; rel=&quot;nofollow&quot;&gt;one&lt;/a&gt;, and &lt;a href=&quot;http://www.wikihow.com/Take-a-Punch&quot; rel=&quot;nofollow&quot;&gt;another&lt;/a&gt;. The reason it seems counter-intuitive is that I'd think if you move in the direction that a force is going to collide into you with, the collision would theoretically be softer. You see that when you catch a baseball barehanded; it hurts much more when you move towards the ball, rather than away from the ball, as it hits your hand. &lt;/p&gt;&#xA;" OwnerUserId="585" LastEditorUserId="585" LastEditorDisplayName="" LastEditDate="2011-07-20T04:12:25.330" LastActivityDate="2011-07-20T04:12:25.330" Title="Should you really lean into a punch?" Tags="&lt;forces&gt;&lt;collision&gt;" AnswerCount="9" CommentCount="2" FavoriteCount="2" />
  <row Id="5244" PostTypeId="2" ParentId="5158" CreationDate="2011-02-16T02:25:17.567" Score="1" ViewCount="" Body="&lt;p&gt;use an &lt;strong&gt;analog clock&lt;/strong&gt; and measure &lt;strong&gt;the angle traveled by the minute hand&lt;/strong&gt; (or the second which is faster but is less time to take notes) over time. It is cyclical and can introduce the functions sine, cosine,.. and uses the same clock for measuring the elapsed time;)&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-02-16T02:25:17.567" CommentCount="1" />
  <row Id="5245" PostTypeId="2" ParentId="5243" CreationDate="2011-02-16T02:29:06.647" Score="6" ViewCount="" Body="&lt;p&gt;Without having heard this argument before I would guess that the plan it to reduce the degree to which the head rattles around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the &lt;em&gt;brain&lt;/em&gt; damage (short term and long term) associate with a punch comes from the brain bouncing off the skull a few times as the head whips back and forth. Minimize the motion, minimize the damage. By leaning in you get a whole complex of muscles pre-tensioned to absorb the impact smoothly and give yourself some room to rock back without losing your balance. Your facial bones will heal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, trying to analyze anything related to fighting in terms of physics is &lt;em&gt;hard&lt;/em&gt; because bio-mechanics is always a complicated subject with a lot of subtle inter-relationships.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-02-17T22:14:28.563" LastActivityDate="2011-02-17T22:14:28.563" CommentCount="3" />
  <row Id="5246" PostTypeId="1" CreationDate="2011-02-16T02:31:38.957" Score="3" ViewCount="145" Body="&lt;p&gt;Now in disordered organics, the band picture is thrown out the window, from what I can tell (due to lack of symmetry). But don't HOMO/LUMO levels basically take the place of conduction/valence bands in molecules? In a organic system (a lot of molecules with no order), then, I am correct to believe that the HOMO and LUMO levels broaden into a Guassian density of states? However, the LUMO 'band' does not act like a conduction band in that the states are still localized. From my reading, it appears that the Fermi Level is in the LUMO 'band'. My question is, how can the LUMO 'band' have filled states since a single molecule of course has no electrons in the LUMO. What am I missing?&lt;/p&gt;&#xA;" OwnerUserId="1939" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-16T10:36:38.313" LastActivityDate="2011-02-16T11:18:34.270" Title="Energy levels in disordered organic semiconductors?" Tags="&lt;condensed-matter&gt;&lt;physical-chemistry&gt;&lt;semiconductor-physics&gt;" AnswerCount="1" />
  <row Id="5247" PostTypeId="2" ParentId="5243" CreationDate="2011-02-16T02:40:31.117" Score="1" ViewCount="" Body="&lt;p&gt;Possibly because the punch has less force behind it, if you meet it earlier. Can't say I ever heard of this conventional wisdom, but then again, I haven't been into many fights or rings. &lt;/p&gt;&#xA;" OwnerUserId="1939" LastActivityDate="2011-02-16T02:40:31.117" CommentCount="1" />
  <row Id="5248" PostTypeId="2" ParentId="5232" CreationDate="2011-02-16T02:45:02.620" Score="2" ViewCount="" Body="&lt;p&gt;Hypercharge assignments in the Standard Model (that is with just the standard leptons and quarks) are determined up to a twofold ambiguity by demanding cancellation of all anomalies. Here's a reference:J.A. Minahan, Pierre Ramond, (Florida U.) , R.C. Warner, (Melbourne U.) . UFIFT-HEP-89-16, Jun 1989. 6pp. &#xA;Published in Phys.Rev.D41:715,1990.&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T04:44:18.060" LastActivityDate="2011-03-02T04:44:18.060" CommentCount="4" />
  <row Id="5250" PostTypeId="2" ParentId="5243" CreationDate="2011-02-16T04:07:12.207" Score="1" ViewCount="" Body="&lt;p&gt;The only possible reason I can see is that the earlier, one faces the punch the less amount of momentum he would face. A punch builds its momentum during its journey towards destination and it reaches its maximum at the end of it. It the target comes close then he would naturally encounter less momentum and hence less force for the change of momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force $F =\frac{(mv - 0)}{\delta t}$ =$\frac{mv}{\delta t}$ if the target lean towards the punch when its seed is v&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force $F\prime = \frac{mV - 0}{\delta t}$ = $\frac{mV}{\delta t}$ if the target does not move and the punch reach its maximum speed V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $ V &amp;gt; v$,   $F\prime &amp;gt; F$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases most of the transferred momentum will be absorbed and carried by the target's muscles and he will feel the pain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It explains why the punch will be less if the target moves towards the punch.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-16T06:45:54.377" LastActivityDate="2011-02-16T06:45:54.377" CommentCount="2" />
  <row Id="5251" PostTypeId="2" ParentId="5194" CreationDate="2011-02-16T05:10:42.640" Score="1" ViewCount="" Body="&lt;p&gt;The title of your question is about &lt;em&gt;calculating&lt;/em&gt; the speed of light ($c$), but the body asks about &lt;em&gt;measuring&lt;/em&gt; $c$. Others have answered you on the measurement issue, but I'd like to include a bit about the calculation of $c$ from principles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light, as an electromagnetic phenomenon, is described by Maxwell's equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{eqnarray}&#xD;&#xA;&#xD;&#xA;\nabla \cdot E &amp;amp;=&amp;amp; \frac{\rho}{\epsilon_0}\\&#xD;&#xA;\nabla \cdot B &amp;amp;=&amp;amp; 0\\&#xD;&#xA;\nabla \times E &amp;amp;=&amp;amp; -\frac{\partial B}{\partial t}\\&#xD;&#xA;\nabla \times B &amp;amp;=&amp;amp; \mu_0 J + \mu_0 \epsilon_0 \frac{\partial E}{\partial t}&#xD;&#xA;&#xD;&#xA;\end{eqnarray} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\rho$ is the charge density, $J$ is the current density, $E$ and $B$ are the electric and magnetic fields, respectively, $\mu_0$ is the magnetic permeability of free space, and $\epsilon_0$ is the electrical permittivity of free space. In the absence of any charges, one solution to these equations is a traveling plane wave with velocity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ c= \frac{1}{\sqrt{\mu_0 \epsilon_0}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this leaves the problem of measuring $\mu_0$ and $\epsilon_0$, but is a great demonstration of the fact that light is truly an electromagnetic phenomenon. As an added bonus, $\mu_0$ and $\epsilon_0$ can be measured in a variety of ways, without requiring very high time resolution.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-02-16T05:10:42.640" />
  <row Id="5252" PostTypeId="1" AcceptedAnswerId="5256" CreationDate="2011-02-16T05:19:54.797" Score="-2" ViewCount="182" Body="&lt;p&gt;Why String theory ?  simple cosmological model is not sufficient for study of nature of universe&lt;/p&gt;&#xA;" OwnerUserId="1843" LastEditorUserId="1843" LastEditorDisplayName="" LastEditDate="2011-02-18T05:47:05.863" LastActivityDate="2011-02-18T05:47:05.863" ClosedDate="2011-02-16T07:41:28.957" Title="Why String theory introduced in cosmological models to study nature of universe?" Tags="&lt;general-relativity&gt;&lt;string-theory&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="5253" PostTypeId="2" ParentId="5243" CreationDate="2011-02-16T05:33:50.053" Score="1" ViewCount="" Body="&lt;p&gt;My guess would be that the objective is to reduce the amount of energy dissipated into a localised area, rather than reducing the force of the strike itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that if you lean into a strike, your muscles will flex and become more rigid. The shockwave from the impact will then travel through the rigid muscle instead of being dissipated. The energy will then be spread out over a larger area (and if your stance is good enough, maybe into the ground). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The get an idea of what I mean, compare a newton's cradle made out of steel and one made from flesh like substance. Also similar to boxers wrapping their hands and wrists.&lt;/p&gt;&#xA;" OwnerUserId="1863" LastActivityDate="2011-02-16T05:33:50.053" />
  <row Id="5254" PostTypeId="2" ParentId="2604" CreationDate="2011-02-16T05:34:42.487" Score="2" ViewCount="" Body="&lt;p&gt;I am looking at this question as a particle physicist and as a reader. I suppose you have explained to your dog about potentials and quantum mechanical solutions which allow electrons to be trapped around nuclei, so the dog is familiar with the quantum nature of the world :).&#xA;You could illustrate with a harmonic oscillator and show that given different strengths the energy levels change accordingly. Then you are ready to do an analogue . Each energy level is a &quot;particle&quot; in potentia, if the right material is there. If you have a hydrogen atom yo have one proton and one electron, and you have only one atom of hydrogen, even though there are many energy levels. If you get a helium you fill two energy levels and the rest of the potential lines are free. You can talk about adding energy to get to an excited state and still have the same atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make an analogue of the standard model, see for example the graph in the  &lt;a href=&quot;http://pdg.lbl.gov/2010/reviews/rpp2010-rev-quark-model.pdf&quot; rel=&quot;nofollow&quot;&gt;particle physics book&lt;/a&gt; figure 14.4. Energy input raises a nucleon (three quarks) to a higher &quot;quasi stable&quot; excited state, that contains  new generations of quarks. This gives the argument  that the quarks and leptons that make up our world are the analogue lowest energy levels filled that create the matter we depend on.  The extra generations are there in the same way that the extra levels are there in the simple quantum mechanical problem and may be filled  and appear given enough energy. They are there because of the form of the &quot;potential&quot; that makes them possible in order and groupings that are necessary given the stable matter solutions we observe, which are still at the frontier of current theoretical studies in physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that &lt;a href=&quot;http://inpa.lbl.gov/blnv2/files/Thursday/Session3/Nath.pdf&quot; rel=&quot;nofollow&quot;&gt;higher order terms in QCD&lt;/a&gt; will include all the generations and it might be that the nucleon solution would not be stable if these higher generations were not there, but maybe somebody else could think of an analogue for that.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-02-16T09:01:32.867" LastActivityDate="2011-02-16T09:01:32.867" CommentCount="1" />
  <row Id="5255" PostTypeId="2" ParentId="5246" CreationDate="2011-02-16T06:45:36.207" Score="1" ViewCount="" Body="&lt;p&gt;Yes, your analogy does work. HOMO/LUMO play the same role for organic semiconductors as the valence band and conduction band play for inorganic semiconductors. In fact, there is a whole discipline of organic electronics, see e.g.:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Organic_electronics&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Organic_electronics&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://en.wikipedia.org/wiki/Organic_semiconductors&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Organic_semiconductors&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, plastic solar panels may usefully become organic in the near future. I guess that it's not a good idea to use organic semiconductors for the state-of-the-art miniature microprocessors because the molecules are too still too large, irregular, and &quot;dirty&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, I don't quite understand what you're missing. Everything you write is true: the LUMO band is unoccupied in the ground state - but it has available states that &lt;em&gt;may&lt;/em&gt; be occupied when the voltage makes it energetically preferred. It's exactly the same thing as the conduction band for other semiconductors. In both cases, these states may be shared by the whole material, so they may become delocalized. I recommend you to search Google Scholar for &lt;a href=&quot;http://scholar.google.com/scholar?q=delocalized+lumo&quot; rel=&quot;nofollow&quot;&gt;delocalized LUMO&lt;/a&gt;. All the best, LUMO&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-16T11:18:34.270" LastActivityDate="2011-02-16T11:18:34.270" CommentCount="1" />
  <row Id="5256" PostTypeId="2" ParentId="5252" CreationDate="2011-02-16T06:49:03.333" Score="1" ViewCount="" Body="&lt;p&gt;A simple cosmological model has nothing to predict about the fate of collision of particles at high energies, or any other question about short-distance physics, for that matter. To describe the Universe not just in a broad macroscopic sketch, but also in the microscopic details, one needs a quantum theory, and string theory is the only known consistent theory that reconciles particle physics (such as gauge theories with chiral fermions) and general relativity (the latter including cosmological models).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-16T06:49:03.333" CommentCount="1" />
  <row Id="5257" PostTypeId="2" ParentId="5232" CreationDate="2011-02-16T07:04:54.770" Score="4" ViewCount="" Body="&lt;p&gt;If you consider particular enough models, the assignment of the allowed hypercharges may be easily calculated. The models themselves are constrained by various additional conditions such as anomaly cancellation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, take an $SO(10)$ Grand Unified Theory. In that theory, all the fermions arise from the chiral spinorial ${\bf 16}$ representations of $SO(10)$ - or $spin(10)$, to be more accurate. It's the simplest case relevant for the Standard Model where it is easy to show that there are no anomalies carried by this ${\bf 16}$ representation: it boils down to the vanishing of &#xA;$$\mbox{Tr}\left(\gamma_{ab} \{\gamma_{cd},\gamma_{ef}\}\right)$$&#xA;for all $a,b,c,d,e,f=1,\dots, 10$ which is trivial to check.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weights of ${\bf 16}$ under $SO(10)$ are&#xA;$$ (\pm \frac12, \pm \frac12, \pm \frac12, \pm \frac12, \pm \frac12) $$&#xA;with an even number of pluses (and odd number of minuses) - or vice versa, for the antifermions. The hypercharge is the inner product of the weight of a state with the generator&#xA;$$ (+2,+2,+2,-3,-3)/3 $$&#xA;The first three entries are linked to the three colors; the last two entries are associated with the weak doublets. The sum of the coordinates above had to vanish because the hypercharge is also a generator of $SU(5)$ where all generators have a vanishing trace (the sum of 5 diagonal elements) because of the $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, take all the inner products. They will be sums of $k_1=-3,-1,+1$, or $+3$ terms $(+2/2)\cdot (+1/2)=1/2$, and $k_2=-2, 0$, or $+2$ times $(+3/2)\cdot(+1/2)=3/4$. List all $4\times 3$ possible values of&#xA;$$ k_1\cdot \frac 12 + k_2\cdot \frac 34 $$&#xA;for the allowed values of $k_1,k_2$ and you will get exactly the 11+1 possibilities you listed (zero will appear twice because $SO(10)$ also predicts the right-handed neutrino).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the derivation of your list from a single representation, the ${\bf 16}$ of $SO(10)$. It's completely inherited by $SU(5)$ and the Standard Model - without the right-handed neutrino. (Some nonzero values of the hypercharge also appear multiply times - for several basis vectors - e.g. quarks appear thrice.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the maximum hypercharge I could have obtained from the inner products was &#xA;$$(2+2+2+3+3)/3/2=6/3$$&#xA;I could have obtained any smaller multiple of $1/3$ because the terms $2/3$ and $3/3$ may conspire in various ways. However, I can't reduce $6/3$ just by $1/3$ because the minimum amounts I can subtract by changing a sign are $2/3$ or $3/3$.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-16T07:25:12.320" LastActivityDate="2011-02-16T07:25:12.320" />
  <row Id="5258" PostTypeId="2" ParentId="4955" CreationDate="2011-02-16T08:30:53.903" Score="0" ViewCount="" Body="&lt;p&gt;I think conformal symmetry is OK, since this open/closed worldsheet is simply a tensor product of CFTs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm not sure how the reparameterization ghosts would work. It seems you can either put the bc ghosts on a disk or on a sphere, and there is no hybrid possibility like there is for the embedding coordinate. One compromise would be to include two bc CFTs, one for the open string boundary conditions and another for the closed string boundary conditions. Nilpotency of the BRST charge would probably then require that there are 26 closed string coordinates and 26 open string coordinates. The worldsheet lagrangian would be a sum of an open string Lagrangian and a closed string Lagrangian, so the resulting theory would be better interpreted as describing two independent strings, rather than one composite string.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be nice if there was a more interesting possibility though :)&lt;/p&gt;&#xA;" OwnerUserId="1856" LastActivityDate="2011-02-16T08:30:53.903" />
  <row Id="5259" PostTypeId="2" ParentId="5195" CreationDate="2011-02-16T08:55:20.787" Score="-2" ViewCount="" Body="&lt;p&gt;I'll tackle the 3D case. I am using the SI system. It should be noted that the electrical resistance of an electrical element (in given case a homogeneous medium) measures its opposition to the passage of an electric current (in given case direct current). The resistance of a homogeneous medium between two electrodes is defined as $$R=\frac{U}{I}=\frac{\rho\epsilon_0}{C}$$ where C is a capacitance between two electrodes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start with the assumption that instead of two points we have two conducting tiny spheres with a radius $r_0$. The distance between the centers of the spheres is $d$ and $r_0&amp;lt;&amp;lt;d$. To simplify the calculation let us assume that the charge on the spheres is distributed spherically symmetric. Then $$C=2\pi\epsilon_0r_0$$ Finally, required resistance:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R=\frac{\rho}{2\pi r_0}$$&lt;/p&gt;&#xA;" OwnerUserId="552" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2011-02-17T10:03:55.643" LastActivityDate="2011-02-17T10:03:55.643" CommentCount="5" />
  <row Id="5260" PostTypeId="2" ParentId="5158" CreationDate="2011-02-16T09:27:21.773" Score="2" ViewCount="" Body="&lt;p&gt;I always get the kids with some sporty things. I would suggest to measure hers or yours &lt;strong&gt;time dependent position while sprinting&lt;/strong&gt;. This is very handy and you can &lt;strong&gt;explain&lt;/strong&gt; to her a lot of physical &lt;strong&gt;concepts&lt;/strong&gt; coming from mechanics like &lt;em&gt;velocity, acceleration, total distance&lt;/em&gt;, ... and also some measurement/error things. I &lt;a href=&quot;http://maximum-maximorum.com/2009/02/13/strength-qualities-of-the-100m-sprinter/&quot; rel=&quot;nofollow&quot;&gt;found&lt;/a&gt; a nice picture illustrating the idea:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/axK9H.gif&quot; alt=&quot;velocity while sprinting&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Greets&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2011-02-16T09:27:21.773" CommentCount="2" />
  <row Id="5261" PostTypeId="2" ParentId="5243" CreationDate="2011-02-16T11:10:16.570" Score="1" ViewCount="" Body="&lt;p&gt;This movement causes the temper of the neck muscles, and strengthens the spine. The punching power is also distributed by the muscles. The soft flesh does not absorb any impact energy and is simply overwhelmed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much more energy is absorbed by the neck less energy will be transmitted to the brain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly a handball goalkeeper hit the ball. Thus the arm is much more tense. From personal experience he is liable to break the arm if he is distracted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best to do: avoid the punch ;)&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-02-16T11:10:16.570" CommentCount="1" />
  <row Id="5262" PostTypeId="1" AcceptedAnswerId="5264" CreationDate="2011-02-16T11:39:29.590" Score="-1" ViewCount="211" Body="&lt;p&gt;A proper time interval $\Delta\tau$ for a given observer is a relativistic invariant. However, the calculation of $\Delta\tau$ requires reference to some arbitrary coordinate time t:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta\tau$ = $\int_{t_1}^{t_2} \! \frac{\mathrm{d}t}{\gamma}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doesn't the invariance of $\Delta\tau$ require a &quot;preferred&quot; reference coordinate time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Proper_time#Mathematical_formalism&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;br&gt;&#xA;&lt;em&gt;Proper time is the pseudo-Riemannian arc length of world lines in four-dimensional spacetime.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doesn't a unique arc-length require a fixed (ie preferred) metric?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, when discussing the twin &quot;paradox&quot;, solutions typically involve drawing world lines against a fixed metric, where the different arc lengths are obvious. Does this not violate the spirit of SR? Is there a better way to look at it that does not involve a preferred reference frame against which arc lengths are defined? &lt;/p&gt;&#xA;" OwnerUserId="1247" LastActivityDate="2011-02-16T17:22:53.973" Title="Why doesn't the existence of proper-time $\tau$ imply a preferred reference frame?" Tags="&lt;special-relativity&gt;" AnswerCount="2" />
  <row Id="5263" PostTypeId="2" ParentId="5236" CreationDate="2011-02-16T11:43:23.663" Score="4" ViewCount="" Body="&lt;p&gt;We have to consider the operator, that doesn't explicitly depends on time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial A}{\partial t} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's apply commutator formula recursively:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d^2 A}{d t^2} = \left(\frac{i}{\hbar}\right)^2[H,[H,A]]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d^3 A}{d t^3} = \left(\frac{i}{\hbar}\right)^3[H,[H,[H,A]]]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.t.c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we combine those derivatives in a series for $A(t)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(t)=A(0)+\frac{d A}{d t}t+\frac{1}{2!}\frac{d^2 A}{d t^2}t^2+\frac{1}{3!}\frac{d^3 A}{d t^3}t^3+...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A(t)=A(0)+\frac{i}{\hbar}[H,A]t+\frac{1}{2!}\left(\frac{i}{\hbar}\right)^2[H,[H,A]]t^2+\frac{1}{3!}\left(\frac{i}{\hbar}\right)^3[H,[H,[H,A]]]t^3+...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then you use &lt;a href=&quot;http://en.wikipedia.org/wiki/Baker-Campbell-Hausdorff_formula&quot; rel=&quot;nofollow&quot;&gt;this formula&lt;/a&gt; to arrive at the result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^XYe^{-X} = Y+\frac{1}{1!}[X,Y]+\frac{1}{2!}[X,[X,Y]]+\frac{1}{3!}[X,[X,[X,Y]]]+...$&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-02-16T11:43:23.663" CommentCount="1" />
  <row Id="5264" PostTypeId="2" ParentId="5262" CreationDate="2011-02-16T11:52:50.233" Score="6" ViewCount="" Body="&lt;p&gt;No, the invariance of proper time $\Delta \tau$ doesn't require any preferred reference frame. It's the very definition of inertial systems that the proper time remains equal to&#xA;$$ \Delta \tau = \sqrt{dt^2-c^{-2}(dx^2+dy^2+dz^2)}.$$&#xA;The set of coordinate systems in which this formula remains valid is continuously infinite - it's the set of all inertial frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proper time has the simple form $\Delta\tau = dt$ in the rest frame of the moving object, but &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;this rest frame is changing as the object accelerates;&lt;/li&gt;&#xA;&lt;li&gt;this rest frame is different for different objects in the Universe;&lt;/li&gt;&#xA;&lt;li&gt;the proper time may be calculated in all other inertial systems, via the formula above.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Indeed, the formula above only takes the simple form in coordinate systems in which the metric has the fixed form &#xA;$$ ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2 $$&#xA;i.e. $\eta_{\mu\nu}=\mbox{diag} (+1,-1,-1,-1)$. Up to the squaring and a multiplication by a power of $c$, this formula is totally equivalent to the first one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's the whole point of the Lorentz symmetry that this form of the metric holds not only in one reference frame but in all (infinitely many) reference frames related by an arbitrary $SO(3,1)$ transformation. So the spirit of STR is that the metric is completely fixed, but it is fixed and equal in all inertial systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general relativity, the metric itself can't be fixed - it is allowed to be curved in any way - but it's still true that locally, physics of general relativity reduces to physics of special relativity where the metric may be fixed to the form above, and it is true in infinitely many inertial frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This statement is totally analogous to the &quot;proper length&quot; in an ordinary Euclidean space. The Pythagorean theorem $ds^2 = dx^2+dy^2(+dz^2)$ for the length of a vector also doesn't require any particular coordinate system; it has the very same form in all coordinate systems related by an arbitrary rotation (and/or translation). The case of the Lorentz symmetry in STR is totally analogous, with an extra coordinate (time) added (with the opposite sign in the Pythagorean theorem).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-16T11:52:50.233" />
  <row Id="5265" PostTypeId="1" AcceptedAnswerId="5510" CreationDate="2011-02-16T12:09:27.020" Score="83" ViewCount="37506" Body="&lt;p&gt;during the breakfast with my colleagues a question popped into my head:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the fastest method to cool a cup of coffee if your only available instrument is a spoon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A qualitative answer would be nice, but if we could find a mathematical model or even better make the experiment (we don't have the means here :-s) for this it would be great! :-D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, the options that we have considered are (any other creative methods are also welcome):&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Stir the coffee with the spoon&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;pros:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the whirlpool has greater surface than the flat coffee, so the better for heat exchange with air.&lt;/li&gt;&#xA;&lt;li&gt;due difference of speed between the liquid and the surrounding air, the Bernoulli effect should lower the pressure and that would cool it too to keep the atmospheric pressure constant.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;cons:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the Joule effect should heat the coffee.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Leave the spoon inside the cup&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As the metal is a good heat conductor (and we are not talking about a wooden spoon!), and there is some part inside the liquid and other outside, it should help with the heat transfer, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A side question about this is what is better, to put it like normal or reversed, with the handle inside the cup? (I think it is better reversed, as there is more surface in contact with the air, as in the CPU heat sinks).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Insert and remove the spoon repeatedly&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The reasoning about this is that the spoon cools off faster when it's outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I personally think it doesn't pay off the difference between keeping it always inside, as as it gets cooler, the lesser the temperature gradient and the worse for the heat transfer).&lt;/p&gt;&#xA;" OwnerUserId="1942" LastEditorUserId="1942" LastEditorDisplayName="" LastEditDate="2011-02-17T10:27:02.823" LastActivityDate="2011-02-22T05:48:15.577" Title="Cooling a cup of coffee with help of a spoon" Tags="&lt;thermodynamics&gt;&lt;experimental-physics&gt;&lt;heat&gt;" AnswerCount="20" CommentCount="9" FavoriteCount="27" />
  <row Id="5266" PostTypeId="1" CreationDate="2011-02-16T12:24:48.150" Score="1" ViewCount="131" Body="&lt;p&gt;What would be the vertical velocity of this ski jumper (ski flyer), after he first touches down, after he breaks the record with a 246.5m jump? What g force would he experience as he slows down?&#xA;&lt;a href=&quot;http://www.wired.com/playbook/2011/02/worlds-largest-skiflying-hill&quot; rel=&quot;nofollow&quot;&gt;http://www.wired.com/playbook/2011/02/worlds-largest-skiflying-hill&lt;/a&gt; (the last video)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think he gets at least 6 seconds of airtime from 26 to 32 second mark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ramp gradient at take off is listed at 11 degrees and the take off speed is about 105km/h.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some statistics are here.&#xA;&lt;a href=&quot;http://berkutschi.com/en/front/hills/show/33-vikersundbakken&quot; rel=&quot;nofollow&quot;&gt;http://berkutschi.com/en/front/hills/show/33-vikersundbakken&lt;/a&gt;&#xA;http://berkutschi.com/en/front/news/show/1347-vikersundbakken-approved---worlds-largest-hill&lt;/p&gt;&#xA;" OwnerUserId="1943" LastActivityDate="2011-02-16T14:03:21.273" Title="Ski Jumper's vertical velocity after 246.5m record?" Tags="&lt;gravity&gt;&lt;forces&gt;&lt;motion&gt;&lt;projectile&gt;" AnswerCount="1" />
  <row Id="5267" PostTypeId="2" ParentId="5265" CreationDate="2011-02-16T12:26:08.577" Score="9" ViewCount="" Body="&lt;p&gt;Well, if you are only allowed to use a spoon, the fastest way to cool the coffee for drinking is to get a spoonful, blow on it, drink it from the spoon, take a next spoonful. Convection does wonders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are allowed a saucer instead of a spoon, pour a bit of coffee in the saucer, blow on it and drink it.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-02-16T12:26:08.577" CommentCount="3" />
  <row Id="5268" PostTypeId="1" CreationDate="2011-02-16T12:29:21.887" Score="7" ViewCount="324" Body="&lt;p&gt;Covariant quantisation in string theory is accomplished by giving the commutator relations &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[X^\mu(\sigma,\tau),P^\nu(\sigma&amp;#39;,\tau)] = &#xD;&#xA;i \eta^{\mu\nu} \delta(\sigma - \sigma&amp;#39;)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although string theory is not a field theory this approach is identical to the quantisation of fields $X^\mu$ in 1+1 dimensions. From this point of view it is also clear how to proceed from here (e.g. getting rid of the negative norm states due to the &quot;wrong&quot; sign of the zeroth component).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are two issues that I do not quite understand. Physical states are supposed to have positive energy, i.e. $P^0 \geq 0$. Especially when supersymmetry comes into the game, the algebra so to say dictates positive energies. But when the energy is bounded from below it is not possible to have a time operator obeying the commutator relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[x^0,p^0] = -i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x$ and $p$ are the above Operators $X$ and $P$ integrated over $\sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the case of quantum mechanics this has been proved by Pauli in 1933. The prove goes roughly like this: For an operator $H$ which is bounded from below let there be a Hermitian time operator $T$ with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[T,H] = i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to shift the spectrum of H because of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^{-i\epsilon T} H e^{i\epsilon T} = H - \epsilon$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for arbitrary $\epsilon$. That means for an eigenstate $|\psi\rangle$ with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H|\psi\rangle = E|\psi\rangle$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there is also an eigenstate &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|\phi\rangle = e^{i\epsilon T}|\psi\rangle$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H|\phi\rangle = (E-\epsilon)|\phi\rangle$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\epsilon &amp;gt; E$ one has to either conclude that H can not be bounded from below or that T is not a Hermitian operator because it would generate states that lie outside of the Hilbert space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible solution could be to allow negative $p^0$ and to apply some kind of Feynman-Stueckelberg approach by reversing the sign of the world-sheet coordinate $\tau$. In this sense the decay of the vacuum into a positive and negative energy state is simply the propagator of a free state. However, this would not be possible in the supersymmetric case where the states have $p^0&amp;gt;0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other issue related to defining the zeroth component of the commutator is the construction of the Hilbert space of the physical states. In his book String Theory Polchinski starts with a &quot;vacuum&quot; state $|0;k\rangle$ with the norm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \langle 0;k|0;k&amp;#39;\rangle \sim \delta^{26}(k-k&amp;#39;)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when integrating over $k$ one ends up with a superficial delta function because of the additional mass shell constraint. Polchinski solves the issue by introducing a reduced product&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \langle 0;k||0;k&amp;#39;\rangle \sim \delta^{25}(k-k&amp;#39;)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to be a logical step because selecting the physical states by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L_0|\psi\rangle = \alpha&amp;#39;(p^2+m^2)|\psi\rangle = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;yields a Hilbert space whose functions depend on 25 momentum components. Still, the full Hilbert space is not normalisable (this differs from what is usually done in QFT during gauge fixing, where the Hilbert space is constructed via an equivalence relation for the gauge and not by selecting a subspace). The question is whether the Hilbert space constructed in such a way does exist at all. Also it seems as if the number of spacetime operators is reduced by one, thus rendering the steps taken for the covariant quantisation needless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(REMARK: I edited my question to make the point a bit clearer. This is not meant to be a criticism against string theory. It is just that during studying string theory I was confronted by someone with this issue. I just want to understand this detail of the basics because I haven't yet been able to resolve it. I hope somebody can help me.)&lt;/p&gt;&#xA;" OwnerUserId="1944" LastEditorUserId="1944" LastEditorDisplayName="" LastEditDate="2011-02-18T22:55:42.733" LastActivityDate="2011-03-14T14:19:34.330" Title="Covariant Quantisation and the Time Operator in String Theory" Tags="&lt;quantum-field-theory&gt;&lt;string-theory&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="3" />
  <row Id="5269" PostTypeId="2" ParentId="5265" CreationDate="2011-02-16T13:11:11.417" Score="13" ViewCount="" Body="&lt;p&gt;With respect to the content in the cup, all Your hampering &#xA;with the spoon is irrelevant. Cooling of a hot coffee is &#xA;achieved by vaporisation of water. At temperatures between &#xA;100 and say 50 °C the vapor pressure is so big, that the heat &#xA;carried away by convection of the hot (and much less dense &#xA;than air!) vapor dominates all other heat transfer mechanisms. &#xA;(ca 540 cal/g heat of vaporisation!) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, because You do not &quot;allow/accept&quot; blowing the surface of the coffee, &#xA;the second best is to wave the spoon over the surface .&#xA;Blowing/waving will enhance the vaporisation due to quick replacement &#xA;of the vapour on top of the surface with fresh, cool, dry air. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1663" LastEditorDisplayName="" LastEditDate="2011-02-16T19:39:04.447" LastActivityDate="2011-02-16T19:39:04.447" CommentCount="2" />
  <row Id="5270" PostTypeId="1" AcceptedAnswerId="5278" CreationDate="2011-02-16T13:26:21.027" Score="1" ViewCount="276" Body="&lt;p&gt;This question arises in a somewhat naive form because I am largely unfamiliar with String Theory. I do know that it incorporates higher space dimensions where I shall take the overall dimensionality to be 10 in this question, for concreteness. Now the traditional Hawking-Penrose Singularity results apply to the the General Relativity manifold of 3+1 dimensions; with the 4D Schwarzchild solution providing an example of a Singularity and Black Hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is: do  singularities (and maybe associated Event Horizons) necessarily form in all 10 dimensions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examining this question for myself I see that &lt;a href=&quot;http://arxiv.org/PS_cache/math/pdf/0603/0603190v3.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper for mathematicians&lt;/a&gt; introduces an N dimensional Schwarzchild metric and in theorem 3.15 an N dimensional Hawking-Penrose singularity theorem. However this cannot answer directly to the intentions of the String theory models. For example it is mathematically possible to extend 4D Schwarzchild to 10D differently by adding a 6D Euclidean metric. So one question is whether this modified 10D Schwarzchild even meets the conditions for the N dimensional Hawking-Penrose theorem. Although such a modification is not likely acceptable as a String Theory extension, it shows that we can consider some cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) All 4D singularities / Event Horizons are actually 10 D ones&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Some/all 4D singularities / Event Horizons are &quot;surface phenomena&quot; in String theory - the underlying Bulk Volume is singularity free&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Expressed a bit more formally this is saying that the String Theory has a singularity free solution $\Phi$ in 10D, but when $\Phi$ is restricted or reduced to 3+1D it is one of the known singular solutions of GR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) Some Singularities in String Theory Bulk (the 6D part) can arise without a corresponding 4D singularity (akin to a &quot;deep earth earthquake&quot; in 10D space-time, perhaps)?&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-02-16T16:26:31.463" LastActivityDate="2011-02-16T16:26:31.463" Title="Black Hole Singularity and String Theory" Tags="&lt;general-relativity&gt;&lt;string-theory&gt;&lt;black-holes&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="5271" PostTypeId="2" ParentId="5268" CreationDate="2011-02-16T13:50:30.197" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Positivity of energy&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability that a single superstring has a negative value of energy is strictly zero, as implied by supersymmetry. There's no contradiction with the commutation relation. It's easy to see why. Just define the wave function of the superstring in the $P^0$ representation. Then you have $\tilde\psi(P^0)$ and you may assume that&#xA;$$\tilde\psi(P^0) = 0\mbox{ for } P^0\leq 0.$$&#xA;The operator $X^0$ may be defined as $X^0=-i \partial/\partial P^0$ and it's manifest that &#xA;$$[X^0,P^0]=-i.$$&#xA;So your favorite anonymous co-father of quantum mechanics, whether it was Heisenberg, Dirac, or anyone else, had to use another assumption that is not satisfied in string theory to argue that it is not possible. (The assumption has to be violated not only in string theory but any particle-like description of a theory similar quantum field theory.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's pretty obvious that you don't quite know what the hypothetical no-go statement is and why it is true, so it may be a sensible idea to forget about this hypothetical statement and discard it as a vague irrational prejudice if you're reading Polchinski's book that is supposed to make everything crystal clear, and I think that it does so. You knowledge of physics can't be built on some vague rumors that you may have heard somewhere. You must actually understand why statements are true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that she or more likely he (the founder of QM) assumed that the Fourier transform of $\tilde \psi(P^0)$, namely the function $\psi(X^0)$, also has to vanish for negative $X^0$. Those two conditions - vanishing of both the functions and its Fourier transform for negative values of the argument - are not compatible conditions. However, there is no condition that $\psi(X^0)$ vanishes for a negative value of $X^0$ for a single superstring (or any single particle, for that matter), so the contradiction disappears and the very construction above proves that there can't be any inconsistency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;String theory is usually treated in the &quot;first-quantized&quot; way, so we obtain one-particle and multi-particle states &quot;directly&quot; and &quot;directly&quot; calculate their scattering amplitudes. That was the treatment I was assuming above; the negative-energy one-particle states are strictly absent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, string theory may also be treated in the &quot;string field theory&quot; formalism that makes it look like a quantum field theory in the spacetime, with infinitely many fields corresponding to the excitations of the string. Then the string fields may be Fourier expanded into components with various values of $P^0$, and the operators in front of the waves with a positive/negative value of $P^0$ are labeled as annihilation/creation operators, just like in any quantum field theory. Only the creation operators may produce nonzero particle states out of the vacuum - which is why particles always carry non-negative or positive values of $P^0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Normalization of eigenstates of momentum&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second question is purely about the normalization of some vectors and their existence. There is no real subtlety here. The off-shell states $|0;k&amp;#39;\rangle$ where $k$ doesn't have to satisfy that $k_\mu k^\mu=m^2$ may be assumed to exist for any $k$. However, the physical - Virasoro - conditions force the $k$ to be on-shell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in mathematical language, if you assume a particular type of a particle and its wave function is written as&#xA;$$|\psi \rangle = \int d^{26} k\,\tilde\psi(k) |0;k&amp;#39;\rangle,$$&#xA;then the Virasoro condition guarantees that $\tilde\psi(k)=0$ for all $k_\mu$ such that $k_\mu k^\mu\neq m^2$. To obtain states of a finite norm that satisfy the on-shell condition, you must pick&#xA;$$\tilde\psi(k) = \sqrt{\delta(k_\mu k^\mu - m^2) } \tilde\psi_{reduced} (\vec k)$$&#xA;where the reduced wave function only depends on the $25$ spatial components of the momentum. I had to add the square root of the delta function because the total norm of the states above will contain the $\delta$-function that, when integrated over $k^0$, will set $k^0$ to the right on-shell value given by the usual square root. In my notation, the vector that only has the 25-dimensional $\delta$-function as the inner product may be obtained from the previous one multiplied by the square root of the $\delta$-function as follows:&#xA;$$||0;k\rangle = \int_0^\infty dk^{\prime 0} \sqrt{ \delta\left( k^{\prime 0} -\sqrt{m^2+\vec k^2} \right) } |0; (k^{\prime 0},\vec k)\rangle $$&#xA;The $\delta$-function in the formula above is the same one as previous one, up to powers of $k^{\prime 0}$ and numerical constants that you should be able to calculate (and the second $\delta$-function also contains $\theta(k^0)$ relatively to the first one - the first one also incorrectly allows negative values of $k^0$). I chose this explicit form here because using the last displayed formula above, you may easily derive the second inner product - with the 25-dimensional $\delta$-function - from the first one. The square root of the $\delta$-function becomes an ordinary one and disappears after the integration over $k^0$, leaving the usual second inner product for the on-shell states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those square roots of the $\delta$-function look awkward and they're not really necessary in any way. It's clear that the right physical space is a simple combination of the vectors $||0;\vec k\rangle$ that only depend on 25 components of the momentum. It's just a physical fact that there is a single on-shell condition, so the number of independent components of the energy-momentum vector is 25 rather than 26. If you pick 25 coordinates out of 26, you inevitably violate the &quot;totally manifest&quot; Lorentz symmetry. Kinematics always does so, even in ordinary quantum field theory where cross sections have the $1/2E$ factors etc. that have to multiply the squared Lorentz-invariant amplitudes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, aside from these kinemetical factors that are universal (independent on the particle type and its interactions) and easy to learn, there's a lot of actual dynamics - the scattering amplitudes that depend on the momenta and on the particle types; and the infinitely many internal &quot;Hagedorn tower&quot; excitations of the string by the stringy oscillators. It's the &quot;difficult&quot; objects from the previous sentence that remain manifestly Lorentz-covariant in the covariant quantization which is why the quantization is called covariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an alternative quantization, such as the light cone gauge, the Lorentz symmetry becomes really hard to prove. In fact, you need to prove that $[J^{i-},J^{j-}]=0$ in the light-cone coordinates and this condition itself, because of various &quot;double commutator terms&quot;, will actually force you to prove that $D=26$ as well. So the calculation of this commutator is not equivalent to the classical Poisson bracket calculations; it knows something about the one-loop behavior of the world sheet theory, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you must learn how the kinematical factors - such as the inner products of the physical states and the extra coefficients that you have to insert to the squared Lorentz-invariant amplitudes to obtain the actual cross sections etc. - work because these are basic and essentially trivial things. The nontrivial physics, i.e. the dynamics, hides in the excitations of the string and the interactions of the strings (or interactions of quantum fields, in similar problems involving quantum field theory), and for those more difficult dynamical questions, it matters whether you want to make unitarity manifest (like in light-cone gauge); or Lorentz symmetry manifest (like in the covariant quantization).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-16T14:13:30.163" LastActivityDate="2011-02-16T14:13:30.163" CommentCount="2" />
  <row Id="5272" PostTypeId="2" ParentId="5266" CreationDate="2011-02-16T14:03:21.273" Score="2" ViewCount="" Body="&lt;p&gt;One would need the angle of the hill the Jumper lands on, the angle of his &quot;flight&quot; &#xA;in the last moment and his velocity. This values can not be calculated without &#xA;knowledge about the aerodynamics of the jump. &#xA;In practice one would measure this figures to calculate the aerodynamics!&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-02-16T14:03:21.273" CommentCount="3" />
  <row Id="5273" PostTypeId="1" AcceptedAnswerId="5274" CreationDate="2011-02-16T14:13:45.920" Score="1" ViewCount="151" Body="&lt;p&gt;First, what do you call this in English?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/YyIyM.gif&quot; alt=&quot;http://www.dieter-heidorn.de/Physik/SS/K06_Elektrisches%20Feld/K01_Ladung_Strom/K11_Ladung/Ladung02.gif&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, how does it work? &#xA;Why do I have not only light at the gas barrier between the two wires? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="534" LastEditorUserId="176" LastEditorDisplayName="" LastEditDate="2011-08-15T06:15:43.093" LastActivityDate="2011-08-15T06:15:43.093" Title="how does a glow lamp work" Tags="&lt;electricity&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="5274" PostTypeId="2" ParentId="5273" CreationDate="2011-02-16T14:25:07.267" Score="1" ViewCount="" Body="&lt;p&gt;This is a &quot;glow lamp&quot;. This special form, btw, is no longer fabricated. &#xA;Zur Arbeitsweise und und Grund für das Vorherrschen des negativen &#xA;Glimmlichts google doch mal nach Glimmentladung. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://de.wikipedia.org/wiki/Glimmentladung&quot; rel=&quot;nofollow&quot;&gt;http://de.wikipedia.org/wiki/Glimmentladung&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electric_glow_discharge&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Electric_glow_discharge&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Der deutsche Artikel ist wesentlich besser.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-02-16T14:56:35.607" LastActivityDate="2011-02-16T14:56:35.607" CommentCount="2" />
  <row Id="5275" PostTypeId="2" ParentId="5270" CreationDate="2011-02-16T14:29:23.410" Score="1" ViewCount="" Body="&lt;p&gt;Yes, the singularities have to form - under similar assumptions - in spacetimes of any dimension. This is not a problem: a singularity just means a place where the quantum-gravity, Planckian effects become important. In particular, the Penrose-Hawking theorem and its generalizations guarantees that black holes with singularities are formed even in 10D vacua.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The singularities that would be, if formed, problematic from the viewpoint of another Penrose's claim - the Cosmic Censorship Conjecture (the CCC) - are naked singularities, i.e. those that are not dressed in an event horizon. In 3+1 dimensions, naked singularities are most likely not formed (the CCC seems to hold) but I think that it has become pretty much established that in higher-dimensional GR, they may get formed (the CCC fails). They're not a real inconsistency - a consistent theory of quantum gravity may predict what happens even in the presence of naked singularities even if Dr Roger Penrose found it counterintuitive or dangerous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Whether the 4D or 10D singularity theorem is relevant for string theory depends on the situation you consider. If there are 10 large dimensions, you need to use the 10D theorem; if there are only 4 large dimensions, physics is approximately equivalent to GR in 4D and the relevant singularity theorem is the 4D theorem. Obviously, if the compact dimensions are as small as the Planck length, we can't resolve them from a cosmological viewpoint, so the cosmological implications are identical. A singularity in a 3+1-dimensional GR description may secretly be extended in additional dimensions - either all the compactified dimensions or just some of them - but this fact doesn't affect the conclusions about the singularities for much longer distances than the Planck scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) I don't fully understand what &quot;surface phenomena&quot; are but it is true that e.g. in the AdS/CFT correspondence, it is extremely difficult if not impossible to &quot;see&quot; inside the black holes, i.e. to the regions inside the event horizon. That's why the singularities inside the black holes remain largely invisible to the boundary CFT description. In some sense, all of physics may be imagined to reside outside the event horizons and the interior of the black hole contains just some subset of this information that is severely reshuffled: this principle rejecting the independent existence of the interior is known as the &quot;black hole complementarity&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) There can be singularities localized in the extra dimensions that are extended over the whole large 3+1-dimensional spacetime. In that case, we simply say that the compactified manifold is singular. Conifolds and orbifolds are two classes of simple examples. In those cases, the 4D effective description may remain non-singular. However, whenever you have any singularities whose character strongly depends on the position in the 3+1-dimensional spacetime (in particular, any singularities that are localized in at least one large dimension among the 3+1 dimensions), they will inevitably manifest themselves as singularities in the 4D description by GR as well. It's simply because one can show that the energy density around those singularities blows up and this energy density has to cause a huge curvature, even according to the effective 4D description. So singularities localized in the large dimensions can't hide themselves - not even if they try to localize themselves in the compact dimensions in any way.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-16T14:29:23.410" CommentCount="1" />
  <row Id="5276" PostTypeId="2" ParentId="5273" CreationDate="2011-02-16T14:36:22.060" Score="2" ViewCount="" Body="&lt;p&gt;This is neon lamp. They are still in production, used for indication of mains power for example. Also, used for decoration - in the form of very long tubes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They work by exciting gas at very low pressure using some 100-200V voltage. &#xA;Color is different for different gases, almost any color is possible.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-02-16T14:36:22.060" CommentCount="4" />
  <row Id="5277" PostTypeId="1" CreationDate="2011-02-16T14:39:21.853" Score="3" ViewCount="685" Body="&lt;p&gt;As I remembered, at the 2 poles of a battery, positive or negative electric charges are gathered. So there'll be electric field existing within the battery. This filed is neutralized by the chemical power of the battery so the electric charges will stay at the poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are electric charges at both poles, there must also be electric fields outside the battery. What happens when we connect a metal wire between the 2 poles of a battery? I vaguely remembered that the wire has the ability to restrain and reshape the electric field and make it within the wire, maybe like a electric field tube. But is that true?&lt;/p&gt;&#xA;" OwnerUserId="1513" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-11T20:55:24.760" LastActivityDate="2011-03-18T06:05:44.520" Title="What happens when we connect a metal wire between the 2 poles of a battery?" Tags="&lt;electricity&gt;&lt;electron&gt;&lt;electrical-engineering&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="5278" PostTypeId="2" ParentId="5270" CreationDate="2011-02-16T15:00:09.873" Score="1" ViewCount="" Body="&lt;p&gt;The Schwarzschild metric for $d$ dimensions is the standard form&#xA;$$&#xD;&#xA;ds^2~=~-e^{2\phi}dt^2~+~e^{2\gamma}dr^2~+~r^2d\Omega^2&#xD;&#xA;$$&#xA;These metric terms in the Einstein field equation gives&#xA;$$&#xD;&#xA;R_{tt}~-~\frac{1}{2}Rg_{tt}~=~G_{tt}~=~ -e^{2\phi}\Big((d~-~1)\frac{e^{2\gamma}}{r} ~+~\frac{(d~+~1)(d~-~2)}{2r^2}(1~-~e^{2\gamma}\Big)&#xD;&#xA;$$&#xA;A multiplication by $g^{tt}$ removes the $-e^{2\phi}$ and we equate this with a pressureless fluid $T^{tt}~=~\kappa\rho$.  So we think of the black hole as composed of “dust.”   Some analysis on this is used to compute the $G_r^r$ gives the curvature term&#xA;$$&#xD;&#xA;G_r^r~=~\Big((d~-~1)(\phi_{,r}~+~\gamma_{,r})\frac{e^{-2\gamma}}{r}~+~\kappa\rho\Big).&#xD;&#xA;$$&#xA;which tells us $\phi~=~-\gamma$, commensurate with the standard Schwarzschild result, and that&#xA;$$&#xD;&#xA;e^{-2\gamma}~=~e^{2\phi}~-~\Big(\frac{r_0}{r}\Big)^{d~-~2}.&#xD;&#xA;$$&#xA;The entropy of the black hole is then computed by writing the density according to these metric elements and computing the Rindler time coordinates $S~=~2\pi(d~-~2)A/\kappa$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The results more or less follow as with the standard $3~+~1$ spacetime result.  The Connection with strings is to work with the entropy of the black hole.  The $1~+~1$ string world sheet  has $d~-~1$ transverse degrees of freedom which contain the field data.  The entropy $S~=~2\pi(d~-~1)T$ may be computed with the string length, which reduces to the holographic results in $d~=~4$ spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The event horizon is $d~-~2$ dimensional, which for $10$ dimension means the horizon is $8$ dimensional.  The singularity is not considered in these calculations.  The factors $e^{-2\gamma}~=~e^{2\phi}$ become extremely large.  The metric approximates&#xA;$$&#xD;&#xA;ds^2~\simeq~\Big(\frac{r_0}{r}\Big)^{d~-~2}dt^2~+~r^2d\Omega^2&#xD;&#xA;$$&#xA;which is a $d~-~1$ dimensional surface where the Weyl curvature diverges for $r~\rightarrow~0$.  For $d~=~4$ this has properties similar to an anti-deSitter space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory of black holes essentially follows in arbitrary dimensions.  It is interesting to speculate on what the singularity is from a stringy perspective.  The event horizon contains the quantum field information which composes the black hole.  This may then have some type of correspondence with the interior singularity, with one dimension larger. For a black hole that is very small $\sim~10^3$ Planck units, the horizon is a quantum fluctuating region, as is the singularity, and the QFT data on the two may have some form of equivalency.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-16T15:00:09.873" CommentCount="2" />
  <row Id="5279" PostTypeId="2" ParentId="5277" CreationDate="2011-02-16T15:17:37.283" Score="5" ViewCount="" Body="&lt;p&gt;When the 2 electrodes are at different potential an electric field will be established. The electric charges will gather at the two poles. Positive charges at the cathode and negative charges at the anode. If the two electrodes are not connected by an external conductor they will not be able to leave the surface of the electrodes and they simply accumulate over there producing an open circuit voltage. As soon as the two electrodes are connected by a conductor the charges will flow by the forces of the electric field in the appropriate direction. If the connecting wire has no resistance or almost zero resistance then it will be a short circuit and a huge current will flow only limited by the internal resistance of the battery. If the electrodes are connected by a conductor through a resistance then the current will be limited according to the Ohm's law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = \frac{V}{R+r}$ where $I$ is the Current, $V$ is the voltage between the electrodes, $R$ is the external resistance and $r$ is the internal resistance of the battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: Electric field is not neutralized by the chemical reaction rather it is maintained by the reaction. No reshaping of the field takes place. The charges move through the conductor since it is the path of least resistance. It is similar to the flow of water through a pipe from a tank. The water flows since there exists a pressure difference. The flow of water does not depend upon the orientation of the pipe. it only depends on the pressure difference at the two ends. Gravitational field does not reshape itself through the pipe. Similarly no electric field reshaping takes place. Only the voltage difference matters between the two terminals.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Additional comments for those who think electric field can reshape in a conductor along the conductor&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/aSP8M.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are simply delta and star connected networks. Now if electric field could reshape then the lines of forces will intersect at three points for the delta and one point for the star networks. But we all know that lines of forces can not intersect. Hence no reshaping of field lines take place.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-03-18T06:05:44.520" LastActivityDate="2011-03-18T06:05:44.520" CommentCount="17" />
  <row Id="5280" PostTypeId="2" ParentId="5265" CreationDate="2011-02-16T15:20:18.617" Score="3" ViewCount="" Body="&lt;p&gt;Pre-cool the spoon first (in the freezer, or in your kitchen thermos of liquid nitrogen :) and put it in the cup. Periodically repeat the process with new spoons. Use a silver spoon (some of us were born with one in our mouths.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: If you allow someone to take the spoon in and out of the coffee, the question allows much too much freedom of action--there is nothing in the bounds of the question to rule out cooling the spoon either then---hence, my answer, which was meant ironically, not seriously, for those who have trouble determining the difference...&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-02-16T18:51:08.890" LastActivityDate="2011-02-16T18:51:08.890" CommentCount="3" />
  <row Id="5281" PostTypeId="2" ParentId="5265" CreationDate="2011-02-16T15:25:11.047" Score="8" ViewCount="" Body="&lt;p&gt;The answer may depend slightly on the humidity in the room (as that will determine the evaporative cooling rate), but basically your best bet is to increase the surface area of your coffee as much as possible and increase the rate of airflow over the coffee as much as possible (so that the local gradient of partial pressure of water vapor is as steep as possible).  I suspect that your best bet would be to pick up spoonfuls of coffee and pour them back into the cup from as high as you can manage without splattering the coffee all over the place--the stream of falling liquid has a high surface to volume ratio and is traveling quickly.&lt;/p&gt;&#xA;" OwnerUserId="1945" LastActivityDate="2011-02-16T15:25:11.047" CommentCount="3" />
  <row Id="5282" PostTypeId="1" AcceptedAnswerId="5289" CreationDate="2011-02-16T15:51:49.777" Score="7" ViewCount="267" Body="&lt;p&gt;That is, are they identical to their anti-particles? (Any results of double beta decay experiments?)&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-02-16T18:55:59.447" LastActivityDate="2011-02-16T18:55:59.447" Title="Are neutrinos Majorana particles?" Tags="&lt;particle-physics&gt;&lt;neutrino&gt;" AnswerCount="3" />
  <row Id="5283" PostTypeId="1" AcceptedAnswerId="5294" CreationDate="2011-02-16T15:53:29.310" Score="3" ViewCount="156" Body="&lt;p&gt;It has been said that our universe is going to eventually become a de sitter universe. Expansion will accelerate until their relative speed become higher than the speed of light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So i want to understand what happens after this point: so from our point of view, we see a progressively shrinking event horizon (each galaxy sees one, arguably each spacetime point sees one). Now, what is the Hawking Radiation expected from this event horizon? It would seems that the event horizon area is shrinking around us, but its actually a black hole turns outwards: the black hole is actually &lt;strong&gt;&lt;em&gt;outside&lt;/em&gt;&lt;/strong&gt; the event horizon, and the visible &quot;well-behaved&quot; spacetime without singularities in the inside of this horizon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, intuitively (i don't have any arguments to support this) i would expect that the Hawking Radiation &lt;em&gt;inside&lt;/em&gt; the horizon would grew larger as this horizon would shrink, but i would love to hear what is actually expected to happen at this point&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-02-16T16:01:54.467" LastActivityDate="2011-02-16T18:54:06.363" Title="de sitter cosmologic limit" Tags="&lt;cosmology&gt;&lt;cosmological-inflation&gt;&lt;hawking-radiation&gt;&lt;de-sitter-spacetime&gt;&lt;event-horizon&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5284" PostTypeId="2" ParentId="5192" CreationDate="2011-02-16T15:57:44.170" Score="0" ViewCount="" Body="&lt;p&gt;Another way of calculating the earth - sun distance is to look at the centrifugal and the gravitational force. This solution assumes that one already knows the mass of the sun, but thats a different problem ;-). One does only need High-School Math and Physics in order to derive a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Newton we know &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_g = -G\frac{Mm}{r^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $G=6,674\quad10^{-11}$ is the gravitational constant. We also know the centrifugal force to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_z =  \frac{mv^2}{r}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting these two equations together one gets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{mv^2}{r} = G \frac{Mm}{r^2}$&#xA;$\Rightarrow r =  \frac{GM}{v^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore we know the duration of a year and therefore we know $v$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v = \omega r = 2 \pi f r = \frac{2 \pi r}{T}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r = \frac{GMT^2}{4 \pi^2 r^2} \Rightarrow r = \sqrt[3]{\frac{GMT^2}{4 \pi^2}} = 149,8 \quad 10^9 \; m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is very close to the real value, which is varying between 147,1 Mio. and 152,1 Mio. km. According to Wikipedia the average distance is 149,6 Mio. km, so our result is actually quite good. &lt;/p&gt;&#xA;" OwnerUserId="393" LastActivityDate="2011-02-16T15:57:44.170" />
  <row Id="5285" PostTypeId="2" ParentId="5282" CreationDate="2011-02-16T16:08:51.697" Score="2" ViewCount="" Body="&lt;p&gt;Neutrinoless double beta decay, a sure sign of Majorana-ness of the neutrino, was claimed to be observed by Klapdor-Kleingrothaus et al. (with 2-3 sigma), see &lt;a href=&quot;http://www.worldscinet.com/mpla/16/1637/S0217732301005825.html&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;. However, their results are far from being accepted by the community.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is certainly not excluded that neutrinos are Majorana particles.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-02-16T16:08:51.697" />
  <row Id="5286" PostTypeId="2" ParentId="5283" CreationDate="2011-02-16T16:09:59.517" Score="2" ViewCount="" Body="&lt;p&gt;We are &lt;em&gt;already&lt;/em&gt; living in a nearly empty de Sitter space - the cosmological constant already represents 73% of the energy density in the Universe - and the Universe won't experience any qualitative change in the future: the percentage will just approach 100%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, once the space may be approximated as an empty de Sitter space, all moments of time are physically equivalent. It's because de Sitter space belongs among the so-called &quot;maximally symmetric spacetimes&quot; - in which each point may be mapped to any other point by a symmetry transformation (isometry).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So nothing will change qualitatively: the radius of the cosmic horizon will converge towards those 100 billion light years or so and never change again; we are not far from that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it is true that de Sitter space is analogous to a black hole except that the interior of the black hole is analogous to the space behind the cosmic horizon - outside the visible Universe. The de Sitter space also emits its thermal radiation, analogous to the Hawking radiation. It's a radiation emitted from the cosmic horizon &quot;inwards&quot;. Because the interior of the de Sitter patch is compact, unlike its black hole counterpart (the exterior of the black hole), the radiation is reabsorbed by the cosmic horizon after some time and the de Sitter space no longer loses energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the right theoretical description of the thermal radiation in de Sitter space is a theoretician's puzzle &lt;em&gt;par excellence&lt;/em&gt; (we are only &quot;pretty sure&quot; about the semiclassical limit, and don't even know whether there exists any description that is more accurate than that), it has absolutely no impact on observable physics because the typical wavelength of the de Sitter thermal radiation is comparable to the radius of the Universe. (Note that it's true for black holes, too: the wavelength of the Hawking radiation is mostly comparable to the black hole radius.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such low-energy quanta are obviously unobservable in practice - and in some sense, they're probably unobservable even in theory. You should imagine that there are just $O(1)$ thermal photons emitted by the cosmic horizons inside the visible cosmos whose energy is $10^{-60}$ Planck energies per photon. From an empirical viewpoint, it's ludicrous.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-16T16:09:59.517" CommentCount="6" />
  <row Id="5289" PostTypeId="2" ParentId="5282" CreationDate="2011-02-16T16:42:47.537" Score="5" ViewCount="" Body="&lt;p&gt;Great question. The experimental situation remains inconclusive. However, theoretically, there exists a damn good reason to think that the neutrinos have Majorana masses - and, consequently, the double beta decays should be possible. It's called the seesaw mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism is justified by an intriguing observation:&#xA;$$ m_{\nu}:m_{Higgs} \approx m_{Higgs}:m_{GUT}$$&#xA;Both ratios are about $10^{-14}$: note that $m_\nu, m_{Higgs}, m_{GUT}$ are about $10^{-12},10^2,10^{16}$ GeV. The light, millielectronvolt-scaled neutrinos, seem to be on the other side of this geometric sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This fact can be explained by a rather robust mechanism. The light neutrino masses - written as a $3\times 3$ mass matrix - may be generated as&#xA;$$ m = -m_D M_R^{-1} m_D^T. $$&#xA;The Dirac masses $m_D$, linking the known left-handed neutrinos with the heavy, unknown, right-handed neutrinos, are comparable to the Higgs mass - the electroweak scale. Meanwhile, the Majorana masses for the right-handed neutrinos, $M_R$, are comparable to the GUT scale. Right-handed neutrinos are automatically predicted e.g. as the &quot;invisible&quot; 16th component of the 16-dimensional spinor multiplet in $SO(10)$ grand unified theories, among others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula above automatically produces the &quot;seesaw&quot;: the resulting masses $m$ are as many times smaller than $m_D$ as $m_D$ is smaller than $m_R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very natural for the right-handed neutrinos to acquire Majorana masses: they're pretty much neutral and nothing prevents them from being coupled in pairs. The Dirac masses comparable to the electroweak scale may also be justified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, one automatically obtains the small masses for the known, left-handed neutrinos: and they're inevitably Majorana masses, too. They can't be Dirac masses simply because Dirac masses would require the right-handed parts of the particles to be as light as the left-handed one. But the right-handed neutrinos are superheavy and may be neglected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The seesaw formula may be obtained by diagonalizing a matrix $((0,m_D),(m_D,M_R))$. One eigenvalue is inevitably close to $M_R$ (which are really three eigenvalues); the other is much smaller than $m_D$. Alternatively, one may view $1/M_R$ in the formula as a suppression factor from more general higher-dimensional operators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be emphasized that the Dirac masses are used in the seesaw formula. However, because they're much smaller than the right-handed Majorana masses, the Dirac masses don't dictate the basic character of the fermions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there may be surprises, for example both Majorana and Dirac masses may occur at low energies, too. That would require some mixing of the three generations of the neutrinos - a scenario that is problematic for many reasons but it is not excluded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the text above, I assumed that the reader understands that the fields needed to describe a Weyl fermion and a Majorana fermion are the same fields: a 2-component complex spinor. Automatically, its Hermitean conjugate (one with the opposite chirality) also exists as an operator on the Hilbert space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether it is more natural to call this field &quot;Weyl&quot; or &quot;Majorana&quot; depends on the mass terms (or the interaction terms). Those that dominate decide about the name. If the Majorana mass terms are much larger than the Dirac masses, the particles are essentially Majorana particles, and the lepton number violation may be substantial while parity is kind of conserved. If the Dirac masses are much larger, the parity is qualitatively preserved - but one needs a doubled number of excitations (left-handed and right-handed particles of the same or similar mass).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can't be any mass terms that would both preserve the lepton number (distinguished particles and particles) and that would also keep the neutrinos purely left-handed.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-16T16:42:47.537" CommentCount="2" />
  <row Id="5290" PostTypeId="2" ParentId="5282" CreationDate="2011-02-16T16:46:20.297" Score="1" ViewCount="" Body="&lt;p&gt;If neutrinos were Majorana particles lepton number conservation would not work. In &lt;a href=&quot;http://hitoshi.berkeley.edu/neutrino/neutrino4.html&quot; rel=&quot;nofollow&quot;&gt;http://hitoshi.berkeley.edu/neutrino/neutrino4.html&lt;/a&gt; they summarize:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Implications of neutrino mass:&#xA;Neutrinos are found to have mass, but the mass is extremely tiny, at least million times lighter than the lighest elementary particle: electron. How do we need to change the Standard Model to explain the neutrino mass? Some argue that our spacetime has unseen spatial dimensions, and we are stuck on three-dimensional &quot;sheets&quot;. Other argue that we need to abandon the sacred distinction between matter and anti-matter.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Majorana neutrinos would require to abandon the distinction between matter and antimatter, and change the Standard Model. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-02-16T16:46:20.297" CommentCount="5" />
  <row Id="5291" PostTypeId="1" CreationDate="2011-02-16T16:53:22.067" Score="0" ViewCount="161" Body="&lt;p&gt;Quantum electrodynamics based upon Euler-Heisenberg or Born-Infeld Lagrangians predict photons to move according to an effective metric, which is dependent on the background electromagnetic field in a given space domain. In other words, photon trajectories are curved in presence of electromagnetic fields, meaning that an effective gravity is acting upon the photons. Charged fermions in the same space domain are obviously electromagnetically interacting; one thus wonder whether the interaction mediating virtual photons are also affected by this effective gravity. If the answer is yes, fermions will likely be affected, according to extended charge models, so that electromagnetic fields will show as another source of gravity for any kind of particles.&lt;/p&gt;&#xA;" OwnerUserId="1823" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-03-06T20:09:07.087" LastActivityDate="2011-03-06T20:09:07.087" Title="Are virtual photons affected by effective gravity in non-linear quantum electrodynamics?" Tags="&lt;quantum-gravity&gt;&lt;quantum-optics&gt;&lt;quantum-electrodynamics&gt;&lt;virtual-photons&gt;" CommentCount="6" />
  <row Id="5292" PostTypeId="2" ParentId="5232" CreationDate="2011-02-16T17:10:48.857" Score="1" ViewCount="" Body="&lt;p&gt;In the hypothetical $SU(5)$ theory, the X and Y bosons have hypercharges of $\pm 5/3$. So do some of the components of the adjoint Higgs needed to break $SU(5)$ down to the Standard Model.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-02-18T17:11:26.573" LastActivityDate="2011-02-18T17:11:26.573" CommentCount="2" />
  <row Id="5293" PostTypeId="2" ParentId="5262" CreationDate="2011-02-16T17:22:53.973" Score="1" ViewCount="" Body="&lt;p&gt;Relativity is not that hard if in classical mechanics some typical examples of &quot;relativity&quot; are considered first. For example, a tree from a distance looks smaller and &lt;strong&gt;needs calculations&lt;/strong&gt; to get the right (proper) tree size. In relativity such calculations involve time intervals too but the principle is the same: one obtains different &quot;raw&quot; experimental data in different RFs and needs recalculations to get the &quot;proper&quot; data. The same is valid for time intervals. They are &quot;seen&quot; differently from different RFs and to get the &quot;proper&quot; periods one needs a simple calculation involving the relative velocity.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-02-16T17:22:53.973" CommentCount="4" />
  <row Id="5294" PostTypeId="2" ParentId="5283" CreationDate="2011-02-16T17:53:11.643" Score="3" ViewCount="" Body="&lt;p&gt;The final fate of the universe can be computed.  The de Sitter spacetime cloaks every region with a cosmological horizon, which is similar in some ways to a Rindler horizon.  $10^{100} years from now there will be nothing going on: Black holes will have decayed away and largely the universe is a de Sitter vacuum.  There may be a smattering of neutron stars around, which as I recall these exist for a very long time.  This horizon does emit quanta, though the flux of this radiation is exceedingly small.  The universe will then eventually decay as the vacuum decays and the cosmological horizon retreats off “to infinity.”  Nothing else much will be going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We consider the decay of the de Sitter vacuum by quantum means and the prospect for this as a mechanism for the production of nascent cosmologies or baby universes.  The observable universe, under eternal inflation from dark energy, will asymptotically evolve to a de Sitter spacetime.  This spacetime is a vacuum configuration with a cosmological constant $\Lambda$.  The stationary metric for this spacetime is &#xA;$$&#xD;&#xA;ds^2~=~Adt^2~–~A(r)^{-1}dr^2~-~r^2d\Omega^2,~ A(r)~=~(1~-~\Lambda r^2/3)&#xD;&#xA;$$&#xA;A radial null geodesic with $ds^2~=~0$ and $d\Omega^2~=~0$ gives the velocity ${\dot r}~=~dr/dt$ $=~A(r)$, where this pertains to both out and in going geodesics near the cosmological horizon $r~=~\sqrt{3/\Lambda}$ as measured from $r~=~0$.  The total action for the motion of a particle is $S~=~\int p_r dr$ $-~\int Hdt$.  Consider the bare action of massless particles, using methods found in [1], across the horizon from $r$ to $r&amp;#39;$,&#xA;$$&#xD;&#xA;S~=~\int_r^{r&amp;#39;}p_rdr~=~\int_r^{r&amp;#39;}\int_0^{p_r}dp_rdr.&#xD;&#xA;$$&#xA;The radial velocity of a particle is ${\dot r}~=~dr/dt$ $=~dH/dp_r$, which enters into the action as,&#xA;$$&#xD;&#xA;S~=\int_r^{r&amp;#39;}\int_0^H{{dH&amp;#39;}\over{\dot r}}dr.&#xD;&#xA;$$&#xA;The field defines $H^\prime~=~\hbar\omega&amp;#39;$.  The integration over frequencies is from $E$ to $E~-~\omega$, for the ADM energy.  The action is properly written as&#xA;$$&#xD;&#xA;S~=~-\hbar\int_r^{r&amp;#39;}\int_E^{E-\omega}{{d\omega&amp;#39;}\over{\dot r}}dr,&#xD;&#xA;$$&#xA;where the negative sign indicates the quanta is tunneling across the horizon to escape the Hubble region with radius  The radial velocity &#xA;$$&#xD;&#xA;{\dot r}~=~\sqrt{\Lambda/3}r&#xD;&#xA;$$&#xA;defines the action &#xA;$$&#xD;&#xA;S~=~-\hbar\int_r^{r’}\int_0^\omega{{d\omega dr}\over{\pm 1~-~\sqrt{\Lambda r^2/3} }}~=~\sqrt{3/\Lambda}tanh^{-1}(\sqrt{\Lambda/3}r)&#xD;&#xA;$$&#xA;The action is then the delay coordinate evaluated as&#xA;$$ &#xD;&#xA;r^*~=~\int {{dr}\over{1~–~\Lambda r^2/3}}~=~\sqrt{3/\Lambda}tanh^{-1}(\sqrt{\lambda/3}r).&#xD;&#xA;$$&#xA;The domain $[0,~\sqrt{3/\Lambda})$ defines a real valued action.  Since, $tanh^{-1}(x)~=~{1\over 2}ln((1~+~x)/(1~-~x))$ for $r~&amp;gt;~\sqrt{3/\Lambda}$ the argument of the logarithm is negative.  In this case the action is&#xA;$$&#xD;&#xA;S~=~\sqrt{3/\Lambda}ln\Big({{\sqrt{\Lambda/3}r~+~1}\over{\sqrt{\Lambda/3}r~-~1 }}\Big)~+~i\pi\sqrt{3/\Lambda}.&#xD;&#xA;$$&#xA;The imaginary part represents the action for the quantum field emission as $r~\rightarrow~\infty$.  The delay coordinate is defined on $[0,~\infty)$ which assures an S-matrix is defined on an unbounded causal domain, and this holds in general as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This action does describe the emission of photons by the cosmological horizon.  A tiny production of bosons occurs which causes the horizon to slowly retreat away.  Eventually the de Sitter spacetime decays away into a Minkowski spacetime as $t~\rightarrow~\infty$.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-02-16T18:54:06.363" LastActivityDate="2011-02-16T18:54:06.363" CommentCount="5" />
  <row Id="5296" PostTypeId="1" CreationDate="2011-02-16T18:20:57.713" Score="2" ViewCount="143" Body="&lt;p&gt;I'm reading this silly Time article: &lt;a href=&quot;http://www.time.com/time/health/article/0,8599,2044517,00.html&quot; rel=&quot;nofollow&quot;&gt;http://www.time.com/time/health/article/0,8599,2044517,00.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And they say &quot;Even at its best, the 20-year-old telescope never had the acuity to peer so far into space, where the rapid expansion of the universe causes light waves to shift to a deep red.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doesn't that imply that things in the far universe are expanding more quickly?  Is that true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't space expanding faster than the movement of objects within space?  Do both cause redshift to be observed?&lt;/p&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2011-05-17T21:28:56.517" Title="Is the &quot;far&quot; universe expanding more quickly?" Tags="&lt;astronomy&gt;&lt;universe&gt;" AnswerCount="3" />
  <row Id="5297" PostTypeId="2" ParentId="5296" CreationDate="2011-02-16T18:25:29.227" Score="4" ViewCount="" Body="&lt;p&gt;this is just an observation of the Hubble's Law which states that the redshift is proportional to the distance of the galaxy. This empirical observation is explained by the isotropic expansion of the universe (that is, a rate of expansion in the metric scale) which implies that observationally, you will see farther objects to speed away faster, since there is more new volume of space per unit of time that separates from those points&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-02-16T18:25:29.227" CommentCount="2" />
  <row Id="5298" PostTypeId="2" ParentId="5265" CreationDate="2011-02-16T18:35:46.827" Score="12" ViewCount="" Body="&lt;p&gt;How to cool a cup of coffee with the help of a spoon. Hmm...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Empty the cup using the spoon, discarding the hot coffee.&lt;/li&gt;&#xA;&lt;li&gt;Strike the cup with the spoon, shattering it and forcing it to release the hot coffee.&lt;/li&gt;&#xA;&lt;li&gt;Drink the coffee with the spoon.&lt;/li&gt;&#xA;&lt;li&gt;Use the spoon to carve a cup-shaped hole in a large block of ice, put the cup in there.&lt;/li&gt;&#xA;&lt;li&gt;Put the spoon in the coffee, attach it to the cold side of a small thermoelectric-effect element.&lt;/li&gt;&#xA;&lt;li&gt;Use the spoon to catapult the cup 200km up. Measure temperature before it comes back down.&lt;/li&gt;&#xA;&lt;li&gt;Use the spoon to add a spoonful of liquid helium to the coffee. Slowly.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm sure you can come up with some good ones (edit away!)&lt;/p&gt;&#xA;" OwnerUserId="1316" LastActivityDate="2011-02-16T18:35:46.827" CommentCount="7" />
  <row Id="5299" PostTypeId="1" CreationDate="2011-02-16T18:58:58.290" Score="3" ViewCount="194" Body="&lt;p&gt;When you look a the Polyakov or Nambu-Goto action, you see a T for tension, but no mass. When you look the action of a relativistic particle, you see the m. See &lt;a href=&quot;http://www.physics.thetangentbundle.net/wiki/String_theory/relativistic_point_particle/action&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's the question. Where is the mass in the String actions? For a classical relativistic string I would expect that the equations of motion have m appearing as a constant. In this case I am thinking that the mass is related to the tension. In the end, if particles are strings in different states, then I would expect the mass to depend on the state.&lt;/p&gt;&#xA;" OwnerUserId="1937" LastActivityDate="2011-02-16T19:09:51.217" Title="Where's the mass?" Tags="&lt;string-theory&gt;" AnswerCount="1" />
  <row Id="5300" PostTypeId="1" AcceptedAnswerId="5389" CreationDate="2011-02-16T18:59:03.593" Score="2" ViewCount="159" Body="&lt;p&gt;From answers to &lt;a href=&quot;http://physics.stackexchange.com/questions/3889/is-a-normal-ordered-product-of-free-fields-at-a-point-a-wightman-field&quot;&gt;a previous question&lt;/a&gt;, a finite degree polynomial in the distribution $i\Delta^+(x)$, with Fourier transform $2\pi\delta(k^2-m^2)\theta(k_0)$, &lt;em&gt;is&lt;/em&gt; a distribution, even though a product of distributions is not in general well-defined. So far I've not been able to find, or be myself sure enough, whether $\exp{\!\left(i\Delta^+(x)\right)}$ is a distribution. It seems that the exponential is the boundary value of the function $\exp{\!\left(\frac{1}{z^2-m^2}\right)}$, but it's not clear to me yet whether that's enough. There seems to be nothing on this in Streater &amp;amp; Wightman (which I own, but don't yet understand well enough to apply the methods), and I don't have heavy lifting books like Reed &amp;amp; Simon, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, what I currently think I'm &lt;em&gt;really&lt;/em&gt; interested in is whether a bounded function of a distribution, such as $\tan{\!{}^{-1}\left(\exp{\left(i\Delta(x)\right)}\right)}$, is a distribution.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-02-17T19:33:57.120" Title="Is the exponential of the distribution $i\Delta^+(x)$, the 2-point function of a free quantized Klein-Gordon field theory, a distribution?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;mathematical-physics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="5301" PostTypeId="2" ParentId="5296" CreationDate="2011-02-16T18:59:38.310" Score="1" ViewCount="" Body="&lt;p&gt;The local Hubble law is the velocity of a galaxy at a distance $d$ moves with a velocity $v~=~Hd$.  So the further out you observe the faster things are moving away.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-16T18:59:38.310" />
  <row Id="5302" PostTypeId="2" ParentId="5299" CreationDate="2011-02-16T19:09:51.217" Score="2" ViewCount="" Body="&lt;p&gt;The mass does depend on the state: it's a basic point of string theory that it gives a deeper derivation of properties of particles such as their masses and charges: they're derived from a deeper starting point. The mass of a particle is no longer a constant: it is an eigenvalue of an operator acting on the string. The set (spectrum) of possible masses of particles - that are fundamentally strings in a state - is calculated in an analogy with the spectrum of the Hydrogen atom or any other bound state in quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classically, the mass of a string would only be given by the tension times the length of the string (which is dynamical). Quantum mechanically, the mass also has contributions from the kinetic energy, because of the $E=mc^2$ equivalence between the mass and energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The squared mass operator for an open string is given by&#xA;$$ m^2 = 2\pi T \sum_{n=1}^\infty \sum_{i=1}^{D-2} \alpha_{-n}^i \alpha_n^i $$&#xA;Here, $T$ is the tension of the string. The constant $2\pi T$ is usually written as $1/ \alpha&amp;#39;$ (read: &quot;one over alpha prime&quot;) where $\alpha&amp;#39;$ is called the Regge slope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term $\alpha_{-n}^i \alpha_n^i$ is equal to $n$ times the integer-valued excitation level of the quantum harmonic oscillator associated with the $n$-th Fourier mode (standing wave) of the coordinate $X^i$ along the open string. As you can see, the squared mass is $2\pi T$ times an integer. That's the spectrum of open (bosonic) strings in the flat space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For closed strings, the formula for $m^2$ has different factors of $2$ and contains sums over left-moving and right-moving oscillators. For superstrings, there are fermions and the integer may also be half-integer, with additional interesting technicalities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/3343/what-is-tension-in-string-theory&quot;&gt;What is tension in string theory?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-16T19:09:51.217" />
  <row Id="5303" PostTypeId="2" ParentId="5265" CreationDate="2011-02-16T19:24:39.267" Score="1" ViewCount="" Body="&lt;p&gt;Strange question :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say use the room temperature, lift the coffee with the soon and drop it back to the cup repeatedly, this will make a part of the coffee be in contact with the &lt;em&gt;cold&lt;/em&gt; room hence getting colder and mixing with the hot one bringing down the overall temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have (obviously??) no mathematical / physical knowledge about how effective this technique might be, and besides I drink the coffee as hot as I can ;P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="1663" LastActivityDate="2011-02-16T19:24:39.267" />
  <row Id="5304" PostTypeId="1" AcceptedAnswerId="5306" CreationDate="2011-02-16T20:08:45.263" Score="3" ViewCount="523" Body="&lt;p&gt;$$\mathbf D = \varepsilon \mathbf E$$&#xA;I don't understand the difference between $\mathbf D$ and $\mathbf E$. &#xA;When I have a plate capacitor different medium inside will change $\mathbf D$, right?&#xA;$\mathbf E$ is only dependent from the charges right?&lt;/p&gt;&#xA;" OwnerUserId="534" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-02-16T21:18:36.550" LastActivityDate="2011-02-17T17:28:11.600" Title="Difference between electric field and electric displacement field" Tags="&lt;electromagnetism&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="5305" PostTypeId="2" ParentId="4226" CreationDate="2011-02-16T21:07:55.483" Score="3" ViewCount="" Body="&lt;p&gt;Hau has demonstrated a memory for classical light pulses (that should also work as a memory for quantum states of light) but her system is - as it stands now - not a viable architecture for a quantum computer.  I'm not familiar with the work you link to from sciencedaily (which is not Hau's work).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure the state of the art for number of qubits (in an architecture that is viable for building a quantum computer) is held by ion traps, with numbers approaching a dozen qubits.  Two groups that are pushing the state-of-the-art are the NIST ion trap group and Rainer Blatt's group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See, ferinstance, one of the latest papers from the Blatt group, with 14 qubits:&#xA;&lt;a href=&quot;http://arxiv.org/pdf/1009.6126&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/1009.6126&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-02-16T21:07:55.483" />
  <row Id="5306" PostTypeId="2" ParentId="5304" CreationDate="2011-02-16T21:16:34.243" Score="10" ViewCount="" Body="&lt;p&gt;$\mathbf E$ is the fundamental field in Maxwell equations, so it depends on all charges. But materials have lots of internal charges you usually don't care about. You can get rid of them by introducing polarization $\mathbf P$ (which is the material's response to the applied $\mathbf E$ field). Then you can subtract the effect of internal charges and you'll obtain equations just for free charges. These equations will look just like the original Maxwell equations but with $\mathbf E$ replaced by $\mathbf D$ and charges by just free charges.&#xA;Similar arguments hold for currents and magnetic fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind, you see that you need to take $\mathbf D$ in your example because $\mathbf E$ is sensitive also to the polarized charges inside the medium (about which you don't know anything). So the $\mathbf E$ field inside will be $\varepsilon$ times lower than for the conductor in vacuum.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-02-16T21:16:34.243" CommentCount="2" />
  <row Id="5307" PostTypeId="1" CreationDate="2011-02-16T21:26:19.290" Score="1" ViewCount="114" Body="&lt;p&gt;Are there any objects out there that have since vanished, but because of their distance and the travel time for light, can be seen &lt;em&gt;with the naked eye&lt;/em&gt;?  Or are those things only visible with magnification?  How much magnification at least?&lt;/p&gt;&#xA;" OwnerUserId="1955" LastActivityDate="2011-02-16T22:06:27.890" Title="Astronomy: Are any vanished objects visible to the naked eye?" Tags="&lt;astronomy&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="5308" PostTypeId="2" ParentId="5307" CreationDate="2011-02-16T21:30:52.970" Score="2" ViewCount="" Body="&lt;p&gt;There's no way to know. The only way to tell whether an object has vanished is by the fact that its light (or other EM radiation) stops reaching us.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-02-16T21:30:52.970" CommentCount="3" />
  <row Id="5309" PostTypeId="2" ParentId="5307" CreationDate="2011-02-16T21:32:52.817" Score="3" ViewCount="" Body="&lt;p&gt;Do supernovas count? They could be visible with naked eye for a few days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/SN_1604&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/SN_1604&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2011-02-16T21:32:52.817" CommentCount="3" />
  <row Id="5310" PostTypeId="2" ParentId="5265" CreationDate="2011-02-16T21:53:31.400" Score="9" ViewCount="" Body="&lt;p&gt;Take the spoon.  Use it to push the coffee over.  It will spill out into a large puddle.  This has maximal surface area and will cool VERY quickly.&lt;/p&gt;&#xA;" OwnerUserId="1209" LastActivityDate="2011-02-16T21:53:31.400" CommentCount="1" />
  <row Id="5311" PostTypeId="2" ParentId="5307" CreationDate="2011-02-16T22:06:27.890" Score="2" ViewCount="" Body="&lt;p&gt;Most famously, the &quot;pillars of creation&quot; part of the Eagle Nebula that appeared in the famous space telescope pictures, are thought to have probably been blown away by now. They are 6000 some lightyears away, and strong interstellar winds are eroding them, so if we could&#xA;be instantly transported there we might find them gone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also high mass stars have short lifetimes of only a few million years. And these are the ones we can see in distant galaxies. So in a galaxy say a hundred million light years away, the brightest stars we see should by now have all died (presumably replaced by new ones).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can detect gamma ray bursts from several billion light years away (and they only last at most a few minutes).....&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-02-16T22:06:27.890" />
  <row Id="5312" PostTypeId="2" ParentId="5304" CreationDate="2011-02-16T22:09:49.273" Score="6" ViewCount="" Body="&lt;p&gt;Like @Marek has said above, the electric field $E$ is the fundamental field, and is in some sense the more physical. However, Maxwell's equations have a neater geometric meaning if you throw in the &quot;auxilary&quot; fields $D$ (and $H$ for $B$). I usually tell my students the following version of electromagnetism:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 4 fields in electromagnetism. We call them $E$, $D$, $B$ and $H$. &lt;em&gt;All of these fields are independent and equally important.&lt;/em&gt; Furthermore, they actually embody geometric concepts which are manifest in the integral equations: $$\oint_S D \cdot dS = Q(S)$$ $$\oint_S B \cdot dS = 0$$ $$\oint_{\partial S} E \cdot dl + \partial_t \int_S B \cdot dS = 0$$ $$\oint_{\partial S} H \cdot dl - \partial_t \int_S D \cdot dS = \int_S j \cdot dS$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$E$ and $B$ form an independent pair, as do $D$ and $H$.&lt;/li&gt;&#xA;&lt;li&gt;$E$ and $B$ do not depend on the sources $Q$ and $j$, but $D$ and $H$ do.&lt;/li&gt;&#xA;&lt;li&gt;$D$ and $B$ are integrated through surfaces, and represent &lt;em&gt;flux&lt;/em&gt; through those surfaces. (The correct mathematical gadget to describe these are actually 2-forms.$&lt;/li&gt;&#xA;&lt;li&gt;$E$ and $H$ are integrated along lines, and end up representing the potential difference across the ends (or circulation in a loop).&lt;/li&gt;&#xA;&lt;li&gt;The latter pair connect the change of flux through surfaces with certain circulations.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These equation form Maxwell's equations. &lt;em&gt;They do not uniquely determine a physical situation.&lt;/em&gt; In particular, they need to be augumented with &lt;em&gt;constitutive relations&lt;/em&gt; which describe (macroscopic) material properties. For example, we might have linear, isotropic, homogeneous (LIH) media, in which case we would have $D = \epsilon E$ and $B = \mu H$. But in general, we might have $\epsilon$ and $\mu$ being tensors, varying as functions of time and space, or even depend on the fields $E$, $B$, etc! These constitutive relations could be arbitrarily complicated, and indeed much of the new field of meta-material engineering is all about creating micro-structures which would yield interesting and useful constitutive relations at the macroscopic scale. More commonly, a scenario where the linearity breaks down is in ferromagnets/ferroelectrics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is usually another constitutive relation, linking current and electric field. In LIH media this is called Ohm's Law: $J = \sigma E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one more equation, which is simply always true, which is the conservation of charge; in the notation above, $\partial_t Q(S) - \int_S j \cdot dS = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: some additional observations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a relativistically covariant form, we can merge $E$ and $B$ together to get the 2-form $F$, and $D$ and $H$ to get its Hodge dual $\star F$. The latter in general depends on the metric we choose. For linear materials it's possible to hide the effects of the material polarisation/magnetisation as a background metric. Incidentally, in this form, the energy is given by $F \wedge \star F$, so it is clear that energy/momentum should be &quot;opposing&quot; pairs, i.e. the Poyntin vector is $N = E \times H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In numerical simulations, it's doubly important that we obey Maxwell's equations --- failure to do so leads to highly unphysical things like superluminal propagation of waves or failure to conserve energy or momentum. It has been found that the key is to be &lt;em&gt;exact&lt;/em&gt; with respect to the integral forms of the equations, and put all of the discretisation error into not meeting the material constitutive properties.&lt;/p&gt;&#xA;" OwnerUserId="1501" LastEditorUserId="1501" LastEditorDisplayName="" LastEditDate="2011-02-17T10:06:26.763" LastActivityDate="2011-02-17T10:06:26.763" CommentCount="3" />
  <row Id="5313" PostTypeId="2" ParentId="5265" CreationDate="2011-02-16T22:11:35.107" Score="1" ViewCount="" Body="&lt;p&gt;By leaving the spoon stationary in the coffee, you are only benefiting from the thermal conduction and dissipation of the spoon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By stirring the coffee, you are not only helping to keep the hottest parts of the coffee in contact with the air &amp;amp; cup, increasing the surface area by introducing abnormalities in the top surface, and helping to move the warm, moist air atop the coffee, but you are also getting all of the benefits of the thermal conduction and dissipation of the spoon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By dunking the spoon repeatedly, you are not only introducing air into the liquid, which will grab lots of heat and be whisked away as soon as it hits the surface, but you are also getting all of the benefits of the thermal conduction and dissipation of the spoon and all of the benefits of stirring the coffee.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring the fact that you will probably splash the coffee and make a mess of the counter, the correct answer is repeatedly dunking the spoon.  Also, you and your coworkers are nerds!&lt;/p&gt;&#xA;" OwnerUserId="1959" LastEditorUserId="1959" LastEditorDisplayName="" LastEditDate="2011-02-16T22:20:02.600" LastActivityDate="2011-02-16T22:20:02.600" CommentCount="1" />
  <row Id="5314" PostTypeId="2" ParentId="5304" CreationDate="2011-02-16T22:24:22.183" Score="0" ViewCount="" Body="&lt;p&gt;To understand what field is &quot;real&quot;, write a charge equation of motion. The force in it is determined with the real field there. In a medium it is still &lt;strong&gt;E&lt;/strong&gt;: $m\vec{a} = q\vec{E}$. In case of magnetic field, it is $\vec{B}$ that determines the force: $m\vec{a} = q \vec{v} \times \vec{B}/c$.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-17T16:38:34.497" LastActivityDate="2011-02-17T16:38:34.497" CommentCount="2" />
  <row Id="5315" PostTypeId="2" ParentId="5265" CreationDate="2011-02-16T22:30:17.407" Score="54" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Stirring will win, hands down, every time.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why physicists need to talk to chemists once in a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Georg correctly remarks, the latent heat of vaporization of water is enormous - but he's wrong about waving the spoon; stirring is the champion here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? Temperature is really the average kinetic energy of the molecules in the bulk substance, which actually have a variety of individual kinetic energies. Stirring is the fastest way to bring high-kinetic-energy outlier water molecules to the surface, where they will overcome the electrostatic bonding mechanisms that keep them in the liquid phase, and jump into the air (vapor phase). This rapid decrease in the high-energy outliers is the quickest way to cool a hot aqueous solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's similar to stirring iced tea. If you just plop ice cubes into a glass of warm tea, it will take quite a while for the warmer tea to cool; if you stir it vigorously, it will reach a cold equilibrium within seconds; the latent heat of fusion absorbed by the ice melting is similarly enormous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This kind of thing has a lot of applications to laboratory and industrial chemical processes, surface catalysis, petroleum cracking, yadda yadda. You learn a lot about it in third-year university physical chemistry, and really must master it before or during graduate work as a chemist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want an even faster way to cool a cup of coffee, here's a tip from my Granddad Parker: &lt;strong&gt;forget the spoon and saucer your coffee&lt;/strong&gt;. In other words, pour the top part of it from the cup into a saucer, and then back again a few times. The large and constantly changing surface area during this process will cause extremely rapid evaporation of those high-energy outliers, much faster than stirring. Saucering was very common up through the Great Depression, which is one of the reasons older coffee sets always included saucers. You also get deep-ish saucers at many restaurants as a holdover from this practice, although I doubt many people do it any more.&lt;/p&gt;&#xA;" OwnerUserId="224" LastActivityDate="2011-02-16T22:30:17.407" CommentCount="16" />
  <row Id="5316" PostTypeId="1" CreationDate="2011-02-16T23:24:25.830" Score="3" ViewCount="194" Body="&lt;p&gt;If Betelgeuse were to go supernova would the sky be twice as bright, or day time extended, depending on what time of year it happened in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically when it does supernova how bright and large will it appear in the sky, and how do you work out the apparent brightness and size for something that hasn't happened yet?&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-02-21T05:06:29.587" LastActivityDate="2011-02-21T05:06:29.587" Title="If Betelgeuse were to go supernova" Tags="&lt;astrophysics&gt;&lt;stellar-physics&gt;" AnswerCount="1" />
  <row Id="5317" PostTypeId="1" AcceptedAnswerId="5338" CreationDate="2011-02-17T00:22:17.150" Score="5" ViewCount="324" Body="&lt;p&gt;I have a numerical method for computing solutions to the Dirac equation for a spin 1/2 particle constrained to an arbitrary surface and am interested in finding applications where the configuration space has a complicated geometry, i.e., not just $R^2$ or the sphere, but a more general curved surface possibly with special boundary conditions.  On the sphere, for instance, one can take advantage of symmetry and simply use the spherical spinors, but for a more general configuration space (or one described only by measurements, for instance) it may not be possible to come up with a nice, closed-form solution.  However, I am not a physicist and would like to better understand where (or indeed, &lt;em&gt;if&lt;/em&gt;) such problems arise.  Any pointers are greatly appreciated.  Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Note that I &lt;em&gt;already have&lt;/em&gt; a numerical method for solving the Dirac equation -- I am not looking for information on how to solve it.  (And, interestingly enough, Clifford algebra / geometric algebra is already the starting point for what we do.)  Also, I am specifically interested in the case of &lt;em&gt;surfaces&lt;/em&gt;, i.e., 2-manifolds, so (many) applications in GR probably don't apply.  Thanks!&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="708" LastEditorDisplayName="" LastEditDate="2011-02-18T00:13:35.700" LastActivityDate="2011-02-18T07:45:36.937" Title="Dirac equation on general geometries?" Tags="&lt;quantum-mechanics&gt;&lt;computational-physics&gt;&lt;dirac-equation&gt;&lt;quantum-computation&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="5318" PostTypeId="2" ParentId="5277" CreationDate="2011-02-17T00:35:26.847" Score="0" ViewCount="" Body="&lt;p&gt;Before you connect the wire, the electric field (at least theoretically, this requires that you ignore everything else in the universe, including the contents of the battery) consists of a + charge and a - charge. With a typical car battery these points are separated by about 10 or 20cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, the charge is distributed in a very complicated manner. For example, if your car battery is rated for 1000 amp-hours, you can compute the &quot;charge&quot; by noting that an amp-second is a coulomb so an amp-hour is 3600 x 1000 coulombs. But actually these charges are held inside the battery chemically and do not appear at the electrodes until you take some of the charge that is there away. So to compute the initial charge you need to know only the voltage and the configuration of the leads. Then it's a matter of electrostatics to compute the charge distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After you connect the wire, current begins flowing according to the resistance of the wire. The wire will have a steadily decreasing voltage from one end to the other. Since the electric field is given by the gradient of the voltage, this means that the electric field along the wire will point down the wire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, this happened accidentally to a battery in the back of his SUV. Unfortunately it happened next to a plastic can of gasoline. The resulting fire destroyed his vehicle seconds after he drove to the side of the road an jumped out of it.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="61" LastEditorDisplayName="" LastEditDate="2011-02-17T02:50:30.900" LastActivityDate="2011-02-17T02:50:30.900" CommentCount="2" />
  <row Id="5319" PostTypeId="2" ParentId="5207" CreationDate="2011-02-17T00:51:58.973" Score="0" ViewCount="" Body="&lt;p&gt;The descent from 24 to 8 seems to happen when the straight sums are substituted by alternating sums. This is known from the theory of the Riemann zeta funtion, whose only pole in $s=1$ is cancelled via a multiplication that produces the Dirichlet eta function,     $$\eta(s) = \left(1-2^{1-s}\right) \zeta(s)$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This function has better analiticity than Zeta. But it is alternating,&#xA;$$    \eta(s) = \frac{1}{1^s} - \frac{1}{2^s} + \frac{1}{3^s} - \frac{1}{4^s} + \cdots$$&#xA;And thus it could be related with the simple expresion I have mentioned above in the comments. But, more important,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta(-1)=\left(1-2^{2}\right) \zeta(-1) = -3 \times {-1\over 12} = {1\over 4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you can suspect that the Eta function does for the superstring the same role that the Zeta does for the bosonic string. And indeed it appears in very similar situations. For instance, Michael B. Green, in his 1986 Trieste lectures &quot;String and Supertring Theory&quot;, section 5.11, calculates the NS sector spectrum and then the normal ordering constant, that appears formally as a difference between the bosonic term and the fermionic term. Such difference can be manipulated to obtain the Eta function as above, times $(D-2)/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if anyone can tell how the Zeta regulator is related to the integer sum up to 24, then we could probably guess how the Eta regulator would be related to the alternating sum up to 8. &lt;/p&gt;&#xA;" OwnerUserId="1335" LastEditorUserId="1335" LastEditorDisplayName="" LastEditDate="2011-02-17T10:24:34.810" LastActivityDate="2011-02-17T10:24:34.810" />
  <row Id="5320" PostTypeId="1" CreationDate="2011-02-17T01:11:16.257" Score="2" ViewCount="281" Body="&lt;p&gt;The size of our observable universe must have grown over time in the early universe.  Conversely with the accelerated expansion, I have heard that eventually our observable universe will shrink down to our local group of gravitationally bound galaxies.  So then the observable universe must go through a maximum at some point in time.  When is that and what is that maximum size?&lt;/p&gt;&#xA;" OwnerUserId="1960" LastActivityDate="2011-08-05T00:04:39.040" Title="Maximum size of the observable universe?" Tags="&lt;cosmology&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="5322" PostTypeId="1" CreationDate="2011-02-17T02:17:36.017" Score="0" ViewCount="215" Body="&lt;p&gt;I'm currently in the final phase preparing for my exam in quantum mechanics, and currently I'm re-doing all the exercises we'd been given over the term and compare my solutions to official one's we were given now for that purpose. And somehow I have the impression that those solutions are rather, well, strange. I'd like to ask my fellow Physics StackExchangers to have a look on those &quot;solutions&quot; and comment on them, because reading some of them I had serious &quot;WTF?!&quot; moments (obviously wrong eigenvalues, but then the correct eigenvectors, missing base vectors for bra-ket spaces, intermediate matrices not orthonormal but further results just like if they were).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the URL to the solution PDFs (also contain the original problems)&#xA;&lt;a href=&quot;http://homepages.physik.uni-muenchen.de/~Wolfgang.Draxinger/t2/&quot; rel=&quot;nofollow&quot;&gt;http://homepages.physik.uni-muenchen.de/~Wolfgang.Draxinger/t2/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd highly appreciate it, if people could have an educated look at them and comment them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastActivityDate="2011-02-17T02:17:36.017" ClosedDate="2011-02-17T03:38:14.853" Title="Request for Review: Quantum Mechanics Exercises - official solutions WTF?" Tags="&lt;quantum-mechanics&gt;&lt;homework&gt;" CommentCount="2" />
  <row Id="5323" PostTypeId="2" ParentId="5320" CreationDate="2011-02-17T02:28:22.557" Score="0" ViewCount="" Body="&lt;p&gt;The observable universe is currently increasing in size at the rate of one light year per year. The universe itself is currently expanding at around 71 km/sec/Mps, so we can still receive incoming light. Even though we are in cosmological constant dominated cosmology and accelerated expansion, the observable universe will not hit a maximum until space is expanding at a rate greater than c. The definition of observable universe is what we can see (or record) because of light that has had time to reach us since the big bang. The size is about 47 billion light years or 14 billion parsecs currently.The future visibility limit is at a co-moving distance of about 19 billion parsecs.&#xA;There does seem to be some variation in the definitions though. The Hubble volume is the volume out to the distance where objects recede from us at greater than the speed of light.&#xA;Because of the accelerating expansion, the Hubble volume is decreasing.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-02-17T17:26:51.163" LastActivityDate="2011-02-17T17:26:51.163" CommentCount="4" />
  <row Id="5324" PostTypeId="2" ParentId="5317" CreationDate="2011-02-17T02:42:41.133" Score="0" ViewCount="" Body="&lt;p&gt;You might take a look at the version of general relativity (GR) done by the Cambridge Geometry Group. They translated GR into &quot;geometric algebra&quot;, which amounts to the gamma matrices. So it's easy for them to do Dirac equation calculations on a black hole. Here's an example papers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In P.G. Bergmann and V. de Sabbata eds, Advances in the Interplay Between Quantum and Gravity Physics, 251-283, Kluwer (2002), Anthony Lasenby and Chris Doran, &lt;em&gt;Geometric Algebra, Dirac Wavefunctions and Black Holes&lt;/em&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.mrao.cam.ac.uk/~clifford/publications/abstracts/anl_erice_2001.html&quot; rel=&quot;nofollow&quot;&gt;http://www.mrao.cam.ac.uk/~clifford/publications/abstracts/anl_erice_2001.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phys.Rev. D66 (2002) 024006, Chris Doran, Anthony Lasenby, &lt;em&gt;Perturbation Theory Calculation of the Black Hole Elastic Scattering Cross Section&lt;/em&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0106039v1&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/gr-qc/0106039v1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dirac equation for the Kerr metric (rotating black hole):&lt;br&gt;&#xA;Phys.Rev. D61 (2000) 067503, Chris Doran, &lt;em&gt;A new form of the Kerr solution&lt;/em&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/9910099v3&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/gr-qc/9910099v3&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These, and papers they cite or are cited by, should be sufficient.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-17T02:42:41.133" />
  <row Id="5325" PostTypeId="2" ParentId="5317" CreationDate="2011-02-17T03:12:55.167" Score="-2" ViewCount="" Body="&lt;p&gt;I've never seen, and i think it would be fun to solve it inside the klein-bottle. Basically its a closed version of the Möebius strip (meaning that is also non-orientable); the boundary conditions are pretty simple&lt;/p&gt;&#xA;&#xA;&lt;p&gt;take a square, with vertices A,B,C,D and edges AB, BC,CD,DA. Now make the following identifications:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) (A, AB, B) =&gt; (D, DC, C)&lt;br&gt;&#xA;this gives you a periodic condition on x, making your square a cylinder&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now if you would do (B, BC, C) =&gt; (A, AD, D) you would get back a torus $T^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you do instead &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) (B , BC, C) =&gt; (D, DA, A) you will get a surface homeomorphic to the klein-bottle&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-02-17T03:12:55.167" CommentCount="2" />
  <row Id="5326" PostTypeId="1" CreationDate="2011-02-17T03:19:37.160" Score="4" ViewCount="285" Body="&lt;p&gt;This is a summary from Physics World of the paper: L J Wang et al. 2000 Nature 406 277--&#xA;&quot;Wang and colleagues begin by using a third continuous-wave laser to confirm that there are two peaks in the gain spectrum and that the refractive index does indeed change rapidly with wavelength in between. Next they send a 3.7-microsecond long laser pulse into the caesium cell, which is 6 centimetres long, and show that, at the correct wavelength, it emerges from the cell 62 nanoseconds sooner than would be expected if it had travelled at the speed of light. 62 nanoseconds might not sound like much, but since it should only take 0.2 nanoseconds for the pulse to pass through the cell, this means that the pulse has been travelling at 310 times the speed of light. Moreover, unlike previous superluminal experiments, the input and output pulse shapes are essentially the same.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realise that the velocity of light in a medium is comprised of the phase velocity, the group velocity, and the front velocity. While the group velocity can exceed the value of c in a vacuum, the front velocity is not supposed to. The way this sounds, the front velocity is exceeding c in a vacuum by 310 times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/2/27/Frontgroupphase.gif&quot; alt=&quot;Frontgroupphase.gif‎&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;blue dot=phase velocity, green=group, red=front (wiki)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wavepacket seems to exit the cell well before it enters, but the negative refractive index &quot;forward shifts&quot; the leading edge of the pulse. It is argued that although this is superluminal, information cannot be transmitted faster than c. &lt;a href=&quot;http://pra.aps.org/abstract/PRA/v67/i6/e063801&quot; rel=&quot;nofollow&quot;&gt;Gauthier and Stenner&lt;/a&gt; introduced a jump discontinuity into the waveform.Its max speed was c Here is a popular account from &lt;a href=&quot;http://www.god-does-not-play-dice.net/speed.html&quot; rel=&quot;nofollow&quot;&gt;NewScientist&lt;/a&gt; . What if you used one photon? What if the photon itself were the information and not a carrier? &lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-02-17T06:08:37.907" LastActivityDate="2011-02-17T14:27:41.060" Title="What does Lijun Wang's experiment about supraluminal speed of light in a medium mean?" Tags="&lt;optics&gt;&lt;special-relativity&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="5327" PostTypeId="2" ParentId="5316" CreationDate="2011-02-17T03:39:15.007" Score="5" ViewCount="" Body="&lt;p&gt;Most models of type II supernovae expect that the explosion has a peak luminosity of around $10^8$ to $10^9$ solar luminosities. This gives it a peak absolute magnitude of -15.16 to -17.16. At a distance of 200 parsecs, this is an apparent magnitude of -8.652 to -11.15.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, so how bright is that? Well, the full moon has an apparent magnitude usually around -12.75. This means the total luminosity from the explosion of Betelgeuse would be something like $1/1000$ times the luminosity of the full moon. But the human eye sees things logarithmically, so we might expect that it looks like its about a third as bright as the full moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This matches up roughly to descriptions of &lt;a href=&quot;http://en.wikipedia.org/wiki/SN_1006&quot; rel=&quot;nofollow&quot;&gt;SN 1006&lt;/a&gt; which was a supernova 1,000 years ago. Observers at the time described a slowly growing ball in the sky about a quarter the brightness of the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your second question, how we predict the brightness: We have our estimates about the brightness of the supernova mostly from computer simulations. Supernovas occur when large stars run out of fuel and their core collapses, causing all the mass to suddenly fall inward. A series of complicated nuclear reactions take place as the matter rebounds in a huge shockwave. The parameters of this whole process are quite nasty to work out in detail and the results depend somewhat (but less than you might think) upon the size and composition of the star. We've figured out those parameters mostly by running lots and lots of computer simulations with different parameters and figuring out which sets of parameters match up best with observations of actual supernovae.&lt;/p&gt;&#xA;" OwnerUserId="1655" LastActivityDate="2011-02-17T03:39:15.007" CommentCount="4" />
  <row Id="5330" PostTypeId="1" CreationDate="2011-02-17T05:19:33.657" Score="9" ViewCount="123" Body="&lt;p&gt;There are lots of models of gravity mediated SUSY breaking with various spectra as well as various general gauge mediation models. Are there any &quot;smoking gun&quot; experimental singnatures that could definitely distinguish between the two scenarios? Would this be possible to do at the LHC or would we need an ILC-type machine to do that?&lt;/p&gt;&#xA;" OwnerUserId="1892" LastEditorUserId="1892" LastEditorDisplayName="" LastEditDate="2011-02-17T05:47:16.043" LastActivityDate="2011-02-17T05:47:16.043" Title="Is there any way to distinguish experimentally gauge mediation from gravity mediation in an unambiguous way?" Tags="&lt;particle-physics&gt;&lt;supergravity&gt;&lt;supersymmetry&gt;" CommentCount="3" FavoriteCount="2" />
  <row Id="5331" PostTypeId="2" ParentId="5265" CreationDate="2011-02-17T05:58:53.670" Score="1" ViewCount="" Body="&lt;p&gt;The fastest method for cooling coffee (or hot chocolate, as I'm more likely to drink) I've discovered when I don't have a saucer or second cup is to ignore the spoon altogether.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without a saucer, spoon, or blowing, I first place my hands around the container without any insulation device and let my hands absorb as much heat as they can stand. I then move my hands to the table to transfer the heat to the table. I then repeat the cupping the cup and transferring to the table as necessary. Within 1-2 minutes I can take a 200+ degree cup down to pleasantly drinkable levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method is all about conducting the heat out of the container to the environment, and has the convenient secondary benefit of warming your hands (and the rest of your body) very quickly while waiting for your drink to cool. Combine it with a swishing of the cup to bring down the overall temperature of the coffee (not just the sides), as well as blowing on the surface (if you are allowed to cheat), and the cup cools even faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works best with glass or metal tables, but wood tables also work well.&lt;/p&gt;&#xA;" OwnerUserId="1971" LastActivityDate="2011-02-17T05:58:53.670" />
  <row Id="5332" PostTypeId="1" CreationDate="2011-02-17T06:18:20.853" Score="19" ViewCount="415" Body="&lt;p&gt;The phenomenon of high temperature superconductivity has been known for decades, particularly layered cuprate superconductors. We know the precise lattice structure of the materials. We know the band theory of electrons and how electronic orbitals mix. But yet, theoreticians still haven't solved high Tc superconductivity yet. What is the obstacle to solving it? What are we missing?&lt;/p&gt;&#xA;" OwnerUserId="1978" LastActivityDate="2011-03-12T05:56:13.253" Title="Why is high temperature superconductivity so hard to solve?" Tags="&lt;lattice-model&gt;&lt;superconductivity&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="5333" PostTypeId="2" ParentId="5265" CreationDate="2011-02-17T06:38:33.013" Score="2" ViewCount="" Body="&lt;p&gt;Using the spoon lift some coffee into the air, and let it pour back into the cup. A nice long slow pour is ideal (as high as you can without splashing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motion of the water through the air will cool it fastest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bigger the spoon the faster it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the real world, to cool coffee fast, get a second cup and pour from one to the other (again, as high as you can). Three pours should be enough to get it from burning hot to drinkable.&lt;/p&gt;&#xA;" OwnerUserId="1983" LastActivityDate="2011-02-17T06:38:33.013" />
  <row Id="5334" PostTypeId="2" ParentId="5317" CreationDate="2011-02-17T06:39:18.483" Score="3" ViewCount="" Body="&lt;p&gt;To write down Dirac's equation over curved spacetime, first express the geometry in terms of vierbeins and spin connections. Spinor bundles are vector bundles over spacetime transforming locally under the local Lorentz gauge group. Using the spin connection, we can write down covariant derivatives for sections of the spinor bundle. To get the Dirac operator, we contract the covariant derivative with the inverse vierbein and contract that with the gamma matrices.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-17T06:39:18.483" />
  <row Id="5335" PostTypeId="2" ParentId="5265" CreationDate="2011-02-17T07:02:57.277" Score="7" ViewCount="" Body="&lt;p&gt;I can't believe the density or material of the spoon hasn't been considered. If the spoon is very dense you can take it and wave it in the air in a 15° arc and say &quot;Dear waiter, if you don't put some cold milk in my coffee I will hit you between the eyes with this abnormally dense spoon&quot;. on the other hand if its made of gold or silver you hold it horizontal to the ground in a north south direction such that it refracts the light and say &quot;dear waiter, here I have a brilliant and valuable spoon which I will give you in exchange for placing my coffee cup in a bath of ice til it cools&quot;&lt;/p&gt;&#xA;" OwnerUserId="1986" LastActivityDate="2011-02-17T07:02:57.277" CommentCount="1" />
  <row Id="5337" PostTypeId="1" CreationDate="2011-02-17T07:59:10.350" Score="2" ViewCount="107" Body="&lt;p&gt;When computing the first order perturbative corrections to string theory over a curved background, we find the background has to be Ricci-flat if the dilaton is constant and we have no fluxes. Such is the case for Calabi-Yau compactifications. However, to fourth order in perturbation theory, we find nonzero contributions to the beta function. But this can be resolved by perturbative modifications to the background metric which cancels the beta function order by order in perturbation theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this procedure work for a generic time-varying background which is Ricci-flat to first order in perturbation theory? If not, does that tell us we can't apply first quantized string theory to such backgrounds?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-17T07:59:10.350" Title="Is string theory over a time varying background a conformal field theory to all orders in perturbation theory?" Tags="&lt;string-theory&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="5338" PostTypeId="2" ParentId="5317" CreationDate="2011-02-17T08:12:11.727" Score="3" ViewCount="" Body="&lt;p&gt;The lagrangian for a massless Dirac fermion in a general background is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ L_D = i (\bar \psi e^{\mu}_I \gamma^I \mathcal{D}_{\mu} \psi + c.c.) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c.c.$ means complex conjugate. $\bar \psi = \gamma^0 \psi^+$, $e^{\mu}_I$ is the tetrad (or more generally n-bien) encoding the background metric via $g_{\mu\nu} = e^I_{\mu} e^J_{\mu} \eta_{IJ}$, where $\eta_{IJ}$ is the minkowski metric $diag(-1,1,1,1)$. $\mathcal{D}_{\mu}$ is the covariant derivative whose action on spinors is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathcal{D}_{\mu} \psi = \partial_{\mu} \psi + g A_{\mu}^{IJ}\gamma_I \gamma_J \psi $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $\gamma_I$ are the Dirac matrices and $\gamma_I\gamma_J$ are the generators of the Lorentz lie algebra. For spinors which transform under the action of a general group the second term in the above expression can be generalized to $A_{\mu}^I T_I \psi$ where $T_I$ are the lie-algebra generators of the relevant group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The n-bien encodes the background metric. The gauge connection encodes the background curvature via $F_{\mu\nu}^{IJ} = \partial_{[\mu}A_{\nu]}^{IJ} + g [A_{\mu}, A_{\nu}]^{IJ} $. The $[..]$ in the first term indicates antisymmetrization over the contained indices. The second term is the commutator of the matrices which constitute the connection and is non-zero only for a non-abelian group. The complete Lagrangian including that for the background geometry and the Dirac fields is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ L_{GR + D} = \frac{1}{8\pi G} e \wedge e \wedge F + L_D $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if the background geometry is or can be taken to be static then only the Dirac term matters. The corresponding equations of motion are easily determined via variation w.r.t $\psi$ and $\bar \psi$. The complete solution must take into account not only the bulk values of the n-bien and connection but also the boundary conditions imposed by the background geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above, what is left unsaid is how the $e \wedge e \wedge F$ term corresponds to the Einstein-Hilbert lagrangian. This is essentially the connection formulation of GR for which an excellent pedagogical reference is the paper by Romano &lt;a href=&quot;http://arxiv.org/abs/gr-qc/9303032&quot; rel=&quot;nofollow&quot;&gt;Geometrodynamics vs. Connection Dynamics&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given all this formalism what is lacking is a concrete example. I have no doubt someone (@Lawrence ?) will soon supply one. But this should get you going. You can also go the Clifford/geometric algebra route @Carl mentioned and gain the benefits of a unified language at the expense of some time spent learning the framework. Hope this helps!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;An explicit example&lt;/strong&gt; of an exact solution can be found in this paper on &lt;a href=&quot;http://arxiv.org/abs/0909.3057&quot; rel=&quot;nofollow&quot;&gt;Graphene Wormholes&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2011-02-18T07:45:36.937" CommentCount="2" />
  <row Id="5339" PostTypeId="1" CreationDate="2011-02-17T08:21:43.210" Score="4" ViewCount="112" Body="&lt;p&gt;In the strong coupling limit of type IIA and heterotic E8 string theory, we get 11 dimensional M-theory in which we have no strings. Instead, we have M2 branes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other backgrounds in string theory which can't be described by first quantized string theory?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-17T08:51:03.450" Title="Are there string backgrounds which can't be described by first quantized string theory?" Tags="&lt;string-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5340" PostTypeId="2" ParentId="5339" CreationDate="2011-02-17T08:51:03.450" Score="3" ViewCount="" Body="&lt;p&gt;the strings &lt;em&gt;always&lt;/em&gt; cease to be the main light degrees of freedom when you go to the strong coupling. In type IIB, you get an isomorphic type IIB string theory at strong coupling but the fundamental strings are what used to be called D1-branes at weak coupling, and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have mentioned three 10-dimensional superstring backgrounds. The remaining 10-dimensional supersymmetric string vacua are those with the $SO(32)$ gauge group - the HO heterotic theory and type I string theory - and they're S-dual to each other. The heterotic strings at the strong coupling become type I D1-branes which are heavy. The type I strings, because they're unoriented, can't carry any string-like conserved charge, become unstable, and they don't leave any &quot;visible traces&quot; in the heterotic string spectrum at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thing holds for lower-dimensional vacua, too. The first quantized string theory is only a good description when the coupling is weak - when the dilaton goes to minus infinity. Whether we use the &quot;first quantized&quot; description of the strings or &quot;string field theory&quot; is just a matter of choice of the formalism; both of them are only valid perturbatively. It used to be believed that string field theory knew &quot;much more&quot; if not &quot;everything&quot; about non-perturbative physics but we no longer believe so. It's just a different way to formulate the methods for perturbative (and transperturbative, I mean calculations of D-brane properties) calculations.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-17T08:51:03.450" />
  <row Id="5341" PostTypeId="1" AcceptedAnswerId="5349" CreationDate="2011-02-17T09:36:45.490" Score="1" ViewCount="260" Body="&lt;p&gt;It might sound stupid but seriously I couldn't find &lt;em&gt;why?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a mass with a constant speed, which is acted on by a unit force which is always at right angles to its direction of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do the mass travels a circular path!?&lt;/p&gt;&#xA;" OwnerUserId="2002" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-17T14:53:54.573" LastActivityDate="2011-02-17T20:24:00.250" Title="Why an object travels a circular path?" Tags="&lt;classical-mechanics&gt;&lt;orbital-motion&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="5343" PostTypeId="2" ParentId="5326" CreationDate="2011-02-17T09:53:26.240" Score="0" ViewCount="" Body="&lt;p&gt;You couldn't use only one photon, because the experiment depends on the light being in a pulse. A pulse is a sum of light waves with a range of different frequencies, which all experience different indices of refraction (i.e. phase velocities) when they travel through the medium. This is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Dispersion_%28optics%29&quot; rel=&quot;nofollow&quot;&gt;dispersion&lt;/a&gt;. It is the precise combination of the frequencies in the pulse, and the dispersion and group-velocity dispersion in the caesium cell, that make this experiment possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same issue as Wang's paper, Nature printed a &lt;a href=&quot;http://www.nature.com/nature/journal/v406/n6793/full/406243a0.html&quot; rel=&quot;nofollow&quot;&gt;&quot;News and Views&quot; article&lt;/a&gt; about it. There, they say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They use smooth, well-defined light pulses, so that the peak of the pulse at the output results from the forward rising edge of the input pulse, which occurs far earlier in time, making it consistent with causality. An abrupt feature in the light pulse would not be able to travel faster than c. This means that even if the 'effect' appears to precede the 'cause', you still can't send useful information — such as news of an impending accident — faster than c.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="107" LastEditorUserId="107" LastEditorDisplayName="" LastEditDate="2011-02-17T14:04:44.417" LastActivityDate="2011-02-17T14:04:44.417" CommentCount="6" />
  <row Id="5344" PostTypeId="1" AcceptedAnswerId="5361" CreationDate="2011-02-17T10:05:12.300" Score="5" ViewCount="338" Body="&lt;p&gt;Will the scale say I weigh more, less, or the same on a carpet as compared to a hard floor? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can assume the scale works via spring mechanism. Free body diagrams encouraged!&lt;/p&gt;&#xA;" OwnerUserId="1090" LastActivityDate="2011-02-17T14:41:37.490" Title="Weight on carpet vs. hard floor" Tags="&lt;measurement&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="5345" PostTypeId="2" ParentId="5332" CreationDate="2011-02-17T10:21:29.113" Score="19" ViewCount="" Body="&lt;p&gt;One problem is that band theory isn't everything! Crucially, band theory completely neglects the interactions between electrons. The fact that often one can do this and obtain near correct results is actually amazing, and worth several lecture courses to flesh out the reasons. However, it cannot always be correct. In many materials the electron-electron interaction dominates --- a good example is the so-called Mott insulator, where by band structure calculations you would think you get a half-filled band and so a conductor, but because the electrons repel each other so strongly you actually get a grid-locked lattice of electrons which cannot move, because moving any of them would put two electrons on top of each other! The cuprates are known to be Mott insulating when they are undoped; this is good evidence that interactions are very important. Unfortunately, without the massively simplifying assumption that electrons are independent (i.e. non-interacting) it is an almost intractable problem to describe their behaviour; indeed, we know from other strongly interacting systems such as fractional quantum hall systems that it's possible to end up with no electrons at all, but fractions of them --- the possibilities for novel electronic structure are really unimaginable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2nd problem which has plagued the field is more technical, which simply that the materials don't behave in universal ways! Although we can point to many superficially similar aspects to many cuprates, it's actually not the case that quantitatively they are the same. For instance, the fabled &quot;linear scaling&quot; is actually incredibly hard to really get --- it's very sensitive on impurities, precise doping levels, etc. The flip-side to this is that if we just look at the qualitative features and ask &quot;what theories predict these?&quot; we actually have quite a few --- marginal Fermi liquid theories, quantum critical theories, strongly coupled gauge theories, Gutzwiller projection theory, etc. All of these will give a superconducting dome, with conducting behaviour at high doping, insulating at low, and some form of anomolous transport. However, experimental signatures are actually very hard to pin down without controversy about what really has been measured, so the debate continues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, the historically long argument has created some unpleasant sociology; some (many?) would claim that actually things are pretty settled, and that their favourite theory is clearly superior. This hasn't helped a consensus to form.&lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-02-17T10:21:29.113" CommentCount="1" />
  <row Id="5347" PostTypeId="1" AcceptedAnswerId="5371" CreationDate="2011-02-17T10:45:02.923" Score="0" ViewCount="90" Body="&lt;p&gt;I need a book dealing in particular with the intermolecular forces and, if possible, the diffusive processes. Someone knows an appropriate one?&lt;/p&gt;&#xA;" OwnerUserId="592" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-02-17T11:31:55.293" LastActivityDate="2011-02-17T15:25:51.300" Title="Soft Condensed Matter" Tags="&lt;books&gt;&lt;soft-matter&gt;" AnswerCount="1" />
  <row Id="5349" PostTypeId="2" ParentId="5341" CreationDate="2011-02-17T10:48:19.673" Score="5" ViewCount="" Body="&lt;p&gt;As Newton did in his Principia Mathematica with an example similar to yours, imagine what happens over a short time $dt$ to the velocity of the mass. Using $F = m\,dv/dt$ gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dv = F\,dt/m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force is at right angles to $v$ and so $dv$ is at right angles to, rather than along $v$. Do the same for the next $dt$ and so on, adding up all the contributions of $dv$ for one complete revolution and you will get $\sum dv = 0$. The additional wobble to the circular motion becomes zero as $dv \rightarrow 0$&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="61" LastEditorDisplayName="" LastEditDate="2011-02-17T13:27:40.597" LastActivityDate="2011-02-17T13:27:40.597" CommentCount="5" />
  <row Id="5350" PostTypeId="1" CreationDate="2011-02-17T11:05:48.567" Score="1" ViewCount="129" Body="&lt;p&gt;The mistery of the mass of the top being in the electroweak scale can be justified by the Higgs mechanism itself; in some sense the top mass is the only &quot;natural&quot; mass, the other masses of fermions being protected by some unknown mechanism so that they are &quot;zero&quot; relative to the electroweak: the typical values of the Yukawas are in the $10^{-3}$ range and even smaller. The masses are still &quot;natural&quot; relative to GUT or Planck Scale cutoffs, because then their corrections are logarithmic and about the same order of magnitude that the mass itself, in fact about a 30%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is about the mass of tau and mu. Is there some reason for them to be in the GeV range, where QCD masses -proton and pion, if you wish, or glue and chiral scales- are?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asking for theories justifying this. For instance, Alejandro Cabo tries to produce first the quark masses from the one of the top quark via QCD, very much as &lt;a href=&quot;http://prd.aps.org/abstract/PRD/v7/i8/p2457_1&quot; rel=&quot;nofollow&quot;&gt;Georgi-Glashow electron-muon&lt;/a&gt; in the early seventies, and then he expects that the leptons should have masses similar to the quarks via the electromagnetic/electroweak coupling (albeit some group theoretical reason could be enough).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hierarchically, and from the point of view that any near zero value for a coupling is a hint for a broken symmetry (so that in the limit where this symmetry is unbroken, the coupling is exactly zero), what I would expect is a symmetry protecting all the yukawas except the top, and then still a subgroup protecting the first generation.&lt;/p&gt;&#xA;" OwnerUserId="1335" LastEditorUserId="1335" LastEditorDisplayName="" LastEditDate="2011-02-17T18:19:44.613" LastActivityDate="2011-02-18T17:09:41.613" Title="Why should the mass of leptons to be near of proton and QCD/chiral scales?" Tags="&lt;particle-physics&gt;&lt;standard-model&gt;&lt;mass&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="5351" PostTypeId="2" ParentId="5243" CreationDate="2011-02-17T11:17:25.357" Score="0" ViewCount="" Body="&lt;p&gt;And don't forget; skull is the hardest bone on your body. When the fist crashes into your skull, some bones are likely to be broken. Hopefully, not the skull. So you can avoid a second punch.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-17T11:17:25.357" />
  <row Id="5352" PostTypeId="2" ParentId="5265" CreationDate="2011-02-17T11:26:26.873" Score="3" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;This is why physicists need to talk to&#xA;  chemists once in a while.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is right, mostly because several classic topics are not really &#xA;thought thoroughly any longer in physics. But: This topic is thaught in chemical technology, not physical chemistry, so most chemists do not know about it as well.&#xA;I had the &quot;fun&quot; to learn chemical technology as a forth main topic to get my diploma, &#xA;due to studing at KIT. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As Georg correctly remarks, the latent&#xA;  heat of vaporization of water is&#xA;  enormous - but he's wrong about waving&#xA;  the spoon; stirring is the champion&#xA;  here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If Bob could read between lines, he might have realized that this &quot;waving&quot; &#xA;contained some irony, with respect to the somewhat artificial kind of problem setting. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why? Temperature is really the average&#xA;  kinetic energy of the molecules in the&#xA;  bulk substance, which actually have a&#xA;  variety of individual kinetic&#xA;  energies. Stirring is the fastest way&#xA;  to bring high-kinetic-energy outlier&#xA;  water molecules to the surface, where&#xA;  they will overcome the electrostatic&#xA;  bonding mechanisms that keep them in&#xA;  the liquid phase, and jump into the&#xA;  air (vapor phase). This rapid decrease&#xA;  in the high-energy outliers is the&#xA;  quickest way to cool a hot aqueous&#xA;  solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No detailed comment to that, this is plainly &quot;popular&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It's similar to stirring iced tea. If&#xA;  you just plop ice cubes into a glass&#xA;  of warm tea, it will take quite a&#xA;  while for the warmer tea to cool; if&#xA;  you stir it vigorously, it will reach&#xA;  a cold equilibrium within seconds; the&#xA;  latent heat of fusion absorbed by the&#xA;  ice melting is similarly enormous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a very bad example, because here Bob deals about two liquid phases, &#xA;bot with viscosity, thus the need to stir. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Bob had learnd the business of heat transfer thoroughly, he knew that &#xA;on the gaseous side of a barrier the big &quot;resistance&quot; is located by &#xA;the laminar boundary layer, which does not conduct heat well. &#xA;Thus the use of &quot;blowing&quot; (forced convection) to make this boundary layer &#xA;as thin as possible. This boundary layer allows heat only by conduction &#xA;(very low for gases) and mass exchange only by diffusion, thus limiting &#xA;the use of vaporisation. Compared to this, the boundary layer of the liquid &#xA;is much less important, because of the higher conductivity of liquid water and &#xA;the high specific heat (= high Thermal diffusivity). &#xA;In general, stirring of the coffee will show some  benefit, &lt;strong&gt;after&lt;/strong&gt; enough &#xA;forced convection (blowing) is applied to the surface, because it is the secomdary &quot;resistance&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to Bobs solutions: very good, but partly outside the rules: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;cup of coffee and a spoon.&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-02-17T11:26:26.873" />
  <row Id="5353" PostTypeId="1" AcceptedAnswerId="5357" CreationDate="2011-02-17T11:46:57.653" Score="1" ViewCount="190" Body="&lt;p&gt;In one dimension -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can one prove that the Hammiltonian and the parity operator commute in the case where the potential is symmetric (an even function)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. that [H, P] = 0 for V(x)=V(-x)&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-07-19T13:50:36.180" LastActivityDate="2011-07-19T13:50:36.180" Title="Symmetric potential and the commutator of parity and hamiltonian" Tags="&lt;quantum-mechanics&gt;&lt;symmetry&gt;&lt;parity&gt;" AnswerCount="1" />
  <row Id="5354" PostTypeId="2" ParentId="5341" CreationDate="2011-02-17T11:56:04.633" Score="1" ViewCount="" Body="&lt;p&gt;Maybe its just because a circle is the only figure with constant 90 ° angle between tangents and radius? &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-02-17T11:56:04.633" CommentCount="3" />
  <row Id="5355" PostTypeId="2" ParentId="5332" CreationDate="2011-02-17T12:08:54.483" Score="2" ViewCount="" Body="&lt;p&gt;Your question implies that things are much better for classic superconductors. &#xA;We know by BCS that some phonon coupling of electrons creates bosonic electron pairs,&#xA;and the theory predicts quantitatively the dependence of transition temperature &#xA;to atomic mass of the metal, veryfied at least one time on aluminium samples of &#xA;different isotopic composition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But afaik  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the BCS does not tell why a certain metal has a high or rather low transition temperature, &#xA;or some metals do not show superconduction down to the lowest temperatures accessible. &#xA;I think, even for the classic superconductors (especially the so called &quot;hard&quot; SCs) &#xA;there is a lot to unravel still. But condensed matter is so complicated :=(&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-02-17T12:08:54.483" />
  <row Id="5356" PostTypeId="2" ParentId="3834" CreationDate="2011-02-17T12:34:39.640" Score="2" ViewCount="" Body="&lt;p&gt;I thought I might just start with an introduction first. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic principal behind Free Electron Laser is that of synchrotron radiation. When electrons or charged particles are made to change momentum (like being bent in a in an arc where the force is radially inwards) they emit electromagnetic radiation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the particles are relativistic then the electromagnetic radiation the lab observer relative to the electron will observe the electromagnetic radiation being emitted in a cone in the direction of motion. (i'll post figures if people really want!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the free electron laser you need to have magnetic arrays which are simply dipole magnets aligned such that the electrons will see as its travelling in a straight line, alternating vertical magnetic fields. This causes the electrons to &quot;undulate&quot; horizontally. With relativistic electrons (which is not hard to do) you will emit synchrotron radiation like a pencil beam. The wavelength from an undulator is given by lambda = undulator_period/(2gamma^2)*(1+a^2/2), where gamma is the relativistic gamma factor, a = e*B*undulator_period/(2*pi*electronmass*speedlight) where B is the strength of the magnetic field and e the electric charge. But the radiation from all the electrons are not very temporally coherent!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now as to why its called a laser! The fundamental feature of why lasers are powerful is because the waves emitted by all sources are emitted in a temporally coherent fashion. I.e. all the electric fields add constructively so the total power scales like &lt;em&gt;pN^2&lt;/em&gt; where &lt;em&gt;p&lt;/em&gt; is the power emitted by a single source and &lt;em&gt;N&lt;/em&gt; is the number of sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a FEL the undulation is small enough that the electrons are irradiated by its own synchrotron radiation and is therefore travelling in an optical field as well as the magnetic field from the dipole magnets. The optical field causes the electrons to micro-bunch and cause further amplification of that particular wavelength (micro-bunching effect) and is referred to as self-amplified spontaneous emission (SASE). This positive feedback generates powers that give an additional gain of &lt;em&gt;N&lt;/em&gt;. Typical bunches of electrons are 10^11 electrons, so you can get orders of magnitude more power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I presume you can reduce the size of the magnetic arrays to just a stretch of material with carefully oriented domains however it will be difficult to reach the fields required. To measure thats the field of spectroscopy which i'm less familiar. For UV you'll have to have some material with dispersion properties like a prism of sorts and a detector that will respond to UV radiation. For x-rays you can use monochromators that use the Bragg principal and determine the energy (calibrated to know emission lines from elements).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy loss along the undulator is a small factor. The emitted photons are very small fractions of its total energy. For example for 10 to 100 nm wavelengths you would need something like 300 MeV electrons (rest mass of an electron is 0.5 MeV) so beta = 0.9999986, and 10 nm photons are only 10s of eV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought the military were focusing on UV FELs like you said, to reduce water absorption. Xrays are much harder to generate with enough power to cause damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That was more than i initially intended to say. Hope its useful.&lt;/p&gt;&#xA;" OwnerUserId="2011" LastActivityDate="2011-02-17T12:34:39.640" CommentCount="2" />
  <row Id="5357" PostTypeId="2" ParentId="5353" CreationDate="2011-02-17T13:17:13.170" Score="3" ViewCount="" Body="&lt;p&gt;You prove the equality of operators by applying them to a function, we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;H = - \frac{\hbar^2}{2 m} \frac{d^2}{dx^2} + V(x)&#xD;&#xA;$$&#xA;Ergo:&#xA;$$&#xD;&#xA;HP f(x) = H f(-x) = (- \frac{\hbar^2}{2 m} \frac{d^2}{dx^2} + V(x)) f(-x) = - \frac{\hbar^2}{2 m} f&amp;#39;&amp;#39;(-x) + V(x) f(-x)&#xD;&#xA;$$&#xA;and&#xA;$$&#xD;&#xA;PH f(x) = P (- \frac{\hbar^2}{2 m} \frac{d^2}{dx^2} + V(x)) f(x) = P (- \frac{\hbar^2}{2 m} f&amp;#39;&amp;#39;(x)) + P (V(x) f(x)) ...&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;... = - \frac{\hbar^2}{2 m} f&amp;#39;&amp;#39;(-x) + V(-x) f(-x)&#xD;&#xA;$$&#xA;When you use &#xA;$$&#xD;&#xA;V(-x) = V(x)&#xD;&#xA;$$&#xA;you see that both expressions are equal.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditorDisplayName="" LastEditDate="2011-02-17T13:44:39.970" LastActivityDate="2011-02-17T13:44:39.970" />
  <row Id="5358" PostTypeId="2" ParentId="5344" CreationDate="2011-02-17T14:27:06.253" Score="1" ViewCount="" Body="&lt;p&gt;Obviously you weight the same. On a soft carpet your weight is distributed to more contact area than a hard floor. Hence pressure, which is force per unit area is greater on a hard floor than soft carpet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $A$ is the area of contact in case of a carpet and $a$ is the same for hard floor. $A$ &gt; $a$ . Your weight is same in both cases. It is $Mg$ where $M$ is your mass and $g$ is the acceleration due to gravity. Pressure in case of a carpet is $p$ $=$ $\frac{Mg}{A}$ and pressure in case of hard floor is $P$ $=$ $\frac{Mg}{a}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $A &amp;gt; a$ it follows $P &amp;gt; p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Newton's third law the same pressure is applied to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the soft carpet feels comfortable but as far as the weight is concerned any &lt;strong&gt;good&lt;/strong&gt; scale will show that your weight being same in both cases. If they deffer then it is manufacturer's fault. (Normally a manufacturer calibrates weight scales on a hard surface)&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-17T14:41:37.490" LastActivityDate="2011-02-17T14:41:37.490" />
  <row Id="5359" PostTypeId="2" ParentId="5326" CreationDate="2011-02-17T14:27:41.060" Score="4" ViewCount="" Body="&lt;p&gt;The front velocity, defined as the propagation speed of the point where the field first differs (by any arbitrarily small amount) from &lt;strong&gt;exactly&lt;/strong&gt; zero, is &lt;strong&gt;always&lt;/strong&gt; no greater than $c$. (In fact, the front velocity is always exactly equal to $c$, no greater or less.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem here is that a Gaussian pulse extends infinitely in both directions, so it simply does not have a &quot;front&quot; to speak of. Of course the amplitude decays super-exponentially on both sides, but that doesn't matter. There is no causality problem with the pulse emerging arbitrarily early in time, because your input pulse made the field start changing long before that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a thought experiment, let's imagine we have a button we can push to start the input pulse going. If any trace of the output pulse comes out before a signal at $c$ has a chance to propagate from when and where the button was pushed, that's a causality violation. But that is impossible, for the following reason:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a theoretically perfect Gaussian pulse has no finite start time, but has a nonzero amplitude at arbitrarily early times, it's impossible to create such a perfect Gaussian pulse by pushing a button. Of course, you can get arbitrarily close to perfection, but there will always be some distortion that gets worse and worse as you try to make the pulse fire &quot;faster&quot;, that is, try to make a shorter separation in time between the button push and the maximum of the pulse. The &lt;strong&gt;theorem&lt;/strong&gt; that the medium has a causal response to the field guarantees that the response to this distortion will always interfere with the response to the perfect Gaussian to cancel out &lt;strong&gt;exactly&lt;/strong&gt; at times earlier than a signal traveling at $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, in a real experiment, the &quot;button push&quot; (actually some electronic signal to the machine that creates the pulse) happens a relatively long time before any trace of the output pulse is detected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a great book chapter I just found about this: &lt;a href=&quot;http://books.google.com/books?id=kE8OUCvt7ecC&amp;amp;pg=PA26&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=kE8OUCvt7ecC&amp;amp;pg=PA26&lt;/a&gt; It has lots more math than my answer (though not too high a level) and might clear up a lot of things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also ask about single photons, but that opens up a huge can of worms I can't really get into (not least because I don't understand it well enough myself). Let me just say that there is always a minimum amount of noise in any mode / degree of freedom of the electromagnetic field, which is equivalent to half a photon. You could make a pulse of the right shape that's so weak there's only a single photon in it (I'm sure people do things like that all the time), but the problem is if you try to &quot;announce&quot; too early that you've detected that photon, you'll be wrong such a large fraction of the time that you can show statistically that no information is being transmitted. It's really difficult to do quantum-limited measurement in the first place (because you have to severely limit the back-reaction of the measurement apparatus on the field you're measuring), and if you try to do it too fast it becomes literally impossible.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-02-17T14:27:41.060" />
  <row Id="5360" PostTypeId="2" ParentId="5332" CreationDate="2011-02-17T14:29:02.227" Score="1" ViewCount="" Body="&lt;p&gt;For strongly correlated systems &amp;mdash; where perturbation theory and mean field theory breaks down &amp;mdash; even if you know the exact form of the Hamiltonian, and regulate it over a lattice, it can still be extremely difficult to compute the ground state, or even figure out its qualitative behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being a quantum system, we can't simulate it using a classical computer. You need a quantum computer for that.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-17T14:29:02.227" CommentCount="1" />
  <row Id="5361" PostTypeId="2" ParentId="5344" CreationDate="2011-02-17T14:30:54.587" Score="6" ViewCount="" Body="&lt;p&gt;According to what I understand you will show as weighing more on a carpet than on a hard floor. From what I understand it is due to the way the hard floor affects the feet of the scales. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a article that explores that question: &lt;a href=&quot;http://www.newscientist.com/article/dn2462-people-weigh-less-on-a-hard-surface.html&quot; rel=&quot;nofollow&quot;&gt;http://www.newscientist.com/article/dn2462-people-weigh-less-on-a-hard-surface.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1914" LastActivityDate="2011-02-17T14:30:54.587" />
  <row Id="5362" PostTypeId="2" ParentId="5265" CreationDate="2011-02-17T14:31:11.887" Score="4" ViewCount="" Body="&lt;p&gt;The fastest and coolest way to cool the coffee, with only a cup and spoon, that is also theoretically possible, is to throw all the coffee up in the air, and with somewhat well-coordinated movement catch it all in the cup as it falls down. This maximises the total surface area of the coffee with the air per time, and thus also the total heat transfer.&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-02-17T14:31:11.887" CommentCount="1" />
  <row Id="5363" PostTypeId="1" AcceptedAnswerId="5365" CreationDate="2011-02-17T14:37:24.230" Score="3" ViewCount="214" Body="&lt;p&gt;Seeing this as an academic community, I hope this question is on-topic. &lt;a href=&quot;http://area51.stackexchange.com/proposals/16617/academia?referrer=dC8p1D-IBbcjazD_NAENyw2&quot;&gt;Academia&lt;/a&gt; is still a long way from beta &lt;code&gt;:(&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a few questions about reading journal papers in the field of engineering/applied physics.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How do you keep and schedule a reading list?&lt;/li&gt;&#xA;&lt;li&gt;From the more recent papers, how do you track down the one (or few) papers that started an idea or technology?&lt;/li&gt;&#xA;&lt;li&gt;Then conversely, how do you move forward in time to trace how that technology evolved? How do you decide which is the next paper to read?&lt;/li&gt;&#xA;&lt;li&gt;(I think being able to do the tasks of 2 and 3 could help me formulate my own research questions in the future)&lt;/li&gt;&#xA;&lt;li&gt;How do you retain the gist of the information you read from an article?&lt;/li&gt;&#xA;&lt;li&gt;How do you do the dirty work of the above? What software do you use, if at all? If you write it down in a notebook, what are the essential data points? Like the date you read the paper, publication date, title, author, then writing down (or illustrating) what you see with your mind's eye the information that the article presented, and... anything else?&lt;/li&gt;&#xA;&lt;li&gt;I envision a notebook with the ideas I learned, then posing my own questions after reading each article. How do you do it?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You may (or may not) answer those questions one-by-one, but they're there to give you an idea of what I want to find out. Offers to make this community wiki are very welcome.&lt;/p&gt;&#xA;" OwnerUserId="293" LastActivityDate="2011-02-17T19:55:12.033" Title="Making and keeping a reading list" Tags="&lt;soft-question&gt;&lt;reseach-level&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="3" />
  <row Id="5364" PostTypeId="1" AcceptedAnswerId="5370" CreationDate="2011-02-17T14:59:35.353" Score="0" ViewCount="173" Body="&lt;p&gt;Hey guys, I'm having a problem in understanding the 1st Law of Thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If i would increase the pressure of a closed system by compressing a gas in a cylinder isothermal, the 1.Law states, that all the work i put in to the system, i get out by heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my 2.state would be the gas at the same temperature (same internal energy), but compressed (at a higher pressure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But isn't this compressed gas able to do more work, than the not compressed?&lt;br&gt;&#xA;If i drilled a hole in the cylinder my high pressure gas could do some work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does this energy come from, since all my input energy i put in by work, came out in form of heat? &lt;/p&gt;&#xA;" OwnerUserId="1789" LastActivityDate="2011-02-18T18:53:52.183" Title="Why isn't pressure a measure of energy?" Tags="&lt;thermodynamics&gt;&lt;energy&gt;&lt;energy-conservation&gt;&lt;pressure&gt;" AnswerCount="3" />
  <row Id="5365" PostTypeId="2" ParentId="5363" CreationDate="2011-02-17T15:02:57.710" Score="3" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Using a wiki: you can write pages about your current knowledge of a specific topic with links to papers and books you'd like to read. You can use either a personal wiki running on your own notebook or an online version (the wiki software that Wikipedia uses is open source, free, and not too difficult to install and get running). An example of a wiki that started as a notebook of a reading list (sort of) is the &lt;a href=&quot;http://ncatlab.org/nlab/show/HomePage&quot; rel=&quot;nofollow&quot;&gt;nLab&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use review papers. Identify the core people of a specialized research community and search for their papers. Besides, finding the papers that really started an idea is difficult, most of the time, there is no simple solution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Usually by interpolation: If there are review papers, these will sometimes list intermediate papers, at least they will give a hint at the historical development of an idea.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do you have to formulate those yourself or do you have an advisor? The easiest way is to ask the experts what they are working on. Another way is to solve problems of increasing difficulty until you think you solved one where you cannot find a solution in the literature.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Again, use a wiki. You can always expand and restructure a wiki, search engines and hyperlinks are your friends when you use a wiki.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can measure your progress by looking at the history of your wiki pages (and become embarrassed if you find a wiki page you wrote a year ago which says that &quot;I need to read this paper soon!&quot;.) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Have an &quot;open questions&quot; paragraph on your wiki pages.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Sorry for being somewhat repetitive :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good example for a reading list for a specific topic that I haven't had the time to pursue is this: &lt;a href=&quot;http://ncatlab.org/nlab/show/AQFT+on+curved+spacetimes&quot; rel=&quot;nofollow&quot;&gt;AQFT on curved spacetime&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-02-17T15:02:57.710" CommentCount="1" />
  <row Id="5366" PostTypeId="2" ParentId="5363" CreationDate="2011-02-17T15:04:38.310" Score="0" ViewCount="" Body="&lt;p&gt;You could try a tool such as &lt;a href=&quot;http://freemind.sourceforge.net&quot; rel=&quot;nofollow&quot;&gt;freemind&lt;/a&gt; for tracking where information is coming from. It allows you to create a topic, then list from there your primary sources. Then any additional sources can be hung from the original source, and so on. This allows you to track and follow your sources to individual facts.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-17T15:04:38.310" />
  <row Id="5367" PostTypeId="2" ParentId="5350" CreationDate="2011-02-17T15:09:12.270" Score="1" ViewCount="" Body="&lt;p&gt;The anthropic principle might explain it.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The difference in mass of the neutron and proton is roughly the difference in mass between that of the down and up quarks, which is comparable to that of the electron mass. If the proton were significantly heavier, they would decay into neutrons &amp;mdash; which are electrically neutral &amp;mdash; leading to no atoms or chemistry. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The instability of neutrons with a half life of 15 minutes is necessary, as otherwise,&#xA;most of the hydrogen and neutrons would fuse together into deuterium during the big bang nucleosynthesis epoch. This means the mass of the neutron should be slightly larger than the sum of proton and electron masses.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For chemistry as we know it to be possible, molecules ought to have definite shapes, as opposed to a delocalized quantum superposition of nuclei positions. This requires the nucleons to be at least 1000 times heavier than the electron, fixing the QCD scale.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The weakness of the weak interaction is due to the smallness of the Yukawa couplings for the first generation of fermions. If the weak interaction were significantly stronger than it is in our universe, the neutrinos generated during the explosive fusion when the core of a massive star collapses will interact too strongly with the stellar matter that they would be absorbed before reaching the outer regions of the star. In our universe, the weak interaction is sufficiently weak that enough neutrinos can reach the outer regions, but yet strong enough to interact sufficiently there to blow the outer region away in a supernova. Without supernovae, we won't have planets with significant quantities of heavier elements needed for life.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;However, I can't think of any anthropic reason why the Yukawa coupling for the tau lepton isn't of order unity.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-02-18T17:09:41.613" LastActivityDate="2011-02-18T17:09:41.613" CommentCount="1" />
  <row Id="5368" PostTypeId="2" ParentId="5364" CreationDate="2011-02-17T15:09:33.373" Score="0" ViewCount="" Body="&lt;p&gt;Pressure is equal to force per unit area which equal to energy per unit volume. Hence the work which is done by the pressure is due to the change in pressure and not pressure itself per say.&lt;br&gt;&#xA;That is given by Bernoulli's equation.&lt;br&gt;&#xA;Here's some more about &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/pber.html#beq&quot; rel=&quot;nofollow&quot;&gt;Bernoulli's equation.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1847" LastActivityDate="2011-02-17T15:09:33.373" />
  <row Id="5369" PostTypeId="2" ParentId="5207" CreationDate="2011-02-17T15:09:59.787" Score="0" ViewCount="" Body="&lt;p&gt;I am separating this answer from the other because it is overly speculative; mainly I wanted to list a few hints about the sequence I named in the comments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The OP names a square sum that happens to be related to the critical dimension of space time of the bosonic string, D-2=24. It seems natural to ask if there is some similar sequence for the critical dimension of the superstring, D-2=8. On other hand, as I said in the other answer, for the open superstring the Zeta regularization is naturally substituted, in some cases, by the Dirichlet Eta, that happens to be an alternating sum. So it is natural for a numerologist to try the alternating square sum and, as said in the comments to the OP, it works:&#xA;$$1^2-2^2+3^2-4^2+5^2-6^2+7^2-8^2= -36 = -6^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main difference, number-wise, with the non alternating sum, is that here the solution is not unique. Still, it is the smallest non trivial one, and all the others can be generated iteratively: the (absolute value of the) sums are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Square_triangular_number&quot; rel=&quot;nofollow&quot;&gt;triangular square numbers&lt;/a&gt;, and it was observed by Colin Dickson (alt.math.recreational March 7th 2004) that such numbers obey a recurrence law $$a_{n+1}={(a_n -1)^2 \over a_{n-1}}$$ with the first two terms being the trivial $a_1=1$ and the above $a_2=36$. For more info, see the OEIS sequences &lt;a href=&quot;http://oeis.org/A001110&quot; rel=&quot;nofollow&quot;&gt;A001110&lt;/a&gt; and &lt;a href=&quot;http://oeis.org/A001108&quot; rel=&quot;nofollow&quot;&gt;A001108&lt;/a&gt;. Note that the sign in the actual solution depends of the number of terms in the sequence, alternating itself, so that actually the sum is $\sigma_n= (-1)^{n-1} a_n$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A way to produce the alternating solutions is to solve Pell equation, and then the sums are also produced from Pell numbers, via $a_n= P_n (P_n+P_{n-1})$ . This could be interesting because the root of the non alternating series, $70$, is a Pell number itself, the 6th, and the next Pell number, $70+(70+29)=169$, is the only Pell number that is an exact square (and the only one that is an exact power). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://oeis.org/A001108&quot; rel=&quot;nofollow&quot;&gt;A001108&lt;/a&gt;, Mohamed Bouhamida mentions some periodicities and some mod 8 relationships for the series. Also the page &lt;a href=&quot;http://www.cut-the-knot.org/do_you_know/triSquare.shtml&quot; rel=&quot;nofollow&quot;&gt;http://www.cut-the-knot.org/do_you_know/triSquare.shtml&lt;/a&gt; gives some hints on some eight factors appearing in a particular &lt;em&gt;subsequence&lt;/em&gt; of square triangular numbers: &quot;Eight triangles increased by unity produce a square&quot;. If these factors can be related to the 8-periodicities of Bott theory or theta functions, I can not tell.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: of course the use of triangular numbers can be telling us that all the business of alternating series is just a decoy: pairing the terms, we can reduce $ (m+1)^2 - m^2 = 2 m + 1 = (m+1) + m $ the alternating squared series to the non alternating non squared series and then simply $$1+2+3+4+5+6+7+8= 36 = 6^2$$ But the goal is to keep at least a formal likeliness with the OP series.  &lt;/p&gt;&#xA;" OwnerUserId="1335" LastEditorUserId="1335" LastEditorDisplayName="" LastEditDate="2011-02-17T15:55:06.580" LastActivityDate="2011-02-17T15:55:06.580" />
  <row Id="5370" PostTypeId="2" ParentId="5364" CreationDate="2011-02-17T15:16:31.473" Score="3" ViewCount="" Body="&lt;p&gt;The pressure is higher, but the volume is less. The kinetic energy per particle stays the same, when compressing isothermally, there are just more particles per volume to &quot;bounce on the walls&quot;, thus the higher pressure. Your higher pressure gas from your drilled cylinder could just do more work per volume, but since there is less volume available the total work stays the same!&lt;/p&gt;&#xA;" LastActivityDate="2011-02-17T15:16:31.473" CommentCount="1" />
  <row Id="5371" PostTypeId="2" ParentId="5347" CreationDate="2011-02-17T15:25:51.300" Score="0" ViewCount="" Body="&lt;p&gt;&quot;Intermolecular and Surface Forces&quot; by Jacob Israelachvili.  A new edition came out recently.  This covers all the types of forces relevant at the molecular scale or nanoscale.&lt;/p&gt;&#xA;" OwnerUserId="266" LastActivityDate="2011-02-17T15:25:51.300" />
  <row Id="5372" PostTypeId="2" ParentId="5364" CreationDate="2011-02-17T15:26:16.637" Score="0" ViewCount="" Body="&lt;p&gt;The first law of thermodynamics says $\delta Q = du + \delta w$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For isothermal compression internal energy of the system remain unaltered. Hence $\delta u = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\delta Q =  \delta w$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the work $\delta w $ that is done on the system has been released to the outside world as $ \delta Q$&#xA;It has no more ability to do work than it had before the compression since all that energy has been already released during the isothermal compression.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-02-17T15:26:16.637" />
  <row Id="5373" PostTypeId="1" CreationDate="2011-02-17T16:13:30.993" Score="2" ViewCount="220" Body="&lt;p&gt;Quantum decoherence is an irreversible process which is the result of interaction of the system with its environment. It prevents interference due to lack of coherence. Environment acts just like a heat bath. Now my question is, is the different branches of the wave function of the universe becoming gradually more and more decoherent so that in a far future no trace of interference can occur? In other words I would like to know whether there will be any quantum death (like heat death of thermodynamics) of the universe if one waits for a sufficiently long time?&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-06-03T12:54:15.270" Title="Quantum death like heat death possible?" Tags="&lt;quantum-mechanics&gt;&lt;cosmology&gt;" AnswerCount="1" CommentCount="13" FavoriteCount="1" />
  <row Id="5374" PostTypeId="2" ParentId="5265" CreationDate="2011-02-17T16:37:48.870" Score="1" ViewCount="" Body="&lt;p&gt;I stir the coffee in a side to side motion with the spoon bowl parallel to the side of the cup. Due to the angle bernoulli's principle moves the coffee through the center of the cup towards the top. This circulates the liquid much more quickly than you'd expect in a controllable fashion that avoids splashing vigorous stirring can cause. It also has the side effect of removing the dry creamer from the bowl of the spoon far more efficiently than any other possible motion. The liquid coming up from the center and being forced across the top to the sides smoothly and without cavitation gaps stirring causes. The liquid also conducts with the sides better and releases heat both via the wave that forms at the top increasing the surface area contact with air and transferring the heat to the body of the cup/heat radiator.&#xA;Actually I do stir that way. It cleans the spoon far better and requires less effort. Other than that the above is merely serendipitous inference.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-17T16:37:48.870" />
  <row Id="5375" PostTypeId="2" ParentId="5265" CreationDate="2011-02-17T16:44:27.130" Score="2" ViewCount="" Body="&lt;p&gt;This question is a classic. On related topics, you can read:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.loreto.unican.es/Carpeta2007/00TorreonCartes2007/CafeyLeche.pdf&quot; rel=&quot;nofollow&quot;&gt;Factors influencing cooling of coffee and tea&lt;/a&gt;, 1988&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://demonstrations.wolfram.com/TheCoffeeCoolingProblem/&quot; rel=&quot;nofollow&quot;&gt;Wolfram simulator&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure I read a paper in Scientific American about this, but I cannot find it back&lt;/p&gt;&#xA;" OwnerUserId="2045" LastActivityDate="2011-02-17T16:44:27.130" CommentCount="2" />
  <row Id="5376" PostTypeId="1" CreationDate="2011-02-17T17:05:55.863" Score="-1" ViewCount="107" Body="&lt;p&gt;With the dawning of the private space industry, if someone was to build a rocket-based craft that powers you up and then glides in for a landing, what would be the quickest you could get across the Atlantic?  Or half way around the world?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Assume a typical flight trajectory - e.g. don't go through the earth or travel at ground level + 1 meter!)&lt;/p&gt;&#xA;" OwnerUserId="491" LastActivityDate="2011-02-17T19:32:48.890" Title="Quickest &quot;flights&quot; across Atlantic and round the world" Tags="&lt;aerodynamics&gt;&lt;rocket-science&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="5377" PostTypeId="2" ParentId="5363" CreationDate="2011-02-17T17:17:46.603" Score="0" ViewCount="" Body="&lt;p&gt;I like to have my reference manager (such as &lt;a href=&quot;http://bibdesk.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;BibDesk&lt;/a&gt;, but there are many others for other platforms) double as a reading list and note taker. This way, I keep references, PDFs, reading status and notes in one place. BibDesk, in particular, stores its database as a BibTeX file, so it is also ready to be referenced by your own LaTeX articles. I now have a pretty sprawling list of articles, neatly organized by tags, where I have my go-to references for each paper I write. Check it out!&lt;/p&gt;&#xA;" OwnerUserId="1252" LastEditorUserId="1252" LastEditorDisplayName="" LastEditDate="2011-02-17T19:55:12.033" LastActivityDate="2011-02-17T19:55:12.033" />
  <row Id="5378" PostTypeId="2" ParentId="5373" CreationDate="2011-02-17T17:18:06.197" Score="1" ViewCount="" Body="&lt;p&gt;This question is a bit strange, and I tend to agree with Anna that this is related to thermodynamics.  The entropy involved here is an entanglement entropy.  Suppose you have system A and system B which form an entanglement.  The entanglement entropy of system A is &#xA;$$&#xD;&#xA;S_A~=~-Tr[\rho_A log\rho_A],&#xD;&#xA;$$&#xA;which equals the entanglement entropy of system B.  If the two systems form a pure state then $S_{A+B}~=~0$.  The entanglement entropy comes from the fact you have access to only one part of the density matrix $\rho~=~\rho_A\otimes\rho_B$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This plays a role in cosmology at large.  During inflation the vacuum energy density was huge.  The cosmological constant $\Lambda~\simeq~(8\pi G/c^4)\rho$, is very large and drives a rapid exponential expansion of space.  There is some theoretical controversy here, but while the energy density of the vacuum was very large, the entropy was not that large.  The entropy is a measure of the number, N, of degrees of freedom in a system that are coarse grained into a macrostate $S~=~k log(N)$.  The other oddball factor is that while the temperature was high, the entropy was low due to the negative heat capacity of event horizons in spacetime.  During inflation the event horizon was smaller than a proton, and the entropy is proportional to the area of the horizon $S~=~k A/4L_p^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bang came about because the exponential expansion rapidly came to a halt, the cosmological constant dropped to a small value (the vacuum energy dropped enormously) and the cosmological horizon adjusted to a very large value.  It is now out about $10^{10}$ light years.  This means a relatively small number of degrees of freedom enter into complicated entanglements which are not accessible in a local region.  The entanglement entropy increases, and these states appear in a highly thermalized form.  This is the bang and fire of the big bang.  It is a form of latent heat of fusion in a phase transition.  The large vacuum energy $\rho~\simeq~10^{100}GeV^4$ crashed into about 10 GeV^4, and the energy gap assumed the form of a thermalized gas of particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was the initial generation of a huge amount of entropy in the early universe.  Subsequently entropy is in the form of black holes, radiation and so forth.  It is interesting to think we can understand this all from the perspective of quantum mechanics.  Into the future the universe will end up as a de Sitter vacuum.  In the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/5283/de-sitter-cosmologic-limit&quot;&gt;de sitter cosmologic limit&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I indicated how the universe will over a vast period of time will decay from the de Sitter vacuum configuration with a small vacuum energy to a Minkowski spacetime.  The horizon will retreat of to “infinity,” which means the entropy becomes infinite.  It might be problematic to think of infinite entropy.  There might be some sort of cut-off in this process.  On the other hand this is just a measure of how the vacuum decays away to zero and there is no energy.  The retreat of the cosmological horizon off to infinity is probably a measure of continued quantum entanglement process, which proceeds almost indefnately.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-17T17:18:06.197" CommentCount="2" />
  <row Id="5379" PostTypeId="2" ParentId="5304" CreationDate="2011-02-17T17:28:11.600" Score="2" ViewCount="" Body="&lt;p&gt;The electrical field $\mathbf E$ is the fundamental one. In principle, you don't need the electrical displacement field $\mathbf D$, everything can be expressed in terms of the field $\mathbf E$ alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works well for the vacuum. However, to describe &lt;strong&gt;electromagnetic fields in matter&lt;/strong&gt;, it is &lt;strong&gt;convenient&lt;/strong&gt; to introduce another field $\mathbf D$. Maxwells original equations are still valid, but in matter, you have to deal with additional charges and currents that are induced by the electric field and that also induce additional electric fields. (More precisely, one usually makes the approximation that the electric field induces tiny dipoles, which are described by the &lt;em&gt;electric polarization&lt;/em&gt; $\mathbf P$.) A little bit of calculation shows that you can conveniently hide these additional charges by introducing the electrical displacement field $\mathbf D$, which then fulfills the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \nabla&amp;#183; \mathbf D = \rho_\text{free} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that this equation involves only the &quot;external&quot; (&quot;free&quot;) charge density $\rho_\text{free}$. Charges that accumulate inside the block of matter have already been taken into account by the introduction of the $\mathbf D$ field.&lt;/p&gt;&#xA;" OwnerUserId="188" LastActivityDate="2011-02-17T17:28:11.600" CommentCount="5" />
  <row Id="5380" PostTypeId="2" ParentId="5243" CreationDate="2011-02-17T17:34:53.963" Score="0" ViewCount="" Body="&lt;p&gt;$m_1,u_1, v_1$ be mass, initial velocity and final velocity of your head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_2,u_2, v_2$ be mass, initial velocity and final velocity of opponents hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there are four cases, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$m_1 &amp;lt; m_2$ AND [ $u_1 &amp;lt; u_2$  OR $u_1 &amp;gt; u_2$ ]&lt;/li&gt;&#xA;&lt;li&gt;$m_1 &amp;gt; m_2$ AND { $u_1 &amp;lt; u_2$  OR $u_1 &amp;gt; u_2$ ]&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$v_1 = \frac{u_1(m_1 - m_2) + 2m_2 (-u_2) }{m_1 + m_2}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_2$ is negative because direction of $u_1$ and $u_2$ opposite.&#xA;Also, we are not worried about $v_2$ because its opponents hands velocity after collision. &lt;em&gt;(Who cares what happens to his hand?)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplifying and not worrying about denominator,&#xA;$v_1 = u_1 ( m_1 - m_2)- 2m_2 u_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if $m_1 &amp;lt; m_2$ then $v_1 &amp;gt; u_1$ if $u_1$ and $u_2$ have opposite directions and $v_1 &amp;lt; u_1$ if $u_1$ and $u_2$ have the same direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if $m_1 &amp;gt; m_2$ then $v_1 &amp;lt; u_1$ if $u_1$ and $u_1$ have opposite directions and $v_1 &amp;lt; u_1$ if $u_1$ and $u_2$ have the same direction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, if mass of your head is less than mass of your opponents hand then move away from the punch otherwise move into the punch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, reason for the recommendation might be that, generally, mass of head is larger than that of hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Assumptions:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Collision is elastic. &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Your head and opponents hand are intact after collision $m_1 , m_2$ not changed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Source:&lt;/em&gt; Wikipedia Elastic Collision&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="126" LastActivityDate="2011-02-17T17:34:53.963" CommentCount="2" />
  <row Id="5381" PostTypeId="2" ParentId="629" CreationDate="2011-02-17T17:50:51.603" Score="2" ViewCount="" Body="&lt;p&gt;There is no pulling, your body is pulling yourself because of the muscles' contraction to counter balance the deceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's model your body by a mass reparteed vertically, plus a your feet.&#xA;No force is applied, you are standing straight.&#xA;Imagine we are at constant speed, the train driving from left to right on the picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/J6SPJ.png&quot; alt=&quot;equilibrium&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the vehicule decelerates, you are the subject of &lt;a href=&quot;http://en.wikipedia.org/wiki/Fictitious_force&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;inertial force&lt;/em&gt;&lt;/a&gt; (if we consider the train as the reference, the inertial force aka &lt;em&gt;fictious force&lt;/em&gt; greatly simplifies the explaination ; you could consider the earth as the reference and study the momentum instead).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's consider you don't wear rollerblades. Your shoes have enough friction on the ground, and an opposite force is applied on your feet. The sum of these two forces is null, and you don't slide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/jqPmb.png&quot; alt=&quot;inertial&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, these two  results in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Couple_%28mechanics%29&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;couple&lt;/em&gt;)&lt;/a&gt; on your body, which begins to rotate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Ur1rb.png&quot; alt=&quot;resulting couple&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'll assume you have a brain. Instinctively, the brain orders your muscles to contract to counteract this couple. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ZHruu.png&quot; alt=&quot;counter couple&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This muscular couple is created by your powerful calf contracting (i.e. pulling your body down), and your strong tibia has an opposite force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of sudden, the train stops, and the deceleration is null. What remains is your own couple. You pull yourself back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/qrDvd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2045" LastActivityDate="2011-02-17T17:50:51.603" CommentCount="4" />
  <row Id="5382" PostTypeId="2" ParentId="5363" CreationDate="2011-02-17T17:52:27.310" Score="1" ViewCount="" Body="&lt;p&gt;Tracking down original paper is easy. You just check the reference list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I print out papers in such a format that leaves space on left and right of the main text. That is where I take my notes and do dirty calculations. Photo:)&#xA;&lt;img src=&quot;http://i.stack.imgur.com/iK4rs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My tools&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Paper&lt;/li&gt;&#xA;&lt;li&gt;Pencil&lt;/li&gt;&#xA;&lt;li&gt;Emacs org-mode&lt;/li&gt;&#xA;&lt;li&gt;Zotero Firefox plug-in&lt;/li&gt;&#xA;&lt;li&gt;Freemind&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="126" LastActivityDate="2011-02-17T17:52:27.310" />
  <row Id="5383" PostTypeId="2" ParentId="5243" CreationDate="2011-02-17T18:04:46.217" Score="0" ViewCount="" Body="&lt;p&gt;Assuming all you want to do is receive a softer punch, it would be better to lean away, since you would diminish the speed at which the blow happens: (speed of punch) - (speed of your head backwards before punch) = (speed at which you get punch). This would result in a slightly less powerful punch.&lt;/p&gt;&#xA;" OwnerUserId="1662" LastActivityDate="2011-02-17T18:04:46.217" />
  <row Id="5384" PostTypeId="2" ParentId="5341" CreationDate="2011-02-17T18:13:06.680" Score="1" ViewCount="" Body="&lt;p&gt;Assume the motion is confined to a plane.  If the force is always at a right angle to the motion, the speed is necessarily constant.  The circular shape follows due to symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the change in speed of an object, we can project the force vector acting on it onto the velocity vector.  For example, the force below will create a large change in speed, because it is pushing mostly the same way an object is going.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://dl.dropbox.com/u/1377494/circular1.png&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This next force is just as big, but it will only create a small change in speed because it is pushing only partially in the direction of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/W3vbc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This final force will create no change in speed because it is perpendicular to the motion.  Its only effect is to change the direction of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/X8Kl8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the problem specifies that the force is always perpendicular to the motion, the object's speed is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we know the speed is constant, it follows from symmetry that the motion is a circle.  If the motion were, say, a parabola, then different parts of the trajectory would &quot;feel&quot; different to the particle.  For example, the very bottom of a parabola is highly curved, but if you go up the sides a way, it is not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/7wlek.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The green part of the parabola, at the bottom, has a very short radius of curvature.  The red part has a very long one.  A ball traveling at constant speed feeling a constant force should not have such different types of motion at different times.  Instead, since the force on it is always the same relative to itself, it should always be doing the same thing.  There should not be any part of the curve that is different than any other part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other shapes, like an ellipse or a spiral, are also different in some parts than in others.  Only the circle has the same curvature everywhere, and so the motion must be a circle.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-02-17T20:24:00.250" LastActivityDate="2011-02-17T20:24:00.250" CommentCount="5" />
  <row Id="5386" PostTypeId="2" ParentId="5363" CreationDate="2011-02-17T18:28:34.233" Score="5" ViewCount="" Body="&lt;p&gt;One Word: &lt;a href=&quot;http://www.tiddlywiki.com/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;TiddlyWiki&lt;/strong&gt;&lt;/a&gt; - personal, light-weight, off-line, extendable, free, cross-platform, tex-enabled wiki cum notebook. Its simply incredible. Everything in &lt;em&gt;one&lt;/em&gt; html file. If you want to use latex &lt;em&gt;jsmath&lt;/em&gt; is required. Just install Bob Mc. Elrath's &lt;a href=&quot;http://bob.mcelrath.org/tiddlyjsmath-2.0.3.html&quot; rel=&quot;nofollow&quot;&gt;jsmath plugin&lt;/a&gt; and you're in tex heaven.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh and did I mention there are hundreds of plugins which you can use to customize your tiddlywiki in endless ways. Here are a few of the prime plugin destinations where you can also download distributions of tiddlywiki with different sets of plugins for different users:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.tiddlytools.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.tiddlytools.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://tiddlyvault.tiddlyspot.com/&quot; rel=&quot;nofollow&quot;&gt;http://tiddlyvault.tiddlyspot.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://tiddlywiki.org/wiki/Plugins&quot; rel=&quot;nofollow&quot;&gt;http://tiddlywiki.org/wiki/Plugins&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also find discussions of this amazing tool on other SE sites such as &lt;a href=&quot;http://superuser.com/questions/87724/which-tiddlywiki-plugins-do-you-use&quot;&gt;this one&lt;/a&gt; on superuser.SE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some very neat plugins that allow you to easily create beautiful vector graphics for your notes. Both the &lt;a href=&quot;http://www.math.ist.utl.pt/~psoares/MathSVG.html&quot; rel=&quot;nofollow&quot;&gt;MathSVG&lt;/a&gt; plugin by Paolo Soares or the &lt;a href=&quot;http://raphaeljs.com/&quot; rel=&quot;nofollow&quot;&gt;Rafael&lt;/a&gt; javascript library are excellent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Happy Wikiing !&lt;/p&gt;&#xA;" LastActivityDate="2011-02-17T18:28:34.233" CommentCount="3" />
  <row Id="5387" PostTypeId="2" ParentId="5376" CreationDate="2011-02-17T19:20:07.530" Score="0" ViewCount="" Body="&lt;p&gt;Well, while some think it's simple x = ct, speed is a question :-)&#xA;Optimal trajectory would be suborbital (but still not higher than ~150-200km), for these distances suborbital speed is gonna be ~5km/sec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as a rough estimate, for 15'000km flights you will need 50 minutes as a rough estimation. &#xA;If we assume that non-prepared humans can handle 2g overload at takeoff/landing, we'll need 5000/19.6 = 4.5*2 = 9 minutes for takeoff and landing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in total it's gonna be about 1 hour. And tonns of fuel ;-)&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-02-17T19:20:07.530" CommentCount="2" />
  <row Id="5388" PostTypeId="2" ParentId="5376" CreationDate="2011-02-17T19:26:12.653" Score="3" ViewCount="" Body="&lt;p&gt;Oh, I see. I may have understood what Alistair is asking. He wants the rocket to accelerate only at the very beginning of the flight, and then fly by free fall, without any force except for Earth's gravity. For short flights, you would imagine that the trajectory will be a parabola.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for significant distances in a spherically symmetric gravitational field of the Earth, the trajectory still has to be an ellipse with the Earth's center as the focus - using Newton's approximation for the law of gravity (Kepler's laws hold, but the Earth is now at the center instead of the Sun). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could calculate the timing more accurately but it's clear that the optimal trajectory will have a pretty constant altitude - so the speed will be close to the &lt;strong&gt;first cosmic velocity&lt;/strong&gt; - the velocity corresponding to the circular orbit. The optimal trajectory will be an ellipse that is very close to the circle around the Earth, of radius $6378+10 km$ or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first cosmic velocity is &#xA;$$ v = \sqrt{gR} $$&#xA;where $g=9.8 ms^{-2}$ and $R=6,378,000 m$. You may see that $v=7.9 km/s$. For 4,000 kilometers between East Coast and Central Europe, you need about 500 seconds which is 10 minutes. Faster than BarsMonster. ;-) It takes about 1 hour and 25 minutes to fly around the Earth (40,000 km) at the orbital speed; the International Space Station is doing it 17 times a day or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand why BarsMonster thinks that &quot;optimal&quot; (for what?) velocity is suborbital. My understanding is that Alistair wants the speed. So the ellipses that are very close to circles - orbital speed - are &quot;optimal&quot; according to my definition of the word. They're faster. Let's hope that you have enough money to pay Richard Branson both expenses and the profit.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-17T19:32:48.890" LastActivityDate="2011-02-17T19:32:48.890" CommentCount="1" />
  <row Id="5389" PostTypeId="2" ParentId="5300" CreationDate="2011-02-17T19:33:57.120" Score="1" ViewCount="" Body="&lt;p&gt;Peter,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since your original post on this topic I have encountered &quot;Causal Perturbation Theory&quot; which does distribution-like calculations using a modification of distributions based on Scaled Test Functions. This paper does lots of interaction and S matrix calculations. Perhaps you are entirely familiar with this, or it is not suitable but here is the link I have been studying:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/hep-th/pdf/9710/9710225v1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/hep-th/pdf/9710/9710225v1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-17T19:33:57.120" CommentCount="1" />
  <row Id="5390" PostTypeId="1" CreationDate="2011-02-17T20:06:44.570" Score="0" ViewCount="121" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/937/how-does-gravity-escape-a-black-hole&quot;&gt;How does gravity escape a black hole?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;This may sound like a stupid question - why do black holes have a gravitational field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light cannot escape the event horizon of a black hole, and it gets redshifted if it travels close by. Gravity travels in waves at the speed of light, so why aren't the gravity waves pulled back into the black hole? Do gravity waves not interact with other gravity waves? Is there some sort of superposition effect going on?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess this is a case of &quot;a little knowledge is a dangerous thing&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-17T20:06:44.570" ClosedDate="2011-02-17T20:40:24.603" Title="Why do black holes have a gravitational field?" Tags="&lt;gravity&gt;&lt;black-holes&gt;" CommentCount="1" />
  <row Id="5392" PostTypeId="2" ParentId="2508" CreationDate="2011-02-17T21:35:03.710" Score="0" ViewCount="" Body="&lt;p&gt;This answer is just for the purpose of discussion. It will be edited sometimes and It may contain wrong conclusions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Martin. I wanted to add another argument that one has a spinning motion as well as an orbital motion. In order to keep the arguments seperated, I choose a new answer. Please consider the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one has a force acting &lt;strong&gt;anywhere&lt;/strong&gt; on a rigid body, this force is causing an acceleration of the center of mass of the body (described by $F=ma$) and it is simultaneously causing an angular acceleration of the body (described by $M=r\times F$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in case of our baby, the force acted upon the pad causes the center of mass of the pad to translate and causes the pad to spin. Using the above principle it should be pretty obvious that both a spinning (around the center of mass of the pad) and a translational motion (an orbital one in our case) of the pad take place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to further clarify the intended motion I have drawn new pictures, illustrating the situation. The red line is marking the starting position of the baby. The curved arrow is indicating the angular velocity of the pad (counterclockwise). In these pictures, the baby itself is always moving clockwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/kQcxn.png&quot; alt=&quot;&quot;&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/Mex4J.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In these pictures, one can see an orbital motion of the center of the pad around the center of masss of the system (COM). This motion is caused by $F=ma$. Furthermore one can see a spinning motion of the pad around the center of the pad, which is caused by $M=r\times F$. As one can see in these pictures, both angular motions contribute to the displacement between the childs current and its original position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Mark. If you are still following this question, feel free to include or modify any of my pictures in your answer.&lt;/p&gt;&#xA;" OwnerUserId="393" LastEditorUserId="393" LastEditorDisplayName="" LastEditDate="2011-02-18T18:09:38.083" LastActivityDate="2011-02-18T18:09:38.083" CommentCount="73" />
  <row Id="5394" PostTypeId="2" ParentId="2604" CreationDate="2011-02-17T22:18:08.683" Score="-1" ViewCount="" Body="&lt;p&gt;This question is related to the problem of family structure.  I will restrict this to QCD, where the gauge group is $SU(3)$, and there are the gauge multiples.  The irreducible representation of $SU(3)$ are ${\bf 3}\times {\bar{\bf 3}}$ and ${\bf 8}~+~{\bf 1}$.  The first irrep describes the quark doublet structure or “flavors,” while the second defines the “color scheme,” which is how quarks may carry two of the gauge coupling charges $r,g,b$ and anti $r,g,b$.  The “one” defines “white” which is color neutral.  So it turns out that the gauge field has a certain group structure and the carriers of these charges also have the same group structure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We might think of the gauge potential $A_\mu$ as written according to $A_\mu~=~A^b_\mu\lambda^b$ where the sum is over the $\bf 8$ color scheme.  The currents for the theory $J_\mu~=~{\bar\psi}^i\gamma_\mu\psi^i$ are determined by the quarks.  Here the index $i$ is with respect to the $\bf 8$ of $SU(3)$.  We have from electromagnetism the Maxwell-Faraday equation&#xA;$$&#xD;&#xA;\nabla\times {\bf H}~=~{\bf J}~+~\frac{\partial {\bf D}}{\partial t}&#xD;&#xA;$$&#xA;where the time derivative of electric displacement vector is the “displacement current.”  In effect the field theory says, “The left hand side sums these up and computes a value for $\bf H$ independently of the particular values of each.”  The two play identical roles.  With QCD we have a similar gauge covariant form of this expression, and the magnetic intensity analogue in QCD obeys a similar rule.  Consequently, the current and gauge field should be interchangeable by changing the irrep --- to put is somewhat loosely I have to admit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does not constitute a proof, but it is suggestive of why the source for the fields has a similar structure as the fields themselves.  The one exception is electromagnetism, which is a $U(1)$ gauge group and all particles carry a charge.  The connection with particle families and electromagnetism is with hypercharge and the Gell-Mann–Nishijima formula that interchanges the centers of higher groups with $U(1)$.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-17T22:18:08.683" CommentCount="4" />
  <row Id="5395" PostTypeId="1" AcceptedAnswerId="5396" CreationDate="2011-02-17T23:48:48.683" Score="0" ViewCount="203" Body="&lt;p&gt;I'm given with the following problem - it's an easy one, but it's nearly 1AM, I'm tired and I need some push into the general direction to get the solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A particle is assumed to be in the state&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(-\sqrt{1 \over 3} Y_1^0(\theta, \phi), -\sqrt{2 \over 3} Y_1^1(\theta, \phi)\right)^T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y$ are spherical harmonics. What are the &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;total angular momentum $\vec{j}^2$&lt;/li&gt;&#xA;&lt;li&gt;total orbital angular momentum $\vec{l}^2$&lt;/li&gt;&#xA;&lt;li&gt;total spin angular momentum $\vec{s}^2$&lt;/li&gt;&#xA;&lt;li&gt;total angular momentum $j_z$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;of the particle&lt;/p&gt;&#xA;" OwnerUserId="1906" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-07T11:05:24.560" LastActivityDate="2011-03-07T11:05:24.560" Title="Angular momentum" Tags="&lt;quantum-mechanics&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="5396" PostTypeId="2" ParentId="5395" CreationDate="2011-02-18T00:38:52.727" Score="4" ViewCount="" Body="&lt;p&gt;Just recall the definitions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;spherical harmonics $Y^m_l$ are eigenvectors of both ${ \hat L}_z$ and $\hat L^2$ with eigenvalues of $m$ and $l(l+1)$ respectivelly (with $\hbar = 1$)&lt;/li&gt;&#xA;&lt;li&gt;if you understand that this is a spin $1 \over 2$ particle then it should be obvious that $\hat s^2 = s(s+1) = {3 \over 4}$. If not, recall how spin matrices look like (e.g. start with Pauli matrices) and compute $\hat s^2$ directly.&lt;/li&gt;&#xA;&lt;li&gt;the total Hilbert space $H$ is a tensor product of the scalar particle system with the spin system $H_{total} = H_s \otimes H_2$. So the operators on this space are $\hat {\mathbf L}_{total} = \hat {\mathbf L} \otimes \hat {\mathbb 1}_2$, $\hat {\mathbf s}_{total} = \hat {\mathbb 1}_s \otimes \hat {\mathbf s}$ (this just means that orbital momentum is still just differential operators and spin is just matrix operator) and $\hat J_{total} = \hat L_{total} + \hat s_{total}$. The rest is direct computation&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-02-18T00:38:52.727" />
  <row Id="5397" PostTypeId="2" ParentId="5320" CreationDate="2011-02-18T00:54:21.813" Score="0" ViewCount="" Body="&lt;p&gt;A related post from arXiv blog.&#xA;Cosmos &lt;a href=&quot;http://www.technologyreview.com/blog/arxiv/26333/&quot; rel=&quot;nofollow&quot;&gt;At Least 250x Bigger Than Visible Universe&lt;/a&gt;, Say Cosmologists.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-02-18T00:54:21.813" />
  <row Id="5398" PostTypeId="1" AcceptedAnswerId="5399" CreationDate="2011-02-18T01:30:45.273" Score="1" ViewCount="138" Body="&lt;p&gt;So I am working on my assignment, and have a question about converting coordinates. I dont know whether I should ask here or the math SE, so lets give it a try here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force in question is $$\vec{F} = -k\,r^{-n}\,\hat{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the conversion equations but I have no idea how to do it. I need to split the $\hat{r}$ into its corresponding i, j, k vectors. I wanna take the curl to make sure that its zero and hence the force is conservative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: the question asks to be converted. I know how to already do it in spherical coordinates.&lt;/p&gt;&#xA;" OwnerUserId="551" LastEditorUserId="869" LastEditorDisplayName="" LastEditDate="2011-02-18T03:01:57.003" LastActivityDate="2011-02-18T03:01:57.003" Title="converting force from spherical to Cartesian coords" Tags="&lt;classical-mechanics&gt;&lt;homework&gt;&lt;coordinate-systems&gt;" AnswerCount="1" />
  <row Id="5399" PostTypeId="2" ParentId="5398" CreationDate="2011-02-18T02:01:07.563" Score="3" ViewCount="" Body="&lt;p&gt;$$r^2 = x^2 + y^2 + z^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does that make $r^{-n}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for converting $\hat{r}$, the position vector can be written&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{r} = r \hat{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in spherical coordinates, but it can also be written&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in rectangular coordinates.  Therefore, the two are equal.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r \hat{r} = x\hat{i} + y\hat{j} + z\hat{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there you can finish up finding a formula for $\hat{r}$ in rectangular coordinates.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-02-18T02:14:49.047" LastActivityDate="2011-02-18T02:14:49.047" CommentCount="4" />
  <row Id="5400" PostTypeId="2" ParentId="5320" CreationDate="2011-02-18T03:43:28.393" Score="-1" ViewCount="" Body="&lt;p&gt;In 10 billion years we will be able to see galaxies in the local group by optical means.  If one were to look with IR it would be possible to see galaxies which have receded further out and are moving out at a faster rate.  If one used millimeter wavelength it is possible then to see very distant galaxies that we can now see in the optical or IR.  The CMB will not be visible in the microwave, but radio wave band.  Every bit of the universe we see today is in principle observable in the future.  Of course as the universe expands and accelerates further it will require radio astronomy to detect any galaxy beyond what then will be the merged Andromeway galaxy and the CMB would only be observable as very long wavelength radio waves.  As time marches on the photons from anything not gravitationally tied to our galaxy will become indefinitely redshifted.  So all of the universe is still available for observation, just ever more redshifted&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also in 100 billion years the galaxy will dimly glow with red dwarf stars.  The universe is generally getting darker and colder.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-18T03:43:28.393" CommentCount="1" />
  <row Id="5402" PostTypeId="1" AcceptedAnswerId="5415" CreationDate="2011-02-18T06:00:47.593" Score="0" ViewCount="98" Body="&lt;p&gt;What is actual meaning of singularity  can we use this term for conclusion in any research paper( related to cosmological models ).what r the types .&lt;/p&gt;&#xA;" OwnerUserId="1843" LastActivityDate="2011-02-18T10:52:50.723" Title="Meaning and types of singularity in case of string or any cosmological model (Mathematical description)" Tags="&lt;general-relativity&gt;&lt;string-theory&gt;&lt;mathematical-physics&gt;" AnswerCount="1" />
  <row Id="5403" PostTypeId="1" AcceptedAnswerId="5406" CreationDate="2011-02-18T06:01:10.017" Score="1" ViewCount="405" Body="&lt;p&gt;Big Bang theory is widely accepted theory when it comes to origin of universe. What other really compelling theories are out there explaining/supporting the origin of universe. I know many people don't believe in Big Bang theory for various reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&#xA;Question was not complete before for some reason.&lt;/p&gt;&#xA;" OwnerUserId="2071" LastEditorUserId="2071" LastEditorDisplayName="" LastEditDate="2011-02-18T09:46:39.700" LastActivityDate="2011-02-18T09:46:39.700" ClosedDate="2011-02-18T09:03:04.373" Title="List of theories supporting origin of universe" Tags="&lt;universe&gt;&lt;theory&gt;&lt;big-bang&gt;" AnswerCount="1" />
  <row Id="5404" PostTypeId="1" AcceptedAnswerId="5405" CreationDate="2011-02-18T07:13:00.023" Score="1" ViewCount="213" Body="&lt;p&gt;What are the weak hypercharge and weak isospin quantum numbers of the helicity states of the $W^\pm$ and $Z^0$ bosons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The W boson is a spin-1 massless particle. Consequently it has three helicity states, negative (left-handed), positive (right-handed), and neutral. As discussed in this paper, these states restrict which orientations of quarks they can interact with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The emitted $b$ quark can be regarded&#xA;  as massless compared to the top quark,&#xA;  and hence expected to be predominantly&#xA;  of negative helicity (left-handed),&#xA;  meaning that its spin points opposite&#xA;  to its line of flight. The emitted $W$&#xA;  boson, being a massive spin-1&#xA;  particle, can assume any of three&#xA;  helicities: one longitudinal ($W_0$)&#xA;  and two transverse states ($W_−$,&#xA;  left-handed and $W_+$, right-handed)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Int.J.Mod.Phys.A24:2899-3037,(2009), Marc-André Pleier, &lt;em&gt;Review of Properties of the Top Quark from Measurements at the Tevatron&lt;/em&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/0810.5226&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0810.5226&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The helicity states of a particle like the electron act very much differently, and they carry different weak hypercharge and/or weak isospin quantum numbers. So really this is two questions. First, are weak hypercharge and weak isospin good quantum numbers for the helicity states of the W and Z bosons? And if they are good states, what are the quantum numbers?&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-18T19:10:52.617" Title="Quantum numbers of the helicity states of the W boson" Tags="&lt;particle-physics&gt;&lt;weak-force&gt;&lt;quark-model&gt;" AnswerCount="1" />
  <row Id="5405" PostTypeId="2" ParentId="5404" CreationDate="2011-02-18T07:22:11.310" Score="3" ViewCount="" Body="&lt;p&gt;Hypercharge and weak isospin are obviously broken generators of $SU(2)\times U(1)$, so they're not conserved which also means that the mass eigenstates are not eigenstates (or representations) of those generators. W-bosons and Z-bosons are states where the breaking of these symmetries is most manifest, so of course, in general, they don't have well-defined values of those quantities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fundamental fields $W^i_\mu$ and $B_\mu$ carry vanishing hypercharge $Y=0$ because $U(1)_Y$ is an Abelian group so its messengers are neutral. However, their helicity components, especially the longitudinal ones, if they're moving quickly, are mixing with components of the Higgs boson (or several Higgs bosons) that carries a nonzero hypercharge $Y$. So especially for the longitudinal component, the value of $Y$ is a superposition of two possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third component of the isospin suffers from the very same uncertainty because the sum of the isospin and the hypercharge is the electric charge - which is well-defined and conserved. One could discuss that the $T_3=-1,0,+1$ triplet (the W-bosons and the $W_3$ combination of the photon and the Z-boson) and the $T_3=0$ singlet states (the $B_0$ combination of the photon and the Z-boson) are getting mixed with the $T_3=\pm 1/2$ doublet of the Higgs field but the conclusion - and the degree of mixing of states with different eigenvalues - would be equivalent to the hypercharge discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(At loop level, the states of the W-bosons and Z-bosons may receive additional contributions from multi-particle states with even different values of the hypercharge and isospin. If $T_3$ may change by $1/2$ by a simple processes, a sequence of processes in a loop diagram may change it by anything.)&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-18T09:59:36.057" LastActivityDate="2011-02-18T09:59:36.057" CommentCount="4" />
  <row Id="5406" PostTypeId="2" ParentId="5403" CreationDate="2011-02-18T07:50:56.943" Score="2" ViewCount="" Body="&lt;p&gt;Cyclic theories (Steinhardt, Turok), Big Bounce theories (LQG), eternal inflation (Linde etc, variant of the Big Bang), Conformal Cyclic Cosmology (Penrose, variant of the Big Bang), &quot;No Boundary&quot; origin (Hawking, Hartle, variant of Big Bang with no singularity and imaginary time), the self-creating universe (&lt;a href=&quot;http://arxiv.org/abs/astro-ph/9712344&quot; rel=&quot;nofollow&quot;&gt;Richard Gott&lt;/a&gt; )-unlikely but fun to consider, Steady State theory--Hoyle Gold and Bondi--not taken seriously now, and, of course various religious creation myths taken seriously by some :)&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-02-18T07:50:56.943" CommentCount="1" />
  <row Id="5407" PostTypeId="1" CreationDate="2011-02-18T08:51:05.837" Score="4" ViewCount="261" Body="&lt;p&gt;Calculations of the shear viscosity of the quark-gluon plasma (QGP), observed in heavy ions collisions, are performed via the AdS/CFT correspondence &lt;a href=&quot;http://arxiv.org/abs/hep-th/0405231&quot; rel=&quot;nofollow&quot;&gt;[ref1]&lt;/a&gt;. One finds that the shear viscosity of the CFT, on the boundary of an AdS spacetime with a black hole in its interior, is equal to the shear viscosity of the fluid theory living on the black hole horizon surface. This quantity in turn can be calculated by considering the cross-section for graviton absorption &lt;a href=&quot;http://arxiv.org/abs/hep-th/9609052&quot; rel=&quot;nofollow&quot;&gt;[ref2]&lt;/a&gt; by the horizon in the limit that the frequency $\omega \rightarrow 0$ (low-energy limit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The end results suggest that the fluid on the horizon (and its dual fluid the QGP) is as close to being a &quot;perfect&quot; fluid as Nature allows. What is found is that the shear viscosity $\eta$ of the horizon is proportional to its surface entropy &lt;em&gt;density&lt;/em&gt; $s$. It is conjectured that there is a lower bound:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\eta}{s} \ge \frac{1}{4\pi} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(in natural units) for the viscosity to entropy ratio for any hydrodynamical system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These results would seem to imply that the evolution of the QGP cannot be given by a unitary theory. Perhaps this is just a naive misinterpretation of the physics on my part. If unitarity in quantum gravity is presumed to be sacrosanct then &lt;strong&gt;how does one reconcile the dissipative behavior of the field theory of a QGP and a black hole horizon with the demands of unitarity?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; @Lubos' answer is fairly complete but I would like to see what others have to say. Thus the bounty. Have at it!&lt;/p&gt;&#xA;" LastActivityDate="2011-03-11T19:31:45.073" Title="Viscosity/Entropy ratio and unitary evolution in quantum gravity" Tags="&lt;quantum-field-theory&gt;&lt;hydrodynamics&gt;&lt;ads-cft&gt;&lt;unitary-evolution&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5408" PostTypeId="2" ParentId="5407" CreationDate="2011-02-18T09:35:18.870" Score="4" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;These results would seem to imply that the evolution of the QGP cannot be given by a unitary theory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Quite on the contrary. Both sides are equivalent so the AdS/CFT is a proof that the black hole dynamics is unitary - because its equivalent to another theory (the CFT) whose dynamics is given by a manifestly and exactly Hermitian Hamiltonian. This also implies that the information is not lost when black holes evaporate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am confused by your statement about &quot;sacrosanct unitarity of quantum gravity&quot;. If it is sacrosanct, why do you claim it is not true? In reality, even if some people could call it sacrosanct today, the unitarity of quantum gravity is much more nontrivial - and has been much more controversial for decades - than the unitarity of the laws governing quark-gluon plasma. The latter are &lt;em&gt;self-evidently&lt;/em&gt; unitary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The AdS/CFT correspondence became the most well-known method - but not the only method - to resolve the much harder problem of unitarity of quantum gravity. It's the key method that has also convinced Hawking that he was wrong and the information is preserved. And the answer is Yes, any evolution in quantum gravity in an asymptotically AdS (or flat) spacetime is unitary even if it involves the birth and disappearance of black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be confused by the term &quot;dissipation&quot;. Dissipation doesn't mean that the theory fails to be unitary. Dissipation is a process in which macroscopic forms of energy are converted to the microscopic forms of energy - heat - which allow the entropy to increase. But the full microscopic theory is still unitary; if you measure the degrees of freedom describing the QGP accurately, they will be shown to evolve in a unitary way (even though, the effective theories we use in practice may display some disappearance of the non-uniformities and information). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the black hole case, people didn't know any reason (e.g. a description) that would indicate that the information was preserved, even in principle. This changed with the AdS/CFT correspondence, Matrix theory, and others. Since the late 1990s, the status of &quot;dissipation&quot; in quantum gravity has been on par with the status in field theories. At the fundamental level, the evolution is unitary and preserves the information even though approximate, macroscopic effective descriptions of the phenomena fail to see this fact.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-18T09:56:42.687" LastActivityDate="2011-02-18T09:56:42.687" CommentCount="6" />
  <row Id="5410" PostTypeId="1" AcceptedAnswerId="5411" CreationDate="2011-02-18T09:55:58.610" Score="0" ViewCount="121" Body="&lt;p&gt;Currently the sun is launching some intense solar flares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.guardian.co.uk/science/2011/feb/17/solar-flares-northern-lights-uk&quot; rel=&quot;nofollow&quot;&gt;http://www.guardian.co.uk/science/2011/feb/17/solar-flares-northern-lights-uk&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Th article I've linked also mentions how a &quot;coronal mass ejection knocked out the power grid in Quebec&quot; in 1989. Some powerful stuff. Thankfully the Earth's magnetic fields protect us rather well from the brunt of the force. But I wondered about those on the International Space Station? How are they protected, if it all? Surely a blackout on the ISS would mean death, so how do they prevent them?&lt;/p&gt;&#xA;" OwnerUserId="2074" LastActivityDate="2011-02-18T17:07:22.520" Title="How does solar activity affect the ISS?" Tags="&lt;astrophysics&gt;&lt;mechanics&gt;&lt;space&gt;&lt;sun&gt;&lt;engineering&gt;" AnswerCount="2" />
  <row Id="5411" PostTypeId="2" ParentId="5410" CreationDate="2011-02-18T10:07:19.863" Score="0" ViewCount="" Body="&lt;p&gt;They are not really protected. Some 1-2mm of metal is nothing.&#xA;So, god bless earth magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prevent blackout there is just proper onboard powergrid engineering with redundancy (which in fact exist in civil power grids too, but fails because someone wanted to save money somewhere). &lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-02-18T10:07:19.863" CommentCount="7" />
  <row Id="5412" PostTypeId="1" CreationDate="2011-02-18T10:26:32.660" Score="0" ViewCount="465" Body="&lt;p&gt;In Mechanics the torque due to a couple is given by $\tau = P\times d$, where $\tau$ is the resulting couple, $P$ is one of the force vectors in the couple and $d$ is the arm of the couple. A couple is made up of two forces of the same magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, a moment is also given by $M = P\times d$. However, there is only one force involved! How can the resulting torque be of the same magnitude if in one case two, in the other case only one force is involved?&lt;/p&gt;&#xA;" OwnerUserId="2075" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-18T10:49:23.950" LastActivityDate="2011-02-18T18:24:19.800" Title="What is the difference between a moment and a couple?" Tags="&lt;mechanics&gt;&lt;torque&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5413" PostTypeId="2" ParentId="2604" CreationDate="2011-02-18T10:30:43.670" Score="4" ViewCount="" Body="&lt;p&gt;Dear Chad, I thought you were an atheist. Most atheists tend to realize that many things that exist in the Universe have no &quot;purpose&quot;. The existence of the Universe has no &quot;purpose&quot; that may be scientifically demonstrated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if life could exist in a Universe with 1 generation of quarks and leptons, which I find plausible (although I couldn't instantly produce any string compactification with 1 generation), one could still ask why the Universe only has 1 of them if it can have several generations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea that 1 generation is inevitably &quot;qualitatively more likely&quot; or &quot;qualitatively more natural&quot; than 3 generations is just flawed. Life could arguably exist somewhere in a Universe with 1 generation. The parameters and molecules relevant for life - and phase diagrams of QCD etc. - would have to be recalculated but no proof is known that would show that life would be impossible in such a Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, the fact that there are 3 generations in a particular Universe can be derived from deeper properties of string theory (half of the Euler character of the Calabi-Yau shape, assuming a conventional heterotic compactification for a while), and as I have hinted, even at this very point, it might be possible to show that the number of generations cannot be one, among other forbidden values. While three-generation models are known, it's not fully known at this moment whether 3 generations is a unique solution to some conditions or whether it's a coincidence, as the anthropic reasoning wants us to immediately believe.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-18T10:30:43.670" />
  <row Id="5414" PostTypeId="2" ParentId="5412" CreationDate="2011-02-18T10:35:43.447" Score="4" ViewCount="" Body="&lt;p&gt;Ingo, when you consider the couple, you may put one of the &quot;spouses&quot; at the origin, so his torque is $P\times d_0$ for $d_0=0$, so his torque vanishes. Meanwhile, she is located at a nonzero $d$ so her contribution is $P\times d$ and nonzero. Because his torque is zero, it doesn't matter whether you add him or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only difference between the whole couple (including the husband at the origin) and the separate wife at nonzero $d$ is that the total &lt;em&gt;force&lt;/em&gt; (not moment) of the whole couple is zero, while the wife separately acts both by torque and by a nonzero ordinary force as well.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-18T10:35:43.447" />
  <row Id="5415" PostTypeId="2" ParentId="5402" CreationDate="2011-02-18T10:52:50.723" Score="2" ViewCount="" Body="&lt;p&gt;We should narrow the question to a concrete theory, like general relativity, to get a definite answer. In general relativity a singularity in a spacetime is characterized by timelike and/or lightlike paths that are not extendable to infinite time in the past and the future: That is, there are paths that end before &quot;time itself comes to an end&quot; (speaking metaphorically) or that begin after &quot;time itself has begun&quot; (again, metaphorically). &#xA;The former describe observers that fall into a singularity, the latter describe observers escaping from a singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naively, one would have expected that &quot;singularity&quot; in general relativity is simply defined as a point of a spacetime = Lorentzian manifold where the curvature diverges. This definition runs into problems, which are very nicely explained in the classic book&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hawking, Ellis: &quot;The large scale structure in space-time&quot;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You may have heard that both Hawking and Ellis are famous experts and that their work is hard to understand: While this may be true, their book is very well and clearly written and should be understandable to anyone with some background in GR and differential geometry. Try it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I'd recommend that you try to become a little bit more specific with your question and narrow it down to a specific physical setting or theory, like: what is a singularity in viscuous flows described by the Navier-Stokes equation, what is the description of a black hole like singularity in a spacetime according to the XY theory of quantum gravity etc. &lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-02-18T10:52:50.723" CommentCount="1" />
  <row Id="5416" PostTypeId="1" AcceptedAnswerId="5433" CreationDate="2011-02-18T11:02:06.497" Score="1" ViewCount="461" Body="&lt;p&gt;In our aerodynamics class we recently discussed the concept of static and dynamic pressure and discussed their application to aircraft instruments. However, I do not understand properly how the altimeter can work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all a small recap of Bernoulli's law. The total pressure is given by $p_t = \frac{1}{2} \rho V^2 + p_s$ and is constant along a streamline. Consequently, the static pressure is given by $p_s = p_t - \frac{1}{2} \rho V^2$.&lt;br&gt;&#xA;Now imagine a low speed windtunnel. In a reservoir the speed is so low that it can be assumed to be $V=0$ m/s. Hence here the static pressure equals the total pressure, $p_t = p_s$. Now the fluid is accelerated along a streamline, and hence the static pressure should drop according to the relation given above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this is where my problem is. I read that the altimeter just measures the static pressure at flight level to obtain the pressure altitude. According to the explanation given above, however, this can not be, as the static pressure in the flow will be lower than the static pressure at the same altitude at zero velocity.&lt;/p&gt;&#xA;" OwnerUserId="2075" LastActivityDate="2011-02-18T18:20:48.190" Title="How does an altimeter work?" Tags="&lt;aerodynamics&gt;" AnswerCount="3" />
  <row Id="5417" PostTypeId="2" ParentId="317" CreationDate="2011-02-18T11:31:08.427" Score="0" ViewCount="" Body="&lt;p&gt;The indeterminism does not originate from Quantum Mechanics. It has a wider philosophical origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider the multi-world interpretation of quantum mechanics. It is a completely deterministic theory which describes unitary, reversible and predictable evolution of a quantum system or the Universe (Multiverse in terms of MWI) as a whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But any actual experiments still will show uncertainty. Why? According the MWI with each act of measurement the observer gets splitted in two copies each of which experience different results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can thus formulate a similar problem but without involving the quantum mechanics: what would one experience if somebody creates an exact copy of him? Will he still experience the old body or the newer one? What happens if the old body is killed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can be formulated several related thought experiments:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;There is a teleportation device that scans your body, sends that information to the receiver which re-creates an exact copy of your body and then the original body is destroyed. Would a reasonable person use such teleporter even if their friends used it and say it's great?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Suppose the medicine of the future became very advanced. Now you are proposed a game: your brain will be splitted in two parts with one of them left in your body and the other transplanted into another body and then the both fully regenerated. The memories of the both parts are completely (or mostly) restored. Now one of the resulting people is given billion of dollars while the other is sent to a life imprisonment. Should you agree to such a game? What is the probability that you will find yourself as a billionaire or as a prisoner after the operation? Should you agree if someone cuts not a half of the brain but a smaller part? What about other parts of the body?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This leads to the yet unresolved philosophical questions which exist from the very ancient times when people knew nothing about quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a list of open philosophical problems that arise in the course of the thought experiment:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hard problem of consciousness (philosophical zombies)&lt;/li&gt;&#xA;&lt;li&gt;Problem of induction&lt;/li&gt;&#xA;&lt;li&gt;Qualia problem&lt;/li&gt;&#xA;&lt;li&gt;Ship of Theseus paradox&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1186" LastEditorUserId="1186" LastEditorDisplayName="" LastEditDate="2011-02-19T03:16:52.683" LastActivityDate="2011-02-19T03:16:52.683" CommentCount="9" />
  <row Id="5418" PostTypeId="2" ParentId="5416" CreationDate="2011-02-18T11:33:06.237" Score="1" ViewCount="" Body="&lt;p&gt;Aircraft altimeter use the pitot-static system&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pitot-static_system&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Pitot-static_system&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In particular, the pitot tube measures the pressure of air ramming into the tube which, under normal circumstances, is equal to the total pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The static pressure is obtained through a static port. The difference is that it is &quot;capped&quot; so the air gets in from the sides. The air isn't rammed into these side holes (by any velocity) so these holes still get balanced with the static pressure outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the Wikipedia text above.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-18T11:33:06.237" CommentCount="2" />
  <row Id="5419" PostTypeId="1" AcceptedAnswerId="5420" CreationDate="2011-02-18T12:32:51.630" Score="2" ViewCount="283" Body="&lt;p&gt;I understand water boils at different temperatures depending on altitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am seeking to get an illustrative explanation for this, including a diagram if possible.&lt;/p&gt;&#xA;" OwnerUserId="1884" LastActivityDate="2011-02-18T14:59:58.943" Title="Explanation for different boiling points of water on different altitudes" Tags="&lt;water&gt;&lt;pressure&gt;" AnswerCount="2" />
  <row Id="5420" PostTypeId="2" ParentId="5419" CreationDate="2011-02-18T13:20:12.630" Score="4" ViewCount="" Body="&lt;p&gt;Boiling point of water changes with altitude because atmospheric pressure changes with altitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how/why boiling point changes with pressure.There is good explanation of this at &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/vappre.html&quot; rel=&quot;nofollow&quot;&gt;hyperphysics (with diagrams)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, why does pressure change with altitude? Imagine you are swimming in water. Deeper you go more pressure you feel, because there is more water above you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P = P_0 + dh$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P$ - pressure, $P_0$- pressure at the surface, $d$ - density of fluid, $h$- height/depth to free surface&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is huge column of air above our heads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rightarrow h$ at sea level &gt; $h$ at hill station. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rightarrow P$ at sea level &gt; $P$ at hill station.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rightarrow$ boiling point at seal level &gt; boiling point at hill station.&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2011-02-18T13:20:12.630" />
  <row Id="5421" PostTypeId="2" ParentId="5265" CreationDate="2011-02-18T14:16:25.883" Score="1" ViewCount="" Body="&lt;p&gt;There's also a significant boundry layer on the liquid side, similar to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sea_surface_microlayer&quot; rel=&quot;nofollow&quot;&gt;Sea surface microlayer&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The organization of the surface of water is an impediment to the diffusion of gases. For example, an unstirred (~5 ml) oximeter cell will take about 10 minutes to equilibrate with atmospheric oxygen, while a stirred cell will equilibrate in under a minute. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface foam, or a monolayer of fatty molecules, such as might be formed by coffee oils or creamer, likely has insulative value. Stirring would minimize that by disrupting the structure.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-18T14:16:25.883" />
  <row Id="5424" PostTypeId="2" ParentId="5416" CreationDate="2011-02-18T14:52:59.037" Score="0" ViewCount="" Body="&lt;p&gt;The location of the static ports on the fuselage are fixed in positions where that impact is minimal (if any).  As an interesting side note, switching to the alternate static (located inside the cockpit of an unpressurized airplane, and used if the outside static ports become obstructed) results in a small increase in indicated altitude as the pressure inside the cockpit is lower than outside.&lt;/p&gt;&#xA;" OwnerUserId="919" LastActivityDate="2011-02-18T14:52:59.037" CommentCount="3" />
  <row Id="5425" PostTypeId="2" ParentId="5419" CreationDate="2011-02-18T14:59:58.943" Score="0" ViewCount="" Body="&lt;p&gt;The boiling point of an element or a substance is the temperature at which the vapor pressure of the liquid equals the environmental pressure surrounding the liquid.&lt;br&gt;&#xA;Atmospheric pressure is due to air above any given point. At greater heights, there is less air above you and hence pressure decreases.&lt;br&gt;&#xA;Lesser pressure hence lower boiling points as you go higher. &lt;/p&gt;&#xA;" OwnerUserId="1847" LastActivityDate="2011-02-18T14:59:58.943" />
  <row Id="5426" PostTypeId="2" ParentId="5265" CreationDate="2011-02-18T15:00:37.913" Score="1" ViewCount="" Body="&lt;p&gt;Move the cup.  One of the main ways to cool something down is to ensure hot particles that have risen off the surface of the drink don't fall back on it again.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-18T15:00:37.913" />
  <row Id="5427" PostTypeId="2" ParentId="5118" CreationDate="2011-02-18T15:09:25.927" Score="1" ViewCount="" Body="&lt;p&gt;Because there are different extensions you can use to give mass to the neutrinos. You can put mass only in left neutrinos, or you can add right neutrinos, and it is even unclear how many species to add. Of course, a GUT-like neutrino, as in SO(10) etc, seems preferable, but it is not the only option.&lt;/p&gt;&#xA;" OwnerUserId="1335" LastActivityDate="2011-02-18T15:09:25.927" />
  <row Id="5428" PostTypeId="2" ParentId="5265" CreationDate="2011-02-18T15:53:10.990" Score="1" ViewCount="" Body="&lt;p&gt;Stir the coffee with the spoon and blow air into the cup with your mouth. Every few seconds change the direction of the stir (clockwise to counterclockwise and vice-versa). This should cool it quickly than just stirring the coffee with the spoon.&lt;/p&gt;&#xA;" OwnerUserId="2084" LastActivityDate="2011-02-18T15:53:10.990" />
  <row Id="5429" PostTypeId="2" ParentId="317" CreationDate="2011-02-18T16:32:31.063" Score="-1" ViewCount="" Body="&lt;p&gt;is not true that this mechanism is mysterious; however most physicists don't have the time to ponder about these philosophical questions and they just prefer to leave the random nature of quantum mechanics as an 'axiom'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to understand how the random happens, first, let's make a Gedanken experiment were our physical bodies (including our brains) are classically described; they all have well-defined position and momenta, and hence any indeterminism in its evolution is completely of practical reasons, and not a matter of principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this hypothetical classical universe, teleportation star trek style is a completely legal operation; you can read all the physical microstate of any person, and write it to another place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in this classical universe, it is also possible to &lt;strong&gt;&lt;em&gt;copy&lt;/em&gt;&lt;/strong&gt; a person microstate: so let's walk on the consequences of such experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So our experimental configuration consists of two separate rooms with big posters on the wall: one of the rooms contains a '+' and the other room contains a '-'. Now we send our test individual into the teleportation chamber, and we will desintegrate this person and create two copies of him into each room&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question arises; if you are the test individual, what is that you are going to experience? well, the truth is that there are not that many options for our experiences after we enter the teleportation chamber:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) we don't experience nothing afterwards, because we have been desintegrated, so we are dead, our goblin soul went away and in the world there are two zombie copies of you that don't have 'soul'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) we experience to appear in a room with a big poster with a '+'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) we experience to appear in a room with a big poster with a '-' &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we discard 1) (i don't wanna argue about religion with anyone here, i'll just say that 1) is preposterous) we are left with two options: 2) and 3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the important thing here, is that, even in this classical, determinist universe, the fact of being able to copy a conscious being, means that some observers/conscious beings, will experience events that are fundamentally random, and are intrinsically non deterministic, even if everything else is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could argue that there are reasons of why the 'real you' went to 2) instead of 3) or viceversa, you could say that the copy in 2) was more 'perfect' than 3) and hence the real you goes there. But the truth is that these arguments are not fundamental; you are basically trying to take to the heart the fact that there is a single 'you'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which makes me go back to your question; How does all this applies to our world?, after all, copying a living being is disallowed (there is even a no-clone theorem on QM) However this is not entirely true. Many-Worlds interpretation of QM is basically taking the determinism of QM to the extreme; if we allow a a quantum superposition to couple to a quantum conscious entity (an observer) the quantum observer will undergo a split and become entangled with the physical system he is measuring; he has physically become two separate copies quantum observers (mutually non-interacting copies, hence the no-clone theorem doesn't apply), experiencing different outcomes, which individually seem random to each one of them.&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-02-18T16:37:54.757" LastActivityDate="2011-02-18T16:37:54.757" CommentCount="10" />
  <row Id="5430" PostTypeId="2" ParentId="5410" CreationDate="2011-02-18T17:07:22.520" Score="0" ViewCount="" Body="&lt;p&gt;The induced EMF from a change in magnetic field strength (note it is a vector quantity so a change in direction also changes things), is the time rate of change in the net magnetic flux going through a closed circuit. Flux is area (normal vector) dotted with the field. So the area covered by a current loop determines the sensitivty to field changes. In the power grid on earth, dimensions are on the order or hundreds to a few thousand kilometers, so a small change in the magnetic field times a large area can generate a high voltage. The ISS is many times smaller, so induced voltages/currents should be orders of magnitude smaller. The real problem for the ISS is the ionizing radiation of the particles, either from the solar wind, or trapped in the earths geomagnetic field. A secondary issue is that the strength of the solar wind interaction determines the temperature and hence the rate of the pressure/density decline with altitude. During periods of high activity the pressure of the earths atmosphere that objects in low earth orbit encounter is higher, so drag forces and orbital lifetimes are affected.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-02-18T17:07:22.520" />
  <row Id="5431" PostTypeId="2" ParentId="3062" CreationDate="2011-02-18T17:40:38.420" Score="1" ViewCount="" Body="&lt;p&gt;In my view, the whole theoretical experiment of the tethered galaxy is intrinsically flawed, because the presence of a tether, rod, or any other implement attached to the two galaxies introduces such an inhomogeneity in the environment that it invalidates the basic assumptions on which the FLRW metric is based.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lest this statement seems unwarranted, I will provide some further explanation.  To note, the inhomogeneity problem lies in the very existence of the rope or rod fastening the galaxies, and not in the masses of the galaxies or the tether.  And to see why this implement flaws the whole experiment, we have first to realize that keeping galaxy B at a fixed proper distance to galaxy A (by any means) results in a symmetrical situation, in which it can also be rightly said that galaxy A is kept at a fixed proper distance to galaxy B. And since neither of the galaxies is &quot;nailed&quot; to the comoving background, the result is that both galaxies are no longer comoving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since the galaxies are no longer comoving, the times ticked by their clocks are no longer comoving time.  Let's assume that both galaxies are of the same mass, so that they are both moving against their surrounding comoving background at the same relative velocity.  Therefore clocks at each galaxy are ticking proper time t' where the relationship between t' and comoving time t is given by simple SR time dilation based on the relative velocity.  On the other hand the middle point of the rope/rod is comoving, and therefore a clock attached to that point is ticking comoving time t.  So we have a &lt;strong&gt;rigid&lt;/strong&gt; object (the rope/rod with the two galaxies attached at its ends) where clocks at different parts thereof tick different proper time!  This logical inconsistency is the necessary consequence of the sheer existence of the rope/rod itself, which violates the basic assumptions on which the FLRW metric is built.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, instead of dealing with a &quot;tethered galaxy&quot; it is much more insightful to study the case of a &quot;stationary rocket&quot; flying towards the origin of coordinates, with the crew continually fine-tuning the impulse in order to compensate exactly for the expansion of space (which they have calculated) so as to keep constant the proper distance of the rocket to the origin, as observed from the origin. At t = tr the crew stop applying impulse so that the rocket starts moving freely along a radial timelike geodesic. This scenario not only does not invalidate the assumptions of the FLRW metric but also is perfectly physical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For t &amp;lt; tr, the proper velocity of the rocket as observed from the local comoving frame of reference (denoted as vsr) is equal to the recession velocity of said FR as observed from the origin, which is the constant proper distance to the origin Dr times the Hubble parameter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;vsr(t) = - Dr H(t)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogue to the tension in the rope is the acceleration that the rocket needs to have with respect to its local comoving FR in order to keep Dr constant, which is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dvsr(t)/dt = - Dr dH(t)/dt     (&quot;needed&quot; acceleration)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If as usual dH(t)/dt &amp;lt; 0, the needed acceleration is positive (away from the origin).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in order to know whether the crew must apply impulse, and in what direction, that &quot;needed&quot; acceleration must be compared to the &quot;free&quot; acceleration that the rocket would have if left free in a timelike geodesic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dvs(t)/dt = - H(t) vs(t) [1 - vs(t)^2/c^2]    (&quot;free&quot; acceleration)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If vs(t) &amp;lt; 0 as in this case, the free acceleration is also positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing both accelerations for different models:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;in the empty linearly expanding model the needed acceleration is greater than the free acceleration, so the rocket must apply impulse away from the origin (i.e. throwing their exhaust towards the origin).  (BTW, the proverbial &quot;rope&quot; would be useless, and what it would be needed would be a rod to keep the galaxy away.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;in the lambda-only exponentially expanding model the free acceleration is greater than the needed (which is 0 as H is constant), so the rocket must apply impulse towards the origin.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For t &gt; tr (engines off), the proper velocity of the rocket starts decreasing according to the equation of the timelike geodesic of the FLRW model in use.  The evolution of the distance of the rocket to the origin depends on the FLRW model:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In the empty linearly expanding &quot;Milne&quot; case, the recession velocity of the rocket's initial comoving background Dr / t decreases faster, so that the rocket starts getting closer to the origin. This in turn starts decreasing Dr(t) and therefore causes the recession velocity Dr(t) / t of the rocket's comoving background to decrease even faster, so that the rocket will keep getting closer to the origin until it reaches and passes&#xA;it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the exponentially expanding &quot;de Sitter&quot; case, the recession velocity of the rocket's initial comoving background Dr H is constant, so that the rocket starts getting away from the origin. This in turn starts increasing Dr(t) and therefore causes the recession velocity Dr(t) H of the rocket's comoving background to increase, so that the rocket will keep getting away from the origin.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2085" LastEditorUserId="2085" LastEditorDisplayName="" LastEditDate="2011-02-19T14:58:25.700" LastActivityDate="2011-02-19T14:58:25.700" CommentCount="2" />
  <row Id="5432" PostTypeId="1" CreationDate="2011-02-18T18:00:27.187" Score="0" ViewCount="94" Body="&lt;p&gt;I am interested in finding out whether the term 'Internal Energy' can be used in both of the following cases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In the case of a macroscopic body upon which light is incident, the internal energy would designate the total energy absorbed by the body, causing it to heat up (in the absence of other heating mechanisms)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the case of a low-density system of atoms upon which light is incident and absorbed, the internal energy would designate the total amount of energy absorbed by the atoms and therefore would relate to the sum of the excitation and kinetic energies of the atoms.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Are both uses of the term 'Internal Energy' correct, or would it only apply to one of the cases? In both cases it would refer to the total energy absorbed by the system from radiation, but how the energy is distributed in the system (kinetic vs excitation) would depend on the type of system.&lt;/p&gt;&#xA;" OwnerUserId="2086" LastActivityDate="2011-02-18T19:54:49.873" Title="Internal Energy Terminology" Tags="&lt;energy&gt;" AnswerCount="2" />
  <row Id="5433" PostTypeId="2" ParentId="5416" CreationDate="2011-02-18T18:08:55.590" Score="4" ViewCount="" Body="&lt;p&gt;The static air pressure seen by the aircraft does not change with the aircraft's velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your confusion is from a common misinterpretation of Bernoulli's principle.   It is not true that a fluid's pressure will decrease simply by virtue of flowing faster.   After all,  this violates the idea that physics should be the same in all inertial frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a simple counterexample to the typical interpretation of the Bernoulli principle. Consider a tube of infinite length and uniform diameter with some gas sitting in it.  Now consider various coordinate systems with a velocity in the direction of the tube.  In these different coordinate systems, the velocity of the gas will be different, but we expect the force on the walls of the tube due to the fluid's pressure to be the same in all cases.   (The tube is not going to rupture simply because of a choice of coordinate system!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, Bernoulli's principle says that, &lt;em&gt;in a given flow&lt;/em&gt; (say, along a streamline), a local increase in velocity is associated with local decrease in pressure.  The canonical example is fluid flow through a tube with a constriction (a &lt;a href=&quot;http://en.wikipedia.org/wiki/Venturi_effect&quot; rel=&quot;nofollow&quot;&gt;venturi&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoting from the Wikipedia article for &lt;a href=&quot;http://en.wikipedia.org/wiki/Bernoulli%27s_principle&quot; rel=&quot;nofollow&quot;&gt;Bernoulli's principle&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bernoulli's principle can be derived from the principle of conservation of energy. This states that, in a steady flow, the sum of all forms of mechanical energy in a fluid along a streamline is the same at all points on that streamline. This requires that the sum of kinetic energy and potential energy remain constant. ... If a fluid is flowing horizontally and along a section of a streamline, where the speed increases it can only be because the fluid on that section has moved from a region of higher pressure to a region of lower pressure; and if its speed decreases, it can only be because it has moved from a region of lower pressure to a region of higher pressure. Consequently, within a fluid flowing horizontally, the highest speed occurs where the pressure is lowest, and the lowest speed occurs where the pressure is highest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note the emphasis on &lt;strong&gt;relative changes&lt;/strong&gt; occurring &lt;strong&gt;on a streamline&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific flaw in your argument is here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now the fluid is accelerated along a streamline, and hence the static pressure should drop according to the relation given above.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the wind tunnel, something has to do work to accelerate the air to the wind tunnel velocity, adding energy to the flow, which violates the conservation of energy assumption in Bernoulli's principle.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-02-18T18:18:45.050" LastActivityDate="2011-02-18T18:18:45.050" />
  <row Id="5435" PostTypeId="1" AcceptedAnswerId="5489" CreationDate="2011-02-18T18:24:02.940" Score="3" ViewCount="157" Body="&lt;p&gt;Been studying hopping conduction and something that everyone is taking for granted is bothering me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we have a bunch of sites that are either unoccupied, singly occupied, or doubly occupied. Due to on-site Coulomb repulsion the two electron levels are separated by U energy at a doubly occupied site. Now everyone is saying that the two electrons on the double site are in the spin singlet state due to, I assume, Pauli exclusion. However the two electrons are not in the same energy level - they are separated by U so why is there a restriction on their spins?&lt;/p&gt;&#xA;" OwnerUserId="1939" LastActivityDate="2011-02-19T17:53:48.447" Title="On-site repulsion and Pauli exclusion" Tags="&lt;condensed-matter&gt;&lt;spin&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="5437" PostTypeId="2" ParentId="5432" CreationDate="2011-02-18T18:27:08.033" Score="1" ViewCount="" Body="&lt;p&gt;Internal energy refers to all the energy you don't care about (or can't distinguish) macroscopically. It's totally unimportant in what form it is present as all forms of energy are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely, by Boltzmann statistics the probability of a microstate with energy $E$ is $p \sim \exp(-E / k_B T)$ law. $E$ here refers to total energy of the microstate which of course consists of both kinetic and interaction terms. The internal energy of a macrostate is then a thermal average of energies of individual microstates.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-02-18T18:27:08.033" />
  <row Id="5440" PostTypeId="1" AcceptedAnswerId="5674" CreationDate="2011-02-18T18:49:06.497" Score="5" ViewCount="246" Body="&lt;p&gt;Power spectra, coherence spectra, and linear transfer functions are ubiquitous tools of experimental physics. However, our instruments often retain small nonlinear effects which can contaminate measurements.  It appears that higher order spectra, in particular the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bispectrum&quot; rel=&quot;nofollow&quot;&gt;bispectrum&lt;/a&gt;, would be ideal tools to investigate nonlinear interactions.  Nonetheless, I've never actually seen them put to use in experimental physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider the (frequency-domain) coherence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{xy}(f) = \frac{\langle X(f)Y(f)^*\rangle}{\sqrt{\langle X(f)X(f)^*\rangle\langle Y(f)Y(f)^*\rangle}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bicoherence considers not two but three signals, and looks for correlations between oscillations at frequencies $f_1$ and $f_2$ combining nonlinearly to produce a signal at $f_1+f_2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{xyz}(f_1,f_2) = \frac{\langle X(f_1) Y(f_2) Z^*(f_1 + f_2)\rangle}{ \sqrt{\langle X(f_1)X^*(f_1)\rangle\langle Y(f_2)Y^*(f_2)\rangle\langle Z(f_1+f_2)Z^*(f_1+f_2)\rangle} } $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...which seems like a useful thing to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are the bispectrum and bicoherence not used more frequently in experimental physics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am specifically thinking about time domain, multi-input/multi-output systems where one is looking for nonlinear couplings between various signals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the top Google hits on the subject is for the Matlab &lt;a href=&quot;http://www.mathworks.com/matlabcentral/fileexchange/3013&quot; rel=&quot;nofollow&quot;&gt;Higher Order Spectral Analysis (HOSA) Toolbox&lt;/a&gt;, which seems like a nice resource (though it appears to be no longer maintained and now suffering from bit-rot).&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-02-22T01:24:52.677" LastActivityDate="2011-02-22T19:45:29.723" Title="Why is the bispectrum not commonly used in experimental physics?" Tags="&lt;experimental-physics&gt;&lt;fourier-transform&gt;&lt;signal-processing&gt;" AnswerCount="3" />
  <row Id="5445" PostTypeId="2" ParentId="5440" CreationDate="2011-02-18T18:55:43.873" Score="3" ViewCount="" Body="&lt;p&gt;It's used a lot in cosmology. Often, to a decent approximation, the quantities we try to measure in cosmology (e.g., CMB temperature and polarization maps, galaxy distributions) are realizations of Gaussian random processes to a decent approximation, but have (or are predicted to have) interesting non-Gaussian features at some low level. People estimate the bispectra of these things all the time to get a handle on various non-Gaussian effects.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-18T18:55:43.873" CommentCount="2" />
  <row Id="5450" PostTypeId="1" AcceptedAnswerId="5451" CreationDate="2011-02-18T19:21:37.613" Score="3" ViewCount="119" Body="&lt;p&gt;I am not someone specialized on physics, I am just curious on why our Body Electrical Resistance measure as shown by a multimeter varies so much.&#xA;When allocating the 2 probes each on one hand, the resistance varies from low 100k's of Ohms up to 700k Ohms and even more.&#xA;Can someone illustrate in a easy way for a physics dummy this phenomena?&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2082" LastActivityDate="2011-08-03T07:31:02.110" Title="Body's Electrical Resistance" Tags="&lt;general-physics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5451" PostTypeId="2" ParentId="5450" CreationDate="2011-02-18T19:47:52.423" Score="7" ViewCount="" Body="&lt;p&gt;The bodies internal electrical resistance is quite low. Bodily fluids have enough ions (dissolved salts mainly) for the conductance to be high. The thin layer of skin provides almost all of the resistance. Once the skin resistance is overcome (by wetting, or for higher voltages arcing), the low internal resistance dominates. So for instance, how good a contact is made, how much surface area, and whether conductive jelly is used makes a large difference. Lie detectors use skin resistance, which largely measures sweating, as one among several measures to try to ascertain the emotional response of the subject.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-02-18T19:47:52.423" CommentCount="1" />
  <row Id="5452" PostTypeId="2" ParentId="5432" CreationDate="2011-02-18T19:54:49.873" Score="0" ViewCount="" Body="&lt;p&gt;At least in mechanics, internal energy covers all forms of energy except bulk kinetic (mass time velocity squared over two (including rotational)). It could be elastic energy, thermal energy, gravitational potential energy, and/or chemical energy. The individual energies of random atomic motion, are usually defined to be part of the thermal energy, which is part of internal, rather than kinetic energy, which applies only to bulk motion.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-02-18T19:54:49.873" />
  <row Id="5453" PostTypeId="1" CreationDate="2011-02-18T20:42:39.853" Score="-6" ViewCount="355" Body="&lt;p&gt;Modern theories of interactions in particle physics are gauge ones. I know how the gauge fields are introduced in equations ($D = \partial + A$). I just do not see any physical motivation in it. I am afraid it is done by analogy with QED and that's it. I conclude that it is not the only possible way of description of interactions. But maybe there is something essential that I am missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; As the main failure of gauge way of &quot;introducing&quot; interactions I can point out its intrinsically perturbative character. One cannot switch off &lt;strong&gt;permanent&lt;/strong&gt; interaction without severe problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dbrane answered to that in comments &quot;... you can treat gauge field theories non-perturbatively and still gather valuable insights ...&quot;  Let us see. First question: With what does $A$ interact - with bare or real electron?&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-30T18:32:56.273" LastActivityDate="2011-08-30T18:32:56.273" ClosedDate="2011-02-20T08:44:28.600" Title="What is physical in the principle of local gauge invariance?" Tags="&lt;particle-physics&gt;&lt;gauge-theory&gt;&lt;interactions&gt;" AnswerCount="3" CommentCount="27" FavoriteCount="1" />
  <row Id="5454" PostTypeId="2" ParentId="5453" CreationDate="2011-02-18T20:53:43.260" Score="1" ViewCount="" Body="&lt;p&gt;Strange, I thought that a perfect example and motivation for gauge fields is the electromagnetic field (the classical one). Isn't that a very good physical motivation!&lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2011-02-18T20:53:43.260" CommentCount="5" />
  <row Id="5455" PostTypeId="2" ParentId="5453" CreationDate="2011-02-18T20:57:24.133" Score="-1" ViewCount="" Body="&lt;p&gt;Of course it was done by analogy with QED. If QED works that way why shouldn't other theories too? Naturally it's not this simple and it took time to make sense of other gauge theories but nevertheless it turns out that it works which is actually the only thing that matters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there is certainly no reason why it should be so. Also, by the very definition of gauge invariance of redundant degrees of freedom, you are free to fix the gauge and forget about gauge theories altogether if you are able to write down your Lagrangian in some other way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in my opinion, there are only two reasons why one would introduce gauge invariance:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;metaphysical one: it's a beautiful theory and the requirement of locality is definitely appealing as well (in the same way people liked moving from instant-action theories to local theories)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;practical: it simplifies things &lt;em&gt;a lot&lt;/em&gt;; both conceptually and computationally. Instead of looking for interaction given by arbitrary terms, you can just specify your field content and the group it's transforming under and you're done. The rest is routine (haha) work of verifying whether theory works.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-02-18T20:57:24.133" CommentCount="18" />
  <row Id="5456" PostTypeId="1" CreationDate="2011-02-18T22:01:44.560" Score="6" ViewCount="311" Body="&lt;p&gt;Sorry for the layman question, but it's not my field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose this thought experiment is performed. Light takes 8 minutes to go from the surface of the Sun to Earth. Imagine the Sun is suddenly removed. Clearly, for the remaining 8 minutes, we won't see any difference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am wondering about the gravitational effect of the Sun. If the propagation of the gravitational force travels with the speed of light, for 8 minutes the Earth will continue to follow an orbit around nothing. If however, gravity is due to a distortion of spacetime, this distortion will cease to exist as soon as the mass is removed, thus the Earth will leave through the orbit tangent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the state of the art of research for this thought experiment? I am pretty sure this is knowledge that can be inferred from observation.&lt;/p&gt;&#xA;" OwnerUserId="217" LastActivityDate="2011-02-20T11:24:46.630" Title="The speed of gravity" Tags="&lt;gravity&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="5457" PostTypeId="2" ParentId="5456" CreationDate="2011-02-18T22:11:49.250" Score="9" ViewCount="" Body="&lt;p&gt;Since general relativity is a local theory just like any good classical field theory, the Earth will respond to the local curvature which can change only once the information about the disappearance of the Sun has been communicated to the Earth's position (through the propagation of gravitational waves).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, the Earth would continue to orbit what should've been the position of the Sun for 8 minutes before flying off tangentially. But I should add that such a disappearance of mass is unphysical anyway since you can't have mass-energy just poofing away or even disappearing and instantaneously appearing somewhere else. (In the second case, mass-energy would be conserved only in the frame of reference in which the disappearance and appearance are simultaneous - this is all a consequence of GR being a classical field theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more realistic situation would be some mass configuration shifting its shape non-spherically in which case the orbits of satellites would be perturbed but only once there has been enough time for gravitational waves to reach the satellite.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-02-18T22:11:49.250" />
  <row Id="5458" PostTypeId="2" ParentId="5456" CreationDate="2011-02-18T22:12:38.357" Score="1" ViewCount="" Body="&lt;p&gt;the fact that distortion travels 'as soon' as a mass is removed or not is not implied in any way by gravity being due to a distortion of spacetime. In fact distortions of spacetime are as limited to travel to the speed of light as any other physical influence.&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-02-18T22:12:38.357" />
  <row Id="5459" PostTypeId="2" ParentId="5456" CreationDate="2011-02-18T22:14:38.313" Score="7" ViewCount="" Body="&lt;p&gt;Gravitational influences do propagate at the speed of light, not instantaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question of what would happen if the Sun instantly disappeared is actually a funny one in general relativity. The equations of general relativity imply as a mathematical consequence that energy must be locally conserved. Therefore, there is no valid solution to the equations that describes the Sun suddenly disappearing (since that scenario violates local energy conservation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A similar statement holds in electromagnetism, by the way: charge conservation is a logical consequence of Maxwell's equations, so if someone asks you what the electric field does when a charge suddenly disappears, there is no correct answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you can sensibly ask what would happen if the Sun suddenly changed its mass distribution -- if it exploded, say, sending its mass in different directions at high speeds. The answer is that the Earth's orbit wouldn't change for 8 minutes.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-02-19T17:24:01.380" LastActivityDate="2011-02-19T17:24:01.380" CommentCount="12" />
  <row Id="5460" PostTypeId="2" ParentId="4289" CreationDate="2011-02-18T22:39:33.077" Score="0" ViewCount="" Body="&lt;p&gt;The entropy of a gravitating particle in Verlinde's theory CANNOT be constant. Consider a screen placed at a distance $r$ from a mass $M$. It has an entropy $S_{screen}(r)$, and this is the maximal entropy the given region surrounded by the screen can have. Now, if you place a particle of mass m at distance $r+\delta r$, the entropy of the screen becomes $S_{screen}(r+\delta r)$. This later entropy is the entropy of the screen placed at $r+\delta r$, where the particle represented by certain microstates. Integrating out this microstates (coarse graining the screen) gives back $S_{sreen}(r)$. Since, the entropy is an additive quantity, the entropy on the screen with a test particle is $S_{screen- without-neutron}(r+\delta r)+S_{neutron}(r+\delta r) = S_{screen}(r)+S_{neutron}(r+\delta r)$. Thus, $S_{neutron}(r+\delta r)=\delta S_{screen}$, that is,  entropy of a test particle $m$ depends on the distance from $M$. Is not that trivial?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second point -- Time evolution of the system seems to be unitary, since the energy eigenvalues obtained are real. There are systems in QM which have such properties, see e.g. &lt;a href=&quot;http://arxiv.org/PS_cache/physics/pdf/9712/9712001v3.pdf&quot; rel=&quot;nofollow&quot;&gt;enter link description here&lt;/a&gt;   &lt;/p&gt;&#xA;" LastActivityDate="2011-02-19T02:49:11.897" />
  <row Id="5461" PostTypeId="2" ParentId="3534" CreationDate="2011-02-18T22:50:41.370" Score="0" ViewCount="" Body="&lt;p&gt;A simple explanation is that it takes more FORCE to ACCELERATE an object of greater MASS.&#xA;A=F/M....or with a constant FORCE, the ACCELERATION is inversely proportional to the MASS. The is a result of inertia. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A larger mass dropped (on a massive objects) requires a GREATER force to accelerate; additionally a GREATER mass exerts a GREATER gravitaional force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting newtons second law to the gravitional force law cancles mass out and therefore renders MASS not related to the ACCELERATION of two gravitationally related objects.&lt;/p&gt;&#xA;" OwnerUserId="2097" LastActivityDate="2011-02-18T22:50:41.370" CommunityOwnedDate="2011-02-18T22:50:41.370" CommentCount="6" />
  <row Id="5462" PostTypeId="1" AcceptedAnswerId="5463" CreationDate="2011-02-18T23:15:08.883" Score="3" ViewCount="156" Body="&lt;p&gt;As a photon leaves a strong gravitational field, it loses energy and redshifts. Is the exchange in potential energy of a photon characterized by energy quanta?&lt;/p&gt;&#xA;" OwnerUserId="2097" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-02-19T08:52:17.590" LastActivityDate="2011-02-19T08:54:11.220" Title="Does the potential energy for a given photon increase or decrease in quanta?" Tags="&lt;gravity&gt;&lt;photons&gt;&lt;energy-conservation&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5463" PostTypeId="2" ParentId="5462" CreationDate="2011-02-18T23:26:18.870" Score="4" ViewCount="" Body="&lt;p&gt;No. A photon of a given frequency $f$ &lt;em&gt;is&lt;/em&gt; exactly one quantum of energy. An electromagnetic wave has a total energy given by $E_\text{total} = \langle N\rangle hf$, where $\langle N\rangle$ is the number of photons in the wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When an EM wave escapes from a gravitational potential well (or falls into one), it's the energy of each individual photon that changes; in other words, the unit in which energy is quantized gets smaller or larger. The number of units (number of photons) stays the same.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-02-18T23:26:18.870" CommentCount="8" />
  <row Id="5464" PostTypeId="2" ParentId="5268" CreationDate="2011-02-18T23:37:18.893" Score="0" ViewCount="" Body="&lt;p&gt;This is an idea I had a while back about time operators, which might work.  I have tried to reconstruct it. Define the time operator &#xA;$$&#xD;&#xA;{\hat T}~=~i\hbar\sum_{j\ne k}{{|E_j\rangle\langle E_k|}\over{E_j~-~ E_k }}. &#xD;&#xA;$$&#xA;that acts upon a ket $|t\rangle~=~N^{1/2}\sum_nexp(iE_nt/\hbar)$ as&#xA;$$&#xD;&#xA;{\hat T}|t\rangle~=~i\hbar\sum_{j\ne k}{{|E_j\rangle\langle E_k|}\over{E_j~-~E_k }}|t\rangle~=~iN^{-1/2}\hbar\sum_{j\ne k}(E_j - E_k)^{-1}|E_j\rangle e^{-iE_kt/ħ}.  &#xD;&#xA;$$&#xA;This is a Fourier summation, which in the continuum limit gives $t|t\rangle$ by the Cauchy integral formula.  Now compute matrix elements of $[T,~H]$ for $|\psi\rangle~=~\sum_ja_j|E_j\rangle$&#xA;$$&#xD;&#xA;\sum_ja_j\langle E_i|[T,~H]|E_j\rangle~=~i\hbar\sum_{j,k,l}a_j\langle E_i|(E_k~-~E_l)^{-1}|E_k\rangle\langle E_l |E_j\rangle&#xD;&#xA;$$&#xA;$$&#xD;&#xA; =~i\hbar\sum_{j,k}a_j\langle E_i| (E_k~-~E_j)^{-1}|E_k\rangle,&#xD;&#xA;$$&#xA;where the matrix element $\langle E_i|(E_k~-~E_j)^{-1}|E_k\rangle~=~\delta_{ik}(E_k~-~E_j)^{-1}$.  $|E_i\rangle$ is not in the projective summation of the time operator for $[{\hat T},~H]~=~i\hbar$, and generally $[{\hat T},~ H]~=~0$.&#xA;\vskip.12in&#xA;A Cauchy sequence of states will converge to a bounded state $|\psi\rangle~=$ $\sum_{j=0}^Na_j(N)|E_j\rangle$, for $N$ the bound on the complete set.  For the coefficient $\sim~(1/j)$ as $N~\rightarrow~\infty$ the accumulation point contains a dense set of points with $E~+~\delta E$ energy eigenvalues which satisfy $\sum_j\langle E_j|\psi\rangle~=~0$.  This means that the commutator $[T,~H]~=~i\hbar$ holds on a measure zero set, and for the function $\psi(t)$ an almost periodic function.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-18T23:37:18.893" CommentCount="1" />
  <row Id="5465" PostTypeId="2" ParentId="2691" CreationDate="2011-02-19T00:32:59.260" Score="-1" ViewCount="" Body="&lt;p&gt;It can definitely generate energy and it's a cool idea!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic principle of a heat engine is that it moves heat between two heat baths. The requirement is that the two heat baths have different temperatures. This is called the Carnot cycle. See the wiki article:&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Carnot_cycle&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Carnot_cycle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual situation is that one heat bath is at ambient (room, say) temperature while the other is heated but it works even better the other way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum theoretical efficiency of a Carnot engine is given by:&lt;br&gt;&#xA;$1 - T_C/T_H$&lt;br&gt;&#xA;where $T_C$ and $T_H$ are the cold and hot temperatures, in degrees Kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you had a source of air at the temperature of boiling water. This could be from an industrial cooling process. That is 373K, a 73C increase from room temp which is around 300K, so the efficiency is about $1-300/373 = 20\%$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose you had a source of cold air at around 73C lower than room temp (i.e. about -46C or -51F). Even though you're still working with a 73C difference in heat bath temperatures the efficiency rises to $1-227/300 = 24\%$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;inefficient&quot; part of the cycle will be the movement of heat from hot to cold without the performance of work. But since that's what they wanted to do anyway, in a certain sense this is a rare example of a 100% efficient process (i.e. compared to how they used to do it).&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-19T00:32:59.260" />
  <row Id="5466" PostTypeId="1" CreationDate="2011-02-19T03:25:18.787" Score="3" ViewCount="249" Body="&lt;p&gt;Is all particles moving and forever?&lt;br&gt;&#xA;Can the movement of particles be stoped ?&lt;/p&gt;&#xA;" OwnerUserId="2101" LastEditorUserId="2101" LastEditorDisplayName="" LastEditDate="2011-02-19T04:16:50.363" LastActivityDate="2011-02-19T20:10:18.023" ClosedDate="2011-02-20T16:46:19.800" Title="Why do particles move?" Tags="&lt;quantum-mechanics&gt;&lt;particle-physics&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="5467" PostTypeId="2" ParentId="5466" CreationDate="2011-02-19T03:31:01.023" Score="1" ViewCount="" Body="&lt;p&gt;I don't know exactly what you are asking. Anyways, it is impossible for any particle to stop moving completely. If it stops the uncertainty principle of quantum mechanics will be violated since in that case both position and its conjugate momentum of the particle will be known with certainty. therefore all particles have to have minimum amount of movement in accordance with the uncertainty principle. For an harmonic oscillator this minimum movement contributes to the zero point energy.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-02-19T03:31:01.023" />
  <row Id="5468" PostTypeId="2" ParentId="5466" CreationDate="2011-02-19T04:28:46.060" Score="0" ViewCount="" Body="&lt;p&gt;Movement itself is a relative quantity. A particle at rest in one frame of reference is moving at a constant velocity in another frame of reference. Rest is never absolute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many ways, particles will never stop moving. Not only is the universe undergoing expansion,but as long as energy is fundamental to our universe, movement should not cease. In another light, without &quot;motion&quot;, the meaning of time would break down, as we would have no way of telling the past from the future. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically speaking, particles move so time can exist, or vice versa.&lt;/p&gt;&#xA;" OwnerUserId="2097" LastActivityDate="2011-02-19T04:28:46.060" />
  <row Id="5469" PostTypeId="2" ParentId="317" CreationDate="2011-02-19T06:46:07.513" Score="1" ViewCount="" Body="&lt;p&gt;My two lepta on this mainly conceptual and semantic problem: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that people have an initial position/desire: those who want/expect/believe that measurements should be predictable to the last decimal point and those who are pragmatic and accept that maybe they are not. The first want an explanation of why there exists unpredictability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An experimentalist knows that measurements are predictable within errors, which errors can sometimes be very large. Take wave mechanics, classical. Try to predict fronts in climate, a completely classical problem. The weather report is a daily reminder how large the uncertainties are in classical problems, in principle completely deterministic. Which leads to the theory of deterministic chaos. So predictability is a concept in the head of the questioner, as far as quantum or classical measurements goes. The difference is that in classical physics we believe we know why there is unpredictability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has physics given up on the predictability of the throw of a dice? Taken to extremes trying to find the physics of the metastable state of the fall of the dice we come again to errors and accuracy of measurement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within errors  in measurements in the order of magnitude we live in, nano to kilometers, quantum mechanics is very predictive, as evinced by all the marvelous ways we communicate through this board . Even in achieving lasing and superconductivity. It is only when probing the very small that the theoretical unpredictability of individual measurements in QM  enters. So small that &quot;intuitions&quot; and beliefs can become dominant to measurement and errors. And there, according to the inherent beliefs of each observer, the desire to have a classical predictability framework or the willingness to explore new concepts plays a role to a physicist, whether he/she will obsess about this conundrum or live with it until TOE..&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-02-19T07:23:34.293" LastActivityDate="2011-02-19T07:23:34.293" />
  <row Id="5470" PostTypeId="1" CreationDate="2011-02-19T07:14:49.140" Score="2" ViewCount="379" Body="&lt;p&gt;Feynman says in his book &quot;QED&quot; that the square root of the fine structure constant is the probability for a charged particle to emit a photon. But for which wavelength? Or is it an average over all wavelengths? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I meant &lt;em&gt;virtual&lt;/em&gt; photon, and I meant a stable charged particle, like the electron. One way to rephrase it would be: how many virtual photons (per unit volume) are there in the Coulomb field around an electron?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-21T19:54:30.923" Title="What does the fine structure constant describe?" Tags="&lt;quantum-field-theory&gt;&lt;quantum-mechanics&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="5471" PostTypeId="1" CreationDate="2011-02-19T07:24:05.887" Score="0" ViewCount="285" Body="&lt;p&gt;Did anyone ever heard about this?I've never seen any serious physicist talk about &quot;mass fluctuations&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the man in his own words &lt;a href=&quot;http://www.intalek.com/Index/Projects/Research/woodward1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.intalek.com/Index/Projects/Research/woodward1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what about this guy:http://aetherwavetheory.blogspot.com/&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He claims his theory can explain virtually every unsolved problem in contemporary physics.&lt;/p&gt;&#xA;" OwnerUserId="2108" LastActivityDate="2011-02-19T10:45:35.807" Title="Is the Woodward effect real?" Tags="&lt;mechanics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5472" PostTypeId="2" ParentId="5470" CreationDate="2011-02-19T08:12:50.593" Score="3" ViewCount="" Body="&lt;p&gt;It's neither - or everything. Feynman means that the &quot;vertex in the Feynman diagram&quot; which has 2 external lines corresponding to a charged particle and 1 external line corresponding to a photon (all three particles have any energy/frequency and momentum/inverse_wavelength you want) is proportional to $e\approx \sqrt{\alpha}$. So he really means the probability amplitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One needs lots of (or at least several) other calculations to calculate the probability. In particular, a stable charged particle can never emit a photon because it would violate the energy or momentum conservation: in its initial rest frame, the energy is minimized, so one can't afford to increase the energy by changing the state of the motion (plus emitting a photon, which would make the final energy even higher). An unstable particle can decay into another particle and a photon. The decay rate will be proportional to the fine-structure constant (without the square root) but the detailed decay rate depends on the mass of the decaying particle as well as the other final decay product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, $\alpha$ slowly depends on a scale - logarithmically. It's the value at $E=0$, or - approximately - anywhere at masses $m&amp;lt;m_e$ where $m_e$ is the electron mass where $\alpha$ has the familiar value $1/137.03604$. Even though I didn't tell you a particular process that is linked to $\alpha$, it's actually a good idea in the process of &quot;renormalization&quot; to define $\alpha$ operationally exactly as a particular probability or amplitude at particular energies, just like you suggested. However, there are many choices how to do that.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-19T08:18:49.700" LastActivityDate="2011-02-19T08:18:49.700" CommentCount="1" />
  <row Id="5474" PostTypeId="2" ParentId="5466" CreationDate="2011-02-19T09:21:31.733" Score="1" ViewCount="" Body="&lt;p&gt;Orbital electron in hydrogen atom is moving forever. Orbital electrons are moving in atoms even at a very low temperature in &lt;a href=&quot;http://en.wikipedia.org/wiki/Bose%E2%80%93Einstein_condensate&quot; rel=&quot;nofollow&quot;&gt;Bose-Einstein condensate&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="238" LastActivityDate="2011-02-19T09:21:31.733" />
  <row Id="5475" PostTypeId="2" ParentId="5471" CreationDate="2011-02-19T10:45:35.807" Score="4" ViewCount="" Body="&lt;p&gt;No, this paper has nothing to do with serious physics. It has two steps&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mach's principle&lt;/li&gt;&#xA;&lt;li&gt;bizarre derivation of varying rest masses while accelerating, based on the previous point&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first point is less problematic - Mach's principle has been inspiring even for Albert Einstein when he was looking for general relativity. However, it turned out that Mach's principle is simply wrong and it hasn't become a part of modern physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, it's the second step that is much more problematic. Mach's principle couldn't be able to imply conclusions about varying rest masses even if this principle were right. The paper makes no sense and whenever the predictions - which don't really follow from the starting assumptions - were treated seriously and tested, they weren't confirmed experimentally. To make it more amusing, the discoverer proposed the hypothetical effect as motors for spaceships - the kind of context where it shouldn't be hard to see that this is probably not serious research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make thing even worse, the blog you mentioned is one by Zephir, a Czech &quot;independent researcher&quot; who is well-known to all physics bloggers in the world, and you don't really want to pay any attention to this stuff.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-19T10:45:35.807" CommentCount="2" />
  <row Id="5477" PostTypeId="2" ParentId="4015" CreationDate="2011-02-19T11:59:46.933" Score="0" ViewCount="" Body="&lt;p&gt;Interestingly, there is still a hypothetical possibility that the opposite might have been the case, i.e. non-empty first and empty then, &lt;strong&gt;pending&lt;/strong&gt; the results of the AEgIS experiment on gravitational behavior of antimatter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://aegis.web.cern.ch/aegis/home.html&quot; rel=&quot;nofollow&quot;&gt;http://aegis.web.cern.ch/aegis/home.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If&lt;/strong&gt; it turns out that matter and antimatter repel each other gravitationally, it will lend support to the theory by Chardin and Benoit-Lévy that the universe may be at large scales an emulsion of matter and antimatter that cancel each out gravitationally, so that the universe is &quot;empty&quot; at large scales (meaning the net gravitational density of large volumes is zero):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/0903.2446&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0903.2446&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this model, the universe &lt;strong&gt;might&lt;/strong&gt; have started with big enough energy density (radiation era) to have closed geometry and finite size and content, and through the matter-antimatter formation and separation process the net energy density (from a gravitational viewpoint, i.e. Rho in Friedmann equations) might have progressively vanished, leading to an open geometry still with finite size and matter content.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just a possibility to keep in mind until the jury is out on AEgIS. BTW, I perceive that some physicists are actually afraid that it finds out that matter and antimatter repel each other, while in contrast some very good wine would be uncorked in France.&lt;/p&gt;&#xA;" OwnerUserId="2085" LastEditorUserId="2085" LastEditorDisplayName="" LastEditDate="2011-02-19T12:47:16.243" LastActivityDate="2011-02-19T12:47:16.243" CommentCount="4" />
  <row Id="5478" PostTypeId="2" ParentId="5435" CreationDate="2011-02-19T12:06:35.613" Score="0" ViewCount="" Body="&lt;p&gt;Maybe I am misunderstanding your question, but let me try to explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity suppose that we have only one site with only one relevant orbital (a big energy gap to other orbitals). Furthermore electrons have internal degrees of freedom, which introduces the index $\sigma = \uparrow, \downarrow$. By Pauli exclusion principle there can only be one electron at each state, so the many-body Hilbert space is four-dimensional&#xA;$ \mathcal H = \{a|0\rangle + b|\uparrow\rangle + c|\downarrow\rangle + d|\uparrow\downarrow\rangle | a,b,c,d\in\mathbb C\}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you are confused because you have forgotten that there is not one, but two singly occupied states. Now suppose you have a Hamiltonian of the form&#xA;$H = \sum_{\sigma}\xi c^{\dagger}_{\sigma}c_{\sigma} + Un_{\uparrow}n_{\downarrow} + B\sum_{\sigma\sigma&amp;#39;}c^{\dagger}_{\sigma}\tau^z_{\sigma\sigma&amp;#39;}c_{\sigma&amp;#39;},$&#xA;where the second term is the &quot;on-site&quot; repulsion and the last term $\sum_{\sigma\sigma&amp;#39;}c^{\dagger}_{\sigma}\tau^z_{\sigma\sigma&amp;#39;}c_{\sigma&amp;#39;} = c^{\dagger}_{\uparrow}c_{\uparrow} - c^{\dagger}_{\downarrow}c_{\downarrow}$ is a Zeeman-term and represents a magnetic field in the $z$-direction. So the on-site repulsion gives a energy penalty $U$ for occupying the orbital with two electrons, while the magnetic field introduces a gap $B$ between single occupation with spin-up and single occupation with spin-down. It is not hard to find the eigenvectors and energies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_{|0\rangle} = 0$,&#xA;$E_{|\downarrow\rangle} = \xi - B$,&#xA;$E_{|\uparrow\rangle} = \xi + B$ and &#xA;$E_{|\uparrow\downarrow\rangle} = 2\xi + U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helped.&lt;/p&gt;&#xA;" OwnerUserId="1469" LastActivityDate="2011-02-19T12:06:35.613" />
  <row Id="5479" PostTypeId="2" ParentId="2604" CreationDate="2011-02-19T14:20:16.160" Score="1" ViewCount="" Body="&lt;p&gt;I've decided to elevate my comment on your question to an answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As a result of dmckee's comment (+1), I Googled &quot;cosmological effects of&#xA;  neutrino mixing&quot;. There are relevant results, though I'm not qualified to&#xA;  sift them for you. Less cosmologically, neutrino mixing might&#xA;  modify the star-forming effects of supernovae in gas clouds nearby.&#xA;  Google for &quot;supernova neutrino mixing&quot; seems to me much more interesting. It&#xA;  may even affect whether there are supernovae at all. If there were none&#xA;  of those, there would be no dogs or bunnies, though, who knows, there&#xA;  might still be squirrels. A more specific question about Supernovae and&#xA;  neutrino mixing might be good.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not to say that neutrino mixing &lt;em&gt;causes&lt;/em&gt; Supernovae, even if it were the case that Supernovae would not happen if there were no neutrino mixing. It is contingently the case that we observe Supernovae, and most models take Supernovae to be a principal source of heavy metals, particularly iron, and it is contingently the case that we observe neutrino mixing. There's bound to be someone on Physics SE who knows straight off whether neutrino mixing plays a significant role in current astrophysics models for Supernovae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of the day, however, this is just to say that effects that are very subtle at small scales &lt;em&gt;may&lt;/em&gt; have manifest consequences at large scales. In Dog World &amp;mdash;which typically doesn't care about butterflies, even if someone speculates that they might cause a hurricane &lt;em&gt;somewhere&lt;/em&gt;&amp;mdash;, if Emmy doesn't eat anything for 2 hours, she might not notice, but if Emmy doesn't eat for three days, &lt;em&gt;everybody&lt;/em&gt; would notice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do want to change a &lt;em&gt;detail&lt;/em&gt; of my comment &amp;mdash; if there might be a metaphysical category of things that behave like &amp;ldquo;squirrels&amp;rdquo;, even without iron in the world, because, counterfactually, we modified the Universe so that there is no metaphysical category of things that behave like &amp;ldquo;neutrino quantum fields that mix&amp;rdquo;, surely there would also be a metaphysical category of things that behave like &amp;ldquo;dogs&amp;rdquo;. I can't pull it off, but I'm envious of your dog trope.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-02-19T14:20:16.160" />
  <row Id="5480" PostTypeId="2" ParentId="4015" CreationDate="2011-02-19T14:51:51.117" Score="0" ViewCount="" Body="&lt;p&gt;If there existed at some point the same amount of matter and antimatter, should there not be a signature in the microwave background radiation? I would expect that the annihilation of electrons and positrons  should still be a separate bump?&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-02-19T14:51:51.117" CommentCount="1" />
  <row Id="5481" PostTypeId="2" ParentId="5453" CreationDate="2011-02-19T15:47:18.047" Score="1" ViewCount="" Body="&lt;p&gt;You sure are having battles here, Vladimir! I find myself, however, having moderate sympathy with this particular question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think your comment, &amp;ldquo;&lt;em&gt;I just do not see any physical motivation in it. I am afraid it is done by analogy with QED and that's it.&lt;/em&gt;&amp;rdquo; can be seen as a large part of its own answer. When we say that something is &quot;Physically Motivated&quot;, I take this to mean that a plausible argument can be given for using, in a new situation, a generalization of a mathematical model that has previously been used successfully as a description for Physical phenomena. The form of a &quot;plausible argument&quot; is not given a priori, it's just a question of what Physicists as a group find plausible. Plausibility has an acid test, which is whether a given Physicist thinks an idea for a new class of mathematical models has enough promise that they &lt;em&gt;spend their own time&lt;/em&gt; developing the mathematics and its relationships with experiment. All that said, QED is physically successful, and enough Physicists found it plausible to consider generalization to non-Abelian gauge fields that, over the course of 15 years, from the mid-1950s to 1972, say, with perhaps a few hundred people working on them, a new, moderately empirically successful class of mathematical models was constructed. The analogy with QED is significant, but it's the arguments for why someone in 1955 might think intensively about such models that I think you're not paying enough attention to. Those arguments are still known in the community, and they play out in various ways in the comments on your question and on your comments, but I think it's fair to say that they are not very &lt;em&gt;clearly&lt;/em&gt; elaborated. There is no &lt;em&gt;axiomatic&lt;/em&gt; QCD, that lays out both the mathematics and why it's especially natural as a Physical model, for a Physicist to point to, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Underlying all your questions, answers, and comments, however, is your railing, as I see it, against renormalization. 40 years ago, you would have been in company with very eminent Physicists, but today the game has largely  moved on. People do talk loosely about bare and real electrons as you do, but the principal modes of discussion are now in terms of the renormalization group and the surrounding mathematics, which is well enough constructed as mathematics that as far as I can tell most Physicists are content with it, and almost all Physicists are content to calculate with it. As far as I've seen, models for real, shielded particles are relatively ad-hoc, and in any case, and insofar as they are not ad-hoc, are ultimately so grounded in the Taylor series mathematics of Feynman diagrams and the renormalization group approach that they are essentially not a new approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All sorts of moves are being made, both in the system and on the periphery, to tighten up the mathematics more, or to construct new non-perturbative methods, but they will &lt;em&gt;remain&lt;/em&gt; peripheral for most Physicists until an approach is constructed that is significantly better as mathematics than the renormalization group approach, which seems more explanatory, and which is, additionally, more usable in an engineering sense than the renormalization group approach. I think it's significant that the renormalization group approach is more-or-less usable as &lt;em&gt;engineering&lt;/em&gt;, but I think it's clearly enough not easy &lt;em&gt;enough&lt;/em&gt; to use the renormalization group approach as engineering that a replacement for this way of thinking and constructing models will emerge in due course. I conclude, therefore, that I agree with you when you write &amp;ldquo;&lt;em&gt;I conclude that it is not the only possible way of description of interactions&lt;/em&gt;&amp;ldquo;, but the hard issue is how to construct something that is better, preferably &lt;em&gt;significantly&lt;/em&gt; better. Gotta do the math, and it's gotta be good math, but, harder, because it requires a certain kind of simplicity that we'll only recognize when we see it, it's gotta be good for use as engineering as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best Wishes, Vladimir!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peter.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-02-19T15:47:18.047" CommentCount="4" />
  <row Id="5482" PostTypeId="1" CreationDate="2011-02-19T16:15:05.967" Score="4" ViewCount="148" Body="&lt;p&gt;I would be glad if someone can explain to me the argument as to why supersymmetry breaking is necessarily accompanied by appearance of a massless fermion, namely the goldstino. (and also why this is a non-perturbative effect) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me quote here the two lines from the third volume of the QFT books by Weinberg where he tries to explain this phenomenon,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;&lt;em&gt;(when supersymmetry is broken)..any n-particle state is accompanied with 2 states of the same energy and momentum and opposite statistics, containing an additional 0-momentum goldstino of spin up or down and with another state of the same energy and momentum and the same statistics, containing two additional 0-momentum goldstinos of opposite spin.&lt;/em&gt;&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;&lt;em&gt;In particular when supersymmetry is spontaneously broken the vacuum state has non-zero energy, so it must be paired with a fermionic state of the same energy and zero-momentum; more precisely, the vacuum and the state containing two zero-momentum goldstinos are paired with the two states of a single zero-momentum goldstino.&lt;/em&gt;&quot;    &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am unable to  understand the above two arguments and will be glad if someone can help. &lt;/p&gt;&#xA;" OwnerUserId="1637" LastActivityDate="2011-02-19T17:27:04.270" Title="Supersymmetry breaking and Goldstino" Tags="&lt;quantum-field-theory&gt;&lt;supersymmetry&gt;" AnswerCount="1" />
  <row Id="5483" PostTypeId="1" CreationDate="2011-02-19T16:43:31.137" Score="4" ViewCount="545" Body="&lt;p&gt;This question was prompted by another question about a paper by Woodward(not mine). IMO Mach's principle&#xA;is very problematic (?wrong) thinking. Mach was obviously influenced by Leibniz. Empty space solutions in GR would result in a Minkowski metric and would suggest no inertia.&#xA;Mach's principle seems incompatible with GR. Gravitational waves could also be a problem.&#xA;I had thought that papers like one by &lt;a href=&quot;http://www.utdallas.edu/physics/faculty/wolfgang.html&quot; rel=&quot;nofollow&quot;&gt;Wolfgang Rindler&lt;/a&gt; had more or less marginalised the Mach Principle, but I see lots of internet discussion of it. Is it correct? Wrong? Is there evidence? (frame dragging experiments)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's use this definition from ScienceWorld.Wolfram.com&#xA;&quot;In his book The Science of Mechanics (1893), Ernst Mach put forth the idea that it did not make sense to speak of the acceleration of a mass relative to absolute space. Rather, one would do better to speak of acceleration relative to the distant stars. What this implies is that the inertia of a body here is influenced by matter far distant.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-02-20T16:09:44.793" LastActivityDate="2011-08-15T05:15:52.540" Title="Is Mach's Principle Wrong?" Tags="&lt;general-relativity&gt;" AnswerCount="7" CommentCount="13" FavoriteCount="2" />
  <row Id="5484" PostTypeId="2" ParentId="5483" CreationDate="2011-02-19T17:03:36.043" Score="0" ViewCount="" Body="&lt;p&gt;The honest answer should be no body knows for sure. Einstein was inspired initially by this principle when he was formulating GR. However it gradually became clear that there are solutions to GR field equations which does not confirm to Mach's principle. Einstein was thoroughly disillusioned in his latter years. As a matter of fact GR has some features which are definitely &quot;Machian&quot; but other features which are &quot;non Machian&quot;. Whether this principle is right or wrong has been always subject of controversy. The present opinion of most experts seem to be totally negative about the Mach principle.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-19T17:15:53.620" LastActivityDate="2011-02-19T17:15:53.620" CommentCount="8" />
  <row Id="5486" PostTypeId="2" ParentId="5466" CreationDate="2011-02-19T17:21:51.320" Score="0" ViewCount="" Body="&lt;p&gt;As long as fields exist, gravitational at any range of distance, and electromagnetic field at short range, the particles will keep moving. &#xA;But the particles themselves are the sources of the fields, and IMO, under modern theoretical grounds the particles 'are' fields. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as &lt;strong&gt;particles exists =&gt; fields exists =&gt; and interaction at a distance exists =&gt; they move&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interaction is a sort of communication between particles: 'come here' or 'get lost' are the messages. The other way of communication is thru exchange of energy packets: 'photons'. &#xA;The accepted theoretical particle for gravitational exchange is the graviton. But GR does not imply it, as space itself is enough to convey the message 'come here'.&#xA;We can only know the momentum or position of any particle thru a measurement, and this is only possible thru an interaction. Actually all measurements that I know of are thru photons exchange that carry away energy from one system/particle to another.&#xA;So, we see that particles are condemned to say, each one to all others, 'come here' or 'get lost'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I will try to see &lt;strong&gt;what can be a immobile world:&lt;/strong&gt; lets put the particles far way one from the others, to consider just the gravitational interaction, and so 'frozen' that we can say : &lt;strong&gt;the temperature is 0&lt;/strong&gt;, equivalent to say that the kinetic energy is zero. &lt;strong&gt;Absolutely homogeneous&lt;/strong&gt; is this world that each particle sense the same attracting force from each direction and would stay imobile forever in the same 'location'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this immobile world, with no definable time, &lt;strong&gt;stable forever? No&lt;/strong&gt;, because the slightest deviation of one particle from his position, due to any quantum fluctuation, will lead, with time, to a circus on all others.&lt;br&gt;&#xA;The Time will start now for the particles in the immediate vicinity of that bastard revolutionary particle, the one that moved away from his correct position, because they will start moving away from the hole left in the vacant position. (what happens to a stocking made of nylon when gets with a loose mesh? the hole widens)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Finally, the only possible conclusion is: particles are condemned to motion,  as long as they  exist.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-02-19T17:21:51.320" />
  <row Id="5487" PostTypeId="2" ParentId="5482" CreationDate="2011-02-19T17:27:04.270" Score="2" ViewCount="" Body="&lt;p&gt;The propositions by Weinberg above are just tautologies. When SUSY is broken it means that&#xA;$$ Q_\alpha|0\rangle \neq 0$$&#xA;the vacuum is not annihilated by the supercharges. It means that $Q_\alpha |0\rangle$ has to be equal to something. Because $Q_\alpha$ is Grassmann-odd, it must be a state with an odd number of fermions. But the momentum is still zero because the momentum of $|0\rangle$ vanishes and $Q_\alpha$ commutes with the energy-momentum vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for broken $N=1$ SUSY, one has four states, &#xA;$$|0\rangle, Q_1 |0\rangle, Q_2 |0\rangle, Q_1 Q_2 |0\rangle $$&#xA;which contain 0,1,1,2 fermions, respectively. This proof is somewhat heuristic because we haven't really proved that there also exist states with a nonzero momentum goldstino. To prove that, one has to define the supergenerators dressed with a momentum. The proof is then totally analogous to the Goldstone theorem for bosonic symmetries.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Goldstone_theorem#Goldstone.27s_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Goldstone_theorem#Goldstone.27s_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-19T17:27:04.270" CommentCount="1" />
  <row Id="5488" PostTypeId="2" ParentId="5483" CreationDate="2011-02-19T17:43:24.310" Score="7" ViewCount="" Body="&lt;p&gt;Mach's principle has influenced Einstein but the final formulation of general relativity as of 1916 clearly invalidates Mach's conjecture. According to Mach's principle, motion - including accelerating and rotating one - may only be defined relatively to other objects. That would imply that there can't exist any gravitational waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, general relativity predicts and experiments confirm that gravitational waves do exist: the relevant observations were awarded by the 1993 physics Nobel prize, too. The waves are vibrations of the space itself. It means that the metric tensor remembers the information about the geometry - and curvature at each point, even in the empty space, something that Mach's principle specifically wanted to prohibit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, the perceptions and other effects of acceleration were supposed to be determined by comparisons with distant objects. This simple fact itself violates locality that has become important already in special relativity, and was simply inherited by general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you care about history, the new cold relationship is mutual: much like general relativity rejected Mach's principle, Mach rejected general relativity - and already special relativity, in fact. ;-) If you care about sociology, there's been a poll among physicists active in relativity, and a vast majority of them would also say that Mach's principle is invalidated by general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people sometimes say that some effects predicted by general relativity, such as frame-dragging, are &quot;Machian&quot; in character. I think it is very misleading because it tries to make the listeners think that Mach's principle may be made compatible with the observations. It's very questionable what Mach's principle would predict about frame-dragging because Mach's principle has never become any viable candidate for a physical theory. But the idea that frame-dragging is Machian is more ideology and hype than a valid observation. Despite the vagueness about such very detailed effects, Mach's principle has said enough for us to be sure that it's incorrect in all of its forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, there's a lot of discussion on the Internet about long-dead ideas in physics - and maybe &lt;em&gt;mostly&lt;/em&gt; about them. However, the Internet has nothing to do with the current state of physics.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-19T17:48:36.553" LastActivityDate="2011-02-19T17:48:36.553" CommentCount="6" />
  <row Id="5489" PostTypeId="2" ParentId="5435" CreationDate="2011-02-19T17:53:48.447" Score="3" ViewCount="" Body="&lt;p&gt;4tnemele's answer is great, but I thought I'd try to give a simpler explanation of the main point of confusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say (as in 4tnemele's answer) that there's only one relevant orbital (or &quot;energy level&quot;) at each site. That means there's two states a single electron can occupy: spin up and spin down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, let's say there's no magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say &quot;Due to on-site Coulomb repulsion the two electron levels are separated by U energy at a doubly occupied site.&quot; This is wrong, because there is only one &quot;level&quot; (orbital), not two. The Coulomb energy penalty U only applies if there are two electrons in that level, one spin up and one spin down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a doubly occupied site, the overall energy is increased by U, but that energy doesn't belong to one electron or the other. It comes from the interaction between the electrons. At a doubly occupied site, the two electrons still have exactly the same spatial wavefunction. The many-body wavefunction is spatially symmetric. This is why the spins have to be opposite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the basic problem is that you shouldn't say &quot;the two electron levels are separated by U&quot;. Instead you should say &quot;the state with two electrons at the same site is increased in energy by U relative to what you'd otherwise expect (twice the single-electron energy)&quot;.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-02-19T17:53:48.447" CommentCount="4" />
  <row Id="5490" PostTypeId="1" CreationDate="2011-02-19T17:59:49.483" Score="5" ViewCount="107" Body="&lt;p&gt;For 4 real SUSY generators, the superspace formalism has been worked out a long time ago. For 8 real SUSY generators, some brilliant theoreticians have worked out the details of harmonic superspace. The major surprise is we have to enlarge superspace by taking the product with $S^2$. When we go down to the original superspace without the $S^2$ factor, we end up with infinitely many harmonics, almost all of which are auxiliary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it worthwhile to construct a manifestly supersymmetric superspace formalism for 16 or 32 real SUSY generators? The number of auxiliary fields will be enormous compared to the number of dynamical fields, but will this effort pay off?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hybrid formalisms like 4D superspace formalism in higher dimensions are unsatisfactory as they don't treat all bosonic dimensions on an equal footing.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-19T18:25:02.747" Title="Would it be worthwhile to work out a manifestly supersymmetric superspace formalism for 16 and 32 real SUSY generators?" Tags="&lt;mathematical-physics&gt;&lt;supersymmetry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="5491" PostTypeId="2" ParentId="5490" CreationDate="2011-02-19T18:19:58.240" Score="2" ViewCount="" Body="&lt;p&gt;Win, it's been tried. For ordinary supersymmetry in flat space with 16 or even 32 supercharges, the superspace description, even if it could work, would surely not be worth the effort. Note that for 32 supercharges, the full multiplet has 4 billion components (4,294,967,296 if you want me to be accurate), much like the number of IPv4 addresses. They managed to run out after 40 years but no one wants to do calculations for 40 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why the super-extended superspaces are not useful is that the superspace formulation of theories with 4 or 8 supercharges is helpful because it&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;makes supersymmetry manifest but also&lt;/li&gt;&#xA;&lt;li&gt;allows us to formulate a large number of physically distinct theories.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you managed to define a superspace for 16 or 32 supercharges, the first condition would still be satisfied. However, the second one would not be. The massless theories with 16 or 32 supercharges are the maximally supersymmetric Yang-Mills theory and maximally supersymmetric supergravity, respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supersymmetry in those two cases is so constraining that only the gauge group may be adjusted in the 16-supercharge Yang-Mills case; and nothing at all can be adjusted in the 32-supercharge case of supergravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's very different from the case of four supercharges, e.g. in $N=1$, where superspace allows you to define a general superpotential and Kähler potential (functions of an arbitrary number of fields - chiral and vector multiplets - you may start with), or $N=2$ where you have a rather adjustable prepotential etc. There's no counterpart of these choices if you have 16 or more supercharges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's not useful although it could be a mathematical curiosity. One would have to decompose the theory in the components, anyway. And it would always be the same theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Counterexamples could involve theories outside flat space - where a high degree of supersymmetry could lead to some surprises. Harmonic superspaces involve bosonic coordinates as well and may produce nontrivial theories, e.g. gauged supergravity. And exceptions could exist for massive multiplets but in those highly supersymmetric theories, massive states are constrained, too. In particular, $N=4$, $d=4$ gauge theory is conformal so it has no preferred mass scale. And massive states in maximal supergravity are inevitably black hole (or black brane) microstates of a sort, so field theory that treats them as excitations is not a good description, anyway.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-19T18:25:02.747" LastActivityDate="2011-02-19T18:25:02.747" />
  <row Id="5492" PostTypeId="1" CreationDate="2011-02-19T18:42:30.097" Score="2" ViewCount="103" Body="&lt;p&gt;At this website:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://heasarc.gsfc.nasa.gov/docs/swift/analysis/threads/uvot_thread_afterglows.html&quot; rel=&quot;nofollow&quot;&gt;http://heasarc.gsfc.nasa.gov/docs/swift/analysis/threads/uvot_thread_afterglows.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The passage at the bottom states that a V-band magnitude of 17.62, with an error $\pm$0.02 is a 49.4-$\sigma$ detection significance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this value calculated? Could you provide the working? I have a similar problem, with a set of magnitudes and errors, and would like to know at what magnitude limit I can claim a statistically significant (6-sigma) detection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-02-19T22:26:25.770" LastActivityDate="2011-02-19T22:26:25.770" Title="Astronomical detection significance from magnitude error" Tags="&lt;experimental-physics&gt;&lt;astronomy&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5493" PostTypeId="1" CreationDate="2011-02-19T19:10:02.867" Score="1" ViewCount="198" Body="&lt;p&gt;Are strange metals described by a quantum critical theory?&lt;/p&gt;&#xA;" LastActivityDate="2011-03-21T23:33:55.387" Title="Are strange metals described by a quantum critical theory?" Tags="&lt;condensed-matter&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="5494" PostTypeId="2" ParentId="5470" CreationDate="2011-02-19T19:19:44.597" Score="0" ViewCount="" Body="&lt;p&gt;The fine structure constant $\alpha$ describes the fine structure of the Hydrogen levels of energy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;I meant a stable charged particle, like the electron.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is the electron really &quot;stable&quot; if &lt;strong&gt;any&lt;/strong&gt; push makes radiation (inelastic &quot;deformation&quot;)? I am afraid it is not. Such radiation occurs always ( probability = 1). The &quot;Coulomb center&quot; picture is obtained by summation of all inelastic pictures. It is called an &quot;inclusive&quot; picture. It is not an &quot;elastic&quot; (non-destructive) one.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-20T00:30:31.717" LastActivityDate="2011-02-20T00:30:31.717" />
  <row Id="5495" PostTypeId="1" CreationDate="2011-02-19T19:39:52.343" Score="3" ViewCount="392" Body="&lt;p&gt;What does it take to become a top physicist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do so many extremely talented young upstarts totally flop as they move to more advanced physics?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-20T05:43:48.640" CommunityOwnedDate="2011-02-19T23:34:40.807" Title="What does it take to become a top physicist?" Tags="&lt;soft-question&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" />
  <row Id="5496" PostTypeId="2" ParentId="5466" CreationDate="2011-02-19T20:10:18.023" Score="-1" ViewCount="" Body="&lt;p&gt;One atom does not move more rapidly than another. In fact, if one only looks at the continued movement of an atom which takes place in an indivisible moment of time, the briefest possible, they all have a movement equally rapid. At the same time, an atom has not, in any moment perceptible to the intelligence, a continued movement in the same direction; but rather a series of oscillating movements from which there results, in the last analysis, a continued movement perceptible to the senses&lt;/p&gt;&#xA;" OwnerUserId="1335" LastActivityDate="2011-02-19T20:10:18.023" />
  <row Id="5497" PostTypeId="2" ParentId="5495" CreationDate="2011-02-19T20:11:26.317" Score="0" ViewCount="" Body="&lt;p&gt;Do not Despair :) .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What does it take to become a top physicist?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start with one has to be a physicist. A physicist is one who studies physics because of a burning curiosity about how the material world works and tries to satisfy it by going to graduate school and accumulating knowledge .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a &quot;middle of the drawer&quot; physicist has in addition to be either very good in math if aiming in theory, or be very good in experiments if aiming to be an experimentalist, and be able to use the accumulated knowledge in research at the frontier of the unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Top physicist is a bit of luck and a bit of attention and dedication to science politics and university politics etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Why do so many extremely talented young upstarts totally flop as they move to more advanced physics?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define &quot;extremely talented&quot; . If they burn to learn physics and they are good in maths why would they flop?  Unless it is the &quot;soft generation syndrome&quot;: lack of direction and persistence, giving up at the first hurdle.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-02-20T05:43:48.640" LastActivityDate="2011-02-20T05:43:48.640" CommunityOwnedDate="2011-02-19T23:34:40.807" CommentCount="2" />
  <row Id="5498" PostTypeId="2" ParentId="5492" CreationDate="2011-02-19T20:32:35.257" Score="3" ViewCount="" Body="&lt;p&gt;I believe those are two different measures. The detection significance is the likelihood that a signal is real; the error is a statement about how precise your measurement is. It's the old &lt;a href=&quot;http://en.wikipedia.org/wiki/Accuracy_and_precision&quot; rel=&quot;nofollow&quot;&gt;accuracy-versus-precision&lt;/a&gt; distinction. The detection significance is measured by comparing how strong your signal is to the background noise of the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the detection significance, it is calculated as $\frac {N_{p} - N_{b}}{\sigma_{b}}$ where &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$N_{p}$ is the number of raw counts at the peak of the star's point spread function, &lt;/li&gt;&#xA;&lt;li&gt;$N_{b}$ is the average number of raw counts in the background near the star, and   &lt;/li&gt;&#xA;&lt;li&gt;$\sigma_{b}$ is the standard deviation in the raw counts of the background near the star.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You'll need the actual raw image to do this calculation.&lt;/p&gt;&#xA;" OwnerUserId="1655" LastActivityDate="2011-02-19T20:32:35.257" CommentCount="2" />
  <row Id="5500" PostTypeId="2" ParentId="5495" CreationDate="2011-02-19T21:09:02.150" Score="0" ViewCount="" Body="&lt;p&gt;A great brain and hard work are required to be a &lt;em&gt;top&lt;/em&gt; physicist. To be an average or below physicist is possible through a deficiency in one and a surplus in the other (I would consider myself to be moderate intelligence but a very hard worker which makes up for it to some degree). As you can see, one of these traits is naturally obtained and the other can be acquired. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is the common perception amongst non-physicist laymen that all physicists are pretty much equally intelligent. Nothing could be further from the truth. We live on the upper end of the bell curve which is hardly constant. &lt;/p&gt;&#xA;" OwnerUserId="1939" LastActivityDate="2011-02-19T21:09:02.150" CommunityOwnedDate="2011-02-19T23:34:40.807" />
  <row Id="5501" PostTypeId="2" ParentId="5495" CreationDate="2011-02-19T21:11:04.983" Score="0" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.phys.uu.nl/~thooft/theorist.html&quot; rel=&quot;nofollow&quot;&gt;How to become a good theoretical physicist by Gerard 't Hooft (Nobel Laureate)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Hang out here :-) &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Always remember the reason why are you doing physics.&#xA;Here is my reason (and it also happens to be Richard Feynman's too :)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Physics is like sex. It may produce some practical results. But that is not why we do it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Becoming the greatest physicist of all the times (yeah... just like dear old Albert) is ambition. Finding out new physics and solving unsolved problems is challenge. But the point is doing physics. Nothing else matters. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Reason for despair of most of the people (including of course me) is that we think is that all the good ideas are already taken. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. When Max Plank went to a professor and said that he wanted to do physics. The professor said physics is finished, there is nothing more to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. After electron was discovered Max Born said physics as we know it will be finished within a week. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;c. Aristotle had invented theories for every other thing. But.. you know what happened when Galileo, Newton turned up. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;To be a top physicist you MUST read hell lot of science fiction along with the serious books.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=HKTSaezB4p8&quot; rel=&quot;nofollow&quot;&gt;And finally, a must watch video for despaired souls ;-)&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Don't take it very seriously. Don't take life seriously. Nobody gets out alive anyway :) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2011-02-19T21:30:23.633" LastActivityDate="2011-02-19T21:30:23.633" CommunityOwnedDate="2011-02-19T23:34:40.807" CommentCount="9" />
  <row Id="5502" PostTypeId="2" ParentId="5456" CreationDate="2011-02-19T21:30:17.033" Score="2" ViewCount="" Body="&lt;p&gt;All observations are consistent with standard GR so far, but I don't think the speed of gravity, in particular, has ever been measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experimental measurements of the speed of gravity was quite a controversy a few years ago when a paper came out claiming that the speed of gravity was very close to $c$ as measured by the Shapiro delay. To see papers on the subject google shapiro+speed+gravity:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.google.com/search?q=speed+of+gravity+site%3Aarxiv.org+shapiro&quot; rel=&quot;nofollow&quot;&gt;http://www.google.com/search?q=speed+of+gravity+site%3Aarxiv.org+shapiro&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clifford Will is an expert in the area and says that there was no measurement. He has a website on the subject that gives links to the various papers:&lt;br&gt;&#xA;&lt;a href=&quot;http://wugrav.wustl.edu/people/CMW/SpeedofGravity.html&quot; rel=&quot;nofollow&quot;&gt;http://wugrav.wustl.edu/people/CMW/SpeedofGravity.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that the Will side won. But academia means &quot;never having to admit you were wrong&quot;. Here's a pair of dueling papers on the subject, published in the same journal at the same time (that date to after Clifford Will last updated his page above):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Class.Quant.Grav. 22 (2005) 5181-5186, Sergei M. Kopeikin, &lt;em&gt;Comment on 'Model-dependence of Shapiro time delay and the &quot;speed of gravity/speed of light&quot; controversy'&lt;/em&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0510048&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/gr-qc/0510048&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Class.Quant.Grav.22 (2005) 5187-5190, S. Carlip, &lt;em&gt;Reply to &lt;code&gt;`Comment on Model-dependence of Shapiro time delay and the&lt;/code&gt;speed of gravity/speed of light' controversy''&lt;/em&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0510056&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/gr-qc/0510056&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-19T21:30:17.033" />
  <row Id="5503" PostTypeId="2" ParentId="5495" CreationDate="2011-02-19T21:45:38.060" Score="4" ViewCount="" Body="&lt;p&gt;What is a top physicist? Can you name a few? If you do, you will see they have little in common except 100% devotion. So that is necessary, but there are no sufficient criteria.&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-02-19T21:45:38.060" CommunityOwnedDate="2011-02-19T23:34:40.807" CommentCount="1" />
  <row Id="5504" PostTypeId="1" CreationDate="2011-02-19T22:13:31.423" Score="4" ViewCount="225" Body="&lt;p&gt;This year, the racetrack at Daytona has been repaved.  The track was always faster than other tracks NASCAR raced at and several cars in a &quot;train&quot; were faster than single cars or smaller trains.  This year is different, however, because two car groups are by far the faster than any other groupings.  When a third car joins a tandem, all three slow down noticeably.  Why is this?  Why wouldn't it still be faster to have more cars pushing?&lt;/p&gt;&#xA;" LastActivityDate="2011-08-19T04:27:04.000" Title="NASCAR drafting at Daytona" Tags="&lt;aerodynamics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5505" PostTypeId="1" AcceptedAnswerId="5685" CreationDate="2011-02-19T23:10:23.933" Score="4" ViewCount="167" Body="&lt;p&gt;Assume an observer sent a beam of photons close to an event horizon, say at some distance x (a distance far enough to avoid the photons falling in.) This light would still be observable, albeit red shifted and with it's path curved appropriately. Now assume the black hole absorbs enough mass to expand it's event horizon beyond the distance x. This stream of photons would stop. Does the observer not have information about a process that occurred exactly at the event horizon, that is it's expansion? Isn't this a region that one should not be able to get any information about due to the fact that nothing could contact the event horizon and return to the observer?&lt;/p&gt;&#xA;" OwnerUserId="1567" LastActivityDate="2011-02-28T12:30:02.457" Title="Information about an expanding event horizon" Tags="&lt;black-holes&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="5506" PostTypeId="2" ParentId="5493" CreationDate="2011-02-19T23:13:24.747" Score="1" ViewCount="" Body="&lt;p&gt;There has been a lot of work on quantum criticality and its CFT equivalence with AdS.  Take a look at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;String Theory, Quantum Phase Transitions and the Emergent Fermi-Liquid&#xA;Authors: Mihailo Cubrovic, Jan Zaanen, Koenraad Schalm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/0904.1993&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0904.1993&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another paper which is less conformal field related is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum Criticality in Heavy Fermion Metals&#xA;Authors: Philipp Gegenwart, Qimiao Si, Frank Steglich&#xA;(Submitted on 13 Dec 2007 (v1), last revised 31 Jan 2008 (this version, v2))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/0712.2045&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0712.2045&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-19T23:13:24.747" />
  <row Id="5508" PostTypeId="1" CreationDate="2011-02-20T01:18:41.383" Score="7" ViewCount="440" Body="&lt;p&gt;How do you effectively study physics? How does one read a physics book instead or just staring at it for hours? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Apologies in advance if the question is ill-posed or too subjective in its current form to meet the requirements of the FAQ; I'd certainly appreciate any suggestions for its modification if need be.)&lt;/p&gt;&#xA;" OwnerUserId="2120" LastActivityDate="2011-08-15T10:41:04.090" Title="How to learn physics effectively and efficiently" Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="5509" PostTypeId="2" ParentId="5508" CreationDate="2011-02-20T01:41:51.410" Score="6" ViewCount="" Body="&lt;p&gt;Read with a pad of paper beside you. When the author says &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;&quot;It follows that ...&quot;&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;&quot;Obviously ...&quot;&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;&quot;From equations (1.137) and (5.17) we can see that ...&quot;&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;don't go on&lt;/strong&gt; until you agree. Sometime it &lt;em&gt;really&lt;/em&gt; will be obvious, other times that may require many sheets of paper (Jackson was particularly bad that was for me).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This will &lt;strong&gt;&lt;em&gt;require&lt;/em&gt;&lt;/strong&gt; undisturbed time, and may be enhanced with some combination of enough sleep, sufficient nutrition, and a IV of the caffeinated beverage of your choice.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-02-20T01:41:51.410" CommentCount="4" />
  <row Id="5510" PostTypeId="2" ParentId="5265" CreationDate="2011-02-20T02:02:07.030" Score="43" ViewCount="" Body="&lt;p&gt;I did the experiment. (dipping wins)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;H2O ice bath  &lt;/li&gt;&#xA;&lt;li&gt;canning jar  &lt;/li&gt;&#xA;&lt;li&gt;thermometer &lt;/li&gt;&#xA;&lt;li&gt;pot of boiling water&lt;/li&gt;&#xA;&lt;li&gt;stop watch&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There were four trials, each lasting 10 minutes.  Boiling water was poured into the canning jar, and the spoon was taken from the ice bath and placed into the jar.  A temperature reading was taken once a minute.  After each trial the water was poured back into the boiling pot and the spoon was placed back into the ice bath.&#xA;&lt;img src=&quot;http://i.stack.imgur.com/zPIDz.jpg&quot; alt=&quot;equipment&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; Method:                  Final Temp.&#xA; 1. No Spoon              151 F &#xA; 2. Spoon in, no motion   149 F&#xA; 3. Spoon stirring        147 F&#xA; 4. Spoon dipping         143 F&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Temperature Readings have an error +-1 F, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/65MiU.png&quot; alt=&quot;chart&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; Red line, no Spoon&#xA; Green line, Spoon in, no motion&#xA; Aqua line, Stirring&#xA; Blue line, Dipping&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Data:&#xA;&#xA;No Spoon       &#xA;&#xA;Min Temp (F)&#xA;0   182&#xA;1   180&#xA;2   174&#xA;3   171&#xA;4   168&#xA;5   164&#xA;6   161&#xA;7   158&#xA;8   155&#xA;9   153&#xA;10  151&#xA;&#xA;Spoon, no stir, no dip       &#xA;&#xA;Min Temp (F)&#xA;0   180&#xA;1   175&#xA;2   172&#xA;3   168&#xA;4   165&#xA;5   162&#xA;6   160&#xA;7   156&#xA;8   153&#xA;9   151&#xA;10  149&#xA;&#xA;Stirring&#xA;&#xA;Min Temp (F)&#xA;0   180&#xA;1   175&#xA;2   171&#xA;3   167&#xA;4   164&#xA;5   161&#xA;6   158&#xA;7   155&#xA;8   152&#xA;9   150&#xA;10  147&#xA;&#xA;Dipping&#xA;&#xA;Min Temp (F)&#xA;0   180&#xA;1   177&#xA;2   173&#xA;3   168&#xA;4   164&#xA;5   160&#xA;6   156&#xA;7   152&#xA;8   149&#xA;9   146&#xA;10  143&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2121" LastEditorUserId="2121" LastEditorDisplayName="" LastEditDate="2011-02-22T05:48:15.577" LastActivityDate="2011-02-22T05:48:15.577" CommentCount="8" />
  <row Id="5511" PostTypeId="2" ParentId="5226" CreationDate="2011-02-20T02:06:45.353" Score="4" ViewCount="" Body="&lt;p&gt;I am a student so please point out in gory detail anything I did wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a process to be quasistatic, the time scales of evolving the system should be larger than the relaxation time. Relaxation time is the time needed for the system to return to equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have an adiabatic process, so equilibrium must be preserved at each point, that is to say&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Working within the validity of the kinetic theory for ideal gases and ignoring friction)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(A L(t))^\gamma P(t)  = (A L_0)^\gamma P(t_0)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Momentum gained by the piston: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta p =  2 m v_x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A molecule would impact the piston every &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta t = \frac{2(L_0+ \delta x) }{v_x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force exerted on the piston is $F =\frac{\Delta p}{\delta t} = \frac{m v_x^2}{L_0+\delta x}$  Pressure would be $P = \frac{P}{A}$ and for $N$ such molecules&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P = \frac{N m &amp;lt;v_x&amp;gt;^2}{A (L_0+\delta x)} = \frac{N m &amp;lt;v&amp;gt;^2}{3A (L_0+\delta x)}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at the instant $t=t&amp;#39;$ where the piston has been displaced by $\delta x$, we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(A L(t))^\gamma P(t) = \frac{N m &amp;lt;v&amp;gt;^2}{3A^{1-\gamma}} (L_0+\delta x)^{\gamma -1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expanding in series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ = \frac{N m &amp;lt;v&amp;gt;^2 L_0^{\gamma-1}}{3A^{1-\gamma}} (1 + \frac{(\gamma-1) \delta x}{L_0}+ O(\delta x^2) )  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituing $\frac{\delta x}{L_0} = \frac{\delta t v_x}{2 L_0} -1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(A L_0)^\gamma P(t_0) = (A L_0)^\gamma P(t_0) (1 + (\gamma-1) (\frac{\delta t v_x}{2 L_0} -1))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want our process to be reversibly adiabitic atleast to first order, we must have from above &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta t = \frac{2 L_0}{&amp;lt;v_x&amp;gt;}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this is time until collision for the starting case. Investigating second orders&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ (A L_0)^\gamma P(t_0) = (A L_0)^\gamma P(t_0) (1 + \frac{(\gamma-1) \delta x}{L_0}+ \frac{1}{2} (\gamma-1)(\gamma-2) (\frac{\delta x}{L_0})^2   +O(\delta x^3) )  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at just the series terms &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 1 + (\gamma-1)\frac{\delta x}{L_0} (1 +\frac{1}{2} (\gamma -2) \frac{\delta x}{L_0}) \approx 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be true for &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta t = \frac{4 L_0}{&amp;lt;v_x&amp;gt;} (\frac{1}{2-\gamma} -1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this is the &quot;time until next collision&quot; for a gas molecule hitting the piston. To maintain reversibility, at least to second order, the piston should be moved from $L_0$ to $L_0 + \delta x$ in time $\tau = \delta t$ so that the system variables follow the adiabatic curve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $&amp;lt;v_x&amp;gt;$ can be calculated from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Maxwell_speed_distribution&quot; rel=&quot;nofollow&quot;&gt;maxwell distribution&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1671" LastEditorUserId="1671" LastEditorDisplayName="" LastEditDate="2011-02-20T02:46:29.127" LastActivityDate="2011-02-20T02:46:29.127" CommentCount="7" />
  <row Id="5512" PostTypeId="2" ParentId="5508" CreationDate="2011-02-20T02:27:00.733" Score="3" ViewCount="" Body="&lt;p&gt;If you are staring into textbooks for hours then you're doing it wrong. A good way is to workout the examples and the problems, first from the text, and then from other sources. Unless you apply what you read to some good problems, imagine what is written and make some of your own examples, you will only have an illusion of understanding. &lt;/p&gt;&#xA;" OwnerUserId="1671" LastActivityDate="2011-02-20T02:27:00.733" CommentCount="1" />
  <row Id="5513" PostTypeId="2" ParentId="5483" CreationDate="2011-02-20T02:41:48.277" Score="2" ViewCount="" Body="&lt;p&gt;Mach's principle is simply a philosophical ancestor of the equivalence principle: &lt;em&gt;matter tells geometry how to curve, geometry tells matter how to move&lt;/em&gt;. So, yes, you can have Minkowski as a solution for the vacuum Einstein equations, but the minute you introduce even the tiniest mass your solution is no longer Minkowski. You might point out that asymptotically the spacetime will still be flat and Minkowski. However, in GR (and in QFT) it is not only the local geometry that matters. Point is that there is no interesting geometry which does not also contain some matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mach's principle has been interpreted in many different ways. I find the following &lt;a href=&quot;http://en.wikipedia.org/wiki/Mach%27s_principle&quot; rel=&quot;nofollow&quot;&gt;Wikipedia definition&lt;/a&gt; to be one I'm most comfortable with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A very general statement of Mach's principle is &quot;&lt;strong&gt;Local physical laws are determined by the large-scale structure of the universe&lt;/strong&gt;.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Depending on what day of the week it is and whatever interpretation is your favorite for that day, Mach's principle could be &quot;right&quot;, &quot;wrong&quot; or &quot;obsolete&quot;. But if stated in the simple manner above, then it is nothing more than a restatement of the equivalence principle and the question of its &quot;correctness&quot; is no longer an issue.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-20T02:41:48.277" CommentCount="6" />
  <row Id="5514" PostTypeId="2" ParentId="5504" CreationDate="2011-02-20T03:10:11.023" Score="0" ViewCount="" Body="&lt;p&gt;I'm waiting for someone who knows the fluid mechanics. It is true that much of the drag is accounted for by the low pressure area in the rear, which is caused by flow separation. What is happening immediately behind does affect this flow, so it is entirely possible that the lead car gets a benefit as well. If not, then the speed wouldn't be any higher than the lead car could do by itself. The following cars could still save on fuel, which might allow them to be more aggressive at another point in the event.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think all the participants in the race have acepted a high degree of risk as part of the cost of participation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In bicycling, the advantage is considered to be to the following riders only. A friendly group will rotate riders so that all can share in the drafting.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-02-20T03:10:11.023" CommentCount="1" />
  <row Id="5515" PostTypeId="2" ParentId="5505" CreationDate="2011-02-20T04:41:45.663" Score="3" ViewCount="" Body="&lt;p&gt;When you look at a black hole, you never see the actual event horizon, because the proper distance to it is infinite and it would take a light signal an infinitely long time to come from it. What you actually see is the matter falling in as the event horizon is &lt;strong&gt;about&lt;/strong&gt; to form. That's why an older name for black hole is &quot;frozen star&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you can't even observe the event horizon form, you definitely can't observe it expand. In your experiment, the closer the photons get to the event horizon the greater the proper length of their path, so the longer you have to wait to see them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you really should think of the event horizon as a 3-dimensional surface in spacetime, not just a 2-dimensional surface in space. The closer the worldline of the light gets to it, the longer you have to wait, so you can't get any information from the 3-surface itself.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-02-20T04:41:45.663" CommentCount="5" />
  <row Id="5516" PostTypeId="2" ParentId="5508" CreationDate="2011-02-20T04:59:00.830" Score="6" ViewCount="" Body="&lt;p&gt;I like the other answers, but here is some more advice:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Learning is an active process, in which you teach yourself. Nobody else can do that for you. So, when reading a textbook your job is to reconstruct it for yourself. Say it differently, make sense out of it in your own language. Ask yourself questions and try to answer them, and don't take anything on authority without making sure it sounds and feels right.  Only then you can be sure you really get it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lots of practice - pianists spend hours stretching and flexing their fingers, doing scales, and all other sort of really boring and not at all glamorous activities. Physics is no different, lots of finger exercises are part of the game, even if they seem trivial.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Also related to other answers - know yourself. Everyone is different, so part of the process is figuring out what works for you and what doesn't and make the necessary adjustments. Make sure you're having fun, at least most of the time, hard to sustain your interest otherwise.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-02-20T04:59:00.830" CommentCount="1" />
  <row Id="5517" PostTypeId="1" CreationDate="2011-02-20T09:06:22.940" Score="8" ViewCount="378" Body="&lt;p&gt;In this question, I would love to hear some independent opinions on an issue I asked Juan Maldacena, Nathan Berkovits, Dan Jafferis, and others, but all the physicists may be missing something. The question has 2 main parts:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is there a dual superconformal symmetry in 2+1-dimensional ABJM theories? In what limit does it apply? (For $N=4$ gauge theory, it's the planar limit, but what is the counterpart in 2+1 dimensions?)&lt;/li&gt;&#xA;&lt;li&gt;Are there spin network states in this extension of Chern-Simons theory? Can their expectation values be related to scattering amplitudes in the ABJM theory, much like the piecewise null Wilson loops are related to scattering amplitudes in $N=4$ gauge theory?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Concerning the first question, there's some confusion what $N$ should go to infinity and what should be kept fixed, especially when it comes to the scaling of the level $k$ in the limit. In string theory, the dual superconformal symmetry (and the Yangian) is very constraining - but that's because it still constrains a lot of the &quot;stringy stuff&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should one necessarily go to the limit with the stringy stuff - e.g. a high-level, a $CP^3$ compactification of type IIA - to see some dual superconformal symmetry? Or is there a dual superconformal symmetry that only works in the planar limit which simply means that only the (leading?) SUGRA amplitudes are constrained?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the second question, the ABJM theory secretly contains membranes, and their discretization - analogous to the piecewise null Wilson loops - could involve faces (pieces of membranes?) surrounded by open Wilson lines. Are they gauge-invariant?&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-20T12:43:32.803" Title="Dual conformal symmetry and spin networks in ABJM" Tags="&lt;quantum-field-theory&gt;&lt;string-theory&gt;&lt;spin-chains&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="5519" PostTypeId="1" CreationDate="2011-02-20T11:06:41.830" Score="3" ViewCount="182" Body="&lt;p&gt;Free field theories are definitely local in .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the interaction picture, we can decompose the fields into creation operator modes and annihilation operator modes. The product of operators can be regulated to finiteness by Wick ordering. In Wick ordering, all creation operators are moved to the left of all annihilation operators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To introduce interactions, we can add a Wick ordered interaction term $:V[\Phi(x)]:$ to the interaction Hamiltonian, but this term is nonlocal in space. To Wick order, we have to Fourier transform to momentum space, subtract some zero point terms, and then Fourier transform back to position space. The end result when expressed in terms of the original spatial fields is a nonlocal convolution with a spatially extended kernel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me clarify a bit. For potentials which are quadratic in the fields, the zero point subtraction is just a c-number, and this is trivially local. But for quartic couplings, for example, the zero point subtraction we get from Wick ordering is still quadratic in the fields with nonlocal point splitting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that mean interacting quantum field theory is nonlocal?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-20T18:01:54.583" Title="If : V(Phi) : is nonlocal in space, does that mean interacting quantum field theory is nonlocal?" Tags="&lt;quantum-field-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" />
  <row Id="5520" PostTypeId="2" ParentId="5456" CreationDate="2011-02-20T11:24:46.630" Score="2" ViewCount="" Body="&lt;p&gt;Your question was first asked by Laplace.   The following is from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Speed_of_gravity&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article on &quot;The speed of gravity&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;strong&gt;Laplace&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first attempt to combine a finite gravitational speed with Newton's theory was made by Laplace in 1805. Based on Newton's force law he considered a model in which the gravitational field is defined as a radiation field or fluid. Changes in the motion of the attracting body are transmitted by some sort of waves.[4] Therefore, the movements of the celestial bodies should be modified in the order v/c, where v is the relative speed between the bodies and c is the speed of gravity. The effect of a finite speed of gravity goes to zero as c goes to infinity, but not as 1/c2 as it does in modern theories. This led Laplace to conclude that the speed of gravitational interactions is at least 7×106 times the speed of light. This velocity was used by many in the 19th century to criticize any model based on a finite speed of gravity, like electrical or mechanical explanations of gravitation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a modern point of view, Laplace's analysis is incorrect. Not knowing about Lorentz invariance of static fields, Laplace assumed that when an object like the Earth is moving around the Sun, the attraction of the Earth would not be toward the instantaneous position of the Sun, but toward where the Sun had been if its position was retarded using the relative velocity (this retardation actually does happen with the optical position of the Sun, and is called annual solar aberration). Putting the Sun immobile at the origin, when the Earth is moving in an orbit of radius R with velocity v presuming that the gravitational influence moves with velocity c, moves the Sun's true position ahead of its optical position, by an amount equal to vR/c, which is the travel time of gravity from the sun to the Earth times the relative velocity of the sun and the Earth. The pull of gravity (if it behaved like a wave, such as light) would then be always displaced in the direction of the Earth's velocity, so that the Earth would always be pulled toward the optical position of the Sun, rather than its actual position. This would cause a pull ahead of the Earth, which would cause the orbit of the Earth to spiral outward. Such an outspiral would be suppressed by an amount v/c compared to the force which keeps the Earth in orbit; and since the Earth's orbit is observed to be stable, Laplace's c must be very large. In fact, as is now known, it may be considered to be infinite, since as a static influence, it is instantaneous at distance, when seen by observers at constant transverse velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a field equation consistent with special relativity (i.e., a Lortentz invariant equation), the attraction between static charges is always toward the instantaneous position of the charge (in this case, the &quot;gravitational charge&quot; of the Sun), not the time-retarded position of the Sun. When an object is moving at a steady speed, the effect on the orbit is order v2/c2, and the effect preserves energy and angular momentum, so that, and orbits do not decay. The attraction toward an object moving with a steady velocity is towards its instantaneous position with no delay, for both gravity and electric charge.&quot;&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2011-02-20T11:24:46.630" CommentCount="1" />
  <row Id="5521" PostTypeId="1" AcceptedAnswerId="5525" CreationDate="2011-02-20T12:41:34.567" Score="3" ViewCount="166" Body="&lt;p&gt;This question has its origin to the reference on the &lt;a href=&quot;http://cdsweb.cern.ch/record/1322447/files/SPSC-SR-079.pdf&quot; rel=&quot;nofollow&quot;&gt;Aegis experiment&lt;/a&gt; at CERN where they aim to produce super cooled antihydrogen and detect whether its reaction to gravity is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It set me thinking that the beams in the Tevatron circulate for more than a second and everything falls about 4.9 meters in a second, so the bunches must be falling too. This of course will be compensated by the fields  that keep the bunches in track among all the other corrections necessary. If though the antiprotons  have a different behavior under gravity, this difference would appear in the orbits of protons and antiprotons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question has two points:&#xA;a) since the beams are travelling equal and opposite paths through the magnetic circuit, a negative gravity effect on antiprotons would disperse the antiproton beam up with respect to the path of  the proton one. Could one get a limit on the magnitude of the gravitational effect difference between protons and antiprotons from this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found one reference where the &lt;a href=&quot;http://cdsweb.cern.ch/record/702738/files/ab-note-2003-009.pdf&quot; rel=&quot;nofollow&quot;&gt;antiproton   beam&lt;/a&gt; has a different behavior in chromaticity than the proton one, and it is explained away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I am completely vague about beam dynamics which I have filed under &quot;art&quot; rather than &quot;physics&quot; but &#xA;b) am wondering whether this observed difference could be interpreted as a gravitational field difference in a dedicated experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe there are beam engineers reading this list. &#xA;My feeling is that if antiparticles  had negative gravity interactions , beam engineers would have detected it since the first e+e- machine, but feelings can be wrong.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-02-20T13:56:27.267" Title="Is there a different behavior between proton and antiproton beams at the Tevatron?" Tags="&lt;experimental-physics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="5522" PostTypeId="2" ParentId="5517" CreationDate="2011-02-20T12:43:32.803" Score="3" ViewCount="" Body="&lt;p&gt;My wild idea would be to look for &lt;a href=&quot;http://arxiv.org/abs/1101.1329&quot; rel=&quot;nofollow&quot;&gt;holomorphic linking&lt;/a&gt; of Wilson loops in the ABJM theory, working from &lt;a href=&quot;http://arxiv.org/abs/1007.4772&quot; rel=&quot;nofollow&quot;&gt;its conjectured Grassmannian integral representation&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1486" LastActivityDate="2011-02-20T12:43:32.803" CommentCount="1" />
  <row Id="5523" PostTypeId="1" CreationDate="2011-02-20T12:46:55.623" Score="5" ViewCount="645" Body="&lt;p&gt;I first saw this on Top Gear, and assumed it was a load of rubbish, but today I tried it out and it actually works, if you walk away from the car continuously unlocking/locking it until it stops, and go a bit further away so that the car will not un/lock. Then out the car keys next to your head and point them directly at the car it will un/lock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does your head/brain/etc increase the range of the keys? And why does it only work next to your head if you point the keys directly at the car, when you do not have to point towards the car when in normal range?&lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2011-04-09T23:08:47.470" Title="Why do car keys have longer range when held next to your head?" Tags="&lt;general-physics&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="5524" PostTypeId="1" CreationDate="2011-02-20T13:04:38.487" Score="1" ViewCount="327" Body="&lt;p&gt;It is well accepted that quantum theory has well adapted itself to the requirements of special relativity. Quantum field theories are perfect examples of this peaceful coexistence. However I sometimes tend to feel little uneasy about some aspects. Consider an EPR pair of particles light years apart. Suppose there are 2 observers moving relative to each other with constant relative velocity. Let us consider, there are spin detection mechanism at both end for each particle. Now suppose one of the observer is at rest w.r.t. the detector for the first particle. As soon as the detection made, the wave function of the 2 particle entangled system will collapse instantaneously and the second particle must realize a definite opposite spin value. Now due to relativity of simultaneity, the second observer may claim that the collapse of the wave function for the two particle system is not simultaneous. He may even claim that the second particle is measured first. In that case a special frame of reference will be privileged, the frame at which the wave function collapsed instantaneously. This will cause a significant strain on the core principle of special relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sure the above reasoning is flawed. My question is where?&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-02-21T09:13:04.280" LastActivityDate="2011-02-21T14:31:52.433" Title="Wavefunction collapse in relativity" Tags="&lt;quantum-mechanics&gt;&lt;special-relativity&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="5525" PostTypeId="2" ParentId="5521" CreationDate="2011-02-20T13:24:17.413" Score="3" ViewCount="" Body="&lt;p&gt;The vertical position of the beams is maintained by quadra-pole (and possibly other high n-pole configurations, I'm not actually familiar with the details of the Tevatron) magnets in the ring as part of the general positioning and focusing mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the difference in coupling strength between gravity and magnetism, any difference due to different behavior in the beams would be too small to measure with the available equipment (or indeed anything you could install in the ring).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There &lt;em&gt;was&lt;/em&gt; a proposal to measure $g$ for anti-matter at Fermilab that's been floating around for some time, but the PAC has turned it down repeatedly and now made the decision to alter some of the hardware it would have relied upon. So, the idea is dead for the time being.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the reasoning went like this&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It ought to work, but no one has used all these principle in one experiment at one time before.&lt;/li&gt;&#xA;&lt;li&gt;It would be a moderately expensive, single measurement experiment&lt;/li&gt;&#xA;&lt;li&gt;It's a good idea, but it doesn't really fit in with our future plans. I mean, it not really &lt;em&gt;&quot;intensity frontier&quot;&lt;/em&gt; physics, is it?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-02-20T13:24:17.413" CommentCount="10" />
  <row Id="5526" PostTypeId="2" ParentId="27" CreationDate="2011-02-20T13:27:56.247" Score="1" ViewCount="" Body="&lt;p&gt;It seems that this question is regarding Copenhagen interpretation (and other related interpretations with collapse and one observer) solely, because it uses the term &quot;measurement&quot; which has no special position in other interpretations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So assuming this is about the Copenhagen interpretation, yes, the measurement is different from any other interaction. The difference is in that the quantum system interacts with the observer, a person who has special physical properties of having ability to trigger the wave function collapse. There is only one such person and the QM provides the theoretical possibility to unambiguously determine who is it based on his special abilities of interacting with matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This constitutes the main problem of Copenhagen interpretation, and the reason why other interpretations were proposed (Relational, MWI) which are observer-agnostic, do not include special, chosen personalities and symmetric across all people. This does not mean however that still there should not be a person apparently having special properties at least in the observable universe. &lt;/p&gt;&#xA;" OwnerUserId="1186" LastActivityDate="2011-02-20T13:27:56.247" CommentCount="1" />
  <row Id="5529" PostTypeId="2" ParentId="5521" CreationDate="2011-02-20T13:56:27.267" Score="0" ViewCount="" Body="&lt;p&gt;I am not good at all in all that but as soon as the particles in accelerators are charged and relativistic, the gravity force is always negligible with respect to other forces present. I think it is like trying to measure the gravity effect on a beam of light.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-02-20T13:56:27.267" CommentCount="2" />
  <row Id="5530" PostTypeId="2" ParentId="5483" CreationDate="2011-02-20T14:08:23.170" Score="1" ViewCount="" Body="&lt;p&gt;The Mach's principle in short is that inertia is not absolute, determined by the matter configuration, and there is no other source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be summed up in two statements:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Massive bodies can affect inertia of other objects (inertia is determined by the matter configuration)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Universe does not rotate and linearly accelerate as a whole (there is no other source of inertia)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The first statement is included in GR and has been proven with the experimental measurement of frame-dragging. In fact inertia even can be screened so people in a rotating (against distant stars) spaceship could not detect the rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second statement is still not proven (in the rotational part) although verified to a high precision. The linear part simply follows from the conservation of impulse.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastEditorUserId="1186" LastEditorDisplayName="" LastEditDate="2011-02-20T14:20:52.433" LastActivityDate="2011-02-20T14:20:52.433" />
  <row Id="5531" PostTypeId="2" ParentId="5524" CreationDate="2011-02-20T14:18:12.943" Score="3" ViewCount="" Body="&lt;p&gt;The reasoning is flawed for multiple reasons. One of them is that there is no physical wave-function collapse so in particular there can't be any problem with relativity. More precisely, entanglement is only about correlations between particles. That is, if you carry out experiments and later (after both observers who measured individual particle meet) they will notice that there was a correlation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this has to do with another flaw: locality. You are simply not allowed to say anything about stuff you don't measure. And that is only the stuff in your past light-cone. You just can't say anything about the other particle which is space-like separated from you (for all you know, it might no longer exist, having been absorbed by a BH).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He may even claim that the second particle is measured first.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sure, this is because there is no notion of causality between space-like separated events. This alone should be enough to convince you that wave-function collapse is completely unphysical. It is just a convenient tool of Copenhagen interpretation (which simply ignores the measurement problem) that has its limitations.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-02-20T14:18:12.943" CommentCount="4" />
  <row Id="5532" PostTypeId="2" ParentId="5524" CreationDate="2011-02-20T14:24:04.550" Score="0" ViewCount="" Body="&lt;p&gt;Let me rephrase your question with the simple, but wery illustrative example &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call the event of detection of the first spin as $A$ and the event of detection of the second spin as $B$. Suppose that $A$ and $B$ are timelike events. Then you can choose such a reference frame, where $t_A&amp;gt;t_B$. But also there exist a reference frame, where $t_B&amp;gt;t_A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion is inavoidable:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In the first reference frame collapse happened before the detection of first spin. &lt;/li&gt;&#xA;&lt;li&gt;In the second reference frame collapse happened before the detection of second spin.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In other words: The moment of collapse of the wave function depends on reference frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any contradiction? -- No. You cannot devise an experiment that &quot;detects&quot; this collapse. You cannot use this collapse to transmit a signal. e.t.c. If you feel like &quot;there is a condtradiction&quot; -- try to get down to the level of a concrete experiment that demostrates this contradiction. You will see that any attempt to formulate such experimet will fail.&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-02-20T14:24:04.550" />
  <row Id="5533" PostTypeId="1" CreationDate="2011-02-20T14:36:34.440" Score="1" ViewCount="103" Body="&lt;p&gt;I'm begining to study  Quantization of field with the second quantization formalism. I've studied phononic field, electromagnetic field in the vacuum and a generic relativistical scalar field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked to me if is possible doing the same thing with the Gravitational Field Hamiltonian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard that we can do it only in the condition of linearized Gravity and we obtain a field with spin 2, but we can't do it in the general condition because we don't have quantization conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are these conditions? And how can we obtain only in linearized gravity the quantized field? And how it has got spin 2? &lt;/p&gt;&#xA;" OwnerUserId="558" LastEditorUserId="558" LastEditorDisplayName="" LastEditDate="2011-02-20T14:47:02.167" LastActivityDate="2011-02-20T14:47:02.167" Title="Quantization of Gravitational Field: Quantization conditions" Tags="&lt;quantum-field-theory&gt;&lt;general-relativity&gt;&lt;gravity&gt;&lt;quantum-gravity&gt;&lt;second-quantization&gt;" CommentCount="4" />
  <row Id="5534" PostTypeId="1" AcceptedAnswerId="5547" CreationDate="2011-02-20T15:11:48.767" Score="1" ViewCount="149" Body="&lt;p&gt;If I wander outside in cold weather for 10-20 minutes my body accumulates charges. I get electric shock if I touch metal, e.g., door knob, car door etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I've two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why does human body accumulate charges in cold weather?&lt;/li&gt;&#xA;&lt;li&gt;What can I do to discharge my body without getting the shock?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks for your time.&lt;/p&gt;&#xA;" OwnerUserId="2125" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-20T16:22:39.383" LastActivityDate="2011-02-20T19:20:46.260" Title="Why does my body accumulate charge?" Tags="&lt;electricity&gt;&lt;theory&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="5535" PostTypeId="1" CreationDate="2011-02-20T15:38:30.033" Score="6" ViewCount="275" Body="&lt;p&gt;There are some interactions that are forbidden by conservation laws, e.g. an electron cannot turn into a positron by conservation of charge and a photon cannot turn into a positron electron pair by conservation of momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is if every interaction (between say up to 3 or 4 particles) that is consistent with all known conservation laws have been observed.&lt;/p&gt;&#xA;" OwnerUserId="37" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-15T16:12:19.593" LastActivityDate="2011-08-15T16:12:19.593" Title="Has every possible interaction between elementary particles been observed?" Tags="&lt;particle-physics&gt;&lt;experimental-evidence&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="5536" PostTypeId="2" ParentId="5535" CreationDate="2011-02-20T15:48:02.027" Score="8" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;proton decay is un-observed, and suspected on the basis of various Beyond the Standard Model theories&lt;/li&gt;&#xA;&lt;li&gt;Nothing directly involving the Higgs has been published as yet&lt;/li&gt;&#xA;&lt;li&gt;neutrinoless double beta decay would be the signature of $\nu + \nu \to \text{::nothing::}$ (off-shell, of course) and would indicate that neutrinos are Majorana particles. There is a report of it, but the significance is limit and it is unconfirmed&lt;/li&gt;&#xA;&lt;li&gt;Evidence for any kind of dark matter interactions outside of gravity is pretty sparse on the ground&lt;/li&gt;&#xA;&lt;li&gt;No super-symmetric partners have been observed&lt;/li&gt;&#xA;&lt;li&gt;&lt;s&gt;I doubt anyone has seen $\nu_\tau + n \to \tau + p$, though this is required by the current electoweak formalism. Or at least, they haven't been able to &lt;em&gt;show&lt;/em&gt; that this is what they saw&lt;/s&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Basically lots of dark corners. Note that much of this is Beyond the Standard Model, so may or may not represent the real state of physics.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-02-20T21:14:22.487" LastActivityDate="2011-02-20T21:14:22.487" CommentCount="4" />
  <row Id="5537" PostTypeId="1" CreationDate="2011-02-20T16:08:00.370" Score="4" ViewCount="120" Body="&lt;p&gt;One reason superstring theory has often been touted as being an improvement over bosonic string theory is that we can impose GSO projections to remove the tachyonic mode. If we insist upon the no tachyon condition, we are left with only five superstring theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if we don't mind tachyonic modes? After all, they only signify the background we are expanding about is unstable. Can we have more string models then? Type 0 string theories are one example, but are there others? I realize we can't have supersymmetry in target space if we have negative energies, but there's no theorem stating that string theories have to be supersymmetric.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-02T04:43:59.260" Title="If we are willing to accept tachyonic string modes, are there valid projections other than the GSO projections?" Tags="&lt;string-theory&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="5538" PostTypeId="2" ParentId="5534" CreationDate="2011-02-20T16:09:20.680" Score="0" ViewCount="" Body="&lt;p&gt;It is a &lt;strong&gt;dry&lt;/strong&gt; weather who prevents you from discharging, not obligatory cold one. Everything &quot;collects&quot; charges while in friction with something else. Any means increasing conductivity helps keep charge low. Discharge yourself with help of a resistance to decrease the current (with a wet stick or something).&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-20T17:02:55.637" LastActivityDate="2011-02-20T17:02:55.637" CommentCount="3" />
  <row Id="5539" PostTypeId="2" ParentId="5535" CreationDate="2011-02-20T16:10:31.797" Score="7" ViewCount="" Body="&lt;p&gt;dmckee has given a good, thorough answer. I'd bet that there are lots of other interactions that haven't been seen. How about $\gamma+\gamma\to Z+\overline Z$, or $\gamma+\gamma\to t\overline t$? To see those reactions, you'd need high-energy photons, and the cross sections should be very low. I doubt that we've produced the right experimental setup for those reactions. Perhaps even better would be a ton of similar $\nu\overline\nu$ reactions: $\nu_\tau+\overline\nu_\tau\to t\overline t$, for instance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these satisfy the conservation laws and are predicted to have nonzero cross sections in the standard model, but I'd be surprised if they've been observed.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-02-20T16:27:31.973" LastActivityDate="2011-02-20T16:27:31.973" />
  <row Id="5540" PostTypeId="1" CreationDate="2011-02-20T16:33:07.407" Score="3" ViewCount="193" Body="&lt;p&gt;For some very special cases of string compactification backgrounds, we get mysterious enhanced gauge symmetries. What is their physical meaning, and why do they arise? Is there any geometrical interpretation for the enhanced gauge symmetries?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-20T20:41:46.183" Title="What is the physical meaning of enhanced gauge symmetries in string compactifications?" Tags="&lt;string-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="5541" PostTypeId="2" ParentId="5519" CreationDate="2011-02-20T16:42:20.003" Score="5" ViewCount="" Body="&lt;p&gt;No, it doesn't mean that quantum field theory is non-local. The fact that there exist operators that exactly (anti)commute at spacelike separation remains exactly valid at the interacting level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your argument is based on a flawed assumption. It is not true that the higher-order interaction term should be normal-ordered in the Hamiltonian. The right interaction Hamiltonian should be written &quot;as it is&quot;, without normal ordering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivation of Feynman rules for scattering amplitudes in the operator language and interaction picture requires us to calculate the matrix elements of $S$ in the initial and final state. The matrix $S$ is the time-ordered exponential:&#xA;$$ S \approx T \left[i \exp\int dt\,H_I \right] $$&#xA;Sorry if there should be $-i$ in front of $H_I$. There is no normal ordering in this formula which is why there's no non-locality of the interaction term; instead, there is time-ordering of everything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Via Wick's theorem, this time-ordered exponential, when expanded via Taylor expansion, may be written via the contractions&#xA;$$ C(x_1,x_2) = \langle 0 | T (\phi_1 \phi_2) | 0\rangle = i\Delta_F(x_1-x_2)$$&#xA;which is the standard Feynman propagator, associated with $i/(k^2-m^2+i\epsilon)$ in the momentum space. But the contraction is the difference of the time-ordered (but not normal-ordered) product of two operators, which always appear in the definition of the S-matrix, and the normal-ordered (but no longer time-ordered) product, which is easy to deal with when evaluating the matrix elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, I think that if you added a higher-order interaction Hamiltonian with the extra normal ordering, you would get a non-local theory, but that's not how QFTs are defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be dissatisfied that without the normal ordering, the interaction Hamiltonian will lead to infinite matrix elements etc. Indeed, it will. But there are many other infinities of a similar kind and all of them have to be dealt with by the process of renormalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea that the interaction Hamiltonian should be normal-ordered is probably a flawed artifact of the intuition to get rid of the infinities &quot;as soon as possible&quot;. For free quantum field theories, one may define finite prescriptions - without renormalization - for quantities such as the total energy and the total charge and normal ordering is helpful to do so easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this treatment of free quantum field theories is not useful to get rid of many other infinities that appear once the interactions are included. To deal with them, one needs renormalization. Normal ordering of the interaction Hamiltonian is not only useless to get rid of the infinities: it would be harmful because, as you correctly observed, it would produce non-localities. (Unless the theory would be equivalent to a local field theory by a field redefinition, and I don't see an obvious way how it could happen.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can see, this trivial mistake - making the interaction Hamiltonian time-ordered - appears at some treatments of quantum field theory in the &quot;axiomatic&quot; or &quot;algebraic&quot; quantum field theory framework, which is why those approaches, at least in some of their versions, are completely incompatible both with renormalization as well as locality.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-20T18:01:54.583" LastActivityDate="2011-02-20T18:01:54.583" />
  <row Id="5542" PostTypeId="2" ParentId="5523" CreationDate="2011-02-20T17:00:59.500" Score="1" ViewCount="" Body="&lt;p&gt;It actually does increase the range. Here is the 'experiment' I performed once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Walk away from the car and find the range at which the key fails to work. Walk away another 5 feet. Turn and put the key against your chest, so that it points towards the car through your body. It worked!! I did some more testing. Apparently the range increases by about ten feet if the key is used through the body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea why.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2011-02-20T17:00:59.500" />
  <row Id="5543" PostTypeId="2" ParentId="5540" CreationDate="2011-02-20T17:27:07.087" Score="0" ViewCount="" Body="&lt;p&gt;If I had to take a guess I'd say these gauge symmetries arise in the same way they do in condensed matter systems as an effective macroscopic description. If you know how this happens then skip ahead to the section marked with (&lt;strong&gt;*&lt;/strong&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a lattice, for instance the honeycomb lattice, perfectly uniform. Consider the case when the lattice is made of carbon atoms at the vertices with $sp^3$ hybridized orbitals forming covalent bonds. Carbon has 4 valence electrons. Three of these are used up in the 3 planar $sp^2$ ($1s+p_x+p_y$)bonds. The fourth electron lives in the $p_z$ orbital which can hold a maximum of two electrons with opposite spins. Each vertex therefore has one electron when it can hold two. The Fermi energy of graphene therefore lies in the center of its conduction band. This setup is referred to as &lt;em&gt;half-filling&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, as has been discussed in a &lt;a href=&quot;http://physics.stackexchange.com/questions/3352/how-do-dirac-fermions-arise-in-graphene-and-what-significance-if-any-does-thi&quot;&gt;previous question&lt;/a&gt;, in the limit that lattice size goes to $\infty$ the effective many-body description of the electron gas is given by the massless Dirac equation. The quasiparticles (distinct from the electrons themselves) are therefore massless fermions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any real graphene lattice, like any other condensed matter system, contains defects and disorder. These can be of various kinds involving doping with other substances to change graphene's electronic properties. We are concerned with lattice defects primarily those where instead of a hexagon we find a pentagon or heptagon. An example is the &lt;a href=&quot;http://newton.ex.ac.uk/research/qsystems/people/latham/papers/1995/C/c1.html&quot; rel=&quot;nofollow&quot;&gt;Stone-Wales defect&lt;/a&gt; where rotation of an edge creates two heptagons and two pentagons where there were four hexagons previously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The presence of defects will obviously modify the long-range effective theory. There are various ways one might imagine generalizing the situation given that initially we have only the massless Dirac equation to work with. An obvious possibility is that the dynamics gets modified in such a way that a gauge field is also required. Then we have massless fermions but now there is a (effective) gauge field which mediates interactions between these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As is shown in this paper (&lt;a href=&quot;http://arxiv.org/abs/cond-mat/9208004&quot; rel=&quot;nofollow&quot;&gt;GGV&lt;/a&gt;) by Gonzales, Guinea and Vozmediano, this is indeed the case and rather surprisingly the emergent gauge field in this example is $SU(2)$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;*&lt;/strong&gt;) Returning to your original question, I would say that that when you take a 10 spacetime and compactify 6 of its dimensions, you can do so such that different regions of the 10D manifold get compactified differently. We are left with a 4D spacetime, with these compactified objects behaving as defects in this (effective) background. The enhanced gauge symmetries then follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully I didn't completely misunderstand your question!&lt;/p&gt;&#xA;" LastActivityDate="2011-02-20T17:27:07.087" CommentCount="2" />
  <row Id="5545" PostTypeId="1" AcceptedAnswerId="5546" CreationDate="2011-02-20T18:08:37.173" Score="2" ViewCount="96" Body="&lt;p&gt;The solar neutrino problem has been &quot;solved&quot; by discovering that neutrinos have mass and they oscillate. So how accurate are &lt;em&gt;now&lt;/em&gt; our predictions about the number and types of solar neutrinos that reach the earth?&lt;/p&gt;&#xA;" OwnerUserId="824" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-06-09T17:08:33.403" LastActivityDate="2011-06-09T17:08:33.403" Title="Solar neutrino predictions" Tags="&lt;experimental-evidence&gt;&lt;neutrino&gt;&lt;stellar-physics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5546" PostTypeId="2" ParentId="5545" CreationDate="2011-02-20T18:18:49.177" Score="5" ViewCount="" Body="&lt;p&gt;The solar data analyzed with current solar models are in good agreement with reactor and atmospheric results for $\theta_{12}$ and $\Delta m_{21}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not, however, a very strong test of the models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The big &lt;a href=&quot;http://prc.aps.org/abstract/PRC/v72/i5/e055502&quot; rel=&quot;nofollow&quot;&gt;SNO paper&lt;/a&gt;. SNO probed the solar neutrino spectrum in detail. On page 34 you'll find their spiffy comparison plot of flux for electron and muon flavor channels between their results; the &quot;standard solor model&quot; with no oscillation; and Super Kamiokande. They are several sigma from the un-oscillated model, and agree with SuperK.&lt;/li&gt;&#xA;&lt;li&gt;The most recent &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v100/i22/e221803&quot; rel=&quot;nofollow&quot;&gt;KamLAND oscillation paper&lt;/a&gt;.  The plot you want here is figure 2, which compares the combined SNO-and-SuperK results to the KamLAND results in $\theta_{12}$--$\Delta m_{21}^2$ space. Note that the allowed regions overlap. Also on the &lt;a href=&quot;http://arxiv.org/abs/0801.4589&quot; rel=&quot;nofollow&quot;&gt;arXiv as 0801.4589&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, we conclude that SNO sees neutrinos fluxes consistent with an oscillated solar model where the oscillation parameters are derived from atmospheric and reactor data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the error bars on the oscillation parameters are non-trivial, so it limits how strong a test of the solar model we have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I presume the SNO paper can be found on the arXiv as well, but I don't know its number.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-02-20T19:24:36.630" LastActivityDate="2011-02-20T19:24:36.630" CommentCount="2" />
  <row Id="5547" PostTypeId="2" ParentId="5534" CreationDate="2011-02-20T18:50:20.973" Score="1" ViewCount="" Body="&lt;p&gt;Here is a similar thread: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/4180/how-to-avoid-getting-shocked&quot;&gt;How to avoid getting shocked&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And: Your bodys &lt;strong&gt;surface&lt;/strong&gt; collects the charge, it is generated by Your shoes &#xA;and the street and Your clothes. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-02-20T18:50:20.973" />
  <row Id="5548" PostTypeId="2" ParentId="5523" CreationDate="2011-02-20T19:02:27.487" Score="13" ViewCount="" Body="&lt;p&gt;It works with a lot of body parts. It will work with a piece of &#xA;wire too. This keys work at about 433 Mhz, a resonant Lamda/4 antenna &#xA;is about 18 cm. Obviously the whole key is shorter, the antanna &#xA;is not tuned for maximum power transmission. By bringing the key &#xA;close to some conductive material, the power transmitted increases.&#xA;A very recondite answer is found &lt;a href=&quot;http://www.youtube.com/watch?v=BoDXyfAUBQc&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &#xA;Same question was asked some weeks ago on &quot;last-word.com&quot;&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-09T23:08:47.470" LastActivityDate="2011-04-09T23:08:47.470" CommentCount="3" />
  <row Id="5552" PostTypeId="1" CreationDate="2011-02-20T19:40:39.350" Score="2" ViewCount="790" Body="&lt;p&gt;Let us take the standard QED ($e^-, e^+, \gamma$) as a model of QFT and ask what is its &quot;short-distance&quot; physics? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They say the UV infinities appear because we do not know the real physics of short distances and initially we introduce it wrong. OK, but after renormalizations, what physics does remain? Do we replace the unknown/wrong physics with certain/right one? Can anybody describe it without appealing to unphysical bare particles? Have we an idea about the real electron from QED? If so, why we cannot use it as the input to construct a reasonable theory from the very beginning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Moderators, please do not close my questions before they are answered, let people answer.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-03-03T09:29:28.503" LastActivityDate="2011-03-06T16:08:59.290" Title="What physics does occur at short distances in QED?" Tags="&lt;quantum-mechanics&gt;&lt;qed&gt;&lt;renormalizations&gt;" AnswerCount="5" />
  <row Id="5554" PostTypeId="2" ParentId="5537" CreationDate="2011-02-20T19:47:21.730" Score="3" ViewCount="" Body="&lt;p&gt;In ten dimensions this question was answered long ago in the three papers listed below.&#xA;The first one found several new possibilities using orbifold projections including a theory without spacetime supersymmetry and with no tree-level tachyon, the latter two papers classified modular invariant spin structures in a fermionic formulation. Below ten dimensions there are many more possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) String Theories in Ten-Dimensions Without Space-Time Supersymmetry. &#xA;Lance J. Dixon, Jeffrey A. Harvey, (Princeton U.) . PRINT-86-0244 (PRINCETON), Feb 1986. 18pp. &#xA;Published in Nucl.Phys.B274:93-105,1986.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Spin Structures in String Theory.&#xA;N. Seiberg, (Weizmann Inst. &amp;amp; Princeton, Inst. Advanced Study) , Edward Witten, (Princeton U.) . Print-86-0218 (PRINCETON), Feb 1986. 28pp. &#xA;Published in Nucl.Phys.B276:272,1986.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) An O(16) x O(16) Heterotic String.&#xA;Luis Alvarez-Gaume, Paul H. Ginsparg, Gregory W. Moore, C. Vafa, (Harvard U.) . HUTP-86/A013, Feb 1986. 15pp. &#xA;Published in Phys.Lett.B171:155,1986.&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T04:43:59.260" LastActivityDate="2011-03-02T04:43:59.260" />
  <row Id="5556" PostTypeId="2" ParentId="5537" CreationDate="2011-02-20T19:59:32.083" Score="3" ViewCount="" Body="&lt;p&gt;First, a disclaimer: expanding around an unstable background is... problematic. The actual dynamics of an unstable background dictates a disappearance of the space within a string time. Spacetime supersymmetry at some level is almost certainly needed for some basic consistency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you care about theories similar to type 0 string theories, we need to focus on the maximally dimensional ones because there are way too many vacua, even the supersymmetric ones, when some dimensions are compactified (the &quot;landscape&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as closed string possibilities go, there is a bosonic string theory in $D=26$. Open strings may be allowed by various D-branes in those theories - the classification of possible D-branes is another step in the classification of backgrounds. And strings may be constrained to be unoriented by the additional of orientifold planes - which are classified analogously to D-branes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for $D=10$ string theories with the world sheet supersymmetry, type 0A and 0B string theories are the only consistent non-supersymmetric theories that use the superstring both for left-movers and right-movers. It's not hard to see why in the RNS formalism. You always absolutely need the &quot;overall&quot; or &quot;diagonal&quot; GSO projection for all the fermions. Type II theories have separate ones for left-movers and right-movers. Type 0 theories only have the shared one and one may distinguish how the GSO operator acts in the PP sector (odd or even forms, again). Type 0 theories are covered as standard material in chapter 10 of Polchinski's book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider heterotic string theories, $D=26$ on one side and $D=10$ on the other, there are extra 10-dimensional theories. The extra left-movers may be fermionized and the actions on the supersymmetric fermions and the left-moving $\lambda$ fields may be coupled in various ways. As explained in section 11.3 of Polchinski's book, you may get seven tachyonic heterotic string theories with $SO(16)\times SO(16)$, $SO(16)\times E_8$, $SO(24)\times SO(8)$, $E_7\times E_7\times SO(4)$, $SU(16)\times SO(2)$, single $E_8$, and $SO(32)$ (without SUSY!) gauge groups.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-20T19:59:32.083" />
  <row Id="5557" PostTypeId="2" ParentId="5540" CreationDate="2011-02-20T20:20:40.640" Score="5" ViewCount="" Body="&lt;p&gt;The enhanced gauge groups look fascinating, indeed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The enhanced continuous gauge symmetry arise because in many cases, one may prove that for particular values of the radii and Wilson lines in various compactifications etc., there exist additional massless spin-one states in the single-string spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because string theory produces consistent physics in spacetime for any value of the radii and Wilson lines, and because Yang-Mills theory is the only consistent way to make spin-one fields interact, it follows that spacetime dynamics has to include the enhanced Yang-Mills symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving away from the enhanced-symmetry point (in the space of the radii and Wilson lines) is equivalent to the Higgs mechanism. The scalar fields that measure the deviation (of the radii and Wilson lines) from the enhanced-symmetry points may be interpreted as Higgs fields that spontaneously break the enhanced symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The emergence of Yang-Mills symmetry was proved from the consistency of string theory which may look like too powerful and abstract a method. However, the emergence may also be showed very explicitly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The new massless gauge-bosons - analogous to W-bosons that become massless at some point, enhancing $U(1)$ to $SU(2)$, if I mention the simplest example - are obtained by a &quot;purely stringy effect&quot;, namely from strings that carry a nonzero winding number. That's why these states can't be seen in the &quot;effective higher-dimensional field theory&quot;: the latter has no wound strings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If those new string states are spin-one states and they are charged under the $U(1)$ and similar &quot;generic&quot; groups, it follows that the physics has to be Yang-Mills theory. But it may be useful to see why there's at least the &quot;global enhanced symmetry group&quot; such as $SU(2)$ at the enhanced point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The appearance of the global symmetry may be seen by the method of &quot;current algebras&quot;. Instead of states of a closed string, it's very useful to switch, by the state-operator correspondence, to the local operators on the world sheet. Much like the total $U(1)$ generator may be written as an integral of a charge density over the string, the remaining $SU(2)$ or other generators may also be written as integrals of a current (its temporal component) over the world sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And indeed, it may be proved that at the enhanced point, the currents have the right OPEs (operator product expansions)&#xA;$$ j^a(z) j^b(0) \sim \frac{ k^{ab}}{z^2} +\frac{ic^{ab}_{\,\,\,\,\,c}}{z}j^c(0) $$&#xA;Because of the general CFT calculus, these OPEs know all about the commutators of the resulting charges, so one may prove that the charges obtained as integrals of those currents have the right commutation relations. One also has to show that they commute with the Hamiltonian - they're conserved i.e. they're symmetries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest and most important triplet of the currents that one should understand are those that generate the $SU(2)$ algebra in a circular compactification on the self-dual radius:&#xA;$$\partial X, \quad :\exp(iX):, \quad :\exp(-iX):$$&#xA;I omit constants, both prefactors and those in the exponent. The $\partial X$ current for a compactified dimension $X$ has the OPEs with others where one simply differentiate with respect to $X$. So it shows that the first current is that of $J_z$ and the other two are $J^\pm$. The most nontrivial commutator or OPE that has to come out right is the OPE of the last two exponentials, but it indeed produces the right result when it should. (Quantum subtleties are critical to calculate the OPE: the commutator is not just the Poisson bracket in this case!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is typical that the enhanced symmetries separately arise from left-movers (holomorphic $z$ dependence) and right-movers (antiholomorphic $\bar z$ dependence) - two CFTs that just happen to be combined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a more overall perspective, one can't say that the enhanced symmetry is an &quot;unnatural&quot; point and the generic one is &quot;natural&quot;. Any point of the moduli space is equally consistent as any other. It sometimes doesn't even make sense to ask what's the &quot;maximum gauge group&quot; that may be broken by the Higgs mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has fascinating consequences. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple example. Take the $SO(32)$ heterotic string: the rank of the gauge group is 16 and the dimension is 496. It looks pretty big. Compactify it on a circle. Turn on generic Wilson lines. It will break the gauge group to $U(1)^{16}$, aside from the additional $U(1)$ from the circle itself (one $U(1)$ from the Kaluza-Klein $U(1)$, another $U(1)$ from a reduction of the B-field along the circle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At an enhanced point, i.e. for proper values of the radius and the Wilson lines, the gauge group becomes $E_8 \times E_8$ - times the same $U(1)^2$. You may decompactify the dimensions differently: if you allow compactifications on a circle, one may interpolate between the $SO(32)$ and $E_8\times E_8$ heterotic string theories by totally consistent theories! Now, notice that $E_8\times E_8$ has rank 16 and dimension 496, much like $SO(32)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In field theory, you would think that there's always the &quot;biggest gauge group&quot; that may perhaps be broken by the Higgs mechanism. In string theory, however, there may be two totally equally big (both rank and dimension), totally different &quot;initial&quot; gauge groups that may be broken to a more generic gauge group. It makes absolutely no sense to ask which of the two heterotic string vacua is more unbroken, more symmetric, or more fundamental. They're clearly equally unbroken, equally symmetric, equally fundamental.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more general cases of enhanced symmetry groups, one has to learn the Dynkin diagrams, weights, roots, Cartan subalgebras etc., but the derivations are de facto just straightforward generalizations of the simplest case of the $SU(2)$ enhanced symmetry. And it is very typical for the CFTs - that are used as world sheet CFTs here in string theory - have several totally equivalent descriptions that use totally different degrees of freedom. For example, 1 boson is equivalent to 2 fermions when the allowed projections and boundary conditions of both kinds of fields are properly adjusted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These were gauge symmetries coming from bulk closed strings. Non-Abelian gauge symmetries in string theory also arise on stacks of D-branes - from open strings - and on singularities - the ADE singularities in M-theory (and their F-theory generalizations) are the key example. In M-theory (and type IIA), the enhanced non-Abelian gauge symmetries on the singularities arise from M2-branes (or D2-branes) wrapped on 2-cycles that shrink to zero size. The gauge bosons are literally tiny membranes, much like they're wound strings or open strings in the other pictures. All these pictures how non-Abelian gauge symmetries may emerge in string theory may be related by various dualities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, I skipped one more method how a non-Abelian group emerges - the $E_8$ group (and the vector supermultiplet) that has to live on all the boundaries of the 11-dimensional spacetime of M-theory.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-20T20:41:46.183" LastActivityDate="2011-02-20T20:41:46.183" />
  <row Id="5558" PostTypeId="2" ParentId="5540" CreationDate="2011-02-20T20:27:04.177" Score="0" ViewCount="" Body="&lt;p&gt;Consider a compactification of M-theory on a $G_2$ holonomy manifold $X$, which in some local neighborhood can be represented as $X=(C^2/Z_N)\times Q$, where $Q\in X$ is a supersymmetric (associative) 3-cycle. The desingularization limit of the orbifold $C^2/Z_N$ is the so-called ALE (asymptotically locally Euclidean) space containing an entire $SU(N)$ root lattice of supersymmetric two-cycles, i.e. one can choose a basis for the two-cycles inside $C^2/Z_N$ such that their intersection matrix is minus the Cartan matrix of the $SU(N)$ Lie algebra. In the smooth limit, i.e. when the two-cycles have a finite volume, one obtains a $U(1)^N$ gauge theory where the N Abelian vector fields come from the first sum in the dimensional reduction of the 11-D supergravity 3-form $C_3=\sum_{i=1}^NA^i_1\wedge \phi_i + axions$, where $\phi_i\in H^2(X,Z)$ are the corresponding basis harmonic two-forms. Now consider the singular limit when all these two-cycles were shrunk to zero size. In this case, there will be new massless gauge degrees of freedom coming from the membranes wrapping the collapsed two-cycles in $C^2/Z_N$ (when a membrane wraps a two-cycle, the mass of the corresponding state is proportional to the volume of the cycle, when the volume shrinks to zero the worldvolume of the membrane effectively turns into a worldline of a massless particle) and when combined with the N vector fields from the KK reduction of the 3-form they fill out the adjoint representation of $SU(N)$. Thus, M-theory compactified on such a space would give rise to a 7-dimensional non-Abelian $SU(N)$ gauge theory living in $R^{3,1}\times Q$. Upon further reduction to 4 dimensions, the volume of supersymmetric tree-cycle $Q$ becomes identified with the inverse of the gauge coupling of the 4-dimensional $SU(N)$ gauge theory (at the KK scale). So, from this perspective the value of the unified gauge coupling in a SUSY GUT, e.g. $SU(5)$ GUT, is directly related to the volume of the three-cycle $Q$ supporting the gauge degrees of freedom in the visible sector $\alpha^{-1}_{GUT}\approx 25=Vol(Q)/{l_M^3}$.&lt;/p&gt;&#xA;" OwnerUserId="1892" LastEditorUserId="1892" LastEditorDisplayName="" LastEditDate="2011-02-20T20:36:56.793" LastActivityDate="2011-02-20T20:36:56.793" />
  <row Id="5559" PostTypeId="2" ParentId="5552" CreationDate="2011-02-20T20:35:15.163" Score="13" ViewCount="" Body="&lt;p&gt;Let us take standard fluid dynamics as a model of continuum media physics. It is obvious what is its &quot;short-distance&quot; physics isn't it? These are atoms and molecules. Let's apply your questions to this case -- suppose that we know nothing about atoms and molecules. in that case we can still use hydrodynamics, don't we?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do we replace the unknown/wrong physics with certain/right one? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is &quot;wrong&quot; and &quot;right&quot; physics in a first place? If your hydrodynamics predicts physical phenomena in it's range of validity then I'm free to call it &quot;right&quot;. Even if it says nothings about atoms. Anything else is a philosophy. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can anybody describe it without appealing to unphysical bare particles?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can anybody describe hydrodynamics without appealing to &quot;unphysical&quot; continuum medium? Yes -- if one knows about atoms. If one doesn't know then one can try to speculate about underlying theory. But it doesn't render hydrodynamics useless and &quot;wrong&quot;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Have we an idea about the real electron from QED?  Why we cannot use it as the input to construct a reasonable theory from the very beginning?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Have we an idea about &quot;real&quot; stream of water? It is made of atoms, right? So in this particular case there is no &quot;real&quot; stream of water at all.  That's why we cannot use it as the input to construct a reasonable theory from the very beginning.&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-02-20T20:35:15.163" CommentCount="6" />
  <row Id="5560" PostTypeId="2" ParentId="5535" CreationDate="2011-02-20T20:57:21.277" Score="6" ViewCount="" Body="&lt;p&gt;There are some extremely important reactions that have never been observed directly. My favorite example is p-p fusion,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{p} + \text{p} \rightarrow \text{d} + \text{e}^+ + \nu_e$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the rate-determining step for the main fusion process in the Sun and all other small stars. But it is utterly impossible to observe this reaction in anything like today's accelerators, because the cross section is so tiny. It's only important in the Sun because there's such a large, dense collection of hot protons. (In fact the smallness of this cross section is the reason all the stars haven't burnt out yet.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other examples would be any reactions of two photons, or two neutrinos, as Ted Bunn said.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-02-20T20:57:21.277" CommentCount="2" />
  <row Id="5561" PostTypeId="1" AcceptedAnswerId="5598" CreationDate="2011-02-20T22:06:22.587" Score="-2" ViewCount="196" Body="&lt;p&gt;I've just moved into my first place, and for reasons I do not condone or recommend I don't have any method of refrigeration right now. Luckily I now live very close to shops, so it isn't a big deal, but it made me think: if I really had to, what would be the most efficient way to keep food cool without using any electrical (or any other means which would be impractical) method of refrigeration?&lt;/p&gt;&#xA;" OwnerUserId="2133" LastActivityDate="2011-02-21T10:28:52.477" Title="Best way to keep food cool indoors without using a fridge/freezer" Tags="&lt;temperature&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="5562" PostTypeId="1" AcceptedAnswerId="5564" CreationDate="2011-02-20T22:12:29.353" Score="2" ViewCount="77" Body="&lt;p&gt;At room temperature....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How long will the sugar stay suspended once dissolved ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What governs the rate of settling ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What part does Brownian motion play ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sugar might be a bad example...please substitute as needed.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2011-02-20T22:25:27.350" Title="What keeps the sugar suspended in the tea?" Tags="&lt;fluid-dynamics&gt;" AnswerCount="1" />
  <row Id="5564" PostTypeId="2" ParentId="5562" CreationDate="2011-02-20T22:25:27.350" Score="3" ViewCount="" Body="&lt;p&gt;Anything dissolved is kept &quot;suspended&quot; by brownian motion. &#xA;Theoreticly there will be a small gradient of concentration &#xA;due to gravity, but this is unmeasurable for sugar. &#xA;(Barometric formula). For much bigger particles (ãbout µm) , this &#xA;gradient can be determined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sugar will stay dissolved until the water evaporates,&#xA;then the sugar will form a solid residue. (if You are lucky, &#xA;as crystals) Settling does not occur with molecules.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-02-20T22:25:27.350" CommentCount="2" />
  <row Id="5565" PostTypeId="2" ParentId="5561" CreationDate="2011-02-20T22:38:44.607" Score="0" ViewCount="" Body="&lt;p&gt;Pasteurized food is preserved for long time (boiled soup without opening the pan, for example).&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-02-20T22:38:44.607" />
  <row Id="5566" PostTypeId="1" AcceptedAnswerId="5578" CreationDate="2011-02-20T23:23:57.760" Score="0" ViewCount="125" Body="&lt;p&gt;Very often the energy released in large-scale events (e.g., earthquakes, meteor impacts, etc) is described in terms of some number of 'tons of TNT'. This may be all well and good for those with an intuitive sense of big explosions, but specifically, how many &lt;em&gt;joules&lt;/em&gt; are there in the 'ton of TNT'? I can work out how many tons of water can be flashed to steam if you give me joules, but 'tons of TNT' means nothing in this regard, without a conversion factor.  What is the conversion factor? &lt;/p&gt;&#xA;" OwnerUserId="717" LastActivityDate="2011-02-21T03:54:00.063" Title="what does 'same energy as a ton of TNT' mean?" Tags="&lt;energy&gt;&lt;si-units&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5567" PostTypeId="1" AcceptedAnswerId="5641" CreationDate="2011-02-20T23:30:03.437" Score="7" ViewCount="193" Body="&lt;p&gt;Classically, &lt;a href=&quot;http://en.wikipedia.org/wiki/Shot_noise&quot; rel=&quot;nofollow&quot;&gt;shot noise&lt;/a&gt; observed in the signal generated by a laser incident on a photodiode is explained as being due to the quantization of light into photons, giving rise to a &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson_process&quot; rel=&quot;nofollow&quot;&gt;Poisson process&lt;/a&gt;.  In quantum optics, on the other hand, the shot noise is said to arise from interference with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Vacuum_state&quot; rel=&quot;nofollow&quot;&gt;vacuum field&lt;/a&gt;, which leaks in at points of optical loss (absorption).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, mechanical oscillators are subject to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fluctuation-dissipation_theorem&quot; rel=&quot;nofollow&quot;&gt;fluctuation-dissipation theorem&lt;/a&gt;, which says, roughly, that the thermal excitement of the various mechanical modes of the oscillator are proportional to the dissipation in that mode.  In the context of electronics, &lt;a href=&quot;http://en.wikipedia.org/wiki/Johnson_noise&quot; rel=&quot;nofollow&quot;&gt;Johnson noise&lt;/a&gt; is an example of this effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently read a thought-provoking statement (in &lt;a href=&quot;http://pra.aps.org/abstract/PRA/v44/i7/p4693_1&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;) that these effects are really &quot;the same&quot;; just as thermal motion arises at points of mechanical dissipation, shot noise arises at points of optical dissipation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is this analogy physically/theoretically significant?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-02-21T16:17:18.797" LastActivityDate="2011-06-11T02:11:40.810" Title="Quantum shot-noise and the fluctuation dissipation theorem" Tags="&lt;optics&gt;&lt;quantum-optics&gt;&lt;fluctuation-dissipation&gt;&lt;oscillators&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="5568" PostTypeId="2" ParentId="2392" CreationDate="2011-02-20T23:50:14.840" Score="1" ViewCount="" Body="&lt;p&gt;There's an unfavourable review of Hydrino theory in New Journal of Physics by A Rathke of ESA Advanced Concepts Team (DG-X), ESTEC Keplerlaan 1, 2201 AZ Noordwijk, The Netherlands:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://iopscience.iop.org/1367-2630/7/1/127/fulltext&quot; rel=&quot;nofollow&quot;&gt;http://iopscience.iop.org/1367-2630/7/1/127/fulltext&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Abstract. Recently, spectroscopic and calorimetric observations of &#xA;  hydrogen plasmas and chemical reactions with them have been &#xA;  interpreted as evidence for the existence of electronic states of the &#xA;  hydrogen atom with a binding energy of more than 13.6 eV. The &#xA;  theoretical basis for such states, which have been dubbed hydrinos, is &#xA;  investigated. We discuss both the novel deterministic model of the &#xA;  hydrogen atom, in which the existence of hydrinos was predicted, and &#xA;  standard quantum mechanics. Severe inconsistencies in the &#xA;  deterministic model are pointed out and the incompatibility of hydrino &#xA;  states with quantum mechanics is reviewed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1345" LastActivityDate="2011-02-20T23:50:14.840" CommentCount="1" />
  <row Id="5569" PostTypeId="1" AcceptedAnswerId="5624" CreationDate="2011-02-20T23:51:37.550" Score="7" ViewCount="201" Body="&lt;p&gt;I would like a tool which allows me to enter some paper citation, and then will begin drawing a graph, where each paper is linked to other papers that cite the original paper or are cited by it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like this would be a handy way to identify influential papers in a field at a glance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anything like this already exist?&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="184" LastEditorDisplayName="" LastEditDate="2011-02-21T18:14:33.270" LastActivityDate="2011-02-21T18:14:33.270" Title="Is there a nice tool to plot graphs of paper citations?" Tags="&lt;soft-question&gt;&lt;software&gt;&lt;publications&gt;&lt;citation-analysis&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" />
  <row Id="5571" PostTypeId="2" ParentId="5483" CreationDate="2011-02-21T01:19:49.350" Score="0" ViewCount="" Body="&lt;p&gt;It is not really a case of Mach principle being wrong.  It is something related to general relativity, and frame dragging or Lense-Thirring effect are similar to Mach’s principle.  However, these are local laws of physics and Mach’s principle is a global hypothesis.  Gravity does tell us that spacetime curvature induces motion of masses, and if these masses are large enough that can in turn change curvature.  So gravity or spacetime curvature acts on a mass locally and the same experiences acceleration as seen from another frame.  This is a geodesic deviation which is defined by the Riemann curvature.  So this tells us the mass which is observed to accelerate is the same mass which responds to the curvature --- the equivalence principle.  This also sounds similar to Mach’s principle: Inertia there determines inertia here.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mach stated the centripetal force on a local rotating frame is equivalent to what happens if the frame is nonrotating and the entire universe rotates around the frame.  Kurt Godel worked out a model of a rotating universe and found a very strange time-looping spacetime.  If the universe rotates it has a structure which is radically different from Mach’s original conjecture.  Further, this universe violates energy conditions.  It is not a physical spacetime, even if it comes from the Einstein field equations.  This does at least suggests there are some radical departures from Mach’s principle and general relativity.  Mach’s principle does not appear to be a consequence of general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mach’s principle is rather vague in some respects.  The inertia of a body is said to be determined by the inertia (or masses) of all other bodies.  Again this sounds like GR, but this extends across the universe.  The actual gravitational interaction a particle here has with a galaxy at z = 8 around12.5 billion light years out is miniscule. So the interaction picture appears funny.  It might be tempting to say the entire universe is some sort of single quantum wave function, and single masses we observe are entangled subsets.  This might sound global and gets away from the $q/r^2$ drop off of gravity.  From there one might be tempted to say inertial is inherited this way.  However, entanglements do not involve forces, and inertia as an entanglement not defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will also say that I suspect Mach had ideas of an ether in space when he suggested this.  So he probably had some picture of a rotating frame moving through this medium, and this is somehow equivalent to the whole medium rotating around the frame.  So the ether vortex in the case of a rotating world is what generated the centrifugal force on the rotating frame.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-21T01:19:49.350" />
  <row Id="5572" PostTypeId="2" ParentId="5569" CreationDate="2011-02-21T01:47:27.010" Score="1" ViewCount="" Body="&lt;p&gt;Centuries ago, there was a Touchgraph demo for Google Scholar. It was the closest thing.&lt;/p&gt;&#xA;" OwnerUserId="1335" LastActivityDate="2011-02-21T01:47:27.010" CommentCount="1" />
  <row Id="5573" PostTypeId="1" CreationDate="2011-02-21T02:57:04.250" Score="5" ViewCount="541" Body="&lt;p&gt;I require only a simple answer. One sentence is enough... (It's for high school physics)&lt;/p&gt;&#xA;" OwnerUserId="2135" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-02-21T04:42:28.540" LastActivityDate="2011-02-23T23:27:49.130" Title="What's the core difference between the electric and magnetic forces?" Tags="&lt;electromagnetism&gt;&lt;homework&gt;&lt;forces&gt;" AnswerCount="8" FavoriteCount="2" />
  <row Id="5574" PostTypeId="2" ParentId="5573" CreationDate="2011-02-21T03:16:55.477" Score="1" ViewCount="" Body="&lt;p&gt;Not to be slick, but starting with Hans Christian Ørsted, then on through James Clerk Maxwell and Albert Einstein we time and again reinforce that &lt;em&gt;there is none&lt;/em&gt; -- they are heads and tails of the same coin.&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2011-02-21T03:16:55.477" CommentCount="4" />
  <row Id="5575" PostTypeId="2" ParentId="5573" CreationDate="2011-02-21T03:30:38.757" Score="3" ViewCount="" Body="&lt;p&gt;For the purposes of high school physics it is reasonable to say that electric fields start and stop on charges but magnetic fields always run in loops.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This fails to be rigorous in at least two ways:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Relative velocity mixes the two types of fields in a manner governed by the Lorentz transform, so the truth is like &lt;a href=&quot;http://physics.stackexchange.com/questions/5573/whats-the-core-difference-between-the-electric-and-magnetic-forces/5574#5574&quot;&gt;wsc says&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;It is an open question if there exist any magnetic monopoles, but theoretical prejudice favors their existence for &lt;a href=&quot;http://physics.stackexchange.com/q/4141/520&quot;&gt;reasons of beauty&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-02-21T03:30:38.757" CommentCount="3" />
  <row Id="5576" PostTypeId="2" ParentId="5573" CreationDate="2011-02-21T03:34:42.843" Score="5" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electron&quot; rel=&quot;nofollow&quot;&gt;Electric monopoles&lt;/a&gt; exist. &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_monopole&quot; rel=&quot;nofollow&quot;&gt;Magnetic monopoles&lt;/a&gt; don't&lt;sup&gt;*&lt;/sup&gt; exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the reason that Maxwell's laws governing electricity and magnetism aren't symmetric. Maxwell's laws say that $\nabla \cdot B = 0$ and $\nabla \cdot E = 4\pi\rho_{e}$, but if magnetic monopoles existed this would be $\nabla \cdot B = 4\pi\rho_{m}$ and $\nabla \cdot E = 4\pi\rho_{e}$. A similar asymmetry in induction would be corrected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experimental lack of evidence for magnetic monopoles can then be taken as the main reason we distinguish between the two forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;*(Probably.)&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1655" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-02-21T04:42:25.673" LastActivityDate="2011-02-21T04:42:25.673" CommentCount="5" />
  <row Id="5577" PostTypeId="2" ParentId="5573" CreationDate="2011-02-21T03:43:52.137" Score="-2" ViewCount="" Body="&lt;p&gt;Electric fields are divergent, magnetic fields are convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why the downvote?  Anyone?&lt;/p&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2011-02-21T03:43:52.137" CommentCount="3" />
  <row Id="5578" PostTypeId="2" ParentId="5566" CreationDate="2011-02-21T03:54:00.063" Score="2" ViewCount="" Body="&lt;p&gt;It is the energy released when TNT detonates - approximately 4.184 gigajoules per ton.  Although TNT is under oxidized and can react further and so it releases about 2x more energy when it mixes with oxygen.  But the bomb makers use 4.184 gigajoules per ton as the conversion factor.&lt;/p&gt;&#xA;" OwnerUserId="1960" LastActivityDate="2011-02-21T03:54:00.063" />
  <row Id="5579" PostTypeId="1" CreationDate="2011-02-21T03:54:12.320" Score="2" ViewCount="199" Body="&lt;p&gt;Magnetic fields are obvious distortions.. of.. something, but what exactly are they distortions of?  Massive objects produce curvatures/gradients in space-time resulting in what we observe as gravity.. what is the equivalent explanation for magnetic/electric fields?&lt;/p&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2011-02-21T23:41:16.887" Title="What is the medium that allows magnetic fields *or any forcefield* to exist?" Tags="&lt;quantum-field-theory&gt;&lt;electromagnetism&gt;&lt;gravity&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" />
  <row Id="5580" PostTypeId="2" ParentId="5573" CreationDate="2011-02-21T03:59:34.563" Score="10" ViewCount="" Body="&lt;p&gt;Electric forces are created by and act on, both moving and stationary charges; while magnetic forces are created by and act on only moving charges.&lt;/p&gt;&#xA;" OwnerUserId="1345" LastActivityDate="2011-02-21T03:59:34.563" CommentCount="3" />
  <row Id="5581" PostTypeId="2" ParentId="5579" CreationDate="2011-02-21T04:16:52.743" Score="10" ViewCount="" Body="&lt;p&gt;The short answer is nothing at all - fields are a fundamental part of the universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What dmckee is teasing you about in the comments is that your intuition saying that there must be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Luminiferous_aether&quot; rel=&quot;nofollow&quot;&gt;luminiferous aether&lt;/a&gt;, which is an idea that was actually conclusively disproven in the late 19th century by Michelson and Morley in one of the most famous experiments of all time. The fact that fields don't need to be permeating through any material was &lt;em&gt;very&lt;/em&gt; disturbing to physicists at the time, because it seems so intuitively bizarre, but that's the way it is - fields just &lt;em&gt;are&lt;/em&gt;. They don't exist &quot;in&quot; or &quot;through&quot; anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some famous physicist - Gell-Mann or someone like that - has a quip about how fields are more real than particles; the idea of a &quot;single electron&quot; should be the idea that's abhorrent to our common sense.&lt;/p&gt;&#xA;" OwnerUserId="1655" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-02-21T04:38:37.657" LastActivityDate="2011-02-21T04:38:37.657" CommentCount="4" />
  <row Id="5582" PostTypeId="1" CreationDate="2011-02-21T04:19:09.733" Score="4" ViewCount="219" Body="&lt;p&gt;When a lambda particle decays into proton and a pion, I am told it does not conserve parity.  Why?&lt;/p&gt;&#xA;" OwnerUserId="1807" LastActivityDate="2011-02-21T07:47:34.633" Title="Why does lambda decay violate parity?" Tags="&lt;particle-physics&gt;&lt;quantum-mechanics&gt;&lt;particles&gt;&lt;parity&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="5583" PostTypeId="2" ParentId="5579" CreationDate="2011-02-21T05:36:28.837" Score="2" ViewCount="" Body="&lt;p&gt;I think that the answer depends on which type of physicist you ask.  First, if you Lorentz transform the ether, then you would expect the Michelson-Morley result and that would not by itself rule out an &quot;ether&quot;. Condensed matter physicists have theories about a medium&#xA;called a spin net liquid and &lt;a href=&quot;http://dao.mit.edu/~wen/&quot; rel=&quot;nofollow&quot;&gt;Xiao-Gang Wen (M.I.T)&lt;/a&gt; thinks that the vacuum is a &lt;a href=&quot;http://dao.mit.edu/~wen/stringnet.html&quot; rel=&quot;nofollow&quot;&gt;quantum liquid of string nets&lt;/a&gt;.&#xA;I don't know much condensed matter physics, but this looks interesting. He has posted the introduction of his book, &lt;a href=&quot;http://dao.mit.edu/~wen&quot; rel=&quot;nofollow&quot;&gt;Quantum Field Theory of Many Body Systems&lt;/a&gt;. Wen studied string theory at Princeton with Ed Witten and then changed fields to condensed matter theory. I don't know enough to make intelligent comments about this theory, but perhaps some string or condensed matter theorists will. This seems to provide the medium you are asking about. I put this answer out there more as what is being suggested by spin networks approach. &lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-02-21T15:57:54.053" LastActivityDate="2011-02-21T15:57:54.053" CommentCount="2" />
  <row Id="5584" PostTypeId="1" AcceptedAnswerId="5589" CreationDate="2011-02-21T05:45:20.787" Score="2" ViewCount="217" Body="&lt;p&gt;What I mean is could it turn out that the world is not described by string theory / M-Theory, but that nevertheless some version of one of these extra-dimensional theories is true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no real background in this area.  I just read Randall and Sundrum's 1999 paper &quot;A Large Mass Hierarchy from a Small Extra Dimension&quot; (http://arxiv.org/PS_cache/hep-ph/pdf/9905/9905221v1.pdf).  Other than the use of the term &quot;brane&quot; and a couple of references to string excitations at TeV scale, I don't see much about string theory, and I notice their theory only requires 1 extra dimension, not 6 or 7.&lt;/p&gt;&#xA;" OwnerUserId="11" LastActivityDate="2011-07-24T03:22:16.293" Title="Do extra-dimensional theories like ADD or Randall-Sundrum require string theory to be true?" Tags="&lt;string-theory&gt;&lt;randall-sundrum&gt;&lt;large-extra-dimensions&gt;" AnswerCount="3" />
  <row Id="5585" PostTypeId="2" ParentId="5573" CreationDate="2011-02-21T06:12:41.397" Score="0" ViewCount="" Body="&lt;p&gt;Unlike electric fields, magnetic fields do no work on charges.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-02-21T06:12:41.397" />
  <row Id="5586" PostTypeId="2" ParentId="3871" CreationDate="2011-02-21T06:26:59.423" Score="-1" ViewCount="" Body="&lt;p&gt;Great question. I'll try to answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum-mechanical world the timeline is completely reversible, the system undergoes unitary evolution and there is no preferred time direction. There is no concept of causalty either. That's why it is possible to move faster than light in the quantum world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, electron (and photon) can travel faster than light with some probability. But it can be interpreted as a virtual electron-positron pair appearing ahead of the propagating electron and then the positron from the pair annihilates with the propagating electron (the trick is that information cannot be transferred such way because you cannot be sure whether you spotted just a virtual particle from a vacuum fluctuation or an actual signal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The causality problem arises from asymmetry of time in classical world. And the asymmetry of time arises from the irreversibility of collapse of the wavefunction. That is the FTL travel can lead to a controversy only when measurements are taken. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is impossible to measure the quantum value and leave it as it is without collapse. Now imagine one can travel in time, measure the value in the future, then return to present. It would appear as if he made a measurement without collapsing the WF. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact the WF collapse is the only existing fundamental law which is not symmetric against the time direction. It is the cause of all other observed irreversible processes, including the second law of thermodynamics.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastActivityDate="2011-02-21T06:26:59.423" CommentCount="5" />
  <row Id="5587" PostTypeId="1" AcceptedAnswerId="5588" CreationDate="2011-02-21T06:59:35.127" Score="3" ViewCount="305" Body="&lt;p&gt;Below is a picture of a simple pendulum clock. Suppose that the bob (a rigid disk) on the end of the pendulum can spin without friction about its geometrical axis and is spinning at an angular velocity $\omega=const$. Will the clock gain or lose time compared to the ordinary pendulum clock. Or it does not have any effect on the time? If it does then what is the percentage? Ignore air drag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ZSwHI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="552" LastActivityDate="2011-03-11T07:20:38.157" Title="A pendulum clock problem" Tags="&lt;classical-mechanics&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="5588" PostTypeId="2" ParentId="5587" CreationDate="2011-02-21T07:16:58.267" Score="6" ViewCount="" Body="&lt;p&gt;The clock runs slightly faster when the disk can spin.  When the disk is attached rigidly to the pendulum, its center translates in space, but it also spins around its own center (compared to gravity).  The disk experiences net acceleration and net angular acceleration.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the disk can spin freely, the center of the disk still accelerates, but the disk has no angular acceleration because there is no torque on it.  (Uniform gravity cannot produce a torque about the center of mass, and the force between the pendulum and the disk has no moment arm.)  This slightly reduces the energy change in the disk for a given movement of the pendulum because there is no energy going in and out of the spin of the disk.  The net result is a lower effective moment of inertia in the pendulum+disk system.  The torque on the pendulum due to gravity is the same, so the clock swings slightly faster when the disk is free to spin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Variable definitions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$l$ length of pendulum&lt;/li&gt;&#xA;&lt;li&gt;$m$ mass of pendulum&lt;/li&gt;&#xA;&lt;li&gt;$\theta$ angle of pendulum with vertical&lt;/li&gt;&#xA;&lt;li&gt;$L$ distance between center of disk and pivot of pendulum&lt;/li&gt;&#xA;&lt;li&gt;$R$ radius of disk&lt;/li&gt;&#xA;&lt;li&gt;$M$ mass of disk&lt;/li&gt;&#xA;&lt;li&gt;$\phi$ angle the disk has rotated relative to gravity&lt;/li&gt;&#xA;&lt;li&gt;$\omega = \dot{\phi}$&lt;/li&gt;&#xA;&lt;li&gt;$g$ gravitational acceleration&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We'll find the Lagrangian of the system and compare the two cases.  When the disk is fixed,  $\phi = \theta$.  When the disk is free, $\phi$ is a new degree of freedom.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to find the kinetic and potential energies of the pendulum and the bob in terms of the variables above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the pendulum is a uniform rod pivoted at one end.  Then its moment of inertia is $ml^2/3$.  Its kinetic energy is $\frac{1}{2} I \dot{\theta}^2 = ml^2\dot{\theta}^2/6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The potential energy of the pendulum is $mg$ times the height of its center of mass above rest, which is $l\cos\theta /2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinetic energy of the disk comes from the motion of its center of mass and from its rotation.  It is $1/2(Mv^2) + 1/2I\omega^2$.  The velocity is $v = L\dot{\theta}$.  The moment of inertia is $I = 1/2 M R^2$.  The total kinetic energy is then $1/2M ( L^2 \dot{\theta}^2 + 1/2 R^2 \dot{\phi}^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The potential energy of the disk is $Mg$ times its height above equilibrium height, or $MgL\cos\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting these together, the Lagrangian for the clock with spinning bob is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{L} = \dot{\theta}^2\left(\frac{ml^2}{6}+ \frac{ML^2}{2}\right) + \dot{\phi}^2\frac{MR^2}{4} + g\cos\theta\left(\frac{ml}{2} + M L\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which we abbreviate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{L} = \dot{\theta}^2 \frac{I_\theta}{2} + \dot{\phi}^2 \frac{I_b}{2} + g\cos\theta M_{eff}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two different cases to explore.  If the bob of the pendulum is fixed, then $\phi = \theta$ because the bottom of the disk points the same way as the pendulum.  That gives us $\dot{\phi} = \dot{\theta}$.  The Lagrangian becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{L}_{fixed} = \dot{\theta}^2 \frac{I_\theta + I_b}{2} + g\cos\theta M_{eff}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or, setting $I_\theta + I_b = I_{fixed}$, it's just&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{L}_{fixed} = \dot{\theta}^2 \frac{I_{fixed}}{2} + g\cos\theta M_{eff}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the disk is free to swing, $\phi$ is a degree of freedom on its own.  The Lagrangian has no cross-terms involving both $\theta$ and $\phi$, meaning their evolutions are independent; these degrees of freedom are decoupled.  $\phi$ is a cyclic coordinate so the angular momentum of the disk is conserved.  The evolution of theta is the same as in a Lagrangian where $\phi$ does not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{L}_{free} = \dot{\theta}^2 \frac{I_\theta}{2} + g\cos\theta M_{eff}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only difference between $\mathcal{L}_{fixed}$ and $\mathcal{L}_{free}$ is that the moment of inertia is slightly larger in $\mathcal{L}_{fixed}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The potential energy terms are the same.  If we make a substitution in the free-bob case, introducing a new time coordinate defined by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t&amp;#39; = t \sqrt{\frac{I_\theta}{I_{fixed}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the Lagrangians take exactly the same form, except that they're expressed in different time coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means the free-bob pendulum will run in the same way, but faster by the factor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\nu_{fixed}}{\nu_{free}} = \frac{t}{t&amp;#39;} = \sqrt{\frac{I_{fixed}}{I_\theta}} = \sqrt{\frac{ml^2/3 + M(L^2 + R^2/2)}{ml^2/3 + ML^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\nu$ indicates a period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For tractability, we might look at some limit, such as $l = L$, $m = M$, $L&amp;gt;&amp;gt;R$.  The factor by which the free-disk clock is faster becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + \frac{1}{3}\left(\frac{R}{L}\right)^2$$&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-03-11T07:20:38.157" LastActivityDate="2011-03-11T07:20:38.157" CommentCount="5" />
  <row Id="5589" PostTypeId="2" ParentId="5584" CreationDate="2011-02-21T07:23:39.217" Score="7" ViewCount="" Body="&lt;p&gt;Extra-dimensional scenarios may be described as &quot;inspired&quot; by string theory but they are independent hypotheses and they may be true even if string theory is not. However, one has to reduce the ambitions and standards of consistency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sociologically, it's surely true that the research of models with extra dimensions has been adopted and pursued by many people who have never take studied proper string theory or taken a course in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite the academic independence, a confirmation of experimentally accessible extra dimensions - which is extremely unlikely to occur, due to their likely tiny size - would be a huge evidence supporting string theory because it's the only framework in which the extra dimensions actually have a justification (many).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-21T07:23:39.217" CommentCount="3" />
  <row Id="5590" PostTypeId="2" ParentId="5582" CreationDate="2011-02-21T07:41:05.167" Score="4" ViewCount="" Body="&lt;p&gt;First, an assignment of the parities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parity of fermions is a bit ambiguous because one may always redefine parity by&#xA;$$ P \to P (-1)^{2J}, P(-1)^L, P(-1)^{3B}, P(-1)^{3Q} $$&#xA;or one may add the product of several factors of this kind because the second factor is a multiplicatively conserved sign. By this definition, one gets another parity that is still conserved (at least in low-energy processes that also conserve $B,L$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there's a convention that assigns a particular parity to fermions. Note that Weinberg has proved that&#xA;$$ P^2 = (-1)^{2J} $$&#xA;so parity behaves much like the rotation by 180 degrees: its square changes the sign of states with an odd number of fermions. In the standard convention, electron, proton, and neutron are set to have $P=+1$: one is allowed to make three choices like that. The parity of the pion is then determined to be negative, $P=-1$, because a deuteron-pion ground state may decay into two neutrons with $L=1$ - the odd orbital momentum changes the sign of the parity, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strong and electromagnetic interactions preserve parity, so one may assign parity to all hadrons, as determined from various strong and electromagnetic interactions. Lambda then turns out to have a positive parity $P=+1$ much like protons and neutrons because it's just another bound state of three quarks and the change of the quarks' identity doesn't change parity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The neutral $\Lambda^0$ decays to a nucleon and a pion, $p+\pi^-$ or $n+\pi^0$, so a $P=+1$ state decays into one particle (nucleon) with $P=+1$ and one (pion) with $P=-1$. That violates the parity because $(+1)\neq (+1)(-1)$. It's because the decay is due to the weak interactions. Weak interactions don't preserve the parity because even the spectrum doesn't: for example, a parity-transformed partner of a left-handed neutrino - the right-handed neutrino - doesn't even exist. This asymmetry is confirmed by the weak interactions that contain two-component (fundamentally left-right asymmetric) spinors or, in the four-component spinor formalism, combinations $(1\pm \gamma_5)$ - scalars and pseudoscalars - that maximally violate the parity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of these interactions, any parity-violating process that respects all the other laws is allowed, although it may be slow because it must rely on the weak interactions which are weak. The decays of the neutral Lambda baryon - and similar decays of the other Lambda particles - belong among the parity-violating ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The violations of parity conservation, discovered half a century ago, was a shock. But once you appreciate the simple fact that fields may be described by a 2-component spinor that prefers one chirality over the other - may describe left-handed particles without the right-handed ones - it's not so shocking. The two-component spinors work because $SL(2,C)=SO(3,1)$, locally. The fundamental representation of $SL(2,C)$ has a spin-1/2, and it also has a small enough number of components that there's no oppositely spinning particle (with the same other charges). The theory is still Lorentz-invariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only parity is violated in some processes: at very high energies, those processes become so common that it isn't even possible to define parity accurately. After all, the parity transformation of the neutrino states is ill-defined. That's why it didn't hurt that I could have redefined parity by the signs coming from the lepton and baryon numbers (which are ultimately violated at very high energies, perhaps with the exception of $B-L$) at the beginning: none of the parity operators is actually fully well-defined on the spectrum of particle states.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-21T07:47:34.633" LastActivityDate="2011-02-21T07:47:34.633" />
  <row Id="5591" PostTypeId="1" CreationDate="2011-02-21T08:31:42.647" Score="1" ViewCount="172" Body="&lt;p&gt;Why is there a consistent theory of continuum mechanics in which one just consider things like differential elements and apply Newtons laws? Is there a deeper reason for it. Is it the nature of newtonian framework that makes it happen or is it somehow related to nature of bodies (topological spaces with borel measure etc)?&lt;/p&gt;&#xA;" OwnerUserId="2138" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-21T09:15:29.150" LastActivityDate="2011-02-21T16:31:33.190" Title="Consistent theory of continuum" Tags="&lt;classical-mechanics&gt;&lt;newtonian-mechanics&gt;&lt;elasticity&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="5592" PostTypeId="2" ParentId="5524" CreationDate="2011-02-21T08:44:50.277" Score="1" ViewCount="" Body="&lt;p&gt;There's no privileged reference frame here.  Just two observers with consistent measurements and a disagreement on the order of events (as is typical in relativity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember the collapse of the wave function isn't something you can observe.  You can observe the spin of one of the particles being measured, and then you can observe the spin of the other particle being measured, and these measurements are guaranteed to be consistent due to the entanglement.  Each observer will conclude that the first measurement (as determined in his reference frame) collapsed the wave function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your observers disagree on which measurement happened first, then perhaps they'll dispute which measurement collapsed the wave function, or when this collapse happened.  But certainly there won't be an observer who thinks the wave function collapse wasn't instantaneous, (unless that observer doesn't understand quantum mechanics).&lt;/p&gt;&#xA;" OwnerUserId="11" LastActivityDate="2011-02-21T08:44:50.277" />
  <row Id="5593" PostTypeId="2" ParentId="5591" CreationDate="2011-02-21T09:06:46.757" Score="-1" ViewCount="" Body="&lt;p&gt;I guess you are talking about the Banach-Tarski paradox, which is not really an inconsistency in mathematics, it is a result following from the axiom of choice, I dont think only Newtons laws implies the negation of the axiom of choice, so I think Banach-Tarski vs mass conservation is resolved (ie can not appear), by the ways of which laws such as conservation of mass is applied to continuum mechanics.&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-02-21T09:06:46.757" CommentCount="8" />
  <row Id="5594" PostTypeId="2" ParentId="5561" CreationDate="2011-02-21T09:15:57.307" Score="0" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Wrapping your food in a moist towel will cool it due to water evaporation. You can use fan to enhance this effect.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For temperatures below zero, ask for some ice in the shop and mix it with salt.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="701" LastActivityDate="2011-02-21T09:15:57.307" />
  <row Id="5595" PostTypeId="2" ParentId="5569" CreationDate="2011-02-21T09:23:18.403" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://isiknowledge.com&quot; rel=&quot;nofollow&quot;&gt;ISI Web of Knowledge&lt;/a&gt; has such tool called &quot;Citation map&quot; but you would probably need a subscription to use this service (most universities have it, I suppose). It is Java-based and sometimes works quite slow. &lt;/p&gt;&#xA;" OwnerUserId="701" LastActivityDate="2011-02-21T09:23:18.403" />
  <row Id="5597" PostTypeId="1" CreationDate="2011-02-21T10:19:11.657" Score="4" ViewCount="216" Body="&lt;p&gt;In all examples that I know, tachyons are described by scalar fields. I was wondering why you can't have a tachyon with spin 1. If this spinning tachyon were to condense to a vacuum, the vacuum wouldn't be Lorentz invariant---seems exotic but not a-priori inconsistent. Is there some stronger consistency requirement which rules out spinning tachyons? If someone could provide a reference that would be helpful too!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's another confusion: I was reading Wikipedia, which claims that tachyons should be spinless and obey Fermi-Dirac statistics(?). (They reference an original paper by G. Feinberg which unfortunately I am not wealthy enough to download). The claim about Fermi-Dirac statistics is baffling---isn't the Higgs field a boson? Does anyone understand what they're talking about?&lt;/p&gt;&#xA;" OwnerUserId="1856" LastActivityDate="2011-02-21T15:54:10.660" Title="Spinning Tachyons" Tags="&lt;quantum-field-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="5598" PostTypeId="2" ParentId="5561" CreationDate="2011-02-21T10:28:52.477" Score="0" ViewCount="" Body="&lt;p&gt;There are cooler boxes . Have a look at &lt;a href=&quot;http://www.minicoolers.co.uk/&quot; rel=&quot;nofollow&quot;&gt;http://www.minicoolers.co.uk/&lt;/a&gt; . I had to make do with one such when my fridge went on the blink till it was repaired. They are as good in keeping food as ice boxes used to be. They just use the normal mains. If you have no electricity you could get one without  used for picnics and buy ice every day to keep things cool.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-02-21T10:28:52.477" />
  <row Id="5599" PostTypeId="1" AcceptedAnswerId="5603" CreationDate="2011-02-21T11:26:45.137" Score="1" ViewCount="136" Body="&lt;p&gt;&lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/1/1f/Feynmann_Diagram_Gluon_Radiation.svg&quot; rel=&quot;nofollow&quot;&gt;http://upload.wikimedia.org/wikipedia/commons/1/1f/Feynmann_Diagram_Gluon_Radiation.svg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the arrow of the positron doesn't show upwards?&lt;/p&gt;&#xA;" OwnerUserId="534" LastActivityDate="2011-02-21T11:42:35.553" Title="feynman diagram - direction of arrows" Tags="&lt;particle-physics&gt;" AnswerCount="1" />
  <row Id="5600" PostTypeId="1" CreationDate="2011-02-21T11:27:34.707" Score="4" ViewCount="311" Body="&lt;p&gt;For example for an isolated system the energy $E$ is conserved. But then any function of energy, (like $E^2,\sin E,\frac{ln|E|}{E^{42}}$ e.t.c.) &#xA;is conserved too. Therefore one can make up infinitely many conserved quantities just by using the conservation of energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why then one can usually hear of &quot;system having N constants of motion&quot;? &quot;System having only one constant of motion&quot;?  &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-03-02T07:21:13.320" Title="What does it mean, when one says that system has N constants of motion?" Tags="&lt;classical-mechanics&gt;&lt;conservation-laws&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="2" />
  <row Id="5601" PostTypeId="1" CreationDate="2011-02-21T11:27:53.670" Score="3" ViewCount="123" Body="&lt;p&gt;How do objects dock in space?&#xA;There's a recent piece of news that Keppler would dock to the ISS sometime this week. I want to know the operational aspects of it - considering the velocity of the objects involved and also the lack of references.&lt;/p&gt;&#xA;" OwnerUserId="2140" LastActivityDate="2011-02-21T19:28:15.527" Title="Object docking in Space" Tags="&lt;space&gt;&lt;motion&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="5602" PostTypeId="2" ParentId="5600" CreationDate="2011-02-21T11:36:17.193" Score="1" ViewCount="" Body="&lt;p&gt;Maybe it is worth distinguishing &lt;strong&gt;independent&lt;/strong&gt; integrals of motion from &lt;strong&gt;functions&lt;/strong&gt; of integrals of motion? See equations (2) and (3) in my &lt;a href=&quot;http://physics.stackexchange.com/questions/4973/do-symmetries-increase-the-number-of-conserved-quantities&quot;&gt;&lt;strong&gt;post&lt;/strong&gt;&lt;/a&gt;. The number of independent constants of motion is limited with the number of &lt;a href=&quot;http://physics.stackexchange.com/questions/4959/can-noethers-theorem-be-understood-intuitively/4960#4960&quot;&gt;independent&lt;/a&gt; initial data (if the problem is well posed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: another way to understand it is to count independent degrees of freedom of the system.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-22T10:57:17.247" LastActivityDate="2011-02-22T10:57:17.247" />
  <row Id="5603" PostTypeId="2" ParentId="5599" CreationDate="2011-02-21T11:42:35.553" Score="5" ViewCount="" Body="&lt;p&gt;The arrow is related to the flow of charge, not the direction of motion.  So for antimatter they go the opposite way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or as some are fond of saying, the positron is like an electron moving backwards in time.&lt;/p&gt;&#xA;" OwnerUserId="11" LastActivityDate="2011-02-21T11:42:35.553" CommentCount="2" />
  <row Id="5604" PostTypeId="1" AcceptedAnswerId="5625" CreationDate="2011-02-21T12:10:15.210" Score="3" ViewCount="231" Body="&lt;p&gt;I'd like to be able to determine the angular acceleration of a system of two rotating masses, which are connected so as to have a variable mechanical advantage between the two. My background with mechanics went as far as a course in statics, so I'm not sure how to proceed with this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have a single mass of some shape and apply a torque to it, I know that the angular acceleration depends on the moment of inertia of that object.  But suppose I have a system of two objects, e.g., gears, and apply torque to one of them, and want to know the angular acceleration. I'd assuming that the effective moment of inertia, at the point where I apply the torque, is the moment of the directly driven mass, plus the moment of the secondary mass multiplied by the mechanical advantage between the gears, and that using this 'effective moment of inertia' with the input torque would tell me how fast the input mass accelerates. (The acceleration of the 2nd mass being implied, as there's only 1 dof here) Not sure if this general approach is even correct, and then there's the real problem . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Introducing the variable mechanical advantage is what's giving me trouble here. If I take the contribution to the effective moment of inertia from the 2nd mass as just its 'intrinsic' moment times the mechanical advantage at any given instant, am I missing something? Calculus instincts tell me that there ought to be a contributing term from the rate-of-change of the mechanical advantage, too.&lt;/p&gt;&#xA;" OwnerUserId="717" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-14T10:14:03.973" LastActivityDate="2011-07-14T17:12:17.447" Title="Dynamics of moment of inertia" Tags="&lt;classical-mechanics&gt;&lt;angular-momentum&gt;&lt;torque&gt;&lt;angular-velocity&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" />
  <row Id="5606" PostTypeId="2" ParentId="5597" CreationDate="2011-02-21T13:11:54.930" Score="2" ViewCount="" Body="&lt;p&gt;It's not completely self-evident but it is true that in consistent theories, tachyons have to be scalar particles - much like the Higgs boson when expanded around the maximum of the potential (zero vev) - which obey, of course, Bose-Einstein statistics. (The claim about Fermi-Dirac is just wrong, or was meant to apply to Faddeev-Popov ghosts or similar fields, not physical tachyons.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In non-interacting string theory, one sees that this conclusion is true because the ground state energy of a single-string Hilbert space is equal to $L_0=-1$ for the scalar tachyon, so any addition of spin - through the string oscillators - raises $L_0$ at least by one, bringing us to the massless or massive level (non-negative $m^2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L_0=-1$ was true for the bosonic string ground state; in the case of the superstring, using the RNS formalism, the ground state has $L_0=-1/2$ but we also have antiperiodic fermions that only raise $L_0$ by $1/2$: that's still enough to show that any addition of spin - which only comes via internal oscillators - brings us to the massless or massive level, above the tachyonic interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scalar character of the tachyon may also be seen in the effective field theory. Dirac or Weyl tachyons are impossible because the Dirac mass term&#xA;$$-m \bar \psi  \psi$$&#xA;has to be Hermitian. It implies that $m$ has to be real, which means that the particle is positively massive. A tachyonic fermion would need an imaginary $m$ but that would produce a non-Hermitian action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true for spin-one particles. Spin-one particles may only get their mass consistently by the Higgs mechanism: the relevant term arises from the covariant version of the kinetic terms for the Higgs fields:&#xA;$$D_\mu \phi^\dagger D^\mu \phi$$&#xA;Again, this has to be Hermitean, and if it is Hermitean, $\phi^\dagger \phi$ that is left if the Higgs field has a nonzero vev, is automatically positively definite, which produces the usual mass term&#xA;$$m^2 A_\mu A^\mu/2$&#xA;with a positive coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning a tachyonic vev, well, any vev of a tachyonic field that solves the equations of motion has to be Lorentz-breaking - because it is a non-constant function of spacetime. The spin of a tachyon would just add one more aspect to this story. But it is intuitively natural that the tachyons have to be scalars - the value of the tachyon away from the maximum of the potential measures &quot;how much the instability has already advanced&quot;, and this quantity is naturally a scalar.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-21T13:11:54.930" CommentCount="13" />
  <row Id="5607" PostTypeId="2" ParentId="5573" CreationDate="2011-02-21T14:23:38.887" Score="0" ViewCount="" Body="&lt;p&gt;Consult this document by Hans de Vries: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://physics-quest.org/Magnetism_from_ElectroStatics_and_SR.pdf&quot; rel=&quot;nofollow&quot;&gt;The simplest, and the full derivation of Magnetism as a Relativistic side efect of ElectroStatics&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He uses only &lt;strong&gt;Electrostactic field and the non-simultaneity to derive the Magnetic Field&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears that the &lt;strong&gt;Magnetic field is a side effect of movement of charges in the electric field&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we pretend that the magnetic field exists beyond a mere artifact then the following question must be answered: &#xA;How can the particles communicate information about its position and velocity so that, in the past, they can adjust the source of the electromagnetic energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) Hans de Vries has a very interesting online book (not yet finished) in his site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;added&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnetic forces are induced by the motion and, much like &lt;strong&gt;Coriolis&lt;/strong&gt; force, is not a real force.  ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;wikipedia ..the Coriolis effect is an &lt;strong&gt;apparent deflection&lt;/strong&gt; of moving objects when they are viewed from a rotating reference frame. (a moving frame induces forces, there is no need to be a rotating one)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ps: it is frustating to be down voted without a comment  :(  it´s an injustice......yes it is - Calimero.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-02-23T23:27:49.130" LastActivityDate="2011-02-23T23:27:49.130" />
  <row Id="5608" PostTypeId="2" ParentId="5601" CreationDate="2011-02-21T14:24:45.720" Score="3" ViewCount="" Body="&lt;p&gt;Well, I am pretty sure Kepler is not docking with ISS, because Kepler is orbiting Sun and ISS is in low Earth orbit. But in general, docking is performed over several days with slow adjustment of velocities so orbits match with final stage docking performed with small thrusters usually called linear RCS. What is important when docking is relative velocities which are usually ~ 10cm/s in final stage of docking. The fact that both vehicle are racing with 7km/s around Earth or 30 km/s around the Sun or 200km/s around the Galaxy OR 600km/s vrt. CMB is quite irrelevant. Also, it may be interesting to note that docking spaceships is quite fun. Do try googling for &quot;orbiter&quot; which is a simulator of newtonian spaceship. There, you can launch your space shuttle and dock with ISS. True fun for any real geek :)  &lt;/p&gt;&#xA;" OwnerUserId="1600" LastEditorUserId="1600" LastEditorDisplayName="" LastEditDate="2011-02-21T15:36:16.520" LastActivityDate="2011-02-21T15:36:16.520" CommentCount="2" />
  <row Id="5609" PostTypeId="2" ParentId="5524" CreationDate="2011-02-21T14:31:52.433" Score="1" ViewCount="" Body="&lt;p&gt;Just add a few points to the other answers. There are two immediate topics that are being conflated in the question. For me the ontology of QM has not been resolved and so I can only discuss some points from my current perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly QFT deals with relativistic fields and so is unlikely to be associated with any Special Relativity paradoxes or oddities. By contrast the EPR experiment was about QM, which &lt;em&gt;is&lt;/em&gt; expressed in a Newtonian manner with phrases like &quot;instantaneous collapse&quot; which are at odds with Special Relativity - whether &quot;collapse&quot; is physical or not the word &quot;instantaneous&quot; should not also be used. Of course it sometimes is because QM is a Newtonian theory - you need QFT for the relativistic aspects!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As other answers have suggested this might be resolved if the &quot;collapse&quot; is not physical. Related to this is the issue as to whether $\Psi(x)$ is defined over physical space (in a field like manner as presented in elementary QM) or whether it should be taken as a member of an abstract Hilbert space only (and thus not subject to spacelike processes). Alternatively a conclusion could be that &quot;collapse&quot; &lt;em&gt;is&lt;/em&gt; physical - but somehow local. GRW (and its variants) is such a physical collapse theory. For such a theory your example then becomes a valid gedanken experiment for that model of collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second point is that one needs to separate the issue of &quot;collapse&quot; - where/when/if and how - from the correlation issues in this EPR experiment. The story of EPR and Bell's theorem is still ongoing. See &lt;a href=&quot;http://arxiv.org/PS_cache/quant-ph/pdf/0703/0703179v3.pdf&quot; rel=&quot;nofollow&quot;&gt; Disproof of Bells Theorem&lt;/a&gt; if you wish to pursue that aspect further.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-21T14:31:52.433" />
  <row Id="5610" PostTypeId="2" ParentId="5597" CreationDate="2011-02-21T15:12:05.440" Score="0" ViewCount="" Body="&lt;p&gt;If the polarizations of a tachyonic vector boson are Lorentz covariant, then its norm has to be indefinite, not positive definite, or positive semidefinite as in the case of massless vector bosons. This is true despite the fact that the polarizations have to be transverse to the 4-momentum because the 4-momentum is spacelike. Negative probabilities don't make any sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tachyonic scalars have to be bosons, not fermions!&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-02-21T15:54:10.660" LastActivityDate="2011-02-21T15:54:10.660" CommentCount="2" />
  <row Id="5611" PostTypeId="2" ParentId="5591" CreationDate="2011-02-21T15:17:05.180" Score="0" ViewCount="" Body="&lt;p&gt;I assume you're referring to condensed matter materials. The continuum limit is only an approximation which breaks down at the atomic/molecular scale. However, for materials many many orders of magnitude larger than the interatomic scale, we may use the continuum limit approximation if the system is in quasiequilibrium thermodynamically. &lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-21T15:17:05.180" />
  <row Id="5612" PostTypeId="2" ParentId="5584" CreationDate="2011-02-21T15:19:58.267" Score="0" ViewCount="" Body="&lt;p&gt;Well, we have to answer another question first: &quot;Is there any complete or consistent theory of quantum gravity which doesn't arise from string theory?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Constructing RS models and ADD models is &quot;phenomenological model-building&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-21T15:19:58.267" />
  <row Id="5613" PostTypeId="1" CreationDate="2011-02-21T15:31:34.783" Score="3" ViewCount="85" Body="&lt;p&gt;I've been looking for questions about dark matter, and I've read some very interesting answers. However, I desire too look into it deeply. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not actually a question. I'm asking the community to recommend interesting references to understanding dark matter and dark energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I accept all sort of references: notes, books, scientific papers etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume some background on classical physics, thermodynamics and basics about quantum theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance :)&lt;/p&gt;&#xA;" OwnerUserId="2145" LastActivityDate="2011-02-22T13:40:21.987" Title="Dark matter references" Tags="&lt;dark-matter&gt;&lt;dark-energy&gt;&lt;reference-request&gt;" AnswerCount="1" />
  <row Id="5614" PostTypeId="1" AcceptedAnswerId="5621" CreationDate="2011-02-21T15:41:08.357" Score="4" ViewCount="250" Body="&lt;p&gt;Can you suggest some references for rigorous treatment of thermodynamics? I want things like reversibility, equilibrium to be clearly defined in terms of the basic assumptions of the framework.&lt;/p&gt;&#xA;" OwnerUserId="2138" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-22T10:20:31.423" LastActivityDate="2011-02-22T10:20:31.423" Title="References about rigorous thermodynamics" Tags="&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="3" />
  <row Id="5615" PostTypeId="1" CreationDate="2011-02-21T15:46:03.683" Score="4" ViewCount="428" Body="&lt;p&gt;Let's say we have one black hole that formed through the collapse of hydrogen gas and another that formed through the collapse of anti-hydrogen gas. What happens when they collide? Do they (1) coalesce into a single black hole or do they (2) &quot;annihilate&quot; into radiation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One would expect (1) to be the case if the No Hair Theorem were to hold. So I guess what I'm really asking for is a modern understanding of this theorem and its applicability given what we know today.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-07-29T02:44:16.630" Title="What happens when a black hole and an &quot;anti-black-hole&quot; collide?" Tags="&lt;general-relativity&gt;&lt;quantum-gravity&gt;&lt;black-holes&gt;" AnswerCount="4" CommentCount="12" FavoriteCount="2" />
  <row Id="5617" PostTypeId="2" ParentId="5601" CreationDate="2011-02-21T16:13:05.307" Score="2" ViewCount="" Body="&lt;p&gt;You might be amused to know that this is the subject of Buzz Aldrin's PhD thesis at MIT, &quot;&lt;a href=&quot;http://dspace.mit.edu/handle/1721.1/12652&quot; rel=&quot;nofollow&quot;&gt;Line-of-sight guidance techniques for manned orbital rendezvous&lt;/a&gt;&quot; (six years before he walked on the moon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a wikipedia page titled &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Space_rendezvous&quot; rel=&quot;nofollow&quot;&gt;space rendezvous&lt;/a&gt;&quot; which will probably lead you to what you are looking for; or perhaps consider the book &quot;&lt;a href=&quot;http://books.google.com/books?id=v_vfn4382YMC&amp;amp;lpg=PA59&amp;amp;ots=me2LdJnfWo&amp;amp;dq=z-bar%20v-bar%20r-bar%20approach&amp;amp;pg=PP1#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Automated rendezvous and docking of spacecraft&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-02-21T19:28:15.527" LastActivityDate="2011-02-21T19:28:15.527" />
  <row Id="5618" PostTypeId="2" ParentId="5591" CreationDate="2011-02-21T16:24:21.390" Score="7" ViewCount="" Body="&lt;p&gt;What evidence do you have that there &lt;em&gt;is&lt;/em&gt; a consistent theory of continuum mechanics? Certainly, when looked at through a macroscope, the universe looks like it behaves according to continuum mechanics, but this completely breaks down on the microscopic level. So you can't justify a consistent theory of continuum mechanics by using the universe. There's no reason that good approximations to our universe have to be consistent theories of physics; physicists currently believe that QED is a very accurate approximation to the electromagnetic force, but that it cannot be made consistent at the smallest scales without adding additional physics. And there are papers showing that if you just define your physics using, say, the wave equation without putting some kind of restrictions on the initial conditions, &lt;a href=&quot;http://www.citeulike.org/user/ehainry/article/7755517&quot; rel=&quot;nofollow&quot;&gt;very funny things can happen&lt;/a&gt;. Also, Newton's laws with point particles have some very unpleasant consequences (I'm not going to look for this reference -- feel free to supply it in the comments). So I would say that, unless you're very clever about how you specify it, a theory of continuum mechanics would very likely &lt;em&gt;not&lt;/em&gt; be consistent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are very clever, you might be able to make a consistent theory of continuum mechanics, but I don't know of anybody who has actually done this.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditorDisplayName="" LastEditDate="2011-02-21T16:31:33.190" LastActivityDate="2011-02-21T16:31:33.190" CommentCount="3" />
  <row Id="5619" PostTypeId="2" ParentId="5579" CreationDate="2011-02-21T16:59:22.877" Score="1" ViewCount="" Body="&lt;p&gt;Well, one should maybe separate the idea of &quot;ether&quot;  from the vacuum of field theory. The vacuum of field theory is there and fields move over it with the appropriate  creation and annihilation operators, but it is a relative frame . So if somebody really wants to hang on to a &quot;medium&quot; picture one can have the vacuum. It just does not have the properties assigned to ether: a universal rest frame. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-02-21T16:59:22.877" />
  <row Id="5620" PostTypeId="2" ParentId="5600" CreationDate="2011-02-21T17:15:36.473" Score="-1" ViewCount="" Body="&lt;p&gt;A system having N constants of motion means having N fundamental, &lt;em&gt;mathematically&lt;/em&gt; derived quantities that are constant for the motion. $F = m dv/dt$ is &lt;em&gt;physically&lt;/em&gt; derived from experiment. You can create the mathematically derived quantity work by forming the scalar product of both sides with a differential displacement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F\cdot dl = m dv/dt \cdot dl = m v \cdot dv$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating over conservative forces, you get a constant of motion, kinetic energy + potential energy = constant. You can create another mathematically derived quantity impulse by multiplying both sides by dt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F dt = m dv/dt dt = m dv$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating over a system where all the forces are internal and action = reaction, you get another constant of motion, total momentum = constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look in any standard text book for the details of the above derivations, and Noether’s theorem which is another way of deriving constants of motion from physically derived quantities. The important point is they’re fundamental, mathematically derived constants of motion from physically derived quantities.&lt;/p&gt;&#xA;" OwnerUserId="1345" LastActivityDate="2011-02-21T17:15:36.473" />
  <row Id="5621" PostTypeId="2" ParentId="5614" CreationDate="2011-02-21T17:18:30.777" Score="7" ViewCount="" Body="&lt;p&gt;1) The pioneer of the rigorous treatment of thermodynamics is Constantin Carathéodory. His aticle (Carathéodory, C., Untersuchung über die Grundlagen der Thermodynamik, Math. Annalen&#xA;67, 355-386) is cited everywhere in this context, but probably you want some newer and more modern things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Buchdahl wrote a lot of papers about this subject in the 40's, 50's and 60's. He summarized these in the book:&#xA;H.A. Buchdahl, The Concepts of Classical Thermodynamics (Cambridge Monographs on Physics), 1966.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) There was a recent series of articles on this subject by Lieb and Yngavason which became famous. You can find the online version of these &lt;a href=&quot;http://arxiv.org/PS_cache/cond-mat/pdf/9708/9708200v2.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, &lt;a href=&quot;http://arxiv.org/PS_cache/math-ph/pdf/9805/9805005v1.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, &lt;a href=&quot;http://arxiv.org/PS_cache/math-ph/pdf/0003/0003028v1.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://arxiv.org/PS_cache/math-ph/pdf/0204/0204007v2.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; :).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Finally, I have come across the book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/9630581701&quot; rel=&quot;nofollow&quot;&gt;T. Matolcsi, &quot;Ordinary Thermodynamics&quot;&lt;/a&gt; (since a few friends of mine went to the class of the author), which treats thermodynamics in a mathematically very rigorous way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope some of these will help you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Greetings,&#xA;Zoltan&lt;/p&gt;&#xA;" OwnerUserId="1855" LastActivityDate="2011-02-21T17:18:30.777" CommentCount="2" />
  <row Id="5622" PostTypeId="2" ParentId="5601" CreationDate="2011-02-21T17:48:35.760" Score="0" ViewCount="" Body="&lt;p&gt;Orbiting mechanics are funny. If I am trailing a satellite in orbit, in order for me to catch up with it I need to slow down. Then I drop into a lower orbit with less orbital period which moves me ahead faster than my target. Then when the time is correct I speed up and jump back to the higher orbit where I can match speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So apply thrust to move back, and apply drag to move forwards. Makes sense!!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back in 1989 I wrote a computer program/game to do just that. Trying to catch up in orbit with something by adding or removing speed. It was loads of fun to try out different techniques for docking.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-02-21T17:48:35.760" />
  <row Id="5623" PostTypeId="2" ParentId="5615" CreationDate="2011-02-21T18:01:24.360" Score="2" ViewCount="" Body="&lt;p&gt;I suspect there is a bit of a difficulty here.  A particle $p$ and its anti-particle $\bar p$ cancel each other, and if they have a mass $m$ this results in the production of massless bosons (photons). Quantum number which identify the particle $p$ are subtracted away by opposite quantum numbers of $\bar p$.  These quantum numbers usually are the lepton or baryon numbers, electric charge, isospin and so forth.  However, anti-matter does not have anti-mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dirac’s original idea was that the Klein-Gordon equation had a square-root according to spinors.  If the particle has a mass there are then two roots for the momentum, which define a surface in $\pm$ portions of the momentum light cone.  The negative portion of the cone defines the so called Dirac sea, where the mass of particles is negative.  All of these negative energy (mass) states are however completely filled up.  This is why this is called a “sea,” for they define a ground state which has no dynamics.  However, if you impart a package of energy to a state in the sea so its energy flips sign you can generate an anti-particle with positive mass-energy.  However, all the other quantum numbers are reversed, including the charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could then construct a black hole from a huge cloud of hydrogen and anti-hydrogen of equal mass.  Based on the final state of the black hole it is not possible to determine whether it was formed by hydrogen or anti-hydrogen.  So if you have two such black holes, one from $H$ and the other from $\bar H$ the two will coalesce into a larger black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Schwarzschild solution in its pure form has a past and future singularity, where the past singularity corresponds to a “white hole.”  The white hole is about the closest thing there is to an “anti-black hole.”  These do not exist in nature, or at least have not been identified astronomically.  They may play some role in the very early universe, but nature is such that there is an asymmetry (asymmetry in their occurrence etc) between the black hole and white hole.  However, the white hole does not have an “anti-mass.”&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-21T18:01:24.360" />
  <row Id="5624" PostTypeId="2" ParentId="5569" CreationDate="2011-02-21T18:13:48.980" Score="3" ViewCount="" Body="&lt;p&gt;Some static ones:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://vue-forums.uit.tufts.edu/posts/list/521.page&quot; rel=&quot;nofollow&quot;&gt;Zotero Firefox plugin for Visual Understanding Environment&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://cluster.cis.drexel.edu/~cchen/citespace/&quot; rel=&quot;nofollow&quot;&gt;CiteSpace&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://longair.net/blog/2009/10/21/thesis-visualization/&quot; rel=&quot;nofollow&quot;&gt;Thesis visualization&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are dynamical ones:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://papercube.peterbergstrom.com/&quot; rel=&quot;nofollow&quot;&gt;PaperCube by Peter Bergstrom&lt;/a&gt; (it's a pity the program is working only on a demo database, the sole arXiv database would be great)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://images.isiknowledge.com/WOK46/help/WOS/h_citation_map.html&quot; rel=&quot;nofollow&quot;&gt;Web of Science - Citation map&lt;/a&gt; (nice, but for seeing not exploring + only for WOS)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(I asked the same question on WebApps SE &lt;a href=&quot;http://webapps.stackexchange.com/questions/10058/visualization-of-citation-data&quot;&gt;http://webapps.stackexchange.com/questions/10058/visualization-of-citation-data&lt;/a&gt; but received no answers. The above is a part of question, as I am still not satisfied.)&lt;/p&gt;&#xA;" OwnerUserId="184" LastActivityDate="2011-02-21T18:13:48.980" />
  <row Id="5625" PostTypeId="2" ParentId="5604" CreationDate="2011-02-21T18:29:33.853" Score="6" ViewCount="" Body="&lt;p&gt;There is indeed a term involving the time derivative of the changing coupling between the masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let's derive the equation for a single mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = \frac{1}{2} I\, \dot{\theta}^2 - V(\theta)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial L}{\partial \dot{\theta}} = I\, \dot{\theta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial L}{\partial \theta} = -\frac{dV}{d\theta} = \tau$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau = \frac{d}{dt} \left( I\, \dot{\theta} \right) = I\, \ddot{\theta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows you that the angular acceleration is proportional to the torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, assume we have two masses. The driven mass has moment of inertia $I_1$ and angular velocity $\dot{\theta}$. The secondary mass has moment of inertia $I_2$ and angular velocity $\dot{\theta_2} = a(t)\, \dot{\theta}$, where $a(t)$ is the changing coupling (for example a changing belt position in a continuously variable transmission).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = \frac{1}{2} I_1\, \dot{\theta}^2 + \frac{1}{2} I_2\, a(t)^2\, \dot{\theta}^2 - V(\theta)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial L}{\partial \dot{\theta}} = \left( I_1 + a(t)^2 I_2 \right) \dot{\theta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial L}{\partial \theta} = -\frac{dV}{d\theta} = \tau$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau = \frac{d}{dt} \left( \left( I_1 + a(t)^2 I_2 \right) \dot{\theta} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau = \left( I_1 + a(t)^2 I_2 \right) \ddot{\theta} + 2 I_2\, a(t) \frac{da}{dt} \dot{\theta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last term, proportional to $a \dot{a} \dot{\theta}$, is the funny term you're looking for. It says that when the coupling is changing, you need to apply some torque just to keep the angular velocity $\dot{\theta}$ constant. Another way to think of it is that in the absence of external torque, $\dot{\theta}$ is no longer constant (as it was for the single mass), but instead $(I_1 + a(t)^2 I_2) \dot{\theta}$ is constant, because that's the real angular momentum.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2011-07-14T17:12:17.447" LastActivityDate="2011-07-14T17:12:17.447" CommentCount="6" />
  <row Id="5626" PostTypeId="2" ParentId="5614" CreationDate="2011-02-21T19:18:16.837" Score="2" ViewCount="" Body="&lt;p&gt;First I'll recommend the &lt;a href=&quot;http://books.google.com/books?id=58SFj9x5RMEC&amp;amp;printsec=frontcover&amp;amp;hl=en#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;book I usually promote&lt;/a&gt;. There are very nice description of thermodynamics from point of view of differential geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you can start from &lt;a href=&quot;http://sgrajeev.com/geometry-of-thermodynamics/&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; for similar treatment.&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-02-21T19:18:16.837" CommentCount="2" />
  <row Id="5627" PostTypeId="1" AcceptedAnswerId="5639" CreationDate="2011-02-21T19:38:09.390" Score="1" ViewCount="161" Body="&lt;p&gt;In the center of our milky way it is assumed that a black hole exists with a mass of approx. 4 times 10^6 our sun's mass. How much light bending (in degrees) would arise for stars that are in perspective near the position of this black hole in the sky?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could the mass of a black holes become so great that it actually distorts the area in the sky we see around it (through telescopes)?&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2011-02-22T14:19:31.770" Title="Light bending by black holes" Tags="&lt;black-holes&gt;&lt;light&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="5628" PostTypeId="1" CreationDate="2011-02-21T19:48:53.537" Score="1" ViewCount="97" Body="&lt;p&gt;Here in this picture &#xA;&lt;a href=&quot;http://de.wikipedia.org/w/index.php?title=Datei:Skineffect_reason.svg&amp;amp;filetimestamp=20080727190721&quot; rel=&quot;nofollow&quot;&gt;http://de.wikipedia.org/w/index.php?title=Datei:Skineffect_reason.svg&amp;amp;filetimestamp=20080727190721&lt;/a&gt;&#xA;you can see I_W which is induced by H.&#xA;But why I_W is not vice versa? &#xA;Because of&#xA;$$rot \, \vec B = \mu_0 \, \left( \varepsilon_0 \frac{\partial E}{\partial t} + \vec j \right)$$ ?&#xA;Maybe I have to use&#xA;$$\oint E \, dr = - \int \frac{\partial B}{\partial t} \, dA = U_{induced}$$&lt;/p&gt;&#xA;" OwnerUserId="534" LastEditorUserId="534" LastEditorDisplayName="" LastEditDate="2011-02-21T21:30:23.037" LastActivityDate="2011-02-21T21:30:23.037" Title="Skin effect and currents" Tags="&lt;electromagnetism&gt;" CommentCount="4" />
  <row Id="5629" PostTypeId="2" ParentId="5470" CreationDate="2011-02-21T19:54:30.923" Score="0" ViewCount="" Body="&lt;p&gt;by Hans de Vries: a numerical elegant approach can be found here: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics-quest.org/fine_structure_constant.pdf&quot; rel=&quot;nofollow&quot;&gt;The ﬁne structure constant: A radiative series leading to it’s exact value&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vries use both 'pi' and 'e' in the formula and reminds me Feynman (wikipedia - Fine-structure_constant ) '...is it related to pi or perhaps to the base of natural logarithms? ...'&#xA;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know the meaning of this serie.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-02-21T19:54:30.923" />
  <row Id="5630" PostTypeId="1" AcceptedAnswerId="5635" CreationDate="2011-02-21T20:35:09.590" Score="3" ViewCount="101" Body="&lt;p&gt;Are there any open source tools for calculating the normal modes of arbitrary objects? Input would be some 3D design format and material properties, output would be movement direction and magnitude for each vertex along with frequency of each normal mode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commercial software such as SYSNOISE does this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Failing that, can anybody provide advice/references on creating such a tool? (I'll publish it, I promise :p)&lt;/p&gt;&#xA;" OwnerUserId="2150" LastActivityDate="2011-02-21T23:28:32.203" Title="Free/open source tools for calculating normal modes, failing that, advice" Tags="&lt;sound&gt;&lt;resonance&gt;" AnswerCount="1" />
  <row Id="5631" PostTypeId="2" ParentId="5604" CreationDate="2011-02-21T20:43:22.340" Score="1" ViewCount="" Body="&lt;p&gt;Consider two rotating masses (&lt;code&gt;1&lt;/code&gt;) &amp;amp; (&lt;code&gt;2&lt;/code&gt;) with a torque $\tau$ applied on (&lt;code&gt;1&lt;/code&gt;) only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you define some sort of coupling between the two, with resulting angular velocities $\omega_2 = \gamma \omega_1 $ then since the power is conserved in the coupling then the two torques through are $T_2 = \frac{1}{\gamma} T_1$ such that the product $T_1 \omega_1 = T_2 \omega_2 $ is the same on both ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiating the angular velocities yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \alpha_2 = \gamma \alpha_1 + \dot{\gamma} \omega_1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of the torques in mass (&lt;code&gt;1&lt;/code&gt;) is $$\tau - T_1 = I_1 \alpha_1$$, and for mass (&lt;code&gt;2&lt;/code&gt;) $$T_2 = I_2 \alpha_2$$. Substituting the above into the second equation yields the reaction torque $T_1 = \gamma I_2 (\gamma \alpha_1 + \dot{\gamma} \omega_1 ) $ and so the grand result is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \alpha_1 = \frac{ \tau - I_2 \omega_2 \dot{\gamma} } {I_1 + \gamma^2 I_2 } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the mechanical advantage counts twice (power of 2), once from the motion, and once from the torque amplification to yield the &lt;em&gt;effective&lt;/em&gt; mass moment of inertia $I_{eff} = I_1 + \gamma^2 I_2 $ (ignoring velocity effects).&lt;/p&gt;&#xA;" OwnerUserId="392" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2011-02-21T21:22:21.153" LastActivityDate="2011-02-21T21:22:21.153" CommentCount="4" />
  <row Id="5633" PostTypeId="2" ParentId="5627" CreationDate="2011-02-21T21:19:58.120" Score="1" ViewCount="" Body="&lt;p&gt;Famously, stars near the sun are deflected 1.75 arc seconds, as predicted by Einstein.&lt;br&gt;&#xA;I calculate that stars on the other side of the galaxy whose light passed very near the galaxy center would be coincidentally be deflected by about the same amount.&lt;br&gt;&#xA;   If you observe in the submillimeter band you could see it, just as Doeleman sees the black hole itself.&lt;br&gt;&#xA;But it would take far more than a lifetime for anything to pass behind the black hole at the center of the galaxy.&lt;br&gt;&#xA;The apparent diameter of the Schwarzschild radius is about 40 microarcseconds, or around 10^-10 radians.&#xA;This is how small you would have to see to detect those stars whose deflection angle is a radian or more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The apparent radius of the Einstein ring in radians is approximately the square root of this i.e. 10^-5 radians or about 2 arc seconds, very similar to Einstein’s number, but this is just a coincidence.&lt;br&gt;&#xA;Look up the Einstein ring formula in wikipedia and you can do a precise calculation if you want.&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-02-21T21:19:58.120" />
  <row Id="5634" PostTypeId="2" ParentId="5614" CreationDate="2011-02-21T22:23:28.573" Score="1" ViewCount="" Body="&lt;p&gt;As some people point out, it depends on what you mean by &quot;rigorous&quot;. If you mean rigorous, in the sense of a mathematician the books by David Ruelle are very nice.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://books.google.com/books?id=mNtWfcj_xhwC&amp;amp;printsec=frontcover&amp;amp;hl=da&amp;amp;source=gbs_ge_summary_r&amp;amp;cad=0#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Statistical mechanics: rigorous results by David Ruelle&lt;/a&gt;.&#xA;This is actually a very nice books, which takes the statistical mechanics point of view (instead of a phenomenological/geometric approach). Everything is defined very rigorously, the existence of the thermodynamic limit is proven in both classical/quantum and lattice/continuous systems (for well-behaved interactions) and the equivalence of different ensembles shown. Also phase transitions are covered where very general versions of the Lee-Yang and the Mermin-Wagner theorems are proven. And much more. The book can be rather dry and formal, so make sure to find the old papers by Michael Fisher which can be helpful.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://books.google.com/books?id=eODJauPWIK4C&amp;amp;printsec=frontcover&amp;amp;hl=da&amp;amp;source=gbs_ge_summary_r&amp;amp;cad=0#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Thermodynamic formalism: the mathematical structures of equilibrium statistical mechanics by David Ruelle.&lt;/a&gt; I don't know this book very well, but it is also a very rigorous (and more advanced than 1.) treatment of thermodynamics.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hope you will find this useful.&lt;/p&gt;&#xA;" OwnerUserId="1469" LastActivityDate="2011-02-21T22:23:28.573" CommentCount="1" />
  <row Id="5635" PostTypeId="2" ParentId="5630" CreationDate="2011-02-21T22:47:23.053" Score="4" ViewCount="" Body="&lt;p&gt;Any physics-oriented FEM solver should do this. I have only done it with COMSOL, which is proprietary and expensive, but searching Ubuntu's repository of free software turns up at least two promising candidates: Elmer and FreeFEM. I'm trying out Elmer now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.csc.fi/english/pages/elmer&quot; rel=&quot;nofollow&quot;&gt;http://www.csc.fi/english/pages/elmer&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Elmer_FEM_solver&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Elmer_FEM_solver&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This example seems to be what you want: &lt;a href=&quot;http://www.csc.fi/english/pages/elmer/examples/eigenmodes&quot; rel=&quot;nofollow&quot;&gt;http://www.csc.fi/english/pages/elmer/examples/eigenmodes&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditorDisplayName="" LastEditDate="2011-02-21T23:28:32.203" LastActivityDate="2011-02-21T23:28:32.203" CommentCount="1" />
  <row Id="5637" PostTypeId="2" ParentId="5579" CreationDate="2011-02-21T23:41:16.887" Score="1" ViewCount="" Body="&lt;p&gt;Ahem, &amp;ldquo;conclusive disproof&amp;rdquo; is &lt;em&gt;perhaps&lt;/em&gt; too strong. I think it's worth pointing out that the reason I wouldn't say you've asked a &quot;good question&quot; is that your question seems not to have focused on the general understanding amongst Physicists that Michelson-Morlety &lt;em&gt;was&lt;/em&gt; conclusive. The conventional position is that there is no aether. To question this is possible, but it has to be done with care, and, significantly, no-one has managed to make an argument for why it might be &lt;em&gt;helpful&lt;/em&gt; when doing Physics to think and do mathematics in terms of an aether. One has to say, for example, that to be compatible with experiment the equations of motion of the aether and of everything else must be Lorentz invariant to quite a high degree of accuracy. That would seem awkward to classical physicists of the late 19th Century, but the advent of liquid Helium experiments makes such an idea seem at least &lt;em&gt;somewhat&lt;/em&gt; more commonplace. Grigori Volovich, for example, has constructed a fairly extended discussion of this analogy (published by OUP, a final draft is available &lt;a href=&quot;http://ltl.tkk.fi/wiki/Grigori_Volovik&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;). [I haven't looked at the book cited by Gordon, but his description makes it sound comparable, albeit not similar.] Harvey Brown has a fairly detailed discussion that takes a comparable point of view from the philosophical side in &quot;&lt;em&gt;Physical Relativity. Space-time structure from a dynamical perspective&lt;/em&gt; (Oxford University Press), 2005&quot;, although this is an entirely classical discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other problem is, as always, how to accommodate quantum mechanics, and quantum field theory in particular. That's much more difficult. The Helium example is certainly not enough to support talking about an analogous aether without far more care than anyone has so far managed to put together in one place (no-one has come very close at all).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your secondary question, &quot;Massive objects produce curvatures/gradients in space-time resulting in what we observe as gravity.. what is the equivalent explanation for magnetic/electric fields?&quot;, does point, loosely, to a mathematical relationship that is much more often given houseroom by Physicists. That is, GR can be constructed in terms of the mathematics of &lt;em&gt;connections&lt;/em&gt;, and so can electromagnetism. This response is, however, at a rather different level from the way in which you have phrased your primary question, and, as just noted, the details of quantum field theory &lt;em&gt;really&lt;/em&gt; gets in the way of any simple understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that said, with apologies for being so direct, &lt;em&gt;WELCOME&lt;/em&gt; to Physics SE! I've learned a lot that would have taken me much longer to learn in other contexts, and I hope you have as good an experience.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-02-21T23:41:16.887" CommentCount="2" />
  <row Id="5638" PostTypeId="2" ParentId="5440" CreationDate="2011-02-22T01:26:33.753" Score="0" ViewCount="" Body="&lt;p&gt;My guesses are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Much more data is needed to produce a low-noise bispectrum than power spectrum(?).&lt;/li&gt;&#xA;&lt;li&gt;The results are confusing to interpret (or at least unfamiliar). Part of this is due to the large amount of redundancy in the output.&lt;/li&gt;&#xA;&lt;li&gt;The sheer amount of data in the bispectrum can be huge.  A bispectrum with the same frequency resolution as a power spectrum requires the &lt;em&gt;square&lt;/em&gt; of the amount of memory!&lt;/li&gt;&#xA;&lt;li&gt;Any good clocks will appear to be bispectrally coherent.  Transients also create spurious bicoherence.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-02-22T01:26:33.753" />
  <row Id="5639" PostTypeId="2" ParentId="5627" CreationDate="2011-02-22T02:38:55.137" Score="3" ViewCount="" Body="&lt;p&gt;The general theory of gravitational lensing shows that a light ray which approaches within a radius $r~&amp;gt;&amp;gt;~2GM/c^2$ will be deflected approximately by an angle $\theta~=~GM/rc^2$.  In a more general setting the deflection of light is given by the Einstein angular radius&#xA;$$&#xD;&#xA;\theta_E~=~\sqrt{\frac{4GM}{c^2}\frac{d_{ls}}{d_ld_{s}}},&#xD;&#xA;$$&#xA;where $d_{ls},~d_l,~d_s$ are the angular diameters to the gravitational lens, the source and the distance between the gravitational lens and the source. For $d_{ls},~d_l,~d_s$ the angular diameters to the gravitational lens, the source and the distance between the gravitational lens and the source. The condition $d_s~=~d_l~+~d_{sl}$ obtains locally where cosmological frame dragging is small.  This theory is the weak gravitational lensing approximation, where the deflection of light is essentially a Newtonian result.  The distance relationships are determined by $\theta d_s~=~\beta d_s~+~\alpha&amp;#39; d_{ls}$, for the angles given in the figure.  The angle of deflection reduced angle of deflection $\alpha(\theta)~=~(d_{ls}/d_s)\alpha&amp;#39;(\theta)$ gives a relationship between the angles of importance $\alpha(\theta)~+~\beta~=~\theta$.  The background which is then distorted scales as the square root of the mass of the black hole or large gravitating body.  So at a fixed distance away the larger the steradian measure of distrotion is observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/RjndW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the position of a source $\vec x$, the propagation of light along the $z$ axis from this source then reduces the visual appearance of the object to $\vec\xi~=~(\xi_x,~\xi_y)$ along the axis of optical propagation.  The weak gravitational lensing of light then indicates that the deflection of the appearance of this object along the axis of optical propagation is given by the &#xA;$$&#xD;&#xA;\Delta{\vec\xi}~=~\nabla\Phi(\xi),&#xD;&#xA;$$&#xA;for $\xi$ the position of the image with the mass present and $\Phi(\xi)$ the gravitational potential.  The difference in the vector position of the image ${\vec\xi}_i~-~{\vec\xi}_s$ is the difference between the position with the mass present and without it being present.  The potential term obeys the Poisson equation so that&#xA;$$&#xD;&#xA;\nabla^2\Phi~=~2\frac{\Sigma(\vec\xi)}{\Sigma_c} &#xD;&#xA;$$&#xA;The integration over the direction of propagation then gives the mass density in the plane of the image, often called the surface mass density $\Sigma(\vec\xi)$.  The angle of deflection $\alpha$ is then determined by the Poisson equation and the potential as&#xA;$$&#xD;&#xA;{\vec\alpha}’(\vec\xi)~=~\frac{4G}{c^2}\int\frac{(\vec\xi~-~\vec\xi&amp;#39;)\Sigma(\vec\xi&amp;#39;)}{|\vec\xi~-~\vec\xi&amp;#39;|^2}d^2\xi&amp;#39;, &#xD;&#xA;$$&#xA;for $\Sigma(\vec\xi)$ a mass/area density distribution in the image.  The function $\Sigma(\vec\xi)$ plays the role of an index of refraction based upon a mass distribution, which for a thin lens will give the angle of deviation.  For a gravitational thin lens, a weak field that is very small compared to the optical path length, and $\Sigma(\vec\xi)$ is a constant.  The deflection angle is simply&#xA;$$&#xD;&#xA;\alpha(\xi)~=~\frac{4\pi G}{c^2}\frac{\Sigma(\xi) d_{ls}\xi}{d_s}&#xD;&#xA;$$&#xA;where for small angles $|\vec\xi|~=\xi~=~d_l\theta$ and&#xA;$$&#xD;&#xA;\alpha(\xi)~=~\frac{4\pi G\Sigma}{c^2}\frac{d_{ls}d_l}{d_s}~=~\frac{\Sigma}{\Sigma_c}\theta&#xD;&#xA;$$&#xA;for the critical mass density $\Sigma_c~=~(c^2/4\pi G)(d_s/d_{ls}d_l)$.  This is the minimal mass density which might be distributed in the area of an Einstein ring.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-02-22T02:46:32.123" LastActivityDate="2011-02-22T02:46:32.123" />
  <row Id="5640" PostTypeId="2" ParentId="5627" CreationDate="2011-02-22T04:02:22.637" Score="0" ViewCount="" Body="&lt;p&gt;For a black hole, light is deflected more and more without limit as a beam approaches the hole. There is no theoretical limit to the angle, but there is a practical limit as the images get closer together.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A sketch of the null geodesics forming&#xA;  the first two relativistic images on&#xA;  the primary image side of the optic&#xA;  axis. The figure on the left&#xA;  represents the first relativistic&#xA;  image, which forms when a photon loops&#xA;  around the black hole once. The figure&#xA;  on the right represents the second&#xA;  relativistic image which forms even&#xA;  closer to the black hole, after&#xA;  looping around twice. In reality, both&#xA;  images are very close together on the&#xA;  lens plane and very close to the black&#xA;  hole’s photon sphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See figure 1.4, page 39/18:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bin-Nun, Amitai Yisrael, &lt;em&gt;Gravitational Lensing with a Large Deflection Angle as a Probe of General Relativity and the Galactic&#xA;Center&lt;/em&gt; (2010). Publicly accessible Penn Dissertations.&#xA;&lt;a href=&quot;http://repository.upenn.edu/cgi/viewcontent.cgi?article=1352&amp;amp;context=edissertations&quot; rel=&quot;nofollow&quot;&gt;http://repository.upenn.edu/cgi/viewcontent.cgi?article=1352&amp;amp;context=edissertations&lt;/a&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/vcmHS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-02-22T14:19:31.770" LastActivityDate="2011-02-22T14:19:31.770" CommentCount="2" />
  <row Id="5641" PostTypeId="2" ParentId="5567" CreationDate="2011-02-22T06:19:12.950" Score="4" ViewCount="" Body="&lt;p&gt;The principle is the same in both pictures.  I'm not sure how to answer &quot;is this analogy significant?&quot;, since I don't think it's an analogy at all.  It's the same phenomena, as explained below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any time you have a mechanism for dissipation (a coupling between the &quot;system&quot; and a &quot;bath&quot;) that coupling mechanism will give rise to back-action of the bath the system (i.e. fluctuations/noise).  This is true whether the bath is in a thermal state or a vacuum state. Quantum optics can treat both cases; the usual fluctuation-dissipation formula only works in the limit that the bath is well approximated by a classical thermal state (i.e. kT &gt;&gt; the energy of 1 quanta of excitation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And just to clarify one point:&#xA;when you say &quot;the shot noise is said to arise from interference with the vacuum field&quot;, that's not always the whole story.  Noise can come from lots of places.  It can be in the original state of your optical field, in which case it will be observed even if you don't have lossy mirrors/detectors.  But the idea is that if you DO have loss, then even if you start with a sub-shot-noise beam, you'll pretty quickly approach the shot noise limit.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-02-22T06:19:12.950" CommentCount="2" />
  <row Id="5642" PostTypeId="1" CreationDate="2011-02-22T06:26:02.427" Score="1" ViewCount="171" Body="&lt;p&gt;What is meant by an &quot;inclined magnetic field&quot;?  How is it different from the usual magnetic field?&lt;/p&gt;&#xA;" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-11T20:09:25.827" LastActivityDate="2011-05-11T20:09:25.827" Title="What is an &quot;inclined magnetic field&quot;?" Tags="&lt;electromagnetism&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="5643" PostTypeId="1" CreationDate="2011-02-22T07:42:43.743" Score="4" ViewCount="228" Body="&lt;p&gt;According to $a = v^2/R$, the circular velocity and radial distance between two attracting objects (such as planets), must remain in perfect proportion in order for orbital motion to take place. How is it possible for objects in nature to achieve this proportion perfectly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only that, to maintain an orbit seems to be impossible. For example, assuming that the moon is orbiting the earth 'perfectly'. Let us say the moon is then hit by a series of meteorites. This would shift the balance slightly, and cause the moon's orbit to decay? Apparently not... How is it possible for the moon to remain in orbit for so long? &lt;/p&gt;&#xA;" OwnerUserId="2156" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-22T10:18:41.393" LastActivityDate="2011-02-22T11:58:16.653" Title="How is it possible for orbits to maintain stability?" Tags="&lt;gravity&gt;&lt;orbital-motion&gt;" AnswerCount="3" />
  <row Id="5644" PostTypeId="2" ParentId="5643" CreationDate="2011-02-22T08:15:44.987" Score="11" ViewCount="" Body="&lt;p&gt;If the velocity of a satellite differs from the right velocity of a circular orbit, Newton's equations imply that the object will simply move along a non-circular orbit, an ellipse. This fact as well as the detailed parameters of this ellipse were already known to Johannes Kepler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All planets and moons in the real world orbit around their stars or planets along ellipses and there is no fine-tuning here whatever. The deviation from a circular orbit is known as &quot;eccentricity&quot; of the ellipse and it is nonzero for all real celestial objects: none of them has a fine-tuned velocity. For any initial position or velocity, one finds an ellipse (which may be a circle if someone, e.g. NASA, fine-tunes the parameters) or a hyperbola or a parabola (if the speed exceeds the escape speed or is equal to it) and the object will move along it, in agreement with Newton's laws of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the elliptical trajectories of the 2-body system are stable (and the elliptical ones are periodic): a small perturbation of the initial state only leads to equally small perturbations of the final state. This proposition has to be modified for 3 bodies and larger numbers (chaotic behavior) and for nearby orbits around very heavy objects in general relativity that may be unstable. But in Newton's theory for 2 bodies, everything is easy.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-22T08:15:44.987" CommentCount="3" />
  <row Id="5645" PostTypeId="2" ParentId="5643" CreationDate="2011-02-22T08:17:01.820" Score="4" ViewCount="" Body="&lt;p&gt;That is the conditions for &lt;em&gt;circular&lt;/em&gt; orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbits have no trouble existing in non-circular (elliptical) varieties. In this case the velocity and radius vary in such a way as to keep the angular momentum constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find the angular momentum at a particular point in time as $L = m * v_t$ where $v_t$ is the velocity &lt;em&gt;transverse&lt;/em&gt; to the line connecting the two bodies. From the considerations you stated for circular orbits you should be able to deduce when the orbit is heading outward and when inward.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-02-22T08:17:01.820" CommentCount="1" />
  <row Id="5646" PostTypeId="2" ParentId="5642" CreationDate="2011-02-22T08:23:01.973" Score="3" ViewCount="" Body="&lt;p&gt;There is no special significance attached to the term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without seeing the context we can't know for certain, but &quot;inclined&quot; almost must refer to the orientation of the field (with respect to what will depend on the problem, but quite possible the horizontal or vertical axis of the system).&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-02-22T08:23:01.973" />
  <row Id="5647" PostTypeId="2" ParentId="5642" CreationDate="2011-02-22T08:24:55.123" Score="5" ViewCount="" Body="&lt;p&gt;An inclined magnetic field is just a magnetic field with a nonzero inclination. Inclination is the angle between the direction of the vector $\vec B$ and another, preferred direction in the problem. In astrophysics, they usually mean the radial direction. By the radial direction, I mean the normal (perpendicular) direction to the surface (of the Sun etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So non-inclined magnetic field is radial; an inclined magnetic field has field lines that are twisted around the Sun (or another object) a little bit. To summarize, the adjective &quot;inclined&quot; doesn't carry any special physical content - it just refers to the geometrical arrangement of the objects relatively to the field.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-02T06:49:11.483" LastActivityDate="2011-05-02T06:49:11.483" />
  <row Id="5648" PostTypeId="1" CreationDate="2011-02-22T08:31:06.617" Score="5" ViewCount="212" Body="&lt;p&gt;It is well known that the Lagrangian of a classical free particle equal to kinetic energy. This statement can be derived from some basic assumptions about the symmetries of the space-time. Is there any similar reasoning (eg. symmetry based or geometrical) why the Lagrangian of a classical system is equal kinetic energy minus the potential energy? Or it is just because we can compare the Newton's equations with the Euler-Lagrange equation and realize how they can match?     &lt;/p&gt;&#xA;" OwnerUserId="1795" LastActivityDate="2011-02-22T14:41:43.133" Title="How the Lagrangian of classical system can be derived from basic assumptions?" Tags="&lt;classical-mechanics&gt;&lt;lagrangian-mechanics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="5649" PostTypeId="2" ParentId="5648" CreationDate="2011-02-22T08:59:00.520" Score="6" ViewCount="" Body="&lt;p&gt;Well, Lagrangian is what one usually starts with, even in the modern physics, and everything else is derived. So I don't believe you can explain it any better than it is the way it is because it matches observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are two points about the above worth taking.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First &lt;strong&gt;symmetries&lt;/strong&gt; can give you a nudge into the right direction of constructing Lagrangian. E.g. in classical mechanics in free space you want to have laws that are translation and rotation invariant, etc., so that excludes lots of strange Lagrangians first-hand. Nevertheless, you could also ask what is the symmetry derived from and for that there is no good answer. At some point you have to stop with the chain of deductions and realize that physics can't really be derived mathematically but has to be observed. It's actually fascinating that the world can be reduced into few simple laws that appear to be purely mathematical.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Second, there can sometimes be a microscopic model of the given system which kind of &quot;explains&quot; why the macroscopic description takes the given form in a limit. One can e.g. derive the principle of least action from quantum mechanics. So you can explain some things. But I don't believe you can really explain Lagrangian.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-02-22T08:59:00.520" />
  <row Id="5650" PostTypeId="1" AcceptedAnswerId="5688" CreationDate="2011-02-22T09:21:18.177" Score="10" ViewCount="446" Body="&lt;p&gt;I am trying to understand graphene as a topological insulator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spin orbital interaction in graphene is very small (~10mK?). But if we consider that, then graphene should be a topological insulator. And at the edge electrons should move in opposite direction according to their spin as shown in the figure for graphene flake with zigzag edges! &lt;img src=&quot;http://i.stack.imgur.com/UM7sq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the map from Brillioun zone to the energy space should be nontrival. The energy space should be topologically equal to a Klein bottle instead of sphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it the right picture?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;update1: Thanks for the comments. Indeed the question is not well defined and I will try to refine it a bit as further as I could.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you glue the 2D Brillouin zone properly you will get a torus. And as I understood there are two cases shown in figure below that create a gap in the 'bulk' graphene. Case 1 show two valleys without gap. Case 2 is the normal insulator case. Case 3 is the TI case.&#xA;&lt;img src=&quot;http://i.stack.imgur.com/h044h.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the energy space I mean the band structure (I don't know its relation with Hilbert space.) Not sure about how to glue it and if a Klein bottle will be formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;update2: Just realized that last update without considering the spin. If we do that, then the momentum space should be 2 torus each for one spin instead of one torus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the problem is how to glue (topologically map) the band structure on that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;update3: Just got some ways of mapping 2 tori to a Klein bottle. (shown below, sorry for the bad drawing)&#xA;&lt;img src=&quot;http://i.stack.imgur.com/pMdoZ.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;The &lt;a href=&quot;http://www.scribd.com/doc/14009839/George-K-Francis-A-Topological-Picture-Book-1988-213p-0387964266&quot; rel=&quot;nofollow&quot;&gt;cover&lt;/a&gt; of the book shows how to map a torus to a sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the way to understand mapping from momentum space to band structure (or topology of the system)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;update4: Well, the next question is how this relates with Berry flux and Chern number. The band structure I mentioned above should be $&amp;lt;\Psi(k)|H|\Psi(k)&amp;gt;$. And here $H$ is proportional to $\nabla_k$. Berry flux ~ $\nabla_k\times &amp;lt;\Psi(k)|\nabla_k|\Psi(k)&amp;gt;$. With it one can calculate the Chern number. Is the right way to link those two pictures? How do this relate with critical points in 2D vector field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;update after 4tnemele's nice answer&lt;/strong&gt;&#xA;Thanks for the answer! I must say it is a very nice one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But may be it is still worth to clear why I asked this question. I am not in the TI field and have no strong background in condensed matter physics, algebra topology, differential geometry etc. Once I read that graphene is a TI. I just cannot &lt;strong&gt;see&lt;/strong&gt; how to construct the calculations of those topological invariants from what I knew at that time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have to start from somewhere and do it in 'my own way'. I intuitively think that the 'energy space' could be a 2D non-orientational close surface. The only thing come into view is a Klein bottle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought it could related with the picture below. If you glue fiber on a torus, you can have no critical point. But it shall be different for the non-orientational surface case (here I call it a Klein bottle.) I thought those fiber could be related with Berry phase. Then I reached the 'energy space', I knew it might be somewhere in between the links. I don't know how to imagine the shape of related Hilbert space.&#xA;&lt;img src=&quot;http://i.stack.imgur.com/S2dbP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hoped I can find a very intuitive way of 'seeing' how to construct the calculation. But now I think I need to read a bit more and it may be other easier ways (like begin from equations) to understand the problem. And I think that to clear the background of the question will help other people to answer it. It is a lesson to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot.&lt;/p&gt;&#xA;" OwnerUserId="1948" LastEditorUserId="1948" LastEditorDisplayName="" LastEditDate="2011-02-23T20:34:39.487" LastActivityDate="2011-02-23T20:34:39.487" Title="Graphene and Klein bottle?" Tags="&lt;solid-state-physics&gt;&lt;topological-insulators&gt;&lt;graphene&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="4" />
  <row Id="5651" PostTypeId="2" ParentId="5226" CreationDate="2011-02-22T09:32:30.470" Score="0" ViewCount="" Body="&lt;p&gt;This is not a direct answer to the question but rather a slightly different perspective on this adiabatic expansion. I am not sure how correct it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let us assume that the piston moves(in direction of $x$-axis) infinitely slowly with velocity $\vec{v}_p$. Let a molecule flies to the piston at a velocity $\vec{v}_k$. With respect to the piston, its velocity will be $\vec{v}_{k_{rel}}=\vec{v_k}-\vec{v_p}$. Normal component(relative piston) of relative velocity is $(v_{k_{rel}})_x=v_{kx}-v_p$. Let's denote by $\vec{v&amp;#39;}_{k_{rel}}$ the velocity of the molecule with respect to the piston, after reflection. Tangential component of relative velocity as a result of the reflection does not change, and the normal changes sign.  $$(v&amp;#39;_{k_{rel}})_x=-(v_{k_{rel}})_x=-v_{kx}+v_p$$ Let's denote by $v&amp;#39;_k$ the velocity of the molecule relative to the fixed cylinder walls after reflection. Its normal component is $v&amp;#39;_{kx}=(v&amp;#39;_{k_{rel}})_x+v_p=-v_{kx}+2v_p$ and the tangential component is the same as that of the velocity $\vec{v}_k$. As a result of the reflection from the piston, the&#xA;kinetic energy of the molecule is incremented:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{2}m(-v_{kx}+2v_p)^2-\frac{1}{2}mv_{kx}^2=-2mv_pv_{kx}+2mv_p^2$$ Let's denote by $n_k$ the number of molecules per unit volume whose velocities are approximately equal to $\vec{v}_k$. Number of hits these molecules on the piston during the time $dt$ is equal to $z_k=An_k(v_{kx}-v_p)dt$ where A is area of the piston. As a result, the kinetic energy of molecules in this group in time dt will increase:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(-2mv_pv_{kx}+2mv_p^2)An_k(v_{kx}-v_p)dt=-2mn_k(v_{kx}^2-v_p^2)dV$$ where $dV=Av_pdt$ is an increase of the volume of gas for the same time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The increment of the kinetic energy of the whole gas:$$dE_{kin}=dU=-dV\sum_{v_{kx}&gt;0}2mn_kv_{kx}^2+2dVv_p^2\sum_{v_{kx}&gt;0}mn_k$$ Here $U$ is the internal energy of an ideal gas. The summation is only for those groups of molecules, which move in the direction of the piston. When summarizing of all groups of molecules, moving like a piston, and from him, then the sum should be divided in half. In this case: $$dU=-dV\sum mn_kv_{kx}^2+dVv_p^2\sum mn_k$$ But by definition, the first sum is the pressure of the gas $P$ and the second sum is simply the density of the gas $\rho$. Thus we get a differential equation:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dU+PdV=\rho v_p^2dV$$ Internal energy of an ideal gas may be expressed as follows: $$U=\frac{f}{2}PV$$ where $f$ is a number of degrees of freedom of molecule. Using the fact that the adiabatic index is $\gamma=\frac{f+2}{f}$, the differential equation can be rewritten:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dP}{dV}+\gamma\frac{P}{V}=(\gamma-1)\frac{\rho}{V}v_p^2$$ If $v_p=0$ then we get from it the adiabatic equation: $PV^{\gamma}=const$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $\rho$ depends on the pressure and temperature it is impossible to integrate the differential equation directly. But for small shifts of the piston we can assume that the density is approximately constant, i think.&lt;/p&gt;&#xA;" OwnerUserId="552" LastActivityDate="2011-02-22T09:32:30.470" />
  <row Id="5652" PostTypeId="1" AcceptedAnswerId="5662" CreationDate="2011-02-22T10:13:19.140" Score="1" ViewCount="162" Body="&lt;p&gt;I have the following setup:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A &amp;lt;---- wire ----&amp;gt; B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;$V_b - V_a = \Delta v = \text{a positive value}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Which end of the wire has a higher potential?&lt;/li&gt;&#xA;&lt;li&gt;What is the direction of the electric field inside the wire?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2157" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-22T10:17:54.733" LastActivityDate="2011-02-22T16:12:13.217" Title="How can I tell which end of a wire will have a higher potential?" Tags="&lt;electromagnetism&gt;" AnswerCount="2" />
  <row Id="5654" PostTypeId="2" ParentId="5643" CreationDate="2011-02-22T11:58:16.653" Score="9" ViewCount="" Body="&lt;p&gt;Let's treat the case of one single particle going around a 1/r potential (i.e. Moon around Earth). In the rotating frame of the Moon's orbit, there is an effective potential given by: $$V(r) = -\frac{GMm}{r} + \frac{J^2}{2mr^2},$$ where $J$ is the angular momentum of the orbit (and is conserved). The problem then reduces to that of a single particle moving in a one-dimensional potential, which has a well-defined minimum: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/hgf88.gif&quot; alt=&quot;wolfram alpha plot&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus any small displacement is stable, and simply results in an oscillation of the orbital radius.&lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-02-22T11:58:16.653" CommentCount="3" />
  <row Id="5655" PostTypeId="1" AcceptedAnswerId="5656" CreationDate="2011-02-22T11:58:52.027" Score="3" ViewCount="178" Body="&lt;ul&gt;&#xA;&lt;li&gt;Let the $i=x,y,z$ components of the angular momentum have the commutation relations with the supersymmetry generators (also called supercharges?) $Q_a$ ($a = \pm \frac{1}{2}$) as, $[J_i , Q_a] = -\frac{1}{2} (\sigma _i )_{ab}Q_b$. Now if I want to calculate the $J^2$ eigenvalue of $Q_a$ then I might have thought of doing the following calculation using properties of Lie bracket, (using the summation convention)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$[J_i^2,Q_a] = J_i[J_i,Q_a] + [J_i,Q_a]J_i = -\frac{1}{2} (\sigma_i)_{ab}\{ J_i, Q_b \} $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is apparently the wrong answer! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right answer is gotten by doing the following calculation where by the action of $J^2$ on $Q_a$ is defined as (using the summation convention)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[J_i,[J_i,Q_a]] = [J_i,-\frac{1}{2} (\sigma _i )_{ab}Q_b] = -\frac{1}{2} (\sigma _i )_{ab} [J_i, Q_b] = \frac{1}{4} (\sigma _i)_{ab} (\sigma _i)_{bc} Q_c = \frac{3}{4}Q_a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know why is the above the correct action and not the former. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;But when lets say one wants to evaluate the action of a supercharge on the square of say the scalar component of a chiral superfield then the right action is of the former kind i.e,&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$[Q,\phi ^2] = \phi [Q,\phi] + [Q,\phi]\phi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand why here what was &quot;wrong&quot; earlier is now the right thing. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When the supercharge acts on the fermion say $\psi$ then in terms of the auxiliary field in the same multiplet (say $F$) the action as derived by Weinberg in his book is,&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$\{ Q_b, \psi_a^*\} =2i \delta _{ab}F^*$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people tell me that the above can't be right because the RHS does not have the same spinorial symmetry as the LHS. Also they say that the above implies that irrespective of the nature of $F$ the RHS will go to $0$ if $a\neq b$. But apparently there are theories with superpotential known where even if the spinor indices don't match the supercharge has non-trivial anti-commutation relation with the fermion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't see anything wrong about Weinberg's argument and nor do I see Weinberg using any assumption about the superpotential in deriving the above. I would like to know what is the right way to think about the above. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $A$ is the gauge field component of a vector superfield and the chiral superfield to which it couples has $\phi$ and $\psi$ as scalar and spinor components then the following seems to be true,&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$[Q,A] = [\phi, \psi]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know if there is a general argument for the above and if it is always true. And also what is the space in which the above commutators on the both sides are defined. &lt;/p&gt;&#xA;" OwnerUserId="1637" LastActivityDate="2011-02-22T16:25:19.213" Title="Supersymmetry algebra" Tags="&lt;quantum-field-theory&gt;&lt;supersymmetry&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="5656" PostTypeId="2" ParentId="5655" CreationDate="2011-02-22T12:35:24.660" Score="5" ViewCount="" Body="&lt;p&gt;The answer for the first question is that you are using wrong definition of the Casimir operator. Casimir operator is defined as $\sum_i X_i^2$ but the $X_i$ have to be understood as generators of the given representation. In your case $J_i \neq X_i$. That's because you are acting on operators, not just vectors. The generators of the action on operators are $X_i = [J_i, \cdot]$. This is the adjoint representation of the usual $J_i$ (which acts the way you suggested only on vectors). So you see that the correct definition of the Casimir operator in this case is the one with doubled commutator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I don't see what relation $[Q,\phi^2]$ has to do with the above. It's completely different computation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't say anything about the last two questions but they seem completely disconnected from the first two. I suggest you ask these things separately for the next time.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-02-22T12:35:24.660" CommentCount="12" />
  <row Id="5657" PostTypeId="2" ParentId="5655" CreationDate="2011-02-22T13:02:16.987" Score="2" ViewCount="" Body="&lt;p&gt;Dear Anirbit, the action of symmetry generators - such as $Q$ - on operators is always given by the commutator (or anticommutator, with both Grassmann-odd) of the symmetry generator with the operator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $Q$ and $J_i$ act on states as&#xA;$$|\psi\rangle \to Q|\psi\rangle, \quad |\psi\rangle \to J_i|\psi\rangle, \quad$$&#xA;but they act on operators $M$ as &#xA;$$M \to [Q,M]_{\pm}, \quad M \to [J_i,M]. \quad$$&#xA;When you ask how $J^2=J_i J_i$ acts on $Q$, you must appreciate that $Q$ is an operator. So the right factor of $J_i$ acts on it as a commutator. The result is $[J_i,Q_a]$ which is still an operator, so the left factor $J_i$ still acts on it as a commutator, so the result is a double commutator summed over $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[Q,\phi^2]$ may be interpreted as an action of symmetry generator $Q$ on the operator $\phi^2$. Note that in this case, it is the &quot;generic operator&quot; - the right player - that is squared while the symmetry generator, $Q$, is not squared. So no double commutators will occur in this case. After all, the identity for $[Q,\phi^2]$ that you wrote is nothing else than the Leibniz rule. The Leibniz rule is always right and it doesn't depend on any definitions &quot;what we mean by the transformation of something by something&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different questions have different answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second question. Weinberg is using a matrix-based formalism in which all 2-valued spinor indices are denoted with the same letter (no dotted letters), and are always written as subscripts. See page xx, Notation, of volume 3 of Weinberg's book. That's very different from the Penrose-like notation I would prefer in which one distinguishes not just two but four different types of spinor indices - lower and upper, dotted and undotted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Penrose's formalism, $\delta_{ab}$ doesn't exist as an invariant. The correct Penrose version of the equation you mention would either have one of the indices $a,b$ raised - so there would be $\delta^a_b$ or vice versa - or it would have $\epsilon_{ab}$ if both indices were kept as subscripts (raising and lowering of 2-valued indices is done by $\epsilon_{ab}$ which is antisymmetric). There is no contradiction when it comes to the antisymmetry in $ab$. In particular, it is not true that $\{Q_b,\psi^*_a\}$ is $ab$-symmetric because $Q$ is not the same letter as $\psi$, so by exchanging them, you don't get the same thing back!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your third question, the identity clearly misses some indices and summations over them - and you haven't mentioned that you may have used the equations of motion at some point (otherwise I have no clue how a completely different chiral superfield could occur in the SUSY variation of a vector superfield). I can't answer this question which is too schematic as stated but I suppose it appeared in your attempt to understand the D-terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, at the schematic level, the supercharge, when written as an operator in terms of fields (integral of its density), contains terms of the type $\int \psi \partial_0 A \phi$ from the covariant derivatives, and when commuted with $A$, $\partial_0 A$ cancels and we're left with the $\psi\phi$ terms - with a dependence on the color indices that was written as a commutator (it only works for the adjoint representation).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-22T16:25:19.213" LastActivityDate="2011-02-22T16:25:19.213" />
  <row Id="5658" PostTypeId="2" ParentId="5613" CreationDate="2011-02-22T13:40:21.987" Score="1" ViewCount="" Body="&lt;p&gt;found in the net:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark Energy : Theory and Observations, &#xA;   491 pages by Luca Amendola and Shinji Tsujikawa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark Sky, Dark matter,&#xA;   216 pages by J M Overduin ans P P Wesson&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does Dark Energy really exist ?,&#xA;   9 pages by Thimoty Clifton and Pedro G Ferreira&#xA;   (... explanation: that our galaxy lies at the center&#xA;of a giant cosmic void...)   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No DE, an artifact of our measuring (a Chimera) :&#xA;   &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0208365&quot; rel=&quot;nofollow&quot;&gt;A relativistic time variation of matter/space fits both local and cosmic data&lt;/a&gt;, &#xA;   14 pages by Alfredo Gouveia Oliveira and Rodrigo de Abreu&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good Luck   &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-02-22T13:40:21.987" CommentCount="3" />
  <row Id="5660" PostTypeId="2" ParentId="5648" CreationDate="2011-02-22T14:41:43.133" Score="0" ViewCount="" Body="&lt;p&gt;There is a nice article by Elisha Huggins, &lt;a href=&quot;http://dx.doi.org/10.1119/1.3502500&quot; rel=&quot;nofollow&quot;&gt;Gravity, Time, and Lagrangians&lt;/a&gt; (Phys. Teach. 48, pp. 512) remarking that, in the gravitational case, the negative sign between kinetic and potential energies in the Lagrangian comes from the fact that moving clocks run slower and higher altitude clocks run faster.  &lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2011-02-22T14:41:43.133" CommentCount="1" />
  <row Id="5661" PostTypeId="2" ParentId="5552" CreationDate="2011-02-22T15:18:52.693" Score="4" ViewCount="" Body="&lt;p&gt;Lets start from experimental data. What is an electron, what do we know about an electron? It is too small to touch or see or smell. Everything we know about an electron comes from several levels of proxies. We end up measuring a track circle in a magnetic field and get e/m, consistently for different &quot;electrons&quot; and we do the millikan oil drop  and get e and then we can assign a mass to these manifestations consistently. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is all we have for the electron, it has a mass m_e measured and a charge e, measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nature has been good to us and a working theory exists for QED. Mathematics is a tool, it can describe and predict measurements but it is not something that creates reality. Reality is what one measures . If the theory predicts, it does not matter if it goes into a yoga position to do so, as long as it can predict consistently. They want to call them bare and dressed mass? Fine. Who can measure anything more than that the measured mass is m_e and and the measured charge e? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better theories/computations  may come up, but to be better they should describe existing measurements and predict more and different ones, that QED cannot explain, for anybody to pay any attention. Or be as overwhelmingly economic and  elegant as the heliocentric is to the geocentric pov. QED works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;added&lt;/strong&gt;: I want to give an example from real physics history that I heard from the horse's mouth back in the 1980s, of how succesful new methods of computation overwhelm tradition and sweep over reluctances once shown to successfully predict faster and accurately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back in the Manhattan day project, a physicist think tank had been set up with the best brains of the time to calculate crossections needed for making the bomb. Feynman was a junior member of the team. They gave the group a problem and a week later people reported the result of their independent calculations, parallel processing. Feynman said that one afternoon he was lying on his bed with his feet on the wall, when the Feynman diagram method came to him, whole ( he had eidetic memory so he probably saw it). He calculated the current problem and waited impatiently for the report of the others. When he gained confidence that his method was as good as the long drawn out s matrix calculations he started playing games with the team. He would get the result in an evening, tell them the next day what they would find, and it would take them the rest of the week to confirm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course Feynman diagrams were universally accepted after that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reminded of this story when I listened to the &lt;a href=&quot;http://streamer.perimeterinstitute.ca/mediasite/viewer/NoPopupRedirector.aspx?peid=ff8f97a4-2848-4b93-98d9-1218de2070af&amp;amp;shouldResize=False&quot; rel=&quot;nofollow&quot;&gt;talk of Nima Arkani-Hamed&lt;/a&gt; which he gave on the &lt;em&gt;twistor revolution&lt;/em&gt;. He finds extremely cumbersome the Feynman diagrams method and is exploring a new one that gives the same results as the thousands of summed QCD feynman diagrams. I was amused, and am sure that Feynman would have been too, if he were still alive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a new computational method is faster, sleeker and as predictive, it will be adopted as surely as God made little cabbages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my experimentalist's opinion of course.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-02-22T18:19:05.153" LastActivityDate="2011-02-22T18:19:05.153" CommentCount="1" />
  <row Id="5662" PostTypeId="2" ParentId="5652" CreationDate="2011-02-22T15:21:49.853" Score="4" ViewCount="" Body="&lt;p&gt;From the given information that $V_b-V_a$ is positive, we know that $V_b&amp;gt;V_a$. The word &quot;potential&quot; means simply $V$, so point B is at a higher potential than point A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know that, you can go and remind yourself of the relationship between electric field and potential (no doubt it's in your textbook). That'll give you the answer to your second question.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-22T15:21:49.853" CommentCount="3" />
  <row Id="5663" PostTypeId="1" CreationDate="2011-02-22T16:01:37.870" Score="1" ViewCount="150" Body="&lt;p&gt;I have been reading the stackexchange questions on enhanced symmetries in string theory, the Leech lattice, monstrous moonshine, etc. , and I have a question to ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An astute commentator pointed out with enhanced symmetries, massless vector bosons have to couple in the Yang-Mills fashion. Only if the massless vectors come in a Lie algebra will that be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine bosonic string theory. Compactify 24 spatial dimensions over the Leech lattice or odd Leech lattice. With respect to the uncompactified dimensions (one spatial and one timelike), we get 196560 massless vector bosons, but they can't be part of any Lie algebra. But that's OK because massless vector bosons in 2 dimensions have zero norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's try again. Compactify 23 spatial dimensions over the shorter Leech lattice. There are 2 spatial and one timelike uncompactified dimensions. Over this space, we have 4600 massless vector bosons, but they're not part of any Lie algebra either. But that's also OK because massless vector bosons in 3 dimensions are physically indistinguishable from massless scalar bosons. The reason for this is there is no helicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we try to find a similar example over at least 4 uncompactified dimensions, where massless vector bosons exist for certain and aren't equivalent to anything else, we discover all unimodular lattices in 22 dimensions or less are part of a root system. Thus, we always have a Lie algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the magical mechanism in string theory enforcing this sort of self-consistency? Is string theory only self-consistent because of accidental coincidences?&lt;/p&gt;&#xA;" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-22T17:43:31.390" LastActivityDate="2011-03-02T04:43:18.607" Title="What mechanism in string theory enforces the consistency of self-couplings of massless vector bosons?" Tags="&lt;string-theory&gt;&lt;mathematical-physics&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="5664" PostTypeId="2" ParentId="5652" CreationDate="2011-02-22T16:12:13.217" Score="1" ViewCount="" Body="&lt;p&gt;Obviously $V_b$ is at a higher potential as Ted already answered. The relation between field and potential is simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec E = - \vec \nabla V$&#xA;In this one dimensional case it is&#xA;$\vec E = - \frac {dV}{dx} \hat x $ where $\hat x$ is the unit vector along the positive x direction. So you see the field is in the $- x$ direction i.e. along $\vec {BA}$&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-02-22T16:12:13.217" />
  <row Id="5665" PostTypeId="1" AcceptedAnswerId="5668" CreationDate="2011-02-22T16:22:13.723" Score="3" ViewCount="188" Body="&lt;p&gt;The mass of an elementary particle in string theory is related with the way the string vibrates. The more frantically a string vibrates the more energy it posses and hence the more massive it is. My question is how is the electric charge of a particle is described in S.T. How is the opposite charges described? More specifically, I would like to know how a particle and its antiparticle are conceptualized?&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-22T16:37:14.170" LastActivityDate="2011-02-22T17:10:50.380" Title="Electric charge in string theory" Tags="&lt;string-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="5666" PostTypeId="2" ParentId="5663" CreationDate="2011-02-22T16:48:47.550" Score="0" ViewCount="" Body="&lt;p&gt;It looks miraculous. However, one may also adopt a less religious attitude to those &quot;coincidences&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, they're not coincidences at all. They're provable mathematical facts. One can explicitly show that the relevant consistency conditions - such as the absence of ghosts - are satisfied by bosonic string theory compactified on any torus of dimension 22 or lower. These no-ghost theorems are even parts of some textbooks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Other, more special consistency conditions, such as stability of the vacuum, clearly fail in bosonic string theory.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it can be showed, using the spacetime perspective - effective field theory - that in 3+1 large dimensions or higher (as you correctly clarified), charged spin-one bosons may only consistently interact if they're a part of Yang-Mills theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this way, combining the two totally mathematical steps above, one can &lt;em&gt;prove&lt;/em&gt; - and indeed, string theorists &lt;em&gt;have&lt;/em&gt; proved - that all unimodular lattices in 22 or less dimensions are a part of a root system. (Jeff Harvey, in the other answer, strengthens the statement of yours: you wouldn't get a contradiction even for 23- or 24-dimensional lattices because the shortest nonzero vectors in the lattice already correspond to massive particles. But I will only talk about your more modest statement that assumes that the dimension of the lattice is 22 or smaller. I would also have a problem with your condition of &quot;unimodularity&quot;. You really want to study lattices with many $p^2=2$ points and no other condition or, if you want to study both left-movers and right-movers, the discussion is about even self-dual lattices in $d_{22}+d_{22}$ dimensions. But let me focus on the philosophical content here, not the technicalities.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could transform this proof into mathematical jargon that will hide all the traces of its stringy origin: but it's still true that string theory is a natural engine that produces remarkable proofs of many amazing facts in mathematics. It tells you how you should optimally proceed if you want to prove an important theorem in mathematics. So what you call is not an &quot;accidental coincidence&quot; - it's a deep, not-at-all coincidental sharp mathematical theorem, and it can be proved by stringy physics-inspired methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this string theory approach also proves a metaphysical statement - namely that this observation is pretty important. It is &lt;em&gt;not&lt;/em&gt; an irrelevant coincidence. It has a deeper meaning. Your opinion that it is just a &quot;coincidence&quot; has been falsified. In the same way, your expectation that this &quot;coincidence&quot; may only be proved by checking the lattice one by one has also been proved incorrect. It's a qualitative statement that has a qualitative, conceptual proof - even though this fact may have failed to be self-evident at the beginning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not necessarily the case that mathematicians must first think of all possible structures, all possible theorems, and all possible proofs, and these results are then used by physicists to describe the physical world. When physicists are working on a theory that is as deep as string theory, the chronology is often the opposite one: they're finding structures, theorems, and/or their proofs that are unknown to mathematicians of their time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Needless to say, such things may occur with physics models that are easier than string theory, too. Witten has used Chern-Simons theory to prove various things about knots and other things. In string theory, those results are omnipresent. String theory produces lots of such results that would look like &quot;accidental coincidences&quot; i.e. &quot;miracles&quot; to a person who doesn't understand any string theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me mention one not-too-quoted but amazing example. There is an identity between sums of products of Bernoulli numbers that was unknown before 2005:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0511286&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0511286&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Look at equation (B.5). It is similar to the Euler-Ramanujan identities. However, it's true and Martin Schnabl discovered this identity - that can be proved (or, at least, certainly checked) once we know it's true while he was solving open string field theory for the disappeared bosonic D25-brane. So this seemingly &quot;miraculous&quot; or &quot;accidentally coincidental&quot; identity follows from the fact that &quot;the space where the D-branes has totally disappeared solves the equations of motion&quot;. By reparameterizations that only look natural with the whole string-theory cannon, one may reduce these identities to some much more obvious, but more formal, identities.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="838" LastEditorDisplayName="" LastEditDate="2011-02-22T17:25:21.477" LastActivityDate="2011-02-22T17:25:21.477" />
  <row Id="5667" PostTypeId="2" ParentId="5663" CreationDate="2011-02-22T17:08:28.080" Score="4" ViewCount="" Body="&lt;p&gt;I think you're confused about the structure of the Leech lattice. The minimum length squared is 4, not 2 as for the root lattices of simply laced Lie algebras. The mass of a state with momentum $p \in \Lambda_{Leech}$ and no oscillator excitations is $p^2/2-1$, so the 196560 states you are talking about are massive, not massless. Before doing an asymmetrical $Z_2$ orbifold projection to get the monster theory there are $24$ massless states, there are just the gauge bosons of $U(1)^{24}$. The $Z_2$ orbifold removes these states and leaves you with a theory which just has massive states. Similar comments apply to your other examples. By the way, although it is not directly relevant to your question, there is an algebraic structure one can define involving the vertex operators for the $196884$ states at the first mass level. It is known as the Griess algebra and its automorphism group is the Monster group.&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T04:43:18.607" LastActivityDate="2011-03-02T04:43:18.607" />
  <row Id="5668" PostTypeId="2" ParentId="5665" CreationDate="2011-02-22T17:10:50.380" Score="2" ViewCount="" Body="&lt;p&gt;In quantum field theory and its extensions including string theory, the electric charge is a generator of a $U(1)$ symmetry which should be promoted to a local symmetry i.e. gauge symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In string theory, the $U(1)$ symmetry and the gauge field often appear as parts of the low-energy effective action. This could be enough to answer the question: we reduce the problem to the same problem in the approximate theory - quantum field theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except that we don't have to end at this point. String theory produces many geometric pictures how to &quot;imagine&quot; or &quot;visualize&quot; the electric charge. Those &quot;visualizations&quot; are often dual to each other: it means that even though these ways to present the charges superficially look totally different, one may actually demonstrate that their physical implications are totally equivalent and indistinguishable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Kaluza-Klein theory&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oldest picture embedded in string theory goes back to 1919 and a discovery by Theodor Kaluza, later refined by brilliant physicist Oskar Klein. Five-dimensional general relativity, with the new dimension compactified on a circle, produces $U(1)$ electromagnetism aside from the four-dimensional general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mixed components of the metric, $g_{\mu 5}$, may be interpreted as the gauge field $A_\mu$ in the large dimensions. The isometry rotating the circle (compact fifth dimension) at each point is interpreted as the $U(1)$ gauge symmetry. And charged particles are particles that carry a momentum in the new, fifth direction. By quantum mechanics, the momentum has to be quantized (for the wave function to be single-valued), $p=Q/R$, where $R$ is the radius of the circle ($2\pi R$ is the circumference) and $Q$ is an integer that may be identified with the electric charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A particle with the opposite charge is simply a particle that moves in the opposite direction along the hidden circular dimension. This works not only for strings but even for point-like particles in higher-dimensional spacetimes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Windings&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;String theory offers a special, more intrinsically stringy origin of the charges, too. Closed strings may wrap around a non-contractible loop in spacetime - such as the circle from the Kaluza-Klein theory. They obey boundary conditions on the string:&#xA;$$ X^5 (\sigma+\pi) = X^5(\sigma)+2 \pi R w. $$&#xA;Those $w$ times wound strings wouldn't exist in a theory without strings. The winding number $w$ - how many times the string is wrapped around the circle - is interpreted as another type of charge. $B_{\mu 5}$, a component of an antisymmetric tensor field, is interpreted as a new gauge field $A_\mu$ for this $U(1)$ symmetry. The oppositely charged particles are strings wrapped in the opposite direction; to be distinct, the closed strings have to be oriented (carry an arrow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This winding number origin of the charge is equivalent to the Kaluza-Klein origin by an equivalence we call T-duality. The gauge groups in the heterotic string theory combine the Kaluza-Klein-like charges and the winding-like charges and promote them to large non-Abelian groups such as $SO(32)$ or $E_8\times E_8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generalizations of wound strings exist for higher-dimensional branes: the total &quot;wrapping number&quot; of some membranes or branes around non-contractible cycles in spacetime (homology) are also manifesting themselves as electric charges. Many non-perturbative dualities exist. Some cycles on which the branes may be wrapped may be shrunk to zero size but they still exist: in those cases, the charged objects are localized in space (the gauge field only lives on a singularity which may be extended just like a brane). That's the case of the ADE singularities. In all cases, oppositely oriented branes correspond to oppositely charged particles. Note that the orientation may be defined for the &quot;whole world volume&quot; so the reversal of the spatial orientation may be mimicked or compensated by the reversal of the world volume in the temporal dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Open strings and D-branes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When open strings are allowed, they can carry charges (historically known as &quot;Chan-Paton factors&quot;) at the end points - these are the points stuck on the D-branes. So the end points behave as quarks: if the string is oriented and carries an arrow from the &quot;beginning&quot; to the &quot;end&quot;, the beginning may be called a quark and the end may be called an antiquark. In this setup, the charges are most analogous to those of point-like particles. The world line &#xA;of the quark and antiquark (going backwards in time) is nothing else than the boundary of the open world sheet as embedded in the spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even this seemingly point-like origin of charges may be dual - exactly equivalent - the purely stringy ways to produce the charges.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-22T17:10:50.380" />
  <row Id="5669" PostTypeId="1" AcceptedAnswerId="5849" CreationDate="2011-02-22T17:58:19.303" Score="3" ViewCount="233" Body="&lt;p&gt;I'd like to make a comparison of prices, availability and properties of semiconductor substrate layers like Si, Ge, GaP, AlN etc. I wonder what are the problems related to getting such data, what are the leading companies that sell such things, what's the quality, waiting time and so on. I'm not going to buy one, It's just a little research, that's why I ask here and not the companies directly. Your own experiences (what problems you've encountered and maybe some sample prices or company names?) are much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1505" LastEditorUserId="1505" LastEditorDisplayName="" LastEditDate="2011-02-25T19:25:33.613" LastActivityDate="2011-02-25T19:25:33.613" CommunityOwnedDate="2011-02-25T18:59:22.507" Title="Purchasing semiconductor substrates" Tags="&lt;semiconductor-physics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5670" PostTypeId="1" CreationDate="2011-02-22T18:51:37.070" Score="5" ViewCount="430" Body="&lt;p&gt;Is this statement correct?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, why is it the case?  If there are two paths available, and one, for example, has a resistor, why would the current run through the other path only, and not both?&lt;/p&gt;&#xA;" OwnerUserId="2165" LastActivityDate="2011-07-03T02:42:32.030" Title="Electricity takes the path of least resistance" Tags="&lt;electricity&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="5671" PostTypeId="2" ParentId="5670" CreationDate="2011-02-22T19:02:32.103" Score="0" ViewCount="" Body="&lt;p&gt;The statement is correct if you interpret it to mean that there is a larger current in the path that has a lower resistance, &lt;em&gt;when both paths have the same voltage across them&lt;/em&gt;. (&lt;strong&gt;This doesn't mean that the path with higher resistance has no current, just less current - as Ted Bunn's example shows&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can understand this by thinking of the analagous situation of a a long pipe that diverges into two branches and re-converges back again. Suppose that the pipe is filled with water and there's a pressure difference (say using a pump) between the two extreme ends of the pipe. One of the branches is just like the rest of the pipe while the other branch is lined with, say, wheels that add to the resistance and make the water flow slower in that branch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure difference across both branches are the same (just like the voltage between two  parallel electrical resistances is the same) but the water flows at a faster speed in the branch without the wheels, just like there's a larger current (rate of flow of electrons) in the path with lower resistance.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-02-22T19:02:32.103" />
  <row Id="5672" PostTypeId="2" ParentId="5670" CreationDate="2011-02-22T19:02:44.737" Score="6" ViewCount="" Body="&lt;p&gt;It's not true. To see this, you can try an experiment with some batteries and light bulbs. Hook up two bulbs of different wattages (that is, with different resistances) in parallel with a single battery:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; ------------------------------------------&#xA;|                     |                   |&#xA;Battery              Bulb 1              Bulb 2&#xA;|                     |                   |&#xA; ------------------------------------------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Both bulbs will light up, although with different brightnesses. That is, current is flowing through the one with more resistance as well as through the one with less resistance.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-22T19:02:44.737" CommentCount="1" />
  <row Id="5674" PostTypeId="2" ParentId="5440" CreationDate="2011-02-22T19:45:29.723" Score="1" ViewCount="" Body="&lt;p&gt;Such objects are used all the time. The mathematics is done in terms of quantum fields, which to some extent conceals what's going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, your &quot;(frequency-domain) coherence&quot; is a correlation &lt;em&gt;coefficient&lt;/em&gt;, which is normalized, whereas Physicists typically work in terms of correlation functions, which typically are not, but they largely amount to the same thing. Your observables $X(f_1)$, etc., are constructed as functions of frequency, however this is a singular object in quantum field theory. In quantum field theory, we instead construct observables $\phi(F_1)$, etc., as &lt;em&gt;functionals&lt;/em&gt; of test functions $F_1(x)$, etc. One singular choice would be $F_1(x)=\exp(if_1\cdot x)$, which makes $\phi(F_1)$ essentially the same object as your $X(f_1)$; it's &lt;em&gt;singular&lt;/em&gt;, however, because $F_1(x)$ is not square integrable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another choice of singular test function is, of course, $\delta(x-y)$, which gives the value of the field at a point, which we might write in your terms as something like $X(y)$. For a quantum field, this is also a rather singular object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, when you say $X(f_1)$, what you really mean is $\int X(f) {\mathrm d}f$, over some small range of frequencies, and in the mathematical and experimental details this has to be taken into account. Making everything precise requires that we know what the frequency ranges of each of the measurements is, which an experimenter either must characterize or must read off from a manufacturer's data sheets. In even more detail, we will have to construct a weight function, saying that frequencies near $f_1$ are still registered by the measurement device, but not as much as near $f_1$. We may well take the weight function, as a first approximation, to be Gaussian. This corresponds to taking the test function $F_1(x)$ to be that Gaussian. Signal analysis usually calls the &lt;em&gt;test function&lt;/em&gt; a &lt;a href=&quot;http://en.wikipedia.org/wiki/Window_function&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;window function&lt;/em&gt;&lt;/a&gt;. Test or Window functions can be difficult to become familiar with, but I believe it's well worth getting there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In these terms, your $C_{xyz}$ is a particular choice of (normalized) 3-point function. The choice of $f_1+f_2$ for the third frequency is of course not necessary, we can consider 3-point correlations between any three frequencies $f_1,f_2,f_3$ (and their vicinities). In quantum field theory, we would represent the 3-point correlation function, in the vacuum state, as $\left&amp;lt;0\right|\phi(F_1)\phi(F_2)\phi(F_3)\left|0\right&amp;gt;$. Replace the vacuum vector by some other state vector, if you like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the particular case when quantum field observables are mutually commutative, it can be understood to generate probability measures that correspond to an equivalent description in terms of probability measures over classical random variables, and hence quite precisely to a stochastic signal analysis. When quantum field observables do not commute, everything gets lots more complicated, but a remnant of the signal processing point of view can be maintained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a mathematics of &lt;em&gt;random fields&lt;/em&gt; that is used in cosmology because it is generally not necessary to worry about measurement incompatibility in that context. Mathematicians generally present signal analysis in Hilbert space terms unless they are writing for an engineering audience.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-02-22T19:45:29.723" CommentCount="2" />
  <row Id="5675" PostTypeId="2" ParentId="5670" CreationDate="2011-02-22T20:09:42.117" Score="1" ViewCount="" Body="&lt;p&gt;No. The statement is not correct. Current will take any path that is available to it. Which means it can even take the path of leaking out of the wire into the surrounding air, which is seen as sparks when dielectric breakdown of air happens. What you intend to mean perhaps is why current distributes itself in the inverse ratio of resistances, given the same potential difference across different resistive elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ohm's law $I=\frac{V}{R}$ would explain what you are asking. Given a common potential, the amount of current flowing through a resistive element is inversely proportional to the resistance. This would mean, and hopefully answer your question, that a path of lower resistance will have more current flowing through it and vice versa. (Normally the resistance of air is so high that current taking &lt;em&gt;that&lt;/em&gt; path and leaking out of the cable is negligibly zero under normal circumstances.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more thorough explanation, currents (and voltages) are distributed to minimize the total power dissipated as heat. This is a consequence of making the action of a disspative system stationary &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{t_1}^{t_2}(L+W)dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here W is the virtual work done by dissipative elements (resistance, capacitance, inductance etc) and L is the dissipation free dynamical system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an alternative, this &lt;a href=&quot;http://books.google.com/books?id=pqJnnPgDOoMC&amp;amp;pg=PA1039&amp;amp;lpg=PA1039&amp;amp;dq=action+principle+electric+current+path+of+least&amp;amp;source=bl&amp;amp;ots=_jFeXEqt4i&amp;amp;sig=PNG6xOrMAQiWDw4r24VcH3trfoY&amp;amp;hl=en&amp;amp;ei=pBJkTavONMPtOfeizb0N&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=5&amp;amp;sqi=2&amp;amp;ved=0CDQQ6AEwBA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; explains how Ohm's Law corresponds to Fermat's Principle of least time.&lt;/p&gt;&#xA;" OwnerUserId="1671" LastEditorUserId="1671" LastEditorDisplayName="" LastEditDate="2011-02-22T21:01:25.650" LastActivityDate="2011-02-22T21:01:25.650" />
  <row Id="5676" PostTypeId="1" CreationDate="2011-02-22T20:11:33.987" Score="3" ViewCount="204" Body="&lt;p&gt;Actually I have few questions. The answer to a question related to string field theory helped me, but let me still ask it as its probably a bit different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When e.g we do bosonic/super string theory we see that the vacuum is $|o,p&amp;gt;$ that is not a true space time vacuua (I mean a unique |0&gt;) (depends on CM momenta of strings). Its okay I thought, as our action to start with, describes a string or even the interaction between them. So, our states can never probably be free of strings. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;So, let me just verify that, is it possible (in first quantize string theory) to get a space time vacuum state as in QFT? Is it because a string never annihilates? Or does it annihilate and can give a unique translationally invariant vacuum state?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I was thinking about this. In QFT (rather in 1 particle case first), we impose space time and momentum commutation relation. In QFT we just apply commutation relation between $\Phi$ and $\Pi$. In string theory we apply between $X^\mu$ and its corresponding $\Pi^\mu$. So, if we don't get a vacuum (like in QFT) in string theory, is it because $X^\mu$ are really not space time coordinate, but only describing strings in Space time and we're quantizing them? Though after learning that a single string action can actually describe the splitting, joining of the string, it seems to me that we can also get a space time vacuum state like in field theory (a state with no strings like a state with no particle). Please verify/clarify?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related to this quantization, let me ask another question. I think when we get a massless spin two particle in string spectrum, we immediately recognize it as graviton. We also consider this gravitons in the background and try to quantize the action (That's the SUGRA action.. right(as in low energy, effective) which we fail to quantize)? &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;So, is it that, in string theory we get the graviton (which will appear also if we can quantize space time metric fluctuation $h_{\mu\nu}$) and that's it from &quot;quantizing gravity&quot; point of view? I mean we actually don't quantize the metric fluctuations in string theory to see the graviton arising. In fact what will I mean by the term Quantizing Gravity at all if not this. Please verify/clarify!&#xA;Of course I am not blaming string theory for this but just want to know our goal if we say &quot;we will try to quantize gravity&quot; and how much string theory does for us. Thanks..&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1349" LastActivityDate="2011-02-23T15:02:58.053" Title="Questions on string theory vacuum and quantizing gravity" Tags="&lt;string-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5677" PostTypeId="1" CreationDate="2011-02-22T20:38:37.887" Score="2" ViewCount="119" Body="&lt;p&gt;In solving different problems in string theory and also in brane theory, we consider this static gauge choice and I think that's due to the local symmetry(Repara. Inv. and Weyl) that we have in our world sheet. So, suppose I start with strings and with a general world sheet metric (and some doable ST metric). Once we use the local symmetries to make the world sheet flat, can we still use them to make the static gauge choice? How do we do it then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly for D branes we also use the static gauge choice ..right? But I guess that's just using the R.I. I don't have any problem in convincing me in this case as I don't have Weyl here so I can't use the local symmetries to make the world volume flat. Anyway, please help me out with my misunderstandings. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1349" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-22T21:31:51.547" LastActivityDate="2011-02-22T21:31:51.547" Title="Static gauge choice in string theory" Tags="&lt;string-theory&gt;" CommentCount="3" />
  <row Id="5678" PostTypeId="2" ParentId="5552" CreationDate="2011-02-22T20:45:33.723" Score="-1" ViewCount="" Body="&lt;p&gt;The benefit of this genre of posts, is that they are challenging. There will be no progress if we speak only of what we were taught.&#xA;Wisdom is the result of testing what we know, and everyone has misconceptions, including myself ;) Honestly: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;on wikipedia: &#xA;&lt;strong&gt;Renormalization&lt;/strong&gt;: ...&#xA; was first developed in quantum electrodynamics (QED) to make sense of infinite integrals in &#xA;perturbation theory. ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What infinities? &lt;strong&gt;I can not conceive of any infinite amount that can be attributed to any object in this universe.&lt;/strong&gt;&#xA;And why not? Because it would take an infinite time or infinite energy or a null or infinite length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example &lt;strong&gt;what is the logic of integrating the EM field from r = 0 to infinity?&lt;/strong&gt; Can be done mathematically, but physically is nonsensical because the short total lifetime of the particle (and universe).&#xA;The wavelength of a photon is limited by the size of the oscillator where it originated. Enormous (or miserably small) wavelengths imply correspondingly long (or short) time of interaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thus the application of cutoffs is natural and the right response.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for perturbative methods, the calculation of a quantity by an infinite sum (not of energy, wavelength, etc) is one approach to another still unknown function. A situation that could raise such representation is the existence of feedback between two components. Any physical process responds within the short interaction time and do not involve computers nor have endless time to do the calculation. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Can someone explain, please, because I am troubled with Vladimir comment: 'At least, Classical Electrodynamics predicts EMW when the &lt;strong&gt;electron is accelerated&lt;/strong&gt;. So this phenomenon (radiation) occurs always. Now look at the Mott, Bhabha, Klein-Nishina formulas - they are the first Born approximations describing elastic processes. &lt;strong&gt;No radiation is obtained&lt;/strong&gt; at'&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-02-22T20:45:33.723" CommentCount="4" />
  <row Id="5679" PostTypeId="2" ParentId="5670" CreationDate="2011-02-22T21:02:01.067" Score="1" ViewCount="" Body="&lt;p&gt;I suspect the statement was meant to be about electrical discharges via dielectric breakdowm. Such as lightening bolts etc. As such it has partial validity, in that say a tall tree is more likely to be hit than a short one. But the reality is that dielectric breakdown is a chaotic process, which is why lightening appears forked, rather than taking a straight pathway. Once you get ionization along a path, more current flows along it causing more ionization, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simple circuits not relying on breakdown, its a simple matter of resistance/impedance, and the current will distribute itself among multiple paths as described above. But for breakdown situation, whatever path gets connected first often takes the entire current.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-02-22T21:02:01.067" CommentCount="1" />
  <row Id="5681" PostTypeId="1" AcceptedAnswerId="7547" CreationDate="2011-02-22T21:27:24.220" Score="1" ViewCount="144" Body="&lt;p&gt;When we observe at close to the cosmological horizon (as close as we currently can) we see the universe as a much younger, denser place. As we feel the gravitational forces from these objects distant, will we measure a tidal force increasing the distance between an object less distant than the cosmological horizon and ourselves?  Or for that matter between any two objects at different distances from earth? If so is it great enough to account for the accelerating expansion of the universe?&lt;/p&gt;&#xA;" OwnerUserId="1567" LastEditorUserId="1567" LastEditorDisplayName="" LastEditDate="2011-02-22T21:33:49.593" LastActivityDate="2011-03-24T23:55:19.760" Title="Gravitational effect of the distant universe" Tags="&lt;cosmology&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="5682" PostTypeId="1" AcceptedAnswerId="5734" CreationDate="2011-02-22T21:50:41.587" Score="9" ViewCount="421" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hopf_algebra&quot; rel=&quot;nofollow&quot;&gt;Hopf algebra&lt;/a&gt; is nice object full of structure (a bialgebra with an antipode). To get some idea what it looks like, group itself is a Hopf algebra, considered over a field with one element ;) usual multiplication, diagonal comultiplication, obvious units and inverse for the antipode. For a less pathological example, a group algebra and universal enveloping algebra can be both quite naturally turned into Hopf algebra. It all obviously relates to representation theory and lots of other neat stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So much for the math-related stuff. Now, I've heard that there should also be some applications to physics.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;For one thing, (and this will be very vague and probably also wrong) Feynman diagrams should somehow carry a structure of Hopf algebra with multiplication given by joining of two lines into a vertex and comultiplication as splitting. It reminds me of cobordisms but I am not sure it really makes sense. Any idea whether something like this works?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Besides that, I heard people try to formalize renormalization using Hopf algebras. I found some papers but I am not sure where to start. Anyone care to give an overview how does this stuff work and whether it's leading anywhere?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Anything else? Sorry, if this is way to vague and there is actually whole industry of Hopfy physics. If so, just try to give some most important examples, preferably with references.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-02-23T21:03:19.983" Title="What are the uses of Hopf algebras in physics?" Tags="&lt;mathematical-physics&gt;&lt;renormalization&gt;&lt;hopf-algebra&gt;" AnswerCount="7" CommentCount="13" FavoriteCount="3" />
  <row Id="5683" PostTypeId="2" ParentId="5669" CreationDate="2011-02-22T21:55:27.647" Score="4" ViewCount="" Body="&lt;p&gt;I was looking for a company who can provide Si (and potentially other) wafers of various types, and I can say the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) You will have a hard time finding 'fixed' pricelist. Companies are not ready to quote you unless you are ready to buy XXX amount. 'I am just checking prices around' not gonna work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Price is drastically different if you are getting standard wafer (which is produced in mass quantity for fabs). If you are getting non-standard wafer you not going to get price/wafer price (or it's going to be ridiculously high). You going to be quoted for crystal growth and slicing+finishing separately. And of course price is negotiable for large quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) You can get almost any needed purity (=more recrystallization passes). It just costs additional money. (so going from 1kOhm to 10kOhm Si resistance adds 30% to the crystal price before slicing)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Delivery of stock wafers depends on your luck &amp;amp; their laziness. Custom crystals take ages to make.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-02-22T21:55:27.647" CommunityOwnedDate="2011-02-25T18:59:22.507" />
  <row Id="5685" PostTypeId="2" ParentId="5505" CreationDate="2011-02-22T22:21:37.967" Score="3" ViewCount="" Body="&lt;p&gt;As I now understand the formulation of this question we have the following situation:(1) an observer is situated near the Event Horizon of a Black Hole;(2) a light source is situated a similar radial distance from the BH and is shining light which is reaching the Observer;(3) the radial distance of both from the BH is approximately x; (4) the Event Horizon expands (reducing the distance between it and the Observer/light source).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several observations about what is likely to happen here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) The Event Horizon is at R=2M for a Schwarzchild solution, but there are unusual photonic effects nearby outside too. At R=3M the photons will go into orbit around the BH. So perhaps if nearby the Observer is picking up a photon from its orbit. Even if a little further out the light bending will be quite extreme, and so the Observer is likely to &quot;see&quot; the light coming from another region of the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) The photons of light from the source may already be in a kind of elliptical orbit which drags them into R=2M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) With an expanded Event Horizon we are assuming greater gravitational force on both Observer and Source. At some point one or both may be overwhelmed by the increasing gravitation near the Event Horizon. Thus they would be unable to maintain any fixed distance, and be drawn in simply because they dont have an infinite power source in their engines. If such movement occured the light from source to Observer could be distorted further as they try to maintain position, but move in different orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) As the Event Horizon approached both would cross the R=3M region, at this point the light source might no longer reach the Observer, depending on exact distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted that the &quot;potential curve&quot; for gravitation near a Black Hole is unlike a standard Newtonian one near a massive object. The exact trajectories of objects is also heavily dependent on a quantity related to the angular momentum of the photon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To complicate matters further the use of the term &quot;Event Horizon&quot; in this expanding example may not be appropriate: other terms like &quot;apparent horizon&quot; may be more appropriate. In a sense the Event Horizon is theoretical construction best applicable to a Black Hole at the end of (its) time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short I am expecting the stream of photons to stop reaching the Observer somewhat &lt;em&gt;before&lt;/em&gt; the Horizon meets the Source, unless the Source and Observer are on some very specially constructed orbits, in which neither are at position x. The rays connecting the two might reform later as both head into the Black Hole, resulting in a light that is intermittent. Finally I believe that anything demonstrated here, is a demonstration of the properties of an apparent horizon.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-02-22T22:44:15.843" LastActivityDate="2011-02-22T22:44:15.843" />
  <row Id="5686" PostTypeId="1" AcceptedAnswerId="5701" CreationDate="2011-02-22T23:37:22.697" Score="5" ViewCount="165" Body="&lt;p&gt;Suppose a very long railway line goes from south africa, to sweden, &#xA;then it's decided to move the entire railway line, sliding it 1 km to north.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(leaving aside the difficulty of moving and the force required)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the railway bend because of coriolis force?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean every point of the railway would have different tangential velocity so when moving, we have different speeds at different time, then that's acceleration, then force, so if those are different forces applied to different points of the rail, then perhaps it would bend. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1916" LastActivityDate="2011-02-23T07:21:19.583" Title="The Coriolis Force bending a railway" Tags="&lt;classical-mechanics&gt;&lt;energy&gt;&lt;coriolis-force&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5687" PostTypeId="2" ParentId="5573" CreationDate="2011-02-23T00:05:31.710" Score="-3" ViewCount="" Body="&lt;p&gt;Electric forces act between charged bodies (charges), magnetic forces act between magnetized bodies (magnets). The latter can be said differently: magnetic forces act between currents of neutralized charges.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-23T15:44:25.337" LastActivityDate="2011-02-23T15:44:25.337" />
  <row Id="5688" PostTypeId="2" ParentId="5650" CreationDate="2011-02-23T00:26:59.840" Score="5" ViewCount="" Body="&lt;p&gt;I am still not sure what you precisely want to be a Klein Bottle, but let me make some comments that might help you clarify want exactly you want to know. (Warning: I am writing this while being very tired, people are invited to correct me.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all one must be careful to distinguish band structure of the &lt;strong&gt;bulk&lt;/strong&gt; from band structure of a &lt;strong&gt;semi-infinite strip&lt;/strong&gt; (with edges). In your second figure, 1 og 2 are bulk band structures while 3 is band structure of a semi-infinite strip. I think you are mixing these.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Bulk picture&lt;/strong&gt;: Both momentum $k_x$ and $k_y$ are good quantum numbers. Assume you start with a trivial insulator (gap in the band structure, number 2 in your picure) and you continuously change the parameters of your system. As long as the bulk gap i open you have a trivial insulator. Suddenly the bulk gap closes, at one value of your parameters (&lt;strong&gt;critical point&lt;/strong&gt;), and quickly opens up again, now you are in the topological phase BUT the bulk band structure still looks like figure 2 above. How to see the difference in these two phases? One way is looking for edge states.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Edge picture:&lt;/strong&gt; Make the system finite along $y$-direction, so now only $k_x$ is a good quantum number. Now you have alot of bands, depending on how many lattice sites you put along $y$. When you are in the trivial phase the system is still gapped and boring. While you change the system parameters and hit the critical point, the gap of all bands close and then &lt;em&gt;most&lt;/em&gt; (but not all) bands open up again (picture 3 above). When one looks at the eigenvectors, one will see that the bands with a gap correspond to bulk states while the band(s) with no gap (with a Dirac point) correspond to states localized along the edge. Furthermore these edge states are robust.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This I mention, in case there were any misunderstanding here. Now to your question, how can one see the connection between band structure, Brillouin zone and topology? The interesting thing is that we only need to analyse the bulk in order to probe the topology of the system, although the interesting physics is along the edge (this is called &lt;strong&gt;bulk-edge correspondence&lt;/strong&gt; in the litterature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume we add a spin-orbit coupling to graphene that preserves $S_z$, meaning that although the spin $SU(2)$ symmetry is not preserved it is still well-defined to speak about spin-up/down (so a $U(1)$ subgroup is still a symmetry). I think that this term will open a gap at the Dirac points at $K$ and $K&amp;#39;$ and the system turns into an insulator. Since $K$ and $K&amp;#39;$ are not important anymore, we have a four-band model (two from spin up/down and two from sub lattice A/B). Since $S_z$ is conserved, the (bulk) Hamiltonian is block diagonal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ H(\mathbf k) = \begin{pmatrix} H_{\uparrow}(\mathbf k) &amp;amp; \\ &amp;amp; H_{\downarrow}(\mathbf k) \end{pmatrix}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $H_{\uparrow}$/$H_{\downarrow}$ are $2\times 2$ matrices. Since the system preserves time-reversal symmetry, the two Hamiltonians are related by a time-reversal transformation $H_{\downarrow} = \Theta H_{\uparrow}\Theta^{-1}$. Since the Brillouin zone is a torus, we need to classify maps $T^2\rightarrow \mathcal H$, where $\mathcal H$ is the appropriate space of $4\times 4$ matrices obeying some constraints (gapped and time-reversal invariance). For this model, however, we don't need to analyze $\mathcal H$. Since the model is block diagonal we can consider each block separately, any $2\times 2$ matrix can be written in the basis of Pauli matrices&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_{\alpha}(\mathbf k) = d_0(\mathbf k) \mathbf I + \mathbf d(\mathbf k)\cdot\mathbf{\sigma}$, where $\alpha = \uparrow, \downarrow$ (the dependence of $d_0$ and $\mathbf d$ on $\alpha$ is suppressed for notational simplicity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the spectrum is $E(\mathbf k) = d_0(\mathbf k) \pm \sqrt{\mathbf d\cdot\mathbf d}$, we can continously deform $d_0$ to zero and deform $\mathbf d\rightarrow \hat{\mathbf d}=\frac{\mathbf d}{|\mathbf d|}$ without closing the gap (and thus still be in the same topological class). Thus we can classify each block of the Hamiltonian, by the map $T^2\rightarrow S^2$, $\mathbf k\rightarrow \hat{\mathbf d}(\mathbf k)$. This is basically classified by the second homotopy group of the 2-sphere $S^2$ (winding number), $\pi_2(S^2) = \mathbb Z$ (well the map is from a torus $T^2$ and not $S^2$, so we need an argument for why we can use the second homotopy group but this I will not delve into right now). The winding number of $\hat{\mathbf d}(\mathbf k)$ is given by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Degree_of_a_continuous_mapping&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;degree of map&lt;/strong&gt;&lt;/a&gt; formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_1^{\alpha} = \frac 1{4\pi}\int_{T^2}d\mathbf k\;\hat{\mathbf d}\cdot\frac{\partial \hat{\mathbf d}}{\partial k_x}\times\frac{\partial \hat{\mathbf d}}{\partial k_y}\in\mathbb Z.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus each block form a Integer Quantum Hall effect separately (see also this &lt;a href=&quot;http://physics.stackexchange.com/questions/3282/simple-models-that-exhibit-topological-phase-transitions/3378#3378&quot;&gt;answer&lt;/a&gt;). Note that $C_1$ is basically the first Chern-number where $\epsilon^{\mu\nu}F_{\mu\nu} = \hat{\mathbf d}\cdot\frac{\partial \hat{\mathbf d}}{\partial k_x}\times\frac{\partial \hat{\mathbf d}}{\partial k_y}$, $F_{\mu\nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$ where $A_{\mu}(\mathbf k) = -i\langle\psi(\mathbf k)|\partial_{\mu}\psi(\mathbf k)\rangle$ is the Berry phase for the filled states (to see how to rewrite the Berry phase in terms of $\hat{\mathbf d}$ see this &lt;a href=&quot;http://pra.aps.org/abstract/PRA/v64/i5/e052101&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt;. The reason I have avoided the Chern number approach is that, one needs to go into discussing vector (or principal) bundles which just would complicate everything even more).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we are almost done. We have found two Chern numbers $C_1^{\downarrow}$ and $C_1^{\uparrow}$, is there a Chern number for the combined system? Since $H_{\downarrow} = \Theta H_{\uparrow}\Theta^{-1}$ are related by time-reversal symmetry, one can show that $C_1^{\uparrow} = -C_1^{\downarrow}$. Thus the sum $C_1^{\uparrow} + C_1^{\downarrow} = 0$, but the difference $C_1^{\uparrow} - C_1^{\downarrow} = 2C_{spin}$ is well defined and sometimes called the &lt;strong&gt;spin Chern-number&lt;/strong&gt;. Thus calculating $C_{spin}\in\mathbb Z$ will tell you if graphene with a spin-orbit coupling is topological or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if one has a perturbation that breaks $S_z$, then our construction breaks down. But it turns out that $C_{spin}$ is still well defined but only modulo 2, $\nu = C_{spin} \text{mod} 2$. This is why people call this a $\mathbb Z_2$-topological insulator, see more details &lt;a href=&quot;http://www.physics.upenn.edu/~kane/pubs/p49.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (pdf-file).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are of course many other ways of seeing the connection to topology and this is probably the most explicit and elementary way to do it. Is this what you wanted? I suspect that you are interested in the map $T^2\rightarrow\mathcal H$, $\mathbf k\rightarrow H(\mathbf k)$ and when you say &quot;energy space&quot; you mean $\mathcal H$. It turns out that this space is &lt;strong&gt;homotopy equivalent&lt;/strong&gt; to $\mathcal H \approx U(8)/Sp(8)$ (called the symplectic class) by a band-flattening procedure. Sadly I am too tired to see if this is diffeomorphic to the &lt;strong&gt;Klein bottle&lt;/strong&gt;. Is this what you are interested in?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer got much longer than originally planned.&lt;/p&gt;&#xA;" OwnerUserId="1469" LastEditorUserId="1469" LastEditorDisplayName="" LastEditDate="2011-02-23T07:43:50.917" LastActivityDate="2011-02-23T07:43:50.917" />
  <row Id="5689" PostTypeId="2" ParentId="5682" CreationDate="2011-02-23T01:24:13.630" Score="2" ViewCount="" Body="&lt;p&gt;I can mention some applications in condensed matter physics, but I must warn you that I know close to nothing about Hopf algebraes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the field of topological order (and topological quantum computation), &lt;strong&gt;Hopf algebras&lt;/strong&gt; has been recently used to construct some very general models with emergent gauge fields, topolgical order and (non-abelian) anyons among other uses. See for example&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/1007.5283&quot; rel=&quot;nofollow&quot;&gt;A hierarchy of topological tensor network states&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/1006.5823&quot; rel=&quot;nofollow&quot;&gt;Electric-magnetic duality and topological order on the lattice&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But I think that Hopf algebras are new thing in this field. &lt;strong&gt;Quantum groups&lt;/strong&gt; on the other hand has been used for longer period, for example to study Fractional Quantum Hall Effect&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/cond-mat/0104035v1&quot; rel=&quot;nofollow&quot;&gt;Quantum groups and nonabelian braiding in quantum Hall systems&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But the main tool in this field is &lt;strong&gt;modular tensor categories&lt;/strong&gt; ($\mathbb C$-additive monoidal categories, with braiding structure and some more). There are many papers to cite, but a random one is&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/1010.1517&quot; rel=&quot;nofollow&quot;&gt;A classification of 2D fermionic and bosonic topological orders&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I think the main connection between these approaches originate from the fact that many topological phases in condensed matter physics are (in the low-energy/long wavelength limit) described by non-Abelian Chern-Simons theories. And there are well-known connection to knot theory, modular tensor categories, representations theory of mapping class groups, Quantum groups and Hopf algebraes (I'm sure you know much much more about this than me).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you can use this, rather vague, answer.&lt;/p&gt;&#xA;" OwnerUserId="1469" LastActivityDate="2011-02-23T01:24:13.630" CommentCount="1" />
  <row Id="5690" PostTypeId="2" ParentId="5682" CreationDate="2011-02-23T02:18:58.840" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/9904014&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9904014&lt;/a&gt; from Brouder and &lt;a href=&quot;http://arxiv.org/abs/q-alg/9707029&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/q-alg/9707029&lt;/a&gt; from Kreimer is, to me, the most puzzling application, doing a coproduct of trees that, when formalised by Connes, provided a different angle to perturbative renormalization. A lot of references can be found in the arxiv, but not a definitive monograph.&lt;/p&gt;&#xA;" OwnerUserId="1335" LastActivityDate="2011-02-23T02:18:58.840" />
  <row Id="5691" PostTypeId="2" ParentId="2900" CreationDate="2011-02-23T02:22:12.463" Score="0" ViewCount="" Body="&lt;p&gt;The Chinese satellite navigation system &lt;em&gt;is&lt;/em&gt; in geostationary orbit.  From th&lt;a href=&quot;http://en.wikipedia.org/wiki/Beidou_navigation_system&quot; rel=&quot;nofollow&quot;&gt;e Wikipedia article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Unlike the American GPS, Russian GLONASS, and European Galileo systems, which use medium Earth orbit(MEO) satellites, BeiDou-1 uses satellites in geostationary orbit(GEO). This means that the system does not require a large constellation of satellites, but it also limits the coverage to areas on Earth where the satellites are visible. The area that can be serviced is from Logitude 70°E to 140°E, and from Latitude 5°N to 55°N.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-02-23T02:22:12.463" />
  <row Id="5692" PostTypeId="1" CreationDate="2011-02-23T02:56:41.180" Score="2" ViewCount="142" Body="&lt;p&gt;This question is kind of inspired in this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/4359/diffm-as-a-gauge-group-and-local-observables-in-theories-with-gravity&quot;&gt;Diff(M) as a gauge group and local observables in theories with gravity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conundrum i'm trying to understand is how is derived the (quite) extraordinary statement that in GR there are no local observables. I just want to stress that this is indeed an extraordinarily counter-intuitive assertion (with extraordinarily dramatic consequences for any compatible theory of quantum reality), and it deserves an extraordinarily robust explanation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among the statements i see are mostly involved in this argument are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;&lt;strong&gt;It is usually argued that Diff(M) is a gauge transformation.&lt;/em&gt;&lt;/strong&gt; This point i don't have any sort of issue; the atlas of reference frames we use to describe space-time are indeed a human convention and physical laws should not depend on such conventions&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;&lt;strong&gt;A physical observable should be invariant under any gauge transformation.&lt;/em&gt;&lt;/strong&gt; whaaa? i mean, this is clearly preposterous non-sense; actually i think this is more dangerous than simple non-sense, is just circular self-justification. This is actually saying that observables should be scalars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can provide you a proof that this is non-sense by saying that the momentum of a particle is an observable in my reference-frame, and in other Diff(M) gauge (that is, another reference frame) i will see a different momentum of the same observable. All eigenvalues and eigenvectors transform according to the vector representation of the Poincare group. You are now of course free of dismiss such 'proof' as non-sense because my assumption that the momentum of a local particle is an observable is non-sense, but then, how are you sustaining that part of the argument, without actually saying again that Diff(M) is a gauge transformation? how do you avoid the circularity in this argument?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;U(1) gauge is a bad example because this is an internal symmetry; Diff(M) are space-time symmetries and i don't think that what is true in there (A eletromagnetic vector potential being unobservable, but B and E being observable, hence observables are invariant under U(1) )&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;&lt;strong&gt;It is usually argued that observables in GR formally exists in the asymptotic boundary of the space-time.&lt;/em&gt;&lt;/strong&gt; Is there an argument for this that is unrelated/independent to the previous point?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-02-23T03:22:01.820" Title="Diff(M) and requirements on GR observables" Tags="&lt;general-relativity&gt;&lt;gauge-theory&gt;&lt;diffeomorphism-invariance&gt;&lt;observables&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="5693" PostTypeId="2" ParentId="5692" CreationDate="2011-02-23T02:58:55.733" Score="-2" ViewCount="" Body="&lt;p&gt;Who says that there are no local observables?  There are two local degrees of freedom in General Relativity, related to the two polarizaiton modes of local gravitational waves.  And they are locally observable--hence the &lt;a href=&quot;http://www.ligo.caltech.edu/LIGO_web/about/&quot; rel=&quot;nofollow&quot;&gt;LIGO experiment&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, general relativity in 2+1 dimensions turns out to be topological--all the degrees of freedom of the theory are determined by the boundary conditions and the matter equation of state.  But that's certainly not true of the 3+1 dimensional world that we inhabit.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-02-23T02:58:55.733" CommentCount="4" />
  <row Id="5695" PostTypeId="2" ParentId="5692" CreationDate="2011-02-23T03:22:01.820" Score="4" ViewCount="" Body="&lt;p&gt;Instead of re-inventing the wheel, here is a bunch of references:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is &lt;a href=&quot;http://diracseashore.wordpress.com/2008/10/09/observables-in-quantum-gravity/&quot; rel=&quot;nofollow&quot;&gt;something I wrote&lt;/a&gt; back when I was blogging, the discussion was kind of fun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a pretty good explanation in this paper by &lt;a href=&quot;http://arxiv.org/abs/hep-th/0512200&quot; rel=&quot;nofollow&quot;&gt;Giddings- Hartle-Marolf&lt;/a&gt; on how this statement doesn't contradict all kind of semi-intuitive ideas people have  (and started commenting on here already) about local observables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, &lt;a href=&quot;http://arxiv.org/abs/0704.1814&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; by Arkani-Hamed and collaborators has very nice introduction explaining why dynamical gravity, and not just Diff invariance, is what prevents the existence of local observables.&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-02-23T03:22:01.820" CommentCount="3" />
  <row Id="5696" PostTypeId="2" ParentId="5682" CreationDate="2011-02-23T03:31:33.687" Score="2" ViewCount="" Body="&lt;p&gt;More references:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science?_ob=ArticleURL&amp;amp;_udi=B6TVC-473DJS2-86&amp;amp;_user=10&amp;amp;_coverDate=02%2F10%2F1992&amp;amp;_rdoc=1&amp;amp;_fmt=high&amp;amp;_orig=search&amp;amp;_origin=search&amp;amp;_sort=d&amp;amp;_docanchor=&amp;amp;view=c&amp;amp;_searchStrId=1652257739&amp;amp;_rerunOrigin=google&amp;amp;_acct=C000050221&amp;amp;_version=1&amp;amp;_urlVersion=0&amp;amp;_userid=10&amp;amp;md5=9f7b06f94783457b95f138507c33c969&amp;amp;searchtype=a&quot; rel=&quot;nofollow&quot;&gt;Mack and Schomerus&lt;/a&gt;: Discussion of Hopf algebras as the general symmetry structure of quantum states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there is quite a bit of work on Hopf algebra for two dimensional chiral CFTs, especially the Wess-Zumino-Witten model. I don't remember references off the top of my head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specific type of Hopf algebra, called the Yangian, plays now an enormous role in understanding the simplification of amplitudes in gauge theories, and potential integrability of N=4 SYM in the planar limit. The literature is too vast to do the subject any justice, but here is &lt;a href=&quot;http://arxiv.org/abs/1012.4005&quot; rel=&quot;nofollow&quot;&gt;one review&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps a bit, looks to me like it is a general structure and you probably want to zoom in before getting into details.&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-02-23T03:31:33.687" CommentCount="1" />
  <row Id="5697" PostTypeId="1" AcceptedAnswerId="5698" CreationDate="2011-02-23T03:35:06.137" Score="1" ViewCount="157" Body="&lt;p&gt;What is the full text (and possibly the source) of the summary of the 3 laws of thermodynamics that goes something along the lines of &quot;Can't break even, can't win and can't even stop playing the game&quot;?&lt;/p&gt;&#xA;" OwnerUserId="846" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-23T04:00:59.603" LastActivityDate="2011-02-23T05:10:46.727" Title="The full entropy quote" Tags="&lt;thermodynamics&gt;&lt;soft-question&gt;&lt;entropy&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5698" PostTypeId="2" ParentId="5697" CreationDate="2011-02-23T04:49:27.963" Score="3" ViewCount="" Body="&lt;p&gt;This comes from the pioneering work of amateur physicist, Michael Jackson:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;You can't win, &lt;br&gt;&#xA;  You can't break even, &lt;br&gt;&#xA;  And you can't get out of the game.&lt;br&gt;&#xA;  People keep sayin', &lt;br&gt;&#xA;  Things are gonna change.&lt;br&gt;&#xA;  But they just look like&lt;br&gt;&#xA;  They're stayin' the same&lt;/em&gt;.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;-Michael Jackson, &quot;&lt;a href=&quot;http://www.youtube.com/watch?v=6YrinCQOxB0&quot; rel=&quot;nofollow&quot;&gt;You Can't Win&lt;/a&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, okay. I think this is one of those jokes that starts between graduate students or something by the water cooler, so it doesn't have an &quot;exact&quot; form. &lt;a href=&quot;http://en.wikiquote.org/wiki/Thermodynamics&quot; rel=&quot;nofollow&quot;&gt;Wikiquotes gives&lt;/a&gt; the original author as &lt;a href=&quot;http://en.wikipedia.org/wiki/C.P._Snow&quot; rel=&quot;nofollow&quot;&gt;C.P. Snow&lt;/a&gt; in the following form:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Zeroth: &quot;You must play the game.&quot; &lt;br&gt;&#xA;  First: &quot;You can't win.&quot; &lt;br&gt;&#xA;  Second: &quot;You can't break even.&quot; &lt;br&gt;&#xA;  Third: &quot;You can't quit the game.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But that seems somewhat dubious. No one can provide a cite of C.P. Snow actually saying or writing it anywhere; I suspect that this history is a little mythical and that the joke has no real precise statement.&lt;/p&gt;&#xA;" OwnerUserId="1655" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-02-23T05:10:46.727" LastActivityDate="2011-02-23T05:10:46.727" />
  <row Id="5701" PostTypeId="2" ParentId="5686" CreationDate="2011-02-23T07:21:19.583" Score="4" ViewCount="" Body="&lt;p&gt;Of course that there would be forces that would try to bend the track but they would be tiny. Each segment of the track would be under the action of $-2m \Omega \times v$ Coriolis force. Note that the Coriolis force only depends on velocities, not accelerations as you stated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, there is the Coriolis acceleration, $-2\Omega\times v$, and you may see that it's tiny for realistic velocities of the moving track. Note that the direction of this acceleration is East-West at each point and its magnitude is zero at the equator where $\Omega$ and $v$ in your experiment have the same direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the Coriolis force would mostly try to rotate the track around the axis going through its equator point. Obviously, the mechanical constraints would compensate this force and it would be largely invisible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you consider a very similar setup to the tracks - namely rivers going to the North - they will eventually make their troughs asymmetric because of the Coriolis force. Albert Einstein was obsessed by this problem and wrote papers about it. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your concern was that the shape of the track - along the length - doesn't change as you move it North, it doesn't matter. The laws of mechanics still recognize that the metal is moving and the Coriolis force depends on the actual velocity, and not just the velocity that is easily seen. ;-)&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-23T07:21:19.583" />
  <row Id="5702" PostTypeId="1" AcceptedAnswerId="5735" CreationDate="2011-02-23T07:55:48.190" Score="6" ViewCount="265" Body="&lt;p&gt;I know that light (or electromagnetic radiation in general) attenuates in intensity as the square of the distance it travels.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A. Why does it attenuate?&lt;br&gt;&#xA;  B. Are the photons being scattered by the medium they pass through?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I also know that the energy carried by light &lt;em&gt;quanta&lt;/em&gt; is a function of their frequency only.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;C. If the answer to B is yes, then can we possibly transmit light energy (and information) over infinite distances through a perfect vacuum?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I may be wrong on some assumptions, so please enlighten me.&lt;/p&gt;&#xA;" OwnerUserId="293" LastEditorUserId="293" LastEditorDisplayName="" LastEditDate="2011-05-18T00:20:07.923" LastActivityDate="2011-05-18T00:20:07.923" Title="Can we transport energy over infinite distances through vacuum using light?" Tags="&lt;light&gt;&lt;energy&gt;&lt;quantum-optics&gt;&lt;frequency&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="5703" PostTypeId="1" CreationDate="2011-02-23T08:11:55.893" Score="-1" ViewCount="119" Body="&lt;p&gt;What is the difference between free energy (over unity) and perpetual motion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please provide some examples, both real world and theoretical.&lt;/p&gt;&#xA;" OwnerUserId="2136" LastActivityDate="2011-02-23T09:05:32.780" ClosedDate="2011-02-23T17:19:43.780" Title="The difference between free energy and perpetual motion" Tags="&lt;quantum-field-theory&gt;&lt;thermodynamics&gt;&lt;vacuum&gt;&lt;polarization&gt;" AnswerCount="1" />
  <row Id="5704" PostTypeId="2" ParentId="2" CreationDate="2011-02-23T08:34:24.153" Score="4" ViewCount="" Body="&lt;p&gt;I've noticed that none of these answers actually answer the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest explanation of string theory I can think of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Particles we currently consider &quot;point particles&quot; (electrons, quarks, photons, etc.) are actually tiny pieces of string with each a characteristic vibration.  They interact in a sort of harmony that results in/manifests as the physical laws we observe.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone with more knowledge in the field can correct me, I ask for improvements.  This is just how I personally explain it to people who ask, and I'd hate to give out false information.&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="71" LastEditorDisplayName="" LastEditDate="2011-02-24T02:15:47.333" LastActivityDate="2011-02-24T02:15:47.333" CommentCount="1" />
  <row Id="5705" PostTypeId="1" CreationDate="2011-02-23T08:44:22.183" Score="1" ViewCount="276" Body="&lt;p&gt;Newton tried to do it with three laws/statements.  While the first can be derived from the second, the three form a pretty nice framework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later on, I've encountered Lagrangian Mechanics, which involves, from what I gather, one statement:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Objects seek a path that minimizes total action.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which, while sort of simple, doesn't sound too natural or fundamental.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard some of these rather elegant statements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The universe is translation-invariant&lt;/li&gt;&#xA;&lt;li&gt;The universe is invariant under different inertial frames of reference&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And from either of these, one may derive Newton's laws and F=ma.  However, I'm not completely sure how this can be done.  Were these claims true?  How is this possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone ever come across a surprisingly/particularly stunningly elegant/fundamental formulation of classical mechanics, and a proof that they are equivalent to Newtonian Mechanics?&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="71" LastEditorDisplayName="" LastEditDate="2011-02-24T02:13:38.990" LastActivityDate="2011-02-24T18:14:19.503" Title="Most elegant/fundamental formulations of the laws of classical mechanics?" Tags="&lt;classical-mechanics&gt;&lt;mechanics&gt;" AnswerCount="3" CommentCount="13" />
  <row Id="5706" PostTypeId="2" ParentId="5702" CreationDate="2011-02-23T08:59:30.333" Score="4" ViewCount="" Body="&lt;p&gt;A. Because the total energy is divided over the whole sphere $4\pi R^2$ of radius $R$ around the source of light which means that the energy density over unit area goes like $1/R^2$. However, this is only true for unfocused light - with chaotic directions. Lasers may create coherent light whose intensity per unit area doesn't drop. The beam may move for kilometers without changing its shape and intensity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. It depends on the environment: photons don't scatter in the vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. No, because the vacuum doesn't interfere with the photons' motion at all, we can transfer energy by photons over any distance. In particular, there's no exponential decrease of the intensity in the vacuum. In this sense, the vacuum is not a &quot;medium&quot; - there is no &quot;luminiferous aether&quot; that fills it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the intensity of energy goes down like $1/R^2$, it's the number of photons per unit area that decreases. The energy of each photon is unchanged.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="377" LastEditorDisplayName="" LastEditDate="2011-02-24T01:33:05.933" LastActivityDate="2011-02-24T01:33:05.933" CommentCount="5" />
  <row Id="5707" PostTypeId="2" ParentId="5702" CreationDate="2011-02-23T08:59:32.220" Score="1" ViewCount="" Body="&lt;p&gt;The diminishing of intensity is best described if you use the wave model of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We make one basic assumption:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Total light energy is conserved over time&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let's say you have a light bulb in a vacuum that makes a flash.  Let's say we enclose that light bulb in a ball, and measure the energy of the light that hits the ball's inside walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we try it again, but this time use a larger ball, with a bigger radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because it's the same &quot;amount&quot; of light that is flashed, both balls, with different radii, should measure the same total energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, Intensity is energy per area (per second).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be clear that with the larger ball, there is less energy &lt;em&gt;per area&lt;/em&gt; than with the smaller ball.  The energy is more concentrated in the smaller ball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this it can be seen that because the surface area of a sphere increases with $r^2$, the intensity of light (energy per area per second) decreases with $r^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light's fading intensity, therefore, is a direct consequence of the conservation of energy.&lt;/p&gt;&#xA;" OwnerUserId="71" LastActivityDate="2011-02-23T08:59:32.220" />
  <row Id="5708" PostTypeId="1" CreationDate="2011-02-23T09:03:03.700" Score="2" ViewCount="127" Body="&lt;p&gt;There might be none.  But I was thinking of links between number theory and physics, and this would seem like an example that would definitely solidify that link.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any known natural systems, or physical systems in general (possibly designed), that test primacy of a number directly using the laws of physics?  (Barring obvious examples such as machines intricately designed with interlocking parts/electronics to do such a job)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are none, I would not be surprised.  But I was just curious.&lt;/p&gt;&#xA;" OwnerUserId="71" LastActivityDate="2011-02-23T09:37:21.867" Title="Natural systems that test primacy of a number?" Tags="&lt;mathematical-physics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="5709" PostTypeId="2" ParentId="5703" CreationDate="2011-02-23T09:05:32.780" Score="2" ViewCount="" Body="&lt;p&gt;&quot;Free energy&quot; is a term used either by science or pseudoscience, and it means totally different things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In pseudoscience, &quot;free energy&quot; is energy for free that can be created out of nothing and it is the same thing as a (source of) perpetual motion, and it's an impossible concept. Conspiracy theories exist that civilizations are suppressing evidence that energy may be created for free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In actual science, free energy is the energy with an extra term, meaning the total energy that may be converted to work. Free energy is&#xA;$$A=U-TS$$&#xA;which is the Helmholtz version or&#xA;$$G=E+pV-TS$$&#xA;which is the Gibbs version. None of those energies is for &quot;free&quot; in the layman's sense - they're just different ways of counting energy and its transfer (including work) in which energy is ultimately conserved.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-23T09:05:32.780" CommentCount="4" />
  <row Id="5710" PostTypeId="2" ParentId="5705" CreationDate="2011-02-23T09:08:57.803" Score="3" ViewCount="" Body="&lt;p&gt;The principle of least action is much more elegant and fundamental than the explicit formulation of the laws of mechanics using three laws of Newton which disproves your intuition that the principle of least action is not elegant or fundamental.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two symmetries you mention are symmetries that constrain the evolution of physical systems but they don't totally determine what the evolution looks like. It is not true that the symmetries you mention are enough to derive the forces among a set of objects. $F=ma$, without specifying how $F$ depends on the positions (e.g. gravitational force), is just a definition of $F$ and it is a vacuous statement so you may derive it from anything - even from no assumptions at all.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-23T09:08:57.803" CommentCount="4" />
  <row Id="5711" PostTypeId="2" ParentId="1717" CreationDate="2011-02-23T09:27:59.913" Score="1" ViewCount="" Body="&lt;p&gt;Read something about Laval nozzles. In a narrow part of the nozzle the air attains the sound velocity and in a divergent part (diffuser) it is accelerated even more. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deceleration of such supersonic flow is only possible in a shock wave (suddenly), not gradually! There are plenty of pictures of supersonic flows from nozzles with visible shock wave disks or &quot;shock diamonds&quot; (in military aircrafts, cosmic jet engines, etc.).&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-24T15:37:23.570" LastActivityDate="2011-02-24T15:37:23.570" />
  <row Id="5712" PostTypeId="2" ParentId="5708" CreationDate="2011-02-23T09:37:21.867" Score="1" ViewCount="" Body="&lt;p&gt;There are connection between some physical system and the Reimann zeta function. Here is one of the related question at mathoverflow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathoverflow.net/questions/54501/riemann-zeta-function-connection-to-quantum-mechanics/54515#54515&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/54501/riemann-zeta-function-connection-to-quantum-mechanics/54515#54515&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I answered there the I recently found the article about this topic which may give you some answer: &quot;Physics of the Riemann Hypothesis&quot; &lt;a href=&quot;http://arxiv.org/abs/1101.3116&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1101.3116&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the Reimann function you can build primality (at least probabilistic) test.&lt;/p&gt;&#xA;" OwnerUserId="1795" LastActivityDate="2011-02-23T09:37:21.867" />
  <row Id="5713" PostTypeId="2" ParentId="5702" CreationDate="2011-02-23T09:59:32.277" Score="1" ViewCount="" Body="&lt;p&gt;Even in perfect/ideal vacuum, each point in a light beam acts as source of light due to its own interactions. So, you can't transport energy over infinite distance.&#xA;&lt;strong&gt;If you use single ray of light (not possible practically), you can - with confidence - transport energy over infinite distance in perfect vacuum.&lt;/strong&gt;&lt;br&gt;&#xA;Practically, even most powerful LASERs don't emit light having wavelengths 100% coherent in space &amp;amp; time. Plus, recorded mass density of best laboratory vacuum is 1.29 × 10^(-20) g/cm^3.&#xA;So, your smart goal (I hope) is just theoretical.&lt;/p&gt;&#xA;" OwnerUserId="2170" LastActivityDate="2011-02-23T09:59:32.277" />
  <row Id="5714" PostTypeId="1" CreationDate="2011-02-23T10:02:23.537" Score="8" ViewCount="654" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;This question exists because it has&#xA;  historical significance, but it is not&#xA;  considered a good, on-topic question&#xA;  for this site, so please do not use it&#xA;  as evidence that you can ask similar&#xA;  questions here.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;More info: &lt;a href=&quot;http://physics.stackexchange.com/faq&quot;&gt;http://physics.stackexchange.com/faq&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What are some failed ideas in physics which had been popular for some time but ultimately turned out to be wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking of something along the lines of Mie theory.&lt;/p&gt;&#xA;" LastEditorUserId="124" LastEditorDisplayName="user2172" LastEditDate="2011-08-28T00:19:43.760" LastActivityDate="2011-08-28T00:19:43.760" CommunityOwnedDate="2011-02-23T10:59:26.097" ClosedDate="2011-08-28T00:19:53.620" Title="Failed ideas in physics" Tags="&lt;soft-question&gt;" AnswerCount="9" CommentCount="5" FavoriteCount="6" />
  <row Id="5715" PostTypeId="1" CreationDate="2011-02-23T10:22:53.760" Score="6" ViewCount="176" Body="&lt;p&gt;Are there any applications of quantum information theory to physics?&lt;/p&gt;&#xA;" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-23T17:21:17.273" LastActivityDate="2011-02-23T18:02:19.293" Title="Are there any applications of quantum information theory to physics?" Tags="&lt;quantum-information&gt;&lt;quantum-computation&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="3" />
  <row Id="5716" PostTypeId="2" ParentId="5714" CreationDate="2011-02-23T10:35:02.003" Score="7" ViewCount="" Body="&lt;p&gt;Hm, strange question because who would be interesting in failed ideas?&#xA;But it turns out that sometimes ideas get recycled. I think the most prominent examples are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Kaluza-Klein_theory&quot; rel=&quot;nofollow&quot;&gt;Kaluza-Klein theory&lt;/a&gt; which attempted to combine general relativity and electromagnetism by introducing additional compact dimension. The theory itself can be said to have failed (mainly because it ignores other forces which were unknown at the time) but the idea itself is very powerful and survives in the form of compact dimensions in modern theories (most prominently string theory).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$SU(2)$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Yang-Mills&quot; rel=&quot;nofollow&quot;&gt;Yang-Mills theory&lt;/a&gt; which was first proposed as a theory of strong interaction. This failed but the theory itself is amazingly versatile and is used everywhere in modern physics. Strong interactions are now understood to be described by a $SU(3)$ YM while $SU(2)$ YM theory itself got recycled as a theory of weak interactions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I can think of other failed ideas (aether, phlogiston, Kelvin's vortices, etc.) but I won't discuss them because I don't see any point in discussing wrong theories. There is quite an infinite number of those out there and it takes enough effort studying the few theories we know are correct...&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-02-23T10:35:02.003" CommunityOwnedDate="2011-02-23T10:59:26.097" CommentCount="3" />
  <row Id="5717" PostTypeId="2" ParentId="5714" CreationDate="2011-02-23T10:47:32.637" Score="5" ViewCount="" Body="&lt;p&gt;There are many failed ideas. The following list is not exhaustive. But these are more famous failed ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Ptolemy's Geocentric model of the universe. (If you considered the model as scientific)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Sir Isaac Newton's theory of light corpuscles. ( Somewhat revived by photons) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Heat was considered to be a fluid called caloric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Luminiferous aether was considered to be a mechanical medium throughout the whole universe with gears and wheels to carry light and heat waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Various failed attempts to explain the Michaelson Morley experiment (aether drag etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) Mie's theory of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7) Thomson's atomic model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8) Einstein's various failed attempts for the &quot;unified field theory&quot; (5 dimesional kaluza Klein theory etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9) &quot;The steady state model&quot; of the universe by F. Hoyle et. al.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10) Various local hidden variable theories of quantum mechanics.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-23T10:56:49.283" LastActivityDate="2011-02-23T10:56:49.283" CommunityOwnedDate="2011-02-23T10:59:26.097" CommentCount="6" />
  <row Id="5718" PostTypeId="2" ParentId="5714" CreationDate="2011-02-23T10:47:43.263" Score="3" ViewCount="" Body="&lt;p&gt;Ether theory is a prime example.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-02-24T19:04:41.733" LastActivityDate="2011-02-24T19:04:41.733" CommunityOwnedDate="2011-02-23T10:59:26.097" />
  <row Id="5719" PostTypeId="2" ParentId="5682" CreationDate="2011-02-23T12:31:51.940" Score="4" ViewCount="" Body="&lt;p&gt;This is an expanded version of the comment I made before. It concerns solely the application of Hopf algebras to renormalization in quantum field theory and the combinatorics of Feynman diagrams. Other applications of quantum groups and Hopf algebras to low-dimensional physics etc. have been mentioned by others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Renormalization of multiloop Feynman graphs brings the problem of disentangling divergences coming from various subgraphs and removing all of them properly by local counterterms. This can be done systematically in perturbation theory by working from one loop up and keeping track of the loop order each counterterm contributes at e.g. by the powers of the Planck's constant. Carrying out this procedure blindly, one cannot make a mistake and at N loops, all subdivergences are removed by counterterms of order up to N-1, leaving only an overall local divergence that can be removed by a new counterterm of order N. Yet, it may be useful to know &lt;em&gt;a priori&lt;/em&gt; e.g. how to renormalize a single given multiloop diagram without having to go all the way from one loop up. The combinatorial recursive solution to this problem was found by Bogoliubov, Parasiuk, Hepp, and Zimmermann (BPHZ). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was originally noticed by Kreimer that the nesting of divergences can be encoded in an oriented rooted tree graph, whose root corresponds to the whole Feynman diagram, the other nodes to its subdiagrams, and a link connecting two nodes expresses the fact that one diagram is a subset of the other. All such rooted trees can be given the structure of a Hopf algebra. Roughly speaking, a coproduct of a tree is given by all splittings into &quot;cut-off branches&quot; and the &quot;remainder of the tree&quot;, connected to the root. In physics terms, this corresponds to all the ways one can shrink different Feynman subdiagrams into points. The antipode of the Hopf algebra then contains exactly the same information as the BPHZ recursion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Kreimer Hopf algebra depends only on the topology of Feynman diagrams. Also, resolution of &lt;em&gt;overlapping&lt;/em&gt; divergences within this framework is subtle. That is why Connes and Kreimer proposed to define a Hopf algebra structure directly on Feynman diagrams. This formulation treats nested and overlapping divergences on the same footing and is designed to allow for additional structure such as tensor structure of the diagrams or dependence on external momentum. Similarly to the previous, the coproduct is defined by all splittings of the Feynman diagram into a subdiagram and a graph where this subdiagram is replaced by a counterterm vertex. The antipode again encodes the BPHZ recursion. Within this approach, one can for example show that the Ward identities in QED generate an ideal in the Hopf algebra, and the renormalization can be carried out on the corresponding factor-Hopf algebra. This only expresses in mathematical terms the well known fact that renormalization preserves gauge invariance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I pointed out before, a mild introduction to this topic can be found in my diploma thesis from 2002, available at my home page. Unfortunately it is written in Czech, so for those non-Czech/Slovak I would recommend e.g. the review paper by Kreimer, &lt;a href=&quot;http://arxiv.org/abs/hep-th/0202110&quot; rel=&quot;nofollow&quot;&gt;hep-th/0202110&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, this was for lovers of neat mathematics. For the others, now comes the disappointment: what is this all actually good for? As @Luboš Motl pointed out in his comment, this idea, no matter how mathematically elegant, had better also contribute something to physics. I doubt it can ever improve our physical understanding of renormalization in quantum field theory; that has been well settled since Wilson. However, it sounds like something that might be helpful in automatizing multiloop calculations. For instance, Broadhurst and Kreimer in &lt;a href=&quot;http://arxiv.org/abs/hep-th/9912093&quot; rel=&quot;nofollow&quot;&gt;hep-th/9912093&lt;/a&gt; used the formalism to resum and renormalize chain-rainbow diagrams in Yukawa theory up to 30 loops with a high numerical precision. Yet, I asked a colleague who is an expert in multiloop calculations and he says that this approach has not become widely used. It seems that it is advocated more or less only by its inventors. Turning my back on a field I worked on myself? That is science :)&lt;/p&gt;&#xA;" OwnerUserId="1641" LastActivityDate="2011-02-23T12:31:51.940" CommentCount="1" />
  <row Id="5720" PostTypeId="1" CreationDate="2011-02-23T13:07:55.813" Score="2" ViewCount="227" Body="&lt;p&gt;Why do we get information about position and momentum when we go to different representations.Why is momentum, which was related to time derivative of position in classical physics, now in QM just a different representation brought about by some unitary transformation..Is Ehrenfest's theorem the only link?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just started studying QM.So please suggest some references explaining the structural aspects and different connections.I dont want to start with noncommutative geometry.I would like something of an introductory nature and motivating.&lt;/p&gt;&#xA;" OwnerUserId="2138" LastActivityDate="2011-02-23T18:05:46.123" Title="Momentum-Representations in Quantum Mechanics" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" />
  <row Id="5721" PostTypeId="1" AcceptedAnswerId="5723" CreationDate="2011-02-23T13:15:39.727" Score="9" ViewCount="385" Body="&lt;p&gt;Let suppose you roll a die, and it falls into a hidden place, for example under furniture.&lt;br&gt;&#xA;Then although the experiment has already been made (the die already has a number to show), that value can not be known, so the experiment was not fully realized.&lt;br&gt;&#xA;Then till you see the die's top side, the probability remain p = 1/6.&lt;br&gt;&#xA;I see no difference between this and the wave function collapse, at least as an analogy.&lt;br&gt;&#xA;Can someone explain a deeper difference?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1916" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-02-24T02:42:23.217" LastActivityDate="2011-02-24T11:14:53.500" Title="A die versus a quantum experiment" Tags="&lt;quantum-mechanics&gt;&lt;statistical-mechanics&gt;" AnswerCount="7" FavoriteCount="1" />
  <row Id="5722" PostTypeId="2" ParentId="5720" CreationDate="2011-02-23T13:25:02.123" Score="1" ViewCount="" Body="&lt;p&gt;Dear Ket, momentum in QM is not &quot;just a different representation brought about by some unitary transformation&quot;. As you probably already know, a physical state in quantum mechanics cannot have simultaneously a well-defined (sharp) value of momentum and position; this the Heisenberg uncertainty principle. However, you can still measure the &lt;em&gt;expectation values&lt;/em&gt; of both momentum and position in the same state. It is at the level of expectation values that momentum and position satisfy exactly the same relation as in classical physics; this is Ehrenfest's theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you talk about representations and unitary transformations, you probably mean the choice of basis in the Hilbert space of physical states. But this is merely a mathematical tool: to be able to work with vectors from the Hilbert space, it is suitable to choose a basis and work with the coordinates in this basis instead of the abstract vectors. Should you choose the basis of eigenstates of the position operator, the &quot;coordinates&quot; will be what is called the wave function. But you can choose any other basis as well. You can work in momentum representation, corresponding to the basis of eigenstates of the momentum operator, which is indeed related to the coordinate representation by a unitary transformation (called the Fourier transform in mathematics). This is because both bases are orthogonal, being formed by eigenstates of self-adjoint (Hermitian) operators. However, you can as well use &lt;em&gt;any&lt;/em&gt; basis, not related to any operator of an observable. What is physical are the expectation values of observables (which are independent of the choice of basis) and relations between them, which are via Ehrenfest's theorem equivalent to classical equations of motion.&lt;/p&gt;&#xA;" OwnerUserId="1641" LastActivityDate="2011-02-23T13:25:02.123" CommentCount="9" />
  <row Id="5723" PostTypeId="2" ParentId="5721" CreationDate="2011-02-23T13:28:25.443" Score="6" ViewCount="" Body="&lt;p&gt;You're absolutely right: the probabilities predicted by quantum mechanics are conceptually fully analogous to probabilities predicted by classical statistical mechanics, or statistical mechanics with a somewhat undetermined initial state - just like your metaphor with dice indicates. In particular, the predicted probability is a &quot;state of our knowledge&quot; about the system and no object has to &quot;collapse&quot; in any physical way in order to explain the measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two main differences between the classical and quantum probabilities which are related to one another:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In classical physics - i.e. in the case of dice assuming that it follows classical mechanics - one may imagine that the dice already has a particular value before we look. This assumption isn't useful to predict anything but we may assume that the &quot;sharp reality&quot; exists prior to and independently of any observations. In quantum mechanics, one is not allowed to assume this &quot;realism&quot;. Assuming it inevitably leads to wrong predictions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The quantum probabilities are calculated as $|c|^2$ where $c$ are complex numbers, the so-called probability amplitudes, which may interfere with other contributions to these amplitudes. So the probabilities of outcomes, whenever some histories may overlap, are not given as the sum over probabilities but the squared absolute value of the sum of the complex probability amplitudes: in quantum mechanics, we first sum the complex numbers, and then we square the result to get the total probability. On the other hand, there is no interference in classical physics; in classical physics, we would surely calculate the probabilities of individual histories, by any tools, and then we would sum the probabilities.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Of course, there is a whole tower of differences related to the fact that the observable (quantities) in quantum mechanics are given by operators that don't commute with each other: this leads to new logical relationships between statements and their probabilities that would be impossible in classical physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A closely related question to yours:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/3158/why-quantum-entanglement-is-considered-to-be-active-link-between-particles/3163&quot;&gt;Why quantum entanglement is considered to be active link between particles?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason why people often misunderstand the analogy between the odds for dice and the quantum wave function is that they imagine that the wave function is a classical wave that may be measured in a single repetition of the situation. In reality, the quantum wave function is not a classical wave and it cannot be measured in a single case of the situation, not even in principle: we may only measure the values of the quantities that the wave function describes, and the result is inevitably random and dictated by a probability distribution extracted from the wave function.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-23T13:37:03.070" LastActivityDate="2011-02-23T13:37:03.070" CommentCount="6" />
  <row Id="5724" PostTypeId="2" ParentId="5721" CreationDate="2011-02-23T13:28:28.263" Score="-1" ViewCount="" Body="&lt;p&gt;There is a huge difference. Firstly for a classical dice, if you could measure the detail initial conditions of the system like position and momentum with arbitrary accuracy you could have predicted the exact result in principle. In principle you can measure but in practice due to practical difficulties you can't. A classical dice always evolve in time in deterministic manner. Since you can't measure you can't predict what will happen. In Q.T. you can't measure the two conjugate variables like position and momentum with equal certainty to arbitrary precision even in principle. Quantum systems are intrinsically indeterministic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, a classical dice always follow a single history in space and time unlike the quantum analogue. In quantum theory, all possible histories do interfere with each other unless measured. That means for a quantum dice all six possible histories will interfere with each other and the state of the system will be given by a linear superposition of all those possibilities. As soon as a measuring device observes the system the interference will disappear and you will see only one of the six possible outcomes according to born rule.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-23T13:42:33.473" LastActivityDate="2011-02-23T13:42:33.473" CommentCount="7" />
  <row Id="5725" PostTypeId="2" ParentId="5682" CreationDate="2011-02-23T13:38:56.150" Score="1" ViewCount="" Body="&lt;p&gt;I think of this kind of math that the hope is that tidying up a complex process like regularization/renormalization by using a relatively neat structure --which a Hopf algebra certainly is-- may give us a hint towards something quite different that we wouldn't have thought of doing if we just left things messy. Different kinds of tidying up may suggest quite different ideas to try next. Feynman says at one point (I'd be glad if someone can tell me where -- a Question, perhaps, if not) that one should try every mathematical way one can think of using to think about a problem, spend lots of time finding the relationships between the different ways, then mull the problem, which may well suggest a new mathematical way to use to think about the problem, then mull again. Repeat until something publishable emerges. From this point of view, I think Hopf algebras are something that anyone who wants to do serious Physics research has to become familiar with, because they are seriously tidy, whether they're useful or not for your present purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Physics, however, I have come to think that Hopf algebras are too tied to perturbation theory and to Feynman diagrams to get us out into a different conceptual playground. Happy to be proved wrong, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The level of abstraction of Hopf algebras is close enough to my math limits that getting them internalized has proved a challenge. My feeling is that no-one has yet found a way to present Hopf algebras at an engineering level of concept and usability. I suspect the 3rd or 4th monograph might get to that point, but we haven't even had the first yet. I also haven't yet seen a review article that has hit the mark. I think an engineering level presentation will find a ready market, because although Tomáš Brauner's expert colleague and others may not have used these methods much, one can be sure that from time to time he and other experts wonder whether it would be a good idea to use them and whether it might be helpful to teach at least some aspects of renormalization using something like Hopf algebras.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-02-23T13:38:56.150" CommentCount="0" />
  <row Id="5726" PostTypeId="2" ParentId="5676" CreationDate="2011-02-23T14:43:02.040" Score="1" ViewCount="" Body="&lt;p&gt;I thought about this last night, when I first read the question.  I am a bit uncertain about the nature of the question.  However, I thought I would offer up a boilerplate answer on the string vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closed string contains two separate sets of modes, left and right (or clockwise and counter-clockwise) modes.  The ground state is a scalar tachyon with $\alpha’M^2~=~-4$.  The $SO(24)$  multiplet of physical states for the open string are doubled, into $SO(24)\times SO(24)$ in the closed string.  The gravitational massless state is the symmetric part of $|\Omega^{\mu\nu}\rangle~=~\alpha^\mu_{-1}{\tilde\alpha}^\nu_{-1}|0\rangle$  in the closed string.  The state ${1\over 2}|\Omega^{\{\mu\nu\}}\rangle$ is a spin $2$ field, which is the graviton.  The antisymmetric portion is a second rank tensor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various combinations or tensor products of left and right movers on the open string can be used to describe the closed string modes.  The for $\alpha’M^2~=~4(N~-~1)$, which from the two modes $N~=~2$ or Young’s Tableaux $\square\square\times\square\square$ you can construct positive states.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-02-23T15:02:58.053" LastActivityDate="2011-02-23T15:02:58.053" CommentCount="3" />
  <row Id="5727" PostTypeId="2" ParentId="2900" CreationDate="2011-02-23T14:55:03.303" Score="3" ViewCount="" Body="&lt;p&gt;This question is &lt;em&gt;slightly&lt;/em&gt; faulty...  Part of the US GPS system &lt;em&gt;is&lt;/em&gt; geostationary (the WAAS component).  It's used in conjunction with the non-geostationary birds for higher precision fixes.  While primarily used for aircraft instrument approaches, there are off the shelf USB GPS computer peripherals that use WAAS.&lt;/p&gt;&#xA;" OwnerUserId="919" LastEditorUserId="919" LastEditorDisplayName="" LastEditDate="2011-02-28T13:25:39.767" LastActivityDate="2011-02-28T13:25:39.767" />
  <row Id="5729" PostTypeId="2" ParentId="5715" CreationDate="2011-02-23T15:02:32.377" Score="3" ViewCount="" Body="&lt;p&gt;There's a &lt;em&gt;huge&lt;/em&gt; body of literature in applying ideas from quantum information to the study of strongly interacting many body systems. If I have time I'll edit this answer with my own personal review (primarily all of the lovely simulation techniques that QIT has given a solid physical and conceptual foundation to) but as a jumping off point, there is a set of lecture notes with excellent references &lt;a href=&quot;http://arxiv.org/abs/1003.3153&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2011-02-23T15:02:32.377" />
  <row Id="5732" PostTypeId="2" ParentId="5682" CreationDate="2011-02-23T15:33:38.147" Score="1" ViewCount="" Body="&lt;p&gt;Hopf algebras are also used in quantum gravity:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In the context of quantum gravity in 2+1 dimensions coupled to particles it is well known that the Poincare (or de Sitter, or anti de Sitter) local symmetry group becomes quantum deformed and turns out to be Drinfeld double of SU(2) (and similarly when the cosmological constant is not zero). The good references are&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Quantum group symmetry and particle scattering in (2+1)-dimensional quantum gravity. hep-th/0205021&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lessons from (2+1)-dimensional quantum gravity arXiv:0710.5844 [gr-qc]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more general picture you may look at the papers of S. Majid like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum geometry and the Planck scale q-alg/9701001&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopf algebras are also used in the context of quantum gravity in 3+1 dimensions, see, eg. my review&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Introduction to doubly special relativity hep-th/0405273&lt;/p&gt;&#xA;" OwnerUserId="1721" LastActivityDate="2011-02-23T15:33:38.147" />
  <row Id="5733" PostTypeId="2" ParentId="5720" CreationDate="2011-02-23T16:26:04.160" Score="2" ViewCount="" Body="&lt;p&gt;You can get information for all observables in any representation. The reason to go to different ones is that it is easier to work with them depending on what you are doing. They are all equivalent by the Stone-von Neumann theorem, so it is a matter of convenience. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a theorem (mathematical) that roughly says that for any operator, that is of interest in QM, there is a representation of the operator as a multiplication operator, in it it acts as a multiplication by a function . In the coordinate space the position operators are multiplication by the coordinates. In momentum space it is the momentum that is represented as a multiplication. It is true for any of the QM observables. Unfortunately (or fortunately) since they do not commute there isn't a single representation for all of them. Hence people use more than one.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; In response to the comment. This is probably written in many books, but here is a reference. Look at Folland's &quot;Quantum Field Theory. A Tourist Guide for Mathematicians&quot;. The first section of chapter 3 gives a nice motivation for the use of self-adjoint operators for modeling QM observables.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2011-02-23T18:05:46.123" LastActivityDate="2011-02-23T18:05:46.123" CommentCount="4" />
  <row Id="5734" PostTypeId="2" ParentId="5682" CreationDate="2011-02-23T16:41:11.810" Score="4" ViewCount="" Body="&lt;p&gt;This is essentially an addition to the list of @4tnemele&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to add some earlier work to this list, namely Discrete Gauge Theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Discrete gauge theory in 2+1 dimensions arises by breaking a gauge symmetry with gauge group $G$ to some lower &lt;strong&gt;discrete&lt;/strong&gt; subgroup $H$, via a Higgs mechanism. The force carriers ('photons') become massive which makes the gauge force ultra-short ranged. However, as the gauge group is not completely broken we still have the the Aharanov-Bohm effect. If H is Abelian this AB effect is essentially a 'topological force'. It gives rise to a phase change when one particle loops around another particle. This is the idea of fractional statistics of Abelian anyons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The particle types that we can construct in such a theory (i.e. the one that are &quot;color neutral&quot;) are completely determined by the residual, discrete gauge group $H$. To be more precise: a particle is said to be charged if it carries &lt;strong&gt;a representation of the group H&lt;/strong&gt;. The number of different particle types that carry a charge is then equal to the number of irreducible representations of the group H. This is similar to ordinary Yang-Mills theory where charged particles (quarks) carry the fundamental representation of the gauge group (SU(3). In a discrete gauge theory we can label all possible charged particle types using the representation theory of the discrete gauge group H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are also other types of particles that can exist, namely those that carry flux. These flux carrying particles are also known as magnetic monopoles. In a discrete gauge theory the flux-carrying particles are labeled by the &lt;strong&gt;conjugacy classes of the group H&lt;/strong&gt;. Why conjugacy classes? Well, we can label flux-carrying particles by elements of the group H. A gauge transformation is performed through conjugacy, where $ |g_i\rangle \rightarrow |hg_ih^{-1}\rangle $ for all particle states $|g_i\rangle$ (suppressing the coordinate label). Since states related by gauge transformations are physically indistinguishable the only unique flux-carrying particles we have are labeled by conjugacy classes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that all then? Nope. We can also have particles which carry both charge and flux -- these are known as dyons. They are labeled by both an irrep and a conjugacy class of the group $H$. But, for reasons which I wont go into, we cannot take all possible combinations of possible charges and fluxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It has to do with the distinguishability of the particle types. Essentially, a dyon is labeled by $|\alpha, \Pi(g)\rangle$ where $\alpha$ is a conjugacy class and $\Pi(N(g))$ is a representation of the associated normalizer $N(\alpha)$ of the conjugacy class $\alpha$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The downside of this approach is the rather unequal setting of flux carrying particles (which are labeled by conjugacy classes), charged particles (labeled by representations) and dyons (flux+compatible charge). A unifying picture is provided by making use of the (quasitriangular) Hopf algebra $D(H)$ also known as a quantum double of the group $H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this language &lt;em&gt;all&lt;/em&gt; particles are (irreducible) representations of the Hopf algebra $D(H)$. A Hopf Algebra is endowed with certain structures which have very physical counterparts. For instance, the existence of a &lt;strong&gt;tensor product&lt;/strong&gt; allows for the existence of &lt;strong&gt;multiple particle configurations&lt;/strong&gt; (each particle labeled by their own representation of the Hopf algebra). The &lt;strong&gt;co-multiplication&lt;/strong&gt; then defines &lt;strong&gt;how the algebra acts&lt;/strong&gt; on this tensored space. the existence of an &lt;strong&gt;antipode&lt;/strong&gt; (which is a certain mapping from the algebra to itself) ensures the existence of an &lt;strong&gt;anti-particle&lt;/strong&gt;. The existence of a unit labels the vacuum (=trivial particle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also go beyond the structure of a Hopf algebra and include the notion of an R-matrix. In fact, the quasitriangular Hopf Algebra (i.e. the quantum double) does precisely this: add the R-matrix mapping. This R-matrix describes what happens when one particle loops around another particle (braiding). For non-Abelian groups $H$ this leads to non-Abelian statistics. These quasitriangular Hopf algebras are also known as quantum groups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays the language of discrete gauge theory has been replaced by more general structures, referred to by topological field theories, anyon models or even modular tensor categories. The subject is huge, very rich, very physical and a lot of fun (if you're a bit nerdy ;)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/9511201&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9511201&lt;/a&gt; (discrete gauge theory)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.theory.caltech.edu/people/preskill/ph229/&quot; rel=&quot;nofollow&quot;&gt;http://www.theory.caltech.edu/people/preskill/ph229/&lt;/a&gt; (lecture notes: check out chapter 9. Quite accessible!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/quant-ph/9707021&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/quant-ph/9707021&lt;/a&gt; (a simple lattice model with anyons. There are definitely more accessible review articles of this model out there though.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/0707.1889&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0707.1889&lt;/a&gt; (review article, which includes potential physical realizations)&lt;/p&gt;&#xA;" OwnerUserId="1287" LastEditorUserId="1287" LastEditorDisplayName="" LastEditDate="2011-02-23T21:03:19.983" LastActivityDate="2011-02-23T21:03:19.983" CommentCount="3" />
  <row Id="5735" PostTypeId="2" ParentId="5702" CreationDate="2011-02-23T16:44:57.333" Score="7" ViewCount="" Body="&lt;p&gt;A. All light sources (even lasers) are subject to &lt;a href=&quot;http://www.rp-photonics.com/diffraction_limited_beams.html&quot; rel=&quot;nofollow&quot;&gt;a diffraction limit&lt;/a&gt;, so any light beam will eventually diverge with an angle $\theta$ given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta \approx \frac{\lambda}{A_T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\lambda$ is the wavelength of the light and $A_T$ is the aperture of the light beam source (and &quot;eventually&quot; means for distances much greater than $A_T$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any beam diverging with a constant angle will have an intensity following an inverse-square law, though the total beam power will be unaffected (if we can neglect light absorption and scattering).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diffraction limit can be seen as a consequence of the Heisenberg uncertainty principle: calling one transverse coordinate $x$ and applying the position-momentum uncertainty relation at the source (assuming that the inequality is saturated and the position uncertainty is equal to the aperture size), we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta p_x^{source} \Delta x^{source} \approx \hbar$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta p_x^{source} \approx \frac{\hbar}{A_T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Far from the source, at a distance $R \gg A_T$, the transverse position uncertainty will be dominated by the transverse momentum uncertainty at the source, giving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta \approx \frac{\Delta x^{far}}{R} \approx \frac{R \Delta p_x^{source}}{p}\frac{1}{R} \approx \frac{\hbar\lambda}{A_T \hbar 2\pi} = \frac{1}{2\pi}\frac{\lambda}{A_T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that differs from the previously given result by a constant, due to the approximations involved and the imprecise nature of the &quot;deltas&quot;. &lt;a href=&quot;http://www.pma.caltech.edu/Courses/ph136/yr2004/book03/chap07/0207.1.pdf&quot; rel=&quot;nofollow&quot;&gt;A more precise treatment&lt;/a&gt; shows that $\theta = \lambda/A_T$ is a better approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. Photons are not scattered in a perfect vacuum. And the intergalactic space, while not a perfect vacuum, is so empty that even photons originated in galaxies at billions of light years can be received.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. Yes, but you will need a receiver with a big aperture to receive this light. In more precise terms, you will need a receiver with an aperture $A_R$ given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_R \approx \frac{\lambda}{A_T}R$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\lambda$ is the wavelength, $A_T$ is the aperture of the transmitter and $R$ is the distance from the transmitter to the receiver.&lt;/p&gt;&#xA;" OwnerUserId="1656" LastEditorUserId="1656" LastEditorDisplayName="" LastEditDate="2011-02-24T12:41:20.987" LastActivityDate="2011-02-24T12:41:20.987" CommentCount="2" />
  <row Id="5736" PostTypeId="1" CreationDate="2011-02-23T16:45:36.397" Score="4" ViewCount="117" Body="&lt;p&gt;How do you derive Noether's theorem when the action combines chiral, antichiral, and full superspace?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-27T10:43:07.337" Title="How do you derive Noether's theorem when the action combines chiral, antichiral, and full superspace?" Tags="&lt;symmetry&gt;&lt;supersymmetry&gt;&lt;noethers-theorem&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="5737" PostTypeId="2" ParentId="5721" CreationDate="2011-02-23T16:54:18.123" Score="10" ViewCount="" Body="&lt;p&gt;A somewhat similar (yet inverted) question was posed, infamously, by Einstein, Podolsky, and Rosen as an argument against the same phenomenon you are challenging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic argument was that such quantum wave collapses are indistinguishable from dice falling under couches, but under opposite grounds as yours -- both must be &quot;determined&quot; by what were later named &quot;hidden variables&quot;.  That a quantum wave collapse, like a falling dice, is completely deterministic by the nature of its initial conditions, and its apparent randomness is due to our current inability to measure these initial conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was John S. Bell who thought up of an experiment to test this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic idea was statistical, and involved quantum entanglement -- if hidden variables existed (and the particles already having a determined, yet unknown state at their inception), statistics should yield a certain correlation between entangled particles.  If there are no hidden variables, then this correlation would not show up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surprisingly, this correlation did not manifest by even a large margin of experimental error.  This proves an inherent difference in the randomness of quantum wave collapse as opposed to simply not knowing the landing of a dice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result culminated in &lt;a href=&quot;http://en.wikipedia.org/wiki/Bell%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Bell's theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-23T17:50:40.167" LastActivityDate="2011-02-23T17:50:40.167" CommentCount="3" />
  <row Id="5738" PostTypeId="2" ParentId="5714" CreationDate="2011-02-23T16:56:54.127" Score="2" ViewCount="" Body="&lt;p&gt;The wikipedia article on &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pathological_science&quot; rel=&quot;nofollow&quot;&gt;Pathological science&lt;/a&gt;&quot; might be good reading. In particular, the &lt;a href=&quot;http://en.wikipedia.org/wiki/N_ray&quot; rel=&quot;nofollow&quot;&gt;N-rays saga&lt;/a&gt; is a great case study in a wrong idea that was even &quot;reproducible&quot; for a while; see the &lt;a href=&quot;http://www.newyorker.com/reporting/2010/12/13/101213fa_fact_lehrer&quot; rel=&quot;nofollow&quot;&gt;decline effect&lt;/a&gt; and Feynman's &lt;a href=&quot;http://www.lhup.edu/~DSIMANEK/cargocul.htm&quot; rel=&quot;nofollow&quot;&gt;cargo cult science&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-02-23T17:04:58.730" LastActivityDate="2011-02-23T17:04:58.730" CommunityOwnedDate="2011-02-23T16:56:54.127" CommentCount="1" />
  <row Id="5739" PostTypeId="1" CreationDate="2011-02-23T16:59:27.750" Score="4" ViewCount="281" Body="&lt;p&gt;Particularly, I can somewhat appriciate the Fermi energy idea - the radius of Fermi surface which is a sphere. But is there any quantitative use of more complicated Fermi surfaces? &lt;/p&gt;&#xA;" OwnerUserId="2177" LastActivityDate="2011-02-23T18:57:54.100" Title="What is Fermi surface and why is this concept so useful in metals research?" Tags="&lt;condensed-matter&gt;&lt;solid-state-physics&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="5740" PostTypeId="1" CreationDate="2011-02-23T17:23:38.467" Score="0" ViewCount="167" Body="&lt;p&gt;What actually is 1 light year? What is the equivalent time in Earth time space?&lt;/p&gt;&#xA;" OwnerUserId="1884" LastActivityDate="2011-02-23T17:27:47.413" ClosedDate="2011-02-23T17:46:38.037" Title="Light years in terms of actual time space" Tags="&lt;spacetime&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="5741" PostTypeId="2" ParentId="5740" CreationDate="2011-02-23T17:27:47.413" Score="1" ViewCount="" Body="&lt;p&gt;&quot;One lightyear&quot; is &quot;the distance that light travels in one year if it goes through empty space.&quot; Since the speed of light is 299,792,458 meters per second, one lightyear is 9,460,730,472,580.8 kilometers.&lt;/p&gt;&#xA;" OwnerUserId="1655" LastActivityDate="2011-02-23T17:27:47.413" />
  <row Id="5742" PostTypeId="2" ParentId="5739" CreationDate="2011-02-23T17:29:35.977" Score="3" ViewCount="" Body="&lt;p&gt;The Fermi surface is simply the surface in momentum space where, in the limit of zero interactions, all fermion states with (crystal) momentum $|k|&amp;lt;|k_F|$ are occupied, and all higher momentum states are empty. Amazingly, Luttinger and Ward showed that the Fermi surface survives even with interactions to all orders in perturbation theory (Oshikawa later showed this nonperturbatively).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of the Fermi surface is that this is where all of the low-lying excitations of the system live -- the Fermi &lt;em&gt;energy&lt;/em&gt; is so much larger than room temperature that for room-temperature experiments, &lt;em&gt;all&lt;/em&gt; of the thermodynamics is dominated by excitations right at the Fermi surface, and thus knowing its structure is very important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more advanced reason for the importance of the Fermi surface is that by knowing its structure (look up &quot;Fermi surface nesting&quot;), we can understand the &lt;em&gt;instabilities&lt;/em&gt; of a metal, for example, for sufficiently low temperature, a normal metal settling into a charge density wave state.&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2011-02-23T17:29:35.977" CommentCount="6" />
  <row Id="5743" PostTypeId="1" CreationDate="2011-02-23T17:29:43.783" Score="-1" ViewCount="335" Body="&lt;p&gt;Reading this PE question&#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/5702/can-we-transport-energy-over-infinite-distances-through-vacuum-using-light&quot;&gt;can-we-transport-energy-over-infinite-distances-through-vacuum-using-light&lt;/a&gt;, a related question arises naturally: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is &lt;strong&gt;energy transported&lt;/strong&gt; (by light)? -- (I did believed in this answer until now)&#xA;or &lt;strong&gt;energy is already 'in site'&lt;/strong&gt; (vacuum) just expecting to be excited by the photons? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This news insinuated the doubt &lt;a href=&quot;http://www.wired.com/wiredscience/2011/02/real-live-antilaser/&quot; rel=&quot;nofollow&quot;&gt;anti-laser&lt;/a&gt;(1)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The antilaser does the reverse: &lt;strong&gt;Two&#xA;  perfect beams of laser light go in,&#xA;  and are completely absorbed.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If vacuum is able to absorb energy then it can do the reverse, and supply energy.&#xA;We are already prepared to accept that &lt;a href=&quot;http://en.wikipedia.org/wiki/Vacuum_energy&quot; rel=&quot;nofollow&quot;&gt;vacuum has energy&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am inclined to accept that &lt;strong&gt;energy do not travel at all&lt;/strong&gt;. What is travelling is the excitation of vacuum, and we call this: &lt;em&gt;photons&lt;/em&gt;.&#xA;It may appear a question of semantics, but I think that the explicit reconaissance of this notion can be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) The two rays entering the slab are in phase opposition when they met, and cancel. The nature of cancelation was obscure to me, until now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;added:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I googled this: &quot;where goes the energy in a destructive interference&quot; and followed past answers to this question. Someone answered &quot;into the surrounding environment.&quot; We are in minority;) Most of the times they said that the total extinguishing is impossible. This anti-laser experiment shows that energy is destroyed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see the same effect with sound cancelation , with boat wake (trailing waves)  cancelation (by double/triple hull or when they sail in formation), and now with light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;added after 2 answers&lt;/strong&gt;&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/xrA2J.jpg&quot; alt=&quot;Destructive interference&quot;&gt; image from &lt;a href=&quot;http://www.astro-canada.ca/_en/a2313.html&quot; rel=&quot;nofollow&quot;&gt;astro-canada.ca&lt;/a&gt;&lt;br&gt;&#xA;What is amazing is that this fact is inside the theory since the begining, quoting from there:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 1801, the British physicist Thomas&#xA;  Young demonstrated that light&#xA;  propagates as waves, like waves on the&#xA;  surface of water. Young understood&#xA;  that when two light waves meet, they&#xA;  interact with each other. Scientists&#xA;  call this “interference”.&#xA;  In the opposite scenario, where the&#xA;  crests of one wave are aligned with&#xA;  the troughs of another, they cancel&#xA;  each other out and the &lt;strong&gt;light&#xA;  disappears&lt;/strong&gt;. This is destructive&#xA;  interference.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As anna says in her answer the actual theorethical framework does not understand a vanishing of energy. &#xA;The question remains open: What happens to energy when '&lt;strong&gt;&lt;em&gt;light disappears&lt;/em&gt;&lt;/strong&gt;' or '&lt;em&gt;light cancels&lt;/em&gt;' or '&lt;em&gt;destructive interference&lt;/em&gt;'?&lt;br&gt;&#xA;Some will invoke extra dimensions as a repository of energy, and will once again ignore the ubiquitous vacuum.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-03-28T10:21:18.897" LastActivityDate="2011-03-28T11:15:03.910" Title="Anti-laser: How sure we are that energy is transported?" Tags="&lt;energy&gt;&lt;photons&gt;&lt;vacuum&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="5744" PostTypeId="2" ParentId="5743" CreationDate="2011-02-23T17:34:30.960" Score="1" ViewCount="" Body="&lt;p&gt;We regularly measure the energy stored in electric and magnetic fields. Think capacitors and inductors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;We do &lt;em&gt;engineering&lt;/em&gt; with energy stored that way.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fields have energy, and the propagation of light generated fields far from the source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So energy was transported. QED.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So where does the energy go in the experiment you linked? Read the article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When the alignment was right, the light waves canceled each other out. The silicon absorbed the light and converted it to another form of energy, like heat or electrical current.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The two beams are being used to drive a single process both directions through it's transition (which will require the same energy input and have the same cross-section by time-reversal symmetry). Presumably these processes couple to the bulk in some way and each dissipates leaks energy through that coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper itself is available as &lt;a href=&quot;http://www.eng.yale.edu/caolab/papers/Sci11b.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.eng.yale.edu/caolab/papers/Sci11b.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-02-23T18:41:41.957" LastActivityDate="2011-02-23T18:41:41.957" CommentCount="9" />
  <row Id="5745" PostTypeId="2" ParentId="5721" CreationDate="2011-02-23T17:35:15.733" Score="2" ViewCount="" Body="&lt;p&gt;In QM one can also deal with probabilities instead of probability amplitudes when one describes the prepared beam in terms of density matrix. For example (I assume there is six states $|n&gt;$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho = (1/6)\sum_n {|n&amp;gt;&amp;lt;n|}$.     (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mixture is generally different from a pure (superposition) state with the same probabilities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\psi = (1/6)^{1/2}\sum_n {|n&amp;gt;}$.   (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dice experiment corresponds to (1). It would be nice if you could read something about difference between mixtures and superpositions in QM (see sources of coherent and non coherent light, for example).&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-24T11:14:53.500" LastActivityDate="2011-02-24T11:14:53.500" />
  <row Id="5746" PostTypeId="2" ParentId="1779" CreationDate="2011-02-23T17:43:05.547" Score="2" ViewCount="" Body="&lt;p&gt;This is where I found most of my opportunities: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://tiptop.iop.org&quot; rel=&quot;nofollow&quot;&gt;http://tiptop.iop.org&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1939" LastActivityDate="2011-02-23T17:43:05.547" />
  <row Id="5747" PostTypeId="2" ParentId="5715" CreationDate="2011-02-23T18:02:19.293" Score="0" ViewCount="" Body="&lt;p&gt;Try this &lt;a href=&quot;http://www.qubit.org/people/david/structure/Documents/Research%20Papers/PPQT.pdf&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; by david deutsch...&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2011-02-23T18:02:19.293" CommentCount="1" />
  <row Id="5748" PostTypeId="1" AcceptedAnswerId="5752" CreationDate="2011-02-23T18:33:03.470" Score="4" ViewCount="80" Body="&lt;p&gt;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Deep_inelastic_scattering&quot; rel=&quot;nofollow&quot;&gt;deep inelastic scattering&lt;/a&gt;, you describe a collision using the &lt;a href=&quot;http://ikpe1101.ikp.kfa-juelich.de/briefbook_part_detectors/node42.html#41&quot; rel=&quot;nofollow&quot;&gt;variables&lt;/a&gt; $Q^2 = -q^2$ (probe virtuality) and $x = Q^2/2p\cdot q$ (Bjorken x, parton momentum fraction). Now, I seem to remember reading somewhere that in the infinite momentum frame, one of these variables primarily depends on momentum in the transverse direction and the other primarily depends on momentum in the longitudinal direction, but I can't remember which, and checking my most likely sources hasn't turned up anything. Is this actually the case, or is my memory playing tricks on me? And if it is true, which is which?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies for asking a vague question, but I'm really drawing a blank on what it is I might have read, and asking here seemed like the quickest way to figure it out ;)&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-02-23T19:38:36.390" Title="Do $x$ and $Q^2$ associate with particular directions in the infinite momentum frame?" Tags="&lt;high-energy-physics&gt;&lt;momentum&gt;&lt;deep-inelastic-scattering&gt;" AnswerCount="1" />
  <row Id="5749" PostTypeId="2" ParentId="5739" CreationDate="2011-02-23T18:57:54.100" Score="1" ViewCount="" Body="&lt;p&gt;I think what you might be interested in are &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_Hove_singularity&quot; rel=&quot;nofollow&quot;&gt;Van Hove singularities&lt;/a&gt; or the critical points of the Fermi surface, where the density of states as given by $dN/dK_{|k=k_f|}$ diverges. Now $dN/dk$ is proportional to the inverse of the gradient of the energy $ dN/dk \propto 1/\nabla E $. The locations with the greatest d.o.s. on the Fermi surface will exhibit singularities in various absorption and emission spectra. These are precisely those locations where the Fermi surface fails to be a smooth, differentiable surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Critical surfaces can be 0D (Fermi point), 1D (line), 2D (Fermi surface). Complicated substances (such as the High $T_c$ superconductors for e.g. which are composed of layers of cuprates) will in general have complicated Fermi surfaces. Because critical surfaces are topological entities they are robust with respect to small perturbations of the microscopic Hamiltonian of the system. In other words the critical surfaces determine the universality class to which the given surface belongs. The universality class determines whether a given material is a superconductor, ferromagnet, Mott insulator etc. Clearly to take a material from one universality class to another requires that one cross a phase boundary. Such a change also requires that the Fermi surface topology undergo a change. Consequently changes in Fermi surface topology can be understood as signs of phase transitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Fermi surface is also important for reasons other than its topology, which describes the &lt;em&gt;global&lt;/em&gt; characteristics of the material. Others will likely mention some of the complementary local aspects of the Fermi surface.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-23T18:57:54.100" CommentCount="5" />
  <row Id="5751" PostTypeId="2" ParentId="5736" CreationDate="2011-02-23T19:02:13.693" Score="3" ViewCount="" Body="&lt;p&gt;The full superspace terms are the &quot;most general ones&quot; but you may convert the chiral and antichiral terms to the full superspace form, too. In particular,&#xA;$$\int d^2\bar\theta = \int d^2\theta\, d^2 \bar\theta\, \theta_1\theta_2$$&#xA;and similarly for its complex conjugate. Sorry if a sign is wrong. Note that $\theta_1\theta_2$ may be written as $\epsilon^{ab}\theta_a \theta_b/2$ up to a sign if you happen to be annoyed by the explicit components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that chiral superfields only depend on $\theta$ but not $\bar\theta$ doesn't matter: it just means that it is a more constrained field. But you may still imagine that it's a function of the full superspace that just happens to have a special form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you convert the action to an integral over the full superspace, you may proceed just like you would proceed for the full superspace, D-term-like terms. I am actually a bit unfamiliar with this thing - so I would convert the action to the components (no superspace at all) and proceeded just like in non-supersymmetric theories. Of course, the resulting conserved quantities wouldn't be nicely organized in supermultiplets - which they can be in a supersymmetric theory.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-23T19:02:13.693" />
  <row Id="5752" PostTypeId="2" ParentId="5748" CreationDate="2011-02-23T19:38:36.390" Score="4" ViewCount="" Body="&lt;p&gt;David, your memory is right. You want to use the Drell-Yan-West frame with $q^+=0$, using the light-cone coordinates. In those coordinates, $Q^2=-q^2$ is clearly $q_\perp^2$ because the $q^+ q^-$ term vanishes and only depends on the transverse components of $q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Drell-Yan limit, the fraction $x$ becomes&#xA;$$x_1 = p_1^- / P_1^-$$&#xA;and only depends on the longitudinal components as in equation (26) of&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0909/0909.4159v2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/arxiv/pdf/0909/0909.4159v2.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where I send you to refresh some memory - as well as provide you with new keywords and references you may search through. I won't try to make my answer perfect, sorry.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-23T19:38:36.390" CommentCount="2" />
  <row Id="5753" PostTypeId="1" AcceptedAnswerId="5754" CreationDate="2011-02-23T19:52:21.317" Score="6" ViewCount="390" Body="&lt;p&gt;This question was inspired by Abstruse Goose :)&#xA;&lt;a href=&quot;http://abstrusegoose.com/342&quot; rel=&quot;nofollow&quot;&gt;http://abstrusegoose.com/342&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's well known that any attempt to describe the spin (of say an electron) in terms of non-internal spatial coordinates is futile since this is just orbital angular momentum and can't take half-integral values. Spin is currently understood to be simply the generator of rotations, which being Hermitian should correspond to an &quot;observable&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I remember hearing about attempts to model the electron as an extended object that literally spins and that in these attempts you run into problems with relativity. (A back of the envelope calculation shows that the &quot;periphery&quot; of the electron should be moving 35 times the speed of light) &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;My first question is whether someone knows of a serious attempt in the literature at doing something as described in the previous paragraph&lt;/li&gt;&#xA;&lt;li&gt;How is spin understood (or incorporated) in the framework of (a) string theory (b) loop quantum gravity?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-02-24T06:27:03.957" Title="The history and modern understanding of spin" Tags="&lt;quantum-mechanics&gt;&lt;history&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="5754" PostTypeId="2" ParentId="5753" CreationDate="2011-02-23T21:05:58.573" Score="3" ViewCount="" Body="&lt;p&gt;First of all, spin, by definition, is the internal angular momentum. And angular momentum is,  by definition, the observable that is conserved as a consequence of the rotational symmetry of the laws of physics. That's the most general definition that follows from Noether's theorem. It follows that in quantum mechanics, the angular momentum is always given by the generator of the rotations. That must be true in any consistent quantum theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of the spin has to be an integer multiple of $1/2$ because the rotation by $4\pi$ is continuously contractible to no rotation at all - so these two transformations can't possibly have qualitatively different eigenvalues. (To see the continuity, represent rotation by $4\pi$ as a rotation around $z$ by $2\pi$, followed by another one. You may take the second rotation by $2\pi$ and continuously change the axis of rotation from $+z$ to $-z$. At the end, you will get a rotation by $2\pi$ and back which has to be an identity, so it proves that the rotation by $4\pi$ has to act trivially on all objects.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, all values of spin that are multiples of $1/2$ are possible. That's because, for example, $Spin(3,1)$ Lorentz group is isomorphic to $SL(2,C)$ that acts on two-component complex spinors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum field theory, the electron's spin arises because every electron is an excitation of a Dirac field that transforms as a spinor. Quantum field theory has to be the right low-energy approximation of any viable theory that has a chance to go beyond quantum field theory, too. So the spin's origin may be reduced to the case of quantum field theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the case of string theory, too. One may also describe the origin of the spin-1/2 particles microscopically. In the RNS formulation of the superstring, there exist zero modes of the $\psi_\mu$ world sheet fermions. The representation of the ground states requires us to quantize them, and because the anticommutators of these $\psi_\mu$ fields&#xA;$$ \{\psi_\mu,\psi_\nu\} = g_{\mu\nu} $$&#xA;are isomorphic to the Dirac algebra of gamma matrices, &#xA;$$ \{\Gamma_\mu,\Gamma_\nu\} = 2g_{\mu\nu} $$&#xA;up to a trivial rescaling by $\sqrt{2}$, the ground state of the single-superstring Hilbert space inevitably transforms as a spinor (in the periodic sector). The GSO projection removes 1/2 of those states that would violate the spin-statistics relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Green-Schwarz formalism, one obtains the ground state - supergravitons and/or gauge supermultiplet - by quantizing fields $\theta^a$ on the world sheet that transform as spinors themselves. In this way, one gets the full supermultiplet - both the fermionic as well as the bosonic states - at the same moment. Also, excitations of non-zero modes of $\theta^a$ are changing the spin of the string. The total angular momentum of a string comes from the spin of the zero modes as well as the nonzero modes - and from both bosonic and fermionic degrees of freedom on the world sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reconciling loop quantum gravity with a half-integral spin remains a matter of wishful thinking. There are many ways to see that one can't get e.g. chiral fermions in lattice-like theories; one can't ever derive the a priori existence of anomalies coming from chiral fermions if the spacetime is discretized; and so on. Half-integral spin is one of the ways to show that discrete models of the reality can't be consistent with basic features of the reality such as the existence of fermions, especially the chiral ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you write about modelling spin's electron as an extended object is refuted in the first part of your very question. A century ago, people would think about electron as a classical object. Obviously, those attempts were totally incompatible with quantum mechanics (and other facts). There has to exist a degree of freedom that has a half-integral spin to start with, and RNS superstring theory only explains its deeper origin, but can't replace it by something completely different.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-24T06:27:03.957" LastActivityDate="2011-02-24T06:27:03.957" CommentCount="11" />
  <row Id="5755" PostTypeId="2" ParentId="5753" CreationDate="2011-02-23T21:15:22.097" Score="2" ViewCount="" Body="&lt;p&gt;The full length answer is a book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0226807940&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;The Story of Spin&lt;/em&gt;&lt;/a&gt; by Sin-itiro Tomonaga which I've only read part of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/wDQFC.jpg&quot; alt=&quot;Cover image for The Story of Spin&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-02-23T21:15:22.097" />
  <row Id="5756" PostTypeId="2" ParentId="5721" CreationDate="2011-02-23T21:18:07.487" Score="2" ViewCount="" Body="&lt;p&gt;The other answers provide an account of quantum theory as a source of the answer. But I would like to consider the question from a more general perspective: that of the meaning of &quot;probability&quot; especially in the classical context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically what is meant by &quot;probability&quot; of the dice value in this question and where has the value 1/6 come from? This may seem like a trivial point, but there are (at least) three different perspectives on the answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) &lt;strong&gt;Frequentist&lt;/strong&gt;. This is the view that probability is a number determined empirically by a series of tests. So this view says that such tests have been conducted on the dice and in 1/6 of cases (subject to some statistical factor) a given number was returned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) &lt;strong&gt;Absolutist&lt;/strong&gt;. This is the view (although I am not sure anyone will promote it) that the probability (with its value here of 1/6) is a derived physical property of the dice (like temperature, say). As a physical property it has been determined by some detailed analysis. One possible such analysis might be on the phase space of all possible motions, which has been appropriately partitioned and it is found that 1/6 of the volume is associated with each resultant value. This analysis might be done for that dice, or for some idealised dice. So the absolutist would claim that physics determines (classical) probabilities, and would view the frequentist data as simply experimental (dis-)confirmation of a specific model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) &lt;strong&gt;Relativist&lt;/strong&gt;. This is really the Bayesian viewpoint as espoused by Jaynes. This is using conditional probability $P(A|C)$ to define a probability in a relative way, with Bayes Theorem - which also allows us to construct $P(C|A)$ - this allows discussion of Inference and Prior Probabilities. The Bayesian view (especially as promoted by Jaynes) would be that there is no absolute probability, but only levels of knowledge, &quot;observer inferencing&quot; and (changing) prior assumptions. So the 1/6 value would be determined subjectively, as a measure of your &quot;expectation&quot; for example. It would have nothing to do with any intrinsic physics of the dice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to return to Quantum Mechanics. It too deals with probabilities, which are calculated from $|\Psi(x)*\Psi(x)|$. So what do these probabilities mean now that we have done some analysis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jaynes liked to promote his Bayesian view (ie 3 above) that he claimed reconciled Bohr with Einstein, thus interpreting $\Psi$ as somewhat epistemological. Phrases I have seen in other Stack answers like &quot;the $\Psi$ collapse is all in your head&quot; when discussing $\Psi$ might suggest that some physicists share an aspect of this view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However an alternative view might be that $\Psi$ really solves the Absolutists problem: namely it provides a well validated physical basis for calculating probabilities in an absolute sense.&#xA;If this were true then the only physical way to get to 1/6 in the dice would be to do some giant quantum calculation from its atoms!&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-23T21:18:07.487" />
  <row Id="5757" PostTypeId="1" AcceptedAnswerId="5781" CreationDate="2011-02-23T22:30:54.470" Score="1" ViewCount="94" Body="&lt;p&gt;Fermilab seems to have ruled out monopoles with mass less than 850 GeV, but I have seen some estimates of the mass thought to be in the order of up to $10^{18}$ GeV, which, of course, would make them undetectable in any accelerators. By 2013, the LHC is scheduled to reach up to 14 TeV. The only disputed sighting of a cosmic ray produced magnetic monopole was in 1982 when Blas Cabrera reported discovering one (Valentine event monopole). This has never been duplicated. CERN has set up the Monopole and Exotics Detector &lt;a href=&quot;http://web.me.com/jamespinfold/MoEDAL_site/Welcome.html&quot; rel=&quot;nofollow&quot;&gt;MOEDAL&lt;/a&gt;. Other experiments set up to detect them are the Antarctic Impulsive Transient Antenna--&lt;a href=&quot;http://arxiv.org/abs/1008.1282&quot; rel=&quot;nofollow&quot;&gt;ANITA&lt;/a&gt;, and the Antarctic Muon And Neutrino Detection Array, aka &lt;a href=&quot;http://www.springerlink.com/content/m46247620hq4u364/&quot; rel=&quot;nofollow&quot;&gt;AMANDA&lt;/a&gt;. While both of these detected neutrinos, neither detected magnetic monopoles (looking for Cerenkov radiation produced by products of monopole interaction).&#xA;Another method of detection is to look for induced current in a superconducting ring when a monopole passes through.&#xA;Joseph Polchinski called the existence of monopoles “one of the safest bets that one can make about physics not yet seen.” This shows some optimism on his part that one will be seen. Is there any reason or way to limit the mass of the monopole on the upside to 14 TEV or less?&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-02-23T22:53:44.370" LastActivityDate="2011-02-24T06:59:31.133" Title="With estimates of mass constraints on magnetic monopoles, how likely is one to be found by the LHC(MoEDAL)?" Tags="&lt;string-theory&gt;&lt;particle-physics&gt;&lt;magnetic-monopoles&gt;&lt;lhc&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5758" PostTypeId="1" AcceptedAnswerId="5761" CreationDate="2011-02-23T22:52:09.660" Score="6" ViewCount="119" Body="&lt;p&gt;Quasiparticles theoretically can form braids on a 2D surface. The braids apparently are quite stable for reasonably long periods, allowing a superposition state more time before it decoheres through environmental interaction. This would be a big step in quantum computing. Has this actually been achieved or is it theory only at this point?&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-02-24T00:02:40.123" Title="Has anyone actually produced any braided anyons (quasi-particles) yet? (quantum computer)" Tags="&lt;particle-physics&gt;&lt;quantum-computation&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5759" PostTypeId="2" ParentId="629" CreationDate="2011-02-23T22:55:32.890" Score="1" ViewCount="" Body="&lt;p&gt;And thus, who can reduce it most: &#xA;   the vehicle designer !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The French car maker Citroen do the compensation since long time ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.citroenet.org.uk/miscellaneous/hydraulics/hydraulics-1.html&quot; rel=&quot;nofollow&quot;&gt;quoting from citroenet&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Anti dive suspension&lt;/strong&gt; is incorporated since the rear brakes&#xA;  take their fluid from the rear&#xA;  suspension &lt;strong&gt;which pulls the tail down&#xA;  under heavy braking.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-02-23T22:55:32.890" />
  <row Id="5760" PostTypeId="1" CreationDate="2011-02-23T23:41:20.590" Score="11" ViewCount="159" Body="&lt;p&gt;Why are so many condensed matter phenomena so sensitive to impurities? In fact, quite a number of them depend upon impurities for their very existence!&lt;/p&gt;&#xA;" OwnerUserId="2180" LastActivityDate="2011-08-17T23:59:33.240" Title="Why are so many condensed matter phenomena so dependent upon impurities?" Tags="&lt;condensed-matter&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="5761" PostTypeId="2" ParentId="5758" CreationDate="2011-02-24T00:02:40.123" Score="3" ViewCount="" Body="&lt;p&gt;So far non-abelian quasiparticles have yet to be observed in quantum hall states. However, as far as theory is concerned, in a &lt;a href=&quot;http://link.aps.org/doi/10.1103/PhysRevB.83.075303&quot; rel=&quot;nofollow&quot;&gt;recent paper&lt;/a&gt; a rigorous proof is provided that such quasiparticles &lt;em&gt;do&lt;/em&gt; exist in quantum hall states. There were many candidate states for this object, starting with the Moore-Read state, however, theoretically there did not yet exist a complete proof of their existence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a really big achievement. Frank Wilzcek describes it as &lt;a href=&quot;http://physics.aps.org/articles/v4/10&quot; rel=&quot;nofollow&quot;&gt;&quot;a landmark proof&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been various incompletely verified &quot;sightings&quot; of the $5/2$ state which is the non-abelian state. For results upto 2008 at least, see the &lt;a href=&quot;http://arxiv.org/abs/0707.1889v2&quot; rel=&quot;nofollow&quot;&gt;RMP paper&lt;/a&gt; on topological quantum computation. At the end of this paper various references are given which claim to have observed this state. I don't know what the latest is, but I would have heard of an definitive experimental observation, or so I like to think :)&lt;/p&gt;&#xA;" LastActivityDate="2011-02-24T00:02:40.123" CommentCount="2" />
  <row Id="5762" PostTypeId="1" CreationDate="2011-02-24T00:03:12.647" Score="5" ViewCount="212" Body="&lt;p&gt;Well,it does according to this preprint: &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1102/1102.3913v1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/arxiv/pdf/1102/1102.3913v1.pdf&lt;/a&gt;&#xA; for certain scales&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would be a simple way to explain MOND to a layman?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it ignore mainstream physics?How much?&lt;/p&gt;&#xA;" OwnerUserId="2108" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-02-24T19:44:49.783" LastActivityDate="2011-02-25T02:56:06.020" Title="Does MOND make good predictions?" Tags="&lt;gravity&gt;&lt;cosmology&gt;&lt;mond&gt;" AnswerCount="3" />
  <row Id="5763" PostTypeId="2" ParentId="5760" CreationDate="2011-02-24T00:08:24.763" Score="5" ViewCount="" Body="&lt;p&gt;A.  You cannot find a condensed matter system &lt;em&gt;without&lt;/em&gt; impurities. Even with the most stringent manufacturing processes you are always left with some (however small) fraction of defects and impurities in any material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B.  The most interesting physics (Kondo Effect, Anderson Localization, Quantum Hall Effect) occurs &lt;em&gt;only in the presence of&lt;/em&gt; defects. If you read Jain's excellent book &quot;Composite Fermions&quot; (or any other review on the QHE) you will discover that in a perfect sample one cannot have the QHE because of Lorentz invariance of the sample. Defects break this symmetry and one can no longer boost to a frame where the induced magnetic field cancels the external one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Defects and impurities allow us extremely fine control over the electronic properties of materials and we would not have things as beautiful as transistors or SQUID devices without them.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-24T00:08:24.763" CommentCount="1" />
  <row Id="5764" PostTypeId="1" CreationDate="2011-02-24T00:24:31.880" Score="4" ViewCount="194" Body="&lt;p&gt;Why do we have so many dualities in string theory? Is there a reason for that?&lt;/p&gt;&#xA;" LastActivityDate="2011-08-14T05:04:30.610" Title="Why do we have so many dualities in string theory?" Tags="&lt;string-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="5765" PostTypeId="1" AcceptedAnswerId="5802" CreationDate="2011-02-24T00:29:03.713" Score="2" ViewCount="173" Body="&lt;p&gt;We suppose for semplicity to have a 1D oscillator, but this is a question abaout the general CCR algebra in oscillators, second quantization, quantum field theory etc. &#xA;We know coherent states are a non orthonormal base and are a overcomplete base.&#xA;We know them are given from $A |\alpha&amp;gt;=\alpha |\alpha&amp;gt;$. where $|\alpha&amp;gt;=e^{-\frac{|\alpha|^2}{2}} e^{\alpha A^+} \Psi_0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we calculate generical scalar product between two different coherent states with eigenvalue $\alpha$ and $\beta$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;e^{-\frac{|\alpha|^2+|\beta|^2}{2}}\Psi_0  e^{\beta A} e^{\alpha A^+} \Psi_0&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="558" LastActivityDate="2011-02-24T21:47:33.437" Title="Scalar product of coherent states " Tags="&lt;quantum-mechanics&gt;&lt;quantum-optics&gt;&lt;second-quantization&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="5766" PostTypeId="2" ParentId="5762" CreationDate="2011-02-24T00:31:04.740" Score="1" ViewCount="" Body="&lt;p&gt;Yes. The would appear to be the point of &lt;a href=&quot;http://arxiv.org/abs/1102.3913&quot; rel=&quot;nofollow&quot;&gt;the paper&lt;/a&gt; you cite. It has been accepted for PRL apparently so that should lend it some credibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MOND - short for MOdified Newtonian Dynamics - is a phenomenological theory that was conceived of by Mordechai Milgrom in order to explain the huge discrepancy between the shapes of galaxy rotation curves as predicted by Newtonian theory and the actual shapes observed. A rotation curve is the plot of the orbital velocity of objects in the galaxy with respect to distance from the center:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/b/b9/GalacticRotation2.svg/250px-GalacticRotation2.svg.png&quot; alt=&quot;Rotation curves&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest statement of MOND is that it is a theory with a minimum acceleration scale $a_0 \sim 10^{-10} m/s^2$. When this scale is reached at some radius $r_0$ in a galaxy, objects appear to cease to respond to gravitational forces. Alternatively one could say that gravitational forces cannot generate an acceleration lower than $a_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many critics of MOND. However, it has had remarkable success over the years. This paper is only the most recent evidence in its favor. MOND makes no claim to explaining the microscopic physics which leads to these effects. Compared to the LCDM model, MOND makes far fewer assumptions (only &lt;em&gt;one&lt;/em&gt;, in fact) and thus has the benefit of being simpler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The challenge for theories of quantum gravity is to either rule out MOND'ian behavior in weak fields or otherwise explain what gives rise to it. So far, none have even tackled the question in part due out of fear of being labeled &quot;fringe&quot; for associating with such rabble as MOND ;]&lt;/p&gt;&#xA;" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-02-24T19:45:17.053" LastActivityDate="2011-02-24T19:45:17.053" CommentCount="2" />
  <row Id="5767" PostTypeId="2" ParentId="5764" CreationDate="2011-02-24T00:56:13.063" Score="6" ViewCount="" Body="&lt;p&gt;That's actually a good question, except I don't think anybody has a good answer. The dualities suggest that there is an underlying structure (sometimes called M theory) behind all the dual theories, and those dualities are a manifestation of that structure. Though we don't have a complete formulation of that structure yet, just the existence of this structure results in some consistency conditions which are those dualities - relationships between naively different theories and calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can have the following rough analogy. Think about the dual theories as different asymptotic expansions of some integral (say around zero and around infinity). Those expansions obey relations that express the fact they come from the same object. The state of affairs now is that we have those asymptotic expansions, and we discovered they obey relations that identify them as expansion of the same object, but we don't really have that object pinned down yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relatedly, look at &lt;a href=&quot;http://physics.stackexchange.com/questions/5064/give-a-description-of-m-theory-your-grandmother-can-understand/5065#5065&quot;&gt;this answer&lt;/a&gt; I gave to someone's grandmother.&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-02-24T00:56:13.063" CommentCount="4" />
  <row Id="5769" PostTypeId="1" CreationDate="2011-02-24T01:09:15.973" Score="6" ViewCount="318" Body="&lt;p&gt;Context is 1D Ising model. Metropolis algorithm is used for simulate that model. Among all possible spins configurations (states) that algorithm generates only states with the desired Boltzmann probability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Algorithm chooses spin at random and makes a trial flip. If trial satisfies certain conditions related to Boltzmann probability, flip is accepted. Otherwise flip is rejected and system is unchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define &quot;acceptance ratio&quot; as a percentage of accepted trials. Simulation shows that acceptance ratio is higher on higher temperature (behave as increasing function of temperature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Why Metropolis algorithm is not efficient at low temperatures? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is efficiency of algorithm related to acceptance ratio?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2179" LastActivityDate="2011-02-24T19:56:12.637" Title="Efficiency of Metropolis algorithm" Tags="&lt;statistical-mechanics&gt;&lt;ising-model&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="5770" PostTypeId="1" CreationDate="2011-02-24T01:28:36.380" Score="-1" ViewCount="124" Body="&lt;p&gt;How much free energy is there in the Earth-ionosphere waveguide resulting from the constant bombardment of lightning strikes all over the Earth, and how do you go about calculating an estimate for it?&lt;/p&gt;&#xA;" OwnerUserId="1670" LastEditorUserId="1670" LastEditorDisplayName="" LastEditDate="2011-02-24T15:19:17.407" LastActivityDate="2011-02-24T15:19:17.407" Title="Amount of free energy in the Earth-ionosphere waveguide" Tags="&lt;thermodynamics&gt;&lt;electrodynamics&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="5771" PostTypeId="2" ParentId="5769" CreationDate="2011-02-24T01:38:57.330" Score="2" ViewCount="" Body="&lt;p&gt;1) The Metropolis algorithm is more general than just the sampling scheme used: the sampling scheme that you describe is a &lt;strong&gt;&quot;local update,&quot;&lt;/strong&gt; however there exist more general sampling schemes (&quot;cluster updates&quot; like Wolff algorithm, Swendsen-Wang, etc.). The point is that local update sampling isn't inefficient at low temperatures, &lt;em&gt;per se,&lt;/em&gt; it's inefficient near phase transitions, since the correlation length diverges (however cluster algorithms do just fine).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) A low acceptance ratio can mean that your algorithm is inefficient, but it can also mean that the probability distribution you'd like to sample is &lt;em&gt;very very narrow&lt;/em&gt;. Strictly speaking the 1D Ising model doesn't order at any $T\neq 0$, but that doesn't mean the correlation length can't get larger than the sample on your computer! In this case Metropolis will quickly anneal to the ferromagnetic state, and the acceptance probability of any change will be $\mbox{exp}(-2J/T)$, quite narrow indeed.&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2011-02-24T01:38:57.330" CommentCount="1" />
  <row Id="5772" PostTypeId="2" ParentId="5769" CreationDate="2011-02-24T01:40:40.040" Score="5" ViewCount="" Body="&lt;p&gt;There are two problems at low temperatures. One is intrinsic to the Ising model, the other to the Metropolis algorithm. The Ising one is a bit more serious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem with the Ising Model at low temperatures&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Ising model of ferromagnetism posits that the electrons in a model can be modeled as little magnetic dipoles, confined to only two quantized spin states, up and down, and that all the electrons in the metal have the same two possible orientations. The Metropolis algorithm counts up the quanta of energy to be gained if a particular dipole were flipped over by counting up the states of the dipole's neighbors and comparing it to the dipole itself. It compares the number it gets with the thermal energy from the Boltzmann distribution to determine whether or not the metal is cool enough that flipping to join the neighbors' alignments is energetically favorable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if the dipoles have more possible orientations than this (ie if their spins don't &lt;em&gt;really&lt;/em&gt; all align on the same plane), that's not the big problem at low temperatures. The real issue is that quantum mechanics is more subtle than this. In reality, it isn't the orientation of each individual which is quantized, but the &lt;em&gt;sum&lt;/em&gt; of their magnetic moments which is quantized. At very low temperatures, this becomes an important failure in the Ising model, and the relevant states of a true ferromagnet are &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Magnon&quot; rel=&quot;nofollow&quot;&gt;magnons&lt;/a&gt;&quot; in which all dipoles are nearly parallel and a single unit of opposite alignment is spread across many dipoles. So, we really shouldn't be comparing the dipole to its neighbors individually at very low temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem with the Metropolis algorithm at low temperatures&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem here is also subtle. The Metropolis algorithm works by randomly picking a single dipole out of the whole magnet, checking it as described above, then checking another - it's just a Monte Carlo method. Now, theoretically, there should be a &quot;critical temperature,&quot; below which it is energetically favorable for the dipoles to side with magnetism instead of temperature. But check out the following plots, which came from a simulation I did of a Metropolis-algorithm Ising magnet in C a while ago. In the units I used, the critical temperature happened to be $T_c=2.27$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Tjz8e.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lowest temperature systems converge much more quickly to a homogeneous magnetization, but even at $T=1.5$ you still see &quot;meta-stable states.&quot; In these cases, large &lt;em&gt;sections&lt;/em&gt; of the magnet are commonly aligned, but the &lt;em&gt;entire&lt;/em&gt; magnet isn't. Change can only occur along the border region - and since you are using a random choice of dipoles, the chance that you randomly select one of those dipoles on the border region is pretty small, so you need a very long timeframe for the system to reach equilibrium using this Monte-Carlo scheme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the efficiency is worse close to the critical point of temperature, you could say that the algorithm's efficiency is worse near the critical point of the acceptance ratio, since the acceptance is related to the temperature.&lt;/p&gt;&#xA;" OwnerUserId="1655" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-02-24T19:56:12.637" LastActivityDate="2011-02-24T19:56:12.637" CommentCount="4" />
  <row Id="5773" PostTypeId="1" AcceptedAnswerId="7592" CreationDate="2011-02-24T01:41:42.280" Score="10" ViewCount="502" Body="&lt;p&gt;In quantum mechanics the state of a free particle in three dimensional space is $L^2(\mathbb R^3)$, more accurately the projective space of that Hilbert space. Here I am ignoring internal degrees of freedom otherwise it would be $L^2(\mathbb R^3)\otimes S$, but let's say it is not that time of the month. The observables are operators in that space and the dynamics is describe by the Schrodinger equation or in other equivalent ways. If we have more particles then it is $L^2(\mathbb R^{3N})$. My question is: are there any examples where one considers a system with configuration space a general manifold $M$, instead of $\mathbb R^3$, say a system of particles (a particle) with some restrictions, so that the state space is $L^2(M)$. There might be physical reasons why this is of no interest and I would be interested to here them. What I am interested in is seeing is specific (or general) examples worked out in detail. For example a system with a given Hamiltonian, where one can explicitly find the spectrum. Or if that is too much to ask for an example where the system has very different properties from the usual case. Say a particle living on the upper half plane with the Lobachevsky geometry, may be some connection to number theory! I am aware that there is quantum field theory on curved spacetime, I am interested in quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Just a small clarification. The examples I would like to see do not have to come from real physics, they can be toy models or completely unrealistic mathematical models. Something along the lines: take your favorite manifold $M$ and pretend this is the space we live in, what can we say about QM in it. The choice of $M$ doesn't have to do anything with general relativity. As I said the upper half plane is interesting or quotients of it by interesting discrete groups or generalizations $\Gamma\backslash G(\mathbb R)/K$ or anything at all. The answers so far are interesting. Hoping to see more.     &lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2011-02-24T16:30:14.857" LastActivityDate="2011-03-26T10:13:38.513" Title="Quantum mechanics on a manifold" Tags="&lt;quantum-mechanics&gt;&lt;mathematical-physics&gt;" AnswerCount="6" CommentCount="4" FavoriteCount="5" />
  <row Id="5774" PostTypeId="2" ParentId="5773" CreationDate="2011-02-24T01:54:55.167" Score="5" ViewCount="" Body="&lt;p&gt;When you study angular momentum in QM, this is a case of particle on a sphere. The wavefunctions are the spherical harmonics, the Hamiltonian is $L^2$, etc. I can think about many other examples where the configuration system of some QM system would be curved (e.g. some group manifold or coset space), so I don't think there is any physical reasons not to look at such examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more sophisticated set of examples, there are many studies on supersymmetric quantum mechanics on various manifolds, starting out with this paper &lt;a href=&quot;http://www.intlpress.com/JDG/archive/1982/17-4-661.pdf&quot; rel=&quot;nofollow&quot;&gt;Supersymmetry and Morse Theory&lt;/a&gt; by Ed Witten. The connection between SUSY QM and QFT on a manifold and the topology (or sometimes even the geometry) of the underlying manifold became a bit of an industry since then.&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-02-24T01:54:55.167" CommentCount="2" />
  <row Id="5775" PostTypeId="1" CreationDate="2011-02-24T04:32:47.157" Score="3" ViewCount="277" Body="&lt;p&gt;Senior physicists constantly complain they spend too much time on administration, teaching, getting grants, serving in committees, peer-reviewing articles, supervising, etc. . Do senior physicists conduct research by getting their post-docs and graduate students to do all the intensive work for them?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-25T02:39:04.340" ClosedDate="2011-02-25T02:49:22.643" Title="Do senior physicists actually conduct research?" Tags="&lt;soft-question&gt;&lt;physics-careers&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="5776" PostTypeId="1" AcceptedAnswerId="5782" CreationDate="2011-02-24T04:39:36.797" Score="3" ViewCount="88" Body="&lt;p&gt;Are there any projects/experiments running for the observation of topological defects (domain walls, monopoles, etc)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any &lt;em&gt;past/canceled&lt;/em&gt; or &lt;em&gt;future&lt;/em&gt; such projects?&lt;/p&gt;&#xA;" OwnerUserId="824" LastActivityDate="2011-02-24T07:31:27.457" Title="Observation of cosmological topological defects" Tags="&lt;cosmology&gt;&lt;experimental-physics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="5777" PostTypeId="2" ParentId="5775" CreationDate="2011-02-24T05:24:28.613" Score="2" ViewCount="" Body="&lt;p&gt;I can answer for experimental particle physics. The answer is yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seniors are instrumental in motivating and, as you say, getting the money, overseeing that the experiment runs in group decisions, since experiments involve many groups. If they are any good  in physics they also have ideas on analysis, but the bulk of the work is carried by the graduate student whose thesis it is. Post docs have freedom of choice once the analysis stage is reached after having worked innumerable hours on hardware or software for the experiment . It is their chance of being original enough to succeed in someday being counted among the seniors.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-02-24T05:24:28.613" />
  <row Id="5778" PostTypeId="2" ParentId="5762" CreationDate="2011-02-24T05:57:40.460" Score="8" ViewCount="" Body="&lt;p&gt;The basic version of MOND pretty much only attempts to explain the rotation curves by postulating a modification of Newtonian gravity at accelerations smaller than a_0, and does a good job at that. However, many other observational facts (cluster dynamics, stability of spiral galaxies) still require large quantities of dark matter, as acknowledged by proponents of MOND themselves. The &lt;a href=&quot;http://apod.nasa.gov/apod/ap060824.html&quot; rel=&quot;nofollow&quot;&gt;Bullet Cluster&lt;/a&gt; is the most dramatic evidence that some form of dark matter must exist and does not necessarily trace the baryonic matter distribution. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In regards to its relationship with physics: MOND does not attempt to reconcile with General Relativity, so it cannot do calculations regarding cosmology, gravitational lensing, etc. There is a theory called &lt;a href=&quot;http://prd.aps.org/abstract/PRD/v70/i8/e083509&quot; rel=&quot;nofollow&quot;&gt;TeVeS&lt;/a&gt; by Bekenstein, which purports to be a relativistic generalization of MOND. I won't pretend I understand it or looked at it in any detail (maybe the value of TeVeS could be a nice question for GR experts). In any case, it has been criticized for yielding unstable solutions for stars, and is still not compatible with the dynamics of the Bullet Cluster.&lt;/p&gt;&#xA;" OwnerUserId="1252" LastEditorUserId="1252" LastEditorDisplayName="" LastEditDate="2011-02-24T15:30:49.207" LastActivityDate="2011-02-24T15:30:49.207" CommentCount="1" />
  <row Id="5779" PostTypeId="2" ParentId="5721" CreationDate="2011-02-24T06:31:16.313" Score="1" ViewCount="" Body="&lt;p&gt;I was going to make this a comment but here it comes, since maybe it is a simpler formulation of a fundamental difference than all the rest of the answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference is that the value of the classical die under the couch can be seen/measured by many people.They will not tell you, and you still will not know,  but  the value that was  thrown will not change. In quantum mechanics &lt;strong&gt;each&lt;/strong&gt; observation/measurement  changes the system, from one measurement to the other . The die's throw will not change  from one look of a hidden to you observer,  to the next.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-02-24T07:02:44.027" LastActivityDate="2011-02-24T07:02:44.027" CommentCount="7" />
  <row Id="5780" PostTypeId="2" ParentId="5721" CreationDate="2011-02-24T06:42:24.460" Score="1" ViewCount="" Body="&lt;p&gt;The deeper difference between the classical and quantum dice is this - quantum dice obey the superposition principle. Consequently you can create entangled states of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\Psi \rangle = \frac{1}{\sqrt{2}}(|x \rangle_{D1} |y \rangle_{D2} \pm |y \rangle_{D1} |x \rangle_{D2} ) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the subscripts denote the particular die (1 or 2), $|x\rangle_{D}$ and $|y\rangle_{D}$ are two different possible states of the die. The simplest case is when the role of the &quot;die&quot; is played by qubits. Then the above expression becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\Psi \rangle = \frac{1}{\sqrt{2}}(|\uparrow \rangle_{D1} |\downarrow \rangle_{D2} \pm |\downarrow \rangle_{D1} |\uparrow \rangle_{D2} ) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Entanglement appears to have no classical analog and it actually plays the role of a usable information resource in quantum computation.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-24T06:42:24.460" CommentCount="1" />
  <row Id="5781" PostTypeId="2" ParentId="5757" CreationDate="2011-02-24T06:54:27.110" Score="2" ViewCount="" Body="&lt;p&gt;Dear Gordon, Polchinski didn't imply anything about the lightness of the monopoles. You are overinterpreting what he said: his statement was an in-principle statement by a theorist. Just to be sure, the MoEDAL collaboration has overinterpreted him, too: he has never suggested that an experiment of their kind will find the beast. Still, it's very important to know whether such qualitatively different particles exist even if we think it's very likely that they won't be directly detected in the near (or far) future. Most of the top contemporary state-of-the-art theoretical physics deals with objects that will probably never be detected by direct experiments, for obvious financial and technological limitations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Directly experimentally, the lower bound on the mass is just 500 GeV or so, but when some theory is wisely added, the mass is pretty much required to be exponentially higher than the TeV scale. In particular, some kind of unification seems necessary to allow the magnetic monopoles, and unification such as GUT can only appear at very high scales, such as the GUT scale of $10^{16}$ GeV: that's where the electromagnetic $U(1)$ may be embedded in a larger group that allows a topologically nontrivial solution (the $SU(2)$ of the weak interactions is not enough - it doesn't contain the hypercharge which has to be twisted as well). So it's unlikely that direct detection succeeds. Inflation has probably diluted the concentration of those massive magnetic monopoles to a very tiny number.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-24T06:59:31.133" LastActivityDate="2011-02-24T06:59:31.133" CommentCount="2" />
  <row Id="5782" PostTypeId="2" ParentId="5776" CreationDate="2011-02-24T07:17:04.337" Score="2" ViewCount="" Body="&lt;p&gt;See e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-ex/0302011&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-ex/0302011&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;MoEDAL at CERN has survived since that time. This is about the magnetic monopoles. Domain walls may be too much stuff too wish for so I don't think that someone is looking for them and nothing else. Cosmic strings are being constantly looked for in the telescopes but so far, the bounds are tightening - they don't seem to be abundant (or they're totally absent).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-24T07:17:04.337" />
  <row Id="5783" PostTypeId="2" ParentId="5776" CreationDate="2011-02-24T07:31:27.457" Score="1" ViewCount="" Body="&lt;p&gt;There is a &lt;a href=&quot;http://axion-wimp.desy.de/&quot; rel=&quot;nofollow&quot;&gt;workshop&lt;/a&gt; on Axions, WIMPs and WISPs which will be held at Mykonos, Royal Myconian &amp;amp; Myconian Imperial Resorts on 26 June - 1 July 2011.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-02-24T07:31:27.457" />
  <row Id="5784" PostTypeId="1" CreationDate="2011-02-24T08:42:10.800" Score="8" ViewCount="151" Body="&lt;p&gt;Why is there a breakdown of Kolmogorov scaling in turbulence? What causes intermittency?&lt;/p&gt;&#xA;" LastActivityDate="2011-07-13T03:52:40.643" Title="Why is there a breakdown in Kolmogorov scaling in turbulence?" Tags="&lt;fluid-dynamics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5785" PostTypeId="1" CreationDate="2011-02-24T08:58:35.723" Score="6" ViewCount="119" Body="&lt;p&gt;If we couple a Yang-Mills theory with a Higgs field and some quarks in the fundamental representation, we can have a Higgs phase and a confining phase. However, they are indistinguishable. The Wilson loops scale according to the perimeter law, not the area law in the confining phase because of hadronization. In the Higgs phase, there is no cluster decomposition for the Higgs field because distant points have to be connected by a Wilson line with exponential falloff as the length of the Wilson line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are both phases indistinguishable?&lt;/p&gt;&#xA;" OwnerUserId="2185" LastActivityDate="2011-02-24T10:50:26.293" Title="Why are the Higgs phase and the confinement phase identical in Yang-Mills-Higgs systems?" Tags="&lt;quantum-chromodynamics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="5786" PostTypeId="1" CreationDate="2011-02-24T09:19:06.307" Score="5" ViewCount="100" Body="&lt;p&gt;Baryons in quantum chromodynamics can be described as a bound state of three quarks. But they can also be described as a topological soliton of the pion field. How can both descriptions be equivalent?&lt;/p&gt;&#xA;" OwnerUserId="2185" LastActivityDate="2011-02-24T09:38:14.643" Title="Why are baryons equivalent to Skyrmions?" Tags="&lt;quantum-chromodynamics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5787" PostTypeId="2" ParentId="5786" CreationDate="2011-02-24T09:38:14.643" Score="2" ViewCount="" Body="&lt;p&gt;The skyrmion description only appears in an effective, non-renormalizable theory where pions are treated as elementary particles. Using a more high-energy viewpoint, however, pions themselves are made out of quarks and gluons, and the short-distance QCD rescription using quarks and gluons is renormalizable and much more accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the pion effective field theory is still marginally applicable at the scale of the proton, so it must inevitably have a description for that. It is not surprising that it must be topologically nontrivial because baryons carry new kinds of charges, the baryon charge, that would vanish in topologically trivial configurations of pion fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no contradiction - the bound state of quarks and gluons that make up the proton is not &quot;any bound state&quot;. It has some specific properties and quantum numbers, and those inevitably mean that the best approximation in an effective pion theory has to be a skyrmion. So this is not a real duality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are very similar descriptions of baryons (among many other objects) that are full-fledged dualities. In AdS/CFT, the baryons are branes wrapped on an internal sphere:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/9805112&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9805112&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In AdS/CFT, the quark-gluon composition of the baryons really becomes invisible in the bulk, so the branes as baryons can't be reduced to quarks and gluons in any direct way. The branes are wrapped, much like the skyrmions are wrapped around some place of the configuration space, but I don't think that there is a direct isomorphism between these two wrappings.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-24T09:38:14.643" />
  <row Id="5788" PostTypeId="2" ParentId="5784" CreationDate="2011-02-24T10:19:45.437" Score="4" ViewCount="" Body="&lt;p&gt;The basic Kolomogorov theory is a mean-field theory -- dissipation rate is considered to be constant over the whole volume of liquid.  While the dissipation rate should depend on position -- that is where intermittency comes from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somewhere in 60's Kolmogorov and Obukhov attempted to account for this effect. But the problem is that there is the &quot;Landau objection&quot; -- the impossibility to devise universal formula, independent on the geometry of the problem (with scale $l$) for relatively small scales (which are $ \ll l$): &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... It may be thought that the possibility exists in principle of obtaining a universal formula, applicable to any turbulent flow, which should give $B_{rr}$ and $B_{tt}$  for all distances $r$ that are small compared with $l$. In fact, however, there can be no such formula, as we see from the following argument. The instantaneous value of $(v_{2i}-v_{1i})(v_{2k}-v_{1k})$ might in principle be expressed as a universal function of the energy dissipation $\epsilon$ at the instant considered. When we average these expressions, however, an important part will be played by the manner of variation of $\epsilon$ over times of the order of the periods of large eddies (with size $\sim l$)  and this variation is different for different flows. The result of averaging therefore cannot be universal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Fluid Mechanics, Landau and Lifshitz, chapter 34.)&lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="829" LastEditorDisplayName="" LastEditDate="2011-07-13T03:52:40.643" LastActivityDate="2011-07-13T03:52:40.643" CommentCount="1" />
  <row Id="5790" PostTypeId="2" ParentId="5785" CreationDate="2011-02-24T10:50:26.293" Score="1" ViewCount="" Body="&lt;p&gt;Dough, in some cases, the Higgs and confinement phases are not separated at all and form the total screening phase, see Banks Rabinovici 1979:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1979NuPhB.160..349B&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/1979NuPhB.160..349B&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and Fradkin Shenker 1979:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://prd.aps.org/abstract/PRD/v19/i12/p3682_1&quot; rel=&quot;nofollow&quot;&gt;http://prd.aps.org/abstract/PRD/v19/i12/p3682_1&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More recently, those and similar relationships were studied in the context of supersymmetric field theories and supersymmetric string vacua. The papers by Seiberg and Witten for $N=2$ were the main progress. See also papers such as this random one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/9408155&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9408155&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some cases, the confinement and Higgs phases may be distinct. The relationships between the two phenomena may be understood as a sort of S-duality - a strong-weak duality or a generalized electro-magnetic duality. For example, the last paper I quoted explains why confinement may be understood as condensation of monopoles (the dual, &quot;magnetically&quot; charged particles, which is kind of the S-dual of the normal Higgs mechanism which gives vevs to &quot;electrically&quot; charged fields).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supersymmetric gauge theories are full of S-dualities, much like stringy vacua.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-24T10:50:26.293" />
  <row Id="5791" PostTypeId="2" ParentId="5773" CreationDate="2011-02-24T12:09:58.550" Score="10" ViewCount="" Body="&lt;p&gt;As far as I understand it there are essentially two ways in which you can study quantum mechanics on a manifold with some curvature. Classically speaking these two ways lead to the same physics but in a quantum mechanical approach they are distinct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first approach is to think of a particle moving &quot;freely&quot; through three-dimensional space, but subject to external forces that confine the particle to some submanifold. The particle lives, in some sense, in a confining potential which defines the manifold. The phase space of the particle is, from the start, the usual phase space associated with the three-dimensional space. However, the external potential limits the particle to some subspace of this phase space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second approach is to work with generalized coordinates, as is done in Lagrangian mechanics. The coordinates of the particle are then a parametrization of the submanifold. What's important here is that there is no reference to the coordinates of the three-dimensional space. An example is the pendulum, which can be described solely in terms of the angle the pendulum makes with the z-axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classically there is no distinction between the two approaches. This no longer holds when you move to quantum mechanics. If you follow the first approach, using some confining potential to keep the particle on the manifold, you will deal with the uncertainty principle that prohibits the exact localization of the particle onto the manifold. Because of this principle the particle will never be fully screened from the larger dimensional space. You can still systematically set up the quantization procedure, though. The advantage of this approach is that quantization works in the usual way (you work with cartesian coordinates, after all). The resolution is to essentially split up the wavefunction and the Schroedinger equation (S.E.) in contributions due to the confining potential and a sort-of effective S.E. for the remaining part of the wavefunction. The effective S.E. then contains two effective potentials due to the &lt;strong&gt;Mean curvature&lt;/strong&gt; and &lt;strong&gt;Gauss curvature&lt;/strong&gt; of the corresponding manifold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very important feature: a cylinder, for instance, has no Gauss curvature, only a mean curvature. In the second approach you will find that there is no distinction between two cylinders with different mean curvatures, because in this approach only the Gauss curvature pops up. Take for instance a particle living on a 1D line. You only require one coordinate to describe this line, so for the the second approach all systems are equivalent. But in the first approach you have to specify in what way the line is embedded in the higher dimensional space, and how the particle is confined to the lower-dimensional space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second approach might feel more natural, if you think like a mathematician. In this approach you require a way to quantize generalized coordinates -- which is a lot more subtle than ordinary quantization. The problem that plagues this approach is the so-called ordering problem. Essentially you want to replace the momentum label by a derivation operator $p \rightarrow -i\hbar\nabla$. Furthermore, there's also the choice of parametrization of the manifold, which should ofcourse have no effect on the underlying physics (similar to general relativity). The ordering problem states that you do not know a priori which way the classical (commuting) variables have to be ordered before you replace them by their quantum mechanical (non-commuting) counterparts. What's even worse, because of the curvature of the space the derivative operator also contains some ambiguity. There is an ambiguity in the choice of your momentum operator and your Hamiltonian (and any other functions). Many quantum mechnical Hamiltonians have the same classical limit, and the equivalence principle (i.e. linking quantum mechanics to classical physics) does not dictate which is best. For instance, the kinetic operator $\nabla^2$ can be defined using the canonical Laplacian or the Laplace-Beltrami operator. Still, there is some work out there which motivates a generalized equivalence principle (see e.g. Kleinert) and results in a consistent quantization procedure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both approaches have interesting feature, but the first one is actually a bit more physical. The reason is that in condensed matter you deal with confining potentials due to some ionic lattice. Take for instance graphene, which is a two-dimensional surface. As it turns out, this surface is not completely flat but will always form some ripples. These deformations of the surface can be interpreted as if the electrons (or Dirac fermions, if you want to use the effective theory) live on a curved manifold embedded in a three dimensional surface. This leads to hilarious applications, such as the existence of wormholes in Graphene. But in the end the curvature has a very physical manifestation in the electronic properties of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;=&#xA;The first of these approaches, which uses a confining potential, is discussed in these papers by Costa:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://link.aps.org/doi/10.1103/PhysRevA.23.1982&quot; rel=&quot;nofollow&quot;&gt;http://link.aps.org/doi/10.1103/PhysRevA.23.1982&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://link.aps.org/doi/10.1103/PhysRevA.25.2893&quot; rel=&quot;nofollow&quot;&gt;http://link.aps.org/doi/10.1103/PhysRevA.25.2893&lt;/a&gt; (many-particle case)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second approach is treated in this review paper by B.S. De Witt:&#xA;&lt;a href=&quot;http://link.aps.org/doi/10.1103/RevModPhys.29.377&quot; rel=&quot;nofollow&quot;&gt;http://link.aps.org/doi/10.1103/RevModPhys.29.377&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also the book by Kleinert, who has a whole chapter on it using a Path Integral approach:&#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/9814273562&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Integrals-Quantum-Mechanics-Statistics-Financial/dp/9814273562&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graphene wormholes:&#xA;&lt;a href=&quot;http://arxiv.org/abs/0909.3057&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0909.3057&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1287" LastEditorUserId="1287" LastEditorDisplayName="" LastEditDate="2011-02-25T09:36:10.353" LastActivityDate="2011-02-25T09:36:10.353" CommentCount="5" />
  <row Id="5792" PostTypeId="1" AcceptedAnswerId="6085" CreationDate="2011-02-24T12:30:00.133" Score="7" ViewCount="546" Body="&lt;p&gt;Well, the Possion bracket:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \{ A(q,p),B(q,p) \} \equiv \sum_{s} \left( \dfrac{\partial A}{\partial q_{s}} \dfrac{\partial B}{\partial p_{s}} - \dfrac{\partial A}{\partial p_{s}} \dfrac{\partial B}{\partial q_{s}} \right) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a coordinate transformation according to the words of this Wikipedia page: (link: &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson_bracket&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Poisson_bracket&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;...&quot;.It places mechanics and dynamics in the context of coordinate-transformations: specifically in coordinate planes such as canonical position/momentum, or canonical-position/canonical transformation.&quot;...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Jacobian of, for example, $ A(q,p) $ and $ B(q,p) $ has the same form of the term on the parenthesis. From what to what coordinates the bracket transform? Does make any sense to make comparison (Poisson bracket and the Jacobian)? &lt;/p&gt;&#xA;" OwnerUserId="1618" LastActivityDate="2011-02-28T10:37:12.590" Title="Meaning of the Poisson bracket as a coordinate transformation" Tags="&lt;classical-mechanics&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="5793" PostTypeId="2" ParentId="5792" CreationDate="2011-02-24T13:06:54.793" Score="5" ViewCount="" Body="&lt;p&gt;There is no specific coordinate transformation being discussed when defining the Poisson bracket of two general observables, $A$ and $B$. Obviously, the phase space has $2N$ coordinates, $q_i$ and $p_j$, so two coordinates $A,B$ wouldn't be enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, the Poisson brackets are relevant for the so-called &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Canonical_transformation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Canonical_transformation&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;canonical transformations&lt;/em&gt; which are transformations that preserve the Poisson bracket, e.g. among the positions and the momenta:&#xA;$$ x_a\to X_a(x_a): \quad \forall a,b, \quad \{X_a,X_b\} = \{x_a,x_b\} $$&#xA;where $X_a$ is either $Q_i$ or $P_j$, and $x_a$ is either $q_i$ or $p_j$. In quantum mechanics, canonical transformations become unitary transformations - because those preserve the commutators between the observables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a kind of coincidence that a single term in the Poisson bracket has the form of a Jacobian. Jacobians occur at many places in this discussion but for many degrees of freedom, the Jacobian would be a determinant of a large matrix. The Poisson bracket is a sum (which is different) of many determinants of $2\times 2$ matrices (only).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it is not a &quot;complete&quot; coincidence that the $2\times 2$ determinant appears here. Mathematically, we say that the canonical transformations preserve the &lt;em&gt;symplectic structure&lt;/em&gt; - given by the antisymmetric prescription for all the &quot;fundamental&quot; Poisson brackets. For 2 variables, a single $q$ and a single $p$, the symplectic structure is given by $\epsilon_{ab}$, an antisymmetric invariant, which may also be thought of as a &quot;volume invariant&quot;. In other words, $Sp(2,R)\equiv SL(2,R)$: the symplectic transformations are the same ones as the unimodular ones. This isomorphism doesn't generalize to many coordinates.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-24T13:12:55.087" LastActivityDate="2011-02-24T13:12:55.087" />
  <row Id="5794" PostTypeId="2" ParentId="5773" CreationDate="2011-02-24T13:10:04.530" Score="3" ViewCount="" Body="&lt;p&gt;One can consider another generalisation of the $L^2(\mathbb R^3)$ model by noting that $\mathbb R^3$ is simply the configuration space, $Q$, of a single particle. There is a field called Geometric Quantization in which the base manifold is extended from a Configuration space onto a full Symplectic Manifold $(M,\omega)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is that all of the Hamiltonian geometry can be encoded in the symplectic 2-form $\omega$. Thus one can talk about Poisson brackets {f,g}, classical observables and so on. The symplectic manifold is the natural geometric space for the study of classical mechanics of any system. Symplectic manifolds might have the form $M=T^*Q$ - as a cotangent bundle (and thus for a single particle be diffeomorphic to $\mathbb R^6$). However they can arise in other cases, for example via constraint reduction, or independently as solutions of field equations. In finite dimensional cases the symplectic manifolds are 2N dimensional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hilbert space is then constructed on top of this. This procedure involves introducing a complex line bundle (locally $U \times \mathbb C$) B over the space M. There are certain topological conditions needed to ensure existence of the sections of this bundle (which are related to the old Bohr quantization conditions). When the sections exist a pairing operator can be introduced and a Hilbert space constructed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The condition that the wave function $\Psi$ be in a representation (say the position representation) is encoded by introducing what is known as a Polarization on M. This is a foliation of M subject to certain conditions. The sections need to be constant along these foliations. This geometric process results in constructing familiar position and momentum expressions for the wave function, and in a sense rebuilds the configuration space if desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However one can often introduce a Complex structure $J$ such that $J^2=-1$ on M which when compatible with the symplectic form $\omega$ results in some further properties. Firstly this introduces a metric on M, and secondly we have $(M,\omega)$ become a Kahler manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now the &quot;phase space&quot; of the classical system is a Kahler manifold. Furthermore the Polarization conditions mentioned above result in $\Psi(z)$ - a holomorphic function of z. As a concrete example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z = x + ip$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;would be the holomorphic coordinate in 2 (real) D. This complex holomorphic representation for elementary examples was introduced by Bergmann in the 1940s, but in the Geometric Quantization context it is the simplest of the Kahler examples. In these Kahler examples the Coherent states play a fundamental part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of non-trivial manifolds another interesting class of examples from Geometric Quantization are from symmetry (Lie group) examples. Here the classical manifold is constructed from the group manifold itself (by examining coadjoint orbits). As a specific example $SU(2)$ has as classical manifold $S^2$. That is the sphere of radius s is the classical phase space for the rotational degrees of freedom of an elementary particle with spin s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this can be a uniform framework for studying the process of quantization and the implications of non-trivial topologies classically (Bohm-Aharanhov, Berry phase, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One text is &lt;a href=&quot;http://books.google.co.uk/books?hl=en&amp;amp;lr=&amp;amp;id=PuhpI98ZkDgC&amp;amp;oi=fnd&amp;amp;pg=PA1&amp;amp;ots=U39Jc_8iLm&amp;amp;sig=Pvgvvtci0HN-BgDHqKKEoEdXAJQ#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Geometric Quantization&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-24T13:10:04.530" />
  <row Id="5795" PostTypeId="2" ParentId="5792" CreationDate="2011-02-24T13:14:10.297" Score="2" ViewCount="" Body="&lt;p&gt;What the Poisson brackets determine is the symplectic structure(&lt;strong&gt;*&lt;/strong&gt;). The symplectic structure in turn determines the time-evolution of the system's trajectory in phase space. Two sets of phase-space co-ordinates describe the same physics as long as they have the same Poisson bracket. These two sets are then said to be related by a &lt;em&gt;canonical transformation&lt;/em&gt; - a transformation that preserves the symplectic structure on the phase space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The R.H.S. of the Poisson bracket is nothing more than the Jacobian of the transformation as you point out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Poisson bracket structure is a powerful guide in transitioning from classical to quantum mechanics. In fact, the simplest prescriptions for &quot;quantizing&quot; involves replacing the phase-space co-ordinates of the classical description (say $A(p,q)$ and $B(p,q)$) with operators ($\hat A(p,q)$, $\hat B(p,q)$), on a Hilbert space, whose commutator is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \{ A(p,q), B(p,q) \} = C(p,q) \rightarrow [\hat A(p,q), \hat B(p,q)] = i \hbar \hat C(p,q) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given two different sets of classical phase-space co-ordinates, which have the same Poisson bracket and are thus related by a canonical transformation, the corresponding sets of operators in the quantum theory are related by a unitary transformation. In other words the symplectic structure determines whether two different representations of the &lt;em&gt;canonical commutation relations&lt;/em&gt; (CCRs) yield the same quantization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a non-trivial problem especially in the context of quantum gravity. For more background on these topics look up the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stone%E2%80%93von_Neumann_theorem&quot; rel=&quot;nofollow&quot;&gt;Stone-von Neumann theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;*&lt;/strong&gt;) Given a function $f$ on a phase space we can obtain a vector field $X(f)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ X(f)_a = \Omega_a^b \nabla_b f $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Omega_{ab}$ is the symplectic anti-symmetric two form. Lie dragging (roughly speaking &quot;parallel transporting&quot;) $\Omega$ along $X(f)$ leaves it ($\Omega$) invariant. $f$ is the &quot;Hamiltonian&quot; and $X(f)_a$ is the &quot;Hamiltonian&quot; vector field which generates time-evolution of a point in phase space.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-24T13:14:10.297" CommentCount="2" />
  <row Id="5796" PostTypeId="2" ParentId="5792" CreationDate="2011-02-24T14:13:42.103" Score="0" ViewCount="" Body="&lt;p&gt;I do not give you an answer to your questions but an addition to the previous answers, so do not judge me too severely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One &lt;strong&gt;may&lt;/strong&gt; perform any coordinate change at one's convenience. Preserving this or that form of equations is not obligatory. The variable change should just be reversible. Often the equations in new coordinates may be made simpler to solve. If so, do them without hesitation - nobody can oblige you to perform only &quot;canonical&quot; transformations ;-). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Description of the same physics is possible in terms of quite different coordinates, not only in terms of &quot;canonical&quot; ones. I do not see any physical sense in &quot;canonical&quot; transformations, to tell the truth. Take, for example, the pairs $p, q$ and $a, a^+$. Their CCRs are different. So what? Who cares?&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-27T22:47:49.847" LastActivityDate="2011-02-27T22:47:49.847" CommentCount="2" />
  <row Id="5797" PostTypeId="2" ParentId="5770" CreationDate="2011-02-24T14:14:41.453" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lightning#Harvesting_lightning_energy&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lightning#Harvesting_lightning_energy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rough estimation, you can take energy of 1 lightening 5 MJ. Multiply that by number of strikes per day/year and you are done.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-02-24T14:14:41.453" CommentCount="1" />
  <row Id="5799" PostTypeId="2" ParentId="5705" CreationDate="2011-02-24T14:30:03.970" Score="1" ViewCount="" Body="&lt;p&gt;The most geometric formulation of classical mechanics is in terms of symplectic geometry. However in terms of the question asked about the principle of least action, the issue is that this formulation &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Objects seek a path that minimizes total action.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is mathematically vague, and that the phrase (and definition of action) can have different variational forms. Some of these are more intuitive and some are more general than others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most general formulation is &lt;strong&gt;Hamilton's Principle&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The motion of the system from time $t_1$ to time $t_2$ is such that the line integral&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = \int_{t1}^{t2} L dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;where $L= T - V$ is a stationary point for the path of motion&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elegance of this is that it identifies one path from amongst all those between initial position at time $t_1$ and final position at time $t_2$ which extremizes the path. The points and paths here belong to the multidimensional configuration space of the system. Not only that but the remaining equations of motion can now be derived from this for all classical systems, with an extension into quantum mechanics as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there are other (older) formulations of similar variational principles which can also refer to &quot;action&quot; and which have use in classical mechanics too. The classic text Goldstein introduces a &quot;Principle of Least Action&quot; which is defined a little differently. If we introduce the Hamiltonian of a system H and assume that it is conserved:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta \int_{t1}^{t2} \Sigma p_i q_{i}^{dot} dt = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term inside the integral is also known as the Action and the variation involved (which is minimized) is denoted by $\Delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand this formulation let us consider only the simplest case of it: assume non-relativistic mechanics with no time dependence and no external forces. Then this expression reduces to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta(t_1 - t_2) =0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This says that of all paths possible between two points, consistent with conservation of energy, the system moves along that particular path for which the time of transit is the least.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other variations of this formulation can have the particle moving along geodesics within configuration space. Some of these formulations are similar to Fermat's principle of Geometric Optics.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-24T14:30:03.970" />
  <row Id="5800" PostTypeId="1" AcceptedAnswerId="5807" CreationDate="2011-02-24T15:00:58.047" Score="8" ViewCount="207" Body="&lt;p&gt;In Physics ordinary terms often acquire a strange meaning, action is one of them. Most people I talk to about the term action just respond with &quot;its dimension is energy*time&quot;. But what is its historial origin?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Action_(physics)#History&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Action_(physics)#History&lt;/a&gt; doesn't really give much insight, as it lacks citations and depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how did &quot;action&quot; become to mean what it means now?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&lt;/p&gt;&#xA;" OwnerUserId="2151" LastActivityDate="2011-02-24T17:27:31.210" Title="What is the historical origin of the term action" Tags="&lt;lagrangian-mechanics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="5801" PostTypeId="1" CreationDate="2011-02-24T15:11:01.823" Score="7" ViewCount="126" Body="&lt;p&gt;What happens when two D-branes annihilate? Do we get a radiation of strings?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" LastActivityDate="2011-06-24T13:49:19.460" Title="What happens when two D-branes annihilate?" Tags="&lt;string-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="5802" PostTypeId="2" ParentId="5765" CreationDate="2011-02-24T15:45:53.310" Score="2" ViewCount="" Body="&lt;p&gt;The computation can be done alternatively in the Bargmann representation of the harmonic oscillator.&#xA;In the following, the required inner product evaluation will be described in this representation. &#xA;In my opinion, this method is computationally superior as well as many other advantages.&#xA;This method is based on the isomorphism between the Hilbert spaces $L^2(\mathbb{R})$ and the Bargmann space of analytical functions on $\mathbb{C}$ with respect to the inner product&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(f,g) = \frac{1}{2\pi}\int_{\mathbb{C}} f(z) \overline{g(z)} exp(-z\bar{z})dz d\bar{z}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The isomorphism is given explicitely by means of the &lt;a href=&quot;http://planetmath.org/encyclopedia/BargmannTransform.html&quot; rel=&quot;nofollow&quot;&gt;Bargmann transform)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Bargmann representation, the creation operator is representaed by the multiplication by $z$&#xA;and the anihilation operator derivative with respect to z and the vacuum state by the constant unit function &#xA;(and, by the way, the energy eigenfunctions of the harmonic oscillator by the monomials $z^n$ - up to a normalization).&#xA;Thus the $\alpha$ coherent state is represented by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\psi_{\alpha}(z) = \exp(-\frac{|\alpha|^2}{2}) \exp(\bar{\alpha}z) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the inner product is therefore given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\psi_{\beta},\psi_{\alpha}) = \frac{\exp(-\frac{(|\alpha|^2+|\beta|^2)}{2})}{2\pi}\int_{\mathbb{C}} \exp(\bar{\beta}z) \exp(\alpha\bar{z}) exp(-z\bar{z}) dz d\bar{z} = \exp(-\frac{(|\alpha|^2+|\beta|^2)}{2}) \exp(\bar{\beta} \alpha)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral is easily evaluated by coordinate translation and square completion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This example is a prototype of quantization in Kahler polarization.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastActivityDate="2011-02-24T15:45:53.310" />
  <row Id="5803" PostTypeId="2" ParentId="5552" CreationDate="2011-02-24T15:48:26.750" Score="2" ViewCount="" Body="&lt;p&gt;If by &quot;short distance physics&quot; you mean &quot;arbitrarily short distances&quot; then the answer to your question is that your question is ill-defined. You cannot consistently describe arbitrarily small distances within QED.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QED (as opposed to QCD) has a Landau pole and thus is inconsistent if viewed as a &quot;fundamental&quot; theory. Like most other quantum field theories QED should be viewed as an effective field theory with its range of validity. You should not trust QED as an appropriate description of Nature at (or beyond) the Planck scale (or even at the GUT scale or at whichever new physics scale may lie between TeV and the GUT scale). &lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-02-24T15:48:26.750" CommentCount="9" />
  <row Id="5804" PostTypeId="1" AcceptedAnswerId="5805" CreationDate="2011-02-24T16:00:55.793" Score="1" ViewCount="257" Body="&lt;p&gt;In a lot of laymen explanations of general relativity it is implied that the four dimensions of the space-time are equivalent, and we perceive time as different only because it is embedded in our human perception to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: is that really how general relativity treats the 4 dimensions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so - what are the implications (if any) this has on causality?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no - can the theory support more than one time dimension?&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-02-24T18:23:46.967" LastActivityDate="2011-03-04T00:48:36.117" Title="Does time have a special status in general relativity?" Tags="&lt;general-relativity&gt;&lt;causality&gt;&lt;dimensionality&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="5805" PostTypeId="2" ParentId="5804" CreationDate="2011-02-24T16:10:51.137" Score="6" ViewCount="" Body="&lt;p&gt;That statement, that space and time are equivalent, is not really correct. In special relativity there is a distinction between spacelike and timelike events, those are events that cannot or can (respectively) be causally connected. This replaces the notion of &quot;simultaneous&quot; and &quot;before or after&quot; to something all inertial observers can agree on. In general relativity, this distinction is made locally - causal influence propagates locally in speeds less than c, but that constraints changes from point to point according to the local metric (which encodes the force of gravity). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this is encoded in the statement that the metric is indefinite, with a specific signature that supports one time direction. While the mathematics of GR can be generalized to other signatures, the physics cannot - the Lorentzian signature is essential in correctly interpreting the theory. Anyone can play all kinds of mathematical games, but those are meaningless unless you are very clear on how the calculations you perform are related (at least in principle) to physically observable quantities.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-02-26T19:54:41.387" LastActivityDate="2011-02-26T19:54:41.387" />
  <row Id="5806" PostTypeId="2" ParentId="5804" CreationDate="2011-02-24T17:13:32.420" Score="5" ViewCount="" Body="&lt;p&gt;The formal statement would be that there is no &lt;em&gt;one&lt;/em&gt; notion of time, and that one persons definition of time may be intermixed with another's definition of space.  What is not correct is that time and space are wholly interchangeable, as Moshe says--there is a distinction between events separated in time and events separated in space.  Causality theory is constructed around this notion--spacetimes that fail to keep the distinct separation between spacelike and timelike separated events will inevitably violate rules of cause and effect.  Also, all observers will agree on which events are timelike and spacelike.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically, it is possible to define a theory with two linearly independent timelike directions.  The physics in such a theory would be extremely novel (for example, if I time evolved in direction one, and then time evolved in direction two, I wouldn't necessarily end up at the same event if I reveresed this order--what would 'time' even end up meaning in such a theory?), and I don't see a particular reason to do so.  But there certainly is existent math to describe a theory with two time directions.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-02-24T17:13:32.420" />
  <row Id="5807" PostTypeId="2" ParentId="5800" CreationDate="2011-02-24T17:27:31.210" Score="8" ViewCount="" Body="&lt;p&gt;See this better introduction to the history:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Principle_of_least_action#Origins.2C_statement.2C_and_the_controversy&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Principle_of_least_action#Origins.2C_statement.2C_and_the_controversy&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course, the notion of &quot;action&quot; only becomes meaningful when one actually knows what the purpose of the action is - to be minimized. Around 1744 and 1746, Pierre Louis Maupertuis figured out that this could be a way to formulate laws of physics when he generalized Fermat's 17th century &quot;principle of least time&quot; (for the trajectory taken by light in any environment with a variable index of refraction) by this proverb:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Nature is thrifty in all its actions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously, some word had to be constructed or borrowed to describe the new quantity whose importance was previously unknown to the humans (and remains to be unknown to most humans even today).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the word &quot;actions&quot; appeared as the only noun of the quote in the context of these minimization problems, so it became known as Wirkung ($W$) in German and action ($S$) in English. I am actually not sure why the letter $S$ was chosen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been claims that Leibniz had found the principle as early as in 1707.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maupertuis also wrote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The laws of movement and of rest deduced from this principle being precisely the same as those observed in nature, we can admire the application of it to all phenomena. The movement of animals, the vegetative growth of plants ... are only its consequences; and the spectacle of the universe becomes so much the grander, so much more beautiful, the worthier of its Author, when one knows that a small number of laws, most wisely established, suffice for all movements.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-24T17:27:31.210" />
  <row Id="5808" PostTypeId="2" ParentId="5804" CreationDate="2011-02-24T18:12:02.187" Score="0" ViewCount="" Body="&lt;p&gt;The only time which has any special status is proper time.  This is the length of a geodesic or path in space.  The interval $ds^2~=~g_{ab}dx^adx^b$ is a measure of proper time.  This is invariant under all Lorentz transformations.  For $U$ an element of the Lorentz group the interval length $s~=~x^T\cdot g\cdot x$ these element transform as $x’~=~Ux$, $x’^T~=~x^TU^{-1}$ and $g’~=~UgU^{-1}$ and it is trivial to see&#xA;$$&#xD;&#xA;s’~=~x^TU^{-1} UgU^{-1}Ux~=~x^Tgx~=~s.&#xD;&#xA;$$&#xA;So the proper time is invariant.  The coordinate time, usually written as $t$ is transform dependent.  This is a manifestation of the coordinate system you chose to work in.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-24T18:12:02.187" />
  <row Id="5809" PostTypeId="2" ParentId="5705" CreationDate="2011-02-24T18:14:19.503" Score="0" ViewCount="" Body="&lt;p&gt;the most elegant form of classical mechanics comes from newton's laws:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;f =ma&lt;/li&gt;&#xA;&lt;li&gt;action = reaction&lt;/li&gt;&#xA;&lt;li&gt;bodys remain at rest unless acted on by other forces.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-02-24T18:14:19.503" />
  <row Id="5810" PostTypeId="2" ParentId="5714" CreationDate="2011-02-24T18:18:31.327" Score="0" ViewCount="" Body="&lt;p&gt;Aristotel's theory of motion where larger masses fell to the earth faster than smaller masses.&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-02-24T18:18:31.327" CommunityOwnedDate="2011-02-24T18:18:31.327" CommentCount="1" />
  <row Id="5811" PostTypeId="2" ParentId="1971" CreationDate="2011-02-24T18:52:58.927" Score="0" ViewCount="" Body="&lt;p&gt;Even a very simple model, say an infinetly thick conductive slab in space so that you only have to deal with radiative effects (solar absorption, and thermal heat loss) gets messy, the solar input is zero at night but follows a curve during the day (easily computable via geometry), but solar input at sunrise and sunset will be discontuous in the first time derivative, which should introduce high order harmonics. Also the Stefan Boltzmann equation is nonlinear in temperature. Then add an atmosphere, it heats the surface until the thermal structure supports convection, so different transport mechanisms are important during different times of day. Then add clouds and variable weather systems....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just want to reasonably reproduce the curves, I'd use only a couple of annual and daily harmonics, otherwise you are probably going to overfit the data. But the statisticians would be the best people to consult about it.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-02-24T18:52:58.927" />
  <row Id="5812" PostTypeId="1" AcceptedAnswerId="5837" CreationDate="2011-02-24T18:53:09.570" Score="6" ViewCount="239" Body="&lt;p&gt;&quot;stability&quot; is invoked as the justification for the axiomatic requirement that the spectrum of the generators of the translation group must be confined to the forward light-cone. The &lt;em&gt;spectrum condition&lt;/em&gt; has pervasive, significant effects in axiomatic QFT. There seems to be no proof, however, that the spectrum condition actually ensures that a quantum field will be stable, partly because there is, AFAIK, no mathematical specification of what stability consists of in QFT. IF there were, I suppose the &lt;em&gt;stability axiom&lt;/em&gt; would be central to axiomatic QFT instead of the positive spectrum condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stability is intimately related with positive energy in classical physics, of course, but the concept of energy is rather different in classical relativistic field physics than in quantum mechanics, being the 00 component of the stress-energy tensor instead of being the 0 component of the 4-vector of generators of translations. The relationship between positive energy and stability in classical physics does not seem enough to justify an uncritical adoption of the spectrum condition in quantum theory &lt;em&gt;as an axiom&lt;/em&gt;, which is supposed to be obvious enough that it is almost beyond question. Negative frequencies are certainly not ruled out for classical field theories, because the energy is not a linear functional of the frequency of the Fourier components of the field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An axiomatic definition of stability would presumably have to specify what deformations would or should not affect the stability of a given construction. A building is only stable, for example, provided a strong enough earthquake does not occur, it is not stable &lt;em&gt;sine die&lt;/em&gt;. Given that the deformations that are possible in quantum field theory are more varied than the deformations that are possible in classical field theory, the spectrum condition seems to require a more substantial justification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Less axiomatically, Feynman integrals include negative frequency/energy components in intermediate calculations, though &lt;em&gt;not&lt;/em&gt; in observables, which seems to bring the spectrum condition into at least some question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Haag discusses the relationship of stability with the spectrum condition only extremely perfunctorily (p.29 of the 2nd edition of &lt;em&gt;Local Quantum Physics&lt;/em&gt;), and I am not aware of an elaborate discussion by other authors. Is there one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Streater &amp;amp; Wightman, in &lt;em&gt;PCT, Spin &amp;amp; Statistics, and all that&lt;/em&gt;, discuss collision states. Their discussion is entirely in terms of perturbation theory, which seems not adequate &lt;em&gt;enough&lt;/em&gt; for an axiomatic discussion. However, because of asking this question I'm starting to see slightly more clearly why a conventional Physicist might be entirely satisfied with what there is on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT(after acceptance of Tim van Beek's Answer): The other aspect of this is that the restriction to positive frequency is apparently not enough to ensure &amp;ldquo;confinement&amp;rdquo;, at least not in an elementary way. That seems to me to be more what &amp;ldquo;stability&amp;rdquo; &lt;em&gt;ought&lt;/em&gt; to mean.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-02-25T16:04:01.930" LastActivityDate="2011-02-25T16:04:01.930" Title="There seems to be no definition of &quot;stability&quot; in axiomatic QFT. Is there? And, if not, is this a problem?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;mathematical-physics&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="5813" PostTypeId="1" AcceptedAnswerId="5814" CreationDate="2011-02-24T19:01:09.370" Score="4" ViewCount="153" Body="&lt;p&gt;Is there a that detection method in use that can distinguish between a neutrino from its anti-neutrino?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-02-24T19:27:05.537" LastActivityDate="2011-02-24T21:00:20.810" Title="Neutrino versus Anti-neutrino Detection" Tags="&lt;particle-physics&gt;&lt;experimental-physics&gt;&lt;neutrino&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="5814" PostTypeId="2" ParentId="5813" CreationDate="2011-02-24T19:07:17.550" Score="3" ViewCount="" Body="&lt;p&gt;Some detection methods only detect neutrinos, as opposed to antineutrinos. For instance, the original solar neutrino experiment in the Homestake mine worked on the basis of the reaction&#xA;$$&#xD;&#xA;\nu+{}^{37}{\rm Cl}\to {}^{37}{\rm Ar}+e^-&#xD;&#xA;$$&#xA;if I recall correctly. Only a neutrino can induce that reaction; an antineutrino can't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other experiments can detect neutrinos through multiple channels, some of which (elastic scattering off an electron, for instance) are sensitive to both $\nu,\overline\nu$ and some of which only detect one.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-24T19:07:17.550" CommentCount="2" />
  <row Id="5815" PostTypeId="1" AcceptedAnswerId="5817" CreationDate="2011-02-24T19:09:46.297" Score="1" ViewCount="302" Body="&lt;p&gt;I would like to expand on what I mean by the title of this question to focus the answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally whenever a theory (e.g. General Relativity) replaces another (e.g. Newtonian Gravity) there is a correspondence requirement in some limit. However there is also normally some experimental area where the new larger theory makes predictions which are different from the older theory which made predictions of the same phenomena. This is ultimately because the newer theory has a deeper view of physics with its own structures which come into play in certain situations that the old theory didnt cover well. Additionally the newer theory will make predictions based on its novel aspects which the older theory did not consider. I know that String Theory is quite rich in this regard, but am not interested in that here. Nor am I concerned as to whether experiment has caught up, as I know that ST (and Quantum Gravity in general) is not easy to test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for the GR to Newtonian example an answer to this question would be: bending of light rays; Mercury perhelion movement - GR had a different results to Newton. What would &lt;em&gt;not&lt;/em&gt; count as an answer would be new structures which GR introduces like Black Holes or even gravitational curvature per se.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So does ST have anything like Mercury perhelion movement waiting to be experimentally verified, and thus &quot;improving&quot; on GR within GR's own back yard?&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-25T00:35:13.403" Title="Does String Theory disagree with General Relativity?" Tags="&lt;general-relativity&gt;&lt;theory&gt;&lt;string&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="5816" PostTypeId="2" ParentId="905" CreationDate="2011-02-24T19:10:04.270" Score="-1" ViewCount="" Body="&lt;p&gt;Very late to this, but entropy is a functional of the phase space DoFs, which, at least for a field theory, is essentially dependent on a choice of a space-like hypersurface (or hyperplane). The thermodynamic dual of entropy is of course temperature, and the all-important $kT$ has dimensions of energy because this determines the probability density $\exp{(-E/kT)}$, and thereby the amplitude of classical thermal fluctuations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 4-dimensional equivalent of $kT$ would be an action. Planck's constant can also be understood to determine the amplitude of fluctuations, which are of course &lt;em&gt;quantum&lt;/em&gt; fluctuations (which are distinct from thermal fluctuations essentially because they are invariant under the action of the Lorentz group). So, a 4-entropy would perhaps be the thermodynamic dual of the amplitude of quantum fluctuations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is essentially allusive, however, since I think this does not allow us, AFAIK, to &lt;em&gt;construct&lt;/em&gt; a 4-entropy. There's something on all this in my &lt;em&gt;Phys. Lett.&lt;/em&gt; &lt;strong&gt;A 338&lt;/strong&gt;, 8-12(2005), or, on arXiv, &lt;a href=&quot;http://arxiv.org/abs/quant-ph/?0411156&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/quant-ph/?0411156&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-02-24T19:10:04.270" />
  <row Id="5817" PostTypeId="2" ParentId="5815" CreationDate="2011-02-24T19:31:16.220" Score="4" ViewCount="" Body="&lt;p&gt;String theory implies new physics in - and only in - the quantum regime. In particular, at distances that are very short - comparable to the string scale or Planck scale - there are new effects. The black holes decay (while they preserve the information), effective actions have higher-derivative terms, e.g. $R^2$, there are strings, branes, fluxes, extra dimensions, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The new physics at the Planck scale - which is very far - is almost certainly not testable by the naive 19th century observations such as the Mercury perihelion's precession. This is not a problem of string theory in any sense: it is a tautological consequence of the questions that string theory addresses - namely the behavior of the Universe at the most fundamental scale - and any other theory that addresses the same questions inevitably shares the inaccessibility of the phenomena by direct tests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you only look at the classical limit or classical physics, string theory exactly agrees with general relativity. In some sense, this is true even in the quantum regime: string theory is the only consistent quantum completion of general relativity.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-24T19:31:16.220" CommentCount="4" />
  <row Id="5818" PostTypeId="2" ParentId="5813" CreationDate="2011-02-24T19:49:51.350" Score="2" ViewCount="" Body="&lt;p&gt;Some more neutrino-antineutrino specific reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nu_e n  \to  p^+ e^-$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\overline\nu_e p^+  \to n e^+$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nu_\mu n  \to  p^+ \mu^-$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\overline\nu_\mu p^+  \to n \mu^+$&lt;/p&gt;&#xA;" OwnerUserId="238" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-02-24T20:01:51.033" LastActivityDate="2011-02-24T20:01:51.033" />
  <row Id="5819" PostTypeId="1" AcceptedAnswerId="5823" CreationDate="2011-02-24T20:24:09.687" Score="2" ViewCount="485" Body="&lt;p&gt;How does one build up an intuitive gut feeling for physics that some people naturally have? Physics seems to be a hodgepodge of random facts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that a sign to quit physics and take up something easier?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for all the answers. On a related note, how many years does it take to master physics? 1-2 years for each level multiplied by many levels gives? &lt;/p&gt;&#xA;" LastActivityDate="2011-08-12T02:37:43.390" Title="How does one build up intuition in physics?" Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="6" FavoriteCount="2" />
  <row Id="5820" PostTypeId="2" ParentId="5819" CreationDate="2011-02-24T20:48:19.053" Score="1" ViewCount="" Body="&lt;p&gt;This question has similarities to at least &lt;a href=&quot;http://physics.stackexchange.com/questions/2219/beginner-physics-resources&quot;&gt;this&lt;/a&gt; previous Stack question on learning physics. I would advise that you check the online links there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An intuition for physics is built by learning the subject and finding out the major theories and their domain of application. So considering a question on Gravity (say working out the gravity on the surface of the Moon) what physics theories apply? As one learns more physics one learns about the limitations of older theories. Perhaps velocities have to be small, or masses below a certain limit: above that limit another related theory applies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are general principles from the major theories which derive much of one's analysis: energy and its conservation and transmission, radiation (of the various kinds). In applied examples a major test of intuition is to include the contributions of factors that are relevant and omit those which are irrelevant. So for example does the chemical composition of the Moon affect its surface gravity, are hollows and bumps on its surface likely to be relevant; could its nearness to the Earth affect the calculation? If we are considering a Pulsar (Neutron star) rather than the Moon do any of these factors change in significance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically there are parallels between many theories which contain fields which decay at $1/r^2$; there are many phenomena describable by similar mathematical models like Harmonic Oscillators (from electronics to quantum physics);then there is the Wave Equation. This allows one to learn to expect similar behaviour in a unfamiliar sub-branch of the subject whenever similar phenomena appear. Indeed one may have already solved a given problem in other variables already.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally there is some evidence to support the hope that all of physics is united under a set of underlying equations and principles. They may not all have been discovered yet, but this suggests that understanding one set of concepts very well could teach us how everything works. We already know of the major branches of physics which are very comprehensive: Thermodynamics; Electromagnetism; Gravitation; Dynamics; Quantum Theory; then there are the subatomic theories.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-24T20:48:19.053" />
  <row Id="5821" PostTypeId="2" ParentId="5813" CreationDate="2011-02-24T20:51:00.293" Score="2" ViewCount="" Body="&lt;p&gt;Yes, as @voix shows in his partial list, neutrinos go either to e- or mu- (or tau-) and antineutrinos go to the postive corresponding particles (conservation of lepton number). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In scattering experiments there may also be any number of hadrons produced depending on the energy of the beam.Because of weak interactions &lt;a href=&quot;http://proj-cngs.web.cern.ch/proj-cngs/PDF_files/ID70_MoriondEW07_CNGS.pdf&quot; rel=&quot;nofollow&quot;&gt;neutrino beams need trillion&lt;/a&gt;s of neutrinos/antineutrinos impinging on the detector in order to see a reasonable number of neutrino/antineutrino interactions by the end of the experiment. Beams by construction are of one type of neutrino anyway, so it is a consistency check.&#xA; Experiments can assign interactions to neutrinos or antineutrinos depending on the charge of the outgoing lepton. The charges of the leptons are found from the curvature of the tracks in the magnetic field of the setup of the experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way one can see if a neutrino or antineutrino is &lt;em&gt;produced&lt;/em&gt;  in an interaction, is from the missing energy, because neutrinos/antineutrinos interact very weakly and cannot be detected within the apparatus. In events where all charged tracks have been measured one can also estimate the missing mass, and a neutrino/antineutrino is assigned by a fitting to a hypothesis  , if the missing mass is small.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-02-24T21:00:20.810" LastActivityDate="2011-02-24T21:00:20.810" CommentCount="2" />
  <row Id="5822" PostTypeId="2" ParentId="5819" CreationDate="2011-02-24T20:51:38.007" Score="6" ViewCount="" Body="&lt;p&gt;Intuition is just pattern recognition. This comes with doing many problems that force you to think and keep your brain completely engaged with the material. I remember when I first started physics, I thought it was impossible to keep track of all of the types of problems. There were balls rolling down inclined planes and masses on pulleys and people pushing blocks up hills with friction. Then there was a certain point when a lightbulb went off and it made sense: &lt;em&gt;these are all the same problem.&lt;/em&gt; All you do is figure out where all the forces are pointing and how strong they are, then add them up  to find the net force, and you've got the equation of motion in hand. Eventually, it becomes second nature, so I can glance at a problem in mechanics and pretty quickly have an intuition for what should happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thing happens in more advanced physics. You start to recognize classes of physical situations and types of symmetry. Let me give you a a more advanced example that was a similar lightbulb for me - it's a bit lengthy, and you don't need to understand every step. The important part is the conclusion. Try evaluating this integral:&#xA;$\int_{-\infty}^{\infty}e^{ip(x-x_0)/\hbar}e^{-|p|/p_0}dp$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks horrible, doesn't it? And yes, if you just charge in and try to integrate it, you'll wind up doing a long series of painful integrations by parts. But there are a bunch of things here that a practiced physics student will recognize that make it easy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $e^{ip(x-x_0)/\hbar}=\cos(p(x-x_0)/\hbar)+i\sin(p(x-x_0)/\hbar)$ (that's clever trick no. 1), this is actually an integral of a cosine plus a sine, multiplied by a decaying exponential, over $(-\infty,\infty)$. But sine is an odd function, so it evaluates to zero over $(-\infty,\infty)$ (clever trick no. 2)! That means the integral is actually&#xA;$\int_{-\infty}^{\infty}\cos(p(x-x_0)/\hbar)e^{-|p|/p_0}dp$ Now, we have an even function inside the integral, so its value over $(-\infty,\infty)$ is twice its value over $(0,\infty)$ (clever trick no. 2, reused). Furthermore, $\cos(p(x-x_0)/\hbar)=Re(e^{ip(x-x_0)/\hbar})$ (going backwards with clever trick no. 1). This makes the integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2Re(\int_{0}^{\infty}e^{ip(x-x_0)/\hbar}e^{-p/p_0}dp)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2Re(\int_{0}^{\infty}e^{(i(x-x_0)/\hbar - 1/p_0)p}dp)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2Re[\frac{1}{i(x-x_0)/\hbar + 1/p_0}]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{2p_0}{1+(x-x_0)^2p_0^2/\hbar^2}$ (Maybe clever trick no. 3 if you aren't very used to complex numbers?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the kicker: &lt;strong&gt;If you do it enough, this sort of integral starts coming intuitively and quickly.&lt;/strong&gt; The even-and-odd tricks become obvious; the conversion back and forth between exponential and trigonometric functions becomes totally natural. An integral like this flies by at the speed of thought, and I don't need to write out all those steps - it just feels sort of obvious at each point. That's just pattern recognition from doing it so many times - or, if you like, &quot;intuition.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to build up your intuition, do many, many problems. As many as you can get your hands on. That's what it takes.&lt;/p&gt;&#xA;" OwnerUserId="1655" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-02-24T22:52:05.063" LastActivityDate="2011-02-24T22:52:05.063" CommentCount="1" />
  <row Id="5823" PostTypeId="2" ParentId="5819" CreationDate="2011-02-24T21:03:47.877" Score="9" ViewCount="" Body="&lt;p&gt;If physics seems to be a hodgepodge of random facts... well, it always seems that way when you're first learning. But physics is all about patterns and relationships, and if you keep working with it you should &lt;em&gt;eventually&lt;/em&gt; get to a point where you see how different facts, formulas, and principles are related to each other. In my experience, for any given &quot;piece&quot; of physics this point comes 1-2 years after you first learn it. :-P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT&lt;/em&gt;: In response to your related note, I don't know that anyone ever really &quot;masters&quot; physics. I'm not even sure exactly what that would mean, since physics as a whole is far too broad for any one person to ever understand in its entirety. But my impression is that, given the usual educational path (college then grad school), it takes anywhere from 3-10 years, depending on the topic and your level of preparation, to learn enough that you can start contributing to research in a single specific topic. Of course you never really stop learning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, to clarify, the 1-2 year figure I mentioned isn't something you can multiply by a number to figure out how long it takes to learn physics ;-) And physics doesn't neatly split itself into levels (well, sort of, although the splitting you'll probably encounter is just a result of the requirements of our educational system). All I meant is that, from the first time you encounter any given physics concept (just to name a few: Newton's laws, the Euler-Lagrange equation, multipole expansion, free energy, quantum entanglement), you may not feel like that concept makes sense to you until 1-2 years later, on average. And that's perfectly normal.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-25T02:35:47.340" LastActivityDate="2011-02-25T02:35:47.340" CommunityOwnedDate="2011-02-24T21:03:47.877" CommentCount="4" />
  <row Id="5824" PostTypeId="1" AcceptedAnswerId="5826" CreationDate="2011-02-24T21:33:18.340" Score="6" ViewCount="835" Body="&lt;p&gt;I just saw this &quot;&lt;a href=&quot;http://www.youtube.com/watch?v=7ctaA2mERzI&amp;amp;feature=player_embedded&quot; rel=&quot;nofollow&quot;&gt;trick&lt;/a&gt;&quot; where a cup of water is turned over onto a table without spilling (using a piece of cardboard.  After removing the cardboard from underneath the cup, the person then removes the cup in a particular way (lifts straight up and twists) and lo and behold, the water stays in it's position as if the cup were still there!?  (watch the video to fully understand)  Is this really possible?  If so, and the real question I'm looking to have answered is, how?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After &lt;a href=&quot;http://www.b3ta.com/links/592518&quot; rel=&quot;nofollow&quot;&gt;further research&lt;/a&gt; it appears that deionized water is needed as well.&lt;/p&gt;&#xA;" OwnerUserId="249" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2011-02-25T15:31:49.327" LastActivityDate="2011-02-25T15:31:49.327" Title="Is it really possible for water to be held in a &quot;cone shape&quot; for a brief period of time? " Tags="&lt;water&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="5826" PostTypeId="2" ParentId="5824" CreationDate="2011-02-24T21:43:47.500" Score="8" ViewCount="" Body="&lt;p&gt;Let's split up the forces, it gives an easier view on what should happen. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Horizontal Forces&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;What we see is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Centrifugal_force&quot; rel=&quot;nofollow&quot;&gt;centrifugal force&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Centrifugal force (from Latin centrum, meaning &quot;center&quot;, and fugere, meaning &quot;to flee&quot;) represents the effects of inertia that arise in connection with rotation &lt;strong&gt;which are experienced as an outward force away from the center of rotation&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, outward force &lt;em&gt;away from the center&lt;/em&gt; of rotation. This means that the water would only be able to appear in place if there is a force in the opposite direction, which is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Friction&quot; rel=&quot;nofollow&quot;&gt;friction&lt;/a&gt;. Where is the friction if you take the glass away? There is no friction anymore, thus that video shows special effects. ;-)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Vertical Forces&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There's also the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_force&quot; rel=&quot;nofollow&quot;&gt;gravitational force&lt;/a&gt;. What would keep the water up so long? Nothing.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Furthermore, if you look at &lt;a href=&quot;http://www.youtube.com/user/dandeentremont&quot; rel=&quot;nofollow&quot;&gt;his other videos&lt;/a&gt;, you can confirm it's CGI... :-)&lt;/p&gt;&#xA;" OwnerUserId="2192" LastEditorUserId="249" LastEditorDisplayName="" LastEditDate="2011-02-24T22:41:59.593" LastActivityDate="2011-02-24T22:41:59.593" CommentCount="5" />
  <row Id="5827" PostTypeId="2" ParentId="5765" CreationDate="2011-02-24T21:47:33.437" Score="1" ViewCount="" Body="&lt;p&gt;Use the Baker-Campbell-Hausdorff identities. They are your best friends for this. Two links, for example, &lt;a href=&quot;http://planetmath.org/encyclopedia/BCHFormula.html&quot; rel=&quot;nofollow&quot;&gt;http://planetmath.org/encyclopedia/BCHFormula.html&lt;/a&gt; , and &lt;a href=&quot;http://www.mathematics.thetangentbundle.net/wiki/Linear_algebra/Baker-Campbell-Hausdorff_formula&quot; rel=&quot;nofollow&quot;&gt;http://www.mathematics.thetangentbundle.net/wiki/Linear_algebra/Baker-Campbell-Hausdorff_formula&lt;/a&gt; . Wikipedia has a page, &lt;a href=&quot;http://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula&lt;/a&gt; , that has an example under the heading &quot;Application in Quantum Mechanics&quot; that is pretty much what you need.&#xA;Best, Peter.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-02-24T21:47:33.437" />
  <row Id="5828" PostTypeId="2" ParentId="5824" CreationDate="2011-02-24T21:58:22.097" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Its a trick.&#xA;There are 2 vessels:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the outer that we saw, and an inner one, transparent and soft.&#xA;Stopt it at &lt;strong&gt;1:48&lt;/strong&gt; you will see the water move out from the bottom. Under atmospheric pression the vessel merge down. The image after this moment is truncated and we can not see the remnants in the table.&#xA;The water was previously set in rotational motion (before the image restarts after those words).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;added:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not needed to explore the inner characteristics of the inner vessel but I will try.&#xA;My first tough : It can be a soft plastic bag.&#xA;My second try:   It can be made of ice. But I suspect it is not so. Rinse the interoir of the vessel, then fridge until ice coated. Rinse again, and repeat,...or two equal vessels one inside the other separated by water and fridge.&#xA;Now, with the tiny ice coating prepared. Rinse the exterior of the vessel with warm water. Wait a little and the exterior will be decoupled from the interior vessel. Start the show&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;added again, my final conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a soft bag. The rotational effect is really need to keep the vertical wall under more tension and will prevent the object from immediate collapse. The centrifugal force will add strength to the vertical wall and the vertical component of force will be lower.&#xA;If it was the icy vessel the rotational effect was not needed.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-02-25T09:20:04.263" LastActivityDate="2011-02-25T09:20:04.263" CommentCount="9" />
  <row Id="5829" PostTypeId="1" CreationDate="2011-02-24T22:04:34.477" Score="7" ViewCount="172" Body="&lt;p&gt;Some topological solitons formed from bosonic fields have fermionic statistics. Why?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-25T06:40:58.030" Title="Why are some solitons formed from bosonic fields fermionic?" Tags="&lt;quantum-field-theory&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="5830" PostTypeId="1" CreationDate="2011-02-24T22:19:07.310" Score="3" ViewCount="133" Body="&lt;p&gt;It is a question about free carrier behavior in graphene flakes. (or may be called charge confinement)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say if we have a perfect hexagonal free standing graphene flake terminated with zigzag edges. Since there are two subatoms (let's call them A,B) in every unit cell. Edges will be terminated, let say, ABABAB (as shown in the figure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/gt1tJ.png&quot; alt=&quot;figure of graphene flake with zigzag edge&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how to calculate (understand) electron wavefunction in this system (only considering relative low energy case, charge carrier can be seen as massless fermions). Another way to ask it what type of &lt;strong&gt;boundary condition&lt;/strong&gt; one shall use for those two types of edges? According to this &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1102/1102.1269v1.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt;(pdf), scattering at the zigzag boundary dosen't couple the two valleys. I think we can only consider one valley picture(is this true?), which means the most simple Dirac equation $-i\hbar \nu_F\nabla \Psi_i=E\Psi_i$, where $i=A,B$ (or Klein-Gordon eq). For simplicity, spin effect here can also be ignored(no magnetic field). I am not only asking about the edge states, also for higher energy states. What confusing me is how to distinguish this from the particle in a box case (relativistic particle, here I mean). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, I think electron will behave as billiard in side of the flake. But there could be two cases, either electrons bouncing between A,B edges, or they do it only in one type of edge, such as only bouncing at edges terminated with subatom A. Can it be understood this way? If yes, which of the two picture is the real one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope it is clear. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1948" LastEditorUserId="113" LastEditorDisplayName="" LastEditDate="2011-02-24T22:40:12.947" LastActivityDate="2011-02-25T06:16:24.797" Title="What is the boundary condition of graphene flake with zigzag edges?" Tags="&lt;condensed-matter&gt;&lt;solid-state-physics&gt;&lt;graphene&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5831" PostTypeId="2" ParentId="5819" CreationDate="2011-02-24T23:02:02.113" Score="1" ViewCount="" Body="&lt;p&gt;Speaking of patterns, you may want to read Shive's &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0852745400&quot; rel=&quot;nofollow&quot;&gt;Similarities in Physics&lt;/a&gt;&lt;/em&gt;.  It discusses common behaviour patterns such as filtering, dispersion, resonance, interference, exponential decay, noise, etc which can be observed across acoustics, optics, mechanics and electronics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a book by Pikovsky et al &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/052153352X&quot; rel=&quot;nofollow&quot;&gt;Synchronization: A Universal Concept in Nonlinear Sciences.&lt;/a&gt; is also interesting in this regard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also take a look at Terence Tao's article on &lt;a href=&quot;http://terrytao.wordpress.com/2010/09/14/a-second-draft-of-a-non-technical-article-on-universality/&quot; rel=&quot;nofollow&quot;&gt;Universality in large dimensions.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1468" LastActivityDate="2011-02-24T23:02:02.113" />
  <row Id="5832" PostTypeId="1" CreationDate="2011-02-25T00:12:35.287" Score="1" ViewCount="167" Body="&lt;p&gt;In my previous question I asked how gravity can be represented as dents in space. It seems that it's a metaphor. But I wondered with spac expanding does the effect of gravity get bigger as well, I know from other questions that mass doesn't expand like the gaps between mass do although I have no idea why. So using the metaphor from before does the expansion of space cause the &quot;dents&quot;(gravity) to increase?  &lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-02-25T12:23:23.117" LastActivityDate="2011-03-28T00:01:32.227" Title="Expansion of space and gravity" Tags="&lt;astrophysics&gt;&lt;space-expansion&gt;" AnswerCount="3" />
  <row Id="5833" PostTypeId="2" ParentId="5815" CreationDate="2011-02-25T00:35:13.403" Score="0" ViewCount="" Body="&lt;p&gt;Roy, the higher derivative terms are there just like in semiclassical gravity simply b/c when you introduce quantum mechanics, any term that is not forbidden by some symmetry principle or other physical criteria must be present, albeit Planck suppressed.  So its not a surprise that string theory predicts them.  The interesting thing is that there are additional objects beyond the usual spectrum of GR.  Things like the dilaton, supersymmetric objects (gravitino et al) and so on.&lt;/p&gt;&#xA;" OwnerUserId="1350" LastActivityDate="2011-02-25T00:35:13.403" />
  <row Id="5834" PostTypeId="2" ParentId="5775" CreationDate="2011-02-25T01:05:24.213" Score="0" ViewCount="" Body="&lt;p&gt;From the way the question is phrased, I take &quot;senior physicist&quot; to mean physicist who are advanced in administration rather than advanced in age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that no, senior physicists do not contribute directly to research. At least, they do not contribute more than everyone else who works at a physics department. And do note that most of the people earn their money because of &quot;physics&quot; do not have an advanced physics degree and certainly are not &quot;physicists&quot;. Instead, for each physicist there are large numbers of undergraduate students, graduate students, janitors, secretaries, various specialties such as plumbers, builders. Large experiments employ huge numbers of technicians, computer programmers, various blue-collar workers, accountants, lawyers, the list goes on and on. Finally, any physics group purchases equipment (example: personal computers, oscilloscopes, raw material) from companies that have little to do with physics. Since physics is an endeavor that is supported by taxes you might want to consider all taxpayers as contributing to physics. In this sense, sure, senior physicists do contribute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The success of the human race is due to its unique organizational skill. Though most of the individuals do not understand this, the species is a classic herd animal. Almost all of the human race's domesticated animals are also herd animals, i.e. dogs, horses, chickens, pigs, ducks, cattle, donkeys, camel, llama, various fish, etc. The only common exception is the cat, hence the common expression, &quot;like trying to herd cats&quot;.1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a herd animal, humans do their activities in packs. They generally form hierarchical structures typically led by an older &quot;alpha&quot; male. This male spends his time keeping the other herd members in line and defending himself from aspirants to his position (which typically comes with improved access to females or various other perks). Their most impressive herd behavior are the large scale conflicts they organize. They use their most advanced available technology to hunt each other in packs. Humans make velociraptors look cuddly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The success of the human species in physics is due to extreme specialization. Any given physicist knows almost nothing from direct observation. Instead, almost all knowledge is obtained from other physicists. This is the ultimate in specialization, there is no other known animal species anything like it. This specialization is extremely useful. For example, the number of physicists who actually measure the mass of the electron is only a tiny fraction of the total but all of them use the same number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stuff that goes on in a physics work groups is is all standard primate behavior that can be observed with more clarity in any jungle (the clarity arises because you will be examining a different species). Any sociologist can tell you these things. And there is at least one sociologist who specializes in physics, Harry Collins. To learn more about the sociology of physics, try picking up some of his fascinating books:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Harry_Collins&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Harry_Collins&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, the sociology of science is not a subject that is appreciated by very many physicists. They would instead prefer to believe that physics is an activity that is governed by a universal inclination to seek the truth. The only reason I ended up reading on the sociology of physics is because I was taught general relativity by Joseph Weber, and Weber was the primary subject of a book by Collins, &quot;Gravity's Shadow: the search for gravitational waves&quot;. Naturally, when I saw the book on sale, I bought it, read it, and was fascinated by the description of physicists in it.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-25T01:05:24.213" CommentCount="5" />
  <row Id="5835" PostTypeId="2" ParentId="5832" CreationDate="2011-02-25T01:09:21.007" Score="0" ViewCount="" Body="&lt;p&gt;Well, no. As the universe expands (note this is an intrinsic expansion. Don't think of it as expanding into anything.), the matter and radiation do not increase, and are present in a much larger space. The vacuum energy (cosmological constant) does not change as the space expands--it is the same per unit volume, and its contribution is to exert a negative pressure, a sort of anti-gravity. Note, however, that the pressure contributes to the right side of Einstein's equation--the Stress-Energy Tensor, so it is important dynamically in the evolution of the gravitational potentials( generating the metric) on the left side of the equation. Most models assume the vacuum energy density is constant, but some consider models where it is slowly decaying or slowly increasing. A huge problem is that the predicted value of the cosmological constant (by QFT) is $10^{120}$ times larger than the observed value, a problem that has not been satisfactorily explained.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditorDisplayName="" LastEditDate="2011-02-25T01:45:17.517" LastActivityDate="2011-02-25T01:45:17.517" />
  <row Id="5836" PostTypeId="2" ParentId="3799" CreationDate="2011-02-25T01:11:28.990" Score="1" ViewCount="" Body="&lt;p&gt;They want to be rich and try to catch funding. ;)&#xA;Two ways: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;They have 'a secret device' that works -&gt; funding, patents, RICH,RICH,RICH ... (I do not believe they have this secret)&lt;/li&gt;&#xA;&lt;li&gt;they have 'a secret lie' - funding,  RICH (it was a risk operation, etc,.. contratual details,...ruined reputation... but RICH)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I will not make a direct judgement on bogus or no bogus. I will wait and see. May be some day, someone, do unexpected experiences like the recent 'anti-laser' that 'destroys' energy.&#xA;To justify my 'operational' position I can tell one story of academic arrogance: The national TV broadcast started here in 1955, in 1957 the russian Sputnic was sent to orbit and, conveniently, an academic appeared in TV and said 'It's impossible, russians are liers'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;em&gt;added : *&lt;/em&gt;&#xA;just found: &lt;a href=&quot;http://www.technologyreview.com/blog/arxiv/26430/&quot; rel=&quot;nofollow&quot;&gt;How to Transmute Elements with Laser Light&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-02-25T03:38:45.950" LastActivityDate="2011-02-25T03:38:45.950" />
  <row Id="5837" PostTypeId="2" ParentId="5812" CreationDate="2011-02-25T02:25:29.303" Score="2" ViewCount="" Body="&lt;p&gt;Stability in this context is usually defined as &quot;there is a vacuum state and there is no state with energy below that of the vacuum state&quot;. For this to be true, you need the energy to be bounded from below. All in all the spectrum condition can be expressed in words as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The spectrum condition is a relativistically invariant way of requiring that the total energy in the theory be nonnegative with respect to every inertial frame of reference and that the quantum system is stable in the sense that it cannot decay to energies below that of the vacuum state.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll find a nice exposition of both this statement and a lot of other interesting results about the vacuum state in AQFT in this paper:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Stephen Summers: &quot;Yet More Ado About Nothing: The Remarkable Relativistic Vacuum State&quot; (&lt;a href=&quot;http://de.arxiv.org/abs/0802.1854v2&quot; rel=&quot;nofollow&quot;&gt;arxiv&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It's maybe worth noting that the spectrum condition can only be formulated in Minkowski spacetime, because it needs the Poincare group for its very formulation. About 17 years ago an alternative formulation was found which is entirely local and can therefore be used to define the &quot;positivity of energy&quot; on any suitable Lorentzian manifold. You can find all necessary references in this paper:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Alexander Strohmaier, Rainer Verch, Manfred Wollenberg: &quot;Microlocal analysis of quantum fields on curved spacetimes: Analytic wavefront sets and Reeh-Schlieder theorems&quot; (&lt;a href=&quot;http://xxx.uni-augsburg.de/abs/math-ph/0202003&quot; rel=&quot;nofollow&quot;&gt;arxiv&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-02-25T02:25:29.303" CommentCount="1" />
  <row Id="5838" PostTypeId="1" CreationDate="2011-02-25T02:29:42.343" Score="2" ViewCount="232" Body="&lt;p&gt;It is well known in racing that driving the car on the ideal &quot;slip angle&quot; of the tire where it is crabbing slightly from the pointed direction produces more cornering speed than a lower slip angle or a higher one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(More explanation as requested) I'm considering two main effects on the tire when in a turn: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The tread of the tire is twisted from the angle of the wheel it is mounted to. There is more force as speed increases, and generally, more twisting. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The tire slides somewhat at an angle on the road surface rather than rolling. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;At low speeds, the angle between the pointed direction of the wheel (90 degrees to the axis of rotation) and the direction of travel is nearly 0.  When the speed increases to the point the angle reaches about 10 degrees, the tire generate more grip and the car goes faster around the turn.  (Higher angles produce lower grip)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the grip is higher at 10 degrees of slip than at 0 or 20 degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the physical effect that causes this increase in grip?&lt;/p&gt;&#xA;" OwnerUserId="2215" LastEditorUserId="2215" LastEditorDisplayName="" LastEditDate="2011-02-25T22:49:55.237" LastActivityDate="2011-02-25T22:49:55.237" Title="Why does a tire produce more traction when sliding slightly?" Tags="&lt;friction&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="5839" PostTypeId="1" AcceptedAnswerId="5852" CreationDate="2011-02-25T02:37:55.493" Score="4" ViewCount="179" Body="&lt;p&gt;Ok, this may seem obvious(I'm not all that experienced in physics)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway. If you had some magnet of incredible strength and attached an iron wire that is one light year long to it, would the other end of the iron wire begin to show a magnetic field in a year, or could it possibly be faster than a year?&lt;/p&gt;&#xA;" OwnerUserId="2217" LastActivityDate="2011-02-25T05:37:00.963" Title="Does magnetic propogation follow the speed of light?" Tags="&lt;magnets&gt;&lt;magnetism&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="5840" PostTypeId="2" ParentId="5775" CreationDate="2011-02-25T02:39:04.340" Score="2" ViewCount="" Body="&lt;p&gt;Physics is top sport. When sports(wo)men age, they generaly discover that it becomes more and more difficult to compete. Some keep training and stay in the game till late age, but sooner or later they discover that the golden years are long past. No problem: these seniors carry a wealth of knowledge, and a position as coach, sports events organizer, etc. provides a natural second career.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does a coach contribute to the results of his/her team? Sure (s)he does!&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-02-25T02:39:04.340" CommentCount="1" />
  <row Id="5841" PostTypeId="2" ParentId="5839" CreationDate="2011-02-25T02:45:09.870" Score="5" ViewCount="" Body="&lt;p&gt;It cannot be faster than a year, because electro-magnetic waves propagate with the speed of light. The magnetization of the wire will take longer, because it depends on the reaction of the iron atoms to the external magnetic field (they align according to the external field).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rule of thumb, one can say that &quot;if it is possible to notify a friend far away with this&quot; then we are talking about the transmission of information, and that can be done with light speed at most according to special relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case you could set of an alarm clock on the other end of the wire once the magnetic field reaches the end :-)&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-02-25T02:45:09.870" />
  <row Id="5842" PostTypeId="2" ParentId="5832" CreationDate="2011-02-25T02:52:11.080" Score="0" ViewCount="" Body="&lt;p&gt;Here is an answer on a more metaphorical level: Think of the &quot;expansion of space&quot; as &quot;new space is created out of nowhere&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know the movie &quot;Poltergeist&quot;? There is a scene where the mother runs down the corridor to reach the door to her daughter's room, who is in trouble, but the corridor expands at almost the same speed she is running at, moving the door away from her. (Don't worry, she accelerates, reaches the door and saves her daughter. I think.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in the case of general relativity, there is no outer space, that is the house is all there is, there is no garden, and if there where, the creation of new space in the corridor does not lead to an expansion of the house that would be observable from the garden. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you replace the door with a star that attracts you (through gravity, not through its beauty), you'll see the star &quot;moving away from you&quot; (only that space is created in between, its not the star that is moving), therefore you'll feel a decrease in the attraction.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-02-25T02:52:11.080" CommentCount="1" />
  <row Id="5843" PostTypeId="2" ParentId="5762" CreationDate="2011-02-25T02:56:06.020" Score="-2" ViewCount="" Body="&lt;p&gt;It is mandatory that MOND is good at predictions. Specially the before mentioned paper (measuring galaxy rotation) will be a success ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Phenomenology_%28science%29&quot; rel=&quot;nofollow&quot;&gt;phenomenological theory&lt;/a&gt; i.e. MOND does not have a model and is accurate because it does data fit, on purpose. As long as galaxies do have a common underlying mechanism MOND will be good.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not really physics, it is statistics, accounting, cheating, and it should be forbidden because it can prevent us from progressing.&#xA;I can say to myself: I do not know the answer, and I will keep looking. Physics, as an institution, can not admit: We dont know. And then Dark matter, MOND, ..., are expressions of our ignorance.&lt;br&gt;&#xA;At least the Ptolomaic epicycles had a model, the Geocentric one.&lt;br&gt;&#xA;Things will change as soon as &lt;em&gt;you&lt;/em&gt; and the  comunity can find the correct answer. &#xA;Try googling &quot;Galáxias e Atractores&quot; and maybe you can find a nice reading. &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-02-25T02:56:06.020" CommentCount="1" />
  <row Id="5844" PostTypeId="1" CreationDate="2011-02-25T03:03:26.247" Score="1" ViewCount="99" Body="&lt;p&gt;I'm a bit rusty on my fluid dynamics.  I'm experimenting with creating my own leaf vacuum.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I determine how much suction a fan creates?&lt;br&gt;&#xA;Bernoulli's equation would apply correct?&lt;br&gt;&#xA;I would also need a fan curve right?&lt;br&gt;&#xA;Does CFM matter much when creating a vacuum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately what I'm looking for is the Force @ inlet (P x Ainlet) &gt; Force on leaves (m x g)...?&lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2011-02-25T03:03:26.247" Title="Creating the proper suction" Tags="&lt;fluid-dynamics&gt;&lt;vacuum&gt;" CommentCount="2" />
  <row Id="5845" PostTypeId="2" ParentId="5819" CreationDate="2011-02-25T03:06:01.470" Score="2" ViewCount="" Body="&lt;p&gt;Intuition is great, but it can also be a trap. I had extreme inutition about physics during highschool and undergraduate days. Probably at the one in a million level. It made much of physics and math almost effortlessly easy. But I basically trainwrecked in grad school. And the reason is that things reached a level of abstraction that was beyond what I could intuit. And all my classmates -none of whom half the intuition I had, had been doing it the hard way, learning symbolically, and mathematically for many years -probably out of necessity. And I suddenly realized I would need to back up several years and relearn things that way. And that was far too much backtrcking to contemplate at that stage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I suspect, that having one area of cognition that you are supergood in at is actually a setup for failure. Train all of the needed abilities with respect to physics. Not just the ones that you are most good at. Otherwise you might find yourself in a col-de-sac.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-02-25T03:06:01.470" CommentCount="2" />
  <row Id="5846" PostTypeId="1" AcceptedAnswerId="5863" CreationDate="2011-02-25T03:37:27.807" Score="2" ViewCount="192" Body="&lt;p&gt;Adding coffee first increases the volume to heat and lowers average temerature, but adding it afterwards seems to have similar effects.  How can you compare the two?&lt;/p&gt;&#xA;" OwnerUserId="2220" LastActivityDate="2011-02-25T10:14:29.713" Title="What is more efficient: Add milk, and then heat up coffee in the microwave, or microwave than milk?" Tags="&lt;thermodynamics&gt;&lt;heat&gt;" AnswerCount="2" />
  <row Id="5847" PostTypeId="1" CreationDate="2011-02-25T04:19:43.717" Score="2" ViewCount="158" Body="&lt;p&gt;I have a physics lab tomorrow and I would appreciate it if you could give me your opinion on whether my reasoning is correct for this situation or not:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i.stack.imgur.com/WMzA7.jpg&quot; rel=&quot;nofollow&quot;&gt;http://i.stack.imgur.com/WMzA7.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The title of the lab is hydrostatic and is roughly described in the attached pic.&#xA;We will take data using different weights to measure the deformation of the spring and the increase in the level of the water (yes, the fluid is water) and the thing that we have to find is the spring constant &quot;k&quot; but (and my professor was very clear about it) for the whole system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I am having some doubts, but I think that he means that we have to find some function of F that uses d (deformation of the spring) and h (increase in water height) and then we'll use least squares (done this in previous labs) to find the constant k, what I'm currently thinking is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;F = restitution force of the spring&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m = mass of the submerged body&lt;br&gt;&#xA;d = deformation of the spring &#xA;h =       increase in water level&lt;br&gt;&#xA;W= weight&#xA;    of    the submerged body&lt;br&gt;&#xA;B =&#xA;    buoyant force&lt;br&gt;&#xA;V= volume of the&#xA;    displaced liquid /      small body&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F=−k*d&lt;br&gt;&#xA;B=d(water density)*V*g &lt;/p&gt;&#xA;&#xA;&lt;p&gt;W=m*g     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking upwards as positive&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F+B−W=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but now I have to find either a) the volume of the displaced fluid or the volume of the body, both of these are not data (or can't be found without the dimensions of the container or the small body as far as I know) at all, the Buoyant force has to be smaller than the weight since it's at equilibrium only after the spring force begins acting&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my current train of thought correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) since the difference in water level changes with different masses I think it's safe to assume that the body to be submerged is not completely inside from the beginning, since if it were that way after adding more masses the total displaced volume would not increase&lt;/p&gt;&#xA;&#xA;&lt;p&gt;II) Still having troubles with the F = W - B part, since I really can't figure how to calculate the Buoyant force without the volume of the object or the water displaced and I can't find the one of the water displaced since I do not know the dimensions of the container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;III) the most smart course of action seems to be to find F using only B and W and then use that F with d (spring deformation) to find&#xA;''k&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IV) I can't grasp what would happen if the mass were to get so big that the body would be completely submerged and &quot;h&quot; would stop growing but it's probably not going to happen in the laboratory&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a feeling I'm missing something really obvious and even a small push in the right direction would be nice&lt;/p&gt;&#xA;" OwnerUserId="2223" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-02-25T06:54:14.783" LastActivityDate="2011-02-25T06:54:14.783" Title="Measuring the spring constant in a system with hydrostatic buoyancy" Tags="&lt;homework&gt;&lt;buoyancy&gt;&lt;spring&gt;" CommentCount="1" />
  <row Id="5848" PostTypeId="2" ParentId="5838" CreationDate="2011-02-25T04:28:33.833" Score="0" ViewCount="" Body="&lt;p&gt;Not familiar with racing but my initial thought is that the vectors are in a more ideal direction.  Maybe the kinetic friction plays a small role (as opposed to static friction)?&lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2011-02-25T04:28:33.833" />
  <row Id="5849" PostTypeId="2" ParentId="5669" CreationDate="2011-02-25T04:28:36.050" Score="8" ViewCount="" Body="&lt;p&gt;As an engineer, I've had to get pricing on stuff I wasn't going to buy many times in the past. I doubt you'll find a website which will have the prices, but you can find them yourself in a few hours work. What you do is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call them up and ask for a &quot;sales rep&quot; (abbreviation for &quot;sales representative&quot; but nobody calls them by the full name) tell him/her that you're working at a physics lab (I presume you are) and that you need &quot;budgetary pricing&quot;. This is pricing that includes no discounts. Budgetary pricing is not an offer to sell. So if the price suddenly goes through the roof because of a revolution in some obscure country budgetary pricing won't get them into trouble. They should give this to you without a lot of trouble as it's not secret information. I've always assumed that they don't put it on their websites because (a) it changes, and (b) it might be construed as being an offer to make a contract.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you can't get information from their website, ask them to give you a &quot;data sheet&quot;. That should be all you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hey, I'll get you started. Here's a company that sells wafers for the valley:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.svmi.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.svmi.com/&lt;/a&gt;&lt;br&gt;&#xA;Here's their extensive non silicon product line:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.svmi.com/non-siliconwafers.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.svmi.com/non-siliconwafers.aspx&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's their contact info:&lt;br&gt;&#xA;Tel:  408.844.7100&lt;br&gt;&#xA;Fax: 408.844.9470&lt;br&gt;&#xA;Email: sales@svmi.com  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now when you talk to them, do not sound like a complete idiot. Make sure you know everything about their product BEFORE you pick up the phone. Make yourself a temporary expert on their product line. Know what size you're interested in and what level of quality, etc. Decide in advance what year you're asking for (new stuff gets cheaper with time), and the approximate quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible, use their website to determine the actual product number for the item you're interested in. By giving them a part number you are making their job very easy. Sales people are not, by nature, inclined to annoy potential customers. Make it easier for them to give you a price than it would be for them to tell you to go away. If you give them a part number it will be very easy to look up a number for you. Do not waste their time by asking for them to speculate about exactly how these prices will change in the future. (But if they're bored they might have time to talk to you about this.) They should be able to tell you that they &quot;expect that these prices will drop as production ramps up&quot;, or give you a vague idea, and maybe you'll get lucky. But remember that you're asking industry to do you a favor for no real good reason.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-25T04:28:36.050" CommunityOwnedDate="2011-02-25T18:59:22.507" CommentCount="2" />
  <row Id="5850" PostTypeId="1" AcceptedAnswerId="5868" CreationDate="2011-02-25T05:03:01.290" Score="2" ViewCount="158" Body="&lt;p&gt;How do I go about finding the pressure exerted on a rectangular surface in a free flowing air stream?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't imagine that this is directly related to the airspeed / surface area, but have no idea where to start. Is there even an equation, or does one need to do some kind of FEA?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance a 1.2m x 2.4m metal sheet suspended some distance above ground level, if I have a gust of wind at 8m/s (directly perpendicular to the sheet), what is the average pressure across the face of sheet?&lt;/p&gt;&#xA;" OwnerUserId="2224" LastActivityDate="2011-02-25T11:34:09.487" Title="Pressure vs wind speed, on a rectangular surface" Tags="&lt;fluid-dynamics&gt;" AnswerCount="2" />
  <row Id="5851" PostTypeId="1" AcceptedAnswerId="5940" CreationDate="2011-02-25T05:18:24.297" Score="9" ViewCount="249" Body="&lt;p&gt;According to the first law of thermodynamics, sourced from wikipedia&#xA;&quot;In any process in an isolated system, the total energy remains the same.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when lasers are used for cooling in traps, similar to the description here: &lt;a href=&quot;http://optics.colorado.edu/~kelvin/classes/opticslab/LaserCooling3.doc.pdf&quot; rel=&quot;nofollow&quot;&gt;http://optics.colorado.edu/~kelvin/classes/opticslab/LaserCooling3.doc.pdf&lt;/a&gt; where is the heat transferred?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I gather of a cursory reading on traps, whether laser, magnetic, etc the general idea is to isolate the target, then transfer heat from it, thereby cooling it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand how sending photons at a(n) atom(s) can cause that structure to shed energy, and this mechanism seems to be the key to these systems. &lt;/p&gt;&#xA;" OwnerUserId="2226" LastActivityDate="2011-02-26T13:13:15.413" Title="When lasers are used to cool atoms or ions, etc where does the heat go?" Tags="&lt;photons&gt;&lt;laser&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" />
  <row Id="5852" PostTypeId="2" ParentId="5839" CreationDate="2011-02-25T05:37:00.963" Score="5" ViewCount="" Body="&lt;p&gt;In your example, the relevant speed isn't the speed of propagation of disturbances in the magnetic field, but rather the speed of the alignment of iron atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are really asking &quot;Does magnetization of a wire/metal propagate at the speed of light?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is &lt;strong&gt;no&lt;/strong&gt;; it propagates at the speed at which each individual iron atom can align its polarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are asking, &quot;Do changes in the magnetic field propagate at the speed of light?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is &lt;strong&gt;yes&lt;/strong&gt;; if a giant, huge, powerful magnet appeared one light year away out of nowhere, then it would take exactly one year for magnets on Earth to feel its pull (however small it may be).  That is, it would take one year for the &quot;magnetic force&quot; to reach the Earth.&lt;/p&gt;&#xA;" OwnerUserId="71" LastActivityDate="2011-02-25T05:37:00.963" CommentCount="1" />
  <row Id="5853" PostTypeId="2" ParentId="5850" CreationDate="2011-02-25T05:54:30.440" Score="1" ViewCount="" Body="&lt;p&gt;Seeing as how I just asked a fluid dynamic question earlier today, I'm probably not the best person to answer this.  I think to get you in the right direction you may want to look at Bernoulli's equation for compressible flow.  Initial pressure is probably atmosphere.  See if that helps.&lt;/p&gt;&#xA;" OwnerUserId="2219" LastActivityDate="2011-02-25T05:54:30.440" />
  <row Id="5854" PostTypeId="2" ParentId="5829" CreationDate="2011-02-25T05:58:10.500" Score="2" ViewCount="" Body="&lt;p&gt;The fermionic statistics has to do with a minus sign acquired when two objects are interchanged. Usually this is unambiguously attributed to statistics, as opposed to interaction between the objects, since any interaction falls off rapidly when they are far apart, but the relative phase acquired due to interchange is the same whether they are close or far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in  low dimensional systems the interaction strength between well-separated objects does not always fall off very fast (or not at all). So, one can get fermionic (or more generally anyonic) behaviour from interaction. For some solitons the interactions is such that they acquire a phase when interchanged, no matter how far apart they are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This arguments only shows it can happen in low dimensional systems, more detailed arguments will show you &lt;em&gt;how&lt;/em&gt; it happens (for example google sine-Gordon and Thirring models, or bosonization). The &lt;em&gt;why&lt;/em&gt; questions I am never sure how to answer.&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-02-25T05:58:10.500" CommentCount="2" />
  <row Id="5855" PostTypeId="2" ParentId="5851" CreationDate="2011-02-25T06:08:55.417" Score="2" ViewCount="" Body="&lt;p&gt;The photon's energy goes into putting the atom into an excited state.  Its momentum goes into slowing down the atom. Both are conserved.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-02-25T06:08:55.417" />
  <row Id="5856" PostTypeId="2" ParentId="5830" CreationDate="2011-02-25T06:16:24.797" Score="1" ViewCount="" Body="&lt;p&gt;Sorry, just a reference. Both Neumann and Dirichlet boundary conditions occur. See &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.springerlink.com/content/h757664275612765/&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/h757664275612765/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Hexagon quantum billiard.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-25T06:16:24.797" />
  <row Id="5857" PostTypeId="2" ParentId="5851" CreationDate="2011-02-25T06:33:24.587" Score="7" ViewCount="" Body="&lt;p&gt;The most widespread type of laser cooling is &quot;Doppler cooling&quot;. The laser light, coming from the sides, is adjusted to a frequency that is slightly below the natural frequency of the atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the atom will only absorb the light if it is moving against the laser beam. Consequently, the velocity will drop. When it absorbs the photon, the energy goes to its excitation. Later, the atom will emit a photon again but in a random direction. So the average magnitude of the atom's velocity after these two steps drops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both steps in the process - emission and absorption - are obviously energy-conserving, much like all other phenomena in the lab. The energy conservation is essential to calculate what happens.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-25T06:33:24.587" CommentCount="1" />
  <row Id="5858" PostTypeId="2" ParentId="5829" CreationDate="2011-02-25T06:40:58.030" Score="2" ViewCount="" Body="&lt;p&gt;There are many examples. Moshe focused on lower-dimensional theories; for example, in 1+1 dimensions, bosonic and fermionic conformal field theories are typically totally equivalent to each other (and one may get fermions as kinks of the bosons, and bosons as bilinear currents of the fermions). I will focus on higher-dimensional theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really know how to get fermions in the $D=26$ bosonic string theory. But in other, $D=10$ string theories that normally contain bosons only, fermions may be obtained - see e.g. this paper by Justin David, Shiraz Minwalla, and Carlos Nunez (whose writing I remember pretty well, back from Santa Barbara 2001 etc.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0107165&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0107165&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The basic trick, known already in 3+1-dimensional bosonic gauge theory (by Jackiw-Rebbi-Hasenfratz-’t Hooft), is &quot;spin from isospin&quot;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://scholar.google.com/scholar?cluster=155257508923742388&amp;amp;hl=en&amp;amp;as_sdt=0,5&quot; rel=&quot;nofollow&quot;&gt;http://scholar.google.com/scholar?cluster=155257508923742388&amp;amp;hl=en&amp;amp;as_sdt=0,5&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The soliton is only invariant under the &quot;diagonal group&quot; of the rotations and some internal isospin group. Noether's theorem dictates that the right definition of the spin of the excitations is associated with whatever symmetry we have. In this case we have the diagonal symmetry, so the new conserved angular momentum will be the diagonal one. Because half-integer-isospin excitations are possible, they will look like half-integer-spin excitations of the soliton.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-25T06:40:58.030" />
  <row Id="5859" PostTypeId="1" CreationDate="2011-02-25T07:53:38.367" Score="6" ViewCount="164" Body="&lt;p&gt;I have a question with regard to probabilistic quantum cloning - see for example &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v80/i22/p4999_1&quot; rel=&quot;nofollow&quot;&gt;http://prl.aps.org/abstract/PRL/v80/i22/p4999_1&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does seems like I can use the proof for no-cloning theorem to disproof the existence of equation (7) in the paper quant-ph/9704020. Do you think this is true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose probabilistic quantum cloning is actually possible, now according to the current papers, if the cloning process fails, the state that we intend to clone would be destroyed. Looking at the equation in the original paper, however, it does seem that a repair or a reverse operation might be possible in the event of a failure. More specifically, in the paper quant-ph/9704020, if I replace the $\left|\phi\right&amp;gt;$ state in Equation (7) with say, $\left|\psi_T\right&amp;gt;\left|0\right&amp;gt;$ where $\left|\psi_T\right&amp;gt;$ is a state orthogonal to the input state $\left|\psi\right&amp;gt;$, the proof would work just as fine. Do you think this is true? If this is true, a repair would be possible in case of cloning failure. The process would still be probabilistic, but I can repeat till it is successful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do hope for some feedback from experts out there. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2233" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-25T15:20:55.543" LastActivityDate="2011-02-26T13:25:06.733" Title="Probabilistic quantum cloning" Tags="&lt;quantum-mechanics&gt;&lt;quantum-information&gt;&lt;quantum-computation&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="5860" PostTypeId="1" CreationDate="2011-02-25T08:05:26.930" Score="1" ViewCount="212" Body="&lt;p&gt;A thin spherical shell made of plastic carries a uniformly distributed negative charge of -Q coulombs. Two large thin disks made of glass carry uniformly distributed positive and negative charges S coulombs and -S coulombs. The radius of the plastic spherical shell is R1, and the radius of the glass disks is R2. The distance from the center of the spherical shell to the positive disk is d.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know how to find two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The potential difference V1-V2. Point 1 is at the center of the plastic sphere, and point 2 is just outside the sphere.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The potential difference V2-V3. Point 2 is just below the sphere, and point 3 is right beside the positive glass disk.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2157" LastActivityDate="2011-02-25T08:05:26.930" Title="How can I find the potential difference of a plastic sphere in a capacitor?" Tags="&lt;electromagnetism&gt;" CommentCount="1" />
  <row Id="5861" PostTypeId="1" CreationDate="2011-02-25T09:45:11.940" Score="2" ViewCount="97" Body="&lt;p&gt;While reading a paper I ran into this particular way of writing a $\cal{N}=3$ fields (in a theory with $N_f$ hypermultiplets) that I couldn't relate to anything I had seen before in the text-books (typically Weinberg's) The claim is that $q^{Aa}$ and $\psi_{\alpha}^{Aa}$ are scalar and fermions such that $a,b$ are $SU(R)_R$ indices, $A,B$ are denoting them to transform in the fundamental representation of $USp(2N_f)$ (why?) and $\alpha, \beta$ are $SO(2,1)$ spinor indices (theory is in $2+1$) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently these fields satisfy some reality conditions given by, the following equations,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(q^\dagger)_{Aa} = \omega_{AB}\epsilon_{ab}q^{Bb}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bar{\psi} ^\alpha _{Aa} = \omega_{AB} \epsilon_{ab} \epsilon ^{\alpha \beta} \psi _{\beta} ^{Bb}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $\omega$ is some symplectic form, which I have no clue about) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find the above equations very unfamiliar. For one thing why do the scalar and the fermion have $R$ charge indices? (I am only familiar with the supercharges having a R-symmetry when the central charges are absent) Also I don't understand the $USp(2N_f)$ index. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently this can be related to usual $\cal{N}=2$ fields $(Q,\tilde{Q})$ as,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q^{A1} = (Q,\tilde{Q})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q^{A2} = (-\bar{\tilde{Q}}, \bar{Q})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am completely unfamiliar with such a relationship or this thinking of $\cal{N}=2$ fields as being thought of as a pair of $Qs$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once I understand the meaning of the above I will post another about its curious way of writing the supersymmetry transformation and the lagrangian with this field content. &lt;/p&gt;&#xA;" OwnerUserId="1637" LastActivityDate="2011-02-25T09:45:11.940" Title="A particular supersymmetry transformation" Tags="&lt;quantum-field-theory&gt;&lt;supersymmetry&gt;" FavoriteCount="1" />
  <row Id="5862" PostTypeId="2" ParentId="5838" CreationDate="2011-02-25T09:58:19.213" Score="0" ViewCount="" Body="&lt;p&gt;The effect sounds the same as braking with ABS. Maximal deceleration is achieved when a wheel rolls slightly slower than the speed of the car. The lateral effect in corners works the same; both are friction effects between rubber and road surface. Both are isotropic materials, i.e. the direction of the friction force doesn't matter.&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-02-25T09:58:19.213" CommentCount="1" />
  <row Id="5863" PostTypeId="2" ParentId="5846" CreationDate="2011-02-25T10:08:01.197" Score="0" ViewCount="" Body="&lt;p&gt;By adding milk first you reduce heat loss during heating due to lower average temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you add milk first, you will get higher temperature at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But personally, I prefer to add milk after, as it's more sensetive to overheating. You can heat cofee to 100C without any damage to it.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-02-25T10:08:01.197" />
  <row Id="5864" PostTypeId="2" ParentId="5846" CreationDate="2011-02-25T10:14:29.713" Score="2" ViewCount="" Body="&lt;p&gt;;-) recipe:&lt;br&gt;&#xA;    0: get a watch, a thermometer, two equal doses of milk and two of coffee and a microwave. As you have a computer to make the annotations, a pen and a stencil are not needed.&lt;br&gt;&#xA;    1 : write down a proper definition of 'efficiency', choose one at your will.&lt;br&gt;&#xA;    2 : do the proposed first experiment, time the results and the temps. Taste ;-)&lt;br&gt;&#xA;    3 : do the proposed second experiment reversing the order, time the results and the temps. Taste ;-)&lt;br&gt;&#xA;    4 : compare the resultant temps and tastes (is a bit subjective but it is important) with the settled goal (efficiency deffinition).&lt;br&gt;&#xA;    5 : kindly let us know your conclusion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may think that I am making fun, but it is not so. The experimental way, when available, is the best solution. You have a courious mind, dont stop. A personnal learning process is good.&lt;br&gt;&#xA; To be more rigourous one can invent a physical model (between point 1 and 2) to advance a candidate outcome. But the point 1 is allways indispensable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;end of recipe ;-)&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-02-25T10:14:29.713" />
  <row Id="5865" PostTypeId="1" AcceptedAnswerId="5867" CreationDate="2011-02-25T10:41:41.890" Score="5" ViewCount="142" Body="&lt;p&gt;I have often seen diagrams, like &lt;a href=&quot;http://en.wikipedia.org/wiki/Lens_%28optics%29#Imaging_properties&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt; on Wikipedia for a thin convex lens that show three lines from a point on the object converging at the image. Do all the other lines from that point on the object that pass through the lens converge at the same point on the image?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;em&gt;Updated question: *&lt;/em&gt; to say, &quot;from that point on the object&quot;&lt;/p&gt;&#xA;" OwnerUserId="119" LastEditorUserId="119" LastEditorDisplayName="" LastEditDate="2011-02-25T11:36:45.707" LastActivityDate="2011-02-25T14:55:38.437" Title="Do all light rays from a point passing through a thin lens converge at the image?" Tags="&lt;optics&gt;&lt;geometric-optics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="5867" PostTypeId="2" ParentId="5865" CreationDate="2011-02-25T11:12:35.660" Score="1" ViewCount="" Body="&lt;p&gt;Yes, at least for functional enough lens or mirrors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as the answer is &quot;Yes&quot;, the lenses or mirrors will produce an image that is totally sharp. If the answer were &quot;No&quot;, a point-like source of light would always look like a fuzzy disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lenses and mirrors and telescopes may be optimized at least for a whole &quot;two-dimensional locus&quot; where the light source may be located to produce sharp images. The shape of the lenses may be constructed so that this condition is satisfied exactly: a real condition (the light ray gets to the right point) has to be satisfied for each distance $y$ from the axis - but one has at least one variable, $x(y)$ (the thickness of the lens), to adjust for each $y$, too. In the first subleading approximation, the shape is always the same: as a function of the vertical coordinate $y$, the thickness of the glass in the horizontal direction goes like $A+By^2$. That approximates a circle, parabola, hyperbola, or anything else, up to errors of order $O(y^4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to totally neglect those terms, it's like neglecting $O(\theta^4)$ terms, quartic in the angle. In this approximation, the angle between the light ray and the horizontal line that is needed for convergence is small and linearly depends on $y$. The linear dependence of the angles on $y$ - how much the direction of the light ray is changed when switching from the air to the glass or back - is equivalent to the quadratic shape of the mirror sketched above (the angle change is a derivative of the shape because this derivative determines the slope of the glass at a given point). This explains why it's not a &quot;miraculous conspiracy&quot;: the change of the angle is a linear function of $y$ so the shape of the glass must be a quadratic function of $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, where the $O(\theta^4)$ terms in the shape can't be neglected, the light sources may be away from the plane where the convergence was guaranteed, and in that case, the answer will be &quot;No&quot; and the image will inevitably be fuzzy. However, it's important to note that e.g. telescopes that observe stars &quot;at infinity&quot; may always be adjusted so that all the images are sharp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the shape of the lens actually has two functions, $x_{left}(y)$ and $x_{right}(y)$ to be adjusted, one may actually guarantee that the condition &quot;light rays converge&quot; is exactly satisfied in a whole region of 3D space, at least in some situations.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-25T11:27:28.990" LastActivityDate="2011-02-25T11:27:28.990" />
  <row Id="5868" PostTypeId="2" ParentId="5850" CreationDate="2011-02-25T11:34:09.487" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.reasonablepower.com/nice_to_know/an_introduction_to_wind_loads.htm&quot; rel=&quot;nofollow&quot;&gt;Wind Load Formula:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_d = \frac{1}{2} \rho v^2 A C_d$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;br&gt;&#xA;$F_d$ is the force of drag (or in this case Force Against the flat plate)&lt;br&gt;&#xA;$\rho$ is the density of the air&lt;br&gt;&#xA;$v$ is the speed of the air against the object&lt;br&gt;&#xA;$A$ is the area of the object which the air is blowing against&lt;br&gt;&#xA;$C_d$ is the drag coefficient&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2011-02-25T11:34:09.487" />
  <row Id="5870" PostTypeId="2" ParentId="5824" CreationDate="2011-02-25T12:19:34.520" Score="1" ViewCount="" Body="&lt;p&gt;We used to do the trick with the glass of water and a piece of paper when I was a child. You fill the glass, to the rim it is more successful, cover with a dry piece of paper or a bit of flat aluminum foil, put your hand flat over it, and upend it slowly. Slowly remove the hand. The water does not empty. The paper stays. The weight of the water is balanced by the ambient pressure and the suction that occurs at the top, as  gravity pulls the water down and a vacuum is formed at the top of the glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first glance I thought that conservation of the angular momentum given by the twist, same with a spinning top, would keep the water there ( it looks as if it is spinning in the video). But the suction from the removal of the glass (goes plop, I am an experimentalist after all) could not produce such an unperturbed shape, imho. So I vote that it is a trick, although the way the water collapses without any debris of bags is realistic, so there must be video manipulation too.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-02-25T12:19:34.520" />
  <row Id="5871" PostTypeId="2" ParentId="5838" CreationDate="2011-02-25T12:23:57.410" Score="3" ViewCount="" Body="&lt;p&gt;At small slip angles, the tire deforms linearly with the side-force demanded of it. This effect saturates when the rubber is essentially fully deformed. The tire then starts to slide over the surface of the road, which reduces the side-force somewhat. Now, this force acts perpendicular to the wheel plane, which is not quite perpendicular to the car's trajectory. This effect reduces the force available to turn the car and causes an &quot;induced drag&quot; that slows it down. The combination of these effects gives rise to the concept of an optimum slip angle.&lt;/p&gt;&#xA;" OwnerUserId="2238" LastEditorUserId="2238" LastEditorDisplayName="" LastEditDate="2011-02-25T12:50:53.773" LastActivityDate="2011-02-25T12:50:53.773" CommentCount="2" />
  <row Id="5872" PostTypeId="1" AcceptedAnswerId="5877" CreationDate="2011-02-25T13:43:12.660" Score="5" ViewCount="300" Body="&lt;p&gt;I would like to migrate this &lt;a href=&quot;http://math.stackexchange.com/questions/514/conjectures-that-have-been-disproved-with-extremely-large-counterexamples&quot;&gt;Math&lt;/a&gt; Question into physics. The question is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Are there conjectures in Physics which have been disproved with extremely large counterexamples? If yes, i would like to know some of them.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="159" LastActivityDate="2011-05-20T19:20:18.373" Title="Conjectures that have been disproved with extremely large counterexamples" Tags="&lt;big-list&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="2" />
  <row Id="5873" PostTypeId="1" CreationDate="2011-02-25T14:28:55.443" Score="-2" ViewCount="210" Body="&lt;p&gt;Maybe these two types of objects do not exist but hypothetically, what would happen if an unstoppable force struck an immovable object. In terms of visually and in Physics terms?&lt;/p&gt;&#xA;" OwnerUserId="2240" LastActivityDate="2011-02-25T15:02:33.443" ClosedDate="2011-02-25T15:11:46.057" Title="What would happen if an immovable object was struck by an unstoppable force" Tags="&lt;forces&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="5874" PostTypeId="2" ParentId="5872" CreationDate="2011-02-25T14:31:38.213" Score="1" ViewCount="" Body="&lt;p&gt;The equivalent in Physics of a counterexample in Mathematics would be a failed experiment. For example: the Michelson Morley experiment is a counterexample to the ether conjecture. But was it big? Can any experiment be &quot;big&quot; in the same sense as Mathematics? Possibly not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I make a conjecture: &quot;any physical conjecture can be disproved with a fairly straightforward experiment.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually it's not a conjecture, it's a simple request that any valid physical theory must be disprovable through experiment (which is pretty much an agreed to principle).&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-02-25T14:31:38.213" CommentCount="2" />
  <row Id="5875" PostTypeId="2" ParentId="5872" CreationDate="2011-02-25T14:33:37.733" Score="4" ViewCount="" Body="&lt;p&gt;Lots of properties that were found to hold locally (in space and time) turned out to be only local approximations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flat earth hypothesis - long journey.&lt;br&gt;&#xA;Galilean transformations - breaks at large velocities.&lt;br&gt;&#xA;Global curvature of spacetime, locally it is flat - large distances.&lt;br&gt;&#xA;Spacetime is not expanding - breaks at large distances (Hubble's law)&lt;br&gt;&#xA;Classical mechanics - breaks also at extremely small scales (QM).&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2011-02-25T15:06:00.027" LastActivityDate="2011-02-25T15:06:00.027" CommentCount="1" />
  <row Id="5876" PostTypeId="2" ParentId="5873" CreationDate="2011-02-25T14:33:58.943" Score="-2" ViewCount="" Body="&lt;p&gt;There may be two issues: or immovable object gets destroyed, or the unstoppable force gets exhausted ;-).&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-02-25T14:33:58.943" CommentCount="3" />
  <row Id="5877" PostTypeId="2" ParentId="5872" CreationDate="2011-02-25T14:34:15.063" Score="7" ViewCount="" Body="&lt;p&gt;One of the better examples of such a reversals is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Steady_State_theory&quot; rel=&quot;nofollow&quot;&gt;&quot;Steady-State Hypothesis&quot;&lt;/a&gt; of Hoyle and Narlikar. Increasing depth and precision in cosmological measurements in the 1960s and 70s, however, emphatically refuted this idea.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-25T14:34:15.063" CommentCount="1" />
  <row Id="5878" PostTypeId="2" ParentId="5873" CreationDate="2011-02-25T14:36:46.847" Score="0" ViewCount="" Body="&lt;p&gt;Movement is relative. If an object is at rest for one observer, it may be moving for another observer. If it is at rest and it has an infinite mass w.r.t. an observer then no finite force can accelerate the object so it will continue to be at rest for that observer.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-25T14:42:35.683" LastActivityDate="2011-02-25T14:42:35.683" />
  <row Id="5879" PostTypeId="1" AcceptedAnswerId="5946" CreationDate="2011-02-25T14:46:16.483" Score="3" ViewCount="229" Body="&lt;p&gt;Can a functional derivative be calculated if we have a function of more than one variable? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The functional derivative of, for example, $F[b(x)]=e^{\int_0^{x&amp;#39;} dx a(x,y) b(x)}$ is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac{\delta F[b(x)]}{\delta b(z)} = a(z,y) e^{\int_0^{x'} dx a(x,y) b(x)}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about if there it's a functional of more than one variable? -  $F[b(x,k)]=e^{\int_0^{x&amp;#39;} \int_0^p  dx dk a(x,y,k) b(x,k) }$? Can we write the following?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac{\delta F[b(x,k)]}{\delta b(z,k)} = a(z,y,k)e^{\int_0^{x'} \int_0^p  dx dk a(x,y,k) b(x,k )}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I just want to note that I've been told that you &lt;strong&gt;cannot&lt;/strong&gt; do this by a lecturer. Therefore, is the answer is yes (or no!), please give some details! (I wasn't convinced that he was correct).&lt;/p&gt;&#xA;" OwnerUserId="873" LastEditorUserId="873" LastEditorDisplayName="" LastEditDate="2011-02-25T16:14:44.523" LastActivityDate="2011-02-26T15:07:37.070" Title="Can a functional derivative be calculated if we have a function of more than one variable?" Tags="&lt;mathematics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="5880" PostTypeId="2" ParentId="5865" CreationDate="2011-02-25T14:55:38.437" Score="5" ViewCount="" Body="&lt;p&gt;No broadly speaking. Yes in a narrow context. &lt;a href=&quot;http://en.wikipedia.org/wiki/Dispersion_%28optics%29&quot; rel=&quot;nofollow&quot;&gt;Chromatic dispersion&lt;/a&gt; guarantees that it is impossible to construct a lens which will be free of caustics and other optical aberrations for &lt;em&gt;all&lt;/em&gt; wavelengths. Individually lenses can be engineered to provide nearly arbitrary precision for narrow bands of wavelengths. However any (traditional, i.e. non-metamaterial) optical device is &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffraction-limited_system&quot; rel=&quot;nofollow&quot;&gt;diffraction limited&lt;/a&gt;, in that the minimum size $d$ of an object that a wavelength $\lambda$ can resolve is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ d = \frac{\lambda}{2n \sin \alpha} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n$ is the index of refraction and $\alpha$ the half-angle subtended by the object at the aperture of the device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this limit also says about the convergence of light-rays is that any &lt;em&gt;pencil&lt;/em&gt; of rays coming from an object of size $ d&amp;#39; \lt d $ will fail to converge at a single point on going through the device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your question is &quot;yes&quot;, as long as the device is used within its operating range!&lt;/p&gt;&#xA;" LastActivityDate="2011-02-25T14:55:38.437" CommentCount="2" />
  <row Id="5881" PostTypeId="2" ParentId="5873" CreationDate="2011-02-25T15:02:33.443" Score="0" ViewCount="" Body="&lt;p&gt;In addition to the relativity of motion, the phrase &quot;unstoppable force&quot; is also ambiguous here. In physics one could have forces that are specific to certain matter types (weak interaction comes to mind). Obviously then the force is not registered by certain matter types, and so has no impact on objects of that type &quot;movable&quot; or not. Neutrinos passing through the Earth also comes to mind here.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-25T15:02:33.443" />
  <row Id="5882" PostTypeId="1" AcceptedAnswerId="5925" CreationDate="2011-02-25T15:47:30.057" Score="0" ViewCount="344" Body="&lt;p&gt;Given a certain running pace uphill, I want to be able to determine an equivalent pace running with no elevation change. Assumptions: similar effort in both cases (say for example running at 90% max heart rate), ignore wind, slope is constant for simplicity, ignore physiological and bio-mechanical factors, weight of the runner is 135 lbs if that matters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: Elevation change +236 feet, distance traveled 1 mile, elapsed time 6 minutes 55 seconds. What could I theoretically run for 1 mile with no elevation change given the same effort?&lt;/p&gt;&#xA;" OwnerUserId="2244" LastEditorUserId="2244" LastEditorDisplayName="" LastEditDate="2011-02-25T19:46:45.003" LastActivityDate="2011-02-26T12:48:18.547" Title="Convert running speed uphill to equivilent speed on flat" Tags="&lt;mechanics&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="5883" PostTypeId="1" AcceptedAnswerId="5891" CreationDate="2011-02-25T15:56:55.210" Score="1" ViewCount="246" Body="&lt;p&gt;a light modifier that is commonly used in studio photography is a honeycomb grid.&lt;br&gt;&#xA;It narrows the beam of light to a circle with soft edges, as it can be seen here:&#xA;&lt;a href=&quot;http://www.honeycombgrids.com/info/grid-diffusion-patterns&quot; rel=&quot;nofollow&quot;&gt;http://www.honeycombgrids.com/info/grid-diffusion-patterns&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: how is this happening?&lt;br&gt;&#xA;A small reporter flash has a rectangular shape, if you place a rectangular shaped grid on it, it produces a &quot;soft&quot; circle of light. How is the light travel modified by the structure of the grid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;Olivier&lt;/p&gt;&#xA;" OwnerUserId="2245" LastActivityDate="2011-02-25T17:24:24.040" Title="How does a honeycomb grid affect the travel of light?" Tags="&lt;light&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="5884" PostTypeId="1" CreationDate="2011-02-25T16:33:00.173" Score="2" ViewCount="167" Body="&lt;p&gt;I have been told (not sure if it is true), that mirror (and glass) do not allow to pass the electromagnetic signals of mobile signals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but for a standard mirror what type of wave can pass through it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="1686" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-25T19:04:43.897" LastActivityDate="2011-02-25T19:06:15.893" Title="What type of electromagnetic wave can pass through silver or aluminium standard mirror?" Tags="&lt;electromagnetic-radiation&gt;&lt;mirror&gt;" AnswerCount="2" />
  <row Id="5885" PostTypeId="2" ParentId="5884" CreationDate="2011-02-25T16:41:08.067" Score="0" ViewCount="" Body="&lt;p&gt;There is only one type of electromagnetic wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hi-frequency electromagnetic wave in the hf radio band can pass through a mirror because the mirror is seen as an antenna of a small size in comparison to wavelenth.&lt;/p&gt;&#xA;" OwnerUserId="2146" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-25T19:06:15.893" LastActivityDate="2011-02-25T19:06:15.893" CommentCount="1" />
  <row Id="5886" PostTypeId="1" CreationDate="2011-02-25T16:47:17.563" Score="3" ViewCount="151" Body="&lt;p&gt;Consider the well known demonstration of diffraction by a narrowing slit.&#xA;(See for example the demonstration at the 30 minute mark of this lecture at MIT by Walter Lewin)&#xA;&lt;a href=&quot;http://ocw.mit.edu/high-school/physics/physical-optics/interference-diffraction/&quot; rel=&quot;nofollow&quot;&gt;http://ocw.mit.edu/high-school/physics/physical-optics/interference-diffraction/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is my (possibly mistaken) understanding that the light emerging after the slit becomes substantially slimmer than one wavelength is polarized.&lt;br&gt;&#xA;This would seem to imply that light of perpendicular polarization would not be transmitted, thus implying a fairly substantial and dramatic difference in the results of the experiment with parallel and perpendicularly polarized light.  That is,  instead of spreading out, the light polarized in the wrong direction would essentially just shut down as the slit narrows below one wavelength.    Is this true? &lt;/p&gt;&#xA;" OwnerUserId="1172" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-25T19:06:45.283" LastActivityDate="2011-02-25T23:01:23.233" Title="What is the effect of polarization on diffraction by a narrow slit?" Tags="&lt;optics&gt;&lt;light&gt;&lt;polarization&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="5887" PostTypeId="2" ParentId="5884" CreationDate="2011-02-25T16:48:57.903" Score="2" ViewCount="" Body="&lt;p&gt;The answer I provided to &lt;a href=&quot;http://physics.stackexchange.com/questions/1094/how-i-can-create-hindrances-to-radio-waves/2307#2307&quot;&gt;this older question&lt;/a&gt; is very applicable here. The degree to which an EM signal is obstructed will depend on the material properties of the object (conductivity, primarily) and the wavelength of the signal. For a mobile phone, the conductive metal coating of the mirror is likely much thinner than the skin depth, so the signal should only be weakened a little bit.&lt;/p&gt;&#xA;" OwnerUserId="869" LastEditorUserId="869" LastEditorDisplayName="" LastEditDate="2011-02-25T18:18:26.520" LastActivityDate="2011-02-25T18:18:26.520" />
  <row Id="5888" PostTypeId="1" AcceptedAnswerId="5890" CreationDate="2011-02-25T16:59:57.987" Score="4" ViewCount="376" Body="&lt;p&gt;I'm new here, so go easy!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I heard on a podcast recently that the supermassive black holes at the centre of some galaxies could have densities less than water, so in theory, they could float on the substance they were gobbling up... can someone explain how something with such mass could float?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please see the below link for the podcast in question...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.universetoday.com/83204/podcast-supermassive-black-holes/&quot; rel=&quot;nofollow&quot;&gt;http://www.universetoday.com/83204/podcast-supermassive-black-holes/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2247" LastEditorUserId="2247" LastEditorDisplayName="" LastEditDate="2011-02-25T17:07:19.207" LastActivityDate="2011-09-03T22:17:13.717" Title="How can super massive black holes have a lower density than water?" Tags="&lt;black-holes&gt;&lt;water&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="5889" PostTypeId="2" ParentId="5886" CreationDate="2011-02-25T17:09:38.880" Score="5" ViewCount="" Body="&lt;p&gt;First, I just want to correct a minor misunderstanding. If anything, it would be the light with its electrical field oscillating in the plane &lt;em&gt;parallel&lt;/em&gt; to the slit which would have more difficulty propagating, and even then, only in specific circumstances. The best way to explain this is to ignore, for a moment, the experiment you've described, and consider a simple polarizing filter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest type of polarizing filter to talk about is simple an array of closely space parallel conductive wires. Such a filter will block light polarized parallel to the wires. Why? Because the E field from this light is parallel to the wires, it can induce currents, which absorb energy from the traveling EM radiation. Perpendicularly polarized light doesn't induce such currents, and is therefore able to pass more easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, going back to your question. If the slit were cut in a conductive material, and the slit width was on the order of the wavelength, then there may indeed be diminished transmission for light polarized parallel to the long axis of the slit because, to a rough approximation, the edges of the slit would act like the conductive wires I described before. &lt;em&gt;However&lt;/em&gt;, while the light transmitted would possibly be dimmer, the form of the resulting diffraction pattern should be unchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is all moot if the slits are cut into a non-conductive material. In any event, the only change would be in the intensity of the pattern. It's up to you to decide if you think this difference qualifies as a &quot;substantial and dramatic difference in the results of the experiment.&quot;&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-02-25T17:09:38.880" CommentCount="3" />
  <row Id="5890" PostTypeId="2" ParentId="5888" CreationDate="2011-02-25T17:18:14.383" Score="10" ViewCount="" Body="&lt;p&gt;Well, it can't (float), since a Black Hole is not a solid object that has any kind of surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When someone says that a super massive black hole has less density than water, one probably means that since the density goes like &#xA;$\frac{M}{R^3}$&#xA;where M is the mass and R is the typical size of the object, then for a black hole the typical size is the Schwarzschild radius which is $2M$, which gives for the density the result&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho\propto M^{-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see from that, that for very massive black holes you can get very small densities (all these are in units where the mass is also expressed in meters). But that doesn’t mean anything, since the Black Hole doesn’t have a surface at the Schwarzschild radius. It is just curved empty space. &lt;/p&gt;&#xA;" OwnerUserId="383" LastEditorUserId="383" LastEditorDisplayName="" LastEditDate="2011-02-25T17:25:01.423" LastActivityDate="2011-02-25T17:25:01.423" CommentCount="9" />
  <row Id="5891" PostTypeId="2" ParentId="5883" CreationDate="2011-02-25T17:24:24.040" Score="3" ViewCount="" Body="&lt;p&gt;From looking at the pictures of the grids provided on that website, I think I understand what they are doing. It's not actually very complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems easiest to explain this if we consider only one of the hexagonal &quot;tubes&quot; formed by the grid. Normally, the light from a flash spreads out over some (relatively wide) angle, but now think of what happens to the light when it is confined to this tube: The light that would normally head off at a steep angle is intercepted by the walls of the tube, while light that was already propagating parallel (or at least close to parallel) to the tube axis is passed without obstruction. This results in a more confined range of angles over which the light can spread. Of course, the confinement is not perfect, as I will explain below, but you can imagine how the angular spread of the beam would be limited to consist only of those rays which can pass cleanly through the tube. This is simple geometry, and you could adjust this angle by varying the length and diameter of the tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the light that is intercepted by the tube walls is not completely eliminated. Instead it scatters into a random direction. It is this scattering that gives the resulting beam a nice soft edge. Instead of being sharply confined to the angular range allowed by the tube, the scattering allows it to &quot;bleed&quot; a little bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should add that the hexagonal shape of the tubes probably has very little effect on the resulting illumination pattern. More likely a hexagonal shape was chosen for some other reason, like easy manufacturing or more mechanical stability. Optically, a square grid or even an array of circular tubes, like a bundle of straws, would work just fine.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-02-25T17:24:24.040" CommentCount="4" />
  <row Id="5892" PostTypeId="2" ParentId="5888" CreationDate="2011-02-25T17:28:21.833" Score="1" ViewCount="" Body="&lt;p&gt;The Schwarzschild radius scales with mass as $r~=~2GM/c^2$.  What might be defined as a Schwarzschild volume would then be $V~=~4\pi r^3/3$ $=~(32/3)\pi(GM/c^2)^3$.  So the density of matter defined by the horizon is $\rho~=~(3/32)(c^2/G)^3M^{-2}$.  So density scales as the inverse square of the mass.  A 10 billion solar mass black hole has a radius about $10^{10}$km, or a volume $V~\sim~10^{30}km^3$ $=~10^{39}m^3$.  A solar mass is $10^{30}$kg and the density defined by the horizon is then $\rho~=~10^{-9}kg/m^3$.  That is actually quite small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if you fall into a black hole of any mass you encounter a region with enormous Weyl curvature and tidal forces.  The source of this is in your future, and eventually you reach it --- it is inescapable.  This region where curvature diverges is a spatial surface of infinite extent.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-25T17:28:21.833" CommentCount="1" />
  <row Id="5893" PostTypeId="1" CreationDate="2011-02-25T17:29:51.103" Score="1" ViewCount="319" Body="&lt;p&gt;Do you know where can I find simple diagram (solid, liquid, gas) for for oxygen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I made a mistake. I wanted oxygen instead of water. sorry.  &lt;/p&gt;&#xA;" OwnerUserId="58" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-25T18:41:45.447" LastActivityDate="2011-02-25T18:41:45.447" Title="Phase diagram for oxygen" Tags="&lt;phase-transition&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="5894" PostTypeId="2" ParentId="5893" CreationDate="2011-02-25T17:44:30.200" Score="2" ViewCount="" Body="&lt;p&gt;For &lt;strong&gt;water&lt;/strong&gt;, see e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.cims.nyu.edu/~gladish/teaching/eao/week2.html&quot; rel=&quot;nofollow&quot;&gt;http://www.cims.nyu.edu/~gladish/teaching/eao/week2.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if you also wanted to see some exotic high-pressure phases of ice, see&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.btinternet.com/~martin.chaplin/phase.html&quot; rel=&quot;nofollow&quot;&gt;http://www.btinternet.com/~martin.chaplin/phase.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For &lt;strong&gt;oxygen&lt;/strong&gt; in the updated question, see&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://images.google.cz/images?q=phase+diagram+oxygen&amp;amp;biw=1031&amp;amp;bih=783&quot; rel=&quot;nofollow&quot;&gt;http://images.google.cz/images?q=phase+diagram+oxygen&amp;amp;biw=1031&amp;amp;bih=783&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.azonano.com/Details.asp?ArticleID=1797&quot; rel=&quot;nofollow&quot;&gt;http://www.azonano.com/Details.asp?ArticleID=1797&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-25T18:32:00.587" LastActivityDate="2011-02-25T18:32:00.587" CommentCount="2" />
  <row Id="5895" PostTypeId="1" AcceptedAnswerId="5897" CreationDate="2011-02-25T19:35:13.430" Score="3" ViewCount="99" Body="&lt;p&gt;Can anyone give a brief overview of what a wimpy particle is?&lt;/p&gt;&#xA;" OwnerUserId="2146" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-26T00:10:06.570" LastActivityDate="2011-02-26T00:10:06.570" Title="What are wimpy particles?" Tags="&lt;astrophysics&gt;&lt;dark-matter&gt;" AnswerCount="2" />
  <row Id="5896" PostTypeId="2" ParentId="5895" CreationDate="2011-02-25T19:37:27.463" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Weakly_interacting_massive_particle&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Weakly_interacting_massive_particle&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-02-25T19:37:27.463" />
  <row Id="5897" PostTypeId="2" ParentId="5895" CreationDate="2011-02-25T19:51:18.510" Score="4" ViewCount="" Body="&lt;p&gt;A little history because I don't like the formulation of the question title.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galaxies exhibit a rotational behavior that is inconsistent with the known law of gravitation and the visibly luminous mass distribution of the galaxy. Two ideas present themselves to explain why:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There is some non-visible mass&lt;/li&gt;&#xA;&lt;li&gt;Gravity doesn't work the way we thought&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Well, option 2 is pretty heady stuff, so lets look first at how option 1 would work. Maybe the so-called &quot;missing mass&quot; is just cold. That way it wouldn't glow and we wouldn't see it. Well, infrared and radio astronomy &lt;em&gt;would&lt;/em&gt; see it if it was spread out in a diffuse gas, and they see a little, but not nearly enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, maybe the missing matter is in big cold lumps. Call them MAssive Compact Halo Objects (MACHOs). Astronomers did extensive microlensing searches for MACHOs in the Milky Way during the 90s, and they found a few but can rule out most scenarios in which MACHOs contribute a large fraction of the missing mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leaves us looking for some mass that simply doesn't glow at any band. That is, we're looking for matter which does not interact with the electromagnetic force, but does interact gravitationally. If we make the conjecture that it also interacts by way of the weak force we can call this stuff Weakly Interacting Massive Particles (WIMPs), which makes a nice counterpart to MACHOs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;wimpy&quot; in the title is just the adjectival form of WIMP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computer models of cosmological evolution that assume the dark matter has the properties expected of WIMPs &lt;em&gt;can&lt;/em&gt; explain the combined lumpiness and large scale uniformity of the universe, so there is considerable hope that this is the explanation we're looking for.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-02-25T20:15:10.517" LastActivityDate="2011-02-25T20:15:10.517" />
  <row Id="5898" PostTypeId="1" CreationDate="2011-02-25T19:55:35.797" Score="2" ViewCount="163" Body="&lt;p&gt;ABS systems work by allowing a tire to continue to rotate rather than &quot;locking-up&quot; (stop turning) due to loss of traction with the road surface.  A rotating tire can influence the direction of the car, a &quot;locked-up&quot; tire has no directional influence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also gives the tire more traction than a fully &quot;locked-up&quot; tire.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a car enters a turn, angular momentum causes weight to shift to the tires on the outside of the turn, reducing weight, and thus traction, on the inside front tire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: does ABS's effect of increasing traction on the inside front tire cause a force on the car that causes it to go more straight (resist turning) compared to a car that has no ABS and whose inside front tire is allowed to stop (or mostly stop) turning?&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2011-07-02T04:40:45.283" Title="Does ABS cause a force that resists turning a car into a corner?" Tags="&lt;angular-momentum&gt;&lt;friction&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="5899" PostTypeId="1" AcceptedAnswerId="5900" CreationDate="2011-02-25T19:56:11.067" Score="3" ViewCount="183" Body="&lt;p&gt;In my country it is a big business now, to sell &quot;water structuring facilities&quot;. Promoters telling that after passing through them the water obtaining fantastic abilities basing on the fact that &quot;water can keep information coded in molecular domains, chained by weak hydrogen bonds&quot;. This &quot;structured&quot; water is a panacea for all diseases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess it is all started from &lt;a href=&quot;http://en.wikipedia.org/wiki/Masaru_Emoto&quot; rel=&quot;nofollow&quot;&gt;this guy&lt;/a&gt;, but now they have a bunch of popular documentary films and many official academics supporting these ideas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which arguments should I use to convince my friends they're charlatans? I don't have a grade in physics, but from my basic university course I believe it is nonsense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there must be some popular and simple facts that can be convincing enough for non-professionals who can understand physics at the high school level.&lt;/p&gt;&#xA;" OwnerUserId="2249" LastActivityDate="2011-02-26T16:14:56.587" Title="Withstanding charlatans: How to protect friends from buying &quot;water structuring facilities&quot;?" Tags="&lt;water&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="5900" PostTypeId="2" ParentId="5899" CreationDate="2011-02-25T20:05:12.830" Score="6" ViewCount="" Body="&lt;p&gt;The lifetime of hydrogen bonds in room temperature water is &lt;a href=&quot;http://www.springerlink.com/content/j82616068358334r/&quot; rel=&quot;nofollow&quot;&gt;picoseconds&lt;/a&gt;. The idea that they could persist long enough to drink is wrong by a factor of trillions.&lt;/p&gt;&#xA;" OwnerUserId="292" LastActivityDate="2011-02-25T20:05:12.830" CommentCount="1" />
  <row Id="5901" PostTypeId="1" CreationDate="2011-02-25T20:07:36.170" Score="6" ViewCount="189" Body="&lt;p&gt;So I wonder if it is any how possible to build laser at home. A powerful one to melt brick.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-25T23:42:51.450" Title="How to build a laser in the garage?" Tags="&lt;laser&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="2" />
  <row Id="5902" PostTypeId="2" ParentId="5901" CreationDate="2011-02-25T20:13:00.893" Score="0" ViewCount="" Body="&lt;p&gt;In principle, yes. Just transform your garage into a laser lab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But.&lt;/strong&gt; Have you computed the power needed to melt brick? Have you factored in the lasers (lack of) efficiency? Do you actually have the skills needed to build circuits handling that much power?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you considered a more modest initial goal?&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-02-25T20:18:55.460" LastActivityDate="2011-02-25T20:18:55.460" />
  <row Id="5903" PostTypeId="2" ParentId="5899" CreationDate="2011-02-25T20:19:27.520" Score="4" ViewCount="" Body="&lt;p&gt;First step: Tell people that according to established physical theories there is no reason to believe that the claimed effect is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does not prove anything, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second step: Ask them if they can detect the difference between processed water and not processed water in any way, if they can, for example &quot;instantly feel&quot; the difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they say they can, propose a &quot;double blindness test&quot; where they get processed and not processed samples of water at random and have to tell you if their sample is processed or if it is not (&quot;double blindness&quot; is the word for word translation from German where neither the people who have to judge the water nor the people bringing them the water know which is which, I don't know the correct idiomatic translation). There are a lot of ways to cheat at such a test, of course. Try to do this in public and with the participation of uncorrupted scientists. There is of course the possibility that they can tell the difference based on an effect that is not related to the claimed properties of the water in any way, that is simply an insignificant side effect of the processing. There is no system and no procedure that saves you from falling into such a trap. Knowing that there can be correlation without causation is important, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe they claim that the difference is only discernible on the long run. Ask for long-term, documented &quot;double blindness&quot; tests. Ask if these tests have resulted in statistically significant results. Run your own tests, if you must. If these prove the correctness of the claimed difference, accept it :-)&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditorDisplayName="" LastEditDate="2011-02-25T20:25:16.050" LastActivityDate="2011-02-25T20:25:16.050" CommentCount="5" />
  <row Id="5904" PostTypeId="2" ParentId="5899" CreationDate="2011-02-25T20:34:31.717" Score="5" ViewCount="" Body="&lt;p&gt;It is not easy to convince someone who wants to believe otherwise. In my experience, no amount of detailed scientific information will help in this regard. People attracted to charlatans often interpret such arguments as &quot;conspiracy of the scientific establishment&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, seeing (or tasting) is believing, and what usually works well is experimental evidence. Make a double-blind test, where you let try your friends ordinary water and &quot;structured water&quot;, and let them evaluate the difference (if any). Double-blind basically means, that neither the tester nor you (nor anyone else in the room) should know which is the &quot;structured water&quot; before the evaluation, to avoid some psychological bias. If you do sufficiently many of these tests (say, 20 probes for each type of water for 5 different persons, so 100 probes in total) you should get nearly a 50% chance that the tap water is identified incorrectly as &quot;structured water&quot; or vice versa. So if in the end only 40-60 of your probes are identified correctly your friends know that the structured water claim is bogus, since flipping coins to decide which water is which would have achieved the same result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beware, however, of slight temperature differences, particularly if the test consists in drinking the water. All physical and chemical conditions should be identical if you really want to test the (non-)effect of giving the water &quot;structure&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some fair the clever guys selling the structured water cooled it slightly below the tap water (about 1 Celsius less). Thus, the &quot;structured water&quot; tasted more fresh than the tap water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a historical aside, in Austria we have a similar issue with so-called Grander-water - funnily, Mr. Grander even received the Austrian cross of honor for scientific merits from former minister Gehrer. So even science ministers are not immune to such charlatans.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-02-25T20:34:31.717" />
  <row Id="5905" PostTypeId="1" AcceptedAnswerId="5906" CreationDate="2011-02-25T20:44:05.967" Score="4" ViewCount="228" Body="&lt;p&gt;The force of gravity is constantly being applied to an orbiting object.  And therefore the object is constantly accelerating.  Why doesn't gravity eventually &quot;win&quot; over the object's momentum, like a force such as friction eventually slows down a car that runs out of gas?  I understand (I think) how relativity explains it, but how does Newtonian mechanics explain it?&lt;/p&gt;&#xA;" OwnerUserId="2252" LastActivityDate="2011-02-25T22:01:17.493" Title="How does Newtonian mechanics explain why orbiting objects do not fall to the object they are orbiting?" Tags="&lt;gravity&gt;&lt;newtonian-mechanics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="5906" PostTypeId="2" ParentId="5905" CreationDate="2011-02-25T20:54:55.287" Score="11" ViewCount="" Body="&lt;p&gt;Newtonian mechanics explains that they &lt;em&gt;do&lt;/em&gt; fall toward the object they're orbiting, the just keep missing.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Quick and dirty derivation for a circular orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the primary have mass $M$ and the satellite mass $m$ such that $m \ll M$ (it can also be done for other cases, but this saves on mathiness).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume we start with an initial circular orbit on radius $r$, velocity $v = \sqrt{G\frac{M}{r}}$. The acceleration of the satellite due to gravity is $a = G\frac{M}{r^2}$ which means we can also write $v = \sqrt{\frac{a}{r}}$. The period of the orbit is $T = \frac{2\pi r}{v} = 2\pi \sqrt{\frac{r}{a}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chose a coordinate system in which the initial position is $r\hat{i} + 0\hat{j}$ and the initial velocity points in the $+\hat{j}$ direction. Chose a short time $t \ll T$ and lets see how far from the primary the satellite ends up after that time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have chosen $t$ short enough, we can approximate gravity as having uniform strength through the time period (and we shall show later that that &lt;em&gt;is&lt;/em&gt; justified).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The new position is $(r - \frac{1}{2}at^2)\hat{i} + vt\hat{j}$ which lies at a distance&#xA;$$ r_2 = \sqrt{r^2 - r a t^2 + \frac{1}{4}a^2 t^4 + v^2 t^2} $$&#xA;pulling our at factor of $r$ we get&#xA;$$ r_2 = r \sqrt{1 - \frac{a}{r} t^2 + \frac{1}{4}\frac{a^2}{r^2} t^4 + \frac{v^2}{r^2} t^2} $$&#xA;and converting all the $\frac{a}{r}$ and $\frac{v}{r}$ terms into expressions of the period we get&#xA;$$ r_2 = r \sqrt{1 - (2\pi\frac{t}{T})^2 + \frac{1}{4}(2\pi\frac{t}{T})^4 + (2\pi\frac{t}{T})^2}$$&#xA;Finally, we drop the $(t/T)^4$ term as negligible and note that the $(t/T)^2$ terms cancel so the result is&#xA;$$r_2 = r$$&#xA;or the radius never changed (which justified the constant magnitude for acceleration, and a small enough $t$ justifies both the constant direction and the dropping of the fourth degree term).&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-02-25T22:01:17.493" LastActivityDate="2011-02-25T22:01:17.493" CommentCount="6" />
  <row Id="5907" PostTypeId="2" ParentId="5905" CreationDate="2011-02-25T21:07:18.493" Score="6" ViewCount="" Body="&lt;p&gt;The force of gravity has little to do with friction. As dmckee says, what is happening is that the body falls, but precisely because it has enough momentum, it falls around the object towards which it gravitates instead of into it. Of course, this is not always the case, collisions do happen. Also systems of astronomical bodies are complicated and the combined effect of the action of several different bodies on one can destabilize trajectories that in a simple 2-body case would be stable ellipses. The result could be collision or escape of the body. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 2-body case however, the crucial aspect of gravity which guarantees the stability of the system is the fact that gravitation is a &lt;strong&gt;centripetal force&lt;/strong&gt;. It always acts towards the center of the other gravitating mass. One can show that this feature implies the &lt;strong&gt;conservation of angular momentum&lt;/strong&gt;, which means that if the 2-body system had some angular momentum to begin with, it will keep the same angular momentum indefinitely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Extra note, even in the 2-body case, there can be collisions and escape to infinity, the first if there is not enough angular momentum (for instance one body having velocity directed towards the other body, like an apple falling from a tree), the other if there is too much angular momentum, resulting in parabolic or hyperbolic trajectories.)&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2011-02-25T21:07:18.493" />
  <row Id="5908" PostTypeId="2" ParentId="5888" CreationDate="2011-02-25T21:10:20.163" Score="1" ViewCount="" Body="&lt;p&gt;I think it is actually misleading to make the claim that is puzzling you. &quot;Density&quot; suggests that the mass is distributed more or less uniformly within the black hole, and this is non-sense. The black hole is mostly empty, and all the mass is concentrated within a tiny region (clasically a point) in the center of the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you ignore this and pretend a black hole of mass $M$ and volume $V\propto r^3$ had a uniform density $\rho$ then you can calculate it, simply using $\rho=M/V$. Since for Schwarzschild black holes the radius of the black hole is proportional to its mass you obtain finally $\rho\propto 1/M^2$, so the heavier a black hole the smaller its density. But again, this provides a highly misleading picture of the mass distribution within the black hole. All its mass is in the center, so classically the density is infinite.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-02-25T21:10:20.163" />
  <row Id="5909" PostTypeId="2" ParentId="5882" CreationDate="2011-02-25T21:54:40.183" Score="0" ViewCount="" Body="&lt;p&gt;Ignoring wind and physiological factors, it requires no energy to run at constant pace with no elevation change. By conservation of energy, your energy has to go somewhere. The only place left (after ignoring wind and physiology) is your kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the same assumptions, running up hill does require energy. In this case your energy goes into potential energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make running up hill equivalent to running with no elevation change if you require that the running with no elevation change be under the condition of constant acceleration. Then you'll be equating the rate of change in potential energy (going up hill) with the rate of change in kinetic energy (accelerating at the same elevation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But none of this is a very good model of human energy.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-25T21:54:40.183" />
  <row Id="5910" PostTypeId="2" ParentId="5898" CreationDate="2011-02-25T22:15:04.587" Score="2" ViewCount="" Body="&lt;p&gt;In terms of forces exerted by the road and the tire on each other, ABS increases the force by keeping your tires in the &quot;static friction&quot; regime instead of the &quot;kinetic friction&quot;. That is, objects tend to have higher friction before they begin to slip then they do after slipping. So when the system detects that a wheel has slipped, it reduces the brake pressure to the point where the wheel returns to the static friction case. (This is said to not work so well on deep gravel, but it's suitable for typical pavement.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other reason ABS is useful is because it allows the driver to control the direction of the force. That is, it helps you use your steering wheel to control the direction that the car moves. So one of the primary purposes of ABS is to help you corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you slam on your brakes without ABS, while going around a corner, you will slide in the direction you're traveling. Since the maximum of rolling friction is larger than sliding friction slamming on your brakes will take you longer to stop. In addition, you will have no control on your direction and you will likely run off the road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you're stopping, your front tires take more of the weight of the car than the rear so I would think that you would lose traction on your rear wheels first. This would definitely make it more difficult to steer and more difficult to stay on the road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if, for one reason or another, only your inside front tire lost traction (without ABS), this would decrease the force on the tire and that would likely make it more difficult for you to turn. That is, you'd have more difficulty getting the car turned without ABS than with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interesting question. I look forward to see what others have to say about this.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-07-02T04:40:45.283" LastActivityDate="2011-07-02T04:40:45.283" CommentCount="3" />
  <row Id="5911" PostTypeId="1" AcceptedAnswerId="5915" CreationDate="2011-02-25T22:24:28.880" Score="2" ViewCount="101" Body="&lt;p&gt;Suppose we have a lens with known focal length and are using the thin lens approximation. Is it possible to determine how an arbitrary beam of light will be deflected by it? Or, is it the case that there are multiple kinds of thin lenses, and so more information is required. I'm asking this question just so I can solve first year university optics problems.&lt;/p&gt;&#xA;" OwnerUserId="119" LastEditorUserId="869" LastEditorDisplayName="" LastEditDate="2011-02-25T22:53:46.033" LastActivityDate="2011-02-26T06:31:34.907" Title="Calculating the deflection of an arbitrary beam of light from a thin lens" Tags="&lt;optics&gt;&lt;homework&gt;" AnswerCount="3" />
  <row Id="5912" PostTypeId="2" ParentId="5901" CreationDate="2011-02-25T22:38:53.263" Score="5" ViewCount="" Body="&lt;p&gt;Yes. 100W CO2 laser is doable at home, and some in fact did that.&#xA;100W one will BURN really well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No other types of power lasers are doable at home. (well, probably there is also killing 200 DVD-RW drives and collimating them all - I am actually doing that, I have 45 RW drives ;-) )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only problems is that you still need few rare things like IR mirrors &amp;amp; transparent window which are very hard to find in garage (there is not much materials transparent at 10um). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the optimal solution would be just buy finished CO2 tube from China (100-200$) and build cooling &amp;amp; power supply system by yourself. This is way way more realistic.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-02-25T22:38:53.263" CommentCount="2" />
  <row Id="5913" PostTypeId="2" ParentId="5911" CreationDate="2011-02-25T22:40:54.883" Score="-1" ViewCount="" Body="&lt;p&gt;Yes it is possible. Read the book while you have time, we all did that and it worked. ;-)&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-02-25T22:40:54.883" />
  <row Id="5914" PostTypeId="1" CreationDate="2011-02-25T22:41:34.343" Score="8" ViewCount="181" Body="&lt;p&gt;Phase transitions are often detected by an order parameter - some quantity which is zero in the &quot;disordered&quot; phase, but becomes non-zero when order is established. For example, if the phase transition is accompanied by a breaking of a global symmetry, an order parameter can be any quantity which transforms non-trivially so that it averages to zero in the disordered phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phases not characterized by their global symmetry are more tricky. For confinement there are well-known order parameters: the area law for the Wilson loop, the Polyakov loop (related to breaking of the center symmetry), and the scaling of the entropy with N in the large N limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is then about the Higgs phase, which is usually referred to (misleadingly in my mind) as spontaneous breaking of gauge &quot;symmetry&quot;. More physical (among other things, gauge invariant) characterization of this phase would be in terms of some order parameter. What are some of the order parameters used in that context?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(One guess would be the magnetic duals to the quantities characterizing confinement, but there may be more).&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-02-26T09:55:42.363" Title="Order Parameters for the Higgs Phase" Tags="&lt;quantum-field-theory&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="3" />
  <row Id="5915" PostTypeId="2" ParentId="5911" CreationDate="2011-02-25T22:55:33.770" Score="2" ViewCount="" Body="&lt;p&gt;The subject you are looking for is called &quot;the ABCD matrix method,&quot; &quot;Matrix Paraxial Optics,&quot; &quot;Matrix Ray Tracing,&quot; or some variation of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A thin lens is represented by a 2x2 matrix and the incident ray is a 2 element vector. Computing the height and angle of the outbound ray is a simple matrix multiplication. &lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-02-25T22:55:33.770" CommentCount="1" />
  <row Id="5916" PostTypeId="2" ParentId="5901" CreationDate="2011-02-25T22:57:26.000" Score="0" ViewCount="" Body="&lt;p&gt;I believe that the answer to your question is no. Of course if you order all special components like resonator mirrors, high voltage power supply, etc., then it doesn't matter where you build the laser. But you cannot build a laser with items from a DIY-shop. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And just a remark - there is no way to melt a brick with a laser. Ceramic plates are very heat-resistant and they are used when you need to block a high-power laser beam (not a trivial task).&lt;/p&gt;&#xA;" OwnerUserId="701" LastActivityDate="2011-02-25T22:57:26.000" />
  <row Id="5917" PostTypeId="2" ParentId="5901" CreationDate="2011-02-25T23:00:58.423" Score="3" ViewCount="" Body="&lt;p&gt;The DIY aluminum foil approach if you have a HV supply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://wiki.4hv.org/index.php/Nitrogen_TEA_laser&quot; rel=&quot;nofollow&quot;&gt;http://wiki.4hv.org/index.php/Nitrogen_TEA_laser&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sparkbangbuzz.com/tealaser/tealaser7.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.sparkbangbuzz.com/tealaser/tealaser7.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.repairfaq.org/sam/lasercn2.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.repairfaq.org/sam/lasercn2.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is based on the TEA approach with nitrogen from the air (or from other sources) as the lasing material.&lt;/p&gt;&#xA;" OwnerUserId="1670" LastEditorUserId="1670" LastEditorDisplayName="" LastEditDate="2011-02-25T23:42:51.450" LastActivityDate="2011-02-25T23:42:51.450" />
  <row Id="5918" PostTypeId="2" ParentId="5886" CreationDate="2011-02-25T23:01:23.233" Score="0" ViewCount="" Body="&lt;p&gt;I tried to make this a comment, but I couldn't do it. I don't know why.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Colin K&#xA;Thank you for your answer, although I’m not sure I understand what you are saying. &#xA;The filter of parallel conducting wires was indeed what I was thinking about.&#xA; Let’s consider the case where the wires are spaced one tenth (or less) of a wavelength apart.  My expectation is that this filter passes the perpendicularly polarized light, and reflects the parallel polarized light.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider a slit cut into a conducting material the same size and shape as the space between two wires in the above filter.  My expectation is that this slit will again reflect parallel polarized light, and pass perpendicularly polarized light.   In addition the passed light will “suffer diffraction” and be spread out significantly.  Do you agree?&lt;br&gt;&#xA;If not please point out exactly where you disagree.&lt;br&gt;&#xA;Thank you. Jim Graber  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, my first guess is that the filter with many wires will have a much smaller, maybe even negligible diffraction angle, an angle based on the width of the entire filter, rather than based on the width of a single slit.&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-02-25T23:01:23.233" CommentCount="1" />
  <row Id="5919" PostTypeId="2" ParentId="5859" CreationDate="2011-02-25T23:02:54.293" Score="1" ViewCount="" Body="&lt;p&gt;The act of measurement of a system which is not an eigenstate of the measurement must destroy information. That is, you cannot return to the initial states after making the measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only information coming into the problem are the states that are being cloned. So by conservation of information it's not possible to clone two states with probability 1 using a sequence of measurements and unitary operations. Maybe you can improve the probabilities this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think that the easiest way to attack the problem would be to take a look at the qubit example they do at the end of the paper. From this you can see just how much information is remaining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is an attractive problem and when I have more time, Sunday, I'll work on it, unless someone solves it before then.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-25T23:02:54.293" CommentCount="1" />
  <row Id="5920" PostTypeId="2" ParentId="5901" CreationDate="2011-02-25T23:17:03.667" Score="1" ViewCount="" Body="&lt;p&gt;You don't need a laser to burn holes in a brick; instead, concentrate the power of the sun using a &lt;a href=&quot;http://www-personal.umich.edu/~bclee/lens.html&quot; rel=&quot;nofollow&quot;&gt;giant fresnel lens&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-02-25T23:17:03.667" />
  <row Id="5921" PostTypeId="2" ParentId="5911" CreationDate="2011-02-25T23:38:09.117" Score="0" ViewCount="" Body="&lt;p&gt;Okay, I've got an idea. I would have liked to draw a diagram, but my graphics software is playing up at the moment. First find where the light ray, &lt;code&gt;R&lt;/code&gt;, crosses the axis of symmetry of the lens - call this &lt;code&gt;X&lt;/code&gt;. Using the paraxial ray approximation: &lt;code&gt;1\s+1\s'=1/f&lt;/code&gt; where &lt;code&gt;s&lt;/code&gt; is the object distance, &lt;code&gt;s'&lt;/code&gt; the image distance and &lt;code&gt;f&lt;/code&gt; is the focal length. Note that &lt;code&gt;s&lt;/code&gt;, &lt;code&gt;s'&lt;/code&gt; are signed variables. Anyway, find where the image, &lt;code&gt;I&lt;/code&gt;, of point &lt;code&gt;X&lt;/code&gt; is by using the formula. All light rays from &lt;code&gt;X&lt;/code&gt; passing through the lens arrive at &lt;code&gt;I&lt;/code&gt;, so we now have the coordinates of two points the light ray must pass through after the lens and hence the angle.&lt;/p&gt;&#xA;" OwnerUserId="119" LastEditorUserId="119" LastEditorDisplayName="" LastEditDate="2011-02-26T06:31:34.907" LastActivityDate="2011-02-26T06:31:34.907" CommentCount="1" />
  <row Id="5922" PostTypeId="1" CreationDate="2011-02-26T00:36:51.323" Score="2" ViewCount="160" Body="&lt;p&gt;This question is inspired by another &lt;a href=&quot;http://physics.stackexchange.com/questions/3282/simple-models-that-exhibit-topological-phase-transitions&quot;&gt;one&lt;/a&gt; about the simplest model of topological insulator, where 4tnemele showed a nice two band model in the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read that and am wondering if we and push that to one dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, by analogy to the graphene case, if we have a Hamiltonian in 1D (say x) as $H(k_x)=(k_x-k_0)+m$ for $k_x&amp;gt;0$. When $k_x=k_0$, one has $m&amp;gt;0$. $H(k_x)=(k_x+k_0)+m$ for $k_x&amp;lt;0$. When $k_x=-k_0$, one has $m&amp;lt;0$. A smooth connection in between, we will have a conductive edge (two ends in the 1D structure), right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I want to make a intuitive picture like below, is it correct?&#xA;&lt;img src=&quot;http://i.stack.imgur.com/n1qkG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestion for real materials show this behavior?&lt;/p&gt;&#xA;" OwnerUserId="1948" LastEditorUserId="1948" LastEditorDisplayName="" LastEditDate="2011-02-27T09:11:54.870" LastActivityDate="2011-02-27T09:11:54.870" Title="1D topological insulator" Tags="&lt;condensed-matter&gt;&lt;topological-insulators&gt;" CommentCount="1" FavoriteCount="2" />
  <row Id="5923" PostTypeId="1" AcceptedAnswerId="5981" CreationDate="2011-02-26T02:27:52.060" Score="2" ViewCount="224" Body="&lt;p&gt;I was interested to read the answers to the other Many-Body Problem questions on this site and was left with one nagging question of my own. What does the Many-Body Problem reveal about reality and our attempt to understand it? To explain: if the solutions we find to this problem are in the form of lengthy iterative expansions may this not signify in some way that nature operates along discrete iterative paths? What does this say about our theories if these problems are so difficult/time-consuming for them but so quickly solved by the real world around us? Is this problematic? Will more complete theories-ie possibly string theory- address this or do they even need to? &lt;/p&gt;&#xA;" OwnerUserId="1567" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-27T01:15:00.250" LastActivityDate="2011-02-27T01:15:00.250" Title="What does the Many-Body Problem say about mathematical physics?" Tags="&lt;mathematical-physics&gt;&lt;theory&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="2" />
  <row Id="5924" PostTypeId="2" ParentId="5851" CreationDate="2011-02-26T03:14:42.267" Score="7" ViewCount="" Body="&lt;p&gt;Just to expand on Dr. Motl's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On average - when laser cooling is working properly and cooling the atoms - the photons emitted by the atoms are slightly higher energy than the photons absorbed.  That's where the energy goes.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-02-26T03:14:42.267" />
  <row Id="5925" PostTypeId="2" ParentId="5882" CreationDate="2011-02-26T03:49:16.577" Score="4" ViewCount="" Body="&lt;p&gt;The easiest (and roughest) way to to do it would be to convert your running &quot;work&quot; into a VO2 score. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The American College of Sports Med's equation is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;VO2= Resting Component + Horizontal Component + Vertical Component&#xA;or VO2= 3.5 + (0.2 x Speed) + (0.9 x Speed x Elevation Gain)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, using your example of 8.67 mph (speed in the equation is in meters per min)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.5 + (.2*232.67) + (0.9*232.67*.045) = 59.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus running on flat ground should give you a speed of 280 m/min or 10.44 mph 5 min 44 sec per mile&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm an exercise scientist, not a physicist)&lt;/p&gt;&#xA;" OwnerUserId="2256" LastActivityDate="2011-02-26T03:49:16.577" CommentCount="2" />
  <row Id="5926" PostTypeId="2" ParentId="5914" CreationDate="2011-02-26T06:08:35.973" Score="1" ViewCount="" Body="&lt;p&gt;Dear Moshe, your question seems to demand the answer to be very complicated and unexpected and the obvious answer seems to be forbidden in between the lines except that I think that the obvious answer &lt;em&gt;is&lt;/em&gt; right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Higgs field's vacuum condensate &lt;em&gt;is&lt;/em&gt; the order parameter of the vacuum in the electroweak theory. And if you define it in a gauge-invariant way,&#xA;$$\sqrt{H^\dagger H},$$&#xA;then it is gauge-invariant, too. Obviously, this quantity is still the same $v$ we all know and love. This fact, that the Higgs field is the order parameter, lies behind the fact that the Higgs mechanism uses the same maths as Landau's theory of phase transitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vev can't be measured &quot;directly&quot; as a mass of something but it may still be measured indirectly - e.g. from the masses it gives to other particles (because the couplings may be measured, too).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-26T06:08:35.973" CommentCount="13" />
  <row Id="5927" PostTypeId="1" CreationDate="2011-02-26T06:10:24.613" Score="9" ViewCount="196" Body="&lt;p&gt;Our observable universe, or a subregion of our universe many times larger than the observable universe, originated from inflating from a very tiny inflationary patch. Being so small, the initial entropy has to be bounded by a very small value. But everything in our universe originated from that tiny patch after inflation, reheating, and further expansion. Does that mean almost all entropy in the observable universe is entanglement entropy if we assume the wavefunction of the universe never collapses and is linear? Because of the peculiarities of quantum mechanics, a subsystem can have far greater entropy than the system as a whole. The many worlds interpretation explains why we subjectively observe a far greater entropy, from the vantage of an Everett branch.&lt;/p&gt;&#xA;" OwnerUserId="2258" LastActivityDate="2011-02-26T16:01:00.837" Title="Is almost all entropy in our universe entanglement entropy?" Tags="&lt;cosmology&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="5928" PostTypeId="1" CreationDate="2011-02-26T07:17:02.627" Score="7" ViewCount="401" Body="&lt;p&gt;This question is migrated from this &lt;a href=&quot;http://mathoverflow.net/questions/8846/proofs-without-words&quot; rel=&quot;nofollow&quot;&gt;MO post&lt;/a&gt;. The question is the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Can you give examples of proofs without words? In particular, can you give examples of proofs without words for non-trivial results?&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="159" LastActivityDate="2011-07-06T10:11:05.163" CommunityOwnedDate="2011-02-26T07:20:21.010" Title="Proofs without words" Tags="&lt;big-list&gt;" AnswerCount="8" CommentCount="6" FavoriteCount="2" />
  <row Id="5929" PostTypeId="1" CreationDate="2011-02-26T08:01:40.157" Score="0" ViewCount="73" Body="&lt;p&gt;A physics formulae sheet I published is available&#xA;&lt;a href=&quot;https://docs.google.com/View?docID=0AcrBEUb_lpPmZGZoOWJkOGZfMTc4aG13M2tjZjc&quot; rel=&quot;nofollow&quot;&gt;https://docs.google.com/View?docID=0AcrBEUb_lpPmZGZoOWJkOGZfMTc4aG13M2tjZjc&lt;/a&gt;&#xA;Though the comments are in Swedish (my native language) I hope and ask if you can deliver any comments upon this page. &#xA;Thank you&#xA;Niklas&lt;/p&gt;&#xA;" OwnerUserId="2260" LastActivityDate="2011-02-26T08:01:40.157" ClosedDate="2011-02-26T08:38:03.363" Title="Can you comment on a page with formulas?" Tags="&lt;quantum-mechanics&gt;&lt;newtonian-mechanics&gt;&lt;formula&gt;" CommentCount="3" />
  <row Id="5930" PostTypeId="1" CreationDate="2011-02-26T08:51:05.653" Score="3" ViewCount="115" Body="&lt;p&gt;If you look at most published articles, or those on the preprint server, you find the vast majority of articles have more than one authors. Are collaborations needed to be productive in physics? In other fields like math, single author articles are the norm.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-26T13:24:20.463" Title="Are collaborations needed to be productive in physics?" Tags="&lt;soft-question&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="5931" PostTypeId="2" ParentId="5914" CreationDate="2011-02-26T09:55:42.363" Score="3" ViewCount="" Body="&lt;p&gt;This is a real, and often encountered problem in general in condensed matter. As the comments in Lubos' answer discusses, it's not easy to invent an &quot;order parameter&quot; which unambiguously yields the correct phase since symmetries can't really be broken, and 2nd order correlators tend to display quantum/thermal fluctuations anyway. In condensed matter theory, a frequently used construct is the Off-Diagonal Long Ranged Order, $\lim_{r \rightarrow \infty} \langle \psi(0)\psi(r) \rangle$, which gives a measure of correlation. In practice however, real systems are finite and it is hard to take the limit physically. Anthony Leggett has though long and hard about this problem in the context of cold atoms and quantum condensation, and suggests something based around the single particle density matrix; it is unclear whether this generalises into things like QCD.&lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-02-26T09:55:42.363" />
  <row Id="5932" PostTypeId="1" CreationDate="2011-02-26T10:41:35.453" Score="4" ViewCount="68" Body="&lt;p&gt;One model for confinement in quantum chromodynamics is the &lt;a href=&quot;http://scholar.google.com/scholar?hl=en&amp;amp;q=savvidy&quot; rel=&quot;nofollow&quot;&gt;Savvidy vacuum&lt;/a&gt;. This is a spontaneous symmetry breaking of color gauge symmetry by the gauge fields. The vacuum is divided into Savvidy domains. Is this process hidden by confinement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;L.Motl: I erased one &quot;d&quot; from Savvidy - everywhere. Link to Savvidy vacuum added.&lt;/p&gt;&#xA;" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-26T11:14:55.953" LastActivityDate="2011-02-26T11:14:55.953" Title="Is the color gauge group spontaneously broken in QCD by the Savvidy vacuum?" Tags="&lt;quantum-field-theory&gt;&lt;quantum-chromodynamics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="5933" PostTypeId="2" ParentId="5932" CreationDate="2011-02-26T11:13:52.337" Score="2" ViewCount="" Body="&lt;p&gt;In each Savvidy domain, the chromomagnetic field has a random direction - both in space and the color space. So of course, in a domain, both the SU(3) gauge symmetry as well as the rotational symmetry is broken. However, the domains change randomly, so the average direction both in the real space and the color space is &quot;nothing&quot;; so one can't say that the situation is equivalent to the Higgs mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Savvidy model is a model for confinement - in particular, for the confinement-deconfinement phase transition. In particular, the Polyakov loop - the monodromy or Wilson loop around the Euclidean thermal time direction - is the order parameter that behaves discontinuously near the phase transition. This behavior may be approximately calculated in the Savvidy model.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-26T11:13:52.337" />
  <row Id="5934" PostTypeId="2" ParentId="5879" CreationDate="2011-02-26T11:30:42.903" Score="1" ViewCount="" Body="&lt;p&gt;I will not do all the math for you -- just the general idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a functional $F[g(x_1,x_2,...,x_n)]$. It takes a function of $n$ variables ( I will not write them further) and returns a number. Now you slightly change the argument $F[g+\delta g]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually the value of this functional for small $\delta g$ is approximately:&lt;br&gt;&#xA;$F[g+\delta g] \simeq F[g]+G[g]\delta g$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $G$ is called the functional derivative $G=\frac{\delta F}{\delta g}$ -- compare this to usual deriavative $f(x+\epsilon)\simeq f(x)+\frac{df}{dx}\epsilon$.  So generally you can take a functional derivative over function of any number of variables. An that is the general answer to your general question.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Concerning your example -- there is something strange happening. The expression $\frac{\delta F[b(x,k)]}{\delta b(y,k)}$ just doesn't make any sense to me. Try going step by step in a way I explained. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-02-26T11:30:42.903" CommentCount="1" />
  <row Id="5935" PostTypeId="2" ParentId="5928" CreationDate="2011-02-26T12:16:20.517" Score="2" ViewCount="" Body="&lt;p&gt;John Baez highlighted a nice geometric proof of the spin statistics theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.ucr.edu/home/baez/spin_stat.html&quot; rel=&quot;nofollow&quot;&gt;http://math.ucr.edu/home/baez/spin_stat.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a nice read: Basically he is saying a worldline of a fermion, that loops in time (ie with intermediate pair creation and annihilation) is topologically the same as a 360° rotated fermion (you stretch the worldline until no loop is left).&lt;/p&gt;&#xA;" OwnerUserId="2151" LastActivityDate="2011-02-26T12:16:20.517" CommunityOwnedDate="2011-02-26T12:16:20.517" CommentCount="4" />
  <row Id="5936" PostTypeId="2" ParentId="5930" CreationDate="2011-02-26T12:30:04.593" Score="3" ViewCount="" Body="&lt;p&gt;Nobody seems willing to try and answer this soft question, so I will have a go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is yes. Experimental High Energy Physics is the worst, with 3000 people signing  the papers in the LHC, even most theoretical papers have at least 2 names.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic reason, imo, is that physics has become a discipline where in order to keep everything in your head you have to have special talents, like an eidetic memory. A middle of the road physicist  has to rely on interactions outside his/her head, and what is better than a colleague to thresh out a problem or a proposition. It is a peer review right there and then, so useless ideas can be discarded fast and new ones explored. Efficiency goes up for success in publications. This holds for theorists and disciplines where one could think that one person could carry off the whole paper. It is more productive to discuss the snags with somebody else, before submitting  a publication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of experimental HEP the added burden is the enormous complexity of the experiments that need teams of people coordinated to build detectors and innumerable shifts to get at the data. It is group work, like building a cathedral.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-02-26T12:30:04.593" />
  <row Id="5937" PostTypeId="2" ParentId="5882" CreationDate="2011-02-26T12:48:18.547" Score="2" ViewCount="" Body="&lt;p&gt;Although this is a biomechanics problem it is of interest to see what kind of physics model would apply here. So I shall take the simplest model of movement against friction and gravity. I shall discuss some more aspects of it after some basics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have translated some of the numbers into SI and also will introduce some notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D = 1609 meters - the distance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T = 415 seconds - the time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H = 71.93 ~ 72 meters - the height increase&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V_{av}$ = 3.877 meters per second - the average velocity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta$ = 2.56 degrees - the elevation angle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$cos \theta = 0.999$ and $tan \theta = 0.045$ (the number used in the V02 formula)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the most basic equivalent model is that of pushing a mass M against a friction force $\mu$ for a distance D: level ($\theta=0$) and inclined ($\theta$&gt; 0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation for moving level is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work Done = $WD^{level}$ = Friction $\times$ D = $\mu mgD$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For moving at incline $\theta$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work Done = $WD^{incline}$ = Friction $\times$ D + WD on gravity over H&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= \mu mgD cos \theta + mgD sin \theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= mgD (\mu cos \theta + sin \theta)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we note that $cos \theta$ is appoximately 1 we can write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$WD^{incline} = \mu mgD (1 + (\mu)^{-1} tan \theta) = WD^{level} \times \alpha^2$ say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus moving a fixed distance against a common value of friction the extra work done in going upward is $\alpha^2 = (1 + (\mu)^{-1} tan \theta)$. This is greater than one (as expected) and in our approximation is work against gravity modified by the inverse coefficient of friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we come to the biomechanics. Can we obtain information about the speed increase in a runner (or any powered object)? The parameter we have available here is the coefficient of friction. We also need to assume some model as to how the extra available energy will be used. Let us assume that all of the extra energy is transfered into kinetic energy: so we assume a linear model (in absence of any more detailed model).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work Done $\propto$ Kinetic Energy = $1/2 m V^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case all the extra energy $\alpha^2$ is being added to the energy for the plane giving an increase of velocity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V_{new}^2 = V_{av}^2(1 + (\mu)^{-1} tan \theta)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thus a proportional decrease of time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_{new} = T (1 + (\mu)^{-1} tan \theta)^{-1/2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know the value of $tan \theta$ but do not know $\mu$. Just to put in a first value for $\mu$ which corresponds to rubber on concrete ie 1 we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_{new} = T \sqrt(1+0.045)^{-1} = 415 \times 0.978 = 406$ seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is somewhat slower than the V02 formula and we have not really a valid model of movement because this assumes a &quot;continuous dragging&quot; but it is a first approximation. A different approximation is to consider rubber tyres rolling. The coefficient of friction is defined differently, but could be as low as 0.01. This ignores all the other friction effects in the vehicle however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inverting the formula and putting in the result from the V02 answer (344 seconds), to derive a value for $\mu$ we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu = 0.045 / ((T/T_{new})^2 -1) = 0.045 / (1.455 - 1) = 0.1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this &quot;generalised friction&quot; model predicts a $\mu$ value smaller than most static friction cases and in the direction of a &quot;rolling model&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-26T12:48:18.547" CommentCount="2" />
  <row Id="5938" PostTypeId="2" ParentId="5859" CreationDate="2011-02-26T13:01:07.280" Score="4" ViewCount="" Body="&lt;p&gt;You can do unambiguous state discrimination (with postselection) in quantum mechanics, and this will let you do probabilistic cloning, so the procedure the authors are claiming to perform is not impossible. Unambiguous state discrimination doesn't give you the optimal probabilistic cloner (with postselection); this paper calculates the optimal probabilistic cloner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is unambiguous state discrimination? Suppose you have a set of $k$ linearly independent pure quantum states. There is a POVM measurement whose outputs are either &quot;state $j$&quot; or &quot;I don't know.&quot; If the input is state $j$, then the output is &quot;state $j$&quot; with some probability $p_j$ and &quot;I don't know&quot; with probability $1-p_j$. On the &quot;I don't know&quot; outputs, the input state is destroyed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unambiguous state discrimination can be used to build a probabilistic cloner: if the output is &quot;state $j$,&quot; prepare an additional copy of state $j$. Postselected on the output not being &quot;I don't know,&quot; you have cloned the input state. However, this is presumably not optimal; the optimal cloner should only be able to produce two copies of a state (the intuition for this is that otherwise it is extracting more quantum information from the state than it really needs to clone). This process can produce arbitrarily many copies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your question about equation 7 in arxiv quant.ph/9704020 (the PRL version has a different numbering), if you replace $|\Phi_{AB}\rangle$ with a state that will let you restore $|\Psi_0\rangle$ and $|\Psi_1\rangle$, as suggested in Frederic's comment, the operator $U$ is no longer unitary. If it were, you would have a unitary operator that doesn't preserve inner products, which is impossible.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditorDisplayName="" LastEditDate="2011-02-26T13:25:06.733" LastActivityDate="2011-02-26T13:25:06.733" CommentCount="2" />
  <row Id="5939" PostTypeId="1" CreationDate="2011-02-26T13:12:23.910" Score="2" ViewCount="155" Body="&lt;p&gt;It seems that virtual photons also exist in vacuum. So the precise question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the additional virtual photon density due to the electric field of a unit charge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or: How many virtual photons per volume are found around a unit charge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer will depend on distance, but what are the exact numbers?&lt;/p&gt;&#xA;" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-03-06T20:08:57.150" LastActivityDate="2011-03-06T20:08:57.150" Title="What is the density of virtual photons around a unit charge?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;electrodynamics&gt;&lt;quantum-electrodynamics&gt;&lt;virtual-photons&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="5940" PostTypeId="2" ParentId="5851" CreationDate="2011-02-26T13:13:15.413" Score="10" ViewCount="" Body="&lt;p&gt;There are two thermodynamic aspects to laser cooling that are worth mentioning. The first, as others have noted, has to do with the frequency of the light that is absorbed and emitted. In Doppler cooling, the laser is tuned slightly below the frequency that the laser wants to absorb. An atom moving toward the laser sees that light shifted slightly up in frequency, and is thus more likely to absorb it than an atom at rest or moving away from the laser is. When it absorbs the light, it also picks up the momentum associated with that photon, which is directed opposite the momentum of the atom (since the atom is headed in the opposite direction from the light), and thus slows down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the atom spontaneously emits a photon a short time later, dropping back to the ground state, the emitted photon has exactly the resonant frequency of the atomic transition (in the atom frame). That means that its frequency is a little higher (in the lab frame) than the frequency of the absorbed light. Higher frequency means higher energy, so the photon has absorbed a low-energy photon and emitted a high-energy photon. The difference in energy between the two has to come from somewhere, and it comes out of the kinetic energy of the moving atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You might reasonably ask what happens with the momentum of the emitted photon; the emission process also gives the atom a kick in the direction opposite the emitted photon's direction. If this happens to be exactly the same as the direction of the laser, the resulting kick sends the atom back to the same initial velocity; however, the direction of spontaneous emission is random, and as a result, it tends to average out over many repeated absorption-emission cycles. On average, the atom loses one photon's worth of momentum with each cycle, and a corresponding amount of kinetic energy.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other aspect, which you didn't ask about, but is worth mentioning, is the entropy. On the surface, it might seem that even though energy is conserved, there would be a thermodynamics problem here because the system moves from a higher entropy (lots of fast-moving atoms) to a lower entropy (lots of slow-moving atoms). The difference is made up in the entropy of the light field. Initially, you have a single monochromatic beam of photons, all with the same frequency, all headed in the same direction, which is about as low-entropy as it gets. After the cooling, those photons have been scattered out into all different directions, and with a range of frequencies, so the entropy of the light is much greater, and more than makes up for the decrease in the entropy of the atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's probably way more detail than you wanted/needed, but there you go.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2011-02-26T13:13:15.413" CommentCount="2" />
  <row Id="5941" PostTypeId="2" ParentId="5930" CreationDate="2011-02-26T13:24:20.463" Score="5" ViewCount="" Body="&lt;p&gt;There are two things at work here: First, physics has advanced to the point where all the interesting problems are sufficiently complex that they almost require multiple people working on them with different areas of expertise. It's not impossible to do publishable physics research as a single individual-- it's more common in theory than experiment, but you do occasionally see single-author experimental papers. (Not in high-energy physics, but high-energy physics isn't the whole of physics by any stretch.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second factor is that modern physics is relatively generous about co-authorship. There are other disciplines where you have to have been far more involved in every aspect of the work to get author credit than you do in physics, where contributing to a sub-part of the data analysis will often get authorship credit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This hasn't always been the case. If you look up some classic papers from the early 20th century, you will often find people thanked in the text for their invaluable contributions; contributions that, when described, make it sound like the person thanked really should've been listed as an author, by modern standards.&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2011-02-26T13:24:20.463" CommentCount="1" />
  <row Id="5942" PostTypeId="1" CreationDate="2011-02-26T14:11:57.020" Score="3" ViewCount="260" Body="&lt;p&gt;Consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_teleportation&quot; rel=&quot;nofollow&quot;&gt;quantum state teleportation protocol&lt;/a&gt; of Bennett et. al.&#xA;How does one prove that this protocol would never violate the conservation of energy? At the face of it, it doesn't seem to be something obvious, as the various measurements, communications, rotations etc. don't seem to be able to account for the differences in the Hamiltonians of Alice's and Bob's labs. The state $\psi$ might have an expected energy $E_1$ in Alice's lab and a completely different expected energy $E_2$ in Bob's. Where does the difference come from?&lt;/p&gt;&#xA;" OwnerUserId="2186" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-26T16:53:16.793" LastActivityDate="2011-08-25T18:34:42.070" Title="Conservation of energy in quantum teleportation" Tags="&lt;energy-conservation&gt;&lt;quantum-information&gt;&lt;entanglement&gt;&lt;teleportation&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="5943" PostTypeId="2" ParentId="5928" CreationDate="2011-02-26T14:14:06.300" Score="7" ViewCount="" Body="&lt;p&gt;Here is a graphical proof that no regular (translation invariant) lattice with pentagonal symmetry (and thus no corresponding crystal) exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/eGTQe.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://web.me.com/paulscott.info/lattice-points/5regular.html&quot; rel=&quot;nofollow&quot;&gt;Source of picture&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="1782" LastEditorDisplayName="" LastEditDate="2011-02-26T16:29:35.753" LastActivityDate="2011-02-26T16:29:35.753" CommunityOwnedDate="2011-02-26T14:14:06.300" CommentCount="3" />
  <row Id="5944" PostTypeId="1" CreationDate="2011-02-26T14:34:12.560" Score="4" ViewCount="248" Body="&lt;p&gt;I will begin my research on ads/cmt,however I find ads/cmt is only a phenomelogical method,&#xA;so I want to know can ads/cmt give some results the condensed matter physicists can not give,or even predeict some unseen exotic matter states? &lt;/p&gt;&#xA;" OwnerUserId="2268" LastActivityDate="2011-03-02T04:43:37.873" Title="How much can ads/cmt tell us?" Tags="&lt;string-theory&gt;&lt;ads-cft&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="5945" PostTypeId="2" ParentId="5939" CreationDate="2011-02-26T14:41:17.197" Score="2" ViewCount="" Body="&lt;p&gt;Apparently you imagine a charge as a point with a Coulomb field around it. Outside the charge there is no charge but there is a field, you say, and it consists of virtual photons. So how many of them are at the distance $r$ from the charge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I let the others answer this question and here I will give you my vision of that. The charge is not point-like but quantum mechanically smeared. It does not stay at a point but &quot;moves&quot; and creates a quantum mechanical &quot;&lt;a href=&quot;http://arxiv.org/abs/0806.2635&quot; rel=&quot;nofollow&quot;&gt;cloud&lt;/a&gt;&quot;. This &quot;cloud&quot; is &quot;large and soft&quot; (like a jelly) - you push the charge and the whole system gets deformed inelastically. You break the original state and the new state is a moving now charge and excited quantum oscillators (oscillating cloud). The latter describes photons.  So normally the initial state is different from the final state. In experiment they usually add together all particular &quot;pictures&quot; of all scattering events and obtain an &quot;inclusive&quot; picture. It is the inclusive picture that corresponds to the point-like image of the charge.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-26T15:36:27.823" LastActivityDate="2011-02-26T15:36:27.823" CommentCount="4" />
  <row Id="5946" PostTypeId="2" ParentId="5879" CreationDate="2011-02-26T15:07:37.070" Score="1" ViewCount="" Body="&lt;p&gt;Jane,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem here might be with the exact form of this expression, which seems to have free and bound variables mixed up. Here is an alternative (the problem is with the k).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac{\delta F[b(x,k)]}{\delta b(z,k')} = a(z,y,k')e^{\int_0^{x'} \int_0^p  dx dk a(x,y,k) b(x,k )}&#xA;\end{equation}&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-26T15:07:37.070" CommentCount="6" />
  <row Id="5947" PostTypeId="2" ParentId="5942" CreationDate="2011-02-26T15:14:07.877" Score="0" ViewCount="" Body="&lt;p&gt;It appears the Bob and Alice have to use identical systems in order to successfully transmit Bob's state $|\phi\rangle$ to Alice. After transmission (I prefer this term over &quot;teleportation&quot;) Alice's state is also $|\phi\rangle$, so I don't see where the question of conservation of energy arises.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does matter is that in order to &quot;teleport&quot; one qubit between two classical observers, one maximally entangled Bell pair is required. This pair is not entangled any more after the state is transmitted. In order to teleport $X$ qubits, one therefore needs $X$ Bell pairs. Work is done every time a entangled pairs are created from a reservoir of unentangled particles and thus an energy cost is associated with each Bell pair and consequently with each qubit that is teleported.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is a consideration of energy but in regards to the cost of the resource which allows the process to occur. But energy is conserved locally at every step of the protocol as @Peter Shor mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also see &lt;a href=&quot;http://arxiv.org/abs/1002.0200&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; by Masahiro Hotta for some nice ideas on how one can teleport energy between two systems in this manner.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-26T15:14:07.877" />
  <row Id="5948" PostTypeId="2" ParentId="5939" CreationDate="2011-02-26T15:20:08.907" Score="2" ViewCount="" Body="&lt;p&gt;I do not think that this question has an answer. A photon is a quantum mechanical object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) there is no conservation of photons, virtual or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b)there is no lower limit to the energy of the photons, so in principle they are infinite ( infrared problem)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c)The energy of the virtual photons will depend on the motion of the charge and or the probe that will be checking for virtual photons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can have an energy spectrum for virtual photons, but not a density estimate . See this caluclation &lt;a href=&quot;http://prc.aps.org/abstract/PRC/v27/i2/p523_1&quot; rel=&quot;nofollow&quot;&gt;for example.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-02-26T15:30:30.943" LastActivityDate="2011-02-26T15:30:30.943" CommentCount="2" />
  <row Id="5949" PostTypeId="2" ParentId="5939" CreationDate="2011-02-26T15:50:04.480" Score="3" ViewCount="" Body="&lt;p&gt;It's infinity. This is the soft photon problem, which requires infrared regularization.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-02-26T15:50:04.480" CommentCount="1" />
  <row Id="5950" PostTypeId="1" CreationDate="2011-02-26T15:56:07.267" Score="5" ViewCount="187" Body="&lt;p&gt;If we can think about the universe as a wave function then many particles should be entangled with many other particles in the universe. The obvious question arises why we don't see those entanglements in everyday circumstances. One standard explanation given is those entanglements average out and cancel so we can ignore those. However, hardly any mathematical justification is given for them to cancel. My question is how much trust one should have on that particular assertion? Is there any mathematical arguments already put forward by anyone?&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-26T16:24:07.317" LastActivityDate="2011-02-27T16:09:26.013" Title="Justification of ignoring large set of entanglements" Tags="&lt;quantum-mechanics&gt;&lt;entanglement&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" />
  <row Id="5951" PostTypeId="2" ParentId="5928" CreationDate="2011-02-26T16:00:39.103" Score="4" ViewCount="" Body="&lt;p&gt;I think that Feynman diagrams are, in a sense, &quot;proofs without words for non-trivial results.&quot; Strictly speaking, they are not proofs, but they are a way of generating true statements about interactions, plus, they are very artistic.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-02-26T16:00:39.103" CommunityOwnedDate="2011-02-26T16:00:39.103" />
  <row Id="5952" PostTypeId="2" ParentId="5927" CreationDate="2011-02-26T16:01:00.837" Score="2" ViewCount="" Body="&lt;p&gt;One also has to take account of the peculiarities of General Relativity as well when discussing the Entropy of the Universe. In fact the Entropy derives from gravitational clumping, and ultimately the formation of Black Holes with all matter fallen in. The early universe could be considered as low Entropy because of a general evenness of matter, which later became more clumped into Galaxies, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantum aspect of this question seems to relate to whether the Universe is primarily &quot;classical&quot; - with lots of decoherence or (if you dont believe that &quot;collapse&quot;) having taken place; or primarily quantum. My understanding was that most physicists considered it to be primarily classical, but there are some unresolved issues of what happened before the time of last scattering.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-26T16:01:00.837" />
  <row Id="5953" PostTypeId="2" ParentId="5899" CreationDate="2011-02-26T16:14:56.587" Score="1" ViewCount="" Body="&lt;p&gt;Another angle to consider here is investigation of these &quot;water structuring facilities&quot; themselves. I dont know whether one is dealing here with something mobile which is itself purchased for household use, or maybe for community use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way I see that there are references in the comments to the water tasting different - citrus-like. This suggests:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Does the machine use electrical power? If so what is powered: a traditional water-cooler powers the valve and fridge to cool the water. Does it have a powered motor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Does it overtly use extra materials? Thus does it need resupplying every so often with something? If so what?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Does it covertly use extra materials? If it is bogus then it may have some filter material (which will eventually run out) which is supplying this extra taste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is no obvious answer to the above, then try to identify what causes the taste difference. Is the water supplied separately (as happens with water coolers)? Also how is the water flowing through the machine?&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-26T16:14:56.587" CommentCount="2" />
  <row Id="5954" PostTypeId="1" AcceptedAnswerId="5966" CreationDate="2011-02-26T16:42:54.110" Score="0" ViewCount="127" Body="&lt;p&gt;My question refers to an overview of the biggest stars we know: &lt;a href=&quot;http://farm5.static.flickr.com/4138/4820647230_faba1c9f3b_o.jpg&quot; rel=&quot;nofollow&quot;&gt;http://farm5.static.flickr.com/4138/4820647230_faba1c9f3b_o.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are some of those blue?&lt;/p&gt;&#xA;" OwnerUserId="2269" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-27T01:27:43.917" LastActivityDate="2011-02-27T01:27:43.917" Title="Why are some of the biggest stars known blue?" Tags="&lt;heat&gt;&lt;color&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5955" PostTypeId="1" CreationDate="2011-02-26T16:49:06.220" Score="5" ViewCount="173" Body="&lt;p&gt;Seiberg traced the nonrenormalization of supersymmetric theories to holomorphy of the superpotential in chiral superspace. However, this overlooks the fact that with a different number of supersymmetric generators, supersymmetry can be real or symplectic, instead of complex. But yet, even for those cases, we still have nonrenormalization. If holomorphy isn't the reason, what is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the reason for nonrenormalizability is totally different for different dimensions and number of supersymmetry generators and choice of superfields, is it then &lt;em&gt;ad hoc&lt;/em&gt; that nonrenormalizability always holds in each case?&lt;/p&gt;&#xA;" LastActivityDate="2011-03-04T16:55:39.830" Title="Is holomorphy the real reason for nonrenormalization in supersymmetry?" Tags="&lt;supersymmetry&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="5956" PostTypeId="2" ParentId="5950" CreationDate="2011-02-26T16:57:30.433" Score="0" ViewCount="" Body="&lt;p&gt;Most of the time, we have decoherence, which suppresses interferences between configurations with different values for the pointer basis. On the other hand, showing a pointer basis always exists is an open problem. According to the Everett interpretation, each pointer basis splits into a parallel world of its own. Or at least, it can be treated as if it did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if some day in the future, we construct a quantum computer implementing Shor's factorization algorithm for integers with a huge number of digits, we can have a complicated multipartite entanglement between many qubits which eventually interfere constructively and destructively to get a nontrivial result.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-02-26T17:07:21.767" LastActivityDate="2011-02-26T17:07:21.767" CommentCount="1" />
  <row Id="5957" PostTypeId="2" ParentId="5950" CreationDate="2011-02-26T17:07:44.037" Score="6" ViewCount="" Body="&lt;p&gt;Dear sb1, the assumptions implicitly contained in your question are upside down. Exactly the fact that the degrees of freedom of the Universe are becoming increasingly entangled &lt;em&gt;means&lt;/em&gt; that we &lt;em&gt;cannot&lt;/em&gt; observe the quantum phenomena in practice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The entanglement that emerges after a relatively short time is the entanglement among a huge number of degrees of freedom. So to detect that it is real, one would have to make an accurate and correlated measurement of &lt;em&gt;all&lt;/em&gt; these entangled degrees of freedom that are relevant, and this is impossible in practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one only studies a subsystem where some degrees of freedom is missing - because they become effectively undetectable - it is enough to &quot;trace over&quot; the unobservable degrees of freedom and to predict the observable system via its density matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Roy Simpson mentioned, the process of &lt;em&gt;decoherence&lt;/em&gt; - which is extremely well understood as of today - guarantees that the density matrix for an observable system with many degrees of freedom - one that interacts with the environment (the unobservable degrees of freedom) - rapidly becomes diagonal, so observable &lt;em&gt;interference&lt;/em&gt; goes to zero and quantum physics de facto reduces to classical physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens with the entanglement? Well, the entanglement between the observable and unobservable degrees of freedom is still there but it is unobservable because the unobservable degrees of freedom are unobservable. And when it comes to the entanglement between various observable degrees of freedom only, the part of it that has decohered disappears (and classical behavior takes over), and the part that hasn't decohered is still there (and quantum mechanics is still important).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think that instead of &lt;em&gt;entanglement&lt;/em&gt;, you could have equally well used the term &lt;em&gt;correlation&lt;/em&gt; because entanglement is nothing else than correlation that may use all the quantum information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even in classical physics, it would be true that the Universe - e.g. all its atoms - are &quot;chaotically&quot; evolving into a state in which the velocities and positions of all the particles are correlated in complicated ways (imagine a description using a distribution function on the phase space; as in reality, you can't quite measure and know the exact point on the phase space). Yes, it is true: evolution makes things complex. But exactly because they are so complex, the correlations become &lt;em&gt;unobservable&lt;/em&gt;; your question incorrectly assumes that the complicated character should make the correlations more &lt;em&gt;observable&lt;/em&gt;). The same is true for the entanglement: the convoluted character of entanglement makes it unobservable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the very end, let me emphasize that a diagonal density matrix is technically &lt;em&gt;entangled&lt;/em&gt;, indeed. It's because this matrix is not a tensor product of a pure state vector with its conjugate: whenever this factorization is impossible, entanglement is nonzero. If the density matrix has many nonzero entries on the diagonal, it is a mixed state containing many independent contributions from pure states. That doesn't mean that this &quot;entangled&quot; character of the density matrix is &quot;highly quantum&quot; i.e. that it should exhibit many striking quantum properties: quite on the contrary, by getting a nearly diagonal density matrix, we get a highly classical situation!&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-26T17:07:44.037" />
  <row Id="5958" PostTypeId="2" ParentId="5954" CreationDate="2011-02-26T17:16:29.037" Score="1" ViewCount="" Body="&lt;p&gt;I guess that some or many planets of this kind are similar to Uranus but I can't guarantee that the chemical composition below is universal for blue planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranus' atmosphere is composed of hydrogen, helium, and methane. Methane in the upper atmosphere is what matters: it absorbs red light from the Sun but reflects the blue light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth is blue mostly because of the oceans of liquid water that have a similar optical property as methane on Uranus.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-26T17:16:29.037" />
  <row Id="5959" PostTypeId="2" ParentId="5955" CreationDate="2011-02-26T17:24:57.500" Score="4" ViewCount="" Body="&lt;p&gt;First of all, when Seiberg's papers say that a non-renormalization theorem may be proved by holomorphy, be sure that Seiberg is right. But it doesn't mean that holomorphy is the only argument that may become crucial in a proof of such theorems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a complex spinor may decompose into real or pseudoreal spinors in a lower dimension but some of the non-renormalization theorems may still apply. Also, the BPS states (annihilated by a subset of supercharges) have masses that are totally dictated by the &quot;central charges&quot; and don't receive any quantum corrections; this argument is independent of holomorphy. It simply follows from the vanishing of some $QQ$ bilinears in the state, which - by SUSY algebra - may be rewritten as a difference between the energy/mass and some central charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other proofs of non-renormalization theorems may involve perturbative arguments that things cancel at each order because SUSY prohibits the &quot;wrong terms&quot; in each case, because of some dimensional analysis. In principle, those arguments - although they are familiar in the holomorphic context - don't depend on holomorphy, either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is misleading to ask about all non-renormalization theorems simultaneously and expect that there is a single word, such as holomorphy, that contains proofs to all of them. Different non-renormalization theorems have different proofs that use different ideas: the required ideas depend not only on the dimension but also on the precise SUSY algebra, precise theory, and even the precise quantity whose non-renormalization we're proving. And in each case, there may exist several highly inequivalent proofs of the same proposition. You would have to ask more specific questions to get more specific answers.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-26T17:24:57.500" />
  <row Id="5960" PostTypeId="2" ParentId="5955" CreationDate="2011-02-26T17:26:25.080" Score="5" ViewCount="" Body="&lt;p&gt;The real reason is, of course, supersymmetry. You can see it in various ways: some quantities don't renormalize since potential divergences cancel between bosonic and fermionic contributions. Or, another way to say it is that potential counterterms are not supersymmetric. This goes beyond renormalizing the action, for example with extended SUSY BPS states get no mass corrections for reasons very similar to non-renormalization theorems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For certain quantities (superpotential, gauge coupling function, prepotential in N=2) &lt;em&gt;holomorphy&lt;/em&gt; is an efficient way to summarize the various restrictions coming from supersymmetry, but it is not the underlying reason for them. Certainly for extended SUSY theories there are more restrictions due to SUSY, not less, but since there may not be a superspace formulation holomorphy is typically not part of the story there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your edited question: with different type of supersymmetry (different number, dimensions etc.) different quantities are simplified in different ways. There is no &quot;unified theory of non-renormalization&quot; in all supersymmetric theories, regardless of context. But, each non-renormalization theorem holds for a whole class of theories and quantities, so I would not describe them as &quot;ad hoc&quot;.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-02-26T18:17:19.130" LastActivityDate="2011-02-26T18:17:19.130" CommentCount="2" />
  <row Id="5961" PostTypeId="1" CreationDate="2011-02-26T17:45:00.517" Score="1" ViewCount="118" Body="&lt;p&gt;How can I find the magnetic interference a stationary 35000 kg block of 100% pure iron would have on a magnetic compass and what the drop off rate of the interference would be.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if said 35000 kg block of iron was 1 meter away from the compass, 100 meters away, or 1000 kilometers way I would like to calculate the rate of drop off of the interference.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This may seem absurd, but it is very important for a conceptual project I am working on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the context of this question, assume everything is perfect, and that we are basically operating in a vacuum and there is no interference from anything else and that all instruments are 100% accurate and infinitely precise. And that I have only a very very basic understanding of physics, mathematics and magnetism.&lt;/p&gt;&#xA;" OwnerUserId="2271" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-02-26T18:28:28.527" LastActivityDate="2011-06-28T07:51:51.980" Title="Find drop-off rate of magnetic interference from a mass of pure iron on a magnetic compass" Tags="&lt;magnetism&gt;&lt;interference&gt;" AnswerCount="2" />
  <row Id="5962" PostTypeId="2" ParentId="5950" CreationDate="2011-02-26T18:15:05.353" Score="4" ViewCount="" Body="&lt;p&gt;The statement that the entanglement with many other objects &quot;cancels out&quot; is a little misleading, giving the impression that the various entanglements somehow undo each other, so they don't affect the final state. In fact, though, all of the various entanglements between a simple system of interest and the objects in its environment will play a role in determining the exact probability of the various possible states, and thus determining the outcome of the measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What relieves us from having to track every single possible entanglement (which would be completely impractical) is that the effects of entanglement with the environment are random and fluctuating. On each repetition of the experiment, the system of interest will interact with a slightly different set of environmental objects, in slightly different ways, so the effects on the wavefunction will be different every time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since quantum mechanics only allows us to predict probabilities, the only way we can check theories by experiment is to make many repeated measurements of our system of interest, and use those repeated measurements to determine the probability of the various measurement outcomes. The random and fluctuating nature of the environmental entanglement means that while the details of the interaction with the environment plays a role in determining the exact probability for any given measurement, the effect of those interactions is different from one measurement to the next. Which means that, when you repeat the measurement many times to get a probability distribution, what you see is a kind of average over all the different ways that the environmental entanglement affects the various different runs of the experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give a highly imperfect analogy, it's a little like statistical mechanics or thermodynamics. While in principle the evolution of a thermodynamic system-- the canonical box full of ideal gas, say-- depends on the motion of every single particle making it up, in practice, the only thing we can measure is the average effect of uncountably huge numbers of moving particles. Thus, we have a well-established mathematical apparatus for figuring out the statistical behavior of all those particles, which gives us the highly successful fields of thermodynamics and statistical mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When people say that the effects of multiple entanglements &quot;average away,&quot; the implied process is sort of similar. It assumes that the experiment will need to be run a very large number of times, and so what you calculate in the end is a sort of average over all the possible effects, in the same way that, say, the calculated temperature change of an expanding gas is an average over the change in speed of the umpty-zillion particles making up that gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is not a precise correspondence, so don't try to push it too far. It's just supposed to get the basic idea of the underlying process.)&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2011-02-26T18:15:05.353" CommentCount="3" />
  <row Id="5963" PostTypeId="1" AcceptedAnswerId="5965" CreationDate="2011-02-26T18:34:55.537" Score="4" ViewCount="123" Body="&lt;p&gt;This is about continuity equation. What does the last integral mean? &#xA;$$\frac{\mathrm{d}Q_V}{\mathrm{d}t}=\iiint_V \mathrm{d}^3x \,\frac{\partial\rho}{\partial t}=-\iiint_V\! \mathrm{d}^3x\,\operatorname{div}\,\mathbf{j}=-\iint\limits_{\partial V}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\;\;\;\subset\!\supset \mathbf j\;\cdot\mathbf n\,{d}S\,,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;And what does $\partial V$ mean? Why not S like surface?&lt;/p&gt;&#xA;" OwnerUserId="534" LastEditorUserId="534" LastEditorDisplayName="" LastEditDate="2011-02-26T18:48:48.640" LastActivityDate="2011-02-26T19:05:21.553" Title="meaning of an integral in the continuity equation" Tags="&lt;electromagnetism&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5964" PostTypeId="2" ParentId="5963" CreationDate="2011-02-26T18:38:14.730" Score="4" ViewCount="" Body="&lt;p&gt;Note that the range of integration has changed from $V$ meaning over the volume to $\partial{V}$ and that the variable of integration has changed from $\text{d}^3x$ to $\text{d}S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It means an integral over the surface that bounds the volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closed curve over the integration signs implies that the surface must be closed.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This expression is an application of what physicist tend to call &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%27s_law&quot; rel=&quot;nofollow&quot;&gt;Gauss's Law&lt;/a&gt;, and mathematicians tend to know as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Divergence_theorem&quot; rel=&quot;nofollow&quot;&gt;divergence theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-02-26T18:48:11.677" LastActivityDate="2011-02-26T18:48:11.677" />
  <row Id="5965" PostTypeId="2" ParentId="5963" CreationDate="2011-02-26T18:40:48.923" Score="8" ViewCount="" Body="&lt;p&gt;The last integral is a surface integral. It is the fluid (current) flux crossing the volume surface integrated over the whole surface. It is how much charge is leaving the volume per second. ($\partial V$ means boundary of V and you sum all local $j_\bot dS$.)&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-26T19:05:21.553" LastActivityDate="2011-02-26T19:05:21.553" CommentCount="2" />
  <row Id="5966" PostTypeId="2" ParentId="5954" CreationDate="2011-02-26T18:52:08.493" Score="6" ViewCount="" Body="&lt;p&gt;dmckee's right: the picture you link to shows stars, not planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The color of a star is almost entirely determined by its temperature. The light coming from a star is, to a good approximation, blackbody radiation (except for absorption lines in its spectrum, which are very important tools for learning about the star but have little effect on its color). The spectrum of a blackbody depends on its temperature, in such a way that it shifts from longer to shorter wavelengths as the temperature increases. So hot stars look blue, and cool stars look red. Even cooler objects, such as you, don't glow significantly at all in the visual part of the spectrum but do in the infrared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment, we know little or nothing about the colors of planets other than those in our solar system. Extrasolar planets are detected indirectly, via their effect on the star they orbit. They are not yet seen directly themselves.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-26T18:52:08.493" CommentCount="0" />
  <row Id="5967" PostTypeId="2" ParentId="5944" CreationDate="2011-02-26T19:11:46.773" Score="0" ViewCount="" Body="&lt;p&gt;I'm not an expert on CMT applications of AdS/CFT aka gauge/string duality so you might get a better answer from such an expert. The idea of applying this formalism to condensed matter systems is quite recent and most of the work has been done in the last 1 1/2 years. So far I don't think there are any spectacularly new results of the type you are asking about, but I think many people hope there will be eventually. Certainly this is a new technique for studying strongly coupled condensed matter systems so in principle it may allow you to understand things that other techniques do not and perhaps even predict some exotic state of matter. Presumably one of the reasons you are starting research in this area is to answer such questions. Perhaps you will report back and answer your own question in a few years.&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T04:43:37.873" LastActivityDate="2011-03-02T04:43:37.873" />
  <row Id="5969" PostTypeId="2" ParentId="5944" CreationDate="2011-02-26T19:38:50.870" Score="2" ViewCount="" Body="&lt;p&gt;I've worked some on AdS/CMT so I can discuss a bit some of my motivation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Jeff points out, we have now a fundamentally new technique to studying strongly coupled systems, so it makes sense to use it to attack any problem under the sun that needs new tools. At our current level of understanding, as you point out, the best we can hope for is constructing phenomenological models for interesting physical phenomena. Part of the price to pay for holography is that the usual description (say in terms of band electrons interacting in a specific way) gets highly scrambled, as it should for a more physical description of strongly coupled systems. But, it is then hard to identify specific models for specific materials,  and one has concentrate on qualitative and semi-quantitative physics. I think this is the nature of the beast and currently there is no way around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I think that CMT is particularly well-suited for this approach, precisely because the phenomenological models are normally the best you can hope for in such inherently complicated problem. For example, when people work on the Hubbard model it is because it is expected to capture some of the features important in high Tc superconductivity, not because it is in any way related specifically to any real material. I can certainly see AdS/CMT reaching this level of usefulness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, along those lines I'd say the best that has come from AdS/CMT so far is a fairly natural description of superconductivity, a large class of quantum phase transitions, and some work on non-Fermi liquids which seems to be related to strange metal phenomenology (e.g all the marginal Fermi surface phenomenology and some indication of the right scaling of transport properties). Probably better results are yet to come, which is why this is a good subject to be working on.&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-02-26T19:38:50.870" CommentCount="4" />
  <row Id="5970" PostTypeId="2" ParentId="5928" CreationDate="2011-02-26T19:52:19.183" Score="6" ViewCount="" Body="&lt;p&gt;The epigraph of Stevinus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/VcQb9.png&quot; alt=&quot;The epigraph of Stevinus&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Vol I of the Feynman lectures.  Since the chain of balls does not go around, we know that a the weights on the following picture are balanced&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Fpzi0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-02-26T19:52:19.183" CommunityOwnedDate="2011-02-26T19:52:19.183" />
  <row Id="5971" PostTypeId="1" CreationDate="2011-02-26T20:04:10.570" Score="4" ViewCount="92" Body="&lt;p&gt;First of all, what is the mathematical relationship between measured linewidth (usually in units of magnetic field) and spin relaxation time? I see papers talk about spin relaxation times in terms of linewidths but I have no idea how to correlate the size of the linewidth to an actual time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, is the measured linewidth in NMR or ESR experiments related to $T_2^{\ast}$ only. If you want just $T_2$, you would have to do spin-echo, right? If that is the case, how is $T_1$ determined?&lt;/p&gt;&#xA;" OwnerUserId="1939" LastActivityDate="2011-02-27T10:20:28.360" Title="Relationship between magnetic resonance linewidth and spin relaxation" Tags="&lt;condensed-matter&gt;&lt;spin&gt;&lt;magnetism&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="5972" PostTypeId="2" ParentId="5961" CreationDate="2011-02-26T20:10:23.610" Score="1" ViewCount="" Body="&lt;p&gt;Specific density of iron is 7.87 times water, so 35,000 kg of iron = 35/7.87 = 4.45 cubic meters. Assuming the block is spherical, this is a sphere with radius 1.02 meters. So I'm assuming you mean 1 meter from the surface of the block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At these distances you can more or less approximate the iron sphere as a bar magnet. In terms of dependency on distance, it acts like a dipole. From the wikipedia article &quot;Dipole&quot; &lt;a href=&quot;http://en.wikipedia.org/wiki/Dipole&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dipole&lt;/a&gt; you have&#xA;&lt;img src=&quot;http://i.stack.imgur.com/BqQs7.png&quot; alt=&quot;dipoleapprox&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;B is the strength of the field, measured in teslas&#xA;r is the distance from the center, measured in metres&#xA;λ is the magnetic latitude (equal to 90° − θ) where θ&#xA;    is the magnetic colatitude, measured in radians or degrees from the dipole axis&#xA;m is the dipole moment (VADM=virtual axial dipole moment),&#xA;   measured in ampere square-metres (A·m2), which equals joules per tesla&#xA;μ0 is the permeability of free space, measured in henries per metre.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the strength depends on the angle $\lambda$ from the axis, and drops off as $1/r^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that at 100 meters away the strength will be about 1,000,000 times less, and at 1000km = $10^6$ meters, it will be reduced by a factor of $10^{18}$.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-26T20:10:23.610" CommentCount="1" />
  <row Id="5973" PostTypeId="1" AcceptedAnswerId="5974" CreationDate="2011-02-26T20:34:42.757" Score="1" ViewCount="285" Body="&lt;p&gt;Why does an object with smaller mass hits the ground at same time compared to object with greater mass? I understand the acceleration due to gravity of earth will be same but won't the object with greater mass will fall faster?&lt;/p&gt;&#xA;" OwnerUserId="2273" LastActivityDate="2011-02-27T06:21:27.730" Title="Free falling of object with no air resistance" Tags="&lt;gravity&gt;&lt;acceleration&gt;&lt;free-fall&gt;" AnswerCount="3" />
  <row Id="5974" PostTypeId="2" ParentId="5973" CreationDate="2011-02-26T20:45:53.887" Score="4" ViewCount="" Body="&lt;p&gt;That is an excellent example for a nice quote I read on the internet: &quot;Common sense may be common, but it certainly isn't sense&quot; :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it is hard to lift heavy objects, we assume that it must be easier for them to drop. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, Newton's laws point out that light and heavy objects will fall with the same velocity. But is there an intuitive reason? Yes!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of an object contributes to two different phenomena: Gravity and inertia.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The heavier an object is, the stronger the gravitational pull it experiences.&lt;/li&gt;&#xA;&lt;li&gt;The heavier an object is, the stronger its resistance to an accelerating force will be: Heavier objects are harder to set in motion, meaning that for the same acceleration you need a larger force.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When people think that heavy objects should fall faster, they only think of the first point. But in reality, the first and second point cancel out each other: Yes, the earth pulls stronger on a heavy object, but the heavy object is more reluctant to get moving.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-02-26T20:45:53.887" CommentCount="3" />
  <row Id="5975" PostTypeId="2" ParentId="5973" CreationDate="2011-02-26T20:50:08.283" Score="2" ViewCount="" Body="&lt;p&gt;If you understand that the acceleration is the same for both objects, then it's the same as saying that they'll fall at the same rate. Acceleration=Rate of Change of Velocity = Rate of Change of Rate of Change of Position. So, if they start at the same point and with the same initial speed and if they have the same acceleration then their velocities and positions change at the same rate and hence they'll hit the ground at the same time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of understanding this is famous thought experiment(by Galileo I think): Assume that there is a difference in the time taken to fall of two bodies with different masses. Now, take two bodies of the same mass and measure the time taken for them to hit the ground. Tie them together. This can be treated as one body of twice the mass. Then that means it will take a different time to hit the ground. But tying them together shouldn't have made any difference as it can thought of as putting the the two bodies very close to each other and then dropping them. Hence, all bodies fall at the same rate.&lt;/p&gt;&#xA;" OwnerUserId="2186" LastActivityDate="2011-02-26T20:50:08.283" CommentCount="1" />
  <row Id="5976" PostTypeId="2" ParentId="5928" CreationDate="2011-02-26T20:53:34.470" Score="7" ViewCount="" Body="&lt;p&gt;The Pythagorean theorem is very simple to prove without words. It might seem trivial now, but I expect that at least 80% of the population knows the theorem but couldn't prove it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/kfJ1N.gif&quot; alt=&quot;pythagorean theorem animation&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1655" LastActivityDate="2011-02-26T20:53:34.470" CommunityOwnedDate="2011-02-26T20:53:34.470" CommentCount="1" />
  <row Id="5977" PostTypeId="2" ParentId="5928" CreationDate="2011-02-26T21:08:17.353" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Newton's drawing proving a central force conserves angular momentum.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big black dot is the center of the central force. The blue triangles are produced at equal steps in time (say 1 second) when there is no central force. They're equal in area, therefore, when there is no central force, angular momentum is conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that a sudden impulse occurs. The red arrow indicates the change in velocity. Therefore at the end of the next second, the red triangle will be produced. But because the two red arrows are parallel (one indicating the force, the other the change in velocity), the red triangle has the same area as the (partially hidden) blue triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore a central force cannot change the angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/R6faJ.jpg&quot; alt=&quot;Newton's drawing&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-02-26T21:08:17.353" CommunityOwnedDate="2011-02-26T21:08:17.353" />
  <row Id="5978" PostTypeId="2" ParentId="5928" CreationDate="2011-02-26T22:44:15.643" Score="3" ViewCount="" Body="&lt;p&gt;What about &lt;a href=&quot;http://en.wikipedia.org/wiki/Method_of_inversion&quot; rel=&quot;nofollow&quot;&gt;method of images&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/DgQks.png&quot; alt=&quot;Wikipedia&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-02-26T22:44:15.643" CommunityOwnedDate="2011-02-26T22:44:15.643" CommentCount="1" />
  <row Id="5979" PostTypeId="1" CreationDate="2011-02-26T22:45:08.513" Score="11" ViewCount="522" Body="&lt;p&gt;I guess it's not a very educated question, but I never quite understood why spacecrafts have to shoot up and can't just reach space by simply continuing an upwards ascent like an airplane.&lt;/p&gt;&#xA;" OwnerUserId="2276" LastActivityDate="2011-02-28T15:52:15.967" Title="Why do space crafts take off with rockets instead of just ascending like an aircraft until they reach space?" Tags="&lt;gravity&gt;" AnswerCount="7" FavoriteCount="1" />
  <row Id="5980" PostTypeId="2" ParentId="5979" CreationDate="2011-02-26T22:59:35.650" Score="12" ViewCount="" Body="&lt;p&gt;Aircraft rely on lift generated by interacting with the atmosphere and on using atmospheric oxygen to burn with fuel they carry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbits aren't stable until you are high enough that there isn't enough atmosphere to interact with, and long before that the oxygen content drops too low to be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to get to a stable orbit, you &lt;em&gt;will&lt;/em&gt; need rockets eventually. Areospace types keep looking at hybrids (air breathing, lift generating while down low then switch to rockets as you get really high) because it &lt;em&gt;does&lt;/em&gt; seem like a good idea, but so far the added complexity hasn't paid. Either the switching gear weighs too much, or the extra equipment adds too much risk of failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe in the future.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2011-02-27T02:34:48.327" LastActivityDate="2011-02-27T02:34:48.327" />
  <row Id="5981" PostTypeId="2" ParentId="5923" CreationDate="2011-02-26T23:20:22.450" Score="3" ViewCount="" Body="&lt;p&gt;The core of the question is about whether Infinite sums and similar expressions which appear in many attempts at an analytical solution to equations of motion can represent &quot;how Nature does it&quot;. Well Feynman once noted something similar in an analysis which ended with the observation that &quot;down at the electron there seemed to be a lot of computation going on in a very small region&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of the answer for classical physics is that we dont always represent the problem that Nature is solving locally: particles move in their local field according to local potentials. In analysing a larger system &lt;em&gt;we&lt;/em&gt; find difficult analytics, but Nature never has to solve these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Quantum Mechanics (and its relatives) we reach the wavefunction ontology question. That is &quot;is the wavefunction just our &lt;em&gt;representation&lt;/em&gt; of the physical reality&quot; or &quot;is the wavefunction &lt;em&gt;actually&lt;/em&gt; physical reality&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly if the wavefunction is just our representation then it is not telling us what Nature actually does; and the analytic issues we have with it (e.g. in the N body case) are just our problem, not Nature's. What Nature actually does would then be a theory which reduces to Quantum Mechanics. The compututational world might describe such a theory as &quot;a theory which &lt;em&gt;Implements&lt;/em&gt; Quantum Mechanics&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the wavefunction corresponds directly to reality, then there probably is still some need to find a good &quot;computational engine&quot; for making it all work. The Quantum Computational school around David Deutsch seem to be in this area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as N-body issues are concerned I am becoming impressed by the &quot;emergent properties&quot; found in that area by the Condensed Matter Physicists. Perhaps this is telling us that Nature has found &quot;higher level descriptions&quot; with which to work with complex systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am still learning about String Theory, but I am getting the impression that it is quite &quot;conventional&quot; in terms of these types of fundamental issues with QM.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-26T23:20:22.450" CommentCount="4" />
  <row Id="5982" PostTypeId="1" CreationDate="2011-02-27T00:45:47.920" Score="1" ViewCount="85" Body="&lt;p&gt;Are there any known uses of modeling with elastic fractals in current physical applications? (Especially uses concerning with self-similarity)&lt;/p&gt;&#xA;" OwnerUserId="2279" LastActivityDate="2011-09-05T04:37:47.567" Title="Current Physical Applications of Elastic Fractals" Tags="&lt;particle-physics&gt;&lt;mathematical-physics&gt;&lt;fun-stuff&gt;" AnswerCount="1" />
  <row Id="5983" PostTypeId="1" CreationDate="2011-02-27T00:46:18.553" Score="2" ViewCount="140" Body="&lt;p&gt;When something fall down, transform potential energy into kinetic one.&#xA;furthermore you can attach a pulley and extract some energy &#xA;(like in an aljibe) for example with a dynamo...&#xA;if the object fall in a deep well, we can extract more energy,&#xA;till the object reach downhole..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but what if we could have a &quot;bottomless pit&quot;..&#xA;Endless energy?.. Well we would need an infinite earth.. &#xA;or like a &quot;black hole aljibe&quot;.. mmm that doesn't seem an option..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;oh wait a minute, we already have one! satellite is like always &quot;falling down&quot;.. &#xA;then is exchanging &quot;potential energy&quot;(who knows from what reference system) into kinetic energy..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what about tied it to a pulley (a long rope I see) to a dynamo, or an earth winding with a magnet satellite, or a satellite winding with the earth magnet, or whatever.. &#xA;the idea is to USE the &quot;bottomless pit&quot; energy..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where is the mistake? ( just in case there be any =P )&lt;/p&gt;&#xA;" OwnerUserId="1916" LastActivityDate="2011-02-27T02:50:18.183" Title="it is possible to extract energy from black holes or..satellites?" Tags="&lt;gravity&gt;&lt;energy&gt;&lt;energy-conservation&gt;" AnswerCount="3" />
  <row Id="5984" PostTypeId="2" ParentId="5983" CreationDate="2011-02-27T00:56:38.550" Score="3" ViewCount="" Body="&lt;p&gt;NASA has already experimented with something along the lines you describe, using earth as a sort of homopolar magnetic generator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electrodynamic_tether&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Electrodynamic_tether&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1670" LastActivityDate="2011-02-27T00:56:38.550" />
  <row Id="5985" PostTypeId="1" CreationDate="2011-02-27T01:07:14.287" Score="8" ViewCount="156" Body="&lt;p&gt;Broadly speaking how do ideas of leading singularities and Grassmanian help in curing infrared divergences when calculating N=4 scattering amplitudes? My understanding is that one gets infra red divergences because the external gluon momenta becomes collinear with the the loop momenta. I am confused as to what Nima's and Freddie's collaboration are doing to avoid this? If people can clarify my confusion and direct me to the appropriate literature I would be grateful.Also, is there a clear way to understand this for theories like planar QCD?&lt;/p&gt;&#xA;" OwnerUserId="2280" LastEditorUserId="429" LastEditorDisplayName="" LastEditDate="2011-02-27T02:56:33.680" LastActivityDate="2011-02-27T06:25:34.763" Title="How do ideas of leading singularities and Grassmanian help in curing infrared divergences when calculating N=4 scattering amplitudes?" Tags="&lt;quantum-field-theory&gt;&lt;string-theory&gt;&lt;supersymmetry&gt;&lt;scattering&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="5986" PostTypeId="2" ParentId="5983" CreationDate="2011-02-27T01:17:36.330" Score="4" ViewCount="" Body="&lt;p&gt;You can, theoretically, extract energy from rotating black holes via the &lt;a href=&quot;http://en.wikipedia.org/wiki/Penrose_process&quot; rel=&quot;nofollow&quot;&gt;Penrose process&lt;/a&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-27T01:17:36.330" />
  <row Id="5988" PostTypeId="2" ParentId="5979" CreationDate="2011-02-27T02:29:29.193" Score="0" ViewCount="" Body="&lt;p&gt;The Wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Spaceplane&quot; rel=&quot;nofollow&quot;&gt;Spaceplanes&lt;/a&gt; should answer all the questions you've asked.&lt;/p&gt;&#xA;" OwnerUserId="2283" LastActivityDate="2011-02-27T02:29:29.193" />
  <row Id="5989" PostTypeId="2" ParentId="5983" CreationDate="2011-02-27T02:40:18.303" Score="5" ViewCount="" Body="&lt;p&gt;A satellite isn't falling down--for a circular orbit, it is maintaining a constant radius.  Since the potential energy is a function of radius, the potential energy of the satellite remains unchanged.  The energy of the orbit is constant--the speed, and thus, the kinetic energy is constant, and so is the position, and thus the potential energy, is also constant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you tied the satellite to a dynamo, then the dynamo would produce a drag on the satellite, which would slow it down, which would cause it to lose kinetic energy, which would then cause the central force to overcome the inertia of the satellite, which would cause the satellite to fall into the Earth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, as Deepak says above, it is possible to have a particle convert the rotational kinetic energy a the black hole (or any rotating body) into energy taken out to infinity.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2011-02-27T02:50:18.183" LastActivityDate="2011-02-27T02:50:18.183" />
  <row Id="5990" PostTypeId="2" ParentId="5979" CreationDate="2011-02-27T03:18:14.573" Score="1" ViewCount="" Body="&lt;p&gt;It's because spacecraft are not generally designed to immediately start returning to the Earth once its fuel has run out, but to have reached the escape velocity of 11.2 kilometers per second at this point. It's then able to orbit the Earth continuously without having to use further fuel in the ideal case, as an example.&lt;/p&gt;&#xA;" OwnerUserId="1345" LastActivityDate="2011-02-27T03:18:14.573" />
  <row Id="5991" PostTypeId="1" AcceptedAnswerId="5998" CreationDate="2011-02-27T03:47:46.707" Score="0" ViewCount="126" Body="&lt;p&gt;Many physics papers now have dozens of authors per paper. Experimental physics may have multi-organizational and multi-country contributing staffs, but I'd guess that most of the names don't contribute a word or equation to a paper, yet they get individual authorship credit.&#xA;My question is who determines the author list, does everybody listed have editing privilages, and perhaps most importantly, who decides on their listed order?   &lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-27T03:49:41.940" LastActivityDate="2011-02-27T15:14:29.133" Title="Publication Authorship Credits" Tags="&lt;soft-question&gt;&lt;general-physics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="5992" PostTypeId="2" ParentId="1019" CreationDate="2011-02-27T04:16:44.617" Score="1" ViewCount="" Body="&lt;p&gt;I highly recommend reading the article &quot;Quantum mechanics: Myths and facts&quot; by H. Nikolic&#xA;&lt;a href=&quot;http://arxiv.org/abs/quant-ph/0609163&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/quant-ph/0609163&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some topics include wave-particle duality, time-energy uncertainty relation and fundamental randomness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've discussed this article in other communities and it seems reliable.&lt;/p&gt;&#xA;" OwnerUserId="301" LastActivityDate="2011-02-27T04:16:44.617" CommunityOwnedDate="2011-02-27T04:16:44.617" />
  <row Id="5993" PostTypeId="2" ParentId="629" CreationDate="2011-02-27T04:42:29.303" Score="2" ViewCount="" Body="&lt;p&gt;I think several answers already pointed correctly to factors. But, at least in cars, I think the dominant factor is the set of front and rear springs. My illustration is very much exaggerated. &#xA;In (a) we have a car with constant speed. When the driver gives the car a deceleration with the breaks the springs system enters in configuration (b). When the vehicle finally stops and there is no acceleration on it, the spring system &lt;strong&gt;oscillates back&lt;/strong&gt; a little (c) before returning to the equilibrium configuration (a). The wrong pull effect happens between (b) and (c)&#xA;&lt;img src=&quot;http://i.stack.imgur.com/kIgt2.jpg&quot; alt=&quot;my illustration&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="301" LastActivityDate="2011-02-27T04:42:29.303" CommentCount="1" />
  <row Id="5994" PostTypeId="1" AcceptedAnswerId="6044" CreationDate="2011-02-27T05:52:58.867" Score="0" ViewCount="237" Body="&lt;p&gt;I seem to be missing something basic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to get a simple orbital simulation working, and my two objects are Earth around the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is this. I placed the Earth at 93M miles away from the Sun, or 155M km. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, the orbital velocity of something at 155M km from the Sun is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = \sqrt{\frac{GM}{r}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging in the numbers for the Sun, I get a velocity of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$29261 \frac{\mathrm{m}}{\mathrm{s}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if I want to get the acceleration that the Sun has upon the earth, I use:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g = \frac{GM}{r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Sun, and 155M km, I get an acceleration of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0.0055\frac{\mathrm{m}}{\mathrm{s}^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I start with a simple body at the proper radius out along the X axis and give it a simply vector of 29261 m/s along the Y axis, then I start applying the 0.0055 m/s^2 acceleration to it. And the acceleration of the Sun is simply not enough to hold the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Earth starts with a vector of (0, 29261 m/s), and after and I add the acceleration vector of (-0.0055 m/s, 0) to it, you can see that after a single second, it doesn't move a whole lot. If I chunk things to days, 86400 seconds, then the acceleration vector is only, roughly, -477 m/day, but the velocity vector is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2,325,974,400 \frac{\mathrm{m}}{\mathrm{day}} = 29,261 \frac{\mathrm{m}}{\mathrm{s}} \times 86,400 \frac{\mathrm{s}}{\mathrm{day}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can imagine, the -477 isn't going to move that much towards the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that better simulations use better techniques than simply adding basic vectors together, but that's not what this is. I seem to be missing something fundamental. I had assumed that given the correct velocity, that the pull of the Sun should keep the Earth in orbit, but the &quot;pull&quot; that I'm using doesn't seem to be having the desired effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'm curious what basic &quot;D'oh&quot; thing I'm missing here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit for Luboš Motl answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps there's something more fundamental I'm missing here. I understand you point, but .0055 m/s * 86,400 is -477. I was doing that math fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply, I have an object with a velocity vector. Then I apply an acceleration at a right angle. I do that for N seconds to come up with a new, right angle velocity vector. I then add that to the original vector to come up with the objects new vector. I then take that vector, apply to the current position of the object, and arrive at a new position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly there is a granularity issue which makes some amount of seconds a better choice for a model than others, but this is high school level simple mechanics, so there's going to be some stepping. I chose one day so that my little dot of a planet on my screen would move. If I update every 1/10th of a second &quot;real time&quot;, and each update is a day, then I should get a rough orbit that's really a 365ish polygon in a little over 30s real time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I choose a step size of 1 second, then my acceleration (0.0055 m/s^2) * 1 s = a right angle velocity vector that's -0.0055 in magnitude. That vector is added to the original vector of 29261 (at right angles), giving me a new vector of (-0.0055, 29261). That's after one second. That's not much of a bump. It's barely a blip. If I apply one days full of acceleration, &quot;all at once&quot;, I am obligated to not only multiply the acceleration by 86,400, but also the original vector (since it's 29261 m/s, and we have 86,400 s), thus giving me, proportionally, the same vector, just longer. And it's still just a bump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'm mis-applying something somewhere here, as I think the numbers are fine. I'm simply &quot;doing it wrong&quot;. Trying to figure out what that wrong part is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2, responding to Platypus Lover&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much for the simple code you posted. It showed me my error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My confusion was conflating updating of the vector with the calculation of the velocity vector. I felt that I had to multiply both the original vector AND the acceleration amount by the time step, which would give me the silly results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was just confused in my head.&lt;/p&gt;&#xA;" OwnerUserId="2285" LastEditorUserId="2285" LastEditorDisplayName="" LastEditDate="2011-02-27T21:29:40.180" LastActivityDate="2011-02-27T21:49:59.680" Title="Missing something basic about simple orbital mechanics" Tags="&lt;newtonian-mechanics&gt;&lt;orbital-motion&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" />
  <row Id="5995" PostTypeId="1" AcceptedAnswerId="7573" CreationDate="2011-02-27T06:02:08.120" Score="4" ViewCount="94" Body="&lt;p&gt;Are Bloch equations (which describe the evolution of macroscopic magnetization) empirical? If so, under what assumptions do they hold?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac {d M_x(t)} {d t} = \gamma (\mathbf{M} (t) \times \mathbf {B} (t) ) _x - \frac {M_x(t)} {T_2}$$&#xA;$$\frac {d M_y(t)} {d t} = \gamma (\mathbf{M} (t) \times \mathbf {B} (t) ) _y - \frac {M_y(t)} {T_2}$$&#xA;$$\frac {d M_z(t)} {d t} = \gamma (\mathbf{M} (t) \times \mathbf {B} (t) ) _z - \frac {M_z(t) - M_0} {T_1} $$&lt;/p&gt;&#xA;" OwnerUserId="824" LastActivityDate="2011-03-26T00:52:19.850" Title="Are Bloch equations empirical?" Tags="&lt;electrodynamics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="5996" PostTypeId="2" ParentId="5973" CreationDate="2011-02-27T06:02:39.010" Score="0" ViewCount="" Body="&lt;p&gt;The real reason is &quot;inertial mass&quot; and &quot;gravitational mass&quot; of an object is same in nature. In Newtonian scheme, the force with which an object is attracted towards earth is $\frac {GMm_G}{R^2}$ $M$ is the mass of the earth, $m_G$ is the gravitational mass of the object and $R$ is the distance between the center of the earth and the object. If the object has an inertial mass $m_I$ and under the force of gravity it has an acceleration  $g$ then the same force is $m_Ig$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $m_Ig = \frac {GMm_G}{R^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m_I = m_G$ then $g = \frac {GM}{R^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the acceleration of the object is independent of its inertial mass.&#xA;Now the deeper reason of why the inertial mass and gravitation mass of an object is same is purely coincidental in Newtonian scheme, but in general relativity it is indispensable. The fundamental postulate on which the edifice of GR is based is the &quot;equivalence principle&quot; which requires these two kinds of masses to be the same.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-02-27T06:21:27.730" LastActivityDate="2011-02-27T06:21:27.730" />
  <row Id="5997" PostTypeId="2" ParentId="5985" CreationDate="2011-02-27T06:25:34.763" Score="5" ViewCount="" Body="&lt;p&gt;I had the same problem - it was one of just two main problems with the motivation of this whole &quot;twistor uprising&quot; (the other, surviving problem for me is the relevance of the whole formalism for the off-shell gauge theory which I find important, especially for AdS/CFT etc.) - but it has been fully answered for me. The solution is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the infrared divergences arise from integrals over momenta or positions - from the parts of the integral with very low momenta or very long distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the normal formalisms and for generic QFTs, the amplitudes are written are the sum of many Feynman diagrams. Each of them integrates over different loop variables but there is no way to reconcile them. The sum is an infrared divergent but there is no &quot;common integrand&quot; that would be convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, at the planar level of gauge theory, one may show that all the terms have a natural common parameterization of the infrared divergences, so the finite piece is totally well-determined. It's because the natural loop variable is what runs around the boundary of the planar diagram, if pictured as a disk, and shifting the internal loop momentum by the external momenta - which is the only ambiguity - doesn't change the form of the &quot;finite piece&quot; of the divergent amplitude. You may also imagine that the planar diagrams are merged into a disk diagram of string theory and you parameterize the infrared divergence in terms of $\tau$ - the length of the disk - in some Schwinger-like parameterization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latest two papers by Nima et al. make this point very manifest because the amplitude for a given process is given by a single integral of a convergent integrand over the Grassmannian (rather than a sum of many similar integrals over different domains). The integration variables and domain may be viewed as &quot;pure kinematics&quot; while the integrand (which is still a sum, but a sum of functions of the universal variables) is the &quot;true dynamics&quot;, so in this sense, the total dynamical result is totally convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't mean that the actual scattering amplitudes &quot;become&quot; convergent. They're still divergent, as in any conformal theory. However, the divergence is fully isolated into some kinematics and there exists a sharp quantity that may be calculated arbitrarily accurately, at any number of normal loops, and that is indisputably related to the scattering amplitudes.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-27T06:25:34.763" CommentCount="3" />
  <row Id="5998" PostTypeId="2" ParentId="5991" CreationDate="2011-02-27T06:39:36.787" Score="5" ViewCount="" Body="&lt;p&gt;OK, this is for experimental high energy physics as I worked in the field for over 40 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are groups in institutions, universities and research ones. There are many such in each country, and there are many countries. The group leaders in the group decide who signs a paper, mainly by the man hours put in the construction and running of the experiment and also considering  contributions in analysis of present and other papers. The order is alphabetical per author within a group, per name of institution. There have been long discussions on changing the credit attribution, but I see that the same holds for LHC papers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are there so many names? In my carreer I worked on one large ( previous had about 50 people) experiment( 350 people) from inception of the idea to taking data and analysis. It took 10 years to build the detector by hundreds of people, years that yielded very few publications from the full work put in, certainly over 8 hours a day. Credit was accumulating from analysis published/worked-on  previous experiment papers. Then another 15 years of data analysis where there are also large numbers of working groups, made up from people from all groups, and many people working on the same subject with their own analysis. The final paper is decided by the working group, a joining of all individual analysis. The names are still the ones the individual institute group leader gives to the working group. The working group proposes a preprint to the collaboration editing board, and if the board approves of the paper it goes to the full collaboration meeting, passed by consensus. Every person who signs can comment and ask for changes. They usually do not, as there is trust that the working groups are doing their job well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system is completely open. Any group member can join in the analysis and comment.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-02-27T06:39:36.787" />
  <row Id="5999" PostTypeId="1" CreationDate="2011-02-27T06:44:11.007" Score="0" ViewCount="54" Body="&lt;p&gt;This question is about the concept of daylight saving. While, this may sound to fall more under geography/meteorology/ etc, but I think, physics is anyway the guiding factor behind these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that daylight saving is done in summer and then the watches are made fast by one hour. The idea behind from what I understand is that in summer days are longer than nights and we want to use the day natural light energy to reduce artificial lights usage, but our working hours in office remain the same. This means that if we are working one hour extra in morning, we are working one hour less in evening. This means that sun light intensity doesn't remain same in morning and in evening and we want to use the higher intensity mornings as against to lower intensity evenings. Is my understanding correct that the basic reason why day light works is because in morning we have more light than in evening in summer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="1093" LastActivityDate="2011-02-27T06:44:11.007" ClosedDate="2011-02-27T06:48:27.950" Title="daylight saving" Tags="&lt;astrophysics&gt;&lt;thermodynamics&gt;&lt;energy&gt;&lt;light&gt;&lt;sun&gt;" CommentCount="3" />
  <row Id="6000" PostTypeId="2" ParentId="5979" CreationDate="2011-02-27T07:01:20.840" Score="1" ViewCount="" Body="&lt;p&gt;In short - A rocket works by exchanging momentum, AND at present airplanes simply do not have the ability to exchange enough momentum. In a rocket - Both the mass of the propellant as well as the high velocity of its exit from the engine system gives the rocket its momentum to propel itself into space. The propellant (fuel)  attains its velocity by burning with an oxidizer in a high-pressure chamber creates a high energy exhaust which then funneled through a nozzle. This velocity, coupled with the right mass properties of the propellant, provides the power, or energy, required to get the vehicle into space.  An airplane on the other hand requires air drag and lift to propel itself off of the ground... and when the atmosphere is to thin ... it cannot exchange momentum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The US Airforce has been working with space planes - however due to the design they still need a ROCKET to lift them up into space.&#xA;Pretty cool stuff - if you want some extra reading...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.spaceflightnow.com/atlas/av026/status.html&quot; rel=&quot;nofollow&quot;&gt;http://www.spaceflightnow.com/atlas/av026/status.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2287" LastActivityDate="2011-02-27T07:01:20.840" />
  <row Id="6001" PostTypeId="2" ParentId="5994" CreationDate="2011-02-27T07:30:47.373" Score="5" ViewCount="" Body="&lt;p&gt;Dear Will, your error is that you think that the acceleration $a$ adds distances $s$ linearly in time $t$. In reality, it adds them quadratically via the formula&#xA;$$ s = \frac{1}{2} at^2 $$&#xA;Many kids know this formula: it's the time $t$ multiplied by the average speed during the interval which is $(v_{initial}+v_{final})/2=at/2$. The graph of the motion is a parabola, a good approximation whenever the acceleration is approximately constant during the short enough time interval. In other words, you have increased the velocity during the first second, but you forgot to increase the velocity by the same amount during the remaining $86,399$ seconds of the first day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the first second, the velocity changes by $0.0055 m/s$ and the total distance the Earth travels during the first second is $0.0055/2 = 0.00275 m$. However, if you study how this distance changes if you increase 1 second to 86,400 seconds, the total distance doesn't jump 86,400 times. Instead, it jumps $(86,400)^2$ times, to&#xA;$$0.0275 \times 86,400^2 = 20,528,000 m.$$&#xA;You may check that 20,000 kilometers is approximately - within the errors that you introduced - the right amount to keep the Earth on a quasi-circular orbit because&#xA;$$\frac{155\times 10^9 m}{2}\left(\frac{2\pi}{365.25}\right)^2 = 20,000,000 m$$&#xA;or so. In the equation above, I calculated the angle of the rotation around the Sun, squared it, divided by two (that's the approximation of $1-\cos\phi$), and multiplied by the radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After 86,400 seconds, the final speed in the transverse direction is not $0.0055 m/s$ as you assumed but $86,400\times 0.0055 m/s$.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-02-27T08:29:45.170" LastActivityDate="2011-02-27T08:29:45.170" />
  <row Id="6002" PostTypeId="1" CreationDate="2011-02-27T08:29:56.263" Score="6" ViewCount="440" Body="&lt;p&gt;What percentage of physics PhDs leave physics to become quantitative analysts, work in computer science/information technology or business? Is physics that bad that so many people leave? Was it worth it?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-27T21:43:39.317" ClosedDate="2011-04-03T09:10:19.300" Title="What percentage of physics PhDs leave physics?" Tags="&lt;soft-question&gt;&lt;physics-careers&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="3" />
  <row Id="6004" PostTypeId="1" AcceptedAnswerId="6023" CreationDate="2011-02-27T09:31:18.660" Score="1" ViewCount="154" Body="&lt;p&gt;Let's consider rotating charged ring. Theoretically mass of this ring  has no limit as rotation speed increases. So what about magnetic moment of the ring? Is it limited by the value of speed of light?&lt;/p&gt;&#xA;" OwnerUserId="238" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-03-05T22:15:43.350" LastActivityDate="2011-03-05T22:15:43.350" Title="Magnetic moment of relativistic rotating ring" Tags="&lt;electromagnetism&gt;&lt;special-relativity&gt;&lt;magnetic-moment&gt;" AnswerCount="1" />
  <row Id="6005" PostTypeId="2" ParentId="6002" CreationDate="2011-02-27T09:41:49.830" Score="4" ViewCount="" Body="&lt;p&gt;A google search would be more help to you. The &lt;a href=&quot;http://www.aip.org/statistics/trends/highlite/emp3/emphigh.htm&quot; rel=&quot;nofollow&quot;&gt;AIP has a page.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More people choose to do a phd in physics then there are tenure spots. Competition is high and monetary rewards not that hot. If one is a real physicist, one hangs on because of the physics interests not the career and yes, it is worth it.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-02-27T09:41:49.830" CommentCount="1" />
  <row Id="6006" PostTypeId="2" ParentId="5971" CreationDate="2011-02-27T10:20:28.360" Score="1" ViewCount="" Body="&lt;p&gt;The theory is rather generic Fourier transform: if you have a perfectly non-decaying oscillation $e^{i\omega_0 t}$ (with real $k$) then the transform gives a perfectly sharp spectrum as a delta function $\delta(\omega - \omega_0)$. But if the excitation decays $e^{i\omega_0 t} e^{-k t}$ then we get $\delta(\omega - \omega_0 - i k)$, which has a real part that is broadened: $1/\left({(\omega-\omega_0)^2 + k^2}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, usually this manifests as some sort of sweep time, as experimentally you would drive the system at some changing frequency $\omega$ and observe the driven amplitude.&lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-02-27T10:20:28.360" />
  <row Id="6007" PostTypeId="2" ParentId="5736" CreationDate="2011-02-27T10:43:07.337" Score="0" ViewCount="" Body="&lt;p&gt;You can think of a generalized innerproduct, where terms get integrated over the &quot;correct&quot; superspace. In my view, the best way of deriving the propagators for chiral fields. It's much cleaner than lifting everything up to full superspace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of this occurs in section 4.8 of &lt;a href=&quot;http://books.google.com.au/books?id=jZ1Y1Z3BoicC&quot; rel=&quot;nofollow&quot;&gt;Ideas and methods of supersymmetry and supergravity&lt;/a&gt; and probably other papers by the authors. In this section only chiral and antichiral spaces are used, but it can be generalized to include a full superspace sector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you've set up your conventions for this type of innerproduct, then everything basically works as you'd expect.&lt;/p&gt;&#xA;" OwnerUserId="429" LastActivityDate="2011-02-27T10:43:07.337" />
  <row Id="6008" PostTypeId="2" ParentId="6002" CreationDate="2011-02-27T11:11:59.030" Score="7" ViewCount="" Body="&lt;p&gt;In Germany the average age of a professor when he/she gets tenure is 42. A professor retires at the age of 65. If you add that the average professor will have one or two students per year finishing a PhD, you can figure out the percentage of physics PhD that can get a tenured position yourself :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a physicist you are trained to do basic research, and there are almost no positions for doing basic research outside of academia, that's how the system works. So the question why one should study physics is legitimate. According to my own experience, physicists are employed as generalists when no specialized engineers are available.  When a company needs a software developer or an electro engineer for example, they will sometimes hire a physicist instead of someone with the matching degree, because &quot;a brilliant physicist is able to learn the necessary skills on the job, so in the long term it is more profitable to have a brilliant physicist instead of a mediocre software scientist&quot;.  &lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-02-27T11:11:59.030" CommentCount="1" />
  <row Id="6009" PostTypeId="1" AcceptedAnswerId="6013" CreationDate="2011-02-27T11:56:05.237" Score="11" ViewCount="368" Body="&lt;p&gt;From the &quot;no hair theorem&quot; we know that black holes have only 3 characteristic external observables, mass, electric charge and angular momentum (except the possible exceptions in the higher dimensional theories). These make them very similar to elementary particles. One question naively comes to mind. Is it possible that elementary particles are ultimate nuggets of the final stages of black holes after emitting all the Hawking radiation it could?&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-02-28T06:52:29.747" Title="Are elementary particles ultimate fate of black holes?" Tags="&lt;general-relativity&gt;&lt;particle-physics&gt;&lt;black-holes&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="4" />
  <row Id="6010" PostTypeId="1" CreationDate="2011-02-27T12:10:48.740" Score="-2" ViewCount="144" Body="&lt;p&gt;There's some constant relating to electrons that also has the same value as the speed of light. What is it, what is the value, and how are they related?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Is it the fine-structure constant?? Are there any other similar constants? If you posted that answer before, you shoulda left it (to the person who deleted their answer)!&lt;/p&gt;&#xA;" OwnerUserId="2291" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-28T00:17:04.617" LastActivityDate="2011-02-28T00:17:04.617" ClosedDate="2011-02-28T00:18:00.230" Title="What is the name and value of the constant that relates to electrons and that coincidentally has the same exact value as the speed of light?" Tags="&lt;speed-of-light&gt;&lt;electron&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="6012" PostTypeId="1" CreationDate="2011-02-27T13:24:45.070" Score="1" ViewCount="230" Body="&lt;p&gt;Will a CFL light bulb and an incandescent light bulb, in separate respective closed systems, produce exactly the same amount of overall temperature increase over time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume you have two identical closed systems with gray walls, with a system input of 20 watts of power each. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT added for clarity: (On the packaging of the CFL light bulb the large print equivalent wattage is irrelevant... the input current of both bulbs is a consistent 20 watts of power each.  The comparison wattage vs. the actual wattage of the CFL is off subject.)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One has a CFL, one has an ordinary incandescent light bulb. Will both systems increase in heat the exact amount, every hour?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to conservation of energy it shouldn't matter if one light source is more efficient, right?... it's the same amount of energy input. One light makes more heat one makes more light, but the light when it hits the gray wall is converted to heat, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no such thing as loss of energy... it's just converted to another form of energy... and light is converted to heat, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The back story of this question is my wondering that if my wife leaves an incandescent light bulb on in the winter time it's not so bad because even though no one is in the room it's still heating up the room.  On the other hand if she leaves on a CFL it's more efficient but it should still add heat to our &quot;system,&quot; i.e., our home.&lt;/p&gt;&#xA;" OwnerUserId="2293" LastEditorUserId="2293" LastEditorDisplayName="" LastEditDate="2011-02-28T11:53:39.530" LastActivityDate="2011-04-18T11:11:47.540" Title="Will a CFL light bulb and an incandescent light bulb produce exactly the same amount of overall temperature increase over time?" Tags="&lt;energy-conservation&gt;&lt;technology&gt;&lt;consumer-packaging&gt;" AnswerCount="2" />
  <row Id="6013" PostTypeId="2" ParentId="6009" CreationDate="2011-02-27T13:28:49.643" Score="5" ViewCount="" Body="&lt;p&gt;This is indeed a tempting suggestion (see also &lt;a href=&quot;http://arxiv.org/abs/hep-th/9202014&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;). However, there is a crucial difference between elementary particles and macroscopic black holes: the latter are described, to a good approximation, by non-quantum (aka classical) physics, while elementary particles are described by quantum physics. The reason for this is simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the classical radius of an object is larger than its Compton wavelength, then a classical description is sufficient. For black holes whose Schwarzschild radius is bigger than the Planck length this is fulfilled. However, for elementary particles this is not fulfilled (e.g. for an electron the &quot;radius&quot; would refer to the classical electron radius, which is about $10^{-13}$cm, whereas its Compton wavelength is about three orders of magnitude larger).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Near the Planck scale your intuition is probably correct, and there is no fundamental difference between black holes and elementary particles - both could be described by certain string excitations.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-02-27T13:28:49.643" />
  <row Id="6014" PostTypeId="2" ParentId="6009" CreationDate="2011-02-27T13:32:31.463" Score="2" ViewCount="" Body="&lt;p&gt;The short answer is no. Have a look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt; on dissipation of black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;quote: &lt;em&gt;Unlike most objects, a black hole's temperature increases as it radiates away mass. The rate of temperature increase is exponential, with the most likely endpoint being the dissolution of the black hole in a violent burst of gamma rays.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possibility of micro black holes from extra dimensions in some string models still has them dissolving thermodynamically into elementary particles as soon as they are formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Herein I have been replying to the question stated clearly in the last sentence:&#xA;&lt;em&gt;Is it possible that elementary particles are ultimate nuggets of the final stages of black holes after emitting all the Hawking radiation it could?&lt;/em&gt;&#xA;Not to the different question that people seem to be replying to: &quot;are black holes like elementary particles.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A yes answer to the latter, does not reply to the former, i.e. whether quarks and leptons are the nugget, what is left over, from a black hole. A yes answer to this last would offer the intriguing model of the snake eating its tail, maybe quite probable in some new more encompassing theory, but not foreseen now, at least from the answers given. If after shedding innumerable quarks leptons and photons and entropy on the way, a black hole ends up as an electron (for example) in an identifiable quantum mechanical history.By this last I mean something similar to a decay chain in nuclear cascades.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-02-28T06:52:29.747" LastActivityDate="2011-02-28T06:52:29.747" CommentCount="5" />
  <row Id="6015" PostTypeId="2" ParentId="6009" CreationDate="2011-02-27T14:06:38.957" Score="1" ViewCount="" Body="&lt;p&gt;The other answers here are fine.  Another point that should be stated is that if you believe in the naïve values of the angular momentum, charge and mass of most elementary particles, and plugged them into the Kerr-Nordstrom solution, you would find that almost every (and probably all of them, I just haven't checked) elementary particle would be a naked singularity, not a black hole--the charge and angular momentum of these objects would be too large to support a horizon.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-02-27T14:06:38.957" CommentCount="4" />
  <row Id="6016" PostTypeId="1" CreationDate="2011-02-27T14:18:25.700" Score="1" ViewCount="532" Body="&lt;p&gt;I am confused about the current and voltage. My intuitive example would be that of a pipe of say water. The diameter of the pipe determines the amount of water flowing per second but the pressure is comparable to voltage. Am I right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what is the difference between voltage source and current source? In what class our electric sockets on the wall do fall?&lt;/p&gt;&#xA;" OwnerUserId="2294" LastActivityDate="2011-02-27T15:57:08.083" Title="Difference between current and voltage sources" Tags="&lt;electricity&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="6017" PostTypeId="2" ParentId="6010" CreationDate="2011-02-27T14:20:39.757" Score="3" ViewCount="" Body="&lt;p&gt;The constant you are looking for is not related to electrons, but to electromagnetics. From the Wikipedia article 'Speed of light': &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The classical behaviour of the electromagnetic field is described by Maxwell's equations, which predict that the speed c with which electromagnetic waves (such as light) propagate through the vacuum is related to the electric constant ε0 and the magnetic constant μ0 by the equation c = 1/√ε0μ0.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-02-27T14:20:39.757" />
  <row Id="6018" PostTypeId="2" ParentId="6016" CreationDate="2011-02-27T14:27:51.437" Score="2" ViewCount="" Body="&lt;p&gt;A voltage source is assumed to deliver energy with a specific terminal voltage which does not depend upon the current from the source. A current source on the other hand is assumed to deliver energy with a specified current through the terminals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both current and voltage sources are ideal. In practice, we represent a real voltage source as an ideal voltage source in series with a resistance and a real current source as an ideal current source in parallel with a resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the best.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-02-27T14:27:51.437" />
  <row Id="6019" PostTypeId="2" ParentId="6016" CreationDate="2011-02-27T14:44:53.917" Score="1" ViewCount="" Body="&lt;p&gt;Your analogy with water flow through a pipe is correct. In that analogy a voltage source corresponds to a pump that generates a specific pressure, and a current source corresponds to a pump that generates a specific water displacement (volume per second).&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-02-27T14:44:53.917" />
  <row Id="6020" PostTypeId="2" ParentId="6016" CreationDate="2011-02-27T14:54:10.460" Score="0" ViewCount="" Body="&lt;p&gt;If you think of it from practical side, it's&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Voltage source = stabilized(fixed) voltage. That is what we usually expect to see as a power source. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current source = stabilized(fixed) current. This kind of source will increase/decrease it's output voltage to match required current. &#xA;This is used to power non-linear devices like LEDs, lasers, gas discharge tubes. &lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-02-27T14:54:10.460" />
  <row Id="6021" PostTypeId="2" ParentId="6012" CreationDate="2011-02-27T14:55:29.460" Score="2" ViewCount="" Body="&lt;h2&gt;No they won't&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;if you're reading the &lt;em&gt;&quot;watts&quot;&lt;/em&gt; off the package.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is because for an incandescent bulb, the &quot;60 W&quot; is a measure of the actual electric power it dissipates.  Most of that power is thermal (i.e. heat) from the get go, and every last Watt ends up heating in the room eventually (when the light energy thermalizes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, CF lamps and LED lamps are labeled with &quot;light output equivalent&quot; wattage, because the first thing people care about when buying a bulb is &lt;em&gt;how much light&lt;/em&gt; they're going to get and we've grown used understanding that in terms of an incandescent bulb's total power dissipation (I suppose that in twenty years this will look like a really stupid convention to the kids). If you read the package closely you'll see that high efficiency lamps generate about the same amount of light, but much less heat. The relationship is not linear in equivalent wattage, but runs about 1/7 for CF bulbs in the 60--120 &quot;Watt&quot; range.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Yes they will&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;if you really mean, that each bulb &lt;em&gt;dissipates&lt;/em&gt; 20 Watts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, that's a really dim incandescent bulb suitable for little more than a night light, but a pretty good CF bulb suitable for keeping a medium sized room well lit and allowing you to read comfortably.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-02-27T15:35:03.687" LastActivityDate="2011-02-27T15:35:03.687" CommentCount="3" />
  <row Id="6022" PostTypeId="2" ParentId="6012" CreationDate="2011-02-27T15:00:47.193" Score="2" ViewCount="" Body="&lt;p&gt;The relevant comparison is between bulbs with similar light power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume for sake of argument that a 20 W CFL light generates as much light as a 100 W incandescent bulb. For instance, both generate 5 W of light. If you leave the CFL light burning in a room, you get 15 W of heat and 5 W of light. If the room has no windows, no light energy leaves the room, and the room heats up with a rate of 15 + 5 = 20 W. In the same room, the incandescent light gives 95 W of heat plus 5 W of light, for a total of 100 W of heating power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is generally not a good idea to heat up your home with incandescent bulbs (or any other form of electrical heating) as it takes much more energy in the form of hydrocarbons to generate a given amount of electrical energy. Efficiency wise, is better to heat up your home by burning the hydrocarbons locally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And where the difference between CFL and incandescent bulbs really starts to matter is during summer when each Watt of power dissipated in your home needs to get pumped to a hot outdoor environment using an AC.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastEditorUserId="1268" LastEditorDisplayName="" LastEditDate="2011-02-27T15:38:04.727" LastActivityDate="2011-02-27T15:38:04.727" CommentCount="4" />
  <row Id="6023" PostTypeId="2" ParentId="6004" CreationDate="2011-02-27T15:11:00.617" Score="4" ViewCount="" Body="&lt;p&gt;The magnetic moment of a rotating charged ring is&#xA;$$&#xD;&#xA;m=IA=\lambda v A.&#xD;&#xA;$$&#xA;Here $I$ is current, $\lambda$ is linear charge density, and $A$ is the enclosed area. This expression is true even relativistically. The quantities $\lambda,A$ don't depend on the rotation speed, so the magnetic moment is limited to $\lambda c A$.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-27T15:11:00.617" />
  <row Id="6024" PostTypeId="1" CreationDate="2011-02-27T15:12:33.970" Score="0" ViewCount="131" Body="&lt;p&gt;How important is unpublished folklore in research? Tidbits of wisdom which can't be proven or explicitly states, but yet so useful.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-27T18:31:49.407" ClosedDate="2011-02-27T18:37:45.057" Title="Importance of folklore in research" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6025" PostTypeId="2" ParentId="5991" CreationDate="2011-02-27T15:14:29.133" Score="1" ViewCount="" Body="&lt;p&gt;It is typical for large collaborations in nuclear and particle physics to have considerable internal bureaucracy, including a &lt;em&gt;Memorandum of Understanding&lt;/em&gt; or similar document that lays out the rules for membership in the collaboration, authorship on papers published as a collaboration, vetting rules for separate publications of  papers on work done on the experiment, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;a href=&quot;http://physics.stackexchange.com/questions/5991/publication-authorship-credits/5998#5998&quot;&gt;anna says&lt;/a&gt; the rules often put the choice into the hands of a institutional leader, but they also might require things like&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Every author must sit at least three days of shift per year (which can result in some grad student having to baby a very senior, but somewhat thumb handed theorist through the shift process (though I learned a &lt;em&gt;lot&lt;/em&gt; when I had that experience))&lt;/li&gt;&#xA;&lt;li&gt;Every author must attend at least one collaboration meeting per year &lt;em&gt;in person&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Some monetary figure that must be contributed toward the collaboration budget (either put into a common fund or spent of capital equipment) for each &quot;author&quot; (this money, of course, generally comes from grants made to the institution, so it goes back to anna's point)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and similar &quot;make sure you're still interested&quot; type qualifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper is usually drafted by committee (the members of which are selected at collaboration meetings, and they generaly want &lt;em&gt;more&lt;/em&gt; not &lt;em&gt;fewer&lt;/em&gt; volunteers), and are made available to the whole collaboration for comments a few times before they are submitted (and usually again when the reviewer comments come back).&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-02-27T15:14:29.133" />
  <row Id="6026" PostTypeId="1" AcceptedAnswerId="6043" CreationDate="2011-02-27T15:42:24.117" Score="7" ViewCount="191" Body="&lt;p&gt;To put it bluntly, weather is described by the Navier-Stokes equation, which in turn exhibits turbulence, so eventually predictions will become unreliable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in a derivation of the time-scale where weather predictions become unreliable. Let us call this the critical time-scale for weather on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could estimate this time-scale if we knew some critical length and velocity scales. Since weather basically lives on an $S^2$ with radius of the Earth we seem to have a natural candidate for the critical length scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I assume that the relevant length scale is the radius of the Earth (about 6400km) and the relevant velocity scale some typical speed of wind (say, 25m/s, but frankly, I am taking this number out of, well, thin air). Then I get a typical time-scale of $2.6\cdot 10^5$s, which is roughly three days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result three days seems not completely unreasonable, but I would like to see an actual derivation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know how to obtain a more accurate and reliable estimate of the critical time-scale for weather on Earth?&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-02-27T18:40:04.303" Title="How to calculate the upper limit on the number of days weather can be forecast reliably?" Tags="&lt;navier-stokes&gt;&lt;weather&gt;&lt;turbulence&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="6027" PostTypeId="1" CreationDate="2011-02-27T15:52:20.920" Score="1" ViewCount="270" Body="&lt;p&gt;I've only learned about to use kinematics equation when solving projectile problems but today i came across the following equations. where does they come from?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Distance travelled&lt;/li&gt;&#xA;&lt;li&gt;Time of flight&lt;/li&gt;&#xA;&lt;li&gt;Angle of reach&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" LastActivityDate="2011-08-01T03:35:48.827" Title="Projectiles problem solving" Tags="&lt;kinematics&gt;&lt;projectile&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6028" PostTypeId="2" ParentId="6024" CreationDate="2011-02-27T15:56:35.430" Score="0" ViewCount="" Body="&lt;p&gt;The credo &lt;em&gt;&quot;It has to be beautiful&quot;&lt;/em&gt; could perhaps be classified as 'physics folklore' (defined as the customs and traditions in physics). But even such a subjective 'tidbit of wisdom' doesn't go unpublished. There is no 'hidden folklore' in physics, everything of (potential) value gets openly discussed and criticized.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastEditorUserId="1268" LastEditorDisplayName="" LastEditDate="2011-02-27T18:31:49.407" LastActivityDate="2011-02-27T18:31:49.407" />
  <row Id="6029" PostTypeId="2" ParentId="6016" CreationDate="2011-02-27T15:57:08.083" Score="0" ViewCount="" Body="&lt;p&gt;Just to put it slightly differently, in high school physics more than fifty years ago we learned that in the water analogy the voltage corresponds to pressure which corresponds to the height of the head in the water system.   Of course this corresponds to DC and our wall sockets are AC.  But the voltage is fixed, so they are voltage sources, just as our water pipes are fixed pressure sources.  We regulate water flow by turning valves (faucets) to change the size of the opening.  We regulate current flow, say in an electric stove or a light with a dimmer switch by changing the resistance.  In each case the control acts like a choke point in the piping system.&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-02-27T15:57:08.083" />
  <row Id="6030" PostTypeId="2" ParentId="5979" CreationDate="2011-02-27T15:57:46.703" Score="3" ViewCount="" Body="&lt;p&gt;For what it's worth, even though a rocket starts its flight going straight up, once it has traveled through most of the atmosphere it soon starts to change its direction so that it spends most of its flight accelerating in the &quot;around the earth&quot; direction (i.e. basically horizontal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, to reach orbit a vehicle either has to reach a high enough speed, or a high enough altitude, or a high enough combination of speed plus altitude.  A gun firing a bullet horizontally at 20,000mph would do the trick.  Or, dropping a ball from 20,000 miles altitude would also work.  Neither of these are &lt;em&gt;practical&lt;/em&gt; altitude/velocity combinations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On to your question.  There isn't any fundamental physical reason why an &quot;airplane&quot; can't fly into space.  A relevant definition here of &quot;airplane&quot; would be a vehicle that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;carries its fuel&lt;/li&gt;&#xA;&lt;li&gt;picks up air from the atmosphere as it flies, and&lt;/li&gt;&#xA;&lt;li&gt;has wings that can generate lift&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note that both an airplane and a rocket generate their thrust from exactly the same physical principle - by accelerating and ejecting some mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reasons that you don't see an aircraft-like vehicle accelerate and climb its way into space are practical engineering reasons, not fundamental physical limitations.  Here is a hypothetical &quot;airplane to orbit&quot; concept that the laws of physics would allow:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Take off from a runway, accelerate and climb to 500mph at 40,000 feet (so far, just like an ordinary transport aircraft)&lt;/li&gt;&#xA;&lt;li&gt;holding 40,000 feet altitude, accelerate from 500mph to 20,000mph&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now here are some of the practical impediments to this actually being a buildable vehicle:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;staying in the atmosphere like this, the faster the vehicle goes the more drag it will experience.  The engine has to generate enough thrust to overcome the drag, plus a bit extra to accelerate the vehicle.  Ordinary jet engines are powerful enough to fly an ordinary jet airplane at 500mph at 40,000 feet.  To go 20,000mph is 40 times as fast, which will require 40*40 times as much thrust, which requires an engine that is 40*40*40 times as powerful.  A particularly unkind law of physics says that the drag increases as the square of the velocity (actually this is a simplification that ignores shock waves, the reality is even more unkind!) and the power required increases as the cube of the velocity.&lt;/li&gt;&#xA;&lt;li&gt;building a conventional airbreathing engine that operates from 500mph to 2,000mph is a major engineering feat (the SR-71 Blackbird can do this).  Getting an airbreathing engine to operate at 4,000mph or 5000mph has not yet been achieved practically (look into current research on &quot;scramjets&quot;).  Getting to 20,000mph is a long, long way further away than any of this.&lt;/li&gt;&#xA;&lt;li&gt;Even if you could accelerate like this, you would face an insurmountable aerodynamic heating problem.  Just like a spacecraft or meteor entering the atmosphere produces white-hot heat, your accelerating aircraft would encounter the same effect.&lt;/li&gt;&#xA;&lt;li&gt;Chemical fuels (hydrocarbons, hydrogen, etc) have a certain amount of energy per unit mass, and engines convert some fraction of this energy into thrust (the amount depends various specifics but one thing you can count on for sure is that 1 Joule of fuel energy will always be converted into less than 1 Joule of propulsive energy (thrust times speed)).  You quickly get silly results from your calculations - what if you need to burn a million pounds of fuel in order to put one pound of payload into orbit?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;EDIT to add: more advanced discussions can be found at the &quot;Selenian Boondocks&quot; blog; start with the post &quot;Part I : Air Launched SSTO&quot; in the &lt;a href=&quot;http://selenianboondocks.com/category/orbital-access-methodologies/&quot; rel=&quot;nofollow&quot;&gt;Orbital Access Methodologies&lt;/a&gt; section.&lt;/p&gt;&#xA;" OwnerUserId="2297" LastEditorUserId="2297" LastEditorDisplayName="" LastEditDate="2011-02-28T15:52:15.967" LastActivityDate="2011-02-28T15:52:15.967" CommentCount="5" />
  <row Id="6031" PostTypeId="2" ParentId="5979" CreationDate="2011-02-27T16:05:46.220" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pegasus_(rocket)&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Pegasus_(rocket)&lt;/a&gt;&lt;br&gt;&#xA;The Pegasus is a hybrid airplane-rocket system in actual current use. &lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-02-27T16:05:46.220" />
  <row Id="6032" PostTypeId="1" CreationDate="2011-02-27T16:07:38.980" Score="0" ViewCount="120" Body="&lt;p&gt;Given that my previous question, namely &quot;What is the number density of virtual photons around a unit charge?&quot; has no precise answer, here is a more precise wording:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the virtual photon spectral density due to the electric field of a unit charge at rest (or, better, its difference with respect to the no-field case)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or: How many virtual photons per volume and per energy are found around a unit charge, at a distance r?&lt;/p&gt;&#xA;" LastActivityDate="2011-02-28T13:59:00.593" Title="What is the spectral energy density of virtual photons around a unit charge at rest?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;quantum-optics&gt;&lt;quantum-electrodynamics&gt;&lt;gauge-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6033" PostTypeId="2" ParentId="5950" CreationDate="2011-02-27T16:09:26.013" Score="1" ViewCount="" Body="&lt;p&gt;In practice, except in very particular circumstances, you can't do experiments to observe highly entangled states. The results of any feasible experiment will be the same as if it was done on an equivalent mixed state. It's the same reason that nobody has observed a cat in the state $\frac{1}{\sqrt{2}}(|$alive$\rangle+|$dead$\rangle)$. What experiment would distinguish between a superposition of an alive and a dead cat and a mixture of these two states?&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2011-02-27T16:09:26.013" />
  <row Id="6035" PostTypeId="2" ParentId="6002" CreationDate="2011-02-27T16:26:34.283" Score="2" ViewCount="" Body="&lt;p&gt;Just to note that people could leave research because they may not have &quot;research grant funding&quot; skills or aptitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A researcher who has invested a few years in a sub-field X may become almost a world expert in an aspect of X  - maybe in one of the experimental/computational/calculational aspects. Unfortunately fashions change in science, in Physics perhaps because experiments are starting to disprove the assumptions/predictions of X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So our researcher finds no more money to work in sub-field X. Our researcher now faces some stark choices: adapt the experimental/calculational techniques to a popular field like Y; physically relocate to an Institution that still pursues X; or develop a new sub-field; or leave research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If our researcher stays on and hopes to change sub-field they will now face a research grant procedure, which although not hostile to them, is hostile to sub-field X which becomes a liability from now on. Some individuals have the skills to negotiate this, some dont.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-27T16:26:34.283" />
  <row Id="6038" PostTypeId="2" ParentId="6009" CreationDate="2011-02-27T16:45:48.623" Score="4" ViewCount="" Body="&lt;p&gt;Yes, black holes are special kinds of elementary particles. That's how they have to be represented in every consistent quantum theory of gravity. This representation of a black hole becomes especially useful and important for small black holes - whose mass is not much larger than the Planck mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And indeed, a black hole evaporates, which is just a form of a decay of a heavy elementary particle, and when it becomes very light, at the end of the Hawking evaporation process, it is literally indistinguishable from a heavy elementary particle that ultimately decays into a few stable elementary particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a difference that you seem to neglect is that black holes actually carry a large entropy &#xA;$$ S = \frac{A}{4A_0} k_B $$&#xA;where $A$ is the area of the black hole's event horizon and $A_0$ is the Planck area $A_0=\hbar G / c^3$. The constant $k_B$ is Boltzmann's constant. This means that there actually exists a huge number of microstates&#xA;$$ N = \exp(S / k_B ) $$&#xA;and a single black hole, with a fixed value of mass, charges, and spin, is just a macroscopic description of the ensemble of $N$ &quot;microstates&quot;. In reality, the black hole carries a huge information - the world distinguishes which of the $N$ microstates is actually present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is these &quot;microstates&quot; that are really analogous to types of elementary particles. But the number of particle species that macroscopically look like the black hole of given mass, charges, and spin is not one: instead, it is huge, approximately $N$.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-27T16:45:48.623" CommentCount="5" />
  <row Id="6039" PostTypeId="2" ParentId="6026" CreationDate="2011-02-27T17:03:04.190" Score="3" ViewCount="" Body="&lt;p&gt;I don't think that such a computation of a theoretical limit of accuracy is possible. There are several sources of uncertainty in weather models:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;initial and boundary data,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;parameterizations,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;numerical instability, rounding and approximation errors of the numerical scheme employed to solve the Navier-Stokes equations for the atmosphere.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The term &quot;parameterization&quot; refers to the approximation of all subgrid processes, these are all processes/influences that happen at a scale that is smaller than the length of a grid cell. This includes effects from the topography, or the local albedo. More sophisticated approximations to subgrid processes can actually lead to less predictivity of a model, because the needed more detailed initial and boundary data are not available. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Navier-Stokes equations themselves are usually approximated up to a minimum length scale that is way larger than the length scales that would be necessary to resolve turbulent flows, these kinds of approximations are called &lt;a href=&quot;http://en.wikipedia.org/wiki/Large_eddy_simulation#Recent_applications&quot; rel=&quot;nofollow&quot;&gt;large eddy simulations&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accuracy of this truncation depends critically on the kind of flow and turbulence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I don't think that it is possible to derive a theoretical limit, what people do instead is performing ensemble runs, where the results of a weather model are compared that are calculated with slightly perturbed initial and boundary data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of such a &quot;twin&quot; experiment can be found here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;David Strauss, Dan Paolino: &lt;a href=&quot;http://en.wikipedia.org/wiki/Large_eddy_simulation#Recent_applications&quot; rel=&quot;nofollow&quot;&gt;Intermediate time error growth and predictability: tropics versus mid-latitudes&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(The result is that the error becomes significant after ca. 15 days of simulated time.)&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-02-27T17:03:04.190" CommentCount="1" />
  <row Id="6040" PostTypeId="2" ParentId="6002" CreationDate="2011-02-27T17:05:39.400" Score="2" ViewCount="" Body="&lt;p&gt;@Tim van Beek To look at it in another way, a lot of people get into physics with the aim of walking away from it after the undergrad, grad or phd. For instance, people may choose to go into environmental science. And a lot of physics phds go into biology because the problems there today are much more interesting than they are in physics. And lets face it. Biology has a lot of catching up to do!! :P&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2011-02-27T17:05:39.400" />
  <row Id="6041" PostTypeId="2" ParentId="6032" CreationDate="2011-02-27T17:40:27.133" Score="3" ViewCount="" Body="&lt;p&gt;It is wrong to think of virtual photons as of some surrounding of the electron. It is &lt;strong&gt;interaction&lt;/strong&gt; and is interaction-dependent thing, if you like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: If you calculate a field of some charge at some distance $\vec{r}$, it has the only physical meaning - as an external filed in the equations of motion of another charge placed at this point.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-02-28T13:59:00.593" LastActivityDate="2011-02-28T13:59:00.593" />
  <row Id="6042" PostTypeId="1" CreationDate="2011-02-27T18:06:39.077" Score="3" ViewCount="200" Body="&lt;p&gt;Assuming the correctness of QM: Would the size of such a computer be smaller than the observable universe? If it were to represent all available information in the universe it seems that it's minimal necessary size would relate to a black hole of a certain mass. Would this lead us to believe this theoretical quantum computer (assuming it didn't represent itself) could be constructed smaller than the universe? Or if the information content of an object is truly expressed in relation to a surface area is the universe the object with the minimal surface required to represent the totality of it's information? &lt;/p&gt;&#xA;" OwnerUserId="1567" LastEditorUserId="1567" LastEditorDisplayName="" LastEditDate="2011-02-27T19:24:24.857" LastActivityDate="2011-02-28T16:43:46.553" Title="Size of a quantum computer to effectively calculate macroscopic reality from quantum mechanics" Tags="&lt;quantum-information&gt;&lt;quantum-computation&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="6043" PostTypeId="2" ParentId="6026" CreationDate="2011-02-27T18:40:04.303" Score="2" ViewCount="" Body="&lt;p&gt;I am not sure how useful this &quot;back of the envelope&quot; calculation of reliability of Numerical Weather Prediction is going to be. Several of the assumptions in the question are not correct, and there are other factors to consider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some correcting points:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The Weather is 3 dimensional and resides on the surface of the planet up to a height of at least 10km. Furthermore the density decreases exponentially upwards. Many atmospheric phenomena involve the third dimension such as rising and falling air circulation effects; jet streams (7-16km).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The equations are fluid dynamics plus thermodynamics. The Navier-Stokes equations are not only too complex to solve, but in a sense inappropriate as well for the larger scales. One problem is that they might introduce &quot;high frequency&quot; effects (akin to every individual gust of wind or lapping of waves), which should be ignored. The earliest weather prediction models were seriously wrong because the high frequency fluctuations of pressure needed to be averaged rather than directly extrapolated. Here is a possible equation for one point of the atmosphere:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Tchange/time = solar + IR(input) + IR(output) + conduction + convection + evaporation + condensation + advection&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The regionality of the model is important too. In a global model there will be larger grid sizes and sources of error from initial conditions and surface and atmosphere top boundary conditions. In a mesoscopic prediction there will be smaller grid sizes but sources of error from the input edges as well. The smallest scale predictions of airflow around buildings and so on might be a true CFD problem using the Navier-Stokes equations however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I dont know that any calculation is done to predict the inaccuracies, although the different types including the numerical analysis (chaos) error sources can be studied separately. Models are tested against historical data for accuracy overall with predictions made 6-10 days out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To assume that the atmosphere &quot;goes turbulent&quot; after 3 days seems to conflate several issues together.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-27T18:40:04.303" CommentCount="3" />
  <row Id="6044" PostTypeId="2" ParentId="5994" CreationDate="2011-02-27T18:59:01.127" Score="3" ViewCount="" Body="&lt;p&gt;I suspect there is something wrong in the way you are adding the acceleration vector to the velocity vector after the first timestep. The simplest way to check you are doing things correctly is to write down your scheme in cartesian coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To point you in the right direction, I wrote a sample orbit integrator for you here: &lt;a href=&quot;https://gist.github.com/846419&quot; rel=&quot;nofollow&quot;&gt;Simplest orbit integrator&lt;/a&gt;, which should be at a level appropriate for a high school student. This is probably simplest integrator you can possibly write. With your code, you should get a nice circle for an x-y plot, and a nice sinusoid for the position and velocity components:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/yxS9k.png&quot; alt=&quot;x-y plot&quot;&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/eWFH3.png&quot; alt=&quot;t-x plot&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that it uses &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%27s_method&quot; rel=&quot;nofollow&quot;&gt;Euler's method&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec{v}(t+\Delta t) = \vec{v}(t) + \Delta t \ \vec{a}(t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec{x}(t+\Delta t) = \vec{x}(t) + \Delta t \ \vec{v}(t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is probably what you have been doing so far without realizing. This is the most inaccurate method to use when integrating, and once you introduce elliptical orbits, this method will give you wrong results after a few orbits. &#xA;There are many simple recommendations I can give you on how to further improve your code once it is fixed (normalizing units, a better integration scheme, etc.).&lt;/p&gt;&#xA;" OwnerUserId="1252" LastEditorUserId="1252" LastEditorDisplayName="" LastEditDate="2011-02-27T19:14:29.840" LastActivityDate="2011-02-27T19:14:29.840" CommentCount="3" />
  <row Id="6045" PostTypeId="1" AcceptedAnswerId="6046" CreationDate="2011-02-27T19:54:08.620" Score="5" ViewCount="218" Body="&lt;p&gt;suppose I have an electrostatic potential which I expand in spherical harmonics via&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{l,m} A^l_m r^n P_l^{|m|}(\cos \theta) e^{im\varphi}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I know that the field has cubic symmetry. Is there something I can say about the coefficients $A^n_m$? I ask because this would be the case for the crystal field in a cubic crystal, where it is known that the $e_g$ and the $t_{2g}$ orbitals are splitted. I am trying to show that the matrix-element is the same for all $e_g$, and the same for all $t_{2g}$. I tried it using the Wigner-Eckart theorem, but that seems not to be enough; I also need some symmetry relations for the $A^n_m$...&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditorDisplayName="" LastEditDate="2011-02-27T20:39:57.443" LastActivityDate="2011-02-27T23:59:22.380" Title="Expansion in spherical harmonics in cubic symmetry" Tags="&lt;quantum-mechanics&gt;&lt;symmetry&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="6046" PostTypeId="2" ParentId="6045" CreationDate="2011-02-27T20:30:24.827" Score="5" ViewCount="" Body="&lt;p&gt;Yes, of course, cubic symmetry is a hugely constraining condition on the coefficients $A_{nlm}$ - I suppose that you meant that the coefficients depend on $l$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cubic symmetry is generated by a few generators such as some rotations by 90 degrees. For example, if the state is symmetric with respect to the 90-degree rotation around the $z$ axis, it means that $\exp(\pi i J_z/2)$ has to keep the state invariant. It implies that $m$ i.e. the eigenvalue of $J_z$ has to be a multiple of four. All coefficients $A_{nlm}$ for which $m$ is not a multiple of four have to vanish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the state is also symmetric with respect to the $z\to(-z)$ reflection (parity), it means that $l$ has to be even: all coefficients with $l$ odd vanish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if the state is invariant under the rotation by 90 degrees around the $x$ axis, there is a similar condition for the coefficients that is slightly more complicated than that $m$ has to be a multiple of four (the condition will constrain some linear combinations of the coefficients), but it is essentially the same thing. If you impose this extra condition, you will find out that only 1/48 of the coefficients may be nonzero because 48 is the order of the group. (Replace 48 by 24 if you don't include the parity-odd transformations.)&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-27T20:30:24.827" CommentCount="7" />
  <row Id="6047" PostTypeId="1" CreationDate="2011-02-27T21:03:26.293" Score="5" ViewCount="698" Body="&lt;p&gt;How and from where does a mathematician learn physics from a mathematical stand point? I am reading the book by Spivak &lt;a href=&quot;http://www.math.uga.edu/~shifrin/Spivak_physics.pdf&quot; rel=&quot;nofollow&quot;&gt;Elementary Mechanics from a mathematicians view point&lt;/a&gt;. The first couple of pages of Lecture 1 of the book summarizes what I intend by physics from a mathematical stand point. I wanted to find out what are the other good sources for other branches of physics. Thanks&lt;/p&gt;&#xA;" OwnerUserId="403" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-08-31T13:05:09.887" LastActivityDate="2011-08-31T13:05:09.887" Title="Physics for mathematicians" Tags="&lt;mathematical-physics&gt;&lt;soft-question&gt;&lt;reference-request&gt;&lt;books&gt;&lt;big-list&gt;" AnswerCount="9" CommentCount="5" FavoriteCount="6" />
  <row Id="6048" PostTypeId="2" ParentId="3243" CreationDate="2011-02-27T21:24:21.277" Score="0" ViewCount="" Body="&lt;p&gt;Here is a paper on using belief population to approximate the partition function &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.143.9334&quot; rel=&quot;nofollow&quot;&gt;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.143.9334&lt;/a&gt; by a former colleague.&lt;/p&gt;&#xA;" OwnerUserId="2306" LastActivityDate="2011-02-27T21:24:21.277" />
  <row Id="6049" PostTypeId="2" ParentId="6002" CreationDate="2011-02-27T21:43:39.317" Score="3" ViewCount="" Body="&lt;p&gt;Not pursuing an academic career does not necessarily mean 'leaving physics'. There are loads of physics PhDs working in industrial R&amp;amp;D. True, their careers don't focus on finding the Higgs boson, but there is a lot of interesting and rewarding work to be done in applied physics. And yes, occasionally industrial scientist scoop up a Nobel. &lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-02-27T21:43:39.317" />
  <row Id="6050" PostTypeId="2" ParentId="5994" CreationDate="2011-02-27T21:49:59.680" Score="2" ViewCount="" Body="&lt;p&gt;Your units are wrong. Your acceleration of 0.0055 is m/s^2, and when you multiply by 86400 s/day, you get m/s/day, and not m/day^2, which is what you think you're getting. That's what confused you. &lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2011-02-27T21:49:59.680" />
  <row Id="6051" PostTypeId="2" ParentId="5979" CreationDate="2011-02-27T22:03:35.070" Score="1" ViewCount="" Body="&lt;p&gt;I looked up this somewhat cheezy Disney clip from the 1950s.  It features a spaceplane idea, but alas the plane takes off like a rocket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=rBgkrhnThek&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=rBgkrhnThek&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some reasons why you generally want rocket flight.  In a vertical ascent you quickly clear the atmosphere.  Within 90 seconds into rocket flight you are above 90% of the air.  Air has friction and is generally not that much help in lifting you far.  Aircraft can fly at a maximum of about 30,000m, and for orbital flight you have to go much further than that.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that ramjets or scramjets could be placed on a rocket.  Once it reaches supersonic these could kick in and provide thrust without the need to carry an oxidant until you reach about 50.000m.  The flight envelopes of modern combat aircraft reflect the safety limit.  If you fire up the aferburners and go near vertical you will reach the “edge of space.”  Of course the engines flame out and your descent will be violent and deadly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The business of spaceplanes is largely a load of BS.  These ideas keep surfacing now and then, and fortunately we have not yet spent billions on such a worthless program.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-02-27T22:03:35.070" />
  <row Id="6052" PostTypeId="1" AcceptedAnswerId="6118" CreationDate="2011-02-27T22:06:45.030" Score="15" ViewCount="955" Body="&lt;p&gt;Decades ago I was in a physics Ph.D. program, made a hash of things, and quite rightly was dropped from the program. Since then I've had a very satisfying career as a computer programmer, and picked up an M.S. in Applied Math. I'm currently working in a bioinformatics research lab, and really enjoying myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I might want to retire in five or six years, and lately I've been daydreaming about returning to my first love, and entering a physics Ph.D. program. This would be entirely for my personal satisfaction. I have no intention of seeking a further career in physics, I'd just like to spend a few years working in a physics research group. I've been out of physics too long to really know what area I'd like to work in, but I'm thinking quantum computing or quantum optics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I wouldn't require any financial support, I recognize that taking on a Ph.D. student is still a major investment of resources for a researcher. My hope is that my years of experience in software development would be enough of an asset to make it worth their while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just curious if there is anybody in the audience who's done something similar, or any faculty who can give an opinion about whether this is a realistic notion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to all for your answers and comments. I think nibot's comment that the thing to do is to establish relationships with some faculty I might like to work with is the most &quot;actionable&quot; answer. I really appreciated all the other anecdotes and advice though.&lt;/p&gt;&#xA;" OwnerUserId="2300" LastEditorUserId="184" LastEditorDisplayName="" LastEditDate="2011-03-01T14:15:55.203" LastActivityDate="2011-03-02T19:23:15.673" Title="Starting a physics Ph.D. at 60?" Tags="&lt;soft-question&gt;&lt;education&gt;&lt;grad-school&gt;" AnswerCount="9" CommentCount="1" FavoriteCount="7" />
  <row Id="6053" PostTypeId="2" ParentId="6047" CreationDate="2011-02-27T22:12:37.580" Score="2" ViewCount="" Body="&lt;p&gt;I personally don't know of particular books dedicated to the subject covering &lt;strong&gt;&lt;em&gt;all&lt;/em&gt;&lt;/strong&gt; areas of physics (maybe &quot;&lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Special%3aBookSources&amp;amp;isbn=9780521679718&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Mathematical Methods for Physics and Engineering&lt;/em&gt;&lt;/a&gt;&quot; by Riley, Hobson and Bence isn't quite what you're looking for), but if you happen to come across the subject of Quantum Field Theory then I suggest you have a look at &quot;&lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Special%3aBookSources&amp;amp;isbn=9780521632652&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Quantum Field Theory for Mathematicians&lt;/em&gt;&lt;/a&gt;&quot; by Ticciati.&lt;/p&gt;&#xA;" OwnerUserId="103" LastEditorUserId="103" LastEditorDisplayName="" LastEditDate="2011-08-25T22:25:56.063" LastActivityDate="2011-08-25T22:25:56.063" CommentCount="1" />
  <row Id="6054" PostTypeId="2" ParentId="6052" CreationDate="2011-02-27T22:16:27.727" Score="8" ViewCount="" Body="&lt;p&gt;After getting a Engineers degree from Naval postgrad in his youth and working twenty-five years as a industry researcher, my father went back to school and got his Piled Higher and Deeper in M.E. a few years after I finished mine in physics{*}, and I had a grad school colleagues in his late fifties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither seems to meet any unusual obstacles or special resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would, however, recommend talking to the department head and some prospective advisers before putting in any paper, just so you don't catch them by surprise.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This generated two funny episodes&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;He called me to ask for help with his homework&lt;/li&gt;&#xA;&lt;li&gt;He was the session chair at a conference to which his adviser was submitting a paper.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;and, of course, I &lt;em&gt;had&lt;/em&gt; to bring my regalia to his graduation for the Doctor and Doctor McKee pictures.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-02-27T22:16:27.727" />
  <row Id="6055" PostTypeId="1" CreationDate="2011-02-27T22:22:11.777" Score="3" ViewCount="281" Body="&lt;p&gt;I am a beginner in this quantum-mechanics stuff. I understand the quantum eraser only from an experimental view. So I didn't understand the formalism that describes the quantum eraser. But what does the experiment tells us? Does the photon know that there is somebody watching it? And this is why it behaves in another way? Does the photon also see the future?&lt;/p&gt;&#xA;" OwnerUserId="534" LastEditorUserId="534" LastEditorDisplayName="" LastEditDate="2011-02-28T10:33:43.463" LastActivityDate="2011-02-28T10:33:43.463" Title="What does the quantum eraser experiment tells us?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="6056" PostTypeId="2" ParentId="6047" CreationDate="2011-02-27T22:24:50.497" Score="3" ViewCount="" Body="&lt;p&gt;For mechanics at the next level (or perhaps skipping a level), you could try &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1441931430&quot; rel=&quot;nofollow&quot;&gt;Jerrold E. Marsden &amp;amp; Tudor S. Ratiu, &quot;Introduction to Mechanics and Symmetry&quot;, Springer, 1994&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For quantum field theory, a recent attempt at a moderately elementary level is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821847058&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Quantum-Theory-Mathematical-Surveys-Monographs/dp/0821847058&lt;/a&gt;, by Gerald B. Folland, which has the subtitle &quot;a tourist guide for mathematicians&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-02-27T22:24:50.497" />
  <row Id="6057" PostTypeId="2" ParentId="6047" CreationDate="2011-02-27T22:29:23.723" Score="4" ViewCount="" Body="&lt;p&gt;It is not just local-patriotism, but because I really think this book series should be very accessible to Mathematicians: Walter Thirring,&#xA;A Course in Mathematical Physics, various volumes.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-02-27T22:29:23.723" />
  <row Id="6058" PostTypeId="2" ParentId="6047" CreationDate="2011-02-27T22:38:26.087" Score="2" ViewCount="" Body="&lt;p&gt;For quantum field theory I like Folland's book &quot;Quantum Field Theory. A Tourist Guide for Mathematicians&quot;, because it is written by a mathematician with mathematicians as readers in mind. It is full of comments and explanations that a mathematician needs and are usually not in the physics books. Also he clearly says which parts a rigorous, from a mathematics point of view, and which are not. &lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2011-02-27T22:38:26.087" />
  <row Id="6059" PostTypeId="2" ParentId="6009" CreationDate="2011-02-27T22:43:10.327" Score="0" ViewCount="" Body="&lt;p&gt;The entropy of a black hole is a measure of the number of microstates, where for $N$ degenerate microstates the entropy is $S~=~k~log(N)$, which is associated with gravity.  The entropy for large $N$ is determined by the area of the event horizon $S~=~kA/4L_p^2$, where for the Schwarzschild black hole $A~=~16\pi M^2$.  The black hole is a system which holds a set of states with energy $E~=~M$ in a degeneracy $g(E)~=~exp(4\pi E^2)$ and the partition function is &#xA;$$&#xD;&#xA;Z(\beta)~=~\sum_E e^{4\pi E^2}e^{-\beta E}.&#xD;&#xA;$$&#xA;This partition function is divergent for $E~\rightarrow~\infty$.  The statistics for the number of degenerate microstates for a black hole is unbounded, and thus the partition function diverges. Black hole entropy is a coarse graining microstate states, which has been accomplished in string theory for large $N$.  The horizon area is a summation of these quantum numbers&#xA;$$&#xD;&#xA;A~=~16\pi\alpha_p\sum_{i=1}^Nn_i,&#xD;&#xA;$$&#xA;for $\alpha_p$ a Planck area.  The quantum numbers $n_i$ determine an element of the horizon area.  The energy is then counted as $E_n~=~\alpha E_p\sqrt{n}$, for $n~=~\sum_{i=1}^Nn_i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The degeneracy for $E_n$ is the number of ways $n~&amp;gt;~0$ is a sum of $N$ or less positive integers $n_i$.  It is the cardinality of the set of elements $\{n_1,~n_2,~,\dots,~n_m\}$, such that $n~=~\sum_{i=1}^m n_i$ for $1~\le~m~&amp;lt;~N$.   The number of ways a positive integer $m$ may be written as a sum of $m$ positive integers is the same problem as computing the number of ways of arranging $n$ balls in $m$ cells in a row.  The result is a degeneracy for the energy $E_n$ &#xA;$$&#xD;&#xA;g(E_n)~=~\sum_{m~=~1}^N\left(\matrix{n~-~1\cr m~-~1}\right),&#xD;&#xA;$$&#xA;for $N~\le~n$.  We also have that $m~\le~n$, which cuts the degeneracy further in &#xA;$$&#xD;&#xA;g’(E_n)~=~\sum_{m~=~1}^n\left(\matrix{n~-~1\cr m~-~1}\right).&#xD;&#xA;$$&#xA;The partition function is a summation of the two degenerate sets, &#xA;$$&#xD;&#xA;Z(\beta)~=~\sum_{n=1}^N\sum{m=1}^n\left(\matrix{n~-~1\cr m~-~1}\right)e^{-E_p\alpha\sqrt{n}}~+~\sum_{n=M+1}^\infty\sum_{m=1}^N\left(\matrix{n~-~1\cr m~-~1}\right)e^{-E_p\alpha\sqrt{n}}.&#xD;&#xA;$$&#xA;The two portions of the partition functions play a role at $n$ small and $n~&amp;gt;&amp;gt;~N$, and may be computed independently.  The convergence occurs for $n~&amp;gt;&amp;gt;N$ with &#xA;$$&#xD;&#xA;Z(\beta)~\simeq~\sum_{n~=~N+1}^\infty(n~-~1)^{N-1}e^{-\beta E_p\alpha\sqrt{n}}&#xD;&#xA;$$&#xA;This is a convergent partition function.  Conversely for a low black hole temperature $n~&amp;lt;&amp;lt;~N$, the degeneracy from the binomial theorem is $g’(E_n)~\simeq~2^{n-1}$ and the black hole entropy is $S~=~k~ln(2^{n-1})$ $=~(n~-~1)ln2$. The area $A~=~16\pi\alpha^2n$ permits us to set $\alpha~=~{1\over 2}\sqrt{{ln2}\over\pi}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above calculation can be looked at according to strings.  By holography the horizon is covered by strings which define all the quantum information which entered the black hole.  A generating function for stringy density of states computes a function which is similar to the above, and in the holographic setting describes the black hole as a string sphere on a stretched horizon.  This part is a bit involved, so I will jump forwards to say that a black hole may be thought of as a statistical state or phase of strings.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These quantum numbers associated with Planck units of area of the event horizon.  This is in keeping with the naturalized units of G = [Area].  These quantum numbers can include a range of quantitites, particular mass, angular momentum and electric charge.  The horizon exists as a radius&lt;/p&gt;&#xA;&#xA;&lt;p&gt;r_\pm~=~m~\pm~sqrt{m^2~-~Q^2~-~J^2}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which corresponds to the outer an inner horizons.  With the term in the square root is zero the two horizons meet and the spacelike region between them is “squashed” into an $AdS_2\times S^2$.  This is an extremal black hole, which has zero Hawking temperature.  In general these charges can be supersymmetric, or supercharges.  In the extremal case these charges are at the BPS bound.  In this case all the quantum numbers associated with those unit areas define an object which is similar to an elementary particle.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-02-28T01:18:55.887" LastActivityDate="2011-02-28T01:18:55.887" CommentCount="1" />
  <row Id="6061" PostTypeId="1" CreationDate="2011-02-27T23:13:16.500" Score="0" ViewCount="124" Body="&lt;p&gt;If I place a compass over a wire(such that the wire is positioned north-south) with charge flowing through it, and it points northeast, how can I determine the direction of the electron current flowing through the wire?&lt;/p&gt;&#xA;" OwnerUserId="2157" LastActivityDate="2011-02-27T23:36:16.330" Title="How is the direction of a compass affected by a charged wire?" Tags="&lt;electromagnetism&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6062" PostTypeId="2" ParentId="6061" CreationDate="2011-02-27T23:36:16.330" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/7/7d/RechteHand.png&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; you can see the magnetic field. This answer your question right?&lt;/p&gt;&#xA;" OwnerUserId="534" LastActivityDate="2011-02-27T23:36:16.330" />
  <row Id="6063" PostTypeId="2" ParentId="6045" CreationDate="2011-02-27T23:59:22.380" Score="2" ViewCount="" Body="&lt;p&gt;This exact problem came up in cosmology years ago, when people first started to consider the possibility of &quot;topologically small&quot; universes. These are models in which the universe is spatially flat, but has the topology of a 3-torus rather than of $R^3$. In other words, the Universe has a finite cell in the shape of a rectangular solid, with periodic boundary conditions. If the size of the cell is smaller than the horizon, then the cosmic microwave background temperature field has the cubic symmetry you describe (in the simplest case of a cubic fundamental cell). Since the connection between CMB observations and theory is generally done in a spherical harmonic expansion, the mathematical problem is identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upshot is that there are a bunch of selection rules of the sort Lubos describes, but no nice simple relations of the sort I think you're looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was a flurry of activity in the CMB community around these models in the 1990s. If you want to see what people came up with, you could try &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v71/i1/p20_1&quot; rel=&quot;nofollow&quot;&gt;Steven, Scott, and Silk (1993)&lt;/a&gt;, &lt;a href=&quot;http://adsabs.harvard.edu/abs/1995ApJ...448..477D&quot; rel=&quot;nofollow&quot;&gt;de Oliveira Costa and Smoot (1995)&lt;/a&gt;, or a bunch of other papers that cite these two.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-27T23:59:22.380" CommentCount="1" />
  <row Id="6064" PostTypeId="2" ParentId="6047" CreationDate="2011-02-28T00:04:36.510" Score="4" ViewCount="" Body="&lt;p&gt;If you're into general relativity, you could try General Relativity for Mathematicians, by Sachs and Wu. I only know general relativity for physicists, so I can't comment on whether this book is any good, but it might be worth a try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sachs is the one known to relativists and cosmologists for the Sachs-Wolfe effect. He taught my E&amp;amp;M class when I was a graduate student at Berkeley in the early 1990s. He was housed in the mathematics department, but I guess he had a joint appointment in physics.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-02-28T00:04:36.510" CommentCount="2" />
  <row Id="6065" PostTypeId="2" ParentId="6052" CreationDate="2011-02-28T00:28:08.533" Score="8" ViewCount="" Body="&lt;p&gt;I would advise to test the waters first. Are you really motivated enough to spend the long hours, days, months? Do you have the stamina to try the next path given that the previous zillion you tried led to nothing? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ArXiv provides a wealth of recent papers on any physics subject (including quantum computing and quantum optics). Make sure you keep up-to-date in a few areas of physics that appeal to you. If you manage to do that in your free hours, chances are you will enjoy your PhD work. It will also help you making an informed choice on the subject of your PhD. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2011-02-28T00:28:08.533" CommentCount="1" />
  <row Id="6066" PostTypeId="2" ParentId="2084" CreationDate="2011-02-28T00:30:08.750" Score="1" ViewCount="" Body="&lt;p&gt;Maybe this one can help too:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rüdiger Paschotta:&#xA;Encyclopedia of Laser Physics and Technology,&#xA;&lt;a href=&quot;http://www.rp-photonics.com/topics.html&quot; rel=&quot;nofollow&quot;&gt;online version&lt;/a&gt; or &#xA;&lt;a href=&quot;http://www.wiley-vch.de/publish/en/books/bySubjectEE00/ISBN3-527-40828-2&quot; rel=&quot;nofollow&quot;&gt;paper version&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can look for &lt;strong&gt;LIA Handbook of Laser Material Processing&lt;/strong&gt;  or &lt;strong&gt;Laser Processing of Engineering Materials&lt;/strong&gt; by John C. Ion.&lt;/p&gt;&#xA;" OwnerUserId="2307" LastActivityDate="2011-02-28T00:30:08.750" />
  <row Id="6068" PostTypeId="1" CreationDate="2011-02-28T02:01:16.680" Score="1" ViewCount="487" Body="&lt;p&gt;What are some good books for learning the concepts of Kinematics, Newton laws, 2D Motion of Object etc.?&lt;/p&gt;&#xA;" OwnerUserId="2273" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2011-02-28T13:52:47.243" LastActivityDate="2011-02-28T18:34:59.263" CommunityOwnedDate="2011-02-28T03:04:13.270" Title="List of good classical physics books" Tags="&lt;newtonian-mechanics&gt;&lt;books&gt;&lt;kinematics&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="6069" PostTypeId="2" ParentId="6055" CreationDate="2011-02-28T02:13:03.867" Score="1" ViewCount="" Body="&lt;p&gt;&quot;Quantum eraser&quot; was first proposed by M. Scully and K. Druhl. In order to understand it, we must know the famous double slit experiment first which I suppose you already know. You must be aware that in the double slit experiment if photons are emitted one at a time an interference pattern forms at the detector screen. As soon as you try to observe which path the photon follows that is which slit it passes through, left or right slit, the interference disappears. That means a knowledge of &quot;which path&quot; information destroys the wave like character of a photon and hence no interference possible. But in 1982, Scully and Druhl suggested a stunning modification of the experiment. They proposed the following on the basis of their quantum mechanical calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose a tagging device is attached by which we can know the &quot;which path&quot; information of the photon. Now if, just before the photon hits the detector screen, we eliminate the possibility of our knowledge of the &quot;which path&quot; information by erasing the mark registered by the tagging device, both possibilities that is the photon passed through the left slit and photon passed through the right slit should come back into play. Both histories should come back once again and interference pattern should reemarge. As if we are kind of shaping the past (warning: it is by no means that future is affecting the past).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experiment carried out by Raymond Chiao, Paul Kwiat and A. Steinberg. Remarkably it worked just as scully and Druhl predicted. Interference pattern indeed reemarged.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-02-28T02:13:03.867" CommentCount="4" />
  <row Id="6070" PostTypeId="2" ParentId="6068" CreationDate="2011-02-28T02:29:23.503" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0805390464&quot; rel=&quot;nofollow&quot;&gt;The Feynman Lectures on Physics Volume 1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Suf1W.jpg&quot; alt=&quot;Image from Amazon.com&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image from Amazon.com&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-02-28T02:29:23.503" CommunityOwnedDate="2011-02-28T03:04:13.270" CommentCount="1" />
  <row Id="6071" PostTypeId="2" ParentId="6068" CreationDate="2011-02-28T02:34:13.403" Score="2" ViewCount="" Body="&lt;p&gt;I guess this was too basic for the Physicists here to answer so this lurking Computer Science guy will take a shot. I read those things in High School using the book by Resnik &amp;amp; Halliday. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were taught in the Russian tradition (like me) with emphasis on math and lot of problem solving, you might like the book by Irodov. Learning Physics by solving problems is IMHO the best way to do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/5030008004&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Problems-general-physics-I-Irodov/dp/5030008004&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2011-02-28T02:34:13.403" CommunityOwnedDate="2011-02-28T03:04:13.270" />
  <row Id="6073" PostTypeId="2" ParentId="6047" CreationDate="2011-02-28T04:59:58.883" Score="7" ViewCount="" Body="&lt;p&gt;You want the book by V.I. Arnold, Mathematical Methods of Classical Mechanics.&lt;/p&gt;&#xA;" OwnerUserId="2311" LastActivityDate="2011-02-28T04:59:58.883" CommentCount="2" />
  <row Id="6074" PostTypeId="1" AcceptedAnswerId="6076" CreationDate="2011-02-28T05:06:15.577" Score="3" ViewCount="407" Body="&lt;p&gt;Do we take gravity = 9.8 m/s² for all heights when solving problems?&lt;/p&gt;&#xA;" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2011-02-28T10:48:08.587" LastActivityDate="2011-02-28T16:10:21.370" Title="Do we take gravity = 9.8 m/s² for all heights when solving problems? Why or why not?" Tags="&lt;gravity&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="1" />
  <row Id="6075" PostTypeId="1" AcceptedAnswerId="6077" CreationDate="2011-02-28T05:13:32.330" Score="2" ViewCount="217" Body="&lt;p&gt;I'm sure a simple question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a video of me jumping off a cliff into a river.  I want to calculate how high it is.  I know my weight, acceleration due to gravity of course, and I can get the time it took to hit the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I calculate (albeit roughly) how high the cliff was?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tim.&lt;/p&gt;&#xA;" LastActivityDate="2011-02-28T22:32:43.493" Title="Can I calculate the height of a cliff from weight of falling object and time taken?" Tags="&lt;gravity&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="6076" PostTypeId="2" ParentId="6074" CreationDate="2011-02-28T05:14:28.507" Score="12" ViewCount="" Body="&lt;p&gt;No, the value $9.8\frac{\mathrm{m}}{\mathrm{s}^2}$ is an approximation that is only valid at or near the Earth's surface. You can go a few miles up or down and it'll still be good enough, but once you get any significant distance away from the surface of Earth, you would need to use a different value for gravitational acceleration. You can calculate the value from Newton's law of gravitation, $F = Gm_1m_2/r^2$, and you'll get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g = \frac{GM}{r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ is the mass of the Earth and $r$ is the distance from the Earth's &lt;strong&gt;center&lt;/strong&gt; to the point for which you are doing the calculation.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-02-28T05:14:28.507" />
  <row Id="6077" PostTypeId="2" ParentId="6075" CreationDate="2011-02-28T05:21:59.097" Score="4" ViewCount="" Body="&lt;p&gt;If I remember correctly, the distance should be $y=\frac{1}{2} g t^2$. This is for if you simply fell off the cliff. Now if you &lt;em&gt;jumped&lt;/em&gt; off the cliff, it will be different because you may have some initial velocity in the y-direction. Also, of course, air friction is not being accounted for. &lt;/p&gt;&#xA;" OwnerUserId="1939" LastActivityDate="2011-02-28T05:21:59.097" CommentCount="4" />
  <row Id="6078" PostTypeId="2" ParentId="6074" CreationDate="2011-02-28T05:25:17.903" Score="8" ViewCount="" Body="&lt;p&gt;To expand a little on &lt;a href=&quot;http://physics.stackexchange.com/questions/6074/why-do-we-take-gravity-9-8-m-s/6076#6076&quot;&gt;David's point&lt;/a&gt; assume we move from the nominal &quot;surface&quot; where $g$ is $9.8\text{ m}/\text{s}^2$ to another point at radius $r + \Delta r$. How much does the acceleration of gravity change?&#xA;$$&#xD;&#xA;g = \frac{GM}{(r+\Delta r)^2} = \frac{GM}{r^2(1 + \Delta r/r)^2}&#xD;&#xA;$$&#xA;and as long as $\Delta r$ is small compared to $r$ we can reasonably approximate this as&#xA;$$&#xD;&#xA;g \approx \frac{GM}{r^2}(1 - 2\frac{\Delta r}{r}) .&#xD;&#xA;$$&#xA;Well, the radius of the Earth is about $6000 \text{ km}$ so the approximation is good at less than 1% error for around $30\text{ km}$ up or down from the nominal surface, which is all the land and sea floor, and a bit up and down from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth noticing that due to variations in the local mass density of the Earth the measured value of $g$ even at the surface can vary by several tenth of a percent.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-02-28T05:25:17.903" CommentCount="1" />
  <row Id="6079" PostTypeId="2" ParentId="6074" CreationDate="2011-02-28T05:25:44.783" Score="4" ViewCount="" Body="&lt;p&gt;$g$ becomes $ g \approx 9.7 \frac{m}{s^2}$ at a height of about 35km, so it would be ok to use the value $9.81$ for &quot;down to earth&quot; problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Earth%27s_gravity&quot; rel=&quot;nofollow&quot;&gt;relevant wikipedia article&lt;/a&gt; has lots of useful information, like for example the following approximation formula for different heights:   $$  g_h=g_0\left(\frac{r_e}{r_e+h}\right)^2 $$&#xA;Where $g_h$, is the gravity measure at height $h$ above sea level; $r_e$, is the Earth's mean radius and $g_0$, is the standard gravity.&lt;/p&gt;&#xA;" OwnerUserId="824" LastActivityDate="2011-02-28T05:25:44.783" CommentCount="2" />
  <row Id="6080" PostTypeId="2" ParentId="6042" CreationDate="2011-02-28T05:46:33.170" Score="1" ViewCount="" Body="&lt;p&gt;Assuming the correctness of QM, you can encode $2^{n}$ complex amplitudes with the computational basis of n entangled qubits. This suggests that the answer to the question is yes, much smaller than the universe in fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately quantum computing technology isn't there yet to maintain useful amount of entangled qubits, or getting that much information out of the qubit system for that matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one of those questions that makes studying quantum computation very interesting.&lt;/p&gt;&#xA;" OwnerUserId="1670" LastEditorUserId="1670" LastEditorDisplayName="" LastEditDate="2011-02-28T16:43:46.553" LastActivityDate="2011-02-28T16:43:46.553" CommentCount="3" />
  <row Id="6081" PostTypeId="2" ParentId="6055" CreationDate="2011-02-28T06:25:58.213" Score="2" ViewCount="" Body="&lt;p&gt;No, the photon doesn't see anyone watching it. And the photon doesn't see its future, either. In fact, the photon doesn't exist in any classical sense prior to its observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of its properties - e.g. which slits it could be taking; whether it behaves more as a particle or a wave etc. - are encoded in the wave function until the very moment of the measurement which is why they may always be &quot;changed back&quot; to the previous answers. For example, in quantum eraser, the photon is ordered to behave as a wave again, even though a premature argument could lead a sloppy person to think that the photon has already decided to behave as a particle forever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you measure the photon, it is finally possible to think of its properties classically and the wave function allows one to calculate all probabilities that the outcome will be something or something else. In the case of the quantum eraser, we restore the interference pattern. But any attempt to &quot;imagine&quot; that the photon has obtained a classical property at any moment before it was measured would lead to wrong predictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is always essential to appreciate that the photon always behaves according to the laws of quantum mechanics and we're never allowed to approximate it by any classical intuition because the classical intuition fails. This strict requirement that classical mechanics is wrong may only be partly circumvented after the photon is actually detected (because then it interacts with a classical object that quickly decoheres) - but not earlier than that. In other words, quantum mechanics always holds: that's the main lesson of this experiment (and many others).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sb1 says that it was remarkable that the experiment behaved as Scully and Druhl predicted. I disagree with this wording. The prediction could have been made by any father of quantum mechanics - no new physics was used whatsoever and they could predict the behavior of any setup of this kind. It could have been remarkable in the 1920s but after the 1920s, all such experiments were mundane physics.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-02-28T06:25:58.213" CommentCount="1" />
  <row Id="6082" PostTypeId="2" ParentId="2084" CreationDate="2011-02-28T08:58:01.300" Score="1" ViewCount="" Body="&lt;p&gt;This is a good all-round book about lasers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Orazio Svelto, &lt;em&gt;Principles of Lasers&lt;/em&gt;, 4th edition. Springer, 1998.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="107" LastActivityDate="2011-02-28T08:58:01.300" />
  <row Id="6083" PostTypeId="2" ParentId="6052" CreationDate="2011-02-28T09:41:04.143" Score="0" ViewCount="" Body="&lt;p&gt;In the UK there is the Open University which is a distance learning organisation with interactive aspects. As well as actual degrees it is possible to sign up for (sets of) modules as part of a longer term MSc (Master of Science) say. So if something similar is available in the US you might want to prepare by choosing some advanced MSc type topics to study in this way. The OU takes students of all ages, as I am sure any equivalent system would.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-02-28T09:41:04.143" />
  <row Id="6084" PostTypeId="1" AcceptedAnswerId="6514" CreationDate="2011-02-28T10:35:38.863" Score="3" ViewCount="246" Body="&lt;p&gt;Let the supersymmetry transformations for the chiral multiplet $(z_k,\psi_{kL},f_k)$ be,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta z_k = 2i \bar{\alpha} \psi_{kL}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta \psi_{kL} = D_\mu z_k \gamma ^\mu \alpha_R + f_k \alpha_L$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta f_k = 2i\bar{\alpha} \gamma^\mu D_{\mu} \psi_{kL} - 2ie\bar{\alpha}\lambda _R z^k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly let the gauge multiplet $(A_\mu,\lambda,D)$ transform as,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta A_\mu = i \bar{\alpha}\gamma_\mu \lambda$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta \lambda = -\frac{1}{2}F^{4}_{\mu \nu}\gamma_4 ^{\mu \nu} \alpha + D \gamma_5 \alpha $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta D = i \bar{\alpha}\gamma_5 \gamma ^ \mu \partial _ \mu \lambda$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here in $F^4$ the $\mu$ and $\nu$ go over indices $0,1,2,3$ and $\gamma_4 ^{\mu \nu} = \frac{1}{2} [\gamma_4^\mu, \gamma_4^\nu]$. (these gamma matrices are defined below) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One does something called &quot;dimensional reduction&quot; of these to $2+1$ dimensional spacetime by assuming that the fields are independent of the third spatial coordinate. In  $2+1$ dimensional spacetime the gamma matrices are defined as,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\gamma ^0_3 = -i\sigma^2$, $\gamma^1_3 = \sigma ^3$ and $\gamma^2_3 = \sigma ^1$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the so called &quot;Majorana representation&quot; the $4-$dimensional Gamma matrices can be written such that $\gamma_4 ^{0\1\2} = \left [ \begin{array}{cc}  &amp;amp; \gamma_3 ^{0\1\2} \\&#xD;&#xA;\gamma_3 ^{0\1\2} &amp;amp;  \end{array} \right ] $ and $\gamma_4 ^3 = \left [ \begin{array}{cc}  I &amp;amp; \\&#xD;&#xA; &amp;amp; -I  \end{array} \right ]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One renames the third component of the gauge field $A_3$ as $C$ and splits the $4-$component fermions into $2-$component spinors in $2+1$ dimensions as, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda = \left [ \begin{array}{c} \lambda_1 \\ \lambda_2 \end{array} \right ]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha = \left [ \begin{array}{c} \alpha_1 \\ \alpha_2 \end{array} \right ]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One substitutes the above in the first set of supersymmetric transformations and sets to $0$ all derivatives with respect to the $3^{rd}$ spatial direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then one is supposed to get, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta A_\mu = i \bar{\alpha}_a \gamma_\mu \lambda _a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta \lambda _a = - \frac{\epsilon ^{\mu \nu \rho}F_{\nu \rho}}{2}\gamma _ \mu \alpha_a + \partial _\mu C \gamma ^\mu \alpha ^a + D \alpha ^a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta C = i\bar{\alpha}^a \lambda _a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta D = i \bar{\alpha}^a\gamma ^\mu \partial_\mu \lambda_a$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a$ runs over $1,2$ and $\alpha^1 = \alpha_2$ and $\alpha^2 = -\alpha_1$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing the above I can get the above transformations for all the fields. (Just that I need to use a fact that is true only for these gamma matrices in the $3$-dimensions that $[\gamma_3^\mu,\gamma_3^\nu]=2\epsilon ^{\mu \nu \rho}\gamma_{3\rho}$)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In this &quot;dimensional reduction&quot; the particular choice of the gamma matrices seemed crucial. Is this true?  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I don't know how but this is supposed to match the transformations of the $\cal{N}=2$ vector multiplet in $2+1$ dimensions which has componets, $A_\mu, \lambda_a,C,D$ (why?) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I don't know how one derives the field content (as above) of the $\cal{N}=2$ vector multiplet in $2+1$ dimensions. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;From the above it is claimed that the following forms a lagrangian respecting the above symmetry and is what gets called the ``supersymmetric Chern-Simons theory&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$L = \frac{\kappa}{2} (\epsilon ^{\mu \nu \rho} A_\mu \partial _\nu A_\rho - i \bar{\lambda_a}\lambda_a + 2CD)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to establish the above claim I need to assume two results,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First is a property which again seems typical of $3$ dimensional gamma matrices that, $\gamma_3^{\mu \dagger} \gamma_3 ^0 = -\gamma_3^0\gamma_3^\mu$ &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Secondly I need to assume the following variation for the fermion self-coupling, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta (\bar{\lambda} \lambda) = 2\delta (\bar{\lambda}) \lambda$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is not clear to me as to why should the above hold. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A crucial difference between the above and the usual Abelian supergauge theory is that the kinetic term for the fermion is missing. How does one understand this?  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1637" LastEditorUserId="1637" LastEditorDisplayName="" LastEditDate="2011-03-08T09:47:09.540" LastActivityDate="2011-03-08T11:47:39.777" Title="Dimensional reduction from $3+1$ to $2+1$ for $\cal{N}=2$ vector superfield" Tags="&lt;quantum-field-theory&gt;&lt;supersymmetry&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="6085" PostTypeId="2" ParentId="5792" CreationDate="2011-02-28T10:37:12.590" Score="3" ViewCount="" Body="&lt;p&gt;Poisson brackets are closely related also to transformations of a system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a &quot;generator&quot; $\delta G$ and some quantity $A$. (All the quantities mentioned are functions of $p_i$, $q_i$, so I'll not write the arguments unless it's necessary). The small transformation, generated by $\delta G$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A \to A+\delta A,\quad\quad \delta A = -\{\delta G, A\} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One have to be careful with the signs in the definition. I use the one &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson_bracket&quot; rel=&quot;nofollow&quot;&gt;from Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example 1 -- momentum&lt;/strong&gt;&lt;br&gt;&#xA;$\delta G = \epsilon_i p_i \quad \delta A = -\epsilon_i\{p_i, A\} = \epsilon_i\frac{\partial A}{\partial q_i},$&lt;br&gt;&#xA;$\quad \Rightarrow \quad A(p_i,q_i)\to A(p_i,q_i)+\epsilon_i\frac{\partial A}{\partial q_i} = A(p_i,q_i+\epsilon_i)$&lt;br&gt;&#xA;The momentum is the generator of translations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example 2 -- angular momentum&lt;/strong&gt;&lt;br&gt;&#xA;$\delta G = \epsilon_i e_{ijk} p_iq_k \quad \delta A = -\epsilon_i e_{ijk} \{p_jq_k, A\}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $e_{ijk}$ -- is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Levi-Civita_symbol&quot; rel=&quot;nofollow&quot;&gt;Levi-Civita symbol&lt;/a&gt;. Expanding:&lt;br&gt;&#xA;$\delta A = \epsilon_i e_{ijk} \sum_\alpha \left(\frac{\partial p_jq_k}{\partial q_\alpha}\frac{\partial A}{\partial p_\alpha}-\frac{\partial p_jq_k}{\partial p_\alpha}\frac{\partial A}{\partial q_\alpha}\right) = \epsilon_i e_{ijk}\left(p_j\frac{\partial A}{\partial p_k}-q_j\frac{\partial A}{\partial q_k}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad \Rightarrow \quad A(p_i,q_i)\to A(p_i,q_i)+\epsilon_i e_{ijk}\left(p_j\frac{\partial A}{\partial p_k} +  q_j\frac{\partial A}{\partial q_k}\right) = A(R_{ij}p_j,R_{ij}q_j)$&lt;br&gt;&#xA;For infinitesimal rotations $R_{ij}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the angular momentum is the generator of rotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example 3 -- energy&lt;/strong&gt;&lt;br&gt;&#xA;By using Hamilton equation one can calculate to which transformations the energy corresponds to (it is essentially a &quot;reverse&quot; of what is written &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson_bracket&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;):  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta G = \epsilon H \quad \delta A = -\epsilon\{H, A\} = \epsilon\frac{d A}{d t},$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\quad \Rightarrow \quad A(p_i(t),q_i(t))\to A(p_i(t),q_i(t))+\epsilon\frac{d A}{d t} = A(p_i(t+\epsilon),q_i(t+\epsilon))$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy is the generator of time evolution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From here one can directly see the relation between symmetries and conservation laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a Hamiltonian is symmetric under certain transformation, then the bracket with the corresponding &quot;generator&quot; must vanish  $\{\delta G, H\} = 0$. But this also means that the value of the  &quot;generator&quot; doesn't change with time.&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-02-28T10:37:12.590" />
  <row Id="6086" PostTypeId="1" AcceptedAnswerId="6383" CreationDate="2011-02-28T12:01:37.503" Score="4" ViewCount="1117" Body="&lt;p&gt;Edit to add: I'm awarding the bounty to dbrane, whose answer will likely retain the lead in votes, even though it's clearly incorrect as I see it. (In short, the correct answer cannot be that an infinitesimally small frame is required to test the EP. It is in fact tested in larger labs. The tidal force need only be small enough that it doesn't affect the outcome. Nor is the horizon a special place in regards to the tidal force, says GR.) I do appreciate the answers. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit to add: this question hasn't been properly answered. The #1 answer below made a false assumption about the question. I've beefed up the question to address the objections in the answers below. I added my own answer to recap the objections and reach a conclusion. Please read the whole post before answering; I may have already covered your objection. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the following brief thought experiment fail to show that general relativity (GR) has a major problem in regards to black holes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The full thought experiment is in &lt;a href=&quot;http://finbot.wordpress.com/2008/03/05/no-black-holes/&quot; rel=&quot;nofollow&quot;&gt;my blog post&lt;/a&gt;. The post claims that GR violates its own equivalence principle at the horizon of a black hole. The principle says that the laws of physics in any sufficiently small, freely falling frame are the same as they are in an inertial frame in an idealized, gravity-free universe. Here's a condensed version of the thought experiment:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In an arbitrarily small, freely falling frame &lt;em&gt;X&lt;/em&gt; that is falling through the horizon of a black hole, let there be a particle above the horizon that is escaping to infinity. A free-floating rod positioned alongside the particle and straddling the horizon couldn't be escaping to infinity as well, or else it'd be passing outward through the horizon. However, if instead the rod didn't extend as far down as the horizon, then in principle it &lt;em&gt;could&lt;/em&gt; be escaping, possibly faster than the particle beside it. &lt;strong&gt;In an inertial frame, unlike in &lt;em&gt;X&lt;/em&gt;, a body's freedom of movement (in principle and if only relative to other free objects in the frame) doesn't depend on the body's position or extent.&lt;/strong&gt; Then a test of the laws of physics can distinguish &lt;em&gt;X&lt;/em&gt; from an inertial frame. If &lt;em&gt;X&lt;/em&gt; was equivalent to an inertial frame, I wouldn't be able to tell whether the rod could possibly be passing the particle in the outward direction, by knowing only whether the rod extends as far down as an imaginary boundary (the horizon) within the frame. If &lt;em&gt;X&lt;/em&gt; was equivalent to an inertial frame, the rod could in principle be passing the particle in the outward direction regardless of its extent within &lt;em&gt;X&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The thought experiment above takes place completely within &lt;em&gt;X&lt;/em&gt;, which is arbitrarily small in spacetime (arbitrarily small both spatially and in duration). That is, the experiment is completely &lt;em&gt;local&lt;/em&gt;. That the particle is escaping to infinity is a process occurring within &lt;em&gt;X&lt;/em&gt;; it tells us that the particle won't cross the horizon during the lifetime of &lt;em&gt;X&lt;/em&gt;. The particle needn't &lt;em&gt;reach&lt;/em&gt; infinity before the experiment concludes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It isn't necessary to be able to detect (by some experiment) that a horizon exists within &lt;em&gt;X&lt;/em&gt;. It's a given (from the givens in the thought experiment) that a horizon is there. Likewise, I am free to specify the initial conditions of a particle or rod in relation to the horizon. For example, I am free to specify that the rod straddles the horizon, and draw conclusions from that. The laws of physics in &lt;em&gt;X&lt;/em&gt; are affected by the presence and properties of the horizon regardless whether an observer in that frame detects the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that the only way the equivalence principle is satisfiable in &lt;em&gt;X&lt;/em&gt; is when in principle the rod can be escaping to infinity regardless of its initial position or extent in &lt;em&gt;X&lt;/em&gt;, which would rule out black holes in a theory of gravity consistent with the principle. Otherwise, it seems the &lt;strong&gt;bolded&lt;/strong&gt; sentence must be incorrect. If so, how? In other words, how can I &lt;em&gt;not&lt;/em&gt; tell whether the rod can possibly be passing the particle in the outward direction, by knowing only whether it extends as far down as the horizon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd appreciate hearing from Ted Bunn or other experts on black holes. A barrier to getting a satisfactory answer to this question is that many people believe the tidal force is so strong at the horizon that the equivalence principle can't be tested there except impossibly, within a single point in spacetime. An equation of GR (see my blog post) shows that a horizon isn't a special place in regards to the tidal force, in agreement with many texts including Ted Bunn's Black Hole FAQ. In fact the tidal force can &lt;em&gt;in principle&lt;/em&gt; be arbitrarily weak in any size &lt;em&gt;X&lt;/em&gt;. To weaken the tidal force in any given size &lt;em&gt;X&lt;/em&gt;, just increase the mass of the black hole. (Or they might believe it's fine to test the principle in numerical approximation in a frame larger than a point, but not fine to test it logically in such frame anywhere. Kip Thorne disagrees, in a reference in my blog post.) Note also that the &lt;a href=&quot;http://chandra.harvard.edu/resources/faq/black_hole/bhole-10.html&quot; rel=&quot;nofollow&quot;&gt;Chandra X-ray Observatory FAQ&lt;/a&gt; tells us that observations of black holes to date aren't confirmations of GR, rather they actually depend on the theory's validity, which is to say the existence of black holes in nature isn't proven.&lt;/p&gt;&#xA;" OwnerUserId="2314" LastEditorUserId="2314" LastEditorDisplayName="" LastEditDate="2011-03-10T02:05:21.040" LastActivityDate="2011-08-23T18:21:35.967" Title="How does this thought experiment not rule out black holes?" Tags="&lt;general-relativity&gt;&lt;gravity&gt;&lt;black-holes&gt;&lt;equivalence-principle&gt;" AnswerCount="8" CommentCount="11" FavoriteCount="1" />
  <row Id="6087" PostTypeId="2" ParentId="970" CreationDate="2011-02-28T12:16:04.293" Score="0" ViewCount="" Body="&lt;p&gt;My understanding is that a freely falling observer who falls into a black hole won't see any future date arrive at a distant point before crossing the event horizon. I think that's true only for an accelerating observer who hovers ever closer to a horizon.&lt;/p&gt;&#xA;" OwnerUserId="2314" LastActivityDate="2011-02-28T12:16:04.293" />
  <row Id="6088" PostTypeId="2" ParentId="5505" CreationDate="2011-02-28T12:30:02.457" Score="0" ViewCount="" Body="&lt;p&gt;Let the light source be pointed directly upward. Assume that the observer (who receives the light) can remain hovering above the black hole as it expands. The observer would notice that the redshift of the light ever increases, until such moment that the last photon is received (subsequent photons fall into the black hole). The last photon could be received at any point in the observer's future: the closer to the horizon (but still above it) that the photon is emitted, the further in the observer's future it is received.&lt;/p&gt;&#xA;" OwnerUserId="2314" LastActivityDate="2011-02-28T12:30:02.457" />
  <row Id="6089" PostTypeId="2" ParentId="2481" CreationDate="2011-02-28T12:39:55.990" Score="7" ViewCount="" Body="&lt;p&gt;I like answers that appeal to symmetry, so I answer this one with a question: If you were at the center, which way would you fall? That tells us you could stay floating there.&lt;/p&gt;&#xA;" OwnerUserId="2314" LastActivityDate="2011-02-28T12:39:55.990" />
  <row Id="6090" PostTypeId="2" ParentId="6086" CreationDate="2011-02-28T13:08:46.050" Score="24" ViewCount="" Body="&lt;p&gt;I just read your blog post and it's clear to me where you've gone wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equivalence principle only allows you to transform to an inertial frame &lt;em&gt;locally&lt;/em&gt;. This means that if your spacetime is curved, then the falling observer can only choose Minkowski coordinates for an infinitesimal region around her. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of a curved surface and having to choose a very small patch on it for it to appear flat. Clearly, you can't extend that flat patch indefinitely and call it an inertial frame of infinite extent (which you require in order to argue that the frame would allow you to send signals out to infinity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizon is a global object that you realize exists when you patch together all the infinitesimal coordinate systems and examine its causal structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, yes, the falling observer can do experiments to realize the horizon exists, but this does not violate the Equivalence Principle because such experiments are not done locally in an infinitesimal region. This applies to the rod that you seem to want to send away to infinity after crossing the horizon too. The infinitesimal flat patch in which you're allowed to play with the EP does not include infinity (or anything beyond the horizon), so you can't throw things outside of the horizon once you've crossed. &lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-02-28T13:16:31.987" LastActivityDate="2011-02-28T13:16:31.987" CommentCount="16" />
  <row Id="6091" PostTypeId="2" ParentId="1916" CreationDate="2011-02-28T13:22:08.607" Score="0" ViewCount="" Body="&lt;p&gt;Since you want to focus on what an individual part of the dead star will &quot;feel&quot; when it becomes a black hole, I think the &quot;no hair&quot; part is outside the scope of that; that's a condition only for someone outside of the hole. (See &quot;externally observable&quot; in that Wikipedia link on &quot;no hair&quot; you gave.) For someone falling with the collapsing body, whether or not it's a star, the body has &quot;hair&quot; all the way short of the singularity. (&lt;em&gt;At&lt;/em&gt; the singularity, general relativity breaks down.) I'd say the singularity is first formed when the first chunk of matter that is below the new horizon reaches the center.&lt;/p&gt;&#xA;" OwnerUserId="2314" LastActivityDate="2011-02-28T13:22:08.607" />
  <row Id="6092" PostTypeId="2" ParentId="6068" CreationDate="2011-02-28T13:26:21.283" Score="2" ViewCount="" Body="&lt;p&gt;Try &lt;strong&gt;Fundamental Laws of Mechanics (by I.E. Irodov)&lt;/strong&gt;. Don't use its problem book as mentioned in a previous answer.&#xA;The book is written in traditional Russian style, but it will give you clear &amp;amp; super advance idea of mechanics. Its useful to a High School guy as well as a PHD student.&#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B0007ASWBW&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Fundamental-laws-mechanics-I-Irodov/dp/B0007ASWBW/&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you feel it somewhat compact &amp;amp; hard-to-learn book, look no further than &lt;strong&gt;Fundamentals of Physics (by Resnik &amp;amp; Halliday)&lt;/strong&gt;.&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/pqq4T.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/047004473X&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Fundamentals-Physics-Chapters-David-Halliday/dp/047004473X/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2170" LastActivityDate="2011-02-28T13:26:21.283" CommunityOwnedDate="2011-02-28T13:26:21.283" />
  <row Id="6093" PostTypeId="2" ParentId="6086" CreationDate="2011-02-28T14:14:20.570" Score="7" ViewCount="" Body="&lt;p&gt;Dbrane's answer contains the essential points. However I should point out that General Relativity is more sophisticated than your models suggest.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The Inertial Frame concept (as used in the Equivalence Principle) is really only valid &lt;em&gt;infinitesimally&lt;/em&gt; (whence it matches Minkowski space and &quot;idealised gravity-free universe&quot;). Some authors have critized the EP for this, and you are too. Most authors accept this and just present the EP &quot;locally&quot; - with &quot;local&quot; meaning no large deviations via curvature. Near the Event Horizon of a Black Hole is not a good place to find such flatness - especially if the BH is rotating - so we would be dealing with small Frames at best. All this makes &quot;Law K&quot; in your post suspect. (EDIT ADD FOR CLARITY) Thus the Blog phrase &quot;Then law K is false in X&quot; needs to say &quot;Then law K is false in General Relativity&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A different problem here is the status of &quot;Event Horizon&quot; (presumed at R=2M in your post). Put simply Event Horizons are difficult to find for an active Black Hole (one that is still eating up matter): its position is actually mobile until the Black Hole finally settles down (at the end of the Universe). This is a very counterintuitive behaviour of Black Holes and of General Relativity and arises because the &quot;M&quot; in &quot;R=2M&quot; hasnt been determined until the Black Hole has stopped growing!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Concerning this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In an inertial frame, unlike in X, a body's freedom of movement (in principle and if only relative to other free objects in the frame) doesn't depend on the body's position or extent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The &quot;freedom of movement&quot; that I think you are referring to whether the object can be accelerated beyond the speed of light, which cannot be done in any Inertial Frame. As it cannot be so accelerated then no physical process in momentarily passing frame X can stop the momentarily straddling rod from entering the Event Horizon (remember that the Frame X is entering the Horizon too).&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-03-02T18:15:56.900" LastActivityDate="2011-03-02T18:15:56.900" CommentCount="24" />
  <row Id="6094" PostTypeId="2" ParentId="6052" CreationDate="2011-02-28T14:32:37.533" Score="8" ViewCount="" Body="&lt;p&gt;I quit computing in the late 1980s as a result of an early mid-life crisis in my mid 30s. After a few years getting myself back together, I ended up in an M.Sc. at the University of Durham, UK, paying my own fees and for my food and lodging. Up to this level is easily manageable, because it's essentially just learning. What can make even this level not work out is what makes the next level more difficult to negotiate &amp;mdash; real research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With some life experience behind me, and having read around quantum theory and having come to naive opinions about what was &quot;wrong&quot; with quantum theory, I almost didn't finish my M.Sc. project, and I was only just adequately prepared for the examination component, because I went off on an almost unguided goose chase after a particular piece of mathematics (Clifford algebras, a pied piper if ever there was one). One academic managed to keep me from going completely off the reservation. I've chased after many other pieces of mathematics in the 20 years of research since, so I have a certain perspective on realizing that I've wasted another year or several on another piece of mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What an older person can bring to research can be a real asset and a real pain for an academic to have to deal with: bull-headedness. If you're older than the academic who is guiding your research, you nonetheless have to take their guidance, at least up to a point. When an academic looks at you, they will be choosing you &lt;em&gt;over&lt;/em&gt; someone younger who is probably also more malleable. An academic has to decide whether they want someone who will be no trouble (and probably also not produce great research of their own in the future, but nonetheless be smart enough to help the academic's research go forward nicely, which is a delicate equation), or be a lot of trouble (but the academic can hope for and look forward to a satisfying experience of seeing their mentoring bear fruit many times over). Of course an academic's ego is a fundamental part of who they choose to take as graduate students. It has always seemed to me that this calculation is generally rather different when an academic considers a mature student. Being a real mentor to a graduate student is exhilarating and exhausting if they're really good at research, but could be worse than pointless if they react badly when they discover they're not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Twenty years ago, I wanted to make some academic let me do things my own way. That didn't work out (!), because I couldn't find an academic who would put me in the place they had instead of taking someone much more promising-looking and just graduated. Putting a brave face on this, I've sometimes thought that it would have been immoral to take those places from the young people who beat me out of those places. Consequently, I've pursued research independently, with an M.Sc. but without a Ph.D., which &lt;em&gt;can&lt;/em&gt; be done if you're persistent &lt;em&gt;and&lt;/em&gt; you can find a way to get good access to an academic library. Although you will make &lt;em&gt;much&lt;/em&gt; better progress in research if you can find a good real-life community, I think it &lt;em&gt;may&lt;/em&gt; be better to spend your money on access to a library and to find effective ways to be part of the on-line community (giving as well as taking, of course).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[My qualification for these observations is somewhat sideways: I'm married to a Classics professor, so I see from quite close up, but still from secondhand, and in an arts subject, the process of becoming an academic researcher, which many promising undergraduates never manage. Second-guessing who will make the jump is a large part of taking someone on as a graduate student. Needless to say, these comments are bashed out in a half-hour, they're incomplete, they're not gold dust, and you're not me. &lt;em&gt;Good luck.&lt;/em&gt;]&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-02-28T14:32:37.533" />
  <row Id="6095" PostTypeId="2" ParentId="6052" CreationDate="2011-02-28T15:00:19.037" Score="4" ViewCount="" Body="&lt;p&gt;Your CV is very similar to mine, although I made my career in supercomputing. I did have one job which was close to physics (doing programming for a plasma physics group). Perhaps if you volunteer yourself to a research group (as free or cheap labour) you can become part of the research effort without having to jump through all the academic hoops. At my age, I'd rather be involved in teaching the next generation, as learning neat stuff, but not having a use for it myself seems a waste.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-02-28T15:00:19.037" />
  <row Id="6096" PostTypeId="1" CreationDate="2011-02-28T15:14:51.440" Score="8" ViewCount="586" Body="&lt;p&gt;I would like to hear the best arguments for and against the Many Worlds interpretation of QM.&lt;/p&gt;&#xA;" OwnerUserId="1658" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-28T21:27:45.850" LastActivityDate="2011-08-11T06:36:10.370" CommunityOwnedDate="2011-02-28T21:27:58.253" Title="Arguments for and against Many Worlds?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-interpretations&gt;" AnswerCount="11" CommentCount="2" FavoriteCount="4" />
  <row Id="6097" PostTypeId="2" ParentId="6096" CreationDate="2011-02-28T15:21:32.877" Score="4" ViewCount="" Body="&lt;p&gt;in my opinion, the best argument is that removes the need to postulate a collapse of wavefunction, and explains the source of the randomness of quantum events as a single observer splitting in multiple version of the same observer entangled to each eigenstate of the observed system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the weak aspect of the many worlds interpretation is that it doesn't give a natural explanation of probabilities as the magnitudes-squared of amplitudes, this is still required to be assumed &lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-02-28T15:21:32.877" CommunityOwnedDate="2011-02-28T21:27:58.253" CommentCount="7" />
  <row Id="6098" PostTypeId="1" AcceptedAnswerId="6156" CreationDate="2011-02-28T15:27:05.150" Score="2" ViewCount="250" Body="&lt;p&gt;I am trying to model a system in which cubes of about 2 cm in size are floating in a circular water thank of about 30 cm in diameter. The cubes move around under the influence of the fluid flow induced by four inlets that point toward the center of the tank, and are located at the positions $0$, $\pi/2$, $\pi$, and $3\pi/2$. The flow velocity ranges from 0 to 10 cm/s, with an average velocity around 6 cm/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What would be the Reynolds number of the system? In particular, should I take as characteristic length the size of the cubes, or that of the tank?&lt;/li&gt;&#xA;&lt;li&gt;For such a system, what is the limit Reynolds number for the turbulent regime?&lt;/li&gt;&#xA;&lt;li&gt;What would be the correct form of the drag force, and do you intuitively think that the orientation of the blocks is negligible from a drag coefficient point of view?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="2320" LastActivityDate="2011-03-01T04:03:20.777" Title="Reynolds number, turbulence regime, and drag force" Tags="&lt;fluid-dynamics&gt;" AnswerCount="2" />
  <row Id="6099" PostTypeId="2" ParentId="6074" CreationDate="2011-02-28T15:33:03.040" Score="1" ViewCount="" Body="&lt;p&gt;Acceleration due to gravity, g is not a universal constant like G. Its calculated by formula mentioned in previous answers. So, for a constant mass system, g depends only on r (distance between center of earth &amp;amp; object in problem). As r = R + h (R is radius of earth &amp;amp; h is height of object from surface) &amp;amp; R is constant, g depends mainly on height.&lt;br&gt;&#xA;&lt;strong&gt;The relation:&lt;/strong&gt; Increase the height, g will become less (as per formula)   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The value 9.8 m/s² is valid for the object at the surface of earth (at sea level).&lt;/strong&gt; When height is small (with respect to radius of earth), the value is slightly less than 9.8 m/s². So, this variation can be neglected for a high school etc problems. When accuracy is important (due to scientific reasons etc), the value of g can't be 9.8 m/s².  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once again, This consideration is valid only for constant mass system. Plus, for relativistic systems, the formula isn't valid with constant space &amp;amp; time scale.&lt;/p&gt;&#xA;" OwnerUserId="2170" LastActivityDate="2011-02-28T15:33:03.040" />
  <row Id="6100" PostTypeId="2" ParentId="6096" CreationDate="2011-02-28T15:34:53.917" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;For:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Collapse is an awkward non unitary concept in the C.I. and likely to be the source of most of the so called paradoxes (like Schrodinger's Cat etc.). Many worlds does not have collapse and thus avoids those paradoxes. It frees one from assuming the &quot;add on&quot; of collapse.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Quantum entanglement finds a &quot;local&quot; meaning.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Removes randomness from quantum theory (although for an observer it does not remove any randomness).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Against:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Huge proliferation of universes for explaining the observations of an observer. In a way it does not respect the Occam's razor. (Many worlders claim MWI respects it since this interpretation has an economy of principles).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The problem of preferred basis.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Defining a suitable measure of probability to achieve Born rule.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Other universes can not be observed. (A variation of saying it does not respect Occam's razor)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is rather a psychological way of thinking about Q.T. rather than a real ontology.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-03T17:26:57.757" LastActivityDate="2011-04-03T17:26:57.757" CommunityOwnedDate="2011-02-28T21:27:58.253" CommentCount="5" />
  <row Id="6101" PostTypeId="2" ParentId="6096" CreationDate="2011-02-28T15:41:00.140" Score="-1" ViewCount="" Body="&lt;p&gt;MWI is an example of misunderstanding of what the wave function is about and it introduces unnecessary &quot;universes&quot; that do not follow from experimental data. It does not solve any conceptual problem. On the contrary. Besides, it is not verifiable. Briefly, it is a funny example of nonsense in science.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-02-28T15:41:00.140" CommunityOwnedDate="2011-02-28T21:27:58.253" CommentCount="23" />
  <row Id="6102" PostTypeId="2" ParentId="6096" CreationDate="2011-02-28T15:44:00.503" Score="0" ViewCount="" Body="&lt;p&gt;A good argument against is Occham's razor. Another is the fact that it is not an experimental testable hypothesis.   The best argument I think against it is the fact that the only reasons &lt;em&gt;for&lt;/em&gt; Many-worlds are based in human language, whose intuition is only developed for a classical setting, as such any human meta-reasoning cannot be expected to apply to non-classical areas, there are no mathemtical or scientific reasoning involved.&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-02-28T15:44:00.503" CommunityOwnedDate="2011-02-28T21:27:58.253" CommentCount="1" />
  <row Id="6103" PostTypeId="2" ParentId="6096" CreationDate="2011-02-28T15:48:44.347" Score="2" ViewCount="" Body="&lt;p&gt;Having studied some &lt;a href=&quot;http://vimeo.com/5490979&quot; rel=&quot;nofollow&quot;&gt;David Deutsch&lt;/a&gt; material recently here are some other points:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A Multiverse framework for considering whether there is &quot;something beyond&quot; traditional quantum theory. EDIT Thus providing a framework for investigating quantum gravity. Deutsch sees the Multiverse as an as yet undiscovered theory, motivated by Everett, Quantum Computation and certain properties of General Relativity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A possible &quot;explanation&quot; for Quantum Computation speedups (EDIT ADDITION) That is for those quantum computations in the BQP Complexity class.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Against&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Those Probability and Preferred Frame problems&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The fact that the theory has split into many sub-theories (e.g. Decoherence, Many Minds, Multiverse, etc) that there is no single model to evaluate any more.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(Related to 2 perhaps) Everett did not publish enough to clarify exactly what the theory actually said in several respects.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-03-01T09:38:34.420" LastActivityDate="2011-03-01T09:38:34.420" CommunityOwnedDate="2011-02-28T21:27:58.253" CommentCount="13" />
  <row Id="6104" PostTypeId="1" CreationDate="2011-02-28T16:00:17.773" Score="2" ViewCount="52" Body="&lt;p&gt;What needs to be the case for a dynamical system with a continuous flow to exhibit chaos? It looks like 1D systems with a continuous flow can't exhibit chaos. Are two dimensions enough or do you need more? I was just thinking about what sort of phase portraits you could have in two dimensions, and it's not immediately obvious if they will always have stable attractors...&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2011-02-28T16:00:17.773" Title="Chaos and continuous flow" Tags="&lt;classical-mechanics&gt;&lt;chaos-theory&gt;" CommentCount="2" />
  <row Id="6105" PostTypeId="1" AcceptedAnswerId="6114" CreationDate="2011-02-28T16:00:52.050" Score="7" ViewCount="148" Body="&lt;p&gt;From the gauge field $A_\mu$ and the QED lagrangian we can derive maxwell's equations in terms of electric and magnetic fields. Are there any situations where similar derivations using the other gauge fields like $W^{\pm}_{\mu}$, $Z_\mu$ or the gluon fields, e.g. $curl W$ could become useful?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: reformulated for clarity.&lt;/p&gt;&#xA;" OwnerUserId="2151" LastEditorUserId="2151" LastEditorDisplayName="" LastEditDate="2011-02-28T16:07:10.560" LastActivityDate="2011-02-28T21:33:04.907" Title="Is there a meaning to the E,B analogues of other gauge fields?" Tags="&lt;quantum-field-theory&gt;&lt;gauge-theory&gt;" AnswerCount="2" />
  <row Id="6106" PostTypeId="2" ParentId="6074" CreationDate="2011-02-28T16:10:21.370" Score="2" ViewCount="" Body="&lt;p&gt;It might also be worth mentioning that $g$ isn't even constant over the earth's surface at sea level. Depending on the mass distribution and the shape (not perfectly spherical!) of the earth, different parts of the world have different $g$. &lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-02-28T16:10:21.370" />
  <row Id="6107" PostTypeId="2" ParentId="629" CreationDate="2011-02-28T16:20:51.453" Score="9" ViewCount="" Body="&lt;p&gt;I spent last weekends making my own realisation of &lt;a href=&quot;http://en.wikipedia.org/wiki/Material_Point_Method&quot; rel=&quot;nofollow&quot;&gt;MPM method&lt;/a&gt; code (just for fun of it). I just had an idea  that I can try to simulate something similar to the problem of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, here is our &quot;car&quot;.&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/B1Y5c.png&quot; alt=&quot;Initial configuration&quot;&gt;&lt;br&gt;&#xA;It is moving to the right with some constant speed. Then I apply some constant external force to the &quot;wheels&quot; to stop them. And that's what I've got:&lt;br&gt;&#xA;1&lt;img src=&quot;http://i.stack.imgur.com/VlMuz.png&quot; alt=&quot;step1&quot;&gt;&#xA;2&lt;img src=&quot;http://i.stack.imgur.com/YagJY.png&quot; alt=&quot;Step2&quot;&gt;&lt;br&gt;&#xA;3&lt;img src=&quot;http://i.stack.imgur.com/qDUle.png&quot; alt=&quot;step3&quot;&gt;&#xA;4&lt;img src=&quot;http://i.stack.imgur.com/ZEe3F.png&quot; alt=&quot;step4&quot;&gt;&lt;br&gt;&#xA;5&lt;img src=&quot;http://i.stack.imgur.com/zOIB6.png&quot; alt=&quot;step5&quot;&gt;&#xA;6&lt;img src=&quot;http://i.stack.imgur.com/l5pSl.png&quot; alt=&quot;step7&quot;&gt;&lt;br&gt;&#xA;The colours denote the amount of stress in the medium. And here is the animation:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/hoG9Q.gif&quot; alt=&quot;Animation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everyone are free to give other ideas for simulations/visualisations... &lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2011-02-28T17:02:18.063" LastActivityDate="2011-02-28T17:02:18.063" CommentCount="3" />
  <row Id="6108" PostTypeId="1" AcceptedAnswerId="6109" CreationDate="2011-02-28T16:28:10.637" Score="10" ViewCount="412" Body="&lt;p&gt;I am looking for a good source on group theory aimed at physicists. I'd prefer one with a good general introduction to group theory, not just focusing on Lie groups or crystal groups but one that covers &quot;all&quot; the basics, and then, in addition, talks about the specific subjects of group theory relevant to physicists, i.e. also some stuff on representations etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is Wigner's text a good way to start? I guess it's a &quot;classic&quot;, but I fear that its notation might be a bit outdated?&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="184" LastEditorDisplayName="" LastEditDate="2011-03-01T14:37:53.790" LastActivityDate="2011-03-01T14:37:53.790" Title="Comprehensive book on group theory for physicists?" Tags="&lt;soft-question&gt;&lt;books&gt;&lt;reference-request&gt;&lt;group-theory&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="7" />
  <row Id="6109" PostTypeId="2" ParentId="6108" CreationDate="2011-02-28T16:31:35.453" Score="5" ViewCount="" Body="&lt;p&gt;there is a book titled &quot;Group theory and Physics&quot; by Sternberg that covers the basics, including crystal groups, lie groups, representations. I think it's a good introduction to the topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521558859&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Group-Theory-Physics-S-Sternberg/dp/0521558859&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to quote a review on Amazon (albeit the only one)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;This book is an excellent introduction to the use of group theory in physics, especially in crystallography, special relativity and particle physics. Perhaps most importantly, Sternberg includes a highly accessible introduction to representation theory near the beginning of the book. All together, this book is an excellent place to get started in learning to use groups and representations in physics.&quot;&lt;/p&gt;&#xA;" OwnerUserId="2151" LastActivityDate="2011-02-28T16:31:35.453" CommentCount="3" />
  <row Id="6110" PostTypeId="2" ParentId="6075" CreationDate="2011-02-28T16:40:35.367" Score="1" ViewCount="" Body="&lt;p&gt;Ignoring air resistance is a good approximation in this case:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;your resolution of the actual time is probably pretty coarse so there's (EDIT TO ADD: no) need to make other parts of your calc significantly more accurate than your least-accurate portion.&lt;/li&gt;&#xA;&lt;li&gt;your speed was quite a a bit lower than your terminal velocity, therefore the drag would have been a very small force relative to gravity's.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To confirm this, do the calculation with air resistance set to zero, and then see if your initial estimate of time and impact velocity are in the &quot;air resistance is negligible&quot; regime, or if it could have been significant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: with no air resistance d=1/2*a*t^2, and in ballpark numbers d=10m and a=10m/s^2 -- so t=1.4seconds; this should match your measured time, if not work backwards from your measured time to get &quot;d&quot; in this no-air-resistance scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work out your highest speed i.e. just prior to impact: v=a*t.  With the above numbers we get v=14m/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now compare this speed to the speed at which air resistance matches your weight.  From skydivers we know that a human body has a terminal velocity of about 120mph in a horizontal position and about 180mph in a diving position.  Going with the lower one, 120mph is about 180ft/sec which is about 60m/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you reached about 1/4 of terminal velocity (14ms / 60m/s).  Air drag scales with the square of velocity so the greatest air drag force you experienced (just prior to hitting the water) was (1/4)^2 or 1/16th of your weight, which is about a 6% error.  Averaged over your fall it would have been much less than 2%, and since your timing probably isn't to 1%-2% precision, it is therefore a sound judgement to neglect air resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT TO ADD: d=1/2*a*t^2 is the standard formula for the distance covered by an object under constant acceleration, starting from rest.  It can be obtained by integrating the formula v=a*t which is the standard formula for velocity attained under constant acceleration, starting from rest.&lt;/p&gt;&#xA;" OwnerUserId="2297" LastEditorUserId="2297" LastEditorDisplayName="" LastEditDate="2011-02-28T22:32:43.493" LastActivityDate="2011-02-28T22:32:43.493" CommentCount="4" />
  <row Id="6111" PostTypeId="2" ParentId="6108" CreationDate="2011-02-28T16:58:41.280" Score="1" ViewCount="" Body="&lt;p&gt;Well, in my dictionary &quot;group theory for physicists&quot; reads as &quot;representation theory for physicists&quot; and in that regard &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387974954&quot; rel=&quot;nofollow&quot;&gt;Fulton and Harris&lt;/a&gt; is as good as they come. You'll learn all the group theory you need (which is just a tiny fragment of all group theory) along the way.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-02-28T16:58:41.280" CommentCount="12" />
  <row Id="6112" PostTypeId="2" ParentId="6096" CreationDate="2011-02-28T17:04:20.043" Score="0" ViewCount="" Body="&lt;p&gt;Has the MWI produced any new physics? What I mean is not people who subscribe to it doing new physics, but people doing new physics because they were thinking within that interpretation. If the answer is nothing, then this, in my eyes, is a very strong argument against it. Of course this could apply to other interpretations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure if I am clear. What I mean is for example, is there a calculation or a derivation or anything like that, which would have not been done (at least for some time) had there been no MWI.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2011-02-28T17:04:20.043" CommunityOwnedDate="2011-02-28T21:27:58.253" CommentCount="7" />
  <row Id="6113" PostTypeId="2" ParentId="6108" CreationDate="2011-02-28T17:11:47.980" Score="3" ViewCount="" Body="&lt;p&gt;I would recommend A. O. Barut and R. Raczka &quot;Theory of Group Representations and applications&quot;. It is about Lie algebras and Lie groups, and you are asking for general group theory, but this book, in my opinion, would be useful to a physicists. The applications are to physics, mainly quantum theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Forgot to comment on the last part of the questions. I think Wigner is a good read. You'll not learn much about general group theory, but you will learn about representation theory of the Poincare group and some general techniques from representation theory like the Mackey machine for induced representations.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2011-02-28T17:26:49.463" LastActivityDate="2011-02-28T17:26:49.463" CommentCount="3" />
  <row Id="6114" PostTypeId="2" ParentId="6105" CreationDate="2011-02-28T17:19:08.550" Score="8" ViewCount="" Body="&lt;p&gt;The decomposition into electric and magnetic fields is valid in a specific reference frame, those fields mix under Lorentz transformations. Based on that, you may expect that this separation will be more useful in the non-relativistic (low energy) limit. There is also the historical fact: Maxwell equations were written down before special relativity was discovered. Since we still, by and large, make the odd choice of teaching things in historical (rather than logical) order, you've probably seen Maxwell equations in components before seeing them in their more compact and symmetric form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, for fields relevant only in very high energies (the electroweak gauge bosons), which were also discovered after SR, the relativistic form of the equations is the default. You can certainly write things in components, but this is not a very useful notation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some exceptions, some corners in high energy physics where this decomposition can give you some intuition. For example if you look at the QFT text by Peskin and Schroeder you'd find some intuitive picture on why the $\beta$ function of non-Abelian gauge theories is negative. This picture is not gauge invariant, but it relies on some &quot;magnetic&quot; as opposed to &quot;electric&quot; effects. But these exceptions are rare and not always all that useful.&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-02-28T17:19:08.550" CommentCount="2" />
  <row Id="6115" PostTypeId="2" ParentId="6098" CreationDate="2011-02-28T17:29:32.610" Score="1" ViewCount="" Body="&lt;p&gt;The characteristic length can be whatever you want it to be.  Characteristic lengths are only prescribed for a limited range of &quot;standard&quot; problems.  For that matter, what do you select as your velocity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, the dependence of flow characteristics on Reynolds number depends on how you define your Reynolds number.  To my knowledge, unlike, say, flow between parallel plates, pipe flow, or flow over an airfoil, this is not a well-studied problem, so there really is no answer anyone can give to your second question.  You'll have to physically do the experiment to see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are referring to the drag force on a specific block, I would ask: does the coefficient of drag for an airfoil change with angle of attack?  There's your answer.&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2011-02-28T17:29:32.610" />
  <row Id="6116" PostTypeId="2" ParentId="6096" CreationDate="2011-02-28T17:59:55.203" Score="-1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://lesswrong.com/lw/r5/the_quantum_physics_sequence/&quot; rel=&quot;nofollow&quot;&gt;http://lesswrong.com/lw/r5/the_quantum_physics_sequence/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the best introductions for MWI out there. Answers queries relating to Occam's razor.. and why it's more than our fantasy..&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2011-02-28T17:59:55.203" CommunityOwnedDate="2011-02-28T21:27:58.253" CommentCount="1" />
  <row Id="6117" PostTypeId="2" ParentId="6052" CreationDate="2011-02-28T18:03:17.417" Score="-10" ViewCount="" Body="&lt;p&gt;Getting a PhD is not easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to attend courses far harder than at undergraduate level and it means being an expert in some area and then adding to that knowledge which no one has done before. Do you honestly believe that you can do this aged 60?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me highly unlikely your 60 year old body, including your brain, being able to compete with that of a virile, agressive, competitive 21 year old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe you could do something easier like a masters, just to see how you do. I'm just concerned that you're not realistic enough, not in touch with what's currently going on in the world. And making a judgement similar to that of A deluded crackpot seeing the world through a brain not as respsonsive as it used to be in their youth.&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-02-28T18:03:17.417" CommentCount="4" />
  <row Id="6118" PostTypeId="2" ParentId="6052" CreationDate="2011-02-28T18:22:12.887" Score="4" ViewCount="" Body="&lt;p&gt;The point of PhD programs is to produce future scientists; it seems unlikely that any decent department would admit someone who indicates that they &quot;have no intention of seeking a further career in physics.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you do have a background in physics, math, and computing, it does seem like you could earn your keep in a physics lab.  Assuming you are interested in experimental physics, I suspect your best bet would be to get a half-time job in a physics lab with an informal agreement that you can also audit the department's courses.   Actually taking a spot in a formal PhD program would probably be seen as unfair to the person you would be displacing, who would actually need the PhD for their career.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretical physics would be more difficult, since it is much more difficult to &quot;earn your keep&quot; as a novice theorist.  I think it would be difficult to find an advisor in theoretical physics who would have the patience to mentor someone who explicitly did not intend to continue in the field--unless that person was manifestly brilliant.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-02-28T18:30:08.000" LastActivityDate="2011-02-28T18:30:08.000" CommentCount="11" />
  <row Id="6120" PostTypeId="2" ParentId="6068" CreationDate="2011-02-28T18:34:59.263" Score="0" ViewCount="" Body="&lt;p&gt;I recommend &quot;Thinking Physics&quot; by Lewis Carroll Epstein.  It's a long accumulation of conceptual problems in introductory physics, for example, &quot;Suppose you drop a ball out a window and it hits the ground at 3 m/s.  Now you throw a ball out the same window straight down at 4 m/s.  How fast will it hit the ground?&quot;  The answer is 5 m/s, and the book illustrates how these numbers are related to the Pythagorean theorem due to kinetic energy depending on the square of velocity.  I learned a lot by working through this book as a supplemental text when I was a college freshman.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0935218084&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Thinking-Physics-Understandable-Practical-Reality/dp/0935218084/ref=sr_1_1?ie=UTF8&amp;amp;qid=1298917867&amp;amp;sr=8-1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-02-28T18:34:59.263" CommunityOwnedDate="2011-02-28T18:34:59.263" />
  <row Id="6122" PostTypeId="1" AcceptedAnswerId="6124" CreationDate="2011-02-28T19:19:14.437" Score="3" ViewCount="107" Body="&lt;p&gt;How can I derive the Einstein's relation $D=k_{b}TB$, where $D$ is the diffusion coefficient and B is the mobility coefficient, from the concept of osmotic pressure?&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2011-02-28T19:56:10.480" Title="Einstein's relation and osmotic pressure" Tags="&lt;diffusion&gt;" AnswerCount="1" />
  <row Id="6123" PostTypeId="2" ParentId="6108" CreationDate="2011-02-28T19:29:16.033" Score="0" ViewCount="" Body="&lt;p&gt;Sternberg's book is excellent and illuminating but perhaps a bit hard for a beginner. I recommend as a first reading &lt;a href=&quot;http://www.springer.com/mathematics/algebra/book/978-0-387-40122-5&quot; rel=&quot;nofollow&quot;&gt;Lie Groups, Lie Algebras, and Representations&lt;/a&gt;. The book deals with representation theory of Lie groups of matrices. After reading this I also recommend the Sternberg's book for physical applications and the topological point of view of group theory.&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2011-02-28T19:29:16.033" />
  <row Id="6124" PostTypeId="2" ParentId="6122" CreationDate="2011-02-28T19:56:10.480" Score="3" ViewCount="" Body="&lt;p&gt;There are two phenomena present&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;diffusion&lt;/em&gt;, which happens due to inhomogeneity in concentration. Particles &quot;want to&quot; go from areas of higher concentration to the lower ones. One can write this in the form of diffusion current &#xA;$$J_{diff}(x) = - D \nabla \rho(x)$$&#xA;where $\rho(x)$ is the concentration. This expression is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Fick%27s_laws_of_diffusion#Fick.27s_First_Law&quot; rel=&quot;nofollow&quot;&gt;Fick's law&lt;/a&gt; but it's actually just the standard linear response to inhomogeneities.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;drift&lt;/em&gt;, which is the terminal velocity particles attain due to presence of some force. E.g. the drift one can observe for balls falling in viscous liquid. One can write &#xA;$$J_{drift} = \rho(x) v(x) = \rho(x) B F(x) = -B \rho(x) \nabla U(x)$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From the requirement of equilibrium we have that $J_{diff} + J_{drift} = 0$ and from Boltzmann statistics we can obtain the concentration $\rho(x) \sim \exp(-{U(x) \over k_B T})$.&#xA;Putting it all together we get&#xA;$$0 = - D \nabla \rho(x) - B \rho(x) \nabla U(x) = - \nabla U(x) \rho(x) (-{D \over k_B T} + B)$$&#xA;and we can see the required relation in the last term.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-02-28T19:56:10.480" />
  <row Id="6125" PostTypeId="2" ParentId="6108" CreationDate="2011-02-28T20:19:13.917" Score="2" ViewCount="" Body="&lt;p&gt;Morton Hamermesh's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486661814&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Group Theory and Its Application to Physical Problems&lt;/em&gt;&lt;/a&gt; is a Dover Press book, so quite inexpensive (though the price seems to be up a bit since I bought it in the '90s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/n4hRt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-02-28T20:19:13.917" CommunityOwnedDate="2011-02-28T20:19:29.847" />
  <row Id="6126" PostTypeId="3" CreationDate="2011-02-28T21:21:50.783" Score="0" ViewCount="" Body="&lt;p&gt;In the context of physics, philosophical questions are those which address the underlying meaning of physical principles. This could include questions about the interpretations of quantum mechanics, although the &lt;a href=&quot;/questions/tagged/quantum-interpretations&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged 'quantum-interpretations'&quot; rel=&quot;tag&quot;&gt;quantum-interpretations&lt;/a&gt; tag is recommended instead for those cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since physics concerns itself with studying nature through observation, &lt;strong&gt;most philosophy questions are off topic for this site&lt;/strong&gt;. An acceptable philosophy question has to be grounded in an established theory of physics.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-28T21:27:04.913" LastActivityDate="2011-02-28T21:27:04.913" />
  <row Id="6127" PostTypeId="3" CreationDate="2011-02-28T21:21:50.783" Score="0" ViewCount="" Body="Philosophical questions in physics concern the underlying meaning or interpretation of physical theories, in ways which cannot be connected to experimental results." OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-28T21:27:04.977" LastActivityDate="2011-02-28T21:27:04.977" />
  <row Id="6128" PostTypeId="5" CreationDate="2011-02-28T21:30:41.043" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-28T21:30:41.043" LastActivityDate="2011-02-28T21:30:41.043" />
  <row Id="6129" PostTypeId="4" CreationDate="2011-02-28T21:30:41.043" Score="0" ViewCount="" Body="Gravity is an attractive force that affects and is produced by all mass and other forms of energy (as well as pressure and stress, in general relativity)." OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-28T21:30:41.043" LastActivityDate="2011-02-28T21:30:41.043" />
  <row Id="6130" PostTypeId="2" ParentId="6105" CreationDate="2011-02-28T21:33:04.907" Score="2" ViewCount="" Body="&lt;p&gt;In the analysis of constraints and canonical/Dirac quantization of gauge fields it is &quot;natural&quot; to choose a time direction. This means that the electric and magnetic components of the field are separated and seem to play different roles. Of course, at the end, Lorentz invariance can be recovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example of this, see Matschull's paper: &lt;a href=&quot;http://arxiv.org/abs/quant-ph/9606031&quot; rel=&quot;nofollow&quot;&gt;Dirac's Canonical Quantization Programme&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="429" LastActivityDate="2011-02-28T21:33:04.907" />
  <row Id="6131" PostTypeId="5" CreationDate="2011-02-28T21:36:38.123" Score="0" ViewCount="" Body="&lt;p&gt;Lagrangian mechanics is a method of analyzing classical mechanical systems. The primary mathematical tool used is the Euler-Lagrange equation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{x}} = \frac{\partial L}{\partial x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each mechanical system is characterized by a Lagrangian $L$, which can be expressed as the difference between the kinetic and potential energies of the system,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = K - U$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging the expression for $L$ for a particular system into the Euler-Lagrange equation yields the equation of motion for the system. This is fundamentally equivalent to solving Newton's second law $\sum F = ma$, but it allows the solution of many problems which would be prohibitively difficult using Newton's laws alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strengths of the Lagrangian approach include handling problems with constraints, and problems which exhibit symmetries in non-Cartesian coordinates.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-28T21:36:38.123" LastActivityDate="2011-02-28T21:36:38.123" />
  <row Id="6132" PostTypeId="4" CreationDate="2011-02-28T21:36:38.123" Score="0" ViewCount="" Body="Lagrangian mechanics is a method of analyzing classical mechanical systems using the Euler-Lagrange equation." OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-28T21:36:38.123" LastActivityDate="2011-02-28T21:36:38.123" />
  <row Id="6133" PostTypeId="1" CreationDate="2011-02-28T21:36:40.510" Score="4" ViewCount="79" Body="&lt;p&gt;I'm looking for contemporary reviews on neutron stars. Seems like this area is pretty active, so even reviews from five or ten years ago are somewhat lacking, though certainly not worthless. Does anyone have recommendations? Newer is better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Books are okay too, but I'd prefer one of those 70-ish page reviews you find in journals.&lt;/p&gt;&#xA;" OwnerUserId="1655" LastActivityDate="2011-03-01T11:48:11.327" Title="Neutron star references?" Tags="&lt;neutron-stars&gt;&lt;reference-request&gt;" AnswerCount="3" />
  <row Id="6134" PostTypeId="5" CreationDate="2011-02-28T21:37:25.860" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-28T21:37:25.860" LastActivityDate="2011-02-28T21:37:25.860" />
  <row Id="6135" PostTypeId="4" CreationDate="2011-02-28T21:37:25.860" Score="0" ViewCount="" Body="The special theory of relativity describes the motion and dynamics of objects moving at significant fractions of the speed of light." OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-02-28T21:37:25.860" LastActivityDate="2011-02-28T21:37:25.860" />
  <row Id="6137" PostTypeId="2" ParentId="2064" CreationDate="2011-02-28T22:49:56.310" Score="2" ViewCount="" Body="&lt;p&gt;well everyone at least knows a mnemonic for the planets in the solar system.&#xA;&quot;My Very Energetic Mother Just Served Us Nine Pizzas&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Planetary_mnemonic&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Planetary_mnemonic&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also to remember where the minus sign goes in $\sigma_2$ of the Pauli Matrices I learned:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;the minus i rides high on the sigma y&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also found this list of some mnemonics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://members.chello.nl/r.kuijt/index-physics.htm&quot; rel=&quot;nofollow&quot;&gt;http://members.chello.nl/r.kuijt/index-physics.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2151" LastEditorUserId="2151" LastEditorDisplayName="" LastEditDate="2011-03-01T12:29:47.960" LastActivityDate="2011-03-01T12:29:47.960" CommunityOwnedDate="2011-02-28T22:49:56.310" CommentCount="4" />
  <row Id="6139" PostTypeId="2" ParentId="6096" CreationDate="2011-02-28T23:56:50.767" Score="0" ViewCount="" Body="&lt;p&gt;It is called &quot;interpretation&quot; for a reason: Ultimately, it all describes the same physics, i.e., so far there is no statement &quot;X&quot; that would be true in the MWI and false in Bohmian QM or the Classical Interpretation. Now, if two ways of explaining something lead to entirely the same results, then for all practical purposes these two ways are identical. In my opinion, it is therefore purely a matter of taste to which of the available interpretations one ascribes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is only when results between different interpretations become inconsistent that we can rule one of them out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which reminds me of a funny text I read on this topic: You can prove to yourself that the MWI is correct by committing &quot;quantum suicide&quot;: Think of Schroedinger's cat, but you are the cat. If the MWI is correct, then there will always be a world in which you aren't dead yet. You should notice that you aren't dead, and after repeating the experiment for a sufficiently long time you can conclude that, with high probability, the MWI is correct. If, however, the classical interpretation is correct, you'll just die with high probability :-( Thus, I say &quot;Don't try this at home!&quot;&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-02-28T23:56:50.767" CommunityOwnedDate="2011-02-28T23:56:50.767" />
  <row Id="6140" PostTypeId="1" AcceptedAnswerId="6185" CreationDate="2011-03-01T00:01:12.523" Score="7" ViewCount="407" Body="&lt;p&gt;Ok, I've stumbled onto what I think is a bit of a paradox. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, say you had some computer in a very fast(near light speed) centrifuge. You provide power to this computer via a metal plate on the &quot;wall&quot; of the centrifuge's container, so it works similar to how subways and streetcars are powered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the computer normally would consume 200 watts, how much power would it consume at say 1/2 of light speed? Would it consume 400 watts from our still viewpoint?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what if you were to be capable of communicating with this computer? Would the centrifuge-computer receive AND transmit messages faster from our still viewpoint? I'm a bit lost in even thinking about it.&lt;/p&gt;&#xA;" OwnerUserId="2217" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2011-03-01T00:26:35.443" LastActivityDate="2011-07-05T12:03:30.723" Title="Special Relativistic Time Dilation -- A computer in a very fast centrifuge" Tags="&lt;relativity&gt;&lt;time&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" />
  <row Id="6141" PostTypeId="2" ParentId="6140" CreationDate="2011-03-01T00:23:06.290" Score="6" ViewCount="" Body="&lt;p&gt;One thing to be aware of is that the principle of relativity would &lt;b&gt;not&lt;/b&gt; apply to this computer--rotating reference frames are not inertial, and therefore, will not be related to 'stationary' reference frames by simple Lorentz transformations.  Also note that if there are any capacitors or anything along those lines in the computer, then they would be accelerating with respect to a rest frame, and this acceleration would produce electromagnetic waves (and even failing this, accelerating current-carrying wires would have a similar effect), which would create a potentially measurable difference between the frame of the computer and the rest frame.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My instinct for this problem, though, is that the power consumption of the computer would go &lt;b&gt;down&lt;/b&gt;, not up--the computer uses energy in its rest frame, which has a slower clock than the lab frame.  So, it would go on using 200 J every second of ITS proper time, which would correspond to more time in the lab frame.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-03-01T00:23:06.290" CommentCount="5" />
  <row Id="6143" PostTypeId="2" ParentId="6133" CreationDate="2011-03-01T00:46:33.130" Score="3" ViewCount="" Body="&lt;p&gt;There are rather new lecture notes (1 year old) by my colleague Andreas Schnmitt focusing on the microscopic aspects of compact stars (formerly known as neutron stars): see &lt;a href=&quot;http://arxiv.org/abs/arXiv:1001.3294&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is more 140-ish than 70-ish, but the layout is rather generous with space, so it might be exactly what you were looking for.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-03-01T00:46:33.130" />
  <row Id="6144" PostTypeId="2" ParentId="6052" CreationDate="2011-03-01T01:21:41.900" Score="4" ViewCount="" Body="&lt;p&gt;The second tier schools offer &quot;terminal masters&quot; physics degrees. Some of these are intended for night students, for example, high school teachers that want an MS to get a better pay rate. For example, U. Washington offers such a program, along with a graduate non-matriculated student admission:&#xA;&lt;a href=&quot;http://www.phys.washington.edu/grad.htm#gnm&quot; rel=&quot;nofollow&quot;&gt;http://www.phys.washington.edu/grad.htm#gnm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you show up as a non matriculated student and then crush the classes, I suspect that they'll let you in as a grad student. If not, you'll have recent grades showing your ability when applying to a different school.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cool thing about not being in physics grad school is that you can study on your own. To do this requires a lot of drive. I suggest looking deeper into those areas of physics which you find absolutely fascinating. Don't make it &quot;work&quot;, make it &quot;play&quot;. Do this a while and soon enough you'll find out that you're a world expert in something. Write papers on it. Publish them. I think this is the right way for older people to edge their way into physics grad school. The idea is that you give yourself a reason for the sacrifices needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless you're completely driven, it's difficult to teach yourself physics on your own. So don't worry about giving yourself a complete graduate physics education. Concentrate on the stuff you're particularly fascinated by. If you decide to go to grad school you can always learn the other stuff in school.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get into US grad school typically requires that you take the GRE. Go ahead and pick up one of the cheap books on the subject and see how you do. The general exam has three parts, quantitative, verbal, and writing. You should get a 750 to 800 on the quantitative as it is just high school algebra. Less than that and you should study / practice the subject. Make it into a video game. Older people do fine on the writing and verbal sections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there's a physics GRE. There are sample tests for that, too. I'm 53 now. I also decided it might be nice to complete a PhD in physics. So I took the Physics GRE last October and maxed it out-990. This is way better than you need to get into physics grad school. So far, I've been rejected at the schools I applied to, but I only applied to the best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as getting a physics PhD as an old man, yes I have no doubt it can be done. But you may want to apply to schools that will be impressed with your credentials.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-01T01:21:41.900" CommentCount="2" />
  <row Id="6145" PostTypeId="2" ParentId="6133" CreationDate="2011-03-01T01:33:33.357" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1102.5735&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1102.5735&lt;/a&gt;&lt;br&gt;&#xA;Coincidentally, this thirty page review appears on tonights ArXiv.&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-03-01T01:33:33.357" />
  <row Id="6146" PostTypeId="1" AcceptedAnswerId="6152" CreationDate="2011-03-01T01:48:59.240" Score="0" ViewCount="98" Body="&lt;p&gt;How to do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Show that if the tension F in a string is changed by a small amount dF, the fractional change in frequency of a standing wave, df/f is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;df/f = 0.5 dF/F&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditorDisplayName="" LastEditDate="2011-03-01T01:57:44.647" LastActivityDate="2011-03-01T02:55:09.153" Title="Standing Wave: How to show that df/f = 0.5 dF/F where f is frequency and F is tension of the string?" Tags="&lt;waves&gt;" AnswerCount="1" />
  <row Id="6147" PostTypeId="1" AcceptedAnswerId="6150" CreationDate="2011-03-01T02:01:53.317" Score="2" ViewCount="256" Body="&lt;p&gt;Okay, I'm asking a question similar to this one here: &lt;a href=&quot;http://physics.stackexchange.com/q/4404/2330&quot;&gt;Time Dilation - what happens when you bring the observers back together?&lt;/a&gt;. Specifically, I am curious about a specific angle on the second part of his question, regarding when two moving frames of reference are brought back together, and how &quot;it&quot; knows which one should be still young.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accepted answer on that question says that it is whichever FoR experienced the forces of acceleration/deceleration. But, isn't that the whole point of relativity, is that it's all ... well, relative; that it is not possible to say with a certainty that it was the traveler in the spaceship who was accelerating/decelerating?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't it the case that it is just as legitimate to say that the universe and people on the planet accelerated/decelerated and the traveler in the spaceship was stationary? This would therefore then lead to that the universe and planet-side people should remain young and the spaceship occupant should be old, no? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does dilation (temporal-spatial) just generally apply to the smaller of the two FoRs, or is there some other system or rule which &quot;decides&quot; which FoR gets dilated?&lt;/p&gt;&#xA;" OwnerUserId="2330" LastEditorUserId="2330" LastEditorDisplayName="" LastEditDate="2011-03-01T02:09:52.220" LastActivityDate="2011-03-01T15:14:59.647" Title="Time Dilation - how does it know which FoR to age slower?" Tags="&lt;special-relativity&gt;&lt;relativity&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="6148" PostTypeId="2" ParentId="6147" CreationDate="2011-03-01T02:15:30.123" Score="1" ViewCount="" Body="&lt;p&gt;In special relativity, acceleration breaks the symmetry between the two observers since SR is applicable only in the inertial frames. If you insists the relativity of &lt;em&gt;all&lt;/em&gt; motions then you have to use GR. If you apply GR to the problem you will get the same answer. However, SR is sufficient to answer the twin paradox. Putting simply, just remember that although all motions are relative according to GR, the equivalence principle demands that acceleration is equivalent to gravitation for the observer accelerating. The symmetry between the two observers break there.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-03-01T02:15:30.123" />
  <row Id="6149" PostTypeId="2" ParentId="6147" CreationDate="2011-03-01T02:20:45.783" Score="4" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Isn't it the case that it is just as legitimate to say that the universe and people on the planet accelerated/decelerated&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, it's not. If you stay on the planet, you know you didn't accelerate (that is, not any more than usual due to the gravitation and rotation of Earth, etc.). But if you are in the spaceship and you turn your engine on, you'll definitely feel the acceleration (same way you feel it in your car). We call the former frames (those that don't feel the force) &lt;em&gt;inertial&lt;/em&gt; and they play very special role in the theory of relativity (both special and general).&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-03-01T02:20:45.783" />
  <row Id="6150" PostTypeId="2" ParentId="6147" CreationDate="2011-03-01T02:44:26.307" Score="7" ViewCount="" Body="&lt;p&gt;&quot;Relativity&quot; is actually a misleading word that Einstein didn't like.  It doesn't mean &quot;every vantage point is equivalent and it's all relative&quot;.  It really means only inertial, non-accelerating vantage points are equivalent.  You could think of it as, prior to relativity, people believed that there was an absolute position/speed to the universe.  Special Relativity shows that there is not, but rather, there is an absolute acceleration to the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is illustrated by the famous rotating bucket thought experiment.  You put a bucket out in the middle of empty space and spin it, and the water in it starts flowing towards the edges.  But if all vantage points were the same, couldn't you also think of it as the universe spinning and the bucket stationary?  But one vantage point is obviously more correct than the other, because only one involves the water flowing towards the edges.  Thus there is a &quot;universal&quot; state of zero acceleration that is unambiguous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is interesting to note that Einstein's original idea for his theory of Special Relativity was a theory of Invariance (of the speed of light)&lt;/p&gt;&#xA;" OwnerUserId="71" LastActivityDate="2011-03-01T02:44:26.307" CommentCount="4" />
  <row Id="6151" PostTypeId="1" AcceptedAnswerId="6179" CreationDate="2011-03-01T02:54:07.817" Score="0" ViewCount="130" Body="&lt;p&gt;I won't even bother to ask about validity as I read the other post on this. So I know how we all feel about that. I hadn't even heard of it till today when I stumbled across it searching for something else. But I'd like to know: How did they get the subatomic particle masses to come out of their equations? Did they just &quot;coach&quot; them in the right directions- meaning are the equations just so vague that you could get any number out of them? I tried to read the papers but just don't get it-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In discussions concerning it's invalidity I've found a lot of arguing over the true accuracy of the mass values. While I've seen tables to compare theoretical and actual values- the sources for such are all a group working on the theory. Has anyone seen the calculations and can answer?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this could be informative as it can show how physics/math can be manipulated to show desired results &lt;/p&gt;&#xA;" OwnerUserId="1567" LastEditorUserId="1567" LastEditorDisplayName="" LastEditDate="2011-03-01T04:31:39.667" LastActivityDate="2011-03-01T13:35:49.733" Title="How did these guys get the masses from Heim Theory?" Tags="&lt;theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6152" PostTypeId="2" ParentId="6146" CreationDate="2011-03-01T02:55:09.153" Score="3" ViewCount="" Body="&lt;p&gt;Let me use a bit more standard notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you pull the string from equilibrium then to the first order there will be a force $F$ pulling it back that is proportional to tension $T$ and the distance $u$ that the string is moved from the equilibrium&#xA;$$F = -A T u$$&#xA;where $A$ is some constant not interesting to the further discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a piece of string with mass $m$ we get to the first order the equation for harmonic oscillator&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0 = m \ddot{u} - F = m \ddot{u} + A T u $$&#xA;which can be solved by $u \sim \exp(i\omega t)$ giving us the relation&#xA;$$\omega \sim \sqrt{T}$$&#xA;and differentiating both sides gives you the relation you wanted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this was actually just a quick and dirty derivation. For a full treatment one needs to account for the infinite number of degrees of freedom of the string. However, this derivation only rests on the fact that the pull-back force is proportional to the tension and that resulting equations are in the form of harmonic oscillator. First point translates to the full string directly while the second is correct because the string is described by a wave equation which can indeed be decomposed into simple harmonic oscillators.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-03-01T02:55:09.153" />
  <row Id="6153" PostTypeId="1" CreationDate="2011-03-01T03:29:35.927" Score="10" ViewCount="364" Body="&lt;p&gt;In the past few days I've become increasingly intrigued by the QHE, mainly thanks to very interesting questions and answers that have appeared here. Unfortunately, I am as of yet very confused by all the (seemingly disparate) stuff I learned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, here are some random points that I've been able to gather&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I(nteger)QHE occurs due to the presence of Landau levels&lt;/li&gt;&#xA;&lt;li&gt;IQHE is an embodiment of topological order and the states are characterized by the Chern number that tells us about topologically inequivalent Hamiltonians defined on the Brillouin zone&lt;/li&gt;&#xA;&lt;li&gt;IQHE requires negligible electron-electron interations and so is dependent on the presence of impurities that shield from Coulomb force&lt;/li&gt;&#xA;&lt;li&gt;F(ractional)QHE occurs because of formation of anyons. In this case Coulomb interaction can't be neglected but it turns out an effective non-interacting description emerges with particles obeying parastatistics and having fractional charge&lt;/li&gt;&#xA;&lt;li&gt;FQHE has again something to do with topology, TQFT, Chern-Simons theory, braiding groups and lots of other stuff&lt;/li&gt;&#xA;&lt;li&gt;FQHE has something to do with hierarchy states&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, here are the questions&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Most importantly, do these points make sense? Please correct any mistakes I made and/or fill in other important observations&lt;/li&gt;&#xA;  &lt;li&gt;How do explanations 1. and 2. of IQHE come together? Landau quantization only talks about electron states while topological picture doesn't mention them at all (they should be replaced by global topological states that are stable w.r.t. perturbations)&lt;/li&gt;&#xA;  &lt;li&gt;How do explanations 4., 5. and 6. relate together&lt;/li&gt;&#xA;  &lt;li&gt;Is there any accessible introductory literature into these matters?&lt;/li&gt;&#xA;  &lt;li&gt;Do IQHE and FQHE have anything (besides last three letters) in common so that e.g. IQHE can be treated as a special case? My understanding (based on 3.) is that this is not the case but several points hint into opposite direction. That's also why I ask about both QHE in a single question&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-03-01T18:55:02.730" Title="Quantum Hall effect for dummies" Tags="&lt;condensed-matter&gt;&lt;quantum-hall-effect&gt;&lt;topological-order&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="3" />
  <row Id="6154" PostTypeId="2" ParentId="1067" CreationDate="2011-03-01T03:36:49.240" Score="0" ViewCount="" Body="&lt;p&gt;what about the probability distribution for these functions?  just because a function is well behaved, that doesn't mean that the probability of finding a particle in a one dimensional box defined by that function is the same from 0 to 1/3 the box size as it is from 1/3 to 2/3 the box size or 2/3 to the length of the box, correct?&lt;/p&gt;&#xA;" LastActivityDate="2011-03-01T03:36:49.240" CommentCount="2" />
  <row Id="6155" PostTypeId="2" ParentId="6153" CreationDate="2011-03-01T03:50:36.077" Score="4" ViewCount="" Body="&lt;p&gt;Oh boy, hard to know where to start. Let me begin and see where I run out of steam. I'll go by the order you wrote your questions and make comments:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When you quantize electrons in a magnetic field, you get Landau levels: discrete energy levels which are highly degenerate. You can visualize each one of them as an electron moving in a circle whose radius is quantized (determined by the Landau level) and whose center can be anywhere (resulting in the degeneracy). Contrary to some discussions you hear sometimes, this by itself does NOT result in quantized Hall conductance.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For the integer QHE, the next crucial step is the presence of a random potential, provided by impurities. Then one can show that each Landau level contributes a fixed value to the Hall conductance, and therefore that conductance counts the number of filled Landau levels. The fact that this is robust is related to the topology, the Chern number and all that good stuff. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;FQHE is a different story, for which the Hall conductance can be fractional. The only thing IQHE and FQHE have in common is the ultimate physical effect, but the mechanism is very different. For the fractional effect you need very pure samples, since it is driven by strong Coulomb intercations between the degenerate electrons in each Landau levels. This is an inherently difficult problem, and in fact it was solved only by a guess - the Laughlin wavefunction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The EFT that describes the low energy excitations is related to the Chern-Simons theory, and those basic excitations obey anyonic statistics. Beyond that, I think all other effects you mentioned (e.g. heirarchy states), could be described as &quot;special topics&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Finally, I am just a humble high energy theorist, so I'll wait for corrections and more complete picture from the experts. Still, that was fun to write.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Incidentally, all of this is well-known stuff appearing in textbooks, though not always in an organized way. One good source: Mike Stone has edited a collection of papers on the subject for which he provided a series of introductions. If you find this book, those introductions are very good.)&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-03-01T03:56:54.263" LastActivityDate="2011-03-01T03:56:54.263" CommentCount="1" />
  <row Id="6156" PostTypeId="2" ParentId="6098" CreationDate="2011-03-01T04:03:20.777" Score="1" ViewCount="" Body="&lt;p&gt;I'd say that you have several regimes that are well defined:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The behavior of the fluid as it exits in inlet jets and enters the bulk without interference from the cubes. [Length scale set by the exit aperture?]&lt;/li&gt;&#xA;&lt;li&gt;Flow of the fluid around isolated cubes when far from the edges of the tank (far being several times the characteristic size &lt;em&gt;of the cube&lt;/em&gt;). [Length scale set by the side of the cube.]&lt;/li&gt;&#xA;&lt;li&gt;Flow of the fluid toward, along and away from the sides of the tank away from the jets and without interference from the cubes. [Length scale set by the boundary behavior?]&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;which is the good news, unfortunately you also have all the cases that mix and match the various length scales:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;case with cubes interacting with the jet near the aperture&lt;/li&gt;&#xA;&lt;li&gt;case with cubes in motion near the walls&lt;/li&gt;&#xA;&lt;li&gt;case with cubes in close proximity to one another&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You can probably find existing treatments for all the former cases, but the latter ones are going to be tricky, and you'll note that &lt;em&gt;they&lt;/em&gt; feature at least two length scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This must be part of why they say CFD is hard.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-03-01T04:03:20.777" CommentCount="1" />
  <row Id="6157" PostTypeId="1" CreationDate="2011-03-01T04:40:54.293" Score="38" ViewCount="2061" Body="&lt;p&gt;I'm trying to amass a list of physics books with open-source licenses, like Creative Commons, GPL, etc. The books can be about a particular field in physics or about physics  in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some freely available physics books on the Internet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a starter:&#xA;&lt;a href=&quot;http://www.phys.uu.nl/~thooft/theorist.html&quot; rel=&quot;nofollow&quot;&gt;http://www.phys.uu.nl/~thooft/theorist.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/6157/list-of-freely-available-physics-books/6242#6242&quot;&gt;jump to list sorted by medium / type&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Table of contents sorted by field (in alphabetical order):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/6157/list-of-freely-available-physics-books/6158#6158&quot;&gt;Chaos Theory&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/6157/list-of-freely-available-physics-books/7690#7690&quot;&gt;Earth System Physics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/6157/list-of-freely-available-physics-books/6159#6159&quot;&gt;Mathematical Physics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/6157/list-of-freely-available-physics-books/6167#6167&quot;&gt;Quantum Field Theory&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1814" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-02T06:43:34.463" LastActivityDate="2011-08-28T06:04:53.857" CommunityOwnedDate="2011-03-01T05:22:45.950" Title="List of freely available physics books" Tags="&lt;education&gt;&lt;books&gt;&lt;reference-request&gt;&lt;big-list&gt;" AnswerCount="19" CommentCount="9" FavoriteCount="32" />
  <row Id="6158" PostTypeId="2" ParentId="6157" CreationDate="2011-03-01T04:46:34.027" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://chaosbook.org/&quot; rel=&quot;nofollow&quot;&gt;Chaos: Classical and Quantum&lt;/a&gt;&lt;/strong&gt;, Cvitanovic´, Artuso, Mainieri, Vattay &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://rack1.ul.cs.cmu.edu/is/newton/&quot; rel=&quot;nofollow&quot;&gt;Mathematical Principles of Natural Philosophy&lt;/a&gt;&lt;/strong&gt;, Newton, translated by Motte&lt;/p&gt;&#xA;" OwnerUserId="1814" LastEditorUserId="2809" LastEditorDisplayName="" LastEditDate="2011-03-28T04:10:45.850" LastActivityDate="2011-03-28T04:10:45.850" CommunityOwnedDate="2011-03-01T05:22:45.950" />
  <row Id="6159" PostTypeId="2" ParentId="6157" CreationDate="2011-03-01T05:03:48.947" Score="9" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.physics.miami.edu/~nearing/mathmethods/&quot; rel=&quot;nofollow&quot;&gt;Mathematical Tools for Physics&lt;/a&gt;, James Nearing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also available in paperback from Dover.  Undergraduate-level math methods book.  Clear writing, many problems and exercises (usually without solution).  IMHO better than Boas.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-03-01T05:03:48.947" CommunityOwnedDate="2011-03-01T05:22:45.950" />
  <row Id="6160" PostTypeId="2" ParentId="6157" CreationDate="2011-03-01T05:07:16.290" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://streetfightingmath.com/&quot; rel=&quot;nofollow&quot;&gt;Street-Fighting Mathematics&lt;/a&gt;, Sanjoy Mahajan&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short book covering dimensional analysis, estimation, and visualizing mathematics.  The goal is to make mathematics easier and more useful for scientists and engineers, specifically physicists.  (Mahajan is a physicist/educator).  Uses mostly undergrad and high school level math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A version is available &lt;a href=&quot;http://mitpress.mit.edu/books/full_pdfs/Street-Fighting_Mathematics.pdf&quot; rel=&quot;nofollow&quot;&gt;free as a pdf&lt;/a&gt; under the creative commons license. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper version is available for purchase from the MIT press.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="2809" LastEditorDisplayName="" LastEditDate="2011-03-28T03:35:19.843" LastActivityDate="2011-03-28T03:35:19.843" CommunityOwnedDate="2011-03-01T05:22:45.950" CommentCount="3" />
  <row Id="6161" PostTypeId="2" ParentId="6157" CreationDate="2011-03-01T05:19:32.520" Score="8" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.pma.caltech.edu/Courses/ph136/yr2008/&quot; rel=&quot;nofollow&quot;&gt;Applications of Classical Physics&lt;/a&gt;, Kip Thorne and Roger Blandford&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unpublished so far, this book covers special relativity, thermodynamics and statistical mechanics, continuum mechanics, and some general relativity at the graduate level.  I've read about a third of it.  It's well-written and surprisingly-polished for being a freely-available preprint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem sets and solutions are available on the website as well.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-03-01T05:19:32.520" CommunityOwnedDate="2011-03-01T05:22:45.950" CommentCount="1" />
  <row Id="6162" PostTypeId="2" ParentId="6157" CreationDate="2011-03-01T05:27:33.087" Score="9" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://mitpress.mit.edu/SICM/&quot; rel=&quot;nofollow&quot;&gt;Structure and Interpretation of Classical Mechanics&lt;/a&gt;&lt;/strong&gt; Sussman, Wisdom, Mayer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/9712019&quot; rel=&quot;nofollow&quot;&gt;A No-Nonsense Introduction to General Relativity&lt;/a&gt;&lt;/strong&gt;, Sean Carroll&lt;/p&gt;&#xA;" OwnerUserId="1814" LastEditorUserId="2809" LastEditorDisplayName="" LastEditDate="2011-03-28T03:56:14.763" LastActivityDate="2011-03-28T03:56:14.763" CommunityOwnedDate="2011-03-01T05:27:33.087" CommentCount="4" />
  <row Id="6163" PostTypeId="2" ParentId="6157" CreationDate="2011-03-01T05:57:05.147" Score="5" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.artcompsci.org/&quot; rel=&quot;nofollow&quot;&gt;The art of computational science&lt;/a&gt;, Piet Hut and Jun Makino&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A computational lab for N-body experiments. Includes books and source code for doing simple N-body simulations. &lt;/p&gt;&#xA;" OwnerUserId="1252" LastActivityDate="2011-03-01T05:57:05.147" CommunityOwnedDate="2011-03-01T05:57:05.147" />
  <row Id="6164" PostTypeId="2" ParentId="6157" CreationDate="2011-03-01T05:59:42.597" Score="5" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/quant-ph/0608140&quot; rel=&quot;nofollow&quot;&gt;1951 Lectures on Advanced Quantum Mechanics Second Edition&lt;/a&gt;, Freeman J. Dyson&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0510040&quot; rel=&quot;nofollow&quot;&gt;Introductory Lectures on Quantum Field Theory&lt;/a&gt;, Luis Alvarez-Gaume, Miguel A. Vazquez-Mozo&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1670" LastActivityDate="2011-03-01T05:59:42.597" CommunityOwnedDate="2011-03-01T05:59:42.597" />
  <row Id="6165" PostTypeId="2" ParentId="6157" CreationDate="2011-03-01T06:29:44.493" Score="5" ViewCount="" Body="&lt;p&gt;Sidney Coleman's QFT &lt;a href=&quot;http://www.physics.upenn.edu/~chb/phys253a/coleman/&quot; rel=&quot;nofollow&quot;&gt;lecture notes&lt;/a&gt;&#xA;and &lt;a href=&quot;http://www.physics.harvard.edu/about/Phys253.html&quot; rel=&quot;nofollow&quot;&gt;videos&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1814" LastEditorUserId="429" LastEditorDisplayName="" LastEditDate="2011-03-01T07:34:30.237" LastActivityDate="2011-03-01T07:34:30.237" CommunityOwnedDate="2011-03-01T06:29:44.493" />
  <row Id="6166" PostTypeId="2" ParentId="6157" CreationDate="2011-03-01T06:55:59.473" Score="1" ViewCount="" Body="&lt;h2&gt;Quantum Mechanics for Engineers&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.eng.fsu.edu/~dommelen/quantum/&quot; rel=&quot;nofollow&quot;&gt;http://www.eng.fsu.edu/~dommelen/quantum/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I stumbled across this book the other day when I was looking for a text on nuclear physics. It seems like a handy resource.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quote the preface:&lt;br&gt;&#xA;&lt;code&gt;&#xA;The book was primarily written for engineering graduate students who find themselves caught up in nano technology. It is a simple fact that the typical engineering education does not provide anywhere close to the amount of physics you will need to make sense out of the literature of your field. You can start from scratch as an undergraduate in the physics department, or you can read this book.&#xA;&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="429" LastEditorUserId="429" LastEditorDisplayName="" LastEditDate="2011-08-28T06:04:53.857" LastActivityDate="2011-08-28T06:04:53.857" CommunityOwnedDate="2011-03-01T06:55:59.473" />
  <row Id="6167" PostTypeId="2" ParentId="6157" CreationDate="2011-03-01T06:57:42.937" Score="10" ViewCount="" Body="&lt;h2&gt;Quantum field theory&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/9912205&quot; rel=&quot;nofollow&quot;&gt;Fields&lt;/a&gt;&lt;/strong&gt;, by W. Siegel &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.physics.ucsb.edu/~mark/qft.html&quot; rel=&quot;nofollow&quot;&gt;Quantum Field Theory&lt;/a&gt;&lt;/strong&gt;, by Mark Srednicki&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0108200&quot; rel=&quot;nofollow&quot;&gt;Superspace, or One thousand and one lessons in supersymmetry&lt;/a&gt;&lt;/strong&gt;&#xA;by S.J. Gates Jr, M.T. Grisaru, M. Rocek and W. Siegel&lt;/p&gt;&#xA;" OwnerUserId="429" LastEditorUserId="2809" LastEditorDisplayName="" LastEditDate="2011-03-28T04:18:41.697" LastActivityDate="2011-03-28T04:18:41.697" CommunityOwnedDate="2011-03-01T06:57:42.937" CommentCount="1" />
  <row Id="6168" PostTypeId="2" ParentId="6151" CreationDate="2011-03-01T07:08:02.257" Score="0" ViewCount="" Body="&lt;p&gt;I googled the subject as I was not familiar with the formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to remind you of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lyman_series&quot; rel=&quot;nofollow&quot;&gt;Rydberg formula&lt;/a&gt; , which fitted the lines of the hydrogen atom long before the Bohr model. In a more recent example I will mention Regge poles, the vector meson exchange model, the eightfold way, all phenomenological formulations that have been absorbed, not discounted,  in the standard model for particle physics. The standard model is on the way of being absorbed by string theory modeling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the relevant question is if there are as many parameters in their formula as output masses. If the formula works, and from what I checked  with a google search there is no proof of such mistakes evident, it is another datum in the search for the TOE. They claim only four parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I were a theorist working full time on a string model I would try to see how this &lt;em&gt;formula&lt;/em&gt; could come out from my model, in analogy  to the way the Rydberg formula came  out when the Bohr model was formulated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Skepticism is good but not to the extent of throwing out the baby with the bathwater, and if the formula, as such, works it is a useful datum. Why, with all the new formulations coming up, as the &lt;a href=&quot;http://streamer.perimeterinstitute.ca/mediasite/viewer/NoPopupRedirector.aspx?peid=ff8f97a4-2848-4b93-98d9-1218de2070af&amp;amp;shouldResize=False&quot; rel=&quot;nofollow&quot;&gt;&quot;twistor uprising&quot;&lt;/a&gt; , who knows maybe even the relevant part to the formula in their  particular spaces will turn out to be some sort of isomorphic subset of the real theory.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-01T07:08:02.257" CommentCount="3" />
  <row Id="6170" PostTypeId="1" AcceptedAnswerId="6172" CreationDate="2011-03-01T09:12:12.757" Score="0" ViewCount="115" Body="&lt;p&gt;how can we relate the inflation and string theory ??  we study some inflationary cosmological model or some string cosmological model  under some specific assumptions or conditions. is their any relation or any difference between these two concepts ... as my knowledge inflation is a theory or concept  to resolve some Standard problems of early universe .and string is ........i already got the answer in more question on this site.i actually want to know the relation between these two terms ?&lt;/p&gt;&#xA;" OwnerUserId="1843" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-01T10:54:38.127" LastActivityDate="2011-03-01T13:07:43.220" Title="Inflation and string theory in general relativity" Tags="&lt;string-theory&gt;" AnswerCount="1" />
  <row Id="6172" PostTypeId="2" ParentId="6170" CreationDate="2011-03-01T10:32:16.440" Score="1" ViewCount="" Body="&lt;p&gt;The theory of cosmic inflation (with the inflaton field) is an example of an effective quantum field theory - much like the Standard Model of particle physics (with quarks, leptons, photons, gauge bosons, etc.). And that's why, if it is correct, it must emerge as an approximation of stringy physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So inflation and string theory are compatible but string theory is much more general a theory, and the inflaton - if it exists - is just one of the many quantum fields predicted by string theory at lower energies (although the inflation scale is probably not too much lower than the string scale).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most &quot;mainstream&quot; incorporation of inflation within string theory is the KKLMMT paper, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0308055&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0308055&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which has 600+ citations now. Of course, with letters such as KKLT in the list of authors, it's not shocking that those people rely on some anthropic tricks to get the slow-roll potential and other things. There may exist better models that predict all the required parameters more automatically. But they won't change the fact that inflation is just a derived concept from string theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exist papers that try to geometrize the inflaton in much more stringy terms - it may be e.g. the distance between a brane and an antibrane that may annihilate. The resulting physics is more or less effectively equivalent to the inflation - it just gives an extra answer to the question &quot;what the inflaton is made out of&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-03-01T13:07:43.220" LastActivityDate="2011-03-01T13:07:43.220" CommentCount="3" />
  <row Id="6173" PostTypeId="2" ParentId="6153" CreationDate="2011-03-01T10:47:55.357" Score="3" ViewCount="" Body="&lt;p&gt;This was too long to fit into a comment, so an answer it will have to be. This is all in supplement to @Moshe R.'s answer, which is excellent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, just to correct your statements (in addition to Moshe's): 3. Impurities do not screen anything. The electrons themselves provide the screening to make an independent electron approximation semi-justified (this is the usual Landau Fermi-liquid argument). Impurities however provide the basic scattering potential to achieve some Anderson localisation, which is crucial for actually getting the plateaus --- otherwise one would never get any resistance at all! Incidentally, understanding this point is crucial for understanding why the longitudinal conductance displays the spikes that it does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In condense matter, we don't get to have exact theories --- everything is a simplified approximation. As such, one will come across in the literature many different theories, which emphasise different aspects of the phenomenon, and have differing amounts of complexity and quantitative accuracy. At this point, it is fair to say that IQHE is well understood, the prevailing theory being a combination of topological states, impurity effects, and 2-parameter scaling theory (of both longitudinal and transverse conductances, ala Khmelnitskii). However, the theory of FQHE has not reached quite the same consensus. In some respects, FQHE is like a IQHE of electrons with extra flux &quot;bound&quot; to them (through an effective interaction due to Coulomb repulsion); in this picture, all the messiness (impurities), etc. are again crucial. This is also related to the hierarchical states because one can imagine binding more flux to the anyonic excitations and getting more IQHE states of those. However, it is clear that since the basic ingredient is the strong Coulomb interaction, without a systematic (the above is very much ad hoc) treatment it is impossible to be confident about the range of validity of the theory. Work on this aspect is on going (but to be fair, somewhat stalled --- it is sufficiently hard theoretically speaking that one is really looking for some fundamental break through in mathematics to finish it off).&lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-03-01T10:47:55.357" CommentCount="9" />
  <row Id="6174" PostTypeId="1" AcceptedAnswerId="6180" CreationDate="2011-03-01T11:15:03.760" Score="7" ViewCount="262" Body="&lt;p&gt;That pretty much says it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I have some powder of $NaCl$. It is kept in contact with itself in vacuum. You are free to remove all the disturbances that bother you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that true that, well, there exist a ($\mbox{very}^\mbox{very}$ large) amount of time $T$ that for every moment $t&amp;gt;T$ you will have a single beautiful crystal with 99% probability.    &lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="184" LastEditorDisplayName="" LastEditDate="2011-03-01T17:43:01.283" LastActivityDate="2011-03-01T17:43:01.283" Title="If I take a handful of salt and wait for an infinite time will it become a single crystal?" Tags="&lt;thermodynamics&gt;&lt;solid-state-physics&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="6175" PostTypeId="2" ParentId="6133" CreationDate="2011-03-01T11:48:11.327" Score="1" ViewCount="" Body="&lt;p&gt;I would suggest these from Living Reviews in Relativity,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://emis.math.tifr.res.in/journals/LRG/Articles/lrr-2003-3/&quot; rel=&quot;nofollow&quot;&gt;Rotating Stars in Relativity&lt;/a&gt;, by Nikos Stergioulas,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://emis.math.tifr.res.in/journals/LRG/Articles/lrr-2008-10/&quot; rel=&quot;nofollow&quot;&gt;Physics of Neutron Star Crust&lt;/a&gt;s, by Nicolas Chamel and Pawel Haensel and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://emis.math.tifr.res.in/journals/LRG/Articles/lrr-2007-1/&quot; rel=&quot;nofollow&quot;&gt;Relativistic Fluid Dynamics: Physics for Many Different Scales&lt;/a&gt;, by Nils Andersson and Gregory L. Comer&lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2011-03-01T11:48:11.327" />
  <row Id="6177" PostTypeId="1" AcceptedAnswerId="6178" CreationDate="2011-03-01T12:37:27.003" Score="2" ViewCount="275" Body="&lt;p&gt;If I apply $1 \text{ V}$ across a $1 \text{ }\Omega$ resistance, I'd get $1 \text{ A}$ flowing. $1 \text{ A}$ is defined as $1 \frac{\text{C}}{\text{s}}$, and $1 \text{ C}$ is equivalent to $6.24150975 &amp;#215; 10^{18}$ electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, Ohm's Law describes only the number of electrons passing a given branch per second. How do we determine the speed of electrons as they move through a conductor? And what is that speed?&lt;/p&gt;&#xA;" OwnerUserId="293" LastActivityDate="2011-03-01T14:52:30.273" Title="How fast do electrons move through a conductor?" Tags="&lt;electricity&gt;&lt;speed&gt;" AnswerCount="1" />
  <row Id="6178" PostTypeId="2" ParentId="6177" CreationDate="2011-03-01T13:05:32.050" Score="4" ViewCount="" Body="&lt;p&gt;You can't measure speed of electrons from these data alone. If the area of the cross section of a cylindrical conductor is A then the formula would be $v = \frac {I}{QeA}$ where $Q$ is the mobile electrons per cc and $e$ is the charge of electron, $v$ is the speed of electron, $I$ is the current. This is for dc. See this &lt;a href=&quot;http://amasci.com/miscon/speed.html&quot; rel=&quot;nofollow&quot;&gt;http://amasci.com/miscon/speed.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you see, you need to know the diameter of the cylindrical conductor. However the electrons don't move in a conductor in an orderly manner. They are constantly colliding with each other and typically have speed components along the conductor about some millimeters per second.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-03-01T14:52:30.273" LastActivityDate="2011-03-01T14:52:30.273" CommentCount="2" />
  <row Id="6179" PostTypeId="2" ParentId="6151" CreationDate="2011-03-01T13:35:49.733" Score="2" ViewCount="" Body="&lt;p&gt;Jaskey13,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have done some googling too on Heim. One problem is just identifying any single formula as &quot;the mass formula&quot;. There are various expressions which are often several dozen symbols in length, for subcomponents of the &quot;formula&quot;. Therein lies part of the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found one link to someone who claimed to have used Mathematica on these formulae, to obtain results which he claimed valid. But this individual could find no reason as to &lt;em&gt;why&lt;/em&gt; they were valid. This kind of thing John Baez called &quot;numerology&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the perspective of &quot;wider lessons&quot; which seems to be part of your question some points:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A scientific theory needs to be able to make predictions too. In the years that these formulae have been around many particles have been predicted by other theories and discovered. So where is the long history of successful predictions in this theory?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This theory is founded on a non-standard mathematics developed by that author. So there is a layer of uncertainty as to what constitutes valid calculations in the formalism. Fortunately (or unfortunately) this characteristic means that the associated theory will not be worked on by anyone else; and if it contains oddities in its physics it will be labeled non-science as has happened here.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It is an old theory though, dating back to the 1950s. Modern scientists have plenty other theories to learn to spend time trying to decipher it all. But do note that there are other such idiosyncratic theories developed over the years in the mathematical sciences connected with a formalism that only their author fully understood. Outsiders might describe these as &quot;crackpot&quot; for understandable reasons.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-03-01T13:35:49.733" />
  <row Id="6180" PostTypeId="2" ParentId="6174" CreationDate="2011-03-01T13:50:40.293" Score="5" ViewCount="" Body="&lt;p&gt;If you wait long enough, it'll become a bunch of iron (assuming it's confined so that the atoms can't evaporate off as Marek pointed out). After all, an iron nucleus is more stable than any other nucleus. The probability of the other nuclei tunnelling together to form a big hunk of iron is absurdly low, but it's not strictly zero. If you wait long enough, it'll happen.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-01T13:50:40.293" CommentCount="1" />
  <row Id="6181" PostTypeId="2" ParentId="6174" CreationDate="2011-03-01T13:59:16.587" Score="-1" ViewCount="" Body="&lt;p&gt;If you wait very, very long then your NaCL will be part of a black hole, for this will be the only entities in the Universe in the far future, at least according to our current understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, eventually, these black holes evaporate and your NaCl will be part of Hawking radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the ensuing thermal ensemble you will have random fluctuations. If you wait long enough then, indeed, you have a probability arbitrarily close to 1 that one of these random fluctuations will be that your original NaCL powder forms a perfect crystal, or that some monkeys type the work of Shakespeare ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in short, the answer seems to be 'yes'.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-03-01T13:59:16.587" CommentCount="3" />
  <row Id="6182" PostTypeId="2" ParentId="6147" CreationDate="2011-03-01T14:01:46.943" Score="5" ViewCount="" Body="&lt;p&gt;Mentions of &quot;acceleration&quot;, or lack of it, to establish the difference between the twins' careers, are confusing the issue.  (This can be proved by concocting roadmaps for two travelers with &lt;em&gt;identical&lt;/em&gt; acceleration or deceleration periods, only placed at &lt;em&gt;different times&lt;/em&gt; of their respective journeys. This way, their watches don't show the same elapsed time when they are reunited.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better mental picture is obtained by considering paths in spacetime for each traveller. Both paths start from the same &quot;event&quot; (the event of the twins saying good bye to each other) and arrive at another event (their getting together again).  Now, just as two different paths in space, with same beginning and same end, need not have the same length, two different paths in space-time need not have the same &quot;four-dimensional length&quot;, so to speak.  The quotes are meant to warn that this expression is not used by relativists:  This &quot;four-dimensional length&quot; is actually what they call &quot;proper time&quot;, and it happens to be (according to Special Relativity), what each twin's wristwatch is measuring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the main issue is this difference in spacetime trajectories, not gravitational effects per se.  That the steady twin finds himself older than his sister at the end comes from the fact that his spacetime trajectory is rectilinear, whereas his twin's is curved.  And contrary to three-dimensional length, &quot;four-dimensional length&quot;, aka proper time, is maximal, not minimal, for straight (i.e., physically, inertial) spacetime trajectories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, one will suffer accelerations if one follows a curved 4-D path.  But,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Meaningful effects (in terms of age difference) can be obtained with small accelerations (of order 1 g) over a few years, so invoking GR is not necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Acceleration per se does not &lt;em&gt;cause&lt;/em&gt; the proper-time discrepancy:  Two traveling twins who suffer the same acceleration-deceleration events (the same in intensity and duration) can age differently if these events are not scheduled the same way.  The real cause is this difference in scheduling, which is logically the same thing as a geometrical difference between the two 4-D paths.&lt;/p&gt;&#xA;" OwnerUserId="1761" LastActivityDate="2011-03-01T14:01:46.943" CommentCount="3" />
  <row Id="6183" PostTypeId="2" ParentId="414" CreationDate="2011-03-01T14:37:16.900" Score="1" ViewCount="" Body="&lt;p&gt;you problably could be interested in this review article &quot;Physics of the Riemann Hypothesis&quot; &lt;a href=&quot;http://arxiv.org/abs/1101.3116&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2338" LastActivityDate="2011-03-01T14:37:16.900" />
  <row Id="6184" PostTypeId="2" ParentId="6147" CreationDate="2011-03-01T14:51:08.527" Score="2" ViewCount="" Body="&lt;p&gt;The other answers all bring out the essential points but I'll try to rephrase things in a way that makes it clear how the two &quot;twins&quot; (of the twin paradox) are different without any reference to General Relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The twin on earth is &lt;strong&gt;always on a geodesic&lt;/strong&gt; of the spacetime (which is flat Minkowski) whereas the round-trip accelerating twin is on a worldine that clearly deviates from a geodesic. So in this picture, it is very clear that the situation is not symmetric with respect to the twins. And you can also calculate the proper time (&quot;length&quot; of the worldines) and see that it is shorter for the non-geodesic path. In fact, &lt;strong&gt;the Minkowski spacetime is such that the proper time is &lt;em&gt;maximized&lt;/em&gt; for geodesics.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-03-01T15:14:59.647" LastActivityDate="2011-03-01T15:14:59.647" CommentCount="1" />
  <row Id="6185" PostTypeId="2" ParentId="6140" CreationDate="2011-03-01T15:56:17.717" Score="3" ViewCount="" Body="&lt;p&gt;If you want your computer to run fastest, leave it at rest ;-).&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-03-01T15:56:17.717" />
  <row Id="6186" PostTypeId="2" ParentId="6174" CreationDate="2011-03-01T16:40:02.673" Score="1" ViewCount="" Body="&lt;p&gt;Technically, if you have infinite time to wait, then yes. This is based on the standard &quot;if it isn't forbidden, it is compulsory&quot; idea of QM. However, as other answers have correctly pointed out, there are other potential quantum fluctuations that are also available and may be more probable. Combined with the unlikelihood of the desired quantum fluctuation, in any practical sense, the answer is no because the universe will end before this occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, with regard to the last line of your question, which is slightly different, the answer is a resounding no. Although you can conceivably get a single crystal at some very large time T, you will not retain a single crystal for all time t&gt;T because another fluctuation will destroy it eventually.&lt;/p&gt;&#xA;" OwnerUserId="1196" LastActivityDate="2011-03-01T16:40:02.673" CommentCount="3" />
  <row Id="6188" PostTypeId="2" ParentId="6153" CreationDate="2011-03-01T18:55:02.730" Score="1" ViewCount="" Body="&lt;p&gt;There is a book that covers exactly the questions you asked:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nathan Goldman, Quantum transport in lattices subjected to external gauge fields: The quantum Hall effect in optical lattices and quantum graphs, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3639163869&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Quantum-transport-lattices-subjected-external/dp/3639163869&lt;/a&gt;&#xA;(I'm reading it now and it's simple and well written.) &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you are short of time (or money) - the book is based on his thesis:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nathan Goldman, Quantum transport and phase transitions in lattices subjected to external gauge fields,&#xA;&lt;a href=&quot;http://theses.ulb.ac.be/ETD-db/collection/available/ULBetd-04012009-152422/&quot; rel=&quot;nofollow&quot;&gt;http://theses.ulb.ac.be/ETD-db/collection/available/ULBetd-04012009-152422/&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="184" LastActivityDate="2011-03-01T18:55:02.730" CommentCount="1" />
  <row Id="6189" PostTypeId="2" ParentId="6157" CreationDate="2011-03-01T19:23:49.480" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.people.fas.harvard.edu/~hgeorgi/new.htm&quot; rel=&quot;nofollow&quot;&gt;Howard Georgi's The Physics of Waves&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish someone would make the (now mostly out of print) Berkeley Physics Course volumes public domain.  Or at least Dover could start cranking out cheap versions.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastEditorUserId="2809" LastEditorDisplayName="" LastEditDate="2011-03-28T03:45:51.523" LastActivityDate="2011-03-28T03:45:51.523" CommunityOwnedDate="2011-03-01T19:23:49.480" />
  <row Id="6191" PostTypeId="2" ParentId="6047" CreationDate="2011-03-01T21:32:39.580" Score="1" ViewCount="" Body="&lt;p&gt;You might try, now in paperback,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Th. Frankel:  &lt;em&gt;The Geometry of Physics,&lt;/em&gt; An Introduction, Cambridge U.P. (Cambridge), 1997.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a course in differential geometry, actually, but one oriented towards physics, with succinct but comprehensive enough developments of physical theories (mechanics, electromagnetism, thermodynamics, Yang-Mills ...).   It's a bit like Burke's &lt;em&gt;Applied Differential Geometry&lt;/em&gt; (which I like too), but longer and  more systematic.&lt;/p&gt;&#xA;" OwnerUserId="1761" LastActivityDate="2011-03-01T21:32:39.580" />
  <row Id="6192" PostTypeId="1" AcceptedAnswerId="6193" CreationDate="2011-03-01T21:53:23.493" Score="1" ViewCount="110" Body="&lt;p&gt;I am trying to solve this simple excercise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You throw a small coin upwards with&lt;/em&gt; $4 \frac{m}{s}$ &lt;em&gt;. How much time does it need to reach the height of&lt;/em&gt; $0.5 m$ &lt;em&gt;? Why do we get two results?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(We get two results for time, because the coin passes the 0.5m mark downwards too.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equtation I used (for constant acceleration):&#xA;$x = x_0 + v_0 t + \frac{1}{2} a t^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These values I know: $x = 0.5m$, $x_0 = 0$, $v_0 = 4 \frac{m}{s}$, $a = -9.81 \frac{m}{s^2}$ So, I get a quadratic equtation which I can solve for $t$. My two results are: $t_1 = 0.15s$ and $t_2 = 0.66s$, however the book has the results 0.804s and 0.013s. What am I doing wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thanks for your help&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2342" LastActivityDate="2011-03-01T22:13:02.027" Title="Simple kinematics excercise, throwing something upwards" Tags="&lt;homework&gt;&lt;kinematics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="6193" PostTypeId="2" ParentId="6192" CreationDate="2011-03-01T21:59:12.943" Score="2" ViewCount="" Body="&lt;p&gt;Nothing, your book did it wrong.  &lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-03-01T21:59:12.943" CommentCount="1" />
  <row Id="6194" PostTypeId="2" ParentId="6192" CreationDate="2011-03-01T22:13:02.027" Score="-2" ViewCount="" Body="&lt;p&gt;You get two results because the particle at this height does not stop but continues flying up. In a while it goes down and passes the same elevation on its way down.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-03-01T22:13:02.027" CommentCount="3" />
  <row Id="6195" PostTypeId="2" ParentId="484" CreationDate="2011-03-01T23:09:24.700" Score="1" ViewCount="" Body="&lt;p&gt;In plasma physics there is a notion of plasma transparency or so. When you heat plasma in TOKAMAK, it is first transparent to radiation and the heat losses are essential. The plasma radiates &quot;from volume&quot;. With temperature increasing the plasma becomes non transparent and radiates only &quot;from surface&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-03-01T23:09:24.700" />
  <row Id="6196" PostTypeId="1" CreationDate="2011-03-01T23:12:11.450" Score="6" ViewCount="181" Body="&lt;p&gt;Can anyone give me a quantum mechanical explanation of the theory of valence? (i.e. why atoms bond just enough to have a complete orbital)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: To clarify, I already have an idea of why atoms bond, and the molecular orbit makes sense to me. The problem is that the valence bond theory provides a simple and fairly accurate description for when a molecule is stable. To make this more concrete: why is CH_4 more stable than CH_3 or CH_5. Looking up the former on wikipedia, I see a better way of phrasing the question might be: why are free radicals reactive? (Should I change the title?)&lt;/p&gt;&#xA;" OwnerUserId="1859" LastEditorUserId="1859" LastEditorDisplayName="" LastEditDate="2011-03-02T22:26:42.443" LastActivityDate="2011-03-02T22:26:42.443" Title="Explaining valence with quantum mechanics" Tags="&lt;quantum-mechanics&gt;&lt;physical-chemistry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="6197" PostTypeId="1" AcceptedAnswerId="6222" CreationDate="2011-03-01T23:18:46.207" Score="9" ViewCount="294" Body="&lt;p&gt;In general relativity light is subject to gravitational pull.  Do light generate gravitational pull, and do two beams of light attract each other?&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditorDisplayName="" LastEditDate="2011-03-02T19:00:46.660" LastActivityDate="2011-03-02T19:00:46.660" Title="Do two beams of light attract each other in general theory of relativity?" Tags="&lt;general-relativity&gt;&lt;light&gt;" AnswerCount="3" FavoriteCount="4" />
  <row Id="6198" PostTypeId="1" CreationDate="2011-03-01T23:28:44.440" Score="1" ViewCount="100" Body="&lt;p&gt;So, the Fermi level in crystals is pretty easy to understand. Been using it and talking about it in terms of the highest occupied level forever. However, I'm now reading about disordered systems. A lot of researchers mention the existence of empty states randomly distributed above and &lt;em&gt;below&lt;/em&gt; the Fermi level. But isn't the Fermi level necessarily the energy at which all energies below are filled states?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: So I suppose chemical potential is the more correct term to use (instead of Fermi energy) because not at T = 0. Anyways, consider this quote:&#xA;&quot;the localized gap states near $E_F$ are spread in energy with singly occupied gap states below $E_F$, doubly occupied states lying above $E_F$, and empty states distributed randomly.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1939" LastEditorUserId="1939" LastEditorDisplayName="" LastEditDate="2011-03-02T02:29:35.103" LastActivityDate="2011-03-02T02:29:35.103" Title="Fermi level in disordered amorphous and/or organic semiconductors" Tags="&lt;condensed-matter&gt;&lt;solid-state-physics&gt;&lt;semiconductor-physics&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="6199" PostTypeId="2" ParentId="6197" CreationDate="2011-03-01T23:37:55.653" Score="7" ViewCount="" Body="&lt;p&gt;Yes. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor#Electromagnetic_stress-energy_tensor&quot; rel=&quot;nofollow&quot;&gt;energy-momentum tensor&lt;/a&gt; (which is on the right hand side of &lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein_field_equations#Mathematical_form&quot; rel=&quot;nofollow&quot;&gt;Einstein's equation&lt;/a&gt;) is non-zero in the presence of any kind of energy density, including radiation. This means that the light beams will curve spacetime (measured by the left hand side of Einstein's equation) and hence affect the path that the light takes. But for typical light beams, this is very small and hence neglectable.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-03-01T23:37:55.653" />
  <row Id="6200" PostTypeId="2" ParentId="6197" CreationDate="2011-03-01T23:45:44.317" Score="8" ViewCount="" Body="&lt;p&gt;According to general relativity, yes, two beams of light would gravitationally attract each other. &lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein_field_equations&quot; rel=&quot;nofollow&quot;&gt;Einstein's equation&lt;/a&gt; says that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R^{\mu\nu} - \frac{1}{2}R g^{\mu\nu} = 8\pi T^{\mu\nu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The terms on the left represent the distortion (&quot;curvature&quot;) of spacetime, and the term on the right represents matter and energy, including light. As long as $T^{\mu\nu}$ is nonzero, there will have to be some sort of induced distortion a.k.a. gravity, since $R^{\mu\nu} - \frac{1}{2}R g^{\mu\nu} = 0$ in flat spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you're interested, the relevant equations are the definition of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Maxwell%27s_equations_in_curved_spacetime#Electromagnetic_stress-energy_tensor&quot; rel=&quot;nofollow&quot;&gt;stress-energy tensor for electromagnetism&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T^{\mu\nu} = -\frac{1}{\mu_0}\biggl(F^{\mu\rho}F_{\rho}^{\ \nu} + \frac{1}{4}g^{\mu\nu}F^{\rho\sigma}F_{\rho\sigma}\biggr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $F$ is the electromagnetic field tensor, and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Maxwell%27s_equations_in_curved_spacetime#Electromagnetic_wave_equation&quot; rel=&quot;nofollow&quot;&gt;electromagnetic wave equation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D_{\alpha}D^{\alpha}F^{\mu\nu} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $D_{\alpha}$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Covariant_derivative&quot; rel=&quot;nofollow&quot;&gt;covariant derivative&lt;/a&gt; operator. In principle, to calculate the gravitational attraction between two beams of light, you would identify the functions $F^{\mu\nu}$ that correspond to your beams (they would have to satisfy the wave equation), and then plug them in to calculate $T^{\mu\nu}$. When you put that into Einstein's equation, it places a constraint on the possible values of the metric $g_{\mu\nu}$ and its derivatives, and you could use that constraint to determine the &lt;a href=&quot;http://en.wikipedia.org/wiki/Geodesic_deviation_equation&quot; rel=&quot;nofollow&quot;&gt;geodesic deviation&lt;/a&gt; between the two beams of light, which, in a sense, corresponds to their gravitational attraction. &lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-03-01T23:45:44.317" CommentCount="3" />
  <row Id="6201" PostTypeId="2" ParentId="5600" CreationDate="2011-03-01T23:52:10.090" Score="3" ViewCount="" Body="&lt;p&gt;Let us assume (for simplicity in this question) that the Hamiltonian H is time independent. Then any function f whose Poisson Bracket with H {f, H}=0 is a constant of motion. Also Poisson's Theorem is that if f,g are both constants of motion then so is {f,g} generating yet more constants of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clarification is that the functions f, g here define canonical changes of variable, so we are only interested in ones that replace the $p_i$ and $q_i$ with new coordinates $(P_i,Q_i)$. There are only going to be at most 2N of these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ideal arrangement is to cause as many of the $Q_i$'s as possible to drop out of the (transformed) Hamiltonian $H=H(P_i,Q_i)$, there only being N maximum available. So yes many &quot;constants of motion&quot; will be functions of one another (akin to having not only x(q),y(p) available as coordinates, but also x+y, $x^2$, etc), but one wants to use only those which lead to a simpler Hamiltonian.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-03-01T23:52:10.090" />
  <row Id="6202" PostTypeId="1" AcceptedAnswerId="6204" CreationDate="2011-03-02T00:16:28.897" Score="1" ViewCount="175" Body="&lt;p&gt;Photons are massless, but if $m = 0$ and $E = mc^2$ then $E = 0c^2 = 0$. This would say that photons have no energy, which is not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, given the formula $E = ℎf$, a photon does have energy as ℎ is a constant not zero and photons with a zero frequency do not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing here, or does the famous formula not apply to photons?&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2011-03-02T00:33:49.153" Title="Does $E = mc^2$ apply to photons?" Tags="&lt;energy&gt;&lt;speed-of-light&gt;&lt;mass&gt;&lt;photons&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6203" PostTypeId="2" ParentId="6202" CreationDate="2011-03-02T00:32:33.863" Score="2" ViewCount="" Body="&lt;p&gt;You dont need quantum mechanics to argue here, you can simply look to the equation with the rest mass, instead the one with the relativistic mass. This is $$E^2= m_0^2 c^4 + p^2 c^2$$&#xA;So, any particle with zero rest mass still has Energy coming from the linear momentum.&lt;/p&gt;&#xA;" OwnerUserId="1335" LastActivityDate="2011-03-02T00:32:33.863" />
  <row Id="6204" PostTypeId="2" ParentId="6202" CreationDate="2011-03-02T00:33:49.153" Score="9" ViewCount="" Body="&lt;p&gt;There are two explanations possible stemming from the fact that the definition of $m$ in the formula is ambiguous. Well, perhaps Einstein had only one of the two meanings in mind in his original paper but I'm afraid I wouldn't know that as I haven't read it. By that as it may one should recall that special relativity mixes space and time and therefore also momentum and energy and the full formula relating all these fundamental quantities has to be&#xA;$$E^2 = m_0^2c^4 + p^2c^2$$&#xA;where $m_0$ is the rest mass of the particle (zero for photons) whose importance lies in the fact that it is invariant w.r.t. Lorentz transformations (rotations and boosts).&#xA;So first explanation is that the famous formula only holds in the object's rest frame ($p=0$). But such a rest frame isn't available for photons so that formula indeed isn't valid for them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other explanation is through the concept of relativistic mass. In that case the $m$ takes on the meaning of apparent mass because the faster the object goes the harder it will be to accelerate it (because of finite speed of light). So formally one can still talk about the photon having a relativistic (or effective) mass $m = E/c^2$. But this concept of mass runs in all kinds of problems so its usage is discouraged.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-03-02T00:33:49.153" CommentCount="1" />
  <row Id="6205" PostTypeId="1" CreationDate="2011-03-02T00:44:32.227" Score="3" ViewCount="98" Body="&lt;p&gt;Can you build an afocal zoom system with only one stage.  It looks like you would have to use more than three lenses.  You would probably have to have all but one of the lenses be fixed; but it might be possible to have more than one moving, as long as they are all moving the same distances (i.e. they are on the same stage).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not; can you prove that you can't do it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some helpful links:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Zoom_lens&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Zoom_lens&lt;/a&gt;&#xA;&lt;a href=&quot;http://spiedigitallibrary.org/oe/resource/1/opegar/v36/i4/p1249_s1&quot; rel=&quot;nofollow&quot;&gt;http://spiedigitallibrary.org/oe/resource/1/opegar/v36/i4/p1249_s1&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2011-07-21T16:09:23.353" Title="Building an Afocal Zoom System with only one stage" Tags="&lt;optics&gt;" AnswerCount="0" CommentCount="1" />
  <row Id="6206" PostTypeId="2" ParentId="6140" CreationDate="2011-03-02T00:44:42.597" Score="2" ViewCount="" Body="&lt;p&gt;A rotating frame has some similarities with special relativity, but since it is an accelerated frame it is not SR.  To describe rotating coordinates consider the transformation between frame $F$ and $F^\prime$&#xA;$$&#xD;&#xA;x^{\mu\prime}~=~{\Lambda^{\mu}}_\nu x^\nu,&#xD;&#xA;$$&#xA;In cylindrical coordinates the radial and azimuthal coordinates are $r^\prime~=~r$ and $z^\prime~=~z$ and&#xA;$$&#xD;&#xA;t^\prime~=~\gamma(t~-~\Omega\chi^t),~\theta^\prime~=~\gamma(\theta~-~\Omega\chi^\theta),&#xD;&#xA;$$&#xA;where $\gamma$ is the Lorentz factor.  The angular velocity $\Omega$ gives the tangential velocity $v~=~\Omega r$ as measured by a nonrotating observer.  The rotation in these coordinates is&#xA;$$&#xD;&#xA;\eta~=~{{d\theta}\over{dt}},~\eta^\prime~=~{{d\theta^\prime}\over{dt^\prime}},&#xD;&#xA;$$&#xA;where in the rotating frame is given by&#xA;$$&#xD;&#xA;\eta^\prime~=~\frac{d\theta~-~\Omega\Big(\frac{\partial\chi^\theta}{\partial t}dt~+~\frac{\partial\chi^\theta}{\partial\theta}d\theta\Big)}&#xD;&#xA;{dt~+~\Omega\Big(\frac{\partial\chi^t}{\partial t}dt~-~\frac{\partial\chi^t}{\partial\theta}d\theta\Big)}.&#xD;&#xA;$$&#xA;Define the velocities $v~=~r\eta$ and $v^\prime~=~r\eta^\prime$,&#xA;for $\chi^t~=~\theta$ and $\chi^\theta~=~t$.  This gives a formula&#xA;for the addition of rotation&#xA;$$&#xD;&#xA;\eta^\prime~=~{{\eta~-~\Omega}\over{1~-~r^2\Omega\eta/c^2}},&#xD;&#xA;$$&#xA;which is analogous to the linear equation for velocity additions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The metric for a rotating frame is then&#xA;$$&#xD;&#xA;ds^2~=~dt^2~-~r^2\Big(d\theta~+~{\Omega\over c} dt\Big)^2~-~dr^2&#xD;&#xA;$$&#xA;or&#xA;$$&#xD;&#xA;ds^2~=~\Big(1~-~\frac{r^2\Omega^2}{c^2}\Big)dt^2~-~r^2d\theta^2~-~2r^2\frac{\Omega}{c} d\theta dt~-~dr^2.&#xD;&#xA;$$&#xA;The horizon for any observer on the rotating frame exists at $r~=~c/\Omega$. A solution to these coordinates with $dr~=~0$ according to the proper time $s$ is found to be&#xA;$$&#xD;&#xA;t~=~ g^{-1}\sinh(gs),~\theta~=~{{g^{-1}}\over r}\cosh(gs)~-~g^{-1}{\Omega\over c} \sinh(gs),&#xD;&#xA;$$&#xA;where $g~=~{{\Omega^2r}\over\sqrt{1~-~\Omega^2r^2/c^2}}$ is the acceleration parameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rotating frame has a sort of event horizon, a Rindler wedge horizon $r~=~c/\Omega$.  However, this is on the frame, so if you send photons along a radially directed fiber optic they get sort of trapped there a bit like a black hole.  If you send them along a free path then there is no horizon.  This is also related to the Sagnac effect, where sending photon pulses along a circular fiber optic in both directions results in one group returning before the other, or in an actual measurement a phase shift. &lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1136" LastEditorDisplayName="" LastEditDate="2011-05-02T05:51:13.163" LastActivityDate="2011-05-02T05:51:13.163" />
  <row Id="6207" PostTypeId="1" CreationDate="2011-03-02T02:33:11.967" Score="3" ViewCount="139" Body="&lt;p&gt;If I am driving a car in the rain, and want to increase the available traction, should I:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Increase pressure in the tires&lt;/li&gt;&#xA;&lt;li&gt;Decrease pressure&lt;/li&gt;&#xA;&lt;li&gt;Leave the pressure set to optimum dry pressure&lt;/li&gt;&#xA;&lt;li&gt;Increase or decrease just the front or back tires&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Does the answer change if the surface has no standing water (i.e. is merely wet)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course, why is your answer correct?&lt;/p&gt;&#xA;" OwnerUserId="2215" LastActivityDate="2011-03-02T07:36:49.827" Title="Best inflation pressure of car tire in the rain- higher, same, lower than normal?" Tags="&lt;friction&gt;" AnswerCount="1" />
  <row Id="6208" PostTypeId="1" CreationDate="2011-03-02T03:36:38.347" Score="7" ViewCount="391" Body="&lt;p&gt;I took an introductory chemistry course long ago, but the rules seemed arbitrary, and I've forgotten most of what I learned.  Now that I have an undergraduate education in physics, I should be able to use physics to learn general chemistry more effectively.  What resources, either books or on-line, are useful for physicists to learn the fundamentals of chemistry?  I'm not enrolled at a university, so official courses and labs aren't realistic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that I am not looking for books on specialized advanced topics, but a general introduction to chemistry that takes advantage of thermodynamics, statistical mechanics, and quantum mechanics while requiring little or no prior knowledge of chemistry.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-10T12:13:43.600" LastActivityDate="2011-07-10T12:13:43.600" CommunityOwnedDate="2011-03-04T18:19:18.983" Title="Where should a physicist go to learn chemistry?" Tags="&lt;soft-question&gt;&lt;books&gt;&lt;reference-request&gt;&lt;physical-chemistry&gt;&lt;big-list&gt;" AnswerCount="4" CommentCount="9" FavoriteCount="2" />
  <row Id="6209" PostTypeId="1" AcceptedAnswerId="6628" CreationDate="2011-03-02T03:54:14.357" Score="6" ViewCount="440" Body="&lt;p&gt;Suppose we have a sphere of radius $r$ and mass m and a negatively charged&#xA;test particle at distance d from its center, $d\gg r$.  If the sphere is electrically neutral, the particle will fall toward the sphere because of gravity.  As we deposit electrons on the surface of the sphere, the Coulomb force will overcome gravity and the test particle will start to accelerate away.  Now suppose we keep adding even &lt;em&gt;more&lt;/em&gt; electrons to the sphere.  If we have n electrons, the distribution of their pairwise distances has a mean proportional to $r$, and there are $n(n-1)/2$ such pairs, so the binding energy is about $n^2/r$.  If this term is included in the total mass-energy of the sphere, the gravitational force on the test particle would seem to be increasing quadratically with $n$, and therefore eventually overcomes the linearly-increasing Coulomb force.  The particle slows down, turns around, and starts falling again.  This seems absurd; what is wrong with this analysis?&lt;/p&gt;&#xA;" OwnerUserId="2072" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-03-08T22:41:47.313" LastActivityDate="2011-03-13T18:21:55.220" Title="Paradoxical interaction between a massive charged sphere and a point charge" Tags="&lt;gravity&gt;&lt;special-relativity&gt;&lt;electrostatics&gt;" AnswerCount="3" CommentCount="13" />
  <row Id="6210" PostTypeId="5" CreationDate="2011-03-02T04:34:16.880" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-03-02T04:34:16.880" LastActivityDate="2011-03-02T04:34:16.880" />
  <row Id="6211" PostTypeId="4" CreationDate="2011-03-02T04:34:16.880" Score="0" ViewCount="" Body="Graphene is a quasi-2D material formed by carbon atoms in a hexagonal lattice. Graphene-based materials are of great interest for Nanoscience and Nanotechnology, mainly for Nanoelectronics." OwnerUserId="-1" LastEditorUserId="1545" LastEditorDisplayName="" LastEditDate="2011-03-02T04:42:41.883" LastActivityDate="2011-03-02T04:42:41.883" />
  <row Id="6212" PostTypeId="5" CreationDate="2011-03-02T04:47:14.187" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-03-02T04:47:14.187" LastActivityDate="2011-03-02T04:47:14.187" />
  <row Id="6213" PostTypeId="4" CreationDate="2011-03-02T04:47:14.187" Score="0" ViewCount="" Body="Topological insulator are materials formed by a insulator bulk and metallic surfaces with topological origin. These materials may be important for developments in Quantum Computing and Spintronics." OwnerUserId="-1" LastEditorUserId="1545" LastEditorDisplayName="" LastEditDate="2011-03-02T05:18:29.527" LastActivityDate="2011-03-02T05:18:29.527" />
  <row Id="6214" PostTypeId="2" ParentId="3349" CreationDate="2011-03-02T06:01:41.053" Score="-3" ViewCount="" Body="&lt;p&gt;I think there is no observation from which we can infer the existence of an &lt;em&gt;actual&lt;/em&gt; horizon. That's the nature of the beast, since the black hole emits nothing at all by definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be kept in mind that it's almost universal for texts to skip over the fact that their detection of the horizon depends on the validity of general relativity. That is, before the conclusion that a black hole (or its horizon) exists is reached, there's a step where the mass determined from &lt;em&gt;some other observation&lt;/em&gt; (something outside the suspected black hole, like the orbit of some gas cloud--anything that's observed is necessarily outside of the horizon) is input into, say, a formula derived from the Schwarzschild metric, and the output tells whether sufficient mass is within a sufficiently small volume, such that the determination that a black hole (or its horizon) is made. The dependence on the metric is a dependence on the validity of general relativity. A different metric for a different theory of gravity could give a different result, such as to the existence of a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Chandra X-ray Observatory FAQ notes that their discoveries of supposedly actual black holes depend on the validity of general relativity. So the black holes they found might not exist after all.&lt;/p&gt;&#xA;" OwnerUserId="2314" LastActivityDate="2011-03-02T06:01:41.053" CommentCount="3" />
  <row Id="6215" PostTypeId="2" ParentId="6196" CreationDate="2011-03-02T06:06:13.617" Score="3" ViewCount="" Body="&lt;p&gt;All these rules are approximate, and all of them follow from Schrödinger's equation that offers a much more accurate, quantitative result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonds in molecules may arise not just from &quot;valence bonds&quot;, which remain localized in atoms, but also from &quot;molecular orbitals&quot; which are localized across the whole molecule. The latter kind of bond is more general and the methods to study it are newer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to the valence bonds. Energetically, the bonds occur because they allow the pair of atoms to reduce its energy. Just like for individual atoms, pairs of atoms are more stable when a whole orbital is filled because the other states that could be filled are separated by a gap and they have a higher energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So whenever it's possible to divide the electrons among the pair of atoms so that both atoms have full shells, then all the electrons with low enough energy have been added, to minimize the ratio of energy per electron, and the addition of an extra electron would increase the energy by a much higher amount, which suggests that the previous full-shell solution is at least a local minimum of the energy per electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the interactions between the electrons themselves are complicated, this superficial wording can't really replace a calculation - quantum mechanics for many electrons, which deals with a $3N$-dimensional wave function. In fact, in this picture, even in the Hartree-Fock approximation, many other emergent phenomena occur such as &quot;orbital hybridization&quot; - the emergence of totally new kinds of orbitals that are appropriate to describe the molecule.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-02T06:06:13.617" />
  <row Id="6216" PostTypeId="1" AcceptedAnswerId="6224" CreationDate="2011-03-02T06:07:48.080" Score="1" ViewCount="169" Body="&lt;p&gt;I saw a paper on arXiv that referenced this approach to an ontology of QM:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phys.Lett. A213 (1996) 1, S. Mancini, V. I. Man'ko, P.Tombesi &lt;em&gt;Symplectic tomography as classical approach to quantum systems&lt;/em&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/quant-ph/9603002&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/quant-ph/9603002&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An introduction to the tomographic picture of quantum mechanics&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/0904.4439&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0904.4439&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-03-02T06:28:00.793" LastActivityDate="2011-03-04T16:06:45.053" Title="What's the worst thing about the tomographic approach to QM?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-interpretations&gt;&lt;ontology&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="6217" PostTypeId="2" ParentId="5600" CreationDate="2011-03-02T07:21:13.320" Score="1" ViewCount="" Body="&lt;p&gt;There are seven fundamental &lt;em&gt;additive&lt;/em&gt; constants of the motion in classical mechanics, these are the energy and the three components of momentum and angular momentum.  In other words, every constant of the motion, $C$, can be written uniquely as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{3cm} C=a_1E+a_2P_x+a_3P_y+a_4P_z+a_5L_x+a_5L_y+a_7L_z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof of this fact can be found in Landau, Vol 1.&lt;/p&gt;&#xA;" OwnerUserId="442" LastActivityDate="2011-03-02T07:21:13.320" />
  <row Id="6218" PostTypeId="2" ParentId="6207" CreationDate="2011-03-02T07:36:49.827" Score="-1" ViewCount="" Body="&lt;p&gt;The tire pressure will not affect the traction in a noticeable way.  The reason the car moves forward is because of friction from the ground, this friction is generated by a downward force (of the spinning tire), which has the effect of pushing the car forward.  This downward force is influenced primarily by the mass of the car, and not the tire pressure; which is why they sell special tires with better traction.  The traction depends not on the pressure, but rather on the material and geometry of the tire surface.&lt;/p&gt;&#xA;" OwnerUserId="442" LastActivityDate="2011-03-02T07:36:49.827" CommentCount="2" />
  <row Id="6219" PostTypeId="1" AcceptedAnswerId="6223" CreationDate="2011-03-02T07:37:04.210" Score="1" ViewCount="91" Body="&lt;p&gt;Does it violate any physical laws to take a portion of the energy out of a system and use it? Specifically I'm referring to heat. (Kinetic energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you have a material which has a lot of energy and has a high temperature, is it possible to take a portion of that energy and use it for something else, and leave the material cooler?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realise it is impossible to cool it to absolute-zero because of the third law of thermodynamics.&lt;/p&gt;&#xA;" OwnerUserId="2348" LastActivityDate="2011-03-02T16:41:30.803" Title="Extracting heat energy from a material" Tags="&lt;thermodynamics&gt;&lt;energy&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="6220" PostTypeId="1" AcceptedAnswerId="7222" CreationDate="2011-03-02T07:41:20.650" Score="5" ViewCount="764" Body="&lt;p&gt;Two towns are at the same elevation and are connected by two roads of the same length.  One road is flat, the other road goes up and down some hills.  Will an automobile always get the best mileage driving between the two towns on the flat road versus the hilly one, if operated by a perfect driver with knowledge of the route?  Is there any hilly road on which a better mileage can be achieved?&lt;/p&gt;&#xA;" OwnerUserId="2072" LastEditorUserId="2072" LastEditorDisplayName="" LastEditDate="2011-03-22T21:44:46.423" LastActivityDate="2011-03-23T12:44:29.463" Title="Can a car get better mileage driving over hills?" Tags="&lt;classical-mechanics&gt;&lt;energy-conservation&gt;" AnswerCount="8" CommentCount="0" FavoriteCount="3" />
  <row Id="6221" PostTypeId="2" ParentId="2670" CreationDate="2011-03-02T07:48:01.673" Score="1" ViewCount="" Body="&lt;p&gt;It's a standard result in the theory of fractals that any set of contraction mappings that don't overlap &quot;too much&quot; will have a unique attractor, and furthermore in principle these attractors have some Hausdorff dimension; I think this is the invariant quantity you are looking for.  See for example &lt;a href=&quot;http://books.google.com/books?id=2Sg3Vug65AsC&amp;amp;printsec=frontcover&amp;amp;dq=princeton+lectures+on+analysis&amp;amp;hl=en&amp;amp;ei=t_ZtTaygLYS6tgeup6WIDw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCgQ6AEwAA#v=onepage&amp;amp;q=princeton%20lectures%20on%20analysis&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Shakarchi and Stein&lt;/a&gt;, Volume 3, Chapter 7, Theorem 2.9.&lt;/p&gt;&#xA;" OwnerUserId="442" LastEditorUserId="442" LastEditorDisplayName="" LastEditDate="2011-03-02T07:53:16.567" LastActivityDate="2011-03-02T07:53:16.567" />
  <row Id="6222" PostTypeId="2" ParentId="6197" CreationDate="2011-03-02T08:12:46.770" Score="8" ViewCount="" Body="&lt;p&gt;The general answer is &quot;it depends.&quot; Light has energy, momentum, and puts a pressure in the direction of motion, and these are all equal in magnitude (in units of c = 1). All of these things contribute to the stress-energy tensor, so by the Einstein field equation, it is unambiguous to say that light produces gravitational effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the relationship between energy, momentum, and pressure in the direction of propagation leads to some effects which might not otherwise be expected. The most famous is that the deflection of light by matter happens at exactly twice the amount predicted by a massive particle, at least in the sense that in linearized GTR, ignoring the pressure term halves the effect (one can also compare it a naive model of a massive particle at the speed of light in Newtonian gravity, and again the GTR result is exactly twice that).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, antiparallel (opposite direction) light beams attract each other by four times the naive (pressureless or Newtonian) expectation, while parallel (same direction) light beams do not attract each other at all. A good paper to start with is: &lt;a href=&quot;http://prola.aps.org/abstract/PR/v37/i5/p602_1&quot; rel=&quot;nofollow&quot;&gt;Tolman R.C., Ehrenfest P., and Podolsky B., &lt;em&gt;Phys. Rev.&lt;/em&gt; &lt;strong&gt;37&lt;/strong&gt; (1931) 602&lt;/a&gt;. Something one might worry about is whether the result is true to higher orders as well, but the light beams would have to be &lt;em&gt;extremely&lt;/em&gt; intense for them to matter. The first order (linearized) effect between light beams is already extremely small.&lt;/p&gt;&#xA;" OwnerUserId="1353" LastActivityDate="2011-03-02T08:12:46.770" CommentCount="2" />
  <row Id="6223" PostTypeId="2" ParentId="6219" CreationDate="2011-03-02T08:20:49.073" Score="3" ViewCount="" Body="&lt;p&gt;Extracting thermal energy is possible, doesn't violate any laws of thermodynamics and is done routinely every day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_engine&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Heat_engine&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1670" LastActivityDate="2011-03-02T08:20:49.073" CommentCount="1" />
  <row Id="6224" PostTypeId="2" ParentId="6216" CreationDate="2011-03-02T08:59:10.240" Score="2" ViewCount="" Body="&lt;p&gt;The worst thing about it is that it is one of a large number of misguided papers that try to deny that the world is not and cannot be classical. The Wigner function may be calculated from a wave function or a density matrix and is approximately related to the probability distribution but not exactly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wigner function is constrained and the precise probabilities  that an observable will have a particular value have to be calculated as the expectation value of the corresponding projection operator (a linear operator on a Hilbert space) - and can never be fully encoded in a distribution function that is a function of several commuting variables simply because the real variables, as quantum mechanics and the experiments verifying it guarantees, do &lt;em&gt;not&lt;/em&gt; commute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several basic facts about quantum mechanics - the probabilistic character and the nonzero commutators of the observables (i.e. the uncertainty principle) are two important examples - and any paper denying those facts is simply wrong and doesn't deserve a further discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To emphasize that they won't be convinced by any evidence, they motivate their paper - in the second sentence of this paper - by a &quot;permanent wish to understand quantum mechanics in terms of classical probabilities.&quot; Sorry, this is not science. This wish has been ruled out and it can never be &quot;unruled out&quot; again. In science, hypotheses are never &quot;permanent&quot;. Hypotheses are only acceptable as long as they haven't been falsified. Falsification ends the debate and it's the case of the hypothesis that the microscopic physics may be described by classical physics.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-02T08:59:10.240" CommentCount="1" />
  <row Id="6226" PostTypeId="1" CreationDate="2011-03-02T09:22:25.680" Score="1" ViewCount="175" Body="&lt;p&gt;It might seem obvious but i can't imagine how is gravitational pull is different from acceleration due to gravity?&lt;/p&gt;&#xA;" OwnerUserId="2273" LastActivityDate="2011-03-02T10:05:41.727" Title="Gravitational pull vs. acceleration due to gravity" Tags="&lt;gravity&gt;&lt;acceleration&gt;" AnswerCount="2" />
  <row Id="6227" PostTypeId="1" CreationDate="2011-03-02T09:27:40.310" Score="4" ViewCount="290" Body="&lt;p&gt;how do the wheels of a train have sufficient grip on a metal track? i mean both of the surfaces are smooth (&amp;amp; not flexible) and it is okay if there is no inclination, but how about on an inclined track?&lt;/p&gt;&#xA;" OwnerUserId="2351" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-02T10:11:03.813" LastActivityDate="2011-03-02T11:14:07.270" Title="Grip of the train wheels" Tags="&lt;friction&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="0" />
  <row Id="6228" PostTypeId="2" ParentId="6226" CreationDate="2011-03-02T09:34:39.897" Score="4" ViewCount="" Body="&lt;p&gt;The term &quot;gravitational pull&quot; is probably meant to convey &lt;em&gt;force&lt;/em&gt;, which is related to but distinct from acceleration. Under a common assumption of constant mass (which may be false in general, e.g., a rocket), F = ma by Newton's second law, so you don't lose much except conceptual precision in very simple situations... but such conceptual difference can become rather important.&lt;/p&gt;&#xA;" OwnerUserId="1353" LastActivityDate="2011-03-02T09:34:39.897" />
  <row Id="6229" PostTypeId="2" ParentId="6226" CreationDate="2011-03-02T10:05:41.727" Score="4" ViewCount="" Body="&lt;p&gt;Gravitational pull is the cause - the force. Acceleration is the consequence of this cause, assuming that the gravitational force is not cancelled by another force. &lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-02T10:05:41.727" />
  <row Id="6230" PostTypeId="2" ParentId="6227" CreationDate="2011-03-02T10:12:55.903" Score="3" ViewCount="" Body="&lt;p&gt;Sliding is prevented by friction and the friction force is equal to the product of the weight - the perpendicular force - and the dimensionless coefficient of static friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coefficient of static friction between steel and steel can be as high as 0.78 so the angle would have to be hugely non-horizontal for the train to slide. And a lot of acceleration may be added, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lowest coefficient of static friction in wet and greasy conditions may be 0.05 which is approximately the angle in radians where one could start to get worried. It is just 3 degrees and if there's lot of oil everywhere on the tracks, the train may get unsafe already for these small angles. However, in reality, the coefficient never drops this low and 15 degrees is usually a safe angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rail_adhesion&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Rail_adhesion&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that the coefficient of static friction is higher than the coefficient of kinetic friction so the hardest thing is to start the sliding. Once the train starts to slide, it is more likely that it will continue to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the text above was about the sliding - the stability in the front-rear direction. The stability in the left-right direction is guaranteed by the shape of the wheels:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/h6brV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-02T11:14:07.270" LastActivityDate="2011-03-02T11:14:07.270" CommentCount="3" />
  <row Id="6231" PostTypeId="1" AcceptedAnswerId="6237" CreationDate="2011-03-02T10:39:07.190" Score="-1" ViewCount="407" Body="&lt;p&gt;The mathematical concept of function is used in physics to represent different physical quantities. For example the air pressure variation with time and space is called an acoustic wave. We use a function to represent a charge distribution (or even electric field strength) in space and time.In gravitation we use it to represent a mass distribution (and momentum distribution) in space and time. In Quantum mechanics we use it to represent a Quantum state and hence the probability distribution of position and momentum of a particle in space and time. Most of the time we do a variety of mathematically well defined operations on these functions to make some conclusions and to answer some questions regarding a physical system.Even the fundamental laws of physics give the mathematical relation between functions representing different physical quantities. Now if i ask a question on the &lt;strong&gt;&lt;em&gt;mathematical nature&lt;/em&gt;&lt;/strong&gt; of a function that represents a physical quantity (say for example a mass distribution in space and time), would it be possible to find an answer ? Does every such question has a physical interpretation ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here i give a sample question ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q. let a function $p(t)$ represent the acoustic pressure variation at a given point in space.Is the function well behaved ? Mathematically speaking &quot;Is $p(t)$ a smooth function ?&quot;&lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-30T18:33:59.453" LastActivityDate="2011-08-30T18:33:59.453" Title="Use of the mathematical concept 'function' in theoretical physics" Tags="&lt;mathematical-physics&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="6233" PostTypeId="2" ParentId="6227" CreationDate="2011-03-02T10:48:27.883" Score="0" ViewCount="" Body="&lt;p&gt;The reason trains stay on track is because the wheels are not cylindrical, but conical.  See Feynman's explanation here:  &lt;a href=&quot;http://www.youtube.com/watch?v=y7h4OtFDnYE&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=y7h4OtFDnYE&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-03-02T10:48:27.883" CommentCount="3" />
  <row Id="6234" PostTypeId="1" CreationDate="2011-03-02T11:04:11.983" Score="12" ViewCount="586" Body="&lt;p&gt;I sometimes hear statements like that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Quantum-mechanically, interference pattern occurs due to quantum interference of wavefunction of a photon. Wavefunction of a single photon only interferes with itself. Different photons (for example from different atoms) do not interfere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First of all -- is this correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is correct -- how do we explain basic classical interference, when we don't care about where do the plane waves came from? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I heard that there are experiments with interference of two different lasers -- is this considered as a refutation of the statement? If it is -- how one should formally describe such process of interference of different photons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally -- such statements are usually attributed to Dirac. Did Dirac really say something like that? &lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-03-02T20:44:29.277" LastActivityDate="2011-08-24T14:01:32.650" Title="Does a photon interfere only with itself?" Tags="&lt;quantum-mechanics&gt;&lt;photons&gt;&lt;quantum-optics&gt;&lt;interference&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="4" />
  <row Id="6235" PostTypeId="2" ParentId="6231" CreationDate="2011-03-02T11:49:06.387" Score="0" ViewCount="" Body="&lt;p&gt;This reply is to the concept illustrated by your example. The answer is that not all functions used in physics are mathematically well behaved/smooth over all the variable field.&#xA;For example some have singularities. Some are described by delta or theta functions. There are branch cut discontinuities in the imaginary plane for certain classes of functions used in physics. etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general boundary conditions can lead to discontinuities in functional representations when solving physics problems.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-02T11:49:06.387" />
  <row Id="6236" PostTypeId="2" ParentId="6231" CreationDate="2011-03-02T11:54:34.920" Score="0" ViewCount="" Body="&lt;p&gt;If your question is whether functions in physics don't necessarily have to be smooth, I can point out a few examples:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;surface charges on a conductor are discontinuous functions in 3-space&lt;/li&gt;&#xA;&lt;li&gt;the corresponding electric field undergoes a discontinuity when it crosses a surface charge&lt;/li&gt;&#xA;&lt;li&gt;first-order phase transitions have a discontinuity in the first derivative of some thermodynamic variable (say, in specific heat)&lt;/li&gt;&#xA;&lt;li&gt;second-order phase transitions have a discontinuity in the second derivative of some thermodynamic variable&lt;/li&gt;&#xA;&lt;li&gt;most of the paths which contribute to the Feynman path integral are not differentiable&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Maybe others can add a few more.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-03-02T11:54:34.920" CommentCount="1" />
  <row Id="6237" PostTypeId="2" ParentId="6231" CreationDate="2011-03-02T11:55:14.800" Score="2" ViewCount="" Body="&lt;p&gt;Dear Rajesh, first of all, different questions have different answers. So the answer below is purely for the actual question about the pressure that you stated and may be wrong for many other, &quot;similar&quot; questions that deal with other functions. You're very naive if you think that all questions about functions in physics have the same answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressure is a macroscopic concept because gases etc. are made out of atoms. So the number of atoms or molecules in a volume of space is an integer. If the volume is really small - if you really want to define the pressure at an &quot;accurate locus&quot; of space - then the integer-valuedness becomes really important and the pressure is behaving as a discrete, discontinuous variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if one averages the pressure over a large enough region of space so that the number of atoms in this region is much greater than one, then the discontinuities in the number of molecules - and their velocity - are suppressed by the statistically large number of the molecules, and the assumption that the pressure is approximately a continuous function of time becomes an acceptable approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classical field theory is a good model of the reality - at least some portions of it - and it is based on continuous (and differentiable) functions of space and time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discreteness above - that arises from the atomic structure of matter - isn't true for other things. If you ask what is the electric field at a given point, it is classically a totally smooth function and there is no disclaimed at all. However, quantum mechanically, the actual values of the quantities don't exist prior to the measurement. There are lots of subtleties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various layers of physics are more or less accurate and neglect various things. Correspondingly, some objects may be continuous, discontinuous, differentiable, or non-differentiable in different descriptions. And in quantum mechanics, there doesn't really exist any objective function that describes the state of the system at each moment at all. Even though the electron is point-like, at least with the accuracy of $10^{-18}$ meters, one can't say what was its trajectory through space, $x(t),y(t),z(t)$, at the same accuracy.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-02T11:55:14.800" CommentCount="5" />
  <row Id="6238" PostTypeId="2" ParentId="6234" CreationDate="2011-03-02T12:28:31.777" Score="4" ViewCount="" Body="&lt;p&gt;A single photon can only interfere with &quot;itself&quot;. However, &quot;itself&quot; is ill-defined because all photons are identical in quantum mechanics. Because of their Bose-Einstein statistics, the wave function of all photons is symmetric - invariant under all permutations of the individual photons. So the states in which some photons are permuted actually do interfere with each other - the symmetry must be preserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two independent atoms emit photons spontaneously and the process is &quot;random&quot;, so there's no correlation between the phases of the two photons. That's really why they can't interfere with one another. Also, one-photon states can't interfere with two-photon states (that would be like adding apples and oranges - one can't define any meaningful &quot;sum of two functions&quot; if the two functions depend on different variables), and photon states with different (perpendicular) polarizations can't interfere with each other, either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A classical electromagnetic wave is a condensate of a large number of photons - essentially all of them are in the same state. The wave function of all the photons is the tensor product of the wave functions of an individual photon - or a tensor power of the state of each photon is the same. So the probabilistic wave function may suddenly be given a classical interpretation.&#xA;$$\psi_n(\vec x_1, \dots, \vec x_n) = \prod_{i=1}^n \psi_1(\vec x_i)$$&#xA;Each single-photon wave function $\psi_1$ essentially evolves independently, so the total state $\psi_n$ for all the photons keeps its factorized form. The average electric field and magnetic field from the many photons carries the same information as $\psi_1$, a single-photon wave function, and they evolve in the same way, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you may still say that the interference of the classical wave with itself is &quot;due to the interference of each photon with itself&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question is non-quantitative - linguistic rather than mathematical - so it's hard to answer it sharply. However, the right description is in terms of mathematics. What you need to understand is that the wave function of the whole system is a function of a maximum set of commuting variables. If those variables include positions, the interference can only occur in between states whose all other quantum numbers take identical values.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-02T12:28:31.777" CommentCount="3" />
  <row Id="6239" PostTypeId="2" ParentId="6234" CreationDate="2011-03-02T13:25:10.427" Score="0" ViewCount="" Body="&lt;p&gt;There are cases when a photon cannot interfere with itself. Take an interferometer with two nearly equal paths and observe interference. As soon as a photon has a limited length itself, it can interfere with itself only if the path difference is inferior to its own length. Otherwise it cannot. The trick is to have a superposition of waves at &lt;strong&gt;the same time&lt;/strong&gt;. If wave trains of photons arrive without &lt;strong&gt;overlapping&lt;/strong&gt;, there is no interference because there is no &lt;strong&gt;superposition&lt;/strong&gt;. Superposition is a local in time notion. It is often implied rather than pronounced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gradually changing the interferometer path difference leads finally to destructing the interference pattern due to finiteness of the photon wave packet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, two distant atoms (or lasers) are considered as a &lt;strong&gt;single&lt;/strong&gt; source and &quot;their&quot; photons do not interfere only if they do not overlap in time (see above). A single source means that one photon interferes with itself, not two different photons from different atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dirac in his book on QM describes interference of a photon with itself but his statement was not original. I read a H. Poincaré article (1912) where he concludes that if we accept quantum nature of the light, then each quantum (photon) interferes with itself, not with other photons. Poincaré arrived at this conclusion by considering a very low intensity beam (flux of one by one quanta). But I am not sure whether it was pronounced for the first time or he wrote what was a &quot;quantum folklore&quot; of that time.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-03-02T17:08:21.537" LastActivityDate="2011-03-02T17:08:21.537" />
  <row Id="6240" PostTypeId="2" ParentId="6216" CreationDate="2011-03-02T13:49:24.133" Score="5" ViewCount="" Body="&lt;p&gt;The answer by Lubos and this unfortunate sentence in page one of the older paper (as quoted by Lubos) &quot;permanent wish to understand quantum mechanics in terms of classical probabilities&quot; gives the impression that this Tomographic approach is trying to do something known to be impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact the field has become the basis of a wide ranging series of practical techniques as discussed in the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_tomography&quot; rel=&quot;nofollow&quot;&gt; Quantum Tomography&lt;/a&gt;. Amongst its applications are quantum optics and practical quantum computation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I will just mention a few points of clarification here. Clearly any simplistic attempt to &quot;reconstruct&quot; $\Psi$ from knowledge of any (say) position (X) probability distribution cannot work. Even knowing the momentum (P) probability distribution as well is not always enough. So a price has to be paid to reformulate $\Psi$ in this kind of way. In summary one needs to know these probability distributions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(\mu, \nu) = \mu X + \nu P$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see this is a parameterised family with 2 dimensions worth of parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Radon transforms, Wigner functions and the like the theory is able to reconstruct (well not $\Psi$) but the density matrix $\rho$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there are still some issues here.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Q&lt;/strong&gt;. How to actually obtain this data? &lt;strong&gt;A&lt;/strong&gt;. By doing multiple measurements on &lt;em&gt;similarly prepared&lt;/em&gt; quantum states. This is how Tomography has found applications in Optics, where there will be plenty of similar source photons to use.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Q&lt;/strong&gt;. Are $\mu$ and $\nu$ not continuous, leading to error? &lt;strong&gt;A&lt;/strong&gt;. Yes. So some form of statistical sampling logic needs to be added in practice.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Q&lt;/strong&gt;. What about other observables apart from X and P? &lt;strong&gt;A&lt;/strong&gt;. Yes the theory can be generalised to other classes of observables. This is one area where the theory may have current limits though.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Q&lt;/strong&gt;. From a Hilbert space point of view what is going on? &lt;strong&gt;A&lt;/strong&gt;. (I think) that what is happening is that alternative representations (to X, P, etc) are being constructed. One example they derive which is familiar to me is the &lt;em&gt;coherent state&lt;/em&gt; representation. In this representation $\Psi$ becomes a function of Z=(X,P) as in Z = X + iP. Every coherent state is defined by a single (complex - hence 2D) point in this representation and the vacuum is 0 = origin. It is well known that the coherent states form an overcomplete basis for $\Psi$ and this is represented in their framework.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;5.&lt;strong&gt;Q&lt;/strong&gt;. Is Tomography really an Interpretation of QM? &lt;strong&gt;A&lt;/strong&gt;. I dont think so, just a set of mathematical tools to be used in practical and maybe theoretical analysis of QM systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the &quot;worst thing&quot;? - lack of a good coherent presentation of the entire area and its research philosophy (at least if those cited papers are anything to go by).&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-03-04T16:06:45.053" LastActivityDate="2011-03-04T16:06:45.053" />
  <row Id="6241" PostTypeId="2" ParentId="6234" CreationDate="2011-03-02T13:52:41.723" Score="3" ViewCount="" Body="&lt;p&gt;No. Even though such claim is in Dirac's classical work it is not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See e.g. &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Hong_Ou_Mandel&quot; rel=&quot;nofollow&quot;&gt;Hong-Ou-Mandel interference&lt;/a&gt;, when exactly two photons interfere (they can be even from different sources). For quotation of Paul Dirac, and some more analysis, see:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Andrzej Dragan, Paweł Ziń, Interference of Fock states in a single measurement, &lt;a href=&quot;http://arxiv.org/pdf/quant-ph/0703273&quot; rel=&quot;nofollow&quot;&gt;arXiv:quant-ph/0703273v1&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, there is some truth in Dirac's claim - photon interferes only with it itself if you have only one detector to measure.&lt;/p&gt;&#xA;" OwnerUserId="184" LastEditorUserId="184" LastEditorDisplayName="" LastEditDate="2011-03-02T14:12:54.647" LastActivityDate="2011-03-02T14:12:54.647" CommentCount="3" />
  <row Id="6242" PostTypeId="2" ParentId="6157" CreationDate="2011-03-02T14:08:29.997" Score="13" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;books&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://physicsmadeeasy.wordpress.com/&quot; rel=&quot;nofollow&quot;&gt;Physics Made Easy&lt;/a&gt;&lt;/strong&gt;  Karura notes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://laserstorm.harvard.edu/files/jsbach/neutron/publications/Other/quantumech.pdf&quot; rel=&quot;nofollow&quot;&gt;Quantum Mechanics For Advanced Beginners. R. Golub and R. Ince&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.physics-quest.org/&quot; rel=&quot;nofollow&quot;&gt;by Hans de Vries: 'Physics Quest' Understanding Relativistic Quantum Field Theory&lt;/a&gt;&lt;/strong&gt;&lt;br&gt;&#xA;- I love this  'book in progress' to understand &lt;strong&gt;Special Relativity&lt;/strong&gt;, and beyond. To see how a real Lorentz contraction do happen (ch. 4) and how magnetic field is &lt;strong&gt;induced&lt;/strong&gt;  by electrostactic field and Non-simultaneity (it is like a Coriollis effect)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.lightandmatter.com/genrel/&quot; rel=&quot;nofollow&quot;&gt;by Benjamin Crowell: 'Light and Matter' - General Relativity&lt;/a&gt;&lt;/strong&gt;&lt;br&gt;&#xA;  explore other physics topics here &lt;a href=&quot;http://www.lightandmatter.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.lightandmatter.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.plasma.uu.se/CED/Book/&quot; rel=&quot;nofollow&quot;&gt;by Bo Thidé: Electromagnetic Field Theory&lt;/a&gt;&lt;/strong&gt;  - advanced Electrodynamics textbook&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://web.mit.edu/6.013_book/www/&quot; rel=&quot;nofollow&quot;&gt;Elecromagnetic Fields and Energy&lt;/a&gt;&lt;/strong&gt; MIT Hypermedia Teaching Facility, by Herman A. Haus and James R. Melcher (with media)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html&quot; rel=&quot;nofollow&quot;&gt;HyperPhysics&lt;/a&gt;&lt;/strong&gt; - &lt;em&gt;everything&lt;/em&gt;, in short.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://physics.info/&quot; rel=&quot;nofollow&quot;&gt;the physics hypertextbook&lt;/a&gt;&lt;/strong&gt; detailed online book, very interesting Work in Progress.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.bartleby.com/173/&quot; rel=&quot;nofollow&quot;&gt;Relativity - The Special and General Theory by Albert Einstein (1920)&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://student.fizika.org/~jsisko/Knjige/Opca%20Fizika/Feynman%20Lectures%20on%20Physics/&quot; rel=&quot;nofollow&quot;&gt;Feynman Lectures (pdfs)&lt;/a&gt; (&lt;a href=&quot;http://student.fizika.org/~jsisko/Knjige/Opca%20Fizika/Feynman%20Lectures%20on%20Physics/Vol%203%20Ch%2022%20-%20Final%20Index.pdf&quot; rel=&quot;nofollow&quot;&gt;final index&lt;/a&gt;)&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Physics&quot; rel=&quot;nofollow&quot;&gt;Wikipedia Physics&lt;/a&gt;&lt;/strong&gt; a portal to start digging. A colaborative gigantic work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikibooks.org/wiki/Special_Relativity&quot; rel=&quot;nofollow&quot;&gt;WikiBooks -SR&lt;/a&gt;&lt;/strong&gt; a textbook on Relativity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikisource.org/wiki/Portal%3aRelativity&quot; rel=&quot;nofollow&quot;&gt;WikiSource - Relativity Portal&lt;/a&gt;&lt;/strong&gt; find here &quot;The Measure of Time&quot; by Henri Poincaré and many other original sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://plato.stanford.edu/contents.html&quot; rel=&quot;nofollow&quot;&gt;Stanford Encyclopedia of Phylosophy&lt;/a&gt;&lt;/strong&gt; a plethora of info related to physics (for ex. &lt;a href=&quot;http://plato.stanford.edu/entries/spacetime-singularities/&quot; rel=&quot;nofollow&quot;&gt;singularities&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://library.wur.nl/WebQuery/edepot/16365&quot; rel=&quot;nofollow&quot;&gt;EXPLORING THE BIOFLUIDDYNAMICS OF SWIMMING AND FLIGHT&lt;/a&gt;&lt;/strong&gt; David Lentink&lt;br&gt;&#xA;&lt;strong&gt;&lt;a href=&quot;http://www.people.fas.harvard.edu/~hgeorgi/onenew.pdf&quot; rel=&quot;nofollow&quot;&gt;The Physics of Waves&lt;/a&gt;&lt;/strong&gt; by HOWARD GEORGI of Harvard&lt;br&gt;&#xA;&lt;strong&gt;&lt;a href=&quot;http://surflibrary.org/wavephysics.pdf&quot; rel=&quot;nofollow&quot;&gt;The Physics of Ocean Waves (for physicists and surfers)&lt;/a&gt;&lt;/strong&gt;, by Michael Twardos at UCI&lt;br&gt;&#xA;&lt;strong&gt;&lt;a href=&quot;http://www.seas.upenn.edu/~tcom503/book/Photonics.pdf&quot; rel=&quot;nofollow&quot;&gt;Photonics - The Basics and Applications&lt;/a&gt;&lt;/strong&gt; 92 pages , University of Pennsylvania&lt;br&gt;&#xA;&lt;a href=&quot;http://ab-initio.mit.edu/book/photonic-crystals-book.pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Photonic Crystals: Molding the Flow of Light&lt;/strong&gt;&lt;/a&gt; 305 pages, Joannopoulos et al, Princeton Univ Press&lt;br&gt;&#xA;&lt;strong&gt;&lt;a href=&quot;http://www.cs.tau.ac.il/~rshamir/algmb/algmb-archive.htm&quot; rel=&quot;nofollow&quot;&gt;Computational Genomics&lt;/a&gt;&lt;/strong&gt;  Algorithms in Molecular Biology, Lecture notes by Ron Shamir  (pdfs)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;journals open access and online collections&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/8942/what-are-the-best-open-access-journals-in-physics-optics&quot;&gt;PSE-list-open-access-journals&lt;/a&gt;&lt;/strong&gt;    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.doaj.org/&quot; rel=&quot;nofollow&quot;&gt;Directory of Open Access Journals&lt;/a&gt;&lt;/strong&gt; free, full text, quality controlled scientific and scholarly journals (6286, been 2735 searchable at article level)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.mathpages.com/&quot; rel=&quot;nofollow&quot;&gt;MathPages&lt;/a&gt;&lt;/strong&gt;  - lectures on various subjects in &lt;strong&gt;physics&lt;/strong&gt; and mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://relativity.livingreviews.org/Articles/subject.html&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;livingreviews journal&lt;/strong&gt;&lt;/a&gt;  articles by invitaton on relativity and beyond  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://blog.relativity.livingreviews.org/&quot; rel=&quot;nofollow&quot;&gt;livingreviews blog&lt;/a&gt;&lt;/strong&gt;       about the journal articles&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.calphysics.org/sci_articles.html&quot; rel=&quot;nofollow&quot;&gt;Calphysics&lt;/a&gt;&lt;/strong&gt; research on the &lt;strong&gt;electromagnetic quantum vacuum&lt;/strong&gt;  (with care, controversial material)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://ocw.mit.edu/index.htm&quot; rel=&quot;nofollow&quot;&gt;MIT - OpenCourseWare&lt;/a&gt;&lt;/strong&gt;  Several courses available&lt;br&gt;&#xA;&lt;a href=&quot;http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-974-fundamentals-of-photonics-quantum-electronics-spring-2006/lecture-notes/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;MIT OCW fundamentals-of-photonics-quantum-electronics&lt;/strong&gt;&lt;/a&gt;  download pdfs &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;sources to use with precaution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;preprints&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://arxiv.org/&quot; rel=&quot;nofollow&quot;&gt;ARXIV&lt;/a&gt;&lt;/strong&gt;    &lt;strong&gt;door to papers that I cannot afford (sometimes good ideas) -- Cornell univ controlled&lt;/strong&gt;&lt;br&gt;&#xA;I follow this archive thru this MIT's blog &lt;a href=&quot;http://www.technologyreview.com/blog/arxiv/&quot; rel=&quot;nofollow&quot;&gt;The Physics arXiv Blog&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://vixra.org/&quot; rel=&quot;nofollow&quot;&gt;VIXRA&lt;/a&gt;&lt;/strong&gt;    &lt;strong&gt;free to post the ugly, the bad, the crazy, and sometimes good ideas&lt;/strong&gt;&lt;br&gt;&#xA;Independent researchers can publish here. The arXiv is usually closed to authors without academic affiliation.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Portals&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.archive.org/details/texts&quot; rel=&quot;nofollow&quot;&gt;Archive.org&lt;/a&gt;&lt;/strong&gt; Access to a &lt;em&gt;world&lt;/em&gt; of original digitized books, and much more.&lt;br&gt;&#xA;&lt;strong&gt;&lt;a href=&quot;http://www.adsabs.harvard.edu/&quot; rel=&quot;nofollow&quot;&gt;NASA ADS&lt;/a&gt;&lt;/strong&gt; Absctract Data Service &lt;a href=&quot;http://www.adsabs.harvard.edu/abstract_service.html&quot; rel=&quot;nofollow&quot;&gt;search&lt;/a&gt;&lt;br&gt;&#xA;&lt;strong&gt;&lt;a href=&quot;http://www.scribd.com/&quot; rel=&quot;nofollow&quot;&gt;Scribd&lt;/a&gt;&lt;/strong&gt; - a generic social publishing site where I find books (scientific/technical) with full or partial access.&lt;br&gt;&#xA;&lt;strong&gt;&lt;a href=&quot;http://scholar.google.com&quot; rel=&quot;nofollow&quot;&gt;scholar.google.com&lt;/a&gt;&lt;/strong&gt;  from the giant that is changing the observable universe of Human beings&lt;br&gt;&#xA;&lt;strong&gt;&lt;a href=&quot;http://www.cosmosportal.org/&quot; rel=&quot;nofollow&quot;&gt;Cosmos Portal&lt;/a&gt;&lt;/strong&gt;  from Digital Library&lt;br&gt;&#xA;&lt;strong&gt;&lt;a href=&quot;http://www.eoearth.org/&quot; rel=&quot;nofollow&quot;&gt;Encyclopedia of Earth&lt;/a&gt;&lt;/strong&gt;  from Digital Library&lt;br&gt;&#xA;&lt;strong&gt;&lt;a href=&quot;http://nanohub.org/home&quot; rel=&quot;nofollow&quot;&gt;NanoHub&lt;/a&gt;&lt;/strong&gt; - A resource for nanoscience and technology    &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;multimedia&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.youtube.com/user/UCBerkeley&quot; rel=&quot;nofollow&quot;&gt;youtube Berkeley Chanel &lt;/a&gt;&lt;/strong&gt; with courses&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.vega.org.uk/video/subseries/8&quot; rel=&quot;nofollow&quot;&gt;Richard Feynman - Science Videos&lt;/a&gt;&lt;/strong&gt; - 4 original videos (recorded at Auckland)  arguably the greatest science lecturer ever.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://theory.tifr.res.in/~minwalla/&quot; rel=&quot;nofollow&quot;&gt;Videos for Shiraz's lectures&lt;/a&gt; on String Theory&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.susskindsblogphysicsforeveryone.blogspot.com&quot; rel=&quot;nofollow&quot;&gt;Leonard Susskind - Modern Theoretical Physics&lt;/a&gt;&lt;/strong&gt; from his &quot;physics for everyone&quot; blog  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://nanohub.org/resources/5346&quot; rel=&quot;nofollow&quot;&gt;Fundamentals of Nanoelectronics&lt;/a&gt;&lt;/strong&gt; NanoHub - Lectures (Purdue Univ ref. ECE 495N)&#xA;  including &lt;strong&gt;&lt;a href=&quot;http://nanohub.org/resources/5522&quot; rel=&quot;nofollow&quot;&gt;Lecture 10: Shrödinger's Equation in 3-D&lt;/a&gt;&lt;/strong&gt; (mp4)   &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Math&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.math.umn.edu/~nykamp/m2374/readings/divcurl/&quot; rel=&quot;nofollow&quot;&gt;Multivariable calculus and vector analysis&lt;/a&gt;&lt;/strong&gt; A set of on-line readings (Interactive click and drag with LiveGraphics3D) ;  &lt;a href=&quot;http://www.math.umn.edu/~nykamp/m2374/readings/topicindex.html&quot; rel=&quot;nofollow&quot;&gt;explore tab Topics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.khanacademy.org/&quot; rel=&quot;nofollow&quot;&gt;KhanAcademy&lt;/a&gt;&lt;/strong&gt; (videos)  mission: to deliver a world-class education to anyone anywhere&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;math and physics online tools:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.codecogs.com/latex/eqneditor.php&quot; rel=&quot;nofollow&quot;&gt;Online Latex Equation Editor&lt;/a&gt;&lt;/strong&gt; (right click the result and apply anywhere)&lt;br&gt;&#xA;&lt;strong&gt;&lt;a href=&quot;http://www.wolframalpha.com&quot; rel=&quot;nofollow&quot;&gt;wolframalpha&lt;/a&gt;&lt;/strong&gt; - computational knowledge engine, do you want to calculate or know about?&lt;br&gt;&#xA; &lt;strong&gt;&lt;a href=&quot;http://sagenb.org/&quot; rel=&quot;nofollow&quot;&gt;sage online&lt;/a&gt;&lt;/strong&gt;  - support a viable open source alternative to Magma, Maple, Mathematica, and MATLAB.&lt;br&gt;&#xA;&lt;strong&gt;&lt;a href=&quot;http://eumat.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;Euler Math Toolbox&lt;/a&gt;&lt;/strong&gt;  free software for numerical and for symbolic computations&lt;br&gt;&#xA;&lt;strong&gt;&lt;a href=&quot;http://www.geogebra.org/cms/&quot; rel=&quot;nofollow&quot;&gt;geogebra&lt;/a&gt;&lt;/strong&gt;     - Free mathematics software for learning and teaching   &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;modeling and simulation&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.openmodelica.org/&quot; rel=&quot;nofollow&quot;&gt;OpenModelica&lt;/a&gt;&lt;/strong&gt; Physical modeling and simulation environment&lt;br&gt;&#xA;&lt;strong&gt;&lt;a href=&quot;http://www.csc.fi/english/pages/elmer&quot; rel=&quot;nofollow&quot;&gt;Elmer&lt;/a&gt;&lt;/strong&gt; Open Source Finite Element Software for Multiphysical Problems  (&lt;a href=&quot;http://www.csc.fi/english/pages/elmer/examples&quot; rel=&quot;nofollow&quot;&gt;examples&lt;/a&gt;)&lt;br&gt;&#xA;&lt;strong&gt;&lt;a href=&quot;http://www.cs.gmu.edu/~eclab/projects/mason/#Screenshots&quot; rel=&quot;nofollow&quot;&gt;Mason&lt;/a&gt;&lt;/strong&gt;  Multiagent based simulation   (IA)&lt;br&gt;&#xA;&lt;strong&gt;&lt;a href=&quot;http://cs.gmu.edu/~eclab/projects/ecj/&quot; rel=&quot;nofollow&quot;&gt;ECJ&lt;/a&gt;&lt;/strong&gt;    Evolutionary Computation      (IA)&lt;br&gt;&#xA;&lt;strong&gt;&lt;a href=&quot;http://www.spiderland.org/&quot; rel=&quot;nofollow&quot;&gt;Breve&lt;/a&gt;&lt;/strong&gt;  A 3d Simulation Environment for Multi-Agent Simulations and Artificial Life &#xA;(IA)&lt;br&gt;&#xA;&lt;a href=&quot;http://nanohub.org/resources/3847&quot; rel=&quot;nofollow&quot;&gt;NanoHub-Periodic Potential Lab&lt;/a&gt; solves the time independent Schroedinger Equation in a 1-D spatial potential variation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://demonstrations.wolfram.com&quot; rel=&quot;nofollow&quot;&gt;demonstrations.wolfram&lt;/a&gt;&lt;/strong&gt; 7050 applets &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Astronomy and astrophysics&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;books and reviews&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/1107.1789&quot; rel=&quot;nofollow&quot;&gt;Cosmology today-A brief review&lt;/a&gt; (arxiv 2011)&#xA;This is a brief review of the standard model of cosmology. We first introduce the FRW models and their flat solutions for energy fluids playing an important role in the dynamics at different epochs. We then introduce different cosmological lengths and some of their applications. The later part is dedicated to the physical processes and concepts necessary to understand the early and very early Universe and observations of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/0809.0631&quot; rel=&quot;nofollow&quot;&gt;review of Big Bang Nucleosynthesis (BBN)&lt;/a&gt;  (arxiv 2008)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;portals&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.astro-canada.ca/_en/plan-site.html&quot; rel=&quot;nofollow&quot;&gt;astro-canada&lt;/a&gt; Introduction to astronomy, light, instruments, etc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;data&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://simbak.cfa.harvard.edu/simbad/sim-fid&quot; rel=&quot;nofollow&quot;&gt;simbad&lt;/a&gt; search data on celestial bodies with the proper tools.&lt;br&gt;&#xA;&lt;a href=&quot;http://pds.nasa.gov/&quot; rel=&quot;nofollow&quot;&gt;NASA PDS: The Planetary Data System&lt;/a&gt; data related to Nasa missions  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;sky viewers&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://skyview.gsfc.nasa.gov/&quot; rel=&quot;nofollow&quot;&gt;Skyview , Nasa &lt;/a&gt; SkyView is a Virtual Observatory on the Net&lt;br&gt;&#xA;&lt;a href=&quot;http://www.worldwidetelescope.org/Home.aspx&quot; rel=&quot;nofollow&quot;&gt;WWT World Wide Telescope, Microsoft&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;simulation and presentations&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.shatters.net/celestia/&quot; rel=&quot;nofollow&quot;&gt;Celestia&lt;/a&gt;  free space simulation that lets you explore our universe in three dimensions.&lt;br&gt;&#xA;&lt;a href=&quot;http://astrolab-parc-national-mont-megantic.org/fr/astronomie.presentations_astronomiques.htm&quot; rel=&quot;nofollow&quot;&gt;astrolab&lt;/a&gt;  presentations_astronomiques (FR)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;other resources&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.physics.princeton.edu/~mcdonald/&quot; rel=&quot;nofollow&quot;&gt;Kirk McDonald page at Princeton.edu&lt;/a&gt;&lt;/strong&gt; a handful of resources on EM,QED,QM (+-5Gb ;-) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.latexsearch.com/&quot; rel=&quot;nofollow&quot;&gt;Springerlink's LaTeXsearch&lt;/a&gt;&lt;/strong&gt; you can search articles by using latex formulas input&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-08-11T17:04:27.813" LastActivityDate="2011-08-11T17:04:27.813" CommunityOwnedDate="2011-03-02T14:08:29.997" CommentCount="8" />
  <row Id="6243" PostTypeId="1" CreationDate="2011-03-02T14:10:41.927" Score="2" ViewCount="208" Body="&lt;p&gt;In Nielsen and Chuang's text on Quantum Information and Computation, the measurement postulate is stated by using a collection of measurement operators and the outcomes are the indices of the measurement operators. See: &lt;a href=&quot;http://books.google.com/books?id=65FqEKQOfP8C&amp;amp;lpg=PA87&amp;amp;ots=Pq9S_kl6GO&amp;amp;dq=measurement%20postulate&amp;amp;pg=PA84#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=65FqEKQOfP8C&amp;amp;lpg=PA87&amp;amp;ots=Pq9S_kl6GO&amp;amp;dq=measurement%20postulate&amp;amp;pg=PA84#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt; . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm little confused over the fact that the eigenvalue postulate for projective measurements (that the outcome of a projective measurement is one of the eigenvalues of the observable), as written in other quantum mechanics texts, does not emerge from the way the postulate is stated in Nielsen &amp;amp; Chuang. While the eigenvalues can serve the same purpose as the indices, they contain more information than just an index: don't they also carry scale information? That is, if one eigenvalue is twice another eigenvalue, then the measuring apparatus should read twice the value for one outcome as compared to the other. So, is this is a weaker form of the measurement postulate? Or are they equivalent and I'm missing something?&lt;/p&gt;&#xA;" OwnerUserId="2186" LastActivityDate="2011-03-03T12:17:37.073" Title="Eigenvalue postulate and experiment outcomes" Tags="&lt;quantum-mechanics&gt;&lt;measurement&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="6244" PostTypeId="1" CreationDate="2011-03-02T14:27:30.833" Score="4" ViewCount="84" Body="&lt;p&gt;The standard textbook treatment of the conformal gauge for a free string with BRST ghosts is fine, but when a string splits into two, we have Gribov ambiguities for the conformal gauge. We can impose the conformal gauge for the initial string before the split, and also the final two strings after the split, but not for the interpolating region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we deal with this?&lt;/p&gt;&#xA;" LastActivityDate="2011-03-02T16:45:13.273" Title="Gribov ambiguities for splitting strings in the BRST conformal gauge" Tags="&lt;string-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6245" PostTypeId="2" ParentId="6216" CreationDate="2011-03-02T14:31:25.323" Score="2" ViewCount="" Body="&lt;p&gt;The second paper is available as &lt;a href=&quot;http://iopscience.iop.org/1402-4896/79/6/065013/&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Phys. Scr.&lt;/em&gt; &lt;strong&gt;79&lt;/strong&gt; (2009) 065013 (29pp)&lt;/a&gt;. An editor and a referee or two think there are not &lt;em&gt;too&lt;/em&gt; many things wrong with it, or that it has enough good things about it. One assumes Luboš was not a referee. The whole community doesn't vote on whether a paper is accepted into a journal, of course, so a pinch of salt, but as of now there are 26 citations to the 2009 paper in published papers, according to Web of Science (which is not shabby at all). The editor required them to cut down the paper significantly, 64pp&amp;rarr;29pp, so one of the worst;-) things about the arXiv version is mitigated in the published paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I take this kind of approach to QM to be on the edge of engineering and interpretation. There is an engineering problem that is addressed by tomography &amp;mdash; how do we determine what quantum observables we are measuring (and the relationships between them, expressed in terms of the mathematical ansatz we use to model them) and what quantum state we have prepared, from the basic classical raw data points (that are written in lab note books or stored in computer memory) and whatever statistics of that raw data we may construct? Note that the engineering problem begins with classical information and ends with quantum observables and a quantum state, and, as such, is intimately concerned with the interpretation of QM, at least in the old-fashioned vaguely formulated terms of the Copenhagen and Bohr's interpretations. For more modern interpretations, which often layer some kind of metaphysics of the quantum state on top of the raw data, the relevance is not as clear-cut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The worst thing about the second paper may not be something for which it can be hung out to dry, just because no one else has achieved definitude: it does not construct a compelling account. Specifically, it does not &lt;em&gt;compel&lt;/em&gt; Luboš to take it seriously. This is an extraordinarily difficult balance to achieve, however it looks to me that if current progress is maintained someone will find an adequate combination of simplicity and mathematical and philosophical sophistication in the next 10 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having written part of this before Roy Simpson's Answer, I have up-voted his Answer and commend it to other people as a counterpoint to the approach I have taken. I commend the papers cited in the Question as useful for engineering purposes and as a valiant attempt at interpretation.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-03-02T14:31:25.323" CommentCount="3" />
  <row Id="6246" PostTypeId="1" AcceptedAnswerId="6297" CreationDate="2011-03-02T15:52:38.703" Score="0" ViewCount="136" Body="&lt;p&gt;In 2000, Nasa conducted an experiment and found that gravity plays an indirect role in flame formation and composition (see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Flame&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt;).  What role does gravity play?  Why does a flame extend in the opposite direction of gravity?  Assuming the experiment could be re-created, are there any other forces that would cause a flame to extend beyond a spherical shape?&lt;/p&gt;&#xA;" OwnerUserId="1115" LastEditorUserId="1115" LastEditorDisplayName="" LastEditDate="2011-03-02T15:57:54.190" LastActivityDate="2011-03-03T14:13:11.993" Title="What role does gravity have on flame composition?" Tags="&lt;gravity&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="6247" PostTypeId="2" ParentId="6196" CreationDate="2011-03-02T16:24:35.413" Score="2" ViewCount="" Body="&lt;p&gt;I will try to give a very qualitative explanation of chemical bonds like the so called covalent bonds. Let's take the example of hydrogen molecule where two hydrogen atoms share two electrons. Now, if the two nuclei of the two hydrogen atoms are separated by a distance, there is a potential barrier between them. Classically the electrons can not cross that barrier, but quantum mechanics allows them to cross it due to tunneling effect. The electrons therefore wander back and forth continuously through the barrier and each of them spends almost equal times in close proximity of the two nuclei. Now quantum mechanical calculation will show that the potential energy of this system is much lower than that of the system if they were far apart. There is an optimum distance between the nuclei when the potential energy of the system is minimum. Naturally the system at that distance will be most stable. This is how covalent bond forms. In other molecules it may be the case that the two atoms are different. In that case electrons spend a little more time to the proximity of the relatively positive nucleus at the most stable (i.e. least potential energy) configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other kinds of bonds as well. The mathematical treatments for them differ but each follows from the rules of quantum mechanics and in all types of bonds the common thing is, the potential energy of the system has to be minimum for the stable configuration.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-03-02T16:29:47.453" LastActivityDate="2011-03-02T16:29:47.453" />
  <row Id="6248" PostTypeId="1" CreationDate="2011-03-02T16:37:53.423" Score="8" ViewCount="212" Body="&lt;p&gt;I have been struggling with these Majorana fermion identities for quite sometime now. I would be grateful if someone can help me with them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\lambda$,$\theta$ and $\psi$ be $4$-component Majorana fermions. Then apparently the following are true,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$(\bar{\theta}\gamma_5 \theta)(\bar{\psi}_L\theta)(\bar{\theta}\gamma^\mu \partial_\mu \psi_L) = \frac{1}{4}(\bar{\theta}\gamma_5 \theta)^2 (\bar{\psi}_L\gamma^\mu \partial_\mu \psi_L)$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$(\bar{\theta}\gamma_5 \theta)(\partial _ \mu \bar{\psi}_L \gamma^\mu \theta)(\bar{\theta}\psi_L) = -\frac{1}{4}(\bar{\theta}\gamma_5 \theta)^2 (\partial_\mu \bar{\psi}_L\gamma^\mu \psi_L)$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$(\bar{\theta}\gamma _5 \gamma _\mu \theta)(\bar{\psi}_L\theta)(\bar{\theta}\psi_L) = \frac{1}{4} (\bar{\theta}\gamma _5 \theta)^2 (\bar{\psi}\gamma ^\mu \psi)$   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$(\bar{\theta}\gamma_5 \theta)(\bar{\psi}_L\theta)(\bar{\theta}\lambda) = \frac{1}{4} (\bar{\theta}\gamma _5 \theta)^2 (\bar{\psi}_L\lambda)$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I guess looking at the above that all the 4 have some generic pattern and hence probably require some same key idea which I am missing. Its not clear to me as to how to &quot;pull out&quot; the $\theta$s between the other fermions to outside and then again repack then into a $(\bar{\theta}\gamma_5 \theta)$. I will be happy to get some help regarding the above. &lt;/p&gt;&#xA;" OwnerUserId="1637" LastEditorUserId="429" LastEditorDisplayName="" LastEditDate="2011-03-02T22:17:37.497" LastActivityDate="2011-03-04T04:49:39.273" Title="Some Majorana fermion identities" Tags="&lt;quantum-field-theory&gt;&lt;fermions&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="6250" PostTypeId="2" ParentId="6244" CreationDate="2011-03-02T16:39:20.020" Score="1" ViewCount="" Body="&lt;p&gt;When we use the BRST formalism for the conformal symmetry, it's because we want to preserve the conformal symmetry as an efficient tool to do the calculations of the interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interactions of strings, which are composed out of splitting and joining interactions whenever you &quot;slice&quot; any history, are described by a world sheet in spacetime. We usually analytically continue to the Euclidean world sheet embedded in a spacetime that may actually be kept Minkowskian. The only manifestly Lorentz-covariant quantities that may be calculated are the scattering amplitudes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scattering amplitudes are calculated as sums of all histories. Each history is a world sheet which has some topology - a genus $h$ Riemann surface with $b$ boundaries and $c$ crosscaps (it has $h$ handles). There exists a &quot;moduli space&quot; of such Riemann surfaces, one must integrate over it to get the amplitude (it's the residual integral from the integration over the auxiliary world sheet metric that can't be fixed by the conformal gauge), and the prescription for the measure is uniquely determined from the BRST formalism. There are no gauge-invariant local observables in quantum theories of gravity:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/4359/diffm-as-a-gauge-group-and-local-observables-in-theories-with-gravity&quot;&gt;Diff(M) as a gauge group and local observables in theories with gravity&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Standard textbooks derive all these formulae and one never encounters anything such as the &quot;Gribov problem&quot; along the way. In fact, this term will remain unknown to any student who learns the thousands of pages of the introductory string theory textbooks. This is no coincidence. Much more generally, it is just an illusion that there is an inevitable problem of this kind. One may always find a way to totally circumvent it. As has been discussed in this question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/4925/why-do-faddeev-popov-ghosts-decouple-in-brst&quot;&gt;Why do Faddeev-Popov ghosts decouple in BRST?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the BRST formalism is just a formalism - not a question about &quot;new physics&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, all of its physical - i.e. in principle measurable - results could be obtained by other formalisms, too. The Gribov problem is a &quot;non-perturbative ambiguity&quot;. However, whenever we have a theory defined perturbatively, we must classify all possible internally consistent (which includes unitary) non-perturbative completions of the theory (it may be unique and it may be non-unique). The previous sentence defines what it means to study the non-perturbative physics properly and this requirement is true regardless of the formalism one uses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one may face a Gribov problem but it is just a problem with a particular formalism. Physical criteria must be imposed to find out whether non-perturbative physics can be consistent at all, and whether it is unique, and with a sufficient patience and skills, one may always find the answer to the physical question - i.e. to classify all possible ways to non-perturbatively complete the theory. If one doesn't use all the physical arguments and clever tricks, he may get stuck with a Gribov-like problem in his calculations - but it's just his problem, not a problem of physics. The latter always has answers.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-02T16:45:13.273" LastActivityDate="2011-03-02T16:45:13.273" />
  <row Id="6251" PostTypeId="2" ParentId="6219" CreationDate="2011-03-02T16:41:30.803" Score="0" ViewCount="" Body="&lt;p&gt;Yes, this is OK.  BTW heat is not generally thought of as being kinetic energy - at a microscale it is, but because the kinetic energy is randomly distributed is cannot be directly, usefully and 100% harnessed as kinetic energy.  So the concept of &quot;heat&quot; is a useful way of making this distinction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember always though that heat only flows from a hotter body to a colder body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you imagine the heat from my warm hands flowing into a hot cup of tea that I am holding, you realize that this is not what actually happens in real life.  This example does not violate any laws about conservation of energy, but it does violate the Second Law of Thermodynamics, which is why it doesn't happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @Janne808 said, extracting heat energy and turning it into work is done all the time (e.g. automobile engine, boiler and steam turbine, thermoelectric junction etc).  Heat engines are governed not only by the conservation of energy, but also by Carnot's Law or the Second Law of Thermodynamics (which places a theoretical limit on the fraction of heat that can be converted into work; the rest must be rejected as waste heat to a lower temperature region)&lt;/p&gt;&#xA;" OwnerUserId="2297" LastActivityDate="2011-03-02T16:41:30.803" />
  <row Id="6252" PostTypeId="2" ParentId="6243" CreationDate="2011-03-02T16:56:35.853" Score="1" ViewCount="" Body="&lt;p&gt;Dear Siddharth, by measuring the projection operators $M_m$, one determines which eigenvalues the corresponding measured observables took. For example, the measurements distinguished by operators $M_m$ may correspond to the measurement of $x$, $p_y$, and $p_z$ of a particle. (I wanted to take a sufficiently complex example to make it clear that we may measure several observables at the same moment - and they don't have to be a set of the &quot;most standard&quot; observables that people usually talk about.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is a one-to-one map between the projection operators and particular values of the three quantities:&#xA;$$M_m |\psi\rangle = 1 |\psi \rangle \quad \Leftrightarrow \quad x |\psi \rangle = \lambda_{x,m} |\psi \rangle, \quad p_y |\psi \rangle = \lambda_{p_y,m} |\psi \rangle,  \quad p_z |\psi \rangle = \lambda_{p_z,m} |\psi \rangle. $$&#xA;Of course, the quantities $x,p_y,p_z$ only take values that are the allowed eigenvalues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when we make a measurement, we assume that the laws describing the physical system are &lt;em&gt;known&lt;/em&gt;. It also means that the precise map between the operators $M_m$ and the corresponding eigenvalues  $\lambda_m$ of the three operators is known, too. The only thing that is unknown is the actual state of an actual physical system, controlled by these laws, at a given moment. That's what we measure. In this context, we don't try to &quot;measure the laws of physics&quot; or &quot;all the allowed eigenvalues which are in principle possible&quot;. It's being assumed that we had known those general things in advance and that we have also chosen a convention what the operators $M_m$ physically represent - we have chosen the set of all the triplets of values $\lambda_m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must understand that this is a textbook on quantum information so the precise values of the eigenvalues that represent the information - which projection operator $M_m$ had eigenvalue equal to one - is irrelevant. A quantum bit may be represented by a Hydrogen atom whose energy is $E=-13.6$ eV or $E=-13.6/4$ eV. But it may also be represented by the spin of the electron, $j_z=\pm 1/2$. The point of quantum information is that the two two-dimensional Hilbert spaces I just presented (and infinitely many other Hilbert spaces, labeled by different eigenvalues of different observables) are isomorphic. One is not interested in how the information is represented in the &quot;hardware&quot;, just in the values (given by wave functions) of the quantum bits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of information you need to determined whether the Hydrogen atom was in the $n=1$ or $n=2$ state, and whether an electron was in $j_z=+1/2$ or $j_z=-1/2$, is the same: it's one qubit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In classical computers, this would be analogous to the statement that we don't care whether bits are stored in the magnetic field on a hard disk or voltages in RAM, or by optical patterns on CD disks, or whatever else: we're only interested in the sequence of bits. Analogously, quantum information is the application of quantum mechanics where analogous technical questions are left to someone else: quantum information physicists only play with the quantum bits and their transformation, not with the hardware that is needed to realize such transformations in practice. But of course, as you correctly observe, every representation of a quantum bit in practice has to be linked to some Hermitean observable with some allowed eigenvalues - and they're usually not &quot;zero&quot; and &quot;one&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-02T17:09:01.007" LastActivityDate="2011-03-02T17:09:01.007" CommentCount="2" />
  <row Id="6253" PostTypeId="1" CreationDate="2011-03-02T16:59:36.430" Score="1" ViewCount="106" Body="&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/jKBIc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have trouble interpreting this illustration. I see why &lt;strong&gt;r&lt;/strong&gt; (position) and &lt;strong&gt;a&lt;/strong&gt; (acceleration) are the way they are, but what happened to &lt;strong&gt;v&lt;/strong&gt;? Why is it smaller than its coordinates? Is this another error in my textbook?&lt;/p&gt;&#xA;" OwnerUserId="2342" LastEditorUserId="2342" LastEditorDisplayName="" LastEditDate="2011-03-02T17:06:45.143" LastActivityDate="2011-03-02T17:06:45.143" Title="Kinematics textbook illustration" Tags="&lt;kinematics&gt;" CommentCount="6" />
  <row Id="6254" PostTypeId="2" ParentId="6248" CreationDate="2011-03-02T17:19:51.607" Score="5" ViewCount="" Body="&lt;p&gt;They're more complicated cousins of the Fierz identities,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fierz_identity&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fierz_identity&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The article above also recommends you Okun's book for the general recipe to prove similar identities. Note that all the identities you wrote except for the third one are just normal Fierz identities because the first factor may be cancelled as it appears (once) both on left-hand side and right-hand side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that it's not trivial to prove those identities doesn't mean that they're not true. If you rewrote them correctly, they are true. You may trust that they're true. In principle, you may verify them by writing the most general values of the spinors $\theta$ and $\psi$ (and $\lambda$, in the last case) - in terms of four complex components each (reduced to two complex by the Majorana condition) - and by calculating the explicit values of the products of the inner products. The identities above will hold. It's kind of inevitable that some identities of a similar form hold because there are just four components in each variable and the number of monomials of the right degree in those components is limited and may be therefore written in different ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spinor identities may be annoyingly technical, especially if one deals with higher dimensions or extended supersymmetry.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-02T17:25:11.243" LastActivityDate="2011-03-02T17:25:11.243" />
  <row Id="6255" PostTypeId="2" ParentId="6234" CreationDate="2011-03-02T17:56:12.467" Score="5" ViewCount="" Body="&lt;p&gt;The photon model of light may be the most frequently over-applied model in physics. Lamb expresses my opinion fairly clearly &lt;a href=&quot;http://www.springerlink.com/content/h16g2307204h5654/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The photon concepts as used by a high&#xA;  percentage of the laser community have&#xA;  no scientific justification.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In my experience, many physicists who answer simple questions about matter without unnecessary reference to quarks or gluons are incapable of answering simple questions about light without unnecessary reference to photons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is puzzling, because &lt;a href=&quot;http://books.google.com/books?id=5t0tm0FB1CsC&amp;amp;lpg=PP1&amp;amp;ots=a7kTItsOzK&amp;amp;dq=the%20quantum%20challenge&amp;amp;pg=PA24#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;very few experiments are capable of distinguishing between the existence and nonexistence of photons&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are genuinely interested in the photon model of light, then be prepared to do an awful lot of math to predict even fairly simple experimental results. You will, of course, be using a more correct model, but one should use the right tool for the right job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, you're interested in the experimentally observable behavior of light, then Maxwell's equations will give you the right answer in the vast majority of cases. For example, you ask if two different lasers can interfere. They can! See this question:&#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/630/is-it-possible-to-observe-interference-from-2-independent-optical-lasers&quot;&gt;Is it possible to observe interference from 2 independent optical lasers?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure the photon model predicts this result, but I suspect not without a fairly strong grasp of the math. If you'd never heard of photons, and all you knew were Maxwell's equations, this result isn't very surprising.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll close my answer with a question: For what type of experimental prediction is the photon model actually relevant? For what types of predictions is the photon model confusing, misleading, or more effort than it's worth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples so far:&lt;br&gt;&#xA;The photon model is relevant to:&lt;br&gt;&#xA;-&lt;a href=&quot;http://dx.doi.org/10.1103%2FPhysRevLett.59.2044&quot; rel=&quot;nofollow&quot;&gt;Hong-Ou-Mandel&lt;/a&gt;&lt;br&gt;&#xA;-&lt;a href=&quot;http://dx.doi.org/10.1209/0295-5075/1/4/004&quot; rel=&quot;nofollow&quot;&gt;The Grangier Experiment&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The photon model is not relevant to:&lt;br&gt;&#xA;-&lt;a href=&quot;http://ajp.aapt.org/resource/1/ajpias/v68/i3/p245_s1&quot; rel=&quot;nofollow&quot;&gt;'Conventional' interference fringes from two independent lasers&lt;/a&gt;&lt;br&gt;&#xA;-&lt;a href=&quot;http://books.google.com/books?id=5t0tm0FB1CsC&amp;amp;lpg=PP1&amp;amp;ots=a7kTItsOzK&amp;amp;dq=the%20quantum%20challenge&amp;amp;pg=PA25#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The photoelectric effect&lt;/a&gt; (despite many &lt;a href=&quot;http://www.einsteinyear.org/facts/photoelectric_effect/&quot; rel=&quot;nofollow&quot;&gt;claims to the contrary&lt;/a&gt;), or the 'clicking' behavior of CCDs and photomultiplier tubes.&lt;br&gt;&#xA;-Any experiment done before the mid-eighties, including &lt;a href=&quot;http://books.google.com/books?id=5t0tm0FB1CsC&amp;amp;lpg=PP1&amp;amp;ots=a7kTItsOzK&amp;amp;dq=the%20quantum%20challenge&amp;amp;pg=PA30#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;the Hanbury-Brown and Twiss experiment&lt;/a&gt;&lt;br&gt;&#xA;-Any commercial technology  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My last two claims are intentionally bold. Prove me wrong!&lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="2359" LastEditorDisplayName="" LastEditDate="2011-03-04T15:18:13.903" LastActivityDate="2011-03-04T15:18:13.903" CommentCount="7" />
  <row Id="6256" PostTypeId="1" AcceptedAnswerId="6261" CreationDate="2011-03-02T18:55:33.873" Score="3" ViewCount="376" Body="&lt;p&gt;I'll break this down to two related questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a fission bomb, Uranium or Plutonium atoms are split by a high energy neutron, thus releasing energy (and more neutrons). Where does the energy come from? Most books I've ever come across simply state e=mc2 and leave it at that. Is it that matter (say a proton or neutron) is actually converted into energy? That would obviously affect the elements produced by the fission since there would be less matter at the end. It would also indicate exactly how energy could be released per atom given the fixed weight of a subatomic particle. I remember hearing once that the energy released is actually the binding energy that previously held the nucleus together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a fusion bomb, two hydrogen isotopes are pushed together to form helium and release energy - same question: where does this energy come from? Is matter actually converted or are we talking about something else here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry is this is rather basic - I haven't done physics since high school.&lt;/p&gt;&#xA;" OwnerUserId="1910" LastActivityDate="2011-03-03T01:59:16.820" Title="Where does the energy from a nuclear bomb come from?" Tags="&lt;nuclear-physics&gt;" AnswerCount="4" />
  <row Id="6257" PostTypeId="2" ParentId="6256" CreationDate="2011-03-02T19:09:32.603" Score="5" ViewCount="" Body="&lt;p&gt;Yes, matter of mass $m$ is directly converted to energy $E=mc^2$ which literally means that the weight of the remnants of the atomic bomb is smaller than the weight of the atomic bomb at the beginning. Fission reduces the mass by 0.1 percent or so; for fusion, you may get closer to 1 percent of mass difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, you may release 100 percent of the $E=mc^2$ energy stored in the mass $m$ - for example by annihilation of matter with antimatter; or by creating a black hole and waiting until it evaporates into pure Hawking radiation - which is pure energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, the differences in the energies - and the corresponding masses because $E=mc^2$ is true universally (in the rest frame), there is really just one independent quantity - may be attributed to the (changing) interaction energy between the nucleons (or quarks). Alternatively, you may imagine that the extra energy was stored in extra gluons and quark-antiquark pairs inside the nuclei. Those two descriptions are actually equivalent although this fact is not self-evident.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-02T19:09:32.603" CommentCount="7" />
  <row Id="6258" PostTypeId="2" ParentId="6052" CreationDate="2011-03-02T19:23:15.673" Score="1" ViewCount="" Body="&lt;p&gt;Whatever you do, do it for fun.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've a similar age, graduate in electronic eng. and I've make a living on IT. I have a passion for physics and cosmology, born 20 years ago, when a friend (also electronic eng.) make me see interesting things untold in the textbooks. IMO new brilliant viewpoints. Now I have a physical explanation to all darkness: DM, DE, BH, even genuine Ufos, as described by pilots. You may smile, doubt and vote down but give me the benefit of the doubt until further explanation. As an example: If a new particle,  Z0 like, makes it appearence near &lt;strong&gt;780 Gev range&lt;/strong&gt; (LHC - 2011 ?)  I will know that the prevision of Douglas Pinnow (no one is reading his book) was right (the next one circa 1450 Gev). For now SUSY seems in trouble ( &lt;a href=&quot;http://www.nature.com/news/2011/110228/full/471013a.html&quot; rel=&quot;nofollow&quot;&gt;&quot;Beautiful theory collides with smashing particle data&quot;&lt;/a&gt; ) Lets wait and see. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another 60yo electronic engineer, and friend, is pursuing an Astronomy graduation. I think that I've more pleasure in studying by myself, asking and searching, without classroom, exams, schedules.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bioinformatics is awesome and a lot of challenging work is yet to be done. I wish I had the time for study it.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-03-02T19:23:15.673" />
  <row Id="6259" PostTypeId="1" AcceptedAnswerId="6492" CreationDate="2011-03-02T19:32:48.963" Score="4" ViewCount="145" Body="&lt;p&gt;I'm using the SOFTSUSY package to generate the sparticle spectrum at the EW scale. One of the input parameters is the ratio of the up and down-type Higgs vevs commonly known as $\tan\beta$. The $\mu$ parameter is computed as an output by constraining the Z-boson mass to be compatible with experiment. I was wondering if there a quick way to figure out the value of the $B\mu$ parameter from $\mu$ and $\tan\beta$. Even an approximate analytic expression would be nice to have.&lt;/p&gt;&#xA;" OwnerUserId="1892" LastActivityDate="2011-03-07T21:00:23.203" Title="$B\mu$ from $\tan\beta$ and $\mu$" Tags="&lt;particle-physics&gt;&lt;supersymmetry&gt;" AnswerCount="1" />
  <row Id="6260" PostTypeId="2" ParentId="6256" CreationDate="2011-03-02T19:34:41.030" Score="1" ViewCount="" Body="&lt;p&gt;Coulomb repulsion between like charges is the biggest contributor to the amount of energy released.&lt;/p&gt;&#xA;" OwnerUserId="1670" LastActivityDate="2011-03-02T19:34:41.030" CommentCount="3" />
  <row Id="6261" PostTypeId="2" ParentId="6256" CreationDate="2011-03-02T19:36:08.910" Score="4" ViewCount="" Body="&lt;p&gt;Energy of a fission nuclear bomb comes from the gravitational energy of the stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Protons and neutrons can coalesce into different kinds of bound states. We call these states atomic nuclei. The ones with the same number of protons are called isotopes, the ones with different number are nuclei of atoms of different kinds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many possible different stable states (that is, stable nuclei), with different number of nucleons and different binding energies. However there are also some general tendencies for the specific binding energy per one nucleon (proton or neutron) in the nuclei. States of simple nuclei (like hidrogen or helium) have the lowest specific nucleon binding energy amongst all elements, but the higher is the atomic number, the higher the specific energy gets. However, for the very heavy nuclei the specific binding energy starts to drop again.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a graph that sums it up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/File:Binding_energy_curve_-_common_isotopes.svg&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/File:Binding_energy_curve_-_common_isotopes.svg&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It means that when nucleons are in the medium-atomic number nuclei, they have the highest possible binding energy. When they sit in very light elements (hidrogen) or very heavy ones (uranium), they have weaker binding. Thus, one can say that for the low &quot;every-day&quot; temperatures, the very heavy elements (like the very light ones) are quasistable in a sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fission bomb effectively &quot;lets&quot; the very heavy atomic nuclei (plutonium, or uranium) to resettle to the atoms with lower number of nucleons, that is, with higher bound energies. The released binding energy difference makes the notorious effect. In terms of the graph cited above, it corresponds to nucleons moving from the right end closer to the peak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet this is not the only way to let nucleons switch to the higher binding energy state than the initial one. We can &quot;resettle&quot; very light elements (like hydrogen) and let nucleons move to the peak from the left. That would be fusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heavy nucleons emerge in the stars. Here the gravitational energy is high enough to let the nucleons &quot;unite&quot; into whatever nuclei they like. Stars usually are formed from the very light elements and the nucleons inside, again, tend to get to the states with lower energies, and form more &quot;medium-number&quot; nuclei. The energy difference powers stars and we see the light emission, high temperatures and all other fun effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, sometimes the temperatures in the stars are so high, that nucleons form the very heavy nuclei from the medium-number nuclei. even though there is no immediate &quot;energy&quot; benefit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These heavy elements then disseminate everywhere with the death of the star. This stored star energy can then be released in the fission bomb.&lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-03-02T20:53:21.150" LastActivityDate="2011-03-02T20:53:21.150" CommentCount="5" />
  <row Id="6262" PostTypeId="1" CreationDate="2011-03-02T19:59:38.067" Score="2" ViewCount="102" Body="&lt;p&gt;I need to find the inaccuracies of this computer generated picture, the picture shows a non existent exoplanet. I need to identify the unrealistic features of the planet and its surroundings and write an essay on some of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image is located here: &lt;a href=&quot;http://i.stack.imgur.com/zXrPy.jpg&quot; rel=&quot;nofollow&quot;&gt;http://i.stack.imgur.com/zXrPy.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you help point out some of these innacuracies?&lt;/p&gt;&#xA;" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-03T00:17:31.153" LastActivityDate="2011-03-03T00:17:31.153" ClosedDate="2011-03-03T00:18:18.733" Title="Identifying unrealistic traits of this exoplanet. There are a few obvious ones, but I cant find them all. Help!" Tags="&lt;homework&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6263" PostTypeId="1" AcceptedAnswerId="12327" CreationDate="2011-03-02T20:02:35.667" Score="3" ViewCount="151" Body="&lt;p&gt;I'm looking for the most common and accurate approximation functions for III-V semiconductor compounds. So far, I found two (InGaAs as the sample compound):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The simplest one: $E(Ga_{1-x}In_xAs)=(1-x)E_{GaAs}+x \cdot E_{InAs}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Binomial: $E(Ga_{1-x}In_xAs)=(1-x)E_{GaAs}+x \cdot E_{InAs} - b \cdot x \cdot (1-x)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For InGaAs, b=1.44&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there more? I mean the methods for which the parameters can be found in articles, not measured by myself. When using the first one is sufficient? Do the scientists bother to calculate next corrections? Are they useful?&lt;/p&gt;&#xA;" OwnerUserId="1505" LastEditorUserId="1505" LastEditorDisplayName="" LastEditDate="2011-03-05T16:45:16.973" LastActivityDate="2011-07-16T13:26:58.467" Title="III-V semiconductor compounds - band gap approximations" Tags="&lt;semiconductor-physics&gt;" AnswerCount="1" />
  <row Id="6264" PostTypeId="2" ParentId="6262" CreationDate="2011-03-02T20:12:45.817" Score="3" ViewCount="" Body="&lt;p&gt;Interesting Homework type exercise - I wonder which Institution is setting this as a physics question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway here are some thoughts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) How many Suns are in this system? Where are they located and what shadows do they cast?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The 3 Body Gravitational problem may have some relevance here. (Look it up if you wonder why.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) What is your view on how weathering works? There is water here, clouds - so do all the landscapes look appropriately weathered?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) What forms Jupiter's cloud belts? Why do I even ask this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enough for now.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-03-02T20:12:45.817" CommentCount="1" />
  <row Id="6265" PostTypeId="2" ParentId="6157" CreationDate="2011-03-02T20:40:36.180" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://arxiv.org/abs/0801.2500&quot; rel=&quot;nofollow&quot;&gt;Making, probing and understanding ultracold Fermi gases&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A review on superfluidity and the BEC-BCS crossover in ultracold Fermi gases.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wolfgang Ketterle (Nobel 2001), Martin W. Zwierlein&lt;/p&gt;&#xA;" OwnerUserId="1814" LastEditorUserId="2809" LastEditorDisplayName="" LastEditDate="2011-03-28T03:59:44.217" LastActivityDate="2011-03-28T03:59:44.217" CommunityOwnedDate="2011-03-02T20:40:36.180" />
  <row Id="6266" PostTypeId="2" ParentId="6157" CreationDate="2011-03-02T20:50:32.133" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.quantiki.org/wiki/Category%3aHandbook_of_Quantum_Information&quot; rel=&quot;nofollow&quot;&gt;Handbook of Quantum Information&lt;/a&gt;&lt;/strong&gt; &quot;an encyclopaedia of everything quantum&quot;&lt;/p&gt;&#xA;" OwnerUserId="1814" LastEditorUserId="2809" LastEditorDisplayName="" LastEditDate="2011-03-28T04:06:12.390" LastActivityDate="2011-03-28T04:06:12.390" CommunityOwnedDate="2011-03-02T20:50:32.133" />
  <row Id="6267" PostTypeId="2" ParentId="6157" CreationDate="2011-03-02T20:59:22.003" Score="2" ViewCount="" Body="&lt;p&gt;A nice set of books have been made available through project Gutenberg, are made available through the &lt;a href=&quot;http://ebooksgo.org/free-physics-ebooks.html&quot; rel=&quot;nofollow&quot;&gt;Open Ebooks Library&lt;/a&gt;, including:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://ebooksgo.org/physics/HandbookOfFormulaeAndConstants.pdf&quot; rel=&quot;nofollow&quot;&gt;Handbook of Formula and Physical constants&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://ebooksgo.org/physics/29782-pdf.pdf&quot; rel=&quot;nofollow&quot;&gt;Space, Time and Gravitation - An Outline of the General Relativity Theory&lt;/a&gt;&lt;/strong&gt;, Sir Arthur Stanley Eddington (beware of this book)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://ebooksgo.org/physics/relat10h.zip&quot; rel=&quot;nofollow&quot;&gt;Relativity : the Special and General Theory&lt;/a&gt;&lt;/strong&gt;, Albert Einstein&lt;/p&gt;&#xA;" OwnerUserId="1814" LastEditorUserId="2809" LastEditorDisplayName="" LastEditDate="2011-03-28T04:16:31.790" LastActivityDate="2011-03-28T04:16:31.790" CommunityOwnedDate="2011-03-02T20:59:22.003" />
  <row Id="6268" PostTypeId="2" ParentId="6208" CreationDate="2011-03-02T21:20:18.237" Score="0" ViewCount="" Body="&lt;p&gt;Videos from Khan Academy is an excellent way to brush up the basics of chemistry&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.khanacademy.org/#Chemistry&quot; rel=&quot;nofollow&quot;&gt;http://www.khanacademy.org/#Chemistry&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One major advantage here is that you can watch only the parts you know you have forgotten and save time on the trivial things. &lt;/p&gt;&#xA;" OwnerUserId="1543" LastActivityDate="2011-03-02T21:20:18.237" CommunityOwnedDate="2011-03-04T18:19:18.983" CommentCount="3" />
  <row Id="6269" PostTypeId="1" CreationDate="2011-03-02T21:59:52.653" Score="0" ViewCount="246" Body="&lt;p&gt;Does the mass of projectile affect its trajectory when it is projected in no air resistance environment and why?&lt;/p&gt;&#xA;" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-03-02T22:13:53.983" LastActivityDate="2011-04-09T16:42:21.663" Title="Projectile motion without air resistance" Tags="&lt;homework&gt;&lt;motion&gt;&lt;projectile&gt;" AnswerCount="3" />
  <row Id="6270" PostTypeId="2" ParentId="6269" CreationDate="2011-03-02T22:30:55.277" Score="2" ViewCount="" Body="&lt;p&gt;No, the mass affects nothing; only the initial velocity, angle of projection and acceleration due to gravity determine the trajectory the projectile takes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because the inertial mass $m_i$ which appears in $F=m_ia$ is equal to the gravitational mass $m_g$ which appears in $F=m_gGM/R^2$ of Newton's gravitational law. These two cancel in all your equations. This property of masses is one way of stating &lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalence_principle&quot; rel=&quot;nofollow&quot;&gt;Einstein's Equivalence Principle&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-03-02T22:30:55.277" CommentCount="1" />
  <row Id="6271" PostTypeId="1" AcceptedAnswerId="6275" CreationDate="2011-03-02T23:03:02.523" Score="2" ViewCount="220" Body="&lt;p&gt;What is a physics law, a theory, and what is the relationship among them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is a very basic question, so any reference to epistemology will be greatly appreciated :)&lt;/p&gt;&#xA;" OwnerUserId="2145" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-02T23:27:09.840" LastActivityDate="2011-03-03T15:10:57.333" Title="laws and theories" Tags="&lt;soft-question&gt;&lt;theory&gt;" AnswerCount="6" CommentCount="5" />
  <row Id="6272" PostTypeId="2" ParentId="6271" CreationDate="2011-03-02T23:07:53.693" Score="4" ViewCount="" Body="&lt;p&gt;In my understanding, laws are the &quot;what&quot; and theories are the &quot;why&quot;. Ohm's law tells you what the relationship between voltage and current in a resistor is. The theory of electrons in solids tells you why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the laws tell you how physical systems behave, and the theories tell you why they do so.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-03-02T23:07:53.693" />
  <row Id="6273" PostTypeId="2" ParentId="6271" CreationDate="2011-03-02T23:17:56.383" Score="4" ViewCount="" Body="&lt;p&gt;A law is a precise statement about a physical relationship, ie Newton's laws. A law is a subset of atleast one theory, which provides a context and framework for the law, ie Classical Mechanics, and can also provide relations between laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A consistent theory normally has a set of axioms, or assumptions, and restrictions on the situation where they hold, and any statement derived from these can be considered laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can ask why of a law, and it can be backtracked inside the theory and back and forth even, between these axioms. Some will call this process an explanation.&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2011-03-02T23:27:22.443" LastActivityDate="2011-03-02T23:27:22.443" />
  <row Id="6274" PostTypeId="1" AcceptedAnswerId="7249" CreationDate="2011-03-02T23:18:33.717" Score="8" ViewCount="371" Body="&lt;p&gt;I am interested to understand how positive entropy Bose Einstein condensation for cold atoms (say) behave. The way I think about it is as follows: We have an ideal pure state where every atom is in the same ground state which depends on a geometry of a certain trap and the atoms do not interact. Now, naively, I thought about the actual positive-entropy state looking as one of the two following alternatives: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Every atom get excited with some small probability p. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A different picture is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) If the type of the intended ideal state puts every atom in the same ground state A (where A, say, depends on the geometry of the trap), then the positive-entropy state is obtained by mixing this ideal state with a state which is Bose Einstein state w.r.t. a different ground state B (or many such B;s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked around a little and what I was told is that the T&gt;0 theory is not part of the original theory discovered in the 1920s but rather a more recent theory that was developed in the 1990s. Moreover, the perturbation for the ideal pure state described by the T&gt;0 state manifests pairwise interaction between the atoms,&#xA;and that there are interesting singularities that occur immediately when T&gt;0 no matter how small. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will be very thankful for some explanations and references. Most helpful for me will be non-technical explanations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; I am aware of one book on a related topic: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3764371161&quot; rel=&quot;nofollow&quot;&gt;The Poincare seminar 2003 on Bose Einstein Condensation - Entropy&lt;/a&gt;. (I did not get a hold of it yet but a few of the papers &lt;a href=&quot;http://www.bourbaphy.fr/mars2003.html&quot; rel=&quot;nofollow&quot;&gt;are here&lt;/a&gt;.) I will appreciate any information regarding the matter. Genneth  recommend the book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0198526431&quot; rel=&quot;nofollow&quot;&gt;Quantum Liquids: Bose Condensation and Cooper Pairing in Condensed-Matter Systems&lt;/a&gt; (Oxford Graduate Texts) by Anthony Leggett.  &lt;/p&gt;&#xA;" OwnerUserId="1360" LastEditorUserId="1360" LastEditorDisplayName="" LastEditDate="2011-03-20T05:15:34.120" LastActivityDate="2011-03-20T05:15:34.120" Title="Bose-Einstein Condensate with T&gt;0  in Theory and Reality" Tags="&lt;bose-einstein-condensate&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="2" />
  <row Id="6275" PostTypeId="2" ParentId="6271" CreationDate="2011-03-02T23:43:47.673" Score="4" ViewCount="" Body="&lt;p&gt;dmckee is right (in the comments), the distinction between &quot;theory&quot; and &quot;law&quot; is quite subjective and varies a lot depending on who you ask and the context in which they are used. Sometimes they can be nearly synonymous. I would advise you to take any information you get about the difference between these two terms (including the remainder of this post) with a grain of salt, and never be afraid to ask someone to clarify what they mean by &quot;theory&quot; or &quot;law&quot; if it matters to your conversation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one difference that I &lt;em&gt;think&lt;/em&gt; most people will agree upon is that in order for something to be called a &quot;law,&quot; there must (or at least should) be experimental evidence supporting it. There is no such requirement to be called a &quot;theory.&quot; So it is possible for a theory to be &quot;upgraded&quot; to a law, once there is enough experimental evidence to make it seem true. However, even when that happens, it doesn't mean people are going to stop calling it a theory; for example, many people still use the terms &quot;theory of gravity&quot; and &quot;theory of relativity&quot; (and many others) even though both those theories have been confirmed by many, many experiments and have unquestionably achieved &quot;law&quot; status.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other difference that I think is common is that &quot;law&quot; often (but not necessarily always) refers to a single principle, typically something that can be expressed as a single equation or a set of closely related equations. A &quot;theory&quot; can be more broad. For example, when someone says &quot;law of gravity,&quot; they're probably talking about the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{F} = -\frac{G m_1 m_2}{r^2}\hat{\mathbf{r}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or a related equation. But I generally don't hear people using the term &quot;law of relativity,&quot; possibly because special relativity involves several equations and a set of related concepts.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-03-02T23:43:47.673" CommentCount="2" />
  <row Id="6276" PostTypeId="2" ParentId="6269" CreationDate="2011-03-02T23:44:52.577" Score="0" ViewCount="" Body="&lt;p&gt;It will only significantly affect the trajectory if the mass of the projectile is significant compared to the mass of the planet or other body providing the gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, Earth's mass is $6 \times 10^{24}$ kg, so even a large aircraft or spacecraft of $10^6$ or $10^7$ kg will not have a measurably different trajectory because of its mass. However, the Moon is $7 \times 10^{22}$ kg, so its trajectory is measurably different from what it would be if it were massless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://physics.stackexchange.com/questions/3534/dont-heavier-objects-actually-fall-faster-because-they-exert-their-own-gravity&quot;&gt;Don't heavier objects actually fall faster because they exert their own gravity?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-03-02T23:44:52.577" CommentCount="1" />
  <row Id="6277" PostTypeId="2" ParentId="6271" CreationDate="2011-03-03T01:43:47.630" Score="2" ViewCount="" Body="&lt;p&gt;Laws are the building blocks of a theory. A theory is a mathematical model which is a collection of one or more laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I don't think, the answer of @Lagerbaer is right. Here is a confusion. &quot;The theory of electrons in solids&quot; are nothing but a set of more general laws. Take chemistry for example. All of the laws of chemistry are derived in principle from more general and fundamental laws of quantum mechanics. Theories are no more than a set of rules relating different quantities of a model or in other words a set of laws. These laws can be explained in terms of deeper and more fundamental laws. It's not that theory explains any &quot;why&quot; of its own laws.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-03-03T15:10:57.333" LastActivityDate="2011-03-03T15:10:57.333" CommentCount="1" />
  <row Id="6278" PostTypeId="2" ParentId="6256" CreationDate="2011-03-03T01:59:16.820" Score="2" ViewCount="" Body="&lt;p&gt;I'm not sure what you're looking for here, but I'll try a different (and simpler) approach than the other answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a &quot;traditional&quot; chemical bomb, the energy comes from the electromagnetic force: you're breaking bonds between atoms and making more stable (lower-energy) bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a fission bomb, the energy comes from the strong nuclear force: you're breaking bonds between nucleons, producing final products that are in a lower-energy state*.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a fusion bomb, the energy is also coming from the strong nuclear force: you're making bonds between nucleons, producing final products that are in a lower-energy state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*To go up one level of complexity (and accuracy), Janne808 is correct: in fission bombs, the electromagnetic force is a significant contributor.  In fact, in the absence of coulomb repulsion between protons, I can't think of any fission reactions that would be exothermic.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-03-03T01:59:16.820" />
  <row Id="6279" PostTypeId="1" AcceptedAnswerId="6284" CreationDate="2011-03-03T02:48:38.260" Score="8" ViewCount="367" Body="&lt;p&gt;If the water were uniform temperature, it would have uniform density, so a bubble should either be all the way at the top (if it's lighter than water) or all the way at the bottom (if heavier). But in reality you don't see this neat separation. Sometimes you see bubbles hovering in the middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this because the water temperature is not uniform? Does this mean that with time, if the room temperature stays constant, all bubbles will neatly separate along the top and bottom?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-03T11:18:18.850" Title="The Galileo thermometer: why do the bubbles float in the middle of the tube?" Tags="&lt;fluid-dynamics&gt;&lt;home-experiment&gt;&lt;buoyancy&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="6281" PostTypeId="1" CreationDate="2011-03-03T04:31:40.237" Score="3" ViewCount="220" Body="&lt;p&gt;An opening paragraph:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The usual approach to introducing interactions in quantum field theory&#xA;  is to make the constraint on the&#xA;  amplitude of the field towards smaller&#xA;  values more forceful than harmonic.&#xA;  For example, instead of the massive&#xA;  Klein-Gordon field's harmonic&#xA;  constraint towards smaller field&#xA;  values,&#xA;  $\partial_\mu\partial^\mu\phi(x)+m^2\phi(x)=0$, we introduce a more forceful&#xA;  constraint,&#xA;  $\partial_\mu\partial^\mu\phi(x)+\lambda_1\phi(x)+\lambda_3\phi(x)^3=0$.&#xA;  We require $\lambda_3&amp;gt;0$; $\lambda_1$&#xA;  may be tuned so that a non-zero value&#xA;  of $\phi(x)$ is preferred, if&#xA;  $\lambda_1&amp;lt;0$. Here, we instead&#xA;  introduce a constraint on the observed&#xA;  field by an alternative, algebraic&#xA;  method. We take the observed field to&#xA;  be a bounded function of the field,&#xA;  $\Phi(x)=\mathsf{gd}(\phi(x))$, where&#xA;  the Gudermannian,&#xA;  $$\mathsf{gd}(z)=2\tan{\!{}^{-1}\!\left(e^z\right)}-\frac{\pi}{2}=\tan{\!{}^{-1}\!\left(\sinh{z}\right)},$$&#xA;  amongst other possible definitions, is&#xA;  complex analytic for $z\not=(2n+1)\pi i$ and has bounded real component&#xA;  between $-\frac{\pi}{2}$ and&#xA;  $\frac{\pi}{2}$. The use of the&#xA;  Gudermannian is inessential to the&#xA;  development, but we will use it as a&#xA;  convenient explicit mapping of the&#xA;  infinite range of measurement values&#xA;  of a free (quantum) field&#xA;  $\hat\phi(x)$ into a bounded range of&#xA;  measurement values&#xA;  $\mathsf{gd}(\hat\phi(x))$, with which&#xA;  we can make some analytic progress. In&#xA;  the usual terms of a potential energy,&#xA;  this construction effectively imposes&#xA;  an infinite potential on the field&#xA;  outside the range&#xA;  $(-\frac{\pi}{2},\frac{\pi}{2})$,&#xA;  because field values outside this&#xA;  range cannot occur, while preserving&#xA;  an attractive harmonic potential when&#xA;  the field is close to zero. It seems&#xA;  possible that the process of&#xA;  renormalization in the context of&#xA;  Hamiltonian or Lagrangian formalisms,&#xA;  by infinite adjustments of parameters,&#xA;  effectively constructs a potential&#xA;  that is comparable to this analytic&#xA;  ideal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My doubts about this paragraph are a little worrying. It emerged from the mathematics that I've been pursuing for the last two months, building on other mathematics that goes back about a year. I'm quite embedded in a way of thinking that is somewhat different from most peoples' conventional. When in this state, something like the above always looks &lt;em&gt;convincing&lt;/em&gt; to me &amp;mdash;I discover six months later why it's not, but right now I'm rather (too) pleased with what you see above. I'm asking here (1) whether people see the paragraph above as plausible enough at least to withhold judgment until they see the Math, or even that it seems enticing; or (2) that it looks hopeless (hopefully including some hint as to why, of course); and (3) whether there's anything that this reminds you of [Citations welcome, in the Math or Physics literature]. Answers that let me shortcut around that six month wait to seeing why this can't work will be welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess this is, as the sidebar recommends, &quot;sharing my research&quot;, though I don't want to take the space to &quot;provide details&quot; beyond a certain point. I'm curious whether it's an effective question. In formulating this question around the above paragraph, it occurs to me to clarify that I'm unhappy with deformation of a Hamiltonian or Lagrangian formulation of a quantum field as the only way we have to construct a new quantum field, particularly because of renormalization, and I'm also unhappy with the nonconstructive tendency in axiomatic quantum field theory. I don't see anything else in the literature, but is there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Luboš rightly focuses on the unlikeliness of the Gudermannian as a preferred choice of transform. As little as possible should depend on the choice of field coordinatization. The working abstract I have calls it a toy model, which I would hope would bring the comment above into focus, that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The use of the Gudermannian is inessential to the development&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, the Gudermannian is mathematically useful in the QFT context I am developing (1) because it has poles only on the imaginary axis, so just two series expansions are enough to evaluate it, effectively around $\pm\infty$, which is of course possible only because the Gudermannian is finite at $\pm\infty$. This makes analysis considerably more manageable. [Explicitly, we can express $\mathsf{gd}(z)=\mp 2\sum\limits_{j=0}^\infty(-1)^j\frac{1-e^{\pm(2j+1)z}}{2j+1}$ for $\Re(z)&amp;lt;0$ and $\Re(z)&amp;gt;0$.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Using $\mathsf{gd}$ instead of $\arctan{}$ is also useful for analysis, because the exponential of the quantum field acts as a displacement operator instead of as a derivation. [Explicitly, $[a(x),f(a^\dagger(y))]=f\mathbf{\,&amp;#39;}(a^\dagger)C(x-y)$, where $[a(x),a^\dagger(y)]=C(x-y)$ is the commutator, acts as a derivation, whereas $e^{\alpha a(x)}f(a^\dagger(y))=f(a^\dagger(y)+\alpha C(x-y))e^{\alpha a(x)}$ acts as a displacement operator.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) $\tanh{}$ is a superficially simpler choice, but it's significantly less convergent. [Explicitly, we have to expand the denominator of $\frac{-1+e^{2z}}{1+e^{2z}}$.] $\mathsf{gd}$ is the simplest bounded function that satisfies these three criteria. I so far can't think of a way to get all this into an opening paragraph, unless there is a precedent I don't know of that sets up a context that is close enough to what I'm doing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find even the little I can do with this toy model gives &lt;em&gt;a little&lt;/em&gt; insight into renormalization, though it's certainly allusive. There is a chaotic regime at time-like separation, for example. Perhaps I have to get into the paragraph above that the move to analysis might be worth trying just because perturbation theory is a non-convergent expansion. I'm already seeing things slightly different because of your comments. Many thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT(2): Luboš asks for clarification of the general idea. At the moment I'm trying to see ways to combine multiple heterodox ways of constructing models. To me, the Hilbert space structure is paramount because it underlies any probability interpretation; positivity of the energy is less important than it is usually taken to be, because it is neither necessary or sufficient to ensure stability; a continuous mass spectrum arises in the Källén-Lehmann representation of the 2-point function of a renormalized theory, so it should not be ruled out as part of other types of models; field renormalization suggests some kind of transformation of the unbounded quantum field to bounded values. Latterly, I've thought that a more analytic approach to models would be better than the series expansions that are universally used. In a few weeks or years I may have dropped any or all of these constituents of the models I have been constructing and characterizing. The published papers that are referred to on my web-site gives some idea of the different kinds of heterodox models I've constructed, understood at some level, and dismissed in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an answer, by the way, to the question &quot;Are there other ways out there?&quot;, which I came across only a few days ago. &lt;a href=&quot;http://dx.doi.org/10.1007/s00220-010-1137-1&quot; rel=&quot;nofollow&quot;&gt;Detlev Buchholz, Gandalf Lechner, Stephen J. Summers, &quot;Warped Convolutions, Rieffel Deformations and the Construction of Quantum Field Theories&quot;&lt;/a&gt;. This and its references claim to construct an interacting QFT in 3+1 in the Haag-Kastler axiomatic context, although I'm just now making my way through it.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-03-07T16:51:18.543" LastActivityDate="2011-03-07T16:51:18.543" Title="An alternative, algebraic way to introduce interactions. Are there other ways out there?" Tags="&lt;quantum-field-theory&gt;&lt;mathematical-physics&gt;&lt;reseach-level&gt;" CommentCount="11" />
  <row Id="6282" PostTypeId="2" ParentId="6248" CreationDate="2011-03-03T05:19:14.307" Score="3" ViewCount="" Body="&lt;p&gt;If you're going to do a lot of work in 4 dimensions, it might be worth learning two-component notation. This is used in the text books, e.g., of &lt;a href=&quot;http://books.google.com.au/books?id=5OepxIG42B4C&quot; rel=&quot;nofollow&quot;&gt;Srednicki&lt;/a&gt;, &lt;a href=&quot;http://books.google.com.au/books?id=jZ1Y1Z3BoicC&quot; rel=&quot;nofollow&quot;&gt;Buchbinder &amp;amp; Kuzenko&lt;/a&gt; or the comprehensive article by &lt;a href=&quot;http://arxiv.org/abs/0812.1594&quot; rel=&quot;nofollow&quot;&gt;Dreiner, Haber and Martin&lt;/a&gt;.&#xA;Two component notation has the advantage that the 4 dimensional objects decompose into the &lt;em&gt;direct sum&lt;/em&gt; (or something similar) of 2 dimensional objects, which have fewer invariants and identities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll use the conventions of B&amp;amp;K (which I think slightly differ from those in the question).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The four component spinors takes the form &#xA;$$ &#xD;&#xA;\psi_a = \begin{pmatrix}\psi_\alpha \\ \bar\psi^{\dot\alpha}\end{pmatrix} \,, \quad&#xD;&#xA;\bar\psi^a = \begin{pmatrix}\psi^\alpha &amp;amp; \bar\psi_{\dot\alpha}\end{pmatrix}&#xD;&#xA;$$&#xA;and the gamma and projection matrices&#xA;$$&#xD;&#xA;\gamma_a = \begin{pmatrix} 0 &amp;amp; \sigma_a \\ \tilde\sigma_a &amp;amp; 0 \end{pmatrix}\,,\quad&#xD;&#xA;\gamma_5 = -i \gamma_0\gamma_1\gamma_2\gamma_3&#xD;&#xA;         =  \begin{pmatrix} 1 &amp;amp; 0 \\ 0 &amp;amp; -1 \end{pmatrix} \,,\quad&#xD;&#xA;$$&#xA;$$&#xD;&#xA;P_L = \frac12 (1 + \gamma_5) \,, \quad P_R = \frac12 (1 - \gamma_5)&#xD;&#xA;$$&#xA;the left and right spinors are then&#xA;$$ &#xD;&#xA;\psi_L = P_L\psi = \begin{pmatrix}\psi \\ 0 \end{pmatrix} \,, \quad&#xD;&#xA;\psi_R = P_R\psi = \begin{pmatrix} 0 \\ \bar\psi \end{pmatrix} \,, \quad&#xD;&#xA;$$&#xA;where there should never be any reason to confuse a 2-component and 4-component spinor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spinor bilinears are&#xA;$$&#xD;&#xA;\bar\theta \theta = \theta^2 + \bar\theta^2 \,,\quad&#xD;&#xA;\bar\theta \gamma_5 \theta = \theta^2 - \bar\theta^2&#xD;&#xA;$$&#xA;etc..., where the contraction convention is &#xA;$\theta^2 = \theta^\alpha \theta_\alpha$ and&#xA;$\bar\theta^2 = \bar\theta_{\dot\alpha} \bar\theta^{\dot\alpha}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, to check your final identity (for example), the left hand side is&#xA;$$ (\bar\theta\gamma_5\theta)(\bar\psi_L\theta)(\bar\theta\lambda)&#xD;&#xA; = (\theta^2-\bar\theta^2)(\bar\psi\bar\theta)(\theta\lambda+\bar\theta\bar\lambda)&#xD;&#xA; = (\theta^2-0)(-\frac12\bar\theta^2)(\bar\psi\bar\lambda)&#xD;&#xA; = -\frac12\theta^2\,\bar\theta^2\,\bar\psi\bar\lambda&#xD;&#xA;$$&#xA;where we used the fact that &#xA;$\theta^3=\theta_{\alpha_1}\theta_{\alpha_2}\theta_{\alpha_3}=0$ &#xA;and similarly for $\bar\theta^3=0$. &#xA;We also used the important identities &#xA;$$ \begin{align} &#xD;&#xA;\theta_\alpha \theta^\beta &amp;amp;= -\frac12\theta^2\delta_\alpha^\beta \,,&amp;amp;&#xD;&#xA;\bar\theta^{\dot\alpha}\bar\theta_{\dot\beta} &amp;amp;=-\frac12\bar\theta^2\delta_{\dot\beta}^{\dot\alpha} \ .&#xD;&#xA;\end{align}$$ &#xA;The right hand side is&#xA;$$ \frac14(\bar\theta\gamma_5\theta)^2(\bar\psi_L\lambda)&#xD;&#xA; = \frac14(\theta^2-\bar\theta^2)(\theta^2-\bar\theta^2)(\bar\psi\bar\lambda)&#xD;&#xA; = -\frac12\theta^2\bar\theta^2\bar\psi\bar\lambda&#xD;&#xA;$$&#xA;so the last identity checks out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest of the identities can be similarly checked.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;  Edit:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Since there seems to be a bit of confusion for the other identities, here's their proofs.&#xA;Because I'm lazy, I've suppressed all indices. For example&#xA;$\bar\theta\partial\psi&#xD;&#xA;=\bar\theta_{\dot\alpha}\tilde\sigma^{\dot\alpha\alpha}_a\partial^a\psi_\alpha&#xD;&#xA;=-(\partial^a\psi_\alpha)\tilde\sigma^{\dot\alpha\alpha}_a\bar\theta_{\dot\alpha}&#xD;&#xA;=-(\partial_a\psi^\alpha)\sigma_{\alpha\dot\alpha}^a\bar\theta^{\dot\alpha}&#xD;&#xA;=-\psi\overleftarrow{\partial}\bar\theta\ .$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Identity 1:&#xA;$$ LHS = (\theta^2-\bar\theta^2)(\bar\psi\bar\theta)(\bar\theta\partial\psi)&#xD;&#xA;= (\theta^2-0)(-\frac12\bar\theta^2)(\bar\psi\partial\psi)&#xD;&#xA;= -\frac12\theta^2\bar\theta^2(\bar\psi\partial\psi)&#xD;&#xA;$$&#xA;$$ RHS = \frac14(\theta^2-\bar\theta^2)^2(\bar\psi\partial\psi) = LHS&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Identity 2 is basically just the complex conjugate result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Identity 3: I think you have a mistake in this one (ignoring the obvious problem of the non-matching $\mu$ index).  This is because $(\bar{\psi}\gamma ^\mu \psi)=0$ if we insert a $\gamma_5$ so it's like the term on the left it's nonvanishing $(\bar{\psi}\gamma_5\gamma_\mu \psi)=2\psi\sigma_\mu\bar\psi$. So &#xA;$$&#xD;&#xA;LHS  &#xD;&#xA;= (\theta\sigma_\mu\bar\theta-\bar\theta\tilde\sigma_\mu\theta)(\bar\psi\bar\theta)(\psi\theta)  &#xD;&#xA;= 2(\psi\theta)(\theta\sigma_\mu\bar\theta)(\bar\theta\bar\psi)&#xD;&#xA;= \frac12\theta^2\bar\theta^2 (\psi\sigma_\mu\bar\psi)&#xD;&#xA;$$&#xA;$$&#xD;&#xA;RHS = \frac{1}{4} (\bar{\theta}\gamma _5 \theta)^2 (\bar{\psi}\gamma_5\gamma_\mu\psi)&#xD;&#xA;= -\theta^2\bar\theta^2(\psi\sigma_\mu\bar\psi)&#xD;&#xA;$$&#xA;so it seems to be out by a factor of $-1/2$. (Although, I might have miscounted the factors of 1/2).&lt;/p&gt;&#xA;" OwnerUserId="429" LastEditorUserId="429" LastEditorDisplayName="" LastEditDate="2011-03-04T04:49:39.273" LastActivityDate="2011-03-04T04:49:39.273" CommentCount="10" />
  <row Id="6283" PostTypeId="2" ParentId="6279" CreationDate="2011-03-03T06:57:10.307" Score="1" ViewCount="" Body="&lt;p&gt;Although I could be wrong, I think the reason is that the density of any fluid, including water, increases very slightly with depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a well-known &lt;a href=&quot;http://en.wikipedia.org/wiki/Pressure#Liquid_pressure_or_pressure_at_depth&quot; rel=&quot;nofollow&quot;&gt;relation between pressure and depth&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta p = -\rho g \Delta y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which says that at the difference in pressure between any two points in the fluid is proportional to the difference in height between those two points. Now, we usually think of a liquid as being incompressible, so that its volume remains the same at any pressure, but in reality this is not true. Liquids do undergo a very slight decrease in volume as the pressure rises. So since the water pressure at the bottom of the tube is higher, the volume of a given amount of water will be slightly smaller at the bottom of the tube than at the top, and therefore the density at the bottom will be a little higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the density of one of the glass bulbs just happens to fall in the range between the density of water at the top and the density of water at the bottom of the tube, then the bulb will come to equilibrium in the middle, at the point where its own density happens to be equal to that of the surrounding water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, &lt;a href=&quot;http://en.wikipedia.org/wiki/Galileo_thermometer&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; reports that the liquid used in a Galilean thermometer is usually not water, but something else that has a more drastic variation of density with temperature. But the argument I've outlined above applies equally well to any normal liquid.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-03-03T11:18:18.850" LastActivityDate="2011-03-03T11:18:18.850" CommentCount="1" />
  <row Id="6284" PostTypeId="2" ParentId="6279" CreationDate="2011-03-03T07:11:35.520" Score="2" ViewCount="" Body="&lt;p&gt;I think there's a temperature gradient in the water.  The other candidate is that the density of water changes due to its compressibility under pressure.  Let's examine the pressure effect first.  From an estimate its size, we can see whether it's a significant factor compared to temperature gradients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bulk modulus of water is about $2*10^9 Pa$ (&lt;a href=&quot;http://en.wikipedia.org/wiki/Bulk_modulus&quot; rel=&quot;nofollow&quot;&gt;Wikipedia source&lt;/a&gt;).  The pressure in a column of water is $\rho g h$ with $\rho$ the density, $g$ gravitational acceleration, and $h$ the height of the column.  For a column of water half a meter tall, this comes to about $5*10^3 Pa$.  This means the water at the density of water at the bottom is slightly greater, and the size of the effect is about $2*10^{-6} \rho$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare this to the effect of temperature.  Water's density varies by about $7*10^{-5} \rho$ for a $1 C$ change in temperature at room temperature (&lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_expansion&quot; rel=&quot;nofollow&quot;&gt;Wikipedia source&lt;/a&gt;).  That means that the density gradient due to pressure and compressibility is equivalent to a temperature difference of about $.03 C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$.03 C$ is tiny - only .01% of the temperature.  There must be temperature fluctuations larger than .01% in a thermometer that's just sitting in your room, so the baubles float mid-way due to a temperature gradient in the water, as you originally suggested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for whether the baubles will separate - yes, I think they will, but I don't have a Galilean thermometer to test this on.  If you've got one and some good ideas on temperature control, make the experiment!&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-03-03T07:11:35.520" />
  <row Id="6285" PostTypeId="2" ParentId="6243" CreationDate="2011-03-03T08:27:36.563" Score="2" ViewCount="" Body="&lt;p&gt;It is not a weaker form. The book quotes &quot;index m &lt;strong&gt;refers&lt;/strong&gt; to the measurement outcome&quot;. As Lubos wrote, the classical measurement outcome could be anything physical, which is not the focus of the postulate, so basically if $\mu(m)$ is a classical measurement outcome and $m$ is your corresponding index then it is implicit that $\mu$ is a mapping&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu :\{1,2,3,...N\} \rightarrow \mathbb{R} \times (\textbf{physical unit of observable}) $&lt;/p&gt;&#xA;" OwnerUserId="1671" LastEditorUserId="1671" LastEditorDisplayName="" LastEditDate="2011-03-03T08:37:19.097" LastActivityDate="2011-03-03T08:37:19.097" CommentCount="2" />
  <row Id="6287" PostTypeId="2" ParentId="6271" CreationDate="2011-03-03T09:40:11.417" Score="0" ViewCount="" Body="&lt;p&gt;As the answers of &lt;a href=&quot;http://physics.stackexchange.com/questions/6271/laws-and-theories/6272#6272&quot;&gt;Lagerbaer&lt;/a&gt; and &lt;a href=&quot;http://physics.stackexchange.com/questions/6271/laws-and-theories/6277#6277&quot;&gt;sb1&lt;/a&gt; show, the answer can seem quite contradictory:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the one hand, you can say a law is like a mathematical axiom, i.e. the starting point for bootstrapping your theories. I guess &lt;em&gt;the&lt;/em&gt; example hear are &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion&quot; rel=&quot;nofollow&quot;&gt;Newton's laws of motion&lt;/a&gt;. There is no explanation why motion should always stem from second order in time ODEs, and yet this assumption has (in the context of classical mechanics) proven to be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, you can say a law is a set of equations (like Ohms law) that can be explained by a theory. But since you can use Ohms law as an axiom as well, I think this latter definition is (IMHO) less correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let's have a look at wikipedia, specifically &lt;a href=&quot;http://en.wikipedia.org/wiki/Theorem#Terminology&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Theorem#Terminology&lt;/a&gt; links law to the article &lt;a href=&quot;http://en.wikipedia.org/wiki/Laws_of_science&quot; rel=&quot;nofollow&quot;&gt;Laws of science&lt;/a&gt;, starting with&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The laws of science are various established scientific laws, or physical laws as they are sometimes called, that are considered universal and invariable facts of the physical universe. Laws of science may, however, be disproved if new facts or evidence contradicts them. A &quot;law&quot; differs from hypotheses, theories, postulates, principles, etc., in that a law is an analytic statement, usually with an empirically determined constant. A theory may contain a set of laws, or a theory may be implied from an empirically determined law.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;not mentioning the word axiom anywhere. However, there's the article &lt;a href=&quot;http://en.wikipedia.org/wiki/Physical_law&quot; rel=&quot;nofollow&quot;&gt;Physical Law&lt;/a&gt;, which states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A physical law or scientific law is a scientific generalization based on empirical observations  of physical behaviour (i.e. the law of nature). Laws of nature are observable. Scientific laws are empirical, describing observable patterns. Empirical laws are typically conclusions based on repeated scientific  experiments and simple observations, over many years, and which have become accepted universally within the scientific community. The production of a summary description of our environment in the form of such laws is a fundamental aim of science. These terms are not used the same way by all authors. Some philosophers e.g. Norman Swartz use &quot;physical law&quot; to mean what others mean by &quot;natural law&quot;/&quot;law of nature&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And &lt;a href=&quot;http://en.wikipedia.org/wiki/Physical_law#Other_fields&quot; rel=&quot;nofollow&quot;&gt;further down&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some mathematical  theorems and axioms are referred to as laws because they provide logical foundation to empirical laws.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Personally, I prefer not using the word law at all (I mean in physics, don't get the wrong idea there ;)&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2011-03-03T09:40:11.417" CommentCount="2" />
  <row Id="6288" PostTypeId="2" ParentId="6271" CreationDate="2011-03-03T09:58:20.867" Score="1" ViewCount="" Body="&lt;p&gt;A classic on this matter is Feynman's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0679601279&quot; rel=&quot;nofollow&quot;&gt;&quot;The Character of Physical Law&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/mofu9.jpg&quot; alt=&quot;Character of Physical Law&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2151" LastActivityDate="2011-03-03T09:58:20.867" />
  <row Id="6289" PostTypeId="2" ParentId="6279" CreationDate="2011-03-03T10:14:07.147" Score="2" ViewCount="" Body="&lt;p&gt;Of course there are temperature gradients, in most rooms if there is no &#xA;forced convection You will find the air cooler at the floor. &#xA;And second, when room temperature changes, the liquid in the cylinder is &#xA;heated or cooled from the wall, which rises a convection current in the liquid. &#xA;Until the column of liquid is in equilibrium to the outside, &#xA;there will be a dynamic gradient on top of the gradient from outside. &#xA;With respect to those calculations of compressibility and thermal expansion, &#xA;one has to include the compressibility and volume expansion of the baubles as well!&#xA;In my youth, calculation of thermometer corrections was standard knowledge, &#xA;but that has gone obviously :=(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water is due to its density anomaly the most useless thermometer liquid. &#xA;Wiki says some hydrocarbon oils are used, and for those (as for all &quot;normal&quot; liquids)&#xA;one can use a volume thermal coefficient of 10⁻3 as a first order approximation. &#xA;Worst case of baubles were a cheap glass with about 30 10⁻6 volume expansion per degree.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-03T10:30:41.880" LastActivityDate="2011-03-03T10:30:41.880" CommentCount="5" />
  <row Id="6290" PostTypeId="1" CreationDate="2011-03-03T12:03:21.693" Score="0" ViewCount="33" Body="&lt;p&gt;I'm looking for a material to release a strong odour as an indicator when wet. Compounds such as used in stink bombs are the inspiration, but I'd like something that can be shaped into a solid practical shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would appreciate any ideas, thanks in advance.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-03T12:03:21.693" ClosedDate="2011-03-04T03:21:39.380" Title="What materials can be used to release odours when wet?" Tags="&lt;material-science&gt;" CommentCount="0" />
  <row Id="6291" PostTypeId="2" ParentId="6243" CreationDate="2011-03-03T12:17:37.073" Score="2" ViewCount="" Body="&lt;p&gt;I have been looking at what is available of the book online, and believe that you are right that the Postulate 3 is (in a sense discussed below) weaker than the usual QM Projection Postulate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly there are some notational issues here. The $M_m$ are a family of operators, called &lt;strong&gt;Measurement Operators&lt;/strong&gt;, indexed by $m$ which is a label for the outcome eigenvectors. However the eigenvalue equation itself does not play a part in Postulate 3, which is partly your point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the book does also discuss &lt;strong&gt;Projection Operators&lt;/strong&gt; as a &lt;em&gt;special case&lt;/em&gt; of Measurement Operators. Here $M = \Sigma m P_m$ where $M$ is a Hermitian operator and m &lt;em&gt;is&lt;/em&gt; now the eigenvalue in its spectral decomposition into Projection operators $P_m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is &quot;special&quot; about this special case is that $P_m$ is idempotent, physically that the &quot;after measurement&quot; state is as an eigenstate of $P_m$ now going to return the same result on immediate measurements. With a Measurement Operator there is the arbitrary state $M_m \Psi$ as after state, which might not be an eigenstate of $M_m$ (hence the notation could get a little confusing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later they introduce a Postulate 4 (on Tensor products and compound systems) and they claim to prove that: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Projective Measurement + Unitary (Postulate 2) + Postulate 4 &quot;implement&quot; General Measurement (Postulate 3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in a sense Postulate 3 is more general than Projective measurement as one could have non-projective measurements, but under all the other Postulates it all comes out the same in the end, if required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose that indeed part of the reason why this has all been done is because the eigenvalue does not do much more than label and distinguish the different outcomes in Quantum Computation and Quantum Information. Its actual physical value is less important in many calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example in the usual qubit situation we might have $|\Psi&amp;gt; = a |0&amp;gt; + b |1&amp;gt;$. Then the eigenvalue equation for $M_0$ might be $M_0 |0&amp;gt; =\alpha |0&amp;gt;$, but the $\alpha$ value is not of interest only its role in indicating that we now have a $|0&amp;gt;$ state.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-03-03T12:17:37.073" />
  <row Id="6292" PostTypeId="1" AcceptedAnswerId="6293" CreationDate="2011-03-03T12:47:06.403" Score="3" ViewCount="119" Body="&lt;p&gt;How could we measure high energy photons, whithout measuring them ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't understand how we can &quot;see&quot; those Gamma Ray Bubbles if they are not reaching here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this graph from Nasa you can see those &quot;bubbles&quot; are not reaching solar system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nasa.gov/mission_pages/GLAST/news/new-structure.html&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://i.stack.imgur.com/PEM2t.jpg&quot; alt=&quot;Art about bubles&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then how could be measure that Gamma Ray without the Gamma Rays&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any answer!&lt;/p&gt;&#xA;" OwnerUserId="1916" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-03-03T18:01:28.887" LastActivityDate="2011-03-03T18:01:28.887" Title="Gamma Ray Bubble at the center of our galaxy seen by Fermi Telescope" Tags="&lt;astrophysics&gt;&lt;astronomy&gt;&lt;universe&gt;&lt;galaxy&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="6293" PostTypeId="2" ParentId="6292" CreationDate="2011-03-03T13:03:17.353" Score="2" ViewCount="" Body="&lt;p&gt;Concerning the 8-shaped bubbles around the galaxy, see&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2010/11/fermi-milky-way-cutting-x-ray-infinity.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/11/fermi-milky-way-cutting-x-ray-infinity.html&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/abs/arXiv:1005.5480&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/arXiv:1005.5480&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They're not pictures of photons - X-rays themselves. The infinity symbol is a picture of X-ray sources: we are observing the X-rays that came from those sources here. Note that the whole structure is smaller than 100,000 light years or so - very tiny when compared to the cosmological distances. So if the 8-shaped sources were created 10 million years ago, the time needed for the photons to get here is negligible. They're here &quot;immediately&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's hard to measure the distance from which an X-ray is approaching us. However, you should understand that the Sun and the Earth are not in the middle of the Milky Way. They're not in the middle of the 8-shaped figure. We're looking at the situation from the &quot;side&quot; (the Solar System is somewhere between the center and the visible edge of our Galaxy) so we literally see something that is 8-shaped in the skies. Assuming that the distribution of the sources is rotationally symmetric - with respect to the Milky Way's axis - one can actually reconstruct the shape of the sources in 3D from the 2D picture we see (because the 3D picture only depends on 2 dimensions, because of the axial symmetry).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-03T13:10:48.470" LastActivityDate="2011-03-03T13:10:48.470" CommentCount="2" />
  <row Id="6294" PostTypeId="1" AcceptedAnswerId="6295" CreationDate="2011-03-03T13:07:36.597" Score="8" ViewCount="158" Body="&lt;p&gt;There is some convention? how is this 3d to 2d mapping done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here an example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://apod.nasa.gov/apod/ap101110.html&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://i.stack.imgur.com/rHRCq.jpg&quot; alt=&quot;Fermi_Data&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1916" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-03-03T13:52:35.187" LastActivityDate="2011-03-03T13:52:35.187" Title="Why are all-Sky images drawn as a filled ellipse?" Tags="&lt;astronomy&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6295" PostTypeId="2" ParentId="6294" CreationDate="2011-03-03T13:11:35.100" Score="9" ViewCount="" Body="&lt;p&gt;The ellipse is a particular way to draw the surface of a sphere - like the sphere of the skies around us, or the surface of the Earth - on a flat piece of paper or screen (because of the curvature of the sphere, it cannot be &quot;flattened&quot; without distortions). This one is called the Mollweide projection and it preserves the areas (and completely sacrifices correct representation of the angles):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mollweide_projection&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mollweide_projection&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://en.wikipedia.org/wiki/Map_projection&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Map_projection&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second link enumerates many other ways how to draw the spherical surface. &lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-03T13:11:35.100" CommentCount="2" />
  <row Id="6297" PostTypeId="2" ParentId="6246" CreationDate="2011-03-03T14:13:11.993" Score="1" ViewCount="" Body="&lt;p&gt;Convection causes the flame to be affected by gravity (hot gas is less dense than colder gas and therefore it rises).  Without gravity there is no convection, so the mixing of the fuel (vaporized candle wax) and oxidizer happens only by diffusion.  Which is a lot slower than convection (so as you can see the flame is a lot less intense) and is also symmetrical (so the flame is spherical, except for where the body of the candle interferes with this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be possible to exert a force on the flame's constituents by using a very strong electric or magnetic field.  If this was done in a zero-g environment and if there were differences in the electric charge or magnetic moment of the flame's particles, in principal some effect could be produced.  For example if hotter bits were a bit more ionized, they would be affected differently by an electric field than colder bits.  In practice I would guess that one would have to choose very carefully the materials used in the flame in order to get an observable effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;W.r.t. your question, a very small air current directed at the flame would alter its shape (that experiment would have to have been carefully done in very still air).&lt;/p&gt;&#xA;" OwnerUserId="2297" LastActivityDate="2011-03-03T14:13:11.993" CommentCount="0" />
  <row Id="6298" PostTypeId="2" ParentId="6208" CreationDate="2011-03-03T14:16:14.373" Score="2" ViewCount="" Body="&lt;p&gt;Quantum mechanics plays an important role in chemistry, as everyone knows, but so does thermodynamics.  I highly recommend Landau, volume 5, chapters 9 and 10; but of course if you haven't read the first 8 chapters yet, you should read them as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think of chemistry sort of as a set of empirical rules born out of experience in the lab.  There are computer programs which use the rules of QM to compute things related to chemistry, and these programs amount to essentially writing down a system of differential equations and solving them with a numerical method.  The point I am trying to make is that trying to understand chemistry from the point of view of QM can only give you a certain type of insight because QM is too fundamental.  I recommend for the QM side of things, you should start with Landau, volume 3, chapters 10-13.  &lt;/p&gt;&#xA;" OwnerUserId="442" LastActivityDate="2011-03-03T14:16:14.373" CommunityOwnedDate="2011-03-04T18:19:18.983" CommentCount="1" />
  <row Id="6299" PostTypeId="1" AcceptedAnswerId="6301" CreationDate="2011-03-03T14:25:26.907" Score="0" ViewCount="118" Body="&lt;p&gt;How should I model the tea concentration as a function of time after a tea bag has been submerged? Is there a simple way of measuring the tea concentration?&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2011-03-04T00:21:05.657" Title="How to model tea diffusion/osmosis?" Tags="&lt;diffusion&gt;" AnswerCount="2" />
  <row Id="6300" PostTypeId="1" AcceptedAnswerId="6319" CreationDate="2011-03-03T14:33:42.530" Score="5" ViewCount="245" Body="&lt;p&gt;So, one thing has been annoying me ever since I learned about &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_hybridisation&quot; rel=&quot;nofollow&quot;&gt;orbital hybridization&lt;/a&gt;: you explain the shape of molecules by postulating that the orbitals of multi-electron atoms are linear combinations of the orbitals of the hydrogen atom.  Fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what gets me, and I've asked several chemistry professors about this, and I haven't gotten an answer.  If I measure the orbital angular momentum of an electron in a hydrogen atom, I get the value $L = \hbar \sqrt{\ell(\ell +1)}$, so an s orbital has angular momentum $0$ and a p orbital has angular momentum $\hbar$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what happens when I measure the orbital angular momentum of a sp${}^{3}$ electron?  Do I get  $\frac{3\hbar}{4}$ with probability 1?  Or do I get $\hbar$ with probability 0.75, and $0$ with probability 0.25?  Are the original orbitals really the fundamental thing, and the hybridized orbitals fundamentally just quantum superpositions of these things?  Or does the multi-electron potential create a state that just looks like a linear combination of one s and 3 p orbitals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is the actual answer more complicated and this a heuristic picture that I'm taking too seriously?&lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2011-03-03T18:26:14.450" LastActivityDate="2011-03-04T16:57:55.687" Title="What is the angular momentum spectrum of an sp${}^3$ electron?" Tags="&lt;quantum-mechanics&gt;&lt;physical-chemistry&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="6301" PostTypeId="2" ParentId="6299" CreationDate="2011-03-03T14:37:07.440" Score="3" ViewCount="" Body="&lt;p&gt;Concentration measurement would be done by colourimetry rather easy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &quot;model&quot; look for 1st and 2nd law of Fick. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-03T14:37:07.440" />
  <row Id="6302" PostTypeId="1" AcceptedAnswerId="6305" CreationDate="2011-03-03T14:39:38.597" Score="4" ViewCount="224" Body="&lt;p&gt;The WIKI Higgs boson site has an interesting diagram illustrating likely Higgs mass intervals that experiments (LEP &amp;amp; Tevatron) or indirect measurements have determined with a 95% confidence level. It posits that the feasible mass ranges are in the intervals of about 115-158 GeV or 175-185 GeV. The Tevatron confirmed the existance of the top quark (172 GeV) in 1995, so it had the energy and time to detect the Higgs in the 115-158 GeV ranges, but could not.&#xA;Given LHC's greater energy and luminosity, why haven't there been anouncements of the Higgs discovery? Do we understand enough about potential particle creation to definitively know they are Higgs decays? &lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-03-03T19:54:07.243" LastActivityDate="2011-03-04T06:01:54.083" Title="Higgs Boson Mass" Tags="&lt;particle-physics&gt;&lt;higgs-boson&gt;" AnswerCount="2" />
  <row Id="6303" PostTypeId="1" CreationDate="2011-03-03T15:16:31.873" Score="3" ViewCount="210" Body="&lt;p&gt;If our universe comes from a flux compactification of string theory over 6 dimensions with a nonzero flux, then it can't be continually deformed to another compactification with zero flux as the total flux is a topological invariant. The cosmological constant of our universe is positive, which in string theory means it is a metastable state. The stable vacuum for a flux compactification in string theory is a BPS compactification with negative cosmological constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if our metastable observable universe is actually embedded in a larger stable anti de Sitter background? An inflating metastable or slowly rolling or chaotically inflating patch of this asymptotically AdS universe inflates to our observable universe. By the AdS/CFT correspondence, our universe can be described by a superconformal field theory in 2+1D with no gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any flaw in my reasoning?&lt;/p&gt;&#xA;" LastActivityDate="2011-03-14T13:52:12.123" Title="Is our universe a 2+1D superconformal field theory?" Tags="&lt;string-theory&gt;&lt;cosmology&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="6304" PostTypeId="2" ParentId="6300" CreationDate="2011-03-03T15:31:33.470" Score="3" ViewCount="" Body="&lt;p&gt;More like the latter than the former, although you've got one detail wrong. An $s$ orbital is $\ell=0$, so it corresponds to $L=0$, not $L=\hbar$. A $p$ orbital is $\ell=1$, corresponding to $L=\hbar\sqrt{2}$. If the electron is in a state that is a superposition of $s$ and $p$, then a measurement of angular momentum will yield either 0 or $\hbar\sqrt 2$, with probabilities given by the squares of the amplitudes of the different parts of the superposition. If it's an equally-weighted superposition (and I confess I don't know enough chemistry to know whether that's correct), then the probabilities are 25% and 75% as you say.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-03T15:31:33.470" CommentCount="1" />
  <row Id="6305" PostTypeId="2" ParentId="6302" CreationDate="2011-03-03T16:04:24.547" Score="5" ViewCount="" Body="&lt;p&gt;There are two considerations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) if you follow a bit the current theories, the manifestation of the Higgs decays can vary, depending on the parameters of the models, and there are many models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Hadron colliders, in contrast to e+e- ones are &quot;dirty&quot;, there are enormous backgrounds that have to be understood, and &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0810/0810.3753v1.pdf&quot; rel=&quot;nofollow&quot;&gt;the few clear decay channels&lt;/a&gt; proposed by theories have small crossections and have to be fished out of these backgrounds. So more data than the ones existing are necessary to be able to answer definitively the Higgs question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LHC is now building up the statistics necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what I foresee will happen, if we are lucky,  is first a new resonance in some of the channels checked will be declared with at least 4 sigma. Then everybody will fall on it, experimentalists to find other decay channels and theorists to fit specific models to decide whether it is the Higgs or not. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-04T06:01:54.083" LastActivityDate="2011-03-04T06:01:54.083" CommentCount="1" />
  <row Id="6306" PostTypeId="1" CreationDate="2011-03-03T16:08:21.107" Score="0" ViewCount="229" Body="&lt;p&gt;Many of the most brilliant ideas of modern physics are very much unpopperian. In view of this, it has become more and more difficult to hold an orthodox Popperian attitude in physics. It simply looks like more of a philosophical dogma nowadays. My question is, should physicists abandon this criterion at least partially for good? If so, what criterion should replace it?&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-03-03T18:51:29.407" ClosedDate="2011-03-03T18:02:44.737" Title="Should physicists abandon Popperian falsification criterion?" Tags="&lt;soft-question&gt;" AnswerCount="3" CommentCount="9" />
  <row Id="6307" PostTypeId="2" ParentId="6306" CreationDate="2011-03-03T16:13:35.103" Score="1" ViewCount="" Body="&lt;p&gt;In my opinion no, they shouldn't. Just because things are more difficult and complicated it does not mean that scientists should &quot;cheat&quot;. That of course does not mean that they should abandon any idea which is not immediately falsifiable. Good ideas should be given time. If you abandon falsifiability and impose rigor then it becomes mathematics. If you don't impose rigor, well then it is &quot;theology&quot;.     &lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2011-03-03T16:13:35.103" CommentCount="5" />
  <row Id="6310" PostTypeId="2" ParentId="6306" CreationDate="2011-03-03T16:39:49.677" Score="7" ViewCount="" Body="&lt;p&gt;Popper's approach to the philosophy of science is of some historical importance, and quite possibly still has some value today, but it's certainly not holy writ. I think that many scientists and philosophers of science have moved beyond Popperianism, and that's fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for what criterion should replace it, my vote is for Bayesian inference. The central idea of science is that evidence matters, and specifically that one must modify one's assessment of the probability of a statement's truth in the light of each additional piece of evidence. A &quot;scientific&quot; statement is one whose probability can be driven arbitrarily close to 1 or to 0 by the accumulation of appropriate evidence. Bayesian inference is the systematic way to think about evidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the statement &quot;there is life on other planets&quot; is scientific, because we can imagine ways of gathering evidence that could support or oppose that statement in such a way as to drive its probability up to nearly 1 or down to nearly 0. Note that this statement is not &quot;falsifiable,&quot; though, since we could never exhaustively check all planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote a longer version of this rant &lt;a href=&quot;http://blog.richmond.edu/physicsbunn/2009/01/22/why-i-am-not-a-popperian/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; quite a while ago. Just to be clear, I'm not trained in philosophy, and my ideas are no doubt quite naive philosophically. In particular, I've never read Popper, so when I criticize Popperianism, I'm really criticizing the cartoon view of it that one seems to encounter among scientists. &lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-03T16:39:49.677" CommentCount="1" />
  <row Id="6311" PostTypeId="2" ParentId="776" CreationDate="2011-03-03T16:59:58.270" Score="-1" ViewCount="" Body="&lt;p&gt;The most likely answer is that probably Jupiter is already ignited: it emits lots of infrared radiation that as far as i know, its largely unexplained. It just doesn't have enough mass to radiate more energy than a extremely dim brown dwarf, so it naively looks to us as a planet&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-03-03T16:59:58.270" CommentCount="1" />
  <row Id="6312" PostTypeId="1" AcceptedAnswerId="6314" CreationDate="2011-03-03T17:13:51.317" Score="5" ViewCount="260" Body="&lt;p&gt;Stop me if i'm wrong but displacement is, for example, when you enter a bathtub the water level rises around you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens when displacement isnt possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll give an example. I have two blocks off steel, one at the top of a chamber and one at the bottom. there is a gap between the two blocks that is filled with water. There is NO gap between the two blocks and the side of the chamber (the water cannot escape)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens when the two block come together, in effect squashing the water? where does the water go?&lt;/p&gt;&#xA;" OwnerUserId="2240" LastActivityDate="2011-03-05T02:24:02.297" Title="When displacement isn't possible" Tags="&lt;forces&gt;&lt;water&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="6313" PostTypeId="2" ParentId="6312" CreationDate="2011-03-03T17:27:41.257" Score="5" ViewCount="" Body="&lt;p&gt;The short answer is that it will compress (increase in density) just as a gas would. But the &lt;a href=&quot;http://en.wikipedia.org/wiki/Properties_of_water#Compressibility&quot; rel=&quot;nofollow&quot;&gt;amount of pressure required is enormous&lt;/a&gt; since most liquids have so small a &lt;a href=&quot;http://en.wikipedia.org/wiki/Compressibility&quot; rel=&quot;nofollow&quot;&gt;compressibility&lt;/a&gt;, that they are for all practical purposes considered to be &lt;a href=&quot;http://en.wikipedia.org/wiki/Incompressible_flow&quot; rel=&quot;nofollow&quot;&gt;incompressible&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Properties_of_water#Compressibility&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; gives more details about the behaviour of compressibility at different temperatures and pressures.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-03-03T18:08:16.340" LastActivityDate="2011-03-03T18:08:16.340" />
  <row Id="6314" PostTypeId="2" ParentId="6312" CreationDate="2011-03-03T17:28:02.130" Score="10" ViewCount="" Body="&lt;p&gt;The blocks of steel and layer of water will just sit there like a sandwich.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is nowhere for the water to go, the blocks of steel will not come together.  Instead, the water will compress slightly, increasing its pressure.  This induced pressure will support the top block of steel.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you make the top block of steel heavier and heavier, the pressure in the water will continually increase.  Eventually you'll see the layer of water shrinking.  The water isn't going anywhere; its density is increasing.  You'll see a significant effect when the pressure on the water (the weight of the steel block divided by its area) becomes on the order of the bulk modulus of water, about 2*10^9 Pascal or 20,000 atmospheres.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure at the bottom of a deep ocean trench is about 5% of this, so your slab of steel will need to be on the order of thousands of meters tall just to see a few percent change in the size of the water layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, you'll burst the sides of the container you're using, or a tiny gap will form.  Then water will come shooting out in a jet.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-03-03T17:28:02.130" CommentCount="3" />
  <row Id="6315" PostTypeId="2" ParentId="6312" CreationDate="2011-03-03T17:33:57.660" Score="1" ViewCount="" Body="&lt;p&gt;It goes through some phases where its temperature and pressure is increasing, until the water molecules break down, forming heavier elements, and eventually you have a black hole.&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-03-03T17:33:57.660" CommentCount="5" />
  <row Id="6316" PostTypeId="2" ParentId="6302" CreationDate="2011-03-03T17:58:25.377" Score="4" ViewCount="" Body="&lt;p&gt;The decay channels were discussed here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/3773/shape-of-the-higgs-branching-ratio-to-zz&quot;&gt;Shape of the Higgs branching ratio to ZZ&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Approximately, the Higgs decays to the heaviest pair of particles that is still consistent with energy conservation. The precise rates are calculable by the Standard Model - if you substitute the so-far unknown Higgs boson mass as input. Of course, if the world isn't described by the ordinary Standard Model, predictions may change even though (for the lightest Higgs, if one exists) the difference is usually not qualitative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It takes some time - more precisely, some number of collisions - for a collider to distinguish the decaying Higgses from other, Higgs-free processes which may produce the same final particles (the so-called background), to be sufficiently sure that some Higgses were produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The required number of collisions depends on the Higgs mass. You get graphs like this one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.symmetrymagazine.org/breaking/wp-content/uploads/2009/02/chance-of-higgs-discovery-at-tevatron.jpg&quot; rel=&quot;nofollow&quot;&gt;http://www.symmetrymagazine.org/breaking/wp-content/uploads/2009/02/chance-of-higgs-discovery-at-tevatron.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The higher the curve is in the $y$-direction for a given mass on the $x$-axis, the easier it is to find the Higgs if the mass is there, and also the easier it is to show that it's not there if it is not there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very light Higgs, near 115 GeV for example, is very hard to be found at the LHC. The work that the Tevatron has done - several inverse femtobarns - will not be surpassed soon. For any substantially higher masses, the LHC energetic advantage is so huge that the LHC, with a much smaller amount of collisions, has already done more to decide whether the Higgs is there than the Tevatron did, or it will soon be the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the whole intervals where the Higgs is conceivable, it should be found by the end of 2012 - at least at 4 sigma - which is why the run was extended so that the Higgs is &quot;almost certainly&quot; found before the upgrade.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-03T17:58:25.377" CommentCount="1" />
  <row Id="6317" PostTypeId="2" ParentId="6306" CreationDate="2011-03-03T18:51:29.407" Score="0" ViewCount="" Body="&lt;p&gt;Physics needs to act the correct way. Physics its important to humanity and can not be fragilized.&lt;br&gt;&#xA;First, I do not agree with the word 'brilliant', and this is only a matter of sensibility: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Many of the most brilliant ideas of&#xA;  modern physics are very much&#xA;  unpopperian.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In time theories may rise up or fall, give the authors time to elaborate, decades if we are willing to pay. Consensus must be avoided. I learn with discussions and past history of science. You can find your evidence, and myself mine, and both of us can be erroneous. I do not want to impose my viewpoint, even if I could, and I expect the same from the comunity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being 'unpopperian' is a problem? In the start of new ideas it is not a problem but, with the progressing of time, one must be prepared to make conclusions. &#xA;An example: SUSY is beautiful, and makes predictions. If experience do not verify the predictions, I can expect one of the two: assume and restart, inspite of the beauty (this is not a physical criterion), or much less correct (cheat is my thought): rephrase the theory and by any kind of ... push previsions to unreacheable ranges of energy, even in the long term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We must have clear what is a 'physical law' (relations obtained empirically and not subject to interpretation), written in stone, and a 'theory'  (a model to interpret the 'why'). Theories come and go, from time to time substituted by revolutions of thought.&lt;br&gt;&#xA;Please compreend that imposing such a thing as 'electrical monopoles' by convenience of some beauty elsewhere can be monstruous. Beauty is reduce the magnetic force to a derived force that vanishes in the appropriate reference frame. Simplify is better (or KISS rule, no offense, Keep It Simple Stupid).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imposing a criterion based on the statistics is ridiculous. This way computers can do the work and the results are written in stone. We'll know tomorrow, the same as today, with greater accuracy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Data should be interpreted because data by itself can be misleading: pls read&#xA;&lt;a href=&quot;http://en.wikisource.org/wiki/The_Measure_of_Time&quot; rel=&quot;nofollow&quot;&gt;Henri Poincaré &quot;The measure of Time&quot;&lt;/a&gt; or his &lt;a href=&quot;http://en.wikipedia.org/wiki/Space#Gauss_and_Poincar.C3.A9&quot; rel=&quot;nofollow&quot;&gt;sphere-world&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Sphere-world&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-03-03T18:51:29.407" />
  <row Id="6318" PostTypeId="2" ParentId="6312" CreationDate="2011-03-03T19:30:19.187" Score="8" ViewCount="" Body="&lt;p&gt;According to the phase diagram at &lt;a href=&quot;http://www.lsbu.ac.uk/water/phase.html&quot; rel=&quot;nofollow&quot;&gt;http://www.lsbu.ac.uk/water/phase.html&lt;/a&gt;, if you compress it slowly enough (so that the temperature remains constant at 300 K), the water will compress slightly and slightly more as you increase the pressure to about 1 GPa (10,000 atm), at which point it will have a density of about 1.18 g/cm$^3$ (18% greater than usual).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, if you continue to apply more pressure, the water will &lt;s&gt;suddenly solidify into&lt;/s&gt; begin forming a crystal known as ice VI. This does not have the same structure as ordinary ice (ice I), which makes sense because ordinary ice is less dense than water, so it will never be formed by compression. Ice VI, on the other hand, is more dense than water (about 1.31 g/cm$^3$), and that's what you get if you apply a pressure greater than 1 GPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you increase the pressure even more, the ice VI changes phase a few more times (forming ice VII, then ice X, it looks like), before getting to the limit of pressures available in the lab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you somehow managed to apply much, much, more pressure than this (for example by throwing the water into a neutron star), then kakemonsteret's answer would begin to apply. The water would first become electron-degenerate matter (like a white dwarf), then the nuclei would start to fuse and it would become like neutron star material. But the pressures required for this are many orders of magnitude higher than anything available outside of a compact star, and at such pressures the steel in your experiment would suffer a similar fate.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditorDisplayName="" LastEditDate="2011-03-05T02:24:02.297" LastActivityDate="2011-03-05T02:24:02.297" CommentCount="2" />
  <row Id="6319" PostTypeId="2" ParentId="6300" CreationDate="2011-03-03T19:59:18.217" Score="2" ViewCount="" Body="&lt;p&gt;You're definitely taking this picture too seriously. Hybridization of atomic orbitals is just an approximation based on an independent particle model created by Pauling to rationalize some structural trends in chemistry with Quantum Mechanics. We can only assign orbitals unambiguously for one-electron systems, though of course independent particle models were widely employed and still are to explain reactivity trends, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, if you prepare an electron in an sp3 state, which is an equally weighted linear combination of the px, py pz and s orbitals, the probability to find and electron at a certain angular momentum might be calculated by taking the square of the projection of this angular momentum on the wave function, as QM tells us to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't worry too much if you find weak spots in general theories developed to explain molecules, reactivity, etc, since they're highly approximated in most cases (unless you do an expensive computation for a system, but then it is not general anymore), and the guys that developed those such as Pauling already knew that. As you know these highly approximated theories (e.g., Molecular Orbital and Valence Bond theory) were and still are very successful, even if they're not very rigorous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It gets really hard to be rigorous in chemistry beyond a certain point...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@TedBunn&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the basic formalism of QM would predict properties of this hypothetical electronic state, I did not say anything that would imply the opposite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, to think about bonding in terms of sp3 hybrid orbitals such as one does in general chemistry is really a very crude picture, unless you're dealing with molecules that have very special properties such as Td symmetry (methane, for example), and still in those cases if you perform a calculation by using VB or MO (Hartree-Fock) theory (and it does not even need to be with a computer, since symmetry is going to make life very easy here) with only the SP3 orbitals of carbon and S of hydrogen in your set of basis functions, you will see a very big quantitative error in predicted ionization energies when you compare to photoelectron spectroscopy measurements of ionization energies. Another good test would be to perform high level quantum chemistry calculation of methane using a software and employing two different basis sets, one containing only s and p functions centered on carbon and s functions centered on hydrogen atoms, and another in which the basis set has functions of several angular momenta centered on each atom. If you compare your results with experiments you will see that the first method will have a much lower accuracy compared to the second one. S and P functions might still be the ones that contribute mostly to the bonding molecular orbitals in the different orbital configurations (set of occupied molecular orbitals) that contribute to a good description of methane, but we can only say that for sure for very special cases. When you go to molecules with very distinct geometries, it is vital to have d functions centered on carbon, for example, in order to determine even qualitatively right chemistry trends. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum chemistry semi-empirical methods widely used in the past often had this problem of only employing valence basis functions centered on an atom in molecule calculations, not giving enough flexibility for the description of molecular geometries, and therefore providing bad results whenever the geometry deviated too much from what the qualitative models (such as the hybridization model) predicted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, I'd really be surprised (but not too much) if any modern paper that discusses chemical bonding, or that qualitatively discusses bonding orbitals to explain computed trends, do that by employing hybridization theory arguments. Those that I have seen almost always do that by looking at molecular orbitals or some refined valence bond treatments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, although hybridization is a deep and important topic that every chemist should dominate, in the way it was developed by Pauling it has severe limitations that people are, or should be, aware of. &lt;/p&gt;&#xA;" OwnerUserId="1695" LastEditorUserId="1695" LastEditorDisplayName="" LastEditDate="2011-03-04T16:57:55.687" LastActivityDate="2011-03-04T16:57:55.687" CommentCount="1" />
  <row Id="6321" PostTypeId="1" CreationDate="2011-03-03T20:31:40.253" Score="6" ViewCount="236" Body="&lt;p&gt;As I hope is obvious to everyone reading this, the universe contains more matter than antimatter, presumably because of some slight asymmetry in the amounts of the two generated during the Big Bang. This raises the question of whether there are any processes short of the Big Bang that produce more matter than antimatter. That is, is there any known process where a particle collider (or whatever) would convert some energy into matter not through the production of particle-antiparticle pairs but through some process that produced more matter than antimatter? This doesn't need to be restricted to current accelerators-- if there's some mechanism for this that requires impractically high energies, but is based on solid theories (i.e., the Standard Model or straightforward extensions thereof), that would be interesting, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm fairly certain that the answer is &quot;no,&quot; because I know that the matter-antimatter asymmetry is related to CP violation, and I also know that existing measurements of CP violation are not enough to explain the asymmetry. If there were a known way to slam protons together and make more quarks than antiquarks, I wouldn't expect this to still be a mystery. My particle physics knowledge is far from comprehensive, though, so it can't hurt to ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I was briefly confused into thinking that there was such an experiment a while back, but it turned out to just be sloppiness about marking the antiquarks on the part of the people writing about it...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is another question prompted by the book-in-progress, on relativity, this time a single word: I wrote that matter created from energy in particle physics experiments is &quot;generally&quot; in the form of particle-antiparticle pairs. Then I started wondering whether that qualifier was really needed, and thus this question.)&lt;/p&gt;&#xA;" OwnerUserId="166" LastActivityDate="2011-03-05T16:48:48.603" Title="Matter-Antimatter Asymmetry in Experiments?" Tags="&lt;particle-physics&gt;&lt;experimental-physics&gt;&lt;theory&gt;&lt;antimatter&gt;&lt;cp-violation&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="6322" PostTypeId="2" ParentId="6321" CreationDate="2011-03-03T20:44:20.743" Score="7" ViewCount="" Body="&lt;p&gt;Dear Chad, you misinterpret the statement that &quot;the known sources of CP-violation are not enough to explain the matter-antimatter asymmetry in the Universe.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You seem to think that the statement means that the known CP-violating parameter (namely the CP-violating phase in the CKM matrix) and the processes based on it are qualitatively insufficient to produce matter-antimatter asymmetry. But they are just &lt;em&gt;quantitatively&lt;/em&gt; insufficient. One simply doesn't get enough of the asymmetry - but qualitatively, the CKM phase would be enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are additional conditions beyond the CP-violation that have (or had) to be satisfied for the Universe to produce matter-antimatter asymmetry. They're known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sakharov_conditions#Sakharov_conditions&quot; rel=&quot;nofollow&quot;&gt;Sakharov conditions&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;CP-violation as well as C-violation&lt;/li&gt;&#xA;&lt;li&gt;Violation of the conservation of the baryon number B (and/or lepton number L)&lt;/li&gt;&#xA;&lt;li&gt;Evolution away from the thermal equilibrium.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All of these &quot;violations&quot; have be present simultaneously to produce quarks and antiquarks asymmetrically. If one of them is absent, the processes remain matter-antimatter symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, lab experiments may deviate from thermal equilibrium but all lab experiments we can perform conserve the baryon number $B$ (as well as the lepton number $L$). That's why we can't imitate the matter-antimatter asymmetry in the lab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The attempted &quot;lab experiments&quot; violating $B$ are the proton decay experiments - those big reservoirs of pure water with sensitive detectors able to see every single proton decay. So far, none of them has been seen (even though the simplest grand unified theories predicted that the proton decay should have been observed rather quickly). For theoretical reasons, it still seems extremely likely that the proton is unstable (although its lifetime is longer than expected in the SU(5) GUT) and $B$ is not conserved. Consequently, $L$ is not conserved, either. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, black holes radiate the Hawking radiation away and the composition of the Hawking radiation carries $B=0$ in average because the event horizon looks the same regardless of the value of $B$ of the initial star that has collapsed into the black hole. This paragraph was meant to be a proof that locality implies that $B$ has to be violated in quantum gravity (or earlier, e.g. in the GUT theory) as long as there are no gauge fields associated with $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the combination $B-L$ may be in principle conserved - it may be a generator of a grand unified group. However, this symmetry is probably broken because there are no long-range forces acting on this combined charge. So all these charges unrelated to gauge symmetries have to be violated (non-conserved) at some level; this reflects the wisdom that quantum gravity doesn't allow any global symmetries. Any symmetry is either explicitly broken by some effects or it is a gauge symmetry.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-03T20:51:04.247" LastActivityDate="2011-03-03T20:51:04.247" />
  <row Id="6323" PostTypeId="2" ParentId="6321" CreationDate="2011-03-03T20:44:48.167" Score="7" ViewCount="" Body="&lt;p&gt;CP violation is necessary but not sufficient condition for matter-anti matter asymmetry. You also need Baryon number violation (sought after but not yet found e.g. in proton decay), and strong deviations from thermal equilibrium. Those three conditions are known as the Sakharov criteria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think it is a realistic goal to achieve all 3 of them in experiment. For example, the reason the standard model does not provide an explanation of the observed asymmetry is partially insufficient CP violation, and partially because the electroweak phase transition is not sufficiently non-equilibrium (i.e. it is second order, or weakly first order transition). By any reasonable criteria, this phase transition is an extremely violent process...&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-03-03T20:44:48.167" CommentCount="2" />
  <row Id="6324" PostTypeId="1" AcceptedAnswerId="6340" CreationDate="2011-03-03T23:00:41.587" Score="3" ViewCount="139" Body="&lt;p&gt;Ok, I was just curious but the electromagnetic force can allow paramagnetism macroscopically in some objects. Can this be done microscopically to the subatomic level? Also, what about other forces (besides gravity I'm guessing) meaning the weak and strong force? Strong force being the strongest, would its para- also be and always be stronger?&lt;/p&gt;&#xA;" OwnerUserId="2279" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-06-09T17:03:32.923" LastActivityDate="2011-06-09T17:03:32.923" Title="Paramagnetism what about Paraweakism or Parastrongism?" Tags="&lt;particle-physics&gt;&lt;forces&gt;&lt;subatomic&gt;" AnswerCount="1" />
  <row Id="6325" PostTypeId="2" ParentId="6299" CreationDate="2011-03-04T00:21:05.657" Score="0" ViewCount="" Body="&lt;p&gt;You could use a first order differentian equation, a Laplacian. You can choose to use x/y/z coordinates, or r/theta/phi depending on the shape of the tea pot will determine which one is simpler to use. I say use a laplacian because I assume the amount of tea in the water spreads out uniformly from the bag, so a tea density gradient would be formed. The solution to this however would likely be fairly complex (or at least not very simple) unless the container was spherical with the tea bag in the middle, and but since a tea pot is a crude approximation of this, it could be used to model it.&lt;/p&gt;&#xA;" OwnerUserId="2350" LastActivityDate="2011-03-04T00:21:05.657" CommentCount="0" />
  <row Id="6326" PostTypeId="2" ParentId="5804" CreationDate="2011-03-04T00:48:36.117" Score="1" ViewCount="" Body="&lt;p&gt;I think I can answer this in more laymen's terms. In relativity theory, space and time are very closely related. For example, when you pass by me, I measure that your wristwatch runs slower than mine, and I measure that you are shorter in your direction of motion than you would be if you stopped to talk to me. Even at walking speed this is true, if only undetectably. From your perspective your wristwatch always runs at its normal rate, and your length in your direction of motion isn't affected by your motion relative to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In relativity theory an observed distortion of time always goes hand-in-hand with an observed distortion of space (length), and by the exact same percentage, so it became convenient to consider that those two concepts are two sides of a coin, so to speak, hence the term &lt;em&gt;spacetime&lt;/em&gt;. It also became convenient to think of observed objects and events in terms of 4 dimensions, the 3 of space and one of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course that's a rough way to put it, but I think it's not much more involved than that. So no, the theory doesn't support more than one time dimension.&lt;/p&gt;&#xA;" OwnerUserId="2314" LastActivityDate="2011-03-04T00:48:36.117" />
  <row Id="6327" PostTypeId="2" ParentId="6086" CreationDate="2011-03-04T01:16:06.583" Score="4" ViewCount="" Body="&lt;p&gt;You're choosing a &quot;freely falling&quot; inertial frame. There's a natural set of coordinates for a non rotating black hole for this called &quot;Gullstrand-Painleve&quot; coordinates. They correspond to the natural coordinates for a particle falling into a black hole from infinity. See the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In these coordinates, the speed of light is different for light trying to move away from the black than light moving towards it. As the little patch enters the black hole, the speed of light moving away from the black hole becomes negative, that is, even light moving away from the black hole still gets sucked into the singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A well written, sort-of introductory and very intuitive paper you might find enlightening on these coordinates, and their generalization to a rotating and/or charged black hole, is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am.J.Phys.76:519-532,2008, Andrew J. S. Hamilton, Jason P. Lisle , &lt;em&gt;The River Model of Black Holes&lt;/em&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0411060&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/gr-qc/0411060&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-04T01:16:06.583" CommentCount="10" />
  <row Id="6328" PostTypeId="2" ParentId="5124" CreationDate="2011-03-04T01:28:00.997" Score="0" ViewCount="" Body="&lt;p&gt;The cyclic action stems from the role of the $3$- form of $G_2$ on a seven dimensional space. In general we may consider the manifold $M^4\times M^7$, where $M^4$ may be any spacetime $(M, g)$, which for explicit purposes is $AdS_4$. The “dual system” with $AdS_7$ can follow accordingly.   A four dimensional field is constructed which for $N$-Killing spinors is $N$ supersymmetric for a $G_2$ holonomy on $M^7$. The exceptional group $G_2$ is the automorphism group of the octonions $o~=~x_0I~+`x_ae_a$ for the basis elements $e_a$  obeying the algebra&#xA;$$&#xD;&#xA;e_ae_b~=~-\delta_{ab}~+~\omega_{abc} e^c&#xD;&#xA;$$&#xA;where the tensor $\Omega_{abc}$ is determined by products of three octonionic elements which are $G_2$ invariant. This is the tensor component of a three-form $\Omega$ which is expanded according to elements on the $M^7$ as&#xA;$$&#xD;&#xA;\Omega~=~{1\over 3!}\omega^{abc}e_a\wedge e_b\wedge e_c&#xD;&#xA;$$&#xA;$$&#xD;&#xA;=~e_1\wedge e_2\wedge e_3~+~e_4\wedge e_3\wedge e_5~+~e_5\wedge e_1\wedge e_6~+~e_6\wedge e_2\wedge e_4~+~e_4\wedge e_7\wedge e_1~+~e_5\wedge e_7\wedge e_2~+~e_6\wedge e_7\wedge e_3 &#xD;&#xA;$$&#xA;$$&#xD;&#xA;=~e_1\wedge e_2\wedge e_3~+~(1/2)e_i\wedge e_m\wedge J_{i mn}e_n &#xD;&#xA;$$&#xA;so the spin tensor $J_{i mn}$ has the element $i~=~1, 2, 3$ and  $m,~n~=~ 4, 5, 6, 7$.  The second line of this equation is equivalent to the projective Fano plane, or the multiplication rule for octonions.  This is an aspect of the alternativity of the octonions which define triplets of quaternions as seen in the index $i$.. The product of the octonionic elements means the product of spin tensors obeys &#xA;$$&#xD;&#xA;J_i\cdot J_j~=~-\delta_{ij}~+~\epsilon_{ijk}J_k. &#xD;&#xA;$$&#xA;By definition the $G_2$ holonomy means the $3$-form is closed $d\Omega~=~0$, and the Hopf fibration  $S^3~\hookrightarrow~S^7~\rightarrow~S^4$ induces a symmetry between elements in seven dimensions so $d*\Omega~=~0$. In addition for the spin connection $\sigma^{ab}$, the projection with the tensor is zero $\omega_{abc}\sigma^{ab}~=~0$. This means that $\Omega$ is covariantly constant, which is a condition it being a Killing spinor. This gives a set of first order differential equations for the metric elements, The existence of additional covariantly constant field-form restricts the $G_2$ holonomy so the Killing spinor equation has more than one solution and the $4$ dimensional field theory has extended $N~&amp;gt;~1$ supersymmetry&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-04T01:28:00.997" />
  <row Id="6329" PostTypeId="1" AcceptedAnswerId="6331" CreationDate="2011-03-04T01:36:03.170" Score="1" ViewCount="257" Body="&lt;p&gt;&quot;Heat rises&quot; is a widely-used phrase (and widely-accepted phenomenon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does hot air really rise?  Or is it simply displaced by colder (denser) air pulled down by gravity?&lt;/p&gt;&#xA;" OwnerUserId="2375" LastActivityDate="2011-03-04T15:10:00.687" Title="Does hot air really rise?" Tags="&lt;thermodynamics&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="6330" PostTypeId="1" AcceptedAnswerId="6334" CreationDate="2011-03-04T01:53:57.180" Score="1" ViewCount="100" Body="&lt;p&gt;Normal-density materials have internal energy, which is the sum of the average energies associated to each of the degrees of freedom. Degrees of freedom can be described as vibrational, translational, and rotational. If you compress this material in such a way that matter cannot escape, you can detect energy escaping in the form of heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine an extremely high-density material such as a neutron star or black hole. I suspect it's not possible to describe the internal energy of (or anything about) an individual atom inside a black hole, since it's impossible to inspect beyond the event horizon, and since I assume that atoms don't exist in an distinguishable form there. But presumably the atoms &lt;em&gt;had&lt;/em&gt; energy when they became part of the thing. How is that energy accounted for?&lt;/p&gt;&#xA;" OwnerUserId="2374" LastActivityDate="2011-03-04T02:48:43.403" Title="What happens to internal energy in high-density materials?" Tags="&lt;astrophysics&gt;&lt;thermodynamics&gt;&lt;soft-question&gt;" AnswerCount="2" />
  <row Id="6331" PostTypeId="2" ParentId="6329" CreationDate="2011-03-04T02:05:18.733" Score="6" ViewCount="" Body="&lt;p&gt;The mechanism responsible for the rising of hot air is floatation: Hot air is less dense than cold air and hence air pressure will exert an upwards force, in the same way air rises in water.&#xA;Now if cold air was magically unaffected by gravity, then it would not be able to exert pressure on the hot air and thus it would not rise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement that &quot;heat rises&quot;, by the way, is not universally true.&#xA;Look at water. Here, it is the cold water that is less dense than warm water (at least in the temperature regime of importance to freezing). In winter, when water gets colder, the cold water raises to the top and eventually will freeze, while the water below remains liquid for the moment.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-03-04T02:05:18.733" CommentCount="3" />
  <row Id="6332" PostTypeId="2" ParentId="6329" CreationDate="2011-03-04T02:06:57.717" Score="2" ViewCount="" Body="&lt;p&gt;Heat does only 1 thing in a closed system, and that is evenly distribute itself about the system as it reaches thermodynamic equilibrium. I dont think this is what you asking about though. I assume you are talking about hot air (hot being a relative term just meaning it is hotter then the surrounding air). This hot air will be less dense then the surrounding air, and will therefore want to be above the more dense, colder air. If you want to actualy see this, get a beaker of water and add some oil, this is same thing that happens with air (as both cases involve 2 liquids of different densities)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the question exactlty, hot air does rise, and it is also displaced by cold air (though often from the side, not directly above it). And yes, gravity is the reason less dense liquids like to sit on top of more dense liquids&lt;/p&gt;&#xA;" OwnerUserId="2350" LastActivityDate="2011-03-04T02:06:57.717" />
  <row Id="6333" PostTypeId="2" ParentId="6330" CreationDate="2011-03-04T02:39:02.760" Score="1" ViewCount="" Body="&lt;p&gt;It is &lt;em&gt;any kind&lt;/em&gt; of constituent particles (and not atoms as such) that move about. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In &lt;em&gt;&quot;ordinary&quot;&lt;/em&gt; gases, liquids, and solids it is the molecules and atoms which move about making up the internal energy.&lt;/li&gt;&#xA;&lt;li&gt;In a plasma that is ions and free electrons. &lt;/li&gt;&#xA;&lt;li&gt;In a degenerate gas (white dwarf material is electron degenerate and neutron star material is neutron degenerate), the bits continue to move, but their choices of energy levels are constrained by Fermi exclusion, resulting in a uniformly filled phase space up to some level called the &quot;Fermi energy&quot; or &quot;Fermi surface&quot;. Note the heavy nuclei are degenerate in protons and neutrons and they have internal degrees of freedom which can be excited, so we have first hand experience of this.&lt;/li&gt;&#xA;&lt;li&gt;In a quark--gluon plasma it is, well, quarks and gluons.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And it doesn't make sense to talk about blackholes having &quot;material&quot; at they represent a physical singularity and by definition we don't know the rule there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting side effect of this is that we have a ready made definition of temperature&#xA;for these systems in terms of the average energy of the internal modes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't talked about what happens at the transitions between these regimes because that can be pretty complicated, but one possible case is the existing constituents get torn apart (as in ionizing a gas).&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-03-04T02:47:40.397" LastActivityDate="2011-03-04T02:47:40.397" CommentCount="2" />
  <row Id="6334" PostTypeId="2" ParentId="6330" CreationDate="2011-03-04T02:48:43.403" Score="2" ViewCount="" Body="&lt;p&gt;The internal energy of a black hole is just its mass.  You can measure the mass of a black hole by its gravitational effects on outside bodies, and then extrapolate an equivalent energy using $E=mc^{2}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An isolated Neutron star also has a well-defined total mass, (the &lt;a href=&quot;http://en.wikipedia.org/wiki/ADM_energy&quot; rel=&quot;nofollow&quot;&gt;ADM mass&lt;/a&gt;), which can be used to define an internal energy.  In the case of a neutron star, since it has X numbers of neutrons, and baryon number is conserved in gravitational collapse, you can even divide the total energy by the number of baryons to get a 'total energy per baryon.'&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-03-04T02:48:43.403" CommentCount="1" />
  <row Id="6338" PostTypeId="2" ParentId="776" CreationDate="2011-03-04T03:45:35.790" Score="0" ViewCount="" Body="&lt;p&gt;This idea seems to crop up now and then.  I will have to confess the thought occurred to me as a kid in grade school.  The AC Clarke “2010” featured Jupiter being imploded by black monoliths and turning into a star.&#xA;Even if you get a fusion bomb into the core of Jupiter, which would be tough to do, the explosion would have no measurable effects on the surface.  There simply is insufficient pressure to fuse nuclei together.  You would need some sort of pressure producing system that is also stable.  You might consider putting a small black hole (an Earth mass BH or so) in the core of Jupiter, which might create a zone of imploding material that has nuclear fusion.  The rub with this idea is you also ultimately turn Jupiter into a 3-4 meter radius black hole.  A collection of Hawking radiating black holes might be better, but their masses would have to be tuned so they don’t absorb or lose mass.  This would mean turning Jupiter into a Hawking radiation source as a &quot;star.&quot;  That is not a stable system.  So to play it safe you tune the Hawking BHs to radiate away and then replenish them later.  Again, none of this is a small feat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the idea is to create a sort of mini-solar system with the various moons.  So even if you could by some means turn Jupiter on as a star the results after a few thousand years would be to turn the Jovian and Saturnian moons into liquid bodies.  These moons would not turn into Earth-like planets.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-04T03:45:35.790" />
  <row Id="6339" PostTypeId="1" CreationDate="2011-03-04T05:13:30.350" Score="3" ViewCount="237" Body="&lt;p&gt;&lt;em&gt;Very unintuitive observation:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I pour myself a Guinness and the bubbles in my glass seem to move down toward the bottom of the glass instead of rising directly to the top of the glass as foam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can this be explained? Why is it that I observe this behavior drinking Guinness and not other carbonated drinks? What system properties (ie, temperature, nature of the solute and solvent) would affect this behavior?&lt;/p&gt;&#xA;" OwnerUserId="173" LastActivityDate="2011-03-05T06:02:38.450" Title="Strange behavior and motion of bubbles in a glass of beer" Tags="&lt;fluid-dynamics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="6340" PostTypeId="2" ParentId="6324" CreationDate="2011-03-04T05:57:58.890" Score="3" ViewCount="" Body="&lt;p&gt;First, there is no universal inequality that would say that materials have to be &quot;paramagnets&quot;. The opposite effects imply that materials may also be &quot;diamagnets&quot; which means that they react oppositely to the magnetic field. I think that atoms and molecules are the smallest objects whose response to the magnetic field may be viewed as the microscopic cause of diamagnetism and paramagnetism. It usually boils down to the electrons' motion but one couldn't analyze the magnetic behavior for the electrons in isolation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, one can't talk about the reaction to uniform weak-magnetic and strong-magnetic fields because there aren't any. At least today, the electroweak symmetry is broken which means that the &quot;weak-electromagnetic waves&quot; are as short as the wavelength of the W-boson or Z-boson which is $10^{-18}$ meters or so; and they never quite move by the speed of light. No atom - and even no actual particle we know - may be squeezed to shorter distances or studied to shorter distances. This prevents us from talking about responses to uniform fields - simply because there are no uniform fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analogously, the &quot;strong electromagnetic fields&quot; are non-uniform and variable because of a different reason in this case: the equations describing them are non-linear. There may be domains of size $10^{-15}$ meters or so where the &quot;strong magnetic fields&quot; may be aligned. See&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/5932/is-the-color-gauge-group-spontaneously-broken-in-qcd-by-the-savvidy-vacuum&quot;&gt;Is the color gauge group spontaneously broken in QCD by the Savvidy vacuum?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But these domains are still too small to be studied as a uniform field in which other particles may peacefully inserted and analyzed. The strong and weak nuclear interactions create such a frantically changing environment that they're almost universally studied and described in terms of particle collisions. There are no peaceful macroscopic uniform fields; the latter is only possible for long-range forces such as electromagnetism and gravity.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-04T05:57:58.890" CommentCount="1" />
  <row Id="6341" PostTypeId="1" CreationDate="2011-03-04T06:52:17.580" Score="3" ViewCount="300" Body="&lt;p&gt;Say you have a bottle of water filled up to a height H.  A small hole is drilled in its side at a height d, so that water squirts out.  The squirting water travels in an arc as it falls, covering some horizontal distance S away from the bottle before it hits the table top that the bottle sits on.  Multi-part question to try to understand this completely:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;At what height d should the hole be placed so that its horizontal travel distance S is maximized?  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Will S or the optimum value of d depend on the hole diameter?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Does the answer change if you assume that the hole presents zero impedance to the water flow?  This might not make sense to ask, or it might be equivalent to asking what happens in the limit of a very large-sized hole (see #2).  Not sure.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks to anyone who can help!&lt;/p&gt;&#xA;" OwnerUserId="1743" LastActivityDate="2011-03-04T19:43:35.277" Title="How far will water squirt out from a hole in a can?" Tags="&lt;fluid-dynamics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="6342" PostTypeId="2" ParentId="6341" CreationDate="2011-03-04T07:27:40.103" Score="3" ViewCount="" Body="&lt;p&gt;If David's right, and this is a homework problem, I'll just give a few hints. Consider a hole at height d. Since there is a flow in the can, there is a streamline; draw it. Now apply Bernoulli's law to this streamline to find the velocity at the point of exit. Finally, using some Newtonian mechanics (parabolic flight etc.), calculate where the water lands on the ground.&#xA;Finally, think about the hole diameter after you've done the above steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[If it wasn't a homework problem or the above hints were not really useful, I'll be happy to elaborate.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I can't figure out all of your notation, but basically it's the correct idea (up to a prefactor, if I'm not mistaken). One thing that people in fluid dynamics tend to do, is to isolate the influence of gravity from pressure. The streamline runs from the top of the can (with P = atmospheric pressure = 0) to the side of the can, where it's also in contact with the air, so at the same pressure, P = 0. You then know that $v^2/2 + gz$ is constant along the streamline, where z is your vertical coordinate. By integrating, you find that $v = \sqrt(2g(H-d))$ (since on the fluid surface, the water has zero velocity). Indeed you find that $S \propto \sqrt{}[d(H-d)]$, which however is maximal at $d = H/2.$&#xA;In terms of hole diameter: we should question whether our assumptions on the flow have been correct. Implicitly, we've taken it to be steady, without divergence, rotation and viscosity. The viscosity one is the most dangerous, since a typical size of viscous flow is given by $\frac{\nu}{v},$ where $\nu$ is a liquid's viscosity. Depending on the liquid you put in and its height/velocity, you can find yourself in a viscous regime, which might alter the flow and violate our assumptions. &lt;/p&gt;&#xA;" OwnerUserId="1471" LastEditorUserId="1471" LastEditorDisplayName="" LastEditDate="2011-03-04T19:43:35.277" LastActivityDate="2011-03-04T19:43:35.277" CommentCount="2" />
  <row Id="6343" PostTypeId="2" ParentId="6339" CreationDate="2011-03-04T07:29:01.980" Score="14" ViewCount="" Body="&lt;p&gt;Yes, this is a real effect. In my hometown of Pilsen which gave the beer its name, every kid knows the physics because they're taught to drink and observe Pilsner Urquell, Gambrinus, and other top brands already when they enter the kindergarten. See e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.stanford.edu/group/Zarelab/guinness/StanfordReport.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.stanford.edu/group/Zarelab/guinness/StanfordReport.htm&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://www.stanford.edu/group/Zarelab/guinness/&quot; rel=&quot;nofollow&quot;&gt;http://www.stanford.edu/group/Zarelab/guinness/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The explanation is that the bubbles in the middle of the glass go up immediately - because it's easier for them to move if they're in the middle. But by going up, they're pushing the liquid in the upward direction, too. Obviously, the liquid has to return to the bottom in some way - to the place that was initially occupied by the rising bubble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the path through which the liquid is returning over there? Well, the path goes near the boundaries of the glass - near the glassy material itself - where the bubbles are easier to be seen. So the rising bubbles at the center create some circulation patterns that go in the opposite direction (down) near the glassy boundary and that's where the bubbles are very easy to be observed (they're not hiding behind other bubbles and opaque liquid). With some exaggeration, the circulation patterns may look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/xpdzK.jpg&quot; alt=&quot;Circulation in Guinness&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect doesn't last long because the bubbles ultimately achieve a higher speed than the circulation speed (note that the bubbles moving up also accelerate, at least for a while). After some time, the liquid is returning to the bottom in between the bubbles, more or less uniformly in the whole horizontal area of the glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that dark beers are more likely to be biased in the direction that the &quot;nearby bubbles, near the glass, are easier to be seen&quot; which is why the visual impression that &quot;the motion down prevails&quot; should be stronger for darker beers such as Guinness.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-04T10:02:43.220" LastActivityDate="2011-03-04T10:02:43.220" CommentCount="2" />
  <row Id="6344" PostTypeId="2" ParentId="6341" CreationDate="2011-03-04T10:41:57.690" Score="2" ViewCount="" Body="&lt;p&gt;Here is the classical apparatus to demonstrate the phenomenons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://shop.corex.de/Sekundarstufe/Physik/Demonstrations_Ger%C3%A4te/Mechanik_der_Fl%C3%BCssigkeiten/Hydrostatik/44660-Auslaufgef%C3%A4%C3%9F&quot; rel=&quot;nofollow&quot;&gt;http://shop.corex.de/Sekundarstufe/Physik/Demonstrations_Ger%C3%A4te/Mechanik_der_Fl%C3%BCssigkeiten/Hydrostatik/44660-Auslaufgef%C3%A4%C3%9F&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In old physics textbooks for high school a picture of the app in action was a must. &#xA;There were versions of this with a rather long narrow pipe coming from the cylinder near the bottom, to demonstrate the pulsating flow due to onset and switch off of turbulence. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-04T11:28:33.047" LastActivityDate="2011-03-04T11:28:33.047" />
  <row Id="6345" PostTypeId="1" CreationDate="2011-03-04T10:47:48.463" Score="2" ViewCount="222" Body="&lt;p&gt;Even if the double slit experiment gives interesting (weird) results, it only concludes that each photon interacts with itself after passing the two slits. I have been thinking about a different experimental setup, where you have two well defined light sources (with specific wave lengths and phase) but no slits. And now to my questions: Has anyone ever done such an experiment, and will there be an interference pattern on the wall?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer to the second question is &quot;no&quot;, light can not be a true wave - it only has some wavelike properties. But if it is &quot;yes&quot;, things become much more interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is an interference pattern on the wall, there has to be an interference pattern even if both light sources are emitting single photons at random, but as seldom as, say, once per minute. That in turn would mean that the photons know about each other, even if they are separated in time with several seconds, and the light sources are independent (not entangled).&lt;/p&gt;&#xA;" LastActivityDate="2011-03-05T11:03:07.453" Title="Do photons interact with each other, or with themselves only?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6346" PostTypeId="1" AcceptedAnswerId="6366" CreationDate="2011-03-04T11:30:54.073" Score="5" ViewCount="177" Body="&lt;p&gt;I'm interested in characterizing the behaviour of measures of quantum information in strongly correlated quantum field theories which admit a gravity dual description, e.g through AdS/CFT duality. In a more broad sense, my question points towards the following: in my limited experience, I use to accept that a duality means that we don't have any right to give a physical meaning to one part of the duality and use the other part as a mere computational tool. I would like to be illuminated by some experts.&lt;/p&gt;&#xA;" OwnerUserId="2237" LastActivityDate="2011-03-04T20:29:38.723" Title="Can the study of the quantum information structure in QFT with holographic duals be relevant to string theory?" Tags="&lt;theory&gt;&lt;ads-cft&gt;&lt;string&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" />
  <row Id="6347" PostTypeId="2" ParentId="6329" CreationDate="2011-03-04T11:41:57.607" Score="0" ViewCount="" Body="&lt;p&gt;The Action force is simultaneous with the Reaction force. One can not happen without the other. &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Archimedes&quot; rel=&quot;nofollow&quot;&gt;Archimedes&lt;/a&gt; settled that a less dense fluid move on top of a denser fluid (&lt;a href=&quot;http://en.wikipedia.org/wiki/Buoyancy&quot; rel=&quot;nofollow&quot;&gt;see &lt;strong&gt;Buoyancy - Archimedes' principle&lt;/a&gt;)&lt;/strong&gt; .&lt;br&gt;&#xA;(and vice-versa: a denser..moves..to bottom..)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Rayleigh%E2%80%93Taylor_instability&quot; rel=&quot;nofollow&quot;&gt;Rayleigh-Taylor instability&lt;/a&gt; describes the evolution of the interface between the two layers. The atomic bomb mushroom cap is due to this effect.&#xA;&lt;img src=&quot;http://i.stack.imgur.com/9Ifvb.jpg&quot; alt=&quot;atomic bomb mushroom cap&quot;&gt;(quoting Wikipedia):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;RT..is an instability of an interface between two fluids of different densities, which occurs when the lighter fluid is pushing the heavier fluid. This is the case with an interstellar cloud and shock system. The equivalent situation occurs when gravity is acting on two fluids of different density — with the dense fluid above a fluid of lesser density — such as water balancing on light oil.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-03-04T11:41:57.607" />
  <row Id="6348" PostTypeId="1" CreationDate="2011-03-04T12:55:48.230" Score="1" ViewCount="172" Body="&lt;p&gt;If double slit experiment is done in a environment with air, then slits could also contain air made up of (approx. 80% nitrogen, 20% oxygen), then there is not empty space inside the slit.  How can it affect the experiment? Has it been taken in account?&#xA;What are specific requirement to define the slit? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were not that way. Why it doesn't affect the measurement?&lt;/p&gt;&#xA;" OwnerUserId="1916" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-03-13T20:16:41.800" LastActivityDate="2011-03-13T20:16:41.800" Title="What's inside the slit in double slit experiment?" Tags="&lt;optics&gt;&lt;experimental-physics&gt;&lt;interference&gt;&lt;double-slit-experiment&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6349" PostTypeId="2" ParentId="6348" CreationDate="2011-03-04T13:19:03.577" Score="2" ViewCount="" Body="&lt;p&gt;Air is everywhere, not only in slit, but also in the optical path before and after the slit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, interaction with transparent media (e.g. air, water, glass) can be easily included: by use of refractive index $n$. Then you know that wavelength in medium is&#xA;$$\lambda = \frac{\lambda_0}{n},$$&#xA;where $\lambda_0$ is the wavelength in the vacuum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If in the slit there were glass - still it is easy to calculate the outcome. (Just you need to include the phase shift caused by it.)&lt;/p&gt;&#xA;" OwnerUserId="184" LastActivityDate="2011-03-04T13:19:03.577" CommentCount="1" />
  <row Id="6350" PostTypeId="1" CreationDate="2011-03-04T13:19:50.683" Score="1" ViewCount="131" Body="&lt;p&gt;I have a doubt about how double slit experiment is made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's think about the perforated wall, what are the requirement for it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can a photographic plate could be used as a wall ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see a problem here, as a photographic plate serve also as a detector, then the single photon experiment could end up with a single point in &lt;em&gt;the wall that contains the slits&lt;/em&gt; and not in the plate located &lt;em&gt;behind the slit's wall.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the &quot;one photon count&quot;, is knowing the emition before the slits (if this count were made after, it will be a detector making a measurement and would simply reflect the particle effect)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wall with the slits could absorb the photon(removing electrons from the material) before something happen on the other side, if that happened, there would be no experiment, the photon won't pass through the slits, just because its energy was already used (supposing all energy is used and the photon doesn't split).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There must be some kind of specification about the material of the wall, need to be special, in the sense of needing a higher energy (than the test photon) to pickup electrons, then the light won't stop there, for that, I think the wall can't be made of a photosensible material, but I don't know, and here I am asking..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks for any answer&lt;/p&gt;&#xA;" OwnerUserId="1916" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-03-13T20:16:36.387" LastActivityDate="2011-03-13T20:16:36.387" Title="Double slit experiment  and perforated walls' properties" Tags="&lt;optics&gt;&lt;experimental-physics&gt;&lt;interference&gt;&lt;double-slit-experiment&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="6351" PostTypeId="2" ParentId="6348" CreationDate="2011-03-04T13:21:19.357" Score="1" ViewCount="" Body="&lt;p&gt;The air will simply introduce a small amount of additional noise into the experiment due to scattering. The majority of the photons will not scatter off the air molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The optical path lengths will also be changed, but not by much. For example, the troughs in the interference pattern will occur when the optical path lengths of the paths to that point are out of phase. Since the optical path lengths in vacuum and air differ by a factor of $n_\mathrm{Air} = 1.00027$ for an arbitrarily chosen visible wavelength of 500 nm, the air has hardly any effect.&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2011-03-04T13:21:19.357" />
  <row Id="6352" PostTypeId="2" ParentId="6350" CreationDate="2011-03-04T13:27:05.423" Score="5" ViewCount="" Body="&lt;p&gt;It doesn't matter whether the wall is photosensitive or not. If the photons' wave functions give some probability of hitting the wall and being absorbed, then some of the photons will hit the wall and be absorbed, whether they produce some chemical reaction in a photographic plate or just heat the wall.&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2011-03-04T13:27:05.423" CommentCount="2" />
  <row Id="6353" PostTypeId="2" ParentId="6300" CreationDate="2011-03-04T13:30:15.437" Score="1" ViewCount="" Body="&lt;p&gt;In the Hydrogen atom $L^2$ commutes with the Hamiltonian so that an eigenstate of $H$ can be at the same time an eigenstate of $L^2$. Additionally, the energy does not depend on the quantum number $l$, so that a superposition of s and p orbitals is still an eigenstate of $H$ with not well-defined angular momentum. In a sp$^3$ state if you measure $L^2$ you will observe $l=0$ 25% of the times, and $l=\sqrt{2}\hbar$ on 75% of the times, as you suggested. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, in a molecule, neither $l_i^2$ (the orbital angular momentum of a single electron) nor $L^2$ (the total orbital angular momentum) commute with $H$. (Molecules are not spherical). You typically write your energy eigenstate as an antisymmetrized product of orbitals, which are eigenstates of the Fock operator (the self-consistent one-electron Hamiltonian). Additionally, each orbital can be written as a linear combination of whatever basis (e.g. atomic orbitals, as is usually done in Chemistry). Obviously $l_i^2$ is not well defined for this orbital, nor it shouldn't since you want to obtain faithful representations of the energy eigenstates, nor of the angular momentum operator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you still want to measure the orbital angular momentum of a single electron in the molecule, you first need to ionize an electron. The angular momentum will depend on the dynamics that led to the ionization. You may still ask what are the angular momentum components of whatever orbital of the molecule, although this information is not physical (I believe there can not be any measurement associated to it as a matter of principle). Then you need to choose an axis first (e.g. whatever bond axis, etc.) There is nothing terribly difficult in doing this, but usually you don't get much interesting information, expect perhaps in diatomic molecules or in studying molecules that are breaking along a particular bond.&lt;/p&gt;&#xA;" OwnerUserId="2379" LastEditorUserId="2379" LastEditorDisplayName="" LastEditDate="2011-03-04T13:46:04.020" LastActivityDate="2011-03-04T13:46:04.020" />
  <row Id="6354" PostTypeId="2" ParentId="6350" CreationDate="2011-03-04T13:40:05.120" Score="1" ViewCount="" Body="&lt;p&gt;1.&#xA;Of course wall sometimes absorbs the photon (light do not evade it in any way). To have interference, you need to have no information (in terms of excitation, creation of photographic grain of whatever) that could be used to say which path the photon 'has chosen'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.&#xA;It is fine to use foil. See e.g. &lt;a href=&quot;http://ultrafast.fuw.edu.pl/publications/ajp_2008.pdf&quot; rel=&quot;nofollow&quot;&gt;http://ultrafast.fuw.edu.pl/publications/ajp_2008.pdf&lt;/a&gt; (so called Fresnel Zone Plates on film or polygraphic foil). Want to test it by yourself? Make an analog photography of two dark lines, and then use film as a diffractive agent. Or - in any DTP studio post a PostScript file e.g. from &lt;a href=&quot;http://migdal.wikidot.com/prazki-moire&quot; rel=&quot;nofollow&quot;&gt;http://migdal.wikidot.com/prazki-moire&lt;/a&gt; -&gt; (don't be afraid it is in Polish; just download a few first ps/pdf files). &lt;/p&gt;&#xA;" OwnerUserId="184" LastEditorUserId="184" LastEditorDisplayName="" LastEditDate="2011-03-04T13:45:17.860" LastActivityDate="2011-03-04T13:45:17.860" CommentCount="1" />
  <row Id="6355" PostTypeId="2" ParentId="6346" CreationDate="2011-03-04T14:11:01.907" Score="0" ViewCount="" Body="&lt;p&gt;The holographic principle in general gives an equivalency between gravity in the interior of a space and the quantum fields on the boundary.  So in the case of the $AdS_n$ its boundary $\partial AdS_n~\simeq~CFT_{n-1}$ is equivalent to a conformal field of one dimension lower.  This is a form of the black hole holographic principle, where the field configuration of a black hole is defined by strings on the horizon.  This means the number of pieces of information required to describe physics in dimensions $N$ is reduced to $N~-~1$ dimensional description.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-04T14:11:01.907" CommentCount="1" />
  <row Id="6356" PostTypeId="2" ParentId="6208" CreationDate="2011-03-04T14:15:41.497" Score="2" ViewCount="" Body="&lt;p&gt;Chemistry is big. Probably your interests are not that wide. For instance one may be interested in molecules/molecular processes such as chemical reactions or in soft stuff (polymers, colloids, membranes, etc.) One may seek for microscopic interpretations of the phenomena or just try to find how complex processes depend on macroscopic variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, without knowing any details, my bet is that as a physicist, if you master Quantum Mechanics, Statistical Mechanics and Thermodynamics, then probably what you would like to read is a general textbook on Physical Chemistry (the application of physics to chemical problems) rather than a book on General Chemistry. The topics mostly overlap except for a lot of descriptive material that you will be happy to skip, probably, but the P-Chem book will be more focused and quantitative while using an understandable language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a particular suggestion, you may try the not-very-well-known book &quot;Principles of Physical Chemistry&quot; by Kuhn and Försterling, rather than the best known P-Chem books of Atkins, Levine and McQuarrie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other books that are worth of looking at in my opinion are: Engel &amp;amp; Reid, Physical Chemistry, and Berry, Rice and Ross with the same title (this one might be a bit too verbose).&lt;/p&gt;&#xA;" OwnerUserId="2379" LastEditorUserId="2379" LastEditorDisplayName="" LastEditDate="2011-03-04T14:31:18.240" LastActivityDate="2011-03-04T14:31:18.240" CommunityOwnedDate="2011-03-04T18:19:18.983" />
  <row Id="6357" PostTypeId="2" ParentId="6329" CreationDate="2011-03-04T15:10:00.687" Score="1" ViewCount="" Body="&lt;p&gt;Another way to think about it is to look at how pressure changes with height. If we place a high density box shaped parcel of fluid immediately next to a lower density fluid parcel, the hydrostatic pressure gradient is greater in the former parcel. So, if say the average pressure of the two parcels is the same, the denser one will have higher pressure at the bottom than the lighter one, and lower pressure than the lighter one at the parcel tops. So the denser parcel will tend to push in at the bottom, and be displaced at the top. To a first order approximation the will rotate, trying to put the lighter parcel on top. If the fluid parcels are of identical composition then the warmer one will be lighter. Of course we a temperature regime in water, where the density versus temperature curve runs backwars between roughly 0C to 4C, and ice is lighter still. But in general warmer fluid is lighter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In anycase the original question is rhetorical. Do we take the mental shortcut and think in terms of bounancy as a lifting force, or try to be more precise and consider the fluids interaction as the cause. In most cases, I'd prefer the former methodology, as that makes it easier to formulate the dynamics.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-03-04T15:10:00.687" />
  <row Id="6358" PostTypeId="2" ParentId="6345" CreationDate="2011-03-04T15:43:32.063" Score="1" ViewCount="" Body="&lt;p&gt;I am answering to the second part of your question, since the first has been covered two days ago. Have a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Optical_lattice&quot; rel=&quot;nofollow&quot;&gt;interfering laser beams&lt;/a&gt;, and read up on &lt;a href=&quot;http://en.wikipedia.org/wiki/Holography&quot; rel=&quot;nofollow&quot;&gt;holography&lt;/a&gt;, the ultimate in interfering beams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;even if both light sources are emitting single photons at random, but as seldom as, say, once per minute.&lt;/em&gt;&#xA;This is wrong. For interference to happen the phases must be kept. Individual photons do not keep the phase with another photon a minute later, so no interference pattern will appear. In contrast to the double slit experiment, where each photon interacts with the slits, in your two beam experiment each photon from one beam has to interact with a photon from the other beam for an interference pattern to appear. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-04T15:48:53.133" LastActivityDate="2011-03-04T15:48:53.133" CommentCount="7" />
  <row Id="6359" PostTypeId="1" CreationDate="2011-03-04T16:36:04.607" Score="1" ViewCount="233" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/76/physics-humor-jokes-and-banter&quot;&gt;Physics Humor - jokes and banter&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Tell me jokes about engineers, physicists and mathematicians.&lt;/p&gt;&#xA;" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-04T18:20:19.000" LastActivityDate="2011-03-04T18:20:19.000" ClosedDate="2011-03-04T18:20:05.317" Title="Jokes about engineers, physicists and mathematicians" Tags="&lt;humor&gt;" CommentCount="4" />
  <row Id="6360" PostTypeId="2" ParentId="5955" CreationDate="2011-03-04T16:55:39.830" Score="2" ViewCount="" Body="&lt;p&gt;Here's my stab at answering your question. We have a supersymmetric quantum field theory (any number of dimensions or supersymmetry generators). In general, it's not a superconformal theory, but we can still analyze it using the AdS/CFT correspondence. To be sure, the geometry deviates from pure anti de Sitter. The couplings characterizing the superpotential coefficients form bulk short supermultiplets. The couplings characterizing the K&amp;aumlaut;hler potential coefficients form bulk long supermultiplets. Being supersymmetric, the dual AdS theory has to be in a BPS configuration. This constrains the VEV of bulk short supermultiplets to vary as a power law in the AdS coordinate $z$ with the exponent fixed. Zero beta function here according to the holographic renormalization group. BPS doesn't constrain the $z$-dependence of the VEV of long multiplets. Nonzero beta functions are permissible.&lt;/p&gt;&#xA;" OwnerUserId="2381" LastActivityDate="2011-03-04T16:55:39.830" />
  <row Id="6361" PostTypeId="2" ParentId="3309" CreationDate="2011-03-04T17:14:49.693" Score="3" ViewCount="" Body="&lt;p&gt;Recall that the LIGO noise limit in some frequency bands is given by &lt;em&gt;the Heisenberg position uncertainty of the two multi kg single crystal sapphire mirrors&lt;/em&gt;, apparently about 10^-18m.  In other words, to lower the noise, the designers are planning to increase the mass of the mirrors from 10 to 40 kg apiece!  This is different from the radiation pressure uncertainties which apparently dominate the error budget in other frequency bands.   See e.g. &lt;a href=&quot;http://elmer.tapir.caltech.edu/cajagwr/pdf/chen.pdf&quot; rel=&quot;nofollow&quot;&gt;elmer.tapir.caltech.edu/cajagwr/pdf/chen.pdf&lt;/a&gt;, for several examples of entanglement, squeezing, etc.   Check out the first slide for a photo of one of the 10 kg mirrors.   To see Heisenberg position uncertainty in an object of this scale is, to me, somehow truly inspiring.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2011-03-04T17:26:15.517" LastActivityDate="2011-03-04T17:26:15.517" />
  <row Id="6362" PostTypeId="2" ParentId="296" CreationDate="2011-03-04T17:58:34.677" Score="-1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Usually, but not always, the Energy is a conserved quantity&lt;/strong&gt; as the others answers have explained.&lt;br&gt;&#xA;But the following clarification is important: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the BB framework where space expands, or in the dual  'shrinking matter' (almost dual)(comoving framework), &lt;strong&gt;the ratio matter/space is not invariant and energy is not conserved&lt;/strong&gt;, i.e.  Nöther's Theorem does not apply. &#xA;It is well known that photons loose energy as they propagate. I can not find an argument to explain why particles shoud not (as they are also matter waves) loose energy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other relevant point is that &lt;strong&gt;energy can be destroyed,&lt;/strong&gt; canceled, annihilated, &lt;strong&gt;already proved by experiment&lt;/strong&gt; as described here: &lt;a href=&quot;http://www.wired.com/wiredscience/2011/02/real-live-antilaser/&quot; rel=&quot;nofollow&quot;&gt;real-live-antilaser&lt;/a&gt;, &lt;a href=&quot;http://www.eng.yale.edu/caolab/papers/Sci11b.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt;, and &lt;a href=&quot;http://physics.stackexchange.com/questions/5743/anti-laser-how-sure-we-are-that-energy-is-transported&quot;&gt;here a tentative discusion&lt;/a&gt;.  Other example: what is the energy radiated by two dipoles centered in the same frequency and in phase opposition ? zero. The same happens with two photons in similar conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;added:&lt;/strong&gt;&lt;br&gt;&#xA;It is well known that the cosmological redshift of light is usually interpreted(*) as a decrease in energy because the photons wavelength is increasing with time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations for the interference of parallel polarized light are :&#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/4862/interference-of-polarized-light/&quot;&gt;see Kostya answer here&lt;/a&gt; and substituting Delta by Pi:&lt;br&gt;&#xA;$\vec{E}=iE_{0}(cos(wt)+cos(wt+\pi))=0$ ; $I\vec{=|\vec{E|}^{2}}=0$.&lt;br&gt;&#xA;Light cancel when the instantaneous &lt;strong&gt;sum&lt;/strong&gt; of the E and B vector field components equals 0. A known fact long time ago (since Maxwell ?). (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Superposition_principle#Wave_interference&quot; rel=&quot;nofollow&quot;&gt;superposition principle&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Interference_%28wave_propagation%29#Constructive_and_destructive_interference&quot; rel=&quot;nofollow&quot;&gt;interference&lt;/a&gt; )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Another two situations&lt;/strong&gt; that should make us think on energy conservation:&lt;br&gt;&#xA;&lt;strong&gt;accelerated charges radiate&lt;/strong&gt; (&lt;a href=&quot;http://www.mathpages.com/home/kmath528/kmath528.htm&quot; rel=&quot;nofollow&quot;&gt;discussion at matpages&lt;/a&gt;)&lt;br&gt;&#xA;&lt;strong&gt;moving bodies&lt;/strong&gt; in a gravitational field radiate &lt;strong&gt;gravitational waves&lt;/strong&gt; (see MotionMountain free e-book, ch 18-Motion in General Relativity)&lt;br&gt;&#xA;(*) I do not share the usual interpretation, but that one is the official.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can not remmember of any other situation where energy is not conserved.&#xA;I'm not including Dark Energy possible issues (&lt;a href=&quot;http://blogs.discovermagazine.com/cosmicvariance/2010/02/22/energy-is-not-conserved/&quot; rel=&quot;nofollow&quot;&gt;see CosmicVariance&lt;/a&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those four exceptional cases should make us think about our concepts. &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-03-05T15:27:38.310" LastActivityDate="2011-03-05T15:27:38.310" CommentCount="2" />
  <row Id="6363" PostTypeId="1" CreationDate="2011-03-04T18:03:01.700" Score="3" ViewCount="139" Body="&lt;p&gt;In flux compactifications to 4D, e.g. Type IIB on a CY orientifold $X$, one uses fluxes to stabilize the axio-dilaton $\tau$ and the complex structure moduli $z_a$ - the periods of the holomorphic three-form $\Omega$ over the basis three-cycles: $z_a=\int_{\alpha_a}\Omega$, by solving a system of equations for a supersymmetric extremum: $\partial_{\tau}W+W\partial_{\tau}K=0$, $\partial_{z_a}W+W\partial_{z_a}K=0$, where $W$ is the flux superpotential $W=\int_X G_3\wedge\Omega$ and $K$ is the Kahler potential. Since in Type IIB the supergravity scalar potential has a no-scale structure, i.e. $V=e^KD^iW{\bar D_i \bar W}$ where $i$ runs over the complex structure and the axio-dilaton, the potential is positive-definite and therefore the supersymmetric extremum appears as an actual minimum, although the Kahler moduli are still unfixed. However, once we include the non-perturbative corrections to $W$ to stabilize the Kahler moduli and break supersymmetry by e.g. an anti-D3 brane, the scalar potential is no longer positive-definite. So, my question is: is there a proof that the eigenvalues of the Hessian of the scalar potential with respect to all the moduli are positive-definite once all the moduli are fixed by the combination of fluxes and non-perturbative terms? In particular, there seems to be no good reason to believe that there are no tachyons in any of the complex structure directions.&lt;/p&gt;&#xA;" OwnerUserId="1892" LastActivityDate="2011-03-11T00:11:01.623" Title="Tachyonic complex structure directions in flux vacua" Tags="&lt;string-theory&gt;&lt;kaluza-klein&gt;&lt;supergravity&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="6364" PostTypeId="2" ParentId="6363" CreationDate="2011-03-04T18:18:33.653" Score="1" ViewCount="" Body="&lt;p&gt;Dear stringpheno, in the KKLT model, the uplifting by the anti-D3-brane means an addition of a positive term $D/\sigma^3$ to the potential. Here, $\sigma$ is what appears in $\exp(-K/3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It creates a new minimum for very small value of $\sigma$ because the coefficient $D$ is also expected (or required) to be small, and because the original second derivative near the zero was positively definite (a supersymmetric AdS vacuum can't have tachyons by supersymmetry) and $D/\sigma^3$ also has a positive second derivative, the second derivative at the new minimum - which is near the original $\sigma=0$ point - remains positively definite, too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if one allows $D$ to become large, the sketched derivation above won't necessarily hold. There can be local maxima somewhere in the configuration space - like the electroweak-symmetry-preserving point of the Higgs potential - which have tachyonic directions but they're not considered &quot;uplifting by anti-D3-branes&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-04T18:18:33.653" CommentCount="1" />
  <row Id="6366" PostTypeId="2" ParentId="6346" CreationDate="2011-03-04T20:29:38.723" Score="4" ViewCount="" Body="&lt;p&gt;This is an interesting question, which quite a few people (myself included) find fascinating, and by and large still remains to be explored. To my knowledge, the farthest this idea has gotten so far is in the work of my colleague, Mark van Raamsdonk: &lt;a href=&quot;http://arxiv.org/abs/arxiv:0907.2939&quot; rel=&quot;nofollow&quot;&gt;Comments on quantum gravity and entanglement&lt;/a&gt; and the essay &lt;a href=&quot;http://arxiv.org/abs/1005.3035&quot; rel=&quot;nofollow&quot;&gt;Building up spacetime with quantum entanglement&lt;/a&gt; which won the GRF essay competition last year. If you have background in many body systems and their description in terms of quantum information, your input on this set of issues could be valuable.&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-03-04T20:29:38.723" CommentCount="2" />
  <row Id="6367" PostTypeId="1" CreationDate="2011-03-04T20:44:30.553" Score="1" ViewCount="116" Body="&lt;p&gt;This is a two-part question...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, models of the specific heat capacity $C$ (i.e. Debye, Einstein) in relation to the temperature $T$ give $C$ as steadily increasing with $T$. I assume that the change in $C$ is due to the heat the system gains being stored in its degrees of freedom- so why doesn't it increase in steps?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, what value does the temperature refer to? If the heat supplied raises the temperature from $T_1$ to $T_2$, does $C$ refer to $T_1$, $T_2$ or somewhere in between? If, in experiment, a large temperature difference was used, can it be assumed that $C$ changes over the range and so the answer is an average/useless?&lt;/p&gt;&#xA;" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-05T00:09:54.827" LastActivityDate="2011-03-05T00:09:54.827" Title="Specific heat and temperature" Tags="&lt;thermodynamics&gt;" AnswerCount="2" />
  <row Id="6368" PostTypeId="2" ParentId="6345" CreationDate="2011-03-04T21:24:49.990" Score="1" ViewCount="" Body="&lt;p&gt;If the energy is large enough, photons interactively create massive particles. So yes photons do interact.&lt;/p&gt;&#xA;" OwnerUserId="1906" LastEditorUserId="1906" LastEditorDisplayName="" LastEditDate="2011-03-05T11:03:07.453" LastActivityDate="2011-03-05T11:03:07.453" CommentCount="1" />
  <row Id="6369" PostTypeId="2" ParentId="6367" CreationDate="2011-03-04T22:04:01.727" Score="6" ViewCount="" Body="&lt;p&gt;When you talk about the heat capacity increasing in steps, I assume you're referring to the equipartition principle, which says that the heat capacity is ${1\over 2}k_B$ times the number of degrees of freedom. As more degrees of freedom become available, the heat capacity goes up in steps of ${1\over 2}k_B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is that equipartition only applies in some circumstances, roughly speaking when the degrees of freedom &quot;look classical,&quot; in the sense that the separation between energy levels is small compared to $k_BT$ (among other conditions). Those assumptions don't hold for the Debye and Einstien models. In particular, as the temperature is raised, more quantum levels become available, and those newly-available ones, more or less by definition, aren't in that classical regime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your second question, the heat capacity only applies to infinitesimal increments. That is, instead of writing&#xA;$$&#xD;&#xA;\Delta Q=C\,\Delta T,&#xD;&#xA;$$&#xA;you have to write&#xA;$$&#xD;&#xA;dQ=C\,dT.&#xD;&#xA;$$&#xA;if $C$ varies significantly over the temperature interval in question, you have to integrate&#xA;$$&#xD;&#xA;\int_{T_1}^{T_2}C(T)\,dT&#xD;&#xA;$$&#xA;to get the required heat input. &lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-04T22:04:01.727" />
  <row Id="6370" PostTypeId="1" AcceptedAnswerId="6371" CreationDate="2011-03-04T22:36:49.903" Score="0" ViewCount="250" Body="&lt;p&gt;How do I find the distance traveled of an object if the speed is not constant?&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2011-03-05T19:04:42.213" Title="I need help with finding distance traveled" Tags="&lt;time&gt;&lt;speed&gt;" AnswerCount="5" />
  <row Id="6371" PostTypeId="2" ParentId="6370" CreationDate="2011-03-04T22:55:51.763" Score="4" ViewCount="" Body="&lt;p&gt;If the speed is a function of time, than the total distance is just the integral with respect to time. For example, the distance traveled $D$ for an object moving at a velocity $v(t)$ over a time interval $t_0$ to $t_f$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D=\int_{t_0}^{t_f}v(t)dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is elementary calculus. If you didn't know this already, then you almost certainly don't know calculus and this is not the place to try to teach you a course in calculus. Either way - you will simply &lt;em&gt;need&lt;/em&gt; calculus to solve this problem.&lt;/p&gt;&#xA;" OwnerUserId="1655" LastActivityDate="2011-03-04T22:55:51.763" CommentCount="2" />
  <row Id="6372" PostTypeId="2" ParentId="6370" CreationDate="2011-03-04T22:56:07.303" Score="3" ViewCount="" Body="&lt;p&gt;Well, you could always lay down a measuring tape between the final position and the initial position and see what it reads ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But seriously though: I'm guessing that all you know is the velocity as a function of time, right? In that case, you'll have to do an integral. Velocity is defined as the time derivative of position,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{v}(t) = \frac{\mathrm{d}\mathbf{x}(t)}{\mathrm{d}t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if you invert that formula (technically: solve the differential equation) to solve for the change in position, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{x}(t) = \int_{t_i}^{t_f} \mathbf{v}(t)\mathrm{d}t$$&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-03-04T22:56:07.303" />
  <row Id="6373" PostTypeId="2" ParentId="6370" CreationDate="2011-03-04T22:56:41.657" Score="2" ViewCount="" Body="&lt;p&gt;You use integral calculus.  The distance traveled is the integral of the speed over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the speed were constant, the distance traveled would be the speed multiplied by the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the speed is changing, we don't know what speed to use.  The solution is to break time up into little chunks - one minute, say.  How fast were you traveling in the first minute?  Multiply that speed by one minute to get the distance traveled in the first minute only.  How fast were you traveling in the second minute?  Multiply that by one minute to get the distance traveled in the second minute.  Add those two up to get the total distance traveled in the first two minutes, and repeat for the entire trip.  Now you have an estimate for the total distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the speed changes significantly inside one minute, this method fails again.  No problem, just break time down into one-second intervals.  Find the speed in each second, multiply by one second, and add them all up.  If the speed is changing significantly in one second, use intervals of .01 seconds, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, as you use smaller and smaller time intervals and calculate the total distance, you'll find that the total distance you calculate converges to some number.  For example, you might find a distance of 10.45m if you calculate in 1-minute chunks, 10.87m in one-second chunks, 10.88m in .01s chunks, and 10.88m in .0001s chunks.  Then you know the true distance traveled is 10.88m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process is called &quot;taking an integral&quot;.  Sometimes it is possible to find the integral exactly without breaking things up into chunks.  For example, if the speed is changing at a constant rate, so speed = acceleration*time for some number &quot;acceleration&quot;, the distance traveled is exactly 1/2*acceleration*time^2.  For more details, read any book on integral calculus.  To learn how to program these algorithms efficiently, look for techniques of numerical integration.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-03-04T22:56:41.657" />
  <row Id="6374" PostTypeId="2" ParentId="2554" CreationDate="2011-03-04T23:19:59.590" Score="2" ViewCount="" Body="&lt;p&gt;Please see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.phys.vt.edu/~takeuchi/relativity/notes/section15.html&quot; rel=&quot;nofollow&quot;&gt;http://www.phys.vt.edu/~takeuchi/relativity/notes/section15.html&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2011-03-04T23:19:59.590" CommentCount="3" />
  <row Id="6375" PostTypeId="2" ParentId="6370" CreationDate="2011-03-04T23:25:32.857" Score="1" ViewCount="" Body="&lt;p&gt;It depends on whether you mean to find the final &lt;em&gt;displacement&lt;/em&gt;,&#xA;$$\mathbf{D} = \int_{t_0}^{t_1}\mathbf{v}\:dt,$$&#xA;or quite literally the &lt;em&gt;distance traveled&lt;/em&gt;. Think of the difference between the two in this way: if you travel from New York to London and back again, do you consider the length of both legs of the journey, or just the difference between your initial and final destination? In words, did you travel (roughly) 11,000 km, there and back, or (roughly) 0 km, since you wound up where you started? The former is the distance you traveled, the latter is the magnitude of your displacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's the total distance traveled you want, then the formula is&#xA;$$S = \int_{t_0}^{t_1}v\:dt,$$&#xA;where $v$ is the magnitude of your velocity velocity vector $\mathbf{v}$. Note that this is in general &lt;em&gt;different&lt;/em&gt; from the magnitude of the displacement $D = |\mathbf{D}|$, unless the motion is always in one direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know the velocity as a function of time, then you're done. But if you're given the trajectory but not the velocity, that becomes a bit more tricky. Consider the Pythagorean theorem or distance formula:&#xA;$$\Delta s^2 = \Delta x^2 + \Delta y^2.$$&#xA;It is also correct in three dimensions for infinitesimal displacements:&#xA;$$ds^2 = dx^2 + dy^2 + dz^2.$$&#xA;Therefore:&#xA;$$\left(\frac{ds}{dt}\right)^2 = \frac{dx^2 + dy^2 + dz^2}{dt^2} = v^2.$$&#xA;Or:&#xA;$$S = \int_{t_0}^{t_1}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2}\:dt.$$&#xA;You can also find lengths of curves that are not given in terms of time, but by some other parameter, even one of the coordinates (just replace $t$ with that parameter above, e.g., if you have a curve as a function of $x$, then replace every $dt$ with $dx$, and be mindful of $dx/dx = 1$).&lt;/p&gt;&#xA;" OwnerUserId="1353" LastActivityDate="2011-03-04T23:25:32.857" />
  <row Id="6376" PostTypeId="2" ParentId="6367" CreationDate="2011-03-04T23:57:24.323" Score="0" ViewCount="" Body="&lt;p&gt;Even if you assume a classical situation, you still have the Boltzmann distribution factor which is exponential, and a smooth function of T. So C won't change in discrete steps.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-03-04T23:57:24.323" />
  <row Id="6377" PostTypeId="1" CreationDate="2011-02-02T22:45:05.737" Score="1" ViewCount="120" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/4506/homework-about-spinning-top&quot;&gt;Homework about spinning top&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I have a top with an unknown mass. It has a moment of inertia of 4.00 * 10^-7 kgm^2&#xA;a string is wrapped around the top and pulls it so that its tension is kept at 5.57 N for a distance of .8 m. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could somebody help me derive some equations to help with this? Or to get me in the right direction?&#xA;I have been trying to derive some sort of equations from KEr = 1/2 * I * w^2 but I cant get anywhere without ending up at radius = radius or mass = mass.&lt;/p&gt;&#xA;" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-03-06T00:43:41.010" LastActivityDate="2011-03-06T00:43:41.010" ClosedDate="2011-03-06T20:08:36.930" Title="Top spun up with string under tension problem" Tags="&lt;homework&gt;&lt;newtonian-mechanics&gt;&lt;rotational-dynamics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6378" PostTypeId="2" ParentId="6377" CreationDate="2011-02-02T22:51:41.300" Score="1" ViewCount="" Body="&lt;p&gt;Hint:  The work done on the top is force times distance.  So given the MOI you can find $\omega$.&lt;/p&gt;&#xA;" OwnerUserId="3257" LastActivityDate="2011-02-02T22:51:41.300" />
  <row Id="6379" PostTypeId="1" AcceptedAnswerId="6380" CreationDate="2011-03-05T01:09:49.720" Score="4" ViewCount="141" Body="&lt;p&gt;It's been claimed that the LHC's 14 TeV energy produces temperatures comparable to that which occurred very soon after the Big Bang. The well-known E=1.5kT formula from classical statistical mechanics predicts LHC produced temperatures in the 2.44E17 K range. This temperature is much higher than I expected.&#xA;Is there an agreed upon method of calculating temperature equivalents of 14 TeV p-p interactions, and if so, what is the result?  &lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="1924" LastEditorDisplayName="" LastEditDate="2011-07-26T16:27:50.657" LastActivityDate="2011-07-26T16:27:50.657" Title="LHC Big Bang Temperatures" Tags="&lt;large-hadron-collider&gt;&lt;big.bang&gt;" AnswerCount="1" />
  <row Id="6380" PostTypeId="2" ParentId="6379" CreationDate="2011-03-05T01:27:57.333" Score="4" ViewCount="" Body="&lt;p&gt;Note that the energy is &lt;em&gt;per degree of freedom&lt;/em&gt;, so you don't use all the 14 TeV (nor are they actually running at that energy yet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how many degrees of freedom? Good question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each proton has three valence quarks, but these are generally agreed to make up a small portion of the mas (and to carry a small portion of the momentum) of the proton. The rest of the mass (or momentum) is carried by particles (qaurks and gluons mostly) from the so-called &quot;sea&quot;; these pop into and out-of existence owing to the uncertainty principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Worse, when the collision happens there is a great deal of energy available to put the virtual particles of the sea on to (or nearly on to) the mass shell, converting them into real particles, each equipped with their own swarm of ghostly hangers on. Then many of these decay in a very short time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the average energy of this multitude of degrees of freedom which you are trying to measure/calculate, and it is non-trivial.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-03-05T01:27:57.333" />
  <row Id="6381" PostTypeId="2" ParentId="6339" CreationDate="2011-03-05T06:02:38.450" Score="0" ViewCount="" Body="&lt;p&gt;It's a beautiful thing to look at, isn't it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I always assumed that it was an observation artifact, basically a visually observable &quot;hole vs. electron motion in semiconductors&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen the same effect in lighter coloured beers too.  It seems to me that a high population density of small bubbles is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(@Lubos I've never been to Pilsen and have aways wanted to try the beer there.  And later today I will be less than 100km away, hmmm...)&lt;/p&gt;&#xA;" OwnerUserId="2297" LastActivityDate="2011-03-05T06:02:38.450" />
  <row Id="6382" PostTypeId="2" ParentId="5066" CreationDate="2011-03-05T07:42:31.753" Score="0" ViewCount="" Body="&lt;p&gt;A fun question for science fiction fans. Ever since strings started to look like the Theory Of Everything I have been fascinated by the extra dimensions. It is the analogy of an ant in a two dimensional maze. If it sees the third dimension it can climb over and be million of ant miles away from his two dimensional path to that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the TOE will be great fun once it is found and established. Even if all extra dimensions are curled, each particle in our four dimensions exists also in those curled dimensions. Who knows what the future may bring in technology. The best bet then is to go full steam ahead and find the TOE, before starting on these journeys worrying about fuel.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-05T07:42:31.753" />
  <row Id="6383" PostTypeId="2" ParentId="6086" CreationDate="2011-03-05T09:59:00.237" Score="-3" ViewCount="" Body="&lt;p&gt;The answers given above are incorrect, I say. I'll cover each type of answer and then give my own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The #1 answer (most votes) above assumes that the existence of the word &quot;infinity&quot; means the thought experiment is global, universe-wide in scope. That's a false assumption. One must look at the context. The particle is &quot;escaping to infinity&quot;, a process that happens locally too. Nothing in principle prevents a local analysis of the particle, and the thought experiment is indeed only a local experiment. Plenty of reputable texts locally analyze things escaping to infinity. A particle can be launched upward in a small (perhaps free-floating) lab at the speed the particle needs to be escaping to infinity, and the movement of that particle can be analyzed locally, within the lab, with the analysis completed before the particle leaves through a hole in the lab. That can be a local experiment despite the word &quot;infinity&quot;. Likewise you can be &quot;escaping to France&quot; even if you haven't left Spain, and your movement can still be analyzed within a lab in Spain. People just jumped the gun here, I say. Note that &quot;escaping to infinity&quot; is a common expression in physics--search the web for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The #2 answer repeats the misconception I noted in the original post, the misconception that the horizon is a special place in regards to the tidal force. It isn't. That is covered further in my blog post, my responses above, and a plethora of reputable texts and sites including &lt;a href=&quot;http://cosmology.berkeley.edu/Education/BHfaq.html#q3&quot; rel=&quot;nofollow&quot;&gt;Ted Bunn's Black Hole FAQ&lt;/a&gt;. It's too bad that this is such a deep-seated misconception that not even direct contradicting evidence by numerous experts sways people. Ideally those answering questions on black holes should be aware that &lt;em&gt;in principle&lt;/em&gt; a horizon of a black hole could sweep over you while you're reading this, with nothing noticed, including by the finest instruments made today or ever in the future. The lights needn't flicker, there needn't be an earthquake or anything else amiss. Only at the singularity is destruction guaranteed--not at the horizon in principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone said that &lt;em&gt;X&lt;/em&gt; needs to be infinitesimal in size, without saying why. That's a variation of the #2 answer. The tidal force is the &lt;em&gt;only&lt;/em&gt; reason there is a size restriction on inertial-equivalent frames in our real, gravity-endowed universe. The equivalence principle (EP) has been tested to high precision in labs that are obviously not infinitesimal in size, and accurately too, because the tidal force in the labs was weak enough that it didn't affect the results at that precision. Likewise, frame &lt;em&gt;X&lt;/em&gt; need only be &lt;em&gt;sufficiently&lt;/em&gt; small. Strictly speaking the EP is true only for a point in spacetime, but that &lt;em&gt;doesn't&lt;/em&gt; mean it can't be tested in a larger frame. We &lt;em&gt;do&lt;/em&gt; test it in labs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thought experiment above applies to an arbitrarily small frame (I let &lt;em&gt;X&lt;/em&gt; be arbitrarily small--it's a given), meaning that no matter how weak the tidal force in &lt;em&gt;X&lt;/em&gt; is, the outcome of the experiment is the same. If a violation of the EP is found in any size lab in which the tidal force is insignificant, that is a bona fide violation. If a violation of the EP is found in an arbitrarily small lab, that is also a bona fide violation, because the tidal force can then definitely be insignificant; just make the lab smaller until the tidal force has no significant effect on the finding of a violation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems there's an undercurrent in the answers above, which goes like this: If you find a violation of the EP, that proves the tidal force in your frame isn't negligible, hence you're not truly local and it wasn't a valid test of the EP. That's just a fancy way of saying the EP isn't testable. But all ideas in the realm of science must be testable; science requires falsifiability. The outcome of the thought experiment, the finding that the EP is violated, is demanded by the property of a horizon that nothing can pass outward through it (that's actually the &lt;em&gt;definition&lt;/em&gt; of a horizon). If the outcome is true, it's true regardless of the strength of the tidal force in &lt;em&gt;X&lt;/em&gt;. To use a silly but valid analogy, the same logic would apply if GR predicted that anything falling through a horizon is eaten by a monster. If someone pointed out that that's a violation of the EP (which it would be), that finding wouldn't be refuted by the tidal force. Like the monster, the property that nothing can pass outward through the horizon isn't because of the tidal force at the horizon. That property is a prediction GR makes regardless of the tidal force there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that covers all the answers above. I've been asking this question for a long time now, not just here. Always the answers are misconceptions or other things that can be shown to be false. The thought experiment is very simple and the logic seems rock-solid. (At its essense it's all of 5 sentences.) I conclude that it validly shows that GR contradicts itself about black holes. I conclude that black holes are a result of a mistake in GR. (The EP isn't the problem. The EP is the heart and soul of GR--you can find 600+ .edu links saying that.) A tweak to the Schwarzschild metric results in a metric that doesn't predict black holes (so the information loss paradox--one of the biggest problems in physics--vanishes) yet is confirmed by all tests of that metric that I know of, and is compatible with quantum mechanics (unlike the Schwarzschild metric) to boot. The vast majority of tests of GR have been tests of the Schwarzschild metric. So a finding that GR has a problem needn't leave us without a seemingly much better metric to replace it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FWIW I'll offer a bounty for this question. It's okay by me if that benefits the #1 answer above. It's all about the science!&lt;/p&gt;&#xA;" OwnerUserId="2314" LastActivityDate="2011-03-05T09:59:00.237" CommentCount="65" />
  <row Id="6384" PostTypeId="1" CreationDate="2011-03-05T10:16:08.930" Score="8" ViewCount="239" Body="&lt;p&gt;Why does a quantum field theory invariant under dilatations also have to be invariant under proper conformal transformations? To show your favorite dilatation invariant theory is also invariant under proper conformal transformations is seldom straightforward. Integration by parts, introducing Weyl connections and so on and so forth are needed, but yet at the end of the day, it can always be done. Why is that?&lt;/p&gt;&#xA;" LastActivityDate="2011-03-09T17:31:39.827" Title="Why does dilatation invariance imply proper conformal invariance?" Tags="&lt;quantum-field-theory&gt;&lt;conformal-field-theory&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="6385" PostTypeId="1" AcceptedAnswerId="6386" CreationDate="2011-03-05T11:32:14.747" Score="3" ViewCount="180" Body="&lt;p&gt;Suppose that a someone decided to calculate, for a given amount of heat added into a planet's atmosphere(&lt;strong&gt;an ideal gas&lt;/strong&gt;), how much is the corresponding temperature rising of the atmosphere, $\Delta T$. Assuming that the whole heat transforms into the internal energy of the gas and distributes uniformly (s)he got $\Delta T=1K$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A question rises: actually, some of the heat is transformed into gravitational potential energy of the atmosphere(we ignore kinetic energy of the atmosphere). If we take also this into account then how much is the temperature rising of the atmosphere, $\Delta T$ now? Maybe the effect due to the gravity is negligible?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity let's assume that the  pressure change is approximately exponential with height, $h$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P=P_0e^{-\frac{h}{H}}$$ where $P_0$ is pressure at $h=0$ and $H$ is called the scale height for pressure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For definiteness, let's take the Earth's atmosphere. In this case $P_0=101$ $kPa$ is average sea-level pressure. $H$ is about $8$ $km$.&lt;/p&gt;&#xA;" OwnerUserId="552" LastActivityDate="2011-03-05T13:24:10.820" Title="A &quot;global warming&quot; problem" Tags="&lt;thermodynamics&gt;&lt;geophysics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="6386" PostTypeId="2" ParentId="6385" CreationDate="2011-03-05T13:07:56.747" Score="5" ViewCount="" Body="&lt;p&gt;The average height of a molecule is $H$ with your exponential distribution, because $\int_0^\infty dt\,t\,\exp(-t)=1$, so its average potential energy is $mgH$ where $m$ is the mass of the molecule. How much does it compare with $3kT/2$ (or $5kT/2$ etc.) in the kinetic energy? And what will happen to $\Delta T$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is given by the virial theorem (so in advance, I am telling you that the reduction of $\Delta T$ will be of the same order as $\Delta T$ itself) but let me use no pre-derived results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, just realize that the exponential decrease in $\exp(-h/H)$ is nothing else than the Maxwell-Boltzmann factor $\exp(-mgh/kT)$ which implies that $H=kT/mg$ and the average potential energy is $mgH=kT$. (At this stage, I am neglecting the dependence of the temperature itself on the height etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you have a monoatomic gas which has $3kT/2$ in the kinetic energy, $kT=2kT/2$ is ultimately given to the potential energy which is $2/5$ of the total energy. As the temperature rises, the scale height will also have to rise, and you can see that $2/5$ of the energy that is initially pumped to the kinetic energy will be converted to the increased potential energy and only $3/5$ of the original increase of the kinetic energy will stay in the kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think that for the monoatomic gas, the answer is $(3/5)\Delta T$. Similarly, it will be $(5/7)\Delta T$ if the kinetic energy of a single molecule is $5kT/2$, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to be sure, this is not equivalent to the &quot;global warming&quot; issue because the greenhouse warming is a permanent change of the energy flows (extra Watts per square meter - Joule in every single second) while the energy needed to change the gravitational potential of the gas when its distribution at different altitudes changes is just a one-time event.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the general observation that the ultimate increase of the temperature (or anything else) will be smaller than the most naively calculated one is valid: the extra heat is ultimately redistributed to and consumed by &quot;many consumers&quot; (in this case, potential energy), which means that one particular consumer ultimately gets less. This is the reason why the feedbacks of stable systems tend to be negative - an insight known as the Le Chatelier's principle (in chemistry) or homeostasis (generally) or Lenz's law (in electromagnetism coupled to mechanics) etc.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-05T13:24:10.820" LastActivityDate="2011-03-05T13:24:10.820" CommentCount="2" />
  <row Id="6387" PostTypeId="1" AcceptedAnswerId="6388" CreationDate="2011-03-05T15:26:25.847" Score="0" ViewCount="58" Body="&lt;p&gt;I'm looking for something to supplement my Physics II class. Last year I started using &lt;a href=&quot;http://www.teach12.com/tgc/courses/course_detail.aspx?cid=1007&quot; rel=&quot;nofollow&quot;&gt;these video lectures&lt;/a&gt; to supplement my Calculus class and it helped tremendously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also turned to &lt;a href=&quot;http://www.thinkwell.com/student/product/physics&quot; rel=&quot;nofollow&quot;&gt;this educational software&lt;/a&gt; when I was stuck in Physics I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't seem to turn up anything similar that deals with Physics II.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty much looking for something that I can watch repeatedly until it sinks in. Perhaps you know of something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2390" LastEditorUserId="2390" LastEditorDisplayName="" LastEditDate="2011-03-05T15:32:31.580" LastActivityDate="2011-03-05T16:03:37.857" Title="Physics II Video Courseware Recommendations" Tags="&lt;electricity&gt;&lt;software&gt;" AnswerCount="1" />
  <row Id="6388" PostTypeId="2" ParentId="6387" CreationDate="2011-03-05T16:03:37.857" Score="0" ViewCount="" Body="&lt;p&gt;Walter Lewin's &lt;a href=&quot;http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/&quot; rel=&quot;nofollow&quot;&gt;lectures&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1671" LastActivityDate="2011-03-05T16:03:37.857" CommentCount="1" />
  <row Id="6389" PostTypeId="1" CreationDate="2011-03-05T16:16:31.803" Score="1" ViewCount="162" Body="&lt;p&gt;I stumbled onto this page&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mylifeisaverage.com/story/1364811/&quot; rel=&quot;nofollow&quot;&gt;http://mylifeisaverage.com/story/1364811/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the post states that they were&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;all making strings and shapes with&#xA;  these sets of 216 really small&#xA;  spherical earth magnets. What did I&#xA;  do? I strung them together and found&#xA;  the string's resonant frequency. It&#xA;  was 15 Hz&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so I'm wondering how the poster could have figured this out?&lt;/p&gt;&#xA;" OwnerUserId="2391" LastActivityDate="2011-03-05T23:29:29.833" Title="How could this person have discovered the resonant frequency from this string of magnets?" Tags="&lt;magnets&gt;&lt;frequency&gt;&lt;resonance&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="6390" PostTypeId="2" ParentId="6321" CreationDate="2011-03-05T16:16:41.523" Score="1" ViewCount="" Body="&lt;p&gt;I'd like to point out that there is a small probability that the assumption on which the question is based:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;As I hope is obvious to everyone reading this, the universe contains more matter than antimatter,&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;may not be true, depending on the result of &lt;a href=&quot;http://aegis.web.cern.ch/aegis/home.html&quot; rel=&quot;nofollow&quot;&gt;the Aegis experiment at CERN&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's because, as Professor Orzel stated in his answer to &lt;a href=&quot;http://physics.stackexchange.com/questions/1165/experimental-observation-of-matter-antimatter-in-the-universe&quot;&gt;this question&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the primary evidence for a lack of vast amounts of antimatter out there in the universe is a lack of radiation from the annihiliation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but if matter and antimatter repel each other gravitationally, there could be whole galaxy clusters of matter, whole galaxy clusters of antimatter, and no particle or antiparticle getting near the boundary region between the matter and antimatter areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is just a low-probability case to keep in mind until the jury is out on Aegis.  I see it as a lifeboat: people keep it just in case, but while all is OK with the main ship, they don't live in it.&lt;/p&gt;&#xA;" OwnerUserId="2085" LastActivityDate="2011-03-05T16:16:41.523" />
  <row Id="6391" PostTypeId="2" ParentId="6086" CreationDate="2011-03-05T16:28:58.167" Score="0" ViewCount="" Body="&lt;p&gt;I also would like to point out that the notion of &quot;escaping to infinity&quot; violates the locality of the equivalence principle, as it requires an infinite amount of time.  That, too, is not a local probe of the gravitational field, as it depends on all the dynamics of the spacetime between the probe point and the time the particle reaches infinity.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-03-05T16:28:58.167" CommentCount="15" />
  <row Id="6392" PostTypeId="2" ParentId="6321" CreationDate="2011-03-05T16:48:48.603" Score="1" ViewCount="" Body="&lt;p&gt;In reply to the second partenthetical question, &lt;em&gt;I wrote that matter created from energy in particle physics experiments is &quot;generally&quot; in the form of particle-antiparticle pairs .&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is too restrictive. Quantum numbers have to be conserved, and they are conserved in pair production, but there can also be associated production of mesons etc:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example a K+and a Lamda (+other particles) may come out of a proton proton collision, conserving strangeness, and similar with other conserved numbers. All sorts of interesting combinations can arise.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-05T16:48:48.603" />
  <row Id="6393" PostTypeId="1" CreationDate="2011-03-05T17:00:22.510" Score="3" ViewCount="118" Body="&lt;p&gt;In the article &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0706/0706.3359v1.pdf&quot; rel=&quot;nofollow&quot;&gt;Three dimensional gravity reconsidered&lt;/a&gt; by Ed Witten, he remarked that the CFT dual to three dimensional quantum gravity has to admit a holomorphic factorization and have a central charge of $c=24k$ for some integer $k$. He argues this by some handwaving from a Chern-Simons crutch even though he admits the nonperturbative theory can't be Chern-Simons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, does this mean if we have a two dimensional conformal field theory which does not factorize into chiral and antichiral parts and/or has a central charge which is not an integer multiple of 24, it does not have an AdS quantum gravity dual?&lt;/p&gt;&#xA;" LastActivityDate="2011-03-05T17:43:01.057" Title="Can a $CFT_2$ which can't be factorized into chiral and antichiral parts and/or have a central charge not a multiple of 24 have AdS duals?" Tags="&lt;string-theory&gt;&lt;conformal-field-theory&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="6394" PostTypeId="2" ParentId="6389" CreationDate="2011-03-05T17:35:38.437" Score="-1" ViewCount="" Body="&lt;p&gt;You shake the arrangement of magnets, then let it sit and wiggle.  You then count the number of wiggles per minute.  Frequency in hertz is then $\frac{{\rm \# of wiggles}}{60 {\rm s}}$.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-03-05T17:35:38.437" CommentCount="2" />
  <row Id="6395" PostTypeId="2" ParentId="6393" CreationDate="2011-03-05T17:43:01.057" Score="5" ViewCount="" Body="&lt;p&gt;This actual reasons behind many statements you mentioned may be a bit different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, one doesn't need too much Chern-Simons handwaving to say that $c_L=24 k_L$ etc. The three-dimensional action of general relativity may be rewritten so that it only depends on $l/G$ where $G$ is Newton's constant (of dimension length) and $l$ is the curvature radius, so that $2/l^2$ is the cosmological constant term added to $R$ in the Einstein-Hilbert action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brown and Henneaux have studied the asymptotic action of the Virasoro group on the three-dimensional spacetime and from the commutators, they could simply determine that $c_L=c_R=3l/2G$. Now, $l/G$ is a multiple of 16 - it's plausible that one needs the Chern-Simons variables to argue this much (one uses the integer-valuedness of the levels). But the isomorphism between CS and GR works perturbatively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if $c_L$ and $c_R$ differ, it's still true that $c_L-c_R$ must be a multiple of 24 for any CFT - not just CFTs that are the duals of some AdS space. This follows from the $\tau\to\tau+1$ modular transformations of the torus - or, equivalently, from the requirement that $L_0-\tilde L_0$ has to be integer-valued for physical states (the implication holds because the ground states have energies $-c_{L,R}/24$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It just happens that when you translate the levels of the Chern-Simons theory to the ground state energies of the boundary CFT, they're equal up to the sign. A necessary condition for holomorphic factorization of a CFT is that the ground state energies for the left-movers and right-movers are two integers - which naturally translates to the integrality of the levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Witten kind of assumed it was also a sufficient condition - i.e. that one can find a dual description with a monster symmetry not only for $k=1$ which is what he did but also for $k=2,3,\dots$. This naturally sounding conjecture was disproved, at least for $k=2$, by Davide Gaiotto by some representation theory of the monster group:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/arXiv:0801.0988&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/arXiv:0801.0988&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems pretty likely that the theories with the monster group symmetry don't exist for any higher $k$, either. It may still be the case that the higher $k$ CFT duals exist and have a smaller sporadic symmetry - much like the ABJM at a general level generalizes the BLG construction but reduces its symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various 2D CFTs may have AdS duals but they won't be pure gravity: they will include some massless bulk states i.e. marginal deformations. The Leech-lattice-based models that Witten studied were special because the Leech lattice doesn't produce any massless states - it only produces massive states which may be interpreted as black hole microstates. Witten hasn't discussed those in his paper. So he was not saying that non-factorizable CFTs can't have an AdS dual: they just can't have a pure gravity AdS dual in three dimensions.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-05T17:43:01.057" CommentCount="5" />
  <row Id="6396" PostTypeId="2" ParentId="6389" CreationDate="2011-03-05T18:00:00.907" Score="2" ViewCount="" Body="&lt;p&gt;There's an interesting question in here if you look hard enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, there's nothing special about the resonant frequency of something made of little magnets. It might as well be a piece of ordinary string, a metal bar, or whatever. In fact I think the fact that it's made of separate little magnets stuck together would give it a much lower Q factor than, for example, a metal bell, which would make the resonant frequency less clear and harder to measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But anyway, let's assume we have some object with a mechanical resonance around 15 Hz. How do we measure this frequency with no other equipment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a human, there are basically two ways you can observe something's vibration frequency. If the vibration is slow enough - below about 5 or maybe 10 Hz - you can just count the individual cycles. If I'm wiggling a jump rope at the resonance frequency and count 300 wiggles per minute, then the frequency is 300/(60 s) = 5 Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other way - which works well between about 50 Hz and 10,000 Hz - is to listen to the pitch of the vibrations and figure out the frequency of that musical pitch. If I listen to a bell and hear the pitch A above middle C, that means the frequency is 440 Hz (A 440). If, like me, you don't have perfect pitch, you'll need some reference pitch to compare it too, such as a piano, but once you figure out the closest note of the musical scale, you know the frequency to within 3%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interesting thing in this case is that 15 Hz is right in the middle of that awkward frequency range which is too fast to count, but too low to hear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do you measure a frequency in this awkward range? If you have some simple lab equipment it becomes easy. For example, just shine a light on a photodiode and arrange it so that the edge of the vibrating object gets in the light path and casts a shadow. Then look at the photodiode signal on an oscilloscope and you'll see the vibrations at 15 Hz. (Interestingly, an ordinary TV camera would be a bad choice here because they take about 30 discrete frames a second, so 15 Hz is right around the Nyquist frequency and aliasing would make it impossible to tell if it were 13 Hz or 17 Hz, for example.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if you don't have an oscilloscope, or video camera or anything like that? I think you would still be able to measure the resonance frequency by hand, you'd just have to be inventive. For example, you can try to excite the vibrations using subharmonics. If you give your string of magnets a sudden shove 3 times a second, for example, then your 3 Hz excitation signal has harmonics at 6 Hz, 9 Hz, 12 Hz, 15 Hz..., because it's not sinusoidal. So if you're able to show that you get a strong resonance at 3 Hz, but not at 2.9 Hz or 3.1 Hz, then you know that the resonant frequency is one of those harmonics. If you repeat that process with some different subharmonic (e.g. 5 Hz), then you can quickly narrow down the possibilities for the resonant frequency.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-03-05T18:00:00.907" CommentCount="1" />
  <row Id="6397" PostTypeId="1" CreationDate="2011-03-05T18:02:31.777" Score="2" ViewCount="205" Body="&lt;p&gt;Some frequencies of electromagnetic waves are used for transmission of information, like radiowaves, microwaves, light, but some not.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about Low Frequencies?:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps low frequencies aren't used for transmision simply because the waves haven't got enough energy to reach destiny, but is well known that if you go near a high voltage transmission tower, you can turn on a fluorescent lamp &lt;a href=&quot;http://www.electric-fields.bris.ac.uk/&quot; rel=&quot;nofollow&quot;&gt;http://www.electric-fields.bris.ac.uk/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the intensity of the high voltage gives more photons to interact there, than in a home installation, but the fact is waves are traveling anyway at 50/60 Hz without need any modulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a limit on how low a frequency can be to put photons into travel? &lt;/p&gt;&#xA;" OwnerUserId="1916" LastEditorUserId="1916" LastEditorDisplayName="" LastEditDate="2011-03-05T18:37:25.570" LastActivityDate="2011-08-11T23:08:14.467" Title="Low frequency electromagnetic waves" Tags="&lt;electromagnetic-radiation&gt;" AnswerCount="3" />
  <row Id="6398" PostTypeId="2" ParentId="6397" CreationDate="2011-03-05T18:14:55.257" Score="2" ViewCount="" Body="&lt;p&gt;&quot;Sending Photons on a Travel&quot; is not the purpose of any radio transmission. &#xA;In fact it is very difficult to show the existence of say, BCB &quot;Photons&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;About frequencies used for transmission, down to 12 or 16 kHz are used &#xA;to contact submersed submarines. In big auditoriums, churches and so on &#xA;there wer/are systems using about 18 kHz to transmit to persons with hearing &#xA;problems. (This is done by IR today in new systems)&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-05T19:02:56.830" LastActivityDate="2011-03-05T19:02:56.830" CommentCount="2" />
  <row Id="6399" PostTypeId="2" ParentId="6384" CreationDate="2011-03-05T18:24:59.727" Score="0" ViewCount="" Body="&lt;p&gt;Maybe this does it:&lt;br&gt;&#xA;$\begin{array}{rccl}&#xD;&#xA;\textrm{Translation:}&amp;amp;P_\mu&amp;amp;=&amp;amp;-i\partial_\mu\\&#xD;&#xA;\textrm{Rotation:}&amp;amp;M_{\mu\nu}&amp;amp;=&amp;amp;i(x_\mu\partial_\nu-x_\nu\partial_\mu)\\&#xD;&#xA;\textrm{Dilation:}&amp;amp;D&amp;amp;=&amp;amp;ix^\mu\partial^\mu\\&#xD;&#xA;\textrm{Special Conformal:}&amp;amp;C_\mu&amp;amp;=&amp;amp;-i(\vec{x}\cdot\vec{x}-2x_\mu\vec{x}\cdot\partial)&#xD;&#xA;\end{array}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the commutation relation gives:&lt;br&gt;&#xA;$[D,C_\mu] = -iC_\mu$&lt;br&gt;&#xA;so $C^\mu$ acts as raising and lowering operators for the eigenvectors of the dilation operator $D$. That is, suppose:&lt;br&gt;&#xA;$D|d\rangle = d|d\rangle$&lt;br&gt;&#xA;By the commutation relation:&lt;br&gt;&#xA;$DC_\mu - C_\mu D = -iC_\mu$&lt;br&gt;&#xA;so&lt;br&gt;&#xA;$DC_\mu|d\rangle = (C_\mu D -iC_\mu)|d\rangle$&lt;br&gt;&#xA;and&lt;br&gt;&#xA;$D(C_\mu|d\rangle) = (d-i)(C_\mu|d\rangle)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But given the dilational eigenvectors, it's possible to define the raising and lowering operators from them alone. And so that defines the $C_\mu$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;P.S. I cribbed this from:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://web.mit.edu/~mcgreevy/www/fall08/handouts/lecture09.pdf&quot; rel=&quot;nofollow&quot;&gt;http://web.mit.edu/~mcgreevy/www/fall08/handouts/lecture09.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-05T18:24:59.727" CommentCount="1" />
  <row Id="6400" PostTypeId="1" AcceptedAnswerId="6403" CreationDate="2011-03-05T18:36:31.110" Score="8" ViewCount="607" Body="&lt;p&gt;Are tidal power plants slowing down Earth's rotation to the speed of the orbiting moon? (1 rotation per 28 cca days) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are they vice versa increasing the speed of moon orbiting by generating some waves in gravitation field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, can you calculate how much energy must be produced by how many tidal power plants (compare it to average nuclear plant please) to slow down the Earth's rotation to 25 hours / day?&lt;/p&gt;&#xA;" OwnerUserId="2394" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-25T15:05:49.337" LastActivityDate="2011-07-25T15:05:49.337" Title="Are tidal power plants slowing down Earth's rotation?" Tags="&lt;orbital-motion&gt;&lt;renewable-energy&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="6401" PostTypeId="2" ParentId="6370" CreationDate="2011-03-05T19:04:42.213" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Integrating&lt;/strong&gt; velocity is OK, but usually  I do simpler things to know the answer.&lt;br&gt;&#xA;It depends on the context. Traveled you said?&lt;br&gt;&#xA;An &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Odometer&quot; rel=&quot;nofollow&quot;&gt;odometer&lt;/a&gt;&lt;/strong&gt; is the ideal instrument. Cars, bycicles, pedestrians can use one.&lt;br&gt;&#xA;I can use a &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/GPS&quot; rel=&quot;nofollow&quot;&gt;GPS&lt;/a&gt;&lt;/strong&gt; in cars, bykes, pedestrian, airplanes and sea turtles, etc, complemented by Google Maps.&#xA;Trucks have a record of the instant speed for audit purposes (I think), this way is more complicated because you will have to integrate.&lt;br&gt;&#xA;A &lt;strong&gt;movie cam&lt;/strong&gt; is sometimes useful to record and keep track of the space traversed. It is used in sports and dancers and to study the body motion.&#xA;In football games on TV sometimes they give us the distance that each player traversed. They have to know the angle of the playfield with the recording camera, identify the player .. and SUM to the previous data. A sumation is more used in the real world than integration because we take measures at time intervals and accumulate to previous data. An integral presume that we have a continuous flux of data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the object is fast compared to light speed then data must be &lt;strong&gt;relativistic corrected&lt;/strong&gt; as the same if you pretend to measure the space traversed when you walk an escalator in relation to the floor of the escalator itself or the outer building. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How interesting that &lt;em&gt;our minds have an automatic complicated answer&lt;/em&gt;.&lt;br&gt;&#xA;Answering 'If you want to know the traversed space you must have to know the velocity' forgets that to know the velocity is more difficult (need to know more: the space and the time consumed at every moment)&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-03-05T19:04:42.213" />
  <row Id="6402" PostTypeId="2" ParentId="6397" CreationDate="2011-03-05T19:51:48.480" Score="7" ViewCount="" Body="&lt;p&gt;Very Low Frequency (VLF) and Extremely Low Frequency (ELF) radio waves &lt;a href=&quot;http://en.wikipedia.org/wiki/Communication_with_submarines&quot; rel=&quot;nofollow&quot;&gt;are used&lt;/a&gt; for communication between submarines, because higher-frequency radiation doesn't propagate well through sea water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the main reason they're not used much outside of the water is simply that higher frequencies are easier to use for a number of reasons. For one thing, the lower the frequency you're trying to use, the longer your antenna needs to be. For another thing, the rate at which you can send information depends on the bandwidth (i.e., the range of frequencies you're using). There just isn't much bandwidth at the low-frequency end.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-05T19:51:48.480" CommentCount="3" />
  <row Id="6403" PostTypeId="2" ParentId="6400" CreationDate="2011-03-05T20:00:19.643" Score="7" ViewCount="" Body="&lt;p&gt;In principle, yes, the ultimate source of energy for a tidal power plant is Earth's rotational energy, so these plants are slowing down the Earth's rotation. By conservation of angular momentum, that means they are pushing the Moon further away as well, although I wouldn't phrase it as being due to &quot;waves in the gravitational field,&quot; as that expression suggests a different phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth's rotational kinetic energy is about $10^{29}$ J, and the world uses something like $10^{22}$ J/year, so you could power the entire world for millions of years before you'd run out of rotational energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your numerical question, you should work out the rotational kinetic energy of the Earth now, and also when the day is 25 hours long. The difference between those is the total energy required. The way to figure out the rotational kinetic energy is ${1\over 2}I\omega^2$.  Here $I$ is the Earth's moment of inertia, which is about $0.4MR^2$ where $M$ and $R$ are Earth's mass and radius. $\omega$ is the Earth's rotation rate in radians per second -- that is, $2\pi$ over the time for one rotation.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-05T20:00:19.643" CommentCount="12" />
  <row Id="6404" PostTypeId="2" ParentId="6400" CreationDate="2011-03-05T20:07:52.867" Score="2" ViewCount="" Body="&lt;p&gt;In order to slow down the rotation of a body angular momentum must transferred off that body.  In the case of Earth and the moon this occurs from the difference in gravity across the Earth, or tidal force.  A tidal power generating system simply converts a tiny fraction of energy in the tidal bulge of the Earth, mostly in the oceans, as it moves around the globe into mechanical or electrical power.  The question is whether that induces a torque on the Earth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These systems might serve to reduce the tidal bulge of the oceans a very tiny amount.  So from the systems perspective the flow of water is impeded, the effective viscosity increased, friction increased and the tidal bulge reduced.  This represents a tiny amount of energy reduced on the Earth in this form.  However, this is a near infinitesimal amount of the Earth’s rotational kinetic energy.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-05T20:07:52.867" />
  <row Id="6405" PostTypeId="2" ParentId="6400" CreationDate="2011-03-05T20:16:59.143" Score="2" ViewCount="" Body="&lt;p&gt;I doubt that tidal power plants change anything. Whether the tidal waves crash on the coast or are channeled in a basin to drive some power generators, does not change dissipation of earths rotational energy. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-05T20:16:59.143" CommentCount="7" />
  <row Id="6406" PostTypeId="1" CreationDate="2011-03-05T21:11:33.267" Score="2" ViewCount="166" Body="&lt;p&gt;The phrase &quot;speed of light&quot; is commonly used for the constant c =3E8 m/s, a feature that's &quot;hardcoded&quot; into the structure of spacetime.  All massless waves and particles move at this speed, and it's a key concept in all fundamental theories in physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light, in vacuum, is one fine example.  But light in air or glass, or microwaves in a waveguide, move slower than c.  The speed of actual light isn't the &quot;speed of light&quot;.   And of course there are plenty of other things besides electromagnetic quanta whose behavior is described by equations involving c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The letter &quot;c&quot; alone isn't good for general writing and conversation, especially with non-scientists.   It is fine as a mathematical symbol in equations, but isn't catchy or self-descriptive (albeit misleading) as &quot;the speed of light&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I have an impression I'm not asking this question in quite the right way.  It's not at all about light, or phase or group velocities of light,  but about our spacetime geometry having a fundamental constant speed &quot;built in&quot; - what to call it?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Historically it was discovered in connection with light, but c isn't specific to electromagnetism.  It's important in all the fundamental forces, and in all situations with high energy particles flying about.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Universal spacetime speed constant&quot; would be an answer, but for that it's long and clumsy to repeat often in writings about physics.  I'm looking for some name or phrase shorter and easier than that, but not so brief as just &quot;c&quot; which is okay in equations and tables of physical contants.&lt;/p&gt;&#xA;" OwnerUserId="353" LastEditorUserId="353" LastEditorDisplayName="" LastEditDate="2011-03-06T06:45:15.810" LastActivityDate="2011-03-06T06:45:15.810" ClosedDate="2011-03-05T21:53:17.083" Title="What's a better phrase than &quot;speed of light&quot; for the universal spacetime speed constant?" Tags="&lt;speed-of-light&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="6407" PostTypeId="2" ParentId="6406" CreationDate="2011-03-05T21:21:18.310" Score="0" ViewCount="" Body="&lt;p&gt;The speed of light, c, actually &lt;em&gt;is&lt;/em&gt; the speed of light. The other thing you mentioned is the phase-speed of light, which should not be confused with c.&lt;br&gt;&#xA;So there are better words for all these thing you can confuse with the speed of light, c.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But not for the speed of light.&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2011-03-05T21:36:36.160" LastActivityDate="2011-03-05T21:36:36.160" CommentCount="1" />
  <row Id="6408" PostTypeId="2" ParentId="2573" CreationDate="2011-03-05T22:40:10.397" Score="0" ViewCount="" Body="&lt;p&gt;I remember a paper, details forgotten, published in Foundations of Physics probably during the late 1970s, where the researchers, wondering about the possibility of some kind of intelligence or awareness of quantum particles, tried to measure if photons had any persisting preference for certain fringes when passing through a grating.  Light in, say, the 2nd fringe would be put through another grating. Photons that prefer the 2nd fringe would make 2nd fringe brighter than the 1st, 3rd or others relative to predictions of basic optics.  They found no such enhancement.  I had a good laugh reading the paper.  But maybe those research are the ones who went on to do whatever work you're wondering about? &lt;/p&gt;&#xA;" OwnerUserId="353" LastActivityDate="2011-03-05T22:40:10.397" />
  <row Id="6409" PostTypeId="2" ParentId="6389" CreationDate="2011-03-05T23:07:50.310" Score="0" ViewCount="" Body="&lt;p&gt;216 magnets suggest &lt;a href=&quot;http://www.theneocube.com/&quot; rel=&quot;nofollow&quot;&gt;Neocube&lt;/a&gt;; I think such string could be just  forced by electromagnet or simply by passing a AC current from a generator. While generator is able to produce arbitrary frequency, it is possible just to try to tune into a resonance and thus obtain the resonance frequency. &lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2011-03-05T23:07:50.310" />
  <row Id="6411" PostTypeId="1" AcceptedAnswerId="6416" CreationDate="2011-03-05T23:52:50.017" Score="3" ViewCount="106" Body="&lt;p&gt;Suppose we have 2 systems with the same partition function, does this mean the 2 systems are the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in 2D CFTs, would the equality of two partition functions imply that the underlying theories are the same (in the CFT sense, I mean same central charge, same OPE, etc). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we take the $\text{N}^{th}$ symmetric product of a mother CFT with a partition function $Z(\tau,\bar{\tau})$ and then I orbifold by the permutation group $S_N$ or any cyclic subgroup $\mathbb{Z}_N$ to get the partition function of the permutation orbifold $\mathbf{Z}$. Now suppose we find another system with the same partition function $\mathbf{Z}$, does this mean that this system should be equivalent to the permutation orbifold? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please feel free to edit or correct my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you all in advance!&lt;/p&gt;&#xA;" OwnerUserId="1531" LastActivityDate="2011-03-06T05:20:00.810" Title="2D CFTs and permutation orbifolds" Tags="&lt;string-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6412" PostTypeId="1" AcceptedAnswerId="6414" CreationDate="2011-03-06T00:49:55.733" Score="4" ViewCount="232" Body="&lt;p&gt;Fierz identities are discussed in the wikipedia article:&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fierz_identity&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fierz_identity&lt;/a&gt;&lt;br&gt;&#xA;but the article doesn't give any derivation. The article implies that they arise from the blade structure of a Clifford algebra.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Some notes to be deleted after we have a solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They seem to arise in QFT calculations and (perhaps as a result) they can always be put into &quot;pure density matrix&quot; form. That is, the spinors can be arranged to appear in an order that makes them natural for a trace operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, with $U,V,W$ as unspecified operators and $\theta,\psi$ as spinors, one might have:&lt;br&gt;&#xA;$(\bar{\theta}U\theta)(\bar{\psi} W\theta)(\bar{\theta} V\psi)$&lt;br&gt;&#xA;where the unbarred and barred spinors do not always match as they do in &quot;$...\theta)(\bar{\theta}...$&quot;. But this can be rearranged to give:&lt;br&gt;&#xA;$(\bar{\theta}U\theta)(\bar{\theta} V\psi)(\bar{\psi} W\theta)$&lt;br&gt;&#xA;One can then turn the $\psi\bar{\psi}$ type elements into (almost) pure density matrices and treat the problem as one of computing traces:&lt;br&gt;&#xA;$\textrm{tr} [ (\theta\bar{\theta}U\theta\bar{\theta} V\psi\bar{\psi} W ]$&lt;br&gt;&#xA;And then one can use various properties such as the fact that with pure density matrices such as $|a\rangle\langle a|^2 = |a\rangle\langle a|$ one has that anything that begins and ends with such a thing has to be a complex multiple of the pure density matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would give the Fierz identities without explicitly looking at the blade structure of the Clifford algebra and that is very suspicious to me. I doubt it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a previous question: &lt;a href=&quot;http://physics.stackexchange.com/questions/6248/some-majorana-fermion-identities&quot;&gt;Some Majorana fermion identities&lt;/a&gt; I intuitively convinced myself that Fierz identities could be established this way (and some very smart people seemed to agree) but when I tried to solve that problem this way I quickly was buried in unsuccessful calculations. Now it could be that that particular problem couldn't easily be so attacked, or my method was misapplied, etc.; I'm not sure.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-06T02:45:12.813" Title="How does one prove Fierz identities?" Tags="&lt;quantum-field-theory&gt;&lt;fermions&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="6413" PostTypeId="2" ParentId="3468" CreationDate="2011-03-06T01:26:09.650" Score="6" ViewCount="" Body="&lt;p&gt;A mis-steered beam at CEBAF simply cut a hole thought the niobium wall of the klystron and flooded half the accelerator with helium (super-conducting klystrons need a liquid helium jacket to work...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We were down for more than a week.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is an electron beam machine, and very high current (up to 400 micro-Amps!), so the details would be rather different than the LHC beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, about $60\text{ }\mu\text{A}$ of $5.5 \text{ GeV}$ beams from that machine partially melted one of my iron targets (thick enough that about 6% of the beam interacted with the target) despite a raster spreading the beam over roughly $2\text{ mm}^2$, because we didn't have good enough thermal contact with the water-cooled frame of the target ladder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;High energy, high current beams can carry a lot of power.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Back to the question as asked:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Treat your hand as water. The particle data booklet puts the energy loss per proton at around $2.5\text{ MeV/g/cm}^2} or something like 4--8 MeV though your hand, depending on how chubby you are. (We're only a few orders of magnitude above the minimum ionization energy, so this is not very sensitive to the actual beam energy.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/3468/what-would-happen-if-you-put-your-hand-in-front-of-the-7-tev-beam-at-lhc/3471#3471&quot;&gt;Phillip says&lt;/a&gt; $1.2 \times 10^{11}$ protons per beam in the ring, about 11,000 passes per second ($3.0 \times 10^8\text{ m/s} / 27\text{ km}$), so $1.3 \times 10^{15}$ protons per second is $8 \times 10^{15}\text{ MeV/s} = 1300\text{ J/s}$ is a fair bit of heat, and results in heating of about $300\text{ K/s/(cubic cm exposed)}$. To finish up here we will have to know something about the beam diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your reflex time to move your hand is on order of 0.1--0.2 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's going to hurt: you will get badly burned, and the damage will extend though the whole depth of the exposed flesh, rather than being limited to the surface as with the contact burns we are all familiar with.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-03-06T01:56:30.417" LastActivityDate="2011-03-06T01:56:30.417" />
  <row Id="6414" PostTypeId="2" ParentId="6412" CreationDate="2011-03-06T02:08:06.853" Score="5" ViewCount="" Body="&lt;p&gt;I always thought of Fierz identities as a kind of completeness relation (*) for products of spinors. To use the bra-ket notation: &#xA;$$|a\rangle\langle b| = \sum k_i \langle b|M_i|a\rangle M_i$$ &#xA;for some convenient trace orthogonal basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the $k_i$ for the specific basis and space that you're working in, you multiply by some $M_j$ and take the trace:&#xA;$$ tr(M_j |a\rangle\langle b|) = \langle b|M_j|a\rangle&#xD;&#xA;  = \sum_i k_i tr( M_i M_j ) \langle b|M_i|a\rangle&#xD;&#xA;$$ &#xA;since the basis is orthogonal with respect to the trace, we see that $k_i^{-1} = tr( M_i M_i )$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is then used to prove the Fierz identities as&#xA;$$ \langle a|U|b\rangle \langle c|V|d\rangle&#xD;&#xA; = \sum_i\langle a|U (k_i \langle c|M_i|b\rangle M_i) V|d\rangle&#xD;&#xA; = \sum_i k_i \langle c|M_i|b\rangle\langle a|U M_i V|d\rangle&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus you get your Fierz rearrangements.  Note that depending on your definitions, if you're deriving the Fierz identities for anticommuting spinors, then you might have a sign factor in the first trace formula and definition of $k_i$ that I gave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard Fierz identities are for 4-dimensional spinors, with the basis of gamma matrices&#xA;$$ 1\,,\quad \gamma_\mu\,,\quad &#xD;&#xA;   \Sigma_{\mu\nu}\;(\mu&amp;lt;\nu)\,,\quad &#xD;&#xA;   \gamma_5\gamma_\mu\,,\quad \gamma_5 $$&#xA;which has $1 + 4 + 6 + 4 + 1 = 16$ elements, which is what you'd expect for 4*4 matrices.&#xA;The basis elements are normally on both the left and right hand side of the Fierz identity. The traces for the above basis can be calculated from the Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_matrices#Trace_identities&quot; rel=&quot;nofollow&quot;&gt;Gamma matrix page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the previous question, &lt;a href=&quot;http://physics.stackexchange.com/q/6248/429&quot;&gt;Some Majorana fermion identities&lt;/a&gt;, I used 2-component notation to check the identities. This is convenient since the completeness relation is really simple. Using the conventions of &lt;a href=&quot;http://books.google.com.au/books?id=jZ1Y1Z3BoicC&quot; rel=&quot;nofollow&quot;&gt;Buchbinder and Kuzenko&lt;/a&gt; the spinor completeness relation is simply the decomposition into an antisymmetric and symmetric part:&#xA;$$ \psi_\alpha \chi_\beta = \frac12\varepsilon_{\alpha\beta}\psi\chi - \frac12(\sigma^{ab})_{\alpha\beta}\psi\sigma_{ab}\chi \,,&#xD;&#xA;$$&#xA;and similarly for the dotted-spinors (complex conjugate representation).&#xA;This means that you don't need a table of coefficients to do Fierz rearrangements, all the steps fit easily in your memory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what the best reference for Fierz identities is. Maybe have a look at &lt;a href=&quot;http://arXiv.org/abs/hep-ph/0306087v1&quot; rel=&quot;nofollow&quot;&gt;Generalized Fierz identities&lt;/a&gt; and references within.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Footnote (*)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You can also think of the completeness relation purely in terms of the matrices.&#xA;This is an alternate approach to derive the Fierz identities and is the one used in &lt;a href=&quot;http://arXiv.org/abs/hep-ph/0306087v1&quot; rel=&quot;nofollow&quot;&gt;Generalized Fierz identities&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="429" LastEditorUserId="429" LastEditorDisplayName="" LastEditDate="2011-03-06T02:45:12.813" LastActivityDate="2011-03-06T02:45:12.813" />
  <row Id="6415" PostTypeId="1" CreationDate="2011-03-06T03:02:59.313" Score="1" ViewCount="170" Body="&lt;p&gt;I have several question regarding axion. Could anyone give me some brief introduction to what is a axion string, axion field and how is this related to fermion zero mode and chiral zero mode?&lt;/p&gt;&#xA;" OwnerUserId="2397" LastActivityDate="2011-03-06T06:20:41.613" Title="Axion related questions" Tags="&lt;quantum-field-theory&gt;&lt;condensed-matter&gt;&lt;qcd&gt;" AnswerCount="1" />
  <row Id="6416" PostTypeId="2" ParentId="6411" CreationDate="2011-03-06T04:09:03.110" Score="4" ViewCount="" Body="&lt;p&gt;The partition function tells you about spectrum of dimensions and degeneracies. It doesn't tell you about OPEs (and higher point functions), so it doesn't determine the theory uniquely. The problem is somewhat reminiscent of the question &quot;can you hear the shape of a drum?&quot;, for which the answer is negative: you can have drums of different shapes that produce the same harmonics. That would be analogous to two different CFTs that have the same spectrum and hence partition function, but differ in more detailed properties.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-03-06T05:20:00.810" LastActivityDate="2011-03-06T05:20:00.810" CommentCount="8" />
  <row Id="6417" PostTypeId="2" ParentId="6400" CreationDate="2011-03-06T05:22:03.867" Score="0" ViewCount="" Body="&lt;p&gt;I am a bit puzzled by the answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earth moon system has a specific distance at a given moment and the gravitational force on the earth a specific magnitude that gives a &quot;spin&quot; to the earth. Does the composition of the earth play a role to the amount of the spin? i.e. water versus land?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may be completely wrong but the bulge of the moon goes at the same rate over land and sea, irrespective of friction. I would expect the angular momentum transfer to exist even if the earth were rock solid and no bulge possible, in the same magnitude, the same way the moon's angular momentum changes by the interaction with the earth's gravitation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus my answer is that the problem with the power plants would be a matter of earth angular momentum conservation that will not be transferred to the moon, and thus the power plants would make no difference because angular momentum of the earth will be conserved whether they are there or not.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-06T05:22:03.867" CommentCount="9" />
  <row Id="6418" PostTypeId="1" CreationDate="2011-03-06T05:57:46.120" Score="5" ViewCount="125" Body="&lt;p&gt;In quantum chromodynamics, long flux tubes will always snap because a quark-antiquark pair gets created from the vacuum, and hadronization results with a quark attached to each new end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In string theory, if we try to stretch a single string out a long distance, will it also snap? If open strings with Neumann boundary conditions are allowed, this can always happen. Suppose this is not the case. If D0-branes exist, we can create a pair of them from the vacuum, and the string will still snap. Even if no D0-branes don't exist, or space filling branes either, as long as we have some Dp-brane for some p in between, the Dp-brane can be wrapped up as finite closed bubble, and we can create a pair of such bubbles. The string will still snap into &quot;tadpoles&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some string theories, strings carry charges and are BPS states. Can they snap? I doubt it.&lt;/p&gt;&#xA;" OwnerUserId="2398" LastActivityDate="2011-03-06T06:52:09.467" Title="Can long strings always snap?" Tags="&lt;string-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6419" PostTypeId="1" AcceptedAnswerId="6425" CreationDate="2011-03-06T06:12:56.583" Score="3" ViewCount="184" Body="&lt;p&gt;If we compare a petrol engine and an electric engine car, it is sufficient for the electric engine car (e.g tesla roadster) just to have 1 gear (forward), whereas petrol engine can have up to 7.&#xA;Why is it sufficient for the electric engine to have a single gear? Is it because it can deliver the required torque always?&lt;/p&gt;&#xA;" OwnerUserId="2351" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-06T13:05:03.820" LastActivityDate="2011-06-20T16:04:21.600" Title="Electric engine transmission" Tags="&lt;torque&gt;" AnswerCount="4" />
  <row Id="6420" PostTypeId="1" CreationDate="2011-03-06T06:20:08.483" Score="6" ViewCount="108" Body="&lt;p&gt;Take two identical closed strings, both tracing out exactly the same path in space. These two strings are coincident everywhere. Call this state I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a single closed string following exactly the same closed path as in the first case, but not closing just yet. The string goes around exactly the same path once again before closing in on itself. Two cycles around the same closed path. Call this state II.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;String field theory tells us unambiguously states I and II are distinct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stretch this closed path to make it much larger than the string scale. Supposedly, stringy nonlocality only happens at the string scale. States I and II still differ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Partition target space into local regions the size of the string scale. The path cuts across a chain of such local regions. If string theory were local, we can reconstruct the state of the entire universe from the restricted states of each subregion if we allow for quantum entanglement between regions. Locally, states I and II ought to be indistinguishable over each local region. For each local region, we always see two string segments passing through it. Thus, states I and II have to be identical?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can't be. Either string theory is inherently nonlocal over scales much larger than the string scale, or it obeys Maxwell-Boltzmann statistics and not Bose-Einstein statistics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't some Aharonov-Bohm effect. Even if we include all the local regions in the &quot;interior&quot; of the closed loop, this doesn't change matters the least bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. Please reread my question more carefully. What you call configuration III is actually my configuration I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PPS: Let me try to understand your explanation. If we have N coincident strings, or a string which winds round the same loop N times or any other combination in between, this can be described by an $S_N$ discrete gauge symmetry. The conjugacy class of the holonomy of this discrete gauge symmetry around the loop distinguishes between the various combinations. Feel free to correct me if I am wrong. This has the flavor of parastatistics, does it not?&lt;/p&gt;&#xA;" OwnerUserId="2398" LastEditorUserId="2398" LastEditorDisplayName="" LastEditDate="2011-03-06T08:16:35.953" LastActivityDate="2011-03-06T08:30:02.383" Title="Statistics and macrolocality in string theory" Tags="&lt;string-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="6421" PostTypeId="2" ParentId="6415" CreationDate="2011-03-06T06:20:41.613" Score="2" ViewCount="" Body="&lt;p&gt;Axionic strings are special cosmic strings defined by the condition that if you look at the axion $a(\phi)$ as a function of the cylindrical coordinate around the cosmic string (which I choose to be along the $z$-axis), it winds around the origin, so&#xA;$$a(\phi) = a_0 \exp(i\phi) $$&#xA;There is a nontrivial &quot;monodromy&quot; around the axionic string, we say. At the very center of the string, $a$ (a complex number) continuously goes to zero. So such an object is protected (at least temporarily, in the case of a gauge $U(1)$ symmetry) by a topological &quot;knot&quot; or &quot;twist&quot;. Note that $a$ is a complex scalar field and it is charged under a $U(1)$ group which explains its lightness; the lightness is necessary for it to play its standard role in QCD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axionic string nevertheless behaves as an object and it has &quot;collective coordinates&quot; - fields that only exist near the locus of the string and are only functions of the coordinate along the length of the string (and time). In particular, the part of the degrees of freedom that doesn't depend on the position along the string are &quot;zero modes&quot;, see&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/2951/what-are-zero-modes&quot;&gt;What are zero modes?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-06T06:20:41.613" />
  <row Id="6422" PostTypeId="2" ParentId="6420" CreationDate="2011-03-06T06:33:50.200" Score="4" ViewCount="" Body="&lt;p&gt;Here, state I is just one closed string going along the path. State III will be used for two strings on the same path - which is called I in the question. Sorry for the permutations: I also discussed a more trivial issue and I don't want to delete it. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The states I and II are always distinct and there is no non-locality implied by this fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, strings usually carry a charge with respect to the B-field - a two-form field in the NS-NS sector; among the 5 ten-dimensional superstring theories, type I theory is the only exception because its string are unoriented. If the path in I,II goes around a non-contractible circle in spacetime, the charge will manifest itself as a &quot;winding number&quot; $w$ - the configurations I,II will have winding numbers $w=1,2$, respectively, and those values of $w$ will behave as different values of electric charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The configuration II simply carries twice as high a charge than the configuration I and is completely different - just like two point-like W-bosons on top of each other are different from one W-boson. In fact, the string example of yours only differs by an added dimension to the strings' shape. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not hard to agree that two W-bosons on top of one another are different from one W-boson. Would you agree that one doesn't need any non-locality to deduce this fact? The configurations of strings I,II differ at each point along the string, not just globally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, your more general point is totally valid: the interactions of strings are totally local in space as long as we look at strings as extended objects, and the right world sheet theory describing these interactions is also completely local on the world sheet. This condition requires that two interactions (processes) where strings are rearranged locally in some region must always have the same amplitude regardless of what the strings in both cases are doing away from the interaction point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you wanted to distinguish in the question was your configuration II and another configuration III (you called I) which has two coincident strings closing around the same path. The configurations II and III may be very hard to distinguish, indeed - in some sense, II is a bound-state version of III. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of matrix string theory, II is a &quot;long string&quot; while III is a pair of strings - and these two configurations may be continuously connected on the configuration space of a Yang-Mills description of perturbative string theory which is what matrix string theory really is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/9701025&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9701025&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/abs/hep-th/9703030&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9703030&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The configurations II, III only differ by a global monodromy of the matrix theory's gauge field along the long path in space that the strings are wrapped around: the configuration III involves a permutation of the two strands while the configuration II doesn't. The permutation group $S_N$ is embedded into the gauge group $U(N)$ with the same value of $N$ in the context of any matrix-theory model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In matrix string theory, all closed-string-like (crossing-over) interactions in string theory are represented by the addition of a transposition to the permutation that remembers the monodromy. Such a transposition may occur whenever the points of several strands are close to each other in the physical spacetime and may be described as the DVV interaction - see the 2nd paper above for details. Once we study it rationally, it is counterproductive to draw the strings in a singular configuration. The same mechanism is really taking place whether or not the remainders of the strings coincide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that the $U(N)$ gauge group arises from D-branes - the fundamental strings as modeled by matrix string theory are really D1-branes which are dual to them, and which appear in a particular extreme kinematic regime: the duality is the pillar of a derivation of matrix string theory (by Seiberg). At any rate, the comments about the differences between II and III really apply to D1-branes, too (or especially to them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matrix string theory is discussed here on physics stack exchange:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/6424/good-introductory-text-for-matrix-string-theory&quot;&gt;Good introductory text for matrix string theory&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-06T08:30:02.383" LastActivityDate="2011-03-06T08:30:02.383" />
  <row Id="6423" PostTypeId="2" ParentId="6418" CreationDate="2011-03-06T06:52:09.467" Score="5" ViewCount="" Body="&lt;p&gt;As you correctly say, QCD flux tubes may snap because (light) quark-antiquark pairs may be created. In string theory, open strings must satisfy the condition that the end points are attached to a D-brane - which may have any dimension as you equally correctly wrote. The endpoints of open strings are points within D-branes that may be identified with quarks or antiquarks in QCD in this (more than) analogy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of creating a D0-brane-anti-D0-brane pair is creative but it doesn't work because D0-branes - and other D-branes - are too heavy and the string simply can't have enough energy to create these new heavy objects (unless the string carries so much vibrational energy on it that the perturbative expansions are hopelessly broken).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The D0-brane mass goes like $m_s/g_s$ which is the same mass as $1/g_s$ times the string length of a string. One would need to use a very long piece of string to get the required energy for the creation of the new D0-brane and its antiparticle - and this won't happen. This is true not just for D0-branes but all D-branes - all their tensions are proportional to $1/g_s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, a D0-brane carries the same mass/energy as a highly excited string mode whose excitation level is $1/g_s$ - well, in fact, it has to be $N\approx 1/g_s^2$ because it's the squared mass that $N$ is proportional to. For such super excited modes, the perturbative approximation of string theory doesn't work. Indeed, the fact that non-perturbative objects such as D0-branes may emerge out of the energy of strings is a way to prove that the perturbative expansions fail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-perturbatively, for large enough values of $g_s$, anything can happen. The perturbative expansions fail, D-branes become as easy-to-be-created as the (no longer) fundamental strings, and there are lots of open strings attached to cheap D0-branes and other D-branes flying in a region with not so high energy.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-06T06:52:09.467" />
  <row Id="6424" PostTypeId="1" CreationDate="2011-03-06T07:30:45.687" Score="11" ViewCount="352" Body="&lt;p&gt;Where can I find a good introductory text for matrix string theory? Most textbooks don't cover it, or only cover it very superficially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the basic idea behind matrix string theory? How can matrices be equivalent to strings?&lt;/p&gt;&#xA;" OwnerUserId="2398" LastActivityDate="2011-03-06T08:50:15.173" Title="Good introductory text for matrix string theory" Tags="&lt;string-theory&gt;" AnswerCount="1" FavoriteCount="5" />
  <row Id="6425" PostTypeId="2" ParentId="6419" CreationDate="2011-03-06T07:35:44.513" Score="5" ViewCount="" Body="&lt;p&gt;Yes, electric motors can always deliver the required torque. What really matters is the efficiency of the conversion. Over a long range of speeds, the motors may convert 90 percent of the energy to its mechanical form. You know, it's just some electromagnetic fields that may become arbitrarily stronger - depending on how much energy is sent into them - and that always do the required work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, petrol engines are only highly efficient over a relatively narrow interval of the number of revolutions per minute (RPM) of the engine. It takes a certain time for the fuel to get burned or explode, for the pressure to spread in the cylinders, and whatever else needs to be done in each cycle. Moreover, at too high frequencies, the motor could get overheated and damaged, among other things. That's why petrol engines want to keep the number of RPM relatively constant and a gearbox is needed to translate the fixed frequency of the motor to a variable speed of the car.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-06T07:35:44.513" CommentCount="0" />
  <row Id="6426" PostTypeId="2" ParentId="6424" CreationDate="2011-03-06T08:02:16.003" Score="13" ViewCount="" Body="&lt;p&gt;Dear Eliza, matrix string theory may be viewed just as a variation of &lt;a href=&quot;http://arxiv.org/abs/hep-th/9610043&quot; rel=&quot;nofollow&quot;&gt;BFSS Matrix Theory&lt;/a&gt;, although arguably an important one, and the original papers are the full introductions at the same moment. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/9701025&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9701025&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/abs/hep-th/9702187&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9702187&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/abs/hep-th/9703030&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9703030&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some of the few hundred followups deal with some more technical issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last paper in the list above, which is the newest one, should be the most optimized one. To say the least, it contains the most detailed treatment of the interactions. One may enumerate a couple of reviews of BFSS Matrix Theory. Some of them dedicate some time to matrix string theory, some of them don't. For example, see&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/9712072&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9712072&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/abs/hep-th/0101126&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0101126&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A derivation of BFSS Matrix Theory was given by Seiberg:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/9710009&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9710009&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;M-theory in 11 dimensions may be compactified on a nearly light-like (slightly space-like) circle - which is still consistent. $X^-$ becomes a periodic variable, $X^-\approx X^-+2\pi R$. (This light-cone treatment was automatically used in my paper above but it was &lt;a href=&quot;http://arxiv.org/abs/hep-th/9704080&quot; rel=&quot;nofollow&quot;&gt;Lenny Susskind&lt;/a&gt; who took credit for it months later - much ado about nothing. The original BFSS paper was using the &quot;infinite momentum frame&quot;.) In the lightlike limit, a Lorentz boost may map the compactification to a compactification of M-theory on a very short spatial circle in 11D Planck units (because the proper length of the nearly light-like circle was tiny) - which is type IIA string theory. Units of momenta along the compact light-like direction become D0-branes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinematic regime guarantees that these D0-branes are non-relativistic. They're well-described by the non-relativistic supersymmetric quantum mechanics - the matrix model - which is the dimensional reduction of the 10D supersymmetric Yang-Mills theory to 0+1 dimensions. The gauge group is $U(N)$. It has 16 non-trivial real supercharges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one can show that all of physics of M-theory, if studied in the light-cone gauge, is equivalent to an ordinary non-gravitational matrix model - a quantum mechanical model with matrix degrees of freedom. The eigenvalues of the $X^i$ matrices may be viewed as the positions of the gravitons (or their superpartners) in 11 dimensions; a threshold (zero-binding-energy) bound state of several such eigenvalues (which can be proved to exist, a remarkable property of $SU(N)$ supersymmetric quantum mechanics) are gravitons that carry a higher number of units of the quantized light-like (longitudinal) momentum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All interactions are encoded in the off-diagonal elements of the matrices which are classically zero but whose virtual quantum effects make the eigenvalues interact so that the resulting picture is indistinguishable from 11D supergravity at low energies; much like AdS/CFT, it is an equivalence of a gravitational theory and a non-gravitational one (in some sense, the compact light-like direction $X^-$ of the matrix model is the holographic direction). The model contains black holes and all other expected objects, too: extended branes may be added. The identical natural of gravitons and gravitinos - with the right Bose-Einstein and Fermi-Dirac statistics - appears because the permutation group is embedded into the $U(N)$ gauge group of the quantum mechanical model, and all physical states must therefore be invariant under this $U(N)$ i.e. also $S_N$. The compact M2-branes (membranes) appear most directly because the whole BFSS matrix model may be viewed as a discretization of the M2-brane world volume theory in M-theory - assuming that the world volume coordinates generate a non-commutative geometry. This equivalence may be derived in a straightforward way, especially for the toroidal and spherical topology of the M2-branes. M5-branes are harder to see but they must be there, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The BFSS Matrix Theory above gave the first complete definition of M-theory in 11 dimensions (the whole superselection sector of the Hilbert space) that was valid at all energies. It's a light-cone-gauge description where sectors with different values of $p^+ = N/R$ are separated and separately described by the $U(N)$ quantum mechanical models. I forgot to say - to really decompactify the $X^-$ coordinate, one needs to send its radius $R$ to infinity. Because $p^+=N/R$ is fixed (physical momentum), $N$ has to be sent to infinity, too. The infinite-space physics is always obtained as the large $N$ limit of calculations in $U(N)$ matrix models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Matrix string theory&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One may apply the same derivation to find the matrix model of other superselection sectors besides the 11D vacuum of M-theory, too. It includes some (simple) compactifications; the right matrix model isn't known for all compactifications. In particular, matrix models for type IIA string theory and heterotic $E_8\times E_8$ string theory have a very simple form. Instead of a quantum mechanical model i.e. 0+1-dimensional field theory arising from the D0-branes, one ends up with a 1+1-dimensional supersymmetric gauge theory originating from D1-branes of type IIB (an extra T-duality is added to the derivation), compactified on a cylinder, the so-called matrix string theory (although the historically more correct name is &quot;screwing string theory&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In matrix string theory, again, the eigenvalues of the $U(N)$ matrices $X^i$ are interpreted as positions of points on strings in the transverse 8-dimensional space (the two light-like directions are treated separately in light-cone gauge: one of them, $X^+$, is the light-like time and the other, $X^-$, is compactified). Those eigenvalues $X^i_{nn}(\sigma)$ still depend on $\sigma$, the spatial coordinate of the cylinder on which the gauge theory is defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, one may obtain strings of an arbitrary length by applying permutations on the eigenvalues: the length determines the light-like longitudinal momentum $p^+=N/R$ which is quantized because $X^-$ is compactified. All these permutations are allowed because $U(N)$ is gauged as a symmetry in the matrix model. Consequently, perturbative type IIA and HE string theory with arbitrary numbers of strings are defined by an orbifold conformal field theory - a single string propagating on the orbifold $R^{8N}/S_N$, if you wish (with the extra fermionic degrees of freedom, too). The permutations now guarantee not only the indistinguishability of strings in the same vibration states but also the existence of strings with higher values of $p^+$ - it looks like your configuration II on the world volume if you wish (but the path in the spacetime is generic) - as well as the validity of the $L_0=\tilde L_0$ condition in the continuum limit, among other things. Interactions work as expected, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The perturbative string theories always emerge in the light-cone gauge Green-Schwarz description. In the heterotic case, the $E_8$ groups arise from the fermionic representation of the $E_8$ current algebra: those extra fermions are fermions transforming in the fundamental representation of $U(N)$; sixteen of them per single Hořava-Witten boundary i.e. per single $E_8$ while the gauge group has to be changed to $O(N)$ and some degrees of freedom (originally Hermitian matrices) become symmetric real tensors of $O(N)$ while others are antisymmetric, see the paper below and its followups:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/9612198&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9612198&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The main advantage of matrix string theory is that while it may be explicitly shown to agree with type IIA or HE string theory at the weak coupling, it provides one with the exact non-perturbative description at any value of the string coupling. In particular, one may see that when the coupling is sent to infinity, matrix string theory reduces to the original BFSS matrix model for M-theory in large 11 dimensions (with an $E_8$ domain wall, in the heterotic case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar matrix models exist for type IIB in ten dimensions, too: one needs the maximally supersymmetric $2+1$-dimensional superconformal field theory which became relevant for the BLG construction (which later transmuted to the ABJM membrane minirevolution). The methods of matrix models become more complicated for backgrounds with additional compact dimensions - by compactifying spacetime dimensions (dimensional reduction), one needs to add dimensions to the matrix model (&quot;dimensional oxidation&quot;) - and no matrix models are known if more than 5 transverse spacetime coordinates are compactified (which is why we can't define matrix models for phenomenologically interesting compactifications, at least as of 2011).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, a long list of introductory literature about all kinds of string-theoretical topics, most recently updated in 2004, is here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0311044&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0311044&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-06T08:50:15.173" LastActivityDate="2011-03-06T08:50:15.173" />
  <row Id="6427" PostTypeId="2" ParentId="6419" CreationDate="2011-03-06T12:27:07.173" Score="2" ViewCount="" Body="&lt;p&gt;There are some types of electric motors who are able to deliver a high torque from 0 rpm on.&#xA;(others do that not at all) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The selection of such a electric motor is for some more features than just efficency. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;high Torque from rest &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;usable as generator (to enable feedback of energy on braking) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;low weight (ideally including the weight of a clutch and gear omitted) to cope for the extra weight of the accumulator. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Maybe as light/cheap to use two of them (close to wheels) saving the differential. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Weight and cost of gear, differential, and clutch are big points. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-06-20T16:04:21.600" LastActivityDate="2011-06-20T16:04:21.600" />
  <row Id="6428" PostTypeId="1" CreationDate="2011-03-06T12:51:52.867" Score="2" ViewCount="192" Body="&lt;p&gt;Can one &quot;interpret&quot; Snell's law in terms of QED and the photon picture? How would one justifiy this interpretation with some degree of mathematical rigour? At the end I would like to have a direct path from qft to snell's law as an approximation which is mathematically exact to some degree and gives a deeper physical insight (i.e. from a microscopic = qft perspective) to Snell's law.&lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorUserId="667" LastEditorDisplayName="" LastEditDate="2011-03-06T16:10:27.840" LastActivityDate="2011-03-06T16:10:27.840" Title="Snell's law starting from qft?" Tags="&lt;quantum-field-theory&gt;&lt;geometric-optics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6429" PostTypeId="2" ParentId="6428" CreationDate="2011-03-06T13:40:28.133" Score="2" ViewCount="" Body="&lt;p&gt;Sure. Start with &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_electrodynamics&quot; rel=&quot;nofollow&quot;&gt;QED&lt;/a&gt;, obtain &lt;a href=&quot;http://en.wikipedia.org/wiki/Maxwell%27s_equations&quot; rel=&quot;nofollow&quot;&gt;Maxwell's equations&lt;/a&gt;, do the &lt;a href=&quot;http://en.wikipedia.org/wiki/Paraxial_approximation&quot; rel=&quot;nofollow&quot;&gt;paraxial approximation&lt;/a&gt; and finally use &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat%27s_principle&quot; rel=&quot;nofollow&quot;&gt;Fermat's principle&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2011-03-06T13:40:28.133" CommentCount="8" />
  <row Id="6430" PostTypeId="2" ParentId="6419" CreationDate="2011-03-06T14:00:49.573" Score="3" ViewCount="" Body="&lt;p&gt;Any kind of motor will have a characteristic relation of torque to speed of rotation. Typically at any speed, there is a maximum torque available from the motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With combustion engines, the shape of the torque-vs-speed curve is fixed (afaik) and the engine must be operated at significant speed in order to deliver enough power to be effective at moving the car. Since the load conditions vary widely from stopped to cruising at top speed, it's necessary to add a device, the multi-gear transmission, to mechanically transform power at the engine's preferred speed/torque ratio to power at the load's required speed/torque ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, the torque-vs-speed curve of electric motors can be dynamically changed, by alteration of the field strength and commutation angle inside the motor. This means that with modern power electronics, the electric motor can be reconfigured at will, so as to grind out large torques at low speeds, or smaller torques at high speed, to efficiently deliver mechanical power that is already at the correct speed/torque ratio, obviating the need for a mechanical transmission.&lt;/p&gt;&#xA;" OwnerUserId="717" LastActivityDate="2011-03-06T14:00:49.573" />
  <row Id="6431" PostTypeId="2" ParentId="6419" CreationDate="2011-03-06T15:26:08.140" Score="1" ViewCount="" Body="&lt;p&gt;Another important point is that the combustion engine not only has a favored RPM range, but also a favoed output range. If the vehicle is moving at a given speed, but needs little or no torque to maintain speed (say on a slight downhill), the engine is still turning, and various dissapative processes (friction of engine parts, and gas being moved in pistoms etc.), so&#xA;efficiency during very low torque conditions is low. At low RPMs, heat loss via conduction through cylinder walls grows as a percentage of the heat of combustion, lowering efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All told, the relative efficiency between these two types of propulsion will be highly dependent upon the driving cycle. Are there lots of starts and stops (or up hill/down hill)? That would favor electric. If the need for is steady continuous power, the internal combustion engine can be tuned for those conditions (load and RPM), and sized for that level of power, rather than for rapid acceleration, and will look much better.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-03-06T15:26:08.140" />
  <row Id="6432" PostTypeId="1" CreationDate="2011-03-06T15:28:20.827" Score="0" ViewCount="113" Body="&lt;p&gt;There is well known gravitational redshift of real photons. What about  gravitational  redshift  of virtual photons of  charged neutron star? Is electrostatic force become weaker while mass of  charged neutron star is growing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can electrostatic field of a charged particle swallowed by a black hole escape from the event horizon? How can charged black holes exist at all?&lt;/p&gt;&#xA;" OwnerUserId="238" LastActivityDate="2011-03-06T15:46:07.467" Title="Gravitational redshift of virtual photons" Tags="&lt;gravity&gt;&lt;black-holes&gt;&lt;gravitational-redshift&gt;&lt;virtual-photons&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6433" PostTypeId="2" ParentId="6432" CreationDate="2011-03-06T15:46:07.467" Score="-1" ViewCount="" Body="&lt;p&gt;There is a difference between the value of a field, and the excitations of that field.  If you have a speherically symmetric charge distribution collapse to a black hole (while maintaining the spherical symmetry), there is no electric nor gravitational radiation--Gauss's Law and Birkhoff's Theorem work in tandem to keep the gravitational and electric forces constant far from the object, and since there is nothing to propogate, nothing needs to propogate.  The field already lives out in the place where information about where the charge distribution needs to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a photon, then?  A photon is a propogating disturbance in the electromagnetic field.  You need to have an accelerating charge in order to produce such a disturbance, and there's no need to violate causality at the black hole horizon in order for a particle to feel force from a field that already exists locally--if, say, an electron were coming close to a charged black hole, it would see the field that existed there, scatter off of the field, and emit a photon.  (and of course, it's new field would be set up and propogate to the black hole, and scatter the black hole--but there's no problem with information falling IN to a black hole, which is allowed to externally display it's mass, charge, angular momentum and linear momentum, all of which get inprinted on the horizon as matter falls in).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in summary, it's not necessary for a photon to be emitted in order for a second particle to feel a force--electrostatic forces in fact don't involve photons at all (at least, until the scattering event begins).    &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-03-06T15:46:07.467" CommentCount="3" />
  <row Id="6434" PostTypeId="2" ParentId="5552" CreationDate="2011-03-06T16:08:59.290" Score="-1" ViewCount="" Body="&lt;p&gt;Before attempting an Answer to this Question, I up-voted Anna V's and Daniel Grumiller's Answers. If we consider that &quot;QED&quot; includes all the calculational apparatus of an expansion as a series of divergent Feynman integrals, their regularization, resummation, and renormalization, then Daniel Grumiller's criticism of the question seems unavoidable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider &quot;QED&quot; to be defined more minimally, however, just by a deformed free field Hamiltonian or Lagrangian, without commitment to a particular elaboration of that definition, then it seems possible that there is a reasonable, analytic answer, but we have not so far constructed any such mathematical approach. In such a case, however, I suppose that any calculational treatment would nonetheless have to make contact with Anna V's request for engineering utility, which I further suppose requires a convergent series expansion of some sort on some finite domain from some &lt;em&gt;different&lt;/em&gt; starting point. It seems unlikely, therefore, that we should start from the &quot;bare particle&quot; free field, since we already know that constructing a series centered on that starting point that is convergent in some sense requires extraordinary, time-consuming measures. To my knowledge, no-one has proved that every possible expansion of QED is everywhere divergent, but functional mathematics is not the relatively elementary playground of complex analytic functions, so such a proof may be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this prelude, I feel myself in a position to give partial answers to the sequence of subsidiary questions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do we replace the unknown/wrong physics with certain/right one? &lt;strong&gt;Maybe, maybe not.&lt;/strong&gt; Can&#xA;  anybody describe it without appealing&#xA;  to unphysical bare particles? &lt;strong&gt;I think the bare particles thing is a problematic starting point, I think we have to start somewhere else.&lt;/strong&gt; Have we&#xA;  an idea about the real electron from&#xA;  QED? &lt;strong&gt;If QED has an alternative series expansion that has some engineering utility, then the current expansion presumably has &lt;em&gt;some&lt;/em&gt; relationship to the new expansion, in which case, yes.&lt;/strong&gt; If so, why we cannot use it as&#xA;  the input to construct a reasonable&#xA;  theory from the very beginning? &lt;strong&gt;It's not necessarily the theory that's problematic, it may be the calculational approach. Just as one example, all quantized free fields already have something similar to the scalar field's property that a 2-point covariance function $C(x-y)$ diverges for $x\rightarrow y$, which makes free fields a questionable starting point.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, however, I acknowledge a fundamental difference from your approach as I understand it, Vladimir, that, to me, &lt;em&gt;algebras of operator-valued distributions&lt;/em&gt; are a &quot;better&quot; starting point than &lt;em&gt;particles&lt;/em&gt;, so that I have had to contort my answers to get near your starting point, a process which may have left the above near to nonsense, though I find the process instructive for my own rethinking of my ideas. For a long time I have thought that QED was itself a poor starting point for constructing mathematically consistent models, whether physical or not, but (for reasons unspoken here) I find myself more open to the idea as a result of attempting to answer this question.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-03-06T16:08:59.290" />
  <row Id="6435" PostTypeId="1" AcceptedAnswerId="6465" CreationDate="2011-03-06T16:39:26.347" Score="2" ViewCount="187" Body="&lt;p&gt;The Pierre Auger Observatory site mentions the detection of a 3E20 eV (48 J) cosmic ray whose energy, well above the GZK cutoff, was based on an analysis of its atmospheric shower. This was equivalent to the kinetic energy of a baseball with a speed of 79.5 m/s or 177 mph. Of course, cosmic rays with such ulta-high energies are extremely rare.&#xA;What kind of damage would occur if an astronaut or a space vehicle encountered such a cosmic ray? How would the damage differ from that from the hypothetical 79.5 m/s baseball?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-07T11:15:55.937" LastActivityDate="2011-03-09T01:10:29.137" Title="Cosmic ray hazards" Tags="&lt;energy&gt;" AnswerCount="3" />
  <row Id="6436" PostTypeId="1" CreationDate="2011-03-06T17:27:36.120" Score="1" ViewCount="154" Body="&lt;p&gt;I'm interested in knowing if ,in the context of AdS_{3}/CFT_{2}, we can (and how to) express the geodesic propagators on the bulk space of the Euclidean AdS_{3} black holes, in terms of  meromorphic functions on its conformal boundary CFT_{2}.&lt;/p&gt;&#xA;" OwnerUserId="2237" LastActivityDate="2011-03-06T22:42:03.457" Title="Can the geodesic propagators in the Euclidean BTZ black hole can be written in terms of meromorphic functions on its conformal boundary?" Tags="&lt;mathematics&gt;&lt;ads-cft&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="6437" PostTypeId="2" ParentId="6435" CreationDate="2011-03-06T17:35:22.437" Score="6" ViewCount="" Body="&lt;p&gt;Ultra-high energy cosmic ray all come from a very, very long way away (anything with the power to create them nearby would constitute a danger to life as we know it). I think the preferred mechanism these days is dynamic acceleration in the jets formed by active galactic nuclei, but don't quote me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, ultra-relativistic though they are, that means they are stable particle. Mostly protons, in fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When A ultra-high energy proton passes though a modest amount of matter, like an astronaut, we can model it's energy loss very simply. The graph in the Particle Physics Data Book, doesn't actually go high enough, but we can extrapolate and say the energy loss is still less than $10\text{ MeV/g/cm}^2$. Our astronaut has a density pretty near $1\text{ g/cm}^3$ and a average thickness (allowing for all aspect ratios) of around 50 cm.&#xA;So the total energy to be expected in only 500 MeV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is ionizing radiation, of course, but not qualitatively different from the rest of the cosmic background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the space craft is a smallish, thin-walled can with a little air in it the situation is only a little worse. There is more chance of showering. But it's just more dose, rather than being a spectacular death.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-03-07T19:09:57.497" LastActivityDate="2011-03-07T19:09:57.497" CommentCount="4" />
  <row Id="6438" PostTypeId="1" AcceptedAnswerId="6439" CreationDate="2011-03-06T19:10:10.927" Score="13" ViewCount="1124" Body="&lt;p&gt;A question in four parts.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What are the main problems which supersymmetry purports to solve? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What would constitute lack of evidence for SUSY at the proposed LHC energy scales (e.g. certain predicted superpartners are not in fact observed)? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are there alternative theoretical approaches which would address the SUSY problem set and which would still be credible in such an LHC no-SUSY-scenario?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Where would LHC-disconfirmation of SUSY leave String Theory?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I would like to think that these four points could be taken together as one question.&lt;/p&gt;&#xA;" OwnerUserId="1394" LastActivityDate="2011-03-06T20:56:18.403" Title="What if the LHC doesn't see SUSY?" Tags="&lt;string-theory&gt;&lt;supersymmetry&gt;&lt;large-hadron-collider&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="6439" PostTypeId="2" ParentId="6438" CreationDate="2011-03-06T19:55:48.940" Score="14" ViewCount="" Body="&lt;p&gt;First, let me emphasize something that is being covered by a thick layer of misinformation in the media these days: it is totally premature to conclude whether the LHC will see SUSY or not. The major detectors have only collected 45/pb (and evaluated 35/pb) of the data. The &quot;slash pb&quot; should be pronounced as &quot;inverse picobarns&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LHC is designed to collect hundreds or thousands times more data than what it has recorded so far, and it should eventually run at a doubled energy (14 TeV total energy instead of the current 7 TeV total energy). Each multiplication of the integrated luminosity (number of collisions) by 10 corresponds to the access of new particles whose masses are approximately 2 times larger or so. It means that the LHC will be able to decide about the existence of new particles at masses that are 4-16 times higher than the current lower bounds (16 also includes the likely upgrade from 2x 3.5 TeV to 7 TeV).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are at least two &quot;mostly independent&quot; parameters with the dimension of mass in SUSY - I mean $m_0$ and $m_{1/2}$. So the number from the previous sentence should really be squared, and in some sensible counting and with a reasonable measure, the LHC has only probed about 1/16 - 1/256 of the parameter space that is accessible to the LHC over its lifetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the only thing we can say now is that SUSY wasn't discovered at an extremely early stage of the experiment - which many people have hoped for but this possibility was never supported by anything else than a wishful thinking. Whether the LHC may see SUSY may remain an open question for several years - unless the LHC will see it much sooner than that. It's an experiment that may continue to 2020 and beyond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't really know where the superpartner masses could be - but they may sit at a few TeV and this would still mean that they're accessible by the LHC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, your questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What SUSY helps to solve&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, SUSY is a natural - and mostly inevitable - consequence of string theory, the only consistent quantum theory that includes gravity as well as the Yang-Mills forces as of 2011. See&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2010/06/why-string-theory-implies-supersymmetry.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/06/why-string-theory-implies-supersymmetry.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this context, supersymmetry is needed for the stability of the vacuum and other things, at least at a certain level. For other reasons, to be discussed below, it's natural to expect that SUSY should be unbroken up to LHC-like energy scales (i.e. that it should be visible at the LHC) - but there's no sharp argument that might calculate the superpartner scale. Some string theorists even say that it should be expected that supersymmetry is broken at a very high scale (near the GUT scale or Planck scale) - because this is a &quot;generic behavior&quot; in the stringy landscape (the &quot;majority&quot; of the minima have a high-scale SUSY breaking which would make SUSY unavailable to any doable experiments) - so these proponents of the anthropic reasoning don't expect SUSY to be seen at the LHC. However, more phenomenological considerations make it more natural for SUSY to be accessible by the LHC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? There are several main arguments: SUSY may offer a very natural dark matter particle candidate, namely the LSP (lightest supersymmetric particle), most likely the neutralino (the superpartner of the photon or Z-boson or the Higgs bosons, or their mixture), that seems to have the right approximate mass, strength of interactions, and other things to play the role of the majority of the dark matter in the Universe (so that the Big Bang theory with this extra particle ends up with a Universe similar to ours after 13.7 billion years). See an article about SUSY and dark matter:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2010/07/susy-and-dark-matter.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/07/susy-and-dark-matter.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, SUSY with superpartners not far from the TeV or LHC energy scale improves the gauge coupling unification so that the strengths of the couplings get unified really nicely near the GUT scale (and maybe incorporated into a single and simple group at a higher energy scale not far from the Planck scale), see:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2010/06/susy-and-gauge-coupling-unification.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/06/susy-and-gauge-coupling-unification.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The unification in the simplest supersymmetric models is only good if the superpartners are not too far from the TeV scale - but if they're around 10 TeV, it's still marginally OK. The same comment with the same value 10 TeV also holds for the dark matter job of the neutralinos discussed above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally and most famously, SUSY with superpartner masses not far from the TeV or LHC scale stabilizes the Higgs mass - it explains why the Higgs mass (and, consequently, the masses of W-bosons and Z-bosons, among other particles) is not driven towards a huge energy scale such as the Planck scale by the quantum corrections (with loops of particle-antiparticle pairs in the Feynman diagrams). Those otherwise expected quantum corrections get canceled at the TeV accuracy if the superpartner masses are near a TeV - and the resulting Higgs mass may then be naturally in the expected 100 GeV - 200 GeV window with an extra 10:1 luck (which is not bad). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lighter the superpartner masses are, the more &quot;naturally&quot; SUSY explains why the Higgs mass remains light. But there is no strict argument that the superpartners have to be lighter than 1 TeV or 10 TeV. It just &quot;sounds strange&quot; if they were much higher than that because a non-negligible portion of the hierarchy problem would remain. See a text on SUSY and the hierarchy problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2010/07/susy-and-hierarchy-problem.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/07/susy-and-hierarchy-problem.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One may say that experiments already do disprove 99.999999999+ percent of the natural a priori interval for a conceivable Higgs mass in the Standard Model. SUSY changes this counting - the probability that the Higgs mass ends up being approximately as low as suggested by the electroweak observations becomes comparable to 100 percent according to a SUSY theory. To agree with other available experiments, SUSY needs to adjust some other parameters but at good points of the parameter space, none of the adjustments are as extreme as the adjustment of the Higgs mass in the non-supersymmetric Standard Model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can we decide whether SUSY is there at the LHC?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SUSY may hide for some time but the LHC is simply scheduled to perform a certain number of collisions at a certain energy, and those collisions may eventually be studied by the most up-to-date methods and the evidence for SUSY will either be there in the data or not. Some phenomenologists often want to stay very modest and they talk about numerous complex ways how SUSY may keep on hiding - or remain de facto indistinguishable from other models. However, sometimes the very same people are capable of reverse-engineering a randomly constructed man-made model (fictitiously produced collision data) within a weekend: these are the games played at the LHC Olympics. So I don't really expect too much hiding. With the data, the fate of the LHC-scale SUSY will ultimately be decided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, if SUSY is there at the LHC scale, the LHC will eventually be discovering fireworks of new effects (SUSY is also the most attractive realistic possibility for the experimenters) - all the superpartners of the known particles, among other things (such as an extended Higgs sector relatively to the Standard Model). Their spins and couplings will have to be checked to (dis)agree with those of the known particles, and so on. All the masses may be surprising for us - we don't really know any of them although we have various models of SUSY breaking which predict various patterns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alternatives in the case of SUSY non-observation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dark matter may be composed of ad hoc particles that don't require any grand structures - but such alternatives would be justified by nothing else than the simple and single job that they should play. Of course that there are many alternatives in the literature but none of them seem to be as justified by other evidence - i.e. not ad hoc - as SUSY. I think that in the case of no SUSY at the LHC, the LHC will remain some distance away from &quot;completely disproving&quot; SUSY particles as the source of dark matter because this role may work up to 10 TeV masses or so, and much of this interval will remain inaccessible to the LHC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the LHC is a great gadget which is stronger than the previous one - but one simply can't guarantee that it has to give definitive answers about all the questions we want to be answered. This fact may be inconvenient (and many laymen love to be promised that all questions will inevitably be answered for those billions of dollars - whether it's true or not) but it's simply a fact that the LHC is not a machine to see every face of God. There are various alternatives how to solve the hierarchy problem - the little Higgs model, the Randall-Sundrum models (which may be disproved at the end of the LHC, too - the LHC is expected to decide about the fate of each solution to the hierarchy problem although they may always remain some uncertainty), etc. - but I am convinced that even in the case that SUSY is not observed at the LHC, superpartners with slightly higher masses than those accessible by the LHC will remain the most well-motivated solution of the problems above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if someone finds some better new models, or some amazing experimental LHC (or other) evidence for some existing models, the situation may change. But right now, away from SUSY, there are really no alternative theories that naturally explain or solve the three problems above at the same moment. This ability of SUSY to solve many things simultaneously is surely no proof it has to be the right solution of all of them - but it is a big hint. It's the reason why particle physicists think it's the most likely new physics at this point - a conclusion that may change but only if new (theoretical or experimental) evidence arrives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it is clear that the absence of SUSY at the LHC would weaken the case for SUSY and all related directions, I am convinced that unless some spectacular new alternatives or spectacular new proofs of other theories are found in the future, SUSY will still remain the single most serious direction in phenomenology. In formal theory, its key role is pretty much guaranteed to remain paramount regardless of the results of LHC or any conceivably doable experiments. The more formal portions of high-energy theory a theorist studies, obviously, the less dependent his or her work is on the LHC findings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have to explain that the absence of SUSY at the LHC would mean a sharper splitting of the particle physics community.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Absence of SUSY and string theory&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, if no SUSY were seen until 2012 or 2015 or 2020, the critics of string theory would be louder than ever before. Within string theory, the anthropic voices and attempts to find a sensible vacuum with the SUSY breaking at a high-energy scale would strengthen. But nothing would really change qualitatively. The LHC is great but it is just moving the energy frontier of the Tevatron at most by 1-1.5 order(s) of magnitude or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is some non-SUSY new physics found at the LHC, most particle physicists will obviously be interested in whatever models that can be relevant for the observations. If the LHC sees no new physics, e.g. if it only sees a single Higgs boson, and nothing else ever appears, the current situation will qualitatively continue and the tensions will only get amplified. Serious physicists will have to continue their predominantly theoretical and ever more careful studies (based on the observations that have been incorporated into theories decades ago) without any guidance about new physics from the available new experiments (simply because there wouldn't be any new data!) - while the not so serious physicists and people around science will strengthen their hostile and utterly irrational claims that physics is no longer science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sociologically, the situation would almost certainly become unpleasant for good physicists and pleasant for populist and uneducated critics of science who are not really interested in the truth about the physical world. But Nature works in whatever way She does. She is not obliged to regularly uncover a part of Her secrets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A paper with the same question in the title&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amusingly, there exists a 2-week-old preprint by 8 authors:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/arXiv:1102.4693&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/arXiv:1102.4693&lt;/a&gt;&lt;br&gt;&#xA;  What if the LHC does not find supersymmetry in the sqrt(s)=7 TeV run?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may see that the question in their title is almost identical to your question at Physics Stack Exchange. Their answer is much like my answer above: if the LHC is not found during the 7 TeV run (that should continue until the end of 2012), SUSY would still remain an acceptable solution to all the problems I mentioned above; just our idea about the masses of the strongly interacting superpartners (gluinos and squarks) would have to be raised above 1 TeV or so. It's pretty natural for those strongly interacting superpartners to be the heaviest ones among the superpartners - which automatically makes them harder to be seen at hadron colliders such as the LHC.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-06T20:33:40.060" LastActivityDate="2011-03-06T20:33:40.060" CommentCount="2" />
  <row Id="6440" PostTypeId="2" ParentId="6438" CreationDate="2011-03-06T20:53:00.183" Score="7" ViewCount="" Body="&lt;p&gt;The most important problem that supersymmetry solves is the &lt;a href=&quot;https://secure.wikimedia.org/wikipedia/en/wiki/Hierarchy_problem&quot; rel=&quot;nofollow&quot;&gt;hierarchy problem&lt;/a&gt;: why is the weak scale, which determines the rate of beta decay or the masses of the W and Z bosons, so much smaller than the Planck scale, which is related to the strength of the gravitational force? In other words, why is the weak force so &lt;em&gt;strong&lt;/em&gt;, compared to gravity? The real problem here is that quantum mechanical corrections tend to raise energy scales close to the largest scale in any given system, unless some mechanism protects them. So even if you arrange the Standard Model to have the weak scale at a particular value, quantum effects want to push it up near the Planck scale. (The Higgs boson mass is the specific parameter that is afflicted with this problem; all the others are protected by symmetries.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supersymmetry solves this problem in an elegant way: in the limit of exact supersymmetry, any correction pushing the Higgs mass scale up is canceled by another correction with the opposite sign. If supersymmetry is broken at around the weak scale, these corrections remain accurate enough to keep the weak scale low. There are essentially only two other good ideas about how to solve the hierarchy problem: one is &lt;a href=&quot;https://secure.wikimedia.org/wikipedia/en/wiki/Technicolor_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;technicolor&lt;/a&gt;, which says the scale arises just as the QCD scale does, because some interaction strength is very weak at very high energies and gradually becomes stronger at lower energies, blowing up precisely at the weak scale; the other is &lt;a href=&quot;https://secure.wikimedia.org/wikipedia/en/wiki/Large_extra_dimension&quot; rel=&quot;nofollow&quot;&gt;large extra dimensions&lt;/a&gt;, which doesn't really solve the problem but translates it into a different, geometrical problem of why certain dimensions are so large. (I'm glossing over most other ideas, like Randall-Sundrum, little Higgs, and composite Higgs models, which are all in some sense particular versions of technicolor that might have extra nice features.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supersymmetry's chief phenomenological advantage over technicolor and large extra dimensions is that it plays nicely with all of our precision observations. Even without directly making new particles, any theory of new physics predicts certain effects that would slightly shift the values and relationships between precisely measured quantities. To date, no such precision measurement unambiguously shows evidence of physics beyond the Standard Model. This means that any new theory has to be tuned somewhat, to cancel all these possible corrections. These tunings in some theories must be done so precisely that we tend to find them implausible. In supersymmetry, relatively little tuning is needed to make the theory consistent with what we've measured so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also has the theoretical advantage of being a deeper idea, certainly than technicolor (which just duplicates physics we already know in QCD) and arguably than extra dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other very compelling reason for SUSY, beyond the hierarchy problem, is gauge coupling unification. This is really an &lt;em&gt;empirical&lt;/em&gt; fact. The measured gauge couplings, if run up to high energies, do not unify in the Standard Model. However, they do unify, quite precisely, in the supersymmetric modification of the Standard Model. This looks like a hint that nature is giving us about very high energy physics, and most of us think it probably isn't an accident. Now, it is possible to make gauge coupling unification work in other scenarios beyond the Standard Model (like some versions of technicolor), but it doesn't seem to &lt;em&gt;want&lt;/em&gt; to happen; it takes some model-building. In the supersymmetric Standard Model, it just comes out immediately. This is pretty convincing, for most of us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark matter is the third point typically mentioned as a reason to believe in SUSY, but I don't find it very compelling. It is easy to build non-SUSY models with reasonable dark matter candidates, and SUSY models without them, and it's very plausible that some other extremely well-motivated particle like an axion is the dark matter. But, dark matter could turn out to be a superpartner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the LHC runs according to plan and finds no evidence of superpartners, I think most of us would expect it will turn up evidence of some other mechanism for solving the hierarchy problem, like technicolor. (There are many ways SUSY could show up, but if superpartners are there at the TeV scale, it's hard to imagine they could be missed entirely. So I think the LHC really could persuade us that they aren't around at the TeV scale.) If it finds evidence of &lt;em&gt;no&lt;/em&gt; such mechanism (just a Standard Model Higgs and nothing else), then we'll be left with a puzzle, and many people would argue that anthropic scenarios are the only answer. That would be pretty unsatisfying, but there's no point in worrying about it much unless it happens.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2011-03-06T20:53:00.183" />
  <row Id="6441" PostTypeId="2" ParentId="6438" CreationDate="2011-03-06T20:56:18.403" Score="3" ViewCount="" Body="&lt;p&gt;There are two things which SUSY brings to physics.  The first is that it is a loophole in the Coleman Mandula “no-go theorem,” which permits internal symmetries of particles to work with external symmetries of spacetime.  SUSY works by transforming bosonic fields into fermionic fields and visa versa.  So particles we know about have their supersymmetric pairs.  This means the vacuum has an equal number of fermionic and bosonic particles, where the zero point energy of these fields is negative and positive respectively.  Unbroken supersymmetry naturally defines a zero energy vacuum.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the LHC does not find SUSY, we are faced with some problems.  It had been thought decades ago that SUSY would show up in the 100GeV range, where the gauge hierarchy problem with light SUSY particles is far simpler to work.  If we do not find SUSY then unification or gauge hierarchy issues are more problematic, and it is possible that nature is simply not supersymmetric.  For various reasons I suspect if we do not get SUSY we may also not find the Higgs field, in which case a vast amount of physics literature is on shaky ground.  Jacque Distler has said he expects nothing out of the LHC, which is the darkest possible outcome for it all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will not get into arguments about whether string theory is science in the face of something like this possible outcome.  The argument would be silly and pointless.  However, a theoretical development in science that consistently fails to make contact with experimental observations is simply in trouble and will eventually wither away.  This is even if it all turns out that if the US had done the super-collider we would have found the Higgs and SUSY and so forth, for the problem is that money and patience on these things is going to be exhausted.  This is particularly with the United States, which is clearly a declining civilization, and Europe is frankly following suit.  Also note that populations around the world and particularly in the US have become highly religious, which bodes ill for the scientific future, particularly big science  If there is a future in this it lies further down this century in China and the other Asian powers.  The western world, from the US across Europe into Russia, is becoming a club of tottering behemoths that are hopelessly adrift and in decline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also means from a personal perspective if nothing is found and you are young enough “Get out of Dodge!”  There will not be much point in trying to keep up with a field that under those conditions is going to wither away.  Switching to newer areas of work, biophysics, bio-informatics, nanotechnology and …, which I did some years ago, is highly advised.  String theory is really cool stuff, but there is a chance the whole thing could implode away.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-06T20:56:18.403" CommentCount="8" />
  <row Id="6442" PostTypeId="2" ParentId="6436" CreationDate="2011-03-06T22:25:12.313" Score="1" ViewCount="" Body="&lt;p&gt;This is an interesting question, which does have a bearing on the problem of qubits and black hole categories, or BPS charges.  I can only lay down some possible ideas for how to think about this.  The anti-de Sitter spacetime is the quotient space&#xA;$$&#xD;&#xA;AdS_n~=~\frac{SO(n-1,2)}{SO(n-1,1)}&#xD;&#xA;$$&#xA;We denote by $g~=~so(n-1,2)$ and $h~=~so(n-1,1)$ the Lie algebras and by   the projection&#xA;$g~\rightarrow~g/h$. An involutive automorphism $\mu:g~\rightarrow~g$ which fixes elements of $h$, and we call $f$ the eigenspace of eigenvalue $−1$ of $\mu$. Thus $g$ decomposes as an algebra as $g~=~h\oplus f$. The compact part of $SO(n-1,2)$ is the decomposition into $K~=~SO(2)\times SO(n − 1)$.  So this defines a set of elements which are in involution with this automorphism.  This defines a certain orbit or path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For points which are nilpotent under the group $AdS_2~\simeq~SL(2,R)$ and $AdS_4~\sim~SL(2,R)^2$ we can define the orbits of the AdS by nilpotent conditions on ${\cal G}~=~SL(2,R)^n$.  A set $S$ which defines a horizon $t^2~-~x^2~=~0$ define the BTZ as $BTZ~\sim~AdS_n\setminus S$, and where the BTZ is given by the fixed or nilpotent points of the group.  So we might then consider looking at these dynamics on the moduli space, and where the Killing vectors are timelike, with no spacelike isometries due to spatial stationary conditions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;SO if I were to set this up I might try the following. Let the $AdS_4$ spacetime with the IR decomposition $AdS_2\times {\cal M}$. The gauge action in $AdS_4$ spacetime is in the STU model is similar to &#xA;$$&#xD;&#xA;S~=~\int d^4x (R~+~G_{ab}\partial{\bar z}^a\partial z^b~-~m_{IJ}F^IF^J~-~n_{IJ}F^I(*F)^J&#xD;&#xA;$$&#xA;The equations of motion may be solved with the metric for a charged black hole, or equivalently a $D$-brane geometry&#xA;$$&#xD;&#xA;ds^2~=~r^2 F(r)(-dt^2~+~dx^idx_i)~+~{1\over{r^2F(r)}}dr^2,~F(r)~=~1~+~{m\over r}&#xD;&#xA;$$&#xA;For the near horizon condition we have the metric reduce to $AdS_2$ on the $t,~r$ coordinates may be removed from the $x^i$ coordinates of the $2$-sphere.  A substitution of variables $R(r)~=~e^{-2U}/r^2$ gives the $AdS_2$ metric $ds^2~=~-e^{-2U}dt^2~-~e^{2U}dr^2$.  The dynamics is then on $D2$-branes.  The portion of the STU action which gives the hyperbolic dynamics on the space ${\cal L}~=~\sqrt{-h}\partial g_{ab}{\bar z}^a z^b$ gives the geodesic equation with $\tau~=~1/r$ where the scalars and gravity decouple:&#xA;$$&#xD;&#xA;{{d^2\phi^i}\over{d\tau^2}}~+~{\Gamma^i}_{jk}{{d\phi^j}\over{d\tau}}{{d\phi^j}\over{d\tau}}&#xD;&#xA;$$&lt;br&gt;&#xA;This geodesic equation the defines the orbits, where the moduli are determined by the nilpotent orbits of the group.  This would then be used to address the question   The condition for the BTZ horizon is then mapped to the boundary of the $AdS_n$ or $CFT_{n-1}$.  This then should be a copy of the conformal group of quantum mechanics $SL(2,R)$.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-03-06T22:42:03.457" LastActivityDate="2011-03-06T22:42:03.457" CommentCount="1" />
  <row Id="6443" PostTypeId="1" AcceptedAnswerId="6444" CreationDate="2011-03-07T00:11:21.690" Score="1" ViewCount="95" Body="&lt;p&gt;In general, it seems cosmological theories that encompass more and more of the phenomena of the universe are expected to be more and more mathematically elegant, in conception if not in detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our experience seems to teach that it is a reasonable expectation to assume more fundamental theories will be more &quot;elegant&quot; or &quot;beautiful&quot; than previous theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It seems like a similar expectation is what triggered Einstein to reject Quantum Mechanics.  We know his experience led him astray.  How do we know our experience isn't causing a similar result in our search for a Theory of Everything?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A correllation between &quot;mathematical elegance&quot; and explanatory power would seem to infer that &quot;elegance&quot; is more than just a human construct.  How can that be?  Why should there be a correllation between what we find pleasing, and how the Universe works?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2226" LastActivityDate="2011-03-07T04:20:21.723" Title="Mathematical Elegance in a Cosmological Theory Considered Necessary?" Tags="&lt;particle-physics&gt;&lt;cosmology&gt;&lt;mathematics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="6444" PostTypeId="2" ParentId="6443" CreationDate="2011-03-07T01:38:43.100" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;We dont.&lt;/li&gt;&#xA;&lt;li&gt;See &lt;a href=&quot;http://physics.stackexchange.com/questions/4141/why-beauty-is-a-good-guide-in-physics&quot;&gt;Why beauty is a good guide in physics?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-03-07T01:38:43.100" CommentCount="2" />
  <row Id="6445" PostTypeId="1" AcceptedAnswerId="6447" CreationDate="2011-03-07T01:42:31.627" Score="3" ViewCount="105" Body="&lt;p&gt;A nonuniform linear charge distribution given by λ = bx, where b is a constant, is located along an x axis from x = 0 to x = L. What is the electric potential at a point   on the y axis? Set potential equal to zero at infinity. Express your answer in terms of the variables given and ε0 (epsilon not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have calculated the answer to be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b * (sqrt(L^2 + y^2) - y) / (4 * pi * ε0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The site I enter this answer onto tells me I am wrong. I have checked this with my dad, who is very good at physics, and he agrees with me on this answer. I would like to know if this answer is correct, and if not, why. Thanks. &lt;/p&gt;&#xA;" OwnerUserId="2404" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-07T02:05:34.857" LastActivityDate="2011-03-07T02:14:16.877" Title="Electric Potential" Tags="&lt;electromagnetism&gt;&lt;homework&gt;" AnswerCount="1" />
  <row Id="6446" PostTypeId="1" AcceptedAnswerId="6454" CreationDate="2011-03-07T01:49:45.873" Score="1" ViewCount="409" Body="&lt;p&gt;I'm a layman without a university background in physics / math.  Since I don't have a background, reading a paper is more of an effort.&#xA;Consequently, when I come across an interesting paper, I can't really just give it a glance, and see if it is science, or psuedoscience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is about &lt;a href=&quot;http://www.theresonanceproject.org/pdf/schwarzschild_proton_a4.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper.&lt;/a&gt;&lt;br&gt;&#xA;If the advent of the internet has taught me anything, its that there is an inverse relationship between cost to publish and the need to vet what is published&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, Lisi's E8 paper has taught me that a paper doesn't necessarily need to be correllated to an establishment like the Perimeter Institute or the Institute for Advanced Study (two examples off the top of my head) to be worth reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the paper worth reading?&lt;/p&gt;&#xA;" OwnerUserId="2226" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-07T13:06:55.877" LastActivityDate="2011-04-25T04:08:45.763" Title="&quot;The Schwarzschild Proton&quot; paper - Is the science valid?" Tags="&lt;particle-physics&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="6447" PostTypeId="2" ParentId="6445" CreationDate="2011-03-07T01:53:59.733" Score="5" ViewCount="" Body="&lt;p&gt;The potential should be symmetric about x-axis, (interchange y-&gt;-y), so you did something  wrong. On closer look that is the only error. Change $y$ to $sqrt(y^2)$&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2011-03-07T02:14:16.877" LastActivityDate="2011-03-07T02:14:16.877" CommentCount="3" />
  <row Id="6448" PostTypeId="2" ParentId="6446" CreationDate="2011-03-07T02:30:02.487" Score="3" ViewCount="" Body="&lt;p&gt;A good guideline is to ask yourself &quot;is this paper published in a reputable journal, or a crackpot website.&quot;  If the former, maybe; the latter, no.  So that would be a no for this and Lisi's paper.&lt;/p&gt;&#xA;" OwnerUserId="1403" LastActivityDate="2011-03-07T02:30:02.487" CommentCount="10" />
  <row Id="6449" PostTypeId="2" ParentId="6384" CreationDate="2011-03-07T03:30:13.893" Score="2" ViewCount="" Body="&lt;p&gt;A nice article about this: &lt;a href=&quot;http://arxiv.org/abs/1101.4886&quot; rel=&quot;nofollow&quot;&gt;Tutorial on Scale and Conformal Symmetries in Diverse Dimensions&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rule-of-thumb is that 'conformal &amp;rArr; scale', but the converse is not necessarily true (some condition(s) needs to be satisfied) — but, of course, this varies with the dimensionality of the problem you're dealing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Polchinski's article: &lt;a href=&quot;http://dx.doi.org/10.1016/0550-3213%2888%2990179-4&quot; rel=&quot;nofollow&quot;&gt;Scale and conformal invariance in quantum field theory&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2011-03-07T03:30:13.893" />
  <row Id="6450" PostTypeId="1" AcceptedAnswerId="6451" CreationDate="2011-03-07T03:38:23.163" Score="3" ViewCount="118" Body="&lt;p&gt;Is there an accepted analogy / conceptual aid for the Higgs field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Physics there are many accepted conceptual aids such as&lt;br&gt;&#xA;   *  Schrödinger's cat&lt;br&gt;&#xA;   *  Maxwell's Demon&lt;br&gt;&#xA;   *  I'm sure I'm missing many, but you get the idea  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an accepted / standard aid for the Higgs?&lt;br&gt;&#xA;I saw a popular treatment of the Higgs boson with Peter Higgs.&lt;br&gt;&#xA;He talked about pearls being dragged through &lt;a href=&quot;http://en.wikipedia.org/wiki/Treacle&quot; rel=&quot;nofollow&quot;&gt;treacle&lt;/a&gt;.&lt;br&gt;&#xA;But the analogy wasn't fully fleshed out, just a two second video clip&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is not an accepted standard aid, what do the various professionals here use to aid explicating to non professionals?  &lt;/p&gt;&#xA;" OwnerUserId="2226" LastActivityDate="2011-03-07T03:47:02.950" Title="Is there an accepted analogy / conceptual aid for the Higgs field?" Tags="&lt;particle-physics&gt;&lt;higgs-boson&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6451" PostTypeId="2" ParentId="6450" CreationDate="2011-03-07T03:47:02.950" Score="4" ViewCount="" Body="&lt;p&gt;This is not as easy as it may sound: in every analogy one has to make a choice between rigor and 'poetic license'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, the one i like better is &lt;a href=&quot;https://www.hep.ucl.ac.uk/~djm/higgsa.html&quot; rel=&quot;nofollow&quot;&gt;Higgs for Waldegrave&lt;/a&gt;: where a crowd-analogy is given. But, as they say, &lt;em&gt;your milage may very&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'd like, you can think in terms of a 'caramel pool', &lt;a href=&quot;https://www.youtube.com/watch?v=7JDXilnpj9Y&quot; rel=&quot;nofollow&quot;&gt;Milky Way Simply Caramel: Pool&lt;/a&gt; : when we say that &lt;em&gt;a particle 'couples' to the Higgs field&lt;/em&gt;, we mean to say that this particle 'sees' this 'caramel pool', and this makes it &quot;harder&quot; for it to move, which we measure as this particle's &quot;mass&quot;. And, as you can imagine, there are particles that do not couple to the Higgs field, meaning to say they do not &quot;see&quot; it: therefore, they move much more easily.  ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, take all this with a grain of salt…&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2011-03-07T03:47:02.950" />
  <row Id="6452" PostTypeId="2" ParentId="6443" CreationDate="2011-03-07T04:20:21.723" Score="0" ViewCount="" Body="&lt;p&gt;What exactly does &quot;elegance&quot; mean? To me, your question seems to answer itself: if elegance is explanatory power, then the best theories are elegant, since explanatory power is what makes them good in the first place. If it's some kind of mystical beauty or simplicity, then we can't count on it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's perfectly possible that people are being limited by expectations or hopes for a simple theory of everything, so I suppose we should all do our best to make sure that any human desire for &quot;beauty&quot; doesn't influence our scientific decisions.&lt;/p&gt;&#xA;" OwnerUserId="156" LastActivityDate="2011-03-07T04:20:21.723" />
  <row Id="6453" PostTypeId="2" ParentId="6435" CreationDate="2011-03-07T04:44:03.457" Score="2" ViewCount="" Body="&lt;p&gt;These are rare, on the order of one per square mile per century. See: &lt;a href=&quot;http://en.wikipedia.org/wiki/Ultra-high-energy_cosmic_ray&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Ultra-high-energy_cosmic_ray&lt;/a&gt; . So a human, with a cross sectional area less than 1 square meter might get hit about once per each 100 million years. I think that the risk of life due to spacecraft malfunction is significantly greater than this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And when they do hit the earth's atmosphere, they expend their energy over the whole depth of the atmosphere. No one single region gets plastered. (Otherwise they'd have been detected long before, by the sound of the air getting overheated.) The thickness of the atmosphere, in terms of equivalent mass of water, is around 10 meters of water. This is much thicker than the equivalent weight distribution of a spacecraft. So my back of the envelope conclusion is that if a spacecraft did get hit by one, most of the energy will exit the far side of the spacecraft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, cosmic rays of all sorts can damage electronics. It can switch ones to zeroes or vice versa, and I would think that such a high energy ray as this could actually hurt a component.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To realistically examine spacecraft in terms of how cosmic rays act in them, it's useful to use numbers for a real spacecraft. In addition to a thin metal shell, a spacecraft also has gas storage, fuel storage, and a place to go to the bathroom. And a spacecraft must be built sturdily even if it doesn't have to land on the earth. They are big, heavy, objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the ISS (International Space Station) weighs 375,000 kg and has a pressurized volume of 907m^3:  &lt;a href=&quot;http://en.wikipedia.org/wiki/International_Space_Station&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/International_Space_Station&lt;/a&gt;&lt;br&gt;&#xA;Thus its density is 375,000/907 = 413 kg/m^3 almost half that of water, and it has a distance between inelastic collisions of about 3.7cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the spacecraft consists of the solar panels, however, like any other nuclear matter, these will generate cascades when they are hit by a cosmic ray. The length of the vessel (along its crew compartment) is 51 meters. With the pressurized volume, if this were a cylinder its diameter would be about 4.8 meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a distance of 51 meters and a density of 0.41 that of water, the long dimension on the ISS is equivalent to 20m of water. This is twice the depth of the earth's atmosphere. The short distance is around 2m of water or 1/5th the earth's atmosphere. This is enough to get a shower going.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These calculations are worst case in that they've imputed all the weight of the spacecraft to the pressurized volume. A more accurate calculation would involve a spacecraft which does not have the solar panels. An example of this would be the US space shuttle orbiter. With payload, it's 93,000 kg (max landing weight 104,000 kg). The length is 37m with a wingspan of 24m. The diameter of the main crew compartment appears to be about 7m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the assumption that the spacecraft is a cylinder of length 37m and diameter 7m, its volume is around 1400 cubic meters, and its density is around 74kg/m^3. Thus a particle traveling the length of the craft will encounter about 37m x 74kg/m^3 = 2.7 meters of water equivalent. At 1/4 the thickness of the earth's atmosphere, this is enough to get a good cascade going.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Anna v notes that the (inelastic) cross-sectional area for an extremely high energy proton-proton collision is only around 1 barn or 1.0E-28m^2. Actually, her (1984) article gives 530mb, but at very high energies, the cross section is unknown and might exceed a barn, see figure (4):&lt;br&gt;&#xA;Nucl.Phys.Proc.Suppl.196:335-340 (2009), Ralf Ulrich, Ralph Engel, Steffen Müller, Fabian Schüssler, Michael Unger &lt;em&gt;Proton-Air Cross Section and Extensive Air Showers&lt;/em&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/0906.3075&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0906.3075&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the proton is about 1.6E-27 kg, and a cubic meter of water weighs 1000 kg (almost entirely nucleons i.e. protons and neutrons). Thus there are 1000/1.6E-27 = 6.25E+29 nucleons per cubic meter of water. Multiplying by the cross-sectional area gives a total cross sectional area/m^3 of 62.5 m^2. Thus the distance at which such a proton begins a cascade, in water, is about 1/62.5 m = 1.6cm, and by the time the particle has exited the human body 50cm later, it has had about 31 such collisions. Thus the size of the human body is sufficient to begin a cosmic ray cascade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a practical long-distance spaceship, it's important to shield the crew from cosmic rays and solar flares. One of the suggestions for doing this is with shields composed of 7kg of aluminum per square meter, which is about 0.7cm water equivalent thickness: &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0701314&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/astro-ph/0701314&lt;/a&gt; Thus there's about a 50% probability for a high energy cosmic ray starting off a cascade on going through this thickness.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-03-09T01:10:29.137" LastActivityDate="2011-03-09T01:10:29.137" CommentCount="5" />
  <row Id="6454" PostTypeId="2" ParentId="6446" CreationDate="2011-03-07T05:16:32.600" Score="13" ViewCount="" Body="&lt;p&gt;I'm going with &quot;Nonsense.&quot; on account of&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mixing physics 101 mechanics with special relativity with no evident effort made to tell which case is applicable. In particular, modeling the proton as a black hole, and asserting that he can use physics 101 circular motion to describe the acceleration of two such objects whose event horizons are in contact!&lt;/li&gt;&#xA;&lt;li&gt;Claiming that the proton's mass arises from &quot;cohering&quot; some of the vacuum, and making no attempt to explain whence the charge comes and why it is always the same.&lt;/li&gt;&#xA;&lt;li&gt;No effort to explain where the neutron comes from, or why &lt;em&gt;it&lt;/em&gt; is chargeless, or why it is unstable.&lt;/li&gt;&#xA;&lt;li&gt;No hint of an explanation of how or why nucleons can bind together, and why some states are stable and some are not.&lt;/li&gt;&#xA;&lt;li&gt;Frankly, I gave up looking at the alleged physics at this point...there is some verbiage that purports to relate the anomalous magnetic moment of the proton, and the author finds it necessary to use scare quotes on anomalous. Not promising.&lt;/li&gt;&#xA;&lt;li&gt;A general sense of &quot;snake oil&quot; in the web site.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A side note that may be of interest. AIP conferences accept a few papers from authors whose theories are...ehm...not well regarded. I don't know why. I &lt;em&gt;do&lt;/em&gt; know that people flock to see these talks as comic relief (and they are generally scheduled very late in the day). Personally I find myself intensely embarrassed on the presenters behalf: I figure they &lt;em&gt;must&lt;/em&gt; know how the audience feels, and don't know why they do it.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-03-07T05:46:37.020" LastActivityDate="2011-03-07T05:46:37.020" CommentCount="3" />
  <row Id="6455" PostTypeId="1" CreationDate="2011-03-07T05:39:33.170" Score="4" ViewCount="154" Body="&lt;p&gt;I am stuck with understanding the following construction. I am breaking it up into segments which I think can be separately answered. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is related to an &lt;a href=&quot;http://physics.stackexchange.com/questions/6084/dimensional-reduction-from-31-to-21-for-caln-2-vector-superfield&quot;&gt;earlier question of mine&lt;/a&gt;. Note that this previous question was edited a lot after the initial comments came in.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Firstly I would like to know how the following field multiplet is determined, &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The gauge multiplet of $\cal{N}=3$ in $2+1$ dimensions is apparently , &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_\mu$, a massive vector field of spin $1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda_a$, $3$ Majorana fermions of spin $\frac{1}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_a$, $3$ neutral scalar bosons&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\chi$, a Majorana fermion of spin $\frac{1}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I guess there should be another scalar field $D_a$ not mentioned in the above list)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Its not clear to me as to what are the superfields one has in mind with the above multiplet structure that helps construct the lagrangians below.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Like in the linked earlier question above one had to &quot;know&quot; what exactly is the meaning of doing the dimensional reduction on a given set of supersymmetric transformations, here its not clear to me as to what are the corresponding starting points. (I wonder if I can start from the $\cal{N}=2$ SYM in $3+1$ lagrangian that is given in say equation $27.9.3$ in Weinberg's QFT book)   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then dimensional reduction of $\cal{N}=2$ SYM in $3+1$ apparently gives the following $\cal{N}=3$ SYM in $2+1$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L_{\cal{N}=3,2+1,SYM} = \frac{1}{g^2}Tr\left [ -\frac{1}{2}F_{\mu \nu} ^2 + (D_\mu C_a)^2 + (D_a)^2 + i \bar{\lambda_a}\gamma^\mu D_\mu \lambda_a + i \bar{\chi}\gamma^\mu D_\mu \chi&#xD;&#xA;&#xD;&#xA;&#xD;&#xA;&#xD;&#xA;+ i\epsilon_{abc}\bar{\lambda_a}[\lambda_b,C_c] -2i\bar{\lambda_a}[\chi,C_a] - \frac{1}{2}[C_a,C_b][C_b,C_a] \right ]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the notation is, $D_\mu = \partial_\mu -iA_\mu$, $ab,c=1,2,3$. The gauge group generators in the fundamental representation satisfy $[T^m,T^n]=if^{lmn}T^l$ with normalization as $Tr{T^mT^n}=\frac{1}{2}\delta ^{mn}$ and $f^{kmn}f^{lmn}=c_v\delta^{kl}$. ($c_v$ being the quadratic Casimir) The fields belong to the adjoint representation and $A_\mu = A_\mu^mT^m$. The metric is given by $\eta_{\mu \nu} = diag(1,-1,-1)$. The purely imaginary gamma matrices satisfy, $\gamma^\mu \gamma ^\nu = \eta^{\mu \nu}-i\epsilon^{\mu \nu \rho}\gamma_\rho$. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It is not clear to me as to how does one argue that the above has $\cal{N}=3$ supersymmetry. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The dimensional reduction of the $\cal{N}=2$ supersymmetry somehow implies that the theory has the not-so-obvious $\cal{N}=4$ supersymmetry. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One also notes that supersymmetrizing the Chern-Simon's terms one is supposed to get the Chern-Simon's lagrangian as, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L_{\cal{N}=3,2+1,SCS} = kTr\left [ \epsilon^{\mu \nu \rho}(A_\mu \partial_\nu A_\rho - \frac{2}{3}iA_\mu A_\nu A_\rho) - \bar{\lambda_a}\lambda_a + \bar{\chi}\chi &#xD;&#xA;&#xD;&#xA;&#xD;&#xA;&#xD;&#xA;+ 2C_aD_a + \frac{i}{3}\epsilon_{abc}C_a[C_b,C_c] \right ]$ &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In all these lagrangians apparently the corresponding $\cal{N}=2$ theories can be obtained by setting $C_1=C_2=D_1=D_2=\lambda_3 = \chi = 0$ and the $\cal{N}=1$ theories can be obtained from the $\cal{N}=2$ theories by further setting $C_3=\lambda_2=0$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(I am unable to get the equations to wrap around properly!) &lt;/p&gt;&#xA;" OwnerUserId="1637" LastEditorUserId="1637" LastEditorDisplayName="" LastEditDate="2011-03-07T05:46:53.240" LastActivityDate="2011-03-08T12:16:38.240" Title="Construction of the $\cal{N}=3$ supersymmetric Yang-Mills Chern-Simons theory in $2+1$ dimensions" Tags="&lt;quantum-field-theory&gt;&lt;supersymmetry&gt;" AnswerCount="1" />
  <row Id="6456" PostTypeId="1" CreationDate="2011-03-07T06:17:58.217" Score="1" ViewCount="94" Body="&lt;p&gt;I am a bio-chemist by profession and recently written an essay for a book which is related to the subject of cosmology. This is not my area of expertise. Is there a cosmologist out there who will be kind enough to review this 8 page essay for me to check it for accuracy, or perhaps give me criticism or even suggestions? I am a member of the APS and its division of Biological Physics. Thanks Paul&lt;/p&gt;&#xA;" OwnerUserId="2407" LastActivityDate="2011-03-07T06:17:58.217" ClosedDate="2011-03-07T08:43:20.987" Title="can someone review a 8 page essay for me for accuracy? essay is in cosmology" Tags="&lt;cosmology&gt;" CommentCount="2" />
  <row Id="6457" PostTypeId="1" AcceptedAnswerId="6459" CreationDate="2011-03-07T06:43:34.963" Score="4" ViewCount="207" Body="&lt;p&gt;Consider a rectangular slab of permanently magnetized material. The slab's dimensions are $L_x$, $L_y$, and $L_z$, and the slab is uniformly magnetized in the $\hat{x}$-direction. The slab is not accelerating or spinning. Does the slab generate an electric field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a frame where the magnet is stationary, we know $\mathbf{E}$ is zero everywhere. In a frame where the magnet is moving, there are at least two ways to attack the problem:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Drop $d\mathbf{M}/d t$ into Maxwell's equations, solve for $\mathbf{E}$ and $\mathbf{B}$  &lt;/li&gt;&#xA;&lt;li&gt;Solve for $\mathbf{B}$ in the slab's rest frame, and use a relativistic boost to transform $\mathbf{B}$ in the slab's frame to $\mathbf{E}$ and $\mathbf{B}$ in the frame where the magnet is moving.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Both these methods give the result that $\mathbf{E}$ is nonzero in a frame where the magnet is moving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's consider a long, thin slab ($L_x, L_y \ll L_z$). In a frame where the slab is moving in the $\hat{z}$-direction, is there an electric field (external to the slab) near the 'center' of the magnet? Both the $d\mathbf{M}/d t$ argument and the Lorentz-boost argument seem unchanged. The magnetic field external to the slab does not vanish near the center of the slab, suggesting there is a nonzero electric field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the backstory laid out, here's my real question: In a frame where the slab is moving in the z-direction, is there still an electric field in the case where $L_z \rightarrow \infty$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lorentz-boost argument seems unchanged, and suggests that there is. However, in the $L_z \rightarrow \infty$ case, $d\mathbf{M}/d t = 0$, suggesting no electric field. Can this case be calculated without Lorentz boosts? How do Maxwell's equations account for moving permanent magnets in the case where $d\mathbf{M}/d t = 0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;Followup questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/6479/whats-a-good-reference-for-the-electrodynamics-of-moving-media&quot;&gt;What&amp;#39;s a good reference for the electrodynamics of moving media?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/6581/what-is-the-electric-field-generated-by-a-spinning-magnet&quot;&gt;What is the electric field generated by a spinning magnet?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="2359" LastEditorDisplayName="" LastEditDate="2011-03-09T23:46:23.550" LastActivityDate="2011-03-09T23:46:23.550" Title="Calculate the electric field of a moving infinite magnet, without boosting" Tags="&lt;electromagnetism&gt;&lt;special-relativity&gt;&lt;magnets&gt;" AnswerCount="1" />
  <row Id="6458" PostTypeId="1" CreationDate="2011-03-07T07:08:51.223" Score="1" ViewCount="112" Body="&lt;p&gt;One of my friends has a problem and we don't know how to get this done,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to to know the mass of a patient who is laying on a bed. One scales was put under the foot of a bed and weighed 232kg and the other under the head of the bed and weighed 220kg. The centre mass of the patient is 0.8 meters from the foot of the bed. What is the mass of the patient?&lt;/p&gt;&#xA;" OwnerUserId="2002" LastEditorUserId="2002" LastEditorDisplayName="" LastEditDate="2011-03-07T12:07:46.760" LastActivityDate="2011-03-07T12:07:46.760" Title="Finding the mass of bed/Patient" Tags="&lt;momentum&gt;&lt;torque&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="6459" PostTypeId="2" ParentId="6457" CreationDate="2011-03-07T07:42:32.740" Score="4" ViewCount="" Body="&lt;p&gt;First, one inevitably gets the same solutions if &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;he solves the problem in the slab's rest frame, and then Lorentz-transforms the result to the frame where the slab is moving; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;or if one solves the problem directly in the frame where the slab is moving. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The reason is that Maxwell's equations are covariant under the Lorentz transformations. So if they're satisfied in one frame, they will be satisfied in any frame related by boosts, too. However, we must properly transform all the magnetizations and material relations etc. and add the corresponding moving sources which will be the main subtlety in the text below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your particular problem, one may say some generic statements about the magnetic (and electric) fields without much thinking. For example, if $\vec M$ is in the $x$-direction, it means that the electrons may be thought of to spin in the $yz$-plane. Take a surface of the slab parallel to the $yz$-plane - i.e. one face that belongs to a $x=x_0$ plane. It's pretty clear that there will inevitably be a component of the magnetic field $\vec B$ in the $x$-direction near the external side of the surface. If one boosts the $B_x$ magnetic field in the $z$-direction, there will inevitably be a nonzero electric field in the $y$-direction, $E_y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the frame where the slab is moving, we seem to have no electric sources $\rho$ of the $\mbox{div}\,\vec D=\rho$ Gauss's law and no right-hand side of the Maxwell-Faraday equation, $\nabla\times \vec E = -\partial \vec B / \partial t$. So because there are no electric sources, you would think that the electric field should vanish. However, this is a flawed argument because the form of Maxwell's equations we're using here are only &quot;Maxwell's equations for materials at rest&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the Gauss's law is optimized for $\vec D$ which we're imagining to be given by $\epsilon \vec E$, and is &quot;purely electric&quot;. However, for a moving material, there should be an extra term of the type $\vec v\times \vec M$ included in $\vec D$. Because the latter has a nonzero $y$-component in the moving frame, there will be a nonzero $E_y$ in this frame, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The precise form of Maxwell's equations in a moving medium may be confusing and unfamiliar so I think it may be a good idea to try to transform the local physics to the rest frame of any material, whenever needed, and perhaps Lorentz-transform back. Whenever subtleties would occur, one would have to revisit the derivation of the &quot;macroscopic Maxwell's equations&quot; (for materials) and redo it with the possibility of moving materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Microscopic Maxwell's equations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you could always try to use the microscopic Maxwell's equations which include the Gauss's law in the form $\vec \nabla\cdot \vec E = \rho / \epsilon_0$. But in this form, $\rho$ includes not only free charges but also the &quot;microscopic charges&quot; related to the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the slab has nonzero values of $j_y$ and $j_z$ (currents inside the material) - recall that the electrons are kind of rotating in the $yz$-plane (to produce the magnetic $x$-field), it's also true that when we boost the system in the $z$ direction, the corresponding multiple of $j_z$ will produce a nonzero value of $\rho$ (microscopic charge density). This will be the source of the $E_y$ field discussed above. In particular, $j_z$ will be proportional to $[\delta(y-y_1)-\delta(y-y_2)]$ in the slab's frame which means that there will be $\rho \sim [\delta(y-y_1)-\delta(y-y_2)]$ in the frame where the slab is moving. It is this $\rho$ that will induce a nonzero value of $E_y$ right outside the material (in the frame where the slab is moving).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-07T07:58:18.793" LastActivityDate="2011-03-07T07:58:18.793" CommentCount="5" />
  <row Id="6460" PostTypeId="2" ParentId="6458" CreationDate="2011-03-07T08:05:19.910" Score="3" ViewCount="" Body="&lt;p&gt;Dear Omeid, this is a homework problem, right? I think we shouldn't be answering the question &quot;explicitly&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let me mention a hint: if you assume that the data are enough to calculate a unique answer, do you think that the information about the length of the bed may be relevant? You want to get a result in kilograms - and there are two pieces of information in kilograms and one in centimeters. To get a result in kilograms, may the figure involving centimeters matter if you require that the result has the right units?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This argument is called &quot;dimensional analysis&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, imagine that you put both scales under the bed simultaneously: one of them is under the foot and one of them is under the head - at the same moment. How much do you think that those scales will show in this case? How can you determine the &lt;em&gt;total&lt;/em&gt; ;-) mass of the bed using this measurement?&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-07T08:05:19.910" CommentCount="1" />
  <row Id="6461" PostTypeId="1" AcceptedAnswerId="6463" CreationDate="2011-03-07T08:35:42.053" Score="4" ViewCount="266" Body="&lt;p&gt;The electron is particle.&#xA;The mass of electron is $9.10938215(45)\times 10^{−31}\, {\rm kg}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why is the mass exactly what it is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What in physics defines the mass of elementary particle?&lt;/p&gt;&#xA;" OwnerUserId="2410" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-03-07T08:43:11.237" LastActivityDate="2011-03-07T14:13:14.250" Title="What defines the mass of elementary particle?" Tags="&lt;particle-physics&gt;&lt;mass&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="6462" PostTypeId="2" ParentId="6461" CreationDate="2011-03-07T08:43:28.420" Score="4" ViewCount="" Body="&lt;p&gt;Right now, the elementary particle masses simply have to be measured. We don't have any explanation for why the masses have the particular values they do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the major goals of unified theories (string theory and the like) is to be able to derive the elementary particle masses, or at least derive the ratios between the masses of different particles. But we're a long way off from doing so.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-03-07T08:43:28.420" />
  <row Id="6463" PostTypeId="2" ParentId="6461" CreationDate="2011-03-07T08:44:34.947" Score="8" ViewCount="" Body="&lt;p&gt;Dear GJ, the mass of every electron is exactly the same - as long as our experiments and theories may say - and the same value of mass manifests itself &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;in the energy/mass conservation law; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;in the source of the gravitational field that the object containing massive particles such as the electron produces; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;in the resistance towards acceleration that the massive objects have.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All those versions of the &quot;mass&quot; are the same quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our modern theories of the world, the mass of the electron is the coefficient of a &quot;mass term&quot; in the Lagrangian or Hamiltonian, $-m\bar\psi\psi$, that is &lt;em&gt;almost&lt;/em&gt; fundamental and can't have any deeper explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, almost. In fact, in the &quot;electroweak theory&quot; - unifying all electromagnetic phenomena (including light) with the weak nuclear interactions (notorious as the origin of the beta-decay) - the mass of electrons (and other particles) arises from the interactions with the Higgs field. So the actual term is actually&#xA;$$-y h \bar\psi\psi$$&#xA;where $\psi$ is the Dirac field for the electron. Here, $h$ is the Higgs field which takes the nonzero value of $h=v$ in the vacuum - the spontaneous symmetry breaking (Higgs mechanism). This value of $v$, about $247$ GeV, is universal for all particles. However, the difficult information about the electron mass is moved to the parameter $y$, the so-called Yukawa coupling, which is dimensionless (no units) and whose value is much smaller than one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the Universe is described by string theory, the value of the Yukawa coupling $y$ and the Higgs value in the vacuum, $v$, can be in principle calculated from first principles (in the natural units, e.g. the Planck units, not in kilograms - the unit of &quot;kilogram&quot; was a randomly chosen social convention that the farmers chose to sell vegetables, without any justification why it's not e.g. $1.3$ times different).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are many (but discretely) possible Universes according to string theory and we can't say which one is right, so in practice, we can't calculate the value of the electron mass from purely theoretical considerations. So far, we need to measure it to know the value. Chances are that the situation won't change for quite some time.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-07T08:44:34.947" />
  <row Id="6464" PostTypeId="1" AcceptedAnswerId="6470" CreationDate="2011-03-07T10:13:55.947" Score="1" ViewCount="716" Body="&lt;p&gt;Added: &lt;strong&gt;5 times down vote for now! Down voter is this religion or physics, please try to explain your decision.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confused about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In physics we know for a &lt;a href=&quot;http://en.wikipedia.org/wiki/Vacuum&quot; rel=&quot;nofollow&quot;&gt;vacuum&lt;/a&gt;, but I think that there is a contradiction in this term. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_fluctuation&quot; rel=&quot;nofollow&quot;&gt;quantum fluctuations&lt;/a&gt; are the phenomena that  contradicts the vacuum existence because, according to them, the vacuum isn't the empty space. &#xA;In vacuum the creation of particle-antiparticle pairs is allowed for small times and this is also proven in practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the other side, the relativistic theory says about space-time that interactions like gravitation bend the vacuum (empty space). There seems to be a contradiction to me. If vacuum is an empty space then we can't bend it, because we can't bend something nonexistent. In other words: we can't bend 'nothing'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can empty space really exist in physics?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT1:&lt;/strong&gt; quote: Luboš Motl&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;By definition, the space without&#xA;  energy is the space whose total value&#xA;  of energy is equal to 0.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But this space is nonexistent, so it is abstract and should exist only in our mind...&#xA;Why is so? My second statement of this post says that in physics we cant bend nothing because such a bending is only thinking and not physics!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is that space is unknown kind of energy, but this is contradiction in modern physics!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can any physics believe that nothing exists?&#xA;By mathematical logic no! And mathematic is elementary tool in physics. Anything other than that is religion!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Nothing (empty space without energy) is only a logical state!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT3:&lt;/strong&gt; quote: Roy Simpson:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The General Relativity Vacuum is a&#xA;  space-time model region without&#xA;  matter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and Luboš Motl says: &quot;By definition, the space without energy is the space whose total value of energy is equal to 0.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Agree... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is only mathematical Euclidean space + time so this is only a mathematic and not physics! With other words: this is only a method of mathematical mapping. But, in real (not theoretic) physics we can't mapping empty things. Empty is only a logical state!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT4:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roy Simpsons argumentation seems to me acceptable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;quote: Roy Simpson:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Einstein struggled with this too, and&#xA;  the problem has come to be known as&#xA;  the &quot;Hole argument&quot; within GR. You&#xA;  have to decide whether you are just&#xA;  interested in GR's vacuum (empty&#xA;  space) or the full physical vacuum&#xA;  which includes quantum aspects as&#xA;  well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2410" LastEditorUserId="2410" LastEditorDisplayName="" LastEditDate="2011-03-09T19:31:20.980" LastActivityDate="2011-05-07T20:13:19.420" Title="Does vacuum (empty space) exist?" Tags="&lt;spacetime&gt;&lt;space&gt;&lt;vacuum&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="1" />
  <row Id="6465" PostTypeId="2" ParentId="6435" CreationDate="2011-03-07T10:26:44.487" Score="3" ViewCount="" Body="&lt;p&gt;One must keep  in mind also that it is the particle, not the shower that goes through the astronaut in  dmckee's estimate above,  where he treats the relativistic particle going through matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shower in your question which gave the energy estimate of the parent particle is generated by cascade/sequential  collisions of deep angle scattering over a long path. The energy is not released in one go unless the astronaut is very unlucky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deep inelastic &lt;a href=&quot;http://www.cosmic-ray.org/papers/physrev52-16.pdf&quot; rel=&quot;nofollow&quot;&gt;scattering crossection&lt;/a&gt; at those energies is still not up to barn values ( a barn is about the size of a uranium nucleus) so the astronaut would have to be very unlucky even to get one energetic scatter let alone to start a shower.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-07T10:26:44.487" />
  <row Id="6466" PostTypeId="2" ParentId="6464" CreationDate="2011-03-07T10:32:50.453" Score="-1" ViewCount="" Body="&lt;p&gt;As strange as this object may feel to you, it is what it is: the empty space. As empty as it can. This is what happens when you take your room and get everything out.&lt;/p&gt;&#xA;" OwnerUserId="100" LastActivityDate="2011-03-07T10:32:50.453" CommentCount="1" />
  <row Id="6467" PostTypeId="2" ParentId="6464" CreationDate="2011-03-07T10:39:50.700" Score="7" ViewCount="" Body="&lt;p&gt;Dear GJ, &quot;vacuum&quot; and &quot;empty space&quot; is always the same thing, but one must always be careful what these two synonymous terms mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General relativity implies that the only &quot;information&quot; that the vacuum carries at each point is the so-called &quot;metric tensor&quot; - a set of numbers that allow one to calculate the distance between any two nearby points. This is enough for the vacuum to be able to bend - much like any material. One doesn't need any atomic constituents to be able to talk about geometry of the space, and to guarantee that the environment is able to get curved (and to distinguish a flat region of the vacuum from a curved one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum field theory implies that the vacuum is full of virtual particles that emerge and quickly disappear. Those virtual particles make their impact on other objects - for example, they make the electromagnetism a little bit weaker at long distances (and stronger at very short distances) than what one expects from the classical Coulomb's law etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, quantum mechanics implies that the vacuum corresponds to a very particular &quot;state&quot; - a vector on the Hilbert space - called $|0\rangle$. It is completely unique and as empty as you can get. In particular, it is the eigenstate of the energy operator with the  minimum allowed energy - essentially zero. (More precisely, the vacuum energy density is nothing else than the magnitude of dark energy but this energy only becomes sizable for huge, cosmological volumes of space.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The uncertainty principle of quantum mechanics implies that when one measures things such as the intensity of the electric field in the vacuum - i.e. when the physical system is found in the state $|0\rangle$ - one may get many random values. It is not allowed for the electric and magnetic fields to be exactly zero, much like a particle can't have a well-defined position and velocity in the quantum mechanics of one particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even though the vacuum has a well-defined (minimal) energy and it is as low as we can get, so the vacuum is as empty as we can get, and there are no particular &quot;atoms&quot; or other particles sitting in the vacuum, there's a lot of activity going on in the vacuum which can be seen by the fact that the measurements of various things, such as the density of energy at a given point, will lead to random results that are not strictly zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the picture of the vacuum as a &quot;literally empty space&quot; that only has the metric tensor at each point; and the quantum picture with all the activity of virtual particles are actually fully compatible with one another. The statement of general relativity that the metric tensor has particular values at a given point should be viewed as a classical approximation, however: when we look at it precisely, the metric tensor is a set of operators, too. They will inevitably have variable and chaotic values if they're measured - that are just &quot;approximately zero&quot; if they're averaged over large enough volumes.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-07T10:39:50.700" CommentCount="7" />
  <row Id="6468" PostTypeId="1" CreationDate="2011-03-07T10:43:33.143" Score="2" ViewCount="85" Body="&lt;p&gt;How is it possible to encrypt multiple images using fresnel transform and inverse the operation to de-multiplex those images?&lt;/p&gt;&#xA;" OwnerUserId="2411" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-07T11:13:23.130" LastActivityDate="2011-03-07T11:13:23.130" Title="Fresnel transform" Tags="&lt;optics&gt;" CommentCount="1" />
  <row Id="6470" PostTypeId="2" ParentId="6464" CreationDate="2011-03-07T12:20:18.350" Score="4" ViewCount="" Body="&lt;p&gt;The concept of vacuum in physics indeed comes from two different theories. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The General Relativity Vacuum is a space-time model region without matter. In General Relativity all of space-time has a &quot;curvature&quot; which relates to the metric which can all have measurable effects, such as the bending of light rays (in the vacuum) near a massive object. One may wish to be a little careful of how one conceptualises the vacuum of empty space however since no events occur there since there is no interacting matter there. As soon as we have interacting matter we no longer just have a vacuum. Also General Relativity has introduced a term, called the Cosmological Constant $\Lambda$, which could be said to measure the curvature of the vacuum at the Cosmological level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Quantum Theory there is the concept of the &quot;Vacuum State&quot; which is a little different: it is the lowest possible energy state of a given quantum system. This lowest possible energy state has quantum fluctuations consistent with the $\Delta E \Delta t &amp;gt; h$ Uncertainty Principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if we straightforwardly apply the &quot;Vacuum State&quot; concept to the space-time vacuum we get a conceptually different model, called the &quot;Curved Space Vacuum&quot;. Some calculations show that this combined &quot;Curved Space Vacuum&quot; is rather different from the vacua of the two component theories: General Relativity and Quantum Mechanics. It has some interesting properties, like Temperature, that the component theories dont have and some calculations do suggest that the &quot;Quantum Vacuum Energy&quot; is different from the corresponding expected $\Lambda$ value by $10^{120}$. These results depend a little on what, if any, quantum fields uniformly pervade space and this aspect is not yet settled.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-03-07T12:20:18.350" CommentCount="7" />
  <row Id="6471" PostTypeId="2" ParentId="6464" CreationDate="2011-03-07T13:17:15.833" Score="-3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Numbers only exist in our minds.&lt;/strong&gt;&#xA;Numbers, any number, only exist in our head and, in physics, must relate to some property of the outer physical world.&lt;br&gt;&#xA;&lt;strong&gt;We measure the properties&lt;/strong&gt; and we use numbers as a tool, or a language.&#xA;We use numbers to measure or classify properties, by comparison with selected references, and we use ratios of measures to obtain adimensional quantities.&lt;br&gt;&#xA;&lt;strong&gt;A property belongs to an Entity&lt;/strong&gt; &#xA;When we measure a property we can not deny the existence of the underlying entity.&lt;br&gt;&#xA;In this case the &lt;strong&gt;Aether, Vacuum, Field, Physical Space, ..., can not be empty&lt;/strong&gt; because it has several properties:&#xA;$\epsilon_0, \mu_0, c^2=1/(\epsilon_0*\mu_0)$   &lt;em&gt;&lt;strong&gt;c&lt;/em&gt; is the Impedance of Vacuum&lt;/strong&gt;. The light speed is a property of space and not of the light. Gravitational waves also propagate at c speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;philosophy or physics&lt;/strong&gt;? Lets see:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Where to goes the energy when exists destructive interference ?&lt;/strong&gt;&lt;br&gt;&#xA;Light can cancel light under appropriate conditions. This statement is backed by theory and by experiment (&lt;a href=&quot;http://physics.stackexchange.com/questions/5743/anti-laser-how-sure-we-are-that-energy-is-transported&quot;&gt;anti-laser discussion&lt;/a&gt; and &lt;a href=&quot;http://www.eng.yale.edu/caolab/papers/Sci11b.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt;). What is the nature of this destructive interference?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;anti-laser experiment&lt;/strong&gt; shows :  The outcome is 0 energy after the destructive interference of two beams of ligth,  each with energy content.&#xA;If energy is to be conserved then the &lt;strong&gt;vacuum must possess that energy&lt;/strong&gt; after we lost it from sight.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Photons are vacuum perturbations that propagate.&lt;/strong&gt; (my friend Alf told me that long time ago)&lt;br&gt;&#xA;This is a possible conclusion to maintain coherence with experiment.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;VACUUM IS ALL&lt;/strong&gt; - is Spinosa(Baruch) view (in &lt;a href=&quot;http://en.wikisource.org/wiki/Ethics_%28Spinoza%29&quot; rel=&quot;nofollow&quot;&gt;Ethics&lt;/a&gt;) and recently Israel Pérez: &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1012/1012.2423v1.pdf&quot; rel=&quot;nofollow&quot;&gt;A physicist’s view of the universe: a philosophical approach&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-03-09T23:54:40.963" LastActivityDate="2011-03-09T23:54:40.963" CommentCount="4" />
  <row Id="6472" PostTypeId="2" ParentId="6086" CreationDate="2011-03-07T13:40:52.460" Score="1" ViewCount="" Body="&lt;p&gt;As this is a Bounty question (and because my other answer has a long set of comments) I have decided to add another answer. This answer is somewhat different from other answers provided, although it is consistent with them. There has been some challenge to understanding the other answers, and there may be some difficulty in understanding this answer too, but I shall record it here for those interested in this physical scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short: there &lt;strong&gt;is&lt;/strong&gt; a contradiction in this physical scenario. The nature of this contradiction and its consequences I shall discuss at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider the two physical assumptions which make up the scenario:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A) A particle on a trajectory starting just above an Event Horizon, this trajectory being a timelike trajectory leading to timelike infinity (in an asymptotic model, say);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(B) A rigid rod (of length L say) straddling ie &quot;half-in&quot; and &quot;half-out&quot; of the Event Horizon at some time t say (in some appropriate coordinates).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scenario continues by discussing Frames and so on, but this answer does not depend on anything else. Let us examine each of these assumptions a little more carefully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A) Is it possible for such a trajectory to exist? It depends on the Black Hole metric, but for the Schwarzchild metric if the particle has energy greater than some minimum E(R), then it may escape the region. If its energy and angular momentum is less than this value, it may orbit the Black Hole, or may directly plunge in. So let us assume that we are dealing with a Schwarzchild metric and that the particle can be given sufficient energy to follow the escaping trajectory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let us consider (B) in more detail:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is (B) possible? I claim that (B) is an inconsistent assumption. I shall outline a proof below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we need to return to the trapped surface property of an Event Horizon: a particle P is inside an Event Horizon if &lt;em&gt;every&lt;/em&gt; trajectory leads to the Singularity. So now consider the rod. This rod is &quot;rigid&quot; in some sense, although we do not require any properties of &quot;rigid&quot; in this proof - so it could be (normal matter) elastic. However it is simpler to assume a regular model of a rigid rod of length L (much smaller than R=2M, say). Consider two points on the rod at time t: P is inside the Event Horizon and Q is outside. Since Q is outside there exists, by definition, some trajectory $\gamma$ such that $\gamma$ does not lead to the Singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let the rod trajectory be such that point Q follows the $\gamma$ trajectory. The point P necessarily will follow a trajectory leading to the Singularity and so the proper distance PQ will extend to at least R = 2M in finite proper time. Thus the rod will break, and so the rod was not rigid as assumed but composed of at least two separate components (this is Hawking radiation!). So we have a contradiction as we assumed that the rod was rigid. So assumption (B) is inconsistent with General Relativity and cannot be used in any thought experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A first objection might be via a &quot;fluid model&quot; intuition of the Event Horizon which allows a rod to be outside, then part straddling, then maybe entirely contained in the &quot;fluid&quot;. But this intuition is not valid here: either the rod &lt;strong&gt;is&lt;/strong&gt; or &lt;strong&gt;is not&lt;/strong&gt; in the Event Horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second objection might be that this implies that the Event Horizon (fluid) has &quot;moved&quot; superluminarily, and this is not possible. The explanation is that the Event Horizon is not a local physical object and not constrained by the restrictions of Special Relativity. In fact it is a Global General Relativity object with counterintuitive properties: discontinuity and achronicity (see Hawking and Ellis 1973) and, as shown here, superluminarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can now understand in outline the paradox that the original question identified with Inertial Frames. These are &lt;em&gt;local&lt;/em&gt; objects in which a &lt;em&gt;Global&lt;/em&gt; GR entity was analysed - but any attempt to account for the behaviour of a global object by purely local analysis will result in the sort of contradictions and paradoxes that the question has uncovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT ADD FOR CLARITY:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a further objection to the stark conclusion of this answer, that can be understood in terms of further thought experiments. I shall discuss these and how they relate to the original question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume that the rod is actually a long spacecraft with a removable capsule at the top. If the lower part of the spacecraft is within the EH, then the capsule &lt;em&gt;might&lt;/em&gt; be fired off, escaping to infinity. Thus in this case it is not true that the entire spacecraft either is or is not contained within the EH, and the EH fluid model somewhat applies. Of course then the rod is not actually rigid as we assumed, so our conclusion is still valid, but only just. We could generalise this scenario to a spacecraft with N modules. A further generalisation simply assumes that the matter of the rod is such that at any distance along its length an explosion can occur (perhaps caused by a striking antiparticle) which would cause the rod to split and the top part to escape (to infinity). In this case (and within the modelling approximation) it would be appropriate to consider a fluid-like  model for the EH and thus talk about the rod &quot;straddling&quot; the Horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this &quot;straddling&quot; model assumes a wider physical scenario than did the original thought experiment, which merely considered the rod as an inert object, and certainly did not consider explosions, the quantum matter in the rod, and colliding particles which might happen (occasionally) in realistic physical situations. When these other factors are present one can discuss &quot;straddling&quot; rods: in the bare model as presented the &quot;straddling&quot; concept becomes inconsistent as discussed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this wider sense the answer to the part of the thought experiment about how the upper straddling rod could in principle be detected in frame X apparently escaping to infinity faster than a nearby particle, is that experiments in X will have detected that the rod has been in some form of explosion near the EH (unlike any non-EH straddling rod which happened also to be escaping). However the original thought experiment was so limited (to a rigid rod idealisation) that &quot;explosions&quot; and the like cannot occur; as a consequence &quot;straddling&quot; cannot occur either.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-03-14T16:36:44.633" LastActivityDate="2011-03-14T16:36:44.633" CommentCount="6" />
  <row Id="6473" PostTypeId="1" CreationDate="2011-03-07T13:48:48.290" Score="4" ViewCount="157" Body="&lt;p&gt;In general relativity, we use the term &quot;time-like&quot; to state that two events can influence one another. In fact, in order for an event to physically interact with another one, they have to be inifnitely close both in time and space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know (correct me if I'm wrong) this principal of &quot;near action/causality&quot; is conserved in all branches of modern physics and that is one of the reasons people are looking for &quot;force carriers&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the case, then would it be accurate to say that causality is simply a measure of continuity in all dimensions - and not only the time dimension?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I don't know anything about continuum mechanics other than its name, but it may have something to do with what I'm asking)&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-03-07T15:41:34.747" LastActivityDate="2011-03-10T16:28:16.843" Title="Is causality synonymous with continuity?" Tags="&lt;general-relativity&gt;&lt;causality&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="6474" PostTypeId="2" ParentId="6461" CreationDate="2011-03-07T14:13:14.250" Score="0" ViewCount="" Body="&lt;p&gt;A not standard answer, i.e. you should not use it in scholar context:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book of &lt;strong&gt;Douglas Pinnow ' (1) The Resonant Universe'&lt;/strong&gt; is a cheap monography of his model of particles.&#xA;It is entirely based in Electromagnetics, and has &lt;strong&gt;only one parameter&lt;/strong&gt; ($m_e$- electron mass) and derives all the relevant particle properties, to the laymen and also with math, &lt;strong&gt;well within 1%  for barionic masses&lt;/strong&gt; without the barionic 'spin crisis'.&lt;br&gt;&#xA;&lt;strong&gt;Concepts like mass, gravitation and charge become visually appealing to my mind.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;He predicted a new particle, $Z_0$ like, near 780 GeV&lt;/strong&gt;, and the next one near 1450 GeV. As SM is probably in trouble, as seen in recent experimental findings, may be we should read the book.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no need of weak or strong force. Only EM, he followed after Kaluza-Klein, Goedecke, Haus (non-radiating condition EM sources). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also by copy_paste ;-) &quot;In addition to generating the correct magnetic moments of the proton and neutron without adjustable parameters, such as required by the standard model, the resonant model predicts a negative value for the magnetic moment of the neutral sigma baryon wich has yet to be measured&quot;.The SM predicts a positive value.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that I'm the only reader of his book. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) - He had a long career in nuclear physics and light engeneering.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-03-07T14:13:14.250" />
  <row Id="6475" PostTypeId="2" ParentId="6464" CreationDate="2011-03-07T14:23:10.790" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;If we have two fundamental particles in space is there empty space between?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am no sure. Take a neutral atom. Outside it the space is empty, no force is acting on a probe particle. But it is because of neutralization of fields of positive and negative charges. Without neutralization the field is a long-range thing. So vacuum is such a combination of external forces that produces zero net effect.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-03-07T14:23:10.790" />
  <row Id="6476" PostTypeId="1" AcceptedAnswerId="6478" CreationDate="2011-03-07T14:56:55.973" Score="6" ViewCount="122" Body="&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/5bscW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like the order behavior shown in the image, is it due to the universality of some fundamental mathematic theory? Is there some general physics explanation for it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-&#xA;edit: This question comes after I read Terence Tao's talk on universality (&lt;a href=&quot;http://terrytao.files.wordpress.com/2011/01/universality.pdf&quot; rel=&quot;nofollow&quot;&gt;pdf&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="1948" LastEditorUserId="1948" LastEditorDisplayName="" LastEditDate="2011-03-07T15:22:46.160" LastActivityDate="2011-03-07T21:24:52.663" Title="How to understand order emerging from many-body system?" Tags="&lt;many-body&gt;" AnswerCount="2" />
  <row Id="6478" PostTypeId="2" ParentId="6476" CreationDate="2011-03-07T15:11:27.437" Score="6" ViewCount="" Body="&lt;p&gt;I don't see much order on this particular chaotic picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I see is some localized mass. The mass is localized because of the attractive force of gravity. Also, what we could see if you included an animation is that the galaxy is spinning. That's also why the shape resembles a disk. This increase of the spinning after the gravitational collapse - or clumping - is due to the conservation of the angular momentum, see&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/2934/why-do-galaxies-and-water-going-down-a-plug-hole-spin/2938&quot;&gt;Why do galaxies and water going down a plug hole spin?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More generally, one could ask why many systems in Nature that contain many degrees of freedom often look simpler - for example why diamond looks so nice and clear even though it's a complicated bound state of many carbon atoms. There's simply not choice because there must always exist an expansion of the behavior of a system around an infinite number of building blocks. The number of mathematically possible, qualitatively different behaviors is limited, so Nature has to recycle them.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-07T15:11:27.437" CommentCount="2" />
  <row Id="6479" PostTypeId="1" AcceptedAnswerId="6490" CreationDate="2011-03-07T15:57:07.357" Score="5" ViewCount="176" Body="&lt;p&gt;The answer to &lt;a href=&quot;http://physics.stackexchange.com/q/6457/2359&quot;&gt;a previous question&lt;/a&gt; suggests that a moving, permanently magnetized material has an effective electric polarization $\vec{v}\times\vec{M}$. This is easy to check in the case of straight-line motion, using a Lorentz boost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect this formula is still correct for motion that is not in a straight line, but I'm not interested in reinventing the wheel. Does anyone know of a textbook or journal article that derives this $\vec{v}\times\vec{M}$ term? Even better, does anyone know of experimental observation of this effect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;br&gt;&#xA;Followup question:&#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/6581/what-is-the-electric-field-generated-by-a-spinning-magnet&quot;&gt;What is the electric field generated by a spinning magnet?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="2359" LastEditorDisplayName="" LastEditDate="2011-03-09T23:48:53.003" LastActivityDate="2011-05-16T08:20:11.813" Title="What's a good reference for the electrodynamics of moving media?" Tags="&lt;electromagnetism&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6480" PostTypeId="2" ParentId="6473" CreationDate="2011-03-07T16:00:54.430" Score="2" ViewCount="" Body="&lt;p&gt;Yes, in the context of special relativity and its extensions, causality is pretty much equivalent to locality - you seem to use the word &quot;continuity&quot; for &quot;locality&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In non-relativistic physics, causality requires that the cause must occur before its effect, $t_{cause}\leq t_{effect}$. Because this condition has to hold in all reference frames - much like all laws of physics - it follows that in relativity, the cause must belong to the past light cone of the effect. In other words, signals etc. may propagate at most by the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This fact also implies that it must be possible to describe the laws of physics in terms of partial differential equations - which contain both derivatives with respect to time as well as those with respect to space. No direct &quot;action at a distance&quot; is possible in relativity. This extrapolation of the properties of the temporal coordinate to the spatial coordinates is a direct consequence of special relativity that introduces a symmetry between space and time, the Lorentz symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See many other articles about causality on this server:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/search?q=causality&quot;&gt;http://physics.stackexchange.com/search?q=causality&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-07T16:00:54.430" />
  <row Id="6481" PostTypeId="2" ParentId="6464" CreationDate="2011-03-07T16:14:46.867" Score="3" ViewCount="" Body="&lt;p&gt;With the risk to repeat someone I'll say what I think. You shouldn't think of vacuum as empty space. Vacuum is the no particle state of a quantum field and no particle doesn't mean empty. The field is still there, it is just in a state we interpret as no particles. And particles are just a very convinced interpretation.   &lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2011-03-07T16:14:46.867" CommentCount="5" />
  <row Id="6482" PostTypeId="1" CreationDate="2011-03-07T16:20:53.073" Score="7" ViewCount="546" Body="&lt;p&gt;Here and on many other forums and blogs people ask the question &lt;a href=&quot;http://physics.stackexchange.com/questions/6438/what-if-the-lhc-doesnt-see-susy&quot;&gt;&quot;What if LHC does not find SUSY?&quot;&lt;/a&gt;. I would like to ask the opposite. What if it finds it? What would the implications be? Is it going to just confirm something understood and expected or is it gong to bring something new? Will there be implications for string theory? String theorists will take it as a good sign, but by itself it is not a confirmation of string theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do realize this is more that one question, but they are on the same topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p.s If this question is not suitable for this site, please delete/close  &lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-03-07T19:05:46.033" LastActivityDate="2011-03-07T21:17:00.450" Title="What if LHC finds SUSY?" Tags="&lt;particle-physics&gt;&lt;supersymmetry&gt;&lt;large-hadron-collider&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="6483" PostTypeId="1" AcceptedAnswerId="6484" CreationDate="2011-03-07T17:00:28.213" Score="9" ViewCount="307" Body="&lt;p&gt;In optics it is widely mentioned real images are projectable onto screens whereas virtual ones can only be seen by a person.  Isn't that contradictory?  I mean in order to see the virtual image it has to be &lt;em&gt;projected&lt;/em&gt; onto the retina (ultimately acting as a screen).  So why can you see virtual images in the first place?&lt;/p&gt;&#xA;" OwnerUserId="2415" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-08T09:18:25.160" LastActivityDate="2011-03-08T09:18:25.160" Title="Why can you see virtual images?" Tags="&lt;optics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="6484" PostTypeId="2" ParentId="6483" CreationDate="2011-03-07T17:10:34.057" Score="11" ViewCount="" Body="&lt;p&gt;Your eye is a second optical system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &lt;b&gt;&lt;i&gt;re&lt;/i&gt;&lt;/b&gt;-focuses the diverging rays to produce a real image on the retina.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process is exactly the same thing it does when looking at a nearby (i.e. not at effective infinity) object.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-03-07T17:10:34.057" CommentCount="2" />
  <row Id="6485" PostTypeId="2" ParentId="6482" CreationDate="2011-03-07T17:29:34.057" Score="6" ViewCount="" Body="&lt;p&gt;An even more excellent question than the opposite one!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Experimenters' job&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first happy group of people for whom the discovery of SUSY at the LHC would be spectacular would be the experimenters. They would experience fireworks of activity, facing the task to find as many superpartners as possible and to measure their properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All their masses - and several other parameters - would be news for the theorists. We would have to learn lots of new, currently unknown numbers from the experiments, the so-called soft SUSY-breaking parameters. It's conceivable that the masses would display some patterns - e.g. mass ratios - that agree with one of the scenarios of supersymmetry breaking. So some of the mechanisms of supersymmetry breaking would be supported; others would ultimately be abandoned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Figuring out the logic of SUSY breaking&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite a lot of data is needed from the first hints of SUSY to the point in the previous paragraph because one needs to see many new particles and some of them are inevitably harder than others. However, if this job is completed, we will know whether SUSY breaking is mediated by gravity mediation, anomaly mediation, gauge mediation, mirage mediation, or something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be a lot of interesting work for the theorists and it is more likely than not that they have already done the essential work for any major type of SUSY breaking that could be observed by the LHC. Obviously, one of the schemes would be studied in much more detail if we knew that it is the correct one. Some knowledge about the mechanism of SUSY breaking - which may be guessed from the superpartner masses - would tell us a great deal about the required underlying compactification of string theory etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the role of the newly found SUSY for all the problems that SUSY is capable of solving - that have been discussed under the question&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/6438/what-if-the-lhc-doesnt-see-susy&quot;&gt;What if the LHC doesn't see SUSY?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(I mean dark matter, gauge coupling unification, and the hierarchy problem) would be studied in much more detail, too. I don't have to enumerate them again because this text would become highly redundant. It's obvious that if the existence of low-energy SUSY were supported by the experiments, all other things that were expected to be solved by SUSY would be studied much more seriously, much more materially, and in much more detail. Some of the advantages of SUSY are only understood superficially - this situation would have to improve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;LHC SUSY discovery and string theory&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discovery of SUSY would be amazing - the first discovery of a new spacetime structure and symmetry since Albert Einstein's relativistic adventures. It would be great, it may happen, and string theorists must be ready to take credit for it. No doubt, at least in the Western (non-Soviet) context, SUSY is a daughter of string theory. First, it was discovered by Pierre Ramond on the 2-dimensional world sheet of a string, before it was exported to the 4D spacetime and other higher-dimensional spacetimes and before it was found in the 10D spacetime of superstring theory as well. Historically, supersymmetry was one of the first amazing ideas that the physicists were forced to discover because string theory led them to discover them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the critics of string theory would be proved to be spectacularly wrong - in fact, narrow-minded folks who wanted to prevent the mankind from discovering one of the most fundamental properties of Nature; everyone could suddenly see that they are on par with the geocentrists and they would hopefully never show up again in the public. I would win a USD 10,000 bet against a phenomenologist who agreed with 100-to-1 odds - this &quot;uneven&quot; number itself is enough to show that some of the enemies of supersymmetry resemble a fundamentalist religious sect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pragmatic and largely non-anthropic phenomenological attitude to string theory would prevail. People would probably agree that a feature of the vacua doesn't have to be &quot;generic&quot; for it to become true. The anthropic principle would fade away. Because low-energy SUSY would become a fact, people would kind of accept that with some extra knowledge about the reality used as assumptions, supersymmetry is a consequence of string theory. See&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2010/06/why-string-theory-implies-supersymmetry.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/06/why-string-theory-implies-supersymmetry.html&lt;/a&gt;&lt;br&gt;&#xA;  Why string theory implies supersymmetry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lots of knowledge about the likely compactification of string theory would become much more accessible. String theorists - which would become a quickly growing group - would very likely converge to some opinion whether heterotic string theory; heterotic M-theory; M-theory on a G2 holonomy manifolds; type IIA intersecting braneworlds; or F-theory on Calabi-Yau four-folds is the most viable approach to phenomenology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have discussed in what sense string theory probably implies supersymmetry. Now, you're obviously interested in the opposite hypothetical implication - whether supersymmetry implies string theory. We can't prove this implication as a mathematical theorem but it would become extremely persuasive. First of all, supergravity (SUGRA) would become an inevitable component of all effective field theories because it follows from general relativity (established) and supersymmetry (hypothetically established in our thought experiment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this text,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2008/07/two-roads-from-n8-sugra-to-string.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2008/07/two-roads-from-n8-sugra-to-string.html&lt;/a&gt;&lt;br&gt;&#xA;  Two roads from N=8 SUGRA to string theory,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I argue that supergravity suffers from two kinds of problems: non-perturbative inconsistencies; and the unacceptable phenomenological limitations of its maximally supersymmetric N=8 version (which is the only perturbatively finite one). Attempts to fix either of those limitations of supergravity inevitably leads to string theory with its more powerful toolkit. If you want to watch a 30-minute lecture by a Dirac Medalist explaining why supergravity can't be decoupled from string theory and why one needs all of string theory to preserve the consistency, see this November 2010 talk by Michael Greene in Trieste:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=UVqCAhLiZDc&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=UVqCAhLiZDc&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://www.youtube.com/watch?v=S8wSl2R3G1o&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=S8wSl2R3G1o&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Similarly, in the text a few paragraphs above, I argue that locality of general relativity implies that there must exist magnetically charged objects - at least black holes - and the Dirac quantization rule implies that the charges must belong to a lattice. The choice of the lattice is equivalent to the point of the moduli space of inequivalent stringy vacua; the noncompact symmetry group of SUGRA is inevitably broken down to its discrete exception subgroup, the U-duality group. In the appropriate limits of the moduli space of the inequivalent vacua, we may derive the existence of objects known from string/M-theory as well as their excitations, and we may pretty much complete the rest of string theory by consistency arguments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Green-Schwarz anomaly cancellation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one more characteristically stringy structure that would become necessary if we add one more assumption: the Green-Schwarz mechanism that mixes tree-level terms with one-loop terms in a very stringy way - originally discovered by Green and Schwarz in 1984 when their discovery sparked the first superstring revolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This mixing of contributions at different orders is extremely unnatural in perturbative field theory. And we would have evidence that it takes place in Nature assuming that there exists at least one axion - or, in SUGRA terms, at least one linear supermultiplet. If there are axions, which may be needed to solve the strong CP-problem, there are also new kinds of anomalies (in particular, the &quot;conformal anomaly&quot; of supergravity) analogous to the 10-dimensional anomalies addressed by Green and Schwarz in 1984. A 4-dimensional version of the stringy Green-Schwarz mechanism would be needed to cancel those anomalies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theorems may be rigorously proved in the highly symmetric vacua only but not in the real world. However, in the real world, the evidence that string theory is right would become overwhelming.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-07T17:53:25.347" LastActivityDate="2011-03-07T17:53:25.347" CommentCount="19" />
  <row Id="6486" PostTypeId="2" ParentId="6473" CreationDate="2011-03-07T17:36:39.353" Score="4" ViewCount="" Body="&lt;p&gt;The answer is &quot;no&quot; if you take precise definitions of &lt;a href=&quot;http://en.wikipedia.org/wiki/Continuity_%28topology%29&quot; rel=&quot;nofollow&quot;&gt;&quot;continuity&quot;&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Causal_structure&quot; rel=&quot;nofollow&quot;&gt;&quot;causality&quot;&lt;/a&gt;. Then these are different concepts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed. The &quot;background&quot; for general relativity is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemannian_manifold&quot; rel=&quot;nofollow&quot;&gt;manifold&lt;/a&gt; which necessarily have some topological structure -- it locally &quot;looks like&quot; an Euclidean space and you just take standard Euclidean topology to your manifold. You can intuitively understand it with the Einsteinian picture of coordinate system, where you have freely falling observer with a clock at every point. That's what I understand under &quot;continuity&quot;. But the structure of a manifold is not enough.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Causal_structure&quot; rel=&quot;nofollow&quot;&gt;causal structure&lt;/a&gt; -- is some &quot;order&quot; that you establish for the points of your manifold. This is the extra structure and it is not the same as the &quot;continuity&quot; in I way one usually understands it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it simply -- you can always continuously move to causally disconnected points (or events) on your manifold.  &lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2011-03-07T17:41:47.507" LastActivityDate="2011-03-07T17:41:47.507" />
  <row Id="6487" PostTypeId="2" ParentId="6482" CreationDate="2011-03-07T18:25:40.517" Score="0" ViewCount="" Body="&lt;p&gt;@MBN I remember lectures in supersymmetry back in the eighties and the justification was that it made computations finite, nothing about strings. I guess it is history in retrospect. The thesis of my doctorate ( experimental) dealt with Regge poles but it is from Lubos I first heard the string connection. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-07T18:25:40.517" CommentCount="2" />
  <row Id="6488" PostTypeId="1" AcceptedAnswerId="6507" CreationDate="2011-03-07T19:21:18.033" Score="2" ViewCount="444" Body="&lt;p&gt;I understand the reason that bicycle pedals are oppositely threaded on either side. What I don't understand is why it works because I'm missing something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the right pedal for example. It's threads are right-handed, so it would be tightened by turning it clockwise (if the pedal were in between you and the bike). Now imagining this from the same perspective, a person pedaling makes the right pedal constantly turn counter-clockwise, relative to the crank. If there's any friction between the pedal and the bolt attaching it to the crank, I would imagine that it would create a force that would try to turn the bolt counter-clockwise along with the pedal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the opposite of what happens, and I've seen the bolts unthread after pedaling backwards a lot myself. What am I missing?&lt;/p&gt;&#xA;" OwnerUserId="2417" LastActivityDate="2011-03-08T10:34:16.537" Title="Why are bicycle pedal threads' handedness left on the left and right on the right?" Tags="&lt;forces&gt;&lt;friction&gt;&lt;torque&gt;&lt;bicycles&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="2" />
  <row Id="6489" PostTypeId="2" ParentId="6488" CreationDate="2011-03-07T19:51:03.420" Score="1" ViewCount="" Body="&lt;p&gt;Edit: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did not understand the explanatins from en.wikipedia really, but the word precession set me on a track. The bolt will roll in the wider bore, its the force from the foot, which holds the pedal down all the circling motion, not only during stroke down: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of a bolt (threaded is not important first) in a bore of slightly higher diameter. &#xA;Then rotate the end of  bolt like a  joy.stick. The bolt will roll in the bore (epicycles!) thus rotating in the opposite direction. If the bolt is threaded, it will scew/unscrew, depending on handedness and direction of &quot;precession&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;de.wikipedia says this effect does overcompensate the friction effect by far., but it is only &#xA;important in case the pedal is loose already! But because under this condition the thread will be damaged soon, they say that this security feature is more theoretical. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-08T10:34:16.537" LastActivityDate="2011-03-08T10:34:16.537" CommentCount="3" />
  <row Id="6490" PostTypeId="2" ParentId="6479" CreationDate="2011-03-07T20:15:05.567" Score="6" ViewCount="" Body="&lt;p&gt;To close the loop, Andrew, the answer to your newest question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best and most famous reference about the electrodynamics of moving bodies is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Einstein, Albert (1905-06-30). &quot;Zur Elektrodynamik bewegter Körper&quot;. &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/andp.200590006/abstract&quot; rel=&quot;nofollow&quot;&gt;Annalen der Physik 17: 891–921&lt;/a&gt;. See also a digitized version at &lt;a href=&quot;http://wikilivres.info/wiki/Zur_Elektrodynamik_bewegter_K%C3%B6rper&quot; rel=&quot;nofollow&quot;&gt;Wikilivres:Zur Elektrodynamik bewegter Körper&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The English translation, &quot;On the Electrodynamics of Moving Bodies&quot;, is here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.fourmilab.ch/etexts/einstein/specrel/www/&quot; rel=&quot;nofollow&quot;&gt;http://www.fourmilab.ch/etexts/einstein/specrel/www/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The content of this paper became known as the special theory of relativity. I am just partly joking because for uniformly moving media, the Lorentz boost to the rest frame is still the most natural way to proceed.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-05-16T08:20:11.813" LastActivityDate="2011-05-16T08:20:11.813" CommentCount="2" />
  <row Id="6491" PostTypeId="2" ParentId="6488" CreationDate="2011-03-07T20:15:17.533" Score="2" ViewCount="" Body="&lt;p&gt;I don't know if it's right or not, but &lt;a href=&quot;http://en.wikipedia.org/wiki/Bicycle_pedal&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; has this to say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The right-side (usually the&#xA;  drive-side) pedal spindle is&#xA;  right-hand threaded, and the left-side&#xA;  (usually the non-drive-side) pedal&#xA;  spindle is left-hand (reverse)&#xA;  threaded to help prevent it from&#xA;  becoming loose by an effect called&#xA;  precession.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Although the left pedal turns&#xA;  clockwise on its bearing relative to&#xA;  the crank (and so would seem to&#xA;  tighten a right-hand thread), the&#xA;  force from the rider's foot presses&#xA;  the spindle against the crank thread&#xA;  at a point which rolls around&#xA;  clockwise with respect to the crank,&#xA;  thus slowly pulling the outside of the&#xA;  pedal spindle anticlockwise&#xA;  (counterclockwise) because of friction&#xA;  and thus would loosen a right-hand&#xA;  thread.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-07T20:15:17.533" CommentCount="1" />
  <row Id="6492" PostTypeId="2" ParentId="6259" CreationDate="2011-03-07T21:00:23.203" Score="5" ViewCount="" Body="&lt;p&gt;At &lt;em&gt;tree level&lt;/em&gt;, the conditions for the electroweak-symmetry breaking vacuum of the MSSM can be found in any of the standard review articles and are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sin(2\beta) = \frac{2B_\mu}{2|\mu|^2 + m_{H_u}^2 + m_{H_d}^2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{2} m_Z^2 = -|\mu|^2 - \frac{m_{H_u}^2 \tan^2\beta - m_{H_d}^2}{\tan^2\beta - 1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the right &lt;em&gt;logical&lt;/em&gt; way to think about this is that a top-down theory determines the values of $\mu$ and the soft-breaking parameters $m_{H_u}^2$, $m_{H_d}^2$, and $B_\mu$. These are really the inputs. On the other hand, we only want to consider the slice of parameter space on which EWSB is realized as in our world, with a particular value of $m_Z$. One can choose other useful coordinates on this slice, like $\tan \beta$, as software inputs to choose only those points on this slice of parameter space, rather than specifying the high-scale input, which in general will fail to realize correct EWSB.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2011-03-07T21:00:23.203" CommentCount="2" />
  <row Id="6493" PostTypeId="2" ParentId="6482" CreationDate="2011-03-07T21:17:00.450" Score="5" ViewCount="" Body="&lt;p&gt;Something analogous happened a couple of decades ago. The discovery of the $W$ and $Z$ bosons simultaneously confirmed the Weinberg/Salam electroweak model &lt;em&gt;and&lt;/em&gt; made &quot;how is electroweak symmetry broken?&quot; the key new unanswered question. It's still unanswered, and it's much more challenging, experimentally and theoretically. The discovery of supersymmetry and a Higgs would go a long way toward answering it (the standard scenario in the MSSM is called &quot;radiative electroweak symmetry breaking.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analogously, the discovery of superpartners would make the key unanswered question in particle physics &quot;how is supersymmetry broken?&quot; Actually, there are two parts to this question: the first is how the symmetry is broken, and the second is how that breaking is &quot;mediated&quot; to the Standard Model. (This is a somewhat technical story, but the point is that supersymmetry breaking must involve fields beyond just the Standard Model, together with some mechanism that allows the breaking to be felt by the Standard Model.) The question of how the breaking is mediated is likely to be easier to answer than the question of how the breaking itself actually happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, though, these will be complicated questions that will depend on the accurate measurement of a number of masses, together with a lot of guesswork and hopefully some luck. There are several good ideas for models, but in the absence of data it's hard to know which ones to believe, and they all have some fairly significant problems.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2011-03-07T21:17:00.450" />
  <row Id="6494" PostTypeId="2" ParentId="6476" CreationDate="2011-03-07T21:24:52.663" Score="3" ViewCount="" Body="&lt;p&gt;Systems containing more than 2 bodies have been proved to be chaotic (Poincarè solved an approach to this problem with perturbation theory, when the king of Sweden proposed a problem: Solve the Solar System movement). Surely, it's been proved that one cannot express the solution of the motion of even a 3-body system in terms of algebraic formulas and integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to classical mechanics, if you know a certain state of the given system, you'll know all the evolution of the system with time, forward and backward. But the problem is that the set of equations one must solve to determine the functions ruling the movement of all the system particles is chaotic. It means that, given an initial state, you CAN predict all the system trajectory, but the system is very sensible to a slight change on the initial state. This is called &lt;strong&gt;deterministic chaos&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to threat the phase space of a system composed with many particles like a Lioville fluid, you'll get the Boltzmann Transport equation. Which also describes the behavior of a distribution of particles. But to derive Boltzmann equation in order to describe the system behavior, one needs to make an assumption about the collisions on the system, where you lose information from one collision to another (it's been a problem for a long time, now...). It results on &lt;strong&gt;random chaos&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The apparent order on the system comes from a dynamical competition between the forces acting among the system's components. You can call it &quot;interaction&quot;. A common characteristic of chaotic systems is the competing of two forces. For example, take the Earth-Sun-Moon system. Acting on each body, you have 2 forces. The order which will arise from the system depends on the tune you make for the forces' intensities and directions. And you should take care with this tune, because the behavior resulting from it may be very different from what you expect, since the system is chaotic. In order to study this kind of problem, one may call upon a lot of recent tools developed since Poincarè approach to it, in the early XX century.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for my bad english!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best regards!&lt;/p&gt;&#xA;" OwnerUserId="2418" LastActivityDate="2011-03-07T21:24:52.663" CommentCount="3" />
  <row Id="6495" PostTypeId="5" CreationDate="2011-03-07T21:41:03.137" Score="0" ViewCount="" Body="&lt;p&gt;Neutrinos are light, uncharged leptons. The neutrino tag should be applied to question relating to neutrino properties or interactions involving neutrinos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutrinos are produced in nuclear reaction involving the weak force. Sources that are useful for experimental efforts include the sun (matter type, electron flavored neutrinos), nuclear fission reactors (anti-matter type, electron flavored neutrinos), the interactions of cosmic rays with the atmosphere and the interactions of man-made particle beams with matter (both matter and anti-matter, and all flavors)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having neither charge nor color, neutrinos interact only by way of gravity and the weak nuclear force. Both of these forces are, well, weak and the neutrinos have relatively low cross-section for interactions with ordinary matter.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Flavor?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Both the charged and the un-charged leptons come in three type which seem to be identical except for mass. The charged leptons are the electron, the muon, and the tau-lepton (often just called &quot;a tau&quot;). For each of these there is a corresponding neutrino, but see the section on mixing below.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Brief History&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A light uncharged particle was first proposed in 1930 by Wolfgang Pauli to solve the problem of the beta decay spectrum. Pauli called his particle a &quot;neutron&quot;, but that name was later adopted for the uncharged nucleon. The name &quot;neutrino&quot; (meaning &quot;little neutral one&quot;) was coined by Enrico Fermi in 1934. Neutrinos were originally modeled as massless for simplicity and in the absence of any measurable mass the assumption was adopted as a given. Neutrinos (actually anti-neutrinos) from a fission reactor were first detected experimentally in 1956 by Cowan and Reines using a delay coincidence technique that remains the standard for reactor neutrinos to this day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting in 1970 Raymond Davis Jr., Kenneth C. Hoﬀman and Don S. Harmer tried to measure the solar neutrino flux using a large tank full of cleaning fluid placed deep in the Homestake mine in South Dakota. They got a figure too low to match theories of stellar structure. This mis-match persisted for two decades, and required a change of theory to resolve: the neutrinos must be considered as massive (albeit light) and allowed to mix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experiments at Sudbury Canada, the Kamioka mine facility in Japan, various nuclear reactor complexes, and at several accelerator sites around the world would eventually show clear evidence of neutrino mixing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current efforts are focused on determining the parameters of the mixing matrix (two mass differences and two mixing angles are known), searching for evidence of CP violation in the neutrino sector, and determining if the neutrinos are Dirac or Majorana particles.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Mixing&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Mixing occurs because the flavor states of the neutrinos, written $\nu_e, \nu_\mu, \nu_\tau$ are not eigenstates of the free Hamiltonian. Those are called the &quot;mass states&quot; and are written $\nu_1, \nu_2, \nu_3$. In a mixing experiment, (anit-)neutrinos are produced in one location (production occurs in &lt;em&gt;flavor&lt;/em&gt; state) and allowed to propagate to another location where they are detected (again, detection is of &lt;em&gt;flavor&lt;/em&gt; states). During the time the neutrinos travel, they are acted upon by the free Hamiltonian which does not keep pure flavor state pure---that is, it mixes them. The result is that the distribution of flavor states detected may not match the distribution of flavor states created.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mixing was actually proposed by Gribov and Pontecorvo in 1968 (even before the Homestake experiment). [Phys. Lett, 28B, vol. 7, p. 493]&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Open Questions&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Measure the remaining parameters of the mixing ($\theta_{13}$ and $\Delta m^2_{31}$) and refine the values of the known parameters.&lt;/li&gt;&#xA;&lt;li&gt;Does neutrino mixing violate CP (i.e. is $\theta_{13} \ne 0$ and $\delta_{CP} \ne 0$?)&lt;/li&gt;&#xA;&lt;li&gt;Mass hierarchy problem.&lt;/li&gt;&#xA;&lt;li&gt;Dirac of Majorana nature?&lt;/li&gt;&#xA;&lt;li&gt;Are there additional neutrinos states?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-03-07T22:28:44.607" LastActivityDate="2011-03-07T22:28:44.607" />
  <row Id="6496" PostTypeId="4" CreationDate="2011-03-07T21:41:03.137" Score="0" ViewCount="" Body="Neutrinos are light, uncharged leptons. The neutrino tag should be applied to question relating to neutrino properties or interactions involving neutrinos." OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-03-07T22:28:46.930" LastActivityDate="2011-03-07T22:28:46.930" />
  <row Id="6497" PostTypeId="1" AcceptedAnswerId="6501" CreationDate="2011-03-07T21:55:04.733" Score="3" ViewCount="250" Body="&lt;p&gt;WARNING: Possibly a naive post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hi guys. I just watched on BBC, [some show] yesterday (sunday) where the host talked about Arrow of Time, where by 2nd law of thermodynamics states that &quot;nature&quot; is always low entropy to high entropy OR &quot;ordered/structured&quot; to &quot;unordered/unstructured&quot; state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;intiutive&quot; examples used, also found elsewhere, are events like iceberg &quot;spontaneously&quot; melting, but never freeze up again; perfume diffuse out of bottle but not entering back in it on its own; etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then what about the other things that appear to naturally create order like formation of cells and life, crystalization, snowflakes etc?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="2419" LastActivityDate="2011-03-09T18:06:03.707" Title="Explaining Arrow of Time with Entropy" Tags="&lt;entropy&gt;&lt;arrow-of-time&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="2" />
  <row Id="6498" PostTypeId="2" ParentId="6497" CreationDate="2011-03-07T22:19:19.080" Score="1" ViewCount="" Body="&lt;p&gt;Raskolnikov's comments are exactly right. If you want more detail, I wrote a &lt;a href=&quot;http://arxiv.org/abs/0903.4603&quot; rel=&quot;nofollow&quot;&gt;pedagogical note&lt;/a&gt; on precisely this subject for the &lt;em&gt;American Journal of Physics&lt;/em&gt; a while back.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-07T22:19:19.080" />
  <row Id="6499" PostTypeId="2" ParentId="6497" CreationDate="2011-03-07T22:19:49.990" Score="0" ViewCount="" Body="&lt;p&gt;Actually, you can maximize the Entropy OR minimize the Energy of a given system, in order to achieve an equilibrium state of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Entropy is not quite understood even in scientific recent works. What we have are only different views of it. In my point of view, a better interpretation of what Entropy really is comes from the derivation of the Boltzmann Transport equation. In order to derive it, you must make an assumption that the momenta of two particles before and after a given collision between them are uncorrelated. In other words, you must assume that you lose spontaneously information on every collision. When you do this (with some other assumptions), you get an reasonable equilibrium solution to the equation and, what is better, the system may has an equilibrium solution. This loss of information one my argue that raises the entropy of the system, so the entropy may be interpreted as a measure of the lack of information about the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &lt;strong&gt;doesn't&lt;/strong&gt; implicate that to have maximum entropy (or minimum information about the system), the system SHOULD be on an unordered state. And to be more precise, this lack of &quot;order&quot; should be interpreted as lack of &quot;information&quot;. Then, you can be happy again and fortunately accept life as an organized state of matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;update: Everything said here should coexist with the comments of the other users...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To read further about entropy as information, take a look at Shannon information theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best regards!&lt;/p&gt;&#xA;" OwnerUserId="2418" LastActivityDate="2011-03-07T22:19:49.990" />
  <row Id="6500" PostTypeId="2" ParentId="3618" CreationDate="2011-03-07T22:27:12.507" Score="2" ViewCount="" Body="&lt;p&gt;I know that Purcell and others have used Lorentz symmetry as a pedagogical device to motivate the introduction of magnetic fields, but I do not recall ever having seen an axiomatic derivation of Maxwell's equations.  It might be an interesting exercise to see precisely what assumptions beyond Lorentz symmetry and Coulomb's Law are necessary to reconstruct Maxwell's equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B fields are not fictitious fields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know the electric and magnetic fields in one inertial frame, you can determine the electric and magnetic fields in any other frame via Lorentz transformation.  If the magnetic field happens to vanish in a given inertial frame, you could think of magnetic effects in other frames as fictitious.  However, it is not always possible to find a frame in which the magnetic fields vanish.  The fastest way to see this is to note that E^2 - B^2 c^2 is a Lorentz invariant quantity (&lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_tensor#Properties&quot; rel=&quot;nofollow&quot;&gt;see Wikipedia&lt;/a&gt;).  If we find that B^2 &gt; E^2/c^2 at a given spacetime point in a given inertial frame, it follows that B^2 &gt; 0 at that point in all inertial frames.  In fact, you could begin in a frame where the electric field vanishes but the magnetic field does not; the electric fields observed in other frames could then be considered fictitious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, neither the electric field nor the magnetic field can be made to vanish under a Lorentz boost.  To see this quickly, note that the dot product of the E field vector with the B field vector at a given spacetime point is a Lorentz invariant quantity (&lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_tensor#Properties&quot; rel=&quot;nofollow&quot;&gt;see Wikipedia&lt;/a&gt;).  If this dot product is nonzero at a given spacetime point in a given inertial frame, the electric and magnetic field vectors will both be nonzero at that  spacetime point in all inertial frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Einstein pointed out, you can understand the motion of a charged particle by referring to the electric field in the rest frame of that particle.  However, if you have multiple particles with different velocities, you need to keep track of the electric field in the instantaneous rest frame of each particle.  Since Lorentz boosts mix the E field with the B field, the only way to keep track of the E field in the rest frame of each of your particles in terms of local quantities in one inertial frame is by reference to the E field and the B field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Locality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if it is possible, it is not clear to me that it would be desirable to use Coulomb's law as an axiom in electromagnetic theory.  Maxwell's equations explain the motion of particles by referring to local degrees of freedom, the fields.  Coulomb's law, on the other hand, is a form of action-at-a-distance, and is manifestly non-local.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is certainly possible to rewrite both the E and B fields in terms of integrals over charge density and current density (I can't post another link, so google &quot;Jefimenko's equations&quot;), and then to use these expressions to interpret electromagnetic forces as a form of retarded action-at-a-distance.  However, to obtain these expressions requires assumptions about the boundary conditions on the E and B fields.  We can always obtain another valid solution of Maxwell's equations by simply changing the boundary conditions on the fields, which demonstrates that the fields have independent existence, and are not mere book-keeping variables to simplify a more fundamental non-local interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Monopoles&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As usually written, Maxwell's equations do not contain terms corresponding to magnetic charge, but it would be consistent to add such terms.  In fact, Dirac showed that the quantization of electric charge could be due to the existence of magnetic monopoles (I can't post another link, so google &quot;magnetic monopole dirac quantization condition&quot;).  Maxwell's equations do not tell us whether magnetic monopoles exist or could exist, but the quantization of electric charge could be evidence that magnetic monopoles exist somewhere in the universe.&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2011-03-07T22:27:12.507" CommentCount="1" />
  <row Id="6501" PostTypeId="2" ParentId="6497" CreationDate="2011-03-07T22:27:25.887" Score="1" ViewCount="" Body="&lt;p&gt;On top of what has already been said before, it's worth noting that there are &lt;em&gt;several&lt;/em&gt; different kinds of &quot;arrows of time&quot; in Physics:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Arrow_of_time&quot; rel=&quot;nofollow&quot;&gt;Arrow of time&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://books.google.com/books?id=4yUu-simAhMC&quot; rel=&quot;nofollow&quot;&gt;The physical basis of the direction of Time&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Sure, Entropy is one of them, but there are others. And the way they &lt;em&gt;interact&lt;/em&gt; is not necessarily something that's very well understood and/or explained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that popular analogies are always missing some of the more rigorous and technical aspects — so, take them with a grain of salt. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Forgot to mention this book earlier, it's quite a gem: &lt;a href=&quot;http://books.google.com/books?id=h7UkMlAbwM0C&quot; rel=&quot;nofollow&quot;&gt;Evolution as entropy: toward a unified theory of biology&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="142" LastEditorUserId="142" LastEditorDisplayName="" LastEditDate="2011-03-08T03:06:43.647" LastActivityDate="2011-03-08T03:06:43.647" />
  <row Id="6502" PostTypeId="1" CreationDate="2011-03-07T22:40:14.770" Score="4" ViewCount="122" Body="&lt;p&gt;For a $c=1$ Boson on a circle at the self-dual rdius, we get an enhanced gauge symmetry $\hat{SU}(2)_1$. It is said that we can orbifold this model by any finite subgroup of $SU(2)$ since $SU(2)$ is a symmetry of the model. But the Lagrangian of a $c=1$ Boson does not have an $SU(2)$ symmetry even at the self-dual radius&#xA;which is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;L=\sqrt{2}\int d^2 z\  \partial \phi \bar{\partial} \phi&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;right?    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that at the self-dual radius we get extra marginal operators which close among each other to form the OPE of $\hat{SU}(2)_1$ but is there another form of the above Lagrangian which makes the $SU(2)$ symmetry manifest? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the $c=1$ Boson at the self-dual radius equivalent to a level 1 WZW model? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1531" LastActivityDate="2011-03-08T06:44:24.760" Title="Orbifolds of the $c =1$ Bosonic theory on a circle" Tags="&lt;string-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6503" PostTypeId="1" AcceptedAnswerId="6556" CreationDate="2011-03-07T23:16:09.133" Score="4" ViewCount="270" Body="&lt;h2&gt;Background&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Calculate how much weight a pole can support without &lt;a href=&quot;http://en.wikipedia.org/wiki/Necking_%28engineering%29&quot; rel=&quot;nofollow&quot;&gt;necking&lt;/a&gt; when a monitor is attached to an articulated arm. This is a &lt;a href=&quot;http://www.efunda.com/formulae/solid_mechanics/beams/casestudy_display.cfm?case=cantilever_endload&quot; rel=&quot;nofollow&quot;&gt;cantilever problem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Problem&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There are three objects involved in this problem: a support pole, a monitor (weight), and an articulated arm (extends the weight away from the post).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Monitor&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;A 12kg &lt;a href=&quot;http://www.samsung.com/nz/consumer/pc-peripherals/monitor/lcd-monitor/LS30HUBCB/XSA/index.idx?pagetype=prd_detail&amp;amp;tab=specification&quot; rel=&quot;nofollow&quot;&gt;monitor&lt;/a&gt; (449.6mm x 690.2mm x 83mm) is mounted to a pole as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/iFRmDl.jpg&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Post&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.omnimountpro.com/Product/Floor_Stands_and_Carts/FS/&quot; rel=&quot;nofollow&quot;&gt;pole&lt;/a&gt; is 10cm in diameter and approximately 1mm thick. The pole is made from &lt;a href=&quot;https://omnimount.box.net/shared/static/v4hvzsg7r1.pdf&quot; rel=&quot;nofollow&quot;&gt;two stainless steel posts&lt;/a&gt;, which attach together at the black plastic ring (which serves to hold up the mounting bracket that wraps around the top post and hide the seam between the posts).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Arm&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.premiermounts.com/product.asp?ProductID=6913&amp;amp;PartID=1775&quot; rel=&quot;nofollow&quot;&gt;articulated arm&lt;/a&gt; (designed for a wall) looks as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/MKrD2l.jpg&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.premiermounts.com/PartData/Docs/LPSA1532_TechSpec.pdf&quot; rel=&quot;nofollow&quot;&gt;arm&lt;/a&gt; weighs approximately 4.1kg and has a maximum extent of 41.1cm.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Problem&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ultimate_tensile_strength#Typical_tensile_strengths&quot; rel=&quot;nofollow&quot;&gt;Stainless steel&lt;/a&gt; has a typical yield strength of 520MPa and ultimate strength of 860MPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An engineer from the pole's manufacturer said that the pole will not support the weight without necking. This conclusion was reached before the engineer knew any of the weights or measurements involved.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Attachment&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The objective is to attach the articulated arm to the bracket attached to the pole. This will either be accomplished using four bolts on the plate, or having the two plates welded together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture of the arm does not show the plate required for wall mounting. It is this plate that will be attached (welded or bolted) to the pole's bracket.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Questions&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is the post strong enough to support the monitor at the maximum extent of the articulated arm without buckling, necking, breaking, or otherwise becoming structurally unsound?&lt;/li&gt;&#xA;&lt;li&gt;How would you perform the calculation?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="2421" LastEditorUserId="2421" LastEditorDisplayName="" LastEditDate="2011-03-09T07:08:21.207" LastActivityDate="2011-03-10T09:51:43.160" Title="Calculate stainless steel pole necking limit" Tags="&lt;classical-mechanics&gt;&lt;mathematical-physics&gt;&lt;forces&gt;&lt;equilibrium&gt;&lt;mass-balance&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="6504" PostTypeId="1" AcceptedAnswerId="6506" CreationDate="2011-03-07T23:32:53.653" Score="2" ViewCount="162" Body="&lt;p&gt;I'm having some problems in understanding the principles of sound propagation. The wave propagates though air (for example) exerts compression, which is followed by rarefaction. I think I got than right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I don't understand: What happens with the particles when the energy passes? Do they oscillate back and forth &quot;in the place&quot; (a pendulum)? Is the oscillation of particles, as shown by some sinusoid, that &quot;back and forth&quot; particle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, does that mean that the whole medium (here air) &quot;vibrates&quot; in some of its volume? Or visually explained: that a sound wave creates a vibrating &quot;tunnel&quot; of particles in air? &lt;/p&gt;&#xA;" OwnerUserId="2422" LastEditorUserId="2422" LastEditorDisplayName="" LastEditDate="2011-03-07T23:47:33.260" LastActivityDate="2011-03-08T00:31:16.113" Title="Sound - what happens with the particles when a wave passes" Tags="&lt;sound&gt;&lt;air&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6505" PostTypeId="1" AcceptedAnswerId="6517" CreationDate="2011-03-08T00:29:50.450" Score="8" ViewCount="288" Body="&lt;p&gt;I'm convinced that water waves for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/4eLQT.jpg&quot; alt=&quot;Water wave jpg&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are a combination of longitudinal and transverse. Any references or proofs of this or otherwise?&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-09T15:04:02.987" Title="Are water waves (i.e. on the surface of the ocean) longitudinal or transverse?" Tags="&lt;classical-mechanics&gt;&lt;waves&gt;&lt;water&gt;" AnswerCount="4" CommentCount="9" />
  <row Id="6506" PostTypeId="2" ParentId="6504" CreationDate="2011-03-08T00:31:16.113" Score="6" ViewCount="" Body="&lt;p&gt;Yes, the particles are just moving back and forth, not moving in any particular direction &lt;strong&gt;on average&lt;/strong&gt;. Here's an animation to make this more clear: &lt;a href=&quot;http://www.acoustics.org/press/151st/Lindwall.html&quot; rel=&quot;nofollow&quot;&gt;http://www.acoustics.org/press/151st/Lindwall.html&lt;/a&gt; (first one on the page)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are at least two weird, unphysical things about that animation. For one, the particles are only moving in response to the sound wave, in perfect synchronization, and they aren't also moving constantly in random directions. If you had some material that would remain a gas at absolute zero, then a sound wave in that absolute-zero gas would look more or less like the animation. But for an everyday-volume sound wave in room-temperature air, the random thermal motion (which is always happening whether or not there is a sound wave) is much stronger than the motion caused by the sound wave. The only reason we notice a sound wave at all is because it is an ordered motion that carries energy in a particular direction. If you followed the motion of a single air molecule, it would look entirely random and there would be no trace of the sound wave. The sound wave only becomes apparent when you look at the large-scale pattern of density variations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other weird thing about the animation is that the molecules stop moving and turn around without colliding with anything! It should be obvious that in a real gas made of electrically neutral particles, there is no long-distance force that would cause this to happen, and a moving particle would not change direction unless it actually collided with another particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your particular questions, yes, the whole medium is &quot;vibrating&quot; in that the density and pressure are increasing and decreasing periodically, as the gas flows back and forth. However, I would not refer to it as a &quot;tunnel&quot;, because, as you can see from the animation, there doesn't have to be any well-defined cross-sectional shape for the sound wave. In fact, the simplest geometry to consider is a plane wave, which extends infinitely in all directions perpendicular to the direction of propagation. It's actually impossible to make a &quot;beam&quot; of sound that will propagate forever without spreading out.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-03-08T00:31:16.113" CommentCount="1" />
  <row Id="6507" PostTypeId="2" ParentId="6488" CreationDate="2011-03-08T01:31:17.617" Score="5" ViewCount="" Body="&lt;p&gt;Take the right pedal as an example.  It uses a right-hand thread, so turning the pedal spindle clockwise (CW) relative to the crank will screw the spindle in, counter-clockwise (CCW) will unscrew it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you put the bike in a repair stand, grab the right pedal and gently simulate the motion of someone riding the bike (always keeping the pedal platform horizontal as if your foot were there).  The body of the pedal will turn CCW relative to the crank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively you would think this might help unscrew the pedal!  But in reality the clamping torque of the threads will be far, far greater than any friction in the bearings could generate.  Even if the bearings were to seize, it would be very difficult to unscrew the pedal (unless it was never tight to begin with).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider someone riding the bike, putting weight on the right pedal.  This applies a force perpendicular to the ground, no matter where the foot is in the pedal stroke.  Relative to the crank arm, this radial force rotates CCW, which creates a CW torque on the pedal spindle (thus tightening it).  The rotations on the left side are all reversed, so it must use the opposite threading to prevent the pedal from coming loose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also see &lt;a href=&quot;http://en.wikipedia.org/wiki/Precession_%28mechanical%29&quot; rel=&quot;nofollow&quot;&gt;&quot;Precession&quot; on Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2420" LastActivityDate="2011-03-08T01:31:17.617" CommentCount="1" />
  <row Id="6508" PostTypeId="2" ParentId="3311" CreationDate="2011-03-08T01:38:49.793" Score="-1" ViewCount="" Body="&lt;p&gt;We assume that under extreme gravitational pressure the star must collapse into a BH because there is no known physical mechanism that can get rid of the star matter too much.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretical models of GRBs, and associated SNe, are struggling against the observations (&lt;a href=&quot;http://arxiv.org/abs/0911.0349&quot; rel=&quot;nofollow&quot;&gt;Gamma Ray Bursts: back to the blackboard&lt;/a&gt; by Maxim Lyutikova).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have for granted that the BHs are an integral part of GR. However, GR survives even if they do not exist. The link referenced in the question make me wonder: &lt;strong&gt;Is it the case that BHs may not exist?&lt;/strong&gt;&lt;br&gt;&#xA;My answer is yes, provided that the energy of the particles becomes radiation (EM X-rays, gamma-rays, ..) before the star collapses. $E=mc^{2}$ allows it but the SM does not. I'm optimistic and I have to me that the LHC will teach us a lot. We must be humble and pursue other routes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue of &lt;strong&gt;singularities&lt;/strong&gt; in space/time is, imo, very difficult to accept &lt;a href=&quot;http://plato.stanford.edu/entries/spacetime-singularities/&quot; rel=&quot;nofollow&quot;&gt;philosophically and physically&lt;/a&gt; and I feel better with the hope that they do not exist.&#xA;The future of physics may be different than we expected.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mankind has the hopes placed in physics because it is a theme central to our collective future. Thankfully, the GUT is not there just around the corner and that's good for the physicists of the future. ;-)&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-03-08T01:38:49.793" CommentCount="8" />
  <row Id="6509" PostTypeId="2" ParentId="6505" CreationDate="2011-03-08T01:41:14.433" Score="7" ViewCount="" Body="&lt;p&gt;In deep waters, the fluid particles describe circles when a wave passes by. So, in a sense, these waves are neither transverse nor longitidinal. For a demonstration, see for example &lt;a href=&quot;http://www.people.fas.harvard.edu/~hgeorgi/new.htm&quot; rel=&quot;nofollow&quot;&gt;Howard Georgi's book&lt;/a&gt; (chapter 11).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In very shallow waters the particles go essentially back and forth. In the intermediate cases they follow eliptical trajectories. &lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="1573" LastEditorDisplayName="" LastEditDate="2011-03-08T02:02:14.067" LastActivityDate="2011-03-08T02:02:14.067" CommentCount="2" />
  <row Id="6510" PostTypeId="2" ParentId="6505" CreationDate="2011-03-08T01:46:46.820" Score="-3" ViewCount="" Body="&lt;p&gt;I will just say what I think I know:&lt;br&gt;&#xA;In the open ocean or great lakes the waves are transverse: the water goes up and down. They are originated by the winds in the surface.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Near the shore the waves become also longitudinal: the small distance from the the surface to the bottom of ocean make the difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;added:&lt;/strong&gt;&lt;br&gt;&#xA;I described the &lt;strong&gt;net transport of energy along the direction of propagation of the waves&lt;/strong&gt;. &#xA;In deep waters it is the vertical motion of the surface that is able to do work (on devices that extend more than a wavelength) against the gravitational field. &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-03-09T15:04:02.987" LastActivityDate="2011-03-09T15:04:02.987" CommentCount="3" />
  <row Id="6511" PostTypeId="2" ParentId="6497" CreationDate="2011-03-08T02:23:48.100" Score="0" ViewCount="" Body="&lt;p&gt;A recent &lt;strong&gt;review&lt;/strong&gt; on the theme &lt;a href=&quot;http://www.mso.anu.edu.au/~charley/papers/LineweaverEgan2008v2.pdf&quot; rel=&quot;nofollow&quot;&gt;Life, gravity and the second law of thermodynamics&lt;/a&gt; is the official position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;;-)&lt;br&gt;&#xA;What do you think about the evolution of this scenario: &#xA;A universe completely homogeneous with T=0, with gravitation and collisionless, all particles are distant one from the others, and by some misfortune one particle move away from his position. Can you imagine what can be the evolution of T? &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-03-08T02:23:48.100" />
  <row Id="6512" PostTypeId="2" ParentId="6502" CreationDate="2011-03-08T05:26:13.093" Score="2" ViewCount="" Body="&lt;p&gt;Yes, a $c=1$ boson at the self-dual radius is exactly equivalent to the SU(2) WZW model at $k=1$. The latter makes the $SU(2)$ symmetry manifest. When one gets used to this and similar equivalences and to the extra marginal momentum/winding operators in the boson description, the $SU(2)$ symmetry becomes &quot;manifest&quot; in both formalisms. There is always some degree of psychology or subjective judgement in what is &quot;manifest&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to be sure about all the $c=1$ CFTs, pages 261-262 of Joe Polchinski's book, volume I, may be helpful. They approximately look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/DzAmp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Special message for Joe: if you wonder why I use an electronic version of the book, it's because Nima Arkani-Hamed borrowed and lost my paper edition. Well, it was actually Volume II that disappeared and I still own this Volume I, but let's ignore those details. ;-)&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-08T06:44:24.760" LastActivityDate="2011-03-08T06:44:24.760" CommentCount="6" />
  <row Id="6513" PostTypeId="2" ParentId="6505" CreationDate="2011-03-08T10:22:05.930" Score="2" ViewCount="" Body="&lt;p&gt;The ocean waves are usually called &quot;surface&quot; waves. Whatever a particle trajectory is, the deeper in water, the smaller its amplitude. Several lengths below surface the water is still.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However deep inside there may be volume waves - from submarines, for example. They are detectable.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-03-08T10:22:05.930" CommentCount="1" />
  <row Id="6514" PostTypeId="2" ParentId="6084" CreationDate="2011-03-08T10:30:29.900" Score="3" ViewCount="" Body="&lt;p&gt;Dear Anirbit, this is a very large number of explicit and implicit, somewhat elementary questions embedded into a copy of a rather technical derivation that doesn't seem to be important for your real &quot;primary&quot; questions at all. It's hard to see what your questions really are because you seem to be hiding them in the convoluted formalism. But I see the following ones:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The particular form of gamma matrices is crucial, isn't it?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, the particular form of gamma matrices is never crucial for physics. Of course, if you want to agree with particular values of components or matrix entries that someone else has derived, you need to use the same convention for the gamma matrices (and other things). But if you want to derive a physical insight, you may choose any basis of the gamma matrices. The derivations will be related by a trivial conjugation of matrices, or sign flips if one chooses different conventions for the metric etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, this is really not a question about representations of 3D supersymmetry algebras. It is a basic question about the difference between physics and social conventions. My answer would be identical in any other context where we talk about conventions. Everyone who has heard or read at least 10 minutes of a lecture about gamma matrices must know that there are different ways to represent them and all of them may be used to do physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How one derives the field content of a supermultiplet?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One takes a state of the multiplet and acts with it by raising and/or lowering supersymmetry generators - those that increase or decrease some $U(1)$ charges by $\pm 1/2$. Some of them are annihilated by the supersymmetry generators. One obtains the spins of all the states in the multiplet. Then one may also attempt to guess Lorentz-covariant fields whose physical polarizations match the desired spectrum of physical particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $d=2+1$, the vector multiplet of $N=2$ algebra - which has 4 supercharges just like $N=1$ in $d=3+1$ - is just the dimensional reduction of the $N=1$ $d=4$ multiplet into 3 dimensions. So the vector multiplet becomes one vector and one scalar - one of the transverse component becomes a scalar - while the Majorana spinor becomes a pair of spinors in 3 dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chiral multiplet in $d=4$ $N=1$ gets directly reduced to three dimensions. In four dimensions, one has a complex scalar and a Majorana spinor. In three dimensions, this gets reorganized as two real scalars and two real two-component spinors - therefore the index $k$ of those fields at the beginning of your text.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How one decomposes multiplets under dimensional reduction?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the supersymmetries themselves transform as spinors, so one decomposes spinors e.g. of $SO(2,1)$ under the $SO(1,1)$ subgroup. If one removes a dimension from spacetime, the higher-dimensional spinor always becomes the lower-dimensional spinor or their pair. Similarly, for a decomposition of a spinor of $SO(d)$ under a maximal $SO(d_1)\times SO(d_2)$ group, a spinor of the former becomes a tensor product of the spinors of the latter smaller groups, or the direct sum of two such tensor products (one has to do a simple exercise concerning the chirality and/or reality of the spinors).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's hard to understand from your text whether you are familiar with these basic facts about spinors or not i.e. whether you want these things to be explained or whether it is just a waste of time because you know them, or whether you are familiar with the very fact that most of these &quot;decomposition&quot; and &quot;dimensional reduction&quot; questions are about decompositions of representations of Lie groups and supergroups under their subgroups. If you're not, you shouldn't have started with obscure technical problems such as $N=2$ vector multiplets in three dimensions; you should have asked what is the procedure to dimensionally reduce a theory. But one should focus at such a question - one question at a time. It just becomes extremely confusing if a physics SE question is composed out of 10 such things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scalars - singlets - remain scalars - singlets. Vectors are decomposed to vectors plus the singlets coming from the reduced directions of the space. When one knows that, it's easy to determine what names you used for those fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why Chern-Simons theory&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the Chern-Simons action may be written using a 3D gauge field and it is gauge-invariant and Lorentz-invariant. However, that's not the only action that one can write down for a gauge field. The Yang-Mills action may also be supersymmetrized. So it's not true that the $N=2$ SUSY implies that the gauge field has to be the Chern-Simons one (which has no physical polarizations since the theory is topological). But the source you are reading simply did study this theory, so it made whatever assumptions are necessary to pinpoint the particular Lagrangian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hermitian conjugation of gamma matrices&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some point, you seem to be puzzled by the relation that you incorrectly claim to &quot;only hold for 3D matrices&quot; that the Hermitian conjugation is equivalent to (minus) the conjugation by the temporal gamma matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This actually holds in any dimension and any (natural!) convention in which the gamma matrices are chosen Hermitian or anti-Hermitian for temporal and spatial directions, respectively. (This choice is needed because the individual matrices must square to plus identity or minus identity matrix, depending on their being spacelike or timelike, because it follows from the anticommutator relations of gamma matrices and the signature of the metric.) It's not hard to see why. The anti-Hermitean ones change the sign by the conjugation, which are exactly those that anti-commute with $\gamma_0$: note that $\gamma_0$ commutes with itself but anticommutes with others, so the conjugation by $\gamma_0$ changes the sign of the &quot;others&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, you shouldn't study obscure supermultiplets in obscure dimensions if you don't understand what gamma matrices do under the Hermitian conjugation. This fact is a basic component of any quantum field theory course. In fact, it is normally discussed in the context of the (single-particle) Dirac equation even &lt;em&gt;before&lt;/em&gt; the students start with quantum field theory. The usual pedagogical treatment may have a $d=4$ bias, for obvious reasons, but it's true that gamma matrices only become worth their name if they can be used above $d=2$ or $d=3$, too: knowing gamma matrices in $d=2$ or $d=3$ only means not to know gamma matrices at all; $d=4$ is already complicated enough so that one could guess how it generalizes to any dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must have missed this whole thing in some way. Again, I recommend you some basic course on quantum field theory because it covers all those questions or, less naturally, an equivalent mathematical introduction to representations of Lie groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The variation of the fermions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other &quot;mysterious&quot; formula for the variation of the fermionic bilinears is just the Leibniz rule for the derivative of a product (now, I would love to believe that you know how to differentiate a product, but this question of yours - in which you didn't do a single step to reduce the question to a more elementary one, i.e. in which you didn't really apply even the Leibniz rule - reduced my certainty even about this point), combined with the fact that the spinors in 3 dimensions are real so that the Dirac conjugation effectively contains &quot;transposition&quot; only and the two terms from the Leibniz rule are equal, producing the factor of two. Again, it's not a big deal; it's just a symmetry of the inner product of two real vectors (spinors) in this case. The only thing one could worry about is the relative sign but because $\bar\lambda \lambda$ doesn't vanish for a fermionic $\lambda$, because they're contracted with an antisymmetric $\epsilon_{\alpha\beta}$ of $SL(2,R)=Spin(2,1)$, it follows that its variation can't be zero, either. So the relative sign has to be plus and the first term from the Leibniz rule has to be doubled rather than canceled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very hard to answer those questions meaningfully because it's a stream of relatively clearly irrelevant technicalities that have nothing to do with the &quot;primary&quot; things you seem to be confused by; combined with your opinions and guesses that are almost always wrong (like guesses that one is not allowed to use different bases); and typographic errors that make a technical discussion impossible (for example, there is no &quot;backslash 1&quot; or &quot;backslash 2&quot; in TeX).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-08T11:47:39.777" LastActivityDate="2011-03-08T11:47:39.777" CommentCount="4" />
  <row Id="6516" PostTypeId="2" ParentId="6455" CreationDate="2011-03-08T12:10:14.937" Score="2" ViewCount="" Body="&lt;p&gt;Dear Anirbit, the field content of a vector multiplet of $N=3$ in $d=3$, if it exists, is obvious. There's one transverse polarization of the gauge boson in $d=3$. The 3 supersymmetries may be used to get 3 different minimal spinorial fermions - Majorana fermions in $d=3$ - out of the vector boson's polarization. That's the $\lambda$ fields. However, in SUSY theories, the multiplets have dimensions that are powers of two. So far we have 3 fermionic fields, we clearly need 4, so one more Majorana spinor, $\chi$, has to be included as well. Now, SUSY has to match bosons and fermions so we must also add 3 extra bosons to the gauge field - they must be spinless e.g. scalars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship of this field content to the Lagrangian you wrote below it is that the Lagrangian you wrote is the Lagrangian that uses the fields you used above, and produces the physical states enumerated in the previous paragraph; a simple check of all the bilinear terms in the Lagrangian is enough to see its free spectrum. The kinetic terms have been supplemented with the interactions with the gauge field to keep the Lagrangian gauge-invariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The starting point in constructing the theory is to decide about its particle spectrum - see the first paragraph. Then one constructs the Lagrangian that produces the required physical states - it's essentially a simple exercise in free field theory. Then one constructs the interacting Lagrangian with all the possible or renormalizable interactions that preserve the pre-determined symmetries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also ask how one &quot;argues&quot; that the Lagrangian is $N=3$ supersymmetric. When, one &quot;argues&quot; in such a way that he must calculate and prove that it is. It means that one needs to find the appropriate SUSY variations of the elementary fields and prove that the action is invariant under those transformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A theory with $N=2$ SUSY, when dimensionally reduced to a lower dimension where the minimal spinor is smaller, automatically has $N=4$ SUSY in the lower-dimensional notation. This simply follows from the spinor's decomposition into two spinors under the smaller group. In your first Lagrangian, $N=3$ SUSY was pretty manifest because the 3-valued index was added to the superpartners of the gauge field. However, such theories often have an automatic higher degree of supersymmetry, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest proof to show the $N=4$ SUSY is to prove the $N=2$ SUSY of the higher-dimensional &quot;mother&quot; theory. You may also check the symmetry under the fourth supersymmetry in $d=3$ explicitly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, I think that your &quot;primary&quot; questions have nothing to do with $N=3$ SUSY in $d=3$ or any of these obscure topics. You shouldn't study these obscure topics before you learn basics of physics and field theory - such as what it means for a theory to have a certain symmetry; certain spectrum; what is the relationship between the Lagrangian and the free spectrum; how one chooses a notation to describe a particular set of fields in particular representations; what it means for objects to transform as representations of symmetries; what happens with the representations if we only look at or preserve a smaller symmetry (subgroup), and so on. As far as I can say, you must be misunderstanding pretty much all these elementary questions which simply makes it inappropriate if not impossible to study vector multiplets in $d=3$, $N=3$ SUSY.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-08T12:16:38.240" LastActivityDate="2011-03-08T12:16:38.240" CommentCount="2" />
  <row Id="6517" PostTypeId="2" ParentId="6505" CreationDate="2011-03-08T12:18:03.460" Score="11" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/U0sU8.gif&quot; alt=&quot;Anim&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each point is moving according to:&lt;br&gt;&#xA;$x(t) = x_0 + a e^{-y_0/l} \cos(k x_0+\omega t)$&lt;br&gt;&#xA;$y(t) = y_0 + a e^{-y_0/l} \sin(k x_0+\omega t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $x_0,y_0$ -- &quot;motion centre&quot; for each particle, $a$ -- the amplitude, $l$ -- decay length with depth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have exact &quot;circular&quot; superposition of longitudinal and transverse waves.&lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2011-03-08T15:10:01.607" LastActivityDate="2011-03-08T15:10:01.607" CommentCount="1" />
  <row Id="6518" PostTypeId="1" AcceptedAnswerId="6519" CreationDate="2011-03-08T13:49:23.813" Score="5" ViewCount="319" Body="&lt;p&gt;I was taught that for the Dirac-equation to &quot;work&quot;, you need matrices of the following form:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$Tr(\alpha^i) = 0$.&lt;/li&gt;&#xA;&lt;li&gt;Eigenvalues +1 or -1&lt;/li&gt;&#xA;&lt;li&gt;2 previous points together: equal number of negative and positive eigenvalues, thus even dimension.&lt;/li&gt;&#xA;&lt;li&gt;$\alpha^i$ and $\beta$ meet the anti-commutator relation.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My Advanced QM professor told me convincingly that there are only 3 linearly independent 2D matrices, the Pauli matrices, which can be used in 2D. That's one short, so you'll need at least 4D matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my QFT professor, who excells at confusing me (and not only me, mind you), stated that it is perfectly possible to use only 2D matrices, by extending the Pauli matrix set with another one. I think he's wrong, but he seems too old (and &quot;wise&quot;) to be wrong, and I didn't remember all conditions above at the time, so I didn't want to argue without arguments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to 2D-ify the Dirac equation? (and not only in a 2D system like graphene, but in general)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="317" LastEditorUserId="317" LastEditorDisplayName="" LastEditDate="2011-03-08T14:58:18.457" LastActivityDate="2011-03-09T20:56:22.177" Title="Why are four-vectors needed in the Dirac equation, when there are 4 linearly independent 2D matrices?" Tags="&lt;quantum-mechanics&gt;&lt;spin&gt;&lt;relativity&gt;&lt;dirac-equation&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="6519" PostTypeId="2" ParentId="6518" CreationDate="2011-03-08T13:58:49.633" Score="7" ViewCount="" Body="&lt;p&gt;Dear rubenb, yes, what your professor says is surely based on solid maths. The reason is that the 4-component Dirac spinor is actually composed of two separate 2-component pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elementary &quot;spinors&quot; for 3+1 dimensions have two complex components. That results from the isomorphism between groups&#xA;$$SL(2,C) \sim Spin (3,1).$$&#xA;Note that both groups have 6 real generators. In particular, in a properly chosen basis, the $\alpha^i$ and $\beta$ matrices may be brought into a block-diagonal form with $2\times 2$ blocks. It follows that the $2\times 2$ blocks themselves satisfy the same algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the four matrices may be written as&#xA;$$(\beta, \alpha^i) = (1_{2\times 2}, \sigma^i)\equiv \sigma^\mu$$&#xA;i.e. as the Pauli matrices supplemented with the identity matrix. Note that the $\alpha^i$ i.e. $\sigma^i$ matrices anticommute with each other while they commute with $\beta$ i.e. $\sigma^0$ and all the matrices square to the identity much like $\beta$, $\alpha^i$ do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The isomorphism above may be viewed as a &quot;noncompact extension&quot; of the usual isomorphism&#xA;$$SU(2) \sim Spin(3).$$&#xA;Note that the group $SU(2)$ is a subgroup of $SL(2,C)$ - it's the same pair as $Spin(3)$ which is a subgroup of $Spin(3,1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two-component spinors are directly relevant for the description of the neutrinos. They only describe a left-handed massless particle (and right-handed massless antiparticle). That's different from the 4-component Dirac spinor that describes a particle that can either left-handed or right-handed. The neutrino is given by a Weyl spinor and the free equation is simply&#xA;$$\sigma^\mu \partial_\mu \chi = 0$$&#xA;which is Lorentz-covariant. However, one must realize that the 4-vector of $2\times 2$ matrices, $\sigma^\mu$, don't transform the 2-dimensional complex space (left-handed Weyl spinors) onto itself but onto another 2-dimensional complex space (of right-handed Weyl spinors) which is the complex conjugate of the first one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Massive charged particles such as the electron require a 4-component spinor - i.e. a pair of two 2-component spinors - but for neutrinos, the minimum amount to describe a single particle is given by one 2-component spinor.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-08T14:10:23.430" LastActivityDate="2011-03-08T14:10:23.430" CommentCount="7" />
  <row Id="6520" PostTypeId="2" ParentId="6518" CreationDate="2011-03-08T16:12:11.213" Score="3" ViewCount="" Body="&lt;p&gt;Luboš' Answer is just fine, and he's adopted a level of presentation that looks just right for you, but I prefer to put this somewhat differently, using the manifestly Lorentz covariant $\gamma^\mu$ operators instead of using the operators $(\beta,\alpha^i)$. For the massive Dirac equation, you need four objects that satisfy the relationships $\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=2g^{\mu\nu}$ to make the presentation manifestly covariant. Using these, you can construct 6 linearly independent objects of the form $\gamma^0\gamma^1$, 4 linearly independent objects of the form $\gamma^0\gamma^1\gamma^2$, and just one linearly independent object of the form $\gamma^0\gamma^1\gamma^2\gamma^3$, 15 in all, which together with the identity gives the same number of dimensions as the algebra of 4x4 matrices. Indeed we can prove that the algebra generated by the $\gamma^\mu$ over the complex field with the relationships above is isomorphic to the algebra of 4x4 complex matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get to 2x2 matrices without breaking Lorentz invariance (almost), we use the object $\gamma^0\gamma^1\gamma^2\gamma^3$, which is invariant under Lorentz transformations &lt;em&gt;except reflections&lt;/em&gt; and which commutes with an 8-dimensional part of the algebra and anti-commutes with a distinct 8-dimensional part of the algebra. We can use this to construct an almost Lorentz-invariant projection $\frac{1}{2}(1+i\gamma^0\gamma^1\gamma^2\gamma^3)$. This projection splits up the 4-dimensional vectors that a 4x4 matrix acts on into two 2-dimensional parts, with eigenvalues 1 and 0. In the chiral representation, this projection operator is $\left(\array{I\ 0\\ 0\ 0}\right)$, where $I$ is a 2x2 identity matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is only a tiny piece of what there is to pick up about Dirac matrices. You will pick up lots of pieces along the way. Some people prefer to work with matrices, some people prefer to work in the slightly more abstract way I've used above, others (mathematicians!) prefer still higher levels of abstraction than you see above. The above is not entirely explicit, but it's much better to work through these things for yourself as far as possible.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-03-08T16:12:11.213" CommentCount="2" />
  <row Id="6521" PostTypeId="2" ParentId="4345" CreationDate="2011-03-08T16:13:53.270" Score="0" ViewCount="" Body="&lt;p&gt;Are we talking about &quot;entanglement&quot; in the sense of one particle controlling the other, or simply about our knowledge of the particles?  As I understand it, one particle must have left spin and the other right spin by conservation of angular momentum; human minds know this.  Then if one particle is observed to have left spin, human minds know that the other one must have right spin.  There is no communication or interaction between the particles themselves.  The connection is made in the human mind, in the understanding of the situation.  Can we say it is the same as counting a bunch of apples?  The bunch is not altered by the counting; only our knowledge of it is altered. &lt;/p&gt;&#xA;" OwnerUserId="2432" LastActivityDate="2011-03-08T16:13:53.270" CommentCount="1" />
  <row Id="6522" PostTypeId="2" ParentId="6503" CreationDate="2011-03-08T16:55:45.137" Score="0" ViewCount="" Body="&lt;p&gt;I used to make robotic hands and the main problem always was the torque generated by the end load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of this arm you have an extremely heavy TV, which increases the force on the poor little column the farther away it is. Since, torque is proportional to the radius. For a small radius the bolts will tolerate it quite well, but there will be a point where the torque exceeds the manufacturers limits of the force per unit cross sectional area (i.e. pressure) of the pole that will lead to the necking...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can actually find this out by using the shear modulus and calculating the amount of weight $m_{\text{TV}}g$ times the perpendicular distance $r$ of the pole to the pole. Using the radius of the bolts you will get something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{Strain} = \frac{\text{stress}}{\text{Shear Modulus for the material}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\delta l}{L} = \frac{\frac{F}{A}}{\text{Shear Modulus}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since, you already know the max pressure you don't have to worry about anything. Just use,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r = \frac{\text{The manufacturer&amp;#39;s pressure rating}}{m_{\text{TV}} g}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result would be the maximum radius possible for your TV without buckling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That aside, what I would do to prevent necking is either get a thicker pole (too expensive) or get another pole and bind it with the first one (just make sure that the restraints for the arm reach around them both. &lt;em&gt;This&lt;/em&gt; is crucial) and then ground it firmly with the first one on a broad base (making sure the entire thing doesn't tip over either).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will work because you've essentially increased the crossectional area of the poll. Kinda like the old stories about how sticks that stick together are harder to break. If this still doesn't work then add a 2 more poles and interlock them with restraints on a broad base. Do note that if I were you I would give em a nice weld and turn this into an awesome home DIY project. So, take everything with a grain of salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that should solve your problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take care,&lt;/p&gt;&#xA;" OwnerUserId="1065" LastEditorUserId="61" LastEditorDisplayName="" LastEditDate="2011-03-08T18:46:55.407" LastActivityDate="2011-03-08T18:46:55.407" CommentCount="9" />
  <row Id="6523" PostTypeId="1" CreationDate="2011-03-08T17:58:15.980" Score="3" ViewCount="149" Body="&lt;p&gt;M-theory wrapped around $S^1$ is type IIA, especially in the limit of zero radius. An M2-brane wrapped around $S^1$ once is a type IIA string. The zero mode anyway. In the limit of very small radius, if we project to states with tiny energy, we get a collection of type IIA strings only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does an M2-brane wrapped around the circle twice correspond to? Its energy is also low. It has to be. Pretend it is an M-theory compactification with twice the radius. Orbifold it over half a periodic translation. It's back to the original. So, is it made up of type IIA strings? Is it two type IIA strings? Or a single type IIA string? Surely it's not a &lt;em&gt;different&lt;/em&gt; kind of type IIA string with $1/\sqrt{2}$ times the string scale and $\sqrt{8}$ times the string coupling?&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2011-03-08T18:46:18.123" Title="M2-brane wrapped twice around $S^1$" Tags="&lt;string-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6524" PostTypeId="1" AcceptedAnswerId="6526" CreationDate="2011-03-08T18:35:16.247" Score="3" ViewCount="187" Body="&lt;p&gt;I am developing an application that uses the magnetometer inside smart-phones to detect orientation w.r.t. the Earth's magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have noticed that when the phone is held close to a metal structure, such as Iron/Steel railings, there is a huge deflection in the magnetometer reading. There is no externally-induced current flowing through the structure; it has &lt;strike&gt;induced&lt;/strike&gt; wrapped the magnetic field by its own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, which types of metals can show this effect? Or more generally which type of elements can show this effect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose Ferromagnetic materials do show this effect. And paramagnetic materials (like Aluminium) will show a very weak effect. What about diamagnetic substances such as copper?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&#xA;Changed the title to mention &lt;em&gt;passively&lt;/em&gt;. I understand that a copper coil forced to carry current would induce a magnetic field that interfers with the Earth's magnetic field. But does the coil distort the Earth's magnetic field when &lt;em&gt;no&lt;/em&gt; EMF is applied?&lt;/p&gt;&#xA;" OwnerUserId="884" LastEditorUserId="884" LastEditorDisplayName="" LastEditDate="2011-03-09T04:57:07.657" LastActivityDate="2011-03-09T04:57:07.657" Title="Which metals can cause magnetic interference (passively)?" Tags="&lt;material-science&gt;&lt;magnetism&gt;" AnswerCount="3" />
  <row Id="6525" PostTypeId="2" ParentId="6523" CreationDate="2011-03-08T18:46:18.123" Score="2" ViewCount="" Body="&lt;p&gt;If this object were the only one - and the original singly wrapped membrane wouldn't exist - then your double wrapped object would really be the fundamental string with twice as high string tension and a different string coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the singly wrapped membrane states also exist in your theory. And they make a lot of difference. In particular, your doubly wrapped membrane - which clearly carries the same winding charge as two ordinary type IIA strings - may decay into the pair of type IIA strings. At least, the charge conservation laws don't prohibit such a decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that your membrane is wrapped on the compact dimension $X^{10}$ and extended in another dimension $X^{9}$. Take a slice for a fixed value of $X^9$. Then the cross section of the membrane looks like a doubly wound string that has been called &quot;state II&quot; in this question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/6420/statistics-and-macrolocality-in-string-theory&quot;&gt;Statistics and macrolocality in string theory&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As discussed in the answer to the question above, this double wound string may self-intersect and a crossing-over interaction may occur at the intersection. This will turn the string into a pair of strings, and the point where this transition occurs may spread in both directions along the $X^9$ axis I neglected, eventually turning the double wrapped membrane into a pair of singly wrapped membranes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some non-minimal values of charges, one must always be careful and check whether a stable state exists, and whether it is a bound state with a nonzero binding energy or a threshold bound state. In this case, I would say it doesn't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A way to see that it doesn't exist is matrix string theory,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/6424/good-introductory-text-for-matrix-string-theory&quot;&gt;Good introductory text for matrix string theory&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the full light-cone gauge model for M-theory on a circle. In the spectrum of this 1+1-dimensional maximally supersymmetric gauge theory, you find the ordinary strings with an arbitrary $P^+$ but you won't find any &quot;doubly wrapped membranes&quot; at all. They would correspond to some block-diagonal matrices that depend on $\sigma$ but those states may be manifestly seen to be equivalent to pairs of ordinary strings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matrix model can't even recognize which strand of the membrane is which. This fact is related to another fact that even in the BFSS matrix model where membranes are expressed by non-commutative geometry, the symmetry under the volume-preserving diffeomorphisms on the M2-brane is actually extended to a full $U(N)$ gauge symmetry that may permute - and mix - arbitrary &quot;cells&quot; of the membrane. So the only physical information is hiding in the approximate positions of the &quot;cells&quot; of the M2-brane and they don't have any other IDs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be emphasized that this conclusion - that no other stable independent states with charges greater than one exist - isn't universal at all. In particular, the threshold bound states of $N$ D0-branes of type IIA string theory &lt;em&gt;do&lt;/em&gt; exist: they're D0-branes with a proportionally higher charge and mass. By T-duality, this is true for any fully toroidally wrapped D-branes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could think that there is a contradiction because by this procedure, one may uncover genuine and independent new threshold bound states of two (or more) D2-branes (wrapped on a two-torus) of type IIA string theory - and D2-branes are the same objects as M2-branes in the corresponding M-theoretical description, aren't they?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this argument is flawed because of one difference: in the type IIA description, the D2-branes correspond to M-theory M2-branes that are localized in the infinitesimal 10th direction of M-theory (that produces type IIA). It's the differences in the location in this tiny new circular direction of space that may distinguish a pair of D2-branes (wrapped on another 2-torus) from a doubly wrapped D2-brane. However, in your example, there's no other &quot;tiny direction&quot; in which the M2-branes are localized, so this extra information doesn't exist. Consequently, the doubly wrapped M2-branes don't produce any new independent objects.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-08T18:46:18.123" />
  <row Id="6526" PostTypeId="2" ParentId="6524" CreationDate="2011-03-08T19:54:19.413" Score="4" ViewCount="" Body="&lt;p&gt;Well, I think that whatever distortion of the magnetic field is caused by paramagnetic materials - such as aluminum - is also caused, with the opposite sign, by diamagnetic materials - such as bismuth (whose effect should be almost exactly opposite to that of aluminum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am convinced you may neglect all those materials. Only paramagnetic materials and various steels and similar metals etc. are relevant. See the relative permeability in this table&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_permeability#Values_for_some_common_materials&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Magnetic_permeability#Values_for_some_common_materials&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and just ignore all the materials whose $\mu/\mu_0$ is very close to one, on either side. The only materials that have the opposite but comparably important effects as ferromagnetic materials are superconductors whose $\mu$ vanishes. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ferromagnets may sustain their field even if the external source disappears: this is probably the biggest source of interference. However, as you point out, even materials without any permanent field could potentially influence the field that you measure by the mobile device.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-08T19:54:19.413" />
  <row Id="6527" PostTypeId="2" ParentId="6524" CreationDate="2011-03-08T20:08:19.193" Score="1" ViewCount="" Body="&lt;p&gt;Maybe not answer but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem should not cause only metals. Any DC or AC current in wires should cause magnetic field. And this magnetic field isn't constant because the power consumption of such device isn't constant. Also any unshielded coil for switcher power supply should cause magnetic interference.&lt;/p&gt;&#xA;" OwnerUserId="2410" LastActivityDate="2011-03-08T20:08:19.193" CommentCount="5" />
  <row Id="6528" PostTypeId="1" AcceptedAnswerId="6529" CreationDate="2011-03-08T20:40:00.247" Score="3" ViewCount="149" Body="&lt;p&gt;You can see Earth in the right most band in this image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/UiTCm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are these bands doing in space?! Can somebody please explain the coloured bands in this image?&lt;/p&gt;&#xA;" OwnerUserId="2391" LastActivityDate="2011-03-09T22:45:14.017" Title="What are the coloured bands in this &quot;Pale Blue Dot&quot; picture?" Tags="&lt;space&gt;&lt;color&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6529" PostTypeId="2" ParentId="6528" CreationDate="2011-03-08T20:52:26.830" Score="3" ViewCount="" Body="&lt;p&gt;From the wikipedia page about the photo: &lt;a href=&quot;http://en.wikipedia.org/wiki/Pale_Blue_Dot&quot; rel=&quot;nofollow&quot;&gt;The Pale Blue Dot&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The light band over Earth is an artifact of sunlight scattering in the camera's optics, resulting from the small angle between the Earth and the Sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="386" LastEditorUserId="2423" LastEditorDisplayName="" LastEditDate="2011-03-09T22:45:14.017" LastActivityDate="2011-03-09T22:45:14.017" />
  <row Id="6530" PostTypeId="1" CreationDate="2011-03-08T20:57:24.657" Score="15" ViewCount="697" Body="&lt;p&gt;Quantum field theory is a broad subject and has the reputation of using methods which are mathematically desiring. For example working with and subtracting infinities or the use of path integrals, which in general have no mathematical meaning (at least not yet) ect. My question is a little vague, but i am interested in hearing what is the status of rigor in QFT. What is known to be mathematically rigorous and consistent, what is known to be not rigorous? Any examples and references are welcome. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added:&lt;/strong&gt; Just to clarify by rigorous I meant anything that a mathematician would find satisfactory. Also my question wasn't for books with rigorous (in some sense) approach, although that was welcomed. It was about specific examples of what is considered mathematically satisfactory and what not. For example the quantization of free fields satisfying the Klein-Gordon equation can be done rigorously. There is no mathematical definition in general of the Feynman path integral and so on.  &lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2011-03-10T21:48:33.920" LastActivityDate="2011-07-13T15:02:11.030" Title="Rigor in quantum field theory" Tags="&lt;quantum-field-theory&gt;&lt;mathematical-physics&gt;" AnswerCount="6" CommentCount="13" FavoriteCount="5" />
  <row Id="6531" PostTypeId="2" ParentId="6530" CreationDate="2011-03-08T21:06:45.563" Score="4" ViewCount="" Body="&lt;p&gt;I'll give a reference that I didn't (yet) managed to finish myself. But it looks really rigorous:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;N. N. Bogoliubov, A. A. Logunov, A. I. Oksak, I. T. Todorov (1990): General Principles of Quantum Field Theory.&lt;/strong&gt;  (ISBN 0-7923-0540-X. ISBN 978-0-7923-0540-8.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's what I call &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nikolay_Bogolyubov&quot; rel=&quot;nofollow&quot;&gt;Bogoliubov&lt;/a&gt; approach&quot; to QFT. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-03-08T21:06:45.563" CommentCount="1" />
  <row Id="6532" PostTypeId="2" ParentId="6524" CreationDate="2011-03-08T21:26:38.953" Score="2" ViewCount="" Body="&lt;p&gt;The main materials that interfere are &lt;em&gt;ferromagnetic&lt;/em&gt; and &lt;em&gt;ferrimagnetic&lt;/em&gt;, especially those based on iron, nickel, cobalt and manganese.&lt;/p&gt;&#xA;" OwnerUserId="1545" LastActivityDate="2011-03-08T21:26:38.953" />
  <row Id="6533" PostTypeId="1" CreationDate="2011-03-08T22:02:33.790" Score="3" ViewCount="155" Body="&lt;p&gt;The dominant method of neutron star cooling is neutrino emission. There are two regimes usually presented, the &quot;direct Urca&quot; and &quot;modified Urca&quot; processes, each of which are sequences of neutron decay and inverse reactions. The direct Urca looks like this:&#xA;$$n\rightarrow p+l+\overline{\nu_l},\quad p + l \rightarrow n + \nu_l$$&#xA;where $l$ is a lepton - either an electron or a muon. These processes cause continuous emission of neutrinos which cools a neutron star relatively quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But below a density of $\rho\approx 10^{15}\mathrm{\,g\,cm^{-3}}$ (about three times the nuclear density) this process is suppressed, which means that the direct Urca process only occurs in the core. This is the reason according to a review of neutron star cooling from Pethick and Yakovlev (2004):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The process can occur only if the&#xA;  proton concentration is sufficiently&#xA;  high. The reason for this is that, in&#xA;  degenerate matter, only particles with&#xA;  energies within ~$k_BT$ of the Fermi&#xA;  surface can participate in reactions,&#xA;  since other processes are blocked by&#xA;  the Pauli exclusion principle. If the&#xA;  proton and electron Fermi momenta are&#xA;  too small compared with the neutron&#xA;  Fermi momenta, the process is&#xA;  forbidden because it is impossible to&#xA;  satisfy conservation of momentum.&#xA;  Under typical conditions one finds&#xA;  that the ratio of the number density&#xA;  of protons to that of nucleons must&#xA;  exceed about 0.1 for the process to be&#xA;  allowed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This makes some sense. But what surprises me is that this process can still work with a slight modification at lower densities. The &lt;em&gt;modified&lt;/em&gt; Urca process can cool the star&#xA;$$n+N\rightarrow p+N+l+\overline{\nu_l},\quad p + N + l \rightarrow n + N + \nu_l$$&#xA;where $N$ is a nucleon - a proton or a neutron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process, I'm told, can work at much lower densities, but produces 7 orders of magnitude less emissivity. As a result, it's the dominant process in the superfluid outer core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why does the additional nucleon permit lower densities? How does an additional neutron or proton get us out of the conservation of momentum problem with the direct Urca process?&lt;/p&gt;&#xA;" OwnerUserId="1655" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-03-11T18:24:47.950" LastActivityDate="2011-04-10T18:40:03.473" Title="What allows the modified Urca process to work at lower density than direct Urca in neutron star cooling?" Tags="&lt;particle-physics&gt;&lt;neutron-stars&gt;" AnswerCount="1" />
  <row Id="6535" PostTypeId="2" ParentId="6530" CreationDate="2011-03-09T00:20:47.257" Score="13" ViewCount="" Body="&lt;p&gt;Your statement &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;working with and subtracting infinities ... which in general have no mathematical meaning&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is not really correct, and seems to have a common misunderstanding in it.  The technical difficulties from QFT do not come from infinities.  In fact, ideas basically equivalent to renormalization and regularization have been used since the beginning of math--see, e.g., many papers by Cauchy, Euler, Riemann, etc. In fact, G.H. Hardy has a book published on the topic of divergent series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821826492&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Divergent-AMS-Chelsea-Publishing-Hardy/dp/0821826492&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is even a whole branch of math called &quot;integration theory&quot; (of which things like Lebesgue integration is a subset) that generalizes these types of issues.  So having infinities show up is not an issue at all, in a sense, they show up out of convenience. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the idea that infinities have anything to do with making QFT axiomatic is not correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real issue, from a more formal point of view, is that you &quot;want&quot; to construct QFTs via some kind of path integral.  But the path integral, formally (i.e., to mathematicians) is an integral (in the general sense that appears in topics like &quot;integration theory&quot;) over a pretty pathological looking infinite dimensional LCSC function space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to define a reasonable measure on an infinite dimensional function space is problematic (and the general properties of these spaces doesn't seem to be particularly well understood).  You run into problems like having all reasonable sets being &quot;too small&quot; to have a measure, worrying about measures of pathological sets, and worrying about what properties your measure should have, worrying if the &quot;$\mathcal{D}\phi$&quot; term is even a measure at all, etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At best, trying to fix this problem, you'd run into an issue like you have in the Lebesgue integral's definition, where it defines the integral and you construct some mathematically interesting properties, but most of its utility is in letting you abuse the Riemann integral in the way you wanted to.  Actually calculating integrals from the definition of the Lebesgue integral is not generally easy.  This isn't really enough to attract the attention of too many physicists, since we already have a definition that works, and knowing all of its formal properties would be nice, and would certainly tell us some surprising things, but it's not clear that it would be all that useful generally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From an algebraic point of view, I believe you run into trouble with trying to define divergent products of operators that depend on renormalization scheme, so you need to have some family of $C^*$-algebras that respects renormalization group flow in the right way, but it doesn't seem like people have tried to do this in a reasonable way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a physics point of view, we don't care about any of this, because we can talk about renormalization, and demand that our answers have &quot;physically reasonable&quot; properties.  You can do this mathematically, too, but the mathematicians are not interested in getting a reasonable answer; what they want is a set of &quot;reasonable axioms&quot; that the reasonable answers follow from, so they're doomed to run into technical difficulties like I mentioned above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally, though, one can define non-interacting QFTs, and quantum mechanical path integrals.  It's probably the case that formally defining a QFT is within the reach of what we could do if we really wanted, but it's just not a compelling topic to the people who understand how renormalization fixes the solutions to physically reasonable ones (physicists), and the formal aspects aren't well-understood enough that it's something one could get the formalism for &quot;for free.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my impression is that neither physicists or mathematicians generally care enough to work together to solve this problem, and it won't be solved until it can be done &quot;for free&quot; as a consequence of understanding other stuff.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should also add briefly that CFTs and SCFTs are mathematically much more carefully defined, and so a reasonable alternative to the classic ideas I mentioned above might be to start with a SCFT, and define a general field theory as some kind of &quot;small&quot; modification of it, done in such a way to keep just the right things well-defined.&lt;/p&gt;&#xA;" OwnerUserId="1403" LastEditorUserId="1403" LastEditorDisplayName="" LastEditDate="2011-03-09T01:43:41.820" LastActivityDate="2011-03-09T01:43:41.820" CommentCount="4" />
  <row Id="6536" PostTypeId="2" ParentId="6530" CreationDate="2011-03-09T01:25:53.973" Score="4" ViewCount="" Body="&lt;p&gt;There are several books that approach QFT (and/or Gauge Theory) from different levels of 'mathematical rigor' (for some definition of &quot;mathematical rigor&quot; — that Moshe would approve ;-).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let me give you some sort of 'preliminary list'… it's by no means complete, and it's in no particular order either, but i think it may pave the way to further work.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540610499&quot; rel=&quot;nofollow&quot;&gt;Local Quantum Physics: Fields, Particles, Algebras&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691070628&quot; rel=&quot;nofollow&quot;&gt;PCT, Spin and Statistics, and All That&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540601422&quot; rel=&quot;nofollow&quot;&gt;Finite Quantum Electrodynamics: The Causal Approach&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3642086357&quot; rel=&quot;nofollow&quot;&gt;Perturbative Quantum Electrodynamics and Axiomatic Field Theory&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521060257&quot; rel=&quot;nofollow&quot;&gt;Quantum Field Theory for Mathematicians&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821847058&quot; rel=&quot;nofollow&quot;&gt;Quantum Field Theory&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521115779&quot; rel=&quot;nofollow&quot;&gt;Mathematical Aspects of Quantum Field Theory&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1107005094&quot; rel=&quot;nofollow&quot;&gt;Quantum Mechanics and Quantum Field Theory: A Mathematical Primer&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540347623&quot; rel=&quot;nofollow&quot;&gt;Quantum Field Theory I: Basics in Mathematics and Physics: A Bridge between Mathematicians and Physicists (v. 1)&lt;/a&gt; and &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540853766&quot; rel=&quot;nofollow&quot;&gt;Quantum Field Theory II: Quantum Electrodynamics: A Bridge between Mathematicians and Physicists&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540769544&quot; rel=&quot;nofollow&quot;&gt;Mathematical Theory of Feynman Path Integrals: An Introduction&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691085463&quot; rel=&quot;nofollow&quot;&gt;Introduction to Algebraic and Constructive Quantum Field Theory&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387905626&quot; rel=&quot;nofollow&quot;&gt;Quantum Physics: A Functional Integral Point of View&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821820125&quot; rel=&quot;nofollow&quot;&gt;Quantum Fields and Strings: A Course for Mathematicians (2 vols)&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821804006&quot; rel=&quot;nofollow&quot;&gt;Geometry and Quantum Field Theory&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0198517734&quot; rel=&quot;nofollow&quot;&gt;Mathematical Theory of Quantum Fields&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In any case… there's much more out there, not only in terms of topics (renormalization, etc) but also in terms of articles, books and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there's plenty of &quot;mathematical rigor&quot; in QFT (and String Theory, for that matter), including different 'levels' of it, that should please and suit varied tastes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: There are other topics here that deal with this topic in a form or another, e.g., &lt;a href=&quot;http://physics.stackexchange.com/questions/3983/haags-theorem-and-practical-qft-computations&quot;&gt;Haag's Theorem and Practical QFT Computations&lt;/a&gt;. So, don't be shy and take a look around. :-)&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2011-03-09T01:25:53.973" />
  <row Id="6537" PostTypeId="1" AcceptedAnswerId="6661" CreationDate="2011-03-09T02:56:51.773" Score="4" ViewCount="120" Body="&lt;p&gt;Suppose we have a rigid token-ring network.  An observer at any node can seemingly determine the angular momentum of the network by measuring the time it takes for a packet to travel around the ring in each of the two directions.  Is it possible by any means for an observer to determine whether the network is knotted?&lt;/p&gt;&#xA;" OwnerUserId="2072" LastEditorUserId="2072" LastEditorDisplayName="" LastEditDate="2011-03-10T23:06:22.943" LastActivityDate="2011-03-11T00:22:57.480" Title="Knotted token-ring network" Tags="&lt;soft-question&gt;&lt;special-relativity&gt;&lt;information&gt;&lt;topological-entropy&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="6538" PostTypeId="2" ParentId="2147" CreationDate="2011-03-09T03:19:20.887" Score="2" ViewCount="" Body="&lt;p&gt;The phase velocity of light in a fully ionized plasma is strictly superluminal; &lt;a href=&quot;http://farside.ph.utexas.edu/teaching/em/lectures/node100.html&quot; rel=&quot;nofollow&quot;&gt;see equation 1156 of Fitzpatrick&lt;/a&gt;. Like all media in which the speed of light is not $c$, plasmas are &lt;a href=&quot;http://en.wikipedia.org/wiki/Dispersion_%28optics%29&quot; rel=&quot;nofollow&quot;&gt;dispersive&lt;/a&gt;, but the phase velocity is still superluminal over a large bandwidth (from the plasma frequency on up).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phase velocity can be tuned by tuning the density of the plasma. My &lt;a href=&quot;http://lasermatter.umd.edu/&quot; rel=&quot;nofollow&quot;&gt;old group&lt;/a&gt; used this denstiy-tuned index of refraction to make transient micro-optics out of plasma.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2011-03-09T03:19:20.887" />
  <row Id="6539" PostTypeId="1" CreationDate="2011-03-09T03:49:59.883" Score="1" ViewCount="158" Body="&lt;p&gt;There’s an interesting &lt;a href=&quot;http://www.scientificamerican.com/article.cfm?id=raizen-entropy-cooling-experiment-interactive&amp;amp;posted=1&quot; rel=&quot;nofollow&quot;&gt;article on Scientific American&lt;/a&gt;&#xA;that tells how to cool individual atoms can bee cooled to within a very tiny fraction of Absolute Zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It uses two laser beams acting on a very cold rarified gas in a magnetic trap.  Initially, all of the atoms are in one state.  The first laser beam (call it red) only affects atoms in a second stable state.  A second laser (call it orange), offset and parallel to the first of a different color is turned on.  This laser switches atoms to the previously mentioned second stable state when it hits an atom.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These excited atoms cannot pass the first laser beam (the excited atoms rebound from this beam).  This leads to a concentration of the excited atoms on the side of the second beam.  All of the atoms will eventually cross the two beams and move into the smaller area, being compressed without raising their temperature (supposedly).  The lasers are then turned off, the gas is allowed to expand (and cool) into its original volume and you end up with a lower temperature gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’ve left the following question there, but haven’t gotten an answer (I keep getting notices that the comments have been added to, but even when I clear my browser cache, or use a different browser, I still don’t see them).  So here is my question.  Can you help me understand?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--Michael&lt;/p&gt;&#xA;" OwnerUserId="2439" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-09T12:19:56.920" LastActivityDate="2011-03-09T22:03:19.887" Title="Maxwell's Demon - laser cooling" Tags="&lt;energy&gt;&lt;photons&gt;&lt;heat&gt;&lt;laser&gt;" AnswerCount="0" CommentCount="11" FavoriteCount="2" />
  <row Id="6540" PostTypeId="2" ParentId="2029" CreationDate="2011-03-09T03:52:50.567" Score="11" ViewCount="" Body="&lt;p&gt;The references in other answers to the Wikipedia pages about &lt;a href=&quot;http://en.wikipedia.org/wiki/Earnshaw%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Earnshaw's theorem&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_levitation&quot; rel=&quot;nofollow&quot;&gt;Magnetic levitation&lt;/a&gt; are right on; you absolutely cannot have stable, static magnetic levitation using only ferromagnets. However, at the time of my posting, &lt;a href=&quot;http://physics.stackexchange.com/questions/2029/can-i-levitate-an-object-without-using-an-electromagnet/2035#2035&quot;&gt;both&lt;/a&gt; other &lt;a href=&quot;http://physics.stackexchange.com/questions/2029/can-i-levitate-an-object-without-using-an-electromagnet/2031#2031&quot;&gt;answers&lt;/a&gt; contain misinformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/2029/can-i-levitate-an-object-without-using-an-electromagnet/2035#2035&quot;&gt;Kakemonsteret's answer&lt;/a&gt; is incorrect*, because Earnshaw's theorem is not the end of the story. If you violate the assumptions of Earnshaw's theorem, its conclusions are no longer valid. There is a special kind of magnet Earnshaw's theorem didn't consider, which you can use for stable, static magnetic levitation: a &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aDiamagnetic_graphite_levitation.jpg&quot; rel=&quot;nofollow&quot;&gt;diamagnet&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Superconductors are strongly diamagnetic, but I suspect using a superconductor is an unsatisfactory solution, unless you have liquid nitrogen on tap. Weaker room-temperature diamagnets like &lt;a href=&quot;http://en.wikipedia.org/wiki/Bismuth&quot; rel=&quot;nofollow&quot;&gt;bismuth&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Pyrolytic_carbon&quot; rel=&quot;nofollow&quot;&gt;pyrolytic carbon&lt;/a&gt; don't support much weight, but they will levitate forever and don't require cooling, electricity, or attention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/2029/can-i-levitate-an-object-without-using-an-electromagnet/2031#2031&quot;&gt;user879's answer&lt;/a&gt; links to the excellent Wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_levitation&quot; rel=&quot;nofollow&quot;&gt;magnetic levitation&lt;/a&gt;. I'd like to emphasize that this page explicitly discusses diamagnetic levitation, which I think is the closest answer to the spirit of your question: a special type of magnet that makes an object levitate without using an electromagnet. However, the answer goes on to state that a bowl of [ferro]magnets would also work. This is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I spent a full summer in high school trying to float a tennis ball covered in fridge magnets above a bowl lined with fridge magnets. I failed; I wasn't violating any of Earnshaw's assumptions. Years later when I finally learned Earnshaw's theorem, I was relieved to realize my failure was predictable, since my goal was &lt;em&gt;provably&lt;/em&gt; impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, buy some strong ferromagnets, buy some strong, light diamagnets, scale down your design's ambitions to something &lt;em&gt;very&lt;/em&gt; light, and start building your decorative item.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't put your credit cards or your hard drives near it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*EDIT: This answer has been updated, and now mentions diamagnets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I can't believe I forgot to mention my favorite aspect of Earnshaw's theorem: it proves there's no such thing as solid objects! This is the single most obvious everyday evidence for quantum mechanics. If you see an apparently static, stable object that isn't held together by gravity, then it can't be modeled as a collection of point charges interacting through Newton's laws and Maxwell's equations.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="2359" LastEditorDisplayName="" LastEditDate="2011-03-15T18:32:11.187" LastActivityDate="2011-03-15T18:32:11.187" CommentCount="5" />
  <row Id="6541" PostTypeId="1" AcceptedAnswerId="6546" CreationDate="2011-03-09T04:00:53.730" Score="6" ViewCount="324" Body="&lt;p&gt;Einstein's general relativity assumes the &lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalence_principle&quot; rel=&quot;nofollow&quot;&gt;equivalence of acceleration and gravitation&lt;/a&gt;. Is there a general class of gravity theories that have this property but disagree with general relativity? Will such theories automatically satisfy any of the tests of general relativity such as the precession of mercury or the bending of light?&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-10T08:55:44.360" LastActivityDate="2011-03-17T16:12:33.567" Title="Gravity theories with the equivalence principle but different from GR" Tags="&lt;general-relativity&gt;&lt;gravity&gt;&lt;equivalence-principle&gt;" AnswerCount="8" CommentCount="1" />
  <row Id="6542" PostTypeId="2" ParentId="6541" CreationDate="2011-03-09T04:22:38.207" Score="9" ViewCount="" Body="&lt;p&gt;The principle of equivalence is built on the observation that motion of matter due to gravity is independent of any of its specific characteristics, it is universal. Contrast that with the electric force, where different particles with different charges will move along different trajectories. Gravity is different because of the equality of the gravitational and inertial masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since gravity affects everything in a universal way, you can declare it to be a property of spacetime instead of being a force. In Einstein's gravity the effect of gravity is summarized in the geometry or more specifically the metric of spacetime. A priori you could imagine that gravity can be manifested in some other universal property of spacetime, but metric is the only choice I know of. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, once you choose the metric to encode the force on test particles due to gravity, I think the principle of equivalence doesn't give any more information. For example, it does not tell you how the metric responds to matter sources - how much spacetime curvature is created by some source of energy-momentum. There could be many answers to this question, all of which would satisfy the principle of equivalence, most of which will differ (potentially dramatically) from GR. For example, if you replace the Einstein-Hilbert action by some more general function of curvature invariants, you'd get a modification of GR which is generally covariant and satisfies the equivalence principle.&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-03-09T04:22:38.207" CommentCount="1" />
  <row Id="6543" PostTypeId="2" ParentId="6541" CreationDate="2011-03-09T04:33:31.760" Score="3" ViewCount="" Body="&lt;p&gt;The real question here is the meaning of the Equivalence Principle: i like to think of in terms of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Frobenius_theorem_%28differential_topology%29&quot; rel=&quot;nofollow&quot;&gt;Fröbenius Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this sense, you could modify the Action by powers of the scalar curvature (Ricci scalar) and get equations of motion different than those of GR — but the Equivalence Principle would still be satisfied (to some approximation). See, for example, &lt;a href=&quot;http://en.wikipedia.org/wiki/F%28R%29_gravity&quot; rel=&quot;nofollow&quot;&gt;$f(R)$ Gravity&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2011-03-09T04:33:31.760" />
  <row Id="6544" PostTypeId="2" ParentId="6537" CreationDate="2011-03-09T05:22:29.650" Score="0" ViewCount="" Body="&lt;p&gt;Measure electric field. This gives you position of electron. And complete topology of wire.&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-03-09T05:22:29.650" CommentCount="4" />
  <row Id="6545" PostTypeId="1" AcceptedAnswerId="6552" CreationDate="2011-03-09T07:35:56.040" Score="8" ViewCount="372" Body="&lt;p&gt;Take a look at this picture (from APOD &lt;a href=&quot;http://apod.nasa.gov/apod/ap110308.html&quot; rel=&quot;nofollow&quot;&gt;http://apod.nasa.gov/apod/ap110308.html&lt;/a&gt;):&#xA;&lt;a href=&quot;http://apod.nasa.gov/apod/ap110308.html&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://apod.nasa.gov/apod/image/1103/titansaturn_cassini_900.jpg&quot; alt=&quot;Saturn's rings at APOD&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I presume that rocks within rings smash each other.  Below the picture there is a note which says that Saturn's rings are about 1 km thick.&lt;br&gt;&#xA;Is it an explained phenomenon?&lt;/p&gt;&#xA;" OwnerUserId="1012" LastEditorUserId="174" LastEditorDisplayName="" LastEditDate="2011-03-09T21:44:56.750" LastActivityDate="2011-03-09T21:44:56.750" Title="Why are Saturn's rings so thin?" Tags="&lt;astrophysics&gt;&lt;newtonian-mechanics&gt;&lt;solar-system&gt;" AnswerCount="2" CommentCount="13" FavoriteCount="1" />
  <row Id="6546" PostTypeId="2" ParentId="6541" CreationDate="2011-03-09T07:37:23.637" Score="4" ViewCount="" Body="&lt;p&gt;Dear Carl, my overall moral answer is the opposite one to the first two answers, so let me write a separate answer. My overall message is that with the right minimal extra assumptions, a correct theory respecting the equivalence principle has to agree with GR at cosmological distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various modifications or competitors of GR, see e.g. this list:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Classical_theories_of_gravitation#Articles_on_specific_classical_field_theories_of_gravitation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Classical_theories_of_gravitation#Articles_on_specific_classical_field_theories_of_gravitation&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First of all, most of the theories tend to be given by an action. This is needed to preserve the Noether conservation laws for symmetries, and so on. No interesting (viable enough and new enough) non-action theory is known, as far as I know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equivalence principle requires that local physics must know how it's mapped to the flat Minkowski space - so there must exist the metric tensor at each point. However, there may also exist other fields or degrees of freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One class you find in the list above is Brans-Dicke theory. It's an example of a broader class of theories that contain new scalar or tensor fields, besides the metric tensor. Torsion is a popular, frequently discussed addition by the people working on &quot;competition to GR&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If such fields are massless, they may modify the long-distance physics. But they effectively destroy the equivalence principle, too. If the extra fields are massless (like moduli, massless scalar fields), their values should be viewed as a part of the gravitational field but these values influence local physics. That's really forbidden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when understood sufficiently strictly, only the metric tensor should form the spectrum of massless (or light) fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you may have many theories which are GR plus additional fields. The additional fields are then treated &quot;matter fields&quot;. They can be anything. An interesting subclass are Kaluza-Klein theories. In them, many new massive fields may be understood as coming from a higher-dimensional general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the equivalence principle fully respected, we really deal with theories defined by an action where the metric tensor is the only massive &quot;gravitational&quot; field. The action for the other fields - matter - is a separate question and there are of course many choices. The action for the metric tensor must be diffeomorphism invariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One may prove that such an action, to have the invariance, must be a function of the Riemann tensor and its covariant derivatives, with properly contracted indices. For example, terms like &#xA;$$\nabla_\alpha R_{\beta\gamma\delta\epsilon} \nabla^\alpha R^{\beta\gamma\delta\epsilon}$$&#xA;are tolerable. However, all such terms contain either extra covariant derivatives or extra powers of the Riemann tensor, relatively to the Ricci scalar $R$. For dimensional reasons, all such extra terms have to be multiplied by a positive power of a distance $L$ and it seems likely that all such distances $L$ that could occur are microscopic or ultramicroscopic. So all such terms become negligible at cosmological (or astrophysical) distances. In this long-distance limit, the gravitational theory has to be given by the Einstein-Hilbert action e.g. Einstein's equations and the predictions for Mercury and bending of light are inevitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It wasn't an accident that Einstein ended up with the right theory. He had to - and many of his &quot;heuristic&quot; arguments (especially &quot;minimality&quot;) may be justified by the more valid RG-like $L$-argument above.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-09T07:37:23.637" CommentCount="7" />
  <row Id="6549" PostTypeId="2" ParentId="6530" CreationDate="2011-03-09T09:15:05.450" Score="6" ViewCount="" Body="&lt;p&gt;First: There is no rigorous construction of the standard model, rigorous in the sense of mathematics (and no, there is not much ambivalence about the meaning of rigor in mathematics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a lot of references that Daniel cited, I'll try to classify them a little bit :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Axiomatic (synonymous: local or algebraic) QFT&lt;/b&gt; tries to formulate axioms for the &lt;b&gt;Heisenberg viewpoint&lt;/b&gt; (states are static, observables are dynamic). There are three sets of axioms known:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the &lt;a href=&quot;http://ncatlab.org/nlab/show/Wightman+axioms&quot; rel=&quot;nofollow&quot;&gt;Wightman axioms&lt;/a&gt;,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the &lt;a href=&quot;http://ncatlab.org/nlab/show/Haag-Kastler+axioms&quot; rel=&quot;nofollow&quot;&gt;Haag-Kastler axioms&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the &lt;a href=&quot;http://ncatlab.org/nlab/show/Osterwalder-Schrader+theorem&quot; rel=&quot;nofollow&quot;&gt;Osterwalder-Schrader axioms&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Roughly, the Wightman axioms describe how fields relate to observables, the Osterwalder-Schrader axioms are the Wightman axioms for Euclidean field theory, and the Haag-Kastler axioms dodge fields entirely and describe the observables per se. All three sets of axioms are roughly equivalent, meaning that the equivalence has been proven, sometimes with additional assumptions that physicists deem to be irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;PCT, Spin and Statistics, and All That&quot; was the first introduction to the Wightman axioms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Local Quantum Physics: Fields, Particles, Algebras&quot; is an introduction to the Haag-Kastler axioms, as is &quot;Mathematical Theory of Quantum Fields&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Perturbative Quantum Electrodynamics and Axiomatic Field Theory&quot; is a description of QED from the viewpoint of the Haag-Kastler axioms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Introduction to Algebraic and Constructive Quantum Field Theory&quot; is about the quantization of given classical equations in the spirit of Haag-Kastler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Quantum Physics: A Functional Integral Point of View&quot; uses the Osterwalder-Schrader axioms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2D conformal field theory can be axiomatized using the Osterwalder-Schrader axioms, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Functorial quantum field theory&lt;/b&gt; axiomatizes the &lt;b&gt;Schrödinger viewpoint&lt;/b&gt;, see e.g. &lt;a href=&quot;http://ncatlab.org/nlab/show/FQFT&quot; rel=&quot;nofollow&quot;&gt;hnLab on FQFT&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This includes for example topological quantum field theories, these describe essentially theories with finite degrees of freedom. This branch has had a lot of impact in mathematics, especially with regard to differential geometry, and here to the theory of 3D and 4D smooth manifolds. I'd put&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Daniel S. Freed (Author), Karen K. Uhlenbeck: &quot;Geometry and Quantum Field Theory&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in this category.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Geometry and Quantum Field Theory&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Quantization of classical field theories&lt;/b&gt;: Note that the axiomatic approaches don't depend on classical field theories that need to be quantized, they open the doors for a direct construction of quantum systems without classical mirror. The Lagrangian approach to QFT is an example of an ansatz that starts with a classical field theory that needs to be quantized, for which different means can be used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ticciati: &quot;Quantum Field Theory for Mathematicians&quot; is actually a quite canonical introduction to Lagrangian QFT, without much ado.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a lot of material about the geometry of classical field theories and variants to quantize them, like &quot;geometric quantization&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book Welington de Melo, Edson de Faria: &quot;Mathematical Aspects of Quantum Field Theory&quot; is an example of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much more advanced is &quot;Quantum Fields and Strings: A Course for Mathematicians (2 vols)&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the path integral there are two points of view:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The path integral - along with the Feynman rules - is a book keeping device for a game called renormalization, that lets you calculate numbers according to arcane rules,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the path integral is a mathematical construct like a &quot;measure&quot; - but not a measure in the sense of measure theory known today - that needs to be discovered and defined appropriately.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;AFAIK there has not been much progress with the second viewpoint, but there are people working on it, for example the authors of the book &quot;Mathematical Theory of Feynman Path Integrals: An Introduction&quot;. You can find a lot more material about the mathematical theory of path integrals on the nLab &lt;a href=&quot;http://ncatlab.org/nlab/show/path+integral&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-03-09T09:15:05.450" />
  <row Id="6550" PostTypeId="1" CreationDate="2011-03-09T11:46:07.307" Score="1" ViewCount="90" Body="&lt;p&gt;If $| \psi \rangle$ is a string mode, how do you compute $\langle \psi | \hat{T}^{\mu\nu}(\vec{x}) | \psi \rangle$ where $\vec{x}$ is a point in target space? This information will tell us the energy distribution of a string. In string theory, the size of a string grows logarithmically as the worldsheet regulator scale. In the limit of zero regulator size, all strings are infinite in size, and this ought to show up in the stress-energy distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about multi-string configurations? The vacuum has virtual string pairs. Only the spatial average $\int d^9x \hat{T}^{00}(\vec{x}) |0\rangle = 0$. $\hat{T}^{00}(0) | 0 \rangle \neq 0$ even though $\langle 0 | \hat{T}^{00}(0) | 0 \rangle = 0$. Here, $|0\rangle$ is the string vacuum. The stress-energy operator can create a pair of strings. It doesn't preserve the total number of strings.&lt;/p&gt;&#xA;" OwnerUserId="2444" LastEditorUserId="2444" LastEditorDisplayName="" LastEditDate="2011-03-09T12:09:37.287" LastActivityDate="2011-03-09T16:49:04.977" Title="What is the stress-energy distribution of a string in target space?" Tags="&lt;string-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6551" PostTypeId="1" CreationDate="2011-03-09T13:28:02.127" Score="0" ViewCount="92" Body="&lt;p&gt;On the web I have found several pages crediting Ernest Rutherford with the invention or at least with discovering the foundation of efficient smoke detectors.  The story often goes like this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The modern smoke detector, responsible for saving so many lives in house fires, can be traced back to 1899 when, at McGill University in Canada, Rutherford blew tobacco smoke into his ionisation chamber and observed the change in ionisation. &lt;a href=&quot;http://www.rutherford.org.nz/biography.htm&quot; rel=&quot;nofollow&quot;&gt;(source)&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many more can be found with a Google search. However, &lt;a href=&quot;http://en.wikipedia.org/wiki/Smoke_detector&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; tells a similar story about the Swiss physicist Walter Jaeger, without mentioning Rutherford at all:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the late 1930s the Swiss physicist Walter Jaeger tried to invent a sensor for poison gas. He expected that gas entering the sensor would bind to ionized air molecules and thereby alter an electric current in a circuit in the instrument. His device failed: small concentrations of gas had no effect on the sensor's conductivity. Frustrated, Jaeger lit a cigarette—and was soon surprised to notice that a meter on the instrument had registered a drop in current. Smoke particles had apparently done what poison gas could not. Jaeger's experiment was one of the advances that paved the way for the modern smoke detector.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tend to be on Wikipedia's (and Jaeger's) side: I find it likely that, Rutherford being such a well known scientist, some discoveries may be falsely attributed to him.  But I wonder if someone can settle the issue with some trustworthy reference.&lt;/p&gt;&#xA;" OwnerUserId="1127" LastActivityDate="2011-03-09T13:28:02.127" ClosedDate="2011-04-08T20:18:52.893" Title="Did Rutherford invent the smoke detector?" Tags="&lt;soft-question&gt;&lt;history&gt;&lt;applied-physics&gt;" CommentCount="2" />
  <row Id="6552" PostTypeId="2" ParentId="6545" CreationDate="2011-03-09T14:05:10.687" Score="6" ViewCount="" Body="&lt;p&gt;There seems to be a known explanation. I quote from &lt;em&gt;Composition, Structure, Dynamics, and Evolution of Saturn’s Rings&lt;/em&gt;, Larry W. Esposito (Annu. Rev. Earth Planet. Sci. 2010.38:383-410):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[The] rapid collision rate explains why each ring is a nearly flat disk. Starting with a set of particle orbits on eccentric and mutually inclined orbits (e.g., the fragments of a small, shattered moon), collisions between particles dissipate energy but also must conserve the overall angular momentum of the ensemble. Thus, the relative velocity is damped out, and the disk flattens after only a few collisions to a set of nearly coplanar, circular orbits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think the key is that particles in a thick ring would not move in parallel planes but would have slanted trajectories, colliding all the time and losing their energy very fast.  &lt;/p&gt;&#xA;" OwnerUserId="1127" LastActivityDate="2011-03-09T14:05:10.687" CommentCount="3" />
  <row Id="6553" PostTypeId="1" CreationDate="2011-03-09T14:08:59.153" Score="-1" ViewCount="257" Body="&lt;p&gt;I know the basics, that by measuring how much x-ray signal reaches each 'pixel' on the receiver we can measure how much has been absorbed. But this gives only a single channel of information, e.g a greyscale image.&#xA;However I'm sure I've seen images where there are opaque white parts (bones) and opaque black parts (metal)... this seems to imply color and opaqueness are two separate channels and I cannot see how that works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone go beyond the basics to describe the &lt;em&gt;real&lt;/em&gt; way a fluoro scan ends up as a picture I can look at? Better references than wikipedia would also be welcome.&lt;/p&gt;&#xA;" OwnerUserId="2447" LastActivityDate="2011-03-09T21:30:54.547" Title="How does fluoroscopy (x-ray) imaging work in practice?" Tags="&lt;x-ray&gt;&lt;medical-physics&gt;" AnswerCount="2" />
  <row Id="6554" PostTypeId="2" ParentId="6545" CreationDate="2011-03-09T14:09:31.680" Score="0" ViewCount="" Body="&lt;p&gt;If one could assume that the ring is a continuous distribution of mass, we could try minimizing the total energy of the system (self energy + energy of interaction with Saturn). These two conditions along with the condition that total mass of the disc is a constant, would (I think) leave us with a unique geometry (inner and outer radius, thickness).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Some Googling gave this paper: &lt;a href=&quot;http://dx.doi.org/10.1016/0019-1035(79)90084-8&quot; rel=&quot;nofollow&quot;&gt;http://dx.doi.org/10.1016/0019-1035(79)90084-8&lt;/a&gt;&lt;/p&gt;&#xA;" LastActivityDate="2011-03-09T15:05:23.780" CommentCount="3" />
  <row Id="6555" PostTypeId="2" ParentId="743" CreationDate="2011-03-09T14:12:21.757" Score="2" ViewCount="" Body="&lt;p&gt;This is not answer to Your question in whole generality, I suggest You take a look in this simple but in my opinion meaningful example of renormalization in  simple situation: &lt;a href=&quot;http://arxiv.org/abs/patt-sol/9709003&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/patt-sol/9709003&lt;/a&gt; &quot;Uses of Envelopes for Global and Asymptotic Analysis; geometrical meaning of renormalization group equation&quot; Teiji Kunihiro and other papers by same author on  arxiv. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it may be generalized to more practical situations it gives also a hint what is it about in elementary way. &lt;/p&gt;&#xA;" OwnerUserId="1012" LastActivityDate="2011-03-09T14:12:21.757" />
  <row Id="6556" PostTypeId="2" ParentId="6503" CreationDate="2011-03-09T14:24:52.193" Score="2" ViewCount="" Body="&lt;p&gt;You have 3 obvious weak points in the system.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Where the (longer) arm is attached to the existing bracket&lt;/li&gt;&#xA;&lt;li&gt;Where the bracket attaches to the pole&lt;/li&gt;&#xA;&lt;li&gt;Where the pole is attached to the ground / base.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Notation I'll use:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$W$ = weight of monitor&lt;/li&gt;&#xA;&lt;li&gt;$L$ = maximum arm length&lt;/li&gt;&#xA;&lt;li&gt;$=&amp;gt; M = WL$ = moment at point of pole attachement&lt;/li&gt;&#xA;&lt;li&gt;$x$ = distance between attachment points. E.g. width of bracket, width of base etc.&lt;/li&gt;&#xA;&lt;li&gt;$\sigma$ = yield stress of material&lt;/li&gt;&#xA;&lt;li&gt;$A$ = area&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1 Arm - bracket&lt;/strong&gt;&#xA;I'm going to assume you'll be able to make this bracket strong enough. If you need to check, you just need to balance moments: $M = x * \sigma * A_{bolts}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2 Bracket attachment to pole&lt;/strong&gt;&#xA;This is the most difficult calculation and I'm not sure if it's possible to solve analytically. It also depends critically on the details of the attachment bracket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moment in the pole is $M$. This is constant through the length of the pole (assuming pole is perpendicular to the arm) from the attachment point to the base. I'm assuming the top of the pole is unattached to anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This moment cause a compressive force along the front of the pole. With a wide, thin-walled pole the most likely failure will be by buckling. The calculations for buckling are difficult and often not possible to solve analytically - normally FEA would be used instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is compounded by the second load on the pole. The moment is conveyed to the pole via the attachment bracket, pulling on the top of the bracket and pushing at the bottom. This means we have a (large) compressive force applied to the front of the pole at the bottom of the bracket - at the point that is already under compression due to the moment. This compressive force would eventually cause failure by buckling the pole at this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the attachment to the pole will fail by causing the pole to buckle under the bottom of the bracket. Calculating the force to do this is very difficult (probably requires FEA) and depends on the precise details of the pole and the bracket itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3 Pole attachment to base&lt;/strong&gt;&#xA;The calculation is exactly as for &lt;strong&gt;1&lt;/strong&gt; - $M$ must be counteracted by forces in the attachment spread over the base width $x$ of the attachment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&#xA;It's going to be difficult (but certainly not impossible) to make the attachments strong enough, due to $L$ being much bigger than $x$, and $W$ being large. Having said that, decent bolts should be plenty strong enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing that'll be difficult to calculate is the weight required to caused buckling of the pole. Since buckling is a sudden failure (switching from one stability mode to another), you won't get much warning of this before failure, so it's difficult to predict just from suck-it-and-see (i.e. it'll be fine, right up to the point where it smashes through your desk).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were going to try this I'd probably try to get an identical pole &amp;amp; bracket and test to destruction. Much more fun, too! Failing that, you might get somewhere by seeing how steady it looks with the monitor not extended too far... (though watch out for sudden failure as above!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can mitigate the buckling effect by:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Making the bracket wider, to spread the load on the pole&lt;/li&gt;&#xA;&lt;li&gt;Adding a sleeve to the pole under the bracket (i.e. make the pole stringer and stiffer)&lt;/li&gt;&#xA;&lt;li&gt;Fill the pole with something strong under compression (eg concrete) that will help to prevent the pole buckling inwards&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;&#xA;I've just looked in a bit more detail at the picture of the bracket itself, and I've realised that's another weak point ...! The bracket appears to be made of a few fairly thin sheets of metal, joined by small bolts (in sliding joints). This isn't designed to carry a large moment (monitor on long arm). I'd expect the sliding joints to slide, and wouldn't be surprised if the bracket itself were to buckle and collapse. Also the height of the bracket is quite small, meaning the forces applied to the pole (to transmit the moment) are large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that, my next recommendation would be to build a new bracket! You've talked about welding some pieces together already, so I'm guessing this should be possible for you. Design a new bracket that is explicitly designed to carry a large moment, and to attach to the specific arm that you have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe something like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Foot-long tube that fits snugly around the pole&lt;/li&gt;&#xA;&lt;li&gt;Foot-long plate, with mounting holes for the arm, welded the full length to the tube (I guess this would be slightly easier to use a C-section to fit more easily to the tube?)&lt;/li&gt;&#xA;&lt;li&gt;If you want to be able to change the height, drill a couple of holes in the tube and screw bolts through to press against the pole. Put these at 90$^o$ to the arm mounting.&lt;/li&gt;&#xA;&lt;li&gt;This should work better, but if the pole itself is too weak it may still fail.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hope this helps. Key thing with all this waffle from me is it's just some general ideas without knowing the full details of the system. I may be being pessimistic about the properties of the system, but without seeing the thing for real it's impossible to say. I just hope I've given you some appropriate things to consider.&lt;/p&gt;&#xA;" OwnerUserId="474" LastEditorUserId="474" LastEditorDisplayName="" LastEditDate="2011-03-10T09:51:43.160" LastActivityDate="2011-03-10T09:51:43.160" CommentCount="4" />
  <row Id="6557" PostTypeId="1" AcceptedAnswerId="6574" CreationDate="2011-03-09T14:27:09.070" Score="0" ViewCount="105" Body="&lt;p&gt;If the space of universe is expanding at fast rates..(even faster than light)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why we do not see the trace of the stars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A possible answer could be that the stars are only moving away but mantaining in geodesic respect earth and each other, that is strange, in a 2D model of an inflating balloon with dots as analogy, one can suspect that, but I can't imagine how that structure should be in 3D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other explanation or possible idea or analogy to explain why we don't see those traces are welcome (or perhaps the information that they are already seen)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1916" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-03-09T21:45:39.560" LastActivityDate="2011-03-09T21:45:39.560" Title="Star trace in the sky " Tags="&lt;astrophysics&gt;&lt;cosmology&gt;&lt;space-expansion&gt;" AnswerCount="2" />
  <row Id="6558" PostTypeId="2" ParentId="6553" CreationDate="2011-03-09T14:32:19.953" Score="0" ViewCount="" Body="&lt;p&gt;By saying that bones and metal are opaque, you actually mean that they are opaque in the visible wavelength of light. They may or may not be opaque in other wavelengths such as X-ray. Thus the color in a X-ray image is not related to the opacity as observed through our eyes. But if your eyes were adapted to X-ray's instead of ordinary light, the X-ray image would kind of look like the negative of what you would see.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-09T20:41:46.023" CommentCount="3" />
  <row Id="6559" PostTypeId="2" ParentId="6557" CreationDate="2011-03-09T14:40:46.383" Score="2" ViewCount="" Body="&lt;p&gt;If by &quot;trace of stars&quot;, you mean the sort of light trail you see when a very fast light source whizzes by you, here is the answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a light trail occurs because of &lt;a href=&quot;http://en.wikipedia.org/wiki/Persistence_of_vision&quot; rel=&quot;nofollow&quot;&gt;persistence of vision&lt;/a&gt;, the phenomenon by which an image on the retina persists for about 1/25 of a second. So a trail will be present only if the image on your retina is moving sufficiently fast. This is determined not just by how fast the object is actually moving, but also by how &lt;em&gt;far away&lt;/em&gt; it is. If two objects are moving at the same speed, the one that is farther away will &lt;em&gt;appear&lt;/em&gt; to move slower because it covers a smaller angular displacement $\theta=l/r$ in the same time. This is of course why an aircraft appears to be moving slowly in the sky even though it is travelling at several hundred kmph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stars are so ridiculously far away that it doesn't matter how fast they are moving - their image will still move so slowly on your retina that you won't see any trail.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-03-09T14:40:46.383" CommentCount="2" />
  <row Id="6561" PostTypeId="1" AcceptedAnswerId="6563" CreationDate="2011-03-09T14:41:13.210" Score="12" ViewCount="319" Body="&lt;p&gt;I'm a statistician with a little training in physics and would just like to know the general consensus on a few things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm reading a book by &lt;a href=&quot;http://en.wikipedia.org/wiki/John_Moffat_%28physicist%29&quot; rel=&quot;nofollow&quot;&gt;John Moffat&lt;/a&gt; which basically tries to state how GR makes failed predictions in certain situations. I know GR is well extremely tested, but I imagine all physicist are aware it doesn't always hold up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book tries to put forth modified theories of gravity that make do without the need of dark matter and dark energy to make GR match real world observations. (ie speed of galaxy rotations etc)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are modified theories of gravity creditable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is dark energy/matter the 'ether' of the 20th/21st century? Is it likely scientists are looking for something that simply doesn't exist and there are unknown fundamental forces at work? What's the best evidence for it's existence other than observations based on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bullet_Cluster&quot; rel=&quot;nofollow&quot;&gt;'bullet' cluster&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any insight would be great.&#xA;Cheers :)&lt;/p&gt;&#xA;" OwnerUserId="1880" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-03-10T10:17:00.933" LastActivityDate="2011-03-10T10:17:00.933" Title="Are modified theories of gravity credible?" Tags="&lt;gravity&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="3" />
  <row Id="6562" PostTypeId="1" AcceptedAnswerId="6577" CreationDate="2011-03-09T15:45:32.757" Score="1" ViewCount="111" Body="&lt;p&gt;Googling yields contradictory results, so here my question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I sing, my vocal chords vibrate, but my whole body is the resonance body, right? So I would say that when I think about standing waves etc. I must think of a resonantor that's closed (i.e. hard boundary conditions) at both ends, because my bottom end is certainly closed, and the top end has the vocal chords which are vibrating similar to woodwind instruments, so it's closed on that end as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am mistaken, could someone point out the physics behind this?&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-03-09T21:26:57.547" Title="Singing: Resonance body open-closed or closed-closed?" Tags="&lt;sound&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6563" PostTypeId="2" ParentId="6561" CreationDate="2011-03-09T15:47:17.053" Score="12" ViewCount="" Body="&lt;p&gt;Excellent question!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, there are two logical possibilities to explain the data:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There is dark matter and a cosmological constant (standard model)&lt;/li&gt;&#xA;&lt;li&gt;Gravity needs to be modified&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Interestingly, both possibilities have historical precedent: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The discovery of Neptune (by  Johann Gottfried Galle and Heinrich Louis d’Arrest) one year after its &lt;a href=&quot;http://en.wikipedia.org/wiki/Neptune#Discovery&quot; rel=&quot;nofollow&quot;&gt;prediction by Urbain le Verrier&lt;/a&gt; was a success-story for the dark matter idea.&lt;br&gt;&#xA;(Of course, after its discovery by astronomers it was no longer dark...)&lt;/li&gt;&#xA;&lt;li&gt;The non-discovery of &lt;a href=&quot;http://en.wikipedia.org/wiki/Vulcan_%28hypothetical_planet%29&quot; rel=&quot;nofollow&quot;&gt;Vulcan&lt;/a&gt; was the a failure of the dark matter idea - instead, gravity had to be modified from Newton to Einstein.&lt;br&gt;&#xA;(Funnily, Vulcan actually WAS observed by Lescarbault a year after its prediction by Urbain le Verrier, but this observation was never confirmed by anyone else.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So basically you are asking: are we in a Neptune or a Vulcan scenario? And could not the Vulcan scenario be more credible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The likely answer appears to be no. Modifications of gravity that seem to explain galactic rotation curves are usually either in conflict with solar system precision tests (where Einstein's theory works extraordinarily well) or they are complicated and less predictive than Einstein's theory (like TeVeS) or they are not theories to begin with (like MOND).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides the gravitational evidence for dark matter, there is also indirect evidence from particle physics. For instance, if you believe in Grand Unification then you must also accept supersymmetry so that the coupling constants merge in one point at the GUT scale. Then you have a natural dark matter candidate, the lightest supersymmetric particle. There are also other particle physics predictions that lead to dark matter candidates, like axions. So the point is, there are no lack of dark matter candidates (rather, there an abundance of them) that may be responsible for the galactic rotation curves, the dynamics of clusters, the structure formation etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that the Standard Model of Cosmology is a rather precise model (at the percent level), and it requires around 23% of dark matter. There are a lot of independent measurements that have scrutinized this model (CMB anisotropies, supernovae data, clusters etc.), so we do have reasonable confidence in its validity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some sense, the best evidence for dark matter is perhaps the lack of good alternatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, as long as dark matter is not detected directly through some particle/astro-particle physics experiment it is scientifically sound to try to look for alternatives (I plead guilty in this regard). It just seems doubtful that some ad-hoc alternative passes all the observational tests.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="1076" LastEditorDisplayName="" LastEditDate="2011-03-09T21:47:20.197" LastActivityDate="2011-03-09T21:47:20.197" />
  <row Id="6564" PostTypeId="2" ParentId="6561" CreationDate="2011-03-09T16:35:53.047" Score="7" ViewCount="" Body="&lt;p&gt;While it is possible that gravity still needs to be modified, it is looking increasingly unlikely that there ISN'T some form of dark matter.  In particular, the observation of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bullet_Cluster&quot; rel=&quot;nofollow&quot;&gt;bullet cluster&lt;/a&gt; is a tall order for the various modified gravity theories (though, arguably, the extra fields in something like bimetric gravity or TeVeS could be self-coupling in a different way than the ordinary fields and this could be what is inducing the seperation, but this is contrived, at best).  The basic MOND idea has be be stretched pretty desperately to explain dark matter in galactic clusters, also.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, looking at cosmological 'freeze-out' scenarios where dark matter falls out of equilibrium with ordinary matter at some high temperature in the early universe produces a model that is consistent with current nucleosynthesis data and with the observed abundance of dark matter.  This is pretty strong evidence that there is something to the beyond the standard model dark matter hypothesis.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But once again, it's not wholly impossible that there is some sort of modified gravity.  After all, quantum effects are going to need to modify gravity at some point.  It would be wrong minded to not look for it just because another hypothesis seems better at the time.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-03-09T16:35:53.047" CommentCount="2" />
  <row Id="6565" PostTypeId="2" ParentId="6550" CreationDate="2011-03-09T16:49:04.977" Score="1" ViewCount="" Body="&lt;p&gt;There's no stress-energy tensor in string theory. If there were, the Weinberg-Witten argument can be applied to the graviton string modes, and that would lead to a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-03-09T16:49:04.977" />
  <row Id="6566" PostTypeId="2" ParentId="6530" CreationDate="2011-03-09T17:24:40.843" Score="1" ViewCount="" Body="&lt;p&gt;I think everything is sufficiently rigorous when you do it according to the math rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheating starts when they say: &quot;The integral of the delta-function squared, although looks as infinity, must be determined from the experimental data&quot;. It is just funny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once I encountered a similar infinity in a simpler but exactly solvable problem. First, I wanted to do renormalizations (determining the integral value from experimental data) but fortunately I managed to choose a better initial approximation and decrease the perturbative corrections. So the problem is in the initial approximation. If it is &lt;strong&gt;good&lt;/strong&gt;, then the perturbative corrections are &lt;strong&gt;small&lt;/strong&gt;. Otherwise they are large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found an explanation why subtractions (discarding corrections) work sometimes. From my current point of view, the QFT needs reformulating since it is badly constructed. Reformulated QFT does not need repairing its solutions on the go.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-03-09T17:24:40.843" />
  <row Id="6567" PostTypeId="1" AcceptedAnswerId="6766" CreationDate="2011-03-09T17:29:20.633" Score="3" ViewCount="262" Body="&lt;p&gt;In &lt;a href=&quot;http://physics.stackexchange.com/questions/6209/paradoxical-interaction-between-a-massive-charged-sphere-and-a-point-charge&quot;&gt;PSE here&lt;/a&gt; electrons are added to a sphere and gravitational modifications are expected.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;br&gt;&#xA;Is there any &lt;strong&gt;experiment&lt;/strong&gt; that show that a negatively charged object is source of a stronger gravitational field than the same uncharged object?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words:&lt;br&gt;&#xA;The active gravitational mass of an electron is equal to his passive gravitational mass &lt;strong&gt;by experiment&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;added, for clarification:&lt;br&gt;&#xA;from here &lt;a href=&quot;http://physics.gmu.edu/~joe/PHYS428/Topic7.pdf&quot; rel=&quot;nofollow&quot;&gt;activ/passive grav mass&lt;/a&gt;:&lt;br&gt;&#xA; - active gravitational mass: establishes the field&lt;br&gt;&#xA; - passive gravitational mass: responds to the field&lt;br&gt;&#xA; - by experiment: a carefully designed setup that evidences how electrons interact in the presence of a gravitational field. Does it exists already? If no, is it doable? etc.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-03-10T01:01:53.400" LastActivityDate="2011-03-12T21:21:10.237" Title="Active gravitational mass of the electron" Tags="&lt;gravity&gt;&lt;experimental-evidence&gt;&lt;electron&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="6568" PostTypeId="2" ParentId="6384" CreationDate="2011-03-09T17:31:39.827" Score="2" ViewCount="" Body="&lt;p&gt;As commented in previous answers, conformal invariance implies scale invariance but the converse is not true in general. In fact, you can have a look at &lt;a href=&quot;http://arxiv.org/abs/hep-th/0006037&quot; rel=&quot;nofollow&quot;&gt;Scale Vs. Conformal Invariance in the AdS/CFT Correspondence&lt;/a&gt;. In that paper, authors explicitly construct two non trivial field theories which are scale invariant but not conformally invariant. They proceed by placing some conformal field theories in flat space onto curved backgrounds by means of the AdS/CFT correspondence. &lt;/p&gt;&#xA;" OwnerUserId="2237" LastActivityDate="2011-03-09T17:31:39.827" CommentCount="2" />
  <row Id="6569" PostTypeId="2" ParentId="6518" CreationDate="2011-03-09T17:47:54.000" Score="1" ViewCount="" Body="&lt;p&gt;There is a recent book by Jean Hladik &quot;Spinors in Physics&quot; that looks good. And there are websites on Clifford Algebra, and articles by Pertti Lounesto and yet more by David Hestenes. Dirac Algebra is a Clifford Algebra, and that is very straightforward stuff. You take any number of anticommuting letters, with any signature, and form a 'number' consisting of all combinations of letters taken r at a time, say: &#xA;  1 + t + x + y + z + tx + ty + tz + xy + yz + zx + txy + tyz + tzx + xyz + txyz  (+ - - -)&#xA;and now you have a 'spacetime algebra' with all the subspaces of spacetime explicitly listed, and the 'even subalgebra' 1 + tx + ty + tz + xy + yz + zx + txyz   can be used to rotate a fourVector - ie, a Lorentz transformation.  Dirac uses complex coefficients. The gamma matrices are just the t,x,y,z above, in a matrix representation, with all the 'bilinear covariants' listed.  The groups and matrices make it look more complicated than it is.&lt;/p&gt;&#xA;" OwnerUserId="1423" LastActivityDate="2011-03-09T17:47:54.000" />
  <row Id="6570" PostTypeId="1" CreationDate="2011-03-09T17:52:41.677" Score="2" ViewCount="64" Body="&lt;p&gt;If a string is unorientable, why is the a string with reversed orientation different from the initial string? Why do we have Kalb-Ramond 2-forms?&lt;/p&gt;&#xA;" OwnerUserId="2452" LastActivityDate="2011-03-09T20:06:17.637" Title="Why are unorientable strings with reversed orientations different?" Tags="&lt;string-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6571" PostTypeId="2" ParentId="6570" CreationDate="2011-03-09T18:05:44.440" Score="2" ViewCount="" Body="&lt;p&gt;the statement in the question is not true. If strings are unorientable, then - by definition - the state with a reversed-orientation string is the same one as the original string, up to a sign (and/or transformation acting on Chan-Paton factors).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also follows that unorientable strings can't carry any charge that could take infinitely many values. Consequently, the Kalb-Ramond 2-form is always removed from the spectrum by the unorientability projection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among the 5 ten-dimensional superstring theories, only type I string theory has unorientable strings. In modern language, type I is type IIB with a spacetime-filling orientifold O9-plane, which makes the strings unorientable, and 16 spacetime-filling D9-branes and their images which allow open (and equally unorientable) strings in the spectrum and give them the $SO(32)$ gauge group carried by the end points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining four 10-dimensional superstring theories - type IIA, IIB, heterotic-O, heterotic-E - contain purely orientable closed strings which are also charged under the NS-NS 2-form B-field. Type II string theories may be supplemented with orientifold planes - type I is an example; type IIA needs even-dimensional orientifold planes and type IIB odd-dimensional ones. Heterotic strings can't be made unorientable (or open) because the left-moving excitations on them are inequivant (bosonic string theory) to the right-moving ones (superstring).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that you have simply confused the words &quot;orientable&quot; and &quot;unorientable&quot; - rotated their meaning by 180 degrees. The &quot;simpler&quot; theories without projections etc. are those with the orientable strings. To make strings unorientable, one has to do extra things and identifications - or, equivalently, add an orientifold plane. It's perhaps linguistically counterintuitive that the &quot;un-...-able&quot; are the objects that require a special treatment while the the &quot;...-able&quot; ones are those that don't. ;-)&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-09T20:06:17.637" LastActivityDate="2011-03-09T20:06:17.637" />
  <row Id="6572" PostTypeId="2" ParentId="6497" CreationDate="2011-03-09T18:06:03.707" Score="0" ViewCount="" Body="&lt;p&gt;G.F.R Ellis and T.Rothman had an article on 'the crystalizing universe' arxiv 0912.0808 with a neat sentence &quot;... the arrow of time arises simply because the future does not yet exist.&quot; That sounds like a pretty straightforward explanation.&lt;/p&gt;&#xA;" OwnerUserId="1423" LastActivityDate="2011-03-09T18:06:03.707" CommentCount="1" />
  <row Id="6573" PostTypeId="2" ParentId="6541" CreationDate="2011-03-09T20:15:49.403" Score="3" ViewCount="" Body="&lt;p&gt;I would like to pick up one class of theories touched apon by Lubos, namely the Torsion extensions of General Relativity. The classic one being the &lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein-Cartan_theory&quot; rel=&quot;nofollow&quot;&gt;Einstein-Cartan Theory&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This theory introduces some form of spin coupling in addition to (metric) gravity. The Torsion tensor is zero in GR and this theory maps Torsion onto a spin connection (both antisymmetric). All this will also cause the Stress-Energy Tensor to not be symmetric (unlike in GR). Note that these spin components are somewhat classical concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So strictly this theory does &lt;em&gt;not&lt;/em&gt; obey the Equivalence Principle (I think) as &quot;spin&quot; components could result in a different measure in a gravitational field. However the Wikipedia article states that any deviation is at $10^{-15}$ and so undetectable (at least on Earth). Some quoted arguments point out that the theory is valid around rotating Black Holes, so effects might be noticable there. Otherwise the Einstein-Cartan theory is not ruled out by observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia also tells us that these ideas formed part of the basis of Loop Quantum Gravity (Ashtekar Variables); and introducing Torsion does provide some element of a mathematical completion to GR. (I believe that Einstein was not aware of the Torsion concept in 1918 and so it is zero in GR simply because he didnt know of it, rather than that it was explicitly considered and rejected. Once Einstein learned of Torsion he became excited by it and even wrote popular News articles explaining it.)&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-03-09T20:15:49.403" CommentCount="4" />
  <row Id="6574" PostTypeId="2" ParentId="6557" CreationDate="2011-03-09T20:19:45.157" Score="3" ViewCount="" Body="&lt;p&gt;Expansion of the universe goes out radially from any observer. So we never see any side-to-side motion from expansion, only the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble%27s_law&quot; rel=&quot;nofollow&quot;&gt;Hubble redshift&lt;/a&gt;. We &lt;em&gt;do&lt;/em&gt; see Hubble redshifting though, which is effectively the &quot;trace&quot; you are looking for. But there's no transverse motion because the expansion of the universe is isotropic.&lt;/p&gt;&#xA;" OwnerUserId="1655" LastActivityDate="2011-03-09T20:19:45.157" CommentCount="2" />
  <row Id="6575" PostTypeId="2" ParentId="6567" CreationDate="2011-03-09T20:37:22.440" Score="2" ViewCount="" Body="&lt;p&gt;I think you'll be unsatisfied with an answer about the gravitational field of the electron--to my knowledge, no one has tested anything involving the gravitational field created by microscopic particles.  The closest we've come is tests of gravitational redshift involving scattering microscopic particles in the external field of the Earth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There, is, however, a known solution to Einstein's equation and Maxwell's equations that represents a black hole with a nonzero charge, known as the Reissner-Nordström metric, given as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ds^{2} = - f dt^{2} + \frac{dr^{2}}{f} + r^{2} d \Omega^{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $f=1-\frac{2M}{r} + \frac{q^{2}}{r^{2}}$.  In the context of this metric, there is a difference in the gravitational field induced by the charge from what it would be without the charge--you get the $r^{2}$ variation in what becomes the gravitational potential function for timelike geodesics.  This is, in principle, measurable (and a failure to measure it would be a contradiction of either Maxwell's equation or General Relativity, which are both stringently tested--the former moreso than the latter).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One, however, must be careful with what one means by 'active gravitational mass' of a system like this--the &lt;a href=&quot;http://en.wikipedia.org/wiki/ADM_energy&quot; rel=&quot;nofollow&quot;&gt;ADM Mass &lt;/a&gt;of this system is still $M$, and is not modified by $Q$, even if particles near the black hole feel different forces due to the presence of the charged particle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, as a bit of a interesting aside, it should be noted that the presence of a charge moves the location of the horizon to the location $r_{\pm}=M \pm \sqrt{M^{2}-Q^{2}}$, so there is no horizon at all if $Q&amp;gt;M$.  It turns out that if you put the values for the electron mass and charge into this equation, you will find that it predicts that the electron should be a naked singularity.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-03-09T20:37:22.440" CommentCount="1" />
  <row Id="6576" PostTypeId="2" ParentId="6518" CreationDate="2011-03-09T20:56:22.177" Score="0" ViewCount="" Body="&lt;p&gt;What has not been explicitly pointed out in the answers above is that the Dirac equation can be expressed in 2-spinor form. Because it is really a 4 spinor object two equations are required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla^A_{B&amp;#39;} \alpha_A = 2^{-1/2}M\beta_{B&amp;#39;}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla^{B&amp;#39;}_{A} \beta_{B&amp;#39;} = 2^{-1/2}M\alpha_{A}$&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-03-09T20:56:22.177" />
  <row Id="6577" PostTypeId="2" ParentId="6562" CreationDate="2011-03-09T21:26:57.547" Score="3" ViewCount="" Body="&lt;p&gt;Both &quot;perfectly open&quot; (zero &lt;a href=&quot;http://en.wikipedia.org/wiki/Acoustic_impedance&quot; rel=&quot;nofollow&quot;&gt;acoustic impedance&lt;/a&gt;) and &quot;perfectly closed&quot; (infinite acoustic impedance) boundary conditions are only idealizations that never occur in practice. For the case of the human vocal tract, they aren't even very good approximations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;bottom end&quot; of the resonating cavity is not, in fact, the lungs, but the vocal folds (as Georg pointed out). This end has some acoustic impedance that's neither extremely low nor extremely high. I'm sure the impedance also changes somewhat with the pitch and volume of the phonation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;top end&quot; of the cavity is of course the mouth, and its impedance changes with the vowel sound you're pronouncing. For aptly-named &quot;open&quot; vowels such as &quot;ah&quot;, &quot;oh&quot;, and so on, the impedance is actually low enough that it might be a good approximation to say it's perfectly open. But for closed vowels (&quot;ee&quot;, &quot;oo&quot;, ...) and especially for humming with closed lips, the acoustic impedance is much higher (but still far from infinite).&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-03-09T21:26:57.547" CommentCount="1" />
  <row Id="6578" PostTypeId="2" ParentId="6553" CreationDate="2011-03-09T21:30:54.547" Score="1" ViewCount="" Body="&lt;p&gt;I'm not sure the assumptions of your question are true. For example, in &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aAnna_Berthe_Roentgen.gif&quot; rel=&quot;nofollow&quot;&gt;this picture&lt;/a&gt;, both the bones and the metal ring appear black. Can you link to examples of the white bones/black metal x-ray images you describe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are ways to obtain contrast with x-rays that do not rely on absorption, such as &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/18672418&quot; rel=&quot;nofollow&quot;&gt;scattering&lt;/a&gt;. However, I think most medical x-rays are based purely on absorption.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2011-03-09T21:30:54.547" CommentCount="4" />
  <row Id="6579" PostTypeId="2" ParentId="6541" CreationDate="2011-03-09T21:56:51.780" Score="0" ViewCount="" Body="&lt;p&gt;Since long ago there exists Relativistic Theory of Gravity (RTG) by A.A. Logunov with co-authors where gravity is a physical field in the Minkowski space-time acting on matter as an effective metric of an effective Riemann space-time. It carries energy-momentum and is different from pure geometry. In this theory there are additive conservation laws. It describes all known experimental data but it does not predict black holes (there is no singularities in it). Instead there are heavy objects of finite size. Some Logunov's papers are now available on arXiv.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-03-09T21:56:51.780" CommentCount="1" />
  <row Id="6581" PostTypeId="1" CreationDate="2011-03-09T23:26:06.033" Score="7" ViewCount="708" Body="&lt;p&gt;Consider a cylinder of permanently magnetized material, with uniform magnetization pointing along the cylindrical symmetry axis (the $z$-direction). The magnet is rotating about its cylindrical symmetry axis with angular velocity $\omega$. What electric field does the rotating magnet generate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Backstory: Moving permanent magnets generally generate an electric field, even in &lt;a href=&quot;http://physics.stackexchange.com/q/6457/2359&quot;&gt;cases where $d\vec{M}/d t = 0$&lt;/a&gt;. In the case of uniform motion, this electric field is straightforward to determine using a Lorentz boost. I'm interested in cases where the simple Lorentz boost does not work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;br&gt;&#xA;As perceived by some of the answers, I am not specifically interested in a cylinder. If your solution is for a ring, a sphere, or pretty much any nontrivial cylindrically symmetric object rotating about its cylindrical symmetry axis, I'm interested, as long as $d\vec{M}/d t = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=FPdAAQAAIAAJ&quot; rel=&quot;nofollow&quot;&gt;Landau and Lifshitz&lt;/a&gt; describe a similar, interesting case where the rotating magnet is also a conductor. I'm interested in the case where the rotating object is not a conductor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Homopolar_generator&quot; rel=&quot;nofollow&quot;&gt;Unipolar induction&lt;/a&gt; is very interesting, but again, involves a rotating conductor, which I am not asking about.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="2359" LastEditorDisplayName="" LastEditDate="2011-03-10T15:33:23.633" LastActivityDate="2011-03-23T01:26:23.963" Title="What is the electric field generated by a spinning magnet?" Tags="&lt;electromagnetism&gt;&lt;magnets&gt;&lt;rotational-dynamics&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="1" />
  <row Id="6582" PostTypeId="1" CreationDate="2011-03-09T23:38:24.680" Score="-1" ViewCount="236" Body="&lt;p&gt;I've been told that in countries like Israel the truth of the matter is that the only available work for B.Sc./M.Sc. in physics/math is only to be high school teacher. To do serious stuff you need to be Ph.D.&lt;/p&gt;&#xA;" OwnerUserId="2120" LastActivityDate="2011-03-10T08:35:57.423" Title="What sort of jobs do physics graduates with B.Sc/M.Sc get?" Tags="&lt;soft-question&gt;&lt;physics-careers&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6583" PostTypeId="1" AcceptedAnswerId="6629" CreationDate="2011-03-09T23:56:42.077" Score="1" ViewCount="125" Body="&lt;p&gt;In the paper &lt;em&gt;The River Model of Black Holes&lt;/em&gt;:&lt;br&gt;&#xA;Am.J.Phys.76:519-532,2008, Andrew J. S. Hamilton, Jason P. Lisle&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0411060&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/gr-qc/0411060&lt;/a&gt;&lt;br&gt;&#xA;The authors give a way of describing the action of a rotating (and or charged) black hole through a collection of local Minkowski frames, that is, as a sort of collection of preferred reference frames, or more accurately, by the tidal effects arising from the movement from one frame to another. Each frame is defined by a &quot;river field&quot; $\omega_{km}$ (See around equation 74). This field is composed as follows:&lt;br&gt;&#xA;$\omega_{0m} = \beta_m$ is the &quot;velocity&quot; of the river, while&lt;br&gt;&#xA;$\mu^i = 0.5 \epsilon^{ikm}\omega_{km}$ gives the &quot;twist&quot; of the river.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the motion of objects due to the black hole can be calculated from the tidal change $\delta\omega^k_m$ which is a local infinitesimal Lorentz transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this: Can this description of a black hole be used to describe general relativity? Note that there is an obvious limitation: since this is based on a flat background metric you can't get worm holes and the like, but I mean, subject to the requirement of trivial topology, can every GR situation be described by a &quot;river field&quot;?&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-10T15:57:29.103" Title="Can Gravity be described in terms of velocities and rotations of local Minkowski space?" Tags="&lt;general-relativity&gt;&lt;gravity&gt;" AnswerCount="2" />
  <row Id="6584" PostTypeId="1" AcceptedAnswerId="6587" CreationDate="2011-03-10T01:58:43.167" Score="10" ViewCount="316" Body="&lt;p&gt;In non-relativistic QM, on one hand we have the following relations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\langle x | P | \psi \rangle = -i \hbar \frac{\partial}{\partial x} \psi(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\langle p | X | \psi \rangle = i \hbar \frac{\partial}{\partial p} \psi(p)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, despite the similarities, the relations cannot be directly applied to energy and time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\langle t | H | \psi \rangle = i \hbar \frac{\partial}{\partial t} \psi(t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\langle E | T | \psi \rangle = -i \hbar \frac{\partial}{\partial E} \psi(E)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just wondering, how can one mathematically prove that the &quot;classical time&quot; (which means no QFT or relativity involved), unlike its close relative &quot;position&quot;, is not a Hermitian operator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask your pardon if you feel the question clumsy or scattered. But to be honest, if I can clearly point out where the core issue of the problem is, I may have already answered it by myself :/&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="749" LastActivityDate="2011-03-10T04:48:24.397" Title="Time as a Hermitian operator in QM?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="4" />
  <row Id="6585" PostTypeId="2" ParentId="2322" CreationDate="2011-03-10T02:48:16.940" Score="4" ViewCount="" Body="&lt;p&gt;Old post, but I just have to make a point. While I completely agree that some of Tesla's beliefs, as pointed out in this thread, do not qualify as science, (you have no argument from me on that point), I think Tesla deserves some credit for other achievements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Jupiter seemed to capture the sentiment of many of your posts when he said that Tesla was considered a crackpot in his time. By whom? If you can look past his crackpot theories from when he was an old man, Tesla was responsible for many, many credible scientific achievements earlier in life. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you ever heard of A/C current? Patent holder: Nikola Tesla. Ever heard of the radio? Patent holder: Nikola Tesla, NOT Marconi, who was erroneously (initially) awarded the patent over Tesla, which landed Marconi a Nobel prize for the achievement. The patent office reversed itself in Tesla's favor in 1942. Thus, the Nobel prize should have been given to Tesla as well. This is a historical fact, yet for some reason people still today refer to Marconi as the inventor of the radio... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't care if this is down rated because it doesn't answer the question. 90% of the posts on this topic are focused on blasting Tesla for some reason. Seriously, get over it people. He got old, lost his edge, and got screwed out of a Nobel prize by Marconi. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, the Tesla mentioned in this post is the same genius who invented A/C and radio... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What have you achieved?&lt;/p&gt;&#xA;" OwnerUserId="2455" LastActivityDate="2011-03-10T02:48:16.940" CommentCount="6" />
  <row Id="6586" PostTypeId="2" ParentId="6363" CreationDate="2011-03-10T03:02:45.717" Score="1" ViewCount="" Body="&lt;p&gt;I think this is already implicit in Lubos's discussion, but to try to be more explicit: the assumption that's being made here is essentially just the usual logic of effective field theory and decoupling. The complex structure moduli will get large masses (which are maybe roughly of order the string scale times some small couplings), and those masses are not tachyonic to the extent that breaking of no-scale structure is small, by the argument you gave in your question. So you should imagine that you can supersymmetrically integrate them out in a consistent way, and then ignore them. The KKLT discussion proceeds after this point: it begins with a superpotential for the remaining Kähler moduli, assuming that there are small numbers involved ($W_0 \ll 1$ by tuning, and $A e^{-aT} \ll 1$ at the minimum dynamically, with similar smallness assumptions on SUSY breaking as Lubos mentioned). It is only to the extent that these remaining moduli fields are light compared to the complex structure moduli that you can consistently assume the CS moduli have been integrated out. (You can always play with a toy model where you keep a heavy modulus around and study how its VEV and mass change after turning on the no-scale breaking $Ae^{-aT}$ term; you will find that, provided you have a large hierarchy of scales in your model, the changes are small.)&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2011-03-10T03:02:45.717" CommentCount="4" />
  <row Id="6587" PostTypeId="2" ParentId="6584" CreationDate="2011-03-10T03:32:10.417" Score="6" ViewCount="" Body="&lt;p&gt;Time is not a variable in Quantum Mechanics (QM), it's a &lt;em&gt;parameter&lt;/em&gt; — much in the same way as it is in Classical (Newtonian) Mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you have a Hamiltonian, e.g., for the harmonic oscillator, you have $\omega$ as a parameter, as well as the masses of the particle(s) involved, say $m$, and you also have time — even though it's not something that shows up explicitly in the Hamiltonian (remember explicit time dependency from Classical Mechanics: Poisson Brackets, Canonical Transformations, etc — in fact, you could get your answer straight from these kinds of arguments).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this sense, just like you don't have a 'transformation pair' between $m$ and $\omega$, you also don't have one between time and Energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you say to convince yourself that $\omega \neq -i\, \partial_m$? Why can't you use this same argument to justify $E \neq -i\, \partial_t$? ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think Roger Penrose makes a nice illustration of how this whole framework works in his book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0679454438&quot; rel=&quot;nofollow&quot;&gt;The Road to Reality: A Complete Guide to the Laws of the Universe&lt;/a&gt;: check chapter 17.&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2011-03-10T03:32:10.417" />
  <row Id="6588" PostTypeId="1" AcceptedAnswerId="6590" CreationDate="2011-03-10T04:00:10.520" Score="5" ViewCount="573" Body="&lt;p&gt;Could someone explain in simple terms (let's say, limited to a high school calculus vocabulary) why decibels are measured on a logarithmic scale?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This isn't homework, just good old fashioned curiousity.)&lt;/p&gt;&#xA;" OwnerUserId="2456" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-13T11:29:30.683" LastActivityDate="2011-07-15T16:15:01.037" Title="Why is the decibel scale logarithmic?" Tags="&lt;sound&gt;&lt;information&gt;" AnswerCount="7" CommentCount="3" FavoriteCount="1" />
  <row Id="6589" PostTypeId="2" ParentId="6588" CreationDate="2011-03-10T04:06:45.817" Score="10" ViewCount="" Body="&lt;p&gt;It's just because sounds that the human ear is capable of hearing range over a very large range of amplitudes.  If you talked about the power delivered to the ear, rather than the log of the power delivered to the ear, you would need to use numbers like $10^{12}$ to talk about airplane engines.  So, rather than deal with that, we use logarithims, so that most of the numbers we deal with when talking about sounds vary over reasonable number ranges.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-03-10T04:06:45.817" CommentCount="1" />
  <row Id="6590" PostTypeId="2" ParentId="6588" CreationDate="2011-03-10T04:07:21.797" Score="11" ViewCount="" Body="&lt;p&gt;I don't know anything about the history of the Bel and related measures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Logarithmic scales--whether for audio intensities, Earthquake energies, astronomical brightnesses, etc--have two advantages:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You can look at phenomena over a wide ranges of scales with numbers that remain conveniently human-sized all the time. An earthquake you can barely detect and one that causes a regional disaster both fit between 1 and 10. Likewise the stillness of an audio-dead room and the pain of an amp turned up to 11 fit between 10 and 130.&lt;/li&gt;&#xA;&lt;li&gt;Fractional measures are converted into &lt;em&gt;differences&lt;/em&gt; which most people find easier to compute quickly. Three decibels reduction is always the same fractional difference; the EEs get a lot of mileage out of this.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These scales may seem very artificial at first, but if you use them they will become second nature.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-03-10T04:12:30.950" LastActivityDate="2011-03-10T04:12:30.950" CommentCount="1" />
  <row Id="6591" PostTypeId="2" ParentId="6584" CreationDate="2011-03-10T04:32:32.163" Score="1" ViewCount="" Body="&lt;p&gt;The energy spectrum is bounded from below. A time operator would contradict the Stone&amp;ndash;von Neumann theorem. This isn't really a problem. All it means is we have limits as to how accurate clocks can be in quantum mechanics. &lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-03-10T04:32:32.163" />
  <row Id="6592" PostTypeId="2" ParentId="6584" CreationDate="2011-03-10T04:48:24.397" Score="0" ViewCount="" Body="&lt;p&gt;Just to add to the answers. I believe that the first time it was shown that time cannot be described by an operator was a theorem of Pauli. You can google it.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2011-03-10T04:48:24.397" />
  <row Id="6593" PostTypeId="2" ParentId="6583" CreationDate="2011-03-10T05:24:09.417" Score="4" ViewCount="" Body="&lt;p&gt;As i said in a response to a previous question of yours, a good way to think about these issues is in terms of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Frobenius_theorem_%28differential_topology%29&quot; rel=&quot;nofollow&quot;&gt;Frobenius theorem (in differential topology)&lt;/a&gt; — or, more generally, in terms of &lt;a href=&quot;http://en.wikipedia.org/wiki/Foliation&quot; rel=&quot;nofollow&quot;&gt;Foliations&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is seems to me that was done in the work you cited is exactly this: a particular 'distribution' (in the sense of Fröbenius's theorem above) was chosen, i.e., a particular collection of flat approximations [to the curved spacetime] — or, if you will, a certain foliation was chosen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that there's a clear analogy between GR and Fluid Mechanics: both are so-called Classical Field Theories. Landau's &quot;Course of Theoretical Physics&quot;, in 10 vols, comes in handy here. My point being that, effectively, it doesn't matter whether you're talking about GR or Fluid Mechanics — and this is the basis behind Unruh's &quot;Sonic Holes&quot;. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in the end of the day, you can use the very same construction in GR, Fluid Mechanics, or any other 'Classical Field Theory'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you can take a look at &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0611154&quot; rel=&quot;nofollow&quot;&gt;MacDowell-Mansouri gravity and Cartan geometry :: arXiv:gr-qc/0611154&lt;/a&gt; and see how the tools of &lt;a href=&quot;http://en.wikipedia.org/wiki/Cartan_connection&quot; rel=&quot;nofollow&quot;&gt;Cartan Geometry&lt;/a&gt; can be used in GR (and, more generally, in Differential Geometry) — in particular, you may be interested in the so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Method_of_moving_frames&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;repère mobile&lt;/em&gt; (moving frame)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I should have said something similar to this in the answer to &lt;a href=&quot;http://physics.stackexchange.com/questions/6541/gravity-theories-with-the-equivalence-principle-but-different-from-gr/&quot;&gt;your previous question&lt;/a&gt;. Note how the Equivalence Principle fits nicely in this framework (moving frames, foliations, 'distributions', etc).&lt;/p&gt;&#xA;" OwnerUserId="142" LastEditorUserId="142" LastEditorDisplayName="" LastEditDate="2011-03-10T05:39:10.187" LastActivityDate="2011-03-10T05:39:10.187" />
  <row Id="6596" PostTypeId="2" ParentId="6588" CreationDate="2011-03-10T08:30:11.543" Score="14" ViewCount="" Body="&lt;p&gt;Human senses, nearly all, works in a manner which obeys Weber–Fechner law: response of the sense machinery is logarithm of an input. It is true at least for hear, but also for eye sensitivity, temperature sense etc. of course in an areas where it works normally, because in extrema there are other processes as pain etc. It is funny that it is not mentioned in Wikipedia...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as in a cause of hear You experience is logarithm of he power of a sound wave, by &quot;biological, natural, hear sense construction&quot;. So i is natural to use logarithmic units because in technical devices You control power of a signal. This is also the cause that in radio potentiometers are logarithmic one in order to obtain &quot;linear experience&quot; of loudness during volume steering.&lt;/p&gt;&#xA;" OwnerUserId="1012" LastEditorUserId="2151" LastEditorDisplayName="" LastEditDate="2011-07-15T16:15:01.037" LastActivityDate="2011-07-15T16:15:01.037" CommentCount="1" />
  <row Id="6597" PostTypeId="2" ParentId="6582" CreationDate="2011-03-10T08:35:57.423" Score="2" ViewCount="" Body="&lt;p&gt;Opinion based on my own experience in central Europe: In order to do research as a physicist (there are jobs, but few), you need a PhD. As a physics master, you'll get hired as a engineer (for software, electronics or similar topics). The same is true if all you got is a bachelor, but you'll have a harder job to convince people to hire you, the master is the norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Germany you'll need a specialized education to get a job as a highschool teacher, so in general you won't be hired as one if all you have is a master in physics, except in very exceptional cases.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-03-10T08:35:57.423" />
  <row Id="6598" PostTypeId="2" ParentId="6561" CreationDate="2011-03-10T09:04:18.753" Score="2" ViewCount="" Body="&lt;p&gt;Keep in mind that while GR is well extremely tested, all those tests have been in a teeny tiny range of what it predicts. For example, GR predicts a gravitational redshift that can range from 0% (no redshift) to 100% (no photons received, a black hole). The largest gravitational redshift detected to date is less than 0.01%.&lt;/p&gt;&#xA;" OwnerUserId="2314" LastActivityDate="2011-03-10T09:04:18.753" />
  <row Id="6599" PostTypeId="2" ParentId="6541" CreationDate="2011-03-10T09:19:28.833" Score="-1" ViewCount="" Body="&lt;p&gt;I don't know that there's any general class of theories that compete with GR. If a competitor doesn't agree with experimental tests to date it wouldn't be viable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vast majority of tests of GR have been tests of the Schwarzschild metric. It's possible to tweak that metric so that it's compatible with quantum mechanics (unlike GR) yet is still confirmed by every experimental test of the Schwarzschild metric to date, including the anomalous perihelion precession of Mercury. Also the black hole information loss paradox vanishes. Tweaking the metric, as long as it's still a smooth curvature, doesn't necessarily deny compatibility with the equivalence principle.&lt;/p&gt;&#xA;" OwnerUserId="2314" LastActivityDate="2011-03-10T09:19:28.833" />
  <row Id="6600" PostTypeId="2" ParentId="2" CreationDate="2011-03-10T09:27:17.140" Score="-2" ViewCount="" Body="&lt;p&gt;Read the book &lt;em&gt;Not Even Wrong&lt;/em&gt;. The theory isn't testable in any practical sense. I'd be wary of investing time in such a theory, since falsifiability is a core requirement of science. It's impossible to distinguish between fantasy and an untestable theory.&lt;/p&gt;&#xA;" OwnerUserId="2314" LastActivityDate="2011-03-10T09:27:17.140" />
  <row Id="6601" PostTypeId="2" ParentId="6588" CreationDate="2011-03-10T09:40:01.277" Score="2" ViewCount="" Body="&lt;p&gt;Not only is the human ear (and other human senses) capable of observing signals over many orders of magnitude, we also perceive these signals more or less on a logarithmic scale. Take the 80 dB of a room full of loud conversation as an example. We don't perceive this as a thousand times louder than the 50 dB of a washing machine, nor a hundred times quieter than the 100 dB of a jackhammer. (Examples from &lt;a href=&quot;http://en.wikipedia.org/wiki/Sound_pressure&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.) Therefore, the decibel scale is not just useful for calculating 3 dB increases or keeping the numbers on human-sized scales, it also approximates the way our senses work.&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2011-03-10T09:40:01.277" />
  <row Id="6602" PostTypeId="2" ParentId="6588" CreationDate="2011-03-10T09:46:22.490" Score="1" ViewCount="" Body="&lt;p&gt;Human ear can detect 10^13 units of intensity. By using logarithmic scale, you get a scale of 130 db.   &lt;/p&gt;&#xA;" OwnerUserId="2422" LastActivityDate="2011-03-10T09:46:22.490" />
  <row Id="6603" PostTypeId="2" ParentId="3477" CreationDate="2011-03-10T10:05:07.413" Score="0" ViewCount="" Body="&lt;p&gt;Hi I need help with to me seems to be a condradiction about the cosmological constant ok here i go.&#xA;I have been reading adrian berry's the giant leap book and in it he talks about oh my god particles or cosmic rays that travel at 99.99999999999999999999999951 per cent of the spped of light he says while it seems unlikely that any civilization could build a ship that would travel this fast were talking about speeds like reaching alpha centuri  in 4.3 thousandths of a second, to the centre of the milkey way 30,000 light years away in 3.2 seconds A voyage to andromada galaxy 2.2 million light years away in 3.5 minutes  to travel to the very edge of the known universe would take just 19 days which is 17 billion light years away, my dilema is if nothing can travel faster then light at 186.000 miles a second like in the cosmological constant how come these can?&lt;/p&gt;&#xA;" OwnerUserId="2463" LastActivityDate="2011-03-10T10:05:07.413" />
  <row Id="6604" PostTypeId="2" ParentId="6581" CreationDate="2011-03-10T10:15:28.193" Score="4" ViewCount="" Body="&lt;p&gt;The electric field is null:  Because of the rotational symmetry assumed there, the magnetic induction  $B$  is constant in time, so $curl\,E = 0$ by Faraday's law.  On the other hand no electric charge is present, so $div E = 0$.  This is enough to make $E = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving permanent magnets do generate an electric field, &quot;even in cases where $\frac{dM}{dt}=0$&quot;, but the  $M$  there is referred to a frame linked with the magnet.  The resulting induction  $B$, being referred to the laboratory frame, changes with time, hence a nonzero $\frac{dB}{dt}$, and a nonzero $curl\,E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One may object that, in the case of the question,  the magnet's frame is moving, too, so a changing  B  should result.  What makes the difference is the rotational symmetry:The field generated by a rotating &lt;em&gt;axisymmetric&lt;/em&gt; magnet is independent of its rotation speed, because any given point of the lab &quot;sees&quot; always the same magnetization, hence the same induction too. So $\frac{dB}{dt} = 0$.  &lt;/p&gt;&#xA;" OwnerUserId="1761" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2011-03-11T14:30:09.477" LastActivityDate="2011-03-11T14:30:09.477" CommentCount="4" />
  <row Id="6607" PostTypeId="2" ParentId="6581" CreationDate="2011-03-10T11:08:02.777" Score="2" ViewCount="" Body="&lt;p&gt;I must say that my previous answer was wrong.&lt;br&gt;&#xA;The correct answer is 0. There is no field outside the rotating cylinder.&#xA;So the answer by @Bossavit is absolutely correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course it was obvious from the very beginning from Gauss law. But I crushed into it &quot;doing it a hard way&quot;. Anyway I setteld down all the details of the problem, so let me present my solution:&lt;br&gt;&#xA;&lt;strong&gt;1. Obtaining the potential inside.&lt;/strong&gt;&lt;br&gt;&#xA;Inside the rotating object we have the Lorentz force that acts on charges (free or bound) inside the medium. The charges redistribute producing the electric field that compensate the force. The electrostatic potential energy, produced by the charge distribution $\rho(r)$ must be equal to the mechanical work against the Lorentz forces:&lt;br&gt;&#xA;$$F_r(r) = \rho(r)\frac{B\omega r}{c}\quad\Rightarrow\quad U(r)=\rho(r)\frac{B\omega r^2}{2c}=\rho(r)\,\phi(r)$$&#xA;Thus, getting the $\phi(r) = \frac{B\omega r^2}{2c}$ inside the cylinder. Let me stress that $\rho(r)$ can be the density of bound charges, free charges or combined density. The result doesn't depend on the nature of these charges.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Obtaining the charge distribution.&lt;/strong&gt;&lt;br&gt;&#xA;Let us first obtain charge density inside the cylinder.  For this I'll just substute $\phi(r)$ into the Poisson equation:&#xA;$$\Delta\phi(r) = -4\pi\rho(r)\quad\Rightarrow\quad&#xD;&#xA;\rho(r)=\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial}{\partial r}\frac{B\omega r^2}{2c}\right) = -\frac{B\omega}{2\pi c}$$&#xA;Inside the cylinder $\rho$ is constant, producing linearly growing field.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also surface charges $\sigma$, responsible for the discontinuity in the electric field. These are obtained from electroneutrality.&#xA;$$2\pi R\sigma = -\pi R^2\rho \quad\Rightarrow\quad \sigma = \frac{B\omega R}{4\pi c}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3. Solving for the potential outside.&lt;/strong&gt;&lt;br&gt;&#xA;Now we have to solve the Laplace equation outside the cylinder. The general solution is:&#xA;$$\Delta\phi = 0 \quad\Rightarrow\quad \phi(r)=A+B\log r$$&#xA;There are two boundary conditions to satisfy: first is the continuity  of the potential $$\phi(R) = \frac{B\omega R^2}{2c}$$ and the second is the discontinuity in the field:&#xA;$$-\phi&amp;#39;(R+0) + \phi&amp;#39;(R-0) = 4\pi\sigma$$&#xA;Obtaining:&#xA;$$ A =  \frac{B\omega R^2}{2c} \quad \text{and}\quad B = 0 $$&#xA;So the potential is constant outside the cylinder. No field.&lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2011-03-11T14:22:54.990" LastActivityDate="2011-03-11T14:22:54.990" CommentCount="5" />
  <row Id="6610" PostTypeId="2" ParentId="3081" CreationDate="2011-03-10T12:20:40.653" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Approximatation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a(n)$ - Area after $n$ folds.&lt;br&gt;&#xA;$t(n)$ - Thickness after $n$ folds.&lt;br&gt;&#xA;$d(n) = k\frac{t(n)}{d(n)}$ - Difficulty after $n$ folds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a(n+1) = \frac{1}{2} a(n) \rightarrow a(n) = c_1 2^{1-n}$&lt;br&gt;&#xA;$t(n+1) = 2 t(n) \rightarrow t(n) = c_2 2^{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d(n) = k4^{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Physics:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You can't fold an atom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Area of atom $= a(N) = c_1 2^{1-N}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve for $N$. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;No paper can be folded more than N times. :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, its very very very very very difficult. After folding a few times. ;)&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2011-03-10T12:20:40.653" />
  <row Id="6613" PostTypeId="1" AcceptedAnswerId="6620" CreationDate="2011-03-10T13:15:44.083" Score="0" ViewCount="79" Body="&lt;p&gt;I am just an space enthusiast and not a physics professional so pardon me if this is not the right place to ask.&lt;br&gt;&#xA;As title says, is there a chance for our Earth to have additional satellite(s) like our Moon? I tried to google it, without luck, however I know here there are very smart people that can perhaps give me some references to read or can give me an answer :)&lt;br&gt;&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2082" LastActivityDate="2011-03-10T15:23:28.890" Title="Is there a chance for Earth to have an additional satellite someday?" Tags="&lt;astronomy&gt;&lt;space&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="6614" PostTypeId="2" ParentId="6613" CreationDate="2011-03-10T13:42:55.613" Score="1" ViewCount="" Body="&lt;p&gt;The chances for this are not terribly good.  There is no current natural satellite around the Earth, unless it is some very small asteroid or meteoroid.  There are some near Earth orbiting asteroids which are in resonant “hunter chaser” orbits with the Earth.  These are orbits which transfers some orbital energy (energy of angular momentum) of the body to the earth and visa versa.  It is a version of the orbital dynamics which permits spacecraft to have a “delta-v” so they can travel to the far outer or inner solar system where our launch craft can’t reach.  There are a couple of upper stages from the Saturn V rocket, which were not directed into the moon, that are in such orbits with the Earth..   These objects may impact the Earth or the moon, or may in the future enter into an orbit around the Earth.  The two small moons around Mars may have been captured this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We humans might decide to adjust the motion of an asteroid so it is in geosynchronous orbit.  If we end up building this rather improbable thing called the space elevator, we may find it preferable to build it from “top down,” rather than lifting things from Earth in a bottom up process.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-10T13:42:55.613" />
  <row Id="6615" PostTypeId="2" ParentId="6613" CreationDate="2011-03-10T14:33:26.413" Score="0" ViewCount="" Body="&lt;p&gt;I would add that the gravitational equations allow it, but humanity will have a collective heart attack if a comet or a large  asteroid headed this way, and would probably scramble to control the path long before it came within capture range.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-10T14:33:26.413" CommentCount="4" />
  <row Id="6616" PostTypeId="1" AcceptedAnswerId="6621" CreationDate="2011-03-10T15:06:13.290" Score="0" ViewCount="171" Body="&lt;p&gt;I get it that there will be a gravitational force between objects attracted towards gravity but can there be a gravitational force between two objects resting on horizontal plane? In other words, does an object experience gravitational force in all directions?&lt;/p&gt;&#xA;" OwnerUserId="2467" LastActivityDate="2011-03-10T15:35:40.283" Title="Gravitational force between two masses" Tags="&lt;gravity&gt;&lt;forces&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="6617" PostTypeId="2" ParentId="6157" CreationDate="2011-03-10T15:13:02.743" Score="4" ViewCount="" Body="&lt;p&gt;Fitzpatrick's &lt;a href=&quot;http://farside.ph.utexas.edu/teaching/plasma/lectures1/index.html&quot; rel=&quot;nofollow&quot;&gt;The Physics of Plasmas&lt;/a&gt; is excellent. He also has notes on &lt;a href=&quot;http://farside.ph.utexas.edu/teaching/jk1/jk1.html&quot; rel=&quot;nofollow&quot;&gt;Classical Electromagnetism&lt;/a&gt;, &lt;a href=&quot;http://farside.ph.utexas.edu/teaching/qm/lectures/index.html&quot; rel=&quot;nofollow&quot;&gt;Quantum Mechanics&lt;/a&gt;, and &lt;a href=&quot;http://farside.ph.utexas.edu/teaching.html&quot; rel=&quot;nofollow&quot;&gt;more&lt;/a&gt;, but I'm less familiar with these works.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2011-03-10T15:13:02.743" CommunityOwnedDate="2011-03-10T15:13:02.743" />
  <row Id="6618" PostTypeId="1" AcceptedAnswerId="6639" CreationDate="2011-03-10T15:17:46.057" Score="1" ViewCount="165" Body="&lt;p&gt;Instead of a flat disc one would expect the centrifugal forces to push it in..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The body is not in a container but suspended in air spinning horizontally (i.e. left to right or vice-versa) - how to do this could be challenging - but that is not the issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt; It is not prolate: &lt;a href=&quot;http://www.youtube.com/watch?v=BxyfiBGCwhQ&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=BxyfiBGCwhQ&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2468" LastEditorUserId="2468" LastEditorDisplayName="" LastEditDate="2011-03-11T03:34:13.313" LastActivityDate="2011-03-11T03:34:13.313" Title="Why is a suspended spinning body of liquid prolate?" Tags="&lt;centrifugal-force&gt;" AnswerCount="2" CommentCount="17" />
  <row Id="6619" PostTypeId="2" ParentId="6616" CreationDate="2011-03-10T15:21:06.520" Score="-2" ViewCount="" Body="&lt;p&gt;If you consider the center of mass of each body to be a point, the gravitational force direction between them are the vectors connecting each center of mass.&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-03-10T15:21:06.520" CommentCount="6" />
  <row Id="6620" PostTypeId="2" ParentId="6613" CreationDate="2011-03-10T15:23:28.890" Score="2" ViewCount="" Body="&lt;p&gt;It's very unlikely. Even if a large body passed near the Earth, it would be on a hyperbolic orbit -- that is, it would have too much kinetic energy to be captured into an elliptical orbit around the Earth. In order to get a new permanent satellite, you'd need a three-body interaction: something would have to come close to the Earth, then, via an interaction with another body, lose enough energy to stay around. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the early days of the Solar System, there was a lot of junk flying around, so events like this could happen. Nowadays, when things have been largely cleaned out, such a three-body interaction is very improbable.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-10T15:23:28.890" CommentCount="1" />
  <row Id="6621" PostTypeId="2" ParentId="6616" CreationDate="2011-03-10T15:27:48.917" Score="4" ViewCount="" Body="&lt;p&gt;The question is confusing, but I think you might possibly mean the following: Are two objects resting on a horizontal table gravitationally attracted to each other? If that's your question, then the answer is yes. The gravitational attraction is very weak for &quot;normal-sized&quot; objects, though. You can use the rule &#xA;$$&#xD;&#xA;F={Gm_1m_2\over r^2}&#xD;&#xA;$$&#xA;to work it out. In this formula, $G=6.67\times 10^{-11}\,{\rm N\,m^2/kg^2}$, $m_1,m_2$ are the masses, and $r$ is the separation between their centers. (Strictly speaking this is only correct if the objects are spheres.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cavendish_experiment&quot; rel=&quot;nofollow&quot;&gt;Cavendish&lt;/a&gt; managed to measure this attraction back in the 1700s, in a truly amazing experiment.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-10T15:27:48.917" />
  <row Id="6622" PostTypeId="2" ParentId="4116" CreationDate="2011-03-10T15:29:28.790" Score="1" ViewCount="" Body="&lt;p&gt;Comment to Spencer Nelson's answer: Consider a spherically symmetric satellite. Let us, for simplicity, consider the idealized situation where all atoms of the satellite are organized into one big crystal held together by covalent bonds, so that all the individual atoms are only allowed to make movements that doesn't break any of their covalent bonds. Even such an idealized satellite can get tidally locked with Earth, because it would still get elastically stretched into an oval shape by tidal forces in the direction of Earth.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastActivityDate="2011-03-10T15:29:28.790" CommentCount="2" />
  <row Id="6623" PostTypeId="2" ParentId="6616" CreationDate="2011-03-10T15:30:39.767" Score="0" ViewCount="" Body="&lt;p&gt;Yes, the question is really about 3 bodies in a Gravitational Field. We can take the masses as m1, m2 and M, where M is the larger planetary body on whose plane surface they both sit. Ted Bunn has just answered whilst I was composing this, so his answer contains the relevant Force formula between m1 and m2.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-03-10T15:30:39.767" />
  <row Id="6624" PostTypeId="1" AcceptedAnswerId="6633" CreationDate="2011-03-10T15:34:59.223" Score="6" ViewCount="248" Body="&lt;p&gt;We all know that there are analytic formulae to solve &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_equation&quot; rel=&quot;nofollow&quot;&gt;quadratic&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_function&quot; rel=&quot;nofollow&quot;&gt;cubic&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Quartic_equation&quot; rel=&quot;nofollow&quot;&gt;quartic&lt;/a&gt; polynomial equations. But it seems to me that the only solution that widely used is physics is the solution of quadratic we all learnt at school.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that when physicist faces quartic or even cubic equation -- he turns to numerical methods to solve them? Have anyone seen the analytic analysis of some real problems in physics, where these formulae were used? &lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="386" LastEditorDisplayName="" LastEditDate="2011-03-10T16:55:58.803" LastActivityDate="2011-03-11T04:32:12.750" Title="Are the solutions in radicals of cubic and quartic of any use in physics?" Tags="&lt;mathematics&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" />
  <row Id="6625" PostTypeId="2" ParentId="6616" CreationDate="2011-03-10T15:35:40.283" Score="0" ViewCount="" Body="&lt;p&gt;Yes.  They still feel the force due to the Earth's gravity on the tabletop.  They also feel an upward force from the tabletop that counteracts the Earth's gravitational pull, and leaves them sitting stationary on the tabletop.  This is why it's harder on your feet when you're carrying a lot of weight than it is when you are not--the ground has to push up on your feet more to cancel out the added gravitational pull on you.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-03-10T15:35:40.283" CommentCount="1" />
  <row Id="6626" PostTypeId="2" ParentId="6624" CreationDate="2011-03-10T15:37:29.517" Score="2" ViewCount="" Body="&lt;p&gt;The problem is that the actual analytic solutions to them are extremely ugly, and it's difficult to see how variation in one of the parameters in your cubic or quartic will affect the location of the roots.  It's much easier to solve the problem numerically, and then plot the location of the roots as a function of the parameters you're varying than it is to carefully analyze the actual analytic solution.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-03-10T15:37:29.517" />
  <row Id="6628" PostTypeId="2" ParentId="6209" CreationDate="2011-03-10T15:53:48.200" Score="3" ViewCount="" Body="&lt;p&gt;The statement that the gravitational attraction will eventually dominate the coulomb repulsion as n increases is correct. &#xA;You probably think the restmass of the electrons will invoke gravitational attraction, but that part is neglible for high electrondensities on the sphere. The gravitational attraction by (the curvature of spacetime caused by) bindingenergy is far greater for high densities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know where the breakpoint is, but suppose I look way past that limit, i.e. an incredible dense sphere(or shell) of electrons. If I make it dense enough, that could very well become a black hole. Note that this would be a strange black hole, since the 'mass' of such a black hole consists almost entirely of the bindingenergy, and not the restmasses of the electrons. From this point of view it might be more easily imaginable that the gravitional attraction will dominate eventually.&lt;/p&gt;&#xA;" OwnerUserId="2470" LastActivityDate="2011-03-10T15:53:48.200" CommentCount="3" />
  <row Id="6629" PostTypeId="2" ParentId="6583" CreationDate="2011-03-10T15:57:29.103" Score="2" ViewCount="" Body="&lt;p&gt;Daniel's answer may be a little bit generic; although I am not exactly sure what you are looking for one similar concept is that of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrad_%28general_relativity%29&quot; rel=&quot;nofollow&quot;&gt;Tetrad&lt;/a&gt;; there are other names like Frame Field for the same concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the introductory paragraph from Wikipedia to see if this is what you are interested in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Frame fields always correspond to a family of ideal observers immersed in the given spacetime; the integral curves of the timelike unit vector field are the worldlines of these observers, and at each event along a given worldline, the three spacelike unit vector fields specify the spatial triad carried by the observer. The triad may be thought of as defining the spatial coordinate axes of a local laboratory frame, which is valid very near the observer's worldline.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Tetrad formalism is closely related to 2-spinors and Tetrads can twist and turn as they move through space-time encoding lots of GR properties more directly than other formalisms.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-03-10T15:57:29.103" />
  <row Id="6630" PostTypeId="1" AcceptedAnswerId="6635" CreationDate="2011-03-10T16:13:33.387" Score="-3" ViewCount="146" Body="&lt;p&gt;I'm not sure how theoretically possible this is but my question is...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we could somehow make a perfect bubble of photons (a massless bubble) and put a spaceship inside it, could it therefore effectively become massless as the perfectly sealed bubble around it has no mass adn therefore travel at lightspeed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if this is just nonsense.&lt;/p&gt;&#xA;" OwnerUserId="2247" LastActivityDate="2011-03-11T11:15:41.997" Title="Massless bubble of photons" Tags="&lt;light&gt;&lt;mass&gt;&lt;photons&gt;&lt;speed&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="6631" PostTypeId="2" ParentId="3906" CreationDate="2011-03-10T16:13:56.157" Score="3" ViewCount="" Body="&lt;p&gt;Marks answer is flawed from the second picture onward. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The major force contributing to the lift of the board is not applied downward on the left axle but instead by pulling parallel to the boards surface (in an up and a litte to the left direction). This is done by the instep of the riders right foot (after rotating the foot around the ankle) as can barely be seen in the video. That is why there is grip tape on the deck in the first place (and why skater's shoes don't last long).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without this (for a beginner rather difficult) motion the board wouldn't lift at all (this is only partly true as any skateboarder can tell you but for now let's assume it's correct).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only applied Mark's suggested downward force on the left axle you would not lift the board but press it right back down to the ground. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall height of the jump then is determined by how high you can jump yourself and how well you can pull the board up with you. There is no necessity in popping (the moment of tail-ground contact) more powerfully to gain height. It is perfectly possible to do an ollie without the tail hitting the ground at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is quite difficult to see this from just watching the video or a full speed skateboarder, so I suggest you try it yourself (while wearing protection of course).&lt;/p&gt;&#xA;" OwnerUserId="2469" LastActivityDate="2011-03-10T16:13:56.157" CommentCount="6" />
  <row Id="6632" PostTypeId="2" ParentId="6630" CreationDate="2011-03-10T16:18:14.497" Score="2" ViewCount="" Body="&lt;p&gt;This is just nonsense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only remotely possible idea that comes to mind in this context is by using the photon's impulse to maintain some sort of balanced spaceship but there is no way to create a bubble of photons. They cannot be confined without some material border (aka mirror)and you would have to carry that weight with you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Ok, not only material borders but also bending of space would allow for photon confinement. But gravitational effects have the side effect that you need some major amount of mass to do that (black holes, galaxy clusters, ...). Unfortunately to accelerate such a body you need a lot of force of some kind since F=ma and thus a=F/m. (a being acceleration, m the mass and F the force)&lt;/p&gt;&#xA;" OwnerUserId="2469" LastEditorUserId="2469" LastEditorDisplayName="" LastEditDate="2011-03-11T11:15:41.997" LastActivityDate="2011-03-11T11:15:41.997" CommentCount="2" />
  <row Id="6633" PostTypeId="2" ParentId="6624" CreationDate="2011-03-10T16:28:07.930" Score="6" ViewCount="" Body="&lt;p&gt;It was a lot of fun to derive the solution for the cubic and quartic equation 22 years ago, when I was a high school kid, but I haven't really used it afterwards. I think the same is true for most physicists I met: they didn't use the complicated formulae, especially not the quartic one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much more generally, the situations in which it would be useful are extremely rare for several reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the situations in which the functions are truncated to a cubic or quartic polynomial are extremely rare&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;known corrections ultimately modify the cubic or quartic polynomial to a more general function&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the position of roots is usually not the key interesting quantity&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the truly interesting and invariant quantities are functions of the roots that can be expressed more easily in terms of the coefficients&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;quite typically, physical problems wouldn't lead to a cubic or quartic equation for one variable but to a set of such equations for many variables which can't be solved in radicals&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;in the interesting cases when a cubic or quartic equation has to be solved, it reduces to a linear or quadratic equation, anyway.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To give some examples, one may consider the most general classical potential for a scalar field which produces a renormalizable theory. This is given by at most a 4th order polynomial of the scalar field. For a general potential, the zeros of such potential would be roots of the polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, zeros of a potential are not too interesting: much more interesting are its stationary points, i.e. extrema, which are the solutions of the cubic equation. Moreover, the coefficients in the polynomial remain more important than the positions of the minima - and observable quantities may be expressed as their functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the canonical shapes of the potential have a $Z_2$ or a related symmetry that effectively reduces the degree of the equation to half of it. Moreover, generally, the theories may include several fields which produces a set of higher-order equations that usually can't be solved in radicals. Equally importantly, quantum corrections add non-polynomial corrections to the potential, anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the set of physical situations in which the problem could be solved using the nontrivial formulae is limited for all the reasons above, and even when the problem may be solved, we usually don't learn anything &quot;qualitative&quot; about the system. In this sense, the existence of the solution for cubic and quartic solutions is a mathematical curiosity that doesn't affect a physicist's life.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-10T16:28:07.930" CommentCount="2" />
  <row Id="6634" PostTypeId="2" ParentId="6473" CreationDate="2011-03-10T16:28:16.843" Score="1" ViewCount="" Body="&lt;p&gt;Just to add to Kostya's answer above, it's worth noting a couple of facts:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1063/1.1704140&quot; rel=&quot;nofollow&quot;&gt;Causality Implies the Lorentz Group&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1017/S030500410004144X&quot; rel=&quot;nofollow&quot;&gt;On the structure of causal spaces&lt;/a&gt; and &lt;a href=&quot;http://dx.doi.org/10.1063/1.522874&quot; rel=&quot;nofollow&quot;&gt;A new topology for curved space–time which incorporates the causal, differential, and conformal structures&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The bottom-line is that the Causal Structure differs from the Metric by a conformal transformation. So, effectively (i.e., modulo a conformal transformation), it doesn't matter whether you supply a Causal Structure or a Metric to your manifold — of course, keep in mind that a point at infinity can cause all sorts of trouble (but usually, in Physics, we'll assume things are &quot;well behaved&quot;... ;-).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To directly answer your question, coupling Luboš's and Kostya's answer: yes, causality and continuity (in the sense given by the metric structure) can be thought of as being the same thing modulo a conformal transformation, i.e., as far as physical events are concerned; having said that, note that acausal regions can be continuously connected, you just can't access those regions physically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to think of this is to 'split' (foliate) your spacetime manifold in three different regions determined by the light-cone, i.e., use your Causal Structure to foliate your manifold into a spacelike region, a timelike region and a lightlike one. Each one of these can be simply-connected (continuous in the metric sense) and understood in terms of the Causal Structure (modulo a conformal transformation).&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2011-03-10T16:28:16.843" />
  <row Id="6635" PostTypeId="2" ParentId="6630" CreationDate="2011-03-10T16:38:22.477" Score="4" ViewCount="" Body="&lt;p&gt;Let me enumerate several reasons why it's indeed nonsense:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Photons are called massless because their rest mass is zero; however, their total mass that is subject to gravity etc. is given by their energy, $E=mc^2=hf$, and it is nonzero&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Photons can't be kept in a shell because they need to move by the speed of light; the closest thing you can get is the &quot;photon orbit&quot; of a black hole - at distance $R=3GM/c^2$ from a neutral black hole center, for example - where photons may move for a long time but this orbit is unstable, so the photons eventually either fall to the black hole or escape from it&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Even if you ignored the problems above, the photons would have no impact on the spaceship inside: analogously, toilet paper is also light but a dumbbell doesn't become light just because you pack it in toilet paper. I have already written enough so if you need to explain why the previous sentence is true, please ask another question and someone else may wrestle with it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-10T16:38:22.477" CommentCount="3" />
  <row Id="6636" PostTypeId="2" ParentId="6209" CreationDate="2011-03-10T17:14:08.693" Score="1" ViewCount="" Body="&lt;p&gt;As a brainstorming answer, lets calculate the binding energy in another way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;suppose we have N electron in the sphere, the electrostatic energy to bring a new electron into the sphere is $Ne/R$. If we add a new electron, only the net electrostatic potential $Ne/R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the net potential on far-away electrons is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N( e -  m_{e} G )/R$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what is actually quadratic in $N$ is the energy required to bind together N electrons in the sphere, it is actually&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e/R + 2e/R + ... Ne/R = N(N-1)e/2R$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also contributes to gravitational weight, but there will be a maximum capacity where the electrons will escape the sphere&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the capacitance of a sphere is given by $4 \pi \epsilon R$. So in this case the binding energy is bounded by the capacitance of the conductor used for your sphere&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-03-10T17:30:12.237" LastActivityDate="2011-03-10T17:30:12.237" />
  <row Id="6637" PostTypeId="1" CreationDate="2011-03-10T17:16:10.753" Score="0" ViewCount="121" Body="&lt;p&gt;Unfortunately I broke the frame of the sunglasses that my wife gave me. I can not weld it with anything. This frame is made ​​from a type of plastic such as polyethylene or polypropylene. Is there a way out?&lt;/p&gt;&#xA;" OwnerUserId="2471" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-10T18:10:28.320" LastActivityDate="2011-03-10T18:10:28.320" ClosedDate="2011-03-10T18:22:21.193" Title="Is there a strong glue for polyethylene or polypropylene?" Tags="&lt;material-science&gt;" AnswerCount="0" CommentCount="1" />
  <row Id="6639" PostTypeId="2" ParentId="6618" CreationDate="2011-03-10T18:28:32.410" Score="0" ViewCount="" Body="&lt;p&gt;It isn't prolate. It's oblate. Centrifugal forces push stuff &lt;em&gt;out&lt;/em&gt; from the center. Nearby surface elements of the liquid are tugged along by surface tension, flattening out the original shape into a disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If something were to take on a prolate shape, there would need to be forces along the axis of rotation, which don't exist for most fluids I can think of, but if you provide a counterexample we could probably explain it. Beyond that, I think you're just confused.&lt;/p&gt;&#xA;" OwnerUserId="1655" LastActivityDate="2011-03-10T18:28:32.410" CommentCount="3" />
  <row Id="6640" PostTypeId="1" CreationDate="2011-03-10T18:32:24.373" Score="1" ViewCount="240" Body="&lt;p&gt;Einstein teaches that as an object gets faster, its relativistic mass increases... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton teaches that as an object's mass increases, so does it's gravitational pull...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... if you a tethered some objects to a wheel, and then accelerated them to a very great speed... Each object would eventually be going fast enough to create sufficient gravity to pull us towards them, correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then just stick this contraption in the centre of a spherical space ship and you've got semi artificial gravity?...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't mind that it might take an unpheasable amount of energy to get the objects to the speed needed... I'd just like to know if the basic theory is correct?&#xA;Or is there something major I'm overlooking...&lt;/p&gt;&#xA;" OwnerUserId="2474" LastEditorUserId="2474" LastEditorDisplayName="" LastEditDate="2011-03-10T18:48:59.943" LastActivityDate="2011-03-10T21:27:14.063" Title="Is This The Answer To Artificial Gravity?" Tags="&lt;gravity&gt;&lt;newton&gt;" AnswerCount="3" />
  <row Id="6641" PostTypeId="2" ParentId="6640" CreationDate="2011-03-10T18:39:24.097" Score="2" ViewCount="" Body="&lt;p&gt;This is correct but because of the wrong reasons; the relativistic mass is only a language concoction created by people that wanted to adapt relativistic mechanics to the newtonian framework. In reality momentum is mass (there is just one fixed mass for a given object) times the 4-velocity $V_{\mu}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a rotational object has a angular momentum energy $I \omega ^2$ asocciated to its spinning, and this energy also contributes to gravity. However, for any macroscopic object for all known materials, the material teardown limit due to centrifugal forces will be reached far before &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the outer velocity reaches any significant portion of $c$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the gravitational energy due to spin reaches any measurable portion of $mG/R$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-03-10T18:39:24.097" CommentCount="1" />
  <row Id="6642" PostTypeId="2" ParentId="6640" CreationDate="2011-03-10T18:50:52.657" Score="2" ViewCount="" Body="&lt;p&gt;The problem is that your tether is made out of matter composed of chemical bonds between atoms. The centrifugal force of your mass at the end of the tether will far outweigh the strength of those chemical bonds, so the tether will snap very early in the process. This would only be feasible if your tether is made out of some unobtainium not made out of normal atoms - to reiterate, this isn't a problem of finding the right material, but one of finding an entirely different &lt;em&gt;type&lt;/em&gt; of materials we've never even considered. That just isn't likely, although you could do it for a science-fiction story, I suppose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note - we've already got a pretty good idea of how to simulate gravity with a spinning space ship. Centrifugal forces push people or things inside the cylindrical shell out towards the walls of the ship, which looks like gravity. This takes way less energy than your method.&lt;/p&gt;&#xA;" OwnerUserId="1655" LastActivityDate="2011-03-10T18:50:52.657" />
  <row Id="6643" PostTypeId="1" CreationDate="2011-03-10T18:54:02.650" Score="2" ViewCount="153" Body="&lt;p&gt;We know,i.e. wood/steel tables in earth will completely broken or disappear in a very future day.&#xA;If we put the table in a perfect/ideal environment (maybe in vacuum), will the disappear/broken still happen again?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Ideal&quot; means every outside factors, temperature, pressure and gravity etc, is just perfect for the table and the table can endure. Table wouldn't move, only the particles of the table move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I always think that the movement of the particles eventually makes the table or other physics object die/decay. &lt;strong&gt;Every objects decay for the internal cause&lt;/strong&gt;. I don't know if this is true, so I ask this question.&lt;/p&gt;&#xA;" OwnerUserId="2475" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-11T08:57:44.600" LastActivityDate="2011-03-11T08:57:44.600" Title="Will physic object in a perfect environment last/exist forever?" Tags="&lt;particle-physics&gt;&lt;mass&gt;&lt;vacuum&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="6644" PostTypeId="2" ParentId="6643" CreationDate="2011-03-10T19:18:05.453" Score="4" ViewCount="" Body="&lt;p&gt;The entropy of a closed system only increases, so if the system is in the state of maximum entropy, it will stay in that state, otherwise it will evolve toward higher entropy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A wood table is probably not maximum entropy, although it would take some work to define exactly what this means.  Wood smithereens spread throughout the entire available volume should have higher entropy.  Chemical changes would result in higher entropy as well.  Eventually, a wood table in a vacuum should evaporate/dissolve/decay.  Doing things like removing all the oxygen from the environment will slow the process down, but the increase of entropy is inevitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a statistically true.  There is a very low probability of entropy decreasing briefly.  For example, see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Poincar%C3%A9_recurrence_theorem&quot; rel=&quot;nofollow&quot;&gt;Poincare recurrence theorem&lt;/a&gt;.  This also ignores cosmological considerations, which seem irrelevant to the question and which I also don't know about.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-03-10T19:18:05.453" />
  <row Id="6645" PostTypeId="2" ParentId="6618" CreationDate="2011-03-10T19:41:50.863" Score="2" ViewCount="" Body="&lt;p&gt;I don't know about liquids 'suspended in air', but a self-gravitating rotating fluid may take many shapes, not only oblate ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jacobi has shown that at high enough spin rates the familiar oblate shape becomes unstable and turns into a triaxial ellipsoid. At even higher spins another instability appears (this one was described by Poincaré) and the system becomes pear-shaped. &lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2011-03-10T19:41:50.863" />
  <row Id="6646" PostTypeId="1" CreationDate="2011-03-10T19:52:38.560" Score="6" ViewCount="232" Body="&lt;p&gt;I'm building a physics simulator for a graphics course, and so far I have it implementing gravitational and Coulomb forces. I want to add collisions next, but I'm not exactly sure how to go about doing it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick summary of how this is working so far is: All objects are spheres of a set radius, mass and charge. The mass's and charges of the spheres are treated as point charge/mass for the calculation. Every step in time (about 1/50th of a second) the forces acting on each object are calculated in a nice big nested for loop that figures out the coloumb and gravitational force between 2 objects, for every set of 2 objects, and then they are summed together. I use this net force to determine the acceleration, and the rest is fairly obvious from there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to add in is collisions. I can deteremine pretty easily if a collision is happening (if the distance between them &amp;lt;= radius of one + radius of other), what I am not so certain of is how I should add in the collision force (and would need to do it component wise). I want elastic collisions, though for now I'd just be happy with getting conservation of momentum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The information I have easily available are: the velocity of each object (and I mean velocity, speed and direction), the position of each object, the mass/charge (charge obviously not needed here) and the Net force calculated so far for each object for the next step in time. I dont need exact formulas (would prefer if they arent exact), just need a nudge in the right direction.&lt;/p&gt;&#xA;" OwnerUserId="2350" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-03-10T20:21:00.687" LastActivityDate="2011-03-10T20:55:18.287" Title="Physics of simple collisions" Tags="&lt;kinematics&gt;&lt;collision&gt;&lt;computational-physics&gt;" AnswerCount="2" CommentCount="10" FavoriteCount="2" />
  <row Id="6647" PostTypeId="2" ParentId="6643" CreationDate="2011-03-10T19:57:54.200" Score="3" ViewCount="" Body="&lt;p&gt;Although the question has some vagueness in its conditions there are some further points which might be of interest. In addition to the Thermodynamic property of Entropy mentioned in the other answer we also have the property of Temperature and its Thermodynamics to consider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The object will (a) approximate a Black Body and will emit thermal radiation reducing its energy over time; and relatedly (b) equalize to the Temperature of its environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take the environment to be somewhere of the same Temperature as the object initially, then that environment will require an infinite amount of energy over time just to remain at that Temperature. So this is arguably unrealistic. The second law of Thermodynamics will also begin to apply in this environment as some atoms are evaporated from the surface of the table, increasing the entropy and destroying the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take the environment to be deep space then (at 3K) the table will equalize to that Temperature. This will affect the steel/wood structure of the table: it will become more brittle over time. Any external energy (ultimately even from photons) will tend to fissure then fracture the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting application of this question is to the formation of the materials surface of space-craft (like Pioneer) which need to stay in space indefinitely. Similar effects will happen here. Generally such spacecraft have an internal power source to keep going, but I dont know how much longer they will survive.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-03-10T19:57:54.200" CommentCount="2" />
  <row Id="6648" PostTypeId="2" ParentId="6646" CreationDate="2011-03-10T20:42:11.073" Score="3" ViewCount="" Body="&lt;p&gt;In an elastic collision between spheres, we can use conservation of energy and momentum, but this alone is not enough.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start by looking in a frame where one ball is stationary (the target) and the other one is coming in from the side (the striker).  The problem is to find the final momenta of the two balls.  Assuming everything stays in a plane, there are 4 degrees of freedom - two for each momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conservation of momentum has two constraints and conservation of energy has one, so applying these still leaves one degree of freedom; one extra parameter is needed to describe the collision.  One choice is the precise point of the collision.  Because the force is being applied at this point and is only a contact force normal to the surface, the target ball will shoot off in the direction of the normal to the surface at the collision point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to summarize the rules is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&lt;li&gt;Boost to a reference frame where the target is initially stationary.  &lt;/li&gt;&#xA;&lt;li&gt; Make the final momentum of the target point in the direction of the vector between the balls' centers at the moment of impact.&lt;/li&gt;&#xA;&lt;li&gt;Apply conservation of kinetic energy.&lt;/li&gt;&#xA;&lt;li&gt;Apply conservation of momentum in both dimensions.&lt;/li&gt;&#xA;&lt;li&gt;Boost back to desired original frame.&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This should uniquely specify where the target and striking balls go after the collision, and the answer is independent of which ball you choose to be the target.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're doing the calculation in 3D, you will have to choose the correct plane.  It's the plane spanned by the displacement vector between the balls and their relative velocity.  (If these vectors are dependent, the motion is only on a line.)&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-03-10T20:42:11.073" CommentCount="3" />
  <row Id="6649" PostTypeId="1" AcceptedAnswerId="6656" CreationDate="2011-03-10T20:46:43.490" Score="7" ViewCount="125" Body="&lt;p&gt;In experimental physics, we often make measurements of linear transfer functions; these are complex-valued functions of frequency.  If the underlying system is causal, then the transfer function must be analytic, satisfying the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kramers%E2%80%93Kronig_relation&quot; rel=&quot;nofollow&quot;&gt;Kramers-Kronig&lt;/a&gt; relations.  Our measurements, however, are corrupted by random (and perhaps systematic) errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is it possible to improve a measurement of a linear transfer function of a causal system in the presence of noise by applying some kind of constraints derived from the Kramers-Kronig relations?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-03-10T21:49:12.597" LastActivityDate="2011-03-10T22:36:27.800" Title="Can the Kramers–Kronig relation be used to correct transfer function measurements?" Tags="&lt;experimental-physics&gt;&lt;signal-processing&gt;&lt;linear-systems&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="6650" PostTypeId="1" CreationDate="2011-03-10T20:53:04.390" Score="6" ViewCount="345" Body="&lt;p&gt;What prevents a static black hole from accumulating more charge than its maximum? Is it just simple Coulomb repulsion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the answer the same for rotating black holes?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Edit&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;What I understand from the answers given so far, is that &lt;strong&gt;maximum charge&lt;/strong&gt; is a &lt;strong&gt;moving traget&lt;/strong&gt;. You can add charge to a black hole but Coulomb repulsion guarantees that you will do so in a manner than will increase &quot;maximum charge value&quot;. Is this correct? &lt;/p&gt;&#xA;" OwnerUserId="824" LastEditorUserId="824" LastEditorDisplayName="" LastEditDate="2011-03-11T13:04:53.837" LastActivityDate="2011-03-11T19:19:37.607" Title="What prevents the accumulation of charge in a black hole" Tags="&lt;black-holes&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="1" />
  <row Id="6651" PostTypeId="2" ParentId="6646" CreationDate="2011-03-10T20:55:18.287" Score="4" ViewCount="" Body="&lt;p&gt;Mark Eichenlaub's answer is 100% correct, but you can also do it without changing reference frames, and I think it's probably easier that way. Here's how I would set it up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that you've determined that two objects are going to collide within the next time step. Determine the positions of the two objects at the moment of collision. Draw a line connecting their two centers. Determine a unit vector $\hat n$ pointing from the center of object 1 to the center of object 2. During the collision, the force will be in this direction, so only this component of the momentum will change. Let the unknown change in momentum be $\vec q$. This vector points in the $\hat n$ direction: $\vec q=q\hat n$. To be specific, if $\hat n$ points from object 1 to object 2, then $\vec q$ is the momentum gained by object 2, and $-\vec q$ is the momentum gained by object 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $\vec p_1,\vec p_2$ be the initial momenta of the two objects, so that the final momenta are $\vec p_1-\vec q,\vec p_2+\vec q$. Energy is related to momentum via the rule $E=p^2/2m$, so conservation of energy says&#xA;$$&#xD;&#xA;{p_1^2\over 2m_1}+{p_2^2\over 2m_2}={(\vec p_1-\vec q)^2\over 2m_1}+{(\vec p_2+\vec q)^2\over 2m_2}&#xD;&#xA;$$&#xA;In this expression, you know everything except the magnitude $q$ of $\vec q$. So you can solve this equation for the unknown $q$. Then you can compute the new momenta, and you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation you solve ends up being a quadratic, but with no constant term -- that is, it's of the form &#xA;$$&#xD;&#xA;aq^2+bq=0.&#xD;&#xA;$$&#xA;One solution is $q=0$, of course, but you don't want that one: it corresponds to the two objects passing right through each other. The nonzero one's the one you want.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-10T20:55:18.287" />
  <row Id="6652" PostTypeId="2" ParentId="6650" CreationDate="2011-03-10T21:02:46.110" Score="5" ViewCount="" Body="&lt;p&gt;Coulomb repulsion it is. Specifically, if a black hole has a lot of charge, then particles with a high charge-to-mass ratio will be repelled. Anything that falls in will contribute &quot;more mass than charge,&quot; heuristically, keeping the charge-to-mass ratio of the black hole from getting too big.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-10T21:02:46.110" CommentCount="12" />
  <row Id="6653" PostTypeId="2" ParentId="6640" CreationDate="2011-03-10T21:27:14.063" Score="4" ViewCount="" Body="&lt;p&gt;I had an idea like this in high school.  The idea was to apply an EM field to an array of ions or charges with some mass.  These masses would oscillate and if the energy applied is large enough it would cause this sort of Lorentz expansion of the mass $m~=~\gamma m_0$.  Then for $gamma$ large enough you would get an increase in gravity.  This idea with a spinning body is essentially the same, where my approach is like replacing the mechanical bits of a clock with a quantum system in an atomic clock.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is wrong with this?  Nothing is wrong in one sense.  If you get this array of ions to absorb this EM field and oscillate it will happen.  Further, imagine that since the charges re-radiate in response there is some high-Q cavity which redirects the EM fields back to the ions.  However, we are not getting gravity “for free.”  All this does is to shift a certain amount of mass-energy from one place to another.  You need some sort of huge generator, which converts mass to energy and drives a dipole antenna that creates the EM fields.  The mass-energy is still present.  In fact before you turn the whole machine on you would have in fuel a huge amount of mass that would by itself have a gravity field equal to, or realistically greater than, the gravity field you are trying to generate.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-10T21:27:14.063" CommentCount="1" />
  <row Id="6654" PostTypeId="1" AcceptedAnswerId="6673" CreationDate="2011-03-10T21:42:25.973" Score="2" ViewCount="138" Body="&lt;p&gt;I've variously heard the shape of the universe being described as multi-dimensional, like a helix or mobius strip, and super string theorem seems to say there are lots of universes all piled up next to each other in vibrating planes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, can the edges of our universe ever join up? Are they already all joined up? as in is our universe really in the shape of a giant donut (yum). or is this shear lunacy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess I am assuming a model of the universe where the &quot;edge&quot; is not necessarily expanding constantly &quot;out&quot; into infinite amounts of empty vacuum, but rather one where the edges are the boundaries to that expansion (so that it seems from an observable position that the universe is expanding, when really its flexing inside these boundaries). As indicated in the comments no doubt an inaccurate description, but I was imagining it like a balloon.&lt;/p&gt;&#xA;" OwnerUserId="582" LastEditorUserId="582" LastEditorDisplayName="" LastEditDate="2011-03-10T21:57:51.360" LastActivityDate="2011-03-11T09:52:20.403" Title="Is it possible for one side of the universe to &quot;meet&quot; the other?" Tags="&lt;universe&gt;&lt;theory-of-everything&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="6655" PostTypeId="1" CreationDate="2011-03-10T22:09:33.497" Score="7" ViewCount="219" Body="&lt;p&gt;A few years ago, when i studied the casimir effect interpretation as the filtering out of vacuum modes with appropiate boundary conditions, i had the following dilemma; supposedly the derivation of the force between the walls was &lt;strong&gt;entirely&lt;/strong&gt; equivalent to just calculating Van der Waals forces. This, as it was argued, compelled us to not take too seriously the idea of a negative energy density of the space between the walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, i was left wondering what would happen if one would take the interpretation based on vacuum modes a bit further, and we could somehow switch on and off the permitivity $\epsilon$ of the conductor; what sort of long range radiation we would expect to see? &#xA;A possible experimental implementation would be if the layers where superconductive, and then a magnetic field would be switched on just to break the superconductive phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We probably need to use the Dirac-Heisenberg-Wigner formalism for these sort of dynamic quantum systems, but what would be intuitively expected to be detected?&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-08-24T18:34:16.397" Title="dynamic casimir effect" Tags="&lt;quantum-field-theory&gt;&lt;dhw-formalism&gt;&lt;casimir-effect&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="6656" PostTypeId="2" ParentId="6649" CreationDate="2011-03-10T22:29:10.177" Score="3" ViewCount="" Body="&lt;p&gt;The problem you describe is (mathematically) similar to &lt;a href=&quot;http://en.wikipedia.org/wiki/Blind_deconvolution&quot; rel=&quot;nofollow&quot;&gt;blind deconvolution&lt;/a&gt;. Given a signal which is the result of blurring an image (a linear operation) and adding noise, blind deconvolution tries to estimate the blur and the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As described &lt;a href=&quot;http://www.mathworks.com/products/demos/image/ipexblind/ipexblind.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, the blind deconvolution process consists roughly of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Guess the blurring function (transfer function)&lt;/li&gt;&#xA;&lt;li&gt;Construct an image consistent with your signal and your guess of the transfer function  &lt;/li&gt;&#xA;&lt;li&gt;Apply physically reasonable constraints to the constructed image. (For example, non-negativity).  &lt;/li&gt;&#xA;&lt;li&gt;Modify your guess of the blurring function to better satisfy these constraints  &lt;/li&gt;&#xA;&lt;li&gt;Goto 2.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It sounds like your idea would apply to step 3. I've never seen the K-K relations used this way, but I imagine they'd work just fine.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="2359" LastEditorDisplayName="" LastEditDate="2011-03-10T22:36:27.800" LastActivityDate="2011-03-10T22:36:27.800" CommentCount="6" />
  <row Id="6657" PostTypeId="2" ParentId="6650" CreationDate="2011-03-10T22:40:18.570" Score="1" ViewCount="" Body="&lt;p&gt;A slightly different answer to this is &quot;neutralisation&quot;. That is the free positive ions around (in gas clouds nearby maybe) will neutralise the charge. This is generally assumed to keep the charge of a Black Hole near to zero in astrophysical contexts.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-03-10T22:40:18.570" />
  <row Id="6658" PostTypeId="1" AcceptedAnswerId="6684" CreationDate="2011-03-10T23:46:46.640" Score="4" ViewCount="158" Body="&lt;p&gt;In a newtonian universe, one can talk of a &quot;time slice&quot;, that is, the state of the universe at a given point in (global) time. In a &quot;typical&quot; classical universe, a time slice would contain enough information to fully compute the state of the universe at any other point in time, backwards or forwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, disregarding what we know of quantum effects and talking purely of a general-relativistic universe, which concept is closest to a time slice?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give a better idea of what I'm after, consider this example: a straight line makes sense in a flat space, but the closest concept in non-flat space is a geodesic. In particular, a straight line &lt;em&gt;is&lt;/em&gt; a geodesic in flat space. Can a &quot;time slice&quot; be generalised meaningfully to general relativity, and if so, how?&lt;/p&gt;&#xA;" OwnerUserId="786" LastActivityDate="2011-03-11T12:29:28.817" Title="What is the closest general-relativistic equivalent of a &quot;time slice&quot;?" Tags="&lt;general-relativity&gt;&lt;time&gt;" AnswerCount="2" />
  <row Id="6659" PostTypeId="2" ParentId="6658" CreationDate="2011-03-10T23:50:45.243" Score="4" ViewCount="" Body="&lt;p&gt;See my answer to &lt;a href=&quot;http://physics.stackexchange.com/questions/2710/what-does-foliation-mean-in-the-context-of-a-foliation-of-spacetime&quot;&gt;this question&lt;/a&gt;, which is complementary to yours. The question there is what is a foliation, which as it turns out is the closest thing in GR to a time slice. In a word, foliation of spacetime is a choice of time slicing of spacetime. Unlike the idea of geodesics which are unique after choosing initial conditions, there is a large amount of freedom in choosing such foliations - at least locally (but globally, such time slicing might not even exist). According to GR, they are all equivalent.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-03-11T01:46:10.807" LastActivityDate="2011-03-11T01:46:10.807" />
  <row Id="6660" PostTypeId="2" ParentId="6363" CreationDate="2011-03-11T00:04:36.433" Score="-1" ViewCount="" Body="&lt;p&gt;The question is also dependent upon supersymmetry.  For the case of $E_6$, relevant for the current “mini-twistor revolution,” positive condition appears straight forwards.  The first Kahler form $K_1~=~ln(W(z)$ is computed by a three-form (or three $1,1$) forms from $h_{1,1}$ in the Hodge diamond, one element for a ${\bf\bar 27}$ of the $E_6$&#xA;$$&#xD;&#xA;W(z)~=~\int\omega\wedge\omega\wedge\omega&#xD;&#xA;$$&#xA;The holomorphic metric on the $E_6$ is given by this potential &#xA;$$&#xD;&#xA;g_{a{\bar b}}~=~-\partial_{z^a}\partial_{{\bar z}^b}W(z)&#xD;&#xA;$$&#xA;The $h_{1,2}$ and $h_{2,1}$, given by $\bf 27$ of $E_6$ determine a second Kahler form&#xA;$$&#xD;&#xA;e^{-iK_2}~=~\int\Omega_3\wedge\Omega^*_3&#xD;&#xA;$$&#xA;and the $\bf 27$ metric in generality is $g’_{a{\bar b}}~=~exp((k_2-K_1)/3) g_{a{\bar b}}$.  The $3,0$ forms determine a superpotential as well, which is related to $K_2$ as the third Betti number is given by $h_{2,1}$, which is the cycle $int\Omega_{3,0}~=~w$  $K_2$ may be computed as well as $log(Im(z^a\partial_aw(z)))$, since $b_3~=~2h_{2,1}~+~2$.  The superpotential is then the holomorphic function&#xA;$$&#xD;&#xA;W(z)~=~\phi^a\phi^b\phi^c\frac{\partial^3w(z)}{\partial z^a\partial z^b\partial z^c}&#xD;&#xA;$$&#xA;Supersymmetry in its unbroken phase will have a zero energy, which will hold for any topology of the CY manifold.  Broken supersymmetry will have positive energy, which is then mapped to the CY manifold.  The holomorphic functions are then given by the topology of the CY manifold.  However, in the Hodge diamond elements are chiral fields of $E_6$ which are positive.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-03-11T00:11:01.623" LastActivityDate="2011-03-11T00:11:01.623" />
  <row Id="6661" PostTypeId="2" ParentId="6537" CreationDate="2011-03-11T00:22:57.480" Score="1" ViewCount="" Body="&lt;p&gt;Actually, an observer can't measure even the angular momentum. For example, suppose the token ring is in the form of a figure 8. Then the times around the two directions are identical even if the token ring is rotating.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-11T00:22:57.480" CommentCount="1" />
  <row Id="6662" PostTypeId="1" AcceptedAnswerId="6670" CreationDate="2011-03-11T00:35:30.073" Score="2" ViewCount="209" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/B-L&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/B-L&lt;/a&gt; says that the difference between baryon number and lepton number is conserved.  Ordinary hydrogen has one of each, but turning it into helium releases only the binding energy.  Complete destruction of hydrogen would satisfy B-L and charge conservation, creating even more energy than fusion.  Is this possible or would it violate a different conservation law?&lt;/p&gt;&#xA;" OwnerUserId="2072" LastEditorUserId="2072" LastEditorDisplayName="" LastEditDate="2011-03-11T03:06:11.107" LastActivityDate="2011-03-11T04:47:28.790" Title="How much energy can be extracted from hydrogen?" Tags="&lt;special-relativity&gt;&lt;energy-conservation&gt;&lt;nuclear-physics&gt;&lt;quantum-chromodynamics&gt;&lt;fusion&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" />
  <row Id="6663" PostTypeId="2" ParentId="6650" CreationDate="2011-03-11T00:35:33.250" Score="3" ViewCount="" Body="&lt;p&gt;The coulomb force cannot be responsible, because given enough energy I can add more charge to the black hole... There was a related question recently:&#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/6209/paradoxical-interaction-between-a-massive-charged-sphere-and-a-point-charge/6628#6628&quot;&gt;Paradoxical interaction between a massive charged sphere and a point charge&lt;/a&gt;&#xA;To summarized my asnwer there: It isn't actually the Coulomb force that prevents the addition of charge to the black hole beyond its maximum, but by adding more charge more electrostatic binding energy is added. A lot! (since the black hole must have an incredibly chargedensity.) Therefore also its mass increases, which in turn leads to an increased maximum charge.....&lt;/p&gt;&#xA;" OwnerUserId="2470" LastActivityDate="2011-03-11T00:35:33.250" />
  <row Id="6664" PostTypeId="2" ParentId="6624" CreationDate="2011-03-11T00:37:03.880" Score="2" ViewCount="" Body="&lt;p&gt;If you &lt;a href=&quot;http://www.google.com/search?q=%22quartic+equation%22+site%3aarxiv.org&quot; rel=&quot;nofollow&quot;&gt;google around&lt;/a&gt; you can find applications. For example, applications of the quartic equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonlinear Landau-Zener Tunnelling in Coupled Waveguide Arrays&#xA;&lt;a href=&quot;http://arxiv.org/abs/1008.1358&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1008.1358&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But they are rare.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-11T00:37:03.880" />
  <row Id="6665" PostTypeId="2" ParentId="6533" CreationDate="2011-03-11T00:46:32.217" Score="1" ViewCount="" Body="&lt;p&gt;I am not an expert on this, but the two cases indicate a change of phase.  The process &#xA;$$&#xD;&#xA;n~\rightarrow~ p+l+\overline{\nu_l},\quad p + l \rightarrow n + \nu_l&#xD;&#xA;$$ &#xA;might be thought of as two steps in a process where a neutron in a high energy state in the Fermi surface, we denote by $n^*$ falls to a low energy by $n^*~\rightarrow~n~+~\nu~+~\bar\nu$.   For a lower mass density $&amp;lt;~10^{15}g/cm^3$ the neutron state, which is a form of liquid as I understand, changes to the cooler phase with nuclear matter.  Using the shorter notion above a neutron scatters off a nucleus, so the $n^*$ neutron scatters off a nucleus $N$ so &#xA;$$&#xD;&#xA;n^*~+~N~\rightarrow~n~+~N~+~\nu~+~\bar\nu&#xD;&#xA;$$&#xA;This process however competes for the process $n^*~+~N~\rightarrow~N’$ which is an isotope of the $N$.  This is a nuclear interaction which is strong than weak interactions, and by that fact probably dominates the neutrino emitting process governed by weak interaction.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-11T00:46:32.217" CommentCount="1" />
  <row Id="6666" PostTypeId="2" ParentId="6624" CreationDate="2011-03-11T01:10:26.843" Score="2" ViewCount="" Body="&lt;p&gt;If you wanna find the time it takes to travel some distance, and your acceleration is increasing linearly in time (constant jerk), then you'd have to solve a cubic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The jerk-equation is also used in electric circuits, jerk circuits.&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-03-11T01:10:26.843" />
  <row Id="6667" PostTypeId="1" CreationDate="2011-03-11T02:35:51.273" Score="0" ViewCount="150" Body="&lt;p&gt;Lets say you have a free particle in a rotating frame of reference with constant angular velocity $\mathbf{\omega}$. By free, I mean there are no real forces on it. Lets call the moving system &quot;primed&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation of the motion is $m\mathbf{a} = \mathbf{F} + \mathbf{F_{fic}}$ but $F = 0$ and the only fictitious forces acting on this particle are the centrifugal force and the coriolis force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to double checked,&#xA;1) there is no azimuthal (or transverse) force since there is no angular acceleration. Correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) And ofcourse there is no translation force since the origin of the primed and nonprimed systems are the same? Right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay so lets assume I am right (which I am since I am following an example from a text :P)&#xA;So we have&#xA;$$m\mathbf{a} = −m \mathbf{\omega} &amp;#215; (\mathbf{\omega} &amp;#215; \mathbf{r&amp;#39;}) − 2m \mathbf{\omega} &amp;#215; \mathbf{v}&amp;#39;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets making the rotating axis fixed with the z axis so $\mathbf{w} = \omega \cdot \mathbf{e_z}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega = \omega \mathbf{e_z} $$ $$\mathbf{r} = x \mathbf{e_x} + y \mathbf{e_y} + z \mathbf{e_z} $$&#xA;and similarly v is pretty much r, but x = x dot, and such. Don't know how to write x dot with latex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my main problem is that for example lets say we are calculating the x component of everything. So then in the equation the coriolis force has a cross product between $e_z$ and $e_x$. This is $e_y$. I get that, but how do I know what sign is it? How come $e_z$ and $e_y$ gives me A NEGATIVE $e_x$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am on page 11 of this pdf, &lt;a href=&quot;http://www.mech.kth.se/~hanno/RelativeMotion.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.mech.kth.se/~hanno/RelativeMotion.pdf&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also confused about how $e_x$ crossed with $e_z$ gives you &quot;y dot&quot; and not &quot;x dot&quot; Look at the pdf (specifically the coriolis force term). &lt;/p&gt;&#xA;" OwnerUserId="551" LastActivityDate="2011-03-12T00:22:49.890" Title="vector cross products" Tags="&lt;vectors&gt;&lt;analyticity&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="6668" PostTypeId="2" ParentId="387" CreationDate="2011-03-11T02:54:01.163" Score="1" ViewCount="" Body="&lt;p&gt;I'm not sure this should be an answer - it is really an anti-answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Quantum Relativity, David Finkelstein has a list of analogies between QM and relativity, detailing &quot;an extended parallel between the structures and developments of relativity and quantum theory&quot;.   (Section 1.4.2)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, he does have GR in mind when he speaks of relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While he points out some deep similarities, the rest of the book explores in depth, in ways that would thrill only a theorist, the underlying nature of each, analogies and differences.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone interested in the relation of QM and GR would benefit from a browse through this book, though it is hardly the only one that should be read.&lt;/p&gt;&#xA;" OwnerUserId="353" LastActivityDate="2011-03-11T02:54:01.163" />
  <row Id="6669" PostTypeId="2" ParentId="6624" CreationDate="2011-03-11T04:32:12.750" Score="2" ViewCount="" Body="&lt;p&gt;FWIIW, coincidentally last month I ran into an algebraic cubic equation when finding the spherically symmetric spacetime equivalent to the flat matter-only FLRW model (the Einstein-de Sitter model).  Specifically when trying to express comoving time in terms of non-comoving time and radial coordinate.  The cubic equation was of the kind with three real roots that must be found by way of intermediate complex numbers, so that none was reducible to a real algebraic expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did learn something &quot;qualitative&quot; from that fact: that expansion of space is the only possible conceptual framework to work with, except for the Milne and de Sitter cases. I had become interested in the topic after reading an exchange of papers mainly between Abramowicz and Chodorowski a few years ago, which had left the issue unsettled.&lt;/p&gt;&#xA;" OwnerUserId="2085" LastActivityDate="2011-03-11T04:32:12.750" />
  <row Id="6670" PostTypeId="2" ParentId="6662" CreationDate="2011-03-11T04:47:28.790" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_atom&quot; rel=&quot;nofollow&quot;&gt;Hydrogen-1&lt;/a&gt; (i.e. hydrogen with no neutrons) has a mass of 1.007825 AMU. To get energy from fusing it you have to preserve baryon number. So you look for the atom that has the lowest mass per nucleon (i.e. lowest mass average over the protons and neutrons that make it up).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This lowest (most stable) atom turns out to be &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron-56&quot; rel=&quot;nofollow&quot;&gt;iron-56&lt;/a&gt;, which has a mass of 55.9349375(7) AMU, or 0.99883817 AMU per nucleon. The difference, $1.007825 - 0.99883817 = 0.008987$ AMU is the energy you get per hydrogen atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To convert &lt;a href=&quot;http://en.wikipedia.org/wiki/Unified_atomic_mass_unit&quot; rel=&quot;nofollow&quot;&gt;AMU&lt;/a&gt; to Joules, first convert the AMU to kg by multiplying by $1.66054\times 10^{-27}$ kg/AMU, then convert to energy (Joules) by $E=mc^2$ with $c$ the speed of light = $3\times 10^8$m/s. The result is $1.343\times 10^{-12}$ Joules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the above numbers, a kilogram of hydrogen-1 has $5.97538\times 10^{26}$ atoms and so the energy per kilogram is $8.025\times 10^{14}$ Joules or 0.1918 Megatons.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-11T04:47:28.790" CommentCount="5" />
  <row Id="6671" PostTypeId="2" ParentId="6667" CreationDate="2011-03-11T05:14:25.473" Score="3" ViewCount="" Body="&lt;p&gt;Some rules for the cross product are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  A \times (\beta B + \gamma C) = \beta (A \times B) + \gamma (A \times C)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (e_x \times e_y) \cdot e_z = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (B \times C)\cdot A = d$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is antisymmetric with respect to transpositions, meaning that if you take any two of those vectors and switch them, you multiply $d$ by $-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These rules uniquely define the cross product.  You can derive other rules from them, for example that the cross product of any vector with itself is zero or more generally that the magnitude of the cross product is the product of the magnitudes times the sine of the angle between the vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the derivation of the specific result you requested:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (e_x \times e_y) \cdot e_z = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is by assumption.  Use rule 3 to switch $e_x$ and $e_z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (e_z \times e_y) \cdot e_x = -1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that the $x$-component of $e_z \times e_y$ is -1.  It remains to show that the $y$ and $z$-components are zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (e_z \times e_y) \cdot e_y = c_y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c_y$ is the unknown $y$-component of $e_z \times e_y$.  Use rule 3 to switch the two appearances of $e_y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (e_z \times e_y) \cdot e_y = -c_y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two expressions are the same, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_y = -c_y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $c_y = 0$, so the $y$-component of $e_z \times e_y$ must be zero.  Similarly for the $z$-component.  Hence &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e_z \times e_y = -e_x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega \times v = - v \times \omega$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Coriolis force, this equation is telling you that the direction of the Coriolis force depends on the velocity.  If you turn around and go the other way, the Coriolis force switches directions.  The Coriolis force pushes things clockwise or counter-clockwise at a certain place, rather than always left or always right.  Hurricanes in the Northern hemisphere all rotate the same way, and the would continue to even if they moved different directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your last question, you can derive from the above that $e_x \times e_z = - e_y$, not $\dot{y}$.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-03-11T05:14:25.473" />
  <row Id="6672" PostTypeId="2" ParentId="5277" CreationDate="2011-03-11T07:00:02.013" Score="4" ViewCount="" Body="&lt;p&gt;Yes Sam, there definitely is electric field reshaping in the wire. Strangely, it is not talked about in hardly any physics texts, but there are surface charge accumulations along the wire which maintain the electric field in the direction of the wire. (Note: it is a surface charge distribution since any extra charge on a conductor will reside on the surface.) It is the change in, or gradient of, the surface charge distribution on the wire that creates, and determines the direction of, the electric field within a wire or resistor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the surface charge density on the wire near the negative terminal of the battery will be more negative than the surface charge density on the wire near the positive terminal. The surface charge density, as you go around the circuit, will change only slightly along a good conducting wire (Hence the gradient is small, and there is only a small electric field). Corners or bends in the wire will also cause surface charge accumulations that make the electrons flow around in the direction of the wire instead of flowing into a dead end. Resistors inserted into the circuit will have a more negative surface charge density on one side of the resistor as compared to the other side of the resistor. This larger gradient in surface charge distribution near the resistor causes the relatively larger electric field in the resistor (as compared to the wire). The direction of the gradients for all the aforementioned surface charge densities determine the direction of the electric fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is very fundamental, and is often misinterpreted or disregarded by people. We are all indoctrinated to just assume that a battery creates an electric field in the wire. However, when someone asks &quot;how does the field get into the wire and how does the field know which way to go?&quot; they are rarely given a straight answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A follow up question might be, &quot;If nonzero surface charge accumulations are responsible for the size and direction of the electric field in a wire, why doesn't a normal circuit with a resistor exert an electric force on a nearby pith ball from all the built up charge in the circuit?&quot; The answer is that it does exert a force, but the surface charge and force are so small for normal voltages and operating conditions that you don't notice it. If you hook up a 100,000V source to a resistor you would be able to measure the surface charge accumulation and the force it could exert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's one more way to think about all this (excuse the length of this post, but there is so much confusion on this question it deserves appropriate detail). We all know there is an electric field in a wire connected to a battery. But the wire could be as long as desired, and so as far away from the battery terminals as desired. The charge on the battery terminals can't be directly and solely responsible for the size and direction of the electric field in the part of the wire miles away since the field would have died off and become too small there. (Yes, an infinite plane of charge, or other suitably exotic configurations, can create a field that does not decrease with distance, but we are not talking about anything like that.) If the charge near the terminals does not directly and solely determine the size and direction of the electric field in the part of the wire miles away, some other charge must be creating the field there (Yes, you can create an electric field with a changing magnetic field instead of a charge, but we can assume we have a steady current and non-varying magnetic field). The physical mechanism that creates the electric field in the part of the wire miles away is a small gradient of the nonzero surface charge distribution on the wire. And the direction of the gradient of that charge distribution is what determines the direction of the electric field there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a rare and absolutely beautiful description of how and why surface charge creates and shapes the electric field in a wire refer to the textbook:&#xA;&quot;Matter and Interactions: Volume 2 Electric and Magnetic Interactions&quot; &#xA;by Chabay and Sherwood,&#xA;Chapter 18 &quot;A Microscopic View of Electric Circuits&quot; &#xA;pg 631-640.&lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="1040" LastEditorDisplayName="" LastEditDate="2011-03-11T07:32:11.657" LastActivityDate="2011-03-11T07:32:11.657" />
  <row Id="6673" PostTypeId="2" ParentId="6654" CreationDate="2011-03-11T09:52:20.403" Score="1" ViewCount="" Body="&lt;p&gt;The short answer is no!&#xA;More about imagination how space of our universe looks (and how expanding) today and in past check here :&lt;a href=&quot;http://map.gsfc.nasa.gov/index.html&quot; rel=&quot;nofollow&quot;&gt;WMAP&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Hidq5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2410" LastActivityDate="2011-03-11T09:52:20.403" CommentCount="3" />
  <row Id="6674" PostTypeId="1" CreationDate="2011-03-11T09:55:58.710" Score="4" ViewCount="353" Body="&lt;p&gt;Basically, how long does it take for electricity to determine there is a closed circuit and how does it know that the circuit exists?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious to know how it knows there is a closed circuit at any length!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For clarification of the thought experiment, picture an electrical source of some type with a 1 light year wire coming from both terminals and connected in the middle.  The wires are of zero resistance and capacitance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If electrons were to start flowing immediately then information (one end of the circuit knowing the ciruit is now closed at the opposite end) has obviously travelled FTL which is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been pointed out that at least some capacitance is required, so if required consider that as a possibility in the hypothetical setup. i'm not quite sure how that affects this situation due to my lack of knowledge.  I'm trying to not get to bogged down in the setup and trying to focus more on the 'how does the circuit know its complete &amp;amp; how long does it take to determine it (particularly when the distance is long)'&lt;/p&gt;&#xA;" OwnerUserId="2482" LastEditorUserId="2482" LastEditorDisplayName="" LastEditDate="2011-03-11T12:33:03.380" LastActivityDate="2011-03-20T03:31:52.483" Title="how long would it take for electricity to begin flowing from one terminal via a 1 light year long wire" Tags="&lt;electricity&gt;&lt;speed-of-light&gt;" AnswerCount="4" CommentCount="9" />
  <row Id="6675" PostTypeId="1" AcceptedAnswerId="6692" CreationDate="2011-03-11T10:10:41.100" Score="2" ViewCount="267" Body="&lt;p&gt;In a scientific article, should I write &quot;3m&quot;, &quot;3 m&quot;, &quot;3 meters&quot;, or &quot;3 [meters]&quot; ? &lt;/p&gt;&#xA;" OwnerUserId="1996" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-11T19:08:03.350" LastActivityDate="2011-03-13T17:55:45.303" Title="How to write units?" Tags="&lt;soft-question&gt;&lt;si-units&gt;&lt;publications&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="6679" PostTypeId="2" ParentId="6675" CreationDate="2011-03-11T10:49:13.230" Score="1" ViewCount="" Body="&lt;p&gt;I'd say it depends on how well established the units are. Commonly used SI units like m, s, kg, eV, etc. can certainly be used without writing their full name. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the distance between vaule and unit is concerned I think it is common to use one space in between (as in 1 m; the space should be unbreakable). For combined units (like kg/m³) there is usually no space in between the units if there is a scale modification (like the k in kg) or a mathematical symbol (the / in kg/m³ or the ³ in m³). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand if there is an ommited multiplication sign (as in kg*m/s) there is usually a little space in between which is smaller than the normal spacing in text.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further info you can read here &lt;a href=&quot;http://www.ece.ucdavis.edu/~jowens/commonerrors.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ece.ucdavis.edu/~jowens/commonerrors.html&lt;/a&gt; . Down at the bottom is a remark to units.&lt;/p&gt;&#xA;" OwnerUserId="2469" LastActivityDate="2011-03-11T10:49:13.230" />
  <row Id="6680" PostTypeId="1" CreationDate="2011-03-11T11:04:19.353" Score="3" ViewCount="196" Body="&lt;p&gt;I want to make a finite element analysis of a cold airflow through warmer pipes. In particular I want to see how the pipes cool down and the air heats up, as it travels through the pipes. Wich are the equations and boundary conditions that I have to consider?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that the velocity field is already given, call it $v$. &#xA;Currently I solve the following equation&#xA;$$&#xD;&#xA;\rho c_p(\frac{\partial T}{\partial t} + v\cdot \nabla T) - k \Delta T = 0&#xD;&#xA;$$&#xA;($T\ $ temperature, $\rho\ $ density, $c_p\ $ heat capacity at constant pressure, $k\ $ heat conductivity) over the whole domain, where inside the pipes I use the given velocity field and in the pipe material I assume $v = 0$. The constants like $\rho\ $ jump on the boundary. I have an initial condition of the same temperature everywhere and then apply a Dirichlet BC at the pipe inlet of a colder temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that this describes convective heat transfer and for $v=0$ the equation reduces to conduction heat transfer. But does it describe my problem correctly? I do not have a boundary condition on the pipe/air boundary. Do I need one? Which one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heatflow inside the metal pipe seems to be very slow, I expected it to be somewhat faster. The thermal diffusivity, i.e. the term $\frac{k}{\rho c_p}$ is of the magnitude $10^{-5}$, I took this value from the literature. However this makes the diffusion part of the equation very slow. Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="2480" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-14T10:08:35.157" LastActivityDate="2011-03-14T15:10:12.647" Title="Mixed conductive and convective heat transfer" Tags="&lt;fluid-dynamics&gt;&lt;computational-physics&gt;&lt;cfd&gt;" AnswerCount="3" />
  <row Id="6681" PostTypeId="2" ParentId="6674" CreationDate="2011-03-11T11:06:24.273" Score="-2" ViewCount="" Body="&lt;p&gt;The &lt;strong&gt;fastest response time&lt;/strong&gt; on the other side of superconductor with no capacity and no inductivity is light speed. And this is valid only for DC (direct current). If we are using alternative current than we have resonant frequency, which is at ideal superconductor equal superconductor length divide c(light speed) . Depend of current source, which supply the circuit, &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrical_impedance&quot; rel=&quot;nofollow&quot;&gt;electrical impedance (which is composed from capacity, inductivity and resistance)&lt;/a&gt; and length of real conductor (wire) we can calculate response time. Current will start to flow immediately if electric potentials are different on both sides of conductors before you do short-circuit.&lt;/p&gt;&#xA;" OwnerUserId="2410" LastEditorUserId="2410" LastEditorDisplayName="" LastEditDate="2011-03-11T13:01:26.130" LastActivityDate="2011-03-11T13:01:26.130" CommentCount="10" />
  <row Id="6682" PostTypeId="1" AcceptedAnswerId="6707" CreationDate="2011-03-11T11:33:36.133" Score="3" ViewCount="223" Body="&lt;p&gt;Is there any good software for construction optical path's in geometrical optics. More specifically I want features like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;draw $k \in \mathbb{N}$ objects $K_1,\dots,K_n$ with indices of refraction $n_1,\dots,n_k$ and light sources $l \in \mathbb{N}$ light sources $L_1,\dots,L_l$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- draw for each some of these light sources light rays and the program constructs the optical path of those rays through $K_1,\dots,K_n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- instead of drawing the light rays, draw a point and the program construction the optical path between the light sources and the given point&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;draw some specific optical elements like lenses, (concave, convex) mirrors and do things like above&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;construct automatically virtual images&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;3D drawings would be fine&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would prefer free software for linux.&lt;/p&gt;&#xA;" OwnerUserId="667" LastActivityDate="2011-03-11T17:15:56.133" Title="Software for geometrical optics" Tags="&lt;software&gt;&lt;geometric-optics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6683" PostTypeId="2" ParentId="6682" CreationDate="2011-03-11T11:47:58.037" Score="2" ViewCount="" Body="&lt;p&gt;I am not aware of a simple program to do what you ask. What comes to mind are either raytracers or finite element methods (FEM).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a list of raytracing software: &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_ray_tracing_software&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/List_of_ray_tracing_software&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not aware of any free FEM software but for an overview of the principle go here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_element&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Finite_element&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2469" LastActivityDate="2011-03-11T11:47:58.037" />
  <row Id="6684" PostTypeId="2" ParentId="6658" CreationDate="2011-03-11T12:29:28.817" Score="2" ViewCount="" Body="&lt;p&gt;The actual equivalent to a Newtonian &quot;slice of time&quot; in GR is a &quot;space-like hypersurface&quot;. It is a 3 dimensional hypersurface in 4D space-time (hence the hypersurface). It is &quot;space-like&quot; because any two points are connected by a &quot;space-like&quot; path in the metric. This means that they are independent from a causality perspective, and can form the basis of forward predictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is a global hypersurface (ie doesnt just stop somewhere, or miss out regions) then it is called a &quot;Cauchy surface&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As remarked elsewhere there is more choice in GR involved. The simplest way to see this is to note that Newtonian theory is $E^4$ with (x,y,z,t) and once t= const is chosen at the space origin, say, the entire hypersurface is now determined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In say Minkowski space (the nearest GR equivalent to Newtonian space) there is a choice of timelike vector to be made first. There are 4 degrees of freedom here, but there is the timelikeness restriction and the restriction on time direction. (Physically this is a choice of Inertial Frame.) This allows a foliation to form. To obtain an actual hypersurface (hyperplane in this case) we can choose one point P on that hypersurface (ie a specific time in the Frame). Now the timelike vector is a normal to that hyperplane and we can determine whether an arbitrary point (x,y,z,t) is on that hyperplane. There will be 3 degrees of freedom as expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GR generalises the Minkowski example in that the hypersurface might be curved and not flat, requiring an infinite number of independent points to identify it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One further generalisation is that in Relativity we have a metric which allows null distances between different points. Thus we can have a &quot;null hyperplane&quot; and &quot;null hypersurface&quot; too, which are quasi-generalisations of the timelike slice Newtonian idea.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-03-11T12:29:28.817" />
  <row Id="6685" PostTypeId="2" ParentId="6680" CreationDate="2011-03-11T12:45:31.870" Score="1" ViewCount="" Body="&lt;p&gt;To describe this correctly, You have to include some realistic fluid dynamics. &#xA;For low air speed You will have a laminar flow with a parabolic velocity profile, &#xA;whereas at higher speeds (Reynolds Number!) You will have a laminar boundary layer &#xA;at the wall of the tube(conduction) , more inside there is turbulence (convection). &#xA;To make things complicated, the boundary layers thickness is a function of air speed and temperature (viscosity). &#xA;A thorough analysis includes &lt;strong&gt;Reynolds&lt;/strong&gt;, &lt;strong&gt;Prandtl&lt;/strong&gt; and &lt;strong&gt;Nusselt&lt;/strong&gt; Numbers. &#xA;Google for &quot;heat transfer&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crucial is the boundary layers thickness, if You know that, the rest is not worth &#xA;modelling, ist very easy to calculate. On the other hand, there is no method to calculate &#xA;when and where You have turbulence, exept the very coarse estimation on base of Reynolds Number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However this makes the diffusion part&#xA;  of the equation very slow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The value is ok, But this value does not make the diffusion &quot;slow&quot;, this value makes the temperature difference low. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, exept for vaporizing or condensing liquids on both sides of a metal wall, the temperature difference across the wall is neglegible.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-14T15:10:12.647" LastActivityDate="2011-03-14T15:10:12.647" CommentCount="3" />
  <row Id="6686" PostTypeId="1" AcceptedAnswerId="6695" CreationDate="2011-03-11T14:01:18.280" Score="0" ViewCount="226" Body="&lt;p&gt;Can we say that we are crystals because just like crystals we are made up of very small unit (cell) making up almost the same shape (our body) everywhere.&lt;/p&gt;&#xA;" OwnerUserId="847" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-12T06:37:47.087" LastActivityDate="2011-03-12T12:07:33.607" Title="Are we crystals?" Tags="&lt;solid-state-physics&gt;&lt;material-science&gt;&lt;definition&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="6687" PostTypeId="1" AcceptedAnswerId="6694" CreationDate="2011-03-11T14:17:32.260" Score="5" ViewCount="269" Body="&lt;p&gt;Why massless particles have zero chemical potential?&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2011-03-12T11:18:11.610" Title="Chemical potential of particles with zero mass" Tags="&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="15" FavoriteCount="1" />
  <row Id="6688" PostTypeId="2" ParentId="6674" CreationDate="2011-03-11T14:21:51.193" Score="2" ViewCount="" Body="&lt;p&gt;You need to analyze the circuit as a transmission line to understand how signals propagate along the wires to other parts of the circuit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Transmission_line&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Transmission_line&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The velocity factor is the ratio of the velocity of a voltage signal traveling along the transmission line, to the velocity of light. A typical value for twin lead is 82%:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Wave_propagation_speed&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Wave_propagation_speed&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On connecting the source to the transmission line, electrons will start to flow from the negative terminal of the source to the wire it's connected to, and from the other wire to the positive terminal of the source. This process propagates down each wire at 82% the speed of light in the case of twin lead. So after 1/82% = 100/82 years, curent will start to flow along that part of the circut 1 light year away from the source, using your example.&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="1345" LastEditorDisplayName="" LastEditDate="2011-03-11T19:22:32.920" LastActivityDate="2011-03-11T19:22:32.920" CommentCount="1" />
  <row Id="6689" PostTypeId="2" ParentId="6674" CreationDate="2011-03-11T14:50:19.427" Score="3" ViewCount="" Body="&lt;p&gt;There is absolutely nothing in any electrical circuits which is nonlocal. Everything is perfectly local and complies with all the local laws of physics. An element of an electrical circuit only &quot;knows&quot; about its immediate vicinity and it can affect or can be affected by the conditions in its immediate vicinity. About your specific question, let me say that if you close a circuit at some point then it takes finite nonzero time for an element of the circuit to &quot;feel&quot; it. The effect of closing at a particular point in the circuit will cause its immediate neighborhood to come to the same potential. That means an elementary length in both sides of the joint now comes to the same potential which in turn implies that current will flow in that elementary length (but not throughout the conductor for a very long conductor). Immediately the next neighborhood elements also come to the same potential. So more parts of the conductor will come to the same potential. All these takes nonzero time. For a one light year long conductor the whole of the conductor will take at least one year to come to the same potential in all of its parts. There is no way you can get an FTL effect for sure.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-03-14T15:02:26.833" LastActivityDate="2011-03-14T15:02:26.833" CommentCount="3" />
  <row Id="6690" PostTypeId="1" CreationDate="2011-03-11T14:51:44.150" Score="1" ViewCount="164" Body="&lt;p&gt;I'm trying to get a grasp of the power within a single hydrogen atom (e.g. the power released when whatever process happens in a hydrogen bomb)... If we could enlarge a hydrogen atom up to the size of your fist, how much energy would it give off?&lt;/p&gt;&#xA;" OwnerUserId="2247" LastActivityDate="2011-05-10T15:44:58.427" Title="Scaled energy output of a hydrogen atom" Tags="&lt;atoms&gt;&lt;power&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6692" PostTypeId="2" ParentId="6675" CreationDate="2011-03-11T15:07:51.030" Score="10" ViewCount="" Body="&lt;p&gt;There is a standard set of rules for writing units in the SI. &lt;a href=&quot;http://physics.nist.gov/cuu/Units/rules.html&quot; rel=&quot;nofollow&quot;&gt;NIST&lt;/a&gt; and &lt;a href=&quot;http://www.aip.org/pt/guide/metric.html&quot; rel=&quot;nofollow&quot;&gt;Physics Today&lt;/a&gt; both have summaries of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, every standard unit has a standard abbreviation, which you're supposed to use. Unit names are set in roman, not italic, type, to distinguish them from variables. That is, $3\,m$ would mean three times the quantity represented by the variable $m$, but $3\,{\rm m}$ means 3 meters. Be careful to avoid ambiguity with compound units. Astrophysicists used to write ${\rm km/s/Mpc}$ a lot for the units for Hubble's constant, but that's bad. Better to write ${\rm km/(s\ Mpc)}$, and even better is ${\rm km\ s^{-1}\ Mpc^{-1}}$. There's a lot more at those Web sites.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-11T15:07:51.030" CommentCount="7" />
  <row Id="6693" PostTypeId="2" ParentId="6690" CreationDate="2011-03-11T15:08:28.950" Score="1" ViewCount="" Body="&lt;p&gt;One can not build a hydrogen bomb with a single hydrogen atom since it requires fusion of more than one nuclei. Neither one can &quot;enlarge&quot; a hydrogen atom. An atom of hydrogen is made of one electron and one proton. There are certain specific orbitals around the nucleus whose sizes are fixed. One just can not rescale it. The size of any atom is fixed by fundamental physical constants. The energy of the hydrogen bomb comes from fusion of hydrogen atoms to form helium atoms. The final mass of the system is less than the initial mass and the lost mass is converted to energy by the equation $E = mc^2$ where $m$ is the mass difference.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-03-11T15:08:28.950" />
  <row Id="6694" PostTypeId="2" ParentId="6687" CreationDate="2011-03-11T15:26:46.053" Score="10" ViewCount="" Body="&lt;p&gt;Massless particles don't always have zero chemical potential. Suppose that you have a box full of photons and other particles, and it's possible for the photons to exchange energy with other particles, but the number of photons cannot change. Then the system will reach a thermal equilibrium in which the photons are described by a Bose-Einstein distribution with (in general) a nonzero chemical potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this doesn't usually happen with photons is that the number of photons is often not conserved in situations like this. If there are photon-number-changing processes, then the equilibrium state for the photons will have zero chemical potential (since otherwise entropy could go up by creating or destroying a photon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, the rules are that the chemical potential must be zero if particle-number-changing interactions are possible, but not otherwise. That distinction often coincides with the massless or massive nature of the particles, but not always.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, there was a period of time in the early Universe when we were in precisely this situation: photons could thermalize via Compton scattering with electrons, but at the temperature and density at the time, photon-number-changing processes essentially did not occur. That means that the cosmic microwave background radiation today could have a nonzero chemical potential. People have tried to measure the chemical potential, but as it turns out it's consistent with zero to quite good precision. This makes sense, as long as the photons and electrons came into thermal equilibrium at an earlier epoch (when photon-number-changing processes did occur), and nothing happened during the later epoch to mess up that equilibrium. Various theories in which particle decays inject energy into the Universe during the constant-photon-number epoch are ruled out by this observation.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-11T15:26:46.053" CommentCount="5" />
  <row Id="6695" PostTypeId="2" ParentId="6686" CreationDate="2011-03-11T15:33:08.033" Score="5" ViewCount="" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, to argue if a definition applies, we must first agree on a definition. &lt;a href=&quot;http://en.wikipedia.org/wiki/Crystal&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; gives this one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A crystal or crystalline solid is a&#xA;  solid material whose constituent&#xA;  atoms, molecules, or ions are arranged&#xA;  in an orderly repeating pattern&#xA;  extending in all three spatial&#xA;  dimensions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Humans are certainly solid-ish, and our constituent molecules are arranged in a somewhat orderly pattern in all three dimensions. However, I think we fail the 'repeating' portion of this definition. If you want to use a more broad definition of a crystal, link to it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, supposing the answer was 'yes'. What are the practical predictions which follow from this assertion? We certainly don't diffract x-rays into a regular grid, for example.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2011-03-11T15:33:08.033" CommentCount="2" />
  <row Id="6696" PostTypeId="2" ParentId="6687" CreationDate="2011-03-11T15:40:49.597" Score="1" ViewCount="" Body="&lt;p&gt;Photons require matter in order to come into thermal equilibrium (if you ignore the negligible contribution from photon-photon scattering). This means that the particle number is not conserved. This further implies that in order for the free energy to be minimum (for a given temperature and volume), you also need $$\partial F/\partial N=0$$ because now $N$ can vary too. Since $$(\partial F/\partial N)_{T,V}=\mu$$ we have $$\mu=0$$.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-03-11T15:40:49.597" />
  <row Id="6697" PostTypeId="2" ParentId="6686" CreationDate="2011-03-11T16:01:10.243" Score="2" ViewCount="" Body="&lt;p&gt;Analogies between crystals and living organisms are a popular thought since centuries. &#xA;The self-organizing of a structureless liquid to a well ordered crystal (with negative deltaS!) leads to such thoughts. The creation of live from &quot;nothing&quot; seemingly adds to this too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW: there are crystalline &quot;pieces&quot; in many animals like shells and bones. And muscle fibers or tendons are likely to produce X-Ray fibre diagrams. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Alchimists had such ideas, miners centurys ago believed that ores (often/mostly crystalline)  &quot;grow&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-12T12:07:33.607" LastActivityDate="2011-03-12T12:07:33.607" />
  <row Id="6698" PostTypeId="1" CreationDate="2011-03-11T16:07:19.683" Score="1" ViewCount="154" Body="&lt;p&gt;Is it true that the expansion of space time cause the CMB to become microwaves from a shorter wavelength. If it is has the amplitude been increased? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing as the amplitude has decreased; why hasn't it increased (/&quot;stretched&quot;) in the same way the wavelength has?&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-03-11T19:06:00.090" LastActivityDate="2011-05-04T13:43:13.560" Title="Effect of expansion of space on CMB" Tags="&lt;astrophysics&gt;&lt;space-expansion&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="6699" PostTypeId="2" ParentId="6680" CreationDate="2011-03-11T16:25:48.567" Score="4" ViewCount="" Body="&lt;p&gt;As &lt;a href=&quot;http://physics.stackexchange.com/questions/6680/cfd-mixed-conductive-and-convective-heat-transfer/6685#6685&quot;&gt;Georg says&lt;/a&gt;, there are a set of dimensionless numbers that control which physics are important under which conditions. To the set he suggested I'll add the Grashof number&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are existing treatment of pipes embedded in a bulk medium (because this problem comes up over and over again...), and they are very complete if the pipe is either horizontal or vertical. Using these will insure that you are using the same definition of the dimensionless numbers as the people who determined (experimentally) where important changes occur (because you've made the same choices of ambiguous length scales (i.e. diameter or radius)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last time I worked thee kind of problems I used a couple of textbooks borrowed from the better half:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Çengel, Y. &lt;em&gt;Heat Transfer: A practical Approach&lt;/em&gt; 1998 &lt;/li&gt;&#xA;&lt;li&gt;Levenspiel, O. &lt;em&gt;Engineering Flow and Heat Exchange&lt;/em&gt; 1998&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and a dusty reference from the bowels of the our university library:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Perry and Chilton &lt;em&gt;Chemical Engineers’ Handbook&lt;/em&gt; 1973&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-03-11T16:40:05.183" LastActivityDate="2011-03-11T16:40:05.183" CommentCount="5" />
  <row Id="6700" PostTypeId="2" ParentId="6698" CreationDate="2011-03-11T16:29:43.970" Score="1" ViewCount="" Body="&lt;p&gt;Yes to the first part of your question. This phenomenon is known as cosmological redshift. Also due to the increase in the volume due to cosmlogical expansion the same amount of photons, as were present during the CMB, now have to occupy a much larger region. Consequently the average temperature (&quot;amplitude&quot; in your words) of such a gas drops from a high of $T \sim 150,000$ deg. Kelvin (corresponding to $T=E/k_B$ with $E$ being the ionization energy of hydrogen (13.6 ev) and $k_B$ is Boltzmann's constant $k_B = 8.6 \times 10^{-5}$ eV/Kelvin) to the currently observed $T&amp;#39;=2.7$ deg. Kelvin.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-11T16:45:49.150" CommentCount="6" />
  <row Id="6701" PostTypeId="2" ParentId="6686" CreationDate="2011-03-11T16:31:13.327" Score="4" ViewCount="" Body="&lt;p&gt;Certainly not. Crystals by definition have some translation invariant lattice structure and humans do not.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-03-11T16:31:13.327" />
  <row Id="6702" PostTypeId="2" ParentId="6698" CreationDate="2011-03-11T16:32:08.373" Score="1" ViewCount="" Body="&lt;p&gt;Just to be clear: by &quot;amplitude&quot; you mean the amplitude of a classical electromagnetic wave -- that is, the peak value of the electric field -- right? In that case, the answer is that the amplitude goes down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For definiteness, let's consider a wave packet of electromagnetic radiation with some fairly well-defined wavelength. At some early time, it has a wavelength $\lambda_1$ and energy $U_1$. (I'm not calling it $E$ because I want to reserve that for the electric field.) After the Universe has expanded for a while, it has a longer wavelength $\lambda_2$ and a smaller energy $U_2$. (Fine print: wavelengths and energies are measured by a comoving observer -- that is, one who's at rest in the natural coordinates to use.) In fact, the ratios are both just the factor by which the Universe has expanded:&#xA;$$&#xD;&#xA;{\lambda_2\over\lambda_1}={U_1\over U_2}={a_2\over a_1}\equiv 1+z,&#xD;&#xA;$$&#xA;where $a$ is the &quot;scale factor&quot; of the Universe. $1+z$ is the standard notation for this ratio, where $z$ is the redshift. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physical extent of the wave packet is also stretched by the same factor. So the energy density in the wave packet goes down by a factor $(1+z)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does that mean about the amplitude of the wave? The energy density in the wave packet is proportional to the electric field amplitude squared. So if the energy density has gone down by $(1+z)^2$, the electric field amplitude must have gone down by $(1+z)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, if the Universe doubles in size, the wavelength of any given wave packet doubles, and the amplitude (peak value of ${\bf E}$) is cut in half.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-11T16:32:08.373" CommentCount="2" />
  <row Id="6703" PostTypeId="2" ParentId="6650" CreationDate="2011-03-11T16:40:43.163" Score="2" ViewCount="" Body="&lt;p&gt;There is actually a nice and simple calculation of gravitational collapse of charged spherical shells, where you can show that Coulomb repulsion is stronger than gravitational attraction if you exceed the critical bound |Q|&gt;M (in convenient units). You find this simple calculation in the &lt;a href=&quot;http://arxiv.org/abs/gr-qc/9707012&quot; rel=&quot;nofollow&quot;&gt;lecture notes by Paul Townsend on black holes&lt;/a&gt; [see chapter 3, in particular eqs. (3.10)-(3.13)].&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-03-11T16:40:43.163" />
  <row Id="6704" PostTypeId="2" ParentId="6686" CreationDate="2011-03-11T16:42:36.553" Score="0" ViewCount="" Body="&lt;p&gt;The answer is No, we are not crystals by any definition of crystal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I think you are asking though can be answered in a larger set of definitions, which includes crystals. &lt;em&gt;&quot; A quantum mechanical state function exists and in principle could be found which mathematically describes crystals and can predict their appearance and  interactions&quot; .&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this sense, one can &lt;em&gt;aim&lt;/em&gt; at finding a quantum mechanical state function describing parts of the living body, neurons for example, or cells in general. In principle it should exist, in practice I do not think anybody has reached that point. People are working with chemical potentials etc, which are a meta level to the underlying quantum mechanical state functions which are  there.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-11T16:42:36.553" />
  <row Id="6705" PostTypeId="1" CreationDate="2011-03-11T16:46:09.507" Score="4" ViewCount="120" Body="&lt;p&gt;motivated by another &lt;a href=&quot;http://physics.stackexchange.com/questions/6655/dynamic-casimir-effect&quot;&gt;question&lt;/a&gt;, i wonder if there are special properties of superconductivity when restricted on 2D or very thin layers related to the effective permittivity in function of the frequency $\epsilon ( \omega )$ near the first-order transitions between the superconducting and normal phases , any references about the main state of the art would be appreciated&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I've noticed that most research papers on superconductivity report on the current transport properties, but they usually don't talk much about the permittivity. Is this because its harder to measure?&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-12T13:19:50.993" LastActivityDate="2011-03-12T13:19:50.993" Title="What do we know of superconductivity in thin layers?" Tags="&lt;superconductivity&gt;&lt;bcs&gt;&lt;boundary-condition&gt;" AnswerCount="1" />
  <row Id="6706" PostTypeId="2" ParentId="6650" CreationDate="2011-03-11T16:55:21.233" Score="3" ViewCount="" Body="&lt;p&gt;the energy required to bind together N electrons in the sphere, it is actually&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e/R + 2e/R + ... Ne/R = N(N-1)e/2R$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is quadratic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each additional electron shoot into the black hole we will have to add energy to it that is quadratic in the current charge (proportional to N, the existing charge), while the charge will just increase linearly (by one)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the (additional) mass-energy of the black-hole will grow quadratically in the charge &lt;/p&gt;&#xA;&#xA;&lt;p&gt;since the maximum charge of a black hole is some function of the mass of the black hole, this implies that this function grow no quicker than a square root of the mass&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-03-11T16:55:21.233" CommentCount="1" />
  <row Id="6707" PostTypeId="2" ParentId="6682" CreationDate="2011-03-11T17:08:29.823" Score="2" ViewCount="" Body="&lt;p&gt;The industrial tool is &lt;a href=&quot;http://www.zemax.com/&quot; rel=&quot;nofollow&quot;&gt;zemax&lt;/a&gt;; however, it is very expensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just want to make diagrams, the TeX package &lt;a href=&quot;ftp://ctan.tug.org/tex-archive/help/Catalogue/entries/pst-optic.html&quot; rel=&quot;nofollow&quot;&gt;pst-optics&lt;/a&gt; might do the trick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the gaussian beam regime, &lt;a href=&quot;http://home.rzg.mpg.de/~ros/optocad.html&quot; rel=&quot;nofollow&quot;&gt;optocad&lt;/a&gt; (free) is a tool often used in the laser interferometer gravitational wave detector community. &lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-03-11T17:15:56.133" LastActivityDate="2011-03-11T17:15:56.133" />
  <row Id="6708" PostTypeId="2" ParentId="6705" CreationDate="2011-03-11T17:18:32.413" Score="2" ViewCount="" Body="&lt;p&gt;The obvious case for &lt;strike&gt;superconductivity&lt;/strike&gt; superflow in 2D is, well, &lt;strike&gt;superconductivity&lt;/strike&gt; superflow in a 2DEG (2D electron gas). One of the many ways this manifests is in the quantum hall effect. There, one observes a step-like behavior for the resistivity of a 2D sample w.r.t externally controllable parameters such as the magnetic field or carrier concentration. The resistivity is quantized in integer or fractional units as shown in the plot below (courtesy of &lt;a href=&quot;http://www.warwick.ac.uk/~phsbm/qhe.htm&quot; rel=&quot;nofollow&quot;&gt;D.R. Leadley, Warwick University 1997&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.warwick.ac.uk/~phsbm/iqhe.gif&quot; alt=&quot;Quantized resistance&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the magnetic field is such that the system lies on the &quot;plateaus&quot;, in between the quantized resistance values, the 2DEG in the bulk of the sample is in a &lt;strike&gt;superconductivity&lt;/strike&gt; state exhibiting dissipationless transport. This is also depicted by the green line in the plot which measures the longitudinal resistance $\rho_{xx}$ of the sample. We see that this quantity vanishes precisely when the system is on a plateau indicating the presence of superflow in the longitudinal direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; As pointed out by @wsc and @4tnmele in the comments below, it is not quite accurate to describe the plateaus as being in a superconducting state. However on the plateaus the hall strip does exhibit dissipationless flow in the longitudinal direction - even though this is primarily due to edge currents. The bulk remains insulating. So while it might be correct to describe this state as exhibiting &quot;superflow&quot; it is incorrect to call it a &quot;superconductor&quot;. I have modified the language in my answer to reflect this change. I am not deleting my answer because I feel that it is still relevant in the context of the OP's question.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-12T13:09:49.267" CommentCount="10" />
  <row Id="6709" PostTypeId="1" AcceptedAnswerId="7039" CreationDate="2011-03-11T17:25:25.800" Score="-1" ViewCount="179" Body="&lt;p&gt;How can I relate mg to R?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to give you a scenario: If a free hanging weight was tied to a motor... I know the voltage (V) and current (I) so I can use V = IR to find R.  What if I reduce the weight by say half?  Will that be the same thing as .5R?&lt;/p&gt;&#xA;" OwnerUserId="2219" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-11T19:03:15.147" LastActivityDate="2011-03-16T21:30:25.630" Title="How does mass relate to resistance (Ohm's Law)" Tags="&lt;homework&gt;&lt;electric-circuits&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="6710" PostTypeId="1" CreationDate="2011-03-11T17:54:05.120" Score="2" ViewCount="106" Body="&lt;p&gt;Is there any sence to say that circular EM wave has angular momentum?&lt;/p&gt;&#xA;" OwnerUserId="2483" LastActivityDate="2011-03-11T18:07:09.040" Title="Angular moment and EM wave" Tags="&lt;electromagnetism&gt;" AnswerCount="1" />
  <row Id="6711" PostTypeId="2" ParentId="6710" CreationDate="2011-03-11T18:07:09.040" Score="3" ViewCount="" Body="&lt;p&gt;Yes. Electromagnetic waves carry energy and momentum, and can carry angular momentum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A linearly polarized wave packet doesn't carry any angular momentum (measured about an axis through its center -- a linearly polarized wave packet moving past you off to one side has angular momentum about an axis located where you are, just as a baseball flying past you does.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A circularly polarized wave packet does carry angular momentum about its center (&quot;spin&quot; as opposed to &quot;orbital&quot; angular momentum, roughly). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Griffiths's book &lt;em&gt;Introduction to Electrodynamics&lt;/em&gt; is a good place to learn about this at the advanced-undergraduate level.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-11T18:07:09.040" CommentCount="4" />
  <row Id="6714" PostTypeId="2" ParentId="6680" CreationDate="2011-03-11T19:00:35.650" Score="1" ViewCount="" Body="&lt;p&gt;Given the stated problem: he says he knows the flow, and with low Reynolds number thats probably a pretty good approximation, it reduces to a 2D PDE (cyllindrical coordinates assuming your flow has the same symmetry) plus time. He almost certainly needs some sort of outflow end boundary condition, although I doubt the details of it will have much effect upstream. So as a start, pick anything reasonable at the outflow boundary.&lt;/p&gt;&#xA;" OwnerUserId="377" LastEditorUserId="377" LastEditorDisplayName="" LastEditDate="2011-03-11T22:41:01.480" LastActivityDate="2011-03-11T22:41:01.480" CommentCount="4" />
  <row Id="6715" PostTypeId="2" ParentId="6687" CreationDate="2011-03-11T19:14:35.737" Score="0" ViewCount="" Body="&lt;p&gt;I thought the real equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E^2 = m^2c^4 + p^2c^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where p is momentum, and photons have momentum. So the equation is not equal to zero, therefore photons do add energy. ianap so I suppose I'm asking as much as saying.&lt;/p&gt;&#xA;" OwnerUserId="2486" LastActivityDate="2011-03-11T19:14:35.737" />
  <row Id="6716" PostTypeId="2" ParentId="6650" CreationDate="2011-03-11T19:19:37.607" Score="2" ViewCount="" Body="&lt;p&gt;There is a way of seeing this more explicitly with the Reissner-Nordstrom (RN) metric&#xA;$$&#xD;&#xA;ds^2~=~-F(r)dt^2~+~F(r)^{-1}dr^1~+~r^2d\Omega^2&#xD;&#xA;$$&#xA;where the $F(r)~=~1~-~r_0/r~+~(Q/r)^2$, $r_0~=~2GM$ and $Q$ the charge in length units.  The metric has two critical points&#xA;$$&#xD;&#xA;r_\pm~=~\frac{r_0}{2}~\pm~\frac{r_0}{2}\sqrt{\frac{4Q^2}{r_0^2}}&#xD;&#xA;$$&#xA;These are the outer and inner horizons for $r_+$ and $r_-$ respectively.  The region between them is a spacelike trapping region, similar to the interior of a Schwarzschild solution.  The extremal condition on the black hole is where $r_+~=~r_-$ which is where the spacelike region between the outer and inner horizons has been “removed,” or in a more subtle way mapped into the spacetime $AdS_2\times S^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the metric components we then compute the Christoffel symbols in the usual straight forwards, though tedious, manner.  The most salient of the connection terms is&#xA;$$&#xD;&#xA;{\Gamma^r}_{tt}~=~F(r)\frac{r_0r~-~2Q^2}{2r^3}&#xD;&#xA;$$&#xA;which gives the geodesic equation&#xA;$$&#xD;&#xA;\frac{d^2r}{ds^2}~+~{\Gamma^r}_{tt}U^tU^t~=~0.&#xD;&#xA;$$&#xA;Far from the black hole We have that $U^t~\simeq~1$ and so $ds~\simeq~dt$ and this is a Newton second law type of equation&#xA;$$&#xD;&#xA;\frac{d^2r}{dt^2}~+~F(r)\frac{r_0r~-~2Q^2}{2r^3}~=~0,&#xD;&#xA;$$&#xA;where for $Q~=~0$ recovers Newton's second law for gravitation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the extremal case.  The connection term is then&#xA;$$&#xD;&#xA;{\Gamma^r}_{tt}~=~\frac{1}{2}\Big(1~-~\frac{r_0}{r}~+~\frac{r_0^2}{4r^2}\Big)\Big(\frac{r_0}{r^2}~-~\frac{r_0}{2r^3}\Big)&#xD;&#xA;$$&#xA;which tells us that a neutral particle is still attracted into the black hole.  Then we consider a charged particle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The field strength 2-form and tensor components is&#xA;$$&#xD;&#xA;F~=~\frac{Q}{r^2}dt\wedge dr&#xD;&#xA;$$&#xA;The geodesic equation is no longer zero, but there is a driving force $F~=~F(r)r_0/2r^2$.  With this Newtonian approximation the total force on the particle can be seen to be zero near the horizon.  So for the extremal black hole a charge near the horizon will experience no net force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other connection terms are also nonzero.  An important one is ${\Gamma^\theta}_{r\theta}~=~-1/r$ .  For the extremal case the radial acceleration of a charge near the horizon approaches zero, but the angular component remains.  Hence if there is a small $U^\theta$ this will move the charged particle off the radial path and ultimately away from the black hole.  This in effect prevents the overcharging of a black hole.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-11T19:19:37.607" CommentCount="5" />
  <row Id="6718" PostTypeId="2" ParentId="6588" CreationDate="2011-03-11T19:53:48.240" Score="0" ViewCount="" Body="&lt;p&gt;It's a historical accident which has left us a lasting pain in the ass.  There's no reason to express sound as decibels: writing the pressure level in Pascals in scientific notation is just as convenient as writing a level in decibels relative to some reference pressure level.  And it's often MORE convenient for calculating certain things (i.e. a noise floor expressed in pascals-per-root-Hz makes sense; dB-per-root-Hz or dB-per-Hz makes no sense)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, there are a few possible reasons for using dB:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) For folks that don't know scientific notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) To match human physiology/psychology, since human perception of sound has a roughly logarithmic response to the pressure level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) For attenuators or amplifiers, you can just add things (this is more common when dealing with RF).  If I put a 10 dB attenuator in series with a 5 dB attenuator I get a 15 dB attenuator.  And if I put a +2 dB power level of RF into a 15 dB attenuators, I get a -13 dB power out.  And even though I don't like dB, I do have to admit that's pretty convenient.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-03-11T19:53:48.240" CommentCount="3" />
  <row Id="6719" PostTypeId="2" ParentId="6588" CreationDate="2011-03-11T20:26:27.890" Score="1" ViewCount="" Body="&lt;p&gt;While most of the answers received so far emphasize the use of decibels as a measure of the loudness of sound, it is important to note that dB are most commonly used in RF engineering and can be used in describing any wave phenomenon that transports energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The utility of dB comes from two properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The logarithmic scale applies conveniently to situations where you have a huge dynamic range.  For example, in an RF system, you might have signals with powers ranging from picowatts to megawatts.  &lt;/li&gt;&#xA;&lt;li&gt;Secondly, the logarithmic scale converts multiplication to addition, which is convenient.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A quantity in dB always represents a &lt;strong&gt;ratio&lt;/strong&gt;.  An absolute power level in &quot;dB&quot; is always given with respect to some reference value.  For example, signal power in RF engineering is often given in &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/DBm&quot; rel=&quot;nofollow&quot;&gt;dBm&lt;/a&gt;&quot;, which are decibels with respect to 1 mW:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm{[dBm]} = 10\ \log_{10} \frac{P}{\left(1 \mathrm{\ mW}\right)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sadly, the &quot;m&quot; in &quot;dBm&quot; refers to &quot;mW&quot; !)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose an initial signal with power 7 dBm is followed by an amplifier with gain 2 (approx 3 dB).   The power at the output will be 7 dBm + 3 dB = 10 dBm.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-03-11T20:26:27.890" />
  <row Id="6720" PostTypeId="1" AcceptedAnswerId="6730" CreationDate="2011-03-11T21:05:49.633" Score="10" ViewCount="1025" Body="&lt;p&gt;A sign of a tsunami is that the water rushes away from the shore, then comes back to higher levels. It seems that waves should be both + and - polarized and that some tsunamis should go in the opposite direction. That is the first indication of them would be that the water begins rising. However, other than situations very close to the source, it seems that the wave always begins with the water drawing away from the coast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tsunami#Warnings_and_predictions&quot; rel=&quot;nofollow&quot;&gt;wikipedia article on tsunamis&lt;/a&gt; states that:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the 2004 Indian Ocean tsunami&#xA;  drawback was not reported on the&#xA;  African coast or any other eastern&#xA;  coasts it reached. This was because&#xA;  the wave moved downwards on the&#xA;  eastern side of the fault line and&#xA;  upwards on the western side. The&#xA;  western pulse hit coastal Africa and&#xA;  other western areas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The above is widely repeated. However, when you search the scientific literature, you find that this is not the case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proc. IASPEI General Assembly 2009, Cape Town, South Africa., Hermann M. Fritza, Jose C. Borrerob, &lt;a href=&quot;http://mahabghodss.net/NewBooks/www/web/digital/nashrieh/Journal%20of%20earth%20qua/Vol.%2022-2006/Vol.22%20Issue%20S3%20June%202006/S219.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Somalia Field Survey after the December 2004 Indian Ocean Tsunami&quot;&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Italian-speaking vice council,&#xA;  Mahad X. Said, standing at the&#xA;  waterfront outside the mosque upon the&#xA;  arrival of the tsunami (Figure 10a),&#xA;  gave a very detailed description of&#xA;  the initial wave sequence. At first, a&#xA;  100-m drawback was noticed, followed&#xA;  by a first wave flooding the beach.&#xA;  Next, the water withdrew again by 900&#xA;  m before the second wave partially&#xA;  flooded the town. Finally, the water&#xA;  withdrew again by 1,300 m offshore&#xA;  before the third and most powerful&#xA;  wave washed through the town. These&#xA;  drawbacks correspond to 0.5-m, 4-m,&#xA;  and 6-m depths. The detailed&#xA;  eyewitness account of the numerous&#xA;  drawbacks is founded on the locations&#xA;  of the offshore pillars.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So is there a physical reason why tsunamis, perhaps over longer distances, tend to be oriented so that the first effect is a withdrawal of the water?&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-03-12T02:45:37.153" LastActivityDate="2011-03-12T06:35:53.783" Title="Why do tsunami waves begin with the water flowing away from shore?" Tags="&lt;waves&gt;&lt;water&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="1" />
  <row Id="6721" PostTypeId="1" AcceptedAnswerId="6724" CreationDate="2011-03-11T21:20:14.533" Score="3" ViewCount="248" Body="&lt;p&gt;After watching the first episode of wonders of the solar system, one question came up which is not explained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bryan Cox says that ultimately the universe will be devoid of matter, so not even a single particle will actually exist. This however, seems to be a state where entropy is zero. There is no entropy since there's no disorder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand the 2nd law of thermodynamics says that the disorder increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I am pretty sure that I am thinking wrong about this in someway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can imagine that the answer might be that an empty universe has an infinite amount of entropy but it feels pretty counter-intuitive for me.&lt;/p&gt;&#xA;" OwnerUserId="2488" LastActivityDate="2011-08-14T19:46:45.390" Title="Entropy of an empty universe" Tags="&lt;cosmology&gt;&lt;universe&gt;&lt;entropy&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="6722" PostTypeId="2" ParentId="6720" CreationDate="2011-03-11T21:51:29.530" Score="2" ViewCount="" Body="&lt;p&gt;Dear Carl, the recession of the sea level is an inevitable consequence of the mass conservation: the extra water in the tsunami has to come from somewhere. It comes from both sides - from the region in front of the wave as well as behind the wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the sea level is most elevated somewhere, it must obviously be lowered at both sides, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that you have a wave packet given by the function&#xA;$$ f(x) = \exp(-ax^2) \cos (kx) $$&#xA;The maximum absolute value of  $f(x)$ is at $x=0$, right? That's the tsunami. However, it's inevitable that the second highest absolute value which can still be seen is at the pair of the nearby local minima. The two maxima at even higher values of $|x|$ are already pretty much negligible.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-11T21:51:29.530" CommentCount="5" />
  <row Id="6723" PostTypeId="2" ParentId="6721" CreationDate="2011-03-11T22:00:00.957" Score="1" ViewCount="" Body="&lt;p&gt;I dont know that episode, but normally in such end of the world scenarios the matter &#xA;is transformed into radiation via Hawking radiation (Black Holes) or decays in some rather speculative way (proton decay). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What remains then is radiation, maybe something analog to CMB. &#xA;There the entropy might &quot;dwell&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-11T22:08:22.823" LastActivityDate="2011-03-11T22:08:22.823" />
  <row Id="6724" PostTypeId="2" ParentId="6721" CreationDate="2011-03-11T22:00:34.343" Score="8" ViewCount="" Body="&lt;p&gt;Dear Paulo, the space will be empty in the sense that the density of particles will be tiny (the particles will include diluted gas as well as the Hawking radiation from the black holes that will have evaporated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the Universe will also be huge because its expansion will continue - and it will become the exponential expansion of the de Sitter space (because of the dark energy, also known as the cosmological constant, that has a constant positive energy density).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's only the total entropy of the system that is increasing and the growth of the total volume of the Universe beats the decrease of the entropy density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's another method to look at the entropy in the future. One may only look inside the &quot;cosmic horizon&quot; of the de Sitter space that our Universe is already becoming. This cosmic horizon - the maximally distant place where you can still see, if you wish - has a fixed radius and indeed, the entropy inside the cosmic horizon will go zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this description, however, the cosmic horizon itself carries a huge entropy equal to&#xA;$$ S = A / 4A_0$$&#xA;where $A_0=\hbar G / c^3$ is the Planck area, about $10^{-70}$ squared meters. The formula is identical to the Bekenstein-Hawking formula for a black hole. You may imagine that this entropy $S$ of the event horizon remembers the entropy of everything that may exist behind the cosmic horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the value of the radius of the cosmic horizon which is determined by the cosmological constant in our Universe, $S$ is approximately $10^{120}$ in natural units (essentially bits of information, also equal to $10^{97}$ J/K or so) that is about 20 orders of magnitude larger than the current entropy $10^{100}$ of the visible Universe - the latter figure is dominated by the entropy of the large black holes in the middle of galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Before people knew about the large black holes and their entropy, they thought that most of the entropy of the current Universe was carried by the cosmic microwave background - but the CMB entropy is roughly 10-20 orders of magnitude lower than the entropy of the black holes.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The entropy of the Universe will continue to increase but in some proper language, it will converge to $10^{120}$ which is the maximum allowed entropy of our Universe. The latter statement - that the entropy of a de Sitter can't exceed this bound equal to the inverse cosmological constant in the Planck units - is a bit of the lore of quantum gravity. I suppose that Brian Cox didn't want to explain this point because the TV program would become really confusing for his typical audiences, much like this comment inevitably will. Whoever is confused by this bound should keep on assuming that $10^{120}$ is infinity and the entropy will continue to grow indefinitely.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-11T22:07:24.057" LastActivityDate="2011-03-11T22:07:24.057" CommentCount="1" />
  <row Id="6725" PostTypeId="1" CreationDate="2011-03-11T23:55:52.913" Score="13" ViewCount="784" Body="&lt;p&gt;and they are on opposite ends of the electromagnetic spectrum, then why can't light travel through walls which is right in the middle of the spectrum?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question has already been asked &lt;a href=&quot;http://www.physlink.com/Education/AskExperts/ae175.cfm?CFID=38736108&amp;amp;CFTOKEN=686d5d990db4e8a0-A73FF6D5-15C5-EE01-B960081244D63FF9&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. However, I am not entirely satisfied by the answer. I am not looking for fanciful analogies and metaphors. I just want to know factual technical reasons why this happens.&#xA;If given a radio source and a gamma source of equal intensity, then will the radio source have more penetrability than the gamma source per formula for &lt;a href=&quot;http://en.wikipedia.org/wiki/Penetration_depth&quot; rel=&quot;nofollow&quot;&gt;penetration depth&lt;/a&gt;? If not, why not?&lt;/p&gt;&#xA;" OwnerUserId="2489" LastEditorUserId="98" LastEditorDisplayName="" LastEditDate="2011-03-15T06:41:27.793" LastActivityDate="2011-03-15T15:39:30.070" Title="If both radio waves and gamma rays can travel through walls" Tags="&lt;electromagnetic-radiation&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="5" />
  <row Id="6726" PostTypeId="2" ParentId="6720" CreationDate="2011-03-11T23:59:31.133" Score="1" ViewCount="" Body="&lt;p&gt;It is a matter of conservation of mass. In order to raise the water a certain height, the extra mass of the water has to come from somewhere. That somewhere is the area directly ahead of the wave. So in the end the average height of the water does not change. Lower the level by 1 meter for 10 meters, and you have enough to make 10 meter wave with 1 meter width.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;clarification:&lt;/strong&gt;&#xA;  This effect travels at the speed of sound underwater, whereas the surface waves travel at a lesser speed, thus the dip precedes the rise.&lt;/p&gt;&#xA;" OwnerUserId="392" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2011-03-12T00:25:09.293" LastActivityDate="2011-03-12T00:25:09.293" CommentCount="3" />
  <row Id="6728" PostTypeId="2" ParentId="6667" CreationDate="2011-03-12T00:22:49.890" Score="1" ViewCount="" Body="&lt;p&gt;I will answer the question of why $\dot{\hat{e}_x} = \hat{e}_y \omega$ when the rotation is $\vec{\omega}=\hat{e}_z \omega$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any instant the components of $e_x$ and $e_y$ are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{e}_{x}=\begin{bmatrix}\cos\theta &amp;amp; \sin\theta &amp;amp; 0\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{e}_{y}=\begin{bmatrix}\mbox{-}\sin\theta &amp;amp; \cos\theta &amp;amp; 0\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\theta$ is the rotation angle. Using the chain rule&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{\hat{e}}_{x}=\begin{bmatrix}\omega\sin\theta &amp;amp; \mbox{-}\omega\cos\theta &amp;amp; 0\end{bmatrix} = \hat{e}_y\,\omega$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{\hat{e}}_{y}=\begin{bmatrix}\mbox{-}\omega\cos\theta &amp;amp; \mbox{-}\omega\sin\theta &amp;amp; 0\end{bmatrix} = \mbox{-}\hat{e}_x\,\omega$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general rule is the for any rotating constant vector $\vec{A}$ riding on a rotating frame (or on a rigid body) with angular velocity $\vec{\omega}$ the rate of change is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dot{\vec{A}} = \vec{\omega}\times\vec{A} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and in your case&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{\hat{e}}_{x} = \left( \hat{e}_z \omega \right) \times \hat{e}_{x} = \hat{e}_{y} \omega $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-03-12T00:22:49.890" />
  <row Id="6730" PostTypeId="2" ParentId="6720" CreationDate="2011-03-12T01:54:32.910" Score="7" ViewCount="" Body="&lt;p&gt;Water waves are rather complicated, and the differential equations which describe them are call Boussenesq equations.  A tsunami is not a transverse wave.  It is a pressure wave with a longitudinal mode.  It also travels very fast at about 700km/hr.  What happens is that this travels as a pressure wave in the open ocean, but when it reaches a continental shelf the wave is reflected partially upwards.  This has the effect of converting it into a transverse wave as water moving along is now pushed upwards.  This is a very nonlinear process and nontrivial to model.  This pushing up of the water does initially cause water at the shore to recede outwards.  The wave which seconds later reaches shore is much more slow moving, and a lot of that wave energy is converted into the towering wave front that sweeps in.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-12T01:54:32.910" CommentCount="2" />
  <row Id="6731" PostTypeId="1" CreationDate="2011-03-12T02:56:59.680" Score="3" ViewCount="154" Body="&lt;p&gt;I am trying to do this question and getting stuck at interpreting what the question is asking for.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$dP/dR = -g(R)\rho(R) = -[GM(&amp;lt;R)/R^2]\rho(R).$$&lt;br&gt;&#xA;  Thus when we descend a distance $\Delta R$ from the surface $R_s$ where $P=0$, the pressure will be approximately:&lt;br&gt;&#xA;  $$P = 0-\Delta P = GM(&amp;lt;R)/R^2\rho(R)\Delta R.$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;By taking $\Delta R = R = R_s$, the radius of the star, and $\rho(R) = \bar{\rho} = 3M/(4\pi R_s^3)$, the mean density of the star, derive a rough estimate for the central pressure of the star and show that the central pressure is independent of mass for the scalings given above.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am substituting the mean density equation instead of $\rho(R)$ into the second equation on the image and simplifying it. I am not sure that $&amp;lt;R$ mathematically represents and when I simplify it I get $P = (3M)/(4\pi R_s)$. I'm not sure if this is correct because my answer still has mass.&lt;/p&gt;&#xA;" OwnerUserId="2491" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-12T11:10:40.777" LastActivityDate="2011-03-14T13:50:24.523" Title="Hydrostatic equilibrium of a star - simplification math error" Tags="&lt;astrophysics&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6732" PostTypeId="2" ParentId="6731" CreationDate="2011-03-12T03:20:19.437" Score="1" ViewCount="" Body="&lt;p&gt;That's some terrible notation in the question, and I think that may be what's tripping you up. Try the more conventional notation for hydrostatic equilibrium $$\frac{dP}{dr}=-\frac{GM(r)\rho(r)}{r^{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are saying&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Make the approximations $dp\approx\Delta P$ and $dr\approx\Delta r$, so instead of a differential pressure between two infinitesimally separated bits of the star, we are comparing broad regions. Then, evaluate this for $\Delta r=R_{s}$, the total radius of the star (what is $\Delta P$ in this case?) and use $\rho(r)=\frac{3M_{total}}{4\pi R_{s}^{3}}$ (the average density).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They drop a hint that you should get an expression which includes the pressure at $r=0$ and which doesn't include a mass term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that this restatement of the problem makes more sense.&lt;/p&gt;&#xA;" OwnerUserId="1655" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-03-14T13:50:24.523" LastActivityDate="2011-03-14T13:50:24.523" CommentCount="1" />
  <row Id="6733" PostTypeId="2" ParentId="6725" CreationDate="2011-03-12T03:35:15.013" Score="13" ViewCount="" Body="&lt;p&gt;The anthropocentric explanation is that if visible light rays could go through wood or plaster we would not build walls with wood or plaster.&lt;/p&gt;&#xA;" OwnerUserId="2492" LastActivityDate="2011-03-12T03:35:15.013" CommentCount="5" />
  <row Id="6734" PostTypeId="2" ParentId="6720" CreationDate="2011-03-12T04:51:22.097" Score="10" ViewCount="" Body="&lt;p&gt;The positive or negative elevation of the leading wave is related to what happened to the ocean floor during the earthquake. If the seabed was raised, a crest should lead the tsunami; if it was lowered then a trough leads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 2004 Sumatra tsunami, a piece of seabed of about 1000 km x 100 km went down a few meters. Another strip, located to the west of the first and with roughly the same size, went up. Because of this, the wave moving east (towards Thailand) was led by a trough. The one moving west, which reached Sri Lanka, India and Africa, was led by a crest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is seen in the simulation of that tsunami shown below (by Kenji Satake, of the Active Fault Research Center in Tsukuba, Japan). In the red areas the water surface is higher than normal, and in the blue ones it is lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/A6Skt.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar animation, with some data added, is at &#xA;&lt;a href=&quot;http://es.ucsc.edu/~ward/indo.mov&quot; rel=&quot;nofollow&quot;&gt;http://es.ucsc.edu/~ward/indo.mov&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2011-03-12T04:51:22.097" CommentCount="1" />
  <row Id="6735" PostTypeId="2" ParentId="5332" CreationDate="2011-03-12T05:56:13.253" Score="1" ViewCount="" Body="&lt;p&gt;In perturbation theory, we split the Hamiltonian into a &quot;free&quot; part, and an &quot;interacting&quot; part, with the magnitude of the interacting part being much smaller than that of the free part. If there are no degeneracies in the free Hamiltonian &amp;mdash; at least for the ground state which we're interested in &amp;mdash; the analysis is relatively straightforward. If we have a mild degeneracy, we can first project the interacting Hamiltonian to the free ground eigenspace, diagonalize it, and work with the diagonalized basis. But what if the ground state is highly degenerate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example would be the fractional quantum Hall effect. In the presence of a uniform background magnetic field, the kinetic part of the Hamiltonian &amp;mdash; i.e the free part &amp;mdash; decomposes into Landau levels, with an infinite degeneracy for the lowest Landau level. With a partially filled lowest Landau level and the Pauli exclusion principle for electrons, we have an extremely large ground eigenspace for the free Hamiltonian. The interaction Hamiltonian is due to the Coulomb repulsion between electrons, which is much weaker. However, diagonalizing the projection of the interaction Hamiltonian onto the lowest Landau level is extremely difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's get on to the more complicated case of a doped Mott insulator. Even with the fermion hopping terms neglected, the dynamics are more complicated. For a slight doping of a Mott insulator, we still get a huge degeneracy.&lt;/p&gt;&#xA;" OwnerUserId="2381" LastActivityDate="2011-03-12T05:56:13.253" />
  <row Id="6737" PostTypeId="2" ParentId="6720" CreationDate="2011-03-12T06:35:53.783" Score="2" ViewCount="" Body="&lt;p&gt;I will simplify the question. When large waves come towards the beach there is always an undertow. Has an &quot;overtow&quot; been seen before the undertow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My explanation while watching waves was that the waves break gathering mass on an incline. Growing in height they draw water symmetrically, but there is less water ahead, because of the incline,  than behind, and the water is sucked up by the wave crest before it arrives. I expect the same mechanism to hold for the much longer wavelength tsunami wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess data from high rocky shores might be enlightening. When I visited Japan I went on tour up the east coast from Tokyo and was impressed that there were no beach facilities and the beaches looked like garbage dumps. I was told this was because of the tsunamis.Long beaches ending up in  precipices of land overlooking them.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-12T06:35:53.783" />
  <row Id="6738" PostTypeId="1" AcceptedAnswerId="6765" CreationDate="2011-03-12T10:23:13.167" Score="4" ViewCount="774" Body="&lt;p&gt;Can you please describe quantum mechanics in &lt;strong&gt;simple&lt;/strong&gt; words? When ever I read this word (quantum computers, quantum mechanics, quantum physics, quantum gravity etc) I feel like fantasy, myth and something very strange that i can never understand.&lt;br&gt;&#xA;So what is quantum mechanics?&lt;/p&gt;&#xA;" OwnerUserId="847" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-12T13:19:55.953" LastActivityDate="2011-03-16T22:50:24.503" Title="Simple explanation of quantum mechanics" Tags="&lt;quantum-mechanics&gt;&lt;definition&gt;" AnswerCount="6" CommentCount="8" FavoriteCount="2" />
  <row Id="6739" PostTypeId="2" ParentId="6738" CreationDate="2011-03-12T11:04:09.267" Score="8" ViewCount="" Body="&lt;p&gt;&quot;Once more unto the breach ...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aspect of quantum mechanics that distinguishes it the most (IMHO) from classical mechanics, is that of &lt;em&gt;superposition of states&lt;/em&gt; - that at any given moment a system is described by a state which can be a linear combination of various, physically realizable, outcomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |\Psi\rangle = c_1 |\psi_1\rangle + c_2 |\psi_2 \rangle + \dots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed one can give a simple visual analogy. Think of that prototypical classical mechanical system - the pool (or &quot;billiards&quot; depending on which time-zone you happen to be in) table. The pool balls obey simple classical trajectories, assuming the absence of non-uniformities on the table, which consist of straight lines interrupted by reflection off of the sides of the table. In classical mechanics one can enumerate all such trajectories and construct the resulting phase space for the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to ask me to characterize the state of the pool table at any instant, I would do so by choosing a point in phase space whose co-ordinates specify the locations and momenta of all the balls on the table at that instant. Given this information and Newton's second law I can determine, &lt;em&gt;to arbitrary accuracy&lt;/em&gt; , the future trajectories of all the pool balls. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum Mechanics introduces the entirely new possibility that I could specify a state which is a linear combination of &lt;em&gt;eigen&lt;/em&gt; (or &quot;physically realizable&quot;) states. In other words a given pool ball could be at location $x$ &lt;em&gt;and&lt;/em&gt; at $x&amp;#39;$ at the same time. Even to a laymen this possibility open up great new vistas. Engineers can only imagine what sort of machines they could construct if they could build a transmission that could operate at more than one speed simultaneously. Computational scientists wonder at the thought that a quantum system could exist in a superposition of $0$ and $1$ simultaneously - this possibility is in fact being realized in the exploding field of quantum computation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In return for all these nice possibilities that superposition opens up we must accept significant trade-offs - there is an inherent uncertainty in whatever physical quantity we measure &lt;strong&gt;and&lt;/strong&gt; we must accept restrictions on which observables can be measured simultaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate the first trade-off (uncertainty in measurement outcomes) I can - in theory - construct a quantum computer which employs &quot;massive parallelism&quot; and solves a problem by considering all possible solutions simultaneously, but the trade-off is that I can never be &lt;em&gt;absolutely&lt;/em&gt; certain that the solution it spits out at the end is the right one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second trade-off is perhaps more significant in terms of its implications for our notions of &quot;reality&quot;. Coming back to the pool ball, superposition allows me to say that it is at location $x$ and $x&amp;#39;$ simultaneously. However, the more accurately I try to localize the ball at these two locations, the less accurately I can localize the ball's momenta. Or as when a cop pulled Heisenberg over for speeding and asked &quot;Do you know how fast you were going?&quot;, he replied &quot;No. But I know exactly where I am&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;strong&gt;Note for the purists and experts:&lt;/strong&gt; The language of this answer is directed at a beginner and not for a conference proceeding, therefore I have avoided technicalities and mathematics. Please keep this in mind when pointing out errors and such.]&lt;/p&gt;&#xA;" LastActivityDate="2011-03-12T11:14:49.803" CommentCount="17" />
  <row Id="6740" PostTypeId="2" ParentId="6738" CreationDate="2011-03-12T11:20:59.503" Score="0" ViewCount="" Body="&lt;p&gt;In the early part of the twentieth century, physics was revolutionized. The foundation of the subject changed once and for all. It was discovered that nature in its deeper scheme of things does not operate according to the deterministic rules of classical Newtonian physics. Heisenberg discovered his famous uncertainty principle in 1926. Physicists came to know that values of certain pairs of classical dynamical variables like position and its conjugate momentum can not be measured to arbitrary precision. The more you are certain about one of them the less you know about the value of the other. The simultaneous precise knowledge of these pairs of variables (at least in principle) are essential to predict the outcome of any experiment in classical physics. Classical mechanics used to assert that in principle nature is completely deterministic. However the uncertainty principle put an end to that dream. The uncertainty principle is a consequence of the fact that there is a wave particle duality for all the so called &quot;particles&quot;(like electron) and &quot;waves&quot; (like electromagnetic wave) in nature. One can say that some times under certain conditions, it is helpful to think a particle like electron as waves or a wave like em wave as particles. Schrödinger discovered his famous equation for particles with non-relativistic speed which describe the evolution of a characteristic function $\psi$ in space with time. Quantum mechanics is based on certain basic postulates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) It is assumed that to every observable physical quantity there exists a hermitian operator. The measurable values of the observable is the various eigen values of the operator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) To every physical system there corresponds an abstract Hilbert space. A state of the system is represented by a vector in this space on which the operator corresponding to the observable acts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The outcome of a measurement of an observable in a particular state is given by the expectation value of the corresponding operator in that state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) The operator corresponding to a dynamical variable is obtained by replacing the classical canonical variables by corresponding quantum mechanical operators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Any pair of canonically conjugate operators will satisfy the Heisenberg commutation rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The edifice of quantum mechanics is based on these basic postulates. What quantum mechanics really implies is that light, electrons, protons etc. can not be described by exclusively in particle terms or wave terms. They are like neither particles nor waves. Particles or waves are approximate classical concepts and to describe quantum reality these classical concepts should be viewed as complimentary concepts rather than contradictory. In the language of Neils Bohr, &quot;However far the phenomena transcend the scope of classical physical explanation, the account of all the evidence must be expressed in classical terms&quot;. This is because &quot;by the word 'experiment' we refer to a situation where we can tell others what we have done and what we have learned&quot; so that &quot;the account of the experimental arrangement and of the results of the observations must be expressed in unambiguous language with suitable application of the terminology of the classical physics&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-03-12T11:20:59.503" CommentCount="1" />
  <row Id="6741" PostTypeId="1" CreationDate="2011-03-12T11:58:02.663" Score="1" ViewCount="289" Body="&lt;p&gt;How can i derive the dynamic of a relativistic charged particle in a uniform magnetic field $B=(0,0,B)$?&lt;/p&gt;&#xA;" OwnerUserId="2501" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-12T22:59:58.873" LastActivityDate="2011-03-13T07:22:01.913" Title="Relativistic charged particles in a constant uniform magnetic field" Tags="&lt;electromagnetism&gt;&lt;special-relativity&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6742" PostTypeId="1" CreationDate="2011-03-12T11:58:49.277" Score="5" ViewCount="365" Body="&lt;p&gt;I am currently studying the motion of relativistic charged particles in electromagnetic fields. More exactly, we first derived the equation of motion in the 4-vector formalism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was a bit confused when my teacher talked about acceleration in special relativity. In which cases are we allowed to talk about it specifically in special relativity? (I mean that I have not studied general relativity yet) What stay constant ? Or are we making an approximation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may of course have a sense, though I have the feeling that in my courses, we only talked about constant speeds in special relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope my question has a sense,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isaac&lt;/p&gt;&#xA;" OwnerUserId="2499" LastEditorUserId="2499" LastEditorDisplayName="" LastEditDate="2011-03-12T12:38:50.607" LastActivityDate="2011-08-18T16:35:51.323" Title="Acceleration in special relativity" Tags="&lt;special-relativity&gt;&lt;acceleration&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="6743" PostTypeId="2" ParentId="6742" CreationDate="2011-03-12T12:49:52.127" Score="7" ViewCount="" Body="&lt;p&gt;Acceleration is an acceptable topic in special relativity.  It is rather easily described.  The metric for flat spacetime is&#xA;$$&#xD;&#xA;ds^2~=~-dt^2~+~dx^2~+~dy^2~+~dz^2~=~g_{ab}dx^adx^b.&#xD;&#xA;$$&#xA;If I divide through by the square of the proper time $ds^2$ this gives unity&#xA;$$&#xD;&#xA;1~=~g_{ab}\frac{dx^a}{ds}\frac{dx^b}{ds}~=~g_{ab}U^aU^b.&#xD;&#xA;$$&#xA;A derivative with respect to the proper time $s$ gives 0 on the left hand side and the spacetime acceleration $A^a$ is clearly seen to be orthogonal to the four velocity $U^a$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I restrict this to two dimensions, where the spatial dimension of importance is the direction the object moves along the metric in a change of signature convention gives us&#xA;$$&#xD;&#xA;1~=~(U^t)^2~-~(U^x)^2&#xD;&#xA;$$&#xA;The equation implies that the two components of the four-velocity are hyperbolic trigonometric functions&#xA;$$&#xD;&#xA;U^x~=~\sinh gs,~U^t~=~\cosh gs,&#xD;&#xA;$$&#xA;for $g$ the acceleration parameter.  The motion of this body asymptotes to a null direction $u~=~t~-~x$ and is a hyperboloid restricted to this one part of the Minkowski spacetime.  This portion is called the Rindler wedge, which has all sorts of surprising structure to it, from Fermat coordinate representation that is a form of Poincare half plane, to Unruh radiation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is some confusion with respect to accelerated frames in special relativity.  It is commonly thought that it is the domain of general relativity.  Special relativity should be thought of as Newton’s laws with the extension of boost symmetries in space plus time.  Then just as with Newton’s laws the symmetries are properly described from an inertial frame, and the dynamics are $F~=~ma$.  The description of accelerated motion for Newton’s second law in special relativity is perfectly acceptable physics.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="3887" LastEditorDisplayName="" LastEditDate="2011-08-18T16:35:51.323" LastActivityDate="2011-08-18T16:35:51.323" CommentCount="2" />
  <row Id="6744" PostTypeId="2" ParentId="6742" CreationDate="2011-03-12T12:50:14.250" Score="4" ViewCount="" Body="&lt;p&gt;It's perfectly possible to have accelerated particles in special relativity in the same way you can have them in classical Newtonian physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to describe them one needs some of the tools and concepts of &lt;a href=&quot;http://en.wikipedia.org/wiki/Basic_introduction_to_the_mathematics_of_curved_spacetime&quot; rel=&quot;nofollow&quot;&gt;differential geometry&lt;/a&gt; used commonly in general relativity. For example, if you'd like to describe physics in the reference frame of the accelerating observer, this is no longer inertial and you can't use the simple math of linear Lorentz transformations anymore because you also need to take into account non-linear effects (some of which are present already in classical case, like centrifugal and Coriolis forces).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you should do is describe the world-line of the particle as a curve $\gamma(\lambda)$ (parametrized by some affine parameter). You can then compute the velocity of the particle ${\mathbf v} = {{\rm \mathbf d}\gamma(\lambda) \over {\rm d} \lambda}$ as a tangent field along the world-line and the acceleration ${\mathbf a} = {{\rm D} {\mathbf v} \over {\rm d} \lambda} $ as a covariant derivative of velocity along the world-line; here we are implicitly using that we have a flat connection available on our manifold. Working backwards, you can postulate that the particle is to have some given acceleration due to forces and figure out its trajectory (same way you do it in classical physics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'd like to figure out additional information you can choose some coordinate system at point $\lambda = \lambda_0$. Then you can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermi%2dWalker_transport&quot; rel=&quot;nofollow&quot;&gt;Fermi-Walker transport&lt;/a&gt; to transport this system along the world-line which gives you information about the way the particle is spinning, etc.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-03-12T12:50:14.250" />
  <row Id="6745" PostTypeId="1" CreationDate="2011-03-12T13:45:54.723" Score="0" ViewCount="434" Body="&lt;p&gt;&lt;a href=&quot;http://www.msnbc.msn.com/id/42044156/ns/world_news-asia-pacific/&quot; rel=&quot;nofollow&quot;&gt;The recent news&lt;/a&gt; says that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Japanese authorities confirmed Saturday that radiation had leaked from a quake-hit nuclear plant after an explosion destroyed a building at the site.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What will be the influence of the nuclear leakage?&lt;/p&gt;&#xA;" OwnerUserId="2503" LastActivityDate="2011-03-12T17:54:38.923" Title="What is the influence of the nuclear leakage in Japan" Tags="&lt;nuclear-physics&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="6747" PostTypeId="2" ParentId="6745" CreationDate="2011-03-12T15:19:02.053" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I hear on the radio that there has been further damage that may effect the containment structure itself. This has the potential (if a series of additional things go wrong) to be rather worse than my original response implies. I haven't heard enough to even guess what might be involved.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I don't like the word &quot;leak&quot; in this context as it implies a passive failure. Steam---steam that had been slightly activated by proximity to the core---was intentionally vented, under controlled conditions to prevent a larger scale failure.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;On the global scale the impact will be trivial: a increase in the extant radiological background that will be difficult or impossible to measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the local scale the increase will presumable be measurably for a time, but will still be small compared to the natural background (see below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the political front the incident will be used as an argument against nuclear power plants by people who don't realize the a coal plant releases more radioactive material than that &lt;em&gt;every day it operates&lt;/em&gt; (simply by dispersing the naturally occurring radiologicals that used to be safely buried with the coal).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The average resident of a technologically developed nation receives a does of roughly $1\text{ mrem/day}$ day-in and day-out for their entire lives. About half of this is the cosmic ray background. Another big chuck is naturally occurring radioactive materials (K-40, U-238 and daughters, C-13...). There is a measurable background related to above-ground nuclear detonations, a little from medical and dental x-rays. When we had CRTs all over the place they leaked a tiny little bit of x-ray.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a few places the natural background is many times the average. Long term residents of these places live just as long as everyone else, and get just the same amount of cancer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The altitude at which you live is usually the single biggest factor in determining your background dose, because the cosmic flux is strongly dependent on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any panic about radiation should always be measured against these basic facts.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-03-12T17:54:38.923" LastActivityDate="2011-03-12T17:54:38.923" CommentCount="7" />
  <row Id="6748" PostTypeId="1" AcceptedAnswerId="6749" CreationDate="2011-03-12T16:10:58.047" Score="1" ViewCount="100" Body="&lt;p&gt;I'm currently working on a physics-based puzzle game with light refraction. I've read about the reflection/refraction incidence ratio, and how it's the proportion of the sine values of the angle of incidences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what formula, if there are any, determines the threshold for % of light reflected and % of light refracted?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the % of light reflected/refracted always either 0% or 100% (that is, they're always either reflected or refracted)? Or is there a curve of some sort for this?&lt;/p&gt;&#xA;" OwnerUserId="2506" LastActivityDate="2011-03-12T16:16:22.920" Title="refraction vs reflection" Tags="&lt;light&gt;&lt;reflection&gt;&lt;refraction&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6749" PostTypeId="2" ParentId="6748" CreationDate="2011-03-12T16:16:22.920" Score="3" ViewCount="" Body="&lt;p&gt;Look here: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fresnel_equations&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fresnel_equations&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is for transparent media, the formulas for metals are somewhat more complicated. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-12T16:16:22.920" />
  <row Id="6750" PostTypeId="1" AcceptedAnswerId="6753" CreationDate="2011-03-12T16:21:22.927" Score="22" ViewCount="2379" Body="&lt;p&gt;One often comes across news articles that claim that an earthquake shifted the earth's axis.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://news.google.com/?q=earthquake%20shifted%20OR%20shifts%20earth%27s%20axis&quot; rel=&quot;nofollow&quot;&gt;http://news.google.com/?q=earthquake%20shifted%20OR%20shifts%20earth%27s%20axis&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you ignore the influence of other celestial bodies, an internal event like an earthquake surely can't change the direction of the angular momentum of the Earth (unless stuff is ejected out of Earth), since angular momentum has to be conserved in the absence of an external torque. So the axis has to remain fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I missing something? Or are geologists trying to say that the resulting movement of tectonic plates causes a change in the point of intersection of the axis (which remains the same) and the plates that include the poles, so that it &lt;em&gt;seems&lt;/em&gt; as if the axis has shifted?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&#xA;Some articles mention the value of the shift in the axis and also the change in the length of the day. If, as Ted Bunn's answer indicates below, the shift in the axis isn't actually real but is because of the movement of tectonic plates with respect to the axis, shouldn't the shift be different at the north and south poles? How are the shifts and the change in day-length calculated?&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="98" LastEditorDisplayName="" LastEditDate="2011-03-13T20:17:13.890" LastActivityDate="2011-03-15T22:48:36.677" Title="How can earthquakes shift the earth's axis?" Tags="&lt;classical-mechanics&gt;&lt;rotational-dynamics&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="7" />
  <row Id="6752" PostTypeId="2" ParentId="6750" CreationDate="2011-03-12T16:34:45.607" Score="17" ViewCount="" Body="&lt;p&gt;Your explanation is right: an earthquake can't change the axis of rotation, relative to a given inertial reference frame -- that is, the axis of rotation doesn't change relative to the &quot;fixed stars&quot; as a result of the earthquake. What the earthquake does is to move material around within the Earth, so that the position of the rotation axis relative to any given marker on Earth's surface changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you prefer, rather than saying the earthquake shifted the Earth's axis, you can say that it shifted all of the stuff on the Earth's surface. Since we find it more convenient (often!) to use points of reference that are fixed to markers on Earth's surface, it looks to us like a shift in the axis.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-12T16:34:45.607" CommentCount="2" />
  <row Id="6753" PostTypeId="2" ParentId="6750" CreationDate="2011-03-12T16:35:37.847" Score="15" ViewCount="" Body="&lt;p&gt;Angular momentum doesn't change, but the angular velocity vector does. This is effectively due to a shift in the body's moment of inertia tensor.&lt;/p&gt;&#xA;" OwnerUserId="2238" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-12T17:13:59.537" LastActivityDate="2011-03-12T17:13:59.537" CommentCount="5" />
  <row Id="6754" PostTypeId="1" AcceptedAnswerId="6756" CreationDate="2011-03-12T17:14:05.277" Score="7" ViewCount="493" Body="&lt;p&gt;The Japanese tsunami, moving at about 700 km/h, affected areas as distant as Chile's coast, 20 hours after the earthquake. How does the Coriolis force affect tsunami? Also, I saw an image of a boat caught within a large whirlpool. Is the whirlpool's rotation due to Coriolis force?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-03-12T18:14:16.630" LastActivityDate="2011-03-13T13:21:32.027" Title="Coriolis effect on Tsunami" Tags="&lt;coriolis-force&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="6755" PostTypeId="2" ParentId="6738" CreationDate="2011-03-12T17:36:27.520" Score="3" ViewCount="" Body="&lt;p&gt;Feynman explained a whole lot in plain language in &quot;QED: the strange theory of light and matter&quot;. But as Zee takes pains to point out, one must pay PARTICULAR attention to what Feynman says, because he is not popularizing or dumbing it down, except for not getting too much into the notion of spin.&lt;/p&gt;&#xA;" OwnerUserId="1423" LastActivityDate="2011-03-16T22:50:24.503" CommentCount="4" />
  <row Id="6756" PostTypeId="2" ParentId="6754" CreationDate="2011-03-12T17:59:10.517" Score="4" ViewCount="" Body="&lt;p&gt;I don't believe the Coriolis force has much effect on a tsunami because it does only affect moving masses.  Coriolis force in fact isn't a force but a movement pattern looking as though a force were involved.   It is a result of inertia &quot;driving&quot; the moving masses towards a constant direction in space and at the same time the earth's rotation taking place.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, while a tsunami travels across the globe there is little water moving, instead what actually is moving is its energy.   By contrast, in hurricanes there is actually a huge amount of air moving which is affected by Coriolis force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE:&#xA;Raskolnikov supplied us with sources that suggest Coriolis force has an effect which I do not question. However, I think it's negligible, although the wave recordings on the images presented to us show a curved trajectory.  Deepak Vaid suggests this is due to ocean currents which I find not very convincing as they move at negligible speeds compared to that one the tsunami moved at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the curvature of the trajectory we see on the images is an illusion due to the map projection just as in the image below showing the geodesic (&lt;strong&gt;straight line&lt;/strong&gt;) between Japan and Chile:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/M8vtY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the aforementioned image mapped onto a sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=yoDFmHn4aLQ&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=yoDFmHn4aLQ&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/878sq.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The NOAA map in &lt;a href=&quot;http://en.wikipedia.org/wiki/Gnomonic_projection&quot; rel=&quot;nofollow&quot;&gt;gnomonic projection&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/eSuVc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2415" LastEditorUserId="2415" LastEditorDisplayName="" LastEditDate="2011-03-13T13:21:32.027" LastActivityDate="2011-03-13T13:21:32.027" CommunityOwnedDate="2011-03-13T13:21:32.027" CommentCount="9" />
  <row Id="6757" PostTypeId="2" ParentId="6754" CreationDate="2011-03-12T18:12:19.923" Score="2" ViewCount="" Body="&lt;p&gt;Your adjective &quot;large&quot; in &quot;large whirlpool&quot; may be very misleading. If you mean hundreds of meters, no effect of the Coriolis force may be visible by a naked eye at this scale. The origin of the whirlpool had to be different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite generally, tsunami is all about waves, and if one has a wave, molecules of water are moving back and forth, in circular patterns - and the radius of the circle doesn't exceed 10 meters - the height of a tsunami.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's very different from the motion of water in rivers which flow along straight lines and the Coriolis force always acts in the same direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the water molecules are making periodic circles, the effect of the Coriolis force is not too different from the case of the Foucault pendulum,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Foucault_pendulum&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Foucault_pendulum&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The plane in which the circle belongs tends to rotate by 360 degrees in 24 hours or so - which is actually comparable to the lifetime of a tsunami on the ocean. However, I think that the internal dynamics of the water tends to redirect all the motion in the waves so that it can only go up and down, or in the same circular patterns correlated with the direction of the wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that the impact won't be measurable at the end - because radially centered waves on the ocean is the only resulting pattern I can easily imagine - but I am not quite sure.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-12T18:12:19.923" CommentCount="1" />
  <row Id="6758" PostTypeId="2" ParentId="6754" CreationDate="2011-03-12T18:20:40.690" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://weatherwatercooler.kcrg.com/2011/03/12/coriolis-force-on-the-pacific-tsunami/&quot; rel=&quot;nofollow&quot;&gt;It does have an effect.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.bioline.org.br/pdf?ja09031&quot; rel=&quot;nofollow&quot;&gt;Also see this paper about modelling tsunami propagation.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted in the paper, the Coriolis force only becomes important over large distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's &lt;a href=&quot;http://mathworld.wolfram.com/news/2005-01-14/tsunamis/&quot; rel=&quot;nofollow&quot;&gt;an article on MathWorld&lt;/a&gt; including many references.&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2011-03-12T18:25:40.973" LastActivityDate="2011-03-12T18:25:40.973" CommentCount="1" />
  <row Id="6759" PostTypeId="2" ParentId="6741" CreationDate="2011-03-12T18:36:47.110" Score="4" ViewCount="" Body="&lt;p&gt;Since the other answer seem to lead to unnecessarily complicated territory, and conflates many different issues (classical versus quantum, particles versus fields, spin and fermionic statistics with charge), let me just sketch the answer whose details can be found in pretty much any text on classical electrodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without relativity you would want to solve Newton's second law $F=ma$, where the force is supplied by the Lorentz force $F = q v \times B$, using standard notation for all the quantities involved. The simplest case is when the motion is in a plane and you want to solve for the trajectory $x(t),y(t)$, which ends up being a circle of radius related to the strength of the magnetic field. Do this exercise first since it is somewhat simpler and the real answer is not that different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the particle start moving fast you'd have to incorporate relativistic effects. The simplest way to proceed is to write Newton's second law as equation for 4-vectors, $f^\mu = m \dot{p^\mu}$, where both sides are 4-vectors and the nearly invisible dot is differentiation with respect to proper time. In a particular reference frame, these 4-vectors include factors of the relativistic $\gamma$ -- look for any exposition of relativistic dynamics to learn about this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you do all that you can write the new equation of motion, which is only slightly modified compared to the non-relativisitic case. The motion is still in a circle (if we choose it to lie on a plane), whose radius now has a factor of $\gamma$ relative to the Newtonian case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, this is just to direct you to the appropriate text, I am skipping all details because this is written very well and in complete detail in many many places. Look at the classical electrodynamics text of your choice, and you'll find it (I like the discussion in Griffiths book which includes all of the above including all details). If you have more specific questions, I'd be happy to answer. In any event, you don't need to discuss either classical or quantum fields, spin and the Dirac equation, or anything like that (which is why this things are not mentioned in those textbooks).&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-03-12T19:37:48.163" LastActivityDate="2011-03-12T19:37:48.163" CommentCount="6" />
  <row Id="6760" PostTypeId="2" ParentId="6738" CreationDate="2011-03-12T18:46:15.350" Score="0" ViewCount="" Body="&lt;p&gt;In classical mechanics the state of a particle is described by a point in phase space. We can directly measure that state. But in the microscopic world this not possible. Then we postulate that we can describe the state by  a ket vector and operator in this space. This operator when acting on this state gives the physical quantities as the eigen value.&lt;/p&gt;&#xA;" OwnerUserId="2508" LastActivityDate="2011-03-12T18:46:15.350" />
  <row Id="6761" PostTypeId="2" ParentId="6754" CreationDate="2011-03-12T19:55:55.760" Score="0" ViewCount="" Body="&lt;p&gt;To supplement @Raskolnikov's answer ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Coriolis force certainly has an important role to play as the image below vividly demonstrates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/deohW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Image courtesy of NOAA, the &lt;a href=&quot;http://weatherwatercooler.kcrg.com/2011/03/12/coriolis-force-on-the-pacific-tsunami/&quot; rel=&quot;nofollow&quot;&gt;weather water cooler&lt;/a&gt; and of course @Raskolnikov who first pointed it out.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @artistoex noted in his answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... while a tsunami travels across the globe there is little water moving, instead what actually is moving is its energy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Exactly! However this energy flow will ultimately follow the underlying oceanic currents. And these &lt;em&gt;are&lt;/em&gt; determined entirely by the Coriolis force. This is one reason why the west coast of the United States is seemingly protected from tsunamis &lt;em&gt;unless&lt;/em&gt; the point of origination is within some finite distance from the coast. From the scale of the map, I'd make a very rough estimate of 500 - 1000 miles for this tsunami &quot;danger-zone&quot; as far the western US is concerned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the &quot;ring of fire&quot; runs from Japan right up around the west coast of the US so the likelihood of a severe earthquake within that zone which would generate a catastrophic tsunami headed for the California coast is high enough to keep emergency planners on their toes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above image and much more incredible information can be obtained at the &lt;a href=&quot;http://www.ngdc.noaa.gov/hazard/tsu.shtml&quot; rel=&quot;nofollow&quot;&gt;NOAA Tsunami Database&lt;/a&gt;. A source for information on ocean currents is &lt;a href=&quot;http://www.windows2universe.org/earth/Water/ocean_currents.html&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-12T19:55:55.760" CommentCount="11" />
  <row Id="6762" PostTypeId="1" CreationDate="2011-03-12T20:16:37.597" Score="3" ViewCount="215" Body="&lt;p&gt;Encouraged by the zeitgeist let me ask the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is it feasible (now or in the future) to build systems a certain distance of a vulnerable coastline which can serve to dampen a tsunami before it reaches the coast itself?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following picture comes to mind. We have some device which is small enough to be mounted on a buoy. An array of such buoys can be anchored some distance off the coastline, controlled by computers which are linked into the global tsunami warning network. This setup is quite feasible given what a cell phone can do today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the detection of an oceanic earthquake or other tsunami generating event (such as a piece of the Rock of Gibraltar falling into the sea) and given present environmental conditions one can calculate the expected path of the tsunami and its intensity. If a TDS (Tsunami Dampening System) of the type I describe above happens to be deployed in range of the tsunami, all relevant information is conveyed to the TDS controllers by the global tsunami warning system. All this happen in a matter of moments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves plenty of time (25min - 2hr) for the devices in the TDS array to be activated and to do their job. The question in this case would be - what form could such devices take? Perhaps mini-turbines which could dissipate the incoming energy into the ocean by first converting it to electrical energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any such undertaking involves physical estimates for the size and shape of such an array, the precise capabilities of each device, etc. Many of these concerns involve simple physical questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you come up with the details of such a system or alternatively outline your own design? If you think that this idea would never work in the first place - for technological, physical or some other reasons - please explain what you think would be the prime obstacle and why?&lt;/p&gt;&#xA;" LastActivityDate="2011-03-13T17:18:29.070" Title="Tsunami dampening mechanisms" Tags="&lt;geophysics&gt;&lt;hydrodynamics&gt;&lt;many-body&gt;&lt;engineering&gt;" AnswerCount="5" />
  <row Id="6764" PostTypeId="2" ParentId="6762" CreationDate="2011-03-12T20:49:35.000" Score="1" ViewCount="" Body="&lt;p&gt;Active systems seem like a engineering nightmare to me: enormously expensive to build and maintain, they would be rarely used and need to be capable of handling huge power loads (or if you over-engineer some continuous use system like a wave power installation, the excess capacity would be rare tested). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A passive system is plausible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A series of artificially shallow regions off-shore would dissipate some of the energy, by causing the wave to partially break several times on the way in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously this is a &lt;em&gt;huge&lt;/em&gt; civil engineering project, and it may have unforeseen consequences such as&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Changing beach erosion patterns. Though &quot;shallow&quot; depends on the wavelength of the wave, and could be set deep enough to not effect day-to-day wave activity. &lt;/li&gt;&#xA;&lt;li&gt;Effects (positive or negative) on the marine eco-system and local fisheries.&lt;/li&gt;&#xA;&lt;li&gt;Is &quot;shallow&quot; deep enough for shipping to pass? If not, how does the added shipping risk (ecological, human, and financial) compare to that posed by the big waves?&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of course, what you get out of this isn't &lt;em&gt;&quot;no tsunami&quot;&lt;/em&gt; it's &lt;em&gt;&quot;less tsunami&quot;&lt;/em&gt;, so you are still looking at building sea walls, paying more for insurance near the shore and running toward high ground if you see the ocean pull back.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-03-12T20:49:35.000" CommentCount="4" />
  <row Id="6765" PostTypeId="2" ParentId="6738" CreationDate="2011-03-12T21:06:01.787" Score="5" ViewCount="" Body="&lt;p&gt;Deepak's answer covers the question, except if &quot;simple&quot; means &quot;in terms an english major would understand&quot;. I have been trying to formulate such an answer, fwiw:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mechanics covers the macroscopic world, the world we see even with glass microscopes and simple telescopes. It explains the motion and interaction of bodies, from large, to quite small. Wave mechanics covers the macroscopic behavior of waves as observed in the sea and the behavior of light. The theories by the beginning of the 20th century were beautifully established so that some physicists thought that real physics was finished, and engineering was all that was left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then came the quanta. They came from various fronts. Chemists studying the elements and measuring atomic weights came up with a table of elements that had discrete numbers. The photo electric effect showed that light was not always behaving as a classical wave, because light could hit material and kick out a single electron. Then lines were found in the light spectra of elements. This forced physicist to think of energy coming in discrete numbers instead of a continuum, &lt;em&gt;quanta&lt;/em&gt; of energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first model of an atom, the Bohr model, had the nucleus  like a miniature sun and the electrons as planets around  it. The discrete lines observed though, meant that the orbits had fixed positions. Electrons could only occupy certain orbits, get kicked to a higher one and release a photon falling back, with a specific line. &lt;em&gt;Classical&lt;/em&gt; mechanics could not solve this conundrum: why there were discrete orbits and why the electrons did not fall into the nucleus anyway since the nucleus is positive and the electrons negative and the attraction inevitable classically. Physicists were forced to postulate &lt;em&gt;quantum&lt;/em&gt; mechanics and developed a whole new set of theories of how the microscopic atomic world worked: in quanta of energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this new mathematical theory the electrons stay in orbit around the nucleus because  they can only change orbits by &lt;em&gt;quanta&lt;/em&gt; of energy. They  cannot fall into the positive nucleus  because there is a lowest stable &lt;em&gt;ground state&lt;/em&gt;, which is the lowest energy an electron can have. An atom can gain a &lt;em&gt;quantum&lt;/em&gt; of energy and the electron can jump to a higher energy state; it can not go lower than the ground state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can never take a micro photo of the electron, only a probability distribution where it might be, can be computed by the new theories. Studying the probability distributions coming out of the solution of quantum mechanical problems, it was found that in the microscopic world particles sometimes behave as waves, and waves (light) sometimes behave as particles, depending on the circumstances under study. Experiments confirmed all this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum mechanics explains beautifully the light spectra of atoms, the periodic table of elements and nuclear interactions and a lot of phenomena, from transistors to lasers. The price payed is a loss of intuitive understanding of  &quot;particle&quot; and &quot;motion&quot; , new intuitions have to be developed to understand the predictions of a quantum mechanical world that need long study and perseverance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;quantum physics&lt;/em&gt; includes &#xA; a) quantum mechanics: solutions of problems with known potential energy using the Schrodinger or Klein Gordon equations. The problems are treated as particles moving in a potential well &#xA;b)second quantization, where particles are treated as creation and annihilation operators acting on the vacuum, and &#xA;c)  quantum strings where one has quantized strings and the particles are energy levels on these strings.&#xA;d) whatever new coming down the theoretical pike.( It is turtles all the way down :) )&#xA; To make any sense of this you have to read further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;quantum computers&lt;/em&gt;,  utilize the knowledge gained by quantum mechanics to create compact computers, and my knowledge is covered by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_computing&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;quantum gravity&lt;/em&gt;  is an attempt to extend quantum mechanics to general relativity, which is a classical theory. There are computational difficulties in doing this. Theorists are aiming at a unified theory of everything, called TOE. To get that, one has to quantize gravity, which means that the gravitational field should be coming in quanta, called gravitons. This is &lt;em&gt;ongoing research&lt;/em&gt; and connected with string theory research, since, up to now, string theories are the only ones that have come up with both the quantum levels needed to describe particles and also to quantize gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deepak's answer is a good  beginning and also sb1's answer. Otherwise start with a quantum mechanics course.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-13T13:24:25.437" LastActivityDate="2011-03-13T13:24:25.437" CommentCount="3" />
  <row Id="6766" PostTypeId="2" ParentId="6567" CreationDate="2011-03-12T21:21:10.237" Score="0" ViewCount="" Body="&lt;p&gt;The question was motivated because I have a suspicion that the electron does not participate to the source of a gravitational field, and eventually not even responds to such a field.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1908 &lt;a href=&quot;http://en.wikipedia.org/wiki/Oil-drop_experiment&quot; rel=&quot;nofollow&quot;&gt;Milikan measured&lt;/a&gt; the charge on a single electron. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Mass-to-charge_ratio#The_electron&quot; rel=&quot;nofollow&quot;&gt;charge-to-electron mass ratio&lt;/a&gt; $q/m_{e}$ was calculated by Thomson in 1897 using the angular momentum and the deflection due to a perpendicular magnetic field. I think that this value reflects the inertial mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this old (1964) doc &#xA;&lt;a href=&quot;http://www.archive.org/details/nasa_techdoc_19650005334&quot; rel=&quot;nofollow&quot;&gt;Gravitational and resonance experiments on very low-energy free electrons&lt;/a&gt; by Fairbank,&#xA;and this entry &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v19/i18/p1049_1&quot; rel=&quot;nofollow&quot;&gt;Experimental Comparison of the Gravitational Force on Freely Falling Electrons and Metallic Electrons&lt;/a&gt; by WittBorn and Fairbank, 1967, with the abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A free-fall technique has been used to&#xA;  measure the net vertical component of&#xA;  force on electrons in a vacuum&#xA;  enclosed by a copper tube. This force&#xA;  was shown to be less than 0.09mg,&#xA;  where m is the inertial mass of the&#xA;  electron and g is 980 cm/sec2. This&#xA;  supports the contention that gravity&#xA;  induces an electric field outside a&#xA;  metal surface, of magnitude and&#xA;  direction &lt;strong&gt;such that the gravitational&#xA;  force on electrons is cancelled&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It seems that the issue remains unsettled.&lt;/strong&gt; - Docs of &lt;a href=&quot;http://rmp.aps.org/abstract/RMP/v64/i1/p237_1&quot; rel=&quot;nofollow&quot;&gt;1992&lt;/a&gt; and &lt;a href=&quot;http://www.sciencedirect.com/science?_ob=ArticleURL&amp;amp;_udi=B6V3S-4N68NMP-7&amp;amp;_user=10&amp;amp;_coverDate=12/31/2007&amp;amp;_rdoc=1&amp;amp;_fmt=high&amp;amp;_orig=gateway&amp;amp;_origin=gateway&amp;amp;_sort=d&amp;amp;_docanchor=&amp;amp;view=c&amp;amp;_acct=C000050221&amp;amp;_version=1&amp;amp;_urlVersion=0&amp;amp;_userid=10&amp;amp;md5=4f55eba3d6981d78151d2caac40b94cb&amp;amp;searchtype=a&quot; rel=&quot;nofollow&quot;&gt;2007 - (Tests of the weak equivalence principle for charged particles in space)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fairbank, as everyone else back then, and even now, believed that the electron must participate in gravity, contrary to my suspicion, preferring to imagine the existence of an imaginary induced electric force, possibly because in the 50s and 60s existed some hype about possible &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrogravitics&quot; rel=&quot;nofollow&quot;&gt;effects relating electricity and gravity&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experimentation is central to advancement of Physics, and the solution of this unsettled issue may prove important in the genesis of a sucessful &lt;strong&gt;theory encompassing both the so called particles and the gravitation field.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can understand that performing the test of Fairbank under microgravity conditions can discriminate if the electron responds to the gravity field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To test if the electron has an active gravitational mass a possible test that may work ('I do not know the tecnological limits...') could be done using a collimated beam of slow neutrons with a path traversing between the plates of a very large high-voltage capacitor. The possible deflection, or not,  of the beam may prove very interesting.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-03-12T21:21:10.237" CommentCount="1" />
  <row Id="6767" PostTypeId="2" ParentId="6738" CreationDate="2011-03-12T22:06:20.730" Score="1" ViewCount="" Body="&lt;p&gt;Consider a cavity or a black body in thermal equilibrium. According to classical notions, every harmonic oscillation mode in that system contains on average the same amount of energy, in equilibrium. But the density of modes rises proportionally to the frequency. Thus the system contains an infinite amount of energy, which obviously doesn't happen in reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantization resolves this. If the modes of the oscillation come at discrete energy levels at much larger steps than the average equilibrium thermal energy, that energy can't occupy every mode equally any more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This however comes with the price of uncertainty. Consider the interaction of a photon and electron in a measurement process. Because of the wave particle duality, energy of the photon is proportional to the frequency of it's wave. If you want a sharp picture of the electron, a point particle, you need high frequency photons to scatter with it. The electron gains kinetic energy in the scattering and so taking an infinitely sharp picture of the electron is practically impossible, you end up with a fuzzy picture no matter what.&lt;/p&gt;&#xA;" OwnerUserId="1670" LastEditorUserId="1670" LastEditorDisplayName="" LastEditDate="2011-03-13T00:30:45.700" LastActivityDate="2011-03-13T00:30:45.700" />
  <row Id="6768" PostTypeId="1" AcceptedAnswerId="6774" CreationDate="2011-03-12T22:45:45.723" Score="10" ViewCount="2029" Body="&lt;p&gt;With all the hubbub over the reactors in Japan and nuclear safety, I was wondering why the insertion of the control rods in Fukushima did not shut down the reactor? Shouldn't they slow the fission reaction so far that cooling is not needed? Why do they need to continue cooling the core when I would expect no more exothermic reaction to take place?&lt;/p&gt;&#xA;" OwnerUserId="435" LastEditorUserId="435" LastEditorDisplayName="" LastEditDate="2011-03-13T18:50:14.470" LastActivityDate="2011-05-12T15:58:27.557" Title="Why didn't the control rods in Fukushima shut down the reactor?" Tags="&lt;nuclear-engineering&gt;" AnswerCount="5" CommentCount="5" />
  <row Id="6769" PostTypeId="1" CreationDate="2011-03-12T22:48:28.373" Score="0" ViewCount="204" Body="&lt;p&gt;Is there an upward force on a object at rest? If yes, where does it come from?&lt;/p&gt;&#xA;" OwnerUserId="2514" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-12T23:00:33.350" LastActivityDate="2011-04-15T20:36:01.730" Title="Upward force on a object at rest" Tags="&lt;gravity&gt;&lt;forces&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="6770" PostTypeId="2" ParentId="6769" CreationDate="2011-03-12T22:54:52.907" Score="6" ViewCount="" Body="&lt;p&gt;Yes, it's called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_force&quot; rel=&quot;nofollow&quot;&gt;normal force&lt;/a&gt;.  It comes from the rigidity of the stuff separating the object from the center of gravitational attraction, i.e. the rigidity of the rocks, dirt, floor, table, etc.  If you'd like, you could think of this stuff as behaving like a spring with a huge spring constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any first-year physics textbook will cover this; there's a very incomplete &lt;a href=&quot;http://physics.stackexchange.com/questions/6068/list-of-good-classical-physics-books&quot;&gt;list of books&lt;/a&gt; in another question.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-03-12T23:06:13.427" LastActivityDate="2011-03-12T23:06:13.427" CommentCount="7" />
  <row Id="6771" PostTypeId="1" AcceptedAnswerId="6772" CreationDate="2011-03-12T23:45:13.973" Score="3" ViewCount="112" Body="&lt;p&gt;I am hoping that someone can clarify this for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these equations. (Boltzmann Law) and radius of Radius of star how does surface temp scale with mass&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$R _x \approx R_\bigodot (\frac{M_x}{M_\bigodot})^.5$ &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$L _x \approx L_\bigodot (\frac{M_x}{M_\bigodot})^3.5$ &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\frac{L}{\pi R^2} = \sigma T^4$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$T_\bigodot = 5800 K$   = Surface Temp of Sun&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I have done is that I have substituted $R_x$ and $L_x$ in the 3rd equation and simplified it in terms of $T$ where my final expression is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$T = (\frac{L_\bigodot M_x^2.5}{\pi\sigma M_\bigodot^2.5 R_\bigodot})^0.25$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Am i done? What am I suppose to do next?&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2491" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-13T20:17:21.923" LastActivityDate="2011-03-13T20:17:21.923" Title="Star Surface Temperature Vs. Mass" Tags="&lt;astrophysics&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="6772" PostTypeId="2" ParentId="6771" CreationDate="2011-03-13T00:13:31.280" Score="2" ViewCount="" Body="&lt;p&gt;Notice that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{L_{\bigodot}}{\pi R_{\bigodot}^2} = \sigma T_{\bigodot}^4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that, dividing through the relation for an arbitrary star and that for the Sun gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{L/L_{\bigodot}}{R^2/R_{\bigodot}^2} = T^4/T_{\bigodot}^4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the other relations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{(M/M_{\bigodot})^{3.5}}{M/M_{\bigodot}} = (T/T_{\bigodot})^4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{M}{M_{\bigodot}} \right)^{2.5} = \left(\frac{T}{T_{\bigodot}}\right)^4$$&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2011-03-13T00:13:31.280" />
  <row Id="6774" PostTypeId="2" ParentId="6768" CreationDate="2011-03-13T01:05:00.843" Score="16" ViewCount="" Body="&lt;p&gt;Per this article on the subject: &lt;a href=&quot;http://theenergycollective.com/nathantemple/53384/how-shutdown-and-core-cooling-japanese-reactors-likely-functions&quot; rel=&quot;nofollow&quot;&gt;http://theenergycollective.com/nathantemple/53384/how-shutdown-and-core-cooling-japanese-reactors-likely-functions&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Even with rods inserted, the reactor&#xA;  continues to produce heat equivalent&#xA;  to about 3% of its full power level.&#xA;  This is not the same as taking a pot&#xA;  off the stove and letting it cool.&#xA;  There are still some atoms splitting&#xA;  and fission products decaying that&#xA;  produce heat. This drops off slowly&#xA;  and is why there needs to be layers of&#xA;  redundant cooling with backup power.&#xA;  During such an earthquake, power from&#xA;  outside the plant would not be&#xA;  expected to be available.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2517" LastEditorUserId="2517" LastEditorDisplayName="" LastEditDate="2011-03-13T08:07:45.037" LastActivityDate="2011-03-13T08:07:45.037" CommentCount="3" />
  <row Id="6775" PostTypeId="1" CreationDate="2011-03-13T01:08:36.457" Score="2" ViewCount="209" Body="&lt;p&gt;Suppose i try to derive the most generic Dirac-like equation (that is, as factors of first-order expression in momenta and mass operator where we allow coefficients that are associative, don't necessarily commute to each other, but they commute with the space-time degrees of freedom)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually the Dirac equation is obtained by taking this expression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E^2 = P^2 + m^2$&#xA; with $E=i \hbar \partial_{t}$, $P=-i \hbar \partial_{x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and getting $E^2 = ( \beta m + \alpha_{i} P_{i} )^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and obtaining a satisfying algebra:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\beta \alpha_{i} + \alpha_{i} \beta = 0$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\alpha_{i} \alpha_{j} + \alpha_{j} \alpha_{i} = 0$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\beta^2 = \alpha_{i}^2 = 1$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However what happens if i don't assume a squared expression and just try to factorize the operator; I note that the most generic expression satisfying the above would be this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E^2 = ( \hat{\beta} m + \hat{\alpha_{i}} P_{i} ) ( \hat{\beta^{-1}} m + \hat{\alpha^{-1}_{i}} P_{i} )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where i use the hat to denote that these are more generic that the usual $\beta, \alpha_{i}$ that we deal with traditionally&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This system satisfies the original expression if we demand the following algebra;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\hat{\beta} \hat{ \alpha^{-1}_{i}} + \hat{\alpha_{i} } \hat{\beta^{-1}} = 0$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\hat{\alpha_{i}} \hat{ \alpha^{-1}_{j}} + \hat{\alpha_{j} } \hat{\alpha^{-1}_{i}} = 0$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The question here is: It seems to me that there is a gauge freedom in specifying these quantities; &lt;strong&gt;what is the meaning of these gauge symmetries and what, if any, interesting invariants we should expected to get from them?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note 1:&lt;/strong&gt; if i assume the following representation of these hatted cliffords in terms of the well-known ones;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\hat{\beta} = \boldsymbol{L} \beta \boldsymbol{M} $&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\hat{\alpha_{i}} = \boldsymbol{L} \alpha_{i} \boldsymbol{M}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;i notice that this satisfies the required algebra &lt;em&gt;(and makes the extra degrees of freedom explicit!)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note 2;&lt;/strong&gt; it seems to me that the most generic solutions look like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$( \hat{\beta} m + \hat{\alpha_{i}} P_{i} ) \boldsymbol{u} = \lambda_{0} \boldsymbol{v} $&lt;/li&gt;&#xA;&lt;li&gt;$( \hat{\beta^{-1}} m + \hat{\alpha^{-1}_{i}} P_{i} ) \boldsymbol{v} = \lambda_{1} \boldsymbol{u} $&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;is this just a rewrite of the left-hand and right-hand Weyl spinors, or is something different?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have purposely not assumed any space-time or spinorial dimensionality in the above, but feel free to make your answers about concrete dimensions&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-03-14T15:14:40.210" LastActivityDate="2011-03-14T15:14:40.210" Title="Dirac equation algebraic derivation, a gauge symmetry" Tags="&lt;dirac-equation&gt;&lt;clifford-algebra&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6776" PostTypeId="1" CreationDate="2011-03-13T01:09:40.750" Score="3" ViewCount="160" Body="&lt;p&gt;Get up off the plane of the ecliptic by a couple of million miles. Look back at the Sun and watch the Earth's orbit in time-lapse for a few centuries. The orbit is an ellipse tilted at 23.5 degrees from the ecliptic. Hold station with the Sun as it rotates about the galactic center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: Does that tilted ellipse -- Earth's orbit -- precess? Over millions of years, does the long axis rotate with the Sun's rotation in the galaxy? Or does it stay fixed in some larger frame of reference, such as the Local Group of galaxies? Or does it exhibit some more complex and subtle motion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the long axis nutate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the angle of the orbit to the ecliptic change in any periodic way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? What are the forces in play?&lt;/p&gt;&#xA;" OwnerUserId="2518" LastActivityDate="2011-03-13T15:05:47.813" Title="Does the long axis of Earth's orbit precess? Does it nutate?" Tags="&lt;astrophysics&gt;&lt;astronomy&gt;&lt;solar-system&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="6777" PostTypeId="1" AcceptedAnswerId="6781" CreationDate="2011-03-13T01:56:46.073" Score="3" ViewCount="188" Body="&lt;p&gt;A positive pion is an up and an anti-down. A negative pion is a down and an anti-up. What's a pion with an electrical charge of 0?&lt;/p&gt;&#xA;" OwnerUserId="2305" LastActivityDate="2011-03-13T08:50:55.417" Title="What Quark and Anti-quark are electrically neutral Pions made out of?" Tags="&lt;particle-physics&gt;&lt;quantum-mechanics&gt;" AnswerCount="2" />
  <row Id="6778" PostTypeId="1" CreationDate="2011-03-13T01:58:34.277" Score="4" ViewCount="157" Body="&lt;p&gt;A purely phase-modulated signal has no power modulation.  This is obvious enough if you look at the time series, but I'd like to &quot;see&quot; it in the frequency domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In physical terms, if we take a laser beam and apply phase modulation via an electro-optic modulator and then send this beam to a photodiode, we will see only a DC term on the photodiode.  I would like to work out how the sidebands created by phase modulation manage to conspire to eliminate any beat frequencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The operation of phase modulation at a frequency $\Omega$ consists of multiplication by $\exp\{i\Gamma\sin\Omega t\}$ which is easily seen to have unit amplitude at all times.  Via the Jacobi-Anger identity, this can be written in terms of sidebands: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\exp\{i\Gamma\sin\Omega t\} = \sum_{n=-\infty}^\infty J_n(\Gamma) \exp\{i n \Omega t\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\exp\{i n \Omega t\}$ is interpreted as creating a sideband at a frequency of $n\Omega$ Hz away from the carrier whose amplitude is given by the nth Bessel function $J_n(\Gamma)$ where $\Gamma$ is the modulation depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A photodiode will see the modulus squared of this signal.   We can write down the result by grouping terms by like powers of $\exp\{i\Omega t\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to see that the energy is conserved at DC due to the relation $ 1 = \sum_{n=-\infty}^\infty |J_n(x)|^2 $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, it's easy to see that there is no signal at frequency $1\Omega$ on the photodiode, because the beat of the $n$th upper sideband with the $(n-1)$st upper sideband always cancels with the beat of the nth lower sideband with the $(n-1)$st lower side band, because of the property of the Bessel functions that $J_{-n}(x)=(-1)^n J_n(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the same sort of nice cancellation does not appear for the $2\Omega$ signal, because the bessel functions always appear in products with two bessel functions of the same parity, i.e. $\cdots + J_{-3}J_{-1} + J_{-2}J_0 + J_{-1}J_1 + J_0 J_2 + J_1 J_3 + \cdots$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can we show that there is no beat signal at frequencies $2\Omega$ (and higher)?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-03-13T20:16:20.427" Title="How does phase modulation conspire to eliminate power variations?" Tags="&lt;optics&gt;&lt;signal-processing&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6780" PostTypeId="2" ParentId="6769" CreationDate="2011-03-13T02:07:14.800" Score="1" ViewCount="" Body="&lt;p&gt;It is all electrostatics. The electrons on the outer shells of the atoms of the object don't want to be anywhere near the electrons of the atoms on the resting surface providing a repelling force which increases with proximity. When this force balances with gravity you have reached &quot;equilibrium&quot;. In fact, everything is somewhat fluid as the atoms move and vibrate nothing is really static. But on a macroscopic scale it is unnoticeable.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-03-13T02:07:14.800" />
  <row Id="6781" PostTypeId="2" ParentId="6777" CreationDate="2011-03-13T02:21:20.140" Score="5" ViewCount="" Body="&lt;p&gt;Up  and anti-up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or down and anti-down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Funny thing is, both of those have the exact same quantum numbers - parity, spin, baryon number and the rest.  So a neutral pion can be a mixture of (u +  anti-u) and (d + anti-d).  There actually result two types of neutral &quot;pion&quot; that decay differently.  One is actually heavier, and we call it the eta meson.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oops I didn't mention yet the strange and anti-strange quark combination, which also gets tangled into the mixes... but it's not important to the neutral pion.&lt;/p&gt;&#xA;" OwnerUserId="353" LastActivityDate="2011-03-13T02:21:20.140" CommentCount="3" />
  <row Id="6782" PostTypeId="1" AcceptedAnswerId="6800" CreationDate="2011-03-13T02:53:33.823" Score="5" ViewCount="224" Body="&lt;p&gt;This concerns the famous two-slit experiment.  Electrons or photons or your favorite particle, doesn't matter.   As we all know, the attempt to detect which slit the quanta pass through leads to loss of the diffraction pattern.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interesting part of the experiment is describing what happens when neither slit has a particle detector.  It is commonly said that each particle goes through both slits at the same time.  At the popular level this makes QM seem weird and mystical, and at the professional level we might cite the Copenhagen Interpretation which boils down to: Don't ask questions about things you can't observe, and so we don't speak of such things, just do the math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How sound this reasoning is at a fundamental level? Can we really conclude each quantum goes through both slits yet as a whole entity?   Can we define a &quot;it went through both slits&quot; observable?  Is there a proper Hermitian operator with (I suppose) eigenvalues 0 and 1, that can distinguish a quantum going through both slits vs. only one slit but without saying which one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps it would make more sense to think about an N-slit experiment, and ask about a Hermitian operator that can report n, the number of slits a quantum takes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect I'm not asking this question quite right, but have an intuition there's something yet to be dug up from this age old gedankenexperiment.  &lt;/p&gt;&#xA;" OwnerUserId="353" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-03-13T20:16:57.277" LastActivityDate="2011-03-14T02:31:02.067" Title="Is it possible to define a &quot;it went through two slits&quot; observable?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-interpretations&gt;&lt;observables&gt;&lt;double-slit-experiment&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="6783" PostTypeId="1" AcceptedAnswerId="6785" CreationDate="2011-03-13T02:53:48.857" Score="2" ViewCount="176" Body="&lt;p&gt;We have two identical massive metal spheres at the same temperature at rest in free space.  Both have an identical charge and the Coulomb force [plus the black-body radiation pressure if the temperature is non-zero] exactly counteracts the gravitational force between them, resulting in no net forces on either object.  They are electrostatically levitating at rest in space.  It is my understanding that the charge distribution in each sphere exists only at the surface, and should be concentrated on the side facing away from the other sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now heat one of the spheres uniformly with an external energy source.  What happens in the instant following?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it be: The increased mass-energy of the hot sphere increases the gravitational force, and the cold sphere starts to fall inward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or else: The increased temperature modifies the charge distribution on the hot sphere by giving it more variance and bringing it closer on average to the cold sphere, increasing the Coulomb force.  The cold sphere starts to fall away.  [Carl's answer says there is no effect on the charge distribution like I describe, but DarrenW points out that there will be an increased black-body radiation pressure between the spheres, similarly causing an outward force.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which direction is correct, and what is the best explanation?  Does it depend on the amount&#xA;of temperature change or the initial conditions?&lt;/p&gt;&#xA;" OwnerUserId="2072" LastEditorUserId="2072" LastEditorDisplayName="" LastEditDate="2011-03-13T23:31:54.270" LastActivityDate="2011-03-13T23:31:54.270" Title="What is the effect of temperature on electrostatic-gravitational balance?" Tags="&lt;gravity&gt;&lt;special-relativity&gt;&lt;temperature&gt;&lt;electrostatics&gt;&lt;equivalence-principle&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="6784" PostTypeId="1" AcceptedAnswerId="6793" CreationDate="2011-03-13T03:21:55.303" Score="6" ViewCount="186" Body="&lt;p&gt;can someone please explain to me what Euler Density is? I encountered it in Weyl anomaly related issues in various articles. Most of them assumes that its familiar, but I couldn't find any accessible paper or a book discussing that. So, it would be nice if I can understand what it is physically and mathematically and also find a reference where I can look it up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also related to that it would be nice to find a reference where people have derived $&amp;lt;T_i^i&amp;gt;$ in curved background which involves Euler Density, $W^i$ etc.&#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1349" LastActivityDate="2011-03-13T06:27:01.837" Title="What is Euler Density?" Tags="&lt;quantum-field-theory&gt;&lt;conformal-field-theory&gt;" AnswerCount="1" />
  <row Id="6785" PostTypeId="2" ParentId="6783" CreationDate="2011-03-13T03:31:24.847" Score="2" ViewCount="" Body="&lt;p&gt;First thing to say is that this system is unstable.  Though the simple gravitational and electrostatic forces may be balanced, there is an additional effect for the electrical interaction that won't exist for gravity: the charge of each sphere polarizes the other.  The charge distributions on each sphere adjust themselves for minimal energy, and that leads to a feeble but non 1/r^2 attractive force.  If anything at all brings the spheres closer together even the tiniest bit, with the plain vanilla Coulomb repulsive force effectively nulled by the attractive gravity (at any distance as both are 1/r^2), this polarizing effect will lead to attraction.  We could try to avoid that by tweaking the charges up for a wee little more repulsion, just enough for balance.  But the system is unstable, like a pencil balanced on its point, since we're balancing forces that follow different power-of-r laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having a nonzero temperature implies random fluctuations. Even at zero temperature, QM says there will be fluctuations.  The net effect of fluctuations over all location on one sphere will lead to random fluctuations in the electric field sensed by the other sphere, and not long you will have to wait for a random extra bit of attraction to start pulling the spheres together.  Will random attractions be cancelled by an equal number of random repulsions?  Short answer: no.  This is like the Van der Waals force between molecules.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making one sphere hotter will add to its time component of energy-momentum, and increase the size and vigor of those random fluctuations.  That would increase the rate at which the random fluctuations and induced polarization do their work.  That, and increased gravity, both increase attraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OTOH, a hotter sphere would emit more thermal radiation - which would push the other sphere away.  How much hotter?  The Stefan-Boltzman law says we can get a lot of bang for the buck, so to speak.  Without a particular physical system to discuss, I imagine that a big increase in temperature would lead to overall repulsion, the thermal radiation winning out over the other effects.  But a small temperature increase?  That might take calculation with particular numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question asks about metal spheres.  So What if the spheres were made of an insulator like ceramic or rubber?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I probably stupidly left off some additional physical effects.  Have at it, commenters!)&lt;/p&gt;&#xA;" OwnerUserId="353" LastActivityDate="2011-03-13T03:31:24.847" CommentCount="4" />
  <row Id="6786" PostTypeId="1" AcceptedAnswerId="6803" CreationDate="2011-03-13T03:44:28.883" Score="2" ViewCount="128" Body="&lt;p&gt;In various articles (I am here talking about specially the ones related to string theory etc.) I have seen the discussion on density and distribution of eigenvalues. I want to know why do we use them (why are they important to consider in some calculation, how to construct them), if there's any physical significances behind them etc. And anything else that you might want to mention. And again a reference could be useful.&#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1349" LastActivityDate="2011-03-27T10:04:26.107" Title="Density/distribution of Eigenvalues" Tags="&lt;quantum-field-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="6787" PostTypeId="1" CreationDate="2011-03-13T03:44:28.963" Score="-1" ViewCount="203" Body="&lt;p&gt;After watching this video:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=j1jIjx0XF_U&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=j1jIjx0XF_U&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experience is made with a speaker that generates a sound wave or mechanic wave. Can you use this to establish a link to HAARP which i believe uses ELF EM radiation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2521" LastActivityDate="2011-03-13T08:27:56.393" Title="Electro Magnetic Waves can cause matter displacement?" Tags="&lt;electromagnetism&gt;&lt;waves&gt;" AnswerCount="3" />
  <row Id="6788" PostTypeId="2" ParentId="6787" CreationDate="2011-03-13T04:02:38.397" Score="2" ViewCount="" Body="&lt;p&gt;I have not watched the video. But the answer of the question in the title is &quot;yes&quot;. Electromagnetic waves carry momentum with it and it can impart momentum to an object. An object at rest in a reference frame can get some momentum from the em waves and start moving unless the object has infinite mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I have watched the video and completely agree with Georg. It is complete crackpottery.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-03-13T07:10:43.947" LastActivityDate="2011-03-13T07:10:43.947" CommentCount="0" />
  <row Id="6789" PostTypeId="1" AcceptedAnswerId="6795" CreationDate="2011-03-13T04:04:00.617" Score="2" ViewCount="178" Body="&lt;p&gt;Is it possible to find the true anomaly of an object in a Kepler orbit given the orbital period of the object, the orbital eccentricity and the time? Assuming a two body system and the mass of the orbiting body is negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm doing this computationally so I'd like to be able to place an object on an orbital ellipse with as few evaluations as possible. I ask about these orbital elements specifically because at the time in the program that I need to calculate the true anomaly, I have these values on hand already.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Edit&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I've been searching extensively for an answer to this, and found &lt;a href=&quot;http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Position_as_a_function_of_time&quot; rel=&quot;nofollow&quot;&gt;these four steps&lt;/a&gt; to solving position as a function of time. The thing I'm having most trouble with is solving Kepler's equation -- it seems like there should be a simpler way to do this -- I can find the velocity at each apsis, and the orbital period relatively easily. As the eccentricity approaches 1 the orbiting object would spend more time near apoapsis and less time near periapsis -- but I can't seem to put that into numbers. &lt;/p&gt;&#xA;" OwnerUserId="417" LastEditorUserId="417" LastEditorDisplayName="" LastEditDate="2011-03-13T05:35:25.297" LastActivityDate="2011-03-13T07:12:43.917" Title="Find true anomaly given period, eccentricity and time" Tags="&lt;astronomy&gt;&lt;orbital-motion&gt;" AnswerCount="2" />
  <row Id="6790" PostTypeId="2" ParentId="6787" CreationDate="2011-03-13T05:31:51.397" Score="1" ViewCount="" Body="&lt;p&gt;This video is complete crackpottery combined with conspiracy babble. &#xA;The link to HAARP is like the link between &quot;Kuchenbacken und Ar...backen&quot;&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-13T05:31:51.397" />
  <row Id="6791" PostTypeId="2" ParentId="6776" CreationDate="2011-03-13T05:54:01.363" Score="1" ViewCount="" Body="&lt;p&gt;The plane of earths orbit &lt;strong&gt;is&lt;/strong&gt; the ecliptic, isn't it? The 23.5 degress are the angle between the earths axis and a normal to the ecliptic. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-13T15:05:47.813" LastActivityDate="2011-03-13T15:05:47.813" />
  <row Id="6792" PostTypeId="1" CreationDate="2011-03-13T06:07:11.277" Score="3" ViewCount="145" Body="&lt;p&gt;The Polyakov action for a string worldsheet has Weyl invariance. In the conformal gauge augmented with Weyl gauge-fixing, we can always impose a flat worldsheet metric in Minkowski coordinates. The residual gauge symmetries take on the form of conformal Virasoro and anti-Virasoro algebras. This is equivalent to a two dimensional conformal field theory. Later, we impose the Virasoro constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, can't we come up with more general worldsheet actions which are invariant under diffeomorphisms but not Weyl transformations? The conformal gauge is still possible, but the worldsheet volume factor is dynamical. Residual diffeomorphisms taking on the same form as conformal transformations, minus the compensating Weyl transformation, still exist. They still take on the form of Virasoro algebras and Virasoro constraints. This is no longer a conformal field theory because we have a characteristic length scale on the worldsheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of these theories are also modular invariant. Do they describe valid string theories?&lt;/p&gt;&#xA;" OwnerUserId="2523" LastActivityDate="2011-03-13T07:52:15.287" Title="Is Weyl invariance absolutely necessary for string worldsheets?" Tags="&lt;string-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6793" PostTypeId="2" ParentId="6784" CreationDate="2011-03-13T06:13:16.393" Score="8" ViewCount="" Body="&lt;p&gt;Euler density is simply the integrand in $2n$ dimensions of the integral that is equal to the Euler characteristic. The Euler characteristic may be written as the integral of the following Euler density in $2n$ dimensions:&#xA;$$E_{2n} = \frac{1}{2^n} R_{i_1 j_1 k_1 l_1} \dots R_{i_n j_n k_n l_n} \epsilon^{i_1 j_1 \dots i_n j_n} \epsilon^{k_1 l_1 \dots k_n l_n} $$&#xA;Note that for $n=1$ i.e. in two dimensions, it is linear in the Riemann tensor - and therefore also in the Ricci scalar (because the Riemann tensor is fully determined by the Ricci scalar in 2D). In four dimensions, the Euler density is quadratic in the Riemann tensor, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Euler character - a &quot;regularized number of points in a manifold&quot; - may also be calculated in many other ways, e.g. for polytopes by adding the number of faces, subtracting edges, adding vertices, etc. For nice manifolds, it's only nonzero for even-dimensional manifolds. For closed orientable two-dimensional Riemann surfaces, it is given by $2-2h$ where $h$ is the number of handles (the genus also known as $g$). One may construct a general open/closed orientable/unorientable two-dimensional manifold by adding $b$ (circular) boundaries i.e. holes and $c$ crosscaps (holes with identified antipodal points, creating an unorientable manifold) and the total Euler characteristic is then&#xA;$$ \chi = 2-2g - b - c.$$&#xA;You should imagine that if there is a function $L(\sigma^i)$ depending on the manifold's coordinates sigma such that $L$ has units of $U$, the path integral $\int DL(\sigma^i)$ has units of $U^\chi$ where $\chi$ is the Euler characteristic: that's what I meant by saying that $\chi$ is the regularized number of points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the Euler characteristic (or character) is arguably the most important and most elementary topological invariant of a manifold. The fact that the integral of $E_{2n}$ is a topological invariant may be seen by calculating its variation which vanishes (for any variation of the metric) - one reduces the derivative to some of the standard identities for the Riemann tensor, especially the two Bianchi identities involving antisymmetrization (and, in one case, one derivative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivation of the trace of the stress-energy tensor is done for $d=2$ in Polchinski's &quot;String Theory&quot;, Volume I. Equation (3.4.31) says&#xA;$$T^a_a (\sigma) = -\frac{C}{12} R(\sigma)$$&#xA;where $R$ is the Ricci scalar, also interpretable as a multiple of the Euler density. The $C$ ends up being a definition of the central charge. I don't know the general form of a similar equation in $d$ dimensions but its exact form - at least the parameters - do depend on the theory. I guess that in general, the trace is equal to some linear combination of the Euler density and perhaps some other generators besides the stress-energy tensor.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-13T06:27:01.837" LastActivityDate="2011-03-13T06:27:01.837" CommentCount="3" />
  <row Id="6794" PostTypeId="2" ParentId="6789" CreationDate="2011-03-13T06:19:10.247" Score="1" ViewCount="" Body="&lt;p&gt;An ellipse can be treated mathematically as a circle and one &quot;epicycle&quot;. &#xA;This might be a method easily (fast) calculated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I remember from the 60ties, that NASA used epicycle methods to calculate &#xA;astronomic data for the moon flights. Still today? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This links seem interesting: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://linuxcommand.org/man_pages/epicycle1.html&quot; rel=&quot;nofollow&quot;&gt;http://linuxcommand.org/man_pages/epicycle1.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dspace-unibg.cilea.it/bitstream/10446/338/1/WPMateRi02(2006)Dixon.pdf&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-13T07:12:43.917" LastActivityDate="2011-03-13T07:12:43.917" CommentCount="3" />
  <row Id="6795" PostTypeId="2" ParentId="6789" CreationDate="2011-03-13T06:36:17.860" Score="1" ViewCount="" Body="&lt;p&gt;Kepler's equation: $M=E-\varepsilon\cdot \sin( E)$, you need to solve for $E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inverse problem may be solved (according to wikipedia) as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E = \begin{cases} \displaystyle \sum_{n=1}^{\infty} {\frac{M^{\frac{n}{3}}}{n!}} \lim_{\theta \to 0} \left( \frac{\mathrm{d}^{\,n-1}}{\mathrm{d}\theta^{\,n-1}} \left( \frac{\theta}{ \sqrt[3]{\theta - \sin(\theta)} } ^n \right) \right) , &amp;amp; \epsilon = 1 \\ \displaystyle \sum_{n=1}^{\infty} { \frac{ M^n }{ n! } } \lim_{\theta \to 0} \left( \frac{\mathrm{d}^{\,n-1}}{\mathrm{d}\theta^{\,n-1}} \left( \frac{ \theta }{ \theta - \epsilon \cdot \sin(\theta)} ^n \right) \right) , &amp;amp; \epsilon \ne 1 \end{cases}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which evaluates to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E = \begin{cases} \displaystyle M + \frac{1}{60} M^3 + \frac{1}{1400}M^5 + \cdots \ | \ x = ( 6 M )^\frac{1}{3} , &amp;amp; \epsilon = 1 \\ \\ \displaystyle \frac{1}{1-\epsilon} M - \frac{\epsilon}{( 1-\epsilon)^4 } \frac{M^3}{3!} + \frac{(9 \epsilon^2 + \epsilon)}{(1-\epsilon)^7 } \frac{M^5}{5!} + \cdots , &amp;amp; \epsilon \ne 1 \end{cases} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is copied &lt;a href=&quot;http://en.wikipedia.org/wiki/Kepler%27s_Equation&quot; rel=&quot;nofollow&quot;&gt;straight out of wikipedia&lt;/a&gt;. It's kind of ugly. Plus, the first equation is written in &quot;x&quot; instead of &quot;M&quot; and you can't be sure it's correct.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So instead, what I'd try is to take the first terms from the above, i.e. put $E_0 = M$ or $M/(1-\epsilon)$, and use &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_method&quot; rel=&quot;nofollow&quot;&gt;Newton's method&lt;/a&gt; to iterate. That is, iterate with:&lt;br&gt;&#xA;$E_{n+1} = E_n - \frac{M - E +\epsilon  \sin E}{-1+\epsilon\cos(E)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To test the method, run a bunch of random data through it, and make sure you test the fence posts (boundary conditions).&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-13T06:36:17.860" CommentCount="3" />
  <row Id="6796" PostTypeId="2" ParentId="6775" CreationDate="2011-03-13T06:38:47.373" Score="2" ViewCount="" Body="&lt;p&gt;There is no &quot;gauge symmetry&quot; in any of these manipulations and you haven't offered any evidence for such a non-existent gauge symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand you well, you have just realized that the exact numerical form of the Dirac matrices depends on conventions. In particular, the matrix entries of the Dirac gamma (and therefore also alpha, beta) matrices depend on the choice of the basis of the complex four-dimensional space of the Dirac spinors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;solution&quot; - a set of matrices $\alpha^k$ which contains the $\beta$ as well (or similarly for $\gamma$) - may also be conjugated by an arbitrary matrix&#xA;$$\{ \alpha^k \} \to \{ U \alpha^k U^{-1} \} $$&#xA;with a universal, $k$-independent matrix $U$. We often want to preserve the hermiticity or antihermiticity of the matrices so we only allow orthonormal bases of the 4D space of Dirac matrices - which means that $U$ is unitary and $U^{-1}$ is the same thing as $U^\dagger$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a well-known thing that one can choose the bases of such spaces differently. For example, one may do it for vectors as well: the coordinates of a vector depend on the choice of the coordinate axes. To some extent, the choice for the spinors is independent of the choice of axes for a vector. But once we choose the gamma matrices (or, equivalently, the beta and alpha matrices), all the conventions for spinors are settled and there is no freedom left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ambiguity in choosing the numerical form of the gamma (or alpha, beta) matrices is not a symmetry of any theory. It's not a gauge symmetry and it's not a global symmetry, either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's because a particular theory describing spinorial degrees of freedom has to depend on the gamma matrices - so the Lagrangian (or the Hamiltonian or other defining quantities) explicitly depends on the form of the Dirac matrices which must be chosen in a particular way. So any well-defined Lagrangian for spinors actually breaks the would-be symmetry and removes the ambiguity. (Of course, we're still allowed to define any theory with any convention, but the precise form of the theory will be different - the theories will be equivalent, but not identical, which is why the change of the gamma matrices is not a symmetry.) Once it says which form of the gamma matrices are used, one must use the same form for all other manipulations with the spinors, otherwise the key identities etc. wouldn't hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I may have completely misunderstand what you're asking. The individual sentences of your question don't really make sense. For example, you say that you find a general &quot;solution&quot;, but the &quot;solution&quot; is actually another unsolved equation. It seems you are confusing the gamma matrices with the spinors themselves, solutions with equations, and so on. Consequently, you also seem to confuse the Dirac equation with the identities that allow one to prove that the Dirac equation implies the Klein-Gordon equation, and so on and so on. It's very hard to read this kind of stuff.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-13T06:45:11.830" LastActivityDate="2011-03-13T06:45:11.830" CommentCount="2" />
  <row Id="6798" PostTypeId="1" CreationDate="2011-03-13T06:49:18.657" Score="4" ViewCount="417" Body="&lt;p&gt;What are the modern marvels in Physics which are crazy but true? Ideas which are ridiculed and dismissed in the beginning but passed the test of time?&lt;/p&gt;&#xA;" OwnerUserId="2524" LastEditorUserId="2192" LastEditorDisplayName="" LastEditDate="2011-03-13T20:17:29.757" LastActivityDate="2011-03-15T06:01:07.813" ClosedDate="2011-03-13T20:18:20.863" Title="What are the modern marvels in Physics that seem crazy but true?" Tags="&lt;soft-question&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="6799" PostTypeId="2" ParentId="6798" CreationDate="2011-03-13T06:58:50.993" Score="2" ViewCount="" Body="&lt;p&gt;Just about all of them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the Earth going around the sun through to relativity and quantum mechanics&lt;/p&gt;&#xA;" OwnerUserId="2525" LastEditorUserId="2525" LastEditorDisplayName="" LastEditDate="2011-03-15T06:01:07.813" LastActivityDate="2011-03-15T06:01:07.813" CommentCount="5" />
  <row Id="6800" PostTypeId="2" ParentId="6782" CreationDate="2011-03-13T07:02:14.193" Score="4" ViewCount="" Body="&lt;p&gt;It's a very good question but the answer is No, there is nothing such as &quot;it went through both slits&quot; observable (i.e. no linear operator that would correspond to this Yes/No question). The reason is that such &quot;information&quot; cannot be observed, not even in principle and not even statistically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much more generally, there don't exist any observables that would say &quot;the physical system did X or Y before it was observed&quot;. It's the very reason why the term &quot;observable&quot; was chosen because things that are, by definition, unobservable because they only occur in the middle of the experiment are simply not observables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these statements are very manifest e.g. in Feynman's path integral formulation of quantum mechanics. One always has to sum the probability amplitudes over all conceivable histories. In this sense, using the double-slit example, the particle &lt;em&gt;always&lt;/em&gt; goes through both slits. For some particular places where the particle may be detected, the first or second slit may contribute zero to the probability density. But the whole probability distribution is always affected by both slits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote that this question is good because these issues are misunderstood by most laymen - and not just those who admit that they are laymen but also many authors of philosophical books about quantum mechanics etc. They usually want to simplify their life and imagine that the particle, after some moment, is guaranteed to &quot;behave as a particle&quot;. When the experimenter does something, he may forget that the particle has quantum and wave-like properties, and he may imagine it has only gone through one of the slits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as the &quot;delayed choice quantum eraser&quot; experiments show, this reasoning is always invalid. The particle may always &quot;recall&quot; that it has wave-like properties, and by modifying the future portions of its journey, the other slit - and the interference between both slits - may always become important once again. There's no reason why Nature should have answers to questions that can't be measured - such as &quot;what was happening before the measurement&quot; - and indeed, whenever we are in the quantum regime and the classical approximation is inapplicable, Nature chooses not to have answers to any of these questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observables are quantities that can be shown as values at actual measuring apparatuses and quantum mechanics can only predict the odds that the value of an observable will be one number or another. Everything else that some people may want to imagine as the &quot;detailed history&quot; that &quot;preceded&quot; the measurement is unphysical.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-13T07:02:14.193" />
  <row Id="6801" PostTypeId="2" ParentId="6783" CreationDate="2011-03-13T07:12:11.840" Score="2" ViewCount="" Body="&lt;p&gt;Heat has no effect on E&amp;amp;M until you get hot enough that the spheres are modified in shape, or hot enough that the charges are able to leave the sphere. So before these sorts of things happen, yes, the increased gravitational attraction will increase the attraction.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In any conceivable practical implementation, these two effects will occur long, long before the gravitational force is changed sufficient to take them significantly out of balance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practical example: 1000kg spheres, 0.5m in radius, nearly touching. The gravitational attraction is $G(1000)^2/1m^2 = 6.67428 \times 10^{-5}$ Newtons, say at 300K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compute the energy change of 1 degree Kelvin, we need to know the substance. Let's use &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron&quot; rel=&quot;nofollow&quot;&gt;iron&lt;/a&gt; so that an increase of 1 degree Kelvin (at 25 °C) gives $25.10$ Joules per mole. At 55.845 grams per mole, 1000kg has 17900 moles so each degree Kelvin increases the energy by about 45,000 Joules. Using $E=mc^2$ so $E/c^2=m$ this gives a mass increase of $5.0\times 10^{-12}$ kg, or about $5\times 10^{-15}$ change in mass. Assuming that both masses are heated up, the change in force is twice this, or $1.0\times 10^{-14}$ per degree Kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current measurements of gravity using spheres gives the gravitational constant to an accuracy of 4 or 5 digits, for example, $6.693(27)(21)\times 10^{-11}$ in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_constant#History_of_measurement&quot; rel=&quot;nofollow&quot;&gt;recent measurement&lt;/a&gt;. In order to get an error in the smallest (6th) digit of 6.67428 you need to have a temperature change of about $10^8$ degrees Kelvin. &lt;a href=&quot;http://en.wikipedia.org/wiki/Plasma_%28physics%29#Ranges_of_plasma_parameters&quot; rel=&quot;nofollow&quot;&gt;At that temperature&lt;/a&gt;, you will have to include fusion (and plasma) effects.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-13T07:12:11.840" CommentCount="2" />
  <row Id="6803" PostTypeId="2" ParentId="6786" CreationDate="2011-03-13T07:22:26.400" Score="3" ViewCount="" Body="&lt;p&gt;Well, whenever there is a large number of &quot;something&quot;, it is always important and relevant to study the &quot;something&quot; by statistical methods. For example, there are many people in China, so it's sensible to ask how many people are there in China, and what is their population density - in the whole China and in its parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density of states $\rho(E)$ whose energy is near a certain value determines the entropy at this value of energy, $S\sim \ln(\rho(E))$: $S$ is typically large so the density of states has to be exponentially large. In string theory, such functions may usually be calculated in several very different ways and compared. The density of high-energy states is, for example, related to the energy of the ground state (by &quot;modular invariance&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like the population density as a function of the latitude and longitude is relevant, so is the whole distribution of the eigenvalues. It may be approximated by various approximate formulae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also important to know how the spacing between two eigenvalues is distributed relatively to the other spacing. If the eigenvalues were numbers that determine the positions of randomly and independently falling needles, the distribution would be &quot;Poisson&quot;: the eigenvalues as functions of energy would display the same patterns as radioactive beeps of a counter as a function of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in quantum systems, the eigenvalues are often distributed differently. Very often, the distribution is similar to that of &quot;random matrix theory&quot; in which the eigenvalues are much less likely to be very close to each other - we say that the &quot;eigenvalues repel&quot;. So the beeping would be much more &quot;rhythmical&quot; than the beeping of the radioactive counters. Ironically enough, the energy levels of large atomic nuclei are the most mundane example of a physical system whose eigenvalues repel - and therefore differ from the beeping of the radioactive counters. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All such statistical tests are doable and they tell us something. So they're important. They're used in every other page of a paper that studies a system with a large number of eigenvalues or eigenstates - which includes any large enough system. One can't enumerate all the situations in which such information is relevant much like one can't give you the references to all papers that use the population density in China.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some introduction to eigenvalues, I may recommend you Jake Barnett, a 12-year-old student of Purdue University in Indianopolis. Here is his tutorial on eigenvalues:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=wShtOJnoeB4&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=wShtOJnoeB4&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-27T10:04:26.107" LastActivityDate="2011-03-27T10:04:26.107" />
  <row Id="6804" PostTypeId="2" ParentId="6787" CreationDate="2011-03-13T07:38:10.063" Score="1" ViewCount="" Body="&lt;p&gt;Dear Luis, in principle, the answer to your question is Yes. Electromagnetic fields carry both energy and momentum - and one may experimentally demonstrate how photons are able to push extraordinarily thin pieces of foil (and make them rotate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the attractive video you included is highly unlikely to correspond to the reality because of the energy conservation. To find out whether the theory in the video is possible, you need not just the Yes/No question whether electromagnetic fields may move objects - they can - but also the quantitative questions e.g. &quot;how much energy you need to cause an earthquake&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to get an idea:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Richter_magnitude_scale&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Richter_magnitude_scale&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Japanese earthquake released energy equivalent to the detonation of 336 megatons of TNT which is 1.41 EJ - or 1.41 times $10^{18}$ Joules. You may calculate it is 392 terawatt-hours. One kilowatt-hour costs about 10 U.S. cents in America so you need to pay about 39 billion dollars for the electricity if you want to use electromagnetic phenomena to create a similar earthquake. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am assuming you will have no losses. This argument is impossible to avoid because of the energy conservation law. So don't expect to be able to produce measurable earthquakes with your microwave oven.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-03-13T08:27:56.393" LastActivityDate="2011-03-13T08:27:56.393" CommentCount="3" />
  <row Id="6805" PostTypeId="2" ParentId="6777" CreationDate="2011-03-13T07:48:07.973" Score="0" ViewCount="" Body="&lt;p&gt;They're made of a combination of up and anti-up, down and anti-down, and strange and anti-strange. And that's just to begin with. To be perfectly accurate you'd have to add all six quarks (okay, maybe not the top).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short reason is that all these quark combinations have the right quantum numbers and so are included. Feynman diagrams of all sorts contribute, for example:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/bXs3d.jpg&quot; alt=&quot;Yet another pion drawing&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above, the wavy lines are virtual vector bosons. I've put two in because, as Anna V notes, the pion is a psuedoscalar and so has zero spin. Of course the heavy quarks don't appear very much in a light meson since they have to be made from borrowed energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People aren't much interested in the heavy quark contribution to the pion. An analogous subject is the heavy quark content of the proton and a reference which will illustrate how this comes about is:&lt;br&gt;&#xA;10.1103/PhysRevD.70.094028, Andrei Kryjevski, &lt;em&gt;Heavy Quark $\bar{q}q$ Matrix Elements in the Nucleon from Perturbative QCD&lt;/em&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/hep-ph/0312196&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-ph/0312196&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the masses of the nucleons are about 1 GeV while the mass of the t-quark is around 171 GeV.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-03-13T08:50:55.417" LastActivityDate="2011-03-13T08:50:55.417" CommentCount="2" />
  <row Id="6806" PostTypeId="2" ParentId="6792" CreationDate="2011-03-13T07:52:15.287" Score="2" ViewCount="" Body="&lt;p&gt;Dear Jenkins, the theories you want to construct are &quot;noncritical string theories&quot; and they're less interesting and less consistent than the &quot;critical string theories&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the Nambu-Goto action - the proper area of the world sheet - is nonlinear. It includes square roots etc. It's much better to introduce an auxiliary metric tensor on the world sheet and the action for the coordinates $X$ becomes nice and bilinear - a free theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we don't want new degrees of freedom to be added. The 2D metric tensor has three independent components. Two of them may be set to a standard form by the 2 degrees of freedom in the 2D coordinate reparameterization symmetry; and the third by the Weyl symmetry if it exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it doesn't exist, it's too bad. The auxiliary world sheet metric may only be brought to the form of $e^\phi \eta_{ab}$. That means that $\phi$, determining the overall scaling, becomes another function of the world sheet coordinates $(\sigma,\tau)$, very analogously to the spacetime coordinates $X(\sigma,\tau)$. In fact, it is really valid to say that the parameter determining the overall scaling of the metric is another spacetime coordinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this coordinate were totally identical to the other coordinates, then there would also be a translation symmetry in the $\phi$ direction - but that's equivalent to the Weyl symmetry (multiplicative scaling of $e^\phi$  is the same thing as additive shifts to $\phi$). Because by assumption, the Weyl symmetry doesn't hold in your theory, the new spacetime coordinate $\phi$ can't have quite the same properties as the other spacetime coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in normal circumstances, you obtain the violations of the Weyl invariance as a disease. In particular, if you try to study string theory in a non-critical dimension, i.e. $D\neq 26$ or $D\neq 10$, you will find out that the field $\phi$ doesn't decouple and the path integral, when calculated including the one-loop accuracy, still depends on $\phi$. So the Weyl symmetry, equivalent to an additive shift of $\phi$ by a function of the world sheet, is not a symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said, this can be interpreted as $\phi$'s becoming a new spacetime coordinate. But if you try to calculate the effective action in the new spacetime that has an additional dimension $\phi$, you will find out that the laws of physics are not invariant under translations in $\phi$ - that's nothing else than the failure of the theory to be Weyl-invariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, you will find out that the dilaton linearly depends on $\phi$: search for papers about &quot;linear dilaton&quot;. The squared gradient of the dilaton is related to the surplus or excess (if it is time-like or space-like) of the spacetime coordinates, relatively to the critical dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the spacetime has two dimensions, one may choose the dilaton to depend on the (only) spacelike coordinate $\phi=X^1$ in such a way that the theory including $\phi$ is Weyl-invariant again. In this case, it's useful to consider not only the right linear dilaton - solving the equations of motion - but also a non-trivial background for the tachyon. One ends up with the so-called &quot;Liouville theory&quot; - a &quot;linear dilaton&quot; theory with some extra tachyonic profile in a non-critical stringy $D=2$ spacetime - which is slightly more consistent than other noncritical string theories. The Liouville theory may also be described by a quantum mechanical model with a large matrix - the old matrix theory.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-13T07:52:15.287" />
  <row Id="6808" PostTypeId="2" ParentId="6754" CreationDate="2011-03-13T08:33:18.647" Score="2" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Coriolis_effect&quot; rel=&quot;nofollow&quot;&gt;Coriolis effect&lt;/a&gt; is proportional to velocity:&lt;br&gt;&#xA;$\boldsymbol{ F}_C = -2 \, m \, \boldsymbol{\Omega \times v}$,&lt;br&gt;&#xA;where $\boldsymbol{\Omega}$ is the angular velocity (of the earth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As waves, tsunamis have very high velocities, you would think it would be enough to see the Coriolis effect, but in fact there is little effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When in deep water, tsunamis move very quickly, but the water of which they are composed move very slowly. &lt;a href=&quot;http://en.wikipedia.org/wiki/Tsunami&quot; rel=&quot;nofollow&quot;&gt;A typical example&lt;/a&gt;, is a wavelength of 200km, amplitude of 1 meter, and a period of around 30 minutes. Therefore, the water only has a speed of on the order of 1 meter per hour. This is a tiny fraction of the speed of clouds in the atmosphere which are effected by the Coriolis effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a tsunami nears land, its height increases dramatically and so the distance the water moves increases as well. Suppose the tsunami moves a distance of 10km with a period of 30 minutes. This implies an average speed of 40km/hour which is fast enough to bring the Coriolis effect into play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting the numbers ($\boldsymbol{v} = 40\; \textrm{km}/\textrm{hour}$ and $\boldsymbol{\Omega} = 2\pi/24\;\textrm{hours}\;\;$ ), into the equation, we find a maximum acceleration of about:&lt;br&gt;&#xA;$\boldsymbol{ F}_C/m = 0.0016\;\textrm{m}/\textrm{s}^2$.&lt;br&gt;&#xA;Since this is small compared to the gravitational acceleration (9.8 m/s$^2$ ) that is governing the last stages of a tsunami, the effect will be completely negligible. Local terrain completely determines the tsunami's behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the Coriolis effect is able to effect clouds because they move for very long times. Over a time of 2 hours, an acceleration of $0.0016 \textrm{m}/\textrm{s}^2$ causes a change in velocity of about $12 \textrm{m}/\textrm{s}$.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-13T08:33:18.647" CommentCount="2" />
  <row Id="6809" PostTypeId="2" ParentId="6768" CreationDate="2011-03-13T08:47:25.113" Score="3" ViewCount="" Body="&lt;p&gt;To be clear: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rods were inserted when shutting down right after the quake. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, the meltdown had occured in much shorter time! &#xA;(I etimate less than an hour)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Here are some figures about the decay heat after shutdown:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Decay_heat&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Decay_heat&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-13T09:52:19.440" LastActivityDate="2011-03-13T09:52:19.440" />
  <row Id="6810" PostTypeId="2" ParentId="6768" CreationDate="2011-03-13T09:41:09.430" Score="2" ViewCount="" Body="&lt;p&gt;Using the Unit 1 electric power rating of 460MW and the 3% figure above give us ~14MW at scram, that is still a lot of power!  So every second 14 million Joules is released(1 Watt = 1 J/S). Now it takes ~4200 joules to heat 1 kilogram of water by 1 degree C. So 14 million joules would heat 1 kg of water to ~3300 degrees C in 1s. You can see why we have a problem! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone check my math, it has been a few years since physics class&lt;/p&gt;&#xA;" OwnerUserId="2528" LastActivityDate="2011-03-13T09:41:09.430" CommentCount="2" />
  <row Id="6811" PostTypeId="1" CreationDate="2011-03-13T09:43:42.540" Score="3" ViewCount="200" Body="&lt;p&gt;Condensed matter physicists have shown using quantum information that in many condensed matter systems, entanglement entropy only scales as the area of the boundary, and not the volume. This is the basis for the density matrix renormalization group and Projected Entangled Pair States (PEPS). Does this also explain the holographic principle in quantum gravity?&lt;/p&gt;&#xA;" OwnerUserId="2529" LastActivityDate="2011-03-13T17:07:25.240" Title="Can PEPS explain the holographic principle in quantum gravity?" Tags="&lt;quantum-gravity&gt;&lt;quantum-information&gt;&lt;entanglement&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="3" />
  <row Id="6812" PostTypeId="2" ParentId="6798" CreationDate="2011-03-13T10:42:59.473" Score="1" ViewCount="" Body="&lt;p&gt;Einstiens theory of relativity was at first considered ridiculous now it is  widely accepted &lt;/p&gt;&#xA;" OwnerUserId="2530" LastActivityDate="2011-03-13T10:42:59.473" CommentCount="4" />
  <row Id="6813" PostTypeId="1" CreationDate="2011-03-13T11:58:59.130" Score="2" ViewCount="169" Body="&lt;p&gt;I just thought of an interesting question.Is the sun actually moving?&#xA;I have learnt that the way the lunar landers and the space ships of the Mercury and Apallo missions moved and controlled their positions through the use of thrusters.NoW apply that same concept to the sun are the constant explosions on the surface of the sun enough to move it?&lt;/p&gt;&#xA;" OwnerUserId="2530" LastActivityDate="2011-03-13T20:16:56.247" Title="is the sun itself moving" Tags="&lt;astrophysics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="6814" PostTypeId="2" ParentId="6798" CreationDate="2011-03-13T12:26:04.187" Score="0" ViewCount="" Body="&lt;p&gt;Relativity and Quantum Theory. That whole division of the waters thing, of the infinite and infinitesimal, leaving our everyday world of beer and sandwiches in the middle. But I still don't understand mass, or gravity or momentum.&#xA;Cataclysm Theory and Continental Drift. Loads of stuff.&#xA;Some of the replies seem a bit shirty, but I'm glad to have found this blog. After all, we have a vicar in every parish and most of them willing to discuss theology; its high time science/scepticism offered the same. I promise not to ask about flying saucers or quantum wormholes, well maybe quantum wormholes...&lt;/p&gt;&#xA;" OwnerUserId="2532" LastActivityDate="2011-03-13T12:26:04.187" CommentCount="1" />
  <row Id="6815" PostTypeId="2" ParentId="6813" CreationDate="2011-03-13T12:30:20.947" Score="2" ViewCount="" Body="&lt;p&gt;All motions are relative and yes the sun is in motion about the center of our Milky Way galaxy, which is itself moving within the local group, which is moving about the center of Virgo cluster, which is a part of Coma supercluster, which is moving towards &quot;the Great Attractor&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constant explosions on the surface of the sun is not causing any significant movement of the sun since explosions are not taking place in any particular direction and more or less taking place on the whole surface. Therefore no significant movement of the center of mass due to these explosions takes place.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="817" LastEditorDisplayName="" LastEditDate="2011-03-13T20:16:56.247" LastActivityDate="2011-03-13T20:16:56.247" />
  <row Id="6816" PostTypeId="1" AcceptedAnswerId="6837" CreationDate="2011-03-13T12:51:56.993" Score="2" ViewCount="156" Body="&lt;p&gt;It would be nice to have a cute method that uses Lorentz transformations of basis vectors by exponential transformation using gamma matrices. To avoid confusion, let's assume -+++ signature. Given $\gamma_\mu$ as gamma matrices that satisfy ${\rm tr}(\gamma_\mu\gamma_\nu) = 4\eta_{\mu\nu}$, then we have ${\rm tr}(\gamma&amp;#39;_\mu\gamma&amp;#39;_\nu) = 4\eta_{\mu\nu}$ if we put:&lt;br&gt;&#xA;$\gamma&amp;#39;_\mu = \exp(-A)\gamma_\mu\exp(+A)$&lt;br&gt;&#xA;where $A$ is any matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boosts and rotations use $A$ as a bivector. For example, with $\alpha$ a real number, $A=\alpha\gamma_0\gamma_3$ boosts in the z direction while $A=\alpha\gamma_1\gamma_2$ gives a rotation around the z axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for the value of $\alpha$ that gives a boost with velocity $\beta = v/c$ appears to be straightforward. But how to do the rotations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And by the way, what happens when you generalize $A$ to be something other than bivectors?&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-03-13T13:13:59.693" LastActivityDate="2011-03-13T17:29:18.487" Title="What is the exponential form gamma matrix for a general rotation and boost?" Tags="&lt;special-relativity&gt;&lt;dirac-equation&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="6817" PostTypeId="1" AcceptedAnswerId="6857" CreationDate="2011-03-13T12:55:16.677" Score="7" ViewCount="173" Body="&lt;p&gt;If an astronomer moves at relativistic speed, the stars and constellations are distorted. He sees the stars towards which he is moving blue shifted, while the ones he's moving away from are red shifted. In addition, the apparent direction of distant stars is modified. I think that the coolest way of representing this is the &quot;relativistic ellipse&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of a spherical shell with the inner surface reflective. Imagine a flashbulb going off at the center. The light is reflected and returns to the source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now think of the same situation observed from a moving frame of reference. The light is still emitted and returns to the center at the same time, but that point has now moved. Since the speed of light is the same in all frames of reference, the distance traveled by each light ray has to be the same. So the paths of the light rays (and the boundary of the spherical shell) describes an ellipse, the &lt;a href=&quot;http://www.relativitybook.com/resources/relativistic_ellipse.html&quot; rel=&quot;nofollow&quot;&gt;&quot;relativistic ellipse&quot;&lt;/a&gt;:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/9PoOJ.png&quot; alt=&quot;RelativisticEllipse&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above drawing is from the book &quot;Relativity in Curved Spacetime&quot; which I highly recommend and can be purchased here:&lt;br&gt;&#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0955706807&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Relativity-Curved-Spacetime-without-relativity/dp/0955706807&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.relativitybook.com/book_cover.html&quot; rel=&quot;nofollow&quot;&gt;http://www.relativitybook.com/book_cover.html&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps when my copy arrives it will have the equation. Should be possible to work it out using Lorentz invariance or contraction or something similar.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Per the answer by &lt;a href=&quot;http://physics.stackexchange.com/users/1257/helder-velez&quot;&gt;Helder Velez&lt;/a&gt;, &lt;a href=&quot;http://physics-quest.org/Book_Chapter_Non_Simultaneity.pdf&quot; rel=&quot;nofollow&quot;&gt;Chapter 4 of Hans de Vries's book&lt;/a&gt; has the following useful diagram:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/gC4Gu.jpg&quot; alt=&quot;HansEllipse&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-03-14T01:29:44.310" LastActivityDate="2011-03-14T01:29:44.310" Title="What is the formula for the relativistic ellipse?" Tags="&lt;special-relativity&gt;&lt;books&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="6818" PostTypeId="2" ParentId="6813" CreationDate="2011-03-13T13:06:09.823" Score="3" ViewCount="" Body="&lt;p&gt;Whether the velocity of the sun changes because of explosive mass ejections depends on whether we include or exclude the ejected mass in the resulting system. The center of mass of &quot;the sun+the ejected mass&quot;, considered as a single system, continues at constant velocity. The velocity of the center of mass of &quot;the sun&quot;, considered as one system, and the velocity of the center of mass of &quot;the ejected mass&quot;, considered as another system, are both changed. So, my classical physics Answer to your Question is no and yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The velocities of &quot;the sun&quot; and &quot;the ejected mass&quot; change so that the total momentum is conserved. &quot;The ejected mass&quot; is very small compared to the mass of &quot;the sun&quot;, so the change of the velocity of &quot;the sun&quot; is very small, though still in proportion to the velocity of &quot;the ejected mass&quot;. A spacecraft's velocity changes quite a lot with its ejection of fuel because the fuel is ejected as fast as we can safely engineer, and the mass of the fuel ejected is ultimately a significant fraction of the initial mass of the spacecraft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various qualifications if we consider the motion of the sun relative to the center of mass of the galaxy. The sun is in orbit around the center of mass of the galaxy, so in classical physics terms the sun is constantly accelerating towards the center of mass of our galaxy, just as the earth is in orbit around and is constantly accelerating towards the center of mass of the solar system. In general relativistic terms, however, a more sophisticated description of the changes of configuration of the earth relative to the sun and of the sun relative to our galaxy is necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two points about how you can think about your question. (1) It's helpful to distinguish between position, velocity, and acceleration, and (2) it's helpful to be as explicit as possible about what the position, velocity, and acceleration are relative to, and to be careful what we choose to use as reference points. I'm moving at a constantly changing velocity relative to you because we are not at the same place on the earth's surface; describing my motion relative to you is considerably more complicated than describing my motion relative to the center of mass and to the rotational axis of the earth. Ultimately, a lot of &lt;em&gt;systematic&lt;/em&gt; care has to be taken. When it comes to a technological application like GPS, a &lt;em&gt;lot&lt;/em&gt; of systematic care is taken to describe and analyze relative positions, velocities, and accelerations.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-03-13T13:06:09.823" CommentCount="1" />
  <row Id="6819" PostTypeId="2" ParentId="6776" CreationDate="2011-03-13T14:27:13.847" Score="2" ViewCount="" Body="&lt;p&gt;The major axis of the Earth's orbit does precess. As Mark Eichenlaub points out, in the case of Mercury this precession provides a classic test of general relativity, but the precession due to general relativity is small compared to the precession due to other, Newtonian effects. In the case of the inner planets of the solar system, the dominant cause of the precession is the gravitational pull of other planets, mainly Jupiter. See the Table &quot;Sources of the precession of perihelion for Mercury&quot; in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tests_of_general_relativity&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; Mark linked to, for instance. Presumably you can follow the reference there for a more scholarly source.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-13T14:27:13.847" />
  <row Id="6820" PostTypeId="1" CreationDate="2011-03-13T14:30:19.133" Score="1" ViewCount="163" Body="&lt;p&gt;Suppose that i have a li-ion battery with voltage 10V (and some capacity, say 1000mAh). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can i charge it completely using 5V voltage? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will hapen if i charge it with 12V voltage?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit:&#xA;Found the &lt;a href=&quot;http://batteryuniversity.com/learn/article/charging_lithium_ion_batteries&quot; rel=&quot;nofollow&quot;&gt;answer&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2534" LastEditorUserId="2534" LastEditorDisplayName="" LastEditDate="2011-03-13T16:02:45.660" LastActivityDate="2011-09-05T18:43:16.017" Title="what happens if i charge a a Li-ion battery with non nominal voltage?" Tags="&lt;electricity&gt;" AnswerCount="1" />
  <row Id="6821" PostTypeId="2" ParentId="6762" CreationDate="2011-03-13T14:41:15.743" Score="2" ViewCount="" Body="&lt;p&gt;Harbors have solved the problem of waves for centuries. It is walls. The same is true of the Netherlands who live below sea level. Solution is walls. A series of artificial islands/walls could be geo-engineered, at great cost, where the tsunami waves could break.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The permanent solution is to forbid any buildings below the height of the maximum recorded tsunamis in the region. Also costly. It is speculated that the minoan civilization disappeared from a tsunami from the explosion of the volcano in &lt;a href=&quot;http://en.wikipedia.org/wiki/Thera&quot; rel=&quot;nofollow&quot;&gt;Thera/Santorini&lt;/a&gt; 3500 years ago. Long time constants. If you are in Greece do not miss a visit to Santorini. It is a breathtaking island with an active caldera still.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-13T14:41:15.743" CommentCount="3" />
  <row Id="6822" PostTypeId="1" CreationDate="2011-03-13T14:45:50.453" Score="3" ViewCount="153" Body="&lt;p&gt;So, when biking, I noticed that when going up hills, it was less tiring if I went up them more quickly.  This is not total Work done as is Force * Distance, as that should be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the longer one is going uphill, the longer gravity is pulling you backwards.  And if you only are providing enough force to counteract the force of gravity (from a stop), you will not make it up the hill, yet you will feel quite tired afterwards. While if one pushes really hard, then one will hardly slow down at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that if you are coasting, then the conservation of energy applies, and $v_i^2 = v_f^2 + C$ where C is the gravitational potential energy at the top of the hill.  But this doesn't explain why it is more taxing to go up a hill slowly than quickly.  It's the same amount of energy transformed into gravitational potential anyways. &lt;/p&gt;&#xA;" OwnerUserId="274" LastActivityDate="2011-03-13T17:08:32.377" Title="&quot;Work&quot; when biking up a hill" Tags="&lt;classical-mechanics&gt;&lt;gravity&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="6823" PostTypeId="2" ParentId="6820" CreationDate="2011-03-13T15:00:26.383" Score="0" ViewCount="" Body="&lt;p&gt;Stop that dangerous tinkering! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lithium batterys can explode/burn violently if charged &#xA;or discharged in a wrong way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To Your questions: Mixing up current and voltage shows You lack the &#xA;simplest basics. A lithium battery is to be charged with the &#xA;equipment You bought along or which is recommended by the manufacterer &#xA;of the battery. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And: This forum is not &quot;Ask Uncle John&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-13T15:00:26.383" CommentCount="3" />
  <row Id="6824" PostTypeId="1" AcceptedAnswerId="6840" CreationDate="2011-03-13T15:02:08.687" Score="4" ViewCount="87" Body="&lt;p&gt;We all know that predicting tsunami and earthquake is difficult, with too many variables involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But with the advent in data collection and computing power and better models, one should be able to predict tsunami better than in the past. How accurate is current tsunami prediction?&lt;/p&gt;&#xA;" OwnerUserId="605" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-03-13T17:15:24.603" LastActivityDate="2011-03-13T17:41:51.687" Title="How good is current tsunami prediction?" Tags="&lt;geophysics&gt;&lt;tsunami&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="6825" PostTypeId="2" ParentId="6762" CreationDate="2011-03-13T15:04:31.657" Score="1" ViewCount="" Body="&lt;p&gt;The cheap (in more than one sense :)) and obvious answer is building restrictions on populated coastlines. Strategic seawalls would also help. These have both been mentioned, but I have framed this as an answer because other methods would be ineffective (predicting long term location of earthquakes is extremely imprecise) and prohibitively expensive. The problem with a tsunami is that until it hits shallow water, it is tiny and the speed is very fast. Our technology is no where near sophisticated enough to stop a tsunami given its spread size and energy.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2011-03-13T15:04:31.657" CommentCount="2" />
  <row Id="6826" PostTypeId="2" ParentId="6822" CreationDate="2011-03-13T15:10:34.893" Score="3" ViewCount="" Body="&lt;p&gt;The same reason applies as You can't hold an arm straight horizontally &#xA;longer than some minutes, although You do not do any &quot;work&quot; in this case. &#xA;Just exerting a force with a muscle (isometric) is work for the muscle!&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-13T16:53:11.870" LastActivityDate="2011-03-13T16:53:11.870" CommentCount="2" />
  <row Id="6827" PostTypeId="1" AcceptedAnswerId="6852" CreationDate="2011-03-13T16:39:46.547" Score="3" ViewCount="250" Body="&lt;p&gt;Why do optical fibers usually have a cladding? Ok, if you put make a bundle of optical fibers this prevents that light leaks from one fiber to another fiber in contact. However, are there other reasons to use claddings? Are there applications of optical fibers without claddings?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am mainly interested in optical fibres which can be described by geometrical optics, i.e. multimode fibers.)&lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-14T15:02:02.393" LastActivityDate="2011-03-19T16:56:40.823" Title="Cladding of optical fibers" Tags="&lt;applied-physics&gt;&lt;geometric-optics&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="6828" PostTypeId="2" ParentId="6822" CreationDate="2011-03-13T16:49:30.877" Score="2" ViewCount="" Body="&lt;p&gt;When your going up a hill at a high rate of speed chances are that you started on a horizontal surface going at a high speed and made your way up the hill at a quicker rate so while you are actually going up the hill you are useing the momentum that you got from the above mentioned horizontal surface so it definitely is easier.&lt;/p&gt;&#xA;" OwnerUserId="2530" LastActivityDate="2011-03-13T16:49:30.877" />
  <row Id="6829" PostTypeId="1" CreationDate="2011-03-13T16:51:03.447" Score="6" ViewCount="306" Body="&lt;p&gt;Why can one bend glass fibers without breaking it, whereas glasses one comes across in real life is usually solid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there also a good high-school level explanation of this?&lt;/p&gt;&#xA;" OwnerUserId="667" LastActivityDate="2011-03-14T10:57:44.367" Title="Why can one bend glass fiber?" Tags="&lt;material-science&gt;&lt;elasticity&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="6830" PostTypeId="2" ParentId="6762" CreationDate="2011-03-13T16:57:45.080" Score="0" ViewCount="" Body="&lt;p&gt;I doubt it makes economic sense. Because tsunamis extend to great depths surface structures to absorb the energy won't be very effective -except perhaps in shallow water. Also they are pretty rare ocurrances, maybe once per decade in the most favorable locations. Various sorts of breakwalls can be made, but like in the case of river levies, if you get an event larger than designed for, they can be overtopped, and because of the false sense of security this might make things worse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As others have said. The key is not to build things in vulnerable areas. I would modify this. You need to take into account the risk. Some structures can be sacrificed to a once a century event, and warning systems can be used for timely evacuation. Also structures can be built to withstand a tsunami, or to minimize damage. The danger needs to be quantified, and the mitigative measures analyzed to determine the most sensible strategies.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-03-13T16:57:45.080" />
  <row Id="6832" PostTypeId="2" ParentId="6824" CreationDate="2011-03-13T17:06:03.680" Score="2" ViewCount="" Body="&lt;p&gt;The problem of determining how much tsunami energy is going to be generated from a given earthquake isn't an easy one. Detailed knowledge of the seafloor motion would be needed. In principle this can probably be recovered from the seismic measurements, but you would need these results in real time. Also sometimes you set off an underwater landslide, i.e. a moderate earthquake can sometimes generate a tsunami because of other nonseismic factors. So the current paradigm is, precautionary evac near the source, then ocean bouy measurements can be used to estimate the magnitide/timing on distant shores. The local effects which can either focus or defocus energy as it interacts with the shoreline aren't easy to understand, so significant safety factors must be used to generate warnings.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-03-13T17:06:03.680" CommentCount="1" />
  <row Id="6833" PostTypeId="2" ParentId="6811" CreationDate="2011-03-13T17:07:25.240" Score="3" ViewCount="" Body="&lt;p&gt;Nope, the very fact that the entanglement entropy is - naturally - proportional to the surface area does &lt;em&gt;not&lt;/em&gt; explain the holographic principle because the holographic principle, reduced to the corresponding entropy bound, implies that the &lt;em&gt;total&lt;/em&gt; entropy of one of the systems can't exceed $A/4G$ where $A$ is the surface. The entanglement entropy is just one tiny term of the entropy, a degree of correlation between two subsystems, so its being proportional to the area is a much weaker and less surprising statement than the holographic principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite generally, non-gravitational physical systems are simply &lt;em&gt;not&lt;/em&gt; holographic in the space (bulk) that they occupy. Only (quantum) gravitational systems obey the holographic principle. In the context of AdS/CFT or AdS/anything, the holographic principle is encoded in the description of the CFT or anything by its having a secret extra (fifth) dimension of spacetime.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-13T17:07:25.240" CommentCount="1" />
  <row Id="6834" PostTypeId="2" ParentId="6822" CreationDate="2011-03-13T17:08:32.377" Score="1" ViewCount="" Body="&lt;p&gt;In can help to consider power, the rate of energy use which is not the same in your two cases.  This can lead to an understanding of how the same total energy use can result in different levels of 'tiredness'.  But as above you also need to take into account biomechanics and recovery time to properly understand the &lt;em&gt;physiological&lt;/em&gt; effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I suggest in direct answer that you are more physically suited to sprinting than endurance physical events, this can be explained best by biologists!&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-03-13T17:08:32.377" />
  <row Id="6835" PostTypeId="1" AcceptedAnswerId="6842" CreationDate="2011-03-13T17:18:10.663" Score="1" ViewCount="160" Body="&lt;p&gt;What are the reasons that optical fibers have to be thin (small radius of the fiber)? Is there a good picture which explains this in detail?&lt;/p&gt;&#xA;" OwnerUserId="667" LastActivityDate="2011-04-01T19:10:02.810" Title="Why should optical fibers be thin?" Tags="&lt;optics&gt;&lt;applied-physics&gt;&lt;geometric-optics&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="6836" PostTypeId="2" ParentId="6762" CreationDate="2011-03-13T17:18:29.070" Score="0" ViewCount="" Body="&lt;p&gt;I agree particularly with the response from anna v.&lt;br&gt;&#xA;Just want to add a possible &lt;strong&gt;measure of last resort&lt;/strong&gt; that can be tested to protect a specific location, for example, the entry of a river.&#xA;You can cause &lt;strong&gt;interference with the tsunami wave&lt;/strong&gt;, picking so that destructive interference is aligned with the entrance of the River.&#xA;The effect could possibly be achieved using &lt;strong&gt;one or more explosions (nuclear?) underwater&lt;/strong&gt;.&#xA;The accuracy of location and timing of the blasts would be key to success. Would have to be adequately studied the cost / benefit on the environment.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-03-13T17:18:29.070" CommentCount="3" />
  <row Id="6837" PostTypeId="2" ParentId="6816" CreationDate="2011-03-13T17:24:06.263" Score="2" ViewCount="" Body="&lt;p&gt;The most general Lorentz transformation that is connected to the identity is given by the conjugation by $\exp(-A)$ where&#xA;$$ A = \frac 12 \omega_{\mu\nu} \gamma^\mu \gamma^\nu $$&#xA;and $\omega_{\mu\nu}$ is an antisymmetric tensor containing $D(D-1)/2$ parameters. The group of all such transformations is isomorphic to $Spin(D-1,1)$. If $\omega$ only contains one component $0\mu$, then it is a boost, and the nonzero numerical value of $\omega$ is the rapidity - the &quot;hyperbolic angle&quot; $\eta$ such that $v/c=\tanh\eta$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If only one doubly spatial component of $\omega$ is nonzero, then this component $\omega_{\mu\nu} = -\omega_{\nu\mu}$ is obviously the angle itself. Note that the spatial-spatial terms in $A$ are anti-Hermitean, producing unitary transformations; the mixed temporal-spatial terms in $A$ are Hermitean and they don't product unitary transformations on the 4-component space of spinors (but they become unitary if they're promoted to transformations of the full Hilbert space of quantum field theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 4 dimensions, a general antisymmetric matrix $4\times 4$ contains 6 independent parameters and has eigenvalues $\pm i a, \pm  i b$, so in 3+1 dimensions, one can always represent a general Lorentz transformation as a rotation around an axis in the 4-dimensional space followed by a boost in the complementary transverse 2-plane. This is the counterpart of the statement that any $SU(2)$ rotations in 3 dimensions is a rotation around a particular axis by an angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you allowed $A$ to contain something else than $\gamma^{\mu\nu}$ matrices which generate the Lorentz group, you could get other groups. Only for a properly chosen subset of allowed values of $\omega$, you would get a closed group from the resulting exponentials (under multiplication). In particular, if you allowed $A$ to be an arbitrary complex combination of any products of gamma matrices, well, then you would allow $A$ to be any complex $4\times 4$ matrix, and its exponentials would produce the full group $GL(4,C)$ - surprising, Carl? ;-) It's not a terribly useful groups in physics because actions are usually not invariant under this &quot;full group&quot;, are they?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there are not too many groups in between $Spin(3,1)$ and $GL(4,C)$ - I guess that there's no proper group of $GL(4,C)$ that has a proper $Spin(3,1)$ subgroup. Obviously, there are many subgroups of $Spin(3,1)$ - such as $Spin(3)$, $Spin(1,1)\times Spin(2)$, and others.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-13T17:29:18.487" LastActivityDate="2011-03-13T17:29:18.487" />
  <row Id="6838" PostTypeId="2" ParentId="6829" CreationDate="2011-03-13T17:29:25.493" Score="3" ViewCount="" Body="&lt;p&gt;The critical parameter for materials under stress is the &lt;em&gt;strain&lt;/em&gt;, defined in a general way{*} as the fraction change some length over that same length:&#xA;$$ \lambda = \frac{\Delta l}{l} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So take a given a fiber of diameter $d$, and bent around a radius of curvature $r$, the strain of either the inside or outside edge is:&#xA;$$ \lambda = \frac{((r \pm d) - r) \theta}{r \theta} = \pm \frac{d}{r} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you need only look up the breaking strain for the glass in question.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;{*} Or fractional change in volume over volume, or displacement as a fraction of length, or... It really depends on the details of the situation you're trying to measure, but the above definition will work for tension and compression.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-03-13T17:29:25.493" />
  <row Id="6839" PostTypeId="1" AcceptedAnswerId="6846" CreationDate="2011-03-13T17:38:23.993" Score="5" ViewCount="58" Body="&lt;p&gt;anyone know some good introductory references on the physics of anyons?&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2011-03-13T18:16:01.470" Title="references on the physics of anyons" Tags="&lt;anyons&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="6840" PostTypeId="2" ParentId="6824" CreationDate="2011-03-13T17:41:51.687" Score="3" ViewCount="" Body="&lt;p&gt;Well, immediately after the earthquake, they created the map showing when the tsunami arrives at different places:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/MczEC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess that it ended up being pretty accurate but I haven't checked. The speed of ocean waves actually depends on the frequency...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your ordering &quot;tsunami and earthquake&quot; is somewhat bizarre. You do acknowledge and realize that the tsunami was a consequence of the earthquake, don't you? ;-) Predicting earthquakes themselves is not really possible. I think that no one knew about the Japanese earthquake until the very moment when it took place.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-13T17:41:51.687" />
  <row Id="6841" PostTypeId="2" ParentId="6827" CreationDate="2011-03-13T17:44:27.330" Score="5" ViewCount="" Body="&lt;p&gt;Do you think a fiber that worked until is got some water on it would be acceptable? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for the fiber to have a reliable, known behavior (transmission speed, attenuation, wavelength acceptance) in all environments, it is necessary to the total internal reflection to happen against a well defined external material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the cladding.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-03-13T17:44:27.330" />
  <row Id="6842" PostTypeId="2" ParentId="6835" CreationDate="2011-03-13T17:47:36.583" Score="2" ViewCount="" Body="&lt;p&gt;The answer comes from the use of optical fibers: information transfer for one. The thinner the more channels in a bundle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the attenuation of the light is smaller in a thin fiber . There is &lt;a href=&quot;http://en.wikipedia.org/wiki/Fiber_optics&quot; rel=&quot;nofollow&quot;&gt;an article&lt;/a&gt; explaining this.   &lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-13T17:47:36.583" />
  <row Id="6843" PostTypeId="2" ParentId="6675" CreationDate="2011-03-13T17:55:45.303" Score="0" ViewCount="" Body="&lt;p&gt;a small aside: In &quot;English-English&quot; (as opposed to &quot;American-English&quot;) a meter is a measuring device, a metre is a unit of length.  This will depend on which dictionary you're following!&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-03-13T17:55:45.303" />
  <row Id="6844" PostTypeId="1" CreationDate="2011-03-13T18:01:12.403" Score="1" ViewCount="231" Body="&lt;p&gt;So, from what I understand: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Special Relativity says that light is always observed moving at the speed of light (c). If some object had a velocity of (3/4)c, and the object had some sort of clock attached to it, it would measure differently from a still clock. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What would a clock moving the speed of light measure compared to a still clock?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Why does it seem that the speed of light is a focal point of time? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I may need to rephrase this slightly. What I want to know is, why is it that observed time changes at high velocities? &lt;/p&gt;&#xA;" OwnerUserId="2536" LastEditorUserId="2536" LastEditorDisplayName="" LastEditDate="2011-03-14T03:05:23.397" LastActivityDate="2011-08-31T03:30:22.720" Title="How are light and time related?" Tags="&lt;special-relativity&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="6845" PostTypeId="2" ParentId="6209" CreationDate="2011-03-13T18:14:49.187" Score="1" ViewCount="" Body="&lt;p&gt;The passive gravitational mass of the electron is, by experiment, &lt;strong&gt;less than 9% of the theoretical expected value&lt;/strong&gt;, as you can see in &lt;a href=&quot;http://physics.stackexchange.com/questions/6567/active-gravitational-mass-of-the-electron&quot;&gt;my answer on this question&lt;/a&gt;.&lt;br&gt;&#xA;The result was so unexpected that no one believed in the correcteness of the result. &#xA;I believe that the active gravitational mass of the electron will also be of the same order (or &lt;strong&gt;zero&lt;/strong&gt;).&lt;br&gt;&#xA;One nice theory could collapse because of one single experiment.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am very much impressed: Almost no question do not include the Black Hole in the response. May be the case that they atract votes up.&#xA;On this site there are already 2000 references to this term.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-03-13T18:21:55.220" LastActivityDate="2011-03-13T18:21:55.220" CommentCount="1" />
  <row Id="6846" PostTypeId="2" ParentId="6839" CreationDate="2011-03-13T18:16:01.470" Score="4" ViewCount="" Body="&lt;p&gt;One of the best recent references is the 2008 RMP article by Nayak et al. &lt;a href=&quot;http://dx.doi.org/10.1103/RevModPhys.80.1083&quot; rel=&quot;nofollow&quot;&gt;Non-Abelian Anyons and Topological Quantum Computation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A somewhat less technical reference is &lt;a href=&quot;http://arxiv.org/abs/hep-th/9209066&quot; rel=&quot;nofollow&quot;&gt;An Anyon Primer&lt;/a&gt; by Sumathi Rao.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many others but these two are good for someone starting out.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-13T18:16:01.470" />
  <row Id="6847" PostTypeId="2" ParentId="6835" CreationDate="2011-03-13T18:34:38.757" Score="2" ViewCount="" Body="&lt;p&gt;If they are thicker then light can take many different paths, bouncing at slightly different angles and still being reflected - this means that the time taken for each photon a pulse to go down a fibre is slightly different which smears out the pulse and so reduces the bits/second that can be sent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thick plastic fibres are used for simply illuminating something because they are cheap, rugged and easy to make, thinner glass fibre is used for short distances and very thin very expensive single mode fibre is used for higher bandwidth long distance&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-03-13T18:34:38.757" CommentCount="2" />
  <row Id="6848" PostTypeId="2" ParentId="6829" CreationDate="2011-03-13T18:41:58.377" Score="10" ViewCount="" Body="&lt;p&gt;In addition to the question of bend radius - there is also an effect of surface scratches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most materials are very strong - they fail because a surface flaw allows a stress concentration - ie a crack to form. glass fibre has a very smooth surface because of the way it is made and can be put under high stress without cracking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can show this with a thick glass rod (or you could before we were banned from doing anything interesting by 'safety'). Clean the surface with acid, or heating and annealing,  so any defects are removed  you can then bend the rod like rubber. But touch the bent part with a metal ruler to create scratches and it will shatter&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-03-13T18:41:58.377" CommentCount="3" />
  <row Id="6849" PostTypeId="2" ParentId="6844" CreationDate="2011-03-13T19:28:10.257" Score="3" ViewCount="" Body="&lt;p&gt;1) It is impossible for matter to go at the speed of light, as it takes an amount of energy equal to $mc^{2}\left(\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}-1\right)$ to accelerate it to a speed $v$, and this energy becomes infinite as $v\rightarrow c$.  In another sense, reference frames moving at the speed of light (or faster) are excluded from Einstein's principle of equivalence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) And this gets us to light.  To the best of our knowledge, light is massless, but still carries energy.  If you look at the equation above, you see that if you allow the mass of the object to go to zero, and still expect the particle to carry energy, then it is necessary that the speed of the object approach $c$, so that you get a $\frac{0}{0}$ indeterminate answer for the energy carried.  The speed of light is the speed at which all massless particles travel.  It is only happenstance that light was the first massless particle that we discovered, which caused $c$ to be named after light, a name which has stuck.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2011-03-13T19:53:02.497" LastActivityDate="2011-03-13T19:53:02.497" CommentCount="1" />
  <row Id="6850" PostTypeId="2" ParentId="6778" CreationDate="2011-03-13T20:16:20.427" Score="2" ViewCount="" Body="&lt;h3&gt;Just to check that we're on the same page:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Multiplication of a constant-amplitude complex signal $s=\exp{\left(i f(t)\right)}$ by another complex function with constant amplitude won't give a time-varying amplitude:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\exp{\left(i f(t)\right)} \cdot \exp{\left(i g(t)\right)} = \exp{\left(i h(t)\right)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $f$ and $g$ are real, $h$ is real, and $\frac{d}{dt}\left|\exp{\left(i h(t)\right)}\right| = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're interested in the particular case where $g(t) = \Gamma \sin{(\Omega t)}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\exp{\left(i f(t)\right)} \cdot \exp{\left(i \Gamma \sin{(\Omega t)}\right)} = \exp{\left(i h(t)\right)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, since $f$ and $\Gamma \sin{(\Omega t)}$ are real, $h$ will be real, and the result will have constant amplitude.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;But, now you want to see the same truth in the frequency domain.&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Starting from the Jacobi-Anger identity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\exp{\left(i \Gamma \sin{(\Omega t)}\right)} = \sum_{n=-\infty}^{\infty}J_{n}(\Gamma)\exp{\left(i n \Omega t\right)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying each side by its complex conjugate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 = \sum_{m=-\infty}^{\infty}\sum_{n=-\infty}^{\infty}J_{m}(\Gamma)J_{n}(\Gamma)\exp{\left(i (m-n) \Omega t\right)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now make the change of variables $p=(m-n)/2$, $q=(m+n)/2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 = \sum_{p=-\infty}^{\infty}\sum_{q=-\infty}^{\infty}J_{q+p}(\Gamma)J_{q-p}(\Gamma)\exp{\left(i 2p \Omega t\right)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rewrite it a little:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 = \sum_{p=-\infty}^{\infty}\exp{\left(i 2p \Omega t\right)}\left(\sum_{q=-\infty}^{\infty}J_{q+p}(\Gamma)J_{q-p}(\Gamma)\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the complex exponentials are orthogonal, I think the sum in parentheses has to vanish unless $p$ is 0:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{q=-\infty}^{\infty}J_{q+p}(\Gamma)J_{q-p}(\Gamma) = \delta(p)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I allowed to assume the Jacobi-Anger identity? If so (and if I haven't made a mistake), I've just proven that this sum in parentheses vanishes. Would you like a separate proof, not based on the Jacobi-Anger identity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclaimer: I'm an experimental physicist. My math is rusty. Please forgive and point out any mistakes.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2011-03-13T20:16:20.427" CommentCount="6" />
  <row Id="6851" PostTypeId="1" CreationDate="2011-03-13T20:40:50.130" Score="-5" ViewCount="197" Body="&lt;p&gt;How to call the states in the superposition $\psi_0 = \sum_n C_n \psi_n^{(0)}$? I mean, whether they are famous virtual states or what?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: $\psi_0$ is the ground state wave function to be calculated by the perturbation theory. $\psi_n^{(0)}$ are the eigenstates of the non perturbed Hamiltonian $H_0$. These are regular notations in QM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 2: My question's arisen because Lubosh called such states &quot;a &lt;a href=&quot;http://js-kit.com/api/static/pop_comments?ref=http://motls.blogspot.com/2010/11/quantum-field-theory-has-no-problems.html&amp;amp;title=The%20Reference%20Frame%3a%20Quantum%20field%20theory%20has%20no%20problems&amp;amp;path=/55672134070675449&amp;amp;standalone=no&amp;amp;scoring=yes&amp;amp;backwards=no&amp;amp;sort=date&amp;amp;thread=yes&amp;amp;permalink=http://js-kit.com/api/static/pop_comments?ref=http%253A%252F%252Fmotls.blogspot.com%252F2010%252F11%252Fquantum-field-theory-has-no-problems.html&amp;amp;path=%252F55672134070675449&amp;amp;skin=echo&amp;amp;smiles=no&amp;amp;editable=yes&amp;amp;thread-title=Echo&amp;amp;popup-title=Echo&amp;amp;page-title=The%20Reference%20Frame%3a%20Quantum%20field%20theory%20has%20no%20problems&quot; rel=&quot;nofollow&quot;&gt;condensate&lt;/a&gt; of virtual particles&quot; or so.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-03-13T22:40:39.297" LastActivityDate="2011-03-14T04:15:36.607" ClosedDate="2011-03-13T23:28:49.137" Title="How to call the states in the superposition $\psi_0 = \sum_n C_n \psi_n^{(0)}$?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="6852" PostTypeId="2" ParentId="6827" CreationDate="2011-03-13T21:05:47.427" Score="7" ViewCount="" Body="&lt;p&gt;For a fiber to guide light, even considering the situation using only geometrical optics, there must be total internal reflection (TIR) at the boundary of the fiber core. For TIR to occur, the angle of incidence of the light must be greater than the critical angle $\theta_c$. If the material before the boundary has index of refraction $n_1$ and the material after the boundary has index $n_2$, then the critical angle is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \theta_c = \arcsin(\frac{n_2}{n_1})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if $\frac{n_2}{n_1} &amp;gt; 1$ then the critical angle is undefined -- TIR cannot occur. This is why TIR is only possible when light encounters an interface with a material of lower index. This should be familiar to you if you think about it: you can see the bottom of a pool from above the water, but the water surface looks silvery and reflective from underneath; the reflective face of a roof prism looks reflective when you try to look through it from the glass-to-air side, but if you look from the air side, you can see clearly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On its own, this doesn't require a fiber to have cladding. Air has an index very near 1, so any other material will have a higher index, and therefore guide light without a cladding. However, this situation would be very unstable. Any material contacting the fiber could produce an area where TIR does not occur, and suddenly the fiber would begin leaking. Consider how easy it would be for this to happen! Any sort of dirt or oil on the fiber would either eliminate TIR completely, or raise the critical angle such that some of the light could escape. The same would be true of most materials you might use to mount the fiber, or even any type of coating you might apply as a mechanical barrier to dirt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, in practice, fibers have a cladding to ensure that the optical guiding characteristics continue to work in real world conditions. They also provide engineers with another way of adjusting the properties of the fiber, so its definitely not something we are unhappy about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do keep in mind though, that there are examples of unclad &quot;fibers&quot; although they are usually just acrylic rods used in teaching demonstrations, so I wouldn't really call them a fiber in the common sense of the word.&lt;/p&gt;&#xA;" OwnerUserId="869" LastEditorUserId="869" LastEditorDisplayName="" LastEditDate="2011-03-19T16:56:40.823" LastActivityDate="2011-03-19T16:56:40.823" CommentCount="0" />
  <row Id="6853" PostTypeId="2" ParentId="6851" CreationDate="2011-03-13T21:13:43.023" Score="0" ViewCount="" Body="&lt;p&gt;Formally they (the $\psi_n^{(0)}$s) form a &lt;a href=&quot;http://en.wikipedia.org/wiki/Basis_%28linear_algebra%29&quot; rel=&quot;nofollow&quot;&gt;basis&lt;/a&gt; of the space and can be called &lt;em&gt;&quot;basis states&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And very likely they are the eigenstates of some operator that is physically important in the problem, so &lt;em&gt;&quot;eigenstates of the Foo operator&quot;&lt;/em&gt; is an option as well.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In light of the edit I can offer only &lt;em&gt;&quot;the unperturbed states of the system&quot;&lt;/em&gt; and similar circumlocutions as I am not aware of any special term for such states.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-03-14T04:15:36.607" LastActivityDate="2011-03-14T04:15:36.607" CommentCount="2" />
  <row Id="6854" PostTypeId="2" ParentId="5188" CreationDate="2011-03-13T21:40:20.490" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The first experiments ever to be conducted in regard to orbits and the earth spinning was done by i believe Galileo he hung a pendulum from the ceiling of his granary it was on a 360 degree hinge of some sort he then  swung it,he coverd the floor in a layer of fine dust he left it for a full day and when he returned the markings in the dust suggested that the earth spins now that was simply to prove that the earth spins on its axis.Issac Newton was the one that suggested that an object in motion must remain in motion until another force acts on it. This force was of course gravity which from the sun on our planet and  all other planets in our solar system is a tremendous force which keeps us in orbit.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The notion that humans had bumped into the other planets is a cute one. The truth is that scientists throughout the ages have all had suspicions that there were other planets in our solar system of course with no way to prove this they were just theories but the truth is astronomers were able with their ancient telescopes to see perhaps as far as mars but up until the last century that was as far as they got.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The first ever satellite was the sputnik created by russian engineers it was sent into orbit in 1957. Now the first man to ever land on the moon was Neil Armstrong followed by Buzz Aldrin the were not afraid to step out of thier ships because NASA did an extremely good job of preparing them for that mission and trust me if you landed on themoon you wouldnt just not go on it was in the words on Armstrong One small step for man,A giant leap for mankind.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Im going to skip part one of this question. Part 2. the speed of light was calculated by interferometer technique this i believe was to shine a beam of light at a multisided mirror and the resulting reflection would show where the light ended up they knew the exact speed that the mirrors were spinning so they could calculate the speed.Not sure about the last part.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Nowadays rockets and satellites provide the bulk of the research.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Humans were very curious to know about mars for several reasons the primary being that it is the only planet besides earth that would be able to support any kinds of life form.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The spacecraft as used in the mercury and apallo missions were rockets with payloads i.e. the satellites, the lunar lander, and space capsules. These rockets would fall back to earth as soon as their purpose was served.&#xA;The space shuttle was to be a ship that would return in full&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2530" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2011-03-13T22:08:04.633" LastActivityDate="2011-03-13T22:08:04.633" />
  <row Id="6855" PostTypeId="1" CreationDate="2011-03-13T21:44:23.557" Score="1" ViewCount="137" Body="&lt;p&gt;What calculations were done to discover the size of all the planets &lt;/p&gt;&#xA;" OwnerUserId="2530" LastActivityDate="2011-03-14T01:39:40.583" Title="The size of the planets" Tags="&lt;astrophysics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="6856" PostTypeId="1" CreationDate="2011-03-13T22:04:57.413" Score="4" ViewCount="71" Body="&lt;p&gt;Given the relevance of this subduction system, I would expect that a wide range of detectors (temperature, vibration, seismometers, whatever) are deployed in the depth of these trenches. Is there some info and or public access to their readings?&lt;/p&gt;&#xA;" OwnerUserId="1335" LastActivityDate="2011-03-13T22:04:57.413" Title="Which are the readings of the detectors at Japan and Izu-Bonin-Mariana Trenchs?" Tags="&lt;geophysics&gt;" CommentCount="5" />
  <row Id="6857" PostTypeId="2" ParentId="6817" CreationDate="2011-03-13T22:10:09.360" Score="3" ViewCount="" Body="&lt;p&gt;Are you asking about the formula for the ellipse, or the formula for how angles in the sky are distorted?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the former case: it's easy to work out the shape and size of the ellipse by considering Lorentz contraction. Suppose that in the rest frame of the source, the sphere has a radius $r$. Since the Lorentz transform has no effect on components of length perpendicular to the motion, then the transverse radius of the ellipsoid remains the same - but that's just the semiminor axis. In the forward direction of motion, the length of the path taken from the source to the sphere is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_1&amp;#39; = \gamma\biggl(r + \frac{vr}{c}\biggr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and from the sphere back to the source,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_2&amp;#39; = \gamma\biggl(r - \frac{vr}{c}\biggr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding these up you get $r_1&amp;#39; + r_2&amp;#39; = 2\gamma r$, but this is the total travel distance which is just twice the semimajor axis of the ellipsoid. So we have an ellipsoid with semimajor axis $\gamma r$ and semiminor axis $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To figure out the formula for the angular distortion - &lt;a href=&quot;http://en.wikipedia.org/wiki/Relativistic_aberration_formula&quot; rel=&quot;nofollow&quot;&gt;relativistic aberration&lt;/a&gt; - consider a photon arriving at the center of the sphere coming at an angle $\theta$ relative to the direction of motion (precisely speaking: relative to an axis which points along the direction of motion in the external observer's frame). From the stationary observer's reference frame, the component of this photon's velocity parallel to the direction of motion is subject to the Lorentz transformation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_{\parallel}&amp;#39; = \frac{u_\parallel - v_\parallel}{1 + \frac{uv}{c^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so if you plug in $u_{\parallel}&amp;#39; = c\cos\theta&amp;#39;$, $u_\parallel = c\cos\theta$, and $v_\parallel = v$ (the relative speed of the source with respect to the observer), you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos\theta&amp;#39; = \frac{c\cos\theta - v}{c + v\cos\theta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives you the angle of an incoming light ray as seen by the external observer, $\theta&amp;#39;$, in terms of the angle as seen by the center of the sphere, $\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I might have mixed up the signs - I'll have to come back to this and recheck it later.)&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-03-13T22:10:09.360" />
  <row Id="6858" PostTypeId="2" ParentId="6855" CreationDate="2011-03-13T22:31:32.617" Score="2" ViewCount="" Body="&lt;p&gt;Planets in our own solar system are measured directly.&#xA;You can measure their angular diameters with a telescope and their distance by measuring the maximum angle they make from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More recently the accurate distance to the inner planets have been measured by radar&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-03-13T22:31:32.617" CommentCount="1" />
  <row Id="6859" PostTypeId="2" ParentId="6817" CreationDate="2011-03-14T01:04:09.990" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://physics-quest.org/Book_Chapter_Non_Simultaneity.pdf&quot; rel=&quot;nofollow&quot;&gt;check Hans de Vries online book&lt;/a&gt;   Chapter 4. Non-simultaneity from the classical wave equation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;chapter 4.9 &lt;strong&gt;The ellipsoids of simultaneity&lt;/strong&gt;&#xA;nice figures 4.13, 4.14, 4.15 and more&lt;br&gt;&#xA;The relevant equations are there.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-03-14T01:04:09.990" CommentCount="2" />
  <row Id="6860" PostTypeId="1" CreationDate="2011-03-14T01:18:45.953" Score="3" ViewCount="27" Body="&lt;p&gt;Suppose you have a distribution function $ f_{\alpha}( \vec{r} , \vec{p} , t)$ obtained from &lt;a href=&quot;http://en.wikipedia.org/wiki/Vlasov_equation#The_Vlasov-Maxwell_system_of_equations_.28cgs_units.29&quot; rel=&quot;nofollow&quot;&gt;Vlasov equations&lt;/a&gt; for a certain $\alpha$ species, say some ions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a rigorous way (in the domain of validity of the model, that is) to estimate a collision density function $ p^{collisions}_{\alpha}( \vec{r} , E_{CM} , t)$ , where $E_{CM}$ is a variable describing collision energy parameters? A link or reference would be great &lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-03-14T02:17:54.233" LastActivityDate="2011-03-14T02:17:54.233" Title="estimating ion collisions from Vlasov equations" Tags="&lt;plasma-physics&gt;&lt;cross-section&gt;&lt;vlasov-equations&gt;" />
  <row Id="6861" PostTypeId="2" ParentId="6782" CreationDate="2011-03-14T01:35:41.070" Score="9" ViewCount="" Body="&lt;p&gt;There are observables corresponding to the light going through both slits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can write down a basis: &quot;it went through slit A + slit B&quot;, and &quot;it went through slit A - slit B&quot;. Although maybe you can't detect these observables easily with an experiment, they're perfectly good observables, they're orthogonal, and a clever enough experiment should be able to measure them (at least approximately). Something like this was done in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Afshar_experiment&quot; rel=&quot;nofollow&quot;&gt;Afshar experiment&lt;/a&gt; and in the quantum eraser experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you can't simultaneously measure both the observables &quot;it went through slit A + slit B&quot;. and &quot;it went through only slit A&quot; because the two observables aren't orthogonal. The observables &quot;it went through slit A&quot; and &quot;it went through slit B&quot; are orthogonal, so they form a basis. Thus, if you measure in this basis, you find it always went through one slit or the other. If you want to measure &quot;it went through both slits&quot; versus &quot;it only went through one slit,&quot; this doesn't correspond to an orthogonal basis, so you can't perform that measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this were polarized light instead of slits, the measurement you want to make would correspond to &quot;this is either vertically polarized or horizontally polarized&quot; versus &quot;this is right-diagonally polarized light&quot;. &lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditorDisplayName="" LastEditDate="2011-03-14T01:57:04.687" LastActivityDate="2011-03-14T01:57:04.687" CommentCount="1" />
  <row Id="6862" PostTypeId="2" ParentId="6855" CreationDate="2011-03-14T01:39:40.583" Score="2" ViewCount="" Body="&lt;p&gt;For exoplanets, the technique used depends on the aspect of &quot;size&quot; being measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Masses of exoplanets are estimated by using doppler spectroscopy to determine the acceleration of the planet's parent star caused by the orbiting planet, while diameters are estimated by observing the percentage of a the parent star's light that is obscured by the planet as it crosses between the star and the observer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Masses are actually minimum masses and diameter can only be determined for planets which &quot;transit&quot; their parents. Both estimates require knowledge of the star's properties—for mass, stellar mass, and for diameter, stellar diameter—both of which are estimated from observation of stellar properties (spectra, known distance, etc.) and stellar evolution. &lt;/p&gt;&#xA;" OwnerUserId="2540" LastActivityDate="2011-03-14T01:39:40.583" CommentCount="2" />
  <row Id="6863" PostTypeId="1" AcceptedAnswerId="6866" CreationDate="2011-03-14T01:43:42.920" Score="1" ViewCount="222" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/6768/why-didnt-the-control-rods-in-fukushima-shut-down-the-reactor&quot;&gt;Why didn&amp;#39;t the control rods in Fukushima shut down the reactor?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;They say that pressure rises and that this can be dangerous. But why does this happen if the reactor is shutdown ?&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2011-03-14T03:26:53.470" ClosedDate="2011-03-14T11:52:40.740" Title="Why does Fukushima pressure rise?" Tags="&lt;nuclear-physics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="6864" PostTypeId="2" ParentId="6829" CreationDate="2011-03-14T01:52:10.290" Score="1" ViewCount="" Body="&lt;p&gt;Having spent a great deal of time working with &quot;normal&quot; glass (one of my many hobbies), I can assure you that, in fact, all glasses can be bent. When cutting large sheets of glass, I always see the sheet bend before it breaks. Every now and then you have to strike it twice; the first time you bend it, it fails to break. Fiber optics differ from the usual glass partly in that it is very thin and so bends much more easily. After you've cut a few thousand pieces of glass you come to recognize differences between different compositions. Some glasses feel very smooth when they bend, others, like the stained art glass shown below, are rather ragged and difficult to bend. (And all breaks or cuts in glass begin with a bend.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a more familiar metaphor, think about wood. It doesn't surprise you when a slat of wood that is 1mm in thickness is bent in a circle. Now imagine doing the same thing with a much thicker piece of wood. You can't do it with the thicker wood because it's too thick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, regular glass differs from fiber optic glass in that fiber optic glass is very pure and this makes it very strong. In the art glass community, thin glass rods is called &quot;stringer&quot;. Let me see if I can find some photos of cold stringer being bent.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Okay, I found some clear stringer. I put it on a sheet of stained glass supported by two pennies. I took two photos, before and after pressing the stringer in the middle. You can see the bending. Straight:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/ZRXhp.jpg&quot; alt=&quot;Unbent&quot;&gt;&lt;br&gt;&#xA;Bent:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/P48sm.jpg&quot; alt=&quot;Bent&quot;&gt;  &lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-03-14T02:10:15.383" LastActivityDate="2011-03-14T02:10:15.383" CommentCount="2" />
  <row Id="6865" PostTypeId="2" ParentId="6782" CreationDate="2011-03-14T02:31:02.067" Score="1" ViewCount="" Body="&lt;p&gt;The first thing I would point out is that the idea of the particle going through both slits at once is flat-out wrong. (Well I suppose you could argue various viewpoints; this is mine) It's a &lt;em&gt;wave&lt;/em&gt; that goes through both slits; specifically, a wave in the quantum field that represents the particle, and there's no confusion about how a wave can go through both slits. Furthermore, the wave &lt;em&gt;always&lt;/em&gt; goes through both slits - that's what waves do, they spread out in every possible direction. A bit of the wave even propagates directly through the wall!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the only reason we have this idea of particles at all is that the interaction between the quantum field of the particle and the quantum field of anything else (e.g. a detector) takes place at a specific point. Essentially, you can only have an interaction with one point on the wave, not the whole wave. When that happens, we see a localized event, so we say &quot;oh, there's the particle.&quot; But all we're really seeing is an interaction between two quantum fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when you detect a particle on the screen in this experiment, what you're really seeing is an interaction involving a point on the wave. This same wave passed through all your slits, and any other routes it could take to get to the screen. It wasn't interacting with anything along the way, so it didn't exhibit any particle-like properties while it was making its way from the emitter to the screen. Thus, it doesn't make sense to talk about which slit the particle passed through, since in the slits, there was no particle (i.e. no interaction), only a propagating wave.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-03-14T02:31:02.067" />
  <row Id="6866" PostTypeId="2" ParentId="6863" CreationDate="2011-03-14T03:26:53.470" Score="1" ViewCount="" Body="&lt;p&gt;Because even if reactor is not-critical, there are lots of radioactive materials in the fuel which decays and produce some heat. You cannot stop that process. Although it gives much less heat than nuclear fission, it is still significant and can cause meltdown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same reason is why spent nuclear fuel is stored under water for few years - it just selfheat and can melt.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-03-14T03:26:53.470" />
  <row Id="6867" PostTypeId="1" AcceptedAnswerId="6870" CreationDate="2011-03-14T03:33:39.110" Score="2" ViewCount="166" Body="&lt;p&gt;In a recent article:&lt;br&gt;&#xA;Physical Review A 83, 032903 (2011), A.Yu. Voronin, P.Froelich, V.V. Nesvizhevsky, &lt;a href=&quot;http://arxiv.org/abs/1103.2150&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Antihydrogen Gravitational Quantum States&lt;/em&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the authors claim that anti-hydrogen has a lifetime of 0.1 seconds when placed in a container with metal walls. They say that the atom is kept from annihilating with the walls because of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Casimir_effect&quot; rel=&quot;nofollow&quot;&gt;Casimir-Polder effect&lt;/a&gt;. This raises in my mind the question of how it is that the atoms annihilate. My first guess is that the positron would annihilate against an electron, and then the proton would be attracted into the metal by the image charge effect. But this would seem to be in contradiction to the Casimir-Polder effect mentioned in the paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how does slow anti-hydrogen decay in the presence of normal matter?&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-25T19:33:21.220" Title="How does slow anti-hydrogen annihilate with normal matter in the lab?" Tags="&lt;antimatter&gt;&lt;casimir-effect&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="6869" PostTypeId="1" CreationDate="2011-03-14T05:57:57.270" Score="1" ViewCount="369" Body="&lt;p&gt;Is there any sort of difference between thermal radiation and infrared radiation at least when detected by a camera? What are the differences between the two?&lt;/p&gt;&#xA;" OwnerUserId="2544" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-16T12:28:02.717" LastActivityDate="2011-03-16T12:28:02.717" Title="What is the difference between thermal and infrared imaging?" Tags="&lt;thermal-radiation&gt;" AnswerCount="3" />
  <row Id="6870" PostTypeId="2" ParentId="6867" CreationDate="2011-03-14T06:03:13.213" Score="2" ViewCount="" Body="&lt;p&gt;Dear Carl, the correct paper to derive the 0.1-second lifetime of the anti-Hydrogen atom in the gravitational field is described after the very sentence you quoted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a &quot;[20]&quot; symbol which means that the sentence is justified in the reference number 20 in the list of literature at the end of the paper you quoted. So the correct paper that answers your question is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Quantum reflection of ultracold antihydrogen from a solid surface&lt;/strong&gt;&lt;br&gt;&#xA;  A Yu Voronin and P Froelich 2005 J. Phys. B: At. Mol. Opt. Phys. 38 L301, doi: 10.1088/0953-4075/38/18/L02&lt;br&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;http://iopscience.iop.org/0953-4075/38/18/L02&quot; rel=&quot;nofollow&quot;&gt;http://iopscience.iop.org/0953-4075/38/18/L02&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://iopscience.iop.org/0953-4075/38/18/L02/pdf/0953-4075_38_18_L02.pdf&quot; rel=&quot;nofollow&quot;&gt;http://iopscience.iop.org/0953-4075/38/18/L02/pdf/0953-4075_38_18_L02.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is fully available online - click the last link for the PDF file. At distances longer than 15 nanometers from the metal, the Casimir-Polder potential has the form $-74/z^4$ in atomic units. At shorter distances between 1 Bohr radius or so to 15 nanometers, the potential becomes $-0.25/z^3$ in atomic units - a van der Waals form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only at distances shorter than 1 Bohr radius or so, the anti-Hydrogen behaves differently than the Hydrogen, and is annihilated. This portion of the potential is not probed in the 0.1-second case: note that to get annihilation, the positron must approach the electron at a shorter distance than the Bohr radius - comparable to the Compton wavelength of the electron (and even shorter, nuclear distances if we want to annihilate the hadrons). The authors calculate the reflection probability as a function of the kinetic energy of the anti-atom. The lower energy/temperature they have, the more likely they will bounce off. For low enough kinetic energy, the reflection probability is as high as 0.99987 (a table).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In those cases, one doesn't probe the short-distance portion of the potential. So the answer to your question why there's no annihilation is the same as the answer to the question why you don't get fusion if you fill a tank with the Hydrogen gas.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-14T06:09:36.700" LastActivityDate="2011-03-14T06:09:36.700" CommentCount="2" />
  <row Id="6871" PostTypeId="2" ParentId="6869" CreationDate="2011-03-14T06:15:42.667" Score="4" ViewCount="" Body="&lt;p&gt;The thermal radiation and the infrared radiation are the same thing if the sources of the radiation have temperatures comparable to the room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the two terms are different: the infrared radiation is defined by having the wavelength in the fixed interval 0.7 - 300 micrometers (a convention)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Infrared_radiation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Infrared_radiation&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;while the thermal radiation of an object depends on its absolute temperature $T$. While any object at nonzero temperature emits at all wavelengths, the wavelength at which the emission is maximized is inversely proportional to the temperature of the source $T$ in Kelvins. See:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_radiation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Thermal_radiation&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For ordinary cold and lukewarm objects, the thermal radiation is mostly emitted in the infrared. For much hotter objects, thermal radiation may be dominated by the visible light (or even ultraviolet light, at even higher temperatures). On the contrary, much colder objects emit radiation at a much longer wavelengths; for example, the empty Universe has temperature 2.7 Kelvin so it is filled by the &quot;cosmic microwave background&quot; which is dominated by much longer electromagnetic waves which are microwaves - similar long waves to those in the microwave oven.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the comments about the &quot;wavelength&quot; of the infrared radiation were new to you, I recommend you to read some basic texts about the electromagnetic spectrum&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_spectrum&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Electromagnetic_spectrum&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which explains all kinds of electromagnetic radiation. In all cases, they only differ by the wavelength (or the energy or frequency which is inversely proportional to it).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-14T06:21:07.183" LastActivityDate="2011-03-14T06:21:07.183" CommentCount="4" />
  <row Id="6872" PostTypeId="1" CreationDate="2011-03-14T08:50:37.953" Score="1" ViewCount="226" Body="&lt;p&gt;What is the exact dependence of the refractive index of air and the temperature? Is there a theoretical derivation of it?&lt;/p&gt;&#xA;" OwnerUserId="667" LastActivityDate="2011-03-14T10:40:23.850" Title="refractive index of air in dependence of temperature" Tags="&lt;optics&gt;&lt;metrology&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="6873" PostTypeId="2" ParentId="6872" CreationDate="2011-03-14T09:23:41.780" Score="0" ViewCount="" Body="&lt;p&gt;The general form of such dependence is known for many different types of substance, but the exact values are not theoretically derivable as far as I know. What you are looking for is the &lt;em&gt;temperature-dependent &lt;a href=&quot;http://en.wikipedia.org/wiki/Sellmeier_formula&quot; rel=&quot;nofollow&quot;&gt;Sellmeier equation&lt;/a&gt;&lt;/em&gt;, but the constants of all Sellmeier equations for any substance are always fit to experimental data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mintaka.sdsu.edu/GF/explain/atmos_refr/air_refr.html&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; is a very good overview of all the available work in this area. Reading this, it looks like Jones' 1981 paper (&lt;a href=&quot;http://nvl.nist.gov/pub/nistpubs/jres/086/1/V86-1.pdf&quot; rel=&quot;nofollow&quot;&gt;freely available&lt;/a&gt;, since it was American government work at what was then called the National Bureau of Standards) contains a formula for the refractive index of air depending on temperature and pressure, among other things, although it doesn't appear to take the form of the Sellmeier equation.&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2011-03-14T09:23:41.780" CommentCount="2" />
  <row Id="6874" PostTypeId="2" ParentId="6872" CreationDate="2011-03-14T09:30:51.010" Score="1" ViewCount="" Body="&lt;p&gt;Refractive index (n - 1 to be exact)  is linearly correlated to density, as long as the material is not &#xA;too different chemically. Reason is, that refraction is caused by the polarizability,&#xA;which is caused solely by electrons in the domain of visible light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is often used in analysis, because refractive index is much easier to measure &#xA;than density. Example: sugar content of grapes, beer etc. Specially adapted Abbe Refractometers are sold to this purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the substance is the same, as in Your case, the correlation to density is exact.&#xA;That all this refers always to the same wavelength, is needless to say. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-14T10:40:23.850" LastActivityDate="2011-03-14T10:40:23.850" />
  <row Id="6875" PostTypeId="1" CreationDate="2011-03-14T12:26:20.657" Score="8" ViewCount="276" Body="&lt;p&gt;My question might be based on a false premise, so here's why I asked.   If you look up the meaning of the moment magnitude scale for measuring earthquake size, the articles usually say that each increase in magnitude by 1 point represents a tenfold increase in the amplitude of the wave, but a 31.6 increase in energy.   Which suggests the energy is going up as the three halves power.  With other waves the energy goes up with the square of the amplitude, so what is going on here?&lt;/p&gt;&#xA;" OwnerUserId="1430" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-14T13:51:33.773" LastActivityDate="2011-03-14T20:55:58.817" Title="Why does energy in earthquake waves seem to go up with the three halves power?" Tags="&lt;waves&gt;&lt;geophysics&gt;" AnswerCount="3" />
  <row Id="6876" PostTypeId="2" ParentId="6875" CreationDate="2011-03-14T12:40:45.347" Score="1" ViewCount="" Body="&lt;p&gt;My guess is the energy does not go up with the three halves power in the first place.  I think it only seems so at 600 km distance from the epicenter: $M_L = M_W \Rightarrow log_{10}(A_{max}) = \frac{2}{3}log_{10}M_0-10.7 \Rightarrow M_0 = \sqrt{A_{max}}^3e^{10.7\frac{3}{2}}$  At different distances there is a correction term in the formula for the Richter scale.: $M_L = log_{10}(A_{max}) - A_0(\delta)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_0(\delta)$ being some empirical function, the answer would be:  what the left part of the $M_L$ term &quot;makes wrong&quot; in terms of wave energy, the right one compensates for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Moment_magnitude_scale&quot; rel=&quot;nofollow&quot;&gt;momentum scale&lt;/a&gt; was designed to yield values similar to those from the Richter scale:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The scale was developed in the 1970s&#xA;  to succeed the 1930s-era Richter&#xA;  magnitude scale (ML). Even though the&#xA;  formulae are different, the new scale&#xA;  retains the familiar continuum of&#xA;  magnitude values defined by the older&#xA;  one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2415" LastEditorUserId="2415" LastEditorDisplayName="" LastEditDate="2011-03-14T15:57:59.377" LastActivityDate="2011-03-14T15:57:59.377" CommentCount="8" />
  <row Id="6877" PostTypeId="1" CreationDate="2011-03-14T12:45:30.070" Score="3" ViewCount="248" Body="&lt;p&gt;As far as i know Zeppelins were abandoned because they were not safe, back in time. Is it still the case in 2011 though? In my opinion it is more economical to operate a Zeppelin than a plane and they are more stable as there is less acceleration so the interior has tables and chairs etc.&lt;/p&gt;&#xA;" OwnerUserId="2351" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-14T13:51:16.630" LastActivityDate="2011-03-15T21:42:43.667" Title="Are Zeppelins still not feasible?" Tags="&lt;soft-question&gt;" AnswerCount="7" CommentCount="1" />
  <row Id="6878" PostTypeId="2" ParentId="6877" CreationDate="2011-03-14T12:55:48.707" Score="4" ViewCount="" Body="&lt;p&gt;They are safe now that they use helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exist zeppelins for special occasions. During the Olympics in Athens in 2004  there was a Zeppelin hovering for the news cover. There existed some by GoodYear for advertisement purposes some years ago. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are not used for transport because of economic choices. I think time is the important factor in plane travel, and they cannot compete there. There are &lt;a href=&quot;http://news.yahoo.com/s/yblog_weekend/20110304/ts_yblog_weekend/up-in-the-air-zeppelin-tours-the-sky&quot; rel=&quot;nofollow&quot;&gt;tourism uses.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-14T13:15:02.947" LastActivityDate="2011-03-14T13:15:02.947" CommentCount="2" />
  <row Id="6879" PostTypeId="2" ParentId="6877" CreationDate="2011-03-14T13:11:13.607" Score="3" ViewCount="" Body="&lt;p&gt;There was a German company &lt;a href=&quot;http://en.wikipedia.org/wiki/CargoLifter&quot; rel=&quot;nofollow&quot;&gt;CargoLifter&lt;/a&gt; that wanted to build commercial zeppelins to transport heavy weight goods. However the company got insolvent. In the german Wikipedia article it is mentioned that Boeing is working on some transportation craft similar to a zeppelin.&lt;/p&gt;&#xA;" OwnerUserId="2480" LastActivityDate="2011-03-14T13:11:13.607" />
  <row Id="6880" PostTypeId="2" ParentId="6877" CreationDate="2011-03-14T13:23:58.930" Score="6" ViewCount="" Body="&lt;p&gt;The problem is we are running out of helium.  Liquid helium is used in the MRI and other applications. It is also profligately wasted on party balloons.  So the prospect for helium lofted dirigibles is not good.  Also the US in the 1930s built some helium lofted dirigibles, which were destroyed by storms.  So they are not entirely safe, as they are pretty vulnerable to weather.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-14T13:23:58.930" CommentCount="2" />
  <row Id="6881" PostTypeId="2" ParentId="6877" CreationDate="2011-03-14T13:38:14.120" Score="2" ViewCount="" Body="&lt;p&gt;There's a company in my home town that's trying to revive the idea for local cargo transport.  Seems to me that there's a lot of technical challenges and very little to gain by doing so.  It'll be interesting to see how it ends up going for them...&lt;/p&gt;&#xA;" OwnerUserId="919" LastActivityDate="2011-03-14T13:38:14.120" />
  <row Id="6882" PostTypeId="2" ParentId="6303" CreationDate="2011-03-14T13:52:12.123" Score="2" ViewCount="" Body="&lt;p&gt;No-one has dared to answer this question yet. So let me get the ball rolling. It seems that there are a large number of problems to be solved for this proposal to work, but I'm not aware that any of these problems are definitely unsolvable (though that might just be my own ignorance). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) This is to be an AdS4/CFT3 duality, then we are dealing with M-theory (a stack of coincident M2-branes) on AdS4 x a 7-manifold, or equivalently, Type IIA string theory (a stack of coincident D2-branes) on AdS4 x a 6-manifold. As pointed out in &lt;a href=&quot;http://www.physicsforums.com/showthread.php?p=555472#23&quot; rel=&quot;nofollow&quot;&gt;a surprising dialogue from 2005&lt;/a&gt;, we immediately face the problems that (i) there's far too much supersymmetry for a realistic model (ii) the 7-manifold is as big as the AdS space. I don't know what to do about the latter problem, but for the better known AdS5/CFT4 duality, we have many variations on the original construction in which supersymmetry is broken a little or a lot, so let us suppose we can do that with AdS4 too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) In order to get the universe we see, a local or temporary patch with de Sitter geometry has to form within the AdS space. I have no idea how that would be represented in terms of the boundary CFT. In fact, I cannot even find a known CFT dual for any AdS space which is used as the basis for a construction of dS space in string theory. The dialogue above mentions a quest for CFT duals of the AdS spaces appearing in &lt;a href=&quot;http://arxiv.org/abs/hep-th/0301240&quot; rel=&quot;nofollow&quot;&gt;the KKLT paper from 2003&lt;/a&gt;; it doesn't seem that they are known yet, even in 2011. Embedding de Sitter space in the CFT framework is an even bigger challenge and figuring out how to do so would be a breakthrough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) The KKLT construction adds anti-D3-branes to the AdS space, to lift it to dS, but they were working in IIB string theory, and it looks like the gravity theory in our AdS4 space will be IIA. Fortunately, &lt;a href=&quot;http://arxiv.org/abs/hep-th/0506181&quot; rel=&quot;nofollow&quot;&gt;there's an alternative mechanism available for IIA&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) We also have to get the fields of the standard model out of this somehow. I'll mention just one problem here, and &lt;a href=&quot;http://arxiv.org/abs/hep-th/0306018&quot; rel=&quot;nofollow&quot;&gt;its potential solution&lt;/a&gt;: &quot;the inclusion of particles in the fundamental representation of the gauge group, i.e. quarks. Since AdS geometries arise as near-horizon limits for coincident branes, the dual field theories have only adjoint matter. The same is true for deformations of AdS. It turns out that fundamental representations can be introduced by including appropriately embedded probe branes.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there you have it. There's a whole raft of problems that would have to be solved to turn AdS4/CFT3 into a source of phenomenological and cosmological models. Some of them may already be demonstrably unsolvable, for technical reasons I don't appreciate, while at least one other basic element of the plan (dS/CFT) is a major unsolved research problem of string theory. Good luck! &lt;/p&gt;&#xA;" OwnerUserId="1486" LastActivityDate="2011-03-14T13:52:12.123" />
  <row Id="6883" PostTypeId="2" ParentId="5268" CreationDate="2011-03-14T14:19:34.330" Score="0" ViewCount="" Body="&lt;p&gt;Since there has been no useful information up to now I figured out some answers myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, the argument could be made even stronger. It is not only the time operator but also the position operators that appear to be problematic because an arbitrary shifted state $|0;k+\epsilon\rangle$ might lie outside the mass shell since generally $k^2 \neq (k+\epsilon)^2$. In usual quantum field theories (e.g. QED) it is possible to construct position operators (if one really has a need for them) which fulfill the Heisenberg algebra with the three-momentum. Unlike the case of string theory they do not commute with the energy operator, i.e. the mass shell condition is preserved. Unfortunately, these position operators do not obey Lorentz transformations and for massless particles they cannot be constructed for a particle spin greater than $1/2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, my original question contains a false assumption which is that the Hilbert space would only contain states being on the mass shell. Actually, the Hilbert space of string theory contains all possible four-momenta, but the quantisation procedure ensures that only states with a momentum fulfilling the mass shell condition contribute to physical amplitudes. I am however not sure about the negative energy states but I guess they also do not contribute to physical amplitudes. &#xA;In quantum theory and standard quantum field theory the Hilbert space (of free fields) is explicitly made up of positive energy states being on the mass shell. There, Pauli's argument applies that it is impossible to construct a time operator. This is not true for string theory as described above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as a consequence the usual norm of its Hilbert space is not well defined. It seems that the only possible solution is a redefinition of the norm. The approach seems to be common also in other approaches to quantum gravity theories with a covariant quantisation mechanism (see e.g. arxiv/1102.1148 and references therein). Not being able to work it out in more detail at the moment, I just take it for granted.&lt;/p&gt;&#xA;" OwnerUserId="1944" LastActivityDate="2011-03-14T14:19:34.330" />
  <row Id="6884" PostTypeId="2" ParentId="6877" CreationDate="2011-03-14T14:31:30.433" Score="5" ViewCount="" Body="&lt;p&gt;First, some nomenclature! (I'm German:=)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Zeppelin is a &lt;strong&gt;rigid&lt;/strong&gt; airship, constucted as a framework of aluminium beams, &#xA;with gas cells in this construction. What You see today from time to time are &#xA;so called &lt;strong&gt;Blimps&lt;/strong&gt;, a balloon-like thing filled with gas with a slight pressure. &#xA;This blimps are much smaller than the Zeppelins of the 30ties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zeppelins were never secure, even not when filled with helium. Almost all crashed in bad weather, fire was not the big problem (but much more sensationel). Zeppelins (and blimps) &#xA;are good in fine weather. I remember an article on psychology of security some years ago, where the author listed an impressive account of Zeppelins fate. All of them failed within some years, the question that author then discussed was: why did they keep on building such useless and insecure airships? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing which lures dreamers and some investors today is the fact that a Zeppelin can hoover &#xA;in a place nearly without energy. So one thinks they would be useful as cranes or to &#xA;monitor an area. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;H2 is not &quot;explosive&quot;! Did the Hindenburg in Lakehurst explode? No it burned, that was all! (And enough) Filling a stratosphere instrument balloon with hydrogen does not bear much risk, because there is no air within such balloons. The &quot;&quot;size of the atom of H2&quot; bigger than He is irrelevant. The figure ruling the effusion of a gas through a sheet of rubber/plastic is molecular weight!&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-15T21:42:43.667" LastActivityDate="2011-03-15T21:42:43.667" />
  <row Id="6886" PostTypeId="2" ParentId="6869" CreationDate="2011-03-14T15:21:59.763" Score="1" ViewCount="" Body="&lt;p&gt;The words &quot;Thermal Imaging&quot; are peferentially used when the thermal radiation of objects is &#xA;used for imaging. E. g. the cameras used to detect heat leakage of buildings, &#xA;or to &quot;see&quot; temperatures on the surface of a machine, or a hot running bearing. &#xA;Police uses thermal imaging from helicopters to search for missing/wanted persons, &#xA;a car which was used recently displays a hot front part etc etc.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-15T22:02:02.550" LastActivityDate="2011-03-15T22:02:02.550" />
  <row Id="6887" PostTypeId="1" CreationDate="2011-03-14T15:26:09.947" Score="2" ViewCount="196" Body="&lt;p&gt;This is a question about deriving effective mass theory for graphene. For the two sub-lattice atoms, the wave equation can be written as the massless Dirac equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \displaystyle -i\hbar v_F \begin{pmatrix} 0 &amp;amp; \partial_x -i\partial_y \\\partial_x +i\partial_y &amp;amp; 0 \end{pmatrix} \left(\begin{array}{c} \Psi_A \\ \Psi_B \end{array}\right)=E \left(\begin{array}{c} \Psi_A \\ \Psi_B \end{array}\right) \ \ \ \ \ (1)$&#xA;where ${A,B}$ are two subatoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivation of the equation went back to &lt;a href=&quot;http://prb.aps.org/pdf/PRB/v29/i4/p1685_1&quot; rel=&quot;nofollow&quot;&gt;1984&lt;/a&gt;, which is the paper I am trying to understand. In the article, they argued that at first order of ${\vec{\kappa}\cdot \vec{p}}$ expansion, the momentum matrix can be written in the form, under group-theoretic arguments (equation (3) in the paper): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \displaystyle \bar{p}\begin{pmatrix} 0 &amp;amp; \hat{x} -i\hat{y} \\ \hat{x} +i\hat{y} &amp;amp; 0 \end{pmatrix} . \ \ \ \ \ (2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the argument behind it? Is there anyone read the paper or know the answer?&lt;/p&gt;&#xA;" OwnerUserId="1948" LastActivityDate="2011-03-14T22:30:02.197" Title="effective theory of graphene" Tags="&lt;condensed-matter&gt;&lt;graphene&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="6888" PostTypeId="2" ParentId="6875" CreationDate="2011-03-14T15:32:34.667" Score="9" ViewCount="" Body="&lt;p&gt;The law looks like a great fundamental law that should have a neat derivation and the power $3/2$ looks like an accurate number. However, you will be disappointed, it is just a phenomenological estimated formula developed by Beno Gutenberg and Charles Richter:&#xA;$$ \log_{10}(E/{\rm erg}) = 11.8 + 1.5 M_S $$&#xA;Note that $M_S$ is calculated from a bandwidth between 18 and 22 seconds. See&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://earthquake.usgs.gov/learn/topics/measure.php&quot; rel=&quot;nofollow&quot;&gt;http://earthquake.usgs.gov/learn/topics/measure.php&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The shaking amplitude shown by the seismometer - and its base-ten logarithm is how the Richter scale is defined - is something between an &quot;energy reading&quot; and an &quot;amplitude reading&quot;. For the latter, you would square the amplitude to get the energy; for the former, the exponent would be one. So in reality, the relationship is something in between and the exponent is $3/2$ but it is just an approximate estimate. It's a messy problem and one would have to make an analysis of the energy that can be released in any allowed frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While we can measure how much seismometers move, we can't really reliably measure the energy released by an earthquake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that Gutenberg's name was ultimately omitted from the name of the Richter scale - it's because Gutenberg didn't like interviews. The term &quot;magnitude&quot; was chosen by Richter because of his self-evident interest in astronomy which uses the same term for a similar logarithmic scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see the actual paper by these two men that introduced the scale in the 1930s, read&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.sciencemag.org/content/83/2147/183&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencemag.org/content/83/2147/183&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-14T15:32:34.667" CommentCount="1" />
  <row Id="6889" PostTypeId="2" ParentId="783" CreationDate="2011-03-14T15:38:00.860" Score="2" ViewCount="" Body="&lt;p&gt;This &quot;looking/watching destroys the interference pattern&quot; is somewhat misleading, because it does not tell that &quot;watching&quot; at quantum level means shooting quanta of similar energy at the slits and the &quot;balls&quot; .&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-14T15:38:00.860" />
  <row Id="6890" PostTypeId="1" AcceptedAnswerId="6891" CreationDate="2011-03-14T15:59:35.133" Score="1" ViewCount="145" Body="&lt;p&gt;The question is hard for me to put into one sentence so please try to completely read the example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I had a stick that is 1000KM long and I would push it forward with 1 millimeter in lets say... 10ms(I have completely no idea how long it would take)but let's say I was about to turn on a switch that is 1000KM + 1 millimeter away from me, how long would it take for me to see the light that is next to the switch to turn on? Basically the answer would be a calculation based on the distance the light needs to travel in order for it to reach me + that one millisecond. But I cannot understand that that calculation does not consider anything about the stick being pushed, if the stick was 1 meter and I would perform the same action the calculation would be the same, resulting in the fact that the stick can have any length no matter how far it would still turn on that light as fast as it would if it was 1 meter long or 1000KM long. Something in my logic way of thinking is telling me that the stick must be slowed down based on the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can provide an answer to the next example I would fully understand this paradox:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have a stick that is 1000KM long and pushing it forward just a little would cause the light in my tower to turn on, 1000KM further away someone is waiting on my push and has the end of the stick placed on the switch of hes light tower. What would the someone see happening if I push that stick ? &#xA;There are 3 answers possible:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;He(the someone, the 2nd person) will see hes light turn on and then the other.&lt;/li&gt;&#xA;&lt;li&gt;He would see both lights turn on simontaniously. (my answer, considering the speed of light delay on the moving stick)&lt;/li&gt;&#xA;&lt;li&gt;He would see the other light turn on 1st and than his. (very unlikely)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The the answer I need is either 1 or 2. I do not understand why it would be 1 but I would love to accept it, in order to accept it I need someone to explain to me why 1 is the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you so much for reading my question and/or leaving a comment or answer !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diede.&lt;/p&gt;&#xA;" OwnerUserId="2547" LastActivityDate="2011-03-20T01:59:20.303" Title="The speed of light also applies for 'distance' materials?" Tags="&lt;speed-of-light&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="6891" PostTypeId="2" ParentId="6890" CreationDate="2011-03-14T16:11:42.227" Score="6" ViewCount="" Body="&lt;p&gt;There is no such thing as an incompressible stick. When you push on your end, a compression wave travels down the stick at the speed of sound, which is much slower than the speed of light. The other end of the stick does not move until this compression wave reaches it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the rest of your problems disappear without the existence of an incompressible stick. If you still have any questions remaining, I'm happy to edit this answer.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2011-03-14T16:11:42.227" />
  <row Id="6892" PostTypeId="2" ParentId="6890" CreationDate="2011-03-14T16:14:27.427" Score="4" ViewCount="" Body="&lt;p&gt;Here's the answer to your first example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though it might take you only 10 ms to push your end of the stick, it would still take a much longer time for the effect of that to reach the other end. This is because there are no perfect rigid bodies. An impulse at one end has to travel at the speed of sound in the material, which is always less than the speed of light. So it will take $l/v_s$ + $l/c$ + 10 ms + 10 ms for you to see the light turn on, where $v_s$ is the speed of sound and $l$ is the length of your stick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't fully understand your second example, but I'm guessing the speed of sound limitation should also make things sensible here.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-03-14T16:14:27.427" CommentCount="2" />
  <row Id="6893" PostTypeId="1" AcceptedAnswerId="6896" CreationDate="2011-03-14T16:14:43.350" Score="1" ViewCount="107" Body="&lt;p&gt;I recently calculated the PSD of the vibration data in a car at constant speed. I would like to know what this means and what if I calculate PSD of the vibration data in a car for the total journey with varying speeds? What do both PSDs signify?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastActivityDate="2011-03-14T17:18:43.120" Title="Power Spectral Density of constant speed data in a car" Tags="&lt;mathematical-physics&gt;" AnswerCount="1" />
  <row Id="6894" PostTypeId="2" ParentId="6875" CreationDate="2011-03-14T16:16:09.617" Score="3" ViewCount="" Body="&lt;p&gt;We have confusion because the Richter scale is very popular, but only measures the energy of the initial second or two of the earthquake. Great quakes like this one take a few minutes to fully rupture and release strain energy over a large area, so the correlation between the sharpness of the initial break and the total energy released wouldn't be expected to be too great. The more modern measure is called seismic moment M, and attempts to account for all the energy. Many journalist hear magnitude, and automatically append &quot;Richter scale&quot; even when it is inappropriate. If you look up the &lt;a href=&quot;http://en.wikipedia.org/wiki/Moment_magnitude_scale&quot; rel=&quot;nofollow&quot;&gt;moment magnitude scale&lt;/a&gt; on Wikipedia, you see that the 3/2 power is explicitly built into the formula, i.e. the log10 of the energy is estimated as M0 and then multiplied by 2/3 to come up with the reported moment. I presume this was done to make the energy/magnitude of the two scales as similar as possible (in the range where they both have reasonable validity).&lt;/p&gt;&#xA;" OwnerUserId="377" LastEditorUserId="377" LastEditorDisplayName="" LastEditDate="2011-03-14T20:55:58.817" LastActivityDate="2011-03-14T20:55:58.817" CommentCount="1" />
  <row Id="6896" PostTypeId="2" ParentId="6893" CreationDate="2011-03-14T17:18:43.120" Score="1" ViewCount="" Body="&lt;p&gt;I'll make a few guesses here.&#xA;  At a first look, I wouln't think speed would have much effect on the normal vibrational effects. At higher speed we have, greater torque and wind resistance, which would look like a slightly different external loading. More significantly, the wheels are rotating faster, and that probably modifies the normal modes spectrum somewhat. Lastly, the excitation comes from tire noise, and road imperfection (and maybe turbulent air flow), which presumably is skewed more towards higher frequencies at higher speeds.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-03-14T17:18:43.120" CommentCount="1" />
  <row Id="6897" PostTypeId="1" CreationDate="2011-03-14T18:13:22.067" Score="7" ViewCount="310" Body="&lt;p&gt;Einstein initially added the Cosmological Constant because (if I get this right) it seemed to him that the universe should be static. I agree that back then this would have been an obvious assumption. I'm curious now, before Hubble, where there any opinions/debates about whether the universe would be expanding or contracting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kind regards,&#xA;Thomas&lt;/p&gt;&#xA;" OwnerUserId="2550" LastActivityDate="2011-04-18T03:08:05.937" Title="Why didn't Newton have a cosmological constant" Tags="&lt;cosmological-constant&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="3" />
  <row Id="6898" PostTypeId="1" AcceptedAnswerId="6902" CreationDate="2011-03-14T18:41:42.667" Score="7" ViewCount="491" Body="&lt;p&gt;I saw an interview with Ed Witten, where he said one way to confirm string-theory is to observe a giant string floating in space, left over from the Big-Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one observe such a string, does it have thickness?&#xA;Is this string any different then the strings that (hypothetically) make up elementary particles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is required to produce such a large string, why is this string so much bigger then others?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happends if one tries to interact with this string, can one stretch/cut it etc?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we have a stable mega-string on earth, and can it be used for anything?&lt;/p&gt;&#xA;" OwnerUserId="1708" LastEditorUserId="1708" LastEditorDisplayName="" LastEditDate="2011-03-15T06:50:38.590" LastActivityDate="2011-03-15T06:50:38.590" Title="Giant string in space" Tags="&lt;string-theory&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="6899" PostTypeId="1" CreationDate="2011-03-14T18:49:06.623" Score="4" ViewCount="73" Body="&lt;p&gt;In a interaction-round-a-face model of $n^2$ particles in a lattice,  a weight $W(a,b,c,d)$ is assigned to each face in the lattice based on the spins $a,b,c,d$ (listed say from the bottom-left corner in counter-clockwise fashion) of the particles at the corners of the face. Based on this, a partition function is formed $Z=\sum\prod W(a,b,c,d)$, where the the sum is over all possible spins of all particles and the product is over all the faces in the lattice. The well-known text &lt;em&gt;Exactly Solved Models in Statistical Mechanics&lt;/em&gt; by Baxter shows how to obtain exact solutions for $Z$ in a couple cases, like the six-vertex model and the eight-vertex model, using &quot;commuting transfer matrices.&quot;  In reading Baxter's book, I am uncertain of the applicability of this method. For instance, I am working on a particular application where particles in a square lattice can have either up spin (+1) or down spin (-1) and the weight of any face $W(a,b,c,d)=1$ unless all four spins around the face alternate, in which case $W(1,-1,1,-1)=W(-1,1,-1,1)=4$. (The boundary condition here is that the all the spins on the boundary must be up.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know if the commuting-transfer-matrix method (or any other method) should yield an exact solution for this particular weighting of the spins? More generally, should the commuting-transfer-matrix method yield exact solutions for &lt;em&gt;any&lt;/em&gt; weighting of the spins (even in the two-spin case)?&lt;/p&gt;&#xA;" OwnerUserId="2542" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-03-14T21:11:51.263" LastActivityDate="2011-03-14T21:11:51.263" Title="Applicability of Baxter's method for IRF models" Tags="&lt;statistical-mechanics&gt;" FavoriteCount="2" />
  <row Id="6900" PostTypeId="2" ParentId="6898" CreationDate="2011-03-14T19:14:54.467" Score="7" ViewCount="" Body="&lt;p&gt;This is the conversion of a superstring into a cosmic string.  A cosmic string is a one dimensional region where the vacuum state is that of the unbroken Lagrangian.  It is similar to the physics of supercooled water that can be colder than the freezing point.  In the phase transition of a gas or liquid to solid there can be boundary zones or grains where the gas or liquid state persists.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The connection to string theory is with F-theory, which takes one into 12-dimensions, or with D-strings that are related to F-strings by S-duality between strong and weak interactions.  One of the corner stones of string theory is the Nambu-Goto action, which is a starting point for the string action determined by the area of the string world sheet.  The one spatial quantum dimension theory also describes the cosmic string.  If a superstring is drawn into a very large filament so that its states remain invariant under the symmetries of the Lagrangian the string can be converted into this &quot;defect&quot; that is a cosmic string.  The string has a huge tension, related to the string parameter $\alpha&amp;#39;$, and if this string is stretched to enormous lengths this can have a large gravitation.  The curvature of space can be thought of as an orbit around the string, but where the area enclosed by the loop is the disk with a wedge cut out of it.  This deficit angle $\theta$ then defines a curvature bounded within that loop as $R~\sim~\theta/2\pi r^2$, where $r$ is the radius of the loop.       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe might have these floating around out there, sort of menacing cosmic bullwhips of sorts, that could have an imprint on the CMB.  It would be bad new if one of these cut through the Earth, for it would gravitationally pancake the Earth some.  As I recall the Earth would be squashed inwards at 4km/sec as it passed through.  We would get a global quake many orders of magnitude worse than the recent Sendai quake.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is hard to know how stable these are, and it could be that interactions with them might break the string up into small strings near the string length.  The deficit angle above would have a gravitational lensing effect, and if one passed between the Earth and a distant object it might be detected that way.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-03-14T19:22:47.327" LastActivityDate="2011-03-14T19:22:47.327" />
  <row Id="6901" PostTypeId="2" ParentId="6869" CreationDate="2011-03-14T19:22:28.460" Score="2" ViewCount="" Body="&lt;p&gt;Infrared imaging can refer to any imaging system that operates in the infrared, which extends from about .7 to 300 microns wavelength, as Lubos stated. Thermal imaging, as the term is commonly used, refers to an infrared imaging system designed specifically for the portion of the infrared range which is emitted by object at or just above room temperature. This is typically in the neighborhood of 10 microns. At this wavelength you will see images that people typically associate with &quot;thermal vision.&quot; People will be brighter than room temperature objects, running car engines will stand out, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Infrared imaging is useful as well, of course. Most &quot;night vision&quot; goggles work by imaging visible and infrared light, and amplifying the resulting image to a level that the user can detect. Using infrared in addition to visible simply means that the goggles have a stronger signal to work with before amplification, because they are sensitive to a broader range of wavelengths.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-03-14T19:22:28.460" CommentCount="0" />
  <row Id="6902" PostTypeId="2" ParentId="6898" CreationDate="2011-03-14T19:33:26.203" Score="11" ViewCount="" Body="&lt;p&gt;The things in question are called &quot;cosmic strings.&quot; They are predicted as a byproduct of some but by no means all theories of the early Universe. There's no experimental evidence that they actually exist, although I suppose the possibility hasn't been ruled out.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How would we observe this?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A long cosmic string would have a couple of observable signatures. The main effect is a sort of gravitational lensing. If a cosmic string lies directly between you and a background object, you will see two copies of the object, one from light that passed around the string to the left, and one from light that went to the right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To go into a bit more detail, a cosmic string produces a &quot;conical singularity&quot; in the space around it. This means that the circumference of a circle surrounding the string will be slightly less than $2\pi$ times the radius. The fractional deficit is independent of radius. It's as if someone had removed a triangular wedge of space, and then &quot;glued together&quot; the faces adjoining the removed bit. To see why this is called a conical singularity, imagine taking a piece of paper, cutting out a triangular wedge (i.e., removing all points with $0&amp;lt;\phi&amp;lt;\alpha$ for some constant $\alpha$), and then gluing together the edges $\phi=0$ and $\phi=\alpha$. The paper will now have the shape of a cone; an ant walking around the singularity will cover a circumference $C=(2\pi-\alpha) r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light travels in straight lines (geodesics) in this conical space, but two such straight lines can pass around opposite sides and meet again. (Draw it on your conical paper if you don't see this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one way to look for cosmic strings is to look for pairs of identical images of something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the cosmic string happens to be moving transverse to your line of sight, then there's another observational signature. Background light that reaches you from the &quot;downstream&quot; side will be blueshifted compared to background light reaching you from the upstream side. In effect, the conical geometry gives a kick to photons coming from one side relative to the other. (That's actually a kind of imprecise way to put it: this is a purely geometrical effect, not an actual force.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does it have a thickness?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Cosmic strings are predicted to have some nonzero thickness, but very small. I don't remember the details.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Are they the same as the strings that (are theorized to) make up elementary particles?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Cosmic strings are not the same as the strings in string theory. There are connections between the two theories, which I don't understand at all. I understand cosmic strings pretty well and string theory not at all. I hope someone else will talk about the connections.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is required to produce such a large string?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Cosmic strings are predicted to result in some theories of the early Universe. They are &quot;topological defects&quot; resulting from a phase transition. During the phase transition, some field tries to &quot;relax&quot; into a lower-energy configuration. There are multiple different versions of the lower-energy state, and the field chooses different versions in different parts of space. A cosmic string results from different regions making incompatible choices, so that the whole system can't relax to the same state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you simulate such a phase transition, in which different points &quot;choose&quot; randomly which state to relax into and then the whole system tries to settle down into a low-energy final configuration, you find that many such strings result. There are large numbers of small loops and smaller numbers of big strings. The very long ones that Witten is talking about are predicted to be rare, but there should be a small number in our observable Universe (if a string-producing phase transition actually occurred).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What happens when you interact with such a string?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The strings are under high tension and have a high energy density confined to a very small cross-sectional area. It'd be hard to cut one, without supplying enough energy to &quot;undo&quot; the phase transition. When two strings (or two sections of the same string) meet, they can &quot;reconnect,&quot; essentially swapping with each other at the intersection point. A loop of string can &quot;pinch off&quot; smaller loops via this process: imagine the string twisting around into a figure-8 shape, and then splitting into two smaller loops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a lot more to the dynamics of these strings. They can lose energy by emitting gravitational waves, for example.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can we have one on Earth?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In principle we could probably produce some small ones, if we could get the energy density in some region up high enough to undo the phase transition, but nobody's done anything remotely like this. If we did have a big one, I don't know how easy it would be to control it.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-14T19:33:26.203" />
  <row Id="6903" PostTypeId="2" ParentId="6897" CreationDate="2011-03-14T20:05:35.230" Score="2" ViewCount="" Body="&lt;p&gt;It's a very good question but the answer is that Newton's Universe actually doesn't have to expand or contract so no cosmological constant is needed. Well, it's a bit more subtle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right non-relativistic gravitational equation where one should add the vacuum energy density is the Poisson equation&#xA;$$ \nabla^2 \phi_g = 4\pi G \rho + \Lambda_{{\rm Newton}}. $$&#xA;I added a Newtonian cosmological constant term. For a Newtonian cosmology with a uniform mass distribution at the cosmological scale (e.g. above hundreds of megaparsecs), you actually have to add this term (with a negative sign), to neutralize the mass density at the very long distance scale. If you omit this term, the $\phi_g$ potential has to have a Laplacian with a constant term, so $\phi_g$ itself will have to contain something like $\vec x^2$ which will inevitably be minimized at some point of the Universe - $\vec x =0$ in my conventions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amusingly enough, one may describe Newton's gravitational forces without any $\phi_g$, by manifestly summing the forces from other point masses in the Universe. It's an equivalent approach to calculate the acceleration but it allows us to avoid the problem with the preferred point in the Universe. I may just claim that the forces acting on the Earth that are caused by very distant objects cancel. This is equivalent to saying that the Earth is the $\vec x = 0$ point - and one may say the same thing for any other object in the Universe (a method to regulate the infrared divergences from the forces caused by very distant objects).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Needless to say, the assumption that we choose a &quot;uniform cutoff&quot; around every probe in the Universe is totally equivalent to adding the neutralizing Newtonian cosmological constant above. Also, you won't be able to invent any non-equivalent yet consistent Newtonian cosmologies with a uniform Universe but a nonzero cosmological constant. That's really because the Newtonian spacetime is flat and the cosmological constant is the curvature of the empty spacetime - which vanishes in Newton's theory by definition.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-14T20:05:35.230" CommentCount="2" />
  <row Id="6904" PostTypeId="1" AcceptedAnswerId="6941" CreationDate="2011-03-14T21:06:35.140" Score="4" ViewCount="300" Body="&lt;p&gt;I asked this question at math.stackexchange but the answers were all too abstract or useless for me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\varphi= \frac{1+\sqrt{5}}{2}\approx 1.6180339887...$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some real world applications of the golden ratio? Does it ever function specifically as a constant in any formulas or theorems? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Original title said Golden Radio... facepalm. &lt;/p&gt;&#xA;" OwnerUserId="2536" LastEditorUserId="2536" LastEditorDisplayName="" LastEditDate="2011-03-14T21:44:05.467" LastActivityDate="2011-03-15T12:09:35.993" Title="Uses of the 'Golden Ratio'" Tags="&lt;soft-question&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="1" />
  <row Id="6906" PostTypeId="1" AcceptedAnswerId="6920" CreationDate="2011-03-14T21:38:05.910" Score="7" ViewCount="209" Body="&lt;p&gt;Given recent events in Japan, this got me wondering. Is it possible to determine radiation levels reliably not having Geiger counters near the possible radiation contaminated zone? According to wikipedia the Chernobyl disaster was first (other than Soviet Union naturally) detected by Swedish via radioactive particles found on clothes of nuclear plant workers. Surely more efficient ways should have been developed by now. &lt;/p&gt;&#xA;" OwnerUserId="2551" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-15T16:32:26.987" LastActivityDate="2011-03-17T20:38:03.683" Title="Is it Possible to Determine Radiation Levels Using Satelites?" Tags="&lt;soft-question&gt;&lt;radiation&gt;" AnswerCount="2" />
  <row Id="6907" PostTypeId="1" AcceptedAnswerId="6950" CreationDate="2011-03-14T21:43:18.467" Score="5" ViewCount="307" Body="&lt;p&gt;So far, the quarks and leptons appear to be fundamental particles. But they're complicated enough that there's always been some speculation that they might be composite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What experimental evidence would be needed to show that a lepton is composite?&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-15T15:41:36.167" LastActivityDate="2011-03-16T19:34:41.267" Title="What Would be the Experimental Signature of Composite Leptons?" Tags="&lt;particle-physics&gt;&lt;experimental-physics&gt;&lt;experimental-evidence&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="1" />
  <row Id="6908" PostTypeId="1" CreationDate="2011-03-14T21:59:55.873" Score="8" ViewCount="197" Body="&lt;p&gt;My question might sound convoluted but my mind is twisting right now so my apologies in advanced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it that when I have one proton and one electron it is Hydrogen a clear flammable gas, and when I have say, twelve, it is carbon the driving force of life as we know it and then we end up with 239 and we have uranium a radioactive element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the amount of protons affect what matter will be? Why does the amount of these particle things magically make one thing air and another cyanide?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have one pebble it is a pebble, if I have 13 its just more of the same pebbles. What makes protons different?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry this is so convoluted, &#xA;any reply is appreciated&#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2552" LastActivityDate="2011-03-15T18:52:50.853" Title="Why does amount of protons define how matter is?" Tags="&lt;atoms&gt;&lt;subatomic&gt;&lt;proton&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="6909" PostTypeId="2" ParentId="6904" CreationDate="2011-03-14T22:03:45.383" Score="1" ViewCount="" Body="&lt;p&gt;Googling arxiv comes up with lots of hits. For example:&lt;br&gt;&#xA;NewJ.Phys.11:063026 (2009), Adisorn Adulpravitchai, Alexander Blum, Werner Rodejohann, &lt;a href=&quot;http://arxiv.org/abs/0903.0531&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Golden Ratio Prediction for Solar Neutrino Mixing&lt;/em&gt;&lt;/a&gt; :&lt;br&gt;&#xA;It has recently been speculated that the solar neutrino mixing angle is connected to the golden ratio $\phi$. Two such proposals have been made, $\cot(\theta_{12}) = \phi$ and $\cos(\theta_{12}) = \phi/2$. We compare these Ansatze and discuss a model leading to $\cos(\theta_{12}) = \phi/2$ based on the dihedral group $D_{10}$. This symmetry is a natural candidate because the angle in the expression $\cos(\theta_{12} = \phi/2$ is simply $\pi/5$, or 36 degrees. This is the exterior angle of a decagon and $D_{10}$ is its rotational symmetry group. We also estimate radiative corrections to the golden ratio predictions.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-14T22:03:45.383" />
  <row Id="6910" PostTypeId="1" CreationDate="2011-03-14T22:06:26.300" Score="4" ViewCount="127" Body="&lt;p&gt;Which type two s-wave superconductor has the smallest London penetration depth? What is l_lamda ?&lt;/p&gt;&#xA;" OwnerUserId="2553" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-14T23:59:12.837" LastActivityDate="2011-03-14T23:59:12.837" Title="London penetration depth" Tags="&lt;superconductivity&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="6911" PostTypeId="2" ParentId="6908" CreationDate="2011-03-14T22:06:38.757" Score="7" ViewCount="" Body="&lt;p&gt;Long answer: Any Chemistry textbook. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short answer: The number of electrons of an atom is the same as the number of protons in the nucleus. This number of electrons (Identical to the position number in PSE!) defines all the chemistry of that atom. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-14T22:06:38.757" CommentCount="2" />
  <row Id="6912" PostTypeId="1" AcceptedAnswerId="6916" CreationDate="2011-03-14T22:12:08.097" Score="4" ViewCount="176" Body="&lt;p&gt;I understand that &lt;em&gt;information&lt;/em&gt; cannot be transmitted at a velocity greater than speed of light. I think of this in terms of the radio broadcast: the station sends out carrier frequencies $\omega_c$ but the actual content is carried in the modulated wave, the sidebands $\omega_c\pm\omega_m$. The modulation envelop has its group velocity and this is the speed at which information is being transmitted. We also know, for example, that x-rays in glass have a phase velocity which is greater than the speed of light in vacuum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what &lt;em&gt;exactly&lt;/em&gt; is it that is travelling faster than the speed of light?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: I know it couldn't be anything physical as its superluminal. My problem is, what is it that has absolutely no information content yet we associate a velocity with it.&lt;/p&gt;&#xA;" OwnerUserId="1671" LastEditorUserId="1671" LastEditorDisplayName="" LastEditDate="2011-03-14T22:24:00.880" LastActivityDate="2011-03-15T00:51:31.353" Title="In superluminal phase velocities, what is it that is traveling faster than light?" Tags="&lt;speed-of-light&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="6913" PostTypeId="2" ParentId="6912" CreationDate="2011-03-14T22:16:17.390" Score="2" ViewCount="" Body="&lt;p&gt;Look at a particular local maximum  on the electromagnetic field.  A small moment of time $dt$ later, there will be a local maximum a small distance $dx$ away.  Superluminal phase velocity means $dx/dt &amp;gt; c$.  The &quot;thing&quot; moving faster than light is not a thing at all, but a point at which the electromagnetic field is a maximum (or has some other, constant phase, i.e. a minimum, inflection point, etc.)&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-03-14T22:16:17.390" CommentCount="1" />
  <row Id="6915" PostTypeId="2" ParentId="6887" CreationDate="2011-03-14T22:30:02.197" Score="1" ViewCount="" Body="&lt;p&gt;In the paper of J.M. McCLURE(1956), he showed how to directly calculate the momentum matrix. (eq.2.5,2.6) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diamagnetism of Graphite,Phys. Rev. 104, 666–671 (1956)&#xA;&lt;a href=&quot;http://prola.aps.org/abstract/PR/v104/i3/p666_1&quot; rel=&quot;nofollow&quot;&gt;http://prola.aps.org/abstract/PR/v104/i3/p666_1&lt;/a&gt;   &lt;/p&gt;&#xA;" OwnerUserId="1948" LastActivityDate="2011-03-14T22:30:02.197" />
  <row Id="6916" PostTypeId="2" ParentId="6912" CreationDate="2011-03-14T22:42:07.553" Score="9" ViewCount="" Body="&lt;p&gt;Shine a flashlight on a wall. Rotate the flashlight so the illuminated spot moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q: How fast does the spot move?&lt;br&gt;&#xA;A: It depends how far away the wall is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q: How fast can the spot possibly move?&lt;br&gt;&#xA;A: There is no limit. Put the wall far enough away, and the spot can move with any speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q: What is moving across the wall?&lt;br&gt;&#xA;A: Nothing. The light that makes up the spot at one instant is unrelated to the light that makes up the spot an instant later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how a wave can be apparently superluminal: we interpret a series of unrelated events as a continuous 'wave'. &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v88/i7/e073901&quot; rel=&quot;nofollow&quot;&gt;Group velocity can also be superluminal&lt;/a&gt;; even though the individual chunks of energy are going at roughly $c$, the region where they superpose constructively (the 'crest of the wave') goes faster than $c$.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="2359" LastEditorDisplayName="" LastEditDate="2011-03-14T22:53:35.750" LastActivityDate="2011-03-14T22:53:35.750" CommentCount="2" />
  <row Id="6917" PostTypeId="2" ParentId="6912" CreationDate="2011-03-14T22:53:20.947" Score="1" ViewCount="" Body="&lt;p&gt;Just one comment. The fact that information cannot travel faster than light means that it cannot travel faster than the speed of light in vacuum. In media were the speed of light is reduced it is still possible that other high energy particles, e.g. electrons, move faster than the speed of light in this medium and thus information speed is faster than the propagation of light there. When the speed of the particle exceeds the phase velocity Cherenkov radiation is produced. This is analogous to the sonic boom of a supersonic aircraft. This effect is used in high energy experiments to determine the velocity of particles. (Look up Cherenkov Radiation at Wikipedia)&lt;/p&gt;&#xA;" OwnerUserId="1944" LastActivityDate="2011-03-14T22:53:20.947" />
  <row Id="6918" PostTypeId="2" ParentId="6907" CreationDate="2011-03-14T22:56:44.927" Score="4" ViewCount="" Body="&lt;p&gt;With the right equipment and enough energy you can look for all the usual naive stuff:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Deviations from the Bhabha scattering cross-section in unpolarized $l + \bar{l}$  or $l + l$ scattering. In particular if there is missing energy in the reaction that might indicate an excited state in the products.&lt;/li&gt;&#xA;&lt;li&gt;Resonance peaks in in $l + \bar{l}+ \to l&amp;#39; + \bar{l}&amp;#39;$. (Of course you can do this with $q + \bar{q} \to l + \bar{l}$ too, but the QCD corrections to the quark vertex makes the theory harder and may hide the signal.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think this is part of the case for a muon collider, but none of it is on the table for experiments running right now.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-03-14T22:56:44.927" CommentCount="2" />
  <row Id="6919" PostTypeId="1" CreationDate="2011-03-14T22:58:45.567" Score="5" ViewCount="195" Body="&lt;p&gt;Hey all, are there any theories out there on the origins (or infinite existence of) the universe beside the big bang that actually adhere to current scientific knowledge and fact?&lt;/p&gt;&#xA;" OwnerUserId="2552" LastActivityDate="2011-03-15T17:51:29.430" Title="Alternative theories to the big bang?" Tags="&lt;theory&gt;&lt;universe&gt;&lt;big-bang&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="6920" PostTypeId="2" ParentId="6906" CreationDate="2011-03-14T23:10:22.523" Score="7" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Short answer:&lt;/strong&gt; no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Longer answer:&lt;/strong&gt; No, excepting neutrinos none of the products of radioactive decay has the penetrating power to pass through the atmosphere, and neutrino detection is not something we can do from satellites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To elaborate, the immediate products of radioactive decay are (some set of, depending on the decay in question) fission fragments, electrons, positrons, alphas, neutrons, photons (gamma rays) and neutrinos. Plus the remnant nucleus. The only secondary product which might be interesting is Cerenkov light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrons and positrons will travel a number of cm in air (at ground level). The gamma might go a few meters. The heavy stuff has no penetrating power at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if lofted to the top of the troposphere, there is just too much air in the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cerenkov light will, of course, go through a lot of atmosphere, but you'd be looking for a pale blue glow against the general light background. For dispersed radionucleides (i.e. contamination), the intensity will be awfully low.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;N.B. I too have seen various TV show and movie where some character from some agency says &lt;em&gt;&quot;We can track the radiation with satellites!&quot;&lt;/em&gt;. I believe this to be misinformed babbling of desperate script writers.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-03-14T23:29:48.583" LastActivityDate="2011-03-14T23:29:48.583" CommentCount="3" />
  <row Id="6921" PostTypeId="2" ParentId="6904" CreationDate="2011-03-14T23:17:24.043" Score="1" ViewCount="" Body="&lt;p&gt;In general, no, the golden ratio is not used often in physics.  As a graduate student in experimental physics, I have never encountered the golden ratio in my work, except in &lt;a href=&quot;http://nibot-lab.livejournal.com/2010/07/02/&quot; rel=&quot;nofollow&quot;&gt;toy problems&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-03-14T23:17:24.043" CommentCount="6" />
  <row Id="6922" PostTypeId="2" ParentId="6908" CreationDate="2011-03-14T23:35:12.773" Score="3" ViewCount="" Body="&lt;p&gt;Your day to day experience of the material world is governed by chemistry.  This is at some level the science of atoms and groups of atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things like hardness, colour, toxicity and others are all largely determined by the interaction of atoms.  In particular the outer coating of atoms, the electrons.  Obviously the details of why element or compound A is harder or softer than compound B is incredibly diverse and complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically the number of electrons is equal to the number of protons (but not always!)  as atoms like to be electrically neutral and so for each positive proton, one negative electron will stick to the atom.  So from this we can see that the number of protons indirectly affects the behaviour of an atom via the number of electrons...&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-03-14T23:35:12.773" />
  <row Id="6923" PostTypeId="2" ParentId="6898" CreationDate="2011-03-14T23:43:22.467" Score="3" ViewCount="" Body="&lt;p&gt;This paper: &lt;a href=&quot;http://arxiv.org/PS_cache/astro-ph/pdf/0302/0302547v1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/astro-ph/pdf/0302/0302547v1.pdf&lt;/a&gt; was a speculative look at a possible double galaxy image. They considered that it had been mirrored down the centre by a cosmic string to repeat the galaxy.  It turned out to be untrue however when looked at in more detail by hubble. &lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-03-14T23:43:22.467" />
  <row Id="6925" PostTypeId="2" ParentId="6904" CreationDate="2011-03-15T00:07:46.623" Score="10" ViewCount="" Body="&lt;p&gt;Actually the golden ratio does show up in at least one interesting physical situation: cobalt niobate is an experimental realization of the 1D Ising model, which -- in a magnetic field transverse to the axis that neighboring spins are coupled along -- has a &lt;strong&gt;quantum&lt;/strong&gt; phase transition. In 1989 Zamoldchikov studied this model and discovered an amazing thing that has quite recently been verified experimentally by Coldea's group: if you look at magnetic excitations (through neutron scattering) right near the quantum critical point, there is structure at a sequence of energies -- the lowest two of these have a ratio that is exactly $\varphi$. [In fact it goes deeper than this, with the excitation energies actually having a connection to the structure of the E8 group... but we'll leave that as homework!]&lt;/p&gt;&#xA;" OwnerUserId="791" LastEditorUserId="791" LastEditorDisplayName="" LastEditDate="2011-03-15T00:42:18.873" LastActivityDate="2011-03-15T00:42:18.873" CommentCount="2" />
  <row Id="6926" PostTypeId="2" ParentId="6912" CreationDate="2011-03-15T00:51:31.353" Score="4" ViewCount="" Body="&lt;p&gt;Let me make an analogy which may be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next time you're at the beach, watch the waves coming in. You will notice that the point where the wave breaks will often move faster than the waves do. This happens when the waves come in close to perpendicular to the beach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect is that if you know the wave is breaking at point A, you can predict that it will break at point B soon afterward. Note that the short vector gives the wave movement while the long arrow shows where on the beach the wave is breaking.  But this is not information that is moving; the wave has the information built into its wave structure:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/6N935.jpg&quot; alt=&quot;Beach&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thing applies to &lt;a href=&quot;http://en.wikipedia.org/wiki/Phase_velocity#Matter_wave_phase&quot; rel=&quot;nofollow&quot;&gt;de Broglie (or quantum matter) waves which are also superluminal&lt;/a&gt;. The short answer is that there isn't anything that is traveling faster than light. It's just an effect of little physical meaning.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-15T00:51:31.353" CommentCount="4" />
  <row Id="6927" PostTypeId="2" ParentId="6750" CreationDate="2011-03-15T01:11:25.133" Score="7" ViewCount="" Body="&lt;p&gt;There is some confusion about the term axis.  The axis about which the Earth rotates of course did not change.  It requires some external torque or transfer of angular momentum for that to happen.  However, the Earthquake changed the shape of the Earth slightly, which changed the figure axis of the Earth.  Calculations also show the Japan quake should have shifted the position of Earth’s figure axis (the axis about which Earth’s mass is balanced) by about 17 centimeters (6.5 inches), towards 133 degrees east longitude. Earth’s figure axis should not be confused with its north-south axis; they are offset by about 10 meters (about 33 feet). This shift in Earth’s figure axis will cause Earth to wobble a bit differently as it rotates, but it will not cause a shift of Earth’s axis in space-only external forces such as the gravitational attraction of the sun, moon and planets can do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jpl.nasa.gov/news/news.cfm?release=2011-080&quot; rel=&quot;nofollow&quot;&gt;http://www.jpl.nasa.gov/news/news.cfm?release=2011-080&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-15T01:11:25.133" CommentCount="2" />
  <row Id="6928" PostTypeId="1" CreationDate="2011-03-15T02:15:32.770" Score="2" ViewCount="850" Body="&lt;p&gt;If lead can absorb or block radiation, would it be possible to pump molten lead into a reactor core which is melting, so that it would eventually cool and contain the radiation?&lt;/p&gt;&#xA;" OwnerUserId="2558" LastEditorUserId="2558" LastEditorDisplayName="" LastEditDate="2011-03-15T02:21:37.757" LastActivityDate="2011-06-10T00:58:42.690" Title="Can a nuclear reactor meltdown be contained with molten lead?" Tags="&lt;nuclear-physics&gt;&lt;power&gt;&lt;nuclear-engineering&gt;" AnswerCount="10" CommentCount="2" />
  <row Id="6929" PostTypeId="2" ParentId="6928" CreationDate="2011-03-15T02:59:43.393" Score="7" ViewCount="" Body="&lt;p&gt;This would guarantee a meltdown. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They're trying to get heat &lt;em&gt;out&lt;/em&gt; of the core because---thought the fission chain reaction has been suppressed---various unstable fission daughters continue to decay. Adding hot lead would add heat to the system and not stop this behavior. Total disaster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the core &lt;em&gt;does&lt;/em&gt; slag out, it will probably end in a hot (thermodynamically and radiologically) heap at the bottom of the primary containment vessel. Presumably you leave it there to cool a bit, they pour on a lot of concrete and post &quot;Keep Away&quot; signs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No need to use anything as expensive lead, as you just want to pile on enough mass to suppressed the primary flux. Dirt would do, but concrete will make it hard for stupid people to go digging in the pile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding some boron to the mix would be helpful because otherwise neutrons get everywhere. Sneaky little bastards that they are.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-03-15T02:59:43.393" CommentCount="2" />
  <row Id="6930" PostTypeId="2" ParentId="6897" CreationDate="2011-03-15T03:08:49.543" Score="1" ViewCount="" Body="&lt;p&gt;This is really a good question and indeed questions were raised at that time about how Newton's universe could give rise to a static universe if gravity is always attractive. Newton argued, if the universe were infinite in extent and matters were distributed more or less uniformly throughout this infinite universe then collapse could be avoided since there would be no center for the universe to collapse. However, this argument is not flawless since one can show that this kind of universe will be highly unstable. A slight perturbation can break the balance and inevitably lead to collapse of the universe. Surprisingly no body argued at that time that the universe could be expanding (or contracting). Instead some people really tried to modify Newton's law of gravity so that gravity may be repulsive at large distances. So you see, a kind of repulsive gravity was indeed proposed much earlier than GR.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-03-15T03:08:49.543" />
  <row Id="6931" PostTypeId="1" AcceptedAnswerId="6932" CreationDate="2011-03-15T03:38:49.037" Score="3" ViewCount="58" Body="&lt;p&gt;The graph of nuclear binding energy is relatively smooth going from H to U, except for He4 (alpha particle). Why is He4's binding energy so anomalously high compared to its neighboring isotopes?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-03-15T04:20:52.200" Title="Binding Energy of He" Tags="&lt;binding-energy&gt;" AnswerCount="1" />
  <row Id="6932" PostTypeId="2" ParentId="6931" CreationDate="2011-03-15T04:20:52.200" Score="5" ViewCount="" Body="&lt;p&gt;It's unusually symmetric.  All four nucleons are in 1s spatial orbitals, in singlet pairs of spin and of isospin.  The more symmetry as system has, the lower its energy.  &lt;/p&gt;&#xA;" OwnerUserId="353" LastActivityDate="2011-03-15T04:20:52.200" />
  <row Id="6933" PostTypeId="2" ParentId="6725" CreationDate="2011-03-15T05:19:10.443" Score="7" ViewCount="" Body="&lt;p&gt;Photons interact with matter if the matter offers quantum transitions that match, or nearly match, the photon's energy in the inertial frame of the matter.  Ordinary matter such as wood, stone, etc offer several groups of possible quantum transitions. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Rotation of molecules (if they are free to rotate, ie not condensed matter)&lt;/li&gt;&#xA;&lt;li&gt;Vibration of molecules - bending, quivering actions&lt;/li&gt;&#xA;&lt;li&gt;Electronic excitations&lt;/li&gt;&#xA;&lt;li&gt;Nuclear excitations (there being various kinds, ignored here for simplicity)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Microwaves have such low energy, they can't do much, though they might excite some types of vibrations on larger floppier molecules - however any type of molecule that could be described as &quot;floppy&quot; probably isn't good for construction materials.  Rotational modes aren't possible in a strong material made of crosslinked polymers or silicates.  So microwaves mostly fly right through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Near infrared and visible light can kick electrons into higher molecular orbitals.  Even if the energies aren't a match, just close, there is interaction as Heisenberg lets them cheat temporarily. Also, having more energy, visible light photons can stir up a greater variety of vibrational modes.  There's nothing in common wall materials to prevent that, and in fact the interaction with photons is so strong that the material, if not super-thin (microns) will be opaque.  Of course, glass is an exception.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gamma rays are of such high frequency, electrons (or ions, or polarized ends of molecules) can't keep up due to inertia.  So no interaction, or only a little.  At the right frequencies, gamma photons can interact with nuclei,  but for a randomly chosen source of gammas, it's photons are unlikely to match close enough with any of the available nuclear excitations, and can't really do much at the molecular level, the material is almost transparent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this is so oversimplified...&lt;/p&gt;&#xA;" OwnerUserId="353" LastActivityDate="2011-03-15T05:19:10.443" CommentCount="1" />
  <row Id="6934" PostTypeId="1" AcceptedAnswerId="6944" CreationDate="2011-03-15T08:17:12.900" Score="4" ViewCount="161" Body="&lt;p&gt;Sometimes old faulty CRT monitors generate nasty high-frequency squeal sound.  What element might be responsible for generating such sound?  I have heard that it might be dry electrolytic capacitor; is it?  What is the physics of generating this sound (how it is generated)?&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2011-03-15T16:12:31.307" Title="What is the source of high-frequency squeal in faulty CRT monitors and TV, and what is physics of generating this sound?" Tags="&lt;electricity&gt;&lt;sound&gt;" AnswerCount="2" />
  <row Id="6935" PostTypeId="2" ParentId="6904" CreationDate="2011-03-15T08:27:47.140" Score="0" ViewCount="" Body="&lt;p&gt;With inspire I obtain 6 hits when searching for &quot;golden ratio&quot; in the title of papers, see&#xA;&lt;a href=&quot;http://inspirebeta.net/search?ln=en&amp;amp;p=golden+ratio&amp;amp;f=title&amp;amp;action_search=Search&amp;amp;sf=&amp;amp;so=d&amp;amp;rm=&amp;amp;rg=100&amp;amp;sc=0&amp;amp;of=hb&quot; rel=&quot;nofollow&quot;&gt;http://inspirebeta.net/search?ln=en&amp;amp;p=golden+ratio&amp;amp;f=title&amp;amp;action_search=Search&amp;amp;sf=&amp;amp;so=d&amp;amp;rm=&amp;amp;rg=100&amp;amp;sc=0&amp;amp;of=hb&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The majority of them is related to neutrino mixing, see &lt;a href=&quot;http://arxiv.org/abs/arXiv:0705.4559&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/arXiv:0705.4559&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is fair to say that none of these papers is well-known.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-03-15T08:27:47.140" />
  <row Id="6936" PostTypeId="1" CreationDate="2011-03-15T08:46:14.493" Score="1" ViewCount="275" Body="&lt;p&gt;Consider a lift, which is at rest in an homogeneous gravity field. There is a thin layer(&lt;strong&gt;with thickness&lt;/strong&gt; $h$) of water on the floor of the lift. At some moment a single cable, supporting the lift, breaks and the lift begins free falling(forever). It is easy to describe qualitatively what happens with water, i think: the formation of a drop begins. During this process the drop jumps up from the floor and once the total kinetic energy of the drop is dissipated into the heat, the center of the drop stands at some height $H$ from the floor. (All the kinetic energy of the drop is coming from the difference in surface tension energy of water between initial and end moments.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: How to determine(approximately) $H$ ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity let's assume that the cross-section of the lift is a disc-shaped with radius $R$.  $(h&amp;lt;&amp;lt;R)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: &lt;strong&gt;A key point is to determine how the total kinetic energy distributes between internal kinetic energy and translational kinetic energy of the drop, i think.&lt;/strong&gt; With this the problem will be solved.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is convenient at the early stage of formation of the drop ignore viscosity effects of water and air resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to clarify that by internal kinetic energy i mean the kinetic energy of macroscopic motion of water inside the drop. At early stage there is no energy loss assumed, for simplicity. &lt;/p&gt;&#xA;" OwnerUserId="552" LastEditorUserId="552" LastEditorDisplayName="" LastEditDate="2011-03-15T12:03:53.463" LastActivityDate="2011-03-15T15:33:18.867" Title="A water drop in a falling lift" Tags="&lt;fluid-dynamics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="6937" PostTypeId="2" ParentId="6928" CreationDate="2011-03-15T09:07:38.373" Score="4" ViewCount="" Body="&lt;p&gt;Hmm, lets see. The melting point of lead is fairly low, 327.46 °C and it is a good absorber of radioactivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the problem with the reactors  is not the heat per se, but the exposure of the fuel rods to the air without cooling because of escaping steam not being replenished by cooling water. The remaining steam etc may blow out the thick container and radioactive material will disperse some distance while the fuel rods will melt.&#xA;I expect that the design is such that even in melt the fuel rods and melted control rods will be sub critical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with the suggestion is, when they have troubles pouring cool seawater in, how ever will they be able to pour hot liquid lead? It is not that the reactor is a pot whose cover can come out!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better question/suggestion would be, since lead has such a low melting point, and the operating temperature of a boiling water reactor is 250C, why don't they set up a sort of cladding to be passively  used in such emergencies: the inner layer of something that can melt when temperatures get higher than 500C, and then lead melts and floods out the system before the fuel rods get exposed, covering them. The answer is similar to why they were not pouring seawater on all reactors from the first minute: economics, trying to save the reactors for production later.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-15T13:06:54.737" LastActivityDate="2011-03-15T13:06:54.737" CommentCount="3" />
  <row Id="6938" PostTypeId="2" ParentId="6936" CreationDate="2011-03-15T09:08:39.373" Score="3" ViewCount="" Body="&lt;p&gt;The only reason the water would form into a drop is the surface tension. You can work out the reduction in the energy stored in the surface tension in being a drop rather than a flat surface using a simple calc of the surface area. There's no particular reason I can see that the drop would float off the floor at all.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-15T09:08:39.373" CommentCount="4" />
  <row Id="6940" PostTypeId="2" ParentId="6936" CreationDate="2011-03-15T10:04:42.517" Score="1" ViewCount="" Body="&lt;p&gt;Unless there is some inital temperature difference, its total kinetic energy is just its translational kinetic energy, which is the energy gained from its change in shape due to surface tension, from which you can get its temperature and also its velocity $v_0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have the surface area, temperatures, conductivities, so you can get the heat transfer per time from fourier's law.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From which you can get the energy E(t), which gives you v(t), a solution to v(t)=0, if it exists, is T, then integrate v(t) from 0 to T to get H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most likely it wont come to a full stop, it could be in thermal equilibrium and still have translation kinetic energy, and even more likely it approaches velocity 0 asymptotically.&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2011-03-15T11:08:38.893" LastActivityDate="2011-03-15T11:08:38.893" CommentCount="1" />
  <row Id="6941" PostTypeId="2" ParentId="6904" CreationDate="2011-03-15T11:06:51.470" Score="9" ViewCount="" Body="&lt;p&gt;A model that has shown some interest in recent years is the &lt;a href=&quot;http://arxiv.org/abs/0902.3275&quot; rel=&quot;nofollow&quot;&gt;golden chain&lt;/a&gt;. In the golden chain you deal with a one-dimensional chain of spin-like particle, similar to the Heisenberg (or Ising) model. But in this model the spin degrees of freedom are replaced by (non-Abelian) anyons (see e.g. &lt;a href=&quot;http://physics.stackexchange.com/questions/6839/references-on-the-physics-of-anyons&quot;&gt;this thread&lt;/a&gt;). The type of anyon used in this model are the Fibonacci anyons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how the golden ratio enters this model we have to look at the Hilbert space, specifically its dimension. In an ordinary spin chain each spin carries a degree of freedom to which we can assign a Hilbert space of dimension 2, $\mathcal{H}$. The tensor product of two spins is spanned by the singlet and triplet combination. The total Hilbert space of a chain with $n$ spins is the tensor product, $\otimes \mathcal{H}$ and it has dimension $2^n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-Abelian anyons on the other hand carry a different kind of spin. When two anyons combine they will form what is known as a fusion product. The fusion product of two anyons depends on the type of anyon you are dealing with. Fibonacci anyons satisfy the relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tau \times \tau = 1 + \tau$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can think of this analogous to spin (with a fundamental difference). When we bring two anyons together they will fuse together and form a composite-like particle. This is similar to the spin of two (s=1/2) particles which combine into a singlet (S=0) or doublet (S=1). In the case of the Fibonacci anyons the particle can form two type of composites: the vacuum particle $1$ (&quot;zero spin&quot;) and the Fibonacci particle $\tau$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens when we bring another $\tau$ \particle to this composite? It will fuse with the composite particle to form another composite. However, the allowed particles which can be formed depend on the fusion channel of the first two anyons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\tau_1 \times \tau_2 \rightarrow 1$, then $(\tau_1 \times \tau_2)\times \tau_3 \rightarrow \tau$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\tau_1 \times \tau_2 \rightarrow \tau$, then $(\tau_1 \times \tau_2)\times \tau_3 \rightarrow 1+\tau$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two ways in which a $\tau$ particle is formed in the end, while there is only one way in which a vacuum particle is formed. All in all:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tau \times \tau \times \tau = 1+ 2\tau$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the factor of two on the right hand side refers to the number of ways in which a $\tau$ particle can be formed. The dimension of the Hilbert space of three tau particles is therefore 3 dimensional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the following conclusion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We set the dimension of the Hilbert space of zero particles equal to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dimension of H for 1 Fib. anyon is also 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dimension of H for 2 Fib. anyons is 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dimension of H for 3 Fib. anyons is 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any guess what the dimension of the 4 anyons will be? It's 5 dimensional. You can derive it yourself: just count the number of ways in which you can fuse the anyons together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence of the dimension of the Hilbert spaces for $n$ anyons is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1,1,2,3,5,8,13,\ldots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, this is the Fibonacci sequence! And the Fibonacci sequence has a very nice feature: It grows roughly as $\phi^n$ where $\phi$ is the golden ratio! The dimension of the Hilbert space of $n$ Fibonacci anyons grows roughly as $\phi^n$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to think of this is that the Fibonacci anyons carry a spin of dimension $\phi$. This statement is wrong though: the Hilbert space always has an integer dimension. It is therefore not referred to as a spin but rather as the quantum dimension of the anyons. The rule is that on average the Hilbert space grows by a factor of $\phi$ every time you add an anyon to the chain (just like the H-space for an Ising chain grows with a factor of two every time you add a spin to the system).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a last note: Fibonacci anyons might be realized in certain quantum Hall systems and they are useful for topological quantum computing, if they are ever found.&lt;/p&gt;&#xA;" OwnerUserId="1287" LastEditorUserId="1287" LastEditorDisplayName="" LastEditDate="2011-03-15T12:09:35.993" LastActivityDate="2011-03-15T12:09:35.993" CommentCount="2" />
  <row Id="6943" PostTypeId="2" ParentId="6907" CreationDate="2011-03-15T11:23:09.313" Score="-2" ViewCount="" Body="&lt;p&gt;In my humble opinion, there are sufficient experimental and theoretical data to consider things to be composite because of their permanent coupling to other things. The problem is in recognizing this permanent coupling and implementing it correctly in our theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider the simplest case of scattering a neutral particle, neutrino, from a charged particle, electron: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nu + e^- \rightarrow \nu + e^-$.     (1) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is, however, unlikely to scatter from a charge elastically because there are thresholdless excitations - photons. In other words, the real charge ($e^-$) is a complicated system including the electromagnetic degrees of freedom and the electron in it is only a part of it. So the true scattering process is written differently: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nu + e^- \rightarrow \nu + e^- + \gamma_1 + \gamma_2 +  ...$      (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, exciting a target (= inelastic processes like (2)) is the first and the principal evidence of the target being compound. And we know from the exact QED equations about this permanent coupling but we do not initially consider the charge to be coupled and write rubbish like (1). This is our grave conceptual error. So inelastic processes like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nu + e^- \rightarrow \nu + e^- + \gamma + $ other neutral stuff      (3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;testify that our target (electron) is not so simple ;-).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We still do not note evident things and decouple &lt;a href=&quot;http://arxiv.org/abs/0806.2635&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;coupled&lt;/strong&gt;&lt;/a&gt; things in our minds and on the paper. Our methodology of &quot;switching the coupling on and off&quot; is wrong - it implies a possibility of perturbative &quot;coupling&quot; as if it were &quot;weak&quot;. It is never weak. When we manage to describe QED correctly, it will be easier to see how other leptons and quarks (and other quasi-particles in composite things) are related to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take an atom as a composite system and scatter from its nucleus or electron. What is a signature of its being composite? Inelastic channels and resonances.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-03-15T22:54:06.403" LastActivityDate="2011-03-15T22:54:06.403" />
  <row Id="6944" PostTypeId="2" ParentId="6934" CreationDate="2011-03-15T11:37:32.760" Score="4" ViewCount="" Body="&lt;p&gt;Such noise is generated by the so called &quot;line&quot; (= horizontal deflection or flyback) transformer.&#xA;This transformer does several things in a CRT or TV at the same time: it couples the horizontal output stage to the horizontal deflection coils and recovers the energy from the coils field during flyback, and generates the high voltage for the CRT (about 20 kV).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transformer has a core made from &lt;a href=&quot;http://en.wikipedia.org/wiki/Flyback_transformer&quot; rel=&quot;nofollow&quot;&gt;ferrite material&lt;/a&gt;. Such materials have a small &#xA;and unwanted magnetostriction effect (There a varieties of ferrites which have a strong effect). In case of failiure the core of such a transformer can be driven into saturation, and the frequency of oscillation comes down to the region where our ears are much more sensitive. Then You hear that vibration of the core. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW in the 50ties, when I was much younger (including my ears) I could hear the 15 625 Hz tone of TV sets out in the street when I passed houses at nighttime (less other noise). &#xA;Whether the transformers were louder in those times, or I just had better ears, I don't know.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-15T11:37:32.760" CommentCount="7" />
  <row Id="6945" PostTypeId="2" ParentId="6928" CreationDate="2011-03-15T12:08:56.417" Score="7" ViewCount="" Body="&lt;p&gt;The melting point of the Zirkonium cladding is about 1850 °C, the Uranium dioxide&lt;br&gt;&#xA;melts at 2176 °C (maybe somewhat lowered by the impurities from fission products) .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lead &lt;strong&gt;boils&lt;/strong&gt; at 1620 °C that means it vaporizes at the temperatures concerned!. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On top of that molten lead will not dissolve Uranium dioxide, but maybe it could reduce &#xA;the oxide to a lover level oxide or to metallic uranium. Who knows? &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-15T12:08:56.417" />
  <row Id="6946" PostTypeId="1" CreationDate="2011-03-15T12:20:21.287" Score="-3" ViewCount="366" Body="&lt;p&gt;I have read that mass of a body increases with the speed and when it reaches the speed of light, it becomes infinity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly does it mean to say that &quot;it becomes infinity&quot;? I am not able to get a picture of &quot;infinite mass&quot; in my mind.&lt;/p&gt;&#xA;" OwnerUserId="1496" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-15T13:53:03.697" LastActivityDate="2011-03-25T11:58:47.280" Title="What does it mean to say &quot;mass approaches infinity&quot;?" Tags="&lt;mass&gt;" AnswerCount="7" CommentCount="6" />
  <row Id="6948" PostTypeId="2" ParentId="6946" CreationDate="2011-03-15T12:30:13.973" Score="3" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am not able to get a picture of&#xA;  &quot;infinite mass&quot; in my mind.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nobody can, and from mathematical use one avoids the wording &quot;is infinite&quot; &#xA;by saying &quot;approaches infinity&quot;. This is to avoid the implication/existence of factual infinity. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-25T09:33:55.173" LastActivityDate="2011-03-25T09:33:55.173" />
  <row Id="6949" PostTypeId="2" ParentId="6946" CreationDate="2011-03-15T12:33:40.170" Score="3" ViewCount="" Body="&lt;p&gt;It means that its inertia (resistance to change in the state of motion) approaches infinity. You probably already know from $F=ma$ that for the same change in speed (acceleration), a larger mass requires a larger force. As the velocity of a body approaches the speed of light, its inertia (i.e. $\gamma m$) becomes so high that you need an impossibly infinite force to accelerate it to exactly the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as Georg says, whenever you see &quot;becomes infinite&quot;, read it as &quot;approaches infinity&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-03-24T23:22:05.897" LastActivityDate="2011-03-24T23:22:05.897" />
  <row Id="6950" PostTypeId="2" ParentId="6907" CreationDate="2011-03-15T12:54:52.617" Score="1" ViewCount="" Body="&lt;p&gt;CMS has  a &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1102/1102.2020v1.pdf&quot; rel=&quot;nofollow&quot;&gt;preprint&lt;/a&gt; out  where they are searching for compositeness in dijet angular distributions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The measured dijet angular distributions can be used to set limits on quark compositeness represented by a four-fermion contact interaction term in addition to the QCD Lagrangian.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They set limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will guess that angular distributions of two lepton events will be in the search of lepton compositeness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering that the compositeness of nuclei and compositeness of nucleons were cleanly found by deep inelastic scattering, I would be very doubtful of interpretations using levels of monte carlo calculations that would give such a drastic conclusion to deviations from QCD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One would have to wait for lepton colliders . From LHC I would need two leptons at a vertex to get the other end of deep inelastic scattering. There is nothing that can beat form factors, imo.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-15T13:23:08.010" LastActivityDate="2011-03-15T13:23:08.010" />
  <row Id="6951" PostTypeId="1" AcceptedAnswerId="6963" CreationDate="2011-03-15T13:28:21.027" Score="2" ViewCount="252" Body="&lt;p&gt;As I understand, the hydrogen is produced by water thermolysis, which need quite high temperature. This temperature is close to previously observed corium temperatures. So:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is the eutectic temperature of the corium of those type of reactors known and publicly accessible?&lt;/li&gt;&#xA;&lt;li&gt;Isn't the hydrogen presence a proof of meltdown?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1715" LastActivityDate="2011-03-16T15:36:31.687" Title="Fukushima - Isn't the hydrogen presence a proof of meltdown?" Tags="&lt;nuclear-engineering&gt;" AnswerCount="1" />
  <row Id="6952" PostTypeId="2" ParentId="6908" CreationDate="2011-03-15T13:45:15.850" Score="0" ViewCount="" Body="&lt;p&gt;The proton is, in some sense, a red herring. Only the total charge of the nucleus (which happens to be the number of protons in certain units) is important to keep the atom electrically neutral. The detailed structure inside the nucleus (like the fact that it consists of protons and neutrons) is not relevant for chemistry. You should be asking to the number of &lt;em&gt;electrons&lt;/em&gt; instead.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastActivityDate="2011-03-15T13:45:15.850" />
  <row Id="6954" PostTypeId="2" ParentId="6877" CreationDate="2011-03-15T13:50:15.603" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In my opinion it is more economical to&#xA;  operate a Zeppelin than a plane&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why would it be more economical? Why would the energy required to propel a large unaerodynamic balloon a certain distance be less than that required to propel an equivalent cargo in an aeroplane?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As far as i know Zeppelins were&#xA;  abandoned because they were not safe,&#xA;  back in time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure they were also abandoned because they're slow and uneconomical.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-15T13:50:15.603" CommentCount="3" />
  <row Id="6955" PostTypeId="1" CreationDate="2011-03-15T14:35:45.720" Score="1" ViewCount="119" Body="&lt;p&gt;Given an uncharged continuum the EM field produces no force on it. Yet in a discrete system we know that atoms are not uncharged locally and momentum can be exchanged due to the electric field(Which Maxwell's equations are suppose to govern).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;viz., The transfer of momentum between atoms is due to the electric field(although it's not as simple as the 1/r^2 version). Maxwell's equations should describe the motion of the atoms exactly when one includes the Lorentz force and Newton's second law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet when one moves up to the continuum these microscopic effects no longer exist and the electric field cannot influence the mechanical motion of density if it is not charged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we include this in Maxwell's equations(or probably instead in the Lorentz force)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking that we simply use and it's conservation law(on the same page)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_stress-energy_tensor&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Electromagnetic_stress-energy_tensor&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but there doesn't seem to be a way of controlling how well the EM energy is coupled to the material(Some materials will experience a larger motion due to the EM field it experiences).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or even&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Maxwell_stress_tensor&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Maxwell_stress_tensor&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in both cases though q = 0 for the continuum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think was thinking that possibly one could temporarily treat the continuum as being charged even though it isn't which will allow the momentum due to the field to propagate and one could compute forces and such. With proper scaling one might get the same effect as if modeling a discrete material. Effectively one is breaking up the material into positive and negative charge, computing the momentum then subtracting... in a continuum this is 0 but for the discrete case it cannot be zero unless there are no particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a better way? How can one model the microscopic collision transfer of momentum that can be applied to the continuous case?&lt;/p&gt;&#xA;" OwnerUserId="2465" LastActivityDate="2011-03-15T23:20:44.200" Title="Mechanical momentum of the EM field in a continuum?" Tags="&lt;particle-physics&gt;" AnswerCount="2" />
  <row Id="6956" PostTypeId="1" AcceptedAnswerId="6958" CreationDate="2011-03-15T14:38:10.833" Score="4" ViewCount="310" Body="&lt;p&gt;The explosions at the Fukushima nuclear reactors were caused by a build of hydrogen.  I was wondering where this hydrogen came from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question &lt;a href=&quot;http://physics.stackexchange.com/questions/6951/fukushima-isnt-the-hydrogen-presence-a-proof-of-meltdown&quot;&gt;Fukushima - Isn&amp;#39;t the hydrogen presence a proof of meltdown?&lt;/a&gt; suggests that the hydrogen is caused by thermolysis of water.&lt;br&gt;&#xA;But surely this can't be the case?  If it were, wouldn't it require the hydrogen and the oxygen to be separated whilst they cooled - else they would recombine into water in a more gradual fashion as the gases cool once they are clear of the scorching hot fuel rods?&lt;/p&gt;&#xA;" OwnerUserId="1187" LastActivityDate="2011-03-16T10:28:18.810" Title="What is generating the hydrogen that has exploded in the Fukushima reactor?" Tags="&lt;nuclear-engineering&gt;" AnswerCount="3" />
  <row Id="6957" PostTypeId="2" ParentId="6946" CreationDate="2011-03-15T14:50:09.693" Score="0" ViewCount="" Body="&lt;p&gt;If you want to imagine an infinite mass just think of a piece of matter which does not response (i.e. accelerate) to any force. Obviously, it is an ideal concept. In reality &quot;the mass approaching infinity&quot; means it is increasing without any finite upper limit.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-03-15T14:50:09.693" />
  <row Id="6958" PostTypeId="2" ParentId="6956" CreationDate="2011-03-15T14:51:15.153" Score="8" ViewCount="" Body="&lt;p&gt;This was (one of the few) facts reported clearly and correct in German news. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Zirconium metal of the fuel tubes reacts with water vapor at elevated temperatures &#xA;(above 1200 °C if  I remember correctly)&#xA;The reaction is simply: Zr + 2 H2O  =&gt; ZrO2 + 2 H2 &#xA;This is proof, that parts (at least) of the fuel rods were not immersed in water, but it does not prove melting. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-15T14:51:15.153" CommentCount="3" />
  <row Id="6959" PostTypeId="1" AcceptedAnswerId="6977" CreationDate="2011-03-15T14:53:55.073" Score="9" ViewCount="388" Body="&lt;p&gt;To focus this question let us consider first classical mechanics (which is time-symmetric). Given a final condition (and sufficient information) one can calculate the system conditions of an earlier time (retrodiction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given Quantum Mechanics (which is time-symmetric) and a final condition what is the status of retrodiction in that theory? Without choosing between them, here are three options:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) An earlier condition can be determined probabilistically exactly as a prediction can be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) An earlier condition can be determined exactly (with enough accessible information) from a final condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) It is inappropriate to use QM for retrodictions: it is a prediction-only theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen some thought experiments on all this in Penrose's books, but it remains inconclusive there, and standard QM texts are not interested in retrodiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT AFTER 6 ANSWERS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the effort on this question, which was challenging. I will continue to investigate it, but some initial thoughts on the Answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) I was expecting to receive comments on decoherence, chaos and something on Interpretations and such are amongst the answers. Perhaps the single most important sentence is Peter Shor:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The time-asymmetry in quantum mechanics beween retrodiction and prediction comes from the asymmetry between state preparation and measurement&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lubos's introduction of Bayesian probability and response (c) is the most useful, although discussion of entropy does not seem immediately relevant. This response though suggests a different framework for retrodiction, with an apriori set of assumptions introduced for the calculation of initial state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A complication that was not clearly enough addressed was the link with Classical Mechanics retrodiction. Statements that Quantum Retrodiction was &quot;impossible&quot; does not square easily with the fact that some will have corresponding classical systems easily retrodicted. Of course the fact that Quantum Prediction is probabilistic, allows Quantum Retrodiction to be probabilistic too (and thus different from classical retrodiction) was not followed up in some answers. As far as references to Chaos is concerned does not &quot;retrodiction Chaos&quot; result in an &lt;em&gt;increase&lt;/em&gt; in ability to classically retrodict given that the trajectories will be converging?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Peter Morgan's points I should say that the question is open to any interpretation of how the experimental apparatus is used - if it is relevant to giving an appropriate answer then so discuss the significance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Deepak's links I should note that these include references to Applications to this idea in Quantum Communication: ie what was the sent state given a received state? I think Lubos's Probability is relevant here too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feel free to EDIT your answers if you think of anything else!&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-03-16T21:48:43.197" LastActivityDate="2011-03-16T21:48:43.197" Title="Retrodiction in Quantum Mechanics" Tags="&lt;quantum-mechanics&gt;" AnswerCount="6" FavoriteCount="0" />
  <row Id="6960" PostTypeId="2" ParentId="6959" CreationDate="2011-03-15T15:07:23.967" Score="-1" ViewCount="" Body="&lt;p&gt;The only (potential, i'm speculating here) issue i see in retrodiction is that you would have to isolate how the system (i.e: the observer and measuring aparatus together) couples to the retrodicted state all the time up to the final measurement which is behaving as your initial condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably if you can guarantee no coupling for a time $\Delta t$, there is still the problem that the state will still be coupled to the system for earlier times, and hence making it not entirely independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of prediction, its is implicitly assumed that your coupling to a future state is zero. Weirdly enough, it doesn't seem you can do the same for postdiction. Definitely a interesting question to ponder&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets elaborate a bit more how an experiment of retrodiction could be done;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;lets take system &lt;strong&gt;A&lt;/strong&gt; to be observed and measure the state at time $t_{0}$, store the measurement result somewhere where it would not decohere with the measuring apparatus. I don't have any evidence to assert that this can always be done consistently, but given the results of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Delayed_choice_quantum_eraser&quot; rel=&quot;nofollow&quot;&gt;delayed quantum erasing experiment&lt;/a&gt;, there is a good chance there is nothing in principle avoiding one to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the measurement is done, if decoherence has happened, the system might be in a (Collapsed) state and if you were to measure the state further, you would find it in a pure-diagonal density matrix (a fully classical state with classical probabilities) so at $t_{1}$ you would observe a certain final state $|\Psi_{final}(t_{1})&amp;gt;$ and the probability of it would be a classical weighted probabilities (of the initial state mixed state classical probabilities after the measurement) &lt;em&gt;times&lt;/em&gt; the probability of each individual initial state in the mixture naturally evolving and finally collapsing into $|\Psi_{final}(t_{1})&amp;gt;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the net probability of this state should be (in the case of decoherence):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum_{ |\Psi_{final}(t_{0}) &amp;gt; } (P(|\Psi_{final}(t_{0}) &amp;gt;) x &amp;lt;\Psi_{final}(t_{1})| e^{-i(t_{1} - t_{0}) H } |\Psi_{final}(t_{0}) &amp;gt; )^2  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in the case of no decoherence, i'm not sure what should happen. I will think on this an update the answer later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, whatever is the resulting probability equation (and how it depends on the fact of measuring or not the delayed measurement state), the experimental procedure to follow (which doesn't depend on the final math that much) would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;experiment template:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;save the $t_{0}$ state &lt;em&gt;somewhere that will not decohere with your apparatus or anything else&lt;/em&gt;. This is probably the hardest part to be achieved, if at all possible&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;let the system evolve to $t_{1}$ and measure the final state&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;now you execute the experiment template many times, and you restrict your dataset to experiments where final state are your $|\Psi_{final}(t_{1})&amp;gt;$. In this way you have properly and consistently defined the problem of retrodicting the state evolution based on a final state&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that the experimental procedure is properly defined, the question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;assumming no decoherence happens in the delayed measurement, would the subset of experiments where final state = $|\Psi_{final}(t_{1})&amp;gt;$, have probabilities amplitudes at $t_{0}$ that are consistent with a propagator of the form $ e^{-i(t_{0} - t_{1}) H }$?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the question is &lt;strong&gt;yes&lt;/strong&gt;, but only if you guarantee that the available physical information of the state at $t_{0}$ is fully encoded in the delayed measurement storage and nowhere else, otherwise you are back in the decoherence scenario&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-03-15T20:48:45.953" LastActivityDate="2011-03-15T20:48:45.953" />
  <row Id="6961" PostTypeId="2" ParentId="6959" CreationDate="2011-03-15T15:08:11.067" Score="-1" ViewCount="" Body="&lt;p&gt;You know QM has two stages (maybe the collapse stage is arbitrary and artificial). The Schrödinger equation is unitary and time symmetric. However the non unitary collapse is time asymmetric in nature. Once measurement is performed, given a final state no one can recover the initial state. However it is difficult to accept that there is any fundamental asymmetry here since collapse is not a physical process in most sensible interpretations. Specially if one accepts that it is decoherence which is giving the appearances of collapse then all traces of non unitarity vanishes and you are left with a perfectly time symmetric theory. But it is the very process of decoherence which is irreversible will make it impossible to retrodict in quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion is therefore it is not possible to retrodict events in quantum mechanics but that does not mean any time asymmetry either.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-03-15T17:41:53.547" LastActivityDate="2011-03-15T17:41:53.547" CommentCount="2" />
  <row Id="6962" PostTypeId="2" ParentId="6955" CreationDate="2011-03-15T15:24:51.830" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Given an uncharged continuum the EM&#xA;  field produces no force on it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This seems to be a root assumption of your question, so let's examine it a little bit. What do you mean by an 'uncharged continuum'? Would you consider a small glass bead to be an uncharged continuum? If so, this statement is disproved by &lt;a href=&quot;http://en.wikipedia.org/wiki/Optical_tweezers&quot; rel=&quot;nofollow&quot;&gt;optical tweezers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean something else, perhaps you could clarify your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;br&gt;&#xA;Chapter 6.6 of &lt;a href=&quot;http://books.google.com/books?id=U3LBQgAACAAJ&amp;amp;dq=jackson+electrodynamics&amp;amp;hl=en&amp;amp;src=bmrr&amp;amp;ei=A_R_TfePKuuz0QG3s4mKCQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CDEQ6AEwAA&quot; rel=&quot;nofollow&quot;&gt;Jackson's Electrodynamics&lt;/a&gt; has a pretty nice derivation of how you go from microscopic to macroscopic. Pay particular attention to polarizability.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="2359" LastEditorDisplayName="" LastEditDate="2011-03-15T23:20:44.200" LastActivityDate="2011-03-15T23:20:44.200" CommentCount="5" />
  <row Id="6963" PostTypeId="2" ParentId="6951" CreationDate="2011-03-15T15:26:48.527" Score="2" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://physics.stackexchange.com/questions/6956/what-is-generating-the-hydrogen-that-has-exploded-in-the-fukushima-reactor&quot;&gt;same question was asked some minutes later&lt;/a&gt;, but I saw that one first. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Eutectic is quite irrelevant here, do You think that the uranium dioxide with some % of &#xA;fission products happens to be a eutectic mix? The probablity of that might be calculatable &#xA;with the aid of a bistromatic drive computer. Uranium dioxide melts around 2300 °C, the fission products will be only partly miscible with it, so I'd say melting point will be &gt;2000°C&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Zirconium metal of the fuel tubes reacts with water vapor at elevated temperatures (above 1200 °C if I remember correctly) The reaction is simply: Zr + 2 H2O =&gt; ZrO2 + 2 H2 This is proof, that parts (at least) of the fuel rods were not immersed in water, but it does not prove melting.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1715" LastEditorDisplayName="" LastEditDate="2011-03-16T15:36:31.687" LastActivityDate="2011-03-16T15:36:31.687" />
  <row Id="6964" PostTypeId="2" ParentId="6897" CreationDate="2011-03-15T15:43:05.513" Score="0" ViewCount="" Body="&lt;p&gt;Newton used this as a proof that the universe was infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all the matter in the universe is attracting every other bit of matter then the universe should collapse into a single point. He reasoned that the only way around this was if the universe was infinite, then every bit of matter would have an equal force in every direction - from the attraction of every other object around it.&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-03-15T15:43:05.513" />
  <row Id="6965" PostTypeId="2" ParentId="6959" CreationDate="2011-03-15T15:52:43.463" Score="-1" ViewCount="" Body="&lt;p&gt;Your question would look different if you consider what a discussion of experimental confirmation of the retrodicted &quot;earlier conditions&quot; would look like. Sure, we can compute what the quantum or classical state must have been at some earlier time, given (1) the quantum or classical state now; and (2) whatever particular dynamics we have good reason (on the basis of &lt;em&gt;past&lt;/em&gt; experience) to believe the system to be well-described by; but this is of limited interest unless we experimentally confirm that the quantum or classical state was in fact what our computation tells us. For the most part, we are just more interested in what the state in the future will be because of the state now than in what the state in the past must have been to cause the state now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible experimental confirmation of a retrodicted state R computed on the basis of a state X would be to look for systems that we measure to be in a similar state to R, and have reason to believe to be subject to the same dynamics, and then verify that those systems in fact evolve into a state similar to X. This kind of discussion, however, seems some distance from the concerns in your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your statements that classical and quantum mechanics are time-symmetric is true enough of those theoretical frameworks in themselves, but in practical use such models of experimental apparatus rarely include &lt;em&gt;all&lt;/em&gt; degrees of freedom that affect the experiment. Even at only moderate levels of accuracy, models usually have to include the effects of, for example, heat flows into and out of the experimental apparatus.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-03-15T15:52:43.463" CommentCount="2" />
  <row Id="6966" PostTypeId="2" ParentId="6946" CreationDate="2011-03-15T15:54:00.543" Score="3" ViewCount="" Body="&lt;p&gt;As the speed of a body approaches the speed of light, it becomes harder and harder to accelerate body. This fact is captured by following formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(v) = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a physicist says that, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;when speed of a body equals the speed of light its mass becomes infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He is not saying $f(v = c)$ but $\lim_{v\to c}f(v) = \infty$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to read &lt;a href=&quot;http://en.wikipedia.org/wiki/Limit_of_a_function&quot; rel=&quot;nofollow&quot;&gt; about limit&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-15T17:44:08.157" LastActivityDate="2011-03-15T17:44:08.157" />
  <row Id="6967" PostTypeId="2" ParentId="6946" CreationDate="2011-03-15T16:05:22.873" Score="1" ViewCount="" Body="&lt;p&gt;It means that if a finite force acts on the mass in its stationary frame, then in all other frames as the measured velocity approaches c, the acceleration approaches 0 which implies the &quot;relativistic mass&quot;, the &quot;apparent mass&quot; the &quot;effective mass&quot;, call it what you will, approaches infinity.&lt;/p&gt;&#xA;" OwnerUserId="2146" LastEditorUserId="2146" LastEditorDisplayName="" LastEditDate="2011-03-15T19:42:49.237" LastActivityDate="2011-03-15T19:42:49.237" />
  <row Id="6968" PostTypeId="2" ParentId="6934" CreationDate="2011-03-15T16:12:31.307" Score="-1" ViewCount="" Body="&lt;p&gt;its down to the pizza-electric effect where faulty electric signals in the audio range cause components to wiggle mechanically in the audio range.&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-03-15T16:12:31.307" CommentCount="3" />
  <row Id="6969" PostTypeId="1" CreationDate="2011-03-15T16:26:41.543" Score="4" ViewCount="174" Body="&lt;p&gt;I live at roughly 52.4,-2.1. On a sunny evenings I've often looked at the moon and the sun and noticed that the light part of the moon does not appear to line up with the sun. For example, at about 17:00 GMT on 13 Mar 2011, I noticed the half moon was facing toward a point roughly 10-20 degrees above where the sun appeared to be. Why?&lt;/p&gt;&#xA;" OwnerUserId="526" LastActivityDate="2011-03-16T04:36:35.927" Title="Why does the light side of the moon appear not to line up correctly with the evening sun?" Tags="&lt;astronomy&gt;&lt;sun&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="6970" PostTypeId="2" ParentId="6969" CreationDate="2011-03-15T16:39:50.083" Score="1" ViewCount="" Body="&lt;p&gt;At sunset the suns is destincly &quot;lower&quot; than it appears. This is due to refraction of the suns rays in atmosphere when low above horizon. Moon is &quot;higer&quot;, that rays come roughly vertical throuh atmosphere,are not deflected. That might explain Your observations at least partly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is from wikipedia &quot;sunset:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Further, because the light from the&#xA;  Sun is refracted, the Sun is still&#xA;  visible after it is geometrically&#xA;  below the horizon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-15T16:45:51.073" LastActivityDate="2011-03-15T16:45:51.073" CommentCount="4" />
  <row Id="6971" PostTypeId="2" ParentId="6959" CreationDate="2011-03-15T17:14:23.360" Score="4" ViewCount="" Body="&lt;p&gt;Doesn't this depend on which interpretation of quantum mechanics you subscribe to? The many-worlds interpretation and the pilot-wave interpretation would both seem to allow for completely accurate retrodiction, given the final state (which the Heisenberg uncertainty principle makes it impossible to actually discover in the many-worlds interpretation, and the inobservability of the pilot wave makes it impossible to discover in the Bohmian interpretation). The Copenhagen interpretation would say that retrodiction is only possible probabilistically (and this opens a can of worms, both classically and quantum mechanically, because putting a Bayesian probability distribution on the prior states of a system needs what might be seemingly irrelevant knowledge about the world -- see Lubos's answer). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time-asymmetry in quantum mechanics beween retrodiction and prediction comes from the asymmetry between state preparation and measurement. It is theoretically possible to prepare a quantum system in an arbitrary state (and experiments can often come very close to doing so) but Heisenberg's uncertainty principle makes it impossible to accurately measure what state a quantum system is in. &lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditorDisplayName="" LastEditDate="2011-03-15T18:42:49.850" LastActivityDate="2011-03-15T18:42:49.850" CommentCount="2" />
  <row Id="6972" PostTypeId="2" ParentId="6969" CreationDate="2011-03-15T17:31:51.893" Score="0" ViewCount="" Body="&lt;p&gt;I am puzzled by you question. When one has two points, the sun and the moon, one can always find a line connecting them, by definition of line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean why the earth is not part of that straight line, it is because the moon has an orbit around the earth, and the angle of the line  earth-moon changes. It is the reason the &lt;a href=&quot;http://en.wikipedia.org/wiki/Moon_phases&quot; rel=&quot;nofollow&quot;&gt;moon has phases&lt;/a&gt;. Earth, moon and sun are on the same line during full moon, and the moon rises while the sun sets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: If as Ted Bun says you mean the bisecting line from the center of the moon, then the drawing given in &lt;a href=&quot;http://64.19.142.10/upload.wikimedia.org/wikipedia/commons/thumb/4/46/Moon_phases_en.jpg/800px-Moon_phases_en.jpg&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; gives an angle because of the motion of the moon around the earth and the motion of the earth  around the sun, except at full moon and new moon (if the rays shown are a correct depiction of the sun's direction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit2: If one looks at the drawing In Carl Brannen's answer in combination with the wiki drawing, I think the parallax arises  depending on the phase of the moon, because  what one sees from the earth is not the total lit up semicircle of the moon. Part of it is hidden from the human observer so the apparent bisector is not the real bisector. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-16T04:36:35.927" LastActivityDate="2011-03-16T04:36:35.927" CommentCount="1" />
  <row Id="6973" PostTypeId="2" ParentId="6877" CreationDate="2011-03-15T17:35:58.800" Score="1" ViewCount="" Body="&lt;p&gt;If I think about the scaling of the aerodynamic drag at a fixed speed with size. A heavier than air airfoil scales linearly with mass, because of the need to generate lift. A Zepplin's area should scale as the 2/3rds power of its size. So for a large enough Zeppline, it should be more efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am agreed, about the high vulnerability to weather events.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-03-15T17:35:58.800" />
  <row Id="6974" PostTypeId="2" ParentId="6946" CreationDate="2011-03-15T17:41:05.147" Score="7" ViewCount="" Body="&lt;p&gt;The answers given so far are fine, but to my surprise nobody's mentioned the most important point: in modern terminology, we generally &lt;em&gt;don't&lt;/em&gt; say that the mass of an object increases with speed. &quot;Relativistic mass increase&quot; is outdated terminology, not used by most physicists anymore. In general, nowadays, &quot;mass&quot; means &quot;rest mass&quot; and is independent of velocity. Igor Ivanov's answer to &lt;a href=&quot;http://physics.stackexchange.com/questions/1686/why-does-the-mass-of-an-object-increase-when-its-speed-approaches-that-of-light&quot;&gt;this question&lt;/a&gt; says it all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't read the &lt;a href=&quot;http://arxiv.org/abs/hep-ph/0602037&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; by Lev Okun that he refers to, but I like the term &quot;pedagogical virus&quot; for this notion.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-15T17:41:05.147" />
  <row Id="6975" PostTypeId="2" ParentId="6919" CreationDate="2011-03-15T17:51:29.430" Score="3" ViewCount="" Body="&lt;p&gt;It depends in part on how much you encompass within the term &quot;big bang.&quot; There is virtually complete consensus among astrophysicists that the Universe is expanding, that it used to be much hotter and denser, and that various observed effects such as the microwave background radiation are artifacts of that hot dense period. If that's what you mean by &quot;the big bang,&quot; then I'd say there are no viable alternatives at present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you think that the term &quot;big bang&quot; encompasses more than that, and especially if you think that it refers to a theory that describes the actual origin of the Universe itself, then the answer is that there are lots of alternatives. We think that we know quite a bit about what the Universe was doing over a wide range of times &quot;after $t=0$,&quot; but we have absolutely no idea what happened &quot;at $t=0$.&quot; I'm using scare quotes here because in fact we have no idea whether &quot;at $t=0$&quot; is even a meaningful thing to talk about!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, there are some models in which, in some sense, times going back to $t=-\infty$ existed (&quot;eternal inflation&quot; being an example). There are others in which there really is a beginning at a finite time in the past. In the somewhat odd terminology of cosmology, we often refer to all of these theories as &quot;big bang models,&quot; because they all involve evolution from a hot dense early stage, even though they're quite different when it comes to the nature or existence of the actual beginning.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-15T17:51:29.430" />
  <row Id="6976" PostTypeId="2" ParentId="6956" CreationDate="2011-03-15T17:55:18.677" Score="0" ViewCount="" Body="&lt;p&gt;If oxidation is the result of hydrogen gas, this means the Zirconium tubes holding the pellets of fissile are corroding. If the core temperature is 2500 C, thermolysis of water could be a significant contributer of hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="2566" LastActivityDate="2011-03-15T17:55:18.677" CommentCount="3" />
  <row Id="6977" PostTypeId="2" ParentId="6959" CreationDate="2011-03-15T18:09:00.783" Score="3" ViewCount="" Body="&lt;p&gt;Dear Roy, (3) is correct. More precisely, retrodictions have to follow completely different rules than predictions. This elementary asymmetry - representing nothing else than the ordinary &quot;logical arrow of time&quot; (the past is not equivalent to the future as far as the logical reasoning goes) - is confusing for a surprisingly high number of people including physicists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this asymmetry between predictions and retrodictions has nothing to do with quantum mechanics per se. In classical statistical physics, one faces the very same basic problem. The asymmetry is relevant whenever there is any incomplete information in the system. The asymmetry occurs because &quot;forgetting is an irreversible process&quot;. Equivalently, the assumptions (=past) and their logical consequences (=future) don't play a symmetric role in mathematical logic. This source of logical asymmetry is completely independent from the CPT-theorem that may guarantee a time-reversal symmetry of the fundamental laws of physics. But whenever there is anything uncertain about the initial or the final state, logic has to be used and logic has an extra asymmetry between the past and the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Predictions: objective numbers&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum mechanics, the probability of a future outcome is calculated from $|c|^2$ where $c$ is a complex probability amplitude calculated by evolving the initial wave function via Schrödinger's equation, or by an equivalent method. The probabilities for the future are completely &quot;objective&quot;. One may repeat the same experiment with the same initial conditions many times and literally &lt;em&gt;measure&lt;/em&gt; the right probability. And this measurable probability is calculable from the theory - quantum mechanics, in this case - too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Retrodictions: subjective choices&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the retrodictions are always exercises in logical inference and logical inference - and I mean Bayesian inference in particular - always depends on priors and subjective choices. There is no theoretical way to calculate &quot;unique&quot; probabilities of initial states from the knowledge of the final state. Also, there is no experimental procedure that would allow us to measure such retrodictions because we are not able to prepare systems in the same &quot;final states&quot;: final states, by definition, are always prepared by the natural evolution rather than by &quot;us&quot;. So one can't measure such retrodictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To estimate the retrodicted probabilities theoretically, one must choose competing hypotheses $H_i$ - in the case of retrodictions, they are hypotheses about the initial states. We must decide about their prior probabilities $P(H_i)$ and then we may apply the logical inference. The posterior probability of $H_i$ is this conditional probability:&#xA;$$ P (H_i|F) = P(F|H_i) P(H_i) / P(F) $$&#xA;This is Bayes' formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, we have observed some fact $F$ about the final state (which may be, hypothetically, a full knowledge of the final microstate although it's unlikely). To know how this fact influences the probabilities of various initial states, we must calculate the conditional probability $P(F|H_i)$ that the property of the final state $F$ is satisfied for the initial state (assumption or condition) $H_i$. However, this conditional probability is not the same thing as $P(H_i|F)$: they are related by the Bayes formula above where $P(H_i)$ is our prior probability of the initial state $P(H_i)$ - our conclusions about the retrodictions will always depend on such priors - and $P(F)$ is a normalization factor (&quot;marginal probability of $F$&quot;) that guarantees that $\sum_i P(H_i|F) = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second law of thermodynamics&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logical asymmetry between predictions and retrodictions becomes arbitrarily huge quantitatively when we discuss the increase of entropy. Imagine that we organize microstates in ensembles - both for initial and final states; and this discussion works for classical as well as quantum physics. What do we mean by the probability that the initial state $I$ evolves to the final state $F$ if both symbols represent ensembles of microstates? Well, we must sum over all microstates in the final state $F$, but average over all microstates in the initial state $I$. Note that there is a big asymmetry in the treatment of the initial and final states - and it's completely logical that this asymmetry has to be present:&#xA;$$ P ( F|I) = \sum_{i,j} P(F_j|I_i) P(I_i) $$&#xA;We sum over the final microstates because $P(F_1 {\rm or } F_2) = P(F_1)+P(F_2)$; &quot;or&quot; means to add probabilities. However, we must average over the initial states because we must keep the total probability of all mutually excluding initial states equal to one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $P(I_i)$ is the prior probability of the $i$th microstate. In normal circumstances, when all the initial states are considered equally likely - which doesn't have to be so - $P(I_i) = 1/N_{I}$ for each $i$ where $N_{I}$ is the number of the initial states in the ensemble $I$ (this number is independent of the index $i$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the formula for $P(F|I)$ is effectively&#xA;$$ P ( F|I) = \frac{1}{N_{I}} \sum_{i,j} P(F_j|I_i) $$&#xA;Note that we only divide by the number of initial microstates but not the final microstates. And the number of the initial states may be written as $\exp(S_I)$, the exponentiated entropy of the initial state. Its appearance in the formula above - and the absence of $\exp(S_F)$ in the denominator - is the very reason why the lower-entropy states are favored as initial states but higher-entropy states are favored as final states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the contrary, if we studied the opposite evolution - and just to be precise, we will CPT-conjugate both initial and final state, to map them to $I&amp;#39;, F&amp;#39;$ - the probability of the opposite evolution will be&#xA;$$ P(I&amp;#39;|F&amp;#39;) = \frac{1}{N_{F}} \sum_{i,j} P(I&amp;#39;_i|F&amp;#39;_j). $$&#xA;Now, the probability $P(I&amp;#39;_i|F&amp;#39;_j)$ may be equal to $P(F_j|I_i)$ by the CPT-theorem: they're calculated from complex amplitudes that are equal (up to the complex conjugation). But this identity only works for the individual microstates. If you have ensembles of many microstates, they're treated totally differently. In particular, the following ratio is not one:&#xA;$$ \frac {P(I&amp;#39;|F&amp;#39;)}{P(F|I)} = \exp(S_I-S_F) $$&#xA;I wrote the numbers of microstates as the exponentiated entropy. So the evolution from $F&amp;#39;$ to $I&amp;#39;$ isn't equally likely as the evolution from $I$ to $F$: instead, they differ by the multiplicative factor of the exponential of the entropy difference - which may be really, really huge because $S$ is of order $10^{26}$ for macroscopic objects. This entropy gets exponentiated once again to get the probability ratio!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This point is just to emphasize the people who claim that the evolution from a high-entropy initial state to a low-entropy final state is &quot;equally likely&quot; as the standard evolution from a low-entropy initial state to a high-entropy final state are making a mistake of a missing or incorrectly added factor of $\exp(10^{26})$ in their formulae, and it is a huge mistake, indeed. Also, there is absolutely no doubt that the inverse processes have these vastly different probabilities - and I would like to claim that I have offered the dear reader a full proof in the text above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their mistake may also be phrased as the incorrect assumption that conditional probabilities $P(A|B)$ and $P(B|A)$ are the same thing: their mistake is &lt;em&gt;this&lt;/em&gt; elementary, indeed. These two conditional probabilities are not the same thing and the validity of the CPT-theorem in a physical theory can't change the fact that these two conditional probabilities are still very different numbers, regardless of the propositions hiding behind the symbols $A,B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to emphasize how shocking it is for me to see that those elementary issues about the distinction of past and future are so impenetrable for so many people in 2011, watch Richard Feynman's The Messenger Lecture number 5, &quot;The Distinction of Past and Future&quot; (Internet Explorer needed):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://research.microsoft.com/apps/tools/tuva/index.html&quot; rel=&quot;nofollow&quot;&gt;http://research.microsoft.com/apps/tools/tuva/index.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The very first sentence - the introduction to this very topic is - &quot;It's obvious to everybody that the phenomena in the world are self-evidently irreversible.&quot; Feynman proceeds to explain how the second law of thermodynamics and other aspects of the irreversibility follow even from the T-symmetric dynamical laws because of simple rules of mathematical logic. So whoever doesn't understand that the past and future play different roles in physics really misunderstands the first sentence in this whole topic - and in some proper sense, even the very title of it (&quot;The Distinction of Past and Future&quot;).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-15T19:46:33.360" LastActivityDate="2011-03-15T19:46:33.360" CommentCount="13" />
  <row Id="6978" PostTypeId="2" ParentId="6955" CreationDate="2011-03-15T18:19:18.937" Score="1" ViewCount="" Body="&lt;p&gt;A continuum distribution of matter, even if it is, in the bulk, electrically neutral, does not need to be electrically inert.  In particularly, it is still able to obtain a &lt;a href=&quot;http://en.wikipedia.org/wiki/Polarization_density&quot; rel=&quot;nofollow&quot;&gt;Polarization density&lt;/a&gt; that is dependent on an external electric and magnetic field.  This enables such materials to interact with electromagnetic waves.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course, in principle, there's nothing that is inherently banned by the laws of physics in having light just pass directly through matter without interacting with it.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-03-15T18:19:18.937" CommentCount="13" />
  <row Id="6979" PostTypeId="2" ParentId="6969" CreationDate="2011-03-15T18:34:13.727" Score="1" ViewCount="" Body="&lt;p&gt;I think it is a parallax effect/optical illusion, and I'm not confident of explaining this clearly but here goes!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The normal vector to the illuminated portion of the moon is pointing generally away from the Earth/moon system towards a point over our horizon.  At low altitudes (evenings) the sun will be close to the horizon and this can lead to the brain interpreting it as closer than it is and messing up the geometry.  This is similar to the enlarged moon illusion when close to the horizon.  Basically the normal vector appears to overshoot the sun as we interpret the sun as closer than it is.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-03-15T18:34:13.727" />
  <row Id="6980" PostTypeId="1" CreationDate="2011-03-15T18:50:50.140" Score="7" ViewCount="430" Body="&lt;p&gt;What I am interested to see are specific examples/reasons why gravity should be quantized. Something more than &quot;well, everything else is, so why not gravity too&quot;. For example, isn't it possible that a quantum field theory on curved space-time would be the way treat QFT and gravity in questions where the effects of neither can be ignored?&lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2011-03-16T16:13:09.830" Title="What are the reasons to expect that gravity should be quantized?" Tags="&lt;gravity&gt;&lt;quantization&gt;" AnswerCount="7" CommentCount="16" FavoriteCount="2" />
  <row Id="6981" PostTypeId="2" ParentId="6908" CreationDate="2011-03-15T18:52:50.853" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.nature.com/news/2011/110127/full/news.2011.48.html&quot; rel=&quot;nofollow&quot;&gt;Heavy and light analogues of hydrogen probe the limits of quantum chemistry&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To make the ultra-light isotope, scientists swapped the proton with a positively charged muon, which has just 11% of the mass of a proton. And to make ultra-heavy hydrogen, they replaced one of the electrons in a helium atom with a negative muon.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The researchers tested the behaviour of these new atoms in a chemical reaction called a hydrogen exchange, in which a lone hydrogen atom plucks another from a two-atom hydrogen molecule — just about the simplest chemical reaction conceivable. In a paper in Science1, they report that both the weedy and the bloated hydrogen atoms behave just as quantum theory predicts they should.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="238" LastActivityDate="2011-03-15T18:52:50.853" />
  <row Id="6982" PostTypeId="2" ParentId="6959" CreationDate="2011-03-15T19:04:03.950" Score="3" ViewCount="" Body="&lt;p&gt;I think what you are looking for is work done by Yakir Aharonov and Lev Vaidman (nice name ;) ) on something called the &quot;two-time&quot; formulation of QM which is incorporates time-symmetry from the get-go and treats prediction and retrodiction (equivalently forward and backward evolution) on an equal footing. I haven't read enough to take a stand but given the credential's of one of the authors (Aharonov as in Aharonov-Bohm effect) I'll take it as an article of faith that there is some intellectual merit therein. Of course the old argument regarding the correlation betwen old-age and &quot;crazy&quot; ideas will likely be raised. I try not to pay heed to such generalizations. Some references are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reznik and Aharonov, &lt;a href=&quot;http://arxiv.org/abs/quant-ph/9501011&quot; rel=&quot;nofollow&quot;&gt;On a Time Symmetric Formulation of Quantum Mechanics&lt;/a&gt;, PRA, 1995&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pregnell &amp;amp; Pegg, &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0510217&quot; rel=&quot;nofollow&quot;&gt;Retrodictive quantum optical state engineering&lt;/a&gt;, J. Mod. Opt., 2004&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vaidman, &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0606208&quot; rel=&quot;nofollow&quot;&gt;Backward Evolving Quantum States&lt;/a&gt;, J. Phys. A, 2006&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the other answers I see reason in @Peter Shor's and @sb1's answer. @sb1's answer would seem to rule out such a picture, or at least any practically useful application of the same. Presumably the &quot;two-time&quot; formulation avoids @sb1's argument in some way. Again I have yet to read the papers and so cannot sensibly comment on this yet.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-15T19:04:03.950" CommentCount="5" />
  <row Id="6984" PostTypeId="2" ParentId="6907" CreationDate="2011-03-15T19:41:35.800" Score="1" ViewCount="" Body="&lt;p&gt;One signature could be similar to that of a parton model.  Suppose leptons are composed of internal particles, preons or rishons or what-ever-ons.  At low energy the lepton will appear to be composed of the valence partons (lepto-partons?), which might just be the lepton itself.  As one transforms to a high energy frame, then in the limit this momentum goes to infinity the Lorentz contraction of the lepton makes other modes, or higher energy partons in excited states, apparent in scattering experiments.  There would then be a Bjorken scaling to scattering amplitudes which act as signatures of the internal constituents of a lepton.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another signature could be some deviation in the magnetic moment of the electron.  The magnetic moment is&#xA;$$&#xD;&#xA;\mu_s~=~-g_s\mu_{bohr}S/\hbar.&#xD;&#xA;$$&#xA;For a Dirac electron with the EM field &quot;turned off&quot; the g-factor is $g_s~=~2$.  In QED this is $g_s~=~2.00231930436$.  If the electron is a constituent particle then there might at some scale be a deviation from the QED expected result.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What might these constituents be?  Most likely any such deviation would to my mind be some stringy physics which due to extra large dimension and related matters is exhibiting an influence on a scale we can detect.  I don’t like the idea of quarks and leptons as composite objects.  This is largely because the energy in binding this system together would be much larger than the masses of the partons.  This would present us with horrendous problems far surpassing those seen with quarks and QCD.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-15T19:41:35.800" />
  <row Id="6985" PostTypeId="1" CreationDate="2011-03-15T20:02:59.657" Score="4" ViewCount="130" Body="&lt;p&gt;This is related to &lt;a href=&quot;http://physics.stackexchange.com/questions/6084/dimensional-reduction-from-31-to-21-for-caln-2-vector-superfield&quot;&gt;this earlier question&lt;/a&gt; I had asked. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using the so called ``Majorana&quot; representation of gamma matrices in $2+1$ dimensions in which everything is real. After doing the dimensional reduction of the $\cal{N}=1$ supersymmetry transformations of the components of the vector superfield in $3+1$ dimensions the supersymmetry transformations of the resulting $\cal{N}=2$ components of the vector superfield in $2+1$ dimensions are, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta F_A = i \bar{\alpha}^a\lambda_{Aa}$$&#xA;$$\delta D_A = i \bar{\alpha}^a\gamma_3^\mu D_\mu \lambda_{Aa}$$&#xA;$$\delta V_{A\mu} = i \bar{\alpha}^a\gamma_{3\mu}\lambda_{Aa}$$&#xA;$$\delta \lambda_{Aa} = -\frac{1}{2}f^3_{A\mu \nu}\gamma_3^{\mu \nu}\alpha_a + D_A\alpha^a + \gamma_3^\mu D_\mu F_A\alpha ^a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A,B,..$ are the gauge group indices, $f^3_{A\mu \nu}$ is the non-Abelian field strength and $\alpha$ is a spinor parameter whose components are raised and lowered as, $\alpha^1 = \alpha_2$ and $\alpha^2 = -\alpha_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the above one can derive the following transformations for the possible terms in the intended super-Chern-Simons' theory,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta(Tr[FD]) = Tr[t_At_B] \{ i\bar{\alpha}^a \lambda _{Aa}D_B -  i \bar{\alpha}^a \gamma_3^\mu\lambda_{Ba}\partial_\mu F_A + i\bar{\alpha}^a \gamma_3^\mu \lambda_{Ca} C_{BB&amp;#39;C}V_{B&amp;#39;\mu}F_A\}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta(Tr[\bar{\lambda}_a\lambda_a]) = 2Tr[t_A t_B]\{ \frac{1}{2}\bar{\alpha}_a\gamma_{3\rho}\lambda_{Ba}f^3_{A\mu\nu}\epsilon ^{\mu \nu \rho} + \bar{\alpha}^a\lambda_{Ba}D_A - \bar{\alpha}^a\gamma_3^\mu \lambda_{Ba}\partial_\mu F_A$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$- \bar{\alpha}^a\gamma_3^\mu \lambda_{Ba}C_{AB&amp;#39;C}V_{B&amp;#39;\mu}F_C \}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta(Tr[\epsilon^{\mu \nu \rho}(V_\mu \partial_\nu V_\rho)]) = -i\epsilon^{\mu \nu \rho}Tr[t_At_B]\bar{\alpha}^a\gamma_{3\rho}\lambda_{Aa}(\partial_\mu V_{B\nu} - \partial_\nu V_{B\mu})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta(Tr[\epsilon ^{\mu \nu \rho}V_\mu V_\nu V_\rho]) = -\frac{3}{2}\epsilon^{\mu \nu \rho}Tr[t_At_D]C_{DBC}\bar{\alpha}^a\gamma_{3\mu}\lambda_{Aa}V_{B\nu}V_{C\rho}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $t_A$ are a chosen basis in the lie algebra of the gauge group such that the structure constants are defined as, $[t_A,t_B]=iC_{DAB}t_D$) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that by choosing a coefficient of $-2$ for the $Tr[FD]$ and $i$ for the $Tr[\bar{\lambda}_a\lambda_a]$ some of the terms the variation of the auxiliary fields can be cancelled and some of the remaining terms of the variation of the fermionic term totally cancel the supersymmetric variation of the kinetic term of the gauge fields. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What remains are,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta(Tr[\epsilon^{\mu \nu \rho}V_\mu\partial _ \nu V_\rho + i \bar{\lambda}_a\lambda_a - 2FD]) = Tr[t_At_B]\{i\bar{\alpha}_a\gamma_{3\rho}\lambda_{Aa}C_{BCD}V_{C\mu}V_{D\nu}\epsilon^{\mu \nu \rho}$$&#xA;$$ - 2i\bar{\alpha}^a\gamma_3^\mu \lambda_{Ba}C_{AB&amp;#39;C}V_{B&amp;#39;\mu}F_C - 2i\bar{\alpha}^a\gamma_3^\mu \lambda_{Ca}C_{BB&amp;#39;C}V_{B&amp;#39;\mu}F_A\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta(Tr[\epsilon ^{\mu \nu \rho}V_\mu V_\nu V_\rho]) = -\frac{3}{2}Tr[t_At_B]\bar{\alpha}^a\gamma_{3\mu}\lambda_{Aa}C_{BCD}V_{C\nu}V_{D\rho}\epsilon^{\mu \nu \rho}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Its not clear that a coefficient can be chosen for the last term so that the supersymmetric variation of the sum of the LHSs go to zero. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The above terms seem to be structurally very different and hence its not clear how they will cancel. Like the variation of the fermionic self-coupling term produces a coupling of the fermionic component, gauge field and auxiliary field. Such a term is not produced by the variation of the gauge field cubed term! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One expects the lagrangian should look something like, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Tr[\epsilon^{\mu \nu \rho}(V_\mu\partial _\nu V_\rho -i\frac{2}{3}V_\mu V_\nu V_\rho     )   + i \bar{\lambda}_a\lambda_a - 2FD ]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to get some help in establishing the above! &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One progress would be if the two terms with the structure constant actually cancel i.e,&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Tr[t_At_B]\{C_{AB&amp;#39;C}\lambda_{Ba} V_{B&amp;#39;\mu}F_C + C_{BB&amp;#39;C}\lambda_{Ca}V_{B&amp;#39;\mu}F_A\} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the above is not clear! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB. My structure constants are defined as $[t_A,t_B]=iC_{DAB}t_D$ &lt;/p&gt;&#xA;" OwnerUserId="1637" LastEditorUserId="1637" LastEditorDisplayName="" LastEditDate="2011-03-16T13:30:56.370" LastActivityDate="2011-03-16T16:57:54.727" Title="Constructing the $\cal{N}=2$ supersymmetric non-Abelian Chern-Simon's theory" Tags="&lt;quantum-field-theory&gt;&lt;supersymmetry&gt;" AnswerCount="2" />
  <row Id="6987" PostTypeId="1" CreationDate="2011-03-15T20:16:55.870" Score="3" ViewCount="140" Body="&lt;p&gt;There are many papers about the running of coupling strength as a function of momentum/energy scale, &#xA;but are there any experimental papers about coupling strength as function of distance? &#xA;Also, are there any good books about this topic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;oops ,really sorry i forgot to specify &quot;experimental&quot;paper&lt;/p&gt;&#xA;" OwnerUserId="1702" LastEditorUserId="1702" LastEditorDisplayName="" LastEditDate="2011-03-16T07:44:14.177" LastActivityDate="2011-03-16T07:44:14.177" Title="running of coupling constant as a function of distance?" Tags="&lt;quantum-field-theory&gt;&lt;renormalization&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6988" PostTypeId="2" ParentId="6980" CreationDate="2011-03-15T20:50:34.530" Score="6" ViewCount="" Body="&lt;p&gt;Reasons for &lt;em&gt;why&lt;/em&gt; gravity should be amenable to &quot;quantization&quot;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Because everything else or as @Marek puts it because &quot;the world is inherently quantum&quot;. This in itself is more an article of faith than an argument &lt;em&gt;per se&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Because QFT on curved spacetime (in its traditional avatar) is only valid as long as backreaction is neglected. In other words if you have a field theory then this contributes to $T_{\mu\nu}$ and by Einstein's equations this must in turn affect the background via:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ G_{\mu\nu} = 8\pi G T_{\mu\nu} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently the QFTonCS approach is valid only as long as we consider field strengths which do not appreciable affect the background. As such there is no technical handle on how to incorporate backreaction for arbitrary matter distributions. For instance Hawking's calculation for BH radiation breaks down for matter densities $\gt M_{planck}$ per unit volume and possibly much sooner. Keep in mind that $M_{planck}$ is not some astronomical number but is $\sim 21 \, \mu g$, i.e. about the mass of a colony of bacteria!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vast majority of astrophysical processes occur in strong gravitational fields with high enough densities of matter for us to distrust such semiclassical calculations in those regimes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Well there isn't really a good third reason I can think of, other than &quot;it gives you something to put on a grant proposal&quot; ;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So the justification for why boils down to a). because it is mandatory and/or would be mathematically elegant and satisfying, and b). because our other methods fail in the interesting regimes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the face of the &quot;inherently quantum&quot; nature of the world we need strong arguments for &lt;em&gt;why not&lt;/em&gt;. Here are a couple:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The world is not only &quot;inherently quantum&quot; but it is also &quot;inherently geometric&quot; as embodied by the equivalence principle. We know of no proper formulation of QM which can naturally incorporate the background independence at the core of GR. Or at least this was the case before LQG was developed. But LQG's detractors claim that in the absence of satisfactory resolutions of some foundational questions (see a recent paper by Alexandrov and Roche, &lt;a href=&quot;http://arxiv.org/abs/1009.4475&quot; rel=&quot;nofollow&quot;&gt;Critical overview of Loops and Foams&lt;/a&gt;). Also despite recent successes it remains unknown as to how to incorporate matter into this picture. It would appear that &lt;a href=&quot;http://arxiv.org/abs/hep-ph/0503213&quot; rel=&quot;nofollow&quot;&gt;topological preons&lt;/a&gt; are the most natural candidates for matter given the geometric structure of LQG. But there does not appear to be any simple way of obtaining these braided states without stepping out of the normal LQG framework. A valiant attempt is made in &lt;a href=&quot;http://arxiv.org/abs/1002.1462&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; but it remains to be seen if this line of thought will bear sweet, delicious fruit and not worm-ridden garbage!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Starting with Jacobson (AFAIK) (&lt;a href=&quot;http://arxiv.org/abs/gr-qc/9504004&quot; rel=&quot;nofollow&quot;&gt;Thermodynamics of Spacetime: The Einstein Equation of State&lt;/a&gt;, PRL, 1995) there exists the demonstration that Einstein's equations arise naturally once one imposes the laws of thermodynamics ($dQ = TdS$) on the radiation emitted by the local Rindler horizons as experienced by any accelerated observer. This proof seems to suggest that the physics of horizons is more fundamental than Einstein's equations, which can be seen as an equation of state. This is analogous to saying that one can derive the ideal gas law from the assumption that an ideal gas should satisfy the first and second laws of thermodynamics in a suitable thermodynamical limit ($N, V \rightarrow \infty$, $N/V \rightarrow$ constant). And the final reason for &lt;em&gt;why not&lt;/em&gt; ...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Because the other, direct approaches to &quot;quantizing&quot; gravity appear to have failed or at best reached a stalemate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;On balance, it would seem that one can find more compelling reasons for &lt;em&gt;why not&lt;/em&gt; to quantize gravity than for &lt;em&gt;why&lt;/em&gt; we should do so. Whereas there is no stand-alone justification for &lt;em&gt;why&lt;/em&gt; (apart from the null results that I mention above), the reasons for &lt;em&gt;why not&lt;/em&gt; have only begun to multiply. I mention Jacobson's work but that was only the beginning. Work by Jacobson's student (?) Christopher Eling (&lt;a href=&quot;http://arxiv.org/find/hep-th/1/au%3a+Eling_C/0/1/0/all/0/1&quot; rel=&quot;nofollow&quot;&gt;refs&lt;/a&gt;) along with Jacobson and a few others &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0602001&quot; rel=&quot;nofollow&quot;&gt;has extended Jacobson's original argument&lt;/a&gt; to the case where the horizon is in a non-equilibrium state. The basic result being that whereas the assumption of equilibrium leads to the Einstein equations (or equivalently the Einstein-Hilbert action), the assumption of deviations from equilibrium yields the Einstein-Hilbert action &lt;em&gt;plus&lt;/em&gt; higher-order terms such as $R^2$, which would also arise as quantum corrections from any complete quantum gravity theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition there are the papers by &lt;a href=&quot;http://arxiv.org/abs/0912.3165&quot; rel=&quot;nofollow&quot;&gt;Padmanabhan&lt;/a&gt; and &lt;a href=&quot;http://arxiv.org/abs/1001.0785&quot; rel=&quot;nofollow&quot;&gt;Verlinde&lt;/a&gt; which set the physics world aflutter with cries of &quot;entropic gravity&quot;. Then there is the holographic principle/covariant entropy bound/ads-cft which also suggest a thermodynamic interpretation of GR. As a simple illustration a black-hole in $AdS_5$ with horizon temperature $T$ encodes a boundary CFT state which describes a quark-gluon plasma at equilibrium at temperature ... $T$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To top it all there is the very recent work Bredberg, Keeler, Lysov and Strominger - &lt;a href=&quot;http://arxiv.org/abs/1101.2451&quot; rel=&quot;nofollow&quot;&gt;From Navier-Stokes To Einstein&lt;/a&gt; which shows an (apparently) exact correspondence between the solutions of the incompressible Navier-Stokes equation in $p+1$ dimensions with solutions of the vacuum Einstein equations in $p+2$ dimensions. According to the abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The construction is a mathematically precise realization of suggestions of a holographic duality relating fluids and horizons which began with the membrane paradigm in the 70's and resurfaced recently in studies of the AdS/CFT correspondence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To sum it all up let me quote from Jacobson's seminal 1995 paper:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since the sound field is only a statistically defined observable on the fundamental phase space of the multiparticle sys- tem, it should not be canonically quantized as if it were a fundamental field, even though there is no question that the individual molecules are quantum mechanical. By analogy, the viewpoint developed here suggests that &lt;strong&gt;it may not be correct to canonically quantize the Einstein equations, even if they describe a phenomenon that is ultimately quantum mechanical.&lt;/strong&gt; (emph. mine)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Standard Disclaimer:&lt;/strong&gt; The author retains the rights to the above work among which are the right to include the above content in his research publications with the commitment to always cite back to the original SE question.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-15T20:50:34.530" CommentCount="12" />
  <row Id="6989" PostTypeId="1" CreationDate="2011-03-15T21:06:40.803" Score="-2" ViewCount="82" Body="&lt;p&gt;If they can not cool the reactors, shouldn't they try to encase them?  I would imagine it should be layered the same way that we would see in nature. I think we should look at volcanoes for an example of how to handle this. Then maybe, lead toward the surface. Just a suggestion to stop this from being the beginning of the end.&#xA;A Biologist/Chemist&lt;/p&gt;&#xA;" OwnerUserId="2569" LastActivityDate="2011-03-15T21:17:54.820" ClosedDate="2011-03-15T22:16:55.097" Title="Reactors in Japan" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="6990" PostTypeId="2" ParentId="6980" CreationDate="2011-03-15T21:12:24.497" Score="2" ViewCount="" Body="&lt;p&gt;I have seen two converging paths as compelling reasons for quantizing gravity, both dependent on experimental observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is the success of gauge theories in particle physics the past decades, theories that organized knowledge mathematically economically and elegantly. Gravitational equations are very tempting since they look like a gauge theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other is the Big Bang theory of the beginning of the universe that perforce has to evolve  the generation of particles and interactions from a unified model, as the microseconds grow. It is attractive and elegant that the whole is unified in a quantum theory that evolves into all the known interactions, including gravity.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-15T21:12:24.497" CommentCount="8" />
  <row Id="6991" PostTypeId="2" ParentId="6989" CreationDate="2011-03-15T21:17:54.820" Score="3" ViewCount="" Body="&lt;p&gt;So you have a heat source that is continuing to generate a lot of heat and contains a lot of unpleasant materials? And the solution is to put a lid on it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gentlemen - I give you the Atomic Pipe Bomb (tm)&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-03-15T21:17:54.820" CommentCount="3" />
  <row Id="6992" PostTypeId="2" ParentId="6980" CreationDate="2011-03-15T21:31:00.717" Score="7" ViewCount="" Body="&lt;p&gt;Gravity has to be subject to quantum mechanics because everything else is quantum, too. The question seems to prohibit this answer but that can't change the fact that it is the only correct answer. This proposition is no vague speculation but a logically indisputable proof of the quantumness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a simple thought experiment. Install a detector of a decaying nucleus, connected to a Schrödinger cat. The cat is connected to a bomb that divides the Earth into two rocks when it explodes. The gravitational field of the two half-Earths differs from the gravitational field of the single planet we know and love.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nucleus is evolving into a superposition of several states, inevitably doing the same thing with the cat and with the Earth, too. Consequently, the value of the gravitational field of the place previously occupied by the Earth will also be found in a superposition of several states corresponding to several values - because there is some probability amplitude for the Earth to have exploded and some probability amplitude for it to have survived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were possible to &quot;objectively&quot; say whether the gravitational field is that of one Earth or two half-Earths, it would also be possible to &quot;objectively&quot; say whether the nucleus has decayed or not. More generally, one could make &quot;objective&quot; or classical statements about any quantum system, so the microscopic systems would have to follow the logic of classical physics, too. Clearly, they don't, so it must be impossible for the gravitational field to be &quot;just classical&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just an explicit proof. However, one may present thousands of related inconsistencies that would follow from any attempt to combine quantum objects with the classical ones in a single theory. Such a combination is simply logically impossible - it is mathematically inconsistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, it would be impossible for the &quot;classical objects&quot; in the hybrid theory to evolve according to expectation values of some quantum operators. If this were the case, the &quot;collapse of the wave function&quot; would become a physical process - because it changes the expectation values, and that would be reflected in the classical quantities describing the classical sector of the would-be world (e.g. if the gravitational field depended on expectation values of the energy density only). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a physicality of the collapse would lead to violations of locality, Lorentz invariance, and therefore causality as well. One could superluminally transmit the information about the collapse of a wave function, and so on. It is totally essential for the consistency of quantum mechanics - and its compatibility with relativity - to keep the &quot;collapse&quot; of a wave function as an unphysical process. That prohibits observable quantities to depend on expectation values of others. In particular, it prohibits classical dynamical observables mutually interacting with quantum observables.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-15T21:31:00.717" CommentCount="25" />
  <row Id="6993" PostTypeId="2" ParentId="6987" CreationDate="2011-03-15T22:26:30.173" Score="1" ViewCount="" Body="&lt;p&gt;I'm not sure if this is what you're asking about, but the energy scale has a one-to-one correspondence with the distance scale. For example, a particle of energy $E$ and mass $m$ has a wavelength&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda = \frac{h}{p} = \frac{hc}{\sqrt{E^2 - m^2c^4}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This wavelength also represents the minimum scale of features that the particle is able to resolve, when it is used as a probe. So if you are analyzing a process that occurs on some characteristic distance scale $d$, you can find the energy associated with that scale by setting $\lambda = d$ and inverting the above equation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = \sqrt{m^2c^4 + \biggl(\frac{hc}{\lambda}\biggr)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tells you the approximate (minimum) energy of particles you will need to probe the process in question, and you can use that energy to evaluate the running coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At high energies, or short distances (specifically, shorter than the Compton wavelength of the probe, $x \ll \frac{h}{mc}$), you can even neglect the mass and just use&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = \frac{hc}{\lambda}$$&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-16T03:14:38.740" LastActivityDate="2011-03-16T03:14:38.740" CommentCount="3" />
  <row Id="6994" PostTypeId="2" ParentId="6985" CreationDate="2011-03-15T22:38:01.647" Score="2" ViewCount="" Body="&lt;p&gt;Without going through and doing the calculation myself, I can only make some general comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your index contractions seem a little weird. In the first term on the RHS of the &#xA;$$\begin{align}&#xD;&#xA;&amp;amp;\delta(Tr[\epsilon^{\mu \nu \rho}V_\mu\partial _\nu V_\rho &#xD;&#xA;          + i \bar{\lambda}_a\lambda_a - 2FD]) \\&#xD;&#xA;= &amp;amp;Tr[t_At_B]\{i\bar{\alpha}_a\gamma_{3\rho}\lambda_{Aa}C_{ABC}V_{B\mu}V_{C\nu} -&#xD;&#xA;   2i\bar{\alpha}^a\gamma_3^\mu \lambda_{Ba}C_{AB&amp;#39;C}V_{B&amp;#39;\mu}F_C \} \,,&#xD;&#xA;\end{align}&#xD;&#xA;$$ &#xA;the Lorentz indices are not contracted, are you missing a $\epsilon^{\mu \nu \rho}$? Also, in the same term you have &lt;em&gt;triply&lt;/em&gt; repeated gauge indices (i.e. the same index appears 3 times), which is unpleasant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you fix the above, and maybe use the symmetry of the $Tr[t_At_D]$ term to move the $D$ index in the RHS of the cubic term $\delta(Tr[\epsilon ^{\mu \nu \rho}V_\mu V_\nu V_\rho])$, then maybe you can get it cancel the first term talked about above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the second term on the RHS of the above displayed equation can not be canceled by anything else. So check your result for $\delta(Tr[\bar{\lambda}_a\lambda_a])$ - maybe the trouble term is meant to vanish... Are you sure that your Susy variations are correct? Have you got a reference (e.g. &lt;a href=&quot;http://arxiv.org/abs/hep-th/9506170&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9506170&lt;/a&gt;) you can check against?&lt;/p&gt;&#xA;" OwnerUserId="429" LastEditorUserId="429" LastEditorDisplayName="" LastEditDate="2011-03-16T03:01:33.307" LastActivityDate="2011-03-16T03:01:33.307" CommentCount="14" />
  <row Id="6995" PostTypeId="2" ParentId="6980" CreationDate="2011-03-15T23:50:55.857" Score="0" ViewCount="" Body="&lt;p&gt;I will answer recasting the question as a thought experiment, based on the example proposed by Lubos;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) a quantum object &lt;strong&gt;A&lt;/strong&gt; in a superposition of two states separated by a distance $X$ somewhere in empty space&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;strong&gt;A&lt;/strong&gt; has an associated gravity, with associated space-time curvature&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) now system &lt;strong&gt;B&lt;/strong&gt;, will approach the region where &lt;strong&gt;A&lt;/strong&gt; is found, and measure space-time curvature, but will not interact directly with &lt;strong&gt;A&lt;/strong&gt; or its non-gravitational fields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) now the system &lt;strong&gt;M&lt;/strong&gt; (aka Measuring Apparatus) approaches the region where both &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; are found, and it will try to measure state correlation between &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; states&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;gravity is quantum&lt;/em&gt;&quot; potential outcome:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; are statistically correlated (entangled), supporting that &lt;strong&gt;B&lt;/strong&gt; coupled with a linear superposition of gravitational fields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;gravity is classical&lt;/em&gt;&quot; potential outcome:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; are uncorrelated quantum mechanically (a direct product of both densities), supporting that any substantial gravity field will collapse (this is basically what Penrose proposes as a mechanism for measurement collapse)&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-03-16T05:17:59.933" LastActivityDate="2011-03-16T05:17:59.933" CommentCount="6" />
  <row Id="6996" PostTypeId="1" CreationDate="2011-03-16T00:21:51.117" Score="3" ViewCount="244" Body="&lt;p&gt;A modern flywheel rotor is suspended in a vacuum by magnetic bearings. This means that nothing touches the rotor as it spins. When time comes that we need to recover that stored kinetic energy, how do we recover it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine a clutch making contact with the rotor, but wouldn't that give quite a jolt, slowing down the rotor considerably?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there other ways of recovering that stored energy?&lt;/p&gt;&#xA;" OwnerUserId="293" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-16T01:06:49.240" LastActivityDate="2011-03-16T10:35:40.627" Title="Recovering energy from a modern, magnetic-levitated flywheel" Tags="&lt;electromagnetism&gt;&lt;energy-conservation&gt;" AnswerCount="4" />
  <row Id="6997" PostTypeId="1" AcceptedAnswerId="7012" CreationDate="2011-03-16T01:12:16.863" Score="3" ViewCount="243" Body="&lt;p&gt;And I ask out of pure curiosity. I know there is the &quot;Vomit Comet&quot; that allows for a non-gravity 'simulation' (for lack of a better word) on Earth, and I was thinking about if there would be a way to reverse that and rather have gravity in space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: Say I was building a space shuttle because I'm dying to go to the moon (true story). Unlimited budget, the smartest people you can fathom helping me build my space shuttle.  I get motion sickness and the whole &quot;anti-gravity&quot; thing wouldn't really work for me.  Would there be a way to have gravity on my space shuttle? Perhaps by using a type of machine or magnets or something of the like? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your patience!&lt;/p&gt;&#xA;" OwnerUserId="2574" LastEditorUserId="2574" LastEditorDisplayName="" LastEditDate="2011-03-16T01:21:58.680" LastActivityDate="2011-05-20T02:55:54.600" Title="Will a machine or a technique ever be possible that allows gravity in space?" Tags="&lt;space&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="6998" PostTypeId="2" ParentId="6997" CreationDate="2011-03-16T01:14:50.593" Score="9" ViewCount="" Body="&lt;p&gt;Turn on your rocket engines or configure your space craft into a rotating cylinder.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-16T01:14:50.593" CommentCount="3" />
  <row Id="7000" PostTypeId="2" ParentId="6996" CreationDate="2011-03-16T01:56:10.570" Score="1" ViewCount="" Body="&lt;p&gt;There has to be a system to speed the rotor up; one imagines that the same system would be used to get the energy back out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose that an electromagnetic system would be the way to do it. There are enough designs out there that it should be possible to google one. Let's see, here's a good link, the &lt;a href=&quot;http://www.energy.ca.gov/reports/2004-04-07_500-04-014.PDF&quot; rel=&quot;nofollow&quot;&gt;California Energy Commission report on the subject of flywheels&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-16T01:56:10.570" CommentCount="1" />
  <row Id="7001" PostTypeId="2" ParentId="6996" CreationDate="2011-03-16T01:56:49.320" Score="1" ViewCount="" Body="&lt;p&gt;I think a homopolar generator setup is used in some situations. Static magnetic field is introduced through the conductive wheel, and a current is induced between the center and outer rim of the disc. You can generate huge pulses of current in this setup, on the order of millions of amperes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Homopolar_generator#Homopolar_generator_development&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Homopolar_generator#Homopolar_generator_development&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1670" LastActivityDate="2011-03-16T01:56:49.320" />
  <row Id="7002" PostTypeId="2" ParentId="6969" CreationDate="2011-03-16T02:06:31.807" Score="0" ViewCount="" Body="&lt;p&gt;This is what you expect, in terms of the moon pointing towards the sun:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/iCXKf.jpg&quot; alt=&quot;half moon illuminated by sun&quot;&gt;&lt;br&gt;&#xA;That is, the line across the moon appears perpendicular to the line towards the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the above is a flat drawing. The sky appears curved (i.e. the dome of the starry sky). So that curve may introduce some apparent distortion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make a drawing that avoids the curvature issue, consider a drawing that only includes the sun, the moon, and a small amount of sky around the line connecting the two. (By the way, even for the curved bowl of the sky, the shortest line connecting the two is well defined except in the case of a new or full moon.) That drawing is approximately flat and will show the above relation.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-16T02:06:31.807" CommentCount="6" />
  <row Id="7003" PostTypeId="1" CreationDate="2011-03-16T03:30:06.900" Score="2" ViewCount="318" Body="&lt;p&gt;If I understand correctly, when an earthquake occurs, energy will be transferred to the water, resulting in water waves. As the waves reach seashore, because the sea depth is getting shallower and wavelength is getting shorter, the height of the wave gets push up, resulting in tsunami. In other words in deep sea, water won't get pushed up as high as the water in shallow seashore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my understanding correct? Is there a quantitative way to express the physics behind all this?&lt;/p&gt;&#xA;" OwnerUserId="605" LastEditorUserId="605" LastEditorDisplayName="" LastEditDate="2011-03-16T04:08:13.327" LastActivityDate="2011-03-16T06:23:11.367" Title="Physics of tsunami: the relationship between wavelength, sea depth and the height of the water" Tags="&lt;geophysics&gt;&lt;tsunami&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="7004" PostTypeId="1" AcceptedAnswerId="7009" CreationDate="2011-03-16T03:39:50.800" Score="1" ViewCount="161" Body="&lt;p&gt;I've come upon Dr. J. Marvin Herndon's theory that the earth's magnetic field is generated by a hot nuclear reactor operating in the center of the earth. This is backed by various papers, some of them peer reviewed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nuclearplanet.com/Herndon%27s%20Nuclear%20Georeactor.html&quot; rel=&quot;nofollow&quot;&gt;http://www.nuclearplanet.com/Herndon%27s%20Nuclear%20Georeactor.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His theory purports to explain various anomalies such as the &lt;a href=&quot;http://www.sciencedirect.com/science?_ob=ArticleURL&amp;amp;_udi=B6V66-487F9KP-3&amp;amp;_user=10&amp;amp;_coverDate=11/30/1987&amp;amp;_rdoc=1&amp;amp;_fmt=high&amp;amp;_orig=gateway&amp;amp;_origin=gateway&amp;amp;_sort=d&amp;amp;_docanchor=&amp;amp;view=c&amp;amp;_searchStrId=1680424709&amp;amp;_rerunOrigin=google&amp;amp;_acct=C000050221&amp;amp;_version=1&amp;amp;_urlVersion=0&amp;amp;_userid=10&amp;amp;md5=9a1068470e01f9f819550010d40e9418&amp;amp;searchtype=a&quot; rel=&quot;nofollow&quot;&gt;high level of He3 in basalt in Hawaii and their dependency with time&lt;/a&gt;. A recent Herndon arXiv paper is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Uniqueness of Herndon's Georeactor: Energy Source and Production Mechanism for Earth's Magnetic Field&lt;/em&gt;&#xA;&lt;a href=&quot;http://arxiv.com/abs/0901.4509&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.com/abs/0901.4509&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His papers suggest that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_nuclear_fission_reactor&quot; rel=&quot;nofollow&quot;&gt;natural reactors sometimes seen in uranium deposits&lt;/a&gt; must also occur at the center of the earth. He says that the conventional explanation for the earth's magnetic field fails because the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rayleigh_number&quot; rel=&quot;nofollow&quot;&gt;Rayleigh number&lt;/a&gt; for the core is inappropriate as far as determining whether convection exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is not accepted geology, what are the problems with the theory?&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-16T18:42:12.983" Title="Can the earth's magnetic field be generated by heat from a natural nuclear reactor?" Tags="&lt;nuclear-physics&gt;&lt;geophysics&gt;&lt;convection&gt;&lt;rayleigh-number&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="7005" PostTypeId="1" CreationDate="2011-03-16T04:11:58.610" Score="3" ViewCount="201" Body="&lt;p&gt;I know that there's no evidence at this point for &quot;white holes&quot; however would it even be mathematically possible for a black hole to be connected to a white hole (total opposite so everything would be expelled, after some really extreme physical conditions)? Maybe with a wormhole connecting them? If this was even possible (if the black hole or connection could actually be created and be stable enough), would that matter be expelled into a different universe, etc? Maybe even a different region of spacetime? Just curious, as it would be a cool idea. &lt;/p&gt;&#xA;" OwnerUserId="2576" LastActivityDate="2011-03-20T03:10:31.923" Title="Is it mathematically possible for a black hole to be connected to a white hole or a worm hole?" Tags="&lt;black-holes&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="7006" PostTypeId="2" ParentId="7003" CreationDate="2011-03-16T04:12:05.957" Score="0" ViewCount="" Body="&lt;p&gt;I think it is just the gravity wave solution. We have a periododic function (sine wave) in the horizontal direction, and an exponential in the vertical direction. I think the wave number and the exponential decay rate are the same number, but hopefully someone in the know can fill in the details. In any case for short waves and deep water you need only consider the exponential term which decays with depth. But for tsunami's the wavelength is greater than the depth, so you have to use both types. Not sure what the boundary conditions are (at the water surface, and at the sea bottom are), but satisfying them would give you the allowable form for the wave at a given wavelength and depth. But in any case, for the tsunami, considerable motion is seen throughout the water column. If the wave is not reflected going into shallower water (I think this means the depth doesn't change much within a horizontal wavelength) then conservation of energy &amp;amp; momentum means the wave amplitude grows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get a hydrolic jump (moving wall of water), because the wavespeed is higher in deeper water, so the higher portion of the wave can catch up with the slower moving portions ahead of it. If that happens, instead of a gradually increasing sealevel, you can get one or several stepfunction type waves coming in.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-03-16T04:12:05.957" />
  <row Id="7008" PostTypeId="1" CreationDate="2011-03-16T04:38:18.560" Score="3" ViewCount="172" Body="&lt;p&gt;Are there any examples of common substances whose decay is not exponential?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're used to thinking about radioactivity in terms of half-lives. This is a concept that makes sense only for a decay that is exponential. However, there are plenty of physics articles on the subject of non exponential decay. It seems to be theoretically ubiquitous. For example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The decay of unstable quantum states&#xA;  is an ubiquitous process in virtually&#xA;  all fields of physics and energy&#xA;  ranges, from particle and nuclear&#xA;  physics to condensed matter, or atomic&#xA;  and molecular science. &lt;strong&gt;&lt;em&gt;The exponential&#xA;  decay, by far the most common type, is&#xA;  surrounded by deviations at short and&#xA;  long times$^{1,2}$.&lt;/em&gt;&lt;/strong&gt; The short-time&#xA;  deviations have been much discussed,&#xA;  in particular in connection with the&#xA;  Zeno effect$^{3,4,5}$ and the&#xA;  anti-Zeno effect$^{6,7,8,9}$.&#xA;  Experimental observations of short$^{10,11}$&#xA;   and long$^{12}$ time deviations&#xA;  are very recent. A difficulty for the&#xA;  experimental verification of long-time&#xA;  deviations has been the weakness of&#xA;  the decaying signal$^{13}$, but also the&#xA;  measurement itself may be responsible,&#xA;  because of the suppression of the&#xA;  initial state reconstruction$^{2,14}$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1) L. A. Khalfin, Zurn. Eksp. Teor.&#xA;  Fiz. 33, 1371 (1957), English&#xA;  translation: Sov. Phys. JETP 6 1053&#xA;  (1958).&lt;br&gt;&#xA;  2) L. Fonda and G. C.&#xA;  Ghirardi, Il Nuovo Cimento 7A, 180&#xA;  (1972).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;10.1103/PhysRevA.74.062102, F. Delgado, J. G. Muga, G. Garcia-Calderon&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/quant-ph/0604005v2&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Suppression of Zeno effect for distant detectors&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So are there any examples of deviations from long time decay? If not, then why not? Is the theory wrong or simply impractical? And is there a simple, intuitive explanation for why long decays should not be exponential?&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-16T11:30:35.867" Title="Long time deviations from exponential decay in radioactivity" Tags="&lt;quantum-mechanics&gt;&lt;nuclear-physics&gt;&lt;radioactivity&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="7009" PostTypeId="2" ParentId="7004" CreationDate="2011-03-16T04:40:39.133" Score="3" ViewCount="" Body="&lt;p&gt;&lt;s&gt;KamLAND&lt;/s&gt; Borexino has set moderately strict limits of the total power of a central geo-reactor. See for instance &lt;a href=&quot;http://geoscience.lngs.infn.it/Program/Pdf_presentations/Maricic.pdf&quot; rel=&quot;nofollow&quot;&gt;Geo-neutrino: Experiments (pdf link)&lt;/a&gt; a talk by one of my colleagues. (Jalena notes that Borexino's limit is the strongest one going, but KamLAND &lt;em&gt;was&lt;/em&gt; the leader for a while.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upper extreme of these limits is less than half the total geological power, but quite non-trivial. The bottom goes all the way to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a recent paper (that I have yet to read) on a variation of this idea: &lt;a href=&quot;http://arxiv.org/abs/1011.3568&quot; rel=&quot;nofollow&quot;&gt;The KamLAND-experiment and Soliton-like Nuclear Georeactor. Part 1. Comparison of Theory with Experiment&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've no idea, how the rest of these ideas stand.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-03-16T18:12:33.363" LastActivityDate="2011-03-16T18:12:33.363" CommentCount="3" />
  <row Id="7010" PostTypeId="2" ParentId="6928" CreationDate="2011-03-16T04:40:49.123" Score="3" ViewCount="" Body="&lt;p&gt;Gamma rays is not a concern at all, just stay 1000m from the reactor, and no gamma rays will reach you from there. So no reason to block gamma rays. (Other types of radiation cannot go even 100m in air).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is release of radioactive gazes &amp;amp; particles, which fly around and end up somewhere in human body. That's why the only thing we could do now is to try to completely isolate reactor from environment using some concrete.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-03-16T04:40:49.123" CommentCount="1" />
  <row Id="7011" PostTypeId="2" ParentId="6996" CreationDate="2011-03-16T04:41:32.147" Score="0" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;but wouldn't that give quite a jolt,&#xA;  slowing down the rotor considerably?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you are trying to extract energy isn't that the point?&lt;br&gt;&#xA;If you apply a force to the flywheel you extract kinetic energy, whether you do this by grabbing it with a clutch, letting in air  or electromagnetically doesn't really matter &lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-03-16T04:41:32.147" />
  <row Id="7012" PostTypeId="2" ParentId="6997" CreationDate="2011-03-16T04:44:56.327" Score="2" ViewCount="" Body="&lt;p&gt;It sounds like you don't want the normal rotating spaceship like in &quot;2001&quot; because you get motion sickness.  No one really gets &quot;motion sickness&quot; just from moving, though.  That's impossible because moving with constant velocity is physically the same as being stationary.  What you get is &quot;acceleration sickness&quot;.  You feel the bumpiness of a car ride.  Even then, it might more-accurately be called &quot;jerk sickness&quot; or &quot;tidal sickness&quot;.  &quot;Jerk sickness&quot; would be when your acceleration is changing over time, as in going over a bump.  &quot;Tidal sickness&quot; is when different parts of your body experience different accelerations at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside a rotating cylinder, what you feel would be identical to normal gravity to first order, so you wouldn't necessarily get motion sick like in a car.  If you get motion sick because you see the stars rotate when you look out the window, don't build any windows and you'll never know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few effects break this illusion of perfect gravity, including tidal forces, Coriolis forces, and non-constant angular velocity of the spaceship.  The bigger your ship, the smaller these effects will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tidal forces arise when the gravitational field is not the same everywhere.  To make tidal forces small, you need the ship to be big.  The strength of the artificial gravity on your ship is $\omega^2 R$ with $\omega$ the angular frequency (how fast the ship is rotating) and $R$ the distance from the rotation axis.  Tidal accelerations are then on the order of $\omega^2\Delta R$, with $\Delta R$ the size of the object we're considering (e.g. your body).  Tidal forces compared to gravity are $\Delta R / R$, or the size of the object compared to the size of the ship.  If you're 1000 times smaller than your ship, the tidal forces you feel will be 1000 times smaller than gravity - too small to notice.  (On Earth they are much smaller still, of course, because you are so small compared to Earth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coriolis forces are forces that makes tossed projectiles appear to bend when you throw them inside the ship, even though nothing is pushing on them.  The size of the Coriolis acceleration is on the order of $\omega v$, with $v$ the speed of the thing being thrown.  In order to make them smaller, you need to decrease $\omega$.  Remember that the strength of gravity is $g = \omega^2 R$, so $\omega = \sqrt{g/R}$.  The make the Coriolis forces small, you need a big ship again.  Suppose you want Coriolis forces to be under some fraction $\alpha$ of $g$ when you go at a certain maximum velocity $v$.  Then the formula for the minimum size of the spaceship is $R &amp;gt; v/(\alpha^2 g)$.  If $v = 10 {\rm m/s}$ and $\alpha = .01$, we need a spaceship 10 kilometers across.  So we're going to have to tolerate fairly high Coriolis acceleration.  But remember you don't feel this at all if you're stationary.  If you're moving it's just a constant force pushing up/down/left/right depending on which way you're going.  Only if different parts of you are moving different directions does it becomes a serious problem.  Your free throws would all be off, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, when stuff moves around in the spaceship, the rotation rate of the entire ship could change because the stuff pushes on the ship.  This acceleration would be on the order of $am/M$, with $a$ the acceleration of the moving object, $m$ the mass of the object, and $M$ the mass of the ship.  Most things on the ship would have a tiny $m/M$, so this wouldn't be a big deal.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an object didn't accelerate, but moved towards the center at a constant speed $v$, the ship would accelerate &lt;a href=&quot;http://physics.stackexchange.com/questions/3611/why-does-a-ballerina-speed-up-when-she-pulls-in-her-arms&quot;&gt;like an ice skater accelerating as they pull their arms in&lt;/a&gt;.   This acceleration would be on the order of $v^2m/RM$, and again would be small for small objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another potential problem is that if the mass distribution of the ship is not symmetric, the ship would wobble.  Suppose that the ship is balanced at some time, but then an object of mass $m$ is moved a distance $d$ along the side of the cylinder towards the end.  Then the frequency of the wobble should be on the order of $\omega md/MR$ and the amplitude should be on the order of $md/M$.  The jerk you feel would then be on the order of $\omega^3 m^2d^2/M^2 R$, which should be small as long as it's a fairly small mass moving a small distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How small can these effects be?  That relies mostly on the size of the ship, as we saw earlier.  If the ship is going to simulate one Earth gravity, it can't be too big.  Eventually, the tension in the sides of the ship would be so great the ship would tear itself apart.  If your ship were a loop, that critical radius would be $T/\lambda g$, with $T$ the tension and $\lambda$ the linear mass density of the wall.  For carbon nanotubes this gets you a ship size of up to $10^7 {\rm m}$, a figure so large (bigger than  Earth) that all effects could be made inconsequential.  For steel it's about $5*10^3 {\rm m}$, meaning that the effects on to human-sized might be in general $.001$ or $.0001$ of gravity.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-03-16T04:44:56.327" CommentCount="2" />
  <row Id="7013" PostTypeId="2" ParentId="6996" CreationDate="2011-03-16T05:07:34.230" Score="2" ViewCount="" Body="&lt;p&gt;We recover the energy in a maglev flywheel in the same way we almost always convert mechanical energy to electrical energy: with a 3 phase electric power generator/motor, also called an alternator, with the rotor &lt;a href=&quot;http://www.aspes.ch/faq.html#How%20energy%20into%20flywheel&quot; rel=&quot;nofollow&quot;&gt;on the same shaft&lt;/a&gt; or &lt;a href=&quot;http://sbir.nasa.gov/SBIR/successes/ss/5-060text.html&quot; rel=&quot;nofollow&quot;&gt;otherwise integrated&lt;/a&gt; with the flywheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In cars with a combined starter/generator, pumped-storage hydroelectric dams, maglev flywheels, etc.,&#xA;3-phase electric power goes into the coils of the stator, and gets converted to mechanical torque to spin the rotor.&#xA;Later, the mechanical energy of the same spinning rotor pushes a magnetic field around that pushes electrons through those same coils of the stator, converting mechanical energy to electrical energy that exits out the wires of the generator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such 3-phase motor/generator electric machines don't need anything to touch the rotor.&#xA;(Unlike some DC motors/generators that require a &quot;brush&quot; to touch the rotor).&#xA;In practice, most of them do have something that touches the rotor --&#xA;a long input shaft to carry mechanical energy into the generator, a long output shaft to carry mechanical energy out of the generator, various bearings to keep the rotor more or less centered end-to-end and radially, or some combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes magnetic levitated &lt;a href=&quot;http://en.wikipedia.org/wiki/flywheel_energy_storage&quot; rel=&quot;nofollow&quot;&gt;flywheel energy storage&lt;/a&gt; a little special is that nothing actually does touch the rotor.&#xA;Some of the coils surrounding the rotor act like the coils of a 3 phase electric machines.&#xA;Those coils convert electric energy to mechanical energy to spin up the rotor in motor mode.&#xA;The same coils later convert the mechanical energy of the rotor back to electrical energy, slowing down the rotor in generator mode.&#xA;Other coils surrounding the rotor are used as a &lt;a href=&quot;http://en.wikipedia.org/wiki/magnetic_bearing&quot; rel=&quot;nofollow&quot;&gt;magnetic bearing&lt;/a&gt; to levitate the flywheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Regenerative_variable-Frequency_drives&quot; rel=&quot;nofollow&quot;&gt;typical regenerative variable-frequency drive&lt;/a&gt; typically uses &lt;a href=&quot;http://www.kilowattclassroom.com/Archive/VFDarticle.pdf&quot; rel=&quot;nofollow&quot;&gt;6 IGBTs to convert DC electric power to mechanical energy and back&lt;/a&gt;; the &lt;a href=&quot;http://www.power-thru.com/22.html&quot; rel=&quot;nofollow&quot;&gt;IGBT arrangement used in maglev flywheels&lt;/a&gt; is no different.&lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2011-03-16T05:07:34.230" />
  <row Id="7014" PostTypeId="1" CreationDate="2011-03-16T05:31:19.633" Score="2" ViewCount="423" Body="&lt;p&gt;For an accelerated charge to radiate, must an electromagnetic field be the source of the force?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it radiate if accelerated by a gravitational field?&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="1345" LastEditorDisplayName="" LastEditDate="2011-03-16T16:52:01.643" LastActivityDate="2011-03-18T21:53:20.850" Title="For an accelerated charge to radiate, is an electromagnetic field as the source necessary?" Tags="&lt;electromagnetic-radiation&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="7015" PostTypeId="2" ParentId="6928" CreationDate="2011-03-16T05:38:20.050" Score="1" ViewCount="" Body="&lt;p&gt;Lead, seawater, or anything with a low boiling point at low pressures would be a waste of time in an actual nuclear core melt down situation. Possibly one could pump sand into the reactor. Hopefully the sand, probably melted by the high temperatures around the pool of melted reactor rods, might dilute the pool enough to eventually stop the nuclear reaction. Certainly looks like the nuclear community has not fully thought out the worst case nuclear scenario; otherwise we would not be having this discussion.  What a mess huh?&lt;/p&gt;&#xA;" OwnerUserId="2577" LastActivityDate="2011-03-16T05:38:20.050" CommentCount="1" />
  <row Id="7016" PostTypeId="2" ParentId="7014" CreationDate="2011-03-16T05:45:26.227" Score="9" ViewCount="" Body="&lt;p&gt;Your question is somewhat abstruse, but here's what I think you're asking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put a charged particle in a uniform external magnetic field.  The particle will move in a circular orbit, but since it's accelerating, it will radiate and its orbit will decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now remove the magnetic field.  Grab the charge and forcibly swing it around in the same circle as before by some other, unknown means.  Does it still radiate in the same way as before?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is yes because Maxwell's equations are linear.  Therefore we can analyze any situation in classical electromagnetism by superposition.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-03-16T05:45:26.227" CommentCount="12" />
  <row Id="7017" PostTypeId="2" ParentId="7014" CreationDate="2011-03-16T06:06:13.833" Score="3" ViewCount="" Body="&lt;p&gt;The radiation, if considered classically, is independent for the reason Mark Eichenlaub gives. But considered quantum mechanically, it is not independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, photons are bosons. So the presence of radiation of a particular polarization and frequency will increase the probability of the particle radiating that polarization and frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a topic I'd not seen before. I'll look around and see if I can find a reference to the effect.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;An accelerated electron produces &lt;a href=&quot;http://en.wikipedia.org/wiki/Synchrotron_radiation&quot; rel=&quot;nofollow&quot;&gt;&quot;synchrotron radiation&quot;&lt;/a&gt;. An example of the electromagnetic field altering the emission of such radiation would be &quot;stimulated synchrotron emission&quot;. &lt;a href=&quot;http://en.wikipedia.org/wiki/Stimulated_emission&quot; rel=&quot;nofollow&quot;&gt;Stimulated emission&lt;/a&gt; was described by Einstein and is the physics behind lasers. An example paper combining these ideas:&lt;br&gt;&#xA;Phys. Rev. Lett. 66, 2312–2315 (1991), J. L. Hirshfield and G. S. Park, &lt;em&gt;Electron-beam cooling by stimulated synchrotron emission and absorption&lt;/em&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://prl.aps.org/abstract/PRL/v66/i18/p2312_1&quot; rel=&quot;nofollow&quot;&gt;http://prl.aps.org/abstract/PRL/v66/i18/p2312_1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-03-16T06:25:34.197" LastActivityDate="2011-03-16T06:25:34.197" CommentCount="1" />
  <row Id="7018" PostTypeId="2" ParentId="7003" CreationDate="2011-03-16T06:23:11.367" Score="4" ViewCount="" Body="&lt;p&gt;A tsunami is basically a shallow-water wave, even in deep seas. This means its velocity is $v=\sqrt{gH}$, where $H$ is the water depth and $g$ is the gravitational acceleration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy of the tsunami scales as the square of its amplitude $A$, and thus the energy flux $S$ goes as $S\sim A^2 \sqrt{H}$. Conservation of energy then implies that the wave amplitude depends on the sea depth as&#xA;$$ A \sim H^{-1/4}$$&#xA;a result known as Green's law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, Green's law predicts that a tsunami with amplitude $A=1$m at $H=5000$m will run up to $A=4$m if the depth becomes $H=20$m.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2011-03-16T06:23:11.367" />
  <row Id="7019" PostTypeId="2" ParentId="7005" CreationDate="2011-03-16T06:48:33.420" Score="3" ViewCount="" Body="&lt;p&gt;Dear Kahtrijn, white holes are microscopically the same objects as black holes, and it's guaranteed - by the second law of thermodynamics - that all macroscopic processes occur in the way as they do in black holes and not white holes (the latter are time-reversed of the former).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you don't care about the second law of thermodynamics that prevents black hole size from shrinking by emitting large objects (which objects?), then you may write down any configuration in general relativity you want. In particular, it's easy to connect white holes to black holes. After all, an &quot;eternal black hole&quot; is a solution that is doing nothing else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The connection between a white hole and a black hole goes in a different direction than you seem to expect, however. It's a connection between the past and the future. It's because the (neutral) white hole singularity is a spacelike singularity in the past, and the (neutral) black hole singularity is a spacelike singularity in the future. To connect those singularities, you need timelike (rather than spacelike) trajectories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One doesn't need any wormhole to write down an eternal black hole. It may sound cool to combine so many different holes but it is totally unnecessary: the wormholes are independent objects from the black holes and white holes.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-16T06:48:33.420" />
  <row Id="7021" PostTypeId="2" ParentId="6980" CreationDate="2011-03-16T08:38:43.730" Score="3" ViewCount="" Body="&lt;p&gt;I am very much surprised to see that apart from all the valid reasons (specially the argument, &lt;em&gt;since everything else is quantum hence gravity should also be the same otherwise many inconsistencies will develop&lt;/em&gt;) mentioned by Lubos et. al. no body pointed out that one of the other main motivations to quantize gravity was that classical GR predicted singularities in extreme situations like big bang or black holes. It was kind of like the the instability of the Ratherford atomic model where electrons should have been spiral inward the nuclus as per the classical electrodynamics. Quantum theory saved physics from this obvious failure of classical physics. Naturally it occured to physicists that quantum theory should be the answer of the singularity problem of classical GR too. However experiences in the last 40 years have been different. Far from removing singularities it appears that our best quantum gravity theory is saying that some of the singularities are damn real. So obviously the motivation of quantization of gravity has changed to an extent and it is unification which is now driving the QG program in my humble opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some additional comments:&lt;/strong&gt;&#xA;@Mbn, There are strong reasons to believe that the uncertainty principle is more fundamental than most other principles. It is such an inescapable property of the universe that all sane physicists imho will try their best, to make every part of their world view including gravity, consistent with the uncertainty principle. All of the fundamental physics has already been successfully combined with it except gravity. That's why we need to quantize gravity.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-03-16T15:39:32.133" LastActivityDate="2011-03-16T15:39:32.133" CommentCount="6" />
  <row Id="7022" PostTypeId="2" ParentId="7008" CreationDate="2011-03-16T10:19:49.983" Score="2" ViewCount="" Body="&lt;p&gt;In German wiki on Zeno effect: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://de.wikipedia.org/wiki/Quanten-Zeno-Effekt&quot; rel=&quot;nofollow&quot;&gt;http://de.wikipedia.org/wiki/Quanten-Zeno-Effekt&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You find this sentence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bisher wurde noch kein Anhalten des&#xA;  radioaktiven Zerfalls durch&#xA;  experimentelle Messungen eines&#xA;  Ensembles von radioaktiven Atomen oder&#xA;  gar eines einzelnen radioaktiven Atoms&#xA;  bestätigt, wie es die Theorie des&#xA;  Quanten-Zeno-Effekts erfordern würde.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Short translation: Not observed in radioactive decay till today. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea on this is, that the Zeno effect affords some observation of &#xA;the nucleus with some measurement touching the nucleus. &#xA;Just registration of the radiation by some counter &lt;strong&gt;after&lt;/strong&gt; the decay won't &#xA;&quot;disturb&quot; the nucleus. Is this an argument in &quot;Schrödingers Cat&quot; too?&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-16T11:30:35.867" LastActivityDate="2011-03-16T11:30:35.867" />
  <row Id="7023" PostTypeId="2" ParentId="6956" CreationDate="2011-03-16T10:28:18.810" Score="0" ViewCount="" Body="&lt;p&gt;Of course, it is not from neutrons going beta to Hydrogen, that should be a hell of neutrons. &lt;/p&gt;&#xA;" OwnerUserId="1335" LastActivityDate="2011-03-16T10:28:18.810" />
  <row Id="7025" PostTypeId="2" ParentId="1184" CreationDate="2011-03-16T10:52:44.573" Score="0" ViewCount="" Body="&lt;p&gt;+1 on the notion that ground speed is the straightforward vector addition of air speed and wind speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Manual flight computers such as the classic Dalton E6B allow you to do this vector math graphically.  From a starting point in the middle of your plastic plate you draw a line in the direction of the wind speed, with a length proportional to the wind speed.  From there you draw a line in the direction of your air speed (which is your aircraft's heading), with a length proportional to your air speed.  Then the vector from start to finish is your ground speed and your ground track (direction you are travelling over the ground).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the extra fuel you need to burn to fly into a headwind or the fuel you save by flying with a tailwind, a most practical consideration to note is that if you fly from pt. A to pt. B into a headwind, and then turn around and fly from B back to A with a tailwind of the same strength, you lose more flying into the headwind than you gain from flying with the tailwind.&lt;/p&gt;&#xA;" OwnerUserId="2297" LastActivityDate="2011-03-16T10:52:44.573" />
  <row Id="7026" PostTypeId="2" ParentId="2377" CreationDate="2011-03-16T11:12:33.060" Score="0" ViewCount="" Body="&lt;p&gt;Ok, what if we were to use some sort of metaphysical observation (no measurement) and then use this observation to choose when to measure thus perhaps partially aligning the colapse with some higher order?  I would assume the waveform would collapse according to the same 'random' order but perhaps because of the partial alignment we could somehow influence the outcome of said measurement.&lt;/p&gt;&#xA;" OwnerUserId="2580" LastActivityDate="2011-03-16T11:12:33.060" />
  <row Id="7027" PostTypeId="1" CreationDate="2011-03-16T11:16:32.073" Score="1" ViewCount="164" Body="&lt;p&gt;Why is the amplitude of the Power spectral density higher for shorter period of time as compared to a longer period of time when calculated for any vibration data?&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-03-17T13:22:09.547" LastActivityDate="2011-03-17T13:22:09.547" Title="Amplitude of power spectral density" Tags="&lt;mathematical-physics&gt;&lt;signal-processing&gt;" AnswerCount="1" />
  <row Id="7028" PostTypeId="2" ParentId="357" CreationDate="2011-03-16T11:58:51.083" Score="0" ViewCount="" Body="&lt;p&gt;Perhaps a Fizeau inteferometer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fizeau_interferometer&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fizeau_interferometer&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of that should be in range of a keen amateur but I'm not sure what you use as a beam splitter without just buying one though.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-03-16T11:58:51.083" />
  <row Id="7029" PostTypeId="2" ParentId="2447" CreationDate="2011-03-16T13:31:10.957" Score="-1" ViewCount="" Body="&lt;p&gt;Where can I find a derivation that proves that Ricci is in fact the deformation of a volume element. By analogy in 2 dimensions I've tried to show that the area works like Ricci; the problem is that the tensor operates on something to get something, like put a vector in to riemann and out pops the delta vector from transport around the specified loop. What does Ricci take?&lt;/p&gt;&#xA;" OwnerUserId="2662" LastActivityDate="2011-03-16T13:31:10.957" CommentCount="4" />
  <row Id="7030" PostTypeId="2" ParentId="357" CreationDate="2011-03-16T13:48:54.500" Score="2" ViewCount="" Body="&lt;p&gt;I think the simplest would be to use an RF oscillator, a receiver to determine its frequency and Lecher wires (i.e. a pair of parallel wires) where the nodes of the standing wave are determined using an RF voltmeter. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Lecher_lines&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lecher_lines&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was one of the experiments in an electronics kit which I had in my youth. The length of the Lecher line was about 5m and the frequency of the oscillator was about 100Mhz using just one transistor in a common base circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a variation it is also possible to change the frequency and measure how much the nodes have moved.&lt;/p&gt;&#xA;" OwnerUserId="1944" LastActivityDate="2011-03-16T13:48:54.500" />
  <row Id="7031" PostTypeId="2" ParentId="7014" CreationDate="2011-03-16T14:38:32.227" Score="1" ViewCount="" Body="&lt;p&gt;If you mean an external EMF, the answer is &quot;The radiation is determined with an external filed&quot;. The charge acceleration is proportional to the external field, and a single  accelerated charge radiates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean the radiated field influence on the charge motion and subsequent radiation, the answer is &quot;No&quot; because the radiation is only expressed via external field. There is no need to invoke the proper field here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In QM radiation of hard single photons happens discontinuously in time so QM is somewhat different.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-03-16T16:08:10.103" LastActivityDate="2011-03-16T16:08:10.103" />
  <row Id="7032" PostTypeId="2" ParentId="6980" CreationDate="2011-03-16T14:42:57.573" Score="0" ViewCount="" Body="&lt;p&gt;There are two questions here. The first is not so much whether we expect a unifying theory to be &quot;quantum&quot; as much as whether we expect a unifying theory to be probabilistic/statistical. I suppose that at or within 5 or 10 orders of magnitude of the Planck scale we can expect that we will still have to work with a statistical theory. Insofar as Hilbert space methods are the simplest effective mathematics for generating probability measures that can then be compared with the statistics of measurements, it's likely we will keep using this mathematics until some sort of no-go theorem proves that we have to use more sophisticated and harder to use mathematical tools (non-associative algebras of observables, etc., etc., etc., none of which most of us will choose to use unless we &lt;em&gt;really&lt;/em&gt; have to).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arguably more characteristic feature of quantum theory is a scale of action, Planck's constant, which determines, &lt;em&gt;inter alia&lt;/em&gt;, the scale of quantum fluctuations and the minimal incompatibilities of idealized measurements. From this we have the Planck length scale, given the other fundamental constants, the speed of light and the gravitational constant. From this point of view, to say that we wish to &quot;quantize&quot; gravity is to assume that the Planck scale is not superseded in dynamical significance at very small scales by some other length scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lack of detailed experimental data and an analysis that adequately indicates a natural form for an ansatz for which we would fit parameters to the experimental data is problematic for QG. There is also a larger problem, unification of the standard model with gravity, not just the quantization of gravity, which introduces other questions. In this wider context, we can construct any length scale we like by multiplying the Planck length by arbitrary powers of the fine structure constant, any of which might be natural given whatever we use to model the dynamics effectively. The natural length for electro-geometrodynamics might be $\ell_P\alpha^{-20.172}$ (or whatever, $\ell_P e^{\alpha^{-1}}$ isn't natural in current mathematics, but something as remarkable might be in the future), depending on the effective dynamics, and presumably we should also consider the length scales of QCD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notwithstanding all this, it is reasonable to extrapolate the current mathematics and effective dynamics to discover what signatures we should expect on that basis. We have reason to think that determining and studying in detail how experimental data is different from the expected signatures will ultimately suggest to someone an ansatz that fits the experimental data well with relatively few parameters. Presumably it will be conic sections instead of circles.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-03-16T14:42:57.573" CommentCount="3" />
  <row Id="7034" PostTypeId="2" ParentId="2377" CreationDate="2011-03-16T15:54:17.467" Score="0" ViewCount="" Body="&lt;p&gt;I think there is another level at which this question is being asked.  Randomness of a symbol string means there is no formal data compression algorithm that reduces the string to some small form.  The shortest description of a string is the level at which its complexity is reduced to an extremum, and that it can be executed by a Turing machine which halts.  This is the Kolmogoroff complexity of a symbol string.  So to emulate the string there exists some Turing machine which has a &quot;tape&quot; and a stack, where the complexity of the string can't be significantly more (longer length, more bits etc) than the Turing machine to guarantee a halting condition.  A set of $n$ coin tosses will produce $N~=~2^n$ possible binary configurations.  For $n$ large a smaller percentage of the $N$ binary strings are likely to satisfy this condition.  The Kolmogoroff complexity is then a form of the Chaitan halting probability, which is itself not computable in general, which does give a bound on the number of strings which are &quot;halting.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to the indefinability of randomness.  To define randomness you need some algorithm which can compute that a string is random.  So given a string $S$ there must exist a Turning machine which determines $Rand(S)~=~T\vee F$.  However, that is not mathematically possible.  This means randomness is not computable.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-16T15:54:17.467" />
  <row Id="7035" PostTypeId="1" CreationDate="2011-03-16T16:10:35.873" Score="2" ViewCount="135" Body="&lt;p&gt;Can you give me examples where the standard physics curriculum in high school or college is inaccurate? When did you realize you had been fed lies?&lt;/p&gt;&#xA;" OwnerUserId="2586" LastActivityDate="2011-03-16T16:10:35.873" ClosedDate="2011-03-16T16:18:31.030" Title="Examples of incorrect physics education" Tags="&lt;education&gt;" CommentCount="10" FavoriteCount="1" />
  <row Id="7036" PostTypeId="2" ParentId="6980" CreationDate="2011-03-16T16:13:09.830" Score="2" ViewCount="" Body="&lt;p&gt;For the sake of argument, I might offer up a plausible alternative.  We might have some quantum underpinning to gravitation, but we might in fact not really have quantum gravity.  It is possible that gravitation is an emergent phenomenon from a quantum field theoretic substratum, where the continuity of spacetime might be similar to the large scale observation of superconductivity or superfluidity.  The AdS/CFT is a matter of classical geometry and its relationship to a quantum field theory.  So the $AdS_4/QFT$ suggests a continuity of spacetime which has a correspondence with the quark-gluon plasma, which has a Bjorken hydrodynamic scaling.  The fluid dynamics of QCD, currently apparent in some LHC and RHIC heavy ion physics, might hint at this sort of connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we might not really have a quantum gravity as such. or if there are quantum spacetime effects it might be more in the way of quantum corrections to fluctuations with some underlying quantum field.  Currently there are models which give quantum gravity up to 7 loop corrections, or 8 orders of quantization.  Of course the tree level of quantum gravity is formally the same as classical gravity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is suggested not as some theory I am offering up, but as a possible way to think about things.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-16T16:13:09.830" CommentCount="2" />
  <row Id="7038" PostTypeId="1" AcceptedAnswerId="7042" CreationDate="2011-03-16T16:41:35.037" Score="5" ViewCount="213" Body="&lt;p&gt;Super string theory establishes that at each point of our 4 dimensional spacetime, there is attached a 6-dimensional Calabi-Yau (CY) manifold which encodes the geometry of the additional compactified spatial dimensions needed to formulate the theory. It is postulated that the size of these compactified dimensions is of the order of the Planck scale. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I've always understood that in quantum gravity, we are confronted with the big problem of wild quantum fluctuations -due to the Heisenberg principle- at the Planck scale.These fluctuations should lead to wild shifts and fluctuations in the metric coefficients and hence, they would make hard to properly talk in the terms of differential geometry (manifold, distance, metric, curvature) that are the terms in which we talk about CY manifolds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know how string theory solves this contradiction that has arisen due to my lack of knowledge of the full technical details of the theory. Is it possible that, the conformal invariance required for the string worldsheet during its propagation, prevents the pathological effects of these quantum fluctuations allowing us to talk in terms of geometry even at the Planck scale?&lt;/p&gt;&#xA;" OwnerUserId="2237" LastActivityDate="2011-03-17T06:03:43.217" Title="Question about Calabi-Yau manifolds and quantum fluctuations at the Planck scale" Tags="&lt;string-theory&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="7039" PostTypeId="2" ParentId="6709" CreationDate="2011-03-16T16:53:47.570" Score="0" ViewCount="" Body="&lt;p&gt;If you want to see how much energy is saved, you can get it in 2 ways, either mechanical, or electric. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If you measure U and I to the motor (directly, using voltmeter and ampermeter, before and afer change), then energy saved is t*U1*i1-t*U2*I2, where t is time, in seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Mechanically. You need to know either total distance traved by the weight, L, or weight speed, v. Evergy saved is (m1-m2)gL, or (m1-m2)gtv. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;V=IR is not the right formula for your purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to use formula to connect the electrical power of the motor (P1) to mechanical energy of raising weight (P2), P1=P2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power P1 (in watts) to/from vertically moving weight is proportional to its speed v and mass m: &#xA;P1 = vmg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electric power P2=$IU$ for direct current, P2=$IU0.77$ for AC, IIRC that constant.&lt;br&gt;&#xA;From P1=P2 you get  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    vmg = IU0.77&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Depending on how the motor is controlled, if weight reduces by half, then&#xA;either I or U need to be reduced by half.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to reduce I or U by connecting resistor R2 in sequence to the motor, then you need, at the very least, to know I and U, and calculate the dissipation power.  You might be surprised.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small resistors are between 0.1W to 1W only. What is power of your motor.  Motor can be kilowatts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resistors are good for several watts only. Above that, resistor will be huge size, it might overheat, and it will waste energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resisters are not how bigger motors are controlled!   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better methods of controlling big AC motor are transformer or semiconductor circuit, not resistors.&lt;/p&gt;&#xA;" OwnerUserId="2585" LastEditorUserId="2585" LastEditorDisplayName="" LastEditDate="2011-03-16T21:30:25.630" LastActivityDate="2011-03-16T21:30:25.630" CommentCount="2" />
  <row Id="7040" PostTypeId="2" ParentId="6985" CreationDate="2011-03-16T16:57:54.727" Score="1" ViewCount="" Body="&lt;p&gt;@Deepak: triply repeated is operationally well-defined as part of Einstein summation convention, but is meaningless in general: contractions are supposed to be group invariants, and in that sense such triple &quot;contractions&quot; are meaningless. &lt;/p&gt;&#xA;" OwnerUserId="2588" LastActivityDate="2011-03-16T16:57:54.727" />
  <row Id="7041" PostTypeId="1" AcceptedAnswerId="7058" CreationDate="2011-03-16T17:04:12.943" Score="7" ViewCount="352" Body="&lt;p&gt;Some threads here touching speed of gravity made me think about that. This lead to some questions. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The speed of gravity was not measured until today (at least there are no undebated papers to that effect).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It seems common knowledge/belief among physicists that the speed of gravity is the same as the speed of light. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And this is my question: Is that speed of light = speed of gravity a result of GR or is that fed into the theory? Or is there some evidence from other sources than GR for this? Does the same speed imply some deep-lying relation between gravity and electromagnetism? &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1778" LastEditorDisplayName="" LastEditDate="2011-06-21T14:56:04.980" LastActivityDate="2011-06-21T14:56:04.980" Title="&quot;Speed&quot; of Gravity and Speed of Light" Tags="&lt;gravity&gt;&lt;speed-of-light&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="7042" PostTypeId="2" ParentId="7038" CreationDate="2011-03-16T17:30:27.020" Score="9" ViewCount="" Body="&lt;p&gt;There are a two levels for the answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, perturbative string theory actually has two different scales: the string scale in which the extended nature of the string comes into play (even classically strings are different than particles), and the Planck scale in which the quantum mechanics of the string (including quantum gravity) becomes important. The string energy scale is lower than the Planck energy scale, their ratio is the string coupling constant (which is proportional to $\hbar$), so it is meaningful to talk about stringy physics in the sub-Planckian regime. Your mental picture then depends on the size of the compact geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the CY manifolds are larger the Planck length you can use perturbative string theory, in which case the notion of a manifold still kind of makes sense. When reaching the Planck scale the string theory is no longer perturbative and large fluctuations in the geometry are part of the story. Since we know quite a bit about non-perturbative string theory (through dualities and non-renormalization theorems) these expected phenomena can be observed and analyzed precisely. For example, topology change of the CY manifold is a smooth process that can be discussed very precisely in this framework. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Incidentally, in this context we see that in the transition region between two well-defined geometries there is no geometrical description at all - contrary to some people's intuition about quantum geometry and spacetime foam and all the rest of that. The geometrical description fails and there is some well-defined procedure to do any well-defined calculation without any reference to any type of geometry whatsoever).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second level of answer is that even when discussing perturbative strings on sub-Planckian CY manifolds, you have to keep in mind that this is a shorthand for something more precise and technical - (2,2) SCFT  which has a limit in which it becomes a sigma model with a CY manifold as a target space. In English this means that even classical strings in some target space probe the geometry in different ways than point particles, and to describe their physics you need more complicated machinery than just differential geometry. As you suspected the fact that the string is extended means that it is more forgiving to non-smooth features of the manifold, even classically. This is encoded in the fact that the SCFT is better behaved than differential geometry on that space, and stays well defined even in the presence of some types of singularities (orbifolds of space are a famous example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in summary &quot;string theory on CY&quot; really means string theory whose classical point particle limit (valid when the manifold is very large and smooth) reduces to CY manifold. Good to remember in general that there is some poetic license taken sometimes when physicists describe their work.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-03-17T06:03:43.217" LastActivityDate="2011-03-17T06:03:43.217" CommentCount="9" />
  <row Id="7043" PostTypeId="1" AcceptedAnswerId="7044" CreationDate="2011-03-16T17:44:47.820" Score="1" ViewCount="140" Body="&lt;p&gt;Imagine three light beams are &quot;sent&quot; to a lens simultaneously, they start at the same position but move towards the lens at different angles. The first light beam passes the lens at its edge, the second light beam passes exactly through its mid whilst the last light beam passes somewhere in between. Behind the lens, there is a screen at the focal point of the light beams. Which of the three light beams reaches the screen first?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume they all reach it at the same moment, but I can't explain why, it's just my intuition. Thus, I hope somebody can help me out here. Thanks for answers in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;//e: Sorry for being inaccurate. This is how the setup described should be like:&lt;br&gt;&#xA;&lt;img src=&quot;http://img6.imagebanana.com/img/ahfdqbrn/IMG_1208.jpg&quot; alt=&quot;This&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="977" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-03-17T03:19:33.170" LastActivityDate="2011-03-17T03:19:33.170" Title="Through a lens, which light beam reaches the screen first?" Tags="&lt;optics&gt;&lt;waves&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="7044" PostTypeId="2" ParentId="7043" CreationDate="2011-03-16T18:10:49.420" Score="5" ViewCount="" Body="&lt;p&gt;If the lens focuses perfectly, so that all rays from the given starting point end up at the same final point, then you're right: they all arrive at the same time. (To be precise, this is true within the limits of applicability of geometrical optics. But if we're not willing to assume that geometric optics applies, then it doesn't make sense to talk about individual rays of light anyway.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is Fermat's principle, otherwise known as the principle of least time. In a situation like this, each ray follows a path of least time joining the given starting and finishing points. If the times weren't the same, then at least one of them wouldn't be a minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might worry that there's a swindle here. Fermat's principle really only says that the path has to be a critical point, not that it has to be a minimum. But that doesn't affect the conclusion. Take two of the rays, and imagine a family of rays smoothly joining one to the other. Fermat's principle says that the derivative of the travel time is zero all along the family of rays, so there's no difference between the time at the beginning and the time at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That last paragraph is a bit awkwardly phrased. Let me restate it precisely. Let $T(r)$ be the travel time corresponding to ray $r$. Let $r(0)$ be one of the rays under consideration, $r(1)$ be another, and $r(s)$ for $0&amp;lt;s&amp;lt;1$ be a smooth function joining the two together. Each $r(s)$ is a ray striking the lens at a different point, but all obeying the correct laws of optics. Fermat's principle says that $\delta T=0$ is constant for all small variations about one of these rays, so $dT(r(s))/ds=0$, and $T(r(0))=T(r(1))$.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-16T18:10:49.420" CommentCount="1" />
  <row Id="7045" PostTypeId="2" ParentId="7043" CreationDate="2011-03-16T18:11:13.600" Score="4" ViewCount="" Body="&lt;p&gt;EDIT&#xA;I wrote this under the assumtion that the screen was placed at the plane where the rays come to focus. I now realize that the question does not state that. If the screen is not at the focal point, then no, the rays will not have equal travel time.&#xA;END EDIT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your intuition is correct, each ray would reach the screen at the same time. In fact, in some sense, you could define the image plane of any imaging system as the plane where all rays from the object plane have an equal travel time. That would be a bit of a sloppy definition and it isn't quite complete in the technical sense, but for an intuitive explanation it will work in most cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speaking in terms of geometrical optics, optical engineers usually refer to the optical path length (OPL) rather than the travel time of the light, but they are equivalent quantities, differing only by a factor of the speed of light (c). The OPL in a medium of refractive index $n$ is defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \text{OPL}=nL$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $L$ is the actual distance traveled in the medium. So, for a ray passing through several diferent materials (like some air, a lens, and then air again) the total OPL can be expressed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \text{OPL}=\sum_in_iL_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This quantity should be equal for each ray when it is properly focused from the object point to the image point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes intuitive sense when you think about it. A lens with positive power is thicker at the center than the edges, and typical glass has $n\approx1.5$. The actual length of the ray through the center of the lens is shorter than the rays that go through the edges of the lens, because it follows a more direct path; but the central ray also goes through the thicker part of the lens, which increases its OPL to match that of the rays through the edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That pretty much sums up the geometrical optics picture.&lt;/p&gt;&#xA;" OwnerUserId="869" LastEditorUserId="869" LastEditorDisplayName="" LastEditDate="2011-03-16T18:32:42.537" LastActivityDate="2011-03-16T18:32:42.537" CommentCount="1" />
  <row Id="7046" PostTypeId="2" ParentId="6827" CreationDate="2011-03-16T18:34:12.240" Score="2" ViewCount="" Body="&lt;p&gt;an optical fibre without cladding exists but is hard to make profit of it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;contrail of jets&lt;/strong&gt; is an optical fiber under special circumstances:&#xA;Ice crystals in the form of hexagonal plates are created in wet and very cold temperatures at high altitudes, typically above 10 km, and form a tube, nearly cylindrical, that can extend for hundreds of kilometres.&#xA;If the atmosphere is quiet the ice crystals will descend slowly, parallel to the ground oscilating up to a maximum of 8 degrees from the vertical, and if It is lit by the Sun or the Moon, not visible at sea level, perpendicular to the cylindrical tube at the opposite site in relation to the observer he will see the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The light will bounce from crystal to crystal under total reflection conditions with very low absortion, and subjected to polarization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The observational evidence:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The perceived usual shapes are like 'discs, cigars, triangles' and correspond to some sort of intersection of a plane with a cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slight change of angle between the sun/moon, the contrail, and observers will make a significant change observed in the lights, which are a decomposition of white light.&#xA;When observing a change in velocity inconsistent with the motion of any known object remember that the light moves at c speed ;-) and the Earth's rotation around its axis contributes to the changing angles.&#xA;The distribution of the observations of UFOs in space and time historically accompanies the evolution of aviation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In colder areas like Alaska and northern Norway, with more candidates to observers and with polar commercial air routes means that there are more reports of observations. Besides the light reflected by the ice cap may also inject more light into the contrail.&#xA;In areas with little commercial air routes as the South Pacific there are hardly any observations of UFOs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;They are what we call UFOs by Night.&lt;/strong&gt;&lt;br&gt;&#xA;This is not the physics involved in the day light observations.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-03-16T18:34:12.240" CommentCount="2" />
  <row Id="7047" PostTypeId="2" ParentId="7004" CreationDate="2011-03-16T18:42:12.983" Score="1" ViewCount="" Body="&lt;p&gt;This theory has been around a while.  The one problem I see is that in order to get a uranium reactor running you need a fair concentration of U235.  The core of the Earth most likely does not have that.  There might be a low level or very subcritical amount of fission which goes on in the core of the Earth.  However, I suspect this is not larger than the amount of energy produced by strong and weak nuclear decay processes.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-16T18:42:12.983" CommentCount="2" />
  <row Id="7048" PostTypeId="2" ParentId="7014" CreationDate="2011-03-16T19:22:54.803" Score="4" ViewCount="" Body="&lt;p&gt;I suspect that two charged objects orbiting one another due to gravitational attraction would radiate, but I can't support that assertion with a citation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question of whether or not a charge radiates when it is uniformly accelerated by gravity is an &lt;a href=&quot;http://www.mathpages.com/home/kmath528/kmath528.htm&quot; rel=&quot;nofollow&quot;&gt;open question&lt;/a&gt;; read the link for an excellent discussion of why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;br&gt;&#xA;The &lt;a href=&quot;http://www.mathpages.com/home/kmath528/kmath528.htm&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; I posted isn't inspiring trust, so I searched for peer-reviewed work. I found two relevant papers:&lt;br&gt;&#xA;&lt;a href=&quot;http://dx.doi.org/10.1088/0143-0807/31/5/006&quot; rel=&quot;nofollow&quot;&gt;Physical interpretation of the Schott energy of an accelerating point charge and the question of whether a uniformly accelerating charge radiates&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://dx.doi.org/10.1119/1.3488985&quot; rel=&quot;nofollow&quot;&gt;The significance of the Schott energy for energy-momentum conservation of a radiating charge obeying the Lorentz–Abraham–Dirac equation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not qualified to comment on the quality of the papers, but both attack the question 'does a falling charge radiate', implying an open question. I didn't find any experimental work on the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;This brings up a topic better suited for meta-discussion: I would like to see &lt;em&gt;even more&lt;/em&gt; external citations in the answers here. I would also like to see more answers clearly indicate their logical foundation. Is your answer based on...&lt;br&gt;&#xA;-  Original research?&lt;br&gt;&#xA;-  Predicted by peer-reviewed theory but unverified?&lt;br&gt;&#xA;-  Indirectly experimentally verified?&lt;br&gt;&#xA;-  Directly experimentally verified?&lt;br&gt;&#xA;Don't make us guess!&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="2359" LastEditorDisplayName="" LastEditDate="2011-03-18T21:53:20.850" LastActivityDate="2011-03-18T21:53:20.850" CommentCount="11" />
  <row Id="7049" PostTypeId="1" CreationDate="2011-03-16T19:30:01.943" Score="12" ViewCount="509" Body="&lt;p&gt;I'm curious to learn how people discovered that electric and magnetic fields could be nicely put into one simple tensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's clear that the tensor provides many beautiful simplifications to the original theory, by applying the abstract theory of tensors to this particular problem. For example, the strange formulas for the transformation of electric and magnetic fields in different reference frames can be explained as the transformation laws of a 2-tensor. The interdependence of the two fields in this transformation, and the fact that electric and magnetic fields are in some ways the same thing in the classical theory, can be explained by this two tensor. The various ad-hoc formulas that make up Maxwell's equations, some of them with curls, some with divergence, can be explained in one beautiful formula by declaring the exterior derivative of the tensor to be 0. The cross product can also be explained as an operation on anti-symmetric tensors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it's clear once someone shows you the tensor formulation that it beautifully weaves together all the parts of the &quot;elementary&quot; (i.e. non-tensorial) theory. My question is, how did people discover this formulation in the first place? What was the motivation, and what is the history?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some thoughts: It's true that the elementary theory provides some hints to the tensor formulation (such as some of the things I list above), but these small hints are not quite enough to motivate all the intricacies of the tensor formula, especially if one has not seen tensors before. Was the theory of tensors already floating around in the time that the field tensor was discovered, and hence did tensor experts simply notice that electromagnetism smelled like a 2-tensor? If this is the case, how did people initially realize that tensors were important in physics? Otherwise, what did happen? Why were people motivated to do it? And, wasn't the original formulation good enough, albeit not quite as mathematically elegant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another related question, did people know the transformation laws for electric and magnetic fields before Einstein came along? Today, you can usually only find those in books on special relativity, or in the very last chapter of a book on electromagnetism, usually a chapter on special relativity. Therefore, if you were reading a book on electromagnetism, then before you got to the chapter on relativity, you would have thought that force vectors and hence electric fields are invariant under change of reference frame, just like forces in Newtonian mechanics.&lt;/p&gt;&#xA;" OwnerUserId="2591" LastEditorUserId="2591" LastEditorDisplayName="" LastEditDate="2011-03-16T20:50:31.063" LastActivityDate="2011-08-15T04:27:10.980" Title="History of Electromagnetic Field Tensor" Tags="&lt;electromagnetism&gt;&lt;history&gt;&lt;tensor-calculus&gt;&lt;field-theory&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="4" />
  <row Id="7050" PostTypeId="2" ParentId="6907" CreationDate="2011-03-16T19:34:41.267" Score="1" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;Composite electrons, muons and taus ought to be much easier to detect in high energy conditions where an electomagnetic field influences decay product paths than composite neutrinos (which are damned hard to observe much of anything about without elaborate and not very statistically powerful purpose built experiments like those being discussed at Neutel11).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If components parts of electon-like leptons had a charge other than -1 (or +1 for antiparticles), the path that even briefly unconfined lepton components took ought to be possible to reverse engineer with great precision (and without a lot of the QCD background issues that make some of the other calculations harder to do -- because you'd be looking at the distribution pattern of where the decay products end up in space relative to the collision point, rather than how many there were).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IIRC, there has been some recent experimental signals that show these kinds of unexpected and unexplained spatial distribution patterns.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Another way to see composite electrons in confined state would be to detect events with signatures that are like mesons or exotic baryons, but much lighter that had previously been screened out of data since we weren't looking for anything like that in that mass range.  For example, suppose that you revised your decay data sorting software and suddenly saw several dozen decays of a particle that was behaving like a Delta plus plus baryon (spin 3/2, charge +2, ddd), but with a mass on the order of 123 eV instead of 1232 MEv.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A third possiblity would be that you could look at processes that appear to show beyond the Standard Model CP violation and do some kind of cluster analysis of the data that show one group of events that closely match the Standard Model and a separate group of events that has some pattern that distinguishes it and then show how a composite lepton model could explain the pattern common to the &quot;excess group&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Strong evidence of B-L non-conservation that seems to be coming from something in the lepton sector.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2592" LastActivityDate="2011-03-16T19:34:41.267" />
  <row Id="7051" PostTypeId="1" CreationDate="2011-03-16T19:47:36.203" Score="1" ViewCount="103" Body="&lt;p&gt;As a general rule of thumb, massive particles (both composite and fundamental) tend to decay rapidly through the weak force, while less massive particles tend to be more stable.  Hence, taus are shorter lived than muons, top quarks are shorter lived than charm quarks, and all mesons and baryons except protons and neutrons are highly unstable.  My understanding is that this relationship is largely captured in the Standard Model electro-weak equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any notable cases where the decay rate and the mass of the particle appear experimentally to deviate from the expected relationship? &lt;/p&gt;&#xA;" OwnerUserId="2592" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-16T19:53:43.270" LastActivityDate="2011-04-30T00:36:54.083" Title="Are there notable cases of anomalies in the mass-decay rate relationship?" Tags="&lt;particle-physics&gt;&lt;experimental-physics&gt;&lt;nuclear-physics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="7052" PostTypeId="2" ParentId="7051" CreationDate="2011-03-16T20:01:31.047" Score="4" ViewCount="" Body="&lt;p&gt;Dear Andrew, first, I edited your &quot;anomoly&quot; which should be &quot;anomaly&quot;. I couldn't watch it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, all the decays you mention use the weak force and the elementary Feynman diagram is always the same: it's a cubic vertex with one W-boson, one decaying fermion, and one fermionic decay product. So the amplitude is essentially $g_{SU(2)} \bar u_{final} \gamma_\mu  u_{initial} \epsilon_W^\mu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what hugely depends on the mass of the fermions are the kinematic factors - the Lorentz-invariant phase space, if you wish. The &quot;universal&quot; amplitude above has to be integrated over all allowed momenta of the final particles, with the $d^3 p / 2E$ measure. Also, there's $1/2E$ for each initial particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most impressive &quot;failure of dimensional analysis&quot; among these weak decays is the decay rate of the ordinary neutron - its half-life is ten minutes! That's an extremely long time scale, especially if you compare it to the half-life of top quark etc. that you mentioned. Both decays are driven by the same elementary process whose Lorentz-invariant amplitude is essentially identical! The neutron is this stable because it's just slightly heavier than the proton, the main decay product, and the phase space for the allowed electron's and antineutrino's momenta in the final state is just extremely small. (There are probably other similarly long half-lives of unstable nuclei that decay via beta-decay - which are just heavier counterparts of the decaying neutron. The neutron decay is also a simple case of a beta-decay.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exists no observed disagreement between any weak decay (of a known particle) and the Standard Model prediction. That's a glimpse of a much more general fact: the Standard Model just universally works. If I am the first one who tells you that it does, it's unfortunate.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-16T20:01:31.047" CommentCount="4" />
  <row Id="7053" PostTypeId="1" AcceptedAnswerId="7061" CreationDate="2011-03-16T20:02:25.987" Score="5" ViewCount="242" Body="&lt;p&gt;So I know that when you collide particles with high enough kinetic energy, (kinetic energy = at least the rest mass of the particles you are making), you get particles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How come potential energy cannot make particles? Say you have an electron held at a potential such that the amount of potential energy it has equals at least the rest mass of a particle you want to make. How come potential energy does not convert into rest mass energy directly? Would this require whatever is creating the potential to disappear because the potential energy turns into a particle? This shouldn't violate any conservation of energy laws does it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question that kind of relates to my previous question is, say we have a 512 keV photon. What prevents this photon from turning into an electron that then has kinetic energy of 1 kev?&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2011-03-16T23:52:29.927" Title="Kinetic Energy vs. Potential energy with regards to creating particles" Tags="&lt;particle-physics&gt;&lt;energy&gt;&lt;photons&gt;&lt;high-energy-physics&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" />
  <row Id="7054" PostTypeId="2" ParentId="6541" CreationDate="2011-03-16T20:15:22.183" Score="2" ViewCount="" Body="&lt;p&gt;While it is a trivial example, any set of equations in the form of the GR equations with different constants would satisfy the equivalence principle while not being confirmed experimentally.  For example, if one took G to be 42 (in conventional units) and the speed of light to be 200 meters per hour, you would have a logically consistent set of equations that met the equivalence principle but did not reproduce GR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A less trivial example is that the value of the cosmological constant is not constrained tightly by anything else in GR.  Other than Hubble constant scale astronomy observations, changing it really doesn't change anything else in GR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bekenstein's TeVeS theory is an example of a non-GR set of equations that satisfy the equivalence principle while producing effects similar to those observed an attributed to dark matter in the weak field that does reproduce the classic GR experimental tests.  (To be clear, I am not stating that TeVeS is the correct theory, particularly after the Bullet Cluster, merely that it has the mathematical character we are discussing.)  More generally, Bekenstein's effort suggests that it is possible as a general matter to craft equations that reduce to GR in most cases and meet the equivalence principle, but differ from it in a systemic way in either the strong field case, or the weak field case, or both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, there is no obvious reason in my mind that one couldn't have a set of equivalence principle compliant, non-GR equations in which gravity was also, for example, a function of overall electro-weak charge, which is neutral in all known cases where GR is observed in astronomy.&lt;/p&gt;&#xA;" OwnerUserId="2592" LastActivityDate="2011-03-16T20:15:22.183" CommentCount="2" />
  <row Id="7055" PostTypeId="2" ParentId="7053" CreationDate="2011-03-16T20:18:15.910" Score="3" ViewCount="" Body="&lt;p&gt;In principle you can create particles with potential, the prototypical case is &lt;a href=&quot;http://en.wikipedia.org/wiki/Color_confinement&quot; rel=&quot;nofollow&quot;&gt;color confinement&lt;/a&gt;. But I believe you'll need field densities that put the necessary energy into a very small space, which is why it works with the strong force, but we don't see it done with others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A (on-shell) photon can't simply turn into an (on-shell) electron because such a process would not be able to conserve both energy and momentum.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-03-16T20:44:14.270" LastActivityDate="2011-03-16T20:44:14.270" CommentCount="5" />
  <row Id="7056" PostTypeId="2" ParentId="7038" CreationDate="2011-03-16T20:24:51.097" Score="-3" ViewCount="" Body="&lt;p&gt;A compactified direction in $10$ dimension has some circle so that $x^9~=~x^9~+~2\pi R$ results in a huge mass.  For the $n^{th}$ quantized momentum state this mass is $m~=~n/R$.  So the large quantum fluctuations compete with the huge mass.  For $R~\simeq~\ell_s$ $=~4\pi\sqrt{\alpha&amp;#39;}$ The tension of the string is $T~\simeq~1/\alpha&amp;#39;$ and the energy of the string is given by the winding number $w$ multiplied bythe energy $E~=~\oint Tdr$ or&#xA;$$&#xD;&#xA;E~=~2\pi wTR~=~\frac{wR}{\alpha&amp;#39;},&#xD;&#xA;$$&#xA;which contributes to the mass in a T-duality which is invariant under $R~\rightarrow~\alpha&amp;#39;/R$.  This mass is the energy associated with the Heisenberg uncertainty principle and the tiny localization of the $x^9$ coordinate direction.  The interchange means this is equivalent to a theory much larger than the string scale.  This exchanges type IIA and type IIB string theories.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-16T20:24:51.097" CommentCount="3" />
  <row Id="7057" PostTypeId="2" ParentId="7041" CreationDate="2011-03-16T20:46:01.067" Score="4" ViewCount="" Body="&lt;p&gt;I think the main argument would be causality. If gravitational disturbances traveled faster than light, presumably you could use them to convey information backwards in time. Although now that I think about it, that argument is based on special relativity. I can't think of a reason it wouldn't carry over to GR, but I don't know whether anyone has explicitly checked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can derive from GR the fact that gravitational waves (disturbances in the weak-field approximation) propagate at speed $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not aware that any experimental measurements have been done, since gravity is a weak force and is difficult to measure. Building a &quot;gravity telescope&quot; is a lot more complicated than a regular telescope, although there are plans in place, e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Laser_Interferometer_Space_Antenna&quot; rel=&quot;nofollow&quot;&gt;LISA&lt;/a&gt;, which might be able to detect gravitational waves for the first time, if it gets funded and built.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-03-16T20:46:01.067" />
  <row Id="7058" PostTypeId="2" ParentId="7041" CreationDate="2011-03-16T20:50:17.180" Score="6" ViewCount="" Body="&lt;p&gt;The special theory of relativity is really enough to see that gravitational signals have to propagate by the speed $c$ which we call &quot;speed of light&quot; because the light is the most commonly understood entity that is moving by this maximum speed. Special relativity is OK to describe infinitesimal deformations of spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All other massless particles also have to propagate by the same speed $c$ because this speed $c$ is needed to enhance the vanishing rest mass to a finite total relativistic energy. And gravitons are inevitably massless because they don't pick any preferred reference frame - or, alternatively, because gravity is a long-range force. Massive particles could only induce short-range forces (similar to the weak nuclear force caused by W,Z bosons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any particle - e.g. neutrino - whose energy is much greater than the rest mass is moving nearly by the speed of light, too. The same thing would hold for massless scalar particles such as the &quot;moduli&quot; (their quanta) if they existed. It's an elementary consequence of the formulae of special relativity. The speed of light is the maximum speed that the information and material objects may pick, by causality, and it's also the typical speed that massless (exactly) and light (approximately) particles actually choose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your last question is No, the appearance of the same speed $c$ doesn't imply any additional dynamical relationship between electromagnetism and gravity - it's a direct and elementary consequence of the special theory of relativity - and its kinematics - that was fully understood in 1905. The importance of the speed $c$ in the scheme of things - because of special relativity - is so high that adult physicists use units in which $c=1$ and they are never ever surprised when $c$ plays an important role - it's exactly the same degree of &quot;surprise&quot; as if the number $1$ appears somewhere in maths.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-16T20:50:17.180" CommentCount="3" />
  <row Id="7059" PostTypeId="2" ParentId="7053" CreationDate="2011-03-16T21:20:55.717" Score="2" ViewCount="" Body="&lt;p&gt;Two electrons race head on, until they are at a distance comparable to their diameter. &#xA;At this distance, they stop for short time, all kinetic energy is tranformed into potential energy (repulsion by charge) if something will happen now, is this driven by potential or kinetic energy? I think its potential. Of course this experiment can be &quot;done&quot; with all other forces keeping particles apart. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your last question: it does not happen, because that would violate lepton number conservation. You have to create a electron/positron pair, then lepton number is not changed. &#xA;But that positron will annihilate with some other electron soon, thus creation of  particles did happen only for some µ or mseconds. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-16T21:41:51.110" LastActivityDate="2011-03-16T21:41:51.110" />
  <row Id="7060" PostTypeId="1" AcceptedAnswerId="7079" CreationDate="2011-03-16T21:29:25.617" Score="7" ViewCount="195" Body="&lt;p&gt;So I have learned in class that light can get red-shifted as it travels through space. As I understand it, space itself expands and stretches out the wavelength of the light. This results in the light having a lower frequency which equates to lowering its energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, where does the energy of the light go? Energy must go somewhere! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the energy the light had before go into the mechanism that's expanding the space? I'm imagining light is being stretched out when its being red-shifted. So would this mean that the energy is still there and that it is just spread out over more space?&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2011-03-23T10:08:17.920" Title="Redshifting of Light and the expansion of the universe" Tags="&lt;light&gt;&lt;energy-conservation&gt;&lt;space-expansion&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="2" />
  <row Id="7061" PostTypeId="2" ParentId="7053" CreationDate="2011-03-16T21:36:28.790" Score="6" ViewCount="" Body="&lt;p&gt;A strong enough static electric field &lt;em&gt;can&lt;/em&gt; create real particle-antiparticle pairs out of the vacuum, so yes, potential energy can contribute to creating particles. This is called the &lt;strong&gt;Schwinger effect&lt;/strong&gt; and was predicted by Schwinger in the 1950s. The original paper is here &lt;a href=&quot;http://prola.aps.org/abstract/PR/v82/i5/p664_1&quot; rel=&quot;nofollow&quot;&gt;http://prola.aps.org/abstract/PR/v82/i5/p664_1&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Schwinger's formula gives $$\frac{N}{VT}=\frac{e^2E_0^2}{4\pi^3}\sum_{n=1}^{\infty}\frac{1}{n^2}\mbox{exp}(-\frac{nm^2\pi}{eE_0})$$ where $N$ e+e- pairs are produced in a volume $V$ and time $T$  by a static uniform electric field $E_0$. This means that the threshold to see any pair production is about $m^2/e$ or approximately $1.3$ x $10^8$ V/m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your last question, a photon cannot turn into an electron because of charge conservation (and several other conserved numbers). It cannot turn into an e+e- pair in vacuum either because energy and momentum cannot be conserved simultaneously - this is a consequence of the masslessness of the photon. A photon &lt;em&gt;can&lt;/em&gt; however decay into e+e- in the presence of some other matter that it scatters off, which allows it to conserve both energy and momentum.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-03-16T21:58:59.753" LastActivityDate="2011-03-16T21:58:59.753" CommentCount="2" />
  <row Id="7062" PostTypeId="2" ParentId="7053" CreationDate="2011-03-16T23:52:29.927" Score="5" ViewCount="" Body="&lt;p&gt;Not that the other answers are wrong or anything, but here's something a different way to think about it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a particle is created, the process that creates the particle is (1) &lt;em&gt;local&lt;/em&gt;, meaning that the particle is created at the same location that the energy used to create it disappears from, and (2) subject to conservation laws for various quantites (generically called &quot;charges&quot;) like charge, total energy, etc. But potential energy is often a function of position and one of these conserved quantities. So, since particle-creating processes don't change either the location or the amount of conserved charge, the end products (including the newly created particle) will have the same amount of potential energy as the initial &quot;reactants.&quot; Therefore, there isn't any extra potential energy you can take to use for particle creation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it's certainly possible to come up with exceptions to this, as included in some of the other answers. It depends to some extent on what you consider a particle. (I think this argument works best in a pure QFT viewpoint where you consider the force carriers as proper particles, but then again in that case there is no such thing as potential energy anyway.)&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-03-16T23:52:29.927" />
  <row Id="7063" PostTypeId="1" CreationDate="2011-03-17T00:03:46.870" Score="5" ViewCount="243" Body="&lt;p&gt;I'm trying to understand what would really happen when large quantities (e.g., 10g) of anti-matter collide with matter. The normal response is that they'd annihilate each other and generate an expanding sphere of gamma ray photons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, thinking about it in more detail, what I see is that the anti-electrons annihilate first against the electrons. Let's assume the energy release in that case is not sufficient to noticeably change the momentum of the projectile. Then the nuclei penetrate the electron-annihilation plasma, and since antiprotons attract protons, their trajectory is changed. However, the nuclei are so small and so widely separated that presumably they just orbit each other as the electron clouds annihilate, and eventually enough energy is generated that the ionic plasma of nuclei and anti-nuclei just expands, with a small fraction of them actually ever combining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, we don't actually see total conversion happening -- only a small fraction of the total mass in an anti-matter/matter collision is turned into gamma rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that what actually would happen? (The ideal answer would be a video!)&lt;/p&gt;&#xA;" OwnerUserId="2595" LastActivityDate="2011-03-17T21:24:54.643" Title="What actually happens when an anti-matter projectile collides with matter?" Tags="&lt;particle-physics&gt;&lt;antimatter&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="7064" PostTypeId="2" ParentId="129" CreationDate="2011-03-17T00:57:49.980" Score="-2" ViewCount="" Body="&lt;p&gt;Every PhD scientist with publications on &lt;a href=&quot;http://en.wikipedia.org/wiki/Technicolor_(physics)#Walking_technicolor&quot; rel=&quot;nofollow&quot;&gt;walking technicolor&lt;/a&gt; (a Higgs boson free approach to many of the problems that the Higgs solves in the Standard Model) gets tenure and has a decent shot at a lateral transfer to a better job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can disprove an observable &quot;light Higgs,&quot; but until somebody came up with a better approach, I suspect that lots of HEP work would simply start calling the Higgs boson a &quot;ghost particle&quot; or &quot;virtual particle&quot; until they came up with something better, in much the same way that QED uses renormalization without having any really rigorous reason to believe that this process reflects a physical reality as opposed to simply functioning as a mathematical shortcut, or complex numbers of used in a wide variety of physics equations that just so happen not to produce any complex observables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, Brian Greene would probably rush to press with a book with a title like &quot;Ghost stories in physics&quot; explaining that and virtual particles and Schroeder's cat and sell lots of books and get an interview on the BBC to talk about it, while some poor post-doc will devise a theory to explain why we should have know that the Higgs was a ghost particle and not an observable one back in the 1970s which will gain general acceptance in the particle physics community.&lt;/p&gt;&#xA;" OwnerUserId="2592" LastActivityDate="2011-03-17T00:57:49.980" CommentCount="2" />
  <row Id="7065" PostTypeId="2" ParentId="1216" CreationDate="2011-03-17T01:23:43.700" Score="0" ViewCount="" Body="&lt;p&gt;In December, Carlo Rovelli summarized the last twenty years of the research agenda of a group of researchers in &lt;a href=&quot;http://arxiv.org/abs/1012.4707&quot; rel=&quot;nofollow&quot;&gt;Loop Quantum Gravity theory&lt;/a&gt;.  In a nutshell, LQG argues that gravity is a property of space-time rather than a quantum field theory mediated by a boson, and that space-time is fundamentally discrete with point-like locations in space-time connected by a network of connections to each other.  In this approach, described by three main equations, the number of dimensions in space-time itself is emergent and neither locality nor the number of dimensions of space-time are well defined concepts at the most fine grained level.  You are at point A and connected to points B, C and D, related by the equations, which when repeated ad infinitum are well approximated by a continuous, four dimensions space that may satisfy the properties of GR in classical approximation.  As he sums up the research agenda:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;There is substantial circumstantial evidence that the large distance limit of the theory is correctly general relativity, from asymptotic analysis and from large distance calculations of n-point functions and in spinfoam cosmology; and there are open directions of investigations to reinforce this evidence. The degrees of freedom are correct and the theory is generally covariant: the low-energy limit is not likely to be much else than general relativity. But there is no solid proof yet.&quot; &lt;/p&gt;&#xA;" OwnerUserId="2592" LastActivityDate="2011-03-17T01:23:43.700" />
  <row Id="7066" PostTypeId="1" CreationDate="2011-03-17T01:41:11.277" Score="3" ViewCount="172" Body="&lt;p&gt;I can't seem to find any info on connected rigid bodies by a joint.  Can someone explain the basics to me?  I'm trying to do a little research to find out how feasible it would be to implement 3d ragdoll physics for my first person shooter game.  &lt;/p&gt;&#xA;" OwnerUserId="2596" LastActivityDate="2011-06-01T23:38:54.450" Title="Rigid body dynamics joints" Tags="&lt;classical-mechanics&gt;&lt;particle-physics&gt;&lt;newtonian-mechanics&gt;&lt;kinematics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="7067" PostTypeId="1" CreationDate="2011-03-17T02:36:47.527" Score="4" ViewCount="145" Body="&lt;p&gt;A recent experimental paper measures a difference between the top quark and anti-top quark masses:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fermilab-Pub-11-062-E, CDF Collaboration, &lt;em&gt;Measurement of the mass difference between $t$ and $\bar{t}$ quarks&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We present a direct measurement of the&#xA;  mass difference between $t$ and&#xA;  $\bar{t}$ quarks using $t\bar{t}$&#xA;  candidate events in the lepton+jets&#xA;  channel, collected with the CDF II&#xA;  detector at Fermilab's 1.96 TeV&#xA;  Tevatron $p\bar{p}$ Collider. We make an&#xA;  event by event estimate of the mass&#xA;  difference to construct templates for&#xA;  top quark pair signal events and&#xA;  background events. The resulting mass&#xA;  difference distribution of data is&#xA;  compared to templates of signals and&#xA;  background using a maximum likelihood&#xA;  fit. From a sample corresponding to an&#xA;  integrated luminosity of 1/5.6 fb, we&#xA;  measure a mass difference,&#xA;  $\mathrm{M}_{t} -&#xD;&#xA;&amp;gt; \mathrm{M}_{\bar{t}}$ $= -3.3 \pm&#xD;&#xA;&amp;gt; 1.4(\textrm{stat}) \pm 1.0(\textrm{syst})$, approximately two standard deviations away from the CPT&#xA;  hypothesis of zero mass difference.&#xA;  This is the most precise measurement&#xA;  of a mass difference between $t$ and&#xA;  its $\bar{t}$ partner to date.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1103.2782&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1103.2782&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to pile on to the recent evidence showing differences between the masses of the neutrinos and anti-neutrinos. But unlike neutrinos, quarks can't be Majorana spinors. So what theoretical explanations for this are possible?&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-17T07:27:59.400" Title="CPT violation and how could quark masses differ from anti-quark masses?" Tags="&lt;particle-physics&gt;&lt;experimental-physics&gt;&lt;cpt-violation&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="7068" PostTypeId="1" AcceptedAnswerId="7072" CreationDate="2011-03-17T03:31:40.110" Score="2" ViewCount="194" Body="&lt;p&gt;Between 1947 and 1962 the US conducted 105 tests of nuclear weapons in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pacific_Proving_Grounds&quot; rel=&quot;nofollow&quot;&gt;&quot;Pacific Proving Grounds&quot;&lt;/a&gt;. I'm wondering how much radiation exposure resulted on the west coast of the US. These were part of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_weapons_and_the_United_States#Nuclear_testing&quot; rel=&quot;nofollow&quot;&gt;1056 nuclear bombs that the US has ignited&lt;/a&gt; over the years (most underground, but two notably in Japanese cities).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how much radiation exposure in the US was caused by the 105 nuke tests in the Pacific? Should the inhabitants of the West coast have taken iodine pills?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd also like to know how much fallout these bombs produced, as compared to the reactor steam releases in Japan, 2011.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-05-12T18:48:09.860" Title="How much radiation exposure in the US was caused by the 105 nuke tests in the Pacific?" Tags="&lt;nuclear-physics&gt;&lt;nuclear-engineering&gt;" AnswerCount="3" />
  <row Id="7069" PostTypeId="1" AcceptedAnswerId="7081" CreationDate="2011-03-17T04:12:45.080" Score="1" ViewCount="144" Body="&lt;p&gt;Suppose we perform a double-slit experiment with a detector placed at a position of minimum intensity (maximum destructive interference), off-center where the path lengths differ by half a wavelength.  The light source is alternately turned on and off (or blocked and unblocked near the source) and the intensity over time is recorded.  I interpret the uncertainty principle to mean that there will be a peak in intensity at the times when the switch is flipped (whether on-to-off or off-to-on).  i.e., it will look something like this (in ASCII art):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;&lt;/em&gt;___&lt;em&gt;&lt;/strong&gt;'-----'&lt;strong&gt;&lt;/em&gt;__&lt;em&gt;_&lt;/em&gt;&lt;/strong&gt;'-----'&lt;strong&gt;&lt;em&gt;_&lt;/em&gt;___&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?  I have had trouble convincingly explaining my reasons for thinking so.  What will be the measured intensity over time and why?&lt;/p&gt;&#xA;" OwnerUserId="2072" LastActivityDate="2011-03-17T10:33:45.120" Title="Double-slit experiment with alternating on-off switch" Tags="&lt;light&gt;&lt;double-slit-experiment&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="7070" PostTypeId="1" AcceptedAnswerId="7071" CreationDate="2011-03-17T04:27:18.713" Score="2" ViewCount="416" Body="&lt;p&gt;I am taking an intro level quantum mechanics class. Our textbook gives a problem like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The deuteron is a nucleus of &quot;heavy hydrogen&quot; consisting of one proton and one neutron. As a simple model for this nucleus, consider a single particle of mass $m$ moving in a fixed spherically -symmetric potential $V(r)$, defined by $V(r)=-V_{0}$ for $r&amp;lt;r_{0}$ and $V(r)=0$ for $r&amp;gt;r_{0}$. This is called a spherical square-well potential. Assume that the particle is in a bound state with $l=0$. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(a) Find the general solution $R(r)$ of the Schrodinger equation for $r&amp;lt;r_{0}$ and $r&amp;gt;r_{0}$. Use the fact that the wave equation must be finite at 0 and $\infty$ to simplify the solution as much as possible. (You do not need to normalize the solutions). &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(b) The deuteron is only just bound: ie., $E$ is nearly equal to $0$. Take $m$ to be the proton mass, $m=1.67*10^{-27}kg$. and take $r_{0}$ to be a typical nuclear radius, $r_{0}=1*10^{-15}m$. Find the value of $V_{0}$ (the depth of the potential well) in MeV (1 MeV=$1.6*10^{-13}$J (Hint: The continuity conditions at $r_{0}$ must be used. The radial wave function $R(r)$ and its dervivative $R&amp;#39;(r)$ must both be continuous at $r_{0}$; this is equivalent to requiring $u(r)$ and $u&amp;#39;(r)$ must both be continuous at $r_{0}$, where $u(r)=rR(r)$. The resulting equations cannot be solved explicitly but can be used to derive the value of $V_{0}$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I found I need to solve an equation of this form when $r&amp;lt;r_{0}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{\hbar^{2}}{2mr}\frac{\partial}{\partial {r}^{2}}(rR(r))-V_{0}R(r)=ER(r)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After expansion this becomes a second order ODE I am not able to solve. But I am not able to use the condition $r=0$ to obtain any good results. At $r=0$ we should have $-V_{0}R(r)=ER(r)$, which implies $E=-V_{0}$. (Is this allowed?). Then I can get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{\hbar^{2}}{2mr}\frac{\partial}{\partial {r}^{2}}(rR(r))=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;yet solving this implies $rR(r)=c_{0}r+c_{1}$, which I &quot;feel&quot; is impossible since it would imply $R(r)=c_{0}+\frac{c_{1}}{r}$, which has a singularity at $r=0$ unless $c_{1}=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand at $r=\infty$ we should have the equation become:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{\hbar^{2}}{2mr}\frac{\partial}{\partial {r}^{2}}(rR(r))=ER(r)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence naming $rR(r)=S(r)$, we should have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial}{\partial {r}^{2}}S(r)=kS(r), k=-\frac{2mE}{{\hbar}^{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and solve it we have $S(r)=e^{\sqrt{k}r}\Rightarrow R(r)=\frac{e^{\sqrt{k}r}}{r}$. This function is obviously non-constant at the origin and since $k$ is negative, it actually &quot;blow up&quot;. I do not know how to reconcile this desperate situation. So I need some help. Hence I cannot continue to part $(b)$ as well. I feel there is something very wrong in my approach. I also want to ask what does it mean &quot;the deuteron is only just bound&quot;, why this implies $E$ is nearly equal to $0$? Also, if I cannot solve the equation explicitly, how can I get $V_{0}$'s value? I am very confused. &lt;/p&gt;&#xA;" OwnerUserId="2597" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-23T18:19:38.830" LastActivityDate="2011-06-23T18:19:38.830" Title="How to solve this Schrödinger equation?" Tags="&lt;quantum-mechanics&gt;&lt;homework&gt;&lt;schrodinger-equation&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="7071" PostTypeId="2" ParentId="7070" CreationDate="2011-03-17T05:00:05.830" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;At $r=0$ we should have $−V_0 R(r)=ER(r)$, which implies $E=−V_0$. (Is this allowed?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nope, not allowed. In any case, that ODE isn't as bad as it looks. Change variables from $R(r)$ to $u(r)$ [which was defined in the problem as $u(r) = r R(r)$], and it will wind up looking very familiar. You'll come up with a second-order ODE that has two linearly independent solutions,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_1(r) = \cdots,\quad u_2(r) = \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can get the two independent solutions to the original Schrodinger equation as $R_i(r) = \frac{u_i(r)}{r}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final &lt;em&gt;physical&lt;/em&gt; solution for $R(r)$ needs to be well-defined at the origin. (In fact you realized this in the second part, but you don't need to worry about it there because $r &amp;gt; r_0$ doesn't include the origin; but you do need to worry about it here) So you need to pick a linear combination of $R_1(r)$ and $R_2(r)$ that is not infinite at $r = 0$. Hint: what needs to be the numerical value of $u(0)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other part, with $r &amp;gt; r_0$, is extremely similar. Again, remember that there are two linearly independent solutions. You only found one of them. Also, the solution you found doesn't blow up if $k$ is negative - but are you sure that $k$ is negative? What do you know about the value of $E$? (Specifically, what does it mean for the particle to be in a bound state?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't need to care about what the solution to the second part does at the origin, because the region you're solving the equation in doesn't include $r = 0$. But it does include $r \to \infty$, so you'll need to pick a linear combination of the solutions that stays finite in that limit.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-03-17T05:00:05.830" CommentCount="11" />
  <row Id="7072" PostTypeId="2" ParentId="7068" CreationDate="2011-03-17T05:02:19.770" Score="3" ViewCount="" Body="&lt;p&gt;It's difficult to compare becuase of different distance and time scales.&#xA;But Iodine 131 fallout from Nevada tests caused doses of 10mGray - 200mGray (1-20Rad) across the central and north eastern US, 10mGray is about 8 X-rays or 1 CT-scan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally bomb fallout contains more of the short lived highly active isotopes which are distributed further by the effects of the bomb. Reactor fires generally disperse fallout more slowly and so many isotopes have decayed before they reach the environment &lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-03-17T05:02:19.770" />
  <row Id="7073" PostTypeId="1" AcceptedAnswerId="7075" CreationDate="2011-03-17T05:37:23.270" Score="4" ViewCount="445" Body="&lt;p&gt;Special relativity was well established by the time the schrodinger equation came out. Using the correspondence of classical energy with frequency and momentum with wave number, this is the equation that comes out, and looks sensible because this is of the form of a wave equation, like the one for sound etc, except with an inhomogeneous term&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla^2 \psi - 1/c^2 \partial_{tt} \psi = (\frac{mc}{\hbar})^2\psi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, we have schrodinger's equation which reminds of the heat equation as it is first order in time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; Some thoughts after seeing the first posted answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is wrong with negative energy solutions? When we do freshman physics and solve some equation quadratic in, say, time... we reject the negative time solution as unphysical for our case. We do have a reason why we get them, like, for the same initial conditions and acceleration, this equation tells usthat given these iniial velocity and acceleration, the particle would have been at the place before we started our clock, since we're interested only in what happens &lt;em&gt;after&lt;/em&gt; the negative times don't concern us.  Another example is if we have areflected wave on a rope, we get these solutions of plane progressive waves travelling at opposite velocities unbounded by our wall and rope extent. We say there is a wall, an infinite energy barrier and whatever progressive wave is travelling beyond that is unphysical, not to be considered, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the same thing could be said about $E=\pm\sqrt{p^2c^2+(mc^2)^2}$ that negative solution can't be true because a particle can't have an energy lower than its rest mass energy $(mc^2)$ and reject that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have a divergent series, and because answers must be finite, we say.. ahh! These are not real numbers in the ordinary sense, they are &lt;em&gt;p-adics&lt;/em&gt;! And in this interpretation we get rid of divergence. IIRC Casimirt effect was the relevant phenomenon here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question boils down to this&lt;/strong&gt;. I guess the general perception is that mathematics is only as convenient as long as it gives us the answers for physical phenomenon. I feel this is sensible because nature can possibly be more absolute than any formal framework we can  construct to analyze it. How and when is it OK to sidestep maths and not miss out a crucial mystery in physics. &lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="2599" LastEditorDisplayName="" LastEditDate="2011-03-17T07:00:23.730" LastActivityDate="2011-03-17T13:19:14.453" Title="Why isn't the wave equation $\nabla^2 \psi - 1/c^2 \partial_{tt} \psi = (\frac{mc}{\hbar})^2\psi$" Tags="&lt;quantum-mechanics&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="2" />
  <row Id="7074" PostTypeId="2" ParentId="7067" CreationDate="2011-03-17T05:42:16.927" Score="5" ViewCount="" Body="&lt;p&gt;There is one simple, obvious, and almost certainly correct theoretical explanation: two-sigma effects show up all the time and, like most of them, this one is not real.&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2011-03-17T05:42:16.927" CommentCount="2" />
  <row Id="7075" PostTypeId="2" ParentId="7073" CreationDate="2011-03-17T05:46:47.927" Score="5" ViewCount="" Body="&lt;p&gt;That's the Klein-Gordon equation, which applies to scalar fields. For fermionic fields, the appropriate relativistic equation is the Dirac equation, but that was only discovered by Dirac years after Schr&amp;ouml;dinger discovered his nonrelativistic equation. The nonrelativistic Schr&amp;ouml;dinger equation is a lot easier to solve too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relativistic equations admit negative energy solutions. For fermions, that was only resolved by Dirac much later with his theory of the Dirac sea. For bosons, the issue was resolved by &quot;second quantization&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with negative energy solutions is the lack of stability. A positive energy electron can radiate photons, and decay into a negative energy state, if negative energy states do exist.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-03-17T08:21:46.257" LastActivityDate="2011-03-17T08:21:46.257" CommentCount="2" />
  <row Id="7076" PostTypeId="2" ParentId="7063" CreationDate="2011-03-17T06:15:55.073" Score="4" ViewCount="" Body="&lt;p&gt;I believe this is where your thought experiment goes wrong:&#xA;&lt;em&gt;However, the nuclei are so small and so widely separated that presumably they just orbit each other as the electron clouds annihilate&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why would they be widely separated?  Solids touch in order of nanometers. The protons and antiprotons would attract each other with the 1/r^2 electric field, which is very strong. Even if the energy of the projectile is small enough to form &quot;protoniums&quot;, ie. have antiprotons in the place of electrons in the quantum mechanical solution, the wave function will be so large that the protons antiproton wave function will overlap and annihilate. The lifetime is of the order of &lt;a href=&quot;http://en.wikipedia.org/wiki/Protonium&quot; rel=&quot;nofollow&quot;&gt;.1 to 10 microseconds.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the projectile has high enough energy, it will be a scattering with the cross section of antiproton on proton, depending on the area, to get the percentage annihilated. It is very difficult to get a plasma of protons and antiprotons under this condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the two bits of matter are in space with only their own gravity, then some will escape a first scatter and then fall back and annihilate or escape if they have escape velocity.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-17T14:50:26.200" LastActivityDate="2011-03-17T14:50:26.200" />
  <row Id="7077" PostTypeId="1" AcceptedAnswerId="7083" CreationDate="2011-03-17T06:38:49.367" Score="3" ViewCount="148" Body="&lt;p&gt;Sorry, I know y'all are probably getting a lot of questions re: Fukushima, but I had a very specific one and no-one has been able to answer it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am specialised in medical radiation, and have been explaining the risks to workers in this accident, but I need to know what are the rough proportions of alpha, beta (+/-) and gamma radiation flying around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have this right, most of the decays in the uranium cycle are alpha or beta+ and beta-. Some isomeric transitions (and obviously the beta+ decays) will release gammas. Obviously there will be a small amount of xrays too, from Bremsstrahlungs and such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what are the percentages of this in general uranium fuel material (I guess 'spent' fuel is the real problem here, not fuel undergoing chain reaction)? Is 50% of the radiation alphas? Is 80% gammas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It makes a huge difference from a biophysics perspective as alphas are easy to protect against (for the workers) but gammas are unshieldable. Beta- decays are in between, pretty easy to deal with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone know the uranium fuel cycle well enough to give a rough estimate here? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Added some thoughts in my response to Bars, who helped a lot. Any other thoughts that follow on from those insights?&lt;/p&gt;&#xA;" OwnerUserId="2600" LastEditorUserId="2600" LastEditorDisplayName="" LastEditDate="2011-03-17T10:45:59.743" LastActivityDate="2011-03-18T14:32:09.180" Title="Ratio of radiation sources in the uranium cycle" Tags="&lt;nuclear-physics&gt;&lt;radiation&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="7078" PostTypeId="2" ParentId="7067" CreationDate="2011-03-17T07:27:59.400" Score="1" ViewCount="" Body="&lt;p&gt;I suppose it will be a painful wait for the 5 sigma. Well, quarks are confined, so maybe it's OK for them to exhibit genuine CPT violation, whereas the neutral particles need renaming according to the mirror picture. But on the other hand, the quark braids have neutral strands, and if we mix (say for the proton) the uud, udu and duu sets, then there are enough neutral strands for component Majorana states.&lt;/p&gt;&#xA;" OwnerUserId="2601" LastActivityDate="2011-03-17T07:27:59.400" />
  <row Id="7079" PostTypeId="2" ParentId="7060" CreationDate="2011-03-17T07:31:49.177" Score="7" ViewCount="" Body="&lt;p&gt;Dear QEntanglement, the photons - e.g. cosmic microwave background photons - are increasing their wavelength proportionally to the linear expansion of the Universe, $a(t)$, and their energy correspondingly drops as $1/a(t)$. Where does the energy go? It just disappears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Energy is not conserved in cosmology.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much more generally, the total energy conservation law becomes either invalid or vacuous in general relativity unless one guarantees that physics occurs in an asymptotically flat - or another asymptotically static - Universe. That's because the energy conservation law arises from the time-translational symmetry, via Noether's theorem, and this symmetry is broken in generic situations in general relativity. See&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2010/08/why-and-how-energy-is-not-conserved-in.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/08/why-and-how-energy-is-not-conserved-in.html&lt;/a&gt;&lt;br&gt;&#xA;  &lt;strong&gt;Why energy is not conserved in cosmology&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cosmic inflation is the most extreme example - the energy density stays &lt;em&gt;constant&lt;/em&gt; (a version of the cosmological &lt;em&gt;constant&lt;/em&gt; with a very high value) but the total volume of the Universe exponentially grows, so the total energy exponentially grows, too. That's why Alan Guth, the main father of inflation, said that &quot;the Universe is the ultimate free lunch&quot;. This mechanism (inflation) able to produce exponentially huge masses in a reasonable time frame is the explanation why the mass of the visible Universe is so much greater than the Planck mass, a natural microscopic unit of mass.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-23T10:08:17.920" LastActivityDate="2011-03-23T10:08:17.920" CommentCount="4" />
  <row Id="7080" PostTypeId="2" ParentId="7077" CreationDate="2011-03-17T07:38:04.773" Score="1" ViewCount="" Body="&lt;p&gt;The thing you have to worry about most (unless you're close to Fukushima) is not the radiation by the reactor itself but the radioactive isotopes of Cs and Iodine, which when consumed expose body cells to direct and no longer shieldable radiation. That's why Iodine pills are distributed (I'm not a doctor, but one shouldn't take some unless instructed to) to saturate the Thyroid so the radioactive particles don't stay. &lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2011-03-17T07:38:04.773" CommentCount="5" />
  <row Id="7081" PostTypeId="2" ParentId="7069" CreationDate="2011-03-17T07:45:22.387" Score="0" ViewCount="" Body="&lt;p&gt;Dear Dan, first of all, you shouldn't use the term &quot;uncertainty principle&quot; if you're talking about &quot;light sources&quot; and light may be explained by ordinary - classical (non-quantum) - electrodynamics where no uncertainty principle applies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just an exercise in the propagation of waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, when you flip the switch, there may be temporary variations of the intensity, but they're not necessary, either. For example, you may find a minimum such that the number of wave peaks on the two trajectories (coming from the two slits) differs by 13.5 - one arm is 13.5 wavelengths longer than the other one. It will mean that the destructive interference only occurs when the beams from both slits are synchronized, and there will always be a period lasting about 13 periods after each flip of the switch when only one beam is coming to the detector. That will indeed eliminate the destructive interference, and give you the &quot;apostrophes&quot; in your ASCII art.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The precise shape of the graph depends on the character of the switches, geometry of the experiment, and other things.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-17T07:45:22.387" />
  <row Id="7082" PostTypeId="1" CreationDate="2011-03-17T08:51:19.857" Score="3" ViewCount="122" Body="&lt;p&gt;How did Nima Arkani-Hamed come up with the following nonzero space-like commutation relation in string theory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\langle \left| \left[ \hat{\mathcal{O}}(\vec{x}), \hat{\mathcal{O}}(\vec{y}) \right] \right|^2 \right\rangle \sim \exp \left( -\frac{k\left| \vec{x} - \vec{y} \right|^{D-2}}{8\pi G} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His explanation isn't too clear.&lt;/p&gt;&#xA;" OwnerUserId="2602" LastActivityDate="2011-03-17T11:43:42.653" Title="Nonzero space-like commutation relation in string theory?" Tags="&lt;string-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="7083" PostTypeId="2" ParentId="7077" CreationDate="2011-03-17T09:41:03.407" Score="3" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;Te-132 - beta &lt;/li&gt;&#xA;&lt;li&gt;I-132 - gamma &lt;/li&gt;&#xA;&lt;li&gt;I-131 - betta &amp;amp; gamma &lt;/li&gt;&#xA;&lt;li&gt;Zr-95 &amp;amp; Nb-95 - beta&lt;/li&gt;&#xA;&lt;li&gt;Ba-140 &amp;amp; La-140 - beta&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/bhM51.png&quot; alt=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/2/2c/AirDoseChernobylVector.svg/600px-AirDoseChernobylVector.svg.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's from wikipedia &amp;amp; &lt;a href=&quot;http://www.matpack.de/Info/Nuclear/Nuclids/&quot; rel=&quot;nofollow&quot;&gt;http://www.matpack.de/Info/Nuclear/Nuclids/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-03-17T09:41:03.407" CommentCount="5" />
  <row Id="7084" PostTypeId="2" ParentId="7073" CreationDate="2011-03-17T09:54:35.093" Score="1" ViewCount="" Body="&lt;p&gt;You can throw away the negative energy solution if you're just doing relativity and not QM, just like in the examples you mentioned. But if you add QM you can no longer do that: remember the energy spectrum are the eigenvalues of the Hamiltonian (operator), so you just can't throw away some eigenvalues without also throwing away the linear algebra the theory is based upon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So no, it is never ok to sidestep math. When you eliminate negative roots of cuadratic equations in classical physics you're not sidestepping math, after all you need to apply the rules of algebra to get the solutions, you're just applying physical considerations to the solutions to get rid of (physiccally) spurious ones that can be ignored without altering the mathematical structure of the theory. In the RQM case the negative energy solutions are unphysical but you can't just ignore them, you have to deal with them.&lt;/p&gt;&#xA;" OwnerUserId="2603" LastEditorUserId="2603" LastEditorDisplayName="" LastEditDate="2011-03-17T10:02:55.743" LastActivityDate="2011-03-17T10:02:55.743" />
  <row Id="7086" PostTypeId="1" AcceptedAnswerId="7403" CreationDate="2011-03-17T10:53:41.920" Score="2" ViewCount="118" Body="&lt;p&gt;In free space, the linear momentum density of an EM wave is given by the Poynting vector $\vec S$ over the speed of light squared,  $\vec g=\frac{\vec S}{c^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a medium, $S$ is generally not directed along the wave vector $\vec k$. If $v_g$ is the group velocity and $W$ is the density of EM wave energy, then $S = v_g W$. On the other hand, it is often assumed that the momentum of a photon is still $\hbar\vec k$. What is the macroscopic momentum density of EM wave in an anisotropic medium such as magnetized plasma ?&lt;/p&gt;&#xA;" OwnerUserId="2560" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-03-17T13:22:05.850" LastActivityDate="2011-03-22T07:53:31.843" Title="What is the linear momentum of an EM wave in a medium?" Tags="&lt;electromagnetic-radiation&gt;&lt;momentum&gt;&lt;dielectric&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7087" PostTypeId="2" ParentId="7066" CreationDate="2011-03-17T10:56:41.810" Score="2" ViewCount="" Body="&lt;p&gt;As a start, look here: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Kinematic_chain&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Kinematic_chain&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is traditionally a part of machine construction business. &#xA;The German page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://de.wikipedia.org/wiki/Getriebelehre&quot; rel=&quot;nofollow&quot;&gt;http://de.wikipedia.org/wiki/Getriebelehre&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is more detailled and contains some reference to computer animation.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-17T10:56:41.810" />
  <row Id="7088" PostTypeId="2" ParentId="6220" CreationDate="2011-03-17T10:59:21.043" Score="0" ViewCount="" Body="&lt;p&gt;I am not sure whether this is primarily a &quot;car&quot; question or a &quot;physics&quot; question. As a physics question there are a few points to note: the fact that the roads are the same length and that one is hilly means that the flat road is necessarily curved between the two towns. Also the time to travel is not relevant here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the flat drive must use petrol/gas for the entire journey between the two towns. However the hilly route only needs to use petrol/gas for the uphill portions: it can glide downhill. So the hilly route will use less fuel if the following equation gets satisfied:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fuel used uphill &amp;lt; Fuel used on Flat (at any speed)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: (With more explanation of the physics.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physics of this situation involves the use of &lt;em&gt;internal power&lt;/em&gt; with which the elementary mechanics examples are misleading. Elementary mechanics examples consider objects moving in frictionless planes. In order to set such an object moving it requires some impulse I for a short time to reach a velocity $v$. Thereafter the object can travel from A to B without further energy. Indeed it can travel to any C (linear with AB) without any further energy. This fact, together with the elementary &quot;conservation of energy&quot; equation in the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;E = T + V&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the basis of both elementary mechanics examples and also fundamental physics examples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is different about the &quot;internal power&quot; scenario is that, for reasons to be discussed below, movement from A to B requires energy. Furthermore movement from A to C will require a different amount of energy in general. In elementary physics examples this energy comes from the external potential V, but in &quot;internal power&quot; systems it comes from some compact fuel. For these reasons, although all the basic laws of physics still apply, simplistic applications of &quot;conservation of mechanical energy&quot; do not apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why movement from A to B requires energy is that the movement is not over a frictionless plane, but over a friction bearing road surface. The friction contributions are from:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Tyre against road - probably a constant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Internal friction in vehicle - proportional to $v$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) Air resistance - proportional to $v^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For these reasons a constant use of fuel is required to continue motion; with no additional use of fuel the vehicle will stop (on a flat road).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the equation has to take account of the amount of fuel used (=amount of energy used), and some approximations need to be made:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if a journey from A to B uses F units of fuel, a journey of twice the length uses 2F fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution proposed here uses 2F fuel for the flat journey and for the uphill journey will use: F + mgH units of fuel. This needs to be less than 2F.  However there is still energy transfer in the downhill part resulting from the mgH potential energy transfered to kinetic energy: but this does not use fuel/ internal power in the extreme case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently I have just seen a TV programme showing &quot;gravity racer&quot; cars: no fuel on the downhill and speeds of 40mph+ were reached without much braking - lots of energy transfer of course.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-03-21T22:51:31.337" LastActivityDate="2011-03-21T22:51:31.337" CommentCount="5" />
  <row Id="7089" PostTypeId="1" AcceptedAnswerId="7504" CreationDate="2011-03-17T11:26:34.453" Score="3" ViewCount="183" Body="&lt;h3&gt;Cost per Power Capacity&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;For a number of energy technologies I'd like to know what the minimum costs are to install a given power capacity.  Are there any such comparisons available?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Research Spendings&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Technologies often own much to publicly funded research efforts so I'm also interested in estimations regarding development cost of the technologies themselves.  Of course foundations to a given technology may have been readily available, so let's limit this to spendings on larger research projects &lt;em&gt;dedicated&lt;/em&gt; to developing a technology and motivated by the need of energy.  I'm aware it's unlikely any exhaustive comparison in full detail exists but maybe something related?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Maybe someone should tag this question 'energy-economics' -- I'm lacking the reputation)&lt;/p&gt;&#xA;" OwnerUserId="2415" LastEditorUserId="2415" LastEditorDisplayName="" LastEditDate="2011-03-18T08:42:03.220" LastActivityDate="2011-03-24T12:56:45.570" Title="Comparing Energy Technologies in terms of: Cost per Power Capacity  and Research Spendings" Tags="&lt;renewable-energy&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="7091" PostTypeId="2" ParentId="7082" CreationDate="2011-03-17T11:43:42.653" Score="2" ViewCount="" Body="&lt;p&gt;This is a cool estimate of the inherent non-locality in quantum gravity but one shouldn't over-interpret it. It is just an estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vacuum expectation value is inserted only to turn the operator into a number. The squaring is only inserted to make it positive and to avoid cancellations; if you omitted the squaring, the order-of-magnitude estimate would only differ by having the constant $k$ from the right hand side divided by two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the formula applies to a hardcore quantum gravity regime - because it depends on $G$, i.e. Newton's constant, rather than $L_{string}$ or something similar that is more specific for perturbative string theory or any other &quot;more well-defined&quot; stringy vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, why it's at least morally correct? Well, it could be justified by a quantum-foam-like calculation. Perturbatively, the commutator vanishes for spacelike separations, by causality. Non-perturbatively, there are instanton-like contributions. One may imagine that some energy sources are inserted into points $\vec x,\vec y$ that allow one to write down a solution - a configuration of general relativity - with things similar to deficit angles at these two points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The typical size of this solution is $|\vec x - \vec y|$, and the Einstein-Hilbert action is equal, according to dimensional analysis, to&#xA;$$ \frac{1}{16 \pi G} |\vec x - \vec y|^D / |\vec x - \vec y|^2 $$&#xA;So the commutator may get a contribution from this gravitational instanton whose action is given by something like the expression above - I wrote $1/16\pi G$ to emphasize that it comes from the gravitational action. The $D$-th power comes from the volume integral over the &quot;body&quot; of the instanton and the second power is from the derivatives in the Ricci scalar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see that my displayed equation is identical to the exponent in your Ansatz. My expression has to be taken with the minus sign and exponentiated, because it's an action - $\exp(-S)$ appears in non-perturbative contributions in the Euclidean path integral calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not 100% certain whether his result or my justification of it is correct in any calculable sense in string theory. If Nima has a more rigorous calculation, then I apologize, but my guess is that it won't differ too much.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-17T11:43:42.653" CommentCount="1" />
  <row Id="7092" PostTypeId="2" ParentId="7027" CreationDate="2011-03-17T12:31:09.997" Score="1" ViewCount="" Body="&lt;p&gt;I suggest you try looking at &lt;a href=&quot;http://en.wikipedia.org/wiki/Periodogram&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Periodogram&lt;/a&gt;. What you're calling the PSD is in fact a &lt;em&gt;periodogram&lt;/em&gt;. What the Wikipedia page I cite above explains is that the periodogram is in general not a good estimator for the PSD, even though, from &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectral_density&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Spectral_density&lt;/a&gt;, &quot;The ensemble average of the average periodogram when the averaging time interval $T\rightarrow\infty$ can be proved (Brown &amp;amp; Hwang) to approach the Power Spectral Density (PSD).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you're asking about the properties of short time period Periodograms, not about the properties of the PSD. PSD is defined as an integral over infinite time. You're probably also talking about a discrete series, not an integral. The Wikipedia Periodogram page also has this to say,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The spectral bias problem arises from a sharp truncation of the sequence,&#xA;  and can be reduced by first multiplying the finite sequence by a&#xA;  window function which truncates the sequence gradually rather than&#xA;  abruptly.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The variance problem can be reduced by smoothing the periodogram. Various&#xA;  techniques to reduce spectral bias and variance are the subject of spectral&#xA;  estimation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not a specialist in signal processing (I attempted to add a signal-processing tag, which with luck might lead someone here who can answer more economically), so this is not a complete Answer to your detailed Question. The word &quot;Periodogram&quot; will lead to more information and towards working out the answer yourself.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-03-17T12:31:09.997" />
  <row Id="7093" PostTypeId="2" ParentId="7068" CreationDate="2011-03-17T12:34:16.830" Score="1" ViewCount="" Body="&lt;p&gt;The steam release disperses the most volatile fission products selectively. &#xA;That is why Iodine and Cesium was talked about mostly when this happened. &#xA;This elements (ie. compound of that elements)  have a certain vapour pressure &#xA;at the steam temperatures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They released the steam into the space between the reactor pressure container and &#xA;the outer wall (sheet metal, deliberately to work as a kind of rupture seal?). &#xA;Maybe they hoped that some of this substances would condense there when the steam cools down?&#xA;But because there was a lot of Hydrogen in the steam, explosions blew away the sheet metal. &#xA;A German reacor safety expert in news said that in a similar case in a German reactor the release would be done through a vent tower.  &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-17T12:34:16.830" />
  <row Id="7094" PostTypeId="2" ParentId="7060" CreationDate="2011-03-17T12:37:41.073" Score="1" ViewCount="" Body="&lt;p&gt;The Friedman-Lemaitre-Robertson-Walker (FLRW) equations can be derived in an elementary way from Newton’s laws, where the energy in the motion of a density of mass-energy in a region is determined by the gravitational potential.  The FLRW energy (so called energy) equation for the evolution of a scale parameter of spatial distance $a$ derived from the metric is,&#xA;$$&#xD;&#xA;\Big(\frac{\dot a}{a}\Big)^2~=~\frac{8\pi}{3}\rho~-~\frac{k}{a^2}&#xD;&#xA;$$&#xA;where $\rho$ is the energy density.  The Hubble parameter or constant with space at each time is $H~=~{\dot a}/a$. We set $k~=~0$ for a flat space $R^3$ to match observations, and which recovers what is derived from Newton’s laws.  Energy density for photons scales inversely with the length of the box.  The box is thought of as a resonance cavity that is equivalent to a situation where the number of photons that leave is approximately equal to the number of photons that enter.  During the radiation dominated period things were in a near equilibrium, so this is not out of line with some physical reasoning.  In a stat-mech course an elementary problem of N-photons in a box uses the same logic, the energy of the photons scales inversely with the size of the box.  So the energy of photons $E~=~hc/\lambda$, and the wave length scales with the scale factor a.  So the density scales as $\rho~\sim~hc/a^4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with this et up let us propose a time dependency on the scale factor a with time $a~\sim~t^n$.  Put this into the &quot;energy equation&quot; and turn the crank and you find that $n~=~1/2$.  The scale factor grows as the square root of time.  This is an energy equation, and the balance tells us that the loss of energy in photons is equal to the gain in gravitational potential energy. This connects well with Newtonian analysis and the Pound-Rebka experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may continue further, for the photons in a box exert a pressure on the sides of the box $p~=~F/a^2$, and the force induces an increment of change in the size of the box $dE~=~Fdx$.  The force is distributed on 3 different directions and so $p~=~ρ/3$.  This may then be used in the equation $pV~=~NkT$ to find that for $p~\sim~a^{-4}$ and $V~\sim~a^3$ with the above $E~\sim~1/\lambda$ that $\lambda~\sim~1/T$, which is Wein's law for the wavelength as the peak of the blackbody curve.  The proportionality of the energy density with scale factor and temperature also gives $E~\sim~T^4$.  So this physics is remarkably in line with laboratory understanding of the basic thermodynamics of radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matter contribution scales as $a^{-3}$, which was smaller than the radiation contribution for a time.  Around 380,000 years into the evolution of the universe the matter density surpassed the radiation density.  The CMB demarks this transition in the mass-energy which dominated the universe. The above dynamics still apply for photons, but radiation is now a minor player in the spacetime structure of the universe.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-03-17T12:45:27.333" LastActivityDate="2011-03-17T12:45:27.333" CommentCount="1" />
  <row Id="7095" PostTypeId="1" AcceptedAnswerId="7144" CreationDate="2011-03-17T12:46:55.183" Score="1" ViewCount="153" Body="&lt;p&gt;I am not sure that this notation is specific to supersymmetry theories but I ran into this while studying that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see people talking of component fields of a chiral superfield as $\phi$ and $\bar{\psi}$ in the adjoint representation of some gauge group. (One sometimes also seems to use the notation of $\bar{\psi}_+$ to denote which component of the spinor is being picked up) One also denotes the covariant derivative operator as $D$ (which is sometimes written as $D_{++}$ to denote which component of it is being chosen..but I don't understand this notation) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now one talks of &quot;single trace operators&quot; of the kind $Tr[\phi \bar{\psi}]$, $Tr[\phi \bar{\psi}^2]$, $Tr[\phi^3D \bar{\psi}^4]$ and stuff like that..basically taking arbitrary combinations of powers of the scalar and the fermion and then taking a &quot;gauge trace&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lot of things that I don't understand.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I am puzzled by the notation of taking powers of a fermionic field. How does one define powers of a spinor? (What is square of a fermion?) Also here in components $\psi$ would have a decomposition like $\psi = \psi_A t_A$ where $\psi_A$ are fermions and $t_A$ runs over the generators of the Lie algebra of a gauge group in the adjoint representation. I guess one is taking a tensor product between the the spinor $\psi_A$ and the matrix $t_A$. But then I don't understand what is a square or any other power of that tensor?  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I am told that $\bar{\psi}^2 \neq 0$ but $Tr[\bar{\psi}^2] = 0$ I don't understand what is this supposed to mean. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;One way I can think of the powers could be that $\bar{\psi}$ are after quantization operators on the Hilbert space of the theory and these are powers of that Hilbert space operator. But in this way of thinking I am confused how to intprete the trace over the gauge indices. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In this language one wants to think of the supersymmetry operators $Q$ to act on the fields in the following way, &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$Q\phi=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q\bar{\psi} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q(DO) = [[\phi,\bar{\psi}],O\} + D(QO)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $O$ is some operator and in the first term on the RHS of the last of the above equations the symbol $[,\}$ means that one will take the commutator or the anticommutator depending on whether $O$ is bosonic or fermionic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above looks like a very different way of thinking about supersymmetry transformations than the language I am familiar with from the books like that of Weinberg where the action of $Q$ on $\phi$ and $\psi$ is through commutator and anti-commutator respectively or one thinks of infinitesimal supersymmetry transformations as $\delta \phi$ etc. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Definitely the last of the above list of commutators is totally unfamiliar to me!  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What the above notion confuses is how I am supposed to think of the action of $Q$ on composite operators like say $Tr[\phi\bar{\psi}^3]$. Apparently one is supposed to think of this as a fermionic operator since the fermion is being raised to an odd power. (I can't see the full argument here!)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then the action of $Q$ is supposed to be (dropping the overall trace),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q(\phi \bar{\psi}^3) = (Q\phi)\bar{\psi}^3 + \phi (Q \bar{\psi})\bar{\psi}^2 - \phi \bar{\psi}(Q \bar{\psi})\bar{\psi} + \phi \bar{\psi}^2(Q \bar{\psi})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The signs alternate depending on how many $\bar{\psi}$s has the $Q$ skipped over. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I would like to know of explanations for the above way of writing powers of fermionic fields and doing supersymmetry transformations on them. I would be glad if I am directed to some reference which explains the above way of thinking which I haven't come across anywhere else in any book.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;{Eventually one interested in calculating the &quot;cohomology of $Q$&quot; on the space of all such single trace operators. (for some reason not clear to me people want to drop from this list of operators which are total covariant derivatives) This depends on the theory one is looking it by what the auxiliary field integrates to and appearing on the RHS. I would like to know references along that too and why this is &quot;cohomology&quot; and why is this calculated.} &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1637" LastEditorUserId="1637" LastEditorDisplayName="" LastEditDate="2011-03-17T12:51:58.970" LastActivityDate="2011-03-18T06:10:07.823" Title="A particular notation about fermions" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;supersymmetry&gt;&lt;fermions&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="7096" PostTypeId="2" ParentId="7063" CreationDate="2011-03-17T13:10:36.647" Score="1" ViewCount="" Body="&lt;p&gt;As an extreme case, one can imagine disappearance of matter and antimatter without noise because the resulting gamma quanta are very penetrating: they will fly away without heating matter in the vicinity. Only chemical reaction heat might be released/consumed.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-03-17T21:24:54.643" LastActivityDate="2011-03-17T21:24:54.643" />
  <row Id="7097" PostTypeId="2" ParentId="7073" CreationDate="2011-03-17T13:19:14.453" Score="4" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;My question boils down to this. I&#xA;  guess the general perception is that&#xA;  mathematics is only as convenient as&#xA;  long as it gives us the answers for&#xA;  physical phenomenon. I feel this is&#xA;  sensible because nature can possibly&#xA;  be more absolute than any formal&#xA;  framework we can construct to analyze&#xA;  it. How and when is it OK to sidestep&#xA;  maths and not miss out a crucial&#xA;  mystery in physics.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your question just took a huge leap sideways. Probably into Soft-question. You don't &quot;sidestep maths&quot; in Physics. You introduce &quot;&lt;em&gt;different&lt;/em&gt; math&quot; and look carefully to see whether it fits the experimental data better. People generally pick a particular mathematical structure and try to characterize its differences from the existing best theories, which for the best alternatives takes people decades. There are other issues, such as whether your different maths is more tractable, whether people think it's beautiful, whether it gives a better explanation, whether it suggests other interesting maths, whether it suggests interesting engineering, whether it's simple enough for it to be used for engineering. The wish-list for a successful theory in Physics is quite long and not very articulated. Philosophy of Physics tries to say something about the process and the requirements for theory acceptance, which I've found it helpful to read but ultimately rather unsatisfying. &quot;miss[ing] out a crucial mystery in physics&quot; would be bad, but it's arguably the case that if it's not Mathematics it's not Physics, which in the end of hard practical use will be because if it's not mathematics you'll be hard pressed to do serious quantitative engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your original question, I've been pursuing why negative frequency components are so bad for about 5 years. If you feel like wasting your time, by all means look at my published papers (you can find them through my SE links) and at my Questions on SE, all of which revolve around negative frequency components (although I think you won't see very clearly why they do in many cases, even if you know more about QFT than your Question reveals). I don't recommend it. I can't at this stage of my research give you a concise Answer to your original Question.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-03-17T13:19:14.453" />
  <row Id="7098" PostTypeId="2" ParentId="7089" CreationDate="2011-03-17T13:28:31.810" Score="3" ViewCount="" Body="&lt;p&gt;Such estimations exist. The big companys in energy, or contractors for power plants or consultants and &quot;green research&quot; do that since long. (There are consultants working  to a lot more of technologies than power plant investment, I remember calculation of rentability of a portlant cement plant as an example in &quot;chemical technology&quot; when I studied chemistry ) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem is, that You won't get such figures free (Exept maybe from Green institutions, but those are often biased, as they blame the others to be :=)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Big problem are the development costs in some cases, when those were partly payed for by state. This is often the case for atomic power plants, because the development often was entangeled with bomb developent. Some reactors (like Tschernobyl) worked (work?) to produce bomb material, energy was(is) a byproduct. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-17T13:35:03.070" LastActivityDate="2011-03-17T13:35:03.070" CommentCount="6" />
  <row Id="7099" PostTypeId="2" ParentId="7060" CreationDate="2011-03-17T13:38:09.360" Score="5" ViewCount="" Body="&lt;p&gt;Other answers have covered the key points correctly, but I'll jump in too, maybe emphasizing a slightly different angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not just that energy is not conserved -- even &lt;em&gt;defining&lt;/em&gt; the total energy of the Universe (or even the total energy in any reasonably large volume) is problematic and, in some sense, unnatural.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What people usually have in mind when they talk about the total energy of the Universe (or a large volume -- from now on I'll stop writing that) is something like the following: Figure out the energy of each particle in that volume, and add 'em up. That's a sensible procedure for figuring out total energy in other contexts: it works great if you want to talk about the energy in all of the air molecules in this room. But it only works if all of the individual energies are determined in &lt;em&gt;the same inertial reference frame&lt;/em&gt;. And in the expanding Universe (or any curved spacetime), there are no inertial reference frames that cover the whole region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When people worry about energy non-conservation as applied to CMB photons, what they're implicitly doing is calculating each photon's energy in the local comoving reference frame (the one that's &quot;at rest&quot; with respect to the expansion). But all of the different comoving frames are in motion with respect to each other, so it's &quot;illegal&quot; to add up those energies and call the result a total energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of a Newtonian analogy: if one person measures the kinetic energy of something on board a moving airplane, and another person measures the kinetic energy of a different object on the ground, you can't add them up to get a total energy. And certainly the sum of those two things won't be a conserved quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to be clear: I know that there are a bunch of contexts (e.g., asymptotically flat spacetimes) in which it does make sense to talk about energy conservation in various forms. But in this specific context, I think that the above is the essence of the issue.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-17T13:38:09.360" />
  <row Id="7100" PostTypeId="2" ParentId="7014" CreationDate="2011-03-17T15:19:06.233" Score="-1" ViewCount="" Body="&lt;p&gt;According to the equivalence principle, a point-like charge in free fall will not radiate EM waves, because its movement its locally equivalent to any other inertial frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;point-like masses will also not radiate gravitational waves in free-fall by the same reason; gravitational wave emission is a function of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_wave#Sources_of_gravitational_waves&quot; rel=&quot;nofollow&quot;&gt;third time derivative of the quadrupole mass&lt;/a&gt; moment; this basically means that only extended, non-symmetric masses will give such radiation (being extended means that significant parts of the mass will NOT be in a inertial free-fall frame)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The only other macroscopic force outside gravity (and the force that is not a force called exclusion principle) is the electromagnetic field. I don't think we have any evidence how accelerated charges radiate in other non-EM forces (i.e: weak and strong) but theoretically acceleration induced on hypotethical macroscopic fields of such should produce the same effect&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-03-17T21:18:07.783" LastActivityDate="2011-03-17T21:18:07.783" CommentCount="9" />
  <row Id="7101" PostTypeId="1" AcceptedAnswerId="7102" CreationDate="2011-03-17T15:19:54.147" Score="3" ViewCount="200" Body="&lt;p&gt;Assume that alpha &amp;amp; beta particles and gamma photons each reach skin at the same energy. It's known that they penetrate most deeply in order by gamma, beta and alpha. How would they compare in terms of total biological impact? In other words, which would be most harmful?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-03-17T22:43:40.470" Title="Alpha Beta Gamma Biological Impact" Tags="&lt;radioactivity&gt;" AnswerCount="4" CommentCount="0" />
  <row Id="7102" PostTypeId="2" ParentId="7101" CreationDate="2011-03-17T15:54:44.283" Score="2" ViewCount="" Body="&lt;p&gt;Gamma is the most harmful - and it's hard to shield it. Alpha get absorbed in dead skin layer. Beta's are in between.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-03-17T15:54:44.283" CommentCount="4" />
  <row Id="7103" PostTypeId="2" ParentId="6541" CreationDate="2011-03-17T16:12:33.567" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://arxiv.org/ftp/gr-qc/papers/9909/9909037.pdf&quot; rel=&quot;nofollow&quot;&gt;Polarizable vacuum representation of general relativity&lt;/a&gt; by Puthoff, 1999&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Abstract:&lt;/strong&gt;&#xA; Standard pedagogy treats topics in general relativity (GR) in terms of tensor &#xA;formulations in curved space-time.  Although mathematically straightforward, &#xA;the curved space-time approach can seem abstruse to beginning students due &#xA;to the degree of mathematical sophistication required.  As a heuristic tool to &#xA;provide insight into what is meant by a curved metric, we present a &#xA;polarizable-vacuum (PV) representation of GR derived from a model by Dicke &#xA;and related to the  &quot;THεµ&quot;  formalism used in comparative studies of &#xA;gravitational theories  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;pag 11 : &lt;strong&gt;Comparison of the Metric Tensors in the GR and PV Approaches&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Having shown by specific calculation that the&#xA;  PV approach to the three classical&#xA;  tests of GR reproduces the traditional&#xA;  GR results&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(the gravitational redshift, the bending of light and the advance of the perihelion of Mercury)&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-03-17T16:12:33.567" />
  <row Id="7104" PostTypeId="1" CreationDate="2011-03-17T16:24:14.473" Score="0" ViewCount="246" Body="&lt;p&gt;1) Why is a hot air balloon stiff? &lt;br /&gt;&#xA;2) Is the pressure inside the balloon higher than the pressure outside (atmospheric pressure)? &lt;br /&gt;&#xA;3) If the pressure inside is higher than the outside, how is it explained by ideal gas law?&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-17T17:02:50.813" LastActivityDate="2011-03-22T07:42:40.033" Title="Why is a hot air balloon  &quot;stiff&quot;?" Tags="&lt;thermodynamics&gt;&lt;kinetic-theory&gt;" AnswerCount="3" />
  <row Id="7106" PostTypeId="2" ParentId="7101" CreationDate="2011-03-17T16:46:38.783" Score="6" ViewCount="" Body="&lt;p&gt;The thing you're trying to get at is the &quot;quality factor&quot; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Sievert#Q_values&quot; rel=&quot;nofollow&quot;&gt;&quot;Q value&quot;&lt;/a&gt; of the radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short short version&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;photons, betas, muons: 1&lt;/li&gt;&#xA;&lt;li&gt;protons, pions, other light mesons: 2&lt;/li&gt;&#xA;&lt;li&gt;neutrons: 1--10 depending on energy&lt;/li&gt;&#xA;&lt;li&gt;alphas and fission fragments: 20&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that alpha have a very high value, but it doesn't count starting from the surface of the skin, as at the usual energies they will be fully stopped before entering live tissue.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-03-17T16:46:38.783" CommentCount="4" />
  <row Id="7107" PostTypeId="2" ParentId="7104" CreationDate="2011-03-17T17:12:41.400" Score="2" ViewCount="" Body="&lt;p&gt;The &quot;hot air&quot; which is the hot flue gas of a flame really (a mixture of nitrogen, water vapour and carbon dioxide) has less density than the surrounding air. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to a situation like water in a bucket, but mirroized upward down:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure at the lower end of that Mongolfiere inside is identical to the pressure of air outside. (necessary condition to be isostatic) The pressure of the lighter column in the ballon higher up is higher, just for the same reason as the pressure of the water is higher down the level from the head. In the old days of &quot;town gas&quot; (hydrogen and carbon monoxide) &#xA;everyone knew that the gas had a higher pressure in upper floors, just reverse than water. &#xA;The natural gas today does not show this effect, because the pressure is much higher, so You do not realize the small difference. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-17T17:12:41.400" CommentCount="2" />
  <row Id="7109" PostTypeId="2" ParentId="6928" CreationDate="2011-03-17T17:50:42.837" Score="1" ViewCount="" Body="&lt;p&gt;The question amounts to whether this thing can be buried away.  Yes, and that will have to happen, and the sooner the better.  The complex of 4 reactors is what is called in the military a cluster f**k --- a total loss and disaster.  Gorbachev in the wave of the Chernobyl meltdown ordered the military to fly helicopters, and servicemen who died of radiation sickness, to the reactor and bury the thing in neutron moderating boron salts, sand and concrete.  The current situation is a slow motion Chernobyl situation, and the “Chernobyl option” seems to be the best shot at stemming this unfolding disaster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molten lead is not a particularly good idea of course.  The radionuclides would dissolve in the liquid lead, and the hot liquid lead would be a source for fires.  Pouring molten lead would make the situation worse.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-17T17:50:42.837" />
  <row Id="7110" PostTypeId="1" AcceptedAnswerId="7113" CreationDate="2011-03-17T17:50:52.800" Score="1" ViewCount="353" Body="&lt;p&gt;As pointed out before, it orginated inside the reactor vessel but exploded outside: how did it went there?  Did it pass the shell by diffusion or was it released by the engineers?&lt;/p&gt;&#xA;" OwnerUserId="2415" LastEditorUserId="2415" LastEditorDisplayName="" LastEditDate="2011-03-17T18:08:27.713" LastActivityDate="2011-03-17T18:56:39.703" Title="Why did the Hydrogen in Fukushima explode *outside* the reactor vessel?" Tags="&lt;nuclear-engineering&gt;" AnswerCount="1" />
  <row Id="7111" PostTypeId="1" CreationDate="2011-03-17T17:53:22.487" Score="2" ViewCount="141" Body="&lt;p&gt;How do we calculate the effective mass of an electron in an Aluminium lattice? Is there any simple analytical way to work it out?&lt;/p&gt;&#xA;" OwnerUserId="304" LastActivityDate="2011-03-17T17:53:22.487" Title="Effective mass in Aluminium lattice?" Tags="&lt;solid-state-physics&gt;" CommentCount="4" />
  <row Id="7112" PostTypeId="1" CreationDate="2011-03-17T17:58:59.970" Score="8" ViewCount="861" Body="&lt;p&gt;I was reading the Feynman lectures in physics and after thinking about it for a while it seems particularly unreasonable to talk about hidden variables. Let us say that the electron has some internal variables as yet unknown which determine its trajectory given a set of initial conditions just like in classical mechanics. But since these hidden variables are unobserved, coupling it with a classical system should make their effect unchanged. This is what Feynman says, I think, in the last paragraph of Ch1 Vol 3, that if in the double slit experiment, if these inner variables dictate that the electron goes through the upper slit and land at a particular place on the opposite screen, and some other place for the lower screen, then the probability must neccesarily be the sum of two Gaussian like peaks, which does not agree with experiment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I concluded that inner workings of an electron had some additional hidden variables, then it should yield, as they should be independent of the classical apparatus, mutually exclusive probabilities that do not quiet add up the way as observed. But then I do a hidden variables search on the archive and a lot of smart guys still write about it, as late as Feb 2011.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the argument I have used might be somehow incomplete, can anyone explain how?&lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="353" LastEditorDisplayName="" LastEditDate="2011-03-18T03:10:26.440" LastActivityDate="2011-05-05T07:49:34.283" ClosedDate="2011-03-19T17:46:43.960" Title="Why do people still talk about bohmian mechanics/hidden variables" Tags="&lt;quantum-mechanics&gt;&lt;quantum-interpretations&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="4" />
  <row Id="7113" PostTypeId="2" ParentId="7110" CreationDate="2011-03-17T18:02:47.440" Score="1" ViewCount="" Body="&lt;p&gt;They released the steam into the space between the reactor pressure container and the outer wall (sheet metal, deliberately to work as a kind of rupture seal?). Maybe they hoped that some of this substances would condense there when the steam cools down? But because there was a lot of Hydrogen in the steam, explosions blew away the sheet metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for the ignition of the hydogen/air mixture is not known. Maybe the temperature of some hot parts of the pressure vessel did that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; &#xA;Inside the reactor vessel the hydrogen cannot explode, there is no oxygen available. &#xA;And:&#xA;All this was dealt already in two treads. Search for Fukushima. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-17T18:56:39.703" LastActivityDate="2011-03-17T18:56:39.703" CommentCount="6" />
  <row Id="7114" PostTypeId="2" ParentId="7112" CreationDate="2011-03-17T18:17:25.040" Score="3" ViewCount="" Body="&lt;p&gt;When we talk about ruling out hidden variable theories, we basically mean non-contextual or local hidden variables, whereas Bohmian mechanics for instance is highly non-local and contextual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the details of contextuality see &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0406166&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/quant-ph/0406166&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or if you're particularly interested in Bohmian mechanics can watch these lectures:&#xA;&lt;a href=&quot;http://pirsa.org/C11001&quot; rel=&quot;nofollow&quot;&gt;http://pirsa.org/C11001&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally think Bohmian mechanics isn't the nicest formalism you can get, but as a physicist, I wouldn't reject a theory only on the aesthetics grounds!&lt;/p&gt;&#xA;" OwnerUserId="304" LastEditorUserId="304" LastEditorDisplayName="" LastEditDate="2011-05-05T07:49:34.283" LastActivityDate="2011-05-05T07:49:34.283" />
  <row Id="7115" PostTypeId="2" ParentId="1775" CreationDate="2011-03-17T18:18:45.657" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Forget SR considerations and lets focus on low velocity/KE particles.&#xA;If I am standing, with my thermometer, in a stream of unidirectional particles going on average 1 mile per second I measure a cetain temperature. If I am acclerated by an outside force to 1 mps then the particles appear stationary except for some wiggle. Has the temperature measured by my thermometer dropped??&#xA;Related to the above does the measured temperature depend on the random spread of energies about the mean or is it soley related to the mean regardless of spread&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="2607" LastActivityDate="2011-03-17T18:18:45.657" />
  <row Id="7116" PostTypeId="2" ParentId="7112" CreationDate="2011-03-17T18:22:55.650" Score="6" ViewCount="" Body="&lt;p&gt;Dear Yayu, I am afraid that this question of yours, while excellent, could belong to a psychology forum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feynman's perspective was already totally sensible; however, since his death, an amazing sequence of ever more straightforward proofs that hidden variables can't exist has been found. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me mention the GHZM state, Hardy's &quot;paradox&quot;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/4040/weak-measurement-and-hardys-paradox&quot;&gt;Weak measurement and Hardy&amp;#39;s paradox.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and various experimentally realized thought experiments disproving various theories of &quot;nonlocal realism&quot; as well, see e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/0704.2529&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0704.2529&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The idea that the probabilistic nature of quantum mechanics is due to the hidden variables was indefensible when Feynman was writing his lectures in the 1960s, and maybe even in the late 1920s when quantum mechanics became settled. But it is much more indefensible today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure whether I fully understand your proof but it &lt;em&gt;is&lt;/em&gt; possible to describe the experimentally validated proof in similarly simple terms. Quantum mechanics remains unacceptable for many people, for quasi-religious reasons. But as Sidney Coleman said at the end of his excellent lecture,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2010/11/sidney-coleman-quantum-mechanics-in.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/11/sidney-coleman-quantum-mechanics-in.html&lt;/a&gt;&lt;br&gt;&#xA;  &lt;strong&gt;Quantum mechanics in your face&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in which he described the conceptual issues of QM very clearly, it's time for those people to seriously consider the hypothesis that quantum mechanics is really how the world works. Well, the lecture is 17 years old by now, too, indicating that the opposition to quantum mechanics will never disappear.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-17T18:22:55.650" CommentCount="17" />
  <row Id="7117" PostTypeId="2" ParentId="7101" CreationDate="2011-03-17T19:12:38.567" Score="0" ViewCount="" Body="&lt;p&gt;Well it's like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;alpha: take a step back, happy days&#xA;beta: cover yourself with a sheet of paper&#xA;gamma: see me running try to keep up.&lt;/p&gt;&#xA;" OwnerUserId="2608" LastActivityDate="2011-03-17T19:12:38.567" CommentCount="1" />
  <row Id="7118" PostTypeId="1" AcceptedAnswerId="7119" CreationDate="2011-03-17T19:36:55.080" Score="1" ViewCount="136" Body="&lt;p&gt;This might be stupid, but is gravity a form of energy? And, if so, couldn't we use it for power?&lt;/p&gt;&#xA;" OwnerUserId="2609" LastActivityDate="2011-03-17T20:08:54.443" Title="Is Gravity Energy?" Tags="&lt;gravity&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="7119" PostTypeId="2" ParentId="7118" CreationDate="2011-03-17T20:08:54.443" Score="3" ViewCount="" Body="&lt;p&gt;Gravitation is something that gives rise to a force. Like most forces, it can be put to work, just like in dams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But gravity is not energy: it's an interaction between physical bodies with mass.&lt;/p&gt;&#xA;" OwnerUserId="1673" LastActivityDate="2011-03-17T20:08:54.443" CommentCount="1" />
  <row Id="7120" PostTypeId="2" ParentId="6906" CreationDate="2011-03-17T20:38:03.683" Score="6" ViewCount="" Body="&lt;p&gt;I hear that the &lt;a href=&quot;http://en.wikipedia.org/wiki/WC-135_Constant_Phoenix&quot; rel=&quot;nofollow&quot;&gt;WC-135 Constant Phoenix&lt;/a&gt; has recently been deployed to monitor radiation in the air around Japan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Vela_%28satellite%29&quot; rel=&quot;nofollow&quot;&gt;Vela satellites&lt;/a&gt; which were operational until 1984, and currently the &lt;a href=&quot;http://en.wikipedia.org/wiki/Defense_Support_Program&quot; rel=&quot;nofollow&quot;&gt;DSP satellites&lt;/a&gt;, are intended to give immediate reports of nuclear bomb detonation and ICBM launches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Vela satellites included gamma ray detectors, which accidentally discovered astronomical gamma-ray bursts (GRBs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since these satellites can detect ICBM launches from the ground and nuclear bomb detonation on the ground, and such detonations produce lots of radiation including gamma rays, and these satellites include gamma ray detectors, it would be easy for people to (incorrectly) jump to the conclusion that these satellites detect gamma rays emitted from the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As dmckee and Martin Beckett pointed out, our atmosphere is far more than thick enough to convert practically all cosmic rays into gamma rays, and practically all gamma rays into visible light and muons.&#xA;That's why all &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma-ray_astronomy&quot; rel=&quot;nofollow&quot;&gt;gamma-ray telescopes&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/X-ray_astronomy&quot; rel=&quot;nofollow&quot;&gt;X-ray telescopes&lt;/a&gt; are designed to be placed outside the earth's atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gamma ray detectors on those satellites can't detect gamma rays emitted by things on the ground; apparently those detectors were originally designed to detect gamma rays emitted by nuclear bombs detonated in space.&#xA;(Has that ever happened?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stuff that happens on the ground is detected by the other sensors on those satellites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can tell, satellites pretty useless for measuring radiation levels around nuclear reactors.&#xA;Radiation levels high enough for them to see (indirectly -- gamma rays converted to visible light) are also bright enough for humans to see, and only nuclear bomb detonation gives radiations levels that high.&#xA;The radiation levels that a damaged nuclear reactor could produce in a worst-case scenario -- lethal but not high enough to indirectly produce enough light for the human eye to see -- that light is also far too dim for these satellites to see.&lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2011-03-17T20:38:03.683" CommentCount="4" />
  <row Id="7121" PostTypeId="1" CreationDate="2011-03-17T20:57:41.683" Score="1" ViewCount="241" Body="&lt;p&gt;According to the reports, the shutdown procedures at all the Fukushima reactors were successful, and all the control rods were fully inserted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So - if there was a meltdown, would the control rods also melt and blend into the resulting material (corium)? If so, would that have the effect of &quot;diluting&quot; the corium in radioactive terms and stabilising it to some extent? (I guess it would depend on the relative melting points of the fuel and the control rods: if the rods are of boron, the melting point is a lot higher than that of uranium.)&lt;/p&gt;&#xA;" OwnerUserId="2611" LastEditorUserId="2611" LastEditorDisplayName="" LastEditDate="2011-03-17T21:15:23.297" LastActivityDate="2011-05-28T19:05:06.633" Title="If there is a meltdown at the Fukushima reactors, would the control rods melt also?" Tags="&lt;nuclear-physics&gt;&lt;nuclear-engineering&gt;" AnswerCount="2" />
  <row Id="7122" PostTypeId="2" ParentId="7121" CreationDate="2011-03-17T21:22:33.400" Score="4" ViewCount="" Body="&lt;p&gt;First:  the Fukushima reactors do not contain uranium, but uranium dioxide. &#xA;The rest is decribed quite extensivly here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Corium_%28nuclear_reactor%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Corium_%28nuclear_reactor%29&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-17T21:22:33.400" />
  <row Id="7123" PostTypeId="1" CreationDate="2011-03-17T21:25:41.557" Score="0" ViewCount="194" Body="&lt;p&gt;Is there any good reference for conceptual problems for students which learn nuclear physics first time? I am not searching problems that involve difficult calculations. Quite the converse, they should be computational rather simple but conceptually difficult. &lt;/p&gt;&#xA;" OwnerUserId="667" LastActivityDate="2011-03-18T05:00:03.823" Title="Conceptual problems in nuclear physics for students?" Tags="&lt;nuclear-physics&gt;&lt;education&gt;&lt;nuclear-engineering&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="7125" PostTypeId="2" ParentId="7112" CreationDate="2011-03-17T22:36:23.957" Score="8" ViewCount="" Body="&lt;p&gt;I agree with Luboš that this question has a lot to do with psychology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the &lt;a href=&quot;http://www.jstor.org/pss/1420555&quot; rel=&quot;nofollow&quot;&gt;tic-tac-toe analogy&lt;/a&gt; is relevant. There are an infinite number of games that are precisely equivalent to tic-tac-toe, but humans are probably terrible at playing most of these games.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chess is even worse. You can teach a child the rules of chess in a few minutes, but imagine doing so without the board! Consider one of the simplest games of chess:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;e4 e5&lt;/li&gt;&#xA;&lt;li&gt;Qh5?! Nc6&lt;/li&gt;&#xA;&lt;li&gt;Bc4 Nf6??&lt;/li&gt;&#xA;&lt;li&gt;Qxf7# 1–0&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A child can grasp the game, but even a grandmaster probably pictures a board as he reads those symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A chess computer does not picture a chess board. It uses a &lt;a href=&quot;http://en.wikipedia.org/wiki/Board_representation_%28chess%29&quot; rel=&quot;nofollow&quot;&gt;different representation&lt;/a&gt; that is more efficient for its calculations. I suspect this formulation of chess would not hold a child's attention for very long, and only a very exceptional child would attain any skill at the game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some representations of chess or tic-tac-toe suit the human mind. Others are efficient to simulate on a computer. Still other representations are utterly useless, despite being precisely equivalent. Surely the same is true of the rules of physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this reason, I like Bohm's efforts to find an &lt;a href=&quot;http://dx.doi.org/10.1103%2FPhysRev.85.166&quot; rel=&quot;nofollow&quot;&gt;experimentally&lt;/a&gt; &lt;a href=&quot;http://dx.doi.org/10.1103%2FPhysRev.85.180&quot; rel=&quot;nofollow&quot;&gt;indistinguishable&lt;/a&gt; reformulation of QM, whether or not he succeeded. Imagine if we've been playing '&lt;a href=&quot;http://www.jstor.org/pss/1420555&quot; rel=&quot;nofollow&quot;&gt;JAM&lt;/a&gt;' all this time, when we could have been playing 'tic-tac-toe'!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that's why people still talk about Bohm.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2011-03-17T22:36:23.957" CommentCount="1" />
  <row Id="7126" PostTypeId="2" ParentId="7101" CreationDate="2011-03-17T22:43:40.470" Score="3" ViewCount="" Body="&lt;p&gt;I am not sure this has been answered clearly enough, despite several good answers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 2 factors to take into account here - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) energy density and likelihood of deposition to tissue&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) penetration and likelihood of reaching tissue&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So, to break that down:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 1)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most energetic particles are alphas. &#xA;That makes sense, as they are essentially helium nuclei (2p,2n). If you think e=mc2, then they are naturally going to have a higher average energy than beta negatives (electrons) or gammas (photons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beta negatives (electrons) are probably next, as they also have mass, and due to the likelihood they also arise with high energy often move relativistically (near light speed). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gammas, as a form of electromagnetic radiation, have a spectrum. Worth noting here that 'gamma' doesnt specify an energy level on the spectrum, but rather than it was generated by a nucelus and not an electron. They can be weaker than most xrays, or contain more energy than most betas. For example you can get a 600keV gamma and a 100keV beta.&#xA;Alphas are more likely to get even further, into the MeV range. The others can too, but it is not common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that is how energetic they are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also relates to how likely they are to deposit into tissue. The more massive a particle is, the more likely it will interact (poor oversimplification, but just accept that for now). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means an alpha, being super duper massive compared to the others, will react quickly. They lose all their energy in less than a mm of tissue. This means, if it is right next to a susceptible piece of tissue, it will give ALL the energy it has to that tissue. Very bad news, &lt;em&gt;cell-killingly bad news&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A beta will probably lose it's energy over several centimetres. This means the damage gets a bit more spread out, which means each section of tissue &lt;em&gt;gets less of a dose&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gamma will get a lot further. A lot of gammas can pass through your body barely interacting at all, and therefore giving each tissue only a fraction of the energy it holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that is how they affect you physically. But then we have the other problem - can they affect you in real life ie. Can the damaging radiation actually reach the cells?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 2)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Penetration. &#xA;This was partially touched upon at the end of the last bit, but is a slightly different topic. We have seen gammas penetrate, betas do less, and alphas barely get anywhere. This means, to tissue, alphas are the biggest baddie.&#xA;&lt;em&gt;But can they get to tissue?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets look at an alpha. These really only come from supermassive nuclei like uranium, and are not relevant in Fukushima as Uranium is not decaying, the daughter products of uranium like cesium, iodine and stronium are (beta and gamma emitters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for alphas, becuase this is a good concept -&#xA;Alphas lose all there energy immediately. This is a bad thing for tissue, to get say 1MeV deposited in a smal(1mm3) space. &#xA;But if the alpha generator is on your skin, then you are totally unharmed! This seems counter intuitive, but because the alpha is so damn good at passing it's deadly energy to tissue, it &lt;em&gt;cannot get through your dead skin layer.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Betas are the next baddie, they can deposit a lot of energy in a small region. But again, for a lot of lower energy betas you can attenuate them out (shield them) with some thick clothes and the dead skin, with an air gap in between. You will only get a fraction of the radioactive dose that is on the outside of your body. To make things even better, your skin and muscle tissue is not very sensitive to radiation anyway, so it has &lt;em&gt;less biological effect&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gammas get through, but can of course pass the whole way through. You will end up with most of the radiation dose, fairly evenly spread throughout your tissues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So thats that right? Gammas get you, high energy betas can, and low energy betas and all alphas are just pussies?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The major population danger is not do to external radiation, but due to the ingestion (swallowing) of a radiation &lt;em&gt;source&lt;/em&gt;. By this I dont mean you swallow an alpha or beta or gamma, but instead swallow radioactive cesium, iodine, stronium etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside your body, the alphas and betas can wreak havok. You have no defenses. All the cesium generated betas will hit small areas in your gut. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The body also processes these compounds naturally whether they are radioactive or not. We use iodine in our thyroid, and stronium is similar enough to calcium that we will put it in our bones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when radioactive versions of these products are absorbed, and decay, the betas and gammas they release have a direct action on the thyoid (for iodine) and the bone marrow (for strontium).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then add to that the problem that the thyroid and blood cells are actually sensitive to radiation, unlike the skin and muscle, and the effect of each Gy absorbed is multiplied by factors of 5 or more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is getting long, so I will just summarise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CONCLUSION: If you can keep them on the outside of your body, alphas and a large percentage of betas are &lt;em&gt;harmless&lt;/em&gt;. Gammas will still get you, even behind several centimeters of lead! &#xA;If &lt;em&gt;sources of alphas and betas&lt;/em&gt; get in your body, via contaminated food, breathing or something else, then they are far, far worse than gammas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the workers at the plant currently, they are worried about gammas, because they have oxygen tanks and shielded suits.&#xA;For the people of Japan, they are worried about alpha and beta generators, because they can get into the food chain, and do far, far more damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps. Sorry for the no pictures!&lt;/p&gt;&#xA;" OwnerUserId="2600" LastActivityDate="2011-03-17T22:43:40.470" CommentCount="6" />
  <row Id="7127" PostTypeId="1" AcceptedAnswerId="7128" CreationDate="2011-03-17T23:50:34.300" Score="1" ViewCount="145" Body="&lt;p&gt;How to find velocity and displacement equations from a given force equation? For instance, it was given the following 1-D equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = b_1(v_1-v) - b_2 v$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_1$, $b_1$ and $b_2$ are constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $F = ma = m\frac{\mathrm{d}v}{\mathrm{d}t}$, but I can't find how to integrate $F$. Is there any technique that can help me or my problem is just calculus?&lt;/p&gt;&#xA;" OwnerUserId="2613" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-17T23:53:35.060" LastActivityDate="2011-03-17T23:57:24.543" Title="Find velocity and displacement from force equation" Tags="&lt;classical-mechanics&gt;&lt;forces&gt;&lt;velocity&gt;" AnswerCount="1" />
  <row Id="7128" PostTypeId="2" ParentId="7127" CreationDate="2011-03-17T23:57:24.543" Score="3" ViewCount="" Body="&lt;p&gt;Start with&#xA;$$&#xD;&#xA;m{dv\over dt}=b_1(v_1-v)-b_2v.&#xD;&#xA;$$&#xA;Move everything involving $v$ to one side of the equation, and everything involving $t$ (in this case, just $dt$) to the other side. Integrate both sides. One side will be just $\int dt$, or $t+C$. The other side will be some function of $v$. Algebraically solve the result to get $v$ in terms of $t$.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-17T23:57:24.543" CommentCount="1" />
  <row Id="7129" PostTypeId="1" AcceptedAnswerId="9223" CreationDate="2011-03-17T23:59:45.447" Score="10" ViewCount="342" Body="&lt;p&gt;Do sound waves in a gas consist of phonons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about a glass? Or other non-crystalline materials such as quasicrystals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the lack of translational symmetry affect the quantization of the displacement field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the answers so far have treated this question at a much more elementary level than I was expecting. I am already quite familiar with the properties of phonons in crystals. Therefore, do not explain the well-known derivations of the dispersion relation and second quantization of phonons in crystal lattices in your answer (and especially don't get them &lt;strong&gt;wrong&lt;/strong&gt;!).&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditorDisplayName="" LastEditDate="2011-04-25T21:14:39.680" LastActivityDate="2011-04-28T05:20:55.087" Title="Phonons in non-crystalline media" Tags="&lt;condensed-matter&gt;&lt;sound&gt;&lt;quantization&gt;" AnswerCount="3" CommentCount="14" FavoriteCount="2" />
  <row Id="7130" PostTypeId="2" ParentId="7112" CreationDate="2011-03-18T03:43:02.753" Score="1" ViewCount="" Body="&lt;p&gt;Bohmian mechanics is a non-local hidden variable theory. It can reproduce the results of ordinary quantum mechanics. Since Bell's theorem and the subsequent experimental verdict ruled out only local hidden variable theory, Bohmian mechanics can still survive in principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality however it survives due to a different reason as pointed out by Lubos. The reason is essentially regressive. People for whom it is hard to swallow the hard reality of quantum mechanics, people who want to understand nature in terms of every day classical intuitions are the ones who often advocate Bohmian mechanics. IMHO it is extremely difficult for a physicist to support this approach. Why? The reasons are as follows.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It violates Occam's razor by introducing extra equation in addition to the Schrödinger's equation.&lt;/li&gt;&#xA;&lt;li&gt;It is explicitly nonlocal and inconsistent with special relativity.&lt;/li&gt;&#xA;&lt;li&gt;It can not be extended to produce a quantum field theory.&lt;/li&gt;&#xA;&lt;li&gt;It is an unnecessary superstructure over ordinary quantum mechanics.&lt;/li&gt;&#xA;&lt;li&gt;The quantum mechanical spin becomes terribly messed up in this approach.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In short it is a cheap alternative to ordinary sane quantum mechanics.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-03-18T04:44:31.530" LastActivityDate="2011-03-18T04:44:31.530" CommentCount="9" />
  <row Id="7131" PostTypeId="1" AcceptedAnswerId="7133" CreationDate="2011-03-18T03:59:18.420" Score="3" ViewCount="146" Body="&lt;p&gt;In all the discussions about how the heavy elements in the universe are forged in the guts of stars and especially during a stars death, I usually hear that once the star begins fusing lighter atoms to produce Iron (Fe) that's the end of the star's life and the whole system collapses onto itself and based on how massive the star was initially it has different outcomes like a white dwarf, a neutron star or a black hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have rarely heard a detailed explanation of how the elements heavier than Iron are produced. I would appreciate a convincing explanation of this process.&lt;/p&gt;&#xA;" OwnerUserId="2616" LastEditorUserId="61" LastEditorDisplayName="" LastEditDate="2011-03-18T05:37:53.907" LastActivityDate="2011-03-18T05:37:53.907" Title="Origin of elements heavier than Iron (Fe)" Tags="&lt;astrophysics&gt;&lt;nuclear-physics&gt;&lt;fusion&gt;" AnswerCount="1" />
  <row Id="7132" PostTypeId="1" AcceptedAnswerId="7136" CreationDate="2011-03-18T04:19:17.080" Score="1" ViewCount="106" Body="&lt;p&gt;The professor asked us to do this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;..Determine all potentials $V(r,\theta, \phi)$ for which it is possible for find solutions of the time independent Schrodinger equation which are also eigenfunctions of the operator $L_{z}$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I try to solve this problem by assuming separation of variables, and I get $$\Phi(r,\theta,\phi)=R(r)F(\theta)e^{im_{l}\phi}$$ Unfortunately I do not know what to do next for this - should I put it into the formula $H\Phi=E\Phi$ in radial form to see whether it works? The professor give us the hint that the commutator $$[L_{z},V(r,\theta,\phi)]=0$$ I do not know how to use this relationship. &lt;/p&gt;&#xA;" OwnerUserId="2597" LastEditorUserId="2597" LastEditorDisplayName="" LastEditDate="2011-03-18T04:33:59.923" LastActivityDate="2011-03-18T15:07:00.390" Title="What is the possible potential?" Tags="&lt;quantum-mechanics&gt;&lt;homework&gt;" AnswerCount="2" />
  <row Id="7133" PostTypeId="2" ParentId="7131" CreationDate="2011-03-18T04:21:55.590" Score="5" ViewCount="" Body="&lt;p&gt;Elements heavier than iron are only produced during supernovae; in these extreme energetic conditions atoms are bombarded by a very large number of neutrons. Rapid successive neutron capture, followed by beta decay, produces the heavier atoms. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Supernova_nucleosynthesis&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Supernova_nucleosynthesis&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2617" LastActivityDate="2011-03-18T04:21:55.590" />
  <row Id="7134" PostTypeId="2" ParentId="7112" CreationDate="2011-03-18T04:22:28.150" Score="1" ViewCount="" Body="&lt;p&gt;For the same reasons as why people still talk about other theories which have been demonstrated to be flawed at some level. Classical mechanics is still useful and gives correct answers in it's regime, even though it's basic assumptions are totally flawed. It's relationship to the 'correct' theory can still be insightful and interesting.&lt;/p&gt;&#xA;" OwnerUserId="1670" LastActivityDate="2011-03-18T04:22:28.150" />
  <row Id="7135" PostTypeId="1" AcceptedAnswerId="7152" CreationDate="2011-03-18T04:42:14.300" Score="3" ViewCount="118" Body="&lt;p&gt;Can anyone point me to a reference (preferably either something online or something a small liberal arts school would be likely to have in its library) that goes through a derivation of the minimum uncertainty wavefunction in more detail than in the Griffiths?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; I've moved the second part of my original question to a separate post: &lt;a href=&quot;http://physics.stackexchange.com/questions/7189/3d-minimum-uncertainty-wavepackets&quot;&gt;3D Minimum uncertainty wavepackets&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditorDisplayName="" LastEditDate="2011-03-18T21:18:13.253" LastActivityDate="2011-03-18T21:18:13.253" Title="Minimum Uncertainty Wavefunction derivation" Tags="&lt;quantum-mechanics&gt;&lt;uncertainty-principle&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="7136" PostTypeId="2" ParentId="7132" CreationDate="2011-03-18T04:45:44.700" Score="2" ViewCount="" Body="&lt;p&gt;Try writing out $[L_z,V(r,\theta,\phi)]=0$. In other words, V has to be a function such that:&lt;br&gt;&#xA;$L_zV \psi = VL_z\psi$&lt;br&gt;&#xA;for absolutely any function $\psi$. Does that tell you anything about V?&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-18T04:45:44.700" CommentCount="2" />
  <row Id="7137" PostTypeId="2" ParentId="7123" CreationDate="2011-03-18T05:00:03.823" Score="1" ViewCount="" Body="&lt;p&gt;Try looking at the nuclear physics questions in the Physics GRE test. These are multiple choice questions that are supposed to be answered in a few minutes each. At this link:  &lt;a href=&quot;http://grephysics.net/ans/type/Advanced%20Topics&quot; rel=&quot;nofollow&quot;&gt;http://grephysics.net/ans/type/Advanced%20Topics&lt;/a&gt;&lt;br&gt;&#xA;You will find a few nuclear physics questions such as these ones from the 8677 test:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://grephysics.net/ans/8677/85&quot; rel=&quot;nofollow&quot;&gt;http://grephysics.net/ans/8677/85&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://grephysics.net/ans/8677/67&quot; rel=&quot;nofollow&quot;&gt;http://grephysics.net/ans/8677/67&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-18T05:00:03.823" />
  <row Id="7139" PostTypeId="2" ParentId="7104" CreationDate="2011-03-18T05:30:46.440" Score="3" ViewCount="" Body="&lt;p&gt;To understand the physics of inflated balloons you have to understand curved membranes under tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there's a membrane under tension $T$ (for simplicity assumed to be isotropic), with fluids on both sides, it will be flat (planar) unless there is a pressure difference between the two fluids. If you notice that it's curved with radius of curvature $R$ (again assumed to be isotropic, like a sphere), then there must be a pressure difference given by&#xA;$$\Delta p = 2 \frac{T}{R}$$&#xA;Of course, the membrane bulges away from the side where the pressure is higher, out into the side where the pressure is lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we have this basic concept, the answers to your questions all come from it because they're all related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hot air balloon is stiff because it is under tension. If a membrane is not under tension it could have folds or wrinkles because there is no energy penalty for having extra area. But since the material of the balloon is under tension it's going to be as smooth as possible. It would be flat if not for the pressure difference, which forces it to have a smooth curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure inside the balloon is indeed higher than the pressure outside, and the difference is given by the formula above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This pressure difference is not really &quot;explained&quot; by the ideal gas law, but it is consistent with it. If you start with a deflated balloon on the ground and gradually fill it up with hot air, you're increasing $N$ (because there are more air molecules inside the balloon than when you started) and $V$, while $p$ and $T$ remain roughly constant (the $T$ inside being higher than the $T$ outside). When $V$ finally gets to the volume that will make the balloon stiff, the pressure begins to go up slightly above the ambient pressure, with the extra pressure being provided by the tension of the balloon. However, the density of the gas inside the balloon is still less than that of the air around it, because although $p$ is greater (tending to increase the density), $T$ is also greater (tending to decrease the density), and the temperature effect dominates.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-03-18T05:30:46.440" CommentCount="3" />
  <row Id="7140" PostTypeId="1" AcceptedAnswerId="7183" CreationDate="2011-03-18T05:35:36.733" Score="8" ViewCount="238" Body="&lt;p&gt;The reply to a question about &lt;a href=&quot;http://physics.stackexchange.com/questions/7131/origin-of-elements-heavier-than-iron-fe&quot;&gt;nucleosynthesis&lt;/a&gt;, that heavier than iron elements are produced in supernovae explosions, raised for me the following question which I could not answer by googling. Partially because the search for planets and stars brings out astrology answers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explosions are dispersive, nevertheless we find minerals in clumps, not only uniformly dispersed in the ground. Is there a coherent presentation that explains how minerals end up in veins and bands?&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-18T08:13:59.130" LastActivityDate="2011-04-18T00:30:20.450" Title="How does uranium from supernovae explosions end up in mineral veins in a planet?" Tags="&lt;geophysics&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="7141" PostTypeId="1" CreationDate="2011-03-18T05:43:39.387" Score="1" ViewCount="88" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/6400/are-tidal-power-plants-slowing-down-earths-rotation&quot;&gt;Are tidal power plants slowing down Earth&amp;#39;s rotation?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Since we have various energy acquiring facilities that harvest energy from earth rotation. (e.g. wind, tidal power planet)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will our earth rotation be slower down? Is there any form of force to accelerate earth rotation speed? &lt;/p&gt;&#xA;" OwnerUserId="1799" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-18T08:05:08.943" LastActivityDate="2011-03-18T08:05:08.943" ClosedDate="2011-03-18T15:27:35.960" Title="Will earth rotation be slowed down?" Tags="&lt;energy&gt;&lt;angular-momentum&gt;" CommentCount="5" />
  <row Id="7142" PostTypeId="2" ParentId="5773" CreationDate="2011-03-18T05:45:30.337" Score="2" ViewCount="" Body="&lt;p&gt;Suppose you wanted to talk about the &quot;quantum mechanical version&quot; of a rotating rigid body which does not &quot;go anywhere&quot; (classically, the center of mass is stationary). Then we would probably consider states in L^2(SO(3,R)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A general rule is that if you have a classically described system and you want to know what the &quot;quantum mechanical version&quot; of it is, you let M be the configuration manifold (ie, that manifold which describes the &quot;position&quot; of the classical system and whose cotangent bundle is the phase space manifold) and you take your states in L^2(M).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the purposes of real physics, this is not always a useful thing.  After all, classical systems (probably) don't exist, so there is not necessarily fundamental value in knowing how to go from &quot;classical to quantum&quot;.  It may however say something very interesting about how the classical limit arises and about the nature of quantum decoherence.&lt;/p&gt;&#xA;" OwnerUserId="2620" LastActivityDate="2011-03-18T05:45:30.337" />
  <row Id="7143" PostTypeId="1" AcceptedAnswerId="7145" CreationDate="2011-03-18T05:51:56.137" Score="5" ViewCount="133" Body="&lt;p&gt;I don't really know anything about physics even though I pretended studying it for years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this explained?: &lt;a href=&quot;http://andrewsullivan.theatlantic.com/the_daily_dish/2011/01/if-you-think-youre-cold-.html&quot; rel=&quot;nofollow&quot;&gt;http://andrewsullivan.theatlantic.com/the_daily_dish/2011/01/if-you-think-youre-cold-.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is there to be said about it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My apologies if you find the question too vague or silly.&lt;/p&gt;&#xA;" OwnerUserId="2619" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-18T08:03:41.380" LastActivityDate="2011-03-18T08:03:41.380" Title="Evaporating water in -30ºC" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="7144" PostTypeId="2" ParentId="7095" CreationDate="2011-03-18T05:56:58.613" Score="4" ViewCount="" Body="&lt;p&gt;Here's my attempt at answering some of your multitude of points...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I know you're often looking at $N=2$ susy. $D_{++}$ type covariant derivatives often appear in &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_superspace&quot; rel=&quot;nofollow&quot;&gt;harmonic superspace&lt;/a&gt;, where the ++ indicates the harmonic charge.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The spinor components of your gauge field are Lie algebra valued fermionic spin-1/2 fields. That means they take a spacetime point and return an element that's in the Grassmann shell of the direct product of the spin-1/2 module and the Lie algebra. Since it is fermionic, it must be an odd element.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the product is a term in the Lagrangian, then everything must be contracted / traced to yield an invariant object.&#xA;If the product is not in the Lagrangian, then it depends on context. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Products of the Lie algebra elements must be either commutators or traces of quadratic terms. In you're case you're in the adjoint rep so the products are secretly commutators. Cubics etc are combinations of the above. $tr(ABC):=tr(A[B,C])=tr([A,B]C)$ etc...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As for &quot;$\bar{\psi}^2 \neq 0$ but $Tr[\bar{\psi}^2] = 0$&quot;... I'm not sure exactly what you mean. Maybe let $\lambda = \lambda^A T_A$ be a anticommuting Lie algebra element. Then the Lie algebra commutator is $[\lambda, \lambda] = \lambda^A \lambda^B [T_A, T_B] \neq 0$ but must have vanishing trace. Or the direct product is $\lambda \lambda = \lambda^A \lambda^B T_AT_B$ which has (assuming an orthogonal basis) the trace &#xA;${\rm tr} (\lambda \lambda) = \lambda^A \lambda^B {\rm tr}(T_AT_B) = \lambda^A\lambda^B \delta_{AB}$&#xA;vanishes because each component is just an odd Grassmann element and so squares to zero.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;After that, I think you leave the realm of standard QFT. You seem to want to make the field become a physical state that is a vacuum and so vanishes under supersymmetry?&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Maybe you should have a quick skim through the first chapter of &lt;a href=&quot;http://books.google.com.au/books?id=jZ1Y1Z3BoicC&quot; rel=&quot;nofollow&quot;&gt;Ideas and methods of supersymmetry and supergravity&lt;/a&gt; that is called (and has a lot of) &quot;mathematical background&quot;. It gives a good description of Grassmann algebras that is often skimmed over in other texts.&lt;/p&gt;&#xA;" OwnerUserId="429" LastEditorUserId="429" LastEditorDisplayName="" LastEditDate="2011-03-18T06:10:07.823" LastActivityDate="2011-03-18T06:10:07.823" CommentCount="7" />
  <row Id="7145" PostTypeId="2" ParentId="7143" CreationDate="2011-03-18T06:24:57.010" Score="3" ViewCount="" Body="&lt;p&gt;This impressive effect has several stages and reason:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Once she throws the water into the air, she hugely increases the surface area because the body of the water becomes lots of small droplets.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Because the water was close to getting vaporized, the large surface of these droplets will be quickly turned to vapor. A low humidity of the air makes this process even more efficient.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The remaining much smaller droplets are much more quickly turned into ice crystals because they have a much higher surface/volume ratio.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The fog that is falling down is made of small pieces of ice - it's a kind of snow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you like unusual effects with water phases, you may also try supercooling. When you open a bottle of Coke taken from the fridge at the right sub-freezing temperature, which was however still liquid, it may fully crystallize within 20 seconds. That's because the suddenly lowered pressure increases the freezing point, so you're suddenly beneath it.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-18T06:24:57.010" CommentCount="3" />
  <row Id="7147" PostTypeId="2" ParentId="7140" CreationDate="2011-03-18T08:19:02.327" Score="1" ViewCount="" Body="&lt;p&gt;p.s. I have found a  &lt;a href=&quot;http://www.csus.edu/indiv/s/slaymaker/Archives/Geochemistry.htm&quot; rel=&quot;nofollow&quot;&gt;course notes precis&lt;/a&gt;  and it  seems there are several models proposed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Completely out of my depth, I would propose that if matter above Fe in the binding energy curve were in a concentration within a &quot;protoplanet&quot; with a high enough pressure, it could form a quark gluon plasma, and then all type of nuclei could precipitate/crystalize as conditions change, according to their quantum mechanical state function, so a supernova explosion might be sufficient to make uranium, but not necessary. This would explain clumping and veins.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-18T16:06:50.627" LastActivityDate="2011-03-18T16:06:50.627" CommentCount="3" />
  <row Id="7148" PostTypeId="2" ParentId="6928" CreationDate="2011-03-18T09:05:26.950" Score="1" ViewCount="" Body="&lt;p&gt;Fingers can be dipped in molten lead at 500 degrees centrigrade and above without getting hurt even trivially. Further Molten lead contains Uranium blasts. After pouring molten lead use common lenses[bvery large ones] to concentrate on the reactors with will send neutrinos and which in turn would destroy all uranium and convert it into a harmless alloy of sorts and will keep molten due to sun's heat. &lt;/p&gt;&#xA;" OwnerUserId="2621" LastActivityDate="2011-03-18T09:05:26.950" />
  <row Id="7149" PostTypeId="1" AcceptedAnswerId="7156" CreationDate="2011-03-18T09:40:25.097" Score="5" ViewCount="672" Body="&lt;p&gt;A BWR reactor core may contain up to 146 tons of uranium.  Why does it not form a critical mass when molten?  Are there any estimates of the critical mass of the resulting zirconium alloy, steel, concrete and uranium oxide mixture?&lt;/p&gt;&#xA;" OwnerUserId="2415" LastEditorUserId="2415" LastEditorDisplayName="" LastEditDate="2011-03-18T09:48:26.833" LastActivityDate="2011-03-24T04:32:26.477" Title="Why does nuclear fuel not form a critical mass in the course of a meltdown?" Tags="&lt;nuclear-physics&gt;&lt;nuclear-engineering&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="2" />
  <row Id="7151" PostTypeId="2" ParentId="7149" CreationDate="2011-03-18T10:57:19.220" Score="0" ViewCount="" Body="&lt;p&gt;My guess would be that the moderator (normally the water, graphite already would have caused the accident) is missing and that is still vital in a compacted mass. The uranium oxide is not enriched to military grade, but I can't speak for MOx-elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should be difficult to say without some decent figures about leftover UO_2, mass, previous distribution.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastActivityDate="2011-03-18T10:57:19.220" CommentCount="3" />
  <row Id="7152" PostTypeId="2" ParentId="7135" CreationDate="2011-03-18T11:05:35.613" Score="3" ViewCount="" Body="&lt;p&gt;First some preliminaries. Suppose you have hermitian operators $A$ and $B$ and some state $\left| \psi \right&amp;gt;$. Denote by $\left&amp;lt;X\right&amp;gt;$ the expectation of $X$ in the state $\psi$, i.e. $\left&amp;lt;\psi\right| X \left| \psi\right&amp;gt;$. Denote by $\bar A := A - \left&amp;lt; A \right&amp;gt;$ and $\bar B := B - \left&amp;lt; B \right&amp;gt;$ the part of of $A$, resp. $B$ with vanishing expectation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's compute $\left&amp;lt; \bar A \right&amp;gt;^2 \left&amp;lt;\bar B \right&amp;gt;^2$. According to Cauchy-Schwarz inequality that this is always greater or equal than&#xA;$\left|\left&amp;lt; \bar A \bar B \right&amp;gt;\right|^2$ (just plug in $\psi$ and interpret these expressions as scalar products). Now, we can express the product as sum of hermitian and antihermitian component&#xA;$$\left|\left&amp;lt; \bar A \bar B \right&amp;gt;\right|^2 = {1\over4}\left&amp;lt; [A,B]/i \right&amp;gt;^2 + {1\over 4}\left&amp;lt; \{\bar A, \bar B\} \right&amp;gt;^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(here we used the fact that $[\bar A, \bar B] = [A,B]$).&#xA;If the commutator is just a number times identity operator then we can discard the expectations and after removing the anticommutator term (because it doesn't have any important interpretation and it doesn't spoil the inequality) we are left with HUP. But we're not interested in this application right now. Instead, we want to minimize the error term and that means we want equalities everywhere (it's not clear that it's possible to attain them, but let's assume this for a while). First, Cauchy-Schwarz inequality becomes equality if the vectors in the scalar product are colinear $$\bar B \left| \psi \right&amp;gt; = c \bar A \left| \psi \right&amp;gt;$$ Second, we want the expectation of anticommutator to vanish $$\left&amp;lt;\psi\right| \{\bar A, \bar B\} \left| \psi \right&amp;gt; = 0$$ So this gives us two equations for $\psi$. Let's see what we can get from them for $A = x$ and $B = p$. For simplicity let's assume that $\left&amp;lt;x\right&amp;gt; = \left&amp;lt;p\right&amp;gt; = 0$ (the general solution doesn't change anything much).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From first condition we obtain&#xA;$$(p - cx) \left | \psi \right&amp;gt; = 0$$ which is a differential equation&#xA;$$ (i \partial_x + cx) \psi(x) = 0$$&#xA;with a solution $\psi(x) = K \exp(-\alpha x^2)$ with ${\rm Re} \alpha &amp;gt; 0$ (so that this is indeed a vector from our Hilbert space) and $K$ being just a normalization constant. Finally from the anticommutator relation we get $\alpha = {1 \over 4(\Delta x)^2}$ and we're done.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-03-18T11:50:02.127" LastActivityDate="2011-03-18T11:50:02.127" />
  <row Id="7153" PostTypeId="1" AcceptedAnswerId="12326" CreationDate="2011-03-18T11:31:22.357" Score="2" ViewCount="129" Body="&lt;p&gt;My aim is to draw a plot of $GaAs_{1-x}Sb_{x}$/InP (GaAsSb on InP substrate) band offset as a function of As/Sb composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first part is easy if I'm doing this correctly: I found VBO (valence band offset) parameters for GaAs and GaSb, then the bowing parameter for $GaAs_{1-x}Sb_{x}$ VBO and finally the bowing parameter for $GaAs_{1-x}Sb_{x}$ band gap. This is what I got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/XOAPX.png&quot; alt=&quot;Unstrained GaAsSb&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I put GaAs or GaSb on InP, strains appear and the valence band is split into heavy hole and light hole subbands. In some articles I found the way to calculate the splitting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta E_{HH}=\partial E_H^V + \partial E_S$&#xA;$\Delta E_{LH}=\partial E_H^V - \partial E_S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\partial E_S=b(1-2\frac{C_{12}}{C_{11}})\epsilon_{||}$&#xA;$\partial E_H^V=2 a^V (1-\frac{C_{12}}{C_{11}})\epsilon_{||}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a, b and Cs are material properties and can be found in this article:&#xA;&lt;a href=&quot;http://jap.aip.org/resource/1/japiau/v89/i11/p5815_s1&quot; rel=&quot;nofollow&quot;&gt;http://jap.aip.org/resource/1/japiau/v89/i11/p5815_s1&lt;/a&gt; (Vurgraftman, Meyer, Ram-Mohan)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm not sure what the $\epsilon_{||}$ is. Is it the direction of growth? Whatever it is, I can't see where I should consider the substrate, as I assume the strain to be substrate-dependent. Is it hidden in the $\epsilon$?&lt;/p&gt;&#xA;" OwnerUserId="1505" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-18T19:17:57.177" LastActivityDate="2011-07-18T19:17:57.177" Title="Band offset and strain-induced valence band splitting in semiconductor compounds" Tags="&lt;semiconductor-physics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7154" PostTypeId="1" CreationDate="2011-03-18T11:32:16.527" Score="-2" ViewCount="174" Body="&lt;p&gt;I'm interested in the extent to which quantum physical effects are seen at a macroscopic level.  I might get some of the physics wrong, but I think I'll get it close enough that I can ask the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's sat that we create a bonfire and let it burn until it burns out.  As the smoke rises from the fire, turbulence takes over and the smoke particles and steam and hot air all mixed together.  By the end of the night when the fire has burned out, the collection of molecules in the system are in some position/velocity X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: Let's assume the multiverse interpretation of quantum physics.  How many possible end state superpositions can there be in this situation?  Ok, that's imprecise and incorrect because it would actually be an uncountable infinitude of possible end states.  How about this: Given the end state that we observed, what percentage of the end state superposition would be &quot;visually&quot; indiscernable from the end state that we observed so that each molecule would be in nearly the same end state across that portion of the multiverse?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or put another way: Do quantum effects sneak into everyday life fast enough that we can observe them?  If we are effected by quantum physics at all, I imagine this is roughly a function of the timescale of the chaos effects.&lt;/p&gt;&#xA;" OwnerUserId="922" LastActivityDate="2011-03-18T14:32:30.833" Title="Chaos and quantum physics: How many ways can a bonfire burn?" Tags="&lt;quantum-mechanics&gt;&lt;determinism&gt;&lt;chaos-theory&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="7155" PostTypeId="2" ParentId="7149" CreationDate="2011-03-18T11:35:30.700" Score="5" ViewCount="" Body="&lt;p&gt;The probable answer is that to make a bomb one needs a very special design and very pure U235, in  a sphere. So even if a critical mass forms in meltdown, a reactor does not have the geometry and purity for a nuclear bomb. No mushrooms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it may continue to heat, acting as a reactor, and the problem is in conveying the heat away without building up steam and hydrogen in order to avoid a chemical explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nuclear engineer should answer whether the amalgam of all the metals does not allow criticallity. Their plans to first cool the melted reactors and then encase them in sand and cement would imply that they do not expect the melted core to be critical. The fact that with the control rods in place the reactor is no longer critical by design, just has  residual radioactivity, argues from conservation of mass that even in melt when the metals will be mixed the same will hold true.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-18T11:35:30.700" CommentCount="2" />
  <row Id="7156" PostTypeId="2" ParentId="7149" CreationDate="2011-03-18T11:52:02.733" Score="11" ViewCount="" Body="&lt;p&gt;If you read Wikipedia page about &lt;a href=&quot;http://en.wikipedia.org/wiki/Corium_%28nuclear_reactor%29&quot; rel=&quot;nofollow&quot;&gt;corium&lt;/a&gt;, they say that critical mass can be achieved locally. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you are concerned about a critical mass allowing a nuclear explosion, the difficulty in nuclear weapon design, as told &lt;a href=&quot;http://en.wikipedia.org/wiki/Critical_mass#Criticality_in_nuclear_weapon_design&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, is to achieve the criticality fast enough. If you do not achieve criticality fast enough, your material heats and its interaction with neutrons decreases, slowing the chain reaction down. And that is with pure ²³⁵U. So basically what happens if criticality happens in a melting nuclear reactor is the release of a lot of heat and radiation, but not in an explosive manner as in an atomic bomb.&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2011-03-18T11:52:02.733" CommentCount="1" />
  <row Id="7157" PostTypeId="2" ParentId="7129" CreationDate="2011-03-18T11:55:58.137" Score="3" ViewCount="" Body="&lt;p&gt;Phonons were named after photons and have the same functionality quantum mechanically. They are both waves and particles. Nobody disputes the wave nature of sound in general. In non ordered materials there is no way that a sound wave will give all its energy to an atom, or a cluster of atoms, as an example. The reason is because all these atoms are in an incoherent state and no pure quantum mechanical state function can be defined. An atom can have a pure state function but the wavelengths of sound are way larger than the wavelengths that an atom can absorb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In ordered materials like crystals this can happen because phases are defined so that there  can be large dimensional coherent scattering: the order allows a quantum mechanical state function for the crystal to be defined, which can interact with a sound wave so that the whole energy of that sound is absorbed by the crystal, thus a phonon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: In the comments to the question it becomes clear that there is a confusion on the use of the term &quot;phonon&quot;. I am using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Phonons&quot; rel=&quot;nofollow&quot;&gt;definition in wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit2: Copying from Carl's comment in the question,  I would add that the quantization is the familiar E=hν=ℏω, and that this applies to fluids as well. But without a translational symmetry, this cannot display quantization but instead can take any value of ω.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we expand the definition of a phonon to a continuous spectrum, it seems that the answer to the question above finally is yes, yes, yes. Though I guess that  in disordered media the particle nature is not manifest. Actually this is also true about photons in ambient light, as an example.&#xA;Maybe somebody should expand the Wikipedia article.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-20T04:57:49.523" LastActivityDate="2011-03-20T04:57:49.523" CommentCount="9" />
  <row Id="7158" PostTypeId="1" CreationDate="2011-03-18T11:59:24.523" Score="7" ViewCount="325" Body="&lt;p&gt;By what mechanism do quantum effects become observable in normal life at the macroscopic level?  For instance, when two molecules &quot;collide&quot; is the momentum a probabilistic event wherein the end state is not unique?  Another example, during a chemical reaction, it is a probabilistic event at the quantum level whether or not any particular molecule within the solution interacts with another molecule?&lt;/p&gt;&#xA;" OwnerUserId="922" LastActivityDate="2011-03-18T22:43:26.627" Title="By what mechanism do quantum effects become observable in normal life at the macroscopic level?" Tags="&lt;quantum-mechanics&gt;&lt;determinism&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="7159" PostTypeId="2" ParentId="7158" CreationDate="2011-03-18T12:07:38.250" Score="-1" ViewCount="" Body="&lt;p&gt;It isnt really an answer, but molecules and chemical reactions are pretty big on a quantum scale, and those interactions I would think are more in the domain of traditional physics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that in medical imaging the constant noise that degrades image quality unrelated to scan parameters is called quantum noise, although I have never actually been sure it was a true quantum event. I suspect it is a quantum thing to do with background energy fluctuations, but it could be related to cosmic rays or something similar too. I would look it up myself but I am off to bed now!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is a quantum event, then there is a good example of how something quantum becomes macro.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Not sure why this was marked down, but if I have to explain why this explanation made sense, here you go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a sleep I found out about this - it is a quantum event observable on a real world (macro) scale. Not sure why anyone would mark this down ... maybe you wanted some esoteric wordy explanation but I thought you might want an example too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mghradrounds.org/index.php?src=gendocs&amp;amp;link=2008_february&quot; rel=&quot;nofollow&quot;&gt;http://www.mghradrounds.org/index.php?src=gendocs&amp;amp;link=2008_february&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the images there show the effect of quantum noise - the grainier right hand side images have a worse signal, and therefore quantum noise is more visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is caused by probabalistic fluctuations in quantum states throughout the system. In fact, this explains very clearly your opening question - as you can see, while there is local variance which is noticable because we are using high energy xrays, the field overall is homogenous. In most systems the heterogeneity would be on such a small scale it would be undetectable without extreme instruments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So each individual event has a probabalistic element, but over macro systems you lose the ability to notice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or just listen to Marek. I think this example is a good one.&lt;/p&gt;&#xA;" OwnerUserId="2600" LastEditorUserId="2600" LastEditorDisplayName="" LastEditDate="2011-03-18T22:43:26.627" LastActivityDate="2011-03-18T22:43:26.627" CommentCount="1" />
  <row Id="7160" PostTypeId="2" ParentId="7158" CreationDate="2011-03-18T12:11:20.707" Score="0" ViewCount="" Body="&lt;p&gt;Quantum properties are manifest everywhere. In the color of objects, structure of crystals, in the electrical, mechanical, chemical properties of different substances and in the very phenomena of life itself. There is no limit how big an object should be to show quantum behavior, only it will be increasingly difficult to maintain coherence. An one mm diameter object (oil droplet) has been &lt;a href=&quot;http://www.physorg.com/news78650511.html&quot; rel=&quot;nofollow&quot;&gt;shown to interfere with itself&lt;/a&gt;. Also &lt;a href=&quot;http://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality#Wave_behavior_of_large_objects&quot; rel=&quot;nofollow&quot;&gt;see this.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-03-18T14:16:23.327" LastActivityDate="2011-03-18T14:16:23.327" CommentCount="2" />
  <row Id="7161" PostTypeId="2" ParentId="7158" CreationDate="2011-03-18T12:23:12.040" Score="4" ViewCount="" Body="&lt;p&gt;One of the most obvious quantum &quot;effects&quot; which are observable at the macroscopic scale is the Pauli exclusion principle for fermions: if it didn't exist matter would collapse as nothing would prevent electrons to radiate and fall into the nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically quantum effects become observable at the macroscopic level when they affect coherently (i.e. in the same way) a large number of quantum particles.&lt;/p&gt;&#xA;" OwnerUserId="1844" LastActivityDate="2011-03-18T12:23:12.040" CommentCount="1" />
  <row Id="7162" PostTypeId="2" ParentId="7158" CreationDate="2011-03-18T12:31:45.710" Score="3" ViewCount="" Body="&lt;p&gt;The mechanism is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_decoherence&quot; rel=&quot;nofollow&quot;&gt;decoherence&lt;/a&gt;. Microscopically, it's true that the system evolves through all allowed configurations (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Path_integral_formulation&quot; rel=&quot;nofollow&quot;&gt;path integral&lt;/a&gt;, which in its more basic description tells you that the particle travels through every trajectory). But in real life we don't observe this. What gives?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turns out that the answer is related to something called &lt;a href=&quot;http://en.wikipedia.org/wiki/Wave_function_collapse&quot; rel=&quot;nofollow&quot;&gt;wave function collapse&lt;/a&gt;. This is a simplification of what happens but it gives you some intuition. When the system is left on its own it evolves as in the first paragraph. But as soon as you observe it (i.e. measure it), it will &quot;collapse&quot; into one of allowed eigenstates of the observable you are measuring. So in the end we are left with pure states without superposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it should be clear that &quot;something&quot; must be observing and collapsing the system even when we are not looking. What is it? Well, if one pauses to think for a second it should be obvious that every system that behaves classically is actually submerged in some bigger system, an environment (the typical example that concerns almost every system on Earth is atmosphere with lots of gas molecules present). In any case, humans are in no way special when dealing with nature and observation. Same role can be played by any big enough system, and environment surely is such a system. We know that according to quantum mechanics systems undergo quantum fluctuations. It is these fluctuations between system and environment that affect any superposition and quickly make it decay into states that are effectively classical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process can actually be imagined very clearly by taking analogy with heat transfer. Heat is transferred because of thermal fluctuations of two objects and basic probability theory dictates that the temperature of the warmer object will go down. Similarly, quantum mechanics and probability dictates that superposed system will gradually lose information about the superposition between its parts. This information is of course not lost but will get transferred to the environment where it will appear as noise and so the result is that we obtain effectively classical system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this means that to observe quantum effects macroscopically, you need to shield your system from decoherence. Either you lower the temperature as much as possible (so that the effect of environment is negligible), or you prepare the system in some clever way so that it is impervious to thermal perturbations. We are of course talking about superfluids, superliquids and similar states of matter.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-03-18T13:30:08.943" LastActivityDate="2011-03-18T13:30:08.943" CommentCount="21" />
  <row Id="7163" PostTypeId="2" ParentId="7154" CreationDate="2011-03-18T12:55:51.097" Score="0" ViewCount="" Body="&lt;p&gt;Deterministic chaos is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Deterministic_chaos&quot; rel=&quot;nofollow&quot;&gt;special branch of physics&lt;/a&gt; that allows estimating the collective  behavior of complex systems which are governed by deterministic differential equations. Your example with the bonfire and the state of the molecules does not fall into chaotic theory, whether classical or quantum. It falls to the chapter of thermodynamics, an axiomatic system of physics that describes the collective behavior of matter  generally without entering on the substrate of the underlying equations, whether classical or quantum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am less familiar with &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_chaos&quot; rel=&quot;nofollow&quot;&gt;Quantum Chaos&lt;/a&gt; Quote: &lt;em&gt;Quantum Chaos is a branch of physics which studies how chaotic classical dynamical systems can be described in terms of quantum theory. The primary question that quantum chaos seeks to answer is, &quot;What is the relationship between quantum mechanics and classical chaos?&lt;/em&gt;&#xA;Since your example is not classically chaotic, again it is not in the regime covered by quantum chaos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe you should rethink what you mean by chaos, since it is not the physics definition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one wades into the many probable outcomes quantum mechanical description, that does not mean that one is dealing with a system that is chaotic, in the definition of chaos in physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for our second formulation: &#xA;We are absolutely affected by quantum mechanics, see the answers to the &lt;a href=&quot;http://physics.stackexchange.com/questions/7158/by-what-mechanism-do-quantum-effects-become-observable-in-normal-life-at-the-macr&quot;&gt;question you raised&lt;/a&gt; on the subject.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-18T12:55:51.097" CommentCount="0" />
  <row Id="7164" PostTypeId="1" CreationDate="2011-03-18T13:29:59.507" Score="2" ViewCount="204" Body="&lt;p&gt;The Dirac operator at 2 dimension can be written as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;D=\sum_{k=1,2}\sigma^{k}D_{k}=\left( \begin{array}{cc}&#xD;&#xA;0 &amp;amp; \partial_{x}-i\partial_{y}-i(A_x-iA_y)\\&#xD;&#xA;\partial_{x}+i\partial_{y}-i(A_x+iA_y) &amp;amp; 0&#xD;&#xA;\end{array} \right) ,&#xD;&#xA;$$&#xA; where $A_i$ is a gauge field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we add some other terms to the operator above, such as $\sigma^{3}\Delta$, where $\Delta$ is a constant, is it still a Dirac operator? Can the last term be understood as some kind of gauge field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clear the background of the question: I am reading Kane and Mele's paper on $Z_2$ topological order (&lt;a href=&quot;http://arxiv.org/PS_cache/cond-mat/pdf/0506/0506581v2.pdf&quot; rel=&quot;nofollow&quot;&gt;pdf&lt;/a&gt;). For my understanding the general question is how the inner structure (in this case is time reversal symmetry) of a Hamiltonian can be used to classify the band structure. They said in this case it is related with twisted Real K theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anyone know how to apply K theory in the graphene case?&lt;/p&gt;&#xA;" OwnerUserId="1948" LastEditorUserId="2545" LastEditorDisplayName="" LastEditDate="2011-03-20T00:18:22.057" LastActivityDate="2011-03-20T00:18:22.057" Title="A question about Dirac operator" Tags="&lt;condensed-matter&gt;&lt;mathematics&gt;&lt;graphene&gt;" CommentCount="4" FavoriteCount="1" />
  <row Id="7165" PostTypeId="2" ParentId="6928" CreationDate="2011-03-18T14:05:41.760" Score="3" ViewCount="" Body="&lt;p&gt;One of the problems is that lead does not &lt;em&gt;absorb&lt;/em&gt; neutrons, it &lt;em&gt;reflects&lt;/em&gt; them. &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutron_reflector&quot; rel=&quot;nofollow&quot;&gt;That has been used to reduce the critical size&lt;/a&gt;, &lt;em&gt;i.e&lt;/em&gt; favour the nucluear reaction by sending the neutron back into the fissile material. And indeed, &lt;a href=&quot;http://en.wikipedia.org/wiki/Criticality_accident#Incidents&quot; rel=&quot;nofollow&quot;&gt;the first deadly criticality accident&lt;/a&gt; was caused by a similar neutron deflector. This is not something someone wants in a nuclear reactor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water is much better and is currently used at Fukushima power plant because :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;it is much easier to manipulate and to get in large quantity than anything else;&lt;/li&gt;&#xA;&lt;li&gt;it is a good coolant, with high vaporization &lt;a href=&quot;http://en.wikipedia.org/wiki/Latent_heat&quot; rel=&quot;nofollow&quot;&gt;latent heat&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;its a good solvant and you an dissolve &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutron_poison&quot; rel=&quot;nofollow&quot;&gt;neutron poisons&lt;/a&gt; in it, like &lt;a href=&quot;http://en.wikipedia.org/wiki/Boric_acid&quot; rel=&quot;nofollow&quot;&gt;boric acid&lt;/a&gt;, which absorbs the neutrons;&lt;/li&gt;&#xA;&lt;li&gt;it is not toxic.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="373" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2011-03-18T14:14:37.230" LastActivityDate="2011-03-18T14:14:37.230" />
  <row Id="7166" PostTypeId="1" AcceptedAnswerId="7170" CreationDate="2011-03-18T14:16:46.213" Score="1" ViewCount="236" Body="&lt;p&gt;Well, reading about &quot;Raman Effect&quot; I saw that when the electron absorb some energy, with frequency $ \omega_{abs} $, that is different from $ \omega_{n} - \omega_{n-1} \neq \omega_{abs1} $, it go to an &quot;virtual energy level&quot; that is unstable, but nevertheless we could measure its lifetime. And more, the electron could stay in this unstable level enough time to absorb another electron with frequency $ \omega_{abs2} $, such that $ \omega_{n} - \omega_{n-1} = \omega_{abs1} + \omega_{abs2} $, and than finally go to an stable energy level, that is foreseen by the Quantum Mechanics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Is this right? An electron could stay a little time in a unstable energy level? &lt;/p&gt;&#xA;" OwnerUserId="1618" LastActivityDate="2011-03-22T17:43:44.337" Title="Unstable energy levels" Tags="&lt;quantum-mechanics&gt;&lt;spectroscopy&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="7167" PostTypeId="2" ParentId="7149" CreationDate="2011-03-18T14:18:44.090" Score="1" ViewCount="" Body="&lt;p&gt;I seriously doubt that any realistic reactor will contain 146 TONS of Uranium at any one time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if it did, it'd be so spaced out as to preclude the creation of a supercritical mass (i.e. a mass that leads to a nuclear explosion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course any operational reactor WILL be critical, if it weren't there'd be no chain reaction going on, no sustained fision reaction, no energy output.&lt;/p&gt;&#xA;" OwnerUserId="2625" LastActivityDate="2011-03-18T14:18:44.090" CommentCount="3" />
  <row Id="7168" PostTypeId="2" ParentId="7077" CreationDate="2011-03-18T14:32:09.180" Score="0" ViewCount="" Body="&lt;p&gt;It is impossible for anyone here to tell you that with any degree of certainty.&#xA;What's emitted depends completely on the state of the system. What materials are contained in it, and at what relative concentrations for example.&#xA;How it's shielded (the shielding itself will get excited and emit both particles and X-rays).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main danger from any radiological incident like this isn't the direct radiation at all (unless you're in the direct vicinity) but fallout from radioactive products dispersed through the atmosphere.&#xA;We can likely expect for example some of that in the form of rain as the water used to cool down the cores evaporates and rains down again later. This will of course be massively diluted depending on distance and time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any distance from the source, even gamma radiation will be low enough that it's no threat.&#xA;What most laymen never seem to grasp (in no small part no doubt due to scare campaigns designed to do just that) is that radiation at low doses is NOT a problem.&#xA;My guess is that the people now panicking in Tokyo (and even the US west coast) aren't getting any more exposure than they'd get during a dental X-ray, at most.&lt;/p&gt;&#xA;" OwnerUserId="2625" LastActivityDate="2011-03-18T14:32:09.180" />
  <row Id="7169" PostTypeId="2" ParentId="7154" CreationDate="2011-03-18T14:32:30.833" Score="0" ViewCount="" Body="&lt;p&gt;Your question is really about the relationship between chaos in classical mechanics and quantum physics.  The many worlds interpretation (MWI) is probably not the best way to look at the problem.  MWI is an interpretation of QM, which like all the others is not effective as a real theory.  Interpretations are not testable or falsifiable in the way a real theory is.  So if you are to use interpretations, and these do have some utility, I would recommend the Bohmian quantum mechanics.  This is not to promote any hidden variable agenda, nonlocal or otherwise, but more because the interpretation involves putting quantum mechanics in this classical-like setting.  In this way you can potentially begin to examine the quantization of chaotic Hamiltonian systems.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-18T14:32:30.833" />
  <row Id="7170" PostTypeId="2" ParentId="7166" CreationDate="2011-03-18T14:57:57.443" Score="0" ViewCount="" Body="&lt;p&gt;All excited energy levels are unstable to this or that extent. So, yes, it can. If below a given energy level $E_n$ there are &lt;strong&gt;many&lt;/strong&gt; lower possible levels $E_{n&amp;#39;} &amp;lt; E_n$, there are usually several channels of transition, each with its own probability and energy difference so we can observe emission of &lt;strong&gt;different&lt;/strong&gt; frequencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, absorbing two different photons, although possible in an intensive EM wave, normally happens rarely. I think it is one-photon absorption (excitation) and many-photon deactivation of atoms and molecules that normally happen (cascade decay).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean existence of unstable energy levels between two neighboring levels, then no, they do not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I remember deriving the atom ionization probability by an intensive EMW of low frequency, following the work by Keldysh. Atom &lt;strong&gt;can&lt;/strong&gt; be excited and even ionized with a strong constant or low-frequency EMF and it is a &lt;strong&gt;simultaneous&lt;/strong&gt; multi-photon process, when a single $\hbar \omega$ is smaller than the energy $E_n - E_{n&amp;#39;}$ required for a given transition.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-03-19T17:14:35.870" LastActivityDate="2011-03-19T17:14:35.870" />
  <row Id="7171" PostTypeId="2" ParentId="7132" CreationDate="2011-03-18T15:07:00.390" Score="0" ViewCount="" Body="&lt;p&gt;I'm not going to answer your homework problem. Instead I'm going to construct a counter example to the hint $[L_{z},V]=0$, which is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider first, for simplicity, a spherically symmetric potential $V(r)$ and a product solution &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Psi(r,\theta,\phi)=R(r)\Theta(\theta)e^{im\phi}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;to the time independent Schroedinger equation $(H-E)\Psi=0$, with Hamiltonian $H:=T+V$, with kinetic term $T:=-\frac{\hbar^2}{2m}\Delta$, and with norm $0&amp;lt; ||\Psi||&amp;lt; \infty$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, you can think of a Hydrogen-like atom, where $V(r)$ is the Coulomb potential, $R(r)$ is a Laguerre polynomial, and $\Theta(\theta)$ is an associated Legendre polynomial. If the principal quantum number $n\neq 1$ is not one, then $R(r)$ will have a radial node, i.e., there exists a radius $r_{*}&amp;gt;0$, such that $R(r_{*})=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now construct a new potential of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tilde{V}(r,\theta,\phi)=V(r)+\delta(r-r_{*})W(\theta,\phi)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution $\Psi$ from before will also solve the new problem $(\tilde{H}-E)\Psi=0$, where $\tilde{H}:=T+\tilde{V}$, since $R(r)\delta(r-r_{*})=0$. But $L_z$ may not commute with $\tilde{V}$. &lt;/p&gt;&#xA;" OwnerUserId="2451" LastActivityDate="2011-03-18T15:07:00.390" CommentCount="10" />
  <row Id="7172" PostTypeId="1" AcceptedAnswerId="7173" CreationDate="2011-03-18T16:00:43.217" Score="6" ViewCount="802" Body="&lt;p&gt;I was not able to find an answer for this question...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some radioactive elements have half-life measured in thousands of years and some others even in millions, but over 4.5 billion years all the radioactive material that was part of the initial material that formed the planet earth should have decayed by now?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is still radioactive material with short half-life to be found in nature. How is this possible and if the answer is that the new radioactive material is constantly being generated somehow, can you explain the mechanism of how this happens? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2628" LastActivityDate="2011-03-20T18:47:19.797" Title="Since radioactive material decays how is it possible that there is any left after 4.5 billion years?" Tags="&lt;radioactivity&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="1" />
  <row Id="7173" PostTypeId="2" ParentId="7172" CreationDate="2011-03-18T16:08:06.143" Score="17" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Uranium-238&quot; rel=&quot;nofollow&quot;&gt;half-life of Uranium 238&lt;/a&gt; is about the age of the Earth, so only about half of the original supply should have decayed by now.  Also, there are some radioactive nuclei that get created by interactions with cosmic rays in the upper atmosphere (carbon-14) or decay from more stable nuclei (all of the daughter nuclei between U-238 and lead, for example).&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-03-18T16:08:06.143" CommentCount="3" />
  <row Id="7174" PostTypeId="2" ParentId="7166" CreationDate="2011-03-18T16:12:23.067" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this right? An electron could stay a little time in a unstable energy level?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Short answer is -- yes, this is a correct standard description of the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me elaborate. In quantum mechanics all observables, including energy, are described with hermitian operators. Eigenvectors of these operators are the states with definite value of the observable. Energy levels are the eigenvectors of the operator of energy -- the Hamiltonian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Virtual level&quot; in the current context is not a state with definite energy. So it's not an energy level in a strict sense. Check &lt;a href=&quot;http://en.wikipedia.org/wiki/Virtual_State_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; on Wikipedia -- the part on &quot;quantum state&quot;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must also add that all that discussion is more about an interpretation than the actual derivation. The math is straightforward and independent on how you call things in your equations. Your question is similar to the question &quot;do &lt;a href=&quot;http://en.wikipedia.org/wiki/Virtual_particle&quot; rel=&quot;nofollow&quot;&gt;virtual particles&lt;/a&gt; exist&quot;. And answer depends on what you mean under &quot;exist&quot;, so you are getting into philosophy instead of physics.  &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-03-18T16:12:23.067" CommentCount="4" />
  <row Id="7175" PostTypeId="2" ParentId="7172" CreationDate="2011-03-18T16:19:37.350" Score="9" ViewCount="" Body="&lt;p&gt;Some elements with short half-lifes, are just decay products of those with long half-lifes.&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-03-18T16:19:37.350" />
  <row Id="7176" PostTypeId="2" ParentId="7172" CreationDate="2011-03-18T16:24:15.760" Score="7" ViewCount="" Body="&lt;p&gt;The short half-life elements ocuring in nature come from the decay of long-half life elements. You can see examples of &lt;a href=&quot;http://en.wikipedia.org/wiki/Decay_chain&quot; rel=&quot;nofollow&quot;&gt;decay chains&lt;/a&gt; on this wikipedia page. For example, ²²⁴Ra (3.6 days half life) is produced by the decay of ²³²Th (14 billion years decay).&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2011-03-18T16:24:15.760" />
  <row Id="7177" PostTypeId="2" ParentId="7172" CreationDate="2011-03-18T16:24:39.160" Score="12" ViewCount="" Body="&lt;p&gt;It's because the half life time is also incredibly long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The half-life of Uranium-238 is $4.5*10^9$ (=4.5 billion) years. Thorium-232 has $1.4*10^{10}$. Potassium-40 has $1.2*10^9$. These are all examples of &lt;a href=&quot;http://en.wikipedia.org/wiki/Primordial_nuclide&quot; rel=&quot;nofollow&quot;&gt;primordial nuclides&lt;/a&gt;. Such half lives are of the order of the age of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also the effect of having a &lt;a href=&quot;http://en.wikipedia.org/wiki/Decay_chain&quot; rel=&quot;nofollow&quot;&gt;decay chain&lt;/a&gt;, since decay products themself can also be radioactive. Although if you look at the tables you'll find that only a few decay products have a significant lifetime -- most are either stable or negligble short. It does explain why you will always find materials with short half lives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One noteworthy mentioning is ofcourse carbon-14, which is used in radiocarbon dating (i.e. estimating the life of certain soils or the remains of plants and animals). Here stable nitrogen-14 is turned into a radioactive carbon-14 by a collission with a cosmic neutron (which replaces one proton). The carbon is then absorbed from the atmosphere by plants or oceans. Carbon-14 has a half-life of roughly 6000 years, which is considrably shorter than the earths lifetime. &lt;/p&gt;&#xA;" OwnerUserId="1287" LastActivityDate="2011-03-18T16:24:39.160" CommentCount="3" />
  <row Id="7179" PostTypeId="1" CreationDate="2011-03-18T16:42:06.760" Score="2" ViewCount="158" Body="&lt;p&gt;According to Hubble's law, light and other kinds of electromagnetic radiation emitted from distant objects are redshifted. The more distant the source, the more intense is the redshift. Now, the expansion of the universe is expected to explain the redshift and its nearly linear dependence on distance between source and observer. But isn't there an other source influencing the redshift? We know that a light beam passing the Sun is deflected by the Sun's gravity in accordance with predictions made by Einstein's general theory of relativity. This deflection is dependant on a gravitational interaction between the Sun and the light beam. Thus, the position of the Sun is affected by the light beam, though by such a tiny amount that it is impossible to detect the disturbance of the Sun's position. Now, during its journey to the Earth a light beam, originating from a distant source in the Universe, is passing a certain amount of elementary particles and atoms. If the light beam interacts gravitationally with those elementary particles and atoms, affecting the microscopic mechanical properties of the individual elementary particles and atoms at issue, can this interaction be detected as a redshift of the light beam? If so, could we use this gravity redshift to measure the mean density of matter and energy in space?         &lt;/p&gt;&#xA;" OwnerUserId="2630" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-03-19T00:31:06.637" LastActivityDate="2011-03-19T00:31:06.637" Title="Gravitation as the source of redshift of light beams" Tags="&lt;gravity&gt;&lt;gravitational-redshift&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="7180" PostTypeId="2" ParentId="490" CreationDate="2011-03-18T17:02:25.020" Score="0" ViewCount="" Body="&lt;p&gt;I we take the initial velocity as zero (u=0) then we have s=1/2at^2 from which a trivial solution provides two displacement s1 and s2 depending on if we take the positive or negative root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drawing a graph of s v t^2 gives a parabola and our two solutions appear somewhere on this curve.  This is where we need to disentangle the mathematical explanation from the physical explanation.  Mathematically they are both valid and it appears our equation is very useful but physically we can choose to prefer one solution to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case I would take the positive root as this describes the position at some positive time, where the negative root gives the position at some negative time.  Taking time to be positive and going forward isn't always correct but it can help differentiate between mathematically equivalent answers.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-03-18T17:02:25.020" />
  <row Id="7181" PostTypeId="2" ParentId="7179" CreationDate="2011-03-18T17:13:48.660" Score="3" ViewCount="" Body="&lt;p&gt;This is similar to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tired_light&quot; rel=&quot;nofollow&quot;&gt;&quot;tired light&quot;&lt;/a&gt; hypothesis, and isn't consistent with current cosmological data.  The best argument against tired light is that if the redshift was caused by an interaction between light and matter, there would be a frequency dependence of this interaction, as light interacts with matter differently when it is at different frequencies.  Astronomical sources, however, don't exhibit any frequency-dependent dispersion in their spectra--they look like they're just redshifted back, as you would expect if they were receding.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-03-18T17:13:48.660" CommentCount="1" />
  <row Id="7182" PostTypeId="2" ParentId="7179" CreationDate="2011-03-18T17:37:53.537" Score="5" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If the light beam interacts gravitationally with those elementary particles and atoms, affecting the microscopic mechanical properties of the individual elementary particles and atoms at issue, can this interaction be detected as a redshift of the light beam? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Consider light passing a particle.  The light comes from a large distance and leaves to a large distance.  From gravity it will very slightly blueshift on approach and redshift when leaving the particle, which is both far too little to measure and the total effect from gravity on the frequency is essentially zero.  Gravity can &lt;em&gt;deflect&lt;/em&gt; the light and this is how galaxies cause gravitational lensing.  So this cannot be detected as redshift.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If so, could we use this gravity redshift to measure the mean density of matter and energy in space?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No.  But the &lt;em&gt;deflection&lt;/em&gt; can be used to measure intervening mass.  Google for the &quot;bullet cluster&quot; to show an example.&lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2011-03-18T17:37:53.537" />
  <row Id="7183" PostTypeId="2" ParentId="7140" CreationDate="2011-03-18T18:33:33.137" Score="2" ViewCount="" Body="&lt;p&gt;Mostly because they are heavy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rocks erode putting their constituents into solution, the heavy stuff settles out in river/sea beds, and metals are heavy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For many metals hydrothermal process are more important. Super hot water deep in the earth  dissolves the rock containing the minerals, it moves along cracks in the rock and cools depositing the salt and metals as lines in the rock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an asteroid with no geological process the metals are found in their raw state having cooled directly from the original ball of primeval gas&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-03-18T18:33:33.137" />
  <row Id="7184" PostTypeId="1" AcceptedAnswerId="7186" CreationDate="2011-03-18T18:36:54.397" Score="4" ViewCount="190" Body="&lt;p&gt;There is no tunneling in the case of infinite potential barrier, but there is when we have a finite well. In the classical analog, in the first case we have a particle bouncing between to infinitely rigid impenetrable walls and there is no tunneling, same as the quantum case. But if we have a finite barrier, means we have walls of finite rigidity, say made of cork or something. Then the particle would just break through some of the cork and it's probability of being found further in the cork wall will decay steadily. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can understand  discrete energy levels being a new thing, because they behave like a wave that's confined and not like particles confined, but why tunneling? &lt;/p&gt;&#xA;" OwnerUserId="2599" LastActivityDate="2011-03-18T19:46:25.680" Title="Why is tunneling not a classical idea?" Tags="&lt;quantum-mechanics&gt;&lt;classical-mechanics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="7185" PostTypeId="2" ParentId="6400" CreationDate="2011-03-18T18:42:55.983" Score="0" ViewCount="" Body="&lt;p&gt;I wonder if hydroelectric or fossil fuel plants slow rotation more?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydroelectric concentrates large masses of water at slightly high elevations, while fossil fuel puts much larger masses of CO2 + H2O much higher in the atmosphere&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-03-18T18:42:55.983" CommentCount="3" />
  <row Id="7186" PostTypeId="2" ParentId="7184" CreationDate="2011-03-18T18:51:01.240" Score="6" ViewCount="" Body="&lt;p&gt;Not a good analogy. Here's why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your cork wall system is &lt;em&gt;very&lt;/em&gt; complicated. That wall isn't just a potential with finite height. Rather, it contains holes where the potential is zero (or near zero), so that the particle can just propagate further and it contains full woody parts where the potential is infinite and the particle bounces back. So it is a very complicated many particle system which you can reduce to one parameter (like penetration depth, possibly depending on energy). But the resulting system will be probabilistic and &lt;em&gt;very&lt;/em&gt; different from just a single particle with finite height potential. Because in this simple system the particle will bounce back with 100% probability if its energy is below the potential and pass with 100% probability if its energy is bigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you take this simple system (with just one degree of freedom) and investigate it from the point of view of quantum mechanics you'll discover that there's non-zero probability that particle will pass through the potential even if it doesn't have sufficient energy to pass it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this strange? Well, the particle has a total energy $E = T + V$. This means that its kinetic energy is ${1\over 2}mv^2 = E - V$. As you can see, this equation makes no sense classically if the quantity on the r.h.s. is negative because you'll obtain imaginary velocity. But in quantum mechanics this still makes sense because the particle is also a wave and that imaginary factor means that this wave will exponentially decay, rather than just propagate freely.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-03-18T18:51:01.240" />
  <row Id="7187" PostTypeId="2" ParentId="7179" CreationDate="2011-03-18T19:06:21.780" Score="2" ViewCount="" Body="&lt;p&gt;As Chris Kuklewicz points out, a photon passing a particle, or a larger clump of matter, is blueshifted as it falls in and equally redshifted as it comes out, so there's no overall &quot;tired-light-like&quot; redshift due to an effect of this sort. On the other hand, for certain choices of coordinates, you actually can regard the observed cosmological redshift as partially or even entirely a gravitational redshift. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Newtonian language, if you imagine the source of the light at the center of a spherically symmetric expanding spacetime, then the light travels &quot;uphill&quot; in a gravitational potential all the way to the observer. The observed redshift is partially due to this redshift and partially due to the observer's motion. Conversely, if you put the observer at the center, the light &quot;falls downhill&quot; all the way to the observer. This gives a different breakdown of the observed redshift into gravitational and Doppler contributions. Of course, the world is relativistic, not Newtonian, but you can think of these two pictures as representations of the general relativistic point of view in two different coordinate systems. The original reference for this stuff is a paper by Bondi, Monthly Notices of the RAS, 107, 410-425 (1947), and it's essentially the point of view taken by Peacock in this &lt;a href=&quot;http://arxiv.org/abs/0809.4573&quot; rel=&quot;nofollow&quot;&gt;set of notes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This just drives home the fact that whether the observed redshift is gravitational or not depends on your choice of coordinates. There is no &quot;fact of the matter.&quot; There are so many confusing statements about this in textbooks that David Hogg and I wrote a &lt;a href=&quot;http://arxiv.org/abs/0808.1081&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; arguing for our own point of view, that the most natural way to think about the cosmological redshift is as a pure Doppler shift -- not a gravitational redshift and &lt;em&gt;certainly&lt;/em&gt; not a whole different animal caused by &quot;the stretching of space.&quot; (The textbook by Harrison, despite many good qualities, is an egregious offender in this regard.) Even if you don't agree with our conclusion, this paper may be useful in putting the issue in context.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-18T19:06:21.780" CommentCount="1" />
  <row Id="7188" PostTypeId="2" ParentId="7184" CreationDate="2011-03-18T19:46:25.680" Score="3" ViewCount="" Body="&lt;p&gt;Marek's answer is rather too &lt;em&gt;non&lt;/em&gt;-statistical classical. Your cork analogy &lt;em&gt;looks&lt;/em&gt; to me at first sight not statistical in a way that gets close enough to quantum theory. As an aside, it's generally better to make your models slightly more abstract-idealized mathematical &amp;mdash; note that Marek has replied with a model that is both much more abstract and much more idealized than your model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're willing to make the potential well have expected depth $E$ with some finite variance because the apparatus that constrains the classical particle is noisy, however, then the analogy is not so bad, if the potential well is then sometimes shallow enough for the classical particle to escape. There is noise in quantum theory &amp;mdash;insofar as one carefully thinks in terms of quantum fluctuations and is careful to distinguish them from thermal fluctuations&amp;mdash;, so it doesn't seem unreasonable to introduce noise into classical models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can alternatively surround the classical particle with some kind of noisy environment, in which case the energy of the particle will fluctuate. If the fluctuations are enough that the energy of the classical particle is sometimes more than the depth of the potential well, then again the particle can escape from the potential well with some finite probability. If you haven't previously looked at Nelson models, get hold of his &quot;&lt;em&gt;Quantum fluctuations&lt;/em&gt; / by Edward Nelson, Princeton University Press, c1985.&quot;&#xA;I don't know of and couldn't find a web resource that gives an adequate description of this approach, however, even though it's arguably closer to being acceptable to Physicists than de Broglie-Bohm type approaches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, your intuition at this level is OK, but it's not in this context that the really difficult issues arise, and also not in the double-slit kind of context. Kochen-Specker and the violation of Bell inequalities are much more challenging. In response to those challenges I have found it much more productive to characterize the differences between classical random fields and quantum fields, both of which are intrinsically statistical, instead of trying to think in terms of classical particles in comparison with quantum particles. The random field context is close enough to the quantum field context that discrete energy levels are not a novelty. I presume you're not &lt;em&gt;getting&lt;/em&gt; my Answers to your Questions, but your Questions are very familiar to me.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-03-18T19:46:25.680" />
  <row Id="7189" PostTypeId="1" AcceptedAnswerId="7192" CreationDate="2011-03-18T21:17:04.987" Score="1" ViewCount="109" Body="&lt;p&gt;Based on the 1D case mentioned in Griffiths, I decided to try looking at the features of 3D Gaussian wavefunctions, i.e. (position basis) wavefunctions of the form $\psi(\mathbf{r}) = Ae^{-\mathbf{r}^\dagger\mathsf\Sigma\mathbf{r}/4}$, where &lt;em&gt;A&lt;/em&gt; is a normalization constant, &lt;strong&gt;r&lt;/strong&gt; is position, &amp;Sigma; is a positive-definite symmetric matrix (which by a suitable change of coordinate basis can be made diagonal), and &lt;sup&gt;&amp;dagger;&lt;/sup&gt; denotes the conjugate transpose. Applying standard results for Gaussian integrals, I was able to get&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\langle \mathbf{r} \rangle = 0$&lt;/li&gt;&#xA;&lt;li&gt;$\langle r^2\rangle = \operatorname{Tr}\mathsf\Sigma$&lt;/li&gt;&#xA;&lt;li&gt;$\langle \mathbf{p} \rangle = 0$&lt;/li&gt;&#xA;&lt;li&gt;$\langle p^2\rangle = \frac{\hbar^2}{4}\operatorname{Tr}\mathsf\Sigma^{-1}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, substituting into Heisenberg's uncertainty principle and rearranging terms, it follows that, in order to get minimum uncertainty with respect to $\mathrm{r}$ and $\mathrm{p}$, we need to have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; $(\operatorname{Tr}\mathsf\Sigma)(\operatorname{Tr}\mathsf\Sigma^{-1})=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where I'm running into a difficulty. As I mentioned before, the matrix &amp;Sigma; can always be assumed to be diagonal. Then the only possible solution for &amp;Sigma; is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathsf\Sigma = \begin{pmatrix} 1 &amp;amp; 0 &amp;amp; 0\\ 0 &amp;amp;-1 &amp;amp;0\\ 0 &amp;amp;0 &amp;amp;1\end{pmatrix}\times\mathrm{constant}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this contradicts the fact that &amp;Sigma; is positive-definite (the -1 would imply that one of the coordinates has negative uncertainty, an absurdity). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming I did all the calculations correctly, this seems to imply that a Gaussian wavefunction is not the minimum uncertainty wavefunction with respect to &lt;strong&gt;r&lt;/strong&gt; and &lt;strong&gt;p&lt;/strong&gt;. On the other hand, it's comparatively trivial to show that it &lt;em&gt;is&lt;/em&gt; the minimum uncertainty wavefunction with respect to &lt;em&gt;x&lt;/em&gt; and &lt;em&gt;p&lt;/em&gt;&lt;sub&gt;&lt;em&gt;x&lt;/em&gt;&lt;/sub&gt;, &lt;em&gt;y&lt;/em&gt; and &lt;em&gt;p&lt;/em&gt;&lt;sub&gt;&lt;em&gt;y&lt;/em&gt;&lt;/sub&gt;, and &lt;em&gt;z&lt;/em&gt; and &lt;em&gt;p&lt;/em&gt;&lt;sub&gt;&lt;em&gt;z&lt;/em&gt;&lt;/sub&gt; individually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a wavefunction which is the minimum unceratinty wavefunction with both respect to the individual coordinates (e.g. &lt;em&gt;x&lt;/em&gt; and &lt;em&gt;p&lt;/em&gt;&lt;sub&gt;&lt;em&gt;x&lt;/em&gt;&lt;/sub&gt;) &lt;em&gt;and&lt;/em&gt; with respect to &lt;strong&gt;r&lt;/strong&gt; and &lt;strong&gt;p&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; It was asked by &lt;a href=&quot;http://physics.stackexchange.com/users/329/marek&quot;&gt;marek&lt;/a&gt; what I meant by  &quot;minimum uncertainty with respect to $\mathbf{r}$ and $\mathbf{p}$&quot;. To answer this, recall that the generalized uncertainty principle takes the form of&#xA;$$ \sigma_A\sigma_B \geq \frac{1}{2}\left|\langle[A,B]\rangle\right|.$$&#xA;Although I'm not entirely sure it's valid to do so, I assumed that to calculate the commutator $[\mathbf{r},\mathbf{p}]$ I could use the formalism of geometric algebra (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometric_algebra&quot; rel=&quot;nofollow&quot;&gt;Geometric algebra&lt;/a&gt;). Then&#xA;$$\begin{align*}&#xD;&#xA; 	[\mathbf{r},\mathbf{p}]f 	&amp;amp;= \frac{\hbar}{i}\mathbf{r}\nabla f - \frac{\hbar}{i}\nabla(f\mathbf{r})\\&#xD;&#xA; 					&amp;amp;= \frac{\hbar}{i}\sum_{jk} \left[x^j\hat{\mathbf{e}}_j\frac{\partial f}{\partial x^k}\hat{\mathbf{e}}^k - \frac{\partial}{\partial x^k}\left(fx^j\hat{\mathbf{e}}_j\right)\hat{\mathbf{e}}^k\right]\\&#xD;&#xA; 					&amp;amp;= \frac{\hbar}{i}\sum_{jk} \left[ x^j\frac{\partial f}{\partial x^k} \hat{\mathbf{e}}_j\hat{\mathbf{e}}^k - \frac{\partial f}{\partial x^k}x^j\hat{\mathbf{e}}_j\hat{\mathbf{e}}^k - f{\delta^j}_k\hat{\mathbf{e}}_j\hat{\mathbf{e}}^k\right]\\&#xD;&#xA; 					&amp;amp;= \frac{\hbar}{i} f,&#xD;&#xA;\end{align*}$$&#xA;where $f$ is an arbitrary function, $x^1,x^2,x^3$ are the position coordinates, and $\hat{\mathbf{e}}_1,\hat{\mathbf{e}}_2,\hat{\mathbf{e}}_3$ are the standard Cartesian basis vectors. Thus, the uncertainty principle for $\mathbf{r}$ and $\mathbf{p}$ takes the form&#xA;$$\sigma_\mathbf{r}\sigma_\mathbf{p} \geq \frac{\hbar}{2},$$&#xA;which means that the minimum uncertainty wavepacket with respect to $\mathbf{r}$ and $\mathbf{p}$ must satisfy &#xA;$$\sigma_\mathbf{r}\sigma_\mathbf{p} = \frac{\hbar}{2}.$$&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditorDisplayName="" LastEditDate="2011-03-18T22:28:29.137" LastActivityDate="2011-03-18T22:28:29.137" Title="3D Minimum uncertainty wavepackets" Tags="&lt;quantum-mechanics&gt;&lt;uncertainty-principle&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7190" PostTypeId="2" ParentId="7112" CreationDate="2011-03-18T21:41:40.923" Score="5" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;First point I want to address is a gross mischaracterization of people working on Bohmian mechanics or other hidden variable interpretations. I quote&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Quantum mechanics remains unacceptable for many people, for quasi-religious reasons.&#xA;  [...]&#xA;  it's time for those people to seriously consider the hypothesis that quantum mechanics is really how the world works.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;People for whom it is hard to swallow the hard reality of quantum mechanics, people who want to understand nature in terms of every day classical intuitions are the ones who often advocate Bohmian mechanics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a straw man. Nobody I know working or having worked on hidden variable interpretations of quantum mechanics, be it Bell, Bohm, Goldstein, Bricmont, Maudlin, de Broglie or other, denies the reality of quantum mechanics. Quite the contrary, I would even say they are/were more aware of the implications than many people otherwise working with quantum mechanics. In particular they fully recognize the non-local nature of quantum mechanics. But I'll come back to that. I can of course not speak for ALL people dealing in hidden variable theories or other alternatives. I'm sure there will be a faire share of crackpots and other cranks. But there are many people who know their stuff.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Second point is the &lt;a href=&quot;http://arxiv.org/abs/0704.2529&quot; rel=&quot;nofollow&quot; rel=&quot;nofollow&quot;&gt;Zeilinger article&lt;/a&gt; that @Luboš Motl mentions. I've only skimmed through the article, but it seems that the only non-local hidden variable theories that are disposed of are ones devised by Leggett. Moreover, on page 3 of the article, the authors recognize that while their work indeed rules out a vast class of non-local hidden variable theories, it does not rule out Bohmian mechanics. So it is disingeneous to use this article to discredit Bohmian mechanics. The work in the article is however very valuable since it puts more strict constraints on the type of non-local hidden variable theories that can reproduce the results of quantum mechanics. This is an interesting question that is also discussed in the excellent book &lt;a href=&quot;http://www.wiley-vch.de/publish/en/books/ISBN3-527-40391-4/?sID=b877331337ea4f5dd071768a451804dd&quot; rel=&quot;nofollow&quot; rel=&quot;nofollow&quot;&gt;Quantum Paradoxes by Yakir Aharonov and Daniel Rohrlich&lt;/a&gt;. In one of the later chapters, they discuss the very special nature of non-locality in quantum mechanics, pointing out that there are many varying degrees of non-locality which are either stronger or weaker than the non-locality of quantum mechanics. They say it is still an open problem to find a precise characterization of the nature of quantum non-locality which could be turned into a postulate from which quantum mechanics could be derived in a way analoguous to the way special relativity is derived from its basic postulates.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now, about Bell's theorem. I'll quote a piece of the &lt;a href=&quot;http://www.bslps.be/meaningWF.pdf&quot; rel=&quot;nofollow&quot; rel=&quot;nofollow&quot;&gt;following article&lt;/a&gt; which itself refers to a thought experiment proposed by Tim Maudlin. Of course, you'll recognize it is a variant of an EPR type experiment.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Here is a puzzle: two persons, call them $X$ and $Y$ , leave a room&#xA;  through opposite doors; at that point, each is asked a question. The&#xA;  precise nature of the questions does not matter, but there are three&#xA;  possible questions (say, $A$, $B$ and $C$). Each person must answer yes&#xA;  or no. This “experiment” is repeated many times, with sometimes&#xA;  the same question, sometimes diﬀerent questions being asked at the&#xA;  two doors. The two persons are allowed to decide, before leaving&#xA;  the room, to follow any strategy they want, but not to communicate&#xA;  with each other, after they have heard the questions.&#xA;  The statistics of answers have some strange properties. First&#xA;  of all, it turns out that when the two people are asked the same&#xA;  question, they always give the same answer. Is that mysterious? Of&#xA;  course not; they simply decide, before leaving the room, to follow a&#xA;  certain strategy: for example, to both say ‘yes’ if the question is $A$,&#xA;  ‘no’ if the question is $B$ and ‘no’ if the question is $C$. Altogether,&#xA;  there are $8 = 2^3$ different such strategies. Before proceeding further, the reader has to answer for himself or herself the following&#xA;  question: Is there any other way? Is there any way to account for&#xA;  the perfect correlations between the results at the two doors without assuming that the answers were predetermined (if we assume&#xA;  that the people cannot have any communication whatsoever with&#xA;  each other once the questions are asked)? I have never seen any&#xA;  suggestion of another possibility and I believe that if Bell’s theorem&#xA;  is arguably the most widely misunderstood result in the history of&#xA;  physics, it is precisely because this question is not answered before&#xA;  proceeding further.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So, let us consider, for the time being, the assumption that the&#xA;  answers are predetermined and let us call $v_i(\alpha) = \pm 1$, $i = X, Y$ ,&#xA;  $\alpha = A, B, C$ those answers. These are “random variables”, namely&#xA;  they may take diﬀerent values when one repeats the “experiment”.&#xA;  However, if one looks at the statistics of answers when diﬀerent&#xA;  questions are asked at the two doors, one ﬁnds that the frequencies&#xA;  of the events in which the same answers are given is 1/4. And this,&#xA;  combined with the perfect correlations is strange. Indeed, a version&#xA;  of the no hidden variable theorems (similar to the one discussed in&#xA;  the previous section), known as Bell’s theorem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll leave out the proof, you can read it for yourself in the &lt;a href=&quot;http://www.bslps.be/meaningWF.pdf&quot; rel=&quot;nofollow&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;. It is just the usual Bell type inequality.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the conclusion of all this? We started from one crucial&#xA;  assumption: absence of “communication” between the two persons&#xA;  once they are asked the questions. I will from now on revert to a&#xA;  less anthropomorphic language and call this assumption “locality”&#xA;  – assume that there is no causal connection whatsoever between the&#xA;  two wings of the experiment. Then, we are led to a contradiction,&#xA;  so that this assumption has to be dropped.&#xA;  It is important to understand the logic of the argument: the&#xA;  perfect correlations plus the absence of communication (i.e. locality) between the two wings of the experiments, leads us to postulate the existence of the variables $v_i(\alpha)$ [...]. However, merely assuming that&#xA;  those variables exist leads to a contradiction with the experimental&#xA;  results obtained when different questions are asked. To put it simply: locality plus perfect (anti)correlation implies hidden variables;&#xA;  however, the latter plus statistics when different angles are measured implies a contradiction. Both the perfect correlations and&#xA;  the statistics for diﬀerent angles are empirical results; the theorem&#xA;  is a theorem, namely a logical deduction; the only assumption was&#xA;  the lack of “communication”, or locality. Hence, locality has to be&#xA;  given up, period.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, I'll stress again the most important point of this text: &lt;strong&gt;hidden variables are postulated to explain the correlations in the absence of communication (i.e. locality assumption). But this leads to a contradiction with the predictions of quantum mechanics. Quantum mechanics is right, we have tested it experimentally, therefore locality isn't right.&lt;/strong&gt; I think that's fairly limpid. Anybody claiming that quantum mechanics doesn't force us to conclude non-locality is the one who doesn't really accept quantum mechanics, contrary to what detractors claimed in point 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In his book &lt;a href=&quot;http://books.google.be/books?id=FGnnHxh2YtQC&amp;amp;dq=speakable+and+unspeakable+in+quantum+mechanics&amp;amp;source=gbs_navlinks_s&quot; rel=&quot;nofollow&quot; rel=&quot;nofollow&quot;&gt;Speakable and unspeakable in quantum mechanics&lt;/a&gt;, which is really a collection of Bell's papers, there are various presentations of the theorem, some of which are clearer than others. Another book which does a good job at explaining the interpretation and consequences of Bell's theorem is Tim Maudlin's &lt;a href=&quot;http://www.chronon.org/reviews/Quantum_Nonlocal.html&quot; rel=&quot;nofollow&quot; rel=&quot;nofollow&quot;&gt;Quantum Non-Locality and Relativity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;And I've barely addressed Bohmian mechanics yet and the question of the OP. Why do people still work on Bohmian mechanics? Well, the original intent of EPR was to show that quantum mechanics is incomplete. EPR showed that assuming locality inevitably leads to hidden variables, which is the first part of the previous argument. But the second part of the argument, provided by Bell shows that this is inconsistent with quantum mechanics. Therefore locality can't be right. One might wonder, even if we have to drop locality, why insist on supplementing the theory with hidden variables?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It's important to understand that the only variables added in Bohm's theory are the positions of particles. There are no hidden variables for spin, momentum, angular momentum, etc... It does however explain how the measured values for spin, momentum, etc... arise from the specific experimental configurations. This is what is called &lt;em&gt;contextuality&lt;/em&gt; and has already been mentioned by @Sina Salek. Non-locality is explained by the fact that the wave-function is a function on configuration space and not in physical space. Hence the possibility of entanglement. You can read more as well as find further references in the &lt;a href=&quot;http://www.bslps.be/meaningWF.pdf&quot; rel=&quot;nofollow&quot; rel=&quot;nofollow&quot;&gt;article I mentioned&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond the fact that Bohmian mechanics makes the non-locality in quantum mechanics more explicit, it also provides a deterministic interpretation of quantum mechanics, showing that quantum mechanics does not force randomness upon us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, wether one likes Bohmian mechanics or not, one can not deny that these are definitely strong points of the theory. However, it does have its weaknesses as well. As @sb1 mentioned, Einstein thought it was cheap and in a way, I do agree that the way the extra equation for the positions is added is rather cheap and even ugly. It does also have its fair share of problems, for instance in trying to generalize into a QFT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I leave this &lt;a href=&quot;http://plato.stanford.edu/entries/qm-bohm/&quot; rel=&quot;nofollow&quot; rel=&quot;nofollow&quot;&gt;link containing a short exposition of Bohmian mechanics and further references.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Finally, I want to address @Matt Reece about QFT and locality. I'm no expert in QFT, but I do remember that locality is imposed on the level of operators by requiring that two observables localized within distinct space-time regions should commute. I'm not entirely sure what this implies w.r.t. entangled states, but I suppose it doesn't rule them out otherwise QFT would be in contradiction with experiment. (I wonder though if there is a complete treatment of entangled states within the context of QFT?) But if entangled states are allowed, then violations of Bell type inequalities still are possible and thus non-locality is a fact. What this means is that whatever the status of the principle of locality in QFT, it's a weaker form of locality than the one required in setting up Bell's inequalities.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2011-03-18T21:53:37.013" LastActivityDate="2011-03-18T21:53:37.013" CommentCount="9" />
  <row Id="7191" PostTypeId="1" CreationDate="2011-03-18T21:42:32.710" Score="3" ViewCount="177" Body="&lt;p&gt;We're finding forces (and whether they're in tension or compression) using the Method of Joints and Method of Sections. I don't understand why sometimes it's necessary to find the support reactions, but sometimes it's not. Please help. Thank you!&lt;/p&gt;&#xA;" OwnerUserId="2638" LastActivityDate="2011-04-07T10:03:14.823" Title="Statics question: When do I need to find support reactions?" Tags="&lt;statics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7192" PostTypeId="2" ParentId="7189" CreationDate="2011-03-18T21:47:49.067" Score="6" ViewCount="" Body="&lt;p&gt;It seems that problem here is with mishandling vector quantities. We want to compute things such as $\left&amp;lt;p^2\right&amp;gt;$ but these are in fact $\sum_i \left&amp;lt;p_i^2\right&amp;gt;$ and so the problem decomposes into components where the standard HUP and minimality conditions can be applied. But what you've done is that you applied one-dimensional HUP to $\left&amp;lt;x^2\right&amp;gt;$ and $\left&amp;lt;p^2\right&amp;gt;$ which just can't be right. The correct form of HUP in this case would be $$\sum_i \left&amp;lt;x_i^2\right&amp;gt;\left&amp;lt;p_i^2\right&amp;gt; \geq 3 {\hbar^2 \over 4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to reiterate, there is really nothing new to solve in more dimensions as the problem decomposes completely and you can write your solution as $\Psi(x,y,z)$ = $\psi_x(x)\psi_y(y)\psi_z(z)$ with each $\psi_{\alpha}$ a Gaussian from the one-dimensional variant of this problem.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-03-18T21:47:49.067" />
  <row Id="7193" PostTypeId="2" ParentId="7172" CreationDate="2011-03-18T22:09:53.870" Score="1" ViewCount="" Body="&lt;p&gt;Since the universe has not yet run out of hydrogen, new stars are forming.  When those stars reach end of life and nova they produce the heavy elements beyond iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So until there are no more stars (large enough to create the heaviest elements such as uranium) and is no more hydrogen to form further stars, there will be a constant creation of further heavy elements.&lt;/p&gt;&#xA;" OwnerUserId="2640" LastActivityDate="2011-03-18T22:09:53.870" />
  <row Id="7194" PostTypeId="2" ParentId="4522" CreationDate="2011-03-18T22:32:54.693" Score="0" ViewCount="" Body="&lt;p&gt;But by crushing the bottle aren't you creating a negative pressure inside the bottle that's just going to suck more CO2 out? The bottle wants to return to its former larger volume..&lt;/p&gt;&#xA;" OwnerUserId="2641" LastActivityDate="2011-03-18T22:32:54.693" CommentCount="1" />
  <row Id="7196" PostTypeId="1" CreationDate="2011-03-19T02:43:39.280" Score="-2" ViewCount="236" Body="&lt;p&gt;Starting with a pot of cold tap water, I want to cook a hard-boiled egg using the minimum amount of energy. Is it more energy efficient to bring a pot to boil first and then put the egg in it, or to put the egg in the pot of cold water first and let it heat up with the water?&lt;/p&gt;&#xA;" OwnerUserId="1462" LastActivityDate="2011-03-19T12:07:18.720" Title="What is the most energy efficient way to boil an egg?" Tags="&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="7197" PostTypeId="2" ParentId="7196" CreationDate="2011-03-19T03:42:35.447" Score="0" ViewCount="" Body="&lt;p&gt;My cookbook says an egg should simmer in water for about 10 minutes. If you wait till the water simmers and then add the egg, the water will start boiling again in a few seconds, then needs to simmer for the 10 minutes. The heat of vaporization loss takes place for about 10 minutes.&#xA;If the egg is placed in cold water first, it'll cook in fewer than 10 minutes of simmering as the pot heats up to boiling, so the heat of vaporization loss will be less than 10 minutes. &lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="1924" LastEditorDisplayName="" LastEditDate="2011-03-19T03:51:40.960" LastActivityDate="2011-03-19T03:51:40.960" />
  <row Id="7198" PostTypeId="2" ParentId="7196" CreationDate="2011-03-19T04:11:00.790" Score="2" ViewCount="" Body="&lt;p&gt;Break egg into vacuum vessel, lower pressure until egg boils (sorry don't  have a phase diagram for eggs handy)&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-03-19T04:11:00.790" CommentCount="6" />
  <row Id="7199" PostTypeId="1" CreationDate="2011-03-19T04:38:12.490" Score="2" ViewCount="243" Body="&lt;p&gt;Let me put things into perspective by comparing with other applications of string theory. Nowadays review papers written by cosmologists about inflation models often discuss string theory scenarios such as brane inflation, among other non-stringy models. Also you often hear about heavy-ion theorists and experimentalists talking about the AdS/CFT bound on QGP viscosity. However, the AdS/CFT approach to high-Tc superconductors, while hailed as a major new achievement of string theory and heavily cited within string theory circles, seems to have received little citation from the condensed matter community. Hardly any physicist with a traditional background in high-Tc superconductors have made references to AdS/CFT in their publications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find this situation to be a bit peculiar. Why do you think this is the case? Does this situation have any implication for the long-term promise of this research field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While being a complete outsider, I'd like to venture some guesses to this question. Please tell me whether they sound sensible:&#xA;1) Condensed matter physicists already had several theories that more or less explain the essential features of high-Tc, so AdS/CFT is not seen as a gift from God. What they really want is a &quot;recipe&quot; for room-temperature superconductors, but AdS/CFT hasn't provided this.&#xA;2) AdS/CFT currently only deals with toy models rather than realistic condensed matter systems, so people whose research interest lies firmly in real-world phenomenology see AdS/CFT as worthless.&#xA;3) Due to very different training, condensed matter physicists simply don't understand this research program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you think?&lt;/p&gt;&#xA;" OwnerUserId="2643" LastActivityDate="2011-03-19T05:15:59.170" Title="Why is the AdS/CFT approach to superconductors rarely cited in condensed matter publications?" Tags="&lt;ads-cft&gt;&lt;superconductivity&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="2" />
  <row Id="7200" PostTypeId="1" CreationDate="2011-03-19T04:46:35.603" Score="4" ViewCount="154" Body="&lt;p&gt;Schwarzschild singularities are described by the Kantowski-Sachs metric with a contracting S&lt;sup&gt;2&lt;/sup&gt;. Of course, T-duality doesn't apply to S&lt;sup&gt;2&lt;/sup&gt;. But what about a Kasner-type singularity with two contracting spatial dimensions compactified over a torus T&lt;sup&gt;2&lt;/sup&gt;, and an expanding spatial dimension? The T-dual of the torus gives rise to a geometry which is expanding in all spatial directions.&lt;/p&gt;&#xA;" OwnerUserId="2645" LastActivityDate="2011-03-19T08:30:22.360" Title="Can T-duality resolve spacelike singularities?" Tags="&lt;string-theory&gt;&lt;quantum-gravity&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="7201" PostTypeId="2" ParentId="7199" CreationDate="2011-03-19T04:50:09.403" Score="-3" ViewCount="" Body="&lt;p&gt;I didn't know that the AdS/CFT correspondence solved high-T&lt;sub&gt;c&lt;/sub&gt; superconductivity. Do you have any references? Most glaringly, high-T&lt;sub&gt;c&lt;/sub&gt; superconductors aren't conformal field theories. How can the AdS/CFT correspondence apply to something which isn't remotely a conformal field  theory?&lt;/p&gt;&#xA;" OwnerUserId="2645" LastEditorUserId="2645" LastEditorDisplayName="" LastEditDate="2011-03-19T05:15:59.170" LastActivityDate="2011-03-19T05:15:59.170" CommentCount="3" />
  <row Id="7202" PostTypeId="1" CreationDate="2011-03-19T05:46:52.657" Score="1" ViewCount="125" Body="&lt;p&gt;I was reading Wikipedia which stated &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;....Voyager 1's current relative velocity is 17.062  km/s, or 61,452 kilometres per hour (38,185 mph).....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It travels away from sun. So sun's gravity must slow it down. What propels Voyager to that speed?&lt;/p&gt;&#xA;" OwnerUserId="1496" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-19T07:29:45.153" LastActivityDate="2011-03-19T07:29:45.153" Title="Velocity of Voyager 1" Tags="&lt;newtonian-mechanics&gt;&lt;velocity&gt;" AnswerCount="1" />
  <row Id="7203" PostTypeId="2" ParentId="7202" CreationDate="2011-03-19T05:51:18.810" Score="5" ViewCount="" Body="&lt;p&gt;Rocket fuels initially, followed by a series of gravitational assists (slingshots):&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gravity_assist&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Gravity_assist&lt;/a&gt;&#xA;The linked article mentions Voyager 1 mission as an example.&lt;/p&gt;&#xA;" OwnerUserId="2643" LastActivityDate="2011-03-19T05:51:18.810" />
  <row Id="7204" PostTypeId="1" CreationDate="2011-03-19T06:21:48.637" Score="-3" ViewCount="164" Body="&lt;p&gt;How are the criteria you use to tell whether or not someone is a crackpot?&lt;/p&gt;&#xA;" OwnerUserId="2647" LastActivityDate="2011-03-19T06:24:31.720" ClosedDate="2011-03-19T07:27:48.930" Title="How do you identify a crackpot?" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="7205" PostTypeId="2" ParentId="7204" CreationDate="2011-03-19T06:24:31.720" Score="4" ViewCount="" Body="&lt;p&gt;Count Baez &lt;a href=&quot;http://math.ucr.edu/home/baez/crackpot.html&quot; rel=&quot;nofollow&quot;&gt;crackpot index.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-03-19T06:24:31.720" CommentCount="2" />
  <row Id="7206" PostTypeId="2" ParentId="6928" CreationDate="2011-03-19T07:19:41.243" Score="1" ViewCount="" Body="&lt;p&gt;The heat of vaporization is not a temperature, but a capacity for holding heat (whose units are btu/lb, or some equivalent to that) which is at the boiling (vaporization) temperature of a substance. Yes, water does have one of the highest heats of vaporization of any substance around, the Fukushima reactors have not melted yet, and so it does make sense to put lots of water on them right now to cool them down to avoid a run-away meltdown. My suggestion deals with a hypothetical, in which say not enough water was put on a reactor, not enough heat could be taken away, and the core actually melted into a pool of molten liquid radioactive nasty. (pray that don't actually happen). My thought was to take something like liquid silica (sand) that might dilute such a liquid mass such as to stop the heat generating and nuclear reactions from taking place. Water could not do that. Once the molten nasty is diluted no longer producing so much heat and radioactivity, one would then entomb it in more sand and cement. &#xA;What I am also wondering about is if anyone in the nuclear community has actually thought out the worst case scenario so far, and what actual plans are out there to deal with a total runaway meltdown?&lt;/p&gt;&#xA;" OwnerUserId="2650" LastActivityDate="2011-03-19T07:19:41.243" />
  <row Id="7208" PostTypeId="2" ParentId="7196" CreationDate="2011-03-19T07:32:39.610" Score="1" ViewCount="" Body="&lt;p&gt;I know the answer :).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most energy efficient way to get a hard boiled egg is to have a pot with a good cover on an electric range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)cover the bottom of pot with 1mm water, put eggs in and pot on the range and and turn it on to the maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)when the cover starts popping, turn off heat completely, leave it on the range, for the residual heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)wait 3 minutes for very soft, 4 to 6 for medium and 8 or more for hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pot with good cover means the cover is not popping once the heat is off.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-19T07:32:39.610" CommentCount="1" />
  <row Id="7209" PostTypeId="2" ParentId="7200" CreationDate="2011-03-19T08:23:36.720" Score="3" ViewCount="" Body="&lt;p&gt;The Kasner solutions were the starting point of the paper that studied whether S,T,U-dualities are enough to de-singularize every possible scaling limit of tori:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Dualities vs singularities&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://motls.blogspot.com/2009/02/dualities-vs-singularities.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2009/02/dualities-vs-singularities.html&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/abs/hep-th/9811194&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9811194&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is that if there are at least 2+1 large dimensions, it's always possible to find a duality that makes all the dimensions larger than the fundamental (Planck/string) scale and that makes the coupling smaller than one, in the case of string theory descriptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moduli space of M-theory on tori is completely covered by regions where M-theoretical, type IIA, or type IIB description is weakly coupled in this sense. One gets pretty beautiful tilings of the moduli space in this way - see the blog post with a Mathematica-generated one. The picture refers to the case of 1 large dimension where the moduli space has a Lorentzian-metric and only the future light-cone may be dualized to well-understood vacua.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/RTHbK.jpg&quot; alt=&quot;A section of the parameter space of rectilinear tori. The three dark colors indicate M, IIA, IIB - the description that becomes weakly coupled.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It just happens that I co-authored the paper above which is most relevant for your question but statistics happens. ;-) We also wrote a heterotic followup. The moduli space is more complicated over there and many more descriptions have to be included. There's no &quot;canonical&quot; subspace of the moduli space, analogous to the rectangular tori above, so the work remained incomplete over there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dualities were performed in the way that would apply to static configurations in general relativity. The time-dependence itself may change many things. For spacelike singularities, the time dependence becomes huge and essential which is why the fate of spacelike singularities remains poorly understood in string theory - very different from the time-like singularities whose physics has been largely cracked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the compact dimensions are spheres - which have no non-contractible loops - the required winding states etc. don't exist and you can't de-singularize the spacetime if it is singular &quot;almost everywhere&quot;. However, you may de-singularize spacetimes where spheres etc. shrink locally - that's what mirror symmetry (a fiber-wise T-duality) often does.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-19T08:30:22.360" LastActivityDate="2011-03-19T08:30:22.360" CommentCount="1" />
  <row Id="7210" PostTypeId="5" CreationDate="2011-03-19T09:07:18.053" Score="0" ViewCount="" Body="&lt;p&gt;Quantum systems have infinitely more information than the corresponding classical system. An example of a two-state classical system is a coin which can be either heads-up or heads-down. This corresponds to a single &lt;em&gt;&quot;bit&quot;&lt;/em&gt; of information; we could code it as &quot;0&quot; or &quot;1&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantum two-state system or &lt;em&gt;&quot;qubit&quot;&lt;/em&gt; is usually described by a normalized complex vector with two elements such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bra-ket_notation&quot; rel=&quot;nofollow&quot;&gt;&quot;ket&quot;&lt;/a&gt;:&lt;br&gt;&#xA;$\left(\begin{array}{c} \alpha \\ \beta \end{array}\right)$&lt;br&gt;&#xA;where $\alpha$ and $\beta$ are two complex numbers subject to $|\alpha|^2+|\beta|^2 =1$. An alternative is the corresponding &lt;a href=&quot;http://en.wikipedia.org/wiki/Bra-ket_notation&quot; rel=&quot;nofollow&quot;&gt;&quot;bra&quot;&lt;/a&gt;, which has the same information content:&lt;br&gt;&#xA;$\left(\begin{array}{cc} \alpha^* &amp;amp; \beta^* \end{array}\right)$&lt;br&gt;&#xA;A less common representation of a qubit is the (pure) &lt;a href=&quot;http://en.wikipedia.org/wiki/Density_matrix&quot; rel=&quot;nofollow&quot;&gt;density matrix&lt;/a&gt; form:&lt;br&gt;&#xA;$\left(\begin{array}{cc}&#xD;&#xA;|\alpha|^2 &amp;amp; \alpha\beta^* \\&#xD;&#xA;\alpha^*\beta &amp;amp; |\beta|^2 \end{array} \right)$,&lt;br&gt;&#xA;in which the extraneous complex phase information has been removed. From this we see that there are only two real informational degrees of freedom in a qubit, for example, $|\alpha|^2$ and the complex phase of $\alpha\beta^*$. Another way of representing these two degrees of freedom is with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bloch_sphere&quot; rel=&quot;nofollow&quot;&gt;Bloch sphere&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-19T09:26:59.483" LastActivityDate="2011-03-19T09:26:59.483" />
  <row Id="7211" PostTypeId="4" CreationDate="2011-03-19T09:07:18.053" Score="0" ViewCount="" Body="Quantum information is the study of the informational content of quantum states. The most common object of study is the &quot;qubit&quot;, the information in a two-state quantum system such as spin-1/2 or photon polarization." OwnerUserId="-1" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-03-19T09:27:07.313" LastActivityDate="2011-03-19T09:27:07.313" />
  <row Id="7212" PostTypeId="1" AcceptedAnswerId="7243" CreationDate="2011-03-19T10:09:58.790" Score="4" ViewCount="215" Body="&lt;p&gt;Who hasn't heard about the &lt;strong&gt;double-slit experiment&lt;/strong&gt;? It figures in any book of quantum physics. But there is something no one can explain to me : I understand why the light cannot be described only as a wave, but I do not understand why it cannot be explained only in terms of a particle, having some trajectory, following other laws of phyisics we may not know. Usually, every book in which I have tried to look for an answer considers that if it cannot be described as a &lt;strong&gt;classical&lt;/strong&gt; trajectory (which is clearly the case in this experiment), then it is not a particle, that is, a material point. Could it not be a particle following new laws?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope I am clear,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your answers,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isaac&lt;/p&gt;&#xA;" OwnerUserId="2499" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-19T18:28:59.620" LastActivityDate="2011-03-20T05:34:50.077" ClosedDate="2011-03-19T17:52:32.017" Title="Young's experiment or why the light can't be described as a particle" Tags="&lt;quantum-mechanics&gt;&lt;particle-physics&gt;&lt;light&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="7218" PostTypeId="1" AcceptedAnswerId="7228" CreationDate="2011-03-19T12:17:02.970" Score="7" ViewCount="321" Body="&lt;p&gt;I understand from basic conservation of energy and momentum considerations, it is clear in classical electrodynamics that the fields should be able to have energy and momentum. This leads to the usual Poynting vector and energy density relations for electromagnetic fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I do not know how to interpret situations where there is a net linear momentum in a static electromagnetic field.  The fields aren't propagating.  It doesn't make sense to me that momentum can be divorced from motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a concrete example to discuss:&#xA;Consider a massless string of length L, with a spherical shell on each end with a magnetic dipole moment m and positive charge q. The radius of the sphere $R&amp;lt;&amp;lt;L$, or alternatively, consider the dipoles to be perfect &quot;point&quot; dipoles. Let the string be along the y axis, with one dipole at the origin and the other at y=+L.  If the magnetic dipole at the origin is oriented in the -z direction, and the other dipole in the +z direction, if you calculate the total linear momentum in the fields, the answer is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_{em} = m q \frac{\mu_0}{2 \pi L^2} \hat{x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this is an unstable equilibrium, it is an equilibrium.  So classically the state can remain static with no need to evoke other external entities, interactions, etc.  So there doesn't seem to be any of the usual potential pitfalls to save us here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please, can someone explain how a static field can have momentum.&lt;/p&gt;&#xA;" OwnerUserId="2652" LastEditorUserId="2652" LastEditorDisplayName="" LastEditDate="2011-03-19T13:48:54.513" LastActivityDate="2011-07-03T22:41:56.190" Title="How can there be net linear momentum in a static electromagnetic field (not propagating)?" Tags="&lt;electromagnetism&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="6" />
  <row Id="7219" PostTypeId="2" ParentId="7112" CreationDate="2011-03-19T12:20:44.427" Score="1" ViewCount="" Body="&lt;p&gt;There's a seemingly 'unknown' argument in this discussion to my eyes, for it seems to always be left out in these discussions about Hidden Variables Theories. I &lt;a href=&quot;https://arsphysica.wordpress.com/2009/03/08/teorema-do-livre-arbitrio/&quot; rel=&quot;nofollow&quot;&gt;wrote about it (in &lt;code&gt;pt_BR&lt;/code&gt;, but Google Translate is your friend ;-)&lt;/a&gt; some time ago: it's the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kochen-Specker_theorem&quot; rel=&quot;nofollow&quot;&gt;Kochen-Specker theorem&lt;/a&gt; and the &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0604079&quot; rel=&quot;nofollow&quot;&gt;The Free Will Theorem&lt;/a&gt; (together with its stronger form, &lt;a href=&quot;http://arxiv.org/abs/0807.3286&quot; rel=&quot;nofollow&quot;&gt;The Strong Free Will Theorem&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, here's the punchline: Hidden Variables Theories hinge on two hypothesis, (1) all hidden variables have a definite value at all times; &amp;amp; (2) these variables values are &quot;intrinsic&quot; and independent of the measuring device used to detect them (aka &quot;observable&quot;). The Kochen-Specker Theorem is a &lt;em&gt;complement&lt;/em&gt; to &lt;a href=&quot;http://math.ucr.edu/home/baez/physics/Quantum/bells_inequality.html&quot; rel=&quot;nofollow&quot;&gt;Bell's Inequalities&lt;/a&gt; and it &lt;strong&gt;proves&lt;/strong&gt; that there is a &lt;em&gt;fundamental contradiction&lt;/em&gt; between the two hypothesis above: in QM, observables do not need to be &lt;em&gt;commutative&lt;/em&gt; (by &lt;em&gt;hypothesis&lt;/em&gt;, a &lt;em&gt;commutative&lt;/em&gt; algebra underlies hidden variables theories). In this fashion, as the Free Will Theorem is a based on the Kochen-Specker one, it invalidates (under very reasonable physical assumptions) all hidden variable theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that i haven't slept in quite some time, it's possible that i haven't been all that clear above. In any case, i'm leaving a list of references that should hopefully patch this. :-)&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;References&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This list of references is by no means comprehensive, but it's extensive enough to give a glimpse into this discussion: the tip of the iceberg, if you will. Also, the 'level' of the refs varies a bit, some links are to Wikipedia and some are to published articles: i tried to use more rigorous refs for the trickier topics (whether or not i succeeded at this is anyone's guess ;-).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully, even if my answer did not connect all the dots, following these breadcrumbs can shed some light in this issue.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Principle_of_locality&quot; rel=&quot;nofollow&quot;&gt;Principle of Locality&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bell%27s_inequalities&quot; rel=&quot;nofollow&quot;&gt;Bell’s Inequalities&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein-Podolsky-Rosen_paradox&quot; rel=&quot;nofollow&quot;&gt;EPR Paradox&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/No_cloning_theorem&quot; rel=&quot;nofollow&quot;&gt;No cloning theorem&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Kochen-Specker_theorem&quot; rel=&quot;nofollow&quot;&gt;Kochen-Specker theorem&lt;/a&gt;, &lt;a href=&quot;http://plato.stanford.edu/entries/kochen-specker/&quot; rel=&quot;nofollow&quot;&gt;The Kochen-Specker Theorem (Stanford Encyclopedia of Philosophy)&lt;/a&gt;, &lt;a href=&quot;http://www.iumj.indiana.edu/IUMJ/dfulltext.php?year=1968&amp;amp;volume=17&amp;amp;artid=17004&quot; rel=&quot;nofollow&quot;&gt;The Problem of Hidden Variables in Quantum Mechanics&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Free_will_theorem&quot; rel=&quot;nofollow&quot;&gt;FWT (Wikipedia)&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.newscientist.com/article/mg19025504.000-free-will--you-only-think-you-have-it.html?full=true&quot; rel=&quot;nofollow&quot;&gt;Free will — you only think you have it&lt;/a&gt;, &lt;a href=&quot;http://www.newscientist.com/article/mg19025530.400-taking-liberties.html&quot; rel=&quot;nofollow&quot;&gt;Taking liberties&lt;/a&gt;, &lt;a href=&quot;http://www.newscientist.com/article/mg19526154.200-free-will--is-our-understanding-wrong.html&quot; rel=&quot;nofollow&quot;&gt;Free will — is our understanding wrong?&lt;/a&gt;, &lt;a href=&quot;http://www.newscientist.com/article/mg19526195.100-quantum-free-will.html&quot; rel=&quot;nofollow&quot;&gt;Quantum free will&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ams.org/notices/200902/rtx090200226p.pdf&quot; rel=&quot;nofollow&quot;&gt;The Strong Free Will Theorem (PDF)&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/quant-ph/0604079&quot; rel=&quot;nofollow&quot;&gt;The Free Will Theorem&lt;/a&gt;, &lt;a href=&quot;http://arxiv.org/abs/0807.3286&quot; rel=&quot;nofollow&quot;&gt;The Strong Free Will Theorem&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/quant-ph/0306072&quot; rel=&quot;nofollow&quot;&gt;Decoherence and the transition from quantum to classical&lt;/a&gt;;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://video.if.usp.br/emaranhamento-realismo-e-n-o-localidade-absurdos-da-mec-nica-qu-ntica&quot; rel=&quot;nofollow&quot;&gt;Emaranhamento, realismo e localidade (video lecture in &lt;code&gt;pt_BR&lt;/code&gt;)&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="142" LastActivityDate="2011-03-19T12:20:44.427" CommentCount="4" />
  <row Id="7220" PostTypeId="1" CreationDate="2011-03-19T12:51:50.730" Score="2" ViewCount="156" Body="&lt;p&gt;Edward Nelson's book &quot;Quantum Fluctuations&quot; (Princeton UP, 1985) gives an alternative way to introduce trajectories, quite different to the trajectories of de Broglie-Bohm type approaches. I've read the book in the past and university libraries generally have copies, but I've been unable to find a good open web reference&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Researching this Question, I came across one of the best graphical presentations I've seen of how a 2-slit interference pattern can be generated by particles, &lt;a href=&quot;http://www.cs.cmu.edu/~lafferty/QF/two-slit.html&quot; rel=&quot;nofollow&quot;&gt;http://www.cs.cmu.edu/~lafferty/QF/two-slit.html&lt;/a&gt; (in Java), with, as explanation, page images and a PDF of chapter 3 of Nelson's book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found plenty of published work. WebOfScience returns 21 review articles that cite Nelson's book. &quot;Research on hidden variable theories: A review of recent progresses&quot;, Genovese M, PHYSICS REPORTS &lt;strong&gt;413&lt;/strong&gt;(6) 319-396, JUL 2005, sent me to what at first glance looks an interesting review article &quot;NON-LOCALITY AND LOCALITY IN THE STOCHASTIC INTERPRETATION OF QUANTUM MECHANICS&quot;, D. BOHM and B.J. HILEY, PHYSICS REPORTS &lt;strong&gt;172&lt;/strong&gt;(3) 93—122 (1989).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT (modified): Does anyone know of &lt;em&gt;other&lt;/em&gt; open access web resources? &lt;i&gt;EDIT (new):If someone has a favorite subscription-only review article, that would be nice to know of as well, but my perception is that Nelson's approach is particularly unknown outside of academia; if someone is asking questions that suggest that thinking about Nelson's approach might widen their horizons, I want an easy place to tell them to look.&lt;/i&gt; Stanford Encyclopedia of Philosophy seems not to have a discussion of Nelson approaches, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I make this request because although it's relatively little known I consider Nelson's approach something that anyone thinking about QM should know about, so I'd like something to be able to point out to people. I consider it significant partly because it demonstrates that de Broglie-Bohm approaches are not a unique way to introduce hidden variables. The way in which stochasticity is introduced is conceptually different in that the trajectories are stochastic instead of the initial conditions, which puts QM in a significantly different light than de Broglie-Bohm approaches.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-03-19T19:25:33.870" LastActivityDate="2011-03-21T17:11:11.863" Title="Web references for Nelson's &quot;Quantum Fluctuations&quot;?" Tags="&lt;quantum-mechanics&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="7221" PostTypeId="2" ParentId="6581" CreationDate="2011-03-19T12:53:37.417" Score="0" ViewCount="" Body="&lt;p&gt;I disagree with the claim &quot;curlE=0 ... divE=0. This is enough to make E=0&quot;.  Consider for instance an electric dipole.  Outside of the dipole, curlE=0 and divE=0, yet E does not equal 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also think the charge distribution considerations are too limited.  For intuitively I expect polarization to show up, but not free charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a simple concrete example showing M constant can still have an electric field:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An infinite cylinder, neutral charge, M constant.  In the rest frame, outside: E=0,B=0, and inside: E=0,B=const.  Now boost to frame moving along the axis, outside: E=0,B=0, and inside B=const, E!=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider a cylinder of length L and radius a, and turn it into a nice symmetric ring (so 'outer' radius = L/2 pi, and 'inner' radius = a).  Inside the ring, in the limit $a&amp;lt;&amp;lt;L$, we need to get back the infinite cylinder case.  So yes, a rotating ring will have non-zero electric field in it.  Also, intuitively, before the infinite limit, the ring will have electric field outside it as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT(yet again)&#xA;I need to think about it, but this can probably be made rigorous as such:&#xA;In the infinite cylinder case, it should be possible to see how M in one frame changes to M and P in another frame.  There may be a simple way to use the symmetry of relativity to explain how these mix.&lt;/p&gt;&#xA;" OwnerUserId="2652" LastEditorUserId="2652" LastEditorDisplayName="" LastEditDate="2011-03-19T13:07:53.097" LastActivityDate="2011-03-19T13:07:53.097" CommentCount="2" />
  <row Id="7222" PostTypeId="2" ParentId="6220" CreationDate="2011-03-19T13:30:17.620" Score="4" ViewCount="" Body="&lt;p&gt;&lt;em&gt;&quot;Is there any hilly road on which a better mileage can be achieved?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is: YES.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's just use three simple reasonable assumptions:&#xA;1) There is rolling friction&#xA;2) There is air friction which increases with speed&#xA;3) In the graph of power vs gas consumption, there is a peak (a maximum)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;note: (power)/(gas/time) has the same units as energy/gas.  If there was no air friction and rolling friction was a constant, we'd want to run the engine at this peak until we could just coast the rest. However, because there is air friction, it may actually be better for gas mileage to run below this power output.  That is the trade off on the flat road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An appropriately shaped hill lets us beat this trade off, because we can go into a gear so that we go very slow essentially giving all our engine power into gravitational energy. We just coast to the hill (or after the hill), with the engine off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Viewed this way, the answer is obvious, since we are essentially using the hill as a gravitational battery.  It lets us beat the flat straight away, because we can give that energy back at 100% efficiency compared to the &amp;lt; 100% efficiency of the motor in the &quot;air friction with speed&quot; vs &quot;power output vs gas consumption&quot; tradeoff we are forced into with the flat road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best road I believe would then be just enough slope to allow rolling against friction all the way until a steep hill right at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the comments and other answers to this question are quite bizarre.  To make it clear, I am not claiming that &lt;em&gt;all&lt;/em&gt; hilly roads are better.  The question asked &quot;Is there &lt;em&gt;any&lt;/em&gt; hilly road on which a better mileage can be achieved?&quot; The answer is YES.  Nor am I claiming that this depends only on the road, as it clearly depends on how the driver decides to run the engine for speeds along the route.  Again, the question seems clear on this, as it says to consider the car &quot;operated by a perfect driver with knowledge of the route&quot;.  So I am not sure where the confusion is coming from.  So I am extending discussion here, in hopes to clear that confusion up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three places where stored or mechanical energy is lost to heat: the engine performance curve, the rolling friction of the tires on the road, and the air friction.  The tire friction is to good approximation a constant, while the air friction increases with speed.  Putting this all together:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the total mechanical energy used to get from A to B:&#xA;$E = mgh|_A^B + \int_A^B (F_{air} + F_{rolling}) ds$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since A and B are at the same elevation in this problem, the gravitational potential energy terms sum to zero and are the same for both routes.  To good approximation the rolling friction is a constant, then if the length of the road is L:&#xA;$E = L F_{rolling}+ \int_A^B F_{air}(v) \ ds$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the rolling friction is the same between the two routes.  The only thing left then is the engine performance curve (the efficiency at which we can get the mechanical energy from the gas) and the air friction.  The answers neglecting the engine performance curves, or air friction, are thus neglecting the real difference between these routes.  I hope I made that abundantly clear now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that if the engine performance curve was flat (a constant), then on a flat road, we'd want to go very slow (the limit v-&gt;0 is the best driving strategy for gas mileage in this unrealistic case).  However, in realistic cases (and as I took as one of my three assumptions in my answer above) the engine performance curve will have a peak.  There is now a trade off between how far off peak performance we run the engine, and how much mechanical energy we waste on air friction.  The details of solving this require detailed knowledge of the engine performance curve, but the general result that there is a trade off is clear regardless.  The issue is that on a flat road: &lt;em&gt;the engine can only generate mechanical energy in the form of kinetic energy&lt;/em&gt;, and kinetic energy in turn causes more energy loss in air friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the case where the road slopes just enough to a hill at the end, that we can just coast down to the hill, and only need to use the engine to get up the hill. (Or, alternatively, as other poster suggested, a hill at the start, and then coast the rest of the way.)  When running the engine now, we can generate mechanical energy in the form of kinetic energy &lt;em&gt;and&lt;/em&gt; gravitational energy.  So a hill allows us to run the engine closer to its peak performance, since we can put the engine output into gravitational energy (which has no loss over the trip) as opposed to just kinetic energy (which we get losses in air friction).&lt;/p&gt;&#xA;" OwnerUserId="2652" LastEditorUserId="2652" LastEditorDisplayName="" LastEditDate="2011-03-22T04:48:32.917" LastActivityDate="2011-03-22T04:48:32.917" CommentCount="9" />
  <row Id="7224" PostTypeId="2" ParentId="7218" CreationDate="2011-03-19T13:53:44.093" Score="5" ViewCount="" Body="&lt;p&gt;It is possible to show that the total momentum of any static system is zero in an inertial frame where nothing is moving. This does not mean that the momenta associated with various components of that system are individually zero. As you point out, there can be finite electromagnetic momentum associated with static charge distributions. Even though there is no obvious motion in the system, the momentum associated with the matter distribution is actually nonzero. It is equal and opposite to the electromagnetic momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is often referred to as the hidden mechanical momentum. It is a special case of a much more general result that the net momentum of an extended object need not be parallel to its center of mass velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrodynamics books like Griffiths or Jackson have a nice microscopic interpretation for this effect in the simple case of a magnetic dipole placed near a charge. Internally, the dipole may be thought of as containing a current loop. The charges in this current loop accelerate and decelerate in response to the external electric field. One may show that this gives them a net momentum that is exactly equal and opposite to the electromagnetic momentum. Note that this is an intrinsically relativistic effect. It does not arise if Lorentz factors are neglected when computing the momenta of the circulating charges.&lt;/p&gt;&#xA;" OwnerUserId="2238" LastActivityDate="2011-03-19T13:53:44.093" CommentCount="2" />
  <row Id="7225" PostTypeId="2" ParentId="7218" CreationDate="2011-03-19T14:03:03.270" Score="-1" ViewCount="" Body="&lt;p&gt;Momentum is a conserved quantity that &lt;em&gt;belongs&lt;/em&gt; to a system.&#xA;When particles are in motion the usual linear and angular moment &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the image of the dipole of the water molecule here &lt;a href=&quot;http://en.wikipedia.org/wiki/Electric_dipole_moment&quot; rel=&quot;nofollow&quot;&gt;Electric_dipole_moment WP&lt;/a&gt;&#xA;you see that the molecule is subject permanently to a tension (red in the image).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electric_dipole_moment&quot; rel=&quot;nofollow&quot;&gt;Electric Dipole moment&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is a measure of the separation of&#xA;  positive and negative electrical&#xA;  charges in a system of charges, that&#xA;  is, a measure of the charge system's&#xA;  overall polarity.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In the simple case of two point&#xA;  charges, one with charge +q and one&#xA;  with charge −q, the electric dipole&#xA;  moment p is:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;dipole moment ~ charge * distance  (of charges)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;compare this with &lt;a href=&quot;http://en.wikipedia.org/wiki/Angular_momentum&quot; rel=&quot;nofollow&quot;&gt;Angular moment&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;angular moment ~  linear moment * distance (to a point)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where linear moment ~ mass * velocity &#xA;as both charge and l.m. are states of the particle in a given moment that represent a potential for action, both formulas of angular and dipole moments are quite similar.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-03-19T14:03:03.270" CommentCount="1" />
  <row Id="7226" PostTypeId="2" ParentId="7212" CreationDate="2011-03-19T14:06:34.687" Score="-3" ViewCount="" Body="&lt;p&gt;As the other answers tell you there have been people trying to explain the two slit experiment with different mathematics than quantum mechanics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Could it not be a particle following new laws?&lt;/em&gt;  ,you ask. Calling a particle a &lt;em&gt;material point&lt;/em&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is no, experiments have confirmed that the quantum mechanical descirption is the  valid one, generally, not just for the two slit experiment. Do look at the link in sb1's answer.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-03-19T15:12:24.353" LastActivityDate="2011-03-19T15:12:24.353" CommentCount="3" />
  <row Id="7227" PostTypeId="2" ParentId="6220" CreationDate="2011-03-19T15:15:07.237" Score="0" ViewCount="" Body="&lt;p&gt;If the hill permits hypermiling to be done. In the most idea case, at the top of the hill the driver would turn off the ignition and shift into neutral, and descend unpowered.[Don't try this as your power brakes and power steering won't work as expected]. I assume the hill is just the right grade to maintain speed unpowered. The advantage comes from avoiding engine braking (losses within the engine/transmission) during the downhill leg. Of course this also requires the engine to not lose much efficiency during the climb. Hybrid cars work like this, charging the battery resembles the uphill climb, and using the battery in &quot;stealth&quot; mode resembles the downhill portion. So hypermiling is simply using gravity as a low tech (but very efficient) hybrid. If you leave the car in gear on the downhill, and thus suffer from the engine braking during the descent I doubt you get the benefit. So if you are a legal driver, you may only see the milage improvement if you drive a hybrid, which is designed to shut the internal combustion motor off when it doesn't need the power. [Of course the hybrid has the additional advantage, that if the hill is too steep, the excess energy (within limits) can be captured by the battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming here that speed is constant. It is almost a trivial excercise to show that we can trade off travel time versus fuel consumption. But modern society rarely gives us the option (especially with other cars on the road). If we assume linear engine braking, then the total engine braking loses for a given trip are proportional to the total number of revolutions the engine made. Running at a fixed RPM, but at higher torque, then in neutral with the engine either idling (or turned off) minimizes the total number of engine revolutions for the trip.&lt;/p&gt;&#xA;" OwnerUserId="391" LastEditorUserId="391" LastEditorDisplayName="" LastEditDate="2011-03-19T19:49:58.643" LastActivityDate="2011-03-19T19:49:58.643" CommentCount="3" />
  <row Id="7228" PostTypeId="2" ParentId="7218" CreationDate="2011-03-19T15:20:01.150" Score="7" ViewCount="" Body="&lt;p&gt;This is a fairly subtle question! Griffiths recently published a paper on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1119/1.3152712&quot; rel=&quot;nofollow&quot;&gt;Hidden momentum, field momentum, and electromagnetic impulse&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Electromagnetic fields carry energy,&#xA;  momentum, and angular momentum. The&#xA;  momentum density, $ϵ_{0}(E\times B)$, accounts&#xA;  (among other things) for the pressure&#xA;  of light. But even static fields can&#xA;  carry momentum, and this would appear&#xA;  to contradict a general theorem that&#xA;  the total momentum of a closed system&#xA;  is zero if its center of energy is at&#xA;  rest. In such cases, there must be&#xA;  some other (nonelectromagnetic)&#xA;  momenta that cancel the field&#xA;  momentum. What is the nature of this&#xA;  “hidden momentum” and what happens to&#xA;  it when the electromagnetic fields are&#xA;  turned off?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.ate.uni-duisburg-essen.de/data/postgraduate_lecture/AJP_2009_Griffiths.pdf&quot; rel=&quot;nofollow&quot;&gt;Free version&lt;/a&gt; of the above link.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="4255" LastEditorDisplayName="" LastEditDate="2011-07-03T22:41:56.190" LastActivityDate="2011-07-03T22:41:56.190" />
  <row Id="7230" PostTypeId="2" ParentId="6220" CreationDate="2011-03-19T15:22:26.077" Score="-1" ViewCount="" Body="&lt;p&gt;absolutely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you assume what Edward assumed, I can give you the optimal path.&#xA;I'm assuming a gas engine, which has an optimal fuel curve and which has a fuel use when its on, even in neutral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Climb a hill immediately, at a speed governed by the ratio of gas used to climb it/gas used by idling.  There will be some optimal climb rate.  Example: if you burn 1 gal/hr, and climbing the hill at 200mph costs you 1 gal in 2 minutes (purely imaginary example) then the right answer is somewhere less than an hour, but more than two minutes, whatever gives you the minimum fuel use.  Then, turn off the engine, and crawl downhill toward the goal.  The optimal shape is the minimum hill that will overcome rolling friction at a crawl.&lt;/p&gt;&#xA;" OwnerUserId="2653" LastActivityDate="2011-03-19T15:22:26.077" CommentCount="4" />
  <row Id="7231" PostTypeId="1" AcceptedAnswerId="7309" CreationDate="2011-03-19T15:51:33.417" Score="12" ViewCount="550" Body="&lt;p&gt;Recently there have been some interesting questions on standard QM and especially on uncertainty principle and I enjoyed reviewing these basic concepts. And I came to realize I have an interesting question of my own. I guess the answer should be known but I wasn't able to resolve the problem myself so I hope it's not entirely trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what do we know about the error of simultaneous measurement under time evolution? More precisely, is it always true that for $t \geq 0$&#xA;$$\left&amp;lt;x(t)^2\right&amp;gt;\left&amp;lt;p(t)^2\right&amp;gt; \geq \left&amp;lt;x(0)^2\right&amp;gt;\left&amp;lt;p(0)^2\right&amp;gt;$$&#xA;(here argument $(t)$ denotes expectation in evolved state $\psi(t)$, or equivalently for operator in Heisenberg picture).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to get general bounds from Schrodinger equation and decomposition into energy eigenstates, etc. but I don't see any way of proving this. I know this statement is true for a free Gaussian wave packet. In this case we obtain equality, in fact (because the packet stays Gaussian and because it minimizes HUP). I believe this is in fact the best we can get and for other distributions we would obtain strict inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to summarize the questions&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Is the statement true?&lt;/li&gt;&#xA;  &lt;li&gt;If so, how does one prove it? And is there an intuitive way to see it is true?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-04-03T06:18:56.670" Title="Does Heisenberg's uncertainty under time evolution always grow?" Tags="&lt;quantum-mechanics&gt;&lt;uncertainty-principle&gt;" AnswerCount="5" CommentCount="19" FavoriteCount="10" />
  <row Id="7232" PostTypeId="1" AcceptedAnswerId="7234" CreationDate="2011-03-19T16:02:16.270" Score="13" ViewCount="462" Body="&lt;p&gt;Does GR provide a limit to the maximum electric field? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've gotten conflicting information regarding this, and am quite confused.  I will try to quote exactly when possible so as not to confuse things more with my paraphrasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author of the Motion Mountain physics textbook claims in his book there is a limit, and clarifies on his site that ( &lt;a href=&quot;http://www.motionmountain.net/wiki/index.php?title=Dislike_Page&quot; rel=&quot;nofollow&quot;&gt;http://www.motionmountain.net/wiki/index.php?title=Dislike_Page&lt;/a&gt; ): &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;electromagnetic fields are limited in&#xA;  magnitude. Now, every electromagnetic&#xA;  field contains energy, and energy&#xA;  density is limited by general&#xA;  relativity: if energy density is too&#xA;  high, a black hole appears. The&#xA;  smallest possible black hole then&#xA;  leads to a field limit. If you deny an&#xA;  upper field limit, you deny general&#xA;  relativity. However, general&#xA;  relativity has been confirmed in every&#xA;  experiment so far.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This sounds very obvious and intuitive to me. However one of my physics TA's got very upset when I used this in a thought experiment when discussing some limits in physics.  When I told him the textbook I got that from, he looked it up, and commented on the Motion Mountain wiki website his argument:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;I'd like to add something here.&#xA;  Suppose for a moment the energy&#xA;  density limit is correct, then if an&#xA;  object is sped up until it length&#xA;  contracts enough such that the energy&#xA;  density is greater than this limit,&#xA;  does it turn into a black hole? No,&#xA;  quoting John Baez &quot;The answer is that&#xA;  a black hole does not form&quot;*&#xA;  &lt;a href=&quot;http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_fast.html&quot; rel=&quot;nofollow&quot;&gt;http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_fast.html&lt;/a&gt;&#xA;  *The claim that &quot;if energy density is too high, a black hole appears&quot; is an&#xA;  incorrect oversimplification. The&#xA;  issue is that in GR, gravity depends&#xA;  on more than just the energy density&#xA;  (component T^00 of the stress energy&#xA;  tensor). So we can't just look at the&#xA;  &quot;relativistic mass&quot; (E/c^2) to judge&#xA;  whether a black hole forms. It&#xA;  actually isn't even enough to look at&#xA;  the &quot;invariant mass&quot;, for while the&#xA;  trace T of the stress energy tensor&#xA;  for a particle is just its invariant&#xA;  mass, for an electromagnetic field it&#xA;  is identically zero even though&#xA;  electromagnetic fields curve spacetime&#xA;  in GR. So none of these concepts of&#xA;  mass are sufficient when discussing&#xA;  gravity using GR (especially when&#xA;  considering electromagnetic fields),&#xA;  because gravity couples to the entire&#xA;  stress energy tensor. I hope this was&#xA;  helpful.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This makes much less sense to me, and I don't understand how energy density tending to infinity could EVER avoid being a black hole. No offense to my TA, but I'm skeptical as he's disagreeing with a textbook author.  Plus, the author's response was that my TA is another Einstein denier, and not worth responding to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'd like a third party's answer on this.  Does GR provide a theoretical limit to the strength of an electric field?  Is it best to just ignore my TA on this one?&lt;/p&gt;&#xA;" OwnerUserId="2655" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-03-19T17:40:39.987" LastActivityDate="2011-08-24T22:28:18.100" Title="Does GR provide a maximum electric field limit?" Tags="&lt;general-relativity&gt;&lt;classical-electrodynamics&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="3" />
  <row Id="7233" PostTypeId="2" ParentId="7232" CreationDate="2011-03-19T16:10:49.257" Score="1" ViewCount="" Body="&lt;p&gt;Trust John Baez over your TA.  Another thing that you need in addition to a large mass density in order to form a black hole is a matter distribution that behaves in a certain way--in particular, you can fill the whole universe with radiation in a certain &lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric&quot; rel=&quot;nofollow&quot;&gt;Robertson-Walker spacetime&lt;/a&gt;, and the density of radiation, and thus the magnitude of the RMS E-field is just a parameter with no upper limit, and the Robertson-Walker spacetime contains no black hole for any value of its parameters.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-03-19T16:10:49.257" CommentCount="10" />
  <row Id="7234" PostTypeId="2" ParentId="7232" CreationDate="2011-03-19T16:16:22.677" Score="8" ViewCount="" Body="&lt;p&gt;Your TA is right that energy density alone does not trigger black hole formation. Consider a ball that's sitting still. Now speed up and look at the ball again. It will have gained (kinetic) energy. Relativistically, you can make the ball's energy density arbitrarily large by moving sufficiently near the speed of light. But the ball hasn't done anything in this process. It's you that has been changing speed. The notion of a black hole is not observer-dependent, so energy density alone cannot make a black hole form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there are senses in which electromagnetic fields can form black holes in general relativity. Two colliding electromagnetic plane waves (which are necessarily gravitational plane waves as well) can do this. The field strength for a single plane wave can be arbitrarily large, however. Any &quot;limits&quot; are highly dependent on the specifics of the configuration.&lt;/p&gt;&#xA;" OwnerUserId="2238" LastActivityDate="2011-03-19T16:16:22.677" CommentCount="6" />
  <row Id="7235" PostTypeId="2" ParentId="7232" CreationDate="2011-03-19T16:20:32.040" Score="0" ViewCount="" Body="&lt;p&gt;I don't know about what Baez is talking about, but i assure you that speeding up an object (relative to my &quot;funny&quot; frame) definitely increases the relative kinetic energy of the object in that frame, but not all energy transforms the same;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;kinetic energy transforms as the 0-component of a cuadrivector, while rest-energy ($mc^2$) transforms as a scalar under Lorentz transformations. Also $T^{00}$ transforms as a tensor component, because it is the integral density of a 0-component vector against a oriented 3-form defining flow over a 3D volume element. Once you integrate $T^{00}$ over a volume you get a scalar density. Gravitational contributions only come from energy &lt;em&gt;tensors&lt;/em&gt; like densities or rest-energies integrated over volumes. A binding energy in an atom would also behave as a scalar so it couples directly to gravity. A high-speed electron has an &lt;strong&gt;unbounded&lt;/strong&gt; kinetic-energy but doesn't not contribute to any tensor energy density in the same unbounded way&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hope this clarifies the issue &lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-03-19T16:20:32.040" CommentCount="5" />
  <row Id="7236" PostTypeId="2" ParentId="1115" CreationDate="2011-03-19T16:31:11.110" Score="1" ViewCount="" Body="&lt;p&gt;The splitting of a vector field $\vec{V}=\vec{V}_{\parallel}+\vec{V}_{\perp}$ into a curl-free part, $\vec{\nabla}\times\vec{V}_{\parallel}=\vec{0}$, and a divergence-free part, $\vec{\nabla}\cdot\vec{V}_{\perp}=0$, is given as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec{V}_{\parallel}:=\vec{\nabla}(\Delta^{-1}(\vec{\nabla}\cdot\vec{V}))$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec{V}_{\perp}:=\vec{V}-\vec{V}_{\parallel}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $\Delta:=\vec{\nabla}\cdot\vec{\nabla}$ is the Laplacian, and $\Delta^{-1}$ is a right inverse $\Delta\circ\Delta^{-1} = {\rm id}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The operators $\Delta$ and $\Delta^{-1}$ take scalars $f:\mathbb{R}^3\to\mathbb{R}$ into scalars. Potential problems are related to zero-modes and whether the inverse $\Delta^{-1}$ is well-defined. For sufficiently well-behaved scalars $f$ (and implicit choice of boundary conditions), the following integral formula applies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\Delta^{-1}f)(\vec{x}) &#xD;&#xA;= - \iiint_{\vec{y}\neq\vec{x}}\frac{d^3y}{4\pi}&#xD;&#xA;\frac{f(\vec{y})}{|\vec{y}-\vec{x}|}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cf. &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson%27s_equation&quot; rel=&quot;nofollow&quot;&gt; Poisson equation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-03-20T11:49:03.640" LastActivityDate="2011-03-20T11:49:03.640" />
  <row Id="7237" PostTypeId="2" ParentId="7212" CreationDate="2011-03-19T16:35:16.793" Score="1" ViewCount="" Body="&lt;p&gt;When we say it doesnt behave as a particle, we mean it doesnt behave as a classical particle. But it does behave as a particle, in the sense of an elementary particle, which share some characterstics with classical particles, and some with classical waves.&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-03-19T16:35:16.793" CommentCount="0" />
  <row Id="7238" PostTypeId="1" CreationDate="2011-03-19T16:42:00.463" Score="2" ViewCount="182" Body="&lt;p&gt;If I calculate the energy contained in the electric field for an electric dipole p in an electric field E, I get (ignoring the terms independent of orientation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U = - \vec{p} \cdot \vec{E}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is as expected.  However, if I do the same for a magnetic dipole m in a magnetic field B, I get (again ignoring the terms independent of orientation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U = \vec{m} \cdot \vec{B}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which clearly has the wrong sign.  Does one have to be careful in interpreting energy in the magnetic field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A colleague pointed out that David Griffiths mentioned this problem in passing in &quot;Dipoles at Rest&quot; (AJP 60, 979, 1992).  But it wasn't very satisfying, as he seems to be referring to the terms which are independent of the orientation to try to argue it fixes the sign somehow.  It didn't make sense to me, maybe because it was only mentioned briefly in passing.  Can anyone expand upon this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if I look at the Lagrangian for electrodynamics, I notice the energy in the electric fields has an overall negative sign (like a potential energy) while the energy in the magnetic fields has an overall positive sign (like a kinetic energy).  So is this a hint I'm not supposed to treat the energy in a magnetic field on the same footing as the energy in an electric field?&lt;/p&gt;&#xA;" OwnerUserId="2652" LastEditorUserId="2652" LastEditorDisplayName="" LastEditDate="2011-03-19T16:47:51.050" LastActivityDate="2011-03-19T21:45:39.487" Title="Energy in magnetic fields" Tags="&lt;electromagnetism&gt;" AnswerCount="1" />
  <row Id="7239" PostTypeId="1" AcceptedAnswerId="7242" CreationDate="2011-03-19T17:09:51.903" Score="5" ViewCount="167" Body="&lt;p&gt;If there is a non-zero expectation value for the Higgs boson even in &quot;vacuum&quot;, since the Higgs boson has a mass unlike photons, then I would expect it to have a rest frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why doesn't a non-zero expectation value for the Higgs boson not only break electroweak symmetry, but also break Lorentz symmetry? &lt;/p&gt;&#xA;" OwnerUserId="2655" LastActivityDate="2011-03-19T17:24:09.650" Title="Is there a rest frame for the Higgs boson?" Tags="&lt;quantum-field-theory&gt;&lt;special-relativity&gt;&lt;higgs-boson&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7241" PostTypeId="2" ParentId="7231" CreationDate="2011-03-19T17:12:39.430" Score="10" ViewCount="" Body="&lt;p&gt;The Schrodinger equation is time-symmetric. The answer is therefore No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From all of the comments, I feel like I must be oversimplifying or missing something, but I can't see what.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-19T17:12:39.430" CommentCount="15" />
  <row Id="7242" PostTypeId="2" ParentId="7239" CreationDate="2011-03-19T17:18:58.597" Score="6" ViewCount="" Body="&lt;p&gt;Dear John, the vacuum doesn't contain any physical Higgs bosons; it only contains a &quot;Higgs condensate&quot;. So the vacuum doesn't break the Lorentz symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The potential energy energy for the Higgs field looks like&#xA;$$ V(h) = \frac{m_H^2}{2} (h-v)^2 + O((h-v)^3) $$&#xA;for $h$ near $v$, so it is minimized by $h(x,y,z,t)=v$. Note that $V(h)$ is a Lorentz scalar so when added to the Lagrangian, it preserves the Lorentz symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nonzero value of $h$ at each point is what breaks the electroweak symmetry because $h$ is a component of a doublet whose only symmetry-invariant point is $h=0$. But because this configuration is constant, it doesn't pick any preferred reference frame. There are no physical particles in the vacuum state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physical Higgs bosons may be added and they're energy quanta that allow $h$ to deviate away from $h=v$ - more than the uncertainty principle requires. Those quanta carry energy and momentum $(E,p)$ that satisfies $E^2 - p^2 = m_H^2$. That's the energy one has to add to the energy of the vacuum state; the latter state is free of particles and its energy may be chosen to be zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Let me not discuss the cosmological constant issues here because they have nothing to do with this question.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to the difference between the vacuum, which has no particles, and states with particles (Higgs bosons), you may define a new field, $h&amp;#39;=h-v$, and this field will be expanded around zero just like the electromagnetic fields or other fields. The vacuum will be at $h&amp;#39;=0$ and it may perhaps make it easier to understand why there's no Lorentz-symmetry breaking in the vacuum. The $SU(2)\times U(1)$ symmetry will act &quot;non-linearly&quot; on $h&amp;#39;$ - in a way that can be easily deduced from $h&amp;#39;=h-v$.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-19T17:24:09.650" LastActivityDate="2011-03-19T17:24:09.650" />
  <row Id="7243" PostTypeId="2" ParentId="7212" CreationDate="2011-03-19T17:19:30.097" Score="3" ViewCount="" Body="&lt;p&gt;I point out that Feynman, in his 1979 Douglass Robb Memorial Lecture, part I (actually entitled &quot;Photons - Corpuscles of Light&quot;) on QED, video freely available at &lt;a href=&quot;http://vega.org.uk/video/programme/45&quot; rel=&quot;nofollow&quot;&gt;http://vega.org.uk/video/programme/45&lt;/a&gt;, asked this very question. He answered definitively that Newton was correct and that light was indeed a particle which, however, propagates according to the usual q.m. laws of adding complex amplitudes, thus reconciling its particle nature with observed interference phenomena. Furthermore see his closely related answers on this question at &lt;a href=&quot;http://www.youtube.com/watch?v=_7OEzyEfzgg&amp;amp;feature=related&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=_7OEzyEfzgg&amp;amp;feature=related&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The determining evidence ruling out waves includes 1. the photoelectric effect since waves can not cause ejection of electrons with the same frequency energy relationship, and the non relationship between the E field amplitude and energy   2.  Compton scattering (not mentioned in his lecture) not explainable under a wave theory  3. An argument he alluded to relating to the equation for scattering multiple particles not corresponding any more to Maxwell's equations (which I don't quite follow).   With the adoption of the probability amplitude propagation theory of QED, apparently no evidence is against the particle hypothesis.  (I myself would also like to see a QED derivation of the old Einstein entropy density of black body radiation in a closed cavity, however, i.e. the one that looks like 1/2 particle and 1/2 wave)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note related to a comment below:I previously (mis)attributed these arguments to his 1964 Messenger Lectures at Cornell, available online (only for PC's?) &lt;a href=&quot;http://research.microsoft.com/apps/tools/tuva/index.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and later published as &quot;The Character of Physical Law&quot;, but the correct ref. is as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS.  I myself do not feel that the question is confused at all, and actually voted it up, but it is the sort of question that often gets many confused answers.   I'm not too sure how we should handle this)&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2011-03-20T05:34:50.077" LastActivityDate="2011-03-20T05:34:50.077" CommentCount="2" />
  <row Id="7244" PostTypeId="1" AcceptedAnswerId="7263" CreationDate="2011-03-19T17:44:06.157" Score="2" ViewCount="209" Body="&lt;p&gt;Facts agreed on by most Physicists -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GR: One can't apply Noether's theorem to argue there is a conserved energy.&#xA;QFT: One can apply Noether's theorem to argue there is a conserved energy.&#xA;String Theory: A mathematically consistent quantum theory of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conclusion -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one can apply Noether's theorem in String Theory to argue there is a conserved energy, String Theory is not compatible with GR. If one can't, it is not compatible with QFT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the conclusion wrong? What is wrong with it? Is there a definition of energy in String Theory? If yes, what is the definition?&lt;/p&gt;&#xA;" OwnerUserId="2656" LastActivityDate="2011-03-19T21:46:21.483" Title="What is energy in string theory?" Tags="&lt;string-theory&gt;&lt;quantum-gravity&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="7245" PostTypeId="1" AcceptedAnswerId="7248" CreationDate="2011-03-19T17:45:28.893" Score="3" ViewCount="103" Body="&lt;p&gt;This should be a very easy question.  If you look at the bottom of &quot;Identical Particles&quot; in Wikipedia, you see Table 1, which gives you the two particle statistics for bosons, fermions and distinguishable particles.  The problem is to extend this table for three, four and five particles, or give an equivalent formula.  I  think I know what the answer is for fermions and distinguishable particles, but what about bosons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first guess is it would always be evenly distributed, but the many body limit does not equal the classical result, so that can't be right.  Or can it?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the formula, and the first few rows of the &quot;Pascals triangle for bosons&quot;? &lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-03-19T18:15:22.363" Title="What are the statistics of three to five bosons?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" />
  <row Id="7246" PostTypeId="1" CreationDate="2011-03-19T17:57:00.243" Score="-1" ViewCount="56" Body="&lt;p&gt;In the last year there has been fuzz regarding &lt;a href=&quot;http://en.wikipedia.org/wiki/Split-ring_resonator&quot; rel=&quot;nofollow&quot;&gt;metamaterials studied for the purpose of cloaking&lt;/a&gt;, and it seem to me they are somewhat glorified 2D waveguides, but in any case it seems a reasonable application of this to funnel radiation captured over a wide area of incidence into smaller area, and i thought that this could be a cheap way to produce ignition radiation for a fusion reactor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reasons why this wouldn't work, or not produce enough energy density? All stuff i've read about &lt;a href=&quot;http://en.wikipedia.org/wiki/National_Ignition_Facility&quot; rel=&quot;nofollow&quot;&gt;fusion ignition with lasers&lt;/a&gt; doesn't seem to depend too heavily on the coherence of the radiation&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-19T18:13:33.160" LastActivityDate="2011-03-20T05:30:28.937" Title="Light bending surfaces and energy funneling properties" Tags="&lt;fusion&gt;&lt;nanoscience&gt;" AnswerCount="2" />
  <row Id="7247" PostTypeId="2" ParentId="7232" CreationDate="2011-03-19T18:08:42.183" Score="4" ViewCount="" Body="&lt;p&gt;The answer is: the Motion Mountain book is wrong, your TA and John Baez are correct&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple answers here are discussing specific solutions in GR.  I think there is a much easier and more general way to answer this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In special relativity, we can choose to use a different coordinate system (we don't need to physically change our motion as some posters seem to claim), and by merely &lt;a href=&quot;http://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity&quot; rel=&quot;nofollow&quot;&gt;changing the coordinate system the electric field changes&lt;/a&gt;.  In fact, the electric field can be made arbitrarily large this way. All that this relies on is the Lorentz symmetry in special relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is possible because the value of the electric field is just a coordinate system dependent component value of a tensor - &lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_tensor&quot; rel=&quot;nofollow&quot;&gt;the electromagnetic tensor&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the global Lorentz symmetry of SR doesn't survive when transitioning to GR, but there is still local Lorentz symmetry.  This means we can still change our coordinate system locally, to have the electromagnetic field tensor change (at a point at least) the same way as before.  So what the Motion Mountain book is claiming it can derive from GR, is actually violating GR.  So we don't even need to look at any specific solutions in GR.  The local Lorentz symmetry of GR is enough to show that the textbook is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As other posters have mentioned, the same goes with energy density.  So the textbook author's starting comment of &quot;energy density is limited by general relativity: if energy density is too high, a black hole appears&quot; is wrong.  In my opinion, if this is an example display of the textbook author's understanding of GR, it is probably best to avoid his books altogether.&lt;/p&gt;&#xA;" OwnerUserId="2652" LastEditorUserId="2652" LastEditorDisplayName="" LastEditDate="2011-03-19T18:15:22.800" LastActivityDate="2011-03-19T18:15:22.800" CommentCount="2" />
  <row Id="7248" PostTypeId="2" ParentId="7245" CreationDate="2011-03-19T18:09:57.470" Score="3" ViewCount="" Body="&lt;p&gt;Dear Jim, it depends on how many different states - boxes - you have for those particles. If you still have 2 states, just like in the Wikipedia example, then for fermions, the probability is 0 everywhere - it's impossible to put more than 2 fermions to 2 states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For distinguishable particles, each particle has 50% odds to be in state 0 and 50% odds to be in state 1. So for $N$ distinguishable particles, there is $(N {\rm\, \, choose\, \,} k)/2^N$ chance for $k$ of them to be in state 0 and the remaining $N-k$ of them to be in state 1. Note that the odds sum to one, if you sum over $k$ from $0$ to $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $N$ bosons, all separations to state 0 ($k$ particles there) and state 1 ($N-k$ particles there) are possible, and each of those separations only corresponds to one multi-particle state. So the odds are $1/(N+1)$ for each possibility - a direct generalization of the table that has $1/3$ in all columns for the $N=2$ bosons. So yes, this is probably what you mean by &quot;equally distributed&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Fermi-Dirac and Bose-Einstein distributions - and both of them, with the same chances - can only be reduced to the Maxwell-Boltzmann distribution if the number of particles is high (it's not really essential) but, more importantly, the number of states in which they can be found is even much higher. Whenever there is a significant chance that the particles want to share a state etc., the classical Maxwell-Boltzmann approximation is inapplicable. So the toy model with two states 0,1 clearly doesn't allow you to approximate the Bose-Einstein or Fermi-Dirac distrtibutions by the Maxwell-Boltzmann one. This is particularly clear for the fermions because the Fermi-Dirac probabilities can't even be calculated because there is &lt;em&gt;no way&lt;/em&gt; to arrange more than two fermions to two states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Maxwell-Boltzmann distribution is applicable when the density of particles (number of particles divided by the number of states) is low. That's why the Maxwell-Boltzmann $\exp(-E/kT)$ is applicable as a replacement for high values of energy $E$ - because this is where a small number of particles will be found, so their arrangement won't be sensitive on their being bosons (collectivists who love siblings in the same state, more than the love between the distinguishable particles) or their being fermions (individualists who absolutely hate to share the state with others). For states at low $E$, that's where the concentration of particles per state is expected to be high, which is why you get the totally new quantum behaviors - such as the Bose-Einstein condensate in the ground state and/or the Fermi liquid - which obviously can't be described by the distinguishable statistics well.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-19T18:15:22.363" LastActivityDate="2011-03-19T18:15:22.363" />
  <row Id="7249" PostTypeId="2" ParentId="6274" CreationDate="2011-03-19T18:27:12.227" Score="3" ViewCount="" Body="&lt;p&gt;First, let me say I'm not sure what is meant by a BEC with $T\gt 0$. Condensation is a finite temperature phenomenon, which occurs due to the presence of pair-wise interactions (generally attractive, but pairing can happen even for repulsive potential) in a many-body system. For instance, in a superconductor below some critical temperature $T_c$, electrons with opposite momenta and spin (s-wave pairing) pair up to form a bound state called a Cooper pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ground state of the unpaired electron gas for $T \gt T_c$ is characterized by the Fermi energy $E_F$. After condensation, the many-body system has a new ground state at energy $E_{bcs} = E_F - \Delta $, where $\Delta \sim k_B T_c$ is the binding energy of a Cooper pair. $\Delta$ is also known as the &lt;em&gt;gap&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $T\lt T_c$ &lt;em&gt;all&lt;/em&gt; the electrons are not paired up due to thermal fluctuations. However, the number of unpaired electrons as a fraction of the total number of electrons (the condensate fraction) goes as $N_{free}/N_{all} = 1- (T/T_c)^\alpha$, where $\alpha\gt 0$. The number of free electrons drops rapidly as $T$ is decreased below $T_c$. In lab setups, BEC's generally undergo some form of evaporative cooling to get rid of particles with energies greater than $\Delta$. At this point the condensate can be treated as a gas of interacting (quasi)particles (cooper pairs) with an approximate hard-core repulsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the gas, before and after condensate formation, is always at finite temperature! This is reflected, for instance, in the dependance of the condensate fraction on $T$ as mentioned above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mean-field solutions for low-energy excitations of the condensate are given by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gross%E2%80%93Pitaevskii_equation&quot; rel=&quot;nofollow&quot;&gt;Gross-Pitaevskii equation&lt;/a&gt;(GPE):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left( - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial r^2} + V(r)  + \frac{4\pi\hbar^2 a_s}{m} |\psi(r)|^2 \right) \psi(r) = \mu \psi(r) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a_s$ is the scattering length for the hard-core boson interaction, with $a \lt 0$ for an attractive interaction and $a \gt 0$ for a repulsive interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably one should be able to construct a canonical ensemble with solutions $\psi(k)$ ($k$ being a momentum label of the above equation), but this is by no means obvious because of the non-linearity represented by the $|\psi(r)|^2$ term. Here a &quot;zero temperature&quot; state would correspond to a perfect BEC with no inhomogeneities, i.e. the vacuum solution of the GPE. However, the entire system is at some finite temperature $T \lt T_c$ as noted above. The resulting thermal fluctuations will manifest in the form of inhomogeneities in the condensate, the exact form of which will be determined by the solutions of the GPE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the GPE's regime of validity is that of dilute bose gases ($l_p \gg a_s$ - the average interparticle separation $l_p$ is much greater than the scattering length). For strong coupling I do not know of any similar analytical formalism. If I had to take a wild guess I'd say that the strong-coupling regime could be made analytically tractable by mapping it to a dual gravitational system, but that's another story altogether.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $l_p$ approaches $a_s$ from above, the GPE breaks down and it will have singular solutions for any given $T$ and these are likely the singularities that you are referring to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference:&lt;/strong&gt; The single best reference I can suggest is &lt;strong&gt;Fetter and Walecka's&lt;/strong&gt; book on many-body physics. I'm sure you can find more compact sources with a little effort. But generally the brief explanations leave one wanting for a comprehensive approach such as the one F&amp;amp;W provides.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;br /&gt;This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-19T19:41:45.860" CommentCount="11" />
  <row Id="7250" PostTypeId="1" AcceptedAnswerId="7259" CreationDate="2011-03-19T18:45:52.963" Score="3" ViewCount="210" Body="&lt;p&gt;&lt;strong&gt;My current understanding of gravitational lensing follows&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a star or other massive body passes between us and another star, the phenomenon&#xA;generally labeled 'gravitational lensing' occurs.  The term gravitational lensing refers to the bending of light rays due to the gravitational influence of the massive body.  Since the bends concentrate light for the observer, they act like a magnifying lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OTOH&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since I was young I have pondered other explanations for such observations.  One thought about what (else) might cause the observed effect is a cloud of gas around the massive body itself.  The assumption being that any change in density between the thin gas of interstellar space and the thicker gas surrounding a massive object should form a refractive boundary.  In the case of an active star, that gas &amp;amp; might be the heliosphere of the star, and for any massive object that is &lt;em&gt;not&lt;/em&gt; creating an &#xA;heliosphere, it could be due to clouds of gas that the massive object is sucking in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I raised this idea with a physics teacher in high school &amp;amp; he assured me that any possible refraction from clouds of gas had been ruled out in observations of gravitational lensing.  I never got to see papers on the subject, so I cannot confirm that (and am not convinced).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Does anybody know of papers on gravitational lensing that discuss the refraction from clouds of gas around the lensing body?&lt;/li&gt;&#xA;&lt;li&gt;How could an observer/experimenter go about quantifying the likely refractive index of such a gas cloud.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2659" LastEditorUserId="2659" LastEditorDisplayName="" LastEditDate="2011-04-06T17:18:43.213" LastActivityDate="2011-04-06T17:18:43.213" Title="Gravitational lensing or cloud refraction?" Tags="&lt;gravity&gt;&lt;light&gt;&lt;refraction&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="7251" PostTypeId="1" AcceptedAnswerId="7258" CreationDate="2011-03-19T18:55:12.000" Score="8" ViewCount="240" Body="&lt;p&gt;If I had a light &quot;diode&quot; - an object that only allowed light (at least for a range of frequencies) to travel through it in one direction, would this &lt;em&gt;necessarily&lt;/em&gt; allow violations of the 2nd Law of Thermodynamics?  Or would it be possible to restrict other properties of the device such that it could be okay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why I think it should violate 2nd law:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I can place such devices at temperature T in a bath of thermal radiation at temperature T, such that it forms a box which only lets the light in or out.  This would appear to always allow one to create a thermal gradient.  Once a thermal gradient is formed, useful energy can be extracted.  This seems to allow useful energy to always be extracted from a thermal bath and thus give perpetual motion.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Release one such device at temperature T in a bath of thermal radiation at temperature T.  Since the photons are absorbed or reflected in one direction, and undisturbed in the other, the device will start moving.  It can use a thermal bath to directly generate useful motion.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Why I think I'm missing something and such a device could potentially co-exist with the 2nd Law:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The above arguments seem valid if the diode was only 'partially' rectifying as well.  However there exist sunglasses which largely reflect on one side, but not from the other side.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Researchers have succeeded in using meta-materials to create a &quot;blackhole&quot; for microwaves.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can anyone help sort this out?&lt;/p&gt;&#xA;" OwnerUserId="2652" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-19T20:06:17.987" LastActivityDate="2011-06-04T20:49:18.733" Title="Light &quot;diode&quot; and 2nd law of thermodynamics" Tags="&lt;thermodynamics&gt;&lt;optics&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="7252" PostTypeId="2" ParentId="1836" CreationDate="2011-03-19T19:01:47.427" Score="1" ViewCount="" Body="&lt;p&gt;Air is mostly composed of first two row small molecules like O2 and N2, and atoms like Argon.  For all these, the absorption is in the deep ultraviolet.   Molecules which absorb in the visible have smaller energy differences in their first absorption bands, although chlorine(2) and Iodine(2) molecules are visible.   This answer is meant to be complementary to and supportive of the evolutionary answer given above&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2011-03-19T19:01:47.427" />
  <row Id="7253" PostTypeId="2" ParentId="7250" CreationDate="2011-03-19T19:41:45.110" Score="2" ViewCount="" Body="&lt;p&gt;Without answering the question I'll note that one can &lt;a href=&quot;http://spiff.rit.edu/classes/phys240/lectures/microlens/microlens.html&quot; rel=&quot;nofollow&quot;&gt;calculate from GR a family  of expected curves&lt;/a&gt; (dependent on the impact parameter) for the apparent brightness of the distance source, and that the micro lensing people matched events against those curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps I'll poke around in search of a paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is something that might be interesting:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.arxiv.com/abs/astro-ph/9504001&quot; rel=&quot;nofollow&quot;&gt;REMOVING DEGENERACY OF MICROLENSING LIGHT CURVES THROUGH NARROW-BAND PHOTOMETRY OF GIANTS&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-03-19T19:48:11.953" LastActivityDate="2011-03-19T19:48:11.953" CommentCount="1" />
  <row Id="7254" PostTypeId="2" ParentId="7251" CreationDate="2011-03-19T19:44:34.453" Score="9" ViewCount="" Body="&lt;p&gt;A light diode that would reflect every photon coming from one side but allowed all photons from the opposite side to be fully transmitted - and that wouldn't absorb anything - would be in contradiction with the second law because that would be nothing else than a photon version of Maxwell's demon: it would be able to collect all particles (photons in this case) in one half of a vessel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such things can't exist because the second law demonstrably holds for all objects with many degrees of freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of the &quot;almost light diodes&quot; from the real world that you mentioned is doing the same thing - mostly because the objects you mention absorb a sufficient fraction of the incoming light from one of the two sides. When they absorb the light, they turn the absorbed photon's energy to heat - a more chaotic form of energy - and increase the total entropy by a sufficient amount so that the overall entropy trend is never negative.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-19T19:44:34.453" CommentCount="2" />
  <row Id="7255" PostTypeId="2" ParentId="7250" CreationDate="2011-03-19T20:05:56.663" Score="6" ViewCount="" Body="&lt;p&gt;Well, in other words, because you don't want to accept that space can get curved, you postulate that there are &quot;real lens&quot; around objects such as the Sun and the galaxies etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think about Eddington's - or any other - experiment, it's clear that the characteristic thickness of the &quot;glass&quot; - let me call it &quot;glass&quot; - surrounding the Sun has to be comparable to the solar radius. By the thickness, I mean the thickness of the region where the refractive index's deviation from one is at least 50% of its maximal deviation from one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Shape of the lens around the Sun&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Eddington's solar setup, the thickness can't be much smaller because that would mean that the stars away from the Sun - visually further than one apparent radius - are not affected but they are affected. And the thickness can't be much greater because of the same reason: too distant stars from the Sun would be affected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fine. The bending observed by Eddington - or at least others who did the same thing more honestly - is 1.37 arc seconds which is 6.6 microradians. By Snell's law, the deviation of $n_{\rm max}$ from one has to be comparable to this angle, so you need a &quot;glass&quot; with something like &#xA;$$ n \approx 1+ 10^{-5} $$&#xA;The air has $1+3\times 10^{-4}$, so if you just dilute air by a factor of 30, you get exactly the &quot;glass&quot; that you need for your lens around the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the starlight has to go through this &quot;glass&quot; and we may ask what is the optical thickness. Well, the solar radius is about 700,000 kilometers so the starlight would have to penetrate through 700,000 kilometers of a material whose density is similar to 1/30 of the air. That's equivalent to a 20,000 kilometers thick layer of air. I assure you that for any conceivable material, no light would survive if it had to go through a 20,000 kilometers thick air-like wall. Note that even the atmosphere which is just 10 kilometers thick (effectively) absorbs a big portion of the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Looking through dozens of miles of water&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you may prefer water with $n=1.33$. If you diluted it about 30,000 times, it would get to $n=1+10^{-5}$. So 700,000 kilometers of your &quot;glass&quot; are equivalent to 20+ kilometers of water. If this &quot;glass&quot; were surrounding the Sun, the stars around the Sun would look similar like if you observe them from the Mariana Trench (times 2-5) - and I assure you that not much starlight can be seen in the Mariana Trench. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This example with the Sun was far too modest. You would have to reproduce the same results for the huge galactic gravitational lens. They bend the light by similar angles but the physical size of the &quot;glass&quot; would have to be many light years. The light in your model would have to penetrate through many light years thick wall of air. Nothing would be left. Moreover, if the density of your &quot;glass&quot; around galaxies resembled the air (over 30), the &quot;glass&quot; would be far more massive than the galaxy itself. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter point may be made spectacular if we estimate the average density of a galaxy. The Milky Way's mass is $6\times 10^{42}$ kg and its radius is $5\times 10^{20}$ meters so $M/R^3$ is $5\times 10^{-20}$ kilograms per cubic meter or so. Compare with the $0.04$ density of the &quot;air over thirty&quot; material. The &quot;glass&quot; you would need to surround the galaxy would be $10^{18}$ times denser than the actual average density of the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you understand why your model is safely dead (by dozens of orders of magnitude) by now? The Universe is - and has to be - remarkably empty. This fact is also being underestimated by proponents of various spin networks and spin foam - they're incompatible with the observations for similar reasons as your &quot;glass&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Designing a better glass&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could try to invent a &quot;better material&quot; for your &quot;glass&quot; - one that doesn't absorb much and one that has the same index of refraction for all colors - an important condition needed to avoid a rainbow-like color pattern of the stars, as explained in another answer by Ted Bunn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were working hard to obey all the required conditions, you would find out that the required material looks nothing like the atomic matter containing electric charges (e.g. electrons and protons). It would look like a sea of virtual gravitons that affect all kinds of light in the same way, and only affect its direction and not its intensity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the very end, if you were doing it right, you would rediscover the &quot;material&quot; of the virtual gravitons in a quantized version of general relativity. ;-) This &quot;material&quot; acts on all photons in the same way (because of the equivalence principle); it maximizes the bending effect relatively to the extra mass it adds to space (which is really zero - the curved spacetime may be fully attributed to the Sun's mass at the center); and it doesn't absorb light.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-19T21:55:03.827" LastActivityDate="2011-03-19T21:55:03.827" CommentCount="1" />
  <row Id="7256" PostTypeId="2" ParentId="7231" CreationDate="2011-03-19T20:36:45.657" Score="1" ViewCount="" Body="&lt;p&gt;Marek,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think in terms of &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_function&quot; rel=&quot;nofollow&quot;&gt;Harmonic Functions&lt;/a&gt; and their &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_function#Maximum_principle&quot; rel=&quot;nofollow&quot;&gt;Maximum Principle&lt;/a&gt; (or Mean Value Theorem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity (and, in fact, without loss of generality), let's just think in terms of a free particle, ie, $V(x,y,z) = 0$. When the Potential vanishes, the Schrödinger equation is nothing but a Laplace one (or Poisson equation, if you want to put a source term). And, in this case, you can apply the Mean Value Theorem (or the Maximum Principle) and get a result pertaining your question: in this situation you saturate the equality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you have a Potential, you can think in terms of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Laplace%E2%80%93Beltrami_operator&quot; rel=&quot;nofollow&quot;&gt;Laplace-Beltrami operator&lt;/a&gt;: all you need to do is 'absorb' the Potential in the Kinetic term via a &lt;a href=&quot;http://arxiv.org/abs/math-ph/0212017&quot; rel=&quot;nofollow&quot;&gt;Jacobi Metric&lt;/a&gt;: $\tilde{\mathrm{g}} = 2\, (E - V)\, \mathrm{g}$. (Note this is just a conformal transformation of the original metric in your problem.) And, once this is done, you can just turn the same crank we did above, ie, we reduced the problem to the same one as above. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps a bit.&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2011-03-19T20:36:45.657" CommentCount="2" />
  <row Id="7257" PostTypeId="2" ParentId="6220" CreationDate="2011-03-19T20:48:10.427" Score="0" ViewCount="" Body="&lt;p&gt;In a simple minded way I would say that the flat road for the same miles and the same speed  would be more economical:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving the car uphill against gravity takes extra energy which is not being gained coming downhill because of braking (constant speed) and not turning off the engine for safety.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each car there is an economic rpm , mine is at 2500 rpm . One can use that on a flat road, but a hill needs lower gears going up , not the economic rpms, and this is not gained back due to braking downhill so I cannot see   how a hilly road can be more economical in any case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: Looking at the preferred answer which claims gains by the method of driving  the car I searched for a curve of rpms to fuel consumption and power output. Surprisingly there are not many out in the links.  &lt;a href=&quot;http://pics.tdiclub.com/members/rwatson/TDIBMEPBSFCHP.jpg&quot; rel=&quot;nofollow&quot;&gt;Here is the only one I found &lt;/a&gt; probably scanned from an engineering textbook. I observe that consumption goes up with rpm. Going uphill needs more rpm. I will also add that most of the inefficiencies are in heating the engine, and the higher rpm the hotter the engine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me simplify the problem: If I pump 100 litres of water up hill and let it run down hill. Will the kinetic energy of the running water be greater than the energy utilized in pumping it up? In the best of conditions, without the inefficiencies of pump, it will break even.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the car analogue then one is playing with inefficiencies of flat versus hill and maybe a computer program with real car data would give a definitive asnwer.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-23T11:55:53.733" LastActivityDate="2011-03-23T11:55:53.733" CommentCount="9" />
  <row Id="7258" PostTypeId="2" ParentId="7251" CreationDate="2011-03-19T21:05:14.530" Score="12" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Optical_isolator&quot; rel=&quot;nofollow&quot;&gt;Light diodes exist&lt;/a&gt;, and yet the 2nd law endures. In general, any &lt;a href=&quot;http://en.wikipedia.org/wiki/Brownian_ratchet&quot; rel=&quot;nofollow&quot;&gt;perfect ratchet&lt;/a&gt; would allow you to violate the 2nd law. Since very few Nobel prizes have been awarded for perpetual motion machines, experimental evidence seems to suggest there is no such thing as a perfect ratchet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When confronted with any type of device that rectifies the flow of energy, the appropriate question, in my opinion, is 'how is this device imperfect'? In the case of a Faraday isolator, Lord Rayleigh &lt;a href=&quot;http://www.nature.com/nature/journal/v64/n1667/abs/064577e0.html&quot; rel=&quot;nofollow&quot;&gt;addressed this question&lt;/a&gt;  a long time ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.usna.edu/Users/physics/mungan/Scholarship/FaradayIsolators.pdf&quot; rel=&quot;nofollow&quot;&gt;The other reference&lt;/a&gt; at the end of the Wikipedia article on Faraday isolators is a pretty nice discussion of this topic.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="2359" LastEditorDisplayName="" LastEditDate="2011-03-20T13:00:44.747" LastActivityDate="2011-03-20T13:00:44.747" CommentCount="4" />
  <row Id="7259" PostTypeId="2" ParentId="7250" CreationDate="2011-03-19T21:14:32.397" Score="8" ViewCount="" Body="&lt;p&gt;One point in addition to the other answers. Refraction is almost always chromatic (i.e., different wavelengths refract differently). Gravitational lensing is achromatic. Some studies of gravitational lensing, particularly the MACHO-like microlensing studies, look specifically for achromaticity to test that what they're seeing really is gravitational.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-19T21:14:32.397" CommentCount="3" />
  <row Id="7260" PostTypeId="1" CreationDate="2011-03-19T21:17:05.293" Score="2" ViewCount="103" Body="&lt;p&gt;I want to heat one side of an aluminum plate enough to hold the other side of that plate $100K$ above ambient.  I'm willing to assume that the heated side of the plate is &quot;well&quot; insulated (along with the edges, if it's significant).  The plate is $500cm^2$ and is open to air in a normal room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the formula is simple enough:  $P = \frac{\Delta T}{R_{pa}}$ Where $R_{pa}$ is the thermal resistance of the plate to air in $\frac{K}{W}$.  The problem I'm having finding the thermal resistance is that most heat-sink-like applications would care about the &lt;em&gt;bottom&lt;/em&gt; of the plate in this case:  the goal would be to &lt;em&gt;cool&lt;/em&gt; the &lt;em&gt;bottom&lt;/em&gt; of the plate.  I don't actually care what the temperature of the bottom is (within reason):  I want to hold the top at a specific temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I attack it from first principles and consider that the the thermal resistance of aluminum is irrelevant (since it's the top surface I care about) then all I need is the thermal resistance of the air above it.  The guessing part comes in when trying to figure out how thick a slab of air to consider.  If I guesstimate 3cm I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_{air} = 0.025 \frac{W}{mK}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T = \frac{0.05m^2}{0.03m}T_{air} = 0.04 \frac{W}{K}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_{pa} = \frac{1}{T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = \frac{100K}{R_{pa}} = 100K \cdot 0.04 \frac{W}{K} = 4W$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 4W result feels about an order of magnitude too low.  Of course if I change my guess of the air thickness to consider to 3mm I get 40W which seems more reasonable.  But instead of calculating the wattage I need I'm just guessing in different units.&lt;/p&gt;&#xA;" OwnerUserId="2660" LastActivityDate="2011-03-19T21:44:48.673" Title="How much power to keep surface of aluminum plate at given temperature?" Tags="&lt;temperature&gt;&lt;power&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7261" PostTypeId="2" ParentId="7238" CreationDate="2011-03-19T21:25:05.347" Score="2" ViewCount="" Body="&lt;p&gt;First of all, energy is energy. It may be converted from any form to any other form and there is absolutely no ambiguity about the sign of any form of energy. If $E$ is correctly defined so that it is conserved, the signs in front of all quantities are well-defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your observations about $E^2$ and $B^2$ having the opposite sign in the Maxwell Lagrangian are correct, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what's incorrect is your assumption that the energy carried by a configuration of electric charges or dipoles or magnetic dipoles may always be reduced to the energy carried by the electromagnetic fields only, $(E^2+B^2)/2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This statement is kind of coincidentally true in the case of the electrostatic energy. The energy between charges, an integral of&#xA;$$\frac 12\int \rho \Phi$$&#xA;where $1/2$ is included to avoid double-counting from the interaction energy of pairs of charges  - may be integrated by parts, using $\rho = \nabla\cdot E$ and $E=-\nabla \Phi$, to get&#xA;$$\frac 12 E^2$$&#xA;That's nice and if one manages to subtract the divergent self-interaction energy of point-like charges (which has been a big puzzle in classical physics and became an equally big puzzle in QED, before it was solved and superseded by renormalization), everything is stored in the electrostatic field energy $E^2/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is simply not the case for the magnetic fields. The actual energy is given by terms like $-\vec m\cdot \vec B$, as you correctly write. In particular, two equally oriented magnetic dipoles, separated by an interval in the same direction as the direction of these dipoles, have a negative interaction energy, see &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_dipole-dipole_interaction&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Magnetic_dipole-dipole_interaction&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's why the force between two such dipoles is attractive - and the very same thing with the same signs would hold for two electric dipoles, too. The term $-\vec m\cdot \vec B$ may be boiled down to the $\pm\vec j\cdot  \vec A$ in the Maxwell Lagrangian with sources. I wrote $\pm$ because I don't want to spend lots of time by fixing this sign because only some relative signs really matter for this question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, mimicking the electrostatic integration by parts here in magnetostatics, $\vec j = {\rm curl} \vec B$ and $\vec B = {\rm curl} \vec A$, won't quite work because&#xA;$$ \vec j\cdot  \vec A = {\rm curl} \vec B \cdot \vec A \neq - \vec B \cdot {\rm curl} \vec A +{\rm curl} (\dots). $$&#xA;Those curl identities don't seem to work in the same way. So the energy of magnetic dipoles can't completely eliminate the non-$B^2$ terms, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in the electrostatic case, we have to be careful how we interpret the different energies to avoid double-counting. Because the $\int\rho\Phi/2$ energy could have been rewritten as $\int E^2/2$, we may imagine that charges are some mysteriously compact lumps of electric field. But aside from the $1/2$ factor that was needed to avoid the double-counting from the electrostatic energy of each pair of charges, there is one more dangerous place where we could double-count: we must either imagine that the energy comes from $\rho\Phi/2$, or from $E^2/2$, but not from both! Again, if we wanted to include both terms, they would have to have the same sign, and we would double the energy relatively to the correct value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnetic counterpart is that we may express the energy density as $B^2/2$ if we don't study how the dipoles and sources are affected, but if we do, we must forget $B^2/2$ and correctly calculate terms such as $-\vec m\cdot \vec B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the problem with the sign would occur even for randomly generated electric dipoles. It's because $\int E^2/2$ is positively definite - for any configuration of dipoles - while the interaction energy between two dipoles may be both positive and negative. So it's just not true that the interaction energy of two ordinary dipoles may be rewritten as $\int E^2/2$. In some sense, the problem does reside in the dipoles' self-interactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\int E^2/2$ is always positively definite - so how it can account for the electric charges' interaction energies that can have both signs? The answer is that $\int E^2/2$ should be identified with $\int \rho\Phi$ plus the &quot;internal mass&quot; of the electric sources - which is basically infinite. However, $\int B^2/2$ is also positively definite while there is significant extra mass (interaction self-energy) associated with the magnetic dipoles themselves. So for the magnetic sources, exactly because the currents may be &quot;light&quot;, the transcription into $\int B^2/2$ cannot work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we have either electric or magnetic dipoles, we must carefully put the right sign in front of the energy, instead of assuming that the energy may always be rewritten as &quot;field-only&quot; energy. It can't.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-19T21:45:39.487" LastActivityDate="2011-03-19T21:45:39.487" />
  <row Id="7262" PostTypeId="2" ParentId="7260" CreationDate="2011-03-19T21:44:48.673" Score="1" ViewCount="" Body="&lt;p&gt;In fact the thermal resistance of the plate can be neglected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That was the simple part. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest: extremely complicated. &quot;From first principles&quot; this affords &#xA;a dimensionless analysis with four Numbers (heat transfer under autoconvection) : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reynolds Number &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prandtl Number &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nusselt Number &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Grashoff Number &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very complicated and rather inaccurate. Results will be not more than a coarse guide to &#xA;some experiments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are rule of thumb formulas often used for heat sinks in electronics, those might &#xA;be as good as a start. &#xA;Be aware that weak air movement or the inclination of the plate changes a lot!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do You need this for? There are rather simple practical solutions. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-19T21:44:48.673" CommentCount="4" />
  <row Id="7263" PostTypeId="2" ParentId="7244" CreationDate="2011-03-19T21:46:21.483" Score="6" ViewCount="" Body="&lt;p&gt;In any theory which includes General Relativity, there is no locally conserved energy. The reason is that energy creates a gravitational field which has energy itself, so &quot;gravity gravitates&quot;. There is a local quantity (the energy-momentum tensor) which is &lt;em&gt;covariantly&lt;/em&gt; conserved, and there are global quantities (like the ADM mass) which expressed the total energy of the system and are conserved. But there are no currents which give locally  conserved quantities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(One more technical way to express that is that the spacetime transformations corresponding to local energy and momentum conservation are now no longer global symmetries but are gauge redundancies). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In field theory (classical or quantum) without gravity spacetime translations can be a global symmetry (if spacetime is flat), and correspondingly energy is locally conserved. Once you couple the theory weakly to gravity (even if gravity is only a background with no dynamics of its own), it is then only approximately conserved. When gravitational effects are large, there is no approximately conserved quantity correpsonding to energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Lubos said in comments, nothing really new happens with respect to this question in string theory. In the most general situation there is no locally conserved quantity, and when the theory reduces to QFT on spacetime which is approximately flat and when gravitational effects are small, then there is an approximately conserved quantity. String theory is &lt;em&gt;compatible&lt;/em&gt; with both QFT and GR, which means it reproduces their results approximately in the appropriate limit. But of course away from those limits it has its own features that are generally different from either one of those theories. For that specific question it is much closer in spirit to GR.&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-03-19T21:46:21.483" />
  <row Id="7264" PostTypeId="1" CreationDate="2011-03-19T21:58:20.250" Score="3" ViewCount="89" Body="&lt;p&gt;Earth's N and S magnetic poles &quot;wander independently of each other and are not at exactly opposite positions on the globe&quot; [from WIKI's &quot;Earth's magnetic field&quot;]. Can these independent motions be consistent with the supposed &quot;dynamo effect&quot; from electric currents of a liquid outer core? Can Earth's spheroidal shape be a factor?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-03-20T00:49:12.440" Title="Why is there independent motion of Earth's magnetic poles?" Tags="&lt;magnetism&gt;" AnswerCount="2" />
  <row Id="7265" PostTypeId="2" ParentId="7250" CreationDate="2011-03-19T22:00:34.337" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The assumption being that any change&#xA;  in density between the thin dust of&#xA;  interstellar space and the thicker&#xA;  dust surrounding a massive object&#xA;  should form a refractive boundary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dust does not lead to refraction. Dust either absorbes light or &#xA;scatters it, depending on particle size and particle properties. &#xA;In both cases that light never will reach us. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, dust is blown away from newformed stars as soon as the &#xA;star lights up. So, the heliospere contains not much dust, as &#xA;long as the star has not left the cloud of his birth, the heliopause &#xA;can be a bubble free of dust in a dust cloud. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-19T22:00:34.337" CommentCount="2" />
  <row Id="7267" PostTypeId="2" ParentId="7264" CreationDate="2011-03-20T00:33:56.173" Score="2" ViewCount="" Body="&lt;p&gt;Yes.  If the earths magnetic field was strictly a dipole field cenetered at the center of the core then they the poles would have to be perfectly symmetric to each other. But the field is much more complicated than that. If you think of the field as being composed of spherical harmonics, you will see that the higher order harmonics fall off faster as a function of radius, so the terms higher than dipole aren't all that strong at the surface, but they are not zero, they so affect the local fields, and can skew the location of the &quot;pole&quot;.&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Spherical_harmonics&quot; rel=&quot;nofollow&quot;&gt;spherical harmonics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-03-20T00:33:56.173" />
  <row Id="7268" PostTypeId="2" ParentId="7264" CreationDate="2011-03-20T00:49:12.440" Score="5" ViewCount="" Body="&lt;p&gt;Well, Earth does not have a giant bar magnet inside, the outer core is an giant mass of conductive fluid in a complex motion guided by magnetohydrodynamics -- the problem is complicated enough to be unsolvable analytically and really gives no hope of simple solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one should rather think of a geomagnetic &lt;strong&gt;field&lt;/strong&gt; which currently happens to be close to a dipole field, thus producing an illusion of poles. &lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2011-03-20T00:49:12.440" />
  <row Id="7269" PostTypeId="2" ParentId="6890" CreationDate="2011-03-20T01:59:20.303" Score="1" ViewCount="" Body="&lt;p&gt;Does&#xA;&lt;a href=&quot;http://amasci.com/miscon/a-rant.html&quot; rel=&quot;nofollow&quot;&gt;&quot;ACOUSTIMAGNETOELECTRICISM!&quot;&lt;/a&gt; by William J. Beaty answer your question?&lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2011-03-20T01:59:20.303" CommentCount="1" />
  <row Id="7270" PostTypeId="2" ParentId="7005" CreationDate="2011-03-20T03:10:31.923" Score="0" ViewCount="" Body="&lt;p&gt;Katrijnn,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lubos Motl's answer is correct.  However, the so-called maximally extended Schwarzschild solution to the equations that govern gravity describe what may also be considered an &lt;em&gt;untraversable&lt;/em&gt; wormhole.   It connects two &quot;universes&quot; (i.e. asymptotically flat regions).  But the connection does not last long enough for a signal to pass through it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a suitable observer this solution essentially describes an object that is a white hole in the past, a short-lived untraversable wormhole in the present, and a black hole in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Note that astrophysical black holes (as opposed to the theoretically permissible black hole solution considered above) result from stellar collapse and do not involve wormholes or white holes.]&lt;/p&gt;&#xA;" OwnerUserId="1726" LastActivityDate="2011-03-20T03:10:31.923" />
  <row Id="7271" PostTypeId="2" ParentId="6674" CreationDate="2011-03-20T03:31:52.483" Score="1" ViewCount="" Body="&lt;p&gt;John McVirgo is correct -- it's a transmission line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you have a 1 light year long pair of wires.&#xA;As soon as you connect a battery to one end of the pair, energy starts racing towards the other end of the pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That energy is mostly in the space between the two wires -- electrical energy does not normally flow inside metal objects.&#xA;&lt;a href=&quot;http://amasci.com/elect/poynt/poynt.html&quot; rel=&quot;nofollow&quot;&gt;Poynting-flow diagrams&lt;/a&gt; show where the energy flows, although they are a bit difficult to draw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various popular cables have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Wave_propagation_speed&quot; rel=&quot;nofollow&quot;&gt;wave propagation speed&lt;/a&gt; that ranges from 42% to 95% of the speed of light.&#xA;Twin lead has a wave propagation speed about 80% of the speed of light.&#xA;If your cable happens to have the same wave propagation speed, then that pulse of energy will hit the end of the 1 light-year-long cable about 1/0.80 years = 15 months later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up until the time that pulse hits at time t=+15 months, it makes no difference whether that far end of the cable is shorted together, open and disconnected, or connected to a lightbulb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the battery is connected for 10 seconds and then disconnected,&#xA;then a light bulb at the far end of the cable will remain dark for about 15 months, then glow for 10 seconds, and then go dark again.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;How long would it take electrical energy to travel 1 light year along a pair of wires?&quot;&#xA;As we just discussed, roughly 15 months.&lt;/li&gt;&#xA;&lt;li&gt;&quot;How long would it take for electrons to begin flowing out the - terminal of a battery after you hook it to a 1 light-year-long cable?&quot; That happens pretty much instantaneously when you connect the battery.&lt;/li&gt;&#xA;&lt;li&gt;&quot;How long would it take an electron to travel 1 light year along a long wire?&quot; That's going to be orders of magnitude longer than a year -- see &lt;a href=&quot;http://amasci.com/miscon/speed.html&quot; rel=&quot;nofollow&quot;&gt;Speed of &quot;Electricity&quot;&lt;/a&gt; by Bill Beaty.&lt;/li&gt;&#xA;&lt;li&gt;&quot;How long would it take for electrons to begin flowing into the + terminal of a battery after you hook it to a 1 light-year-long cable?&quot; That happens pretty much instantaneously when you connect the battery.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="965" LastActivityDate="2011-03-20T03:31:52.483" />
  <row Id="7272" PostTypeId="2" ParentId="7246" CreationDate="2011-03-20T04:25:01.407" Score="0" ViewCount="" Body="&lt;p&gt;Assuming that you have a source whose temperature is T, the best any optics system can do (in applying the energy from the source to a target) is in such a way that the target may be heated up to the temperature T. This is because of a basic law of thermodynamics: heat flows from hot things to cold things and never vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's another way of saying this that applies to imaging optics; that is, to optical systems that produce images. I'll edit it in if I find it. But the mechanism you're describing is more general than an optics system.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-20T04:25:01.407" CommentCount="1" />
  <row Id="7273" PostTypeId="2" ParentId="7246" CreationDate="2011-03-20T05:30:28.937" Score="1" ViewCount="" Body="&lt;p&gt;If you read the laser fusion article in &lt;a href=&quot;http://en.wikipedia.org/wiki/Laser_fusion&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; you will see that white light is used in some designs to feed the lasers, not very efficiently. Better wave guides might improve efficiency but certainly cannot substitute for the energy  concentration  of lasers on the fuel pellets.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-20T05:30:28.937" />
  <row Id="7274" PostTypeId="2" ParentId="6581" CreationDate="2011-03-20T05:38:47.340" Score="0" ViewCount="" Body="&lt;p&gt;It would seem that this problem could be re-expressed in a manner that makes the form of the answer intuitively clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a cylinder oriented vertically.  Its top (disk-shaped) surface is coated with a thin layer of &quot;North&quot; magnetic monopoles.  Its bottom surface is similarly coated with &quot;South&quot; magnetic monopoles.  The spinning of the cylinder along its vertical axis creates rings of magnetic current due to the resulting circular paths traced out by the magnetic monopoles.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These magnetic currents appear in the Maxwell equation corresponding to Faraday's Law in a manner exactly analogous to the appearance of electric currents in the Maxwell equation corresponding to Ampere's Law. [This term in the Faraday's Law equation is normally zero, because there are no magnetic monopoles, hence no magnetic currents.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rings of magnetic current will produce toroidal electric fields (concentrated at the ends of the cylinder).  These fields are analogous to the toroidal magnetic fields produced by rings of electric current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The validity of this answer (that E is not zero, but a toroidal configuration at both ends) depends on whether a magnetic dipole formed from the aforementioned distribution of magnetic monopoles is equivalent to your cylindrical magnet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, although magnetic monopoles could be used to create a magnetic field identical to that of your magnetic (whose field result from electric charges circulating about nuclei), the situations are not equivalent.  There are only two ways to generate an electric field -- electric charges or magnetic currents.  Spinning the magnet does not spontaneously generate electric charge.  Nor, in the absence of mononoples, does it create a magnetic current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, Gauss' Law does &lt;em&gt;not&lt;/em&gt; imply that the electric field is zero.  It only implies that the integral of the electric field on a surface enclosing the cylinder is zero.  [As pointed out by Edward, for an electric dipole this integral is zero, but the electric field itself is not.]&lt;/p&gt;&#xA;" OwnerUserId="1726" LastEditorUserId="1716" LastEditorDisplayName="" LastEditDate="2011-03-22T18:21:55.490" LastActivityDate="2011-03-22T18:21:55.490" />
  <row Id="7275" PostTypeId="1" CreationDate="2011-03-20T06:31:58.040" Score="0" ViewCount="33" Body="&lt;p&gt;i need some information (link , ebook , etc ) about 'Modeling medical phenomenon with mechanics fluid'&#xA;i don't know where can i search about this subject.&#xA;regards shabnam.&lt;/p&gt;&#xA;" OwnerUserId="2667" LastActivityDate="2011-03-20T06:31:58.040" ClosedDate="2011-03-20T06:36:16.693" Title="Modeling medical phenomenon with mechanics fluid" Tags="&lt;mechanics&gt;" CommentCount="1" />
  <row Id="7276" PostTypeId="1" CreationDate="2011-03-20T07:24:52.033" Score="11" ViewCount="352" Body="&lt;p&gt;Bell's inequality theorem, along with experimental evidence, shows that we cannot have both realism and locality.  While I don't fully understand it, Leggett's inequality takes this a step further and shows that we can't even have non-local realism theories.  Apparently there are some hidden variable theories that get around this by having measurements be contextual.  I've heard there are even inequalities telling us how much quantum mechanics does or doesn't require contextuality, but I had trouble finding information on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all confusing to me, and it would be helpful if someone could explain precisely (mathematically?) what is meant by: realism, locality (I assume I understand this one), and contextuality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What combinations of realism, locality, and contextuality can we rule out using inequality theorems (assuming we have experimental data)?&lt;/p&gt;&#xA;" OwnerUserId="2655" LastActivityDate="2011-03-20T14:53:58.993" Title="What combinations of realism, non-locality, and contextuality are ruled out in quantum theory?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;quantum-interpretations&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="7277" PostTypeId="2" ParentId="7276" CreationDate="2011-03-20T08:16:21.667" Score="8" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Realism&lt;/strong&gt; refers to a philosophical position that says that certain attributes of the world of experience are independent of our observations. Let's take a physics example. In classical physics we used to say that a particle has a definite position and a definite momentum at a certain instant of time. These are represented by real numbers and they have those definite numbers independent of any observation. This seemed to be the only sane position one can take about the objective world. However the uncertainty principle of quantum mechanics tells us that a particle can not have both a well defined value of position and well defined value of momentum at the same time along the same direction independent of measurement. The more accurately one tries to measure one the less accurately one can have the knowledge of the other. Philosophically it means that position and its conjugate momentum can not have simultaneous reality. This realization had led the founding fathers of quantum theory to reformulate mechanics into a new theory called quantum mechanics. In QM a system is represented by a state vector in an abstract space. The length (norm) of this vector remain unchanged but with time its direction changes (for simplicity I am discussing Schrödinger picture). The various components of this state vector along the axes are various eigenstates with definite value of certain observables. Obviously the state vector is the linear combination of these eigenstates. Whenever a measurement is performed the state vector collapses to one of the eigenstates with certain probability determined by the Schrödinger's equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The so-called realists claim that the system was already in a definite state characterized by some additional hidden parameters before the measurement and since we are not aware of those hidden parameters we have an incomplete knowledge of the system. The random outcome reflects our incomplete knowledge of the system. There are number of hidden variable theories developed which reproduced the results of ordinary quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then surprisingly Bell discovered the famous Bell's inequality and showed that not all results are identical for both qm and local hidden variable theories. Experiment carried out and the verdict was clear. QM won. Nature supported QM. Therefore local hidden variable theories were ruled out. However there are nonlocal hidden variable theories which still survived like Bohmian mechanics. (I would also like to emphasize that MWI is an interpretation which is to some extent realist in spirit and it is by no means ruled out)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is &lt;strong&gt;locality&lt;/strong&gt;?&#xA;&lt;strong&gt;Locality&lt;/strong&gt; is the assumption that an object can be influenced only by its immediate surroundings by the events which took place in its immediate past. All classical and quantum field theories depends on this assumption in an essential way. Non locality implies that two events which are separated from each other by space-like separation can affect each other. Some people demand (imho) falsely that EPR type entanglement violates locality. In reality in never does. All one need to abandon is realism. Entanglement just shows that there exists quantum correlations between particles which were in past had some common origin. It also shows that &lt;em&gt;if it were a classical world then the EPR entanglements effects were nonlocal&lt;/em&gt;. But we live in a quantum world and there is no non locality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore in a nutshell, locality is certainly &lt;em&gt;not&lt;/em&gt; ruled out. Realism is ruled out to a large extent.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-03-20T09:00:54.733" LastActivityDate="2011-03-20T09:00:54.733" CommentCount="8" />
  <row Id="7278" PostTypeId="1" CreationDate="2011-03-20T08:23:06.250" Score="0" ViewCount="152" Body="&lt;p&gt;What sort of academic environment appeals to you the most? How would you describe your ideal academic environment?&lt;/p&gt;&#xA;" OwnerUserId="2668" LastActivityDate="2011-03-20T08:23:06.250" ClosedDate="2011-03-20T08:34:09.903" Title="How would you describe your ideal academic environment?" Tags="&lt;soft-question&gt;" />
  <row Id="7279" PostTypeId="1" AcceptedAnswerId="7295" CreationDate="2011-03-20T08:33:18.173" Score="4" ViewCount="128" Body="&lt;p&gt;The title question would be too long if I tried to specify it clearly.  So let me be more clear.  Consider the class of theories having the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The langrangian density is only dependent of scalars created from the curvature tensors (for example $R, R^{ab}R_{ab}, R^{abcd}R_{abcd}, C^{abcd}C_{abcd}, (R)^2, R^{ab}R_{bc}R^{cd}R_{da}$, etc.) + generic matter using the usual minimal coupling.&lt;/li&gt;&#xA;&lt;li&gt;The field theory is generated from the lagrangian by considering variations in the metric (ie. we will not consider Palatini or torsion theories)&lt;/li&gt;&#xA;&lt;li&gt;The field equation automatically provides a covariantly conserved stress energy tensor (this means like in GR, the covariant derivative of the Ricci terms are collectively zero just from geometric requirements and not an extra postulate. I've seen papers claim this follows automatically from 1, but I have not seen proof, so I'm specifying it just in case.)&lt;/li&gt;&#xA;&lt;li&gt;Let's stay with 3+1 dimensions.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My question is then: GR is the theory starting from using $R$ as the lagrangian density, and the vacuum equation in GR reduces to $R^{ab}=0$.  Is there another function of curvature scalars in this class which would yield the same vacuum equations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For clarity: Yes, for this question I don't care if the field equations differ in non-vacuum.  I'm just curious if they can possibly match in vacuum.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One may be tempted to immediately say it is unique, but consider the theories from $(R)^2$, $R^{ab}R_{ab}$, and $R^{abcd}R_{abcd}$.&lt;a href=&quot;http://arxiv.org/abs/astro-ph/0410031&quot; rel=&quot;nofollow&quot;&gt;(1)&lt;/a&gt;  One might intuitively expect that $C^{abcd}C_{abcd}$ would then add yet another theory, but it turns out this theory can be written as a combination of the previous three. &lt;a href=&quot;http://en.wikipedia.org/wiki/Curvature_invariant_%28general_relativity%29&quot; rel=&quot;nofollow&quot;&gt;(2)&lt;/a&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/gr-qc/pdf/9912/9912060v1.pdf&quot; rel=&quot;nofollow&quot;&gt;(3)&lt;/a&gt;  And it turns out that in 4-dimensions, there are actually only two linearly independent combinations of these four theories. &lt;a href=&quot;http://liu.diva-portal.org/smash/get/diva2:244008/FULLTEXT01&quot; rel=&quot;nofollow&quot;&gt;(4)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Playing with it to try to find if there is a theory equivalent to GR in vacuum, I was looking for a theory which in vacuum after I contract the indices, I can show the only possible result is R=0  (For an example if the contraction gave a polynomial like $R + R^3 + R^{abcd}R_{abcd}R = 0$, then the only solution with real curvature is $R=0$).  And then hoping that applying this to the field equations reduces everything to $R^{ab}=0$.  This hunt and peck method is failing me, as I haven't found anything yet and clearly won't let me prove GR is unique if that is the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone give an example which yields the GR vacuum equation or prove mathematically no such theory exists and thus the GR vacuum equation is unique in this class?&lt;/p&gt;&#xA;" OwnerUserId="2652" LastEditorUserId="2652" LastEditorDisplayName="" LastEditDate="2011-03-20T13:10:28.907" LastActivityDate="2011-03-20T14:35:14.963" Title="Is GR vacuum equation unique?" Tags="&lt;general-relativity&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="7280" PostTypeId="1" CreationDate="2011-03-20T08:54:28.143" Score="4" ViewCount="232" Body="&lt;p&gt;I still don't really understand what contextuality means in reference to quantum mechanics.  If someone could give a clear definition that would be great.  It sounds like it means you can't always relate the state of a system to its wavefunction or something?  That confuses me since it seems to violate quantum mechanics.  So a clear explanation would be great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is then, does any interpretation of quantum mechanics require us to accept contextuality?  This paper seems to claim yes, but I don't know enough to evaluate it, so would appreciate a professional's eye: &lt;a href=&quot;http://arxiv.com/abs/1006.0500&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.com/abs/1006.0500&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is a follow up on a previous question: &lt;a href=&quot;http://physics.stackexchange.com/questions/7276/what-combinations-of-realism-non-locality-and-contextuality-are-ruled-out-in-qu&quot;&gt;What combinations of realism, non-locality, and contextuality are ruled out in quantum theory?&lt;/a&gt; )&lt;/p&gt;&#xA;" OwnerUserId="2655" LastActivityDate="2011-03-27T04:46:57.920" Title="Is contextuality required in quantum mechanics?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;quantum-interpretations&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="7281" PostTypeId="2" ParentId="7276" CreationDate="2011-03-20T09:00:33.520" Score="5" ViewCount="" Body="&lt;p&gt;This is a truly excellent question in my opinion.  It is still being worked on.&#xA;Here are some professional references that will somewhat clarify the issue, or perhaps even confuse you further:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1102.4467&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1102.4467&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/abs/1007.5518&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1007.5518&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/1006.3680&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1006.3680&lt;/a&gt;&lt;br&gt;&#xA;Michael J.W. Hall&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/0808.2178&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0808.2178&lt;/a&gt;&lt;br&gt;&#xA;Travis Norsen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/quant-ph/0209123&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/quant-ph/0209123&lt;/a&gt;&lt;br&gt;&#xA; Laloe, Franck&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/0711.4650&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0711.4650&lt;/a&gt;    Adam Brandenburger, Noson Yanofsky&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/1102.0264&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1102.0264&lt;/a&gt;   Samson Abramsky, Adam Brandenburger&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-03-20T09:00:33.520" CommentCount="1" />
  <row Id="7282" PostTypeId="2" ParentId="7280" CreationDate="2011-03-20T09:47:02.213" Score="1" ViewCount="" Body="&lt;p&gt;I think this should clear it up that quantum mechanics in it's all models or interpretations is indeed contextual. &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0406166&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/quant-ph/0406166&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="304" LastActivityDate="2011-03-20T09:47:02.213" CommentCount="1" />
  <row Id="7283" PostTypeId="2" ParentId="7276" CreationDate="2011-03-20T09:55:22.190" Score="0" ViewCount="" Body="&lt;p&gt;You may find these detailed lectures helpful. Particularly lecture 3 will answer your question. However, if you don't have the background you might want to go through all the three. &lt;a href=&quot;http://qi10.ca/summerschool/speakers.html#FQM&quot; rel=&quot;nofollow&quot;&gt;http://qi10.ca/summerschool/speakers.html#FQM&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="304" LastActivityDate="2011-03-20T09:55:22.190" />
  <row Id="7284" PostTypeId="1" CreationDate="2011-03-20T11:04:03.257" Score="2" ViewCount="86" Body="&lt;p&gt;If water is introduced in a container maintained at 20C in vacuum conditions, a gaseous phase will appear and the pressure will stabilize at the vapour pressure for the given temperature inside the container. So far so good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine the experiment is repeated but instead of vacuum conditions, the water is presurized with nitrogen at 1 atm. According to the phase diagram of water, liquid is the stable form of water in these conditions. Yet it is commonly observed that the water molecules with the highest kinetic energy will escape and form a gaseous phase. The partial pressure of gaseous water will be equal to the saturation pressure at this temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do we say that this is a liquid/vapour equilibrium in this case, since the liquid phase is not at pressure required for this equilibrium to appear?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I just realized that my question seems to be answered by Raoult's law.&lt;/p&gt;&#xA;" OwnerUserId="1559" LastEditorUserId="1559" LastEditorDisplayName="" LastEditDate="2011-03-20T11:22:01.683" LastActivityDate="2011-03-20T11:22:01.683" Title="Thermodynamics of evaporation" Tags="&lt;thermodynamics&gt;" />
  <row Id="7285" PostTypeId="1" AcceptedAnswerId="7287" CreationDate="2011-03-20T11:07:47.950" Score="2" ViewCount="94" Body="&lt;p&gt;Classically, we obtain the equations of motion by finding a path which has an action that is stationary with respect to small changes in the path.  That is the path for which:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta S =0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scaling the action by a constant should therefore do nothing.  However, some books seem to consider the overall sign of the action important (since if we changed the sign we could find a path with arbitrarily negative action).  Does a scaling factor or change in sign actually matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For quantum mechanics we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K(x,y;T) = \langle x;T|y;0 \rangle = \int_{x(0)=x}^{x(T)=y} e^{i S /\hbar } Dx $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it looks like scaling the action $S$ will cause changes in how paths will interfere, but the overall sign of the action looks like it still won't matter.  So now we can somehow measure the absolute scale of the action?&lt;/p&gt;&#xA;" OwnerUserId="2655" LastActivityDate="2011-03-20T11:51:32.293" Title="Are there measurable effects to scaling the action by a constant?" Tags="&lt;quantum-mechanics&gt;&lt;classical-mechanics&gt;&lt;lagrangian-mechanics&gt;" AnswerCount="3" />
  <row Id="7286" PostTypeId="2" ParentId="7285" CreationDate="2011-03-20T11:26:27.653" Score="2" ViewCount="" Body="&lt;p&gt;For classical mechanics, the magnitude of a scaling factor does not matter, but the sign may depend on how you formulate your action principle.  The &lt;a href=&quot;http://www.scholarpedia.org/article/Principle_of_least_action&quot; rel=&quot;nofollow&quot;&gt;least action principle&lt;/a&gt; can be taken a bit too literal by some, but as you note it is whether the path is stationary that is the more rigorous definition.  For short enough time differences, in classical mechanics for particles the action is &lt;em&gt;always&lt;/em&gt; a minimum (well, or a maximum depending on sign choice).  See &lt;a href=&quot;http://www.eftaylor.com/pub/Gray&amp;amp;TaylorAJP.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;When action is not least&quot;&lt;/a&gt; by Taylor and Gray. However even for short times, the &quot;path&quot; of fields in classical mechanics are saddle points in the action.  So at some point one needs to give up taking the &quot;least&quot; in least action literally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Quantum mechanics, the scaling will cause an effect.  This in one of the root reasons why classical actions that lead to the same equations of motion, can actually lead to different quantum theories.  So this effect is indeed measurable in principle.  The sign however is still not measurable, as the sign choice is just a convention.&lt;/p&gt;&#xA;" OwnerUserId="2652" LastEditorUserId="2652" LastEditorDisplayName="" LastEditDate="2011-03-20T11:34:24.403" LastActivityDate="2011-03-20T11:34:24.403" CommentCount="2" />
  <row Id="7287" PostTypeId="2" ParentId="7285" CreationDate="2011-03-20T11:29:30.500" Score="6" ViewCount="" Body="&lt;p&gt;Well, you have basically answered you question yourself. Rescaling of the action is the same thing as rescaling of the Planck's constant. Obviously this can't have any effect classically. But on the quantum level $\hbar$ measures the non-commutativity of observables and in the extreme limit $\hbar \to 0$ you recover classical mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the sign, this doesn't matter either classically or quantum mechanically. We are not interested in only minimizing the action but in all extrema. Reverting the sign only means that we swap the meaning of maximum and minimum but the solutions won't change at all.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-03-20T11:29:30.500" CommentCount="2" />
  <row Id="7289" PostTypeId="2" ParentId="7285" CreationDate="2011-03-20T11:51:32.293" Score="0" ViewCount="" Body="&lt;p&gt;Your argument is absolute correct; a change of sign/scale factor won't change the solutions to the 'classical' equation $\delta S = 0.$ However, observables do change. If you've seen some field theory, you know an observable F is calculated like&#xA;$$\langle F \rangle = \frac{\int\mathrm{d}\phi F[\phi] e^{iS[\phi]/\hbar}}{\int\mathrm{d}\phi e^{iS[\phi]/\hbar}};$$ changing the scale of S &lt;em&gt;does&lt;/em&gt; change what you measure. It is indeed equivalent to a change of scale in $\hbar,$ so practically any observable will somehow change its value!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, when authors say that the sign does matter, it helps to think about the link between quantum and classical field theories. For example, if you 'Wick rotate' the free QFT, you get $$S[\phi] = \int\mathrm{d}^Dx \frac{1}{2}(\nabla\phi(x))^2 + \frac{1}{2}m^2\phi(x)^2, $$ which is by construction positive everywhere: $S[\phi] \geq 0.$ In that case, expressions like&#xA;$$\langle F \rangle = \frac{\int\mathrm{d}\phi F[\phi] e^{-\beta S[\phi]}}{\int\mathrm{d}\phi e^{-\beta S[\phi]}}$$ make sense, since the integrals converge. If you change the sign of $S$, the integrals suddenly diverge horrendously - you suddenly don't have a nice field theory anymore. This is a very general property: large field values and large deviations should be 'punished' by being exponentially supressed; if you change the sign, large values and deviations are being exponentially favoured.&lt;/p&gt;&#xA;" OwnerUserId="1471" LastActivityDate="2011-03-20T11:51:32.293" CommentCount="3" />
  <row Id="7290" PostTypeId="1" AcceptedAnswerId="7304" CreationDate="2011-03-20T13:26:49.297" Score="9" ViewCount="258" Body="&lt;p&gt;No hair theorem says that a black hole can be characterized by a small number of parameters that are visible from distance - mass, angular momentum and electric charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me it is puzzling why local quantities are not included, i.e. quantum numbers different from electrical charge. Lack of such parameters means breaking of the conservations laws (for a black hole made of baryons, Hawking radiation then is 50% baryonic and 50% anti-baryonic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If lack of baryonic number as a black hole parameter is a well established relation?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;OR&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is (or may be) only an artifact of lack of unification between QFT and GR?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="184" LastActivityDate="2011-03-20T21:23:22.473" Title="No hair theorem for black holes and the baryon number" Tags="&lt;quantum-field-theory&gt;&lt;general-relativity&gt;&lt;black-holes&gt;&lt;antimatter&gt;&lt;conservation-laws&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="7291" PostTypeId="1" CreationDate="2011-03-20T13:51:04.423" Score="3" ViewCount="78" Body="&lt;p&gt;In analogy to the electromagnetic tensor, with the components defined as the electric field $E$ and magnetic field $B$ as such:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F^{ab} = \begin{bmatrix}&#xD;&#xA;0 &amp;amp; -E_x/c &amp;amp; -E_y/c &amp;amp; -E_z/c \\&#xD;&#xA;E_x/c &amp;amp; 0 &amp;amp; -B_z &amp;amp; B_y \\&#xD;&#xA;E_y/c &amp;amp; B_z &amp;amp; 0 &amp;amp; -B_x \\&#xD;&#xA;E_z/c &amp;amp; -B_y &amp;amp; B_x &amp;amp; 0&#xD;&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can one form a relativistic tensor from the electric displacement field $D$ and magnetizing field $H$ like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bar{F}^{ab} = \begin{bmatrix}&#xD;&#xA;0 &amp;amp; -D_x/c &amp;amp; -D_y/c &amp;amp; -D_z/c \\&#xD;&#xA;D_x/c &amp;amp; 0 &amp;amp; -H_z &amp;amp; H_y \\&#xD;&#xA;D_y/c &amp;amp; H_z &amp;amp; 0 &amp;amp; -H_x \\&#xD;&#xA;D_z/c &amp;amp; -H_y &amp;amp; H_x &amp;amp; 0&#xD;&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thus obtain an &quot;electromagnetic tensor&quot; that can be easily used to handle the effects of fields in materials?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $D = \epsilon_0 E + P$ and $H = \frac{1}{\mu_0}B - M$, this is equivalent to asking if $P$ and $M$ can be used to form a relativistic tensor.&lt;/p&gt;&#xA;" OwnerUserId="2652" LastActivityDate="2011-03-20T18:43:22.547" Title="Can D and H form an 'in materials' version of the electromagnetic tensor?" Tags="&lt;special-relativity&gt;&lt;electrodynamics&gt;" AnswerCount="2" />
  <row Id="7292" PostTypeId="1" AcceptedAnswerId="7312" CreationDate="2011-03-20T13:53:37.143" Score="6" ViewCount="439" Body="&lt;p&gt;I recently realised that quaternions could be used to write intervals or norms of vectors in special relativity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(t,ix,jy,kz)^2 = t^2 + (ix)^2 + (jy)^2 + (kz)^2 = t^2 - x^2 - y^2 - z^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it useful? Is it used? Does it bring anything? Or is it just funny?&lt;/p&gt;&#xA;" OwnerUserId="2499" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-03-20T20:47:45.137" LastActivityDate="2011-03-23T17:07:16.993" Title="Quaternions and 4-vectors" Tags="&lt;relativity&gt;&lt;vectors&gt;" AnswerCount="6" CommentCount="1" />
  <row Id="7293" PostTypeId="2" ParentId="7292" CreationDate="2011-03-20T14:03:42.287" Score="1" ViewCount="" Body="&lt;p&gt;First note that $q = t + xi + yj +zk$ and $q\prime = t - xi - yj - zk$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $qq\prime = t^2 + x^2 + y^2 + z^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, it should be $ t^2 + x^2 + y^2 +z^2$ and hence it is not the right signature of special relativity. Secondly since quaternion does not have an analog of a notion like holomorphic function (since its conjugate is not independent) it is not physically ideal or useful as far as I know. You can generalize it and have excellent usefulness but you have to give up its division algebra property. Clifford algebra is an example.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-03-23T17:07:16.993" LastActivityDate="2011-03-23T17:07:16.993" CommentCount="5" />
  <row Id="7294" PostTypeId="1" CreationDate="2011-03-20T14:11:50.857" Score="-1" ViewCount="301" Body="&lt;p&gt;Do you really need to be a math whiz to succeed in physics? Can someone with lousy math skills but good physics intuition succeed? After all, Michael Faraday sucked at math, but turned out to be one of the best physicists of his time.&lt;/p&gt;&#xA;" OwnerUserId="2674" LastActivityDate="2011-03-20T15:58:02.077" ClosedDate="2011-03-20T16:11:26.573" Title="Do you really need to be a math whiz to succeed in physics?" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="7295" PostTypeId="2" ParentId="7279" CreationDate="2011-03-20T14:17:39.213" Score="2" ViewCount="" Body="&lt;p&gt;You can add a Gauss-Bonnet term $R^2 - 4R^{\mu\nu}R_{\mu\nu} + R^{\mu\nu\rho\sigma}R_{\mu\nu\rho\sigma}$. It's purely topological.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-03-20T14:34:40.347" LastActivityDate="2011-03-20T14:34:40.347" CommentCount="1" />
  <row Id="7296" PostTypeId="2" ParentId="7290" CreationDate="2011-03-20T14:25:05.237" Score="0" ViewCount="" Body="&lt;p&gt;One way out of this--if you believe that Hawking radiation is real, and that black holes completely evaporate, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) the Hawking radiation violates the positive energy conditions.  These are assumptions in the no-hair theorem.  So, we don't expect the no-hair theorem to be valid for black holes if Hawking radiation is important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If the evaporation profile evolves in a certain way, you can prevent the formation of a true event horizon, only having an &lt;a href=&quot;http://blog.relativity.livingreviews.org/tag/apparent-horizons/&quot; rel=&quot;nofollow&quot;&gt;apparent horizon&lt;/a&gt; appear for a finite amount of time.  This will enable the information about the input baryon number to escape the black hole.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But classically, you are right.  It is well-established that you lose the information about what you put into the black hole--classically, a black hole formed from neutrino collapse is identical to a black hole formed from the collapse of a neutron star or photons or gravitational waves or whatever.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-03-20T14:25:05.237" CommentCount="21" />
  <row Id="7297" PostTypeId="2" ParentId="7294" CreationDate="2011-03-20T14:29:36.747" Score="9" ViewCount="" Body="&lt;p&gt;Note when Faraday lived : Michael Faraday, FRS (22 September 1791 – 25 August 1867) . .He was called a &quot;natural philosopher&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was still a time when higher education meant learning a lot of Latin and Greek and maybe geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the year 2011 higher education has become much more demanding and being a physicist, let alone a successful one, requires passing a lot of courses in university with fairly good grades, and mathematics is a prerequisite. In addition, the transmission of knowledge in physics depends crucially on mathematical formulations. One has to develop a mathematical intuition for physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if somebody is really interested in physics he/she should devote some extra effort to get good/passable grades in mathematics too. Tutoring by a good tutor.  It is all a matter of persistence and perseverance and elbow grease, if maths do not come easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is perseverance and persistence  that are lacking, then it is better not to aim at becoming a physicist anyway, because a lot of that is needed in any physics field chosen.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-20T14:29:36.747" />
  <row Id="7299" PostTypeId="2" ParentId="7276" CreationDate="2011-03-20T14:53:58.993" Score="3" ViewCount="" Body="&lt;p&gt;What I take to be elementary significant papers on this question pre-date arXiv, so they are unfortunately usually available only behind paywalls. I've always found the simplicity of Willem de Muynck's argument in Physics Letters A 114, 65 (1986), &quot;THE BELL INEQUALITIES AND THEIR IRRELEVANCE TO THE PROBLEM OF LOCALITY IN QUANTUM MECHANICS&quot;, somewhat compelling. I can reproduce the basic argument here under fair use, from the first page,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In his original derivation Bell &lt;a href=&quot;http://www.mth.kcl.ac.uk/~llandau/Homepage/Math/bell.html&quot; rel=&quot;nofollow&quot;&gt;3&lt;/a&gt;&#xA;  assumed his hidden variables theory to&#xA;  satisfy a locality condition which he&#xA;  deemed to be a&quot;vital assumption&quot;.&#xA;  Presumably due to this fact there&#xA;  still exists a widespread belief- also&#xA;  among specialists - that the Bell&#xA;  inequalities can not be derived for&#xA;  nonlocal hidden variables theories.&#xA;  This would leave open the possibility&#xA;  that quantum mechanics might be&#xA;  reproduced by a nonlocal hidden&#xA;  variables theory. From the following,&#xA;  however, it should be clear that the&#xA;  mere existence of hidden variables is&#xA;  sufficient to yield the Bell&#xA;  inequalities. Hence not only local but&#xA;  also nonlocal hidden variables&#xA;  theories are incompatible with quantum&#xA;  mechanics. &lt;em&gt;Local and nonlocal theories&#xA;  being on an equal footing it also&#xA;  follows that the Bell inequalities are&#xA;  completely irrelevant to the problem&#xA;  of (non)-locality in hidden variables&#xA;  theories.&lt;/em&gt; (my emphasis here)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After I extracted the above, I found &lt;a href=&quot;http://www.phys.tue.nl/ktn/Wim/dMIrrBI86.pdf&quot; rel=&quot;nofollow&quot;&gt;a PDF of the paper&lt;/a&gt; on &lt;a href=&quot;http://www.phys.tue.nl/ktn/Wim/muynck.htm&quot; rel=&quot;nofollow&quot;&gt;de Muynck's web-page&lt;/a&gt;, I'm pleased to say (it's elementary math, and only 4 pages). A similar but rather more algebraic construction, which I think is mathematically quite a bit nicer, can be found in Lawrence J. Landau, Physics Letters A 120, 54, 1987, &quot;ON THE VIOLATION OF BELL'S INEQUALITY IN QUANTUM THEORY&quot;, without, however, making anything like de Muynck's claim for its significance (I don't believe it, but &lt;a href=&quot;http://www.mth.kcl.ac.uk/~llandau/Homepage/Math/bell.html&quot; rel=&quot;nofollow&quot;&gt;I found it here&lt;/a&gt;). IMO, this simple algebra underlies the question of locality/nonlocality to this day -- one takes this argument seriously, or one does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, locality is very closely tied to measurement compatibility because measurement compatibility is required for measurements that are at space-like separation in quantum field theory. The implication does not apply in reverse, however, so space-like separation of measurements is &lt;em&gt;not&lt;/em&gt; equivalent to measurement compatibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spanner in the works, a big one, is that measurement compatibility (and hence by implication space-like separation) does &lt;em&gt;not&lt;/em&gt; imply no-correlation. There &lt;em&gt;are&lt;/em&gt; correlations at space-like separation in quantum (field) theory, &lt;em&gt;but&lt;/em&gt; one can prove (I realize here that I don't know precisely what additional assumptions are needed, but conventional QM frameworks are enough) that one can't use those correlations to send messages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to point out that you shouldn't take too much for granted your parenthetical comment &quot;(I assume I understand this one)&quot;. If you look at the other Answers here, you'll see that locality is far from simple. I particularly draw your attention to sb1's introduction of &quot;influence&quot; as part of his last paragraph's discussion, which I suggest is not simple at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important, IMO, to understand that this argument has been gradually changing over the last 50 years. It's not clear when or whether a novel argument will appear that makes it worthwhile to think in terms outside quantum (field) theory for practical purposes, but novel arguments are constantly emerging. The fact that Michael J. W. Hall (cited by Jim Graber above) has managed to publish his novel argument in Physics Review Letters is enormously impressive, particularly when one sees the robust tone he adopts, because PRL sets the bar very high indeed for foundations papers, but time will tell whether the argument can be used constructively in a quantum field theory context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, reading through my Answer, I realize that it does not &lt;em&gt;directly&lt;/em&gt; address &quot;realism&quot; and &quot;contextuality&quot;. That's because I associate mutual measurement compatibility of all observables directly with classical realism, all measurements commute, and the sometime presence of measurement incompatibility with contextuality. Measurements may have an &quot;influence&quot; (hee!) on some other measurements, and not on others. I might expand on this already overlong Answer later.&#xA; Best wishes,&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-03-20T14:53:58.993" CommentCount="0" />
  <row Id="7300" PostTypeId="2" ParentId="7292" CreationDate="2011-03-20T14:55:58.963" Score="3" ViewCount="" Body="&lt;p&gt;Cornelius Lanczos has a chapter on quaternions and special relativity in his &quot;The Variational Principles of Mechanics&quot;.  So, is has been used. But it seems more straightforward to consider the multivector algebra of spacetime so t,x,y,z really are on the same footing.&lt;/p&gt;&#xA;" OwnerUserId="1423" LastActivityDate="2011-03-20T14:55:58.963" />
  <row Id="7301" PostTypeId="2" ParentId="7290" CreationDate="2011-03-20T15:42:18.217" Score="2" ViewCount="" Body="&lt;p&gt;A standard answer is that the nuclear force obeys a Yukawa potential $V(r)~=~-e^{-ar}/r$ which drops off with a very short range.  A charge dropped on a black hole will have a long range electric field which connects the horizon with conformal infinity.  The electric charge becomes distributed on the horizon and remains apparent.  The Yukawa type of force that drops off rapidly does not appear in this manner.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is the holographic content of quantum fields or strings which fall onto a black hole.  The baryon number, and in fact all quantum numbers, are preserved on the stretched horizon.  They are annihilated in the far future as they are replaced by Hawking radiation modes.  However, this Hawking radiation is just some “recoding” of the quantum information.  Without the master decyption system what emerges appears to be noise.  Yet if quantum information is preserved then baryon number, or more fundamentally quark family types and color charges, is simply transformed into some other form of quantum information.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-20T15:42:18.217" CommentCount="2" />
  <row Id="7302" PostTypeId="2" ParentId="7294" CreationDate="2011-03-20T15:58:02.077" Score="1" ViewCount="" Body="&lt;p&gt;First, Farady did NOT suck at math. He never went to school but he could have done well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can NOT become physicist without learning math. Why? &#xA;Aristotle, one of the smartest men known to mankind, thought that heavier bodies fall faster than lighter ones. Do you see what happens when intuition is not guided by math and experiment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Math is the language of modern physics. Without it you will understand NOTHING. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though math required is fairly advanced, you don't have to be a mathematician to become a physicist. &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2011-03-20T15:58:02.077" CommentCount="2" />
  <row Id="7303" PostTypeId="1" CreationDate="2011-03-20T17:21:31.120" Score="1" ViewCount="353" Body="&lt;p&gt;We can see the moon in the night because it reflects sunlight. But the light is incident on the opposite side of moon with respect to the observer in the night. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, how does the moon reflect light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure if I have articulated my question correctly.&lt;/p&gt;&#xA;" OwnerUserId="1496" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-03-20T20:59:52.970" LastActivityDate="2011-03-21T10:46:52.567" Title="How does the moon reflect light?" Tags="&lt;astronomy&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="7304" PostTypeId="2" ParentId="7290" CreationDate="2011-03-20T17:21:41.120" Score="6" ViewCount="" Body="&lt;p&gt;The no hair theorem is proven in classical gravity, in asymptotically flat 4 dimensional spacetimes, and with particular matter content. When looking at more general circumstances, we are starting to see that variations of the original assumptions give the black hole more hair. For example, for asymptotically AdS one can have scalar hair (a fact which is used to build holographic superconductors). For five dimensional spaces black holes (and black rings) can have dipole moments of gauge charges. Maybe there are more surprises.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, the basic intuition behind the no hair theorem is still valid. The basic fact used in all these constructions is that when the object falls into a black hole, it can imprint its existence on the black hole exterior  only if it associated with a long ranged field. So for example an electron will change the charge of the black hole which means that black hole will have a Coulomb field. You'd be able to measure the total charge by an appropriate Gauss surface. Note that gravity has no conserved local currents (see &lt;a href=&quot;http://physics.stackexchange.com/questions/7244/what-is-energy-in-string-theory/7263#7263&quot;&gt;this discussion&lt;/a&gt;), the only thing you'd be able to measure is the &lt;em&gt;total&lt;/em&gt; charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for baryon number, it is not associated with long ranged force, when it falls into  black hole there is nothing to remember that fact, and the baryon number is not conserved. This is just one of the reasons there is a general belief that global charges (those quantities which are not accompanied by long ranged forces) are not really conserved. For the Baryon number we know that for a fact: our world has more baryons than anti baryons, so the observed baryon number symmetry must only be approximate. It must have been violated in the early universe when all baryons were generated (look for a related discussion &lt;a href=&quot;http://physics.stackexchange.com/questions/6321/matter-antimatter-asymmetry-in-experiments/6323#6323&quot;&gt;here&lt;/a&gt;), a process which is referred to as baryogenesis.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-03-20T21:23:22.473" LastActivityDate="2011-03-20T21:23:22.473" CommentCount="2" />
  <row Id="7305" PostTypeId="2" ParentId="7303" CreationDate="2011-03-20T17:36:58.887" Score="1" ViewCount="" Body="&lt;p&gt;I think Marek's comment is the correct answer. I will rephrase it here slightly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think one of your assumptions is faulty. Look at &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aMoon_phases_en.jpg&quot; rel=&quot;nofollow&quot;&gt;this picture&lt;/a&gt;. Do you still have a question, after seeing this picture?&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2011-03-20T17:36:58.887" CommentCount="5" />
  <row Id="7306" PostTypeId="2" ParentId="7291" CreationDate="2011-03-20T18:14:52.613" Score="3" ViewCount="" Body="&lt;p&gt;Yes, you can form such a tensor with $D$ and $H$ but it is not as useful and natural as the tensor with $E$ and $B$ because $E$ and $B$ are the actual physical fields that naturally transform into each other by Lorentz transformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, you may declare that $D$ and $H$ transform into each other by the Lorentz transformations but it is not too useful because these fields are only useful to describe fields inside a material, and the presence of this material spontaneously breaks the Lorentz symmetry, anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is related to some recent questions on this server about electrodynamics of moving media. $D$ and $H$ are normally used, together with the well-known equations they satisfy, only in the rest frame of a material. If the material is moving, the whole analysis has to be redone. To be sure that we do it right, we should return to the actual $B$ and $E$ field and if we define new ones to simplify our life, we shouldn't guess the equations for these new fields: we should derive them from the equations for the vacuum fields $E,B$ and the actual charge densities and currents, including the bound ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting Maxwell's equations for a medium, which include $B,E,D,H$, are not Lorentz-covariant, anyway - because the material picks a preferred frame. To introduce new fields $D,H$ is just a bookeeping device to simplify some calculations - and it only works nicely in the medium's rest frame.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-20T18:43:22.547" LastActivityDate="2011-03-20T18:43:22.547" />
  <row Id="7307" PostTypeId="2" ParentId="7292" CreationDate="2011-03-20T18:18:57.180" Score="10" ViewCount="" Body="&lt;p&gt;It's just funny. Note that your equation doesn't actually use any single general quaternion. You only use the $i,j,k$ imaginary units in an &lt;em&gt;ad hoc&lt;/em&gt; way to get three minus signs whenever you need them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were using an actual quaternion&#xA;$$ q = t + xi + yj + zk,$$&#xA;then the only semi-natural real bilinear invariant you may construct out of it is&#xA;$$ q\bar q =  (t + xi + yj + zk ) ( t - xi - yj - zk) =  t^2+ x^2 +y^2 +z^2 $$&#xA;so the 4 real components in a quaternion still have the Euclidean, rather than Minkowskian, signature. But even for a 4-dimensional Euclidean space, the quaternions are actually just a game because we haven't really used the main nontrivial structure of the quaternions, their multiplication, in any nontrivial way. Quaternions are not genuine quaternions if you never use the relations $ij=-ji = k$ and its cyclic permutations - and we haven't used them above. We only used the fact that $i,j$ etc. anticommute with each other, but we didn't really care what their product is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because we haven't really used those relations, we haven't used full quaternions - except as a meaningless bookkeeping device. In the same way, one may organize 8 real numbers under the umbrella of a single &quot;octonion&quot; except that if the complicated and cool octonion multiplication table - with the $G_2$ automorphism group - is never employed, it's clear that the &quot;octonion&quot; interpretation was just a game to give a name to a collection of 8 numbers. But not every collection of 4 or 8 numbers deserves to be called &quot;quaternion&quot; and &quot;octonion&quot;, even though, of course, one may get the individual components out of the &quot;quaternion&quot; and &quot;octonion&quot;, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the very same way, a general pair of two real numbers is simply not a complex number. By its very essence, a complex number must act as one number - so there must be a notion of holomorphy required somewhere or everywhere in the formalism - rather than two numbers. The references linked in the other answers don't understand the purpose and relevance of all those mathematical structures, so they lead to incorrect answers to the fundamental question whether the trick is real or just a fun. The right answer is that it is just a fun, and your fun even used a wrong signature that differs from a somewhat more natural fun.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-20T18:26:52.600" LastActivityDate="2011-03-20T18:26:52.600" CommentCount="6" />
  <row Id="7308" PostTypeId="2" ParentId="7291" CreationDate="2011-03-20T18:22:39.377" Score="2" ViewCount="" Body="&lt;p&gt;I'm not 100% sure, but I think the answer is yes. Here's my reasoning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physical effects of $\mathbf{P,M}$ result from the bound charges and currents they produce: $\rho_b=-\nabla\cdot{\bf P}$ and ${\bf J}_b=\nabla\times{\bf M}$. I'd expect $j\equiv (\rho_b,{\bf J}_b)$ to transform as a 4-vector (after all, they are &quot;real&quot; charges and currents). If you define an antisymmetric tensor $P^{\mu\nu}$ out of $\mathbf{P,M}$ in the natural way, then the definitions of $\rho_b,{\bf J}_b$ reduce to $j^\mu={P^{\mu\nu}}_{,\nu}$. If $P^{\mu\nu}$ is covariant, then $j^\mu$ is automatically covariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's just a plausibility argument, not a proof, but I find it pretty convincing. Let me see if I can tighten up the argument a bit. In the local rest frame of the matter, &lt;em&gt;define&lt;/em&gt; $P^{\mu\nu}$ to be the tensor built out of $\mathbf{P,M}$ in the natural way. In any other frame, define $P^{\mu\nu}$ via Lorentz transformations. Finally, define $j^\mu={P^{\mu\nu}}_{,\nu}$. Then the four-vector $j$ gives the correct bound charges and currents in the rest frame, and hence (since everything's covariant) it gives the correct bound charges and currents in all frames. Everything that matters about $\mathbf{P,M}$ is contained in the bound charges and currents, so we're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct answer is presumably contained in Landau &amp;amp; Lifschitz, &lt;em&gt;Electrodynamics of Continuous Media&lt;/em&gt;, which I've never read. If what's in there contradicts what I've said, then I'm sure that they're right and I'm wrong.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-20T18:22:39.377" />
  <row Id="7309" PostTypeId="2" ParentId="7231" CreationDate="2011-03-20T18:46:23.753" Score="16" ViewCount="" Body="&lt;p&gt;The question asks about the time dependence of the function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(t) := \langle\psi(t)|(\Delta \hat{x})^2|\psi(t)\rangle&#xD;&#xA; \langle\psi(t)|(\Delta \hat{p})^2|\psi(t)\rangle,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta \hat{x} := \hat{x} - \langle\psi(t)|\hat{x}|\psi(t)\rangle, \qquad&#xD;&#xA;\Delta \hat{p} := \hat{p} - \langle\psi(t)|\hat{p}|\psi(t)\rangle, \qquad&#xD;&#xA;\langle\psi(t)|\psi(t)\rangle=1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will here use the Schroedinger picture where operators are constant in time, while the kets and bras are evolving. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Spurred by remarks of Moshe R. and Ted Bunn let us add that (under assumption (1) below) the Schroedinger equation itself is invariant under the time reversal operator $\hat{T}$, which is a conjugated linear operator, so that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{T} t = - t \hat{T}, \qquad \hat{T}\hat{x} = \hat{x}\hat{T}, \qquad \hat{T}\hat{p} = -\hat{p}\hat{T}, \qquad  \hat{T}^2=1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we are restricting ourselves to Hamiltonians $\hat{H}$ so that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\hat{T},\hat{H}]=0.\qquad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, if &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\psi(t)\rangle = \sum_n\psi_n(t) |n\rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a solution to the Schroedinger equation in a certain basis $|n\rangle$, then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{T}|\psi(t)\rangle := \sum_n\psi^{*}_n(-t) |n\rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;will also be a solution to the Schroedinger equation with a time reflected function $f(-t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if $f(t)$ is non-constant in time, then we may assume (possibly after a time reversal operation) that there exist two times $t_1&amp;lt;t_2$ with $f(t_1)&amp;gt;f(t_2)$. This would contradict the statement in the original question. To finish the argument, we provide below an example of a non-constant function $f(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a simple harmonic oscillator Hamiltonian with the zero point energy $\frac{1}{2}\hbar\omega$ subtracted for later convenience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{H}:=\frac{\hat{p}^2}{2m}+\frac{1}{2}m\omega^{2}\hat{x}^2&#xD;&#xA;-\frac{1}{2}\hbar\omega=\hbar\omega\hat{N},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\hat{N}:=\hat{a}^{\dagger}\hat{a}$ is the number operator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us put the constants $m=\hbar=\omega=1$ to one for simplicity. Then the annihilation and creation operators are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{a}=\frac{1}{\sqrt{2}}(\hat{x} + i \hat{p}), \qquad&#xD;&#xA;\hat{a}^{\dagger}=\frac{1}{\sqrt{2}}(\hat{x} - i \hat{p}), \qquad&#xD;&#xA;[\hat{a},\hat{a}^{\dagger}]=1,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or conversely,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{x}=\frac{1}{\sqrt{2}}(\hat{a}^{\dagger}+\hat{a}), \qquad&#xD;&#xA;\hat{p}=\frac{i}{\sqrt{2}}(\hat{a}^{\dagger}-\hat{a}), \qquad &#xD;&#xA;[\hat{x},\hat{p}]=i,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{x}^2=\hat{N}+\frac{1}{2}\left(1+\hat{a}^2+(\hat{a}^{\dagger})^2\right), \qquad &#xD;&#xA;\hat{p}^2=\hat{N}+\frac{1}{2}\left(1-\hat{a}^2-(\hat{a}^{\dagger})^2\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider Fock space $|n\rangle := \frac{1}{\sqrt{n!}}(\hat{a}^{\dagger})^n |0\rangle$ &#xA;such that $\hat{a}|0\rangle = 0$. Consider initial state &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\psi(0)\rangle := \frac{1}{\sqrt{2}}\left(|0\rangle+|2\rangle\right), \qquad &#xD;&#xA;\langle \psi(0)|  = \frac{1}{\sqrt{2}}\left(\langle 0|+\langle 2|\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\psi(t)\rangle = e^{-i\hat{H}t}|\psi(0)\rangle&#xD;&#xA; = \frac{1}{\sqrt{2}}\left(|0\rangle+e^{-2it}|2\rangle\right),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle \psi(t)| = \langle\psi(0)|e^{i\hat{H}t}&#xD;&#xA; = \frac{1}{\sqrt{2}}\left(\langle 0|+\langle 2|e^{2it}\right),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle\psi(t)|\hat{x}|\psi(t)\rangle=0, \qquad &#xD;&#xA;\langle\psi(t)|\hat{p}|\psi(t)\rangle=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle\psi(t)|\hat{x}^2|\psi(t)\rangle=\frac{3}{2}+\frac{1}{\sqrt{2}}\cos(2t), \qquad &#xD;&#xA;\langle\psi(t)|\hat{p}^2|\psi(t)\rangle=\frac{3}{2}-\frac{1}{\sqrt{2}}\cos(2t),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because $\hat{a}^2|2\rangle=\sqrt{2}|0\rangle$. Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(t) = \frac{9}{4} - \frac{1}{2}\cos^2(2t),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is non-constant in time, and we are done. Or alternatively, we can complete the counter-example without the use of above time reversal argument by simply performing an appropriate time translation $t\to t-t_0$.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-03-21T18:19:43.740" LastActivityDate="2011-03-21T18:19:43.740" CommentCount="6" />
  <row Id="7310" PostTypeId="2" ParentId="7172" CreationDate="2011-03-20T18:47:19.797" Score="-1" ViewCount="" Body="&lt;p&gt;the concept of half life is that it follows an exponential curve... so you can not predict there exact depletion stage. As for the case of Naturally occurring Isotopes like Uranium etc. they have half life's in billion of years... we all are exposed to radioactive materials.. RADON is a gas which is present in our environment which is constantly generated from the earth itself...then we have Artificially occurring Isotopes like Iridium 192, Cobalt 60, Cs. etc which are produced by activation of stable elements like Ir 191, Co 59 resp....there half lives varies from few days to some hundred of years only. They are used for industrial applications.&lt;/p&gt;&#xA;" OwnerUserId="2681" LastActivityDate="2011-03-20T18:47:19.797" CommentCount="1" />
  <row Id="7311" PostTypeId="1" AcceptedAnswerId="7313" CreationDate="2011-03-20T18:55:25.980" Score="4" ViewCount="175" Body="&lt;p&gt;What does the quantum canonical momentum operator corresponding to arbitrary canonical position. For example, in Cartesian coordinates ($x^i$), the canonical momentum operator with respect to each $x^i$ is $-i\hbar \frac{\partial}{\partial x^i}$. For arbitrary canonical coordinates $q^i$, would the corresponding canonical momentum operators just be $-i\hat{\mathbf q}^i\cdot\hbar \nabla$?&lt;/p&gt;&#xA;" OwnerUserId="2618" LastActivityDate="2011-03-20T19:56:08.540" Title="Canonical momentum operators in curvilinear coordinates" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="7312" PostTypeId="2" ParentId="7292" CreationDate="2011-03-20T19:47:01.847" Score="8" ViewCount="" Body="&lt;p&gt;The object you're talking about is called, in mathematics, a Clifford algebra. The case when the algebra is over the complex field in general has a significantly different structure from the case when the algebra is over the real field, which is important in Physics. &lt;em&gt;In&lt;/em&gt; Physics, in the specific case of 4 dimensions, using the Minkowski metric as you have in your Question, and over the complex field, the algebra is called the Dirac algebra. Once you have the name &lt;em&gt;Clifford algebra&lt;/em&gt;, you can look them up in Google, where the first entry is, unsurprisingly, Wikipedia, &lt;a href=&quot;http://en.wikipedia.org/wiki/Clifford_algebra&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Clifford_algebra&lt;/a&gt;, which gives you a reasonable flavor of the abstract construction methods that mathematicians prefer. The John Baez page that is linked to from the Wikipedia page is well worth reading (if you spent a year learning everything that John Baez has posted over the years, almost always with unusual clarity and engagingly, you would know most of the mathematics that might be useful for Physics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not so much that the Clifford algebras are funny. Their quadratic construction is interrelated, often closely, with many other constructions in mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are people who are enthusiastic about Clifford algebras, sometimes very or too much so, and a &lt;em&gt;lot&lt;/em&gt; of ink has been spilled (Joel Rice's and Luboš Motl's Answers are rather inadequate to the literature, except that I think they chose to interpret your Question narrowly where I've addressed what your construction has led to in Mathematics more widely), but there are many other fish in the sea to admire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Particularly in light of Marek's comments below, it should be said that I interpreted Isaac's Question generously. There is a somewhat glaring mistake in the OP that is pointed out by Luboš (which I hope you see, Isaac). Nonetheless there is a type of construction that is closely related to what I chose to take to be the idea of the OP, Clifford algebras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isaac, this is how I think your derivation ought to go, if we just use quaternions, taking $q=t+ix+jy+kz$, &#xA;$$q^2=(t+ix+jy+kz)(t+ix+jy+kz)=t^2-x^2-y^2-z^2+2t(ix+jy+kz).$$&#xA;The $xy,yz,zx$ terms cancel nicely, but the $tx,ty,tz$ terms don't, unless we do as Luboš did and introduce the conjugate $\overline{q}=t-ix-jy-kz$. This, however, doesn't do what I take you to be trying to do. So, instead, we introduce a fourth object, $\gamma^0$, for which $(\gamma^0)^2=+1$, and which anti-commutes with $i$,$j$, and $k$. Then the square of $\gamma^0t+ix+jy+kz$ &lt;em&gt;is&lt;/em&gt; $t^2-x^2-y^2-z^2$. The algebra this generates, however, is more than just the quaternions, it's the Clifford algebra $C(1,3)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT(2): Hi, Isaac. I've thought about this way too much overnight. I think now that I was mistaken, you didn't make a mistake. I think you intended your expression $(a,b,c,d)^2$ to &lt;em&gt;mean&lt;/em&gt; the positive-definite inner product $a^2+b^2+c^2+d^2$. With &lt;em&gt;this&lt;/em&gt; reading, however, we see three distinct structures, the positive-definite inner product, the quaternions, and the Minkowski space inner product that emerges from using the first two together. Part of what made me want to introduce a &lt;em&gt;different&lt;/em&gt; construction is that in yours the use of the quaternions is redundant, because you'd get the same result that you found remarkable if you just used $(a,ib,ic,id)^2$ (as Luboš also mentioned). Even the positive-definite inner product is redundant, insofar as what we're really interested in is just the Minkowski space inner product. Also, of course, I know something that looks similar and that has been mathematically productive for over a century, and that can be constructed using just the idea of a non-commutative algebra and the Minkowski space inner product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To continue the above, we can write $\gamma^1=i$, $\gamma^2=j$, $\gamma^3=k$ for the quaternionic basis elements, together with the basis element $\gamma^0$, then we can define the algebra by the products of basis elements of the algebra, $\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=2g^{\mu\nu}$. Alternatively, for any vector $u=(t,x,y,z)$ we can write $\gamma(u)=\gamma^0u_0+\gamma^1u_1+\gamma^2u_2+\gamma^3u_3$, then we can define the algebra by the product for arbitrary 4-vectors, $\gamma(u)\gamma(v)+\gamma(v)\gamma(u)=2(u,v)$, where $(u,v)$ is the Minkowski space inner product. Hence, we have $[\gamma(u)]^2=(u,u)$. Now everything is getting, to my eye, and hopefully to yours, rather neat and tidy, and nicely in line with the conventional formalism.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-03-21T11:02:42.630" LastActivityDate="2011-03-21T11:02:42.630" CommentCount="11" />
  <row Id="7313" PostTypeId="2" ParentId="7311" CreationDate="2011-03-20T19:56:08.540" Score="2" ViewCount="" Body="&lt;p&gt;If you use &lt;a href=&quot;http://en.wikipedia.org/wiki/Curvilinear_coordinates&quot; rel=&quot;nofollow&quot;&gt;curvilinear coordinates&lt;/a&gt; you basically need to transform your derivatives likewise, ie, you need to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant&quot; rel=&quot;nofollow&quot;&gt;Jacobian&lt;/a&gt; of the coordinate transformation you used to go from Cartesian coords to the curvilinear system you have at hands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, you can think in terms of Quantum Mechanics over a curved manifold, in which case you would simply use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Covariant_derivative&quot; rel=&quot;nofollow&quot;&gt;covariant derivative&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2011-03-20T19:56:08.540" CommentCount="5" />
  <row Id="7314" PostTypeId="2" ParentId="7292" CreationDate="2011-03-20T20:07:50.673" Score="2" ViewCount="" Body="&lt;p&gt;There is a book:&#xA;&quot;Quaternions, Clifford Algebras and Relativistic Physics.&quot; by Patrik R. Girard. Find this if you want to learn more -- very good reading, not very complex and not very long. I'll just cite the first paragraph of chapter 3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From the very beginning of special relativity, &lt;strong&gt;complex&lt;/strong&gt; quaternions have been used&#xA;  to formulate that theory [45]. This chapter establishes the expression of the Lorentz&#xA;  group using complex quaternions and gives a few applications. Complex quaternions&#xA;  constitute a natural transition towards the Clifford algebra H ⊗ H.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well and the reference:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[45] L. Silberstein, The Theory of Relativity, Macmillan, London, 1914.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-03-20T20:07:50.673" CommentCount="1" />
  <row Id="7315" PostTypeId="1" AcceptedAnswerId="7365" CreationDate="2011-03-20T21:19:52.110" Score="0" ViewCount="156" Body="&lt;p&gt;Which diode (semiconductor material) has the lowest voltage in the direction of conductivity?&lt;/p&gt;&#xA;" OwnerUserId="58" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-21T13:54:28.910" LastActivityDate="2011-03-24T18:30:37.283" Title="The diode of the lowest voltage" Tags="&lt;electronics&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="7316" PostTypeId="2" ParentId="7231" CreationDate="2011-03-20T21:41:25.163" Score="-2" ViewCount="" Body="&lt;p&gt;A physical way of seeing this is that the phase space volume of a system is preserved.  Hamiltonian mechanics preserves the volume of a system on its energy surface H = E, which in quantum mechanics corresponds to the Schrodinger equation.  The phase space volume on the energy surface of phase space is composed of units of volume $\hbar^{2n}$ for the momentum and position variables plus the $\hbar$ of the energy $i\hbar\partial\psi/\partial t~=~H\psi$.  This is then preserved.  Any growth in the uncertainty $\Delta p\Delta q~=~\hbar/2$ would then imply the growth in the phase space volume of the system.  This would then mean there is some dissipative process, or the quantum dynamics is replaced by some master equation with a thermal or environmental loss of some form.  For a pure unitary evolution however the phase space volume of the system, or equivalently the $Tr\rho$ and $Tr\rho^2$ are constant.  This means the uncertainty relationship is a Fourier transform between complementary observables which preserve an area $\propto~\hbar$.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-03-21T12:55:32.250" LastActivityDate="2011-03-21T12:55:32.250" CommentCount="2" />
  <row Id="7317" PostTypeId="1" CreationDate="2011-03-20T22:09:36.233" Score="3" ViewCount="221" Body="&lt;p&gt;In meteorology, the atmosphere is considered to be divided into &lt;a href=&quot;http://en.wikipedia.org/wiki/Air_mass&quot; rel=&quot;nofollow&quot;&gt;air masses&lt;/a&gt;, regions of relatively uniform temperature and humidity with &lt;a href=&quot;http://en.wikipedia.org/wiki/Weather_front&quot; rel=&quot;nofollow&quot;&gt;fronts&lt;/a&gt; on their borders. But why doesn't the air from different air masses mix together when they meet? I can't find the answer in any weather book, and nobody seems to ask. People just accept the answer that different densities don't readily mix.&lt;/p&gt;&#xA;" OwnerUserId="2683" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-20T22:16:07.240" LastActivityDate="2011-03-21T16:33:47.307" Title="Why don't different air masses mix immediately?" Tags="&lt;air&gt;&lt;atmosphere&gt;&lt;atmospheric-science&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="7318" PostTypeId="2" ParentId="7303" CreationDate="2011-03-20T22:14:14.043" Score="6" ViewCount="" Body="&lt;p&gt;The problem with most of the earth-moon pictures is that they show the earht and moon very close which suggests that the moon is in the earth's shadow for almost half of the time. So in the picture linked to above - it looks like  a full moon should be dark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/cmfTj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real picture is more like this &#xA;&lt;img src=&quot;http://i.stack.imgur.com/PiNUT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-03-20T22:14:14.043" CommentCount="1" />
  <row Id="7319" PostTypeId="1" CreationDate="2011-03-20T22:46:28.047" Score="2" ViewCount="128" Body="&lt;p&gt;&lt;img src=&quot;https://docs.google.com/drawings/pub?id=1cm2nq5OKck4zKphMYgMIJuCREZDO9JlNriRA0_4pUTc&amp;amp;w=400&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This diagram is of a block (b) moving down a ramp unaided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking gravity (g) as $9.8ms^{-2}$, and the mass of the block (b) as $0.2kg$, how can I find the force acting along the slope (f).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW: the ramp is inclined 1.8 degrees, not radians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought I could use basic trigonometry, and $F=ma$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Sin(1.8) = \frac{0.2*9.8}{f} $$&#xA;$$ f = \frac{0.2*9.8}{sin(1.8)} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this results in: $f = 62.40$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing as force is a vector I thought I could use trig to split into it's components but obviously I'm doing something wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the answer should be ~0.06N)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the noob question, but it's got me clueless.&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="4" LastEditorDisplayName="" LastEditDate="2011-03-21T15:58:21.787" LastActivityDate="2011-03-21T16:04:21.250" Title="Forces on a simple flat slope" Tags="&lt;homework&gt;&lt;forces&gt;&lt;newtonian-mechanics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="7320" PostTypeId="2" ParentId="7319" CreationDate="2011-03-21T00:24:23.650" Score="2" ViewCount="" Body="&lt;p&gt;Here's something that might be worth remembering: when you split a vector into components (as you were trying to do), the original vector is &lt;em&gt;always&lt;/em&gt; the hypotenuse. See if you can use that fact to find the error you made in writing the formula &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Sin(1.8) = \frac{0.2*9.8}{f}$$&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-03-21T00:24:23.650" CommentCount="4" />
  <row Id="7321" PostTypeId="1" AcceptedAnswerId="7345" CreationDate="2011-03-21T00:43:56.317" Score="3" ViewCount="97" Body="&lt;p&gt;I am writing a lab report for class on PNMR experiment that we did. &lt;em&gt;How come in this experiment we don't worry about the electron spins in our sample?&lt;/em&gt; Aren't the electrons affected by the PNMR machine as much as the protons? Isn't there electron proton spin interactions? For some reason, the lab only talks about the proton spins and does not mention the electron spins at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we did this at room temperature, does this mean that the electrons have so much energy that their spins doesn't even align with the permanent magnetic field, or are they all paired up with other electrons and therefore do not contribute much? I am very confused. Please help!!!&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2011-03-21T10:46:56.763" Title="PNMR, Pulsed Nuclear Magnetic Resonance, Experiment" Tags="&lt;experimental-physics&gt;&lt;nuclear-physics&gt;&lt;magnetism&gt;&lt;electron&gt;" AnswerCount="1" />
  <row Id="7322" PostTypeId="1" AcceptedAnswerId="7330" CreationDate="2011-03-21T01:10:57.430" Score="4" ViewCount="197" Body="&lt;p&gt;What paper(s) or theory(s) describe or prove that the elementary particles that we have determined today cannot be made up of smaller more fundamental particles?&lt;/p&gt;&#xA;" OwnerUserId="2578" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-03-21T09:48:49.190" LastActivityDate="2011-03-27T03:58:34.253" Title="References on the non-compositeness of the known elementary particles" Tags="&lt;quantum-mechanics&gt;&lt;string-theory&gt;&lt;particle-physics&gt;&lt;nuclear-physics&gt;&lt;particles&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="7323" PostTypeId="2" ParentId="7322" CreationDate="2011-03-21T01:41:02.100" Score="3" ViewCount="" Body="&lt;p&gt;There is no such theory. We treat the elementary particles as elementary simply because we have never seen any evidence of them having a substructure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose someone might have published a paper claiming that they must be fundamental (after all, there are a lot of papers out there), but the vast majority of physicists do not take such claims seriously.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-03-21T01:41:02.100" />
  <row Id="7324" PostTypeId="1" AcceptedAnswerId="7325" CreationDate="2011-03-21T02:55:54.817" Score="4" ViewCount="213" Body="&lt;p&gt;I apologize if this is a naive question, but I never really learned about this. I'm curious as to what happens to sound waves after they are &quot;used&quot;? For example, if I say something to you verbally, then a sound wave is transmitted and picked up by your ears, but what happens to the wave after that? Does it float away into space? And if sound waves are never destroyed (not saying that is the case, but if it is), does this theoretically mean we can reproduce or attain all sound waves from the history of man, including every word ever spoken?&lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2011-03-21T09:48:04.490" Title="What happens to sound waves?" Tags="&lt;waves&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="7325" PostTypeId="2" ParentId="7324" CreationDate="2011-03-21T03:01:03.173" Score="3" ViewCount="" Body="&lt;p&gt;Ha Ha :D No mohabitar :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not possible. Sound ultimately transforms into heat energy. You can not reproduce all the useful and useless sounds from history. In general all energy is ultimately converted to heat energy and heat energy flows from an object at higher temperature to an object at lower temperature in order to attain thermal equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-03-21T03:01:03.173" CommentCount="3" />
  <row Id="7326" PostTypeId="1" AcceptedAnswerId="7332" CreationDate="2011-03-21T03:43:31.950" Score="7" ViewCount="204" Body="&lt;p&gt;I have a couple of naive questions from the topic of the title.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;We know&#xA;\begin{eqnarray}&#xA;\Omega-1=\frac{k}{a^2H^2}-\frac{\Lambda}{3H^2}&#xA;\end{eqnarray}&#xA;Now I read that from SBB model $\frac{1}{aH}$ increases with time and in recent time $\Omega\approx 1$. So, $\Omega$ has to be fine tuned to stay close to 1. I guess its because as we go back in time $\frac{1}{aH}$ decreases in SBB, and for fixed $\Lambda$ this drives $\Omega$ away from 1. I guess that's o.k as $aH$ decreases faster than $H$ alone in the second term above. But why do we want $\Omega\approx 1$ in earlier times? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My confusion deepens after reading how inflation solves this problem. In ``Inflation'', it is argued that $\frac{1}{aH}$ decreases with time. So I thought to myself As before going back in time resulted in decrease in Hubble length (comoving), so in the above equation the 1st term was large that is effectively with non-zero $k$. But now as going back in time increases $\frac{1}{aH}$, the first term decreases and it acts as if $k$ is nearly zero which is the criterion for flat universe. So my confusion boils down to the question on Why do we want $\Omega\approx 1$ in earlier times? Also how does Inflationary era solve the whole flatness problem? What about when inflation ends and SBB begins? Moreover, the above equation is valid for every case (inflation and SBB). So, even if inflation kills the problem, SBB should revive it yielding non unity of $\Omega$ now!&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Also I came to know that as long as $\dot{\phi}^2&amp;lt;V(\phi)$, inflation takes place and this is generally the case when potentials are flat enough (with $\phi$ of course). How to see that?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Related to that, I learned in Hybrid Inflation, since $\phi$ is driven to zero for $\psi&amp;gt;1$ (Why??), the potential in $\psi$ direction is flat and satisfies slow-roll conditions, so that $\psi$ is considered inflaton. How can we say potential in $\psi$ direction is flat?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="1349" LastActivityDate="2011-03-21T06:47:32.047" Title="On Flatness problem, Inflation etc" Tags="&lt;cosmology&gt;" AnswerCount="1" />
  <row Id="7327" PostTypeId="1" CreationDate="2011-03-21T04:50:48.923" Score="4" ViewCount="203" Body="&lt;p&gt;I would begin by apologizing for a soft question. Reading the FAQ, I have tried to formulate my  question precisely in the last paragraph and you may skip the following paragraph where I make subjective observations, which I think may be relevant and any comments would be more than helpful. By formal education I mean registering in an academic department in a university as a grad/Phd Student.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subjective belief: I feel I am personally not upto the mark of high-quality physics students. I need to learn a lot more, and do a lot more problems to be any good. I've noticed that Physics particularly is a very selective field dominated by hard-working and primarily young people who accomplish a lot at an early age (olympiads/contests, early phd's from very good places, etc). Even among this bunch a lot get filtered out and wouldn't get faculty positions. Theoretical Physicists tend to be outliers in a field almost entirely composed of outliers. And to learn from such a community, I feel an academic environment is essential, and not just reading books in isolation. So I beleive academia is the only legitimate environment that functions to produce sane physicists who know their field. However, if one has been set-back for some reason but carries the self-beleif that he can be smart enough to do it, would ageing be a parameter. I read a statement in Kleppner's undergrad mechancis book that &quot;by finishing all this at the age of 26, Newton set the trend which follows today, that great advances are made by young physicists&quot;. I haven't found a counterexample so I would trust academic departments have the same criterion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q: Do people who missed out all the action in their 13-20 ages miss out on having a good career? What are the possible reasons to take a hiatus of &quot;self-study&quot; for a 22-year old so that he may be a good candiate for a respectable physics career (Phd/postdoc etc) later. &lt;/p&gt;&#xA;" OwnerUserId="1671" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-21T05:13:09.360" LastActivityDate="2011-03-21T07:52:12.030" Title="Is it possible to restart formal higher education in physics at a later age?" Tags="&lt;soft-question&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="7328" PostTypeId="2" ParentId="7327" CreationDate="2011-03-21T04:58:40.870" Score="4" ViewCount="" Body="&lt;p&gt;Remember the famous people are outliers. Superstars - just like in Hollywood - are not the norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of science is done by very bright, hard working but ordinary people, the same very bright ordinary people that work in industry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes there are fields, particulalrly maths, where there may be a biological reason that people do all their best work before their brains hit 25 - but it's not true in a lot of fields. It takes a lot of years to get the experience to know how to do things in many branches, especially experimental ones, and even more years to get up the tree to a position where you can get do to the experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BUT taking 10years out because at 18 you don't think you know as much as everyone else is equally silly, most of your peers don't know as much as they pretend either!&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-03-21T04:58:40.870" />
  <row Id="7329" PostTypeId="2" ParentId="7327" CreationDate="2011-03-21T05:02:11.440" Score="4" ViewCount="" Body="&lt;p&gt;It depends on the country. In Greece there are so called &quot;open university&quot; academic programs,  programs geared for older people wanting a higher education. Last week a 91 year old man was interviewed on TV because he is pursuing studies for a degree in &quot;art appreciation&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would agree with Martin Becket , that if one is serious enough in wanting to study physics one should not put it off for &quot;not good enough in background&quot; reasons.  Yes, it is a competitive field, still, like any race, it is best to run it and see the results than not to and regret it. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-21T05:13:02.493" LastActivityDate="2011-03-21T05:13:02.493" />
  <row Id="7330" PostTypeId="2" ParentId="7322" CreationDate="2011-03-21T05:19:40.723" Score="6" ViewCount="" Body="&lt;p&gt;One of the questions under investigation in the data being gathered at LHC is the search for compositeness of quarks and leptons. &lt;a href=&quot;http://arxiv.org/pdf/1102.2020.pdf&quot; rel=&quot;nofollow&quot;&gt;They gave limits&lt;/a&gt; for quark compositeness from the data of 2010.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is, it is an open question under investigation, though not popular with the theorists.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-21T05:19:40.723" />
  <row Id="7331" PostTypeId="2" ParentId="7324" CreationDate="2011-03-21T05:43:36.287" Score="0" ViewCount="" Body="&lt;p&gt;To supplement sb1's answer I would like you to think the same question about waves of the sea, even if directional, like the wake of a boat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sound is waves of motion of all the particles that compose the air, similar to the waves of the sea. The shape and energy that the waves of the sea carry become finally thermal motion of the water, the air, and the sand if they break on a beach.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-21T05:43:36.287" />
  <row Id="7332" PostTypeId="2" ParentId="7326" CreationDate="2011-03-21T06:47:32.047" Score="3" ViewCount="" Body="&lt;p&gt;the normal cosmological evolution dominated by matter or radiation - which was the case for billions of years after the Big Bang - makes $|\Omega-1|$ &lt;em&gt;increase&lt;/em&gt; with time because of changes captured by the Friedmann equations. However, WMAP and others show that $|\Omega-1|$ is not greater than 0.01 today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that $|\Omega-1|$ had to be even much smaller a minute - and a fraction of second - after the Big Bang - something like $10^{-{\rm dozens}}$. This is called the flatness problem because there's no reason for a generic Universe to have such a small number of a quantity such as $|\Omega-1|$ which can a priori be anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cosmic inflation solves the flatness problem because it reverses the evolution $|\Omega-1|$: as time goes to the future, $|\Omega-1|$ is (more precisely, was) &lt;em&gt;decreasing&lt;/em&gt; in this case. It's because the evolution is driven by the last, temporary cosmological constant term in your equation. So because of inflation, one may start with a generic value of $|\Omega-1|$ before the inflation, and one ends up with $|\Omega-1|$ close to zero at the end, anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that I don't have to explain that if a quantity increases/decreases with time, it will decrease/increase if you read the time in the opposite direction. Nevertheless, it seems that 1/3 of your question is focusing on this triviality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot\phi^2 &amp;lt; V (\phi)$ is just a condition saying that $V(\phi)$, the cosmological constant term, is the dominant term in the Friedmann equation during inflation. If that's so, you may neglect the kinetic term $\dot \phi^2$. I am not sure whether the coefficient in the equation is one - I doubt so. You should only view the inequality as an estimate. If the kinetic term is much smaller, it's guaranteed that you can neglect it and the exponential inflationary expansion dictated by $V(\phi)$, the temporary cosmological constant, follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning hybrid inflation, obviously, you can't determine that the potential for $\psi$ is a slow-rolling one just from the assumptions you have shared with us: it's just an assumption that the potential has the property. In particular, the potential is supposed to resemble the potentials in the theory of second order phase transitions - which occurs very generically in string theory models etc. Imagine that $\phi$ is the temperature, more precisely $T-T_c$, and $\psi$ is the magnetic field. For a positive $\phi$, the minimum is at $\psi=0$, so only $\phi$ is changing. However, once $\psi$ reaches zero, you get a phase transition and the right magnetic field is at a nonzero value, so one rolls the channel at $\psi=\pm M$. Near the $\phi=0$ point, the potential is inevitably changing slowly because it's a stationary point. See e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://nedwww.ipac.caltech.edu/level5/Liddle/Liddle5_7_2.html&quot; rel=&quot;nofollow&quot;&gt;http://nedwww.ipac.caltech.edu/level5/Liddle/Liddle5_7_2.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The purpose of the other fields in hybrid inflation is to stop the inflation.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-21T06:47:32.047" CommentCount="2" />
  <row Id="7333" PostTypeId="2" ParentId="7317" CreationDate="2011-03-21T06:58:28.307" Score="2" ViewCount="" Body="&lt;p&gt;There's always a reason although for different weather fronts, the discussion has to be different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different densities have to have a reason - different pressure and/or humidity etc. If there is a different pressure, there is a mechanical force the preserves the pressure difference: think about the cyclones that have a lower pressure in the center. The cyclones rotate in the right direction and the cyclone may be preserved by the Coriolis force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the two air masses differ by humidity, the mixing will almost always lead to precipitation - which includes a phase transition for water etc. It's because the vapor from the more humid air mass gets condensed under the conditions of the other. You get some rain. In general, intense precipitation, thunderstorms, and other visible isolated weather events are linked to weather fronts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, a mixing of two air masses is a nontrivial, violent process in general. That's why the boundary is called a &quot;front&quot;. In the military jargon, a &quot;front&quot; is the contested frontier of a conflict. So your idea that the air masses could mix quickly and peacefully - whatever you exactly mean quantitatively - either neglects the inertia of the air, a relatively low diffusion coefficient, a low thermal conductivity, and/or high latent heat of water vapor. A front is something that didn't disappear within minutes so pretty much tautologically, there must be forces that make such a quick disappearance impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape of the clouds etc. is complicated and meteorological models are not capturing them nicely. And I surely don't plan to defend long-term climate models whose predictions are uncorrelated to the reality, and they usually turned out to be negatively correlated, in fact. But I assure you that the meteorological models used to forecast weather for the next weak deal with the weather fronts - they're among the main entities that these models have to deal with - and predict them in a satisfactory way, at least at a qualitative level, while using well-known laws of physics only.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-21T06:58:28.307" CommentCount="2" />
  <row Id="7334" PostTypeId="2" ParentId="7327" CreationDate="2011-03-21T07:52:12.030" Score="3" ViewCount="" Body="&lt;p&gt;Taking a short hiatus should &lt;em&gt;not&lt;/em&gt; prove fatal to your futur career, whatever that may be. I believe the theorist Dijkgraaff went to art school halfway through his studies! You're still very young, and the idea that all good work in physics and math is done before someone's 25th birthday seems farfetched; any major experiment takes years to set up, and good idea's in theory also take some time (the name of Mischa Gromov springs to mind). However, &lt;em&gt;if&lt;/em&gt; you take a break, do it for the right reasons. It's not very likely that after one year, you'll return as the head of your class, full of wisdom about Chern classes and fermion doubling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, 'ageing' also differs per country. For example: in France, people tend to finish PhDs really early (around their 25th), but in Germany it's quite common to still be working on your Master's at that age.&lt;/p&gt;&#xA;" OwnerUserId="1471" LastActivityDate="2011-03-21T07:52:12.030" />
  <row Id="7336" PostTypeId="1" CreationDate="2011-03-21T09:35:15.937" Score="5" ViewCount="128" Body="&lt;p&gt;I googled in vain and found nothing. I am surprised. Maybe somebody knows about videos of Julian Schwinger available on Internet?&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-03-21T09:35:15.937" ClosedDate="2011-04-13T23:21:16.580" Title="Julian Schwinger videos, is there any?" Tags="&lt;soft-question&gt;" CommentCount="1" />
  <row Id="7337" PostTypeId="2" ParentId="3635" CreationDate="2011-03-21T09:36:44.797" Score="0" ViewCount="" Body="&lt;p&gt;The problem you have posed is very interesting.I made a similar trial some time ago.&#xA;If you like you can go to www.anconanotizie.it where i published a note on the possibility &#xA;to explain the stability of  the hidrogen atom on a classical basis.&#xA;The crucial point, according to my opinion, is the following: the proton is a very strong&#xA;source of magnetic field mainly due to its spin. So the electron will be submitted to the Lorentz force which presumably will oppose to the Coulomb force.If the electron try to collapse on the proton a strong repulsive force will be established between the two particles explained by the induction electromagnetic law of Faraday-Lenz.&#xA;So it's possibile that will exist motion conditions that prevent the system from radiation&#xA;just because the system should not collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ronchini Riccardo&lt;/p&gt;&#xA;" OwnerUserId="2685" LastActivityDate="2011-03-21T09:36:44.797" />
  <row Id="7338" PostTypeId="2" ParentId="7324" CreationDate="2011-03-21T09:48:04.490" Score="3" ViewCount="" Body="&lt;p&gt;Just one more aspect to consider is the comparison with the electromagnetic waves from radio and TV broadcasts. One could ask the analogous question here:&quot;what happens to them?&quot; The answer is that some were broadcast also into space (when transmitters were entirely undirectional) and so &lt;em&gt;are&lt;/em&gt; still available for reception, albeit only by aliens light years away and very attenuated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sound waves stay on Earth in the atmosphere, and become part of that atmospheric motion, which is gaseous motion. This motion exists because there is a non-zero temperature for the atmosphere (say 300K). So the sound wave attenuation will reduce the given wave intensity, and corresponding particle motion, and corresponding air pressure difference (which is sound) below that of the ambient atmosphere. In a good (outdoor or indoor) auditorium the sound can last for longer and travel further, but eventually it will reach the general level of atmospheric noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I suppose that with sound (unlike with EM waves in space) one could say that the medium ultimately destroys the message.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-03-21T09:48:04.490" />
  <row Id="7339" PostTypeId="1" AcceptedAnswerId="7354" CreationDate="2011-03-21T10:16:41.183" Score="3" ViewCount="202" Body="&lt;p&gt;If a cable used to power something is exposed to a radioactive source will it over time make the entire cable radioactive? &lt;/p&gt;&#xA;" OwnerUserId="2686" LastEditorUserId="2686" LastEditorDisplayName="" LastEditDate="2011-03-21T10:29:54.333" LastActivityDate="2011-05-19T20:39:37.767" Title="Can electricity transfer radioactivity?" Tags="&lt;electricity&gt;&lt;radioactivity&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="7340" PostTypeId="1" CreationDate="2011-03-21T10:29:48.027" Score="3" ViewCount="384" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Water_rocket#World_record&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt;, the current record for greatest altitude achieved by a water and air propelled rocket is 2044 feet (623 meters).&lt;br&gt;&#xA;In this connection, i had one question: How the  greatest altitude depends of the density of the liquid used in the rocket?&lt;br&gt;&#xA; Suppose that we have the opportunity to choose a liquid with an arbitrary density and with the same hydrodynamic properties as water. Then there must be a density at which the greatest altitude has its absolute maximum, i think.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more specific let's consider a simpler problem:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let a two-liter soda bottle rocket, filled with some amount of a liquid mentioned above and air at the atmospheric pressure, is launched in deep space($g=0$)(say an astronaut is experimenting). What is the absolute maximum of the speed the rocket can achieve? For simplicity let's assume the expansion of the air to be isothermal (which it is not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/RVXX7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="552" LastActivityDate="2011-08-03T10:20:57.603" Title="Water rocket physics" Tags="&lt;classical-mechanics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="7341" PostTypeId="1" CreationDate="2011-03-21T10:40:41.340" Score="2" ViewCount="66" Body="&lt;p&gt;what is the meaning of expansion ,shear and viscosity in context of universe? how can we  conclude a result after getting a numerical value of above terms?&lt;/p&gt;&#xA;" OwnerUserId="1843" LastActivityDate="2011-03-21T10:40:41.340" Title="shear , viscosity and expansion of universe" Tags="&lt;cosmology&gt;" CommentCount="5" />
  <row Id="7342" PostTypeId="2" ParentId="7339" CreationDate="2011-03-21T10:41:35.363" Score="2" ViewCount="" Body="&lt;p&gt;Electricity is the movement of electrons. Electrons cannot become radioactive. So a current in the cable will not change the radioactivity of the cable.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-21T10:41:35.363" CommentCount="3" />
  <row Id="7344" PostTypeId="2" ParentId="7303" CreationDate="2011-03-21T10:46:52.567" Score="1" ViewCount="" Body="&lt;p&gt;The surface of the moon is rough rather than shiny so it scatters light in all directions.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-21T10:46:52.567" />
  <row Id="7345" PostTypeId="2" ParentId="7321" CreationDate="2011-03-21T10:46:56.763" Score="2" ViewCount="" Body="&lt;p&gt;1st all electrons in &quot;normal&quot; matter occur in pairs with opposite spins, thus cancelling each other. &#xA;Exeptions like stable radicals or O2, which has a triplet ground state, interact with the nuclear spins. Look for CIDNP, &quot;Chemical Induced Dynamic Nuclear Polarisation&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general electron magnetic contribution eg paramagnetism act to shorten the lifetime of the &#xA;nuclear spin states, this is often used to enhance the sensitivity of NMR, when You get some rare earth salt infusion for brain scan. But electron spin/angular momentum can also make NMR impossible by line broadening. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd there &lt;strong&gt;is&lt;/strong&gt; interaction between electrons and nulei, the coupling of the protons in different &#xA;atoms of a molecule is mediated by the electrons of the bonds in between. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-21T10:46:56.763" />
  <row Id="7346" PostTypeId="1" CreationDate="2011-03-21T11:01:31.283" Score="2" ViewCount="302" Body="&lt;p&gt;And you jumped in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would happen when you got to the middle of the earth? Would you gradually slow down, until you got to the middle and once you were in middle would every direction feel like it was up? &lt;/p&gt;&#xA;" OwnerUserId="406" LastActivityDate="2011-03-21T17:15:55.803" Title="If it was possible to dig a hole that went from one side of the earth to the other" Tags="&lt;gravity&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="7347" PostTypeId="1" CreationDate="2011-03-21T11:12:26.103" Score="-2" ViewCount="221" Body="&lt;p&gt;I saw this question in a test. I would have answered kinectic energy due to rotation and translation. It that correct. Else what is the answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh no, i forgot to mention it was objective type question.There were  three other options.I just looked them up&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)kinetic energy due to rotation&#xA;2)kinetic energy due to translation&#xA;3)none of the above&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the question looked confusing and the choices looked misleading at that time.&#xA;Thanks for all your resopnses&lt;/p&gt;&#xA;" OwnerUserId="2687" LastEditorUserId="2687" LastEditorDisplayName="" LastEditDate="2011-03-23T09:00:41.953" LastActivityDate="2011-03-23T19:06:12.123" Title="what energies do the wheels of a moving car posses?" Tags="&lt;mechanics&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="0" />
  <row Id="7348" PostTypeId="2" ParentId="7346" CreationDate="2011-03-21T11:25:50.863" Score="8" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;And you jumped in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, I'd refuse, You should do that Yourself. :=(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do You want to &quot;solve&quot; this with or without friction by air? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without friction, You would fall and reach maximum speed in the center of earth, &#xA;going on until You reach the antipods, where You would stand still for a fraction of a second, &#xA;then You would go down again. Very boring indeed and You should take some sandwiches &#xA;and some drink with You. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With friction You wold accelerate up to a speed around 300 km/h, then fall down &#xA;with this constant speed, slowing down when approaching the center, and go on some &#xA;distance, fall down to center again and do that oscillation until all Your potential/kinetic  energy is consumed by friction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because about a dozen of such holes were drilled here in physics.SE alone, You might find some company at the center of earth. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-21T12:05:18.677" LastActivityDate="2011-03-21T12:05:18.677" CommentCount="3" />
  <row Id="7349" PostTypeId="2" ParentId="7347" CreationDate="2011-03-21T11:36:08.297" Score="0" ViewCount="" Body="&lt;p&gt;This is correct, there is no &quot;else&quot;. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-21T11:36:08.297" CommentCount="7" />
  <row Id="7350" PostTypeId="2" ParentId="7347" CreationDate="2011-03-21T11:37:20.583" Score="1" ViewCount="" Body="&lt;p&gt;They also posses internal energy consisting of chemical energy (from interaction between molecules), kinetic energy of individual molecules, mass energy of individual elementary particles, strong interaction energy between quarks and gluons et cetera et cetera. There will also be some interaction energy coming from the interface of wheels with materials it is in contact with (be it air or road). And of course also interaction energy between wheel and its axis (relevant for shock absorption, etc.)&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-03-21T13:11:17.097" LastActivityDate="2011-03-21T13:11:17.097" CommentCount="15" />
  <row Id="7351" PostTypeId="2" ParentId="7347" CreationDate="2011-03-21T11:37:55.840" Score="1" ViewCount="" Body="&lt;p&gt;They are in a gravitational field, so they have some gravitational potential energy.&lt;br&gt;&#xA;The atoms etc in the wheels have nuclear binding energy.&lt;br&gt;&#xA;There is some stretched rubber which carry elastic potential energy.&lt;br&gt;&#xA;They have temperature which amounts to kinetic energy.&#xA;They have electric charge, so there is some electric potential energy.&#xA;As you mentioned, they have translational and rotational kinetic energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take the union (some are partly overlapping) of all these energies you find their total energy E=mc^2.  &lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-03-21T11:37:55.840" CommentCount="4" />
  <row Id="7353" PostTypeId="1" CreationDate="2011-03-21T11:59:36.093" Score="0" ViewCount="153" Body="&lt;p&gt;Is precision a &quot;quality&quot; of a measurement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a better (accepted by the literature) word?&lt;/p&gt;&#xA;" OwnerUserId="2688" LastActivityDate="2011-03-21T13:07:38.810" ClosedDate="2011-03-21T16:11:41.960" Title="Metrology: What is precision for a measurement?" Tags="&lt;statistics&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7354" PostTypeId="2" ParentId="7339" CreationDate="2011-03-21T12:25:03.577" Score="7" ViewCount="" Body="&lt;p&gt;Are you worried that the cables that go to the Fukushima reactors will carry radioactivity out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is No. You should &lt;a href=&quot;http://en.wikipedia.org/wiki/Radioactivity&quot; rel=&quot;nofollow&quot;&gt;read up a bit&lt;/a&gt; on radioactivity and educate yourself, since it is one of the facts of life. In the article you will see that it is atoms that are responsible for radioactivity whereas the current in the cables is due to electrons. The parts of the cables that are in a radiation environment will become radioactive, but that activity will remain in the locality, and the length of cable that was exposed. It in no way can be transmitted away from the region the way the current is transmitted.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-21T12:25:03.577" CommentCount="2" />
  <row Id="7355" PostTypeId="2" ParentId="7353" CreationDate="2011-03-21T13:07:38.810" Score="1" ViewCount="" Body="&lt;p&gt;No, precision = detail, it's &lt;strong&gt;accuracy&lt;/strong&gt; that could be related to quality (and it implies precision too)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Accuracy_and_precision&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Accuracy_and_precision&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1916" LastActivityDate="2011-03-21T13:07:38.810" CommentCount="8" />
  <row Id="7356" PostTypeId="2" ParentId="7089" CreationDate="2011-03-21T13:40:29.010" Score="1" ViewCount="" Body="&lt;p&gt;The question is complicated and related to something called ERoEI (Energy Return on Energy Investments).  This is the general question on how much energy one must put into getting a source of energy balanced with how much energy you get out.  The direct case can be seen with fusion and the break even point where you get net energy out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Georg indicates the nuclear power industry has been hugely subsidized in connection with nuclear weaponry.  The industry is largely a spin off from plutonium production for weaponry.  The stable Pu isotope for weapons is Pu-239, while the other isotopes tend to spontaneously fission.  These isotopes are farmed out to the nuclear power industry.  So the nuclear power industry is a spin off from the weapons industry.  The net financial investment by the United States in nuclear technology is over $10 trillion.  We might of course question whether we have recovered this cost, either in financial terms or energy terms.  Nuclear accidents are also seriously costly and dangerous, and the industry has some appearances of being like the dirigible industry before 1936.  At some point there may be the big nuclear “Hindenburg” incident which might put the kibosh on the whole thing.  Events of late come somewhat close to that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The industry, in part due to its connections with weaponry, has a huge political connection.  Unfortunately this has been politics of the far ends of the spectrum, where in the USSR this was with the extreme left communists and in the US this has been associated with the far right that is moving further to the right.  The further on either spectrum politics becomes the more it degenerates into a filthy pile of lies.  Unfortuantely the economics and politics surrounding nuclear power, mainly with those who run the system, have a long history of being surrounded and mired in mendacities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue with research costs is of course important.  The fusion research programs, such as ITER, seem to be a long process of getting closer to a goal, but where the goal also seems itself to recede further away.  Renewable energy research tends to get at best second billing, for it is not as “sexy” as “big science” programs with impressive expensive facilities.  The funding and political interest is highly skewed in some ways.  For instance there is the whole thing about “clean coal,” which is a way of subsidizing the future of the coal industry.  However, clean coal makes about as much sense as saying “partially pregnant.”  In spite of what the mendacious right wing in the US says, we do have a little problem with $CO_2$ in the atmosphere and the climate perturbation it is generating.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ERoEI issues then comes into play, where there are people who spend considerable time on this problem.  As petroleum becomes less accessible and approaches this peak situation the ERoEI declines.  This is a ratio of energy out/energy in, and for a large oil field that is “virgin” the ERoEI is about 25.  Currently ERoEI is estimated for oil production globally to be around 10 to 15.  Nuclear power is estimated to be a number of values, where about the highest I have seen is 5.  The solar, wind, and renewable power systems have a similar ERoEI.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-21T13:40:29.010" CommentCount="1" />
  <row Id="7357" PostTypeId="2" ParentId="7317" CreationDate="2011-03-21T13:40:58.223" Score="1" ViewCount="" Body="&lt;p&gt;I would like to add a few points to the other answer. The question could be taken to be asking why the Earth's atmosphere consists of the thermodynamic equivalent of &quot;domains&quot;. That is why are there different regions (of homogeneous parameter values) with discontinuities between those regions? These regions here being the &quot;air masses&quot; - defined as having a homogeneous temperature and humidity - which are listed on the Wiki articles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted that any question about the atmosphere is a question covering many areas of physics: thermodynamics, turbulence, chemical processes. As a turbulent system the atmosphere is not in equilibrium and so, for example, does not have a constant temperature, or even a constant temperature gradient across either time or surface. It is in somewhat of a steady state with the Solar energy, however, and this results in the atmospheric equivalent of large eddies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the Wiki article explains the &quot;air mass&quot; regions are actually maintained at their temperature and humidity by their underlying ground structure and vegetation. According to that diagram too, the air mass regions are actually physically separated from each other. So they will only interact when the air above one of them moves near to the territory of another. Part of the cause of this movement will be the Earth's rotation and its consequences in circulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now by definition an air mass region has a given Temperature and Humidity. But Humidity is really a measure of the air-water mix and so the &quot;gas&quot; (ie gaseous state) of each region is different. Put differently the chemical composition of the two regions is different, as well as their temperature. Now one factor that arises is that more humid air is less dense than less humid air (originally discovered by Newton apparently). Another factor is that humidity (absolute and relative) can double with a mere $10^o$ temperature change, resulting in saturation or near saturation. We also have that lower temperature air can contain less water than higher temperature air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in regions with different (T, H) levels it has to be thermodynamically attractive for them to mix immediately, and the scenario described above is not looking promising... Basically the warmer air (being less dense and maybe with more water) will rise above the colder air rather than just mix. Unfortunately this has consequences for those on the ground nearby. The momentum of the colder denser air also keeps it moving in its original direction somewhat intact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully on other planets they do things differently.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-03-21T13:40:58.223" />
  <row Id="7358" PostTypeId="1" CreationDate="2011-03-21T13:49:22.567" Score="0" ViewCount="385" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicates:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/3158/why-quantum-entanglement-is-considered-to-be-active-link-between-particles&quot;&gt;Why quantum entanglement is considered to be active link between particles?&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/317/why-cant-the-outcome-of-a-qm-measurement-be-calculated-a-priori&quot;&gt;Why can't the outcome of a QM measurement be calculated a-priori?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Why do some (the majority of?) physicists conclude non-determinism from quantum uncertainty?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we can't measure something, it seems to me like it's just a reflection of our ignorance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, from what I've read and seen, physicists actually interpret that as a reflection of the underlying non-determinism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't understand why this is. It seems to violate the fundamental axiom of science. Specifically causality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia's page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Uncertainty_principle&quot; rel=&quot;nofollow&quot;&gt;Uncertainty Principle&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(..) Certain pairs of physical properties, such as position and momentum, cannot be simultaneously known to arbitrarily high precision. (...) The principle implies that it is impossible to determine simultaneously both the position and the momentum of an electron or any other particle with any great degree of accuracy or certainty. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This seems like a statement about a limit of our instruments and knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see no reason to conclude &quot;the underlying nature of reality is inherently nondeterministic&quot; from &quot;we cannot make precise measurement&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like Einstein was pretty much the only prominent physicist who rejected it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, to quote Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Albert Einstein believed that randomness is a reflection of our ignorance of some fundamental property of reality.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I find myself naturally agreeing with Einstein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see no logical pathway from &quot;we can't measure nature&quot; to &quot;nature is random&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why do physicists adopt this view?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please provide an answer that is clear and simple, and not smothered by formality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note: this question is not for debating. I'm genuinely asking &lt;em&gt;why&lt;/em&gt; is this the prevalent point of view? I honestly see no logical pathway to that conclusion at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider the question &quot;Is it day or night in Cairo right now?&quot;, and assume we don't know what causes day and night. If we try to find out the answer to this question at any point in time, there's 50% chance that it's day and 50% chance that it's night. This doesn't mean that there's no answer to the question until we check, it simply means we don't really know what causes day and night to occur in Cairo at any given point in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why, oh why, would anyone conclude from this thought experiment, that Cairo has the fundamental property that day and night are non-deterministic in it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems patently clear that there's an awful lot about sub-atomic particle that we don't know. If we knew more, perhaps we could come up with better ways to measure things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify what I mean by non-determinism:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a particle has 30% chance of being &quot;here&quot;, and 70% chance of being &quot;there&quot;, then I would assume that there's some underlying reason that determines where the particle is. But the prevalent view (as I understand it) is that there's no underlying reason, the particle just happens to sort of &quot;choose&quot; to be &quot;here&quot; 30% of the time and &quot;there&quot; 70% of the time with no particular reason. (I find this view absurd)&lt;/p&gt;&#xA;" OwnerUserId="2690" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-23T00:18:40.523" LastActivityDate="2011-03-23T00:18:40.523" ClosedDate="2011-03-23T00:17:52.350" Title="Causality and Quantum uncertainty" Tags="&lt;quantum-mechanics&gt;&lt;uncertainty-principle&gt;" AnswerCount="4" CommentCount="10" FavoriteCount="1" />
  <row Id="7359" PostTypeId="1" CreationDate="2011-03-21T14:18:07.210" Score="3" ViewCount="238" Body="&lt;p&gt;If the universe is expanding, what is it expanding into?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly when the big bang happened where and how did it occur? - where did the energy come from? energy can not be created or destroyed does that mean, energy has existed before the universe was here?&lt;/p&gt;&#xA;" OwnerUserId="2691" LastActivityDate="2011-03-23T16:16:37.727" Title="origins of the universe questions" Tags="&lt;universe&gt;&lt;big-bang&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="7360" PostTypeId="2" ParentId="7319" CreationDate="2011-03-21T14:21:44.350" Score="0" ViewCount="" Body="&lt;p&gt;When in doubt, &lt;em&gt;always&lt;/em&gt; draw a force diagram.  Remember, $g$ will always point straight down, $f$ will always run parallel to the plan upon which you're moving, and your third vector will always be perpendicular to the plane.  With that in mind, your problem will have the following force diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/JUNbz.png&quot; alt=&quot;Force Diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, $a$ is the force component of gravity perpendicular to the plane.  If you were working with friction, it would be important ... but you're not, so feel free to ignore it.  The angle directly opposite $f$ is $1.8^{\circ}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the law of sines (basic trig) to calculate f at this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{f}{sin 1.8} = \frac{g}{sin 90}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f = g * sin 1.8$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll let you plug in the numbers (gravity and the mass of the block) to solve for $f$.&lt;/p&gt;&#xA;" OwnerUserId="2197" LastEditorUserId="2197" LastEditorDisplayName="" LastEditDate="2011-03-21T16:04:21.250" LastActivityDate="2011-03-21T16:04:21.250" CommentCount="7" />
  <row Id="7361" PostTypeId="2" ParentId="7358" CreationDate="2011-03-21T14:37:06.757" Score="-5" ViewCount="" Body="&lt;p&gt;Firstly let me mention you have a wrong definition of causality. Causality does not mean determinism. It means simply, no signal can travel faster than the speed of light and future is always influenced by the past not the reverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let us come to your main question. Are physicists so stupid that they can't understand that if they fail to measure a quantity with 100% accuracy that does not mean a precise value does not exist. No they are not and let me tell you they are the smartest of people in many respect. Ask yourself what is position or momentum? They are nothing but concepts invented by human mind for the sole purpose of describing our world of experiences, to make reliable mental models to relate different sense data. If uncertainty principle shows that even in principle you can not have the knowledge of both position and its conjugate momentum with 100% accuracy then it shows that the uncertainty should be ascribed to nature rather than ourselves. Our mental construct should be such that it can adapt to nature or more precisely the laws of nature. Since there is uncertainty in the very nature of nature there is an element of randomness too. You may consider that there exists a &quot;real&quot; universe where the elements of reality may have definite features and values. But then you are in metaphysics. You are no different than those who discuss and argue how many angels can dance on the head of a pin. As far as physics is concerned nature does not posses a definite value of position and its conjugate momentum and there are many such conjugate pairs which are equally subject to the uncertainty principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Albert Einstein was a genius. It was possible, in his time, to consider that some hidden variable theory could manage to replace these essential ideas. But the history has proved him wrong in this regard. It showed that nature can be stranger than what could be thought by a genius of Einstein's caliber.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-03-21T14:37:06.757" CommentCount="1" />
  <row Id="7362" PostTypeId="2" ParentId="7358" CreationDate="2011-03-21T14:37:43.260" Score="6" ViewCount="" Body="&lt;p&gt;A short answer (which has just become somewhat longer) is that there is &lt;em&gt;more&lt;/em&gt; than just the Uncertainty Principle thought experiment involved in the non-determinism deduction. If it were only that then physicists might just conclude that it was some classical wave phenomenon (which it is in a way) that gave rise to the HUP. The key other factor at work is the mysterious object called $\Psi$ which obeys an equation and is said to describe all of (quantum) physics. That is all the atomic experiments (spectroscopy etc) can be calculated from it and its equation (called the Schrodinger equation). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The factor then is that $\Psi$ is not like other objects in traditional physics - these give a specific answer to a specific question. Thus a clock measures the time, and a calculation based on that can determine whether it is or is not night time in Cairo (the date and other physics input might be needed for this in general). But for physics involving $\Psi$ all that happens is a probability! So there is maybe a 80% probability of result X and 20% result not-X. EDIT: I shall have more to say on Einstein's position on this topic after some references at the end of the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly physicists want to know whether this is some kind of approximation (as Einstein viewed) or whether it is a final result. In some way the maximum possible answer being just a probability. You will find many Stack questions which take this to the next level involving subtle tests of this probability (like Bell's Theorem). Many things are known from these theorems and corresponding experiments, and $\Psi$ clearly gives non-deterministic results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the HUP is simply one consequence of the basic properties of $\Psi$ giving somewhat of an explanation of that result. If you want to study this more note that words like &quot;determinism&quot; and &quot;non-determinism&quot; would need to be analysed a little bit carefully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Some references:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary topic to study further is Einstein's thought experiment to advance his perspective. This is known as the Einstein-Podolsky-Rosen (EPR) experiment. It has become famous in exactly this area: &lt;a href=&quot;http://en.wikipedia.org/wiki/EPR_paradox&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/EPR_paradox&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Derived from this have been further theorems primarily:&#xA;Bell's Theorem : &lt;a href=&quot;http://en.wikipedia.org/wiki/Bell_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Bell_theorem&lt;/a&gt;&#xA;This theorem is subject to &quot;debate&quot; from several directions. It rules out what are called &quot;hidden variables&quot;, subject to certain conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an old stack question related to yours:&#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/7/will-determinism-be-ever-possible&quot;&gt;Will Determinism be ever possible?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more recent one related to the Bell Theorem topics is (which I havent studied yet):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/7280/is-contextuality-required-in-quantum-mechanics&quot;&gt;Is contextuality required in quantum mechanics?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another theorem related to this question is the Kocken-Specker (Conway) theorem, also discussed here: &lt;a href=&quot;http://physics.stackexchange.com/questions/4172/what-does-john-conway-and-simon-kochens-free-will-theorem-mean&quot;&gt;What does John Conway and Simon Kochen&amp;#39;s &amp;quot;Free Will&amp;quot; Theorem mean?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT (after some clarification on the Question):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there are two parts to this: (a) Einstein's reaction to the points mentioned above and (b) why &quot;most physicists&quot; dont accept that reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Einstein's reaction was to develop some thought experiments, essentially the EPR thought experiment, which has become famous. This makes the claim that even although QM is accurate and even complete in its own terms, it is incomplete as a physical theory. The general idea was to show that uncertainty in a value (or even the impossibility to obtain a value) was somehow &lt;em&gt;just&lt;/em&gt; a consequence of Quantum Mechanics as it then stood (and still stands). The EPR thought-experiment was intended to show that there was other &quot;physical information&quot; out there which was not being captured by $\Psi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially I think that physicists just ignored this paper, as no experiments were done or further development was provided. There was also a theorem by John Von Neumann which supported the view that there could be no &quot;hidden variables&quot; - which is what the Einstein position seemed to imply. (A hidden variable might be if $\Psi(x) = \Psi(x,h)$ where somehow we only measure $\Psi(x)$ probabilistically, but $\Psi(x,h)$ is more deterministic - but we never see or determine h, maybe. A few options here as one sees.) This wasnt quite what EPR claimed, but Bohm did develop a version of QM which had a hidden field (not quite a variable) which stochastically determined things. This was a counterexample, of a sort, to the von Neumann theorem. This is now called the &quot;Bohmian Interpretation&quot; of QM and showed that &quot;Interpretations&quot; might exist which were different from the ones known in earlier days of QM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bell examined this argument in the 1960s with a Theorem to update von Neumann and clarify what was excludable by the maths of QM itself. The Kochen-Specker theorem also derived similar results in a special finite case. When the Bell theorem relations were tested experimentally it showed that the Bell theorem was correct experimentally. This is why most physicists consider this topic closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some philosopher-physicists and some mathematical physicists like to study remaining loopholes in these theorems and their experiments, and so the topic is under subtle discussion from that perspective. Finally it is known that Quantum Mechanics is not Quantum Gravity so it is possible that some modification of QM might be required before it becomes QC. Maybe something in that modification will also add some subtletly to the discussion.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-03-21T16:28:06.380" LastActivityDate="2011-03-21T16:28:06.380" CommentCount="8" />
  <row Id="7363" PostTypeId="2" ParentId="7292" CreationDate="2011-03-21T14:47:25.933" Score="1" ViewCount="" Body="&lt;p&gt;You've stumbled on a fertile area.  Though not strictly what you were asking about, I can tell you that perhaps the most interesting relationship between orthogonal groups and quaternions comes from looking at spinors.  As you may know, the symmetry group called $Spin$ double covers the group of rotations, and is the more relevant group for physics since spinors transform under this larger group.  A useful example is the double cover $SU(2) \rightarrow SO(3),$ that is, $SU(2) = Spin(3)$ in the Euclidean signature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Topologically, $SU(2)$ is a 3-sphere, which we can think of as the unit quaternions (remember, the norm is Euclidean, as pointed out by others).  To understand the map $SU(2) \rightarrow SO(3),$ let $v$ be an &lt;em&gt;imaginary&lt;/em&gt; quaternion (which we can think of as a 3-vector), and let $q$ be in $SU(2).$  Then since multiplication of quaternions preserves the norm,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\overline{q} v q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has the same norm as v, and you will note that it is still imaginary.  Thus, the action $v\stackrel{q}{\mapsto} \overline{q}vq$ of $SU(2)$ on $R^3$ (the imaginary quaternions) is by a &lt;em&gt;rotation.&lt;/em&gt;  Furthermore, $q$ and $-q$ act the same way.  So we have described a double cover of $SU(2)$ onto 3-dimensional rotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this is not what you immediately inquired about or discovered, but it's probably worth knowing.&lt;/p&gt;&#xA;" OwnerUserId="20" LastActivityDate="2011-03-21T14:47:25.933" />
  <row Id="7365" PostTypeId="2" ParentId="7315" CreationDate="2011-03-21T15:35:44.143" Score="3" ViewCount="" Body="&lt;p&gt;In fact, Schottky Diodes &lt;/p&gt;&#xA;&#xA;&lt;p&gt;have the lowest &lt;strong&gt;forward voltages&lt;/strong&gt;. This means, that this is not a question of band gap &#xA;&quot;voltage&quot; (this is a energy difference originally!) but of technology. &#xA;Second are Germanium point contact diodes with gold wires. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Diodes&quot; made from Galena maybe are very low too, but due to the wiggely properties &#xA;I would not dare to write about the &quot;impressions&quot; I had some times for some seconds on a scope. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-21T15:35:44.143" />
  <row Id="7367" PostTypeId="2" ParentId="7358" CreationDate="2011-03-21T15:43:02.217" Score="2" ViewCount="" Body="&lt;p&gt;Hasen j hacker-not-engineer. I suspect the reason is that causality in the classical sense hasn't been found to be useful &lt;em&gt;enough&lt;/em&gt; for Physics and Engineering to justify the extra baggage. It's enough for practical purposes to construct good models for the statistics of experimental data, and the most effective mathematics for doing this is the mathematics of Hilbert spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interpretations of quantum theory exist &amp;mdash;such as the de Broglie-Bohm and the Nelson stochastic interpretation, but there are various others&amp;mdash; that will more-or-less fulfil what looks rather like a wish for classical causality in your Question, but they add a relatively awkward layer of mathematics that doesn't, for most Physicists, add enough insight or ability to do better Physics, Mathematics, or Engineering. There are other detailed reasons why not many physicists use these interpretations, particularly concerning Special Relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can model an experiment in a way that is as close to classical as we feel like having, while staying in the quantum theory fold, by introducing increasingly detailed models of every part of an experimental apparatus, instead of just modeling small numbers of electrons or photons, etc. This is, very loosely, called &quot;moving the Heisenberg cut&quot;. This is quantum theory's version of &quot;hidden variables&quot;; model refinement happens, it's just not done by introducing classical hidden variables, it's done by introducing quantum mechanical variables that weren't in the less refined models. Consequently, there's no strong need to introduce classical hidden variables, even though it's not impossible to do so. This is loosely tied to a currently popular interpretation of what happens in quantum mechanical measurements, &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_decoherence&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;decoherence&lt;/em&gt;&lt;/a&gt;, which says that classical properties emerge because of the huge numbers of degrees of freedom that surround quantum systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should understand that I've insinuated some unconventional views in this presentation of the conventional view. Also, there are more people in the Physics community than you might think who have committed their whole working lives trying to find good reasons why the Physics community should do Physics more in a way that, crudely, Einstein would approve of, so far without much success. New approaches and arguments emerge fairly regularly, are considered seriously by people who would jump if they could see practical advantage in it, and are found not good enough. The de Broglie-Bohm approach is essentially a product of the 50s, the Nelson approach is essentially a product of the 60s, from the 80s we have the GRW approach (apologies to anyone whose favorites I've missed out); it's more difficult to give a name to an approach from the 90s and 00s that people might cite in 20 years time, but there are many candidates, however none of them is obviously more &lt;em&gt;useful&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Try looking at all the interpretations listed on the Wikipedia page &lt;a href=&quot;http://en.wikipedia.org/wiki/Interpretations_of_quantum_mechanics&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Interpretations_of_quantum_mechanics&lt;/a&gt;, and think whether there's a way in which any of them helps to make the construction of models more practical. The Wikipedia presentations are not necessarily the best available, but the best presentations are probably not in a different ballpark as far as the level of complexity involved is concerned. It's OK to be annoyed that there's not anything better available, but constructing something better is real hard. Not a lot different from the situation for large-scale software, perhaps, where the balance of features and simplicity is also hard to master. Wanna displace Microsoft? We have to go ahead and try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT(2): In response to your EDIT,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If a particle has 30% chance of being &quot;here&quot;, and 70% chance of being&#xA;  &quot;there&quot;, then I would assume that there's some underlying reason that&#xA;  determines where the particle is. But the prevalent view (as I understand&#xA;  it) is that there's no underlying reason, the particle just happens to&#xA;  sort of &quot;choose&quot; to be &quot;here&quot; 30% of the time and &quot;there&quot; 70% of the time&#xA;  with no particular reason.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As I see it, this introduces complications, because what the conventional view is depends on whether you put a question in terms of particles or in terms of quantum fields. Consider this restatement, which changes the single reason that you invoke to cause events, a particle, into a confluence of multiple or a continuum of reasons, which is much more appropriate to a field,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If there is a 30% chance of observing an event &quot;here&quot;, and a 70% chance of&#xA;  observing an event &quot;there&quot;, then I would assume that there are underlying&#xA;  reasons that determine whether we observe an event &quot;here&quot; or &quot;there&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is rather too little interpretation of quantum field theory, but it's to QFT that Physicists retreat when people press for details. The state of the quantum field describes where we can expect to see &lt;em&gt;events&lt;/em&gt; (OK, to stay in the conventional I should say particles, so you can stop reading now, but it's fairly widely understood that the concept of a particle in QFT is theoretically problematic, whereas statistics of recorded events are measured) when we use a particular experimental apparatus. Now, if you want a reason why events happen &quot;here&quot; or &quot;there&quot;, you can, if you're careful to understand that a quantum field is an operator-valued distribution, not a classical field, say that the statistics we observe are because of the quantum field, which is everywhere between the preparation apparatus and the events in the measurement apparatus. Art Hobson is the only person I know who has published on this, although some of my papers falteringly hint towards this kind of approach. Art's papers are not perfect, but they're available &lt;a href=&quot;http://physics.uark.edu/hobson/papers.html&quot; rel=&quot;nofollow&quot;&gt;on his web-site&lt;/a&gt;, try &quot;Teaching quantum physics without paradoxes&quot; and try others if you like that. Brigitte Falkenburg's book, &lt;a href=&quot;http://www.springerlink.com/content/978-3-540-33731-7&quot; rel=&quot;nofollow&quot;&gt;Particle Metaphysics&lt;/a&gt; may be too much Philosophy for most people, but I find it a very good counterpoint to Art Hobson.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This says only that the &lt;em&gt;statistics&lt;/em&gt; of the events are caused by the quantum field. I'm ambivalent about this, I'd prefer to say only that the quantum field describes or models the statistics of the events, but you do what you like. If you want to know what causes &lt;em&gt;individual&lt;/em&gt; events, then I'm going to leave you on your own. I &lt;em&gt;think&lt;/em&gt; something more satisfactory than deBB-type trajectories for fields can be done in the field context, but I haven't done it. Actually, I've been trolling around here at PhysicsSE for a week or so while I recoup my spirits for a new thrash at issues that halted me a few weeks ago (I've been 20 years at it, so don't hold your breath). My approach is certainly not conventional, and you can find so many other approaches out there that there is no reason at all to read what I have to say about it. If we stay with conventional Physics, however, as you see in some of the other Answers, we can hardly engage with your Question at all.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-03-22T03:25:22.147" LastActivityDate="2011-03-22T03:25:22.147" CommentCount="6" />
  <row Id="7368" PostTypeId="2" ParentId="7317" CreationDate="2011-03-21T15:51:37.423" Score="0" ViewCount="" Body="&lt;p&gt;They do in fact mix, although because of the time scale it may be reasonable mental simplification to treat them as if they don't. Meterologists speak of frontogenesis, and frontolysis, which is the creation and death of weather fronts. Fronts as ideal stepfunction changes in air properties don't actually exist, but steep gradients do. Usually there is some dynamic inhibiting mechanical mixing, such as denser air underneath lighter air. But even in that case, you would still have diffusion working to lessen the gradients.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-03-21T15:51:37.423" CommentCount="2" />
  <row Id="7369" PostTypeId="2" ParentId="7359" CreationDate="2011-03-21T15:54:37.797" Score="3" ViewCount="" Body="&lt;p&gt;There are much better people to answer this question, but I will answer it from the point of view of an experimental particle physicist, the way I have filed this knowledge in my head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all the &lt;a href=&quot;http://en.wikipedia.org/wiki/Big_Bang&quot; rel=&quot;nofollow&quot;&gt;big bang is a particular solution&lt;/a&gt; within the theory of &lt;a href=&quot;http://en.wikipedia.org/wiki/General_relativity&quot; rel=&quot;nofollow&quot;&gt;general relativity,&lt;/a&gt; which treats gravity as a deformation of space time around massive bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This theory differs from the Newtonian and special relativity world we have explored and are familiar with. For  these  we have definitions of energy and also conservation of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the general relativity framework there exists energy but the concept of conservation of energy is not valid. Energy can appear according to the appropriate equations. General relativity is necessary in the description  of observations of large dimensions in space and time and very massive bodies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the big bang in this theory happens from one point, as an explosion as shown in the links I gave you. All points in our universe are this point. Practically metaphysical but true. So there is no meaning on &quot;what it is expanding to&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A useful for me analogy is the surface of a balloon. Our universe, in smaller dimensions, is like the &lt;em&gt;surface&lt;/em&gt; of an expanding   balloon. All points on the surface are receding from each other and as far as the surface is concerned, it just expands. Others like the analogy of a cake with fruit interspersed. As it rises in the oven each piece of fruit recedes from the others and if we could think that all of it could be originating from  a point, all points in the expanded cake would  be the &lt;em&gt;beginning&lt;/em&gt; point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for when the big bang happened, read the links I gave you.Astronomy has measured this.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-21T16:14:58.310" LastActivityDate="2011-03-21T16:14:58.310" />
  <row Id="7370" PostTypeId="2" ParentId="7317" CreationDate="2011-03-21T16:07:17.890" Score="0" ViewCount="" Body="&lt;p&gt;To answer this question, one had to give You a comprehensive treatment of common sense &#xA;and basic physics of gases and some metereology &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with a counter-question: Why do You use a spoon to mix the milk with Your coffee?&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-21T16:33:47.307" LastActivityDate="2011-03-21T16:33:47.307" CommentCount="3" />
  <row Id="7371" PostTypeId="2" ParentId="7359" CreationDate="2011-03-21T16:07:54.490" Score="0" ViewCount="" Body="&lt;p&gt;Energy conservation is funny with general relativity.  To have a conservation law you need something which is a constant of the motion.  In the case of energy in spacetime this requires there be a Killing vector that is constant along all timelike geodesics.  This isometry is what defines the conservation of energy.  Spacetime metrics for cosmologies do not admit a Killing vector which is constant along timelike geodesics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe expands as a manifold.  The spacetime is foliated by spatial surfaces, where points on that surface increase their distance along the time direction of the foliation.  If the universe is flat these spatial surfaces are just $R^3$ spaces, and the curvature of spacetime is defined by their extrinsic curvature in the space plus time = spacetime.  The universe is not particularly expanding into anything outside of itself. &lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-21T16:07:54.490" CommentCount="1" />
  <row Id="7372" PostTypeId="1" CreationDate="2011-03-21T16:09:59.590" Score="7" ViewCount="280" Body="&lt;p&gt;Is there a classification of &lt;em&gt;all&lt;/em&gt; possible BPS configurations in string theory? It has to include Calabi-Yau orbifolds, intersecting D-branes, coincident D-branes, etc. . To simplify matters, take the limit of zero string coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2693" LastActivityDate="2011-09-01T15:38:51.477" Title="Classifying all possible BPS configurations in string theory" Tags="&lt;string-theory&gt;&lt;supersymmetry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="7373" PostTypeId="2" ParentId="7359" CreationDate="2011-03-21T16:44:34.227" Score="0" ViewCount="" Body="&lt;p&gt;Only a physics undergrad here, so I only know of this stuff cause ive done a bit of extra reading, perfectly possible that I misunderstood it. But you can &quot;create&quot; energy/matter, the catch being that you must also make a matching amount of anti matter/energy, resulting in a net gain of 0 energy/matter. Something I saw before I learned this that kind of started my understanding of it was this video on Hawking radiation &lt;a href=&quot;http://www.youtube.com/watch?v=S6srN4idq1E&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=S6srN4idq1E&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understood the reading correctly, this is a possible mechanism of creating the universe (particle/anti particle mechanism). This apparently happens all the time in Quantum Mechanics, but on a much smaller scale, and apparently the only way to use this effect to create something that lasts more then a fraction of a second is to have it create a universe worth of matter/energy and matching anti matter/energy. What makes this so hard to get a definite understanding is the fact that you have to apply general relativity as well as Quantum field theory, and the 2 can sometimes disagree (I don't have enough understanding of either to say exactly what the disagreement is, I just know there is apparently one)&lt;/p&gt;&#xA;" OwnerUserId="2350" LastActivityDate="2011-03-21T16:44:34.227" CommentCount="3" />
  <row Id="7374" PostTypeId="2" ParentId="7220" CreationDate="2011-03-21T17:11:11.863" Score="2" ViewCount="" Body="&lt;p&gt;I shall build up links piece by piece as I find them in this answer. This paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.springerlink.com/content/v07n5l128613x5g0/&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/v07n5l128613x5g0/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from 1989 is suggesting that there is an inadequacy in the Nelson approach. I will need to follow that up now.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-03-21T17:11:11.863" CommentCount="1" />
  <row Id="7375" PostTypeId="2" ParentId="7346" CreationDate="2011-03-21T17:15:55.803" Score="2" ViewCount="" Body="&lt;p&gt;I seem to remember Dr Karl on the Triple J radio station (Karl Kruszelnicki) explaining this one. In a zero friction environment, you would oscillate back and forth. It would take about 40 minutes to travel from one side of the earth to the other. An interesting side note was that if you were to dig the hole from any point in the Earth to any other point, it would still take the same amount of time to travel from one side to the other.&lt;/p&gt;&#xA;" OwnerUserId="1910" LastActivityDate="2011-03-21T17:15:55.803" CommentCount="1" />
  <row Id="7376" PostTypeId="2" ParentId="6220" CreationDate="2011-03-21T17:42:31.040" Score="-1" ViewCount="" Body="&lt;p&gt;This is a question which might be best for “Click and Clack” on “Car Talk.”  The question is a bit unclear on a couple of things.  So I will assume that the path length on the straight road and the hilly road is the same.  I will also assume that the average velocity of the drive on the two roads is the same.  Clearly if the straight road is much longer then the hilly road would be more economical.  So suppose you drove the hilly road and you take your foot off the fuel on the descending portions of the drive.  The idea is then that the potential energy you “banked” in driving up is converted to kinetic energy, and your gas mileage on the trip down is near “infinity.”  If you use the brakes a lot on the way down, which might be advised for safety, then some of that energy is dissipated as heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the two be equal?  I don’t think so.  The reason is thermodynamics, for a good chuck of the energy used to climb the hills is lost as heat, so you recover a rather small amount of that energy as kinetic energy on the way down.  We might think of the thermodynamics of the hilly drive as thermodynamic losses of the straight drive plus thermodynamic losses which are incurred in raising the potential energy of the car on the various hills.  This is even more so if the engine idles on the way down, which is advised with modern cars.  An old VW bug might save a bit on saving this energy loss with old fashioned brakes.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-03-21T19:08:10.387" LastActivityDate="2011-03-21T19:08:10.387" CommentCount="5" />
  <row Id="7377" PostTypeId="1" CreationDate="2011-03-21T17:51:19.553" Score="0" ViewCount="489" Body="&lt;p&gt;While not directly a physics question, I can't think of forum better capable of answering my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In discussions over Japan's nuclear reactor situation the observation was made that reactors there and in the United States are built on shorelines or close to large bodies of water. Is this purely for the &quot;backup backup backup&quot; cases where immediate water is necessary or are there more reasons justifying these locations?&lt;/p&gt;&#xA;" OwnerUserId="1586" LastActivityDate="2011-03-24T18:18:11.350" Title="Why build nuclear reactors on shorelines?" Tags="&lt;nuclear-physics&gt;&lt;nuclear-engineering&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="7378" PostTypeId="2" ParentId="7358" CreationDate="2011-03-21T17:55:02.107" Score="1" ViewCount="" Body="&lt;p&gt;In order to test the speed and position of a particle, you need to hit it with a photon. In doing so, you change the outcome of where the particle is, where it's going and its speed. Only by not measuring it does the particle maintain its true characteristics. So, in this way, it is not merely the inaccuracy of the detector which causes the uncertainty, but the act of measuring in itself.&lt;/p&gt;&#xA;" OwnerUserId="2694" LastActivityDate="2011-03-21T17:55:02.107" CommentCount="4" />
  <row Id="7379" PostTypeId="2" ParentId="7377" CreationDate="2011-03-21T18:08:35.183" Score="1" ViewCount="" Body="&lt;p&gt;As Deepak Vaid said above:&lt;br /&gt;&#xA;&lt;b&gt;WATER&lt;/b&gt; especially with older BWR reactors having lots of water nearby proves to be very useful for keeping it running cool.&lt;/p&gt;&#xA;" OwnerUserId="2590" LastActivityDate="2011-03-21T18:08:35.183" />
  <row Id="7380" PostTypeId="2" ParentId="7377" CreationDate="2011-03-21T18:21:13.550" Score="2" ViewCount="" Body="&lt;p&gt;You need a reliable suppy of cooling water to operate the turbine cycle. &#xA;(Think of Carnot-cycle, and for emergency cooling as well). &#xA;Steam is not the problem, the water used to make steam is condensed and recirculated. &#xA;This boiler water has to be very &quot;pure&quot; in a sense. &#xA;This means You need a real river or sea nearby. &#xA;Japan as a kind of island(s) nowwhere has rivers big enough. &#xA;To feed rivers You need some landmass, where enough rain falls and &#xA;where water is stored in the underground for the dry season.   &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-21T18:37:02.820" LastActivityDate="2011-03-21T18:37:02.820" />
  <row Id="7381" PostTypeId="2" ParentId="7377" CreationDate="2011-03-21T18:40:08.880" Score="4" ViewCount="" Body="&lt;p&gt;In addition to the above it might be noticed that coal fired power stations are often inland despite their similar requirement for lots of water.  The difference being in the weight of fuel required to run the power station.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A significant desirable fact of nuclear power is the very high energy density of the fuel compared to coal.  So we have more flexibility in where it is economic to build, so the second factor of water supply becomes dominant and siting by a coast helps with this expense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I guess that in the worst case scenario there is a ~50% chance any radioactive cloud will blow away from civilian populations.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-03-21T18:40:08.880" CommentCount="4" />
  <row Id="7382" PostTypeId="2" ParentId="6157" CreationDate="2011-03-21T19:05:35.790" Score="3" ViewCount="" Body="&lt;p&gt;The &lt;strong&gt;&lt;a href=&quot;http://freescience.info/Physics.php&quot; rel=&quot;nofollow&quot;&gt;FreeScience&lt;/a&gt;&lt;/strong&gt; website links to a large listing of free books on a wide range of physics, math, and other fields of science. &lt;/p&gt;&#xA;" OwnerUserId="2696" LastEditorUserId="2809" LastEditorDisplayName="" LastEditDate="2011-03-28T04:00:55.543" LastActivityDate="2011-03-28T04:00:55.543" CommunityOwnedDate="2011-03-21T19:05:35.790" />
  <row Id="7383" PostTypeId="2" ParentId="7377" CreationDate="2011-03-21T19:22:33.217" Score="0" ViewCount="" Body="&lt;p&gt;Also, access for shipping heavy components is easier. This is probably important for minimizing construction costs. Properly sited, a plant could still be near the shore, but be high enough up to be out of reach of any tsunami.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-03-21T19:22:33.217" CommentCount="5" />
  <row Id="7384" PostTypeId="2" ParentId="6581" CreationDate="2011-03-21T19:39:20.207" Score="-3" ViewCount="" Body="&lt;p&gt;The &lt;strong&gt;Earth&lt;/strong&gt; is rotating and has a permanent Magnetic field.&#xA;It does not exhibits a correspondent Electric field.&#xA;The answer is 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnetic field is due to the presence of a relative motion between the charges inside the Earth (electric dipole) and the observer. Thus, it is a derived quantity much like the Coriolis force.&#xA;Instead of saying magnetic &lt;em&gt;field&lt;/em&gt; I think more appropriate to say magnetic &lt;em&gt;force&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by Hans de Vries:&#xA;&lt;a href=&quot;http://physics-quest.org/Magnetism_from_ElectroStatics_and_SR.pdf&quot; rel=&quot;nofollow&quot;&gt;The simplest, and the full derivation of Magnetism as a Relativistic side efect of ElectroStatics&lt;/a&gt;&#xA;He uses only Electrostactic field and the non-simultaneity to derive the Magnetic Field.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-03-23T01:26:23.963" LastActivityDate="2011-03-23T01:26:23.963" CommentCount="8" />
  <row Id="7385" PostTypeId="1" AcceptedAnswerId="7389" CreationDate="2011-03-21T20:30:46.873" Score="5" ViewCount="115" Body="&lt;p&gt;I have the Particles Physics Booklet and I noticed that the resonances that decay into a nucleon and pion are indicated by an abbreviation. For example $P_{33}$ is associated to the $\Delta (1232)$ resonance. What is the meaning of this symbol?&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2011-03-21T22:18:26.407" Title="Nomenclature of hadronic resonances" Tags="&lt;nuclear-physics&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="7386" PostTypeId="1" AcceptedAnswerId="7387" CreationDate="2011-03-21T20:57:54.340" Score="2" ViewCount="185" Body="&lt;p&gt;Do they create energy? Or do they just disappear with zero energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they create energy when disappearing, that means it takes energy to create them, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they disappear into zero energy, could that be a clue about how to create matter from nothing?&lt;/p&gt;&#xA;" OwnerUserId="2690" LastActivityDate="2011-03-22T12:29:26.230" Title="When matter and anti-matter collide" Tags="&lt;energy&gt;&lt;energy-conservation&gt;&lt;antimatter&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="7387" PostTypeId="2" ParentId="7386" CreationDate="2011-03-21T21:04:22.460" Score="8" ViewCount="" Body="&lt;p&gt;The do &lt;strong&gt;not&lt;/strong&gt; disappear with zero energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their energy (both that originating from rest mass and any kinetic energy) appears &lt;em&gt;somehow&lt;/em&gt;. As photons, a spray of other (lighter) particles, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, when an &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron%E2%80%93positron_annihilation&quot; rel=&quot;nofollow&quot;&gt;electron meets a positron (that is, a anti-electron)&lt;/a&gt; the most common result is a pair of gamma rays each of 511 keV (in the center of momentum (CoM) frame of the $e$--$e^+$ pair). It has to be at least two because you have to conserve both energy and momentum (and angular momentum, too, but that is a complication we will ignore for the moment), and that also tells us that there have opposite momenta in the CoM frame.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-03-21T21:04:22.460" />
  <row Id="7388" PostTypeId="1" AcceptedAnswerId="7394" CreationDate="2011-03-21T21:36:11.310" Score="6" ViewCount="168" Body="&lt;p&gt;With the current QED framework, If an electric field is strong enough (say, near a nucleus with $Z &gt; 140$) , pair production will occur spontaneously? Is this a real effect or an artifact before renormalization is carried out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how can energy be conserved in such scenario?&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-03-22T13:29:58.943" Title="critical electric field that spontaneously generates real pairs" Tags="&lt;qed&gt;&lt;pair-production&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="7389" PostTypeId="2" ParentId="7385" CreationDate="2011-03-21T22:15:19.730" Score="7" ViewCount="" Body="&lt;p&gt;$\Delta$ represents the constituents (up and down quarks, three in total) in the spin and isospin 3/2 state, the number (1232) represents its the mass of the resonance in MeV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the notation $P_{33}$ -- this is an example of the notation $L_{2I,2J}$. $I$ stands for the isospin ($I=3/2$ in this case) and $J$ stands for the total angular momentum ($J=3/2$) of the resonance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The letter $L$ takes the 'values' $S$, $P$, $D$, $F$, $G$, $H$,.. (which corresponds to $L=0,1,2,...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It represents the angular momentum state of the nucleon-pion pair to which the resonance decays. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://jazz.physik.unibas.ch/~krusche/resonance_basics.html&quot; rel=&quot;nofollow&quot;&gt;http://jazz.physik.unibas.ch/~krusche/resonance_basics.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1287" LastActivityDate="2011-03-21T22:15:19.730" />
  <row Id="7390" PostTypeId="2" ParentId="7385" CreationDate="2011-03-21T22:18:26.407" Score="3" ViewCount="" Body="&lt;p&gt;For baryons, the notation $P_{33} = L_{2I\;2J}$ indicates the symmetry of this set of particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;L&quot; gives the angular momentum. This counts up as &quot;S&quot;, &quot;P&quot;, &quot;D&quot;, &quot;F&quot;. In the case of $P_{33}$ we have a P which indicates the angular momentum is 1 (or $\hbar$ if you prefer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;I&quot; values is 3/2. This is the &quot;isospin&quot; quantum values. The 3/2 says that there are four values possible, 3/2, 1/2, -1/2, and -3/2. Isospin is the symmetry that relates up and down quarks. Up quarks have charge +2/3 while down quarks have charge -1/2. So the particle comes in four isospin states: $\Delta^{++},\Delta^+,\Delta^0,\Delta^-$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;J&quot; is for total angular momentum. So if you measured the total angular momentum you could get +3/2, +1/2, -1/2, or -3/2.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-21T22:18:26.407" />
  <row Id="7391" PostTypeId="1" AcceptedAnswerId="7392" CreationDate="2011-03-22T00:40:42.190" Score="3" ViewCount="182" Body="&lt;p&gt;I'm assuming the LHC can create a Higgs and an anti-Higgs boson. If so, would their fields be identical with respect to mass effects? How would LHC detectors distinguish between the two bosons?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-03-22T01:39:11.037" Title="Difference between Higgs and anti-Higgs Fields" Tags="&lt;particle-physics&gt;" AnswerCount="1" />
  <row Id="7392" PostTypeId="2" ParentId="7391" CreationDate="2011-03-22T01:32:14.990" Score="6" ViewCount="" Body="&lt;p&gt;The Higgs is a &lt;strong&gt;real&lt;/strong&gt; scalar field, so there's &lt;strong&gt;no&lt;/strong&gt; &quot;anti-Higgs&quot; particle. All imaginary part of initial complex doublet are absorbed by the weak gauge bosons (Ws and Z), only a real scalar field remains after this.&lt;/p&gt;&#xA;" OwnerUserId="1545" LastEditorUserId="1545" LastEditorDisplayName="" LastEditDate="2011-03-22T01:39:11.037" LastActivityDate="2011-03-22T01:39:11.037" CommentCount="4" />
  <row Id="7393" PostTypeId="2" ParentId="7372" CreationDate="2011-03-22T02:35:36.667" Score="-1" ViewCount="" Body="&lt;p&gt;This presentation is based on a reading of &quot;Four-qubit entanglement from string theory&quot; L. Borsten, D. Dahanayake, M. J. Duff, A. Marrani, W. Rubens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1005/1005.4915v2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/arxiv/pdf/1005/1005.4915v2.pdf&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exist mathematical correspondences between black holes and entanglement theory.  3-Q-bit theory is related to BPS black holes and 4-Q-bit entanglements are related to extremal black holes.  And this leads to STU theory of coset QFTs with entanglement and entropy correspondences.  The Stochastic Local Operations and Classical Communication (SLOCC) is a relationship between states in entanglement and teleportation.  Two states are SLOCC related by a teleporation if they can be inter-converted to each other in a reversible manner with some probability of success.  This uses group theory, where the group $G_{SLOCC}$ for this process is an $N$-partite system of q-bits with some group $GL(2,~C)$.  The states further transform as a $(2~2,~...,~2)$.&#xA;$$&#xD;&#xA;G_{SLOCC}~=~SL(2,~C)_1\otimes SL(2,~C)_2\otimes\dots\otimes SL(2,~C)_N,&#xD;&#xA;$$&#xA;which is the group that acts on the moduli space of black holes and is the U duality group.  A composite state is then,&#xA;$$&#xD;&#xA;|\psi_{12...N}\rangle~=~SL(2,~C)_1\times SL(2,~C)_2\times\dots\otimes SL(2,~C)_N|\phi_{12...N}\rangle&#xD;&#xA;$$&#xA;So this is an $N$-partite quantum information system where the entanglements are determined by the group element $G_{SLOCC}$ and polynomials of this group.  This is the moduli space for black holes composed of Q-bits and the U-duality group.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a $2$ Q-bit system this construction is apparent.  You have a state of the form $\sum_{ij}a_{ij}|i,j\rangle$ for $i$ and $j$ running form $0$ to $1$.  The elements $a_{ij}$ transform as $(2,~2)$ of the $G_{SLOCC}$.  The invariant element is the determinant of these matrices so $det(a_{ij})$ transformed under the $G_{SLOCC}$ into&#xA;$$&#xD;&#xA;det(a_{ij})~\rightarrow~det(a&amp;#39;_{ij})~=~det(U_{i&amp;#39;i}a_{ij}U&amp;#39;^\dagger_{j&amp;#39;j})~=~det(a_{ij})&#xD;&#xA;$$&#xA;with the obvious result on the determinant of a product that the transformation elements have unit determinant.  The entanglement entropy is given by this measure so $S_{ij}~=~4|det(a_{ij})|^2$.  For multipartite systems the same rule generally applies, but the matrix interpretation is different. For an $N$-partite system the entanglement entropy is given by a $2\times 2\times\dots\times 2$ (N times) set of elements.  This leads to the entangled states $|00\rangle~+~|11\rangle$ and $|01\rangle~+~|10\rangle$ (without normalization) for singlet and triplet entangled states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a $3$ q-bit system the amplitudes are elements $a_{ijk}$ --- a $2\times 2\times 2$ element that is not a matrix, which has no diagonalization procedure.  We then have to focus on invariants.  The determinant is replaced by a hyperdeterminant that transforms as a $(2,~2,~2)$, and there are elements $\sigma_i,~\sigma_j$, and $\sigma_k$, co-invariants, that transform as $(3,~1,~1),~ (1,~3,~1)$ and $(1,~1,~3)$ of the $G_{SLOCC}$.  So these four then construct entanglement measures $S_{ijk},~ S_{ij}$, $S_{ik}$ and $S_{jk}$, for the GHZ state and the bipartite elements which for a W-state.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a $3$ Q-bit system we focus on invariants.  The determinant is replaced by a hyperdeterminant that transforms as a $({\bf 2},~{\bf 2},~{\bf 2})$, and there are elements $\sigma_i,~ \sigma_j$ and $\sigma_k$, co-invariants, that transform as $({\bf 3},~{\bf 1},~{\bf 1})$ $({\bf 1},~{\bf 3},~{\bf 1})$  and $({\bf 1},~{\bf 1},~{\bf 3})$ of the $G_{SLOCC}$.  These four construct entanglement measures $S_{ijk}$, $S_{ij}$, $S_{ik}$ and $S_{jk}$.  $S_{ABC}$ is a tri-partite entanglement entropy with&#xA;$$&#xD;&#xA;S_{ABC}~=~S_{A(BC)}~-~S_{AB}~-~S{AC}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bipartite elements are pair entanglements and the tripartite involves a triplet which is entanglement.  So $B~=~Bob$ is maximally entangled with $A~=~Alice$ and $C~=~Carl$ in the tripartite state. Where if I trace out $A$ $B$ or $C$ there is a complete mixed state, classical information due to tracing, and no correlation, while with the three bipartite entanglements Bob's entanglement is with $A$ and $C$, and a tracing out of  his quantum state entanglement continues to have the $A-C$ entanglement.  The tripartite entanglement corresponds to a large black hole, while the set of bipartite entangled states a small black hole.  The tripartite state is also called the GHZ state for Greenberger, Horne, Zeilinger formulation of a three particle entanglement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The N-partite entanglement is different from the standard bipartite entanglement.  These correspond to three separable state, bipartite states plus one separated, nonseparable bipartite states and a tripartite GHZ state.  These states in black hole logic correspond to small black holes $1/2$ supersymmetric, small black holes with two BPS charges and $1/4$ supersymmetric, two black holes with $3$ charges and $1/4$ supersymmetric and, the GHZ state is $4$ charges and $1/8$ supersymmetric.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-03-22T02:45:03.210" LastActivityDate="2011-03-22T02:45:03.210" CommentCount="8" />
  <row Id="7394" PostTypeId="2" ParentId="7388" CreationDate="2011-03-22T03:00:00.960" Score="9" ViewCount="" Body="&lt;p&gt;The Schwinger pair production can be understood as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a constant electric field E in some region of space, pointing in the x direction. This is created for example by a large capacitor. Inside this capacitor, the energy of an electron-positron pair separated by some distance $\Delta x$ is $V(\Delta x)= 2m- q E \Delta x$, where $m$ is the mass of the electron (and positron) and $q$ is its charge. The first term is the rest energy of two massive particles, and the second is their potential energy in the presence of the electric field. You can see that for large enough separation the total energy is negative - it becomes energetically favourable to have a pair present instead of an empty capacitor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process can be thought of as tunnelling: the configuration of empty space and an electron-positron pair are separated by a barrier $V(\Delta x)$. You therefore have an exponentially small probability to create pairs at the classical turning point $V(\Delta x)=0$, or $\Delta x = \frac{2m} {qE}$. As always in tunnelling process, energy is conserved - it is zero before and after the tunnelling in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the particles are created, they accelerate away from each other, and eventually end up neutralizing in part the capacitor, in other words reducing the electric field. Note that there is no critical electric field - pair creation occurs for arbitrarily small electric field, though the probability is exponentially suppressed, roughly as $\exp(- \pi m^2/ q E)$. The derivation of this formula by Schwinger (and corrections to all orders) is a real joy to see, I'd recommend at least for theorists to take a look at the &lt;a href=&quot;http://prola.aps.org/abstract/PR/v82/i5/p664_1&quot; rel=&quot;nofollow&quot;&gt;original paper&lt;/a&gt;. This may well be the first use of instanton methods in quantum mechanics, though I'm far from being an expert on the history.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no real connection of this to renormalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Variants of this calculation are useful in cosmology and QFT in curved spacetime, for example for particle production by time varying backgrounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, the effect has never been observed due to difficulties creating large enough electric field. I may be wrong on that as well though.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-03-22T03:20:40.203" LastActivityDate="2011-03-22T03:20:40.203" CommentCount="2" />
  <row Id="7395" PostTypeId="1" CreationDate="2011-03-22T04:32:34.867" Score="3" ViewCount="236" Body="&lt;p&gt;I've heard that regular light bulbs with a filament are better for the eyes. Is the spectrum of one worse than the other? If so, are there any regulations for their use in industrial settings for worker safety?&lt;/p&gt;&#xA;" OwnerUserId="2700" LastActivityDate="2011-07-14T01:11:22.583" Title="Are regular light bulbs better for the eyes than CFLs or &quot;tube lights&quot;?" Tags="&lt;light&gt;&lt;electromagnetic-radiation&gt;&lt;vision&gt;&lt;biology&gt;" AnswerCount="1" />
  <row Id="7396" PostTypeId="1" AcceptedAnswerId="7405" CreationDate="2011-03-22T04:41:42.987" Score="5" ViewCount="262" Body="&lt;p&gt;Einstein's field equations are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_{ab} - {1 \over 2}g_{ab}\,R + g_{ab} \Lambda = {8 \pi G \over c^4} T_{ab}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And since the Ricci curvature tensor is &quot;less information&quot; than the Riemann curvature tensor because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_{ab} = R^c{}_{acb}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the Riccie curvature scalar is even &quot;less information&quot; than the Ricci curvature tensor because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R = R^a{}_{a}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This appears to indicate the GR field equations don't contain enough information to specify how the full Riemann curvature tensor evolves.  So it looks like something missing.  Is it even possible to say how the full Riemann curvature tensor evolves in GR?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Riemann curvature tensor can be obtained from the metric, if we can solve for how the metric evolves using the Ricci curvature tensor, we can then get the Riemann curvature tensor evolution.  So the equation is equivalent to the title question, what all do we need to solve for the metric in GR?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In books where they obtain the solution outside a static spherical object, they seem to always refer back to the Newtonian limit and compare to Newtonian gravity to fully fix the answer.  This seems a bit scary, as it seems to suggest indeed that GR needs some other equations specified to get the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I choose a coordinate system and specify the metric everywhere at some initial time, is that initial metric + the GR field equations enough to solve for the metric everywhere in spacetime? Or is there some way to use GR to get the metric without any prior geometry put in?&lt;/p&gt;&#xA;" OwnerUserId="2655" LastActivityDate="2011-03-22T09:08:35.400" Title="What all is needed to solve for the metric in GR?" Tags="&lt;general-relativity&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="7398" PostTypeId="1" CreationDate="2011-03-22T05:41:32.153" Score="3" ViewCount="324" Body="&lt;p&gt;So in physics, one smart teacher told me, &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;V = IR&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I = V / R&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;but it is not always true, because &lt;code&gt;P = VI&lt;/code&gt; and each power supplying device (such as battery) has a &quot;maximum power&quot; it can provide.  So &lt;code&gt;I = V / R&lt;/code&gt; is true as long as &lt;code&gt;P = VI&lt;/code&gt; doesn't exceed the maximum power the battery can supply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So will old style AA batteries, Alkaline AA batteries, and NiMH, rechargeable alkaline, Lithium AA (Energizer) all have similar MAXIMUM POWER?  I am suspecting my Mario Kart Remote (fit in a wheel) for playing Mario Kart Wii not behave the same when using different batteries.  The best seems to be the Duracell Ultra Alkaline, and then the Duracell rechargeable pack for Wii remote (2.4V only) works also flawlessly (but can it be, because it is very snug?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, sometimes I turn left and do a drifting and the console thinks that I am drifting right... so it does seem that different batteries have different outcome...  and I am suspecting can the Wii need more power than some batteries can provide?  and so how do they work really?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I cannot add a tag for &lt;code&gt;batteries&lt;/code&gt; or &lt;code&gt;ohm's law&lt;/code&gt; because I am new.  Can somebody add it for me?)&lt;/p&gt;&#xA;" OwnerUserId="2701" LastEditorUserId="2701" LastEditorDisplayName="" LastEditDate="2011-05-01T16:33:05.530" LastActivityDate="2011-05-01T16:33:05.530" Title="Do Alkaline batteries, NiMH, rechargeable alkaline, Lithium AA, all have similar MAX POWER?" Tags="&lt;electricity&gt;&lt;power&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="7399" PostTypeId="1" AcceptedAnswerId="7400" CreationDate="2011-03-22T06:20:30.603" Score="3" ViewCount="235" Body="&lt;p&gt;When you come to a stop normally, the brakes don't pulse when you stop.  Since the car can only know its speed by the rotation of the wheels, how can it distinguish between the car is stopped normally and the car wheel's stopped rotating because they locked up?&lt;/p&gt;&#xA;" OwnerUserId="2702" LastActivityDate="2011-03-22T06:23:54.687" Title="How do anti-lock brakes know when to brake?" Tags="&lt;mechanics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="7400" PostTypeId="2" ParentId="7399" CreationDate="2011-03-22T06:23:54.687" Score="3" ViewCount="" Body="&lt;p&gt;You are correct that the car can't use the speedometer output for this.  Instead there are sensors on the brakes that measure how quickly the wheel rotation is decelerating.  If this is faster than the brakes can actually slow the car down, then it means the breaks are locking up and it is only slowing the wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information and images, check out this link: &lt;a href=&quot;http://auto.howstuffworks.com/auto-parts/brakes/brake-types/anti-lock-brake.htm&quot; rel=&quot;nofollow&quot;&gt;http://auto.howstuffworks.com/auto-parts/brakes/brake-types/anti-lock-brake.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2652" LastActivityDate="2011-03-22T06:23:54.687" CommentCount="2" />
  <row Id="7401" PostTypeId="1" AcceptedAnswerId="7429" CreationDate="2011-03-22T06:42:58.697" Score="8" ViewCount="179" Body="&lt;p&gt;Someone described to me the difficulty of numerically simulating turbulence as that as you look at smaller length scales you see more structure like you do in a fractal.  Searching on google for 'fractal turbulence' does seem to bring up quite a few hits.  But since fractals are self repeating as you go to smaller and smaller scales, doesn't this mean that we would already know all about the turbulence if we calculate it at one scale, and then add in the all the other scales assuming they are self similar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not a physicist, so please answer with simple language and not overly mathematical.&lt;/p&gt;&#xA;" OwnerUserId="2702" LastActivityDate="2011-03-22T22:07:50.267" Title="Fractal nature of turbulence" Tags="&lt;classical-mechanics&gt;&lt;turbulence&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="7402" PostTypeId="2" ParentId="7104" CreationDate="2011-03-22T07:42:40.033" Score="0" ViewCount="" Body="&lt;p&gt;I have an explanation for this. I would like to verify if that makes sense. &#xA;Basically my hypothesis is that whenever there is a temperature gradient, there will be a pressure gradient. In our case there is a temperature gradient between outside the balloon and inside the balloon. This leads to a pressure gradient. This pressure gradient (read pressure difference) causes the balloon to be stiff.&lt;br /&gt;&#xA;Please refer to &lt;a href=&quot;http://rejeev.blogspot.com/2011/03/temperature-and-pressure-gradient-in.html&quot; rel=&quot;nofollow&quot;&gt;http://rejeev.blogspot.com/2011/03/temperature-and-pressure-gradient-in.html&lt;/a&gt; for details.&lt;br /&gt;&#xA;I would like to seek feedback from the community on my hypothesis.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2011-03-22T07:42:40.033" />
  <row Id="7403" PostTypeId="2" ParentId="7086" CreationDate="2011-03-22T07:53:31.843" Score="1" ViewCount="" Body="&lt;p&gt;Best description of the problem of those I found is given by Nobel winner Vitaly Ginzburg. See, e.g. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ufn.ru/en/articles/1973/3/k/&quot; rel=&quot;nofollow&quot;&gt;http://ufn.ru/en/articles/1973/3/k/&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or (even better)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;VL Ginzburg, Theoretical Physics and Astrophysics (Pergamon, Oxford, 1979); other editions with somewhat different titles also available.&lt;/p&gt;&#xA;" OwnerUserId="2560" LastActivityDate="2011-03-22T07:53:31.843" />
  <row Id="7404" PostTypeId="2" ParentId="7398" CreationDate="2011-03-22T09:00:57.750" Score="0" ViewCount="" Body="&lt;p&gt;All batteries have &lt;a href=&quot;http://en.wikipedia.org/wiki/Internal_resistance&quot; rel=&quot;nofollow&quot;&gt;internal resistance&lt;/a&gt; which must be added to the R term for any circuit.  This value depends on the type of battery as well as its level of charge.  I found &lt;a href=&quot;http://www.ka7oei.com/nicds.html&quot; rel=&quot;nofollow&quot;&gt;this page about batteries&lt;/a&gt; which explains the effects in detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is NO, because P=V^2/R and R varies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Wii_Remote#Power_source&quot; rel=&quot;nofollow&quot;&gt;wiki page about the Wii Remote&lt;/a&gt; says that a &quot;3300µF capacitor provides a temporary source of power during quick movements of the Wii Remote&quot;.  This suggests that the device may have power requirements which exceed those of the battery; however even if the internal resistance of the battery is as high as 1 Ohm, this capacitor can be charged in 3.3ms, which is too small to account for your perceptions.  Maybe the errors are caused by low power but I don't think t = R C is telling the whole story.&lt;/p&gt;&#xA;" OwnerUserId="2072" LastEditorUserId="2072" LastEditorDisplayName="" LastEditDate="2011-03-23T07:57:20.463" LastActivityDate="2011-03-23T07:57:20.463" CommentCount="1" />
  <row Id="7405" PostTypeId="2" ParentId="7396" CreationDate="2011-03-22T09:08:35.400" Score="4" ViewCount="" Body="&lt;p&gt;Dear John, let me post the same thing that Marek has said as a standard answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Einstein's equations are not equations for the Riemann tensor because the Riemann tensor's components are not independent fields. Instead, Einstein's equations are differential equations for the metric tensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 4 dimensions, the metric tensor has 10 components - a symmetric tensor - and Einstein's equations have 10 components - a symmetric tensor - too. It doesn't matter that the Riemann tensor has 20 components because these 20 functions of space and time are calculated from the 10 component functions of the metric tensor and its (first and second) derivatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the 10-10 counting is oversimplified. Four &quot;differential combinations&quot; of Einstein's equations vanish identically because $\nabla_\mu R^{\mu\nu}=0$ is an identity (that always holds, even if the equations of motion are not satisfied). The same identity holds for the corresponding other tensors that are added to the Ricci tensor in Einstein's equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So instead of 10 equations, the Riemann equations are, in some sense, just 6 independent equations. That means that they don't determine the metric completely: they leave 4 functions undetermined and these are exactly the 4 functions that you may choose arbitrarily to specify a diffeomorphism, mapping one solution into another (equivalent) solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up to the coordinate transformations which are always allowed to be made, initial conditions for the metric and its first derivative determine the metric tensor - and therefore the whole Riemann tensor - everywhere in the future. One doesn't need any Newtonian equations as a &quot;mandatory supplement&quot; in general relativity. That doesn't mean that the Newtonian limit is unimportant: of course, it is one of the most important approximate consequences of general relativity.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-22T09:08:35.400" CommentCount="7" />
  <row Id="7407" PostTypeId="2" ParentId="7386" CreationDate="2011-03-22T12:29:26.230" Score="1" ViewCount="" Body="&lt;p&gt;Interaction like this produce anything that is allowed by conservation laws, which track the sum totals of various values (including quantum numbers).  An anti-matter version of a particle has the opposite/negative value of most conserved quantities.  So the result of the interaction might be almost any &lt;em&gt;other&lt;/em&gt; particle-antiparticle pair, so long as the total energy is available.  In particular the result can be a pair of photons: electron-positron collisions produce a pair of gamma rays carrying off the energy from the rest mass of the particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You questions about total energy are confused.  The total energy before and after a collision will always be the same.  Thus some particles have to exist after the collision.  For particle plus its anti-particle collisions there will be more than one resulting product because a single product by itself would have some non-zero quantum numbers.&lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2011-03-22T12:29:26.230" CommentCount="2" />
  <row Id="7409" PostTypeId="2" ParentId="7388" CreationDate="2011-03-22T13:29:58.943" Score="1" ViewCount="" Body="&lt;p&gt;I would like to complement rather than to answer: one can see the pair creation as a relaxation mechanism. For example, in a usual capacitor the charges are artificially separated and there is a potential energy of their interaction. In an ideal case of infinite dielectric resistance, the system is stable but in reality there is always a current (leakage) that serves to diminish the system potential energy. Similarly in QED: pair creation serves to neutralize the separated charges. Of course, the energy is conserved. In reality it is transformed into heat due to resistance of the dielectric, for example.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-03-22T13:29:58.943" CommentCount="2" />
  <row Id="7410" PostTypeId="1" CreationDate="2011-03-22T13:12:32.273" Score="3" ViewCount="246" Body="&lt;p&gt;Currently accepted scientific theory says that all elements heavier than hydrogen, helium, and a little lithium have been created in supernova explosions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions, specifically, is has anyone done the math to see if the observed amounts and the distribution of, the heavier elements in the universe agrees with the 13.7 billion year age of the universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the birth, life, and death of stars at the currently observed and accepted rates... does the math work out? Can you please refer me to a resource where I can review the material or get further clarification. (I freely admit that my maths aren't up to the task, but I'm hoping for a deeper understanding of the phenomenon.)&lt;/p&gt;&#xA;" LastActivityDate="2011-03-23T02:26:54.217" Title="Does the math work out for there being enough time for the formation of the heavier elements and their distribution as seen in today's universe?" Tags="&lt;astronomy&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="7411" PostTypeId="2" ParentId="7410" CreationDate="2011-03-22T13:53:20.930" Score="5" ViewCount="" Body="&lt;p&gt;Of course people have &quot;done the math&quot; on this. The chemical evolution of the Universe is a significant research area in astrophysics. There are tons of observations charting the evolution over time (by looking at different distances, we can measure things at different times), and tons of work on numerical models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could try &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521559588&quot; rel=&quot;nofollow&quot;&gt;this book&lt;/a&gt; for an entry point into the voluminous literature, if you're really interested. (It's about 15 years old, so there are probably more recent review articles, but your question of whether the timing can work out has been understood for far longer than that, so this book should be OK.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, and Edward's comment is quite right, although I think he should have said &quot;iron&quot; rather than &quot;lead.&quot; Elements from beryllium to iron are mostly produced during &quot;normal&quot; stellar evolution, not supernovae; elements heavier than iron are produced in supernovae.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-03-22T16:53:10.463" LastActivityDate="2011-03-22T16:53:10.463" CommentCount="3" />
  <row Id="7412" PostTypeId="1" CreationDate="2011-03-22T14:01:30.237" Score="6" ViewCount="369" Body="&lt;p&gt;Can anyone explain to me why we have had no major breakthroughs in the theory of quantum computation in the past 15 years?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shor's algorithm set the standard, since then we've had Grover's algorithm and an improvement on the Deutsch–Jozsa algorithm. What quantum algorithms of practical value have been created since Grover's?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes it so hard to create a quantum algorithm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2707" LastActivityDate="2011-07-15T21:19:18.210" Title="Recent breakthroughs in quantum computing?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-information&gt;&lt;quantum-computation&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="4" />
  <row Id="7413" PostTypeId="1" CreationDate="2011-03-22T14:27:24.647" Score="0" ViewCount="118" Body="&lt;p&gt;The &lt;a href=&quot;http://www.iso.org/sites/JCGM/VIM/JCGM_200e.html&quot; rel=&quot;nofollow&quot;&gt;ISO VIM&lt;/a&gt; defines them as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;measurement method:&lt;/strong&gt;&#xA;  generic description of a logical organization of operations used in a&#xA;  measurement.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;measurement procedure:&lt;/strong&gt; detailed description of a measurement according to one or more measurement principles and to a given measurement method, based on a measurement model and including any calculation to obtain a measurement result.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I'm still not sure what their difference is. Can anyone explain this to me, maybe with some examples?&lt;/p&gt;&#xA;" OwnerUserId="2688" LastActivityDate="2011-03-22T16:06:40.573" ClosedDate="2011-03-22T16:35:22.163" Title="What's the difference between &quot;measurement method&quot; and &quot;measurement procedure&quot;?" Tags="&lt;terminology&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7414" PostTypeId="2" ParentId="7398" CreationDate="2011-03-22T14:54:52.120" Score="0" ViewCount="" Body="&lt;p&gt;You can estimate that ALkaline ones have resistance 0.1 Ohm, NiMh 0.01 and LiIon 0.005, so in short pulses you should be able to get 33W, 144W and 450W of power in short pulses, but your mario cart should not eat THAT much. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BUT half-discharged Alkalines became 'weak' much faster(their internal resistance increases rapidly) than NiMh/LiIon, and LiIon is the very best by far.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-03-22T14:54:52.120" />
  <row Id="7415" PostTypeId="2" ParentId="7398" CreationDate="2011-03-22T15:47:42.603" Score="-1" ViewCount="" Body="&lt;p&gt;The maximum power of a battery depends on the internal resistance (which gives the maximum current you can draw) and the chemistry which determines how long you can draw that current for before the battery is damaged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You probably aren't interested in how much short circuit current you can draw from a battery for a millisec before it's destroyed - the maximum continuos current is more complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly NiCad highest, then Lithium ion, then NiMH then Alkaline.&lt;br&gt;&#xA;But in terms of stored capacity (how long you can use them for) then Li-Ion, Alkaline then NiMH/Nicad&lt;/p&gt;&#xA;" OwnerUserId="2525" LastEditorUserId="2525" LastEditorDisplayName="" LastEditDate="2011-03-22T16:03:31.010" LastActivityDate="2011-03-22T16:03:31.010" />
  <row Id="7416" PostTypeId="2" ParentId="7413" CreationDate="2011-03-22T16:06:40.573" Score="1" ViewCount="" Body="&lt;p&gt;There is no real difference - it's just an ISO thing.&lt;br&gt;&#xA;Under the &quot;measurement method&quot; you write out the overall reason for what you are testing and in &quot;measurement procedure&quot; the detailed steps, test rig settings etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you print these out put them in a nice binder and bring them out once a year to show the ISO inspector.&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-03-22T16:06:40.573" />
  <row Id="7417" PostTypeId="2" ParentId="7149" CreationDate="2011-03-22T16:10:05.800" Score="3" ViewCount="" Body="&lt;p&gt;One of the questions often being answered is “why don’t meltdowns go supercritical?” ie. “Why don’t they create a nuclear explosion?”  A good understanding of why they don’t provides a good framework to understand why corium should not be critical, and why subcritical can be fairly bad.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Critical mass is more a description of neutron flux than actual mass of material.  Perhaps a better term would be critical concentration, but I think “critical mass” sounds better; don’t you?  In the supercritical mass the fuel can act as moderator and neutron generator.  Fission produces mostly fast neutrons which are not absorbed at high probability by fuel nuclei, but if you have enough fuel nuclei this problem is obviated.  Some nuclei will absorb fast neutrons and give off slow neutrons, and a gamma ray.  Some of the first reactor designs were slurry reactors where finely ground moderator and fuel were mixed together.   I suppose if a slurry reactor melted down the corium would be critical. Modern reactors use fuel rods and both solid and liquid moderators.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fuel in the rods has both an internal and a surface neutron flux.  The perfect nuclear reactor would have surface flux where all the fast neutrons escaping the fuel rod were reflected back as slow neutrons.  This would allow the fuel to “burn” most efficiently.  By arraigning the fuel rods and moderators in an optimal configuration the closest to perfect flux is achieved.  If you remove the moderator the fuel is not critical by itself.  If you change the arrangement the fuel should go subcritical also.  If it were not for the fact that much more fuel is present for the minimal critical configuration it would be so statistically unlikely to even have local areas of critical flux that it would really be impossible.  With the overabundance of fuel it is only highly improbable that a local slurry reactor zone could form in the corium blob.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subcritical neutron flux does not mean that fission has stopped; it only means that fission will eventually drop to the level expected with only spontaneous fission.  Depending on how subcritical the reactor is this could take a while.  &lt;/p&gt;&#xA;" OwnerUserId="2708" LastEditorUserId="2708" LastEditorDisplayName="" LastEditDate="2011-03-24T04:32:26.477" LastActivityDate="2011-03-24T04:32:26.477" />
  <row Id="7418" PostTypeId="1" AcceptedAnswerId="7422" CreationDate="2011-03-22T17:29:21.433" Score="6" ViewCount="360" Body="&lt;p&gt;scalars (spin-0) derivatives is expressed as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla_{i} \phi = \frac{\partial \phi}{ \partial x_{i}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;vector (spin-1) derivatives are expressed as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla_{i} V^{k} = \frac{\partial V^{k}}{ \partial x_{i}}  + \Gamma^k_{m i} V^m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what is the expression for covariant derivatives of spinor (spin-1/2) quantities?&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-03-23T08:32:33.340" Title="covariant derivative for spinor fields" Tags="&lt;covariance&gt;&lt;spinors&gt;" AnswerCount="4" />
  <row Id="7419" PostTypeId="2" ParentId="7166" CreationDate="2011-03-22T17:43:44.337" Score="0" ViewCount="" Body="&lt;p&gt;Firstly, Raman scattering, photoelectric effect, Compton scattering, etc., all produce change in energy levels, due to absorption/release of photon energy $h\Delta\nu$, typically in a molecule, in an atom, in an electron, etc., respectively. Secondly, the &lt;em&gt;virtual energy level&lt;/em&gt; is the (sometimes somewhat artificial) adding of the photons energy to the energy level of the system that is hit by the incoming photon. It may (but not always) only be a mathematical abstraction if the system releases an outgoing photon as well. Thirdly, in quantum mechanics we are typically allowed to ask about probabilities, half times, etc, of various energy levels of the system. It does not make sense to ask how much time did the system spend with specific energy $E&amp;lt;0$, with no uncertainty in $E$, if $E$ does not correspond to an actual energy level of the system.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastActivityDate="2011-03-22T17:43:44.337" />
  <row Id="7420" PostTypeId="2" ParentId="6220" CreationDate="2011-03-22T17:43:45.573" Score="0" ViewCount="" Body="&lt;p&gt;Answer: YES &lt;br /&gt;&#xA;It is possible under the following circumstances. &lt;br /&gt;&#xA;1) The energy spent on itself by the engine (internal friction of engine) is constant (this is true when the engine running at constant speed regardless of the speed of the vehicle (may be by using a infinitely variable gear). Assume the energy spent per second by internal friction is A and the work done (external) is Y. Total fuel consumed is (X + Y). &lt;br /&gt;&#xA;2) The efficiency of the engine is very low under normal working conditions. That is Y/(X+Y) is very low (say 10%) at 50 km per hour speed. &lt;br /&gt;&#xA;3) Total frictional force is very high at higher speed. Lower at low speed. &lt;br /&gt;&#xA;Condition (1) and (3) means that there will be point of maximum efficiency - say at 20 km/hour speed. &lt;br /&gt;&#xA;Now construct a road with a downward slop such that the vehicle just rolls down without engine (engine off) till a an intermediate point. From that intermediate point the road starts upward slop with maximum steep (depends on the grip of the tyres). That means intermediate point is more close the upward sloping end. &lt;br /&gt;&#xA;Now vehicle runs till the intermediate point without any fuel cost and the remaining section with marginal increase in fuel cost than a vehicle running at plane road (as most of the energy spent is on internal friction). &lt;br /&gt;&#xA;In this case the efficiency of the downhill/uphill road is higher than a plane road. &lt;br /&gt;&#xA;I assume the mathematical calculation behind my arguments are obvious. If not let me know I can elaborate that further.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2011-03-22T17:43:45.573" />
  <row Id="7421" PostTypeId="1" AcceptedAnswerId="7423" CreationDate="2011-03-22T18:19:36.833" Score="7" ViewCount="237" Body="&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/7346/if-it-was-possible-to-dig-a-hole-that-went-from-one-side-of-the-earth-to-the-othe&quot;&gt;This recent question&lt;/a&gt; about holes dug through the Earth led me to wonder: if I wanted to dig out a tube from the north pole to the equator and build a water slide in it, which shape would be the fastest?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're assuming a frictionless tube, of course.  Let's ignore centrifugal forces.  Coriolis forces do no work, and so shouldn't matter.  Also, let's assume the Earth is a sphere with uniform density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to solve this problem by writing down an integral in polar coordinates for the time, then applying the Euler-Lagrange equations.  However, I didn't make any progress on the resulting differential equation.  Is there an analytical expression for the curve?&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-03-22T18:25:14.723" LastActivityDate="2011-03-22T19:12:55.120" Title="Brachistochrone Problem for Inhomogeneous Potential" Tags="&lt;classical-mechanics&gt;&lt;brachistochrone-problem&gt;&lt;variational-calculus&gt;&lt;minimization&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="7422" PostTypeId="2" ParentId="7418" CreationDate="2011-03-22T18:22:14.477" Score="7" ViewCount="" Body="&lt;p&gt;For the covariant spinor derivative we need to introduce a connection which can parallel transport a spinor. Such a connection takes values in the Lie-algebra of the group the spinor transforms under. Then we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ D_i \psi = \partial_i \psi + g A_i^I T_I \psi $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $T_I$ are the generators of the lie-algebra and are matrix valued. We have suppressed spinorial indices. Writing them out explicitly we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ D_i \psi_a = \partial_i \psi_a + g A_{i\,I} T^I{}_a{}^b \psi_b $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For eg, for $SU(2)$ the lie-algebra generators are given by the three pauli matrices $\sigma_x,\sigma_y, \sigma_z$ which then act on two component spinors. If you wish to work with four-component spinors $\psi_A$, transforming under the Lorentz group, the relevant generators are those of $SO(3,1)$. You can find these in Peskin and Schroeder, page 41.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are relations between the spin connection, the christoffel connection and the metric but this is the definition of the spin connection.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-22T18:22:14.477" CommentCount="3" />
  <row Id="7423" PostTypeId="2" ParentId="7421" CreationDate="2011-03-22T19:12:55.120" Score="6" ViewCount="" Body="&lt;p&gt;Yes there is, the curve is a a hypocycloid.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See for instance:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hypocycloid&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hypocycloid&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://demonstrations.wolfram.com/SphereWithTunnelBrachistochrone/&quot; rel=&quot;nofollow&quot;&gt;http://demonstrations.wolfram.com/SphereWithTunnelBrachistochrone/&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.physics.unlv.edu/~maxham/gravitytrain.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.physics.unlv.edu/~maxham/gravitytrain.pdf&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-03-22T19:12:55.120" CommentCount="5" />
  <row Id="7424" PostTypeId="2" ParentId="7418" CreationDate="2011-03-22T20:55:09.477" Score="-3" ViewCount="" Body="&lt;p&gt;I would like you to pay your attention that this way of introducing &quot;interaction&quot; is only good for describing external fields (that may be switched on and off physically). This way of coupling with the proper field (that can never be switched off) is not good and needs resolving IR and UV divergences if implemented. After renormalizations and IR diagram summation the true coupling with the proper field is different from the &quot;covariant derivative&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-03-22T21:04:16.037" LastActivityDate="2011-03-22T21:04:16.037" />
  <row Id="7425" PostTypeId="2" ParentId="2229" CreationDate="2011-03-22T21:01:54.820" Score="-1" ViewCount="" Body="&lt;p&gt;Maybe if you guys defined the 'stress energy tensor' a little more? &#xA;To me that seems as the answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A photons momentum is not not a property solely of the object/photon, as far as I understand? It's more of a shared property of the object (the photons 'energy &amp;amp; speed') relative a SpaceTime/detector? So discussing the momentum of a photon seems, to me, to become a description of a SpaceTime/detector interaction with the photons energy (and 'speed'), or am I bicycling in the great younder here? Without any interaction taking place I would assume the momentum to be undefined as light quanta can come in different energies as observed by the photoelectric effect and black body radiation. That should mean that the interaction defines the 'momentum'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And seen that way I assume that the same can be said about any object moving relative something else? As described by Louis de Broglie who said that the wavelength of a particle is inversely proportional to the momentum of the particle. And 'relativistic mass' is a useful concept, but photons do not have a mass, neither relativistic nor invariant, well, as far as I know that is? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The proper equation to use is E2 = m2c4 + p2c2 So in the case of a photon, m=0 so E = pc or p = E/c. On the other hand, for a particle with mass m at rest (i.e., p = 0), you get back the famous E = mc2. &quot; But a photon has no 'rest frame' that I know off? 'And according to Coulombs law the photon 'mass' must be less than 10^{-14}eV - AND - theoretically zero as defined by the gauge symmetries of Maxwell's equations.' &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it becomes confusing again as light quanta can be off different energies? That should then mean that they will 'bend' differently, shouldn't it? It would have been easier for me if I could assume that all light quanta had the same 'energy'? But that only seems to hold true relative their 'frequency' (waves) which in the case of a photon then should mean that different 'photons' must contain different 'energies'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope I made some sense here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(That photons can import momentum or 'energy' to a box and so create a higher invariant mass is a dissipating effect, isn't it? Only when handling matter aka 'the compressed spring' does it seem to me that you can conserve that energy?)&lt;/p&gt;&#xA;" OwnerUserId="2709" LastEditorUserId="2709" LastEditorDisplayName="" LastEditDate="2011-03-22T21:26:48.543" LastActivityDate="2011-03-22T21:26:48.543" CommentCount="4" />
  <row Id="7426" PostTypeId="1" AcceptedAnswerId="7427" CreationDate="2011-03-22T21:06:15.830" Score="7" ViewCount="394" Body="&lt;p&gt;I've on occasion seen when a cashier has trouble getting a credit card to read, will wrap it in a thin plastic grocery bag, and try swiping it again. Much to my surprise, it seemed to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the card uses a magnetic strip, I can't see how the plastic would help. Is there any reason it should, or is this a case where it worked, but not because of the plastic?&lt;/p&gt;&#xA;" OwnerUserId="1638" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-22T21:28:08.183" LastActivityDate="2011-03-23T12:03:01.763" Title="If a credit card won't read when swiped, does wrapping it in plastic actually help and if so, why?" Tags="&lt;magnetism&gt;" AnswerCount="1" />
  <row Id="7427" PostTypeId="2" ParentId="7426" CreationDate="2011-03-22T21:20:48.470" Score="18" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://cr4.globalspec.com/thread/2432/plastic-bags-improve-credit-card-swiping&quot; rel=&quot;nofollow&quot;&gt;best explanation I have seen for this&lt;/a&gt; is that the plastic bag increases the distance between the read head and the magnetic strip.  As the magnetic strip is used, small particles break off and become embedded elsewhere on the stripe.  These have very little magnetic force, but enough that they throw off today's more sensitive readers, even though they're embedded in a larger area of stronger magnetism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By increasing the distance between the read head and the mag stripe, even though it's just the thickness of the plastic bag, the smaller &quot;noisy&quot; bits impart a smaller field on the sensor than the larger areas of magnetic material surrounding them. It increases the signal to noise ratio by placing a low pass filter between the strip and the read head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, it's similar to putting on earmuffs so you can only hear the bass line of a piece of music, muffling the midrange, and almost muting the high range frequencies.&lt;/p&gt;&#xA;" OwnerUserId="2710" LastEditorUserId="2710" LastEditorDisplayName="" LastEditDate="2011-03-23T12:03:01.763" LastActivityDate="2011-03-23T12:03:01.763" CommentCount="12" />
  <row Id="7428" PostTypeId="2" ParentId="7418" CreationDate="2011-03-22T21:42:03.713" Score="3" ViewCount="" Body="&lt;p&gt;There is an interesting way to look at Christoffel connections with spinor fields.  The usual Dirac operator is written as $\gamma^\mu\partial_\mu$.  It is interesting to change this to $\partial_\mu(\gamma^\mu\psi)$.  This then becomes&#xA;$$&#xD;&#xA;\partial_\mu(\gamma^\mu\psi)~=~ \gamma^\mu\partial_\mu~+~(\partial_\mu\gamma^\mu)\psi.&#xD;&#xA;$$&#xA;The anticommutator $\{\gamma^\mu,~\gamma^\nu\}~=~2g^{\mu\nu}$ and the covariant constancy of the metric gives $\partial_\mu\gamma^\mu~=~\Gamma^\mu_{\mu\sigma}\gamma^\sigma$.  So we may then write the Dirac operator in this different form as&#xA;$$&#xD;&#xA;\delta_\nu^\mu\partial_\mu(\gamma^\nu\psi)~=~ \delta^\mu_\nu \gamma^\nu\partial_\mu\psi~+~\delta^\mu_\nu \Gamma^\nu_{\mu\sigma}\gamma^\sigma\psi.&#xD;&#xA;$$&#xA;Now if you peel off the Kronecker delta you have a covariant derivative of the spinor field.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means is that in general the Clifford algebra $CL(3,1)$ representation of the Dirac matrices is local.  The connection coefficient can then be seen as due to transition functions between these representations, so the differential produces connection coefficients.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-03-22T21:53:49.160" LastActivityDate="2011-03-22T21:53:49.160" CommentCount="1" />
  <row Id="7429" PostTypeId="2" ParentId="7401" CreationDate="2011-03-22T22:07:50.267" Score="6" ViewCount="" Body="&lt;p&gt;The statement that &quot;fluids are fractal&quot; is not quite correct (or at the very least is not precise).  Instead what really happens is that energy in fluids transitions to higher and higher frequencies via a recursive formula which looks slightly fractal-like (called the &quot;Selection rule&quot;).  This is one of the most famous results in fluid mechanics and is due to Kolmogorov.  Here is a slightly mathematical, but very eloquent exposition of this beautiful discovery:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sjsu.edu/faculty/watkins/kolmo.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.sjsu.edu/faculty/watkins/kolmo.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note however that this only tells us the approximate structure of a fluid flow; ie its power spectrum.  It does not give us enough information to say exactly what any given fluid will do at fine scales, or even if a solution exists!  These latter two problems remain unsolved even today and are the subject of one of the famous millennium problems:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.claymath.org/millennium/Navier-Stokes_Equations/&quot; rel=&quot;nofollow&quot;&gt;http://www.claymath.org/millennium/Navier-Stokes_Equations/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2712" LastActivityDate="2011-03-22T22:07:50.267" CommentCount="2" />
  <row Id="7430" PostTypeId="1" CreationDate="2011-03-22T22:26:30.943" Score="8" ViewCount="397" Body="&lt;p&gt;At the beginning of quantum computation, David Deutsch made a strong claim that the Many Worlds interpretation of quantum theory was at the foundation of his ability to do what he did. There was a lot of interest in his claim, and Many Worlds had something of a resurgence at the time. Amongst other strong claims, I'm particularly aware that Chris Fuchs has pressed a Bayesian perspective that is rather different from the Many Worlds interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My perception, however, is that quantum computation/information/communication is now much less in thrall to any particular interpretation, unless we say there is such a thing as the minimalist interpretation.&#xA;It seems that most quantum computation work is reported in journals in as close to the unadorned Fock space formalism of quantum optics as possible, often cut down to a small, finite number of harmonic oscillators, in one of the many formalisms and ansätze of statistical or condensed matter Physics, or, of course, the discussion revolves around &lt;em&gt;qubits&lt;/em&gt;.&#xA;Nonetheless, quantum information is not called quantum &lt;em&gt;particle&lt;/em&gt; information &amp;mdash;it's not part of &lt;em&gt;particle&lt;/em&gt; physics&amp;mdash; and the Bell and other inequalities that form the basis of the general rejection of classical particle physics are central instead of being as peripheral as they often seem in particle physics, so it seems, to me, that there is at least something of a change of perception, of zeitgeist, if you will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum computation for a long time was a subject of keen interest in the academic quantum foundations conferences that I went to back in the day. &lt;del&gt;Is that interest still there?&lt;/del&gt; As Peter Shor points out in a comment, this interest &lt;em&gt;is&lt;/em&gt; still there, but Is any of that interest felt in the wider Physics community that has relatively little practical interest in quantum foundations? [Peter Shor Answered that too &amp;mdash;in a way that agrees with my prejudices, so it must be right&amp;mdash; while I was writing the EDIT below, which tries to address one way in which I think the zeitgeist is being affected.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I suppose a significant number of Physicists will occasionally read PRL articles on quantum information that are highlighted by the editors, and it would be hard to avoid all Scientific American, New Scientist, and Nature articles on the subject. Someone considering grant sources would presumably at least briefly consider whether their work could be applied to QI (I'm bored already with the comp../inf../comm..) and occasionally talk to the QI group at their Institution (almost everybody has one of those by now, right?). Insofar as they do, then it seems inevitable that some QI viewpoints must be increasingly out there. Bell's theorem is surely more a topic of informed discussion than it was 20 years ago. To make the question more specific, and to include my bias in a more explicitly leading way, &lt;em&gt;Do people agree with the implication of my second paragraph that &amp;ldquo;particles&amp;rdquo; are less in the air than they were in the heyday of Particle Physics?&lt;/em&gt; [If you not unreasonably think that's too specific, feel free to Answer the original and more general Question of the title. The more specific Question still admits at least evidence of the sort given by Peter Shor, and shouldn't be thought of or addressed as only a Yes/No question.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hat Tip to &lt;a href=&quot;http://physics.stackexchange.com/users/1460/roy-simpson&quot;&gt;Roy Simpson&lt;/a&gt; for this Question (his is the credit, mine is the blame).&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-03-24T13:11:29.403" LastActivityDate="2011-03-24T13:11:29.403" Title="How much is quantum computation changing the interpretation of quantum theory, and, if at all, how?" Tags="&lt;quantum-information&gt;&lt;quantum-interpretations&gt;&lt;quantum-computation&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="7432" PostTypeId="2" ParentId="7340" CreationDate="2011-03-23T01:26:21.063" Score="5" ViewCount="" Body="&lt;p&gt;This problem is a combination of fluid mechanics and the physics of the rocket equation from Newton’s laws.  The two need to be put together properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To derive the rocket equation suppose there exists a mass $M$ that fragments into $M~-~\delta M$ and $\delta M$, where $\delta M~&amp;lt;&amp;lt;~M$. Further, suppose that $\delta M$ flies away at a velocity $V$.  The large mass will then experience a change in velocity $-\delta v$ so that&#xA;$$&#xD;&#xA;0~=~\delta MV~-~(M~-~\delta M)\delta v.&#xD;&#xA;$$&#xA;If these increments are small the term $\delta M\delta v$ may be ignored.  This is the conservation of momentum for a brief increment in the rocket flight. From here in the calculus limit these increments become infinitesimal&#xA;$$&#xD;&#xA;0~=~VdM~-~Mdv,~\rightarrow~\int_0^v dv^\prime~=~V\int_M^m {{dM^\prime}\over{M^\prime}}.&#xD;&#xA;$$&#xA;This leads to a final velocity of the rocket with mass $m$ after expending a mass $M~-~m$ of burned fuel plus oxidant $v~=~Vln(M/m)$.  This equation is not adequate for the velocity of the reaction mass is assumed to be constant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We consider the physics internal to the rocket bottle.  Air pressure pushes the water out, where this pressure depends on the volume of the air.  In fact this is the natural gas law $pV~=~NkT$, and $p~=~NkT/V$.  We also need the Navier Stokes equation in one dimension.  We then have&#xA;$$&#xD;&#xA;\rho\frac{dv}{dt}~=~\frac{dp}{dx}&#xD;&#xA;$$&#xA;The pressure is dependent on the volume with $V~=~2\pi R^2 x$ and we have that&#xA;$$&#xD;&#xA;\frac{dv}{dt}~=~-\frac{NkT}{2\pi R^2\rho x^2}&#xD;&#xA;$$&#xA;The volume of water decreases as $x$ increases, as the air expands.  This $x~=~ut$ will be related to the velocity of the water jet out the nozzle by the ratio of the bottle area and nozzle area $2\pi r^2$ so $u~=~v(r/R)^2$.  Thus &#xA;$$&#xD;&#xA;\frac{dv}{dt}~=~-\frac{NkTR^2}{2\pi v^2\rho r^4t^2}&#xD;&#xA;$$&#xA;and thus by rearranging things and solving the integrations&#xA;$$&#xD;&#xA;\frac{v^3}{3}~=~\frac{NkTR^2}{6\pi r^4\rho t}&#xD;&#xA;$$&#xA;and so the water velocity out the nozzle is&#xA;$$&#xD;&#xA;v~=~t^{-1/3}\Big(\frac{NkTR^2}{2\pi r^4\rho}\Big)^{1/3}&#xD;&#xA;$$&#xA;I have been a bit cavalier here, for one needs to take care of the end points of this integration to cut off the integral at some time t.  Clearly the water velocity does not asymptote to zero as time goes to infinity.  I leave that detail to the reader.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the job is to rework the rocket equation with this as the plume velocity $V$.  One also expresses the $dM$, the change in the mass, according to the rate of change in the column of water in the bottle.  At this point this should be reduced to further calculations of this sort.  I will halt this at this point, for this is already rather TeX intensive and pages tend to be slow.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-07-31T04:37:48.517" LastActivityDate="2011-07-31T04:37:48.517" CommentCount="3" />
  <row Id="7433" PostTypeId="1" CreationDate="2011-03-23T02:06:48.087" Score="3" ViewCount="166" Body="&lt;p&gt;I would like to find out how much energy is required to liquify gases (oxygen or nitrogen, for example). &lt;strong&gt;What are some relevant equations?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious whether creating small quantities of dry ice or cryogenic fluids is feasible at home. If James Dewar was able to liquify hydrogen in 1898 surely it must not be that difficult!&lt;/p&gt;&#xA;" OwnerUserId="2716" LastEditorUserId="2716" LastEditorDisplayName="" LastEditDate="2011-03-25T00:14:15.580" LastActivityDate="2011-06-12T07:05:41.140" Title="Equations describing the liquifaction of gases" Tags="&lt;physical-chemistry&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="7434" PostTypeId="2" ParentId="7410" CreationDate="2011-03-23T02:26:54.217" Score="1" ViewCount="" Body="&lt;p&gt;For the lighter elements the standard model nucleosynthesis works pretty good. But all is not quite perfect in the nucleosynthesis world, particularly with lithium. Google &quot;lithium problem&quot; to find lots of papers on this. For example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;JCAP 0811:012 (2008), Richard H. Cyburt, Brian D. Fields, Keith A. Olive, &lt;em&gt;A Bitter Pill: The Primordial Lithium Problem Worsens&lt;/em&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/0808.2818&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0808.2818&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The lithium problem arises from the&#xA;  significant discrepancy between the&#xA;  primordial 7Li abundance as predicted&#xA;  by BBN theory and the WMAP baryon&#xA;  density, and the pre-Galactic lithium&#xA;  abundance inferred from observations&#xA;  of metal-poor (Population II) stars.&#xA;  This problem has loomed for the past&#xA;  decade, with a persistent discrepancy&#xA;  of a factor of 2--3 in 7Li/H. Recent&#xA;  developments have sharpened all&#xA;  aspects of the Li problem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's a paper suggesting that there's problems even with 4He:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Int.J.Mod.Phys.E15:1-36 (2006), Gary Steigman, &lt;em&gt;Primordial Nucleosynthesis: Successes And Challenges&lt;/em&gt;&lt;br&gt;&#xA;&lt;strong&gt;Summary and Conclusions:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Perhaps the $^4$He challenge to SBBN&#xA;  is a signal of new physics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/astro-ph/0511534&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/astro-ph/0511534&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this is good news for physicists; there jobs are safe.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-23T02:26:54.217" />
  <row Id="7435" PostTypeId="1" CreationDate="2011-03-23T05:54:00.663" Score="0" ViewCount="176" Body="&lt;p&gt;How did the universe begin ?  how far back can we probe?can string theory determine the initial condition ? Was there a time before big bang crunch or big bang????time may be taken as negative .i mean if at big bang we take time is as 0 or time is just started ...then when t is negative, is it possible ???and if possible what form of universe was there. &lt;/p&gt;&#xA;" OwnerUserId="1843" LastEditorUserId="1843" LastEditorDisplayName="" LastEditDate="2011-03-23T10:39:34.897" LastActivityDate="2011-03-23T10:39:34.897" ClosedDate="2011-03-23T07:41:59.580" Title="How did the universe begin ? before big bang what was there ?? time may be taken as negative " Tags="&lt;string-theory&gt;&lt;astrophysics&gt;&lt;cosmology&gt;&lt;spacetime&gt;" CommentCount="1" />
  <row Id="7436" PostTypeId="2" ParentId="7430" CreationDate="2011-03-23T06:21:35.680" Score="3" ViewCount="" Body="&lt;p&gt;The overlap between &quot;interpretations of quantum mechanics&quot; and &quot;quantum computing&quot; is still imprinted onto the descriptions of various research groups and grant proposals but as far as I can say, pretty much everyone realizes that there is no actual overlap in the research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's been clear from the beginning which interpretations are viable and which interpretations are not - a few basic experiments (and their theoretical descriptions) are enough to test an interpretation. An interpretation must be &quot;de facto equivalent&quot; to e.g. the Copenhagen interpretation or any other equivalent interpretation, otherwise it's wrong. And everyone realizes that the selection of the &quot;personally preferred interpretation&quot; from the equivalence class is a matter of personal taste and physicists usually don't talk about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the quantum computation business has become a specialized and advanced &quot;engineering field&quot; which is standing on totally firm foundations and doesn't need to ask any questions about them. On the other hand, much of the continuing research of &quot;foundations of quantum mechanics&quot; has become a philosophical field whose purpose is to unlearn certain things that have already been learned, or solve the solved questions again and again. The two disciplines have totally diverged in practice.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-23T06:21:35.680" CommentCount="4" />
  <row Id="7437" PostTypeId="1" AcceptedAnswerId="7439" CreationDate="2011-03-23T06:50:19.120" Score="7" ViewCount="534" Body="&lt;p&gt;Once I asked this question from my teacher and he replied &quot;because it passes light&quot;, &quot;and why it passes light&quot; I asked and he said &quot;because it is transparent&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same question again, Why glass is transparent. Why it passes light while opaque objects do not?&lt;/p&gt;&#xA;" OwnerUserId="847" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-23T17:58:06.827" LastActivityDate="2011-07-08T14:31:28.720" Title="Why glass is transparent?" Tags="&lt;electromagnetism&gt;&lt;optics&gt;&lt;light&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="7438" PostTypeId="2" ParentId="7437" CreationDate="2011-03-23T07:21:03.843" Score="5" ViewCount="" Body="&lt;p&gt;Essentially because of absorption. When photon flies into the material it interacts with its constituents. This interaction can be divided into two contributions. One of them is elastic and is the source of the index of refraction (because effectively it just slows the photon down) while the other one is inelastic. Photon gets absorbed by an atom (say) and later it is emitted as thermal radiation in random direction thereby losing the original information it carried.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you look at this macroscopically, this process will be described by some parameter like penetration depth and intensity w.r.t. depth will decay exponentially. So if you made opaque objects thin enough, they would still be transparent (although the outgoing light would be weaker depending on thickness). Of course, this discussion completely avoids surface effects (reflection, refraction, scattering, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that all of this depends on frequency of the incident light. Atoms (let's just talk about them for simplicity; in reality there will be contribution also from molecules, lattice, free electrons and whatnot) have something called absorption spectrum. This arises because for certain frequencies electron can catch the photon and get excited to the higher energy state. So, while a material can be transparent in certain range of frequencies (like glass is for visible light) it can be quite opaque in others.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-03-23T07:21:03.843" CommentCount="6" />
  <row Id="7439" PostTypeId="2" ParentId="7437" CreationDate="2011-03-23T08:27:21.310" Score="6" ViewCount="" Body="&lt;p&gt;Photons pass through glass because they are not absorbed. And they are not absorbed because there is nothing which &quot;absorbs&quot; light in visual frequencies in glass. You may have heard that ultraviolet photons are absorbed by glass, so glass is not transparent for them. Exactly the same happens with X-rays for which our body is nearly transparent whilst a metal plate absorbs it. This is experimental evidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any photon has certain frequency - which for visible light is related to the colour of light, whilst for lower or upper frequencies in the electromagnetic spectrum it is simply a measure of the energy transported by photon. A material's absorption spectrum (which frequencies are absorbed and how much so) depends on the structure of the material at atomic scale. Absorption may be from atoms which absorb photons (remember - electrons go to upper energetic states by absorbing photons), from molecules, or from lattices.&#xA;There are important differences in these absorption possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Atoms absorb well-defined discrete frequencies. Usually single atoms absorb only a few frequencies - it depends on the energetic spectrum of its electrons. Regarding atomic absorption, the graph of absorption (plotted as a function of frequency of light) contains well-defined peaks for frequencies when absorption occurs, and no absorption at all between them.&lt;/li&gt;&#xA;&lt;li&gt;Molecules absorb discrete frequencies but there are many more absorption lines because even a simple molecule has many more energetic levels than any atom. So molecules absorb much more light.&lt;/li&gt;&#xA;&lt;li&gt;Crystalline lattices may absorb not only discrete frequencies but also continuous bands of frequencies, mainly because of discrepancies in the crystalline structure.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As glass is a non-crystalline, overcooled fluid, consisting of molecules, its absorption occurs in the 1st and 2nd ways, but because of the matter it is composed of, it absorbs outside our visible spectrum.&lt;/p&gt;&#xA;" OwnerUserId="1012" LastEditorUserId="4283" LastEditorDisplayName="" LastEditDate="2011-07-08T14:31:28.720" LastActivityDate="2011-07-08T14:31:28.720" CommentCount="4" />
  <row Id="7440" PostTypeId="2" ParentId="7418" CreationDate="2011-03-23T08:32:33.340" Score="2" ViewCount="" Body="&lt;p&gt;Before you can even introduce spinor bundles in curved spacetime, we need to introduce vierbeins first. This defines a local orthonormal frame. If you wish, you can introduce a principle frame bundle with $Spin(d,1)$ as the gauge group. Spinors can be defined with respect to this frame. The key is that spinors are representations of $Spin(d,1)$, a double cover of $SO(d,1)$, but not of the general linear group $GL(d+1,\mathbf{R})$. The affine connection is a connection over the latter group, but assuming metricity, we may map that into a spin connection over the former principle bundle.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-03-23T08:32:33.340" />
  <row Id="7441" PostTypeId="1" AcceptedAnswerId="7545" CreationDate="2011-03-23T09:18:33.013" Score="2" ViewCount="444" Body="&lt;p&gt;When I was a boy I used to daydream about building a tower so tall that the top of it would &lt;strike&gt;stick out of the top of Earth's atmosphere&lt;/strike&gt; project into near space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There would perhaps be a zero gravity area in the penthouse where my friends and I could bounce around and play space versions of various earth-based games and sports in most excellent zero-g conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much to my continued disappointment and despite all the technological advances of the last thirty or so years, no one has built such a structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain the physical limitations/constraints that are preventing someone from realising my fantasy of a 'Space Tower'?&lt;/p&gt;&#xA;" OwnerUserId="2639" LastEditorUserId="2639" LastEditorDisplayName="" LastEditDate="2011-03-23T09:55:11.017" LastActivityDate="2011-05-24T15:01:33.697" Title="What are the Constraints on Building a Tower to Space?" Tags="&lt;classical-mechanics&gt;&lt;gravity&gt;&lt;mechanics&gt;&lt;forces&gt;&lt;space&gt;" AnswerCount="8" CommentCount="1" />
  <row Id="7442" PostTypeId="2" ParentId="7441" CreationDate="2011-03-23T09:36:52.850" Score="6" ViewCount="" Body="&lt;p&gt;First of all, it is an elementary misconception that there would be a &quot;zero gravity&quot; environment in a tower that would only reach the top of the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the air molecules exist at a height smaller than 10 kilometers - and above 100 kilometers from the Earth's surface, the air is so diluted that it becomes undetectable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the height of 10 kilometers - where the atmospheric pressure is almost zero - the gravitational acceleration is just 0.3% weaker than it is on the surface and even at 100 kilometers, it is just 3% weaker. So forget about &quot;lunar games&quot;. The gravitational forces over there are pretty much indistinguishable by humans from those we know on the surface. At 100 kilometers, a 75-kilogram person may feel 2 kg lighter but it may be compensated by the suit he needs to avoid suffocation. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absence of air has nothing to do with the absence of the gravitational force. The air tries to be at low attitudes in order to minimize its potential energy; how much it wants to minimize the energy is given by the molecular mass and the temperature. However, the air density is something totally different than the gravitational acceleration - they're surely not proportional to each other in any sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The air density is proportional to $\exp(-\Phi m / kT)$ where $\Phi$ is the gravitational potential, $m$ is the molecule's mass, $k$ is Boltzmann's constant, and $T$ is temperature in Kelvin's degrees. However, the gravitational acceleration is $d\Phi / dh$. These two functions depend totally differently on the height $h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tall buildings&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tallest building in the world is Burj Khalifa in Dubai - it has 828 meters. It's about 10% of the thickness of the atmosphere in the &quot;narrow sense&quot;. It's hard to build tall buildings - one must guarantee that they're stable despite the immense weight of the material above each floor and despite the wind and vibrations of the Earth's surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are no &quot;strictly physical&quot; limitations that would prevent one from building a tower that reaches 10 kilometers above the surface. One may say that all such limitations are of engineering character. Tall mountains such as Mount Everest may be viewed as &quot;natural tall buildings&quot; and their height isn't far from the top of the atmosphere (in the narrow sense). The design of very tall buildings would probably have to be a bit hierarchical - with a solid base made out of a heavier material and lighter floors near the top, just like in the case of mountains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One would surely start to face problems to find reasonable materials if he wanted buildings that substantially reduce the gravity on the roof - buildings that are thousands of kilometers tall. For example, one of my kindergarten visions was to build an elevator that could take one to the Moon and that could convert the kinetic energy of the Moon's motion  around the Earth. That's a really challenging task for engineers. One will run into problems with conventional materials etc. - but one may still say that the limitations are of an engineering (and budgetary) character rather than fundamental physics limitations.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-23T09:43:08.807" LastActivityDate="2011-03-23T09:43:08.807" CommentCount="3" />
  <row Id="7443" PostTypeId="2" ParentId="7437" CreationDate="2011-03-23T11:07:08.367" Score="3" ViewCount="" Body="&lt;p&gt;I watch lately Sixty Symbols Video about Energy Gap. And there is explanation why material are transparent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=Omr0JNyDBI0&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=Omr0JNyDBI0&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.sixtysymbols.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.sixtysymbols.com/&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2718" LastActivityDate="2011-03-23T11:07:08.367" CommentCount="1" />
  <row Id="7444" PostTypeId="2" ParentId="6220" CreationDate="2011-03-23T12:44:29.463" Score="1" ViewCount="" Body="&lt;p&gt;Unfortunately the answer very much depends on the poorly specified &quot;an automobile&quot; portion of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll use a riding lawnmower to prove the hilly case:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Assume the flat route is flat.&lt;/li&gt;&#xA;&lt;li&gt;Assume the &quot;hilly&quot; route is not more than perhaps a dozen degrees going up to the midpoint, then similar angle going down to the endpoint - just barely steep enough that gravity is sufficient to overcome the rolling friction of the lawnmower.&lt;/li&gt;&#xA;&lt;li&gt;Assume that at the speed the lawnmower is traveling, about 5-10MPH, there is no significant air friction.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A riding lawnmower's engine and gear train is so inefficient that it will consume the same amount of gas riding on a the specified slight incline as it will riding on a flat surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first half of both routes, the same amount of fuel is consumed.  For the second half, the flat route requires more fuel, but the hill route does not.  Therefore with the efficiency of the engine assumed, the hill route could take up to 50% less fuel than the flat route.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we scale this up to a regular car, then we need only pay attention to the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does air resistance come into play&lt;/li&gt;&#xA;&lt;li&gt;Is the engine efficiency different between the two routes&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Internal combustion engines have a lower bound for energy output.  You can't get an automobile engine to put out some fraction of this lower bound when it's running - it's spending the same amount of fuel whether you're drawing 100 watts or 500 watts.  Once you get into the higher end, the engine consumes fuel at a rate that corresponds to output energy.  Fuel consumption only increases from this lower bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why the riding lawnmower is an easy case - the total range of efficiency of the riding lawnmower is so small, that there's no point in that range where going faster (therefore spending less fuel due to time spent driving) will use less total fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, some cars &lt;em&gt;will&lt;/em&gt; have such low air resistance, and such high efficiency at higher speeds, that the flat route will consume less than the hilly route, because even the slight slope we're positing will make a 2x difference in the energy required to drive the whole route.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the question is inadequately specified to answer for the general case.&lt;/p&gt;&#xA;" OwnerUserId="2710" LastActivityDate="2011-03-23T12:44:29.463" />
  <row Id="7445" PostTypeId="1" CreationDate="2011-03-23T12:49:12.507" Score="6" ViewCount="355" Body="&lt;p&gt;I was watching a BBC documentary on space last night.  It was talking about gravity, and it said that the reason we only ever see one side of the moon, is because the earths gravity is strong enough to actually stretch the moon into a longer shape, creating a 7m bulge on the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This bulge travelled around the moon.  It was described on the documentary as something along the lines of, if you were there, you would see the ground make a 7m high wave as it rotated.  The bulge then acted a brake, and the moons spin slowly ground to a halt until where we are today when it's spin is very very slow so it looks like we only see one side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So where is this giant death bulge tidal wave on earth?  Shouldn't we been seeing the ground rising up a few meters in relation to the suns position, demolishing and killing everything in its path?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume the documentary grossly simplified what actually happened, or I misunderstood!  I'm a physics noob, please go easy on me, chances are I'm not going to understand any formulas that use anything but addition or subtraction.&lt;/p&gt;&#xA;" OwnerUserId="2720" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-23T16:40:06.613" LastActivityDate="2011-03-23T18:44:31.557" Title="Where's earths death bulge, destroying everything in it's path?" Tags="&lt;gravity&gt;&lt;astrophysics&gt;&lt;tidal-effect&gt;" AnswerCount="4" CommentCount="6" />
  <row Id="7446" PostTypeId="1" AcceptedAnswerId="7457" CreationDate="2011-03-23T12:59:51.027" Score="2" ViewCount="439" Body="&lt;p&gt;Let's say I fire a bus through space at the speed of light. If I'm inside the bus (sitting on the back seat) and I run up the aisle of the bus will I in fact be traveling faster than the speed of light? Relative to earth that I just took off from.&lt;/p&gt;&#xA;" OwnerUserId="2721" LastActivityDate="2011-08-15T13:59:06.390" Title="Travelling faster than the speed of light" Tags="&lt;speed-of-light&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="3" />
  <row Id="7447" PostTypeId="2" ParentId="7445" CreationDate="2011-03-23T13:07:31.630" Score="12" ViewCount="" Body="&lt;p&gt;In the ocean, it's called the tides, and it happens twice a day. You can see it if you're standing on the seashore, but on the deep ocean it happens so gradually you wouldn't even notice it. On land the tidal bulge only rises up a third of a meter or so (not seven meters, because the moon's gravity is much less than the earth's), and, again, it happens so gradually we don't even notice it. I suspect that even the seven meter bulge on the moon would happen so gradually you wouldn't notice it, although I suspect (this is pure speculation) it would put some extra strain on the crust, but I don't know what effects that would cause.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't even notice that you said &quot;sun&quot; and not &quot;moon&quot;. The sun causes tides, too, but because the sun is so much farther than the earth, they're weaker than the moon's. From wikipedia, the moon causes 70% of the tidal forces and the sun causes 30% of the tidal forces. When they're lined up, you have especially high tides, called &quot;spring tides,&quot; and when they're 90&amp;deg; from each other, you have especially low tides, called &quot;neap tides.&quot; Since &quot;neap&quot; comes from old English, and is a word that only applies to tides, it's clear that sailors and fishermen have been noticing this for a long time.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditorDisplayName="" LastEditDate="2011-03-23T13:46:22.797" LastActivityDate="2011-03-23T13:46:22.797" CommentCount="4" />
  <row Id="7448" PostTypeId="2" ParentId="7441" CreationDate="2011-03-23T13:07:52.403" Score="4" ViewCount="" Body="&lt;p&gt;The level at which this question is being asked is uncertain.  I thought I would mention the idea of the space elevator, which some people take seriously.  However, Lubos is correct in saying that the edge of the atmosphere does not mean the end of gravity.  A spacecraft orbits the Earth because it is falling towards the Earth.  However, it is just moving fast enough so that it keeps missing the Earth which curves away under the spacecraft’s trajectory.  The apparent loss of gravity, as seen with space shuttle and ISS astronauts floating around, is due to the fact the astronauts and everything in the spacecraft is falling and moving with the rest of the spacecraft.  Remember that Galileo demonstrated that different masses fall with the same acceleration, and so everything in a spacecraft falls with the same acceleration.  However, the whole thing is moving fast enough to keep missing the Earth, and this acceleration of gravity provides the centripetal force which maintains a circular orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is this “Jack and the Beanstalk” idea of the space elevator.  I seriously question whether this will ever be built, but the idea is possible in principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Space_elevator&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Space_elevator&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea has some problems of course.  In particular it is tough to stack up mass elements without it falling over.  If the gravity force at the center of mass deviates from its foundation the stack fall over.  So I think the idea of building the tower from the ground up is probably wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prospect for this lies in building from the top down.  There are ideas about manipulating the orbits of asteroids.  The Russians want to change the orbit of Apophis asteroid, which will come close to the Earth in 2029.  Suppose we get good at doing this, and we manipulate the orbit of an asteroid into geosynchronous orbit around the Earth.  A geosynchronous orbits is at a radius of 37,000 km where the orbital period is equal to the rotational period of the Earth.  As the Wiki page shows one must then have a counter weight beyond geosynchronous orbital radius.  So if one had an asteroid of sufficient mass and with the proper material constitution one could then build the tower downwards from this point.  This would be accompanied by building upwards with an amount of mass so the center of mass of the emerging structure remains at geosynchronous orbit.  Eventually this would then be constructed into this tower.  The gravity gradient on this emerging structure would have to be carefully monitored and the vibrations on this controlled.  It would not be at all trivial to do this.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-23T13:07:52.403" CommentCount="7" />
  <row Id="7449" PostTypeId="2" ParentId="7446" CreationDate="2011-03-23T13:10:03.263" Score="1" ViewCount="" Body="&lt;p&gt;Hmm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that your bus is approaching the speed of light, because if it had reached it, its mass would be infinite and the question becomes metaphysical as far as the contents and passengers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, momentum conservation insures that the bus would drop back from the speed it has to compensate for your momentum,as long as you are airborn but when you hit the front glass, it will gain it back.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-23T13:10:03.263" CommentCount="2" />
  <row Id="7452" PostTypeId="2" ParentId="7446" CreationDate="2011-03-23T13:18:22.027" Score="2" ViewCount="" Body="&lt;p&gt;I will have to answer this quickly, for I suspect this question will be closed.  However, this thought experiment is similar to what Einstein thought about 10 years before he published his paper on special relativity.  The problem is this.  If you were on a reference frame moving at the speed of light you would observe that light, or any electromagnetic wave, as a wave of oscillating electric and magnetic fields.  However, this would be stationary, which contradicts the Maxwell equations for the propagation of electromagnetic radiation.  Einstein worked to fix this contradiction, which lead to special relativity.  The conclusion is that you can’t place yourself on a frame where light is observed to have any velocity other than the speed of light $c~\simeq~300,000km/sec$.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-23T13:18:22.027" CommentCount="1" />
  <row Id="7453" PostTypeId="2" ParentId="7445" CreationDate="2011-03-23T13:21:27.733" Score="5" ViewCount="" Body="&lt;p&gt;It would be good if you read an &lt;a href=&quot;http://en.wikipedia.org/wiki/Tides&quot; rel=&quot;nofollow&quot;&gt;article about tide&lt;/a&gt;s, and the bulge the sun/moon combination makes on the earth, seas and crust. Earth tides are interesting,  the whole crust moves reaching 55 cms at places. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-23T13:21:27.733" />
  <row Id="7454" PostTypeId="1" AcceptedAnswerId="7456" CreationDate="2011-03-23T13:24:11.917" Score="2" ViewCount="219" Body="&lt;p&gt;Is there any reason why Earth's magnetic poles are so close to its axis of rotation, or is it a coincidence? Why does the Earth have a somewhat uniform magnetic field with two poles? If you take a bunch of random particles and put them together, why should a pole form on each side of this collection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it in practice possible to create a device capable of canceling the Earth's magnetic field in a region the size of the North Sea? This could help us control the wandering of certain unwanted crabs.&lt;/p&gt;&#xA;" OwnerUserId="1708" LastEditorUserId="2710" LastEditorDisplayName="" LastEditDate="2011-03-23T15:55:59.830" LastActivityDate="2011-03-24T02:59:32.717" Title="Cancel the Earth's magnetic field" Tags="&lt;electromagnetism&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="7455" PostTypeId="2" ParentId="7454" CreationDate="2011-03-23T13:44:09.193" Score="4" ViewCount="" Body="&lt;p&gt;I keep referring to wiki articles, and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_field_of_the_earth&quot; rel=&quot;nofollow&quot;&gt;earth magnetic field one&lt;/a&gt; is quite ok.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is that the rotation of the earth plays a role, but it is not as simple as one to one. &#xA;&lt;em&gt;The dynamo theory proposes a mechanism by which a celestial body such as the Earth or a star generates a magnetic field. The theory describes the process through which a rotating, convecting, and electrically conducting fluid can maintain a magnetic field over astronomical time scales.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now about the crabs, and the north sea: though in principle one can change the magnetic field in small  regions changing the field on such magnitudes is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you described the problem with magnetic fields and the crabs, maybe a solution could be designed generating strong magnetic fields to attract them to special points where they can be harvested/eliminated. Strong magnetic fields, orders of magnitude stronger than the earth can be made without too much expense.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-23T13:54:38.687" LastActivityDate="2011-03-23T13:54:38.687" CommentCount="3" />
  <row Id="7456" PostTypeId="2" ParentId="7454" CreationDate="2011-03-23T13:46:33.640" Score="4" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If you take a bunch of random particles and put them together, why should a pole form on each side of this collection?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some particles already have a magnetic field.  Many particles are polar, such that they will orient themselves in a magnetic field.  If you jumble them all together, they will self align, and eventually one strong field will be externally detectable even though their individual fields were small and unorganized at the start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perform this experiment:  Drop a bunch of magnetic powder and dirt into a bag.  Shake vigorously.  What is the resulting clump's magnetic signature?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it in practice possible to create a device capable of canceling the earth's magnetic field in a region the size of the north sea? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No.  What you want is a Helmholtz coil, adjusted electronically to react to the earth's changing field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the area of the field required, even though it would be relatively low magnetic force, would require entirely too much energy to be practical.  Further, an ideal Helmholtz coil, where the field is uniformly 0 everywhere inside the coils, requires essentially a cubic structure.  The North Sea is 970 KM long, and thus the coils would need to be 970KM in diameter, vertically oriented, buried a significant portion of that depth into the ground on either side of the north sea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, it would really mess up the compasses of people traveling anywhere near the coils, not to mention other animals that appear to depend on magnetic fields, such as some migrating birds.&lt;/p&gt;&#xA;" OwnerUserId="2710" LastEditorUserId="2710" LastEditorDisplayName="" LastEditDate="2011-03-23T14:24:28.240" LastActivityDate="2011-03-23T14:24:28.240" />
  <row Id="7457" PostTypeId="2" ParentId="7446" CreationDate="2011-03-23T13:50:25.210" Score="16" ViewCount="" Body="&lt;p&gt;Your question has to do with addition of velocities in special relativity. For objects moving at low speeds, your intuition is correct: say the bus move at speed $v$ relative to earth, and you run at speed $u$ on the bus, then the combined speed is simply $u+v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, when objects start to move fast, this is not quite the way things work. The reason is that &lt;em&gt;time&lt;/em&gt; measurements start depending on the observer as well, so the way you measure time is just a bit different from the way it is measured on the bus, or on earth. Taking this into account, your speed compared to the earth will be $\frac{u+v}{1+ uv/c^2}$. where $c$ is the speed of light. This formula is derived from special relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some comments on this formula provide direct answer to your question:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If both speeds are small compared with the speed of light, they approximately add up as your intuition tells you.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If one of the speeds is the speed of light $c$, you can see that adding any other speed to it does not in fact change it: the speed of light is the same in all reference frames.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you add up any two speeds below $c$, you end up still below the speed of light. So, any material object which has a mass (unlike light, which doesn't), moves at a speed less than $c$. Adding to it according to the correct rule makes it closer to the speed of light, but you can never exceed it, or in fact not even reach it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'd recommend Wheeler and Taylor's &quot;Spacetime Physics&quot; to read about this. Unlike the reputation of the subject it is actually pretty intuitive (I learned that formula in high school).&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="125" LastEditorDisplayName="" LastEditDate="2011-04-28T05:51:03.630" LastActivityDate="2011-04-28T05:51:03.630" CommentCount="7" />
  <row Id="7458" PostTypeId="2" ParentId="7398" CreationDate="2011-03-23T14:12:27.517" Score="1" ViewCount="" Body="&lt;p&gt;For the purpose of this question, I'm assuming by MAX POWER you mean the maximum peak wattage - or in other words the intersection of voltage and current draw where power is at its peak.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So will old style AA batteries, Alkaline AA batteries, and NiMH, rechargeable alkaline, Lithium AA (Energizer) all have similar MAXIMUM POWER? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No.  They have different chemistry, and the power curve is different.  But that doesn't matter, because the root of your question doesn't depend on maximum power:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am suspecting my Mario Kart Remote (fit in a wheel) for playing Mario Kart Wii not behave the same when using different batteries.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...sometimes I turn left and do a drifting and the console thinks that I am drifting right... so it does seem that different batteries have different outcome... and I am suspecting can the Wii need more power than some batteries can provide? and so how do they work really?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Wii remote should consume between 50 and 200mA, or 150 to 600mW.  All these battery chemistry types are perfectly capable of meeting these requirements while they are new and /or fully charged.  As the power is consumed, the batteries will eventually be unable to meet the 600mW power requirement, and the wiimote will generally shut off to prevent players from using the remote when it is unable to operate correctly due to low power conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, changing the type of battery will not change the operation of the remote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you believe this to be the case, there are instructions for connecting your wii remote to a computer and logging the data stream.  You can therefore construct tests to verify your theory that different battery types will affect the performance of the wiimote.&lt;/p&gt;&#xA;" OwnerUserId="2710" LastActivityDate="2011-03-23T14:12:27.517" />
  <row Id="7459" PostTypeId="1" CreationDate="2011-03-23T15:08:52.783" Score="0" ViewCount="117" Body="&lt;p&gt;The adiabatic index of a gas changes with temperature, because the degrees of freedom change. For example take Argon. At 20 C it has $\gamma = 1.67$, but at -180 C it is $\gamma = 1.76$ (values taken from Wikipedia). My question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a gas, where the adiabatic index changes considerably between 0C and 100C? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll let &quot;considerably&quot; undefined, but I am thinking about a jump of maybe 0.1 similar to the Argon example, which is unfortunately at lower temperature.&lt;/p&gt;&#xA;" OwnerUserId="2480" LastEditorUserId="2710" LastEditorDisplayName="" LastEditDate="2011-03-23T15:56:04.760" LastActivityDate="2011-03-23T17:25:51.413" Title="gases with changing adiabatic index" Tags="&lt;thermodynamics&gt;&lt;statistical-mechanics&gt;" AnswerCount="1" />
  <row Id="7461" PostTypeId="2" ParentId="7437" CreationDate="2011-03-23T15:40:53.163" Score="4" ViewCount="" Body="&lt;p&gt;Strange,&#xA;all three predecessors (&quot;KKM&quot;) did not mention scattering. There are a lot of materials which do not absorb light by any of the ways mentioned, but are intransparent by scattering at crystal imperfections or grain boundarys. (Like froth on beer or clouds are intransparent)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lump of eg plaster of paris or a hand full of salt are intransparent without any absorption. So a second condition of transparency other than absoption is abcence of scattering, in detail all crystallites/grains have to be much smaller than the wavelength one wants to pass the material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are filters by this principle, one lets grow crystals in a glass matrix similar to the process of zero dilatation glass. The process is stopped when, lets say, the crystals are 300 nm. This glass will be transparent from red to 300 nm, intransparent to shorter wavelangths. Just by stopping sooner or later You control the &quot;edge&quot; of intransparency.&lt;br&gt;&#xA;(Numbers of crystal diameters only symbolic!)&#xA;This principle is rather old, &quot;Christiansen filters&quot; are known since about &lt;a href=&quot;http://de.wikipedia.org/wiki/Christiansen-Filter&quot; rel=&quot;nofollow&quot;&gt;1900&lt;/a&gt;. &#xA;There is no English version of this article :=(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS Of couse there is the anthropocentric answer: if it wasn't, we would not use it for windows, some other material would be used and would be named &quot;glass&quot;. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-23T18:08:36.380" LastActivityDate="2011-03-23T18:08:36.380" CommentCount="5" />
  <row Id="7462" PostTypeId="1" AcceptedAnswerId="7463" CreationDate="2011-03-23T15:48:00.307" Score="7" ViewCount="524" Body="&lt;p&gt;When trying to find the Fourier transform of the Coulomb potential&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V(\mathbf{r})=-\frac{e^2}{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one is faced with the problem that the resulting integral is divergent. Usually, it is then argued to introduce a screening factor $e^{-\mu r}$ and take the limit $\lim_{\mu \to 0}$ at the end of the calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This always seemed somewhat ad hoc to me, and I would like to know what the mathematical justification for this procedure is. I could imagine the answer to be along the lines of: well, the Coulomb potential doesn't have a &lt;em&gt;proper&lt;/em&gt; FT, but it has a &lt;em&gt;weak&lt;/em&gt; FT defined like this …&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Let me try another formulation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would a mathematician (being unaware of any physical meanings) do when asked to find the Fourier transform of the Coulomb potential?&lt;/p&gt;&#xA;" OwnerUserId="2726" LastEditorUserId="2726" LastEditorDisplayName="" LastEditDate="2011-03-24T21:38:07.877" LastActivityDate="2011-03-24T21:38:07.877" Title="Fourier transform of the Coulomb potential" Tags="&lt;mathematical-physics&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="7463" PostTypeId="2" ParentId="7462" CreationDate="2011-03-23T15:54:09.223" Score="8" ViewCount="" Body="&lt;p&gt;It can be justified using distribution theory.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-03-23T15:54:09.223" CommentCount="12" />
  <row Id="7464" PostTypeId="2" ParentId="7433" CreationDate="2011-03-23T16:04:44.120" Score="1" ViewCount="" Body="&lt;p&gt;This question first requires some &quot;sorting&quot;. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;As long as You want to do that with nitrogen or oxygen, so called &quot;permanent gases&quot; &#xA;in this context, the answer is no. Linde machines are &lt;strong&gt;plants&lt;/strong&gt;! The first liquefied air was &#xA;made in lab (Krakau) with simpler apparatus, but nevertheless I doubt You could tinker something like that, high pressures and precooling to rather deep temperatures are necessary. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If You content with &quot;not permanent&quot; gases, then just take a flask for refill of lighters &#xA;(Butane) and open the nozzle pointing it down into a glass. some liquid and cold and boiling butane will collect. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;dry&quot; ice is simple too, just take a fire extinguisher filled with liquid carbon dioxide &#xA;and open the nozzle and blow into a bag made from coarse linen. Dry ice (snow) will collect in the bag.  Depending on inner construction of the gas cylinder (raiser or not) &#xA;You have to hold it upside down maybe. In general do the way prescribed for the extinguisher in use.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Edit, I forgot Your asking for &quot;equations&quot;. Look for &quot;Joule-Thompson-effect&quot; (Wikipedia) &#xA;the only thing You will need, is the inversion temperature. Calculations maybe will show You the difficulty to make liquid air. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit/PS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If You try to tinker in this domain, please be aware that steel (and other metals) &#xA;will be very brittle at liquefied air temperatures.(Danger of explosions) Selection of materials is a very important part of cryogenics. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-24T10:46:12.610" LastActivityDate="2011-03-24T10:46:12.610" />
  <row Id="7465" PostTypeId="2" ParentId="7462" CreationDate="2011-03-23T16:16:21.620" Score="3" ViewCount="" Body="&lt;p&gt;I'm not sure that a justification is required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't feel the need to justify solutions obtained by making a change of variables, nor those obtained by differentiating under the integral sign, nor those obtained by analytic continuation and application of contour integrals and so on for a host of other tricks of symbol manipulations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you insist...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can interpret this as &lt;a href=&quot;http://en.wikipedia.org/wiki/Yukawa_potential&quot; rel=&quot;nofollow&quot;&gt;the potential due to a massive scaler boson (the Yukawa potential)&lt;/a&gt;, and then take the limit of vanishing mass.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-03-23T16:16:21.620" />
  <row Id="7466" PostTypeId="2" ParentId="7359" CreationDate="2011-03-23T16:16:37.727" Score="1" ViewCount="" Body="&lt;p&gt;Your assumption that there has to be something before, after, outside or even in between, some sequence of events, while natural to classical Physics, is not necessary. Google for &amp;ldquo;Lucretius arrow &quot;edge of the universe&quot;&amp;rdquo; finds various web references, none of which I particularly want to cite, to the fact that Lucretius asked what happens at the edge of the universe around 100BC. With various similar search terms, Google leads to various Philosophical forums that have had similar Questions asked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your Question also has slight similarity to the ancient &lt;a href=&quot;http://plato.stanford.edu/entries/paradox-zeno/#Arr&quot; rel=&quot;nofollow&quot;&gt;Zeno's paradox of the Arrow&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AFAIK, there is no settled Answer to this. It's often convenient in mathematics to assume that there's a continuum, and such models can be quite effective, but there's no necessity. Taking things to be without end, infinite, in endless ascents or descents, etc., while often useful, also requires care when constructing mathematical models, because without care things can end up being badly defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To step into the detailed Question you put, you are asking about before, after, and outside what we can confidently infer from fitting experimental data to models of current physical theory. There does not &lt;em&gt;have&lt;/em&gt; to be anything at all there. A different possibility, perhaps equally disconcerting, is that we may be using models of the scientific theories that we find to be very accurately consistent close to Earth and the Sun in a domain in which the the experimental data does not fit any single consistent model. In particular, there is no guarantee that there is a consistent atlas of the big bang (and yes, for those who know of these things, I also mean this technically, as of a manifold).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I fear that this is rather general, but insofar as I see the Question blows Philosophical so blow I.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-03-23T16:16:37.727" />
  <row Id="7467" PostTypeId="2" ParentId="7445" CreationDate="2011-03-23T16:59:25.643" Score="3" ViewCount="" Body="&lt;p&gt;We ARE/Have experienced slow down in the rotation of the earth on its axis just as the moon has.  We used to have ~18 hour days, now it is ~24 hours (give or take the remainder amount).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CITATION Colorado State Univ. (with other citations including notes on neap tides) &lt;a href=&quot;http://asa.chm.colostate.edu/archive/evolution/199607/0052.html&quot; rel=&quot;nofollow&quot;&gt;http://asa.chm.colostate.edu/archive/evolution/199607/0052.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the &quot;death bulge&quot; there is also about a 1 meter tidal bulge in the rock/solid part of the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The height of a tidal bulge on a planet is proportional to the inverse cube of the distance between the planet and the object causing the tidal bulge. The torque which slows down the planet is proportional to the inverse sixth power of the distance.&quot;  from here: &lt;a href=&quot;http://www.exo.net/~pauld/physics/tides/tidalevolution.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.exo.net/~pauld/physics/tides/tidalevolution.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is the citation from Ohio State Univ. &lt;a href=&quot;http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit4/tides.html&quot; rel=&quot;nofollow&quot;&gt;http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit4/tides.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2729" LastActivityDate="2011-03-23T16:59:25.643" />
  <row Id="7468" PostTypeId="2" ParentId="7459" CreationDate="2011-03-23T17:25:51.413" Score="1" ViewCount="" Body="&lt;p&gt;I think values of specific heat of gases are more common, &#xA;at least there is a table in CRC Handbook. &#xA;Specific heats reflect the increase of degrees of freedom in the same way. &#xA;(Argon and all other inert gases do not change degrees at all, no necessity to &#xA;look for them) &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-23T17:25:51.413" CommentCount="2" />
  <row Id="7469" PostTypeId="2" ParentId="7430" CreationDate="2011-03-23T17:32:18.003" Score="2" ViewCount="" Body="&lt;p&gt;Since I am mentioned in the question, I shall make a few points about what I understand about the history of Quantum Computation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Originally Deutsch was proposing quantum computation as a &quot;thought experiment&quot; which would prove the Many Worlds Interpretation, or at least provide a theoretical basis for constructing experiments. Several claims were made at the launch of his quantum computation approach, one of which related to the &lt;em&gt;efficiency&lt;/em&gt; of the quantum algorithms that could be constructed, versus the &lt;em&gt;efficiency&lt;/em&gt; of the corresponding Turing algorithms. This mismatch in efficiency was presented as an argument that the quantum case was demonstrating some form of parallelism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What has happened in the years since is that the field of Quantum Complexity Theory has grown, and all the quantum algorithms have been shown to belong to a class called &lt;a href=&quot;http://en.wikipedia.org/wiki/BQP&quot; rel=&quot;nofollow&quot;&gt;BQP&lt;/a&gt; which is larger than the serial &lt;strong&gt;P&lt;/strong&gt; class, but yet smaller than the larger parallelism classes known classically (like &lt;strong&gt;NP&lt;/strong&gt;). This leads to some puzzlement as to what quantum algorithms actually prove about physics reality. Had a more dramatic result been proven (which theoretically could still happen), such as &lt;strong&gt;BQP&lt;/strong&gt; = &lt;strong&gt;NP&lt;/strong&gt;, then physicists might have taken note and started to wonder why this was true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the meantime interest in the Many Worlds Interpretation has continued outside of the Quantum Computation framework arising from the Quantum Gravity perspective; and I suppose that few Quantum Computation Engineers see their work as relating to cosmological issues, although Deutsch likely remains the exception with his Multiverse ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One aspect of the Deutsch model which seemed incongruous to me, was the restriction to finite qubits in the &quot;Universal Quantum Computer&quot; model. [Part of the reason for this being that one could directly match the Turing Machine which was explicitly finite.] From a practical perspective having a finite number of qubits might make sense (then again perhaps not...), but as a thought experiment one should perhaps extend the basic notion of quantum computation to a (countably) infinite set of states. I am not sure whether this idea has been followed up, but it seems more foundational than the engineering challenges currently considered in papers, and might lead to more useful discussions on QM Interpretations from the Quantum Computation perspective.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-03-23T17:42:41.240" LastActivityDate="2011-03-23T17:42:41.240" CommentCount="1" />
  <row Id="7470" PostTypeId="1" AcceptedAnswerId="7474" CreationDate="2011-03-23T18:13:37.123" Score="3" ViewCount="222" Body="&lt;p&gt;This is something probably very basic but I was led back to this issue while listening to a recent seminar by Allan Adams on holographic superconductors. He seemed very worried to have a theory at hand where the chemical potential is negative. (why?)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For fermions, isn't the sign of the chemical potential a matter of definition?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The way we normally write our equations for the Fermi-Dirac distribution the chemical potential happens to that value of energy at which the corresponding state has a occupation probability of half. And within this definition the holes in a semiconductor have a negative chemical potential.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It would be helpful if someone can help make a statement about the chemical potential which is independent of any convention.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;{Like one argues that negative temperature is a sign of instability of the system.}&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Also isn't it possible for fermions in an interacting theory to have a negative chemical potential?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Also if there is a &quot;physical argument&quot; as to why bosons can't have a positive chemical potential? (Again, can an interacting theory of bosons make a difference to the scenario?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And how do these issues change when thinking in the framework of QFT?&#xA;     (No one draws the QCD phase diagram with the chemical potential on the negative X-axis!)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In QFT does the chemical potential get some intrinsic meaning since relativistically there is a finite lower bound of the energy of any particle given by its rest mass?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1637" LastEditorUserId="1637" LastEditorDisplayName="" LastEditDate="2011-03-23T18:34:23.750" LastActivityDate="2011-03-23T19:10:55.913" Title="Chemical potential" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;statistical-mechanics&gt;&lt;solid-state-physics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="7471" PostTypeId="2" ParentId="7462" CreationDate="2011-03-23T18:30:09.007" Score="4" ViewCount="" Body="&lt;p&gt;Dear Emerson, the &quot;weak Fourier transform&quot; of the potential, which is simply $4\pi / q^2$, is the &quot;proper physical&quot; value of the Fourier transform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In physics, it simply makes no sense to say that a Fourier transform &quot;doesn't exist&quot;: the Coulomb potential clearly does exist, and when we're calculating something that depends on its Fourier transform, the answer also has to exist because the world has to behave in some way. It's just generally impossible in physics to say that a valid theory produces an &quot;ill-defined answer&quot;. We must always struggle to get an actual, well-defined answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And quantum electrodynamics, even though it has a superficially divergent Fourier transform of the Coulomb potential, is an excellent theory. Still, we need the Fourier transform to answer many questions. So the only question can be how to find out the right answer - not whether an answer exists. Of course that it does. The physically correct answer is most easily obtained by regulating the potential - as a limit of the Yukawa potential - and by exchanging the limit with the integration in the middle of the calculation, to guarantee that the result is sensible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is like giving a mass to the photon - we call it an infrared regulator. Such things are omnipresent in quantum field theory. If an expression doesn't naively make sense, one tries to express it as a limit of a more general expression, and calculate a more general calculation, which is more well-defined, and try to take the required limit at the very end. This is the right physicist's approach to seemingly ill-defined expressions and it always leads us out of the trouble and it always tells us something more meaningful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A failure - a refusal to calculate the answers - is just not acceptable in physics.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-23T18:30:09.007" CommentCount="2" />
  <row Id="7472" PostTypeId="2" ParentId="7445" CreationDate="2011-03-23T18:44:31.557" Score="0" ViewCount="" Body="&lt;p&gt;Even if there was a 7 meter bulge that traveled around the earth. I don't think you could perceive it.  7 meters tapering off over 1000s of miles would not be visible, and each square mile of crust wont change in location much to the next square mile.&lt;/p&gt;&#xA;" OwnerUserId="503" LastActivityDate="2011-03-23T18:44:31.557" />
  <row Id="7473" PostTypeId="2" ParentId="7347" CreationDate="2011-03-23T18:55:01.337" Score="1" ViewCount="" Body="&lt;p&gt;I think there's an important lesson for the OP lurking in this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything as large as a car wheel is almost certainly extremely complicated as well.  The wheel has more atoms than a beach has grains of sand.  It's made of long, convoluted polymer molecules that were formed in complicated chemical processes, as well as from air and metals, themselves interesting materials to understand.  The wheel deforms and bounces as it rotates.  It dissipates energy into the road and its surroundings, and takes energy in from the motor.  If you wanted to understand everything about the wheel, by the time you are done, you'll have a learned a great deal from all over the natural sciences.  I once read an entire book called &quot;The Bicycle Wheel&quot; that really only covered a few select topics, like the tension in the spokes and the alignment.  Everything that can be said would take many books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, physicists do not want to know everything about the wheel.  Part of the art of physics is figuring out the bare minimum amount of information about the wheel to take into consideration.  What to consider depends on what sort of question you're answering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you are testing the brakes, and you want the car to stop very suddenly.  Then the brakes must dissipate all the kinetic energy of the car.  In that case, you might wonder, &quot;Can I calculate this energy using simply $KE = 1/2 mv^2$, or do I need to consider the extra energy of the rotating wheels?&quot;  Here we would ignore most sorts of energy, and focus only on the translational and rotational kinetic energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In another situation, we might ask, &quot;How will my gas mileage improve if I inflate my tires another 5 psi?&quot; (In the US, we use absurd units for everything.)  In that case, we would be concerned with how the energy of the wheel is being turned into heat, and we'd have to consider the energy involved in deforming the wheel repeatedly as it goes around - stretching the rubber and slipping against the road, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's true that there are many forms of energy in the wheel, and so the full picture is complicated.  However, the goal is to ignore as many complications as is reasonable at any given time.  Figuring out which things to consider and which to ignore is a difficult skill that physicists gain only by dint of long and diligent practice.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-03-23T19:06:12.123" LastActivityDate="2011-03-23T19:06:12.123" />
  <row Id="7474" PostTypeId="2" ParentId="7470" CreationDate="2011-03-23T19:05:04.483" Score="5" ViewCount="" Body="&lt;p&gt;consider the grand canonical ensemble,&#xA;$$ \rho \sim \exp[-\beta (E-\mu N)] $$&#xA;In the exponent, the inverse temperature $\beta = 1/kT$ is the coefficient in front of one conserved quantity, the (minus) energy, while another coefficient, $\beta\mu$, is in front of the number of particles $N$. The chemical potential is therefore the coefficient in front of the number of particles - except for the extra $\beta$. The number of particles has to be conserved as well if $\mu$ is non-zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as the sign of $N$ is well-defined, the sign of the chemical potential is well-defined, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for bosons, $\mu$ can't be positive because the distribution would be an exponentially increasing function of $N$. Note that in the grand canonical ensemble - which is really the ensemble in which $\mu$ is sharply defined - the dual variable to $\mu$, namely $N$, is not sharply defined. However, the probability for ever larger - infinite $N$ would be larger, so the distribution would be peaked at $N=\infty$. Such a distribution couldn't be well-defined. We want, in the thermodynamic limit, the grand canonical ensemble, while assuming a fixed $\mu$, also generate a finite and almost well-defined $N$, within an error margin that goes to zero in the thermodynamic limit. That couldn't happen for bosons and a positive $\mu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This catastrophe would be possible because $N=\sum_i n_i$, a summation over microstates $i$, and every $n_i$ can be an arbitrarily high integer for bosons. For fermions, the problem doesn't occur because $n_i=0$ or $1$ for each state $i$. So for fermions, we can't argue that $\mu$ has to be positive. Note that $E-\mu N$ in the exponent is the sum  $\sum_i N_i (e_i-\mu)$. For bosons, the problem occurred for states for which $e_i-\mu$ was negative i.e. $e_i$ was low enough. For fermions, however, the number of such states - and therefore the maximum number of fermions in them - is finite, so the divergence doesn't occur if the chemical potential is positive. For fermions, positive $\mu$ is OK.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, for fermions, both positive and negative $\mu$ is OK. Also, it is easy to see that if both particles and antiparticles exist, $\mu$ of the antiparticle has to be minus $\mu$ of the particle because only the difference $N_{\rm particles} - N_{\rm antiparticles}$ is conserved; this is true both for bosons and fermions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the potential for electrons is positive, the potential for positrons or holes (which play the very same role) has to be negative, and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing changes about the meaning of the chemical potential when one switches from classical physics to quantum physics: in fact, above, I was assuming that there are &quot;discrete states&quot; for the particles, just like in quantum physics - otherwise we wouldn't be talking about bosons and fermions which are only relevant in the quantum setup. Classical physics is a limit of quantum physics in which the number of states is infinite because $\hbar$ goes to zero, so a finite number of particles never ends up in &quot;exactly the same state&quot;. In some sense, Ludwig Boltzmann, while working in the context of classical statistical physics, was inherently using the thinking and intuition of quantum statistical physics - he was a truly ingenious &quot;forefather&quot; of quantum physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In relativity, one has to be careful how we define the energy of a state. Note that the physically meaningful combination that appears in the exponent is $e_i-\mu$, so if one shifts $e_i$ e.g. by $mc^2$, the latent energy, one has to shift $\mu$ in the same direction by the same amount. The notions of chemical potential obviously work in relativity, too. Relativistic physics is not a &quot;completely new type of physics&quot;. It is just a type of the old physics that happens to respect a symmetry - the Lorentz symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, in quantum field theory which combines both quantum mechanics and relativity, statistical physics including the notion of the chemical potential also works but one must be careful that particle-antiparticle pairs may be created with enough energy. That implies $\mu_{p}=-\mu_{np}$, as I said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can't be any general ban on a negative chemical potential of fermions: fermionic $\mu$ can have both signs. However, in the particular theory that Allan wanted to describe, he could have had more detailed reasons why $\mu$ should have been positive for his fermions. I am afraid that this would be an entirely different, more specific question - one about superconductors. As stated, your question above was one about statistical physics and I tend to believe that the text above exhausts all universal facts about the sign of the chemical potential in statistical physics.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-23T19:10:55.913" LastActivityDate="2011-03-23T19:10:55.913" CommentCount="7" />
  <row Id="7475" PostTypeId="1" CreationDate="2011-03-23T21:13:59.073" Score="7" ViewCount="194" Body="&lt;p&gt;In &lt;a href=&quot;http://arxiv.org/abs/0909.2207v4&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; paper the authors refer to&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;transverse susceptibility $\chi_{ \perp}$ [meV $^{−1}$]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was taught that the magnetic susceptibility is dimensionless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I get $\chi$ in the above units??&lt;/p&gt;&#xA;" OwnerUserId="2399" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-03-24T01:50:42.487" LastActivityDate="2011-07-03T08:15:30.130" Title="Magnetic susceptibility in 1/eV" Tags="&lt;electromagnetism&gt;&lt;unit-conversion&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="7476" PostTypeId="1" CreationDate="2011-03-23T21:30:49.473" Score="0" ViewCount="97" Body="&lt;p&gt;I'm wondering if the water lifting secret method of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hanging_Gardens_of_Babylon&quot; rel=&quot;nofollow&quot;&gt;Hanging Gardens of Babylon&lt;/a&gt; still unresolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to find anything related to it at &lt;a href=&quot;http://www.google.com/patents&quot; rel=&quot;nofollow&quot;&gt;Google Patent&lt;/a&gt; but nothing was there related to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that means that no one have made an explanation of the way that water was lifted? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PS.&lt;/strong&gt; I guess that it's mystery is the water lifting with the nothing of the available technology. Isn't it?&lt;/p&gt;&#xA;" OwnerUserId="2731" LastActivityDate="2011-03-23T21:30:49.473" ClosedDate="2011-03-23T22:00:48.647" Title="Is the Hanging Gardens of Babylon mystery still unresolved?" Tags="&lt;gravity&gt;&lt;water&gt;&lt;lift&gt;" CommentCount="3" />
  <row Id="7477" PostTypeId="1" CreationDate="2011-03-23T22:11:46.117" Score="2" ViewCount="154" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/6582/what-sort-of-jobs-do-physics-graduates-with-b-sc-m-sc-get&quot;&gt;What sort of jobs do physics graduates with B.Sc/M.Sc get?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I've been told over and over again...&quot;I would not major in physics unless you are looking to get a Ph.D.&quot; What is it that physics majors do with B.S./M.S. degrees? I'm currently studying engineering physics (though most of my courses lean toward engineering theory and not pure physics, really)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Apologies in advance if the question is ill-posed or too subjective in its current form to meet the requirements of the FAQ; I'd certainly appreciate any suggestions for its modification if need be.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, I'd greatly appreciate some responses  &lt;/p&gt;&#xA;" OwnerUserId="2120" LastActivityDate="2011-03-23T22:11:46.117" ClosedDate="2011-03-23T23:08:59.200" Title="Is there any reason to major in physics if you're certain you will never do a Ph.D?" Tags="&lt;soft-question&gt;&lt;education&gt;&lt;physics-careers&gt;" CommentCount="3" />
  <row Id="7478" PostTypeId="1" CreationDate="2011-03-23T23:46:27.403" Score="1" ViewCount="82" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicates:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/679/learning-physics-online&quot;&gt;Learning physics online?&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/2219/beginner-physics-resources&quot;&gt;Beginner Physics Resources?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I never formally studied physics.  My background is in engineering and I have taken many basic physics courses, but never beyond the basics of classical physics (very minor visiting of subjects such as relativity and quantum mechanics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am very interested in the subject and often find myself reading about specific topics. For instance, I have always been fascinated by the results of special relativity such as the constancy of the speed of light, time dilation etc.  Another example is the the subject of unifying general relativity and quantum mechanics.  I read about these subjects but always end up frustrated at my lack of complete and beyond the surface understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Granted I probably sound very naive, but my question is this: what would you (as a formal student of physics) recommend to help me gain a deeper understanding of such subjects? Going back to school for a phd is out of the question. I am not trying to trivialize an extraordinarily complex subject, I just want to be able to follow discussions such as the quantum mechanical origins of magnetism (when I am pondering the essence how the heck magnets works). I feel like there is a whole world I am missing out on by not having a true deep understanding of such subjects and would appreciate any advice on how to be more enlightened.&lt;/p&gt;&#xA;" OwnerUserId="2732" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-23T23:58:41.450" LastActivityDate="2011-03-23T23:58:41.450" ClosedDate="2011-03-24T00:01:59.030" Title="Independent physics study" Tags="&lt;soft-question&gt;&lt;education&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="7479" PostTypeId="1" CreationDate="2011-03-23T23:51:52.313" Score="3" ViewCount="155" Body="&lt;p&gt;Why is it that if I'm sitting on a seat on a bus or train and its moving quite fast, I am able to throw something in the air and easily catch it? Why is it that I haven't moved 'past' the thing during the time its travelling up and down?&lt;/p&gt;&#xA;" OwnerUserId="2639" LastActivityDate="2011-03-24T03:38:27.223" Title="A Basic Question about Gravity, Inertia or Momentum or something along those lines" Tags="&lt;gravity&gt;&lt;momentum&gt;&lt;velocity&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="7480" PostTypeId="2" ParentId="7479" CreationDate="2011-03-24T00:01:13.727" Score="2" ViewCount="" Body="&lt;p&gt;Barring the affects of air resistance and such, just think of the object as also &quot;part&quot; of the train moving at the same velocity as you.  Essentially there is no way to distinguish between an object at rest or moving at a constant velocity.&lt;/p&gt;&#xA;" OwnerUserId="2732" LastActivityDate="2011-03-24T00:01:13.727" CommentCount="1" />
  <row Id="7481" PostTypeId="2" ParentId="7479" CreationDate="2011-03-24T00:01:48.263" Score="0" ViewCount="" Body="&lt;p&gt;Momentum is conserved.  If you are on a frame (the bus) moving at a velocity that is constant, then everything else is as well.  The momentum of every object is $p~=~mv$.  This is whether or not there is something holding to the frame.  In the absence of some force a body maintains a constant momentum.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-24T00:01:48.263" CommentCount="3" />
  <row Id="7482" PostTypeId="2" ParentId="7479" CreationDate="2011-03-24T01:32:44.523" Score="3" ViewCount="" Body="&lt;p&gt;The thing you throw in the air is also traveling at the same speed you are, in the same direction.  When you throw it up, it doesn't matter that the earth below is moving backwards at speed, nor that the moon is moving past even more quickly, nor that the earth itself is spinning and moving relative to the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ball has a speed and direction and currently that matches your speed and direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you throw the ball up, you have &lt;em&gt;added&lt;/em&gt; force in a new direction, which alters its speed and direction, but only with respect to your speed and direction.  In other words, to you the ball appear to go up and down, but to the earth it's falling like a projectile - forward up and down.  Since you are traveling forward at the same speed as the projectile, it appears to you that it only goes up, then down, even though during that time you &lt;em&gt;both&lt;/em&gt; moved forward.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'm not actually going to break out the math, but here's the short version:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You and the object are moving at a speed and in a direction that we'll call vector P and B, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently your two vectors match.  Relative to some other reference frame you are both moving, but relative to you, since your vectors match, the object appears to be motionless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You apply a force on vector B, which alters its trajectory.  Now this force results in additional speed and direction described by vector T.  The object, therefore, is now moving according to the vector B + T.  However, again, since B = P, it appears to you that the object is only moving according to vector T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity is applying a force to the object, which will eventually reverse T in the down direction, unless the ball is acted upon by another force, such as your hand catching the ball again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So regardless of what vector you apply to it, it will be in &lt;em&gt;addition&lt;/em&gt; to the vector you are already traveling at, and therefore it will appear to you as though it is only traveling along its new vector.&lt;/p&gt;&#xA;" OwnerUserId="2710" LastActivityDate="2011-03-24T01:32:44.523" />
  <row Id="7483" PostTypeId="1" CreationDate="2011-03-24T02:56:28.463" Score="2" ViewCount="377" Body="&lt;p&gt;Inflation seems to solve many of the problems of cosmology like horizon problem, flatness problem etc. Now suppose, I am a devil's advocate and tries to find holes in this beautiful theory. I argue that the early universe were having a fractal geometry. For any generic initial conditions suppose it was indeed a fractal. Then obviously no amount of stretching can make it smooth and flat locally. How would one reconcile this picture with inflation?&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-08-18T12:26:47.743" Title="If the universe were a fractal" Tags="&lt;cosmology&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" />
  <row Id="7484" PostTypeId="2" ParentId="7441" CreationDate="2011-03-24T02:57:33.917" Score="1" ViewCount="" Body="&lt;p&gt;There's an interesting review article on the subject:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Review of New Concepts, Ideas and Innovations in Space Towers&lt;/strong&gt;&lt;br&gt;&#xA;Mark Krinker, (2010)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A lot of new concepts, ideas and&#xA;  innovation in space towers were&#xA;  offered, developed and researched in&#xA;  last years especially after 2000. For&#xA;  example: optimal solid space towers,&#xA;  inflatable space towers (include&#xA;  optimal space tower), circle and&#xA;  centrifugal space towers, kinetic&#xA;  space towers, electrostatic space&#xA;  towers, electromagnetic space towers,&#xA;  and so on. Given review shortly&#xA;  summarizes the[s]e researches and gives&#xA;  a brief description [of] them, note some [of]&#xA;  their main advantages, shortcomings,&#xA;  defects and limitations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1002.2405&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1002.2405&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is sufficiently interesting that I would recommend it as a place to look for interesting problems for your undergraduate classes in mechanics.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-24T02:57:33.917" CommentCount="0" />
  <row Id="7485" PostTypeId="2" ParentId="7454" CreationDate="2011-03-24T02:59:32.717" Score="2" ViewCount="" Body="&lt;p&gt;The magnetic poles are apparently defined by the movement of material deep inside the earth's liquid core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the core is liquid, there are Coriolis effects which align the currents in the core. The effect is similar to that seen in the earth's atmosphere, but with a far longer time scale. So while the structure of the flows in the earth's core are unknown, we expect that they will have a structure vaguely reminiscent of the earth's atmosphere. See the wikipedia article on &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Prevailing_winds&quot; rel=&quot;nofollow&quot;&gt;prevailing winds&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus I suspect that a literature search will reveal that the earth's magnetic north pole is approximately aligned parallel (or anti-parallel) with the planet's rotation axis because this is a symmetry of the forces that create the magnetic field.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-03-24T02:59:32.717" CommentCount="3" />
  <row Id="7486" PostTypeId="2" ParentId="7475" CreationDate="2011-03-24T03:17:16.493" Score="1" ViewCount="" Body="&lt;p&gt;Magnetic susceptibility is a static constant relating the amount of magnetization to the applied magnetic field; it is dimensionless.  Transverse susceptibility is a measure of the magnetic response of a material to an alternating field; it is not dimensionless. The transverse susceptibility is often used when talking to the ferromagnetic resonance (like I think they are doing in the paper you cite) of a material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the short answer is that MS is dimensionless and TS is not.&lt;/p&gt;&#xA;" OwnerUserId="2708" LastActivityDate="2011-03-24T03:17:16.493" CommentCount="4" />
  <row Id="7487" PostTypeId="1" AcceptedAnswerId="7489" CreationDate="2011-03-24T03:38:13.933" Score="2" ViewCount="2471" Body="&lt;p&gt;There is currently a lot of discussions regarding radioactivity in Japan, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Iodine-131&quot; rel=&quot;nofollow&quot;&gt;iodine-131&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iodine-131 has a half-life of 9 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it mean that people drinking radioactive water (with iodine-131) will lose half of it withing 9 days? &lt;br/&gt;&#xA;And thus the health risk would be lower and lower over time?&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="174" LastEditorDisplayName="" LastEditDate="2011-03-26T16:19:38.397" LastActivityDate="2011-04-08T14:21:25.750" Title="Iodine-131 half-life and reality" Tags="&lt;radioactivity&gt;" AnswerCount="6" CommentCount="8" />
  <row Id="7488" PostTypeId="2" ParentId="7479" CreationDate="2011-03-24T03:38:27.223" Score="2" ViewCount="" Body="&lt;p&gt;All physics that we know obeys the &lt;a href=&quot;http://en.wikipedia.org/wiki/Principle_of_relativity&quot; rel=&quot;nofollow&quot;&gt;principle of relativity&lt;/a&gt;, which states that it is impossible to tell whether the train is moving at a constant speed or not without looking outside.  In a real train, you can tell, but only because the train ride is bumpy and the train changes speeds.  In a perfectly-smooth train, it would be fundamentally impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you can throw the ball up and catch it again when the train isn't moving over the tracks, you can do it when the train is moving over the tracks.  If the ball behaved differently, you'd be able to tell when the train is moving.  Since there can be no way to tell, the ball must behave the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no deeper justification of this principle.  We can take the various physical theories we have and prove that they obey the principle (which several other answerers have done), but ultimately many physicists have come to believe that relativity is simply built into the universe, and that future theories will obey it, and that if we ever find an exception to it, it will apply only in extreme situations (like, for example, tiny distances much smaller than an atomic nucleus).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also want to see &lt;a href=&quot;http://physics.stackexchange.com/questions/2230/why-and-how-is-the-speed-of-light-constant&quot;&gt;this earlier question&lt;/a&gt; about relativity and the speed of light, where I said pretty much the same thing in more words.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-03-24T03:38:27.223" />
  <row Id="7489" PostTypeId="2" ParentId="7487" CreationDate="2011-03-24T03:54:40.253" Score="5" ViewCount="" Body="&lt;p&gt;The Iodine-131 undergoes beta decay.  This is what makes it radioactive.  The faster it decays the more radiation it gives off per unit time.  Yes the material disappears faster than something with a longer half-life, but its speed of decay means that it is hotter.  &lt;/p&gt;&#xA;" OwnerUserId="2708" LastActivityDate="2011-03-24T03:54:40.253" CommentCount="3" />
  <row Id="7490" PostTypeId="2" ParentId="7487" CreationDate="2011-03-24T03:57:50.893" Score="3" ViewCount="" Body="&lt;p&gt;Iodine-131 &lt;a href=&quot;http://en.wikipedia.org/wiki/Iodine_131#Radioactive_decay&quot; rel=&quot;nofollow&quot;&gt;decays&lt;/a&gt; with a half-life of 8.02 days into stable &lt;a href=&quot;http://en.wikipedia.org/wiki/Isotopes_of_xenon&quot; rel=&quot;nofollow&quot;&gt;xenon-131&lt;/a&gt; which is &lt;a href=&quot;http://www.lenntech.com/periodic/elements/xe.htm#Health%20effects%20of%20xenon&quot; rel=&quot;nofollow&quot;&gt;not considered carcinogenic&lt;/a&gt;.  Xenon certainly does not belong in your body, but the amount produced would be chemically &lt;a href=&quot;http://home.earthlink.net/~sweetwind7/thyca/thyca.html#xips&quot; rel=&quot;nofollow&quot;&gt;insignificant&lt;/a&gt;.  Iodine has a &lt;a href=&quot;http://www.epa.gov/rpdweb00/radionuclides/iodine.html#inbody&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;biological&lt;/em&gt; half-life&lt;/a&gt; of 100 days in the body so any iodine-131 will probably decay before being excreted.  So the answer is YES to both questions, because the radiation is much more dangerous than the presence of xenon and this risk is incurred at the time of decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The above answer assumes iodine-131 has been consumed and some time has passed, and it is incorrect unless there is some way of removing the radioactive iodine (KI therapy?).  The implication is that such treatment should be started as soon as possible, but it may not be worth the &lt;a href=&quot;http://en.wikipedia.org/wiki/Potassium_Iodide#Adverse_reactions&quot; rel=&quot;nofollow&quot;&gt;cost&lt;/a&gt; if months have already passed.  Without this possibility, the risk is incurred at the time of &lt;em&gt;ingestion&lt;/em&gt; and the answer to the question of health risk is NO.  The above answer is also incorrect if we allow for other possibilities (e.g. a cure for cancer).  If as Henry suggests in the comments we have a glass of water contaminated with iodine-131, and we keep it stored in the fridge for 90 days, then more than 99.9% of the isotope will have decayed to stable and harmless xenon.  But I would drink it only if there was no doubt that it was free of &lt;em&gt;other&lt;/em&gt; radioactive or poisonous chemicals (such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Cesium_137#Health_risk_of_radioactive_caesium&quot; rel=&quot;nofollow&quot;&gt;caesium-137&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2072" LastEditorUserId="174" LastEditorDisplayName="" LastEditDate="2011-03-26T16:25:56.037" LastActivityDate="2011-03-26T16:25:56.037" CommunityOwnedDate="2011-03-25T04:18:50.050" CommentCount="2" />
  <row Id="7491" PostTypeId="1" CreationDate="2011-03-24T05:46:03.657" Score="5" ViewCount="442" Body="&lt;p&gt;The traditional physics career is an academic job at some university, with the eventual goal of becoming a tenured professor. Is it possible for a mostly self-educated outsider working outside academia to come up with significant results in physics? Let's take significant to mean accepted in a high ranking peer review journal with a high citation count. Let's just say due to external life circumstances, the academic path is unfeasible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any examples of notable physics results coming from outsiders? e.g. a third class patent clerk coming up with light quanta, Brownian motion and relativity.&lt;/p&gt;&#xA;" OwnerUserId="2738" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-24T17:11:18.117" LastActivityDate="2011-05-08T08:29:23.040" Title="Is it possible to work on physics independently outside academia?" Tags="&lt;soft-question&gt;&lt;physics-careers&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="3" />
  <row Id="7492" PostTypeId="2" ParentId="7491" CreationDate="2011-03-24T06:21:25.260" Score="7" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Short answer:&lt;/strong&gt; Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Slightly longer answer:&lt;/strong&gt; Yes, but you will have to prove your chops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Discussion:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the physics &quot;career path&quot; of many garden variety crackpots, so you may meet some initial skepticism. Take care to understand and speak the language. Don't overstate your preparation or competence; nor the significance of any particular findings. Be prepared to show that you work is consistent with known physics. Establish some connections with people on more traditional career paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another concern: it may be hard to find a problem that can be given meaningful work without a lot of time devoted to it. They are out there, but you need to know the landscape to find them.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-03-24T06:21:25.260" />
  <row Id="7493" PostTypeId="2" ParentId="7491" CreationDate="2011-03-24T06:34:13.043" Score="6" ViewCount="" Body="&lt;p&gt;Most probably, no. This opinion is based on two considerations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)Being unable to join the academic stream means that other work decisions have been taken. Inevitably this leads to inability to focus 16/24 on a physics problem ( I put some time in for sleep). People who have left their mark in the history of physics were very focused on it, and  focusing I would consider  a prime prerequisite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My no is not a complete one  because in this day and age we have the internet. If the reason one is unable to join academia is something like Hawking would have suffered if his disease caught up early with him, or if one is taking care of a very sick loved one, there is still the internet to learn the latest and discuss the most important, currently. As long as time is not a problme . Which brings me to the second consideration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I had a professor in ancient greek who imparted the following wisdom: Knowledge is a circle. There is a frontier where explorations go on, but the more you know, the larger the radius, the greater the frontier. Unfortunately now there is an enormous body of knowledge continuously expanding and continuously needing new tools. Within academia a sifting takes place through discussions, coffee in the cafeteria, informal lectures, conferences, long before papers for  peer review are ready. Ideas are threshed out and collaborations set up that allow for faster knowledge accumulation. Some of this may start to happen on the internet  and it would need a particular type of personality to be able to survive without the face to face interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if he/she has the focus and the time to pursue a physics problem that is not contradictory to known experimental facts and proposes new experiments ( i.e. it is falsifiable) then  I wish them luck. I think the peer review process in physics is not too biased to look at a paper on its merits, but in any case there exist the archive locations where a paper can be put up on the internet and discussed on sites like this . &lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-05-08T08:29:23.040" LastActivityDate="2011-05-08T08:29:23.040" CommentCount="8" />
  <row Id="7494" PostTypeId="1" AcceptedAnswerId="7497" CreationDate="2011-03-24T07:40:34.353" Score="1" ViewCount="153" Body="&lt;p&gt;$\alpha$ radiation consist of positive charged helium nuclei, $\beta$ radiation of negative charged electrons. So why don't the $\alpha$ particles take those electrons to get neutral?&lt;/p&gt;&#xA;" OwnerUserId="2740" LastActivityDate="2011-03-24T11:29:38.360" Title="$\alpha$ and $\beta$-decay, why don't they neutralize each other?" Tags="&lt;nuclear-physics&gt;&lt;radiation&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="7495" PostTypeId="2" ParentId="7491" CreationDate="2011-03-24T07:47:57.700" Score="1" ViewCount="" Body="&lt;p&gt;It is certainly possible, given the many who have done so in the past.&#xA;Problems however quickly arise in modern physics, and that's the enormous cost of the equipment needed for a lot of fundamental research. The cost of this is beyond the means of most individuals, and even most corporations.&lt;/p&gt;&#xA;" OwnerUserId="2625" LastActivityDate="2011-03-24T07:47:57.700" CommentCount="11" />
  <row Id="7496" PostTypeId="2" ParentId="7494" CreationDate="2011-03-24T07:51:03.690" Score="0" ViewCount="" Body="&lt;p&gt;when they meet, it might happen. But calculate the chances of 2 particles meeting that have the right energy levels to actually combine. You'll find those chances are miniscule.&lt;/p&gt;&#xA;" OwnerUserId="2625" LastActivityDate="2011-03-24T07:51:03.690" CommentCount="1" />
  <row Id="7497" PostTypeId="2" ParentId="7494" CreationDate="2011-03-24T08:23:57.917" Score="2" ViewCount="" Body="&lt;p&gt;I agree with jwenting but in some sense, I feel that he is not answering the question: why there's no &quot;combined $\alpha$ plus $\beta$ decay in which a nucleus emits e.g. a helium atom?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, let me start with the $\beta$-decay. Nuclei randomly - after some typical time, but unpredictably - may emit an electron because a neutron inside the nuclei may decay via&#xA;$$ n\to p+e^- +\bar\nu$$&#xA;which may be reduced to a more microscopic decay of a down-quark,&#xA;$$ d\to u + e^- +\bar\nu.$$&#xA;This interaction, mediated by a virtual W-boson, is why a nucleus - with neutrons - may sometimes randomly emit an electron. So the $\beta$-decay is due to the weak nuclear force. On the other hand, the $\alpha$-decay is due to the strong nuclear force: the nucleus literally breaks into pieces, with a very stable combination of 2 protons and 2 neutrons appearing as one of the pieces (helium nucleus).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two processes above are independent, and each of them can kind of be reduced to a single elementary interaction whose origin is different. This independence and different origin is why the &quot;combined&quot; decay, with an emission of both electron (or two electrons) and a Helium nucleus, is extremely unlikely. Such an emission of a whole atom (which is electrically neutral but it is surely not &quot;nothing&quot;!) could only occur if several of the elementary decay interactions would occur at almost the same time which is extremely unlikely.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-24T08:23:57.917" />
  <row Id="7498" PostTypeId="2" ParentId="7487" CreationDate="2011-03-24T09:58:01.950" Score="3" ViewCount="" Body="&lt;p&gt;Hmmm,&#xA;those answers are ok, but they miss two critical points:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the faster the decay, the stronger stronger (decay/time) the radiation. &#xA;To include that, one should calculate something like amount/(half life). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;extremely important is, that land animals have chemical means to extract and &#xA;store incredible low amounts of iodine from drinks and what You eat. &#xA;Thus extremely small amounts of iodine in water and soil will be concentrated &#xA;1 by millions (or billions) in our thyroid gland. That is the big danger &#xA;and that is reason to give iodine tablets to endangered people, to &quot;saturate&quot; &#xA;the iodine uptake system. &#xA;All this is a heritage from the &quot;fish&quot; which changed their lifestyle to dwell &#xA;on land, where iodine is rare. We still preseve that piece of metabolism &quot;Invented&quot; &#xA;under marine conditions. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-24T10:48:50.063" LastActivityDate="2011-03-24T10:48:50.063" />
  <row Id="7500" PostTypeId="1" AcceptedAnswerId="7503" CreationDate="2011-03-24T11:02:37.370" Score="1" ViewCount="104" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;$HCl(g) + H_2O(l) -&gt; H_3O^{+}(aq) + OH^{-}(aq)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming $HCl$ completely ionizes in water, if we prepare a $HCl$ solution, will the $[HCl]=[H_3O]$ ($HCl$ concentration = $H_3O$ concentration)? That is, the $HCl$ before the reaction, and the $H_3O$ after reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, then if we observe the following equilibrium reaction for the auto-ionization of water:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$2H_2O \rightleftharpoons H_3O^{+} + OH^{-}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it can be concluded that $[H^{+}][OH^{-}]$ is a constant value, that is, the product of their equilibrium concentrations is a constant value. However, if the $H_3O$ concentration increases by the ionization of $HCl$, wouldn't the reaction shift to the left to counteract the change (by Le Chatelier's principle)? Consequently, this would decrease the concentration of $H_3O$ ions in solution, and contradict the equality of $HCl$ concentration = $H_3O$ concentration, so was the original proposition incorrect?&lt;/p&gt;&#xA;" OwnerUserId="1846" LastEditorUserId="1846" LastEditorDisplayName="" LastEditDate="2011-03-24T11:30:49.497" LastActivityDate="2011-03-24T11:50:37.907" Title="Auto-ionization of water" Tags="&lt;chemical-compounds&gt;" AnswerCount="1" />
  <row Id="7502" PostTypeId="2" ParentId="7462" CreationDate="2011-03-24T11:24:13.373" Score="1" ViewCount="" Body="&lt;p&gt;If the final result is &lt;em&gt;unique and independent&lt;/em&gt; of the regularization method, then it is a mathematically correct answer.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-03-24T14:26:47.407" LastActivityDate="2011-03-24T14:26:47.407" />
  <row Id="7503" PostTypeId="2" ParentId="7500" CreationDate="2011-03-24T11:37:59.633" Score="2" ViewCount="" Body="&lt;p&gt;No, the autoprotolysis equation equilibrium makes the concentration of OH- &#xA;to decrease appropriately. Say, the H3O+ is 1 mole/ltr, the OH- will be 10exp-14 moles per liter. If You calculate that precisely, You will find that the OH- and H3O+ will differ a tiny bit from that values. But compared to the deviations of the quilibrium due to temperature &#xA;and some other reasons, that is way below any acuracy of chemical measurement. &#xA;(and the calculus becomes somewhat ugly) &#xA;Chemists as I live very happily with such simplifications. pH differences of 0.1 are can measured quite reliantly, 0.01 affords a lot of experience and some tricky preparation and apparatus (unknown to say, 95 % of chemistry PHDs), and that is the end of all we can do. &#xA;Imagine, that that logarithmic scale (for equilibrium constants as well!) is not just some trick of chemists, nature behaves like that! Chemical potential depends linearly on log of concentration. Summa: that is why chemistry works so well on base of thumb rules. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-24T11:50:37.907" LastActivityDate="2011-03-24T11:50:37.907" CommentCount="4" />
  <row Id="7504" PostTypeId="2" ParentId="7089" CreationDate="2011-03-24T12:13:09.790" Score="2" ViewCount="" Body="&lt;p&gt;I have found &lt;a href=&quot;http://en.wikipedia.org/wiki/Relative_cost_of_electricity_generated_by_different_sources&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; in Wikipedia on relative energy costs. It contains a lot of data and some discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarise parts of it we note that the official UN bodies have developed a formula called &quot;Levelised Energy Cost&quot; (&lt;strong&gt;LEC&lt;/strong&gt;) which is a lifetime accounting formula which includes the following cost factors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Investment (per year) - the totals are&#xA;over the n years of the system's&#xA;life.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Maintenance and Operations (per year)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Fuel Use (per year) [minimal for some&#xA;sources like Nuclear, obviously high &#xA;for anything Coal/oil based.]&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With the benefit factor being the Electricity generation itself (per year).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the other Answers have reminded us, with Nuclear in particular we need to be careful about including the &quot;hidden subsidy&quot; and dual use (civilian/military) of these technologies, so maybe one should use the Nuclear columns in the data as a reference point, and investigate it in more detail if required. For the less military fuels the data may be more accurate (at least from that perspective).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the data tables in these lists give the LEC values, for the different energy types as estimated by various national bodies: US, UK , Australia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the &lt;a href=&quot;http://en.wikipedia.org/wiki/File:Levelized_energy_cost_chart_1,_2011_DOE_report.gif&quot; rel=&quot;nofollow&quot;&gt;US DOE Table&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a recent (commercial) study for the UK:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technology --------------------------------------                                 Cost range (£/MWh)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;New nuclear --------------------------------                                   80-105&lt;/li&gt;&#xA;&lt;li&gt;Onshore wind ------------------------------                                        80-110&lt;/li&gt;&#xA;&lt;li&gt;Biomass -------------------------------------                                  60-120&lt;/li&gt;&#xA;&lt;li&gt;Natural gas turbines with CO2 capture---                                           60-130&lt;/li&gt;&#xA;&lt;li&gt;Coal with CO2 capture --------------------                                    100-155&lt;/li&gt;&#xA;&lt;li&gt;Solar farms        -------------------------------                                125-180&lt;/li&gt;&#xA;&lt;li&gt;Offshore wind    -----------------------------                                150-210&lt;/li&gt;&#xA;&lt;li&gt;Natural gas turbine, no CO2 capture------                                          55-110&lt;/li&gt;&#xA;&lt;li&gt;Tidal power     -------------------------------                                   155-390&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In all these data the &quot;research costs&quot; have not been separated out, but seem to be included in the &quot;investment costs&quot; associated with the given estimates. As the articles also point out there are other costs associated with each type in addition to those considered in this question including any transmission and connection costs which any &quot;geographically remote&quot; technologies might incur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few extra points about reading the data from Commercial organisations. If company A has expertise in technology a then obviously their assessment (no matter how genuine) of the costs of technologies a, b, c will price the R&amp;amp;D and other costs for b and c much higher than a. (An oil company estimating solar versus oil costs, might be an example.) Whereas a specialist company B in technology b (which might be much smaller than A if dealing with newer technologies) will have a lower cost figure for b (versus anything else for them, and lower than A's estimate of b too if their product is good.) Of such are commercial and political arguments made.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-03-24T12:56:45.570" LastActivityDate="2011-03-24T12:56:45.570" CommentCount="1" />
  <row Id="7505" PostTypeId="1" CreationDate="2011-03-24T12:46:16.290" Score="2" ViewCount="151" Body="&lt;p&gt;I'm trying to determine the divergence angle of light from a single lens that is completely illuminated by a high power LED. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most optics textbooks only deal with imaging optics and I'm having a hard time wrapping my head around what happens when I don't care about the image on the focal plane - this is for a non-imaging application (I'm just projecting the light), and I just want to be able to determine the spot size at any distance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LED will not necessarily be at the focal point of the optic (though it will be along its axis). Furthermore, the LED isn't a point-source, as its size is approximately 1/10th of the focal length of the lens.&lt;/p&gt;&#xA;" OwnerUserId="2735" LastActivityDate="2011-03-25T15:15:58.903" Title="How to determine divergence of a LED source from a single biconvex lens" Tags="&lt;optics&gt;&lt;light&gt;" AnswerCount="1" />
  <row Id="7506" PostTypeId="1" CreationDate="2011-03-24T13:58:42.773" Score="7" ViewCount="1283" Body="&lt;p&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1103/1103.1879v1.pdf&quot; rel=&quot;nofollow&quot;&gt;The half-page arxiv doc by Joy Christian&lt;/a&gt; of Oxford Uni, UK has the Title and Abstract:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Disproof of Bell’s Theorem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We illustrate an explicit&#xA;  counterexample to Bell’s theorem by&#xA;  constructing a pair of dichotomic&#xA;  variables that exactly reproduce the&#xA;  EPR-Bohm correlations in a manifestly&#xA;  local-realistic manner.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As The Bell's theorem is often cited it seems important to clarify this short doc.&lt;br&gt;&#xA;Is the doc OK?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT add  2011/07/12&lt;br&gt;&#xA;this recent paper &#xA;&lt;a href=&quot;http://arxiv.org/abs/1107.1007&quot; rel=&quot;nofollow&quot;&gt;Comments on &quot;Disproof of Bell's theorem&quot;&lt;/a&gt; by Florin Moldoveanu (July 06,2011)&lt;br&gt;&#xA;can shed light in the interpretation and importance of the work of Joy&#xA;and may be the case that the answers could benefit from it.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One conclusion that seems relevant, imo, is that Geometric Algebra is very promising to model the physical world.&lt;br&gt;&#xA;EDIT add  end&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-07-12T18:27:30.143" LastActivityDate="2011-07-12T18:27:30.143" Title="Disproof of Bell’s Theorem" Tags="&lt;quantum-mechanics&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="7" />
  <row Id="7507" PostTypeId="2" ParentId="7483" CreationDate="2011-03-24T14:00:28.567" Score="0" ViewCount="" Body="&lt;p&gt;This question is quite speculative.  What I will write below is speculative as well, and doubtless there will be down voters on this.  However, here goes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there may be a role for fractal or fractal-like geometry in spacetime physics with respect to discrete groups.  This might lie with the use of discrete groups with hyperbolic spaces, such as the anti de Sitter spacetime.  The AdS in 2-dimensions has a structure similar to the Poincare disk.  The Escher print below illustrates a particular discrete group, eg $PSL(2,Z)~\subset~PSL(2,R)$, which tessellates the space.  This is a Euclideanized version of the AdS, and the boundary of this disk corresponds to a conformal structure that is equivalent to the isometries of the AdS.  $AdS_n~=~SO(n-1,2)/SO(n-1,1)$ with the isometry group $SO(n-1,2)$  for $n~=~2$ this is the $SO(2,1)~\simeq~SL(2,R)$, which is the elementary group of conformal quantum mechanics.  For an $AdS_4$ that contains a black hole, where $AdS_4$ is the perfect box to hold a black hole due to its negative Gaussian curvature, the structure of the spacetime becomes $AdS_2\times S^2$ near the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is mentioned because it is possible that the tessellation of the $AdS_2$, which may extend to $AdS_3$ or even higher dimensions, has a quasi-periodic structure if there is a fibration over the AdS of a particular nature.  This tessellation is a case of quasi-crystals which are five-fold tilings in $2$ and $3$ dimension have pentagonal or isosahedral symmetry.  The $AdS_2$ may then exist in as the result of compactification and the near horizon condition on larger $AdS_n$, so that the resulting fibration over the $AdS_2$ gives a quasi-periodic tiling of the hyperbolic space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are similarities in behavior between quasi-crystals and the zeros of the Riemann zeta function.  It is possible that quasi-crystal structure might play a role in some future proof of the Riemann zeta function conjecture.  Where this might play a role in physics is if the quantum states of the universe turn out to have some correspondence to the zeros of the Riemann zeta function.  So there is a prospect that quasi-crystals and tessellations are an important part of the understanding the quantum states of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one dimensional ‘fractal” or quasi-periodic structure has already played a role in number theory.  The integer partition and Ramanujan’s congruencies have been found to be due to a one dimensional fractal or quasi-periodic structure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.aimath.org/news/partition/folsom-kent-ono.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.aimath.org/news/partition/folsom-kent-ono.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is then not unreasonable to think there are even deeper relationships between these types of structures and the quantum states of gravitation and the universe.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $AdS$ spacetime is of course not the spacetime of the universe. It really is more of a mathematical gadget.  So this may be somewhat indirect with respect to this question.  Further, there are elements to what I wrote here that are not known.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-24T14:00:28.567" />
  <row Id="7508" PostTypeId="2" ParentId="7315" CreationDate="2011-03-24T14:07:25.433" Score="1" ViewCount="" Body="&lt;p&gt;EDIT&lt;br&gt;&#xA; Georg's note make me see that is not an answer to the OP.&lt;br&gt;&#xA;     I leave the rest of the text only to make notice of a very promising new techonoly.&lt;br&gt;&#xA;end EDIT  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.technologyreview.com/blog/arxiv/26560/?ref=rss&quot; rel=&quot;nofollow&quot;&gt;Superconducting Niobium Chip Smashes Silicon Power Consumption Standards&lt;/a&gt;&lt;/strong&gt;&#xA;A superconducting logic chip with a clock speed of 6 GHz &lt;strong&gt;&lt;em&gt;beats silicon energy efficiency by two orders of magnitude&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ref: &lt;a href=&quot;http://arxiv.org/abs/1103.4269&quot; rel=&quot;nofollow&quot;&gt;Ultra-Low-Power Superconductor Logic&lt;/a&gt;&lt;br&gt;&#xA;Abstract:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We have developed a new&#xA;  superconducting digital technology,&#xA;  Reciprocal Quantum Logic, that uses AC&#xA;  power carried on a transmission line,&#xA;  which also serves as a clock. Using&#xA;  simple experiments we have&#xA;  demonstrated zero static power&#xA;  dissipation, thermally limited dynamic&#xA;  power dissipation, high clock&#xA;  stability, high operating margins and&#xA;  low BER. These features indicate that&#xA;  the technology is scalable to far more&#xA;  complex circuits at a significant&#xA;  level of integration. On the system&#xA;  level, Reciprocal Quantum Logic&#xA;  combines the high speed and low-power&#xA;  signal levels of Single-Flux- Quantum&#xA;  signals with the design methodology of&#xA;  CMOS, including low static power&#xA;  dissipation, low latency combinational&#xA;  logic, and efficient device count.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-03-24T18:30:37.283" LastActivityDate="2011-03-24T18:30:37.283" CommentCount="5" />
  <row Id="7509" PostTypeId="1" CreationDate="2011-03-24T14:10:11.383" Score="1" ViewCount="155" Body="&lt;p&gt;I recently read up about gyroscopes, angular momentum and mechs (the big Cockpit controlled robots) and was wondering if it would be possible to get a stable walking mech (only as example, not meant to be a serious attempt &lt;a href=&quot;http://autopixx.de/bilder/lM23mgzq/battle-mech.jpg&quot; rel=&quot;nofollow&quot;&gt;http://autopixx.de/bilder/lM23mgzq/battle-mech.jpg&lt;/a&gt;) by adding a big enough gyroscope which stabilizes the mech as it lifts a leg in the walk cycle. The mech is about 4 meters high, but I can't speculate on how much it probably would weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just a question out of curiosity. Thanks in advance :)&lt;/p&gt;&#xA;" OwnerUserId="2743" LastActivityDate="2011-08-16T11:49:21.200" Title="Mech stability through gyroscope" Tags="&lt;angular-momentum&gt;&lt;stability&gt;&lt;gyroscopes&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7510" PostTypeId="2" ParentId="7506" CreationDate="2011-03-24T14:14:49.547" Score="9" ViewCount="" Body="&lt;p&gt;Oh crud, this!  I read this paper a month ago.  What Joy Christian does is to write out the Bell inequalities, and then effectively identifies quantum states with the variables in the terms of the inequality.  This is silly, for the whole point is to erect the inequalities and then demonstrate how quantum states violate them.  Joy identifies the quantum states with the elements in the inequality.  The whole thing is what I would call tautologically false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing worse that somebody trying to disprove nonlocal properties of quantum mechanics are knuckleheads who try to show that relativity is all wrong.  There seems to be an endless stream of this sort of nonsense.  These types of papers are simply best ignored.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;{\bf Addendum}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joy Christian tacitly equates the elements in the inequality with the quantum states because there is no sign value to the outcome of a quantum measurement.  There is this 2-1 relationship with a Bloch sphere and the $R^3$ sphere.  The lack of this sign, say a spin up or down, involves equating the average of noncommuting variables with average of commuting variables.  It is for this reason that the quantum variable effectively slip into place with the classical probability variables in the inequalities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do understand Bell's theorem well enough.  It is a demonstration that quantum mechanics does not obey classical set theory. The corresponding case classically involve projecting onto subspaces of an entangled state &#xA;$$&#xD;&#xA;|\psi\rangle~=~1/\sqrt{2}(|+,-\rangle~+~|-.+\rangle)&#xD;&#xA;$$&#xA;for the singlet state configuration. So the Pauli matrices for the two are $\sigma_i~\tau_i$, the set of projector operators on the 1 and 2 states are employed&#xA;$$&#xD;&#xA;P(1)_z~=~(1/2)(1~+~\sigma_z), P(2)_z~=~(1/2)(1~+~\tau_z) &#xD;&#xA;$$&#xA;and for the 45 degree case &#xA;$$&#xD;&#xA;P(1)_{45}~=~(1/2)(1~+~(\sigma_z~+~\sigma_x)/\sqrt{2}), ~P(2)_{45}~=~(1/2)(1~+~(\tau_z~+~\tau_x)/\sqrt{2}) &#xD;&#xA;$$&#xA;and &#xA;$$&#xD;&#xA;P(1)_x~=~(1/2)(1~+~\sigma_x), P(2)_x~=~(1/2)(1~+~\tau_x). &#xD;&#xA;$$&#xA;The projections onto the entangled state which correspond to the classical probability rules is&#xA;$$&#xD;&#xA;Prob(|, /)~=~P(1)_z*P(2)_{45}&#xD;&#xA;$$&#xA;$$&#xD;&#xA;Prob(/, \_ )~=~P(1)_{45}*P(2)_x&#xD;&#xA;$$&#xA;$$&#xD;&#xA;Prob(|,\_ )~=~P(1)_z*P(2)_x&#xD;&#xA;$$&#xA;Some calculations with the matrices and the states leads to the Bell result that this does violate the inequality $Prob(|,\_)~\ge~Prob(|,/)~+~Prob(/,\_)$ expected classically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In effect the classical inequality is derived from union and intersection rules $\cup,~\cap$, with OR and AND logical meaning, while the quantum analogue involves additions and products of operators to construct “spans” on a vector space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are these extraordinary claims which come about now and then.  Of course 100 years ago the claims of quantum mechanics would have been seen as extraordinary.  Even Einstein did not like it, though he laid down some initial groundwork on it.  However, the physics of quantum mechanics and its unusual implications have a vast data base of experimental support behind them.  There is no “crisis” with quantum mechanics, where the potential deviations may lie near the Planck scale.  However, for most purposes these are not important.  Claims that quantum physics is wrong, or that there are hidden variables or that Bell’s theorem is wrong or violated have come and gone.  Much the same is likely to happen here as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have spent far more time on this than I wanted to.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-03-24T17:36:26.067" LastActivityDate="2011-03-24T17:36:26.067" CommentCount="26" />
  <row Id="7511" PostTypeId="1" AcceptedAnswerId="7520" CreationDate="2011-03-24T14:24:01.033" Score="5" ViewCount="286" Body="&lt;p&gt;This new &lt;a href=&quot;http://www.newscientist.com/article/mg20928054.000-bestever-quantum-measurement-breaks-heisenberg-limit.html&quot; rel=&quot;nofollow&quot;&gt;Best-ever quantum measurement breaks Heisenberg limit&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;PHYSICISTS have made the most accurate&#xA;  quantum measurement yet, breaking a&#xA;  theoretical limit named for Werner&#xA;  Heisenberg.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nature, DOI: 10.1038/nature09778  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When an experiment breaks a theoretical limit we have to reassess what we are accustomed to know.&#xA;Is it the case with this experiment? &#xA;At what level ?&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-03-24T16:10:07.187" Title="The Heisenberg limit is not a limit?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7512" PostTypeId="2" ParentId="7491" CreationDate="2011-03-24T14:31:10.633" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.hhmi.org/research/groupleaders/betzig_bio.html&quot; rel=&quot;nofollow&quot;&gt;Eric Betzig&lt;/a&gt; did this... kinda. He was a trained physicist who worked in academia, but produced some &lt;a href=&quot;http://dx.doi.org/10.1126/science.1127344&quot; rel=&quot;nofollow&quot;&gt;significant results&lt;/a&gt; during a break from professional science. Notice, though, that when Betzig got finished dreaming in isolation, he motivated a team of excellent scientists to help make his dream reality. Very few papers in Science and Nature have an author list of one person.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you don't get a job at a university, consider associating yourself with one, somehow. If you're planning to work for free anyway, you might as well volunteer in someone's lab. In my personal experience, I work much more productively when surrounded by other scientists, especially if they care about my work. It's hard to have truly original thoughts without at least &lt;em&gt;some&lt;/em&gt; outside stimulus, and it can be hard to stay motivated without the social reward of 'showing off'.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2011-03-24T14:31:10.633" />
  <row Id="7513" PostTypeId="1" CreationDate="2011-03-24T14:51:02.807" Score="0" ViewCount="73" Body="&lt;p&gt;Suppose one knows almost nothing about the nature of radioactivity (like the discoverers of this phenomenon). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the detailed/rigorous logical steps/arguments to show experimentally that radioactive radiation ionizes air?&lt;/p&gt;&#xA;" OwnerUserId="2740" LastActivityDate="2011-03-24T16:24:42.260" Title="Detailed logical steps to show experimentally that radioactivity ionizes air?" Tags="&lt;nuclear-physics&gt;&lt;radioactivity&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="7515" PostTypeId="2" ParentId="7513" CreationDate="2011-03-24T15:15:00.727" Score="1" ViewCount="" Body="&lt;p&gt;Have a look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cloud_chamber&quot; rel=&quot;nofollow&quot;&gt;cloud chamber&lt;/a&gt; article in wikipedia. It shows that there is ionisation of the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the discovery of the harmful effects of &lt;a href=&quot;http://en.wikipedia.org/wiki/Radioactivity#Discovery&quot; rel=&quot;nofollow&quot;&gt;radioactivity.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.scienceinschool.org/2010/issue14/cloud&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is how to build your own cloud chamber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sort of measurements are repeated in student labs the world over.&#xA;These are the first researchers in &lt;a href=&quot;http://www.accessexcellence.org/AE/AEC/CC/historical_background.php&quot; rel=&quot;nofollow&quot;&gt;radiation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-24T16:24:42.260" LastActivityDate="2011-03-24T16:24:42.260" CommentCount="2" />
  <row Id="7516" PostTypeId="2" ParentId="7513" CreationDate="2011-03-24T15:25:47.300" Score="0" ViewCount="" Body="&lt;p&gt;I would read the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Smoke_detector#Ionization&quot; rel=&quot;nofollow&quot;&gt;smoke detectors&lt;/a&gt;. Then I would buy a smoke detector, crack one open, and try to build a similar device myself.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2011-03-24T15:25:47.300" CommentCount="2" />
  <row Id="7519" PostTypeId="2" ParentId="7505" CreationDate="2011-03-24T15:55:27.013" Score="3" ViewCount="" Body="&lt;p&gt;Determine = design or measure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A dark-ish room, a white piece of paper and a ruler should do for a bright LED.  Modelling it accurately, especially if the LED has a plastic lens built-in and you don't know the precise details of the die position, is trickier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the LED has a builtin lens then about the best you can do is just match the angle coming out of the led to a lens size and put the LED at the focal point (actually just beyond the focal point is better)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/0aMnx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since your lens isn't a single point you will end up with something like this - because the edges of your LED can't both be at the focal point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/gJC8B.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://alumnus.caltech.edu/~leif/infratag/lens_choice.html&quot; rel=&quot;nofollow&quot;&gt;http://alumnus.caltech.edu/~leif/infratag/lens_choice.html&lt;/a&gt; for details - it's specifically about IR leds but that doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: for free space optical comms. If you can choose the LED then a bare surface mount device with no built-in lens is much closer to a point source. A longer focal length will give the best colimation - for experiments a 500mm camera mirror lens or telescope is a good start&lt;/p&gt;&#xA;" OwnerUserId="2525" LastEditorUserId="2525" LastEditorDisplayName="" LastEditDate="2011-03-25T15:15:58.903" LastActivityDate="2011-03-25T15:15:58.903" CommentCount="6" />
  <row Id="7520" PostTypeId="2" ParentId="7511" CreationDate="2011-03-24T16:10:07.187" Score="11" ViewCount="" Body="&lt;p&gt;The use of the term &quot;Heisenberg Limit&quot; is somewhat misleading for outsiders (that is non-quantum Interferometers). If we recall the Heisenberg Uncertainty Principle is a limit on simultaneous measurement of two complementary variables. In the case of (quantum) metrology one is only interested in the measurement of a &lt;em&gt;single&lt;/em&gt; variable to high accuracy, and this does not (directly) conflict with the HUP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of interferometry the variable of interest is $\Delta \Phi$ the phase difference between two waves detected in two arms. In basic interferometry there were some limits as to how accurately this could be measured:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum Shot Noise : $\Delta \Phi = 1/N^{1/2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heisenberg Limit : $\Delta \Phi = 1/N $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the N corresponds to how many quanta are required for the given accuracy, so the second is more accurate when it can be achieved, as was eventually done using entangled states, and perhaps squeezed light. If you cannot use these features of QM one gets just the Quantum Shot Noise accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well a few years ago it was noticed that the assumption behind the Heisenberg limit calculation was that the Hamiltonian was quadratic in its (key) variables: this corresponded to the assumption of linearity amongst the measuring quanta. If the Hamiltonian could be made non-linear then an improvement on the Heisenberg limit would be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This interaction between the measuring photons is discussed in the given paper, in Arxiv form &lt;a href=&quot;http://arxiv.org/abs/1012.5787&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-03-24T16:10:07.187" CommentCount="3" />
  <row Id="7522" PostTypeId="1" AcceptedAnswerId="7524" CreationDate="2011-03-24T16:46:20.607" Score="1" ViewCount="139" Body="&lt;p&gt;For a given black body radiation curve, would the changes to the spectrum resulting from cosmological expansion and those from Doppler effects be distinguishable on the basis of the shapes of the resulting curves alone? Or, put another way, starting from the same spectrum, can both processes produce the same observation (for suitably chosen magnitudes of expansion or velocity)?&lt;/p&gt;&#xA;" OwnerUserId="2540" LastActivityDate="2011-03-25T00:42:42.560" Title="Do cosmological and Doppler redshift produce different patterns?" Tags="&lt;spectroscopy&gt;&lt;space-expansion&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="7524" PostTypeId="2" ParentId="7522" CreationDate="2011-03-24T17:14:58.053" Score="3" ViewCount="" Body="&lt;p&gt;The redshift due to cosmological expansion is identical to a Doppler shift in its effect on the spectrum of any source. To be specific, both phenomena &quot;stretch&quot; all wavelengths by the same factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a very good reason for this: in a suitable coordinate system, the cosmological redshift &lt;em&gt;is&lt;/em&gt; a Doppler shift. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll find statements in some textbooks saying that this isn't true, but a weak version of this statement, which is nonetheless strong enough to explain why the effects on the spectra are identical, is uncontroversially true. To be specific, the redshift of a distant galaxy can be thought of as the accumulation of many infinitesimal Doppler shifts along the line of sight. (Each member of a family of comoving observers is in motion relative to her neighbor, and each can &quot;watch&quot; the redshift build up gradually due to these relative velocities.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One perspective on this subject (mine, to be precise) can be found in &lt;a href=&quot;http://arxiv.org/abs/0808.1081&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;. Even if you don't like our point of view in this paper, our description of and references to other treatments may be of interest.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-24T17:14:58.053" CommentCount="6" />
  <row Id="7526" PostTypeId="1" CreationDate="2011-03-24T17:44:54.270" Score="1" ViewCount="181" Body="&lt;p&gt;Gravity is ignored in the SM. The proton rest mass is ~0.938 GeV/$c^2$. LHC protons will move with 7 TeV energy, presumably with a relativistic mass about 7,450 times rest mass. A cosmic ray with the highest energy was detected at about $6\times 10^{21}$ eV. If it was a proton, its relativistic mass would have been about $6.4\times 10^{12}$ times rest mass. My question is, at what energy levels would it be necessary to include gravity in the SM? I recognize this isn't an issue for forseeable particle accelerators, but astronomical events like supernova are capable of generating extremely high energy particles.&lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-03-24T18:25:37.213" LastActivityDate="2011-03-25T03:03:07.480" Title="Gravity and the Standard Model" Tags="&lt;gravity&gt;&lt;standard-model&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="7527" PostTypeId="2" ParentId="7377" CreationDate="2011-03-24T18:18:11.350" Score="-1" ViewCount="" Body="&lt;p&gt;If it use &lt;a href=&quot;http://www.crossfirefusion.com/nuclear-fusion-reactor/overview.html&quot; rel=&quot;nofollow&quot;&gt;aneutronic fusion&lt;/a&gt; instead nuclear fission, it will produce electricity directly with an efficiency exceeding 90%, requiring less water for cooling.&lt;/p&gt;&#xA;" OwnerUserId="2748" LastActivityDate="2011-03-24T18:18:11.350" CommentCount="2" />
  <row Id="7528" PostTypeId="2" ParentId="7526" CreationDate="2011-03-24T18:19:25.063" Score="1" ViewCount="" Body="&lt;p&gt;This question is a bit open ended.  There are theories and maybe some experimental hints of extra large dimensions and black hole or AdS amplitudes with heavy ion collisions.  The jury is of course still out, but there might be some gravity physics creeping into the standard model + QCD in the TeV energy range and above.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-24T18:19:25.063" />
  <row Id="7529" PostTypeId="2" ParentId="5066" CreationDate="2011-03-24T18:26:19.537" Score="-3" ViewCount="" Body="&lt;p&gt;I think the most realistic way of high speed space propulsion to get close to the speed of light is by using &lt;a href=&quot;http://tinyurl.com/nuclear-fusion-starship&quot; rel=&quot;nofollow&quot;&gt;phase-shifted electrodynamic propulsion&lt;/a&gt; instead of expelling-mass rockets.&lt;/p&gt;&#xA;" OwnerUserId="2748" LastActivityDate="2011-03-24T18:26:19.537" />
  <row Id="7530" PostTypeId="1" CreationDate="2011-03-24T18:33:24.597" Score="4" ViewCount="262" Body="&lt;p&gt;Due to certain reasons, I don't have the chance for going to college for under-graduation, but that doesn't mean I can't choose to walk the hard path and learn. Over the past month or so a lot of people have asked about resources for physics online or offline, but what I want isn't just the resources, but advice on what to study.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have taught myself the basic undergrad stuff (essentially classical mechanics and electromagnetism with a bit of modern physics with things like the one dimension time in/dependent Schrödinger wave equation) through a combination of MIT OCW, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0132431068&quot; rel=&quot;nofollow&quot;&gt;Giancoli&lt;/a&gt; and asking around, but what boggles me is how to proceed ahead. I want to understand things instead of just knowing their names and there is so much to physics beyond this that I don't know where to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm reaching out like this because I'm hoping that someone will give me a basic idea on where to start tackling these things. For example, I have a deep interest in thermodynamics to learn it what I would love is an answer that actually talks about; what's the next step (or book)? What do you recommend to get a solid grinding for any special math behind it? Any good experiments to actually see what I'm learning? And so on...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this might be a tall order, but I just had to ask. I'm sorry if this detracts from the main purpose of this site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh and as far as where my interest lies. Well I just like to create things and what I'm fascinated about is creating systems no matter what form they take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; I know that this question may seem broad and vague, but it's just that I see things in a sort of connected way in my mind. So, although I want to create things and it might appear to be an engineering question it really isn't. I think that traditionally what comprises of engineering is just one view at the same problem. Physics is another. So, if you're wondering what topic to answer on then pick anyone that you know well and you think is connected to the world and helps to expose some of our day to day realities. Things I can use to gain perspective while creating designs. (Particle physics for example, doesn't really make the cut, but fluid dynamics does and so on.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update 2&lt;/strong&gt;: I'm sorry for all of the confusion, but I'm essentially asking how to learn applied physics.&lt;/p&gt;&#xA;" OwnerUserId="1065" LastEditorUserId="1065" LastEditorDisplayName="" LastEditDate="2011-03-25T17:47:36.947" LastActivityDate="2011-03-26T01:05:58.673" Title="Going Beyond Introductory Physics" Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="4" CommentCount="16" FavoriteCount="1" />
  <row Id="7531" PostTypeId="2" ParentId="7506" CreationDate="2011-03-24T18:36:57.730" Score="2" ViewCount="" Body="&lt;p&gt;Mr. Crowell has yet again demonstrated his total lack of understanding of Bell’s theorem, as well as my argument against it. To begin with, Bell’s theorem has nothing to do with quantum mechanics per se, or classical set theory for that matter. It is a theorem about any possible future theory of physics---without prejudice or preconception displayed by Mr. Crowell concerning what that theory could be---and involves only some very basic assumptions about the completeness of that theory, and whether or not it could be a locally causal theory. My papers too have nothing to do with quantum mechanics per se (so please stop misrepresenting them without ever having read them). They have to do with how Bell’s argument against a certain class of possible locally causal theories is simply wrong, and more importantly how we can better understand the origin and strength of the observed correlations in nature. Mr. Crowell, you DO NOT understand Bell’s theorem, or my work in particular. Your opinion about my work is based on prejudice and ignorance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in response to the original question raised by Helder Velez let me describe the essence of my argument in the short paper. Given the background described above, the basis of Bell’s theorem is the claim that no local and realistic model can reproduce the experimental data observed in the EPR-Bohm type bipartite experiments. Therefore, all one has to do to refute Bell’s theorem is to produce such a local-realistic model. In the above paper I show that half-a-page is all it takes to produce such a model. No elaborate arguments are needed, since Bell’s argument explicitly rests on the impossibility of such a local model. Thus, a correct response to Helder Velez’s question should have been an attempt by the participants to assess whether or not I have produced the local model I claim I have produced. But that is not what we are getting so far.&lt;/p&gt;&#xA;" OwnerUserId="2755" LastEditorUserId="2755" LastEditorDisplayName="" LastEditDate="2011-03-24T23:04:58.283" LastActivityDate="2011-03-24T23:04:58.283" CommentCount="9" />
  <row Id="7532" PostTypeId="1" CreationDate="2011-03-24T19:09:53.067" Score="-2" ViewCount="83" Body="&lt;p&gt;Let's suppose there is a bulding $10m$ tall and the earth's gravity is rounded to $10m/s^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I dropped a ball from the top of the building. It takes 1 second to hit the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final velocity I:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V = g.t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V = 10.1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V = 10m/s$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S - S0 = 10.1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S - S0 = 10m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless: Final Velocity II&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V = \sqrt{2g.(S - S0)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V = \sqrt{2.10.10}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V = \sqrt{200 m/s}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$10  ≠  \sqrt{200}$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Please, could someone point out why I'm making this mistake ? I'm trying to understand it better but a need some help to figure out why $S - S0$ doesn't work for the second example. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2750" LastActivityDate="2011-03-24T19:15:50.393" Title="Kinematics on dropping a ball from the peak" Tags="&lt;classical-mechanics&gt;&lt;kinematics&gt;" AnswerCount="2" />
  <row Id="7533" PostTypeId="2" ParentId="7532" CreationDate="2011-03-24T19:14:27.980" Score="1" ViewCount="" Body="&lt;p&gt;The first incorrect statement is &quot;It takes 1 second to hit the ground.&quot; If that were true, it would have an &lt;em&gt;average&lt;/em&gt; speed of 10 m/s over the entire 1-second journey (since it traveled 10 m in 1 s). In fact, though, 10 m/s is the &lt;em&gt;final&lt;/em&gt; speed at the end of the 1-second interval. The average speed over the entire journey is less than the final speed -- in fact, for something starting from rest and moving with constant acceleration, the average speed is half the final speed.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-24T19:14:27.980" CommentCount="4" />
  <row Id="7534" PostTypeId="2" ParentId="7532" CreationDate="2011-03-24T19:15:50.393" Score="1" ViewCount="" Body="&lt;p&gt;s = 1/2 * g * t² &lt;/p&gt;&#xA;&#xA;&lt;p&gt;s = 1/2 * 10 * 1 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;s = 5 m&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-24T19:15:50.393" />
  <row Id="7535" PostTypeId="2" ParentId="7530" CreationDate="2011-03-24T19:45:41.753" Score="0" ViewCount="" Body="&lt;p&gt;I like these video-lectures very much&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.newpackettech.com/Resources/Susskind/&quot; rel=&quot;nofollow&quot;&gt;http://www.newpackettech.com/Resources/Susskind/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statistical physics course explains some of the microscopic basic principles underlying thermodynamics...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Professor Susskind is an awfully good and patient teacher. &#xA;I`ve never seen someone else giving such an easily accessible explanation of the meaning of the curvature tensor in GR for example.&lt;/p&gt;&#xA;" OwnerUserId="2751" LastEditorUserId="2751" LastEditorDisplayName="" LastEditDate="2011-03-24T20:11:16.973" LastActivityDate="2011-03-24T20:11:16.973" />
  <row Id="7536" PostTypeId="1" AcceptedAnswerId="7537" CreationDate="2011-03-24T20:12:42.977" Score="3" ViewCount="109" Body="&lt;p&gt;Some food or medical items, like rasberry juice concentrate I just bought, have a &quot;protect from light&quot; warning on the label. How can light influence products like milk and what does this warning mean in practice?&lt;/p&gt;&#xA;" OwnerUserId="2752" LastActivityDate="2011-03-24T20:46:18.143" Title="&quot;Protect from light&quot; warning" Tags="&lt;light&gt;&lt;physical-chemistry&gt;&lt;everyday-life&gt;" AnswerCount="1" />
  <row Id="7537" PostTypeId="2" ParentId="7536" CreationDate="2011-03-24T20:46:18.143" Score="4" ViewCount="" Body="&lt;p&gt;The photons of light, especially ultra violet photons from sunlight, carry energy that can break chemical bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bonds in chemicals that make up colours, ie. dye molecules, are most affected  - since they create the colour effect by absorbing some photons (of one colour) and not others. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So they have chemical bonds of the correct energy to absorb light, these molecules will eventually be destroyed by strong light and the color will fade - which is what happens to clothes left in strong light.  In the case of the fruit juice it will probably lose some of it's strong colour, it may also destroy some of the vitamins - but it's unlikely to make the juice poisonous!&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-03-24T20:46:18.143" CommentCount="1" />
  <row Id="7538" PostTypeId="2" ParentId="7441" CreationDate="2011-03-24T20:47:23.457" Score="2" ViewCount="" Body="&lt;p&gt;Re: the original question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted &quot;zero gravity&quot; at the top of the tower, you'd have to build a tower tall enough to reach the height of geostationary orbit: a point at which the orbital period of an object in freefall matches the time it takes the earth to rotate once.  As other commenters have pointed out, that'll occur at a height of roughly 35000 km above ground.  Good luck!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Re: The claim that &quot;but one may still say that the limitations are of an engineering (and budgetary) character rather than fundamental physics limitations.&quot;  I disagree.  One fundamental physical limitation is the fact that matter is held together by chemical bonds.  This limits the ultimate strength of any material (although AFAIK for all macroscopic materials mankind currently produces, the ultimate strength is far shy of what one would get from a &quot;perfect&quot; material).  The ultimate strength will limit how tall a tower one can build.  This is, for example, the reason why small asteroids can be quite aspherical, but large asteroids (and planets) cannot: an highly aspherical planet would collapse under its own gravitational force and become near-spherical due to the finite strength of the materials it's made of.  The ultimate strength limit also means that you can't build the &quot;space elevator&quot; other folks mention using any currently available materials that I know of.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-03-24T20:47:23.457" />
  <row Id="7539" PostTypeId="1" CreationDate="2011-03-24T22:05:42.370" Score="3" ViewCount="350" Body="&lt;p&gt;Nuclear fusion research appears to promise sustainable, practically unlimited energy, and according to &lt;a href=&quot;http://www.iter.org/environment&quot; rel=&quot;nofollow&quot;&gt;ITER&lt;/a&gt; it will be a perfectly safe, non-polluting source of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is if there are any plausible scenarios under which operating a thermonuclear reactor could cause a major environmental disaster?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Products of the fusion process are Helium, which is inert and harmless, and neutrons, which will lodge in the vessel walls and produce heat and activation of materials.&quot; This obviously assumes normal operating conditions, but what would happen if some man-made or natural catastrophe was to impact the reactor and create abnormal operating conditions?&lt;/p&gt;&#xA;" OwnerUserId="2628" LastEditorUserId="2710" LastEditorDisplayName="" LastEditDate="2011-03-28T04:29:02.700" LastActivityDate="2011-03-28T07:56:55.523" Title="Is there any inherent danger in operating a Thermonuclear Reactor?" Tags="&lt;nuclear-engineering&gt;&lt;fusion&gt;" AnswerCount="6" CommentCount="6" FavoriteCount="1" />
  <row Id="7540" PostTypeId="1" AcceptedAnswerId="7597" CreationDate="2011-03-24T22:06:51.843" Score="7" ViewCount="279" Body="&lt;p&gt;A question needed for a &quot;solid&quot; sci-fi author: How to detect a strong magnetic monopole? (yes, I know no such thing is to be found on Earth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of basic construction details, principles of operation and necessary components of a device capable of detecting/recognizing a macroscopic object emitting magnetic field of equivalent of order ~0.1-10 Tesla near its surface, but with only one pole, reliably distinguishing it from normal (2-pole) magnets, preferably at a distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Preferably a robust method, not involving extremely advanced technology. Detect the presence, possibly distance (or field strength) and direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know of SQUIDs, but these concentrate on extreme sensitivity. I'm thinking of something less sensitive but more robust (like, no need for the monopole to fall through the loop) and still able to recognize a monopole against a magnet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how would such a macroscopic object behave practically? Such a &quot;one-pole magnet&quot; about the size and strength of a refrigerator magnets - how would it behave around ferromagnetics, normal magnets and so on?&lt;/p&gt;&#xA;" OwnerUserId="2754" LastActivityDate="2011-03-26T16:14:42.820" Title="How would I go about detecting monopoles?" Tags="&lt;electromagnetism&gt;&lt;soft-question&gt;&lt;measurement&gt;&lt;magnetic-monopoles&gt;" AnswerCount="5" FavoriteCount="2" />
  <row Id="7541" PostTypeId="2" ParentId="7539" CreationDate="2011-03-24T22:14:04.397" Score="5" ViewCount="" Body="&lt;p&gt;A heck of a lot of energy in one place always represents a local danger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question of a more wide spread danger depends on a lot of details. Does the core reaction produce neutrons? How bad is the activation rate in the plant? Would a disaster event spread activated material over a wide area?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the absence of data I would guess &quot;less dangerous than fission&quot;, but it is just a guess.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-03-24T22:14:04.397" CommentCount="8" />
  <row Id="7542" PostTypeId="2" ParentId="7441" CreationDate="2011-03-24T22:15:35.160" Score="1" ViewCount="" Body="&lt;p&gt;Towers supported from the bottom are a bit tricky.&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Column#Equilibrium.2C_instability.2C_and_loads&quot; rel=&quot;nofollow&quot;&gt;Buckling&lt;/a&gt;&#xA;  limits how tall a column can be. One needs to additional lateral stiffness to overcome this, usually by putting up guy wires. Even so there are going to be real limits, as Anonymous Coward has mentioned above, solids obtain their stiffness from chemical interactions between molecules and atoms, and the strength to weight ratio is limited. There are some plans for some structures up to about a kilometer, but the cost per unit volume of building goes up for tall buildings. We could probably go a lot higher by the use of carbon nanotubes, but we are years away from being about to construct practical guy wires from them.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-03-24T22:15:35.160" />
  <row Id="7543" PostTypeId="2" ParentId="7506" CreationDate="2011-03-24T22:35:32.760" Score="5" ViewCount="" Body="&lt;p&gt;Helder,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of comments surround this question and we are awaiting some further responses from the Author of the papers. However this is my understanding of the conclusion of these papers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Every theory - even classical physics - violates the Bell Inequalities&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in a sense there is no dispute with the Bell calculation as a demonstrable result of Quantum Mechanics. All of that and the Aspect experiments are fine and to be expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem, it is claimed, lies in a mathematical (topological) assumption at the root of the Bell calculation about the geometric nature of classical spin. Once this is taken account of then one can show that a corresponding classical system would have similar properties. Indeed Joy has even proposed a classical experiment involving an exploding ball to test this result (I cannot find the link right now.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are mathematical technical counter challenges in the Papers quoted by Peter Morgan, and there is the question as to what this all would really mean for the other Classical-Quantum questions that exist (determinism, non-locality, etc) and I am not too clear about that as yet. I would hope that further answers could clarify these aspects. Otherwise after I study these other papers I might be able to finalise an overall opinion.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-03-24T22:35:32.760" CommentCount="4" />
  <row Id="7545" PostTypeId="2" ParentId="7441" CreationDate="2011-03-24T22:38:26.997" Score="3" ViewCount="" Body="&lt;p&gt;First off, the limitation is a material that would not collapse under the weight - earth crust is not quite hard enough. Buckling and other instabilities, nope. Generally, forget a tower built on earth. Not a chance, no such material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start building from geostationary orbit and extend the &quot;rope&quot; both inside and outside the orbit. The outside may be just heavy counterweights, as the inside will begin to pull towards earth. Make the orbital part thicker to support extra weight, as you extend the lower part, until it reaches earth surface below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the problem is the material. The only material in existence with sufficient weight-strength ratio is buckytubes. These are currently centimeters long at most, extremely expensive and you'd not only need thousands of kilometers of them... the rope to sustain its own weight would have to be about 1km thick in the thickest place (near the geostationary orbit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;earth carbon supply, I don't think all coal mines combined could mine that much carbon&lt;/li&gt;&#xA;&lt;li&gt;construction craft fuel. This all would have to be lifted high enough. A LEO rocket takes many times more fuel than its payload weight. A geostationary orbit rocket - much more. The good news is the fuel can be hydrogen+oxygen which is water, and we have that aplenty. The bad news is you need at least as much energy to separate them as you gain from burning them, so the power consumption for fuel production would exceed whole world's power production.&lt;/li&gt;&#xA;&lt;li&gt;environmental impact of that much steam released into atmosphere&lt;/li&gt;&#xA;&lt;li&gt;account for micrometeorites that can really rain over your parade. And this thing being that big, collisions WILL happen. Also account for space junk.&lt;/li&gt;&#xA;&lt;li&gt;account for winds and storms once you reach the atmosphere. Also, upper atmosphere is pretty hot... not nice work environment, also nanotubes aren't extremely fire-proof.&lt;/li&gt;&#xA;&lt;li&gt;cost and impact on economy. Coal becomes super-expensive and we look for alternate sources of carbon.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And when you finally build it, calculate how long a lift travelling some 300km/h would take to reach 37,000km of the 0-gravity orbit...&lt;/p&gt;&#xA;" OwnerUserId="2754" LastActivityDate="2011-03-24T22:38:26.997" CommentCount="4" />
  <row Id="7546" PostTypeId="2" ParentId="7540" CreationDate="2011-03-24T23:06:37.163" Score="2" ViewCount="" Body="&lt;p&gt;Blas Cabrera designed and built a magnetic monopole detector.  Here's how he did it (more or less):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.slac.stanford.edu/cgi-wrap/getdoc/ssi82-025.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.slac.stanford.edu/cgi-wrap/getdoc/ssi82-025.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-03-24T23:06:37.163" CommentCount="2" />
  <row Id="7547" PostTypeId="2" ParentId="5681" CreationDate="2011-03-24T23:55:19.760" Score="1" ViewCount="" Body="&lt;p&gt;I think the question here asks what the role of distant gravity fields are.  This can be addressed with Newtonian gravity.  The motion of a body in Newtonian gravity has the total energy&#xA;$$&#xD;&#xA;E~=~\frac{1}{2}mv^2~-~\frac{GMm}{r}.&#xD;&#xA;$$&#xA;We may use this in cosmology for the scale factor $a$.  For a coordinate system with coordinates $x_i$, the distance between objects with these coordinates expands by a scale factor.  The relevant coordinate is the radius, and for $r~=~0$ our origin any radial distance then scales according to $r~=~ar_0$.  The radial velocity of objects away from the origin is then $dr/dt~=~(da/dt)r_0$.  We may then write this equation with $r_0~=~1$ (normalized) as&#xA;$$&#xD;&#xA;E~=~\frac{1}{2}\Big(\frac{da}{dt}\Big)^2~-~\frac{GM}{a}&#xD;&#xA;$$&#xA;So far this has assumed a homogeneous distribution of matter, and the motion of this mass appears isotropic.  So this mass $M$ corresponds to everything “out there” out to some radius given by the scale factor a.  So this mass is determined by a constant density $\rho$, and $M~=~\rho\times vol$ for the volume $vol~=~(4\pi/3)r^3$ so that&#xA;$$&#xD;&#xA;E~=~\frac{1}{2}{\dot a}^2~-~\frac{4\pi G\rho a^2}{3}~\rightarrow~\frac{\dot a}{a}~=~\frac{8\pi G\rho}{3}~+~\frac{2E}{a^2}.&#xD;&#xA;$$&#xA;We may then simplify all of this.  The energy $E$ is set to zero, where we assume the space if flat.  This is the $k~=~0$ case in the Friedman-Lemaitre-Robertson-Walker (FLRW) spacetime.  We have derived a central part of it without using general relativity!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this equation describes the motion of particles, or galaxies, for a coordinate origin.  That origin is chosen to be where we are, and we do indeed observe galaxies moving outwards.  Hubble empirically “derived” or demonstrate an equation.  The equation says that a galaxy at a distance $d$ from the Earth is moving outwards with a velocity $v~=~Hd$.  The distance out was measured with Cepheid variable luminosity and the velocity measured by Doppler shifting.  The Hubble factor (constant on a spatial surface) is then $H~=~{\dot a}/a$, which is our scale factor “velocity.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let us consider matter that is moving or expanding.  The density will scale as $a^3$ and we assume that the scale factor evolves with some power of time, $a(t)~=~bt^n$, for $b$ a constant.  We put this into both sides of the FLRW equation and we find that the scale factor evolves as $a(t)~=~bt^{-2/3}$.  The calculation is a bit more complicated, but a universe with radiation, such as the time prior to the CMB limit, the universe evolved as $a(t)~=~bt^{-1/2}$.  This is the primary role of matter in the universe upon the structure and dynamics of the universe.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does have to be mentioned that it has been found that the scale factor began to evolve in an exponential manner, where the dominance of matter began to give away to the dynamics of dark energy.  Dark energy is likely due to the vacuum energy density of the universe.  Prior to about 4 billion years ago the universe evolved according to a matter dominated dynamics.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-24T23:55:19.760" />
  <row Id="7548" PostTypeId="2" ParentId="7522" CreationDate="2011-03-25T00:42:42.560" Score="1" ViewCount="" Body="&lt;p&gt;This addresses the comment to Professor Bunn's answer more than the original question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any spacetime described by the FLRW metric, where objects are static in comoving coordinates and redshift is due to expansion of space, can in principle be also described, via an appropriate change of coordinates, by a spherically-symmetric (SS) metric, where objects move along radial timelike geodesics and redshift is due to positional (gtt) and Doppler factors. Since both metrics describe the same physical system, they are observationally equivalent in all respects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To note, the equivalent SS metric is static, meaning that gtt and grr do not depend on time, only in the empty (Milne) and lambda-vacuum (de Sitter) cases. Moreover, even in the next simplest case, the flat matter-only FLRW model known as Einstein-de Sitter model, gtt and grr cannot be expressed algebraically in terms of non-comoving time, as I mentioned in &lt;a href=&quot;http://physics.stackexchange.com/questions/6624/are-the-solutions-in-radicals-of-cubic-and-quartic-of-any-use-in-physics/&quot;&gt;my answer to this question&lt;/a&gt;.  I take it as a strong hint that the FLRW model with expansion of space &lt;strong&gt;is&lt;/strong&gt; the real thing, but even if you take a purely utilitarian position, you can at least say that expansion of space is the only conceptual framework with which it is possible to work.&lt;/p&gt;&#xA;" OwnerUserId="2085" LastActivityDate="2011-03-25T00:42:42.560" CommentCount="1" />
  <row Id="7549" PostTypeId="2" ParentId="7539" CreationDate="2011-03-25T02:45:38.790" Score="7" ViewCount="" Body="&lt;p&gt;As far as a catastrophic plasma disruption is concerned, it might be a problem for the first wall, and hence the ability to use the reactor. But despite the high energy per nucleon, the desity of the plasma is extremely low, the total energy is dominated by the energy in the magnetic fields, and thats not tremendous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you do have the radioactivity produced by the absorbption of high energy neutrons. In some hybrid schemes these neutrons are to be absorbed by fissile material, which multiplies the net energy per fussion by a factor of about 10. Once you actually produce a decent amount of fusion in a thermo reactor, then its environs become radioactive and are no longet easy to work in.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-03-25T02:45:38.790" />
  <row Id="7550" PostTypeId="2" ParentId="7526" CreationDate="2011-03-25T03:03:07.480" Score="1" ViewCount="" Body="&lt;p&gt;At or near Planck energy gravity has to be taken seriously and directly. But even at much lower energy scale number of extra ordinary phenomena should take place which should be &quot;beyond standard model&quot;. However no body knows for sure about the exact energy scale of this &quot;beyond standard model&quot; physics. It is likely that some &quot;beyond S.M.&quot; physics should be observed at the LHC energy scale.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-03-25T03:03:07.480" />
  <row Id="7551" PostTypeId="2" ParentId="7540" CreationDate="2011-03-25T04:53:42.293" Score="0" ViewCount="" Body="&lt;p&gt;Of course one could go about trying to search for monopoles with sophisticated apparatus in a similar vein as searches for dark matter and supersymmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, one could follow the path taken by Castelnovo, Moesnner and Sondhi who &lt;strong&gt;demonstrated&lt;/strong&gt; the existence of monopoles as emergent excitations in spin-ice: &lt;a href=&quot;http://arxiv.org/abs/0710.5515&quot; rel=&quot;nofollow&quot;&gt;Magnetic Monopoles in Spin Ice&lt;/a&gt; (Nature, 2007). This answers quite definitely the question of &lt;em&gt;do magnetic monopoles exist in Nature ?&lt;/em&gt;.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-25T04:53:42.293" CommentCount="7" />
  <row Id="7552" PostTypeId="1" AcceptedAnswerId="7567" CreationDate="2011-03-25T05:21:17.863" Score="7" ViewCount="377" Body="&lt;p&gt;I have come across a limit of Gaussian integrals in the literature and am wondering if this is a well known result.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The background for this problem comes from the composition of Brownian motion and studying the densities of the composed process.  So if we have a two sided Brownian motion $B_1(t)$ we replace t by an independent Brownian motion $B_2(t)$ and study the density of $B_1(B_2(t))$. If we iterate this composition  n times we get the iterated interal in (**) below as an expression for the density of the n times iterated Brownian motion.  The result I am interested in is derived in the following paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original reference is &quot;Fractional diffusion equations and processes with randomly varying time&quot; Enzo Orsingher, Luisa Beghin &lt;a href=&quot;http://arxiv.org/abs/1102.4729&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1102.4729&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Line (3.14) of Orsingher and Beghins paper reads&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(**) $\lim_{n \rightarrow \infty} 2^{n} \int_{0}^{\infty} \ldots \int_{0}^{\infty} \frac{e^{\frac{-x^2}{2z_1}}}{\sqrt{2 \pi z_1}} \frac{e^{\frac{-{z_1}^2}{2z_2}}}{\sqrt{2 \pi z_2}} \ldots  \frac{e^{\frac{-{z_n}^2}{2t}}}{\sqrt{2 \pi t}} \mathrm{d}z_1 \ldots \mathrm{d}z_n = e^{-2 |x|}   $&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How do you prove this result without using probability?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is this a type of path integral (functional integral)? Or is this integrand some kind of kinetic plus potential term arsing in quantum mechanics?  Do expressions like (**) ever come up in physics literature?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(I tried using the change of variable theorem for Wiener measure to transform (**) into a Wiener integral with respect a specific integrand and have had some success with this.. I think this shows how to compute a Wiener integral with respect to a function depending on a path and not just a finite number of variables but did not see how to take this any further - The change of variable theorem for Wiener Measure was taken from &quot;The Feynman Integral and Feynman's Operational Calculus&quot; by G. W. Johnson and M. L. Lapidus.)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I have been studying a slight generalization of ** from the probability side of things and have been trying to use dominated convergence to show the LHS of ** is finite but I am having problems finding a dominating function over the interval $[1,\infty)^n$.  Is dominated convergence the best way to just show the LHS of (**) is finite?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2757" LastActivityDate="2011-03-29T16:15:21.453" Title="Question about a Limit of Gaussian Integrals and how it relates to Path Integration (if at all)?" Tags="&lt;quantum-mechanics&gt;&lt;path-integral&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" />
  <row Id="7553" PostTypeId="2" ParentId="7540" CreationDate="2011-03-25T06:41:56.023" Score="2" ViewCount="" Body="&lt;p&gt;In this I am replying to the question stated about macroscopic monopoles, as you describe them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A magnetic monopole would attract magnetized matter,  falling in strength over distance by 1/r^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would know it is a monopole if you go with your normal everyday compass and your spaceship all around it and see that the compass is pointing always north, or always south, all around. This could be done quite a distance away if it is a monopole in the range of Tesla, as you seem to ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as the dimensions are macroscopic, as simple compass will tell you if is a monopole using it to map it all around. A macroscopic monopole would be attracted to the opposite side pole of a dipole magnet  and to ferromagnetic materials just as the dipole magnets attract them. There are &lt;a href=&quot;http://www.lessemf.com/dcgauss.html&quot; rel=&quot;nofollow&quot;&gt;commercial instruments&lt;/a&gt; for measuring magnetic fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see I am a science fiction fan. My favorite is Terry Pratchett, where he talks of magnetism as &quot;the love of iron&quot; :).&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-25T07:31:37.850" LastActivityDate="2011-03-25T07:31:37.850" CommentCount="5" />
  <row Id="7554" PostTypeId="2" ParentId="7539" CreationDate="2011-03-25T08:15:52.730" Score="2" ViewCount="" Body="&lt;p&gt;A fusion reactor tries to harness the energy that was explosive in the Hydrogen bomb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question of danger then goes as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible for ITER to turn into a hydrogen bomb?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer for radiation catastrophies in a large area is no, it cannot even remain highly radiative in the sense that the Japanese reactors are now and cannot be controlled except by time and cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same reason that makes building a fusion engine so hard,it is over 50 years when the stellatron was being discussed, is the reason that makes it safe for the larger environment to have a fusion reactor in the vicinity. They needed an atomic bomb to trigger the hydrogen bomb. ITER is generating a plasma in a tokamak and a plasma is something that has to be nursed and is destroyed if disrupted. In addition the feeding of new fuel is done on the same principles as feeding gas to a car engine, incrementally. There is no way the unburned fuel will become critical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Locally, as others have said, walls and metals will become radioactive and if an explosion happens for some unforeseen reason, war, terrorism etc, the debris will be local. No iodine and cesium etc byproducts in bulk to be sent to the atmosphere since the plasma has very little mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is why countries are spending resources to support ITER. It is the ultimate free  clean energy .&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-28T05:53:52.807" LastActivityDate="2011-03-28T05:53:52.807" />
  <row Id="7555" PostTypeId="1" AcceptedAnswerId="7558" CreationDate="2011-03-25T09:52:38.823" Score="3" ViewCount="81" Body="&lt;p&gt;if i understood &lt;a href=&quot;http://en.wikipedia.org/wiki/Galvanic_cell&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; correctly, the determination of voltage for a specific voltaic (gallvanic) cell is determined only by the chemical correlation between the two metals. is this true? &#xA;for an example , if i use iron instead of zinc in the example from wikipedia, i will get a different voltage?&lt;/p&gt;&#xA;" OwnerUserId="2758" LastActivityDate="2011-03-25T10:55:54.993" Title="determination of voltaic cell voltage" Tags="&lt;electricity&gt;&lt;physical-chemistry&gt;" AnswerCount="1" />
  <row Id="7556" PostTypeId="2" ParentId="7530" CreationDate="2011-03-25T10:06:37.533" Score="4" ViewCount="" Body="&lt;p&gt;Creativity in physics exists, even in experimental physics, in trying to devise, design, build  experiments to test/falsify theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the way you are forming your question it seems to me you want to learn physics in the way somebody wants to learn french, or chinese. As a tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is fine if you have the stamina to go through the grind of mastering the tools just to design a different mousetrap. Guest gave you a link to a series of lectures. It could be a start to see whether your desire is satisfiable or you will be biting more than you can chew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;an elder Anna&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-25T10:06:37.533" CommentCount="3" />
  <row Id="7557" PostTypeId="2" ParentId="7441" CreationDate="2011-03-25T10:16:50.863" Score="1" ViewCount="" Body="&lt;p&gt;Some &quot;attachment&quot; to the questiones/thoughts re &quot;the highest tower possible&quot; and the &#xA;constraints. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A tower build to reach the highest altitude possible (and no other &quot;use&quot; is intended) &#xA;will be not towerlike, but more a kind of pyramid. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The tower will be very &quot;thick&quot; at its base, &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;and will get gradually thinner upward. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is for the very same reasons as space rockets are made with &quot;stages&quot;. &#xA;An ideal rocket would contain an infinit number of infinitely small stages. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That ideal tower might have a form like a cone or pseudosphere or somewhat in between. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a design will lessen the need for those &quot;guy&quot; cables. Strength of materials against pressure is generally higher than to strain. The limits then will be plastic flow under pressure or &quot;Querzugspannung&quot; (I dont find a English word for that). Of course all this does not include questions of &quot;fundaments&quot;, thus this limit still would not be realistic.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-25T11:12:17.140" LastActivityDate="2011-03-25T11:12:17.140" CommentCount="2" />
  <row Id="7558" PostTypeId="2" ParentId="7555" CreationDate="2011-03-25T10:55:54.993" Score="3" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;for an example , if i use iron instead&#xA;  of zinc in the example from wikipedia,&#xA;  i will get a different voltage?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is determined only by the chemical&#xA;  correlation between the two metals. is&#xA;  this true?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The voltage of a galvanic cell is the sum of the voltages of the &quot;half cells&quot;, &#xA;wich can (not must!) be made from a metal and a solution around that &quot;electrode&quot;. &#xA;Both the metal (or what else is there as &quot;active&quot; material) and the solutions (&quot;electrolyte&quot;)&#xA;quality and concentration determine the voltage of this &quot;electrode&quot;.&lt;br&gt;&#xA;There are cases, in which there is a single electrolyte for both &quot;electrodes&quot; and &#xA;where the influence of the electrolyte on both electrodes cancel out. &#xA;(Eg the oldfashioned acid Zink/Manganese dioxide &quot;dry&quot; cells and the more recent &#xA;&quot;alkaline&quot; Zink/mangese dioxide cells, both have about 1.5 Volts). But in general, &#xA;all components have an influence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS&#xA;Be aware that the popular use of &quot;electrodes&quot; (up to physics textbooks!) is wrong. &#xA;The carbon rod in a Zink/&quot;carbon&quot; cell is not an electrode at all. This is a contact &#xA;material able to withstand the corrosive environment. The electrode is a mass of manganese dioxide.  The latter then is &quot;explained&quot; as &quot;depolarisator&quot;, a theory obsolete &#xA;(and known to be wrong) since Nernsts law was published more than 100 years ago. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-25T10:55:54.993" />
  <row Id="7559" PostTypeId="2" ParentId="7530" CreationDate="2011-03-25T11:21:25.103" Score="1" ViewCount="" Body="&lt;p&gt;Anna,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another aspect of any physics education is the role of mathematics in physics. As you will know there is lots of mathematics underlying physics. Whether and to what extent you want or perhaps need to learn the underlying maths depends on your objectives too. Now I see this line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Things I can use to gain perspective while creating designs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which suggests that you have another purpose in learning physics, perhaps the &quot;artistic appeal&quot; of certain structures in physics? If something like this is your reason, then some suggestions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Chaos theory and Fractals. This subject is slightly underneath several parts of physics, and may relate to Turbulence and similar topics in Fluid Dynamics. There are certain common structures in this theory, like strange attractors.&lt;/li&gt;&#xA;&lt;li&gt;Common solutions. Feynman Vol1 discusses how many unrelated physical processes (electrostatics, neutron diffusion, etc) happen to obey exactly the same equations. So the similar behaviour is repeated in their solutions as well.&lt;/li&gt;&#xA;&lt;li&gt;Theory of crystals. These are very regular objects, but there is a surprisingly extensive mathematical theory of the different types of 3D shapes which are possible.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There will be other topics too, but I can envisage how study of these topics could provide some input when creating physics based designs, or looking at some similarities between different topics.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-03-25T11:21:25.103" />
  <row Id="7560" PostTypeId="1" CreationDate="2011-03-25T11:26:29.080" Score="2" ViewCount="203" Body="&lt;p&gt;How to change the Fermi-Energy level  for extrinsic Semiconductor with the temperature change?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="2760" LastEditorUserId="2760" LastEditorDisplayName="" LastEditDate="2011-03-25T11:55:36.743" LastActivityDate="2011-03-25T11:55:36.743" Title="Fermi-Energy Level for Extrinsic Semiconductor" Tags="&lt;semiconductor-physics&gt;" CommentCount="8" />
  <row Id="7561" PostTypeId="1" CreationDate="2011-03-25T11:44:25.017" Score="6" ViewCount="264" Body="&lt;p&gt;It's a well known fact that an observer that accelerates at a constant rate from $-c$ at past infinity to $+c$ at future infinity sees a horizon in flat Minkowski spacetime. This is easy to see from a spacetime diagram once you realize that the union of past light cones on such a trajectory has a boundary that divides the spacetime into two regions - one inaccessible by the accelerating observer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to the classic result of Unruh radiation when one looks at the quantum field theory for such an observer. The horizon plays a crucial role here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one go about determining whether a horizon is seen by a general class of worldines? In particular, is there any reason to believe that a horizon would exist for an observer that is stationary for all time except for a finite period of acceleration and deceleration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any other class of worldlines other than an indefinitely accelerating observer for which a horizon is known to exist in flat spacetime?&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-03-25T21:19:02.393" LastActivityDate="2011-03-26T22:05:54.530" Title="Horizon and Unruh radiation for a finite period of acceleration" Tags="&lt;quantum-field-theory&gt;&lt;special-relativity&gt;&lt;event-horizon&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="7562" PostTypeId="2" ParentId="6946" CreationDate="2011-03-25T11:58:47.280" Score="-1" ViewCount="" Body="&lt;p&gt;The mass will never be infinite because it would require infinite work to reach required speed. It's just a figure of speech.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of trying to imagine &quot;infinite mass&quot; try imagining the process: as you apply force (&quot;press the accelerator pedal&quot;) the object instead of going faster, gets heavier. The more work you provide, the heavier the object gets.&lt;/p&gt;&#xA;" OwnerUserId="2754" LastActivityDate="2011-03-25T11:58:47.280" CommentCount="4" />
  <row Id="7563" PostTypeId="2" ParentId="7561" CreationDate="2011-03-25T12:32:50.003" Score="4" ViewCount="" Body="&lt;p&gt;It seems to me that this question is quite trivial, so I hope I am not overlooking anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a horizon to form, the observer needs to be &quot;as light-like as possible&quot; (otherwise the slope of his world-line will be more than 45 degrees and union of his past light-cones will cover all of universe). And because there is no other way of achieving that than just asymptotically by accelerating indefinitely I think your question is resolved.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-03-25T12:32:50.003" CommentCount="12" />
  <row Id="7564" PostTypeId="2" ParentId="7506" CreationDate="2011-03-25T13:04:26.123" Score="5" ViewCount="" Body="&lt;p&gt;My reading of Joy's paper &amp;mdash;just as it is, without having carefully read the arXiv paper I cited, nor all of Joy's responses to critics that I also mentioned&amp;mdash; is, so far: the left and right hand sides of eq(1) and eq(2), without the central interpolations, state that $A(\mathbf{a},\lambda)=\lambda$ and $B(\mathbf{b},\lambda)=-\lambda$, where $\lambda$ takes the values $\pm 1$. $A(\mathbf{a},\lambda)$ and $B(\mathbf{b},\lambda)$ are independent of $\mathbf{a}$ and $\mathbf{b}$, respectively, hence the expected value of the product is $-1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The central interpolations introduce nine algebraic objects, each of which is the basis of and satisfies the algebraic relations of a quaternion algebra, $\beta_i$ and $\beta_{i&amp;#39;}(\lambda)$, with $\lambda=\pm 1$. For $\lambda=+1$, the $\beta_{i&amp;#39;}(+1)$ satisfy the same algebra as the $\beta_i$; for $\lambda=-1$, $-\beta_{i&amp;#39;}(-1)$ satisfy the same quaternion algebra, with the sign change to be noted. To fix the algebraic structure further, which is absolutely necessary so we know how to handle products like $\beta_i\beta_{i&amp;#39;}(+1)$, Joy states that $\beta_{i}(\lambda)=\lambda\beta_i$, so we are in fact dealing with a purely quaternion algebra, of real dimension 4. The whole of the prelude to eq(5-7) could be stated using only $\beta_i$; for me the $\beta_i(\lambda)$ just obscures things. I would like to see a mathematical justification for introducing the $\beta_i(\lambda)$ instead of just using $\lambda\beta_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The notation of eq(5-7) is problematic because it seems to play fast and loose with the non-commutative structure of the quaternions. One cannot in general write $\frac{p}{q}$ for two quaternions $p$ and $q$, because in general $pq^{-1}$ is different from $q^{-1}p$. Since eq(5-7) obtains a different result from the result that I get in my first paragraph, I'd want to see the whole thing rewritten using inverses so that the order of the multiplications is kept under control. Unless there is a potent reason for using the $\beta_{i&amp;#39;}(\lambda)$ notation, I'd like to see everything written out using only the $\beta_i$. If the answer is still $-\mathbf{a.b}$ I'd want to check that it does not make any &lt;em&gt;unwarranted&lt;/em&gt; reversal of the quaternions $a_i\beta_i$ and $b_i\beta_i$, even one of which would be &lt;em&gt;exactly&lt;/em&gt; enough to get the result $-\mathbf{a.b}$ instead of $-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I currently cannot see any way to justify the jump from the left hand expression of eq(6) to the right hand expression. Perhaps someone can show me how to get from one to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my discussion above is OK, this leaves questions about Joy's earlier papers. My impression is that Joy tried to make the argument of his earlier papers as succinct as possible. He may have made a mistake in doing so, in which case, if he claims the earlier papers do not make any mistake, then they have to be considered on their own merits. On the other hand, before I would consider checking that I think I would want to see Joy withdraw or replace this paper on the arXiv with something that at least made play at addressing my discussion here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I look forward to comments.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-03-25T13:04:26.123" CommentCount="2" />
  <row Id="7565" PostTypeId="2" ParentId="7561" CreationDate="2011-03-25T14:05:34.717" Score="2" ViewCount="" Body="&lt;p&gt;The question seems to point to the question of whether or not one can “turn on” the Unruh effect.  This of course has some experimental relevance, for if we test it with highly accelerated particles then we do have to prepare a system where the acceleration is initiated.  So this question is related to other funny problems, such as the physics which generates a black hole.  The exact solution is an eternal solution, but an astrophysical black hole is generated by the implosion of a stellar core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An accelerated observer will measure a thermal background of radiation on the Rindler chart, which defines wedges in flat spacetime.  The Rindler wedge is a region an accelerated observer may completely causally interact with.  Consider the Cartesian coordinate chart&#xA;$$&#xD;&#xA;ds^2~=~-dT^2~+~dX^2~+~dY^2~+~dZ^2,&#xD;&#xA;$$&#xA;for $0~&amp;lt;~X~&amp;lt;~\infty$ and $-X~&amp;lt;~T~&amp;lt;~X$.  This slice of spacetime is a Rindler wedge, and the accelerated observer is on the transformed coordinates,&#xA;$$&#xD;&#xA;T~=~x~\mathrm{sinh}(t),~X~=~x~\mathrm{cosh}(t),~y~=~y,~Z~=~z,&#xD;&#xA;$$&#xA;so the Minkowski line element is&#xA;$$&#xD;&#xA;ds^2~=~-x^2dt^2~+~dx^2~+~dy^2~+~dz^2,~~0~&amp;lt;~x~&amp;lt;~\infty,~-\infty~&amp;lt;~t,~y,~z~&amp;lt;~\infty.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the Rindler chart projections of null geodesics onto a spatial hypersurface are semicircular arcs.  A highly accelerated observer will observe a laser beam bent into an arc, just as a projectile on this frame will move on a curve.  A Killing vector defines a foliation of spatial surfaces orthogonal to static frames.  This is a metric on these spatial surfaces according to a conformal transformation of the original metric, defined as the Fermat metric. For a fixed time $t~=~0$ the line element&#xA;$$&#xD;&#xA;ds^2~=~g_{tt}dt^2~+~g_{ij}dx^idx^j,&#xD;&#xA;$$&#xA;gives the Fermat line element&#xA;$$&#xD;&#xA;d\sigma^2~=~{{g_{ij}}\over{g_{tt}}}dx^idx^j.&#xD;&#xA;$$&#xA;The Killing vector $\xi_t~=~\partial_t$ defines a foliation of these surfaces so the static spacetime is a trivial vacuum solution to the Einstein field equation.  The Fermat metric for Rindler observers is&#xA;$$&#xD;&#xA;d\sigma^2~=~x^{-2}(dx^2~+~dy^2~+~dz^2),~0~&amp;lt;~x~&amp;lt;~\infty.~-\infty~&amp;lt;~\{y,~z\}~&amp;lt;~\infty,&#xD;&#xA;$$&#xA;which is the hyperbolic three space ${\cal H}^3$ in the upper half of the space.  The projection on $z~=~0$ is the Poincare half place ${\cal H}^2$ with semicircular geodesics. This space is a conformal map from the plane ${\mathbb R}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hyperbolic geometry of the spacetime reduced to two dimensions gives the propagator&#xA;$$&#xD;&#xA;G(X,~X^\prime)~=~-{1\over{4\pi^2}}{1\over{(X^\prime_0~-~X_0~-~i\epsilon)^2~-~(X^\prime_i~-~X_i)^2}}&#xD;&#xA;$$&#xA;for coordinate index $i$ chosen. This propagator has the value&#xA;$$&#xD;&#xA;G(X,~X^\prime)~=~-{1\over {16\pi^2(\alpha~+~1)}}\mathrm{cosech}^2(\Delta&#xD;&#xA;t/\alpha~-~i\epsilon/\alpha),&#xD;&#xA;$$&#xA;for $\alpha$ determined by the hyperbola $X^2~-~T^2~=~\alpha^2$. This propagator satisfies a harmonic condition $(\Delta~-~\lambda)G(X,~X^\prime)$ $=~\delta(X,~X^\prime)$, for $\lambda$ an eigenvalue.  This propagator has a thermal or heat kernel interpretation.  The thermal radiation is what the accelerated observer detects in the Rindler wedge.         &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This spacetime is very interesting, for it has an analogue with de Sitter and anti de Sitter spacetimes.  Another characteristic is that this is eternal.  The exact solution assumes a Rinder wedge for an accelerated mass or observer in Minkowski spacetime that is perpetual.  This is analogous in a way to the eternal black hole solution, which is somewhat fictional.  However, the eternal solution approximates very closely a final state of a black hole formed from a collapse.  In a similar manner if one accelerates a particle to $a$ at a time much earlier than one performs and experiment then the exact theory above will serve as an approximation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;{\bf addendum}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel somewhat compelled to write an addendum to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unruh radiation is something which is observed only by the accelerated observer.  An inertial observer watching an accelerated mass would detect none of this radiation at all.  The accelerated mass starts out at $v~=~-c~+~-\epsilon$ approaches the origin where the split horizon of the Rindler wedge exists at $v~-~0$ and then accelerates off to $v~=~c~-~\epsilon$ in the other direction.  The horizon exists at a distance $d~=~a/c^2$, for $a$ the acceleration, from the accelerated mass.  If you were on this frame this horizon would “appear” behind your direction of motion.  It would be a region from which no information can be received.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume that the mass starts accelerating at a time $t~=~0$.  This would be a case where only the upper half of the Rindler wedge exists.  We now assume that the particle accelerates for a long period of time, so it approaches $v~=~c~=~\epsilon$.  The motion asymptotes to a null line, and the path is half of a hyperbola.  The absence of the bottom null asymptote will have less and less influence on what happens later on.  This is a physical statement.  First off if an inertial observer is moving at a velocity $u$ in the direction of the accelerated mass, this mass will appear to approach this observer and then recede.  For $u$ large enough this accelerated mass would appear to come at this observer with a velocity $v’~=~-c~+~\delta$ where $\delta~&amp;gt;~\epsilon$, but not by “that much.”  The accelerated mass will then approach this observer, stop and accelerate outwards.  So “modulo” this small error, this situation is a good approximation to the eternal accelerated case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The comparison to a black hole is with the Penrose diagram for the Schwarzschild metric.  The physical black hole removes the bottom half of the diagram which corresponds to a “white hole.”  The Rindler wedge is a near horizon approximation to a black hole.  We do not expect any strange loss of Hawking radiation from a physical black hole to result from this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The particle horizon partitions regions of the Minkowski spacetime so the accelerated observer receives signals from only a portion of the spacetime.  This causal partition means the accelerated observer measures the vacuum state as being a mixed state of thermal bosons.  The partition removes a unitary description of the vacuum for the accelerated observer.  So what the inertial observer witnesses as a pure vacuum states with zero temperature, the accelerated observer witnesses as a vacuum plus thermal distribution of radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now have the problem of detecting this.  The inertial observer does not see the Unruh radiation.  The accelerated observer does.  The temperature is $T~\simeq~10^{-21}K/cm-sec^{-2}$  So you need a pretty serious acceleration to observe anything on that frame.  Clearly a real observer can’t be on the frame.  So not only do we need a start for the acceleration, we need a stop as well.  The object on the accelerated frame will absorb this Unruh radiation, and if the inertial observer can “catch” the object its temperature may be measured.  If this object is a Bose Einstein condensate of Ba ions that are accelerated enormously through a high $\vec E$ field region, say in a capacitor, the Unruh radiation might be enough to change the phase of the BEC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote a paper a fair number of years ago which got “honorable mention” for the Gravity Essay Prize where I proposed just this.  Unfortunately nobody seemed interested in actually doing the experiment --- which is what I wrote this paper for!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly we can’t have an eternal accelerated object, so we must work with approximate situations.  However, some physical reasoning should strongly suggest that a removal of the $\infty$ or eternal aspects of the Unruh results can be made into a small perturbation.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-03-25T19:11:16.393" LastActivityDate="2011-03-25T19:11:16.393" CommentCount="8" />
  <row Id="7566" PostTypeId="2" ParentId="7539" CreationDate="2011-03-25T14:50:04.257" Score="-2" ViewCount="" Body="&lt;p&gt;The main risk of thermonuclear energy is unlimited production.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If humanity is not limited by fuel supply or production capacity, energy production will rise exponencially untill this would change earth climate directly, instead of indirect change due to CO2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll just start having 10kW CPUs, personal jets and oasises at north pole.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-03-25T14:50:04.257" CommentCount="8" />
  <row Id="7567" PostTypeId="2" ParentId="7552" CreationDate="2011-03-25T17:24:43.343" Score="6" ViewCount="" Body="&lt;p&gt;Our plan of an answer is as follows. Firstly, we will introduce Planck's constant $\hbar$ so that the particular value $\hbar=1$ corresponds to the original problem. Secondly, we mention a connection to (what physicists often calls) the group property of Feynman path integrals. Thirdly, we will show that the sought-for formula happens to be the classical &quot;instanton&quot; contribution in a saddle point/steepest descent asymptotic expansion, which becomes valid as $\hbar\to 0$. We are currently unaware if semi-classical localization methods can be applied to justify the saddle point/steepest descent expansion, and we shall not attempt to make a justification here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let us get to business. Define endpoints $x_0\equiv x&amp;gt;0$ and $x_{n+1}\equiv t&amp;gt;0$. We start by introducing Planck's constant $\hbar$ into the $u_n$ function in eq. (1.9) of &lt;a href=&quot;http://arxiv.org/abs/1102.4729&quot; rel=&quot;nofollow&quot;&gt; arXiv:1102.4729&lt;/a&gt;, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_n(x,t,\hbar) &#xD;&#xA;:=\left[\prod_{j=1}^n 2 \int_0^{\infty} \frac{\mathrm{d}x_j}{\sqrt{\hbar}}\right]&#xD;&#xA;\prod_{i=1}^{n+1}\frac{e^{-\frac{x_{i-1}^2}{2\hbar x_i}}}{\sqrt{2 \pi x_i}}. &#xD;&#xA;\qquad (1.9\hbar)&#xD;&#xA;$$&#xA;In particular, for $n=0$, we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ u_{n=0}(x,t,\hbar)=\frac{e^{-\frac{x^2}{2\hbar t}}}{\sqrt{2 \pi t}}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $u_n$ function enjoys various scaling/homogeneity properties,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ u_n( x,t,\hbar) = \sqrt{\lambda} u_n(\lambda x,\lambda t,\lambda\hbar) &#xD;&#xA;= \lambda u_n(\lambda x,\lambda^{2^{n+1}} t,\hbar) $$ &#xA;$$= \sqrt{\lambda} u_n( x,\lambda^{2^n} t,\frac{\hbar}{\lambda}),&#xD;&#xA;\qquad \lambda&amp;gt;0. \qquad (H) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the help of the first homogeneity property in eq. ($H$), we can immediately deduce the corresponding $\hbar$ generalization of eq. (3.14) in arXiv:1102.4729,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lim_{n\to \infty}u_n(x,t,\hbar)&#xD;&#xA;=\frac{1}{\sqrt{\hbar}}e^{-\frac{2x}{\hbar}}. \qquad (3.14\hbar) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is basically &lt;em&gt;how do we derive, understand, motivate, etc.,  eq. (3.14$\hbar$) physically?&lt;/em&gt; To get to a path integral interpretation, we note that the $u_n$ function has (what physicists often call) a group property,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ u_{n+1+m}(x,z,\hbar) = 2 \int_0^{\infty} \frac{\mathrm{d}y}{\sqrt{\hbar}} u_{n}(x,y,\hbar)u_{m}(y,z,\hbar),  \qquad (G)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in close analogy with the Feynman propagator $K(x_f,t_f;x_i,t_i)$ with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ K(x_3,t_3;x_1,t_1) = \int_{-\infty}^{\infty}\mathrm{d}x_2 \ K(x_3,t_3;x_2,t_2) K(x_2,t_2;x_1,t_1).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the &quot;sum of histories&quot; from $x$ to $z$ can be calculated by integrating over an intermediate point $y$. The $n$ in the $u_n$ function plays the role of a discretized time variable. As a consistency check, it is easy to see (by performing some elementary integrals) that the right-hand side of eq. (3.14$\hbar$),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_{n=\infty}(x,t,\hbar)&#xD;&#xA;=\frac{1}{\sqrt{\hbar}}e^{-\frac{2x}{\hbar}} \qquad \left(\to \sqrt{2\pi x} \delta(x)  \quad \mathrm{for} \quad\hbar \to 0\right),   $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;does indeed solve the group equation $(G)$ in the particular cases $n,m=0,\infty$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ u_{\infty}(x,z,\hbar) = 2 \int_0^{\infty} \frac{\mathrm{d}y}{\sqrt{\hbar}} u_{\infty}(x,y,\hbar) u_{\infty}(y,z,\hbar) $$&#xA;$$= 2 \int_0^{\infty} \frac{\mathrm{d}y}{\sqrt{\hbar}} u_{\infty}(x,y,\hbar) u_{0}(y,z,\hbar)$$&#xA;$$= 2 \int_0^{\infty} \frac{\mathrm{d}y}{\sqrt{\hbar}} u_{0}(x,y,\hbar) u_{\infty}(y,z,\hbar).&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next introduce Gaussian momenta $p_1, \ldots,p_{n+1},$ with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-\infty}^{\infty}\frac{\mathrm{d}p_i}{2\pi\sqrt{\hbar}}e^{-\frac{1}{2\hbar}x_ip_i^2}&#xD;&#xA;= \frac{1}{\sqrt{2 \pi x_i}}, \qquad x_i&amp;gt;0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the $u_n$ function becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_n(x,t,\hbar) = \left[\prod_{j=1}^n 2\int_0^{\infty}\frac{\mathrm{d}x_j}{\sqrt{\hbar}}\right]\left[\prod_{i=1}^{n+1}\int_{-\infty}^{\infty}\frac{\mathrm{d}p_i}{2\pi\sqrt{\hbar}}\right]e^{-\frac{S}{\hbar}},&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with Euclidean phase space action &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S:=\frac{1}{2}\sum_{i=1}^{n+1}\left(\frac{x_{i-1}^2}{x_i}+x_i p_i^2\right).$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let us turn to the saddle point/steepest descent asymptotic expansion.&#xA;The classical equations of motion are &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0 \approx \frac{\partial S}{\partial p_i} = x_i p_i, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0 \approx \frac{\partial S}{\partial x_i} &#xD;&#xA;= \frac{x_i}{x_{i+1}}-\frac{x_{i-1}^2}{2x_i^2} + \frac{p_i^2}{2}, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where we use $\approx$ sign instead of $=$ sign to emphasize when classical equations of motion have been applied. The classical solution is  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_i \approx 0, \qquad q_i \approx q_{i-1}^2,$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where we have defined $q_i :=\frac{x_i}{2x_{i+1}}$. &#xA;So $q_i \approx q_{i-1}^2 \approx q_{i-2}^4 \approx \ldots \approx q_0^{2^i}$. Now the telescopic product&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\prod_{i=0}^n 2q_i =\prod_{i=0}^n\frac{x_i}{x_{i+1}}=\frac{x_0}{x_{n+1}}=\frac{x}{t},$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is fixed by boundary conditions $x$ and $t$. So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q_0^{2^{n+1}-1}=q_0^{\sum_{i=0}^n 2^{i}}\approx\prod_{i=0}^n q_i =\frac{x}{2^{n+1}t},$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and therefore the unique classical solution is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q_i \approx \left( \frac{x}{2^{n+1}t} \right)^{\frac{2^i}{2^{n+1}-1}} \to 1 \qquad&#xD;&#xA;\mathrm{for} \qquad n \to \infty. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence classically $x_i \approx 2^{-i}x$ for $n=\infty$. The classical value of the action is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S_{\mathrm{cl}} \approx \sum_{i=0}^{n} x_i q_i \to x\sum_{i=0}^{\infty}2^{-i} =2x \qquad&#xD;&#xA;\mathrm{for} \qquad n \to \infty, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the classical &quot;instanton&quot; contribution $e^{-\frac{S_{\mathrm{cl}}}{\hbar}}$ happens to be the right-hand side of eq. (3.14$\hbar$), up to a $\sqrt{\hbar}$ factor. This is our main observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more complete treatment would now calculate the one-loop Van Vleck determinant  $\det(\partial^2S)$ in the saddle point/steepest descent asymptotic expansion. Here we will only make a couple of further remarks. The Hessian $\partial^2 S$ of the action is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\partial^2 S}{\partial x_i\partial x_j} =&#xD;&#xA;\delta_{i,j}\left(\frac{1}{x_{i+1}}+\frac{x_i^2}{x_{i+1}^3}\right)-\delta_{i+1,j}\frac{x_i}{x_j^2}-\delta_{i-1,j}\frac{x_j}{x_i^2}, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\partial^2 S}{\partial p_i\partial x_j} &#xD;&#xA;= \frac{\partial^2 S}{\partial x_i\partial p_j} =  \delta_{i,j}p_i\approx 0, \qquad&#xD;&#xA;\frac{\partial^2 S}{\partial p_i\partial p_j} = \delta_{i,j}x_i. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we have $n+1$ momenta $p_i$, but only $n$ positions $x_i$, we would naively expect the Van Vleck determinant $\det(\partial^2S) \sim x$ to be proportional to $x$ on-shell. This would mean a $1/\sqrt{x}$ factor in the expansion. It would be interesting to see a detailed calculation of the Van Vleck determinant $\det(\partial^2S)$.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-03-29T16:15:21.453" LastActivityDate="2011-03-29T16:15:21.453" CommentCount="6" />
  <row Id="7568" PostTypeId="1" AcceptedAnswerId="7569" CreationDate="2011-03-25T18:22:34.637" Score="1" ViewCount="164" Body="&lt;p&gt;Someone told me that reading glasses (&lt;em&gt;a priori&lt;/em&gt; with a magnifying glass effect only) improve one's eyesight of objects lying in the long range distance. I am really sceptic about it since everything is blurry if I look at objects through those glasses! Is it possible (for this particular person at least)?&lt;/p&gt;&#xA;" OwnerUserId="545" LastActivityDate="2011-03-25T18:32:49.453" Title="Can reading glasses improve one's eyesight of objects lying in the long range?" Tags="&lt;optics&gt;&lt;geometric-optics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="7569" PostTypeId="2" ParentId="7568" CreationDate="2011-03-25T18:32:49.453" Score="3" ViewCount="" Body="&lt;p&gt;I don't think this is possible. Suppose that your unaided eye can focus on objects over some range of distanced from $d_1$ (closest) to $d_2$ (furthest). Putting a converging lens in front of your eye &lt;em&gt;reduces&lt;/em&gt; both $d_1$ and $d_2$. That is, it shifts the range of distances over which you can focus closer to you, not further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the geometric-optics proof of this. Varying the distance at which you can focus is equivalent to varying the effective focal length of your eye. If $D$ is the diameter of your eye, and $d$ is the distance of the object you're focusing on, then the focal length is given by&#xA;$$&#xD;&#xA;{1\over f}={1\over d}+{1\over D}.&#xD;&#xA;$$&#xA;When you put on your reading glasses, the effective $f$ is decreased, or to put it another way, the effective power of the lens, $1/f$, is increased. (To be specific, $1/f_{\rm new}=1/f_{\rm old}+1/f_{\rm lens}$.) So the left side of the equation becomes larger. $D$ doesn't change, so $1/d$ must become larger. So $d$, the distance to which you focus, becomes less.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-25T18:32:49.453" CommentCount="7" />
  <row Id="7570" PostTypeId="2" ParentId="6867" CreationDate="2011-03-25T19:33:21.220" Score="1" ViewCount="" Body="&lt;p&gt;Let me say that antihydrogen is never &quot; slow&quot; in the process of annihilation, when it comes to matter it is already accelerated by Casimir-Polder potential to the energies of few eV, which corresponds to temperature of dozens of thousands of degrees. The attraction forces between particles and antiparticles tear antihydrogen into antiproton, which is captured by matter atom, and positron, which most likely forms pair with electron. Antiproton falls down on nuclei, ejecting electrons from the atom. Then it&#xA;Usually destroys nuclei and annihilates. This process usually takes time of order of 10^-9 s or so, electron-positron in best case could survive for 10^-7 s. &lt;/p&gt;&#xA;" OwnerUserId="2761" LastActivityDate="2011-03-25T19:33:21.220" />
  <row Id="7571" PostTypeId="1" AcceptedAnswerId="7581" CreationDate="2011-03-25T22:33:54.870" Score="4" ViewCount="129" Body="&lt;p&gt;I read this artice: &lt;a href=&quot;http://web5.cns.utexas.edu/news/2010/05/brownian-particles/&quot; rel=&quot;nofollow&quot;&gt;Physicists Prove Einstein Wrong with Observation of Instantaneous Velocity in Brownian Particles&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;“We’ve now observed the instantaneous&#xA;  velocity of a Brownian particle,” says&#xA;  Raizen. “In some sense, we’re closing&#xA;  a door on this problem in physics. But&#xA;  we are actually opening a much larger&#xA;  door for future tests of the&#xA;  equipartition theorem at the quantum&#xA;  level.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;What did Einstein propose and what did they actually prove?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;What door does this experiment open?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="425" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-25T22:54:23.750" LastActivityDate="2011-03-26T06:38:40.767" Title="What does this observation of instantaneous velocity in Brownian particles mean?" Tags="&lt;statistical-mechanics&gt;&lt;velocity&gt;&lt;brownian-motion&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="7572" PostTypeId="2" ParentId="4068" CreationDate="2011-03-25T23:35:18.863" Score="2" ViewCount="" Body="&lt;p&gt;I think if you want a rigorous approach to QFT that incorporates the ideas of renormalization and effective field theory, you might want to take a long look at the work of the 'constructive QFT' school.  Their work on rigorous Euclidean functional integration is very much in the spirit of Wilson.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two examples, chosen randomly among many:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Euclidean functional measures are usually constructed as cylinder measures on spaces of distributions.  (These spaces of distributions arise as duals of nuclear spaces, which are vector spaces of functions which get their topology from families of norms which quantify &quot;how sharply concentrated is this function?&quot;.)   These measures are constructed as limits of cutoff measures, as the cutoff scale goes to zero, and the cutoff measures are measures are constructed in such a way that they approximate a renormalization group flow.  For example, the construction of continuum Yang-Mills theory on a torus is done by rather explicitly block-spinning lattice Yang-Mills theory.  Same thing for 2d Maxwell-Higgs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) In order to be an honest measure, a cylinder measure must satisfy one additional property:  it must be countably additive.  In Glimm &amp;amp; Jaffe's work, this property is obtained a consequence of a property called 'vanishing at infinity', which states, basically, that the measure is insensitive to the region of field space probed by test functions which are large and very sharply localized.  This is almost exactly what one means by an 'effective' field theory.&lt;/p&gt;&#xA;" OwnerUserId="1504" LastActivityDate="2011-03-25T23:35:18.863" CommentCount="2" />
  <row Id="7573" PostTypeId="2" ParentId="5995" CreationDate="2011-03-26T00:52:19.850" Score="3" ViewCount="" Body="&lt;p&gt;I wouldn't call them empirical but rather phenomenological. These equations relate our model of how spins behave in magnetic fields to observables such as macroscopic magnetization. One of the things required then is the ability to define a sensible M. What you usually do is to say &#xA;M = 1/V SUM mu_i, where M is the magnetization made up of the sum of magnetic moment of all spins i in the volume V. What you want is a V large enough to contain a few spins but small enough for the external fields over it to be constant. (Amongst other things external B0 is defining the magnitude of M0). The set of spins in the volume is called a spin isochromat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can derive transverse and longitudinal relaxation from a QM basis. (Slichter: 'Principles or MR' or worse and less digestible: Bloembergen Purcell Pound: PhysRev 73 p670 (1948)) but beyond the validity of QM theory and the ones above no further assumptions have to be satisfied. [I mean, yes, if you vary temperature, M0 will change but let's not be silly]&lt;/p&gt;&#xA;" OwnerUserId="1825" LastActivityDate="2011-03-26T00:52:19.850" />
  <row Id="7574" PostTypeId="2" ParentId="7530" CreationDate="2011-03-26T01:05:58.673" Score="0" ViewCount="" Body="&lt;p&gt;'Learn applied physics' is a tricky question to answer because to 'apply' 'physics' (apologies for the quote marks but i feel they are necessary) you often need to reduce the general equations taught at undergrad level to specific situations with specific limits/boundaries.  To paraphrase an old quote: 'you need to learn more and more about less and less'.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't really answer your question so much as prepare you for the fact that learning the detailed explanation of any experiment often requires a very narrow interpretation of more general and complete laws.  This means that learning the details can often be a not very good way to learn 'physics' and concentrating on the text book higher level is in some way more satisfying for the amateur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to avoid be patronisng so apologies if this all comes across as obvious.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-03-26T01:05:58.673" />
  <row Id="7575" PostTypeId="2" ParentId="7571" CreationDate="2011-03-26T01:24:51.003" Score="1" ViewCount="" Body="&lt;p&gt;This is a bit strange.  The Langevin equation &#xA;$$&#xD;&#xA;\frac{dv}{dt}~+~\beta v~=~\frac{F}{m}&#xD;&#xA;$$&#xA;for the motion of a free particle under a stochastic force $F$ evaluates the velocity as an average or in an interval.  The stochastic force has a Gaussian probability distribution $\langle F(t)F(t’)\rangle~=$ $2\beta kT\delta(t~-~t’)$, which is also a Markov process.  The Langevin equation has an associated Fokker-Planck equation.  This is a diffusion type of equation,&#xA;$$&#xD;&#xA;\frac{\partial P}{\partial t}~=~\frac{\partial}{\partial v}{\beta v P}~+~\frac{D}{m^2}\frac{\partial^2P}{\partial v^2},~ D~=~m\beta kT&#xD;&#xA;$$&#xA;This looks a bit like a heat equation for the evolution of a probability function $P$ for a particle to be found in some region of space.  Under a Fourier transform this is converted into a first order differential equation in $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This experiment is measuring the “cut off” in the one dimensional fractal path of the particle.  On a small enough of a scale and with time slowed down the motion is due to individual atoms recoiling off of the small particle.  This is a domain where the thermal-classical world transitions into the quantum world.  So in this transition place the above Fokker-Planck equation gives way to a Schrodinger equation.  The above stochastic force in this scale becomes replaced by small change in momentum $\delta p$ which defines an operator ${\hat p}~=~\langle p\rangle~+~\delta p$, where  $\delta p$ will have the Markovian properties above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an interesting domain to think about, for it appears to be in the netherworld between thermal-classical physics and quantum mechanics.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-26T01:24:51.003" />
  <row Id="7576" PostTypeId="1" CreationDate="2011-03-26T02:19:47.207" Score="0" ViewCount="240" Body="&lt;p&gt;There is always a debate around &lt;a href=&quot;http://www.earthhour.org&quot; rel=&quot;nofollow&quot;&gt;Earth Hour&lt;/a&gt; every year, and the opposite side of Earth Hour usually claims that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The (sudden) decrease and increase of the power usage in the start and end of Earth Hour will cause much more power loss (than the save of power), and even do damage to the power supply system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is this statement true? To what extent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="2503" LastActivityDate="2011-03-26T13:09:09.450" Title="Will Earth Hour do damage to power supply system?" Tags="&lt;power&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" />
  <row Id="7577" PostTypeId="2" ParentId="7576" CreationDate="2011-03-26T03:29:41.967" Score="2" ViewCount="" Body="&lt;p&gt;Probably not, lighting uses a decreasing proportion of the power load these days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the UK the grid used to carefully monitor TV, during the commercial breaks in popular soaps the power draw would go up by giga-watts as everybody made tea! This was in the days when 20M people watched the same show and in a country with electric kettles.&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-03-26T03:29:41.967" />
  <row Id="7578" PostTypeId="2" ParentId="7576" CreationDate="2011-03-26T04:44:28.283" Score="2" ViewCount="" Body="&lt;p&gt;The grids can take turning the lights off, by experimental observation: everyday all over the world the lights come up at about the same time for each geographical region, and turn off at about the same time due to the similar sleep schedule of millions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total black out might overload the system, but the percentage of people who take part in the game is small so even then dangers are minimal.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-26T04:44:28.283" CommentCount="3" />
  <row Id="7579" PostTypeId="1" AcceptedAnswerId="7582" CreationDate="2011-03-26T06:13:33.540" Score="2" ViewCount="134" Body="&lt;p&gt;I was reading &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0305045&quot; rel=&quot;nofollow&quot;&gt;Quantum Computation explained to my mother&lt;/a&gt;, which makes the following claim&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Postulate 2 A closed physical system&#xA;  in state V will evolve into a new&#xA;  state W , after a certain period of&#xA;  time, according to W = UV where U is a&#xA;  n × n unit matrix of complex numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, &lt;code&gt;V&lt;/code&gt; is a column matrix with &lt;code&gt;n&lt;/code&gt; rows. Can anybody justify this assumption?&lt;/p&gt;&#xA;" OwnerUserId="119" LastActivityDate="2011-03-26T06:55:49.467" Title="Representation of quantum transformations as matrices" Tags="&lt;quantum-information&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="7580" PostTypeId="2" ParentId="7579" CreationDate="2011-03-26T06:35:29.397" Score="3" ViewCount="" Body="&lt;p&gt;If we assume time evolution preserves the Hilbert space norm, then Wigner had shown it can only be a linear unitary transformation, or an antilinear antiunitary transformation. If time evolution is continuous in time, it can't possibly be antiunitary as there's no continuous deformation of the identity operator to an antiunitary transformation.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-03-26T06:35:29.397" CommentCount="1" />
  <row Id="7581" PostTypeId="2" ParentId="7571" CreationDate="2011-03-26T06:38:40.767" Score="4" ViewCount="" Body="&lt;p&gt;The article makes no sense. Einstein realized that matter was composed out of atoms, so the number of collisions of a Brownian particle with the surrounding molecule is finite in a finite period of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for times $t$ much longer than the typical scale between the collisions, the particle moves by a distance scaling like $\sqrt{t}$. It follows that the velocity measured in time interval $t$ goes like &#xA;$$v=s/t\sim\frac{\sqrt{t}}{t} =\frac{1}{\sqrt{t}}$$&#xA;and it diverges in the $t\to 0$ limit, that of the instantaneous velocity. This fact is described by the statements that the instantaneous velocity &quot;cannot exist&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the scaling law above can't really be extrapolated for time scales shorter than the time between two collisions and what the people you mentioned may have succeeded in is to increase the time resolution so that the individual collisions may be distinguished - an advance that was impossible when Einstein wrote his paper 100+ years ago. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, it is not a big deal and it doesn't contradict anything Einstein believed - even though he may have failed to write down the explanation above explicitly.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-26T06:38:40.767" CommentCount="2" />
  <row Id="7582" PostTypeId="2" ParentId="7579" CreationDate="2011-03-26T06:55:49.467" Score="3" ViewCount="" Body="&lt;p&gt;Dear Casebash, the statement above is a &quot;postulate of quantum mechanics&quot; (note that it is even called in this way) - a fundamental &quot;axiom&quot; or &quot;assumption&quot; of any theory that wants to be called &quot;a quantum mechanical theory&quot;. There are about 5-10 general postulates of quantum mechanics and the logic of quantum mechanics only starts to make sense to someone once he understands all of them. So it is counterproductive to take individual postulates out of the context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This particular postulate, referred to as the &quot;linearity of the wave functions&quot;, guarantees that the wave functions can only be constrained by linear equations. The set of allowed values of the wave functions is known as the &quot;Hilbert space&quot; and the postulate above guarantees that the set (the Hilbert space) must be a linear vector space and all physically meaningful operations, including the values of observables and evolution, have to be given by linear operators on this Hilbert space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For infinitesimal evolution in time, the evolution of a wave function or, more generally, a &quot;state vector&quot; is dictated by Schrödinger's equation,&#xA;$$i\hbar \frac{d}{dt} |\psi\rangle =  H | \psi\rangle.$$&#xA;If the linear equation above is right, one can also mathematically prove that &#xA;$$|\psi(t_1)\rangle = U | \psi(t_0)\rangle.$$&#xA;In fact, we may exactly calculate what $U$ is in terms of the Hamiltonian $H$: it is&#xA;$$ U = \exp(H(t_1-t_0) / i \hbar) $$&#xA;where we have to exponentiate an operator - which is possible. More generally, if the norm $\langle \psi| \psi\rangle$ is conserved, and it should be because it may be interpreted as the total probability (100 percent) that the physical system is found in any of the mutually excluding states, it must be true that the operator $H$ has to be Hermitian and, equivalently, $U$ has to be unitary or antiunitary. (Only the unitary option is possible if $U$ is obtained from $H$ by a continuous evolution.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There doesn't exist a single experiment in which the linearity of quantum mechanics would be violated. In fact, by pure thought, one may see that such a deviation from linearity would require one to readjust the equations for $|\psi\rangle$ or their interpretations so that the total probability is returned back to 100 percent. Such readjustments would inevitably violate locality. Moreover, they would make the &quot;collapse&quot; of a wave function (and its precise moment) observable, at least in principle. These are de facto physical inconsistencies because they contradict the special theory of relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, as of 2011, the postulates of quantum mechanics including linearity of the Hilbert space (of allowed wave functions) seem to be 100 percent valid and there doesn't even exist a logically plausible proposed way to &quot;deform&quot; them that wouldn't lead to other problems that are viewed as physical inconsistencies.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-26T06:55:49.467" />
  <row Id="7583" PostTypeId="1" CreationDate="2011-03-26T08:00:16.157" Score="-2" ViewCount="150" Body="&lt;p&gt;Where do you get your good physics ideas from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your colleagues, your boss? From reading articles and textbooks? When scratching notes on a notepad? In front of the computer? In &quot;bed, bath and bus&quot;? From discussions with others?&lt;/p&gt;&#xA;" OwnerUserId="2767" LastActivityDate="2011-03-26T16:21:32.897" ClosedDate="2011-03-26T16:25:30.983" Title="Where do you get your good physics ideas from?" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="7584" PostTypeId="1" AcceptedAnswerId="7586" CreationDate="2011-03-26T08:05:13.347" Score="4" ViewCount="413" Body="&lt;p&gt;If a radioactive material takes a very long time to decay, how is its half life measured or calculated? Do we have to actually observe the radioactive material for a very long time to extrapolate its half life?&lt;/p&gt;&#xA;" OwnerUserId="2766" LastActivityDate="2011-03-30T10:46:43.257" Title="How do we know that some radioactive materials have a half life of millions or even billions of years?" Tags="&lt;nuclear-physics&gt;&lt;radioactivity&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="7585" PostTypeId="1" CreationDate="2011-03-26T08:05:49.447" Score="3" ViewCount="232" Body="&lt;p&gt;As I learned in my string theory course, you can describe the quantum spin-0 particle by quantizing the arclength Lagrangian of a relativistic classical particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is whether you can get a photon, or arbitrary particle with spin, by quantizing a particle Lagrangian. Since we usually call our quanta in quantum field theory &quot;particles,&quot; it seems we should have a Lagrangian describing said particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have heard of some constructions involving worldline supersymmetry. What is the relation between worldline and spacetime supersymmetry in such Lagrangians? Is it possible to have a Lagrangian without worldline or spacetime supersymmetry? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone knows a good reference or review article I would be most happy!&lt;/p&gt;&#xA;" OwnerUserId="1856" LastActivityDate="2011-03-27T14:08:46.097" Title="Particle Lagrangians" Tags="&lt;quantum-field-theory&gt;" AnswerCount="3" />
  <row Id="7586" PostTypeId="2" ParentId="7584" CreationDate="2011-03-26T08:10:30.957" Score="9" ViewCount="" Body="&lt;p&gt;No, one doesn't need to measure the material for years - or even millions or billions of years. It's enough to watch it for a few minutes (for time $t$) and count the number of atoms $\Delta N$ (convention: a positive number) that have decayed. The lifetime $T$ is calculated from&#xA;$$ \exp(-t/T) = \frac{N - \Delta N}{N}$$&#xA;where $N$ is the total number of atoms in the sample. This $N$ can be calculated as &#xA;$$N={\rm mass}_{\rm total} / {\rm mass}_{\rm atom}.$$&#xA;If we know that the lifetime is much longer than the time of the measurement, it's legitimate to Taylor-expand the exponential above and only keep the first uncancelled term:&#xA;$$ \frac{t}{T} = \frac{\Delta N}{N}.$$&#xA;The decay of the material proceeds atom-by-atom and the chances for individual atoms to decay are independent and equal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get some idea about the number of decays, consider 1 kilogram of uranium 238. Its atomic mass is $3.95\times 10^{-25}$ kilograms and its lifetime is $T=6.45$ billion years. By inverting the atomic mass, one sees that there are $2.53\times 10^{24}$ atoms in one kilogram. So if you take one kilogram of uranium 238, it will take $2.53\times 10^{24}$ times shorter a time for an average decay, e.g. the typical separation between two decays is&#xA;$$t_{\rm average} = \frac{6.45\times 10^9\times 365.2422\times 86400}{2.53\times 10^{24}}{\rm seconds} = 8.05\times 10^{-8} {\rm seconds}. $$&#xA;So one gets about 12.4 million decays during one second. (Thanks for the factor of 1000 fix.) These decays may be observed on an individual basis. Just to be sure, $T$ was always a lifetime in the text above. The half-life is simply $\ln(2) T$, about 69 percent of the lifetime, because of some simple maths (switching from the base $e$ to the base $2$ and vice versa).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we observe $\Delta N$ decays, the typical relative statistical error of the number of decays is proportional to $1/(\Delta N)^{1/2}$. So if you want the accuracy &quot;1 part in 1 thousand&quot;, you need to observe at least 1 million decays, and so on.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-30T10:46:43.257" LastActivityDate="2011-03-30T10:46:43.257" CommentCount="15" />
  <row Id="7587" PostTypeId="2" ParentId="7585" CreationDate="2011-03-26T08:17:25.730" Score="1" ViewCount="" Body="&lt;p&gt;As spin has no classical analog, you either need to start with a classical field theory, or a quantum wave function.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-03-26T08:17:25.730" />
  <row Id="7588" PostTypeId="2" ParentId="7585" CreationDate="2011-03-26T08:31:18.030" Score="3" ViewCount="" Body="&lt;p&gt;Yes, you may obtain spin from first-quantized Lagrangians as well. However, such a particle with spin must have some extra degrees of freedom - which don't need to be field-theory degrees of freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In supersymmetric theories, one may consider the propagation of a point-like particle in a superspace instead of a regular bosonic space. The world line of such a &quot;superparticle&quot; is then given by functions&#xA;$$X^\mu (\tau), \theta^a (\tau)$$&#xA;where $\theta$ are additional Grassmannian (fermionic) functions of the parameter $\tau$ along the world line. Once you consider the quantum version of this theory, you will find out that the wave function of the particle depends not only on $X$ but also the fermionic directions $\theta^a$. The wave function may be Taylor-expanded in these $\theta$ variables - one only gets $2^N$ terms where $N$ is the number of these $\theta$ variables. And the individual coefficients in front of products of $\theta$ describe the wave function for the particle's being in a particular state of spin, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, you may consider a superparticle in 11 dimensions of M-theory. Then you effectively have 8 $\theta$ variables that matter - after you eliminate most of the covariant 32 components of the corresponding spinor. The Taylor-expanded function of the 8 physical $\theta$ variables has 256 components and they describe various components of the graviton field, gravitino, and the $3$-form generalized gauge potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of string theory is another major example of what you're asking. It's because perturbative string theory is usually studied in the first-quantized form. Aside from the &quot;center of mass&quot; degrees of freedom $X^\mu$ and their possible fermionic (supersymmetric) counterparts $\theta^a$ I discussed above, a string is a particle that also has the other Fourier modes $X^\mu_n$ and $\theta^a_n$ where $n$ is an integer - they can be obtained by Fourier-transforming $X^\mu(\sigma)$ and $\theta^a(\sigma)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $n\neq 0$ modes of $X$ and $\theta$ are internal degrees of freedom of the particle and by exciting them, in the same way as we do for the harmonic oscillator etc., we may change the spin of the particle as well. Some fermionic degrees of freedom are always necessary for the spin to be allowed to take half-integral values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, string theory may be formulated as a generalized field theory, the so-called &quot;string field theory&quot;. One may only calculate physically equivalent answers from this &quot;second-quantized approach&quot; to string theory and because it doesn't bring us too many new things, it's only studied by a rather limited group of string theorists who are fascinated by the formalism of string field theory itself.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-26T08:31:18.030" CommentCount="4" />
  <row Id="7590" PostTypeId="2" ParentId="7583" CreationDate="2011-03-26T08:54:23.917" Score="2" ViewCount="" Body="&lt;p&gt;I see you are new here, so you could edit your question and give more details on what you are really after. You have answered yourself with all the alternatives you offer for the case of a &quot;eureka&quot; moment. The same would be true for any human endeavor which needed sudden inspiration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most research physicists get their ideas from following consistently the questions posed from the research they have done before, by colleagues and further studies. They pursue them if they believe they are good ideas that will lead to a peer reviewed publication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes change of research direction can be stimulated by talks and discussions in conferences or at the institute, and the possibility of collaborating with a colleague on a problem.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-26T08:54:23.917" />
  <row Id="7591" PostTypeId="2" ParentId="5773" CreationDate="2011-03-26T10:01:03.620" Score="2" ViewCount="" Body="&lt;p&gt;You may have a look at deformation quantization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bayen, F. ; Flato, M.; Frønsdal, C.; Lichnerowicz, A. ; Sternheimer, D.: Quantum Mechanics as a Deformation of Classical Mechanics.&#xA;In: Lett. Math. Phys. 1 (1977), S. 521–530&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bayen, F. ; Flato, M. ; Frønsdal, C. ; Lichnerowicz, A. ; Sternheimer, D.: Deformation Theory and Quantization. In: Ann. Phys. 111&#xA;(1978), S. 61–151&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the original papers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See for example &lt;a href=&quot;http://omnibus.uni-freiburg.de/~sw12/Download/intro.pdf&quot; rel=&quot;nofollow&quot;&gt;http://omnibus.uni-freiburg.de/~sw12/Download/intro.pdf&lt;/a&gt; for a short elementary introduction. &lt;a href=&quot;http://iopscience.iop.org/1742-6596/103/1/012002&quot; rel=&quot;nofollow&quot;&gt;http://iopscience.iop.org/1742-6596/103/1/012002&lt;/a&gt; may be also interesting as introduction.&lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorUserId="667" LastEditorDisplayName="" LastEditDate="2011-03-26T10:06:26.567" LastActivityDate="2011-03-26T10:06:26.567" />
  <row Id="7592" PostTypeId="2" ParentId="5773" CreationDate="2011-03-26T10:13:38.513" Score="4" ViewCount="" Body="&lt;p&gt;Here is an overview of quantization methods: &lt;a href=&quot;http://arxiv.org/abs/math-ph/0405065&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/math-ph/0405065&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of this article deals with QM on manifolds.&lt;/p&gt;&#xA;" OwnerUserId="667" LastActivityDate="2011-03-26T10:13:38.513" />
  <row Id="7593" PostTypeId="2" ParentId="7576" CreationDate="2011-03-26T11:07:10.680" Score="2" ViewCount="" Body="&lt;p&gt;It is unlikely that the Earth Hour will cause substantial damage to the power supply system. However, it will require a lot of energy and manpower to adjust the power plants to the changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, generators are automatically taken off the grid if it is detected that the power supply is higher than the current demand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the Earth Hour ends, power generation companies will have to turn the generators on (which also consumes time and energy) is such a way that they can supply energy to all those who turn their lights on. But they can't do this too rapidly, or the generators will automatically shut down again.&lt;/p&gt;&#xA;" OwnerUserId="2752" LastActivityDate="2011-03-26T11:07:10.680" CommentCount="2" />
  <row Id="7594" PostTypeId="2" ParentId="7576" CreationDate="2011-03-26T11:22:01.120" Score="2" ViewCount="" Body="&lt;p&gt;Not at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Germany illumination is about 3 % of total electric power. I think this &#xA;the same value as in most countries. &#xA;When You look at the power consumption curve  on a typical day, You will realize that &#xA;switching lights on and off is peanuts.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-26T11:22:01.120" />
  <row Id="7595" PostTypeId="2" ParentId="7540" CreationDate="2011-03-26T12:23:52.250" Score="1" ViewCount="" Body="&lt;p&gt;Chapter 6.11 of Jackson's Electrodynamics mentions that Dirac's quantization argument fixes the strength of a magnetic monopole, were one to exist. Jackson goes on to say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...Their coupling strength is enormous,&#xA;  making their extraction from matter&#xA;  with dc magnetic fields and their&#xA;  subsequent detection very simple in&#xA;  principle. For instance, the energy&#xA;  loss in matter by a relativistic Dirac&#xA;  monopole is approximately the same as&#xA;  that of a relativistic heavy nucleus&#xA;  with Z=137n/2. It can presumably be&#xA;  distinguished from such a nucleus if&#xA;  it is brought to rest because it will&#xA;  not show an increase in ionization at&#xA;  the end of its range...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The method suggested wouldn't work at a distance, but it seems to cover the heart of your question.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2011-03-26T12:23:52.250" />
  <row Id="7596" PostTypeId="2" ParentId="7576" CreationDate="2011-03-26T13:09:09.450" Score="0" ViewCount="" Body="&lt;p&gt;As the vast majority of people aren't stupid enough to fall for the scam/hoax, the effects will be miniscule.&#xA;And of those who do, as said, their electric lights are only a small percentage of their total power consumption (and most likely they'll compensate by turning to other electrical appliances that consume more electricity than those lights).&lt;/p&gt;&#xA;" OwnerUserId="2625" LastActivityDate="2011-03-26T13:09:09.450" CommentCount="4" />
  <row Id="7597" PostTypeId="2" ParentId="7540" CreationDate="2011-03-26T13:38:36.203" Score="2" ViewCount="" Body="&lt;p&gt;Consider the motion of a magnetic monopole in a completely symmetric Maxwell system, where &#xA;$$&#xD;&#xA;\nabla\cdot {\vec B}~=~4\pi\rho_{mag},&#xD;&#xA;$$&#xA;and &#xA;$$&#xD;&#xA;\nabla\times{\vec E}~=~4\pi{\vec J}_{mag}~-~\frac{\partial{\vec B}}{\partial t}&#xD;&#xA;$$&#xA;The first equation is then a Gauss’ law for magnetic monopole charge, and the second is a magnetic current form of the Maxwell-Faraday equation.  For the occurrence of a magnetic monopole flying through space this will act as a transient current.  The last term on the right hand side is a displacement monopole current in this case.  The left hand side will by Stokes’ law $\int\nabla\times{\vec E}\cdot da~=$ $\int{\vec E}\times d{\vec l}$, produce an electric current in a loop.  So the right hand side could be measured by the torque this magnetic field induces on an ordinary magnetic dipole.  The right hand side measured in a solenoid.  If the left hand side and the last right hand side term do not equal each other in the standard form of the Maxwell equation with ${\vec J}_{mag}~=~0$, this would be a signal for the detection of a magnetic monopole.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-03-26T16:14:42.820" LastActivityDate="2011-03-26T16:14:42.820" CommentCount="1" />
  <row Id="7598" PostTypeId="2" ParentId="7584" CreationDate="2011-03-26T14:05:55.580" Score="2" ViewCount="" Body="&lt;p&gt;Lubos has already answered your question comprehensively. I would like to add just one thing. It is one of the funny things of the theory of probability that you need not have to wait for that long to confirm the half life. For example it has been estimated that proton has a half life of approximately $10^{32}$ years in some grand unified theories. Now in order to observe proton decay one does not have to wait for $10^{32}$ years. You just take some material which has $10^{32}$ numbers of protons. According to the theory of probability you should observe at least one proton decay per year (if a proton doest indeed take $10^{32}$ years to decay). For an easier example, suppose you throw a dice 6 times. The probability of a particular outcome is same or equivalent to throwing 6 dices at once.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-03-26T14:05:55.580" CommentCount="1" />
  <row Id="7599" PostTypeId="1" CreationDate="2011-03-26T14:37:05.320" Score="1" ViewCount="294" Body="&lt;p&gt;What are some important examples of paradoxes in physics which are still currently unresolved?&lt;/p&gt;&#xA;" OwnerUserId="2770" LastEditorUserId="2770" LastEditorDisplayName="" LastEditDate="2011-03-26T16:19:35.357" LastActivityDate="2011-03-26T16:19:35.357" CommunityOwnedDate="2011-03-26T16:21:16.400" Title="List of physics paradoxes" Tags="&lt;soft-question&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="7600" PostTypeId="2" ParentId="7599" CreationDate="2011-03-26T14:58:36.743" Score="0" ViewCount="" Body="&lt;p&gt;In quantum gravity, we have the black hole information paradox. If information is thrown into a black hole past the event horizon and is destroyed at the singularity inside, and the black hole subsequently decays away via Hawking radiation, where does the information go to? If information can't travel faster than light, it can't escape from inside the black hole to outside it. But if information is truly lost, this would lead to violations of unitarity in quantum mechanics, with terrible consequences.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-03-26T15:05:30.857" LastActivityDate="2011-03-26T15:05:30.857" CommunityOwnedDate="2011-03-26T16:21:16.400" />
  <row Id="7601" PostTypeId="2" ParentId="7599" CreationDate="2011-03-26T15:01:08.857" Score="2" ViewCount="" Body="&lt;p&gt;Let me assure you that all paradoxes are false paradoxes if you understand the physics. There are many misunderstandings for naive mortals like us for a period which are called &quot;paradoxes&quot; until resolved. Some familiar examples are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Twin paradox&lt;/li&gt;&#xA;&lt;li&gt;EPR paradox&lt;/li&gt;&#xA;&lt;li&gt;Schrödinger's cat paradox&lt;/li&gt;&#xA;&lt;li&gt;Faradays rotating disc paradox (quite an old one)&lt;/li&gt;&#xA;&lt;li&gt;Black hole information paradox (a recent one) (solution of this is still controversial)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-03-26T15:01:08.857" CommunityOwnedDate="2011-03-26T16:21:16.400" CommentCount="5" />
  <row Id="7602" PostTypeId="2" ParentId="7599" CreationDate="2011-03-26T15:58:35.183" Score="0" ViewCount="" Body="&lt;p&gt;I will put in my two cents of the euro. Paradoxes arise when two  axiomatic systems are mixed up. Paradoxes are resolved with the meta (often content versus context)concept in general cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Famous example: All Cretans are liars, said the Cretan.  It is resolved by the meta language level concept, where one level is the Cretan speaking, and the other is the statements coming out of his/her mouth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One cannot mix, for example, classical thermodynamics and quantum statistical mechanics concepts without a thorough investigation of underpinnings, because both are axiomatic systems valid in certain regions of phase space and one has to be very careful of overlaps in order not to double count, the energy, as an example. ( which happens in  climate studies).&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-26T15:58:35.183" CommunityOwnedDate="2011-03-26T16:21:16.400" />
  <row Id="7603" PostTypeId="1" CreationDate="2011-03-26T16:01:40.940" Score="0" ViewCount="254" Body="&lt;p&gt;As you all know, journal subscriptions are extremely expensive, and many colleges and libraries don't subscribe to them. Many important articles aren't available on the preprint server arxiv. My question is, where can I find pirated copies of journals? Your help means so much to me.&lt;/p&gt;&#xA;" OwnerUserId="2771" LastActivityDate="2011-03-26T22:02:11.273" ClosedDate="2011-03-26T16:22:39.473" Title="Where can I find pirated copies of journals?" Tags="&lt;soft-question&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="7604" PostTypeId="2" ParentId="7603" CreationDate="2011-03-26T16:13:04.067" Score="1" ViewCount="" Body="&lt;p&gt;I highly doubt anyone would go through the trouble of pirating journals, simply because most people who need the articles are affiliated to some university or the other that subscribes to the journals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My advice is to use a library for the older articles that aren't on the arxiv and at the same time - make friends at universities who can email you articles once in a while!&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-03-26T22:02:11.273" LastActivityDate="2011-03-26T22:02:11.273" />
  <row Id="7605" PostTypeId="2" ParentId="7603" CreationDate="2011-03-26T16:14:39.187" Score="1" ViewCount="" Body="&lt;p&gt;I will tell you what I do if a paper whose title I know is behind a pay wall and I am away from a library. I take the title and put it in a search engine. If it is interesting enough  somebody has put up a pdf version someplace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whole journals are a tall order and really not legal to copy them.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-26T16:14:39.187" />
  <row Id="7606" PostTypeId="2" ParentId="7603" CreationDate="2011-03-26T16:18:28.030" Score="0" ViewCount="" Body="&lt;p&gt;Most libraries will order articles for you for a small fee. You need to bear in mind that it takes money to sustain these journals and you're fundementally a naughty person expecting something for nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about getting a job so you can pay for the articles?&lt;/p&gt;&#xA;" OwnerUserId="2146" LastEditorUserId="2146" LastEditorDisplayName="" LastEditDate="2011-03-26T19:47:27.473" LastActivityDate="2011-03-26T19:47:27.473" CommentCount="4" />
  <row Id="7607" PostTypeId="2" ParentId="7583" CreationDate="2011-03-26T16:21:32.897" Score="-2" ViewCount="" Body="&lt;p&gt;By getting to know the problem like the inside of my mouth&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-03-26T16:21:32.897" />
  <row Id="7608" PostTypeId="2" ParentId="7487" CreationDate="2011-03-26T19:10:16.147" Score="2" ViewCount="" Body="&lt;p&gt;I131 is a gamma and beta emitting isotope. As such, the beta emission is the destructive component. Beta emission is an electron ejected from the nucleus and is ionizing. This aspect makes beta electrons very destructive to thyroid tissue which concentrates iodine ~200 times more than any tissue in order to make thyroxine, an endocrine hormone which regulates the rate of metabolic processes. The important consideration is absorbed dose. This is the total dose absorbed while the I131 is present. So, the health risk is not 1/2 as much in one half life because the dose is internal and 1st half has already been absorbed.&lt;/p&gt;&#xA;" OwnerUserId="2773" LastActivityDate="2011-03-26T19:10:16.147" CommentCount="1" />
  <row Id="7610" PostTypeId="1" CreationDate="2011-03-26T20:57:10.290" Score="3" ViewCount="302" Body="&lt;p&gt;Differences? They are both an electron and a proton, since the neutron decays to a proton and an electron, what's the difference between a neutron and proton + electron? so is it just a higher binding energy between the two?&lt;/p&gt;&#xA;" OwnerUserId="2775" LastEditorUserId="622" LastEditorDisplayName="" LastEditDate="2011-06-11T02:00:42.530" LastActivityDate="2011-06-11T02:00:42.530" Title="What is the difference between a neutron and hydrogen?" Tags="&lt;particle-physics&gt;&lt;electron&gt;&lt;neutron&gt;&lt;proton&gt;&lt;binding-energy&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="7611" PostTypeId="1" AcceptedAnswerId="7624" CreationDate="2011-03-26T20:58:14.503" Score="4" ViewCount="303" Body="&lt;p&gt;I am your nightmare poster - a layman trying to learn special relativity.  (I'm also a refugee from stackoverflow.com: trust me, it's only going to get worse).  Apologies in advance if my question is nonsense or too open-ended.  I won't be offended if you vote to close it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm working through Taylor &amp;amp; Wheeler 'Spacetime Physics'.  The book starts with a nice analogy of different teams of surveyors mapping out a world with different sized measuring rods (and even the same teams using different sized rods for different coordinate axes), and coming to realize that there is an underlying structure which is independent of the coordinate system.  The idea is that spacetime intervals likewise give us the objective measuring stick by which we can map out spacetime (presumably by some process analogous to triangulation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just not able to intuit spacetime intervals like this, especially that distances in time and space have opposite sign. I can draw the diagrams and answer the exercises by plugging numbers into essentially the same equations over and over.  Nor do I have any trouble working with the different inertial observers with all their meter rules and clocks.  It's just the God's eye spacetime view which is eluding me.  If intervals are real and objective, it ought to be possible to project this lattice of intervals onto the world and see it independently of any frame of reference (just like I can imagine the earth without the surveyors' grid lines superimposed on it).  I accept that it might be some funny shaped space like a tennis ball turned inside out, but still...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my questions are: do you ever get a proper intuitive model for Minkowski spacetime, so that you can visualize the intervals, move objects about relativisticly in your mind's eye, and so on?  If so, are there any tricks or insights which helped you get there?  Do things 'really' move in your model, or is time just another static dimension like in the Minkowski diagrams?&lt;/p&gt;&#xA;" OwnerUserId="2772" LastEditorUserId="2772" LastEditorDisplayName="" LastEditDate="2011-03-26T23:14:10.130" LastActivityDate="2011-03-28T04:22:29.243" Title="How do I develop an intuitive model of spacetime?" Tags="&lt;special-relativity&gt;&lt;spacetime&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="7612" PostTypeId="2" ParentId="5743" CreationDate="2011-03-26T20:59:51.360" Score="1" ViewCount="" Body="&lt;p&gt;From the experiment with the silicon wafer we get the information that heat was generated. When two coherent beams of photons interfered destructively, heat appeared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is heat in this particular case? Heat results when a bunch of infrared incoherent photons interact with matter. The destructive interference of the waves of photons became a burst of incoherent infrared photons heating the wafer..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same will happen in vacuum. The disappearance of the laser coherence  will be accompanied by a burst of infrared photons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can think of it as the particle nature of light  dominating, with multiple photons scatterings on virtual electron loops from each other. The energy would be all there in the infrared burst. Because the momenta of the two beams are in the same  x direction, they can only scatter within the Heisenberg uncertainty allowance in space, and thus turn into photons of very low energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could be tested experimentally in vacuum , by detecting the infrared from the destructive interference. I think it is the correct view because the claim that &quot;energy disappears&quot; is not viable within our local physics framework . If the experiment shows no infrared correlated with the interference, then one would have proof of extra dimensions into which the energy was dissipated.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-28T11:15:03.910" LastActivityDate="2011-03-28T11:15:03.910" CommentCount="11" />
  <row Id="7613" PostTypeId="2" ParentId="7610" CreationDate="2011-03-26T21:19:29.597" Score="13" ViewCount="" Body="&lt;p&gt;A neutron is &lt;em&gt;not&lt;/em&gt; a proton and an electron lumped together (as your question seems to suggest you think)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hydrogen atom is a bound state of an electron and a proton (bound by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_force&quot; rel=&quot;nofollow&quot;&gt;electromagnetic force&lt;/a&gt;) whereas a neutron is a bound state of three quarks (bound by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Strong_force&quot; rel=&quot;nofollow&quot;&gt;strong force&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be tempted to think that a neutron is also a bound state of an electron and a proton because a neutron can decay into an electron and a proton and the neutron is also slightly more massive than a proton. But you'd be wrong. Here's why:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both a neutron and a proton are bound states of three quarks. &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_decay&quot; rel=&quot;nofollow&quot;&gt;Beta decay&lt;/a&gt; can convert a neutron to a proton like this $$udd \rightarrow uud + e^- + \bar{\nu_e}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, one of the down quarks in the neutron gets converted into an up quark, an electron and an antineutrino through the mediation of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weak_interaction&quot; rel=&quot;nofollow&quot;&gt;weak force&lt;/a&gt;. Since the up quark has a charge of $+2/3$ and the down quark $-1/3$, that explains the difference in charge of a neutron and a proton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/kIp2f.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's what turns a neutron into a proton, not some kind of ejection of an electron out of the neutron, but a genuine transmutation of a fundamental particle (a down quark) through the weak force.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-03-26T21:33:59.210" LastActivityDate="2011-03-26T21:33:59.210" CommentCount="3" />
  <row Id="7614" PostTypeId="2" ParentId="7610" CreationDate="2011-03-26T21:19:47.753" Score="4" ViewCount="" Body="&lt;p&gt;The neutron decays into a proton, an electron and an antineutrino. So even the end components are different from Hydrogen which is just a proton with an electron orbiting around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The binding forces are also different. The proton and the electron are bound by the electromagnetic force. The neutron by the strong to the rest of the nucleons in a nucleus. When it decays, it decays  by the weak force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at this  &lt;a href=&quot;http://teachers.web.cern.ch/teachers/archiv/HST2002/webgroup/mcclean/Introduction%20to%20Particle%20Physics.ppt&quot; rel=&quot;nofollow&quot;&gt;presentation for teachers&lt;/a&gt;  where also you will see what is inside a neutron and a proton and you will understand better the nature of particles.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-27T07:21:52.543" LastActivityDate="2011-03-27T07:21:52.543" CommentCount="2" />
  <row Id="7615" PostTypeId="1" CreationDate="2011-03-26T20:07:14.323" Score="3" ViewCount="317" Body="&lt;p&gt;There is a common analogy about the structure of an atom, such as the nucleus is a fly in the centre of a sports stadium and the electrons are tiny tiny gnats circling the stadium (tip of the hat to 'The Greatest Show on Earth') but what is in the space between the 'fly' and circling 'gnats'?&lt;/p&gt;&#xA;" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-26T22:28:17.567" LastActivityDate="2011-03-27T11:13:24.763" Title="What is in the space between a nucleus of an atom and its electrons?" Tags="&lt;quantum-mechanics&gt;&lt;atoms&gt;&lt;subatomic&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="7617" PostTypeId="2" ParentId="7561" CreationDate="2011-03-26T22:05:54.530" Score="0" ViewCount="" Body="&lt;p&gt;Since you're really interested in the horizon as part of an approach to understanding the Unruh effect, I will address that instead of the initial Question. What experimental resources are we to consider an accelerating observer A to have? If A can make only local measurements in the laboratory &lt;strong&gt;La&lt;/strong&gt; in the diagram, then (in the Wightman or Haag-Kastler axiomatic backgrounds) A can make exactly the same deductions about the state from their measurements as can an observer who can use the laboratory &lt;strong&gt;L0&lt;/strong&gt;.&lt;img src=&quot;http://i.stack.imgur.com/KYOvw.png&quot; alt=&quot;An inertial laboratory L0 and an accelerated laboratory La&quot;&gt; Because of the Reeh-Schlieder theorem, what is at space-like separation from the two laboratories is equally accessible to both (in the sense that both can get as close as they like to knowing what the statistics of measurement results would be everywhere in Minkowski space by making precise enough measurements in their local laboratory, on the assumption that the Wightman axioms hold or that the Haag-Kastler axioms hold). [It would have been better not to have drawn the two laboratories the same shape, because the shapes are not relevant to the Reeh-Schlieder argument, but it's uploaded now.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A local laboratory, however, &lt;em&gt;cannot tell&lt;/em&gt; whether the state of the universe is in the &lt;em&gt;folium&lt;/em&gt; of the vacuum state, in the folium of a thermal state, or in the folium of some other less symmetrical state; the folium of a highly symmetric state is the set of states that are in the Hilbert space that is constructed by the GNS construction from the vacuum state (resp. from a thermal or some other state) over the net of algebras of observables associated with regions of Minkowski space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I take your discussion about the Unruh effect that is prompted by the earlier Answers to boil down to the fact that it's only if our laboratory is not confined to a finite region of space-time that we can make any &lt;em&gt;global&lt;/em&gt; claims about the state of the universe, despite the Reeh-Schlieder theorem. We could only possibly determine whether the state is in the folium of a thermal state instead of in the folium of the vacuum state if our laboratory is infinite in extent. Having available as our experimental resource a particle that interacts with the vacuum state in an ad-hoc way for the whole of an accelerating trajectory, as is proposed in Unruh and Wald's &lt;em&gt;Phys. Rev.&lt;/em&gt; &lt;strong&gt;A&lt;/strong&gt; 29,1047(1984), even though it's infinite in extent, seems unlikely to be enough to make such a fine-grained determination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't plan on putting the terse comments that I made earlier on the Question into this Answer, however they do inform the perspective a little if you choose to think about the Unruh effect in something like the way I have done here. The above is essentially tentative, something I've been mulling for several years, so I'd be glad to have suggestions for how anyone here would fill the gaps or of what I haven't taken into account that will shoot it down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, thanks @dbrane for the Question. I held off giving it +1 because although it seemed definitely useful it didn't seem quite clear enough, but when I realized how much it was &lt;em&gt;stimulating&lt;/em&gt;, I decided that it was better for me than had it been clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. Hans Halvorson's paper on &quot;Algebraic Quantum Field Theory&quot; (with Michael Mueger), &lt;a href=&quot;http://arxiv.org/abs/math-ph/0602036&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/math-ph/0602036&lt;/a&gt;, is a good on-line resource, although in many ways it takes you one step higher than Haag. His treatment of the concept of a folium is more sophisticated than Haag introduces in his Local Quantum Physics, in terms of quasi-equivalence of states. Hans is one of the few Philosophers of Physics who can really come up to scratch with Mathematicians on AQFT.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-03-26T22:05:54.530" CommentCount="4" />
  <row Id="7618" PostTypeId="1" CreationDate="2011-03-26T22:37:09.730" Score="3" ViewCount="188" Body="&lt;p&gt;I would like to know about the larger picture, current state and future prospects of the sequence of papers that were written by Sheldon Katz and Cumrun Vafa on F-theory. (Freddy Cachazo was also a co-author in many of these papers)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the same is also known as &quot;geometric engineering&quot;. &#xA;(Kindly explain if that is not the same)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been &lt;a href=&quot;http://arxiv.org/pdf/1103.3287&quot; rel=&quot;nofollow&quot;&gt;recent works&lt;/a&gt; on F-theory by Cumrum Vafa, Jonathan Heckman and others. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know of how this recent work fits in with the earlier work by Katz and Vafa and where do people see this pursuit to be going and what does the community think of its future prospects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these Katz-Vafa works a prospective field for beginning grad students?  &lt;/p&gt;&#xA;" OwnerUserId="2778" LastActivityDate="2011-03-27T06:01:46.327" Title="Katz and Vafa's work on F-theory" Tags="&lt;quantum-field-theory&gt;&lt;string-theory&gt;&lt;supersymmetry&gt;&lt;supergravity&gt;&lt;super-strings&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="7619" PostTypeId="2" ParentId="7615" CreationDate="2011-03-26T22:47:23.840" Score="0" ViewCount="" Body="&lt;p&gt;You might enjoy pictures from &lt;a href=&quot;http://vladimirkalitvianski.wordpress.com/2010/12/02/zoom-in-atom-or-unknown-physics-of-short-distances/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;here&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-03-26T22:47:23.840" CommentCount="1" />
  <row Id="7621" PostTypeId="1" AcceptedAnswerId="7622" CreationDate="2011-03-27T00:22:49.490" Score="3" ViewCount="190" Body="&lt;p&gt;Something that has bothered me for a while regards the interpretation of chemical potential for different statistics. While I understand its meaning in metals (and its relation with the Fermi surface), I cannot quite relate this definition with the thermodynamic chemical potential, &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_potential&quot; rel=&quot;nofollow&quot;&gt;defined as&lt;/a&gt; the change in energy of the system when one particle is added to it (or according to &lt;a href=&quot;http://demonstrations.wolfram.com/BoseEinsteinFermiDiracAndMaxwellBoltzmannStatistics/&quot; rel=&quot;nofollow&quot;&gt;Wolfram Demonstrations&lt;/a&gt;, &quot; It can be interpreted, for example, as the ability of the system to perform phase transitions or chemical reactions, or its tendency to diffuse&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Are those concepts (thermodynamic vs Fermi-Dirac chemical potential) related or they should be thought as different things?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Am I missing something trivial or this cited demonstration is misleading? It is mentioned there that the x coordinate corresponds to (E-&lt;em&gt;u&lt;/em&gt;). Shouldn't the function blow up for x = 0 independent of temperature, then? I do not understand the shift for negative values as temperature increases.     &lt;/p&gt;&#xA;" OwnerUserId="2765" LastActivityDate="2011-03-27T12:47:15.903" Title="Chemical potential interpretation" Tags="&lt;statistical-mechanics&gt;&lt;condensed-matter&gt;&lt;physical-chemistry&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="7622" PostTypeId="2" ParentId="7621" CreationDate="2011-03-27T00:36:19.030" Score="4" ViewCount="" Body="&lt;p&gt;Of course they are related, the Fermi-Dirac chemical potential is &lt;em&gt;defined&lt;/em&gt; to enforce the exclusion principle &lt;em&gt;inside&lt;/em&gt; a thermodynamic framework rather than having to apply it with a separate mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider three systems, an ideal gas, a van der Waals type gas, and a Fermi gas. At a fixed volume and energy, ask &lt;em&gt;&quot;How much energy is required to add another particle?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In an ideal gas the presence of a new particle has no effect on the resident particles.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the van der Waals case, introducing a new particle pushed the particles on average closer together, and so adds to the total potential energy, but the new particle can be introduced with little or no kinetic energy. The amount of energy required (i.e. the chemical potential) depends on the density and the strength of the van der Waals force.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the Fermi case, if the low lying states are filled then you &lt;em&gt;can not&lt;/em&gt; introduce a new particle with low KE, but must introduce it with a lot of energy (how much being set by the current level of the Fermi surface). The chemical potential is at least the level of the Fermi surface.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="520" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-27T12:47:15.903" LastActivityDate="2011-03-27T12:47:15.903" CommentCount="1" />
  <row Id="7623" PostTypeId="2" ParentId="7280" CreationDate="2011-03-27T01:41:38.843" Score="1" ViewCount="" Body="&lt;p&gt;Some helpful answers: &lt;a href=&quot;http://thisquantumworld.com/wp/?s=contextuality&quot; rel=&quot;nofollow&quot;&gt;http://thisquantumworld.com/wp/?s=contextuality&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2784" LastActivityDate="2011-03-27T01:41:38.843" />
  <row Id="7624" PostTypeId="2" ParentId="7611" CreationDate="2011-03-27T03:17:26.857" Score="8" ViewCount="" Body="&lt;p&gt;Here's a little story comparing visualizing distances with visualizing spacetime intervals.&lt;/p&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;p&gt;In plane Euclidean geometry, the shape that's invariant with respect to rotations is a circle.  If you rotate a circle any amount, no one can tell.  This is in contrast to other curves, which have at most discrete symmetries with respect to rotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have a Cartesian coordinate system, the equation for a circle of radius $R$ located at the origin is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 + y^2 = R^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantity $x^2 + y^2$ uniquely picks out which circle you're on, and everywhere on a given circle has the same value for that quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you draw a circle on a piece of graph paper, then put your finger down somewhere on the circle, then rotate the paper around the center of the circle while leaving your finger in the same spot, by the end of the rotation your finger is still on the same circle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, although the x-coordinate and y-coordinate under your finger change, the value of $x^2 + y^2$ does not.  $x^2  + y^2$ is given the name &quot;distance&quot;, and it is invariant with respect to rotations.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;In 1+1 dimensional Minkowski spacetime, the shape that's invariant with respect to Lorentz boosts is a right hyperbola.  If you boost a right hyperbola by any amount, no one can tell.  This is in contrast to other curves, which have at most discrete symmetries with respect to boosts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Wikipedia image illustrates the action of boosts, which are less intuitive than rotations.  The diagonal lines shown are essentially the special hyperbola $x^2 - y^2 = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Z1x83.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you watch a single point, it drifts downward when the observer is not accelerating because time is moving forward.  When the observer accelerates, the point will quickly move along a hyperbola, then start drifting downwards again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we introduce a coordinate system, the general equation for a right hyperbola is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 - t^2 = s^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $s^2$ can be positive or negative.  The quantity $x^2 - t^2$ uniquely picks out which hyperbola you're on, and everywhere on a given hyperbola has the same value for that quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you draw a hyperbola on a piece of graph paper, then put your finger down somewhere on the hyperbola, then somehow made the paper undergo a hyperbolic rotation (i.e. Lorentz boost), by the end of the hyperbolic rotation your finger would still be on the same hyperbola.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hyperbolic rotation can't be done with a normal sheet of paper, but it looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/b/b4/Animated_Lorentz_Transformation.gif&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, although the x-coordinate and t-coordinate under your finger change, the value of $x^2 - t^2$ does not.  $x^2 - t^2$ is given the name &quot;spacetime interval&quot;, and it is invariant with respect to Lorentz boosts.&lt;/p&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;p&gt;This story begins by positing a transformation, then looking at what sort of shape is invariant under that transformation.  Physically, I think it's a little more insightful to begin with the invariant - distance or spacetime interval - and then ask what transformations leave it invariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is historically how the Lorentz transformations were discovered.  They are linear transformations that leave Maxwell's equations invariant.  Lorentz found them by searching for such transformations before Einstein published his first paper on relativity.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-03-27T03:23:07.617" LastActivityDate="2011-03-27T03:23:07.617" CommentCount="3" />
  <row Id="7626" PostTypeId="2" ParentId="7322" CreationDate="2011-03-27T03:58:34.253" Score="1" ViewCount="" Body="&lt;p&gt;This answer, though somewhat dated, comes with the best credentials: &quot;The question is still open experimentally, but theory and experiment are pointing more than ever before to the &#xA;possibility that we have discovered the 'ultimate constituents'.&quot; — National Research Council (U.S.), Elementary-Particle Physics Panel (1998), &lt;em&gt;Elementary-Particle Physics&lt;/em&gt;, National Academy Press, Washington, D.C., p. 23.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But are the &quot;ultimate constituents&quot; pointlike entities, or are they &lt;a href=&quot;http://thisquantumworld.com/wp/?page_id=1568#shapes&quot; rel=&quot;nofollow&quot;&gt;formless&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="2784" LastActivityDate="2011-03-27T03:58:34.253" CommentCount="3" />
  <row Id="7627" PostTypeId="1" CreationDate="2011-03-27T04:04:58.440" Score="1" ViewCount="190" Body="&lt;p&gt;This question came about from a side discussion that arose on this:&#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/7232/does-gr-provide-a-maximum-electric-field-limit/7240#7240&quot;&gt;Does GR provide a maximum electric field limit?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we change our choice of coordinate system completely independent of physical motion, and still refer to coordinate dependent components as observations?  For example if we have an inertial coordinate system in which we measure the electric field to be E, and now I want to boost the coordinate system to a different inertial coordinate system and state what the electric field E' is in this new coordinate system, do I need to physically accelerate to justify this change in coordinate system (and therefore need to worry about effect such as Unruh, etc. during &quot;changing&quot; the coordinate system)?&lt;/p&gt;&#xA;" OwnerUserId="2655" LastEditorUserId="2655" LastEditorDisplayName="" LastEditDate="2011-03-27T07:23:10.420" LastActivityDate="2011-03-27T07:23:10.420" Title="Does change of coordinate system require acceleration?" Tags="&lt;general-relativity&gt;&lt;special-relativity&gt;&lt;coordinate-systems&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="7628" PostTypeId="2" ParentId="7627" CreationDate="2011-03-27T04:22:43.540" Score="-1" ViewCount="" Body="&lt;p&gt;It is not entirely clear what you are getting at but by whatever I have managed to understand the answer is in order to go from one frame to another you always need to take gravity into account. Gravity plays the role of a bridge between different reference frames which are accelerating w.r.t. each other.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-03-27T05:56:11.090" LastActivityDate="2011-03-27T05:56:11.090" />
  <row Id="7629" PostTypeId="2" ParentId="7627" CreationDate="2011-03-27T04:41:26.347" Score="4" ViewCount="" Body="&lt;p&gt;This is a common confusion students have with coordinate systems when first learning SR. They hear us say things like &quot;the time according to observer Bob&quot; or &quot;in the reference frame of Alice&quot; or &quot;boost to the rest frame of Charlie&quot;.  To be clear, these are short hand.  Something isn't literally/physically &quot;in&quot; one frame and not &quot;in&quot; another frame.  When we refer to the coordinate system &quot;of&quot; an observer, or something &quot;in&quot; someone's frame, it is usually short hand for discussing values according to the instantaneous co-moving inertial coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So do not take such short hand literally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A coordinate system is just a systematic labeling of spacetime points.  We can choose any coordinate system we want.  As an example that will hopefully get past this intuition barrier, choose a coordinate system in which you are at rest.  Now change the coordinate system by adding a constant (shifting the origin).  You don't need to physically move to use this new coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, nature doesn't care how we label spacetime points. So we should be able to explain the physics independent of choice of coordinate systems.  This is indeed possible, so you can really use &lt;em&gt;whatever coordinate system you choose&lt;/em&gt;.  A particular choice may be simplier to work with based on your motion, but no choice is more mathematically correct than any other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As should be apparent from the other question you asked, the Motion Mountain book is not reliable.  Just searching in Google brings up the phrase crackpot in link titles on the first page of results.  Please stick with reputable sources.&lt;/p&gt;&#xA;" OwnerUserId="2652" LastEditorUserId="2652" LastEditorDisplayName="" LastEditDate="2011-03-27T05:10:49.020" LastActivityDate="2011-03-27T05:10:49.020" CommentCount="9" />
  <row Id="7630" PostTypeId="2" ParentId="7280" CreationDate="2011-03-27T04:46:57.920" Score="3" ViewCount="" Body="&lt;p&gt;To understand contextuality in QM &lt;a href=&quot;http://scienceblogs.com/pontiff/2008/01/contextuality_of_quantum_theor.php&quot; rel=&quot;nofollow&quot;&gt;see this helpful example&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-03-27T04:46:57.920" CommentCount="1" />
  <row Id="7631" PostTypeId="2" ParentId="7618" CreationDate="2011-03-27T06:01:46.327" Score="3" ViewCount="" Body="&lt;p&gt;I find these questions about the &quot;current state&quot; of old papers confusing because such questions about &quot;current state&quot; seem to assume that the validity of papers is changing like the weather. Well, it's not. The current state of the Katz-Vafa papers is that they are correct and important papers on F-theory that are almost 15 years old. It's very likely that they will always be correct and important papers from the late mid 1990s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the recent (five years or so) F-theoretical bottom-up phenomenology by Vafa, Heckman, and others (let me mention Marsano, Saulina, Schafer-Nameki, Dolan, Wecht, Cecotti, Cheng, Seo, Kane) is still a relatively new research direction that continues to be very active. It sometimes uses older papers such as the Katz-Vafa ones but it is surely not &quot;just about them&quot;. The newer work is focusing much more on the physics of singular cycles and phenomenological applications, among other things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The particular questions raised by the papers in the mid 1990s have been largely answered so there's not too much room for completely new fundamental research; however, that doesn't mean that you can't discover something completely new over there. The F-theory bottom-up phenomenology remains full of open questions and it's surely an excellent area of research if someone can get familiar with everything he needs. This F-theory phenomenology has become one of the 5-6 major &quot;scenarios&quot; how string theory could be realized in the real world.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-27T06:01:46.327" CommentCount="2" />
  <row Id="7632" PostTypeId="2" ParentId="7610" CreationDate="2011-03-27T06:48:56.140" Score="5" ViewCount="" Body="&lt;p&gt;A neutron is a fermion, a hydrogen atom is a boson. This is related to the fact that a neutron decays into three fermions rather than two which is what you seem to think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A neutron is composed of three valence quarks, $u,d,d$, while a hydrogen atom is made out of $u,u,d,e^-$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The internal size of a neutron is about $10^{-14}$ meters while the internal size of the hydrogen atom is about $10^{-10}$ meters, about 10,000 times larger which gives the hydrogen a trillion times greater volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The excited (but still bound) states of a hydrogen atom have energies that differ by at most 13.6 electronvolts. On the other hand, the excitations of a neutron have to add many MeVs, a million times higher energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydrogen atom is a stable particle - neglecting the proton decay for a while. The lifetime of a neutron is just 15 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The neutron's behavior is studied by nuclear physics; the atoms' behavior is studied by atomic physics which is something very different, referring to different length scales and energy scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much more generally, it is a fallacy to imagine that if a particle X decays to YZ, then X is the &quot;same thing&quot; as YZ. One may say that it has the same conserved charges. But conserved charges don't fully define the identity of an object.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-27T06:48:56.140" />
  <row Id="7633" PostTypeId="1" AcceptedAnswerId="7637" CreationDate="2011-03-27T06:54:22.517" Score="0" ViewCount="221" Body="&lt;p&gt;What is the direction is the magnetic force vectors pointing from a coil of wire that has current running through it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ndt-ed.org/EducationResources/CommunityCollege/MagParticle/Graphics/coil1.gif&quot; rel=&quot;nofollow&quot;&gt;http://www.ndt-ed.org/EducationResources/CommunityCollege/MagParticle/Graphics/coil1.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above link is a picture of a wire with current running through it. I see the blue arrows indicating the magnetic field lines, but I am having trouble visualizing the magnetic force lines. Where are they pointing? Please help.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2011-03-27T08:11:15.277" Title="Direction of Magnetic force from a current running through a coil of wire" Tags="&lt;electromagnetism&gt;&lt;vectors&gt;&lt;electric-circuits&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="7634" PostTypeId="2" ParentId="7633" CreationDate="2011-03-27T07:01:24.213" Score="1" ViewCount="" Body="&lt;p&gt;The tangent to any magnetic field lines at a point gives the direction of the force at that point on an imaginary isolated north pole whose own magnetic field strength is negligible to change the original field. the direction of magnetic lines of force for a current carrying conductor can be known by an easy method. just imagine to grab a current carrying conductor with your right hand with the thumb pointing towards the direction of the flow of current. The alignment of the other fingers will point to the direction of the magnetic lines of force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Q8cSj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case if you would like to know the law which is called the Biot-savart law is given by&#xA;&lt;img src=&quot;http://i.stack.imgur.com/LYTUG.gif&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/4kDwy.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-03-27T07:28:47.780" LastActivityDate="2011-03-27T07:28:47.780" />
  <row Id="7635" PostTypeId="2" ParentId="7633" CreationDate="2011-03-27T07:06:26.537" Score="1" ViewCount="" Body="&lt;p&gt;the magnetic field is given by Ampere's law and the direction is determined by the right-hand rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Ua5WX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the current goes in the direction of the right thumb (convention for the direction of current is from plus to minus), the remaining fingers show the direction of $B$ around the wire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ampere_law&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Ampere_law&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a bar magnet, the magnetic moment is directed from South to North; it's also the direction in which the magnet will try to rotate in an external $B$ field of the same direction. See e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Magnet#Two_models_for_magnets:_magnetic_poles_and_atomic_currents&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Magnet#Two_models_for_magnets:_magnetic_poles_and_atomic_currents&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course, those rules - and why direction is chosen rather than the opposite one - depend on a couple of conventions that physicists ultimately adopted globally. Some of the &quot;signs&quot; in the conventions are independent, others are not.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-27T07:06:26.537" CommentCount="3" />
  <row Id="7636" PostTypeId="2" ParentId="7615" CreationDate="2011-03-27T07:24:46.417" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Short answer:&lt;/strong&gt;&#xA;The space between the nucleus and the electron is not empty space, it is filled with an electron cloud. (You will understand this answer better if you read the long answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long answer:&lt;/strong&gt;&#xA;Firstly, physics is a description of what we can observe. Depending on the scale of which you are describing, physicists, over the years, have different mathematical descriptions describing at different scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you ask about the space between the electron and the nucleus, confusion arises when trying to use a classical description to describe things on the atomic scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the atomic scale, physicists have found that quantum mechanics describes things very well on that scale. And when we use quantum mechanics to describe particles, electrons, protons, neutrons, etc, they are no longer thought of as point like particles whizzing around like the description that you are confused about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particle locations in quantum mechanics are not at an exact position, they are described by a probability density function. The previous answers other people gave before me are links to pictures of a probability density function of Hydrogen at different energy levels in 3D space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These probability density function's shapes are often known as &quot;electron clouds&quot;. They show where it's likely to find the electron. The darker regions are places of higher probability of finding the electron and the lighter regions are lower probability of finding the electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important thing to note is that the pictures show &quot;white space&quot;. Its actually a smooth picture with the probability getting smaller and smaller so fast that those locations look &quot;white&quot;. This means that the probability of finding the electron in those seemingly white spaces are extremely low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what does it mean to &quot;find and electron?&quot; This is another question you will have to ask. Its not related to this question.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2011-03-27T07:24:46.417" />
  <row Id="7637" PostTypeId="2" ParentId="7633" CreationDate="2011-03-27T08:11:15.277" Score="2" ViewCount="" Body="&lt;p&gt;Let me start from your comment on Lubos' answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If we have a electron near a coil of wire that has current running through it, certainly the electron will move a certain direction right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, it's not that simple. For a given coil of wire producing a given magnetic field, the electron can experience a force in &lt;em&gt;any&lt;/em&gt; direction that is perpendicular to the field. It depends on which way the electron is moving. (The force is always perpendicular to both the field and the electron's velocity.) In fact,  if the electron is just sitting at rest, or is moving parallel to the magnetic field, it experiences &lt;em&gt;no force at all&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be confused because you're thinking of the electrostatic force. That one is always parallel to the electric field; it doesn't matter how the particle is moving, and that's why you can draw electrostatic force lines. But that doesn't work with the magnetic force.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-03-27T08:11:15.277" CommentCount="4" />
  <row Id="7638" PostTypeId="1" CreationDate="2011-03-27T08:18:22.837" Score="0" ViewCount="172" Body="&lt;p&gt;How would I go about showing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{A}^{\dagger} + \hat{B}^{\dagger} = \left( \hat{A} + \hat{B} \right) ^{\dagger}$$&lt;/p&gt;&#xA;" OwnerUserId="2789" LastActivityDate="2011-03-27T10:10:20.223" Title="Proving Operator identities (Quantum Physics)" Tags="&lt;homework&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="7639" PostTypeId="1" CreationDate="2011-03-27T09:00:18.963" Score="1" ViewCount="173" Body="&lt;p&gt;In quantum physics we've defined:&#xA;$$ \psi (x) = \sqrt{ \dfrac{1}{2 \pi \hbar} } \int^{ \infty }_{-\infty } \phi (p) \exp \left( i \dfrac{px}{ \hbar} \right) dp $$&#xA;Now, &#xA;$$a(k) \equiv \sqrt{ \hbar } \phi (p)\quad  {\rm and}\quad  k = \dfrac{p}{ \hbar } $$&#xA;Where,&#xA;$$ a(k) = \left\{ \begin{array}{cccc} 0 &amp;amp; k &amp;lt; - \dfrac{ \epsilon }{2} \\ \sigma + \dfrac{2 \sigma }{ \epsilon } k &amp;amp; - \dfrac{ \epsilon }{2} &amp;lt; k &amp;lt; 0 \\ \sigma - \dfrac{2 \sigma }{ \epsilon } k &amp;amp; 0 &amp;lt; k &amp;lt; \dfrac{ \epsilon }{2} \\ 0 &amp;amp; k &amp;gt; \dfrac{ \epsilon }{2} \\ \end{array} \right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normalizing $a(k)$ I get $\sigma$ to be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sigma = \sqrt{ \dfrac{3}{ \epsilon } } $$&#xA;But I can't get anything reasonable from the Fourier Integral.&lt;/p&gt;&#xA;" OwnerUserId="2789" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-27T09:30:03.957" LastActivityDate="2011-04-26T10:41:41.557" Title="Finding $\psi(x)$ from Fourier modes" Tags="&lt;homework&gt;&lt;fourier-transform&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="7640" PostTypeId="2" ParentId="7639" CreationDate="2011-03-27T09:37:12.723" Score="1" ViewCount="" Body="&lt;p&gt;This isn't really physics, just math. Just perform straightforward piecewise integrations (I guess we can't really help you if you don't at least understand basics of integration).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you can observe that you can obtain your function (which is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Triangular_function&quot; rel=&quot;nofollow&quot;&gt;triangular&lt;/a&gt;, by the way, for obvious reasons) as a convolution of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rectangular_function&quot; rel=&quot;nofollow&quot;&gt;rectangular function&lt;/a&gt;. That means that its Fourier transform will be square of the Fourier transform of the rectangular function.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-03-27T09:37:12.723" />
  <row Id="7641" PostTypeId="2" ParentId="7639" CreationDate="2011-03-27T09:40:42.407" Score="1" ViewCount="" Body="&lt;p&gt;If I understand well, you're asking what is the Fourier transform of the &quot;triangle&quot; function that is piecewise linear, continuous, nonzero in the interval $\{-\epsilon/2,+\epsilon/2\}$, and having maximum $\sigma$ at $k=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, this is a purely mathematical question but the answer is &#xA;$$ \psi(x) = \frac{\sigma\epsilon}{\sqrt{8\pi\hbar}}  \, {\rm sinc}^2(\epsilon x/4\hbar) $$&#xA;where &#xA;$${\rm sinc}(y) = \frac{\sin(y)}{y}$$&#xA;and you may substitute your value of $\sigma$. Marek suggests deriving the result as the product of the Fourier transform of two &quot;rectangle&quot; functions because the &quot;triangle&quot; function is the convolution of two &quot;rectangle&quot; functions and the Fourier transform maps convolution to a product. That's why we simply obtained the square of the ${\rm sinc}$ function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integrals may be calculated without any convolutions, directly, using the integration by parts. The required techniques are explained e.g. by Jacob Barnett, a 12-year-old student of Purdue University in Indianopolis, here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=YFmrlIEpJOE&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=YFmrlIEpJOE&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And don't worry: Jake is here to help you with all your math phobias. The followup video by Jacob dealing with the trigonometric integrals may also be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to be sure: if the original poster wanted to implicitly claim that the particular value of $a(k)$ is &quot;canonical&quot; or &quot;special&quot; or &quot;more important than any other $L^2$-normalizable function $a(k)$&quot;, then he is wrong. It's just one function in the Hilbert space and there are infinitely many very different functions over there, too. The triangular function, in the $\epsilon\to 0$ limit, may be viewed as an approximation of the $\delta$-function, but even in this limited category, it is extremely far from being unique.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-27T10:21:25.007" LastActivityDate="2011-03-27T10:21:25.007" />
  <row Id="7642" PostTypeId="2" ParentId="7638" CreationDate="2011-03-27T10:00:35.183" Score="2" ViewCount="" Body="&lt;p&gt;$$\left\langle \chi | \left( A + B \right)^\dagger | \psi \right\rangle = \left\langle \psi | (A +B) | \chi \right\rangle^* =$$ &#xA;$$=\left\langle \psi | A | \chi \right\rangle^* + \left\langle \psi | B | \chi \right\rangle^* = \left\langle \chi | A^\dagger | \psi \right\rangle + \left\langle \chi | B^\dagger | \psi \right\rangle = \left\langle \chi | (A^\dagger +B^\dagger) | \psi \right\rangle$$&#xA;for all $|\chi\rangle$, $|\psi\rangle$.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-27T10:10:20.223" LastActivityDate="2011-03-27T10:10:20.223" CommentCount="4" />
  <row Id="7643" PostTypeId="1" CreationDate="2011-03-27T10:20:11.530" Score="-3" ViewCount="92" Body="&lt;p&gt;Are you free to choose your own research topics, or do you let your supervisor, grant agencies, hiring committees, fads, peer pressure, etc. tell you what to work on?&lt;/p&gt;&#xA;" OwnerUserId="2790" LastActivityDate="2011-03-27T11:35:08.560" ClosedDate="2011-03-27T13:12:30.463" Title="Do you let others determine your research topics?" Tags="&lt;physics-careers&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" />
  <row Id="7644" PostTypeId="1" CreationDate="2011-03-27T10:39:51.910" Score="2" ViewCount="252" Body="&lt;p&gt;Quick question for the nuclear engineers/physicists out there&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does I-134 come from? I cant find it in any of the charts of standard decay products of Uranium fission, but there is tons of the stuff in Fukushima reactor 2 building right now (2900 MBq/ml of water! Nasty!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;half life is 52 minutes, so either it is being made by fission (which would be bad news) or something from weeks ago is still decaying into it in large quantities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&lt;/p&gt;&#xA;" OwnerUserId="2600" LastActivityDate="2011-05-04T04:23:14.203" Title="Iodine isotope question" Tags="&lt;nuclear-physics&gt;&lt;radioactivity&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="7646" PostTypeId="2" ParentId="7615" CreationDate="2011-03-27T11:13:24.763" Score="1" ViewCount="" Body="&lt;p&gt;Maybe one should add to the analysis of @QEntanglement and the nice electron probability clouds in the illustrations in the other answers,  that also the space between the nucleus and the electrons is teaming with the exchange of virtual particles between the electrons and the nucleus, necessary to create the potential which determines the energy levels of the particles. These will mostly be virtual photons with loops and stuff.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-27T11:13:24.763" />
  <row Id="7647" PostTypeId="2" ParentId="7643" CreationDate="2011-03-27T11:35:08.560" Score="2" ViewCount="" Body="&lt;p&gt;You might look at the answers to &lt;a href=&quot;http://physics.stackexchange.com/questions/7583/where-do-you-get-your-good-physics-ideas-from/7590#7590&quot;&gt;a similar question&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you are new here, I will try to answer this question, from a particle physicist's perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short answer is &lt;em&gt;seniority&lt;/em&gt; determines the freedom of choosing research topics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A thesis is given by the adviser a professor or research scientist in some institute. So there is very little freedom there, except for the choice of professor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Post docs start acquiring seniority and may be given the freedom to choose a subject or a direction within a larger group. Future career opportunities play a large role here, particularly if one is aiming at a tenure track post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a tenure track post there is more freedom in initiating research and joining groups working on problems of interest to the individual. Still the objective of tenure will weigh in the decisions of what to pursue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once on tenure, freedom is the maximum possible within the constraints of financing, convincing others of the value of the research ( from thesis students to tenure track colleagues) . Also for large experiments, depending on being accepted as a member with one's group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This said, individuals with strong views and excellent abilities can move within this system and pursue/initiate their physics interests with success, if their main desire is  to do good physics and  guides the decisions.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-27T11:35:08.560" CommentCount="3" />
  <row Id="7648" PostTypeId="2" ParentId="7644" CreationDate="2011-03-27T11:40:19.667" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;First, I need to point out that the news about the &quot;10 million times higher radiation&quot; that spread across the world media today were just &lt;a href=&quot;http://www.kjct8.com/news/27334529/detail.html&quot; rel=&quot;nofollow&quot;&gt;an error&lt;/a&gt;. TEPCO revealed that the error came from a misinterpretation of the concentration of cobalt-56 as iodine-134. Because of this fixed mistake, it's completely plausible that iodine-134 was never detected over there, and there was just cobalt-56. However, let me continue with the bulk of the answer that was written before the mistakes in the reporting began to be appreciated.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chernobyl and similar accidents have been usually associated with iodine-131 (half-life of 8 days) as the primary source of thyroid cancer. However, iodine-134 (and other isotopes) is usually produced in greater quantities. Its immensely short lifetime makes it quickly irrelevant, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, when one studies &quot;acute&quot; (and therefore also short-lived) problems with high radiation, such as the &quot;10 million times increased radiation in the water&quot; today, the quickly decaying isotopes such as iodine-134 are just dominant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where do nuclei such as iodine-134 come from? Well, there is fission going on in the reactor, so fissiles such as uranium-235 are broken into pretty much arbitrarily smaller pieces and iodine-134 is often among them. See&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_fission&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Nuclear_fission&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;for the typical products of fission - the mass is often divided in the 3:2 ratio etc. so isotopes such as iodine-134 are common. Yes, iodine-134 is a direct product of fission but most likely not ordinary neutron-driven fission but probably only photofission (fission initiated by gamma rays hitting the large nucleus), see these papers for details:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1982PhRvC..25.1546T&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/1982PhRvC..25.1546T&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;http://ieeexplore.ieee.org/Xplore/login.jsp?url=http%3A%2F%2Fieeexplore.ieee.org%2Fiel5%2F4816218%2F4823592%2F04823601.pdf%3Ftp%3D%26arnumber%3D4823601%26punumber%3D4816218&amp;amp;authDecision=-203&quot; rel=&quot;nofollow&quot;&gt;http://ieeexplore.ieee.org/Xplore/login.jsp?url=http%3A%2F%2Fieeexplore.ieee.org%2Fiel5%2F4816218%2F4823592%2F04823601.pdf%3Ftp%3D%26arnumber%3D4823601%26punumber%3D4816218&amp;amp;authDecision=-203&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the uranium-235 splits into a $Z=53$ iodine isotope and one more, this &quot;one more&quot; has to be yttrium with $Z=39$. The half-life of yttrium-99 and nearby isotopes you may get (after emitting a few neutrons) is comparable to a second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is exactly iodine-134 so much more represented among the fission products than other iodine isotopes? Well, because it has the rigth proton-neutron ratio. The fissile, uranium-235, has $Z/A$ equal to $92/235$. And because iodine's $Z=53$, the same ratio of protons and neutrons - a democratic fission - is obtained for the total nuclear mass&#xA;$$ 235 / 92 \times 53 = 135.4.$$&#xA;So it's most likely to produce iodine isotopes whose $A$ is close to 135. Iodine-135 has 6.6 hours of half-life, so it's not that important &quot;acutely&quot;, while iodine-136 has 83 seconds and disappears too quickly. It just happens that the contamination of water, as was measured, is dominated by half-lives close to 1 hour, and iodine-134 is the key representative in this realm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other elements, different from iodine, have typically much longer lifetimes for the same $Z/A$ ratio. Still, it's plausible that much of iodine occurs from decay of other fission products, e.g. from beta-decay of tellurium-134 (see the other answer): fission directly prefers to produce even $Z$ elements while iodine has $Z=53$, odd. But this even-$Z$ rule is not &quot;strict&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, fission is a messy process and it can produce &quot;almost anything&quot; in the list of isotopes. Again, just to be sure, fission is not about getting incremental alpha and beta decays only; it's the process of splitting the large nucleus into two comparable pieces.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-27T17:15:47.647" LastActivityDate="2011-03-27T17:15:47.647" CommentCount="3" />
  <row Id="7649" PostTypeId="2" ParentId="6768" CreationDate="2011-03-27T11:57:52.207" Score="-2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Even with rods inserted, the reactor&#xA;  continues to produce heat equivalent&#xA;  to about 3% of its full power level.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That seems an extraordinary engineering design oversight - even negligence, given that the control rods are the primary security for the nuclear reaction.&#xA;Just the most basic precautionary principle would lead to install 10 times more control rods than needed to close the reaction down completely; and then even better to make ten separate sets of rods, each set differently powered.&lt;/p&gt;&#xA;" OwnerUserId="2792" LastActivityDate="2011-03-27T11:57:52.207" CommentCount="1" />
  <row Id="7650" PostTypeId="2" ParentId="7585" CreationDate="2011-03-27T14:08:46.097" Score="2" ViewCount="" Body="&lt;p&gt;I want to add to Lubos answer and give some old and new references on the classical representation of spinning particles. In addition, I'll try to explain, the main idea of the construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classical description of spin by means of Grassmann variables was introduced by F.A. Berezin and M.S. Marinov (for the nonrelativistic case) in: F. A. Berezin and M. S. Marinov, Classical spin and Grassmann algebra, Sov. Phys. JETP Lett. 21 (1975) 320-321. (This &lt;a href=&quot;http://physics.technion.ac.il/~site/wisw/memoriam/marinov/pub/Mar35.pdf&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; can be found in Prof. Marinov memorial page).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later, this work was extended to the relativistic case in: F.A.Berezin and M. S. Marinov, Particle spin dynamics as the Grassmann variant of classical mechanics, Ann. Phys. (NY) 104 (1977) 336-362. (Here is the version of this &lt;a href=&quot;http://physics.technion.ac.il/~site/wisw/memoriam/marinov/pub/Mar40.pdf&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relativistic Lagrangian is given in this paper is in the arc length (square root) form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll try to briefly explain the logic behind their construction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The translational phase space is parameterized by the position $q_i$ and momentum $p_j$ coordinates which are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Commuting as functions on the phase space $q_jq_i=q_iq_j,  p_jq_i=q_ip_j,  p_jp_i=p_ip_j$&lt;/li&gt;&#xA;&lt;li&gt;Satisfying the canonical commutation relation: $ \{q_i,q_j\} =0, \{p_i,p_j\} =0, \{q_i,p_j\} = \delta_{ij}$&lt;/li&gt;&#xA;&lt;li&gt;Their quantization operators satisfy, (the Weyl algebra) commutaion relation $[\hat{q}_i,\hat{p}_j] = i \hbar \hat{1} \delta_{ij}$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In analogy, the spin phase space is parameterized by the Grassmann variables $\xi_i$, which are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Anti-commuting as functions on the phase space $\xi_j\xi_i=-\xi_i\xi_j$&lt;/li&gt;&#xA;&lt;li&gt;Satisfying the commutation relation: $ \{\xi_i,\xi_j\} = \frac{1}{2i}\delta_{ij}$&lt;/li&gt;&#xA;&lt;li&gt;Their quantization operators satisfy, (the clifford algebra) commutaion relation $[\hat{\xi}_i,\hat{\xi}_j] = \frac{\hbar}{2} \delta_{ij}$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thus, the main idea is that as the Weyl algebra is the quantization of the classical translational phase space, the Clifford algebra is the quantization of the classical spin phase space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As known from the theory of Clifford algebras, the spin generators are the bivectors in the basic &#xA;Clifford generators, for example in the three dimensional case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hat{S}_i = i \epsilon_{ijk} \hat{\xi}_j\hat{\xi}_k$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the infinite dimensional case, one obtains, the usual the canonical anti-commutation algebra (CAR).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A review &lt;a href=&quot;http://arxiv.org/PS_cache/hep-th/pdf/9601/9601020v1.pdf&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; by Andrzej Frydryszak of the Berezin-marinov model can be found in: LAGRANGIAN MODELS OF PARTICLES WITH SPIN: THE FIRST SEVENTY YEARS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be mentioned that the Berezin-Marinov model describes only spin $\frac{1}{2}$ particles. &#xA;(This property is desirable from a quantization theory, because it results an &#xA;irreducible representation of the operator algebra). In order to describe photons, one should add additional noncommuting coordinates to the phase space, See for example the following solution by : &lt;a href=&quot;http://fangdian.net/PS_cache/hep-th/pdf/9401/9401132v1.pdf&quot; rel=&quot;nofollow&quot;&gt;Gitman and Goncalves&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the Berezin-Marinov construction is a prototype geometric quantization model of &#xA;a phase space which is symplectic superspace, namely the supercotangent bundle of the bosonic phase space. &#xA;The following &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1004/1004.1595v1.pdf&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;: Conformal geometry of the supercotangent and spinor bundles:&#xA; by J.P. Michel describes the model from this point of view and its generalization to &#xA;the non-flat case, where the configuration space is a Riemannian manifold. Here, the quantization of the supercotangent bundle&#xA;of a spin manifold is its spinor bundle.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastActivityDate="2011-03-27T14:08:46.097" CommentCount="1" />
  <row Id="7651" PostTypeId="1" AcceptedAnswerId="7656" CreationDate="2011-03-27T14:19:25.283" Score="1" ViewCount="168" Body="&lt;p&gt;I am struggling with a problem from classical mechanics. Imagine a massless wheel (to make it simpler) with a mass $m$ fixed to it rolling without slipping on a horizontal ground. If we now try to find the equations of motion of the wheel (for instance the angle $\alpha$ it turns) we will find that all the forces are independent of velocity, so $\alpha &amp;#39;&amp;#39;=f(\alpha)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After doing that I decided to solve this problem using Euler-Langrange equations (since friction does no work). I came up with $L=\frac{1}{2} m R^{2} (\frac{d\alpha}{dt})^{2} (1+\cos \alpha)-mgR(1+\cos \alpha)$ which, upon solving, gives $\alpha &amp;#39;&amp;#39;$ as a function of both $\alpha$ and $\alpha &amp;#39;$. What is my problem?&lt;/p&gt;&#xA;" OwnerUserId="2752" LastEditorUserId="2752" LastEditorDisplayName="" LastEditDate="2011-03-27T16:34:06.400" LastActivityDate="2011-03-28T18:13:40.470" Title="Mass fixed to a wheel - mechanics problem" Tags="&lt;classical-mechanics&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="7652" PostTypeId="1" CreationDate="2011-03-27T14:56:09.370" Score="6" ViewCount="512" Body="&lt;p&gt;What practical application can we expect from particle physics a century or two from now? What use can we make of quark-gluon plasmas or strange quarks? How can we harness W- and Z-bosons or the Higgs boson? Nuclear physics has given us nuclear plants and the promise of fusion power in the near future. What about particle physics? If we extend our timeframe, what promise does string theory give us? Can we make use of black holes?&lt;/p&gt;&#xA;" OwnerUserId="2794" LastActivityDate="2011-08-28T06:44:09.400" Title="What future technologies does particle physics and string theory promise?" Tags="&lt;string-theory&gt;&lt;particle-physics&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="0" />
  <row Id="7653" PostTypeId="2" ParentId="7652" CreationDate="2011-03-27T15:24:40.977" Score="10" ViewCount="" Body="&lt;p&gt;Quantum Chromodynamics, the electroweak theory, or general theory of relativity - or quantum gravity and string theory - are not methods to obtain new devices; they're theories meant to understand the truth about the Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find it unlikely that any of those things will become practically useful. It may still hypothetically happen, but if it will, no one can predict how this could happen at present - but even if those things happen to have practical applications sometime in the future, that's not why they've been and why they are being studied.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-27T15:24:40.977" CommentCount="7" />
  <row Id="7654" PostTypeId="1" AcceptedAnswerId="7721" CreationDate="2011-03-27T15:52:58.387" Score="8" ViewCount="270" Body="&lt;p&gt;This question &lt;a href=&quot;http://physics.stackexchange.com/questions/7421/brachistochrone-problem-for-inhomogeneous-potential&quot;&gt;Brachistochrone Problem for Inhomogeneous Potential&lt;/a&gt; has the obvious extension. Namely the same question, when gravity is treated according to general relativity. To make it specific let's consider the case of Schwarzschild metric. Given to points, outside the event horizon, what is the trajectory (or the worldline) of a test body, which goes from the first point to the second for least time. All considered from the point of view of a distant observer at rest (with respect to the star/back hole). Of course other metrics or the general case are also interesting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't find much by searching on the net. There is one article with the tittle suggesting relevance (actually more but this one seems very relevant), but it is over 60 pages of technicalities. So it would be nice to see a shorter answer.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-27T16:27:17.747" LastActivityDate="2011-08-16T03:23:22.777" Title="Brachistochrone problem in general relativity" Tags="&lt;general-relativity&gt;&lt;brachistochrone-problem&gt;" AnswerCount="2" CommentCount="10" FavoriteCount="1" />
  <row Id="7655" PostTypeId="2" ParentId="7652" CreationDate="2011-03-27T16:37:17.260" Score="5" ViewCount="" Body="&lt;p&gt;The experimental testing of these theories, and further assuming we might be able to test string theory sometime in the next decades, will result in new methods of detection, new techniques in data management and filtering signal from noise, and push other developments.  These may then be transferable to other applications.  The Apollo lunar program had a similar effect, though we brought nothing back from the moon which had any great utility.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-27T16:37:17.260" />
  <row Id="7656" PostTypeId="2" ParentId="7651" CreationDate="2011-03-27T17:18:04.417" Score="2" ViewCount="" Body="&lt;p&gt;(Updated to reflect my better understanding of the problem, based on our discussion in the comments.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my comments above, I didn't understand the question. Now that I do, it seems to me that your Lagrangian is correct, except that the factor $1\over 2$ in the kinetic energy term doesn't belong. The equation of motion does involve $\dot\alpha$. I don't think I understand why that's a problem. I can't think of a general argument that proves that there shouldn't be any first derivatives in the Euler-Lagrange equations in a system like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, consider a force-free particle moving in two dimensions. If we express things in polar coordinates, the Lagrangian is&#xA;$$&#xD;&#xA;L={1\over 2}m(\dot r^2+r^2\dot\phi^2).&#xD;&#xA;$$&#xA;The Euler-Lagrange equation for $\phi$ reduces to &#xA;$$&#xD;&#xA;2\dot r\dot\phi+r\ddot\phi=0.&#xD;&#xA;$$&#xA;This involves first derivatives, even though there's no force at all (hence &lt;em&gt;a fortiori&lt;/em&gt; no velocity-dependent force).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In subsequent comments (below) you explained your reason for not expecting there to be $\dot\alpha$ terms: calculating the torque about the instantaneous point of contact and setting that equal to $I\ddot\alpha$ yields an equation with no $\dot\alpha$'s. The reason that doesn't work is that the torque is equal to $d{\bf l}/dt=d(I\dot\alpha)/dt$. Since $I$ is a function of time, this is not equal to $I\ddot\alpha$. There's a bit more detail in the comments below. Anyway, the final equation of motion, derived either from the Euler-Lagrange equation or from torque considerations, comes out to&#xA;$$&#xD;&#xA;2(1+\cos\alpha)\ddot\alpha=(\dot\theta^2+g/R)\sin\alpha.&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-03-28T18:13:40.470" LastActivityDate="2011-03-28T18:13:40.470" CommentCount="5" />
  <row Id="7657" PostTypeId="1" AcceptedAnswerId="7658" CreationDate="2011-03-27T17:51:57.840" Score="5" ViewCount="311" Body="&lt;p&gt;I'm a computer programmer that never studied physics in school and now it's coming back to bite me a bit in some of the stuff I'm being asked to program.  I'm trying to self study some physics and I've got a few open source intro physics books and understanding it for the most part but I'm a bit confused on this statement I've stumbled upon in a section about static friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's already gone over the formula for static friction and so forth.  It gets into a section explaining that the weight of a train increases static friction between the wheels and the tracks.  Alright, makes sense.  But then it says this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;The reason locomotives are built to be so heavy is for traction.&#xA;The upward normal force of the rails on the wheels, FN, cancels&#xA;the downward force of gravity, FW, so ignoring plus and minus&#xA;signs, these two forces are equal in absolute value, FN = FW.&#xA;Given this amount of normal force, the maximum force of static&#xA;friction is Fs = sFN = sFW. This static frictional force, of the&#xA;rails pushing forward on the wheels, is the only force that can&#xA;accelerate the train, pull it uphill, or cancel out the force of air&#xA;resistance while cruising at constant speed.  The coefficient of&#xA;static friction for steel on steel is about 1/4, so no locomotive can&#xA;pull with a force greater than about 1/4 of its own weight. If the&#xA;engine is capable of supplying more than that amount of force, the&#xA;result will be simply to break static friction and spin the wheels.&#xA;&lt;br /&gt;&lt;br /&gt;&#xA;- &quot;Newtonian Physics&quot;, Light and Matter - Book 1, p158 B. Crowell&#xA;&lt;a href=&quot;http://www.lightandmatter.com/bk1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.lightandmatter.com/bk1.pdf&lt;/a&gt;&#xA;&lt;/blockquote&gt;  &#xA;&#xA;&lt;p&gt;I'm confused as to how static friction is the only thing that can move the train forward.  I thought static friction was what kept it in place in the first place.  There's another force - that I can't think of the name of, but I've heard of somewhere - that I thought was more what they're describing here, where the weight of the wheels pushing down and forward slightly on the tracks causes the tracks to push up and forward (from the opposite side).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain to me what this is saying?&lt;/p&gt;&#xA;" OwnerUserId="2798" LastActivityDate="2011-03-28T11:17:34.200" Title="Static Friction - Only thing that can accelerate a train?" Tags="&lt;newtonian-mechanics&gt;&lt;friction&gt;&lt;motion&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="7658" PostTypeId="2" ParentId="7657" CreationDate="2011-03-27T17:59:05.233" Score="5" ViewCount="" Body="&lt;p&gt;This is all a complicated (and confusing, or just plain confused) way to say that, if you want the locomotive to pull the train, you don't want its wheels to slip.  It's friction that prevents the wheels from slipping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest you simply delete this sentence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This static frictional force, of the rails pushing forward on the wheels, is the only force that can accelerate the train, pull it uphill, or cancel out the force of air resistance while cruising at constant speed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The paragraph makes a lot more sense without it.  The author is trying to get at Newton's third law (equal and opposite reaction) but this way of putting it provides more confusion than insight. &lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-03-27T21:44:29.163" LastActivityDate="2011-03-27T21:44:29.163" CommentCount="5" />
  <row Id="7659" PostTypeId="2" ParentId="7657" CreationDate="2011-03-27T18:00:05.287" Score="1" ViewCount="" Body="&lt;p&gt;The thing you need to think about here is that the wheels &lt;em&gt;roll&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there was no friction between the wheels and the track, firing up the locomotive would just cause its drive wheel to spin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Friction acts to prevent or resist &lt;em&gt;relative motion&lt;/em&gt; between the two surfaces. So, if there is a torque on the wheels and the point of contact can not move relative the rail (just where it touches) because of static friction, the only way the wheel can turn is if the train moves relative the wheel.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-03-27T18:00:05.287" CommentCount="6" />
  <row Id="7660" PostTypeId="2" ParentId="7657" CreationDate="2011-03-27T18:32:32.930" Score="3" ViewCount="" Body="&lt;p&gt;There is a question of what you are trying to learn about friction from this convoluted example. What may be slightly confusing is that there are two types of friction &lt;em&gt;static friction&lt;/em&gt; and &lt;em&gt;kinetic friction&lt;/em&gt;. Kinetic friction is the friction associated with two substances sliding against one another, which can only happen when one is moving relative to another. Kinetic friction is therefore associated with motion. Static friction is the friction associated with keeping an object stationary. However when one considers rolling wheel motion the rolling wheel friction is static friction, despite the motion of the wheel, because the wheel never actually slips on the track (or at least it is not meant to).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force name your are looking for in the last paragraph is called the Normal Force. Look at the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Friction&quot; rel=&quot;nofollow&quot;&gt;Friction&lt;/a&gt; if that helps as it gives more equations and diagrams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT (after a comment below) From the original question and comments we can see that an additional component of this question was to have a better understanding of a more complex case of elastic deformation. To understand this consider a massive object on a springy mattress: then there will be a U or V shaped deformation in the material. Modelling this would require some mathematics and &lt;a href=&quot;http://en.wikipedia.org/wiki/Hooke%27s_law&quot; rel=&quot;nofollow&quot;&gt;Hooke's Law&lt;/a&gt; (F=-kx) would seem the most appropriate as an approximation. Now if this object were moving forward at some velocity v then the corresponding deformation will travel at this velocity too. It would also be necessary to model the surface tension of the material (which could rip if the strain was too large). Together all this will cause leading and lagging deformations (and possibly oscillations) in the material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this is for a computer model I am not clear whether the physics of this material oscillation has to be exactly correct according to some specific parameters and equations, or whether it is just an effect for demonstration on some computer game. In either case some further modelling is required to determine the equation for such motion, based on the principles above. If we are dealing with motion of cars on roads or trains on tracks then this extra modelling should be unnecessary and such elastic deformation can be ignored.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-03-27T21:29:57.757" LastActivityDate="2011-03-27T21:29:57.757" CommentCount="6" />
  <row Id="7661" PostTypeId="2" ParentId="2347" CreationDate="2011-03-27T19:22:01.173" Score="2" ViewCount="" Body="&lt;p&gt;I'm going to have to throw out a vote for &lt;a href=&quot;http://en.wikipedia.org/wiki/X-Plane_%28simulator%29&quot; rel=&quot;nofollow&quot;&gt;X-Plane&lt;/a&gt;, approved by the FAA for flight simulation.&lt;/p&gt;&#xA;" OwnerUserId="2785" LastActivityDate="2011-03-27T19:22:01.173" CommunityOwnedDate="2011-03-27T19:22:01.173" />
  <row Id="7662" PostTypeId="2" ParentId="7652" CreationDate="2011-03-27T20:01:15.860" Score="7" ViewCount="" Body="&lt;p&gt;Allow me to answer your question with some quotes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;The energy produced by the atom is a very poor kind of thing. Anyone who expects a source of power from the transformation of these atoms is talking moonshine.&quot; &amp;mdash;Ernst Rutherford&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;There is not the slightest indication that nuclear energy will ever be obtainable. It would mean the atom would have to be shattered at will.&quot; &amp;mdash;Albert Einstein&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;There is no likelihood that man can ever tap the power of the atom. The glib supposition of utilizing atomic energy when our coal has run out is a completely unscientific Utopian dream, a childish bug-a-boo.&quot; &amp;mdash;Robert Millikan&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Radio has no future.&quot; &amp;mdash;Lord Kelvin&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;The more important fundamental laws and facts of physical science have all been discovered, and these are now so firmly established that the possibility of their ever being supplanted in consequence of new discoveries is exceedingly remote.... Our future discoveries must be looked for in the sixth place of decimals.&quot; &amp;mdash;Albert. A. Michelson, &lt;strong&gt;1894&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;There is nothing new to be discovered in physics now. All that remains is more and more precise measurement.&quot; &amp;mdash;Lord Kelvin, &lt;strong&gt;1900&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also, I couldn't find a good quote for this one, but it was widely believed that number theory was an abstract mathematical discipline with no practical application. Now it is the basis of all modern cryptography.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically the idea I'm trying to get across is that it's &lt;strong&gt;impossible to predict&lt;/strong&gt; the future applications of basic research.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-03-27T20:01:15.860" CommentCount="3" />
  <row Id="7663" PostTypeId="1" CreationDate="2011-03-27T20:44:41.870" Score="4" ViewCount="319" Body="&lt;p&gt;Or to put the question another way - what is the result of a proton-positron collision, or an up quark-charm antiquark collision, etc.? As far as I know, annihilation happens only between particles of opposite charge and same mass, but perhaps I am wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if the types of annihilation mentioned above cannot happend, what are the reasons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2800" LastActivityDate="2011-04-29T16:04:02.937" Title="Does a particle annihilate only with its antiparticle? If yes, why?" Tags="&lt;particle-physics&gt;&lt;antimatter&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="7664" PostTypeId="1" CreationDate="2011-03-27T20:51:20.287" Score="1" ViewCount="151" Body="&lt;p&gt;I'm trying to determine the electric field at a point P (located on the +Z axis) due to a square of side length [L] and centered at the x-y plane origin. The square has a constant surface density [s]. I'm thinking that I should go about splitting up the square into four smaller squares (one in each quadrant) and then calculate the field from each on the point P. Is this the correct way of solving this type of problem, or should I be splitting the big square up into infinitesimal strips and calculating the field that way?&lt;/p&gt;&#xA;" OwnerUserId="2801" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-03-28T09:44:03.203" LastActivityDate="2011-03-28T09:44:03.203" Title="Electric field at a point from a square surface" Tags="&lt;homework&gt;&lt;vector-fields&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="7665" PostTypeId="2" ParentId="7663" CreationDate="2011-03-27T21:03:19.077" Score="4" ViewCount="" Body="&lt;p&gt;Some &quot;Numbers&quot; have to be conserved. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Charge&lt;/li&gt;&#xA;&lt;li&gt;Baryon Number &lt;/li&gt;&#xA;&lt;li&gt;Lepton number&lt;/li&gt;&#xA;&lt;li&gt;More numbers alike (see dmckee's comment)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These Numbers are $+1$ or $-1$ for most particles, and when two leptons annihilate the lepton number in summation is zero before and after. If you &quot;mix&quot;, this is not the case. This explains not much, of course, that decribes just the way such particles behave.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="103" LastEditorDisplayName="" LastEditDate="2011-04-29T16:04:02.937" LastActivityDate="2011-04-29T16:04:02.937" CommentCount="1" />
  <row Id="7666" PostTypeId="2" ParentId="7664" CreationDate="2011-03-27T21:10:05.637" Score="1" ViewCount="" Body="&lt;p&gt;I don't see any advantage of cutting up the plate into four (you are basically asking the same question again: the field of an individual plate). The other way you suggest would work. Cut the square plate into tiny 1-D strips and integrate over.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-27T21:10:05.637" CommentCount="4" />
  <row Id="7668" PostTypeId="1" AcceptedAnswerId="7673" CreationDate="2011-03-27T21:21:57.577" Score="7" ViewCount="358" Body="&lt;p&gt;Let me admit beforehand that this is quite possibly a very stupid question. I was also uncertain of where to post this question, as it doesn't fit cleanly into either physics or math stackexchange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In dimensional analysis, it does not make sense to, for instance, add together two numbers with different units together. Nor does it make sense to exponentiate two numbers with different units (or for that matter, with units at all) together; these expressions make no sense:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(5  m)^{7 s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(14  A)^{3 A}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is plainly this: &lt;em&gt;why&lt;/em&gt; do they not make sense? Why does only multiplying together numbers with units make sense, and not, for instance, exponentiating them together? I understand that raising a number with a unit to the power of another number with a unit is quite unintuitive - however, that's not really a good reason, is it?&lt;/p&gt;&#xA;" OwnerUserId="2802" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-28T01:06:35.887" LastActivityDate="2011-03-28T01:06:35.887" Title="Fundamental question about dimensional analysis" Tags="&lt;units&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="7669" PostTypeId="2" ParentId="7668" CreationDate="2011-03-27T21:28:14.437" Score="8" ViewCount="" Body="&lt;p&gt;Because of the way an exponential is defined. By an expression like $a^b$ we mean to say that the quantity $a$ is multiplied $b$ times with itself. Thus an expression like $(5m)^{7s}$ would mean $5m$ multiplied &quot;7 seconds&quot; times with itself, which is meaningless.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-27T21:28:14.437" CommentCount="1" />
  <row Id="7670" PostTypeId="2" ParentId="7663" CreationDate="2011-03-27T21:31:26.963" Score="5" ViewCount="" Body="&lt;p&gt;It depends on your definition of annihilation. But microscopically all processes are described by Feynman diagrams such as these&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/76Ejv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of which last one describes electron positron annihilation (if it weren't for the typo in the out-going photon). But as you can see it's all a simple matter of how you turn your head around and the very same diagram represents emission (or absorption) of photon by electron (or positron). Does electron annihilate with photon and create a brand new electron? You can certainly interpret it that way. In other words, it's just a question of terminology and interpretation. Actual physics doesn't depend on how you call the process. It is encoded in the underlying math of quantum field theory (QFT).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the punchline is that annihilation (in the strict sense of particle-antiparticle inelastic collision) doesn't have a special place in one's vocabulary once they learn their QFT and particle physics. It's just one particular kind of interaction. So you might as well ask which arbitrary interactions are allowed. And answer to that is: there's quite a lot of them and they are described by Standard Model. But the basic picture is that particles can be charged under certain charges: either the familiar electromagnetic, or less familiar weak and strong charges. Or in more modern language whether some families of particles form a multiplet under some gauge group. For weak force with group SU(2) you get lepton (e.g. electron-neutrino) and quark (e.g. up-down) doublets. For strong force with group SU(3) you need triplets and these are precisely the red green blue colors of quarks that you probably heard about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, for every multiplet there is a diagram like the ones above where you have two charged particles and one mediating particle between them (photon, weak bosons or gluons). Besides this you can also get funnier diagrams with e.g. three or four gluon lines. But that's it, these are all of allowed interactions of Standard Model.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-03-27T21:53:34.893" LastActivityDate="2011-03-27T21:53:34.893" CommentCount="1" />
  <row Id="7671" PostTypeId="1" AcceptedAnswerId="7739" CreationDate="2011-03-27T21:40:19.200" Score="5" ViewCount="241" Body="&lt;p&gt;To take the simplest example, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Phi(\vec{r}) = \begin{cases} V &amp;amp;  r\leq a \\ 0 &amp;amp;  r\geq a \end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a plane conducting surface that extends all the way. Choosing an appropriate $G(\vec{r},\vec{r&amp;#39;})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Phi(\vec{r}) =\frac{1}{4\pi\epsilon_0} \int_V G(\vec{r},\vec{r&amp;#39;}) \rho(\vec{r&amp;#39;})d^3 r&amp;#39; -\frac{1}{4\pi}\oint_S \Phi(\vec{r&amp;#39;}) \frac{\partial G(\vec{r},\vec{r&amp;#39;})}{\partial n&amp;#39;} da&amp;#39;  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I can find all the stuff by making a rigged guess of a Green's function (from having solved the simpler 0 potential problem). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is really elementary though. The $\rho(\vec{r&amp;#39;})$ that appears in the problems only takes into account the initial charges present but not any induced charges on the plane. I cannot understand that convincingly. There are some issues floating in my mind:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Induced charges appear as an effect of the inital charge distributions present, and are not part of the initial value problem. However, this suggests there is some kind of time evolution. Wrong reasoning, move on. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Induced charges are an unknown for some known  $\rho(\vec{r&amp;#39;})$. Our effort is to investigate the effect on the system. So we have the time when the effect due to the  $\rho(\vec{r&amp;#39;})$ is not present, that is the instant we assume to solve the problem. Whatever we get however appears to apply to later time when effects have distributed themselves and accounts for the induced charges present.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All this sounds fudgy when I use the Green's function method to solve BVPs. The more elementary usage of method of images seemed highly intuitive and a natural consequences of the uniqueness theorem. We take an image charge, and that &lt;em&gt;accounts&lt;/em&gt; for all induced charges. Can someone tell me how to reason in the correct way and why the two are equivalent. How do I intuitively think of the Green's function so that I naturally arrive at coherent conclusions like in the simplistic approach.  &lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="2599" LastEditorDisplayName="" LastEditDate="2011-03-27T22:00:36.110" LastActivityDate="2011-04-28T09:42:06.523" Title="In Method of images type problems, why isn't the induced charge taken into account while solving it via Green's functions?" Tags="&lt;electrostatics&gt;" AnswerCount="1" />
  <row Id="7672" PostTypeId="1" CreationDate="2011-03-27T21:44:50.803" Score="2" ViewCount="120" Body="&lt;p&gt;I need to create an online service displaying latest Kp index.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where I can take the data? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data should be in machine-readable format, i.e. text files, XML, or CGI gateways, for instance. No graphical plots!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I found this:&#xA;&lt;a href=&quot;http://www.swpc.noaa.gov/wingkp/wingkp_list.txt&quot; rel=&quot;nofollow&quot;&gt;http://www.swpc.noaa.gov/wingkp/wingkp_list.txt&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is Est. Kp what I need? I compared the data with plots covering several last days and found that kp values in this file are lower that on the charts.&lt;/p&gt;&#xA;" OwnerUserId="2803" LastActivityDate="2011-03-27T21:44:50.803" Title="live Kp index data" Tags="&lt;geophysics&gt;&lt;sun&gt;&lt;computation&gt;" CommentCount="3" />
  <row Id="7673" PostTypeId="2" ParentId="7668" CreationDate="2011-03-27T21:52:40.417" Score="16" ViewCount="" Body="&lt;p&gt;Another way to look at it &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ e^x = \sum_n \frac{x^{n}}{n!} = 1 + x +\frac{x^2}{2} + \dots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which comes down to adding quantities with different dimension, which you have already accepted makes no sense. This is why you can't exponentiate values with units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we can do a similar thing with most transcendental functions.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-03-27T22:18:59.047" LastActivityDate="2011-03-27T22:18:59.047" CommentCount="2" />
  <row Id="7674" PostTypeId="1" CreationDate="2011-03-27T22:02:55.687" Score="5" ViewCount="117" Body="&lt;p&gt;I'd like to create a very rough animation of a wave crashing on a beach. I'm guessing it would have to be a particle simulator, where you code in the forces between the particles and then integrate forward in time. I've done similar things, like simulations of charged particles, but there the forces are pretty straightforward, whereas here I guess I'd have to account for 1) tides 2) gravity 3) water surface tension. These seem like widely different forces acting on different scales. I don't even know where to begin. Any hints or links to papers related to this topic? &lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-03-27T22:29:09.220" Title="How to simulate a crashing wave?" Tags="&lt;waves&gt;&lt;water&gt;&lt;simulation&gt;&lt;computation&gt;&lt;tidal-effect&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7675" PostTypeId="2" ParentId="7668" CreationDate="2011-03-27T22:24:53.577" Score="6" ViewCount="" Body="&lt;p&gt;One further point to note, is that strictly one is just saying that the exponent is dimensionless, not that it does not contain expressions with dimension. So for example we could have some expression like $X=a^{(E/E_0)}$ where the exponent for a is a ratio of energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several restrictions on the space (sometimes viewed as a vector space) of dimensional quantities: for example units are raised to rational, but not irrational values. This allows a theorem: &lt;a href=&quot;http://en.wikipedia.org/wiki/Buckingham_%CF%80_theorem&quot; rel=&quot;nofollow&quot;&gt;The Buckingham $\Pi$ Theorem&lt;/a&gt; to form.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-03-27T22:24:53.577" />
  <row Id="7676" PostTypeId="2" ParentId="7674" CreationDate="2011-03-27T22:29:09.220" Score="5" ViewCount="" Body="&lt;p&gt;This is, no doubt, one of the biggest challenges for realistic simulations: waves crashing, hair moving under wind and whatever other movement involving turbulence will be hard to solve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though it is true that one can solve the equations of motion for each individual particle in a 'molecular dynamics' fashion, that is just infeasible for a system that goes beyond a few million particles. What can be done instead is to solve the equations of motion for the fluid by coarse graining the whole liquid as a homogeneous material (solving Navier-Stokes equations) but, again, if the simulation involves turbulence the equations become very time consuming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big question is how accurate you need your results to be: if the idea is to make it look pretty and somewhat real there are a couple of models - mainly for graphical processors (gpu) - that can be used to simulate waves. There is a really nice (and recent) overview on the subject at Science called &quot;&lt;a href=&quot;http://www.sciencemag.org/content/330/6012/1756.full&quot; rel=&quot;nofollow&quot;&gt;Computation Physics in Film&lt;/a&gt;&quot;. I think it will answer most of your remaining questions.&lt;/p&gt;&#xA;" OwnerUserId="2765" LastActivityDate="2011-03-27T22:29:09.220" />
  <row Id="7677" PostTypeId="2" ParentId="2347" CreationDate="2011-03-27T22:41:41.453" Score="1" ViewCount="" Body="&lt;p&gt;Half-Life 2 has what appears to be nice physics, which are also involved in puzzles, where you, for example, have to move floating barrels under a broken bridge so it rises and supports your vehicle.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-03-27T22:41:41.453" CommunityOwnedDate="2011-03-27T22:41:41.453" />
  <row Id="7679" PostTypeId="1" AcceptedAnswerId="7697" CreationDate="2011-03-27T22:54:51.337" Score="7" ViewCount="304" Body="&lt;p&gt;Most textbooks on basic quantum mechanics tell you that when your initial Hamiltonian $H_0$ has degenerate states, then before you can do (time independent) perturbation theory with a perturbation matrix $V$ on it, you have to first diagonalize $H_0 + V$ in the subspace of the degenerate states. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sounds nice enough, but what if $V$ is already diagonal in that subspace?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example I have in mind is a hydrogen molecule in the tight-binding + Heitler-London limit, e.g. $U \gg t$. The Hamiltonian is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\begin{array}{cccc} U &amp;amp; 0 &amp;amp; t &amp;amp; t\\ 0 &amp;amp; U &amp;amp; t &amp;amp; t \\ t &amp;amp; t &amp;amp; 0 &amp;amp; 0 \\ t &amp;amp; t &amp;amp; 0 &amp;amp; 0\end{array}\right)$$&#xA;where the first two basis states are the ionized states (both electrons on the same atom) and the other two basis states are the covalent states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this Hamiltonian is easy enough to diagonalize directly, and if the new ground-state energy is expanded up to second order in $t$, we get $-4t^2/U$ as the drop in energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now try to picture this in simple terms of second order perturbation theory, and there I only see a drop of $-2t^2/U$: Because from either of the unperturbed ground states (the covalent ones), I can have two second-order processes, with amplitudes $t^2$ each and with energy denominator $U$ each. Since I basically have a two-fold degenerate ground state, is this where the missing factor $2$ comes from? If so, what's the mechanism behind this?&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-04-09T19:12:01.087" Title="Perturbation theory with degeneracy even after 1st order" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="7680" PostTypeId="1" AcceptedAnswerId="7687" CreationDate="2011-03-27T23:10:34.393" Score="3" ViewCount="251" Body="&lt;p&gt;There have been a lot of sci-fi shows recently using the &quot;rotating space station&quot; explanation for gravity on space stations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After watching these videos:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=49JwbrXcPjc&amp;amp;NR=1&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=49JwbrXcPjc&amp;amp;NR=1&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.youtube.com/watch?v=_36MiCUS1ro&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=_36MiCUS1ro&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering what would happen, if you were facing the direction of rotation, while standing on the &quot;floor&quot; of a rotating space station and tossed a ball &quot;up&quot;. From the video it looks like the ball should land in front of you. Is this in fact what would happen and if you dropped a ball would it land behind you?&lt;/p&gt;&#xA;" OwnerUserId="2805" LastActivityDate="2011-03-28T02:00:56.027" Title="&quot;Artificial gravity&quot; on a rotating space stations" Tags="&lt;gravity&gt;&lt;space&gt;&lt;coriolis-force&gt;&lt;centrifugal-force&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="7681" PostTypeId="1" CreationDate="2011-03-27T23:28:02.523" Score="1" ViewCount="115" Body="&lt;p&gt;From my preliminary research I've learned that desiccants are a good (best?) method to absorb water from air. Assuming I'm using silica gel (or kitchen salt, or rice) then what is the best way to extract the water out of the desiccant?&lt;/p&gt;&#xA;" OwnerUserId="2807" LastEditorUserId="2807" LastEditorDisplayName="" LastEditDate="2011-03-31T15:37:30.577" LastActivityDate="2011-03-31T15:37:30.577" ClosedDate="2011-03-28T07:46:40.653" Title="Extracting water out of silica gel" Tags="&lt;water&gt;&lt;material-science&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="7682" PostTypeId="2" ParentId="7652" CreationDate="2011-03-27T23:29:54.673" Score="0" ViewCount="" Body="&lt;p&gt;I have long dreamed of the applicability of Quantum entanglement. However I find now that observing a particle presumably affected by Quantum entanglement, destroys it's entanglement. Thus making it's affects unobservable. The odd thing of it is that we can mathematically prove that things work a certain way, but we can never prove this through observation, and similarly the phenomenon is impossible to harness. There exists a conjecture that even if we can theoretically prove that something works a certain way through our maths, yet this proof is completely unobservable and inapplicable in any way, does the phenomenon exist at all or is it simply an artefact of the mathematical constructs we have since used to describe the universe? Long ago we thought the universe consisted of nothing but ideal platonic solids. Perhaps the beautiful simple equations we have produces are similarly too ideal. Perhaps our understanding of what we think we understand is flawed in that we choose to favour simple solutions over more complicated solutions. Quantum Entanglement, Dark matter, string theory, and so on are all ideas and concepts bred from mathematical oddities which we have no observable experience or applicable use. Do they exist at all? How can we be so sure that the equations we have constructed that describe the physical forces we observe, accurately describe them throughout the universe and not just in our small neighbourhood? Einstein proved that the equation we use for gravity is utterly flawed at great distances due to space and time curvature. But we so love the simple equation that we use it as a basis to prove that dark matter exists. These things are illusionary and it's more likely that they don't exist than they do. And if they do exist, what is the point in knowing them if they have no application and cannot be observed? In the end, it becomes nothing different from Religion. If you believe in dark matter, string theory, or quantum entanglement you have no ground to be cynical of people of religious faiths. For these beliefs are cousins.&lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2011-03-27T23:29:54.673" CommentCount="4" />
  <row Id="7683" PostTypeId="1" AcceptedAnswerId="7689" CreationDate="2011-03-28T00:00:58.280" Score="1" ViewCount="107" Body="&lt;p&gt;Here &lt;a href=&quot;http://en.wikipedia.org/wiki/Internal_energy&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Internal_energy&lt;/a&gt; we can read &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U= TS-PV+\sum_i \mu_i N_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's suppose i=1 and a ideal gas. We know:&#xA;$$U=N/N_A c_v T$$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$PV=NKT$$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu=\frac{\partial U}{\partial N}= c_v T/N_A$$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so $S=\frac{U+PV-\mu N}{T}$ would be $&#xD;&#xA;NK&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but it's not true. Where is the mistake?&lt;/p&gt;&#xA;" OwnerUserId="558" LastActivityDate="2011-03-28T02:33:25.593" Title="Internal Energy and entropy in a open system" Tags="&lt;thermodynamics&gt;&lt;heat&gt;&lt;chemical-potential&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7684" PostTypeId="2" ParentId="5832" CreationDate="2011-03-28T00:01:32.227" Score="2" ViewCount="" Body="&lt;p&gt;The view of spacetime as sliding point of space in a foliation of spatial 3-dim surfaces is helpful.  Gravity commoves particles with the sliding of points.  In the case of a black hole this commoves particles with points sliding into the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/8zHFo.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The de Sitter spacetime is similar to this, but with the flow in the opposite direction, without the singularity and the coordinate dependency with the radius.  The cosmological horizon $\Lambda$ determine the Hubble parameter $H~=~ 8\pi G\rho/3$, for $\rho$ the vacuum density of the universe. This determines the cosmological constant as&#xA;$$&#xD;&#xA;\Lambda = (3H^2\Omega/c^2)&#xD;&#xA;$$&#xA;where we take $\Omega~=~1$. The cosmological constant $\Lambda~\simeq! 10^{54}cm^{-2}$, and the cosmological horizon is at $r~=~\sqrt{3/\Lambda}$ which is at about $10^{10}$ light years. The CMB limit is beyond the cosmological horizon. Multiplying this by the relationship $v~\sim~Hd$ gives $v~=~c$. The cosmological Doppler shift is then at the cosmological horizon $z~=~1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some departures from this, for the de Sitter metric involves an exponential and there are deviations from $v~=~Hd$ for large enough $d$, but this is close enough. The most distant galaxies observed with z = 7 or 8, these galaxies are being commoved by the expansion of space at about 7 or 8 times the speed of light! The CMB with $z~\sim~ 1000$ is moving out at about that many times the speed of light. This is terribly confusing, and I see professionals, even some rather lionized ones, get tripped up on this. An observer who passes into a black hole has entered a region where they are being dragged by the isometries of space faster than light relative to things outside. Yet this observer will still be able to observe things outside the black hole before they crash into the singularity. What the observer is not able to do is to send a message to things beyond the black hole horizon. The same thing works with the cosmological horizon. We will never be able to send a message to any galaxy beyond it. We can however see everything into the past of the universe as far as our past light cone reaches.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-28T00:01:32.227" />
  <row Id="7685" PostTypeId="2" ParentId="7680" CreationDate="2011-03-28T00:14:55.403" Score="3" ViewCount="" Body="&lt;p&gt;What you indicate appears to be the case.  You want to minimize the Coriolis effect in any volume of space, which requires making the rotating station large.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-28T00:14:55.403" />
  <row Id="7686" PostTypeId="1" AcceptedAnswerId="7692" CreationDate="2011-03-28T01:59:38.697" Score="2" ViewCount="107" Body="&lt;p&gt;I am wondering: If I have a regular octahedron as my starting point, oriented along the x-y-z axis, and now Jahn-Teller suggest I elongate or compress along the $z$-axis, what happens along the other axis? I expect that these move in the opposite direction, but by how much? Say my displacement in $z$-direction is $\delta$. Is the displacement in $x$- and $y$-direction then $-\delta$, or do I have to find $\delta_x = \delta_y$ so that the total Volume is conserved?&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-03-28T03:06:31.010" Title="Nature of tetragonal distortion in Jahn-Teller effect" Tags="&lt;solid-state-physics&gt;&lt;crystals&gt;" AnswerCount="1" />
  <row Id="7687" PostTypeId="2" ParentId="7680" CreationDate="2011-03-28T02:00:56.027" Score="6" ViewCount="" Body="&lt;p&gt;Yes, the ball would land in front of you.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you watch from outside the space station, the ball moves in a straight line at constant speed while you move in a circle at constant speed.  That means the distance the ball takes to get from point A (where you release it) to point B (where it hits the floor) is shorter than the distance you take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, since you threw the ball, it's going faster than you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ball is going a shorter distance at a faster speed, so it gets there ahead of you.  From your point of view, the ball must curve forward in order to get ahead of you, so that's what you see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some time ago I made a Mathematica Demonstration where you can see the curved path the ball takes, varying the ship's radius and rotation rate as  well as how hard you throw the ball.  See here:&#xA;&lt;a href=&quot;http://demonstrations.wolfram.com/PathOfAProjectileInARotatingFrame/&quot; rel=&quot;nofollow&quot;&gt;http://demonstrations.wolfram.com/PathOfAProjectileInARotatingFrame/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read a full derivation of the equations on a doc I put on Scribd around that time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.scribd.com/doc/3880728/rotating-ship&quot; rel=&quot;nofollow&quot;&gt;http://www.scribd.com/doc/3880728/rotating-ship&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-03-28T02:00:56.027" CommentCount="1" />
  <row Id="7688" PostTypeId="2" ParentId="7664" CreationDate="2011-03-28T02:19:24.060" Score="3" ViewCount="" Body="&lt;p&gt;Hint: The square is symmetric with respect to the transformation $x \to -x$.  That means the x-component of the electric field must be equal minus itself i.e. it must be zero.  Similarly for y.  So the electric field only points up, and you can simply calculate the z-component for a strip and integrate in one dimension.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-03-28T02:19:24.060" />
  <row Id="7689" PostTypeId="2" ParentId="7683" CreationDate="2011-03-28T02:33:25.593" Score="4" ViewCount="" Body="&lt;p&gt;The problem here is that the thermodynamic potentials are functions of three thermodynamic variables each. Now, each thermodynamic potential has a set of natural variables. For the internal energy $U$, these are S, V and N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, your partial derivatives should explicitly state which other variables are held  constant. For example,&#xA;$$ \mu = (\frac{\partial U}{\partial N})_{S,V} $$&#xA;i.e. you keep $S$ and $V$ constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are, of course, free to express the internal energy as a function of other quantities, such as temperature instead of entropy. This would be achieved by expressing $S$ as a function of $V$, $N$ and $T$, so $S = S(T,V,N)$ or $T = T(S,V,N)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if you want to use the equation $U = N/N_A c_v T$ to compute $\mu$, you get $1/N_A c_v T$ from the $N$-dependency, but also some contribution from&#xA;$$ (\frac{\partial T}{\partial N})_{S,V}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's where your error comes from.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-03-28T02:33:25.593" />
  <row Id="7690" PostTypeId="2" ParentId="6157" CreationDate="2011-03-28T02:41:52.943" Score="3" ViewCount="" Body="&lt;h1&gt;Earth System Physics&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://dust.ess.uci.edu/facts/rt/rt.pdf&quot; rel=&quot;nofollow&quot;&gt;Radiative Transfer in the Earth System&lt;/a&gt;&lt;/strong&gt;, Charlie Zender (All of Zender's books are &lt;a href=&quot;http://www.gnu.org/copyleft/fdl.html&quot; rel=&quot;nofollow&quot;&gt;GNU FDL&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://dust.ess.uci.edu/facts/aer/aer.pdf&quot; rel=&quot;nofollow&quot;&gt;Natural Aerosols in the Climate System&lt;/a&gt;&lt;/strong&gt;, C. Zender&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://dust.ess.uci.edu/facts/psd/psd.pdf&quot; rel=&quot;nofollow&quot;&gt;Particle Size Distributions: Theory and Application to Aerosols, Clouds, and Soils&lt;/a&gt;&lt;/strong&gt;, C. Zender&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://oceanworld.tamu.edu/resources/ocng_textbook/contents.html&quot; rel=&quot;nofollow&quot;&gt;Introduction to Physical Oceanography&lt;/a&gt;&lt;/strong&gt;, Robert Stewart open source, accepting contributions&lt;/p&gt;&#xA;" OwnerUserId="2809" LastEditorUserId="2809" LastEditorDisplayName="" LastEditDate="2011-03-28T03:49:30.290" LastActivityDate="2011-03-28T03:49:30.290" CommunityOwnedDate="2011-03-28T02:41:52.943" />
  <row Id="7691" PostTypeId="2" ParentId="7681" CreationDate="2011-03-28T02:57:50.470" Score="1" ViewCount="" Body="&lt;p&gt;Heating the silica will drive out the water, I forget it's melting point but you aren't goign to get there!&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-03-28T02:57:50.470" />
  <row Id="7692" PostTypeId="2" ParentId="7686" CreationDate="2011-03-28T03:06:31.010" Score="3" ViewCount="" Body="&lt;p&gt;Very generally, Jahn-Teller distortions are indeed volume conserving, and there is an &quot;elastic&quot; cost associated with moving the atoms that make up the octahedron from their equilibrium positions, so if you stretch along the $z$-axis, say, then you have to compress along the $x$- and $y$-axes in such a way as to minimize an elastic energy that goes as $\delta x^2+\delta y^2$. Symmetrically compressing the axes is the best way to minimize that.&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2011-03-28T03:06:31.010" CommentCount="1" />
  <row Id="7693" PostTypeId="2" ParentId="7663" CreationDate="2011-03-28T03:21:52.540" Score="1" ViewCount="" Body="&lt;p&gt;“In the fall of 1940, Feynman received a tele­phone call from John Wheeler [Feynman’s thesis advisor] at the Grad­uate Col­lege in Princeton, in which he [Wheeler] said that he knew why all elec­trons have the same charge and the same mass. ‘Why?’ asked Feynman, and Wheeler replied, ‘Because they are all one and the same elec­tron.’” — Jagdish Mehra&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What Wheeler meant was that the annihilation of an electron and a positron can be described by saying that the electron gets scattered backward in time. An electron going pastward looks like a positron going futureward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An  anal retentive comment on Marek's statement that &quot;microscopically all processes are described by Feynman diagrams such as these...&quot; Make this: the amplitude associated with a scattering event (with given numbers and species of incoming and outgoing particles) can be calculated by summing over an infinite series of complex numbers, each represented by a Feynman diagram.&lt;/p&gt;&#xA;" OwnerUserId="2784" LastActivityDate="2011-03-28T03:21:52.540" CommentCount="1" />
  <row Id="7694" PostTypeId="2" ParentId="7663" CreationDate="2011-03-28T04:20:01.150" Score="1" ViewCount="" Body="&lt;p&gt;All the answers given are great answers once one has acquired some modern physics background and mathematics. For an aspiring young physicist they  might be a big mouthful. So once more I will offer the simpler &quot;everyday&quot; interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When particles interact we talk of it as scattering of one particle with the other and have built machines that do that. Scatterings are described well with the theories developed, as in the answer by Marek, and if one means by &quot;annihilation&quot; that a particle disappears and becomes another particle, that is the way it can happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Real&quot; annihilation happens, is measured,  experimentally when all the quantum numbers describing the two incoming interacting particles become 0 in the interaction region and the output is photons, as in the case of electron positron annihilation at low energy, and/or a bunch of particle whose quantum numbers sum up to 0 at energies that allow their creation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when the two incoming particles have equal and opposite quantum numbers that describe them in the Standard Model,  experimentalists call the interaction  annihilation if it is not an elastic scattering. Elastic scattering retains the quantum numbers of the two incoming particles, either wholly or partially. It can happen that an antiproton becomes  an antineutron, retaining baryon number, for example. The processes can only be disentangled studying further with diagrams of the type shown by Marek.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-28T04:44:05.497" LastActivityDate="2011-03-28T04:44:05.497" />
  <row Id="7695" PostTypeId="2" ParentId="7611" CreationDate="2011-03-28T04:22:29.243" Score="0" ViewCount="" Body="&lt;p&gt;You can't describe the world without describing it. An inertial frame is a language. You can describe a physical system or process by using any language (inertial frame) you like, but you cannot describe it by using no language (inertial frame) at all. To be able to visualize a 4-vector, which is invariant under Lorentz transformations (translations from one language or frame to another), you would have to visualize at once all its frame-dependent decompositions into a spatial and a temporal component. The &#xA;&lt;A HREF=&quot;http://upload.wikimedia.org/wikipedia/commons/b/b4/Animated_Lorentz_Transformation.gif&quot; rel=&quot;nofollow&quot;&gt;animation provided by Mark&lt;/A&gt; does this to some extent.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...do you ever get a proper intuitive model for Minkowski spacetime...?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Allow me to deepen you mystification. For any two events A,B there exist two reference frames F&lt;SUB&gt;A&lt;/SUB&gt; and F&lt;SUB&gt;B&lt;/SUB&gt; and a third event C such that C is simultaneous with A in F&lt;SUB&gt;A&lt;/SUB&gt; and simultaneous with B in F&lt;SUB&gt;B&lt;/SUB&gt;. This &quot;simultaneity by proxy&quot; of A with B compels us to conceive of all parts of the spatiotemporal whole as coexistent and as equally real. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or does it? The coexistence of the spatiotemporal whole cannot be a &lt;em&gt;simultaneous&lt;/em&gt; existence — that would be self-contradictory. It can only be a a &lt;em&gt;tenseless&lt;/em&gt; or &lt;em&gt;atemporal&lt;/em&gt; coexistence. Can you imagine an expanse whose character is neither spatial nor temporal? If you can, then you got a proper intuitive model for Minkowski spacetime.&lt;/p&gt;&#xA;" OwnerUserId="2784" LastActivityDate="2011-03-28T04:22:29.243" CommentCount="2" />
  <row Id="7696" PostTypeId="2" ParentId="7539" CreationDate="2011-03-28T06:45:58.770" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://video.google.com/videoplay?docid=1996321846673788606#&quot; rel=&quot;nofollow&quot;&gt;Should google go nuclear.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should all watch that video. It is EXTREMELY informative. It completely trounces the tokamak style reactor. It also mentions that there are different combinations of hydrogen atoms that could be utilized to produce slightly less power output, but have zero radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it is far less dangerous than fission. Yes it produces neutrons, but as I mentioned, using a different type of hydrogen drastically reduces the energy (and thus damage potential) of those neutrons. Neither fusion or fission reactors could ever &quot;explode&quot; like an atomic bomb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with fission is that they use a physical core of uranium or plutonium. Fusion is done with only atoms. The fissile material undergoes a chain reaction, and is submerged in water to generate heat. Once the chain reaction begins, you can only stop it by inserting graphite rods to absorb the excess neutrons. All of the major nuclear incidents, Three Mile Island, Chernobyl, and now Japa, have involved the rods not being inserted, and the core somehow gets exposed to the air, and radiation pours out. Chernobyl for example, was a explosion caused by steam that blew the reactor core apart, exposing the radioactive material inside. At this point it becomes far to hot and radioactive to get anywhere near it. The crisis in Japan is the first time it has been the result of some catastrophic event. The others were human error. This could obviously never happen with fusion. If there was a disaster, the reactor would just shut off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://emc2fusion.org/&quot; rel=&quot;nofollow&quot;&gt;http://emc2fusion.org/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Website with some information and pictures of working prototypes of &quot;Electric fusion&quot; reactors. &lt;/p&gt;&#xA;" OwnerUserId="2812" LastActivityDate="2011-03-28T06:45:58.770" />
  <row Id="7697" PostTypeId="2" ParentId="7679" CreationDate="2011-03-28T07:33:26.610" Score="6" ViewCount="" Body="&lt;p&gt;First, just to be sure about the answers to this particular problem: the eigenvalues of the $4\times 4$ matrix are &#xA;$$0,\quad U\quad {\rm and}\quad U/2\pm \sqrt{(U/2)^2+4t^2}$$&#xA;When expanded to the first nontrivial order, the last two eigenvalues are &#xA;$$ 0 - \frac{4t^2}U \quad {\rm and} \quad U+\frac{4t^2}U. $$&#xA;Note that the corrections to the energy arise at order $t^2$ so the first-order perturbation theory is not enough in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the problem that prevents us from choosing the right eigenstates by the simple method is - as you correctly notice - that the matrix $V$, i.e. the matrix multiplied by $t$, has a vanishing upper left $2\times 2$ block as well as the right lower $2\times 2$ block - both of these blocks vanish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $V$ doesn't lift the degeneracy &quot;inside the degenerate subspaces&quot; only. This is, of course, related to the fact that the first-order $O(t)$ corrections to the energy eigenvalues vanish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard formula of perturbation theory for the second-order corrections to energy is&#xA;$$ E_n = E_n^{(0)} + t \langle n^{(0)} |V|n^{(0)}\rangle + t^2 \sum_{k\neq n}\frac{\left| \langle k^{(0)} |V| n^{(0)} \right|^2}{E_n^{(0)}-E_k^{(0)}} +O(t^3)  $$&#xA;Now, the $t^2$ term should give us $\pm 4t^2/U$ if it works. And of course, it does as long as we choose the right superpositions as the zeroth-order eigenvectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, let's just reveal the eigenvectors. The eigenvalue $U$ comes with the eigenvector $(1,-1,0,0)^T / \sqrt{2}$ and similarly the eigenvalue $0$ comes with the eigenvector $(0,0,1,-1)^T / \sqrt{2}$. The $U+4t^2/U$ comes from $(1,1,0,0)^T/\sqrt{2}$ and similarly $0-4t^2/U$ comes from $(0,0,1,1)^T/\sqrt{2}$. The transposition means that the vectors should be written in the conventional column form. I added the $1/\sqrt{2}$ factor to make all of them normalized - and they're orthogonal, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each calculated state (and its second order correction to energy), there's just one nonzero term in the $\sum_{k\neq n}$ sum. It has the denominator - the energy difference - $U$ if we calculate $E_n$ near $U$ or $-U$ if we calculate $E_n$ near $0$. And it connects the states $(1,1,0,0)^T/\sqrt{2}$ with $(0,0,1,1)^T/\sqrt{2}$ or vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the matrix element of the $V$ matrix with the two $((1,1),(1,1))$ off-diagonal blocks between the two states I mentioned at the end of the previous paragraph is $1/2$ (from the two $1/\sqrt{2}$ normalization factors) times $8$ (because there are eight nonzero entries $t$ in your matrix, or $1$ in mine, and each of them contributes the same one to the matrix element). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the matrix element is simply $4$ and the perturbation theory gives you the right $\pm 4t^2/E$ correction to the energy, from a single term, with the proper sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the only step I haven't quite justified was the right choice of the eigenvectors - such as $(1,1,0,0)^T  / \sqrt{2}$. How could I have seen this was the right one? Well, in this case, it was the intuitively right choice. While the $2\times 2$ blocks on the diagonal preserved the degeneracy, the energy shift came from the $2\times 2$ off-block-diagonal blocks and those made it natural to use this basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if we're in a similar situation that the degeneracy isn't lifted by first-order corrections in $V$ in the subspaces, what we really have to diagonalize is $V (H_0-E_0)^{-1\prime} V$ where the prime indicates that one has to omit the (divergent) terms from the vanishing energy differences (of course, we need a sensible result). This is the operator whose expectation value de facto gives us the second-order energy correction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this particular case, this operator will have a block diagonal form, $U^{-1} {\rm diag} [ ((+1,+1),(+1,+1)),  ((-1,-1),(-1,-1)) ]$, and by diagonalizing, you get the right initial eigenvectors to deal with. Once you have the right eigenvectors to start with, their perturbations are infinitesimal at each order of the perturbation theory and the standard formulae of perturbation theory work without any extra subtleties, as the example above showed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, the only thing one has to be careful about are the right zeroth-order initial eigenvectors. In a more generic case, they're given as eigenvectors of $V$ because $V$ lifts the degeneracy in each subspace. If it doesn't lift the degeneracy, only higher-order terms do so. But the operator $V (H-E_0)^{-1\prime} V$ plays the same role as $V$, and by diagonalizing it, we get the right initial eigenvectors.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-28T07:33:26.610" CommentCount="3" />
  <row Id="7698" PostTypeId="1" AcceptedAnswerId="8551" CreationDate="2011-03-28T07:47:49.100" Score="4" ViewCount="110" Body="&lt;p&gt;Extending my previous question &lt;a href=&quot;http://physics.stackexchange.com/questions/6710/angular-moment-and-em-wave&quot;&gt;Angular moment and EM wave&lt;/a&gt;, does it make sense to talk about the angular momentum of electromagnetic waves in an anisotropic medium? It is not obvious that the angular momentum is conserved in this case. However, if the anisotropy is introduced by the external magnetic field (eg, magnetized plasmas), the projection of the angular momentum of a wave packet on the direction of the magnetic field might be conserved.&lt;/p&gt;&#xA;" OwnerUserId="2560" LastEditorUserId="2560" LastEditorDisplayName="" LastEditDate="2011-04-15T02:21:06.110" LastActivityDate="2011-05-15T02:46:18.910" Title="Is it true that the angular momentum of electromagnetic waves in an anisotropic medium is an integral of motion?" Tags="&lt;condensed-matter&gt;&lt;electrodynamics&gt;&lt;quantum-electrodynamics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="7699" PostTypeId="2" ParentId="7539" CreationDate="2011-03-28T07:56:55.523" Score="0" ViewCount="" Body="&lt;p&gt;You need an aneutronic energy cycle to make fusion secure. Examples: p+B11, D+He3. TriAlpha is a running commercial project aimed on p+B11 fusion.&lt;/p&gt;&#xA;" OwnerUserId="2560" LastActivityDate="2011-03-28T07:56:55.523" />
  <row Id="7700" PostTypeId="1" CreationDate="2011-03-28T10:23:04.933" Score="1" ViewCount="206" Body="&lt;p&gt;Just wondering about the definitions and usage of these three terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my understanding so far, &quot;covariant&quot; and &quot;form-invariant&quot; are used when referring to physical &lt;em&gt;laws&lt;/em&gt;, and these words are synonyms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Invariant&quot; on the other hand refers to physical &lt;em&gt;quantities&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would you ever use &quot;invariant&quot; when talking about a law? I ask as I'm slightly confused over a sentence in my undergrad modern physics textbook:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;In general, Newton's laws must be replaced by Einstein's relativistic laws...which hold for all speeds and are &lt;strong&gt;invariant&lt;/strong&gt;, as are all physical laws, under the Lorentz transformations.&quot; [emphasis added]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;~ Serway, Moses &amp;amp; Moyer. Modern Physics, 3rd ed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Did they just use the wrong word?&lt;/p&gt;&#xA;" OwnerUserId="2557" LastActivityDate="2011-05-07T18:26:59.077" Title="Definitions and usage of Covariant, Form-invariant, Invariant?" Tags="&lt;definition&gt;&lt;covariance&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="7702" PostTypeId="1" CreationDate="2011-03-28T11:02:30.937" Score="-2" ViewCount="179" Body="&lt;p&gt;There is a known thought experiment, connected to quantum immortality: a duel between physicist and a philosopher. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each turn physicist and philosopher fire at each other with a pistol. The quantum immortality predicts that each of the participants will find themselves alive and the opponent dead after a number of shots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can modify the experiment in the following ways:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;They do not fire at each other, but it is lightning that shots them. The outcome should not change.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are not two but more participants.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They are located at separate islands in the ocean.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It seems that even if there are hundreds of participants, each of them will eventually find all others killed by lightning. If he could not observe other isles directly, after he discovers and explores other isles, he will find them uninhabitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now pretend that the participants actually live on separate planets, and in each million years there happens disaster that kills all inhabitants on 1% of all planets. It seems that after years the physicist that lives on one of the planets should discover that all other planets are uninhabitable. Even more: there is even no chance that organic matter on each other planet could actually evolve into anything resembling actual life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said the physicist will find out that there is no other inhabitable planet in the Uinverse. On the other hand, this does not mean that extraterrestrial life does not exist. It actually cannot not to exist. But it exists in parallel universes (in terms of MWI) and thus unobservable. Both physicist on planet A and philosopher on planet B exist, but will never meet in the same universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, QI it seems predicts one concrete observable consequence that extraterrestrial life can be never detected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is your opinion?&lt;/p&gt;&#xA;" OwnerUserId="1186" LastEditorUserId="1186" LastEditorDisplayName="" LastEditDate="2011-03-28T11:10:12.083" LastActivityDate="2011-03-28T11:36:16.933" ClosedDate="2011-03-28T15:17:28.360" Title="Does quantum mechanics prohibit extraterrestrial life?" Tags="&lt;quantum-mechanics&gt;&lt;cosmology&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="7703" PostTypeId="2" ParentId="7657" CreationDate="2011-03-28T11:17:34.200" Score="1" ViewCount="" Body="&lt;p&gt;Some remark on &lt;/p&gt;&#xA;&#xA;&lt;p&gt;friction of rolling wheel = static friction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This true theoretically, in practice the friction of a wheel will &#xA;be somewhat less. Depending on details of smoothness of the track and &#xA;the details of the wheels suspension (springs, mass and damping) &#xA;the normal force will &quot;oscillate&quot;. In extreme cases the wheel can &#xA;jump for a piece of way, for that then there is no friction at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW &#xA;some  Swedish people in superconductor research praised much lighter &#xA;lokomotives with superconductor motors  :=) &#xA;I do not find the link anymore.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-28T11:17:34.200" />
  <row Id="7704" PostTypeId="2" ParentId="7700" CreationDate="2011-03-28T11:25:45.490" Score="2" ViewCount="" Body="&lt;p&gt;These words do have different meanings, this is a general guide to their differences. In different fields they may have slightly varying definitions. I would recommend looking them up to be certain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Invariant means does not change at all. Everything is the same (whether physical law, quantity or anything). In terms of vectors, invariant is a scalar which does not transform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Form-invariant means the form does not change, for example the inverse square law, will always be inverse square but the constants may differ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Covariant, has a specific meaning when relating it to vectors, as it specifies the transformation rules. (This is as opposed to contravariant which is the other one).&#xA;For more information &lt;a href=&quot;http://en.wikipedia.org/wiki/Four-vector#Mathematics_of_four-vectors&quot; rel=&quot;nofollow&quot;&gt;see wikipedia&lt;/a&gt;, towards the end of the Mathematics of four vectors section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To specifically answer your question on the phrase, Einsten's relativistic laws are invariant under Lorentz transformations, the laws don't change at all. The constants don't change, neither does the form.&lt;/p&gt;&#xA;" OwnerUserId="133" LastActivityDate="2011-03-28T11:25:45.490" CommentCount="2" />
  <row Id="7705" PostTypeId="2" ParentId="7702" CreationDate="2011-03-28T11:36:16.933" Score="1" ViewCount="" Body="&lt;p&gt;Dear Anixx, we know very well from the observations - and theory - here on Earth that repeated lightnings are nowhere close to be able to exterminate the life on the planet. Because it's true here and because the laws of physics - including the emergent ones - are at least approximately translationally invariant, it's almost certainly true on similar other planets in the Universe: lightnings are not enough to exterminate life. Your &quot;conclusion&quot; - actually an &quot;assumption&quot; - that the life on other planets has to go extinct because of lightnings is therefore invalid. It is invalid in the whole Universe and this paragraph is a proof that can't be &quot;unproved&quot; by any silly logical game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very obvious that none of those thought experiments can actually determine the real odds - or our rational expectations about them - of having some extraterrestrial life. It's because the emergence of life (whatever its definition is) and its extinction on every celestial body is controlled by the objective laws of physics rather than subjective feelings of an observer who is likely to die soon. So all calculations of an individual that simply &quot;erase&quot; the possibility that he will die from the calculation of odds are subjective and can't be used by other individuals as an objective calculation. So they cannot be used to count the extraterrestrial civilizations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this reason, the &quot;quantum immortality&quot; itself is a logical fallacy. From an objective viewpoint, whether an observer dies during a period of time or not is just a well-defined question that can have both answers and eliminating possibilities where he dies is objectively unjustifiable. By the way, the philosophies and religions that believe in &quot;life after death&quot; obtain much more valid figures for the probabilities than Max Tegmark's philosophy that simply denies the possibility that observers die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No &quot;interpretation&quot; of quantum mechanics or physics, as long as it is consistent with the standard physical predictions, can change anything about the paragraphs above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, if you find my comment that the fate of life is given by the laws of physics and not games in philosophy controversial, let me mention that this battle between physics and philosophy has been settled since 1687. A general in this physics-philosophy war, philosopher Isaac Newton, published his aptly named &quot;Mathematical Principles of Natural Philosophy&quot; and declared a surrender. Since that time, it was clear that all natural phenomena have to be studied by physics (that he helped to establish) and not by meaningless and logically flawed philosophical games you have presented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I would respectfully disagree that the answer above is &quot;my opinion&quot;. It is the objective answer to your question. Physics is not about personal opinions; it is the scientific method to search (and often find) the objective truth. Also, this server is meant to offer the right physics answers to questions that can be answered as of 2011, rather than an assorted flow of chaotic personal opinions, and yours was surely one of the questions that can be answered today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't claim that the extraterrestrial civilizations have to exist or have to be absent - but I surely do claim that there exists no rational argument based on &quot;quantum immortality&quot; that could influence the odds.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-28T11:36:16.933" CommentCount="19" />
  <row Id="7707" PostTypeId="1" CreationDate="2011-03-28T12:30:16.077" Score="7" ViewCount="186" Body="&lt;p&gt;I have a problem with energy conservation in case of interfering waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine two harmonic waves with amplitudes $A$. They both carry energy that is proportional to $A^2$, so the total energy is proportional to $2A^2$. When they interfere, the amplitude raises to $2A$, so energy is now proportional to $4A^2$ and bigger than before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equivalent question is what happens to the energy with the superposition of two waves that interfere destructively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if someone could comment on the statement about this problem in my physics book (Bykow, Butikow, Kondratiew): the sources of the waves work with increased power during the interference because they feel the wave from the other source.&lt;/p&gt;&#xA;" OwnerUserId="2752" LastActivityDate="2011-03-29T15:24:03.947" Title="Energy conservation and interference" Tags="&lt;energy-conservation&gt;&lt;interference&gt;" AnswerCount="3" FavoriteCount="4" />
  <row Id="7708" PostTypeId="2" ParentId="409" CreationDate="2011-03-28T12:31:17.473" Score="3" ViewCount="" Body="&lt;p&gt;Strange, the &quot;stars orbiting black holes&quot; method was mentioned, &#xA;but not the fact that this method was applied to infrared images &#xA;from the Saggitarius region, (= the center of our own galaxy). &#xA;The movement of some stars over some years there is the most direct &#xA;proof for a black hole that exists till now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link with a beautiful animation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.scienceblogs.de/weitergen/2008/12/das-schwarze-loch-im-zentrum-unserer-milchstrasse.php&quot; rel=&quot;nofollow&quot;&gt;http://www.scienceblogs.de/weitergen/2008/12/das-schwarze-loch-im-zentrum-unserer-milchstrasse.php&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-03-28T12:31:17.473" CommentCount="1" />
  <row Id="7709" PostTypeId="1" AcceptedAnswerId="7711" CreationDate="2011-03-28T12:47:06.493" Score="-1" ViewCount="244" Body="&lt;p&gt;For the last few years, My friend worked on figuring out the &quot;theory of everything&quot;. She is afraid of sharing her theory with well known physicists, because she doesn't want other people to take credit for her idea. So, my here's what I'm wondering:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1) What is the best way for her to validate her theory with experienced physicists&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2) If the theory ends up being worth publishing, how can she go about publishing it considering the fact that she has no reputation in the field at the moment?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2815" LastEditorUserId="2815" LastEditorDisplayName="" LastEditDate="2011-03-28T14:15:47.350" LastActivityDate="2011-03-28T14:51:45.957" ClosedDate="2011-03-28T14:54:49.877" Title="How to validate a new physics theory and keep the credit for it" Tags="&lt;theory&gt;" AnswerCount="1" CommentCount="22" />
  <row Id="7710" PostTypeId="2" ParentId="7707" CreationDate="2011-03-28T13:00:25.100" Score="4" ViewCount="" Body="&lt;p&gt;It is guaranteed that finite wave packets always create places where the interference is constructive as well as places where it is destructive: the energy simply flows from the maxima to the minima.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your convention, it's guaranteed that the total energy at the end is always &quot;in between&quot; the energy from the constructive interference and the energy from the destructive one, which is simply the average&#xA;$$ (4 A^2 + 0 ) / 2 = 2A^2,$$&#xA;exactly as the original energy. Otherwise, the energy conservation can be proved even locally - as a continuity equation - directly from Maxwell's equations so it always holds. This is particularly easy to prove for vacuum Maxwell's equation - enough for propagation and interference of light,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_stress-energy_tensor&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Electromagnetic_stress-energy_tensor&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-28T14:26:33.000" LastActivityDate="2011-03-28T14:26:33.000" CommentCount="3" />
  <row Id="7711" PostTypeId="2" ParentId="7709" CreationDate="2011-03-28T13:33:16.147" Score="2" ViewCount="" Body="&lt;p&gt;Even though  I tend to be of the same mind as the comments, I would like to answer, before the question is closed, since comments are limited in character counts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1st) validation. Validation means that somebody,, in this case the investigator proposing a new theory comes up with hard numbers from the proposed theory and validates them by checking with the accurate data available from elementary particle tables to astronomy ones. i.e. checks predictions against data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, for example, the standard model of particle physics does not come out naturally from the new theory, it is invalidated as a theory of everything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) if the first stage is passed, and the theory of your friend by some miraculous manner is validated by the existing data up to now,she can publish with her name and address, her theory  and its validation in &lt;a href=&quot;http://arxiv.org/help/faq/index&quot; rel=&quot;nofollow&quot;&gt;the archives&lt;/a&gt; so there is no way somebody will take the credit from her. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-28T14:51:45.957" LastActivityDate="2011-03-28T14:51:45.957" CommentCount="8" />
  <row Id="7712" PostTypeId="2" ParentId="7707" CreationDate="2011-03-28T13:43:37.513" Score="2" ViewCount="" Body="&lt;p&gt;The energy does not double.  This is prevent in part by the Born rule.  Consider the state vector $|\psi\rangle~=~\sum_n c_n|n\rangle$, where it is easy to see $c_n~=~\langle\psi|n\rangle$.  The energy eigenvalues are computed as $H|n\rangle~=~E_n|n\rangle$.  For the single wave function the amplitudes are found by the modulus square $\langle\psi|\psi\rangle~=~1$ of the wave, where&#xA;$$&#xD;&#xA;\langle\psi|\psi\rangle~=~\sum_nc_n^*c_n~=~\sum_nP_n~=~1&#xD;&#xA;$$&#xA;Now consider the expectation of the Hamiltonian&#xA;$$&#xD;&#xA;\langle\psi|H|\psi\rangle~=~\sum_nP_nE_n.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address this question directly we then consider the sum of two waves in a superposition with the state vectors $|\psi_1\rangle$ and $|\psi_2\rangle$, &#xA;$$&#xD;&#xA;|\psi_1\rangle~=~\sum_n c_{n,1}|n\rangle,~|\psi_2\rangle~=~\sum_n c_{n,2}|n\rangle,&#xD;&#xA;$$&#xA;where the two waves are expanded in their own coefficients which are normalized for the sum of the two waves.  Consider the modulus square of the sum $|\psi_t\rangle~=~|\psi_1\rangle~+~|\psi_2\rangle$ of the two waves&#xA;$$&#xD;&#xA;\langle\psi_t|\psi_t\rangle~=~\sum_n(c_{n,1}^*c_{n,1}~+~ c_{n,2}^*c_{n,2}~+~ c_{n,1}^*c_{n,2}~+~ c_{n,2}^*c_{n,1})&#xD;&#xA;$$&#xA;At this point you can now see the answer to the question, for the amplitude coefficients $c_{n,1}$ and $c_{n,2}$ are normalized to counter the growth by four of the modulus square of the amplitudes.  This is particularly easy to see if $c_{n,1}~=~c_{n,2}$, where now the amplitudes for the two waves are normalized to $1/\sqrt{2}$ their independent values.  One might have $c_{n,1}~=~exp(ikx_1)c_n$ and $c_{n,2}~=~exp(ikx_2)c_n$, where the interference term &#xA;$$&#xD;&#xA;c_{n,1}^*c_{n,2}~+~ c_{n,2}^*c_{n,1}~=~c_n(e^{ik(x_1~-~x_2)}~+~ e^{ik(x_2~-~x_1)})&#xD;&#xA;$$&#xA;which gives a oscillating term with respect to the difference $ x_1~-~x_2$.  Then further the expectation of the  Hamiltonian $\langle\psi_t|H|\psi_t\rangle$ is similarly expanded as&#xA;$$&#xD;&#xA;\langle\psi_t|\psi_t\rangle~=~\sum_n(c_{n,1}^*c_{n,1}~+~ c_{n,2}^*c_{n,2}~+~ c_{n,1}^*c_{n,2}~+~ c_{n,2}^*c_{n,1})E_n&#xD;&#xA;$$&#xA;where there is clearly no increase in the energy.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-28T13:43:37.513" />
  <row Id="7713" PostTypeId="1" AcceptedAnswerId="7716" CreationDate="2011-03-28T15:58:48.713" Score="3" ViewCount="170" Body="&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/q/7584/520&quot;&gt;Question 7584&lt;/a&gt; illustrated a procedure to forecast the decay rates of isotopes with known long average lifetimes. Lifetimes of the many U isotopes vary from micoseconds to gigayears. F has only one stable isotope while Sn has 10. Can Standard Model principles be used to predict the stability of isotopes and the average lifetimes for unstable isotopes, or can this only be done by measurement?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-03-28T16:11:38.193" LastActivityDate="2011-03-28T19:03:01.883" Title="Predicting Decay Rates via the Standard Model" Tags="&lt;standard-model&gt;&lt;radioactivity&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="7714" PostTypeId="2" ParentId="409" CreationDate="2011-03-28T16:00:57.683" Score="3" ViewCount="" Body="&lt;p&gt;For a stellar mass black hole the quick answer is there is no hard surface from which infalling material can crash on and splash or scatter back out.  The event horizon is not a hard or material substance.  So a black hole has some distinct signature differences from a neutron star, where material in an accretion disk strikes a hard surface.  It is in this way one can identify an object as a black hole.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-28T16:00:57.683" />
  <row Id="7716" PostTypeId="2" ParentId="7713" CreationDate="2011-03-28T16:39:01.650" Score="3" ViewCount="" Body="&lt;p&gt;The nuclear forces are a complex amalgam of primarily Quantum Chromodynamics forces and electromagnetic ones, but to deal with the diagramatic way of calculating in Quantum Field Theory, is not possible. The weak force  responsible for beta decays should also be in the calculations.Too many diagrams and too convoluted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum mechanical models with a potential well to estimate the collective forces are used for this.  Nuclear physics has been using various models successfully, like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_shell_model&quot; rel=&quot;nofollow&quot;&gt;shell model.&lt;/a&gt; to predict energy levels in nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ways  of estimating  lifetimes are taught in nuclear engineering, &lt;a href=&quot;http://www.nuc.berkeley.edu/NE-101&quot; rel=&quot;nofollow&quot;&gt;for example:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Course Outcomes: Students must be&#xA;able to...&#xA;calculate the consequences of radioactive&#xA;growth and decay and nuclear reactions.&#xA;calculate estimates of nuclear masses and energetics&#xA;based on empirical data and nuclear models.&#xA;calculate estimates of the lifetimes of nuclear&#xA;states that are unstable to alpha-,beta- and gamma decay and internal&#xA;conversion based on the theory of simple nuclear models.&#xA;use nuclear models to predict low-energy level&#xA;structure and level energies.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-03-28T17:54:56.903" LastActivityDate="2011-03-28T17:54:56.903" />
  <row Id="7717" PostTypeId="1" AcceptedAnswerId="7719" CreationDate="2011-03-28T17:55:18.293" Score="4" ViewCount="132" Body="&lt;p&gt;With this being a potentially subjective question I know I'm threading on thin ice, but this is an actual &quot;physics&quot; problem I face from time to time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation is this: I hear about some phenomenon. For the sake of definiteness, let's say it is the Jahn-Teller-Theorem. I'd then like to find a good source that has a graduate level &lt;em&gt;introduction&lt;/em&gt; to this phenomenon. The problem is that all Google turns up is either a very simplified explanation of the thing at hand (think Wikipedia), or a research level text that assumes its readers already have a perfect understanding of what the thing is. And original papers often do not provide such a clear answer (Legend has it that Born and Oppenheimer's paper on the adiabatic approximation is quite inaccessible and the relevant thing, the BO-Expansion, appears hidden somewhere in the appendix...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that I am not alone with this problem and I wonder if there is a straightforward approach to finding good introductory (but NOT superficial) texts on a given phenomenon. &lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-03-28T18:22:51.000" Title="How to find 'good' literature on a phenomenon?" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="7718" PostTypeId="2" ParentId="7717" CreationDate="2011-03-28T18:22:11.157" Score="1" ViewCount="" Body="&lt;p&gt;As already said, there is probably no universal answer. Probably best and most obvious options are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-asking here :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-searching directly in Google Books. You will probably find both introductory, but still advanced texts, and very technical ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-checking references. This is obvious, but it is usually worthy to a) find any advanced paper on the subject b) look up all the papers mentioned in the introduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way: google has lauched beta of 'Google scholar', which allows to search papers by key-words, citations etc&lt;/p&gt;&#xA;" OwnerUserId="2752" LastActivityDate="2011-03-28T18:22:11.157" />
  <row Id="7719" PostTypeId="2" ParentId="7717" CreationDate="2011-03-28T18:22:51.000" Score="9" ViewCount="" Body="&lt;p&gt;If I'm trying to learn about, say, geometric phases, what I tend to do is fire up &lt;strong&gt;&lt;a href=&quot;http://scholar.google.com&quot; rel=&quot;nofollow&quot;&gt;scholar.google.com&lt;/a&gt;&lt;/strong&gt; and search for&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;geometric phases&quot; review&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;geometric phases&quot; resource letter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That usually does the trick. And I tried this for { &quot;jahn-teller&quot; review } and I found an AJP article titled &quot;The Jahn–Teller effect: An introduction and current review&quot;. I presume that's the kind of thing you're looking for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the broader your search topic, the more likely you are to find a good review or resource letter.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-03-28T18:22:51.000" CommentCount="3" />
  <row Id="7720" PostTypeId="2" ParentId="7713" CreationDate="2011-03-28T18:55:47.517" Score="1" ViewCount="" Body="&lt;p&gt;In principle, the Standard Model contains the answer to all these questions about half-lives. In practice, it is impossible to calculate at this point - as far as I know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is very easy to explain why the lifetimes span such an exponentially huge interval of possible time scales. Here's why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alpha-decay - and only alpha-decaying nuclei with &quot;really long-lived&quot; half-lives will be explained by the simple argument below - may be approximated as the confinement and quantum tunneling of an alpha particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that a big nucleus $M$ decays to a smaller one $N$ and an alpha-particle:&#xA;$$ M \to N + \alpha$$&#xA;In this context, $M$ may be viewed as a bound state of $N$ and $\alpha$.  The alpha-particle is confined by the attractive potential of the nucleus $N$. However, we know that this bound state isn't quite stable - it decays. It follows that a free $N$ plus a free $\alpha$ must have a lower total mass/energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, the bound state of $N$ and $\alpha$, and that's what we called $M$, must be metastable and the alpha-particle is confined by a potential wall. Classically, it would stay there forever. Quantum mechanically, there is quantum tunneling. There's a nonzero probability that the alpha-particle tunnels through the potential wall and gets out. However, the rate of this event is exponentially small because the wave function, having an imaginary momentum inside the potential wall (a negative kinetic energy needed to go through the wall), exponentially decreases in the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability or decay rate $\Gamma$ is then comparable to&#xA;$$ \Gamma = dP/dt \sim \exp (-V/V_0) \times (10^{-24}{\rm seconds})^{-1} $$&#xA;where $V$ is the total &quot;thickness&quot; of the potential created by the nucleus $N$ inside the bound state $M$, and $V_0$ is some typical QCD-scale unit for such thickness. To get the right dimensions, I also included $10^{-24}$ seconds as the typical &quot;QCD time scale&quot;. However, as you can see, the half-lives of the isotopes will be exponentially larger. Relatively moderate changes of $V/V_0$ - from 20 to 100 - may change the lifetime from a fraction of a second to billions of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This qualitative pattern is surely well understood. In fact, approximate estimates of the lifetimes can be done by models in nuclear physics, and those models can be at least approximately be showed compatible with the Standard Model. But no one has calculated the half-life of uranium-235 via a bound state of many quarks and gluons - as far as I know. Also, however, there is not a single glimpse of a demonstrable discrepancy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, I was cheating a bit: even beta-decay may lead to incredibly long lifetimes.  An example is potassium-40 which can decay by any three types of the beta-decay - and its lifetime is still over a billion of years. Microscopically, the beta-decay is a bit different than the alpha-decay because it requires a real transformation of the types of particles - not just a different state of protons' and neutrons' motion and position. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same moment, beta-decay is somewhat simpler because its &quot;microscopic process&quot; is just a single weak interaction vertex with a W-boson - and the rest of the nucleus is less changed (only 1 nucleon is added/killed, instead of 4). In some sense, the beta-decays combine the quantum tunneling above with the elementary W-boson-induced weak interaction and they may have very diverse lifetimes, too.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-28T19:03:01.883" LastActivityDate="2011-03-28T19:03:01.883" CommentCount="2" />
  <row Id="7721" PostTypeId="2" ParentId="7654" CreationDate="2011-03-28T19:06:38.817" Score="8" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: I made a number of mistakes in the original version of this post, although I think all the big ideas are right. I tried to fix everything, but I wouldn't be at all surprised if I've made additional errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure that this is a very hard problem! I think I know how to get started, but I doubt I can finish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll work in Schwarzschild coordinates, with $c=1$, Schwarzschild radius $R$, and $(+---)$ signature, so the metric is&#xA;$$&#xD;&#xA;ds^2=(1-R/r)dt^2-(1-R/r)^{-1}dr^2-r^2(d\theta^2+\sin^2\theta\,d\phi^2).&#xD;&#xA;$$&#xA;All the action can be taken to lie in the equatorial plane $\theta=\pi/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a particle traveling on a geodesic in this geometry, the energy-like conserved quantity (i.e., the one arising from time-translation invariance of the metric) is $u_0$ where $u$ is the 4-velocity. I'll call this quantity $E$ (it's really energy per unit mass):&#xA;$$&#xD;&#xA;E=\left(1-{R\over r}\right)\left(dt\over d\tau\right).&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to &lt;em&gt;assume&lt;/em&gt; that this quantity is conserved even for our particle whizzing along in the tube. I'm pretty sure this is the correct generalization of the assumption that the constraint forces due to the tube do no work. I think you could prove this by looking at the physics in a local inertial reference frame in which the tube is at rest as the particle whizzes by. In that frame, the above energy conservation law is equivalent to the statement that the particle's speed is constant, which follows from a special-relativistic analysis in that frame, because the tube pushes in a direction perpendicular to the velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next we use the fact that the four-velocity has unit norm:&#xA;$$&#xD;&#xA;1=u_\mu u^\mu=(1-R/r)\dot t^2-(1-R/r)^{-1}\dot r^2-r^2\dot\phi^2,&#xD;&#xA;$$&#xA;where dots mean $d/d\tau$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide through by $\dot t^2$:&#xA;$$&#xD;&#xA;\dot t^{-2}=1-{R\over r}-\left(1-{R\over r}\right)^{-1}\left({dr\over dt}\right)^2-r^2\left({d\phi\over dt}\right)^2.&#xD;&#xA;$$&#xA;Express $\dot t$ in terms of the energy, and rearrange:&#xA;$$&#xD;&#xA;\left(1-{R\over r}\right)^{-1}\left(dr\over dt\right)^2+r^2\left(d\phi\over dt\right)^2&#xD;&#xA;=1-{R&#xD;&#xA;\over r}-\left(1-R/r\over E\right)^2.&#xD;&#xA;$$&#xA;Say the particle starts from rest at position $r_0$. Then $E=(1-R/r_0)^{1/2}$. So&#xA;$$&#xD;&#xA;dt=\sqrt{(1-R/r)^{-1}(dr/d\phi)^2+r^2\over 1-R/r-(1-R/r)^2/(1-R/r_0)}\,d\phi.&#xD;&#xA;$$&#xA;If our initial points are $(r_0,0)$ and $(r_0,\alpha$), then the quantity we want to minimize is&#xA;$$&#xD;&#xA;t=\int_0^\alpha \sqrt{(1-R/r)^{-1}(dr/d\phi)^2+r^2\over 1-R/r-(1-R/r)^2/(1-R/r_0)}\,d\phi.&#xD;&#xA;$$&#xA;You can in principle use standard calculus of variations techniques from here to get $r(\phi)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's enough for me! You said in the comments that you'd be happy with just the functional. Are you happy?&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-03-28T22:12:53.773" LastActivityDate="2011-03-28T22:12:53.773" CommentCount="2" />
  <row Id="7722" PostTypeId="1" AcceptedAnswerId="8170" CreationDate="2011-03-27T06:55:36.930" Score="1" ViewCount="1938" Body="&lt;p&gt;SWITL - what could this be? Ultra thin paper with some adhesive that binds to fat? (or as I originally thought, a hoax)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A robotic hand to scoop gel-like substances...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://japantechniche.com/2009/06/09/the-world-first-robotic-hand-scoop-switl/&quot; rel=&quot;nofollow&quot;&gt;http://japantechniche.com/2009/06/09/the-world-first-robotic-hand-scoop-switl/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Truly weird video - looks like some kind of super gluey paper &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://furukawakiko.com/tech/page285.html&quot; rel=&quot;nofollow&quot;&gt;http://furukawakiko.com/tech/page285.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=MQECrcZs6W&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=MQECrcZs6W&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: I did do a patent search. I think it might be this patent I found via &lt;a href=&quot;http://worldwide.espacenet.com/publicationDetails/biblio?CC=JP&amp;amp;NR=2003070411A&amp;amp;KC=A&amp;amp;FT=D&amp;amp;date=20030311&amp;amp;DB=EPODOC&amp;amp;locale=en_EP&quot; rel=&quot;nofollow&quot;&gt;Espacenet&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2795" LastActivityDate="2011-04-23T08:36:28.237" Title="What could make the SWITL materials behave like this?" Tags="&lt;biology&gt;" AnswerCount="4" CommentCount="10" FavoriteCount="2" />
  <row Id="7723" PostTypeId="2" ParentId="7722" CreationDate="2011-03-27T08:10:34.483" Score="1" ViewCount="" Body="&lt;p&gt;Even though it seems quite impressive, it appears to be exactly what the company's site says it is. It is a product in testing intended to transfer dough in assembly line-situations.  Apparently they have taken a demo to food service industry &lt;a href=&quot;http://www.foomajapan.jp/fooma2010/english/exhibition/exhibitor_list.html&quot; rel=&quot;nofollow&quot;&gt;conventions&lt;/a&gt; in Japan for the past few years.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously they are not divulging the specifics of this particular piece of equipment, because it states that a patent is pending. It is reasonable to theorize that the machine is utilizing some property, material, or process that is uniquely suited to moving non-newtonian or high viscosity liquids such as ketchup, mayonaise or dough which all behave differently depending on composition and where and how force is applied to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you have the patience and the time, you could attempt a patent search to find out the specifics. While there is not yet enough information available to determine if it will function as well as it seems to in the ad (and what ever does?) there is no reason to think it is a hoaxed video.      &lt;/p&gt;&#xA;" LastActivityDate="2011-03-27T08:10:34.483" CommentCount="3" />
  <row Id="7724" PostTypeId="1" AcceptedAnswerId="7726" CreationDate="2011-03-28T19:27:03.463" Score="2" ViewCount="187" Body="&lt;p&gt;I'm interested in the size of atoms (extent of the outermost electron cloud), in particular its cross-sectional area and how it scales by increasing Z.  (I trust that this won't be affected much by the number of neutrons?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recognize that for a given atom/ion one can simply look up the data, but I'm interested in a more 'first-principles' kind of view.  The distances of electrons in given orbitals is not constant across different atoms, since as the nucleus gains protons the orbitals tend to get crushed somewhat closer I believe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In fact my final goal will be to determine scattering for charged particles, which is based on much more than the geometry, but I became interested in this particular subproblem apart from that ultimate question.)&lt;/p&gt;&#xA;" OwnerUserId="2818" LastActivityDate="2011-03-29T08:49:55.460" Title="Size of atoms based on atomic number" Tags="&lt;particle-physics&gt;&lt;measurement&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="7725" PostTypeId="2" ParentId="7724" CreationDate="2011-03-28T19:33:04.947" Score="2" ViewCount="" Body="&lt;p&gt;Atomic sizes are surprisingly close to $a_0$ when in the ground states. It is the negative/positive ions that are significantly larger/smaller.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-03-28T19:43:09.090" LastActivityDate="2011-03-28T19:43:09.090" />
  <row Id="7726" PostTypeId="2" ParentId="7724" CreationDate="2011-03-28T19:44:14.343" Score="5" ViewCount="" Body="&lt;p&gt;Atoms have no sharp outer boundaries. All these things and clouds are probabilistic. The electrons always have a nonzero chance to be arbitrarily far from the nucleus, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, the visual size of an atom will depend on the frequency of light one uses to &quot;see&quot; the atom, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that you understand all these disclaimers and you only want some expectation value, and only an estimate of it because no analytic formula exists beyond Hydrogen, here is why all atomic radii are comparable to the Bohr radius $a_0$, independently of $Z$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bohr_radius&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Bohr_radius&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's neglect the electron-electron interactions. In that case, we have $Z$ electrons in the radial electrostatic potential.  The Bohr radius itself is&#xA;$$ a_0 = \frac{4\pi\epsilon_0\hbar^2}{m_e e^2} .$$&#xA;If the nucleus has charge $Ze$, there will be an extra $Z$ in the denominator i.e. $Ze^2$ instead of $e^2$. However, we also need to fill $Z$ electrons in the Hydrogen-like potential. That will fill the shells up to $n=Z^{1/2}$ or so because there are $2n_{\rm max}^2$ of electrons with $n\leq n_{\rm max}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, just like in the Bohr model of the atom that gets the scaling right, the radius of the orbit for the quantum number $n$ - that is somewhat misinterpreted as $l$ in Bohr's model - goes like $n^2$. So if $n\sim Z^{1/2}$, then $n^2\sim Z$, and this enhancement of the radius by $Z$ cancels the reduction of the size caused by the stronger potential induced by the stronger nucleus, as discussed at the beginning of the calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the radius will be close to $a_0$ regardless of the $Z$. The electrons' self-interactions will only change the typical outer shell's radius by &quot;dozens of percent&quot;, to put it verbally. As Vladimir says, negative ions will include more electrons so they will be substantially larger than $a_0$ while the positive ions will remove some electrons, so the positive ions will be smaller than $a_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because various elements differ in electronegativity, the atoms such as Li, Na, K, Rb etc. will be larger - with the extra electron in the new shell - while the atoms such as He, Ne, Ar, Kr, Xe, Rn will be smaller, just having filled the last shell - much like positive ions. See a visual table of the size of atoms here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://intro.chem.okstate.edu/1314F00/Lecture/Chapter7/Lec111300.html&quot; rel=&quot;nofollow&quot;&gt;http://intro.chem.okstate.edu/1314F00/Lecture/Chapter7/Lec111300.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One can see that when things are evaluated, the &quot;covalent radii&quot; still increase with $Z$ a little bit. It's because the central potential, which I assumed to be $Ze$, is actually growing less quickly with $Z$ once the electron-electron interactions are taken into account (because from the outer electron's viewpoint, the inner electrons screen much of the electric charge of the protons in the nucleus).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-29T08:49:55.460" LastActivityDate="2011-03-29T08:49:55.460" />
  <row Id="7727" PostTypeId="1" AcceptedAnswerId="7728" CreationDate="2011-03-28T19:47:48.907" Score="3" ViewCount="113" Body="&lt;p&gt;usually on physical theories one would have Lagrangians or Hamiltonians with multiple fields; say, a vector $A_{\mu}$ and a scalar $\phi$ and one would postulate ad hoc a coupling between the fields which will be used to derive some physical states and then, retroactively support the coupling from a match with the physics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually one reads about 'minimal couplings' and for scalars with vectors on might see a factor like  $\frac{ \partial{\phi}}{ \partial{ x_{\mu}} } A_{\mu}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my question goes on in what general or abstract sense such factors represent a minimal coupling? minimum of what? if is the total degree of powers of the fields? do we have some variational principle for lagrangian couplings that these terms can be called a stable point in such variations?&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-03-29T12:52:31.133" Title="formal framework for talking about 'minimal couplings'" Tags="&lt;lagrangian-mechanics&gt;&lt;hamiltonian-mechanics&gt;&lt;field-theory&gt;&lt;minimal-coupling&gt;" AnswerCount="1" />
  <row Id="7728" PostTypeId="2" ParentId="7727" CreationDate="2011-03-28T19:53:03.763" Score="3" ViewCount="" Body="&lt;p&gt;They're the most renormalizable (or least non-renormalizable) couplings. If the term in the Lagrangian has a coefficient of dimension ${\rm length}^\Delta$, we want to minimize $\Delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equivalently, we want to minimize the mass dimension of the operator. That means reducing a combination of the powers of the fields, as well as the number of derivatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, $(\partial_\mu \phi) A^\mu$ is the minimal coupling between $\phi$ and $A_\mu$ (among the Lorentz-invariant terms) - this bilinear term is usually banned by gauge invariance or conservation laws (symmetries).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dimension of the operator is 3 - one from $\phi$, one mass from $\partial_\mu$, one mass from $A_\mu$. Every factor of $A_\mu$, every new derivative, or every new copy of $\phi$ would increase $\Delta $ by one away from the minimality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The points above are kind of modern - they are related to the Renormalization Group of the 1970s. The minimal couplings are those that are most important at long distances (by power laws) and least problematic (divergent) at short distances. During Einstein's times, the justification of the minimality would be understood just heuristically, based on feelings of beauty etc. That's no longer needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, one could consider non-polynomial or even non-local interactions for which the discussion above would break down - but such terms would almost certainly not be called &quot;minimal&quot;, anyway.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-28T19:53:03.763" CommentCount="6" />
  <row Id="7729" PostTypeId="2" ParentId="7724" CreationDate="2011-03-29T00:19:03.250" Score="2" ViewCount="" Body="&lt;p&gt;Wikipedia has two periodic tables with &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_radius&quot; rel=&quot;nofollow&quot;&gt;atomic radii&lt;/a&gt;, one empirically measured covalent radii and the other theoretically modelled for single atoms. They give different numbers and patterns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the modelled table, atomic radius has a fairly smooth pattern, growing as Period increases and as Group falls (i.e. moving down and left across the table). The reason given for this is that extra shells are larger, but more electrons (and protons) bind more tightly.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the empirical table there is no data for elements which do not have easily measurable covalent compounds, so no figures for noble gases.  For the data it does have, there are peaks near Caesium and Tellurium; non-metals tend to be noticeably smaller.  The patterns are not monotonic: for example Molybdenum is larger than either Chromium or Tungsten while Osmium is smaller that either Rhenium or Iridium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross-sectional area will be proportional to the square of the radius.&lt;/p&gt;&#xA;" OwnerUserId="2714" LastActivityDate="2011-03-29T00:19:03.250" CommentCount="1" />
  <row Id="7730" PostTypeId="1" AcceptedAnswerId="7735" CreationDate="2011-03-29T02:05:40.433" Score="4" ViewCount="291" Body="&lt;p&gt;The curve for a chain hanging between two poles in a uniform gravitational field is known as the catenary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there known an expression for the curve of a hanging chain on a planet of mass M which has a gravitational field of $M/r^2$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the rope have length L, uniform massdensity $\rho$ and negligible thickness.&#xA;The poles are a distance D apart, each a distance R from the planets centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in solutions where the gravitational field may be taken to point straight down aswell as when it point towards the center of the planet.&lt;/p&gt;&#xA;" OwnerUserId="1708" LastEditorUserId="1708" LastEditorDisplayName="" LastEditDate="2011-03-29T04:09:50.337" LastActivityDate="2011-03-30T03:17:46.990" Title="Hanging chain under gravity" Tags="&lt;classical-mechanics&gt;&lt;gravity&gt;&lt;mathematical-physics&gt;&lt;lagrangian-mechanics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="7731" PostTypeId="1" AcceptedAnswerId="7747" CreationDate="2011-03-29T02:59:19.993" Score="1" ViewCount="153" Body="&lt;p&gt;How can we calculate the change in distance over time between (say) two galaxies that they are at rest in respect to each other? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, how do we modify the &quot;classical&quot; equation:&#xA;$$ \ddot{\mathbf{r}} = -G\frac{\mu}{|r|^2}\hat{\mathbf{r}} $$&#xA;to take into account expansion, G changes over time, etc?&lt;/p&gt;&#xA;" OwnerUserId="824" LastEditorUserId="824" LastEditorDisplayName="" LastEditDate="2011-03-29T07:38:47.583" LastActivityDate="2011-03-30T06:45:01.890" Title="Distance in cosmology" Tags="&lt;cosmology&gt;" AnswerCount="3" />
  <row Id="7732" PostTypeId="1" AcceptedAnswerId="7734" CreationDate="2011-03-29T04:58:39.093" Score="2" ViewCount="187" Body="&lt;p&gt;I am interested in theoretical and practical considerations.&lt;/p&gt;&#xA;" OwnerUserId="2823" LastActivityDate="2011-03-29T05:07:13.350" Title="If the Sun were to suddenly become a black hole of the same mass, what would the orbital periods of the planets be?" Tags="&lt;general-relativity&gt;&lt;black-holes&gt;&lt;solar-system&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="7733" PostTypeId="1" CreationDate="2011-03-29T05:00:53.537" Score="3" ViewCount="267" Body="&lt;p&gt;Is it something unexpected?Why universality in cold atomic gases is important?What researches are looking for?Can this be useful for topological quantum computers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we expect a whole myriad of these states?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1012.2698&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1012.2698&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks                          &lt;/p&gt;&#xA;" OwnerUserId="2108" LastActivityDate="2011-08-02T04:36:54.317" Title="What's the relevance of Efimov physics?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="7734" PostTypeId="2" ParentId="7732" CreationDate="2011-03-29T05:07:13.350" Score="3" ViewCount="" Body="&lt;p&gt;It would be exactly the same, (atleast in Newtionian picture), no gravitational fields outside planets radius would change. The easiest way to see this I think is to use the gravitational analogoue of Gauss law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we have spherical symmetry in both cases Int G dA = G*4*pi*r^2 ~ M&lt;br&gt;&#xA;So G is constant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Gauss&lt;/a&gt;'s_law  &lt;/p&gt;&#xA;" OwnerUserId="1708" LastActivityDate="2011-03-29T05:07:13.350" CommentCount="3" />
  <row Id="7735" PostTypeId="2" ParentId="7730" CreationDate="2011-03-29T05:28:10.520" Score="6" ViewCount="" Body="&lt;p&gt;This should be possible to solve in the same way we do the ordinary catenary problem, by variational calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the angular separation between the endpoints is $\Delta$, where we could define $\Delta = \frac{D}{R}$ if I understand the problem correctly. Let the shape of the rope be given by a function $r(\phi)$, and write the potential energy of the rope as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U_G = \int_0^\Delta \frac{GM}{r}\lambda\sqrt{r&amp;#39;^2 + r^2}\mathrm{d}\phi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $\lambda$ is the linear mass density of the rope, and I've left the $\phi$ dependence of $r(\phi)$ implicit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also need a constraint that enforces the total length of the rope remain fixed at $L$. According to the method of Lagrange multipliers, we'll have to add a term&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U_L = \int_0^\Delta l\sqrt{r&amp;#39;^2 + r^2}\mathrm{d}\phi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for a total of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U = \int_0^\Delta \biggl(\frac{GM}{r}\lambda + l\biggr)\sqrt{r&amp;#39;^2 + r^2}\mathrm{d}\phi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we define $T$ to be the integrand,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T(r, r&amp;#39;, \phi) = \biggl(\frac{GM}{r}\lambda + l\biggr)\sqrt{r&amp;#39;^2 + r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this can be solved by the Euler-Lagrange equation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}}{\mathrm{d}\phi}\frac{\partial T}{\partial r&amp;#39;} = \frac{\partial T}{\partial r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(r&amp;#39;&amp;#39;r - r&amp;#39;^2)(GM\lambda + lr) = lr(r&amp;#39;^2 + r^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There does appear to be an analytic solution to this equation, but Mathematica makes it look rather icky to say the least... I'll have to come back to this and try to find some sensible solution and/or physical interpretation since I'm out of time right now.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-03-29T05:28:10.520" CommentCount="6" />
  <row Id="7736" PostTypeId="2" ParentId="7731" CreationDate="2011-03-29T05:33:44.990" Score="2" ViewCount="" Body="&lt;p&gt;If the two objects were the only particles in the Universe, it would obviously not be our Universe which has something like $10^{80-90}$ particles (this number counting mostly CMB photons) in its visible portion. Because it would be a different Universe, you would have to explain its rules of the game, especially the cosmological constant, scales of inflation, etc. The answer to your question obviously depends on all these things. There's no universal concept of &quot;the Universe&quot; in which you may change absolutely everything but you still expect the events to be unique. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, in any Universe with matter species, it is impossible for the Universe to only contain 2 particles after the inflation because the kinetic energy from the rolling inflaton field is always converted to gadzillions of particles when inflation is ending.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you meant our Universe, with all the extra matter in it, and you're just asking how the distance between two objects has been changing, well, it's been changing proportionally to $a(t)$ that enters the Friedmann equations. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Friedmann_equations&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want to compare the present and a particular moment in the past, you need to know what the moment is. Even if you say that it is the end of inflation, it depends on the energy scale where inflation occurred and we don't exactly know that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could calculate the answer for a GUT-scale inflation but the vagueness of your question isn't a good argument that such a calculation would be more than a waste of time. At any rate, such ratios of distances will be of the form &quot;ten to a few dozens&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-29T05:33:44.990" CommentCount="2" />
  <row Id="7737" PostTypeId="1" CreationDate="2011-03-29T06:34:39.683" Score="5" ViewCount="234" Body="&lt;p&gt;I am going to learn some math about functionALs (like functional derivative, functional integration, functional Fourier transform). Just looking forward to any good introductory text for this topic. Any idea will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="749" LastEditorUserId="749" LastEditorDisplayName="" LastEditDate="2011-03-30T05:40:22.180" LastActivityDate="2011-03-30T11:47:49.547" Title="Introductory texts for calculus of Functionals" Tags="&lt;quantum-field-theory&gt;&lt;mathematical-physics&gt;&lt;books&gt;" AnswerCount="1" CommentCount="13" />
  <row Id="7738" PostTypeId="1" AcceptedAnswerId="7740" CreationDate="2011-03-29T08:49:29.793" Score="2" ViewCount="359" Body="&lt;p&gt;At first I thought it's because of Coriolis, but then someone told me that at the bathtub scale that's not the predominant force in this phenomenon.&lt;/p&gt;&#xA;" OwnerUserId="2826" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-29T11:23:14.830" LastActivityDate="2011-03-31T17:31:54.320" Title="Why there's a whirl when you drain the bathtub?" Tags="&lt;fluid-dynamics&gt;&lt;coriolis-force&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="1" />
  <row Id="7739" PostTypeId="2" ParentId="7671" CreationDate="2011-03-29T08:58:13.730" Score="0" ViewCount="" Body="&lt;p&gt;I remember that during the mathematical physics course at the university, the Green functions method seemed to me &lt;em&gt;exactly&lt;/em&gt; like the method of images. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that you have to solve the equation for Green function:&#xA;$$\Delta \phi(\vec{r}-\vec{r}&amp;#39;) = -2\pi \delta(\vec{r}-\vec{r}&amp;#39;) $$ &lt;strong&gt;With corresponding boundary conditions.&lt;/strong&gt; And to account for the boundary conditions you have to do  virtually the same method-of-images-stuff.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is some reference I found, that supports my memory: &lt;a href=&quot;http://ocw.mit.edu/courses/mathematics/18-303-linear-partial-differential-equations-fall-2006/lecture-notes/greensfn.pdf&quot; rel=&quot;nofollow&quot;&gt;http://ocw.mit.edu/courses/mathematics/18-303-linear-partial-differential-equations-fall-2006/lecture-notes/greensfn.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-03-29T08:58:13.730" />
  <row Id="7740" PostTypeId="2" ParentId="7738" CreationDate="2011-03-29T10:09:36.087" Score="4" ViewCount="" Body="&lt;p&gt;The whirl is due to the net angular momentum the water has before it starts draining, which is pretty much random.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the circulation were due to Coriolis forces, the water would always drain in the same direction, but I did the experiment with my sink just now and observed the water to spin different directions on different trials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Coriolis force is proportional to the velocity of the water and the angular velocity of Earth.  Earth's angular velocity is $2\pi/24\ {\rm hours}$, or about $10^{-4}\ s^{-1}$.  If water's velocity as it drains is $v$ the Coriolis acceleration is about $10^{-4} v\ s^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water moves about a meter while draining, which takes a time $1\ m/v$, so the total velocity imparted by Coriolis forces could be at most $10^{-4} v\ s^{-1} * 1\ m/v = 10^{-4} \ m/s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the Coriolis effect is quite a small effect.  But this first-order Coriolis effect does not cause the water to rotate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The direction of Coriolis force depends on your direction of motion.  All the water in your tub is moving the same direction, so the Coriolis force pushes it all the same direction.  The effect is that if the bathtub starts out perfectly flat and begins draining (and it points north), all the water will get pushed east.  The two edges of the tub will have very slightly different depths of water, because the Coriolis force is pushing sideways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Coriolis force could create &quot;spinning&quot; on uniformly-moving water, but only as a second-order effect.  As you move away from the equator, the Coriolis force changes.  This change in the Coriolis force is because the angle between &quot;north&quot; and the angular velocity vector of Earth changes as you move around; as you go further north (in the Northern Hemisphere) the &quot;north&quot; direction gets closer and closer to making a right angle with the angular velocity vector, so the Coriolis force increases in strength.  The size of this effect would be proportional to the ratio of the size of your tub to the radius of Earth.  That ratio is $10^{-7}$, so this effect is completely negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Coriolis force could also create some &quot;spinning&quot; if different parts of the water are moving different speeds.  If the tub is draining to the north in the northern hemisphere, and water near the drain is moving faster than water far away, then the water near the drain would be pushed east more than water far away is.  If you subtracted out the average effect of the Coriolis force, what remained would be an easterly push near the drain and a westerly push far away.  This gives a clockwise spin as viewed from above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've already estimated the typical velocities as $\omega L$, so the angular momentum per unit mass induced this way would be on the order of $\omega L^2$ (but maybe smaller by a factor of 10).  That's only $10^{-4}\ m^2/s$.  To get an equivalent effect, in a tub of $100\ L$, you could give just one liter of water on the edge of the pool a velocity of a few cm/s, something you surely do many time over when removing your body from the tub.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This effect is too small to affect your bathtub, but it's still observable under the right conditions.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Coriolis_effect#Draining_in_bathtubs_and_toilets&quot; rel=&quot;nofollow&quot;&gt;According to Wikipedia&lt;/a&gt;, Otto Tumlirz conducted several experiments in the early 20th century that demonstrated the effects of the Coriolis forces on a draining tub of water.  The tub was allowed to settle for 24 hours in a controlled environment before the experiment began.  This was enough to damp out the residual angular momentum left over from filling the tub up to the point where Coriolis effects were dominant.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-03-29T10:09:36.087" CommentCount="3" />
  <row Id="7741" PostTypeId="2" ParentId="7738" CreationDate="2011-03-29T10:19:12.717" Score="0" ViewCount="" Body="&lt;p&gt;The main effect is angular momentum (rotational inertia) in the water set up by various movements before you start observing, such as getting out of your bath. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This results in the water level being lower near the centre of rotation than further away, setting up centripital forces which maintain the rotation. When the difference in levels is significant relative to the average water level, you notice the typical whirl effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other things happening too, including the Coriolis force.&lt;/p&gt;&#xA;" OwnerUserId="2714" LastActivityDate="2011-03-29T10:19:12.717" />
  <row Id="7742" PostTypeId="2" ParentId="7738" CreationDate="2011-03-29T10:35:19.420" Score="1" ViewCount="" Body="&lt;p&gt;A discussion by 'The Straight Dope' website&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.straightdope.com/columns/read/149/do-bathtubs-drain-counterclockwise-in-the-northern-hemisphere&quot; rel=&quot;nofollow&quot;&gt;http://www.straightdope.com/columns/read/149/do-bathtubs-drain-counterclockwise-in-the-northern-hemisphere&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;references experimental work carried out but Ascher Schapiro in 1962, which concluded something like it all depends on the shape of container and how its stirred before being left to empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is Schapiro's paper but I feel you will need academic access via a university or library to read the full PDF:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://journals.cambridge.org/action/displayAbstract?fromPage=online&amp;amp;aid=368912&quot; rel=&quot;nofollow&quot;&gt;http://journals.cambridge.org/action/displayAbstract?fromPage=online&amp;amp;aid=368912&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-03-29T10:35:19.420" CommentCount="2" />
  <row Id="7743" PostTypeId="1" AcceptedAnswerId="7744" CreationDate="2011-03-29T12:04:33.200" Score="2" ViewCount="132" Body="&lt;p&gt;If I have an ocean with constant temp 10 degrees celcius and I throw in an 8 gram iron ball that's 50 degrees celcius, I'd like to find how to calculate the amount of energy that flows into the water before equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about doing this?&lt;/p&gt;&#xA;" OwnerUserId="2829" LastActivityDate="2011-03-29T12:14:23.563" Title="Cooling a ball of Iron" Tags="&lt;heat&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="7744" PostTypeId="2" ParentId="7743" CreationDate="2011-03-29T12:14:23.563" Score="4" ViewCount="" Body="&lt;p&gt;In order to lower the temperature of the iron, it will have to transfer heat to the water.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat transferred per degree of cooling (per gram) is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_capacity&quot; rel=&quot;nofollow&quot;&gt;heat capacity&lt;/a&gt; of the iron.  This number is about 0.45 J/(gK) (same link).  In general, the heat capacity can depend on the temperature, but for many solids it is pretty much constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the total heat transferred, multiply the mass, heat capacity, and temperature change.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-03-29T12:14:23.563" />
  <row Id="7746" PostTypeId="1" CreationDate="2011-03-29T14:22:41.663" Score="2" ViewCount="194" Body="&lt;p&gt;There are many stories about radioactivity and the relative danger of it in the news lately, but very little actual information. The radioactivity levels around Fukushima Daiichi are high, but seem negligible in just somewhat removed locations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real danger seems to stem from ingesting radioactive particles. Just how likely is it for that to happen in any considerable distance from the reactor, say in Tokyo, and how dangerous for the human body is it really? How far can these particles travel in any dangerous concentration?&lt;/p&gt;&#xA;" OwnerUserId="2830" LastActivityDate="2011-03-31T13:12:38.773" Title="What needs to happen for one to ingest radioactive particles and how likely is this?" Tags="&lt;radioactivity&gt;&lt;particles&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="7747" PostTypeId="2" ParentId="7731" CreationDate="2011-03-29T14:57:41.530" Score="3" ViewCount="" Body="&lt;p&gt;The answer depends on the exact details of the physical model you're considering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest model for the expanding Universe is a homogeneous one, in which the density is uniform in space (but varying in time, of course). If we imagine the two galaxies as &quot;test particles&quot; of negligible mass, living in such a Universe, then it's possible to work out the paths of both particles through space. If they are &quot;at rest&quot; with respect to each other at some initial time (meaning, to be precise, that their physical distance, measured at fixed cosmic time, is initially unchanging), then, roughly speaking, they start to fall towards each other if the Universe is dominated by &quot;ordinary&quot; matter (which pulls), and they start to fall away from each other if the Universe is dominated by &quot;dark energy&quot; (which pushes). &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0104349&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; gives details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it looks to me like you want to consider a more complicated model, in which the galaxies themselves have non-negligible mass. That is, I think you're imagining that the density of the Universe is not uniform but has &quot;lumps&quot; at the locations of the galaxies. In that case, life gets complicated -- indeed, you can make it pretty much as complicated as you like! There's a large industry in astrophysics in simulating many-body gravitational systems like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's one case I know of that can be treated analytically and that might give you a feel for how the answer goes. Suppose that the Universe has uniform density, except for &lt;em&gt;one&lt;/em&gt; large galaxy, which can be treated as an extra pointlike &quot;lump&quot; of mass. Suppose that that galaxy is at rest in comoving coordinates (i.e., it's going with the flow of the expansion). Suppose that the other galaxy is a test particle of negligible mass. Finally, suppose that it's not too far away -- to be precise, its distance is small compared to $c/H$ where $H$ is the Hubble constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that situation, the equation of motion of the test galaxy can be found from a roughly Newtonian analysis. The force on the test galaxy is determined by the gravitation of all of the stuff in a sphere centered on the big galaxy. To be precise, it's&#xA;$$&#xD;&#xA;\ddot r = -{GM_{\rm eff}\over r^2},&#xD;&#xA;$$&#xA;where $r$ is the physical distance of the test galaxy (not the comoving distance), and the &quot;effective mass&quot; is&#xA;$$&#xD;&#xA;M_{\rm eff}=M+(\rho+3p/c^2){4\over 3}\pi r^3.&#xD;&#xA;$$&#xA;Here $M$ is the mass of the big central galaxy, and $\rho,p$ are the average  density and pressure. Without the $p$ term, this would be exactly the Newtonian result: $M_{\rm eff}$ would be the total mass in the sphere. The only general-relativistic correction is that pressure also gravitates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a Universe with lots of &quot;dark energy,&quot; as ours appears to be, the pressure is negative and contributes more strongly than $\rho$, by the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for a bit of self-promotion, but here it is: John Baez and I wrote an &lt;a href=&quot;http://math.ucr.edu/home/baez/einstein/&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; for the American Journal of Physics explaining Einstein's equation in relatively simple terms, and one of the main examples we worked out was the dynamics of the expanding Universe. If you want to know how to get from the exact general-relativity treatment to the quasi-Newtonian picture, you might want to check it out.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-29T14:57:41.530" CommentCount="1" />
  <row Id="7748" PostTypeId="6" CreationDate="2011-03-29T15:00:24.103" Score="0" ViewCount="" Body="&lt;p&gt;I have been one of the pro tem moderators and I would like to continue my mission as a regular mod.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I certainly admit that not everything is going smooth and well now and it is also my fault, but I hope you'll give me a chance for redemption ;-) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sort-of-a-my-programme:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I believe it can be and I think it should be a many-level, broad and rich site&lt;/li&gt;&#xA;&lt;li&gt;I think that discursive questions should be killed with fire and redirected to chat (probably we need not-constructive close reason as on Programmers.SE)&lt;/li&gt;&#xA;&lt;li&gt;I consider limited level of teasing others normal to intelligent people&lt;/li&gt;&#xA;&lt;li&gt;I hope to finally implement chat journal club here&lt;/li&gt;&#xA;&lt;li&gt;I'll be resolving flags and fixing typography as usual&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm from CET zone.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-04-04T15:45:40.933" LastActivityDate="2011-04-04T15:45:40.933" CommunityOwnedDate="2011-03-29T15:00:24.103" CommentCount="5" />
  <row Id="7751" PostTypeId="2" ParentId="7707" CreationDate="2011-03-29T15:24:03.947" Score="1" ViewCount="" Body="&lt;p&gt;I think that you are dealing with classical electrodynamics and I will answer within this domain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/4862/interference-of-polarized-light&quot;&gt;In Kostya answer in PSE&lt;/a&gt; about interference look at the equations of parallel polarization that I copy here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Total field: $\vec{E} =&#xD;&#xA;&amp;gt; \vec{i}E_0\left(\cos\omega&#xD;&#xA;&amp;gt; t+\cos(\omega t+\Delta)\right)$.&lt;br&gt;&#xA;  Intensity: $I_\parallel\sim&#xD;&#xA;&amp;gt; E_0^2\langle\cos^2\omega t+2\cos\omega&#xD;&#xA;&amp;gt; t\cos(\omega t+\Delta)+\cos^2(\omega&#xD;&#xA;&amp;gt; t+\Delta)\rangle=E_0^2(1+\cos\Delta)$,&#xA;  which nicely depends on the phase&#xA;  shift between the waves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When $\Delta= \pi$ we got $\vec{E} = \vec{0}$ and $I = 0$&lt;/strong&gt;. If the antennas are headed on the Poyinting vector also cancels.&lt;br&gt;&#xA;Two waves and zero net field. And this means trouble as I will point out later and justifies the naive words of Bykov.&#xA;This image from &lt;a href=&quot;http://web.sbu.edu/chemistry/wier/electrons/young.html&quot; rel=&quot;nofollow&quot;&gt;sbu.edu&lt;/a&gt;  &lt;img src=&quot;http://i.stack.imgur.com/xZpyv.jpg&quot; alt=&quot;constructive and destructive interference of light waves&quot;&gt; we see in the image at left a constructive pattern and at right a destructive pattern. This game to 'shape the field' in space is played with antenna arrays always.  What means the  obvious diference in the intensities? &lt;strong&gt;Physically one radiator is moved apart from the other half wavelength&lt;/strong&gt;, and the feed (current intensity, frequency,phase) of both radiators was kept invariant. It appears that when the field cancels one (strange property) must say to both radiators: Stop radiating! Or as Bykov '&lt;strong&gt;they &lt;em&gt;feel&lt;/em&gt; the wave from the other source&lt;/strong&gt;'. But this is complete nonsence.&lt;br&gt;&#xA;Think of this: Does the electrons in the radiators have sensor to all space to say : radiate in this direction and distance; stop radiating because the other radiator is changing position; then add N radiators,... We can use light originated in the stars and do they play this game?&lt;br&gt;&#xA;&lt;strong&gt;There is no theory to cover Bykov words&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, as you say, 'the conservation of energy'. When I studied EM field (is light), radiation, antennas, I never used the concept of 'photons' as particles, but only the 'EM field'. We can cancel the field, as equations show, but can not cancel the 'particles' and lost the energy.&#xA;A recent experiment on &lt;strong&gt;&lt;a href=&quot;http://www.eng.yale.edu/caolab/papers/Sci11b.pdf&quot; rel=&quot;nofollow&quot;&gt;'anti-laser'&lt;/a&gt;&lt;/strong&gt; motivated me &lt;a href=&quot;http://physics.stackexchange.com/questions/5743/anti-laser-how-sure-we-are-that-energy-is-transported&quot;&gt;to ask a question&lt;/a&gt; that is  similar to your's. The experiment is a head-on photon-photon cancelation consistent with the above EM equations. &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-03-29T15:24:03.947" CommentCount="1" />
  <row Id="7752" PostTypeId="2" ParentId="7738" CreationDate="2011-03-29T15:27:26.923" Score="-5" ViewCount="" Body="&lt;p&gt;It's because of the Coriolius effet. In the Northen hemisphere, it goes around one direction, the Southern hemisphere around the opposite direction, and goes straight down bang on the equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black natives demostrate the effect to White tourists in this clip:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water flow at the equator, Coriolis effect&#xA;&lt;a href=&quot;http://www.youtube.com/watch?v=ZU1EB0z-h7w&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=ZU1EB0z-h7w&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-03-29T15:27:26.923" CommentCount="7" />
  <row Id="7753" PostTypeId="2" ParentId="7746" CreationDate="2011-03-29T15:27:50.640" Score="0" ViewCount="" Body="&lt;p&gt;How far they travel depends mostly on weather. How much you ingest also depends on chemistry - you are more likely to ingest something that is water soluble then a heavy metal atom.&#xA;Then how much damage it does depends on biochemistry, a heavy metal atom that doesn't have any biological role may just go straight through you while something like iodine or strontium that are used in the body will be absorbed and may stay in you long enough to do some damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember a few particles don't do much harm, we can detect radioactivity at amazingly low concentrations, down to single atoms, so simply detecting radioactivity on the other side of the pacific doesn't mean much from a health perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most dangerous radioisotope for health is Radon (in terms of killing most people/year). It's a gas that comes out of the ground and gets trapped in houses, and because it's a gas you breathe it in so it's easily ingested. &lt;/p&gt;&#xA;" OwnerUserId="2525" LastEditorUserId="2525" LastEditorDisplayName="" LastEditDate="2011-03-29T16:03:15.617" LastActivityDate="2011-03-29T16:03:15.617" CommentCount="7" />
  <row Id="7755" PostTypeId="2" ParentId="7700" CreationDate="2011-03-29T15:37:26.493" Score="0" ViewCount="" Body="&lt;p&gt;This is a good question because I think physicists nowadays don't understand the difference between form invariant and covariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations of physics are form invariant under a Lorentz transformation, but they're not co-variant as in they don't vary &lt;em&gt;with&lt;/em&gt; the Lorentz transformation.&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-03-29T15:37:26.493" />
  <row Id="7756" PostTypeId="2" ParentId="817" CreationDate="2011-03-29T16:22:14.550" Score="1" ViewCount="" Body="&lt;p&gt;There is a recent paper about Noether theoreom on discrete systems which i found pretty interesting, i thought to share;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1103.4785&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1103.4785&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-03-29T16:22:14.550" CommentCount="2" />
  <row Id="7757" PostTypeId="1" AcceptedAnswerId="7766" CreationDate="2011-03-29T16:32:44.433" Score="3" ViewCount="73" Body="&lt;p&gt;on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalence_principle#Tests_of_the_weak_equivalence_principle&quot; rel=&quot;nofollow&quot;&gt;wikipedia article about the equivalence principle&lt;/a&gt; there is a mention about testing the EP against parity-violating masses;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The equivalence principle is untested&#xA;  against opposite geometric parity&#xA;  (chirality in all directions) mass&#xA;  distributions. A parity Eötvös&#xA;  experiment contrasting solid single&#xA;  crystal spheres of identical&#xA;  composition α-quartz in enantiomorphic&#xA;  space groups P3121 (right-handed screw&#xA;  axis) versus P3221 (left-handed screw&#xA;  axis) is appropriate. Equivalence&#xA;  principle parity violation validates a&#xA;  chiral vacuum background forbidden&#xA;  within general relativity but allowed&#xA;  within Einstein-Cartan theory; affine,&#xA;  teleparallel, and noncommutative&#xA;  gravitation theories.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't understand any of this, is this correct? why the crystal layout has anything to do with a parity-violating fields? (AFAIK electro-weak force is the only known force to be parity-violating, but this force is short range) so why the alignment of the by-comparison-macroscopic crystal atomic layout does matter? is this crackpot physics or i just don't understand it?&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-29T16:42:04.803" LastActivityDate="2011-03-29T21:38:03.693" Title="Weak equivalence principle tests" Tags="&lt;equivalence-principle&gt;&lt;parity&gt;&lt;chirality&gt;" AnswerCount="1" />
  <row Id="7758" PostTypeId="2" ParentId="7738" CreationDate="2011-03-29T18:09:28.177" Score="1" ViewCount="" Body="&lt;p&gt;The whirl happens in the draining tube, whose optimal solution to drain the bathtub is a laminar flow allowing for some rotation in the tube. What you see in the surface is the match between the solution of flow in the tube and the solution of flow in the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Angular momentum of the flow gets modified a lot as the tube twists and twists, sometimes even siphoning up and down. &lt;/p&gt;&#xA;" OwnerUserId="1335" LastActivityDate="2011-03-29T18:09:28.177" />
  <row Id="7760" PostTypeId="1" CreationDate="2011-03-29T18:52:42.657" Score="1" ViewCount="165" Body="&lt;p&gt;If you take apart a syringe you are left with one piece that is a tube with a tiny hole in one end and the open end of a tube at the other. If you then submerge this underwater and cover the small hole, then lift out of the water, the water remains in the tube/syringe until you release your finger from the small hole. Why is that?&lt;/p&gt;&#xA;" OwnerUserId="2840" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-29T19:52:01.550" LastActivityDate="2011-03-29T19:52:01.550" Title="Tube under water and air pressure" Tags="&lt;water&gt;&lt;pressure&gt;&lt;air&gt;" AnswerCount="2" />
  <row Id="7761" PostTypeId="2" ParentId="7760" CreationDate="2011-03-29T19:06:05.843" Score="0" ViewCount="" Body="&lt;p&gt;I assume the small hole is at the top and the open end is down?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weight of 10 miles of air above you is pressing down and then pressing up on the surface of the water. This air weighs around 10tons / square meter!&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-03-29T19:06:05.843" />
  <row Id="7762" PostTypeId="2" ParentId="7760" CreationDate="2011-03-29T19:06:30.480" Score="1" ViewCount="" Body="&lt;p&gt;This is because, when you close the end with the tiny hole in it, you prevent air from exerting any pressure on the water column. Thus the air pressure from the open end on the other side prevents it from falling. Once you take your hand away, you let air get into the syringe as a result of which the water falls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting thing happens when you close the tiny hole as the water is flowing down. The water doesn't stop instantaneously after you close the hole. It flows down a little more and stops. This is because even after you close the hole, the air pressure above the water is same as the air pressure below it. But soon, because of the flowing down water, the air column above expands resulting in a pressure drop stopping the water.&lt;/p&gt;&#xA;" OwnerUserId="966" LastActivityDate="2011-03-29T19:06:30.480" CommentCount="1" />
  <row Id="7763" PostTypeId="2" ParentId="7746" CreationDate="2011-03-29T19:14:04.530" Score="2" ViewCount="" Body="&lt;p&gt;I would like to add that radioactive isotopes are a constant background in all natural environments, and more so in stone  and concrete buildings. The relevant question is how much more than the natural background is the artificial radiation induced by human activities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a recent &lt;a href=&quot;http://www.bbc.co.uk/news/world-12860842&quot; rel=&quot;nofollow&quot;&gt;viewpoint in BBC news&lt;/a&gt;  professor Alison gives relevant numbers. For example, the human body has about 50 becquerel per kilogram in the natural state. An interesting chart that puts radiation in perspective is &lt;a href=&quot;http://xkcd.com/radiation/&quot; rel=&quot;nofollow&quot;&gt;here,&lt;/a&gt; and this is &lt;a href=&quot;http://people.reed.edu/~emcmanis/radiation.html&quot; rel=&quot;nofollow&quot;&gt;a graph&lt;/a&gt; from the scientist who provided the numbers for the chart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when you see animated maps of how the radiation is spread from Japan, read the scales. You will see that the numbers are within natural variations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The immediate danger of death come for huge doses, look at the chart. Even for people next to the reactors there has been no such exposure.  There are long term effects of ingesting or breathing in isotopes for the people in the region, which have to do with a larger cancer rate over twenty or thirty years. The rest of the world is nowhere close to such levels of exposure.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-29T19:14:04.530" />
  <row Id="7764" PostTypeId="1" CreationDate="2011-03-29T19:14:48.580" Score="2" ViewCount="157" Body="&lt;p&gt;Suppose you have two thermodynamical systems $X_1$ and $X_2$ (for example water and air) with different temperatures ($T_1 &gt; T_2$). Now put them into thermal contact. Is there a formula which describes, how $T_1$ and $T_2$ changes with time?&lt;/p&gt;&#xA;" OwnerUserId="2740" LastActivityDate="2011-03-31T09:30:25.477" Title="time dependence of temperature equalization" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="7765" PostTypeId="2" ParentId="7764" CreationDate="2011-03-29T20:29:18.597" Score="3" ViewCount="" Body="&lt;p&gt;In a simple heat conduction theory (no convection) the corresponding time is determined with the lowest eigenvalue of the Sturm-Liouville problem: $\Delta T(t) = A\cdot e^{-t/\tau}$ (regular regime), see, for example, my &lt;strong&gt;&lt;a href=&quot;http://arxiv.org/abs/0906.3504&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: For any time $t &amp;gt; 0$ the temperature difference is a series like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta T(t) = \sum_{n=0}^\infty A_n \cdot e^{-t/\tau_n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tau_{n+1} &amp;lt; \tau_n$, for example, $\tau_n \propto \frac{1}{\pi^2(n+1)^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $t &amp;gt;&amp;gt; \tau_1$, the fast decaying exponentials $e^{-t/\tau_n}$ are small with respect to the slowest one $e^{-t/\tau_0}$ so only one term remains in this sum. The latter regime is called a regular regime of the heat exchange. For certain one-layer 1D systems $\tau \propto \frac{\rho c L^2}{\kappa \pi^2(n+1)^2}$ Here $\kappa$  is the heat conductivity, $\rho$  is the material density, $c$  is the specific heat capacity, and $L$  is the layer thickness (exact $n$-dependence  is determined with the boundary conditions).&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-03-31T09:30:25.477" LastActivityDate="2011-03-31T09:30:25.477" CommentCount="6" />
  <row Id="7766" PostTypeId="2" ParentId="7757" CreationDate="2011-03-29T20:41:36.840" Score="1" ViewCount="" Body="&lt;p&gt;Dear lurscher, the quote is the kind of C-physics described by the C-word which is a favorite word of mine but is discouraged on this server, so I won't use it - but you have used it. You don't misunderstand anything - quite on the contrary, you're right on the money.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These comments about a non-existent test of parity in the equivalence principle are due to &quot;Uncle Al&quot; Schwartz, see  e.g. the first comment under&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.iac.es/galeria/masc/Outreach_files/Dark-energy_particle_spotted_NatureNews290609.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.iac.es/galeria/masc/Outreach_files/Dark-energy_particle_spotted_NatureNews290609.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to see that even extremely microscopic details about the &quot;idea&quot; coincide with the paragraph you quoted. &quot;Uncle Al&quot; is known to most physics bloggers - as well as many other physics forums.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a physics viewpoint, the claims are completely preposterous. First of all, one needs pretty special objects to get a parity-violating physics in 4 spacetime dimensions: either chiral (Weyl) spinors or self-dual 2-forms (and their interactions). None of those things is included in GR; Einstein-Cartan theory; conventional affine, teleparallel, and noncommutative gravitational theories - which is why &lt;em&gt;all&lt;/em&gt; these gravitational theories automatically preserve parity. Moreover, self-dual 2-forms don't exist in 3+1 dimensions (although they do exist in 4+0 dimensions) because $*^2=-1$ in 3+1 dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you correctly say, only the weak interactions as expressed by the electroweak theory violate parity which is because the gauge field interacts with chiral (Weyl) spinors. The gauge fields $A_\mu$, when coupled to spinors, get naturally multiplied by $\gamma^\mu$ which switches the chirality (because it anticommutes with $\gamma_5$); on the other hand, $g_{\mu\nu}$ with an even number of indices preserves the chirality, so the spinors interactions with the metric have to be parity-preserving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, the existing tests of the equivalence principle de facto eliminate the possibility of substantially different interactions of left-handed and right-handed helixes, too. For example, Newton's apple contains vitamin C (L-ascorbic acid) which is parity-asymmetric, and it would accelerate differently if the laws of gravity cared about the mirror images. There's really no way compatible with an effective field theory how to make the two mirror objects accelerate differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some semi-serious paper by authors from well-known places who talk about gravitational parity violation, see e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/arXiv:1005.3310&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/arXiv:1005.3310&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the papers mentioned by it (by Stephon Alexander and others). It would still be impossible for me to say that any scenario in those papers is defensible but at least those papers don't sound as a self-evident nonsense of the kind that &quot;Uncle Al&quot; Schwartz has managed to incorporate into the Wikipedia page.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-29T21:38:03.693" LastActivityDate="2011-03-29T21:38:03.693" CommentCount="4" />
  <row Id="7767" PostTypeId="1" AcceptedAnswerId="7768" CreationDate="2011-03-29T21:40:53.167" Score="7" ViewCount="108" Body="&lt;p&gt;What does it mean to &quot;weakly gauge&quot; a global symmetry in a gauge theory?  I have seen this term used in a number of papers, but have not seen it defined.&lt;/p&gt;&#xA;" OwnerUserId="2841" LastActivityDate="2011-03-29T22:07:54.470" Title="Weakly gauge a symmetry?" Tags="&lt;quantum-field-theory&gt;" AnswerCount="1" />
  <row Id="7768" PostTypeId="2" ParentId="7767" CreationDate="2011-03-29T21:54:19.653" Score="5" ViewCount="" Body="&lt;p&gt;it is a favorite construction by model-builders. To weakly gauge a global (continuous) symmetry $G$ means to add a gauge field $A_\mu$ in the adjoint of $G$, its kinetic term $-{\rm Tr} F_{\mu\nu}F^{\mu\nu}/2$, and the interaction potential term in the Lagrangian&#xA;$$L_{\rm int} = g\int {\rm Tr} (j^\mu A_\mu)d^{d-1} x$$&#xA;where $j^\mu$ is the Noether current of the global symmetry - or its supersymmetric generalizations, depending on the amount of SUSY that is being expected  - for example ${\mathcal L}_{\rm int} = 2g \int d^4\theta  {\mathcal J}\, {\mathcal V}$ for ${\mathcal N}=1$ SUSY. A new gauge symmetry is added in this way: it is created out of the original global symmetry. That's where the word &quot;gauged&quot;, describing the transformation of the character of the symmetry, comes from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The adverb &quot;weakly&quot; means that it is assumed that $g\ll 1$. Because of this assumption, it is hard to observe the new force mediated by the new gauge field (it is very weak), and the new gauge symmetry is, to a large extent, physically indistinguishable from the original global symmetry. As an example, imagine that the baryon number, or at least $B-L$, may be &quot;weakly gauged&quot; in this way; the coupling constant has to be vastly smaller than one, otherwise the theory would predict unobserved new forces between baryons and leptons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just a warning: it is unnatural, marginally incompatible with ideas about the &quot;grand unification&quot;, and possibly downright inconsistent with quantum gravity, to have gauge groups with excessively tiny values of $g$ (relatively to masses of charged objects expressed as multiples of the Planck mass); for the claim about quantum gravity problems, see the &quot;weak gravity conjecture&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0601001&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0601001&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-29T22:07:54.470" LastActivityDate="2011-03-29T22:07:54.470" CommentCount="7" />
  <row Id="7769" PostTypeId="1" AcceptedAnswerId="7788" CreationDate="2011-03-29T22:46:22.787" Score="7" ViewCount="142" Body="&lt;p&gt;In discussion of LHC searches for SUSY particles, physicists seem to assume they will decay quickly to the lightest SUSY particle which then remains stable (at least within the time it takes to leave the detector).  As an example: &lt;a href=&quot;http://physics.aps.org/articles/v4/27&quot; rel=&quot;nofollow&quot;&gt;http://physics.aps.org/articles/v4/27&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stability of the lightest SUSY particle is desired as it provides an explanation of dark matter.  This stability is usually given by postulating R-parity, the conservation of super-symmetric particles (kind of like lepton, or baryon number).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully my understanding up to this point is mostly correct. My question is then in two parts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we assume R-parity, then wouldn't the lightest electrically charged SUSY particle also be stable? However we don't see charged dark matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if it can decay, into the lightest SUSY particle + some standard model particles to get rid of the charge, AND theorists expect this decay to be so very rapid that only the decay products will be visible to detectors (so &quot;high cross section&quot; for decay reaction), THEN wouldn't the reverse reaction also have to have a huge cross section?  That is, if colliding  (SM stuff) + (dark matter) --&gt; (other SM stuff) + (charged SUSY particle), has such a HUGE cross section, and dark matter is everywhere, then we don't need the energy to create a SUSY particle to see SUSY particle, we only need roughly the energy &lt;em&gt;difference&lt;/em&gt; between lightest SUSY and lightest charged SUSY particle.  In which case, if the cross section is really that big, it seems strange we haven't already seen this.&lt;/p&gt;&#xA;" OwnerUserId="2655" LastEditorUserId="2655" LastEditorDisplayName="" LastEditDate="2011-03-30T03:50:17.093" LastActivityDate="2011-03-30T07:25:12.243" Title="Decay of SUSY particles" Tags="&lt;particle-physics&gt;&lt;supersymmetry&gt;&lt;large-hadron-collider&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" />
  <row Id="7770" PostTypeId="1" AcceptedAnswerId="7790" CreationDate="2011-03-29T23:08:52.663" Score="6" ViewCount="89" Body="&lt;p&gt;In 2006 the IAU deemed that Pluto was no longer a planet because it fails to &quot;clear&quot; the neighborhood around its Kuiper Belt orbit. Presumably, this is because Pluto (1.305E22 kg) has insufficient mass to do the job. How massive must a body in Pluto's orbit (semi-major axis 39.5 AU) be to &quot;clear&quot; its orbit? Would Mars (6.24E23 kg) or Earth (5.97E24 kg) be declared planets if they were in Pluto's orbit?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-03-30T07:38:01.630" Title="Mass Needed to Clear an Orbital Neighborhood" Tags="&lt;gravity&gt;&lt;astronomy&gt;&lt;newtonian-mechanics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="7771" PostTypeId="1" AcceptedAnswerId="7773" CreationDate="2011-03-30T01:10:12.553" Score="8" ViewCount="389" Body="&lt;p&gt;I am speaking about GR with classical fields and energy. One question, spread over three increasingly strict situations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an energy density limit in GR? (literally, can the energy density have an arbitrarily large value at some point in space at some point in time)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an energy density limit beyond which a blackhole will always form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's choose a small volume, for here I'll just choose the Planck volume.  Is there an average energy density limit over this volume beyond which a blackhole will always form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarification:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In light of &lt;a href=&quot;http://en.wikipedia.org/wiki/Mass_in_general_relativity&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mass_in_general_relativity&lt;/a&gt; , can those that are answering that the energy density is limited and referring to a mass $M$ in some equations please specifically state how you are defining the $M$ in terms of the energy density, or defining $M$ in terms of $T^{\mu\nu}$ the stress-energy tensor.  Does your $M$ depend on coordinate system choice?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, reading some comments, it sounds like there is confusion on what energy density means.  Based on wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/File:StressEnergyTensor.svg&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/File:StressEnergyTensor.svg&lt;/a&gt; , it sounds like we can consider energy density = $T^{00}$ of the stress-energy tensor.  If you feel this is not correct terminology, please explain and I'll edit the question if necessary.&lt;/p&gt;&#xA;" OwnerUserId="2655" LastEditorUserId="2655" LastEditorDisplayName="" LastEditDate="2011-03-31T07:05:35.057" LastActivityDate="2011-04-06T20:16:31.767" Title="Is there an energy density limit in GR?" Tags="&lt;general-relativity&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" />
  <row Id="7773" PostTypeId="2" ParentId="7771" CreationDate="2011-03-30T02:22:41.070" Score="9" ViewCount="" Body="&lt;p&gt;The answer is NO.  There is no energy density limit (for all three questions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest way to see this is that the energy density is just the $T^{00}$ component of the stress energy tensor.  The solution in GR depends on the &lt;em&gt;full&lt;/em&gt; stress energy tensor, so it is not enough to just talk about the energy density.  Furthermore, because the energy density is just a component of a tensor, it is a coordinate system dependent quantity.  So starting from a solution that doesn't become a blackhole, and has some energy somewhere, we can always choose the coordinate system to make the energy density arbitrarily large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More clearly stated: Local Lorentz symmetry alone is enough to show that the energy density is not limited in GR.  And furthermore since there exist non-zero energy solutions that don't become blackholes, this also answers your second question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make the answer to the third question more clear, let's discuss an exact solution.  Consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann-Lema%C3%AEtre-Robertson-Walker_metric&quot; rel=&quot;nofollow&quot;&gt;Robertson-Walker&lt;/a&gt; solution with a perfect fluid.  Here's an example stress energy tensor for a perfect fluid in the comoving frame:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T^{ab} =\left( \begin{matrix} \rho &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; p &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; p &amp;amp; 0 \\               0 &amp;amp; 0 &amp;amp; 0 &amp;amp; p \end{matrix} \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we change to a different coordinate system, using the coordinate transformation:&#xA;$\Lambda^{\mu}{}_{\nu} =\left( \begin{matrix} \gamma &amp;amp;-\beta \gamma &amp;amp; 0 &amp;amp; 0 \\ -\beta \gamma&amp;amp;\gamma&amp;amp;0&amp;amp;0\\ 0&amp;amp;0&amp;amp;1&amp;amp;0\\ 0&amp;amp;0&amp;amp;0&amp;amp;1\\ \end{matrix} \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see the energy density will transform as:&#xA;$\rho&amp;#39; = \gamma^2 \rho + p \beta^2 \gamma^2 = \gamma^2 (\rho + p \beta^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So not only can the energy density be arbitrarily large, but even over a finite volume.&lt;/p&gt;&#xA;" OwnerUserId="2652" LastEditorUserId="2652" LastEditorDisplayName="" LastEditDate="2011-03-30T23:00:11.887" LastActivityDate="2011-03-30T23:00:11.887" CommentCount="19" />
  <row Id="7774" PostTypeId="2" ParentId="7733" CreationDate="2011-03-30T02:37:48.937" Score="2" ViewCount="" Body="&lt;p&gt;Efimov physics is interesting because it's a neat bit of fundamental physics describing something we have a tendency to neglect: 3-body quantum scattering problems. As for being unexpected -- I'm not sure what you want; if Efimov states were expected then V.N. Efimov wouldn't have had to discover them (already more than 40 years ago!). The experimental verification was not unexpected, but did have to wait for the right tools to emerge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which brings me to my next point -- I've seen the paper that you linked to; I don't have the ability to judge its merits as a topology paper, but it is &lt;em&gt;not&lt;/em&gt; a physics paper. If you are interested in learning about Efimov &lt;em&gt;physics,&lt;/em&gt; then &lt;a href=&quot;http://physics.aps.org/articles/v3/9&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; is a gentle introduction with solid references if you are serious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While Efimov states share an interesting &quot;feature&quot; of three-body bound states even when no two of the bosons would be bound (analogous to the Borromean rings) this has nothing to do with nontrivial topological properties that would be used for quantum computation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; It's also worth noting, in retrospect, that there exists an analogous phenomenon in 4-body physics: recently it was noticed that for every Efimov trimer, there are two corresponding tetramers. I think some people conjecture that there are similar any $N&amp;gt;2$-body Efimov states, which seems to me still rather different from the logic of the paper in the original question.&lt;/p&gt;&#xA;" OwnerUserId="791" LastEditorUserId="791" LastEditorDisplayName="" LastEditDate="2011-03-31T00:22:52.243" LastActivityDate="2011-03-31T00:22:52.243" />
  <row Id="7775" PostTypeId="2" ParentId="7730" CreationDate="2011-03-30T03:17:46.990" Score="1" ViewCount="" Body="&lt;p&gt;We have had a related problem with the brachistochrone problem in Schwarzschild coordinates here.  This problem here has struck me as rather odd, for this seems to refer to geodesics, where to me the .  However, it has occurred to me there might be a problem here of some relevancy.  Suppose we have a string in some orbit around a black hole.  We then impose some condition that this string has end points attached to a fixed surface of some kind --- a Dp-brane?  This string is then in an orbit around the black hole and will assume some configuration in the curved spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to set this up we start with the line element, which gives the unit 4-velocity,&#xA;$$&#xD;&#xA;1~=~\Big(1~-~\frac{R}{r}\Big) \Big(\frac{dt}{ds}\Big)^2~-~ \Big(1~-~\frac{R}{r}\Big)^{-1} \Big(\frac{dr}{ds}\Big)^2~-~r^2\Big(\frac{d\theta}{ds}\Big)^2~-~r^2sin^2\theta\Big(\frac{d\phi}{ds}\Big)^2,&#xD;&#xA;$$&#xA;where $R~=~2GM/r$.  We identify the $dt/ds$ with the energy $E$ and do some algebra to obtain an equation of motion&#xA;$$&#xD;&#xA;\Big(\frac{dr}{ds}\Big)^2~=~\frac{E}{m}~-~1~-~V(r)&#xD;&#xA;$$&#xA;for the potential function&#xA;$$&#xD;&#xA;V(r)~\simeq~\frac{1}{2}\Big(\frac{R}{r}~-~\frac{L^2}{2mr^2}~+~\frac{R}{mr^2}\Big)&#xD;&#xA;$$&#xA;The $L$ is the angular momentum of the object in orbit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now use this in the potential for the computation of the shape of the string&#xA;$$&#xD;&#xA;U_G~=~\int_0^\Delta V(r)\lambda\sqrt{r&amp;#39;^2~+~r^2}d\phi&#xD;&#xA;$$&#xA;$\Delta~=~D/r$ and $r’~=~dr/d\phi$ and employ the Lagrange multiplier $U_L$.  The Euler-Lagrange equations are applied to the&#xA;$$&#xD;&#xA;{\cal L}~=~(V(r)~+~L)\sqrt{r’^2~+~r^2}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is somewhat artificial for it assumes the string endpoints are held apart by some fixed distance.  This could be relaxed by treating the endpoints are particle geodesic or orbits.  In that case the angular separation of the fixed points varies and the problem is time dependent.  This might be an interesting tool to use in some problems with strings and black holes.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-30T03:17:46.990" />
  <row Id="7776" PostTypeId="2" ParentId="7771" CreationDate="2011-03-30T03:22:32.760" Score="-2" ViewCount="" Body="&lt;p&gt;The Schwarzschild radius is $R~=~2GM/c^2$, where if you pack a mass $M$ into a volume with a radius $R$ you get a black hole.  The term mass-energy limit is not standard language usage, but if you push sufficient mass into a volume it will becomes a black hole and causally sealed off from the outside world.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-30T03:22:32.760" CommentCount="3" />
  <row Id="7777" PostTypeId="1" AcceptedAnswerId="7792" CreationDate="2011-03-30T03:28:12.727" Score="5" ViewCount="584" Body="&lt;p&gt;This is not a question pertaining to interpretations, after the last one I realized I should not open Pandora's Box ;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For theories to be consistent, they must reduced to known laws in the classical domains. The classical domain can be summed up as: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hbar=0 ; c=\infty$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is OK. I need to know, however, is that if QM is an independent and fundamental &lt;em&gt;theory&lt;/em&gt;, why does it rely so heavily on the classical formalism. Is it necessary for a classical formalism to exist in order to have a quantum formalism? From as far as I have read, it does not seem so, and I find this puzzling. Suppose you have a dissipative system, or an open system when you cannot write an autonomous Hamiltonian in the classical case, how then do we approach these quantum mechanically, when we cannot even write down the corresponding Hamiltonian. &lt;/p&gt;&#xA;" OwnerUserId="2599" LastActivityDate="2011-03-31T23:30:17.233" Title="Why is a classical formalism necessary for quantum mechanics?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="8" CommentCount="3" FavoriteCount="5" />
  <row Id="7778" PostTypeId="1" CreationDate="2011-03-30T03:33:01.337" Score="5" ViewCount="163" Body="&lt;p&gt;Here's a question inspired by &lt;a href=&quot;http://physics.stackexchange.com/questions/7771/is-there-an-energy-density-limit-in-gr/7773#7773&quot;&gt;Edward's answer&lt;/a&gt; to &lt;a href=&quot;http://physics.stackexchange.com/questions/7771/is-there-an-energy-density-limit-in-gr&quot;&gt;this question&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's my understanding that the average energy density of a black hole in its rest frame is $\rho_\text{BH}(A)$, a function of surface area. I calculated $3c^2/2GA$ for a Schwarzschild black hole, but that's presumably not applicable here since I'm talking about an extended energy distribution. Anyway, suppose you are in a space filled with some sort of energy, matter, or whatever, which produces a potentially time-dependent stress-energy tensor. And further suppose that there is some finite, spherical region of surface area $A$ in this space, over which you measure the average energy density to be $\rho_\text{BH}(A)$, the net charge to be zero, and the total angular momentum of the matter within the region to be zero. (I'm assuming there is a measurement procedure available which can be carried out without entering the region, if it matters.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, Edward's argument in the other question shows that there are at least two ways to produce that (average) energy density:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The energy density arises from a fluid or other material at rest, such that the region is a black hole and there are no timelike geodesics exiting it&lt;/li&gt;&#xA;&lt;li&gt;The energy density has been &quot;augmented&quot; by a Lorentz boost from some other frame, implying that there are timelike geodesics exiting the region&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is it possible, in general, to distinguish between case 1 and case 2 by only looking at the other components of the stress energy tensor, without actually calculating the geodesics? If so, how? What would be the &quot;signature&quot; of a black hole in the components of $T^{\mu\nu}$ (or rather, their averages over the region in question)?&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-03-31T00:02:37.467" LastActivityDate="2011-03-31T00:02:37.467" Title="How can you tell if a critical energy density is actually a black hole?" Tags="&lt;general-relativity&gt;&lt;energy&gt;&lt;black-holes&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="7779" PostTypeId="2" ParentId="7777" CreationDate="2011-03-30T03:36:01.600" Score="1" ViewCount="" Body="&lt;p&gt;This is an excellent question and the answer is it follows from the crucial postulates of quantum mechanics. Namely, the operator corresponding to a dynamical variable is obtained by replacing the classical canonical variables by corresponding quantum mechanical operators and any pair of canonically conjugate operators will satisfy the Heisenberg commutation rules. Physically speaking, it has definite relevance to these remarks of the great Neils Bohr, &quot;However far the phenomena transcend the scope of classical physical explanation, the account of all the evidence must be expressed in classical terms&quot;. This is because &quot;by the word 'experiment' we refer to a situation where we can tell others what we have done and what we have learned&quot; so that &quot;the account of the experimental arrangement and of the results of the observations must be expressed in unambiguous language with suitable application of the terminology of the classical physics&quot;. (Note that for some observables like spin there is no classical analog and we take different approach.)&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-03-30T03:36:01.600" />
  <row Id="7780" PostTypeId="1" CreationDate="2011-03-30T03:48:44.963" Score="6" ViewCount="224" Body="&lt;p&gt;To maintain the surface tension which formed our original bubble (in order to keep the bubble from breaking), we may change the temperature/pressure of air on both sides of the bubble varyingly, with different rates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is such a configuration possible? Examples of any such system?&lt;/p&gt;&#xA;" OwnerUserId="2844" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-03-31T07:24:02.053" LastActivityDate="2011-03-31T08:57:46.050" Title="Can a water bubble be frozen into ice bubble?" Tags="&lt;thermodynamics&gt;" AnswerCount="2" />
  <row Id="7781" PostTypeId="1" AcceptedAnswerId="7798" CreationDate="2011-03-30T03:49:47.037" Score="3" ViewCount="273" Body="&lt;p&gt;Whenever I read about the curvature of spacetime as an explanation for gravity, I see pictures of a sheet (spacetime) with various masses indenting the sheet to form &quot;gravity wells.&quot; Objects which are gravitationally attracted are said to roll down the curved sheet of spacetime into the gravity well. This is troubling to me, because, in-order for objects on the locally slanted spacetime sheet to accelerate, gravity must be assumed. Therefore I ask; does the explanation of gravity as the curvature of spacetime assume gravity? If yes, what is the point of the theory? If No, what am I missing?&lt;/p&gt;&#xA;" OwnerUserId="2843" LastEditorUserId="2843" LastEditorDisplayName="" LastEditDate="2011-03-30T03:55:29.890" LastActivityDate="2011-04-03T18:20:31.187" Title="Does the curvature of spacetime theory assume gravity?" Tags="&lt;gravity&gt;&lt;spacetime&gt;&lt;theory&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="7782" PostTypeId="2" ParentId="7777" CreationDate="2011-03-30T03:52:29.263" Score="0" ViewCount="" Body="&lt;p&gt;As far as I can see, this &lt;EM&gt;is&lt;/EM&gt; an interpretational issue. The testable part of quantum mechanics is a probability calculus. It serves to assign probabilities to possible outcomes of measurements on the basis of actual outcomes of other measurements. In other words, it concerns correlations between measurement outcomes. The rest (for instance, claims about what happens &lt;em&gt;between&lt;/em&gt; measurements) is &quot;not even wrong&quot; in Wolf­gang Pauli’s felic­i­tous phrase. In other words, it is a matter of interpretation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The piv­otal role played by mea­sure­ments in all stan­dard for­mu­la­tions of quantum mechanics is an embar­rass­ment known as the “mea­sure­ment problem.” As an anony­mous ref­eree for a phi­los­ophy of sci­ence journal once put it to me, “to solve this problem means to design an inter­pre­ta­tion in which mea­sure­ment processes are not dif­ferent in prin­ciple from ordi­nary phys­ical inter­ac­tions.” Yet how can reducing mea­sure­ments to “ordi­nary phys­ical inter­ac­tions” solve this problem, con­sid­ering that quantum mechanics describes “ordi­nary phys­ical inter­ac­tions” in terms of cor­re­la­tions between the pos­sible out­comes of mea­sure­ments per­formed on the inter­acting sys­tems? This kind of “solu­tion” merely sweeps the problem under the rug. The way I see it, to solve the mea­sure­ment problem means to design an inter­pre­ta­tion in which the cen­tral role played by mea­sure­ments in all stan­dard axiom­a­ti­za­tions of quantum mechanics is under­stood. (For my two cents, see the links on &lt;A HREF=&quot;http://thisquantumworld.com/wp/?page_id=197&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/A&gt; at my personal website.&lt;/p&gt;&#xA;" OwnerUserId="2784" LastEditorUserId="2784" LastEditorDisplayName="" LastEditDate="2011-03-31T02:14:57.897" LastActivityDate="2011-03-31T02:14:57.897" CommentCount="3" />
  <row Id="7783" PostTypeId="2" ParentId="7777" CreationDate="2011-03-30T04:12:27.977" Score="4" ViewCount="" Body="&lt;p&gt;Note that all the experiments we did prior to discovering quantum weirdness are still out there. Your complete theory has to be able to explain all the demos we do in a introductory freshman mechanics class. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you can show that a new theory evolves to the old one in the realm where the old one was correct, you &lt;em&gt;don't&lt;/em&gt; have to re-prove everything you already knew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can build the new theory on the footing of the old one, your halfway to that point already.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-03-30T04:12:27.977" />
  <row Id="7784" PostTypeId="1" CreationDate="2011-03-30T04:24:07.677" Score="2" ViewCount="292" Body="&lt;p&gt;I don't know how to ask this more clearly than in the title.&lt;/p&gt;&#xA;" OwnerUserId="2843" LastActivityDate="2011-08-30T07:36:08.040" Title="How does the curvature of spacetime induce gravitational attraction?" Tags="&lt;gravity&gt;&lt;spacetime&gt;&lt;theory&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="7786" PostTypeId="2" ParentId="7731" CreationDate="2011-03-30T06:34:31.103" Score="1" ViewCount="" Body="&lt;p&gt;In line with what professor Bunn said, the galaxies being &quot;at rest&quot; with each other implies that either both of them (if their masses are similar) or one of them (if its mass is much smaller than the other's) is &lt;strong&gt;not&lt;/strong&gt; at rest with respect to the local comoving background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A metric commonly used to model a central gravitational mass in a flat FLRW spacetime with no accretion onto the central object is the McVittie metric, whose meaning and characteristics are still undergoing discussion. See Faraoni &amp;amp; Jacques (2007) &quot;Cosmological expansion and local physics&quot; &lt;a href=&quot;http://arxiv.org/abs/0707.1350&quot; rel=&quot;nofollow&quot;&gt;arXiv:0707.1350&lt;/a&gt; and Carrera &amp;amp; Giulini (2008) &quot;Influence of global cosmological expansion on local dynamics and kinematics&quot; &lt;a href=&quot;http://arxiv.org/abs/0810.2712&quot; rel=&quot;nofollow&quot;&gt;arXiv:0810.2712&lt;/a&gt;. To note, if the FLRW spacetime is de Sitter (just cosmological constant with no matter), the McVittie solution is equivalent, through an appropriate coordinate transformation, to the Schwarzschild–de Sitter (Kottler) metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In words of the abstract of the second paper: &quot;However, the general problem of gaining a qualitative and quantitative understanding of how the cosmological dynamics influences local systems remains challenging, with only partial clues being so far provided by exact solutions to the field equations of General Relativity.&quot;&lt;/p&gt;&#xA;" OwnerUserId="2085" LastEditorUserId="2085" LastEditorDisplayName="" LastEditDate="2011-03-30T06:45:01.890" LastActivityDate="2011-03-30T06:45:01.890" />
  <row Id="7787" PostTypeId="2" ParentId="7781" CreationDate="2011-03-30T07:08:11.770" Score="3" ViewCount="" Body="&lt;p&gt;Those sheets with dips having stars at their centers illustrate gravity by illustrating that freely falling objects move along geodesics, and that geodesics are curved by the dips in a way that looks as if freely falling objects were attracted by the stars. (Needless to say, the 2-dimensional sheets remain poor substitutes for the real McCoy, which is a (3+1)-dimensional pseudo-Riemannian manifold.)&lt;/p&gt;&#xA;" OwnerUserId="2784" LastEditorUserId="2710" LastEditorDisplayName="" LastEditDate="2011-03-30T12:23:29.717" LastActivityDate="2011-03-30T12:23:29.717" />
  <row Id="7788" PostTypeId="2" ParentId="7769" CreationDate="2011-03-30T07:18:57.100" Score="4" ViewCount="" Body="&lt;p&gt;Dear John, your basic logic is valid but the way you &quot;estimate&quot; the cross section for the LSP production by the time reversal of the decay is way too emotional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction by which you expect charged dark matter to be created everywhere, namely&#xA;$${\rm SM stuff} + {\rm LSP}_{\rm from\,dark\,matter} \to {\rm charged\, superpartners}$$&#xA;is indeed possible and it is essentially what the underground detectors of dark matter are trying to find. However, this reaction is extremely rare because dark matter, while everywhere, is very diluted - less than half a GeV per cubic centimeter&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/0907.0018&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0907.0018&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With this - relatively to nuclear densities - tiny mass density (even smaller than the cosmological constant), it takes some time to detect the collisions with the dark matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also misapply the time reversal. While the laws of physics produce CPT-invariant - and approximately T-invariant - Lorentz-invariant amplitudes, the actual probabilities of the opposite processes contain different kinematic factors: that's implied by the time asymmetry of mathematical logic, an elementary fact that - as I found out - a huge number of people misunderstand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the decay of the charged superpartner to the LSP and a charged standard model particle is described by the &quot;decay rate&quot; while the opposite process is described by a &quot;cross section&quot;. The cross section and the decay rate don't even have the same dimension (units). So it's obvious that they can't be &quot;equal&quot;. And saying that one of them is &quot;small&quot; if the other is &quot;small&quot; is only true if you properly &quot;normalize&quot; what you mean by the vague adjective &quot;small&quot;. You apparently didn't do it right. The sentence is surely not right if &quot;small&quot; is meant as &quot;small for purposes I find practical&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;decay rate&quot; and the &quot;cross section&quot; of the two processes related by the time reversal differ by kinematic factors comparable to powers of the total energy of a particle in the process. If the decay rate (width) is e.g. $1 {\rm GeV}$, which makes the particle decay almost immediately in the detector, and if its mass is $100 {\rm GeV}$, the cross section, the corresponding cross section might be comparable to $1 {\rm GeV}/ (100 {\rm GeV})^3$ which is a tiny cross section. It was made tiny by the huge denominator, and the denominator is huge because the mass of the superpartner is high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two processes differing by &quot;time reversal&quot; have different prior probabilities. There is simply no &quot;simple&quot; way to relate the opposite probabilities. In a similar way, if glass &quot;certainly&quot; breaks down if it falls to the floor, it doesn't mean that it's likely that the pieces will &quot;often&quot; collect themselves and create a perfect unbroken glass. The opposite process is suppressed, relatively to the first one, by $\exp(-\Delta S)$, the exponential of the entropy difference between the initial and final state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even for microscopic processes, the relevant &quot;probabilities&quot; of the process have - aside from the shared $|{\mathcal M}|^2$, the squared Lorentz-invariant amplitude, lots of kinematic factors - namely $1/2E$ from both initial and final particles (which is symmetric with respect to the time reversal); and the integration over $d^3 p$ of final particles (not initial ones, which is time-reversal-asymmetric). So expecting that the probabilities of the time-reversed evolutions are &quot;equal&quot; is a trivial logical fallacy. They're not equal - that's not what T-invariance or CPT-invariance mean. &lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-30T07:25:12.243" LastActivityDate="2011-03-30T07:25:12.243" CommentCount="1" />
  <row Id="7789" PostTypeId="2" ParentId="7784" CreationDate="2011-03-30T07:24:50.560" Score="6" ViewCount="" Body="&lt;p&gt;The simple answer is that the curvature of spacetime does not induce gravitational attraction. It &lt;em&gt;describes&lt;/em&gt; it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all of classical physics (not only general relativity), the motion of bodies can be described as motion along the geodesics of a suitable differential geometry. Body A acts on body B by influencing the curvature of the geodesic along which body B moves. See the illustrations on &lt;a href=&quot;http://thisquantumworld.com/wp/?page_id=1035&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt; of my personal website for both classical electrodynamics and general relativity.&lt;/p&gt;&#xA;" OwnerUserId="2784" LastEditorUserId="2784" LastEditorDisplayName="" LastEditDate="2011-03-31T02:49:13.153" LastActivityDate="2011-03-31T02:49:13.153" CommentCount="5" />
  <row Id="7790" PostTypeId="2" ParentId="7770" CreationDate="2011-03-30T07:38:01.630" Score="4" ViewCount="" Body="&lt;p&gt;Dear Michael, there is an article dedicated to this criterion:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Clearing_the_neighbourhood&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Clearing_the_neighbourhood&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The relevant quantity is &#xA;$$\Lambda / \Lambda_E = \frac{M^2 / P}{M_E^2 / P_E} $$&#xA;where $M$ is the mass of the would-be planet and $P$ is the orbital period. The subscript $E$ means that the value is for the Earth. The definition above guarantees the ratio is $1$ for the Earth and $1.08$ for its evil sister, Venus. Uranus and Neptune are close with 2.51 and 1.79, respectively. It's 8,500 and 300 for Jupiter and Saturn, 0.01 and 0.006 for Mercury and Mars. Pluto, much like Ceres, Eris, Makemake, and Haumea have figures between $10^{-9}$ and $10^{-7}$ - too small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;threshold&quot; was chosen to be $1/153,000$ (by Stern and Levison in 2000, I don't understand where the threshold came from - probably some model of how the planets are clearing the neighborhood as a function of $\Lambda$) so the non-planets are beneath it. All the explanations should be in this paper:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.boulder.swri.edu/~hal/PDF/planet_def.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.boulder.swri.edu/~hal/PDF/planet_def.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-30T07:38:01.630" CommentCount="2" />
  <row Id="7791" PostTypeId="6" CreationDate="2011-03-30T07:50:47.123" Score="0" ViewCount="" Body="&lt;p&gt;While I'm not very active in terms of &lt;a href=&quot;http://physics.stackexchange.com/questions/629/why-does-one-experience-a-short-pull-in-the-wrong-direction-when-a-vehicle-stops&quot; title=&quot;just to name&quot;&gt;questions&lt;/a&gt;/&lt;a href=&quot;http://physics.stackexchange.com/questions/197/how-are-forces-mediated/200#200&quot; title=&quot;my two favorites&quot;&gt;answers&lt;/a&gt; on the main site, I did collect quite some experience over at &lt;a href=&quot;http://meta.stackoverflow.com/users/146482/tobias-kienzler&quot;&gt;meta.SO&lt;/a&gt; concerning the SE mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to a lack of time I unfortunately had to turn down my &lt;a href=&quot;http://meta.physics.stackexchange.com/questions/100/nominate-and-elect-our-temporary-moderators/111#111&quot;&gt;pro-tempore nomination&lt;/a&gt;, but since I'm regularly &lt;a href=&quot;http://xkcd.com/303/&quot; rel=&quot;nofollow&quot; title=&quot;XKCD: Compling&quot;&gt;waiting&lt;/a&gt; for simulations to finish I still spend some time here acting in the background doing some &lt;a href=&quot;http://blog.stackoverflow.com/2009/05/a-theory-of-moderation/&quot; rel=&quot;nofollow&quot;&gt;janitor work&lt;/a&gt;. Since I originally &lt;a href=&quot;http://area51.stackexchange.com/proposals/1908/physics&quot;&gt;proposed&lt;/a&gt; this site I obviously want to take good care of it, too.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I live in Germany, so (until &lt;a href=&quot;http://en.wikipedia.org/wiki/Daylight_saving_time&quot; rel=&quot;nofollow&quot; title=&quot;Daylight saving time&quot;&gt;DST&lt;/a&gt; ends) the time is UTC +2 (see also &lt;a href=&quot;http://en.wikipedia.org/wiki/Time_in_Germany&quot; rel=&quot;nofollow&quot; title=&quot;Wikipedia&quot;&gt;Time in Germany&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-04-04T08:39:54.583" LastActivityDate="2011-04-04T08:39:54.583" CommunityOwnedDate="2011-03-30T07:50:47.123" CommentCount="4" />
  <row Id="7792" PostTypeId="2" ParentId="7777" CreationDate="2011-03-30T07:50:59.317" Score="16" ViewCount="" Body="&lt;p&gt;Quantum mechanics and quantum mechanical theories are totally independent of the classical ones. The classical theories may appear and often do appear as limits of the quantum theories. This is the case of all &quot;textbook&quot; theories - because the classical limit was known before the full quantum theory, and the quantum theory was actually &quot;guessed&quot; by adding the hats to the classical one. In a category of cases, the full quantum theory may be &quot;reverse engineered&quot; from the classical limit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, one must realize that this situation is just an artifact of the history of physics on Earth and it is not generally true. There are classical theories that can't be quantized - e.g. field theories with gauge anomalies - and there are quantum theories that have no classical limits - e.g. the six-dimensional $(2,0)$ superconformal field theory in the unbroken phase. Moreover, it's typical that the quantum versions of classical theories lead to new ordering ambiguities (the identity of all $O(\hbar^k)$ terms in the Hamiltonian is undetermined by the classical limit in which all choices of this form vanish, anyway), divergences, and new parameters and renormalization of them that have to be applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the predictions of quantum mechanics don't need any classical crutches. Quantum mechanics works independently of its classical limits, and the classical behavior may be deduced from quantum mechanics and nothing else in the required limit. Historically, people discussed quantum mechanics as a tool to describe the microscopic world only, assuming that the large objects followed the classical logic. The Copenhagen folks divided the world in these two subworlds, in an ad hoc way, and that simplified their reasoning because they didn't need to study quantum physics of the macroscopic measurement devices etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But these days, we fully understand the actual physical mechanism - decoherence - that is responsible for the emergence of the classical logic in the right limits. Because of decoherence, which is a mechanism that only depends on the rules of quantum mechqnics, we know that quantum mechanics applies to small as well as large objects, to all objects in the world, and the classical behavior is an approximate consequence, an emergent law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To know the evolution in time, one needs to know the Hamiltonian - or something equivalent that determines the dynamics. The previous sentence is true both in classical physics and quantum mechanics, for similar reasons, but independently. If a classical theory is a limit of a quantum theory, it of course also means that its classical Hamiltonian may be derived as a limit of the quantum Hamiltonian. Of course, if you don't know the Hamiltonian operator, you won't be able to determine the dynamics and evolution with time. Guessing the quantum Hamiltonian from its classical limit is one frequent, but in no way &quot;universally inevitable&quot;, way to find a quantum Hamiltonian of a quantum theory.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-30T16:57:02.387" LastActivityDate="2011-03-30T16:57:02.387" CommentCount="15" />
  <row Id="7793" PostTypeId="2" ParentId="7784" CreationDate="2011-03-30T08:12:11.847" Score="1" ViewCount="" Body="&lt;p&gt;If You are in flat spacetime and move from A to B, the easiest way is to travel by fragment of an straight line between points A and B. When there is gravitation, spacetime is bending. So the easiest way is to follow the line which is the shortest, but it is not straight line, because of the curvature. So another choice is the line which is shortest but remain within spacetime, and is &quot;minimal&quot;. And this is exactly physical trajectory of small particle in such curved space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As gravitation mass of a body and inertial mass of a body are exactly equal, there is equivalence between such geometrical description (when curvature is dependent of gravitational mass) with dynamical one which uses forces to describe particle movement ( because inertial mass is equal to gravitational one - so the two pictures are perfectly equivalent). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when You go from point A towards point B you follow no stright line but more complicated curve such as parabola or ellipse -  because it is has &quot;minimal feature&quot; among accessible trajectories. Of course You may run some kind of engine, and run on different - non minimal trajectory - for example straight line. This cost You some bill for fuel, and is not minimal at all...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As planets, meteorites, comets, thrown rocks etc. do not have its own engines, they have to move along &quot;minimal&quot; trajectories.. As &quot;minimal&quot; lines in curved spacetime are exactly trajectories of body within gravitational fields, this two pictures describes the same move with different language. Of course attraction has its causes in curvature when You use geometrical language - many trajectories goes close and close each other. So You may say - gravitation is represented by curvature.&lt;/p&gt;&#xA;" OwnerUserId="1012" LastActivityDate="2011-03-30T08:12:11.847" />
  <row Id="7794" PostTypeId="2" ParentId="7777" CreationDate="2011-03-30T08:43:28.620" Score="2" ViewCount="" Body="&lt;p&gt;I want to quote for you F.A. Berezin's point of view regarding this question as given in his seminal papers&#xA;in &quot;General concept of quantization&quot; Commun. math. Phys. 40, 153-174(1975) and &quot;Quantization&quot; Math. USSR Izv. 8 (1974) 1109-1165.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In these papers, Berezin presented his theory of quantization having the correspondence principle as one of its postulates (at $\hbar \rightarrow 0$). The quantum and classical theories describe the same mechanical reality; the quantum description of reality is more detailed than the classical one, but once we have this detail, we have the freedom of giving up on some of it. &#xA;Then we should reach the experimentally well established classical limit. This is like the case of one having a telescope but intentionally defocusing its lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should also be mentioned that this interpretation is consistent with the nonfunctoriality point of view of (first) quantization, because (as emphasized in Berezin's articles) a lot of different detailed pictures of the reality may correspond to the same &quot;defocused&quot; limit of the classical theory.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Update&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Referring to Peter's comment. I should clarify that the quantum theory converges at the limit ($\hbar \rightarrow 0$) &#xA;to a classical theory only for a subset of the states and observables of the system. For example Weyl operators on coherent&#xA;states have a well defined limit. A detailed discussion of the classical limit (and why other states do not have a well defined limit) &#xA;may be found in the  following &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0506082&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; by N.P. Landsman: &quot;Between classical and quantum&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I think that Berezin's view as a general principle is still valid as a basis of our understanding the quantum phenomena.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorUserId="2190" LastEditorDisplayName="" LastEditDate="2011-03-30T16:12:49.027" LastActivityDate="2011-03-30T16:12:49.027" CommentCount="3" />
  <row Id="7796" PostTypeId="2" ParentId="7778" CreationDate="2011-03-30T11:35:46.460" Score="3" ViewCount="" Body="&lt;p&gt;David,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see from Lubos's comment there is a serious flaw in the formulation of this question, so I will use this answer simply to explain some aspects of the topic. First the flaw:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the energy density of a Schwarzchild black hole&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Schwarzchild solution is a solution of the Vacuum Einstein equations ie $R^{u,v}=0$ and so $T^{u,v}=0$ ie the Stress Energy Tensor is zero throughout the Schwarzchild solution (removing the origin from this manifold avoids the undefinedness there). With zero stress-energy there are no fluids or local energy densities to measure or examine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is meant in the question (as it arises from the earlier Stack questions) has some relationships with the Hoop conjecture as Edward points out, but just for clarification I shall add more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us not consider any more an Einstein vacuum and assume that some form of matter (or radiation - I will just say matter below) must have been present &quot;originally&quot;. This matter is part of a &lt;em&gt;non-vacuum&lt;/em&gt; solution of the Einstein equations, and so there will be a corresponding &lt;em&gt;non-zero&lt;/em&gt; Stress-Energy Tensor whose matter is destined to become the Black Hole. So now the question begins to make sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is really about what determines whether a given non-vacuum solution of Einstein's equations forms a Black Hole and whether this fact is measurable locally. The sentence that asks this is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What would be the &quot;signature&quot; of the Black Hole in the components of $T^{u,v}$?$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I dont believe that this answer is known, partly because the space of all solutions of Einstein's equations is not yet known. If one considers the points made below, one might also conclude that GR alone was inadequate to predict the BH formation - matter properties are central too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are the classical Hawking-Penrose theorems which give a topological-geometric answer to this question by positing the existence of &quot;closed trapped surfaces&quot;, along with certain properties of $T^{u,v}$. In that sense there is an answer, but it doesnt tell us &lt;em&gt;when&lt;/em&gt; the closed trapped surfaces will form (generically).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black Holes arose as physically plausible solutions to Einstein's equations because of the early work of Oppenheimer et al. Here there are two metrics combined to form the matter leading to a Black Hole:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Friedmann Dust (interior) + Schwarzchild (exterior)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a clever trick to consider &quot;massless dust&quot; allows one to use only the Friedmann Dust solution). These two solutions need to be &quot;glued together&quot; to form the surface of the star. The $T^{u,v}$ (in the comoving frame and its translation into other frames) for this was given in Edward's earlier answer, and is non-zero in the star interior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What causes another layer of complication in discussing the formation of Black Holes and Event Horizons is the teleological nature of their formation in General Relativity. This arises because &quot;time&quot; is just a parameter in the theory and the formation of the Black Hole is determined by the overall solution (thus in a time independent way). Now it has been concluded that for stellar objects of mass &gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_limit&quot; rel=&quot;nofollow&quot;&gt; TOV limit&lt;/a&gt; a Black Hole will form. But as remarked above translating this condition into a condition on the Stress-Energy Tensor alone may not be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physically one might expect that there is some local condition despite all these issues, such as the Hoop conjecture which includes the object's mass in its formulation. There are several subtleties connected with this unproven conjecture and one problem here is that &quot;mass&quot; is not a local property in GR (because mass = energy and the gravitational field contributes energy too, not just the Stress-Energy Tensor - hence we again may need the full solution of $G^{u,v}=T^{u,v}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I shall add &lt;a href=&quot;http://williewong.wordpress.com/2010/01/28/the-hoop-conjecture-of-kip-thorne-and-spherically-symmetric-space-times/&quot; rel=&quot;nofollow&quot;&gt;this link from Willie Wong&lt;/a&gt; for anyone interested in the latest on the Hoop Conjecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally my answer to the &lt;a href=&quot;http://physics.stackexchange.com/questions/7771/is-there-an-energy-density-limit-in-gr&quot;&gt;linked question&lt;/a&gt; might be of interest.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-03-30T19:04:48.377" LastActivityDate="2011-03-30T19:04:48.377" CommentCount="1" />
  <row Id="7797" PostTypeId="2" ParentId="7737" CreationDate="2011-03-30T11:47:49.547" Score="1" ViewCount="" Body="&lt;p&gt;The standard encyclopedic treatise of nonlinear functional analysis is the 5 volume opus of Eberhard Zeidler, &quot;Nonlinear Functional Analysis and Its Applications&quot;. It covers a lot of material about variational calculus, for example, in volume III &quot;Variational Methods and Optimization&quot;. The applications are usually applications from physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is too much material, there is also a two volume version including some topics of linear functional analysis as well, &quot;Applied functional analysis. Main principles and their applications.&quot; and &quot;Applied functional analysis. Applications to mathematical physics.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-03-30T11:47:49.547" />
  <row Id="7798" PostTypeId="2" ParentId="7781" CreationDate="2011-03-30T12:03:13.717" Score="3" ViewCount="" Body="&lt;p&gt;I greatly sympathize with your question. It is indeed a very misleading analogy given in popular accounts. I assure you that curvature or in general, general relativity (GR) describe gravity, they don't assume it. As you appear to be uninitiated I shall try to give you some basic hints about how gravity is described by GR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &lt;em&gt;absence&lt;/em&gt; of matter/energy the spacetime  (space and time according to the relativity theories are so intimately related with each other it makes more sense to combine them in a 4 dimensional object called space-time) is flat like a table top. This resembles closely with (not completely) Euclidean geometry of plane surfaces. We call this spacetime, Minkowski space. In this space the shortest distance between any two points are straight lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However as soon as there is some matter/energy the geometry of the surrounding spacetime is affected. It no longer remains Minkowski space, it becomes a (pseudo) Riemannian manifold. By this I mean the geometry is no longer like geometries of a plane surface but rather like geometries of a curved surface. In this curved spacetime the shortest distance between any two points are not straight lines in general, rather they are curved lines. It is not very hard to understand. Our Earth is a curved surface and the shortest distance between any two points are great circles rather than straight lines. Similarly the shortest distance between any two points in the 4 dimensional spacetime are curved lines. An object like sun makes the geometry of spacetime curved in such a way that the shortest distance between any two points are curved. This is called a geodesic. A particle follows this curved geometry by moving along this geodesic. Einstein's equations are mathematical descriptions of the relation of the geometry to the matter/energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how gravity is described in general relativity.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-04-03T18:20:31.187" LastActivityDate="2011-04-03T18:20:31.187" CommentCount="6" />
  <row Id="7799" PostTypeId="2" ParentId="7777" CreationDate="2011-03-30T13:32:22.097" Score="1" ViewCount="" Body="&lt;p&gt;As an addition to the previous answers, let me give my own point of view on this subject. Briefly, the classical picture is an &lt;strong&gt;inclusive&lt;/strong&gt; quantum picture - it includes many different quantum events in one classical &quot;object&quot; by definition. It is just like taking a photograph with photon by photon,  point by point, pixel by pixel. Each point is not classical but their ensemble taken together and associated with an &quot;average&quot; or &quot;inclusive&quot; image (&quot;object&quot;) is usually described with the classical physics (deterministic point-like or wavy classical behavior). The brightest example - an interference picture. It is a multi-photon stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally one can see it from the Ehrenfest equations if one remembers that an &quot;average&quot; means &quot;inclusive&quot; and &quot;normalized&quot;. There is no need to tend here $\hbar$ to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is funny but the very well known Coulomb potential is also an inclusive rather than &quot;elastic&quot; picture, see my opus &lt;strong&gt;&lt;a href=&quot;http://arxiv.org/abs/0806.2635&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Seeing downvotes, I conclude that too few understand or know the Ehrenfest theorems and ascribe the inclusive picture to my personal &quot;speculations&quot;. Too bad!&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-03-31T23:30:17.233" LastActivityDate="2011-03-31T23:30:17.233" CommentCount="3" />
  <row Id="7800" PostTypeId="2" ParentId="7784" CreationDate="2011-03-30T14:10:13.657" Score="7" ViewCount="" Body="&lt;p&gt;I'm a bit worried about getting a reputation for citing myself too much, but I'll go for it anyway. (In my defense, I always admit it when I'm doing it!) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;John Baez's and my pedagogical paper &lt;a href=&quot;http://math.ucr.edu/home/baez/einstein/&quot; rel=&quot;nofollow&quot;&gt;The Meaning of Einstein's Equation&lt;/a&gt; aims to address exactly this question. We describe the meaning of spacetime curvature and the way that Einstein's equation connects spacetime curvature to the matter content of a region of space. As one example, we use this description to heuristically &quot;derive&quot; Newtonian gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the most important point is that, in &quot;ordinary&quot; situations involving particles moving at speeds much less than $c$, the &quot;time&quot; part of spacetime curvature is by far the most important part. Intuition about curved &lt;em&gt;space&lt;/em&gt; (as opposed to spacetime) only gets you so far. &lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-03-30T14:10:13.657" CommentCount="5" />
  <row Id="7801" PostTypeId="2" ParentId="7777" CreationDate="2011-03-30T16:03:34.187" Score="-1" ViewCount="" Body="&lt;p&gt;It's the other way around, a quantum formalism is necessary for classical mechanics.&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-03-30T16:03:34.187" />
  <row Id="7802" PostTypeId="2" ParentId="885" CreationDate="2011-03-30T16:38:39.190" Score="1" ViewCount="" Body="&lt;p&gt;Here is my answer, which is basically an expanded version of Greg Graviton's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question of why one can treat position and velocity as independent variables arise in the &lt;em&gt;definition&lt;/em&gt; of the Lagrangian $L$ itself, &lt;em&gt;before&lt;/em&gt; one thinks about varying the action $S:=\int_{t_i}^{t_f}dt \ L$, and has therefore &lt;em&gt;nothing&lt;/em&gt; to do with calculus of variation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On one hand, let us consider first the role of the Lagrangian.&#xA;Let there be given an arbitrary but fixed instant of time $t_0\in [t_i,t_f]$. The (instantaneous) Lagrangian $L(q(t_0),v(t_0),t_0)$ is a function of both the instantaneous position $q(t_0)$ and the instantaneous velocity $v(t_0)$ at the instant $t_0$. &#xA;Here $q(t_0)$ and $v(t_0)$ are &lt;em&gt;independent&lt;/em&gt; variables. Note that the (instantaneous) Lagrangian $L(q(t_0),v(t_0),t_0)$ does not depend on the past $t&amp;lt;t_0$ nor the future $t&amp;gt;t_0$. (One may object that the velocity profile $\dot{q}\equiv\frac{dq}{dt}:[t_i,t_f]\to\mathbb{R}$ is the derivative of the position profile $q:[t_i,t_f]\to\mathbb{R}$, so how can $q(t_0)$ and $v(t_0)$ be truly independent variables? The point is that one is still entitled to make two &lt;em&gt;independent&lt;/em&gt; choices of initial conditions.) We can repeat this argument for any other instant $t_0\in[t_i,t_f]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, let us consider calculus of variation.&#xA;The action functional $S[q] := \int_{t_i}^{t_f}dt \ L(q(t),\dot{q}(t),t)$ depends on the whole (perhaps virtual) path $q:[t_i,t_f]\to\mathbb{R}$. Here the time derivative $\dot{q}\equiv\frac{dq}{dt}$ do &lt;em&gt;depend&lt;/em&gt; on the function $q:[t_i,t_f]\to \mathbb{R}$. Extremizing the action functional &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0=\delta S = \int_{t_i}^{t_f}dt\left[\left.\frac{\partial L(q(t),v(t),t)}{\partial q(t)}\right|_{v(t)=\dot{q}(t)} \delta q(t) +\left.\frac{\partial L(q(t),v(t),t)}{\partial v(t)}\right|_{v(t)=\dot{q}(t)}\delta \dot{q}(t)\right] $$ &#xA;$$ = \int_{t_i}^{t_f}dt\left[\left.\frac{\partial L(q(t),v(t),t)}{\partial q(t)}\right|_{v(t)=\dot{q}(t)} \delta q(t) +\left.\frac{\partial L(q(t),v(t),t)}{\partial v(t)}\right|_{v(t)=\dot{q}(t)}\frac{d}{dt}\delta q(t)\right] $$&#xA;$$ = \int_{t_i}^{t_f}dt\left[\left.\frac{\partial L(q(t),v(t),t)}{\partial q(t)}\right|_{v(t)=\dot{q}(t)} - \frac{d}{dt}\left(\left.\frac{\partial L(q(t),v(t),t)}{\partial v(t)}\right|_{v(t)=\dot{q}(t)} \right)\right]\delta q(t) $$ &#xA;$$+ \int_{t_i}^{t_f}dt\frac{d}{dt}\left[\left.\frac{\partial L(q(t),v(t),t)}{\partial v(t)}\right|_{v(t)=\dot{q}(t)}\delta q(t)\right] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with appropriate boundary conditions leads to Euler-Lagrange equation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\frac{d}{dt}\left(\left.\frac{\partial L(q(t),v(t),t)}{\partial v(t)}&#xD;&#xA;\right|_{v(t)=\dot{q}(t)} \right)&#xD;&#xA;= \left.\frac{\partial L(q(t),v(t),t)}{\partial q(t)}&#xD;&#xA;\right|_{v(t)=\dot{q}(t)} &#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\frac{d}{dt}$ is a total time derivative, not an explicit time derivative $\frac{\partial}{\partial t}$, so that Euler-Lagrange equation is really a second-order ordinary differential equation (ODE),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\ddot{q}(t)\frac{\partial}{\partial v(t)}+\dot{q}(t)\frac{\partial}{\partial q(t)}+\frac{\partial}{\partial t}\right) \left. \frac{\partial L(q(t),v(t),t)}{\partial v(t)}&#xD;&#xA;\right|_{v(t)=\dot{q}(t)} &#xD;&#xA;= \left.\frac{\partial L(q(t),v(t),t)}{\partial q(t)}&#xD;&#xA;\right|_{v(t)=\dot{q}(t)} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve for the path $q:[t_i,t_f]\to \mathbb{R}$, one should specify two initial conditions, e.g., $q(t_i)=q_i$ and $\dot{q}(t_i)=v_i$.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-04-12T15:04:18.003" LastActivityDate="2011-04-12T15:04:18.003" />
  <row Id="7803" PostTypeId="2" ParentId="7778" CreationDate="2011-03-30T17:02:07.843" Score="0" ViewCount="" Body="&lt;p&gt;It really is not a density.  The critical spatial quantity is the Schwarzschild radius $R~=~2GM/c^2$, where the mass $M$ is contained in a volume with a radius $R$.  The critical factor is not the amount of matter per unit volume, or density.  A super massive black hole has a mass of 10 billion solar masses, and a radius 10 billion times the solar mass radius of 1.5 km.  The volume in standard coordinates is $3.4\times 10^{30}km^3$. The sun has a volume of $1.4\times 10^{18}km^3$.  !0 billion suns have a volume of  $1.4\times 10^{28}km^3$.  So 10 billion suns could be packed into the volume of a 10 billion solar mass black hole with room to spare.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-30T17:02:07.843" CommentCount="2" />
  <row Id="7804" PostTypeId="1" AcceptedAnswerId="7808" CreationDate="2011-03-30T18:08:12.260" Score="2" ViewCount="70" Body="&lt;p&gt;So, there is a sound at $S$, whose intensity $I$ obeys the inverse square law ($I \sim \frac{1}{x^2}$). At point $P$, at a distance $r$ from $S$, the air molecules oscillate with an amplitude of $8μm$. Point $Q$ is at a distance of $2r$ from $S$. What is the amplitude of the air molecules at $Q$? What is the relationship between amplitude and distance?&lt;/p&gt;&#xA;" OwnerUserId="2853" LastEditorUserId="2853" LastEditorDisplayName="" LastEditDate="2011-03-30T18:18:26.340" LastActivityDate="2011-03-30T19:03:47.223" Title="Amplitude at distance from source" Tags="&lt;classical-mechanics&gt;&lt;waves&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7805" PostTypeId="1" AcceptedAnswerId="7806" CreationDate="2011-03-30T18:29:59.110" Score="2" ViewCount="110" Body="&lt;p&gt;I'm currently reading a nice introductory book (german, could be translated as &quot;Physics with a pencil&quot;). The author works a lot with differential calculus and antiderivatives (integrals will be used later). I'm stuck at a kind of mathematical question inmidst the easier physics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So he takes the equation of force:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m \ddot{\vec{r}} = \vec{F}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and dot-multiplies it with the speed $\dot{\vec{r}}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m \dot{\vec{r}} \cdot \ddot{\vec{r}} = \dot{\vec{r}} \cdot \vec{F}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I do not understand how he gets to the next step:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta_t \left(\frac{1}{2}m\dot{\vec{r}}^2\right) = \frac{d\vec{r} \cdot \vec{F}}{dt}\ \ \ $ (with $\frac{mv^2}{2} = E_{kin}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because if I resolve the last equation again I find something different than what we began with, because the derivation of the left part leaves me no single $\dot{\vec{r}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{2}{2} m \ddot{\vec{r}} = \dot{\vec{r}} \cdot \vec{F}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe I misinterpret the application $\delta_t$ to both $\vec{r}$ or the $\frac{d\vec{x}}{dr}$-notation not applying to $\vec{F}$ here? (The author states before that he use the partial differentiation symbol $\delta_t$ in the same way as the total $\frac{d\vec{x}}{dr}$ because the partial states exactly what is to be derived.)&lt;/p&gt;&#xA;" OwnerUserId="1089" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-11T20:05:46.143" LastActivityDate="2011-05-11T20:05:46.143" Title="Problem deriving kinetic energy from work" Tags="&lt;newtonian-mechanics&gt;&lt;mathematics&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="7806" PostTypeId="2" ParentId="7805" CreationDate="2011-03-30T18:39:59.837" Score="4" ViewCount="" Body="&lt;p&gt;I think you're just making a calculus error. &#xA;$$&#xD;&#xA;{d\over dt}\left({1\over 2}m\dot{\bf r}^2\right)={1\over 2}m{d(\dot{\bf r}\cdot\dot{\bf r})\over dt}={1\over 2}m(\ddot{\bf r}\cdot\dot{\bf r}+\dot{\bf r}\cdot\ddot{\bf r})=m\dot{\bf r}\cdot\ddot{\bf r}.&#xD;&#xA;$$&#xA;You can tell that something has gone wrong in your calculation, because the units of $(d/dt)({1\over 2}m\dot{\bf r}^2)$ aren't the same as those of $m\ddot{\bf r}$. Also, the first one is a scalar, and the second is a vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, I wrote vectors in boldface rather than with arrows. Both notations are fine. I don't like having multiple things (arrows and dots) piled on top of symbols, but that's just my taste.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-03-30T18:49:27.440" LastActivityDate="2011-03-30T18:49:27.440" CommentCount="1" />
  <row Id="7808" PostTypeId="2" ParentId="7804" CreationDate="2011-03-30T19:03:47.223" Score="1" ViewCount="" Body="&lt;p&gt;Yes, Electro, the intensity $I$ - and energy density $T_{00}$ and similar quantities - is proportional to the squared amplitude,&#xA;$$ I \sim A^2 $$&#xA;Because the intensity must go as&#xA;$$ I \sim \frac{1}{r^2} $$&#xA;as the energy gets spread over the sphere of area $4\pi r^2$, it follows that&#xA;$$ A \sim \frac{1}{r}.$$ &#xA;See e.g. the $1/r$ factor in this formula:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dipole_antenna#Elementary_doublet&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dipole_antenna#Elementary_doublet&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-30T19:03:47.223" />
  <row Id="7809" PostTypeId="1" AcceptedAnswerId="7811" CreationDate="2011-03-30T20:47:26.583" Score="11" ViewCount="280" Body="&lt;p&gt;I.e. online discussion with your friends. A forum is probably too overkill in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet so far nothing can beat direct communication.&#xA;Important feature: the ability to archive discussions. We don't want to retell our story to people who just missed the &quot;conference&quot; (let the newcomers dig through the archives by themselves).&lt;/p&gt;&#xA;" OwnerUserId="1814" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-04-01T11:51:40.757" LastActivityDate="2011-04-01T16:38:46.097" CommunityOwnedDate="2011-03-30T21:57:04.000" Title="Online physics collaboration tools" Tags="&lt;soft-question&gt;&lt;software&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="5" />
  <row Id="7810" PostTypeId="2" ParentId="7809" CreationDate="2011-03-30T21:23:32.897" Score="3" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.mathim.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.mathim.com/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Latex based chatroom partially fulfills the requirement . The best scenario I can imagine is a mathjax/latex based wiki/website with an embedded $\LaTeX$ chat like M$\alpha$thIM. &lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="2599" LastEditorDisplayName="" LastEditDate="2011-03-30T21:54:16.053" LastActivityDate="2011-03-30T21:54:16.053" CommunityOwnedDate="2011-03-30T21:57:04.000" />
  <row Id="7811" PostTypeId="2" ParentId="7809" CreationDate="2011-03-30T22:26:10.223" Score="5" ViewCount="" Body="&lt;p&gt;I should mention the wonderful userscripts written by &lt;a href=&quot;http://alexeev.org/display-latex2.html&quot; rel=&quot;nofollow&quot;&gt;Valery Alexeev&lt;/a&gt; which utilize mathjax and allow you to render latex on &lt;em&gt;any&lt;/em&gt; webpage, even if it does not support such rendering internally. Currently supported webpages include arXiv and gmail, though it should be easy to add your own to the list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Used in combination with mathim.com and &lt;strong&gt;Tiddlywiki&lt;/strong&gt; we have a powerful platform for online tex-enabled collaborations.&lt;/p&gt;&#xA;" LastActivityDate="2011-03-30T22:26:10.223" CommunityOwnedDate="2011-03-30T22:26:10.223" CommentCount="11" />
  <row Id="7813" PostTypeId="2" ParentId="7777" CreationDate="2011-03-30T22:31:05.833" Score="3" ViewCount="" Body="&lt;p&gt;I would like to question some of the assumptions of the question itself here. Firstly the title:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is a Classical Formalism necessary for Quantum Mechanics?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is this premise actually true? One could imagine setting up a quantum structure (Hilbert Spaces, $\Psi$, Operators) without any reference to anything classical. The closest one comes is in formulating the experimental meanings of the eigenvalues.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For theories to be consistent, they must reduced to known laws in the classical domains.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I dont agree that this is a requirement for theory consistency. Of course it depends what &quot;reduced&quot; means here, but some quantum theories live quite happily without an (apparent) classical counterpart (spin?). Anyway theory consistency is about whether if P(x) is deduced then not P(x) is not also deducible ie a logical requirement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point that I suspect is intended here is that there is expected to be &quot;theory correspondence&quot; from Quantum to Classical: this is weaker than requiring &quot;reduction&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The classical domain can be summed up as $\hbar=0$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well not quite, it is now known that h was discovered in Classical Statistical Mechanics by Gibbs as an area proportionality term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as writing Hamiltonians is concerned the Classical theory could be said to simply provide a heuristic guide - in a sense it is not necessary since it may not exist. That is &lt;em&gt;if&lt;/em&gt; we have the classical Hamiltonian H(p,q), then we know that the quantum one is unordered as $H(-i\hbar\partial /{\partial x},x)$, but we still dont know what the ordered (correct) form is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real question that is being asked here is how to write a quantum Hamiltonian when there is no (apparent) classical theory. Perhaps the answer lies in Feynman's saying &quot;Guess, then try it out&quot; or perhaps there is some greater theory of Quantum Hamiltonians yet to be found...&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-03-30T22:31:05.833" CommentCount="6" />
  <row Id="7814" PostTypeId="2" ParentId="7809" CreationDate="2011-03-30T22:47:01.337" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://convore.com&quot; rel=&quot;nofollow&quot;&gt;convore&lt;/a&gt;: $\frac1{\sqrt3}(|IRC\rangle + |mailing list\rangle + |chat\rangle)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No built-in latex support yet, but this can be remedied by installing the &lt;a href=&quot;http://alexeev.org/display-latex2.html&quot; rel=&quot;nofollow&quot;&gt;display latex&lt;/a&gt; userscript (thank you Deepak Vaid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nice complement to convore: online whiteboard (e.g. scriblink, flockdraw).&lt;/p&gt;&#xA;" OwnerUserId="1814" LastEditorUserId="1814" LastEditorDisplayName="" LastEditDate="2011-04-01T16:38:46.097" LastActivityDate="2011-04-01T16:38:46.097" CommunityOwnedDate="2011-03-30T22:47:01.337" CommentCount="3" />
  <row Id="7815" PostTypeId="2" ParentId="7809" CreationDate="2011-03-30T22:50:09.167" Score="1" ViewCount="" Body="&lt;p&gt;alternative to tiddlywiki: &lt;a href=&quot;http://www.instiki.org/show/HomePage&quot; rel=&quot;nofollow&quot;&gt;instiki&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a simple wiki with latex support out of the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't really explored it yet, so please share your experience.&lt;/p&gt;&#xA;" OwnerUserId="1814" LastActivityDate="2011-03-30T22:50:09.167" CommunityOwnedDate="2011-03-30T22:50:09.167" CommentCount="2" />
  <row Id="7816" PostTypeId="2" ParentId="7778" CreationDate="2011-03-30T23:40:33.443" Score="2" ViewCount="" Body="&lt;p&gt;Also, note that there are exact solutions of Einstein's equation that correspond to naked singularities.  For an easy example, if you have a charged black hole with $Q&amp;gt;M$, there is no event horizon at all, but you still have a mass $M$ measured by an observer at asymptotic infinity.  So, these solutions are a class of solutions that have &quot;infinite&quot; density at the singularity, but at the same time, aren't black holes.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-03-30T23:40:33.443" CommentCount="6" />
  <row Id="7817" PostTypeId="1" CreationDate="2011-03-31T02:52:36.373" Score="8" ViewCount="278" Body="&lt;p&gt;Firstly, I understand that we have no observational evidence for 'wormholes'. They are theorised solutions to general relativity equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, if macroscopic wormholes do exist---how do they work? I've been thinking about this for the past couple of days, and I have some questions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What does the mouth of a wormhole look like? I've seen &lt;a href=&quot;http://www.spacetimetravel.org/wurmlochflug/wurmlochflug.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;; is that more or less accurate?&lt;/li&gt;&#xA;&lt;li&gt;If I place one end of a wormhole in a high magnetic field, and the other end on my desk, will all my ferrous objects fly into the wormhole?&lt;/li&gt;&#xA;&lt;li&gt;If so, with what force? What does the magnetic field look like at each end of the wormhole?&lt;/li&gt;&#xA;&lt;li&gt;If the distance between the mouths of the wormhole was comparable to the force that the magnets exert, would the magnet 'pull' on itself and draw itself into the wormhole?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Do any of these questions even make sense?&lt;/p&gt;&#xA;" OwnerUserId="2824" LastActivityDate="2011-04-01T00:33:18.557" Title="How do wormholes work?" Tags="&lt;general-relativity&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="7818" PostTypeId="1" CreationDate="2011-03-31T04:35:57.667" Score="4" ViewCount="249" Body="&lt;p&gt;In the book &quot;Kinetic theory in the expanding Universe&quot; (J. Bernstein, 1988, Camb. Univ. Press), it was stated that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;for nonstationary Robertson-Walker&#xA;  matrixes [sic] there is no spacelike&#xA;  Killing vector.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(page 6, footnote.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we know this is not true, since one can see that the generators of spatial translations/rotations are manifestly space-like. (For example, see R. Maartens and S.D. Maharaj, Class. Quantum Grav. 3 (1986) 1005-1011, equation (1.7)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was the author of the book wrong, or was it a slip of the hand, where &quot;spacelike&quot; should have been &quot;timelike&quot; (but that would amount to tautology anyway, since &quot;nonstationary&quot; is defined in terms of the absence of a time-like Killing vector).&lt;/p&gt;&#xA;" OwnerUserId="2859" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-03-31T07:41:53.200" LastActivityDate="2011-03-31T13:05:01.490" Title="Space-like Killing vector of Robertson-Walker metric?" Tags="&lt;general-relativity&gt;&lt;spacetime&gt;" AnswerCount="2" />
  <row Id="7819" PostTypeId="1" CreationDate="2011-03-31T05:27:57.013" Score="7" ViewCount="325" Body="&lt;p&gt;Given: Law of Conservation of Angular Momentum.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Reverse spinning with dense atmosphere (92 times &gt; Earth &amp;amp; CO2 dominant sulphur based).&lt;/li&gt;&#xA;&lt;li&gt;Surface same degree of aging all over.&lt;/li&gt;&#xA;&lt;li&gt;Theoretical large impact is not a sufficient answer.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Assuming any object large enough to alter a planets rotation or even orbit would likely destroy most of its shape, yet Venus has retained a spherical property with a seemingly flat, even terrain indicating no volcanoes,and few if any visible meteor impacts. It would be fragmented and dispersed for billions of years.  Even the question of what meteor, comet, asteroid composition could survive traveling that close to the sun's temperature, radiation, electromagnetic energy, solar flares, or gravity to equal a mass reactionary change as to alter it's spin.&lt;/p&gt;&#xA;" OwnerUserId="2860" LastActivityDate="2011-03-31T17:49:19.610" Title="Why does Venus rotate the opposite direction as other planets?" Tags="&lt;gravity&gt;&lt;astrophysics&gt;&lt;astronomy&gt;&lt;spacetime&gt;&lt;rotational-dynamics&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="7820" PostTypeId="2" ParentId="7819" CreationDate="2011-03-31T06:34:01.923" Score="2" ViewCount="" Body="&lt;p&gt;Well, I Binged and found some &lt;a href=&quot;http://www.astronomycafe.net/qadir/q50.html&quot; rel=&quot;nofollow&quot;&gt;references.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seems that a collision is most probable, if it happened at a time when the whole system was malleable. But there is no solid explanation.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-31T06:34:01.923" CommentCount="6" />
  <row Id="7821" PostTypeId="2" ParentId="7780" CreationDate="2011-03-31T06:52:17.170" Score="4" ViewCount="" Body="&lt;p&gt;The answer is &lt;a href=&quot;http://en.wikipedia.org/wiki/Soap_bubble#Freezing&quot; rel=&quot;nofollow&quot;&gt;YES&lt;/a&gt;.  Note that &lt;a href=&quot;http://en.wikipedia.org/wiki/Surface_tension#Surfactants&quot; rel=&quot;nofollow&quot;&gt;pure water bubbles are unstable&lt;/a&gt; but it should be possible to freeze one without it popping as long as it is thick enough and the pressure is correctly regulated.  (&lt;a href=&quot;http://www.youtube.com/watch?v=ddST_7n9peg&quot; rel=&quot;nofollow&quot;&gt;This video&lt;/a&gt; shows in detail a soap bubble partially freezing and then bursting on contact, and &lt;a href=&quot;http://www.youtube.com/watch?v=OM9WXrGftXE&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt; shows more completely frozen soap bubbles rolling around and eventually disintegrating.)  If completely frozen, the bubble will no longer be a perfect sphere because the ice crystal will nucleate at certain positions and this will distort the surface.&lt;/p&gt;&#xA;" OwnerUserId="2072" LastEditorUserId="2072" LastEditorDisplayName="" LastEditDate="2011-03-31T07:05:57.103" LastActivityDate="2011-03-31T07:05:57.103" />
  <row Id="7822" PostTypeId="1" CreationDate="2011-03-31T07:48:46.783" Score="7" ViewCount="221" Body="&lt;p&gt;This paper was published in a peer review journal, and claims the answer is yes.&#xA;&lt;a href=&quot;http://arxiv.org/abs/physics/0607090&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/physics/0607090&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivation in the paper seems more like dimensional analysis hand-waving in the beginning.  But then uses some specific results from differential geometry to uniquely get to Einstein's field equations.  I can't tell if he is assuming something along the way during the hand-waviness that accidentally garauntee's the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the article states that this can be worded also as postulating a maximum total energy flow/flux through a surface.  But in discussion from &lt;a href=&quot;http://physics.stackexchange.com/questions/7771/is-there-an-energy-density-limit-in-gr&quot;&gt;Is there an energy density limit in GR?&lt;/a&gt;, it sounds like components of the stress energy tensor can be arbitrarily large by changing coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can GR be derived by postulating a maximum force, or a maximum total energy flow through a surface?&lt;/p&gt;&#xA;" OwnerUserId="2655" LastActivityDate="2011-04-07T03:00:56.407" Title="Can GR be derived by postulating a maximum force?" Tags="&lt;general-relativity&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="2" />
  <row Id="7823" PostTypeId="1" CreationDate="2011-03-31T08:34:52.880" Score="8" ViewCount="662" Body="&lt;p&gt;Time travel -- often featured in movies, works of fiction, or facetiously in conversation. There are of course theories treating time as simply another dimension, which to the layperson might imply forward and backward movement is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what do we scientifically know with respect to the possibility or impossibility of time travel? Are there any testable theories on the horizon (or theories already tested) that support a likelihood or unlikelihood for time travel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: by &quot;time travel&quot; I mean the ability to (some day?) willingly transport a complex object or being through time (perhaps only to the past, or a copy of the past, which is an answer too).&lt;/p&gt;&#xA;" OwnerUserId="2496" LastEditorUserId="2496" LastEditorDisplayName="" LastEditDate="2011-04-01T02:44:27.793" LastActivityDate="2011-05-20T02:37:52.547" Title="Is time travel possible?" Tags="&lt;time-travel&gt;" AnswerCount="4" CommentCount="13" FavoriteCount="2" />
  <row Id="7824" PostTypeId="2" ParentId="7818" CreationDate="2011-03-31T08:39:50.920" Score="2" ViewCount="" Body="&lt;p&gt;From the book by Kolb and Turner &quot;The Early Universe&quot;, chapter 3.5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the strictest mathematical sense it is not possible for the Universe to be in thermal equilibrium, as the FRW cosmological model does not possess  a &lt;strong&gt;time-like&lt;/strong&gt; Killinig vector. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Emphasis is mine). So it confirms the &quot;slip of the hand&quot; hypothesis.   &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-03-31T08:39:50.920" CommentCount="1" />
  <row Id="7825" PostTypeId="2" ParentId="7780" CreationDate="2011-03-31T08:57:46.050" Score="0" ViewCount="" Body="&lt;p&gt;No. Thermodynamic functions of the phase transitions of water explain the science behind this question.&#xA;Long term, a definite No.  The question does have a valid common sense reference answer: Short term, Yes. But only for a short time.  Also its not a perfect sphere and most wouldn't refer to it as a bubble. Think of simple ice cube tray in the freezer that is removed before fully frozen with some liquid still in the center. Temperature at the core hasn't reached freezing point yet. Technically a pocket of liquid water and air within a solid mass of water. &#xA;Thinking also in the other direction, the same could be said of a water vapor bubble momentarily existing in a container of boiling water. Long term, a definite No.&#xA;Using a water phase diagram chart used to show the physical states of water at different temperatures and pressures, the phase boundaries reveal the singular conditions of temperature and pressure where liquid water, gaseous water and hexagonal ice stably coexist, there is a 'triple point' if you look closely.  (chart can be found here)&gt; &lt;a href=&quot;http://www.lsbu.ac.uk/water/phase.html&quot; rel=&quot;nofollow&quot;&gt;http://www.lsbu.ac.uk/water/phase.html&lt;/a&gt;  However it is impossible to create a water bubble within an ice bubble for any long term time period.&lt;/p&gt;&#xA;" OwnerUserId="2860" LastActivityDate="2011-03-31T08:57:46.050" />
  <row Id="7826" PostTypeId="1" CreationDate="2011-03-31T08:59:00.013" Score="2" ViewCount="85" Body="&lt;p&gt;I'm working with a megapixel sensor with a 1/2&quot; format. I would like to know which parameters have to be taken into account in order to design a lens that can be fit with 1/2&quot; format.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The light source wavelenght is 850 nm produced by high power leds. &#xA;The working distance is about 15 meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Raphael&lt;/p&gt;&#xA;" OwnerUserId="2861" LastEditorUserId="869" LastEditorDisplayName="" LastEditDate="2011-03-31T12:37:00.293" LastActivityDate="2011-03-31T12:37:00.293" Title="Which lens parameters determines it's format?" Tags="&lt;optics&gt;&lt;geometric-optics&gt;" AnswerCount="1" />
  <row Id="7828" PostTypeId="2" ParentId="7746" CreationDate="2011-03-31T09:37:02.337" Score="0" ViewCount="" Body="&lt;p&gt;There are a myriad of answers to your question available on every TV and website, but most have vague mass suggestions on how to stay safe.  Its better to focus on the size of the particulates of contamination. See a video referring to a form of radiation here: &lt;a href=&quot;http://www.youtube.com/watch?v=WgQ79-oDX2o&amp;amp;feature=related&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=WgQ79-oDX2o&amp;amp;feature=related&lt;/a&gt; Also not to let rain contact your skin directly when theres a report of radioactivity in an area. Radioactive substances are in our everyday life and some reports may be false positives, but always better to practice caution.  Also the length of time of exposure is just as critical. With the Japan leakage, the exposure to the rest of the world is minimal so far. Keep in mind that its an airborne particulate breathed in that is worrisome and anything to reduce that exposure is best.&#xA;As for food, we irradiate it already with small doses.&lt;/p&gt;&#xA;" OwnerUserId="2860" LastActivityDate="2011-03-31T09:37:02.337" />
  <row Id="7829" PostTypeId="1" AcceptedAnswerId="7830" CreationDate="2011-03-31T10:10:11.403" Score="3" ViewCount="232" Body="&lt;p&gt;What is the minimum thickness of an aggregate concrete membrane to prevent radioactive particles from escaping a broken reactor?&#xA;I am interested in calculating the amount of concrete required to entomb a reactor.&lt;/p&gt;&#xA;" OwnerUserId="582" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-17T05:52:47.160" LastActivityDate="2011-04-17T14:38:16.537" Title="What is the required thickness of concrete needed to entomb a leaking reactors emissions?" Tags="&lt;nuclear-physics&gt;&lt;radiation&gt;" AnswerCount="2" />
  <row Id="7830" PostTypeId="2" ParentId="7829" CreationDate="2011-03-31T10:52:39.730" Score="4" ViewCount="" Body="&lt;p&gt;Some amount of radiation always penetrates through a barrier of thickness $T$ - the proportion is given by $(1/2)^{(T/T_0)}$ where $T_0$ is the halving thickness. The decrease is exponential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The halving thickness depends both on the material as well as the type of the radiation. Generally, gamma rays - heavier cousins of X-rays (photons) - are the most penetrating ones. See e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Radiation_protection#Shielding_design&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Radiation_protection#Shielding_design&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The halving thickness for lead and concrete is 1 and 6 centimeters, respectively. However, the gamma radiation obviously escapes in random directions and is only a concern in the very vicinity of the source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beta-radiation - electrons that fly nearly by the speed of light - is somewhat less penetrating than gamma-rays, so the halving thickness is a few millimeters for solids:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.fas.harvard.edu/~scdiroff/lds/QuantumRelativity/PenetrationandShielding/PenetrationandShielding.html&quot; rel=&quot;nofollow&quot;&gt;http://www.fas.harvard.edu/~scdiroff/lds/QuantumRelativity/PenetrationandShielding/PenetrationandShielding.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The least penetrating ones are alpha particles and their heavier sisters, heavy radioactive nuclei, that are physically screened by dozens of micrometers of a solid; see the page above. So a minimal layer of a solid - with no holes - is always enough to protect a tomb against the release of some radioactive material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When thinking about the tomb, it may be more important to guarantee that it will be mechanically stable against excessive pressure inside etc. that could lead to explosions etc. It's not quite trivial to decide how to lower the pressure in the hermetically isolated tomb because the stuff that creates the pressure is radioactive and harmful.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-31T10:52:39.730" CommentCount="2" />
  <row Id="7832" PostTypeId="2" ParentId="7826" CreationDate="2011-03-31T12:36:25.410" Score="3" ViewCount="" Body="&lt;p&gt;When a lens designer is asked to make a lens for a specific sensor size, there are a couple of important factors that come in to play. The most obvious of course, is that the image must fill the sensor. This can be verified by a simple equation which calculates the height of an optical ray as it strikes the image plane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ h=f \tan{\theta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f$ is the effective focal length of the lens, and $\theta$ is the field of view. It is important to note that the sensor may be 1/2&quot;, so you might expect that the maximum ray height above the optical axis would be 1/4&quot;, but because the array is usually not circular, the corner of the array will have a height above the optical axis of more the 1/4&quot;. For a square array it would be $\frac{\sqrt{2}}{4}$ inches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the focal length and field of view are themselves important lens parameters, so appropriate choices must be made for those values. IF the requirements of the lens being designed specify a particular sensor size and field of view though, this equation exactly selects the focal length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a case like yours, where the object distance is known and fixed, designers also often talk about the &quot;Reduction Ratio&quot; or magnification of the lens, which is the ratio of the size of the image of an object, to the size of the object itself. If you were trying to have your object fill the image plane, this would set your reduction ratio. The reduction ratio is not independent of the equation I stated above, it is just another way to express a design specification for a given field of view at a given object distance. It is obviously undefined for a lens which images objects at infinity, such as a telescope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a pretty good picture of how first order lens parameters are chosen to give you a specific image size, but in designing a good lens, there is a lot more to consider. Most of the work in lens design happens after these first order parameters are chosen. Then the designer must make countless choices in order to produce a lens which has good optical performance over the full field of view. This is a process that takes a number of graduate level courses and years of experience to master, and is in some ways still as much an art as a science.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-03-31T12:36:25.410" CommentCount="2" />
  <row Id="7833" PostTypeId="2" ParentId="7818" CreationDate="2011-03-31T13:05:01.490" Score="-1" ViewCount="" Body="&lt;p&gt;The de Sitter metric &#xA;$$&#xD;&#xA;ds^2~=~dt^2~-~e^{\sqrt{\Lambda/3}t}(dr^2~+~r^2d\Omega^2)&#xD;&#xA;$$&#xA;has this time dependent factor.  This prevents a time-like Killing vector, for any vector formed by $\sqrt{g_{rr}^{-1}}\partial/\partial t$, or something similar, will not constant on a timelike spacetime vector.  The same holds for spacelike directions, for space and time are interchangeable in $g_{rr}$.  This means global conservation laws are not defined.  Energy conservation in general relativity is a funny thing, and is not something which can be established.  In the case of cosmologies, these metrics FLRW etc, are type O Petrov-Penrose-Pirani solutions, which have no isometries or Killing vectors.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-31T13:05:01.490" CommentCount="6" />
  <row Id="7834" PostTypeId="2" ParentId="7746" CreationDate="2011-03-31T13:12:38.773" Score="0" ViewCount="" Body="&lt;p&gt;The big danger is less from eating radioactive substances than from breathing them.  The gut is a continuous “tube,” or in other words our body plans are a torus.  As a result you are more likely to pass it out within 24 hours or so.  The respiratory system is an “in and out” system, where the lungs form a fractal branching pattern that reaches to the cellular level.  So a piece of dust with radioactive substances can get lodged in the lung.  So an alpha emitter can sit there and do lots of damage.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-31T13:12:38.773" />
  <row Id="7835" PostTypeId="2" ParentId="7738" CreationDate="2011-03-31T13:13:09.357" Score="1" ViewCount="" Body="&lt;p&gt;Since you want to explain it to your daughter, take a plastic bottle, cut the bottom open, turn it upside town, hold the top closed and fill it with water. Give her that bottle and have her release the top (which is on the bottom now, sorry for the bad phrasing). The water will whirl in different orientations whenever you repeat this (if it whirls at all) and she can influence it by accelerating the bottle in a circular motion to understand that an initial disturbance is responsible for the whirl orientation.&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2011-03-31T13:13:09.357" CommentCount="4" />
  <row Id="7837" PostTypeId="2" ParentId="7823" CreationDate="2011-03-31T14:03:01.760" Score="6" ViewCount="" Body="&lt;p&gt;Well, this is a migrated question and it deserves an answer from a physicist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, within the physics framework we have developed up to now, i.e. the totality of the accumulated theories which are based on experimental results, time travel in the sense of a human controlling his/her position in time as one can control it in space, is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course theories change and evolve, so this is the present answer. If, as @Luke in the comment said, a proven time traveler had appeared, that would have been experimental evidence that would have forced a change in theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only traveling we do in time is from the present to the future, willy nilly.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-03-31T14:03:01.760" CommentCount="15" />
  <row Id="7838" PostTypeId="1" CreationDate="2011-03-31T14:15:14.080" Score="2" ViewCount="210" Body="&lt;p&gt;With the notion of causality, firmly fixed by GR, we derived the concept of a singular point from where space-time begun. Causality alone gives us the possibility to talk about a known past (i.e. every effect has a cause) and by this we can trace back in time every event. To reach the moment of the Big Bang we only have to extrapolate backwards the Hubble Law.(I see this like a reversal of effect with cause in every point of space-time.)&#xA;With all this said, my question is about the moment of Big Bang; if even time started &lt;em&gt;there&lt;/em&gt;, we have an effect without a cause and causality breaks. In this perspective, when did causality started to play a role in the evolution of the universe? Does this path allow for a pre-Big Bang theory? Like a universe, where all its mass has been turned into energy (and in this case, with no mass present, the ideas of small and large are equivalent. And an infinite universe would be equivalent with a point sized one. Hence, the possibility of point like starting place for the Big Bang, but with pre existing &lt;em&gt;time&lt;/em&gt;). From the point of view of causality we reach a logical nonsense at the moment of the BB. How can we modify the BBT to avoid this problem? (or is has been done already?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks! &lt;/p&gt;&#xA;" OwnerUserId="2473" LastActivityDate="2011-07-29T04:01:53.660" Title="on causality and The Big Bang Theory" Tags="&lt;cosmology&gt;&lt;big-bang&gt;" AnswerCount="2" />
  <row Id="7839" PostTypeId="1" AcceptedAnswerId="7853" CreationDate="2011-03-31T15:00:52.323" Score="0" ViewCount="314" Body="&lt;p&gt;According to some source or other (I forget which now) it is theoretically impossible for bumblebees to fly by virtue of their size/bulk/aerodynamic properties. Is this old adage apocraphyal or true? And if its true, how come they do fly ...?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for reading.&lt;/p&gt;&#xA;" OwnerUserId="2639" LastActivityDate="2011-03-31T20:31:25.460" Title="Is it really Impossible for Bumblebees to fly?" Tags="&lt;theory&gt;&lt;aerodynamics&gt;&lt;nature&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="7840" PostTypeId="2" ParentId="7823" CreationDate="2011-03-31T15:06:13.570" Score="-3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following analogy I have failed to take into account the Relativistic Doppler Effect which would cause &quot;me&quot; to age more slowly than &quot;you&quot;. Since the speed of light is said to be independent of the motion of the observer the Doppler Effect, which may imply the opposite, is hard to accept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I have may lie in my idea of what &quot;independent of the motion of the observer&quot; means. But I digress...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Disclaimer: Someone else may want to&#xA;  expand on this in a more formal,&#xA;  correct, and scientific answer; but&#xA;  here is &lt;strong&gt;my&lt;/strong&gt; layman's idea of&#xA;  theoretical time travel (well,&#xA;  technically what i'm describing is&#xA;  time dilation). It is probably wrong&#xA;  but this &quot;understanding&quot; appeases my&#xA;  curious brain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Light is tricky. And relative. It &quot;travels&quot; at a constant speed (180000 mi/s) relative to the OBSERVER.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my home-brewed analogy on the strangeness of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets just pretend that light moves at 10 miles per hour (or km/h if that's your thing) for simplicity's sake (I have a hard time comprehending just how fast 186000 mi/s really is).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm driving my car toward you, with my lights on, at the speed of light&lt;sup&gt;1&lt;/sup&gt; (10 mph) how fast would the the light from my headlights be moving toward you? 20 mph, right? That would make sense...but it is not the case. The light from my headlights would travel at the constant 10 mph - relative to your perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, relative to my speed (10 mph), at what speed would I perceive the light moving away from my car? 10 mph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hey, its light... it can do whatever it wants!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm going to drive right on past you at my speedy 10 mph. And lets say instead of tail lights I've got a big digital clock shining back at you (its a 24-hour clock, where 00:00 = 12:00AM). Tick Tock. You never want to be late and don't want to rely on my clock so you have your own just to make sure mine is correct at all times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, think about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tick Tock... I'm now 10 miles away from you still travelling at the speed of light. The hour hand on my clock changes from 00:00 to 01:00. The light zooms toward you at 10 mph. Meanwhile, I'm still driving away from you. Another hour passes for me and my clock changes again from 01:00 to 02:00; I am now 20 miles away. At that exact moment you perceive that first change from 00 to 01. Now you think to yourself, &quot;Man, his clock must be slow!!! Its an hour off according to my clock!&quot; Your clock says it is 02:00 (and so does mine, but you haven't perceived that yet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you wait another hour and look back up at my clock expecting a change. But nothing happens. You wait another hour and BAM, my clock finally changes from 01:00 to 02:00. But it is now 04:00 on your clock (it's 04:00 on mine too, but again... you haven't perceived that yet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suddenly and instantaneously turn around (impossible, yes; but I'm traveling at the speed of light so I'll do what I want). I'm coming towards you at 10 mph. My light (which beaming back at you telling you it is 04:00), from your perspective, is also traveling at you at 10mph. Since I am 40 miles away it should take me, and my &quot;04:00&quot; light, 4 hours to get back to you. So, 4 hours later, when you see my &quot;04:00&quot; light, I'll show up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I appeared to either have a slow clock, or be 4 hours in the past (or 4 hours younger depending on how you look at it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the catch. If I had been monitoring your clock I would have witnessed the same time dilation. &lt;strong&gt;Your&lt;/strong&gt; clock would have appeared to be slow and &lt;strong&gt;you&lt;/strong&gt; would have appeared to be moving through time more slowly than me. (That's the &quot;relativity&quot; thing kicking in)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is brain thumper: what if I could go &lt;em&gt;faster&lt;/em&gt; than the speed of light&lt;sup&gt;2&lt;/sup&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it is not really time travel... but it is about as close as we are going to get until we figure out those darned wormholes&lt;sup&gt;3&lt;/sup&gt;. :-p&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;[1]&lt;/sup&gt; Einstein pretty much said we can't really travel at the speed of light under our current understanding of physics. (&lt;a href=&quot;http://thefastertimes.com/tech/2011/03/30/boy-genius-12-challenges-einsteins-theory-of-relativity/&quot; rel=&quot;nofollow&quot;&gt;This kid&lt;/a&gt; may have something to say on the matter, though.)&lt;br&gt;&#xA;&lt;sup&gt;[2]&lt;/sup&gt; AKA &quot;FTL&quot; according to Stargate&lt;br&gt;&#xA;&lt;sup&gt;[3]&lt;/sup&gt; AKA &quot;Hyperdrive&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, about wormholes: If we could travel to the other side of the universe and then look back on ourselves the light we would see would be from a long long time ago (in a galaxy far far away).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if we got in our super-speedy FTL spaceship after having been instantly transported via wormhole to the other side of the universe? Could we show up back on earth before we left?&lt;/p&gt;&#xA;" OwnerUserId="1031" LastEditorUserId="125" LastEditorDisplayName="" LastEditDate="2011-05-20T02:37:52.547" LastActivityDate="2011-05-20T02:37:52.547" CommentCount="6" />
  <row Id="7841" PostTypeId="2" ParentId="7839" CreationDate="2011-03-31T15:14:41.337" Score="8" ViewCount="" Body="&lt;p&gt;No it's an urban myth. It's impossible for them to fly using a very simple and inappropriate model of wing behaviour - possibly closer to say that bumble bees can't glide like albatrosses&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-03-31T15:14:41.337" CommentCount="2" />
  <row Id="7842" PostTypeId="2" ParentId="7838" CreationDate="2011-03-31T15:34:09.737" Score="2" ViewCount="" Body="&lt;p&gt;You are confusing causality with determinism. Causality implies, nothing can travel faster than light and future can not affect the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the singularity theorems of GR there was a singularity and a singularity means certain physical quantities were infinite. In other words GR is not capable to handle such badly behaved point. The notion of time does not have well defined meaning at this point and even if there were events &lt;em&gt;before&lt;/em&gt; the big bang those events can not have any observational consequence in the post bang universe. Hence any such prebang events will be causally disconnected from the post bang universe. One can therefore say that time indeed began at the big bang. There was always this belief that in a proper quantum theory of gravity, somehow this singularity can be avoided. But it is far from clear now and any prebang physics is pure speculation at this stage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that even in the classical GR, two events which are spacelike separated are causally disconnected. They are outside of each other's light cone and therefore they can't have any causal link between them.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-03-31T15:34:09.737" />
  <row Id="7843" PostTypeId="1" AcceptedAnswerId="7847" CreationDate="2011-03-31T15:40:42.770" Score="6" ViewCount="327" Body="&lt;p&gt;We’re told that ‘all forces are gauge forces’. The process seems to start with the Lagrangian corresponding to a particle-type, then the application of a local gauge symmetry leading to the emergence of the force bosons via the associated symmetry group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But where did &lt;em&gt;exactly four forces&lt;/em&gt; come from? Could new, perhaps supersymmetric particles hint at new fundamental forces? Is there a deeper theory which predicts what final set of forces we’ll eventually end up with?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, is the concept of force in unified physics really that fundamental at all?&lt;/p&gt;&#xA;" OwnerUserId="1394" LastActivityDate="2011-04-01T11:21:27.913" Title="How many fundamental forces could there be?" Tags="&lt;forces&gt;&lt;gauge-theory&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="7844" PostTypeId="2" ParentId="7843" CreationDate="2011-03-31T16:02:14.527" Score="2" ViewCount="" Body="&lt;p&gt;The answer from an experimentalist is that depending on the model that fits best the data, yes, there will be more interpreted as forces. Whenever vector bosons can be exchanged, like W and Z and gamma a corresponding force can be defined. The hypothetical graviton is a spin 2 exchanged particle and is associated macroscopically with gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a sense, any Feynman diagram with the exchange of a particle might be interpreted as the carrier of a force, since it gives an impulse, a change in momentum. It is traditional to keep the focus on the forces that are known macroscopically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is that these new &quot;forces&quot; will be irrelevant to the level of matter as we see it. The important one is gravity, so we stick on the earth, then electromagnetic, so we navigate and don't fall to the center of the earth, strong, for the creation of nuclei, and weak because we observe decays. By the time we reach weak the exchanged bosons are heavy, and the newer ones will be heavier still, and have a very small effect as forces and tiny contributions to amplitudes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, forces are fundamental on human sized problems. Not in the theories the theorists are building. They are an effect, imo.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-04-01T11:21:27.913" LastActivityDate="2011-04-01T11:21:27.913" CommentCount="3" />
  <row Id="7845" PostTypeId="2" ParentId="7823" CreationDate="2011-03-31T16:20:14.617" Score="3" ViewCount="" Body="&lt;p&gt;You can travel to the future by accelerating near the speed of the light. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In physics, the twin paradox is a thought experiment in special relativity, in which a twin makes a journey into space in a high-speed rocket and returns home to find he has aged less than his identical twin who stayed on Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Twin_paradox&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Twin_paradox&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia discusses various theoretical possibilities for backward time traveling, which include by walking near high density cosmic masses, such as black holes that deform the space-time fabric, in a particular direction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Residing just outside of the event horizon of a black hole, or sufficiently near an object whose mass or density causes the gravitational time dilation near it to be larger than the time dilation factor on Earth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Time_travel&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Time_travel&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2868" LastEditorUserId="2868" LastEditorDisplayName="" LastEditDate="2011-03-31T17:12:23.400" LastActivityDate="2011-03-31T17:12:23.400" CommentCount="1" />
  <row Id="7846" PostTypeId="2" ParentId="7817" CreationDate="2011-03-31T16:30:46.713" Score="2" ViewCount="" Body="&lt;p&gt;I am not an expert by any means (just starting my studies in these topics) but i will share my understanding of the topic so to complement any further discussion of your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1)&lt;/strong&gt; The example of joining two points in a sheet of paper makes good intuition but i think its physically very misleading; there is no such operation in GR as joining two space-time points, because in the case of the sheet of paper the points are joined by being close in an embedding space (euclidean 3-dimensional space), while classical GR does not have any notion of an embedding space in which space-time exists. Other theories (not sure, but probably string theory?) may have further assumptions in this regard, but as far as i can tell, such embedding spaces are not allowed to produce any direct, physical consequences&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2)&lt;/strong&gt; GR allows solutions in space-time topologies that are non-trivial (where trivial means homeomorphic to asymptotically flat minkowski space), but the dynamics of GR doesn't allow by itself any dynamical transition between &lt;em&gt;space&lt;/em&gt; topologies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What there exists in the literature is postulating &lt;em&gt;ad-hoc&lt;/em&gt; the geometry in question, in this case two asymptotically flat minkowski spaces connected by the identification of two $S^3$ mouths on each space. &lt;em&gt;Then&lt;/em&gt; one uses GR equations to derive the required $T_{uv}$ tensor that makes the geometry stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What there is &lt;strong&gt;not&lt;/strong&gt; in the literature to my knowledge is taking flat minkowski space, and obtain a consistent evolution of trivial, flat space topology in a given time foliation parametrized as $t_{0}$ into another non-trivial space topology in another time $t_{1}$, even assuming exotic stress-energy densities, i.e: $T_{00} &amp;gt; T_{11} + T_{22} + T_{33}$. And the reason is that GR as a dynamical theory, is strictly a &lt;em&gt;local theory&lt;/em&gt; (please, correct me if this is wrong, this just represents my knowledge of the subject) and as a local theory, doesn't really tell much about &lt;em&gt;how the manifold behaves globally&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for a much more in-depth answer, please see &lt;a href=&quot;http://physics.stackexchange.com/questions/1787/what-is-known-about-the-topological-structure-of-spacetime&quot;&gt;What is known about the topological structure of spacetime?&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-03-31T16:43:39.463" LastActivityDate="2011-03-31T16:43:39.463" />
  <row Id="7847" PostTypeId="2" ParentId="7843" CreationDate="2011-03-31T16:41:37.550" Score="7" ViewCount="" Body="&lt;p&gt;The very claim that there are &quot;four fources&quot; is an approximation. We know that the electromagnetic and the weak force have to be unified to an electroweak theory. So counting the electroweak theory as one force, there are just three known elementary forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electroweak theory is based on the $SU(2)\times U(1)$ group which has two factors, but these two factors are not in one-to-one correspondence with the electromagnetism and the weak force, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strong force with its $SU(3)$ group is another seemingly independent factors, except that there is evidence that all three non-gravitational forces get unified into a grand unified force of a GUT theory at high energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;String theory unifies the non-gravitational forces with gravity, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every vacuum of string theory predicts gravity described by GR plus extra non-gravitational forces. The number of factors and their Higgs-like breaking patterns are essentially random properties of the string vacua. According to the anthropic picture of the world, the number of low-energy forces is an accidental property of our world that could be different in different parts of the multiverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to non-anthropic reasoning, the precise selection of our vacuum - including the fact that it has 4 low-energy forces - could be derivable from some more unique theoretical principles. However, this research program remains a wishful thinking as of 2011.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-31T16:41:37.550" />
  <row Id="7848" PostTypeId="2" ParentId="7843" CreationDate="2011-03-31T17:19:12.123" Score="3" ViewCount="" Body="&lt;p&gt;Currently there have been searches for 'fifth forces' in the literature, although exactly how you search depends on the details of the force in question.  The punchline is that no such thing has yet been detected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it acts somewhat like gravity, you could say look for departures from the equivalence principle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively if it was more particle physics based, you would expect that on general grounds the force was either spontaneously broken (making it very short range), or confining.  Thus you would need a detector of sufficient energy to produce the mediating particle, or alternatively to look for substructure in existing particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, all such searches have come up negative and although there are some theories that produce these objects, they tend to be less favored than simpler models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for 'why' 4 forces question (and not 3 or 5 or 100).  Well, 'why' questions in physics are very hard to answer, and typically lead you inexorably all the way to whatever fundamental theory lies at the heart of everything.  So see Lubos's post for what eg string theory might say, and how they are probably all unified anyway. &lt;/p&gt;&#xA;" OwnerUserId="1350" LastActivityDate="2011-03-31T17:19:12.123" CommentCount="1" />
  <row Id="7849" PostTypeId="1" CreationDate="2011-03-31T17:19:23.617" Score="1" ViewCount="124" Body="&lt;p&gt;say that i have 2 qubits - 2 spin half fermions.&#xA;my initial condition is |00&gt; in the spin-wave function and some anti-symmetrical spacial wave function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering about what happens when applying a NOT gate. The fermions total wave function still has to be inverted when swapping the particles. so does it mean that a NOT gate has to effect the spacial wave function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y.a.&lt;/p&gt;&#xA;" OwnerUserId="2869" LastActivityDate="2011-04-30T19:42:30.423" Title="fermions and quantum gates" Tags="&lt;quantum-information&gt;&lt;fermions&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="7850" PostTypeId="2" ParentId="7843" CreationDate="2011-03-31T17:21:58.920" Score="4" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But where did exactly four forces come from? Could new, perhaps supersymmetric particles hint at new fundamental forces?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sure, conceivably. After all, we didn't even know about the strong force and the weak force until the 1950s/1960s, so it's not at all unreasonable that the list of fundamental forces (or, if you prefer, the set of known gauge bosons a.k.a. &quot;force carriers&quot;) might be expanded again as experiments are able to probe higher energies. Right now, the theoretical models only involve the particular forces they do because there isn't any evidence of others.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-03-31T17:21:58.920" CommentCount="3" />
  <row Id="7852" PostTypeId="2" ParentId="7849" CreationDate="2011-03-31T17:39:54.700" Score="0" ViewCount="" Body="&lt;p&gt;It is not exactly clear which of the quantum &quot;NOT gates&quot; you meant - there are many quantum counterparts of the classical &quot;NOT gate&quot;. But you clearly meant an operation that changes the antisymmetry with respect to the spins, assuming that the qubits are encoded in the spins. Such operations obviously exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, if the spin-part of the wave function changes from an antisymmetric one to a symmetric one, and if the electrons fully carry the information about the qubits, then the spatial part of the wave function has to change from a symmetric to an antisymmetric one. That's because the total wave function is antisymmetric for identical fermions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's not &quot;difficult&quot; in any sense because in a quantum computer, if the information is stored by the electrons, they're supposedly located at different places - or different states - so the position (or another observable) effectively makes them distinguishable particles: you may identify the individual electrons employed by the computer by their (discrete) position, by the &quot;gate&quot; where they belong etc. Their identical character only becomes important if you try to physically exchange the fermions - their positions - or if you're trying to squeeze them to the very same place. But these conditions don't hold for electrons used in a quantum computer so the fact that they're identical fermions is pretty much irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, the wave function of $N$ qubits in a quantum computer have to be allowed to be symmetric, antisymmetric, or having no symmetry, otherwise it wouldn't be $N$ independent qubits of information and they couldn't be used to calculate anything of value.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-03-31T17:39:54.700" />
  <row Id="7853" PostTypeId="2" ParentId="7839" CreationDate="2011-03-31T17:41:41.800" Score="4" ViewCount="" Body="&lt;p&gt;This story may have originated with August Magnan and &lt;a href=&quot;http://en.wikipedia.org/wiki/Andr%C3%A9_Sainte-Lagu%C3%AB&quot; rel=&quot;nofollow&quot;&gt;André Sainte-Laguë&lt;/a&gt;. In the forward to his book &lt;em&gt;Le vol des insectes&lt;/em&gt;, August Magnan wrote &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Tout d'abord poussé par ce qui se fait en&#xA;  aviation, j'ai appliqué aux&#xA;      insectes les lois de la résistance de l'air, et je suis arrivé avec M.&#xA;  Sainte-Laguë a cette conclusion que&#xA;  leur vol est impossible.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;First, prompted by what is done in&#xA;  aviation, I&#xA;  applied the laws of air resistance to&#xA;  insects, and with M. Sainte-Laguë came to the conclusion&#xA;  that their flight is impossible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(my translation ... please correct if it's wrong).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He is talking about insects, and not specifically bumblebees.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditorDisplayName="" LastEditDate="2011-03-31T18:59:48.970" LastActivityDate="2011-03-31T18:59:48.970" />
  <row Id="7854" PostTypeId="2" ParentId="7819" CreationDate="2011-03-31T17:49:19.610" Score="0" ViewCount="" Body="&lt;p&gt;IMO there is no solid explanation, as anna said. &lt;a href=&quot;http://en.wikipedia.org/wiki/Venus#Orbit_and_rotation&quot; rel=&quot;nofollow&quot;&gt;Only clues (WP)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this simulation (2002) &lt;a href=&quot;http://www.imcce.fr/Equipes/ASD/preprints/prep.2002/venus2.2002.pdf&quot; rel=&quot;nofollow&quot;&gt;Long term evolution of the spin of Venus- II, Numerical simulations&lt;/a&gt; we find a mix of: 'chaotic zone', instability, large impact, close encounter, tidal effects, planetary perturbations,... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;There is room for speculation:&lt;/strong&gt;&lt;br&gt;&#xA;I think that the heavy atmosphere is not a significant factor of slowing the rotation.&lt;br&gt;&#xA;I discarded any violent event because it can easily change more than one parameter; in this case we have only one (the direction).&lt;br&gt;&#xA;The planet's minute axial tilt (less than three degrees, compared with 23 degrees for Earth) make me think to keep only the &lt;strong&gt;tidal perturbation&lt;/strong&gt;, although the present configuration Sun-Venus-Earth is not able to justify it.&lt;br&gt;&#xA;As the Venusian surface rotates as low as 6.5 km/h (on Earth is about 1,670 km/h) we can think that Venus may have changed the direction of rotation not long time ago.&lt;br&gt;&#xA;The solar system angular momentum problem is not solved (the planets strangely have almost all of it) and I think that the solution of this does one is not related to the present question.&lt;br&gt;&#xA;The &lt;strong&gt;equation 35&lt;/strong&gt; of this paper (&lt;a href=&quot;http://arxiv.org/PS_cache/astro-ph/pdf/0208/0208365v1.pdf&quot; rel=&quot;nofollow&quot;&gt;a new model, undiscussed&lt;/a&gt;) allows the slow evolution of the configuration Sun-Venus-Earth.  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-03-31T17:49:19.610" />
  <row Id="7856" PostTypeId="1" CreationDate="2011-03-31T18:26:32.307" Score="6" ViewCount="239" Body="&lt;p&gt;Suppose You have semigroup instead of typical group construction in Noether theorem. Is this interesting? In fact there is no time-reversal symmetry in the nature, right? At least not in the same meaning as with other symmetries ( rotation, translation etc). So why we construct energy as invariant of such kind of symmetry, not using semigroup instead? Is is even possible? There is a plenty of references about Lie semigroups - i looks like it is active field of mathematics. Is there any kind of Noether theory constructed within such theories? Could You give any references to, ideally introductory texts?&lt;/p&gt;&#xA;" OwnerUserId="1012" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-04-01T11:51:07.123" LastActivityDate="2011-04-01T11:51:07.123" Title="Noether theorem with semigroup of symmetry instead of group" Tags="&lt;lagrangian-mechanics&gt;&lt;symmetry&gt;&lt;reference-request&gt;&lt;noethers-theorem&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="7857" PostTypeId="2" ParentId="7856" CreationDate="2011-03-31T18:31:12.523" Score="1" ViewCount="" Body="&lt;p&gt;The main problem with semi-groups is that they are not in general invertible, so it doesn't apply well to the notion of a symmetry, which is something that behaves the same after performing an operation that transforms the states into each other&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-03-31T18:48:52.430" LastActivityDate="2011-03-31T18:48:52.430" CommentCount="3" />
  <row Id="7858" PostTypeId="2" ParentId="7856" CreationDate="2011-03-31T18:41:26.853" Score="6" ViewCount="" Body="&lt;p&gt;As lurscher says, semigroups are not too interesting in physics. They're not really &quot;symmetries&quot;. An important physics example of a semigroup in physics is the (misleadingly called) &quot;Renormalization Group&quot; that allows us to derive effective laws for long distances from the short-distance ones, but this &quot;integrating out&quot; or &quot;flowing&quot; is irreversible - which is exactly why the RG operations deserve to constitute a &quot;semigroup&quot; rather than a &quot;group&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However, more importantly, you are totally confused about the case of Noether's observable called &quot;energy&quot;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy is associated with the time-translational-invariance and it is a standard symmetry, expressed by a group and not &quot;just a semigroup&quot;. This symmetry says that the laws of physics don't change if the events occur $t$ seconds later. Also, they didn't change if the processes occurred $t$ minutes earlier. The time translations go in both directions, they can be both positive and negative, and they're always symmetries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has nothing to do with the irreversibility of the processes or the asymmetries associated with the arrow of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you are actually wrong even about the time reversal. Even the time reversal symmetry, which is a symmetry of the microscopic laws of physics (if CP holds; or at least CPT is such a symmetry universally) is a symmetry, and it is again expressed by a $Z_2$ group. There is no non-group semigroup anywhere in those spacetime-related operations!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Back to why &quot;semigroups&quot; are not symmetries&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite generally, objects such as the action cannot be symmetric under semigroups that are not groups. The example of the Renormalization Group clearly shows what happens if the would-be symmetry transformations were irreversible. Irreversibility always means that some information is getting lost, so the contributions to the action from the degrees of freedom whose information is getting lost are inevitably being suppressed, so that action cannot stay constant. If one &quot;flows&quot; to long distances via the Renormalization Group flows, he actually &quot;integrates out&quot; the shortest-distance degrees of freedom which means that they cannot be excited in the new effective theory. So the operation cannot provide us with a one-to-one map for all states which, as explained in the previous sentences, is necessary for a symmetry, and symmetries therefore have to groups and not just semigroups.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-03-31T18:47:54.447" LastActivityDate="2011-03-31T18:47:54.447" CommentCount="5" />
  <row Id="7859" PostTypeId="2" ParentId="7849" CreationDate="2011-03-31T19:24:28.623" Score="1" ViewCount="" Body="&lt;p&gt;First of all, you need to be careful what you mean by a NOT gate. If you have only one qubit, you could use the gate (typically called X) $$\begin{pmatrix}0 &amp;amp; 1\\1&amp;amp; 0\end{pmatrix},$$ which flips the two basis states of the qubit. Since your system here consists of only one electron, this won't affect the 'spatial wave function' very much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're talking about two or more qubits, you need to keep in mind that quantum operations are always done whilst (physically) identifying each of the qubits. If, for example, you'd have the following operation:&#xA;$$|00\rangle \mapsto \frac{1}{\sqrt{2}}\left[|01\rangle - |11\rangle\right],$$ that means that the 'left' bit is mapped to 0 or 1, whilst the 'right' one is mapped to 1. You're therefore never really 'swapping particles'. In fact, if you were to do quantum computing on an entangled two-fermion system, (say: a deuterium atom), how would you probe the system, i.e., how would you experimentally distinguish between $|01\rangle$ and $|10\rangle$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you &lt;em&gt;could&lt;/em&gt; do, is swap the basis states of the two qubits, using a gate like&#xA;$$\begin{pmatrix}1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0\\0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 \\ 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 \\0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1\end{pmatrix},$$ where the basis states are labeled $|00\rangle,\quad |10\rangle,\quad|01\rangle,\quad|11\rangle.$ [The matrix above is indeed unitary.] In fact, this quantum operation would leave the spin-symmetric states $|00\rangle, \quad |11\rangle$ alone.&lt;/p&gt;&#xA;" OwnerUserId="1471" LastActivityDate="2011-03-31T19:24:28.623" />
  <row Id="7860" PostTypeId="2" ParentId="7839" CreationDate="2011-03-31T20:31:25.460" Score="4" ViewCount="" Body="&lt;p&gt;Insect flight is different than bird flight.  The rapidly moving wings, which do a figure 8 sort of motion, generates a vortex tube over the wings.  This vortex by Bernoulli principle has less pressure, which permits the larger air pressure underneath to lift the animal up.  If one is trying to understand insect flight according to the mechanics of bird or aircraft flight you are then pounding a square peg into a round hole, which makes you conclude bees can’t fly.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-03-31T20:31:25.460" />
  <row Id="7861" PostTypeId="2" ParentId="7856" CreationDate="2011-03-31T22:22:44.750" Score="2" ViewCount="" Body="&lt;p&gt;Here is an argument why &quot;a Noether Theorem with Lie monoid symmetry&quot; essentially wouldn't produce any new conservation laws. Noether's (first) Theorem is really not about Lie groups but only about Lie algebras, i.e., one just needs $n$ &lt;em&gt;infinitesimal&lt;/em&gt; symmetries to deduce $n$ conservation laws. If one is only interested in getting the $n$ conservation laws one by one (and not so much interested in the fact that the $n$ conservation laws together form a representation of the Lie algebra), then one may focus on a $1$-dimensional Abelian subgroup of symmetry. The corresponding Lie subalgebra then becomes just $u(1)\cong\mathbb{R}$. Now returning to the question, one may, of course, artificially truncate a Lie group into a Lie monoid, say, if $q$ is a cyclic variable for a Lagrangian $L$, then artificially declare that the symmetry monoid is $q \to q + a$ for only non-negative translations $a\geq 0$, while artificially denying all negative $a&amp;lt;0$. On the other hand, one needs at least access &quot;from one side&quot; because Noether's Theorem is about &lt;em&gt;continuous&lt;/em&gt; symmetry. But in practice, one can then always extend, at least infinitesimally, to &quot;the other side&quot; as well, and then one is back to a standard $u(1)$ Lie algebra and a standard Noether Theorem.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-04-01T10:52:58.470" LastActivityDate="2011-04-01T10:52:58.470" CommentCount="4" />
  <row Id="7862" PostTypeId="1" CreationDate="2011-03-31T22:47:59.993" Score="6" ViewCount="159" Body="&lt;p&gt;Does salt water conduct mostly by the ions travelling through the solution, or by electrons collectively flowing or hopping through the solution like in metals?&lt;/p&gt;&#xA;" OwnerUserId="2702" LastActivityDate="2011-04-01T21:01:39.740" Title="Salt water conduction" Tags="&lt;electricity&gt;" AnswerCount="2" />
  <row Id="7863" PostTypeId="1" CreationDate="2011-03-31T23:12:52.980" Score="6" ViewCount="452" Body="&lt;p&gt;I have seen in popular media, claims that Hawking does not believe the Higgs boson exists due to microscopic black holes and even made a bet against it.  This is based on something published in journal Physical Review D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have access to journal Physical Review D, and I can't find a clear detailed explanation what his claim is, and what his proposed alternative is.  Can someone explain it for this curious layman?&lt;/p&gt;&#xA;" OwnerUserId="2702" LastActivityDate="2011-04-01T03:59:36.300" Title="Hawking's alternative to Higgs Boson" Tags="&lt;black-holes&gt;&lt;large-hadron-collider&gt;&lt;higgs-boson&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7864" PostTypeId="1" CreationDate="2011-03-31T23:23:57.140" Score="4" ViewCount="424" Body="&lt;p&gt;I am trying to reconcile data that I have found in &lt;a href=&quot;http://www.xn--rheinischesmuseumfrphilologie-2bd.com/fileadmin/content/dokument/archiv/silvaegenetica/18_1969/18-1-2-40.pdf&quot; rel=&quot;nofollow&quot;&gt;one publication (Allen 1969)&lt;/a&gt; with data that I found in &lt;a href=&quot;http://www.jstor.org/stable/1514306&quot; rel=&quot;nofollow&quot;&gt;another publication (George 2003)&lt;/a&gt;  that synthesized this data. The &#xA;data is root respiration rate, it was originally measured at $27\ ^\circ C$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Approach&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I am trying to convert a rate of oxygen consumed as volume per mass of root per time to carbon dioxide produced as mass per unit mass per time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the appendix table, George 2003 reports the range of root respiration rates, converted to $15\ ^\circ C$ and standard units:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[11.26, 22.52]  \frac{\mathrm{nmol CO}_2}{\mathrm{g}\ \mathrm{s}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the original publication Allen (1969), root respiration was measured at  $27\ ^\circ $C. &#xA;The values can be found in table 3 and figure 2. &#xA;The data include a minimum (Group 2 Brunswick, NJ plants) and a maximum (Group 3 Newbery, South Carolina), which I assume are the ones used by George 2003:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[27.2, 56.2] \frac{\mu\mathrm{L}\ \mathrm{O}_2}{10\mathrm{mg}\ \mathrm{h}}$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Step 1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Transformed George 2003 measurements back to the measurement temperature using a rearrangement of equation 1 from George, the standardized temperature of  $15\ ^\circ $C stated in the Georgeh table legend, and  Q$_{10} = 2.075$ from George 2003, and the measurement temperature of $27\ ^\circ $C reported by Allen 1969: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_T = R_{15}[\exp(\ln(Q_{10})(T- 15))/10]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[11.26, 22.52] * exp(log(2.075)*(27 - 15)/10)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have the values that we would have expected to find in the Allen paper, except that the units need to be converted back to the original:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[27.03,54.07] \mathrm{nmol CO}_2\ \mathrm{g}^{-1}\mathrm{s}^{-1}$$ &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Step 2: convert the units&lt;/h2&gt;&#xA;&#xA;&lt;h3&gt;Required constants:&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;inverse density of $\mathrm{O}_2$ at $27^\circ C$: $\frac{7.69 \times 10^5\ \mu\mathrm{L}\ \mathrm{O}_2}{\mathrm{g}\ \mathrm{O}_2}$ first assume that Allen converted to sea level pressure (101 kPa), although  maybe they were measured at elevation (Allen may have worked at \~{} 900 kPa near Brevard, NC)&lt;/li&gt;&#xA;&lt;li&gt;molar mass of $\mathrm{O}_2$: $\frac{32\mathrm{g}\ \mathrm{O}_2}{\mathrm{mol}}$&lt;/li&gt;&#xA;&lt;li&gt;treat 10mg, which is in the unit of root mass used by Allen, as a unit of measurement for simplicity &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now convert $$[27.03,54.07] \mathrm{nmol CO}_2\ \mathrm{g}^{-1}\mathrm{s}^{-1}$$ to units of $\frac{\mu\mathrm{L}\ \textrm{O}_2}{10\mathrm{mg}\ \mathrm{root}\ \mathrm{h}}$.&#xA;The expected result is the original values reported by Allen: $[27.2, 56.2] \frac{\mu\mathrm{L}\ \mathrm{O}_2}{10\mathrm{mg}\ \mathrm{h}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[27.03, 54.07]\ \frac{\mathrm{nmol}\ \mathrm{CO}_2}{\mathrm{g}\ \mathrm{root}\ \mathrm{s}} \times \frac{1\ \mathrm{g}}{100\times10\mathrm{mg}} \times \frac{3600\ \mathrm{s}}{\mathrm{h}} \times \frac{3.2 \times 10^{-8}\ \mathrm{g}\ \mathrm{O}_2}{\mathrm{nmol}\ \mathrm{O}_2}\times \frac{7.69\times10^5\ \mu\mathrm{L}\ \mathrm{O}_2}{\mathrm{g}\ \mathrm{O}_2}$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Result:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$$[23.8, 47.8]&#xD;&#xA;  \frac{\mu\mathrm{L}\ \textrm{O}_2}{10\mathrm{mg}\ \mathrm{h}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the units reported in the Allen paper, but they appear to be underestimates . Since the ratio of observed:expected values are different, it is not likely that Q$_{10}$ or the atmospheric pressure at time of measurement would explain this error.  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Am I doing something wrong?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.xn--rheinischesmuseumfrphilologie-2bd.com/fileadmin/content/dokument/archiv/silvaegenetica/18_1969/18-1-2-40.pdf&quot; rel=&quot;nofollow&quot;&gt;Reference 1&lt;/a&gt;: Allen, 1969, Racial variation in physiological characteristics of shortleaf pine roots., Silvics Genetics 18:40-43&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.jstor.org/stable/1514306&quot; rel=&quot;nofollow&quot;&gt;Reference 2&lt;/a&gt;: George et al 2003, Fine-Root Respiration in a Loblolly Pine and Sweetgum Forest Growing in Elevated CO2. New Phytologist, 160:511-522&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Footnote 1: The values from reference 2 are adjusted from the $15^\circ C$ reference temperature to the $27^\circ C$ in reference 1 using the Ahhrenius equation, but I am off by an order of magnitude so I do not think that this is relevant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_T = R_{15}[\exp(\ln(Q_{10})(T- 15))/10]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[26.9, 54.0] = [11.2, 22.5] * exp(log(2.075)*(27 - 15)/10)$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;note:&lt;/em&gt; I have been updating the equation based errors pointed out by Mark and rcollyer, but the problem remains&lt;/p&gt;&#xA;" OwnerUserId="2809" LastEditorUserId="2809" LastEditorDisplayName="" LastEditDate="2011-04-07T07:26:27.617" LastActivityDate="2011-04-08T17:43:18.890" CommunityOwnedDate="2011-04-07T07:04:57.143" Title="What am I doing wrong when I try to make this unit conversion?" Tags="&lt;unit-conversion&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="7865" PostTypeId="1" CreationDate="2011-03-31T23:49:24.197" Score="8" ViewCount="226" Body="&lt;p&gt;Please note, I am only interested in classical mechanics discussion on this.  Please do not involve quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inspired by this question: &lt;a href=&quot;http://physics.stackexchange.com/questions/822/is-angular-momentum-truly-fundamental/4045&quot;&gt;Is Angular Momentum truly fundamental?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can there be a concept of angular momentum separate from &quot;orbital&quot; angular momentum in classical mechanics?  For example, can there be a thing such as &quot;intrinsic&quot; angular momentum in a classical theory that could be distinguished from the limit of shrinking a spinning ball to size zero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seemed obvious to me that the answer should be no. -- No such distinction could be made in a classical theory.  However searching on related topics and reading more brought me to this article in wikipedia:&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein%E2%80%93Cartan_theory&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Einstein%E2%80%93Cartan_theory&lt;/a&gt;&lt;br&gt;&#xA;While I don't follow the math details, the overview comments seem to claim there is a classical notion of intrinsic spin, but GR cannot handle intrinsic angular momentum.  And a different classical theory is presented which does, and differs from GR.  For instance, some important tensors can be non-symmetric now, which can't happen in GR no matter the size we make a spinning ball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore there seems to be a real distinction between &quot;intrinsic&quot; angular momentum, and the limit of shrinking a spinning ball to size zero, even in classical mechanics.  This blows my mind.  So if the answer to the above questions are YES!, can someone help explain what this classical &quot;intrinsic&quot; angular momentum is?&lt;/p&gt;&#xA;" OwnerUserId="2702" LastActivityDate="2011-04-04T14:41:43.197" Title="Intrinsic angular momentum in classical mechanics" Tags="&lt;general-relativity&gt;&lt;classical-mechanics&gt;&lt;angular-momentum&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="7866" PostTypeId="2" ParentId="7862" CreationDate="2011-04-01T00:18:00.827" Score="5" ViewCount="" Body="&lt;p&gt;Let's look at pure Water with pure Sodium Chloride in it. We know that in Sodium Chloride, the Sodium atoms give one electron to the Chlorine. This results in both atoms having closed shells. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For electrons to collectively flow/hop through a material, you need to have electron states that are delocalized over many atoms. This is what happens in metals: The individual atomic orbitals overlap and form bands. Now, band theory also tells you that for conduction to occur, you need a band that is partially filled. If all your bands are completely empty or completely full, you cannot have conduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since, as I have described at the top, your Sodium and Chloride ions have closed shells, there will be no transport of electrons among the atoms. If you think about it in terms of hopping: Both Sodium and Chloride will be very happy with the number of electrons they currently possess. Hence, they will be very reluctant to either give up or accept hopping electrons. I will not go as far as claiming that these processes are entirely unlikely, just strongly suppressed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, it'll be the ions itself doing the (bulk) of the transport.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: My reasoning above rules out that the electrons will behave the same as in a metal, i.e. forming a Fermi gas. As pointed out in the comments, chemical solutions exist where electrons are free to move around, behaving like ions. Another answer explains why this cannot happen in water. &lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditorDisplayName="" LastEditDate="2011-04-01T18:01:42.227" LastActivityDate="2011-04-01T18:01:42.227" CommentCount="9" />
  <row Id="7867" PostTypeId="2" ParentId="7817" CreationDate="2011-04-01T00:33:18.557" Score="0" ViewCount="" Body="&lt;p&gt;Wormholes are similar to black holes, but where the even horizon is replaced by a membrane of some type of quantum field which causes geodesics to diverge.  The Hawking-Penrose energy condition, in particular the weak energy condition, gives geodesics which focus inwards in a spacetime diagram.  In order to get a wormhole it requires that inward focusing geodesics be “defocused” near or at the region where the event horizon of a black hole would otherwise exist.  This means the geodesics are defocused into some other region of spacetime.  The wormhole is then two 3-balls cut out of spacetime and where the boundaries of the two balls have points identified with each other.  The special field which violates the Hawking-Penrose energy conditions defines a junction in spacetime where the curvature exhibits an abrupt change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We might think of a wormhole as a cut in space, say the removal of a ball $B^3$ with a spherical surface $S^2~=~\partial B^3$, which is matched by another ball, where the surfaces of the two balls are identical.  The two spheres $S^2$ and $S’^2$ then have points on them which are identified with each other.  The two ball then in effect when sutured together define a three sphere.  The junction is then a region at this surface, or just above where the curvature of spacetime jumps with a negative sign.  This results in a defocusing of geodesics across the $S^2$, which forces the geodesic to emerge at $S^2$.  This is a multiply connected space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty comes if one of these openings is accelerated in a “send and return” process.  This in special relativity results in the so called “twin paradox,” which for the case of a wormhole means the clock on the accelerated opening ends up in the past of the clock on the unaccelerated opening.  This is a form of time machine.  The multiply connected space is embedded in spacetime and by this means is transformed into a multiply connected space plus time.  This is the primary difficulty with the whole idea of wormholes, for the symmetries of spacetime mean one can transform this into the funny form of a time machine.  Now it might be that at the point the two openings in this send and return process have the same proper time and are connected by a null ray that something disastrous might happen.  As this condition is met the vacuum will wind through the wormhole in a “pile up” or asymptotic condition approaching this null condition.  This might violently destroy the wormhole before it can be transformed to a time machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the negative energy of the exotic quantum field may do that anyways.  This quantum field is not bounded below, which means the ladder of states descends endlessly to $E~=~-\infty$.  This means a vast amount of radiation would flow out from this strange quantum field.  In fact this will always happen to the extent the negative energy field is “cancelled out.”  So this negative energy or exotic quantum field seems to be self-canceling and simply can’t exist.  Wormholes on a macroscopic scale most likely do not exist.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-04-01T00:33:18.557" />
  <row Id="7868" PostTypeId="1" CreationDate="2011-04-01T02:36:25.830" Score="4" ViewCount="402" Body="&lt;p&gt;Why does the air we blow/exhale out from our mouths change from hot to cold depending on the size of the opening we make with our mouth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not just a subtle difference, but significant in my opinion. I'm inclined to discredit the notion that it's just a matter of speed because I can blow fast with an open mouth and still, it's hot; and blow slow with an almost closed (tighter) mouth and again, it's cold.&lt;/p&gt;&#xA;" OwnerUserId="2874" LastActivityDate="2011-04-03T16:34:21.147" Title="Why does the air we blow/exhale out from our mouths change from hot to cold depending on the size of the opening we make with our mouth?" Tags="&lt;heat&gt;&lt;temperature&gt;&lt;air&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="7869" PostTypeId="2" ParentId="7868" CreationDate="2011-04-01T02:44:42.073" Score="5" ViewCount="" Body="&lt;p&gt;It is mainly due to air entrainment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you blow through a tight mouth, there is smaller volume of air but a higher velocity. This pulls in and mixes with a lot of ambient air (Venturi or Bernoulli effect) - in fact typically the air stream is only 40% body warmth and 60% ambient so it will be markedly colder. As an experiment you pucker and blow through a tube held to your mouth, this excludes the ambient air and you will get reduced airflow but at the higher temperature again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a wide mouth there is hardly any air entrainment. Its temperature will be almost same as its temperature in your lungs which is higher than the ambient temperature.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-04-03T16:34:21.147" LastActivityDate="2011-04-03T16:34:21.147" CommentCount="5" />
  <row Id="7871" PostTypeId="1" AcceptedAnswerId="7876" CreationDate="2011-04-01T03:11:30.590" Score="4" ViewCount="115" Body="&lt;p&gt;Since strings are extended objects, is all angular momentum in string theory essentially &quot;orbital&quot; angular momentum?  Or is there still a kind of intrinsic angular momentum assigned to a string?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, is there anything that prevents the &quot;intrinsic spin&quot; of a particle represented by a string from being arbitrarily large?&lt;/p&gt;&#xA;" OwnerUserId="2702" LastActivityDate="2011-04-01T06:06:12.577" Title="Angular momentum in string theory" Tags="&lt;string-theory&gt;&lt;angular-momentum&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="7872" PostTypeId="1" CreationDate="2011-04-01T03:16:38.100" Score="7" ViewCount="277" Body="&lt;p&gt;What are the major recent findings and their corresponding contributions to an overall picture?&#xA;How well explained are the various regions of the dome, is there any thing that is pretty well understood?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is left and how are people trying to figure it out?&lt;/p&gt;&#xA;" OwnerUserId="2875" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-30T18:31:17.070" LastActivityDate="2011-08-30T18:31:17.070" Title="How far away are we from resolving high temperature superconductivity?" Tags="&lt;condensed-matter&gt;&lt;superconductivity&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="7873" PostTypeId="2" ParentId="7868" CreationDate="2011-04-01T03:23:09.613" Score="0" ViewCount="" Body="&lt;p&gt;I don't think it's quite that physical. The pressures involved aren't that high&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you breathe slowly on your hand the air is war, moist and 37C so feels warm compared to the surroundings, if you blow through a small opening the flow of air increases the cooling and evaporation from your skin.&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-04-01T03:23:09.613" CommentCount="5" />
  <row Id="7874" PostTypeId="1" CreationDate="2011-04-01T03:34:26.140" Score="2" ViewCount="86" Body="&lt;p&gt;While the microscopic mechanism leading to type II superconductivity is still debated, do scientists at least have a handle on the phenomonology?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, are the experimental properties of type II superconductivity understood well enough that it is possible in principle for the properties of a newly discovered material to lead to declaring a type III superconductor?  Or is type II so poorly understood, that it is best described as &quot;not type I&quot; superconductivity, and thus leaving no room for a type III superconductor?&lt;/p&gt;&#xA;" OwnerUserId="2702" LastActivityDate="2011-04-01T03:34:26.140" Title="Type II Superconductors" Tags="&lt;superconductivity&gt;" CommentCount="2" />
  <row Id="7875" PostTypeId="2" ParentId="7863" CreationDate="2011-04-01T03:52:19.690" Score="2" ViewCount="" Body="&lt;p&gt;There is a &lt;a href=&quot;http://blogs.plos.org/badphysics/2010/10/25/br_sample/&quot; rel=&quot;nofollow&quot;&gt;book review&lt;/a&gt; which mentions this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Famously, Stephen Hawking made his bet for $100 with Gordy Kane that neither LEP nor any other particle collider would ever see the Higgs boson because of virtual black holes.  Perhaps unsurprisingly, Peter Higgs was very doubtful of the calculations that lead Hawking to this conclusion (and Peter certainly wasn’t alone) and this lead to a little public drama between the two scientists in the media.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe the book has details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://motls.blogspot.com/2008/09/why-lhc-will-see-higgs-peter-higgs-vs.html&quot; rel=&quot;nofollow&quot;&gt;Lubos Motl's blog&lt;/a&gt; on the subject where also the &lt;a href=&quot;http://arxiv.org/abs/hep-th/9510029&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt; to the paper of Hawking is discussed, rather unfavorably. :;&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-04-01T03:59:36.300" LastActivityDate="2011-04-01T03:59:36.300" />
  <row Id="7876" PostTypeId="2" ParentId="7871" CreationDate="2011-04-01T06:06:12.577" Score="3" ViewCount="" Body="&lt;p&gt;The orbital momentum of a string may be arbitrarily large. Whether it should be called &quot;orbital&quot; or &quot;intrinsic&quot; depends on the perspective. The right answer is the formula, such as&#xA;$$J_{ij} = \int d\sigma [x_i (\sigma) p_j (\sigma) - p_i(\sigma) x_j(\sigma) + \gamma_{ij}^{ab} \theta_a(\sigma) \theta_b(\sigma)]$$&#xA;I added some superstring term, too. The $xp$ terms may be viewed as a density of the orbital angular momentum on the string; the fermionic term is its most direct fermionic generalization. However, both of these terms, and especially the latter, become &quot;intrinsic angular momentum&quot; when you expand the fields $x,p,\theta$ to Fourier modes and interpret the string as a particle with internal oscillations. In particular, the intrinsic spin-1/2 always comes from the quantization and/or excitations of the fermionic degrees of freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula says much more than just some confusing dogmatic word &quot;intrinsic&quot; vs &quot;orbital&quot;, at least to a person who wants to understand the terms accurately enough - at the level of maths.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-01T06:06:12.577" />
  <row Id="7877" PostTypeId="1" AcceptedAnswerId="7878" CreationDate="2011-04-01T07:33:55.473" Score="5" ViewCount="228" Body="&lt;p&gt;Consider a pendulum, consisting of a string of length $l$ tied to a ball of negligible mass and radius $r$. The bob is filled with water, which has density $d$, and the pendulum is given a small push and left to oscillate. There is no friction or damping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an analysis I've heard during a high school class. Qualitatively, when the water level is nearly full, the CG of the bob is at its centre, so the length of the pendulum is effectively $l+r$, and it oscillates with the corresponding period. As water drains out, the CG is lowered, until it reaches $l+2r$ , and the period of the pendulum correspondingly increases. Once all the water has drained out,  the pendulum resumes oscillating with a length of $l+r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions: Is this analysis correct? I'd really like to know the quantitative model for this.&lt;/p&gt;&#xA;" OwnerUserId="2878" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-04-01T11:49:16.093" LastActivityDate="2011-04-08T05:36:45.853" Title="Pendulum with water dripping out" Tags="&lt;classical-mechanics&gt;&lt;fluid-dynamics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="7878" PostTypeId="2" ParentId="7877" CreationDate="2011-04-01T08:57:23.490" Score="4" ViewCount="" Body="&lt;p&gt;Not quite. But before getting into the math, you'd need to fix up the problem a bit: if the string and the ball are &lt;em&gt;both&lt;/em&gt; massless, then there will be nothing left to oscillate once the water drains out. It probably makes the most sense to assume that the ball is a uniform spherical shell with a mass $m$. For simplicity, I'll make two more assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The rate at which the water drains is slow, so that the change in mass over each oscillation is negligible (i.e. the process is adiabatic)&lt;/li&gt;&#xA;&lt;li&gt;The momentum/energy of the stream of water escaping is also negligible&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm also using the result which Georg brought up, that the surface of the water remains perpendicular to the pendulum's axis even as it swings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I'll compute the volume of water in the ball when it's filled up to a height $h$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V = \int_{-r}^{h-r} \pi(r^2 - z^2)\mathrm{d}z = \pi h^2\biggl(r - \frac{h}{3}\biggr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then the moment of that water,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M z_{w} = \int_{-r}^{h-r} zd\pi (r^2 - z^2)\mathrm{d}z = \pi d h^2 \biggl(hr - r^2 - \frac{h^2}{4}\biggr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the overall center of mass, you add this to the moment of the sphere (which is zero since its center of mass is at $z = 0$) and divide by the total mass. Since $M = dV$, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z_{CM} = \frac{\pi d h^2\bigl(hr - r^2 - h^2/4\bigr)}{(d) \pi h^2\bigl(r - h/3\bigr) + m} = \frac{hr - r^2 - \frac{h^2}{4}}{r - \frac{h}{3} + \frac{m}{\pi h^2 d}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing that you're calling effective length in the problem is $l + r - z_{CM}$. Using $m = 1$, $r = 1$, and $d = 1$ as sample values (e.g. if you are using CGS units with lengths in centimeters and masses in grams), a plot looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/9mHLF.png&quot; alt=&quot;plot of first approximation to effective length&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph does start and end at zero, which means that at the beginning and at the end, the center of mass is at the center of the sphere at those times, as you said. But in the middle, it never goes below about -0.24, which means that the radius of the swing of the center of mass never gets longer than $l + 1.24 r$ (in this example). It won't reach all the way out to $l + 2r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as Georg also mentioned, approximating the pendulum as a point mass on a string of varying length isn't quite accurate. To do better, we can use the rotational equation $\vec\tau = I\vec\alpha$ to account for the pendulum's finite extent perpendicular to the axis. The torque is produced by the force of gravity acting downward through the center of mass, and after a little rearranging the equation becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ddot\theta + \frac{mg r_g}{I}\sin\theta = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $I$ is the moment of inertia of the pendulum and $r_g$ is the radius at which the gravitational force acts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the behavior of the pendulum in this model is determined entirely by the value of the constant $\omega^2 \equiv \frac{mg r_g}{I}$. For a point pendulum of length $l$, you have $r_g = l$ and $I = ml^2$, so $\omega^2 = \frac{g}{l}$. We can therefore define the &quot;effective length&quot; of a more complex pendulum as the length that gives the same value of $\omega^2$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$l_\text{eff} = \frac{g}{\omega^2} = \frac{I}{m r_g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the dripping pendulum, we already have $r_g$: it's just the radius of the swing of the center of mass, which is what you were calling &quot;effective length&quot; in the problem. But the new definition also incorporates the moment of inertia of the pendulum bob around the pivot point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = \bigl(I_\text{shell,CM} + m(l + r)^2\bigr) + \bigl(I_\text{water,CM} + M(l + r - z_w)^2\bigr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term is the moment of inertia of the spherical shell around its own center of mass, $I_\text{shell,CM} = \frac{2}{3}mr^2$. The third term is the moment of inertia of the &lt;em&gt;water&lt;/em&gt; around its own center of mass. It takes the shape of a &quot;chopped&quot; sphere so we have to do the integral to calculate it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_\text{water,CM} = \iiint x^2\,\mathrm{d}^3 M = \int_{-r}^{h-r}\int_{0}^{2\pi}\int_{0}^{\sqrt{r^2 - z^2}} \bigl((s\cos\phi)^2 + (z - z_w)^2\bigr) d\,s\,\mathrm{d}s\mathrm{d}\phi\mathrm{d}z$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second and fourth terms in $I$ come from the parallel axis theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting it all together, we get a long but straightforward expression which can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$l_\text{eff} = l + \frac{12mrl + 20mr^2 +\pi dh^2\bigl[-\frac{9}{5}h^3 + 24r^2(l+2r)+3h^2(l+5r)-4hr(4l+11r)\bigr]}{12m(l+r) + \pi dh^2\bigl[4l(3r-h)+3h^2-16hr+24r^2\bigr]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(yes, I used Mathematica for the algebra :-P) The graph for $l=10$ looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/VT0hK.png&quot; alt=&quot;plot of more accurate effective length&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red line is $l_{eff}$, the blue line is the swing radius of the center of mass (the same thing that was in the last graph) just for comparison. Notice that for these values they're pretty similar, and as $r$ shrinks relative to $l$ the curves get even closer together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, you can show this from the analytical analysis as well. I won't show the complete details, but if you eliminate $h$ and $r$ in favor of $\epsilon = \frac{h}{r}$ and $\beta = \frac{r}{l}$, and do a series expansion in $\beta$, if Mathematica is to be trusted, you wind up with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{l_{eff}}{l} = 1 + \beta + \frac{2}{3}(\beta^2 - \beta^3) + P^4(\epsilon)\mathcal{O}(\beta^4)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the amount by which the sphere is filled doesn't even matter until you reach the fourth-order correction in $\frac{r}{l}$.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-08T05:36:45.853" LastActivityDate="2011-04-08T05:36:45.853" CommentCount="5" />
  <row Id="7879" PostTypeId="1" AcceptedAnswerId="7883" CreationDate="2011-04-01T09:48:28.553" Score="2" ViewCount="209" Body="&lt;p&gt;When I look at pictures taken by a modern spacecraft (like that one showing the surface of Titan: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/b/bc/Huygens_surface_color.jpg&quot; alt=&quot;Titan&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and taken by &lt;a href=&quot;http://en.wikipedia.org/wiki/Cassini%E2%80%93Huygens&quot; rel=&quot;nofollow&quot;&gt;Cassini&lt;/a&gt;) I kinda understand how they might be produced. There might be some kind of CCD sensor installed and the data is later transmitted to Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then I think about it I start to wonder how cameras worked on older spacecrafts like &lt;a href=&quot;http://en.wikipedia.org/wiki/Mariner_program&quot; rel=&quot;nofollow&quot;&gt;Mariner&lt;/a&gt;, for example this one taken by &lt;a href=&quot;http://en.wikipedia.org/wiki/Viking_1&quot; rel=&quot;nofollow&quot;&gt;Viking 1&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/a/ae/Mars_Viking_12a001.png&quot; alt=&quot;Mars&quot;&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know there were no digital cameras back then. How did they get around it and were able to produce photos and more importantly sent imaging data back to Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can imagine that there was an analog camera installed along with some scanning facility but that sounds &lt;em&gt;sooo&lt;/em&gt; unlikely as it would require a rather complex machinery which isn't feasible for an operation in space. But other than that I'm out of ideas. Wikipedia is surprisingly silent about it, too.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Thanks, everybody, for your great answers! I upvoted everything I could and accepted &lt;a href=&quot;http://physics.stackexchange.com/users/1370/georg&quot;&gt;Georg&lt;/a&gt;'s answer for a great Wiki link.&lt;/p&gt;&#xA;" OwnerUserId="541" LastEditorUserId="541" LastEditorDisplayName="" LastEditDate="2011-04-02T00:17:47.800" LastActivityDate="2011-04-02T00:17:47.800" Title="How do spacecraft cameras work?" Tags="&lt;space&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="7880" PostTypeId="1" CreationDate="2011-04-01T10:12:17.613" Score="0" ViewCount="138" Body="&lt;p&gt;What makes zinc so desperate to give up its electrons? Silver has one extra electron, and zinc has two, but why is zinc more reactive?&lt;/p&gt;&#xA;" OwnerUserId="2881" LastActivityDate="2011-09-05T05:19:45.563" Title="What makes zinc reactive?" Tags="&lt;homework&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7882" PostTypeId="2" ParentId="7862" CreationDate="2011-04-01T11:49:52.193" Score="2" ViewCount="" Body="&lt;p&gt;No, &#xA;free electrons in water cannot &quot;survive&quot;. In water those electrons would react to form &#xA;hydrogen atoms and then molecules. Water is just a little bit too &quot;acidic&quot;. &#xA;There is no influence from the salt ions, in purest water the conduction is by &#xA;H+ and OH- exclusively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Free electrons&quot; are possible in liquid ammonia and in some (aprotic) ethers. &#xA;You can see that electrons in ammonia due to their intense blue colour. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-01T21:01:39.740" LastActivityDate="2011-04-01T21:01:39.740" CommentCount="4" />
  <row Id="7883" PostTypeId="2" ParentId="7879" CreationDate="2011-04-01T11:59:15.743" Score="4" ViewCount="" Body="&lt;p&gt;There were a lot of (TV) Cameras from the first &lt;a href=&quot;http://www.oebl.de/Fernseh/Kamera/index.html&quot; rel=&quot;nofollow&quot;&gt;Ikonoskops&lt;/a&gt; &#xA;(by Zworykin) in the 30ties up to Orthicons or Plumbicons and what &lt;a href=&quot;http://en.wikipedia.org/wiki/Video_camera_tube&quot; rel=&quot;nofollow&quot;&gt;-cons&lt;/a&gt; ever &#xA;were invented.  Just google for that names. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS The word &quot;digital&quot; is misused a lot today. The CCDs &#xA;are not digital, there is some analog/digital stage coupled to the CCD or &#xA;maybe &quot;onboard&quot; the CCD chip. But nevertheless the CCD works &#xA;analog.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-01T15:39:50.317" LastActivityDate="2011-04-01T15:39:50.317" CommentCount="5" />
  <row Id="7884" PostTypeId="1" AcceptedAnswerId="7890" CreationDate="2011-04-01T13:12:20.657" Score="3" ViewCount="149" Body="&lt;p&gt;My wife's kindergarten class asked, &quot;What would happen to a tree planted on the moon?&quot; Aside from the obvious that it would die from lack of water/air, what physical effects would happen to a tree exposed to the vacuum of Space?&lt;/p&gt;&#xA;" OwnerUserId="2882" LastEditorUserId="2882" LastEditorDisplayName="" LastEditDate="2011-04-01T13:17:56.500" LastActivityDate="2011-04-01T15:01:34.993" Title="Effects on a tree exposed to the vacuum of Space" Tags="&lt;space&gt;&lt;vacuum&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7885" PostTypeId="2" ParentId="7823" CreationDate="2011-04-01T13:47:52.437" Score="-2" ViewCount="" Body="&lt;p&gt;The real problem is whether it's possible to move &lt;em&gt;back&lt;/em&gt; in time (we all move forward in time)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our whole understanding of the world is based on basic concepts, one of the is causality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;causality is a property of our &lt;em&gt;mind&lt;/em&gt; (not the physical world) - we believe that one event causes a second event when some conditions hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The causality links events such the cause preceded the effect&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(this a very brief and partial definition of causality, but it should be enough to make the point)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a visitor from the future is like a ghost, or other paranormal phenomena&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;We'll either find a different explanation, which is consistent with the basic ways our mind works&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We'll ignore it &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So even if time travel is physically possible, we won't interpret it's implications as time travel,&#xA;So our mind (and science, which is a product of our mind) wont be able to  &lt;/p&gt;&#xA;" OwnerUserId="2883" LastActivityDate="2011-04-01T13:47:52.437" />
  <row Id="7886" PostTypeId="2" ParentId="1601" CreationDate="2011-04-01T14:03:03.780" Score="2" ViewCount="" Body="&lt;p&gt;I think that &lt;strong&gt;Geometric Algebra&lt;/strong&gt; suits Classical Mech without Coords, and more.&lt;br&gt;&#xA;There are a lot of free resources in the net. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The approach is computational easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from the &lt;strong&gt;Hestenes&lt;/strong&gt; book &lt;a href=&quot;http://geocalc.clas.asu.edu/html/NFCM.html&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;New Foundations for Classical Mechanics&lt;/strong&gt;&lt;/a&gt;&lt;br&gt;&#xA;quoting:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...introduction to geometric algebra&#xA;  as a unified language for physics and&#xA;  mathematics... introduces new,&#xA;  &lt;strong&gt;coordinate-free methods&lt;/strong&gt; for rotational dynamics and orbital&#xA;  mechanics, developing these subjects&#xA;  to a level well beyond that of other&#xA;  textbooks. These methods have been&#xA;  widely applied in recent years to&#xA;  biomechanics and robotics, to computer&#xA;  vision and geometric design, to&#xA;  orbital mechanics in governmental and&#xA;  industrial space programs, as well as&#xA;  to other branches of physics. The book&#xA;  applies them to the major&#xA;  perturbations in the solar system,...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or &lt;strong&gt;&lt;a href=&quot;http://www.mrao.cam.ac.uk/~clifford/publications/ps/chris_thesis.pdf&quot; rel=&quot;nofollow&quot;&gt;Geometric Algebra and its Application to Mathematical Physics&lt;/a&gt;&lt;/strong&gt; by Chris J. L. Doran (Chris Thesis) free download&lt;br&gt;&#xA;....&lt;br&gt;&#xA;or &lt;a href=&quot;http://geocalc.clas.asu.edu/pdf/OerstedMedalLecture.pdf&quot; rel=&quot;nofollow&quot;&gt;Oersted Medal Lecture 2002: Reforming the Mathematical Language of Physics (GA)&lt;/a&gt;&#xA;free download  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... This has produced a&#xA;  comprehensive language called&#xA;  &lt;strong&gt;Geometric Algebra&lt;/strong&gt;, which I introduce&#xA;  with emphasis on how it simplifies and&#xA;  integrates classical and quantum&#xA;  physics.&lt;br&gt;&#xA;  ... After explaining the &lt;strong&gt;utter&#xA;  simplicity of the GA grammar&lt;/strong&gt; ..&#xA;  unique features of the mathematical&#xA;  language: &#xA;  (1) GA seamlessly integrates&#xA;  the properties of vectors and complex&#xA;  numbers to enable a completely&#xA;  coordinate-free treatment of 2D&#xA;  physics.&lt;br&gt;&#xA;  (2) GA articulates&#xA;  seamlessly with standard vector&#xA;  algebra to enable easy contact with&#xA;  standard literature and mathematical&#xA;  methods.&lt;br&gt;&#xA;  (3) GA &lt;strong&gt;Reduces “grad, div,&#xA;  curl and all that” to a single vector&#xA;  derivative&lt;/strong&gt; that, among other things,&#xA;  combines the standard set of four&#xA;  Maxwell equations into a single&#xA;  equation and provides new methods to&#xA;  solve it.&lt;br&gt;&#xA;  (4) The GA formulation of&#xA;  spinors facilitates the treatment of&#xA;  rotations and rotational dynamics in&#xA;  both classical and quantum mechanics&#xA;  &lt;strong&gt;without coordinates or matrices&lt;/strong&gt;.&lt;br&gt;&#xA;  (5) GA provides fresh insights into the&#xA;  geometric structure of quantum&#xA;  mechanics with implications for its&#xA;  physical interpretation.&lt;br&gt;&#xA;  All of this &lt;strong&gt;generalizes smoothly to a completely coordinate-free language&lt;/strong&gt; for&#xA;  spacetime physics and general&#xA;  relativity to be introduced in&#xA;  subsequent papers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-04-01T14:03:03.780" CommentCount="1" />
  <row Id="7887" PostTypeId="2" ParentId="7879" CreationDate="2011-04-01T14:04:04.107" Score="2" ViewCount="" Body="&lt;p&gt;Viking1 used &lt;a href=&quot;http://en.wikipedia.org/wiki/Vidicon#Vidicon&quot; rel=&quot;nofollow&quot;&gt;vidicons&lt;/a&gt; basically the same thing that studio TV cameras used into the 80s&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-04-01T14:04:04.107" />
  <row Id="7888" PostTypeId="2" ParentId="7879" CreationDate="2011-04-01T14:40:24.840" Score="1" ViewCount="" Body="&lt;p&gt;Others answered about the old cameras, but a piece of trivia about current space craft cameras to clarify a small misconception: the primary cameras used in the Mars landers (like Spirit and Opportunity) are high-resolution B&amp;amp;W cameras equipped with switchable color filters for R, G and B. The craft sends 3 b&amp;amp;w images filtered through the 3 filters and they only get composed into one color image by earth crew.&lt;/p&gt;&#xA;" OwnerUserId="2754" LastActivityDate="2011-04-01T14:40:24.840" CommentCount="2" />
  <row Id="7889" PostTypeId="1" AcceptedAnswerId="7891" CreationDate="2011-04-01T14:58:57.710" Score="0" ViewCount="140" Body="&lt;p&gt;Because of the apparent discrepancy of how some CFT and GR books define conformal transformation unlike in string theory area, I wanted to get rid of all the confusion from McGreevy's lecture notes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://ocw.mit.edu/courses/physics/8-821-string-theory-fall-2008/lecture-notes/lecture10.pdf&quot; rel=&quot;nofollow&quot;&gt;http://ocw.mit.edu/courses/physics/8-821-string-theory-fall-2008/lecture-notes/lecture10.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have understood and happy with the definition of conformal group and conformal transformation now, but I have the following two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In page 2, where he defined conformal invariance, did he mean invariance under the diffeomorphism part only (which gives the conformal group)? The reason I think so is that he changed the metric which according to his definition of conformal transformation don't change. So, the reason I am asking is that I thought conformal invariance will be a suitable name for a theory, invariant under conformal transformation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Just to make sure: the transformation in page 2: $\delta g_{\mu\nu}=2\lambda g_{\mu\nu}$ and $\delta g_{\mu\nu}=\delta\Omega(x)\delta g_{\mu\nu}$, these are results of diffeomorphism ..right? and not Weyl in any way. In other words, what are these transformations?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All my questions are for general dimension and general space time (classically of course).&lt;/p&gt;&#xA;" OwnerUserId="1349" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-01T15:16:44.743" LastActivityDate="2011-04-01T15:37:05.467" Title="conformal, Weyl transformations, apparent discrepancies and confusions" Tags="&lt;conformal-field-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="7890" PostTypeId="2" ParentId="7884" CreationDate="2011-04-01T15:01:34.993" Score="2" ViewCount="" Body="&lt;p&gt;This is a question for a biologist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start with, I guess it would freeze dry, by the water evaporating due to the lack of atmosphere, as Ted observed. If you have forgotten an unwrapped piece of food in the freezer you would know what it means. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then also the rest of the molecules  which are made up mostly of Carbon and Hydrogen would slowly sublimate. If in the sun, much faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if one assumes a bubble of atmosphere and appropriate water and fertilizer, and the question concerns if the moon soil could support a plant, the answer is , yes, if earth soil microorganisms are introduced into the moon soil where the plant is planted. The &lt;a href=&quot;http://aem.asm.org/cgi/content/full/71/12/8714&quot; rel=&quot;nofollow&quot;&gt;technique exists&lt;/a&gt; used for improving poor soils.&#xA;(I recently learned about it from discussions about planting a sour cherry tree in poor soil.)&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-01T15:01:34.993" />
  <row Id="7891" PostTypeId="2" ParentId="7889" CreationDate="2011-04-01T15:37:05.467" Score="2" ViewCount="" Body="&lt;p&gt;It's somewhat unclear how you &quot;understood&quot; and are &quot;happy&quot; about the definition of the conformal transformations because your questions, while referring to page 2 and other things, are nothing else than misunderstandings about the definition of a conformal transformation which is explained on page 1, not 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;John wrote his equation 1 which states that conformal transformations &quot;are&quot; diffeomorphisms that only change the metric by a local scalar coefficient - by a Weyl rescaling. However, the invariance of a theory under these &lt;em&gt;diffeomorphisms&lt;/em&gt; is a trivial property. If one is allowed to change the terms coupled to the &quot;metric&quot;, a diffeomorphism-transformed theory has a different action in general, and it is always possible to rewrite the original action in the conformally transformed coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what's physically nontrivial is the condition that the action, in its original form, is actually invariant under the operations - that's what we mean by the theory's being conformally invariant. If a theory is conformally invariant, we don't allow any &quot;change of the coefficients&quot; in the integral of the Lagrangian density. This condition of conformal invariance, as he shows, is equivalent to the invariance of the theory under the &quot;Weyl rescaling&quot; only: we just completely eliminate the diffeomorphisms from the picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Again, the invariance under the combined &quot;diffeomorphism&quot; and &quot;Weyl rescaling&quot; (the latter changes the form of the action) is a tautology. Obviously, by conformal invariance, we don't mean a tautology, so by conformal invariance, we mean the invariance of the action under the diffeomorphism separately, without changing the form of the metric in the action. Because the invariance under the &quot;combo&quot; is tautological, conformal invariance is equivalent to the invariance under the &quot;Weyl rescaling part&quot; of the transformation only.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No, on page 2, the transformations are exactly what the equations say: point-dependent transformations of the metric tensor itself i.e. a Weyl rescaling. There is no diffeomorphism at this stage. The invariance under those Weyl transformations is equivalent to the invariance under some diffeomorphisms (with the  modification of the metric erased), as explained in the previous point.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The formalism may depend on classical physics but your comment that it is &quot;classically only of course&quot; is incorrect, too. All those facts about conformal transformations are completely valid quantum mechanically as well - and indeed, this background is &lt;em&gt;primarily&lt;/em&gt; mean to understand some quantum theories.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-01T15:37:05.467" CommentCount="0" />
  <row Id="7892" PostTypeId="2" ParentId="7864" CreationDate="2011-04-01T17:27:07.700" Score="5" ViewCount="" Body="&lt;p&gt;As far as I can tell, you went wrong in a couple of places: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;instead of converting from $\mathrm{g}$ to $\mathrm{mg}$ your converting to $\mathrm{kg}$ in your first term, and&lt;/li&gt;&#xA;&lt;li&gt;in your fourth term, the units should be $\mathrm{g\, O_2}$ and $\mathrm{ nmol\, O_2}$, for consistency. This would've let you know that you've inverted your conversion from $\mathrm{ nmol\, CO_2}$ to $\mathrm{ nmol\, O_2}$ in your third term.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;When I'm performing unit conversions, I do a couple of things to make my life easier.  First, like it is displayed in your question, I write everything on two lines so that I can easily distinguish between the numerator and denominator, but I doubt that is at issue here.  Second, like you, I tend to do them all at once, but occasionally, like this case, it may be worthwhile to do it piecemeal.  Last, I convert systematically, working from left to right.  For instance, in this case, I'd convert from $\mathrm{ nmol\, CO_2}$ to $\mathrm{\mu L\, O_2}$ first, and then add in the conversions from $\mathrm{g}$ to $\mathrm{mg}$ and $\mathrm{s}$ to $\mathrm{h}$ after that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: in your most recent revision, you've removed you conversion from $\mathrm{CO_2}$ to $\mathrm{O_2}$, so the units you get are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{ \mu L\, \mathrm{CO_2}}{10 g h}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I remember correctly, the conversion was a power of 2, so it will not fix the deviation between the two units.&lt;/p&gt;&#xA;" OwnerUserId="770" LastEditorUserId="770" LastEditorDisplayName="" LastEditDate="2011-04-08T17:43:18.890" LastActivityDate="2011-04-08T17:43:18.890" CommentCount="1" />
  <row Id="7894" PostTypeId="1" AcceptedAnswerId="7897" CreationDate="2011-04-01T19:01:58.430" Score="5" ViewCount="177" Body="&lt;p&gt;I'm working through Polchinski's book on string theory, and I ran into something that I don't think I understand. I'm hoping that someone who knows this stuff can help me out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before calculating the Dp-brane tension in Chapter 8, Polchinski says that we could have obtained the same result by calculating the amplitude for graviton emission from the D-brane (instead of calculating closed string exchange between two D-branes). It seems like we would do this by placing a graviton vertex operator on the disk with Dirichlet boundary conditions on (25-p) coordinates and Neumann on the rest. But doesn't the amplitude with only one vertex operator vanish, since there aren't enough ghost insertions to get a nonzero result? I must be misunderstanding something, because it seems like if we only fix the real and imaginary parts of the position of the vertex operator, then we have to divide by the volume of the rest of the CKG of the disk, which is infinity. Any ideas or hints? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2888" LastEditorUserId="2888" LastEditorDisplayName="" LastEditDate="2011-04-01T19:12:10.873" LastActivityDate="2011-04-01T20:44:40.277" Title="Graviton Emission from D-Branes" Tags="&lt;string-theory&gt;" AnswerCount="1" />
  <row Id="7895" PostTypeId="2" ParentId="6835" CreationDate="2011-04-01T19:10:02.810" Score="1" ViewCount="" Body="&lt;p&gt;Just to add a little more info, there are also &lt;a href=&quot;http://en.wikipedia.org/wiki/Single-mode_optical_fiber&quot; rel=&quot;nofollow&quot; title=&quot;example&quot;&gt;monomode optical fibers&lt;/a&gt; that are even thinner (8 to 20 µm). These are useful to transport light pulses, and only allow one spatial mode within the fiber. That means that instead of the typical picture of light bouncing in the walls of the fiber, you get that light can only propagate forward with a certain intensity profile. Check &lt;a href=&quot;http://www.answers.com/topic/optical-fiber&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; or just google for a comparison of the three main types of fibers: multimode, graded-index and monomode.&lt;/p&gt;&#xA;" OwnerUserId="2890" LastActivityDate="2011-04-01T19:10:02.810" />
  <row Id="7896" PostTypeId="2" ParentId="7864" CreationDate="2011-04-01T20:05:34.403" Score="1" ViewCount="" Body="&lt;p&gt;After making the corrections suggested by rcollyer, you were very close. Looking back to the original paper, note that the units were reported in the unexpcted units of  $$\frac{\mu\mathrm{L}\ \mathrm{O}_2}{10\mathrm{mg}\ \mathrm{h}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a 10 in the denominator, so multiply your answer by 10 and your calculations are pretty close.&#xA;&lt;img src=&quot;http://i.stack.imgur.com/4N9QC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2809" LastEditorUserId="2809" LastEditorDisplayName="" LastEditDate="2011-04-02T00:11:31.267" LastActivityDate="2011-04-02T00:11:31.267" />
  <row Id="7897" PostTypeId="2" ParentId="7894" CreationDate="2011-04-01T20:44:40.277" Score="2" ViewCount="" Body="&lt;p&gt;After you fix the closed string vertex operator, the remaining group of isometries is one dimensional and its volume is finite. For example if the vertex operator is fixed at the center of the disk, the remaining isometries are rotations. The volume of that group in some units is $2\pi$. Figuring out the precise constants involved is a bit messy though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another subtlety with this calculation - the on-shell graviton is pure gauge and strictly speaking this amplitude is zero. But with appropriate limiting procedure you can extract the tension as the proportionality constant involved in a slightly off-shell amplitude (which is slightly ill-defined). This is another reason why the annulus calculation is cleaner.&lt;/p&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-04-01T20:44:40.277" CommentCount="7" />
  <row Id="7898" PostTypeId="1" CreationDate="2011-04-01T21:43:24.187" Score="6" ViewCount="128" Body="&lt;p&gt;I tried this sight before to find filtering options and got fantastic results, so let's try again!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am setting up an experiment that requires light of two different frequencies (445nm and 350nm).  The light ultimately needs to be focused on a small area.  I can think of two good ways to make this happen:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Get a broad spectrum light source that emits over this range - maybe a halogen lamp? - and buy two filters:  A high pass that cuts off at ~420nm and a low pass that cuts off at ~380nm.  For the low pass, I may also include a high pass in order to make a band pass (or just flat out buy a band pass filter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will then simply wrap the filters in a black paper cone to &quot;focus&quot; the light (for this experiment, light spread after hitting my target doesn't seem to important).  We would use a power meter to tell us what flux is actually hitting our experiment. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My questions for this approach:  What kind of broad spectrum source would you recommend?  Can you recommend a retail outlet for this source?  Our grant ain't so grand, so afford ability is important.  Can you think of a better way to focus the light after it passes through the filters?  Again, any retail sources would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Get two more highly focused sources.  Given our small area focusing requirements, we are looking at lasers (Again, 340nm and 445nm).  I have found a couple of retail sources that have such lasers, but they are fairly cost prohibitive.  Do you know of any sources that sell such lasers - power sources and all - for &amp;lt;500?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance for any help and recommendations.  If a paper is produced from this - which I very much am counting on - I will thank the guy who provides the best answer in it (if that is an incentive).&lt;/p&gt;&#xA;" OwnerUserId="1037" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-20T23:14:30.650" LastActivityDate="2011-07-20T23:14:30.650" Title="Fairly Broad Spectrum Light Source Options" Tags="&lt;optics&gt;&lt;experimental-physics&gt;&lt;laser&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="7899" PostTypeId="1" AcceptedAnswerId="7949" CreationDate="2011-04-01T22:23:35.143" Score="4" ViewCount="284" Body="&lt;p&gt;Ok, from astronomical observations we can tell that the observable matter is separating - so rewind the clock about 13.7 billion years and it was all at a single point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, how do we distinguish between the following two options:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Universe is expanding  &lt;/li&gt;&#xA;&lt;li&gt;Matter distribution is increasing into infinite void&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarification 1:&lt;/strong&gt;&lt;br&gt;&#xA;(My notion of) The traditional notion is that all time/space/matter was created at the instant of the big bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. BB was inital conditions of: $t=0$, $V=0$, $E=very big$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People say &quot;the universe is expanding&quot;, rather than &quot;observable matter is separating&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we know that the big bang event wasn't started by all matter condensed at a single point within a larger (otherwise empty) universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we know that BB wasn't: $t=0$, $V(universe)&amp;gt;0$ but $V(matter)=0$, $E=very big$?&lt;/p&gt;&#xA;" OwnerUserId="937" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-03T15:29:00.403" LastActivityDate="2011-04-03T15:29:00.403" Title="How do we know the size of the universe?" Tags="&lt;universe&gt;&lt;big-bang&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="7900" PostTypeId="2" ParentId="7899" CreationDate="2011-04-01T22:32:21.547" Score="1" ViewCount="" Body="&lt;p&gt;We can't know about what lies beyond the cosmological horizon. (roughly, any region more than 13.7 bn light years from us, though that's not completely right).  It is impoosible for any signal from there to travel to where we are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we can say is that all observations that we've made are consistent with an infinite universe everywhere filled with matter.  It turns out that this is very close to some very simple exact solutions of Einstein's equation.  So, it would make sense to take this model as our first guess/approximation as to what the universe, as a whole looks like.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-04-01T22:32:21.547" CommentCount="5" />
  <row Id="7901" PostTypeId="2" ParentId="7899" CreationDate="2011-04-01T23:26:54.347" Score="2" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/UoqrK.png&quot; alt=&quot;expanding universe&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you overlap a point from figure A with corresponding point from figure B, you will see that every other point has moved away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You go to any point in the universe you will see that things are moving away from you. It is like you put ink dots on a balloon and when you blow the balloon distance between the ink dots will increase. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If matter distribution were expanding into infinite void [point (b)] there is a question what is causing that expansion? As gravity pulls things there must be some force that is stronger than gravity that is pushing things away. But we don't know any such force. So it must be universe expanding.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Watch &lt;a href=&quot;http://www.youtube.com/watch?v=7ImvlS8PLIo&quot; rel=&quot;nofollow&quot;&gt;&quot;A universe from nothing&quot; by Lawrence Krauss.&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;NOTE: I would have done better job at explaining but I don't know much GR. :(&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2011-04-01T23:26:54.347" CommentCount="2" />
  <row Id="7902" PostTypeId="2" ParentId="7872" CreationDate="2011-04-02T00:53:14.833" Score="-1" ViewCount="" Body="&lt;p&gt;Possibly very close.  Peer review needed: &lt;a href=&quot;http://www.superconductors.org/20C.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.superconductors.org/20C.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2780" LastActivityDate="2011-04-02T00:53:14.833" CommentCount="3" />
  <row Id="7903" PostTypeId="1" CreationDate="2011-04-02T01:07:29.087" Score="-1" ViewCount="145" Body="&lt;p&gt;In regards to the right hand rule, given Earth's electric and magnetic field, in which &quot;direction&quot; would a particle go?&lt;/p&gt;&#xA;" OwnerUserId="2780" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-11T20:05:31.417" LastActivityDate="2011-05-11T20:05:31.417" Title="What direction does the force vector point in regards to Earth's E and B fields?" Tags="&lt;electromagnetism&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="7904" PostTypeId="2" ParentId="7822" CreationDate="2011-04-02T01:34:47.797" Score="2" ViewCount="" Body="&lt;p&gt;The maximum force / power are obtained in the paper as:&#xA;$$\textrm{Force} \le \frac{c^4}{4G}\;\;\;\;\;\;\;\;\;\textrm{Power} \le \frac{c^5}{4G}$$&#xA;These are just 1/4 the Planck force or Planck mass one would obtain from the Planck length, mass, and time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First noticeable weirdness is that he keeps talking about momentum and energy in high gravity situations when these concepts are not very well defined in the context of GR. I guess that's okay in that he's deriving GR rather than making a statement about GR. Later he explains that force and power are highly observer dependent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Page 3 has another weirdness, could be some sort of typo or oversight by the author:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As a result, a maximum force&#xA;  (respectively power) value in nature&#xA;  implies the following statement: one&#xA;  imagines a physical surface and&#xA;  completely covers it with observers;&#xA;  then the integral of all momentum&#xA;  (respectively energy) values flowing&#xA;  through that surface per time,&#xA;  measured by all those observers, never&#xA;  exceeds the maximum value.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The above is weird in that the maximum force should depend on the size of the surface. That is, a surface of twice the size should allow twice the force. Maybe this is incorrect intuition on my side, but it could at least use a better description.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the same page, he notes that force and power are parts of a 4-vector in relativity. Kind of cool. I never thought about that. Might be nice to see a reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading further, he makes it clear that the &quot;maximum forces&quot; he's talking about are on the event horizons of black holes. That seems reasonable. In the Schwarzschild coordinates objects freeze on the event horizon but in the coordinates I prefer, Gullstrand-Painleve (GP), objects fall through and one can compute an acceleration at the event horizon. The amusing thing is that I wrote the definitive paper on this subject:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Int.J.Mod.Phys.D18:2289-2294 (2009) Carl Brannen, &lt;em&gt;The force of gravity in Schwarzschild and Gullstrand-Painlevé coordinates&lt;/em&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/0907.0660&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0907.0660&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at equation (11) of the above, the acceleration is zero at the event horizon for stationary observers in Schwarzschild or Gullstrand-Painleve coordinates. For the Schwarzschild case this is intuitively obvious. That's part of why particles get stuck on the event horizon. I think his force doesn't make sense.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Ah hah! He has some counterarguments to critiques of his theory. The first one is the &quot;mountatin example&quot;. In it, he writes:  &lt;em&gt;Nuclei are quantum particles and have an indeterminacy in their position; this indeterminacy is essentially the nucleus–nucleus distance.&lt;/em&gt; This is clearly untrue; the electron wave functions define the nucleus-nucleus distance and this is far greater than the size of the nucleus wave function. One's measured in barns, the other's measured in Angstroms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The counter-argument for &quot;multiple neutron stars&quot; fails to take into account how special relativity deals with simultaneous events. He's claiming that (in a given rest frame) it's impossible to have simultaneous events that are widely separated. This is a confusion of spacelike separation and simultaneity. With his use it eliminates any set of observers to be simultaneous but it seems that was an intrinsic assumption of his argument.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So my verdict is that it's junk, but better junk than most in that I didn't see any violations of units. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He uses arguments that all rely on the Planck units to argue that the Planck units are fundamental. For example, the diameter of a black hole has a close relationship between the mass of the black hole and the gravitational constant. So when you consider the time light takes to cross this distance, of course you're going to get more Planck constant related things. There's plenty of opportunities to get the right constants as there's an infinite number of ways of choosing arbitrary lengths and distances from the geometry of a black hole. No surprises.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-04-02T01:34:47.797" CommentCount="1" />
  <row Id="7905" PostTypeId="1" AcceptedAnswerId="7909" CreationDate="2011-04-02T03:25:10.767" Score="11" ViewCount="393" Body="&lt;p&gt;Are there any &lt;strong&gt;massless&lt;/strong&gt; (zero invariant mass) particles carrying electric &lt;strong&gt;charge&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, why not? Do we expect to see any or are they a theoretical impossibility?&lt;/p&gt;&#xA;" OwnerUserId="824" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-04-02T17:22:29.663" LastActivityDate="2011-08-11T18:06:56.197" Title="Massless charged particles" Tags="&lt;particles&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="3" />
  <row Id="7906" PostTypeId="1" AcceptedAnswerId="7928" CreationDate="2011-04-02T03:44:34.553" Score="2" ViewCount="143" Body="&lt;p&gt;How does an antenna behave when it is cooled so that its black-body radiation is emitting energy at its resonant frequency?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: To clarify, its not how they're related in general, but how might thermal radiation and resonance interact with each other when their spectra are aligned well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Also, I'm sure that the thermal radiation spectra that have a significant peaks are associated with incredibly high temperatures, and peak at incredibly small wavelengths, rendering such an antenna completely impractical to build. Still, I'm still interested in the theoretical concept.&lt;/p&gt;&#xA;" OwnerUserId="2896" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-02T10:24:14.470" LastActivityDate="2011-04-02T10:24:14.470" Title="How might a resonant antenna and black body radiation interact?" Tags="&lt;thermodynamics&gt;&lt;electromagnetic-radiation&gt;&lt;thermal-radiation&gt;&lt;resonance&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="7907" PostTypeId="1" AcceptedAnswerId="7926" CreationDate="2011-04-02T03:54:32.647" Score="5" ViewCount="326" Body="&lt;p&gt;I was reading Landau and Lifschitz's &quot;Classical Field Theory&quot; and came across a comment that the action for electromagnetism must be a scalar, not a pseudoscalar (footnote in section 27). So I was wondering, is it possible/interesting to construct a classical field theory with a pseudoscalar action? If not, why not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: I was motivated to look at this because of a question from Sean Carroll's &quot;Spacetime and Geometry&quot;, which asks us to show that adding a pseudoscalar term ($\vec{E} \cdot \vec{B}$) to the Lagrangian doesn't change Maxwell's equations.)&lt;/p&gt;&#xA;" OwnerUserId="2897" LastActivityDate="2011-04-02T18:07:11.617" Title="Pseudoscalar action in classical field theory" Tags="&lt;classical-electrodynamics&gt;" AnswerCount="2" />
  <row Id="7908" PostTypeId="2" ParentId="7905" CreationDate="2011-04-02T03:59:38.450" Score="2" ViewCount="" Body="&lt;p&gt;None found so far - but they may be possible apparently (don't understand how!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.springerlink.com/content/d2457278u3m78173/&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/d2457278u3m78173/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-04-02T03:59:38.450" CommentCount="1" />
  <row Id="7909" PostTypeId="2" ParentId="7905" CreationDate="2011-04-02T04:05:58.040" Score="3" ViewCount="" Body="&lt;p&gt;There's no problem in writing down a theory that contains massless charged particles. Simple $\mathcal{L} = \partial_{\mu} \phi \partial^{\mu} \phi^*$ for a complex field $\phi$ will do the job. You might run into problems with renormalization but I don't want to get into that here (mostly because there are better people here who can fill in the details if necessary).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disregarding theory, those particles would be easy to observe assuming their high enough density. Also, as you probably know, particles in Standard Model compulsorily decay (sooner or later) into lighter particles as long as conservation laws (such as electric charge conservation law) are satisfied. So assuming massless charged particles exist would immediately make all the charged matter (in particular electrons) unstable unless those new particles differed in some other quantum numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you didn't mention electric charge in particular, the answer would be simpler as we have massless (color-)charged gluons in our models. So it's definitely nothing strange to consider massless charged particles. It's up to you whether you consider electric charge more important than color charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another take on this issue is that Standard Model particles (and in particular charged ones) were massless before electrosymmetric breaking (at least disregarding other mechanisms of mass generation). So at some point in the past, this was actually quite common.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-04-02T04:05:58.040" CommentCount="4" />
  <row Id="7910" PostTypeId="1" CreationDate="2011-04-02T04:14:34.763" Score="1" ViewCount="198" Body="&lt;p&gt;&lt;strike&gt;Science and science fiction alike&lt;/strike&gt;Science fiction describes black holes as these amazingly different entities in space that don't behave according to the same laws of physics that the rest of the universe is bound to. I've heard them described as wormholes to other universes, singularities, or tears in space-time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, are they really that special? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thought is that all matter has event horizons. However, most of it isn't dense enough for that event horizon to be large enough to affect the way the matter interacts with the rest of the universe. So, a black hole is simply just a dense star that is simply dark. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, if black holes &lt;em&gt;must&lt;/em&gt; have a singularity at their center because of general relativity, can't we conclude from that that all matter creates singularities, since the gravitational field intensifies into infinity the closer you get to the point-mass? What is the difference being really close to a point-mass and being very close to a black hole?&lt;/p&gt;&#xA;" OwnerUserId="2896" LastEditorUserId="2896" LastEditorDisplayName="" LastEditDate="2011-04-02T05:45:16.423" LastActivityDate="2011-04-02T05:45:16.423" Title="Are black holes really that special?" Tags="&lt;black-holes&gt;" AnswerCount="2" />
  <row Id="7911" PostTypeId="1" AcceptedAnswerId="7935" CreationDate="2011-04-02T04:42:23.673" Score="6" ViewCount="348" Body="&lt;p&gt;According to an answer in &lt;a href=&quot;http://skeptics.stackexchange.com/questions/1664/it-is-not-the-voltage-that-kills-you-it-is-the-current/1667#1667&quot;&gt;this thread&lt;/a&gt; on Skeptics:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you take one of the little 12V&#xA;  garage door opener batteries and short&#xA;  out (directly connect) the two&#xA;  terminals with a piece of wire or&#xA;  something else. You'll get a light&#xA;  current flow through the wire or&#xA;  metal. It may get a little warm. This&#xA;  battery is only capable of supplying a&#xA;  small amount of current.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If you take a 12V car battery and&#xA;  short out the two terminals (don't do&#xA;  it, it's not fun), you will be met&#xA;  with a huge current arc that will&#xA;  likely leave a burn mark on whatever&#xA;  was used to short it. This is because&#xA;  the car battery is capable of&#xA;  discharging a large amount of current&#xA;  in a very short period of time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how this could work given Ohm's law V=IR. If we assume the resistance of toucher is constant, then we'd expect the current to be the same as well.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Could it be that a car battery has&#xA;less internal resistance than a&#xA;garage battery?&lt;/li&gt;&#xA;&lt;li&gt;Does it have anything to do with contact area? If it does, then how would you model it? Generally resistances add in series, but if I only half touch a contact, then neither the battery nor I change, so you'd expect our resistances to stay the same, but somehow our total resistance changes. So which objects resistance would change - mine or the batteries?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="119" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-02T15:42:34.107" LastActivityDate="2011-05-26T21:39:51.917" Title="Why can different batteries with the same  voltage send different currents through the same object?" Tags="&lt;electricity&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" />
  <row Id="7912" PostTypeId="2" ParentId="7910" CreationDate="2011-04-02T04:43:15.880" Score="0" ViewCount="" Body="&lt;p&gt;&quot;Science and science fiction alike describe black holes as these amazingly different entities in space that don't behave according to the same laws of physics that the rest of the universe is bound to. I've heard them described as wormholes to other universes, singularities, or tears in space-time.&quot; - this is not quite correct! Black holes obey the same physical laws as the rest of the observable universe. They are neither worm holes nor gateway to another universe, nor any tears in space-time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classical GR does not allow any tear in spacetime. In GR, geometry of the space-time can change but topology can't change. Only string theory allow such tears but those are not necessarily related with black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes are special in the sense that they contain a singularity at their core and nothing can come of it including light except Hawking radiation which are emitted from the vicinity of a black hole. They are clothed. That means black holes can affect the causal structure of spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing that is special is black holes are ripe for studying quantum gravity since gravity is extreme in that case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;My thought is that all matter has event horizons.&quot; - sorry, but this is not right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think, you should study physics a little more carefully. Cheers.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-04-02T05:08:18.013" LastActivityDate="2011-04-02T05:08:18.013" CommentCount="11" />
  <row Id="7913" PostTypeId="2" ParentId="7910" CreationDate="2011-04-02T04:46:11.983" Score="5" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Science and science fiction alike describe black holes as these amazingly different entities in space that don't behave according to the same laws of physics that the rest of the universe is bound to. I've heard them described as wormholes to other universes, singularities, or tears in space-time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Um... no. Science fiction may do that, but not science, at least not when it's properly described. Admittedly, popular science writers do have a tendency to treat black holes as mystical objects, but they're really not. The general theory of relativity, our best description of gravity, works just as well for black holes as it does for anything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be clear, let me give out a couple of definitions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A &lt;em&gt;black hole&lt;/em&gt; is a region which, once entered, is impossible to escape. This has a precise mathematical definition in general relativity. The theory allows us to calculate the set of all possible paths that a particle can travel on; they're called &lt;em&gt;geodesics&lt;/em&gt;. When you have a black hole, all geodesics which pass inside the black hole remain inside the black hole.&lt;/li&gt;&#xA;&lt;li&gt;An &lt;em&gt;event horizon&lt;/em&gt; is a surface that can only be crossed in one direction. The boundary of a black hole is one example.&lt;/li&gt;&#xA;&lt;li&gt;A &lt;em&gt;singularity&lt;/em&gt; is a point (or a set of points) at which a theory predicts that some physical quantity becomes infinite. In the case of a black hole, the quantity is the spacetime curvature. In reality, it doesn't mean that the quantity actually becomes infinite, it just means that you need a better theory.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My thought is that all matter has event horizons. However, most of it isn't dense enough for that event horizon to be large enough to affect the way the matter interacts with the rest of the universe. So, a black hole is simply just a dense star that is simply dark.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;And, if black holes must have a singularity at their center because of general relativity, can't we conclude from that that all matter creates singularities, since the gravitational field intensifies into infinity the closer you get to the point-mass? What is the difference being really close to a point-mass and being very close to a black hole?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're partly right, partly possibly right, and partly wrong. Under the current models which treat elementary particles as points of zero size, then yes, they would have tiny event horizons. People have actually &lt;a href=&quot;http://adsabs.harvard.edu/abs/2006tima.conf..270T&quot; rel=&quot;nofollow&quot;&gt;investigated&lt;/a&gt; the implications of this, whether elementary particles could actually be black holes, and &lt;a href=&quot;http://arxiv.org/abs/hep-th/9202014&quot; rel=&quot;nofollow&quot;&gt;vice versa&lt;/a&gt;. As far as I know, it seems unlikely, but we don't particularly care for now because these black holes would be far tinier than anything we can currently detect. (One of the appeals of string theory is that it spreads out the mass of an elementary particle along an extended object, so there isn't necessarily a high enough density to be a black hole.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a black hole is &lt;em&gt;not&lt;/em&gt; just a dense, dark star. Yes, it is dense and dark, but the key difference is that a black hole does not allow anything to escape from inside it, i.e. it has an event horizon. That is not the case with a normal star, at least not on a macroscopic scale.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-04-02T04:46:11.983" CommentCount="2" />
  <row Id="7915" PostTypeId="1" AcceptedAnswerId="7942" CreationDate="2011-04-02T04:51:36.533" Score="3" ViewCount="329" Body="&lt;p&gt;I understand that in nature wind would never get high enough, but I am just curious as to whether physics would allow this to occur or not.&lt;/p&gt;&#xA;" OwnerUserId="2544" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-02T05:14:35.537" LastActivityDate="2011-04-03T22:36:09.463" Title="Is it possible for wind to break the sound barrier?" Tags="&lt;fluid-dynamics&gt;&lt;sound&gt;&lt;air&gt;" AnswerCount="6" CommentCount="4" />
  <row Id="7918" PostTypeId="1" AcceptedAnswerId="8006" CreationDate="2011-04-02T05:33:28.403" Score="3" ViewCount="224" Body="&lt;p&gt;The many-worlds interpretation of quantum physics is built around a configuration space, where the position of a particle is three components of the position of that universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens with particle-antiparticle creation or annihilation? It can't just change the number of dimensions, can it?&lt;/p&gt;&#xA;" OwnerUserId="2898" LastEditorUserId="2898" LastEditorDisplayName="" LastEditDate="2011-04-03T06:08:58.877" LastActivityDate="2011-04-03T19:59:01.870" Title="What happens when particle-antiparticle pairs annihilate in MWI?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-interpretations&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="7919" PostTypeId="2" ParentId="7905" CreationDate="2011-04-02T05:43:33.883" Score="5" ViewCount="" Body="&lt;p&gt;Massless charged particles can't exist in Nature because they would be easily produced by the colliders, and they haven't been. Such a production would simply arise from the Feynman diagram with an intermediate photon that &quot;splits&quot; into the new charged massless particle and its antiparticle. The cross section of this process would be calculable, and not small in any way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the fine-structure constant $\alpha=1/137.036$, one expressing the strength of the electrostatic interactions in the natural units, is not a real constant. It's running. However, it's only running at energy scales such that there exist lighter charged particles. In Nature, it means that the constant is only running above the mass of the electron or positron - the lightest charged particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there were massless charged particles, the electron and positron would become unstable - one problem - and the fine-structure constant would run to $\alpha=0$ at very long distances - another problem, and it obviously doesn't. So massless charged particles are theoretically impossible in our world - assuming that we empirically know some things such as the fact that there is a limiting Coulomb force at long distances.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-02T05:43:33.883" CommentCount="6" />
  <row Id="7920" PostTypeId="2" ParentId="7906" CreationDate="2011-04-02T05:52:22.170" Score="1" ViewCount="" Body="&lt;p&gt;They're not related. The black-body radiation as well as the resonance curve may look like &quot;bumps&quot; but they are very different bumps mathematically. The black body radiation gets emitted at all frequencies, and the &quot;uncertainty of the frequency&quot; is maximized, in some sense. On the other hand, resonances are peaked around a particular frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resonance curves are about matrix elements between pure states; thermal curves are traces over the whole Hilbert spaces so they arise from mixed stated. That's why the exponentials only appear in the thermal curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the only thing they share is that they produce intensities as a function of frequency - but many other things in physics do the same thing - and in both cases, complex numbers are useful ($E_0-i\Gamma/2$ for resonances and imaginary time $i\beta$ in the thermal case) - but complex numbers are useful across physics.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-02T05:52:22.170" CommentCount="2" />
  <row Id="7922" PostTypeId="1" CreationDate="2011-04-02T06:11:18.513" Score="0" ViewCount="362" Body="&lt;p&gt;The question assumes the standard formalism with projector-valued measures rather than POVMs. Suppose a measurement has two possible outcomes, and the corresponding probabilities are greater than 0 and less than 1. Neither outcome is therefore certain. Then why is it certain that either outcome is obtained (as it seems, if the probabilities add up to 1)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added after four answers&lt;/strong&gt;: All the answers provided so far elaborate on the comment by @Vladimir: &quot;It is not a 'quantum mechanical' feature but a consequence of probability definition.&quot; @Lubos and @Mark cast the question into a quantum-mechanical form, e.g., why do the absolute squares of the amplitudes associated with the possible outcomes  of a measurement add up to 1? They also explain why the sum remains equal to 1. (However, for a decaying particle the probability of finding it decreases, while the probability of finding its decay products increases. So the &quot;conservation of probability&quot; has something to do with the proper conservation laws.) @David makes it clearest why these answers are insufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that no actual measurement is perfect. While theorists may ignore this, experimenters know well enough that in many runs of a given experiment no outcome is obtained. (The efficiency of many real-world detectors is rather low.) This means that in order to make the probabilities add up to 1, one discards (does not consider) all those experiments in which no outcome is obtained. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let me follow up with &lt;a href=&quot;http://physics.stackexchange.com/questions/7969/does-quantum-mechanics-allow-us-to-formulate-causally-sufficient-conditions-for-t&quot;&gt;another question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2784" LastEditorUserId="2784" LastEditorDisplayName="" LastEditDate="2011-04-03T05:50:16.243" LastActivityDate="2011-04-03T19:30:07.650" Title="In quantum mechanics, why do the probabilities of the possible outcomes of a measurement add up to 1?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="6" CommentCount="6" FavoriteCount="1" />
  <row Id="7923" PostTypeId="2" ParentId="7922" CreationDate="2011-04-02T06:40:22.550" Score="7" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose a measurement has two possible outcomes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's why. One of the possible outcomes has to occur - it's the definition of &quot;possible outcomes.&quot; Probability theory takes this as an axiom, &lt;a href=&quot;http://en.wikipedia.org/wiki/Probability_axioms&quot; rel=&quot;nofollow&quot;&gt;the axiom of unit measure&lt;/a&gt; according to Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we often have to normalize the probabilities so that they add up to 1.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-04-02T06:40:22.550" />
  <row Id="7924" PostTypeId="2" ParentId="7922" CreationDate="2011-04-02T06:59:07.130" Score="4" ViewCount="" Body="&lt;p&gt;The probabilities of individual outcomes are given by the squared absolute values of the complex probability amplitudes $a_i$ associated with the individual outcomes. Their sum&#xA;$$ {\rm Total\,\,probability} = \sum_{i} |a_i|^2 $$&#xA;is therefore nothing else than a formula for the squared length of the state vector $|\psi\rangle$: note that it is a complexified version of the Pythagorean theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum mechanics, if the state vector (wave function) has the length equal to one at the beginning, it will have the length equal to one at all times - because of the so-called &quot;unitarity&quot;. Unitarity means that the evolution according to Schrödinger's equation is essentially just a rotation around some &quot;axes&quot; in the Hilbert space - an element of $U(N)$ or $U(\infty)$, a complex and/or infinite-dimensional generalization of $O(N)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unitarity means that the evolution of $|\psi\rangle$ - by Schrödinger's equation - preserves the length of the vector, and the squared length of the vector is nothing else than the sum of probabilities of all mutually exclusive outcomes of experiments, regardless of the orthogonal basis of the Hilbert space that we choose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unitarity condition is $U U^\dagger={\bf 1}$ for the evolution operators $U$ and may be reduced to $H=H^\dagger$, the hermiticity of the Hamiltonian, whenever the evolution is encoded in a Hamiltonian via $U=\exp(Ht/i\hbar)$, as Mark's answer shows explicitly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important that in quantum mechanics, we don't have to normalize the probabilities in a way that would depend on the evolution: the preservation of the total probability is guaranteed by the equations of motion, namely by the hermiticity of the Hamiltonian that enters these equations of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an ad hoc normalization were required, quantum mechanics would be spoiled by a new source of non-locality and the &quot;wave function collapse&quot; would become observable. Both of these problems would lead to inconsistencies with the observations - and, from a theorist's viewpoint, internal logical inconsistencies in the theory.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-02T07:48:28.447" LastActivityDate="2011-04-02T07:48:28.447" />
  <row Id="7925" PostTypeId="2" ParentId="7922" CreationDate="2011-04-02T07:10:40.267" Score="8" ViewCount="" Body="&lt;p&gt;Suppose we have a quantum state that is normalized at some time.  Then it will remain normalized if the Hamiltonian is Hermitian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial}{\partial t}\langle \Psi \mid \Psi \rangle = \left(\frac{\partial}{\partial t} \langle \Psi \right) \mid \Psi \rangle + \langle \Psi \mid \left(\frac{\partial}{\partial t} \Psi \rangle \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Schrodinger's equations says&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial}{\partial t}\mid \Psi \rangle = \frac{-i}{h} H \mid \Psi \rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial}{\partial t} \langle \Psi \mid = \frac{i}{h} \langle H \Psi \mid = \frac{i}{h} \langle \Psi \mid H^*$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting in gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial}{\partial t}\langle \Psi \mid \Psi \rangle = \frac{i}{h}\left( -\langle \Psi \mid H \Psi \rangle + \langle \Psi \mid H^* \Psi \rangle\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as $H = H^*$, this is zero.  The initial normalization is a postulate, as David said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer is pretty much an explicit way of saying what Lubos wrote - I was half way done when he posted, so check  there for more detail.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-04-02T07:10:40.267" CommentCount="1" />
  <row Id="7926" PostTypeId="2" ParentId="7907" CreationDate="2011-04-02T07:11:05.953" Score="6" ViewCount="" Body="&lt;p&gt;Electromagnetism is parity-symmetric. Because all other terms in the action - such as $mv^2-V(x)$ for particles - are parity-even, the electromagnetic contribution has to be parity-even, too. Otherwise the different terms would transform differently and the combined theory would violate parity. &quot;Parity-even&quot; simply means that the Lagrangian density is a scalar, not a pseudoscalar. It's the same thing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actions' being invariant under &#xA;$$(x,y,z)\to(-x,-y,-z),$$&#xA;including the sign ($S\to +S$), is simply what we mean by the parity symmetry, and scalars (as opposed to pseudoscalars) are the objects that preserve their signs under this operation, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E\cdot B$ is a term that doesn't affect equations of motion because it is a total derivative (which gets integrated to a constant, unaffected by variations of the fields, as long as the variations of the fields at $t=\pm \infty$ vanish):&#xA;$$ E\cdot B \sim \epsilon_{\alpha \beta\gamma\delta} F^{\alpha\beta} F^{\gamma\delta} \sim \partial^\alpha( \epsilon_{\alpha \beta\gamma\delta} A^\beta F^{\gamma\delta}) $$&#xA;You see that it is a total derivative; the $\partial^\alpha F^{\gamma\delta}$ term contracted with the $\epsilon$ symbol vanishes identically because it is the Bianchi identity (in the language of forms, $d^2 A\equiv 0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In non-Abelian theories, however, terms of the type&#xA;$${\rm Tr }\, F_{\mu\nu} F^{*\mu \nu}$$&#xA;change the physics even though they are total derivatives. It's because they get integrated to a nontrivial integral in the Euclidean spacetime where the configuration of the gauge field is topologically nontrivial - an instanton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In its Feynman's path-integral formulation, quantum mechanics calculates the transition amplitudes as the sum over the normal histories as well as the instantons, and the additive shifts in instantons matters. Because the instanton action above is integer - after a proper normalization - the coefficient $\theta$ in front of it is defined modulo $2\pi$ - as an angle - because a change of the action $S$ by $2\pi i$ doesn't matter since the path integral only depends on $\exp(iS)$. For example, in QCD, the term&#xA;$$ \theta{\rm Tr }\, F_{\mu\nu} F^{*\mu \nu}$$&#xA;is known to affect the physics but experimentally, the coefficient $\theta$ is smaller than $10^{-9}$ which is surprising and unnatural: we would expect $\theta$ to be of order one. The $\theta$-term above, if nonzero, is also P- (pseudoscalar) and CP-odd, and it would lead to new sources of CP-violation which is not observed (the only CP-violation that has been observed comes from the phase of the CKM matrix mixing quark masses).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This smallness of the $\theta$-angle, which is apparently not explained and not needed, not even for life (so even the anthropic principle fails to help), is called the strong CP-problem. The main candidate explanation why the observed $\theta$ is small, even though it doesn't have to be, is the Peccei-Quinn mechanism using the axions. $\theta$ gets promoted to a light scalar field in a way...&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-02T07:59:52.530" LastActivityDate="2011-04-02T07:59:52.530" />
  <row Id="7927" PostTypeId="2" ParentId="7911" CreationDate="2011-04-02T07:47:50.813" Score="2" ViewCount="" Body="&lt;p&gt;First an aside, before getting directly to the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many standard battery types that output 1.5V, with some example common ones being AA,AAA,C,D.  You can even buy one of each type with the same exact chemistry inside.  So what is the difference between these, and why would a designer require a couple big D cell batteries for a flashlight when a couple AAA have the same voltage rating?  There are two main reasons.  One is the total energy contained (a D cell is larger in volume, and thus has more room to store chemical energy, and thus can run a device longer than a AAA could).  The second is what you are getting at here, and that is the short circuit current.  A D cell battery should be able to output more current than a AAA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads us to your question, of how this can possibly fit with Ohm's law since the battery voltage rating is the same.  From an electrical engineer's standpoint, the open circuit voltage of a battery is only half the story.  One needs to know the open circuit voltage as well as the internal resistance rating, to be able to predict the actual output of a battery into some device.  You could consider the &quot;equivalent circuit&quot; of a real voltage source as an ideal voltage source in series with a resistor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, you can look at the technical documents for typical batteries here:&lt;br&gt;&#xA;&lt;a href=&quot;http://www1.duracell.com/oem/productdata/default.asp&quot; rel=&quot;nofollow&quot;&gt;http://www1.duracell.com/oem/productdata/default.asp&lt;/a&gt;&lt;br&gt;&#xA;and here's one for a D cell alkaline&lt;br&gt;&#xA;&lt;a href=&quot;http://www1.duracell.com/oem/Pdf/new/MN1300_US_CT.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www1.duracell.com/oem/Pdf/new/MN1300_US_CT.pdf&lt;/a&gt;&lt;br&gt;&#xA;notice it lists the &quot;nominal voltage&quot; and the &quot;impedance&quot;. For a D cell the impedance is listed as 136 mOhm.  If we look at a AAA using the same chemistry we see&lt;br&gt;&#xA;&lt;a href=&quot;http://www1.duracell.com/oem/Pdf/new/MN2400_US_CT.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www1.duracell.com/oem/Pdf/new/MN2400_US_CT.pdf&lt;/a&gt;&lt;br&gt;&#xA;the impedance is about 250 mOhm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this two parameter description of power sources is intuitive and the first order approach to describing a device, or course the real life devices can be a bit trickier (the impedance can depend on frequency as batteries can only respond so quickly, draining the battery quickly is usually less efficient, etc.).  So the technical documents often provide real test curves if you want to see even more detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a little 12V battery in a remote control will not be designed to have the low impedance required to start a car engine.  The 12V car battery should definitely be able to put out more current when shorted.&lt;/p&gt;&#xA;" OwnerUserId="2652" LastEditorUserId="2652" LastEditorDisplayName="" LastEditDate="2011-04-02T08:14:43.510" LastActivityDate="2011-04-02T08:14:43.510" />
  <row Id="7928" PostTypeId="2" ParentId="7906" CreationDate="2011-04-02T07:50:31.877" Score="2" ViewCount="" Body="&lt;p&gt;OK, the simple answer:&#xA;When there is  a resonance in the antenna  you have a coherent phenomenon. All the bands of electrons of the antenna are marching in tune. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black body radiation is an incoherent phenomenon coming from the individual atoms of the antenna. Even if the peak of the black body radiation were sitting on the resonance of the antenna it is still an incoherent phenomenon that cannot couple to the coherent behavior of the electrons in the current that resonate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of a single violin tune and a crowd of people talking. The noise of the people does not cover the clarity of the violin even in high volumes.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-02T07:50:31.877" CommentCount="4" />
  <row Id="7929" PostTypeId="2" ParentId="7907" CreationDate="2011-04-02T08:03:04.170" Score="7" ViewCount="" Body="&lt;p&gt;In recent years, it has become apparent that a class of materials called &lt;em&gt;topological insulators&lt;/em&gt; can be described by an action where the term $E\cdot B$ is added.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The action is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S_{top} = S_{em} + \frac{\theta}{2\pi}\frac{e^2}{\hbar c&amp;#183;2\pi} \int d^3xdt\, E&amp;#183;B.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ordinary insulators, we have $\theta=0$ while for topological insulators, we have $\theta=\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the $E&amp;#183;B$ term is a total derivative, Maxwells equations are unchanged inside and outside the insulator. But the point is that $\theta$ has a gradient at the surface, and that's where something interesting happens. Namely, an outside electric field can induce surface currents and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One might think that the action is not invariant under time-reversal, because then we would have to map $\theta\to-\theta$. But with periodic boundary conditions, it turns out that the value $\theta$ is &lt;em&gt;only well-defined modulo $2\pi$&lt;/em&gt;. Thus, both $\theta=0$ and $\theta=\pi=2\pi-\pi$ are possible. Note that you need quantum mechanics to understand that $\theta$ can only be defined up to $2\pi$, it is not possible to see that classically.&lt;/p&gt;&#xA;" OwnerUserId="188" LastActivityDate="2011-04-02T08:03:04.170" CommentCount="5" />
  <row Id="7930" PostTypeId="2" ParentId="7911" CreationDate="2011-04-02T08:18:11.587" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;&quot;Could it be that a car battery has less internal resistance than a garage battery?&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes! Let's denote a car battery internal resistance by $r_c$ , a garage battery internal resistance by $r_g$ , resistance of a toucher by $R$ ,voltage of a battery by $U$. Then current through the toucher:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I=\dfrac{U}{R+r}$$ The toucher is releasing power: $$P=I^2R= \dfrac{U^2R}{(R+r)^2}$$ Let's introduce the ratio:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu=\dfrac{P_c}{P_g}= \left(\dfrac{R+r_g}{R+r_c}\right)^2$$ $P_c$ is the power the toucher is releasing on a car battery and $P_g$ is the power the toucher is releasing on a garage battery. Internal resistance of a car battery is estimated about $r_c=0.001$ ohms and internal resistance of a garage battery around $r_g=0.1$ ohms.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity let's suppose that $R&amp;lt;&amp;lt;r_c$. Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu=\dfrac{P_c}{P_g}= \left(\dfrac{r_g}{r_c}\right)^2=100^2$$ &lt;/p&gt;&#xA;" OwnerUserId="552" LastActivityDate="2011-04-02T08:18:11.587" />
  <row Id="7931" PostTypeId="2" ParentId="7915" CreationDate="2011-04-02T08:28:33.603" Score="3" ViewCount="" Body="&lt;p&gt;Wind is caused by differences in pressure. Wikipedia's discussion of atmospheric Wind is &lt;a href=&quot;http://en.wikipedia.org/wiki/Wind&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I suspect that what is requested is whether the velocity can ever exceed the speed of sound for air &gt; Mach 1 ie supersonic wind. Well there is such a thing as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Supersonic_wind_tunnel&quot; rel=&quot;nofollow&quot;&gt;Supersonic Wind Tunnel&lt;/a&gt; which is generated by pressure difference ratios of 10 or so. This corresponds to the use of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Nozzle&quot; rel=&quot;nofollow&quot;&gt;nozzle&lt;/a&gt; with those pressure ratios between input and output streams. There are also &lt;a href=&quot;http://www.youtube.com/watch?v=U_9VNDS79Nk&amp;amp;feature=related&quot; rel=&quot;nofollow&quot;&gt;videos &lt;/a&gt; of objects in these tunnels.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-04-03T22:14:09.603" LastActivityDate="2011-04-03T22:14:09.603" CommentCount="4" />
  <row Id="7932" PostTypeId="2" ParentId="7898" CreationDate="2011-04-02T09:15:49.927" Score="2" ViewCount="" Body="&lt;p&gt;I haven't checked exhaustively, but I don't think you'll get a laser for under 500 (dollars?). I would recommend a LED light source. &lt;a href=&quot;http://www.oceanoptics.com/Products/ls450.asp&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a page with an example; they're more than &lt;code&gt;$&lt;/code&gt;500 but less than &lt;code&gt;$&lt;/code&gt;1000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I seem to remember that our lab got a cheap fiber-coupled white light LED source for 300 euros, but I can't remember the brand right now and I can't find it online.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps &quot;black light&quot; fluorescent tubes would meet your needs - the article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_light&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; has a table listing typical emission peaks depending on what kind of phosphor is used in the tube. I have no idea where the best place to buy those is, but a Google search seems to show that they're available everywhere, perhaps with varying quality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for a better way of focusing, it's hard to say without knowing more about your requirements - apparently you don't need temporally coherent light, since you're considering not using a laser, but do you need spatially coherent light? You say light spread doesn't matter, so I assume not. Do note that by &quot;focusing&quot; with a black paper cone, you'll be throwing a lot of light away; perhaps you might use lenses. Microscope objectives are expensive, but perhaps if you can &quot;borrow&quot; one from a microscope in the area, you could use one to focus.&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2011-04-02T09:15:49.927" CommentCount="1" />
  <row Id="7933" PostTypeId="1" CreationDate="2011-04-02T09:36:18.383" Score="4" ViewCount="127" Body="&lt;p&gt;A condensate of open strings with both ends attached to the same D-brane can be equivalent to a displacement of the D-brane with no open string condensate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A solution to the D-brane Born-Infeld action gives rise to a semi-infinite string extension which is entirely equivalent to an open string with one end attached to the D-brane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why aren't D-branes and strings independent degrees of freedom?&lt;/p&gt;&#xA;" OwnerUserId="2899" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-04-02T10:06:42.570" LastActivityDate="2011-04-02T12:55:04.877" Title="Why aren't D-branes and strings independent degrees of freedom?" Tags="&lt;string-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="7934" PostTypeId="2" ParentId="7915" CreationDate="2011-04-02T09:40:55.203" Score="1" ViewCount="" Body="&lt;p&gt;No.&#xA;Look at max pressure differences on isobaric (weather) charts. &#xA;With such low differences You cant procuce winds of such speed. &#xA;There are very fast winds in stratosphere (&quot;jet streams&quot;)&#xA;but those never exceed 650 km/h &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Jet_stream&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Jet_stream&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(max speed in discussion page) &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-02T09:40:55.203" />
  <row Id="7935" PostTypeId="2" ParentId="7911" CreationDate="2011-04-02T10:02:44.817" Score="5" ViewCount="" Body="&lt;p&gt;The answers of Martin and Edward &#xA;are quite right, but theý lack a componet which plays a role &#xA;in the case of that &quot;garage opener battery&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If current is not limited in the circuit outside the battery, &#xA;first internal resistance can be the limiting factor. &#xA;But, often in such small (the measure for &quot;small&quot; is the area/volume of &#xA;active material at the battery poles) batteries &#xA;the current is limited by the speed of chemical reaction at the &#xA;electrodes. This is not a linear function of current, thus&#xA;it cannot viewed as a part of internal resistance. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does it have anything to do with&#xA;  contact area? If it does, then how&#xA;  would you model it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, contact area is important, but this is a problem at high &#xA;current densities mostly. &#xA;Depending on voltage in such a circuit, small contact area &#xA;(= high current density) &#xA;can result in contact welding, sparking, or ignition of an arc,&#xA;or in some strange semiconductor effects of the contact point. &#xA;(metal surfaces are nearly always coated by some oxide)&#xA;Did You ever watch when an electric arc is started by a welder?&#xA;This is main business when designing contacts in relays, swiches in &#xA;home, or the switches in 230 kV lines. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-02T11:37:31.240" LastActivityDate="2011-04-02T11:37:31.240" CommentCount="3" />
  <row Id="7936" PostTypeId="1" AcceptedAnswerId="8002" CreationDate="2011-04-02T10:11:48.733" Score="3" ViewCount="245" Body="&lt;p&gt;My physics teacher explained the difference between voltage and current using sandwiches. Each person gets a bag full of sandwiches when they pass through the battery.  Current = the number of people passing through a particular point per unit time.  Voltage = the (change in) number of sandwiches per person.  In a parallel circuit the number of people (current) is divided between the two paths, but the number of sandwiches per person (voltage) remains the same.  In a series circuit the number of people passing through a particular point remain the same, but they drop off a certain percentage of their sandwiches at every resistor.   Therefore, there is a voltage drop that occurs between the points before and after every resistor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This analogy naturally leads to the question: how do the electrons &quot;know&quot; that they are going to have to share their voltage between two resistors before they reach the second one?  (In other words, not drop off all their sandwiches at the first resistor they find)&lt;/p&gt;&#xA;" OwnerUserId="2900" LastEditorUserId="2900" LastEditorDisplayName="" LastEditDate="2011-04-02T10:19:29.833" LastActivityDate="2011-04-03T18:57:41.187" Title="How do electrons &quot;know&quot; to share their voltage between two resistors?" Tags="&lt;electromagnetism&gt;&lt;electricity&gt;&lt;electron&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="7937" PostTypeId="2" ParentId="7936" CreationDate="2011-04-02T10:54:03.943" Score="1" ViewCount="" Body="&lt;p&gt;Well, they don't know that actually. The picture you are familiar with is quite simplified. To explain the apparent emergent properties you need statistics and knowledge of microscopic physics. So what happens with that electron as it passes through the circuit? Well, let's start with resistors. Why is there any resistance at all? It's because electrons collide with the crystal lattice of the conductor and thereby lose energy (let me use this more familiar term instead of your sandwiches). How much? Well, it depends on precise realization of the collision. Some electrons scatter elastically (losing no energy at all, just changing direction of motion), some don't. But if you average this over all electrons (because there is so many of them in the material), you'll obtain the familiar Ohm's law for the heat losses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gaining of energy by electrons is pretty much same, but in reverse. There is a possibility that the electron will absorb a photon (which is a quantum of electromagnetic field) and this will boost it. Again, whether the collision is elastic or not doesn't matter. Once you average over all electrons and photons you'll be left with macroscopic effect of current produced by classical EM field.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-04-02T10:54:03.943" CommentCount="4" />
  <row Id="7938" PostTypeId="1" CreationDate="2011-04-02T11:13:05.853" Score="2" ViewCount="132" Body="&lt;p&gt;What do you guys think of alternative career paths in physics away from academia free from the usual academic shackles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples: Garrett Lisi who spends his time surfing and skiing while not working on E&lt;sub&gt;8&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or Daniel Bedingham who is able to support himself part-time as an investment banker when not working on quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or Julian Barbour who supports himself by translating Russian journal articles into English when not working on quantum gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it feasible to count upon Foundational Questions Institute to support such unconventional career paths using Templeton money? &lt;/p&gt;&#xA;" OwnerUserId="2902" LastActivityDate="2011-04-02T11:43:58.367" ClosedDate="2011-04-02T12:51:10.870" Title="Alternative career paths in physics?" Tags="&lt;physics-careers&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="7939" PostTypeId="2" ParentId="7936" CreationDate="2011-04-02T11:32:28.073" Score="1" ViewCount="" Body="&lt;p&gt;Continuing Marek's reply, I will add that in  electrical circuits the value of the current depends on the elements of the circuit: the  &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrical_network&quot; rel=&quot;nofollow&quot;&gt;voltage , the resistors&lt;/a&gt; summed when in series, summed inversely if in parallel. The whole defines the value of the current, and currents in the parallel parts, according to the electricity laws. The individual electrons participating in the current follow the flow, it is the current that &quot;knows&quot; where the resistances are . &lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-02T11:32:28.073" CommentCount="3" />
  <row Id="7940" PostTypeId="2" ParentId="7938" CreationDate="2011-04-02T11:43:58.367" Score="1" ViewCount="" Body="&lt;p&gt;Before closure, thanks for informing me of the &lt;a href=&quot;http://en.wikipedia.org/wiki/FQXi&quot; rel=&quot;nofollow&quot;&gt;foundational question institutes '&lt;/a&gt; existence :) .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the link :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It has run two worldwide grant competitions. The first competition provided US $2M to 30 projects&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any researcher who can compete is free to try for those 30 grants. On the other hand academic and research posts run into the hundreds.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-02T11:43:58.367" />
  <row Id="7941" PostTypeId="2" ParentId="7933" CreationDate="2011-04-02T12:49:06.830" Score="3" ViewCount="" Body="&lt;p&gt;D-branes and fundamental strings are not independent degrees of freedom because D-branes are solitons or topological defects constructed out of closed string fields. At weak coupling, the closed string degrees of freedom are the only genuinely elementary fields; &lt;em&gt;everything else&lt;/em&gt;, including D-branes, may be thought of as being constructed out of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equivalently, it's because the string world sheets that determine all the dynamics of the D-branes are the same world sheets that also determine other things, and these world sheets - histories of strings - can't be double-counted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why the degrees of freedom are not independent of each other is the same as the reason (known as the &quot;modular invariance&quot;) why the torus with the complex structure $\tau$ is equivalent to a torus with the complex structure $-1/\tau$: they only differ in the way how we &quot;read them&quot; or &quot;slice them&quot;. One is not allowed to double-count $\tau$ and $-1/\tau$ because they represent two configurations related by a large world sheet diffeomorphism, and all world sheet diffeomorphisms - including the large ones - have to be counted as gauge symmetries (for consistency).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way, a cylinder with circumference of the circular boundaries $2\pi t$ and with the distance $\pi$ between the two boundaries may be thought of as a loop of an open string - if we slice the cylinder into line intervals, i.e. &#xA;$${\rm Tr}_{\rm open} \exp(-\beta H_{\rm open}),$$&#xA;or as a closed string propagation between two boundary states (closed string states encoding the full information about the D-brane), &#xA;$$\langle B_1|\exp(-\beta H_{\rm closed})|B_2\rangle.$$&#xA;This &quot;failure to be independent&quot; between the degrees of freedom is the basis of the UV-IR connection and gauge-gravity or open-closed dualities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any configuration of a D-brane or D-branes that solves the stringy equations of motion may be represented by the &quot;boundary state&quot;, the closed string state determining the allowed boundary conditions of a world sheet. It must obey various conditions. On the other hand, every perturbation of a D-brane state is also generated by the quanta of open strings attached to this D-brane. These descriptions are mathematically different but physically equivalent and the corresponding degrees of freedom coincide - they shouldn't be double counted.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-02T12:55:04.877" LastActivityDate="2011-04-02T12:55:04.877" />
  <row Id="7942" PostTypeId="2" ParentId="7915" CreationDate="2011-04-02T13:08:35.077" Score="2" ViewCount="" Body="&lt;p&gt;I think that some confusion is here.&#xA;Wind speed is determined by differences in the air pression between two points. The max. limits that we observe must set the max wind speed. Wind is the movement of a mass of air. The Sound speed determines the sound barrier. Sound speed is determined by considerations about density, temperature... and is a property about the relative speed of the wavefront of a sound event (a perturbation of te medium - air, that propagates) in relation to te the air (considered at rest). The speed of sound in relation to the ground is the vector addition/subtraction of the sound speed (+-340m/s) with the speed of wind in relation to the ground. The max wind speed is observed in the jet streams in altitude and in the tornados and hurricanes at surface. Somewhere we can find the max speed of the exhausted air in the jets and determine if air moves trhu the air with a speed superior to the speed barrier. Nothing prevent this from happening with enough thrust. I've no time to search now, sorry.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT add:&lt;br&gt;&#xA;There is is need to search. The exhausted mixture of jet engines, not beeing wind but mostly a mass of air, can travel thru the atmosphere at speeds much superior of sound barrier. If it is was no so the jet planes could not cross the sound barrier.&lt;br&gt;&#xA;So, inspite of the confusion in the question, I the short answer is a Yes (a mass of air can travel faster than max sound speed). &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-04-03T22:36:09.463" LastActivityDate="2011-04-03T22:36:09.463" />
  <row Id="7944" PostTypeId="2" ParentId="7915" CreationDate="2011-04-02T14:31:26.050" Score="1" ViewCount="" Body="&lt;p&gt;The hypothetical &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypercane&quot; rel=&quot;nofollow&quot;&gt;Hypercane&lt;/a&gt; only has winds of around 800kph, so even this extreme situation is well below the soundspeed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not so sure that some gas giants might not have local winds in excess of the sound speed, here is wikipedia on Neptune&#xA;&quot;Neptune has extremely dynamic weather systems, including the highest wind speeds in the solar system, thought to be powered by the flow of internal heat. Typical winds in the banded equatorial region can possess speeds of around 350 m/s, while storm systems can have winds reaching up to around 900 m/s, nearly the speed of sound in Neptune's atmosphere. Several large storm systems have been identified, including the Great Dark Spot, a cyclonic storm system the size of Eurasia, the Scooter, a white cloud group further south than the Great Dark Spot, and the Wizard's eye/Dark Spot 2, a southern cyclonic storm.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect Gas Giants can have high local wind speed with respect to the local speed of sound, because wind systems may extend to great depth/pressures, where the sound speed is much greater, so that the ratio of soundspeed at depth, versus soundspeed high in the atmosphere can become quite large.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-04-02T14:31:26.050" />
  <row Id="7947" PostTypeId="2" ParentId="7922" CreationDate="2011-04-02T15:17:59.890" Score="0" ViewCount="" Body="&lt;p&gt;Lets look at simple probability theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you toss a coin $N$ times. And you get heads $H$ times and tails $T$ times. &#xA;When you don't get heads you get tails. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means $T = N - H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, probability of getting heads $P_h = \frac{H}{N}$ (by definition)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and probability of getting tails $P_t = \frac{T}{N} = \frac{N-H}{N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_h + P_t = \frac{H}{N} + \frac{N-H}{N} = \frac{H + N - H}{N} = \frac{N}{N} = 1$&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2011-04-02T15:17:59.890" />
  <row Id="7948" PostTypeId="1" CreationDate="2011-04-02T15:47:42.723" Score="1" ViewCount="96" Body="&lt;p&gt;Please Can anyone explain the concepts of electromagnetic induced transparency? I am having problem with the technicality of the explanation on wikipedia. Please I am an engineer with a physics background though and would love if this can be explained with little to no jargons . I am working on optical buffers and studying recent technologies and would love and explanantion on this . Thanks &lt;/p&gt;&#xA;" OwnerUserId="2904" LastActivityDate="2011-04-02T16:26:18.500" Title="What is electromagnetic Induced Transparency" Tags="&lt;quantum-optics&gt;&lt;quantum-electrodynamics&gt;" AnswerCount="1" />
  <row Id="7949" PostTypeId="2" ParentId="7899" CreationDate="2011-04-02T16:02:31.363" Score="1" ViewCount="" Body="&lt;p&gt;What I think is being asked here is about the alternatives theoretically (and experimentally) consistent with the cosmological solutions in GR. Also some history of terminology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First it would be useful to distinguish between the &lt;a href=&quot;http://en.wikipedia.org/wiki/Observable_universe&quot; rel=&quot;nofollow&quot;&gt;Observable Universe&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Universe&quot; rel=&quot;nofollow&quot;&gt;Universe&lt;/a&gt;. There are so many theories around that this distinction will help clarify communication (and understanding). Of course the Observable Universe is what is observed and deductions about what is true outside of that (from some theory) are not going to be astronomically measurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to parts of the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;People say that &quot;the universe is expanding&quot; rather than &quot;observable matter is separating&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the Hubble galactic expansion had been discovered in the decades before Einstein's General Relativity (GR), then perhaps the second phrase would have been more common initially. However GR also contains Cosmological solutions intended therefore to describe the Universe (and not just the Observable one). At around the time of Hubble's discoveries the Lemaitre-Friedmann solution was gaining ground in which GR is seen to provide a specific expansion of the Universe from a t=0 singularity. So the early and essentially continual match between theory and astronomical observation has resulted in the first phrase being used as well.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do we know that BB wasn't: t=0, V(universe)&gt;0 but V(matter)=0, E=very big?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Strictly what describes the t=0 solution of GR is outside of GR. However leaving that point aside there are some comments.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;V(universe) =0 or &gt;0&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This really asks what the singularity can be modelled as in GR. In the days when the closed universe model was prominently discussed it was natural to talk about universe radius and origin at a point for the singularity. However, and especially when the flat or open models seem more observationally accurate, those earlier assumptions do need to be re-considered in accounts. For a flat model say, then the singularity could be an &lt;strong&gt;E&lt;/strong&gt;^3 - much different from the &quot;point&quot; often discussed (although other geometries may be possible too).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the density of the &lt;em&gt;observable&lt;/em&gt; universe's matter at this point would still tend toward infinity at t=0 and thus occupy V(matter)=0 as you have written it. So this leaves a big question about what might be true of the Universe as a whole. Obviously unknown, but a range of theories abound, including some based on Inflation ideas.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;E= very big?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Surprisingly perhaps this need not be true. The reason is that gravitational energy (a kind of binding energy) counts negatively. So many theorists prefer the idea that E= small, with E=0 also being popular because then there is no net energy in the universe at t=0.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-04-02T16:02:31.363" CommentCount="3" />
  <row Id="7950" PostTypeId="2" ParentId="7948" CreationDate="2011-04-02T16:26:18.500" Score="1" ViewCount="" Body="&lt;p&gt;As you may know, electromagnetic induced transparency (or EIT) is a technique for eliminating the effect of a medium on a propagating beam of electromagnetic radiation. To obtain transparency or to improve transmission, you have to apply two laser wavelengths whose frequencies differ by a Raman transition of a medium. From a classical point of view, you can eliminate the effect of a medium on a propagating beam by stopping the electrons to move at the frequencies of the applied field (i.e. if the electrons do not move, they do not contribute to the dielectric constant). This non-movement will happen if at each applied frequency, the electron is subjected to two sinusoidal forces of opposing phase. &#xA;For a quantum mechanical treatment of the problem and other details I would suggest this article &lt;a href=&quot;http://www.unl.edu/amop/pdf_files/RMP_EIT_Fleischhauer.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.unl.edu/amop/pdf_files/RMP_EIT_Fleischhauer.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.    &lt;/p&gt;&#xA;" OwnerUserId="2473" LastActivityDate="2011-04-02T16:26:18.500" />
  <row Id="7951" PostTypeId="2" ParentId="1961" CreationDate="2011-04-02T16:52:54.093" Score="2" ViewCount="" Body="&lt;p&gt;Summary: A good rule of thumb is roughly 2/3 the speed of light in a vacuum for all these transmission media. Optic fiber is slower than some and faster than others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Details:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://amasci.com/elect/poynt/poynt.html&quot; rel=&quot;nofollow&quot;&gt;Electrical energy normally doesn't flow inside of metals&lt;/a&gt;.&#xA;Poynting-flow diagrams show where the energy flows, and the path(s) traveled by the information carried by that energy, although they are a bit difficult to draw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As conqenator pointed out,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Transparent_materials#Optical_waveguides&quot; rel=&quot;nofollow&quot;&gt;refractive index&lt;/a&gt; ... Typical values for core and cladding of an optical fiber are 1.48 and 1.46, respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which gives a wave propagation speed of 1/1.48 and 1/1.46 of the speed of light in a vacuum (0.67 c and 0.68 c)  in these regions of a typical optical fiber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various popular cables have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Wave_propagation_speed&quot; rel=&quot;nofollow&quot;&gt;wave propagation speed&lt;/a&gt; that ranges from 42% to 95% of the speed of light. In particular, 100 Ohm &lt;a href=&quot;http://en.wikipedia.org/wiki/Category_5_cable&quot; rel=&quot;nofollow&quot;&gt;Cat5e cable&lt;/a&gt; has a typical propagation speed of 0.64 c. The 300 Ohm Twin-lead has a typical propagation speed of 0.82 c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dmckee reminded us that it's usually adequate to approximate all of these as around 2/3 c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you use Cat5e cable, bits will arrive slightly later than bits sent through fiber optic. If you use 300 Ohm Twin-lead, bits will arrive slightly before bits sent through fiber optic. &lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2011-04-02T16:52:54.093" />
  <row Id="7952" PostTypeId="2" ParentId="7905" CreationDate="2011-04-02T16:54:12.833" Score="0" ViewCount="" Body="&lt;p&gt;I retract the following answer (see comments): &lt;em&gt;emphasized textA massless charged particle would violate special relativity. It could be accelerated by an electric field beyond the speed of light.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1567" LastEditorUserId="1567" LastEditorDisplayName="" LastEditDate="2011-04-02T19:15:54.800" LastActivityDate="2011-04-02T19:15:54.800" CommentCount="16" />
  <row Id="7953" PostTypeId="1" CreationDate="2011-04-02T18:21:11.057" Score="3" ViewCount="139" Body="&lt;p&gt;I have no prior experience of using matlab. My teacher want me to solve this question. I have been trying for a couple of hours now with no luck, please help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of 100 g hanging in a spring with spring constant 40 N / m. The mass is set into vibration by the spring is stretched 14 mm. Neglecting any energy loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use MATLAB and plot a graph showing the distance from the equilibrium position as&#xA;function of time with initial conditions x(0) = 14 mm and v (0) = 0 mm/s&lt;/p&gt;&#xA;" OwnerUserId="2905" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-02T23:23:04.280" LastActivityDate="2011-04-02T23:23:04.280" Title="Plotting a SHO in matlab" Tags="&lt;homework&gt;&lt;simulation&gt;&lt;harmonic-oscillator&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="7954" PostTypeId="2" ParentId="7915" CreationDate="2011-04-02T19:37:20.927" Score="1" ViewCount="" Body="&lt;p&gt;Pure air sound velocity is very high and is not attainable for the atmosphere pressure differences. However in a two-phase mixture (air-water, for example) the sound velocity can be very low and comparable with the wind velocity.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-04-02T19:37:20.927" />
  <row Id="7955" PostTypeId="2" ParentId="7953" CreationDate="2011-04-02T20:03:55.960" Score="5" ViewCount="" Body="&lt;p&gt;Addressing just the physics part (go to stack overflow for the programming), and using the equation that you've been given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x(t) = A \cos \left( \omega t + \delta \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at the form of the solution.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is sinusoidal&lt;/li&gt;&#xA;&lt;li&gt;The curve will have a maximum value of $A$ (because cosine has a maximum value of 1)&lt;/li&gt;&#xA;&lt;li&gt;When $\delta$ is $0, \pm 2\pi, \pm 4\pi \dots$ the maximum will occur at $t = 0, \pm \frac{2\pi}{\omega}, \pm \frac{4\pi}{\omega} \dots$, and will move around by $-\delta$. So we conclude that $\delta$ controls the phase of the oscillation.&lt;/li&gt;&#xA;&lt;li&gt;We call $\omega$ the angular frequency of the oscillation and find that the period is $T = \frac{2\pi}{\omega}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With that, lets look at your initial conditions and see if we can fill in some of those variables...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The string starts stretched down from the equilibrium position. I'm going to call that the $-x$ direction. So $x(0) = -14$ rather than $+14$ as you wrote (but that is just a detail  of the coordinate system I've chosen).&lt;/li&gt;&#xA;&lt;li&gt;Will the weight ever move farther down that the starting position? What does that tell you about $A$ and $\delta$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That leaves us with $\omega$ as an unknown. I presume you've been shown a derivation of simple harmonic oscillation, and already know that for a mass on a spring&#xA;$$ \omega = \sqrt{ \frac{k}{m} }$$. If you have not look at my answer to &lt;a href=&quot;http://physics.stackexchange.com/questions/1018/explanation-simple-harmonic-motion/1020#1020&quot;&gt;Explanation: Simple Harmonic Motion&lt;/a&gt; (or indeed several of the other answers to that question).&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-04-02T20:03:55.960" CommentCount="1" />
  <row Id="7956" PostTypeId="2" ParentId="7936" CreationDate="2011-04-02T21:02:35.303" Score="2" ViewCount="" Body="&lt;p&gt;The analogy used by the teacher is quite inappropiate, as pointed out by Georg. Do the electrons have to stop running when they are out of sandwiches? Or are they allowed to run on credit? Then: resistance is not part of the analogy (at least it is not mentioned) which is impossible because the analogy is about Ohms law: $V=IR$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then: also when using better models it is always a problem to think about individual electrons: how does one particular electron know that the voltage is lower at the other end of the loop? Individual electrons do not exist in QM, electrons are indistinguishable, they are a collective, so in your analogy, should all sandwhich eaters exist of some kind of clairvoyant species with a collective conscience?&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2011-04-02T21:02:35.303" />
  <row Id="7957" PostTypeId="2" ParentId="7918" CreationDate="2011-04-02T21:35:08.487" Score="-1" ViewCount="" Body="&lt;p&gt;As I look upon the MWI: every event is a branch point of a particular universe (i.e. the universe before the event) to &lt;em&gt;a set of universes&lt;/em&gt; after the event. Every universe in the set has another outcome of the event as it's start point. All universes are to be considered as objectively real, all outcomes of the event do occur, all in another universe and the different universes cannot interact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You descrive an event with one particular outcome: the annihilation of a particle-antiparticle pair. The MWI is not related to this situation, your universe is the same before and after this event because you do not consider different outcomes of the event, you assume an event with only one result (annililation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some further remarks: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Your phrase &quot;position of the universe&quot; does not make sense, because position is measured &lt;em&gt;ín&lt;/em&gt; some universe. For a universe to have a position we have to consider another universe that contains it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The MWI is extremley absurd. When there are 10^80 paricles in the universe and each particle has 10^40 events per second, then after 10^17 seconds there are (10^80^40^17 = 10^54400) universes. MWI defies common sense also in other respects.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2011-04-02T21:35:08.487" CommentCount="6" />
  <row Id="7958" PostTypeId="2" ParentId="7899" CreationDate="2011-04-02T22:19:38.633" Score="2" ViewCount="" Body="&lt;p&gt;I will assume that E in the question is energy density.  First we have to distinguish between two sizes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;size of the observable universe = current proper distance to particle horizon = current proper distance to infinite redshift, and&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;size of the whole universe.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Assuming trivial global topology, the size of the entire universe is infinite if space is flat or open (hyperbolic), and finite if space is closed (spherical).  To note, if the magnitude of the current density parameter for curvature $\Omega_k(to)$ is smaller than a certain threshold, it is impossible to know whether the universe is flat and infinite or closed and finite. (&lt;a href=&quot;http://arxiv.org/abs/0901.3354&quot; rel=&quot;nofollow&quot;&gt;Vardanyan et al (2009) &lt;em&gt;How flat can you get?&lt;/em&gt;&lt;/a&gt;)  Clearly the total size of an infinite universe was infinite at t=0, while the total size of a finite universe was zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the alternative view of matter travelling into pre-existing empty space (Minkowski spacetime) is &lt;strong&gt;precisely&lt;/strong&gt; the model proposed by Edward Milne in 1932.  For that &lt;strong&gt;precise&lt;/strong&gt; model to be valid we have to assume:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;either that GR is not applicable to cosmology (Milne's original position),&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;or that the universe is gravitationally empty at large scales (&lt;a href=&quot;http://physics.stackexchange.com/questions/4015/empty-universe-in-the-past-non-empty-in-the-future/5477#5477&quot;&gt;an unlikely but still open possibility&lt;/a&gt;).  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, if you stay within GR, then matter distribution determines spacetime, so that a non-neglibible matter density implies a spacetime different from Minkowski (so that if the matter moves, the spacetime changes over time). Having cleared that, as I said in &lt;a href=&quot;http://physics.stackexchange.com/questions/7522/do-cosmological-and-doppler-redshift-produce-different-patterns/7548#7548&quot;&gt;this answer&lt;/a&gt;, any spacetime described by the FLRW metric, where objects are static in comoving coordinates and redshift is due to expansion of space, can also be described, via an appropriate change of coordinates, by a spherically-symmetric (SS) metric, where objects move along radial timelike geodesics and redshift is due to positional (gtt) and Doppler factors. Since both metrics describe the same physical system, they are observationally equivalent in all respects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two notes on the equivalent SS metric:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;usually (but not in the Milne case) has a cosmological horizon (gtt &amp;amp; grr -&gt; $\infty$), located at the Hubble distance in flat FLRW models,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;is static (gtt &amp;amp; grr do not depend on time), &lt;strong&gt;only&lt;/strong&gt; in the empty (Milne) and lambda-vacuum (de Sitter) cases,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;even in the next simplest case, the flat matter-only FLRW model known as Einstein-de Sitter model, gtt &amp;amp; grr cannot be expressed algebraically in terms of non-comoving time (t' below), as I mentioned in &lt;a href=&quot;http://physics.stackexchange.com/questions/6624/are-the-solutions-in-radicals-of-cubic-and-quartic-of-any-use-in-physics/6669#6669&quot;&gt;this answer&lt;/a&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I expanded this below to practice a bit of LaTex:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any homogeneous and isotropic spacetime described by the FLRW metric in terms of a set of coordinates $(t, r, \theta, \phi)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{equation}&#xD;&#xA;ds^2 = - c^2 dt^2 + \frac {a(t)^2 dr^2} {(1 - k r^2)} + a(t)^2 r^2 d\Omega^2&#xD;&#xA;\end{equation}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;t = comoving time = proper time for all observers at constant $(r, \theta, \phi)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;r = comoving radial coordinate = radial coordinate enclosing constant proper mass&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d\Omega^2 = d\theta^2 + sin^2(\theta) d\phi^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be described by a spherically symmetric (SS) metric in terms of a different set of coordinates $(t&amp;#39;, r&amp;#39;, \theta, \phi)$, where the ' does &lt;strong&gt;not&lt;/strong&gt; mean a derivative:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{equation}&#xD;&#xA;ds^2 = - c^2 gtt(t&amp;#39;, r&amp;#39;) dt&amp;#39;^2 + grr(t&amp;#39;, r&amp;#39;) dr&amp;#39;^2 + r&amp;#39;^2 d\Omega^2&#xD;&#xA;\end{equation}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&#xA;$\Omega$, $\theta$ and $\phi$ are the same as in the FLRW metric, and obviously&#xA;$r&amp;#39; = a(t) r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expressing $t&amp;#39; = f(t, r)$ and its partial derivatives as $ft$ and $fr$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{equation}&#xD;&#xA;grr = \frac {1} {[1 - k r^2 - \left( \frac {r} {c} \frac {da(t)} {dt} \right)^2]}&#xD;&#xA;\end{equation}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{equation}&#xD;&#xA;gtt = \frac {grr} {ft^2} (1 - k r^2)&#xD;&#xA;\end{equation}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem lies in expressing gtt &amp;amp; grr in terms of $(t&amp;#39;, r&amp;#39;)$, which can be done only in the Milne and de Sitter cases.&lt;/p&gt;&#xA;" OwnerUserId="2085" LastEditorUserId="2085" LastEditorDisplayName="" LastEditDate="2011-04-02T23:11:21.937" LastActivityDate="2011-04-02T23:11:21.937" />
  <row Id="7959" PostTypeId="1" AcceptedAnswerId="7967" CreationDate="2011-04-02T23:46:37.200" Score="3" ViewCount="168" Body="&lt;p&gt;This is a follow up question from: &lt;a href=&quot;http://physics.stackexchange.com/questions/7905/massless-charged-particles&quot;&gt;Massless charged particles&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since by definition such a particle would interact with photons- resulting in some change in momentum- would the particle emit Bremsstrahlung during this interaction? If it does, it seems that there is a problem as the particle cannot decelerate, yet the Bremsstrahlung would necessarily carry away some of the energy. Or perhaps the fact the particle is massless precludes such emissions. Is this a real problem? What would the physical effects of Bremsstrahlung radiation be for this particle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: for some reference- this question came out of a discussion in the comments after my answer to   &lt;a href=&quot;http://physics.stackexchange.com/questions/7905/massless-charged-particles&quot;&gt;Massless charged particles&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1567" LastEditorUserId="1567" LastEditorDisplayName="" LastEditDate="2011-04-03T03:12:24.313" LastActivityDate="2011-04-03T17:50:21.520" Title="Is there Bremsstrahlung radiation for a charged massless particle?" Tags="&lt;particles&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="7960" PostTypeId="2" ParentId="7865" CreationDate="2011-04-03T00:01:59.100" Score="2" ViewCount="" Body="&lt;p&gt;I can give some indication of what is going on here, but I have not studied all the references as yet. This topic is complicated by the fact that there are several (possibly independent) ingredients to this issue. You can follow up on some or all aspects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;General Relativity and Einstein-Cartan Theory&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic equations of GR we can write as $G^{ab}=P^{ab}$. Here the LHS is the geometric content of the theory and the RHS is the matter (stress-energy tensor) content of a given solution. Basic properties show that both Tensors (and several others) are symmetric. The geometric solutions are described by a metric which generates &lt;em&gt;curvature&lt;/em&gt;. Cartan discovered later that this was not the most general form of such a geometry. Essentially the geometry could contain non-symmetric Tensors too, say $T_{a}^{bc}$ as the &lt;em&gt;Torsion&lt;/em&gt; tensor. Thus a manifold could have two independent types of non-Euclideanness: Curvature and Torsion. The curvature in GR is well known, but the Torsion is set to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to have a manifold which has zero curvature (so all parallel lines stay parallel), but has non-zero Torsion. So how does this manifest itself in the manifold? Well basically given say a unit vector (eg pencil with a point) moving it along a curve will result in the pencil having rotated from its original direction. Parallel lines stay parallel, but the space has a sort of built-in helix structure causing an intrinsic - or geometric - rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interesting mathematical idea, but does it relate to anything physical? Einstein became excited by the idea for a time, developing Teleparallel gravity - lots of torsion but no curvature (hence the Parallel aspect).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This theory was developed further in the 1960s by coupling (ie equating) the Torsion to the &quot;Spin Tensor&quot; $S_{a}^{bc}$. This added a second equation to the usual Einstein Tensor equation, generating EC (KS) Theory. The spin tensor is classically defined as a generalisation of angular momentum to tensors. So in the EC theory the angular momentum of an object also has gravitational effect (calculations show that it would only be significant in high spin neutron stars or maybe early universe.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Spin Density&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple-seeming generalisation of the idea of &lt;em&gt;density&lt;/em&gt; from mass density ie $\rho(x)$ to linear momentum density to spin density $s(x)$ is associated with the name of Wessenhoff in 1947 who was trying to develop a &quot;classical&quot; Dirac theory. Here a spin value is associated with each point in space - at least in the mathematics. Later there is discussion of a &quot;Wessenhoff fluid&quot; - which is a strange quasi-classical fluid with vortices at all length scales. Since then the phrase &quot;spin density&quot; has become used in some parts of the literature without explicitly including quantum effects, but corresponds to a use of this mathematical idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some authors seem to map the idea directly to quantum spin without considering such quasi-classical models - so the ideas can do double-duty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Geometric Algebra&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;David Hestenes has expressed all the above concepts in Geometric Algebra (known as STA) which also allows discussion of spin density in this sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A paper on Spinning (Wessenhoff-like) fluids in Cosmology (under EC theory): &lt;a href=&quot;http://deepblue.lib.umich.edu/bitstream/2027.42/49195/2/cq940917.pdf&quot; rel=&quot;nofollow&quot;&gt;http://deepblue.lib.umich.edu/bitstream/2027.42/49195/2/cq940917.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A paper by David Hestenes:&#xA;&lt;a href=&quot;http://geocalc.clas.asu.edu/pdf-preAdobe8/Decouple.pdf&quot; rel=&quot;nofollow&quot;&gt;http://geocalc.clas.asu.edu/pdf-preAdobe8/Decouple.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Intrinsic Angular Momentum&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also one should make a more elementary comment about the possible use of this phrase in Classical Mechanics. A given system of particles will have an angular momentum &lt;strong&gt;M'&lt;/strong&gt; in a given frame K' in which it is otherwise at rest. From the perspective of a different frame K there will be a translation formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;M&lt;/strong&gt; = &lt;strong&gt;M'&lt;/strong&gt; + &lt;strong&gt;R&lt;/strong&gt; x &lt;strong&gt;P&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;strong&gt;R&lt;/strong&gt; is the radius to centre of mass in K and &lt;strong&gt;P&lt;/strong&gt; is the momentum in K. Thus from the perspective of K the angular momentum consists of the &quot;intrinsic angular momentum&quot; &lt;strong&gt;M'&lt;/strong&gt; plus the angular momentum of the whole system. [Landau/Lifshitz Vol1 (9.6) actually use the expression &quot;intrinsic angular momentum&quot;.]&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-04-03T17:11:39.713" LastActivityDate="2011-04-03T17:11:39.713" />
  <row Id="7961" PostTypeId="1" AcceptedAnswerId="7975" CreationDate="2011-04-03T00:05:45.950" Score="3" ViewCount="122" Body="&lt;p&gt;During solar system formation, many bodies achieved hydrostatic equilibrium, a spherical shape where their self gravitational force was balanced by internal pressure. Many also achieved differentiation, where a body is seperated into layers of different density (core, mantle and crust). Differentiation was mainly the result of the melting heat from isotopes like Al 26 and Fe 60, with half lives of ~10^6 years, coupled with the body's gravity. My question is, can differentiation occur without the presence of radioactive isotopes? Can a body's mass alone create sufficient gravitational force to induce melting, say by the attraction of meteoroids, and develop differentiation?&#xA;Of course stars are differentiated, but I'm interested in their orbiting bodies.&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-04-03T07:53:13.803" Title="Gravity and Planetary Differentiation" Tags="&lt;gravity&gt;&lt;astronomy&gt;&lt;radioactivity&gt;" AnswerCount="1" />
  <row Id="7962" PostTypeId="1" AcceptedAnswerId="7965" CreationDate="2011-04-03T01:23:46.427" Score="3" ViewCount="590" Body="&lt;p&gt;From Michael on &lt;a href=&quot;http://skeptics.stackexchange.com/questions/1664/it-is-not-the-voltage-that-kills-you-it-is-the-current/1677#1677&quot;&gt;Skeptics Stackexchange&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How about a wire that's grounded? &#xA;  Safe to touch, right?   WRONG.&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;   ________________ 30 amps -&amp;gt; ________________&#xA;  |                                            |&#xA;  +                                            |&#xA; 220V                                        Load&#xA;  -                                            |&#xA;  |______(YOU ARE HERE)______&amp;lt;- 30 amps________|&#xA;  |&#xA;Ground&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;The wire you touched was not only at 0&#xA;  volts, but also grounded, and yet, you&#xA;  are feeling pretty shitty in this&#xA;  diagram.  You have ceased to be as a&#xA;  human, and you are now a part of a&#xA;  circuit, functioning as part of a&#xA;  return leg (pictured above) or as a&#xA;  &quot;parallel path to ground&quot; (not&#xA;  pictured above.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't get how this can work. If the wire is at 0 volts and you are at 0 volts, then there is no potential difference and hence I'd expect no current. Is this physics correct?&lt;/p&gt;&#xA;" OwnerUserId="119" LastActivityDate="2011-05-26T21:50:49.670" Title="A circuit with no voltage difference, but current flowing" Tags="&lt;electricity&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="3" />
  <row Id="7963" PostTypeId="1" AcceptedAnswerId="8061" CreationDate="2011-04-03T02:00:49.173" Score="2" ViewCount="120" Body="&lt;p&gt;Searching for this on google proved to be quite tedious, but I reckon that someone working with crystals a lot might know this off the top of his head:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a good source that lists the Madelung constants for a variety of geometries? I'd be particularly interested in that for a NaCl-type 110 surface.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-03T02:25:41.013" LastActivityDate="2011-04-05T03:45:59.130" Title="Madelung constant list (for surfaces as well)" Tags="&lt;solid-state-physics&gt;&lt;reference-request&gt;&lt;crystals&gt;" AnswerCount="2" />
  <row Id="7964" PostTypeId="2" ParentId="7962" CreationDate="2011-04-03T02:08:55.993" Score="3" ViewCount="" Body="&lt;p&gt;If you are in the circuit at that point then there is a 240V potential different across you!  One side of you is at 0v and there is no potential difference between you and the 240V line (except for the wire resistance) and so that side of you is at 140V&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean you are standing on the ground and touch the circuit at that point then, no the line is at approximately 0 and there is no potential difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously in the real world don't do this anyway - just in case!&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-04-03T02:08:55.993" />
  <row Id="7965" PostTypeId="2" ParentId="7962" CreationDate="2011-04-03T02:24:26.053" Score="18" ViewCount="" Body="&lt;p&gt;The issue is that by touching a wire, you're augmenting the circuit with yourself as a resistor. (At first I wrote &quot;inserting yourself&quot; but as mmc's comment pointed out, that is a misleading phrase to use.) And whenever you change the layout of an electrical circuit, all the potentials and currents are subject to change. So the wire that is at ground potential before you touch it won't necessarily still be at the same potential after you touch it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this specific case, I'd guess that the $30\ \mathrm{A}$ value is the current that flows with only the battery and the load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/OuOvw.png&quot; alt=&quot;circuit without person&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can calculate that the resistance of the load has to be $7.3\ \Omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could actually insert yourself into the circuit as shown in the diagram, you would be adding to the resistance of the circuit, which reduces the current. The resistance of the human body varies greatly depending on several factors, but a &quot;typical&quot; value might be on the order of $10\ \mathrm{k\Omega}$ which reduces the current to a small fraction of an Ampere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/0hmUG.png&quot; alt=&quot;circuit with person in series&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even with that small current, though, you're going to experience a large voltage drop because your resistance is so large compared to the rest of the circuit. In the diagram above, there would be a voltage drop of $219.8\ \mathrm{V}$ across your body. By inserting yourself in the circuit, you've prevented the wire coming out of the load from being grounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if you're trying to determine whether this is a dangerous situation to be in, the value you should be looking at is the &lt;em&gt;current&lt;/em&gt; of $0.022\ \mathrm{A}$, not the $219.8\ \mathrm{V}$ voltage drop. That's what it means to say that it's the current that kills you, not the voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you instead grab on to the wire while standing on the ground (which seems like a more realistic situation), you're not actually inserting yourself in the circuit. Instead you wind up with a setup more like this,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Rbya6.png&quot; alt=&quot;circuit with person in parallel&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the wire coming out of the load resistor is still at potential zero because it's connected to ground via a zero-resistance path. (Keep in mind that this is an ideal model; real wires do have some small nonzero resistance so in reality the wire wouldn't quite be at zero volts.) So the voltage difference across your body is going to be basically zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides, any current that flows between the circuit and the ground can do so by one of two paths, either through you or through the open wire. Since your bodily resistance is much higher than that of the wire, essentially all the current will go through the wire, not through your body.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-03T04:22:56.737" LastActivityDate="2011-04-03T04:22:56.737" CommentCount="13" />
  <row Id="7966" PostTypeId="2" ParentId="7962" CreationDate="2011-04-03T02:27:59.373" Score="0" ViewCount="" Body="&lt;p&gt;There will  be voltage across you since you are part of the circuit. How much? Well, apply Kirchoff's law,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$220 = IR + v$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $I$ is the current in the circuit and also through your body which is $30 amp$ in this case, $R$ is the resistance of the load. $v$ is the voltage across you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v = 220 - IR$&#xA;or $v = 220 - 30R$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the voltage at a point just right of (you are here). You see it is not at 0 volt.&#xA;The voltage at a point just left of (you are here) is 0 however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence 30 amps current will flow through your body which &lt;em&gt;is&lt;/em&gt; dangerous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't try this experiment.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-04-03T04:19:16.940" LastActivityDate="2011-04-03T04:19:16.940" CommentCount="15" />
  <row Id="7967" PostTypeId="2" ParentId="7959" CreationDate="2011-04-03T04:37:27.933" Score="4" ViewCount="" Body="&lt;p&gt;From an experimental point of view, we know one mass less particle, the photon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We cannot describe the photon relativistically by E=mc^2. Its energy is E=h*nu, When it interacts and loses energy, it is the frequency that changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus I would expect, if a massless charged particle could exist on shell, a corresponding energy definition would give it a change in its wave &quot;frequency&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is simpler to say that it would loose energy and go from E1 to E2 where E1-E2 would be the brehmstrahlung energy. That is the way interactions of neutrinos were treated when they were thought to be massless.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-03T04:37:27.933" CommentCount="3" />
  <row Id="7968" PostTypeId="2" ParentId="7922" CreationDate="2011-04-03T04:58:47.737" Score="1" ViewCount="" Body="&lt;p&gt;An outcome is anything that can happen. You might not want to fail to detect a particle, but if you do, it's an outcome. It still causes quantum decoherence (and waveform collapse if you believe the Copenhagen interpretation), in that there will not be interference between waves that are there if the particle is detected and waves that are there if it isn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it's entirely possible to predict the probability of not detecting the particle when it's there, and accidentally detecting the particle when it isn't. In case you're wondering, the decoherence is essentially from the detector. There will be interference between part of a waveform that missed the detector and part that just didn't manage to set it off.&lt;/p&gt;&#xA;" OwnerUserId="2898" LastEditorUserId="2898" LastEditorDisplayName="" LastEditDate="2011-04-03T06:44:42.997" LastActivityDate="2011-04-03T06:44:42.997" CommentCount="1" />
  <row Id="7969" PostTypeId="1" CreationDate="2011-04-03T05:06:37.420" Score="0" ViewCount="172" Body="&lt;p&gt;This is a follow-up question to &quot;&lt;a href=&quot;http://physics.stackexchange.com/questions/7922/in-quantum-mechanics-why-do-the-probabilities-of-the-possible-outcomes-of-a-meas&quot;&gt;In QM, why do the probabilities ... add up to 1?&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No actual measurement is perfect. While theorists may ignore this, experimenters know well enough that in many runs of a given experiment no outcome is obtained. (The efficiency of many real-world detectors is rather low.) This means that in order to make the probabilities of the possible outcomes of a measurement add up to 1, one discards (does not consider) all those experiments in which no outcome is obtained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum mechanics thus allows us to predict the probabilities of measurement outcomes &lt;em&gt;on condition that&lt;/em&gt; there is an outcome. But is there anything in quantum mechanics — after all, the theoretical framework of contemporary physics — that allows us to predict that a measurement, which is about to be made, will have an outcome? Does quantum mechanics allow us to formulate causally sufficient conditions for the occurrence of an outcome (no matter which) or the success of a measurement?&lt;/p&gt;&#xA;" OwnerUserId="2784" LastEditorUserId="2784" LastEditorDisplayName="" LastEditDate="2011-04-03T05:23:54.987" LastActivityDate="2011-04-06T23:04:40.647" Title="Does quantum mechanics allow us to formulate causally sufficient conditions for the occurrence of an outcome?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="5" CommentCount="6" />
  <row Id="7971" PostTypeId="2" ParentId="6096" CreationDate="2011-04-03T06:07:34.090" Score="1" ViewCount="" Body="&lt;p&gt;There's no experimental proof one way or the other. The only way to tell is Occam's razor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's two interpretations of Occam's razor. One says that the number of laws should be small. The other says the number of entities should be. I can't prove one of them is correct philosophically, but given that the laws could fit readably on a single note card, and there are 10^80 particles (which are distinct entities), I'd go with the former.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The many-worlds interpretation has strictly fewer laws. That is to say, the laws that make it up exist entirely within the Copenhagen interpretation, where they govern entangled particles. As such, no matter what language you use, the many-worlds interpretation will come out smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: My mistake. I can prove one is correct philosophically. The proof of Occam's razor is basically that all the prior probabilities have to add to one, and if it didn't decrease as complexity increased, they'd add to infinity. This gives a lower limit to how fast probability must decrease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It applies to number of laws and number of entities, but it just means that exactly 10^80 particles is 80 orders of magnitude less likely than 1. 10^80 to 10^81 is still about as likely as 1 to 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you deal with laws, you're not talking about how many bits the theory is. You're talking about the theory itself. If it takes 267 bits to express, and it's an exact set of 267 bits, that's analogous to there being exactly 10^80 particles.&lt;/p&gt;&#xA;" OwnerUserId="2898" LastEditorUserId="2898" LastEditorDisplayName="" LastEditDate="2011-04-03T17:13:55.620" LastActivityDate="2011-04-03T17:13:55.620" CommunityOwnedDate="2011-04-03T06:07:34.090" />
  <row Id="7972" PostTypeId="2" ParentId="7231" CreationDate="2011-04-03T06:18:56.670" Score="1" ViewCount="" Body="&lt;p&gt;No. Here's a simple example where it shrinks:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a particle that has a 50% chance of being on the left going right, and a 50% chance of being on the right going left. This has a macroscopic error in both position and momentum. If you wait until it passes half way, it has a 100% chance of being in the middle. This has a microscopic error in position. There will also only be a microscopic change in momentum. (I'm not entirely sure of this as the possibilities hit each other, but if you just look right before that, or make them miss a little, it still works.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, the error in position decreased significantly, but the error in momentum stayed about the same.&lt;/p&gt;&#xA;" OwnerUserId="2898" LastActivityDate="2011-04-03T06:18:56.670" />
  <row Id="7973" PostTypeId="2" ParentId="7963" CreationDate="2011-04-03T06:35:34.433" Score="0" ViewCount="" Body="&lt;p&gt;It's between 1.748 and 1.75, see&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=GnCWJF6TQUwC&amp;amp;pg=PA228&amp;amp;dq=Madelung+constant+NaCl+110+1.748&amp;amp;hl=cs&amp;amp;ei=1xSYTYS4C8Xl4wbD9tDUAQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=4&amp;amp;ved=0CDkQ6AEwAw#v=onepage&amp;amp;q=Madelung%20constant%20NaCl%20110%201.748&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=GnCWJF6TQUwC&amp;amp;pg=PA228&amp;amp;dq=Madelung+constant+NaCl+110+1.748&amp;amp;hl=cs&amp;amp;ei=1xSYTYS4C8Xl4wbD9tDUAQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=4&amp;amp;ved=0CDkQ6AEwAw#v=onepage&amp;amp;q=Madelung%20constant%20NaCl%20110%201.748&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See some other promising sources&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://books.google.com/books?q=Madelung-constant+nacl+110+surface&amp;amp;btnG=Search+Books&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?q=Madelung-constant+nacl+110+surface&amp;amp;btnG=Search+Books&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The figure 1.748 is confirmed by Wikipedia - in the thick table in the middle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Madelung_constant&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Madelung_constant&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-03T06:35:34.433" CommentCount="1" />
  <row Id="7974" PostTypeId="2" ParentId="7969" CreationDate="2011-04-03T06:59:54.423" Score="2" ViewCount="" Body="&lt;p&gt;Let us take a simple example of a theory predicting something, Ohms Law from Electricity and Magnetism&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I=V/R&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am given the value of R and asked to measure I. I take a voltmeter, measure V and have a value for I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same circuit, same conditions, I take and ampmeter and measure I. The value differs by Delta(I), inevevitably.   This is because my measuring  instruments have by construction a measurement error, that should always be taken into account when reporting a measurement. Ohm's law is validated within errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above comes from a simple theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quantum mechanical prediction comes from more complicated solutions of quantum mechanical equations describing a system. The measuring apparatus is much more complicated than in my simple example. The measurements are always given with the measurement error estimated. This &lt;a href=&quot;http://pdg.lbl.gov/2010/reviews/rpp2010-rev-cross-section-plots.pdf&quot; rel=&quot;nofollow&quot;&gt;pdf from the particle data group&lt;/a&gt; shows the meticulous estimate and recordings of measurement errors. These have nothing to do with the quantum mechanical prediction, except to allow limits wherein a specific quantum mechanical model can play  and be consistent with experiment.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-03T14:36:57.010" LastActivityDate="2011-04-03T14:36:57.010" CommentCount="2" />
  <row Id="7975" PostTypeId="2" ParentId="7961" CreationDate="2011-04-03T07:53:13.803" Score="2" ViewCount="" Body="&lt;p&gt;Yes, a body's mass is sufficient to cause melting. For example, suppose the moon, instead of orbiting the earth, fell onto the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moon weighs about 7.35E22 kg, and is a distance of about 3.8E8 m from the earth which weighs about 6E24 kg and has a radius of around 6.4E6 m. So the difference in potential energy $U = -Gm_1m_2/R$ for the earth-moon system at 3.8E8m and 6.4E6m is.&#xA;$$ -G(7.35\times 10^{22})(6\times 10^{24})(1/3.8\times 10^8-1/6.4\times 10^6) = 4.5\times 10^{30}\; \textrm{Joules}$$&#xA;where $G= 6.7\times 10^{-11}$ is the gravitational constant. The combined mass is still about 6E24 kg so the collision creates heat of&#xA;$$ 4.5\times 10^{30}/6.0\times 10^{24} = 750,000\;\textrm{Joules/kg}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To heat a kilogram of iron up by 1 degree kelvin requires about 460 Joules. Granite needs about 790 while basalt is 840. Other materials making up most of the earth have &lt;a href=&quot;http://www.engineeringtoolbox.com/specific-heat-solids-d_154.html&quot; rel=&quot;nofollow&quot;&gt;similar specific heats&lt;/a&gt;. So the collision will heat up the material by something around 1000 degrees kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that was just for the moon whose mass is only a tiny fraction of the earth's. If a larger mass collided with the earth the temperature would be proportionally larger. If the body were about 4x the mass of the moon, the temperature increase would be around 4000 K and that would certainly thoroughly melt the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, it is possible for gravity alone to melt planets.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-04-03T07:53:13.803" CommentCount="4" />
  <row Id="7976" PostTypeId="1" CreationDate="2011-04-03T08:55:28.470" Score="0" ViewCount="106" Body="&lt;p&gt;What are the chances of becoming a tenured professor in physics? Is that a goal worth striving for? If a person is denied tenure, would it all have been a waste of time?&lt;/p&gt;&#xA;" OwnerUserId="2915" LastActivityDate="2011-04-03T08:55:28.470" ClosedDate="2011-04-03T09:09:48.553" Title="What are the chances of becoming a tenured professor in physics?" Tags="&lt;physics-careers&gt;" CommentCount="2" />
  <row Id="7977" PostTypeId="2" ParentId="7959" CreationDate="2011-04-03T09:07:52.113" Score="4" ViewCount="" Body="&lt;p&gt;It is not so hypothetical situation if you consider very relativistic electrons, for example. The velocity is always $c$ but the energy-momentum changes. Quantum mechanically it means a change in the De Broglie wave frequency. So a less energetic relativistic electron is like a less energetic photon - it has lower frequency. Classically it correspond to changes of energy-momenta that are determined not only with particle velocity.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-04-03T09:07:52.113" />
  <row Id="7978" PostTypeId="1" AcceptedAnswerId="7998" CreationDate="2011-04-03T11:59:55.113" Score="2" ViewCount="57" Body="&lt;p&gt;Currently the charge-to-mass ratio of the electron is known to &lt;a href=&quot;http://en.wikipedia.org/wiki/Mass-to-charge_ratio#The_electron&quot; rel=&quot;nofollow&quot;&gt;10 orders of magnitude&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, i'm curious if:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Are there any experiments trying to bound the anisotropy of this ratio for different directions of space? Over what time does the data needs to be averaged to get to our current ten-figure precision? Less than a day? Less than a second? Less than a month?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Does an spatial anisotropy (or an spatial inhomogeneity) of this ratio break Poincare symmetry, or can that be restored with replacing the mass scalar with a tensor object? Also, what theoretical ideas explicitly disallow a non-scalar fundamental mass? What about a non-scalar electromagnetic charge?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Does mass spectroscopic experimental data bound any anisotropic variations? Do we have any measurements of this ratio on small cavities where van der waals forces are dominant?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="955" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-03T15:07:09.167" LastActivityDate="2011-04-03T18:24:51.650" Title="Experimental limits on anisotropies in the $e/m_{e}$ ratio" Tags="&lt;quantum-field-theory&gt;&lt;mass-spectrometry&gt;&lt;poincare-symmetry&gt;&lt;symmetry-breaking&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="7981" PostTypeId="1" AcceptedAnswerId="7990" CreationDate="2011-04-03T13:04:08.720" Score="4" ViewCount="247" Body="&lt;p&gt;If I search online for how radio telescopes work, the found articles talk about how RF is on the spectrum, etc, how the parabolic collector is the aperture which contributes to the sensitivity and reflects the signals into the detector at the focal point.  Then they skip to the signal processing circuitry the leads to a rendered image.  They also mention that the furthest nodes in a telescope array determine the virtual aperture of the array.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few parts I don't understand:  How does the detector work?  From my reading, it is merely an RF antenna.  I can't think how you'd get but one pixel per &quot;moment&quot; of detection.  That the telescope is positioned and begins listening, and as the earth turns one scan-line of image is captured.  For high resolution 2D image this would take thousands of passes.  This seems impractical as the earth is orbiting and time marches on, otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps multiple telescopes each take a scan line and interlace the results ... but especially at such high resolutions, I can't see how you'd engineer the huge machines with such precision.  Please explain this to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I have the 1D scan-line-antenna idea wrong?  Is it more like a 2D CCD sensor in a digital camera with the antenna just behind the focus of the dish?  If so, I assume the sensor would have to be a 2D array of nanometer scale antennas?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may be clear in the explanation, but in layman's terns, how are do telescope arrays work together to create the virtual aperture?  What is lost, if any between the virtual and a theoretical &quot;real&quot; aperture of equal size?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I consider this a physics question because understanding the answer to this real-world application will help correct whatever theoretical misconceptions I have about RF, antennas, and radio telescopes)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2910" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-03T14:42:05.993" LastActivityDate="2011-04-05T16:39:50.373" Title="How do radio telescopes work?" Tags="&lt;radio&gt;&lt;antennas&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="7982" PostTypeId="2" ParentId="7969" CreationDate="2011-04-03T13:59:39.487" Score="3" ViewCount="" Body="&lt;p&gt;I suggest that you need to modify your &quot;Quantum mechanics thus allows us to predict the probabilities of measurement outcomes on condition that there is an outcome.&quot; I suggest instead that quantum mechanics allows us to use probability measures to model observed statistics of actual raw data. As an aside, whether this is prediction of future statistics or fitting of past statistics seems to me not relevant to the question you're asking here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experimental raw data &lt;em&gt;can&lt;/em&gt; take the form of a list of times and whether a thermodynamic transition occurred in an experimental device within some given time tolerance of each time in the list. That could be a list of some analogue signal value (averaged over some time period and converted to n bits), recorded, say, each picosecond. The same list is more-or-less equally well represented, however, by a list of times at which thermodynamic transitions occurred. I say more-or-less because the list of times at which events occurred essentially filters features of the experimental apparatus such as noise and &lt;a href=&quot;http://en.wikipedia.org/wiki/Dead_time&quot; rel=&quot;nofollow&quot;&gt;dead-times&lt;/a&gt; out of the raw list of signal values, which may well be done in hardware. The rawest possible data is often not recorded because it is too voluminous. Some choices have to be made about what is and is not interesting to record; every now and then someone records something that wasn't expected to be interesting but turns out to be unexpected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The list of signal conditioning processes that are applied by experimentalists is quite long. To me, qua theorist, the methods used by experimentalists to describe and/or model the signal conditioning processes they use in a particular experiment often seem quite ad-hoc, or at least not clearly unified, which is, I think, a way of more generally formulating the concern you seem to me to be voicing in your Question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT (after noticing your postscript on your previous Question): Not only is no measurement perfect, nor is any preparation. The process of &lt;em&gt;characterizing&lt;/em&gt; both what the preparation apparatus prepares and what the measurement apparatus measures is &lt;em&gt;nonlinear&lt;/em&gt;, insofar as we know neither the density operators $\hat\rho_i$ nor the measurement operators $\hat O_j$ in the linear equations that determine the expected values $\mathsf{Tr}\!\left[\hat\rho_i\hat O_j\right]=V_{ij}$, where $V_{ij}$ is the mean of a raw dataset (a statistic) that we obtain when we use the $i$-th state preparation apparatus with the $j$-th measurement apparatus. Both all the density operators $\hat\rho_i$ and all the measurement operators $\hat O_j$ have to be determined from the finite set of experimentally determined statistics $V_{ij}$. Of course things get considerably more nonlinear when we start to model second and higher moments of raw datasets, $W_{ij}$, say, so we also have equations such as $\mathsf{Tr}\!\left[\hat\rho_i\hat O_j^2\right]=W_{ij}$, and still more nonlinear when we introduce unitary or more general state transformations. Note that the experimental statistics also have to determine the dimensionality of the Hilbert space we use. The nonlinearity of the process of characterization is superficially avoided because we know from experience what dimensionality of Hilbert space usually constructs effective models for a given state preparation and we may even have datasheets from the manufacturer that tell us what a given preparation or measurement device prepares or measures to a good approximation.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-04-03T15:53:13.460" LastActivityDate="2011-04-03T15:53:13.460" />
  <row Id="7983" PostTypeId="6" CreationDate="2011-04-03T14:15:52.170" Score="0" ViewCount="" Body="&lt;p&gt;I have been an active user in the first few months of Physics and I've helped build up the community until the go live phase. I would very much like to help our community grow and improve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why do I want your vote?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I love Physics and I love StackExchange, therefore I see this site as one with a great &lt;em&gt;potential&lt;/em&gt;!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I am a pro-tempore moderator on Skeptics.SE, which is a very young, but promising site: I do know what moderating a &lt;del&gt;tough&lt;/del&gt; active site looks like.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I am also a long term user of StackOverflow, as I participated since the beta. I know why moderation is in place, and I've lived through a great experience in building a community from zero up to very high numbers of participants.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I believe in a &lt;em&gt;broad spectrum&lt;/em&gt; site: the focus should be towards having the highest quality for questions and answers. I believe that &lt;strong&gt;bad questions&lt;/strong&gt; should be fixed or closed, and &lt;strong&gt;good questions&lt;/strong&gt; should be rewarded with &lt;strong&gt;great answers&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I am very active in various chat rooms, and I would like to help making our &lt;a href=&quot;http://chat.stackexchange.com&quot;&gt;chat&lt;/a&gt; a great one!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;About me&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a moderator I think I would be a good &lt;a href=&quot;http://blog.stackoverflow.com/2009/05/a-theory-of-moderation/&quot; rel=&quot;nofollow&quot;&gt;human exception handler&lt;/a&gt;, cleaning up messy situations and helping the community be &lt;em&gt;harmonic&lt;/em&gt;. I believe my current experience as a moderator would really help this site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a physicist I am university trained to ~grad level, and as such I believe I could be most effective in dealing with lower level questions -- which is something that more skilled physicists don't have in their heart to deal with, most of the times.&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-03T18:16:32.837" LastActivityDate="2011-04-03T18:16:32.837" CommunityOwnedDate="2011-04-03T14:15:52.170" CommentCount="0" />
  <row Id="7984" PostTypeId="2" ParentId="7969" CreationDate="2011-04-03T14:24:22.463" Score="2" ViewCount="" Body="&lt;p&gt;Looking at both this question and your previous one, I think there's a bit of talking at cross purposes. Theoretically, &lt;em&gt;something&lt;/em&gt; has to happen with regard to the outcome of a measurement. Therefore, the total probability of all outcomes must sum to one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving to a real-world scenario, instead of the idealized equations most likely worked up by theorists, one of the possible outcomes of the measurement attempt is to find a value of zero. As you state, because of inefficiencies in a detector, one of the quantum mechanical probabilities in a properly formulated equation that takes into account both the detector and the thing to be measured should be that there is no interaction, i.e. the detector detects nothing. Although the measurement value is zero, the probability for that (non)measurement is non-zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the disconnect you're experiencing is from thinking about measuring the particle, for example, rather than the interaction between the particle and the detector. The theoretical equations used to teach QM are simplified by assuming a 100% efficient detector so that the interaction does not need to be computed. In the real world, however, this is necessary. A good example might be neutrino detection, where almost all the probability is for no detection.&lt;/p&gt;&#xA;" OwnerUserId="1196" LastEditorUserId="1196" LastEditorDisplayName="" LastEditDate="2011-04-06T23:04:40.647" LastActivityDate="2011-04-06T23:04:40.647" CommentCount="2" />
  <row Id="7985" PostTypeId="1" CreationDate="2011-04-03T14:26:17.633" Score="-2" ViewCount="195" Body="&lt;p&gt;Does the rotational speed of a planet consistently become faster and faster given that there are no conflicting events?&lt;/p&gt;&#xA;" OwnerUserId="2874" LastActivityDate="2011-04-05T06:51:08.727" ClosedDate="2011-04-05T16:02:50.900" Title="Does the rotational speed of a planet consistently become faster and faster given that there are no conflicting events?" Tags="&lt;astronomy&gt;&lt;rotational-dynamics&gt;&lt;momentum&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="1" />
  <row Id="7986" PostTypeId="2" ParentId="7985" CreationDate="2011-04-03T14:38:15.340" Score="4" ViewCount="" Body="&lt;p&gt;No, it will continue to rotate at the same rate forever. This is a consequence of angular momentum conservation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Angular_momentum&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Angular_momentum&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1708" LastEditorUserId="1708" LastEditorDisplayName="" LastEditDate="2011-04-03T14:57:25.393" LastActivityDate="2011-04-03T14:57:25.393" />
  <row Id="7987" PostTypeId="2" ParentId="7969" CreationDate="2011-04-03T14:39:17.713" Score="1" ViewCount="" Body="&lt;p&gt;I am not sure whether I fully understand this question as it contains some ambiguities I shall discuss below. However in the most simplistic reading of the Title the answer is: YES.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I am taking the phrase &quot;the occurrence of an outcome&quot; to mean a non-zero eigenvalue for a given measurement operator. To make this work one needs to mention an axiom of QM, namely that of the existence of a complete set of observables: A, B, C,... As a result the basis of the Hilbert space can be described in terms of the eigenfunctions $\Psi_i$ (usually written in Dirac notation) of that complete set, giving a decomposition $\Psi=\Sigma{_i}a_i\Psi_i$. Now a given operator $P_j$ may have zero and non-zero eigenvalues; if the only eigenvalues are {0,1} then it is a Projection operator. The Spectral decomposition theorem may also be used for Measurement operators to define the outcomes, their probabilities and the resultant eigenspaces too. In particular a zero outcome (for a given measurement $P_j$) can have a non-zero probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly it is necessary in an experimental context to ensure that the complete set has been defined and that the algebra is worked through correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question however introduces more topics than just this simple interpretation, as there is discussion of non-ideal measurement instruments, and any implication from that fact. There will be implications in terms of measurement in practice, and measurement procedures in general as well. However to discuss this further one would need to introduce more clarification into the Question as to exactly what one is concerned about experimentally. To focus on one point. Even although one can calculate that eigenvalues {a1,a2,a3,a4,a5,...} might be observed with different probabilities with an ideal measuring instrument; it might well be that the &lt;em&gt;resolution&lt;/em&gt; of the measuring apparatus cannot distinguish between {a1,a2},{a3,a4}, etc. Clearly calculation of quantum probabilities need to take into account this non-ideal resolution. So perhaps your question has to do with whether a complete theory of how to do this exists?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should also mention, following on from the last point and remarks in the question, that one may need to consider extending the framework of quantum mechanics as described in these questions to deal with classical uncertainty arising from uncertainty in the experimental apparatus or setup. This additional set of probabilities combined with the quantum probabilities results in something called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Density_matrix&quot; rel=&quot;nofollow&quot;&gt;Density Matrix&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can ask a range of questions about the formalism of Density Matrices (e.g. what sums to 1 and why) as well as ask about their interpretation and use. However Density Matrices have not been mentioned in these questions so I will leave that topic for further questions.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-04-03T14:39:17.713" CommentCount="3" />
  <row Id="7989" PostTypeId="2" ParentId="7981" CreationDate="2011-04-03T15:32:40.637" Score="2" ViewCount="" Body="&lt;p&gt;A single dish aperture can operate a single pixel detector but this pixel projected onto the sky can be quite large compared to optical and infrared telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resolution is proportional the wavelength and therefore this can be 1000's of times the optical resolution.  Therefore a radio telescope can take its image in a reasonable time simply because they aim at larger targets and take 'worse' images than you might imagine.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-04-03T15:32:40.637" />
  <row Id="7990" PostTypeId="2" ParentId="7981" CreationDate="2011-04-03T15:42:38.893" Score="4" ViewCount="" Body="&lt;p&gt;In the microwave band here are multi-element detectors, but at longer wavelengths the telescope is a single pixel.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes it does take a while to build up an image, but radio pictures aren't usually very large - not the millions  of pixels of an optical/IR image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One big advantage of radio telescopes is that you can combine telescopes 1000s of km apart to create an image with the resolution of a single dish that large ( you can just about do this in the optical with 100m separation now)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know about the interference pattern you get with two slit? If you picture the two telescopes as the two slits and you interfere (electrically) the signals to form the fringe pattern. You can then calculate the shape of the light source - a single point will produce the classic fringe pattern, 2 close points will produce a slightly different pattern etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: Building an optical 'radio' telescope - the &lt;a href=&quot;http://www.mrao.cam.ac.uk/telescopes/coast/handout.html&quot; rel=&quot;nofollow&quot;&gt;original English solution&lt;/a&gt; and the rather more &lt;a href=&quot;http://en.wikipedia.org/wiki/VLT&quot; rel=&quot;nofollow&quot;&gt;impressive Teutonic result&lt;/a&gt; (if you have a lot more $$$)&lt;/p&gt;&#xA;" OwnerUserId="2525" LastEditorUserId="2525" LastEditorDisplayName="" LastEditDate="2011-04-04T15:36:46.643" LastActivityDate="2011-04-04T15:36:46.643" CommentCount="1" />
  <row Id="7991" PostTypeId="2" ParentId="222" CreationDate="2011-04-03T15:58:06.083" Score="3" ViewCount="" Body="&lt;p&gt;I'd like to add that GNU Octave is a very good free alternative to Matlab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contrary to Scilab which does not aim at being compatible with Matlab, you can practically run your Matlab scripts with Octave with very few modifications (at least with their latest version).&lt;/p&gt;&#xA;" OwnerUserId="2917" LastActivityDate="2011-04-03T15:58:06.083" CommunityOwnedDate="2011-04-03T15:58:06.083" />
  <row Id="7992" PostTypeId="2" ParentId="222" CreationDate="2011-04-03T16:04:14.483" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.ginac.de/&quot; rel=&quot;nofollow&quot;&gt;GiNaC&lt;/a&gt; is a c++ symbolic manipulation framework oriented to high-energy physics computations. It has a couple of interactive frontends, although its main usage is as part of the Root framework at CERN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A derivative of GiNaC is &lt;a href=&quot;http://pynac.sagemath.org/&quot; rel=&quot;nofollow&quot;&gt;Pynac&lt;/a&gt;, which forms the backend for symbolic expressions in &lt;a href=&quot;http://www.sagemath.org/&quot; rel=&quot;nofollow&quot;&gt;Sage&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="429" LastEditorDisplayName="" LastEditDate="2011-06-25T06:57:31.883" LastActivityDate="2011-06-25T06:57:31.883" CommunityOwnedDate="2011-04-03T16:04:14.483" CommentCount="3" />
  <row Id="7993" PostTypeId="1" AcceptedAnswerId="7994" CreationDate="2011-04-03T17:11:34.040" Score="2" ViewCount="178" Body="&lt;p&gt;Suppose we have an atom with several energy levels (e.g. an hydrogen), and it is hit by photons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that in order to have the atom change energy levels, the photon must have an energy level precisely equal to the energy difference of the two levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens if the photon has more energy than the ionization energy for the ground state (i.e., the sum of the series of differences between consecutive levels)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not clear to me. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2918" LastActivityDate="2011-04-03T21:24:11.583" Title="Photon hitting an atom with higher energy than needed to ionize" Tags="&lt;quantum-mechanics&gt;&lt;photons&gt;&lt;atoms&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="7994" PostTypeId="2" ParentId="7993" CreationDate="2011-04-03T17:30:51.937" Score="5" ViewCount="" Body="&lt;p&gt;If the photon has more energy than the ionization energy, it can (and often does) ionize the atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One nice way to think about this is to put the various energy levels of the atom in an energy level diagram like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ew1tW.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(taken from &lt;a href=&quot;http://web.njit.edu/~gary/728/Lecture2.html&quot; rel=&quot;nofollow&quot;&gt;this web site&lt;/a&gt;). All of the bound-state energy levels lie below the line marked $E_\infty=0$ eV. Above this line is the region called &quot;free electrons&quot; in the diagram. This corresponds to an ionized atom. In this range, the energy levels are not quantized -- that is, any energy is possible. So any photon with enough energy to put the total energy somewhere in that shaded region is capable of ionizing the atom.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-03T17:30:51.937" CommentCount="2" />
  <row Id="7995" PostTypeId="2" ParentId="7959" CreationDate="2011-04-03T17:43:57.547" Score="1" ViewCount="" Body="&lt;p&gt;The Dirac current obeys a continuity equation&#xA;$$&#xD;&#xA;\partial_\mu({\bar\psi}\gamma^\mu\psi)~=~0,&#xD;&#xA;$$&#xA;which for electromgnetism we generalize $\partial_\mu~\rightarrow~\partial_\mu~+~ieA_\mu$ in a covariant manner.  The $J^0~=~{\bar\psi}\gamma^0\psi$ $\psi^\dagger\psi~=~\rho$, which is the density. The spatial part is&#xA;$$&#xD;&#xA;{\vec j}~=~\psi^\dagger{\vec\alpha}\psi&#xD;&#xA;$$&#xA;The application of the momentum operator ${\hat p}~=~-i\hbar\nabla~-~ie{\vec A}$ gives a divergence of the 3-current &#xA;$$&#xD;&#xA;{\hat p}\cdot{\vec j}~=~i\hbar\big((\nabla\psi^\dagger)\cdot{\vec\alpha}\psi~+~\psi^\dagger{\vec\alpha}\cdot\nabla\psi\big)~-~ie{\vec A}\cdot\psi^\dagger{\vec\alpha}\psi,&#xD;&#xA;$$&#xA;where the spatial component $\alpha^i~=~\gamma^0\gamma^i$.  This continuity equation tells us this equals $\partial\rho/\partial t$ and the quantum current divergence through a volume is equal to the rate the quantum density changes in that volume.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To consider the Bremsstrahlung question we need to address the force ${\vec F}~=~d{\vec p}/dt$ with respect to the current flow.  The time derivative is evaluated with the Dirac Hamiltonian&#xA;$$&#xD;&#xA;H~=~{\vec\alpha}(\cdot{\hat p}~-~ie{\vec A}),&#xD;&#xA;$$&#xA;where the absence of mass should be noted.  The time derivative is evaluated according to the commutator $[{\hat p},~H]$ and the calculation result in&#xA;$$&#xD;&#xA;\frac{d}{dt}{\hat p}\cdot{\vec j}~=~\langle F\rangle~=~e\langle{\vec E}~+~{\hat\alpha}\times\nabla\times{\vec A}\rangle.&#xD;&#xA;$$&#xA;The vector ${\vec\alpha}$ plays the role of a velocity of the particle, where the matrices are units in space with unit magnitude or $v~=~c$ in the Lorentz equation $({\vec v}/c)\times{\vec B}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have now that a massless particle which carries a charge will obey electrodynamics.  Then what about the Bremsstrahlung?  We can now use this to compute the energy of work of this force $W~=~\int\langle F\rangle\cdot d{\vec r}$  The power is the time derivative of this &#xA;$$&#xD;&#xA;P~=~dW/dt~=~\int\langle F\rangle\cdot d{\vec v}&#xD;&#xA;$$ &#xA;We then consider the “c velocity” given by $\vec\alpha$ as a rotation given by the $SO(3)$ matrix elements $M$ that ${\vec\alpha}’~=~M{\vec\alpha}M^T$, where the time derivative on the velocity is due to a rotational $dM/dt$ and $dM^T/dt$.  So the velocity is a particle at the speed of light moving in a circular path --- ignoring the electric field.  This will then result in an acceleration $\dot a$ from the time derivatives of these matrices.  At this point I am largely sketching this out but it is clear that a form of the Abraham-Lorentz force and Bremsstrahlung can be derived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rats, I see this has slit me from my previous identity here --- am back to 1.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-04-03T17:50:21.520" LastActivityDate="2011-04-03T17:50:21.520" CommentCount="6" />
  <row Id="7996" PostTypeId="2" ParentId="7985" CreationDate="2011-04-03T18:10:00.787" Score="4" ViewCount="" Body="&lt;p&gt;The question in ill-stated and open to multiple interpretations. I'm going to answer two of them.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Assuming that by planet you mean a more-or-less solid object (like the Earth), then no. Conservation of angular momentum means that unless acted on by a non-central outside force the speed of rotation remains forever the same.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Assuming that the planet is an warm, gaseous object (like Jupiter shortly after the formation of the solar system), than the system is supported in part by &lt;a href=&quot;http://en.wikipedia.org/wiki/Virial_theorem&quot; rel=&quot;nofollow&quot;&gt;Virial heating&lt;/a&gt;, and will continue to contract as it cools. Because Angular momentum is conserved in this system too, as the body contracts it's speed of rotation &lt;em&gt;will&lt;/em&gt; increase. Eventually however, the contraction will stop as processes other than thermal pressure in the gas begin to dominate. Once that happens we're back to the first case: the rate of rotation is fixed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-04-03T18:10:00.787" />
  <row Id="7997" PostTypeId="2" ParentId="7915" CreationDate="2011-04-03T18:10:39.013" Score="1" ViewCount="" Body="&lt;p&gt;Wind speed is how fast the air is moving. The laws of physics don't have a privileged rest frame, so it can't have a bounded speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you're in a supersonic airplane. From your point of reference, you're stopped and the air is moving. Hence, supersonic wind.&lt;/p&gt;&#xA;" OwnerUserId="2898" LastActivityDate="2011-04-03T18:10:39.013" />
  <row Id="7998" PostTypeId="2" ParentId="7978" CreationDate="2011-04-03T18:11:23.287" Score="0" ViewCount="" Body="&lt;p&gt;You ask in the comments to the main question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;how sure are we (quantitatively, and theoretically) that mass and charge are scalar constants&quot; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our concept of mass has developed from classical mechanics, the same with charge and classical electrodynamics. If charge and/or mass behaved differently at different times in different locations we would have noticed experimentally and would have developed different classical mechanics and electromagnetism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass and charge are assumed to be scalar in QM as  continuation of the classical mechanics attributes. The many and accurate measurements of e/m allow very little leeway to a different interpretation, within measurement errors.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-03T18:11:23.287" />
  <row Id="8000" PostTypeId="2" ParentId="7969" CreationDate="2011-04-03T18:39:34.917" Score="2" ViewCount="" Body="&lt;p&gt;Yes. A detector isn't ideal, so a model that assumes it is will be flawed. It does, however, obey the laws of physics. Just model it as part of the system, and you can predict the probabilities for any result the detector is capable of giving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, rather than checking if a photon makes it to where the detector is, actually model the detector and see if it ends up going off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any error about the probabilities is due to lack of knowledge about the detector, not the detector not being good enough.&lt;/p&gt;&#xA;" OwnerUserId="2898" LastActivityDate="2011-04-03T18:39:34.917" />
  <row Id="8001" PostTypeId="1" CreationDate="2011-04-03T18:56:45.217" Score="1" ViewCount="124" Body="&lt;p&gt;Given d(slit separation)= $0.158mm$, $\lambda _{red}= 665nm$, $\lambda _{g/y}= 565nm$, L(distance from screen)= $2.24m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the distance between the third order red and green/yellow fringes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using the equation $X _{3}= \frac{m \lambda L}{d}$ (just to find the distance to the third red fringe). My teacher told me this equation is approximate, but I think this is beyond simple approximation error... I keep getting the answer $2.8284 \times 10^{7}m$. This seems unreasonable; what am I doing wrong? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m = mm\times10^{-13}, m= nm\times10^{-9}$, right? &lt;/p&gt;&#xA;" OwnerUserId="2536" LastEditorUserId="2536" LastEditorDisplayName="" LastEditDate="2011-04-03T19:08:54.030" LastActivityDate="2011-04-03T19:08:54.030" ClosedDate="2011-04-03T21:13:12.723" Title="Double Slit Problem, Waves and Optics" Tags="&lt;optics&gt;&lt;homework&gt;&lt;light&gt;&lt;waves&gt;" CommentCount="5" />
  <row Id="8002" PostTypeId="2" ParentId="7936" CreationDate="2011-04-03T18:57:41.187" Score="2" ViewCount="" Body="&lt;p&gt;These laws are based on a circuit in equilibrium. If you made a circuit where you had +1 volt on the left, 1 ohm in the middle, and +2 volts on the right, and you started out with the resistor not under any voltage, electricity would start out moving towards the center of the resistor until it builds up a voltage of about 1.5 volts. (It would gradually change from 1 volt to two based on which source you're closer to.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to extend the sandwich analogy, imagine, for some reason, that the strength of people is proportional to their sandwiches. They also have no idea where they're going, and they push at random. And the number of sandwiches they drop is proportional to their speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my circuit, at first people will be pushed by the people behind them, until they get to the center. At this point, the guys on the right have more sandwiches, so they shove the people on the left back, until you end up with just a group of people going from the right to the left. They're slowed down exactly enough to have one sandwich at the end, since if they went slower, they'd have more sandwiches than the guys they're pushing against, speeding them up. If they went faster, they'd have fewer, slowing them down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason you always end up with everyone having the same number of sandwiches when they meet at a node is if they don't, the guys with more sandwiches push back on the guys with less, slowing them down, until they end up having the same number of sandwiches. It might be off for a little bit, but it will quickly enter an equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="2898" LastActivityDate="2011-04-03T18:57:41.187" />
  <row Id="8003" PostTypeId="1" CreationDate="2011-04-03T19:18:45.937" Score="2" ViewCount="203" Body="&lt;p&gt;One of the weirdest things about quantum mechanics (QM) is the exponential growth of the dimensions of Hilbert space with increasing number of particles.&#xA;This was already discussed by Born and Schrodinger but here's a recent reference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0711/0711.4770v2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/arxiv/pdf/0711/0711.4770v2.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and another one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.scottaaronson.com/papers/are.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.scottaaronson.com/papers/are.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another difference between quantum mechanics and classical physics is that QM is discrete.&lt;br&gt;&#xA;A good example is the line spectrum of hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any good observational example which demonstrates the exponential growth of Hilbert space dimensions?&lt;br&gt;&#xA;Even for low numbers?&lt;br&gt;&#xA;(Does it perhaps show up in the line spectrum  of a three or four valent atom?)&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-04-04T15:11:12.417" Title="Observing the exponential growth of Hilbert space?" Tags="&lt;quantum-mechanics&gt;&lt;hilbert-space&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="8004" PostTypeId="2" ParentId="7922" CreationDate="2011-04-03T19:30:07.650" Score="0" ViewCount="" Body="&lt;p&gt;What &lt;em&gt;operationally&lt;/em&gt; is &quot;no outcome&quot;?   As mentioned above this has nothing to to with quantum mechanics, since every quantum experiment must have a classical reading (Bohr's dictum).   Perhaps this is the part of which you are unaware.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have thought about it myself since yesterday.   One definition of the absence of outcome is &quot;no stability&quot; or reproducibility.  Something like, you think you see blue.  You blink or wish or something and the outcome changes, you see green.  Blink again and it changes again.   Another possibility might be that the experimental apparatus is destroyed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, merely apparent measurements of unphysical quantities may have no outcome, like trying to measure our absolute x,y,z coordinates in the universe, but even this is not operationally so well defined.   Some purely statistical experiments like telepathy appear to have no outcome, as they change on reinterpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have other operational possibilities in mind for &quot;no outcome&quot;?   Please focus on the meaning of &quot;operational&quot;.   This was the giant advance in 20th century physics.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2011-04-03T19:30:07.650" CommentCount="4" />
  <row Id="8005" PostTypeId="2" ParentId="7985" CreationDate="2011-04-03T19:58:30.943" Score="4" ViewCount="" Body="&lt;p&gt;A planet with no forces acting on it will rotate at the same speed for ever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However a real planet will experience tidal forces which will eventually slow it's rotation until the same side always faces the sun (or possibly until it reaches a resonance with the sun's influence and other nearby large planets) &lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-04-03T19:58:30.943" />
  <row Id="8006" PostTypeId="2" ParentId="7918" CreationDate="2011-04-03T19:59:01.870" Score="2" ViewCount="" Body="&lt;p&gt;In the many-worlds interpretation of quantum mechanics, there is at all times just one state vector for the entire Universe. It is a vector in a certain (infinite-dimensional) vector space, and that vector space is always the same. So there's nothing in the theory whose dimension changes when the number of particles in the Universe changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be a bit more precise, the space in which the state vector of the Universe lives is (something like) a &lt;a href=&quot;http://en.wikipedia.org/wiki/Fock_space&quot; rel=&quot;nofollow&quot;&gt;Fock space&lt;/a&gt;. Vectors in that space include states with all possible different numbers of particles, as well as superpositions containing different numbers of particles. So if a particle-antiparticle pair is created, the state vector simply &quot;wanders&quot; from one part of that space to another; the space itself needn't get any bigger.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-03T19:59:01.870" CommentCount="10" />
  <row Id="8007" PostTypeId="1" CreationDate="2011-04-03T20:13:36.027" Score="0" ViewCount="71" Body="&lt;p&gt;I'm taking an (obligatory...) course in &quot;Scientific history&quot; at university level. It mainly consists of a written project: ~20 pages and oral presentation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have planned to study the status of fundamental physics research throughout history, or more exactly from ~1900, by picking a number of events/areas (Einstein's relativity, the formulation of QM, nuclear research prior to WW2, QFT, string theory, the search for new fundamental particles) and see how society influenced them, and how they influenced society. The question I pose concern e.g. how governments/people in general/scientists have viewed the funding of directly non-applicable research, what geographical differences can be seen and how these findings have evolved during the 20th century and onwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the plan, at least. With this question I'm looking for input in the subject, and I'm very interested in literature suggestions (papers/blogs/traditional books). I am currently writing my MSc thesis in fundamental physics, and the course is given at BSc level, to give an idea of which level it concerns.&lt;/p&gt;&#xA;" OwnerUserId="949" LastActivityDate="2011-04-03T20:13:36.027" ClosedDate="2011-04-03T21:10:18.077" Title="What has been the status of basic physics research throughout history?" Tags="&lt;history&gt;" CommentCount="3" />
  <row Id="8008" PostTypeId="2" ParentId="7993" CreationDate="2011-04-03T20:14:33.467" Score="1" ViewCount="" Body="&lt;p&gt;The photoelectric effect (http://en.wikipedia.org/wiki/Photoelectric_effect) is a phenomenon whereby the surface atoms are ionised through liberation of some of their electrons.  These electrons can then leave with a kinetic energy equal to the incident photon energy minus the 'work function' which was the energy required to ionise the atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The work function is the energy required to remove the electron from the surface of a material.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastEditorUserId="2535" LastEditorDisplayName="" LastEditDate="2011-04-03T20:56:41.887" LastActivityDate="2011-04-03T20:56:41.887" CommentCount="2" />
  <row Id="8009" PostTypeId="2" ParentId="7993" CreationDate="2011-04-03T20:29:01.763" Score="1" ViewCount="" Body="&lt;p&gt;If your photon is sufficiently energetic, in the final state you can find any configuration allowed with the energy-momentum conservations laws, i.e., an excited atom and a lower-energy photon(s): $\gamma (\omega) + A_0 \rightarrow \gamma &amp;#39;(\omega&amp;#39;) + A_n,$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \omega&amp;#39;&amp;lt;\omega$.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-04-03T21:24:11.583" LastActivityDate="2011-04-03T21:24:11.583" />
  <row Id="8010" PostTypeId="6" CreationDate="2011-04-03T20:29:07.673" Score="0" ViewCount="" Body="&lt;p&gt;There has been &lt;a href=&quot;http://meta.physics.stackexchange.com/questions/601/moderator-nominee-debates/605#605&quot;&gt;some complaint that we haven't enough candidates&lt;/a&gt;, so I'll put myself forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently one of the &lt;em&gt;pro tem&lt;/em&gt; moderators on &lt;a href=&quot;http://codegolf.stackexchange.com/&quot;&gt;codegolf.SE&lt;/a&gt;, &lt;a href=&quot;http://stackoverflow.com/users/2509/dmckee&quot;&gt;I have been around since the Stack Overflow private beta&lt;/a&gt;, and have &lt;a href=&quot;http://meta.stackoverflow.com/users/2509/dmckee&quot;&gt;accumulated a bit of reputation on the mother meta&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;do&lt;/em&gt; have opinions about what content does and does not belong (including a dislike of Big Lists, a dislike of keeping elementary homework problem on the site (I try to provide a pointer in the comments, instead), and having &lt;a href=&quot;http://meta.physics.stackexchange.com/questions/340/discouraging-the-use-of-personal-attacks-and-epithets-such-as-crackpot/344#344&quot;&gt;little truck with...er...non-standard theories&lt;/a&gt;), but have tried not to move too preemptively on CodeGolf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can be convinced to change my mind at times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The demands of real life come and go and I may not always be able to devote a lot of time to moderation around here, but can generally check in every day (with the rare exception of travel days on overseas trips).&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-04-03T21:29:32.170" LastActivityDate="2011-04-03T21:29:32.170" CommunityOwnedDate="2011-04-03T20:29:07.673" CommentCount="4" />
  <row Id="8011" PostTypeId="1" CreationDate="2011-04-03T20:46:18.030" Score="3" ViewCount="131" Body="&lt;p&gt;If one starts with a flat piece of paper, removes a wedge, and tapes the paper together, you get a cone.  The angle of the removed wedge is called the &quot;angular deficit&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if this is done in 3 spatial dimensions, removing a 'wedge' along an entire line, there is similarly an angular deficit around the line defect. Using cylindrical coordinates we can write the spacetime metric as:&lt;br&gt;&#xA;$g^{\mu\nu} = dt^2 - dz^2 - dr^2 - r^2 d\theta^2$&lt;br&gt;&#xA;with $0 \leq \theta \leq 2 \pi - \Delta \phi$ and the boundary condition that $\theta=0$ is equivalent to $\theta = 2 \pi - \Delta \phi$, where $\Delta \phi$ is the angular deficit.  This line defect is sometimes referred to as a 'cosmic string'.  In GR this can also be used to describe the spacetime outside of a cylindrically symmetric object in certain conditions ( &lt;a href=&quot;http://prd.aps.org/abstract/PRD/v39/i4/p1084_1&quot; rel=&quot;nofollow&quot;&gt;http://prd.aps.org/abstract/PRD/v39/i4/p1084_1&lt;/a&gt; ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides the change in boundary condition for $\theta$, this looks exactly like flat spacetime.  So the spacetime is locally flat (the Riemann curvature vanishes) everywhere except at the line defect where it is undefined.  So without observing the defect directly, its gravitational presence can only be seen in a topological sense as it requires measurements of paths going around the defect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make sure my intuition is okay up to this point, first I'd like to verify that last line:&lt;br&gt;&#xA;In case I'm missing something... In such a spacetime, is it possible to somehow measure the angular deficit &quot;locally&quot; or using paths that have a zero winding number around the defect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This intuitive understanding of angular deficit seems to implicitly require that spacetime is locally flat and have the symmetry of an infinite cylinder.  For example, if I broke translation symmetry along the line defect by making it finite in length, one can't even use intuitive topology to state whether a path went 'around' the line defect or not.  However, physical intuition states that spacetime around the center of the finite line defect, at distances much less than the finite length, should still approximate the infinite line case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can &quot;angular deficit&quot; along a path be rigorously defined for arbitrary metrics somehow?  If not, what symmetries are necessary to do so?&lt;/p&gt;&#xA;" OwnerUserId="2652" LastActivityDate="2011-04-04T08:37:13.120" Title="Angular deficit" Tags="&lt;general-relativity&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="8012" PostTypeId="2" ParentId="8003" CreationDate="2011-04-03T21:14:48.387" Score="6" ViewCount="" Body="&lt;p&gt;There is more to the picture than just the superposition principle applied to long state vectors. @Lubos notes that &lt;em&gt;This number of basic configurations is exponentially large because the spins are independent&lt;/em&gt;, right. But there are certain sectors of the Hilbert space which are distinguished from others in the amount of entanglement among the constituent spins of state vectors in the respective sectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if I have two spins, then I can have states such as: $(|0\rangle|1\rangle + |0\rangle|0\rangle)/\sqrt{2} = |0\rangle(|1\rangle+|0\rangle)/\sqrt{2}$ which are factorizable and those which are not such as $(|0\rangle|1\rangle \pm |1\rangle |0\rangle)/\sqrt{2}$. This is the simplest example of an entangled state and is called &lt;em&gt;bipartite&lt;/em&gt; entanglement. In general one can have a state with entanglement among more than 2-spins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a state with $n$ spins as $n$ grows so does the dimension and complexity of the subspace describing sequences with bipartite, tripartite and so on to k-partite entanglement ($k\le n$) among the constituent spins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from the possible ways to exploit multipartite entanglement for quantum computation, there is the question that @Jim asks:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why don't we observe multipartite entanglement in the real world?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course one response is simply that such states are very fragile and decay incredibly fast into non-entangled states due to decoherence and are thus hard to make and harder to observe. Though one might see them in microscopic systems they have no utility for describing phenomena at larger scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question as to whether such states are more stable than this naive picture suggest is still wide open. It is possible, and I strongly believe, that multipartite entanglement will turn out to play a far more important role in our daily lives than we realize at present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One very compelling prospect is that life, which has been evolving for billions of years and which has developed sophisticated mechanisms which we are only now discovering, has evolved methods to exploit quantum information processing. Here I will just list some references:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Apoorva Patel has been working on the relation between &lt;a href=&quot;http://arxiv.org/find/quant-ph/1/au:+Patel_A/0/1/0/all/0/1&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/find/quant-ph/1/au:+Patel_A/0/1/0/all/0/1&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Seth Lloyd summarizes these developments beautifully in &lt;a href=&quot;http://www.youtube.com/watch?v=wcXSpXyZVuY&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Again, neither of these are explicit demonstrations of &lt;em&gt;multipartite&lt;/em&gt; entanglement in action but it would be a shame if life failed to exploit such a wonderful resource.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Let me answer the question with a &lt;em&gt;concrete&lt;/em&gt; example of multipartite entanglement in action in the &quot;real-world&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are certain insulating magnets whose magnetic susceptibility as a function of temperature scales not according to the $1/T$ Curie law, but according to a power law $T^{-\alpha}$ ($\alpha \lt 1$). Ghosh et al, demonstrate with numerical simulations that such power law behavior can be explained only is long-range entanglement of dipoles in the spin-system is taken into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quote from the concluding section of &quot;&lt;a href=&quot;http://arxiv.org/abs/cond-mat/0402456&quot; rel=&quot;nofollow&quot;&gt;Entangled Quantum States of Magnetic Dipoles&lt;/a&gt;&quot; by Ghosh et al. (Nature, 2003):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a growing realization that entanglement is a useful concept for understanding quantum magnets, thus unifying two rapidly evolving areas, quantum information theory and quantum magnetism. The discussions to date have focused on one-dimensional magnets and measures of entanglement with clear theoretical meaning but no simple experimental implementation. &lt;strong&gt;Our experiments and simulations represent a dramatic illustration of how entanglement, rather than energy level redistribution, can contribute significantly to the simplest of observables&lt;/strong&gt; – the bulk susceptibility – in an easily stated model problem. (emph. mine)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" LastActivityDate="2011-04-04T06:01:28.787" CommentCount="2" />
  <row Id="8013" PostTypeId="1" CreationDate="2011-04-03T22:15:21.350" Score="5" ViewCount="189" Body="&lt;p&gt;For example it is stated that if supersymmetry breaking is &lt;em&gt;soft&lt;/em&gt; then stability of gauge hierarchy can be still maintained.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2011-04-04T09:07:03.433" Title="What does &quot;soft&quot; in &quot;soft symmetry breaking&quot; mean?" Tags="&lt;quantum-field-theory&gt;&lt;supersymmetry&gt;&lt;symmetry&gt;" AnswerCount="2" />
  <row Id="8016" PostTypeId="2" ParentId="8011" CreationDate="2011-04-04T08:37:13.120" Score="1" ViewCount="" Body="&lt;p&gt;The Riemann tensor is exactly zero everywhere except for the locus of this cosmic string, so any open set in this locally flat space will behave in the same way as the corresponding open set in the flat space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is a nonzero Riemann tensor proportional to the delta-function right at the origin of the two-dimensional $xy$ plane transverse to the cosmic string.&#xA;$$R_{xyxy} = \Delta\phi \cdot \delta^{(2)}(x,y) $$&#xA;All other components of the Riemann tensor vanish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that if you study what happens with a vector $\vec v$ under parallel transport around a curve that encircles the $(x,y)=(0,0)$ point, i.e. the locus of the cosmic string, you will find out that that this vector $\vec v$ will be rotated by $\Delta \phi$ in the $xy$ plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a cosmic string may therefore act as a kind of gravitational lens. Light rays from the same original source may be coming to our telescopes from two directions that differ by $\Delta \phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that if the cosmic string were an open string, one couldn't invariantly say whether a path in space encircles this cosmic string or not. Consequently, the &quot;monodromy&quot; would be undetermined: should it be zero, or the deficit angle? The resolution to this paradox is simply that it cannot happen that a cosmic string of this kind is &quot;open&quot;. One can't write down a geometry that would look like a flat geometry at long distances from the open cosmic string - in all directions - but that would also exhibit a deficit angle, but that would also be locally flat almost everywhere. The observation that such a geometry would lead to inconsistent, ambiguous predictions for the relative angle between light rays is a &lt;em&gt;proof&lt;/em&gt; that such a geometry cannot exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, there are many reasons why a cosmic string can't be open. If a cosmic string could easily create end points, it would be unstable much like the QCD flux tube. And each open segment of such a cosmic string would be rapidly shrinking in order to minimize its length - which also minimizes the energy. The actual cosmic strings in physics have to be at least locally stable. Also, many cosmic and not only cosmic strings carry various &quot;charges&quot; that may be interpreted as &quot;winding numbers&quot; (infinite if the strings extend across the Universe) - however, in that case, there is also a generalized &quot;electromagnetic field&quot; around them which is also a nonzero energy density and sources the gravitational field. Such things are heavily constrained, too. But even in the absence of such conserved charges, the cosmic strings tend to be stable and they don't admit end points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies if there are sign errors or errors in the factors of two above.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-04T08:37:13.120" />
  <row Id="8017" PostTypeId="2" ParentId="8013" CreationDate="2011-04-04T08:48:47.700" Score="9" ViewCount="" Body="&lt;p&gt;In general, the terminology &quot;soft&quot; in particle physics refers to &quot;low energy&quot; or &quot;low frequency&quot; while &quot;hard&quot; refers to &quot;high energy&quot; or &quot;high frequency&quot;. It's because hard radiation can penetrate and/or modify matter. Even in medicine, people are familiar with &quot;soft X-rays&quot; between 0.12 and 12 keV and &quot;hard X-rays&quot; between 12 and 120 keV - the latter can penetrate matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term &quot;soft SUSY breaking&quot; has to be interpreted as one term. Of course, it is related to the softness explained above. The technical definition is that &quot;soft SUSY breaking&quot; is a term added to the Lagrangian that breaks supersymmetry, but only softly. This adverb means that the modification of physics at high energies is so small that we don't create any new divergent contributions to the mass of the scalars such as the Higgs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the advantages of supersymmetry in phenomenology is that the bosons and fermions cancel their quadratically divergent loop corrections to the Higgs mass; soft SUSY breaking are non-supersymmetric terms that still preserve this cancellation. However, soft SUSY breaking obviously allows - and does cause - &lt;em&gt;finite&lt;/em&gt; loop corrections to the Higgs mass. However, the quadratically divergent corrections are still absent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one looks which terms in the Lagrangian have this property, he finds out that it's generally the terms whose coefficients have a positive power of the mass - especially various mass terms for the superpartners of the known fermions and bosons. However, the terms that are generated from dimensionless (e.g. Yukawa or gauge) interactions with the Higgs or the gauge field can't be included among soft SUSY breaking terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the minimal supersymmetric standard model (MSSM), there are about 105 new soft SUSY-breaking parameters. All of them influence the low-energy physics but preserve the cancellation of the divergences for the Higgs mass. In principle, the values of all these coefficients may be fully calculated from an underlying UV theory - such as an explicit compactification of string theory; in the full picture, the breaking of SUSY is always spontaneous. However, in the absence of knowledge about such a UV theory or vacuum, people partially give up and parameterize the low-energy manifestations of the supersymmetry breaking by the soft SUSY-breaking terms which are breaking SUSY &quot;explicitly&quot; and may be viewed as an effective description of the spontaneous SUSY breaking mechanism whose precise character in Nature - if any - is not understood at this moment.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-04T08:48:47.700" />
  <row Id="8019" PostTypeId="2" ParentId="8013" CreationDate="2011-04-04T09:07:03.433" Score="0" ViewCount="" Body="&lt;p&gt;I saw the words &quot;soft breaking&quot; used without relation to SUSY. I doubt that there is a strict definition of this, but naively it seems to mean the following thing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A symmetry of your model can be broken either&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Spontaneously -- with some of your fields receiving non-invariant vacuum expectation value.&lt;/li&gt;&#xA;&lt;li&gt;Or explicitly -- by some terms in the Lagrangian that is not invariant under your symmetry transformation.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Explicit symmetry breaking is called &quot;soft&quot; if the terms introduced are of the dimension less than four.&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-04-04T09:07:03.433" />
  <row Id="8020" PostTypeId="1" CreationDate="2011-04-04T09:26:56.997" Score="1" ViewCount="119" Body="&lt;p&gt;This is a serious question from someone engaged in evaporating large quantities of water to turn sap into syrup at this time of year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably some background will help. When sap boils vigorously it creates quite a bit of foam, which will overflow the evaporator (incidentally filling the building with a pleasing maple caramel smell as it burns on the side of the evaporator). When the foam gets too high we touch it with a bit of lard and the foam level drops (surface tension - I know). However, it is tempting for me to give a good swipe so that the foam almost disappears (instead of just dropping). The old-timers however contend that I should just reduce the foam to the point where it isn't overflowing any more. They say that it will take longer to boil away the water if I eliminate the foam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I fail to see how the foam will improve evaporation (although it seems to me that it might slow it down).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: by request ( @georg ) , a link to the evaporator in question &lt;a href=&quot;https://sites.google.com/site/lindsayssugarbush/_/rsrc/1240515239201/Home/2005-03-30--12-25-21.jpg&quot; rel=&quot;nofollow&quot;&gt;https://sites.google.com/site/lindsayssugarbush/_/rsrc/1240515239201/Home/2005-03-30--12-25-21.jpg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2923" LastEditorUserId="2923" LastEditorDisplayName="" LastEditDate="2011-04-04T20:12:01.090" LastActivityDate="2011-04-05T04:35:54.357" Title="Does the foam on top of boiling maple sap affect the rate of evaporation?" Tags="&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="8022" PostTypeId="2" ParentId="8020" CreationDate="2011-04-04T10:20:34.507" Score="0" ViewCount="" Body="&lt;p&gt;I am too lazy to compute anything, but consider what is the foam?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The foam is a bubble of steam covered with a layer of water and sugar solution.  If there is no foam the area of the pot exposed to air or evaporation  is two dimensional, with maybe some big bubbles from the boil. Each little bubble in the multitude  contributes more to the area exposed to the air where the surface cover water can evaporate, burst and release trapped steam, making way for the next bubble. It is a fractal problem,but the dimensions of the available evaporation surface are larger than 2 by far when there is foam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets put it another way. The foam raises a film of hot water to contact with the air,  without obstructing the trapped steam, so when a foam bubble breaks there is water out of the pot that would have been in the pot if there were no foam,  the extra surface of the foam increases the  evaporation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the oldtimers are probably correct.This argument depends on the surface of the foam bubbles being a solution with water.Of course an experiment should be carried out with the specific liquid to decide on this.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-04-05T04:35:54.357" LastActivityDate="2011-04-05T04:35:54.357" CommentCount="10" />
  <row Id="8023" PostTypeId="2" ParentId="4506" CreationDate="2011-04-04T10:27:33.600" Score="1" ViewCount="" Body="&lt;p&gt;This question and ones like it are trying to get at the 'transferability' of energy between different frames of reference.  You begin in a linear domain and move to a rotational domain.  Whether you are considering the problem in the linear or rotational sense the inherent physics remains the same (at least in elementary examples such as this.), and so you can transfer physical quantities from one to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point I learned was that it helps to select a regime in which it is easy to calculate some figure of merit and which can be transfered simply to a regime in which the answer exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy is often useful in this case as it is scalar and invariant in the transform.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-04-04T10:27:33.600" CommentCount="2" />
  <row Id="8024" PostTypeId="2" ParentId="8020" CreationDate="2011-04-04T10:28:39.620" Score="1" ViewCount="" Body="&lt;p&gt;Yes and No,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The evaporation is controlled by the heat flow into that evaporator. &#xA;(How does that evaporator look like? Could You post/link a picture?)&#xA;So foam is not the control, but the limiting factor, because You have &#xA;to reduce heat flow to avoid the foam spilling over. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To &quot;fight&quot; that foam, there are several possibilities: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Kind of stirrer/cam agitating above the surface of liquid. &#xA;Whether this is helpful, You know better than me, just &#xA;try the foams reaction to some agitation by a stick/paddle. &#xA;Coating the upper part of the evaporator and that foam breakers &#xA;with some hydrophobic plastic (eg PTFE) might help.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Big diameter flat evaporation vessel with ample space above liquid. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Antifoam agents: more theoretically, because those might leave &#xA;some unpleasant smell/taste in the syrup. You needed a food &#xA;grade defoamer, I don't know whether such are available. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some hints to defoaming: most defoamers are some hydrophobic &#xA;substances like petroleum (that used for lamps) or fat. &#xA;This fatty substances  will disperse in the foaming liquid as &#xA;fine droplets, which act as rupture initiation points on the foam &#xA;lamellae. &#xA;If the syrup was for my own use only, I'd try to add some paraffine wax &#xA;(food grade) for a trial. (one gram on 100 ltr of syrup, not more!)&#xA;The wax will separate after cooling down, (and swim on top I assume?). &#xA;Just try that in a pan with a small amount of syrup. The critical &#xA;factor is this removal, maybe the paraffin remains dispersed as tiny &#xA;droplets in the syrup. Removal by adsorption filtration is possible, but &#xA;that is more messy/lossy than slow enough evaporation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point: some of the taste of maple syrup comes from heating &#xA;for some time (caramel reaction). In case You reduce this time by &#xA;reduced foaming, You might get a very light (color) syrup missing the &#xA;&quot;right&quot; flavor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS &#xA;foam is not just a question of surface tension, think of water, which has &#xA;the highest surface tension of all liquids (exept molten metals). &#xA;Foaming is a very complicated process, viscosity, not too high &#xA;surface tension are the main &quot;incredients&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &#xA;Thank You for the picture! A evaporator more flat than this one &#xA;is not possible, I think. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In chemical technology there are a number of machines which were developed &#xA;to evaporate/concentrate viscous and foamy liquids. But those are big machines,&#xA;not for such a small scale business. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-04T21:24:52.947" LastActivityDate="2011-04-04T21:24:52.947" />
  <row Id="8025" PostTypeId="2" ParentId="1286" CreationDate="2011-04-04T14:09:45.343" Score="0" ViewCount="" Body="&lt;p&gt;The stone creates a contact patch with the ice. In this case it is roughly circular in shape with approximate radius of $a=0.721\,\sqrt[3]{W\,D\left(\dfrac{1-\nu_{1}}{E_{1}}+\dfrac{1-\nu_{2}}{E_{2}}\right)}$ where $D$ is the diameter of the spherical contact surface (the crown diameter of the stone), $W$ is the weight of the stone, and $\nu$, $E$ are the Poisson's ratio and Youngs modulus for the two materials contacting (ice and stone).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now because the surface is uneven, the pressure on the contact is not necessarily symmetric, and may have an pressure center of axis from the spin axis of the stone. At most the eccentricity is equal to $a$, yielding the maximum possible torque applied to the stone to be $T=\mu\,W\,a$ where $\mu$ is the coefficient of friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torque the is responsible for altering the spin the stone. Once the stone has spin, then the hydrodynamic contact properties alter the path of the stone.The friction is &lt;em&gt;not&lt;/em&gt; constant as it depends on the pressure applied on each spot. Think of the top layer of ice as a lubrication layer, and the stone has to plow through it. As far as the mechanism that this complex interplay creates a side force (tracking force) I am not sure I can explain it. Maybe someone else can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ref: &quot;Formulas for Stress &amp;amp; Strain&quot;, Roark, 3rd ed. 1954&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-04-04T14:09:45.343" CommentCount="2" />
  <row Id="8026" PostTypeId="2" ParentId="7985" CreationDate="2011-04-04T14:17:30.537" Score="-2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;The answer is 'No'&lt;/strong&gt;. If you have to answer at school then the previous answers are fine, consistent whith mainstream.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But  ...&lt;br&gt;&#xA;&lt;strong&gt;The answer is 'Yes'&lt;/strong&gt; under &lt;strong&gt;&lt;a href=&quot;http://arxiv.org/abs/astro-ph/0208365&quot; rel=&quot;nofollow&quot;&gt;this undiscussed model&lt;/a&gt;&lt;/strong&gt; (page 11, Rotation, eq 38) where the variation of LOD (Earth's Length Of Day) is discussed. A less known fact is that the Earth is increasing the long term rate of rotation, i.e. shorter days (&lt;a href=&quot;http://maia.usno.navy.mil/whatiseop.html&quot; rel=&quot;nofollow&quot;&gt;image from here&lt;/a&gt;). The angular momentum redistribution due to tidal effects is not enough to account for the measures, because it implies a slowing instead of an acceleration of rotation. The overall redistribuition of water/ice and movement of the solid crust is also not sufficient to account for the measures.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My personal answer is 'Yes' because I can not find a single reason to disproof that model .&lt;br&gt;&#xA;The extreme rotational speed of millisecond pulsars can be easily explained by model, as suggested there.&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/8rVi4.png&quot; alt=&quot;$\Delta$LOD - Length Of Day&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-04-04T14:17:30.537" CommentCount="5" />
  <row Id="8027" PostTypeId="1" CreationDate="2011-04-04T14:18:37.020" Score="4" ViewCount="272" Body="&lt;p&gt;I'm trying to explain to someone learning elementary physics (16 year old) that linear momentum and energy are conserved independently. I'm not a professional physicist and haven't tried to explain this stuff for years, and I can't think of any convincing elementary argument to show that this is the case. Does anyone know of an elementary approach to this ? (i.e. one that doesn't contain the expressions &quot;Lagrangian&quot; and &quot;Noether's Theorem&quot;)&lt;/p&gt;&#xA;" OwnerUserId="2925" LastEditorUserId="2925" LastEditorDisplayName="" LastEditDate="2011-04-04T14:46:00.477" LastActivityDate="2011-04-05T12:28:07.323" Title="How to explain independence of momentum and energy conservation?" Tags="&lt;classical-mechanics&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="2" />
  <row Id="8028" PostTypeId="2" ParentId="6096" CreationDate="2011-04-04T14:18:56.320" Score="0" ViewCount="" Body="&lt;p&gt;I don't understand why nobody has mentioned the apparent violation of conservation of energy. Am I naive here for thinking that a quantum mechanical trigger that allows the bifurcation of world lines, violates the conservation of energy?&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-04-04T14:18:56.320" CommunityOwnedDate="2011-04-04T14:18:56.320" CommentCount="3" />
  <row Id="8029" PostTypeId="2" ParentId="8027" CreationDate="2011-04-04T14:35:45.993" Score="-1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;edited&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conservation of energy is almost self-evident. Conservation of linear momentum comes from the fact the in an isolated system of particles there is no preferred location. To try to make a point that the two are related seems that would invoke some really convoluted arguments. One deals with invariance of energy and the other with invariance of location.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.eng.fsu.edu/~dommelen/quantum/style_a/physsym.html&quot; rel=&quot;nofollow&quot;&gt;Reference on conservations and symmetry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="392" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2011-04-04T16:50:56.057" LastActivityDate="2011-04-04T16:50:56.057" CommentCount="6" />
  <row Id="8030" PostTypeId="2" ParentId="7865" CreationDate="2011-04-04T14:41:43.197" Score="2" ViewCount="" Body="&lt;p&gt;The classical Dirac and EM fields have intrinsic angular momentum. For the Dirac field, for example (see $\S$2-1-3 in Itzykson&amp;amp;Zuber, &lt;em&gt;Quantum Field Theory&lt;/em&gt;, McGraw-Hill), there is a component $-i\sigma_{\mu\nu}\omega^{\mu\nu}/4$ of the infinitesimal generators of Lorentz transformations as well as the orbital component $x_\mu\partial_\nu\omega^{\mu\nu}$. That is to say, the Dirac equation is invariant under infinitesimal transformations&#xA;$$\psi&amp;#39;(x)=\left(I-\frac{i}{4}\sigma_{\mu\nu}\omega^{\mu\nu}+x_\mu\partial_\nu\omega^{\mu\nu}\right)\psi(x).$$&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-04-04T14:41:43.197" CommentCount="5" />
  <row Id="8031" PostTypeId="2" ParentId="8027" CreationDate="2011-04-04T14:47:29.167" Score="1" ViewCount="" Body="&lt;p&gt;For a particle in a 1D external time-dependent field there are no energy and momentum conservation laws, yet there are two independent conserved quantities. It is because the differential equation is of the second order and it is accompanied with two independent initial data - the initial position and initial velocity. See an example &lt;strong&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/4973/do-symmetries-increase-the-number-of-conserved-quantities&quot;&gt;here&lt;/a&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-04-04T14:47:29.167" />
  <row Id="8032" PostTypeId="2" ParentId="8003" CreationDate="2011-04-04T15:11:12.417" Score="0" ViewCount="" Body="&lt;p&gt;This exponential growth is what allows us to have multipartite entanglement, and exponentially many Everett branches in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If in the near future, we can construct a quantum computer implementing Shor's algorithm, this will be a good observational example.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-04-04T15:11:12.417" />
  <row Id="8033" PostTypeId="2" ParentId="8027" CreationDate="2011-04-04T15:34:22.610" Score="1" ViewCount="" Body="&lt;p&gt;All the following explanation is for elementary students.&#xA;First ensure that he/she understands that momentum is a vector quantity and energy is scalar. Also s/he might be knowing Newton's 2nd law of motion. Momentum of a system is unique in given direction.  For a given system if net external force (on the system) is zero then net momentum of the system &lt;strong&gt;does not change&lt;/strong&gt;. He can argue that so what energy will also not change. Yes, but we can have situation where we apply force on a body and its &lt;strong&gt;momentum is changed but energy is not&lt;/strong&gt;. (A force applied perpendicular to the direction of motion all times). I hope this sets the sail for discussion.&lt;/p&gt;&#xA;" OwnerUserId="2927" LastActivityDate="2011-04-04T15:34:22.610" CommentCount="2" />
  <row Id="8034" PostTypeId="1" AcceptedAnswerId="8056" CreationDate="2011-04-04T16:03:15.760" Score="1" ViewCount="302" Body="&lt;p&gt;Given a stream of random binary numbers(*)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to differentiate if they came from a Truly Random or from a formula/algorithm ? how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if there is no way to decide this, then, I can't find any basis, to keep denying that behind the &quot;truly random&quot; of quantum mechanics can be a hidden algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I am talking about the posibility of a &quot;hidden variables&quot; theory, but I can't find any other explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) Is known that is possible to create &lt;a href=&quot;http://www-2.dc.uba.ar/profesores/becher/turing.pdf&quot; rel=&quot;nofollow&quot;&gt;computable normal numbers&lt;/a&gt; from which first is possible to extract an infinite random binary stream, second, it tell us that a finite logic expression (again a binary stream) can contain a rational, and even an irrational number, so there is no big difference within a bit stream and the measurement of a random outcome, (and even less if we consider the accuracy) I say this, because the argument that obtaining bits is not the same as answer the complex question that can be made to quantum experiments, I think that &quot;random&quot; results can be perfectly read from a random stream of bits&lt;/p&gt;&#xA;" OwnerUserId="1916" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-04-04T19:15:07.887" LastActivityDate="2011-04-05T05:54:18.547" Title="How to tell if QM is really random?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="3" CommentCount="14" />
  <row Id="8035" PostTypeId="2" ParentId="408" CreationDate="2011-04-04T16:04:22.643" Score="1" ViewCount="" Body="&lt;p&gt;There are two levels at which this is being asked I think.  The animations in the link below illustrates what black holes appear as in front of stars or the galaxy.  It also portrays the black surface which demarks this as a black hole.  Will we ever see a black in this form?  Maybe, or we might get some better signature of this.  Much higher resolution could demonstrate angular dependencies with accretion disks around black holes.  Similarly it is possible that the super massive black hole in this galaxy could be resolved in a similar way, at least to measure optical lensing of distant stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://jila.colorado.edu/~ajsh/insidebh/&quot; rel=&quot;nofollow&quot;&gt;http://jila.colorado.edu/~ajsh/insidebh/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is of course the prospect that black holes exhibit some quantum properties around the 1-10 TeV range of energy.  This is the large extra-dimension prospect, where maybe Calabi-Yau manifolds “unfold” to larger scales at higher energy.  In doing so physics at the TeV range of energy may have some small amplitudes corresponding to graviton and black hole physics.  At low energy the physics is QFT, while at higher energy it assumes some gravitational aspects.  We might think of this as doing QFT on the surface of the AdS, but where at sufficiently high energy we get some small sampling of gravity in the interior of the AdS.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extent to which we probe the interior of the AdS might be modeled with a simple field.  We might think of this as the manifestation of an auxiliary field which converts one sort of physics at low energy into another at high energy.  The transverse modes of a string on the boundary define the CFT there, and these assume a certain value on this boundary.  In the interior the longitudinal modes are nonvanishing, but vanish on the boundary.  The noncommutative coordinate uncertainty principle $\Delta X^+\Delta X^-~\simeq~{\ell}_s$ $=~4\pi\sqrt{\alpha’}$ is then off quadrature near the boundary of the AdS.  This auxiliary force is then analogous to the squeezing operator in quantum optics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a toy model consider a $n+1$dimensional “spacetime plus R” space. This additional dimension is a space with a gauge connection or potential $A~=~\phi$ which defines a force $F~=~-d\phi/dx_5$, for $x_5$ a parameter on this fifth dimension.  An obvious simple case for a field in a one dimensional chord of length $L$ would be where $\phi~=~\pi x_5/L$ as a model similar to a constant gravity on Earth.  This “force” then adjusts the noncommutative uncertainty principle, which is in a complete squeezed state on the AdS boundary.  We then at high energy adjust this slightly by pushing the conformally flat spacetime on the boundary into an additional dimension in the interior of the AdS.  We might then gets some signatures or small amplitudes of quantum gravity, or quantum black hole physics, in LHC scale experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prospect is not entirely absurd, for at the EW symmetry recovery level QFT may exhibit a renormalization group flow to high energy.  This flow is then logarithmic in energy, and so the difference in scale from $10 TeV$ to $10^{16}TeV$ is not that dramatic.  It then might be the case that small amplitudes for quantum black hole physics could turn up in the LHC.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-04-04T16:04:22.643" />
  <row Id="8036" PostTypeId="1" AcceptedAnswerId="8047" CreationDate="2011-04-04T16:12:36.787" Score="1" ViewCount="373" Body="&lt;p&gt;I recently bought some &lt;a href=&quot;http://www.getbuckyballs.com/&quot; rel=&quot;nofollow&quot;&gt;buckyballs&lt;/a&gt;, considered to be the world's best selling desk toy. Essentially, they are little, spherical magnets that can form interesting shapes when a bunch of them are used together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After playing around with these buckyballs for a while, I wondered: &quot;Can these guys ever lose their magnetism?&quot; Then I went a step further and thought, &quot;How are magnets affected by the 2nd law of thermodynamics?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how are magnets affected by the 2nd law of thermodynamics? Do they break down and lose their magnetism over time (like iron rusts over time)?&lt;/p&gt;&#xA;" OwnerUserId="2609" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-04T20:30:59.037" LastActivityDate="2011-04-09T19:13:39.247" Title="Do magnets lose their magnetism?" Tags="&lt;electromagnetism&gt;&lt;entropy&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="8037" PostTypeId="2" ParentId="7864" CreationDate="2011-04-04T16:17:58.097" Score="1" ViewCount="" Body="&lt;p&gt;There are a few problems with this formula and calculation. I shall present those that I have found below and hopefully this will help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we note that the intended conversion ratios from the data: old = [27.2,56.2]; new = [11.26,22.52] give a conversion factor between new/old = [0.414,0.401]. So perhaps because of rounding errors one is seeking a conversion factor between these two values, of perhaps around 0.41.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This conversion is done in two logically distinct steps: Unit Conversion (U say) - your step 2- and Temperature Conversion (R say) - your step 1. Both of these conversions contain issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 1 - Temperature Conversion.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem here is not so much the formula (as far as I know), but the assumption of $27^0$C. The Allen paper does do plant tests at $27^0$C and also $30^0$C, $35^0$C and $40^0$C. So which value to choose? The main issue for me however is figure 2 in that paper where the data (from table 3) are plotted against mean annual temperature in degrees Fahrenheit ie the true natural ground temperature. This range is $53^0$F to $67^0$F ie [11.6,19.4] Centrigrade range with median at $15.5^0$C. So it is possible that George has taken $15.5^0$C - which gives an R value of 1.04 or even just $15^0$C giving an R value of 1. If that were true it would largely cancel out the error you have found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2 - Unit Conversion.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some problems here however which complicates things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First the density of $O_2$ would need to change to that at (say) $15^0$C. Now the figure quoted there is not the density but the inverse density (and there is a typo in the microunit). Using the ideal gas law the inverse density would change to 7.37 microunits from your value of 7.69 microunits - a decrease in this term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem are the errors in the &quot;respiration&quot; formula. This formula does not balance on the Oxygen: LHS=4, RHS=6. Furthermore the LHS chemical is Formaldehyde! So there must be at least a typo here - or perhaps the whole idea of using respiration formulae is wrong here? As this is just a unit conversion it may not rely on the details of plant respiration but just be based on atomic weights (8 mol $CO_2$ = 11 mol $O_2$)?? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT I have just seen what is wrong with this chemical formula: it is intended to be an &quot;empirical formula&quot; reducing from the glucose respiration formula (in Wikipedia). So it is 2$O_2$ on the LHS to balance. In that case it is 1 mol $O_2$ = 1 mol $CO_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The George paper does not explain how the $O_2$ to $CO_2$ conversion works and so I dont know how to take this any further, but hopefully the above is helpful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Incidentally to &quot;reverse engineer&quot; the formula under such circumstances you might also check the conversion factor implied by the other cited papers in Appendix 1 where it was maybe clearer what was being converted.)&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-04-05T18:36:50.563" LastActivityDate="2011-04-05T18:36:50.563" CommentCount="2" />
  <row Id="8038" PostTypeId="2" ParentId="8036" CreationDate="2011-04-04T16:22:49.153" Score="4" ViewCount="" Body="&lt;p&gt;The second law of thermodynamics - about the increasing entropy - which is apparently what you're talking about - holds for any system. Permanent magnets are no exception.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ferromagnet may look &quot;more ordered&quot; than a non-magnetic material because the spins are oriented in the same direction, rather than random directions. But physical systems may only try to maximize their entropy among configurations that conserve energy (much like the momentum, charge, and other conserved quantities). For ferromagnets, the configuration with spins oriented in random directions would have a much higher energy - because one reduces the energy by orienting the spins, elementary magnets, in the same direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the spontaneous disappearance of the uniform electrons' spin would violate the energy conservation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among the configurations with the same energy, the magnet still tries to maximize its entropy. In particular, the heat is flowing from warmer pieces of the material to colder ones, and so on. More generally, the entropy never goes down, and that's the only general statement that follows from the second law of thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ferromagnets are not special among physical objects that could have a higher entropy if you allowed the energy to increase. For example, any object would raise its entropy - the amount of disorder - if its temperature increased. But a higher temperature requires a higher energy, too. One can't violate the first law of thermodynamics (energy conservation) just because it would make it more straightforward to satisfy the second law. Both of them hold in Nature.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-04T16:37:12.563" LastActivityDate="2011-04-04T16:37:12.563" CommentCount="2" />
  <row Id="8039" PostTypeId="2" ParentId="8027" CreationDate="2011-04-04T16:43:01.953" Score="4" ViewCount="" Body="&lt;p&gt;Give him an example of inelastic collisions and explain why momentum is conserved but kinetic energy is not. If you explain the reasoning, (all forces are internal hence momentum is conserved) and that there are losses so KE of the system is lost in heat/sound/other forms of energy.. he should get the idea that the two are different beasts. At such a level it is best to illustrate with counterexamples. &lt;/p&gt;&#xA;" OwnerUserId="2599" LastActivityDate="2011-04-04T16:43:01.953" CommentCount="4" />
  <row Id="8040" PostTypeId="2" ParentId="8034" CreationDate="2011-04-04T16:52:56.793" Score="0" ViewCount="" Body="&lt;p&gt;No, you cannot prove that a string of numbers is random, and no, this is not a problem for quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you are observing integers between 0 and 9 and you observe the 12-number sequence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 6 7 1 9 4 0 8 5 2 8 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it random?  It looks like it, but there's always a possibility it was generated by the following algorithm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) print 4&#xA;2) print 6&#xA;3) print 7&#xA;4) print 1&#xA;...&#xA;12) print 8&#xA;13) go back to 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can test whether this was the correct algorithm by getting the next digit.  Say you get &quot;3&quot; when the previous algorithm said you should get &quot;4&quot;.  That's no problem, you can amend the previous algorithm to add printing &quot;3&quot; at the end of it before it repeats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there's an algorithm for any sequence of numbers, but it's vacuous.  The point isn't so much that quantum measurements are random.  According to the Copenhagen interpretation they are, but what matters is whether they can be predicted, not whether they can be post-dicted.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-04-04T16:52:56.793" CommentCount="1" />
  <row Id="8041" PostTypeId="1" CreationDate="2011-04-04T17:06:18.497" Score="2" ViewCount="185" Body="&lt;p&gt;I wonder how one can show that general relativity is ghost-free? By ghost I mean the negative norm state that breaks the unitarity. I think it is a well-known &quot;fact&quot; but I just couldn't find any derivation or proof in any reference. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2928" LastEditorUserId="2928" LastEditorDisplayName="" LastEditDate="2011-04-05T01:36:59.890" LastActivityDate="2011-04-05T01:36:59.890" Title="Why is GR ghost-free?" Tags="&lt;quantum-field-theory&gt;&lt;general-relativity&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8042" PostTypeId="2" ParentId="8041" CreationDate="2011-04-04T17:29:07.283" Score="1" ViewCount="" Body="&lt;p&gt;Heng, the answer is a part of the &quot;linearized general relativity&quot; that is used when one derives the polarizations of the gravitational waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result of the derivation is that only positively-normed &quot;purely transverse&quot; polarizations exist. In $D=4$ spacetime dimensions, there are two physical positive-normed polarizations which may be represented as the $\pm 2$ helicity eigenstates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The metric tensor may be chosen to be &#xA;$$g_{\mu\nu} = \epsilon_{\mu \nu} \exp(i k \cdot x).$$&#xA;At the linearized level, Einstein's equations act as the wave equation or massless equation for the components, implying that $k^2=0$ i.e. the momentum vector has to be light-like. However, the constraint part of the equations - the combination of Einstein's equations that contains no second time derivatives - says that&#xA;$$ k_\mu \epsilon^{\mu\nu} = 0 $$&#xA;while the general covariance makes another identification for the polarization tensor,&#xA;$$\epsilon_{\mu\nu} \equiv \epsilon_{\mu\nu} + k_\mu \delta x_\nu + k_\nu \delta x_\mu.$$&#xA;He, I suggestively used the symbol $\delta x^\mu$ for the infinitesimal coordinate variations - the parameters of gauge symmetry of general relativity. Those conditions and identifications allow us to choose the most general $\epsilon_{\mu\nu}$ to be in the purely transverse form,&#xA;$$\epsilon_{ij}, \quad i,j=1,2,\dots D-2.$$&#xA;On this space, there is an obvious positively definite metric for the states. Because a residual gauge symmetry may also be used to eliminate the trace (sum of $(D-2)$ diagonal components) of the tensor $\epsilon_{ij}$, this tensor has&#xA;$$\frac{(D-2)(D-1)}{2} - 1$$&#xA;components which is equal to two components in $D=4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the last step - the elimination of the trace (corresponding to the overall volume form) - may be a bit subtle. In the Einstein-Hilbert action, this component of the metric actually does correspond to a bosonic field with the wrong sign of the kinetic term, a ghost. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This mode still ends up being unphysical. But if all spatial coordinates were compactified on a torus, one could actually see the wrong-sign scalar. That's e.g. where the Lorentzian structure in the paper&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/9811194&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9811194&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;comes from.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-04T17:29:07.283" CommentCount="2" />
  <row Id="8043" PostTypeId="1" AcceptedAnswerId="8046" CreationDate="2011-04-04T17:34:13.723" Score="6" ViewCount="124" Body="&lt;p&gt;According to Rindler the geodetic effect can be considered as consisting of Thomas precession combined with the effect of moving through curved space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wolfgang Rindler (2006) Relativity: special, general, and cosmological (2nd Ed.) p234&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However according to Misner, Thorne, and Wheeler, Gravitation, p. 1118, Thomas precession does not come into play for a freely moving satellite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See: &lt;a href=&quot;http://en.wikipedia.org/wiki/Talk:Geodetic_effect&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Talk:Geodetic_effect&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that although a freely moving satellite doesn't feel gravity, it's relation to an observer is still subject to lorentz transformations and hence Thomas precesssion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So who is right ? Rindler or Misner, Thorne, and Wheeler ?&lt;/p&gt;&#xA;" OwnerUserId="2929" LastActivityDate="2011-04-04T18:23:09.550" Title="Decomposing geodetic/de Sitter effect into Thomas precession and spatial curvature" Tags="&lt;relativity&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="8044" PostTypeId="2" ParentId="7487" CreationDate="2011-04-04T17:59:44.007" Score="1" ViewCount="" Body="&lt;p&gt;Beta rays, a product of I131 decay,have a very short range. If concentrated in the thyroid, it will for all practical puirposes, not be biologically eliminated before it decays. This will ionize a lot of the surrounding tissue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, gamma rays have a low absorbtion cross-section in the body and so do very little damage. Finally, Xenon is inert and should cause no damage at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accordingly, the trick is for people to minimize the uptake of I131. There is some atmospheric in dust I131 near the reactor. There is contaminated ground and sea water. There could also be uptake in marine life and vegetation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would seem to me, that except for people who cannot avoid being close to the plant and its environs, Iodide tablets are not called for. Avoidance would be much more practical. &lt;/p&gt;&#xA;" OwnerUserId="2930" LastActivityDate="2011-04-04T17:59:44.007" />
  <row Id="8045" PostTypeId="1" CreationDate="2011-04-04T18:08:06.873" Score="1" ViewCount="126" Body="&lt;p&gt;I apologize for this simple question, but I lost a factor of 2 and can't find it anymore, so now I'm looking on the internet, perhaps one of you has some information about its whereabouts. :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the electromagnetic field tensor $F_{\mu\nu}$ which corresponds to a differential form $F =\frac12 F_{\mu\nu} dx^\mu \wedge dx^\nu$. The wedge product of $F$ with itself can be expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F \wedge F = \frac14 F_{\mu\nu}\mathcal F^{\mu\nu} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathcal F^{\mu\nu} = \frac12 \epsilon^{\mu\nu\sigma\tau}F_{\sigma\tau}$ is the dual field tensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Writing $A = A_\mu dx^\mu$ for the vector potential, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_V F \wedge F = \int_V (d A) \wedge F = \int_V d(A \wedge F) = \int_{\partial V} A \wedge F$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it appears to me that the latter integral is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dots = \int_{\partial V} dx^\mu\ A_\nu \mathcal F^{\mu\nu} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, at the same time, we can write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_V dx^\alpha \frac12 F_{\mu\nu}\mathcal F^{\mu\nu} = \int_V dx^\alpha (\partial_\mu A_\nu)\mathcal F^{\mu\nu} = \int_V dx^\alpha \partial_\mu (A_\nu\mathcal F^{\mu\nu}) = \int_{\partial V} dx^\mu\ A_\nu \mathcal F^{\mu\nu} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, that cannot be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have lost a factor of two. Where is it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="188" LastActivityDate="2011-04-04T18:08:06.873" Title="I lost a factor of two in the electromagnetic field tensor" Tags="&lt;electromagnetism&gt;&lt;differential-geometry&gt;&lt;tensor-calculus&gt;" CommentCount="7" />
  <row Id="8046" PostTypeId="2" ParentId="8043" CreationDate="2011-04-04T18:17:31.143" Score="3" ViewCount="" Body="&lt;p&gt;The difference between Rindler's wording and the MTW wording is just a difference in the choice of coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thomas precession in STR&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, what is the Thomas precession? It is a special relativistic effect so the original derivation of the Thomas precession only applies in flat spacetimes. In other words, Rindler's application of the Thomas precession in the context of general relativity requires one to specify how the flat special relativistic spacetime is identified with, or embedded into, the curved spacetime in general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Thomas precession is a change of the angular momentum that a gyroscope undergoes when it is attached to another object whose velocity is changing and going along a curve in the space of velocities. Why does it occur?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, in special relativity, the velocity is a vector $u^\mu$ normalized so that $u^\mu u_\mu=1$. The space of such vectors is a two-part hyperboloid in a Minkowski space (its intrinsic signature is purely spacelike). Now, the angular momentum of a gyroscope moving together with an object is given by an antisymmetric tensor $J_{\mu\nu}$ that satisfies $J_{\mu\nu}u^\nu=0$. Now, if you try to parallel transport this tensor $J_{\mu\nu}$ along a path inside the hyperboloid, it will not return to itself because the hyperboloid is a curved submanifold. Instead, it will rotate by an angle around an axis (one that depends on the path in the space of velocities) - and this is what we mean by the Thomas precession.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;General relativity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a gyroscope attached to a satellite in a gravitational field described by general relativity, we may describe the vicinity of the world line of the satellite as a piece of flat Minkowski spacetime. After a whole orbit, we return to the same place in space. However, the orbit won't be quite periodic in our &quot;flat, thin, and long Minkowski cylinder&quot; surrounding the world line. Instead, we will induce both a rotation of the velocity space as well as a rotation of the ordinary position space, caused by the actual curvature of the space. Rindler probably calculates the full effect by summing these two contributions - from the monodromy in the uniformly curved velocity space and from the monodromy in the actual spacetime curved by the presence of matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am pretty confident that Rindler did his job correctly and avoided any sign errors. Gravity Probe B has confirmed the result, after all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MTW are arguably able to calculate the total result correctly, too. But they organize their calculation differently, attributing the whole effect to the curvature of space. Effectively, their coordinates for the &quot;cylinder surrounding the satellite's world line&quot; differ by a &quot;twist&quot; (a time-dependent rotation) from Rindler's cylinder - the two groups of relativists use different coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The MTW proposition seems to &lt;em&gt;directly&lt;/em&gt; contradict Rindler's calculational strategy and I think that the MTW, in the very satellite context that is relevant and with the Rindler's choice of coordinates, is incorrect. A statement that would be true and similar to the MTW proposition is that if the satellite were freely moving in space and had &quot;no intrinsic rotation&quot; relatively to a chosen system of coordinates, then the velocity could be viewed as a &quot;constant&quot; and the Thomas effect would be exactly zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the coordinates chosen by Rindler contradict the assumption because the satellite itself - not just the gyroscope - is rotating in this system of coordinates (the gyroscope is also rotating, and differently). So the velocity itself is non-constant even in the flattened cylinder surrounding the world line, and the Thomas precession contribution is nonzero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the existence or absence of the Thomas precession depends on whether or not we consider the velocity of the satellite to be &quot;constant&quot; and this question depends on whether or not our &quot;natural&quot; (there is no natural one!) coordinate system is rotating relatively to other people's natural systems. If there is no Thomas precession in one system, it's because we embedded the special relativistic spacetime in such a way that the velocities stay &quot;constant&quot;. However, if we choose a &quot;twisted&quot; coordinate system that is rotating with respect to the first one, the velocities will be non-constant and the precession will receive contributions from the Thomas precession.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By changing the coordinate systems, arbitrary parts of the overall precession after one period - that everyone has to agree with - may be moved from the Thomas precession contribution to the curvature-of-space contribution.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-04T18:23:09.550" LastActivityDate="2011-04-04T18:23:09.550" CommentCount="2" />
  <row Id="8047" PostTypeId="2" ParentId="8036" CreationDate="2011-04-04T18:40:45.927" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Do magnets lose their magnetism?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This question still may be not the thing You mean. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A permanent magnet will loose its macroscopic  magnetisation, &#xA;but not it magnetism. :=)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read  Ferromagnetism in Wikipedia. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-04T18:40:45.927" CommentCount="1" />
  <row Id="8048" PostTypeId="2" ParentId="8027" CreationDate="2011-04-04T19:18:46.513" Score="0" ViewCount="" Body="&lt;p&gt;Momentum and energy are both different depending on what I compare the motion of an object with. If I'm in a train, I have no momentum or kinetic energy relative to the train. Relative to the fields outside the train, however, I have lots of momentum and energy. If I jump off the train, I will come to a stop relative to the fields, so the momentum and energy relative to the fields have to go somewhere. On the other hand, I will then be moving relative to the train, so the momentum and energy to make that happen also have to come from somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between the momentum and the energy comes from the fact that the forces that change my speed have to act both for a certain amount of time &lt;em&gt;and&lt;/em&gt; over a certain distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the force that slows me down is constant, and that the force doesn't make me spin or break me up into pieces. This is the sort of wild idealization that gets Physics a bad name with 16 year olds, but it's a first approximation from which we can go on to a second approximation that's better, and no-one has yet thought up a better first approximation. If the train travels twice as fast, the force has to act for &lt;em&gt;twice as much time&lt;/em&gt; to bring me to stop, that's the change of momentum, &lt;em&gt;but&lt;/em&gt; the force has to act for &lt;em&gt;four times the distance&lt;/em&gt; to bring me to a stop, that's the change of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gets very tricky, because someone in an airplane that's moving really fast sees the force acting for the same amount of time as someone standing in the field sees the force acting for, but the person in the airplane sees the force acting for much more distance, because when the force started I was right next to the airplane, say, but when the force ended I was a &lt;em&gt;long&lt;/em&gt; way behind. So, the change of energy from the point of view of the person in the airplane was much bigger than someone in the fields thinks it was, even though everyone agrees that the change of momentum was the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To switch to a different analogy, the energy is important because it determines how far it takes me to stop a car using the brakes, so it determines whether I hit the brick wall that I suddenly see in front of me. The momentum determines how much time it takes to come to a stop, but I can't immediately think of a really graphic situation when that's important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can construct different situations endlessly. It can be done in equations, of course, but you'll have to decide whether that's appropriate. I'll be interested if there's anything about this Answer that you think could be made clearer. It certainly isn't complete. &lt;em&gt;Welcome&lt;/em&gt; from an Englishman in the USA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Of course overnight I realize that I mention &lt;em&gt;conservation&lt;/em&gt; not once. From the point of view of the above it's enough to note that both can be understood to be because of Newton's third law, which is, from Wikipedia, &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion&quot; rel=&quot;nofollow&quot;&gt;&quot;The mutual forces of action and reaction between two bodies are equal, opposite and collinear&quot;&lt;/a&gt;. As a result, we can say that the energy added to an object is taken away from the other object, and the same for momentum. The independence of the two conservation laws is essentially just because the two quantities are independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've decided to add a few simplified equations,&#xA;$$Force = Mass \times Acceleration,$$&#xA;$$Kinetic\ Energy = \mathsf{The\ Sum\ Of}\ The\ Forces\ Applied \times The\ Distance\ each\ Force\ is\ Applied\ For,$$&#xA;$$Momentum = \mathsf{The\ Sum\ Of}\ The\ Forces\ Applied \times The\ Length\ of\ Time\ each\ Force\ is\ Applied\ For,$$&#xA;or, as vector equations, almost certainly beyond what you need,&#xA;$$\underline{F}=m\underline{a},&#xD;&#xA;  \qquad E=\int \underline{F}(t,\underline{s}(t))\cdot\frac{d\underline{s}(t)}{dt}dt,&#xD;&#xA;  \qquad \underline{P}=\underline{F}(t,\underline{s}(t))dt.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really, this stuff should be left to specialist educators, the best of whom will take time not only to create new ways to explain ideas, but also to study how well different strategies of explanation work for different kinds of student, but I've always been interested occasionally to put myself in this mindset. It's always humbling to discover how much creativity is needed to do it well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As often, trawling around in Wikipedia, starting from the page on Newton's laws that I mention above, will render up some gems amongst the too-much-information for the purposes of your Question. I particularly like the tail-end comment that &quot;Conservation of energy was discovered nearly two centuries after Newton's lifetime, the long delay occurring because of the difficulty in understanding the role of microscopic and invisible forms of energy such as heat and infra-red light.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-04-05T12:28:07.323" LastActivityDate="2011-04-05T12:28:07.323" CommentCount="2" />
  <row Id="8049" PostTypeId="1" AcceptedAnswerId="8052" CreationDate="2011-04-04T19:23:25.310" Score="15" ViewCount="450" Body="&lt;p&gt;Suppose Stanford Research Systems starts selling a two-level atom factory. Your grad student pushes a button, and bang, he gets a two level atom. Half the time the atom is produced in the ground state, and half the time the atom is produced in the excited state, but other than that you get the exact same atom every time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;National Instruments sells a cheap knockoff two-level atom factory that looks the same, but doesn't have the same output. In the NI machine, if your grad student pushes a button, he gets the same two-level atom the SRS machine makes, but the atom is always in a 50/50 superposition of ground and excited states with a random relative phase between the two states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;random relative phase between the two states&quot; of the NI knockoff varies from atom to atom, and is unknown to the device's user.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these two machines distinguishable? What experiment would you do to distinguish their outputs?&lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-05T10:17:04.883" LastActivityDate="2011-08-14T21:42:21.050" Title="Are these two quantum systems distinguishable?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="6" CommentCount="4" FavoriteCount="4" />
  <row Id="8052" PostTypeId="2" ParentId="8049" CreationDate="2011-04-04T20:04:22.767" Score="17" ViewCount="" Body="&lt;p&gt;These systems are &lt;em&gt;not&lt;/em&gt; distiguishable. The average density matrix is the same, and the probability distribution obtained by performing any measurement depends only on the average density matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first system, the density matrix is&#xA;$$\frac{1}{2} \left[\left(\begin{array}{cc}1&amp;amp;0\cr 0&amp;amp;0\end{array}\right)+ \left(\begin{array}{cc}0&amp;amp;0\cr 0&amp;amp;1\end{array}\right)\right].$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second system, the density matrix is&#xA;$$\frac{1}{2\pi} \int_\theta \frac{1}{2}\left(\begin{array}{cc}1&amp;amp;e^{-i\theta}\cr e^{i \theta}&amp;amp;1\end{array}\right) d \theta.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easily checked that these are the same.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditorDisplayName="" LastEditDate="2011-04-04T20:09:45.363" LastActivityDate="2011-04-04T20:09:45.363" CommentCount="2" />
  <row Id="8054" PostTypeId="2" ParentId="8049" CreationDate="2011-04-04T20:20:31.273" Score="9" ViewCount="" Body="&lt;p&gt;Case 1: $\frac{1}{2}\left[\left|0\right&amp;gt;\left&amp;lt;0\right|+\left|1\right&amp;gt;\left&amp;lt;1\right|\right]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2, average over phases $0$ to $2\pi$: $$\frac{\int\left[(\left|0\right&amp;gt;+e^{i\theta}\left|1\right&amp;gt;)&#xD;&#xA;                         (\left&amp;lt;0\right|+e^{-i\theta}\left&amp;lt;1\right|)\right]d\theta}&#xD;&#xA;          {\int\left[(\left&amp;lt;0\right|+e^{-i\theta}\left&amp;lt;1\right|)&#xD;&#xA;                         (\left|0\right&amp;gt;+e^{i\theta}\left|1\right&amp;gt;)\right]d\theta}.$$&#xA;The cross terms average to zero because $\int\limits_0^{2\pi} e^{i\theta}d\theta=0$, so it's the same density matrix. If this is really what the different manufacturers deliver, it's not a cheap knock-off.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-04-04T20:20:31.273" CommentCount="6" />
  <row Id="8055" PostTypeId="2" ParentId="8020" CreationDate="2011-04-04T22:20:16.433" Score="0" ViewCount="" Body="&lt;p&gt;It could go either way. If you are heating the liquid to the boiling point, then the foam will not limit boiling (unless it raises the pressure), but will limit convection/advection of air near the surface. Note that latent heat of water vapor is not the only method of heat loss from your pot. If air advects/convects over the surface, you are also heating air molecules. Also some heat is being lost by the surface via thermal radiation (probably roughly a kilowatt per meter squared). So the bubbles provide insulation, so that the heat loses other than into latent heat of water vapor are reduced..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, if it is not actually boiling, but the temperature is controlled to be some value below boiling, then it loses water via evaporation, and that requires fluid to flow to and away from the surface, and the foam would seriously inhibit that.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-04-04T22:20:16.433" CommentCount="1" />
  <row Id="8056" PostTypeId="2" ParentId="8034" CreationDate="2011-04-04T22:44:39.097" Score="7" ViewCount="" Body="&lt;p&gt;This question involving &quot;randomness&quot; and Quantum Mechanics introduces some subtleties. Firstly we have the definition of &quot;randomness&quot; to consider. It turns out that there are various ways to define this term: the two I shall consider here are (the digit sequence of a) normal number (mentioned in the Question) and Martin-Lof Randomness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As explained in the Wikipedia article a &quot;Normal Number&quot; is &quot;Finite State Machine Random&quot;. So it looks random to a FSM. However such numbers can be computable by a Turing Machine (as the link shows that Turing proved).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Martin-Lof random sequence is based on the familiar notion of incompressibility and cannot be computed by a Turing Machine. Phrased alternatively because it is an infinite sequence it will not have a finite compression onto a Turing Machine - so there can be no program for it (which has to be finite).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logical link between the two is that every Martin-Lof random sequence is normal (but not conversely - as shown by Turing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that every finite sequence can be generated by an algorithm (and also it will be the solution of a polynomial). The reason why the random sequences work is that although the first N digits can be replicated via a program P, a different program (in general) is required for the first N+1 digits of the sequence ie P will fail to &quot;predict&quot; N+1. To get the entire sequence requires an infinite series of programs so we are back where we started with the random sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we have some definitions available we can examine connection to Physics. The problem in a experimental based theory is that we only ever have a finite amount of data. Thus the claim that a given series is &quot;random&quot; is strictly not empirically provable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this puts the Copenhagen claim that &quot;QM sequences are random&quot; into an awkward semi-scientific status. Such claims cannot be proved experimentally, yet Copenhagen asserts this. So where is the proof? Indeed what constitutes a proof? Also we have seen several definitions of randomness (there are more) - so which type of randomness does QM have exactly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To return to specific points in the OP question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there any way to differentiate if they came from a Truly Random or from a formula/algorithm ? how?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, because one only ever has a finite amount of stream data to analyse, which can always be explained algorithmically (as discussed above).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if there is no way to decide this, then, I can't find any basis, to keep denying that behind the &quot;truly random&quot; of quantum mechanics can be a hidden algorithm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &quot;truly random&quot; of quantum mechanics &lt;em&gt;might&lt;/em&gt; be Martin-Lof randomness, for which there is no algorithm; however if it is &lt;em&gt;really&lt;/em&gt; normal number randomness then there &lt;em&gt;might&lt;/em&gt; be an algorithm. I suspect that most physicists take the view that QM is as &quot;random as it gets&quot;, hence would prefer the Martin-Lof option.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I know I am talking about the possibility of a &quot;hidden variables&quot; theory, but I can't find any other explanation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The link between algorithmic underlying structure and &quot;hidden variables&quot; would appear to be close. If its a non-algorithmic type of randomness then we have to decide whether it belongs to one of &quot;oracle classes&quot; associated with Martin-Lof randomness, and what that would mean in terms of &quot;hidden variables&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some readers might recall that in his book &quot;The Emperor's New Mind&quot; in 1989 Roger Penrose proposed that aspects of quantum mechanics were &quot;non-computable&quot;. Although that argument was not formulated as I have here, it is consistent (I believe) with the idea of Martin-Lof randomness too.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-04-04T22:44:39.097" CommentCount="2" />
  <row Id="8058" PostTypeId="2" ParentId="8049" CreationDate="2011-04-05T01:28:45.730" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Reread it some hours later and found my error. I figured I was doing something wrong. I was applying operations out of order when calculating the conditional probability.  It is 1/2 in each case.  I'll leave the commentary untouched.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;I think the answer is Yes, or at least I'm not entirely convinced the answer is no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will provide an example below, but I don't find it very convincing since I just ad-hoc approached it, and don't have a nice &quot;overarching&quot; principle to take away from this.  Basically, consider this more as a comment to get discussion going, than a full fledged answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other answers show that the expectation value of measuring the system to be in a particular state is the same.  Basically the density matrix of the ensemble is the same, but the density matrix of the first machine only has two possible outputs while the second has an infinite number.  Focusing immediately on the ensemble average seems to be throwing away any possibility we have of distinguishing them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;Here's an attempt at distinguishing them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Machine 1 possible output, only pure states&lt;br&gt;&#xA;$|0\rangle$&lt;br&gt;&#xA;$|1\rangle$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Machine 2 possible output, any state&lt;br&gt;&#xA;$\frac{1}{\sqrt{2}}(|0\rangle + p |1\rangle)$&lt;br&gt;&#xA;where $p = e^{i\theta}$ with $0 \le \theta &amp;lt; 2\pi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take some other qubit B (it doesn't matter here physically what it is) of prepared state $\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$ to get the product states:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;machine 1&lt;br&gt;&#xA;$\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)|0\rangle = \frac{1}{\sqrt{2}}(|00\rangle +|10\rangle)$&lt;br&gt;&#xA;$\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)|1\rangle = \frac{1}{\sqrt{2}}(|01\rangle +|11\rangle)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;machine 2&lt;br&gt;&#xA;$\frac{1}{2}(|0\rangle+|1\rangle)(|0\rangle + p |1\rangle) = \frac{1}{2}(|00\rangle+p|01\rangle + |10\rangle + p |11\rangle)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's introduce an interaction which can cause some interference:&lt;br&gt;&#xA;$|00\rangle \rightarrow |00\rangle$&lt;br&gt;&#xA;$|01\rangle \rightarrow \frac{1}{\sqrt{2}}(|01\rangle + |10\rangle)$&lt;br&gt;&#xA;$|10\rangle \rightarrow \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)$&lt;br&gt;&#xA;$|11\rangle \rightarrow |11\rangle$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;now we have&lt;br&gt;&#xA;machine 1&lt;br&gt;&#xA;$\frac{1}{\sqrt{2}}(|00\rangle +|10\rangle) \rightarrow \frac{1}{\sqrt{2}}|00\rangle +  \frac{1}{2}(|01\rangle-|10\rangle)$&lt;br&gt;&#xA;$\frac{1}{\sqrt{2}}(|01\rangle +|11\rangle) \rightarrow \frac{1}{\sqrt{2}}|11\rangle +  \frac{1}{2}(|01\rangle+|10\rangle)$&lt;br&gt;&#xA;machine 2&lt;br&gt;&#xA;$\frac{1}{2}(|00\rangle+p|01\rangle + |10\rangle + p |11\rangle) \rightarrow \frac{1}{2}(|00\rangle + p\frac{1}{\sqrt{2}}(|01\rangle+|10\rangle) + \frac{1}{\sqrt{2}}(|01\rangle-|10\rangle) + p |11\rangle)$&lt;br&gt;&#xA;$ \ \ \ = \frac{1}{2}(|00\rangle + (p+1)\frac{1}{\sqrt{2}}|01\rangle+(p-1)\frac{1}{\sqrt{2}}|10\rangle + p |11\rangle)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's do two measurements.  First measure the state of B to be 0 or 1, then measure the sate of the atom to be 0 or 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conditional probability on the ensemble:&lt;br&gt;&#xA;Given that we find B in state 1, what is the probability of finding the atom in state 0?&lt;br&gt;&#xA;machine 1&lt;br&gt;&#xA;(1/2) x 1 + (1/2) x (1/3) = 4/6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;machine 2&lt;br&gt;&#xA;$\frac{\frac{1}{2}(p-1)^2}{\frac{1}{2}(p-1)^2 + p^2} = \frac{\frac{1}{2}(2 - 2\cos\theta)}{\frac{1}{2}(2 - 2\cos\theta) + 1} = \frac{1 - \cos\theta}{2 - \cos\theta}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now averaging over $\theta$&lt;br&gt;&#xA;$\mathrm{Prob} = \frac{1}{2\pi}\int_0^{2\pi}\frac{1 - \cos\theta}{2 - \cos\theta} d\theta = 1 - \frac{1}{\sqrt{3}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;Now, it is quite possible I've made a mistake here.  But my main point is that the other answers seem to be throwing away the useful information to obtain solely an average of the initial output states.  As the answers stand now, they do not mathematically convince me that we can never obtain an effect by adding interactions and multiple measurements with conditional probability or maybe 'weak' measurements, since individually the states have much different density matrices.  Hopefully I didn't make a mistake above, but even if I did, I'd still very much like to hear more in the other answers beyond what is currently written.  This is a fascinating question, so I'm quite interested in discussing this further.&lt;/p&gt;&#xA;" OwnerUserId="2652" LastEditorUserId="2652" LastEditorDisplayName="" LastEditDate="2011-04-05T08:11:46.973" LastActivityDate="2011-04-05T08:11:46.973" CommentCount="4" />
  <row Id="8061" PostTypeId="2" ParentId="7963" CreationDate="2011-04-05T03:37:59.077" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.virginia.edu/ep/SurfaceScience/xps.html&quot; rel=&quot;nofollow&quot;&gt;This page&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, the surface Madelung constant computed in the (100) plane is 96 % of the bulk value for crystals with a sodium chloride structure. In the (110) planes in crystals with sodium chloride and cesium chloride structures, the surface Madelung constant is 86% and 90% respectively, of the bulk value.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seems high, but I don't have access to the cited reference (&lt;a href=&quot;http://jcp.aip.org/resource/1/jcpsa6/v57/i10/p4446_s1?isAuthorized=no&quot; rel=&quot;nofollow&quot;&gt;P. H. Citrin and T. D. Thomas, J. Chem. Phys. 57, 4446 (1972)&lt;/a&gt;) to check.&lt;/p&gt;&#xA;" OwnerUserId="1656" LastEditorUserId="1656" LastEditorDisplayName="" LastEditDate="2011-04-05T03:45:59.130" LastActivityDate="2011-04-05T03:45:59.130" CommentCount="2" />
  <row Id="8062" PostTypeId="1" AcceptedAnswerId="8147" CreationDate="2011-04-05T04:25:39.527" Score="8" ViewCount="739" Body="&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the wave function complex? I've collected some layman explanations but they are incomplete and unsatisfactory. However in the book by Merzbacher in the initial few pages he provides an explanation that I need some help with: that the de broglie wavelength and the wavelength of an elastic wave do not show similar properties under a galilean transformation. He basically says that both are equivalent under a Guage transform and also, separately by lorentz transforms. This, accompanied with the observation that $\psi$ is not observable, so there is no &quot;reason for it being real&quot;. Can someone give me an intuitive prelude by what is a Guage transform and why does it give the same result as a Lorentz tranformation in a non-relativistic setting? And eventually how in this &quot;grand scheme&quot; the complex nature of the wave function becomes evident.. in a way that a dummy like me can understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A wavefunction can be thought of as a scalar field (has a scalar value in every point ($r,t$) given by $\psi:\mathbb{R^3}\times \mathbb{R}\rightarrow \mathbb{C}$ and also as a ray in Hilbert space (a vector). How are these two perspectives the same (this is possibly something elementary that I am missing out, or getting confused by definitions and terminology, if that is the case I am desperate for help ;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way I have thought about the above question is that the wave function can be equivalently written in $\psi:\mathbb{R^3}\times \mathbb{R}\rightarrow \mathbb{R}^2 $ i.e, Since a wave function is complex, the Schrodinger equation could in principle be be written equivalently as coupled differential equations in two real functions which staisfy the Cauchy Reimann conditions. ie, if $$\psi(x,t) = u(x,t) + i v(x,t)$$ and $u_x=v_t$ ; $u_t = -v_x$ and we get $$\hbar \partial_t u  = -\frac{\hbar^2}{2m} \partial_x^2v + V v$$ $$\hbar \partial_t v  = \frac{\hbar^2}{2m} \partial_x^2u - V u$$&#xA;(..in 1-D) If this is correct what are the interpretations of the $u,v$.. and why isn't it useful. (I am assuming that physical problems always have an analytic $\psi(r,t)$).&lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-06T06:58:44.037" LastActivityDate="2011-04-08T03:50:15.510" Title="About the complex nature of the wave function?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="2" />
  <row Id="8063" PostTypeId="2" ParentId="7981" CreationDate="2011-04-05T04:43:45.003" Score="4" ViewCount="" Body="&lt;p&gt;It depends on what kind of radio telescope you're talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're talking about a single dish, with say, a horn feed at the focus, it behaves like a regular 'dish antenna' by focussing radio waves to the focus where it is amplified, processed in some manner (such as downconversion), digitised, and so on. A single dish can also have a feed array at the focus to act analogous to a camera sensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're talking about a radio telescope &lt;em&gt;array&lt;/em&gt;, it can operate in one of two ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first method, the antennas in the array are pointed at the source, and receive signals that are at different phases due to the path difference associated with the relative locations of the antennas and the source. These phase differences are corrected for, the signals are combined, and then processed as before. This is usually employed when observing a point source (such as a pulsar) with a telescope array.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second way is as an interferometer. This method is used to create radio images of (usually) extended sources such as radio galaxies. In this method, the phase differences are not removed, but utilised along with the rotation of the Earth to create a Fourier-synthesised image. The measured signal from each antenna is correlated with those from all other antennas to get what is called the 'visibility'. Taking the Fourier Transform of the measured visibilities gives you the brightness distribution in the sky (a.k.a. the image). Following the behaviour of the Fourier Transform, long baselines (antennas that are farther apart) measure small spatial variations (high spatial frequency) in the brightness distribution, while short baselines measure large spatial variations (low spatial frequency). In other words, it neither behaves like a camera with a CCD sensor, nor does it do raster-scanning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 'Antennas &amp;amp; Radiometers' chapter of this course should tell you more in detail:&#xA;&lt;a href=&quot;http://www.cv.nrao.edu/course/astr534/ERA.shtml&quot; rel=&quot;nofollow&quot;&gt;http://www.cv.nrao.edu/course/astr534/ERA.shtml&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These lectures from the Sixth NRAO/NMIMT Synthesis Imaging Summer School might also be helpful:&#xA;&lt;a href=&quot;http://adsabs.harvard.edu/cgi-bin/nph-abs_connect?version=1&amp;amp;warnings=YES&amp;amp;partial_bibcd=YES&amp;amp;sort=BIBCODE&amp;amp;db_key=ALL&amp;amp;bibstem=Synthesis+Imaging+in+Radio+Astronomy+II&amp;amp;year=&amp;amp;volume=&amp;amp;page=&amp;amp;nr_to_return=100&amp;amp;start_nr=1&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/cgi-bin/nph-abs_connect?version=1&amp;amp;warnings=YES&amp;amp;partial_bibcd=YES&amp;amp;sort=BIBCODE&amp;amp;db_key=ALL&amp;amp;bibstem=Synthesis+Imaging+in+Radio+Astronomy+II&amp;amp;year=&amp;amp;volume=&amp;amp;page=&amp;amp;nr_to_return=100&amp;amp;start_nr=1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2931" LastEditorUserId="2931" LastEditorDisplayName="" LastEditDate="2011-04-05T16:39:50.373" LastActivityDate="2011-04-05T16:39:50.373" CommentCount="2" />
  <row Id="8064" PostTypeId="2" ParentId="8034" CreationDate="2011-04-05T05:06:25.280" Score="0" ViewCount="" Body="&lt;p&gt;An individual measurement from a quantum mechanical wave function describing the system is random within the probability distribution described by the whole function. A sufficient number of measurements will show the probability distribution and not randomness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have a Psi that has expectation +1 and -1. Randomness means 50% of the time the measurement will find 1 and 50% -1. One can make an algorithm to mimic this. The only way the algorithm could have a connection with measurements is if, by some method, the experimenter could control it, and get either 1 or -1 at will. Immediately one gets to extra interactions, outside the solution Psi, by the word &quot;control&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There could be a million algorithms. The existence of algorithms &lt;strong&gt;is not sufficient&lt;/strong&gt; to demand the existence of hidden variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hidden variables would be necessary&lt;/strong&gt; if the data did not agree with the 50/50 calculated expectation value for this Psi and data in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If hidden variables exist, there will be a mathematical description for them, which you could call an algorithm, that might even  give 50/50 in this simple case, and something spectacularly different for scattering, for example, that had necessitated their supposition.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-04-05T05:54:18.547" LastActivityDate="2011-04-05T05:54:18.547" />
  <row Id="8065" PostTypeId="2" ParentId="8062" CreationDate="2011-04-05T05:27:27.627" Score="8" ViewCount="" Body="&lt;p&gt;Among other things, the OP reprinted a page of a textbook, asking what &quot;it is all about&quot;. I think it is impossible to answer this kind of questions because what is the OP's problem all about is totally undetermined, and the people who offer their answers could be writing their own textbooks, with no results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wave function in quantum mechanics has to be complex because the operators satisfy things like &#xA;$$ [x,p] = xp-px = i\hbar.$$&#xA;It's the commutator defining the uncertainty principle. Because the left hand side is anti-Hermitian,&#xA;$$ (xp-px)^\dagger = p^\dagger x^\dagger - x^\dagger p^\dagger = (px-xp) = -(xp-px),$$&#xA;it follows that if it is a $c$-number, its eigenvalues have to be pure imaginary. It follows that either $x$ or $p$ or both have to have some non-real matrix elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, Schrödinger's equation&#xA;$$i\hbar\,\, {\rm d/d}t |\psi\rangle = H |\psi\rangle$$&#xA;has a factor of $i$ in it. The equivalent $i$ appears in Heisenberg's equations for the operators and in the $\exp(iS/\hbar)$ integrand of Feynman's path integral. So the amplitudes inevitably have to come out as complex numbers. That's also related to the fact that eigenstates of energy and momenta etc. have the dependence on space or time etc.&#xA;$$\exp(Et/i\hbar)$$&#xA;which is complex. A cosine wouldn't be enough because a cosine is an even function (and the sine is an odd function) so it couldn't distringuish the sign of the energy. Of course, the appearance of $i$ in the phase is related to the commutator at the beginning of this answer. See also&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2010/08/why-complex-numbers-are-fundamental-in.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/08/why-complex-numbers-are-fundamental-in.html&lt;/a&gt;&lt;br&gt;&#xA;  &lt;strong&gt;Why complex numbers are fundamental in physics&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Concerning the second question, in physics jargon, we choose to emphasize that a wave function is &lt;em&gt;not&lt;/em&gt; a scalar field. A wave function is not an observable at all while a field is. Classically, the fields evolve deterministically and can be measured by one measurement - but the wave function cannot be measured. Quantum fields are operators - but the wave function is not. Moreover, the mathematical similarity of a wave function to a scalar field in 3+1 dimensions only holds for the description of one spinless particle, not for more complicated systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the last question, it is not useful to decompose complex numbers into real and imaginary parts exactly because &quot;a complex number&quot; is one number and not two numbers. In particular, if we multiply a wave function by a complex phase $\exp(i\phi)$, which is only possible if we allow the wave functions to be complex and we use the multiplication of complex numbers, physics doesn't change at all. It's the whole point of complex numbers that we deal with them as with a single entity.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-05T05:27:27.627" CommentCount="3" />
  <row Id="8066" PostTypeId="1" AcceptedAnswerId="8067" CreationDate="2011-04-05T05:53:16.507" Score="3" ViewCount="129" Body="&lt;p&gt;If you rubbed a balloon with a towel, where would the electrons go: the balloon or the towel? Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing the electrons would go to the object with a larger mass, but it's just a guess. :)&lt;/p&gt;&#xA;" OwnerUserId="125" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-05T05:57:26.723" LastActivityDate="2011-04-05T15:35:52.913" Title="Balloon rubbing; where do the electrons go?" Tags="&lt;electricity&gt;&lt;electrostatics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8067" PostTypeId="2" ParentId="8066" CreationDate="2011-04-05T06:20:17.153" Score="5" ViewCount="" Body="&lt;p&gt;This article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Triboelectric_series&quot; rel=&quot;nofollow&quot;&gt;triboelectricity&lt;/a&gt; gives a triboelectric  series, listing materials according to whether they will expel electrons or acquire them , when rubbed or in contact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Towel is not in there :) but cotton is, with 0 but it is more positive than rubber. So the answer is that the electrons will move to the balloon, if you manage to get cotton to give up its electrons :) . Did you do the experiment? A woolen towel would be much better. Maybe it was your hand that gave up the electrons after all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what wikipedia has to say on the causes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After coming into contact, a chemical&#xA;  bond is formed between some parts of&#xA;  the two surfaces, called adhesion, and&#xA;  charges move from one material to the&#xA;  other to equalize their&#xA;  electrochemical potential. This is&#xA;  what creates the net charge imbalance&#xA;  between the objects. When separated,&#xA;  some of the bonded atoms have a&#xA;  tendency to keep extra electrons, and&#xA;  some a tendency to give them away,&#xA;  though the imbalance will be partially&#xA;  destroyed by tunneling or electrical&#xA;  breakdown (usually corona discharge).&#xA;  In addition, some materials may&#xA;  exchange ions of differing mobility,&#xA;  or exchange charged fragments of&#xA;  larger molecules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hence the need of a table.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="125" LastEditorDisplayName="" LastEditDate="2011-04-05T07:58:24.513" LastActivityDate="2011-04-05T07:58:24.513" CommentCount="3" />
  <row Id="8068" PostTypeId="1" CreationDate="2011-04-05T06:40:30.753" Score="4" ViewCount="93" Body="&lt;p&gt;Can someone point to a good review article covering contemporary developments in big bang nucleosynthesis, beyond what's covered in Kolb and Turner's &lt;em&gt;The Early Universe&lt;/em&gt;? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2643" LastActivityDate="2011-05-07T04:43:57.810" Title="Looking for review articles on big bang nucleosynthesis" Tags="&lt;big-bang&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="8069" PostTypeId="2" ParentId="7985" CreationDate="2011-04-05T06:51:08.727" Score="1" ViewCount="" Body="&lt;p&gt;Your question does lack a few specifics, but it's not at all incomprehensible?  Common sense indicates that your initial premise must come from a basic example taught in public schools.  That is that if you launch a rocket or spacecraft into outer space, it will continue to accelerate faster and faster. But unlike an orbiting planet which is defined by conservation of angular momentum, a rocket has a propulsion system with certain differing variables in each application.  Although the immediate answer is no. I would recommend rephrasing your question after reviewing angular moment&lt;/p&gt;&#xA;" OwnerUserId="2860" LastActivityDate="2011-04-05T06:51:08.727" />
  <row Id="8071" PostTypeId="2" ParentId="2901" CreationDate="2011-04-05T07:08:24.567" Score="2" ViewCount="" Body="&lt;p&gt;A black hole at the center of the Milky Way is an observation that is dependent on the validity of the general theory of relativity (GR). An input into an equation of GR is one of the steps to determine the existence of a black hole (anywhere). If GR is invalid, no black hole may exist there. No black hole has been directly observed. For confirmation of this see the &lt;a href=&quot;http://chandra.harvard.edu/resources/faq/black_hole/bhole-10.html&quot; rel=&quot;nofollow&quot;&gt;Chandra X-ray Observatory FAQ&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2314" LastActivityDate="2011-04-05T07:08:24.567" CommentCount="1" />
  <row Id="8072" PostTypeId="1" AcceptedAnswerId="8073" CreationDate="2011-04-05T07:40:34.053" Score="1" ViewCount="100" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Brian_Cox_%28physicist%29&quot; rel=&quot;nofollow&quot;&gt;Prof Brian Cox&lt;/a&gt; mentioned on &lt;a href=&quot;http://en.wikipedia.org/wiki/Wonders_of_the_Universe&quot; rel=&quot;nofollow&quot;&gt;Wonders of the Universe&lt;/a&gt; when discussing chemical elements that Carbon 12 consists of 6 protons and 6 neutrons, he also mentioned that Helium consists of 2 protons and 2 neutrons. He then went on to say that by combining certain elements we can create others - can anyone explain what he means by this with a practical example?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I get myself some Carbon and some Helium and somehow combine them? :)&lt;/p&gt;&#xA;" OwnerUserId="2840" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-05T10:21:52.243" LastActivityDate="2011-04-05T10:21:52.243" Title="Combining chemical elements" Tags="&lt;nuclear-physics&gt;" AnswerCount="1" />
  <row Id="8073" PostTypeId="2" ParentId="8072" CreationDate="2011-04-05T08:01:56.643" Score="1" ViewCount="" Body="&lt;p&gt;Brian Cox was talking about &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_nucleosynthesis&quot; rel=&quot;nofollow&quot;&gt;stellar nucleosynthesis&lt;/a&gt;.  In theory you could combine the nucleus of a Helium atom (also called an &lt;em&gt;alpha particle&lt;/em&gt;) and the nucleus of a Carbon atom to produce an Oxygen nucleus and some energy in the form of gamma radiation&lt;br&gt;&#xA;$$ ^4_2He + ^{12}_6C \rightarrow ^{16}_8O +\gamma$$&#xA;This is an &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Alpha_process&quot; rel=&quot;nofollow&quot;&gt;alpha process&lt;/a&gt;&lt;/em&gt;, and does happen inside certain stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, you are unlikely to be able to do this personally, as you probably cannot get the nuclei moving fast enough or accurately enough to enable fusion.&lt;/p&gt;&#xA;" OwnerUserId="2714" LastActivityDate="2011-04-05T08:01:56.643" CommentCount="1" />
  <row Id="8074" PostTypeId="1" AcceptedAnswerId="8088" CreationDate="2011-04-05T08:13:37.690" Score="20" ViewCount="1257" Body="&lt;p&gt;I made a naive calculation of the height of Earth's equatorial bulge and found that it should be about 10km.  The true height is about 20km.  My question is: why is there this discrepancy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculation I did was to imagine placing a ball down on the spinning Earth.  Wherever I place it, it shouldn't move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational potential per unit mass of the ball is $hg$, with $h$ the height above the pole-to-center distance of the Earth (call that $R$) and $g$ gravitational acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity wants to pull the ball towards the poles, away from the bulge.  It is balanced by the centrifugal force, which has a potential $-\omega^2R^2\sin^2\theta/2$ per unit mass, with $\omega$ Earth's angular velocity and $\theta$ the angle from the north pole.  This comes taking what in an inertial frame would be the ball's kinetic energy and making it a potential in the accelerating frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ball doesn't move, this potential must be constant, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U = hg - \frac{(\omega R \sin\theta)^2}{2} = \textrm{const}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we might as well let the constant be zero and write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h = \frac{(\omega R \sin\theta)^2}{2g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Earth, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R = 6.4*10^6m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega = \frac{2\pi}{24\ \textrm{hours}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g = 9.8\ m/s^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives 10.8 km when $\theta = \pi/2$, so the equatorial bulge should be roughly that large.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Equatorial_bulge&quot; rel=&quot;nofollow&quot;&gt;According to Wikipedia&lt;/a&gt;, the Earth is 42.72 km wider in diameter at the equator than pole-to-pole, meaning the bulge is about twice as large as I expected.  (Wikipedia cites diameter; I estimated radius.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is the extra bulge coming from?  My simple calculation uses $g$ and $R$ as constants, but neither varies more than a percent or so.  It's true the Earth does not have uniform density, but it's not clear to me how this should affect the calculation, so long as the density distribution is still spherically-symmetric (or nearly so).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Wikipedia also includes &lt;a href=&quot;http://en.wikipedia.org/wiki/Equatorial_bulge#Mathematical_expression&quot; rel=&quot;nofollow&quot;&gt;an expression&lt;/a&gt;, without derivation, that agrees with mine.)&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-04-22T19:07:58.300" Title="Why is the Earth so fat?" Tags="&lt;rotational-dynamics&gt;&lt;geophysics&gt;&lt;newtonian-gravity&gt;&lt;centrifugal-force&gt;&lt;equilibrium&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="6" />
  <row Id="8076" PostTypeId="1" CreationDate="2011-04-05T08:58:44.783" Score="2" ViewCount="324" Body="&lt;p&gt;This is a question for which I've found it surprisingly hard to find a good answer. Biology texts talk mystically about the ATP-&gt;ADP reaction providing energy to power other reactions. I'd like to know some more details. Is this following roughly right ?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Each reaction in a cell has a specific enzyme.&lt;/li&gt;&#xA;&lt;li&gt;Each enzyme has binding sites for, say, two molecular species AND for an ATP molecule.&lt;/li&gt;&#xA;&lt;li&gt;When a reaction takes places, the two species bind to the enzyme, and a little later, an ATP molecule binds.&lt;/li&gt;&#xA;&lt;li&gt;For some reason (why ?), the ATP-&gt;ADP reaction is now energetically favourable, so the high-energy bond breaks.&lt;/li&gt;&#xA;&lt;li&gt;This releases electromagnetic energy at some characteristic frequency.&lt;/li&gt;&#xA;&lt;li&gt;Certain bonds in the enzyme have a resonant frequency that allow them to absorb this electromagnetic energy (the EM energy disturbs molecular dipoles ?).&lt;/li&gt;&#xA;&lt;li&gt;The 3D structure of the enzyme is disturbed (i.e. it bends) in such a way that the 2 molecular species are mechanically forced together, providing sufficient activation energy for the reaction in question.&lt;/li&gt;&#xA;&lt;li&gt;The newly formed species no longer binds nicely to the enzyme (why ?) so it detaches, as does the ADP, which also doesn't bind as nicely as ATP.&lt;/li&gt;&#xA;&lt;li&gt;The End.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, is that an accurate summary ? Anyone care to add some more physical details ? (like all the thermodynamics and quantum chemistry I have no idea about) &lt;/p&gt;&#xA;" OwnerUserId="2925" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-04-05T12:54:14.413" LastActivityDate="2011-08-27T08:56:49.400" Title="How does ATP transfer energy to a reaction?" Tags="&lt;energy&gt;" AnswerCount="6" CommentCount="9" />
  <row Id="8077" PostTypeId="1" CreationDate="2011-04-05T09:19:09.823" Score="5" ViewCount="66" Body="&lt;p&gt;I would like to test my hardware under vibration that can appear on a highway gantry.&#xA;If someone has a model of such vibration. i.e period and amplitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my lab I have a motor that can be regulated according to its RPM.&lt;/p&gt;&#xA;" OwnerUserId="2861" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-05T10:39:07.493" LastActivityDate="2011-04-05T17:31:08.030" Title="How to reproduce highway vibration?" Tags="&lt;experimental-physics&gt;&lt;waves&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8079" PostTypeId="2" ParentId="8077" CreationDate="2011-04-05T09:44:16.100" Score="2" ViewCount="" Body="&lt;p&gt;All (some of) the items fastened to such gantries must have been tested somewhere.&lt;br&gt;&#xA;For that reason there might exist a test lab to do that.&lt;br&gt;&#xA;I think your street authority wants to see tests from such a lab &#xA;anyway. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, climb on such a gantry equiped with some accelerometer or &#xA;&quot;structure-borne noise microphone&quot; and record the vibrations. &#xA;I think this will be nearly 100% of the natural vibration of that gantry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The designer of that gantry maybe knows that frequency. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-05T10:40:58.133" LastActivityDate="2011-04-05T10:40:58.133" />
  <row Id="8080" PostTypeId="2" ParentId="8076" CreationDate="2011-04-05T10:01:03.097" Score="1" ViewCount="" Body="&lt;p&gt;Biochemistry is not my profession, &#xA;but as far as I had to learn it for my Diploma, Ill try to &#xA;help You. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 to 3 seem ok. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4: ATP =&gt; ADP is always &quot;favourable&quot;. That it happens right when &#xA;the energy is needed, is due to catalysis. How this happens in molecular detail,&#xA;is unknown as far as I know. (Would earn You a Noble price, if You unraveled that :=)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 and 6 No. No such radiation is known (or possible) &#xA;To name one obstacle: You can't &quot;channel/concentrate electromagnetic&#xA;radiation in that way. Compare wavelength of the energy quanta related to &#xA;the dimensions of the molecules in that process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;7 and 8  Yes, that is what happens, to Your Why?: that is the final &#xA;work of that enzyme, as wonderful as all it does for a chemist. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general: only very few of the &quot;elementary steps&quot; in the acting of &#xA;enzymes are known/understood. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe some day there will be &quot;microscopes&quot; working in such dimensions &#xA;and fast enough :=(&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-05T10:31:02.410" LastActivityDate="2011-04-05T10:31:02.410" CommentCount="8" />
  <row Id="8081" PostTypeId="1" AcceptedAnswerId="8084" CreationDate="2011-04-05T10:02:18.400" Score="5" ViewCount="83" Body="&lt;p&gt;Say you have a chemical compound made up of one or more radioactive nuclei. If the nucleus decays, does the compound? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possible outcomes I can think of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;the compounds continues to exist if a bonding is still possible between the decay product and the rest of the original compound.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the compound just ceases to exist.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the decay product forms a new compound with some fraction of the remainder of the original one but discards some other part.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Generally, it will do whatever results in the least energetic state, but is there some kind of regularity to it?&lt;/p&gt;&#xA;" OwnerUserId="2151" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-05T10:14:03.387" LastActivityDate="2011-04-05T11:03:24.560" Title="What happens to chemical compunds that include radioactive nuclei, when those decay?" Tags="&lt;physical-chemistry&gt;&lt;radioactivity&gt;" AnswerCount="1" />
  <row Id="8082" PostTypeId="1" CreationDate="2011-04-05T10:03:06.537" Score="2" ViewCount="192" Body="&lt;p&gt;(little background: I'm trying to develop a small, quick 'n dirty static physics engine to determine whether a stacking of boxes is stable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have a 3D rigid box (with the bottom in the horizontal plane), resting on n points (at [xn, yn]), and we apply a downward force F at [xF, yF], how can I calculate the resulting forces Fn at these n points?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the system is in equilibrium then sum(Fn) should be equal to F.&#xA;However, I must also consider that the system might not be in equilibrium (for instance, if all xn &amp;lt; xF), so I probably can't use equilibrium equations. I'd still like to know the forces in that case, though, so that I can calculate the resulting torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an easy formula for this? &lt;/p&gt;&#xA;" OwnerUserId="2933" LastActivityDate="2011-05-07T10:44:03.573" Title="Distribution of forces" Tags="&lt;mechanics&gt;&lt;forces&gt;&lt;statics&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="8083" PostTypeId="2" ParentId="2901" CreationDate="2011-04-05T10:21:25.943" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;and powered by a black hole??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is Your idea of &quot;powered by&quot; ? Do You think about a galaxy like a car &#xA;who needs fuel even when going at constant speed? &#xA;A galaxy in toto does not need to be &quot;powered&quot; at all. The energy of the stars &#xA;revolution around the center was brought along by the gas which collected &#xA;to form the galaxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And last but not least: A black hole does not permit anything to evade by definition.&#xA;So , how should it power something? &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-05T10:21:25.943" CommentCount="2" />
  <row Id="8084" PostTypeId="2" ParentId="8081" CreationDate="2011-04-05T11:03:24.560" Score="5" ViewCount="" Body="&lt;p&gt;To deduce this, You have to specify the kind of decay and the nature of the &quot;compund&quot;&#xA;is it a crystal, a small molecule in gas phase, a organic material? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beta decay shifts the nucleus one position upward in PSE, &#xA;thus any &quot;compound&quot; will be transformed into a cation by &#xA;loss of an electron, and whre say a iodide Ion had been, &#xA;there will be an Xe atom. (which will not &quot;fit&quot; chemically of course) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There will be some recoil in this process, which can cause the nucleus &#xA;to leave its place.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electron will ionize everything along its path, those products &#xA;of ionisation can alter/ destroy the molecule (compound) where the&#xA;electron was emitted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar is the case of alpha, with a strong recoil and severly ionisation. &#xA;The decaying nucleus is shifted two &quot;down&quot; in PSE. The alpha particle will stay &#xA;in the crystal, if it is big enough. Think of Helium gassing out of &#xA;Pechblende when heated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Gamma, recoil will be less, ionisation is distributed along a long path &#xA;(maybe meters)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are special cases in crystals, when the recoil is taken not by the emitting &#xA;nucleus alone, but collectively ba the crystal lattice. (Mößbauer effect) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, radioactive decay is so energetic, that any chemical bonds/lattice forces &#xA;are broken. What happens then is very complicated and not to be answered by a simple scheme. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-05T11:03:24.560" />
  <row Id="8085" PostTypeId="2" ParentId="8062" CreationDate="2011-04-05T11:05:06.353" Score="-4" ViewCount="" Body="&lt;p&gt;EDIT add:&lt;br&gt;&#xA;My Answer is GA centric and after the comments I felt the need to say some words about the beauty of &lt;strong&gt;Geometric Algebra:&lt;/strong&gt;&lt;br&gt;&#xA;On 2nd page of Oersted Medal Lecture (link bellow):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;(3) GA Reduces “grad, div, curl and all&#xA;  that” to a single vector derivative&#xA;  that, among other things, combines the&#xA;  standard set of four Maxwell equations&#xA;  into a single equation and provides&#xA;  new methods to solve it.&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Geometry Algebra (GA) encompasses in a single framework for all this:&lt;br&gt;&#xA;Synthetic Geometry, Coordinate Geometry, Complex Variables, Quaternions, Vector Analysis, Matrix Algebra, Spinors, Tensors, Differential forms. It is one language for all physics.&lt;br&gt;&#xA;Probably Schrödinger, Dirac, Pauli, etc ... would have used GA if it existed at the time.&lt;br&gt;&#xA;To the Question: WHY is the wave function complex? This Answer is not helpful: because the wave function is complex (or has a &lt;strong&gt;i&lt;/strong&gt; on it). We have to try something different, not written in your book.&lt;br&gt;&#xA;In &lt;strong&gt;the abstracts I bolded the evidence that the papers are about the WHYs&lt;/strong&gt;. If someone begs a fish I'll try to give a fishing rod.&lt;br&gt;&#xA;I'm an old IT analyst who would be unemployed if I had not evolved. Physics is evolving too.&lt;br&gt;&#xA;end EDIT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently I've found the &lt;strong&gt;Geometric Algebra&lt;/strong&gt;, Grassman, Clifford, and David Hestenes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will not detail here the subject of the OP because each one of us need to follow paths, find new ideas and take time to read. I will only provide some paths with part of the abstracts:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://geocalc.clas.asu.edu/html/Overview.html&quot; rel=&quot;nofollow&quot;&gt;Overview of Geometric Algebra in Physics&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://geocalc.clas.asu.edu/pdf/OerstedMedalLecture.pdf&quot; rel=&quot;nofollow&quot;&gt;Oersted Medal Lecture 2002: Reforming the Mathematical Language of Physics&lt;/a&gt;  (a good start)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In this lecture Hestenes is arguing&#xA;  for a reform of the way in which&#xA;  mathematics is taught to physicists.&#xA;  He asserts that using Geometric&#xA;  Algebra will make it easier to&#xA;  understand the fundamentals of&#xA;  physics, because the mathematical&#xA;  language will be clearer and more&#xA;  uniform.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://geocalc.clas.asu.edu/pdf/SnarkPaper.pp.pdf&quot; rel=&quot;nofollow&quot;&gt;Hunting for Snarks in Quantum Mechanics&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Abstract. A long-standing debate over&#xA;  the interpretation of quantum&#xA;  mechanics has centered on the  meaning&#xA;  of Schroedinger’s wave function ψ for&#xA;  an electron. Broadly speaking, there&#xA;  are two major  opposing schools. On&#xA;  the one side, the  Copenhagen school&#xA;  (led by Bohr, Heisenberg and Pauli) &#xA;  holds that  ψ provides a  complete&#xA;  description of a single electron&#xA;  state; hence the probability &#xA;  interpretation of ψψ* expresses an&#xA;  irreducible uncertainty in electron&#xA;  behavior that is intrinsic in  nature.&#xA;  On the other side, the realist school&#xA;  (led by Einstein, de Broglie, Bohm and&#xA;  Jaynes) holds  that  ψ represents a&#xA;  statistical ensemble of possible&#xA;  electron states; hence it is an &#xA;  incomplete  description of a single&#xA;  electron state. I contend that the&#xA;  debaters have overlooked crucial facts&#xA;  about  the electron revealed by Dirac&#xA;  theory. In particular, analysis of&#xA;  electron zitterbewegung (first noticed&#xA;  by Schroedinger) opens a window to&#xA;  particle substructure in quantum&#xA;  mechanics &lt;strong&gt;that explains the  physical&#xA;  significance of the complex phase&#xA;  factor in  ψ&lt;/strong&gt;. This led to a  testable&#xA;  model for particle  substructure with&#xA;  surprising support by recent&#xA;  experimental evidence. If the&#xA;  explanation is upheld by  further&#xA;  research, it will resolve the debate&#xA;  in favor of the realist school. I give&#xA;  details. The perils of  research on&#xA;  the foundations of quantum mechanics&#xA;  have been foreseen by Lewis Carroll in&#xA;  The  Hunting of the Snark!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://geocalc.clas.asu.edu/pdf/Kinematic.pdf&quot; rel=&quot;nofollow&quot;&gt;THE KINEMATIC ORIGIN OF COMPLEX WAVE FUNCTION&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Abstract. A reformulation of the Dirac&#xA;  theory &lt;strong&gt;reveals that i¯h has a&#xA;  geometric meaning relating it to&#xA;  electron spin&lt;/strong&gt;. This provides the basis&#xA;  for a coherent physical interpretation&#xA;  of the Dirac and Sch¨odinger theories&#xA;  wherein &lt;strong&gt;the complex phase factor&#xA;  exp(−iϕ/¯h) in the wave function&#xA;  describes electron zitterbewegung, a&#xA;  localized, circular motion generating&#xA;  the electron spin and magnetic moment.&lt;/strong&gt;&#xA;  Zitterbewegung interactions also&#xA;  generate resonances which may explain&#xA;  quantization, diﬀraction, and the&#xA;  Pauli principle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://geocalc.clas.asu.edu/html/UGC.html&quot; rel=&quot;nofollow&quot;&gt;Universal Geometric Calculus&lt;/a&gt; a course, and follow:&lt;br&gt;&#xA;&lt;a href=&quot;http://geocalc.clas.asu.edu/html/Impl_QM.html&quot; rel=&quot;nofollow&quot;&gt;III. Implications for Quantum Mechanics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The Kinematic Origin of Complex Wave Functions&lt;/strong&gt;&lt;br&gt;&#xA;  Clifford Algebra and the Interpretation of Quantum Mechanics&lt;br&gt;&#xA;  The Zitterbewegung Interpretation of Quantum Mechanics&lt;br&gt;&#xA;  Quantum Mechanics from Self-Interaction&lt;br&gt;&#xA;  Zitterbewegung in Radiative Processes&lt;br&gt;&#xA;  On Decoupling Probability from Kinematics in Quantum Mechanics&lt;br&gt;&#xA;  Zitterbewegung Modeling&lt;br&gt;&#xA;  Space-Time Structure of Weak and Electromagnetic Interactions  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;to keep more references together:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.mrao.cam.ac.uk/~clifford/publications/ps/chris_thesis.pdf&quot; rel=&quot;nofollow&quot;&gt;Geometric Algebra and its Application to Mathematical Physics&lt;/a&gt; (Chris Thesis)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(what lead me to this amazing path was a paper by Joy Christian '&lt;a href=&quot;http://physics.stackexchange.com/q/7506/1257&quot;&gt;Disproof of Bell Theorem&lt;/a&gt;')&lt;br&gt;&#xA;'Bon voyage', 'good journey', 'boa viagem'   &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-04-06T15:20:20.743" LastActivityDate="2011-04-06T15:20:20.743" CommentCount="11" />
  <row Id="8086" PostTypeId="2" ParentId="8076" CreationDate="2011-04-05T11:14:20.243" Score="3" ViewCount="" Body="&lt;p&gt;This is rather a strange model; I'll took a chance to explain it from scratch. Basically all the chemical reactions look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_1 + S_2 + \cdots + S_n  \leftrightarrows P_1 + P_2 + \cdots + P_n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what means that a bunch of atoms forming few substrate molecules may rearrange into few product molecules. Both states have some energies and one postulates that the reaction involves some intermediate state when everything is mixed up (this is a simple view; since there may be few such states, few hidden subreactions, stuff -- I will not discuss it).  This can be plotted as:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/dzUAc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;where S are the substrates state, with energy $E_S$, than we heave an intermediate state of a higher energy $E_I$ and finally the products state of energy $E_P$. Now, the core of chemistry:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The balance between atoms in substrate state and product state after infinite time is only governed by the energy difference $E_S-E_P$ (and temperature, pressure, and other stuff that we assume constant)&lt;/li&gt;&#xA;&lt;li&gt;The speed of the reaction depends on the intermediate state energy $E_I$, because we assume that the substrates must gather $E_I-E_S$ from the thermal fluctuations and collide do form intermediate state.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;An enzyme forms a proper environment and usually improves spatial organisation of substrates to decrease $E_I$ and greatly improve the reaction speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in a cell, we have dozens of different molecules and thus huge amount of possible reactions. Moreover, the cell is under a constant matter exchange with environment, so the situation is dynamic and far from static equilibrium defined by $E_S-E_P$ differences. Because of that, the enzyme ability to change reaction speed is enough to form a dynamic equilibrium which we know as a metabolism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I can finally go back to the ATP role. Let's consider a situation where we have a reaction $S  \leftrightarrows P$ but $E_S&amp;lt;E_P$; this reaction will of course rather go the opposite way we want and the equilibrium will be shifted into substrates. Now, let's say we have also a reaction $S_X  \leftrightarrows P_X$ for which $E_{S_X}&amp;gt;&amp;gt;E_{P_X}$. One can expect that it is possible that the both reactions may co-occur creating reaction $S+S_X  \leftrightarrows P+P_X$ which would have positive $E_{S}-E_{P}$ difference, however in most cases this would be very improbable (corresponding to large $E_I$). Yet, this obstacle could be easy omitted -- with a help of an appropriate enzyme. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reassuming, enzyme role is to temporarily shift the balance of endoenergetical reaction by enlarging the probability of its co-occurrence with exoenergetic ATP to ADP+P transition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is of course a trivial and very wrong picture =) You can start learning more &lt;a href=&quot;http://en.wikipedia.org/wiki/Transition_state_theory&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2011-04-05T11:14:20.243" CommentCount="4" />
  <row Id="8087" PostTypeId="2" ParentId="8076" CreationDate="2011-04-05T12:01:55.863" Score="0" ViewCount="" Body="&lt;p&gt;It's usually more indirect than that.   You will need to know biochemical statistical thermodynamics of irreversible processes rather better.  I'll look for some references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To go to your example of an Enzyme E catalyzing the exothermic reaction summarized by X + Y + E + ATP -&gt; X-Y + E + ADP + P(i), I agree with the answer above that (4) the ATP -&gt; ADP + P(i) reaction in water at standard lab conditions is always energetically favorable, i.e. free energy is always released, and entropy increased as a result of this reaction.  It is this increase in entropy which will drive the reaction in the forward direction.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, the free energy released will be in the form of many far away infrared photons or phonons, which is much more probable (many more such states) than having it all localized in the single bond in ATP, so the reaction will go statistically in one direction.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your points 5 and 6 are overly specific.  The specific mechanism you refer to, resonant energy transfer can happen in some rather special cases, but it is unusual. Instead, more often enzymes form a temporary labile phosphorylated intermediate E-P which changes the charge in the active site and therefore the shape of the active site via &#xA;a)E + ATP -&gt; E-P + ADP  b)H2O + E-P -&gt;E + P(i) as part of the overall reaction summarized above.  No resonance is necessary, since charge effects do all the work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other different enzymes work by forming labile intermediates with one of the substrates as part of the overall reaction summary.   E + X -&gt; E-X  .  Many other mechanisms are also possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;main point&lt;/em&gt; is that everything does not need to happen at once, since the statistics will determine a net direction for the coupled reactions which comprise the overall reaction to go, based upon the overall free energy change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point is that the enzyme is always changing shape somewhat from the ordinary room temperature fluctuations it undergoes, according to the Boltzmann formula.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reiterate, reactions can be driven forward overall by statistics even if the early steps are energetically unfavorable.   So the whole situation is much more diverse than you described (which is why it is chemistry).  A good example of this last mechanism is in the synthesis of nucleic acids from nucleotide triphosphates, the overall reaction being driven forward by the much later release of free energy associated with the hydrolytic splitting of the previously released pyrophosphate molecule&lt;br&gt;&#xA;H2O + P-P -&gt; 2 P(i).    &lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2011-04-05T12:10:03.593" LastActivityDate="2011-04-05T12:10:03.593" CommentCount="7" />
  <row Id="8088" PostTypeId="2" ParentId="8074" CreationDate="2011-04-05T12:42:55.697" Score="20" ViewCount="" Body="&lt;p&gt;The error is that you assume that the density distribution is &quot;nearly spherically symmetric&quot;. It's far enough from spherical symmetry if you want to calculate first-order subleading effects such as the equatorial bulge. If your goal is to compute the deviations of the sea level away from the spherical symmetry (to the first order), it is inconsistent to neglect equally large, first-order corrections to the spherical symmetry on the other side - the source of gravity. In other words, the term $hg$ in your potential is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just imagine that the Earth is an ellipsoid with an equatorial bulge, it's not spinning, and there's no water on the surface. What would be the potential on the surface or the potential at a fixed distance from the center of the ellipsoid? You have de facto assumed that in this case, it would be $-GM/R+h(\theta)g$ where $R$ is the fixed Earth's radius (of a spherical matter distribution) and $R+h(\theta)$ is the actual distance of the probe from the origin (center of Earth). However, by this Ansatz, you have only acknowledged the variable distance of the probe from a spherically symmetric source of gravity: you have still neglected the bulge's contribution to the non-sphericity of the gravitational field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you include the non-spherically-symmetric correction to the gravitational field of the Earth, $hg$ will approximately change to $hg-hg/2=hg/2$, and correspondingly, the required bulge $\Delta h$ will have to be doubled to compensate for the rotational potential. A heuristic explanation of the factor of $1/2$ is that the true potential above an ellipsoid depends on &quot;something in between&quot; the distance from the center of mass and the distance from the surface. In other words, a &quot;constant potential surface&quot; around an ellipsoidal source of matter is &quot;exactly in between&quot; the actual surface of the ellipsoid and the spherical $R={\rm const}$ surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will try to add more accurate formulae for the gravitational field of the ellipsoid in an updated version of this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update: gravitational field of an ellipsoid&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have numerically verified that the gravitational field of the ellipsoid has exactly the halving effect I sketched above, using a Monte Carlo Mathematica code - to avoid double integrals which might be calculable analytically but I just found it annoying so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took millions of random points inside a prolate ellipsoid with &quot;radii&quot; $(r_x,r_y,r_z)=(0.9,0.9,1.0)$; note that the difference between the two radii is $0.1$. The average value of $1/r$, the inverse distance between the random point of the ellipsoid and a chosen point above the ellipsoid, is $0.05=0.1/2$ smaller if the chosen point is above the equator than if it is above a pole, assuming that the distance from the origin is the same for both chosen points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Code:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{xt, yt, zt} = {1.1, 0, 0};&#xA;&#xA;runs = 200000;&#xA;totalRinverse = 0;&#xA;total = 0;&#xA;&#xA;For[i = 1, i &amp;lt; runs, i++,&#xA; x = RandomReal[]*2 - 1;&#xA; y = RandomReal[]*2 - 1;&#xA; z = RandomReal[]*2 - 1;&#xA; inside = x^2/0.81 + y^2/0.81 + z^2 &amp;lt; 1;&#xA; total = If[inside, total + 1, total];&#xA; totalRinverse = &#xA;  totalRinverse + &#xA;   If[inside, 1/Sqrt[(x - xt)^2 + (y - yt)^2 + (z - zt)^2], 0];&#xA;]&#xA;&#xA;res1 = N[total/runs / (4 Pi/3/8)]&#xA;res2 = N[totalRinverse/runs / (4 Pi/3/8)]&#xA;res2/res1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Description&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the Mathematica code above: its goal is to calculate a single purely numerical constant because the proportionality of the non-sphericity of the gravitational field to the bulge; mass; Newton's constant is self-evident. The final number that is printed by the code is the average value of $1/r$. If {1.1, 0, 0} is chosen instead of {0, 0, 1.1} at the beginning, the program generates 0.89 instead of 0.94. That proves that the gravitational potential of the ellipsoid behaves as $-GM/R - hg/2$ at distance $R$ from the origin where $h$ is the local height of the surface relatively to the idealized spherical surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the code above, I chose the ellipsoid with radii (0.9, 0.9, 1) which is a prolate spheroid (long, stick-like), unlike the Earth which is close to an oblate spheroid (flat, disk-like). So don't be confused by some signs - they work out OK.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus from Isaac&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mariano C. has pointed out the following solution by a rather well-known author:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=ySYULc7VEwsC&amp;amp;lpg=PP1&amp;amp;dq=principia%20mathematica&amp;amp;pg=PA424#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=ySYULc7VEwsC&amp;amp;lpg=PP1&amp;amp;dq=principia%20mathematica&amp;amp;pg=PA424#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-06T05:40:43.950" LastActivityDate="2011-04-06T05:40:43.950" CommunityOwnedDate="2011-04-06T05:40:43.950" CommentCount="18" />
  <row Id="8089" PostTypeId="2" ParentId="8076" CreationDate="2011-04-05T12:55:16.343" Score="-1" ViewCount="" Body="&lt;p&gt;there are some good answers here and I'll agree it is important to understand the thermodynamical aspects of enzyme function, but I sympathize with the OP on this point: down at the lowest level you still want to describe what happens with the bonds and energy even if the macroscopics of the system (temperature, pressure, configuration) controls the reaction speeds. After all, chemistry is low-energy physics, and thermodynamics is big-number physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is linked a paper (from 2005) which summarizes the then-current knowledge about ATP hydrolysis in the F1-ATPase, one of the most strikingly mechanical enzyme complexes in nature (see the linked YouTube-video!). It's basically a stator and a rotor, which can convert between ATP&amp;lt;-&gt;ADP+P and mechanical rotation of the rotor. It is one of the crucial proteins that power respiration in the mitochondrias in human cells in a very mechanical way - protons fall down a gradient through the mitochondrial membrane, pulling the rotor around, which turns ADP+P into ATP. In the bacterial flagelli (tails) it is used the other way, ATP powers it as a motor which enables the bacteria to swim. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3D-structure of the F1-ATPase has been known for some time and there are some beautiful images of it on the web but the process is not completely understood. The paper discusses results from experiments and atomic simulations (one of my research interests) which give clues to exactly how the ATP bond energy is transferred. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in most protein functions, you have a very mechanical function (as the OP suggests in pt. 7) in combination with redistribution of Coulumbic charge as well as hydrogenic and covalent bonds. Some speculate that elastic storage of mechanical energy can participate (and hence the possibility of phonons I guess) - but I have never heard of any suggestions of intra-protein photon exchange as the OP suggests in pt. 5-6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried posting the link to the researcher's homepage of the paper linked below but this site didn't allow me posting more than 2 links in a post :) Sorry..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1509602/&quot; rel=&quot;nofollow&quot;&gt;Zooming in on ATP hydrolysis in F1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=J8lhPt6V-yM&quot; rel=&quot;nofollow&quot;&gt;ATP F1 ATPase youtube video&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-04-05T12:55:16.343" CommentCount="5" />
  <row Id="8090" PostTypeId="1" CreationDate="2011-04-05T13:01:39.050" Score="4" ViewCount="169" Body="&lt;p&gt;I was searching a lot and could only find dosages for curing cancer and allowed emission, but no Iodine-131 dose that could be connected with increased thyroid cancer risk (like, 10mSv is the radiation dose resulting with measurable increase of cancer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Central Laboratory of Radiological Protection publishes Iodine-131 content in air measured in $\mu$Bq/m$^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number has grown by 3 orders of magnitude since the Fukushima accident, and while I strongly believe it's still good 6 orders of magnitude below dangerous levels, there's no convincing of some people without solid numbers - and finding these has proven quite hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how can I then get either the safe levels of I-131 in air in ($\mu$Bq/m$^3$) , or a way to (indirectly - this gonna be convoluted) convert the content in air ($\mu$Bq/m$^3$) to dose equivalent radiation ($\mu$Sv) from absorbed Iodine?&lt;/p&gt;&#xA;" OwnerUserId="2754" LastActivityDate="2011-04-05T14:48:50.113" Title="Dangerous dose of I-131?" Tags="&lt;radioactivity&gt;&lt;unit-conversion&gt;&lt;isotope&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="8091" PostTypeId="2" ParentId="8076" CreationDate="2011-04-05T13:11:13.123" Score="0" ViewCount="" Body="&lt;p&gt;The double welled potential diagram mbq posted is an operative diagram for the energetic state of a molecule, call it M, which is phosphorylated ATP + M --&gt; ADP + M-P.  The phosphorous atom binds onto an amino acid residue group in a polypeptide chain, often threonine or tyrosine in the case of kinases.  Kinases are enzymes which initiate biochemical pathways.  The amino acid residue which is phosphorylated has dihedral bond angles with adjacent amino acid residues in the chain.  The unphosphorylated and phosphorylated bond angles are substantially different, reflecting the two minima in the double welled potential.  The result is the enzyme exhibits two conformational shapes.  These shapes can act as binary conditions, literally on and off, for turning on and off a biochemical pathway.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may be compared to Landauer’s principle which relates a binary information condition to thermodynamics.  ATP contains 76j/mole of energy, which we can identify as the E.  The work or useful energy extracted is the change in the energy from the lower portion of the well to the higher.  Also there is energy in the entropy due to a chemical change, which for the two states is $k~ln(2)$.  Therefore, the free energy per mole $F~=~ 76j/mole$ must be greater than the entropy for a change in the configuration or shape of the kinase or other molecule.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-04-05T13:11:13.123" />
  <row Id="8092" PostTypeId="2" ParentId="8090" CreationDate="2011-04-05T13:32:13.397" Score="1" ViewCount="" Body="&lt;p&gt;It's difficult to get hard figures without exposing people and seeing how many die!  Then the problem is that for low levels of exposure you have to workout how many extra people have died &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.cancer.gov/cancertopics/causes/i131&quot; rel=&quot;nofollow&quot;&gt;Nation Cancer Institute&lt;/a&gt; has some calculators and papers based on US nuclear testign exposure&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-04-05T13:32:13.397" CommentCount="3" />
  <row Id="8094" PostTypeId="1" CreationDate="2011-04-05T13:51:45.493" Score="2" ViewCount="219" Body="&lt;p&gt;what about a theory where no time exists, only discrete space, steady states and occasion chains. what would speak against this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;;-)&lt;/p&gt;&#xA;" OwnerUserId="2937" LastActivityDate="2011-04-05T15:53:22.733" ClosedDate="2011-04-05T16:06:37.073" Title="what about there is no time, only space" Tags="&lt;space&gt;&lt;time&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="8095" PostTypeId="1" AcceptedAnswerId="8109" CreationDate="2011-04-05T14:00:56.063" Score="3" ViewCount="168" Body="&lt;p&gt;Every orbiting of a satellite around a mass is nothing else but a constant fall - and therefore acceleration - towards this mass. In a way it is a &quot;falling around&quot; that mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question&lt;/strong&gt;&lt;br&gt; Is it possible to measure this acceleration on earth due to its &quot;falling around&quot; the sun?&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2011-04-06T01:02:56.970" Title="Measuring acceleration of earth due to its fall around the sun" Tags="&lt;gravity&gt;&lt;measurement&gt;&lt;orbital-motion&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="8097" PostTypeId="2" ParentId="8094" CreationDate="2011-04-05T14:18:17.300" Score="0" ViewCount="" Body="&lt;p&gt;a physical theory is meant to describe observations. It may be an important pastime of an idle, crazy pseudo philosopher to ponder over imaginary speculations, but the job of physicists are much more prosaic and constrained. You have asked what would be against your &quot;hypothetical theory&quot;. Nothing much, except that it is a good for nothing.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-04-05T14:18:17.300" CommentCount="7" />
  <row Id="8100" PostTypeId="1" AcceptedAnswerId="8101" CreationDate="2011-04-05T15:12:29.817" Score="1" ViewCount="237" Body="&lt;p&gt;I find plane waves are uncompatible with light cone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps plane waves are &quot;virtual&quot; and can never be measured&#xA;in that case, shouldn't we call plane waves as &quot;virtual plane waves&quot;?&#xA;(other option could be that plane waves allows waves travel faster than c)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to clarify this point through this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If plane waves would exist(as measurable), then higher than c speed &#xA;could be reached like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A wave going from X to Y at a speed c, it will reach Z at higher&#xA;than c speed, because it will reach at same time, traveling more distance.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                   (X).&#xA;                    |   &#xA;                    v   &#xA;                  ________________________________________________  &#xA;plane waves       ________________________________________________  &#xA;going X to Y      ________________________________________________&#xA;                   (Y).                                       (Z).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In a real situation the wave will be a circle (or a sphere in 3d) &#xA;so it will get Z later then that's not a plane wave.&lt;/p&gt;&#xA;" OwnerUserId="1916" LastActivityDate="2011-04-06T00:38:57.450" Title="Are electromagnetic &quot;plane&quot; waves measurable or just a virtual concept?" Tags="&lt;electromagnetism&gt;&lt;electromagnetic-radiation&gt;&lt;quantum-electrodynamics&gt;&lt;plane-wave&gt;" AnswerCount="4" />
  <row Id="8101" PostTypeId="2" ParentId="8100" CreationDate="2011-04-05T15:22:27.827" Score="10" ViewCount="" Body="&lt;p&gt;A &quot;plane wave&quot; generally refers to an infinite and perfectly flat wavefront, which cannot exist in reality, of course. However, there is nothing at all impossible about a plane wave of finite extent. Such a wave will experience diffraction at its edges, of course, but can still propagate over long distances before losing its planar nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with your question about faster than light information transmission, is that if X were the only point emitting a wave, then it would be a spherical wave, not a plane wave. A plane wave can be thought of as being composed as a superposition of spherical waves emitted in phase from every point on an infinite plane. So in your example, information would not be arriving at Z from X, but instead from some other source point within the causal past of Z. In your diagram that point could be at the same height as X, but directly above Z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering a wavefront (planar or otherwise) as a superposition of spherical waves is the central feature of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Huygens%E2%80%93Fresnel_principle&quot; rel=&quot;nofollow&quot;&gt;Huygens-Fresnel Principle&lt;/a&gt;, which would be a good reading on the topic.&lt;/p&gt;&#xA;" OwnerUserId="869" LastEditorUserId="869" LastEditorDisplayName="" LastEditDate="2011-04-05T16:00:14.147" LastActivityDate="2011-04-05T16:00:14.147" CommentCount="10" />
  <row Id="8102" PostTypeId="2" ParentId="8049" CreationDate="2011-04-05T15:32:41.543" Score="2" ViewCount="" Body="&lt;p&gt;The density matrices in both cases are identical. If quantum mechanics is exactly linear, both states ought to be indistinguishable. But if there are some slight nonlinearities in the time evolution, we ought to be able to distinguish between them in principle. But you have to realize nonlinearities in quantum mechanics lead to all sorts of problems, which is why most people assume quantum mechanics is exactly linear.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastEditorUserId="1039" LastEditorDisplayName="" LastEditDate="2011-04-05T15:49:27.337" LastActivityDate="2011-04-05T15:49:27.337" />
  <row Id="8104" PostTypeId="2" ParentId="8049" CreationDate="2011-04-05T15:47:37.637" Score="4" ViewCount="" Body="&lt;p&gt;Let me give some reference that might be useful to make things clear.&lt;br&gt;&#xA;It's Landau-Lifshitz, book 5, chapter 5:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The averaging by means of the statisitcal matrix ... has a twofold nature. &#xA;  It comprises both the averaging due to the probalistic nature of the quantum description (even when as complete as possible) &#xA;  and the statistical averaging necessiated by the incompleteness of our information concerning the object considered....&#xA;  It must be borne in mind, however, that these constituents cannot be separated; the whole averaging procedure is carried &#xA;  out as a single operation, and cannot be represented as the result of succesive averagings, &#xA;  one purely quantum-mechanical and the other purely statistical.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This &quot;twofold averaging&quot; is exactly the reason why the two states cannot be distinguished in any way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me add another nice citation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It must be emphasised that the averaging over various $\psi$ states, which we have used in order to &#xA;  illustrate the transition from a complete to an incomplete quantum-mechanical description has only &#xA;  a very formal significance. In particular, it would be quite incorrect to suppose that the description &#xA;  by means of the density matrix signifies that the subsystem can be in various $\psi$ states with various &#xA;  probabilities and that the average is over these probabilities. Such a treatment would be in conflict &#xA;  with the basic pronciples of quantum mechanics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-04-05T15:47:37.637" />
  <row Id="8105" PostTypeId="2" ParentId="8094" CreationDate="2011-04-05T15:53:22.733" Score="3" ViewCount="" Body="&lt;p&gt;I am not sure this is a completely nonsensical question, though it is phrased in an odd way.  Julian Barbour is pretty much into the notion that time does not exist.  This is based largely on the Wheeler DeWitt equation ${\cal H}\Psi[g]~=~0$, which is a quantum version of the Hamiltonian constraint in ADM relativity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could well enough imagine presenting how time exists, but space does not. We could presume there is some one dimensional space, a line or curve, and there is a fibration on that space by a three dimensional space.  This internal space is a symmetry of the dynamics of this one dimensional parameterized space we label as time.  This then connects to relativity when we consider the metric line element&#xA;$$&#xD;&#xA;ds^2~=~-c^2dt^2~+~g_{ij}dx^idx^j,&#xD;&#xA;$$&#xA;where mixed time-space metric components are not included.  We have here two notions of time.  The first is the proper time $\tau~=~s/c$, which is the invariant of relativity.  The other time is a coordinate time $t$, which is not an invariant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious question to ask is whether $ds$ is real.  We can multiply it by $mc^2$ and define an action according to the extremal principle of the proper interval&#xA;$$&#xD;&#xA;S~=~mc\int ds,&#xD;&#xA;$$&#xA;which appears real in some sense. It has units of action, or angular momentum, which is a measurable quantity. Yet there is something a bit troublesome about all of this. How does the observer on this world line actually measure this interval? A clock is employed which must have some system of oscillations, such as a spring. Yet this is measuring the invariant interval according to something carried on that world line that deviates from the world line. Hence some sort of nongeodesic motion is being used to define or measure an interval along a geodesic path. Of course I am thinking primarily of a mechanical clock, but an atomic one still appears to hold for an EM field must be applied to knock electrons in the Ce atoms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Lagrangian is measured according to something which is not invariant.  So we might then consider that action as $dS~=~pdq~-~Hdt$. Now we have some Hamiltonian, which might include a part for the dynamics of the clock. Hamiltonians must be specified on some Cauchy surface of data with a coordinate time direction. Yet this has gotten us into some funny issue, for to define an invariant interval it appears that we need a coordinate defined clock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far we have some identification of $Hdt$, or the square of this, with the $c^2dt^2$ in the interval above.  We then have that the bare action term $pdq$ is identified with $mc\sqrt{g_{ij}dx^idx^j}$.  So we have a bare action given by our fibration, but we also have some constraint, where $H$ acts as a Lagrange multiplier.  So we then have our one dimensional curve defined in a spacetime, where the space is the space of fibration and the Lagrange multiplier determines the symmetry of that fibration which is the Lorentz group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to make things curious, we could imagine this picture as dual in some ways to the picture where time does not exist, but space does.  The duality might then have a noncommutative coordinate geometric content in quantum gravity.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-04-05T15:53:22.733" CommentCount="6" />
  <row Id="8106" PostTypeId="2" ParentId="8095" CreationDate="2011-04-05T17:12:50.917" Score="2" ViewCount="" Body="&lt;p&gt;The acceleration can be found by a number of means.  $F~=~ma$ with the Newtonian law of gravity and the centripetal acceleration $a~=~v^2/r$,&#xA;$$&#xD;&#xA;m\frac{v^2}{r}~=~-\frac{GMm}{r^2}.&#xD;&#xA;$$&#xA;This gives $v~=~\sqrt{GM/r}$, which is $v~=~29.5km/s$ or $v~=~2.95\times 10^4m/s$, for the mass of the sun and $r~=~1.5\times 10^8km$.  The acceleration is then $5.8m/s^2$.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-04-05T17:12:50.917" />
  <row Id="8107" PostTypeId="1" CreationDate="2011-04-05T17:31:00.283" Score="0" ViewCount="105" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/6646/physics-of-simple-collisions&quot;&gt;Physics of simple collisions&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I have looked around the internet and gradually built up a better understanding. However I'm still not 100% clear on the physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have 2 identical circles (in a 2D world, so they are on the same plane) and they collide, from any 2 directions, how do I calculate which direction they will travel in after the collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial though was to get the tangent between the 2 circles, and each circle would bounce off the tangent in the same a ball bounces off a flat object (angle into the collision equals the angle out, but reversed around the normal):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where the red lines are parallel, and the line between the center of the circles is perpendicular to the red lines. And the angles x1 and x2 are equal in value, as are y1 and y2, but x1 and y1 are not [necessarily] equal)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://docs.google.com/drawings/pub?id=13Kw1gxNjd_7Pd181QdVFzoOAtWXw4LO1mJerdnFrVj4&amp;amp;w=521&amp;amp;h=497&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But after reading articles on the internet, it doesn't appear this is right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(NOTE: This is not homework, but rather learning by example)&lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2011-04-05T17:31:00.283" ClosedDate="2011-04-05T17:44:33.037" Title="The physics of two circles colliding" Tags="&lt;forces&gt;&lt;newtonian-mechanics&gt;" CommentCount="6" FavoriteCount="0" />
  <row Id="8108" PostTypeId="2" ParentId="8077" CreationDate="2011-04-05T17:31:08.030" Score="-1" ViewCount="" Body="&lt;p&gt;put it on a big board, supported at either end with brick say, and then get some people to stamp on the board in the middle while you test it. To make the vibration heavier, just add more people.&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-04-05T17:31:08.030" />
  <row Id="8109" PostTypeId="2" ParentId="8095" CreationDate="2011-04-05T17:34:17.293" Score="4" ViewCount="" Body="&lt;p&gt;The answer depends in part on what you mean by &quot;measure&quot;. You can certainly &lt;em&gt;calculate&lt;/em&gt; the acceleration using the laws of kinematics and dynamics, as Lawrence B. Crowell points out in his answer. Does that, in your mind, count as &quot;measuring the acceleration&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a much more direct recipe that I think would uncontroversially count as measuring the acceleration. Pick a set of fixed stars, and measure the velocity of the Earth relative to them using the Doppler effect. Any one measurement only gives you one component of ${\bf v}$, but if you measure a few, you can get the full vector. Do this twice, at two different times, and calculate $\Delta{\bf v}/\Delta t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With current technology it would be very easy to do this to the required accuracy. In fact, astronomers are making precisely these required measurements all the time -- not with the purpose of measuring the acceleration, but for a wide variety of other reasons. In fact, astronomers often deliberately subtract out the time-varying Doppler shifts due to Earth's changing velocities, because what they're interested in are (often much smaller) changes due to the actual motions of the stars. &lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-05T17:34:17.293" CommentCount="3" />
  <row Id="8110" PostTypeId="6" CreationDate="2011-04-05T17:55:46.937" Score="0" ViewCount="" Body="&lt;p&gt;I've finally decided to throw my hat into the ring :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been one of the pro-tem moderators for the site for the past few months, so people who have been around here probably know my style. I tend to be a little hesitant about &quot;doling out punishment&quot; (suspensions etc.) and a little aggressive about closing questions, but only because I really care about keeping the site focused on proper physics (not philosophy, pseudoscience, etc.) and I think it is important to act quickly on questions which are inappropriate here. I don't support restricting the topics we cover to a certain level, the way Math Overflow does, but I do support restricting based on &lt;em&gt;quality&lt;/em&gt;: the questions we get should be reasonably precise and answerable using established physics. I believe it's possible to ask a good question, or a bad question, at any level of study.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll freely admit that I may not have done the best job as a moderator so far, but at least &lt;em&gt;some&lt;/em&gt; people seem to think I've been decent at it, so I think it's fair to give the community the option of having me continue in the role.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now I'm in the middle of a busy period that keeps me from being fully available for moderating (or participating in) the site, but that should end in about 3 weeks. This means I wouldn't be fully available for the first couple weeks of the new moderators' term, but if I'm elected, I'm hoping that won't be a serious problem. I've had some chats with Rebecca (Chernoff) and she seems to think it won't be too big of a deal.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-06T06:26:12.310" LastActivityDate="2011-04-06T06:26:12.310" CommunityOwnedDate="2011-04-05T17:55:46.937" CommentCount="1" />
  <row Id="8112" PostTypeId="1" AcceptedAnswerId="8148" CreationDate="2011-04-05T18:26:12.473" Score="-1" ViewCount="294" Body="&lt;p&gt;The other three forces' mediating particles (photons etc.) are absorbed by their appropriate charge-carrying particles, but I can't seem to find a clear answer that applies to the gravitational force in a quantified scenario. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The different answers to this most basis question seem to elude me in frameworks of String theory and LQG, though it becomes more intuitive in the latter.&lt;/p&gt;&#xA;" OwnerUserId="2942" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-05T20:16:45.380" LastActivityDate="2011-04-06T15:01:32.453" Title="What do gravitons interact with?" Tags="&lt;quantum-gravity&gt;" AnswerCount="1" CommentCount="13" />
  <row Id="8113" PostTypeId="2" ParentId="8100" CreationDate="2011-04-05T18:33:37.417" Score="1" ViewCount="" Body="&lt;p&gt;As soon as your source is X, there is no wave propagation from Y to Z, whatever wave front looks like (spherical, plane, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About &quot;virtuality&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A wave is &lt;strong&gt;real&lt;/strong&gt; if it is of sufficient intensity, i.e., it consists of many-many photons. Then your detector at Y will work as well as your detector at Z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a low-intensity wave it is not the case and the wave describes the &lt;strong&gt;probability amplitude&lt;/strong&gt;. If you have only one photon in it, only one detector will register it.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-04-05T20:28:24.950" LastActivityDate="2011-04-05T20:28:24.950" CommentCount="3" />
  <row Id="8114" PostTypeId="2" ParentId="8074" CreationDate="2011-04-05T18:41:25.273" Score="8" ViewCount="" Body="&lt;p&gt;There was some doubt about Lubos' answer (which I've accepted), so this is just a verification.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I copied the method Lubos described and found the potential difference for an ellipsoid with different eccentricities.  Sure enough, for an oblate spheroid, if you make the center-equator distance a fraction $e$ larger than the center-pole distance, the potential is roughly a fraction $e/2$ smaller at the equator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do the whole problem, we'd have to take into account the varying density of the Earth, but as a rough estimate this seems to do the job.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for the oblate spheroid &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{x}{1.01}\right)^2 + \left(\frac{y}{1.01}\right)^2 + z^2 &amp;lt; 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the average value of $1/r$ at $(0,0,1)$ is about .996, and the average at $(1.01,0,0)$ is about .991.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Python code below (please excuse the amateur-ness)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import random&#xA;import math&#xA;&#xA;points = 10000000&#xA;e = .01&#xA;rad = 1+e&#xA;diam = 2*rad&#xA;&#xA;pot= 0&#xA;count = 0&#xA;for i in range(1,points):&#xA;    x = diam*random.random()-rad&#xA;    y = diam*random.random()-rad&#xA;    z = diam*random.random()-rad&#xA;    r = math.sqrt((x-rad)*(x-rad)+y*y+z*z)&#xA;    if x*x/(rad*rad)+y*y/(rad*rad) + z*z &amp;lt; 1:&#xA;        pot = pot + 1/r&#xA;        count = count + 1&#xA;&#xA;print pot/count&#xA;&#xA;pot2 = 0&#xA;count = 0&#xA;for j in range(1,points):&#xA;    x = diam*random.random()-rad&#xA;    y = diam*random.random()-rad&#xA;    z = diam*random.random()-rad&#xA;    r = math.sqrt(x*x+y*y+(z-1.0)*(z-1.0))&#xA;    if x*x/(rad*rad)+y*y/(rad*rad) + z*z &amp;lt; 1:&#xA;        pot2 = pot2 + 1/r&#xA;        count = count + 1&#xA;&#xA;print pot2/count&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="74" LastEditorUserId="1656" LastEditorDisplayName="" LastEditDate="2011-04-06T02:41:06.290" LastActivityDate="2011-04-06T02:41:06.290" CommentCount="6" />
  <row Id="8115" PostTypeId="1" CreationDate="2011-04-05T18:43:51.103" Score="2" ViewCount="177" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/7359/origins-of-the-universe-questions&quot;&gt;origins of the universe questions.&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;i am a software engineer and not an astrophysicist but i want to know if anyone is working on sorting out what exists outside the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what about outside space ? what is there ? or shall i assume there are infinite galaxies ?&#xA;and even that is not a satisfying answer. &#xA;I know this question might get closed soon ... but guys please tell me what is there outside the space as we know it ? since childhood i have wondered abt this . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its really a simple question. If we say that we live &quot;inside&quot; a Universe .. then we are objectively saying that there IS something &quot;outside&quot; the universe as well ? am i crazy ? what exists there ? is it just empty space ? or what ? &lt;/p&gt;&#xA;" OwnerUserId="2943" LastEditorUserId="2943" LastEditorDisplayName="" LastEditDate="2011-04-05T18:52:36.337" LastActivityDate="2011-04-05T19:56:45.100" ClosedDate="2011-04-05T20:24:56.757" Title="If the Big Bang theory suggests that the Universe is &quot;expanding&quot; then what is it expanding INTO?" Tags="&lt;astrophysics&gt;&lt;astronomy&gt;&lt;spacetime&gt;&lt;space&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="8116" PostTypeId="2" ParentId="8115" CreationDate="2011-04-05T18:54:11.107" Score="0" ViewCount="" Body="&lt;p&gt;you are not crazy. its a very good (and deep) question but current physical theories don't have any compelling evidence to work about what sort of space, space-time itself would be embedded on, or if such a embedding space could have any physical consequences on our space-time. There is brane-world scenarios where such models are studied, but no evidence  to support anything of this sort so far&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from the current factual perspective, current General Relativity assumes space-time is curved due to gravity sources, but to understand curvature mathematically you don't need to assume any outer space, so an expanding space-time in itself is a totally consistent idea mathematically. There is no requirement to postulate an outside, ambient space&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-04-05T18:59:53.377" LastActivityDate="2011-04-05T18:59:53.377" CommentCount="2" />
  <row Id="8117" PostTypeId="2" ParentId="8115" CreationDate="2011-04-05T19:12:18.943" Score="0" ViewCount="" Body="&lt;p&gt;There exists a mathematical space into which our universe is embedded, but it is mathematical because nothing as we know it can get out of the universe or enter the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the  big bang the universe is expanding in three dimensions analogous to the way that the surface of a balloon expands in two dimensions: all points on the balloon surface recede from each other and the surface gets large. The balloon surface is not expanding into anything, as far as the points on the balloon go.  In three dimensions all points in our universe recede from each other, having all started from the one point of the big bang.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-05T19:12:18.943" CommentCount="3" />
  <row Id="8118" PostTypeId="2" ParentId="8115" CreationDate="2011-04-05T19:56:45.100" Score="0" ViewCount="" Body="&lt;p&gt;The universe is expanding into Time.  At least that's one way to think about it.&lt;/p&gt;&#xA;" OwnerUserId="2945" LastActivityDate="2011-04-05T19:56:45.100" CommentCount="1" />
  <row Id="8119" PostTypeId="1" CreationDate="2011-04-05T19:57:50.043" Score="2" ViewCount="120" Body="&lt;p&gt;From an earlier version of the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Thomas_precession&quot; rel=&quot;nofollow&quot;&gt;Thomas precession&lt;/a&gt; concerning TP and LP=&lt;a href=&quot;http://en.wikipedia.org/wiki/Larmor_precession&quot; rel=&quot;nofollow&quot;&gt;Larmor precession&lt;/a&gt;, regarding the paper: G B Malykin, &quot;Thomas precession: correct and incorrect solutions&quot;, Physics-Uspekhi 49 (8) 837-853 (2006)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a 2006 survey of the literature Malykin notes that there are numerous conflicting expressions for Thomas precession. This is partly explained by the fact that different authors use &quot;Thomas precession&quot; to refer to different things, often without saying what they are referring to and subsequent authors then misinterpret the results and apply them to other things, but, even taking this into account, some of the expressions in the literature are just plain wrong.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Malykin explains the source of some of these errors: &lt;i&gt;&quot;We emphasize that Thomas considered the rotation of the axes of the coordinate system accompanying the electron in its motion rather than the electron spin rotation. Subsequently, this led to a misunderstanding and the emergence of incorrect work on the TP problem. &lt;b&gt;It is possible to introduce three different reference frames accompanying the electron motion&lt;/b&gt; around a circular orbit and, in the most general case, along a curvilinear trajectory: (i) a reference frame whose coordinate axes remain parallel or retain their angular position relative to the axes of a laboratory IRF, (ii) a reference frame one of whose coordinate axes is always coincident with the electron velocity vector, and (iii) a reference frame in which the electron spin vector retains its orientation relative to the coordinate axes. It is evident that the electron spin vector precesses relative to the coordinate axes of the two first systems, but the angular velocity of its precession is different in these systems. ... In several papers concerned with the TP, calculations are performed in the first approximation in v^2/c^2, where v is the speed of an elementary particle in the laboratory IRF and c is the speed of light. In this case, all authors arrive at the same expression first derived by Thomas, this being so irrespective of whether they consider the relativistic rotation of the particle spin or the relativistic rotation of the axes of the coordinate system comoving with the particle. In the most general case, however, the expressions for the TP obtained by different authors are radically different. As noted above, the problem is complicated by the fact that different authors assign different meaning to this expression: some imply the relativistic rotation of the particle spin in the laboratory IRF, some in the comoving reference frame (in this case, as noted above, the rotation law for the axes of this system may be defined in three ways), while others refer to the relativistic rotation of the axes of the coordinate system accompanying the particle in motion. ... As noted above, the expression for the TP in Thomas's first paper was obtained in the first approximation in v^2/c^2 and is always correct when this condition is fulfilled. In his subsequent work, on performing calculations for an arbitrary electron velocity v, Thomas derived an expression that correctly describes the relativistic rotation of the axes of the comoving coordinate system relative to the rest-frame (laboratory) system. However, because the majority of authors use the term TP in reference to the precession of the spin of an elementary particle, this subsequently led to several errors and misunderstandings. ... &lt;b&gt;In 1952, in his famous monograph [78], the Danish scientist C Møller (1904-1980), an acknowledged expert on the theory of relativity, derived an expression for the TP that coincides, up to a sign, with the corresponding Thomas expression and is correct in the comoving frame of reference. However, it was stated in Ref. [78] that this expression was written for the laboratory IRF, which is incorrect. Møller's immense scientific prestige played a negative role in this case: since then, the majority of authors of papers, monographs, and lecture courses started using the expression for the TP from Ref. [78] or, in the derivation of suchlike expressions, tried to make them coincident with that given in Ref. [78].&lt;/b&gt; ... At the same time, because the experimentally observed particle spin precession is caused by the sum of two effects, the TP and the LP, it is possible to choose different expressions for each of these effects, only provided that these expressions add up to correspond to expression (14). we conclude that the preferred method is the above-discussed method of recording the TP with the aid of mechanical gyroscopes in their orbital motion because quantum mechanical effects in experiments on charged elementary particles partly complicate the interpretation of experimental data.&quot;&lt;/i&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Question: Is Malykin's criticism of the literature correct ?     &lt;/p&gt;&#xA;" OwnerUserId="2929" LastActivityDate="2011-04-05T19:57:50.043" Title="Criticism of the Thomas precession literature" Tags="&lt;special-relativity&gt;" CommentCount="4" />
  <row Id="8120" PostTypeId="2" ParentId="8082" CreationDate="2011-04-05T21:31:48.693" Score="0" ViewCount="" Body="&lt;p&gt;It's not a &lt;em&gt;very&lt;/em&gt; clear Question. I &lt;em&gt;think&lt;/em&gt; you're saying that the center of mass of the 3D box is above the point $[xF,yF]$, at $[xF,yF,zF]$. I'm going to run with that because the background is clearly a few days work or more to get under control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this assumption, the box is stable if and only if the point $[xF,yF]$ is contained within the maximal convex polygon defined by the $N$ points $[x_n,y_n]$. Any points $[x_i,y_i]$ that are contained within the maximal convex polygon can be removed from the list as irrelevant. To make the algorithms go more quickly, it will be useful to construct a list of the edges of the convex polygon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the box is not stable, only one of the edges of the convex polygon matters, which is the edge that is closest to the point $[xF,yF]$. Now you have just two points to worry about, the center of mass of the 3D box $[xF,yF,zF]$, and the closest point on the closest edge to $[xF,yF]$, at a distance $D$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torque acting on the box is its weight times the distance $D$. As the box tips over, the torque that acts will increase as the position of the center of mass of the box moves away from the closest point of the convex polygon, so the box will tip increasingly rapidly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need to know the forces that act at the points $[x_n,y_n]$ then you need to know what the distribution of the mass inside the 3D box is. Knowing the Moment of Inertia Tensor of the 3D box is necessary to calculate the angular velocity of the box as a result of the torque just calculated, for which you would also have to know what the distribution of the mass inside the 3D box is. You probably want to assume that the mass of the box is not distributed in a non-trivial way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the full stacking of the boxes problem that you give as a background, good luck. If you're commercial it may be cheaper to buy software that can calculate this stuff than develop it yourself, or there may be open source available. Perhaps try &lt;a href=&quot;http://gamedev.stackexchange.com/&quot;&gt;http://gamedev.stackexchange.com/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-04-05T21:31:48.693" />
  <row Id="8121" PostTypeId="2" ParentId="8100" CreationDate="2011-04-05T21:44:55.333" Score="2" ViewCount="" Body="&lt;p&gt;A plane wave $e^{-i(kx~-~\omega t)}$ is a case where the Fourier transform of the wave is delta function of momentum.  The physical reality of a plane wave is then somewhat more in line with a mathematical “fiction.”  Usually there is some envelope with the wave, such as a Gaussian packet, which attenuates the wave off to “infinity.”  However, the plane wave is useful for waves with well known momentum, or a small spread or uncertainty in $\Delta k$ or $\Delta p$.  Further, one can write a plane wave as an infinite sum of Bessel functions.  The wave as it interacts with a hole in a screen selects out some of these components by Kirchoff’s rule, and the wave at the other side of the screen is a spherical wave front. &lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-04-05T21:44:55.333" />
  <row Id="8122" PostTypeId="1" CreationDate="2011-04-05T22:59:24.520" Score="8" ViewCount="187" Body="&lt;p&gt;I am a graduate student in a large experimental physics collaboration.  Newcomers to the collaboration invariably complain about excessive use of jargon and insufficient documentation as barriers to their understanding and quick integration into the collaboration.  Of course, jargon is unavoidable in any technical pursuit, and one should not expect to find comprehensive, polished documentation of what is essentially a work-in-progress, but there must be some effective techniques to communicate this stuff to new students with a minimum of frustration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What are some &quot;best practices&quot; in communicating institutional knowledge to new (graduate student and post-doc) members of a large scientific collaboration?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, what is the best way to prepare students for interaction with a big collaboration?  Perhaps a few bits of advice at the beginning could help them become integrated and acquire institutional knowledge much more quickly(?).&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-04-06T00:24:30.280" LastActivityDate="2011-04-06T23:03:22.553" Title="How to communicate institutional knowledge in a big physics collaboration?" Tags="&lt;soft-question&gt;&lt;experimental-physics&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="1" />
  <row Id="8123" PostTypeId="1" AcceptedAnswerId="8127" CreationDate="2011-04-05T23:45:16.307" Score="4" ViewCount="108" Body="&lt;p&gt;Inspired by this question: &lt;a href=&quot;http://physics.stackexchange.com/questions/8049/are-these-two-quantum-systems-distinguishable&quot;&gt;Are these two quantum systems distinguishable?&lt;/a&gt; and discussion therein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an ensemble of states, the randomness of a measurement outcome can be due to classical reasons (classical probability distribution of states in ensemble) and quantum reasons (an &lt;em&gt;individual&lt;/em&gt; state can have a superposition of states).  Because a classical system cannot be in a superposition of states, and in principle the state can be directly measured, the probability distribution is directly measurable.  So any differing probability distributions are distinguishable.  However in quantum mechanics, an infinite number of different ensembles can have the same density matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What assumptions are necessary to show that if two ensembles initially have the same density matrix, that there is no way to apply the same procedure to both ensembles and achieve different density matrices? (ie. that the 'redundant' information regarding what part of Hilbert space is represented in the ensemble is never retrievable even in principle)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To relate to the referenced question, for example if we could generate an interaction that evolved:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) an ensemble of states $|0\rangle + e^{i\theta}|1\rangle$ with a uniform distribution in $\theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) an ensemble of states $|0\rangle + e^{i\phi}|1\rangle$ with a &lt;em&gt;non-uniform&lt;/em&gt; distribution in $\phi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;such an mapping of vectors in Hilbert space can be 1-to-1. But it doesn't appear it can be done with a linear operator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it hints that we can probably prove an answer to the question using only the assumption that states are vectors in a Hilbert space, and the evolution is a linear operator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone list a simple proof showing that two ensembles with initially the same density matrix, can never evolve to two different density matrices?  Please be explicit with what assumptions you make.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: I guess to prove they are indistinguishable, we'd also need to show that non-unitary evolution like the projection from a measurement, can't eventually allow one to distinguish the underlying ensemble either.  Such as perhaps using correlation between multiple measurements or possibly instead of asking something with only two answers, asking something with more that two so that finally the distribution of answers needs more than just the expectation value to characterize the results.&lt;/p&gt;&#xA;" OwnerUserId="2702" LastEditorUserId="2702" LastEditorDisplayName="" LastEditDate="2011-04-06T00:17:15.767" LastActivityDate="2011-04-06T00:56:47.677" Title="Distinguishability in Quantum Ensembles" Tags="&lt;quantum-mechanics&gt;&lt;quantum-information&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="8124" PostTypeId="1" CreationDate="2011-04-05T23:51:06.570" Score="2" ViewCount="56" Body="&lt;p&gt;The average energy we receive from the Sun is 1,366 w/m^2, and this only varies by 0.1% from the activity peak to trough of its 11 year cycle. About 9% of the energy comes from wavelengths less than 400 nm. The 200-300 nm region varies on the order of a few percent, 150-200 nm by 10-20% and shorter regions by over 50% with maximums at the cycles peak activity. Energy above 400 nm is almost constant over the cycle. Why does solar energy variation over the Sun's activity cycle vary inversely with wavelength?&#xA;&lt;a href=&quot;http://astro.ic.ac.uk/research/solar-irradiance-variation&quot; rel=&quot;nofollow&quot;&gt;http://astro.ic.ac.uk/research/solar-irradiance-variation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-04-06T08:38:35.017" Title="Solar Cycle UV Variation" Tags="&lt;astronomy&gt;&lt;radiation&gt;&lt;sun&gt;" AnswerCount="1" />
  <row Id="8125" PostTypeId="2" ParentId="8123" CreationDate="2011-04-06T00:22:08.863" Score="1" ViewCount="" Body="&lt;p&gt;Density matrices are an alternative description of quantum mechanics. Consequently, if two ensembles have the same density matrix, they are not distinguishable.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Example, consider the unpolarized spin-1/2 density matrix which can be modeled as a system that is half pure states in the +x direction and half in the -x direction, or alternatively, as half pure states in the +z direction (i.e. spin up) and half in the -z direction (i.e. spin down):&lt;br&gt;&#xA;$$\begin{pmatrix}0.5&amp;amp;0\\0&amp;amp;0.5\end{pmatrix} = 0.5\rho_{+x}+0.5\rho_{-x} = 0.5\rho_{+z}+0.5\rho_{-z}$$&#xA;Now compute the average value of an operator $H$ with respect to these ensembles. Let&#xA;$$H = \begin{pmatrix}h_{11}&amp;amp;h_{12}\\h_{21}&amp;amp;h_{22}\end{pmatrix}$$&#xA;then the averages for the four states involved are:&#xA;$$\begin{array}{rcl}&#xD;&#xA;\langle H\rangle_{+x} &amp;amp;=&amp;amp; 0.5(h_{11}+h_{12}+h_{21}+h_{22})\\&#xD;&#xA;\langle H\rangle_{-x} &amp;amp;=&amp;amp; 0.5(h_{11}-h_{12}-h_{21}+h_{22})\\&#xD;&#xA;\langle H\rangle_{+z} &amp;amp;=&amp;amp; h_{11}\\&#xD;&#xA;\langle H\rangle_{-z} &amp;amp;=&amp;amp; h_{22}&#xD;&#xA;\end{array}$$&#xA;From the above, it's clear that taking the average over $\pm x$ will give the same result as taking the average over $\pm z$, that is, in both cases the ensemble will give an average of&#xA;$$\langle H\rangle = 0.5(h_{11}+h_{22})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any preparation of the system amounts to an operator acting on the states and so $H$ can stand for a general operation. Therefore there is no way of distinguishing an unpolarized mixture of +- x from an unpolarized mixture of +-z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument for general density matrices is similar, but I think this gets the point across.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-04-06T00:56:47.677" LastActivityDate="2011-04-06T00:56:47.677" CommentCount="3" />
  <row Id="8126" PostTypeId="2" ParentId="8100" CreationDate="2011-04-06T00:38:57.450" Score="1" ViewCount="" Body="&lt;p&gt;The questioner is considering point sources. If you consider a source that is planar, you will find that, at least near the source, it produces plane waves. Of course this is all just an approximation, but, since all experiment has accuracy limitations, everything in physics is an approximation.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-04-06T00:38:57.450" />
  <row Id="8127" PostTypeId="2" ParentId="8123" CreationDate="2011-04-06T00:56:46.480" Score="6" ViewCount="" Body="&lt;p&gt;You only need to assume&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the Schrödinger equation (yes, the same old linear Schrödinger equation, so the proof doesn't work for weird nonlinear quantum-mechanics-like theories)&lt;/li&gt;&#xA;&lt;li&gt;the standard assumptions about projective measurements (i.e. the Born rule and the assumption that after you measure a system it gets projected into the eigenspace corresponding to the eigenvalue you measured)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Then it's easy to show that the evolution of a quantum system depends only on its density matrix, so &quot;different&quot; ensembles with the same density matrix are not actually distinguishable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, &lt;a href=&quot;http://www.nyu.edu/classes/tuckerman/stat.mech/lectures/lecture_13/node3.html&quot; rel=&quot;nofollow&quot;&gt;you can derive from the Schrödinger equation&lt;/a&gt; a time evolution equation for the density matrix. This shows that if two ensembles have the same density matrix and they're just evolving unitarily, not being measured, then they will continue to have the same density matrix at all future times. The equation is&#xA;$$\frac{d\rho}{dt} = \frac{1}{i\hbar} \left[ H, \rho \right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, when you perform a measurement on an ensemble, the probability distribution of the measurment results depends only on the density matrix, and the density matrix after the measurement (of the whole ensemble, or of &lt;strong&gt;any sub-ensemble&lt;/strong&gt; for which the measurement result was some specific value) only depends on the density matrix before the measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, consider a general observable (assumed to have discrete spectrum for simplicity) represented by a hermitian operator $A$. Let the diagonalization of $A$ be&#xA;$$A = \sum_i a_i P_i$$&#xA;where $P_i$ is the projection operator in to the eigenspace corresponding to eigenvalue (measurement outcome) $a_i$. Then the probability that the measurement outcome is $a_i$ is&#xA;$$p(a_i) = \operatorname{Tr}(\rho P_i)$$&#xA;This gives the complete probability distribution of $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density matrix of the full ensemble after the measurment is&#xA;$$\rho&amp;#39; = \sum_i P_i \rho P_i$$&#xA;and the density matrix of the sub-ensemble for which the measurment value turned out to be $a_i$ is&#xA;$$\rho&amp;#39;_i = \frac{P_i \rho P_i}{\operatorname{Tr}(\rho P_i)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since none of these equations depend on any property of the ensemble other than its density matrix (e.g. the pure states and probabilities of which the mixed state is &quot;composed&quot;), the density matrix is a full and complete description of the quantum state of the ensemble.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-04-06T00:56:46.480" CommentCount="2" />
  <row Id="8129" PostTypeId="2" ParentId="8062" CreationDate="2011-04-06T01:56:36.150" Score="0" ViewCount="" Body="&lt;p&gt;From the Heisenberg Uncertainty Principle, if we know a great deal about the momentum of a particle we can know very little about its position. This suggests that our mathematics should have a quantum state that corresponds to a plane wave $\psi(x)$ with a precisely known momentum but entirely unknown position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A natural definition for the probability of finding the particle at the position $x$ is $|\psi(x)|^2$. This definition makes sense for both a real wave function and an imaginary wave function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a plane wave to have no position information is to imply that $|\psi(x)|$ does not depend on position and so is constant. Therefore we must have $\psi$ complex; otherwise there would be no way to store the information &quot;what is the momentum of the particle&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in my view, the complex nature of wave functions arises from the interaction between the necessity for (1) a probability interpretation, (2) the Heisenberg uncertainty principle, and (3) plane waves.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-04-06T01:56:36.150" CommentCount="3" />
  <row Id="8130" PostTypeId="2" ParentId="8122" CreationDate="2011-04-06T02:47:19.527" Score="2" ViewCount="" Body="&lt;p&gt;Wiki - start with something informal and simple that you and fellow grad students/new-comers use then expand it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combined with informal bag-lunch technical talks by people about their part - preferably about the equipement/software/experiment rather than the science it will do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some massive all encompassing documentation project will either never get started or develop more jargon and convoluted understanding of it's own than it ever explains.&lt;/p&gt;&#xA;" OwnerUserId="2525" LastEditorUserId="2525" LastEditorDisplayName="" LastEditDate="2011-04-06T04:46:48.050" LastActivityDate="2011-04-06T04:46:48.050" CommentCount="3" />
  <row Id="8132" PostTypeId="1" AcceptedAnswerId="8324" CreationDate="2011-04-06T03:39:20.227" Score="9" ViewCount="405" Body="&lt;p&gt;I have been studying scattering theory in Sakurai's quantum mechanics.  The phase shift in scattering theory has been a major conceptual and computational stumbling block for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How (if at all) does the phase shift relate to the scattering amplitude?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does it help you calculate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, any literature or book references that might be more accessible than Sakurai would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1807" LastActivityDate="2011-04-13T21:44:09.120" Title="Phase shifts in scattering theory" Tags="&lt;quantum-mechanics&gt;&lt;particle-physics&gt;&lt;particles&gt;&lt;scattering&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" />
  <row Id="8133" PostTypeId="1" AcceptedAnswerId="8138" CreationDate="2011-04-06T04:21:56.880" Score="1" ViewCount="126" Body="&lt;p&gt;Can someone please help me explaining how to convert units in problems like the one below? I've done many things, but I am very unsure of all my methods. Could you tell me what steps should I take?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$5.66\ \frac{\mathrm{hm}}{\mathrm{min.}} \frac{\hspace{2cm}}{}\frac{\mathrm{km}}{\mathrm{hr.}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2950" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-06T06:35:29.490" LastActivityDate="2011-04-06T07:11:17.183" Title="Unit conversion help (school)" Tags="&lt;homework&gt;&lt;unit-conversion&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="8134" PostTypeId="1" AcceptedAnswerId="9299" CreationDate="2011-04-06T04:45:43.333" Score="11" ViewCount="345" Body="&lt;p&gt;Seeing the news about 14 entangled states today @ Innsbruck:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't found a clear guide online to how many qubits we are aiming for a first practical quantum computer,&#xA;e.g. &#xA;Factorization, &#xA;Search or &#xA;re-implementing large scale computing problems?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Given the relatively few algorithms we have, and the fact that algorithms don't necessarily have to map 1:1 with the size of the domain (i.e. be multi-step), can we make any reasonable guesses for the above use cases?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia entry for Shor's algorithm seems to state &quot;The input and output qubit registers need ... twice as many qubits as necessary ...&quot; so we would need 1024 qubits for common encryption in use today e.g. AES.&#xA;Is this a correct understanding?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;references:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Shor%27s_algorithm&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Shor%27s_algorithm&lt;/a&gt;&#xA;http://en.wikipedia.org/wiki/Key_size &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are more info on any quantum algorithms suitable for large scale computing problems yet?&#xA;e.g. 50-100 qubits for 'useful' (1999) eigen* calculations &#xA;&lt;a href=&quot;http://prl.aps.org/abstract/PRL/v83/i24/p5162_1&quot; rel=&quot;nofollow&quot;&gt;http://prl.aps.org/abstract/PRL/v83/i24/p5162_1&lt;/a&gt; - only have access to the abstract&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&lt;/p&gt;&#xA;" OwnerUserId="2951" LastEditorUserId="2951" LastEditorDisplayName="" LastEditDate="2011-04-06T05:30:59.797" LastActivityDate="2011-04-30T14:57:11.593" Title="How many qubits are needed for useful computation?" Tags="&lt;quantum-computation&gt;&lt;computation&gt;&lt;quantum-computer&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="7" />
  <row Id="8135" PostTypeId="2" ParentId="8133" CreationDate="2011-04-06T05:01:52.600" Score="-1" ViewCount="" Body="&lt;p&gt;One hectometer is equal to .1 kilometers, thus you multiply 5.66Hm/m by 1km/10Hm.  The hectometers cancel out and your left with .566km/m.  You know that there are 60 minutes in an hour so you multiply by 56.6km/m by 60min/1h.  Your minutes will cancel out and you'll be left with 33.96 km/h.  If you're using significant figures this rounds to 34.0 km/h.  The only way to eliminate fractions would be to set up your answer as 33 and 96/100 which multiplied out to a simple fraction becomes 849/25.  However if you're using significant figures you'll see that the answer rounds to a whole number.&lt;/p&gt;&#xA;" OwnerUserId="2952" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-06T06:04:17.860" LastActivityDate="2011-04-06T06:04:17.860" CommentCount="3" />
  <row Id="8136" PostTypeId="2" ParentId="8122" CreationDate="2011-04-06T05:21:13.160" Score="4" ViewCount="" Body="&lt;p&gt;Although this is phrased as a management-type question, I have some experience with such a scenario when I was a summer student in CERN as an undergrad. My advisor met me once a day for an hour and most of that time was spent in debugging/checking the elementary simulations I was making in ROOT, which hardly left me with much time to know about all the physics jargon that is prominent there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes the issue with professors and senior researches in labs is that they know too much and forget where they learnt the elementary stuff three decades ago. I think it is best to find someone with not too much of an age difference who's been there longer than you, and make them your confidante. People in your age group would tend to explain stuff to you in a much more accessible and informal way and guide you through the introductory literature. The learning curve is usually steep in the beginning, but you'll get past it. Everyone does. &lt;/p&gt;&#xA;" OwnerUserId="1671" LastActivityDate="2011-04-06T05:21:13.160" />
  <row Id="8138" PostTypeId="2" ParentId="8133" CreationDate="2011-04-06T06:01:46.770" Score="7" ViewCount="" Body="&lt;p&gt;The way you convert between units is really just multiplying by several factors of 1. But it's 1 written in a slightly unusual way. Think about this: you're probably familiar with conversion factors in the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\text{number})(\text{unit}) = (\text{other number})(\text{other unit})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course, you can divide both sides of any equation by the same thing, and the equation will continue to hold. So you can also write the conversion factor as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{(\text{number})(\text{unit})}{(\text{other number})(\text{other unit})} = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{(\text{other number})(\text{other unit})}{(\text{number})(\text{unit})} = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same equation, just rearranged a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's see what happens when you have some value, expressed in $(\text{unit})$, that you want to convert to $(\text{other unit})$. As you know, you can always multiply anything by 1, and that doesn't change the value at all. So, you can go look up the conversion relationship between $(\text{unit})$ and $(\text{other unit})$, change it into one of the forms above, and use it like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\text{given value})(\text{unit})\times 1 = (\text{given value})(\text{unit})\times \frac{(\text{other number})(\text{other unit})}{(\text{number})(\text{unit})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you have $(\text{unit})$ in both the numerator and denominator, you can cancel those out. Note that it's up to you to pick the right conversion factor to use so that you can cancel out units! Anyway, you'll be left with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{(\text{given value})(\text{other number})}{(\text{number})}(\text{other unit})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything except for $(\text{other unit})$ is just a number, so now you can actually figure out the numerical value (e.g. plug things into your calculator if necessary).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your given value has a compound unit (one made as a combination of other units, as in your examples), then you will need to use more than one unit conversion factor - basically, you will have to multiply by 1 more than once, using a different conversion factor each time. In your example, you would have to multiply by a conversion factor to go from $\mathrm{hm}$ to $\mathrm{km}$, and another one to go from $\mathrm{min.}$ to $\mathrm{hr.}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how that applies to your example, assuming Mark Eichenlaub is right about the mistake:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$(\text{given value}) = 5.66$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For one part of the unit conversion:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$(\text{unit}) = \mathrm{hm}$&lt;/li&gt;&#xA;&lt;li&gt;$(\text{other unit}) = \mathrm{km}$&lt;/li&gt;&#xA;&lt;li&gt;The conversion factor is $10\text{ hm} = 1\text{ km}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the other part of the unit conversion:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$(\text{unit}) = \mathrm{min.}$&lt;/li&gt;&#xA;&lt;li&gt;$(\text{other unit}) = \mathrm{hr.}$&lt;/li&gt;&#xA;&lt;li&gt;The conversion factor is $60\text{ min.} = 1\text{ hr.}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that the time units are in the denominator! But that doesn't really change anything  - you still just need to arrange the conversion factor in the way that makes the original units cancel out.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-04-06T06:01:46.770" />
  <row Id="8139" PostTypeId="2" ParentId="8133" CreationDate="2011-04-06T07:11:17.183" Score="2" ViewCount="" Body="&lt;p&gt;I think &lt;a href=&quot;http://physics.stackexchange.com/questions/8133/unit-conversion-help-school/8138#8138&quot;&gt;David's answer&lt;/a&gt; is the right one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the wrong one, but it's the way I actually think about unit conversions when I do them.  (I mean &quot;wrong&quot; in the sense of &quot;unofficial&quot;, not that it gives the wrong answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I tell you there are 5280 feet in a mile, and ask how many feet are in two miles.  You don't need to think about what to do.  It's obvious there are twice as many feet in two miles as in one, so it's 2*5280 = 10,560 ft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you want to convert 15 miles per hour to meters per second.  Do it in two steps.  Go from miles per hour to meters per hour, and then from meters per hour to meters per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are about 1609 meters in a mile, so how many meters per hour are you going?  In one hour you go 15 miles, so you go 15*1609 = 24135 meters in an hour.  You have to multiply because you go a lot more meters (which are small) than miles (which are big).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, there are 3600 seconds in an hour.  If you go 24135 meters in an hour, how many meters do you go per second?  Seconds are short, while hours are long.  You clearly go a much shorter distance in one second than in one hour, so divide by 3600 to get 24135/3600 = 6.7 m/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you can always check your answer using Google like this:&#xA;&lt;a href=&quot;http://www.google.com/search?&amp;amp;q=15+mph+in+m/s&quot; rel=&quot;nofollow&quot;&gt;http://www.google.com/search?&amp;amp;q=15+mph+in+m/s&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or using Wolfram Alpha like this:&#xA;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=15+mph+in+m%2Fs&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=15+mph+in+m%2Fs&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-04-06T07:11:17.183" CommentCount="1" />
  <row Id="8140" PostTypeId="2" ParentId="8122" CreationDate="2011-04-06T07:16:56.243" Score="4" ViewCount="" Body="&lt;p&gt;To orient oneself in a large group, now reaching the 3000 number, is formidable. Fortunately it is broken in smaller groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My advice for a new student would be to concentrate on the specific area of his/her task, and slowly, step by step, by asking questions of people on shift, searching on the net, get the necessary vocabulary for that task. A small notebook might help,for new jargon and concepts.&#xA;After about a month, expand the circle of &quot;knowledge&quot; to a similar group: if you have been working on the hadron calorimeter, find coffee company with electromagnetic calorimetry people, or the trigger group etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data base  will be expanding step by step. By the end of the year reading the technical report will not be an enormous task.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-06T07:16:56.243" />
  <row Id="8141" PostTypeId="2" ParentId="8124" CreationDate="2011-04-06T08:38:35.017" Score="1" ViewCount="" Body="&lt;p&gt;This must be due to the fact, that majority of visible spectra is emitted by whole sun surface, which have stable size &amp;amp; temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While short-wave UV must be emitted by hotspots &amp;amp; bursts, number &amp;amp; size of which is not constant and varies in large range.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-04-06T08:38:35.017" CommentCount="1" />
  <row Id="8142" PostTypeId="1" CreationDate="2011-04-06T09:57:55.807" Score="3" ViewCount="233" Body="&lt;p&gt;First of all I am not a professional physicist. I was curious as to how a particle of light can reach the maximum speed of light c given that particles experience an increase in mass as they approach c ?&lt;/p&gt;&#xA;" OwnerUserId="2954" LastActivityDate="2011-08-23T14:14:33.643" Title="How does a particle of light reach the max speed of light?" Tags="&lt;special-relativity&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8143" PostTypeId="1" AcceptedAnswerId="8160" CreationDate="2011-04-06T10:00:20.253" Score="4" ViewCount="180" Body="&lt;p&gt;I'm studying basic physics. I'm using the text available at &lt;a href=&quot;http://www.anselm.edu/internet/physics/cbphysics/downloadsI.html&quot; rel=&quot;nofollow&quot;&gt;http://www.anselm.edu/internet/physics/cbphysics/downloadsI.html&lt;/a&gt;. It develops the universal law of gravitation by postulating the existence of a vector at each point of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;g_P = \sum G\frac{m_i}{|r|_i^3}r_i,&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $m_i$ and $r_i$ are the mass of and separation vector from $P$ for all particles that aren't at the point $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It examines the effect of one particle on another. If the separation vector is $r$ then from the above equation, we see that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;g = G\frac{m_1}{|r|^3}r&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that when a particle of mass $m_2$ is placed at the given point, the force will be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;&#xD;&#xA;F = G\frac{m_1m_2}{|r|^3}r&#xD;&#xA;&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors then go on to claim that we can either repeat the development to see the effect of the second particle on the first one or apply Newton's third law. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does Newton's third law apply through a field? If it's the field exerting the force, then Newton's third law would require a force on the field and not the object &quot;generating&quot; the field, correct?&lt;/p&gt;&#xA;" OwnerUserId="2953" LastActivityDate="2011-04-06T17:08:38.633" Title="Fields and Newton's Third Law" Tags="&lt;classical-mechanics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8144" PostTypeId="1" AcceptedAnswerId="8146" CreationDate="2011-04-06T10:05:35.753" Score="4" ViewCount="138" Body="&lt;p&gt;Imagine, if you will, a string background with an extremely tiny value for the string coupling. The Planck scale is many orders of magnitude smaller in size than the string scale. Is the UV/IR mixing scale the Planck scale, or the string scale? What happens at intermediate scales?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitons are string modes, and as such, their spatial extent is the string scale. They can never resolve anything smaller than that. On the other hand, D-branes are more sensitive.&lt;/p&gt;&#xA;" OwnerUserId="2955" LastActivityDate="2011-04-06T10:29:41.067" Title="At what scale does UV/IR mixing happen in string theory?" Tags="&lt;string-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="8145" PostTypeId="2" ParentId="8142" CreationDate="2011-04-06T10:12:20.797" Score="10" ViewCount="" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Photon&quot; rel=&quot;nofollow&quot;&gt;photon&lt;/a&gt; does not &quot;reach&quot; light speed, it is &quot;born&quot; with &#xA;that velocity. &#xA;And because the photon has no rest mass, there is no problem &#xA;inherent. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-04-06T10:49:41.900" LastActivityDate="2011-04-06T10:49:41.900" />
  <row Id="8146" PostTypeId="2" ParentId="8144" CreationDate="2011-04-06T10:29:41.067" Score="1" ViewCount="" Body="&lt;p&gt;In perturbative string theory, for $g\ll 1$, it's always the string scale where the UV/IR mixing happens first. It's the lightest mass scale associated with the lightest (possibly extended) objects - and in perturbative string theory regimes, it's always the fundamental strings. For example, the low-energy spectrum at masses below $M_{\rm string}$ is always related to the high-energy spectrum at masses above $M_{\rm string}$ by the modular invariance (of the toroidal world sheet) - and similarly for open strings and cylindrical world sheets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different effects may see different forms of UV/IR mixing for which different scales are relevant, however. The term UV/IR mixing is a somewhat vague, umbrella term that covers many effects in string theory or quantum gravity. D-branes may resolve shorter distances than the string scales but there is actually always some sense in which even e.g. D0-branes are linked to the string scale. If they resolve distances $\Delta x$ and times $\Delta t$, then $\Delta x \cdot \Delta t &amp;gt; L_{\rm string}^2$. Yoneya would play with insights like this. D0-branes may resolve sub-stringy distances but only if their velocity is very small, in which case they can only measure the time with much-worse-than-stringy resolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also - less understood - aspects of the UV/IR mixing that arise at the Planck scale. But some of them are known so incompletely that the arguments are only valid up to the assumption that $g$ is considered to be of order one. The relevant scale where geometry - with degrees of freedom at each &quot;layer&quot; of scales being independent of each other - fails to be applicable is always given by the scale associated with the first higher-order corrections to the Einstein-Hilbert action etc. and it's always the string scale i.e. the scale associated with the tension of the lightest fundamental objects.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-06T10:29:41.067" CommentCount="2" />
  <row Id="8147" PostTypeId="2" ParentId="8062" CreationDate="2011-04-06T12:26:30.147" Score="7" ViewCount="" Body="&lt;p&gt;More physically than a lot of the other answers here (a lot of which amount to &quot;the formalism of quantum mechanics has complex numbers, so quantum mechanics should have complex numbers), you can account for the complex nature of the wave function by writing it as $\Psi (x) = |\Psi (x)|e^{i \phi (x)}$, where $i\phi$ is a complex phase factor.  It turns out that this phase factor is not directly measurable, but has many measurable consequences, such as the double slit experiment and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Aharonov%E2%80%93Bohm_effect&quot; rel=&quot;nofollow&quot;&gt;Aharonov-Bohm effect&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are complex numbers essential for explaining these things?  Because you need a representation that both doesn't induce time and space dependencies in the magnitude of $|\Psi (x)|^{2}$ (like multiplying by real phases would), AND that &lt;b&gt;DOES&lt;/b&gt; allow for interference effects like those cited above.  The most natural way of doing this is to multiply the wave amplitude by a complex phase.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2011-04-06T15:32:34.653" LastActivityDate="2011-04-06T15:32:34.653" />
  <row Id="8148" PostTypeId="2" ParentId="8112" CreationDate="2011-04-06T12:51:38.440" Score="2" ViewCount="" Body="&lt;p&gt;Well, the question as stated : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What do gravitons interact with?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is naive enough . It assumes gravitons exist in the way photons and gluons and Z exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In such a framework, the answer is : they interact with everything for which a Feynman diagram can be imagined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The other three forces' mediating particles (photons etc.) are absorbed by their appropriate charge-carrying particles,&lt;/em&gt;  is not quite correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force mediating quanta interact with all matter, even if not at first order, through higher order loop Feynman diagrams , is the correct formulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;but I can't seem to find a clear answer that applies to the gravitational force in a quantified scenario.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mean a quantized scenario.&#xA;In string theory, where the Feynman diagram formalism is appropriately modified for the extension to strings, a graviton will be a mediating particle on par with the other  &quot;forces&quot; mediating interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As was discussed in &lt;a href=&quot;http://physics.stackexchange.com/questions/7843/how-many-fundamental-forces-could-there-be&quot;&gt;another question,&lt;/a&gt; the forces are an artificial carry over from the macroscopic classical world. What exist are interactions that are characterized by coupling constants which distinguish  if the specific interaction is strong, gravitational, supersymmetric or...&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-04-06T15:01:32.453" LastActivityDate="2011-04-06T15:01:32.453" CommentCount="1" />
  <row Id="8149" PostTypeId="1" AcceptedAnswerId="8293" CreationDate="2011-04-06T14:00:05.800" Score="4" ViewCount="370" Body="&lt;p&gt;Let's assume two spherical particles $p_1$ and $p_2$ of finite radius $r_1$ and $r_2$, which are at locations $(\pm\frac{d}{2},0,0)$ a distance $d$ apart at initial time $t$. These particles diffuse with coefficients $D_1$ and $D_2$, respectively. How can I obtain the probability distribution of the collision time (that is, if the time at which they collide is $t + \Delta t$, I would like to know the probability density function of the random variable $\Delta t$), or at least some of its moments, in these two cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;when the domain is infinite, e.g., $\mathbb{R}^3$, and&lt;/li&gt;&#xA;&lt;li&gt;when it is bounded (inside a sphere with surface area $A$ and the origin $(0,0,0)$ as its center)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I tried to look up for an answer in Van Kampen's book, but I couldn't really come up with an answer.&lt;/p&gt;&#xA;" OwnerUserId="2320" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-04-08T20:37:47.943" LastActivityDate="2011-05-07T18:01:48.443" Title="Collision time of Brownian particles" Tags="&lt;statistical-mechanics&gt;&lt;diffusion&gt;" AnswerCount="2" CommentCount="13" FavoriteCount="5" />
  <row Id="8150" PostTypeId="1" CreationDate="2011-04-06T14:50:40.467" Score="9" ViewCount="233" Body="&lt;p&gt;A flop transition changes the second homotopy group of a Calabi-Yau compactifation, but not the fundamental group or the number of connected components. Can the number of connected spatial components change in string theory? Can a part of space pinch off never to interact with the rest of space?&lt;/p&gt;&#xA;" OwnerUserId="2955" LastActivityDate="2011-04-11T05:38:35.343" Title="Can the connectivity of space change in string theory?" Tags="&lt;string-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" />
  <row Id="8151" PostTypeId="1" AcceptedAnswerId="8152" CreationDate="2011-04-06T15:12:01.313" Score="6" ViewCount="288" Body="&lt;p&gt;In mathematics, we talk about tangent vectors and cotangent vectors on a manifold at each point, and vector fields and cotangent vector fields (also known as differential one-forms). When we talk about tensor fields, we mean differentiable sections of some tensor power of the tangent or cotangent bundle (or a combination).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various natural differentiation operations, such as the exterior derivative of anti-symmetric covariant tensor fields, or the Lie derivative of two vector fields. These have nice coordinate-free definitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In physics, there is talk of &quot;covariant derivatives&quot; of tensor fields, whose resulting objects are different kind of tensor fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering, what is the abstract interpretation of the general notion of a covariant derivative in terms of (tensor products of) tangent vectors and vector fields.&lt;/p&gt;&#xA;" OwnerUserId="2591" LastActivityDate="2011-04-06T19:59:02.093" Title="What is the covariant derivative in mathematician's language?" Tags="&lt;mathematics&gt;&lt;tensor-calculus&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="8152" PostTypeId="2" ParentId="8151" CreationDate="2011-04-06T15:19:45.613" Score="6" ViewCount="" Body="&lt;p&gt;The notion of &lt;em&gt;covariant derivative&lt;/em&gt; is equivalent to the notion of &lt;a href=&quot;http://en.wikipedia.org/wiki/Connection_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;connection&lt;/a&gt;. More precisely, for every connection $\nabla$ and vector field $X$, the operation $\nabla_X$ is a covariant derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Connections $\nabla$ on the tangent bundle $TM$ of a manifold are usually induced by a metric, this is the so called &lt;a href=&quot;http://en.wikipedia.org/wiki/Levi-Civita_connection&quot; rel=&quot;nofollow&quot;&gt;Levi-Citiva connection&lt;/a&gt;. It is essentially determined by the requirement that the scalar product fulfills the product rule&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ d_X\langle v,w\rangle = \langle \nabla_X v, w\rangle + \langle v, \nabla_X w \rangle .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this does not yet determine the connection completely, one has to add the additional requirement that there is no torsion,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \nabla_X Y - \nabla_Y X = [X,Y] .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;This is a coordinate-free specification of a connection. Note, however, that unlike the Lie-derivative or the derivatives of differential forms, which are defined in terms of the manifold alone, connections represent additional data. Many different connections are possible on a single manifold, whereas the other two notions of derivative are unique.&lt;/p&gt;&#xA;" OwnerUserId="188" LastEditorUserId="188" LastEditorDisplayName="" LastEditDate="2011-04-06T17:53:49.907" LastActivityDate="2011-04-06T17:53:49.907" CommentCount="3" />
  <row Id="8153" PostTypeId="2" ParentId="8122" CreationDate="2011-04-06T15:22:08.137" Score="0" ViewCount="" Body="&lt;p&gt;i think the solution is to identify a single individual that will serve as a information manager that will ask people if they ought to create a new wiki or contribute/modify an existing one. Since that role only involves redirecting people to where they should contribute, &lt;em&gt;should&lt;/em&gt; not be very time consuming&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will take a bit of time to reroute the information this way into a more coherent structure, but it will pay off&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-04-06T15:22:08.137" />
  <row Id="8154" PostTypeId="2" ParentId="8143" CreationDate="2011-04-06T15:46:22.357" Score="0" ViewCount="" Body="&lt;p&gt;The force field makes sense when it is an external force in a probe mass equation motion. It is too strong claim that the field exists &quot;everywhere&quot;. In fact, it is a wrong claim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a classical source of light at a point A, it may emit a spherical wave that &quot;exists&quot; everywhere at a given R. But for a low-intensity, quantum source, this wave describes not the wave amplitude but the probability amplitude. Not all probe bodies placed around a quantum source can receive a single photon.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-04-06T15:46:22.357" />
  <row Id="8155" PostTypeId="2" ParentId="7191" CreationDate="2011-04-06T16:05:52.323" Score="1" ViewCount="" Body="&lt;p&gt;I have examined some material on Statics and having recently discovered this unanswered question I can provide some indications from a physics perspective of what is going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fundamental issue in statics is to provide the values for the forces acting on a rigid construction (usually of rods) with everything in equilibrium. (The examples I have looked at are 2 dimensional, so I will assume that in what follows; but one could add similar equations for 3 D although the essence of the OP question can be discussed in 2D.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations available are from the Newtonian conservation laws:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Sigma F_x = 0$ - the sum of forces in the x-axis is zero&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Sigma F_y = 0$ - the sum of forces in the y-axis is zero&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Sigma M_a = 0$  - the moment (torque) about a point (hence any point) is zero&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So these equations will be true at any point, especially connection points on the construction (often called a &quot;truss&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variables that form in these equations are the Forces at given points on the system: more accurately the x- and y- components of the Forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system (truss) will be placed on a ground at N points; at each of these points we have a &quot;support reaction&quot; onto the truss. Thus there will be N unknown support reactions (in the y-axis say) to find in principle: labeled $V_1$, $V_2$,....,$V_N$ say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There will also likely to be some form of external force in the problem. This external Force $F$ is normally assumed applied at a specific point on the structure, and may have a horizontal (x-axis) component and a downward vertical (-ve y-axis component). I should remark here that in a wider class of physics and engineering problems the force F is assumed to be &lt;em&gt;distributed&lt;/em&gt; across the structure and not located just at a point: the primary example of such a force would be gravity. However the mathematical techniques to solve this would involve calculus and so are outside of the basic Statics theory I have seen and that is covered here. So assume a single force is acting at a point externally. Can this problem be solved for all the forces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well it turns out that there is a mathematical problem here, which I shall explain with some simple examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example 1: Beam with 2 supports and external force&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the length of the beam be L units, supports (A,B) at each end and an external force F acting purely vertically on the beam at a point distance $a$ from A, $b$ from B (hence $L=a+b$). Then we have to solve (signs are important in general):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] $\Sigma F_y$ = $V_A + V_B - F = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] $\Sigma M_A$ = $aF - (a + b) V_B = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $F$, $a$ and $b=L-a$ are known and $V_A$ and $V_B$ are unknown. The key point here is that we have two equations with two unknowns. The solution is easily seen to be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V_A = (b/a+b) F = (b/L)F$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V_B = (a/a+b) F = (a/L)F$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is called a &lt;strong&gt;statically determinate&lt;/strong&gt; system. However the simplest of modifications can result in a &lt;strong&gt;statically indeterminate&lt;/strong&gt; system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example 2: Beam with 3 supports&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wont go through the details which are in this &lt;a href=&quot;http://en.wikipedia.org/wiki/Statically_determinate&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;. The point is that there are the same equations except that there is now a third reaction force $V_C$. Therefore there are too many variables and this problem cannot be solved (by Statics alone). In a sense this is quite a general case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the engineering challenge remains to model the forces on a connected truss. The &lt;strong&gt;Method of Joints&lt;/strong&gt; does this going through the Truss iteratively: at each connection point with just a few forces and variables such that the problem is determinate and such that a new unknown force is determined. Eventually by repeated application of this technique all the forces on all the beams of the truss can be determined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A different question can always be asked about any physical system however: namely what are the forces acting on a particular point? This problem does not always need the full solution of the forces acting on every other point determined before it can be answered. So the idea is to construct a &lt;a href=&quot;http://en.wikipedia.org/wiki/Free_body_diagram&quot; rel=&quot;nofollow&quot;&gt;Free Body Diagram&lt;/a&gt; which extracts the key physics (in this case Statics) of the problem. Doing so in Static Truss like problems is called the &lt;strong&gt;Method of Sections&lt;/strong&gt;. The key equations are the Newton equations of Force and Torque used in Statics, but applied in a selective way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This technique of determining where to &quot;cut&quot; the truss looks like a bit of an art, although there do seem to be some principles used in the Tutorial material I have seen. Here is one such &lt;a href=&quot;http://www.ce.memphis.edu/3121/notes/notes_03c.pdf&quot; rel=&quot;nofollow&quot;&gt;Trusses - Method of Sections &lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-04-07T10:03:14.823" LastActivityDate="2011-04-07T10:03:14.823" />
  <row Id="8156" PostTypeId="1" AcceptedAnswerId="8193" CreationDate="2011-04-06T16:08:51.510" Score="2" ViewCount="333" Body="&lt;p&gt;When finding the discrete energy states of a operator I have been taught to use the time-independent Schrodinger equation which restates the definition of eigenvalues and eigenvectors. What I don’t understand is why the eigenvalues are the energy states, is there firstly a mathematical reason and secondly a physical reason?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this arise from Hamiltonian or Lagrangian mechanics which I am not familiar with?&lt;/p&gt;&#xA;" OwnerUserId="2958" LastEditorUserId="2958" LastEditorDisplayName="" LastEditDate="2011-04-06T17:01:23.230" LastActivityDate="2011-04-08T16:43:13.283" Title="Eigenvalues of an operator correspond to energy states in quantum mechanics, why?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="8158" PostTypeId="2" ParentId="8156" CreationDate="2011-04-06T16:31:33.940" Score="3" ViewCount="" Body="&lt;p&gt;Both eigenvalues and eigenstates belong to some &lt;strong&gt;operator&lt;/strong&gt;. In your case, this is the Hamiltonian operator $\hat H$. It's fundamental because of many reasons. First is that it is indeed an operator that represents energy in a sense that possible energy levels are encoded in its spectrum (i.e. a set of eigenvalues). The second important reason is that it is the operator that can be found in Schrodinger equation $i \hbar \partial_t \left | \psi(t) \right &amp;gt; = \hat H \left | \psi(t) \right &amp;gt;$. This equation can then be solved by writing $\left | \psi(t) \right &amp;gt;$ as superposition of eigenstates of $\hat H$: $\left | \psi(t) \right &amp;gt; = \sum_n c_n(t) \left | \psi_n \right &amp;gt;$. If we can find these states, we are done as $c_n(t) = \exp({-iE_n t \over \hbar}) c_n(t=0)$ solves the equation (and it also shows the importance of these eigenstates because they are preserved by time evolution).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this means the problem of time evolution in quantum mechanics can be reduced to the problem of finding the eigenvalues and eigenstates of $\hat H$, the equation for that being $\hat H \left | \psi_n \right&amp;gt; = E_n \left| \psi_n \right&amp;gt;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt; the above assumes that $\hat H$ is time-independent. If it's not (as is the case in basically all practical applications, using perturbation theory) then we use different techniques, e.g. of path integration, or various scattering formulas.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-07T09:05:16.697" LastActivityDate="2011-04-07T09:05:16.697" />
  <row Id="8159" PostTypeId="2" ParentId="8150" CreationDate="2011-04-06T16:52:01.330" Score="1" ViewCount="" Body="&lt;p&gt;There is a very interesting way in which the process of pinching off a spacetime (or geometry) to create two disconnected regions. This picture is due to Mark Van Raamsdonk which is outlines in the essay &lt;a href=&quot;http://arxiv.org/abs/1005.3035v1&quot; rel=&quot;nofollow&quot;&gt;Building up spacetime with quantum entanglement&lt;/a&gt;. Incidentally this paper was also the winner of the &lt;a href=&quot;http://www.gravityresearchfoundation.org/announcements.html&quot; rel=&quot;nofollow&quot;&gt;2010 Gravity Research Foundation&lt;/a&gt; essay contest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture itself is fairly straightforward if you have some familiarity with the basic notions of the duality between (super)gravity on an (asymptotically) Anti-DeSitter (AdS) spacetime and a CFT (conformal field theory) on living on the boundary of this spacetime. However, I just read the paper a day or two ago so please excuse any stupid mistakes in my explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Raamsdonk's reasoning builds upon a &lt;a href=&quot;http://arxiv.org/abs/hep-th/0603001&quot; rel=&quot;nofollow&quot;&gt;result&lt;/a&gt; according to which given a CFT on a boundary $\partial M$ of a spacetime $M$, the entanglement entropy between the degrees of freedom of a subset ($A$ in the figure below taken from Raamsdonk's paper) of $\partial M$ is given by the area of the static minimal surface ($\tilde A$ in figure) in the bulk AdS geometry whose boundary is given by $\partial A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/J4cxw.jpg&quot; alt=&quot;Minimal Surface&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the situation when, as shown in the figure above, the boundary $\partial A$ of the minimal surface $\tilde A$ divides the boundary of the bulk spacetime into two regions $A$ and $B$. Then the argument goes as follows - the area of $\tilde A$ is a measure of the entanglement entropy between the degrees of freedom in $A$ and in $B$. Shrinking the minimal surface $\tilde A$ will then have the effect of reducing the amount of entanglement between the d.o.f. in these two regions. The physical effect of decreasing the area of $\tilde A$ in the &lt;em&gt;bulk&lt;/em&gt; geometry is to &quot;pinch off&quot; the two regions of the AdS spacetime separated by this minimal surface, until eventually we are left with two disconnected copies of AdS spacetime, each of whose respective boundary CFT's are no longer entangled with each other. This is illustrated in the following figure also taken from Raamsdonk's paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/T0DnM.jpg&quot; alt=&quot;Spacetime Pinching&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this line of reasoning suggests is that the emergence of a classical spacetime is intricately linked with the entanglement of the d.o.f of a quantum many-body system - in this case the boundary CFT. In other words by entangling the CFT's on the boundaries of two disconnected spacetimes we can join the two into &lt;em&gt;one&lt;/em&gt; connected spacetime and vice versa by disentangling the d.o.f of the boundary CFT of a given spacetime we create conditions for it to separate into disconnected components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arguments as presented here are necessarily lacking in details for lack of space and time. For these I suggest reading Raamsdonk's GRF essay winning paper linked to above and the &lt;a href=&quot;http://arxiv.org/abs/hep-th/0603001&quot; rel=&quot;nofollow&quot;&gt;Ryu-Takayanagi&lt;/a&gt; PRL which shows the relation between minimal surfaces in AdS and entanglement entropy of the CFT on the boundary of the AdS spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;strong&gt;Nb:&lt;/strong&gt; Since the AdS/CFT duality is usually understood as having a string-theoretic basis, the above answer satisfies the OP's requirement that string theory be made use of in explaining the connectivity of spacetime.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;strong&gt;Edit:&lt;/strong&gt; For more on the relation between the geometry of AdS and entanglement in spin-systems, see @genneth's very interesting answer to the question &lt;a href=&quot;http://physics.stackexchange.com/questions/8162/whats-so-special-about-ads&quot;&gt;What's so special about AdS?&lt;/a&gt; where he talks about the so-called Multi-scale Entanglement Renormalisation Ansatz (MERA) and its relationship to AdS geometry]&lt;/p&gt;&#xA;" LastActivityDate="2011-04-11T05:38:35.343" />
  <row Id="8160" PostTypeId="2" ParentId="8143" CreationDate="2011-04-06T17:08:38.633" Score="3" ViewCount="" Body="&lt;p&gt;In Newtonian physics, the field is not really something physical that has an independent existence. Particularly in Newtonian gravity, the gravitational force is really action at a distance with nothing mediating the force in between. The field $g$ defined here is simply for mathematical convenience and is not the usual field that you talk about in a fully relativistic classical field theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as long as you're not doing any relativistic calculations and asking questions like &quot;Does the force act instantaneously? In what frame?&quot;, it's perfectly fine to just treat the particles as obeying the Third Law, with equal and opposite forces between them that act at a distance, and use the field $g$ only as a convenience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you want a relativistic theory, you have to introduce a real physical field that can carry momentum and energy at every point in space. This is done in classical electrodynamics and general relativity, for example. You will see there that particles do not obey a simple Newton's Third Law since you have to take into account the &lt;em&gt;dynamics&lt;/em&gt; of the field too. (There is no dynamics for the field itself in Newtonian physics, because like I said, the field is not something physical in Newtonian physics) This is what gives rise to electromagnetic and gravitational waves in those two field theories, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary: Don't take the &quot;field&quot; $g$ in a physical sense.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-04-06T17:08:38.633" />
  <row Id="8161" PostTypeId="2" ParentId="8082" CreationDate="2011-04-06T17:45:37.903" Score="1" ViewCount="" Body="&lt;p&gt;Following on from Peter's answer I will add some more suggestions although I also recognise some ambiguity (or underdetermination) in the problem - some will be identified at the end of the discussion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There seems to be two aspects to the physics of this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Determine the forces at $[x_i,y_i]$ for all the N points on the box&lt;/li&gt;&#xA;&lt;li&gt;Determine whether there will be some non-zero moment to the box.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The equations of statics that are basic to this are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Sigma F_x = 0$ - the x axis forces sum to zero&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Sigma F_y = 0$ - the y axis forces sum to zero&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Sigma F_z = 0$ - the z axis forces sum to zero&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Sigma M_a = 0$ - total moment about any axis is zero for equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other forces include the reaction $V_i$ forces at each $[x_i,y_i]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One problem is that this scenario is &lt;em&gt;statically indeterminate&lt;/em&gt; - meaning that the distribution of forces using &lt;em&gt;only&lt;/em&gt; statics will not give a complete answer. Some other form of approximation or data is also required. The method of sections (briefly in Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Structural_analysis#Strength_of_materials_methods_.28classical_methods.29&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) and perhaps the &lt;a href=&quot;http://en.wikipedia.org/wiki/Moment_distribution_method&quot; rel=&quot;nofollow&quot;&gt; moment distribution method&lt;/a&gt; are worth studying. These methods are for solving general engineering problems of this type (like the truck itself maybe). I do not know whether there is freedom to simplify some of these properties in your model, thus resulting in a much simpler &quot;force distribution model&quot; inside the box. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One suggestion might be to consider a &quot;truss approximation&quot; to the floor of the box (even the entire box) with the truss joints corresponding to the reaction points of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the perspective of the non-zero moment of the box it is not clear where the freedom of movement is coming from. For example if each of the N points were considered a high stilt then the configuration might not be stable, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the simplest possible example of this case we have a(n essentially massless) beam of length L sitting on two (high) structs at A and B with each point interior to the beam, maybe A is distance a, B distance b from the LHS. Let a force $F_1$ be acting on the LHS downwards, and let $F_2$ be acting on the RHS downwards (perhaps from edges etc). Then the &lt;/p&gt;&#xA;&#xA;&lt;p&gt;LHS moment = $aF_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RHS moment = $(L-a)F_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For stability we require RHS&gt;LHS ie $(L-a)F_2 &amp;gt; aF_1$ ie $F_2 &amp;gt; (a/(L-a))F_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly these equations can and must be generalised further depending on the exact level of assumptions one includes in the full model (massive box, height/strength of structs, interior mass distribution in box, strength of box walls, side forces, friction, ...).&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-04-07T09:59:42.703" LastActivityDate="2011-04-07T09:59:42.703" CommentCount="2" />
  <row Id="8162" PostTypeId="1" CreationDate="2011-04-06T19:19:28.993" Score="11" ViewCount="481" Body="&lt;p&gt;This question is coming from someone who has very little experience with M-Theory but is intrigued by the AdS/CFT correspondence and is beginning to study it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the gauge/gravity duality discussed almost always in the context of anti-deSitter space? What is unique about it? What are the difficulties in studying it in Schwarzschild, deSitter, etc.? References to work done on the gauge/gravity duality in these more physical spacetimes would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-04-07T15:33:10.347" Title="What's so special about AdS?" Tags="&lt;string-theory&gt;&lt;ads-cft&gt;" AnswerCount="5" FavoriteCount="2" />
  <row Id="8163" PostTypeId="1" AcceptedAnswerId="8165" CreationDate="2011-04-06T19:51:11.833" Score="0" ViewCount="230" Body="&lt;p&gt;What happens if you would ride on top of a light stream and you would look into a mirror that is in front of you, could you actually see your own face? I am asking this because I heard that nothing can be faster than light, this should also be the case here, shouldn't it?&lt;/p&gt;&#xA;" OwnerUserId="2968" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-06T20:26:03.530" LastActivityDate="2011-04-06T20:26:03.530" Title="Can you see yourself in a mirror when you are riding on top of a light stream?" Tags="&lt;special-relativity&gt;&lt;light&gt;&lt;speed-of-light&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8164" PostTypeId="2" ParentId="8151" CreationDate="2011-04-06T19:59:02.093" Score="4" ViewCount="" Body="&lt;p&gt;A covariant derivative on some manifold is a map which takes each differentiable tensor field of type (n,m) to a tensor field of type (n,m+1) with the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Linearity&lt;/li&gt;&#xA;&lt;li&gt;Leibnitz rule&lt;/li&gt;&#xA;&lt;li&gt;Commutativity with contraction&lt;/li&gt;&#xA;&lt;li&gt;Consistency with notion of tangent vectors&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Additionally, one may wish to impose the condition of vanishing torsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using abstract index notation the above conditions translate to the following conditions for the covariant derivative $\nabla_a$:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\nabla_a (A T^{a_1 \dots a_n}{}_{b_1 \dots b_m} + B \hat T^{a_1 \dots a_n}{}_{b_1 \dots b_m} ) = A \nabla_a T^{a_1 \dots a_n}{}_{b_1 \dots b_m} + B \nabla_a \hat T^{a_1 \dots a_n}{}_{b_1 \dots b_m}$&lt;/li&gt;&#xA;&lt;li&gt;$\nabla_a (T^{a_1 \dots a_n}{}_{b_1 \dots b_m}\hat T^{\hat a_1 \dots \hat a_n}{}_{\hat b_1 \dots \hat b_m}) = (\nabla_a T^{a_1 \dots a_n}{}_{b_1 \dots b_m}) \hat T^{\hat a_1 \dots \hat a_n}{}_{\hat b_1 \dots \hat b_m} + T^{a_1 \dots a_n}{}_{b_1 \dots b_m} \nabla_a \hat T^{\hat a_1 \dots \hat a_n}{}_{\hat b_1 \dots \hat b_m}$&lt;/li&gt;&#xA;&lt;li&gt;$\nabla_a \big(T^{a_1 \dots c \dots a_n}{}_{b_1 \dots c \dots b_m}\big) = \nabla_a T^{a_1 \dots c \dots a_n}{}_{b_1 \dots c \dots b_m}$&lt;/li&gt;&#xA;&lt;li&gt;For any tangent vector $v^a$ and any sufficiently smooth scalar function $f$ you have $v(t)=v^a\nabla_a f$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The condition of vanishing torsion requires the covariant derivative to commute on sufficiently smooth scalar functions $f$, $[\nabla_a,\nabla_b]f=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take the commutator of the covariant derivative acting on vector fields you get a nice algebraic interpretation of torsion $T$ and curvature $R$:&#xA;$&#xD;&#xA;[\nabla_a,\nabla_b]v^c = T^d{}_{ab}\nabla_d v^c + R^c{}_{dab} v^d&#xD;&#xA;$&#xA;Thus, torsion coefficients are like structure constants in a Lie algebra and curvature resembles a central charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general relativity torsion is required to vanish. Together with the condition of metric compatibility, $\nabla_a g_{bc}=0$, where $g_{ab}$ is the metric, this defines uniquely your covariant derivative. The ensuing connection is then the Christoffel connection.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-04-06T19:59:02.093" />
  <row Id="8165" PostTypeId="2" ParentId="8163" CreationDate="2011-04-06T20:07:17.863" Score="4" ViewCount="" Body="&lt;p&gt;This question has a great pedigree! Supposedly it's what Einstein asked himself when he was first thinking about relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that you can't &quot;ride on top of a light stream&quot; -- that is, you can't go as fast as the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed of light is invariant -- same for all observers. So at any sub-light speed you can see yourself in a mirror just fine. No matter how fast you're going, a beam of light going past you still moves at the speed of light, relative to you. No matter how hard you fire your rocket engines, you never catch up with it.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-06T20:07:17.863" CommentCount="3" />
  <row Id="8167" PostTypeId="2" ParentId="8162" CreationDate="2011-04-06T20:20:49.103" Score="5" ViewCount="" Body="&lt;p&gt;AdS$_d$ in any space-time dimension $d\geq 2$ is maximally symmetric with isometry group so(d-1,2) (for Minkowski signature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This group coincides with the conformal group in d-1 dimensions (again for Minkowski signature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, for AdS$_5$ you obtain the isometry group so(4,2), which is the conformal group in 4 dimensions. This matching is a slightly trivial consistency check that something like an AdS$_5$/CFT$_4$ correspondence can work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another special feature of AdS as opposed to dS is that it provides a stable vacuum in most theories (whereas dS is only meta-stable), and that it is compatible with SUSY (whereas dS is not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spaces that asymptote to AdS have very special properties, too. Brown and Henneaux showed in d=3 that any consistent quantum theory of gravity must be dual do a 2-dimensional conformal field theory, in the sense that the Hilbert space must fall into irreducible representations of two copies of the Virasoro algebra, with central charge determined by Newton's constant and the cosmological constant. This was an important precursor of the AdS/CFT correspondence, where such a duality is realized explicitly (but in higher dimensions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Minkowski space is also maximally symmetric and stable, but not as susceptible to holography as AdS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, AdS spaces are simple and have interesting physical properties, which is why they are used quite frequently.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-06T20:41:21.513" LastActivityDate="2011-04-06T20:41:21.513" />
  <row Id="8168" PostTypeId="2" ParentId="7722" CreationDate="2011-04-06T20:49:58.007" Score="1" ViewCount="" Body="&lt;p&gt;If you look at it closely, it seems as though that the film on the platform is being rotated inwards as the assembly sticks out its platform. That's why the platform doesn't have to be absolutely level with the surface the condiments are spread on.&lt;/p&gt;&#xA;" OwnerUserId="2960" LastActivityDate="2011-04-06T20:49:58.007" CommentCount="1" />
  <row Id="8169" PostTypeId="1" AcceptedAnswerId="8203" CreationDate="2011-04-06T20:57:50.167" Score="6" ViewCount="262" Body="&lt;p&gt;I understand what a qubit-based quantum computer by the current definition is and how they are constructed. I read another &lt;a href=&quot;http://physics.stackexchange.com/questions/3412/using-delayed-choice-interference-experiments-as-a-computing-device&quot;&gt;thread&lt;/a&gt; where someone suggested encoding a computation into a dual-slit experiment and reading out the answer as interference vs non-interference, and this was mostly shot down since it's not &quot;useful&quot;, because the system ceased to be a quantum computer when decoherence occurred.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this reminded me of an old idea I had on other ways to encode computations in physics than the &quot;classical&quot; qubit-based schemes. As a striking example, consider one of the canonical uses for a quantum-computer - to efficiently simulate quantum-level systems. Well - to do this without a quantum-computer, you could actually assemble the system itself and let it &quot;run&quot;, and then read out what you were looking for to simulate in the first place in many cases, in a vastly more simple way than digitally calculating it with qubits. Is this then not also a &quot;quantum computation&quot;, really? The closest analogue I can come up with is the way analog computers were designed before digital computers took over, and that they were indeed much more efficient at a certain class of problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even a system of 3 particles performs a calculation (of sorts) which is difficult to simulate traditionally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, if you could encode your problem into a hamiltonian which you subsequently map onto a physical structure wouldn't this be very useful? You might not be able to run &lt;em&gt;algorithms&lt;/em&gt; on it, but you might be able to format the problems the algorithms were meant to solve in the first place. I'd be interested in hearing if this approach is pursued and I have missed it or if it is deemed to be useless for some reason, or if it's useful but &quot;digital&quot; quantum computation is simply more attractive in the long-run so most of the focus is on that. Maybe there is a measure on the computational complexity that can be mapped or performed by a general quantum structure (non-qubit-based)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I seem to remember a Canadian company called D-Wave which I think do this more or less, and there was (is) much heated discussions on if it was &quot;real quantum computation&quot; or not.&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-04-07T13:26:30.697" Title="Is &quot;analog&quot; quantum-computation not useful?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-information&gt;&lt;quantum-computation&gt;&lt;quantum-computer&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="8170" PostTypeId="2" ParentId="7722" CreationDate="2011-04-06T21:03:05.543" Score="1" ViewCount="" Body="&lt;p&gt;The main thing here is the clever belt system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To avoid some traces adhering to this belt, I'd use belts made from PTFE.&#xA;PTFE is hydrophobic and oleophobic as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only substances which could make problem were fluorinated detergents. &#xA;(But those are not used in food technology, afaik) &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-06T21:03:05.543" />
  <row Id="8171" PostTypeId="2" ParentId="8169" CreationDate="2011-04-06T21:56:10.920" Score="2" ViewCount="" Body="&lt;p&gt;Well, i am just not sure that it would be useless, and just meant as an example (and not as a full answer to the more general aspects of your question), you could have super-FFT if you could encode the inputs of your FFT as dark/transparent pixels in a 2D light filter and then recover the FFT output as an 2D interference pattern with CCD. If this is actually scalable or interesting from a cost-benefit perspective its another, engineering-related matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would of course be counted by many as a classical-analog computer, because it doesn't rely on entanglement of qubits, just good old wave huyghens interference&lt;/p&gt;&#xA;&#xA;&lt;p&gt;best FFT algorithms have complexities like $O( n \log(n^2) )$, so i don't know why no one has come up with an application for this yet. As i said, probably the reasons are entirely economical&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-04-06T22:06:22.857" LastActivityDate="2011-04-06T22:06:22.857" CommentCount="5" />
  <row Id="8172" PostTypeId="2" ParentId="8169" CreationDate="2011-04-06T22:21:04.973" Score="2" ViewCount="" Body="&lt;p&gt;The difference between a traditional analogue computer and a quantum computer is at least that the analogue computer is not stochastic (at least in concept; there's still noise, but the idea is to eliminate the noise as far as possible, not to use it). In particular, from the field theoretic perspective I usually work in, the mathematical structure of a classical field is of a lower order than the mathematical structure of a probability measure over values of a classical field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one goes to the higher order structure of probability measures over classical analogue structures instead of deterministic analogue structures, one is then close to the computational power of quantum mechanics. In particular, in a recent paper, &quot;Equivalence of the Klein-Gordon random field and the complex Klein-Gordon quantum field&quot;, &lt;a href=&quot;http://dx.doi.org/10.1209/0295-5075/87/31002&quot; rel=&quot;nofollow&quot;&gt;EPL 87 (2009) 31002&lt;/a&gt;, &lt;a href=&quot;http://arxiv.org/abs/0905.1263&quot; rel=&quot;nofollow&quot;&gt;arXiv:0905.1263v2&lt;/a&gt;, I show that a quantized complex Klein-Gordon field is &lt;em&gt;empirically equivalent&lt;/em&gt; to a real Klein-Gordon random field, which is essentially at the mathematical level of probability measures over classical fields. The traditional Klein-Gordon field, with no stochastic or probabilistic structure, however, is not of the same mathematical order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, a classical stochastic model can be presented in a Koopman-von Neumann formalism, which essentially just applies Hilbert space mathematics to classical probabilities. To return to the parenthesis of my first paragraph, for a classical analogue computer to approach the effectiveness of a quantum computer, my intuition (without enough justification) is that it would have to use some kind of noise in an essential way to improve the effectiveness of an algorithm. It may or may not be significant that &lt;a href=&quot;http://en.wikipedia.org/wiki/Monte_Carlo_method&quot; rel=&quot;nofollow&quot;&gt;Monte Carlo methods&lt;/a&gt; are a well-known example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real crux, however, is that a quantum mechanical way of thinking about the experimental apparatuses that are currently being used to push forward quantum computing as a technology, essentially in terms of qubits and closely similar finite-dimensional Hilbert space mathematics, is more productive than thinking in terms of classical stochastic signal processing, even though the concept of incompatible measurements is as natural in the classical mathematics of processing stochastic analogue signals as it is in quantum theory. [You go a long way towards providing this reason in your Question.] The particular advantage, as far as I can tell from outside the field, is that the mathematics of Hilbert spaces allows a quasi-digital way of algebraic thinking that bridges the gap between analogue and digital thinking more effectively than thinking just in terms of probability measures over analogue mathematical structures. Because almost no-one works in stochastic-analogue terms, there is a dearth of established results and a consequent lack of leverage for people who do work with these ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a good Question. I look forward to seeing a more conventional Answer that I hope will be forthcoming from someone else.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-04-06T22:21:04.973" />
  <row Id="8173" PostTypeId="2" ParentId="8162" CreationDate="2011-04-06T22:27:12.507" Score="8" ViewCount="" Body="&lt;p&gt;I'm going to provide a possible controversial answer, in an attempt to provoke some discussion. I do this in good faith, and in the belief that what I state are true, and back with references (comments of &quot;citation needed&quot; will be very handy). I context this with: I'm a condensed matter theorist, and I think the usual exposition of AdS/CFT has the cart before the horse. I will take a long detour, but hopefully I will come back at the end and answer the actual question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start with a spin 1/2 chain on a 1D lattice, infinite in extent. The Hilbert space is a product of 2 dimensional spaces. Let the Hamiltonian be anti-ferromagnetic Ising with an external magnetic field, so that at a critical field strength we will get a quantum phase transition from anti-ferromagnetic to ferromagnetic. We deal only with the ground state (i.e. at zero temperature). Let's then make a couple of observations: away from the phase transition, the correlation length is finite, and the entanglement entropy of any given block of length $L$ is asymptotically a constant (as $L \rightarrow \infty$); at the phase transition, the correlation length is infinite, and the entanglement entropy goes as $\log(L)$. Note that these are quite special features of the ground state, since the typical (defined as average over the canonical Haar measure) state has entanglement entropy which scales as $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, instead of writing the ground state with full generality $$\left| \Omega \right\rangle = \sum_{s_1,s_2,\ldots} c_{s_1,s_2,\ldots} \left|s_1\right\rangle\otimes\left|s_2\right\rangle\otimes\ldots$$ where we would have to specify the matrix $c$ with an exponentially large number of dimensions (spanning the full Hilbert space), we're going to restrict our attention to so-called Matrix Product States (MPS) with the form: $$\left|\Omega\right\rangle = \sum_{s_1,s_2,\ldots} \mathrm{Tr}\left(\hat A^{s_1} \hat A^{s_2} \ldots \right) \left|s_1\right\rangle\otimes\left|s_2\right\rangle\otimes\ldots$$ where the matrices $\hat A^{s_i}$ are arbitrary matrices of dimension $m$. Essentially, we're staring in the corner of Hilbert space which is spanned by a &lt;em&gt;linearly&lt;/em&gt; increasing number of dimensions. Now, as $m \rightarrow \infty$ we recover the full Hilbert space, but away from the critical point, a finite $m$ suffices to fully (exactly) describe the ground state, because of the prior point about finite entanglement entropy; essentially, the dimension $m$ controls how much entanglement is possible between adjacent sites, and the MPS ansatz fully spans all such states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, as mentioned, the entanglement in a critical state is not bounded. In this case, we can use a different ansatz, the Multi-scale Entanglement Renormalisation Ansatz (MERA). The construction is difficult to describe in words, but easier in pictures. If we use tensor network diagrams (first identified by Penrose and called spin networks), we depict each tensor as a blob with a number of legs equal to its rank. Treating the matrices $\hat A^{s_i}$ as 3-rank tensors (one extra due to the spin index), we can draw the MPS as: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/TAGF3.png&quot; alt=&quot;MPS&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the lower legs are the spin indices. The MERA is then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/QMXct.jpg&quot; alt=&quot;MERA&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(but imagine that the &quot;tree&quot; continues upwards without end). The essence is that we reify coarse graining (i.e. renormalisation) into the ground state description by a tree of disentanglers and coarse graining. Again, if we do this right, this can describe the ground state with perfect accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These tensor network diagrams also give a picturesque reason for why the entanglement entropy scales as a constant and as $\log(L)$ respectively. The argument is that the entanglement is localised at the boundary of a block (as it has to, since each connection in that network can only support a finite amount of entanglement), but the &quot;boundary&quot; actually scales differently in the two cases: the the non-critical case, it is just the edges of a 1D chain, which clearly don't care about the bulk; in the critical case, it needs to include not only the bottom layer, but all the layers above it, and there are $\log(L)$ layers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, everything is basically (up to corner cases) true. Let's now turn to more conjectural/interpretational stuff. Focus on MERA. Notice that if we treat it as a space, then a natural distance measure is the number of &quot;hops&quot; we need to do from one vertex to another; notice also, that in the continuum limit this is a homogeneous hyperbolic space, i.e. AdS. In the original Ising model, at the critical point, the field theory should be conformally invariant, and thus be a CFT. This is all but AdS/CFT, except we haven't specified that the MERA coefficients are computed by a quantum gravitational theory (it probably can't be, I think... the central charge is 1, and nothing is supersymmetric). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, at this point, you might think &quot;Aha! See? AdS/CFT is of primary importance to even mundane things like condensed matter!&quot; However, I'd like to present some evidence that actually, AdS/CFT is a mundane consequence of a very clever idea, which is to geometrically interpret the information in a ground state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider instead an interacting fermion system in 1D. The usual electrons with Coulomb repulsion will do. It is known that the physical ground state is that of non-interacting solitons of fractionalised electrons: holons (carrying the charge) and spinons (carrying the spin). Our ansatz will then be that of MERA, but at a certain depth in the tree, we duplicate everything above it --- so that we end up with two 1D systems, one for holons and one for spinons. In the geometric picture above, it will be as if we glued an extra AdS space onto the usual one, so that we get a fork. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this suggests that actually the ground state should come first and the holography principle second is two fold:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Holography only holds for special states like the ground state, where the entanglement entropy scales sub-bulk.&lt;/li&gt;&#xA;&lt;li&gt;The internal AdS space might not be AdS, or even admit any sort of nice geometrical picture, and even if it does, it might not be given by some sort of Lagrangian based field theory.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, back to the question: &quot;what is special about AdS?&quot; Other answers will no doubt focus on the special geometry that makes the maths work, but I would answer that the key is never the inner space, but the boundary: the (super-)CFT. The inner space, in this case, AdS, just comes along for a ride. If we had some other kind of boundary theory, we'd have some other kind of inner space, or not a space at all!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seminal (?) paper on correspondence between MERA and holography: &lt;a href=&quot;http://arxiv.org/abs/0905.1317&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0905.1317&lt;/a&gt;&#xA;Branching MERA as exotic holography: &lt;a href=&quot;http://pirsa.org/10110076&quot; rel=&quot;nofollow&quot;&gt;http://pirsa.org/10110076&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1501" LastEditorUserId="1501" LastEditorDisplayName="" LastEditDate="2011-04-07T05:37:22.053" LastActivityDate="2011-04-07T05:37:22.053" CommentCount="12" />
  <row Id="8174" PostTypeId="2" ParentId="8122" CreationDate="2011-04-06T23:03:22.553" Score="1" ViewCount="" Body="&lt;p&gt;My suggestion would be to take the next new student and give them the task of compiling a list of the jargon (&quot;institutional knowledge&quot;) that they come across, with definitions. Don't have them go out of their way to do so, since that would detract from their getting &quot;real&quot; work done, but have them make sure to jot down things as they come up (asking more senior members for definitions as necessary).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the next new student joins, they inherit the list and do the same. In order to add new things, they'll have to learn what's already on the list, which will be handily compiled already (and which they can reference). After a few iterations, if implemented sufficiently across the collaboration, a complete list will exist and can, if necessary, by synthesized into one overarching list. If only subsections interact regularly, the decentralized lists may suffice.&lt;/p&gt;&#xA;" OwnerUserId="1196" LastActivityDate="2011-04-06T23:03:22.553" />
  <row Id="8175" PostTypeId="2" ParentId="7898" CreationDate="2011-04-06T23:58:13.340" Score="2" ViewCount="" Body="&lt;p&gt;Can you be more specific about what you are trying to do, especially your power/focusing requirements, and the wavelength sensitivity of your experiment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;350nm LEDs are available from this company: &lt;a href=&quot;https://dcomponents.com/?keyword=350nm&amp;amp;content=results&quot; rel=&quot;nofollow&quot;&gt;https://dcomponents.com/?keyword=350nm&amp;amp;content=results&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;445nm LEDs are also widely available, and at this wavelength high powered laser diodes which can be focused tightly can also be obtained: &lt;a href=&quot;http://www.lasersurplusparts.com/shop/80-1w-445nm-laser-diodes.html&quot; rel=&quot;nofollow&quot;&gt;http://www.lasersurplusparts.com/shop/80-1w-445nm-laser-diodes.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 350nm laser light, your options are much more limited. HeCd has a line at 325, and N2 lasers emit around 337. Pulsed N2 lasers can be built fairly cheaply and easily but are not very coherent.&lt;/p&gt;&#xA;" OwnerUserId="2963" LastActivityDate="2011-04-06T23:58:13.340" />
  <row Id="8176" PostTypeId="1" AcceptedAnswerId="8212" CreationDate="2011-04-07T01:47:38.840" Score="3" ViewCount="153" Body="&lt;p&gt;Newton's theory of gravity supports &quot;gravity waves&quot; in that moving objects cause changing gravitational fields. For example, two bodies rotating around their center of mass will have a stronger gravitational field when they are longitudinally oriented than when they are transverse oriented. Given two masses of mass $M$ orbiting on a circle of radius $r$, at a distance $d$ from an observer, the strength of the attraction is:&#xA;$$\begin{matrix}&#xD;&#xA;F_{min} &amp;amp;=&amp;amp; \frac{2GM}{d^2},\\&#xD;&#xA;F_{max} &amp;amp;=&amp;amp; \frac{2GM}{d^2}\frac{1+r^2/d^2}{(1-r^2/d^2)^2}&#xD;&#xA;\end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/dIjHy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be detectable at long distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this: With that sort of gravity, would the laser interferometer based gravity wave detectors be able to detect the gravity wave? An example is the &lt;a href=&quot;http://en.wikipedia.org/wiki/LIGO&quot; rel=&quot;nofollow&quot;&gt;LIGO, Laster Interferometer Gravitational-wave Observatory&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-04-07T18:13:38.457" Title="Are Newton's gravity waves detectable by a laser interferometer?" Tags="&lt;gravity&gt;&lt;newtonian-gravity&gt;&lt;interferometry&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="8177" PostTypeId="2" ParentId="8156" CreationDate="2011-04-07T01:49:28.043" Score="1" ViewCount="" Body="&lt;p&gt;A transformation from one set to another can be regarded as a matrix if we define a particular basis. Likewise, an operator can be thought of as a matrix. What is the matrix equation that relates eigenvalues and eigenvectors? You are solving an eigenvalue problem when you are solving the time independent Schrodinger equation.&lt;/p&gt;&#xA;" OwnerUserId="2965" LastActivityDate="2011-04-07T01:49:28.043" />
  <row Id="8178" PostTypeId="2" ParentId="4654" CreationDate="2011-04-07T02:14:22.663" Score="0" ViewCount="" Body="&lt;p&gt;Pulsars are easiest to detect if some of the time they're pointed at us (and so are bright) and some of the time they're pointed away. This can happen because the pulsar emits radiation in a direction different from the direction it rotates. So as the pulsar rotates, it beams to different directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pulsars are thought to run out of energy quite quickly and then become silent. So 99% of the pulsars that have ever lived are thought to now be silent. Astronomers and astrophysicists are surprisingly intelligent and take these facts into account when estimating the amount of matter contained in pulsars. If it were possible for sufficient matter to be contained in pulsars so as to significantly effect dark matter I'm sure they would have thought of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's always fun to look in the literature to see some light curves:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The optical light curve of the LMC pulsar B0540-69 in 2009&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/1012.0738&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1012.0738&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ejlY1.jpg&quot; alt=&quot;LMC pulsar B0540-69 in 2009&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the above you can see that (for this particular pulsar observed in a particular range of light frequencies at this particular time) there was no significant beaming effect. It could also be that this pulsar is pointed directly at us, but wobbling slightly. Other effects can smear out a pulsar's light curve. The pulsar could be seen by its effect on nearby dust or other matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a paper showing a lot of light curves, some of which do form nice beams, see:&#xA;&lt;strong&gt;High-Resolution Timing Observations of Spin-Powered Pulsars with the AGILE Gamma-Ray Telescope&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/0810.1516&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0810.1516&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a paper analyzing the offset angles and the angle between the pulsar rotation axis and the earth, see:&#xA;&lt;strong&gt;Modeling of Gamma-ray Millisecond Pulsar Light Curves Revealed by Fermi-LAT&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/0911.4890&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0911.4890&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-04-07T02:14:22.663" />
  <row Id="8179" PostTypeId="1" AcceptedAnswerId="8353" CreationDate="2011-04-07T02:15:00.760" Score="5" ViewCount="355" Body="&lt;p&gt;I have trouble imagining how two point-particles can have different properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And how can finite mass, and finite information (ie spin, electric charge etc.) be stored in 0 volume?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only that, but it can also detect all fields without having any structure.&#xA;Maybe it can check curvature of spacetime to account for gravity, but how can a point contain the information of what the other fields-vectors are? This seems to mean that also the information/volume in space is infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, a point cant have any intrinsic structure, so how does physics which is a mathematical theory explain this?&lt;/p&gt;&#xA;" OwnerUserId="1708" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-09T23:05:57.830" LastActivityDate="2011-04-10T17:03:27.443" Title="How can a point-particle have properties?" Tags="&lt;particle-physics&gt;&lt;mathematical-physics&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="8180" PostTypeId="2" ParentId="8179" CreationDate="2011-04-07T02:27:48.603" Score="1" ViewCount="" Body="&lt;p&gt;My feeling is that, at the foundations, the information about a particle needs to be encoded in its propagator. That is, all the coupling constants, mass, spin, and other information needs to show up in the propagator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that instead of the information being contained in a point, instead the information is contained essentially in how the particle moves from point to point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of saying this is to note that what we call a &quot;particle&quot; is an effect seen by our experiments, where nature does something that seems to be repeated in a consistent manner. Relativity requires that an object be able to move, hence the propagator.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-04-07T02:27:48.603" CommentCount="2" />
  <row Id="8181" PostTypeId="2" ParentId="8176" CreationDate="2011-04-07T02:52:25.980" Score="2" ViewCount="" Body="&lt;p&gt;I don't think so. To first order, the maximum variation in your Newton's gravity force is  $$F_{min}-F_{max}=\frac{6GMmr^2}{d^4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume that the effects of gravity propagate at light speed, then in order for one mirror to experience $F_{max}$ while the other experiences $F_{min}$, the objects must rotate 90 degrees in the time $l/c$ where $l$ is the path length of the interferometer (4 kilometers for LIGO). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we choose the speed of light as an upper bound for the speed of the rotating objects, this means that they can be separated by no more than about 5 kilometers ($r=2.5$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lower bound on $d$ can be 4 lightyears, the distance to the nearest star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's suppose the mass of a mirror is $m = 100 kg$, and the mass of the rotating objects is 20 billion solar masses, the largest known mass in the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Punching this into google calculator gives a force difference of about $10^{-28}$N, which is 4 orders of magnitude smaller than any force yet measured - and of course most of these assumptions are unphysically generous.&lt;/p&gt;&#xA;" OwnerUserId="2963" LastEditorUserId="2963" LastEditorDisplayName="" LastEditDate="2011-04-07T02:58:28.640" LastActivityDate="2011-04-07T02:58:28.640" CommentCount="4" />
  <row Id="8182" PostTypeId="1" CreationDate="2011-04-07T00:02:50.027" Score="3" ViewCount="147" Body="&lt;p&gt;Let's say I have an infinite line source in 3d wave medium, like a pulsating cylinder, emitting a broadband noisy signal. I know that the pressure falls off at 3 dB per doubling in distance, unlike a point source which falls off at 6 dB per doubling in distance, because the cylindrical wavefront is only spreading out along one direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am unclear on whether this has a frequency-dependent effect, and on the relationship of this situation to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stationary_phase_approximation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Stationary_phase_approximation&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the signal received by a listener some distance from the line frequency-dependent? Does it roll-off at high frequencies?? And am I posting at the wrong site?&lt;/p&gt;&#xA;" OwnerUserId="3043" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-07T03:58:27.923" LastActivityDate="2011-08-05T07:39:08.063" Title="Frequency response of infinite line source" Tags="&lt;waves&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8183" PostTypeId="2" ParentId="8068" CreationDate="2011-04-07T03:52:44.897" Score="2" ViewCount="" Body="&lt;p&gt;Since there hasn't been any answer from experts, I did some search myself. This well cited article seems to be very comprehensive: &lt;a href=&quot;http://arxiv.org/abs/0809.0631&quot; rel=&quot;nofollow&quot;&gt;Primordial Nucleosynthesis: from precision cosmology to fundamental physics&lt;/a&gt;&#xA;It painstakingly walks through the implications of BBN on all kinds of physics models, including exotic ones like large extra dimensions and brane inflation.&lt;/p&gt;&#xA;" OwnerUserId="2643" LastActivityDate="2011-04-07T03:52:44.897" />
  <row Id="8184" PostTypeId="2" ParentId="8156" CreationDate="2011-04-07T04:03:20.023" Score="2" ViewCount="" Body="&lt;p&gt;The reason why it is the eigenvalues of the &lt;strong&gt;Hamiltonian&lt;/strong&gt; and not some other operator that will give you the energy states is that in &lt;em&gt;classical&lt;/em&gt; Mechanics, the Hamiltonian function is just the energy of your system, expressed as a function of position $x$ and momentum $p$. As a simple example, the Hamiltonian for a harmonic oscillator is &#xA;$$H(x,p) = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2$$&#xA;Note that this really is just the sum of kinetic and potential energy, so we could write &#xA;$$H(x,p) = E$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get to quantum mechanics, one now performs what is called &lt;em&gt;canonical quantization&lt;/em&gt;. There is no mathematically rigorous reason why this will give you a correct quantum mechanics. Since quantum isn't classical, we cannot really expect to find a seamless and watertight derivation of the former from the latter. To my knowledge, this approach has, however, always given correct results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in canonical quantization, what one does is to replace the variables of the Hamiltonian, i.e., $x$ and $p$, with their operator versions, $\hat x$ and $\hat p$. Now we cannot simply write $H(x,p) = E$ anymore, since the energy is a scalar, but the Hamiltonian $H$ is now an operator. Operators are functions that take a wavefunction and modify it in some way and give you a new wavefunction. Now, another postulate of quantum mechanics is that you get the expectation value of an operator $\hat A$ in a given state $\Psi$ by calculation the integral&#xA;$$ \int dx \Psi^*(x) \hat A \Psi(x) $$&#xA;Hence, we get the expectation value of the energy by calculating&#xA;$$ \int dx \Psi^*(x) H \Psi(x)$$&#xA;Obviously, if $H\Psi(x) = E\Psi(x)$, then the expectation value yields $E$, and it is not hard to show that for such an eigenstate, the variance of $E$ will be $0$, i.e. &lt;em&gt;every&lt;/em&gt; measurement of the energy in state $\Psi$ will yield the same value $E$.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-04-07T04:03:20.023" />
  <row Id="8185" PostTypeId="2" ParentId="8162" CreationDate="2011-04-07T04:51:22.497" Score="6" ViewCount="" Body="&lt;p&gt;There are some extensions of the AdS/CFT correspondence, so it is more accurate to call the more general set of dualities the gauge/gravity correspondence. In all such dualities one has a gravity dual of some quantum field theory, and the theory in question determines the boundary conditions on all the bulk fields (including the asymptotics of the bulk geometry). States of those theories correspond to small fluctuations (normalizable modes) moving in the bulk of that spacetime, and the vacuum state usually corresponds to the maximally symmetric (&quot;empty&quot;) space. There are many such examples which are not AdS even asymptotically, though sadly asymptotically dS is not yet one of them (partially because it is not clear what the expression &quot;asymptotically dS&quot; really means). Asymptotically flat examples, alas with linear dilaton background, do exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, within the set of all holographic dualities there is something special in spaces which are asymptotically AdS. Those correspond to theories which become conformal at short distances. Using Wilsonian language, those are theories whose renormalization group flow can be continued to all energy scales, so they are completely well-defined quantum field theories without a cutoff. Such field theories are defined as relevant deformations of fixed points in the UV, and the holographic translation of that statement is that the gravity dual is asymptotically AdS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing with the Wilsonian language, usually one needs to use QFT only as an effective field theory, and it is then inherently defined with an UV cutoff. Such more general quantum field theory (defined only up to certain energy scale) correspond to instances of gauge-gravity duality which are not asymptotically AdS. The correspondence between EFT with a cutoff and non-asymptotically-AdS space (at least on some occasions dubbed the &quot;non-AdS/non-CFT correspondence&quot;) is less well-understood than AdS/CFT (with the amount of work on AdS/CFT, this applies to many other subjects...). But, it is a very useful and interesting subject, in some ways more so than the original AdS/CFT correspondence, the one that opened the flood gates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any event, the type of boundary conditions imposed on the space is only restrictive when discussing global questions. Any local process you are interested in (say, the formation and evaporation of black holes) can be embedded in asymptotically AdS space with an arbitrarily small cosmological constant. I wouldn't then think about the set of examples provided by AdS/CFT as &quot;unphysical&quot; in any way - it may not address all the possible questions one may be interested in, but it the best way to address of whole bunch of fascinating ones.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-04-07T05:29:33.740" LastActivityDate="2011-04-07T05:29:33.740" />
  <row Id="8186" PostTypeId="2" ParentId="8162" CreationDate="2011-04-07T05:28:05.333" Score="3" ViewCount="" Body="&lt;p&gt;AdS space is basically just hyperbolic space, with a time direction. Here's a nice geometrical fact. Consider 2d hyperbolic space in the Poincaré disc model. (Generalizing to higher dimensions is straightforward.) The metric is $ds^2 = \frac{dr^2 + r^2 d\theta^2}{(1-r^2)^2}$. The corresponding area element is $\frac{rdrd\theta}{(1-r^2)^2}$. So consider the circle at fixed $r = r_0$. This has circumference $2\pi r_0\frac{1}{1-r_0^2}$, and area $2\pi \int_0^{r_0} \frac{rdr}{(1-r^2)^2}$. For $r_0 = 1 - \epsilon$, with $\epsilon \ll 1$, these are $\frac{\pi}{\epsilon} - \frac{\pi}{2} + {\cal O}(\epsilon)$ and $\frac{\pi}{2\epsilon} - \frac{\pi}{2} + {\cal O}(\epsilon)$, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this mean? It means that, for circles large compared to the curvature radius of the space, perimeter and area scale in the same way as you make the circle larger. (As opposed to flat space, where one scales like the square of the other.) I think this is a hint about holography; in some sense, AdS is the space in which holography becomes almost trivial, because $d$ and $d-1$ dimensional volumes are almost identical, which is why we understand holography much better in AdS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, this isn't unrelated to the ideas about conformal symmetry, etc. But I think this geometric fact sheds some light and is easy to understand without getting into details of physics.)&lt;/p&gt;&#xA;" OwnerUserId="19" LastActivityDate="2011-04-07T05:28:05.333" CommentCount="1" />
  <row Id="8187" PostTypeId="2" ParentId="8182" CreationDate="2011-04-07T06:06:28.417" Score="1" ViewCount="" Body="&lt;p&gt;If you let me interpret 'broadband, noisy', 'pressure', and 'pulsating cylinder' creatively enough, sure, you can get behavior that deviates from $1/r$, at least for a little while. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can construct a carefully rigged source such that the waves it emits fall off in &lt;a href=&quot;http://www.opticsinfobase.org/ol/abstract.cfm?&amp;amp;uri=ol-30-13-1686&quot; rel=&quot;nofollow&quot;&gt;funny ways&lt;/a&gt;. For example, suppose your waves are light, your wave medium is vacuum, and your line source emits two short pulses of light simultaneously. One pulse propagates away perpendicular to the line source; the other pulse propagates at a small angle $\theta$ away from the line source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two pulses will both look like expanding cylindrical shells, but they will have different velocities in the radial direction, since one pulse is also carrying energy in the axial direction. Each individual pulse will fall off like $~1/r$, but they will eventually cease to overlap, so their summed intensity won't fall off like $1/r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your broadband, noisy source is quasi-cw and incoherent, and your wave medium is not absorptive or reflective, however, I think it will have to emit waves with a $1/r$ intensity falloff to conserve energy. I don't see how this would depend on frequency, and I don't see what this has to do with the stationary phase approximation. Is this the specific case you're interested in? If so, I can try to justify these statements.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="2359" LastEditorDisplayName="" LastEditDate="2011-04-07T06:30:00.690" LastActivityDate="2011-04-07T06:30:00.690" />
  <row Id="8188" PostTypeId="1" AcceptedAnswerId="8190" CreationDate="2011-04-07T08:20:26.967" Score="2" ViewCount="135" Body="&lt;p&gt;Is it possible to measure the spin of a single electron? What papers have been published on answering this question? Would the measurement require a super sensitive SQUID, Superconductive Quantum Interference Device?&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2011-05-21T08:31:02.800" Title="Measuring the spin of a single electron" Tags="&lt;electromagnetism&gt;&lt;measurement&gt;&lt;superconductivity&gt;&lt;electron&gt;&lt;measurement-problem&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="8189" PostTypeId="1" AcceptedAnswerId="8202" CreationDate="2011-04-07T08:50:11.720" Score="3" ViewCount="123" Body="&lt;p&gt;So I learned that after a measurement of, lets say the position of the wavefunction of a particle is made, if another measurement of the position of the particle is made right away, you should get the same result because the wavefunction has collapsed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you tell the difference between the observer measuring the wavefunction which causes the collapse vs. something else that measured the wavefunction and collapsed it already and you are measuring it the second time around?&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2011-04-07T13:05:36.970" Title="Knowing when wavefunction collapses" Tags="&lt;quantum-mechanics&gt;&lt;waves&gt;&lt;measurement&gt;&lt;measurement-problem&gt;" AnswerCount="3" />
  <row Id="8190" PostTypeId="2" ParentId="8188" CreationDate="2011-04-07T08:51:55.517" Score="2" ViewCount="" Body="&lt;p&gt;Dear QEntanglement, the spin of a single electron has been measured since the very first moment when the people understood that every electron possesses a spin. A Stern-Gerlach experiment - a magnetic field - is enough to measure the spin:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stern-Gerlach_experiment&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Stern-Gerlach_experiment&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-07T08:51:55.517" CommentCount="5" />
  <row Id="8191" PostTypeId="1" AcceptedAnswerId="8199" CreationDate="2011-04-07T08:59:48.123" Score="1" ViewCount="85" Body="&lt;p&gt;Is it possible to measure the magnitude of the magnetic field of a single electron due to its spin? The electron's intrinsic magnetic field is not dependent upon the amount of energy it has does it? Does this require a SQUID, a Superconducting Quantum Interference Device? What published papers or experiments have been done to answer this question?&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2011-04-07T12:41:45.060" Title="Measuring the magnitude of the magnetic field of a single electron due to its spin" Tags="&lt;quantum-mechanics&gt;&lt;measurement&gt;&lt;electron&gt;&lt;measurement-problem&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="8192" PostTypeId="2" ParentId="8189" CreationDate="2011-04-07T09:45:07.987" Score="0" ViewCount="" Body="&lt;p&gt;As you correctly say in first paragraph, measuring twice is the same thing as measuring just once. Therefore you &lt;em&gt;can't&lt;/em&gt; distinguish these two alternatives. But there is no need to tell. If you are an experimentalist, all you care about is that the wavefunction has collapsed and there is some number shown on your measuring apparatus. You don't care how precisely was the collapse realized.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-04-07T09:45:07.987" CommentCount="2" />
  <row Id="8193" PostTypeId="2" ParentId="8156" CreationDate="2011-04-07T10:06:31.923" Score="2" ViewCount="" Body="&lt;p&gt;You seem to be confusing two things, namely the eigenstates of an operator and Schrödinger's equation. &lt;em&gt;A priori&lt;/em&gt;, these two have nothing to do with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Quantum Mechanics, measurable quantities are represented by (hermitian) &lt;strong&gt;operators&lt;/strong&gt; on a Hilbert space. For instance there is an operator $P$ corresponding to the momentum. In general, when measuring the momentum of a state $|\psi \rangle$, the result will not be deterministic. However, the &lt;em&gt;average&lt;/em&gt; over several measurements will be equal to the expection value&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \langle \psi | P | \psi \rangle $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when $|\psi\rangle$ is an &lt;em&gt;eigenvector&lt;/em&gt; of the operator, $P|\psi\rangle = \lambda |\psi\rangle$, then the measurement will always be the same value $\lambda$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, there is an operator corresponding to the total energy, the &lt;strong&gt;Hamiltionian&lt;/strong&gt; $H$. The form of this operator can be obtained from classical physics if you replace momentum and location by their corresponding operators. For instance, the Hamiltonian of an electron in an electric potential $V$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H = \frac1{2m} P^2 + eV(X) .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the expectation value for the energy of a state $|\psi\rangle$ is $\langle \psi|H|\psi\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;Now, the Hamiltionian is a very interesting operator because it features prominently in the equation of motion, the Schrödinger equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ i\hbar \partial_t |\psi(t)\rangle = H |\psi(t)\rangle .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this have to do with the eigenvalues of the Hamiltionian? &lt;em&gt;A priori&lt;/em&gt; nothing, but the point is that knowing the eigenvectors and -values of $H$ allow you to &lt;em&gt;solve&lt;/em&gt; this equation. Namely, if you have an eigenvator $|\psi_n\rangle$, then you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ i\hbar \partial_t |\psi_n(t)\rangle = H |\psi_n(t)\rangle = E_n|\psi(t)\rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be solved to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |\psi_n(t)\rangle = e^{-\frac{i}{\hbar} E_nt} |\psi(0)\rangle $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, the eigenvalues of an operator tells you something about what happens when you perform &lt;em&gt;measurements&lt;/em&gt;, but in addition, the eigenvalues of the energy operator help you &lt;em&gt;solve the equations of motion&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="188" LastActivityDate="2011-04-07T10:06:31.923" />
  <row Id="8194" PostTypeId="1" AcceptedAnswerId="8218" CreationDate="2011-04-07T10:26:31.247" Score="4" ViewCount="180" Body="&lt;p&gt;I'd like to extend this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/8176/are-newtons-gravity-waves-detectable-by-a-laser-interferometer&quot;&gt;Are Newton&amp;#39;s gravity waves detectable by a laser interferometer?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but with some changes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has somebody ever tried to measure the gravity oscillations nearby (&amp;lt; some meters) of such rotating masses (as in Carl Brannens question)?&#xA;Say some kgs or grams rotating on a beam with the highest rpm technically possible? &#xA;(In a vacuum chamber of course!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I dont know what will &quot;radiate&quot; stronger , &#xA;some grams at 100 000 rpm or some kg at say, 5000 rpm (and some cm distance), therefore, has anybody ever tried something like that in lab? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware, that such a experiment is plagued by mechanical (elastic) coupling through &#xA;air, supporting structures etc, but I think that one can make use of mechanical resonance &#xA;in the detecting device, helping a lot for selectivity. The fact that the oscillations come from a quadrupole(frequency doubling) , whereas mechanical coupling is not at double frequency, might help to &quot;see&quot; something. As often, its a question of signal/noise ratio :=(&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-07T11:44:33.163" LastActivityDate="2011-04-08T20:58:19.343" Title="Has somebody ever tried to measure the gravity oscillations of nearby rotating masses?" Tags="&lt;gravity&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="8195" PostTypeId="2" ParentId="8176" CreationDate="2011-04-07T10:29:23.757" Score="-1" ViewCount="" Body="&lt;p&gt;I will try to &lt;strong&gt;amplify&lt;/strong&gt; the effect based in angles to an ensemble of distant stars. I assume that the observer has a large mass or a sensitive device (the all planet where he inhabits or a Near Large Star, and all the other parameters are known. In the figure any change in position of the observer or the  'Device Star' in a periodic movement will be theoretically amplified in the measure of the angles &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;b&lt;/strong&gt;. &#xA;&lt;img src=&quot;http://i.stack.imgur.com/h9bQP.png&quot; alt=&quot;amplifyer ;)&quot;&gt;&#xA;I am measuring the gravitational influence but I do not think you are measuring 'gravity waves'. The problem with LIGO is that the device is subject to dynamic changes in length, and the laser light undergoes the same changes, making impossible the detection of any variation. After all LIGO is measuring 0.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-04-07T10:29:23.757" />
  <row Id="8196" PostTypeId="2" ParentId="8194" CreationDate="2011-04-07T11:03:43.893" Score="4" ViewCount="" Body="&lt;p&gt;There are no gravitational waves in the Newtonian gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A wave is defined by its shape: it is something that depends on space like $\cos(kx)$ but nothing like that can be produced by Newton's gravity whose action is instantaneous. In the standard Ansatz for a wave,&#xA;$$\exp(-i\omega t +i k x),$$&#xA;one has $k=\omega/c$ but if $c$ is sent to infinity, $k$ goes to zero and the space dependence disappears. So you can't really be talking about waves - something that has a periodic dependence on the spatial coordinates. There aren't any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, for equivalent reasons, according to non-relativistic mechanics, an orbiting binary star isn't losing any energy by the emission of gravitational waves because there aren't any. If there are extra objects in space besides the binary star, they will influence the internal motion of the binary star (a three-body problem) but this is in no way equivalent to the effect of gravitational waves. In particular, there's no &quot;guaranteed sign&quot; that would ensure that the internal energy of the binary stars decreases. It would go up and down equally often. The energy is only carried by the &quot;point masses&quot; and not by &quot;gravitational waves&quot; because there aren't any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In non-relativistic mechanics, you may only talk about time-dependent gravitational fields. But it's only the quadrupole moment that may be periodically changing if the source of gravity is changing. And the gravitational force from an oscillating quadrupole goes like&#xA;$$ \cos(\omega t)/r^4.$$&#xA;I added two powers of $1/r$ to the Newton's $1/r^2$ inverse square law. On the other hand, if there are gravity waves, they - e.g. the magnitude of $\delta g_{\mu\nu}(x,y,z,t)$ - depend on the location as&#xA;$$\cos(\omega t-kr) / r$$&#xA;because the density of energy goes like the square of the amplitude above and $1/r^2$ is the right way how energy spreads in space (gets distributed over the spherical area of $4\pi r^2$). Note that the multiplicative difference between the true gravitational waves in general relativity - that go like $1/r$ - and the time-dependent quadrupole in Newtonian gravity - that goes like $1/r^4$ - is $1/r^3$. That's a huge difference, especially if the distances are very large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, the effect of the quadrupole decreases much more quickly with the distance from the source than the intensity of the gravitational waves. Obviously, one can't measure the former if we can't even measure latter - the latter, gravitational waves, is incomparably larger than its Newtonian artifact, but we still haven't managed to measure it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the whole idea that there is something like &quot;gravitational waves&quot; in non-relativistic gravity is totally misguided. This comment is not meant to say that one can't measure time dependence of gravitational fields. Of course that one can. Ocean tides that change twice a day are exactly an example of a periodic change of the higher moments from other celestial bodies - the Moon and the Sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But they're not &quot;waves&quot; in any sense because the energy isn't carried away by those would-be waves, and it isn't distributed over the sphere as the energy of waves would. It's up to you whether you consider ocean tides to be an example of &quot;measured gravity oscillations of nearby rotating masses&quot; - I think that you should. But this observation only means a tautologically trivial thing - that changing configurations of matter produce changing gravitational forces - and this trivial thing is in no way analogous to gravitational waves which are independent objects that exist according to general relativity.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-07T11:17:22.810" LastActivityDate="2011-04-07T11:17:22.810" CommentCount="5" />
  <row Id="8197" PostTypeId="2" ParentId="8189" CreationDate="2011-04-07T12:33:32.790" Score="3" ViewCount="" Body="&lt;p&gt;What Marek wrote is accurate with for example a photon polarization measurement (the canonical example is that an additional polarizer after a first polarizer won't change the intensity if they are oriented the same way) but in the OPs example, actually if you measure the position accurately the first time, you will collapse the wavefunction into an uncertainty of the momentum so the next position measurement will not be entirely predictable and hence you will not measure the same thing. This by itself does not allow you to realize that the wavefunction already collapsed though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you allow your position measurements to be done in different places, you can exploit the fact that your collapsed wavefunction has lost some of its components which can be crucial for subsequent interference and hence detect the difference you are asking about, at a certain efficiency at least.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the typical dual-slit experiment and the fact that a position measurement of which slit the particle goes through causes a loss of interference at the detection screen. You can put a detector at a fringe which should be completely dark and if you do detect a particle there, you can infer that the wavefunction collapsed before the both slits could interfere (and therefore conclude that a measurement was done at certain places in the experiment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be considered a form of &quot;interaction-free measurement&quot; of which the &lt;a href=&quot;http://en.wikipedia.org/wiki/Elitzur-Vaidman_bomb-testing_problem&quot; rel=&quot;nofollow&quot;&gt;bomb-detection experiment&lt;/a&gt; is the most popular I think - again in that, you can detect the presence of a position-measurement (a photon-sensitive bomb going off) in a certain place in the experiment by the loss of interference effects in front of later detectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these examples are examples of &lt;a href=&quot;http://en.wikipedia.org/wiki/Counterfactual_definiteness&quot; rel=&quot;nofollow&quot;&gt;counterfactual measurements&lt;/a&gt;, where you draw conclusions about something you don't explicitely measure, and hence provide a loophole to your question.&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-04-07T12:33:32.790" />
  <row Id="8199" PostTypeId="2" ParentId="8191" CreationDate="2011-04-07T12:41:45.060" Score="0" ViewCount="" Body="&lt;p&gt;The following &lt;a href=&quot;http://www.sciencedaily.com/releases/2007/11/071101144942.htm&quot; rel=&quot;nofollow&quot;&gt;publication&lt;/a&gt; measures single electrons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Researchers at the Delft University of Technology's Kavli Institute of Nanoscience and the Foundation for Fundamental Research on Matter (FOM) have succeeded in controlling the spin of a single electron merely by using electric fields. This clears the way for a much simpler realization of the building blocks of a (future) super-fast quantum computer.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose they have taken as given the value of the electron magnetic field, but the apparatus could be used to measure it, I am sure.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-07T12:41:45.060" />
  <row Id="8200" PostTypeId="1" CreationDate="2011-04-07T12:56:56.333" Score="0" ViewCount="59" Body="&lt;p&gt;Is it feasible that time could depend on scale? We generally think because of relativity that time is velocity dependent... but maybe it's the other way around? Maybe time is dependent on scale and/or velocity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do I ask such a question? Because we generally think of &quot;small&quot; things as moving fast and large things as moving slow. The universe seems to evolve on a much slower time scale than atomic particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe the reason is simply there is a different time scale due to the size(or maybe gravity or whatever).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any logic or thought experiment that could be used to prove that time much be the same at all &quot;scales&quot;? By scale I don't mean simply scale of length because that is too arbitrary to define properly. It would have to be some type of quantity that makes sense as scale of length as we typically comprehend it actually increases. The only thing I can think of that really scales like this is gravity.&lt;/p&gt;&#xA;" OwnerUserId="1551" LastActivityDate="2011-04-07T12:56:56.333" ClosedDate="2011-04-07T13:55:36.243" Title="Time dependent on scale?" Tags="&lt;mathematical-physics&gt;" CommentCount="1" />
  <row Id="8201" PostTypeId="1" CreationDate="2011-04-07T13:04:30.700" Score="9" ViewCount="379" Body="&lt;p&gt;I started reading about Conformal Field Theory a few weeks ago. &#xA;I'm from a more mathematical background. I do know Quantum Mechanics/Classical Mechanics, &#xA;but I'm not really an expert when it comes to Quantum Field Theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while reading through several papers, some questions appeared and&#xA;I would be more than happy if someone could answer them: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;First I'm refering to &lt;a href=&quot;http://www.phys.ethz.ch/~mrg/CFT.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.phys.ethz.ch/~mrg/CFT.pdf&lt;/a&gt; (which is a german document)&#xA;Chapter 3.2.1 (starting at page 29). &#xA;The main topic here is the Free Boson as an example of a CFT. &#xA;In this chapter he tries to proof the conformal invariance of the quantized theory. &#xA;He's doing this by constructing the generators of conformal symmetry. &#xA;Now here comes the first question: Why does this show conformal invariance? &#xA;In the classical case you show conformal invariance by showing that the integrand of&#xA;the action functional doesn't change under a conformal group action (which seems reasonable). &lt;strong&gt;But why does constructing generators of conformal symmetry imply conformal invariance in the quantum case?&lt;/strong&gt; &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Next I'm refering to equation (3.2.26) in the same chapter. Here he states &#xA;that the equation (3.2.26) $[L_{m},\phi]=z^{m+1}\partial_{z}\phi(z,\bar{z})$&#xA;proofs that the operators L_{m} actually do implement conformal transformations &#xA;of the type $L_{n}=-z^{n+1} \partial_{z}$. Why is that so? Why does this proof that&#xA;the L_{m} actually implement these conformal transformations $L_{n}=-z^{n+1} \partial_{z}$? Or better: &lt;strong&gt;What does he mean by &quot;implementing a transformation&quot;?&lt;/strong&gt; &#xA;(What's the defintion if you want). The equation $[L_{m},\phi]=z^{m+1}\partial_{z}\phi(z,\bar{z})$ kind of looks like an eigenvalue-equation (which of course it isnt). &#xA;But if you think that there's no Lie-Bracket. It looks like $z^{m+1}\partial_{z}$ being an &quot;eigenvalue&quot; of $L_{m}$ of the eigenvector $\phi$. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I'm now refering to the paper of David Tong: &lt;a href=&quot;http://www.damtp.cam.ac.uk/user/tong/string/four.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.damtp.cam.ac.uk/user/tong/string/four.pdf&lt;/a&gt; (which is english). In Chapter 4.5.1 (on page 92) he adresses radial quantization. So he has a Theory living on a cylinder and maps this theory with &#xA;the exponential map to the complex plane. So first of all, this seems a bit restrictive&#xA;to me. I mean &lt;strong&gt;why should we only consider theories living on a cylinder as the general case?&lt;/strong&gt; What I've heard is that that since you have a finite spatial direction on the cylinder you avoid singularities. But since I don't know Quantum Field Theory well enough I'm not so sure about this. N*&lt;em&gt;ext it's not so obvious where the actual quantization appears in this process.&lt;/em&gt;* For me quantization means imposing certain commutation relations on operators/fields. And I don't see where this is happening here? Maybe it happens implicitly when you transform the energy momentum tensor from the cylinder to the plane by using the Schwartzian derivative?? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'd really be more than happy if someone can provide me with some detailed explanations. &#xA;(because I'm really not an expert yet in this subject)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!!&lt;/p&gt;&#xA;" OwnerUserId="2969" LastEditorUserId="1287" LastEditorDisplayName="" LastEditDate="2011-04-07T15:04:21.470" LastActivityDate="2011-04-07T17:27:35.347" Title="Beginners questions concerning Conformal Field Theory" Tags="&lt;quantum-field-theory&gt;&lt;string-theory&gt;&lt;conformal-field-theory&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="8202" PostTypeId="2" ParentId="8189" CreationDate="2011-04-07T13:05:36.970" Score="3" ViewCount="" Body="&lt;p&gt;There's a more detailed model of the experiment in which you aren't making the same measurement twice. You say it yourself, the two measurements are made at different times. It's better, IMO, to get out of thinking in terms of the first approximation that is quantum mechanics, instead moving to thinking in terms of the second approximation that is quantum field theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a quantum field theory, there is a difficult question of just what you might be measuring. If we think in terms of QFT being a &lt;em&gt;field&lt;/em&gt; theory, it's not at all clear that you're measuring the position of a single particle, because in field terms you're just measuring a statistic of the field in two different places in space-time (constructing an ensemble of data points or some other way of relating probabilities to statistics needs some care in this conceptual background, of course). Even if we think in terms of QFT being a particle theory, which IMO, and in the opinion of an increasing number of papers in the literature, goes somewhat against the current of the formalism, the two measurements may be measurements of the positions of different particles, not of two positions of the same particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would further reduce Marek's comment that you don't need to tell when there is a collapse. If you're an instrument-level experimentalist, aka a lab technician, all you care about is that there is an analogue signal. You might notice that the signal has more-or-less discrete transitions between different levels or different types of behavior, and because of that you might start to record when those discrete events occur. Only if you care about theory, and want to give an explanation for the discrete events, do you mention collapse (although &quot;collapse&quot; is more an encoded description of the observed discrete structure than an explanation). I think it's only if you care about theory that you have particular wishes as to what features of the analogue signal you will convert to logbook or digital representation or put into a [published paper, which we can suppose will include particular interest in any identifiable discrete features there might be in the analogue signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bjorn's Answer appeared while I wrote this. It's a good way to start discussing more detailed models of what measurements do (it's Useful!) in quantum mechanical terms. I can get carried away by QFT ways of thinking. In the QM sort of vein, you might try the book &quot;Operational Quantum Physics&quot; by Paul Busch, Marian Grabowski, and Pekka J. Lahti (don't be misled by the &quot;Operational&quot;, it's far more theory than experimental, and also not elementary).&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-04-07T13:05:36.970" />
  <row Id="8203" PostTypeId="2" ParentId="8169" CreationDate="2011-04-07T13:10:25.083" Score="5" ViewCount="" Body="&lt;p&gt;It's been proposed, and there are &lt;a href=&quot;http://www.sciencemag.org/content/326/5949/108.abstract&quot; rel=&quot;nofollow&quot;&gt;a variety of possible experimental systems&lt;/a&gt; people are working on. I expect this to be a very useful way of simulating quantum systems, which will work well for many questions long before large-scale universal quantum computers can be built.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditorDisplayName="" LastEditDate="2011-04-07T13:26:30.697" LastActivityDate="2011-04-07T13:26:30.697" CommentCount="3" />
  <row Id="8204" PostTypeId="2" ParentId="8176" CreationDate="2011-04-07T13:24:23.640" Score="1" ViewCount="" Body="&lt;p&gt;This does not imply gravity waves.  This can be seen by the following calculation. The mass configuration is that of two masses in a circular orbit around each other.  This is a dipole term, which defines ${\vec p}~=~m{\vec d}$, for $\vec d$ the distance between the two masses.  So this dipole moment projected along the field determines a dipole potential&#xA;$$&#xD;&#xA;\phi_d(r)~=~{\vec p}\cdot\nabla\frac{GMm}{r}&#xD;&#xA;$$ &#xA;If this generates power in waves that escape the system, then a time derivative is dependent on $md{\vec d}/dt$.  However, momentum conservation tells us this must be zero in the initial frame of the two masses.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in order for gravity waves to exist one must have quadrupole moments and higher.  However, this contributes to gravity waves in the post Newtonian expansion to second order.  The &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gravity wave is a perturbation on a background metric $\eta_{ab}$ with the total metric&#xA;$$&#xD;&#xA;g_{ab}~=~\eta_{ab}~+~h_{ab}.&#xD;&#xA;$$&#xA;The flat background metric has zero Ricci curvature so that to first order in the perturbation expansion $R_{ab}~=~\delta R_{ab}$, which enters into the Einstein field equation $R_{ab}~-~\frac{1}{2}Rg_{ab}~=~\kappa T_{ab}$, where $\kappa~=~8\pi G/c^4$ is the very small coupling constant between the momentum-energy source and the spacetime configuration or field. The Ricci curvature to first order is then&#xA;$$&#xD;&#xA;R_{ab}~=~{1\over&#xD;&#xA;2}\Big(\partial_c\partial_a{h^c}_b~+~\partial_c\partial_b{h^c}_a~-~\partial_a\partial_bh)~-~\partial_c\partial^ch_{ab}\Big).&#xD;&#xA;$$&#xA;The harmonic gauge $g^{bc}\Gamma^a_{bc}~=~0$, to first order as $\partial_c{h^c}_a~=~\frac{1}{2}\partial_\mu h$ the Einstein field equation gives&#xA;$$&#xD;&#xA;\partial^c\partial_ch_{ab}~-~\frac{1}{2}\eta_{ab}\partial^c\partial_ch~=~{{16\pi G}\over {c^4}}T_{ab},&#xD;&#xA;$$&#xA;which is well defined for the traceless metric term ${\bar h}_{ab}~=~h_{ab}~-~\frac{1}{2}\eta_{ab}h$ with the simple wave equation&#xA;$$&#xD;&#xA;\partial^c\partial_c{\bar h}_{ab}~=~{{16\pi G}\over {c^4}}T_{ab}.&#xD;&#xA;$$&#xA;This is then a basic wave equation for a tensor field with two independent components $h_{\times\times}$ and $h_{++}$ for polarization directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is of interest is the source of the gravity waves $T_{ab}$.  The metric perturbation far from the source $r~=~|{\bf x}|~&amp;gt;&amp;gt;~|{\bf y}|$ is approximated by,&#xA;$$&#xD;&#xA;{\bar h}_{ab}~=~{{4G}\over{rc^4}}\int_V d^3yT_{ab}(t~-~r/c,~{\bf y}).&#xD;&#xA;$$&#xA;The covariant constancy of the momentum-energy tensor $\nabla\cdot {\bf T}~=~0$ results in a number of relationships. The energy $T_{tt}~=~\rho U_0U_0$ component of with these relationships gives that for the quadrupole moment $I_{ij}~=~\int_Vx_ix_j\rho(t,~{\bf x})$ the metric wave is given by&#xA;$$&#xD;&#xA;{\bar h}_{ij}~=~{{2G}\over {rc^4}}{{d^2}\over{dt^2}}I_{ij}(t,~{\bf x}).&#xD;&#xA;$$&#xA;This term is different from a naïve idea of gravity waves suggested here.  The coupling constant is $~\sim~G/c^4$ which comes from the Einstein field equation and is extremely small.  The post Newtonian terms to $c^{-4}$ are second order, and this is a departure from a Newtonian gravity.&lt;/p&gt;&#xA;" OwnerUserId="2970" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-07T18:13:38.457" LastActivityDate="2011-04-07T18:13:38.457" CommentCount="1" />
  <row Id="8206" PostTypeId="1" CreationDate="2011-04-07T14:23:02.540" Score="2" ViewCount="159" Body="&lt;p&gt;Perturbatively, open strings with both ends attached to the same D-brane definitely exist. But D-branes have nonzero p-form charges. These charges give rise to a nonzero flux with nonzero stress-energy tensor. This backreaction leads to curved spacetime. Being BPS, the metric becomes that of an extremal black brane with the event horizon at an infinite distance (logarithmic divergence) away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to apologize for asking bad questions. Please put up with a newbie like me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy of a string due to its tension is proportional to the length of the string and also the time factor aka the gravitational redshift factor. A string is a worldsheet after all. An open string extending radially from the event horizon has an infinite length, but the gravitational redshift factor cancels that, and the total energy of the open string remains finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no question after all.&lt;/p&gt;&#xA;" OwnerUserId="2955" LastEditorUserId="2955" LastEditorDisplayName="" LastEditDate="2011-04-09T15:47:40.063" LastActivityDate="2011-04-09T15:47:40.063" Title="Do open strings with both ends attached to the same D-brane really exist?" Tags="&lt;string-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="8207" PostTypeId="1" CreationDate="2011-04-07T14:54:58.423" Score="2" ViewCount="88" Body="&lt;p&gt;We just (20mn ago) suffered from another strong aftershock (of the initial March 11th earthquake) in Japan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some questions related to seismic science. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Why is there so much time between the initial quake and the following strong aftershocks (initial was March 11, the stronger aftershock happened 20mn ago, April 7)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why does it seem that progress in seismic science (and quakes predictions) is so poor? It seems people have only raw probabilities, based on historical data.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Would anybody have any graphical / imagery data of the physical fault move on March 11?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-07T17:12:29.087" LastActivityDate="2011-04-07T17:12:29.087" Title="Seismic science questions" Tags="&lt;geophysics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="8208" PostTypeId="2" ParentId="8162" CreationDate="2011-04-07T15:33:10.347" Score="1" ViewCount="" Body="&lt;p&gt;The $AdS$ in a Euclidean form is a is a hyperbolic space.  In two dimensions hyperbolic plane ${\cal H}^2$ is a simply connected manifold with constant Gaussian curvature $-1$.  The two dimensional $AdS_2$ holds near the event horizon of a black hole, which is the Poincare disk ${\cal D}^2~=~\{z:|z|~&amp;lt;~z_0\}$, and related to the Rindler spacetime the upper half plane ${\cal H}^2~=~\{z:Im(z)~&amp;gt;~0\}$.  The group of isometries $Iso({\cal H}^2)$ is the set of smooth transformations $z^\prime~=~gz$ which satisfy the hyperboloid metric for $s~=~s(z,~gz)$.  The half plane and the Poincar{\'e} disk are related by a conformal transformation, so the transformation of coordinates are given by&#xA;the same group. In the half plane the isometries are the fractional linear transformations, or modular group&lt;br&gt;&#xA;$$&#xD;&#xA;g:{\cal H}~\rightarrow~{\cal H},~z:~\mapsto~gz:~=~{{az~+~b}\over{cz~+~d}},~\left(\matrix{a &amp;amp;&#xD;&#xA;b\cr c &amp;amp; d}\right)~\in~SL(2,~{\bf R}).&#xD;&#xA;$$&#xA;The matrices $g$ and $-g$ are the same fractional linear transformations, so the isometries  $Iso({\cal H}^2)$ may be identified with the projective Klein group $PSL(2,~{\mathbb R})~=$ $SL(2,~{\mathbb R})/\{\pm 1\}$.  The group is restricted further by the isotropy subgroup which leaves elements of ${\cal H}^2$ unchanged $z~=~gz$.  Any such $g~\in~PSL(2,~{\mathbb R})$ defines the $SO(2)$ rotation group.  Then ${\cal H}^2~=~PSL(2,~{\mathbb R})/SO(2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discrete structure, or $PSL(2,{\mathbb Z})$ is manifested in the tessellation symmetry of the hyperbolic half-plane or disk.  These symmetries are seen in the Escher prints called limit circles.  These discrete structures give the MERA structure which Genneth references.  The “piling up” of structure towards the boundary is a renormalization of the Ising spin system, for spins at the vertices in the tessellation.  What follows is in part a brief outline of the discrete coset $AdS$ completion due to Charles Frances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.u-psud.fr/~frances/&quot; rel=&quot;nofollow&quot;&gt;http://www.math.u-psud.fr/~frances/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boundary space $\partial AdS_{n+1}$ is a Minkowski spacetime, or a spacetime $E_n$ that is simply connected that with the $AdS$ is such that $AdS_{n+1}\cup E_n$ is the conformal completion of $AdS_{n+1}$ under the discrete action of a Kleinian group.  For the Lorentzian group $SO(2,~n)$ there exists the discrete group $SO(2,n,Z)$ which is a Mobius group.  For a discrete subgroup $\Gamma$ subset $SO(2,~n,~Z)$ that obeys certain regular properties for accumulation points in the discrete set $AdS_{n+1}/\Gamma$ is a conformal action of $\Gamma$ on the sphere $S_n$.  This is then a map which constructs an $AdS/CFT$ correspondence.  Given that $AdS_n~=~O(n,2)/O(n,1)$ this coset structure is a Clifford-Klein form, or double coset structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lightlike geodesics in $E^n~=~M^n$, the Minkowski spacetime, are copies of $RP^1$, which at a given point p define a set that is the lightcone $C(p)$.  The point p is the projective action of $\pi(v)$ for $v$ a vector in a local patch $R^{n,2}$ and so $C(p)$ is then $\pi(P\cap C^{n,2})$, for $P$ normal to $v$, and $C^{n,2}$ the region on $R^{n,2}$ where the interval vanishes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The space of lightlike geodesics is a set of invariants and then due to a stabilizer on $O(n,2)$, so the space of lightlike curves $L_n$ is identified with the quotient $O(n,2)/P$, where $P$ is a subgroup defined the quotient between a subgroup with a Zariski topology, or a Borel subgroup, and the main group $G~=~O(n,~2)$.  This quotient $G/P$ is a projective algebraic variety, or flag manifold and $P$ is a parabolic subgroup.  The natural embedding of a group $H~\rightarrow~G$ composed with the projective variety $G~\rightarrow~G/P$ is an isomorphism between the $H$ and $G/P$.  This is then a semi-direct product $G~=~P~\rtimes~ H$.  For the $G$ any $GL(n)$ the parabolic group is a subgroup of  upper triangular matrices. This is the Heisenberg group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This connection to Heisenberg groups and parabolic groups is particularly interesting.  The structure here has a $\theta$-function realization, and is related to the density of states in string theory.  This leads one in my opinion into some very deep structure which is not at all completely explored.&lt;/p&gt;&#xA;" OwnerUserId="2970" LastActivityDate="2011-04-07T15:33:10.347" CommentCount="1" />
  <row Id="8209" PostTypeId="2" ParentId="8207" CreationDate="2011-04-07T15:36:26.330" Score="1" ViewCount="" Body="&lt;p&gt;I also live in an earthquake prone region,in Greece, and we got a 6.3 a week or so ago, fortunately in the middle of the sea between Crete and Carpathos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We got two stong ones in  Athens during my life time: one in 1981 and one in 1990, from different fault lines, though all of them finally are due to the fact that the African plate is pushing against the European one. At the time of the shocks all physicists in my region read up on seismology. It is not an advanced science. Some tried to predict aftershocks with dubious results : large error for epicenter, large error on time, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that the pressure builds at the fault until it breaks. Then the system settles with &quot;small&quot; after shocks, going on for months. Large after shocks can also happen. Predictions are very uncertain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://earthquake.usgs.gov/earthquakes/step/explain.php&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; may help.  &lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-07T15:36:26.330" CommentCount="3" />
  <row Id="8210" PostTypeId="2" ParentId="8206" CreationDate="2011-04-07T15:43:36.770" Score="2" ViewCount="" Body="&lt;p&gt;Dear Shankar, a good question. Andy Strominger would be telling us stories how, around 1995, Andy would speak about (black) $p$-branes during lunches in Santa Barbara. Then he would stop and Joe Polchinski would speak about D-branes during the rest of the lunch. They wouldn't notice that they were speaking about the same object - up to some moment when they did realize that. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $p$-branes are physically the same objects as D-branes. However, you can't use both of these descriptions at the same moment. If you describe the object as a D-brane, you're treating it as a light perturbation of the background and you assume that the backreaction is small. The backreaction is effectively proportional to $g_{\rm string} N$ where $N$ is the number of D-branes in the stack. So at weak coupling, for a single (or a few) D-branes, there is no backreaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the backreaction is large, then $g_{\rm string} N\gg 1$, or a similar inequality holds. In that case. the perturbative expansion based on open strings attached to the D-brane(s) and their world sheet loops wouldn't converge and it is better to switch to the dual description which sees the curved geometry of the extremal black $p$-brane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you can never use both of these descriptions at the same moment! The picture of D-branes as sheets of paper inserted into a pre-existing space is valid or useful if $g_{\rm string} N\ll 1$ while the curved geometry with the event horizon is valid, useful, and nicely convergent in the opposite limit $g_{\rm string} N\gg 1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &quot;duality of regimes&quot; also becomes decisive in the AdS/CFT correspondence. If 't Hooft's coupling $g_{\rm Yang-Mills}^2 N \ll 1$, the Yang-Mills perturbative expansion - inherited from the open string perturbative physics attached to D-branes - is convergent. When $g_{\rm Yang-Mills}^2 N \gg 1$, the new fifth holographic direction becomes visible and you should switch to the AdS higher-dimensional gravitational description. In fact, the AdS space is just the near-horizon limit of the backreacted geometry of the stack of D-branes, and in a similar way, Yang-Mills theory is the corresponding low-energy limit of the open &quot;string field theory&quot; living on the D-branes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is unacceptable to calculate D-brane loops and backreaction of the geometry at the same moment. It is as invalid a double counting as if you take the toroidal world sheet with the complex structure $\tau$ and you would double count the thin and fat tori with $\tau$ and $\tau&amp;#39;=-1/\tau$. Only the effects from one of the descriptions may be counted as independent ones, otherwise you're double-counting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you consider the energy of open strings at the background of D-branes, you always calculate their energy from perturbative string theory, assuming the background without backreactions. If you also include higher-loop corrections, all the effects of the backreactions will be correctly accounted for and they will be relatively low. The physical open strings with Chan-Paton factors never terminate on the event horizon etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is the holographic duality between the two descriptions - one that is also visible in AdS/CFT (which is just the near-horizon/low-energy limit of what you're considering). In the CFT description, there is an $SU(N)$ gauge symmetry - in the most canonical example of the AdS/CFT correspondence - and no gravity and no diffeomorphism symmetry. In the dual AdS gravitational description, the natural gauge group includes the bulk diffeomorphism of general relativity but there is no $SU(N)$ Yang-Mills symmetry. The two pictures have the same physical Hilbert spaces - only the physical ones, after you eliminate the gauge-variant states etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is also untrue that open strings attached to D-branes always want to become closed strings because they're shorter. In fact, many open string states attached to D-branes with both ends may be BPS - which proves that they can't decay to anything lighter, because of energy conservation. When counting their energy at weak 't Hooft coupling, you shouldn't include any event horizons or infinite enhancement of the energy: you must use the pre-existing metric. Alternatively, you may use the &quot;closed string&quot; description with the backreacted metric. But if you do so, you don't see open strings going through the complicated space. These open-string degrees of freedom get reinterpreted as closed-string degrees of freedom. You should never double-count them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the oscillations of the shape of a D3-brane in the transverse 6 dimensions are represented by scalar open string modes at weak coupling. At the strong coupling, the stack of D3-branes becomes a black $p$-brane and the variations of the shape of the brane within the 10-dimensional spacetime are represented by changes of the black-hole-like solution, which is clearly carried by closed-string modes such as the graviton. Double counting would be a mistake: it would be literally equivalent to taking the $N=4$ $SU(N)$ gauge theory in 3+1 dimensions and adding the five-dimensional curved bulk metric as well. That's not how the physics works: the dynamics is fully determined by the gauge theory, or fully determined by the gravitating (closed) string theory in the bulk - and these two descriptions are equivalent and you don't have to (and you shouldn't) combine them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Chan-Paton factors become literally invisible if you switch to the closed-string description (with nonzero backreaction). They're some hidden degrees of freedom, much like the bulk diffeomorphisms are hidden in the boundary CFT. There's no contradiction because the symmetry arising from the Chan-Paton factors is a gauge symmetry, so all physical states have to be singlets under this symmetry, anyway. The same holds in the opposite way, for the gauge symmetry in the bulk - e.g. the diffeomorphisms in $AdS_5$.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-07T15:48:44.600" LastActivityDate="2011-04-07T15:48:44.600" CommentCount="4" />
  <row Id="8211" PostTypeId="2" ParentId="8201" CreationDate="2011-04-07T15:47:01.870" Score="0" ViewCount="" Body="&lt;p&gt;I'm not really qualified to answer, but I think I can give a hint on your first question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you have to think of the field $\phi$ as an operator. In particular, I tend to think of $\phi$ as a linear combination&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \hat\phi(z,\bar z) = \int dz d\bar z\ \phi(z,\bar z) \Psi^\dagger(z,\bar z)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of operators $\Psi^\dagger(z,\bar z)$ that create particles at positions $(z,\bar z) = (x+iy,x-iy)$ (where $x$ and $y$ are the independent coordinates in the plane). The construction in terms of normal modes is similar, except that you use a different basis - not positions but modes (mainly because creating a particle that it localized at precisely one point has its mathematical problems).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The following is not very accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what does it mean for the operator $\mathcal L(\phi, \partial\phi)$ to be invariant under a symmetry operator $T$? It simply means that both operators commute,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ [T,L(\phi, \partial\phi)] = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the symmetry operator $T=L_n$ fulfills the commutator relation in question, and the Lagrangian &lt;em&gt;as a function&lt;/em&gt; is invariant under conformal transformations, then it follows that the Lagrangian &lt;em&gt;as an operator&lt;/em&gt; commutes with $T$. The reason is that the commutator relation for $L_n$ makes $[L_n,&amp;#183;]$ act on the Lagrangian operator exactly like the corresponding conformal symmetry acts on the Lagrangian function. (I think this can be shown by using the above picture of $\phi$ as an operator.)&lt;/p&gt;&#xA;" OwnerUserId="188" LastEditorUserId="188" LastEditorDisplayName="" LastEditDate="2011-04-07T17:27:35.347" LastActivityDate="2011-04-07T17:27:35.347" CommentCount="3" />
  <row Id="8212" PostTypeId="2" ParentId="8176" CreationDate="2011-04-07T15:48:30.530" Score="4" ViewCount="" Body="&lt;p&gt;This is exactly the approach taken in Bernard Shutz's note &quot;&lt;a href=&quot;http://ajp.aapt.org/resource/1/ajpias/v52/i5/p412_s1?&quot; rel=&quot;nofollow&quot;&gt;Gravitational waves on the back of an envelope&lt;/a&gt;&quot; (Am. J. Phys. &lt;strong&gt;50&lt;/strong&gt; vol 5 pp 412).  The abstract reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Using only Newtonian gravity and a little special relativity we calculate most of the important effects of gravitational radiation, with results very close to the predictions of full general relativity theory&lt;/strong&gt;. Used with care, this approach gives helpful back‐of‐the‐envelope derivations of important equations and estimates, and it can help to teach gravitational wave phenomena to undergraduates and others not expert in general relativity. We use it to derive the following: the quadrupole approximation for the amplitude h of gravitational waves; a simple upper bound on h in terms of the Newtonian gravitational field of the source; the energy flux in the waves, the luminosity of the source (called the ‘‘quadrupole formula’’), and the radiation reaction in the source; order‐of‐magnitude estimates for radiation from supernovae and binary star systems; and the rate of change of the orbital period of the binary pulsar system. Where our simple results differ from those of general relativity we quote the relativistic ones as well. We finish with a derivation of the principles of detecting gravitational waves, and we discuss the principal types of detectors under construction and the major limitations on their sensitivity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(If you don't have access to Am J Phys, &lt;a href=&quot;http://www.gw-indigo.org/tiki-download_wiki_attachment.php?attId=45&quot; rel=&quot;nofollow&quot;&gt;this talk&lt;/a&gt; seems to recapitulate the details.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A major difference in this Newtonian scalar theory from the real GR theory of gravitational waves is in the effect of the waves on an inertial test particle.   This Newtonian theory predicts that the waves would appear as an oscillating force along the direction between the source and the test particle.  By contrast, GR predicts an oscillating differential tidal effect in the plane &lt;em&gt;perpendicular&lt;/em&gt; to the line connecting the source and the test mass.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, while I think LIGO would still detect the &quot;Newtonian&quot; form of gravitational waves, the antenna pattern of the detector would be different.  The L-shaped LIGO detector has optimal sensitivity to a source located directly overhead in the GR case (allowing the gravitational wave to stretch one arm while it is compressing the other).  There would be no sensitivity to a &quot;Newtonian&quot; source directly overhead.  However, you could detect it if the &quot;Newtonian&quot; source were aligned with either arm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, &quot;Newtonian noise&quot; (the near-field action of Newtonian gravity arising from density waves in the material near the detector) is a real concern for terrestrial gravitational wave detectors!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. To be pedantic, it is best to avoid the term &quot;gravity wave&quot; (as opposed to &quot;gravitational wave&quot;), since a &quot;gravity wave&quot; (&quot;Newtonian gravity wave&quot; even!) is something &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravity_wave&quot; rel=&quot;nofollow&quot;&gt;completely different&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-04-07T16:22:58.843" LastActivityDate="2011-04-07T16:22:58.843" CommentCount="1" />
  <row Id="8213" PostTypeId="2" ParentId="8201" CreationDate="2011-04-07T16:43:18.160" Score="6" ViewCount="" Body="&lt;p&gt;concerning the first question, related to the German text by Matthias Gaberdiel (greetings to him):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Closed algebra implies symmetry&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is enough to construct the generators of an algebra - in this case, conformal algebra - and calculate their commutators $[L_m,L_n]$ and so on. If the commutators are linear combinations of other generators, we say that the generators form a closed algebra. Now, you are right that we also want to use some &quot;dynamical information about the theory&quot;. You wrote that the action should be invariant under the transformations that these generators generate - and you worry that the action - all the dynamical information about the theory - has been completely removed from the proof, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a good point but the dynamical information hasn't been removed because a particular generator (or, for a general basis of generators, a linear combination of generators) is the Hamiltonian that determines the dynamics itself. For conformal symmetry, it is $L_0+\tilde L_0$ that plays the role of the Hamiltonian. It generates translations of the cylinder or, equivalently (as we will discuss below), multiplicative translations in the radial coordinate. (Maybe additive shifts such as $c/24$ should be added on one of the backgrounds.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $L_0+\tilde L_0$ is a linear combination of some generators and you can show that the set of generators is closed under the operation of taking the commutator, it proves that the whole algebra generated by this set of generators is a dynamical symmetry. In this case, the commutator $[L_m, L_0+\tilde L_0]$ is not strictly zero - so the generators $L_m$ don't commute with the Hamiltonian. Instead, the commutator is equal to another combination of the symmetry generators. But we still say that $L_m$ is a symmetry of the system and we know this situation from other contexts as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in special relativity, the angular momentum $J_{12}=J_z$ commutes with the energy $p_0$. However, the Lorentz boost generator $J_{03}$ doesn't commute with the Hamiltonian $p_0$: their commutator is proportional to $p_3=p_z$, a component of the momentum. It's nonzero but it's another symmetry generator. It's normal for symmetry generators that act nontrivially on time - such as the $J_{03}$ boost generator in relativity or $L_m$ in conformal symmetry - to have nonzero commutators with the Hamiltonian $p_0$ or $L_0+\tilde L_0$, respectively. What's important is that the commutator is another operator we know to be a generator of a symmetry, and the symmetry algebra is fully described by group theory - by the structure constants $f$ in $[L_m,L_n]=f_{mn}^k L_k$ - and doesn't need us to know any detailed dynamical information about the fields etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You proposed that one should be verifying that the action is invariant under symmetry generators. That sounds good except that the action is only good for a classical description - or a quantum description that is obtained by a direct quantization of a classical theory. Such a way to obtain a quantum theory is only smooth or useful if the quantum theory is &quot;close enough&quot; to a classical theory. The most general CFT, especially in 2 dimensions, is so strongly &quot;quantum&quot; that there is no natural notion of an action and classical degrees of freedom. One must directly work with the quantum operators, their commutators, and they don't have any natural or helpful classical limit. Take the Ising model CFT as an example. You will find lots of fields, such as spin fields and twist fields, whose (mass) dimensions are fractional numbers such as $1/16$, something that would be unthinkable in a classical theory: the whole dimension comes from quantum effects. That's why the structure of the CFT science tries to be as independent of classical concepts such as the action as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Implementing a symmetry on operators&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a generator $G$ of a Lie symmetry, it (infinitesimally) acts on ket states $|\psi\rangle$ and bra states $\langle \varphi|$ as&#xA;$$\delta |\psi \rangle = i \epsilon G  |\psi \rangle, \quad \delta \langle \varphi| = i \epsilon \langle \varphi| G$$&#xA;If you also define the action of the generator $G$ on a general operator $M$ as &#xA;$$\delta M = i \epsilon [G,M]$$&#xA;then you may prove that all matrix elements will be invariant under the symmetry,&#xA;$$\delta \langle \varphi | M | \psi \rangle = 0$$&#xA;by the Leibniz rule. So the natural action of symmetry generators such as $L_m$ on operators such as $\phi(z,\bar z)$ is&#xA;$$\delta \phi(z,\bar z) = i \epsilon [L_m,\phi].$$&#xA;So if the commutator of $L_m$ with some fields - operators - is the same as the appropriate $(n+1)$-st derivative of these operators, then the generators $L_m$ implement the symmetry whose infinitesimal form involves $\delta \phi\sim z^{n+1} \partial^{n+1} \phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your comments about &quot;eigenvalues&quot; are conceptually misguided because a defining property of an &quot;eigenvalue&quot; is that it must be a &quot;value&quot; - a $c$-number - but $\partial_z$ is not a value - it is an operation. (I avoided the word &quot;operator&quot; because $\partial_z$ is not an operator acting on the Hilbert space of the CFT; only operators such as $\phi(z,\bar z)$ and $\partial_z \phi(z,\bar z)$  or $L_m$ are operators acting on the CFT Hilbert space. Instead, $\partial_z$ itself is just a rule to produce one operator from another one. It would be an operator if the wave functions - state vectors - were equivalent to functions of $z,\bar z$ but in a two-dimensional CFT, they surely aren't.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why cylinder is important&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the question based on David Tong's text (greetings to David!), the cylinder is important exactly because a CFT on a cylinder is exactly equivalent to a CFT on the infinite plane. If $w=\sigma+i\tau$ lives on a cylinder - with $\sigma$ being $2\pi$-periodic - and if $z=\exp(-iw)$, then the infinite cylinder will be fully mapped to the plane, in a one-to-one way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I actually think that David is being very clear about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the analysis of the CFT defined on a full plane in general, and its behavior near the $z=0$ origin in particular, is totally equivalent to an analysis of a CFT defined on a cylinder in general, and in the limit $w\to -i\infty$ in particular. The two problems are exactly equivalent, exactly because of the conformal symmetry. The cylinder has a periodic spatial coordinate but this periodicity is not postulated for &lt;em&gt;ad hoc&lt;/em&gt; reasons. It's postulated because if you write $z$ in the radius/phase form,&#xA;$$ z= \exp(-i\sigma+\tau), $$&#xA;then the points with $\sigma\sim \sigma+2 \pi$ are identified with one another. The coordinate $\sigma$ is periodic. This is the basic fact of the exponential function - or its inverse function, the logarithm, if viewed as a function of a complex variable. And the exponential conformal map is very useful which can be seen if you follow what David is doing with that. You could ban the exponential function because you don't like it (or you think that other functions are being discriminated against) - but then you couldn't learn much of the CFT calculus because the CFT calculus largely depends on this clever exponential conformal map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because a small piece of any two-dimensional world sheet - regardless of the topology - looks like the flat plane and because the flat plane is equivalent to the infinite cylinder, the infinite cylinder is important for the understanding of local physics of the CFT at any Riemann surface - of any topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is just true that I may just describe the infinite plane in coordinates such that one of them is periodic. This is what makes the analysis of the states defined on the cylinder - closed string states - automatically useful for the analysis of any properties of CFT, including its operators on the plane. In fact, the states of a closed string - obtained by quantizing the CFT on a cylinder - are in one-to-one correspondence with the local operators $\phi_K(0)$ at the origin (or any other point), because of the very same conformal map from the plane to the cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you ask, where is the quantization?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of the formulae would work for a classical (non-quantum) conformal field theory, too. However, there are is no Hilbert space of &quot;states&quot; of a closed string, obtained from the quantization. So many of the interesting things, including the state-operator correspondence discussed two paragraphs above this one, only arise in the quantum theory. Pretty much all the objects such as $H$, $T_{\rm cylinder}$, and so on that David lists on page 86 or almost any other page are operators, so he deals with a quantum theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some equations, David surely also uses commutators, to prove that it is a quantum theory, but it's not necessarily page 86 or another page you could find that has no commutators on it. ;-) But your complaint that David doesn't play with commutators of some fields exactly on some page where you would expect it surely not a sensible complaint, is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure that if you listen carefully, you will also understand that the commutators of operators in a CFT may be obtained from the OPEs, the operator product expansions. Just place two operators $T_k(z)$ and $T_l(0)$ to two nearby points $0$ and $z$ and calculate their product. The product will typically include a singularity that diverges as $z\to 0$ - as the two operators are very close to each other. (The singularity will be visible in any sensible expectation value.) The coefficient of $1/z$ or $1/z^2$ or $1/z^4$ - the leading singularity - is either a $c$-number or another operator. From that operator, you may determine the commutator of Fourier modes of $T_k$ and $T_l$ expanded over the cylinder, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum mechanics has many effects that you wouldn't encounter in classical physics. For example, as you correctly mention, it influences the transformation from the cylinder to the plane, and so on. However, I don't know what to do with questions such as &quot;And I don't see where this is happening here?&quot; What should be happening here? Well, what's happening is probably something else than what you were expecting to be happening - but that's the very reason why you're trying to learn new things from David Tog, isn't it? If you were only learning old things that you knew, you would be wasting your time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The things that you have to learn to understand two-dimensional conformal field theories are not &quot;the same things&quot; that you already learned for a generic quantum field theory in a generic flat space (such as a four-dimensional one). It is a new topic with new special features such as the exponential maps, OPEs, state-operator correspondence, and so on, and you shouldn't insist that the physics of OPEs has to be composed of the same insight that you already knew from QED in $d=4$. It is not the same thing - if it were the same thing, people wouldn't teach it twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would suggest that you ask about some particular statements that David makes and that you don't understand. A necessary assumption is that you actually try to listen what David is saying, instead of trying to force him to say things that you wanted to hear in the first place. ;-) When you switch to this mode of learning, the discussion could become a little bit more constructive. At any rate, I assure you that David is talking mostly about quantum mechanical systems, so all observables are operators on a Hilbert space that can get multiplied and whose expectation values may be calculated. The previous sentence could help if you misunderstood every single formula in David's lectures that includes an operator - which would be pretty much every formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I can't explain you all other details about David's text (and not even all effects of quantum mechanics - because pretty much everything in the text is quantum mechanical) unless you say exactly what's your problem. I would have to take 107 of his pages, inflate them by a factor of 10, and you could still end up being dissatisfied because your dissatisfaction could have some totally different causes. ;-)&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-07T17:05:52.410" LastActivityDate="2011-04-07T17:05:52.410" CommentCount="2" />
  <row Id="8214" PostTypeId="2" ParentId="8207" CreationDate="2011-04-07T17:01:28.720" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does it seem that progress in seismic science (and quakes predictions) is so poor? It seems people have only raw probabilities, based on historical data.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's because the actual earthquake depends on tiny initial events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine standing on a glass so it breaks. We can calculate how much force it takes for a glass to break. But it would be very hard to predict in advance at which microscopic atomic scale flaw in the material the crack will start and at exactly what nano-second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An earthquake is just the same on a larger scale.&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-04-07T17:01:28.720" CommentCount="2" />
  <row Id="8215" PostTypeId="1" AcceptedAnswerId="8292" CreationDate="2011-04-07T17:28:20.510" Score="4" ViewCount="132" Body="&lt;p&gt;In many articles and web pages I found people claiming that the maximum critical temperature for superconductivity in the BCS framework is about 40K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometime, more accurate writers report that the above temperature is the maximum critical temperature achievable with electron-phonon coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases I never sow a reliable proof of these statements, so I'm asking if they are correct and where I can find a reference (a paper or simply few physical considerations)&#xA;leading to this result.&lt;/p&gt;&#xA;" OwnerUserId="2399" LastActivityDate="2011-05-16T08:20:06.573" Title="Maximum electron-phonon coupling SC temperature" Tags="&lt;superconductivity&gt;" AnswerCount="1" />
  <row Id="8218" PostTypeId="2" ParentId="8194" CreationDate="2011-04-07T18:38:09.860" Score="5" ViewCount="" Body="&lt;p&gt;The near-field effect (of Newtonian gravity) can be measured by a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravimeter&quot; rel=&quot;nofollow&quot;&gt;gravimeter&lt;/a&gt; (which you might say was invented by &lt;a href=&quot;http://en.wikipedia.org/wiki/Cavendish_experiment&quot; rel=&quot;nofollow&quot;&gt;Cavendish&lt;/a&gt; at the end of the 18th century).  Here's a quote from the gravimeter page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The superconducting gravimeter achieves extraordinary sensitivities of one nanogal, one thousandth of one billionth (10-12) of the Earth surface gravity. In a dramatic demonstration of the sensitivity of the superconducting gravimeter, Virtanen (2006), describes how an instrument at Metsähovi, Finland, detected the gradual increase in surface gravity as workmen cleared snow from its laboratory roof.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is left as an exercise for the reader to calculate what system of rotating masses will generate 1 nano&lt;a href=&quot;http://en.wikipedia.org/wiki/Gal_%28unit%29&quot; rel=&quot;nofollow&quot;&gt;gal&lt;/a&gt; at the sensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, suppose you want to measure the far-field effect (gravitational waves).  This turns out to be completely impractical.  See section 2.1 (&quot;Laboratory generators of gravitational waves&quot;) of Peter Saulson's note &quot;&lt;a href=&quot;http://www.slac.stanford.edu/gen/meeting/ssi/1998/media/saulson.ps.gz&quot; rel=&quot;nofollow&quot;&gt;Physics of Gravitational Wave Detection&lt;/a&gt;&quot;, or take a look at &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/9810218206&quot; rel=&quot;nofollow&quot;&gt;his book&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-04-07T18:38:09.860" CommentCount="2" />
  <row Id="8219" PostTypeId="1" AcceptedAnswerId="8228" CreationDate="2011-04-07T18:47:13.640" Score="1" ViewCount="147" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/7346/if-it-was-possible-to-dig-a-hole-that-went-from-one-side-of-the-earth-to-the-othe&quot;&gt;If it was possible to dig a hole that went from one side of the earth to the other&amp;hellip;&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;This question may not be suitable here and it could be closed in seconds but I wanna give it a try. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;is it possible to dig a hole to the&#xA;other side of the Earth?&lt;/li&gt;&#xA;&lt;li&gt;If we did somehow manage to dig a&#xA;hole to the other side of the Earth,&#xA;would we fall through?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2976" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-07T21:17:06.547" LastActivityDate="2011-04-07T21:17:06.547" ClosedDate="2011-04-07T21:39:17.410" Title="Dig a hole to the other side of the Earth" Tags="&lt;gravity&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="8221" PostTypeId="1" AcceptedAnswerId="8225" CreationDate="2011-04-07T18:59:12.773" Score="8" ViewCount="344" Body="&lt;p&gt;When determining the gravitational attraction between 2 solid bodies, we can simplify computations by taking their masses to be concentrated at their respective centres of mass. However, had they been electrically charged bodies and we needed to compute electrostatic attraction, there is no &quot;centre of charge&quot; notion available to us. ( We would evaluate a 2-dimensional integral, or equivalently, apply Gauss's law. )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any simple reason for the absence of such a centre of charge? Frequently, I've got the answer: &quot;charge just doesn't work that way&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2878" LastActivityDate="2011-08-19T10:04:57.593" Title="Why isn't there a centre of charge?" Tags="&lt;electrostatics&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="2" />
  <row Id="8222" PostTypeId="2" ParentId="8221" CreationDate="2011-04-07T19:12:12.723" Score="1" ViewCount="" Body="&lt;p&gt;Charges for electrostatic interaction determine the interaction force. In this respect they are similar to masses that determine the gravitational force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However when we look at the particle dynamics, it is the particle mass that determines the particle inertia whatever the force nature is. In this respect the masses are different from charges. According to the Newton equations, the center of mass makes sense as a quasi-particle position (the system as a whole).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, charge distribution can be described and is often described as system charge form-factors.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-04-07T21:40:37.723" LastActivityDate="2011-04-07T21:40:37.723" />
  <row Id="8223" PostTypeId="2" ParentId="8221" CreationDate="2011-04-07T19:14:14.047" Score="0" ViewCount="" Body="&lt;p&gt;You can define something like the center of mass for charge, but it can be undefined, when the total charge is equal to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at the definition of the dipole moment as well.&lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2011-04-07T19:14:14.047" />
  <row Id="8224" PostTypeId="2" ParentId="8221" CreationDate="2011-04-07T19:27:30.737" Score="8" ViewCount="" Body="&lt;p&gt;In the same way that bodies that are far away from you can be approximated as point masses, charge distributions that are sufficiently far away can be approximated as point charges, with total charge $Q$, and then you would, indeed, compute something like the center of charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the first part of the so-called multipole expansion: You begin by approximating a charge distribution $\varrho(x)$ by a point charge of charge $Q = \int dx \varrho(x)$. The next step would be to look at the dipole moment, $p = \int dx x\varrho(x)$, followed by the Quadrupole moment etc&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-04-07T19:27:30.737" CommentCount="1" />
  <row Id="8225" PostTypeId="2" ParentId="8221" CreationDate="2011-04-07T19:32:27.330" Score="9" ViewCount="" Body="&lt;p&gt;There is no &quot;center of charge&quot; that simplifies calculations completely because there is no center gravity that does this, either.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two rigid bodies feel gravity that is quite similar to that of point masses, but not exactly the same.  Unless the objects are perfect spheres, they feel &lt;a href=&quot;http://en.wikipedia.org/wiki/Tidal_force&quot; rel=&quot;nofollow&quot;&gt;tidal forces&lt;/a&gt;, which depend in second-order and higher derivatives of the potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can only calculate gravitational interactions as points at the center of mass if we want to ignore these second-order effects.  On a day-to-day basis, using gravity for things like playing baseball with your nephew, that's fine.  The tidal forces are usually very small in applications that we encounter because the phenomena are small compared to the characteristic length scales involved (i.e. a baseball's trajectory is small compared to the radius of Earth, Earth's diameter is fairly small compared to the Earth-Sun distance, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can indeed make the same first-order approximation in electrostatics that we do in Newtonian gravity.  In practice, a slightly different procedure is normally used, called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Multipole_expansion&quot; rel=&quot;nofollow&quot;&gt;multipole expansion&lt;/a&gt;.  First, a reference point is fixed.  The total charge, called the monopole moment, is independent of this reference point, so if you want your first-order approximation to be good, it is wise to choose a reference point near the center of your charge distribution.  Higher-order effects are then calculated relative to the reference point by computing &lt;a href=&quot;http://en.wikipedia.org/wiki/Electric_dipole_moment&quot; rel=&quot;nofollow&quot;&gt;dipole&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadropole#Electric_quadrupole&quot; rel=&quot;nofollow&quot;&gt;quadrupole&lt;/a&gt;, and higher moments of the charge distribution.  These higher orders are more likely to come into play in everyday life in electrostatics than in gravity because the size of the charge distributions is similar to the separation between charged bodies (i.e. two balloons you rubbed on your head and are playing with are not very far separated from each other, compared to their diameters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also be interested in the hard sci-fi novel &quot;Incandescence&quot;, in which a species of beings living inside a large asteroid orbiting a black hole observe tidal effects to discover general relativity without ever seeing the outside world.  It's an interesting demonstrations of what higher-order gravity effects look like beyond the center-of-mass approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author, Greg Egan, has a web page explaining the tidal effects described in his novel &lt;a href=&quot;http://gregegan.customer.netspace.net.au/INCANDESCENCE/Orbits/Orbits.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="4760" LastEditorDisplayName="" LastEditDate="2011-08-19T10:04:57.593" LastActivityDate="2011-08-19T10:04:57.593" CommentCount="9" />
  <row Id="8226" PostTypeId="2" ParentId="8221" CreationDate="2011-04-07T19:35:27.747" Score="0" ViewCount="" Body="&lt;p&gt;Defining a centre of mass relies on the symmetry of particles with a mass charge.  Namely that they are all of the same sign.  This often allows the problem to be reduced to a geometrically simple description.  For electric charge there are two signs and therefore this simple symmetry doesnt exist at short distances, though as mentioned above it can be approximated to at large distance.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-04-07T19:35:27.747" />
  <row Id="8227" PostTypeId="1" AcceptedAnswerId="8231" CreationDate="2011-04-07T19:39:39.457" Score="51" ViewCount="19992" Body="&lt;p&gt;Why is it that when you look in the mirror left and right are flipped, but not the up and down?  &lt;/p&gt;&#xA;" OwnerUserId="2978" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-07T21:18:49.353" LastActivityDate="2011-07-26T08:10:08.287" Title="A mirror flips left and right, but not up and down" Tags="&lt;light&gt;&lt;geometric-optics&gt;&lt;mirror&gt;" AnswerCount="11" CommentCount="7" FavoriteCount="16" />
  <row Id="8228" PostTypeId="2" ParentId="8219" CreationDate="2011-04-07T19:54:48.503" Score="5" ViewCount="" Body="&lt;p&gt;No, it is not feasible to dig a hole to the center of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deepest such hole appears to be the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kola_Superdeep_Borehole&quot; rel=&quot;nofollow&quot;&gt;Kola Superdeep Borehole&lt;/a&gt;, which goes down about 12km, or about 0.2% of the way to the center of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure at the center of the Earth is so high that rocks are about five or six times denser there than at the crust; they have literally been crushed down to a fraction of their original size.  The temperature there is &lt;a href=&quot;http://en.wikipedia.org/wiki/Geothermal_gradient&quot; rel=&quot;nofollow&quot;&gt;believed to be around 7000K&lt;/a&gt;, hot enough to &lt;a href=&quot;http://www.muggyweld.com/melting.html&quot; rel=&quot;nofollow&quot;&gt;melt just about any metal&lt;/a&gt;.  Clearly, this is not a reasonable engineering goal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be interested in a 2003 paper by Dave Stevenson proposing to study the deep interior of Earth by dropping molten iron down a crack in the crust and allowing it to sink all the way to the core.  I can't seem to find the original paper, but here's &lt;a href=&quot;http://eands.caltech.edu/articles/LXVI/modest.html&quot; rel=&quot;nofollow&quot;&gt;a readable summary&lt;/a&gt;.  The reference is&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A87. Stevenson, David J. Mission to Earth’s Core -A Modest Proposal. Nature, 423, 239-240, 2003.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-04-07T20:38:09.523" LastActivityDate="2011-04-07T20:38:09.523" CommentCount="3" />
  <row Id="8229" PostTypeId="2" ParentId="8227" CreationDate="2011-04-07T19:57:08.687" Score="3" ViewCount="" Body="&lt;p&gt;First, lets separate the concepts; there is nothing that is &quot;flipped&quot; in the mirror image regarding one orientation more than others. the full group of transformations $O(3)$ includes transformations where $det(R) = -1$. You can consider the following transformations examples of this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) they have one random direction flipped in sign, or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) for the special case where $D= 2n+1$, you can flip all direction signs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;moreover, all these rotations are equivalent to each other, that is, any of them is equivalent to any other by a normal rotation $det(R) = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;explained in other way, the mirror image is also equivalent to your image with up and down flipped. You only think in the equivalent with the right-left flipped version because its easier to build in your mind. That might be related with the fact that our bodies are almost right-left symmetrical, but very poorly up-down symmetrical&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-04-07T19:57:08.687" />
  <row Id="8230" PostTypeId="2" ParentId="8227" CreationDate="2011-04-07T20:01:18.807" Score="18" ViewCount="" Body="&lt;p&gt;Because they don't flip left with right (or up with down), they flip the 3D space you're standing in &quot;inside out&quot;, so far from the mirror becomes far away inside the mirror and vice versa. A hand 1 meter from the mirror seems like it's 1 meter on the other side of the mirror but in the same spot with regards to left/right so nothing is flipped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wiggle your left hand - you'll see the hand which is to the left in the mirror wiggle. Wiggle your toes and the toes in the mirror image wiggle etc.&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-04-07T20:01:18.807" CommentCount="3" />
  <row Id="8231" PostTypeId="2" ParentId="8227" CreationDate="2011-04-07T20:02:02.457" Score="54" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=msN87y-iEx0&quot; rel=&quot;nofollow&quot;&gt;Here's a video&lt;/a&gt; of physicist Richard Feynman discussing this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a blue dot and a red dot.  They are in front of you, and the blue dot is on the right.  Behind them is a mirror, and you can see their image in the mirror.  The image of the blue dot is still on the right in the mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's different is that in the mirror, there's also a reflection of you.  From that reflection's point of view, the blue dot is on the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the mirror really does is flip the order of things in the direction perpendicular to its surface.  Going on a line from behind you to in front of you, the order in real space is &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Your back&lt;/li&gt;&#xA;&lt;li&gt;Your front&lt;/li&gt;&#xA;&lt;li&gt;Dots&lt;/li&gt;&#xA;&lt;li&gt;Mirror&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The order in the image space is&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mirror&lt;/li&gt;&#xA;&lt;li&gt;Dots&lt;/li&gt;&#xA;&lt;li&gt;Your front&lt;/li&gt;&#xA;&lt;li&gt;Your back&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Although left and right are not reversed, the blue dot, which in reality is lined up with your right eye, is lined up with your left eye in the image.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is that you are roughly left/right symmetric.  The eye the blue dot is lined up with is still your right eye, even in the image.  Imagine instead that Two-Face was looking in the mirror. (This is a fictional character whose left and right side of his face look different.  His image on Wikipedia looks like this:)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/JqDty.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two-face looked in the mirror, he would instantly see that it was not himself looking back!  If he had an identical twin and looked right at the identical twin, the &quot;normal&quot; sides of their face would be opposite each other.  Two-face's good side is the right.  When he looked at his twin, the twin's good side would be to the original two-face's left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, the mirror Two-face's good side is also to the right.  Here is an illustration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/G9jgK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ot5BH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So two-face would not be confused by the dots.  If the blue dot is lined up with Two-Face's good side, it is still lined up with his good side in the mirror.  Here it is with the dots:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/dyMrw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two-face would recognize that left and right haven't been flipped so much as forward and backward, creating a different version of himself that cannot be rotated around to fit on top the original.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-04-07T20:37:05.640" LastActivityDate="2011-04-07T20:37:05.640" CommentCount="7" />
  <row Id="8232" PostTypeId="1" CreationDate="2011-04-07T21:13:12.223" Score="4" ViewCount="170" Body="&lt;p&gt;In the letter of introduction to Einstein's 1916 paper on General Relativity, he writes, &quot;The mathematical tool sthat are necessary for general relativity were readily available in the 'absolute different calculus,' which is based upon the research on non-Euclidean manifolds by Gauss, Riemann, and Christoffel, and which has been systematized by Ricci and Levi-Civita and has already been applied to problems of theoretical physics.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In what way had it already been applied to problems of theoretical physics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.alberteinstein.info/gallery/pdf/CP6Doc30_English_pp146-200.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.alberteinstein.info/gallery/pdf/CP6Doc30_English_pp146-200.pdf&lt;/a&gt; for the quote.&lt;/p&gt;&#xA;" OwnerUserId="2591" LastActivityDate="2011-04-08T08:50:13.833" Title="How were non-Euclidean manifolds applied to physics before Einstein?" Tags="&lt;general-relativity&gt;&lt;history&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="8233" PostTypeId="2" ParentId="8232" CreationDate="2011-04-07T21:51:38.997" Score="1" ViewCount="" Body="&lt;p&gt;It is difficult to know exactly what Einstein had in mind, but the following is the first application of non-Euclidean geometry to our physical world. From &lt;a href=&quot;http://en.wikipedia.org/wiki/Space&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;: &lt;em&gt;Carl Friedrich Gauss, a German mathematician, was the first to consider an empirical investigation of the geometrical structure of space. He thought of making a test of the sum of the angles of an enormous stellar triangle and there are reports he actually carried out a test, on a small scale, by triangulating mountain tops in Germany.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-04-08T08:50:13.833" LastActivityDate="2011-04-08T08:50:13.833" CommentCount="2" />
  <row Id="8234" PostTypeId="1" CreationDate="2011-04-07T21:52:05.750" Score="1" ViewCount="218" Body="&lt;p&gt;I have read through the wikipedia page and several lecture notes/arxiv papers from my google search (and several related P.SE questions), but I'm still hopelessly confused.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Consider a 'classical Schrodinger field in a box' problem. Since the field evolves just like a normal wavefunction, I could still extract the x-projection of any operators I want from it, in addition to the field momentum $\frac{\partial\mathcal{L}}{\partial\dot{\phi}}$ or any other field variables.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;From my understanding, there are two ways to arrive at QFT:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;from QM/RQM: we change our basis into particle-number basis (fock space).&lt;/li&gt;&#xA;&lt;li&gt;from CFT: we &quot;2nd quantize&quot; the field and field-momentum.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Looks like the &quot;particle creation&quot; interpretation of the field in QFT can be deduced only from the first approach. Which means, the interpretation of the latter approach can't be completely independent from the former.&#xA;But they don't give me any hint on how to interpret the classical field. Or am I missing something? Or should I just give it up because the quantum field is not an observable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Random comment: I could still see $\hbar$ lurking inside CFT (some people might argue that it's there to keep the dimension consistent, but I could apply the same argument to Schrodinger equation, too).&lt;/p&gt;&#xA;" OwnerUserId="1814" LastActivityDate="2011-08-12T02:39:54.697" Title="What is a &quot;classical Schrodinger field&quot;, really?" Tags="&lt;quantum-field-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="8236" PostTypeId="2" ParentId="8234" CreationDate="2011-04-07T22:28:15.517" Score="4" ViewCount="" Body="&lt;p&gt;I am not sure I am getting the source of the difficulty, but let me try:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The usual object described by a Schrodinger equation is the wavefunction, which is not observable. There are some classical systems which involve continuous media (= fields) which evolve according to classical equations similar in form to the Schrodinger equation. Be that as it may, the physics is all in interpreting the results of the mathematical framework, and classical fields are different objects than quantum wave functions, even if they obey similar equations.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When you quantize a &lt;em&gt;free&lt;/em&gt; classical field theory you obtained results that are isomorphic to a multi-particle Fock space. But, quantum field theory is more general than that - the physics of strongly coupled QFTs can be very different from that of weakly interacting particles, for example they may well not have any sort of particles involved. The distinction between the two ways of approaching QFT is historical: the first way you quote is the older and less general way to approach the subject, and the second way is more precise and general (reproducing the results of the first one in a specific limit of free fields). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In general field theory the field operators are observable, but not always easy to interpret as results of concrete measurements, one usually constructs correlation functions which are more directly related to simple experiments.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, as for the random comment: if you phrase you classical theory in terms of quantities which stay finite in the classical limit, the Planck constant $\hbar$ ought to completely disappear. If you keep an inherently quantum mechanical quantity (say the mass or charge of individual particle) fixed, there will be (misleading) factors of $\hbar$ left.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="566" LastActivityDate="2011-04-07T22:28:15.517" CommentCount="4" />
  <row Id="8237" PostTypeId="1" CreationDate="2011-04-07T22:59:37.030" Score="1" ViewCount="185" Body="&lt;p&gt;Referencing this news article:&#xA;&lt;a href=&quot;http://www.smh.com.au/technology/sci-tech/cosmic-burst-in-far-away-galaxy-puzzles-nasa-20110408-1d6kz.html&quot; rel=&quot;nofollow&quot;&gt;http://www.smh.com.au/technology/sci-tech/cosmic-burst-in-far-away-galaxy-puzzles-nasa-20110408-1d6kz.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also references an event id:&lt;br&gt;&#xA;(GRB) 110328A&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article seems to imply that the reason for the unusually long period of gamma bursts being visible is because the remnants of the exploding star is near the centre of that galaxies nucleaus, and hence probably near the supermassive black hole. It goes on to say that the observable event is a stream of this energy towards the sm-bh.&lt;br&gt;&#xA;Is this the most logical hypothesis for this event?&lt;br&gt;&#xA;What other possibilities could explain such a lengthy burst of gamma radiation being visible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;br&gt;&#xA;If it is even speculation that it is a star being ripped apart, then what other possibilites could explain the explosion that lead to the unusually long bursts?&lt;/p&gt;&#xA;" OwnerUserId="582" LastEditorUserId="582" LastEditorDisplayName="" LastEditDate="2011-04-09T10:27:48.573" LastActivityDate="2011-04-20T02:34:14.070" Title="Is the conclusion that the cosmic burst observed in the Draco constellation from gamma rays that are streaming towards a massive black hole correct?" Tags="&lt;soft-question&gt;&lt;astronomy&gt;&lt;cosmic-rays&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8238" PostTypeId="2" ParentId="723" CreationDate="2011-04-07T23:00:53.690" Score="2" ViewCount="" Body="&lt;p&gt;It should be stressed that electrons are very very easy to trap in the magnetic confinement region created by the arrangement of magnetic coils in Polywell.  The electrons in the trap interact with the magnetic fields and actually bend the lines of force into a spherical magnetic chamber with small regions where the lines of force lead outwards. This happens because electrons by themselves are diamagnetic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some positive ions (protons, deuterium nuclei, B11 nuclei, etc.) will drift into the well also.  Electrons and ions can escape the well, but travel along the magnetic lines of force and get fed back into the well.  The high voltage static electric field continuously draws electrons into the trap, and the ions are attracted to them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pocket of trapped plasma is called quasi-neutral because there are always just a bit more electrons than ions.  Ions are also injected from far outside the well and are accelerated to a high speed by the electric field toward the quasi-neutral (on the negative side of neutral) gas in the pocket.  Because these ions are fed in from far outside the well, their speed is so high at the center of the trap, that if a collision does occur, then fusion will happen.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gas pressure used is so very low that &quot;thermalization&quot; does &lt;strong&gt;not&lt;/strong&gt; happen.  &quot;Thermalization&quot; is when gas particles randomly collide and create a Boltzmann distribution.  Because the gas pressure is very very low, hardly any non-fusion, high speed, accelerated, incoming ion with trapped ion collisions occur.  Therefore the accelerated ions maintain their speed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ions that do not fuse will just fly out the other side of the potential well, slow down, and fall back in again.  The see-saw effect of falling in, flying out and falling back into the well can be maintained by adding a small oscillation to the generally static electric field which resonates with the particular species of ion that is desired to be accelerated into the cloud of &quot;target&quot; ions.  If the population of gas particles in the trap gets too dense, &quot;thermalization&quot; will occur.  The accelerated ions get slowed down by too many non-fusion collisions, and fusion stops.&lt;/p&gt;&#xA;" OwnerUserId="2982" LastEditorUserId="2982" LastEditorDisplayName="" LastEditDate="2011-04-12T13:04:13.000" LastActivityDate="2011-04-12T13:04:13.000" />
  <row Id="8240" PostTypeId="1" CreationDate="2011-04-07T23:41:40.567" Score="6" ViewCount="175" Body="&lt;p&gt;The Coleman-Mandula theorem states that under certain seemingly-mild assumptions on the properties of the S matrix (roughly: one particle states are left invariant and the amplitudes are analytic in external momenta) the largest possible Lie algebra of symmetries of a (non-trivial) S matrix is given by Poincaré times an internal symmetry.&lt;br&gt;&#xA; On the other hand, there are (interacting) field theories whose Lagrangians are symmetric under the conformal extension of the Poincaré group, and in some rare case this property is retained even at the quantum level.&lt;br&gt;&#xA; Why (interacting) conformal invariant QFTs do not contradict the theorem? Is it possible to define an S matrix in these theories? I have read somewhere that they do not admit a particle interpretation, what does it mean exactly?&lt;/p&gt;&#xA;" OwnerUserId="2981" LastActivityDate="2011-04-08T00:06:03.570" Title="CFT and the Coleman-Mandula Theorem" Tags="&lt;quantum-field-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="8241" PostTypeId="2" ParentId="8240" CreationDate="2011-04-07T23:56:45.080" Score="6" ViewCount="" Body="&lt;p&gt;You sort of answered the question yourself. In CFT there is no Notion of &quot;particles&quot; - individual lumps of energy which exists independently of each other when sufficiently far away from each other. Other ways to say the same thing - The Hilbert space of the theory does not organize naturally into a Fock space, or there is no cluster decomposition. All of this follows from lack of scale in the theory. As a result LSZ reduction which depends on contraction of appropriate asymptotic states does not work (if you try to force it you find IR divergences which cannot be resumed). So, as you say the S matrix does not exist, which is the loophole to the Coleman Mandula theorem.  &lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-04-08T00:06:03.570" LastActivityDate="2011-04-08T00:06:03.570" CommentCount="5" />
  <row Id="8242" PostTypeId="2" ParentId="8232" CreationDate="2011-04-08T02:43:56.807" Score="1" ViewCount="" Body="&lt;p&gt;The Stanford Encyclopedia entry on &lt;a href=&quot;http://plato.stanford.edu/entries/geometry-19th/&quot; rel=&quot;nofollow&quot;&gt;&quot;Nineteenth Century Geometry&quot;&lt;/a&gt; is not quite what you're asking for, but it seems nonetheless likely to be of interest. It's always worth trying this resource for History as well as for Philosophy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found this in Google books, &lt;a href=&quot;http://books.google.com/books?id=r9C-SCXymPoC&amp;amp;lpg=PA93&amp;amp;ots=LhLRqi5Fkh&amp;amp;dq=geometry%20before%20einstein&amp;amp;pg=PA109#v=onepage&amp;amp;q=geometry%20before%20einstein&amp;amp;f=true&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=r9C-SCXymPoC&amp;amp;lpg=PA93&amp;amp;ots=LhLRqi5Fkh&amp;amp;dq=geometry%20before%20einstein&amp;amp;pg=PA109#v=onepage&amp;amp;q=geometry%20before%20einstein&amp;amp;f=true&lt;/a&gt;, which is considerably closer to an Answer, but still not quite what your Question asks for (you can see in this link that my search term into Google was pretty simple-minded, &quot;geometry before Einstein&quot;). This reference makes it seem as if there was a lot of speculation in the air, perhaps enough that Einstein and others, particularly including Hilbert, would have taken it for granted that geometry was where the story would be, but there's still no sign of the smoking gun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: It is perhaps noteworthy that in my pursuit of a more definitive Answer to this Question I find myself stumbling into the substantial literature on one of the great controversies of the last 15 years in the History of Physics, the question of priority for GR, between Hilbert and Einstein. Reading a review of this sometimes acrimonious debate in &lt;a href=&quot;http://arxiv.org/abs/physics/0504179&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/physics/0504179&lt;/a&gt; (it's a colloquium talk), one is further struck by the extent to which geometry is absolutely in the air for these guys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT(2): What I think is a better citation (by searching the PhilSci archive): &lt;a href=&quot;http://philsci-archive.pitt.edu/4377/&quot; rel=&quot;nofollow&quot;&gt;http://philsci-archive.pitt.edu/4377/&lt;/a&gt;. The discussion here includes the following, on page 42: &quot;Early in 1914, in a joint paper with Lorentz's former student Adriaan D. Fokker, Einstein reformulated Nordström's theory using Riemannian geometry (Einstein and Fokker 1914).&quot; That it wasn't yet successful Physics doesn't matter, it &lt;em&gt;has&lt;/em&gt; already been &lt;em&gt;applied&lt;/em&gt; to problems of Physics. Good enough?&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-04-08T03:35:07.890" LastActivityDate="2011-04-08T03:35:07.890" />
  <row Id="8243" PostTypeId="2" ParentId="8062" CreationDate="2011-04-08T03:41:06.577" Score="7" ViewCount="" Body="&lt;p&gt;Alternative discussion by Scott Aaronson:&#xA;&lt;a href=&quot;http://www.scottaaronson.com/democritus/lec9.html&quot; rel=&quot;nofollow&quot;&gt;http://www.scottaaronson.com/democritus/lec9.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;From the probability interpretation postulate,&#xA;we conclude that the time evolution operator $\hat{U}(t)$ must be &lt;strong&gt;unitary&lt;/strong&gt; in order to keep the total probability to be 1 all the time. Note that the wavefunction is not necessarily complex yet.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;From the website: &quot;Why did God go with the complex numbers and not the real numbers?&#xA;Answer: Well, if you want every unitary operation to have a square root, then you have to go to the complex numbers... &quot; $\hat{U}(t)$ must be complex if we still want a continuous transformation. This implies a complex wavefunction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hence the operator should be: $\hat{U}(t) = e^{i\hat{K}t}$ for hermitian $\hat{K}$ in order to preserve the norm of the wavefunction.&lt;/p&gt;&#xA;" OwnerUserId="1814" LastEditorUserId="1814" LastEditorDisplayName="" LastEditDate="2011-04-08T03:50:15.510" LastActivityDate="2011-04-08T03:50:15.510" CommentCount="1" />
  <row Id="8244" PostTypeId="2" ParentId="8232" CreationDate="2011-04-08T03:57:54.143" Score="0" ViewCount="" Body="&lt;p&gt;The work of Ricci and Levi-Civita, what is also called absolute differential calculus, was less than ten years old when Einstein started working on general relativity. There isn't much room for other applications. I suppose what he meant was that it was accessible to everyone, not that it had been around for ages. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reminds me of another quote by Hilbert. &quot;Every schoolboy in Gottingen knows more about higher dimensional geometry than Einstein, but chalk is cheaper than gray matter.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2011-04-08T03:57:54.143" CommentCount="2" />
  <row Id="8245" PostTypeId="1" CreationDate="2011-04-08T04:45:08.330" Score="1" ViewCount="66" Body="&lt;p&gt;With all the sudden discussion related to Nuclear power, safety, fuel, waste, etc, there's one question whose answer has so far eluded me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How big (height and diameter) is the actual pressure vessel of a conventional BWR or PWR?  I've seen, now, about a thousand images depicting the structures of Fukushima, Chernobyl, and TMI, but none of them include units of measurement for the size of what I'm looking at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&#xA;Eric B&lt;/p&gt;&#xA;" OwnerUserId="2988" LastActivityDate="2011-04-08T04:45:08.330" ClosedDate="2011-04-08T05:37:27.820" Title="How big is a nuclear reactor pressure vessel?" Tags="&lt;nuclear-engineering&gt;" CommentCount="1" />
  <row Id="8246" PostTypeId="1" CreationDate="2011-04-08T06:04:10.593" Score="6" ViewCount="337" Body="&lt;p&gt;Some background info on what an evaporator is:&#xA;It is a system of metal pans set over a heat source.  Sap constantly enters the first pan controlled by a float valve to keep a constant depth.  The pans have channels which are open to each other at alternating ends, so that the sap flows back and forth in a serpentine pattern.  At the end of the pan system there is a manual valve where you periodically draw off finished syrup.  From the beginning of the pan system to the end there is a ever-increasing concentration of sugar.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question is, how does the sap get more concentrated towards the end of the pan?  Or conversely, how is it that the sap doesn't maintain a homogenous sugar concentration throughout?  (If you let the fire go out, once the sap cools it will redistribute itself until it is homogenous.)&lt;/strong&gt;    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think somebody will probably tell me it has to do with the distance the sap has flowed through the pan by the time it gets to the other end (it would be a linear distance of about 24 feet in my system) but I don't think that answers the question because when you first start a batch, you flood the whole system with cold, fresh sap, all of uniform sugar content, and then you start the fire.  It doesn't take long before there is a visible color difference between each channel of the final pan, with the darkest one being next to the draw-off valve.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And no, it's not flow caused by the pans being angled downwards either (a BS answer posted on yahoo!answers to a similar question).  The pans are perfectly level.  And as far as I can tell the heat intensity from the fire below is uniform at least for the final four channels, which are where you really begin to see visible color graduation as it becomes syrup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/4NVQS.jpg&quot; alt=&quot;here is a quick drawing of the setup that may help with visualization&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one detail that I didn't describe at first because I couldn't figure out how to without a drawing, so now that I've added the drawing I'll elaborate on that.  The first pan, where the sap goes in, has flues, which are deep, narrow vertical pockets in the bottom of the pan, which reach all the way down into the stove itself.  The flames actually pass between the flues on their way to the stovepipe.  This (obviously) greatly increases the surface area being heated.  I don't know if this changes anything but I thought I should add it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/klCOD.jpg&quot; alt=&quot;evaporator (cold)&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="2987" LastEditorUserId="2987" LastEditorDisplayName="" LastEditDate="2011-04-11T01:55:07.603" LastActivityDate="2011-08-28T21:36:18.283" Title="How does a maple syrup evaporator work?" Tags="&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="8247" PostTypeId="1" AcceptedAnswerId="8257" CreationDate="2011-04-08T09:21:46.787" Score="2" ViewCount="221" Body="&lt;p&gt;Apologies if too specific.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Watched a documentary &lt;a href=&quot;http://natgeotv.com/nl/mega-factories-2/videos/bugatti-veyron&quot; rel=&quot;nofollow&quot;&gt;National Geographic Megafactories Bugatti Veyron&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I told an colleague (engineer) about the need to warm the car up and then unleash it on the long stretch to be able to make it to 405km/h and mused that the curve before the straight probably had to be taken at 250km/h to do so. He immediately replied that the G-force of going around a(ny) curve at 250km/h would kill the driver... I find that very hard to believe but do not have the skills nor knowledge to refute it. It would be cool to get him this app &lt;a href=&quot;http://www.dynolicious.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.dynolicious.com/&lt;/a&gt; and a trip in a Veyron ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the &lt;a href=&quot;http://googlesightseeing.com/2007/03/super-secret-volkswagen-test-track/&quot; rel=&quot;nofollow&quot;&gt;Ehra-Lessien track &lt;/a&gt; and here is the &lt;a href=&quot;http://googlesightseeing.com/maps?p=&amp;amp;c=&amp;amp;t=k&amp;amp;hl=en&amp;amp;ll=52.595202,10.787587&amp;amp;z=16&quot; rel=&quot;nofollow&quot;&gt;actual curve&lt;/a&gt; where some of the clever people here could glean the radius - I think 350m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone show some math and a graph or so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been looking at &lt;a href=&quot;http://en.wikipedia.org/wiki/Formula_One_car#Lateral_force&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Formula_One_car#Lateral_force&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Turn 8 at the Istanbul Park circuit, a&#xA;  190° relatively tight 4-apex corner,&#xA;  in which the cars maintain speeds&#xA;  between 265 and 285 km/h (165 and 177&#xA;  mph) (in 2006) and experience between&#xA;  4.5g and 5.5g for 7 seconds—the longest sustained hard cornering in&#xA;  Formula 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which supports my claim - but would love some more information on this specific matter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link on what an &lt;a href=&quot;http://www.drivingfast.net/techniques/racing-line.htm&quot; rel=&quot;nofollow&quot;&gt;Apex means in motoring&lt;/a&gt;&lt;br&gt;&#xA;And here are some &lt;a href=&quot;http://forums.autosport.com/lofiversion/index.php/t53791.html&quot; rel=&quot;nofollow&quot;&gt;formulas&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2795" LastEditorUserId="2795" LastEditorDisplayName="" LastEditDate="2011-06-16T11:39:46.553" LastActivityDate="2011-06-17T15:28:57.183" Title="How many Gs would a driver of the Bugatti Veyron experience on the Ehra-Lessien track when cornering before record attempt?" Tags="&lt;classical-mechanics&gt;&lt;centrifugal-force&gt;&lt;sports&gt;" AnswerCount="2" />
  <row Id="8248" PostTypeId="1" AcceptedAnswerId="8249" CreationDate="2011-04-08T09:42:30.103" Score="10" ViewCount="955" Body="&lt;p&gt;please consider the publication&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1104.0699&quot; rel=&quot;nofollow&quot;&gt;Invariant Mass Distribution of Jet Pairs Produced in Association with a W boson in $p\bar{p}$ Collisions at $\sqrt{s} = 1.96$ TeV&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Collider_Detector_at_Fermilab&quot; rel=&quot;nofollow&quot;&gt;CDF-Collaboration&lt;/a&gt;, already with &lt;a href=&quot;http://www.google.de/search?q=fermilab%20new%20force&quot; rel=&quot;nofollow&quot;&gt;huge media attention&lt;/a&gt;. I must admit that I am &lt;strong&gt;puzzled, astonished and excited&lt;/strong&gt; at the same time. Not beeing an expert, it would be really nice if someone could shed light on&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h3&gt;Did we find a fifth fundamental force that cannot be explained by the standard model?&lt;/h3&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What do you think?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS.: A part of figure 1 in the publication that is re-printed over and over again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/JOgV1.jpg&quot; alt=&quot;Fig 1&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2011-04-08T09:47:43.170" LastActivityDate="2011-04-10T18:28:46.220" Title="Did the guys at Fermilab find a fifth force?" Tags="&lt;forces&gt;&lt;standard-model&gt;&lt;accelerator-physics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8249" PostTypeId="2" ParentId="8248" CreationDate="2011-04-08T09:55:45.943" Score="13" ViewCount="" Body="&lt;p&gt;The likely answer is No, the whole signal is just an artifact of the difficult statistical manipulations. A Weizmann Institute physicist has noted that the peak has been shifted by one bin and the whole discovery could therefore be due to a pile-up effect or jet energy calibration; a small shift of the jet energy removes the effect. See also&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2011/04/fermilab-cdf-new-force-press-conference.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2011/04/fermilab-cdf-new-force-press-conference.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;for more details. Additional doubts about the valid statistical procedures were mentioned on Tommaso Dorigo's blog, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/3AziQ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;An adjustment of jet energy by 3 percent is enough to make the signal totally insignificant. Animation by Tommaso Tabarelli de Fatis, a member of CMS.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The text above also reviews some theoretical literature and clarifies that if the signal happens to be real - and the D0 Collaboration is going to release its own verdict about the phenomenon in a few weeks - the most likely explanations are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;a new, fifth force - one mediated by a &lt;strong&gt;Z' boson&lt;/strong&gt; (with the mass of 144 GeV or so) which is the particle that was decaying to two jets in those events. It must be leptophobic - (almost) no interactions with the leptons - and such Z' bosons, messengers of new $U(1)$ groups similar to the electroweak Z' boson but independent from them, are predicted by a very large fraction of grand unified theories and/or string theory models; articles by Alon Faraggi and many others are listed above; the simplest (but not the only) way to get a new leptophobic $U(1)$ is to obtain it as a piece of a color $SU(4)$ broken to $SU(3)\times U(1)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a &lt;strong&gt;technipion&lt;/strong&gt;, a particle analogous to a pion in technicolor theories that break the electroweak symmetry by similar composite particles, and not necessarily scalar ones (so there is no true Higgs particle in those theories); a paper by Kenneth Lane et al. is linked above, too; technicolor has been thought to be nearly dead for years - the optimistic proposal to interpret the bump as a technicolor effect doesn't solve all the detailed problems with technicolor theories&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a &lt;strong&gt;stop squark&lt;/strong&gt; in a supersymmetric theory - but it must be an R-parity-violating version of supersymmetry (e.g. because the new superpartners were not produced in pairs) which is unattractive for many other reasons (in simple terms, R-parity-violating theories are only consistent with proton stability given immense new assumptions, and they don't produce natural well-behaved dark-matter candidates); an article is linked to by the weblog above, too&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-10T08:10:10.117" LastActivityDate="2011-04-10T08:10:10.117" CommentCount="8" />
  <row Id="8250" PostTypeId="2" ParentId="723" CreationDate="2011-04-08T09:59:45.640" Score="3" ViewCount="" Body="&lt;p&gt;Question 1: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You confuse Energy and Force. You need a force to keep the electrons within &#xA;the inner grid, provided by the voltage on the grid. As long as no electrons are lost &#xA;(and must be replenished, this needs energy), the situation will stay for ever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 2: this is rather confusing to me too, I don't know.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="176" LastEditorDisplayName="" LastEditDate="2011-04-08T17:52:36.290" LastActivityDate="2011-04-08T17:52:36.290" />
  <row Id="8251" PostTypeId="2" ParentId="8247" CreationDate="2011-04-08T10:16:18.420" Score="2" ViewCount="" Body="&lt;p&gt;The problem is within Your text:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and mused that the corner probably had&#xA;  to be taken at 250km/h to do so. He&#xA;  immediately replied that the G-force&#xA;  of 250km/h in a corner would kill the&#xA;  driver..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First &quot;the corner&quot; then &quot;a corner&quot;. Such corners use to have different radius &#xA;of curvature. The less this radius, the higher the &quot;gs&quot; at same speed. &#xA;(I assume that &quot;apex4&quot; mentioned in that link is some classification in &#xA;this sense) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, You can answer Your question Yourself, just get the curvature radius of the curve &#xA;You have in mind and calculate according to the formulas in Wiki:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Centrifugal_force&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Centrifugal_force&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-08T10:16:18.420" CommentCount="7" />
  <row Id="8254" PostTypeId="1" CreationDate="2011-04-08T13:22:36.013" Score="3" ViewCount="114" Body="&lt;p&gt;We know, string theory, from its inception, has a technical problem of the instability of the extra dimensions. I have heard of the partial solution of the problem quite a few years ago. My question is, has it been completely solved now?&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-04-08T14:52:50.730" Title="Is the &quot;moduli problem&quot; completely solved in string theory?" Tags="&lt;string-theory&gt;" AnswerCount="1" />
  <row Id="8255" PostTypeId="1" CreationDate="2011-04-08T13:27:44.890" Score="-1" ViewCount="221" Body="&lt;p&gt;body is missing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;please delete it&lt;/p&gt;&#xA;" OwnerUserId="1916" LastEditorUserId="1916" LastEditorDisplayName="" LastEditDate="2011-04-10T20:59:25.253" LastActivityDate="2011-04-10T20:59:25.253" ClosedDate="2011-04-08T17:48:26.933" Title="When are we now? Time coordinate system" Tags="&lt;spacetime&gt;&lt;time&gt;&lt;coordinate-systems&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8256" PostTypeId="1" AcceptedAnswerId="8270" CreationDate="2011-04-08T14:06:22.013" Score="4" ViewCount="189" Body="&lt;p&gt;Books such as &lt;a href=&quot;http://books.google.com/books?id=Pd8-s6rOt_cC&amp;amp;printsec=frontcover&quot; rel=&quot;nofollow&quot;&gt;Mathematical methods of classical mechanics&lt;/a&gt; describe an approach to classical (Newtonian/Galilean) mechanics where Hamiltonians turn into a theory of symplectic forms on manifolds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering why it's at all interesting to consider such things in the classical case. In more advanced theories of physics, manifolds become relevant, but in classical mechanics, everything is Euclidean and homogenous, so this isn't really needed. There, I'm wondering what motivated people to develop the theory of Hamiltonians through the theory of symplectic manifolds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One idea: There wasn't really and physical motivation, but someone noticed it, and it was mathematically nice. If this is the case, then how did people notice it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, maybe this theory was not developed until general relativity or quantum mechanics came about. That is, manifolds became relevant to more advanced physics, and Arnold thought it reasonable to teach students this manifold formalism in classical mechanics in order to better prepare them for more advanced physics. But, if this is true, no one thought of using symplectic manifolds purely for reasons of classical physics.&lt;/p&gt;&#xA;" OwnerUserId="2591" LastEditorUserId="2591" LastEditorDisplayName="" LastEditDate="2011-04-10T02:35:52.963" LastActivityDate="2011-04-10T02:35:52.963" Title="Why is the symplectic manifolds version of Hamiltonians used in Newtonian mechanics?" Tags="&lt;classical-mechanics&gt;&lt;history&gt;&lt;hamiltonian-mechanics&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" />
  <row Id="8257" PostTypeId="2" ParentId="8247" CreationDate="2011-04-08T14:17:22.947" Score="4" ViewCount="" Body="&lt;p&gt;Neglecting friction, the force experienced is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Centrifugal_force&quot; rel=&quot;nofollow&quot;&gt;Centrifugal Force&lt;/a&gt; $F=\frac{mv^2}{r}$ (it would be less if you included friction since the car actually slips) vectorially added to the orthogonal gravitational force $F_g=mg$, i.e. $F = m\sqrt{\left(\frac{v^2}r\right)^2 + g^2}$ where $g = 9.81 \frac{m}{s^2}$&lt;sup&gt;1&lt;/sup&gt;. Divide this by $F_g$ to obtain a result in Gs. Remember to use SI-units, i.e. divide a km/h speed by 3.6 (1000 km/m / 3600 s/h) to obtain m/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the 350 m curve and 285 km/h that yields about 2.08 gs only, to obtain the 5.5 gs mentioned a radius of about 118 m is required, or some higher velocity (490 km/h for the 350 m curve).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1)&lt;/sup&gt; Thanks &lt;a href=&quot;http://physics.stackexchange.com/users/394/j-h&quot;&gt;J.H.&lt;/a&gt; for this important correction!&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-06-17T15:28:57.183" LastActivityDate="2011-06-17T15:28:57.183" CommentCount="4" />
  <row Id="8258" PostTypeId="2" ParentId="7487" CreationDate="2011-04-08T14:21:25.750" Score="0" ViewCount="" Body="&lt;p&gt;To this discussion should be added that apparently radioactive iodine-131 has what is known as an &quot;effective half life&quot; when in the human body....now physically, the 131 isotope has a half-life of 8.02 days and this remains precisely the same whether it is internal or external to the body---the helf-life of any radioactive isotope is a physical constant that depends strictly on the structure of the nucleus and its location within or without the human body is of no consequence whatsoever. Nonetheless, for some physiological reason, when radioactive iodine-131 is introduced into a human (usually as NaI) and a radiation meter, Geiger counter or some instrument that reads counts per minute (CPM) is placed close to the thyroid gland to which the iodine is most strongly gravitated and held, and the time for the count to drop to exactly half its original highest value is taken and averaged....the resulting figure is other than 8.02 days. This could result from migration of the &quot;hot&quot; iodine to other parts of the body or whatever but this new &quot;half-life&quot; as measured is referred to as the &quot;effective half-life&quot; in the body...but this term can mislead people into believing that once ingested, radioactive iodine's physical properties change---this is not so.&lt;/p&gt;&#xA;" OwnerUserId="2998" LastActivityDate="2011-04-08T14:21:25.750" />
  <row Id="8259" PostTypeId="2" ParentId="8256" CreationDate="2011-04-08T14:35:21.673" Score="2" ViewCount="" Body="&lt;p&gt;The action principle, stemming from Fermat and Maupartuis, and developed by Lagrange tells us that a classical system extremizes the action $S~=~\int L(q,{\dot q})dt$ so that the following equation is obeyed&#xA;$$&#xD;&#xA;\frac{d}{dt}\frac{\partial L}{\partial{\dot q}}~-~\frac{\partial L}{\partial q}~=~0.&#xD;&#xA;$$&#xA;This defines a differential equation for each Lagrangian $L(q,{\dot q})$, which is equivalent to Newton’s second law.  The action is a bare action plus a Hamiltonian under a Legendre transformation $Ldt~=~pdq~-~Hdt$, where this tells us that the Lagrangian is a form of constraint which sets the dynamics on some manifold of solution.  This manifold is called a contact manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hamiltonian can be showed with several steps of mathematics to obey the equations&#xA;$$&#xD;&#xA;{\dot q}~=~\frac{\partial H}{\partial p},~{\dot p}~=~-\frac{\partial H}{\partial q},&#xD;&#xA;$$&#xA;which are the Hamiltonian dynamical equations.  In a general setting the evolution of quantity $z$ is given by ${\dot z}~=~\{z,~H\}$, which is a Poisson bracket.  The antisymmetric structure of this theory can be encoded according to Hamiltonian flows determined by a symplectic matrix $\Omega_{ij}$ which is antisymmetric.  If we let $z~=~(q_1,\dots, q_n,~p_1,\dots,p_n)$ be a $2n$ vector, then for $\Omega_{ij}$ a $2n\times 2n$ matrix $dz_i/dt~=~\Omega_{ij}z_j$ is a general form of dynamics in symplectic form.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an outline of how this is developed.  Arnold’s book on Math Methods of Classical Mech is a good source for a deeper study of this.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-04-08T16:14:01.050" LastActivityDate="2011-04-08T16:14:01.050" CommentCount="4" />
  <row Id="8260" PostTypeId="2" ParentId="8254" CreationDate="2011-04-08T14:47:48.547" Score="1" ViewCount="" Body="&lt;p&gt;This question has many aspects because there are many groups of stringy vacua, some of them have been showed to have (unstabilized) moduli, the status of others was (or is) unknown. Those that were (or are) almost certainly stabilized may be stabilized at the classical level, or by calculable non-perturbative effects, or by not-quite-understood non-perturbative effects which may required fluxes or antibranes, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most classes of stringy vacua, the boundaries have been understood which means that it is known whether the vacua have some unstabilized moduli or not, and if they don't, the relevant potentials became largely understood, at least qualitatively. See papers such as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0505160&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0505160&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/abs/hep-th/0201028&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0201028&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/abs/hep-th/0611001&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0611001&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/abs/hep-th/0602120&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0602120&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/abs/hep-th/0506090&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0506090&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/abs/hep-th/0604087&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0604087&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/abs/hep-th/0508171&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0508171&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://arxiv.org/abs/hep-th/0303016&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0303016&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So if you ask whether the question has been solved for all vacua, the answer is surely No. But for many big classes etc., the answer is Yes, at least morally. And yes, there are fully stabilized vacua.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-08T14:52:50.730" LastActivityDate="2011-04-08T14:52:50.730" />
  <row Id="8262" PostTypeId="2" ParentId="8255" CreationDate="2011-04-08T15:05:34.437" Score="2" ViewCount="" Body="&lt;p&gt;The Universe doesn't come equipped with any &quot;clocks at each moment&quot; that would immediately recognize different &quot;moments&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, the laws of physics - except for cosmology - are totally invariant with respect to translations in time, which is just a different way of saying that there is no way, even in principle, to find out whether an event occurred at time $t_1$ or $t_2$: the events are guaranteed to proceed in an identical way if the initial conditions are identical, whether the events occur at $t_1$ or $t_2$. By Noether's theorem, this symmetry is inseparably connected with energy conservation. Still, spacetime exists and events occur at different times, and the distances between events in time - duration of processes etc. - are well-defined numbers and can be measured by various tools.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the proper time of world lines may be measured by all kinds of clocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of cosmology, the Universe is expanding and the time-translational symmetry is broken by the expansion (and so is the corresponding energy conservation law). It follows that one may define a &quot;cosmic time&quot; - the proper time of the &quot;static&quot; (a frame in which the density of momentum of the CMB radiation vanishes) world line stretched between a point of the Big Bang and a present event. This cosmic time is related to the current CMB temperature and many other things. Still, the CMB temperature is obviously not accurate enough to measure the time separations of some very recent events from each other. In that case, we prefer things like atomic clocks that may achieve an unbelievable precision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Archeologists use radioactive isotopes to determine the age of various things because it works and they don't use a more accurate method because no method that would be more accurate is available. Pharaohs could have put (and reset) digital clocks into the pyramids when they built them except that they failed to do so and it's very hard to prosecute them for their negligence today. ;-) At any rate, those questions are for geologists, historians, and archeologists - not physicists. If a historian asked whether physicists may come up with a better method, the answer is almost certainly No. But this is mostly a question about the creative engineering tricks and inventions that are possible given the known laws of physics; it is not a question about physics itself. If we measure the age of objects and materials, we must look at the time-dependent changes of these objects and materials. Chemistry doesn't bring us too far and nuclear physics is the way to go - we're back to various types of radioactive dating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, I probably don't understand the question - and I would bet that I will end up being in the majority - so I have no way to predicting whether the text above will be found satisfactory by HDE.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-08T15:12:32.797" LastActivityDate="2011-04-08T15:12:32.797" />
  <row Id="8263" PostTypeId="2" ParentId="3634" CreationDate="2011-04-08T15:12:30.433" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Pinnow model does not suffer of the barionic 'spin crisis':&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;Douglas Pinnow&lt;/strong&gt; book '&lt;strong&gt;&lt;em&gt;Our Resonant Universe&lt;/em&gt;&lt;/strong&gt;' (2003) is a monography of a new model of particles, based only on ElectroMagnetism, that has only one parameter: the $m_{e}$ - electron mass, and he &lt;strong&gt;derives the particle properties to within 1% of their values&lt;/strong&gt; (barion masses well bellow 1%), masses, angular momenta, spin. He makes predictions and his model is very simple based in the  &lt;strong&gt;non-radiation condition and he worked on top of Haus, Goedecke and Kaluza-Klein theory&lt;/strong&gt;. The author is an 'old' nuclear engineer, and NASA fellow, with a large career in photonics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quote from pag 9 (in the Overview):  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Chaper 2, &lt;em&gt;Composite particles&lt;/em&gt;, is&#xA;  dedicated to the interactions among&#xA;  the standing-wave resonances&#xA;  (elementary, massive particles) which&#xA;  form the composite barions. Models of&#xA;  such structures lead to mass&#xA;  calculations that are accurate to&#xA;  within 1% of observation for all &#xA;  spin$-\frac{1}{2}$ baryons.&#xA;  The magnetic moments of the proton and&#xA;  neutron are also easily calculated&#xA;  from such first principles. &lt;strong&gt;An&#xA;  estimate of the magnetic moment of the&#xA;  neutral sigma barion ($\Sigma^{0}$) is&#xA;  offered as a test of this approach&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Overview &lt;a href=&quot;http://www.authorhouse.com/BookStore/BookDetail.aspx?Book=233588&quot; rel=&quot;nofollow&quot;&gt;from here&lt;/a&gt; (more to read: Free Preview and About the Author)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Our Resonant Universe presents the&#xA;  resonance model, a ground-breaking&#xA;  vision of the nature of the universe.&#xA;  By applying a &lt;strong&gt;simple electromagnetic&#xA;  approach to both elemental matter and&#xA;  nuclear reactions, Pinnow and Miller&#xA;  unite mass, spin, and charge with the&#xA;  four fundamental forces of the&#xA;  universe, without invoking the&#xA;  arbitrary constants of the standard&#xA;  model or the multi-dimensional&#xA;  mathematics of string theory&lt;/strong&gt;. The&#xA;  results are striking in their accuracy&#xA;  and in their descriptive power. The&#xA;  model also provides a series of&#xA;  predictions that will soon put this&#xA;  conception of the universe to the&#xA;  test.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sidenote:&lt;br&gt;&#xA;From this &lt;a href=&quot;http://www.citebase.org/abstract?id=oai%3aarXiv.org:0904.1735&quot; rel=&quot;nofollow&quot;&gt;citebase link&lt;/a&gt; we can access the sources of all the references cited in the Thomas  article as G-scholar or pre-print ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book Pinnow, low cost, was probably too little read and has not been discussed!&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-04-08T15:12:30.433" CommentCount="2" />
  <row Id="8264" PostTypeId="1" CreationDate="2011-04-08T15:23:12.937" Score="3" ViewCount="202" Body="&lt;p&gt;I have 2 questions, as a expat and new parent residing in Shanghai (2000km from Fukushima) where we are now experiencing radioactive fallout from the Fukushima Nuclear Plant (Japan), we can not get any 'real' infomation about what are safe limits for food and air. We simply do not understand the reports we are given, nor the poor explanations that accompany the reports, which are written by design to try make people feel safe and not panic. In fact, we have no idea if the numbers being released are even true. What we are being told is that yes there is fallout and some vegetables have had 1-3 Bq/Kg detected on them, i.e. spinach. And for Air in Shanghai today (mBq/m(3)) 131-I (.58), 137Cesium (.17) and 134Cesium (.16). Now I know -from what i have learned -no one can give me specifics and certainties with all this, but what are we looking at here? Can any of you geniuses in this community - and I say this respectfully-- tel me if these levels are dangerous? Would appreciate your advice...&lt;/p&gt;&#xA;" OwnerUserId="3001" LastActivityDate="2011-04-09T21:33:32.223" Title="Are these radioactive particle matter and air emmissions dangerous, 2000KM from Fukushima?" Tags="&lt;radiation&gt;" AnswerCount="3" />
  <row Id="8265" PostTypeId="1" AcceptedAnswerId="8271" CreationDate="2011-04-08T15:36:04.000" Score="4" ViewCount="139" Body="&lt;p&gt;I have found several nice versions available for purchase, but all of the free ones are either Latin, old English, or pixillated. I searched google and was only able to locate the free ones linked to from the wikipedia page on &quot;Philosophiæ Naturalis Principia Mathematica.&quot; &lt;/p&gt;&#xA;" OwnerUserId="2843" LastActivityDate="2011-04-08T16:12:42.407" Title="Is there a good Modern English translation of Newton's &quot;Philosophiæ Naturalis Principia Mathematica&quot; available for free viewing/download?" Tags="&lt;mathematical-physics&gt;&lt;newton&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="1" />
  <row Id="8266" PostTypeId="2" ParentId="8264" CreationDate="2011-04-08T15:38:09.227" Score="5" ViewCount="" Body="&lt;p&gt;First, just recall that one becquerel is one radioactive decay per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your spinach has had 1-3 Bq/kg, then you should just completely forget about any threats - there is no danger whatsoever. The safe limits are 100-500 Bq/kg, see e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.anarchyjapan.com/2011/04/for-now-no-changes-in-food-radiation-safety-levels/&quot; rel=&quot;nofollow&quot;&gt;http://www.anarchyjapan.com/2011/04/for-now-no-changes-in-food-radiation-safety-levels/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which are 100+ times higher than what you report. Of course, I can't completely eliminate the risk that your next package of spinach will be unchecked and will contain 1/3 of the weight stored in plutonium :-) but it is unlikely because things are being checked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/VvMIg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compare, an average (healthy) banana - because of its potassium content (radioactive potassium-40 is &lt;a href=&quot;http://en.wikipedia.org/wiki/Potassium#Isotopes&quot; rel=&quot;nofollow&quot;&gt;0.01% of natural potassium&lt;/a&gt;; it decays to argon-40 with half-life 1.25 billion years) - has about 15 becquerels which is like 5-15 kilograms of your (not so) &quot;radioactive&quot; spinach. That's despite the fact that the average banana's mass is just 120 grams! Because of its measurable radioactivity content, a &quot;banana equivalent dose&quot; is sometimes used to compare radioactivity with the exposure in normal life - and this is an example where the comparison is useful. Potassium-40 is the main source of radioactivity in humans, beating even carbon-14. The conclusion is that the spinach is not radioactive in any practical sense of the word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radioactivity in the air is negligible as well. The U.S. limits are about 150 becquerels per cubic meters, so if you tell us about some isotopes contributing less than one millibecquerel per cubic meter, which is one million times less, there is really nothing to talk about - as long as they mention isotopes that are relatively short-lived (and assuming that none of the data and units have been misinterpreted).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one got the same amount of becquerels from plutonium which has a much longer half-life, one would need much more plutonium (by mass) to explain the data, and this plutonium would stay in your lungs forever - it is dangerous. But millibecquerels per cubic meter stored in short-lived isotopes are totally irrelevant and chances are that the numbers don't differ from the natural background in Shanghai in any recognizable way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of those quantities are actually within the safe levels even a few miles away from Fukushima, so if your distance is thousands of miles, be sure that the effect of the hassles in Fukushima is between inconsequential and undetectable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretty much all the hysteria is completely irrational and scientifically unjustifiable.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-09T19:46:25.770" LastActivityDate="2011-04-09T19:46:25.770" CommentCount="1" />
  <row Id="8268" PostTypeId="2" ParentId="8264" CreationDate="2011-04-08T15:45:15.573" Score="3" ViewCount="" Body="&lt;p&gt;To supplement Lubos' answer, this &lt;a href=&quot;http://xkcd.com/radiation/&quot; rel=&quot;nofollow&quot;&gt;graph&lt;/a&gt; gives numbers and put things in perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that just sleeping next to another human you are exposed to some radiation. The table uses sieverts, which is the way the cumulative harm of radiation is measured on human health.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exists always a background level of radiation, from the ground and the air and the cosmic rays. The numbers you are quoting are within the statistical error in your region, so no, it is not dangerous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One Bq is defined as the activity of a quantity of radioactive material in which one nucleus decays per second.  You have to realize that as we sit or sleep or do anything, &lt;a href=&quot;http://cosmic.lbl.gov/SKliewer/Cosmic_Rays/Muons.htm&quot; rel=&quot;nofollow&quot;&gt;one cosmic ray per minute per cm**2&lt;/a&gt; passes through us at sea level and is about half the natural radiation count level. That is what I mean when I say the numbers you are quoting are within the error of the ambient radiation. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-04-09T06:35:22.003" LastActivityDate="2011-04-09T06:35:22.003" CommentCount="1" />
  <row Id="8269" PostTypeId="2" ParentId="8264" CreationDate="2011-04-08T15:46:59.487" Score="4" ViewCount="" Body="&lt;p&gt;A Bq is a Becquerel or decay rate of 1 event per second (on average, because this is different from Hz which is events coming 1 second apart like, well, clockwork). Note in the chart &lt;a href=&quot;http://physics.stackexchange.com/questions/8264/are-these-radioactive-particle-matter-and-air-emmissions-dangerous-2000km-from-f/8268#8268&quot;&gt;anna posted&lt;/a&gt; the dose from cosmic rays, that comes from an event rate rather more than Bq.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much radioactive dose that is depends on how it is delivered. From Iodine-131 it is $\beta^-$, which (a) does not have a lot of penetrating power (good while it is outside you, not so good once you ingest it) and (b) how a &lt;a href=&quot;http://en.wikipedia.org/wiki/Sievert#Q_values&quot; rel=&quot;nofollow&quot;&gt;quality factor&lt;/a&gt; of 1 (a nice low value). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iodine however, poses a special risk if you ingest it: it gets collected in the thyroid glands, which means that the dose (and the risk) is not spread out all over, it is concentrated in one place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one quick comparison: banana, spinach, etc. naturally have a lot of Potassium, and that means some Potassium-40 which is also radioactive. A typical banana carries several Bq of K-40. This is the source of the dose from eating a banana in the XKCD link that anna gave you. (I've used this fact to demonstrate high sensitivity gamma detectors to visiting VIPs because it does not require any of the paperwork associated with radioactive test sources: just put your lunch in the detector.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the risk you're hearing about now is not large. Pay attention to the reports (to learn if things change), consider eating a little less of the affected foods, and don't fret it.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-04-09T21:33:32.223" LastActivityDate="2011-04-09T21:33:32.223" />
  <row Id="8270" PostTypeId="2" ParentId="8256" CreationDate="2011-04-08T16:04:08.720" Score="1" ViewCount="" Body="&lt;p&gt;This question is somewhat of a physics history question about when and why. I dont have a specific reference as to who first introduced symplectic geometry into physics, but there are some broader indications as to what has happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The symplectic manifold idea is a formalisation (in terms of differential geometry) of the phase space idea. Looking up phase space in Wikipedia gives this useful reference and explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In mathematics and physics, a phase space, introduced by Willard Gibbs in 1901, is a space in which all possible states of a system are represented, with each possible state of the system corresponding to one unique point in the phase space. For mechanical systems, the phase space usually consists of all possible values of position and momentum variables.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A second factor that would have been relevant in this case was &lt;a href=&quot;http://en.wikipedia.org/wiki/Liouville%27s_theorem_%28Hamiltonian%29&quot; rel=&quot;nofollow&quot;&gt;Liouville's Theorem&lt;/a&gt;. This theorem asserts that the density of points in phase space is conserved. It is about volumes in phase space, thus suggesting a differential geometry treatment of phase space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once it became clear (I dont know whether this was Gibbs or later) that the symplectic form exists as a geometric object similar in some ways to a metric, and closely connected with the Poisson bracket and the rest of classical mechanics, then it would be natural to describe the whole of classical mechanics in this geometric way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could be another topic that was first noticed by mathematicians, but only noticed by physicists after the advent of quantum mechanics (1920s or 1930s) when explicit connections between the two theories (Classical and Quantum) were searched for. Names associated with this are Weyl and Moyal (in 1949). More on these early phase-space to QM ideas in this &lt;a href=&quot;http://en.wikipedia.org/wiki/Weyl_quantization&quot; rel=&quot;nofollow&quot;&gt;Wikipedia link&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also needs to be remembered as we discuss post General Relativity and post Quantum Mechanics eras, that phase space formulations of GR have been sought as an aid to quantization. All this is a highly geometric approach to Quantum Gravity, associated with many names although &lt;a href=&quot;http://en.wikipedia.org/wiki/Abhay_Ashtekar&quot; rel=&quot;nofollow&quot;&gt;Ashtekar&lt;/a&gt; worked on this overlap. Eventually this led to his (co-) formulation of Loop Quantum Gravity.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastActivityDate="2011-04-08T16:04:08.720" />
  <row Id="8271" PostTypeId="2" ParentId="8265" CreationDate="2011-04-08T16:07:30.287" Score="4" ViewCount="" Body="&lt;p&gt;You overestimate the amount of evolution that the English language has undergone in the last 300 years. I recommend you the perfectly readable English translation by my great great grand uncle Motte, as revised by Cajori:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=ySYULc7VEwsC&amp;amp;printsec=frontcover&amp;amp;hl=en&amp;amp;source=gbs_ge_summary_r&amp;amp;cad=0#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=ySYULc7VEwsC&amp;amp;printsec=frontcover&amp;amp;hl=en&amp;amp;source=gbs_ge_summary_r&amp;amp;cad=0#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let me just be a linguist for a while. ;-) Modern English has existed since 1550&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Modern_english&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Modern_english&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;when the Great Vowel Shift took place, so not only Motte's 1729 translation linked above is written in Modern English but Isaac Newton was speaking Modern English throughout his life, too. Of course, Newton would write in Latin.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-08T16:12:42.407" LastActivityDate="2011-04-08T16:12:42.407" CommentCount="5" />
  <row Id="8272" PostTypeId="1" CreationDate="2011-04-08T16:33:34.780" Score="2" ViewCount="107" Body="&lt;p&gt;reading the paper about &lt;a href=&quot;http://ftp.ics.uci.edu/pub/wayne0/papers/belgrade/Wisdom-swimming03.pdf&quot; rel=&quot;nofollow&quot;&gt;spacetime swimming by Wisdom&lt;/a&gt;&lt;br&gt;&#xA; (something related to this has been &lt;a href=&quot;http://physics.stackexchange.com/questions/886/swimming-in-spacetime-apparent-conserved-quantity-violation&quot;&gt;previously asked here&lt;/a&gt;) can't help but think that there is more to this than what is on the paper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically the paper just proves that inertially, an object can get a net displacement by using the fact that phases are path-dependent due to non-commutativity in space paths, and he uses the rotational argument to properly ambient his point. The punchline here is that this same argument seems to be applicable when the masses have also for instance, a magnetic dipole, and there is a geometric phase in the magnetic field due to curvature&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you think that the same argument are applicable to obtain a similar effect due electric charges or magnetic dipoles in a curved manifold? i'm aware the question is vague but i hope this can entice (or stir) at least some deep discussion on the subject&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side (but very important) question; is not clear from the paper if the displacement is constrained to go only down in the gravitational potential well, but most discussions and papers referencing &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0510054&quot; rel=&quot;nofollow&quot;&gt;it says it should be so&lt;/a&gt; (check the conclusions section and comments)&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-04-08T17:02:16.617" LastActivityDate="2011-04-08T17:02:16.617" Title="net displacement and path dependence" Tags="&lt;general-relativity&gt;&lt;geometric-phase&gt;&lt;spacetime-swimming&gt;" />
  <row Id="8273" PostTypeId="2" ParentId="8156" CreationDate="2011-04-08T16:43:13.283" Score="1" ViewCount="" Body="&lt;p&gt;The basic experimental fact the inventors of QM had to deal with was the uncertainty principle.  The mathematics behind this principle has two major parts, one involving linear algebra and another involving Fourier analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the operator algebra of QM is necessary in order to have a theory which obeys the uncertainty principle, and if you want to know why this is true, you have to study the mathematics.&lt;/p&gt;&#xA;" OwnerUserId="442" LastActivityDate="2011-04-08T16:43:13.283" />
  <row Id="8274" PostTypeId="1" CreationDate="2011-04-08T16:47:51.943" Score="0" ViewCount="164" Body="&lt;p&gt;Can you give me some examples of recreational physics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thx&lt;/p&gt;&#xA;" OwnerUserId="3004" LastActivityDate="2011-04-08T17:11:45.827" ClosedDate="2011-04-08T17:55:53.197" Title="Can you give me some examples of recreational physics?" Tags="&lt;soft-question&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="8275" PostTypeId="2" ParentId="8274" CreationDate="2011-04-08T17:06:38.950" Score="3" ViewCount="" Body="&lt;p&gt;There is nothing such as &quot;recreational physics&quot;. The term &quot;recreational mathematics&quot; is a standard umbrella term for mathematical puzzles and mathematical games. The term &quot;recreational physics&quot; may only be used by individuals who want to say that they find physics - whether theory or experiments - amusing. But they realize that no one will exactly know what they mean because &quot;recreational physics&quot; is not a standard term in the dictionary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, this blog&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.recreationalphysics.blogspot.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.recreationalphysics.blogspot.com/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;was named &quot;Recreational physics&quot; because its author thought that it is an amusing title that combines physics and attractiveness. But the title doesn't mean anything specific and you won't find anything special on the blog that you wouldn't find on other physics blogs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For physics to be &quot;recreational&quot; in the same way as recreational mathematics, there would have to be some &quot;game&quot; about them - a large number of possibilities that may always train our brains. But when it is so, it is mathematics. Physics, pretty much by definition, is all about &quot;fully solving&quot; problems. So if one has an amusing physics experiment, it may be viewed as &quot;amusing physics&quot;, but if one does it - or understands it - once, there is really no point in doing the same thing again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That makes a difference. For example, people may solve Sudoku thousands of times because it's always a bit different and they have always some &quot;new fun&quot;. But if someone finds out that water may be frozen to ice e.g. in this way&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=wzHXiGdMvkU&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=wzHXiGdMvkU&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;it's not terribly recreational to repeat the same procedure one thousand times because it's always the same hard work. ;-) The part of &quot;physics puzzles&quot; that may get modified deserves to be classified as &quot;maths&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-08T17:11:45.827" LastActivityDate="2011-04-08T17:11:45.827" />
  <row Id="8276" PostTypeId="1" CreationDate="2011-04-08T17:33:14.430" Score="3" ViewCount="168" Body="&lt;p&gt;My daughter's coloring book has perforated pages, so you can easily rip out the pages for easier coloring action. Great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trouble is, the perforation isn't enough, and it's almost impossible to rip out the pages without tearing the paper. It is difficult to grab the paper near the perforation since it is so close to the binding of the book.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember having this problem as a child, too. Funny, you'd think after 30 years coloring book page perforation would be a &quot;solved problem&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found myself sometimes the tearing the page at a shallow angle as possible helps, but other times not, and sometimes a sharp angle works, and sometimes not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if there is any optimal angle or other technique to tear the paper, to maximize the stress applied to the perforation while minimizing the stress on the surrounding area, as to not rip the paper. The goal here is something that takes the minimum amount of time and effort, and maximizing chances of a successful tear.&lt;/p&gt;&#xA;" OwnerUserId="3006" LastEditorUserId="3006" LastEditorDisplayName="" LastEditDate="2011-04-08T18:13:27.700" LastActivityDate="2011-04-08T18:32:06.063" ClosedDate="2011-04-09T04:16:30.140" Title="Optimal technique to tear perforated paper along the perforation" Tags="&lt;forces&gt;&lt;material-science&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="8277" PostTypeId="2" ParentId="8276" CreationDate="2011-04-08T17:38:41.540" Score="3" ViewCount="" Body="&lt;p&gt;I know creasing the paper along the perforation first helps.&lt;/p&gt;&#xA;" OwnerUserId="2898" LastActivityDate="2011-04-08T17:38:41.540" CommentCount="1" />
  <row Id="8278" PostTypeId="2" ParentId="8276" CreationDate="2011-04-08T17:46:22.050" Score="2" ViewCount="" Body="&lt;p&gt;The easiest fix which is also an option for a child is to take an unsharpened pencil and take the edge of the pencil and run it down the perforation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may have to bend the book back at the binding, opening the book beyond the 180 degree angle the book was designed for, but this will not create a usability issue with the book itself. :)&lt;/p&gt;&#xA;" OwnerUserId="3007" LastActivityDate="2011-04-08T17:46:22.050" />
  <row Id="8279" PostTypeId="1" AcceptedAnswerId="8280" CreationDate="2011-04-08T18:18:57.457" Score="3" ViewCount="83" Body="&lt;p&gt;If a stationary charge is placed in a changing uniform magnetic field to measure the induced E field at some point, what would be the direction of E?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think E must equal zero normal to B to maintain the symmetry of the problem, leaving E entirely along B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-04-08T19:14:07.423" Title="What is the direction of the induced E field from a changing uniform magnetic field?" Tags="&lt;electromagnetism&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8280" PostTypeId="2" ParentId="8279" CreationDate="2011-04-08T18:29:54.400" Score="3" ViewCount="" Body="&lt;p&gt;Faraday's law of induction - one of Maxwell's equations - says&#xA;$$\nabla \times E = - \frac{\partial B}{\partial t}.$$&#xA;So the time derivative of $B$ induces an electric field. However, what you incorrectly assumed was that the time derivative of $B$ will directly be proportional to the induced electric field. In reality, it is the &quot;curl&quot; of the electric field that is equal to the time derivative of $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if $B$ and its time derivative is uniform in space, the field $E$ will &lt;em&gt;not&lt;/em&gt; be uniform. For example, if $-\partial B / \partial t = (0,0,\beta)$ in the $z$-direction, the most natural $E$ will depend linearly on spatial coordinates, and may be e.g.&#xA;$$ E = (-y\beta/2, x\beta/2, 0) $$&#xA;This electric field is rotating around the $z$-axis - around the direction of the magnetic field (its change in time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that it is constructed so that its curl is equal to $(0,0,\beta)$. However, this configuration is also nicely rotational symmetric because up to the extra factor of $\beta$, the vector $E$ is nothing else than the vector $r$ projected to the $xy$-plane and rotated by 90 degrees around the $z$-axis - and this definition doesn't depend on the choice of axes in the $xy$-plane, so the configuration is rotationally symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not translationally symmetric in the $x$ and $y$ directions, however. There exists a specific axis - in my case, the $z$-axis - where the induced $E$ vanishes. The solution to Maxwell's equations is not unique. I may add any other uniform $E$ field (or, more generally, any electromagnetic wave solution of Maxwell's equations with no right hand side) without spoiling the validity of the Maxwell's equations - this is a kind of trivial artifact of the linearity of these equations.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-08T19:14:07.423" LastActivityDate="2011-04-08T19:14:07.423" CommentCount="2" />
  <row Id="8281" PostTypeId="2" ParentId="8276" CreationDate="2011-04-08T18:32:06.063" Score="1" ViewCount="" Body="&lt;p&gt;Could you could run a toothpick or something similar down the perforated edge to enlarge the holes in the perforation?&lt;/p&gt;&#xA;" OwnerUserId="3008" LastActivityDate="2011-04-08T18:32:06.063" />
  <row Id="8282" PostTypeId="1" AcceptedAnswerId="8296" CreationDate="2011-04-08T18:56:14.307" Score="3" ViewCount="328" Body="&lt;p&gt;My evidence is entirely anectodal and non-scientific, but I've noticed food gets cold faster when it's been heated in a microwave instead of a stove. Is this true? And if it is, why does it happen?&lt;/p&gt;&#xA;" OwnerUserId="3009" LastActivityDate="2011-04-09T00:52:49.037" Title="Why does microwaved food get cold faster" Tags="&lt;thermodynamics&gt;&lt;heat&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="8283" PostTypeId="2" ParentId="8282" CreationDate="2011-04-08T19:12:43.117" Score="6" ViewCount="" Body="&lt;p&gt;There are differences in the amount of moisture which affects the cooling time as well but I think that most of the commonly &quot;observed&quot; effect is simply due to the fact that the regular oven heats up the container - the dishes - while the microwave oven ideally heats up the content only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the dishes are hot in the classical oven, their heat is keeping the food warm for a longer time than the relatively cooler dishes taken from a microwave.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-08T19:12:43.117" CommentCount="1" />
  <row Id="8284" PostTypeId="1" AcceptedAnswerId="8297" CreationDate="2011-04-08T19:28:22.110" Score="4" ViewCount="114" Body="&lt;p&gt;The hypothetical Oort cloud is an explaination for long period comets. It presumably is made up of icy bodies orbiting up to 50,000 AU from the Sun. If so, shouldn't the Spitzer IR telescope have  detected the radiation generated by solar energy from such a spherical cloud?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="1924" LastEditorDisplayName="" LastEditDate="2011-04-08T22:11:49.520" LastActivityDate="2011-04-09T01:35:58.547" Title="Oort Cloud Detection" Tags="&lt;astronomy&gt;&lt;radiation&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="8285" PostTypeId="2" ParentId="8282" CreationDate="2011-04-08T19:35:30.717" Score="1" ViewCount="" Body="&lt;p&gt;Another effect is that instructions for heating in a microwave are of the order of minutes. Mostly the water molecules get heated to 100C during that time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When heating in the  oven one may set it   180C which is normal cooking temperatures, and one leaves it a longer time; of course then there is more heat content and it takes longer to cool.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-08T19:35:30.717" CommentCount="5" />
  <row Id="8286" PostTypeId="1" AcceptedAnswerId="8288" CreationDate="2011-04-08T19:52:41.820" Score="4" ViewCount="119" Body="&lt;p&gt;I suppose this is a not-too-bright question but, in a nutshell: why are nuclear fuels radioactive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this I mean, which is the connection between being a fissile (or fertile, for that matter) material and being radioactive? Does the former imply the latter? Does the latter imply the former? Why?&lt;br&gt;&#xA;Would it be thinkable for an isotope to be able to capture a neutron, become instable and split without it being radioactive in the first place?&lt;/p&gt;&#xA;" OwnerUserId="3011" LastActivityDate="2011-04-09T07:40:17.950" Title="Radioactive nuclear fuel?" Tags="&lt;nuclear-physics&gt;&lt;radioactivity&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" />
  <row Id="8287" PostTypeId="1" CreationDate="2011-04-08T20:13:40.657" Score="6" ViewCount="110" Body="&lt;p&gt;So I have some follow up questions from this thread: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/6419/electric-engine-transmission&quot;&gt;Electric engine transmission&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I am looking for a more precise mathematical statement that makes this true. Why is it that the electric motor can provide torque as necessary and the gas engine can't? What can we do to the gas engine to make it behave like an electric engine? &#xA;I have only taken 2nd year analytical mechanics so keep that in mind. &lt;/p&gt;&#xA;" OwnerUserId="551" LastActivityDate="2011-04-08T20:13:40.657" Title="Torque of an electric engine" Tags="&lt;electrostatics&gt;&lt;electrical-engineering&gt;" CommentCount="4" FavoriteCount="1" />
  <row Id="8288" PostTypeId="2" ParentId="8286" CreationDate="2011-04-08T20:24:47.827" Score="4" ViewCount="" Body="&lt;p&gt;Qualitatively, isotopes that carry a lot of unnecessary extra energy are unstable - and it makes them more unstable with respect to alpha or beta decay as well as fission (the latter is useful for nuclear fission energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the question why fissiles are radioactive (unstable) has a trivial answer. Only elements up to $Z=82$, $N=126$ i.e. $A=208$ may be stable nuclides, see&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/File:Table_isotopes_en.svg&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/File:Table_isotopes_en.svg&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the other hand, fissiles are heavier - with $A=233,235$ (uranium), and $A=239,241$ (plutonium). Obviously, all those nuclides must be unstable - even though the half-life may be close to billions of years (very long).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most cases, it's much more favorable for a nucleus to alpha-decay than to undergo fission because the alpha particle has a very high binding energy per nucleon, around 7 MeV, which is close to the maximum value around 8.8 MeV per nucleon achieved by nickel and iron:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin.html#c2&quot; rel=&quot;nofollow&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin.html#c2&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin2.html#c1&quot; rel=&quot;nofollow&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin2.html#c1&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In particular, a fission to fragments with $2&amp;lt;Z&amp;lt;30$ is almost impossible because those have a relatively low binding energy per nucleon. If one can energetically win by fusion, it's pretty much guarantee that one may also find an energetically favored alpha decay because the alpha particle has a pretty high binding energy per nucleon, allowing the other decay product to have a similar binding energy per nucleon if not lower, thus making the decay more likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The opposite implication obviously doesn't hold: most of the known radioactive isotopes are the light ones that can't split by fission (alpha decay is not counted even though it is a sort of &quot;minifission&quot;). In general, alpha and beta radioactivity is much more common among the known nuclides than fission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important to realize that the link between fissiles and radioactivity is a bit indirect: the main isotopes that bring radioactivity to spinach etc. and that represent a public health concern are not the heavy nuclei themselves: it's the decay products such as the isotopes of iodine, cesium, strontium etc. Sometimes they're products of long chains of radioactive decay; the fission of the heavy nuclei is just the beginning of the chain, a tip of an iceberg.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-09T07:40:17.950" LastActivityDate="2011-04-09T07:40:17.950" CommentCount="1" />
  <row Id="8289" PostTypeId="1" AcceptedAnswerId="8306" CreationDate="2011-04-08T20:45:49.093" Score="9" ViewCount="257" Body="&lt;p&gt;Start with a set of points $x_1, x_2, \ldots$ that are connected by wires with some resistance. Represent the resistance by a conductance matrix (conductance being one over the resistance), where $\mathbf{C}_{ij}$ is the conductance between points $i$ and $j$, if the point are connected by a wire, otherwise the $\mathbf{C}_{ij}=0$. Can one solve for the equivalent resistance between two points by some matrix transform of $\mathbf{C}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The comments bring up some interesting points - and suggest an alternate phrasing: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can you compute the resistance distance for a graph when the resistances are not all unit values using matrix operations?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="846" LastEditorUserId="846" LastEditorDisplayName="" LastEditDate="2011-04-09T05:35:23.957" LastActivityDate="2011-04-09T07:35:29.233" Title="Matrix solution of an equivalent resistance circuit problem" Tags="&lt;electric-circuits&gt;&lt;linear-systems&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" />
  <row Id="8290" PostTypeId="2" ParentId="8194" CreationDate="2011-04-08T20:58:19.343" Score="2" ViewCount="" Body="&lt;p&gt;The nearby effect of rotating masses are detectable and Joe Weber (who I took general relativity from) measured this effect in the 1950s. It's how he calibrated his gravity wave detectors. This is old technology and largely forgotten now. For example, see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jpn. J. Appl. Phys. 19, pp. L123-L125, (1980), Katsunobu Oide, Kimio Tsubono and Hiromasa Hirakawa, &lt;em&gt;The Gravitational Field of a Rotating Bar&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The dynamic gravitational field around&#xA;  a rotating bar, and the forced&#xA;  oscillation of a resonant antenna&#xA;  located in this field, are treated in&#xA;  terms of the quadrupole-quadrupole&#xA;  interaction. Some of these&#xA;  calculations have been confirmed by&#xA;  observation of the resonant&#xA;  oscillation of a quadrupole antenna in&#xA;  several orientations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://jjap.jsap.jp/link?JJAP/19/L123/&quot; rel=&quot;nofollow&quot;&gt;http://jjap.jsap.jp/link?JJAP/19/L123/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proceedings of Space Technology and Applications International Forum 2000, STAIF2000, Giorgio Fontana, &lt;em&gt;Gravitational Radiation and its Application to Space Travel, principles and required&lt;/em&gt; scientific developments&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Historically, the emission of&#xA;  gravitational radiation has been&#xA;  studied in astrophysical systems, as&#xA;  reported in the second section of this&#xA;  paper, and a quite large literature&#xA;  exists on the subject. Instead, the&#xA;  laboratory generation of gravitational&#xA;  radiation is still in its theoretical&#xA;  stage of development except, perhaps,&#xA;  the rotor employed for the calibration&#xA;  of a GW antenna (Astone, 1991), which&#xA;  has indeed produced experimental&#xA;  results for a near field detection.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The above were found by googling &quot;Weber bar&quot;+calibration&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-04-08T20:58:19.343" CommentCount="1" />
  <row Id="8291" PostTypeId="1" AcceptedAnswerId="8304" CreationDate="2011-04-08T22:03:54.507" Score="3" ViewCount="342" Body="&lt;p&gt;I presume the answer is that it depends on the mass and size of the star and black hole, but I was wondering if somebody could provide some rough bounds  (e.g. hours vs thousands of years).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;eating&quot; I mean the time it takes a star to go through the event horizon of the black hole, although maybe there are other easier-to-detect or more relevant events one can measure?&lt;/p&gt;&#xA;" OwnerUserId="1527" LastEditorUserId="1527" LastEditorDisplayName="" LastEditDate="2011-04-09T01:37:58.390" LastActivityDate="2011-04-09T14:14:49.863" Title="How long does it take a black hole to eat a star?" Tags="&lt;black-holes&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="8292" PostTypeId="2" ParentId="8215" CreationDate="2011-04-08T22:20:57.517" Score="2" ViewCount="" Body="&lt;p&gt;The thermal energy required to break the Cooper pairs determines the value of Tc in BCS superconductors.  In the work of McMillan, (&lt;a href=&quot;http://link.aps.org/doi/10.1103/PhysRev.167.331&quot; rel=&quot;nofollow&quot;&gt;Physical Review 167 (2), 1968 p 331-344&lt;/a&gt;), he shows that with considerations to phonon excitations in materials, the limitation of phonon vibrational energies is the limiting factor in BCS transition temperatures. In his work, McMillan estimated the upper Tc for electron-phonon superconductors is 28K.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The superconductivity in MgB2, with a Tc of ~40K is a form of BCS electron phonon coupling, but is isolated to only a few phonon modes, which enhances the effect of the coupling.&lt;/p&gt;&#xA;" OwnerUserId="2433" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-05-16T08:20:06.573" LastActivityDate="2011-05-16T08:20:06.573" />
  <row Id="8293" PostTypeId="2" ParentId="8149" CreationDate="2011-04-08T23:23:02.050" Score="2" ViewCount="" Body="&lt;p&gt;Marek suggested I post my comment (which doesn't completely answer the question) as an answer. Here it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have two Brownian motions with diffusion coefficients $D_1$ and $D_2$, which start at the same point at $t=0$. Let $x_i$ be the average displacement vector for particle $i$ after time $t$. Then, $\langle x_i^2 \rangle  = 2D_it$, where $x_i^2$ means the square of the length of vector $x_i$. Now, consider the relative position of these two random walks after time $t$ (assuming they start out at the same place). The expected value of the relative displacement will be $$\langle (x_1 - x_2)^2 \rangle = \langle x_1^2 \rangle + \langle x_2^2 \rangle = 2 (D_1 + D_2)t,$$ where you can ignore the cross terms since the expected inner product of the two displacement vectors is 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully, this argument convinces you that the relative position of two points undergoing Brownian motion is still Brownian. You can make it completely rigorous if you want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your question boils down to: what is the probability distribution for the time it takes Brownian motion starting at distance $d$ from the origin to first reach a disc of radius $r=r_1+r_2$ around the origin. It's straightforward to find the probability that Brownian motion is in that disc after time $t$, but computing the probability distribution for the first time it's reached is harder. I am sure this has been done, but right now I don't have time to do a literature search for this.&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2011-04-08T23:23:02.050" CommentCount="1" />
  <row Id="8294" PostTypeId="1" CreationDate="2011-04-08T23:29:59.823" Score="2" ViewCount="141" Body="&lt;p&gt;&lt;a href=&quot;http://blogs.discovermagazine.com/badastronomy/2011/04/05/astronomers-may-have-witnessed-a-star-torn-apart-by-a-black-hole/&quot; rel=&quot;nofollow&quot;&gt;http://blogs.discovermagazine.com/badastronomy/2011/04/05/astronomers-may-have-witnessed-a-star-torn-apart-by-a-black-hole/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of the star in the disc, a lot of the star in the jets, precisely how much of the star actually falls into the black hole?&lt;/p&gt;&#xA;" OwnerUserId="2780" LastActivityDate="2011-04-08T23:29:59.823" Title="How much of a star falls into a black hole?" Tags="&lt;black-holes&gt;" />
  <row Id="8295" PostTypeId="1" AcceptedAnswerId="8298" CreationDate="2011-04-09T00:20:48.333" Score="10" ViewCount="303" Body="&lt;p&gt;I was reading a book on the history of Quantum Mechanics and I got intrigued by the &lt;em&gt;gendankenexperiment&lt;/em&gt; proposed by Einstein to Bohr at the 6th Solvay conference in 1930.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For context, the thought experiment is a failed attempt by Einstein to disprove Heisenberg's Uncertainty Principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/en/6/65/Ebohr4.gif&quot; alt=&quot;Einstein's box as drawn by Bohr&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Einstein considers a box (called Einstein's box; see figure) containing electromagnetic radiation and a clock which controls the opening of a shutter which covers a hole made in one of the walls of the box. The shutter uncovers the hole for a time Δt which can be chosen arbitrarily. During the opening, we are to suppose that a photon, from among those inside the box, escapes through the hole. In this way a wave of limited spatial extension has been created, following the explanation given above. In order to challenge the indeterminacy relation between time and energy, it is necessary to find a way to determine with adequate precision the energy that the photon has brought with it. At this point, Einstein turns to his celebrated relation between mass and energy of special relativity: $E = mc^2$. From this it follows that knowledge of the mass of an object provides a precise indication about its energy.&lt;br&gt;&#xA;  --&lt;a href=&quot;http://en.wikipedia.org/wiki/Bohr%E2%80%93Einstein_debates#Einstein.27s_second_criticism&quot; rel=&quot;nofollow&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bohr's response was quite surprising: there was uncertainty in the time because the clock changed position in a gravitational field and thus it's rate could not be measured precisely.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bohr showed that [...] the box would have to be suspended on a spring in the middle of a gravitational field. [...] After the release of a photon, weights could be added to the box to restore it to its original position and this would allow us to determine the weight. [...] The inevitable uncertainty of the position of the box translates into an uncertainty in the position of the pointer and of the determination of weight and therefore of energy. On the other hand, since the system is immersed in a gravitational field which varies with the position, according to the principle of equivalence the uncertainty in the position of the clock implies an uncertainty with respect to its measurement of time and therefore of the value of the interval Δt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How can Bohr invoke a General Relativity concept when Quantum Mechanics is notoriously incompatible with it? Shouldn't HUP hold up with only the support of (relativistic) quantum mechanics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarifying a bit what my doubt is/was: I thought that HUP was intrinsic to QM, a derived principle from operator non-commutability. QM shouldn't need GR concepts to be self consistent. In other words - if GR did not exist, relativistic QM would be a perfectly happy theory. I was surprised it's not the case.&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-09T09:25:02.783" LastActivityDate="2011-04-09T09:25:02.783" Title="Einstein's box - unclear about Bohr's retort" Tags="&lt;quantum-mechanics&gt;&lt;history&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="8296" PostTypeId="2" ParentId="8282" CreationDate="2011-04-09T00:52:49.037" Score="3" ViewCount="" Body="&lt;p&gt;There are two physical processes that are important in a microwave oven. First is &lt;a href=&quot;http://en.wikipedia.org/wiki/Dielectric_heating&quot; rel=&quot;nofollow&quot;&gt;dielectric heating&lt;/a&gt;: basically, the oven sets up an electromagnetic wave inside itself, and certain molecules in the food absorb energy from the wave. Specifically, an absorber molecule needs a nonzero &lt;a href=&quot;http://en.wikipedia.org/wiki/Electric_dipole_moment&quot; rel=&quot;nofollow&quot;&gt;electric dipole moment&lt;/a&gt;, and it has to be free to rotate. Liquid water is a very efficient microwave absorber, fats and oils are somewhat less so, but a lot of molecules are not. So only some of the molecules in the food can actually absorb energy from the microwave oven.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other important process is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermalization&quot; rel=&quot;nofollow&quot;&gt;thermalization&lt;/a&gt;. All that means is that the hotter molecules (which in this case happen to be the absorbers) will move around and bump into other molecules, and transfer some of their energy away. This is how heat spreads from the absorbers throughout the food.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the thermalization process takes some time, typically more time than the food actually spends in the microwave oven. So when you take it out, it's unevenly heated: some of the molecules still have more energy than others. The temperature does even out after some time, but that means that the energy from the absorbers has to be shared out among all the molecules in the food (and its container), so they're not as hot as they were when they came out of the oven. This is the cooling process you've noticed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, when you heat food on a stove or in a conventional oven, the only process involved is thermalization, and every molecule in the food (and pot) is equally capable of receiving energy by that process. So when you take your food off the stove, &lt;em&gt;all&lt;/em&gt; the molecules have been completely heated up. The energy doesn't have to be &quot;shared out&quot; any more. For that reason, the food has more total thermal energy which keeps it hot longer.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-04-09T00:52:49.037" CommentCount="1" />
  <row Id="8297" PostTypeId="2" ParentId="8284" CreationDate="2011-04-09T01:35:58.547" Score="1" ViewCount="" Body="&lt;p&gt;Wikipedia puts the &lt;a href=&quot;http://en.wikipedia.org/wiki/Spitzer_Space_Telescope&quot; rel=&quot;nofollow&quot;&gt;Spitzer Cold mission imaging capability in the band 3 -- 180 micrometers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I calculate the blackbody temperature that would have a peak emission at the extreme of that band like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The frequency is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \nu = \frac{3.0 \times 10^8 \text{ m/s}}{180 \times 10^{-6} \text{ m}} = 1.7 \times 10^{12} \text{ Hz} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And from that a &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_body#Wien.27s_displacement_law&quot; rel=&quot;nofollow&quot;&gt;corresponding temperature&lt;/a&gt; of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T = \frac{1.7 \times 10^{12} \text{ Hz}}{59 \times 10^9 \text{ Hz}} = 28 \text{ K} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK. What to compare that to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=surface%20temperature%20of%20Pluto&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha puts the surface temperature of Pluto&lt;/a&gt; at around $44\text{ K}$, so Spitzer Cold should have been able to image Pluto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, &lt;a href=&quot;http://www.wolframalpha.com/input/?i=Radius%20of%20the%20orbit%20of%20Pluto&quot; rel=&quot;nofollow&quot;&gt;Wolfram also puts the radius of Pluto's orbit&lt;/a&gt; at around 40 AU. The Oort cloud objects would somehow need to be 2/3 or Pluto's temperature to have been (just barely!) visible to Spitzer Cold. This is not my specialty, but I kinda doubt it.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-04-09T01:35:58.547" CommentCount="3" />
  <row Id="8298" PostTypeId="2" ParentId="8295" CreationDate="2011-04-09T02:09:43.990" Score="3" ViewCount="" Body="&lt;p&gt;Bohr realized that the weight of the device is made by the displacement of a scale in spacetime.  The clock’s new position in the gravity field of the Earth, or any other mass, will change the clock rate by gravitational time dilation as measured from some distant point the experimenter is located.  The temporal metric term for a spherical gravity field is $1~-~2GM/rc^2$, where a displacement by some $\delta r$ means the change in the metric term is $\simeq~(GM/c^2r^2)\delta r$.  Hence the clock’s time intervals $T$ is measured to change by a factor &#xA;$$&#xD;&#xA;T~\rightarrow~T\sqrt{(1~-~2GM/c^2)\delta&#xD;&#xA;r/r^2}~\simeq~T(1~-~GM\delta r/r^2c^2),&#xD;&#xA;$$&#xA;so the clock appears to tick slower.  This changes the time span the clock keeps the door on the box open to release a photon. Assume that the uncertainty in the momentum is given by the $\Delta p~\simeq~\hbar/\Delta r~&amp;lt;~Tg\Delta m$, where $g~=~GM/r^2$. Similarly the uncertainty in time is found as $\Delta T~=~(Tg/c^2)\delta r$.  From this $\Delta T~&amp;gt;~\hbar/\Delta mc^2$ is obtained and the Heisenberg uncertainty relation $\Delta T\Delta E~&amp;gt;~\hbar$.  This demands a Fourier transformation between position and momentum, as well as time and energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This argument by Bohr is one of those things which I find myself re-reading.  This argument by Bohr is in my opinion on of these spectacular brilliant events in physics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This holds in some part to the quantum level with gravity, even if we do not fully understand quantum gravity.  Consider the clock in Einstein’s box as a blackhole with mass $m$.  The quantum periodicity of this blackhole is given by some multiple of Planck masses.  For a blackhole of integer number $n$ of Planck masses the time it takes a photon to travel across the event horizon is $t~\sim~Gm/c^3$ $=~nT_p$, which are considered as the time intervals of the clock. The uncertainty in time the door to the box remains open is &#xA;$$&#xD;&#xA;\Delta T~\simeq~Tg/c(\delta r~-~GM/c^2),&#xD;&#xA;$$&#xA;as measured by a distant observer.  Similary the change in the energy is given by $E_2/E_1~=$ $\sqrt{(1~-~2M/r_1)/(1~-~2M/r_2)}$, which gives an energy uncertainty of &#xA;$$&#xD;&#xA;\Delta E~\simeq~(\hbar/T_1)g/c^2(\delta&#xD;&#xA;r~-~GM/c^2)^{-1}.&#xD;&#xA;$$&#xA;Consequently the Heisenberg uncertainty principle still holds $\Delta E\Delta T~\simeq~\hbar$.  Thus general relativity beyond the Newtonian limit preserves the Heisenberg uncertainty principle.  It is interesting to note in the Newtonian limit this leads to a spread of frequencies $\Delta\omega~\simeq~\sqrt{c^5/G\hbar}$, which is the Planck frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The uncertainty in the $\Delta E~\simeq~\hbar/\Delta t$ does have a funny situation, where if the energy is $\Delta E$ is larger than the Planck mass there is the occurrence of an event horizon.  The horizon has a radius $R~\simeq~2G\Delta E/c^4$, which is the uncertainty in the radial position $R~=~\Delta r$ associated with the energy fluctuation.  Putting this together with the Planckian uncertainty in the Einstein box we then have&#xA;$$&#xD;&#xA;\Delta r\Delta t~\simeq~\frac{2G\hbar}{c^4}~=~{\ell}^2_{Planck}/c.&#xD;&#xA;$$&#xA;So this argument can be pushed to understand the nature of noncommutative coordinates in quantum gravity.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-04-09T02:09:43.990" CommentCount="1" />
  <row Id="8299" PostTypeId="2" ParentId="8237" CreationDate="2011-04-09T02:32:55.217" Score="-1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;https://www.llnl.gov/str/December05/gifs/vanbreugel3.jpg&quot; rel=&quot;nofollow&quot;&gt;Diagram&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.llnl.gov/str/December05/vanBreugel.html&quot; rel=&quot;nofollow&quot;&gt;Link to diagram&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3014" LastActivityDate="2011-04-09T02:32:55.217" CommentCount="1" />
  <row Id="8301" PostTypeId="1" AcceptedAnswerId="8318" CreationDate="2011-04-09T03:18:36.923" Score="8" ViewCount="208" Body="&lt;p&gt;Some scenarios describing the fate of the matter vs dark energy tug of war on the universe involve the acceleration of the universe increasing to the point that it ends up ripping apart even atoms. This is called the &lt;a href=&quot;http://hubblesite.org/hubble_discoveries/dark_energy/de-fate_of_the_universe.php&quot; rel=&quot;nofollow&quot;&gt;Big Rip&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes no sense to me.  It looks like all of these GR models of the universe assume it has a uniform isotropic distribution of matter and energy.  This works great at long scales, but is also clearly wrong even at the length scale of the separation of galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density of atomic nuclei remains the same even though the universe is expanding even as we speak.  I don't think the &quot;scale&quot; in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann_equations&quot; rel=&quot;nofollow&quot;&gt;Friedmann equation&lt;/a&gt; can be interpretted so literally.  Or maybe said better, it's got to break down when the cosmological horizon distance gets to the scale where the isotropic assumption breaks down, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can scientists be claiming that run away expansion will rip apart atoms?&lt;/p&gt;&#xA;" OwnerUserId="2655" LastActivityDate="2011-04-09T14:46:56.743" Title="Can &quot;big rip&quot; rip apart an atomic nucleus?" Tags="&lt;general-relativity&gt;&lt;cosmology&gt;&lt;cosmological-inflation&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="8302" PostTypeId="2" ParentId="8295" CreationDate="2011-04-09T05:03:08.603" Score="3" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;How can Bohr invoke a General Relativity concept when Quantum Mechanics is notoriously incompatible with it?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You may have misheard, Sklivvz. General relativity is perfectly compatible with quantum mechanics. If you want the full and completely accurate theory that answers questions that depend on both GR and QM, in any regime, it is called string theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But obviously, you don't need the sophisticated cannon of string theory to answer these Bohr-Einstein questions. String theory is only needed when the distances are as short as the Planck length, $10^{-35}$ meters, or energies are huge, and so on. Whenever you deal with ordinary distance scales, semiclassical GR is enough - a simple quantization of general relativity where one simply neglects all loops and other effects that are insanely small. And indeed, string theory does confirm (and any other hypothetical consistent theory would confirm) that those effects are small, suppressed by extra powers of $G$, $h$, or $1/c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in this Bohr-Einstein case, you don't even need semiclassical general relativity. You don't really need to quantize GR at all. This is just about simple quantum mechanics in a pre-existing spacetime, and Bohr's correct answer to Einstein is just a simple comment about the spacetime geometry. The extreme phenomena that make it hard to unify QM and GR surely play no detectable role in this experiment. They don't even play much role in &quot;quantum relativistic&quot; phenomena such as the Hawking radiation: all of their macroscopic properties may be calculated with a huge accuracy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Shouldn't HUP hold up with only the support of (relativistic) quantum mechanics?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nope. The Heisenberg uncertainty principle is a principle that holds for all phenomena in the Universe. Moreover, it's a bit confusing why you wrote &quot;only&quot; in the context of relativistic quantum mechanics - relativistic quantum mechanics is the most universally valid framework to describe the reality because it includes both the quantum and relativistic &quot;refinements&quot; of physics (assuming that we do the relativistic quantum physics right - with quantum field theory and/or string theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Einstein, in his claim that he could violate the uncertainty principle, used gravity, so it's not surprising that the error in Einstein's argument - one that Bohr has pointed out - has something to do with gravity, too. Because we talk about the uncertainty principle, you surely didn't want to say that we should be able to describe it purely in non-quantum language. If you wanted to say that &lt;em&gt;non-relativistic&lt;/em&gt; quantum mechanics should be enough to prove Einstein wrong, then it's not true because photons used in the experiment are &quot;quantum relativistic&quot; particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the mass of a photon that he wants to measure is $m=E/c^2 = hf/c^2$. Because photons and electromagnetic waves in any description are produced at a finite frequency $f$, we cannot let $c$ go to infininity because the change of the mass $m=hf/c^2$ that Einstein proposed to measure (by a scale) would vanish so he couldn't determine the change of the mass - and he wanted to calculate the change of the energy from the mass so he wouldn't be able to determine the energy, either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Einstein's strategy to show that $E,t$ may be determined simultaneously uses effects that depend on the finiteness of both $h$ and $c$, so he is using relativistic quantum phenomena. To get the right answer or the right predictions what will happen and what accuracy can be achieved, he should do so consistently and take into account &lt;em&gt;all&lt;/em&gt; other relevant phenomena &quot;of the same order&quot; that also depend both on relativity and the quanta. The time dilation, as pointed out by Bohr, is one such effect that Einstein neglected, and if it is included, not surprisingly, HUP gets confirmed again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such proofs are somewhat redundant in theories that we know. Whenever we construct a quantum theory, whether the gravity is described relativistically or not, with fields or not, the uncertainty principle is automatically incorporated into the theory - by the canonical commutators - so it is never possible to find a measurement for which the theory predicts that HUP fails. This conclusion is safer than details of Bohr's particular &quot;loophole&quot; - I am not going to claim that Bohr's observation is the only (or the main) effect that Einstein neglected. There are probably many more.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-09T06:13:53.523" LastActivityDate="2011-04-09T06:13:53.523" CommentCount="1" />
  <row Id="8303" PostTypeId="1" CreationDate="2011-04-09T05:19:54.040" Score="5" ViewCount="245" Body="&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/7358/causality-and-quantum-uncertainty/7361#7361&quot;&gt;This answer of mine&lt;/a&gt; has been strongly criticized on the ground that it is no more than a philosophical blabbering. Well, it may well be. But people seem to be of the opinion that HUP alone does not ensure randomness and you need Bell's theorem and other features for the randomness in QM. However, I still believe it is the HUP which is all one needs to appreciate the probabilistic feature of QM. Bell's theorem or other such results reinforces this probabilistic view only. I am very much curious to know the right answer.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-09T05:29:34.013" LastActivityDate="2011-04-10T16:00:29.733" Title="Does the HUP alone ensure the randomness in QT?" Tags="&lt;quantum-mechanics&gt;&lt;uncertainty-principle&gt;" AnswerCount="6" CommentCount="12" />
  <row Id="8304" PostTypeId="2" ParentId="8291" CreationDate="2011-04-09T05:21:26.067" Score="1" ViewCount="" Body="&lt;p&gt;One of the main ways black holes are noticed is by looking at a solar system where the star appears to move as though it were a binary star system (e.i. two stars) when only one is seen. In these situations, depending on the distances, the black hole &quot;feeds&quot; off the original star, and a stream of the stellar plasma is slowly pealed off the star into the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This matter can sometimes form a very vivid accretion disk, that can be observed using telescopes (see Herbig–Haro object). This process can take a very long time, on the order of millions of years. However, of course, a rough black hole could enter a star system head on and collide right with the sun and &quot;suck it up,&quot; which would happen rather quickly (to an outside observer).&lt;/p&gt;&#xA;" OwnerUserId="3018" LastActivityDate="2011-04-09T05:21:26.067" CommentCount="1" />
  <row Id="8305" PostTypeId="2" ParentId="8303" CreationDate="2011-04-09T05:53:02.590" Score="2" ViewCount="" Body="&lt;p&gt;It's very strange for someone to say that &quot;Bell's theorem ensures something in quantum mechanics&quot;. Bell's theorem is a theorem - something that can be mathematically proved to hold given the assumptions. It's valid in the same sense as $1+1=2$. Is $1+1=2$ needed for something in physics? Maybe - but the question clearly makes no sense. Mathematics is always valid in physics - and everywhere else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, even the assumptions of Bell's theorem surely can't be &quot;necessary building blocks&quot; for some results in quantum mechanics because Bell's theorem is not a theorem about quantum mechanics at all. It is a theorem (an inequality) about local realist theories - exactly the kind of theories that quantum mechanics is &lt;em&gt;not&lt;/em&gt;. Whether someone needs $1+1=2$ doesn't matter because this fact is imposed upon him, anyway. Any proof may be modified so that $1+1=2$ is needed and any proof may be modified so that $1+1=2$ is not needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even if one ignores the comment about &quot;Bell's theorem and other such results&quot; that can't possibly have anything to do with the question, it's nontrivial to make the question precise. The uncertainty principle is normally formulated as a part of quantum mechanics - we say that $\Delta x$ and $\Delta p$ can't have well-defined sharp values at the same moment. What it means for them not to have sharp values? Well, obviously, it means that one measures their values with an error margin, and the fluctuations or choice of the measured value from the allowed distribution has to be random.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were not random, there would have to be another quantity for which one should do the same discussion. Again, if the uncertainty principle applied to this hidden variables (and a complementary one), it would imply that its values have to be random. Do you allow me to assume that the HUP holds for whatever variables we have? If you do, obviously, there has to be random things in the Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even the term &quot;random&quot; is too ill-defined. Do you require some special vanishing of correlations etc.? If you do, shouldn't you describe what those requirements are?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I don't think it's possible to fully answer vague questions of this kind. I would say a related comment that quantum mechanics - with its random character - is the only mathematically possible and self-consistent framework that is compatible with certain basic observations of the quantum phenomena. The outcomes in quantum mechanics take place randomly, with probabilities and probability distributions that can be calculated from the squared probability amplitudes, and all other attempts to modify the basic framework of quantum mechanics have been ruled out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's so, and it is so, there's really no point in trying to decompose the postulates of quantum mechanics into pieces because the pieces only combine into a viable theoretical structure, able to explain the behavior of important worlds such as ours, when all these postulates are taken seriously at the same moment.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-09T05:53:02.590" />
  <row Id="8306" PostTypeId="2" ParentId="8289" CreationDate="2011-04-09T07:35:29.233" Score="4" ViewCount="" Body="&lt;p&gt;Well, surely you can compute it using matrix operations. But it won't be very natural. Let me instead provide you with a very similar solution (based on a similar matrix) that you'll hopefully find useful. It's not new at all (Kirchhoff, 1847) but I think it's not very well known. I first learned about it in this &lt;a href=&quot;http://rmp.aps.org/abstract/RMP/v54/i1/p235_1&quot; rel=&quot;nofollow&quot;&gt;Wu's review paper of Potts model&lt;/a&gt;, p. 252. Let me reproduce the main points of the derivation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write $U_i$ for the potential at the site $i$ and $I_i$ for the external current flowing into the site $i$. Then continuity equation gives us&#xA;$I_i = \sum_{j \neq i} C_{ij} (U_i - U_j)$&#xA;which can be rewritten as $I_i = \sum_j A_{ij} U_j$ with&#xA;$$A_{ij} = \begin{cases} \sum_{k \neq i} C_{ij} &amp;amp; i = j \\ -C_{ij} &amp;amp; i \neq j \end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now one can proceed directly to solve for $U_i$, given external current flows. But it turns out that thanks to special properties of the matrix $A$ (notice that sum entries of each row gives zero) more can be said. It turns out (read the paper for details) one can express equivalent resistance between points $k$ and $l$ as&#xA;$$R_{kl} = {{\rm det}A^{(kl)} \over {\rm det} A^{(l)}}$$ where the indexed matrices are obtained by removing the said rows and columns from the matrix $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last remark (not related directly to your question but it would be shame not to mention it now) is that those determinants can be interpreted naturally as spanning tree polynomials in $C_{ij}$ on the given graph $G$ (with or without $(kl)$ edge) and this in turn can be computed directly from the partition function of $q \to 0$ limit of the $q$-state Potts model on the said graph $G$ with weights on the edges related to their resistances.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-04-09T07:35:29.233" CommentCount="1" />
  <row Id="8307" PostTypeId="1" AcceptedAnswerId="8309" CreationDate="2011-04-09T07:47:15.747" Score="7" ViewCount="181" Body="&lt;p&gt;If we have a perfect Schwarzschild  black hole (uncharged and stationary), and we &quot;perturb&quot; the black hole by dropping in a some small object. For simplicity &quot;dropping&quot; means sending the object on straight inward trajectory near the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly the falling object will cause some small (time dependent) curvature of space due to its mass and trajectory, and in particular, once it passes the even horizon, the object will cause some perturbation to the null surface (horizon) surrounding the singularity (intuitively I would think they would resemble waves or ripples).  Analogously to how a pebble dropped in a pond causes ripples along the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to calculate (i.e. approximate numerically) the effect of such a perturbation of the metric surrounding the black hole?, and specifically to calculate the &quot;wobbling&quot; of the null surface as a result of the perturbation,maybe something analogous to quantum perturbation theory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or more broadly, does anyone know of any papers or relevant articles about a problem such as this?&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-09T12:19:41.897" LastActivityDate="2011-04-09T12:19:41.897" Title="Perturbation of a Schwarzschild Black Hole" Tags="&lt;general-relativity&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="8308" PostTypeId="2" ParentId="8303" CreationDate="2011-04-09T07:56:19.380" Score="0" ViewCount="" Body="&lt;p&gt;Well, you misinterpreted what I (and others) said at least in two important ways.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Bell's theorem surely isn't responsible for randomness in QM. That's because it doesn't actually tell you anything about QM itself, only about other theories trying to reproduce the same results that QM (and nature) produces. The reason I mentioned it is that it (severely) restricts the class of non-random theories that can describe the nature. Without such a theorem one might hope (and people still do) that it is possible to construct a &lt;em&gt;deterministic&lt;/em&gt; framework that could be compatible with observations. So HUP certainly doesn't imply intrinsic randomness. You need further work to establish that &lt;em&gt;no&lt;/em&gt; viable theory (and not just QM) is deterministic. Measurement of violation of Bell's inequalities is what does it (at least if one assumes locality).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;QM is based on lots of principles. HUP is fundamental (and is built-in by including non-commutative operators into the framework) but no less fundamental than other postulates. Trying to isolate one particular feature of a theory doesn't always make sense. You could try to obtain deterministic QM by removing HUP but that essentially means letting $\hbar \to 0$ and obtaining classical physics, thereby losing all the other special effects of QM.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In other words, your statement &quot;HUP which is all one needs to appreciate the probabilistic feature of QM&quot; couldn't be more far removed from reality. To appreciate this probabilistic aspect, one needs to master the mathematical formalism of QM, the way it connects to experiment and the way measurements are interpreted. HUP is only a small part in it and actually, the one thing you almost never care for as it is built-in into the theory from the start.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-04-09T07:56:19.380" CommentCount="4" />
  <row Id="8309" PostTypeId="2" ParentId="8307" CreationDate="2011-04-09T08:06:56.113" Score="3" ViewCount="" Body="&lt;p&gt;Your intuitive picture is basically correct. If you perturb a black hole it will respond by &quot;ringing&quot;. However, due to the emission of gravitational waves and because you have to impose ingoing boundary conditions at the black hole horizon, the black hole will not ring with normal-modes, but with quasi-normal modes (QNMs), i.e., with damped oscillations. These oscillations depend on the black hole parameters (mass, charge, angular momentum), and are therefore a characteristic feature for a given black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Historically, the field of black hole perturbations was pioneered by Regge and Wheeler in the 1950ies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;p&gt;For a review article see &lt;a href=&quot;http://arxiv.org/abs/gr-qc/9909058&quot; rel=&quot;nofollow&quot;&gt;gr-qc/9909058&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the specific case of the Schwarzschild black hole there is a very nice analytical calculation of the asymptotic QNM spectrum in the limit of high damping by Lubos Motl, see &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0212096&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. See also his &lt;a href=&quot;http://arxiv.org/abs/hep-th/0301173&quot; rel=&quot;nofollow&quot;&gt;paper with Andy Neitzke&lt;/a&gt; for a generalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise usually you have to rely on numerical calculations to extract the QNMs.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-04-09T08:06:56.113" CommentCount="4" />
  <row Id="8310" PostTypeId="1" AcceptedAnswerId="8313" CreationDate="2011-04-09T08:50:13.623" Score="5" ViewCount="349" Body="&lt;p&gt;I believe I understand how tuning a radio with an analog tuner works: turning the dial physically changes the length of the antenna, which determines which broadcast wavelength will resonate in the antenna and get picked up. In contrast, what is the mechanism that makes a digital tuner work? My guess is it'll be some clever little circuit that somehow selects a resonant frequency by changing the amount of current going through it (or something like that). &lt;/p&gt;&#xA;" OwnerUserId="2966" LastActivityDate="2011-06-15T23:11:10.550" Title="How does a digital radio tuner work?" Tags="&lt;electromagnetism&gt;&lt;engineering&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="8311" PostTypeId="2" ParentId="8301" CreationDate="2011-04-09T09:42:33.963" Score="-2" ViewCount="" Body="&lt;p&gt;I looked at the link, and it is a general scenaria link, so I will concentrate on the isotropic part of your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The density of atomic nuclei remains the same even though the universe is expanding even as we speak.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Define density of a piece of metal. It is the number of atoms*(atomic mass) per cubic centimeter.&#xA;Number is constant. Consider atomic mass constant too. Cubic centimeter was defined as a cm cubed,  drawn on a piece of metal at some central bank. Now they take the wavelength of some atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe is expanding, from the time of big bang, &lt;em&gt;at each point of space&lt;/em&gt;. It means as an example that the distance between these two points :  is expanding at the rate of expansion of the universe , and the dots are expanding, and this page is expanding and ... The same is true for matter composed of atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would not  notice the expansion, even if our senses were sensitive enough to be able to see such small cosmological changes, because of our definition of distance and units in general. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the isotropy and uniformity do not have a role to play in an argument whether the big Rip, as expounded in the link you gave, can happen or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some theory has to enter, see for example the &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0508052&quot; rel=&quot;nofollow&quot;&gt;link provided by Jim Graber above&lt;/a&gt;, where results are conditional to the strength of forces.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-04-09T14:46:56.743" LastActivityDate="2011-04-09T14:46:56.743" CommentCount="15" />
  <row Id="8312" PostTypeId="1" CreationDate="2011-04-09T09:58:33.503" Score="1" ViewCount="121" Body="&lt;p&gt;According to experts, in order to become an international scientist, you have to put in 80 hour weeks for a decade. Is becoming an international scientist incompatible with a work/life balance?&lt;/p&gt;&#xA;" OwnerUserId="3019" LastActivityDate="2011-04-09T09:58:33.503" ClosedDate="2011-04-09T11:01:12.770" Title="Is a work/life balance incompatible with becoming an international scientist?" Tags="&lt;physics-careers&gt;" CommentCount="4" />
  <row Id="8313" PostTypeId="2" ParentId="8310" CreationDate="2011-04-09T11:04:22.240" Score="12" ViewCount="" Body="&lt;p&gt;In the early days of radio, the resonance of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Antenna_%28radio%29&quot; rel=&quot;nofollow&quot;&gt;antenna&lt;/a&gt; in combination with its associated inductive and capacitive properties was indeed the item which &quot;dialed in&quot; the frequency you wanted to listen to. You didn't actually change the length of the antenna, but by changing the inductor (a coil) or capacitor connected to the antenna you tuned the resonance. The output signal is an alternating voltage, and by rectifying it with a diode (called a &quot;crystal&quot; then..) you could extract a signal modulated as a varying amplitude of the carrier wave. &lt;a href=&quot;http://en.wikipedia.org/wiki/Crystal_radio_receiver&quot; rel=&quot;nofollow&quot;&gt;All this without any battery&lt;/a&gt;! :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But actually the antenna in a normal modern radio is not the component that &quot;dials in&quot; the selected broadcast frequency. The antenna circuit should indeed have a resonance within the &lt;strong&gt;band&lt;/strong&gt; of frequencies you are interested in but this wide-band signal is then &lt;em&gt;mixed&lt;/em&gt; with an internally generated sinusodial signal in the radio in an analog component, this subtracts the frequencies and lets the rest of the radio operate on a much easily handled frequency band (called the intermediate frequency). It is in the mixer you &lt;em&gt;tune&lt;/em&gt; the reception in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Superheterodyne_receiver&quot; rel=&quot;nofollow&quot;&gt;modern superheterodyne radio receiver&lt;/a&gt;. It is much easier to synthesize an exact mixing frequency to tune with than to change the resonance of the antenna circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest is not really physics, but the difference between an analog and a digital radio comes in the circuits after this and basically an analog radio extracts a modulation from the intermediate frequency which is amplified and sent to the speakers or radio output. In a digital radio, the signal represents a digital version of the audio, just like a WAV or MP3-file on a computer is a digital representation which can be turned back into an analog signal you can send to a speaker. The benefit of this is that the digital signal requires (potentially) less bandwidth in the air so you can fit more signals in the same &quot;airspace&quot; and that the digital signal can be less susceptible to noise. I write &quot;can&quot;, because unfortunately many commercial digital radio/TV stations don't do this to improve the viewing or listening quality but just to fit in more content.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me reiterate that in a &quot;digital&quot; radio, the component that selects the reception frequency is still analog but the mixing (tuning) frequency is digitally controlled and selected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a very interesting thing called &lt;a href=&quot;http://en.wikipedia.org/wiki/Software-defined_radio&quot; rel=&quot;nofollow&quot;&gt;Software Defined Radio&lt;/a&gt;, SDR, which is the principle where the intermediate frequency (or in some cases the antenna frequency directly) is turned into a digital signal and demodulated by a signal processor which is completely software-upgradeable. Since it is much easier to program new software than to solder electronic components around, this created large interest in the radio hobby community where you can completely change the properties of a radio receiver just by downloading someone else's software from the net or write a new one yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you include SDR, and apply it without any intermediate frequency (take the antenna directly to an analog/digital converter and into a signal processor), you do indeed have a purely software-way of tuning your source like you ask for, although this is not how the most common digital radios work currently.&lt;/p&gt;&#xA;" OwnerUserId="2936" LastEditorUserId="2936" LastEditorDisplayName="" LastEditDate="2011-04-09T11:09:58.393" LastActivityDate="2011-04-09T11:09:58.393" CommentCount="1" />
  <row Id="8314" PostTypeId="1" AcceptedAnswerId="8330" CreationDate="2011-04-09T11:14:20.340" Score="5" ViewCount="305" Body="&lt;p&gt;From Surely you must be joking, Mr Feynman.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You blast off in a rocket which has a&#xA;  clock on board, and there's a clock on&#xA;  the ground. The idea is that you have&#xA;  to be back when the clock on the&#xA;  ground says one hour has passed. Now&#xA;  you want it so that when you come&#xA;  back, your clock is as far ahead as&#xA;  possible. According to Einstein, if&#xA;  you go very high, your clock will go&#xA;  faster, because the higher something&#xA;  is in a gravitational field, the&#xA;  faster its clock goes. But if you try&#xA;  to go too high, since you've only got&#xA;  an hour, you have to go so fast to get&#xA;  there that the speed slows your clock&#xA;  down. So you can't go too high. The&#xA;  question is, exactly what program of&#xA;  speed and height should you make so&#xA;  that you get the maximum time on your&#xA;  clock?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This assistant of Einstein worked on&#xA;  it for quite a bit before he realized&#xA;  that the answer is the real motion of&#xA;  matter. If you shoot something up in a&#xA;  normal way, so that the time it takes&#xA;  the shell to go up and come down is an&#xA;  hour, that's the correct motion. It's&#xA;  the fundamental principle of&#xA;  Einstein's gravity--that is, what's&#xA;  called the &quot;proper time&quot; is at a&#xA;  maximum for the actual curve.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How is this result proven?&lt;/p&gt;&#xA;" OwnerUserId="119" LastActivityDate="2011-04-13T01:43:47.207" Title="Maximum time difference between clocks in a gravity field" Tags="&lt;general-relativity&gt;&lt;gravity&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="8315" PostTypeId="2" ParentId="8291" CreationDate="2011-04-09T11:20:00.037" Score="-2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;By &quot;eating&quot; I mean the time it takes a&#xA;  star to go throug&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your question is very unclear, apparently You lack knowledge &#xA;on Newtons mechanics for celestial bodys. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Benjamin wrote about the usual and &quot;slow&quot; process.&lt;br&gt;&#xA;In case the star hits the black hole &quot;head on&quot; &#xA;(very unlikely, but not impossible) the star would be &#xA;swallowed in seconds or milliseconds. Only a question of &#xA;relative velocity.   &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-09T11:26:56.190" LastActivityDate="2011-04-09T11:26:56.190" />
  <row Id="8316" PostTypeId="2" ParentId="8314" CreationDate="2011-04-09T11:58:39.123" Score="6" ViewCount="" Body="&lt;p&gt;As Raskolnikov says, it's the principle of least action (more precisely &lt;em&gt;extremal&lt;/em&gt; action but let's put that off till later). Consider the task of finding the shortest path between two points in Euclidean space. We call such a special path &lt;strong&gt;geodesic&lt;/strong&gt; and it's obvious that here it is just plain old line segment between the two points. But to prove this statement is not so trivial (one needs calculus of variations at hand), so let's concentrate on just the intuitive picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try to deform the line a little to make the total path shorter, you'll find it's not possible. Reason is that infinitesimally, each such deformation will correspond to a triangle with longest side $c$ being the part of the line and $a$ and $b$ lines comprising the perturbed path. As per basic Euclidean geometry we have $c &amp;lt; a + b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Minkowski space-time, the picture is very similar. Straight lines are again geodesics although now they extremize &lt;em&gt;space-time intervals&lt;/em&gt; or equivalenty &lt;em&gt;proper time&lt;/em&gt; along the path (at least if the particle is massive). Again, if you repeat the above argument you will obtain an inequality but with the sign reversed because time component has different signature than the space components. That means that time will be maximized along the geodesic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above can be generalized to curved spaces and space-times. E.g. on a sphere you'll find the shortest path between two points are parts of great circles. And in fact, our argument will work too because infinitesimally sphere looks like a plane (as I am sure you know from your day to day experience living on the surface of Earth) so our arguments holds. And similarly for space-times and proper times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A note on the word &lt;em&gt;extremal&lt;/em&gt; is in order. If you inspect the sketches of the proofs I made, you'll see that we always make only local (infinitesimal) perturbations around the extremal paths to see if there is something shorter/longer around. That means that we only find local extrema. Also, these paths need not exist or be unique between every pair of points (consider the sphere again and all the geodesics between north and south poles). This is good to keep in mind but I won't delve into general mathematical theory as it's messy and not needed usually.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-04-09T12:10:45.357" LastActivityDate="2011-04-09T12:10:45.357" CommentCount="3" />
  <row Id="8318" PostTypeId="2" ParentId="8301" CreationDate="2011-04-09T12:31:16.297" Score="6" ViewCount="" Body="&lt;p&gt;Look at this paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/0508052&quot; rel=&quot;nofollow&quot;&gt;&quot;In an expanding universe, what doesn't expand?&quot; by Richard Price&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The expansion of the universe is often viewed as a uniform stretching of space that would affect compact objects, atoms and stars, as well as the separation of galaxies. One usually hears that bound systems do not take part in the general expansion, but a much more subtle question is whether bound systems expand partially. In this paper, a very definitive answer is given for a very simple system: a classical &quot;atom&quot; bound by electrical attraction. With a mathemical description appropriate for undergraduate physics majors, we show that this bound system either completely follows the cosmological expansion, or -- after initial transients -- completely ignores it. This &quot;all or nothing&quot; behavior can be understood with techniques of junior-level mechanics. Lastly, the simple description is shown to be a justifiable approximation of the relativistically correct formulation of the problem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Short summary&lt;br&gt;&#xA;All or nothing behavior:  If the binding force is greater, the object does not expand significantly. If the cosmological costant is stronger, the object expands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, as the cosmological constant grows to infinity (or minus infinity, according to the usual convention), more and more strongly bound systems are ripped apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this analysis assumes the binding force (i.e. gravity or electrodynamics) decreases with distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strong force, however, increases with distance.&lt;br&gt;&#xA;So the cosmological constant versus the strong force is still an interesting question.&lt;/p&gt;&#xA;" OwnerUserId="1172" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-09T13:25:40.547" LastActivityDate="2011-04-09T13:25:40.547" CommentCount="5" />
  <row Id="8319" PostTypeId="1" AcceptedAnswerId="8321" CreationDate="2011-04-09T14:06:12.183" Score="1" ViewCount="138" Body="&lt;p&gt;Let's imagine this in an situation where a person jumps down from a smaller building, a garage for example. The exact numbers aren't really important here, since I'm not aiming for an exact answer, but we can call it 2 metres in height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To have both situations as similar to each other as possible, let's define them like this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In first &quot;try&quot;, the person runs on the building, the jumps down. He is not jumping up to increase his jumping distance, he simply 'drops down', but considering that he's running with a certain speed forward, he will be falling in about 45 degrees angle. He drops at the distance &lt;strong&gt;X&lt;/strong&gt; from the building.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In second &quot;try&quot;, the person runs on the building, but does a forward somersault during the flight. As before, he doesn't necessarily jumps up, as the building height is already enough for him to complete the sault and as with the previous situation, he drops at the same distanes &lt;strong&gt;X&lt;/strong&gt; from the building.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In which scenario does this person hits the ground at the lower fall speed?&lt;/p&gt;&#xA;" OwnerUserId="2834" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-09T23:33:55.783" LastActivityDate="2011-04-09T23:33:55.783" Title="Is the speed of a person falling down different if doing a forward somersault during the fall?" Tags="&lt;kinematics&gt;&lt;free-fall&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="8320" PostTypeId="2" ParentId="8291" CreationDate="2011-04-09T14:14:49.863" Score="1" ViewCount="" Body="&lt;p&gt;If you are asking about a star being &quot;eaten&quot; by a supermassive black hole at the center of a galaxy, as is being &#xA;in another question here on Physics SE due to a recent event covered in the popular press (GRB 110328A), &#xA;this relatively authoritative paper says that they stay bright for one month to one year.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This long period is due to the orbiting of the torn up parts of the star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probing Quiescent Massive Black Holes:&lt;br&gt;&#xA;Insights from Tidal Disruption Events&lt;br&gt;&#xA;A Whitepaper Submitted to the Decadal Survey Committee&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0903/0903.1107v1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/arxiv/pdf/0903/0903.1107v1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-04-09T14:14:49.863" CommentCount="1" />
  <row Id="8321" PostTypeId="2" ParentId="8319" CreationDate="2011-04-09T14:45:38.047" Score="5" ViewCount="" Body="&lt;p&gt;The person hits the ground at the same speed in both scenarios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you're in the air, you fall towards the ground with a constant acceleration of about 10 m/s^2.  Everything falls the same way - rocks, cannonballs, people - regardless of size or somersaulting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be some small effects from air resistance, but not enough to be noticeable.  The sommersaulter may also land with a different orientation so that his center of mass falls a further distance.  Then he could hit the ground with slightly higher speed.  But basically, everything falls the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out &lt;a href=&quot;http://www.youtube.com/watch?v=5C5_dOEyAfk&quot; rel=&quot;nofollow&quot;&gt;this Youtube video&lt;/a&gt; of a feather and hammer falling simultaneously on the moon, which shows that heavy and light objects fall the same way absent air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also see &lt;a href=&quot;http://www.youtube.com/watch?v=D9wQVIEdKh8&quot; rel=&quot;nofollow&quot;&gt;this Youtube video&lt;/a&gt; of the TV show Mythbusters dropped a bullet straight down and firing one from a gun.  The bullets fall in the same amount of time, regardless of their horizontal speed.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-04-09T14:45:38.047" />
  <row Id="8322" PostTypeId="2" ParentId="8319" CreationDate="2011-04-09T15:48:07.910" Score="0" ViewCount="" Body="&lt;p&gt;There will be points on a rotating body moving faster and slower than the CoM rate, however I assume your gymnast has pulled out of the spin and lands on their feet, then they land at the same rate, as already mentioned. &lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-04-09T15:48:07.910" CommentCount="1" />
  <row Id="8323" PostTypeId="2" ParentId="8303" CreationDate="2011-04-09T15:52:01.347" Score="0" ViewCount="" Body="&lt;p&gt;The title question is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the HUP alone ensure the randomness of QM?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I claim that the answer to this question is:No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The HUP has the basic forms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta E.\Delta t \ge \hbar$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta x.\Delta p \ge \hbar$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore quantum mechanics books prove that for non-commuting observables:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[P,Q] \neq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta P.\Delta Q \ge \hbar$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the HUP is proven generally as a consequence of the non-commutativity of the observables. In order to understand &lt;em&gt;why&lt;/em&gt; there are non-commuting observables in QM takes us to the rest of the postulates of QM and so explains why the other answers say that HUP is a consequence of QM &lt;em&gt;in toto&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there is more to the topic of &quot;QM randomness&quot; than this, and we have not yet responded to your remarks about Bell's Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first point to note is that in classical engineering, there is a concept of time-domain and frequency domain (for a wave) and the associated law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta \omega.\Delta t \ge 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This law is a consequence of the Fourier transform between these domains and therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{-i\omega t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the HUP formula is more widespread than just quantum mechanics. Of course if one puts &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E=\hbar \omega$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then one obtains&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta E.\Delta t \ge \hbar$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;once again! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So where does quantum randomness (assuming for the moment, that that is the correct term) come from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One published book that makes this point explicitly is Roger Penrose &quot;The Emperor's New Mind&quot;, p297&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[In quantum collapse..] these &lt;em&gt;real&lt;/em&gt; numbers play a role as actual &lt;em&gt;probabilities&lt;/em&gt; for the alternatives in question. Only &lt;em&gt;one&lt;/em&gt; of the alternatives survives into the actuality of physical experience.. &lt;em&gt;It is here, and only here, that the non-determinism of quantum theory makes its entry.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The italics is mine (and this is where Penrose introduces his &lt;strong&gt;R&lt;/strong&gt; definition for describing quantum wave function reduction). Thus if you are familiar with quantum mechanics then this is the reduction postulate (in words).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have several different concepts in play here: HUP, QM Postulates, Bell's Theorem, randomness.&lt;/p&gt;&#xA;" OwnerUserId="1460" LastEditorUserId="1460" LastEditorDisplayName="" LastEditDate="2011-04-10T16:00:29.733" LastActivityDate="2011-04-10T16:00:29.733" CommentCount="5" />
  <row Id="8324" PostTypeId="2" ParentId="8132" CreationDate="2011-04-09T16:17:22.810" Score="5" ViewCount="" Body="&lt;p&gt;Suppose you treat scattering of a particle in a central potential. This means that the Hamiltonian $H$ commutes with the angular momentum operators $L^2$ and $L_z$. Hence, you can find simultaneous eigenfunctions $\psi_{k,l,m}$. &#xA;You might know, for example from the solution of the hydrogen atom, that these functions can be expressed in terms of the spherical harmonics:&#xA;$$\psi_{k,l,m}(x) = R_{k,l}(r) \Psi_m^l(\theta, \varphi)$$&#xA;where the radial part satisfies&#xA;$$\frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dR_{k,l}}{dr}\right)&#xD;&#xA;+\left(n^2 - U(r) - \frac{l(l+1)}{r^2}\right) R_{k,l} = 0$$&#xA;with $U(r) = 2m/\hbar^2 V(r)$, your central potential, and $k$ is the particle's wavenumber, i.e., $E = \frac{\hbar^2 k^2}{2m}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step is to look for a special case with simple solutions. This would be the &lt;strong&gt;free particle&lt;/strong&gt;, with $U(r) = 0$. Then, the radial equation is a special case of Bessel's equation. The solutions are the &lt;em&gt;spherical Bessel functions&lt;/em&gt; $j_l(kr)$ and $n_l(kr)$, where the $j_l$ are &lt;em&gt;regular&lt;/em&gt; at the origin whereas the $n_l$ are &lt;em&gt;singular&lt;/em&gt; at the origin. Hence, for a free particle, the solutions are superpositions of the $j_l$:&#xA;$$\psi(x) = \sum_{l,m} a_{l,m} j_l(kr) Y^l_m(\theta, \varphi)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we also have axial symmetry, only $m = 0$ is relevant. Then we can rewrite the spherical harmonics using Legendre polynomials. This will lead to&#xA;$$\psi(x) = \sum_{l,m} A_{l} j_l(kr) P_l(\cos \theta)$$&#xA;One important special case of such an expansion is the &lt;strong&gt;Rayleigh plane wave expansion&lt;/strong&gt;&#xA;$$e^{ikz} = \sum_l (2l+1) i^l j_l(kr) P_l(\cos\theta)$$&#xA;which we will need in the next step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We move away from free particles and consider scattering from a potential with a &lt;strong&gt;finite range&lt;/strong&gt; (this excludes Coulomb scattering!). So, $U(r) = 0$ for $r &amp;gt; a$ where $a$ is the range of the potential. For simplicity, we assume axial symmetry. Then, &lt;em&gt;outside&lt;/em&gt; the range, the solution must be again that of a free particle. But this time, the origin is not included in the range, so we can (and, in fact, must) include the $n_l(kr)$ solutions to the Bessel equations:&#xA;$$\psi(r) = \sum_l (a_l j_l(kr) + b_l n_l(kr)) P_l(\cos \theta)$$&#xA;Note how the solution for a given $l$ has two parameters $a_l$ and $b_l$. We can think of another parametrization: $a_l = A_l \cos\delta_l$ and $b_l = -A_l \sin \delta_l$. The reason for doing this becomes apparent in the next step:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spherical Bessel functions have &lt;em&gt;long range approximations&lt;/em&gt;:&#xA;$$j_l(kr) \sim \frac{\sin(kr - l\pi/2)}{kr}$$&#xA;$$n_l(kr) \sim \frac{\cos(kr - l\pi/2)}{kr}$$&#xA;which we can insert into the wavefunction to get a long range approximation. After some trigonometry, we get&#xA;$$\psi(r) \sim \sum_l \frac{A_l}{kr} \sin(kr - l\pi/2 + \delta_l) P_l(\cos \theta)&amp;#39;$$&#xA;So, this is how our wavefunction looks like for large $r$. &lt;strong&gt;But&lt;/strong&gt; we already know how it &lt;em&gt;should&lt;/em&gt; look like: If the incoming scattered particle is described as a plane wave in $z$-direction, it is related to the scattering amplitude $f$ via&#xA;$$\psi(\vec{x}) \sim e^{ikz} + f(\theta) \frac{e^{ikr}}{r}.$$&#xA;Obviously, both forms for writing down a long-range approximation for $\psi$ should give the same, so we use the Rayleigh plane wave expansion to rewrite the latter form. We also rewrite the $\sin$ function using complex exponentials. The ensuing calculations are a bit tedious, but not complicated in itself. You just insert the expansions. What we can do afterwards is comparing the coefficients in both expressions for the same terms, e.g. equation the coefficients for $e^{-ikr}P_l(\cos\theta)$ will give you&#xA;$$A_l = (2l+1)i^l e^{i\delta_l}$$&#xA;whereas equating coefficients for $e^{ikr}$ gives you&#xA;$$f(\theta) = \frac{1}{2ik} \sum_l (2l+1) \left( e^{2i\delta_l} - 1 \right) P_l(\cos \theta).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Interpretation of the Phase Shift:&lt;/strong&gt; Remember the long range limit of the wavefunction. It lead to an expression for the $l$-th radial wavefunction in the long-range of&#xA;$$u_l(r) = kr\psi_l(r) \sim A_l \sin(kr - l\pi/2 +\delta_l).$$&#xA;For a free particle, the phase shift $\delta_l$ would be $0$. One could therefore say that the phase shift measures how far the asymptotic solution of your scattering problem is displaced at the origin from the asymptotic free solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Interpretation of the Partial Wave Expansion:&lt;/strong&gt; In the literature, you will often come across terms such as $s$-wave scattering. The partial wave expansion decomposes the scattering process into the scattering of incoming waves with definite angular momentum quantum number. It explains in which way $s$-, $p$-, $d$-waves etc. are affected by the potential. For low energy scattering, only the first few $l$-quantum numbers are affected. If all but the first term are discarded, only the $s$-waves take part in the scattering process. This is an approximation that is, for example, made in the scattering of the atoms in a Bose-Einstein condensate. &lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditorDisplayName="" LastEditDate="2011-04-13T21:44:09.120" LastActivityDate="2011-04-13T21:44:09.120" CommentCount="1" />
  <row Id="8325" PostTypeId="1" CreationDate="2011-04-09T16:36:01.173" Score="-2" ViewCount="92" Body="&lt;p&gt;I need to get the x/y values of a fired projectile given the angle of the initial power, and air resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formulas I have are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_x=V_o \cos\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X=V_o \cos \theta t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_y=V_o \sin \theta -gt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Y=V_o \sin \theta t-(1/2)gt^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the initial velocity $V_o$, I did&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_o= \sqrt{2E/m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E$ = power with which the ball was fired, &#xA;$m$ = mass of ball&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I need is how to add air resistance to this equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides, are my equations correct?&lt;/p&gt;&#xA;" OwnerUserId="3023" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-09T23:25:18.010" LastActivityDate="2011-04-09T23:25:18.010" ClosedDate="2011-04-09T23:29:59.533" Title="projectile power with air resistance" Tags="&lt;homework&gt;&lt;projectile&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="8326" PostTypeId="2" ParentId="6269" CreationDate="2011-04-09T16:42:21.663" Score="0" ViewCount="" Body="&lt;p&gt;check here for more explanation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.physicstutorials.org/home/mechanics/1d-kinematics/riverboat-problems/23-projectile-motion?showall=1&quot; rel=&quot;nofollow&quot;&gt;http://www.physicstutorials.org/home/mechanics/1d-kinematics/riverboat-problems/23-projectile-motion?showall=1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3023" LastActivityDate="2011-04-09T16:42:21.663" />
  <row Id="8327" PostTypeId="1" AcceptedAnswerId="8339" CreationDate="2011-04-09T16:55:32.557" Score="2" ViewCount="107" Body="&lt;p&gt;Let's imagine a plastic container completely closed. We put alcohol inside it filling it at half. Now we make a hole to introduce 2 little Conductor one near the other so that they don't let the air pass. We make eletricity pass trough one of the conductor so that a Spark being created. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: does the air+alcohol-gas burn creating a big pressure or does the container need of a little hole to burn?&lt;/p&gt;&#xA;" OwnerUserId="3024" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-09T21:20:00.723" LastActivityDate="2011-04-09T21:20:00.723" Title="Alcohol inside a container" Tags="&lt;pressure&gt;&lt;air&gt;&lt;explosions&gt;" AnswerCount="2" />
  <row Id="8328" PostTypeId="2" ParentId="8327" CreationDate="2011-04-09T17:09:51.537" Score="1" ViewCount="" Body="&lt;p&gt;'Homework' answer - depending on the amount of air and the temperature the alcohol-air may form a ratio of alcohol-oxygen that will burn until all the oxygen is consumed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'real world' - the spark might set fire to the plastic, the alcohol may get hot and spray out of the electrode hole as a flame-thrower. = Don't try this at home. &lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-04-09T17:09:51.537" CommentCount="4" />
  <row Id="8329" PostTypeId="2" ParentId="8303" CreationDate="2011-04-09T17:57:53.183" Score="1" ViewCount="" Body="&lt;p&gt;In my opinion HUP is not a &quot;principle&quot; but a consequence of the mathematical framework of QM  - it is derived rather than &quot;postulated&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Randomness or uncertainty in measuring some variable in some state is not strictly related to the uncertainty of its canonically conjugate variable. HUP establishes some limitation on them and that's it. What I want to underline is that, say, uncertainty in momentum is determined with the given QM state itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About randomness, it is easy to understand if we remember that the information is gathered with help of photons. When the number of photons in one &quot;observation&quot; is large, their average is well determined and it is what the classical physics deals with. If the number of photons is small, the uncertainty makes an impression of a strong randomness in measuring, say, position of a body. Even the Moon position is uncertain if based on few-photon measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uncertainty in measurements is a fundamental feature of states in physics. Determinism is possible only for &quot;well-averaged&quot; measurements. Look at the Ehrenfest equations - they involve average (expectation) values. It implies many-many measurements. In other words, the classical determinism is due to its &lt;strong&gt;inclusive&lt;/strong&gt; character.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-04-09T19:07:59.813" LastActivityDate="2011-04-09T19:07:59.813" />
  <row Id="8330" PostTypeId="2" ParentId="8314" CreationDate="2011-04-09T18:11:58.977" Score="6" ViewCount="" Body="&lt;p&gt;To elaborate on Marek's (correct) answer, since it seems that math is the issue that @Casebash is having:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with an integral representing the time elapsed on the moving observer's clock in terms of the stationary observer's coordinates (I'm supressing G and c below.  Feel free to replace all $t$'s with $c\,t$'s, and all $M$'s with $\frac{GM}{c^{2}}$'s):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{equation}&#xD;&#xA;\Delta \tau = \int d\tau \sqrt{{\dot t}^{2}(1-\frac{2M}{r})- \frac{{\dot r}^{2}}{1-\frac{2M}{r}}- r^{2} {\dot \theta}^{2} - r^{2}\sin^{2}\theta {\dot \phi}^{2}}&#xD;&#xA;\end{equation}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $(t,r,\theta,\phi)$ are all considered to be functions of $\tau$ denoting the moving observer's position at time $\tau$.  Our problem amounts to looking for the path beginning at $(t_{0},R,\Theta,\Phi)$ and ending at $(t_{f},R,\Theta,\Phi)$ that maximizes this integral, subject to the constraint that the quantity under the square root will be equal to $1$ when the calculation is complete.  To make our calculations easier, note that the square root is a monotonic function over all its domain, so we might as well maximize&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{equation}&#xD;&#xA;\frac{1}{2}\int d\tau \left({\dot t}^{2}(1-\frac{2M}{r})- \frac{{\dot r}^{2}}{1-\frac{2M}{r}}- r^{2} {\dot \theta}^{2} - r^{2}\sin^{2}\theta {\dot \phi}^{2}\right)&#xD;&#xA;\end{equation}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is considerably simpler.  It would take a whole page to carefully vary (meaning that we basically, but not quite, take the derivative of the function with respect to) this with respect to the four independent functions.  So, I'm going to just give you a taste of the task by varying with respect to $\theta$.  The path of maximum &quot;moving clock&quot; time (heretofore called 'proper time') will be the one for which these variations are zero.  The other four functions follow just as easily.  The variation of this integral with respect to $\theta$ gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;\delta \Delta\tau |{}_{\theta}&amp;amp;=\int d\tau\left(-r^{2}{\dot \theta}\delta {\dot \theta}-r^{2}\sin\theta \cos \theta {\dot \phi}^{2}\delta \theta\right)\\&#xD;&#xA;0&amp;amp;=\int d \tau \left(-\frac{d}{d\tau}\left({\dot \theta} r^{2}\delta \theta\right) + \left({\ddot \theta}r^{2} + 2 r {\dot r}{\dot \theta}-r^{2}\sin\theta \cos \theta {\dot \phi}^{2}\right)\delta \theta\right)\\&#xD;&#xA;&amp;amp;=\left(\dot \theta(t_{0})r(t_{0})^{2}\delta \theta (t_{0})-\dot \theta(t_{f})r(t_{f})^{2}\delta \theta (t_{f})\right)\\&#xD;&#xA;&amp;amp;\quad + \int d\tau\; r^{2}\left({\ddot \theta} + \frac{1}{r}2{\dot r} {\dot \theta}- \sin \theta \cos \theta {\dot \phi}^{2}\right)\delta \theta&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term, that we obtained by integrating the total derivative, vanishes since the variation of $\theta$ is zero at the endpoints $t_{0}$ and $t_{f}$ (our space of states is paths beginning and ending at a fixed value of $\theta$.  If the variation will be zero, then the remaining stuff under the integral must turn out to be zero if it is going to be a minimum for an arbitrary fixed-point variation (if the integrand isn't zero, just imagine making the variation a hundred kajillion only at the point where it is nonzero--clearly this isn't an extremum).  Therefore, the $\theta$ variable of the maximum proper time path must satisfy ${\ddot \theta} + \frac{1}{r}2{\dot r} {\dot \theta}- \sin \theta \cos \theta {\dot \phi}^{2}=0$.  It turns out that this equation is exactly the same equation you get for geodesic motion of a particle, which is the path followed by an observer being acted upon only by gravity.  A simple calculation of the first integral for any other path (and the fact that it will come out higher) will then show you that this, in fact, is the maximum time path to travel.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2011-04-10T17:06:24.707" LastActivityDate="2011-04-10T17:06:24.707" CommentCount="3" />
  <row Id="8331" PostTypeId="1" AcceptedAnswerId="8335" CreationDate="2011-04-09T18:52:29.867" Score="2" ViewCount="159" Body="&lt;p&gt;This is my first question here, so I hope it's on-topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today my younger brother asked me from where does the 1 Pa  = 0.00750061683 mmHg formula for mercury barometer come. He needs a way to derive it or an academic source which can be cited. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After doing some calculations we got the formula for standard U manometer: $P=\frac{h_2}{h_1}P_0$ where $P_0$ is atmospheric pressure, $P$ is pressure being measured, $h_1$ is height is mercury column exposed to atmospheric pressure and $h_2$ is height of the column exposed to pressure being measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that in the case of barometer, the $h_2$ is exposed to vacuum and I don't know how to use that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've searched the Internet, and got countless sites which explain how a mercury column barometer works, but I was unable to find a site which explain which forces are acting there and how the number was derived. To make things even worse, none of the physics books I have access to don't have a detailed explanation.&lt;/p&gt;&#xA;" OwnerUserId="398" LastEditorUserId="398" LastEditorDisplayName="" LastEditDate="2011-04-09T19:17:25.267" LastActivityDate="2011-04-09T19:35:39.113" Title="How is formula for converting pressure from mmHg to Pa derived?" Tags="&lt;pressure&gt;" AnswerCount="1" />
  <row Id="8332" PostTypeId="2" ParentId="8325" CreationDate="2011-04-09T18:52:57.853" Score="2" ViewCount="" Body="&lt;p&gt;In order to add air resistance (drag), you have to have some specific model of air resistance in mind. Which model you choose depends on how accurate you want to be and which different fluid dynamics phenomena you want to describe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often people model the drag in the simplest possible way, as a force proportional to the velocity but in the opposite direction: $\mathbf{F} = -b \mathbf{v}$ This is known as Stokes drag and is only approximately true at low velocities (low Reynolds numbers). At higher velocities the drag force becomes proportional to $v^2$ rather than $v$, and if the velocity ever approaches the speed of sound, the situation becomes more complicated with the possibility of shocks forming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your equations all appear correct, but it will probably be difficult to generalize them to include drag, even with as simple a model as Stokes drag. Depending on what you're planning to use this for, it might be much more straightforward to integrate the equation of motion numerically. Look up &quot;numerical ODE solving&quot; or &quot;Euler method&quot; for a specific example.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-04-09T18:52:57.853" CommentCount="1" />
  <row Id="8333" PostTypeId="2" ParentId="7679" CreationDate="2011-04-09T19:12:01.087" Score="0" ViewCount="" Body="&lt;p&gt;Some remarks:&#xA;1) A 3-dimensional example is the matrix&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 2 Εo         ΑΕ                     0&#xA;&#xA;  Α*Ε          Εo                    2ΑΕ&#xA;&#xA;    0            2Α*Ε                2 Εo&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This also has two degenerate states that are not directly connected, only indirectly via the nondegenerate second state. In fact one of the degenerate states&#xA;is unshifted even with the full diagonalization&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)Diagonalising to find the eigenvalues: This is ok as a post-mortem, but kind of cheating when one wants to do the problem in perturbation theory: If one has done the&#xA;analysis to get the eigenvalues and eigenvectors of the full matrix, why do perturbation theory? Of course in the examples shown, it's actually easier  and faster to do the problem exactly than to do &quot;post-mortem&quot; analysis on what did not work and why&#xA;In other words the problem is: How do you go about it if you do absolutely want to use pert theory without first finding eigenvalues and eigenvectors of the full  matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) What one can easily see is that because the degenerate states are not connected by V or H' matrix elements, one needs to  involve states outside the degenerate block. I am not sure what the second order propagator term  solves in this case-the issue is not the operator, but the states used. Anyway, the point is that perturbation theory will not do  here. In my example, if one writes&#xA;|1'&gt;=c11|1&gt;+c21|2&gt;+c31|3&gt; and correspondingly for the rest&#xA;(Ε1 –2 Ε0 ) c11  = ΑΕ c21     (1)&#xA;(Ε3 –2 Ε0 ) c33  = 2ΑΕ* c23    (2)&#xA; (Ε1 -ε2)c21=Η21’c11 +Η23’c33+  Η22’ c21 =&gt;(Ε1 -Ε0 )c21=Η21’c11 +Η23’c33=&gt;&#xA;c21= [ Η23’c33 +Η21’c11 ] /(Ε1 –Ε0 )  (3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Ε3 -ε2)c23=Η23’c33 +Η21’c11+  Η22’ c23 =&gt;(Ε3 -Ε0)c23=Η23’c33 +Η21’c11=&gt;&#xA;c23= [Η23’c33 +Η21’c11 ] /(Ε3 –Ε0 )  (4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and eliminating c23 and c21 in (1) and (2) we get a homogeneous system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Ε1 –2 Ε0 ) c11  = ΑΕ[ AEc33 + Α*Εc11 ] /(Ε1 –Ε0 )]&#xA;(Ε3 –2 Ε0 ) c33  = 2Α*Ε[ AEc33 + Α*Εc11 ] /(Ε3 –Ε0 ) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting the determinant to 0 gets an equation connecting E1 and E3, but these are two unknowns. So bottom line: in such a case where degenerate states are not connected by the perturbation,  diagonalize the part consisting of these states plus the states that are &quot;most connected&quot; to them, i.e. have the strongest matrix elements&lt;/p&gt;&#xA;" OwnerUserId="3025" LastActivityDate="2011-04-09T19:12:01.087" />
  <row Id="8334" PostTypeId="2" ParentId="8036" CreationDate="2011-04-09T19:13:39.247" Score="0" ViewCount="" Body="&lt;p&gt;Where does the energy go when a magnet looses energy, magnetism over time in an isolated system&lt;/p&gt;&#xA;" OwnerUserId="3026" LastActivityDate="2011-04-09T19:13:39.247" CommentCount="1" />
  <row Id="8335" PostTypeId="2" ParentId="8331" CreationDate="2011-04-09T19:21:10.600" Score="2" ViewCount="" Body="&lt;p&gt;If the height difference between the mercury level in the two arms is $h$ (it's called $\delta h$ on the figure), then &#xA;$$P_1 - P_2 = h\rho g$$&#xA;where $P_1,P_2$ are the pressures in both wings (called $P,P_{\rm ref}$ on the figure). One of them is the measured atmospheric pressure. The two pressures are being subtracted because the air pushes the liquid from the two sides in two opposite directions. You may also move $P_2$ to the right hand side, so that the two sides exactly express the pressure in both directions (to be specific, you may think about forces acting on a special separator inserted to the point $B$ at the bottom of the figure - most of the mercury cancels, only the height difference doesn't).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic-school formula $h\rho g$ for the pressure may be derived as the force of the mercury column per unit area of the base. The mass is $V\rho = A h\rho$, the force is $g$ times larger i.e. $A h \rho g$, and the force per unit area is therefore $h\rho g$ because $A$ cancels. My derivation is only valid for &quot;cylindrical&quot; shapes but the $h\rho g$ formula is actually true for any shape - the pressure only depends on the depth $h$ beneath the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/SQuue.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Restricting our attention to the pressure and height differences only, it's clear that $h=1$ millimeter of mercury corresponds to the pressure difference&#xA;$$ \delta P = h \rho g = 0.001 \,{\rm m} \times  13,595.1\, {\rm kg}/{\rm m}^3 \times 9.80665\,{\rm m}/{\rm sec}^2 = 133.332 \,{\rm Pa} $$&#xA;The inverse relationship is 1 Pascal equivalent to $1/133.332 = 0.0075006$ mm Hg. The exact values of the densities are a little bit conventional - the densities depend on temperature and pressure and the gravitational acceleration depends on the place. In the past, 1 mm Hg wasn't needed that accurately. In modern era, we define 1 mm Hg by your relationship, and 1 Pa is much more accurately defined in terms of &quot;fundamental physics&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-09T19:35:39.113" LastActivityDate="2011-04-09T19:35:39.113" CommentCount="1" />
  <row Id="8336" PostTypeId="2" ParentId="102" CreationDate="2011-04-09T19:40:23.493" Score="-1" ViewCount="" Body="&lt;p&gt;Something everyone missed, gravity is a wave tension from all matter and energy induced by the mechanism of wave front formation as a back action. Time and space are visible actions of this process of matter and energy decay via a gravitational wave. Time is a measure of change and is relative as sound measures an energy change. Time is relative as sound is relative. Space is the volume of change. Gravity is the reaction of the tension created by constructive wave interference from continuously generated waves.&#xA;   Time,space, and gravity are generated actions,by-products of matter and energy decay. This concept is different than what is accepted currently but it takes all observations which have unknown explains and explains them with one new concept which is easy to understand. - Michael Turner&#xA;&quot;We don't need to think more, we need to think differently&quot;- Albert Einstein&#xA;So to answer your question, matter decay into a wave and wave front formation is the force and gravity is the opposite force, the reaction to an action.&lt;/p&gt;&#xA;" OwnerUserId="3026" LastEditorUserId="3026" LastEditorDisplayName="" LastEditDate="2011-04-09T19:47:43.490" LastActivityDate="2011-04-09T19:47:43.490" CommentCount="1" />
  <row Id="8337" PostTypeId="2" ParentId="7049" CreationDate="2011-04-09T19:54:24.300" Score="0" ViewCount="" Body="&lt;p&gt;Regarding whether people figured out the Lorentz transformation of E&amp;amp;M fields before Einstein, the answer (of course) is &quot;sort of.&quot;  According to my physics professor (Columbia U), people realized that the fields transformed according to the Lorentz transformation (although of course it wasn't yet called that), shortly before Maxwell came up with his laws.  My professor roughly said that people had an idea that there was this strange dependence (Lorentz) on relative velocity of reference frames, but they didn't know what its significance was, or its relation to, for example, the not-yet-discovered position/time transformation or other Lorentz transformations.&lt;/p&gt;&#xA;" OwnerUserId="3027" LastActivityDate="2011-04-09T19:54:24.300" CommentCount="1" />
  <row Id="8338" PostTypeId="2" ParentId="8303" CreationDate="2011-04-09T20:10:37.740" Score="4" ViewCount="" Body="&lt;p&gt;I'm not sure if an undergrads perspective would be useful here- but I'll give it a shot (at worst I'll learn something new.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;David Griffith's &quot;Introduction to Quantum Mechanics&quot; takes great care to motivate the uncertainty principle from more basic founding postulates of Q.M. First Hilbert space and the  state vector, as the description of the particle, are defined. Next classical observables are formulated as operators on the state vector. Eigenvalues and the basis of the operators are explored and it is revealed that for certain (conjugate) operators, the state vector cannot be written in the same basis if a unique value for those operator's corresponding observable is desired. It is shown that such operators do not commute. It is finally shown that from non-commuting the uncertainty principle can be mathematically derived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the point of this summary (all of which I'm sure you already know well) is the order that things are done. Griffiths is so far my favorite text book author, and I'm sure there is a reason he laid things out so explicitly. He stresses the classical nature of the observables and how the state vector is truly fundamental. It always seemed to me (and thus how I understand it) that what he was getting at is that observables like position and momentum are classical and what we are doing is trying to perform classical observation on a quantum system. When we attempt to do this, we are putting limitations on the state vector that nature simply doesn't do on her own. The result of this is that we end up with  non-comeasurable observables, simply because of our classical bias in &quot;translating&quot; the true state of the particle, which is simply not completely expressible solely in terms of classical observables. To me this, what Q.M. is actually doing, seems more fundamental than the HUP. Perhaps it borders on metaphysics- but it seems to be the logical conclusion of the math/algorithms.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And because Bell's Theorem is mentioned: the inputs for this theorem are already there in Q.M.- the theorem simply tells us how to properly combine them and then conclude the character of the correlations between observables. In a way (once again seeming to me) it &quot;measures&quot; what kind of probabilities we are expressing in our theory.  &lt;/p&gt;&#xA;" OwnerUserId="1567" LastActivityDate="2011-04-09T20:10:37.740" CommentCount="9" />
  <row Id="8339" PostTypeId="2" ParentId="8327" CreationDate="2011-04-09T20:19:06.323" Score="0" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The question is: does the&#xA;  air+alcohol-gas burn creating a big&#xA;  pressure or does the container need of&#xA;  a little hole to burn?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Strange &quot;need &lt;strong&gt;of&lt;/strong&gt; a little hole&quot;, What does that mean? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The question was: do we need of hole&#xA;  to start the ratio of oxygen+alcohol&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another time strange &quot;need &lt;strong&gt;of&lt;/strong&gt; hole&quot;. And on the whole two rather different questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing I can say: In a container with alcohol vapor (I assume ethyl alcohol is meant) &#xA;and air one will have &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;40 mmHg partial pressure of ethanol at 19 °C&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;720 mmHg partial pressure of air  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This makes 40/760 = 5.3  vol-% of ethanol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A view in &lt;a href=&quot;http://en.wikipedia.org/wiki/Flammability_limit&quot; rel=&quot;nofollow&quot;&gt;tables&lt;/a&gt; says that ethanol has &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lower explosive limit (LEL) 3.3 vol% &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upper explosive limit (UEL) 19 vol% &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer is: Your container will explode, the liquid ethanol &#xA;will be sprayed around, burning. I'd prefer to be several&lt;br&gt;&#xA;dozens of meters away from such &quot;experiments&quot;, at least. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-09T20:19:06.323" CommentCount="3" />
  <row Id="8340" PostTypeId="1" AcceptedAnswerId="8345" CreationDate="2011-04-09T20:20:42.337" Score="9" ViewCount="193" Body="&lt;p&gt;An asteroid was recently discovered that is in a &lt;a href=&quot;http://arxiv.org/abs/1104.0036&quot; rel=&quot;nofollow&quot;&gt;horseshoe orbit with respect to the earth&lt;/a&gt;. Is there an intuitive explanation for these orbits? It seems that the earth acts as a repulsive force where one would expect gravity to be attractive. From the above linked article:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accepted in MNRAS, Apostolos A. Christou, David J. Asher, 1104.0036, &lt;em&gt;A long-lived horseshoe companion to the Earth&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/lesXN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better description of the horseshoe orbits, along with the tadpole orbits and Lagrange points:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/FrWhl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-04-09T21:09:02.907" LastActivityDate="2011-04-09T21:37:11.113" Title="How do horseshoe orbits work?" Tags="&lt;gravity&gt;&lt;celestial-mechanics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="8341" PostTypeId="1" AcceptedAnswerId="8378" CreationDate="2011-04-09T20:38:15.930" Score="1" ViewCount="167" Body="&lt;p&gt;Suppose I'm on a non rotating planet. I have two identical, perfect watches. I synchronize them. Then I throw one of them into the air and catch it. Does the one I throw into the air gain or lose time with respect to the one I was holding?&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-04-10T18:08:13.290" Title="Does throwing a watch into the air cause it to gain or lose time?" Tags="&lt;general-relativity&gt;&lt;special-relativity&gt;" AnswerCount="2" CommentCount="10" />
  <row Id="8342" PostTypeId="2" ParentId="8340" CreationDate="2011-04-09T20:54:03.317" Score="4" ViewCount="" Body="&lt;p&gt;Here everything is explained: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Horseshoe_orbit&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Horseshoe_orbit&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lagrange points  are the &quot;return&quot; attractors. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-09T20:54:03.317" />
  <row Id="8343" PostTypeId="1" CreationDate="2011-04-09T20:57:32.207" Score="1" ViewCount="90" Body="&lt;p&gt;The image distance is positive.&#xA;The image distance is negative.&#xA;The image is inverted.&#xA;The image is upright.&#xA;The image height is positive.&#xA;The image height is negative.&#xA;The magnification of the image is positive.&#xA;The magnification of the image is negative.&#xA;It was made by a converging optic.&#xA;It was made by a diverging optic.&#xA;The focal length of the optic is positive.&#xA;The focal length of the optic is negative.&lt;/p&gt;&#xA;" OwnerUserId="3029" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-09T21:17:31.837" LastActivityDate="2011-04-09T23:54:21.703" Title="If you are told an image is real and you know it was made by a single optic (either a lens or a mirror), which of the following must be true?" Tags="&lt;optics&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8344" PostTypeId="2" ParentId="8343" CreationDate="2011-04-09T21:22:41.657" Score="0" ViewCount="" Body="&lt;p&gt;All those statements are potentially true.  None of them is logically necessary.  This can be seen from the lens equation $\frac{1}{d_{o}}+ \frac{1}{d_{i}} = \frac{1}{f}$, remembering the convention that the focal length is positive for a converging lens and negative for a diverging lens.  All that remains is to draw a few &lt;a href=&quot;http://www.google.com/search?sourceid=chrome&amp;amp;ie=UTF-8&amp;amp;q=ray+diagram&quot; rel=&quot;nofollow&quot;&gt;ray diagrams&lt;/a&gt; to get a sense for the inversion of the image.  (or using the derived result${}^{1}$ $M=-\frac{d_{i}}{d_{o}}$, where M is the magnification of the lens.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{1}$It's easiest to get THIS result from noting that the ray from the top of the object will pass through the center of the lens undeflected, and also gets mapped to the top of the image.  So the ray going from the object to the lens creates a similar triangle to the one formed going from the lens to the image.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-04-09T21:22:41.657" />
  <row Id="8345" PostTypeId="2" ParentId="8340" CreationDate="2011-04-09T21:37:11.113" Score="7" ViewCount="" Body="&lt;p&gt;The horseshoe orbit shape does occur only in the reference frame of the Earth’s orbit. It is a manifestation of a third body problem, and the orbit is in an accelerated reference frame. The loop, which is this distended horseshoe shape, has no central gravitational source inside the loop. As a result the orbit is a “pseudo-orbit.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the perspective of an inertial frame in heliocentric coordinates this asteroid is in a circular orbit (topologically a circle) around the sun. When the orbit is closer to the sun than the Earth’s orbit the asteroid has a smaller orbital period, or equivalently it has a larger velocity. It will eventually catch up with the Earth, but it is not necessarily gravitationally drawn into the Earth. It interacts with the Earth’s gravity field in its frame with an effective and repulsive potential $L^2/2mr$, $L$ = angular momentum $r$ = distance from Earth.  The gravitational potential plus this effective potential pushes the asteroid into a higher orbital radius.  The Lagrange points $L_4$ and $L_5$ act then as attractor points in the rotational frame of the asteroid.  Its orbital velocity is now smaller and recedes away from the Earth. Eventually the Earth approaches the asteroid and the process is repeated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a “hunter-chaser” type of orbit. The thinner the horseshoe is the smaller the angular momentum L is with respect to the Earth at close approach. This means the gravitational attraction can become larger.  This is a physical way of thinking about this.  Technically this problem requires using Hill’s method for the 3 body problem.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-04-09T21:37:11.113" />
  <row Id="8346" PostTypeId="2" ParentId="8343" CreationDate="2011-04-09T23:42:29.737" Score="1" ViewCount="" Body="&lt;p&gt;A &lt;em&gt;real&lt;/em&gt; image comes from a concave mirror or a converging lens, and for a single optic the image will be inverted.  Whether the real image is the same side of the optic as the source depends on whether you have a mirror or a lens.  The source must be further away than the focal length of the optic to produce a real image; the image will be reduced if the source is more than two focal lengths away from the optic, and magnified if the source is between one and two focal lengths away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is homework and some of the questions depend on the conventions you have been taught, I will leave the rest to you.&lt;/p&gt;&#xA;" OwnerUserId="2714" LastEditorUserId="2714" LastEditorDisplayName="" LastEditDate="2011-04-09T23:54:21.703" LastActivityDate="2011-04-09T23:54:21.703" />
  <row Id="8347" PostTypeId="1" AcceptedAnswerId="8350" CreationDate="2011-04-10T00:13:51.430" Score="3" ViewCount="193" Body="&lt;p&gt;Given a solid whose interior is a hollow sphere with perfectly reflecting mirrors. A small hole is drilled in the sphere and a photon is sendt in at some angle. Will it always eventually exit through the hole it entered?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any arrangement of mirrors one can place inside the sphere such that the photon will never escape?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, is there a principle that explains it, or some way to prove it?&lt;/p&gt;&#xA;" OwnerUserId="1708" LastEditorUserId="1708" LastEditorDisplayName="" LastEditDate="2011-04-10T01:30:10.683" LastActivityDate="2011-04-11T17:22:26.757" Title="Trapping a lightray" Tags="&lt;optics&gt;&lt;geometric-optics&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="1" />
  <row Id="8348" PostTypeId="2" ParentId="8347" CreationDate="2011-04-10T00:25:18.047" Score="-1" ViewCount="" Body="&lt;p&gt;Yes, it will always exit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to trap it - you will need to to be able to move mirror a little while photon is somewhere in the middle of your trap.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-04-10T00:25:18.047" />
  <row Id="8349" PostTypeId="2" ParentId="8347" CreationDate="2011-04-10T00:55:24.503" Score="0" ViewCount="" Body="&lt;p&gt;If you're asking a mathematical question: does a line starting at a point on the inside of a sphere, and reflecting off the sides of a sphere, ever come back to its starting point, then the answer is yes if and only if the angle the line makes with the tangent to the sphere at the starting point is 90/n degrees, for integral n &gt; 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're asking a physical question: does a photon going through a finite-size hole and bouncing around inside ever come out, the answer is yes -- even if you close the hole up after you shoot the photon inside.&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="2595" LastEditorDisplayName="" LastEditDate="2011-04-10T01:02:21.470" LastActivityDate="2011-04-10T01:02:21.470" CommentCount="3" />
  <row Id="8350" PostTypeId="2" ParentId="8347" CreationDate="2011-04-10T00:58:49.053" Score="7" ViewCount="" Body="&lt;p&gt;For the first problem, yes. Because you shoot in the photon from the boundary of the sphere, the trajectory of the photon, using elementary geometry, will stay within a fixed plane (the plane is defined by the center of the sphere, the point of entry, and the initial direction of the photon). So you reduce the problem to an essentially two-dimensional problem. It is well known that the billiards trajectory in a circle is integrable, and the trajectory is either periodic or hits a dense set of points on the boundary. In the former case the photon will quickly exit the sphere, even in the limit as the size of the initial hole tends to zero. In the latter case for every non-zero sized hole, you can find some finite time $T$ such that the photon will be guaranteed to exit the hole after $T$. But as you shrink the size of the hole, $T$ increases unboundedly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second problem, it is possible to set up trapping obstacles inside your sphere. For an example, see &lt;a href=&quot;http://mathoverflow.net/questions/38307/trapped-rays-bouncing-between-two-convex-bodies&quot; rel=&quot;nofollow&quot;&gt;this question/answer on MathOverflow&lt;/a&gt;. However, in general you can only trap trajectories for a small set of initial angles; I am inclined to say that you cannot do so for all angles simultaneously, though I don't have a proof for that. &lt;/p&gt;&#xA;" OwnerUserId="829" LastActivityDate="2011-04-10T00:58:49.053" CommentCount="3" />
  <row Id="8351" PostTypeId="2" ParentId="8347" CreationDate="2011-04-10T01:49:59.773" Score="0" ViewCount="" Body="&lt;p&gt;A line entering a sphere will bounce around in the plane containing the original ray and a ray from the entering point through the center of the sphere. It will bounce around until it hits the entrance, then exit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your first example with theoretically perfect surfaces would work of course if you could perfectly close the entrance hole after you inject the photon, but I suspect this is not what you were looking for. &lt;strong&gt;Edit&lt;/strong&gt;: noticed the other answers with the trapping arrangements so I won't try to answer that part here :)&lt;/p&gt;&#xA;" OwnerUserId="2936" LastEditorUserId="2936" LastEditorDisplayName="" LastEditDate="2011-04-10T01:57:06.307" LastActivityDate="2011-04-10T01:57:06.307" />
  <row Id="8352" PostTypeId="1" AcceptedAnswerId="8354" CreationDate="2011-04-10T01:55:38.580" Score="6" ViewCount="235" Body="&lt;p&gt;When one turns on the tap in the kitchen, a circle is observable in the water flowing in the sink. The circle is the boundary between laminar and turbulent flow of the water (maybe this is the wrong terminology?). On the inside the height of the water is lower than on the outside. I'm sure that you have seen it many times, if not you can try it out yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found out by experiment that the effect is most likely independent of the curvature of the sink, as it works in strongly curved bathroom sinks as well as in flat kitchen sinks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is there a circle, why not a gradual change and what is happening qualitatively and/or quantitatively in this system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I have not found a satisfying answer to this problem yet, every time I turn on a tap, I am reminded of my own ignorance. It is really annoying.&lt;/p&gt;&#xA;" OwnerUserId="3031" LastActivityDate="2011-04-10T14:07:40.287" Title="water flow in a sink" Tags="&lt;water&gt;&lt;hydrodynamics&gt;&lt;continuum-mechanics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8353" PostTypeId="2" ParentId="8179" CreationDate="2011-04-10T02:40:44.780" Score="6" ViewCount="" Body="&lt;p&gt;When one says that an elementary particle is point-like, one is referring to the fact that theoretically, there's no limit to how small a region a detector can localize a particle to. For the sake of argument, let's imagine two wrong things (a) that such an ideal detector is possible and (b) complications arising from Planck scale physics don't change anything conceptually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you allow for that, your worry that information is being stored in a zero volume region is still unfounded. It would be a legitimate worry in pre-relativistic-QFT physics. But we know particles aren't pellets that move around carrying information. They can disappear and be spontaneously created out of the vacuum. What this is hinting at (though some might prefer a different picture) is that particles aren't fundamental - fields are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantum fields for various particles are defined everywhere in space. Once you specify what kind of structure the quantum field is (a scalar, vector, spinor, etc.) and what its other properties are (say, the symmetry group under which it has local gauge invariance), you have specified what spin, charge, mass etc. its particle excitations will carry. Since the field is defined everywhere in space, there's plenty of room for all that information. So in a certain sense, the information that a detector detects is encoded everywhere in space (because the field is everywhere) - and the specific structure of the detector just picks out the right information you asked for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, two point-particles (say an electron and a muon) have different properties because they are excitations of two different fields defined everywhere in space - and the detector you build specifically for the electron will pick out the &quot;signal&quot; from the electron.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-04-10T16:46:09.663" LastActivityDate="2011-04-10T16:46:09.663" CommentCount="1" />
  <row Id="8354" PostTypeId="2" ParentId="8352" CreationDate="2011-04-10T03:28:57.997" Score="6" ViewCount="" Body="&lt;p&gt;You are observing a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydraulic_jump&quot; rel=&quot;nofollow&quot;&gt;hydraulic jump&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia article is very good, so I won't try to out-do it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In brief summary, when the water starts running out from the place where it hits the sink, it the same flow is spread out over a larger and larger circumference as you move out.  This means the flow gets shallower and moves more slowly as you move further out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a wave propagates in this flow, its wave speed depends on the height of the water.  Its speed relative to the flow also depends on the flow speed.  So the propagation of wave changes as we move further out as the flow underneath the wave changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wave itself changes the height of the water - the water is deep at the peak of the wave and shallow at a trough.  Different parts of the wave moves at different speeds.  This is clearly a non-linear phenomenon.  Similar to waves crashing at the beach, eventually the propagating wave crashes over on itself.  This causes a &quot;hydraulic jump&quot;.  The main effects are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The speed of the flow goes down.&lt;/li&gt;&#xA;&lt;li&gt;The height of the water increases, converting some kinetic energy to potential.&lt;/li&gt;&#xA;&lt;li&gt;Some energy is lost to heat through turbulence.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The physics of water in your sink is not very easy - since the flow is so shallow, surface tension has considerable importance.  You can learn more details in the Wikipedia article linked above.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-04-10T03:28:57.997" CommentCount="7" />
  <row Id="8355" PostTypeId="1" CreationDate="2011-04-10T04:06:23.027" Score="1" ViewCount="195" Body="&lt;p&gt;&lt;a href=&quot;http://astrophysics.fic.uni.lodz.pl/100yrs/pdf/07/063.pdf&quot; rel=&quot;nofollow&quot;&gt;In Hartle-Hawking &quot;no boundary&quot; proposal&lt;/a&gt; it is proposed that Riemannian spacetimes rather than Lorentzian dominated the path integral near the big bang. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moments after the big bang however spacetimes with Lorentzian metrics started to dominate over the Riemannian. The dominance of the Riemannian spacetimes is characterized by positive definite metrics obtained by applying Wick rotation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to compute the path integral some approximations are made and the process of analytic continuation is applied. What bothers me is analytic continuation of an approximate holomorphic function is not guaranteed to remain analytic in another region. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why then this unreliable process applied&amp;mdash;can one rely on this scheme which seems to be mathematically dubious? &lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-10T09:51:37.350" LastActivityDate="2011-09-03T23:07:24.997" Title="Mathematical justification of Hartle-Hawking &quot;no boundary&quot; proposal" Tags="&lt;cosmology&gt;&lt;quantum-gravity&gt;" AnswerCount="1" CommentCount="11" />
  <row Id="8356" PostTypeId="1" AcceptedAnswerId="8730" CreationDate="2011-04-10T05:10:19.337" Score="9" ViewCount="250" Body="&lt;p&gt;Are the basic postulates of QM, such as complex Hilbert space, unitary evolution, Hermitian operator observables, projection hypothesis etc., the &lt;strong&gt;unique and only&lt;/strong&gt; set of postulates that gives rise to a semi-deterministic and semi-probabilistic theory, in which the time evolution is non-degenerate? By non-degenerate, I mean different initial states never produce the same final state probabilities, which in QM is guaranteed by unitarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phrased in another way, is it possible to prove from some general principles, such as semi-determinism, semi-reversibility (not for collapse), causality, existence of non-compatible observables etc., that a physical theory must satisfy these postulates? In particular, is it possible to prove that complex numbers, or a mathematical equivalent, must be fundamental to the theory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't studied anything about foundational issues of QM, so feel free to point out if I'm being a crackpot. I suppose this question may be similar to something like &quot;can you prove that gravity must be a metric theory entirely from the equivalence principle?&quot;, whose answer is no, but I'll be glad if it turns out to be otherwise.&lt;/p&gt;&#xA;" OwnerUserId="2643" LastActivityDate="2011-05-12T15:22:01.600" Title="Are the basic postulates of QM the only set of postulates that can give rise to a sensible semi-probabilistic physical theory?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="2" />
  <row Id="8357" PostTypeId="2" ParentId="8356" CreationDate="2011-04-10T06:34:07.767" Score="-1" ViewCount="" Body="&lt;p&gt;This is not a bad question in my opinion and indeed some smart people like Anton Zeilinger (I can remember his article but at the moment I can't find it to give a reference. But &lt;a href=&quot;http://scienceblog.com/community/older/2001/C/200113660.html&quot; rel=&quot;nofollow&quot;&gt;this is an indirect reference&lt;/a&gt;) have thought about it before. However, no body as yet come close to formulate it. There is no guarantee however that some such simple principles exists from which the postulates of QM can be derived. The mainstream idea is that the postulates of QM are fundamental and they are here to stay with us. In case you are interested, my personal humble opinion belongs to the mainstream group and I don't believe any such simple principles exist.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-04-10T06:39:38.573" LastActivityDate="2011-04-10T06:39:38.573" CommentCount="4" />
  <row Id="8358" PostTypeId="1" CreationDate="2011-04-10T07:36:20.820" Score="2" ViewCount="195" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/8248/did-the-guys-at-fermilab-find-a-fifth-force&quot;&gt;Did the guys at Fermilab find a fifth force?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Did the LHC just find a Z' boson, or is it just your typical media overblown sensationalism with no basis in fact?&lt;/p&gt;&#xA;" OwnerUserId="3034" LastActivityDate="2011-04-10T07:36:20.820" ClosedDate="2011-04-13T00:05:53.897" Title="Did the LHC just find a Z' boson?" Tags="&lt;particle-physics&gt;" CommentCount="2" />
  <row Id="8359" PostTypeId="1" CreationDate="2011-04-10T07:48:20.737" Score="8" ViewCount="264" Body="&lt;p&gt;Sorry if this is really naive, but we learned in Newtonian physics that the total energy of a system is only defined up to an additive constant, since you can always add a constant to the potential energy function without changing the equation of motion (since force is negative the gradient of the potential energy). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then in Quantum Mechanics we showed how the ground state of a system with potential energy $V(x) = \frac{1}{2} m \omega^{2} x^{2}$ has an energy $E_{0}=\frac{1}{2} \hbar \omega $. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we add a constant to $V(x)$ won't that just shift the ground state energy by the same constant? So in what sense can we actually say that the ground state energy has an absolute value (as opposed to just a relative value)? Is there some way to measure it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask this in part because I have heard that Dark Energy might be the ground state energy of quantum fields, but if this energy is only defined up to a constant, how can we say what it's value is?&lt;/p&gt;&#xA;" OwnerUserId="3035" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-10T09:43:18.370" LastActivityDate="2011-04-10T09:43:18.370" Title="If energy is only defined up to a constant, can we really claim that ground state energy has an absolute value?" Tags="&lt;energy&gt;&lt;energy-conservation&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="8360" PostTypeId="2" ParentId="8359" CreationDate="2011-04-10T08:24:25.667" Score="10" ViewCount="" Body="&lt;p&gt;In non-relativistic and non-gravitational physics (both conditions have to be satisfied simultaneously for the following proposition to hold), energy is only defined up to an arbitrary additive shift. In this restricted context, the choice of the additive shift is an unphysical, unobservable convention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Special relativity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in special relativity, energy is the time component of a 4-vector and it matters a great deal whether it is zero or nonzero. In particular, the energy of the empty Minkowski space has to be exactly zero because if it were nonzero, the state wouldn't be Lorentz-invariant: Lorentz transformations would transform the nonzero energy (time component of a vector) to a nonzero momentum (spatial components).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;General relativity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general relativity, the additive shifts to energy also matter because energy is a source of spacetime curvature. A uniform shift of energy density in the Universe is known as the cosmological constant, and it will curve the vacuum. So it's important to know what it is - and it is not just a convention. Also, in general relativity, the argument from the previous paragraph may be circumvented: dark energy, regardless of its value, preserves the Lorentz (or de Sitter or anti de Sitter, which are equally large) symmetry because the stress energy tensor is proportional to the metric tensor (because $p=-\rho$). However, as long as there is gravity, the additive shift matters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, we don't measure the zero-point energy by its gravitational effects, and the value of the cosmological constant remains largely mysterious. So I surely have a different, more observationally relevant answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Casimir energy, comparison of situations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The additive shifts to the energy are also important when one can compare the energy in two different situations. In particular, the Casimir effect may be measured. The Casimir force arises because in between two metallic plates, the electromagnetic field has to be organized to standing waves - because of the different boundary conditions. By summing the $\hbar\omega/2$ zero-point energies of these standing waves (each wavelength produces a harmonic oscillator), and by subtracting a similar &quot;continuous&quot; calculation in the absence of the metallic plates, one may discover that the total zero-point energy depends on the distance of the metallic plates if they're present, and experiments have verified that the corresponding force $dE/dr$ exists and numerically agrees with the prediction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many other contexts in which the zero-point energy may be de facto measured. For example, there exist metastable states that behave like the harmonic oscillator for several low-lying states. The energy of these metastable states may be compared with the energy of the free particle at infinity, and the result is $V_{\rm local\,minimum}-V_{\infty}+\hbar\omega/2$. This is somewhat analogous to calculating the energies of the bound state in a Hydrogen atom - which may be measured (think about the ionization energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, whenever one adds either special relativity or gravity or comparisons of configurations where the structure and frequencies of the harmonic oscillators differ, the additive shift becomes physical and measurable.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-10T08:24:25.667" CommentCount="1" />
  <row Id="8361" PostTypeId="2" ParentId="8359" CreationDate="2011-04-10T08:24:44.760" Score="2" ViewCount="" Body="&lt;p&gt;It's quite correct that you can additively shift energy, even in quantum mechanics, and one can always make the ground state carry zero energy. Nevertheless, you can still measure some other energy even in the ground state: the kinetic energy. Because $T = {p^2 \over 2m}$ the expectation of kinetic energy in a given energy state is essentially its uncertainty in momentum (because the average value of momentum is zero). So even in the ground state of the oscillator there is some intrinsic movement present (of course only in this sense, the state is still stationary w.r.t. evolution), notwithstanding that it has zero energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From another point of view, consider your potential $V(x) = {1\over 2} m\omega^2 x^2 - E_0$: it will intersect the $x$-axis. But the ground state energy lies at $E=0$. So it's not found at the bottom of your potential (as one would expect for a ground state in classical physics). This relative position of $E_0$ and $V(x)$ is independent of any shifts in energy.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-04-10T08:24:44.760" CommentCount="6" />
  <row Id="8362" PostTypeId="2" ParentId="8248" CreationDate="2011-04-10T08:42:25.627" Score="13" ViewCount="" Body="&lt;p&gt;Let me add an experimentalist's opinion . From the plots shown by Lubos above, one sees that two distributions are being subtracted in order to bring up the signal. A monte carlo background from expected interactions with a number like 500 events/8GeV, and a similar number for the experimental data. As the Monte Carlo background has no error bars, I presume the statistics are much higher and the histogram is just normalized to the number of events in the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The  statistical error in number of events for each data bin is about sqrt(500)=22.4 events. If I measure the error in the difference plot you show above, it is not larger than this number for each bin. This means they have not included systematic errors in their error estimates. One such is the error produced by the shift in energy as discussed by Lubos. This should have been added to the errors by varying the Monte Carlo background according the the 1 sigma error of the energy of the jet , and added to the error bars. There are other systematic errors one can think about in subtracting data from Monte Carlo events and in cut decisions. The effect of each cut, should be in a systematic error. Each variable used in the cuts, (including the 8 Gev binning above), should be varied in the Monte Carlo within the error bars of the variable and the error estimated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that systematic errors are added linearly and not in quadrature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this bump is not a statistical fluctuation but a result of underestimation of systematic errors, even if CDF doubles the statistics the problem will remain. It is independent experiments that will  inform us of whether it is a statistical fluctuation, an analysis artifact or a true signal. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-04-10T18:28:46.220" LastActivityDate="2011-04-10T18:28:46.220" CommentCount="5" />
  <row Id="8364" PostTypeId="2" ParentId="8352" CreationDate="2011-04-10T10:57:10.393" Score="2" ViewCount="" Body="&lt;p&gt;When You use a flat glass plate and a fine nozzle for the water jet attached &#xA;to tap by a hose, You can vary the angle between  the water jet and the plate. &#xA;The interesting part of this is, that the circle stays a circle, even when &#xA;the jet impinges rather oblique. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some &lt;strong&gt;recreational physics&lt;/strong&gt; article (SCIAM?) I read some decades ago, &#xA;the author said that the velocity at same distance from jet is the same &#xA;irrespective of angle, but the thickness of the laminar film is less in &#xA;the &quot;backward&quot; direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rather big  sink, a glas plate and a 20 to 50 ml Pipette as a nozzle &#xA;is all You need for that &quot;recreation&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-10T11:42:45.207" LastActivityDate="2011-04-10T11:42:45.207" CommentCount="1" />
  <row Id="8365" PostTypeId="1" CreationDate="2011-04-10T11:06:20.070" Score="3" ViewCount="149" Body="&lt;p&gt;Can a trapped surface be formed without any massive bodies &lt;strong&gt;inside&lt;/strong&gt; that trapped surface, but only by a configuration of massive bodies surrounding the trapped volume?&lt;/p&gt;&#xA;" OwnerUserId="1186" LastEditorUserId="1186" LastEditorDisplayName="" LastEditDate="2011-04-10T17:07:13.490" LastActivityDate="2011-04-10T17:07:13.490" Title="Can a trapped surface be formed by a mass configuration outside of that trapped surface?" Tags="&lt;general-relativity&gt;&lt;black-holes&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8366" PostTypeId="2" ParentId="8365" CreationDate="2011-04-10T11:30:27.440" Score="-1" ViewCount="" Body="&lt;p&gt;By definition future directed geodesics emanating orthogonally from a trapped surface either outwards or inwards have to converge in finite time. Important thing to note is that this affine distance in which this convergence has to happen goes inversely with the curvature of the trapped surface and doesn't depend on any finer detail of the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If mass is existing outside such a surface then one has to ask the question of what is keeping it stable there? Why are they not falling in? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't see how I can rule out such a case if say the mass is placed such that it is outside this converging distance (which is intrinsically determined by the trapped surface only). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure how good an example this is..but one can always have stable orbits of small amounts of matter around a Schwarzschild black hole and every spherical surface inside the event horizon is a trapped surface in this case. So in some sense there is no matter inside any of the trapped surfaces. (but this orbiting small object is a matter outside all the trapped surfaces) &lt;/p&gt;&#xA;" OwnerUserId="1637" LastActivityDate="2011-04-10T11:30:27.440" />
  <row Id="8367" PostTypeId="1" CreationDate="2011-04-10T11:54:08.887" Score="7" ViewCount="208" Body="&lt;p&gt;This question is in relation to this &lt;a href=&quot;http://arxiv.org/pdf/1104.0680&quot; rel=&quot;nofollow&quot;&gt;recent paper&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I would like to know how the so called supersymmetry transformations at the start of page 27 or at the end of page35 (equation 8.4) or at the end of page 37 (equations 9.8,9.9,9.10) of the above paper are obtained. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I have never seen any supersymmetry transformation look like that! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be glad if someone can help me derive these transformation equations. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Also if anyone can explain the meaning of the conjectures being made about some sort of &quot;cohomology&quot; in the tables on page 28 and 31. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I have tried to check these conjectures on my own in some simple cases. (by almost blind-foldedly using the supersymmetry transformations!) &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Can anyone suggest possible methods of trying to prove these conjectures or if there exists other papers with similar content? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is this what is called &quot;quantum cohomology&quot; ? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1637" LastActivityDate="2011-04-12T12:30:13.343" Title="An unfamiliar way of writing supersymmetry transformations" Tags="&lt;quantum-field-theory&gt;&lt;supersymmetry&gt;" FavoriteCount="1" />
  <row Id="8368" PostTypeId="2" ParentId="8365" CreationDate="2011-04-10T12:57:16.117" Score="5" ViewCount="" Body="&lt;p&gt;If I understand your question correctly, the answer is yes, by virtue of the fact that &lt;a href=&quot;http://arxiv.org/abs/0805.3880&quot; rel=&quot;nofollow&quot;&gt;trapped surfaces can form without there being any massive bodies in the universe whatsoever&lt;/a&gt;. In other words, trapped surfaces can form purely from the focusing of gravitational waves. Now take some massive bodies all very far apart from each other, and shake them vigorously. If the shaking is hard enough, the profile of the gravitational waves would be sharp enough that the results in the linked paper would (in principle) apply. &lt;/p&gt;&#xA;" OwnerUserId="829" LastActivityDate="2011-04-10T12:57:16.117" CommentCount="1" />
  <row Id="8369" PostTypeId="2" ParentId="8179" CreationDate="2011-04-10T13:57:53.863" Score="2" ViewCount="" Body="&lt;p&gt;The assumption of particles being points is a modelling assumption used extensively in mathematical physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the orbits of the planets are calculated, the center of mass points are used in the first order calculations and assigned the mass of the whole planet. The sun too is a point at first order. Nobody has trouble with that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly when solving Schrodinger's equation for the hydrogen atom , as a potential well for the electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In second quantization the &quot;particle&quot; has a structure of Feynman diagram loops of all orders accompanying its probable position as particle. In strings even the Feynman vertex is no longer a point but has extension. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-10T13:57:53.863" />
  <row Id="8371" PostTypeId="2" ParentId="8303" CreationDate="2011-04-10T14:36:28.420" Score="-1" ViewCount="" Body="&lt;p&gt;I think I'm largely going to repeat what Roy, Vladimir, and Jaskey13 have already said, but perhaps, I hope, not so totally that this won't be Useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I take it that HUP, despite its grandiose title, is &lt;em&gt;not&lt;/em&gt; a principle; it's &lt;em&gt;derived&lt;/em&gt; as a consequence of the various mathematical structures of QM. As such, HUP is a part of a characterization of the properties of QM. HUP is, however, something of a lesser part of that characterization because it is not enough to characterize all the differences between classical stochastic physics and QM. It is possible, as Roy says, to construct local classical models for which, under a reasonable physical interpretation of the mathematics, HUP is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not completely sure what you mean by &quot;HUP alone does not ensure randomness&quot;? I suppose the interpretation of QM is all probability all the time. In various comments you protest, and I believe, that you know the axioms of QM and their basic interpretation well enough. What I &lt;em&gt;take&lt;/em&gt; you to mean is that &quot;HUP alone does not ensure &lt;em&gt;intrinsic&lt;/em&gt; randomness&quot;. This qualification, which is fairly commonly used, makes sense, to me, of your following comment, with my qualification inserted, that &quot;you need Bell's theorem and other features for the [intrinsic] randomness in QM&quot;, whereas the relevance of Bell inequalities to your Question seems to have troubled other people here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I take &amp;ldquo;intrinsic&amp;rdquo; to be a rather coded way to say that a classical probability theory is not isomorphic to quantum probability theory. I've previously cited on Physics SE the presentations of Bell-CHSH inequalities that I think best make this clear, due to Landau and to de Muynck, &lt;a href=&quot;http://physics.stackexchange.com/questions/7276/what-combinations-of-realism-non-locality-and-contextuality-are-ruled-out-in-qu/7299#7299&quot;&gt;here&lt;/a&gt;, where I note that &lt;em&gt;you&lt;/em&gt; also left a notably Useful(8) Answer. Their derivations use the CCRs in a way that is not significantly more obscure than does the derivation of the HUP. I take the Bell-CHSH inequalities to be a reasonable lowest-order characterization of the difference. There is of course confusion concerning the relevance of locality to the Bell inequalities, which I think could get in the way of my discussion here, but I see that you have a relatively sophisticated view of that confusion.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-04-10T14:36:28.420" CommentCount="3" />
  <row Id="8372" PostTypeId="2" ParentId="723" CreationDate="2011-04-10T16:15:16.273" Score="2" ViewCount="" Body="&lt;p&gt;I guess some more remarks are necessary with regard to electrons being inside and outside the &quot;grid&quot;.  If you studied Physics in High School and/or College, you may recall one exercise where you integrate the electric force inside a charged hollow sphere.  The net force on a charged particle inside that sphere integrates out to zero.  In Polywell, the &quot;grid&quot; is a set of 6 charged tori in a cubical arrangement.  While that is not exactly a hollow sphere, some distance inward from the tori, the static electric field will tend to blend toward spherical symmetry.  Even with a cubical charged shell, the forces on a charged particle inside will be small.  What happens is that the potential inside a charged hollow container is a constant level everywhere inside, so forces are the &quot;same&quot; in all directions.  The more spherical the container, the more uniform that potential level will be. In the case of Polywell, the closer to center inside the &quot;grid&quot;, the more spherical symmetry the electric field will exhibit.  Outside the &quot;grid&quot;, and the farther away from the center, the more the electric field appears as a point charge.  So anywhere inside the &quot;grid&quot;, there will be a very lowered electric force at least, and of that, less and less force due to lack of perfect spherical symmetry the closer to center an ion may come. Outside the grid, there will be some force acting more like a point charge the farther away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that there are positive ions inside the electrostatic trap as well.  The quasi-neutral plasma having positive ions in the mix, will also  tend to smooth out whatever electric forces remain.  What keeps the plasma in the trap are mostly the magnetic fields.  The plasma is held in the trap as a &quot;target&quot; so high speed ions coming from far outside the magnetic trap and accelerated by the electric field can collide with an ion in the trap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the other question, the magnets are not used to &quot;repel&quot; electrons exactly.  Rather, in a magnetic field, electrons are constrained to travel along magnetic lines of force.  The electrons are prevented from grounding on the positively charged metal &quot;grids&quot; by simply being constrained to travel along magnetic lines of force that never intersect the &quot;grids&quot;.  That is why the &quot;grids&quot; were made to conform to the contour of the magnetic lines of force.  The use of magnetic coils inside circular cross-section toroidal static charged metal &quot;grids&quot; was Bussard's solution to electron loss.  If you read about his earlier experiments, you will see that the EMC2 team was having great difficulties with electron loss because they were using tori with a square cross-section. There are even pictures on the internet that show the burn marks on the shell where electrons following magnetic lines of force were channeled right into the toroid casing and &quot;grounded&quot;.  So the solution was to create a torus with a cross-section that is more conformal to the magnetic lines of force.  The electrons then travel along magnetic lines of force that never (well, at lest fewer) intersect the metal shell of the charged torus, and having less of a path to ground, the electron loss was greatly reduced.  I would guess that future toroid shells may evolve into a sort of teardrop or maybe egg-shaped cross-section to conform even more closely to the contour of the magnetic lines of force.&lt;/p&gt;&#xA;" OwnerUserId="2982" LastEditorUserId="2982" LastEditorDisplayName="" LastEditDate="2011-04-12T01:17:12.903" LastActivityDate="2011-04-12T01:17:12.903" CommentCount="2" />
  <row Id="8373" PostTypeId="1" AcceptedAnswerId="8376" CreationDate="2011-04-10T16:22:19.003" Score="14" ViewCount="527" Body="&lt;p&gt;Forgive me if this topic is too much in the realm of philosophy. John Baez has an interesting perspective on the relative importance of dimensionless constants, which he calls fundamental like alpha, versus dimensioned constants like G or c [ &lt;a href=&quot;http://math.ucr.edu/home/baez/constants.html&quot; rel=&quot;nofollow&quot;&gt;http://math.ucr.edu/home/baez/constants.html&lt;/a&gt; ]. What is the relative importance or significance of one class versus the other and is this an area that physicists have real concerns or expend significant research?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="3037" LastEditorDisplayName="" LastEditDate="2011-04-10T17:19:55.123" LastActivityDate="2011-04-11T16:11:24.270" Title="Dimensionless Constants in Physics" Tags="&lt;physical-constants&gt;" AnswerCount="4" FavoriteCount="3" />
  <row Id="8374" PostTypeId="2" ParentId="8179" CreationDate="2011-04-10T16:57:50.490" Score="1" ViewCount="" Body="&lt;p&gt;Particles are never detected as points. The more precisely you want to measure the coordinates the more energy you would need. This means that to determine a particle's coordinate exactly you'd need infinite energy. In all other cases the particle appears as a wave package stretched in space, you only know the volume area where it is located.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When someone say the particle is point-like, it means that the dependence of the measurement precision on energy is smooth and logarithmic, and potentially you can localize the particle to any given however small volume given enough energy.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastEditorUserId="1186" LastEditorDisplayName="" LastEditDate="2011-04-10T17:03:27.443" LastActivityDate="2011-04-10T17:03:27.443" />
  <row Id="8375" PostTypeId="2" ParentId="8341" CreationDate="2011-04-10T17:03:18.953" Score="2" ViewCount="" Body="&lt;p&gt;What astounds me is there is considerable quibbling over the nature of the question, but nobody answers it!  This is comparatively simple to address.  Let us consider the Schwarzschild metric in a weak gravity field&#xA;$$&#xD;&#xA;ds^2~=~-(1~-~2\phi/c^2)dt^2~+~dr^2~+~r^2d\Omega^2&#xD;&#xA;$$&#xA;for $\phi~=~GM/r$ the Newtonian gravity potential.  The unit velocity is then&#xA;$$&#xD;&#xA;1~=~-(1~-~2\phi/c^2)u_t^2~+~u_r^2~-~\dots&#xD;&#xA;$$&#xA;where we can consider the motion in the radial direction for simplicity.  The derivative of this with respect to the proper time $s$ is then&#xA;$$&#xD;&#xA;0~=~-(1~-~2\phi/c^2)u_ta_t~+~u_ra_r.&#xD;&#xA;$$&#xA;If the gravity potential is zero the solutions are $t~=~g^{-1}sinh(gs)$ $r~=~g^{-1}cosh(gs)$, for $g$ the acceleration parameter.  Here $g$ counters the gravitation of the Earth.  If the gravity potential is turned on we can the write the time solution solution as $t~=~g^{-1}sinh(gs~+~\gamma)$, which we input into the third equation&#xA;$$&#xD;&#xA;0~=~-(1~-~2\phi/c^2)g~cosh(gs)sinh(gs)~+~g~cosh(gs~+~\gamma) sinh(gs~+~\gamma)&#xD;&#xA;$$&#xA;$$&#xD;&#xA;=~(1~-~2\phi/c^2)\frac{g}{2}sinh(2gs~+~2\gamma) ~+~sinh(2gs)&#xD;&#xA;$$&#xA;If we consider weak fields, small accelerations and small proper time $s$ we have&#xA;$$&#xD;&#xA;0~\simeq~-(1~-~2\phi/c^2)(g^2s~+~g\gamma)~+~g^2s,&#xD;&#xA;$$&#xA;where $\gamma~\simeq~-2\phi gs/c^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coordinate time is reduced with the turning on of the acceleration.  This implies that the watch on the accelerated frame will mark off a shorter interval of time than the watch which is placed on a geodesic motion in the local gravity field with acceleration $g$.  This is a gravitational version of the twin paradox.  The twin which travels outwards and back is on an accelerated frame, which is a path in spacetime that is non-extremal, or maximal.  As a result the proper time marked off is shorter.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-04-10T17:03:18.953" />
  <row Id="8376" PostTypeId="2" ParentId="8373" CreationDate="2011-04-10T17:08:00.963" Score="21" ViewCount="" Body="&lt;p&gt;Dear Michael, first of all, the question you are asking is very important and you may master it completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dimensionful constants are those that have units - like $c, \hbar, G$, or even $k_{\rm Boltzmann}$ or $\epsilon_0$ in SI. The units - such as meter; kilogram; second; Ampere; kelvin - have been chosen partially arbitrarily. They're results of random cultural accidents in the history of mankind. A second was original chosen as 1/86,400 of a solar day, one meter as 1/40,000,000 of the average meridian, one kilogram as a mass of a randomly chosen prototype, one Ampere so that $4\pi \epsilon_0 c^2$ is a simple power of 10 in SI units, one Kelvin as 1/100 of the difference between the melting and boiling points of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, the circumference of the Earth, the solar day, a platinum prototype brick in a French castle, or phase transitions of water are not among the most &quot;fundamental&quot; features of the Universe. There are lots of other ways how the units could be chosen. Someone could choose 1.75 meters - an average man's height - to be his unit of length (some weird people in the history have even used their feet to measure distances) and he could still call it &quot;one meter&quot;. It would be his meter. In those units, the numerical values of the speed of light would be different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exactly the products or ratios of powers of fundamental constants that are &lt;em&gt;dimensionless&lt;/em&gt; are those that don't have any units, by definition, which means that they are &lt;em&gt;independent&lt;/em&gt; of all the random cultural choices of the units. So all civilizations in the Universe - despite the absence of any interactions between them in the past - will agree about the numerical value of the proton-electron mass ratio - which is about $6\pi^5=1836.15$ (the formula is just a teaser I noticed when I was 10!) - and about the fine-structure constant, $\alpha\sim 1/137.036$, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Standard Model of particle physics, there are about 19 such dimensionless parameters that &quot;really&quot; determine the character of physics; all other constants such as $\hbar,c,G,k_{\rm Boltzmann}, \epsilon_0$ depend on the choice of units, and the number of independent units (meter, kilogram, second, Ampere, Kelvin) is actually exactly large enough that all those constants, $\hbar,c,G,k_{\rm Boltzmann},\epsilon_0$, may be set equal to one which simplifies all fundamental equations in physics where these fundamental constants appear frequently. By changing the value of $c$, one only changes social conventions (what the units mean), not the laws of physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The units where all these constants are numerically equal to 1 are called the Planck units or natural units, and Max Planck understood that this was the most natural choice already 100 years ago. $c=1$ is being set in any &quot;mature&quot; analysis that involves special relativity; $\hbar=1$ is used everywhere in &quot;adult&quot; quantum mechanics; $G=1$ or $8\pi G=1$ is sometimes used in the research of gravity; $k_{\rm Boltzmann}=1$ is used whenever thermal phenomena are studied microscopically, at a professional level; $4\pi\epsilon_0$ is just an annoying factor that may be set to one (and in Gaussian 19th century units, such things are actually set to one, with a different treatment of the $4\pi$ factor); instead of one mole in chemistry, physicists (researchers in a more fundamental discipline) simply count the molecules or atoms and they know that a mole is just a package of $6.022\times 10^{23}$ atoms or molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 19 (or 20?) actual dimensionless parameters of the Standard Model may be classified as the three fine-structure constants $g_1,g_2,g_3$ of the $U(1)\times SU(2)\times SU(3)$ gauge group; Higgs vev divided by the Planck mass (the only thing that brings a mass scale, and this mass scale only distinguishes different theories once we also take gravity into account); the Yukawa couplings with the Higgs that determine the quarks and fermion masses and their mixing. One should also consider the strong CP-angle of QCD and a few others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you choose a modified Standard Model that appreciates that the neutrinos are massive and oscillate, 19 is lifted to about 30. New physics of course inflates the number. SUSY described by soft SUSY breaking has about 105 parameters in the minimal model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original 19 parameters of the Standard Model may be expressed in terms of more &quot;fundamental&quot; parameters. For example, $\alpha$ of electromagnetism is not terribly fundamental in high-energy physics because electromagnetism and weak interactions get unified at higher energies, so it's more natural to calculate $\alpha$ from $g_1,g_2$ of the $U(1)\times SU(2)$ gauge group. Also, these couplings $g_1,g_2$ and $g_3$ run - depend on the energy scale approximately logarithmically. The values such as $1/137$ for the fine-structure constant are the low-energy values, but the high-energy values are actually more fundamental because the fundamental laws of physics are those that describe very short-distance physics while long-distance (low-energy) physics is derived from that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mentioned that the number of dimensionless parameters increases if you add new physics such as SUSY with soft breaking. However, more complete, unifying theories - such as grand unified theories and especially string theory - also imply various relations between the previously independent constants, so they reduce the number of independent dimensionless parameters of the Universe. Grand unified theories basically set $g_1=g_2=g_3$ (with the right factor of $\sqrt{3/5}$ added to $g_1$) at their characteristic &quot;GUT&quot; energy scale; they may also relate certain Yukawa couplings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;String theory is perfectionist in this job. In principle, all dimensionless continuous constants may be calculated from any stabilized string vacuum - so all continuous uncertainty may be removed by string theory; one may actually prove that it is the case. There is nothing to continuously adjust in string theory. However, string theory comes with a large discrete class of stabilized vacua - which is at most countable and possibly finite but large. Still, if there are $10^{500}$ stabilized semi-realistic stringy vacua, there are only 500 digits to adjust (and then you may predict everything with any accuracy, in principle) - while the Standard Model with its 19 continuous parameters has 19 times infinity of digits to adjust according to experiments.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-10T17:47:06.703" LastActivityDate="2011-04-10T17:47:06.703" CommentCount="3" />
  <row Id="8377" PostTypeId="2" ParentId="8373" CreationDate="2011-04-10T17:25:32.470" Score="5" ViewCount="" Body="&lt;p&gt;Only dimensionless quantities are important. They are just pure numbers and there can't be any ambiguity about their value. This is not so with dimensionful quantities. E.g. if I tell you my speed $v$ relative to you is $0.5\, \rm speedons$ that doesn't give you much information as I have a freedom to define my $\rm speedon$ units any way I want. Only way I &lt;em&gt;can&lt;/em&gt; give you some information is if I give you dimensionless quantity like $v/c = 0.5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what we need to make dimensionful quantities dimensionless is some reference scale (in previous example it was $c$). We can in principle choose any scale we want but usually it will be something from day to day experience. E.g. you choose meter to be what it is so that stuff you usually encounter (other people, houses, trees, etc.) is of the order $\sim 1$ with respect to meter. These is how all our units originated. Naturally, there's nothing particularly special about humans and the scales they usually work with. We know there are lots of important scales as we go down to atomic and nuclear sizes. We also know there is more important speed scale (namely, ultra-relavistic $v/c \to 1$). And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, we need to choose some units to work with to be able to compute anything and it would be nice to choose some units that wouldn't suffer from the above-mentioned arbitrariness. It turns out we are in luck because Nature has given us few special constants. Each of them is related to some fundamental theory ($c$ in special relativity, $G$ in gravity, $\hbar$ in quantum mechanics, etc.). It would be silly not to exploit this generous gift. So we can talk about speeds being 0.9 (meaning actually $v/c$), action of 20 ($=S/\hbar$) and so on. This system of units is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck_units&quot; rel=&quot;nofollow&quot;&gt;Planck's&lt;/a&gt; and while it's not used in day to day life for obvious reasons, it's very useful anytime we deal with fundamental physics.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-04-10T17:25:32.470" />
  <row Id="8378" PostTypeId="2" ParentId="8341" CreationDate="2011-04-10T18:08:13.290" Score="2" ViewCount="" Body="&lt;p&gt;Lawrence has the details, but this question can actually be determined exactly without assuming anything about mass distribution, etc. A geodesic (which is what the watch in free-fall follows) in GR has maximal proper time along it. The &quot;stationary&quot; watch, which is actually accelerated, is following some other path and so must experience a shorter proper time. &lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-04-10T18:08:13.290" CommentCount="1" />
  <row Id="8379" PostTypeId="1" AcceptedAnswerId="8380" CreationDate="2011-04-10T18:14:20.620" Score="2" ViewCount="174" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/2392/have-the-rowan-university-hydrino-findings-been-replicated-elsewhere&quot;&gt;Have the Rowan University &amp;ldquo;hydrino&amp;rdquo; findings been replicated elsewhere?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I came across this controversial company: &lt;a href=&quot;http://blacklightpower.com/&quot; rel=&quot;nofollow&quot;&gt;Black Light Power&lt;/a&gt;&#xA;I want to know if there is any validity of the scientific claims made by this company and its founder Dr. Mills - especially concerning the existence of &quot;hydrinos.&quot; &lt;/p&gt;&#xA;" OwnerUserId="2843" LastEditorUserId="2843" LastEditorDisplayName="" LastEditDate="2011-04-10T18:20:27.153" LastActivityDate="2011-04-10T18:25:01.490" ClosedDate="2011-04-10T20:57:53.820" Title="Do Hydrinos Exist?" Tags="&lt;fusion&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="8380" PostTypeId="2" ParentId="8379" CreationDate="2011-04-10T18:25:01.490" Score="1" ViewCount="" Body="&lt;p&gt;No, this stuff is complete balderdash.  It is nonsense of the first degree.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-04-10T18:25:01.490" CommentCount="3" />
  <row Id="8381" PostTypeId="1" CreationDate="2011-04-10T18:38:38.217" Score="9" ViewCount="408" Body="&lt;p&gt;It has been proposed that AdS/CFT arises because of the entanglement structure of quantum field theories, e.g. &lt;a href=&quot;http://physics.stackexchange.com/questions/8162/whats-so-special-about-ads/8173#8173&quot;&gt;see the discussion which occurred right here&lt;/a&gt;. Until now I have been skeptical of the idea, on the grounds that so much else is going on in the duality. A CFT only has a gravity dual if it has &lt;a href=&quot;http://arxiv.org/abs/0907.0151&quot; rel=&quot;nofollow&quot;&gt;the right properties&lt;/a&gt;, or so I thought. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I've just started looking into Vasiliev's &quot;higher spin gravity&quot; or &quot;higher spin gauge theory&quot;, and eventually I ran across &lt;a href=&quot;http://arxiv.org/abs/1011.4926&quot; rel=&quot;nofollow&quot;&gt;&quot;Holographic dual of free field theory&quot;&lt;/a&gt;, which claims to construct a dual Vasiliev-type theory for &lt;em&gt;every&lt;/em&gt; free QFT! Looking at the paper's only figure, I began to wonder if the &lt;a href=&quot;http://physics.stackexchange.com/questions/8162/whats-so-special-about-ads/8173#8173&quot;&gt;tensor-network MERA diagrams&lt;/a&gt; could somehow be interpreted as Witten diagrams (which are the boundary-to-bulk Feynman diagrams used in AdS/CFT). Or perhaps the &lt;a href=&quot;http://arxiv.org/abs/0905.1317&quot; rel=&quot;nofollow&quot;&gt;&quot;discrete version of anti de Sitter space&quot;&lt;/a&gt; which the MERA diagrams can provide, could be the background space on which a &quot;discrete Witten diagram&quot; is drawn. Suddenly I'm not so sure about my skepticism. Does anyone have an argument either way? &lt;/p&gt;&#xA;" OwnerUserId="1486" LastActivityDate="2011-04-27T01:24:27.033" Title="Vasiliev gravity and &quot;holographic&quot; entanglement" Tags="&lt;entanglement&gt;&lt;ads-cft&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="8382" PostTypeId="2" ParentId="8373" CreationDate="2011-04-10T19:20:59.633" Score="-3" ViewCount="" Body="&lt;p&gt;Speaking of the electron-proton mass ratio (which is about 1/1836), Lubosh found it might be connected with $\pi$, and I think it is kind of a coupling constant in the Hydrogen atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atom has the center of inertia variables and internal motion variables. When an external force is applied to the atomic nucleus, the atom is accelerated as a whole and its internal motion can also be excited. The ratio $m_e/m_p$ determines efficiency of &quot;pumping&quot; the internal degrees of freedom of an atom with an external force acting on the nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Seeing so many downvotes, I changed my mind. I agree with Lubosh: $m_p/m_e = 6\pi^5$ and has nothing to do with physics :-(.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-04-11T16:11:24.270" LastActivityDate="2011-04-11T16:11:24.270" CommentCount="2" />
  <row Id="8383" PostTypeId="2" ParentId="8347" CreationDate="2011-04-10T19:24:20.910" Score="3" ViewCount="" Body="&lt;p&gt;A  &lt;em&gt;hollow sphere with perfectly reflecting mirrors&lt;/em&gt; is a particular case of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Resonator#Cavity_resonators&quot; rel=&quot;nofollow&quot;&gt;cavity resonator&lt;/a&gt; &#xA;The others answers presume that the light ray (or a single photon if you prefer) travel along a plane that contains the centre of the sphere and the question set no constraint on this. Also the dimension of the sphere (radius) must be adequate to the wavelenght of the photon to trap it (total reflection). &#xA;There are certain configurations of angles where the ray will return to the same initial point, and I suspect that in any situation it will always return to the origin. Probably it is already proved.&lt;br&gt;&#xA;&lt;em&gt;Is there any arrangement of mirrors one can place inside the sphere such that the photon will never escape?&lt;/em&gt;&lt;br&gt;&#xA;Yes - The number of mirrors, size, the shape and orientation are important and some constraints on these must be previously known.  The solution is trivial if we can change the mirrors configuration after the ray is inside or with a proper rotation of the sphere. I am not visualizing a solution if we can not change the configuration. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A cavity resonator is a hollow&#xA;  conductor blocked at both ends and&#xA;  along which an electromagnetic wave&#xA;  can be supported. It can be viewed as&#xA;  a waveguide short-circuited at both&#xA;  ends (see Microwave cavity).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arstechnica.com/science/news/2009/08/physicists-trap-light-in-a-bottle.ars&quot; rel=&quot;nofollow&quot;&gt;physicists-trap-light-in-a-bottle&lt;/a&gt;  and &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v103/i5/e053901&quot; rel=&quot;nofollow&quot;&gt;bottle microresonators&lt;/a&gt;&lt;br&gt;&#xA;Even a perfect reflection can not avoid the &lt;a href=&quot;http://en.wikipedia.org/wiki/Evanescent_wave#Total_internal_reflection_of_light&quot; rel=&quot;nofollow&quot;&gt;Evanescent_wave&lt;/a&gt; &#xA;so, there are no perfect mirrors. Some of the energy will evade.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In optics and acoustics, evanescent&#xA;  waves are formed when waves traveling&#xA;  in a medium undergo &lt;strong&gt;total internal&#xA;  reflection&lt;/strong&gt; at its boundary because&#xA;  they strike it at an angle greater&#xA;  than the so-called critical&#xA;  angle. The physical explanation&#xA;  for the existence of the evanescent&#xA;  wave is that the electric and magnetic&#xA;  fields (or pressure gradients, in the&#xA;  case of acoustical waves) cannot be&#xA;  discontinuous at a boundary, as would&#xA;  be the case if there were no&#xA;  evanescent wave field. In quantum&#xA;  mechanics, the physical explanation is&#xA;  exactly analogous—the Schrödinger&#xA;  wave-function representing particle&#xA;  motion normal to the boundary cannot&#xA;  be discontinuous at the boundary.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In optics, evanescent-wave coupling is&#xA;  a process by which electromagnetic&#xA;  waves are transmitted from one medium&#xA;  to another by means of the evanescent,&#xA;  exponentially decaying electromagnetic&#xA;  field.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/quant-ph/pdf/0210/0210039v1.pdf&quot; rel=&quot;nofollow&quot;&gt;dielectric microsphere resonators&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The whispering gallery modes (WGMs) of&#xA;  quartz &lt;strong&gt;microspheres&lt;/strong&gt; are investigated&#xA;  for the purpose of strong &lt;strong&gt;coupling between single photons and atoms in cavity&lt;/strong&gt; quantum &#xA;  electrodynamics (cavity QED). ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In optics the treatment is done using only the wave properties of the light, and as a particle when it is absorbed by some atom/electron/...  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT add: The first sentence of this answer is not very interesting. In one situation the study uses 'geometric optics' because the wavelength is &amp;lt;&amp;lt; sphere Radius and we must use EM pure treatment in a waveguide (the wavelength is in the order of transversal dimension of the waveguide). &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-04-11T17:22:26.757" LastActivityDate="2011-04-11T17:22:26.757" />
  <row Id="8384" PostTypeId="1" CreationDate="2011-04-10T20:42:09.480" Score="10" ViewCount="187" Body="&lt;p&gt;I'm having a hard time finding information on how to model/simulate this.  I attached a couple files, both of which show an example tank &amp;amp; pump network.  It's just nonsense that I made up for this post, but it shows what kind of system I'm trying to simulate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I just put it on the web instead)&#xA;&lt;a href=&quot;http://gneissshirt.com/Example%20for%20pipe%20model.pdf&quot; rel=&quot;nofollow&quot; rel=&quot;nofollow&quot;&gt;http://gneissshirt.com/Example%20for%20pipe%20model.pdf&lt;/a&gt;&#xA;http://gneissshirt.com/Example%20for%20pipe%20model.vsd&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been looking on the web to find information on how to do this, but I keep running into formulas that are way more complex than what I need to do.  I just want to know what the pressure and flow is at any node in the system, with 80% or so accuracy.  There's no need to consider fluid compression, and we can just use a constant for the resistance experienced through valves or narrowing pipes.  We don't need to consider friction of fluid against piping.  We can ignore heat.  It's important to note, again, that I don't need complete accuracy, it's much more the behavior of the simulated system that is important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried this going in both directions, that is following a flow path down from the tanks/sources and propogating pressure changes through the system, and I've also tried it directly from the node in question, and considering that each node has a number of source nodes in terms of pressures and flows.  Both of these simulated SW systems become highly erratic the further downstream from the pumps that you get, and the more on/off valves that I put into the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you just told me where to look or what book to read, that would be a huge help.  If there's example source code out there somewhere that does this, I will buy you a 60 oz beer of your choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&#xA;~~~~&#xA;(post edit 4/12)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the answers so far, I've ended up spending quite a bit of time looking around for a solution that already exists, because I was just &lt;em&gt;convinced&lt;/em&gt; that many have run into this problem before, and knew that it was solvable in a general sense.  Part of the reason I was so interested in a general solution is that I can rip through 12 more of these simulations that I have to do quickly and accurately if that's the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found another piece of software called EPAnet, which seems like a simpler solution than what was listed below, and I'm looking through their SDK at the moment and playing around with it.  I'll keep you posted, the few that are interested ... this might not be a big deal to most, but to the few that it matters to it will be huge, so I'll write up what I find.&lt;br&gt;&#xA;~~~~&#xA;(edit 4/13)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've spent the last two days building a solid wrapper around the EPAnet programmer's toolkit, and making a scratch program that does some of the things I want to do.  (Incidentally, I've broken a new record for the # of times in a day I can make LabVIEW crash quickly and silently.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's really quite complex underneath, and I'm delighted by how much mathematical power I can stand on.  I don't think they've fleshed out all the issues, so I've been working around some of them.  But it's looking like it's 80% chance so far that it's going to work for what I want.&lt;/p&gt;&#xA;" OwnerUserId="3038" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-04-14T07:21:51.297" LastActivityDate="2011-04-18T17:26:55.350" Title="How to model/simulate pressures and flows in a network of pipes" Tags="&lt;classical-mechanics&gt;&lt;fluid-dynamics&gt;&lt;pressure&gt;&lt;simulation&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="8385" PostTypeId="2" ParentId="8384" CreationDate="2011-04-10T21:21:44.517" Score="4" ViewCount="" Body="&lt;p&gt;My advice is to look into the Lattice Boltzmann Method, which is robust and simple for athermal non-compressible flows. Try out this open source GPL3 software for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.lbmethod.org/palabos/&quot; rel=&quot;nofollow&quot;&gt;http://www.lbmethod.org/palabos/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1670" LastActivityDate="2011-04-10T21:21:44.517" CommentCount="4" />
  <row Id="8386" PostTypeId="2" ParentId="2347" CreationDate="2011-04-10T21:32:57.830" Score="2" ViewCount="" Body="&lt;p&gt;Quantum Soccer:&#xA;&lt;a href=&quot;http://gregegan.customer.netspace.net.au/BORDER/Soccer/Soccer.html&quot; rel=&quot;nofollow&quot;&gt;http://gregegan.customer.netspace.net.au/BORDER/Soccer/Soccer.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbiter:&#xA;&lt;a href=&quot;http://orbit.medphys.ucl.ac.uk/&quot; rel=&quot;nofollow&quot;&gt;http://orbit.medphys.ucl.ac.uk/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plasma Pong:&#xA;&lt;a href=&quot;http://www.youtube.com/watch?v=NDjseVmruH8&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=NDjseVmruH8&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1670" LastActivityDate="2011-04-10T21:32:57.830" CommunityOwnedDate="2011-04-10T21:32:57.830" />
  <row Id="8387" PostTypeId="1" CreationDate="2011-04-10T21:46:34.663" Score="2" ViewCount="140" Body="&lt;p&gt;If we know the universe is expanding in whatever direction we look, can't we reasonably estimate where the 'center' of the Universe is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the rate of expansions in all directions the same?&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2011-04-10T22:36:34.547" Title="Expanding universe" Tags="&lt;universe&gt;&lt;space-expansion&gt;&lt;gravitational-redshift&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8388" PostTypeId="2" ParentId="8387" CreationDate="2011-04-10T22:02:27.730" Score="2" ViewCount="" Body="&lt;p&gt;The speed that a galaxy moves away from you is larger the further away the galaxy already is. This is most consistent with an expanding universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find the following mental picture helpful: Imagine a balloon, with lots of little dots painted on it. If you inflate the balloon, then every point is moving away from every other point. Where would the &quot;center&quot; be? Pretty much everywhere.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-04-10T22:02:27.730" />
  <row Id="8389" PostTypeId="2" ParentId="8387" CreationDate="2011-04-10T22:36:34.547" Score="2" ViewCount="" Body="&lt;p&gt;Wikipedia kindly presents four models you can use to visualize the expansion of the universe. All of these are meant to explain why we think the universe is expanding, yet we think there is no center. Everything&lt;sup&gt;&amp;dagger;&lt;/sup&gt; moves away from everything else, with expanding velocity proportional to the distance.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;In the &quot;ant on a rubber rope model&quot; one imagines an ant (idealized as pointlike) crawling at a constant speed on a perfectly elastic rope which is constantly stretching. If we stretch the rope in accordance with the ΛCDM scale factor and think of the ant's speed as the speed of light, then this analogy is numerically accurate—the ant's position over time will match the path of the red line on the embedding diagram above.&lt;/li&gt;&#xA;  &lt;li&gt;In the &quot;rubber sheet model&quot; one replaces the rope with a flat two-dimensional rubber sheet which expands uniformly in all directions. The addition of a second spatial dimension raises the possibility of showing local perturbations of the spatial geometry by local curvature in the sheet.&lt;/li&gt;&#xA;  &lt;li&gt;In the &quot;balloon model&quot; the flat sheet is replaced by a spherical balloon which is inflated from an initial size of zero (representing the big bang). A balloon has positive Gaussian curvature while observations suggest that the real universe is spatially flat, but this inconsistency can be eliminated by making the balloon very large so that it is locally flat to within the limits of observation. This analogy is potentially confusing since it wrongly suggests that the big bang took place at the center of the balloon. In fact points off the surface of the balloon have no meaning, even if they were occupied by the balloon at an earlier time.&lt;/li&gt;&#xA;  &lt;li&gt;In the &quot;raisin bread model&quot; one imagines a loaf of raisin bread expanding in the oven. The loaf (space) expands as a whole, but the raisins (gravitationally bound objects) do not expand; they merely grow farther away from each other.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;&amp;mdash;&lt;a href=&quot;http://en.wikipedia.org/wiki/Metric_expansion_of_space#Other_models_of_expansion&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&amp;dagger; Bound objects&amp;mdash;in particular galaxies&amp;mdash;are held together by binding forces like gravitation and therefore do not expand internally.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-04-10T22:36:34.547" />
  <row Id="8390" PostTypeId="1" AcceptedAnswerId="8419" CreationDate="2011-04-10T23:24:59.017" Score="4" ViewCount="314" Body="&lt;p&gt;Researchers at Michigan State University recently invented the &lt;a href=&quot;http://www.physorg.com/news/2011-03-msu-prototype-video.html/&quot; rel=&quot;nofollow&quot;&gt;Wave Disk Generator&lt;/a&gt; that is supposed to get 60% fuel efficiency. What allows it to be so much more efficient than a traditional Internal Combustion Engine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware that there is better mixing of fuel and air, but surely this alone does not produce the extreme efficiency. &lt;/p&gt;&#xA;" OwnerUserId="2843" LastEditorUserId="2843" LastEditorDisplayName="" LastEditDate="2011-04-10T23:33:38.560" LastActivityDate="2011-05-06T21:45:29.570" Title="What allows the Wave Disk Generator to be so efficient?" Tags="&lt;waves&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="8391" PostTypeId="1" CreationDate="2011-04-11T00:01:54.013" Score="4" ViewCount="115" Body="&lt;p&gt;If we look at the atomic positions in a single crystal sample with a diamond like lattice, there exist directions along which there are long hexagonal &quot;tubes&quot; (I'm not sure if these have a proper term).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/a4Y2v.jpg&quot; alt=&quot;hexagonal tubes in diamond like lattice&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happens in silicon for example.  Now if a hydrogen atom was inserted in the material, it would be too large and just get trapped in the 'matrix' of the silicon and present a defect in the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If instead I took a proton and muon in a bound &quot;muonic hydrogen&quot; state, the bohr radius which goes like 1/m would be ~ 200x's smaller.  Due to being neutral, small, and tightly bound (a binding energy &gt; 2.5 keV), I am tempted to view it almost like a large neutron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;br&gt;&#xA;Due to its much smaller size, could such a 'large neutron' pass freely through a crystal along one of these tubes?&lt;br&gt;&#xA;Could one eliminate trapped hydrogen in a crystal by making it &quot;mobile&quot; in this fashion by putting the sample in a muon beam of low enough energy that it won't damage the crystal structure but still create 'muonic hydrogen'?&lt;/p&gt;&#xA;" OwnerUserId="2655" LastActivityDate="2011-04-11T16:27:09.120" Title="Mobile &quot;muonic hydrogen&quot;" Tags="&lt;condensed-matter&gt;&lt;solid-state-physics&gt;" AnswerCount="3" />
  <row Id="8392" PostTypeId="1" CreationDate="2011-04-11T00:25:55.357" Score="2" ViewCount="212" Body="&lt;p&gt;I would like to know what it is.&lt;/p&gt;&#xA;" OwnerUserId="3041" LastActivityDate="2011-04-11T15:25:38.360" Title="What is virtual ground?" Tags="&lt;electricity&gt;&lt;electric-circuits&gt;" AnswerCount="2" />
  <row Id="8393" PostTypeId="2" ParentId="8392" CreationDate="2011-04-11T00:33:00.933" Score="6" ViewCount="" Body="&lt;p&gt;&quot;Ground&quot; refers to a particular voltage, generally taken to be &quot;zero&quot;, or the voltage of the earth. A &quot;virtual ground&quot; is a wire in a circuit whose voltage is held to be zero not because it is directly connected to the true ground, but instead because it is actively driven to that voltage typically by feedback mechanisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example of a virtual ground circuit:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/Zlj9O.jpg&quot; alt=&quot;Virtual ground circuit&quot;&gt;&lt;br&gt;&#xA;In the above, the large triangle represents an OP-AMP, that is, an &lt;a href=&quot;http://en.wikipedia.org/wiki/Operational_amplifier&quot; rel=&quot;nofollow&quot; rel=&quot;nofollow&quot;&gt;operational amplifier&lt;/a&gt;. It drives its output (marked VGND) to be + voltage when its + input has a higher voltage than its - input, and to be - voltage when its - input has a higher voltage than its + input. Since the + input is connected to true ground (marked GND), the op-amp will keep it output VGND at the GND level (to the best of its ability). Since the VGND wire is held at ground but is not connected to ground, it is called a &quot;virtual ground&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, in the usual &quot;electric-circuits&quot; limit, circuits are composed of &quot;wires&quot; where we make the assumption that the voltage of a wire is the same no matter where on the wire we look. That is, we ignore the process by which a signal is sent down a wire.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-04-11T01:47:00.087" LastActivityDate="2011-04-11T01:47:00.087" />
  <row Id="8394" PostTypeId="2" ParentId="8246" CreationDate="2011-04-11T00:57:25.617" Score="1" ViewCount="" Body="&lt;p&gt;What's going on is that there are two flows here, a gas flow (produced by boiling) and a liquid flow. The liquid flow is from the entrance to the exit while the gas flow is to the air. The gas flow is controlled by allowing only one exit for the gas, near the place where you add syrup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The longer the fluid spends in the contraption the more water is boiled out of it and the more sugar that is left. So by the time the liquid gets to the output spigot it's spent a lot of time boiling and has lost a lot of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process is similar to distillation in that the steam preferentially carries off water rather than sugar. This suggests that a more efficient method of processing maple syrup is distillation, and that designing an evaporator so that it mimics the distillation process as much as possible is a good idea. The US Forest Service agrees:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.fs.fed.us/ne/newtown_square/publications/research_papers/pdfs/scanned/ne_rp385p.pdf&quot; rel=&quot;nofollow&quot;&gt;Processing Maple Syrup with a Vapor Compression Distiller: An Economic Analysis&lt;/a&gt; Lawrence D. Garrett, Forest Service Research Paper NE-385, 1977&lt;br&gt;&#xA;Forest Service, U.S. Department of Agriculture,&lt;br&gt;&#xA;Northeastern Forest Experiment Station&lt;br&gt;&#xA;6816 Market Street, Upper Darby, PA 19082&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A test of vapor compression distillers&#xA;  for processing maple syrup revelaed&#xA;  that (1) vapor compression equipment&#xA;  tested evaporated 1 pound of water&#xA;  with 0.047 pounds of steam equivalent&#xA;  (electrical energy); open-pan&#xA;  evaporators of similar capacity&#xA;  required 1.5 pounds of steam&#xA;  equivalent (oil energy) to produce 1&#xA;  pound of water; (2) vapor compression&#xA;  evaporators produced a syrup of equal&#xA;  quality to that from a conventional&#xA;  open-pan evaporation plant; and (3) a&#xA;  central plant producing 8,000 gallons&#xA;  of syrup per year should yield a&#xA;  return of 16 percent on investment.&#xA;  Increasing annual product output&#xA;  should increase the return on&#xA;  investment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-06-13T04:17:51.540" LastActivityDate="2011-06-13T04:17:51.540" CommentCount="7" />
  <row Id="8395" PostTypeId="2" ParentId="8391" CreationDate="2011-04-11T02:21:10.897" Score="-1" ViewCount="" Body="&lt;p&gt;This is a case of particle &lt;a href=&quot;http://en.wikipedia.org/wiki/Channelling_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;channelling&lt;/a&gt;.  A particle can pass through planes or “tubes” in crystals of this nature.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-04-11T02:21:10.897" CommentCount="1" />
  <row Id="8396" PostTypeId="1" CreationDate="2011-04-11T03:54:57.263" Score="1" ViewCount="163" Body="&lt;p&gt;This is in continuation to my &lt;a href=&quot;http://physics.stackexchange.com/questions/8303/does-the-hup-alone-ensure-the-randomness-in-qt&quot;&gt;previous question&lt;/a&gt;. It is not a duplicate of the previous one. This question arises because of the answers and discussions in that question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we call a theory, quantum theory, if it is consistent with HUP? For example, suppose there is a finite and self consistent theory of gravity which incorporates the uncertainty principle. Can we at once call this theory a quantum theory of gravity or does it have to satisfy other conditions too?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question may be too basic but it is intriguing my mind.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-04-11T07:11:07.407" Title="What are the conditions to be satisfied by a theory in order to be a quantum theory?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-gravity&gt;" AnswerCount="4" />
  <row Id="8397" PostTypeId="2" ParentId="8396" CreationDate="2011-04-11T05:14:42.593" Score="3" ViewCount="" Body="&lt;p&gt;A quantum theory is a theory that obeys the postulates of quantum mechanics, see e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Postulates_of_quantum_mechanics#Postulates_of_quantum_mechanics&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Postulates_of_quantum_mechanics#Postulates_of_quantum_mechanics&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One may choose a slightly different wording of the conditions, add some detailed ones or unify the conditions above to a smaller list, but it must always be effectively the same thing. There are no exceptions from this definition. In particular, a theory of quantum gravity is a theory predicting gravity that is compatible with the postulates of quantum mechanics. (We usually want a theory of quantum gravity to be compatible with the principles of relativity, too.)&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-11T05:14:42.593" CommentCount="0" />
  <row Id="8398" PostTypeId="2" ParentId="8396" CreationDate="2011-04-11T06:00:29.197" Score="2" ViewCount="" Body="&lt;p&gt;In mathematical reasoning, and not only, there are what are defined as necessary conditions, and others that are defined as sufficient conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Necessary means that without this condition the theory cannot be proven/ be consistent. Sufficient means that just from this condition the theory is consistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The QM postulates are necessary &lt;strong&gt;and&lt;/strong&gt; sufficient for a quantum mechanical theory to be fully developed.  The HUP is a theorem arising from the postulates of the QM theory, so it is a necessary attribute. But it is not sufficient to define the postulates, they cannot be derived from it,  and thus not sufficient to define a theory as a quantum theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the answer by @Roy Simpson in your previous question. I remember this &quot;uncertainty principle&quot; from my classical electrodynamics course. That it exists, it does not make classical electrodynamics a quantum theory.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-11T06:00:29.197" CommentCount="1" />
  <row Id="8399" PostTypeId="2" ParentId="8396" CreationDate="2011-04-11T07:04:02.793" Score="0" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;.NET is a framework&lt;/li&gt;&#xA;&lt;li&gt;Software that is built using languages supported by .NET is called .NET software. &lt;/li&gt;&#xA;&lt;li&gt;That software may or may not break if some part of .NET breaks. &lt;/li&gt;&#xA;&lt;li&gt;It is necessary to have .NET in place for the software to work.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now replace .NET with QM and software with theory. :-D&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2011-04-11T07:04:02.793" />
  <row Id="8400" PostTypeId="1" AcceptedAnswerId="8406" CreationDate="2011-04-11T07:10:36.833" Score="5" ViewCount="158" Body="&lt;p&gt;In textbooks the characteristic length scale of an exciton, or an electron bound to dopant atom, in silicon is calculated by analogy to the vacuum case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bohr radius in vacuum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_0 = \frac{4 \pi \varepsilon_0 \hbar^2}{m_0 e^2} = 0.53\ {\rm &amp;#197;}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\varepsilon_0$ is the vacuum  permittivity, $m_0$ is the electron mass, and $e$ is the electron charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exciton bohr radius in material:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a = \frac{4 \pi \varepsilon_{\mathrm{r}} \varepsilon_0 \hbar^2}{m_e e^2} = a_0 \varepsilon_{\mathrm{r}} \frac{m_0}{m_e}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\varepsilon_{\mathrm{r}}$ is the relative permittivity, and $m_e$ is the effective mass of the electron in the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For silicon, these can be looked up in &lt;a href=&quot;http://inmmc.org/ftp/material/silicon-electrical.html&quot; rel=&quot;nofollow&quot;&gt;a properties table&lt;/a&gt; and are:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\varepsilon_{\mathrm{r}} =  11.7$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_e = 0.26 m_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a = \frac{11.7}{0.26} 0.53\ {\rm &amp;#197;} = 23.85\ {\rm &amp;#197;}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, experimental evidence suggests an exciton radius of $4.3\ \mathrm{nm}$ in silicon (according to Yoffe, AD. Advances in Physics, 42(2) pg 173, 1993).  This seems quite a bit off to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the idea of effective mass not entirely appropriate for this analogy?&lt;br&gt;&#xA;And more specifically, what about in heavy fermion systems where the effective mass of the electron can be even larger than a muon!? This would give an incredibly small scale for the exciton or defect bound radius.&lt;/p&gt;&#xA;" OwnerUserId="2655" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-11T07:48:47.403" LastActivityDate="2011-04-11T09:01:50.973" Title="When is use of the 'effective mass' concept appropriate?" Tags="&lt;condensed-matter&gt;&lt;solid-state-physics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="8401" PostTypeId="2" ParentId="8396" CreationDate="2011-04-11T07:11:07.407" Score="2" ViewCount="" Body="&lt;p&gt;Apparently, &lt;strong&gt;No&lt;/strong&gt;. To quote &lt;a href=&quot;http://plato.stanford.edu/entries/qt-uncertainty/&quot; rel=&quot;nofollow&quot;&gt;SEP&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A second point is the question whether the theoretical structure or the quantitative laws of quantum theory can indeed be derived on the basis of the uncertainty principle, as Heisenberg wished. Serious attempts to build up quantum theory as a full-fledged Theory of Principle on the basis of the uncertainty principle have never been carried out. Indeed, the most Heisenberg could and did claim in this respect was that the uncertainty relations created &quot;room&quot; (Heisenberg 1927, p. 180) or &quot;freedom&quot; (Heisenberg, 1931, p. 43) for the introduction of some non-classical mode of description of experimental data, not that they uniquely lead to the formalism of quantum mechanics. A serious proposal to construe quantum mechanics as a theory of principle was provided only recently by Bub (2000). But, remarkably, this proposal does not use the uncertainty relation as one of its fundamental principles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All the mystique surrounding the Heisenberg Uncertainty principle vanishes if you take into account the deBroglie hypothesis. I remember from the Feynman lectures, '...what follows is something that is familiar to anyone who works with waves'. It might be worthwhile also to look at Terry Tao's &lt;a href=&quot;http://terrytao.wordpress.com/2009/02/18/hardys-uncertainty-principle/&quot; rel=&quot;nofollow&quot;&gt;post&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1671" LastActivityDate="2011-04-11T07:11:07.407" />
  <row Id="8402" PostTypeId="1" AcceptedAnswerId="8405" CreationDate="2011-04-11T07:13:49.620" Score="-1" ViewCount="211" Body="&lt;p&gt;What variables affect the damping of a spring executing simple harmonic motion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the independent variables, and what variables would need to be controlled in an experiment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm attempting to complete an investigation where I measure the decrease in amplitude of a damped spring, and to prove the relationship between variables in the motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3044" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-04-11T07:56:49.833" LastActivityDate="2011-04-11T17:22:10.550" Title="What affects the damping of a spring?" Tags="&lt;homework&gt;&lt;harmonic-oscillator&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8403" PostTypeId="2" ParentId="2347" CreationDate="2011-04-11T07:21:34.043" Score="1" ViewCount="" Body="&lt;p&gt;You should definitely look into Algodoo (formerly Phun) at &lt;a href=&quot;http://www.algodoo.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.algodoo.com/&lt;/a&gt;&#xA;This is not a game per se, just a &quot;physics sandbox&quot; - a game with no goal other than one you set yourself. Rich mechanics, fluid dynamics, air friction, motors, ability to change object mass, friction, flexibility, collision layers and so on...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another game you might be interested in is the Powder Game, &lt;a href=&quot;http://dan-ball.jp/en/javagame/dust/&quot; rel=&quot;nofollow&quot;&gt;http://dan-ball.jp/en/javagame/dust/&lt;/a&gt; much more simple but with rich particle physics simulation - various materials of specific properties: water(freezable, evaporable), explosives, magma, sand, rock, oil (burnable) and so on.&lt;/p&gt;&#xA;" OwnerUserId="2754" LastActivityDate="2011-04-11T07:21:34.043" CommunityOwnedDate="2011-04-11T07:21:34.043" />
  <row Id="8404" PostTypeId="2" ParentId="8392" CreationDate="2011-04-11T07:26:46.423" Score="6" ViewCount="" Body="&lt;p&gt;Virtual ground refers to a circuit element not directly connected to ground, held at a reference voltage.  This reference voltage need not be the same voltage as ground either.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example many op-amp circuits were originally designed for dual power supplies (say +12V and -12V) and could handle filtering or modification of a signal that was oscillating about ground.  This need for two power supplies is annoying, so it is possible to remake these circuits with a single supply, and a &quot;virtual ground&quot; held at half the power supply.  As long as signal voltage are compared to this virtual ground, filtering or modification of the oscillating signals can be handled just fine as they only go negative relative to the virtual ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way, it is fairly straight forward to convert dual supply op-amp circuits with single supply circuits by replacing the ground reference voltage in the circuit to a virtual ground to which the signal is referenced.  This is the main context in which I have seen 'virtual ground' have a useful application.&lt;/p&gt;&#xA;" OwnerUserId="2652" LastActivityDate="2011-04-11T07:26:46.423" CommentCount="1" />
  <row Id="8405" PostTypeId="2" ParentId="8402" CreationDate="2011-04-11T07:45:19.787" Score="2" ViewCount="" Body="&lt;p&gt;You should already be familiar with damping.  It simply refers to the fact that if you set a spring going, it eventually stops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Damping&quot; rel=&quot;nofollow&quot;&gt;The wikipedia article&lt;/a&gt; should cover most of what you want to know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any particular spring may be damped for all sorts of reasons.  Any way the spring can lose energy contributes to damping, so it could be lost internally to heat due to stressing the material, or externally to heat via friction, or externally to an electromagnetic field, to some sort of mechanical &lt;a href=&quot;http://en.wikipedia.org/wiki/Dashpot&quot; rel=&quot;nofollow&quot;&gt;dashpot&lt;/a&gt;, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were designing an experiment to study damping, you could be interested in a number of different particular things.  You will have to make your own choice about what the most interesting thing to study is.  For example, would you like measure the &lt;a href=&quot;http://en.wikipedia.org/wiki/Damping_ratio&quot; rel=&quot;nofollow&quot;&gt;damping ratio&lt;/a&gt;?  The overall magnitude of the damping effect?  The time it takes your spring to reduce its amplitude to 1/2 its previous value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a particular effect picked out, do you want to know how it depends on the load on the spring, or the material of the spring, or whether the spring is in a vacuum, etc.  There are many possible things to study, so your question has no definite answer.  I'm sure you can find something particular you find worthwhile.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-04-11T07:45:19.787" CommentCount="9" />
  <row Id="8406" PostTypeId="2" ParentId="8400" CreationDate="2011-04-11T08:01:40.377" Score="3" ViewCount="" Body="&lt;p&gt;Dear John, note that 23.85 Å is equal to 2.385 nm, while the observed 4.3 nm is approximately two times larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a simple error in your calculation that exactly fixes the factor of two. Note that the actual calculation you should have done has a radius proportional to $1/m$ and the correct $m$ that you should substitute is the reduced mass of the two-body problem governing the relative position of the two particles.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Reduced_mass&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Reduced_mass&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reduced mass is $m_1 m_2 / (m_1+m_2)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the important point is that an exciton is not a bound state of the effective electron and a superheavy nucleus: instead, it is a bound state of an effective electron and an effective hole - a larger counterpart of the positronium (an electron-positron bound state). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Exciton&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Exciton&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming that both the electron and hole masses are equal, 0.26 $m_0$, the reduced (and still also effective) mass is 0.26/2 $m_0$ = 0.13 $m_0$, and the resulting $a$ is twice as big as your result, 4.77 nm - assuming that your arithmetics is right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deviation from 4.3 nm is not too large but I can only handwave if I were trying to pinpoint the most important source of the discrepancy. It could be a different effective mass of the hole; finite-size effects caused by the fact that the silicon atoms were not quite uniformly distributed inside the exciton, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, in fact, I noticed that your properties table does include a special figure of the effective hole's mass and it differs from the electron mass: 0.38 $m_0$. So the reduced mass is &#xA;$$\frac{0.38\times 0.26}{0.38+0.26} m_0 = \frac{0.0988}{0.64} m_0 = 0.154 m_0 $$&#xA;and the calculated radius is&#xA;$$ \frac{11.7}{0.154} \times 0.53\ &amp;#197; = 40.3\ &amp;#197;. $$&#xA;Well, this is 7 percent too small, much like the previous one was 7 percent too big. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hydrogen atoms with composite heavy fermions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning your second question, as you clearly realize, the calculated radius of the &quot;atom&quot; with such &quot;heavy electrons&quot; would be much smaller than the ordinary atom. This also proves that the assumptions of such a calculation fail: the heavy fermions (in condensed matter physics) are the result of the collective action of many atoms on the electron and its mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the large mass of the heavy fermions is only appropriate for questions about physics at long distances - much longer than the ordinary atom. If you look at very short distances - a would-be tiny atom with the heavy fermion - you cannot use the long-distance or low-energy effective approximations of condensed matter physics. You have to return to the more fundamental, short-distance or high-energy description which sees electrons again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, you will find out that there can be no supertiny atoms created out of the effective particles such as heavy fermions. The validity of all such phenomenological effective theories - such as those with heavy fermions - is limited to phenomena at distances longer than a certain specific cutoff and highly sub-atomic distances surely violate this condition, so one must use a more accurate theory than this effective theory, and in those more effective theories, most of the fancy emergent condensed matter objects disappear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Non-relativistic effective theories&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just a disclaimer for particle physicists: in this condensed matter setup, we are talking about non-relativistic theories so the maximum allowed energy $E$ of quasiparticles doesn't have to be $pc$ where $p$ is the maximum allowed momentum in the effective theory. In other words, we can't assume $v/c=O(1)$. Quite on the contrary, the validity of such effective theories in condensed matter physics typically depends on the velocities' being much smaller than the speed of light, too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the mass of the heavy fermions is much greater than $m_0$ which would make $m_e c^2$ much greater than $m_0 c^2$; however, the latter is not a relevant formula for energy in non-relativistic theories. Instead, $p^2/2m_e$, which is (for heavy fermions) much smaller than the kinetic energy of electrons, is relevant. The maximum allowed $p$ of these quasiparticles is much larger than $\hbar/r_{\rm Bohr}$ - the de Broglie wavelength must be longer than the Bohr radius. That makes $p^2/2m_e$ really tiny relatively to the Hydrogen ionization energy.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-11T09:01:50.973" LastActivityDate="2011-04-11T09:01:50.973" CommentCount="2" />
  <row Id="8407" PostTypeId="1" CreationDate="2011-04-11T08:04:15.083" Score="-1" ViewCount="279" Body="&lt;p&gt;Why do electrons (and other very small particles) sometimes behave as particles (i.e. when we are not looking at them) where as other times they behave as waves?&lt;/p&gt;&#xA;" OwnerUserId="3046" LastEditorUserId="3048" LastEditorDisplayName="" LastEditDate="2011-04-11T13:23:50.767" LastActivityDate="2011-04-11T13:23:50.767" Title="Why electrons behave as a particle and also as a wave?" Tags="&lt;quantum-mechanics&gt;&lt;particle-physics&gt;&lt;waves&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="8408" PostTypeId="1" CreationDate="2011-04-11T08:06:00.993" Score="-1" ViewCount="278" Body="&lt;p&gt;This questions started with a question I had about gravity.  If two objects of different weights fall to the earth at the same rate of acceleration, then it seems to me that gravity is in some ways 'calculating' the weight of each item and applying the appropriate force to each item so as to have it fall at the same rate of acceleration.  Is this true (or at least close to the truth)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This got me think that perhaps this is what all of the mathematical equations of physics are really saying - namely that there are mathematical equations that are getting applied to the real world in one way or another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this right?  If not, why not?&lt;/p&gt;&#xA;" OwnerUserId="3045" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-11T13:19:44.767" LastActivityDate="2011-04-11T15:09:58.557" ClosedDate="2011-04-11T17:00:01.000" Title="Are the laws of physics applied mathematics?" Tags="&lt;mathematics&gt;&lt;laws-of-physics&gt;" AnswerCount="6" CommentCount="5" />
  <row Id="8409" PostTypeId="2" ParentId="8407" CreationDate="2011-04-11T08:25:55.920" Score="7" ViewCount="" Body="&lt;p&gt;I love a quote from my QM teacher &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An electron is what it is [...] words like wave or particle are coined by us to paraphrase its properties, and these properties are properties of objects with which we have daily experience. [...] its a semantic failure. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So you might want to discuss at &lt;a href=&quot;http://english.stackexchange.com/&quot;&gt;English.SE&lt;/a&gt; ;)&lt;/p&gt;&#xA;" OwnerUserId="1671" LastActivityDate="2011-04-11T08:25:55.920" />
  <row Id="8410" PostTypeId="2" ParentId="8407" CreationDate="2011-04-11T08:27:42.010" Score="6" ViewCount="" Body="&lt;p&gt;Electrons behave like the same thing all the time.  That thing is a solution to a &lt;a href=&quot;http://en.wikipedia.org/wiki/Partial_differential_equation&quot; rel=&quot;nofollow&quot;&gt;partial differential equation&lt;/a&gt;, called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation&quot; rel=&quot;nofollow&quot;&gt;Schrodinger equation&lt;/a&gt;.  The solution to the Schrodinger equation is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wave_function&quot; rel=&quot;nofollow&quot;&gt;wave function&lt;/a&gt; of the electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the behavior is always in accord with the same equation, in different circumstances the qualitative aspects of the wave function may be different.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical wave functions include &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_function&quot; rel=&quot;nofollow&quot;&gt;gaussians&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Sine_wave&quot; rel=&quot;nofollow&quot;&gt;sinusoids&lt;/a&gt; (perhaps inside some sort of &lt;a href=&quot;http://en.wikipedia.org/wiki/Wave_packet&quot; rel=&quot;nofollow&quot;&gt;wave packet&lt;/a&gt;).  These functions can either be very narrow or very broad.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the wave function is broadly-spread compared to the thing it's interacting with, we see significant &lt;a href=&quot;http://en.wikipedia.org/wiki/Interference_%28wave_propagation%29&quot; rel=&quot;nofollow&quot;&gt;interference&lt;/a&gt; effects.  This is frequently called &quot;behaving like a wave&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the wave function is narrow compared to the thing it's interacting with, we don't see much interference, and this is called &quot;behaving like a particle&quot;. Mathematically, though, the wave function (and hence the electron) is behaving in the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many applets online that let you play around with wave functions to see how they behave.  The first one I found by clicking around is &lt;a href=&quot;http://www.physics.orst.edu/~rubin/nacphy/CPapplets/QMWAVE/Potbar/Potbar.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-04-11T08:34:39.223" LastActivityDate="2011-04-11T08:34:39.223" />
  <row Id="8411" PostTypeId="2" ParentId="8408" CreationDate="2011-04-11T08:40:47.500" Score="0" ViewCount="" Body="&lt;p&gt;From my point of view, the equations are &quot;only&quot; used to model the observed reality. You can fine-tune your equations and constants, construct new models and try predicting phenomena beyond current observations, but since one probably will never be able to examine the real &quot;sourcecode&quot; of the universe, you can only iterate and never tell when convergence is actually achieved.&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2011-04-11T08:40:47.500" CommentCount="2" />
  <row Id="8412" PostTypeId="2" ParentId="8408" CreationDate="2011-04-11T08:51:25.210" Score="0" ViewCount="" Body="&lt;p&gt;As Tobias wrote in his answer, the equations we have are just models. They describe the &lt;em&gt;observed&lt;/em&gt; relationships between measured (or measurable) quantities, that's all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're asking whether nature actually goes through the process of doing calculations to figure out, say, how much the force between two objects is, that question is outside the scope of physics.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-04-11T08:51:25.210" CommentCount="8" />
  <row Id="8413" PostTypeId="2" ParentId="8408" CreationDate="2011-04-11T08:59:55.457" Score="1" ViewCount="" Body="&lt;p&gt;This is a &lt;em&gt;very&lt;/em&gt; philosophical question. Some pointers from giants:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Timothy Gowers' has an excellent &lt;a href=&quot;http://books.google.com/books?id=kC25qnVj4doC&amp;amp;lpg=PP10&amp;amp;dq=gowers%20mathematics%20a%20very%20short&amp;amp;pg=PA1#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;description&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As does &lt;a href=&quot;http://www.youtube.com/watch?v=1SrHzSGn-I8&quot; rel=&quot;nofollow&quot;&gt;Feynman&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1671" LastActivityDate="2011-04-11T08:59:55.457" CommentCount="2" />
  <row Id="8414" PostTypeId="2" ParentId="8408" CreationDate="2011-04-11T09:12:03.343" Score="0" ViewCount="" Body="&lt;p&gt;A mathematical description of an observation does not say  how  the observed was created.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Draw a perfect circle of radius a. Do the ink points &quot;know&quot; that a compass and your hand was used to make this mathematical description: r=a , into a perfect circle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many referential levels, meta levels, in any mathematical description. Sometimes known by construction as in the circle example, sometimes as in your gravity question  still a point of research. The referential level of two objects dropping are different from the level of the gravitational equations of general relativity which are the ultimate at the moment mathematical model for gravity. The earth is not calculating anything,  in the same sense that the points on the circle were not calculating the r=a. The equations of motion of bodies is a mathematical model of observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: When one says &quot;Nature does this or that&quot; a level of anthropomorphism enders. It certainly belongs to  metaphysics questions  and not to physics. My almost metaphysical view is that mathematical forms  exist , the way 2+2 exists, and matter as we know it, given the appropriate conditions &quot;crystallizes&quot; out following the energy levels etc. of the mathematical equations. Our mathematical models are successive levels of approximations to these ideal equations. No calculations by nature or anybody are involved, imo.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-04-11T11:17:18.507" LastActivityDate="2011-04-11T11:17:18.507" CommentCount="4" />
  <row Id="8415" PostTypeId="2" ParentId="2347" CreationDate="2011-04-11T09:12:59.720" Score="2" ViewCount="" Body="&lt;p&gt;Mafia II (2010) currently has the world's best physics model for gravity, acceleration, driving and destruction of parts of cars (exact models of physical characteristics of dozens of types of historical cars), buildings, piling up of debris, ashes, broken glass on the floor, people, friction, fire, smoke, optics including complicated absorption and reflection models off different types of metals on the cars' surface, and everything else. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all those respects, it was a huge winner, as expected from a similarly groundbreaking Mafia (I) in 2002. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the play fun itself is limited as people often think that there are not too many things to nontrivially interact with in the city and the tasks form a completely linear sequence which is often considered boring - even though this was the Mafia II's plan from the beginning because this leads to much more organized, movie-like experience. Mafia II - and dozens of other games that exist today - uses NVIDIA's PhysX enhancement of the physics calculations by graphics cards, see e.g. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://youtube.com/watch?v=GMHeIGjLZ9I&quot; rel=&quot;nofollow&quot;&gt;http://youtube.com/watch?v=GMHeIGjLZ9I&lt;/a&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want a more impressive, financially unrealistic but physically valid scene with thousands of barrels that are falling and exploding, see Crysis Physics&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=YG5qDeWHNmk&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=YG5qDeWHNmk&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which still remains a benchmark.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-11T09:19:37.937" LastActivityDate="2011-04-11T09:19:37.937" CommunityOwnedDate="2011-04-11T09:12:59.720" />
  <row Id="8417" PostTypeId="2" ParentId="8408" CreationDate="2011-04-11T09:34:42.330" Score="0" ViewCount="" Body="&lt;p&gt;The fact that some of the laws describing very complicated and chaotic systems are in fact very short and 'elegant', and that any other description of the phenomenon in equations must be many orders of magnitude longer, seems to me to be a very good indicitor that Nature does infact follow some of the laws we currently know exactly. And that these equations are not just some human invented thing to explain phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ofcourse we can never prove which laws Nature follows and if it does so, but likewise we can never prove there aren't exact laws governing Nature, which some people here seem to think.&lt;/p&gt;&#xA;" OwnerUserId="1708" LastEditorUserId="1708" LastEditorDisplayName="" LastEditDate="2011-04-11T10:10:16.993" LastActivityDate="2011-04-11T10:10:16.993" CommentCount="1" />
  <row Id="8419" PostTypeId="2" ParentId="8390" CreationDate="2011-04-11T10:30:11.633" Score="2" ViewCount="" Body="&lt;p&gt;That Prof Müller said it: Shock waves in addition of the usual things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture in Your link is rather different from the thing shown &#xA;in the video, for the time being, I do not understand really what is &#xA;the new thing. From thermodynamics it is clear, that they (hope?) &#xA;to have a higher &lt;strong&gt;effective&lt;/strong&gt; DeltaT, obviously without having &#xA;higher temperatures at machiney parts (which makes them expensive and/or short-lived) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that those shock waves can be transformed into working pressure &#xA;without the high temperatures of the combustion shock wave touchng machine parts. &#xA;I hope we will hear more  from Prof Müller in near future. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This link is somewhat more detailed and less press-release-silly. &#xA;in general it says what I surmised (by application of thermodynamics basics) &#xA;&lt;a href=&quot;http://www.zdnet.com/blog/emergingtech/wave-disk-engines-to-make-hybrid-vehicles-cheaper-more-efficient/1887&quot; rel=&quot;nofollow&quot;&gt;http://www.zdnet.com/blog/emergingtech/wave-disk-engines-to-make-hybrid-vehicles-cheaper-more-efficient/1887&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-11T11:02:34.800" LastActivityDate="2011-04-11T11:02:34.800" CommentCount="1" />
  <row Id="8420" PostTypeId="1" AcceptedAnswerId="8421" CreationDate="2011-04-11T11:20:24.200" Score="2" ViewCount="279" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/4243/what-does-it-mean-to-say-gravity-is-the-weakest-of-the-forces&quot;&gt;What does it mean to say &amp;ldquo;Gravity is the weakest of the forces&amp;rdquo;?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;It is said nuclear force is the strongest force in nature..&#xA;But it is not true near a black hole where gravitational force exceeds nuclear force..&#xA;So which is the strongest force in nature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and Regards&#xA;Athreya&lt;/p&gt;&#xA;" OwnerUserId="3047" LastActivityDate="2011-04-11T12:47:10.677" Title="Strongest force in nature" Tags="&lt;gravity&gt;&lt;black-holes&gt;&lt;nuclear-physics&gt;&lt;forces&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" />
  <row Id="8421" PostTypeId="2" ParentId="8420" CreationDate="2011-04-11T12:12:30.290" Score="3" ViewCount="" Body="&lt;p&gt;The gravitational force exerted by a black hole is strong because the force is proportional to the mass and the black hole as a large mass. In the same way, the gravitational field of the Earth may beat a weak enough magnet - because the Earth is large and the magnet is small and we're comparing apples and oranges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But fundamentally speaking, among particles with masses that correspond to elementary particles, gravity is the weakest force. For example, the gravitational force between two electrons is $10^{43}$ times weaker than the electrostatic one. The weak nuclear force is as strong as the electromagnetic one at very short distances - but exponentially drops at distances much longer than the W-boson wavelength. The strong nuclear force is the strongest among the four forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that the weakness of gravity relatively to all other forces, when evaluated at the level of elementary particles, is a general principle that has to hold in any consistent quantum theory of gravity, see&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0601001&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0601001&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-11T12:12:30.290" />
  <row Id="8423" PostTypeId="2" ParentId="8420" CreationDate="2011-04-11T12:47:10.677" Score="3" ViewCount="" Body="&lt;p&gt;There are two elements to this question.  The first has to do with the strength of gravitation relative to other forces of nature.  It is not hard to see that gravity is extremely weak.  One can see this mathematically in the ratio of the magnitude of the gravitational and electromagnetic forces&#xA;$$&#xD;&#xA;F_g~=~\frac{GMm}{r^2},~F_e~=~\frac{q_1q_2}{4\pi\epsilon_0 r^2},&#xD;&#xA;$$&#xA;for $M$ and $m$ the masses of a proton and electron and $q_1~=~-q_2~=~e$.  The ratio is about $F_e/F_g~\simeq~10^{39}$.  A more direct semi-empirical test is to drop an elastic ball from a height $h$.  The ball accelerates by gravity to the floor, where upon hitting the floor it very quickly reverses the direction of motion and returns to some height $h’~&amp;lt;~h$.  The electromagnetic forces which hold the molecules together in the ball and floor in a very small time period induce a change in momentum $\Delta p~=~ma\Delta t$, so the force $ma~&amp;gt;&amp;gt;~mg$ for $g~=~9.8m/s^2$ the acceleration of gravity near the Earth’s surface.  It also takes the entire mass of the Earth to give rise to the acceleration of Earth’s gravity, while it take a few grams or maybe a kilogram of mass to induce the acceleration of the ball through molecular forces, which are ultimately electromagnetic.  The nuclear force is about $100$ times the electromagnetic force, though it is short ranged and confined to the scale of hadron length $\sim~10^{-13}cm$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity becomes large if a huge amount of mass is compacted into a small volume.  The mass equivalent of the sun compressed into a volume of about $1.5km$ radius becomes a black hole.  The electromagnetic force does not tend to do this.  A concentration of lots of charge is repulsive and requires another force to bind it together.  In the case of the nucleus this is the emergent nuclear force of baryons and mesons, which is a low energy scale aspect of QCD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second aspect of this problem has to do with the variation of the relative strength of forces at different energy scale of interaction.  The function $G$, such as a propagator function, for the interaction varies with interaction coupling parameter $g$ as $G(\partial G/\partial g)^{-1}$ and where the coupling strength depends upon the energy scale $e$, often denoted by $\mu$, as a beta function&#xA;$$&#xD;&#xA;\frac{\partial g}{\partial ln(\mu)}~=~\beta(g).&#xD;&#xA;$$&#xA;This equation emerges from the dependency of terms on the renormalization cut-off, which would require a long discussion to break out.  The connection to the propagator function above is the Callin-Symanzik equation for an n-point propagator&#xA;$$&#xD;&#xA;\Big(\lambda\frac{\partial}{\partial\lambda}~+~\beta(g)\frac{\partial}{\partial g}~+~n\gamma\Big)G(x_1,x_2,\dots,x_n,g,m)~=~0,&#xD;&#xA;$$&#xA;for $\gamma$ a field scaling factor, and $\lambda$ the scaling bound or reference --- often the cut-off.  This equation is similar to a Navier-Stokes equation, and it is generalized into renormalization flow equations.  The $\beta$-function connects this most effectively with supersymmetry and strings.  The running coupling constants $g$ which depend on the energy scale converge in a fairly natural way near the Planck scale of energy.  This means the electromagnetic, or electroweak coupling, and gravitational coupling converges with the strong nuclear force at extremely high energy.  In this way the forces of nature tend to “fuse” into a single interaction with a large symmetry group and with a single interaction strength.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-04-11T12:47:10.677" />
  <row Id="8424" PostTypeId="2" ParentId="8407" CreationDate="2011-04-11T13:04:55.477" Score="0" ViewCount="" Body="&lt;p&gt;Because it is a wave, and a wave sometimes may look as a particle.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-04-11T13:04:55.477" />
  <row Id="8425" PostTypeId="1" AcceptedAnswerId="8446" CreationDate="2011-04-11T13:18:35.910" Score="2" ViewCount="128" Body="&lt;p&gt;What are the equations that describe the flow of a fluid in a porous medium? Is there a variation of the Navier-Stokes equations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to model the flow of air through a sponge-like structure. Instead of modelling the structure of the sponge, I would like to treat it like a homogenous area, where I solve appropriate equations for conservation of momentum, mass and energy.&lt;/p&gt;&#xA;" OwnerUserId="2480" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-04-14T10:28:00.007" LastActivityDate="2011-04-14T10:55:42.277" Title="Laws of fluid flow in porous medium" Tags="&lt;fluid-dynamics&gt;" AnswerCount="2" />
  <row Id="8426" PostTypeId="1" AcceptedAnswerId="9283" CreationDate="2011-04-11T13:36:34.247" Score="8" ViewCount="283" Body="&lt;p&gt;There have been some &lt;a href=&quot;http://physics.stackexchange.com/questions/5075/is-anyone-studying-how-the-topology-of-space-arises-from-more-fundamental-notions&quot;&gt;very&lt;/a&gt; &lt;a href=&quot;http://physics.stackexchange.com/questions/6980/what-are-the-reasons-to-expect-that-gravity-should-be-quantized&quot;&gt;nice&lt;/a&gt; &lt;a href=&quot;http://physics.stackexchange.com/questions/8150/can-the-connectivity-of-space-change-in-string-theory/8159#8159&quot;&gt;discussions&lt;/a&gt; &lt;a href=&quot;http://physics.stackexchange.com/questions/1216/could-gravity-be-an-emergent-property-of-nature&quot;&gt;recently&lt;/a&gt; centered around the question of whether gravity and the geometry and topology of the classical world we see about us, could be phenomena which emerge in the low-energy limits of a more fundamental microscopic theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among these, @Tim Van Beek's reply to the question on &quot;&lt;a href=&quot;http://physics.stackexchange.com/questions/5075/is-anyone-studying-how-the-topology-of-space-arises-from-more-fundamental-notions&quot;&gt;How the topology of space [time] arises from more fundamental notions&lt;/a&gt;&quot; contains the following description of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Reeh%E2%80%93Schlieder_theorem&quot; rel=&quot;nofollow&quot;&gt;Reeh-Schlieder theorem&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It describes &quot;action at a distance&quot; in a mathematically precise way. According to the Reeh-Schlieder theorem there are correlations in the vacuum state between measurements at an arbitrary distance. The point is: The proof of the Reeh-Schlieder theorem is independent of any axiom describing causality, showing that quantum entanglement effects do not violate Einstein causality, and don't depend on the precise notion of causality. Therefore a change in spacetime topology in order to explain quantum entanglement effects won't work.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is also preceded with an appropriate note of caution, saying that the above paragraph:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... describes an aspect of axiomatic quantum field theory which may become obsolete in the future with the development of a more complete theory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I had a bias against AQFT as being too abstract an obtuse branch of study to be of any practical use. However, in light of the possibility (&lt;a href=&quot;http://physics.stackexchange.com/questions/8162/whats-so-special-about-ads&quot;&gt;recently&lt;/a&gt; &lt;a href=&quot;http://physics.stackexchange.com/questions/8381/vasiliev-gravity-and-holographic-entanglement&quot;&gt;discussed&lt;/a&gt; on physics.SE) that classical geometry arises due to the entanglement between the degrees of freedom of some quantum many-body system (see Swingle's paper on &lt;a href=&quot;http://arxiv.org/abs/0905.1317&quot; rel=&quot;nofollow&quot;&gt;Entanglement Renormalization and Holography&lt;/a&gt;) the content of the Rees-Schilder theorem begins to seem quite profound and far-sighted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question therefore is: Does the Rees-Schlieder theorem provide support for the idea of building space-time from quantum entanglement? or am I jumping the gun in presuming their is some connection between what the theorem says and the work of Vidal, Evenbly, Swingle and others on &quot;holographic entanglement&quot;?&lt;/p&gt;&#xA;" LastActivityDate="2011-04-30T22:38:56.860" Title="The Reeh-Schlieder theorem and quantum geometry" Tags="&lt;quantum-gravity&gt;&lt;entanglement&gt;&lt;renormalization&gt;&lt;geometry&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="8427" PostTypeId="2" ParentId="2347" CreationDate="2011-04-11T13:41:09.250" Score="2" ViewCount="" Body="&lt;p&gt;Real-life is pretty good (at most scales - it gets pretty unrealistic in the very small details)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/xfXnC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(http://xkcd.com/880/)&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-04-11T13:41:09.250" CommunityOwnedDate="2011-04-11T13:41:09.250" />
  <row Id="8428" PostTypeId="1" CreationDate="2011-04-11T14:11:11.597" Score="-4" ViewCount="684" Body="&lt;p&gt;P. Dirac was worried with the infinities and their discarding in QED. He wanted us to reformulate the theory in order to eliminate infinities and renormalizations from the very beginning. Is there anybody who works professionally on this subject? I mean, who earns his salary for that. I know several amateurs but no one professional. &lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-04-14T07:50:33.247" LastActivityDate="2011-04-14T07:50:33.247" Title="Who works professionally on reformulation of QFT?" Tags="&lt;quantum-field-theory&gt;&lt;soft-question&gt;&lt;renormalization&gt;&lt;reformulation&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="3" />
  <row Id="8429" PostTypeId="2" ParentId="8373" CreationDate="2011-04-11T14:15:28.690" Score="1" ViewCount="" Body="&lt;p&gt;I'm nitpicking here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quote from Lubos's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;... one meter as 1/40,000,000 of the average meridian, one kilogram as a mass of a randomly chosen prototype ...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost!  Yes, 1 metre was 1/40,000,000th of the average meridian, then 1 litre = 10cm x 10cm x 10cm, and 1 kilogram is the mass of 1 litre of water.&lt;/p&gt;&#xA;" OwnerUserId="3049" LastActivityDate="2011-04-11T14:15:28.690" CommentCount="1" />
  <row Id="8430" PostTypeId="2" ParentId="8428" CreationDate="2011-04-11T14:20:28.190" Score="8" ViewCount="" Body="&lt;p&gt;Kenneth Wilson. He won a Nobel prize for it.&lt;/p&gt;&#xA;" OwnerUserId="3050" LastActivityDate="2011-04-11T14:20:28.190" CommentCount="4" />
  <row Id="8431" PostTypeId="1" CreationDate="2011-04-11T14:26:00.527" Score="1" ViewCount="82" Body="&lt;p&gt;In theories with extended supersymmetry, both short and long multiplets exist. For some reason or other, short multiplets are studied more often. Why? What's wrong with long multiplets?&lt;/p&gt;&#xA;" OwnerUserId="3050" LastActivityDate="2011-04-11T18:07:28.657" Title="Why is there more analysis of short multiplets compared to long multiplets?" Tags="&lt;supersymmetry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8432" PostTypeId="1" CreationDate="2011-04-11T14:26:05.580" Score="2" ViewCount="156" Body="&lt;p&gt;Suppose that an electron with spin up emits a photon in the field of an ion (&lt;a href=&quot;http://en.wikipedia.org/wiki/Bremsstrahlung&quot; rel=&quot;nofollow&quot;&gt;bremsstrahlung&lt;/a&gt;). What is the spin of the emitted photon? Is it correct to say that the photon is circularly polarized if the spin of the electron flips down and linearly polarised if it remains up?&lt;/p&gt;&#xA;" OwnerUserId="2560" LastEditorUserId="174" LastEditorDisplayName="" LastEditDate="2011-04-11T20:00:42.053" LastActivityDate="2011-04-11T20:00:42.053" Title="Is spin is a conserved entity?" Tags="&lt;photons&gt;&lt;spin&gt;" AnswerCount="2" />
  <row Id="8433" PostTypeId="1" CreationDate="2011-04-11T14:55:38.937" Score="-2" ViewCount="160" Body="&lt;p&gt;Eventually, we are going to reach our limits in particle physics, or maybe find the theory of everything. Particle physics isn't easy by any means. It requires at the minimum years of graduate school and more to master. The main motivation of learning is to come up with original research later. But if there is no more research to do in the future, hardly anyone would be motivated to learn particle physics given its intrinsic difficulty. What would happen then? Would knowledge of particle physics disappear because no one would be around to remember it? Or would it become a scholastic tradition passed down dogmatically from generation to generation.&lt;/p&gt;&#xA;" OwnerUserId="3052" LastActivityDate="2011-04-11T15:54:27.257" ClosedDate="2011-04-12T16:45:43.310" Title="The future of particle physics" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8434" PostTypeId="1" CreationDate="2011-04-11T15:01:03.750" Score="2" ViewCount="140" Body="&lt;p&gt;I'm working on a 2D He superfluid system with vortices. I was asked to calculate the kinetic energy of vortex-(anti-)vortex pairs and compare the two situations. One finds in literature that the vortex-vortex situation is unstable, and the vortex-antivortex pair can be stable. This is a crucial observation in what follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the calculation of the energies, I came across a striking fact, which may or may not have any physical implications: the (kinetic) energy of the vortex-vortex situation had an imaginary component, while in the vortex-antivortex energy, everything imaginary nicely cancelled. It seems my results are consistent with what is expected, and certainly scale as they should for the system (energy logarithmically in the size of the system, and in the distance between the vorteces).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For now, I just &quot;forgot&quot; about the imaginary part of the energy, but my question is this:&#xA;Can the imaginary part of the energy be interpreted as the lifetime of an unstable state in the case of a vortex-vortex situation (comparable to the lifetime of resonances in S-wave scattering, which also comes into play as the complex part of the energy of the resonance). Or is it merely a remnant of my approximate calculation and the result of logarithms in the complex plane and have I been staring at this too long?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: for those interested, the integration I performed can be viewed as a question on math.stackexchange.com. Please ignore the question itself, because it is not really an example of my intellect in this field, just some conjectures/&quot;fleeting hope&quot; &lt;code&gt;;)&lt;/code&gt; &lt;a href=&quot;http://math.stackexchange.com/questions/32286/choosing-the-branch-of-a-logarithm&quot;&gt;http://math.stackexchange.com/questions/32286/choosing-the-branch-of-a-logarithm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="317" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-11T16:44:52.130" LastActivityDate="2011-05-11T17:45:11.563" Title="Complex valued energy" Tags="&lt;resonance&gt;&lt;stability&gt;&lt;superfluidity&gt;&lt;vortex&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8435" PostTypeId="2" ParentId="8408" CreationDate="2011-04-11T15:09:58.557" Score="1" ViewCount="" Body="&lt;p&gt;Plato pondered the relationship between physical reality and mathematics 2500 years ago.  He used the allegory of his cave to show that we see traces of structure in the physical world, but we do not see the full picture.  I don’t particularly hold to the idea of Platonia, but the idea is food for some thought.  It certainly suggests that we may never know the answer to whether reality is ultimately some sort of mathematics which is reified by some means.  Penrose has suggested this reification is the existence of mind, where a mind is a way that this Platonia has to become aware of itself and observes this system as “reality.”  Again, I don’t know whether it is worth embracing this as Truth, for I see no way it can ever be verified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tegmark has proposed some ideas along these lines.  He argues that mathematics, or that set of it which is first order and halting and non-halting if it converses to some describable set, in his mathematical universe hypothesis defines the entire set of reality.  The ensemble has some statistical weight to its elements which defines the probability the mathematics exists in some “universe” or exists in some reality.  Of course the problem is that this set is infinite.  An infinite set is one which admits a bijective map from the set to any nontrivial (finite) subset.  Such a set of maps will include the Cantor diagonalization and as a consequence the Godel incompleteness issue.  So it would appear that Tegmark’s system is computed within the Chaitan halting probability function, which is itself not computable.  We may then never know if this system has any bearing on reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that a determination of the truth value of these conjectures require that we somehow access information or knowledge which is outside of physical reality.  Without doing that this question might amount to chasing one’s own tail endlessly.  It is very unlikely this question will be satisfactorily addressed, even as a hypothesis supported by some data set, by ordinary means of science.  So in general this question is best something thought about in evening over scotch and cigars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To conclude one might ask whether reality is fully based on mathematics, or whether on some deep level it is based on magic.  By magic we might mean supernaturalism or some theological imposition of an infinite will.  If physics is not fully based on mathematics, then there is ultimately some sort of magic deep down in the “rivers of hidden funk.”  The question might not be resolvable, but the negative seems far more disturbing.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-04-11T15:09:58.557" CommentCount="1" />
  <row Id="8436" PostTypeId="1" CreationDate="2011-04-11T15:13:40.077" Score="2" ViewCount="117" Body="&lt;p&gt;Related to my other misguided question, D-branes are equivalent to p-branes. D-branes are described by a sheet in topologically trivial spacetime. p-branes are extremal black branes. They have an event horizon which is infinitely far away along spatial geodesics. However, we may cross the event horizon in a finite affine time for null and timelike geodesics. So, we need to extend spacetime, and if we do that, the Penrose diagram consists of an infinite chain of universes, ordered by time. Each universe has its own null infinity. Causal information is &quot;created&quot; in the past null infinity of each universe. So, a universe in the &quot;future&quot; contains information which is causally independent from and inaccessible to &quot;our universe&quot;. All this is invisible if we work with D-branes. Where did all the other universes go to? Where is all the information contained within them encoded?&lt;/p&gt;&#xA;" OwnerUserId="2955" LastActivityDate="2011-04-11T15:13:40.077" Title="If D-branes are the same as p-branes, and p-branes open out to &quot;another universe&quot;, why is the other universe invisible in D-branes?" Tags="&lt;string-theory&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="8437" PostTypeId="2" ParentId="8433" CreationDate="2011-04-11T15:17:07.053" Score="3" ViewCount="" Body="&lt;p&gt;There are huge challenges left in particle physics. There are number of free parameters which are put by hand to make proper predictions. Physicists will continue to struggle to find a theory which will predict those parameters. There are other compelling reasons to consider that there must be some post standard model physics. There is still the problem of a completely satisfactory quantum theory of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;String theory is one answer which is an work in progress. It has not predicted those parameters from the theory till now but who knows what surprises may come in this journey. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Huge fields of study in particle physics lay before the physicists and the investigation is by no means over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As someone says ‎&quot;Confusion and disorientation reign…. Very likely we are still confused beginners with very wrong mental pictues, and ultimate reality remains far beyond our grasp… The more we discover, the less we seem to know. That’s physics in a nutshell...&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-04-11T15:48:30.893" LastActivityDate="2011-04-11T15:48:30.893" CommentCount="1" />
  <row Id="8439" PostTypeId="2" ParentId="8433" CreationDate="2011-04-11T15:54:27.257" Score="1" ViewCount="" Body="&lt;p&gt;A &lt;a href=&quot;http://smsc.cnes.fr/html-images/basic_physics.htm&quot; rel=&quot;nofollow&quot;&gt;view of history&lt;/a&gt; helps:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Lord Kelvin (whose name was used for a unit of temperature) also said at the end of the 19th century that physics was finished, that everything had been understood, that our theories worked so well that they could not be wrong, but that there were perhaps two small clouds in the blue sky (in fact these small clouds were to lead to the theories of relativity and quantum physics!!!).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will repeat the trite that knowledge is a circle with an expanding radius. Within the circle is the known, without is the unknown. The more the known is enlarged the greater the contact with unknowns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now specifically for particle physics, maybe it will go the way of thermodynamics, or electromagnetism :) . These disciplines are still being studied by physicists because new frontiers also open new questions in old disciplines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particle physics is no harder than other physics branches when one tries to describe and learn as much as possible. Now suppose that strings do turn up to be the theory of everything. Have you looked at what this everything is? at least six more dimension and God knows what this means as far as our reality goes. Rather future generations of physicists will try to see what this means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition the tools available to physicists for calculations and experiments could not even be dreamed about during Lord Kelvin's time. Why should we be able to dream successfully about how physics  and experiments will be 100 years hence?&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-11T15:54:27.257" />
  <row Id="8440" PostTypeId="2" ParentId="8391" CreationDate="2011-04-11T15:55:08.237" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;This happens in silicon for example.&#xA;  Now if a hydrogen atom was inserted in&#xA;  the material, it would be too large&#xA;  and just get trapped in the 'matrix'&#xA;  of the silicon and present a defect in&#xA;  the material.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The problem is,: how should the hydrogen be &quot;inserted&quot;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either You use &quot;brute force&quot; (ion implantation) or the &#xA;hydrogen will not &quot;insert&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very common feature of many crystals: &#xA;&quot;channels&quot; where molecules can diffuse in and out. &#xA;(Look for sulfur chains in ultramarine, or molecular sieves in general) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether the atom/molecules fit in such channels depends &#xA;on relative size and temperature. In the case of a hydrogen atom in &#xA;silicon I think that is not possible without damaging the &#xA;lattice, so by ion implantation and annealing one would get &#xA;a lattive where a silicon atom misses, and a hydrogen atom is there, &#xA;presumably bound to one of the adjacent silicone atoms like in a silane. &#xA;(Or some ther lattice imperfection, there are a lot of possibilities)&#xA;Another possibility were that the hydrogen after ion implantation &#xA;remains a H+ ion, which is able to diffuse in the lattice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to Your question: Look for data of crystal dimensions and diffusion &#xA;of atomar spezies in such lattices. This gives You an idea, what atom size&lt;br&gt;&#xA;to void size relation is necessary to allow diffusive movement in those voids. &#xA;An example: at about 800 °C hydrogen atoms diffuse easily through &#xA;palladium metal. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-11T15:55:08.237" />
  <row Id="8441" PostTypeId="1" AcceptedAnswerId="8449" CreationDate="2011-04-11T16:10:55.547" Score="4" ViewCount="357" Body="&lt;p&gt;I am searching for a complete and comprehensive book for QFT. What is, in your opinion, a good one?&lt;/p&gt;&#xA;" OwnerUserId="592" LastEditorUserId="1782" LastEditorDisplayName="" LastEditDate="2011-04-11T16:30:34.897" LastActivityDate="2011-04-11T18:37:29.767" CommunityOwnedDate="2011-04-11T18:11:23.047" Title="What is a complete book for quantum field theory?" Tags="&lt;quantum-field-theory&gt;&lt;books&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="1" />
  <row Id="8442" PostTypeId="2" ParentId="8441" CreationDate="2011-04-11T16:21:57.927" Score="1" ViewCount="" Body="&lt;p&gt;I would recommend Peskin and Schroeder book. Srendnicki book is also good for the starters.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-04-11T16:21:57.927" CommunityOwnedDate="2011-04-11T18:11:23.047" />
  <row Id="8443" PostTypeId="2" ParentId="8391" CreationDate="2011-04-11T16:27:09.120" Score="0" ViewCount="" Body="&lt;p&gt;On the specific &quot;muonic hydrogen&quot; part of the question: yes it will be much more compact than hydrogen and would probably fit but I see difficulties in generating such a beam that will last long enough  at the necessary low energy and be incorporated in the crystal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A muon decays in about 10**-6 seconds. So a beam of muonic hydrogen could tunnel through a crystal along the axis  and not scatter destructively, but I find it improbable it will be captured within. Anyway the muon will decay and the proton will be an imperfection in the lattice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beams of particles have been &lt;a href=&quot;http://cdsweb.cern.ch/record/366497/files/sl-98-059.pdf&quot; rel=&quot;nofollow&quot;&gt;tunneled through crystals&lt;/a&gt;. The problem will be in targetting interstitial hydrogen because nothing guarantees it will be along the crystal axis.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-11T16:27:09.120" />
  <row Id="8444" PostTypeId="2" ParentId="8441" CreationDate="2011-04-11T16:31:13.523" Score="1" ViewCount="" Body="&lt;p&gt;Anthony Zee's book &lt;a href=&quot;http://press.princeton.edu/titles/7573.html&quot; rel=&quot;nofollow&quot;&gt;QFT in a Nutshell&lt;/a&gt; is remarkably complete, and yet rather small. It provides a good intro to the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A complete, thorough and good book is Steven Weinberg's &lt;a href=&quot;http://books.google.com/books?id=OoR0lZdqxqIC&amp;amp;printsec=frontcover&amp;amp;dq=steven+weinberg+quantum+theory+of+fields&amp;amp;source=bl&amp;amp;ots=wT_eo4qdrh&amp;amp;sig=HPDVKP0SV7cFzgeN1OPQ7I-9VH8&amp;amp;hl=en&amp;amp;ei=JCyjTei2OoPzsgbzndD3AQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=7&amp;amp;ved=0CEUQ6AEwBg#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;3-volume opus on QFT&lt;/a&gt;. It is perhaps &lt;i&gt;too&lt;/i&gt; complete for beginners.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-04-11T16:31:13.523" CommunityOwnedDate="2011-04-11T18:11:23.047" />
  <row Id="8445" PostTypeId="2" ParentId="8428" CreationDate="2011-04-11T16:32:13.570" Score="0" ViewCount="" Body="&lt;p&gt;Anyone working on non-perturbative QFT really. (temporary) infinities arise from our mathematical approach.&lt;/p&gt;&#xA;" OwnerUserId="3053" LastActivityDate="2011-04-11T16:32:13.570" CommentCount="5" />
  <row Id="8446" PostTypeId="2" ParentId="8425" CreationDate="2011-04-11T16:50:23.640" Score="0" ViewCount="" Body="&lt;p&gt;Situationes where Stokes (without Navier) rules, are well known,&#xA;eg hydrogeology (ground water) or oil geology (secondary or tertiaty recovery) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A word which might be usefull to search for: percolation. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-14T09:56:04.047" LastActivityDate="2011-04-14T09:56:04.047" />
  <row Id="8447" PostTypeId="2" ParentId="8434" CreationDate="2011-04-11T16:59:18.237" Score="1" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Can the imaginary part of the energy be interpreted as the lifetime of an unstable state in the case of a vortex-vortex situation&lt;/em&gt; ... Yes. Vortices have repulsive interactions between them and cannot form bound states. So the result is physically reasonable. And complex energy eigenvalues do measure decay rates of resonances. This is shown in this article: &quot;&lt;a href=&quot;http://arxiv.org/abs/quant-ph/0201091&quot; rel=&quot;nofollow&quot;&gt;A Pedestrian Introduction to Gamow vectors&lt;/a&gt;&quot;. Gamow vectors are the eigenvectors corresponding to the complex energy eigenvalues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[George Gamow was also a famous physics writer with books such as &lt;em&gt;One, Two, Three Infinity&lt;/em&gt; and &lt;em&gt;Physics: Foundations &amp;amp; Frontiers&lt;/em&gt;. Respect.]&lt;/p&gt;&#xA;" LastActivityDate="2011-04-11T16:59:18.237" />
  <row Id="8448" PostTypeId="2" ParentId="8432" CreationDate="2011-04-11T17:10:14.370" Score="3" ViewCount="" Body="&lt;p&gt;Total angular momentum is always conserved.  If there's nothing else changing (like the orbital angular momentum of the electron around the ion), then what you said sounds correct, neglecting details like the direction of propagation of the light vs. the axis of quantization of the spin.  But in general, there WILL be other stuff going on as well (such as the aforementioned orbital angular momentum, etc. etc.).&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-04-11T17:10:14.370" />
  <row Id="8449" PostTypeId="2" ParentId="8441" CreationDate="2011-04-11T17:13:05.980" Score="2" ViewCount="" Body="&lt;p&gt;Each book reflects the author's taste on this or that matter in QFT. It is better to have many books to have a comprehensive picture of the subject. If you are going to study QFT seriously, many &quot;incomplete&quot; books is better than one &quot;complete&quot; course.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-04-11T17:13:05.980" CommunityOwnedDate="2011-04-11T18:11:23.047" />
  <row Id="8450" PostTypeId="2" ParentId="8432" CreationDate="2011-04-11T17:13:13.380" Score="1" ViewCount="" Body="&lt;p&gt;Spin of the photon is always 1 as it is a spin 1 particle. If you are asking about projection of spin into some direction in the center of mass reference frame then this has no meaning (because photon is massless). Instead, one needs to project into the direction of motion (by forming a scalar product of the spin operator with momentum) and this is called helicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, by conservation of angular momentum it must always be the case that the electron's spin flips in the first order (because $1/2 = -1/2 + 1$ is the only possible solution) and that means the photon is circularly polarized. In the second order, it could stay the same if you emitted two photons with opposite helicities one after the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you can actually measure the projection of the spin of the electron by measuring the polarization of the photon (which should be circular). If this experiment is however done over the sample which has electrons in both states, then one needs to average over that and you obtain (circularly) unpolarized photons.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-04-11T17:13:13.380" />
  <row Id="8451" PostTypeId="2" ParentId="8441" CreationDate="2011-04-11T17:27:29.277" Score="1" ViewCount="" Body="&lt;p&gt;I personally like Folland's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821847058&quot; rel=&quot;nofollow&quot;&gt;Quantum Field Theory&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2011-04-11T17:27:29.277" CommunityOwnedDate="2011-04-11T18:11:23.047" />
  <row Id="8452" PostTypeId="1" CreationDate="2011-04-11T17:30:23.767" Score="9" ViewCount="466" Body="&lt;p&gt;The equation describing the force due to gravity is $F = G \frac{m_1 m_2}{r^2}$. Similarly the force due to the electromagnetic force is $F = k \frac{q_1 q_2}{r^2}$. Is there a similar equation that describes the force due to the strong nuclear force? What are the equivalent of masses/charges if there is? Is it still inverse square or something more complicated?&lt;/p&gt;&#xA;" OwnerUserId="2908" LastActivityDate="2011-04-12T11:18:22.823" Title="Is there an equation for the strong nuclear force?" Tags="&lt;high-energy-physics&gt;" AnswerCount="5" FavoriteCount="3" />
  <row Id="8453" PostTypeId="1" CreationDate="2011-04-11T17:30:56.310" Score="4" ViewCount="435" Body="&lt;p&gt;D'Alembert's principle suggests that the work done by the internal forces for a virtual displacement of a mechanical system in harmony with the constraints is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is obviously true for the constraint of a rigid body where all the particles maintain a constant distance from one another. It's also true for constraining force where the virtual displacement is normal to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone think of a case where the virtual displacements are in harmony with the constraints of a mechanical system, yet the total work done by the internal forces is non-zero, making D'Alembert's principle false?&lt;/p&gt;&#xA;" OwnerUserId="2146" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-04-11T22:19:06.600" LastActivityDate="2011-07-11T04:09:57.413" Title="Are there examples in classical mechanics where D'Alembert's principle fails?" Tags="&lt;classical-mechanics&gt;" AnswerCount="2" />
  <row Id="8454" PostTypeId="1" AcceptedAnswerId="8458" CreationDate="2011-04-11T17:39:32.603" Score="5" ViewCount="109" Body="&lt;p&gt;I am having a hard time in finding out what exact light medium(s) do laser shows use. I am trying to build a laser show myself. I know that the laser light is reflected off these particles in such a way that that it makes the laser line &lt;strong&gt;&quot;viewable&quot; in all directions&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody explain to me how exactly do the collection of particles make it viewable in all directions and what exact conditions are necessary? Does the angle of the incoming laser light matter? Does the size of the particles matter? Does the uniformity of how the particles are dispersed matter? Would water vapor work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried using a fog machine, but the red laser that I am using only reflects off of the fog particles in a way that makes it &lt;strong&gt;viewable in only at a certain perspective!!&lt;/strong&gt; This would not be a good show to the people standing in one side of the room vs. another.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2011-06-05T10:11:33.590" Title="What reflective medium(s) do laser shows use?" Tags="&lt;experimental-physics&gt;&lt;light&gt;&lt;water&gt;&lt;laser&gt;&lt;information&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8455" PostTypeId="2" ParentId="8452" CreationDate="2011-04-11T17:51:07.403" Score="6" ViewCount="" Body="&lt;p&gt;No, there is none such equation. Reason is that these equations are highly classical and invalid in both relativistic (there is an action at a distance, incompatible with finite speed of light) and quantum mechanical regime (distances strong force is important at are quite microscopic). Also, strong force is confining, meaning you can't ever observe individual color charged particles (color is a property associated with strong force), so there can't really be a macroscopic equation for them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You obviously need at least quantum mechanics to account for strong force, because distances are so tiny (on the scale of nucleus or smaller). But it turns out you need relativity too. The complete theory which incorporates both QM and relativity is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_field_theory&quot; rel=&quot;nofollow&quot;&gt;quantum field theory&lt;/a&gt; and individual forces are described by &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrangian#Lagrangians_in_quantum_field_theory&quot; rel=&quot;nofollow&quot;&gt;QFT Lagrangians&lt;/a&gt; which essentially tell you which particles interact with which other particles (e.g. photons with electrically charged particles, gluons with color charged particles, etc.). This is the fundamental theory and the electric force equation you described can be derived from it in classical (both non-QM and nonrelativistic) limit. As for gravitation law, that too can be derived but from different theory, namely &lt;a href=&quot;http://en.wikipedia.org/wiki/General_relativity&quot; rel=&quot;nofollow&quot;&gt;general relativity&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-04-11T18:00:02.157" LastActivityDate="2011-04-11T18:00:02.157" CommentCount="6" />
  <row Id="8456" PostTypeId="1" CreationDate="2011-04-11T17:55:00.353" Score="0" ViewCount="166" Body="&lt;p&gt;The concept of relativistic energy comes from it's conservation in relativistic mechanics for an elastic collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that another possible derivation could equate the energy of a single particle before, with the kinetic energy after of N particles after the energy is distributed amongst them in the form of the Boltzman distribution, as well as individually being of the form $mv^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any problems in doing this?&lt;/p&gt;&#xA;" OwnerUserId="2146" LastEditorUserId="2146" LastEditorDisplayName="" LastEditDate="2011-04-12T14:45:20.210" LastActivityDate="2011-04-12T14:45:20.210" Title="Derivation of relativistic energy" Tags="&lt;special-relativity&gt;&lt;energy&gt;&lt;statistical-mechanics&gt;" CommentCount="7" FavoriteCount="1" />
  <row Id="8457" PostTypeId="2" ParentId="8431" CreationDate="2011-04-11T18:07:28.657" Score="3" ViewCount="" Body="&lt;p&gt;Long multiplets are bad because they have an unnecessarily high mass. This fact usually makes them&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;unstable because lighter objects with the same conserved charges exist&lt;/li&gt;&#xA;&lt;li&gt;decouple from low-energy physics which often comes from short multiplets&lt;/li&gt;&#xA;&lt;li&gt;contribute zero to the indices that can be calculated in various ways (they always contribute zero!)&lt;/li&gt;&#xA;&lt;li&gt;impossible to be exactly calculated with - because their mass or energy is unconstrained by symmetries and, in this sense, arbitrary&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Short multiplets have the corresponding virtues. Invert the four points above.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-11T18:07:28.657" CommentCount="3" />
  <row Id="8458" PostTypeId="2" ParentId="8454" CreationDate="2011-04-11T18:19:43.583" Score="0" ViewCount="" Body="&lt;p&gt;Note that there is a huge difference between standard particles (elementary, atoms, or molecules) and droplets. Droplets are quite macroscopic and having a definite spherical shape, they would act as nontrivial optical medium (think about rainbows caused by water droplets) which could selectively prefer some directions. So fog is definitely not recommended.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you have some gas at room temperature the scattering should be effectively classical and in all directions. So my recommendation would probably to use some non-lethal ;) gas.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-04-11T18:19:43.583" CommentCount="2" />
  <row Id="8459" PostTypeId="2" ParentId="8452" CreationDate="2011-04-11T18:25:38.067" Score="1" ViewCount="" Body="&lt;p&gt;The strong force as seen in &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_force&quot; rel=&quot;nofollow&quot;&gt;nuclear matter&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The nuclear force, is now understood as a residual effect of the even more powerful strong force, or strong interaction, which is the attractive force that binds particles called quarks together, to form the nucleons themselves. This more powerful force is mediated by particles called gluons. Gluons hold quarks together with a force like that of electric charge, but of far greater power.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Marek is talking of the strong force that binds the quarks within the protons and neutrons. There are charges, called colored charges on the quarks, but protons and neutrons are color neutral.  Nuclei are bound by the interplay between the  residual strong force , the part that is not shielded by the color neutrality of the nucleons, and the electromagnetic force due to the charge of the protons. That also cannot be simply described. Various potentials are used to calculate nuclear interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplicity and similarity of form for all forces comes not in the formalism of forces, but as Marek said the formalism of quantum field theory.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-11T18:25:38.067" />
  <row Id="8460" PostTypeId="2" ParentId="8441" CreationDate="2011-04-11T18:37:29.767" Score="6" ViewCount="" Body="&lt;p&gt;See a list here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2006/01/qft-didactics.html&quot; rel=&quot;nofollow&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2006/01/qft-didactics.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-11T18:37:29.767" CommunityOwnedDate="2011-04-11T18:37:29.767" />
  <row Id="8461" PostTypeId="2" ParentId="8428" CreationDate="2011-04-11T18:48:45.090" Score="2" ViewCount="" Body="&lt;p&gt;What about the &lt;a href=&quot;http://motls.blogspot.com/2011/01/twistor-minirevolution-goes-on.html&quot; rel=&quot;nofollow&quot;&gt;Twistor uprising&lt;/a&gt; of Arkani Hamed discussed in Lumos' blog?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Nima explains that the goal is to think different - to eliminate the word &quot;Feynman&quot; from the QFT calculations as completely as possible. ;-) In particular, the new description and calculational methods proudly make locality in the ordinary spacetime obscure while they succeed in making many other, more exotic properties of the N=4 gauge theory manifest&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do not know what twistors do to renormalization, quite out of my depth. But the claim in the lecture is that they simplify QCD calculations, which is one of your worries.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-11T18:48:45.090" CommentCount="5" />
  <row Id="8462" PostTypeId="2" ParentId="8453" CreationDate="2011-04-11T18:49:02.610" Score="1" ViewCount="" Body="&lt;p&gt;Well, if the internal forces, e.g., include sliding friction, then they produce a non-zero negative work. You may want to distinguish between constraining internal forces and non-constraining internal forces.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-04-11T18:57:14.237" LastActivityDate="2011-04-11T18:57:14.237" />
  <row Id="8463" PostTypeId="2" ParentId="8381" CreationDate="2011-04-11T19:34:33.380" Score="0" ViewCount="" Body="&lt;p&gt;There are several proposals about the emergence of holographic gravity duals to QFT, based on the spatial structure of quantum correlations such as the discussion that you mention as well as the very interesting ideas posed by &lt;a href=&quot;http://arxiv.org/abs/0907.2939&quot; rel=&quot;nofollow&quot;&gt;Van Raamsdonk&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related to your last question, in &lt;a href=&quot;http://arxiv.org/abs/0905.1317&quot; rel=&quot;nofollow&quot;&gt;Entanglement Renormalization and Holography&lt;/a&gt; what is postulated is that the full structure of the MERA tensor network (defined by the structure of quantum entanglement in the boundary theory) happens to define &quot;a discrete version&quot; of AdS space. In this proposal, the points in the MERA bulk, represent points in a discretized version of the Anti de Sitter space, while the links between these points account for the structure of the causal cones of operators defined on sites on the boundary. The causal cone of a block of sites is defined as the set of sites, disentanglers, and isometries that can affect the chosen block. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, if one analyses how MERA computes the entanglement entropy of a block of L sites on the boundary of the tensor network (the original quantum many body system), one finds an striking similarity with a proposal for the computation of entanglement entropy in theories with a &quot;right&quot; gravity dual posed by &lt;a href=&quot;http://arxiv.org/abs/hep-th/0603001&quot; rel=&quot;nofollow&quot;&gt;Ryu &amp;amp; Takayanagi&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2237" LastActivityDate="2011-04-11T19:34:33.380" CommentCount="1" />
  <row Id="8464" PostTypeId="2" ParentId="8452" CreationDate="2011-04-11T19:35:26.150" Score="5" ViewCount="" Body="&lt;p&gt;At the level of quantum hadron dynamics (i.e. the level of nuclear physics, not the level of particle physics where the real strong force lives) one can talk about a Yukawa potential of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ V(r) = - \frac{g^2}{4 \pi c^2} \frac{e^{-mr}}{r} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ is roughly the pion mass and $g$ is an effective coupling constant. To get the force related to this you would take the derivative in $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a semi-classical approximation, but it is good enough that Walecka uses it breifly in his book.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-04-11T19:35:26.150" />
  <row Id="8465" PostTypeId="1" CreationDate="2011-04-11T20:11:43.550" Score="0" ViewCount="168" Body="&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/2RWJU.gif&quot; alt=&quot;pendulum&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help me visualise the forces found in a pendulum. I know weight can be decomposed into $F_x$ and $F_y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Centripetal force, acceleration and velocity are three words that I'm confused. Centripetal force is in L and is equal to acceleration ? it's the gravity that move the pendulum, the weight, but there is a rope, so it's not the same situation that happens to a satellite.&lt;/p&gt;&#xA;" OwnerUserId="3056" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-11T20:36:03.677" LastActivityDate="2011-04-11T22:30:45.483" Title="Forces in a Pendulum" Tags="&lt;gravity&gt;&lt;forces&gt;&lt;newtonian-mechanics&gt;" AnswerCount="2" />
  <row Id="8466" PostTypeId="1" AcceptedAnswerId="8470" CreationDate="2011-04-11T20:17:54.417" Score="2" ViewCount="164" Body="&lt;p&gt;If a Lagrangian containing an N-multiplet of fields is invariant under global $\mathbf{SU}(N)$ transformations, does that necessarily imply it is invariant under $\mathbf{SU}(N-1)$, $\mathbf{SU}(N-2)$,... all the way down to $\mathbf{U}(1)$ as well, where the representations in each case are $N$x$N$ matrices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, given a symmetry group $\mathbf{SU}(N)$, do representations involving $M$x$M$ matrices exist for all $M \geq N$?&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-04-11T21:11:53.373" Title="SU(N) symmetry and its representations" Tags="&lt;symmetry&gt;&lt;group-representations&gt;&lt;lie-algebra&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8467" PostTypeId="1" AcceptedAnswerId="8475" CreationDate="2011-04-11T20:17:59.130" Score="0" ViewCount="136" Body="&lt;p&gt;So I have learned in my QM classes that you can't tell one electron from another electron. They are indistinguishable. I also learned that the wavefunction of a particle includes the spacial part and the spin part. (There might be other parts I don't know! Please don't fault me for not knowing.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also learned that it is possible to use the quantum teleportation protocol to retrieve exactly, 100%, the spin state of one particle in another particle after particular measurements.&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_teleportation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Quantum_teleportation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, if you can teleport the spatial state and the spin state and whatever other state left there is that describes the particle, then wouldn't you be &lt;strong&gt;effectively teleporting the original particle?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sub question directly related to this is; Is matter/particles just a changeable glob of stuff that acts differently, thus being a different particle, depending upon what quantum state its in? (This kinda sounds like string theory to me... Oh no!!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what's never answered is, &lt;strong&gt;What state does the original particle assume when its state is teleported to the new particle?&lt;/strong&gt; Does the two particles, the original particle and the particle retrieving the quantum information, switch states?&lt;/p&gt;&#xA;" OwnerUserId="2578" LastEditorUserId="2578" LastEditorDisplayName="" LastEditDate="2011-04-11T20:30:31.083" LastActivityDate="2011-04-11T22:50:50.430" Title="Is the quantum state or information of a particle is all that differentiates it from another particle?" Tags="&lt;quantum-mechanics&gt;&lt;string-theory&gt;&lt;particle-physics&gt;&lt;spin&gt;&lt;teleportation&gt;" AnswerCount="1" />
  <row Id="8468" PostTypeId="2" ParentId="8452" CreationDate="2011-04-11T20:45:17.427" Score="7" ViewCount="" Body="&lt;p&gt;From the study of the spectrum of quarkonium (bound system of quark and antiquark) and the comparison with positronium one finds as potential for the strong force&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V(r) = - \frac{4}{3} \frac{\alpha_s(r) \hbar c}{r} + kr$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the constant $k$ determines the field energy per unit length and is called string tension. &#xA;For short distances this resembles the Coulomb law, while for large distances the $kr$ factor dominates (confinement). It is important to note that the coupling $\alpha_s$ also depends on the distance between the quarks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula is valid and in agreement with theoretical predictions only for the quarkonium system and its typical energies and distances. For example charmonium: $r \approx 0,4 fm$. So it is not as universal as eg. the gravity law in Newtonian gravity.  &lt;/p&gt;&#xA;" OwnerUserId="3031" LastActivityDate="2011-04-11T20:45:17.427" CommentCount="2" />
  <row Id="8470" PostTypeId="2" ParentId="8466" CreationDate="2011-04-11T21:11:53.373" Score="6" ViewCount="" Body="&lt;p&gt;Well, if something is invariant under the action of some group it means that every element leaves it invariant. In particular, every element that belongs to some subgroup. Also, you can get an obvious representation of the subgroup by restriction. If $\rho: G \to {\rm Aut}(V)$ is a representation then so is $\rho_{H}: H \to {\rm Aut}(V)$ for $H &amp;lt; G$. Naturally, this need not be irreducible anymore (obvious example being representation of a trivial subgroup which is reducible to $N$ components).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, they surely do because one can take $T\oplus \dots \oplus T$ where $T$ is trivial one dimensional representation. Let's pretend that you meant irreducible representations so that this question is actually non-trivial. Groups of $A_l$ type (with $l = N-1$) are of rank $l$. This means that finite-dimensional representations (as is the case for the irreducible representations of compact $\mathbf {SU}$ forms of these groups) are indexed by $l$ non-negative integers $\{\lambda_i\}_{i=1}^l$. E.g. for $l=1$ (${\mathbf {SU}(2)}$) this integer is connected to spin by $\lambda_1 = 2s$ (and dimension of these representations is $\lambda_1 +1 = 2s+1$). So we are interested in determining dimensions of the irreducible representations indexed by $\lambda$. This is not so straight-forward; in principle one can determine these dimensions by looking at the weight space of the given representation and counting the number of ways one can arrive at any given weight starting from highest weight and applying annihilation operators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, one can exploit the fact that $\mathbf {SL}$ is basically the same thing as $&#xD;&#xA;\mathbf {GL}$ and use &lt;a href=&quot;http://en.wikipedia.org/wiki/Schur-Weyl_duality&quot; rel=&quot;nofollow&quot;&gt;Schur-Weyl duality&lt;/a&gt; between $\mathbf {GL}(N,\mathbb C)$ and $\mathfrak{S}_k$ (i.e. the symmetry group of $k$-element set) which states that the representation of the above groups on $(\mathbf C^n)^{\otimes k}$ (with obvious actions) decomposes as sum over partitions $\lambda \in Par(k,N)$ (e.g. $(3,2,1) \in Par(6,3)$) of products of irreducible representations $F^{\lambda} \otimes G^{\lambda}$ (with $F$ resp. $G$ being a representation of $\mathbf {GL}(N,\mathbb C)$, resp. $\mathfrak{S}_k$). The dimensions of these representations can then be computed by fancy combinatorial formulas derived from &lt;a href=&quot;http://en.wikipedia.org/wiki/Young_tableau&quot; rel=&quot;nofollow&quot;&gt;Young tableaux&lt;/a&gt; (which enter due to partitions, of course). Basically one writes down some numbers related to $n$ and number of rows under and columns to the right into each box of the tableaux and multiplies all those numbers together. As for whether every integer greater or equal to $N$ can be obtained in this way, I don't know. I suppose by shear amount of these tableaux this should be true but I'll come back to check it later.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-04-11T21:11:53.373" />
  <row Id="8471" PostTypeId="2" ParentId="8465" CreationDate="2011-04-11T21:11:58.520" Score="2" ViewCount="" Body="&lt;p&gt;In the picture you presented weight (mg) is decomposed into $F_t$, that is tangential force, and $F_r$, that is the radial force (along the radius). Problems from rotational dynamics are usually solved in either laboratory (fixed to earth) or inertial (fixed to the rotating mass) reference frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider inertial reference frame. Forces acting on the mass at the end of the rope are weight and rope tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pendulum has a &lt;strong&gt;velocity&lt;/strong&gt; that is tangential, because it is moving around the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its &lt;strong&gt;acceleration&lt;/strong&gt; can be considered to have both tangential and radial components. Tangential component says that the pendulum is accelerating (so the value of the velocity is changing). Radial component says that the direction of velocity is changing (because the body is moving around the circle, for instance for $\theta=90 \deg$ velocity vector points down and for $\theta=0 \deg$ is points to the left).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Centipetal force&lt;/strong&gt; is the force that is responsible for radial acceleration. Its value is equal to $mV^2/R$ and is a sum of the rope tension and $F_r$&lt;/p&gt;&#xA;" OwnerUserId="2752" LastActivityDate="2011-04-11T21:11:58.520" CommentCount="1" />
  <row Id="8472" PostTypeId="1" AcceptedAnswerId="8473" CreationDate="2011-04-11T21:19:52.210" Score="1" ViewCount="110" Body="&lt;p&gt;Say we start with an entangled electron positron pair and we separate them. I want to take the entangled electron and bound it to a proton or the outer shell of an atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this possible to do this while having the entangled electron to still remain entangled with the positron? Would the electromagnetic interaction with the proton &quot;make a measurement&quot; of the entangled electron spin because of its magnetic field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming my above example is possible, say we then measure the spin of the positron that's far away. Whatever it may be, the entangled electron bound to the atom will now assume the opposite spin state. The total angular momentum of the atom will now be at a different value then what it was before, and therefore something must happen to conserve this, emit a photon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, if my above example has a measurable effect whether it emits a photon or changes in some way, then this suggest to me that if we have an ensemble of these special atoms, we will be able to send information via quantum entanglement. By having these special atoms in one station and the entangled positrons in another station, we can measure or not measure, 1 or 0,  the spin of the entangled positron and thus produce a measurable effect at the other station via the special atoms.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2011-04-12T15:22:14.143" Title="Is it possible to bound an entangled electron to the outer shell of an atom or just a proton?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-information&gt;&lt;entanglement&gt;&lt;spin&gt;&lt;atoms&gt;" AnswerCount="2" />
  <row Id="8473" PostTypeId="2" ParentId="8472" CreationDate="2011-04-11T22:26:19.710" Score="2" ViewCount="" Body="&lt;p&gt;I'm not going to answer your electron&amp;lt;-&gt;atom+photon interaction question because my atomic physics is veeery rusty..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But regardless, suppose it works like you write, what you are proposing is simply a spin-measurement device. By measuring the positron in the &quot;other end&quot;, you collapse the wavefunction consisting of the positron and electron+atom+photon system. Likewise if your atom ejects a photon which you detect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your first question, the electron would involve the atom in the spin superposition at first, but if it includes the emission and non-emission of a photon as well, this is where the fun ends since the photon emission and detection is a &quot;collapsing&quot; act.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you are not better off than simply measuring the electron spin the same way you measure the positron spin, with equally zero chance of information transmission (as have doubtlessly been discussed in many other threads here).&lt;/p&gt;&#xA;" OwnerUserId="2936" LastEditorUserId="2936" LastEditorDisplayName="" LastEditDate="2011-04-11T22:32:55.850" LastActivityDate="2011-04-11T22:32:55.850" />
  <row Id="8474" PostTypeId="2" ParentId="8465" CreationDate="2011-04-11T22:30:45.483" Score="2" ViewCount="" Body="&lt;p&gt;Hopefully you know that velocity is how fast an object's position changes, and acceleration is how fast an object's velocity changes. Both of these are vectors, so you can split them up into components along any set of perpendicular axes. For example, you could split them up into $x$ and $y$ components, or into tangential and radial components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true of the force - or rather, forces. There are multiple forces acting on the pendulum bob, both its weight and a force from the string (tension). Each of those forces is a vector, so you can split it up into components along any set of perpendicular axes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also add up all the forces (as vectors) and produce something called the &lt;em&gt;net force&lt;/em&gt; (or total force). This net force is related to the acceleration by Newton's second law,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{F}_\text{net} = m\vec{a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, since the net force is a vector, you can split it up into components along any set of perpendicular axes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose you choose one axis to point toward the center of the circle. The component of the net force that points along that axis is called the &lt;em&gt;centripetal force&lt;/em&gt;. The component of the acceleration that points along that axis is called the &lt;em&gt;centripetal acceleration&lt;/em&gt;. And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having chosen one axis to point toward the center of the circle, the only way you can choose the other axis to be perpendicular is to have it point parallel to the edge of the circle. The component of, say, the acceleration along that axis is called the &lt;em&gt;tangential acceleration&lt;/em&gt;. Similarly for velocity, force, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you get out of introductory physics it's common to replace the centripetal axis (the one that points toward the center) with an axis that points in the exact opposite direction, away from the center. The component of the net force that points along this axis could be called the &lt;em&gt;radial force&lt;/em&gt; (or just &quot;radial component of the net force&quot;). It's the negative of the centripetal force, because you switched the direction of the axis. And of course, the same goes for acceleration, velocity, etc.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-04-11T22:30:45.483" />
  <row Id="8475" PostTypeId="2" ParentId="8467" CreationDate="2011-04-11T22:50:50.430" Score="1" ViewCount="" Body="&lt;p&gt;the quantum teleportation experiments teleports for example the spin-state of the particle to be teleported with its superpositioned state intact, as you write. But the experiment is designed on the premise that the recipient has his/her own particle to imprint this superpositioned state on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the &quot;spatial state&quot; of the particle involves the location of the senders lab, there would be no point in teleporting it. As you yourself suggest, it would be a null operation. On the other hand, if you want the particle to appear in the &lt;strong&gt;recipient&lt;/strong&gt; lab, the spatial state obviously is NOT the same and hence you can't talk about teleportation at all in that case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subquestion: yes, the particles of the same type are interchangeable (given compatible quantum states) and actually this is always taken into consideration when calculating the superpositioned interactions in Quantum Field Theory - you need to consider all the ways the particles might get exchanged with each other during a process you are calculating, there are no &quot;particle A&quot; and &quot;particle B&quot; labels if their states are identical otherwise. With some stretch of words and imagination, the &quot;glob&quot; you are talking about is essentially the quantum field of the particle type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Never answered question: The original particle's superpositioned state is destroyed in the quantum teleportation process. This is mentioned in the beginning of the document you link, actually. You can't clone the state without destroying the source - this is actually a theorem called the &lt;a href=&quot;http://en.wikipedia.org/wiki/No-cloning_theorem&quot; rel=&quot;nofollow&quot;&gt;No-cloning Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-04-11T22:50:50.430" />
  <row Id="8476" PostTypeId="1" CreationDate="2011-04-11T23:08:26.587" Score="7" ViewCount="132" Body="&lt;p&gt;Following on Jim Graber's answer to: &lt;a href=&quot;http://physics.stackexchange.com/questions/8301/can-big-rip-rip-apart-an-atomic-nucleus&quot;&gt;Can &amp;quot;big rip&amp;quot; rip apart an atomic nucleus?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the cosmological constant is large enough, even the ground state of a hydrogen atom can be affected.  So why is the energy scale for quantum gravity set by the planck mass and not by the cosmological constant?  Is it because the cosmological constant can be associated with other theories (inflatons, or vacuum energies of the matter fields, etc.) and thus is just considered an ingredient and not gravity itself?  If this does come down to a semantics issue, I'd still be interested to hear if the scale set by the cosmological constant suggests we may be able to see interesting quantum effects at that scale depending on what the cosmological constant 'is'.&lt;/p&gt;&#xA;" OwnerUserId="2655" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-04-11T23:11:03.810" LastActivityDate="2011-04-12T05:02:15.467" Title="Scale set by cosmological constant" Tags="&lt;quantum-mechanics&gt;&lt;general-relativity&gt;&lt;quantum-gravity&gt;" AnswerCount="1" />
  <row Id="8477" PostTypeId="1" AcceptedAnswerId="8481" CreationDate="2011-04-12T02:33:04.910" Score="2" ViewCount="205" Body="&lt;p&gt;Photons do not have mass (that's why they can move at speed of &quot;light&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is how the gravity of black hole can stop light from escaping?&lt;/p&gt;&#xA;" OwnerUserId="3057" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-12T07:07:39.517" LastActivityDate="2011-04-12T07:07:39.517" Title="Why can't light escape from a black hole?" Tags="&lt;black-holes&gt;" AnswerCount="2" />
  <row Id="8478" PostTypeId="2" ParentId="8477" CreationDate="2011-04-12T02:39:42.070" Score="3" ViewCount="" Body="&lt;p&gt;Even though photons have no mass, they are still affected by gravity. That's how we can see black holes - by the way they &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_lens&quot; rel=&quot;nofollow&quot;&gt;distort the light&lt;/a&gt; going near them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason nothing can escape a black hole is because within the event horizon, space is curved to the point where all directions are actually &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_hole#Event_horizon&quot; rel=&quot;nofollow&quot;&gt;pointing inside&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Escape_velocity&quot; rel=&quot;nofollow&quot;&gt;escape velocity&lt;/a&gt; from within a black hole's event horizon is &lt;a href=&quot;http://en.wikipedia.org/wiki/Escape_velocity#List_of_escape_velocities&quot; rel=&quot;nofollow&quot;&gt;faster than the speed of light&lt;/a&gt;, hence light cannot go at that speed and thus cannot escape.&lt;/p&gt;&#xA;" OwnerUserId="2348" LastActivityDate="2011-04-12T02:39:42.070" CommentCount="2" />
  <row Id="8479" PostTypeId="1" CreationDate="2011-04-12T02:48:32.083" Score="6" ViewCount="149" Body="&lt;p&gt;What is highest water pressure at which electrolysis can be performed to derive hydrogen and oxygen? Does the dielectric constant of water, which which lowers as pressure increases, have an effect on this?&lt;/p&gt;&#xA;" OwnerUserId="3058" LastActivityDate="2011-05-12T13:14:45.673" Title="What is highest water pressure at which electrolysis can be performed?" Tags="&lt;water&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="8480" PostTypeId="1" AcceptedAnswerId="8485" CreationDate="2011-04-12T02:54:01.497" Score="6" ViewCount="225" Body="&lt;p&gt;According to the third law of motion, you van't have an mass move in a particular direction unless there is a proportional opposite mass/acceleration ratio in the opposite direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No-one has been able to provide a convincing argument otherwise, but the best one to date is Shawyer's &lt;a href=&quot;http://en.wikipedia.org/wiki/EmDrive&quot; rel=&quot;nofollow&quot;&gt;EM Drive&lt;/a&gt;. He claims some fancy relativistic effects are what allows his engine to work, but I have read some papers which claim he is a fraud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, &lt;i&gt;why&lt;/i&gt; is it impossible to move a mass in a given direction without a proportional change in the opposite direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not talking about a perpetual motion machine, or anything. Sure, the device would need to consume at least the amount of energy proportional to the energy required to accelerate the mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a highly hypothetical example:&#xA;Say we either can project a gravity well in front of our vehicle, and/or project a gravity hill behind. In empty space, the effects of the gravity will be near-negligible by the time they reach any other object, however close to the vehicle they will be more significant. The end result would be the vehicle would move in the given direction, and nothing else around would really move at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An even cruder example would be to shine a bright torch out the back of your vehicle. Even though the photons have no mass, wouldn't the vehicle move forward?&lt;/p&gt;&#xA;" OwnerUserId="2348" LastActivityDate="2011-04-12T08:39:38.267" Title="Reactionless Drives" Tags="&lt;motion&gt;&lt;relativity&gt;&lt;acceleration&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="8481" PostTypeId="2" ParentId="8477" CreationDate="2011-04-12T03:09:34.867" Score="5" ViewCount="" Body="&lt;p&gt;Black holes affect the causal structure of spacetime in such a manner that all future light cones within a black hole lie within the event horizon of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although photons are massless they have energy and have to obey the geometry of a curved spacetime. Since all future lies within the event horizon, photons are trapped inside the black hole.&#xA;&lt;img src=&quot;http://i.stack.imgur.com/RUtH4.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-04-12T03:09:34.867" CommentCount="2" />
  <row Id="8482" PostTypeId="2" ParentId="8480" CreationDate="2011-04-12T03:20:29.787" Score="3" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My question is, why is it impossible&#xA;  to move a mass in a given direction&#xA;  without a proportional change in the&#xA;  opposite direction?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because otherwise the principle of conservation of linear momentum will be violated (A more accurate answer is, it is a law of nature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction less drive is a complete fraud.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-04-12T03:20:29.787" CommentCount="1" />
  <row Id="8483" PostTypeId="1" AcceptedAnswerId="8491" CreationDate="2011-04-12T03:53:04.533" Score="6" ViewCount="239" Body="&lt;p&gt;Someone sent me this link to a talk by &lt;a href=&quot;http://www.ks.uiuc.edu/~kschulte/&quot; rel=&quot;nofollow&quot;&gt;Prof. Klaus Schulten&lt;/a&gt; from the University of Illinois: (my emphasis)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Quantum Computing and Animal Navigation&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Quantum computing is all the rage nowadays. But this type of computing may have been discovered and used by living cells billion of years ago. Nowadays migratory birds use a protein, Cryptochrome, which absorbs weak blue light to produce &lt;strong&gt;two quantum-entangled electrons in the protein&lt;/strong&gt;, which by monitoring the earth's magnetic field, allows birds to navigate even in bad weather and wind conditions. The lecture tells the story of this discovery, starting with chemical test tube experiments and ending in the demonstration that the navigational compass is in the eyes and can be affected by radio antennas. The story involves theoretical physicists who got their first paper rejected as &quot;garbage&quot;, million dollar laser experiments by physical chemists to measure the entangled electrons, and ornithologists who try to 'interrogate' the birds themselves. &lt;strong&gt;This work opens up the awesome possibility that room-temperature quantum mechanics may be crucial in many biological systems.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now here's my question: What's the big deal with entangled electrons? I mean, if I do not neglect electron-electron interaction, then pretty much all electrons in a condensed matter system are entangled, are they not? Electrons in the same angular momentum multiplet are entangled  via Hund's rule, electrons on neighboring sites in a tight-binding (or, in the interacting case, Hubbard) model can all be entangled due to an antiferromagnetic exchange coupling, etc. etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sure, for a quantum computer I'd like to have physically separated electrons maintain their entanglement, and I'd like to have fine-grained control over which of the electrons are entangled in which way etc, but for chemical processes in molecules such as these earth-magnetic-field receptors, is it not a bit sensationalist to liken such a process to quantum computing?&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-04-15T13:52:16.610" Title="Quantum Computing and Animal Navigation" Tags="&lt;quantum-information&gt;&lt;entanglement&gt;&lt;biophysics&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" />
  <row Id="8484" PostTypeId="1" AcceptedAnswerId="8488" CreationDate="2011-04-12T03:54:02.083" Score="3" ViewCount="181" Body="&lt;p&gt;I am reading Ludwig's paper &quot;&lt;a href=&quot;http://arxiv.org/abs/1010.0936&quot; rel=&quot;nofollow&quot;&gt;Electromagnetic and gravitational responses and anomalies in topological insulators and superconductors&lt;/a&gt;&quot;, and in this paper, although I am clear how they get the descent equation which introduced the relationship between anomaly and the existence of topological insultor, I am confused about the theta-term they mentioned on section VA5, they said that the integral of anomaly polynomial $\Omega_{2n+2}$ corresponds to the $\theta$ term. Could anyone help and explain to me what is this theta-term referring to and where can I find information about it?&lt;/p&gt;&#xA;" OwnerUserId="2397" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-12T05:49:19.343" LastActivityDate="2011-04-12T05:54:06.707" Title="$\theta$ term of anomaly related with topological insulators" Tags="&lt;condensed-matter&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8485" PostTypeId="2" ParentId="8480" CreationDate="2011-04-12T03:57:47.533" Score="9" ViewCount="" Body="&lt;p&gt;It appears to me the issue is understanding momentum conservation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An even cruder example would be to shine a bright torch out the back of your vehicle. Even though the photons have no mass, wouldn't the vehicle move forward?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You also refer to mass in this manner in the paraphrasing of Newton's third law &quot;proportional opposite mass/acceleration ratio in the opposite direction&quot;.  It is not the mass that matters here.  Newton's third law can be thought of as a statement of conservation of momentum.  Despite having no invariant mass, photons do have momentum.  So a light sail, or a flash light in your case, works because to conserve momentum if light is reflected off your craft or emitted from your craft, your momentum must change to compensate for the change in momentum of the light or the momentum the light carried away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Noether%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Noether's theorem&lt;/a&gt; is an even stronger statement, which shows that if we can describe the physics in a manner that does not depend on position (ie. you could redo this experiment 1km to the right and it wouldn't effect the results), then momentum must be conserved.  So this forbids a reactionless drive in special relativity as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spatial translation symmetry becomes a bit messy in GR, since the spacetime itself is dynamic.  So your gravity wave idea could work in principle (&lt;a href=&quot;http://en.wikipedia.org/wiki/Alcubierre_drive&quot; rel=&quot;nofollow&quot;&gt;Alcubierre drive&lt;/a&gt;).  However it is possible to formulate a type of energy and momentum conservation with pseudo-tensors in GR (this is how Einstein discussed it).  In this view we can keep track of the energy and momentum of gravity waves as well, and so we can still use momentum conservation for these scenarios as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your question:&lt;br&gt;&#xA;&lt;em&gt;Why&lt;/em&gt; is it impossible to move a mass in a given direction without a proportional change in the opposite direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the answer is:  Momentum conservation forbids a reactionless drive, and momentum conservation itself follows from the translational symmetry of the physics describing a closed system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So like free energy machines, we don't even need to see the details to know something is a misrepresentation or a scam.  We can reject such &quot;inventions&quot; on very general grounds.  And indeed, the U.S. patent office explicitly will not review a patent on a perpetual motion machine, or a device that could be used to build such a machine.  They may have a similar restriction on &quot;reactionless drives&quot;, but I am not sure of that.&lt;/p&gt;&#xA;" OwnerUserId="2652" LastEditorUserId="2652" LastEditorDisplayName="" LastEditDate="2011-04-12T04:08:25.243" LastActivityDate="2011-04-12T04:08:25.243" CommentCount="3" />
  <row Id="8486" PostTypeId="2" ParentId="8476" CreationDate="2011-04-12T05:02:15.467" Score="4" ViewCount="" Body="&lt;p&gt;A very good question. The cosmological constant has been taken to be the natural scale in various not-quite-mainstream papers such as those written by my ex-adviser Tom Banks, starting from e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0007146&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0007146&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More generally, the cosmological constant scale is extremely long in our Universe - it's the millimeter scale when you calculate the energy scale from the energy density; and the Hubble scale when you calculate it from the curvature of the vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particle physics has this habit of thinking about short-distance physics as the fundamental physics; the long-distance physics is the derived discipline. It's because one can derive how big objects behave if he knows how the small ones do; but it is not possible in the opposite direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, the most fundamental scale has to be the shortest distance scale where &quot;interesting things happen&quot;. Because the Planck scale - the radius of the smallest black holes and many other things - is much shorter than the cosmological constant scale, it has to be more fundamental.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-12T05:02:15.467" />
  <row Id="8488" PostTypeId="2" ParentId="8484" CreationDate="2011-04-12T05:47:50.280" Score="3" ViewCount="" Body="&lt;p&gt;The $\theta$-term is also known as the axion term and it's simply the $F\wedge F$ term known to particle physicists. In a more condensed-matter-friendly language,&#xA;$$\Delta {\mathcal L} = \theta\left( \frac{e^2}{2\pi h}  \right) \vec B \cdot \vec E $$&#xA;I don't know the optimum starting point but you may begin with &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1001.3179&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1001.3179&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and its followups and references. More generally, the $\theta$-term means the integral of the anomaly polynomial. Note that the anomaly polynomial is a nice gauge-invariant expression - but in higher dimensions. The actual anomaly in the original spacetime is related to it by several operations.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-12T05:54:06.707" LastActivityDate="2011-04-12T05:54:06.707" CommentCount="2" />
  <row Id="8489" PostTypeId="2" ParentId="8483" CreationDate="2011-04-12T06:13:54.840" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What's the big deal with entangled electrons?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you had even mentioned the possibility that entanglement would play &lt;em&gt;any&lt;/em&gt; role in biological systems, ten years ago or so you would have been greeted with eye-rolling and finger-wagging. Fortunately since then we have become accustomed to seeing mentions of &quot;entanglement&quot; all over the place. This is justifiably so. Entanglement is a new resource, much as coal, oil or nuclear power were at one point in history, that holds almost unlimited possibilities.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I mean, if I do not neglect electron-electron interaction, then pretty much all electrons in a condensed matter system are entangled, are they not?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. Not in general. Entanglement is, of course, present in any many-body quantum system to some extent. One example where this is quite explicit is in some proposed mechanisms of high-T_c superconductivity built around the the Resonance Valence Bond (RVB) state [see this &lt;a href=&quot;http://physics.stackexchange.com/questions/2964/what-is-a-resonating-valence-bond-rvb-state&quot;&gt;question&lt;/a&gt; on physics.SE and the accompanying answer] where the &quot;valence bond&quot; between two neighboring spins is nothing more than a spin-singlet entangled bell state:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |\Psi\rangle = \frac{1}{\sqrt{2}}\left( |\uparrow\rangle |\downarrow\rangle - |\downarrow\rangle |\uparrow\rangle \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an example of &lt;em&gt;bipartite&lt;/em&gt; entanglement, i.e. entanglement between two subsystems. In general one can have *multi*partite engtanglement between $N$ number of quantum systems. Again, the presence of bipartite or pairwise entanglement in a RVB state or due to the sort that lead to a contribution of the specific heat of a material (&lt;a href=&quot;http://arxiv.org/abs/cond-mat/0402456&quot; rel=&quot;nofollow&quot;&gt;ref:Sondhi et al.&lt;/a&gt;), &lt;em&gt;does not&lt;/em&gt; imply by any means that &quot;all electrons are entangled&quot;. I hope this point is very clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Entanglement on &lt;em&gt;large-scales&lt;/em&gt; (i.e. scales comparable to the size of the system itself, only then can you justify the statement that &quot;all the particles&quot; in a system are entangled) is a non-generic phenomenon that is a signature of so-called quantum phase transitions. It has been suggested [for references and background see the &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0702225v2&quot; rel=&quot;nofollow&quot;&gt;RMP&lt;/a&gt; by the Horodecki clan - in particular col. 2 first para. page 4.] that the entanglement entropy of any susbsystem with the rest of the system diverges near a quantum phase transition and can considered (see for eg. &lt;a href=&quot;http://arxiv.org/abs/cond-mat/0510613&quot; rel=&quot;nofollow&quot;&gt;Levin and Wen, PRL, 2006&lt;/a&gt; and  &lt;a href=&quot;http://arxiv.org/abs/hep-th/0510092&quot; rel=&quot;nofollow&quot;&gt;Kitaev and Preskill, PRL, 2006&lt;/a&gt;) as an order parameter for characterizing topological phases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum, (AFAIK) it is only near the critical point associated with a quantum phase transition (when the entanglement entropy of a &quot;test&quot; subsystem diverges) where all subsystems of the given system can be said to be entangled with all other subsystems thus justifying the phrase &quot;all electrons ... are entangled&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;strong&gt;Nb:&lt;/strong&gt; See also remarks on entanglement in the Slater determinant state at the end of this answer.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The presence of entanglement in 1D spin-chains and such might be ubiquitous but its presence in the (approx.) 1D biomolecular chains such as DNA, RNA and various proteins would certainly be nothing less than extraordinary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If entanglement is actively exploited in any biological processes - avian navigation, photosynthesis or others - the implications for biology are phenomenal. We've already seen what evolution has managed to come up with as far as nano-technology (ATP synthesis, bacterial navigation, myosin motors, etc.)  and command-and-control systems (nervous system, endocrine system, etc.) are concerned. What more can be explained by throwing entanglement into the mix?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sure, for a quantum computer I'd like to have physically separated electrons maintain their entanglement, and I'd like to have fine-grained control over which of the electrons are entangled in which way etc, but for chemical processes in molecules such as these earth-magnetic-field receptors, &lt;strong&gt;is it not a bit sensationalist to liken such a process to quantum computing?&lt;/strong&gt; [emph. mine]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Perhaps a little but the enthusiasm is understandable. First of all the greatest challenge of quantum computing is to be able to design systems that do not lose their entanglement due to decoherence. For this generally one tries to keep the system at low temperatures. But it would not be feasible for a biological system which operates at room temperatures and above to maintain sub-zero temperature. Despite this obstacle, the fact that quantum systems stable against decoherence &lt;em&gt;already&lt;/em&gt; exist in the warmest, slushiest of all regimes - biology - is therefore a big shock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might still say that comparing the situation to a quantum computer is a bit of a stretch. Then again what lengths will people go to in order to attract the funding agencies' good graces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cryptochrome, at its most basic, is a detection device such as a Geiger counter. On its own a single such device does not constitute a circuit, &lt;em&gt;per se&lt;/em&gt;. The question is: &lt;em&gt;do biological systems utilize quantum computation in a non-trivial manner?&lt;/em&gt; (i.e. with circuits which perform a more complex computation than just the detection of the direction of a magnetic field).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PS:&lt;/strong&gt; The answer by @Edward &lt;em&gt;appears&lt;/em&gt; to suggest that the Slater determinant state is a &lt;em&gt;de facto&lt;/em&gt; entangled state. If this was the case then &lt;em&gt;any&lt;/em&gt; (even a non-interacting) many-body fermionic system would be considered entangled. There is the suggestion (&lt;a href=&quot;http://arxiv.org/abs/quant-ph/0508078&quot; rel=&quot;nofollow&quot;&gt;T.A. Kaplan&lt;/a&gt;) that according to the traditional definition of entanglement as being present in any non-factorizable state, the Slater determinant should be considered entangled. However, the notion of entanglement as discussed in &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0205069v5&quot; rel=&quot;nofollow&quot;&gt;Shi&lt;/a&gt;, &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0701081v2&quot; rel=&quot;nofollow&quot;&gt;Wang &amp;amp; Kais&lt;/a&gt; and &lt;a href=&quot;http://arxiv.org/abs/cond-mat/0009083&quot; rel=&quot;nofollow&quot;&gt;Schliemann, Loss &amp;amp; MacDonald&lt;/a&gt; explicitly rules out consideration of the Slate state as a a measure of entanglement.&lt;/p&gt;&#xA;" LastActivityDate="2011-04-15T13:52:16.610" CommentCount="4" />
  <row Id="8490" PostTypeId="2" ParentId="5031" CreationDate="2011-04-12T07:28:00.490" Score="0" ViewCount="" Body="&lt;p&gt;What astronomers claim to be black holes are objects that &quot;progressively make themselves more similar to [a black hole] without actually reaching the stage of having an event horizon&quot;, &lt;em&gt;as they reckon&lt;/em&gt;. That's assuming that GR is valid, since all such claims depend on GR's equations. Plenty of books on GR note that black holes are perhaps better named &quot;frozen stars&quot; from a distant observer's perspective.&lt;/p&gt;&#xA;" OwnerUserId="2314" LastActivityDate="2011-04-12T07:28:00.490" />
  <row Id="8491" PostTypeId="2" ParentId="8483" CreationDate="2011-04-12T07:31:20.500" Score="5" ViewCount="" Body="&lt;p&gt;I think the issue is we need to separate the 'expected'/obvious quantum effects, from the unexpected ones.  For instance, some of your questions refer to quantum mechanics in molecular structure.  In the most trivial sense, we couldn't even have molecular structure or even stable atoms if $\hbar \rightarrow 0$ such that there are no quantum effects.  So in a trivial sense, every protein structure is due to quantum mechanics.  Following up on this...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I mean, if I do not neglect electron-electron interaction, then pretty much all electrons in a condensed matter system are entangled, are they not?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes.  In the simplest picture of matter, chemists refer to electrons occupying molecular orbitals.  This is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hartree%E2%80%93Fock_method&quot; rel=&quot;nofollow&quot;&gt;Hartee-Fock approximation&lt;/a&gt; in quantum chemistry, and since the wavefunction is written as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Slater_determinant&quot; rel=&quot;nofollow&quot;&gt;slater determinant&lt;/a&gt; of the occupied molecular orbitals, in this approximation there is no electron-electron position correlation.  Obviously in the real world this is not the case.  The wavefunction will be correlated to have two electrons further apart on average than in the simplified molecular orbital picture.  This means the state is not a product state and there is indeed entanglement of the electron positions.  So you are correct on this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And as an aside, more advanced multi-electron computational methods beyond Hartree-Fock can of course be used to account for most of this correlation energy in theoretical calculations.  The molecular orbit picture works very well though, which is why this is how it is introduced to students and how chemists colloquially discuss and visualize the quantum mechanics of a molecule.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, again this is in some sense the 'trivial' effects of quantum mechanics, as it is just &quot;chemistry&quot;.  But as we get to larger structures, due to interaction with the environment and decoherence, we can increasingly well describe everything with phenomenological parameters and a classical theory.  True the phenomenological parameters are due to the quantum mechanics, but this is not the exciting part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exciting part is when we &lt;em&gt;can't&lt;/em&gt; describe the protein interactions with phenomenological models due to the quantum coherence playing an important part &lt;em&gt;at the macroscopic level&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it is somewhat arbitrary where we draw this line.  But few really thought biology would end up using something so fragile as the coherence of entangled states to actually improve &lt;em&gt;biological function&lt;/em&gt;.  The better studied example I've seen regarding this is energy transfer in a particular photosynthesis step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether something is amazing or not isn't really a scientific question, but hopefully I clarified enough regarding your question of entanglement of electrons in molecules to see how the use of coherence to improve biological function is at the least unexpected, and hopefully exciting to you as well.&lt;/p&gt;&#xA;" OwnerUserId="2652" LastEditorUserId="2652" LastEditorDisplayName="" LastEditDate="2011-04-12T07:38:37.933" LastActivityDate="2011-04-12T07:38:37.933" />
  <row Id="8492" PostTypeId="2" ParentId="8314" CreationDate="2011-04-12T08:02:39.527" Score="2" ViewCount="" Body="&lt;p&gt;There's a way to see the answer in the OP logically, no math. In SR, clocks that &quot;feel&quot; acceleration, like the ground clock does, run slower than floating clocks (ones in free fall) do. Therefore the clock that floats during the entire experiment will elapse the most time. Only a clock &quot;shot up in a normal way&quot;, to use your words, floats during the entire experiment. Search for the &lt;em&gt;Non-Mathematical Proof of Gravitational Time Dilation&lt;/em&gt; for an elaboration. The falling clock in that experiment is one half of a symmetrical experiment involving a clock &quot;shot up in a normal way&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clock on the ground is analogous to the traveling twin that thrusts its rocket engines in the twin paradox. The clock &quot;shot up in the normal way&quot; is analogous to the stay-at-home twin that floats &quot;stationary&quot; in space, waiting for the traveling twin to return. Any clock shot up, but not in the normal way, will be a bit less like the stay-at-home twin and a bit more like the traveling twin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, the reason that a floating clock elapses more time than a clock that &quot;feels&quot; acceleration, when they start and end at the same location, is also given by SR: The clocks move relative to each other, completing trips between when they start and end at the same location. The trip travel distance of the clock that &quot;feels&quot; acceleration is shortened by length contraction predicted by SR. (A ruler that freely falls downward and near the clock on the ground shortens as it falls, as measured in that clock's frame, because the ruler's speed is increasing as measured in that frame. The ruler measures a portion of the remaining distance the clock has to travel.) Therefore that clock travels less distance at the same average speed relative to the other clock, elapsing less time than the floating clock does.&lt;/p&gt;&#xA;" OwnerUserId="2314" LastEditorUserId="2314" LastEditorDisplayName="" LastEditDate="2011-04-13T01:43:47.207" LastActivityDate="2011-04-13T01:43:47.207" CommentCount="1" />
  <row Id="8493" PostTypeId="2" ParentId="409" CreationDate="2011-04-12T08:21:24.447" Score="1" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://chandra.harvard.edu/resources/faq/black_hole/bhole-10.html&quot; rel=&quot;nofollow&quot;&gt;Chandra X-Ray Observatory FAQ&lt;/a&gt; answers this question, about the supposed black hole at the center of the Milky Way. One of the steps is to input value(s) observed into GR's equations. Therefore any black hole observed depends on GR's validity, as that link confirms. If GR is invalid, whatever we think is a black hole could instead be an object with a physical surface, with an escape velocity at its surface that is very close to the speed of light.&lt;/p&gt;&#xA;" OwnerUserId="2314" LastActivityDate="2011-04-12T08:21:24.447" />
  <row Id="8494" PostTypeId="2" ParentId="8480" CreationDate="2011-04-12T08:39:38.267" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/8480/reactionless-drives/8485#8485&quot;&gt;Edward's&lt;/a&gt; and &lt;a href=&quot;http://physics.stackexchange.com/questions/8480/reactionless-drives/8482#8482&quot;&gt;sb1's&lt;/a&gt; answers already state this well in words, but I prefer formulae:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motion is determined by the total force $\vec F_j$ acting on a particle $j$, for which you can find a potential&lt;sup&gt;1&lt;/sup&gt; $V$ such that $\vec F_j = \vec\nabla_j V$ (with $\nabla_j$ I mean derivation with respect to the $j$&lt;sup&gt;th&lt;/sup&gt; particle's coordinates). Since even in non-relativistic mechanics there is no reason to emphasize one coordinate system over the other, this means that $V$ can only depend on the positions of a particle relative to the others, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V=V(\vec x_1 - \vec x_2, \vec x_1-\vec x_3, ... , \vec x_{N-1} - \vec x_N)$ for $N$ particles&lt;sup&gt;2&lt;/sup&gt;. The result is that the sum of all forces vanishes, $\vec F = \sum\limits_j\vec F_j = \sum\limits_j\vec\nabla_j V = 0$, which for $N=2$ particles yields the famous &lt;em&gt;actio = reactio&lt;/em&gt;. This also means that, in general, a force acting on one particle (e.g. $j=1$) is cancelled out by the total force acting on all other particles, i.e. $\sum\limits_{j\neq 1}\vec F_j = -\vec F_1$ - that is why it is impossible to move a mass in a given direction without a proportional change in the opposite direction.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1)&lt;/sup&gt; for friction etc. generalized potentials can still be used in the Lagrangian formulation&lt;br&gt;&#xA;&lt;sup&gt;2)&lt;/sup&gt; I assume no external forces are present - they are caused by (maybe many) particles as well and just because you don't notice earth being attracted in your direction as well doesn't mean it doesn't happen&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2011-04-12T08:39:38.267" CommentCount="2" />
  <row Id="8495" PostTypeId="1" AcceptedAnswerId="8499" CreationDate="2011-04-12T08:41:18.130" Score="-1" ViewCount="188" Body="&lt;p&gt;I'm going to conduct an investigation into the dampening of a spring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, what specific factors could be investigated? Currently I'm planning to investigate the effect of changing the mass on the end of the spring, the effect of using different spring constants, and the effect of changing the initial extension of the spring on the amount of dampening.&lt;/p&gt;&#xA;" OwnerUserId="3044" LastEditorUserId="3044" LastEditorDisplayName="" LastEditDate="2011-05-03T09:41:17.300" LastActivityDate="2011-05-03T09:41:17.300" Title="Investigating shm in a spring?" Tags="&lt;harmonic-oscillator&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="8496" PostTypeId="1" AcceptedAnswerId="8497" CreationDate="2011-04-12T08:53:47.173" Score="1" ViewCount="105" Body="&lt;p&gt;I've observed this many times when I spill coffee, It gets attracted towards the cup or the vessel. why is this so? Which is the force acting? &#xA;(Hope you understood my question).&lt;/p&gt;&#xA;" OwnerUserId="3047" LastActivityDate="2011-04-12T09:04:20.603" Title="Spilling a cup of coffee" Tags="&lt;experimental-physics&gt;&lt;forces&gt;&lt;experimental-evidence&gt;" AnswerCount="1" />
  <row Id="8497" PostTypeId="2" ParentId="8496" CreationDate="2011-04-12T09:04:20.603" Score="1" ViewCount="" Body="&lt;p&gt;I'm reasonably certain it's surface tension&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Surface_tension&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Surface_tension&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3060" LastActivityDate="2011-04-12T09:04:20.603" CommentCount="3" />
  <row Id="8498" PostTypeId="2" ParentId="8483" CreationDate="2011-04-12T09:54:53.800" Score="2" ViewCount="" Body="&lt;p&gt;The chemistry behind that navigation is production of a biradical. &#xA;Simplified:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A--B + photon  =&gt;  A. +  B.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A and B are maybe identical halves of some bigger molecule)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is very common, a lot of photochemistry relies on such reaction, &#xA;eg photoinduced chlorination of hydrocarbons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This radical pair is a singulet, the spins are antiparallel. &#xA;(Because of total spin multiplicity conservation) &#xA;Magnetism comes into play, when the lifetime of the radical pair is &#xA;long enough to undergo &quot;intersystem crossing&quot;. This is the word for &#xA;the forbidden (ie slow) reaction of the singulett to a triplett. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that moment on magnetic fields can play a role for the outcome &#xA;of the chemical reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An examle for such things is &quot;CIDNP&quot;, where this effect drives a kind of &#xA;chemical maser. (There is a wiki on CIDNP) &#xA;This is &lt;strong&gt;quantum chemistry&lt;/strong&gt; of radical pairs. &#xA;This is common organic chemical knowledge since the 70ties, any (good) &#xA;chemist will think of that when magnetic fields and some chemistry are &#xA;&quot;entangeled&quot; :=)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What quantummechanical entanglement has to do in this context, I dont know. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-12T09:54:53.800" />
  <row Id="8499" PostTypeId="2" ParentId="8495" CreationDate="2011-04-12T10:05:33.850" Score="3" ViewCount="" Body="&lt;p&gt;and welcome to the site. I suppose that you have to do this as some kind of homework so I would not try to invent new things and just look what &lt;strong&gt;mathematical description(s)&lt;/strong&gt; of such a system exists. The governing equations will have some &lt;strong&gt;constants&lt;/strong&gt; in it and it would be already a nice thing to look if the given framework can explain the dynamics and if you can determine those constants. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before we start, you can be assured that the harmonic oscillator is most likely the physical model with the highest impact on life in general - applications from molecule spectra to airplane design.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Mathematical description: Damped harmonic oscillator&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_oscillator#Damped_harmonic_oscillator&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; has a really nice collection of articles for these kind of things. There, we can see that a damped harmonic oscillator can be described by the equation of motion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m\ddot{x} + c\dot{x} = -k x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where now $x$ is the position of the mass depending on time, dots denote derivative(s) with respect to it, $k$ is the spring &quot;constant&quot; and $c$ is the damping coefficient. All you have to do is to solve this equation by an &lt;strong&gt;exponential ansatz&lt;/strong&gt; of the form &#xA;$$x(t) = x_0 e^{\lambda t}$$&#xA;and discuss the several cases as given in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_oscillator#Damped_harmonic_oscillator&quot; rel=&quot;nofollow&quot;&gt;mentioned article&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motion in the underdamped case will look like an exponentially decaying oscillation&#xA;$$x(t) = x_0 e^{-\frac{c}{2m}}\cos (\omega t + \phi_0)\,:$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/On3II.gif&quot; alt=&quot;animation&quot;&gt;&lt;img src=&quot;http://i.stack.imgur.com/6SpMz.png&quot; alt=&quot;x(t) damped&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture was taken from the very good &lt;a href=&quot;http://es.wikipedia.org/wiki/Oscilador_arm%C3%B3nico&quot; rel=&quot;nofollow&quot;&gt;spanish wikipedia site&lt;/a&gt; concerning the topic, $c \equiv b$. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Experiment&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For a damped oscillation, you will have to determine two constants, $k$ and $c$. They are linked to the &lt;strong&gt;angular frequency&lt;/strong&gt; of the system by&#xA;$$\omega = \sqrt{\frac{k}{m}-\left( \frac{c}{2m}\right) ^2}\, .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, can &lt;strong&gt;you&lt;/strong&gt; tell me, with respect to the given form of the movement, how to measure the constants just with a stopwatch and a ruler?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After you figured everything out, you can go into a lot of different directions like how the spring constant depends on $m$ (nonlinear motion), aperiodicity, Fourier space (&lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy_distribution&quot; rel=&quot;nofollow&quot;&gt;Lorentzian&lt;/a&gt;) etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely &lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2011-04-12T10:05:33.850" CommentCount="11" />
  <row Id="8500" PostTypeId="2" ParentId="8479" CreationDate="2011-04-12T11:01:47.103" Score="3" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is highest water pressure at&#xA;  which electrolysis can be performed to&#xA;  derive hydrogen and oxygen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In practice this is a question of reactor construction/material selection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theroretically there is no limit, You just have to &quot;pay&quot; by higher potential &#xA;You need. Any tenfold of pressure  (=concentration) costs about 52 mV &#xA;at each electrode.(Maybe even more, in case that higher orders of reactants &#xA;concentrations are in the equilibrium constant of the primary reaction)&lt;br&gt;&#xA;This extra potential times current  is transformed into volume work (compression).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the dielectric constant of water,&#xA;  which which lowers as pressure&#xA;  increases, have an effect on this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is something only very distantly related. This might happen at extremely &#xA;(out of practical use) pressures by changing the water structure. &#xA;The dielectric constant per se in this context is not relevant. &#xA;Under such pressures the water can be a solid, no electrolysis any longer :=(&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-05-12T13:14:45.673" LastActivityDate="2011-05-12T13:14:45.673" CommentCount="2" />
  <row Id="8501" PostTypeId="2" ParentId="8452" CreationDate="2011-04-12T11:18:22.823" Score="3" ViewCount="" Body="&lt;p&gt;Let me add one obvious thing: There is an exact equation for the strong force.  It is what Gross, Politzer and Wilczek got the Nobel prize for.&#xA;It is called quantum chromodynamics (QCD).  Google it or look it up in Wikipedia, and you can see the Lagrangian for QCD, and  compare it to the Lagrangian for electrodynamics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you could argue about the similarities and differences of a Lagrangian, and a force equation, such as your two examples.&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-04-12T11:18:22.823" />
  <row Id="8502" PostTypeId="1" AcceptedAnswerId="8509" CreationDate="2011-04-12T12:41:12.087" Score="7" ViewCount="259" Body="&lt;ol&gt;&#xA;&lt;li&gt;Most images you see of the solar system are 2D and all planets orbit in the same plane. In a 3D view, are really all planets orbiting in similar planes? Is there a reason for this? I'd expect that the orbits distributed all around the sun, in 3D.&lt;/li&gt;&#xA;&lt;li&gt;Has an object made by man (a probe) ever left the Solar System?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="3064" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-12T20:41:19.797" LastActivityDate="2011-04-12T20:41:19.797" Title="Questions about the Solar System" Tags="&lt;astronomy&gt;&lt;orbital-motion&gt;&lt;solar-system&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="8503" PostTypeId="2" ParentId="8428" CreationDate="2011-04-12T12:56:21.330" Score="5" ViewCount="" Body="&lt;p&gt;I think you need to look for the following book, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387510583&quot; rel=&quot;nofollow&quot;&gt;Finite Quantum Electrodynamics&lt;/a&gt;: this is not something &quot;fringe&quot; nor some &quot;crackpot&quot; off-shoot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The name of the game is &lt;a href=&quot;http://en.wikipedia.org/wiki/Causal_perturbation_theory&quot; rel=&quot;nofollow&quot;&gt;Causal Perturbation Theory&lt;/a&gt;, and was pioneered by &lt;a href=&quot;http://www.numdam.org/numdam-bin/item?id=AIHPA_1973__19_3_211_0&quot; rel=&quot;nofollow&quot;&gt;Epstein, Glaser: &quot;The role of locality in perturbation theory&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as i understand your question (in the context of your comments, etc), this answer should suffice. However, maybe it'd be a good idea to take a look at what's known as &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540610499&quot; rel=&quot;nofollow&quot;&gt;Local Quantum Physics&lt;/a&gt;: i think this can give you a broader view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Another route that seems worth pursuing is that of &lt;a href=&quot;http://en.wikipedia.org/wiki/Vertex_operator_algebra&quot; rel=&quot;nofollow&quot;&gt;Vertex Operator Algebra (VOAs)&lt;/a&gt; (sometimes known just as 'Borcherds Algebra'). Note that these &quot;infinities problems&quot; appear in QFT because, ultimately, we're multiplying distributions (in the sense of generalized functions — in fact, operator-valued distributions) at the same point, which is not an allowed operation mathematically speaking. Thus, the way out is to do what people have suggested above and follow Ken Wilson's approach of &quot;point-splitting&quot;: this is just the physicist's version of the idea behind VOAs. The bottom line is that &lt;em&gt;every&lt;/em&gt; QFT is valid &lt;em&gt;only&lt;/em&gt; on some energy scale, so when you cross such line, you're bound to get meaningless results (think of it as the sites of your lattice — defined by this energy scale — merging: completely distinct points gradually become the same). In this fashion, you can take the following path: &lt;em&gt;define&lt;/em&gt; your QFT via a certain VOA that &lt;em&gt;explicitly&lt;/em&gt; sets your energy scale in its &lt;a href=&quot;http://en.wikipedia.org/wiki/Operator_product_expansion&quot; rel=&quot;nofollow&quot;&gt;Operator product expansion&lt;/a&gt; — which is nothing but the 'multiplication rule' of the VOA you have at hands. In principle, this is mathematically rigorous, although &lt;strong&gt;very&lt;/strong&gt; difficult. In any case, this also answers your question.&lt;/p&gt;&#xA;" OwnerUserId="142" LastEditorUserId="142" LastEditorDisplayName="" LastEditDate="2011-04-12T13:08:59.143" LastActivityDate="2011-04-12T13:08:59.143" CommentCount="15" />
  <row Id="8504" PostTypeId="2" ParentId="8502" CreationDate="2011-04-12T13:22:24.043" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The orbital planes of different planets has small inclinations to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ecliptic#Ecliptic_and_planets&quot; rel=&quot;nofollow&quot;&gt;ecliptic&lt;/a&gt; plane. The corresponding wikipedia diagram should give a better view than the 2D images you've seen.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Both &lt;a href=&quot;http://en.wikipedia.org/wiki/Voyager_program&quot; rel=&quot;nofollow&quot;&gt;Voyager 1 and 2&lt;/a&gt; are beyond the Solar System&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1671" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-04-12T13:49:42.010" LastActivityDate="2011-04-12T13:49:42.010" CommentCount="1" />
  <row Id="8505" PostTypeId="2" ParentId="8502" CreationDate="2011-04-12T13:23:13.197" Score="5" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;More or less yes.  The planets mostly orbit in the same plane but with small deviations compared to the size of the system.  The largest relative tilt is around 4 to 6 degrees. This 'flatness' is due to orbital mechanics, where the solar system started spherical it now has 'decayed' into a flat disc.  This disc is essentially a stable low energy configuration and therefore the system has become more and more disc like over time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Voyager 1 probe is 'leaving' the solarsystem as we speak but there is no real firm boundary and so it's hard to give a definitive answer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2535" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-12T20:40:41.623" LastActivityDate="2011-04-12T20:40:41.623" CommentCount="2" />
  <row Id="8506" PostTypeId="1" AcceptedAnswerId="8508" CreationDate="2011-04-12T13:27:47.223" Score="3" ViewCount="107" Body="&lt;p&gt;I've been looking around about LCD monitors, and how they polarise light. When a pixel needs to be black, the light is twisted so that it can't go through the polarising sheet in front.&#xA;What happens to this light? Does it relfect back into the screen? Surely that would mean the screen would get quite hot if you leave it on a black image for a while? (but I've never noticed that happen).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens to the energy (from the electricity used to create the light) when it's block by polarisation.&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="107" LastEditorDisplayName="" LastEditDate="2011-04-12T14:16:28.350" LastActivityDate="2011-04-12T14:16:28.350" Title="Where does the energy go, when light is blocked by polarisation" Tags="&lt;optics&gt;&lt;light&gt;&lt;energy-conservation&gt;&lt;polarization&gt;" AnswerCount="1" />
  <row Id="8507" PostTypeId="2" ParentId="8495" CreationDate="2011-04-12T13:33:02.197" Score="0" ViewCount="" Body="&lt;p&gt;Is this a real spring? If it is, then damping is affected by a) The number of active coils (mass that is in motion), b) The cross section of the wire, c) Any rubbing or clashing between the coils, and d) the rate at which loads are applied on the coil (frequency dependency).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest putting strain gauges on each 1/4 turn and hitting it with a hammer. You will see the response of the spring clearly and how the dynamics of the coils really affect the overall behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest find a good engineering book an spring design and analysis and you will be amazed at the complexity that springs exhibit in real life.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-04-12T13:33:02.197" CommentCount="0" />
  <row Id="8508" PostTypeId="2" ParentId="8506" CreationDate="2011-04-12T13:39:29.007" Score="9" ViewCount="" Body="&lt;p&gt;It is possible to &quot;block&quot; light based on its polarization in a number of ways. In the situation you are describing, where the light hits a polarizing filter, it is simply absorbed by the filter. The filter does indeed heat up, and in fact if you put your hand near the screen you can usually feel that it is quite warm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also possible to have a polarized mirror, which either reflects or transmits light based on its polarization. In this case the mirror doesn't absorb any* of the energy, so it doesn't heat up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*In the ideal case. In reality, of course, there is some small amount of absorption, but it can usually be ignored.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-04-12T13:39:29.007" />
  <row Id="8509" PostTypeId="2" ParentId="8502" CreationDate="2011-04-12T14:27:12.863" Score="7" ViewCount="" Body="&lt;p&gt;Nic and Approximist's answers hit the main points, but it's worth adding an additional word on the reason the orbits lie roughly in the same plane: &lt;em&gt;Conservation of angular momentum.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Solar System began as a large cloud of stuff, many times larger than its current size. It had some very slight initial angular momentum -- that is, it was, on average, rotating about a certain axis. (Why? Maybe just randomly! All of the constituents were flying around, and if you add up those random motions, there'll generically be some nonzero angular momentum.) Because angular momentum is conserved, as the cloud collapsed the rotation rate sped up (the usual example being the figure skater who pulls in her arms as she spins, and speeds up accordingly). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further collapse in the direction perpendicular to the plane of rotation doesn't change the angular momentum, but collapse in the other directions would change it. So the collapse turns the initial cloud, whatever its shape, into a pancake. The planets formed out of that pancake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, you can see the signs of that initial angular momentum in other things too: not only are all of the planets orbiting in roughly the same plane, but so are most of their moons, and most of the planets' rotations about their axes as well.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-04-12T14:52:05.097" LastActivityDate="2011-04-12T14:52:05.097" CommentCount="5" />
  <row Id="8510" PostTypeId="1" CreationDate="2011-04-12T15:21:12.990" Score="5" ViewCount="90" Body="&lt;p&gt;Inspired by &lt;a href=&quot;http://physics.stackexchange.com/questions/8502/questions-about-solar-system&quot;&gt;this question&lt;/a&gt;, are there any known planetary systems with largely varying planes of orbit? For example a system where two planets have perpendicular planes?&lt;/p&gt;&#xA;" OwnerUserId="3067" LastActivityDate="2011-04-12T20:27:32.063" Title="Are there planetary systems where the planes of orbits vary greatly?" Tags="&lt;astronomy&gt;&lt;orbital-motion&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8511" PostTypeId="2" ParentId="8472" CreationDate="2011-04-12T15:22:14.143" Score="1" ViewCount="" Body="&lt;p&gt;You say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I want to take the entangled electron and bound it to a proton or the outer shell of an atom. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The act of binding the electron with a proton, or an outer shell of an atom, means that some energy in the form of electromagnetic radiation will be given up by the electron in order to fall in an available energy level. A photon takes away spin 1 and any mathematical relationship that your electron's spin had with the world outside the atom world is lost. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is no. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-12T15:22:14.143" />
  <row Id="8513" PostTypeId="1" AcceptedAnswerId="8600" CreationDate="2011-04-12T15:49:03.513" Score="9" ViewCount="213" Body="&lt;p&gt;If you blow a bunch of soap bubbles outside, and a gust of wind hits them, will the bigger ones be more or less accelerated by the wind than the smaller ones?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, and maybe from remembered experience, I expect the small ones to be accelerated more. But I'm not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on theory I would think they'd be affected about the same, because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a = \frac{F}{m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F$ proportional to cross sectional area, which is proportional to $r^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m$ proportional to surface area, which is also proportional to $r^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming that bigger bubbles tend to have about the same thickness as smaller ones, but I don't know that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, damping due to air friction should affect big and small bubbles about the same, right?&lt;/p&gt;&#xA;" OwnerUserId="70" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-04-15T12:47:19.990" LastActivityDate="2011-04-15T12:47:19.990" Title="Are smaller soap bubbles more accelerated by wind?" Tags="&lt;fluid-dynamics&gt;&lt;acceleration&gt;" AnswerCount="6" CommentCount="4" />
  <row Id="8514" PostTypeId="1" CreationDate="2011-04-12T15:52:15.580" Score="9" ViewCount="89" Body="&lt;p&gt;Take $N$ parties, each of which receives an input $s_i \in {1, \dots, m_i}$ and produces an output $r_i \in {1, \dots, v_i}$, possibly in a nondeterministic manner. We are interested in joint conditional probabilities of the form $p(r_1r_2\dots r_N|s_1s_2\dots s_N)$. Bell polytope is the polytope spanned by the probability distributions of the form $p(r_1r_2\dots r_N|s_1s_2\dots s_N) = \delta_{r_1, r_{1, s_1}}\dots\delta_{r_N, r_{N, s_N}}$ for all possible choices of numbers $r_{i,s_i}$ (in other words, each input $s_i$ produces a result $r_{i,s_i}$ either with probability 0 or 1, regardless of other players' inputs). Polytopes of this kind are of interest in quantum information theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every Bell polytope has a certain amount of trivial symmetries, like permutation of parties or relabelling of inputs or outputs. Is it possible to give an explicit Bell polytope with nontrivial symmetries? (e.g. transformations of the polytope into itself that takes faces to faces and is not trivial in the above sense) In other words, I'm interested whether a specific Bell scenario can possess any &quot;hidden&quot; symmetries&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bell polytopes in literature are usually characterized by their faces, given by sets of inequalities (Bell inequalities), which, however, usually do not have any manifest symmetry group. &lt;/p&gt;&#xA;" OwnerUserId="307" LastEditorUserId="307" LastEditorDisplayName="" LastEditDate="2011-04-12T16:17:54.540" LastActivityDate="2011-04-12T16:17:54.540" Title="Bell polytopes with nontrivial symmetries" Tags="&lt;quantum-information&gt;&lt;bells-inequality&gt;" CommentCount="0" FavoriteCount="2" />
  <row Id="8515" PostTypeId="2" ParentId="8510" CreationDate="2011-04-12T15:55:01.573" Score="1" ViewCount="" Body="&lt;p&gt;I'm not sure we know. I think that the methods used to detect planets orbiting other stars don't allow a determination of the plane of the orbit. One method involves noticing eclipses of the star by the planet, but planets orbiting in different planes, which both eclipse the star, would look the same as planets orbiting in the same plane. The other main method is looking at the Doppler shift of the star as it &quot;wobbles&quot; in response to the planet's gravity. Since that only allows a determination of one component of the velocity, I think it also doesn't allow a determination of the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to tell would be to watch a system over time and measure perturbations of one planet's orbit due to another planet. Those perturbations would be different depending on whether the planets were in the same plane. But I think it'd be easier to get a &quot;null&quot; result (the planes are close to the same) than a positive result from this method: if they're in different planes, I think the perturbations would be too weak to observe, and it'd be hard to draw a firm conclusion from your failure to see them.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-12T15:55:01.573" CommentCount="3" />
  <row Id="8516" PostTypeId="2" ParentId="8513" CreationDate="2011-04-12T15:55:47.767" Score="2" ViewCount="" Body="&lt;p&gt;Yes this is right. Friction is also proportional to cross-section which is r^2. So the acceleration should be approximately invariant. (The friction here is nothing but the wind in the opposite direction)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another argument that makes the result much more clear:&lt;br&gt;&#xA;Since these things are floating in the air, the mass of the bubble is irrelevant (ie the density is approximately that of air).&lt;br&gt;&#xA;Second, any of these bubbles are very large compared to the microscopic properties of air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we can treat the air as a continuous medium, and you have a completely scale-invariant situation where the size of the sphere have no influence.&lt;/p&gt;&#xA;" OwnerUserId="1708" LastEditorUserId="1708" LastEditorDisplayName="" LastEditDate="2011-04-13T13:07:09.860" LastActivityDate="2011-04-13T13:07:09.860" CommentCount="1" />
  <row Id="8517" PostTypeId="1" CreationDate="2011-04-12T16:05:12.220" Score="4" ViewCount="412" Body="&lt;p&gt;Inspired by &lt;a href=&quot;http://physics.stackexchange.com/questions/8510/are-there-planetary-systems-where-the-planes-of-orbits-vary-greatly&quot;&gt;this question&lt;/a&gt;. Can a 3 body problem, starting with two planets orbiting a larger one (so massive it may be taken to stand still) in perpendicular planes, be stable?&lt;br&gt;&#xA;Is there known an analytical solution to this 3 body problem?&lt;br&gt;&#xA;Or a qualitative description of the evolution?&lt;br&gt;&#xA;Will the two bodies approach coplanar orbits?  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Bounty for this:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is Classical Mechanics (that is Newtons 1/r^2 law) sufficient for some large random collection of point-particles (with nonzero net angular momentum) in orbit around a larger one, sufficient to explain that they approach coplanar orbits?&lt;br&gt;&#xA;Or is it needed to add collisions and energy loss etc..&lt;/p&gt;&#xA;" OwnerUserId="1708" LastEditorUserId="1708" LastEditorDisplayName="" LastEditDate="2011-04-16T06:04:47.620" LastActivityDate="2011-04-17T10:43:48.997" Title="Two orbiting planets in perpendicular planes" Tags="&lt;gravity&gt;&lt;astrophysics&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="8518" PostTypeId="1" AcceptedAnswerId="8521" CreationDate="2010-08-24T08:21:23.537" Score="17" ViewCount="598" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Noether%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Noether's theorem&lt;/a&gt; states that, for every &lt;em&gt;continuous&lt;/em&gt; symmetry of a system, there exists a conserved quantity, e.g. energy conservation for time invariance, charge conservation for U(1). Is there any similar statement for &lt;em&gt;discrete&lt;/em&gt; symmetries?&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-04-12T16:06:38.460" LastActivityDate="2011-04-14T09:54:22.223" Title="Is there something similar to Noether's theorem for discrete symmetries?" Tags="&lt;mathematical-physics&gt;&lt;group-theory&gt;&lt;discrete-symmetry&gt;" AnswerCount="7" CommentCount="6" FavoriteCount="3" />
  <row Id="8519" PostTypeId="2" ParentId="8518" CreationDate="2010-08-24T20:26:14.867" Score="5" ViewCount="" Body="&lt;p&gt;No, because discrete symmetries have no infinitesimal form which would give rise to the (characteristic of) conservation law. See also &lt;a href=&quot;http://dx.doi.org/10.1119/1.18109&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; for a more detailed discussion.&lt;/p&gt;&#xA;" OwnerUserId="348" LastActivityDate="2010-08-24T20:26:14.867" CommentCount="2" />
  <row Id="8520" PostTypeId="1" CreationDate="2011-04-12T16:40:47.820" Score="0" ViewCount="62" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/1193/why-does-the-atmosphere-rotate-along-with-the-earth&quot;&gt;Why does the atmosphere rotate along with the earth?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;suppose a helicopter  rise vertically from the ground and remain in a fixed point, when it comes to land, whether in the same point of land comes down(Do things that are in the atmosphere rotates with the Earth)&lt;/p&gt;&#xA;" OwnerUserId="3069" LastActivityDate="2011-04-12T16:40:47.820" ClosedDate="2011-04-12T16:42:07.213" Title="Do things that are in the atmosphere rotates with the Earth" Tags="&lt;astronomy&gt;&lt;newtonian-mechanics&gt;&lt;space&gt;" CommentCount="0" />
  <row Id="8521" PostTypeId="2" ParentId="8518" CreationDate="2011-04-12T17:14:33.233" Score="12" ViewCount="" Body="&lt;p&gt;For continuous global symmetries, Noether theorem gives you a locally conserved charge density (and an associates current), whose integral over all of space is conserved (time independent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For global discrete symmetries you have to distinguish between the cases where the conserved charge is continuous or discrete. For &lt;em&gt;infinite&lt;/em&gt; symmetries like lattice translations the conserved quantity is continuous, albeit a periodic one. So  in such case momentum is conserved modulo vectors in the reciprocal lattice. The conservation is local just as in the case of continuous symmetries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of &lt;em&gt;finite&lt;/em&gt; group of symmetries the conserved quantity is itself discrete. You then don't have local conservation laws because the conserved quantity cannot vary continuously in space. Nevertheless for such symmetries you still have a conserved charge which gives constraints (selection rules) on allowed processes. For example, for parity invariant theories you can give each state of a particle a &quot;parity charge&quot; which is simply a sign, and the total charge has to be conserved for any process, otherwise the amplitude for it is zero.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="566" LastEditorDisplayName="" LastEditDate="2011-04-12T18:51:29.117" LastActivityDate="2011-04-12T18:51:29.117" CommentCount="3" />
  <row Id="8522" PostTypeId="1" AcceptedAnswerId="8548" CreationDate="2011-04-12T17:24:56.377" Score="8" ViewCount="529" Body="&lt;p&gt;Is it energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it energy per unit volume?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it energy per unit time i.e power?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is it?&lt;/p&gt;&#xA;" OwnerUserId="2146" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-04-14T04:32:32.913" LastActivityDate="2011-05-17T11:15:23.167" Title="What exactly is heat?" Tags="&lt;thermodynamics&gt;&lt;heat&gt;&lt;temperature&gt;" AnswerCount="6" CommentCount="5" FavoriteCount="3" />
  <row Id="8523" PostTypeId="2" ParentId="8513" CreationDate="2011-04-12T17:26:22.427" Score="1" ViewCount="" Body="&lt;p&gt;No wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Friction of a sphere is proportional to radius, &#xA;as Mr Stokes found out, and mass of bubble is prop  r³ &#xA;as far as entrapped air is concerned, surface is prop r² &#xA;and now thickness of the soap lamella is the only unknown &#xA;remaining. &#xA;From colors of interference I assume the thickness is in the magnitude &#xA;of some µmeters. Take 10 µm and calculate.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-12T17:26:22.427" CommentCount="11" />
  <row Id="8524" PostTypeId="2" ParentId="8522" CreationDate="2011-04-12T17:30:10.643" Score="-1" ViewCount="" Body="&lt;p&gt;Its largely a historical artifact.  But it is often used to refer to energy.  In any physics text book it will be meant as analagous to energy and so measured in joules.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-04-12T17:30:10.643" CommentCount="1" />
  <row Id="8525" PostTypeId="7" CreationDate="2011-04-12T17:33:53.237" Score="0" ViewCount="" Body="&lt;p&gt;Physics - Stack Exchange is for active researchers, academics and students. We welcome questions of all levels, but please stick to the following topics:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;b&gt;Explanations of observed physical phenomena&lt;/b&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/629/why-does-one-experience-a-short-pull-in-the-wrong-direction-when-a-vehicle-stops&quot;&gt;Why does one experience a short pull in the wrong direction when a vehicle stops?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;b&gt;Experimental designs and results&lt;/b&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/185/what-is-needed-to-claim-the-discovery-of-the-higgs-boson&quot;&gt;What is needed to claim the discovery of the Higgs boson ?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;b&gt;Accepted and/or actively researched theories&lt;/b&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/2051/why-do-we-think-there-are-only-three-generations-of-fundamental-particles&quot;&gt;Why do we think there are only three generations of fundamental particles?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;b&gt;Mathematics and physics&lt;/b&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/414/number-theory-in-Physics&quot;&gt;Number theory in physics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You can expect a range of answers from very basic to highly technical, depending on how you phrase your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we do not allow certain kinds of questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;b&gt;&quot;Do my homework&quot;-type physics questions&lt;/b&gt;&lt;br&gt;&#xA;&lt;i&gt;&quot;A 4kg ball is traveling at 8m/s in the x direction, how do I find...&quot;&lt;/i&gt;&lt;br&gt;&#xA;Physics - Stack Exchange is not a homework help site. If you have a question about a homework problem, or any problem of an educational nature, narrow it down to the specific concept that is giving you trouble and ask about that. You can find more &lt;a href=&quot;http://meta.physics.stackexchange.com/questions/714/how-do-i-ask-homework-questions-on-physics-stack-exchange&quot;&gt;information about acceptable homework questions&lt;/a&gt; on our meta site.&lt;/li&gt;&#xA;&lt;li&gt;&lt;b&gt;Pitches for your own personal theories or work&lt;/b&gt;&lt;br&gt;&#xA;We deal with mainstream physics here. Anything that couldn't be published in a reputable journal is probably not appropriate at this site.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-14T17:59:06.440" LastActivityDate="2011-07-14T17:59:06.440" />
  <row Id="8526" PostTypeId="2" ParentId="8522" CreationDate="2011-04-12T17:38:57.060" Score="8" ViewCount="" Body="&lt;p&gt;The word &quot;heat&quot; is used to apply to energy &lt;em&gt;transfer&lt;/em&gt; from a high-temperature body to a low-temperature body. You should never say that a body &lt;em&gt;contains&lt;/em&gt; a certain amount of heat; you should only say that a certain amount of heat &lt;em&gt;flowed&lt;/em&gt; from one body to another. As Nic correctly says, it comes in units of energy (joules, or sometimes calories).&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-12T17:38:57.060" CommentCount="9" />
  <row Id="8528" PostTypeId="2" ParentId="8518" CreationDate="2011-04-12T17:58:18.193" Score="5" ViewCount="" Body="&lt;p&gt;You mentioned crystal symmetries. Crystals have a discrete translation invariance: It is not invariant under an infinitesimal translation, but invariant under translation by a lattice vector. The result of this is conservation of momentum &lt;em&gt;up to a reciprocal lattice vector&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an additional result: Suppose the Hamiltonian itself is time independent, and suppose the symmetry is related to an operator $\hat S$. An example would be the parity operator $\hat P|x\rangle = |-x\rangle$. If this operator is a symmetry, then $[H,P] = 0$. But since the commutator of an operator with the Hamiltonian also gives you the derivative, you have &#xA;$\dot P = 0$.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-04-12T17:58:18.193" />
  <row Id="8529" PostTypeId="2" ParentId="8522" CreationDate="2011-04-12T17:59:48.830" Score="2" ViewCount="" Body="&lt;p&gt;Three meanings:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;amount of energy (in joules) transfered from a &quot;hot body&quot; to &quot;cold body&quot; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the process of 1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a non exact 1-form Q such that $dE=W+Q$ or $Q=TdS$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Heat is not a substance. Heat cannot be &quot;extracted&quot; from a body. Microscopically heat is &quot;the flow of kinetic energy&quot; during the process of achieving thermal equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="1377" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-12T20:14:28.093" LastActivityDate="2011-04-12T20:14:28.093" CommentCount="4" />
  <row Id="8530" PostTypeId="1" AcceptedAnswerId="8533" CreationDate="2011-04-12T18:03:45.177" Score="2" ViewCount="208" Body="&lt;p&gt;If a charged particle with charge $q$ and mass $m$ has spin $s \neq 0$ we can measure an intrinsic magnetic moment $\mu = g \frac{q}{2m}\hbar \sqrt{s(s+1)}$. This is how spin was discovered in the first place in the Stern-Gerlach Experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for a neutral particle $\mu = 0$, so we cannot measure the spin of the particle in the same manner. But it is said, that e.g. the Neutron or the Neutrino both have a spin $s=1/2$. How was or can this be measured? &lt;/p&gt;&#xA;" OwnerUserId="1648" LastActivityDate="2011-04-13T05:11:56.670" Title="How to measure the spin of a neutral particle?" Tags="&lt;quantum-mechanics&gt;&lt;spin&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="8531" PostTypeId="2" ParentId="8530" CreationDate="2011-04-12T18:14:46.103" Score="3" ViewCount="" Body="&lt;p&gt;Also the neutron has a magnetic moment. &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutron_magnetic_moment&quot; rel=&quot;nofollow&quot;&gt;Check this out&lt;/a&gt;. The reason is that the neutron is not an elementary particle but built up from quarks which have charge...&lt;/p&gt;&#xA;" OwnerUserId="2962" LastActivityDate="2011-04-12T18:14:46.103" CommentCount="7" />
  <row Id="8532" PostTypeId="2" ParentId="8517" CreationDate="2011-04-12T18:18:18.573" Score="1" ViewCount="" Body="&lt;p&gt;There are &lt;a href=&quot;http://www.seasky.org/astronomy/astronomy-software.html&quot; rel=&quot;nofollow&quot;&gt;free programs&lt;/a&gt; for playing with gravitational simulations  for example &lt;a href=&quot;http://www.sat-net.com/winorbit/&quot; rel=&quot;nofollow&quot;&gt;satellite orbits&lt;/a&gt; , so there is no problem of calculating numerically any kind of  orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not see why two perpendicular orbit planes would not be stable in principle. It would depend on the parameters of course, masses and distances. If the satellite  bodies or the central one have oceans, tides will affect the orbits and long term stability will be lost. Otherwise, once the orbits are set up angular momentum conservation should keep them in the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are really interested you could try downloading and playing around with some program to make sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit/correcting : in this &lt;a href=&quot;http://en.wikipedia.org/wiki/N-body_problem&quot; rel=&quot;nofollow&quot;&gt;wiki article&lt;/a&gt; the three body problem is examined and it is stated that the orbits are in general chaotic, in fact the three body gravitational problem led to the study of deterministic chaos. In the gif provided for a three  small mass particle problem I cannot see what goes on with angular momentum conservation. Some stability exists in special solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the problem is not as simple as observing the sun moon and earth, and depends on specific initial assumptions. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-04-13T11:48:36.527" LastActivityDate="2011-04-13T11:48:36.527" CommentCount="6" />
  <row Id="8533" PostTypeId="2" ParentId="8530" CreationDate="2011-04-12T18:30:14.157" Score="6" ViewCount="" Body="&lt;p&gt;Conservation of angular momentum is invoked for the neutrinos because beams of neutrinos cannot be collimated   for an experimental measurement. Neutron spin  can be measured in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Stern_gerlach&quot; rel=&quot;nofollow&quot;&gt;Stern Gerlach&lt;/a&gt; setup. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interactions and decays were carefully examined in various experiments and the only consistent spin values are the ones assigned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I see that the question should be formulated as : why the neutron has a Dirac magnetic moment, although it is neutral, which is the formula that is displayed above, and does the neutrino have a Dirac magnetic moment? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The neutron, and other baryons,  has a magnetic moment because the quarks that compose it have a Dirac magnetic moment. See for example &lt;em&gt;Perkins, Introduction to High Energy Physics, section: baryon magnetic moments&lt;/em&gt; for the derivation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether the neutrino has a magnetic moment due to higher order loop diagrams is a research question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, though spin in charged point like particles is connected to  magnetic moment with the formula above, analogous to classical charges circulating in a loop having a magnetic moment, , charge is &lt;strong&gt;not necessary&lt;/strong&gt; for spin to appear. There is intrinsic spin which for the neutrino comes from the angular momentum balance in the interactions where it appears. The neutrino is a spinor  in the Dirac formalism.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-04-13T05:11:56.670" LastActivityDate="2011-04-13T05:11:56.670" CommentCount="5" />
  <row Id="8534" PostTypeId="2" ParentId="8518" CreationDate="2011-04-12T18:32:27.493" Score="7" ViewCount="" Body="&lt;p&gt;Put into one sentence, Noether's first Theorem states that a continuous, global, off-shell symmetry of an action $S$ implies a local on-shell conservation law. By the words &lt;em&gt;on-shell&lt;/em&gt; and &lt;em&gt;off-shell&lt;/em&gt; are meant whether Euler-Lagrange equations of motion are satisfied or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question asks if &lt;em&gt;continuous&lt;/em&gt; can be replace by &lt;em&gt;discrete?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should immediately be stressed that Noether Theorem is a machine that for &lt;em&gt;each&lt;/em&gt; input in form of an appropriate symmetry produces an output in form of a conservation law. To claim that a Noether Theorem is behind, it is &lt;em&gt;not&lt;/em&gt; enough to just list a couple of pairs (symmetry, conservation law). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, where could a discrete version of Noether's Theorem live? A good bet is in a discrete lattice world, if one uses &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_difference&quot; rel=&quot;nofollow&quot;&gt;finite differences&lt;/a&gt; instead of differentiation. Let us investigate the situation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our intuitive idea is that finite symmetries, e.g., time reversal symmetry, etc, can not be used in a Noether Theorem in a lattice world because they don't work in a continuous world. Instead we pin our hopes to that discrete infinite symmetries, that become continuous symmetries when the lattice spacings go to zero, can be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine for simplicity a 1D point particle that can only take discrete positions $q_t\in\mathbb{Z}a$ on a 1D lattice $\mathbb{Z}a$  with lattice spacing $a$, and that time $t\in\mathbb{Z}$ is discrete as well. (This was, e.g., studied in J.C. Baez and J.M. Gilliam, Lett. Math. Phys. 31 (1994) 205; hat tip: Edward.) The velocity is the finite difference&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_{t+\frac{1}{2}}:=q_{t+1}-q_t\in\mathbb{Z}a,$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and is discrete as well. The action $S$ is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S[q]=\sum_t L_t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with Lagrangian $L_t$ on the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L_t=L_t(q_t,v_{t+\frac{1}{2}}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define momentum $p_{t+\frac{1}{2}}$ as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ p_{t+\frac{1}{2}} := \frac{\partial L_t}{\partial v_{t+\frac{1}{2}}}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naively, the action $S$ should be extremized wrt. neighboring virtual discrete paths $q:\mathbb{Z} \to\mathbb{Z}a$ to find the equation of motion. However, it does not seem feasible to extract a discrete Euler-Lagrange equation in this way, basically because it is not enough to Taylor expand to the first order in the variation $\Delta q$ when the variation $\Delta q\in\mathbb{Z}a$ is not infinitesimal. At this point, we throw our hands in the air, and &lt;em&gt;declare&lt;/em&gt; that the virtual path $q+\Delta q$ (as opposed to the stationary path $q$) does not have to lie in the lattice, but that it is free to take continuous values in $\mathbb{R}$. We can now perform an infinitesimal variation without worrying about higher order contributions,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 =\delta S := S[q+\delta q] - S[q]&#xD;&#xA;= \sum_t \left[\frac{\partial L_t}{\partial q_t} \delta q_t + p_{t+\frac{1}{2}}\delta v_{t+\frac{1}{2}} \right] $$ &#xA;$$ =\sum_t \left[\frac{\partial L_t}{\partial q_t} \delta q_{t} + p_{t+\frac{1}{2}}(\delta q_{t+1}- \delta q_t)\right] $$&#xA;$$=\sum_t \left[\frac{\partial L_t}{\partial q_t} - p_{t+\frac{1}{2}} + p_{t-\frac{1}{2}}\right]\delta q_t + \sum_t \left[p_{t+\frac{1}{2}}\delta q_{t+1}-p_{t-\frac{1}{2}}\delta q_t \right].$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the last sum is telescopic. This implies (with suitable boundary conditions) the discrete Euler-Lagrange equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial L_t}{\partial q_t} = p_{t+\frac{1}{2}}-p_{t-\frac{1}{2}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the evolution equation. At this point it is not clear whether a solution for $q:\mathbb{Z}\to\mathbb{R}$ will remain on the lattice $\mathbb{Z}a$, if we specify two initial values on the lattice. We shall from now on restrict our considerations to such systems for consistency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, one may imagine that $q_t$ is a cyclic variable, i.e., that $L_t$ does not depend on $q_t$. We therefore have a discrete global translation symmetry $\Delta q_t=a$. The Noether current is the momentum $p_{t+\frac{1}{2}}$, and the Noether conservation law is that momentum $p_{t+\frac{1}{2}}$ is conserved. This is certainly a nice observation. But this does &lt;em&gt;not&lt;/em&gt; necessarily mean that a Noether Theorem is behind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that the enemy has given us a global vertical symmetry $\Delta q_t = Y(q_t)\in\mathbb{Z}a$, where $Y$ is an arbitrary function. (The words &lt;em&gt;vertical&lt;/em&gt; and &lt;em&gt;horizontal&lt;/em&gt; refer to translation in the $q$ direction and the $t$ direction, respectively. We will for simplicity not discuss symmetries with horizontal components.) The obvious candidate for the bare Noether current is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$j_t = p_{t-\frac{1}{2}}Y(q_t).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is unlikely that we would be able to prove that $j_t$ is conserved merely from the symmetry $0=S[q+\Delta q] - S[q]$, which would now unavoidably involve higher order contributions. So while we stop short of declaring a no-go theorem, it certainly does not look promising.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps, we would be more successful if we only discretize time, and leave the coordinate space continuous? I might return with an update about this in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A general remark: It is a misnomer to attribute all pairs of (symmetry, conservation law), that one encounters, to Noether Theorem. For instance, it is a misnomer to call consequences of Hamilton's equations of motion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\{A,H\}=0 \wedge \frac{\partial H}{\partial t}=0 \wedge \frac{\partial A}{\partial t}=0   \Rightarrow  \frac{d A}{dt}=0,  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or equivalently, in quantum mechanics, consequences of Heisenberg's equations of motion &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\hat{A},\hat{H}]=0 \wedge \frac{\partial \hat{H}}{\partial t}=0 \wedge \frac{\partial \hat{A}}{\partial t}=0   \Rightarrow  \frac{d \hat{A}}{dt}=0,  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for &quot;a Noether Theorem&quot;, because the argument never involved an &lt;em&gt;off-shell&lt;/em&gt; symmetry, which is the input for the Noether Theorem machine. Moreover, some quantum-mechanical&#xA;pairs (symmetry, conservation law) do not have classical counterparts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example from the continuous world, that is good to keep in mind: Consider a &lt;a href=&quot;http://en.wikipedia.org/wiki/Pendulum&quot; rel=&quot;nofollow&quot;&gt;simple gravity pendulum&lt;/a&gt; with Lagrangian&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L(\varphi,\dot{\varphi}) = \frac{m}{2}\ell^2 \dot{\varphi}^2 + mg\ell\cos(\varphi).$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has a global discrete periodic symmetry $\varphi\to\varphi+2\pi$, but the (angular) momentum $p_{\varphi}:=\frac{\partial L}{\partial\dot{\varphi}}= m\ell^2\dot{\varphi}$ is &lt;em&gt;not&lt;/em&gt; conserved if $g\neq 0$.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-04-14T09:54:22.223" LastActivityDate="2011-04-14T09:54:22.223" CommentCount="2" />
  <row Id="8536" PostTypeId="2" ParentId="8530" CreationDate="2011-04-12T18:33:10.880" Score="3" ViewCount="" Body="&lt;p&gt;If that neutral particle is a neutron, You find a negative &#xA;value -1.91 nuclear magnetons with a apparatus by Rabi &#xA;developed on basis of Stern-Gerlach method. Bloch measured the &#xA;neutrons magnetic moment first. &#xA;(This was one of the hints that a neutron is not a pointlike &#xA;or homogenous particle, quarks and so on ...) &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-12T18:33:10.880" />
  <row Id="8537" PostTypeId="2" ParentId="8522" CreationDate="2011-04-12T19:19:23.817" Score="0" ViewCount="" Body="&lt;p&gt;Energy can be defined as the capacity to do &lt;em&gt;work&lt;/em&gt; or produce &lt;em&gt;heat&lt;/em&gt; (among others) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can define a system through defining its boundaries, &#xA;then energy flowing across system boundaries could be heat or work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although Energy can exist &lt;em&gt;contained&lt;/em&gt; in a system as potential ( in many forms) to exchange it from one system to another it takes (or it is named as) these specific forms: heat and work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the difference between heat and work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mechanical work is a force acting through a distance in certain direction, it's relevant and useful for our daily tasks, like moving an arm, a piston, a machine, then a car, etc.. it's in a given preferential direction, this energy can be used to respond to some intention over the -whole system-, that's one of the reasons it's often called &lt;em&gt;useful&lt;/em&gt; energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand at microscopic scale particles are too many, and not so easy to control, they have no preferential direction of movement, particles vibrates, and move randomly, they do not move in a coherent direction although they have a mean speed (related to system temperature). It's a kind of symmetry in the sense of having no preferential direction, but this kinky energy, can be transfered, so that isotropy, with dissipated amount of &quot;surprises&quot; and information, crosses the boundaries under the name of  Heat, turning the interacting systems to a whole boring dead state (to witch systems usually evolve). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat flows from High to Low temperature systems. Changes in the systems made through heat are strongly path dependent (its effects depend on pressure, volume, time, temperature differences, order of events ..)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, mechanical work, besides to produce the moving of large ammount of particles in a coherent direction, this &quot;intended action&quot; generate of course at microscopic scale a lot of &quot;unintended actions&quot; too (random), and that's heat again, then &lt;em&gt;work is never alone.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work can be converted completely into heat, but heat can not be converted completely into work. That's why heat is usually seen as energy loss, and even the definition of energy is often : the ability a physical system has to do work (leaving The Heat out)&lt;/p&gt;&#xA;" OwnerUserId="1916" LastEditorUserId="1916" LastEditorDisplayName="" LastEditDate="2011-04-13T12:39:17.727" LastActivityDate="2011-04-13T12:39:17.727" />
  <row Id="8538" PostTypeId="1" CreationDate="2011-04-12T19:35:09.607" Score="1" ViewCount="180" Body="&lt;p&gt;The following image is taken from teaching physics lecture &lt;a href=&quot;http://www.univie.ac.at/pluslucis/FBW0/FBW2008/Material/Muckenfuss_Wien_2008/Workshop_Optik/03Aufsaetze_Optik/spiegelbilder_DPG_Vortrag.pdf&quot; rel=&quot;nofollow&quot;&gt;Was man aus virtuellen Bildern lernen kann&lt;/a&gt; (in German):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Dzq9r.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the cited paper claims that the left hand side is the correct picture to explain the virtual image if you look with your eyes, the right hand side is correct if you use a camera (but incorrect if the camera is replaced with the eye). In the case of a plane mirror both constructions lead to the same result. However in the case of curved mirrors they lead to different results (which is also discussed in the cited paper). Now I am a bit confused since the right picture (with an eye instead of the camera) is that one typically drawn in physics textbooks (for example  Pedrotti &quot;Introduction to optics&quot; p.38 or Halliday Resnick &quot;Fundamentals of Physics&quot; p.1140). So my question is, which picture is the correct one? Do you know further references of papers or books which discuss this problem more in detail? (I have only found this cited paper in German).&lt;/p&gt;&#xA;" OwnerUserId="667" LastActivityDate="2011-06-27T16:13:44.207" Title="Virtual images in (plane) mirrors?" Tags="&lt;optics&gt;&lt;education&gt;&lt;geometric-optics&gt;&lt;mirror&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" />
  <row Id="8539" PostTypeId="1" AcceptedAnswerId="8546" CreationDate="2011-04-12T19:42:52.033" Score="2" ViewCount="113" Body="&lt;p&gt;I have just started looking into special relativity and I have come up with an intriguing &lt;em&gt;gedanke&lt;/em&gt;, as Einstein himself called such theoretical thought experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a space shuttle traveling through space at a constant velocity close to $c$. As the shuttle passes earth, a previously set-up camera starts broadcasting from earth to the shuttle. Since radio waves travel at the speed of light, the shuttle is receiving a constant transmission feed, assuming the camera is broadcasting 24/7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, from what I have understood of special relativity so far, time will flow slower for  the astronaut than for the earthlings. Hence, assuming $v=0.8c$, the astronaut will after 30 years have received a video transmission 50 years long!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my reasoning correct, that even though the transmission is live, the astronaut would actually be watching things that happened many years ago, while still receiving the &quot;live&quot; feed, which would be stored/buffered in the shuttles memory, thus making it possible for the astronaut to fast-forward the clip to see what happened more than 30 years after passing the earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second question is, what happens when we consider the space shuttle to be at rest and the earth to be moving instead? If that would imply that it has been 50 years from the astronauts point of view, while only 30 years have passed on earth, then the astronaut would run out of video material after the first 30 years of watching the broadcast. Then what?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope it makes sense. Thank you!&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditorDisplayName="" LastEditDate="2011-04-12T19:57:38.557" LastActivityDate="2011-04-12T21:09:23.230" Title="A video conference between earth and a space shuttle" Tags="&lt;special-relativity&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8540" PostTypeId="1" CreationDate="2011-04-12T19:50:53.200" Score="4" ViewCount="275" Body="&lt;p&gt;Since I don't know how to add another question to an already existing topic, &#xA;I'm opening a new threat. However I'm refering to:&#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/8201/beginners-questions-concerning-conformal-field-theory&quot;&gt;Beginners questions concerning Conformal Field Theory&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted, a few weeks ago I started reading about Conformal Field Theory. &#xA;I'm actually from a more mathematical background, however I'm not very familiar &#xA;with Quantum Field Theory. Though I'm quite familiar with Quantum Mechanics/Classical Mechanics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now again some questions again turned up: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Think of a theory with an energy-momentum-tensor that is given &#xA;on the plane. Let's assume the most general form $T(z)=\sum z^{-n-2} L_{n}$ and &#xA;$L_{n} = \frac{1}{2 \pi i} \oint dz z^{n+1} T(z)$. &#xA;Now some of my reference (such as David Tong in the reference question above) point&#xA;out that $L_{0}$ generates scalings/rotations and $L_{1},L_{-1}$ generate translations. &#xA;So let's consider the example of a rotation. The generator of a rotation is $z \frac{\partial}{\partial z}$. Now in order to show that $L_{0}$ actualyy generates &#xA;this rotation one needs to show that $[L_{0},\phi]$ =  $z \frac{\partial}{\partial z}$ \phi.&#xA;I've shown this for the example of the free boson, however I'm not 100% sure how to prove it in the general case. Can someone help me? (Maybe it's related to Operator Product Expansions...) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The second question goes a bit deeper into the theory. &#xA;It concern Current Algebras. I've read some articles on the Sugawara construction &#xA;and there Mr Sugawara proposes and Energy-Momentum-Tensor of the form&#xA;$T(z)  = \gamma \sum_{a=1}^{dim g} : j^{a}(z) j^{a}(z):$ . However I don't really see &#xA;how he comes up with it or why this seems to be a &quot;natural choice&quot; of an Energy-Momentum tensor. I've heard that it includes the Energy Momentum Tensor of the free boson &#xA;(given by $T(z)=\partial_{z} \phi \partial_{z} \phi$) as a special case. &#xA;For me this is not so obvious. Can someone please explain to me how he comes up that&#xA;in an easy way. I don't think it's necessary to show me all the calculations. &#xA;Just the basic idea would be usefull to get some intuition. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm having some troubles on understanding the intuition behind current algebras. &#xA;(I haven't read about WZW Models yet). The Virasoro algebra appeared to me in a kind&#xA;of natural way in the example of the free boson. The generalization is then pretty much&#xA;straight forward. However I don't have that kind of intuition for current algebras. &#xA;I've read that they provide some &quot;additional symmetry structure&quot; which reduces the number of possible correlation functions. But I don't know any details. &#xA;I'd be more than happy if someone could comment on that.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!!&lt;/p&gt;&#xA;" OwnerUserId="2969" LastActivityDate="2011-06-14T12:58:25.320" Title="Some questions on Conformal Field Theory" Tags="&lt;quantum-field-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="8541" PostTypeId="2" ParentId="8538" CreationDate="2011-04-12T19:57:22.117" Score="1" ViewCount="" Body="&lt;p&gt;The left picture shows the light rays from foot and top of the candle, those rays will end at the corresponding points of the real picture on the retina. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This allows the estimation of the distance by &lt;strong&gt;experience&lt;/strong&gt; . You just know how big such a candle is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second picture shows the the projection of one point of the candls (the tip) to &#xA;the camera. The camera has to be focused to that distance to get a point instead of a &#xA;contourless stain. This is estimation of distance by focus, which is done by Your eyes &#xA;as well but only at close distances, says the text. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The word &quot;Bildlage&quot; is inapropriate for me, I do not understand it in this context. &#xA;I think this an error. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, there is distance &quot;seeing&quot; by stereoscopic effect as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the new links the word &quot;Entfernungseindruck&quot; appears, that makes sense!.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-12T20:27:34.853" LastActivityDate="2011-04-12T20:27:34.853" CommentCount="4" />
  <row Id="8542" PostTypeId="1" CreationDate="2011-04-12T20:12:08.167" Score="1" ViewCount="150" Body="&lt;p&gt;As compared to when the coffee is just hot from brewing.  I suspect it has something to do with the way the microwaves are affecting the molecules of the coffee.&lt;/p&gt;&#xA;" OwnerUserId="3074" LastActivityDate="2011-04-12T20:33:54.380" ClosedDate="2011-04-12T20:42:53.930" Title="Why is it when you microwave cold coffee and then add milk it creates a foam head?" Tags="&lt;thermodynamics&gt;" AnswerCount="0" CommentCount="11" FavoriteCount="0" />
  <row Id="8544" PostTypeId="2" ParentId="8510" CreationDate="2011-04-12T20:27:32.063" Score="0" ViewCount="" Body="&lt;p&gt;If the planets are all formed from a simple accretion disk, they should end up very close to the plane of the accretion disk. A planet with a different orientation must have had a different formation history, or to have suffered a severe deflection of its motion. IIRC one observed disk around a yound star was thought to be orbitting in opposite directions, i.e. the close in part going one direction, and an outer disk going in the opposite direction. But if you read Ted's answer about detection, the chances are very high that we would only see the planets from a single plane, and be unaware of any out of plane objects. The chances are that to get two accretion disks with different orientations (possibly at different times), you would have to incorportate different gas clouds with different relative velocities during planetary formation.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-04-12T20:27:32.063" />
  <row Id="8546" PostTypeId="2" ParentId="8539" CreationDate="2011-04-12T21:09:23.230" Score="4" ViewCount="" Body="&lt;p&gt;The astronaut will not have received 50 years worth of transmissions at the time you specify. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start by stating precisely what time dilation means in this instance. Consider two &quot;events&quot;: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The people on Earth have a party to celebrate 50 years of radio broadcasting.&lt;/li&gt;&#xA;&lt;li&gt;The astronaut on board the ship has a party to celebrate 30 years of travel.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In a reference frame in which the Earth is at rest, those two events are simultaneous. That's what we mean when we say that the astronaut's clocks run slow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the astronaut won't receive the 50th year of radio broadcasting until much later -- to be precise, 200 more years of Earth-time will elapse before he receives this signal. (Let's check this: in 200 years at $0.8c$, he will travel 160 light-years. Add that to the 40 light-years he's already gone, and you find that he's 200 light-years from Earth at the time. The radio signal will just be reaching that distance at that time.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, because of time dilation, that's only 120 additional years of astronaut time. Still, it means the astronaut can't see into the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To say it another way, at the moment in question (when the astronaut throws his 30-year party), the radio signal he is receiving is one that left Earth merely 10 years after he passed Earth. In Earth's reference frame, that radio signal traveled a distance of 40 light-years (since that's how far away the astronaut is at that moment), and took 40 years to do it (50 years minus 10 years).&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-12T21:09:23.230" CommentCount="7" />
  <row Id="8547" PostTypeId="2" ParentId="8522" CreationDate="2011-04-12T22:35:18.483" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_energy&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Thermal_energy&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;read about molecular / statistical description of thermal energy (which is not heat) u would really get the feel what it really is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as far as the rigorous description of heat is in terms of entropy, let me denote it by q and entropy by s&#xA;(dq is total differential of q)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dq&amp;lt;=Tds  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;equality being true for reversible process ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you may come across many intuitive(mostly wrong) definitions and descriptions of heat .. which may still work if you are a mechanical engineer dealing with simple (linear) thermal/fluid systems but if you are a physicist, you really need to know entropy and heat in terms of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dimensionally q (which i have said heat) is dimensionally equal to energy&lt;/p&gt;&#xA;" OwnerUserId="3077" LastActivityDate="2011-04-12T22:35:18.483" />
  <row Id="8548" PostTypeId="2" ParentId="8522" CreationDate="2011-04-12T22:43:24.093" Score="19" ViewCount="" Body="&lt;p&gt;I'll try to give an answer in purely classical thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat is a way of accounting for energy transfer between &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_system&quot; rel=&quot;nofollow&quot;&gt;thermodynamic systems&lt;/a&gt;.  Whatever energy is not transferred as work is transferred as heat.  If you observe a thermodynamic process and calculate that system A lost $Q$ calories of heat, this means that if the environment around system A were replaced with $Q$ grams of water at 14C and the process were repeated, the temperature of that water would rise to 15C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Energy&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy is a number associated with the &lt;a href=&quot;http://en.wikipedia.org/wiki/State_function&quot; rel=&quot;nofollow&quot;&gt;state&lt;/a&gt; of a system.  It can be calculated if you give the state variables - things like mass, temperature, chemical composition, pressure, and volume.  (These state variables are not all independent, so you only need to give some combination of them.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes the energy can be accounted very simply.  For an ideal gas, the energy is simply proportional to the temperature, number of molecules, and number of dimensions.  For a system with interesting chemistry, internal stresses and deformation, gravitational potential, etc. the energy may be more complicated.  Essentially, we get to invent the formulas for energy that are most useful to us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a nice overview of energy, summarizing Richard Feynman, &lt;a href=&quot;http://www.chipbrock.org/?p=98&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  For a more theoretical point of view on where these energy formulas come free, see Lubos Motl's answer &lt;a href=&quot;http://physics.stackexchange.com/questions/3014/what-is-energy-where-did-it-come-from/3017#3017&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Energy Conservation&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as we make the right definitions of energy, it turns out that energy is conserved.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have an isolated system.  If it is not in equilibrium, its state may change.  Energy conservation means that at the end of the change, the new state will have the same energy.  (For this reason, energy is often treated as a constraint.  For example, an isolated system will maximize its entropy subject to the constraint that energy is conserved.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves the question of what an isolated system is.  If we take another system (the environment) and keep it around the isolated system, we find no observable changes in the environment as the state of the isolated system changes.  For example, changes in an isolated system cannot change the temperature, pressure, or volume of the environment.  Practically, an isolated system should have no physical mechanisms for interacting with the rest of the universe.  Matter and radiation cannot leave or enter, and there can be no heat conduction (I'm jumping the gun on that last one, of course, but take &quot;heat conduction&quot; as a rough term for now).  A perfectly isolated system is an idealization only. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next we observe systems A and B interacting.  Before the interaction, A has 100 joules of energy.  After interacting, A has 90 joules of energy, so it has lost 10 joules.  Energy conservation says that if we measure the energy in system B before and after the interaction, we will always find that system B has gained 10 joules of energy.  In general, system B will always gain exactly however much system A loses, so the total amount is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are nuances and caveats to energy conservation.  See &lt;a href=&quot;http://physics.stackexchange.com/questions/296/is-energy-really-conserved&quot;&gt;this question&lt;/a&gt;, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Work&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work is defined by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\textrm{d}W = P\textrm{d}V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Pressure&quot; rel=&quot;nofollow&quot;&gt;pressure&lt;/a&gt;; $V$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Volume&quot; rel=&quot;nofollow&quot;&gt;volume&lt;/a&gt;, and it is fairly easy to give operational definitions of both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this equation, we must ensure that $P$ is the pressure the environment exerts on the system.  For example, if we took a balloon into outer space, it would begin expanding.  However, it would do no work because the pressure on the balloon is zero.  However, if the balloon expands on Earth, it does work given by the product of its volume change and the atmospheric pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That example treats the entire balloon as the system.  Instead, we might think of only the air inside the balloon as a system.  Its environment is the rubber of the balloon.  Then, as the balloon expands in outer space, the air inside does work against the pressure from the elastic balloon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote more about work in &lt;a href=&quot;http://physics.stackexchange.com/questions/2451/why-does-a-car-engine-not-do-work-if-the-wheels-dont-slip/2474#2474&quot;&gt;this answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Adiabatic Processes&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work and energy, as described so far, are independent ideas.  It turns out that in certain circumstances, they are intimately related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some systems, we find that the decrease in energy of the system is exactly the same as the work it does.  For example, if we took that balloon in space and watched it expand, the air in the balloon would wind up losing energy as it expanded.  We'd know because we measure the temperature, pressure, and volume of the air before and after the expansion and calculate the energy change from a formula.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, the air would have done work on the balloon.  We can calculate this work by measuring the pressure the balloon exerts on the air and multiplying by the volume change (or integrating if the pressure isn't constant).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remarkably, we could find that these two numbers, the work and the energy change, always turned out to be exactly the same except for a minus sign.  Such a process is called adiabatic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, adiabatic processes are approximations.  They work best with systems that are almost isolated, but have a limited way of interacting with the environment, or else occur too quickly for interactions beside pressure-volume ones to be important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our balloon, the expansion might fail to be adiabatic due to radiation or conduction between the balloon and the air.  If the balloon were a perfect insulator and perfectly white, we'd expect the process to be adiabatic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sound waves propagate essentially adiabatically, not because there are no mechanisms for one little mass of air to interact with nearby ones, but because those mechanisms (diffusion, convection, etc.) are too slow to operate on the time scale of the period of a sound wave (about a thousandth of a second).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads us to thinking of work in a new way.  In adiabatic processes, work is the exchange of energy from one system to another.  Work is still calculated from $P\textrm{d}V$, but once we calculate the work, we know the energy change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Heat&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Real processes are not adiabatic.  Some are close, but others are not close at all.  For example, if I put a pot of water on the stove and turn on the burner, the water's volume hardly changes at all, so the work done as the water heats is nearly zero, and what work is done by the water is positive, meaning the water should lose energy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water actually gains a great deal of energy, though, which we can discover by observing the temperature change and using a formula for energy that involves temperature.  Energy got into  the pot, but not by work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that work is not a sufficient concept for describing energy transfer.  We invent a new, blanket term for energy transfer that is not done by work.  That term is &quot;heat&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat is simply any energy transferred between two systems by means aside from work.  The energy entering the boiling pot is entering by heat.  This leads to the thermodynamic equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\textrm{d}E = -\textrm{d}W + \textrm{d}Q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E$ is energy, $W$ work, and $Q$ heat.  The minus sign is a convention.  It says the if a system does work, it loses energy, but if it receives heat, it gains energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Interpreting Heat&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used to be very confused about heat because it felt like something of a deus ex machina to say, &quot;all the leftover energy must be heat&quot;.  What does it mean to say something has &quot;lost 30 calories through heat&quot;?  How can you look at it and tell?  Pressure, temperature, volume are all defined in terms of very definite, concrete things, and work is defined in terms of pressure and volume.  Heat seems too abstract by comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to get a handle on heat, as well as review everything so far, is to look at the experiments of &lt;a href=&quot;http://en.wikipedia.org/wiki/James_Prescott_Joule&quot; rel=&quot;nofollow&quot;&gt;James Joule&lt;/a&gt;.  Joule put a paddle wheel in a tub of water, connected the wheel to a weight so that the weight would drive the wheel around, and let the weight fall.  Here's the Wikipedia picture of the set up:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/mCSRL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the weight fell, it did work on the water; at any given moment, there was some pressure on the paddles, and they were sweeping out a volume proportional to their area and speed.  Joule assumed that all the energy transferred to the water was transferred by work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weights lost energy as they fell because their gravitational potential energy went down.  Assuming energy is conserved, Joule could then find how much energy went into the water.  He also measured the temperature of the water.  This allowed him to find how the energy of water changes as its temperature changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next suppose Joule starting heating the water with a fire.  This time the energy is transferred as heat, but if he raises the temperature of the water over exactly the same range as in the work experiment, then the heat transfer in this trial must be the same as the work done in the previous one.  So we now have an idea of what heat does in terms of work.  Joule found that it takes 4.2 joules of work to raise the temperature of one gram of water from 14C to 15C.  If you have more water than that, it takes more work proportionally.  4.2 joules is called one calorie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At last we can give a physical interpretation to heat.  Think of some generic thermodynamic process.  Imagine it happening in a piston so that we can easily track the pressure and volume.  We measure the energy change and the work during the process.  Then we attribute any missing energy transfer to heat, and say &quot;the system gave up 1000 joules (or 239 calories) of heat&quot;.  This means that if we took the piston and surrounded it with 239 grams of water at 14C, then did exactly the same process, the water temperature would rise to 15C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Misconceptions&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I discussed in this post is the first law of thermodynamics - energy conservation.  Students frequently get confused about what heat is because they mix up its definition with the role it plays in the second law of thermodynamics, which I didn't discuss here.  This section is intended to point out that some commonly-said things about heat are either loose use of language (which is okay as long as everyone understands what's being said), or correct use of heat, but not directly a discussion of what heat is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things do not have a certain amount of heat sitting inside them.  Imagine a house with a front door and a back door.  People can come and go through either door.  If you're watching the house, you might say &quot;the house lost 3 back-door people today&quot;.  Of course, the people in the house are just people.  The door only describes how they left.  Similarly, energy is just energy.  &quot;Work&quot; and &quot;heat&quot; describe what mechanism it used to leave or enter the system.  (Note that energy itself is not a thing like people, only a number calculated from the state, so the analogy only stretches so far.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We frequently say that energy is &quot;lost to heat&quot;.  For example, if you hit the brakes on your car, all the kinetic energy seems to disappear.  We notice that the brake pads, the rubber in the tires, and the road all get a little hotter, and we say &quot;the kinetic energy of the car was turned into heat.&quot;  This is imprecise.  It's a colloquialism for saying, &quot;the kinetic energy of the car was transferred as heat into the brake pads, rubber, and road, where it now exists as &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_energy&quot; rel=&quot;nofollow&quot;&gt;thermal energy&lt;/a&gt;.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat is not the same as temperature.  Temperature is what you measure with a thermometer.  When heat is transferred into a system, its temperature will increase, but its temperature can also increase because you do work on it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship between heat and temperature involves a new state variable, &lt;a href=&quot;http://en.wikipedia.org/wiki/Entropy&quot; rel=&quot;nofollow&quot;&gt;entropy&lt;/a&gt;, and is described by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Second_law_of_thermodynamics&quot; rel=&quot;nofollow&quot;&gt;second law of thermodynamics&lt;/a&gt;.  Statements such as &quot;heat flows spontaneously from hot bodies to cold bodies&quot; are describing this second law of thermodynamics, and are really statements about how to use heat along with certain state variables to decide whether or not a given process is spontaneous; they aren't directly statements about what heat is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat is not &quot;low quality energy&quot; because it is not energy.  Such statements are, again, discussion of the second law of thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Reference&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This post is based on what I remember from the first couple of chapters in Enrico Fermi's &lt;em&gt;Thermodynamics&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-05-17T11:15:23.167" LastActivityDate="2011-05-17T11:15:23.167" CommentCount="2" />
  <row Id="8549" PostTypeId="2" ParentId="5714" CreationDate="2011-04-12T22:45:43.623" Score="0" ViewCount="" Body="&lt;p&gt;Heat being considered as a fluid... the most disastrous one .. which shows poor scientific understanding of very fundamental (and ordinarily perceivable) phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Caloric_theory&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Caloric_theory&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but laws of thermodynamics (0 to 2) really enlightened about not just the nature of heat but also behaviour of material in bulk (statistical description of thermal and continuum phenomena) &lt;/p&gt;&#xA;" OwnerUserId="3077" LastActivityDate="2011-04-12T22:45:43.623" CommunityOwnedDate="2011-04-12T22:45:43.623" CommentCount="1" />
  <row Id="8550" PostTypeId="1" AcceptedAnswerId="8553" CreationDate="2011-04-12T23:08:52.563" Score="3" ViewCount="103" Body="&lt;p&gt;I was reading about this asteroid (apparently, it has a moon, isn't that awesome?) and I started thinking about if I was on this asteroid, and I jumped, would I fall off?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's been a while since I did something like this, so I went to Wikipedia to get some formulas. There I found&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_e = \sqrt{\frac{2GM}{r}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$G$ is the gravitational constant ($6.673 × 10^{-11}\ \mathrm{m}^3\mathrm{kg}^{-1}\mathrm{s}^{-2}$)&lt;/li&gt;&#xA;&lt;li&gt;$M$ is mass of the body (found to be $4.2 × 10^{16}\ \mathrm{kg}$)&lt;/li&gt;&#xA;&lt;li&gt;$r$ distance from center of mass (found to be $53.6\ \mathrm{km}$)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$v_e = \sqrt{\frac{2(6.673\times 10^{-11})(4.2\times 10^{16})}{53600}} = \sqrt{104} = 10.22\,\frac{\mathrm{m}}{\mathrm{s}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a fairly fit guy, about 73kg and .6m vertical leap here on earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a formula for the speed of a free falling object to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = \sqrt{2gd}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I assume that my speed when I hit the ground is the same as when I jump up, that means my jump speed is about 3.4 m/s, which is significantly less than the required 10.4 m/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it would seem I would not be able to jump off of this asteroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are my formulas, assumptions, and calculations correct here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how would I use this information to determine how high I would jump (would it be enough to freak me out? Well I'd be jumping off an asteroid, so it would freak me out anyway... although I imagine I would be going slow enough for (my ship?) to pick me back up.)&lt;/p&gt;&#xA;" OwnerUserId="1954" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-13T00:01:11.937" LastActivityDate="2011-04-13T17:31:58.253" Title="Escape Velocity of Asteriod 243 Ida" Tags="&lt;gravity&gt;&lt;space&gt;&lt;velocity&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="8551" PostTypeId="2" ParentId="7698" CreationDate="2011-04-12T23:49:27.880" Score="2" ViewCount="" Body="&lt;p&gt;As was pointed out in an answer to your previous question, the polarization of your beam can be thought of as the &quot;spin&quot; angular momentum being carried by it. Circularly polarized light carries angular momentum, linearly polarized light does not (not counting the &quot;orbital&quot; angular momentum carried by the spatial profile of the beam).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Viewed in this light, I would say that angular momentum is not conserved when light propagates through an arbitrary anisotropic medium. A simple example is a half-waveplate, a device which can convert right-hand circular polarized into left-circular. Waveplates are very simple -- they are just a small plate of anisotropic material, cut to a certain width and with their &quot;fast axis&quot; oriented at some set angle relative to the input polarization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing up, the angular momentum of light propagating in an anisotropic medium is not necessarily conserved. The excess angular momentum is, presumably, taken up by the medium.&lt;/p&gt;&#xA;" OwnerUserId="2979" LastActivityDate="2011-04-12T23:49:27.880" />
  <row Id="8552" PostTypeId="2" ParentId="8518" CreationDate="2011-04-13T00:05:44.163" Score="2" ViewCount="" Body="&lt;p&gt;Maybe,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.technologyreview.com/blog/arxiv/26580/&quot; rel=&quot;nofollow&quot;&gt;http://www.technologyreview.com/blog/arxiv/26580/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am by no means an expert, but I read this a few weeks ago. In that paper they consider a 2d lattice and construct an energy analogue. They show it behaves as energy should, and then conclude that for this energy to be conserved space-time would need to be invariant. &lt;/p&gt;&#xA;" OwnerUserId="3080" LastActivityDate="2011-04-13T00:05:44.163" />
  <row Id="8553" PostTypeId="2" ParentId="8550" CreationDate="2011-04-13T00:22:58.273" Score="6" ViewCount="" Body="&lt;p&gt;Looks as if your arithmetic is correct, and you could not escape 243 Ida. Given your leap velocity capability of 3.4 m/s, you would be able to leap 593 m above 243 Ida's surface, since its surface gravitational acceleration is only about 9.75E-3 m/s^2.&lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="1924" LastEditorDisplayName="" LastEditDate="2011-04-13T00:33:53.837" LastActivityDate="2011-04-13T00:33:53.837" CommentCount="2" />
  <row Id="8554" PostTypeId="2" ParentId="8538" CreationDate="2011-04-13T00:45:09.120" Score="-2" ViewCount="" Body="&lt;p&gt;I think it isw possible that one is a cocave lens and the other is a convex lens that is also why in the right picture the angles converge&lt;/p&gt;&#xA;" OwnerUserId="2530" LastActivityDate="2011-04-13T00:45:09.120" />
  <row Id="8555" PostTypeId="1" CreationDate="2011-04-13T01:03:29.970" Score="0" ViewCount="96" Body="&lt;p&gt;If the object has an acceleration 9.81 m/s^2, and is 100 meters from the ground with a starting velocity of 10 meters per second, how much time will it take to reach the ground, not considering air friction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula is:&#xA;$d=v_it+\frac{at^2}2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then what does t equal? I would like a formula for time.&lt;/p&gt;&#xA;" OwnerUserId="3041" LastActivityDate="2011-04-13T01:03:29.970" ClosedDate="2011-04-13T01:04:44.160" Title="How much time will it take to reach the ground?" Tags="&lt;gravity&gt;&lt;time&gt;&lt;velocity&gt;&lt;acceleration&gt;&lt;speed&gt;" CommentCount="1" />
  <row Id="8556" PostTypeId="2" ParentId="8518" CreationDate="2011-04-13T01:29:06.360" Score="2" ViewCount="" Body="&lt;p&gt;As was said before, this depends on what kind of 'discrete' symmetry you have: if you have a &lt;em&gt;bona fide&lt;/em&gt; discrete symmetry, as e.g. $\mathbb{Z}_n$, then the answer is in the negative in the context of Nöther's theorem(s) — even though there are conclusions that you can draw, as &lt;em&gt;Moshe R.&lt;/em&gt; explained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you're talking about a discretized symmetry, i.e. a continuous symmetry (global or local) that has been somehow discretized, then you do have an analogue to Nöther's theorem(s) à la Regge calculus. A good talk introducing some of these concepts is &lt;a href=&quot;http://ccom.ucsd.edu/~astern/pdfs/cambridge20081121.pdf&quot; rel=&quot;nofollow&quot;&gt;Discrete Differential Forms, Gauge Theory, and Regge Calculus (PDF)&lt;/a&gt;: the bottom line is that you have to find a Finite Difference Scheme that preserves your differential (and/or gauge) structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a big literature on Finite Difference Schemes for Differential Equations (ordinary and partial).&lt;/p&gt;&#xA;" OwnerUserId="142" LastActivityDate="2011-04-13T01:29:06.360" />
  <row Id="8557" PostTypeId="1" CreationDate="2011-04-13T04:36:44.363" Score="5" ViewCount="257" Body="&lt;p&gt;Doing a Google search i found a paper called &lt;a href=&quot;http://www.sciencedirect.com/science?_ob=ArticleURL&amp;amp;_udi=B6TJ4-4M21SWY-4&amp;amp;_user=10&amp;amp;_coverDate=06%2F30%2F2007&amp;amp;_rdoc=1&amp;amp;_fmt=high&amp;amp;_orig=gateway&amp;amp;_origin=gateway&amp;amp;_sort=d&amp;amp;_docanchor=&amp;amp;view=c&amp;amp;_searchStrId=1716211562&amp;amp;_rerunOrigin=google&amp;amp;_acct=C000050221&amp;amp;_version=1&amp;amp;_urlVersion=0&amp;amp;_userid=10&amp;amp;md5=25e00e3ba9bac4c17fe64b8a9de5bc6c&amp;amp;searchtype=a&quot; rel=&quot;nofollow&quot;&gt;The maximum number of elementary particles in a super symmetric extension of the standard model&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It claims in the abstract that the upper bound is 84 (i don't have access to the article)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Is there a max number of types of elementary particles predicted in advanced physics theories such as string theory? What are the reasons for this?Are the arguments purely mathematical?&lt;/p&gt;&#xA;" OwnerUserId="2108" LastEditorUserId="2108" LastEditorDisplayName="" LastEditDate="2011-04-13T04:59:31.760" LastActivityDate="2011-04-13T07:25:00.550" Title="Is there a maximum number of types of elementary particles?" Tags="&lt;string-theory&gt;&lt;particle-physics&gt;&lt;supersymmetry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="8558" PostTypeId="1" CreationDate="2011-04-13T04:56:18.957" Score="3" ViewCount="206" Body="&lt;p&gt;I'm just curious.  I figure atoms fuse occasionally just by chance, like quantum tunneling or rogue waves.  Is this true?  If so, any idea how often?&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2011-04-13T15:45:37.097" Title="How often does nuclear fusion occur within the human body?" Tags="&lt;nuclear-physics&gt;&lt;fusion&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8559" PostTypeId="2" ParentId="8557" CreationDate="2011-04-13T05:29:00.360" Score="0" ViewCount="" Body="&lt;p&gt;The number of particles depends on the theory assumed. Symmetries, such as super symmetry impose limits, but who knows what the theory of everything is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, the arguments are purely mathematical, until some experiment at a future date will chose among the multiplicity of theoretical models.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-13T05:29:00.360" CommentCount="2" />
  <row Id="8560" PostTypeId="2" ParentId="1030" CreationDate="2011-04-13T05:45:47.687" Score="2" ViewCount="" Body="&lt;p&gt;As pointed out by Weinberg in his book Cosmology (note, this is NOT Gravitation and Cosmology. He also has a book of that name), inflation was proposed to explain 3 problems:&#xA;1)Horizon problem&#xA;2)Flatness problem&#xA;3)Monopole problem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)Horizon problem: The evolution of the scale factor before and after decoupling is $\sqrt{t}$ and $t^{\frac{2}{3}}$. We compute the linear dimension of the forward and backward lightcones at the time of decoupling in the hot big bang model. The radius of this light cone is the physical size of the region on the last scattering surface from which we receive the CMB. The backward lightcone is $l_{B} \approx 3(t_{dec}^{2}t_{0})^{1/3}$ ($t_{0}$ is present time.). The forward lightcone radius is $l_{F} = 2t_{dec}$. The ratio $R \equiv \frac{l_{B}}{l_{F}} \approx 70$. The physical wavelength associated with cosmological perturbations grows faster than the Hubble radius as we go back in time. If, a causal mechanism is responsible for the inhomogenities, then these scales should be inside Hubble scale in very early universe. This is possible if, the perturbation associated wavelength decreases faster than Hubble radius as we go back in time. So, $-\frac{d}{dt}\left( \frac{\lambda}{d_{H}}\right) &amp;lt;0 $ ($d_{H}$ is the Hubble radius). This leads to $\ddot{a} &amp;gt;0$. In most cases, we model it as a single scalar field which causes this inflation in a de Sitter background ($\Lambda$ dominated universe)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)Flatness problem: A less convincing argument of inflation. Experimentally, we observe a vanishing spatial curvature parameter $\Omega_{K} = -\frac{K}{a^2 H^2} = -\frac{K}{a^2}$. In solving this problem, we assume that nothing much happens to the cosmic scale factor and expansion rate from the end of inflation to the beginning of the radiation dominated era i.e $a_{Inflation}H_{Inflation} \approx a_{rad. domination}H_{rad. domination}$. The small value of $|K|/\dot{a}^2$ could be explained by taking $K=0$ i.e a spatially flat universe. However, inflation opens up the possibility that the universe is not at all homogenous and isotropic and that its apparent flatness of the cosmic metric is just the result of inflation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)Monopole problem: Standard Model predicts that in a hot early universe, a large number of monopoles must be produced by symmetry breaking from some single gauge theory since it is at an energy scale of about $M = 10^{16}$ GeV. Those monopoles should have persisted even to the present days. However, that is not the case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amongst all the above problems, the horizon problem is the most serious one. Since, the other two can be explained by other mechanisms. Also, any number of $e$-foldings not only solves the horizon problem but also the flatness problem and the monopole problem.&lt;/p&gt;&#xA;" OwnerUserId="2973" LastActivityDate="2011-04-13T05:45:47.687" CommentCount="1" />
  <row Id="8562" PostTypeId="1" CreationDate="2011-04-13T06:00:09.577" Score="0" ViewCount="104" Body="&lt;p&gt;Where exactly can I find some reading material on ADM Formalism? Actually, Maldacena has used it in one of his papers (http://arxiv.org/abs/astro-ph/0210603) in computing the the three point correlation function (i.e the non-Gaussianity) parameter for primordial cosmological perturbations. Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="2973" LastActivityDate="2011-04-13T06:00:09.577" Title="Reading material on ADM Formalism" Tags="&lt;general-relativity&gt;&lt;cosmology&gt;" CommentCount="3" />
  <row Id="8564" PostTypeId="1" AcceptedAnswerId="8565" CreationDate="2011-04-13T06:10:57.990" Score="4" ViewCount="123" Body="&lt;p&gt;Movies and TV shows frequently show buildings being bombed, cars blowing up, etc.  Frequently these are really explosions of miniatures filmed up close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from the speed that the explosion expands relative to the size of the object (which can be adjusted by slowing down the film), are there any features of an explosion that clue us in to the scale on which it occurs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be definite, let's imagine two tank trucks of gasoline, one 10m long and one 10cm long, but otherwise with the same proportions and made from the same materials.  We film the big one exploding from 100m away and film the little one exploding from 1m away.  How would we tell which footage is which?&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-04-13T10:36:01.830" Title="How can you tell a model explosion from the real thing?" Tags="&lt;classical-mechanics&gt;&lt;fluid-dynamics&gt;&lt;explosions&gt;&lt;scaling&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8565" PostTypeId="2" ParentId="8564" CreationDate="2011-04-13T06:21:47.357" Score="4" ViewCount="" Body="&lt;p&gt;Note that by changing the overall scale by a factor k, you are changing the volume of the gasoline by k^3, and the area you are viewing by k^2. So the overall size of the explosion (ie visible flames etc) is not invariant. To find out what is k, you'd have to know lots of details about the fuel etc I suppose.&lt;/p&gt;&#xA;" OwnerUserId="1708" LastEditorUserId="1708" LastEditorDisplayName="" LastEditDate="2011-04-13T06:27:21.987" LastActivityDate="2011-04-13T06:27:21.987" CommentCount="1" />
  <row Id="8566" PostTypeId="2" ParentId="8513" CreationDate="2011-04-13T06:53:47.470" Score="-1" ViewCount="" Body="&lt;p&gt;This is not a trivial question and it makes sense to consider at first the simplest case where a bubble falls freely in static air to find the bubble's terminal velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R$-radius of a bubble&lt;br&gt;&#xA;$\rho_w$-density of water&lt;br&gt;&#xA;$\rho_a$-density of air&lt;br&gt;&#xA;$l$-thickness of the bubble &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force on a bubble due to gravity:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_g=4\pi R^2l\rho_wg$$ Because it is difficult to decide when the actual drag force is proportional to $v$ or $v^2$ then to avoid Reynolds numbers lets choose a different way.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $v$ be the terminal velocity of the bubble. If the bubble falls distance $h$ then it transmits to the air mass $m=\pi R^2h\rho_a$ amount of energy $E=F_dh=\frac{mv^2}{2}$. From this drag force:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_d=\pi R^2\rho_a\frac{v^2}{2}$$ So actual drag force is proportional rather to $v^2$, i think.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally from $F_g=F_d$ we get the terminal velocity of freely falling bubble:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v=\sqrt{\frac{8\rho_wgl}{\rho_a}}$$ Thus as a first approximation, terminal velocity of a bubble does not depend of it's radius but depends of it's thickness. So in general, answer to the question in title, i think, is: &lt;strong&gt;No&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="552" LastActivityDate="2011-04-13T06:53:47.470" CommentCount="9" />
  <row Id="8567" PostTypeId="1" CreationDate="2011-04-13T07:18:34.397" Score="2" ViewCount="317" Body="&lt;p&gt;I am trying to understand energy levels of electron configurations.  I visited the &lt;a href=&quot;http://physics.nist.gov/PhysRefData/ASD/levels_form.html&quot; rel=&quot;nofollow&quot;&gt;NIST web site&lt;/a&gt; and discovered that the notation used here are called term symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading corresponding &lt;a href=&quot;http://en.wikipedia.org/wiki/Term_symbol&quot; rel=&quot;nofollow&quot;&gt;wikipedia entry&lt;/a&gt; I worked through the carbon example in the &quot;Term symbols for an electron configuration&quot; section.  So it appears to me that each of the $\binom{t}{e}$ microstates [(in this case $\binom{6}{2}=15$), where $t$ is number of slots in the outside subshell, and $e$ is the number of electrons] will each have an assigned energy level, or term symbol - since in the end there are 15 1s distributed in three different matrices, but only 5 possible term symbols (1 term symbol for 5 microstates, 3 term symbols for 9 microstates, and 1 term symbol for 1 microstate).  However, I don't see how a given microstate maps to a specific term symbol.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, $M_L = 0$ and $M_S = 0$ is true for 3 microstates, but how to determine which term symbols correspond to them? Does it even make sense to do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks gigacyan, for the detailed answer. By the wording I am not sure at the end if you mean that $M_L = 0$ and $M_S = 0$ can only have the $^1S$ term. If that is true, then the following must be totally off, but I will take a shot anyway:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So are you saying that ANY given microstate cannot be assigned a term symbol (energy level), or that just certain microstates (such as the $M_L = 0$ and $M_S = 0$ case above) cannot be assigned since there are more than one term symbol possibility?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, it seems that when there is only one microstate for a given $M_L$ and $M_S$ combination, it is uniquely determined as long as it falls into a matrix that has only one term symbol - for carbon, say $M_L = 2$ and $M_S = 0$ (row 7 of 15 in the microstate table) then it falls into the 5x1 table, which must be $^1D_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when there is only one microstate for a given $M_L$ and $M_S$ combination but it falls into a matrix with more than one term symbol - for carbon, say $M_L = 1$ and $M_S = 1$ (row 1 of 15 in the microstate table), it falls into the 3x3 matrix, which means that this microstate must be one of $^3P_2$, $^3P_1$, or $^3P_0$, but it is not known which.  Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="3082" LastEditorUserId="3082" LastEditorDisplayName="" LastEditDate="2011-04-14T04:57:23.077" LastActivityDate="2011-04-15T02:50:52.263" Title="How do electron configuration microstates map to term symbols?" Tags="&lt;quantum-mechanics&gt;&lt;energy&gt;&lt;spectroscopy&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8568" PostTypeId="2" ParentId="8557" CreationDate="2011-04-13T07:25:00.550" Score="7" ViewCount="" Body="&lt;p&gt;The paper you cite employs Mohammed El Naschie's &quot;E-infinity theory&quot; of physics, which is one big exercise in what physicists call &quot;numerology&quot;. Numerology is where you match up numbers - e.g. the three generations of particles in the standard model, and the three dimensions of space - and then you state or insinuate that there is a connection; but you cannot justify the connection logically (deductively). Another common example is where people find formulae for particle masses and other unexplained quantities, using combinations of transcendental numbers, other particle masses, and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &quot;numerology&quot; sometimes does work in physics and mathematics. That is, the search for quantitative coincidences sometimes does stumble upon relationships which have a deeper origin. Balmer's formula for the emissions of the hydrogen atom was explained by quantum mechanics; the coincidence in mathematics known as &quot;monstrous moonshine&quot; was proven to be true by Richard Borcherds; there are many other examples. But it is also possible to make extremely contrived relationships - e.g. you can approximate any real number arbitrarily closely, using combinations of &lt;em&gt;e&lt;/em&gt; and $\pi$, if you use enough of them. You can also pile up lots of deductively unjustified &quot;connections&quot;, and claim to have a theory of everything. &quot;E-infinity theory&quot; is in the latter category. These papers don't contain even the moderately difficult sorts of calculation that you see in real particle physics papers - I mean scattering amplitudes, particle lifetimes, and all the other detailed quantities which come from employing a theory with a proper equation of motion. Instead, these papers are full of basic algebra equations in which various known quantities are &quot;explained&quot; in a meaningless way. But these papers don't actually explain anything, nor do they predict anything, and the journal which publishes most of them is considered low-quality for this reason. &lt;/p&gt;&#xA;" OwnerUserId="1486" LastActivityDate="2011-04-13T07:25:00.550" CommentCount="7" />
  <row Id="8569" PostTypeId="1" AcceptedAnswerId="8572" CreationDate="2011-04-13T09:37:12.697" Score="4" ViewCount="150" Body="&lt;p&gt;I am studying how animals (including the human beings) can perceive distances thanks to their eyes and their ears. I am focusing on the fact that they always go in pairs: two eyes, two ears, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the sight, I think our eyes use the parallax. But is this enough? Is this the only way we can perceive distances with them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the sounds, I think it is merely by interferometry, but I am not convinced, because the distances between our ears and the wavelength are not always comparable to one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you think about it?&#xA;Thanks in advance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isaac&lt;/p&gt;&#xA;" OwnerUserId="2499" LastActivityDate="2011-04-13T13:05:46.997" Title="How do animal perceive distances with their eyes and ears" Tags="&lt;sound&gt;&lt;biophysics&gt;&lt;interferometry&gt;&lt;eye&gt;" AnswerCount="3" CommentCount="9" />
  <row Id="8570" PostTypeId="1" AcceptedAnswerId="8585" CreationDate="2011-04-13T09:47:04.043" Score="1" ViewCount="156" Body="&lt;p&gt;You have three separate glass rods, and you know one is positive, one is negative and one is neutral. You also have an electroscope that is positively charged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you determine which rod is positive, negative and neutral?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you will be able to determine which rod is negatively charged, as that rod will cause the electroscope leaf to rise. I'm not sure how to differentiate between the positive+neutral though, as both will cause the leaf to collapse. Could you place it next to the negative rod and see which repels?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially it would be great if somebody could explain the actual physics behind the situation. Why will the negative rod cause the leaf to rise and vice versa?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3044" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-04-13T15:31:56.377" LastActivityDate="2011-05-13T16:45:58.273" Title="Determine Charge With Electroscope?" Tags="&lt;homework&gt;&lt;electrostatics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="8571" PostTypeId="2" ParentId="8564" CreationDate="2011-04-13T10:31:01.667" Score="1" ViewCount="" Body="&lt;p&gt;This wiki: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Deflagration&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Deflagration&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gives You an impression and further links into the business &#xA;of fire/deflagration/explosion/detonation. Important is, &#xA;that the things in filmes are almost always fires or deflagrations. &#xA;(Everything more violent/faster does not make &quot;good&quot; pictures.) &#xA;Those trucks would cause a big fire, not more (bad enough).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bombed buildings might be pictures from buldings tored down (with explosives),&#xA;but both give You tumbeling down brickwalls and a lot of dust then, &#xA;flames much later or not at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For cars, I have seen reports (&quot;making of&quot;), where they did that 1:1. &#xA;For two tank trucks, this will depend on budget:=)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One easy way of betrayal is to combine sound of explosions with the &#xA;pictures of a fire/deflagration, which do not sound much.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course, a lot is done today on computers, &quot;light and magic&quot;. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-13T10:36:01.830" LastActivityDate="2011-04-13T10:36:01.830" />
  <row Id="8572" PostTypeId="2" ParentId="8569" CreationDate="2011-04-13T12:07:03.323" Score="2" ViewCount="" Body="&lt;p&gt;With vision depth is determined by parallax.  This largely works for objects out to 100m or less.  Depth beyond that distance is assessed by familiarity with these objects, say large mountains or vistas, and experience with them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depth with hearing is determined by two means.  A tone which is perceived louder in one ear than the other is usually perceived as having a direction based on that.  For lower frequency sounds phase differences between left and right auditory perception can be used by the brain to detect angular direction relative to the facial direction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some species of whales are thought to generate a mental map of the ocean, where a blue whale in the Atlantic maps the outlay of the ocean bottom and coasts through echo location.  Clearly this indicates a huge amount of neural processing which uses these acoustical data.  Dogs generate maps of their immediate world through olfactory means.  Life forms generate maps of their world by a variety of means.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-04-13T12:07:03.323" CommentCount="2" />
  <row Id="8573" PostTypeId="2" ParentId="8569" CreationDate="2011-04-13T12:56:42.343" Score="1" ViewCount="" Body="&lt;p&gt;Adding to Lawrence's answer, apart from the monocular cues like depth from shading, lighting, haze, geometrical structure and familiarity of objects, you also have binocular cues from the eye motor convergence system (eyes point at the spot you are looking at in 3d space) as well as the feedback from the lens focusing system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these systems work together when you look at objects in the real world, but in for example 3d movies and tvs only some are active, which can cause various sorts of dizziness or headache among viewers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly viewing plain old 2d photos of 3d objects/worlds does not infer this sickness at all..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regards to hearing, apart from loudness and phase the auditory system also uses timing of onset, this obviously relates to phase discrimination depending on the audio frequency and size of the head. Furthermore, the spectral filtering of the signals at both ears is shaped by the asymmetrical shape of the outer ears, this allows the auditory system to gauge sound localization in front of, behind, above and below. Monaural cues include, as with vision, familiarity of certain sounds or speakers and how these sounds degrade by passing through air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several hundred million years of evolution has been pretty good at extracting every bit of useful information from all our senses :)&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-04-13T12:56:42.343" CommentCount="2" />
  <row Id="8574" PostTypeId="2" ParentId="8569" CreationDate="2011-04-13T13:05:46.997" Score="1" ViewCount="" Body="&lt;p&gt;My &quot;Stereotelemeter&quot; (Swiss Army) from late 30ties, built &#xA;by Wild Heerbrugg and used to determine distances of aircraft &#xA;from ground (Flak), has a scale up to 10 000 meters. &#xA;The Army Handbook I got with it says that depth vision &#xA;works up to 1000 meters, and the Telemeter works up to &#xA;10 000 meters due to 10 fold magnification and a basis of 1.25 meters. &#xA;Of course there are big personal differences in such abilities. &#xA;I read in a textbook on Photography (Mutter, about 1960) that &#xA;about 10 % of normal population is not able to see stereoscopically.&#xA;(Given as one reason why 3D failed regularly in photography and cinema) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, depth sense works as well with knowledge &#xA;of real and apparent size or on basis of accomodation &#xA;(at small distances, some meters) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depth of sound sensing is weak, some sounds can be determined &#xA;by the alteration by distance. (Thunder is more &quot;rolling&quot; if &#xA;far away, close to the strike it sounds like a beam of wood is &#xA;ruptured) &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-13T13:05:46.997" CommentCount="3" />
  <row Id="8576" PostTypeId="2" ParentId="8558" CreationDate="2011-04-13T13:49:40.370" Score="5" ViewCount="" Body="&lt;p&gt;This'll be a &lt;em&gt;very&lt;/em&gt; rough order of magnitude estimate, but as you'll see it's good enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that two hydrogen atoms bump into each other. In order to fuse, the nuclei have to tunnel to within about a nuclear distance of $10^{-15}$ m of each other. The tunneling probability is something like $e^{-(2mE)^{1/2}L/\hbar}$, where $E$ is the energy gap, $m$ is the particle mass, and $L$ is the distance. The distance is of order $10^{-10}$ m (a Bohr radius) and the energy is about an MeV (the electrical potential energy of two protons right next to each other. I work out the numbers to get a probability of about $e^{-20000}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'd next have to multiply that by the number of &quot;chances&quot; (number of times two atoms collide with each other). That's a large number by ordinary standards, but it's not exponentially large in the same way that the probability is exponentially small. Say you've got $10^{29}$ atoms in you, and each one collides with something else $10^{10}$ times per second. Then the number of chances per second is a mere $10^{39}$. I made up that number $10^{10}$ out of nowhere, but whatever it is, it's not $10^{10^4}$, which is what it'd have to be for there to be any significant probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it never happens.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-13T13:49:40.370" CommentCount="6" />
  <row Id="8577" PostTypeId="5" CreationDate="2011-04-13T14:05:21.220" Score="0" ViewCount="" Body="&lt;p&gt;Superconductivity is the transmission of current with no resistive losses, and is one of the most active areas of condensed matter physics research.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Superconductivity was first observed by Kamerlingh Onnes in 1911 when he discovered that the resistivity of mercury went to zero at 4.2K.  Since that first discovery, superconductivity has been found in a wide variety of material classes, from pure elements to complex structures like cuprates, organometallics, and heavy fermions.  The current record for high temperature superconductivity is close to 150K, which is found in a cuprate compound at high pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Important outstanding questions in superconductivity research remain the origination of the pairing mechanism for unconventional superconductors, and whether room temperature superconductivity is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many web resources for superconductivity.  General introductions to the topic may be found in the following web sites: &#xA;Hyperphysics: &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/solids/scond.html&quot; rel=&quot;nofollow&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/solids/scond.html&lt;/a&gt;&#xA;Wikipedia: &lt;a href=&quot;http://en.wikipedia.org/wiki/Superconductivity&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Superconductivity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current research edge of the field may be found on the arXiv preprint server, which has an entire subsection devoted to superconductivity research: &lt;a href=&quot;http://arxiv.org/list/cond-mat.supr-con/recent&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/list/cond-mat.supr-con/recent&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="2433" LastEditorDisplayName="" LastEditDate="2011-04-13T14:19:25.080" LastActivityDate="2011-04-13T14:19:25.080" />
  <row Id="8578" PostTypeId="4" CreationDate="2011-04-13T14:05:21.220" Score="0" ViewCount="" Body="Superconductivity is the transmission of current with no resistive losses, and is one of the most active areas of condensed matter physics research." OwnerUserId="-1" LastEditorUserId="2433" LastEditorDisplayName="" LastEditDate="2011-04-13T14:19:26.250" LastActivityDate="2011-04-13T14:19:26.250" />
  <row Id="8579" PostTypeId="2" ParentId="8517" CreationDate="2011-04-13T14:18:58.667" Score="1" ViewCount="" Body="&lt;p&gt;If you have two orbiting bodies, with angular momentum ${\vec L}_1$ and ${\vec L}_2$ the total angular momentum of the system is ${\vec J}~=~{\vec L}_1~+~{\vec L}_2$.  The angular momentum of the system has a pseudo-potential $C(r_1,r_2){\vec J}\cdot{\vec J}$, for $C(r_1,r_2)$ a function of the orbital radii and masses of two bodies.  The relevant portion is the angular momentum squared which is&#xA;$$&#xD;&#xA;{\vec J}\cdot {\vec J}=~{\vec L}_1\cdot {\vec L}_1~+~{\vec L}_2\cdot {\vec L}_2~+~2{\vec L}_1\cdot {\vec L}_2&#xD;&#xA;$$ &#xA;This is related to the orbit-orbit interaction in QM, but we can physically see that the total angular momentum $\vec J$ is largest when the two ${\vec L}_i$ are parallel, and minimal when they are anti-parallel.  The potential, or pseudo-potential term is negative, and so the potential is minimal (most negative) when the orbits are aligned with each other.  This is the opposite of what happens in QM with electrons, due to the magnetic poles associated with orbital and spin directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is somewhat descriptive, for a full presentation requires working with the 3-body problem in classical mechanics.  That would require a much more extensive coverage of the problem.  However, this roughly answers the basic question.  It is also one reason that solar systems (stellar systems) have planets which orbit in approximately the same plane and with angular momentum approximately aligned.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-04-13T14:18:58.667" />
  <row Id="8580" PostTypeId="1" AcceptedAnswerId="8595" CreationDate="2011-04-13T15:03:53.360" Score="0" ViewCount="143" Body="&lt;p&gt;I read that the magnetic moment of the proton can be expressed as the sum of the magnetic moments of the three quarks like $\mu_p = \frac{4}{3}\mu_u - \frac{1}{3}\mu_d$. But I couldn't find a table listing the measured magnetic moments of the quarks. Can you point me to a reference?&lt;/p&gt;&#xA;" OwnerUserId="1648" LastActivityDate="2011-04-13T18:52:35.847" Title="Where can I find the measured magnetic moments of the quarks?" Tags="&lt;particle-physics&gt;&lt;experimental-physics&gt;&lt;quark-model&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="8581" PostTypeId="2" ParentId="2239" CreationDate="2011-04-13T15:37:26.853" Score="3" ViewCount="" Body="&lt;p&gt;Although I agree that some of the physical phenomena described in the posts above are real effects, I think they are tertiary.  If you are you are observing this from a commercial airplane, two main things explain the lion's share of the higher lift near ground: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) The airplane maintains a higher angle of attack.  This means the nose (and therefore the wing) tilts up a few degrees, greatly increasing the lift at a given airspeed.  This also greatly increases drag, which is why you will hear the engines INCREASE thrust, just before landing.  So three observable things happen:  The nose tilts up; the engines increase thrust (you hear increased engine noise) and the airplane slows down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.)  In addition to the nose tilt, modern large commercial airplanes have adjustable flaps and slats, which extend to greatly increase the wing area; which greatly increases lift at slower airspeeds (that is their purpose for existence).  These are extended on approach, to give the airplane the ability to maintain lift at the slow airspeeds desired during landing.  &lt;/p&gt;&#xA;" OwnerUserId="3087" LastActivityDate="2011-04-13T15:37:26.853" />
  <row Id="8582" PostTypeId="2" ParentId="8558" CreationDate="2011-04-13T15:45:37.097" Score="1" ViewCount="" Body="&lt;p&gt;Do not hold your breath. The probability is infinitesimally small, though I do not think anybody has calculated it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In simpler systems, like crystals, people have attempted tunneling like calculations in trying to propose cold fusion. Crystals have the possibility of coherent behavior  and some models  have been proposed, but imo hand waving. There was a discussion a while &lt;a href=&quot;http://physics.stackexchange.com/questions/3799/why-is-cold-fusion-considered-bogus&quot;&gt;ago here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-13T15:45:37.097" />
  <row Id="8584" PostTypeId="2" ParentId="8580" CreationDate="2011-04-13T16:09:09.870" Score="0" ViewCount="" Body="&lt;p&gt;Quarks are confined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any &quot;measurements&quot; come from the measurements of seen particles, utilizing  the Dirac equation, Quantum Field Theory  and the symmetries of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_model&quot; rel=&quot;nofollow&quot;&gt;standard model&lt;/a&gt;  to constrain the quantities of the quarks, be it masses or moments or charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard model was extremely well verified by the measurements at &lt;a href=&quot;http://en.wikipedia.org/wiki/LEP&quot; rel=&quot;nofollow&quot;&gt;LEP&lt;/a&gt; up to the energies of LEP.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-04-13T18:52:35.847" LastActivityDate="2011-04-13T18:52:35.847" />
  <row Id="8585" PostTypeId="2" ParentId="8570" CreationDate="2011-04-13T16:24:56.437" Score="1" ViewCount="" Body="&lt;p&gt;I did a google search and found that: When a glass rod is rubbed with a paper towel, the glass becomes positively charged. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.)ground the electroscope through touching it with you fingers (removing any excess charge); one can assume it is neutral/has no net charged at this point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) touch one of the charged rods to the electroscope. The leaves should rise with an unknown charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) Vigorously rub the uncharged glass rod with a paper towel to give it a positive charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.) bring the positively charged rod close to the Electroscope. If the leaves are repulsed, then you know that the glass rod of unknown charge was/is of the same charge, and therefore is positively charged. Opposite charges attract. Therfore if the leaves attract one knows that the glass rod of unknown charge was negatively charged. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5.) [check] repeat steps 1-4 to verify that that last rod is of opposite charge.&lt;/p&gt;&#xA;" OwnerUserId="2843" LastActivityDate="2011-04-13T16:24:56.437" />
  <row Id="8586" PostTypeId="1" AcceptedAnswerId="8588" CreationDate="2011-04-13T16:29:32.540" Score="4" ViewCount="105" Body="&lt;p&gt;I was speaking with some friends of mine, one of whom was an aerospace engineer. He posited the infeasibility of a hypothetical &quot;Margaritaville Satellite&quot; that orbited earth in such a way that wherever it was above, it was six-o'clock (the idea being, the Margaritaville Satellite is in Happy Hour 24/7), matching Earth's rotation exactly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there was disagreement on whether or not the satellite could actually be put in such an orbit, as if it were rotating at the exact speed of earth, it would effectively just be &quot;falling&quot; straight towards Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to sync a satellite's orbiting with the time of day of the earth without some sort of propulsion to keep it aloft?&lt;/p&gt;&#xA;" OwnerUserId="3088" LastEditorUserId="871" LastEditorDisplayName="" LastEditDate="2011-04-13T16:52:01.923" LastActivityDate="2011-04-13T18:29:13.103" Title="The Feasibility of a satellite orbiting at a fixed time" Tags="&lt;classical-mechanics&gt;&lt;orbital-motion&gt;&lt;newtonian-gravity&gt;" AnswerCount="3" />
  <row Id="8588" PostTypeId="2" ParentId="8586" CreationDate="2011-04-13T16:48:38.960" Score="3" ViewCount="" Body="&lt;p&gt;There is definitely satellites that achieve this. The type of orbit is called &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sun_synchronous&quot; rel=&quot;nofollow&quot;&gt;Sun synchronous&lt;/a&gt;&quot;. It's impossible to always be 6:00 PM, but it can always be either 6:00 PM or 6:00 AM. The trick is that it's a slight retrograde polar orbit, that allows for this feat to be achieved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very common for earth observing (And for that matter, Mars, Moon, Venus, and Mercury observing) satellites to follow this type of orbit, due to the illumination remaining consistent for all photos.&lt;/p&gt;&#xA;" OwnerUserId="871" LastEditorUserId="871" LastEditorDisplayName="" LastEditDate="2011-04-13T16:55:48.737" LastActivityDate="2011-04-13T16:55:48.737" CommentCount="3" />
  <row Id="8589" PostTypeId="1" AcceptedAnswerId="8601" CreationDate="2011-04-13T16:50:46.077" Score="0" ViewCount="119" Body="&lt;p&gt;taking rough numbers from &lt;a href=&quot;http://en.wikipedia.org/wiki/Muon&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Muon-catalyzed_fusion&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, it seems that with natural cosmic radiation sources, we could sustain $10^6$ muon-catalyzed fusion reactions per square-meter per minute. This would be $14 \times 10^6$ MeV of neutron energy (for D-T reactions, assuming 14 MeV per event) per square-meter per minute. In Watts-hour this would be $4 \times 10^{-12}$ KWh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is the right estimate, but wanted to check if someone spots a problem with my estimate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cheers&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-04-13T17:35:35.287" LastActivityDate="2011-04-22T04:20:46.653" Title="cosmic radiation muons as a source for muon-catalyzed fusion" Tags="&lt;fusion&gt;&lt;cosmic-rays&gt;&lt;cold-fusion&gt;&lt;muon-catalyzed-fusion&gt;" AnswerCount="4" />
  <row Id="8591" PostTypeId="2" ParentId="8589" CreationDate="2011-04-13T17:12:32.430" Score="0" ViewCount="" Body="&lt;p&gt;I have a problem with this :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In watts this would be 4×10−12 KWh&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-13T17:12:32.430" CommentCount="3" />
  <row Id="8592" PostTypeId="2" ParentId="8518" CreationDate="2011-04-13T17:12:36.320" Score="0" ViewCount="" Body="&lt;p&gt;Sobering thoughts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conservation laws are not related to any &lt;strong&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/4973/do-symmetries-increase-the-number-of-conserved-quantities&quot;&gt;symmetry&lt;/a&gt;&lt;/strong&gt;, to tell the truth. For a mechanical system with N degrees of freedom there always are N conserved quantities. They are complicated combinations of the dynamical variables. Their existence is provided with existence of the problem solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When there is a symmetry, the conserved quantities get just a simpler look.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I do not know how they teach you but the conservation laws are not related to Noether theorem. The latter just shows how to construct some of conserved quantities from the problem Lagrangian and the problem solutions. Any combination of conserved quantities is also a conserved quantity. So what Noether gives is not unique at all.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-04-13T21:49:29.847" LastActivityDate="2011-04-13T21:49:29.847" CommentCount="11" />
  <row Id="8593" PostTypeId="2" ParentId="8586" CreationDate="2011-04-13T17:23:17.177" Score="2" ViewCount="" Body="&lt;p&gt;There's another possibility, aside from the sun-synchronous orbit mentioned by @Pearsonartphoto. If you want your satellite to maintain a constant position relative to both Earth and Sun, you can put it at one of the Earth-Sun &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrangian_point&quot; rel=&quot;nofollow&quot;&gt;Lagrange points&lt;/a&gt;. It's always noon at L1 and always midnight at L2. L4 is always directly above a point on Earth corresponding to some time in the morning, and L5 is always directly above a point corresponding to some time in the afternoon, although in the latter cases the points are so far away that it's a bit funny to identify those terrestrial times of day with times on board the satellite.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-13T17:23:17.177" />
  <row Id="8594" PostTypeId="2" ParentId="8550" CreationDate="2011-04-13T17:31:58.253" Score="3" ViewCount="" Body="&lt;p&gt;&quot;So it would seem I would not be able to jump off of this asteroid.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That APPEARS to be correct.  There are two adjustments that need to be made to the formula you used, plus other corrections; but I don't think they would get you permanently off 243 Ida.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Are my formulas, assumptions, and calculations correct here?&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculations appear to be good, but there are two 'mis-assumptions' which invalidate the formulas you used, plus one error for constants (radius), plus plausibly one error in initial velocity:&#xA;1.) The geometry of 243 Ida is not even close to spherical.&#xA;2.) The formula you chose for speed of a falling object assumes a constant gravitational field, which is far from the truth. I think the easiest to use formula would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(v^2)/2 = Integral of (g(h))dh, where 'h' is height above asteroid surface, and 'g(h)' is a function of h.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.)  the dimesions of 243 Ida are roughly 54km x 24km x 15km; so the largest value for 'r' would be 54km/2 = 27km.  'r' could be as small as 7.5km.&#xA;4.)  Your jump velocity of 3.4m/s is in Earth's gravity.  That means you are fighting your own weight.  The excess force OVER and above your weight is what is used to generate your jump velocity through the mechanism of F=ma.  What do you think your initial jump velocity would be if you were hooked up to a bungie cord which lifted you with a force of (your body weight minus 1 ounce)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Also, how would I use this information to determine how high I would jump (would it be enough to freak me out? Well I'd be jumping off an asteroid, so it would freak me out anyway... although I imagine I would be going slow enough for (my ship?) to pick me back up.)&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The near field gravity on the surface of 243 Ida varies greatly, depending on where you are.  To understand this better, look at a pic of the rock.  It is roughly barbell-shaped.  Try to imagine if we took this to the extreme, and 243 Ida was shape like two identical spheres, with a very small connecting waist.  What would be the gravity at that waist?  Very close to zero, since the forces of gravity of the two spheres would cancel each other out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not going to try to work out the integral, here; but will make a few assumptions which will show how much the answer given by the standard formula could be off:  The standard answer for surface gravity on Ida 243 is 1.1 cm/s^2 down to 0.3 cm/s^2.  Assuming it is 0.3 cm/s^2 at the waist and remains constant (which is not true, but wait...) AND you jumped with an inital velocity of 5m/s (a conservative estimate), you would attain a height of 4167 meters.  Now, assuming r = 7500m at the waist, you might be tempted to think that gravitational pull would fall off by a factor of (4167/7500)^2 = 0.31, and so your jump would actually be higher; BUT, because of the geometry of the problem (two point sources of gravity instead of one), I think gravity would actually increase with height, as the two lobes of the asteroid started appearing more as a point source.&lt;/p&gt;&#xA;" OwnerUserId="3087" LastActivityDate="2011-04-13T17:31:58.253" CommentCount="2" />
  <row Id="8595" PostTypeId="2" ParentId="8580" CreationDate="2011-04-13T17:56:20.767" Score="1" ViewCount="" Body="&lt;p&gt;Since a quark $q_i$ is a spin $\frac{1}{2}$ Dirac particle with charge $z_i e$ and constituent mass $m_i$ with $i=u,d,s...$ theory predicts its magnetic moment as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \mu_i = \frac{z_i e \hbar}{2 m_i}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the formula in your question is only valid in the constituent quark model where a baryon consists only of three constituent quarks. Therefore you have to use the constituent quark mass in the formula for the magnetic moment too. A list of constituent quark masses is eg. found &lt;a href=&quot;http://en.wikipedia.org/wiki/Constituent_quark&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. These masses are obtained from hadron spectroscopy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not aware of any direct measurements of quark magnetic moments.&lt;/p&gt;&#xA;" OwnerUserId="3031" LastActivityDate="2011-04-13T17:56:20.767" CommentCount="4" />
  <row Id="8596" PostTypeId="2" ParentId="8586" CreationDate="2011-04-13T17:59:20.197" Score="2" ViewCount="" Body="&lt;p&gt;Agreed with Ted Bunn's answer, that it is possible to keep a satellite over a fixed time point on earth by using the Lagrange points.  One of those L-points is on the opposite side of the sun, which doesn't help much.  There are two problems with the other four.  One is that there are exactly four solutions, corresponding to exactly 4 local times on Earth: high noon; midnight; ROUGHLY 7am; and ROUGHLY 5pm.  The other problem is that all such L-points are far, FAR outside of LEO or geosynchronous orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility, which would take very little station-keeping fuel, would be to send the satellite into exactly the same orbit around the Sun as Earth.  Thus, it would appear to an Earth-bound observer as if the satellite was always above the dividing point of day and night (6:00am or 6:00pm, depending on whether the satellite was leading or lagging earth in the shared orbit around the Sun).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The satellite would have to be  much, much farther away than geosynchronous orbit in order to be able to fight the constant pull of Earth.  It might be possible to place it far enough out, and close enough inside lunar orbit, to gain enough periodic lunar pull to aid in stationkeeping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, six solutions have surfaced, for stationkeeping over a fixed local Earth time with little or no stationkeeping fuel burn: ~7:00 am; 6:00 am; noon; ~5:00 pm; 6:00 pm; and midnight. Happily, one of those solutions coincides with &quot;Margaritaville time.&quot;&lt;/p&gt;&#xA;" OwnerUserId="3087" LastEditorUserId="3087" LastEditorDisplayName="" LastEditDate="2011-04-13T18:29:13.103" LastActivityDate="2011-04-13T18:29:13.103" />
  <row Id="8597" PostTypeId="2" ParentId="8567" CreationDate="2011-04-13T18:00:04.103" Score="1" ViewCount="" Body="&lt;p&gt;Energy levels are assigned not to microstates but to terms. Terms are electron configurations with well-defined orbital and spin quantum numbers and in case of carbon there will be &lt;strong&gt;three&lt;/strong&gt; terms: $^1D$, $^3P$ and $^1S$. Due to spin-orbit interaction $^3P$ term is split into three closely spaced energy levels $^3P_2$, $^3P_1$ and $^3P_0$ (you cannot say which microstates belong to each of these levels).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true – you cannot start with $M_S$ and $M_L$ equal zero and determine which term it belongs to. Instead you start with the largest values of $M_L$ and $M_S$. Note that maximal $M_L$ is +2 which means that there should be a term with $L=2$ (denoted as $D$). All possible $M_L$ values for this term are +2,+1,0,-1,-2 and you should exclude these five microstates from your pool. The next highest $M_L=1$ for which $M_S=1$. This means that there &lt;em&gt;have&lt;/em&gt; to be $^3P$ term which is represented by 9 microstates. If you cross them out, the only combination you are left with is $M_L=0$, $M_S=0$ which can only be $^1S$ term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was this your question? &lt;/p&gt;&#xA;" OwnerUserId="701" LastActivityDate="2011-04-13T18:00:04.103" CommentCount="0" />
  <row Id="8598" PostTypeId="2" ParentId="8589" CreationDate="2011-04-13T18:27:24.713" Score="3" ViewCount="" Body="&lt;p&gt;Moun catalyzed fusion requires very specific kinematics, but cosmic muons come in all energies from stopped to tens of GeV at any particular spot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Care to work out the cross-section for having the &lt;em&gt;right&lt;/em&gt; kinematics? If you're having trouble I know a graduate student who is familiar with several of the common cosmic muon Monte Carlo generators.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-04-13T18:27:24.713" CommentCount="2" />
  <row Id="8599" PostTypeId="1" CreationDate="2011-04-13T18:44:27.550" Score="5" ViewCount="180" Body="&lt;p&gt;For the CS theory on a three manifold $M$ with a gauge group $G$, it is said that the gauge field $A$ is a connection on the trivial bundle over $M$. Why the bundle should be trivial? I know that space of classical solutions is made of flat connections but why should we assume the bundle is trivial from the beginning? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question: Can anyone explain the meaning of farming of a three manifold in simple terms? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1531" LastActivityDate="2011-04-14T14:33:16.030" Title="Chern-Simons Theory in 3D" Tags="&lt;string-theory&gt;" AnswerCount="1" />
  <row Id="8600" PostTypeId="2" ParentId="8513" CreationDate="2011-04-13T19:12:05.977" Score="0" ViewCount="" Body="&lt;p&gt;F=ma.  Which means $a = \frac{F}{m}$.  To find out any variations in 'a' versus bubble size, we need to find how F/m varies with bubble size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force on the bubble due to wind is somewhere between $k r^2 v $ and $k r^2 v^2 $.  In either case, for any given wind velocity of interest, it is proportional to $r^2$: $F = k_1 r^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the bubble is made of two components: The mass of the bubble film (MF) and the mass of the enclosed air (MA).  The mass of the film (MF) is proprtional to bubble surface area which is proportional to $r^2$: $MF = k_2 r^2$.  The mass of the entrapped air (MA) is proportional to bubble volume which is proportional to $r^3$: $MA = k_3 r^3$.  So...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F/m = \frac{k_1 r^2 }{MF + MA} = \frac{k_1 r^2 }{ k_2 r^2+k_3r^3} = \frac{k_1}{k_2+k_3 r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, acceleration due to wind is proportional to $\frac{k_1/k_3}{(k_2/k_3)+r}$, which means acceleration is INVERSELY proportional to $(k2/k3)+r$.  So, as bubble size goes up acceleration of the bubble goes down: The velocities of bigger bubbles are less affected by the wind.&lt;/p&gt;&#xA;" OwnerUserId="3087" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-04-15T10:37:25.907" LastActivityDate="2011-04-15T10:37:25.907" CommentCount="10" />
  <row Id="8601" PostTypeId="2" ParentId="8589" CreationDate="2011-04-13T19:22:00.343" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Muon-catalyzed_fusion&quot; rel=&quot;nofollow&quot;&gt;Muon catalyzed fusion&lt;/a&gt; needs the muons to be low enough energy to replace an electron and stay in a stable orbit. Since the reason the catalysis happens is because the atom is much smaller and  two protons can  get close together enhancing the probability of overlap and fusion, one needs a large number of low energy muons so that the probability of two muonic hydrogens to be in contact  would be high. Fusion has been observed in  molecules that have an orbital electron replaced by a muon, binding the atoms of the molecule into closer distances, so that the nuclear force can have a probability of fusing the two nuclei, of H2 for example, and releasing the muon for a next round.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 10.000 muons per minute per square meter are mainly above GeV energies whereas for atomic sizes one needs  keV at best, so your calculations cannot be correct. In addition the muon decays in 10^-6 seconds, so it cannot accumulate in a volume to keep on catalyzing.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-04-22T04:20:46.653" LastActivityDate="2011-04-22T04:20:46.653" />
  <row Id="8602" PostTypeId="1" AcceptedAnswerId="8605" CreationDate="2011-04-13T19:52:45.607" Score="6" ViewCount="251" Body="&lt;p&gt;in the context of $$a_-  |0\rangle =0$$&lt;/p&gt;&#xA;" OwnerUserId="3239" LastActivityDate="2011-04-13T21:46:19.833" Title="What is the difference between $|0\rangle $ and $0$" Tags="&lt;quantum-mechanics&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="8603" PostTypeId="1" CreationDate="2011-04-13T19:55:59.040" Score="0" ViewCount="65" Body="&lt;p&gt;I'm looking for a simple schematic or instructions to build a unipolar motor with inexpensive parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps something like this: &lt;a href=&quot;http://bit.ly/rolling-unipolar-motor&quot; rel=&quot;nofollow&quot;&gt;http://bit.ly/rolling-unipolar-motor&lt;/a&gt; (1.5MB AVI Video)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="2390" LastEditorUserId="2390" LastEditorDisplayName="" LastEditDate="2011-04-13T20:01:48.567" LastActivityDate="2011-04-13T20:01:48.567" ClosedDate="2011-04-13T20:16:05.870" Title="Simple Unipolar Motor Plans" Tags="&lt;electromagnetism&gt;&lt;magnetic-monopoles&gt;" />
  <row Id="8604" PostTypeId="2" ParentId="8602" CreationDate="2011-04-13T20:05:16.110" Score="12" ViewCount="" Body="&lt;p&gt;$|0\rangle$ is a particular nonzero vector in the Hilbert space associated with this system. That vector is nonzero -- in fact, it's usually normalized to have magnitude 1. The 0 on the right refers to the zero vector in the Hilbert space. So they're quite different. For one thing, $|0\rangle$ is a possible state for a particle to be in. 0 isn't (since only unit-magnitude vectors are possible states).&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-13T20:05:16.110" CommentCount="3" />
  <row Id="8605" PostTypeId="2" ParentId="8602" CreationDate="2011-04-13T20:05:56.550" Score="12" ViewCount="" Body="&lt;p&gt;$|0\rangle$ is just a quantum state that happens to be labeled by the number 0. It's conventional to use that label to denote the ground state (or vacuum state), the one with the lowest energy. But the label you put on a quantum state is actually kind of arbitrary. You could choose a different convention in which you label the ground state with, say, 5, and although it would confuse a lot of people, you could still do physics perfectly well with it. The point is, $|0\rangle$ is just a particular quantum state. The fact that it's labeled with a 0 doesn't have to mean that anything about it is actually zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, $0$ (not written as a ket) is &lt;em&gt;actually zero&lt;/em&gt;. You could perhaps think of it as the quantum state of an object that doesn't exist (although I suspect that analogy will come back to bite me... just don't take it too literally). If you calculate any matrix element of some operator $A$ in the &quot;state&quot; $0$, you &lt;em&gt;will get 0&lt;/em&gt; as a result because you're basically multiplying by zero:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle\psi| A (a_-|0\rangle) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for any state $\langle\psi|$. In contrast, you can do this for the ground state without necessarily getting zero:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle\psi| A |0\rangle = \text{can be anything}$$&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-04-13T20:05:56.550" />
  <row Id="8606" PostTypeId="2" ParentId="8602" CreationDate="2011-04-13T20:24:28.610" Score="4" ViewCount="" Body="&lt;p&gt;You may consider 0 as an eigenvalue and write $a|0\rangle = 0|0\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any eigenvector $a|\alpha \rangle = \alpha |\alpha \rangle$ is of different &quot;length&quot; than the corresponding normalized vector $|\alpha \rangle$. In your particular case the vector $0|0\rangle$ is of zeroth length.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-04-13T21:46:19.833" LastActivityDate="2011-04-13T21:46:19.833" />
  <row Id="8607" PostTypeId="1" AcceptedAnswerId="8608" CreationDate="2011-04-13T21:46:24.937" Score="2" ViewCount="112" Body="&lt;p&gt;I'm need to understand the following: to keep the room at confortable temperature (70 degree, for example), how does the amount of energy consumed by the AC grow as the outside air temperature rises in the summer over the same length of time (a day, say)? Is it more or less linear? quadratic? exponential? or anything else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I posted the same question on &lt;a href=&quot;http://electronics.stackexchange.com/questions/13015/temperature-vs-ac-energy-consumption&quot;&gt;another StackExchange site&lt;/a&gt;, but I figured that physicist may provide more help because it seems to me that it is related to both heat conduction and the working mechanism of AC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3093" LastActivityDate="2011-04-13T22:15:00.127" Title="Temperature vs AC energy consumption" Tags="&lt;heat&gt;" AnswerCount="1" />
  <row Id="8608" PostTypeId="2" ParentId="8607" CreationDate="2011-04-13T22:06:38.033" Score="1" ViewCount="" Body="&lt;p&gt;Without getting too complicated I'd say it's linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two bodies in contact exchange heat at a rate that's proportional to the temperature difference between them.  The rate of heat transfer in with an outside temperature of 80 degrees will be twice the rate at 75 degrees (assuming your 70-degree room temperature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your AC needs to pump this heat out as fast as it comes in to maintain 70 degrees, so it will need to do it twice as fast when it's 80 degrees outsite than when it's 75 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a linear relationship.&lt;/p&gt;&#xA;" OwnerUserId="379" LastActivityDate="2011-04-13T22:06:38.033" CommentCount="9" />
  <row Id="8610" PostTypeId="1" AcceptedAnswerId="8616" CreationDate="2011-04-14T01:00:58.790" Score="6" ViewCount="164" Body="&lt;p&gt;I have learned in my physics classes about three different types of masses and I am confused about the differences between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the difference between the three masses: inertial mass, gravitational mass and rest mass?&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2011-04-14T03:39:12.763" Title="What's the difference between the three masses: inertial mass, gravitational mass and rest mass?" Tags="&lt;mass&gt;" AnswerCount="3" />
  <row Id="8611" PostTypeId="2" ParentId="8610" CreationDate="2011-04-14T01:22:35.193" Score="5" ViewCount="" Body="&lt;p&gt;A body's inertial mass is the mass measured by its resistance to motion. Its gravitational mass is the mass measured by its attraction by gravitational force. Its rest mass is the mass when it's at rest with respect to an observer, and is then equivalent to its inertial mass.&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-04-14T01:22:35.193" />
  <row Id="8612" PostTypeId="1" CreationDate="2011-04-14T01:24:23.353" Score="5" ViewCount="140" Body="&lt;p&gt;This question: &lt;a href=&quot;http://physics.stackexchange.com/questions/8567/how-do-electron-configuration-microstates-map-to-term-symbols&quot;&gt;How do electron configuration microstates map to term symbols?&lt;/a&gt;&lt;br&gt;&#xA;And the discussion of multielectron effects here: &lt;a href=&quot;http://physics.stackexchange.com/questions/8483/quantum-computing-and-animal-navigation&quot;&gt;Quantum Computing and Animal Navigation&lt;/a&gt;&lt;br&gt;&#xA;Inspired me to try to understand quantum mechanics of multi-electron atoms better.  As it is aparrently difficult to solve anything in closed form, I found a numeric multi-electron quantum mechanics calculation package that is free for students! (Called &lt;a href=&quot;http://www.msg.ameslab.gov/gamess/&quot; rel=&quot;nofollow&quot;&gt;GAMESS&lt;/a&gt;)  My goal is to understand the basics at least at the level of the 'molecular orbital' picture, and be able to calculate some energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the discussion in the other question and reading wikipedia I think I understand the basic idea behind &lt;a href=&quot;http://en.wikipedia.org/wiki/Hartree-Fock_method&quot; rel=&quot;nofollow&quot;&gt;Hartree Fock&lt;/a&gt; and the wavefunction made from a &lt;a href=&quot;http://en.wikipedia.org/wiki/Slater_determinant&quot; rel=&quot;nofollow&quot;&gt;Slater determinant&lt;/a&gt;.  And I was able to use the program to calculate the orbital energies and ground state energy for a  ${}^1S_{0}$ term symbol atom like Neon.  These make sense to me as each orbital filled with two electrons:&#xA;$$ \begin{array}{ccc}&#xD;&#xA;\underline{\uparrow \downarrow} &amp;amp; \underline{\uparrow \downarrow} &amp;amp; \underline{\uparrow \downarrow} \ \underline{\uparrow \downarrow} \ \underline{\uparrow \downarrow} \\&#xD;&#xA;\underset{1s}{\ } &amp;amp; \underset{2s}{\ } &amp;amp;  \underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{2p} \end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main question is regarding atoms with open shells:&lt;br&gt;&#xA;&lt;strong&gt;How is the wavefunction represented in the orbital picture for open shell atoms?&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example Lithium with term symbol ${}^2S_{1/2}$&#xA;$$ \begin{array}{ccc}&#xD;&#xA;\underline{\uparrow \downarrow} &amp;amp; \underline{\uparrow \ } &amp;amp; \underline{\quad \ } \ \underline{\quad \ } \  \underline{\quad \ } \\&#xD;&#xA;\underset{1s}{\ } &amp;amp; \underset{2s}{\ } &amp;amp;  \underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{2p} \end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;versus Boron with term symbol ${}^2P_{1/2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{array}{ccc}&#xD;&#xA;\underline{\uparrow \downarrow} &amp;amp; \underline{\uparrow \downarrow } &amp;amp; \underline{\uparrow \ } \ \underline{\quad \ } \  \underline{\quad \ } \\&#xD;&#xA;\underset{1s}{\ } &amp;amp; \underset{2s}{\ } &amp;amp;  \underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{2p} \end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like these should be described exactly the same way in molecular orbitals: one open shell below which all the remaining shells are closed.  However this apparently isn't the case.  The Lithium atom is a &quot;doublet&quot; and can be described by ROHF with GAMESS, while the Boron case is apparently not just a doublet since somehow the degeneracy needs to be taken into account (does this 'kind of doublet' have a name besides the term symbol?) although this GAMESS document says they are both 'technically ROHF' states ( &lt;a href=&quot;http://haucke-jena.de/gamess/gam_grund_1/gvb_ref.htm&quot; rel=&quot;nofollow&quot;&gt;http://haucke-jena.de/gamess/gam_grund_1/gvb_ref.htm&lt;/a&gt; ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question &quot;part 2&quot;:&lt;br&gt;&#xA;&lt;strong&gt;Is this trying to say we need more than one slater determinant for these degenerate states?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, something like:&#xA;$$ \frac{1}{\sqrt{2}} \left[ \left(&#xD;&#xA;\begin{array}{ccc}&#xD;&#xA;\underline{\uparrow \downarrow} &amp;amp; \underline{\uparrow \downarrow } &amp;amp; \underline{\uparrow \ } \ \underline{\quad \ } \  \underline{\quad \ } \\&#xD;&#xA;\underset{1s}{\ } &amp;amp; \underset{2s}{\ } &amp;amp;  \underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{2p} \end{array}&#xD;&#xA;\right)&#xD;&#xA;+&#xD;&#xA;\left(&#xD;&#xA;\begin{array}{ccc}&#xD;&#xA;\underline{\uparrow \downarrow} &amp;amp; \underline{\uparrow \downarrow } &amp;amp; \underline{\quad \ } \ \underline{\quad \ } \  \underline{\uparrow \ } \\&#xD;&#xA;\underset{1s}{\ } &amp;amp; \underset{2s}{\ } &amp;amp;  \underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{2p} \end{array}&#xD;&#xA;\right) \right]&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this would be called a &quot;multi configuration&quot; state I think (MCSCF?), but the quantum calculation documentation (previous link) says MCSCF isn't necessary for either of these states.  So are those two &quot;pieces&quot; of the wavefunction somehow not different slater determinants?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm very confused and would appreciate any help.  Please clearly explain the terminology you use in an answer, as the difference between RHF, ROHF, GVB, MCSCF, etc. seem to be a large part of my confusion here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;UPDATE:&lt;br&gt;&#xA;Some example calculation results for an Oxygen atom.&lt;br&gt;&#xA;energy = -74.8002167288 Hartrees --- triplet (ROHF with MULT=3)&lt;br&gt;&#xA;energy = -74.6780126893 Hartrees --- singlet (ROHF with MULT=1)&lt;br&gt;&#xA;energy = -74.6780126893 Hartrees --- singlet (RHF), should be same as ROHF MULT=1&lt;br&gt;&#xA;energy = -74.7986255288 Hartrees --- ??? (GVB with setting specified by GAMESS doc for ${}^3P_2$ term symbol)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The GVB output also says:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;     MULLIKEN ATOMIC POPULATION IN EACH MOLECULAR ORBITAL&#xA;&#xA;                      1          2          3          4          5&#xA;&#xA;                  2.000000   2.000000   1.333333   1.333333   1.333333&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which seems to me like a uniform mixing of&#xA;$$&#xD;&#xA;\left(&#xD;&#xA;\begin{array}{ccc}&#xD;&#xA;\underline{\uparrow \downarrow} &amp;amp; \underline{\uparrow \downarrow } &amp;amp; \underline{\uparrow \downarrow } \ \underline{\uparrow \ } \  \underline{\uparrow \ } \\&#xD;&#xA;\underset{1s}{\ } &amp;amp; \underset{2s}{\ } &amp;amp;  \underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{2p} \end{array}&#xD;&#xA;\right),&#xD;&#xA;&#xD;&#xA;\left(&#xD;&#xA;\begin{array}{ccc}&#xD;&#xA;\underline{\uparrow \downarrow} &amp;amp; \underline{\uparrow \downarrow } &amp;amp; \underline{\uparrow \ } \ \underline{\uparrow \downarrow } \  \underline{\uparrow \ } \\&#xD;&#xA;\underset{1s}{\ } &amp;amp; \underset{2s}{\ } &amp;amp;  \underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{2p} \end{array}&#xD;&#xA;\right),&#xD;&#xA;&#xD;&#xA;\left(&#xD;&#xA;\begin{array}{ccc}&#xD;&#xA;\underline{\uparrow \downarrow} &amp;amp; \underline{\uparrow \downarrow } &amp;amp; \underline{\uparrow \ } \ \underline{\uparrow \ } \  \underline{\uparrow \downarrow } \\&#xD;&#xA;\underset{1s}{\ } &amp;amp; \underset{2s}{\ } &amp;amp;  \underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{2p} \end{array}&#xD;&#xA;\right)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how can that center one be a ${}^3P_2$ state? Shouldn't it be a ${}^3S_1$ state?  This all seems very conflicting.  Especially that trying to include the effects of the degeneracy actually raised the energy instead of lowering it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the GAMESS input I used if someone else wants to play with it:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA; $CONTRL UNITS=ANGS COORD=unique RUNTYP=energy $END&#xA;! p**4 3-P state&#xA; $CONTRL SCFTYP=GVB MULT=3 $END&#xA; $SCF NCO=2 NSETO=1 NO=3 COUPLE=.TRUE.&#xA;    F(1)= 1.0 0.66666666666667&#xA;    ALPHA(1)= 2.0 1.33333333333333 0.83333333333333&#xA;    BETA(1)= -1.0 -0.66666666666667 -0.50000000000000 $END&#xA; $BASIS GBASIS=N311 NGAUSS=6 NPFUNC=1 NDFUNC=1 $END&#xA; $DATA&#xA; Oxygen atom&#xA; C1&#xA; O           8.0   0.0000000000   0.0000000000   0.0000000000&#xA; $END&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;EDIT: My latex is kludgy.  I had trouble making things line up.  If someone could edit this so I can see the proper way to do things, that would be great.&lt;/p&gt;&#xA;" OwnerUserId="2655" LastEditorUserId="2655" LastEditorDisplayName="" LastEditDate="2011-04-14T08:03:58.430" LastActivityDate="2011-04-15T03:32:56.983" Title="Open shells in Quantum mechanics of multielectron atoms" Tags="&lt;quantum-mechanics&gt;&lt;atoms&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="8613" PostTypeId="2" ParentId="90" CreationDate="2011-04-14T01:44:08.320" Score="1" ViewCount="" Body="&lt;p&gt;For a neutron star to be called a pulsar, we need to detect a periodic signal pulse from the object. The &quot;lighthouse model&quot; explains this as a spinning object, with a magnetic field off-kilter from the spin axis, beaming radiation out from the poles.  So, there are certainly some neutron stars where lighthouse beams are spinning around but never point at the earth, and we don't see them. In some cases, we observe a pulsar in a binary with another neutron star, but can't detect any radiation from the companion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, neutron stars (and thus pulsars) do emit other thermal radiation, it's just hard to detect that if they are far away. The surfaces are really small. Or we see some surface radiation, plus a brighter blip from the &quot;lighthouse&quot;. For example, several nearby pulsars ( &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Magnificent_Seven_(neutron_stars&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/The_Magnificent_Seven_(neutron_stars&lt;/a&gt;) ) are detected mostly from their constant thermal radiation in x-rays.  But they also have small periodic pulsations on top of the constant emission - the &quot;pulsed fraction&quot; is 1% to about 20% of the total ( &lt;a href=&quot;http://arxiv.org/abs/0801.1143v1&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0801.1143v1&lt;/a&gt; ) - so they are still called pulsars.&lt;/p&gt;&#xA;" OwnerUserId="3094" LastActivityDate="2011-04-14T01:44:08.320" CommentCount="2" />
  <row Id="8614" PostTypeId="2" ParentId="8610" CreationDate="2011-04-14T01:56:00.460" Score="6" ViewCount="" Body="&lt;p&gt;Inertial mass is the mass that appears in Newton's Second Law $$F=ma$$ Gravitational mass is what appears in Newton's Law of Gravity $$F=\frac{GMm}{r^2}$$ Einstein's Equivalence Principle requires that inertial mass and gravitational mass are equal so that all masses react the same way to a given gravitational field ($m$ cancels in the above two equations to give the same acceleration). This equality has been established to great precision in many experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rest mass is a somewhat obsolete term for what is referred to more commonly today as the invariant mass, proper mass or simply just mass in relativistic physics. Given the 4-momentum $p^\mu$ of a particle, a scalar invariant can be obtained from it which is the square of its mass. $$p^\mu p_\mu=m^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it is an invariant, it holds true in any reference frame. However, in the frame in which the particle is at rest, $m$ equals the total energy of the particle (in units of $c=1$), hence the old name of &quot;rest mass&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton's second law takes a different form when relativity is taken into account, so it's not helpful to compare rest mass and inertial mass, except of course in the rest frame of the particle. The important conceptual difference in relativity however is that &lt;em&gt;inertia&lt;/em&gt;, defined as the resistance to motion, depends on the velocity of the particle so that the higher the velocity, the harder it is to accelerate it. You can read more about that in the Wikipedia article &lt;a href=&quot;http://en.wikipedia.org/wiki/Mass_in_special_relativity&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mass_in_special_relativity&lt;/a&gt; or a nice textbook on Special Relativity.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-04-14T01:56:00.460" />
  <row Id="8615" PostTypeId="1" AcceptedAnswerId="8638" CreationDate="2011-04-14T02:28:23.507" Score="5" ViewCount="145" Body="&lt;p&gt;Could you please help me understand what is the difference between rocket engines designed to work in air (first stage) and vacuum (later stages)? &lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="1965" LastEditorDisplayName="" LastEditDate="2011-04-14T04:27:30.597" LastActivityDate="2011-04-14T16:28:23.870" Title="Rocket engines: air &amp; vacuum" Tags="&lt;rocket&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="8616" PostTypeId="2" ParentId="8610" CreationDate="2011-04-14T03:26:40.940" Score="6" ViewCount="" Body="&lt;p&gt;Let us define the inertial mass, gravitational mass and rest mass of a particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Inertial mass:&lt;/strong&gt; - To every particle in nature we can associate a real number with it so that the value of the number gives the measure of inertia (the amount of resistance of the particle to accelerate for a definite force applied on it) of the particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Newton's laws of motion,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_i = F/a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Gravitational mass&lt;/strong&gt; - (This is defined using Newton's law of universal gravitation i.e. the gravitational force between any two particle a definite distance apart is proportional the product of the gravitational masses of the two particles.) To every particle in nature we can associate a real number with it so that the value of the number gives the measure of the response of the particle to the gravitational force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = \frac{Gm_{G1}m_{G2}}{R^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All experiments carried out till date have shown that $m_G = m_i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the reason why the acceleration due to gravity is independent of the inertial or gravitational mass of the particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_ia = \frac{Gm_{G1}m_{G2}}{R^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $m_{G1} = m_i$ then $a = \frac{Gm_{G2}}{R^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is acceleration due to gravity of the particle is independent of its inertial or gravitational mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rest mass&lt;/strong&gt; - This is simply called the mass and is defined as the inertial mass of a particle as measured by an observer, with respect to whom, the particle is at rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was an obsolete term called relativistic mass which is the inertial mass as measured by an  observer, with respect to whom, the particle is at motion. The relation between the rest mass and the relativistic mass is given as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m = \frac{m_0}{\sqrt{1-v^2/c^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $v$ is the speed of the particle and $c$ is the speed of light, $m$ is the relativistic mass and $m_0$ is the rest mass.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-04-14T03:39:12.763" LastActivityDate="2011-04-14T03:39:12.763" />
  <row Id="8617" PostTypeId="1" AcceptedAnswerId="8618" CreationDate="2011-04-14T03:50:22.017" Score="3" ViewCount="45" Body="&lt;p&gt;Where would I look to get the most accurate experimental values of &lt;a href=&quot;http://en.wikipedia.org/wiki/Parton_%28particle_physics%29&quot; rel=&quot;nofollow&quot;&gt;parton distribution functions&lt;/a&gt; for the proton? I know these functions aren't measured &lt;em&gt;directly&lt;/em&gt;, but I'd basically like to find a fit that involves recent experimental data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading &lt;a href=&quot;http://arxiv.org/abs/hep-lat/9609018&quot; rel=&quot;nofollow&quot;&gt;a paper&lt;/a&gt; on PDFs which briefly discussed how the theoretical functions are fit to data and included some sample figures, but it dates from 1996 and I'm looking for something more recent.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-04-14T04:13:16.520" Title="Where can I get the most accurate measurements of parton distribution functions?" Tags="&lt;particle-physics&gt;&lt;experimental-physics&gt;&lt;quantum-chromodynamics&gt;" AnswerCount="1" />
  <row Id="8618" PostTypeId="2" ParentId="8617" CreationDate="2011-04-14T04:13:16.520" Score="4" ViewCount="" Body="&lt;p&gt;The two most well-known parton distribution function tables are &lt;a href=&quot;http://arxiv.org/abs/1007.2241&quot; rel=&quot;nofollow&quot;&gt;CTEQ from the US&lt;/a&gt; and &lt;a href=&quot;http://arxiv.org/abs/0901.0002&quot; rel=&quot;nofollow&quot;&gt;MSTW from Europe&lt;/a&gt;. Some people claim that the European one is of higher quality, but I don't know enough to judge.&lt;/p&gt;&#xA;" OwnerUserId="2643" LastActivityDate="2011-04-14T04:13:16.520" CommentCount="1" />
  <row Id="8619" PostTypeId="2" ParentId="8615" CreationDate="2011-04-14T04:23:46.190" Score="1" ViewCount="" Body="&lt;p&gt;While it does have a small affect on the way a rocket works, the pressure of the atmosphere is trivial in comparison to the inertia of the rocket fuel. The atmosphere is easily pushed out of the way in the vicinity of the rocket due to the force on the rocket fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary force exerted on a rocket is that of the pressure that forces the fuel to move in the only direction it can, the other way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the rocket has so much inertia due to its single solid mass, it's relative velocity is a fraction of the fuel's velocity, but in the end, the pressure of the ignition puts the same force on both systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why both the vacuum and the atmosphere have negligable effect on the way a rocket works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's all in the hips.&lt;/p&gt;&#xA;" OwnerUserId="1965" LastActivityDate="2011-04-14T04:23:46.190" CommentCount="2" />
  <row Id="8620" PostTypeId="1" AcceptedAnswerId="8621" CreationDate="2011-04-14T06:06:51.987" Score="4" ViewCount="329" Body="&lt;p&gt;How does Newton's color disk work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton's disk - Take a circular white color disk, make 7 equal intersections and paint section with respective VIBGYOR colors, now when you spin the disk in certain speed, it appears white.&lt;/p&gt;&#xA;" OwnerUserId="3047" LastEditorUserId="3047" LastEditorDisplayName="" LastEditDate="2011-04-15T06:02:22.073" LastActivityDate="2011-04-15T06:02:22.073" Title="Newton's color Disk" Tags="&lt;optics&gt;&lt;light&gt;&lt;spectroscopy&gt;&lt;biophysics&gt;&lt;color&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8621" PostTypeId="2" ParentId="8620" CreationDate="2011-04-14T06:31:58.997" Score="3" ViewCount="" Body="&lt;p&gt;The perception of a picture or a color stays in the human brain for a fraction of a second. Newton's color disk is a mechanical device that rotates an array of colors arranged as petals or gradients around an axis and when the disc rotates fast enough the perception of the color changes to white. This is because white light is nothing but the mixture of light of all wavelengths in the visible range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sir Isaac Newton used this device to prove the above assertion that white color is the mixture of all colors.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-04-14T06:31:58.997" CommentCount="14" />
  <row Id="8622" PostTypeId="1" AcceptedAnswerId="8630" CreationDate="2011-04-14T07:24:05.880" Score="12" ViewCount="629" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Summary&lt;/strong&gt;: This turned out to be a rather trivial one indeed. As Marek mentioned in the comment, the continuity equation is trivial. And it indeed turns out be so. Godfrey Miller elaborates on this, showing that the continuity equation merely shows that the transition amplitude is constant with time. All this is only confusing as long as you stick to the belief that  the continuity equation is some exclusive indication of probability density. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\rho = \psi_1^* \psi_2$ is the probability density for transition from state 2 to 1.  I am having trouble interpreting the probability though $$dP_{1\leftarrow 2}=\psi_1^* \psi_2 d^3 r $$ does this mean the probability in a region of volume $d^3 r$ to go from state 2 to state 1? That does not make sense at all!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I more or less agree with Marek's answer. However, that makes it seem this is not the probability density. &lt;strong&gt;How do we reconcile that with the fact that it satisfies the continuity equation.&lt;/strong&gt; If we have two states:&lt;br&gt;&#xA;$$ i \hbar \partial_t \psi_{1,2}= -\frac{\hbar^2}{2m} \nabla^2 \psi_{1,2} + V\psi_{1,2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After making standard manipulations (taking conjugate, multiplying by conjugate, subtracting) we would get&#xA;$$\partial_t ( \psi_1^*\psi_2) +\frac{\hbar}{2mi} \nabla\cdot \left( \psi_1^* \nabla \psi_2 - (\nabla \psi_1^* )\psi_2\right)=0$$&#xA;This seems consistent with the continuity equation: $$\partial_t \rho +\nabla.\mathbb{j} =0$$ However, we seem to be changing its relation to the probability by making the probability $P=\int \int\rho^*(r,r&amp;#39;) \rho(r,r&amp;#39;) d^3r d^3r&amp;#39; $ instead of the regular $P=\int \rho(r) d^3r$ For a kind of intuitive &quot;dimensional consistency&quot; can we instead say $P=\int \int \; \sqrt{\rho^* \rho} \; d^3r d^3r&amp;#39;$&#xA;$$****$$&#xA;&lt;strong&gt;Full steps:&lt;/strong&gt; &#xA;$\psi_1$ and $\psi_2$ are two states of the same system, i.e both satisfy the SE of the same potential:&#xA;Taking the SE for $\psi_2$  and multiplying by $\psi_1^*$ :&#xA;$$ \psi_1^*\times\left( i \hbar \partial_t \psi_2 = -\frac{\hbar^2}{2m} \nabla^2 \psi_2 + V\psi_2\right)$$ Taking the SE for $\psi_1$ complex conjugating the entire equation and multiplying by $\psi_2$ :&#xA;$$\left( -i \hbar \partial_t \psi_1^*= -\frac{\hbar^2}{2m} \nabla^2 \psi_1^* + V\psi_1^*\right)\times \psi_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($V$ is real, no decay processes etc) On subtracting the second equation from the first &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\partial_t (\psi_1^*\psi_2 )= -\frac{\hbar}{2mi} \left( \psi_1^* \nabla^2 \psi_2 - (\nabla^2 \psi_1^* )\psi_2\right)$$&#xA; which is the same as $$ \partial_t (\psi_1^*\psi_2 )= -\frac{\hbar}{2mi} \nabla\cdot \left( \psi_1^* \nabla \psi_2 - (\nabla \psi_1^* )\psi_2\right)$$&lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="2599" LastEditorDisplayName="" LastEditDate="2011-04-25T05:28:03.730" LastActivityDate="2011-04-25T05:28:03.730" Title="How do I correctly interpret $\rho = \psi_1^* \psi_2$?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" />
  <row Id="8623" PostTypeId="1" AcceptedAnswerId="8624" CreationDate="2011-04-14T07:29:52.463" Score="2" ViewCount="71" Body="&lt;p&gt;What is the resistivity coefficient of  graphene?&lt;/p&gt;&#xA;" OwnerUserId="3097" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-04-14T07:52:43.450" LastActivityDate="2011-04-14T07:54:52.327" Title="What is the resistivity coefficient of  graphene?" Tags="&lt;graphene&gt;" AnswerCount="1" />
  <row Id="8624" PostTypeId="2" ParentId="8623" CreationDate="2011-04-14T07:54:52.327" Score="1" ViewCount="" Body="&lt;p&gt;Graphene has a high electron mobility at ordinary room temperature. Experiments also has shown that the conductivity of electron and holes are almost the same. Between the temperature 10 to 100 k it is independent of temperature. At room temperature the resistivity is $10 ^{-6}$ ohmcm. There is a good &lt;a href=&quot;http://en.wikipedia.org/wiki/Graphene&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-04-14T07:54:52.327" CommentCount="1" />
  <row Id="8625" PostTypeId="1" CreationDate="2011-04-14T08:18:22.787" Score="1" ViewCount="28" Body="&lt;p&gt;Assume I have a bounded material with heat sources inside. The material is known (i.e. I know heat capacity and all relevant data) and the temperature of the boundary is fixed. I solved the (steady state) heat equation and hence know the temperature distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I compute the heat flow through the boundary? I expect something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\int_\Gamma k\cdot \nabla T&#xD;&#xA;$$&#xA;where $\Gamma$ is the boundary and $T$ the temperature. Is this correct? If so, what value for $k$ do I have to use?&lt;/p&gt;&#xA;" OwnerUserId="2480" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-04-14T08:57:48.757" LastActivityDate="2011-04-14T09:53:11.693" Title="How to compute the heat flow for a specific material for some given boundary temperature?" Tags="&lt;heat&gt;" AnswerCount="1" />
  <row Id="8626" PostTypeId="1" AcceptedAnswerId="9250" CreationDate="2011-04-14T08:35:54.127" Score="9" ViewCount="538" Body="&lt;p&gt;If a system has $N$ degrees of freedom (DOF) and therefore $N$ independent&lt;sup&gt;1&lt;/sup&gt; &lt;s&gt;conserved quantities&lt;/s&gt; integrals of motion, can continuous symmetries with a total of $N$ parameters be found that deliver these conserved quantities by means of &lt;a href=&quot;http://en.wikipedia.org/wiki/Noether%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Noether's theorem&lt;/a&gt;? I think this is not exactly the opposite of Noether's theorem since I don't ask whether for &lt;em&gt;each&lt;/em&gt; conserved quantity &lt;em&gt;a&lt;/em&gt; symmetry can be retrieved, I ask about a connection between the &lt;em&gt;whole set&lt;/em&gt; of conserved quantities and symmetries.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1)&lt;/sup&gt; or $2N-1$, or $N$, depending on definition and details that are irrelevant here. But let me expand on it anyway... I consider the number of DOFs equal to the number of initial conditions required to fully describe a system in Classical Mechanics. That means, velocities (or momenta) are considered individual DOFs, and not that each pair of coordinate + velocity make up only one DOF. Time is no DOF however, it is a &lt;em&gt;parameter&lt;/em&gt;. Please discuss this in &lt;a href=&quot;http://physics.stackexchange.com/q/8860/97&quot;&gt;this question&lt;/a&gt; if you disagree.&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-04-20T18:14:58.577" LastActivityDate="2011-04-28T21:06:26.180" Title="If all conserved quantities of a system are known, can they be explained by symmetries?" Tags="&lt;mathematical-physics&gt;&lt;symmetry&gt;&lt;noethers-theorem&gt;&lt;conservation-laws&gt;&lt;integrals-of-motion&gt;" AnswerCount="4" CommentCount="21" FavoriteCount="1" />
  <row Id="8627" PostTypeId="2" ParentId="2690" CreationDate="2011-04-14T08:41:57.773" Score="0" ViewCount="" Body="&lt;p&gt;The quantity that &lt;em&gt;is&lt;/em&gt; conserved is the squared momentum $p^2$ of the &lt;em&gt;entire&lt;/em&gt; universe (or any isolated system), caused by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Poincar%C3%A9_group&quot; rel=&quot;nofollow&quot;&gt;Poincaré invariance&lt;/a&gt;. It also results in the conservation of total spin by the way. &lt;a href=&quot;http://en.wikipedia.org/wiki/Wigner%27s_classification&quot; rel=&quot;nofollow&quot;&gt;Wigner's classification&lt;/a&gt; provides some interesting further reading, too.&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2011-04-14T08:41:57.773" />
  <row Id="8628" PostTypeId="2" ParentId="8625" CreationDate="2011-04-14T08:53:15.230" Score="1" ViewCount="" Body="&lt;p&gt;Don't reinvent wheels:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Conduction_%28heat%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Conduction_%28heat%29&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fourier solved that 150 years ago. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-14T09:53:11.693" LastActivityDate="2011-04-14T09:53:11.693" CommentCount="2" />
  <row Id="8629" PostTypeId="2" ParentId="8567" CreationDate="2011-04-14T09:26:14.223" Score="1" ViewCount="" Body="&lt;p&gt;The Hamiltonian of an atom commutes with the square of the total angular momentum $\hat{J} = \hat{L} + \hat{S}$ and one component, disregarding very small hyperfine splittings. Knowing the energy $E$, $J$ and $J_z$ amounts to solving the fine structure of the atom. Since $\hat{J}$ is an angular momentum there are 2J+1 degenerate components in the absence of an external magnetic field. For non heavy atoms the Hamiltonian also approximately commutes with $\hat{L}^2$, $\hat{L}_z$, $\hat{S}^2$ and $\hat{S}_z$, the orbital and spin angular momenta, so you can have approximately well defined energy and $L$, $M_L$, $S$ and $M_S$ quantum numbers. These states have the right (or approximately right) symmetry, so you want to have your eigenstates with all those labels, say $|E,L,M_L,S,M_S,J,M_J\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In typical Quantum calculations, however, you write the n-electron wave function as a Slater determinant of {\it atomic orbitals}, which have well defined one-electron orbital and spin angular momenta, $l_j$, $m_{l,j}$, $s_j$ and $m_{s,j}$. The number of micro-states that you can find specifying the lower case quantum numbers $l_j$, $m_{l,j}$, etc. (you only need to consider electrons that don't fill a &quot;shell&quot;) is the same as the number of &quot;real&quot; states (or microstates) that you find specifying the upper case quantum numbers $L$, $M_L$, etc. {\bf but} the wave functions are not the same. However you can transform from one representation to the other by a unitary operator. In the case of two electrons (or holes) in the outer open shell, the coefficients of this transformation are the well known Clebsch-Gordan coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular the micro-states specification is related to a single Slater determinant of atomic spin-orbitals, but open shell quantum states with well defined angular momentum typically require a sum of several Slater determinants. There are several simple techniques that show you how to find the right linear combination of Slater determinants. Gigacyan pointed to the most usual one. Since for the highest possible $M_L$ there is only one possible single determinant, you can generate the linear combination for the other $M_L$ states with the same $L$ by using ladder angular momentum operators. The same procedure can be used to generate the different $M_S$ states sharing equal $S$. Somewhat more generally, you can also use projection operators. See for instance L\&quot;owdin, Adv. Phys. 5, 1 (1956).&lt;/p&gt;&#xA;" OwnerUserId="3100" LastEditorUserId="3100" LastEditorDisplayName="" LastEditDate="2011-04-15T02:50:52.263" LastActivityDate="2011-04-15T02:50:52.263" />
  <row Id="8630" PostTypeId="2" ParentId="8622" CreationDate="2011-04-14T10:44:52.930" Score="6" ViewCount="" Body="&lt;p&gt;That's probably (no pun intended) because it's not a probability density at all. Wave functions by themselves don't have any physical interpretation. Only their squares do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of transition is computed like this&#xA;$$P_{1 \to 2} = {\left \Vert \left &amp;lt; \psi_1 | \psi_2 \right &amp;gt; \right \Vert^2 \over \left \Vert \psi_1 \right \Vert^2 \left \Vert \psi_2 \right \Vert^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you've written is just a density of the probability amplitude $\left&amp;lt;\psi_1|x\right&amp;gt;\left&amp;lt;x|\psi_2\right&amp;gt;$. It can be complex so it certainly cannot be probability in any sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit regarding your additional comments:&lt;/em&gt; assuming that $\left \Vert \psi_1 \right \Vert^2 = \left \Vert \psi_2 \right \Vert^2 = 1$ you could write $P_{1 \to 2}$ like this &#xA;$$P_{1 \to 2} = {\rm Tr} \left( \left|\psi_1\left&amp;gt; \right&amp;lt;\psi_1\right|  \left|\psi_2\left&amp;gt; \right&amp;lt;\psi_2\right| \right) = \int \psi_1(y) \psi_1^*(x)  \psi_2(x) \psi_2^*(y) {\rm d} x {\rm d} y$$&#xA;interpreting it as a trace of the product of projection operators. But this integral representation is basically never done, it's much easier to only compute the amplitude and square it when you're finished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a moral, you shouldn't insist on restricting your attention to some portion of position space. Quantum theory &quot;doesn't like it&quot; (except for few special constructions).&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-04-16T07:44:52.640" LastActivityDate="2011-04-16T07:44:52.640" CommentCount="11" />
  <row Id="8631" PostTypeId="2" ParentId="8425" CreationDate="2011-04-14T10:55:42.277" Score="1" ViewCount="" Body="&lt;p&gt;The keyword for porous flows is Darcy's flow, which is based on the Darcy law guiding the field of mass flux instead of velocity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{q}=-K\nabla p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec{q}:=\vec{u} \cdot \text{porosity}$, so it is equivalent to normal flow only for homogeneous media.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation comes from averaging the NS over porous volume; however, as you can see, it is very different from NS -- and this is just inevitable that such averaging will change the mathematical form of the problem. And, as in your case, when some more complex theory is needed, one should get through the averaging of NS with a set of assumptions that suit your problem (clues for instance &lt;a href=&quot;http://shyylab.engin.umich.edu/files/papers/AIAA-2007-5549-526.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2011-04-14T10:55:42.277" />
  <row Id="8632" PostTypeId="2" ParentId="8626" CreationDate="2011-04-14T11:10:59.610" Score="7" ViewCount="" Body="&lt;p&gt;Yes, this is the opposite of Noether's theorem. So let's call our conserved quantity $A$ (we will consider just one conserved quantity for starters) and begin with $\left \{H, A \right \} = 0$ law for conservation. Because of the connection between Poisson bracket with flows on the phase space this tells you both that $\mathcal{L}_{V_H} A$ = 0 ($A$ is conserved in time evolution) and $\mathcal{L}_{V_A} H = 0$ (Hamiltonian is conserved under symmetry that has fundamental field $V_A$) as long as the vectors fields associated to the functions on phase space as $V_X = ({\rm d} X)^{\sharp}$ are non-degenerate in certain sense. &lt;em&gt;Note that raising&lt;/em&gt; $(\cdot)^{\sharp}$ &lt;em&gt;operator is defined obviously using the inverse symplectic form&lt;/em&gt;. This e.g. means that for example genuine constants (which are certainly also constants of motion) won't work because ${\rm d} C = 0$ and we obtain zero vector field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, as long as all the conserved quantities are non-degenerate we can always find the associated symmetry flows by integration of the above mentioned vector fields. But note that what we receive in the end are symmetries of the &lt;strong&gt;phase space&lt;/strong&gt;. Whether these are also directly linked to some symmetry of the underlying position space (if there is such a space, that is; in general there need not be one but in usual applications we take the phase space of the position manifold $M$ as the cotangent bundle $T^*M$) is a question for further investigation. I'll try looking into it later, if I find some time.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-04-14T11:10:59.610" CommentCount="8" />
  <row Id="8633" PostTypeId="1" AcceptedAnswerId="8651" CreationDate="2011-04-14T13:09:12.317" Score="3" ViewCount="151" Body="&lt;p&gt;In the photoelectric effect there is a threshold frequency that must be exceeded, to observe any electron emission, I have two questions about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I) &lt;strong&gt;Lower than threshold&lt;/strong&gt;:&#xA;What happen with lesser frequency/energy photons? I mean there is no energy transferred? If some energy were transferred, (and it must be transfered quantized), how can be experimentally known it was quantized, if there is no electron emission?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;II) &lt;strong&gt;Intensity dependence&lt;/strong&gt;: There is any known dependency with intensity? (I mean a dependency not about the number of electrons &lt;em&gt;(I am already aware of this)&lt;/em&gt; but about the energy of them) I thought energy of electron emited was independent with intensity, but then I found this link &lt;a href=&quot;http://www.sciencedirect.com/science?_ob=ArticleURL&amp;amp;_udi=B6TVM-3WN77X7-16&amp;amp;_user=10&amp;amp;_coverDate=06/17/1996&amp;amp;_rdoc=1&amp;amp;_fmt=high&amp;amp;_orig=gateway&amp;amp;_origin=gateway&amp;amp;_sort=d&amp;amp;_docanchor=&amp;amp;view=c&amp;amp;_acct=C000050221&amp;amp;_version=1&amp;amp;_urlVersion=0&amp;amp;_userid=10&amp;amp;md5=4d0612bde6536d59f730cd066af81c23&amp;amp;searchtype=a&quot; rel=&quot;nofollow&quot; title=&quot;dependency with intensity&quot;&gt;dependency with intensity&lt;/a&gt; (is a paper to buy that I haven't read) but relates energy with intensity.&lt;/p&gt;&#xA;" OwnerUserId="1916" LastEditorUserId="1916" LastEditorDisplayName="" LastEditDate="2011-04-14T20:02:50.947" LastActivityDate="2011-04-14T20:18:58.347" Title="Is energy exchange quantized?" Tags="&lt;quantum-mechanics&gt;&lt;photons&gt;&lt;electron&gt;&lt;photoelectric-effect&gt;" AnswerCount="3" />
  <row Id="8634" PostTypeId="1" CreationDate="2011-04-14T13:19:44.123" Score="4" ViewCount="150" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How does one derive the superconformal algebra? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Especialy how to argue the existence of the operator $S$ which doesn't exist either in either the supersymmetric algebra or the conformal algebra? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the explanation of the convention of choosing If $J _ \alpha ^\beta$ as a generator of $SU(2) \times SU(2)$ and $R^A _B$ and $r$ are the generators of the $U(m)$ R-symmetry ( $m \neq 4$)? How does one get these $R$ and $r$? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;One further defines $R_1, R_2$ and $R_3$ as the generators of the Cartan subalgebra of $SU(4)$ (the $N=4$ R-symmetry group) where these are defined such that $R_k$ has $1$ on the $k^{th}$ diagonal entry and $-1$ on the $(k+1)^{th}$ diagonal entry. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Using the above notation apparently one of the $Q-S$ commutation looks like,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{ S^A_\alpha , Q^\beta _B\} = \delta ^A_B J^\beta _\alpha + \delta ^\beta _\alpha R^A_B + \delta ^A_B \delta ^\beta _\alpha (\frac{H}{2} + r \frac{4-m}{4m})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know how the above comes about and why the above equation implies the following equation (..each of whose sides is defined as $\Delta$) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2\{ Q_4 ^- , S^4_ - \} = H - 3J_3 + 2\sum _{k=1} ^ 3 \frac{k}{4} R_k$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't understand how to get the second of the above equations from the first of the above and why is this $\Delta$ an important quantity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;{Is there any open source online reference available which explains the above issues? &lt;/p&gt;&#xA;" OwnerUserId="1637" LastEditorUserId="429" LastEditorDisplayName="" LastEditDate="2011-04-15T07:48:16.877" LastActivityDate="2011-04-15T14:48:41.540" Title="The superconformal algebra" Tags="&lt;quantum-field-theory&gt;&lt;supersymmetry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="8635" PostTypeId="2" ParentId="8633" CreationDate="2011-04-14T13:24:59.697" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Lower than threshold:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The light will be reflected, scattered or absorbed. &#xA;Look at the optical properties of the material used &#xA;as cathode. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Intensity dependence:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is/was used in millions of photocells, camera tubes &#xA;and multipliers. The number of electrons is strictly &#xA;proportional to intensity. The link You gave is about &#xA;some &lt;strong&gt;circular polarized&lt;/strong&gt; light. Presumably not the &#xA;basics You want to understand. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-14T13:24:59.697" CommentCount="7" />
  <row Id="8636" PostTypeId="1" AcceptedAnswerId="8639" CreationDate="2011-04-14T13:29:23.450" Score="6" ViewCount="127" Body="&lt;p&gt;On calm days, the water in the toilet looks completely still. But when it's rainy and windy out, the water looks like it moves and pulsates. Why is this?&lt;/p&gt;&#xA;" OwnerUserId="1000" LastActivityDate="2011-04-14T14:04:36.817" Title="Why does the water in the toilet move around so much on stormy days?" Tags="&lt;water&gt;&lt;household-physics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8637" PostTypeId="2" ParentId="8633" CreationDate="2011-04-14T13:32:35.537" Score="2" ViewCount="" Body="&lt;p&gt;If the frequency of the incident radiation is lower than the threshold frequency then either a photon is fully absorbed or not absorbed at all. It is absorbed only if it has an energy which is just enough to excite the electron to the higher state but not enough (less than the work  function) to make the electron to leave the surface of the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way we know that the energy of the electron is quantized in an atom is by analyzing its spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a particular frequency above the threshold, the number of the electrons emitted will linearly increase with the intensity of the incident radiation.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-04-14T13:32:35.537" CommentCount="1" />
  <row Id="8638" PostTypeId="2" ParentId="8615" CreationDate="2011-04-14T14:02:42.397" Score="3" ViewCount="" Body="&lt;p&gt;The basic design difference between atmospheric and vacuum rocket design is its nozzle.&#xA;The rocket thrust equation is &lt;a href=&quot;http://www.braeunig.us/space/sup1.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.braeunig.us/space/sup1.htm&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F = q \times V_e + (P_e - P_a) \times A_e $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{eqnarray}&#xD;&#xA;&#xD;&#xA;F &amp;amp;= &amp;amp;\mbox{Thrust} \\&#xD;&#xA;q &amp;amp;= &amp;amp;\mbox{Propellant mass flow rate} \\&#xD;&#xA;V_e &amp;amp;= &amp;amp;\mbox{Velocity of exhaust gases} \\&#xD;&#xA;P_e &amp;amp;= &amp;amp;\mbox{Pressure at nozzle exit} \\&#xD;&#xA;P_a &amp;amp;= &amp;amp;\mbox{Ambient pressure} \\&#xD;&#xA;A_e &amp;amp;= &amp;amp;\mbox{Area of nozzle exit} \\&#xD;&#xA;&#xD;&#xA;\end{eqnarray} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $V_e$ and $P_e$ are inversely proportional, maximum thrust occurs when $P_e$ = $P_a$. Nozzle exit pressure ($P_a$) can be increased/decreased by smaller/larger exit nozzle areas ($A_e$). Unless one designs a variable exit area nozzle, a single stage rocket's fuel usage can be optimal only either on Earth or in space. Fortunately however, as Nick pointed out, there's little advantage for nozzle optimization. &lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="869" LastEditorDisplayName="" LastEditDate="2011-04-14T16:28:23.870" LastActivityDate="2011-04-14T16:28:23.870" CommentCount="2" />
  <row Id="8639" PostTypeId="2" ParentId="8636" CreationDate="2011-04-14T14:04:36.817" Score="7" ViewCount="" Body="&lt;p&gt;Any siphon in Your house will be affected in this way. &#xA;The inside surface (that you can see) is affected by the air pressure in Your house, &#xA;the other level is connected to the sewer piping system, wich gets &#xA;his pressure from a vent of the sewer system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both pressures can be affected in many ways by the stormy wind, &#xA;depending where some passages (like windows not really tight) &#xA;for the air are located , eg at the windward or leeward side of Your house. &#xA;The vent of the sewer system can develop pressure variations &#xA;depending on wind speed by some water-aspirator-like action. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the siphon in Your toilet works as a pressure differential indicator. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-14T14:04:36.817" CommentCount="1" />
  <row Id="8640" PostTypeId="1" CreationDate="2011-04-14T14:31:56.173" Score="2" ViewCount="171" Body="&lt;p&gt;UFOs as shown in movies are shown as disk like objects with raised centers that emit some sort of light from bottom. Can such a thing fly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My very limited knowledge in physics tell me that a disk like object may not be able to maneuver unless it has thrusters on sides and simple light can not be enough to make any object go up in the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible?&lt;/p&gt;&#xA;" OwnerUserId="3101" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-15T10:04:11.330" LastActivityDate="2011-04-15T10:04:11.330" Title="Can a disk like object (like UFO's) really fly?" Tags="&lt;gravity&gt;&lt;newtonian-gravity&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="8641" PostTypeId="2" ParentId="8599" CreationDate="2011-04-14T14:33:16.030" Score="6" ViewCount="" Body="&lt;p&gt;Answering the first question. The Chern-Simons functional expressed in terms of the vector potential $A$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(A) = \frac{k}{8 \pi^2} \int_M tr(A \wedge dA + \frac{2}{3} A \wedge A \wedge A )$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is well defined only for a trivial principal bundle, because in this case the gauge connection is &#xA;a Lie algebra one form over $M$. When $M$ is an oriented three manifold and the gauge group $G$ is compact connected and&#xA;simply connected, then every $G-$ bundle over $M$ is trivializable (see, for example the following &lt;a href=&quot;http://arxiv.org/PS_cache/hep-th/pdf/9206/9206021v1.pdf&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; by D. Freed)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case when the principal bundle is not trivial (for example when the above restrictions on the &#xA;gauge group are not satisfied) was treated by &lt;a href=&quot;http://citeseer.ist.psu.edu/viewdoc/summary?doi=10.1.1.128.1806&quot; rel=&quot;nofollow&quot;&gt;Dijgraaf and Witten&lt;/a&gt;&#xA;, using the standard construction by Witten: One can choose a 4-manifold $B$ such that $M$ is its boundary.&#xA;The CS action is defined: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(A) = \frac{k}{8 \pi^2} \int_B tr(F \wedge F)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This action is the proper definition of the CS action for the following two reasons&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It reduces to the usual CS action when the bundle is trivial (Using Stoke's theorem).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If k is an integer, the general form is independent modulo 1 of the extension $B$ of $M$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2190" LastActivityDate="2011-04-14T14:33:16.030" CommentCount="1" />
  <row Id="8642" PostTypeId="1" CreationDate="2011-04-14T14:36:21.007" Score="6" ViewCount="163" Body="&lt;p&gt;Nuclearity is a postulate in algebraic quantum field theory (AQFT). Basically, it says thermal states at any temperature always have a thermodynamic limit with extensive quantities. This is violated by string theory at the Hagedorn temperature. Does this mean anything?&lt;/p&gt;&#xA;" OwnerUserId="3102" LastActivityDate="2011-08-12T03:06:47.427" Title="Does the lack of modular nuclearity in string theory mean anything?" Tags="&lt;quantum-field-theory&gt;&lt;string-theory&gt;" AnswerCount="2" />
  <row Id="8643" PostTypeId="1" AcceptedAnswerId="8649" CreationDate="2011-04-14T14:55:46.343" Score="14" ViewCount="1121" Body="&lt;p&gt;I am not a physiologist, but whatever little I know about human eyes always makes me wonder by its details of optical subtleties. A question always comes to mind. Are human eyes the best possible optical instrument evolved by natural selection? Is there any possible room for further major improvement as per the laws of physics in principle? Can there be any camera prepared by human beings which can be superior to human eyes?&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-04-14T15:16:29.663" LastActivityDate="2011-04-15T00:57:52.467" Title="Are human eyes the best possible camera?" Tags="&lt;optics&gt;&lt;light&gt;&lt;biology&gt;&lt;technology&gt;" AnswerCount="7" CommentCount="5" FavoriteCount="3" />
  <row Id="8644" PostTypeId="2" ParentId="8643" CreationDate="2011-04-14T15:09:39.090" Score="5" ViewCount="" Body="&lt;p&gt;If You &lt;/p&gt;&#xA;&#xA;&lt;p&gt;think of our eyes as &quot;cameras&quot; they are not good at all, compared to &#xA;even the most simple camera. &#xA;Whether &quot;evolution&quot; could not do better, I don't know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture on the retina is very distorted and rather blurry, &#xA;exept for a very small portion in the center, and there is a big &#xA;&quot;blind spot&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quality of our visual perception is very high nevertheless, &#xA;but this is due to a lot of data processing in retina and brain. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-14T15:09:39.090" CommentCount="2" />
  <row Id="8645" PostTypeId="2" ParentId="8640" CreationDate="2011-04-14T15:15:38.660" Score="2" ViewCount="" Body="&lt;p&gt;Beam propulsion has already been done here on Earth.  :-)  &lt;a href=&quot;http://en.wikipedia.org/wiki/Beam-powered_propulsion&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Beam-powered_propulsion&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="919" LastActivityDate="2011-04-14T15:15:38.660" CommentCount="2" />
  <row Id="8646" PostTypeId="2" ParentId="8640" CreationDate="2011-04-14T15:15:46.213" Score="1" ViewCount="" Body="&lt;p&gt;You are right of course. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But please keep mind that those UFOs in cinema are theme of a kind of fairy tales. &#xA;Do You think about reality of the tent of Paribanu or the reality of &#xA;Arthurs sword? &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-14T18:18:36.437" LastActivityDate="2011-04-14T18:18:36.437" CommentCount="2" />
  <row Id="8647" PostTypeId="2" ParentId="8643" CreationDate="2011-04-14T15:36:37.407" Score="12" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;This is a common misconception -- evolution has not stopped a million years ago leaving all the creatures in the &quot;best possible state&quot;; it is a continuous pursuit of adapting to the current environmental conditions, with only aim in reproductive success. Moreover most of the population stays in even more suboptimal surroundings due to random mutations (the advantage of that is to achieve faster adaptation).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes; even a huge one, see Georg's answer. Squids have eyes without blind spot and with no risk of retina detach. Bees see light polarization and some UV. Cat's eye has a mirror that doubles the signal strength. Etc, etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Human eye has about 2Mpixel resolution -- my phone's camera is better.  Of course this is not a fair comparison, since the most usability of a sight stays in the brain, but if we talk about the eye itself...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-04-14T15:59:58.800" LastActivityDate="2011-04-14T15:59:58.800" CommentCount="11" />
  <row Id="8648" PostTypeId="2" ParentId="8643" CreationDate="2011-04-14T15:38:04.630" Score="11" ViewCount="" Body="&lt;p&gt;[Updated to correct a couple of mistakes pointed out in comments. Thanks!]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At my age, it's clear that there's room for at least one major improvement: more accommodation. Accommodation, in this context, means the ability of the eye to focus at different distances. This is accomplished by changing the shape, and hence the focal length, of the lens. The lens material gets stiffer over time, so that most people lose the ability to focus on nearby objects as they age. In addition, of course, many people, including me, lack the ability to focus on faraway objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were in charge of natural selection, I'd make those muscles strong enough to let me focus on nearby objects. Presumably, back in the savannah, the decreasing ability of middle-aged people (or, as they were known back then, very old people!) to focus at objects less than a meter away had little effect on the number of descendants those people produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another natural figure of merit to use to assess any optical system is how close it is to the diffraction limit -- that is, how close it comes to attaining the best possible resolution for a system of its size. I'm not an expert, it looks like the eye is within a factor of a few of the diffraction limit (&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/raylei.html&quot; rel=&quot;nofollow&quot; rel=&quot;nofollow&quot;&gt;example&lt;/a&gt;), so there's a bit of room for improvement there, but not much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to look at is the quantum efficiency, effectively the fraction of all incident photons that an optical system uses. Again, things depend a lot on conditions, but in general the eye's very good on this score, reaching levels of about &lt;a href=&quot;http://astronomy.ua.edu/keel/techniques/eye.html&quot; rel=&quot;nofollow&quot; rel=&quot;nofollow&quot;&gt;10%&lt;/a&gt;. (That may not sound so great, but I think it's hard to imagine ways of improving it dramatically.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to improve the eye would be to boost the range of frequencies it could see. Going into the ultraviolet probably wouldn't do much good, but if you added some infrared receptors on your retina, your night vision would be much better.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-04-14T18:43:29.080" LastActivityDate="2011-04-14T18:43:29.080" CommentCount="6" />
  <row Id="8649" PostTypeId="2" ParentId="8643" CreationDate="2011-04-14T15:42:46.443" Score="28" ViewCount="" Body="&lt;p&gt;This question is sort of difficult to answer in an objective way, because it depends very strongly on your definition of &quot;best.&quot; Natural selection favors traits which provide a reproductive advantage; no more, no less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could our eyes be better by the standards of modern optical design, in terms of precision and features? Sure. I could easily design a camera with a larger aperture, better resolution, less abberation, a broader wavelength sensitivity, etc. There are even some precedents for this in nature. Some animals can see ultraviolet or infrared; some can even detect polarization (mostly birds I think?). I believe there are some fish that can swivel their eyes around through nearly a full circle in any direction. Cats and dogs have higher sensitivity in low light due to the reflective layer behind their retina, and squid have eyes with truly huge apertures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worth pointing out though that the human eye, despite what I've said above, is still a pretty good camera. It can re-focus pretty quickly, it's got a fantastic image processing computer attached to it that can do incredible amounts of pattern recognition, noise reduction, and image stabilization. While its wavefront aberration is not perfect, it is pretty good for a biological system. They have very low chromatic aberration, and are quite compact. Additionally, our eyes have a curved focal plane (our retina) which is a lens designers dream -- it eliminates the problem of field curvature, which is an inherent aberration in any optical system that is nearly impossible to eliminate.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-04-14T15:42:46.443" CommentCount="9" />
  <row Id="8651" PostTypeId="2" ParentId="8633" CreationDate="2011-04-14T17:16:52.553" Score="3" ViewCount="" Body="&lt;p&gt;Energy exchange is quantized when moving a electron from one &lt;em&gt;bound&lt;/em&gt; state to another &lt;em&gt;bound&lt;/em&gt; state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't because the exchange is inherently quantized, but because the states the electron may occupy are quantized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the standard photo-electric effect in which a photon can not excite an atom unless it has a minimum energy.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;However,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There &lt;em&gt;are&lt;/em&gt; multi-photon processes by which sub-threshold light can excite transitions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cross-section for them go by intensity-squared (or worse) and are very small for any reasonable light intensity. To study or employ them you get powerful, short pulsed laser systems. Where short pulsed means nano-second or faster pulses and powerful means &lt;em&gt;&quot;Do not look into beam with remaining eye&quot;&lt;/em&gt;. Even then you don't get a lot of rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;These processes are &lt;strong&gt;utterly&lt;/strong&gt; negligible for the kind of benchtop experiment we use to teach the photoelectric effect:&lt;/em&gt; you just can't get enough intensity. (See below for how negligible.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conceptual model here is that the first photon bumps the electron to a short-lived, unstable state without well defined quantum numbers, and the second comes along &lt;em&gt;before&lt;/em&gt; that state decays and finishes the job. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're currently exploring the application of such a process to calibrating light yields, opacities in a large volume of scintillating material.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From &lt;em&gt;&lt;a href=&quot;http://arxiv.org/abs/1011.6001&quot; rel=&quot;nofollow&quot;&gt;New J.Phys.12:113024, 2010&lt;/a&gt;&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For gases the one-photon absorption cross-section $\sigma_1$ is typically of the order of $10^{−17}\text{ cm}^2$, &#xA;  whereas the two-photon and the three-photon cross-sections are of the order of &#xA;  $\sigma_s = W/F_2 \approx 10^{−50}\text{ cm}^4\text{ s}$ &#xA;  and &#xA;  $\sigma_3 = W/F_3 \approx 10^{−83}\text{ cm}^6\text{ s}^2$, &#xA;  respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where $F$ is intensity in photons/second and W is excitation rate in reciprocal seconds.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-04-14T20:18:58.347" LastActivityDate="2011-04-14T20:18:58.347" CommentCount="5" />
  <row Id="8653" PostTypeId="1" CreationDate="2011-04-14T17:47:21.950" Score="4" ViewCount="187" Body="&lt;p&gt;Is there the smallest particle that can be guaranteed to be unable to be broken down into smaller particles?&lt;/p&gt;&#xA;" OwnerUserId="3106" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-14T17:56:47.463" LastActivityDate="2011-04-14T18:07:28.650" Title="Is there the smallest particle that can be guaranteed to be unable to be broken down into smaller particles?" Tags="&lt;particle-physics&gt;&lt;atoms&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="8654" PostTypeId="2" ParentId="8653" CreationDate="2011-04-14T17:57:48.167" Score="2" ViewCount="" Body="&lt;p&gt;A particle can only decay into particles with smaller (rest) masses than itself. So the lowest-mass particle can't decay. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are zero-mass particles, that argument by itself doesn't suffice to show that there are any massive particles that can't decay, but we can strengthen the argument a bit to show that there are. Because charge is conserved, the decay products of a charged particle must include at least one charged particle. So the lowest-mass charged particle can't decay. That's why electrons are stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same argument applies to any conserved quantity, not just electric charge. In (some? all?) supersymmetric theories, there is a conserved quantity called R-parity. The lightest particle with nontrivial R-parity must be stable. That means that such theories always have an additional stable particle, besides the ones we know about. That particle is often called the lightest supersymmetric particle (LSP). In these theories, particles like this should have been produced in the early Universe. Since the LSP can't decay, it should still be around today. The LSP is one of the most popular candidates to be the dark matter in the Universe.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-14T17:57:48.167" />
  <row Id="8655" PostTypeId="2" ParentId="8653" CreationDate="2011-04-14T17:58:15.187" Score="6" ViewCount="" Body="&lt;p&gt;In particle physics, you have to be specific about what you mean by &lt;em&gt;breaking down&lt;/em&gt; a particle: A neutron in an atom can decay into a proton, an electron, and an electron-antineutrino. &lt;strong&gt;But&lt;/strong&gt; this does not mean that a neutron is &lt;em&gt;made of&lt;/em&gt; a proton, an electron and an electron-antineutrino. What the neutron &lt;em&gt;is&lt;/em&gt; made of are three quarks (one up and two down). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If by &lt;em&gt;unable to break down into smaller particles&lt;/em&gt; you mean that a particle has no internal structure, then our current understanding is that electrons, muons, taus, their corresponding neutrinos, all the quarks, as well as the photon, the gluon, the Z- and the W-boson, together with all the anti-particles, are all &lt;strong&gt;elementary particles&lt;/strong&gt;. ,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this be guaranteed? No. Nothing really can be &lt;em&gt;guaranteed&lt;/em&gt; in physics. Ever finer measurements mean that we are ever more sure about the elementary nature of, e.g., the electron. We can conclude that at the length-scales currently accessible for probing, the electron has no internal structure. This does not rule out that a structure might be found at tinier scales. This is what happened with protons and neutrons, where scattering experiments gave results that were inconsistent with assuming a point-particle with no internal structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, although the aforementioned particles are counted as elementary particles, they &lt;em&gt;can&lt;/em&gt; decay: See &lt;a href=&quot;http://en.wikipedia.org/wiki/Muon#Muon_decay&quot; rel=&quot;nofollow&quot;&gt;Muon Decay&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-04-14T17:58:15.187" />
  <row Id="8656" PostTypeId="1" AcceptedAnswerId="8660" CreationDate="2011-04-14T18:03:41.627" Score="4" ViewCount="74" Body="&lt;p&gt;My favorite physics 101 textbook (Giancoli) explains precession in terms of a spinning top whose axis is tilted from the vertical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way the book sets things up, $L$ (angular momentum) points along the top's axis, and $\operatorname{d}\!L$ is perpendicular and horizontal to $L$, pointing in the direction of precession (in their figure, $\operatorname{d}\!L$ ends up pointing to the right). It then goes on to state that since $\operatorname{d}\!L$ is always perpendicular to $L$, $L$'s magnitude is constant, and only its direction changes. Finally, it concludes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since $L$ points along the axis of the top, we see that this axis moves to the right... (Giancoli, 2008)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I just don't understand why you can assume that it will always be true that $L$ and the axis of the top will always lie on top of one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book doesn't address this, but is there only one way in which the top can move in order to produce the required motion of the angular momentum vector? If that is the case, then I guess it would make more sense? I dunno. Help me out here folks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I know that there is a deviation in $L$ due to the precessional motion itself, but another one of the book's assumptions is that this is irrelevant as long as the top is spinning quickly enough.)&lt;/p&gt;&#xA;" OwnerUserId="2820" LastEditorUserId="1136" LastEditorDisplayName="" LastEditDate="2011-04-14T22:07:26.150" LastActivityDate="2011-04-14T22:16:25.810" Title="Why can you assume that the angular momentum vector of a top will always track its axis of rotation?" Tags="&lt;classical-mechanics&gt;&lt;mechanics&gt;&lt;precession&gt;" AnswerCount="1" />
  <row Id="8657" PostTypeId="2" ParentId="8653" CreationDate="2011-04-14T18:07:28.650" Score="1" ViewCount="" Body="&lt;p&gt;I would like to add to the previous answers that compositeness of the elementary particles of the standard model is an ongoing research question, not favored by theoriticians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CMS has a &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1102/1102.2020v1.pdf&quot; rel=&quot;nofollow&quot;&gt;preprint out&lt;/a&gt; where they are searching for compositeness in dijet angular distributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The measured dijet angular distributions can be used to set limits on quark compositeness represented by a four-fermion contact interaction term in addition to the QCD Lagrangian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have set limits.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-14T18:07:28.650" />
  <row Id="8658" PostTypeId="2" ParentId="8640" CreationDate="2011-04-14T18:19:14.547" Score="1" ViewCount="" Body="&lt;p&gt;Have a look at the &lt;a href=&quot;http://news.bbc.co.uk/2/hi/6970031.stm&quot; rel=&quot;nofollow&quot;&gt;flying car&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is even a &lt;a href=&quot;http://www.liveleak.com/view?i=b2e_1188457915&quot; rel=&quot;nofollow&quot;&gt;video.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-14T18:19:14.547" />
  <row Id="8659" PostTypeId="1" AcceptedAnswerId="8662" CreationDate="2011-04-14T18:44:58.487" Score="7" ViewCount="279" Body="&lt;p&gt;I understand that the closer something travels to the speed of light, that time will stretch by a factor, and distance will compress by the same factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, if something travels in a circle, close to the speed of light, what does the distance of the journey look like to them? They measure that the trip took them 10 minutes. And an outside observer says that the journey took 20 minutes, and the outside observer measured that they did, for example, 1000 laps of a circle circumference 1000 km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the plane had a distance trip counter, what would it read? And if they were looking out of the window, would the circle still look like it had a circumference of 1000km?&lt;/p&gt;&#xA;" OwnerUserId="3109" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-11T05:11:09.223" LastActivityDate="2011-05-11T05:11:09.223" Title="Invariant spacetime - distance - Circular Motion" Tags="&lt;general-relativity&gt;&lt;special-relativity&gt;&lt;spacetime&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" />
  <row Id="8660" PostTypeId="2" ParentId="8656" CreationDate="2011-04-14T19:11:44.793" Score="3" ViewCount="" Body="&lt;p&gt;Essentially it is because the top is a rigid body with a well defined symmetry axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examine the definition of angular momentum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec{L} = \int_V dV  \left( \vec{r} \times \vec{v} \right) \rho(\vec{r}) $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choosing a (instantaneously) cylindrical coordinate system with $\vec{z}$ coincident with the axis of the top, we see that $\vec{v}$ is always in the $\phi$ direction so the angular momentum &lt;em&gt;due to the spinning&lt;/em&gt; always points in the azimuthal direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec{L} = \int_V dV r^2 \omega \left( \hat{r} \times \hat{\phi} \right)  \rho(\vec{r})&#xD;&#xA;           = -\omega \int_V dV  r^2 \rho(\vec{r})  \hat{\theta} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;by taking advantage of the symmetry of the top, we find that only the $\hat{z}$ component contributes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec{L} = -\omega \hat{z} \int_V dV  r^2 \sin^2(\theta) \rho(\vec{r})   $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have already identified the first correction, and as the book says if the top spins fast enough (and precesses slowly enough) we can neglect that.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-04-14T22:16:25.810" LastActivityDate="2011-04-14T22:16:25.810" CommentCount="2" />
  <row Id="8661" PostTypeId="2" ParentId="8643" CreationDate="2011-04-14T19:15:57.040" Score="4" ViewCount="" Body="&lt;p&gt;This question can’t be answered exactly properly.  The eyes we humans have evolved to be “about” the best available to us hominids.  There are other species of life with eyes better adapted for their econiches.  Eagles and hawks have eyes capable of far better resolution at great distances.  They can soar 500 meters above the ground and spot a small animal for prey, where upon they dive from that height to snatch it.  Our eyes are nowhere near that sharp.  On the other hand their eyes may not be as adapted for fine resolution of very close vision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Human eyes have some funny aspects to them which illustrate how evolution selects for adaptive feature built from previously existing structures.  For instance, the optical neuron innervates the eye from the front, which is quite the opposite of what one might expect.  It would be as if one built a CCD with the pixel connections connected in the optical path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best answer which can be given is no, but the eye is evolved to work within the limits of what is biologically adaptable.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-04-14T19:15:57.040" />
  <row Id="8662" PostTypeId="2" ParentId="8659" CreationDate="2011-04-14T19:46:42.560" Score="7" ViewCount="" Body="&lt;p&gt;Circular motion in special relativity is somewhat tricky: Note that for circular motion, the acceleration in the spaceship travelling in a circle is &lt;em&gt;not&lt;/em&gt; zero, so the spaceship is not in a single frame of inertia. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an interesting thought: Distances &lt;em&gt;perpendicular&lt;/em&gt; to the direction of motion are not subject to contraction. Hence, if the observes on earth see the spaceship going on a circle with radius $R$, then in their frame of inertia the spaceship is always a distance $R$ away from the earth. Since the line from earth to spaceship is perpendicular to the direction of flight, the people on the spaceship will &lt;em&gt;also&lt;/em&gt; believe that they are always a distance $R$ away from earth, so they will also fly on a circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They will, nonetheless, experience a different circumference! The best way to solve this is consider a polygon with $N$ sites and then letting $N$ go to infinity. If people on the earth measure each side of the polygon as $L_0/N$ where $L_0$ is the circumference of the polygon in the earth's frame of inertia, then the spaceship-people will measure each side to be &#xA;$L_0/(N\gamma)$. Hence, for $N \rightarrow \infty$, the polygon becomes a circle. Measured from earth, it has circumference $L_0$, but for the spaceship it has circumference $L_0/\gamma$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This suggests that the spaceship moves through non-Euclidean geometry, because it travels on a circle whose ratio between circumference and diameter is less than $2\pi$. This is a hint that accelerating frames have non-Euclidean geometry, which is excessively treated in General Relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Reference&lt;/em&gt;: &lt;a href=&quot;http://abacus.bates.edu/~msemon/WortelMalinSemon.pdf&quot; rel=&quot;nofollow&quot;&gt;http://abacus.bates.edu/~msemon/WortelMalinSemon.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-04-14T19:46:42.560" CommentCount="4" />
  <row Id="8663" PostTypeId="1" AcceptedAnswerId="8705" CreationDate="2011-04-14T20:16:43.303" Score="6" ViewCount="279" Body="&lt;p&gt;In Feynmans book &quot;Quantum Mechanics and Path Integrals&quot; Feynman states that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the probability $P(b,a)$ to go from point $x_a$ at time $t_a$ to the point $x_b$ at the time $t_b$ is $P(b,a) = \|K(b,a)\|^2$ of an amplitude $K(b,a)$ to go from $a$ to $b$. This amplitude is the sum of contributions $\phi[x(t)]$ from each path. $$ K(b,a) = \sum_{\text{paths from $a$ to $b$}} \phi[x(t)]$$ The contributions of a path has a phase proportional to the action $S$: $$ \phi[x(t)] = \text{const}\ e^{(i/\hbar)S[x(t)]}$$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why must the contribution of a path be $\sim e^{(i/\hbar)S[x(t)]}$? Can this be somehow derived or explained? Why can't the contribution of a path be something else e.g. $\sim \frac{S}{\hbar}$, $\sim \cos(S/\hbar)$, $\log(S/\hbar)$ or $e^{- (S[x(t)]/\hbar)^2}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I have to admit that in the first version of this question, I didn't exclude the possibility to derive the contribution of a path directly from Schrödingers equation. So answers along this line are valid although not so interesting. I think when Feynman developed his formalism his goal was to find a way to quantize systems, which cannot be treated by Schrödingers equation, because they cannot be described in terms of a Hamiltonian (e.g. the Wheeler-Feynman absorber theory). So I think a good answer would explain Feynmans Ansatz without referring to Schrödingers equation, because I think Schrödingers equation can only handle a specific subset of all the systems that can be treated by Feynmans more general principle. &lt;/p&gt;&#xA;" OwnerUserId="1648" LastEditorUserId="1648" LastEditorDisplayName="" LastEditDate="2011-04-15T19:42:32.660" LastActivityDate="2011-04-15T19:42:32.660" Title="Why is the contribution of a path in Feynmans path integral formalism $\sim e^{(i/\hbar)S[x(t)]}$" Tags="&lt;quantum-mechanics&gt;&lt;path-integral&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="2" />
  <row Id="8664" PostTypeId="2" ParentId="506" CreationDate="2011-04-14T20:18:30.267" Score="5" ViewCount="" Body="&lt;p&gt;A report appeared in &lt;em&gt;Science&lt;/em&gt; today which addresses this exact question: Kooijman et al., &lt;em&gt;Science&lt;/em&gt; &lt;strong&gt;332&lt;/strong&gt; (6027): 339-342, &quot;&lt;a href=&quot;http://www.sciencemag.org/content/332/6027/339.abstract&quot; rel=&quot;nofollow&quot;&gt;A Bicycle Can Be Self-Stable Without Gyroscopic or Caster Effects&lt;/a&gt;.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The abstract reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A riderless bicycle can automatically steer itself so as to recover from falls. The common view is that this self-steering is caused by gyroscopic precession of the front wheel, or by the wheel contact trailing like a caster behind the steer axis. We show that neither effect is necessary for self-stability. Using linearized stability calculations as a guide, we built a bicycle with extra counter-rotating wheels (canceling the wheel spin angular momentum) and with its front-wheel ground-contact forward of the steer axis (making the trailing distance negative). When laterally disturbed from rolling straight, this bicycle automatically recovers to upright travel. Our results show that various design variables, like the front mass location and the steer axis tilt, contribute to stability in complex interacting ways.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is also a &lt;a href=&quot;http://news.sciencemag.org/sciencenow/2011/04/how-to-keep-a-riderless-bike-fro.html&quot; rel=&quot;nofollow&quot;&gt;blurb in ScienceNOW&lt;/a&gt; that you can read without subscription.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://www.scienceblogs.de/geograffitico/StableBicyclev34Revised.pdf&quot; rel=&quot;nofollow&quot;&gt;free-to-read preprint&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-04-15T20:30:17.287" LastActivityDate="2011-04-15T20:30:17.287" CommentCount="4" />
  <row Id="8665" PostTypeId="2" ParentId="8643" CreationDate="2011-04-14T20:40:13.017" Score="3" ViewCount="" Body="&lt;p&gt;a few different answers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;no, it's easy to imagine small improvements: more regular lenses, wider spectrum sensibility, more than three color sensors...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;not only not the best possible, but also have some fortuitous bad choices, the most commented is the retina orientation that makes it easy to detach with a big impact.  the very similar but independently evolved squid eye has the retina backwards, making it a lot more resistant.  unfortunately there seems no way for evolution to flip it without many blind generations in between. obviously not a good evolution path...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;best for what?  for producing images that will be seen by whom?  IOW: if the end observer is a pair of human eyes, then it will always be better to observe directly than to use an intermediary, no matter how high the quality/resolution/contrast/etc of such an intermediate reproduction.  (unless you want to see things not visible to the naked eye)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2034" LastActivityDate="2011-04-14T20:40:13.017" />
  <row Id="8666" PostTypeId="2" ParentId="8663" CreationDate="2011-04-14T21:15:16.123" Score="0" ViewCount="" Body="&lt;p&gt;I think the justification goes like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;first a couple observations from the classical limit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) paths that are far from the classical solution are not near an extremum value of the action, which means that the action will have a non-zero variance in all paths that are neighbour to this path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) the classical solution itself is an extremum value of the action (either a minimum or a maximum)&#xA;in the space of classical paths. in the neighbourhood of this path, the variance of the paths will approach zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so, an approach to construct a quantum limit would be to think on the double-slit experiment and see that the interference pattern is constructed from taking two plane waves paths that go from the source to each slit, and then from the slit to the a point in the interference screen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, none of the paths match exactly the classical path. if you write a plane wave you'll see that the argument $p x - E t$ and you'll notice that this is actually the action of a free particle. So you can think of the De Broglie wave as a plane wave with the action of a free particle $e^{i(kx - \omega t)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, is just a small step to just infer that in general, when paths are not restricted by a double slit, you need to allow for all classical paths possible, and when the action is more complex than a free particle, you need to replace the wavefunction argument by the action of the path&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-04-14T21:15:16.123" />
  <row Id="8667" PostTypeId="2" ParentId="8517" CreationDate="2011-04-14T22:22:20.287" Score="1" ViewCount="" Body="&lt;p&gt;In the limit of infinitesimal secondary masses the problem is clearly stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So consider the problem of how the secondary (planetary) masses interact with each other. For simplicity, let's make the situation symmetric about the plane midway (45 degrees) between the two planetary orbits. To minimize their interaction, let's make them 180 degrees out of phase so that one is at apogee while the other is at perigee.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we treat the interaction between the planets as a perturbation, the system is stable before we include the perturbation (of course). Since the orbital periods are the same, the perturbation is stable. The question is &quot;can we arrange for the perturbation to be zero?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an interesting problem and one that it appears I will have to simulate in Java.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-04-14T22:22:20.287" />
  <row Id="8668" PostTypeId="2" ParentId="8663" CreationDate="2011-04-14T23:10:52.847" Score="4" ViewCount="" Body="&lt;p&gt;You start by writing down the probability to find a particle at $y$ at time $t$ when it was at $x$ at time $0$, denoted as $K(y,t;x,0)$. &#xA;You get this by solving the Schrödinger equation with the initial condition $\psi(y,0) = \delta(y-x)$. Then, $K(y,t;x,0) = \psi(y,t)$.&#xA;Thus, to solve this, we need to know the time development of $\psi(y,t0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us start with the simple example of a &lt;strong&gt;free particle&lt;/strong&gt;. This is easiest solved in momentum-representation, obtained by Fourier-transforming $\psi(y,t)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi(y,t) = \frac{1}{\sqrt{2\pi\hbar}} \int dp exp(ipy/\hbar) \tilde \psi(p,t)$$&#xA;For $\tilde \psi$, the Schrödinger equation gives &#xA;$$\tilde \psi(p,t) = \frac{1}{\sqrt{2\pi\hbar}} \exp \left(-\frac{i}{\hbar} &#xD;&#xA;\left[\frac{p^2 t}{2m} - px\right]\right)$$&#xA;This can be inserted back into the equation for $\psi(y,t)$. The integral over $p$ can be solved exactly. The final result is &#xA;$$K_\text{free}(y,t;x,0) = \sqrt{\frac{m}{2\pi i\hbar t}} \exp\left(\frac{im(x-y)^2}{2\hbar t}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next step: The solution of the Schrödinger equation can generally be written as &#xA;$$|\psi, t\rangle = \exp\left(-\frac{iHt}{\hbar}\right) |\psi,0\rangle$$&#xA;with $H$ being the Hamiltonian of your system. &#xA;Writing $H = T+V$, the general formula for $K$ becomes&#xA;$$K(y,t;x,0) = \langle y \mid \exp(-\frac{i(T+V)t}{\hbar}) \mid x \rangle$$&#xA;We use the &lt;strong&gt;Trotter-Kato Formula&lt;/strong&gt; (which holds under certain conditions which I won't go into detail at this point. It allows us to write &#xA;$$K(y,t;x,0) = \lim_{N\rightarrow \infty} \langle y \mid \left[ \exp(-\frac{iTt}{N\hbar})&#xD;&#xA;\exp(-\frac{iVt}{N\hbar})\right]^N \mid x\rangle$$&#xA;We insert the unity operator, decomposed as $1 = \int dx | x \rangle \langle x |$ $N-1$ times, which gives us &#xA;$$K(y,t;x,0) = \int dx_1 dx_2 \dots dx_{N-1} \prod_{j=0}{N-1}&#xD;&#xA;\langle x_{j+1} \mid \exp(-iTt/N\hbar) \exp(-iVt/N\hbar) \mid x_j \rangle$$&#xA;Note that $V$ as an operator acting on $|x\rangle$ gives just $V(x) |x\rangle$. And &#xA;$\langle x_{j+1} | \exp(-iTt/N\hbar) | x_j \rangle$ gives us just the contribution of a free particle, i.e. $$\sqrt{\frac{mN}{2\pi i\hbar t}} \exp(\frac{imN}{2\hbar t}(x_{j+1} - x_j)^2$$.&#xA;If we abbreviat $\tau = t/N$, we can write:&#xA;$$K(y,t;x,0) = \lim_{N\rightarrow \infty} \int dx_1 dx_2 \dots dx_{N-1} &#xD;&#xA;\left( \frac{mN}{2\pi i\hbar t}\right)^{N/2} \times$$&#xA;$$\exp \left(\frac{i\tau}{\hbar} \sum_{j=0}^{N-1} \left[ \frac{m}{2}\left(\frac{x_{j+1}-x_j}{\tau}\right)^2 - V(x_j)\right]\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is to see the values $x_j$ as points of a certain path $x(t&amp;#39;)$ evaluated at points &#xA;$t&amp;#39; = t_j = j\tau = jt/N$. If $\tau$ is small, we write &#xA;$$\sum_{j=0}^{N-1} \tau f(t_j) \rightarrow \int f(t&amp;#39;) dt&amp;#39;$$&#xA;$$\frac{x_{j+1} - x_j}{\tau} \rightarrow \dot x(t&amp;#39;)$$&#xA;where the dot denotes the time-derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument of the exponential then becomes &#xA;$$\frac{i}{\hbar} \int_0^t dt&amp;#39; \left( \frac{m\dot x(t&amp;#39;)^2}{2} - V(x(t&amp;#39;))\right)$$&#xA;You will have no trouble identifying the integrand as the &lt;strong&gt;Lagrangian&lt;/strong&gt; $L = T-V$. The integral itself, therefore, is the &lt;strong&gt;classical action&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the formula we have for $K$ can be interpreted as the sum over all possible paths from $(x,0)$ to $(y,t)$ of the function $\exp(\frac{i}{\hbar} S(t,0)$ of the classical action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;interpretation&lt;/em&gt; of this was given in other answers: The classical path is that which minimizes the action, i.e. the action is stationary for the classical path. In your path-integral formula, this path will have a large contribution, as all paths that vary only slightly from the classical path will still have pretty much the same phase factor as the classical one, leading to constructive interference of those paths. For paths far from the classical path, the action will vary greater among the paths, so that there all possible phases occur, which will ultimately cancel each outer out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference&lt;/strong&gt; A lecture on advanced quantum mechanics given by Prof. Crispin Gardiner. Lecture notes are, unfortunately, not freely available. It was a good lecture :)&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-04-14T23:10:52.847" CommentCount="9" />
  <row Id="8669" PostTypeId="2" ParentId="8663" CreationDate="2011-04-14T23:13:28.133" Score="3" ViewCount="" Body="&lt;p&gt;If you accept that Quantum Mechanics is built upon the fact that you sum complex amplitudes of processes &lt;a href=&quot;http://physics.stackexchange.com/questions/8062/about-the-complex-nature-of-the-wave-function&quot;&gt;(see this previous Question/Answers about this fact)&lt;/a&gt; you would expect that a sum over multiple paths behaves like a sum of different complex phases: $$M \sim \sum e^{i*phase}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the variational principle to the phase, you see that the paths which vary their phase the least will contribute the most to the sum (because the others will average each other). Add the fact that you want the classical path to be the main contribution (because we want to match classical physics, this is the correspondence princple), and that the classical path is the path where the &lt;em&gt;action&lt;/em&gt; $S$ varies the least, you can identify the phase with the action and get $phase \sim S[x(t)]$. Then you get $1 / \hbar$ as an experimental constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if this is a satisfactory answer, but most of the &quot;strangeness&quot; here comes from the QM superposition principle in the first place anyway. Note that the variational principle in classical mechanics was known and used before QM was invented and had the teleological property of &quot;sniffing out&quot; paths of least action. In the QM path-integral method this is at least explained from a more local point of view.&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-04-14T23:13:28.133" />
  <row Id="8670" PostTypeId="2" ParentId="7395" CreationDate="2011-04-14T23:47:06.500" Score="1" ViewCount="" Body="&lt;p&gt;The &quot;warmer&quot; white-color of the filament (incandescent) bulb is usually considered more relaxing than the &quot;colder&quot; white of tubes and certain white LEDs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the main ergonomic drawback of tubes is that many of them flicker. Some people are more sensitive to this than others. Some tubes have high-frequency modulators which greatly reduce or remove this effect. There is a very slight emittance in the UV-spectrum from tubes as well which could affect people wiht a photosensitive condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since tube-lighting is used in virtually all offices and industries I don't think there are any particular regulations limiting their use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the wikipedia entries for &lt;a href=&quot;http://en.wikipedia.org/wiki/Fluorescent_lamp&quot; rel=&quot;nofollow&quot;&gt;tube lighting&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Color_temperature&quot; rel=&quot;nofollow&quot;&gt;color temperature&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-04-14T23:47:06.500" CommentCount="2" />
  <row Id="8671" PostTypeId="1" CreationDate="2011-04-15T00:11:05.800" Score="0" ViewCount="57" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/8659/invariant-spacetime-distance-circular-motion&quot;&gt;Invariant spacetime - distance - Circular Motion&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I understand that the closer something travels to the speed of light, that time will stretch by a factor, and distance will compress by the same factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, if something travels in a circle, close to the speed of light, what does the distance of the journey look like to them? They measure that the trip took them 10 minutes. And an outside observer says that the journey took 20 minutes, and the outside observer measured that they did, for example, 1000 laps of a circle circumference 1000 km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the plane had a distance trip counter, what would it read? And if they were looking out of the window, would the circle still look like it had a circumference of 1000km?&lt;/p&gt;&#xA;" OwnerUserId="3120" LastActivityDate="2011-04-15T00:11:05.800" ClosedDate="2011-04-15T00:29:26.057" Title="A question about the relativity of time" Tags="&lt;general-relativity&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" />
  <row Id="8672" PostTypeId="2" ParentId="8663" CreationDate="2011-04-15T00:38:44.463" Score="1" ViewCount="" Body="&lt;p&gt;An approach similar to Lagerbaer’s may be formulated without reference to the probability function.  The overlap between states at different times $\langle\psi_{t^\prime}|\psi_t\rangle$ may be written according to a product of $|q_{t+n\delta t}\rangle\langle q_{t+n\delta t}$ at different time slices.  The wave overlap is then&#xA;$$&#xD;&#xA;\langle\psi_{t^\prime}|\psi_t\rangle~=~\lim_{\delta t\rightarrow 0}\lim_{N\rightarrow\infty}&#xD;&#xA;\prod_{n=0}^N \int dq_{t+n\delta t}\langle\psi_{t+n\delta t}|q_{t+n\delta t}\rangle\langle q_{t+(n-1)\delta t}|\psi_{t+(n-1)\delta t}\rangle.&#xD;&#xA;$$&#xA;This description of the overlap is then according to snapshots determined by projectors, where in the limit the time increment vanish they recover the density matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now focus on a product defined on one particular time slice. Each infinitesimal overlap is written as&#xA;$$&#xD;&#xA;\psi^*(q,t)\psi(q,t~-~\delta t)~=~ \psi^*(t)\Big(\psi(q,t)~+~\delta t{\frac{d\psi}{dt}}(q,t)~+~O(\delta t^2)\Big).&#xD;&#xA;$$&#xA;The term to $O(\delta t)$ is easily seen to be&#xA;$$&#xD;&#xA;\delta t {\frac{d\psi}{dt}}(q,t) ~=~\delta t \Big(\frac{\partial\psi}{\partial t}(q,t)~+~ \frac{dq}{dt}\nabla \psi(q,t)\Big)~=~\frac{i\delta t}{\hbar}\Big(\frac{dq}{dt} p~-~ H\psi(t)\Big).&#xD;&#xA;$$&#xA;The integrand of the infinitesimal overlap is&#xA;$$&#xD;&#xA;\psi^*(q,t)\psi(q,t~-~\delta t)~=~ e^{\frac{i\delta t}{\hbar} \big({\dot q} p~-~H\big)}\psi^*(t)\psi(t)&#xD;&#xA;$$&#xA;This is a way of deriving the probability function above.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-04-15T00:50:53.313" LastActivityDate="2011-04-15T00:50:53.313" />
  <row Id="8673" PostTypeId="1" AcceptedAnswerId="8690" CreationDate="2011-04-15T00:42:53.973" Score="-3" ViewCount="243" Body="&lt;p&gt;In discussing relativity with a (somewhat mathematical) friend the other day, I ran into a problem showing why special and/or general relativity could be considered as exact descriptions of reality rather than just approximations that are working okay so far. Here's his argument:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Special relativity applies to a situation where there is no curvature of spacetime. Since both matter and energy cause curvature, the only situation where special relativity would apply would be one which has no matter and energy, and therefore one which is not achievable in any experiment. And even as a thought experiment, a clock has to have mass and energy and so there can be no exact predictions in such a theory.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He went on to say that therefore the general theory of relativity rests on a very shaky foundation. This is as compared, for example, to thermodynamics which rests comfortably on statistical mechanics and quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A related question: &lt;a href=&quot;http://physics.stackexchange.com/questions/2926/special-relativistic-approximation-to-gr&quot;&gt;Special Relativistic approximation to GR&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-04-16T08:03:03.790" Title="Is special relativity an exact description of reality?" Tags="&lt;general-relativity&gt;&lt;special-relativity&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="8674" PostTypeId="2" ParentId="8643" CreationDate="2011-04-15T00:57:52.467" Score="3" ViewCount="" Body="&lt;p&gt;The human eye is an indifferent camera. Compared to a hawk's, it is laughable. A hawk can sit on top of the Empire State Building and make out a dime on the sidewalk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes the human eye any good at all is the software behind it: what our brains provide. You do realize that what really focuses on our retinas is an upside-down, tunnel-vision image that our brains sort into the images that we &lt;em&gt;think&lt;/em&gt; we see, right? And that in experiments, people who were made to wear glasses that made everything look upside-down adapted in a few days so that upside-down looked right-side up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We recognize patterns, colors, shapes where cameras only record differences in color and tone. It is not the eye that is remarkable. It is the brain.&lt;/p&gt;&#xA;" OwnerUserId="3121" LastActivityDate="2011-04-15T00:57:52.467" CommentCount="2" />
  <row Id="8675" PostTypeId="1" AcceptedAnswerId="8679" CreationDate="2011-04-15T01:05:39.320" Score="0" ViewCount="132" Body="&lt;p&gt;If you connected the positive terminal of a battery to the negative terminal to a battery with a wire with (hypothetically) no resistance, and are asked to give the voltage drop of a segment of wire in the circuit, how would you determine this?&lt;/p&gt;&#xA;" OwnerUserId="3123" LastActivityDate="2011-04-15T03:43:03.623" Title="Voltage drop in a circuit" Tags="&lt;electric-circuits&gt;" AnswerCount="3" />
  <row Id="8676" PostTypeId="1" AcceptedAnswerId="9123" CreationDate="2011-04-15T01:12:05.887" Score="2" ViewCount="147" Body="&lt;p&gt;How big would a hollow rigid object need to be to float, (not in water but in air) if all of the air was vacuumed out and the container sealed?&lt;/p&gt;&#xA;" OwnerUserId="3124" LastActivityDate="2011-04-25T23:22:52.053" Title="Making a &quot;heavier-than-air&quot; craft float" Tags="&lt;gravity&gt;&lt;forces&gt;&lt;pressure&gt;&lt;aerodynamics&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="8677" PostTypeId="2" ParentId="8675" CreationDate="2011-04-15T01:13:58.827" Score="2" ViewCount="" Body="&lt;p&gt;If you are asking what happens when a constant-voltage source with infinite current capability is short-circuited by 0 resistance, it's not a well-defined question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, the source has an intrinsic resistance &gt; 0 and the wire has a resistance &gt; 0 and if that is your question, Ohms law (or a voltmeter) and basic circuit theory will give you answers but I don't know exactly what you are asking currently.&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-04-15T01:13:58.827" />
  <row Id="8678" PostTypeId="2" ParentId="8676" CreationDate="2011-04-15T01:23:36.443" Score="1" ViewCount="" Body="&lt;p&gt;Since air weighs about 1.2 kilograms per cubic meter, if you could make your Magdeburg hemispheres strong enough and they enclosed one cubic meter and weighed less than 1.2 kilograms, they ought to float like a helium balloon&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-04-15T01:23:36.443" CommentCount="2" />
  <row Id="8679" PostTypeId="2" ParentId="8675" CreationDate="2011-04-15T01:35:41.687" Score="0" ViewCount="" Body="&lt;p&gt;Since you specified the wire, but not the battery, as being ideal, the result will be that there will be no voltage on the wire. Instead, the battery will be shorted out.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-04-15T01:35:41.687" />
  <row Id="8680" PostTypeId="2" ParentId="8676" CreationDate="2011-04-15T02:49:15.290" Score="2" ViewCount="" Body="&lt;p&gt;Just to flesh out Jim's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume you made a cube (makes the math easier!)  1m x 1m x 1m&lt;br&gt;&#xA;Aluminium has a density of 2700kg/m^3&lt;br&gt;&#xA;And you need the cubeto have a mass of less than 1.2Kg&lt;br&gt;&#xA; - so can use (1.2/2700) m^3 of material&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface area is 6m^2 then you can have a thickness of (1.2/2700)m^3 / 6m^2 = 70um&lt;br&gt;&#xA;Or about half the thickness of a hair!&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-04-15T02:49:15.290" />
  <row Id="8681" PostTypeId="2" ParentId="8540" CreationDate="2011-04-15T02:50:10.293" Score="-1" ViewCount="" Body="&lt;p&gt;This question is pretty open ended.  The second part of this involves the $SL(2,{\mathbb R})$ subgroup of the Virasoro algebra.  So I thought that at the risk of giving answers which might not be relevant I thought I would try to connect this with Lie theory.  The Lie algebra g has a maximal set of commuting matrices that define the Cartan center $H^i$, $i~=~1,\dots,~ rank(g)$.  These operators act upon the remaining operators $E^\alpha$ as $[H^i,~E^\alpha]~=~\alpha^i E^\alpha$, where $\alpha^i$ are the roots of the algebra.  The Jacobi theorem &#xA;$$&#xD;&#xA;[[H^i,~E^\alpha],~E^\beta]~+~[[E^\beta,~H^i],~E^\alpha]~+~[[E^\alpha,~E^\beta],~H^i]~=~0&#xD;&#xA;$$&#xA;permits us to compute&#xA;$$&#xD;&#xA; [E^\alpha,~E^\beta]~=~\matrix{ C(α,β)E^{α+β}~&amp;amp; :~\alpha~+~\beta~a~root \cr&#xD;&#xA;2\alpha\cdot H/\alpha^2~&amp;amp;	   ~: \alpha~+~\beta~=~0\cr&#xD;&#xA;0~&amp;amp;		   ~: otherwise}&#xD;&#xA;$$&#xA;The structure constant $|C(\alpha,\beta)~=~\pm 1$ and in the second of these the contraction of $H^i$ with the root $\alpha^i$ is a trace of $H^i$ and is used in a normalization $E^\alpha E^{-\alpha}~= 2/\alpha^2$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The operators for the string modes obey a Virasoro algebra,&#xA;$$&#xD;&#xA;{[L^{a_j},~L^{b_j}]~=~(a_j~-~b_j)L^{a_j + b_j}~+~c(a_j ,b_j ).} &#xD;&#xA;$$&#xA;The Virasoro generators are expanded according to the Laurent expansion&#xA;$$&#xD;&#xA;L^{a_n}~=~\oint\frac{dz}{2\pi iz}z^{a_n+2}T(z),~T(z)~=~-\sum_{a_n=\infty}^\infty\frac{L^{a_n}}{z^{m+2}}.&#xD;&#xA;$$&#xA;Commutators of the Virasoro generators $L^{-1},~L^0,~L^1$ produce the $SL(2,{\mathbb R})$ algebra&#xA;$$&#xD;&#xA;[L^0,~L^{-1}]~=~L_{-1},~[L^0,~L^1]~=~-L^1,~[L^1,~L^{-1}]~=~2L^0.&#xD;&#xA;$$&#xA;This is the same in form as the $SU(2)$ algebra for the angular momentum operators $L_\pm,~L_z$, but is noncompact.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A general commutator of an element $T^a~=~T^a(z)$ in the vector space of a Lie algebra obeys $[T^a,~T^b]~=~iC^{ab}_cT^c$.  The inner product of these elements defines a positive element $\langle T^a,~T^b\rangle~=~h^{ab}$.  This serves as a metric in the vector space of the Lie algebra.  This defines a rule &#xA;$$&#xD;&#xA;\langle[T^a,~T^b],~T^c\rangle~+~\langle T^b,~[T^a,~T^c]\rangle~=~0.&#xD;&#xA;$$&#xA;So the metric $h^{ab}$ defined in some representation, $r$, of matrix element $t^a_r$ then gives the Schur’s lemma result $tr(t^a_rt^b_r)~=~T_rh^{ab}$.  This further gives the definition of the Coxeter number cox(g)&#xA;$$&#xD;&#xA;-\sum_{cd}C^{ac}_dC^{bd}_c~=~cox(g)(α_L)^2h^{ab}&#xD;&#xA;$$&#xA;for $\alpha_L$ any long root.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With some of these Lie algebraic basics down operator produce expansions (OPE) can be found.  The bosonic vertex operator for the heterotic string is of the form $j(z)\phi^i({\bar z})exp(ik\cot X)$, for $X$ the string world sheet.  A gauge bosonic vertex operator is similarly $j(z){\bar\partial}X^i({\bar z})exp(ik\cdot X)$.  The current is holomorphic in the complex $z$, and stress-energy constructed from currents in order to be conformal must also be holomorphic.. The most basic form of a an OPE is the $(1,0)$ holomorphic current is&#xA;$$&#xD;&#xA;j^aj^b~\sim~ k^{ab}/z^2~+~i(c^{ab}_c/z)f^c(0).&#xD;&#xA;$$&#xA;The algebraic content is found by taking the Laurent expansion of the current&#xA;$$&#xD;&#xA;j^a(z)~=~\sum_{m=-\infty}^∞\frac{j^a_m}{z^{m+1}},&#xD;&#xA;$$&#xA;where the current coefficients satisfy a Lie algebra&#xA;$$&#xD;&#xA;[j^a_m,~j^b_n]~=~mk^{ab}\delta_{m,-n}~+~iC^{ab}_cj^c_{m+n},&#xD;&#xA;$$&#xA;which is a Virasoro algebra.  The coefficients $k^{ab}~=~kh^{ab}$.  For $m = 0,\pm 1$ the Virasoro algebra obeys a closed algebra of commutators&#xA;$$&#xD;&#xA;[j^a_0,~j^b_{\pm1}]~=~ic^{ab}_cj^c_{\pm 1},~ [j^a_1,~ j^b_{-1}]~=~2J_0,&#xD;&#xA;$$&#xA;which is an $SU(2)$ algebra of the elements $2\alpha\cdot H/\alpha^2$, $E^\alphaα_0$, $ E^{-\alpha}_0$, or the elements $(2\alpha\cdot H~+~k)/\alpha^2$, $E^\alpha_1 E^{-\alpha}_{-1}$.  So we connect with the Lie algebraic construction above.  The Coxeter number above defines an OPE stress-energy&#xA;$$&#xD;&#xA;T~=~[(k~+~cox(g))(\alpha_L)^2]^{-1}:jj(z):&#xD;&#xA;$$&#xA;With :  : meaning a normalization.  With additional work the current algebra of the system constructs OPE expansions for relevant terms.  In this way a conformal consistent stress-energy can be constructed.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-04-15T12:45:40.100" LastActivityDate="2011-04-15T12:45:40.100" />
  <row Id="8682" PostTypeId="2" ParentId="8675" CreationDate="2011-04-15T03:11:18.270" Score="1" ViewCount="" Body="&lt;p&gt;If you have an ideal voltage source with 0 internal resistance and 0 external resistance (0 resistance of the wire) then the voltage drop is indeterminate since it is of the form $\infty \times o$. The question then is not well defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in practice, we represent a practical voltage source by an ideal voltage source in series with a resistance $r$ (the internal resistance of the practical voltage source). This will be in series with an external practical wire of resistance $r\prime$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying Kirchhoff's law, we have,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I \times (r + r\prime) = V$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or, $$I\times r\prime = V - I\times r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V$ is the voltage of the source, $I$ current through the wire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the voltage drop is, $V - I \times r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or $I \times r\prime$&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-04-15T03:43:03.623" LastActivityDate="2011-04-15T03:43:03.623" />
  <row Id="8683" PostTypeId="2" ParentId="8673" CreationDate="2011-04-15T03:20:01.510" Score="3" ViewCount="" Body="&lt;p&gt;Newton's laws of motion are idealized laws as well. You can't have $0$ force acting on a particle in reality since gravity is a long range force and is present everywhere. So these laws are limiting cases as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SR is a good description of reality in the above sense. In the limiting case when gravity may be ignored it is a valid theory of reality.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-04-15T03:20:01.510" CommentCount="1" />
  <row Id="8684" PostTypeId="2" ParentId="8612" CreationDate="2011-04-15T03:32:56.983" Score="1" ViewCount="" Body="&lt;p&gt;I am not so certain about the terminology because I do not use electronic structure software, but since nobody has answered so far, I will try to see if my answer clarifies something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a closed-shell configuration you know that a single Slater determinant built from atomic orbitals (or for that matter, any function with angular dependence in terms of spherical harmonics) will be an eigenstate of $\hat{L}^2$ and $\hat{S}^2$ with quantum numbers L = S = 0. Additionally in this case there is no spin-orbit coupling, so that J = 0 as well. Therefore the angular momentum properties of this wave function will be well defined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the simplest ab initio molecular orbital calculation you obtain your atomic orbitals solving the restricted Hartree-Fock equations (RHF). Why restricted? Because if you have n electrons and the dimension of your basis (whatever STO, GTO or more advance basis, assuming you expand your atomic orbitals in a basis) is L &gt; n/2, you populate the first n/2 orbitals and assume each orbital is doubly occupied by two electrons with spins $\uparrow$ and $\downarrow$. You could use the unrestricted Hartree-Fock (UHF) method, populating the lowest n spin-orbitals, but then the density of electrons with spin $\uparrow$ will be in general different than the density of electrons with spin $\downarrow$. Thus, although the energy of your system will be lower (some restrictions in the variational procedure are lifted) $S$ will not be well defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, for an open shell configuration the angular momentum will not be well defined by a single Slater determinant wave function. How do you create the proper wave function? You can use several strategies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more general strategy is to use a Multi-Configurational Self Consistent Field (MCSCF) approach, in which you generate several Slater determinants with the L orbitals, in which both the orbitals and the coefficients multiplying the Slater determinants are variational parameters that you have to determine consistently within the MCSCF approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simpler method including correlation (among the many that exists) is the Configurational Interaction (CI) in which you optimize only once the orbitals and then generate the Slater determinants and then at the end of the day obtain the coefficients of these Slater determinants by diagonalizing the Hamiltonian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The previous methods are called method with dynamic correlation or beyond the HF approach. More simply, you can use the RHF scheme and after obtaining your n/2 populated orbitals &#xA;force the symmetry by suitable projector operators that generate the other Slater determinants needed so that your overall wave function is an eigenstate of $\hat{L}^2$ and $\hat{S}^2$ (and $\hat{L}_z$ and $\hat{S}_z$). This is good only in the so-called L-S coupling scheme when spin-orbit coupling is assumed small (small relativistic corrections). But of course you can later diagonalize the Hamiltonian with the spin-orbit coupling terms to resolve the fine-structure of the atom. I believe this is what GAMESS calls the ROHF method but I am not sure. To be sure you have to look at the manual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More commonly, you can use the UHF method and after obtaining you n populated spin-orbitals you force the right symmetry by applying projector operators to filter the &quot;wrong&quot; components that give an incorrect spin symmetry. This method is often called the PHF or Projected Hartree-Fock. Finally, you can also use the RHF equations but start from a number of Slater determinants with the coefficients fixed by the projector operator (that is, they are no longer variational parameters) and then variationally optimize the orbitals for this set of determinants. The equations become more difficult and I believe this method, called EHF or Extended Hartree-Fock is not often applied. If you want to work with several Slater determinants you typically use CI or MCSCF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now going back to your examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Lithium you have two microstates for the ground state that you can write assigning the 4 quantum numbers $n_j$, $l_j$, $m_j$ and $m_{s,j}$ for each electron $j$, assuming the usual Aufbau principle. However these two states are degenerate. Therefore you can write two Slater determinants which will give you the same ground state energy. In the absence of an external magnetic field, any of these Slater determinants (or whatever linear combination of the two) is a good description of your ground state. It will be an eigenstate of the four angular momentum operators (or of $\hat{J}^2$ and $\hat{J}_z$ too).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your second example, Boron, you can write 6 microstates, because you can put your last electron in $|2,1,0,\pm 1/2 \rangle$ or $|2,1,\pm 1,\pm 1/2 \rangle$ where in the nice Dirac nomenclature the third quantum number corresponds to $m_j$ and the last to $m_{s,j}$. Although all the Slater determinants that you can built with these microstates assignments are degenerate eigenstates with well defined $\hat{L}^2$, $\hat{L}_z$, $\hat{S}^2$ and $\hat{S}_z$, the degeneracy will be broken when you include the spin-orbit coupling (fine structure). Then some states will be written in terms of a combination of several Slater determinants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your last example, Oxigen, you have two holes in the outer (open) shell. If you use the ROHF method (assuming I got the terminology right) then it suffices to use the Clebsch-Gordon coefficients to know the right combination of Slater determinants that determines well defined values of $\hat{L}^2$, $\hat{L}_z$, $\hat{S}^2$ and $\hat{S}_z$ for the set of 4 lowest-energy populated orbitals found by the RHF method.&lt;/p&gt;&#xA;" OwnerUserId="3100" LastActivityDate="2011-04-15T03:32:56.983" />
  <row Id="8685" PostTypeId="2" ParentId="8517" CreationDate="2011-04-15T03:41:43.180" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Henri_Poincar%C3%A9#The_three-body_problem&quot; rel=&quot;nofollow&quot;&gt;Poincaré&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The version finally printed contained&#xA;  many important ideas which lead to the&#xA;  theory of chaos. The problem as stated&#xA;  originally was finally solved by Karl&#xA;  F. Sundman for n = 3 in 1912 and was&#xA;  generalised to the case of n &gt; 3&#xA;  bodies by Qiudong Wang in the 1990s.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Karl_F._Sundman&quot; rel=&quot;nofollow&quot;&gt;Karl_F._Sundman&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;used analytic methods to prove the&#xA;  existence of a convergent infinite&#xA;  series solution to the three-body&#xA;  problem in 1906 and 1909.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Qiudong_Wang&quot; rel=&quot;nofollow&quot;&gt;Qiudong_Wang&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Wang is best known for his paper The&#xA;  &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1991CeMDA..50...73W&amp;amp;data_type=PDF_HIGH&amp;amp;whole_paper=YES&amp;amp;type=PRINTER&amp;amp;filetype=.pdf&quot; rel=&quot;nofollow&quot;&gt;global solution of the n-body&lt;/a&gt;&#xA;  problem (*), in which he generalised&#xA;  Karl F. Sundman's results from 1912 to&#xA;  a system of more than three bodies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(*)With Zero Angular Momentum, it seems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a large colection of N-Body codes available from the net, and some of them work with GPUs (graphics hardware)&#xA;&lt;a href=&quot;http://www.scholarpedia.org/article/N-body_simulations&quot; rel=&quot;nofollow&quot;&gt;a SoftPedia list of opensource codes&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've downloaded &lt;strong&gt;Gravit&lt;/strong&gt; from the site of &lt;a href=&quot;http://gravit.slowchop.com/&quot; rel=&quot;nofollow&quot;&gt;Gerald Kaszuba&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've choosed his work because it is loaded with options, even if it is NOT physically correct:&lt;br&gt;&#xA;I've included the Velocity Verlet Integrator to solve this problem.  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;// Velocity Verlet integrator algorithm&#xA;// r(i+1) = r(i) + v(i)*dt + a(i)*dt^2 / 2&#xA;// v(i+1/2) = v(i) + a(i) * dt/2&#xA;//     1st -- act. pos, 2nd calc accel = a(i+1), 3rd vel, 4th acc&#xA;// a(i+1= sum of accels on i+1  (accel)&#xA;// v(i+1) = v(i+1/2) + a(i+1)*dt/2  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I've also included code to work with my GPU using BROOK+ of AMD/ATI (320 parallel processors, 60000 objects)&#xA;In the site of CUDA/OpenCL of NVIDEA can be found NBODY code to work in OpenCL.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ast.cam.ac.uk/research/nbody/&quot; rel=&quot;nofollow&quot;&gt;Sverre Aarseth codes&lt;/a&gt; are a strong reference (&lt;a href=&quot;https://gist.github.com/mtrenti/NBODY6&quot; rel=&quot;nofollow&quot;&gt;code in Git&lt;/a&gt;).&lt;br&gt;&#xA;&lt;a href=&quot;http://www.artcompsci.org/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Theory&lt;/strong&gt; in The Art of Computational Science&lt;/a&gt; and &lt;a href=&quot;http://www.artcompsci.org/kali/index.html&quot; rel=&quot;nofollow&quot;&gt;Maya Project&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.grav-sim.com/simulation.html&quot; rel=&quot;nofollow&quot;&gt;Grav-Sim&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.cm.ph.bham.ac.uk/cpp/projects/physics/nbody.html&quot; rel=&quot;nofollow&quot;&gt;types of code, and links resources&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are solutions stables with 3 body as we can see &lt;a href=&quot;https://sites.google.com/site/ufoexplained/fotos_ufo/3stableNBody.bmp&quot; rel=&quot;nofollow&quot;&gt;in this image&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question with two bodys in perpendicular orbits, I think that in general the orbits will tend to become planar.  I didnt tested yet about the stability but I think that for aproximate radius they will be unstable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT add&lt;/strong&gt;:&lt;br&gt;&#xA;'The three-body problem with close encounters is notoriously ill-conditioned because it admits chaotic solutions that manifest extreme sensitivity to initial conditions.'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My code is good for a large ensemble of bodies like a galaxy, a star cluster or a disk of matter.&lt;br&gt;&#xA;The code for a few bodies must be of 'direct' type (and non GPU) and make use of 'adaptive time step'. In the short range distances the time step must be finer then in the long range and the use of a different reference frame (a local) could benefit the solution.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Picard and Parker-Sochacki based Methods&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;If I were you I'd explore this very interesting paper &lt;a href=&quot;http://tphex.hep.by/Journal_of_Computational_Physics/Volume%20187/1/16.pdf&quot; rel=&quot;nofollow&quot;&gt;An adaptive N-body algorithm of optimal order&lt;/a&gt; (2002 ) and &lt;a href=&quot;http://www.jmu.edu/hyper/visualization/nbody/expository.pdf&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt; and &lt;a href=&quot;http://www.jmu.edu/hyper/visualization/nbody/&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt; and &lt;a href=&quot;http://www.jmu.edu/hyper/visualization/nbody/ps.pdf&quot; rel=&quot;nofollow&quot;&gt;THIS ONE with 47 lines of code inside&lt;/a&gt;&lt;/strong&gt;&lt;br&gt;&#xA;(I must say 'thank you' because you made me find this new, interesting way) &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Table I: &lt;strong&gt;Basic Source Code for Solving&#xA;  the N-body Problem&#xA;  The preceding 47 lines of code&lt;/strong&gt; ...&#xA;  What is stunning is&#xA;  the simplicity of the solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT add 2&lt;/strong&gt;&lt;br&gt;&#xA;I've done some simulations &lt;strong&gt;&lt;a href=&quot;http://heldervelez1.tiddlyspace.com/#MySimulations&quot; rel=&quot;nofollow&quot;&gt;available here&lt;/a&gt;&lt;/strong&gt; where a xls can be downloaded.&#xA; &lt;img src=&quot;http://i.stack.imgur.com/M6sir.jpg&quot; alt=&quot;example image&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-04-17T10:43:48.997" LastActivityDate="2011-04-17T10:43:48.997" CommentCount="3" />
  <row Id="8686" PostTypeId="1" AcceptedAnswerId="8699" CreationDate="2011-04-15T03:43:32.040" Score="2" ViewCount="257" Body="&lt;p&gt;If you're writing about a theory with Yang-Mills/Gauge fields for an arbitrary reductive gauge group coupled to arbitrary matter fields in some representation,&#xA;is it best to call it a Yang-Mills theory or a Gauge theory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard that one is more likely to refer to a theory with no matter sector - but I can't remember which one!&#xA;Or are the terms basically interchangeable in the context of quantum field theory?&lt;/p&gt;&#xA;" OwnerUserId="429" LastActivityDate="2011-04-15T18:43:36.277" Title="Nomenclature: Yang-Mills theory vs Gauge theory" Tags="&lt;quantum-field-theory&gt;&lt;nomenclature&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="8687" PostTypeId="2" ParentId="8686" CreationDate="2011-04-15T04:55:28.563" Score="1" ViewCount="" Body="&lt;p&gt;&quot;Yang-Mills theory&quot; is pretty much equivalent to &quot;gauge theory&quot;. You could say gauge theories are more general, since we usually think of Yang-Mills theories as being gauge theories of SO or SU groups, or sometimes one of the other classical lie groups. But you could make a distinction by saying something like &quot;gravity is a gauge theory but not a Yang-Mills theory,&quot; since its gauge group would be local Poincare transformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term you're thinking of without matter is probably &quot;pure Yang-Mills theory&quot; or &quot;pure gauge theory.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1403" LastActivityDate="2011-04-15T04:55:28.563" CommentCount="3" />
  <row Id="8688" PostTypeId="1" CreationDate="2011-04-15T04:58:00.950" Score="2" ViewCount="136" Body="&lt;p&gt;I understand that matter will gravitate toward matter. (ex: Earth gravitates a satellite toward it, and the satellite toward Earth.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this always apply, regardless of distance? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take two atoms, the farthest apart in the universe. Do they gravitate toward each other, but at an infinitesimal rate? (And actual position displacement is likely in a different direction, due to stronger/closer forces)&lt;/p&gt;&#xA;" OwnerUserId="2836" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-04-15T12:48:08.257" LastActivityDate="2011-04-15T12:48:08.257" Title="Do all atoms in the universe gravitate each other?" Tags="&lt;gravity&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="8689" PostTypeId="2" ParentId="8688" CreationDate="2011-04-15T05:08:57.320" Score="5" ViewCount="" Body="&lt;p&gt;Any mass/energy will interact with  the gravitational field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational field at an atoms position will be the combined result of all the mass/energy of the universe at that position, but the strength falls with distance from the gravitational sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coupling to gravity is so very small with respect to other forces acting on an atom, it is ignored when solving problems, except gravitational ones for aggregates of atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, atoms interact with other atoms gravitationally, as a function of the distance, but the strength of the interaction is infinitesimally small.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-15T05:08:57.320" CommentCount="7" />
  <row Id="8690" PostTypeId="2" ParentId="8673" CreationDate="2011-04-15T05:36:11.913" Score="5" ViewCount="" Body="&lt;p&gt;From an experimentalist's point of view, data trumps theory and mathematics too :) .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Special relativity is a measured fact, within the accuracy of our measuring systems, and has not been falsified. It rests on data, as all physical theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Your friends&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;the general theory of relativity rests on a very shaky foundation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;reflects the necessity of special relativity in any theory of gravity. His argument that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the only situation where special relativity would apply would be one which has no matter and energy, and therefore one which is not achievable in any experiment &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is at fault, because he is not including experimental limits and data. Any mathematical curvature is orders of magnitude smaller than our measuring instruments can measure and validate, and physical theories are judged on that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physical theories go mathematically as far as the data can take them. Before  the observation of particles special relativity was an interesting observation of Lorenz in the Maxwell electromagnetic theory. Before the photoelectric effect classical mechanics sufficed. If ever experiments can go down to the curvatures of gravity for elementary particles, maybe special relativity will be modified to fit the data. It is theories that change to fit the data, not vice versa.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-04-16T08:03:03.790" LastActivityDate="2011-04-16T08:03:03.790" CommentCount="2" />
  <row Id="8691" PostTypeId="2" ParentId="8686" CreationDate="2011-04-15T05:47:09.000" Score="2" ViewCount="" Body="&lt;p&gt;What I've usually heard in practice is that a Yang-Mills theory is a gauge theory with a non-Abelian gauge group. That would disqualify classical E&amp;amp;M, for example, where the gauge group is $U(1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Yang%E2%80%93Mills_theory&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; has a slightly more restrictive definition, saying that a Yang-Mills theory is a gauge theory based on $SU(N)$ specifically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not quite sure which definition to go with. It probably depends on who you're talking to. Regardless, the only distinction I'm familiar with is based on the nature of the gauge group and has nothing to do with whether there happen to be matter fields or not.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-04-15T05:47:09.000" CommentCount="3" />
  <row Id="8692" PostTypeId="1" CreationDate="2011-04-15T06:28:54.790" Score="0" ViewCount="63" Body="&lt;p&gt;I've got a Jupiter moon series telescope and I've a got 2x Barlow lens along with the set.&#xA;Which lens(Ramsden,Plossl,Nagler.. or any other) should I go for next to buy, as a novice astronomer.&lt;/p&gt;&#xA;" OwnerUserId="3047" LastActivityDate="2011-04-15T06:28:54.790" ClosedDate="2011-04-15T10:41:23.780" Title="Telescope lens for basic sky watching" Tags="&lt;optics&gt;&lt;astronomy&gt;" CommentCount="2" />
  <row Id="8693" PostTypeId="1" AcceptedAnswerId="8696" CreationDate="2011-04-15T06:51:04.360" Score="1" ViewCount="202" Body="&lt;p&gt;1.How does a Telescope work? &#xA;2.What factors increase the magnification of the lens?&lt;/p&gt;&#xA;" OwnerUserId="3047" LastEditorUserId="3047" LastEditorDisplayName="" LastEditDate="2011-04-18T11:12:27.090" LastActivityDate="2011-04-18T11:12:27.090" Title="How does Telescope lens work?" Tags="&lt;optics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8694" PostTypeId="2" ParentId="8673" CreationDate="2011-04-15T08:07:27.100" Score="5" ViewCount="" Body="&lt;p&gt;The general theory of relativity &lt;em&gt;does not rest&lt;/em&gt; on special relativity, so it does not make sense to say the foundation of GR is shaky because SR is shaky. One merely requires consistency, i.e., we want general relativity to reduce to special relativity in the limit.&lt;/p&gt;&#xA;" OwnerUserId="2299" LastActivityDate="2011-04-15T08:07:27.100" CommentCount="2" />
  <row Id="8695" PostTypeId="2" ParentId="8673" CreationDate="2011-04-15T08:24:06.617" Score="2" ViewCount="" Body="&lt;p&gt;Special relativity accurately applies to experiments in a lab or reference frame that is small enough that the tidal force (exactly the same thing as spacetime curvature) in it doesn't alter the outcome of the experiment at the given precision. Search the web for: &quot;For sufficiently small regions, the special theory of relativity is correct!!&quot;&lt;/p&gt;&#xA;" OwnerUserId="2314" LastActivityDate="2011-04-15T08:24:06.617" />
  <row Id="8696" PostTypeId="2" ParentId="8693" CreationDate="2011-04-15T09:14:20.687" Score="6" ViewCount="" Body="&lt;p&gt;It's not quite clear what you mean by &quot;telescope lens&quot; - do you mean the system of lenses that make up a telescope? If so, there are two basic types. The actual lenses in your telescope are probably more complicated and correct for all kinds of &lt;a href=&quot;http://en.wikipedia.org/wiki/Spherical_aberration&quot; rel=&quot;nofollow&quot;&gt;aberrations&lt;/a&gt;, but they work like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Keplerian telescope (top one in the diagram) consists of two positive lenses, with different focal distances, with their foci at the same point, in between the lenses. Imagine your eye on the left. Two parallel rays will converge to a point at the focus of the right-hand lens, and since this point is also at the focus of the left-hand lens, they will become parallel again, but inverted. Of course you are usually not looking at parallel rays with your telescope, but picturing it this way is what helped me to understand why we see a magnified image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might think at first glance from the diagram that this makes the image smaller; but what it does is take parallel rays traveling at different angles to the optical axis, and make them parallel again on the other side of the telescope, but traveling at a &lt;em&gt;larger&lt;/em&gt; angle. This increases the apparent size of the object. (Google Docs isn't very good for drawing detailed diagrams - you could take a look at the more complicated ones on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Galilean_telescope#Refracting_telescope_designs&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; on telescopes.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Galilean telescope (bottom one in the diagram) consists of a negative and a positive lens, again with their foci at the same point. This time the point is on the outside of the telescope, at where your eye is. The positive lens focuses the parallel rays to that point, and the negative lens takes the converging rays and makes them parallel again. This time, the image is not inverted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/NE1EW.png&quot; alt=&quot;Illustration of Galilean and Keplerian telescopes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the second part of your question is about magnification. The focal lengths of the lenses are the only factors that influence the magnification: it is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M=-\frac{f_2}{f_1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minus sign seems counter-intuitive, but think about it - we fill in a negative focal length for the negative lens in the Galilean telescope, so the magnification comes out positive. For the Keplerian telescope, the magnification comes out negative - this indicates that the image is magnified, but also inverted.&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2011-04-15T09:14:20.687" CommentCount="0" />
  <row Id="8698" PostTypeId="2" ParentId="8513" CreationDate="2011-04-15T09:33:29.643" Score="0" ViewCount="" Body="&lt;p&gt;This is a response to Georg: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The condition of applicability of the Stokes formula $F_d=6\pi\mu Rv$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Re=\frac{\rho_aRv}{\mu}&amp;lt;&amp;lt;1$$ $\mu$ is the dynamic viscosity of air  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From $mg=6\pi\mu Rv$;  $m=4\pi R^2l\rho_w$ we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v=\frac{2\rho_wglR}{3\mu}$$ and $$Re=\frac{2\rho_a\rho_wglR^2}{3\mu^2}&amp;lt;&amp;lt;1$$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R&amp;lt;&amp;lt;\sqrt{\frac{3\mu^2}{2\rho_a\rho_wgl}}$$ Now $\mu=1.8*10^{-5}$ Pa*s, $\rho_a=1.2$ kg/m^3, $\rho_w=10^3$kg/m^3, $g=10$m/s^2, $l=10^{-7}$m  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives: $$R&amp;lt;&amp;lt;0.1cm$$ We're not talking about such bubbles, i think.  &lt;/p&gt;&#xA;" OwnerUserId="552" LastEditorUserId="552" LastEditorDisplayName="" LastEditDate="2011-04-15T09:51:14.620" LastActivityDate="2011-04-15T09:51:14.620" CommentCount="4" />
  <row Id="8699" PostTypeId="2" ParentId="8686" CreationDate="2011-04-15T10:14:40.770" Score="3" ViewCount="" Body="&lt;p&gt;Very briefly, a classical theory is a &lt;em&gt;gauge theory&lt;/em&gt; if its field variables $\varphi^i(\vec{x},t)$ have a non-trivial local gauge transformation that leaves the action $S[\varphi]$ gauge invariant. Usually, a gauge transformation is demanded to be a continuous transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Gauge theory is a &lt;em&gt;huge&lt;/em&gt; subject, and I only have time to give some explanation here, and defer a more complete answer to, e.g., the book &quot;Quantization of Gauge Systems&quot; by M. Henneaux and C. Teitelboim. By the word &lt;em&gt;local&lt;/em&gt; is meant that the gauge transformation in different space-time point are free to be transformed independently without affecting each others transformation (as opposed to a &lt;em&gt;global&lt;/em&gt; transformation). By the word &lt;em&gt;non-trivial&lt;/em&gt; is meant that the gauge transformation does not vanish identically on-shell. Note that an infinitesimal gauge transformation does &lt;em&gt;not&lt;/em&gt; have to be on the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta_{\varepsilon}A_{\mu}(\vec{x},t) = D_{\mu}\varepsilon(\vec{x},t),$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;nor&lt;/em&gt; does it have to involve a $A_{\mu}$ field. More generally, an infinitesimal gauge transformation is of the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta_{\varepsilon}\varphi^i(x) = \int d^d y \ R^i{}_a (x,y)\varepsilon^a(y),$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R^i{}_a (x,y)$ are Lagrangian gauge generators, which form a gauge algebra, which, in turn, may be &lt;em&gt;open&lt;/em&gt; and &lt;em&gt;reducible,&lt;/em&gt; and $\varepsilon^a$ are infinitesimal gauge parameters. Besides gauge transformations that are continuously connected to the identity transformation, there may be so-called &lt;em&gt;large&lt;/em&gt; gauge transformations, which are &lt;em&gt;not&lt;/em&gt; connected continuously to the identity transformation, and the action may not always be invariant under those. Ultimately, physicists want to &lt;em&gt;quantize&lt;/em&gt; the classical gauge theories using, e.g., &lt;a href=&quot;http://en.wikipedia.org/wiki/Batalin%E2%80%93Vilkovisky_formalism&quot; rel=&quot;nofollow&quot;&gt;Batalin-Vilkovisky formalism&lt;/a&gt;, but let's leave quantization for a separate question. Various subtleties arise at the quantum level as, e.g., pointed out in the comments below. Moreover, some quantum theories do not have classical counterparts.] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yang-Mills theory is just one example out of many of a gauge theory, although the most important one. To name a few other examples: &lt;a href=&quot;http://en.wikipedia.org/wiki/Chern-Simons_theory&quot; rel=&quot;nofollow&quot;&gt;Chern-Simons theory&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/BF_model&quot; rel=&quot;nofollow&quot;&gt;BF theory&lt;/a&gt; are gauge theories. Gravity can be viewed as a gauge theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yang-Mills theory without matter is called &lt;em&gt;pure Yang-Mills theory.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-04-15T18:43:36.277" LastActivityDate="2011-04-15T18:43:36.277" CommentCount="3" />
  <row Id="8700" PostTypeId="2" ParentId="8513" CreationDate="2011-04-15T11:51:16.340" Score="0" ViewCount="" Body="&lt;p&gt;This question going on and on made me think about : &quot;&lt;strong&gt;where is the error?&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather simple, (and my own error/fault!), the mass of air inside must not be &#xA;considered. For stokes flow only the effective mass (mass minus buoyancy) &#xA;is relevant.  This , of course is zero for a air filled bubble in air. &#xA;What remains is the weight of the soap lamella. &#xA;This thickness is in the magnitude of µmeters (by colors), but often &#xA;big drops at the &quot;bottom&quot; of the bubble are included. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does somebody know about values of speed of falling bubbles?&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-15T12:39:02.197" LastActivityDate="2011-04-15T12:39:02.197" CommentCount="1" />
  <row Id="8701" PostTypeId="1" CreationDate="2011-04-15T13:03:32.887" Score="6" ViewCount="129" Body="&lt;p&gt;There are proofs in the literature that QFT including microcausality is &lt;em&gt;sufficient&lt;/em&gt; for it not to be possible to send signals by making quantum mechanical measurements associated with regions of space-time that are space-like separated, but is there a proof that microcausality is &lt;em&gt;necessary&lt;/em&gt; for no-signaling?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I take microcausality to be trivial commutativity of measurement operators associated with space-like separated regions of space-time. In terms of operator-valued distributions $\hat\phi(x)$ and $\hat\phi(y)$ at space-like separated points $x$ and $y$, $\left[\hat\phi(x),\hat\phi(y)\right]=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An elementary observation is that classical ideal measurements satisfy microcausality just insofar as all ideal measurements are mutually commutative, irrespective of space-like, light-like, or time-like separation. For a classical theory, ideal measurements have no effect either on the measured system or on other ideal measurements, so ideal measurements cannot be used to send messages to other ideal experimenters. Whether a classical theory admits space-like signaling &lt;em&gt;within&lt;/em&gt; the measured system, in contrast to signaling by the use of ideal measurements applied by observers who are essentially outside the measured system, is determined by the dynamics. Applying this observation to QFT, I note that microcausality is not sufficient &lt;em&gt;on its own&lt;/em&gt; &amp;mdash;without any dynamics, which I take to be provided in QFT by the spectrum condition&amp;mdash; to prove no-signaling at space-like separation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an auxiliary question, what proofs of sufficiency do people most often cite and/or think most well-stated and/or robust?&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-04-15T13:03:32.887" Title="Is microcausality *necessary* for no-signaling?" Tags="&lt;quantum-field-theory&gt;&lt;quantum-information&gt;&lt;reference-request&gt;" FavoriteCount="2" />
  <row Id="8702" PostTypeId="2" ParentId="8634" CreationDate="2011-04-15T14:48:41.540" Score="3" ViewCount="" Body="&lt;p&gt;I'm not sure what you mean by &quot;derving&quot; the algebra, but one way of obtaining a superconformal algebra (let me consider $N=4$) is to start with the four-dimensional conformal algebra and the $N=4$ supersymmetry algebra and consider the closure of the combined algebra. The conformal algebra consists of the Lorentz generators $J^\alpha{}_\beta$ and $\dot{J}^{\dot{\alpha}}{}_{\dot{\beta}}$ (I write the Lorentz group as $SL(2)\times SL(2)$, which is what I think you do in your question), the translations $P^{\alpha\dot{\beta}}$, the special conformal transformations $K_{\alpha\dot{\beta}}$ and the dilatation $H$. The $N=4$ supersymmetry group is generated by the $SU(4)$ R-symmetry generators $R^A{}_B$, the translations $P^{\alpha\dot{\beta}}$ and the supercharges $Q^A{}_\alpha$ and $\dot{Q}_{\dot{\alpha}A}$. These two algebras are both closed, but if we want to consider the combined symmetry the above generators are not enough to close the algebra. In particular the commutator between $K^{\alpha\dot{\beta}}$ and $\dot{Q}_{\dot{\gamma}A}$ gives rise to a new generator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[K^{\alpha\dot{\beta}},\dot{Q}_{\dot{\gamma}C}] = \delta^{\dot{\beta}}_{\dot{\gamma}} S^{\alpha}{}_C$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By also including the superconformal charges $S^{\alpha}{}_A$ (and a correpsonding charge $\dot{S}^{A\dot{\alpha}}$) we ge a closed algebra. The other (anti-)commutation relations follow if we require the the Jacobi identity is satisfied. In particular we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{S^{\alpha}{}_a,Q^b{}_\beta\} = \delta^b_a J^{\alpha}{}_\beta + \delta^{\alpha}_{\beta} R^B{}_A + \frac{1}{2} \delta^b_a \delta^{\alpha}_{\beta} H$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is your next-to-last equation (in somewhat different conventions when it comes to indices). To obtain your last equation you just have to take the particular case of $A=B=4$ and $\alpha=\beta=-$ and express the result in terms of your Cartan generators $R_1$, $R_2$, $R_3$ and $J_3$. The importance of this last expression really depends on in which context you have encountered the algebra. If you give some more details maybe someone can answer.&lt;/p&gt;&#xA;" OwnerUserId="3131" LastActivityDate="2011-04-15T14:48:41.540" CommentCount="5" />
  <row Id="8703" PostTypeId="2" ParentId="8517" CreationDate="2011-04-15T15:12:42.830" Score="1" ViewCount="" Body="&lt;p&gt;Strictly speaking, no, if you are asking whether the two smaller planets can stay in orbits in perpendicular planes.  Looking down on the system along the line where the two planes intersect, at some stage the two will have an angle at the the megaplanet less than 180 degrees or $\pi$ radians between them, and at that point the gravitational attraction between the small planets will start to move them out of their original planes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does not preclude a more complicated pattern where the orbits of the two smaller planets are not in planes but are something more complicated with substantially different average inclinations relative to each other.  I don't know, but perhaps it is possible for these to be perpedicular on average.  &lt;/p&gt;&#xA;" OwnerUserId="2714" LastActivityDate="2011-04-15T15:12:42.830" />
  <row Id="8704" PostTypeId="1" CreationDate="2011-04-15T16:03:12.940" Score="5" ViewCount="196" Body="&lt;p&gt;Physically, quantum entanglement is ranged from full long-range entanglement (Bose-Einstein condensate), described by a basis of states that look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |\Psi\rangle = |\phi_{i_{0} i_{1} ... i_{N}}\rangle $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to full-decoherence (a Maxwell-Boltzmann ideal gas) which a basis of states that look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |\Psi\rangle = \prod_{i}{ |\phi_{i}\rangle }  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in the middle of this range we have 2-particle entanglement terms, 3-particle entanglement, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it seems natural to arrange the wavefunction as a series looking like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |\Psi\rangle = \sum{ \prod_{i}{ |\phi_{i}\rangle } } + \sum{ \lbrace \prod_{i_{0} ,i_{1} &amp;gt; i_{0}}{ |\phi_{i_{0} i_{1}}\rangle } \rbrace } + \sum{ \lbrace \prod_{i_{0} ,i_{1} &amp;gt; i_{0} , i_{2} &amp;gt; i_{1}}{ |\phi_{i_{0} i_{1} i_{2}}\rangle } \rbrace } + \dotsb  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BEGIN EDIT&lt;/strong&gt; I feel that i need to put in a bit more clear footing the mathematics behind this separation. Let's take an arbitrary state vector $|\Psi\rangle$, we might write it like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |\Psi\rangle = \prod_{i}{ c^{0}_{i} |\phi_{i}\rangle } + |\Psi_{Remainder}\rangle  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, we write the state vector as a vector that is completely separable and a remainder that is not. &lt;em&gt;This decomposition is unique&lt;/em&gt;. Proof: take another decomposition with $\widehat{c^{0}_{i}}$, take the difference between both decompositions and verify that both remainders are completely separable, which is against the definition&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is that one should be able to further this decomposition of the remainder state vector, for instance lets take a vector with zero completely separable part (that is, we are on the equivalence class of our remainder above) and attempt to write it like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |\Psi_{R_0}\rangle = \prod_{i_{0} ,i_{1} &amp;gt; i_{0}}{ c^{1}_{i_0 i_1} |\phi_{i_{0} i_{1}}\rangle } + |\Psi_{R_1}\rangle  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analogously, one can prove that the $c^{1}_{i_0 i_1}$ are unique and depend only on the $|\Psi_{R_0}\rangle$ vector. And we don't need to do any symmetrization operation to get this result, so this is an universal decomposition of the state vector (for bosons and fermions)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(note: i'm aware that this leaves out a lot of products with mixed entanglement, i.e: some single-particle states multiplied by two-particle entangled states, but since they don't add anything to this particular argument i choosed to leave them aside)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;END EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BEGIN 2ND EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;separability of states is a property that is invariant under unitary transformations, so a separable state is not equivalent to a separable one in any basis you choose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've investigated a bit more and the problem in general of knowing if a state is separable or not is known as QSP (quantum separability problem). For a definition please look at &lt;a href=&quot;http://arxiv.org/abs/cs/0603047&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;END 2ND EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:&#xA;Where can i read more about this sort of expansion and do you know if there is a computational framework to estimate the relative magnitudes of each term (for instance, i would expect that for Bose-Einstein condensates you need to keep all the terms of the expansion, while for relatively high-temperature solids you would be able to get away with 3 or 4 terms)&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-04-17T16:33:01.000" LastActivityDate="2011-04-17T16:33:01.000" Title="Expansion of multi-particle state vector as a sum of n-entangled states" Tags="&lt;quantum-mechanics&gt;&lt;condensed-matter&gt;&lt;entanglement&gt;&lt;bose-einstein-condensate&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="8705" PostTypeId="2" ParentId="8663" CreationDate="2011-04-15T16:13:48.433" Score="2" ViewCount="" Body="&lt;p&gt;There are already several good answers. Here I will only answer the very last question, i.e., if the Boltzmann factor in the path integral is $f(S(t_f,t_i))$, with action $S(t_f,t_i)=\int_{t_i}^{t_f} dt \ L(t)$, why is the function $f:\mathbb{R}\to\mathbb{C}$ an exponential function, and not something else?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, since the Feynman &quot;sum over histories&quot; propagator should have the group property&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ K(x_3,t_3;x_1,t_1) = \int_{-\infty}^{\infty}\mathrm{d}x_2 \ K(x_3,t_3;x_2,t_2) K(x_2,t_2;x_1,t_1),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one must demand that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(S(t_3,t_2)f(S(t_2,t_1)) = f(S(t_3,t_1)) = f(S(t_3,t_2)+S(t_2,t_1)),$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(S(t_1,t_1)) = 1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question boils down to: How many continuous functions $f:\mathbb{R}\to\mathbb{C}$ satisfy $f(s)f(s^{\prime})=f(s+s^{\prime})$ and $f(0)=1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: The exponential function! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof (ignoring some mathematical technicalities): If $s$ is infinitesimally small,  then one may Taylor expand&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(s) = f(0) + f^{\prime}(0)s +{\cal O}(s^{2}) = 1+cs+{\cal O}(s^{2}), $$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;with some constant $c:=f^{\prime}(0)$. Then one calculates&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f(s)=\lim_{n\to\infty}f(\frac{s}{n})^n&#xD;&#xA;=\lim_{n\to\infty}\left(1+\frac{cs}{n}+o(\frac{1}{n})\right)^n&#xD;&#xA;=e^{cs}, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e., the exponential function. &lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-04-15T18:55:12.470" LastActivityDate="2011-04-15T18:55:12.470" CommentCount="3" />
  <row Id="8706" PostTypeId="1" AcceptedAnswerId="8958" CreationDate="2011-04-15T16:14:05.253" Score="2" ViewCount="110" Body="&lt;p&gt;If you go to the Atlas experiment &lt;a href=&quot;http://atlas.ch/&quot; rel=&quot;nofollow&quot;&gt;http://atlas.ch/&lt;/a&gt; and click the status button, there's an AFS reading at the bottom with a current value&lt;/p&gt;&#xA;&#xA;&lt;p&gt;50ns_228b+1small_214_12_180_36bpi_8inj&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 50ns seems to refer to the bunch spacing, the 228b the number of bunches, but what do the other numbers refer to?&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-04-21T08:42:04.987" Title="What are the AFS values in the Atlas experiment?" Tags="&lt;large-hadron-collider&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="8708" PostTypeId="1" CreationDate="2011-04-15T18:51:39.990" Score="5" ViewCount="128" Body="&lt;p&gt;There are two experiments that are often used to explain Quantum Mechanics: the two-slit experiment and the EPR paradox.  I am curious what would happen if you combined them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine an experiment where you fire pairs of entangled particles at two simultaneous two-slit setups.  If you used detectors, you could find out how the entangled particles' paths correlate.  Perhaps you'd be able to deduce, based on the result of one detector, which slit the other particle went through.  Now, if you were to run the experiments with a detector on one side and no detector on the other side, would the unobserved particles still form an interference pattern, even though you know which slit they would have went through?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition is that the answer is yes.  Despite being entangled, the particles should not have correlated actions, otherwise we would have invented faster-that-light communication.  You could create a device that constantly fired entangled particles toward two far-apart worlds, and if one world suddenly started observing the particles on their side, the particles arriving at the other world would instantly cease to create an interference pattern.&lt;/p&gt;&#xA;" OwnerUserId="3136" LastEditorUserId="3136" LastEditorDisplayName="" LastEditDate="2011-04-15T19:10:34.793" LastActivityDate="2011-04-16T01:02:17.547" Title="Can observations of entangled particles affect their unobserved counterparts?" Tags="&lt;quantum-mechanics&gt;&lt;entanglement&gt;&lt;epr-experiment&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="8709" PostTypeId="1" CreationDate="2011-04-15T19:57:19.390" Score="1" ViewCount="107" Body="&lt;p&gt;Does the mass of both the parent object, and the child object affect the speed at which the child object orbits the parent object?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I though it didn't (something like T^2 = R^3) until I saw an planet on the iphone exoplanet app, that is closer to it's star than a planet in another system, yet takes longer to complete one orbit. Both of the planets were a similar mass, as were the stars. &lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-04-15T20:07:51.930" LastActivityDate="2011-04-15T21:33:30.100" Title="Does mass affect speed of orbit at a certain distance?" Tags="&lt;classical-mechanics&gt;&lt;orbital-motion&gt;" AnswerCount="2" />
  <row Id="8710" PostTypeId="2" ParentId="8709" CreationDate="2011-04-15T20:04:48.500" Score="3" ViewCount="" Body="&lt;p&gt;In the limit where $m_2 \ll m_1$, only the mass of the &lt;em&gt;heavy&lt;/em&gt; body matters (along with the semi-major axis of the orbit, of course).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where that limit does not apply, varying the mass of either body changes the reduced mass:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mu = \frac{m_1 m_2}{m_1 + m_2} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the system acts as if a negligibly massive object was moving in the field of one having the reduced mass, this &lt;em&gt;does&lt;/em&gt; alter the period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that in the limit above $\mu \to m_2$ and we recover the expected behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Marion and Thorton give the full expression for the period $\tau$ in the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \tau^2 = \frac{4 \pi}{G} \frac{a^3}{m_1 + m_2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a$ is the length of the semi-major axis of the orbit and $G$ is the gravitational constant. It should be obvious that in the limit of a heavy primary this reduces to $\tau^2 = \frac{4 \pi}{G} \frac{a^3}{m_1}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Side comment: The rule you recall is the one Kepler found for planets in &lt;em&gt;our&lt;/em&gt; Solar System. In this case the mass of the sun dominates in every case. Jupiter is about 0.001 solar masses, so the largest correction in at the tenth of a percent level. Observable, but not at all large.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-04-15T20:21:45.183" LastActivityDate="2011-04-15T20:21:45.183" CommentCount="1" />
  <row Id="8711" PostTypeId="2" ParentId="6769" CreationDate="2011-04-15T20:36:01.730" Score="0" ViewCount="" Body="&lt;p&gt;The upward force on an object at rest is called the Normal force and is always perpendicular to the surface. If you recall from Newton's Third law, &quot;Every action has an equal and opposite reaction.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So an example is a block sitting on a table. The block is exerting a force DOWN on the table from the gravitational force, its weight. By Newton's Third law, there is an equal an opposite reaction due to this downward force. The block is &quot;pushing&quot; down on the table, so the table must also push UP on the block. This pushing from the table is the normal force. If this force were not present, the block would accelerate right through the table due to Newton's Second Law.&lt;/p&gt;&#xA;" OwnerUserId="3137" LastActivityDate="2011-04-15T20:36:01.730" CommentCount="3" />
  <row Id="8712" PostTypeId="2" ParentId="8709" CreationDate="2011-04-15T21:33:30.100" Score="1" ViewCount="" Body="&lt;p&gt;As @dmckee's answer says, in the limit where the mass of the planet is much less than the mass of the star, the mass of the planet does not have a significant effect on the period. I just want to add a more explicit comment on this part of your question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I saw an planet on the iphone&#xA;  exoplanet app, that is closer to it's&#xA;  star than a planet in another system,&#xA;  yet takes longer to complete one&#xA;  orbit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason for this is almost certainly not the masses of the planets but rather the masses of the two stars. The systems you're looking at in the app almost certainly satisfy the rule $m_{\rm planet}\ll m_{\rm star}$, so that the planet's mass is unimportant. You say that the masses of the stars are &quot;similar,&quot; but I bet they're different enough that that's the explanation for what you're seeing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way of writing Kepler's third law, as it applies to planets orbiting other stars, is&#xA;$$&#xD;&#xA;T^2={R^3\over M},&#xD;&#xA;$$&#xA;which is valid only in a certain choice of units: periods in years, radii in astronomical units, masses in solar masses. dmckee gives the more general formula. This version corresponds to a choice of units that makes the combination of constants $4\pi/G$ come out to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the app give you specific information about the numerical values of the various quantities? If so, you could check this. If not, are you sure about your statement that the masses are &quot;similar&quot;?&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-15T21:33:30.100" />
  <row Id="8713" PostTypeId="1" CreationDate="2011-04-15T21:48:40.540" Score="3" ViewCount="60" Body="&lt;p&gt;Maximum tidal heights vary widely across the globe, from 16 m in the Bay of Fundy to mere centimeters elsewhere. These variations are due to coastline and shoreline differences. This makes it difficult to determine a global average ocean tidal maximum height. Let's assume the Earth is a landless planet covered throughout by an average one km of ocean. How can we calculate the maximum ocean tidal height from a new moon or full moon syzygy, assuming semi-major Moon &amp;amp; Sun distances? &lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-04-16T11:32:40.917" Title="Determining Average Tidal Effects" Tags="&lt;gravity&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8714" PostTypeId="1" AcceptedAnswerId="8723" CreationDate="2011-04-15T21:55:12.113" Score="1" ViewCount="103" Body="&lt;p&gt;I was doing some reading on wikipedia and found it interesting that one page says the Grand Canonical Ensemble does not allow for exchange of particles, however another page says it does. So I went on google books and tried to look for a more trust worthy source, again the same happens one source says it allows the other says it doesn't so which is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Book that says it does allow exchange:&#xA;&lt;a href=&quot;http://www.scribd.com/doc/52426748/46/Grand-Canonical-Ensemble&quot; rel=&quot;nofollow&quot;&gt;http://www.scribd.com/doc/52426748/46/Grand-Canonical-Ensemble&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another that says it isn't allowed:&#xA;&lt;a href=&quot;http://books.google.co.uk/books?id=5sd9SAoRjgQC&amp;amp;pg=PA62&amp;amp;dq=grand+canonical+ensemble+exchange+particles&amp;amp;hl=en&amp;amp;ei=hL6oTZ2THc_p4wbEh-3DCg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CC0Q6AEwAA#v=onepage&amp;amp;q=grand%20canonical%20ensemble%20exchange%20particles&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.co.uk/books?id=5sd9SAoRjgQC&amp;amp;pg=PA62&amp;amp;dq=grand+canonical+ensemble+exchange+particles&amp;amp;hl=en&amp;amp;ei=hL6oTZ2THc_p4wbEh-3DCg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CC0Q6AEwAA#v=onepage&amp;amp;q=grand%20canonical%20ensemble%20exchange%20particles&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What gives?&lt;/p&gt;&#xA;" OwnerUserId="2057" LastEditorUserId="661" LastEditorDisplayName="" LastEditDate="2011-04-16T02:23:53.940" LastActivityDate="2011-04-16T02:23:53.940" Title="Does the Grand Canonical Ensemble allow for exchange of particles or not?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8715" PostTypeId="1" AcceptedAnswerId="8719" CreationDate="2011-04-15T22:19:06.303" Score="2" ViewCount="255" Body="&lt;p&gt;I will probably prove that I am an ignoramus by asking this question...&#xA;When answering questions about light, I see that we conveniently shift between wave and particle duality of light to match the answer: isn't this really cheating? &#xA;Or, is it the principle that the observation changes the nature of the object?&lt;/p&gt;&#xA;" OwnerUserId="3140" LastActivityDate="2011-04-16T04:55:18.960" Title="Isn't wave particle duality of light actually cheating?" Tags="&lt;light&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="8716" PostTypeId="2" ParentId="8713" CreationDate="2011-04-15T22:20:44.430" Score="0" ViewCount="" Body="&lt;p&gt;I think you would have to get into the fluid dynamics. A uniform ocean approximation would help a lot. With any luck rather than solving for a fully 3D time dependent flow, an assumption that the result is an expansion of a few well choosen spherical harmonics, with the same periodicity as the tidal frequency might yield a closed form solution (if you are lucky).&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-04-15T22:20:44.430" CommentCount="2" />
  <row Id="8717" PostTypeId="2" ParentId="8715" CreationDate="2011-04-15T22:30:36.553" Score="7" ViewCount="" Body="&lt;p&gt;The truth is that light (like other things) isn't a wave or a particle. It's another kind of a thing, namely an excitation of a quantum field. Under some circumstances, excitations of quantum fields behave like waves; under other circumstances, they behave like particles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not right to say that we switch descriptions &quot;to match the answer.&quot; Rather, we switch descriptions to match the question! Certain experimental setups and measurements are more naturally analyzed by thinking about things as waves; others are better analyzed in the particle framework. There are good reasons for using one or the other in any given situation. In effect, both wave and particle pictures are approximations that arise as limiting cases of the correct, exact theory. If you study how those approximations arise from the exact theory, then you can understand when one or the other is appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One final note. I &lt;em&gt;hate&lt;/em&gt; language like &quot;observation changes the nature of the object.&quot; The nature of light is always the same: excitation of a quantum field. Coincidentally, I ranted about this &lt;a href=&quot;http://blog.richmond.edu/physicsbunn/2011/04/13/particle-and-wave/&quot; rel=&quot;nofollow&quot;&gt;elsewhere&lt;/a&gt; a bit recently.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-15T22:30:36.553" CommentCount="3" />
  <row Id="8718" PostTypeId="2" ParentId="8659" CreationDate="2011-04-15T22:42:11.220" Score="1" ViewCount="" Body="&lt;p&gt;You seem to be forgetting that the distance (ie. the laps) will contract in length as the rocket ship approaches the speed of light. From the perspective of the pilot, EACH lap that he takes will be shorter in distance than the one that is seen from the point of view of the bystander. In order for this to be balanced out, time will dilate for the pilot.&lt;/p&gt;&#xA;" OwnerUserId="3120" LastEditorUserId="3120" LastEditorDisplayName="" LastEditDate="2011-04-16T04:53:03.140" LastActivityDate="2011-04-16T04:53:03.140" CommentCount="2" />
  <row Id="8719" PostTypeId="2" ParentId="8715" CreationDate="2011-04-15T22:56:07.753" Score="4" ViewCount="" Body="&lt;p&gt;You wouldn't think it's cheating if you understood the &quot;wave&quot; involved to be NOT a physical wave in some ether-like medium, but a measure of the information we have about the system (technically, a wavefunction -- the complete description of the system).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When there is no detector to access that information, the wave is spread out in space, more wave-like because of its spatial extent and displays all of the typical wavy behaviour of interference and diffraction. If you now have a detector that measures position, it will localize the wave in space to such a small region that it will &lt;em&gt;appear&lt;/em&gt; to behave like a particle, but really still is a wavepacket -- albeit very sharply peaked around the measured position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The details of how detectors and measurement devices are able to do this is partially described by the theory of quantum decoherence (some would argue, fully described). But I hope it's clear to you that the wave-particle duality is not some kind of ad hoc prescription; it's a natural consequence of asserting that the wavefunction is a complete description of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality#Treatment_in_modern_quantum_mechanics&quot; rel=&quot;nofollow&quot;&gt;Modern treatment of the wave-particle duality&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_decoherence&quot; rel=&quot;nofollow&quot;&gt;Quantum Decoherence&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-04-16T00:40:33.507" LastActivityDate="2011-04-16T00:40:33.507" CommentCount="2" />
  <row Id="8720" PostTypeId="1" CreationDate="2011-04-15T23:08:29.173" Score="0" ViewCount="115" Body="&lt;p&gt;I'm working on a simulation of water drops falling into a pool. I'm specifically interested in the waves generated by the impact of the drops. In order to calculate the vertical motion of the waves, I believe that I need to know the &quot;stiffness&quot; and damping constants of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if anyone can give me, or can point me to a book that has, those values? Would they be related to viscosity?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;I'm no physics expert, so if this is something that doesn't make sense to ask, please let me know and I'll edit or delete the question. My thinking is that I can treat the drop, whose mass and velocity I know, as an impulse input to the water which (at the point of impact) can be treated like a &quot;mass-on-a-spring&quot;. I am doing some deliberate oversimplification of the model; my goal is aesthetic, not engineering.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="3141" LastActivityDate="2011-04-16T11:18:59.037" Title="Damping and stiffness constants of water" Tags="&lt;simulation&gt;&lt;harmonic-oscillator&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8721" PostTypeId="2" ParentId="8708" CreationDate="2011-04-16T00:45:49.817" Score="1" ViewCount="" Body="&lt;p&gt;An experiment very close to the proposed experiment has been done by Zeilingers grad-student Dopfler in 1998. She used a down-conversion crystal to produce pairs of entangled photons in a quantum-eraser type of experiment involving a dual-slit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the members went through the dual-slit and was detected by a detector A which is scanning the space behind the slits to see if there is an interference pattern or not. The other member passes through a lens to another detector B whose distance to the lens can be varied (angle is fixed). By moving the detector B into or out of focus with the lens, the other pair member can be detected as passing through one of the slits (in focus, it &quot;sees&quot; the slit holes) or the which-way information is erased (out of focus, information from both slits are merged).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A useful way of looking at these experimental setups is to pretend that the photon is &lt;strong&gt;emitted&lt;/strong&gt; by one of the detectors, passing backwards through the experiment, through the down-conversion crystal with momentum intact, and finally being absorbed by the other detector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This experiment is thus described simply as the lens + detector B either watching the which-slit information or not at the other section of the experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Dopfler and &lt;a href=&quot;http://www.hep.yorku.ca/menary/courses/phys2040/misc/foundations.pdf&quot; rel=&quot;nofollow&quot;&gt;Zeilinger&lt;/a&gt; the experiment worked, but I have not read anything about it since and the original dissertation has been pulled from the web, but a &lt;a href=&quot;http://web.archive.org/web/20070714025355/www.quantum.univie.ac.at/publications/thesis/bddiss.pdf&quot; rel=&quot;nofollow&quot;&gt;copy&lt;/a&gt; can be found at the internet archive. Since it used a coincidence counter to increase the signal-to-noise between both detectors, they had not really demonstrated any FTL-signalling, however the speculation is that the experiment could run without the counter. Zeilinger calls the conceptual alternative to spacelike signalling &quot;retrocausality&quot; I think, where the cause is timelike in both paths after the down-conversion, but runs backwards in one of them (like in the &quot;pretend&quot; tool mentioned above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J. &lt;a href=&quot;http://en.wikipedia.org/wiki/John_G._Cramer&quot; rel=&quot;nofollow&quot;&gt;Cramer&lt;/a&gt; apparently currently works on refining this experiment.&lt;/p&gt;&#xA;" OwnerUserId="2936" LastEditorUserId="2936" LastEditorDisplayName="" LastEditDate="2011-04-16T01:02:17.547" LastActivityDate="2011-04-16T01:02:17.547" CommentCount="1" />
  <row Id="8722" PostTypeId="2" ParentId="8704" CreationDate="2011-04-16T01:22:26.500" Score="5" ViewCount="" Body="&lt;p&gt;Lagerbaer gets it right in the comments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of all (properly [anti-]symmetrized) product states is a complete basis. That's it, game over. So the first sum in your series represents every possible state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could also make (generally overcomplete) bases that consist entirely of products of maximally entangled pairs -- for example the valence bond basis is the natural description for resonating valence bond (or valence bond solid) states. Thus the second term in your series, by itself, could also represent a complete many-body basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there is a flaw implicit in the notation, and this has to do with the monogamy of entanglement -- if two particles in my system are maximally entangled, they cannot be entangled with anything else! My point being: you have to be more clear with what you mean by &quot;entanglement in a BEC.&quot; Yes a BEC has long-range &lt;em&gt;correlations,&lt;/em&gt; but the ideal BEC &lt;em&gt;is still a product state&lt;/em&gt;!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, as far as estimating relative contributions, writing down the best possible product ground state for a certain interacting hamiltonian is essentially mean-field theory. The true ground state will have corrections that will be superpositions of states, but still in the product basis. As a side-note, for a particularly cute way of going slightly beyond MFT in a language that helps to understand what entanglment entails in many-body states, look up Steve White's &quot;Minimally Entangled Typical Thermal States&quot; algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT: after edit in the OP...&lt;/strong&gt;&#xA;Okay, so the point is that non-separable states are not such different objects from separable states (at least when talking about pure states). The structure of Hilbert space is such that when I take two objects and combine them, I get a complete basis from all possible tensor products of the two original bases! As an example, suppose I mix two qubits: I can write &lt;em&gt;any state at all&lt;/em&gt; as the vector&#xA;$$|\Psi\rangle=\sum_{ij}c_{ij}|i\rangle|j\rangle$$&#xA;Nonseparable states are states like Bell states with, for example, $c_{01}=c_{10}=0, c_{11}=c_{00}=1/\sqrt{2}$. There is no separable part versus non-separable part, the state is a superposition of product states, and is non-separable because it cannot be written as a single product without the sum. From your edit it seems like you understand this, but I feel the need to be explicit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally we might imagine writing states in the Bell basis; no one can stop me from writing&#xA;$$|\Psi\rangle=c_{1}(|00\rangle+|11\rangle)+c_{2}(|00\rangle-|11\rangle)+c_{3}(|10\rangle+|01\rangle)+c_{4}(|10\rangle-|01\rangle)$$&#xA;It seems like this is what you imagine as the second term in your series, but it's nothing more than a rotation of my original basis. I don't gain anything at all by mixing my descriptions -- the original basis already provided a complete description of nonseparable states.&lt;/p&gt;&#xA;" OwnerUserId="791" LastEditorUserId="791" LastEditorDisplayName="" LastEditDate="2011-04-17T14:10:22.093" LastActivityDate="2011-04-17T14:10:22.093" CommentCount="10" />
  <row Id="8723" PostTypeId="2" ParentId="8714" CreationDate="2011-04-16T02:12:56.247" Score="6" ViewCount="" Body="&lt;p&gt;There are, essentially, three types of ensembles used in thermodynamics:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Microcanonical Ensemble:&lt;/strong&gt; This is used to describe closed systems. The number of particles and the total energy are constant (since no interaction with the environment takes place). In such a system, the entropy will eventually be maximized.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Canonical Ensemble:&lt;/strong&gt; Now, your system is in contact with a big reservoir and allowed to exchange &lt;em&gt;energy&lt;/em&gt; with the system. This means that your system will be in thermal equilibrium with the bath. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Grand Canonical Ensemble&lt;/strong&gt; Now, your system can exchange both energy and particles with the system. Hence, it will be in thermal equilibrium with the bath &lt;em&gt;and&lt;/em&gt; in chemical equilibrium, i.e. the chemical potential for adding a particle to your system equals the chemical potential for adding a particle to the bath.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If exchanging particles wasn't allowed in the grand canonical ensemble, it would become the canonical ensemble.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-04-16T02:12:56.247" />
  <row Id="8725" PostTypeId="2" ParentId="8715" CreationDate="2011-04-16T04:01:11.360" Score="7" ViewCount="" Body="&lt;p&gt;Yes, it's cheating. Treating light as &lt;em&gt;either&lt;/em&gt; waves or particles is cheating. If you don't want to cheat, you should do everything from first principles in quantum optics and never mention either waves or particles (both of which are only approximations to the true nature of light). Unfortunately, to do that you have to pretty much throw all your intuition out the window. Cheating is much easier.&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2011-04-16T04:01:11.360" />
  <row Id="8726" PostTypeId="2" ParentId="8384" CreationDate="2011-04-16T04:12:36.950" Score="1" ViewCount="" Body="&lt;p&gt;Turns out, EPAnet can actually do what I want.  It's a little hairy to try to plug into, as I don't think any real updates have come for 5 years or more, but there's several commercial products that base their underlying engine on EPAnet, even if they extend or tweak it a bit.  This little fact gave me more confidence plowing forward that I wasn't jumping down a rabbit hole.&lt;/p&gt;&#xA;" OwnerUserId="3038" LastEditorUserId="3038" LastEditorDisplayName="" LastEditDate="2011-04-18T17:26:55.350" LastActivityDate="2011-04-18T17:26:55.350" />
  <row Id="8727" PostTypeId="1" CreationDate="2011-04-16T04:41:29.193" Score="1" ViewCount="63" Body="&lt;p&gt;Or are they in all subshells??&lt;/p&gt;&#xA;" OwnerUserId="3135" LastActivityDate="2011-04-16T16:36:44.313" Title="Are valence electrons located solely in the s and p subshells?" Tags="&lt;physical-chemistry&gt;" AnswerCount="1" />
  <row Id="8728" PostTypeId="2" ParentId="8715" CreationDate="2011-04-16T04:55:18.960" Score="1" ViewCount="" Body="&lt;p&gt;To add a lighter note, a reminder of the four blind men and the elephant story:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What is an elephant like&lt;/em&gt;, asks the first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It is like a snake&lt;/em&gt; says the second, holding the trunk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;You are  wrong&lt;/em&gt;, says the third, &lt;em&gt;It is like the trunk of a tree&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Ha ha ha&lt;/em&gt; says the fourth,  &lt;em&gt;what a great imagination you all have, to see this great big wall as snakes and trees&lt;/em&gt;  leaning on the elephants side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Strange! you are all confused&lt;/em&gt;, said the first, &lt;em&gt;I have caught a long piece of string&lt;/em&gt; (wink, wink)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least we have  a dual ambiguity  only:) .&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-16T04:55:18.960" />
  <row Id="8729" PostTypeId="1" CreationDate="2011-04-16T05:13:59.810" Score="0" ViewCount="101" Body="&lt;p&gt;What are the currently relatively unexplored areas of physics which are promising in terms of potential?&lt;/p&gt;&#xA;" OwnerUserId="3144" LastActivityDate="2011-04-16T05:13:59.810" Title="What are the currently relatively unexplored areas of physics which are promising?" Tags="&lt;soft-question&gt;" CommentCount="6" />
  <row Id="8730" PostTypeId="2" ParentId="8356" CreationDate="2011-04-16T05:37:18.880" Score="3" ViewCount="" Body="&lt;p&gt;There are some recent efforts in trying to derive the mathematical structure of quantum mechanics from some reasonable and/or operational axioms. You may want to give a look, for example, at &lt;a href=&quot;http://arxiv.org/abs/1011.6451&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1011.6451&lt;/a&gt; and references therein.&lt;/p&gt;&#xA;" OwnerUserId="3145" LastActivityDate="2011-04-16T05:37:18.880" />
  <row Id="8731" PostTypeId="1" CreationDate="2011-04-16T06:09:44.263" Score="0" ViewCount="79" Body="&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Wa have a constant force $F_1$, always perpendicular to the course of an object. At time $t_0$, the momentum is 0.2 kg m/s and after 0.05 sec we find that the change in momentum &lt;strong&gt;VECTOR&lt;/strong&gt; was 0.2 kg m/s. What is the net force?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the help&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: fixed the text of the excercise (in &lt;strong&gt;bold&lt;/strong&gt;), and deleted my proposed (wrong) solution&lt;/p&gt;&#xA;" OwnerUserId="2342" LastEditorUserId="2342" LastEditorDisplayName="" LastEditDate="2011-04-16T13:40:28.877" LastActivityDate="2011-04-16T13:40:28.877" Title="Excercise: impulse &amp; circular motion" Tags="&lt;newtonian-mechanics&gt;&lt;momentum&gt;" CommentCount="6" />
  <row Id="8732" PostTypeId="1" CreationDate="2011-04-16T09:55:03.677" Score="2" ViewCount="143" Body="&lt;p&gt;At the current LHC luminosities, will it take years to detect the Higgs boson, superpartners, or any other forms of new physics at the LHC? What should particle physicists do in the meantime?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We particle physicists have been waiting too long now for experimental input.&lt;/p&gt;&#xA;" OwnerUserId="3146" LastEditorUserId="3146" LastEditorDisplayName="" LastEditDate="2011-04-16T10:11:40.650" LastActivityDate="2011-04-16T11:39:19.843" Title="When do we expect to get results from the LHC?" Tags="&lt;particle-physics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8733" PostTypeId="1" CreationDate="2011-04-16T10:09:17.503" Score="0" ViewCount="225" Body="&lt;p&gt;I am interested in examples of crackpots coming up with correct results in physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do mainstream physicists look down so much upon &quot;crackpots&quot;?&lt;/p&gt;&#xA;" OwnerUserId="3147" LastActivityDate="2011-04-16T13:28:42.230" ClosedDate="2011-04-16T13:41:09.593" Title="Give me examples of crackpots who were right after all" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="8734" PostTypeId="2" ParentId="8706" CreationDate="2011-04-16T10:23:46.567" Score="2" ViewCount="" Body="&lt;p&gt;The LHC filling-scheme coding is described &lt;a href=&quot;http://lpc.web.cern.ch/lpc/fillingschemes.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Found after some digging :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very cool live displays, I must say! It's almost like having your own backyard accelerator...&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-04-16T10:23:46.567" CommentCount="6" />
  <row Id="8735" PostTypeId="1" CreationDate="2011-04-16T11:05:12.827" Score="2" ViewCount="171" Body="&lt;p&gt;Since every question has to be asked in a seperate topic, &#xA;I'm asking a question refering to the following topic:&#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/8201/beginners-questions-concerning-conformal-field-theory&quot;&gt;Beginners questions concerning Conformal Field Theory&lt;/a&gt;&#xA;In particula I'm refering to the subsection &quot;Implementing a symmetry on operators&quot; of the answer given by Lubos Motl. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear to me that we obtain $\delta\phi=i\epsilon [L_{m},\phi]=i \epsilon z^{m+1} \partial_{z} \phi$ and that $z^{m+1} \partial_{z}$ are the generators of conformal symmetry.&#xA;What confuses me though is that these generators fulfill the Witt-Algebra and not&#xA;the Virasoro Algebra. Since we're in the quantum case and we want to show that the theory is invariant on a quantum level, shouldn't we get generators that fulfill the Virasoro Algebra?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for the responses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best regards.&lt;/p&gt;&#xA;" OwnerUserId="2969" LastActivityDate="2011-04-18T18:39:05.240" Title="Question on Conformal Field Theory" Tags="&lt;conformal-field-theory&gt;" AnswerCount="2" />
  <row Id="8736" PostTypeId="2" ParentId="8720" CreationDate="2011-04-16T11:18:59.037" Score="1" ViewCount="" Body="&lt;p&gt;Water waves are rather complicated, because the span very different &#xA;things: so called capillary waves (shortest wavelenghts, surface tension is important) &#xA;to gravity waves, which are a complicated movement (elliptical) of water, &#xA;longitudional and transversal at the same time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This wiki is a good point to start:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Wind_wave&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Wind_wave&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If You want to simulate the primary action of a drop falling into a liquid,&#xA;that is much more complicated, because absolutely outside of waves physics. &#xA;Maybe You have seen pictures from high-speed cameras when a &#xA;kind of &quot;crown&quot; is formed? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.clipfish.de/video/2001849/slow-motion-wassertropfen/&quot; rel=&quot;nofollow&quot;&gt;http://www.clipfish.de/video/2001849/slow-motion-wassertropfen/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Stiffness&quot; is not a property of any liquid. maybe You thought of viscosity? &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-16T11:18:59.037" CommentCount="1" />
  <row Id="8737" PostTypeId="2" ParentId="8713" CreationDate="2011-04-16T11:32:40.917" Score="1" ViewCount="" Body="&lt;p&gt;The wiki on tides has a line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;, but also because wave propagation&#xA;  across the ocean has a natural period&#xA;  of the same order of magnitude as the&#xA;  rotation period: if there were no land&#xA;  masses, it would take about 30 hours&#xA;  for a long wavelength surface wave to&#xA;  propagate along the equator halfway&#xA;  around the Earth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This implies, that somebody calculated that idealized ocean-earth &#xA;already. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-16T11:32:40.917" CommentCount="2" />
  <row Id="8738" PostTypeId="2" ParentId="8732" CreationDate="2011-04-16T11:39:19.843" Score="1" ViewCount="" Body="&lt;p&gt;As particle physicists know, the detection of Higgs at the LHC  will depend on the cross section for producing it as well as at what mass it is. There exists a &lt;a href=&quot;http://cdsweb.cern.ch/record/1318996/files/cern-2011-002.pdf&quot; rel=&quot;nofollow&quot;&gt;working group&lt;/a&gt; &lt;em&gt;Report of the LHC Higgs Cross Section Working Group&lt;/em&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculations show that the higher the mass of the Higgs the smaller the cross section of inclusive production at pp scattering in the LHC energy range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some limits already &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1104/1104.1619v1.pdf&quot; rel=&quot;nofollow&quot;&gt;given by CMS&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is patience and keeping fingers crossed that nature has a low mass Higgs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://personalpages.to.infn.it/~bolognes/docu/HiggsLCWS.pdf&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is an analysis for low luminosity detection.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-16T11:39:19.843" CommentCount="2" />
  <row Id="8739" PostTypeId="2" ParentId="8733" CreationDate="2011-04-16T12:27:14.047" Score="2" ViewCount="" Body="&lt;p&gt;There's some nice articles on crackpots who were right at viXra blog:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://blog.vixra.org/category/crackpots-who-were-right/&quot; rel=&quot;nofollow&quot;&gt;http://blog.vixra.org/category/crackpots-who-were-right/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://blog.vixra.org/2010/03/06/%E2%80%9Ccrackpots%E2%80%9D-who-were-right-1-boris-belousov/&quot; rel=&quot;nofollow&quot;&gt;http://blog.vixra.org/2010/03/06/%E2%80%9Ccrackpots%E2%80%9D-who-were-right-1-boris-belousov/&lt;/a&gt;&#xA;&lt;a href=&quot;http://blog.vixra.org/2010/03/12/%E2%80%9Ccrackpots%E2%80%9D-who-were-right-2-alfred-wegener/&quot; rel=&quot;nofollow&quot;&gt;http://blog.vixra.org/2010/03/12/%E2%80%9Ccrackpots%E2%80%9D-who-were-right-2-alfred-wegener/&lt;/a&gt;&#xA;&lt;a href=&quot;http://blog.vixra.org/2010/03/17/crackpots-who-were-right-3-ernst-stuckelberg/&quot; rel=&quot;nofollow&quot;&gt;http://blog.vixra.org/2010/03/17/crackpots-who-were-right-3-ernst-stuckelberg/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading through the list they all appear to have been highly educated in their particular field anyway and so more likely to succeed compared to the majority that are uneducated and deluded, yet think they're some new Einstein.&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="1345" LastEditorDisplayName="" LastEditDate="2011-04-16T12:33:24.383" LastActivityDate="2011-04-16T12:33:24.383" CommentCount="4" />
  <row Id="8740" PostTypeId="2" ParentId="8733" CreationDate="2011-04-16T13:28:42.230" Score="-1" ViewCount="" Body="&lt;p&gt;I think there is a genre of physicists writing on the internet, where good manners mean not using four letter words, who use the term &quot;crackpot&quot; to  describe physicists who disagree with their beliefs about physics current orthodoxy. It is as if a consensus model of physics is sought rather than an exploration of truth.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do mainstream physicists look down so much upon &quot;crackpots&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There exist true crackpots in physics, people who think they have proven Einstein wrong with two lines of incoherence, people who have found a perpetual motion machine etc. They are like all those Napoleons in the asylums, just not so deep in the delusion. It is easy to look down upon such claims.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It becomes more difficult if the  candidate for crack pottery is well versed in mathematics, because they develop convoluted mathematical arguments that one has to spend time with to really dismiss. That is why in physics candidates for the crackpot label are theoreticians. I would draw the line there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To call crackpots people seeking alternative quantum field theories for gravity, or different ways of describing quantum mechanics, is abuse of the concept. It displays the herd mentality, of which mammals  partake and physicists are mammals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Progress in physics happens at the fringes . If the researchers are well versed, sincere and hard working they should be allowed to develop their theories on equal footing with mainstream science, to be tested and rejected if necessary by experiments. Otherwise physics will become fossilized.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-16T13:28:42.230" CommentCount="6" />
  <row Id="8741" PostTypeId="1" AcceptedAnswerId="8742" CreationDate="2011-04-16T13:57:44.920" Score="0" ViewCount="99" Body="&lt;p&gt;Introductory texts and popular accounts of why we see the &quot;once hot&quot; CMB as microwaves nearly always say something about the photons &quot;cooling&quot; since the Big Bang. But isn't that misleading? Don't those photons have long (&quot;cool&quot;) wavelengths because space expanded since they were emitted. There's no separate &quot;cooling&quot; process, is there?&lt;/p&gt;&#xA;" OwnerUserId="2540" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-04-16T23:37:26.807" LastActivityDate="2011-04-16T23:37:26.807" Title="Have CMB photons &quot;cooled&quot; or been &quot;stretched&quot;?" Tags="&lt;cosmology&gt;&lt;photons&gt;&lt;space-expansion&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8742" PostTypeId="2" ParentId="8741" CreationDate="2011-04-16T14:17:51.227" Score="3" ViewCount="" Body="&lt;p&gt;Cooling and stretching essentially mean the same thing here. The temperature of any blackbody radiation is related to the peak wavelength by &lt;a href=&quot;http://en.wikipedia.org/wiki/Wien%27s_displacement_law&quot; rel=&quot;nofollow&quot;&gt;Wien's Law&lt;/a&gt; $$\lambda_{\mathrm{max}} = \frac{b}{T} $$ Therefore as the universe expands, all of the photon wavelengths get stretched out and so does the peak wavelength. $$\lambda \propto a(t)$$ This decreases the temperature of the radiation by the same factor that expands the wavelength of the photons. $$ T \propto \frac{1}{a(t)} $$ This factor $a(t)$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Scale_factor_%28cosmology%29&quot; rel=&quot;nofollow&quot;&gt;scale factor&lt;/a&gt; of the universe and increases with time for an expanding universe.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-04-16T14:17:51.227" />
  <row Id="8743" PostTypeId="2" ParentId="8741" CreationDate="2011-04-16T14:19:31.447" Score="2" ViewCount="" Body="&lt;p&gt;I think the best way to think about it is that the sentence &quot;the photons have cooled&quot; is simply describing a fact, not explaining that fact. At early times, the photons at any given location had a thermal (blackbody) distribution corresponding to a high temperature (as measured by observers at rest in the natural, comoving reference frame). At later times, the photons at any given location had a thermal distribution corresponding to a lower temperature. That's what we mean when we say that they &quot;cooled.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it's then very natural to ask &lt;em&gt;why&lt;/em&gt; they cooled. That's where the &quot;stretching of space&quot; explanation comes in. I think that that explanation is problematic, as I wrote at great length &lt;a href=&quot;http://arxiv.org/abs/0808.1081&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, but others disagree.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-16T14:19:31.447" CommentCount="4" />
  <row Id="8744" PostTypeId="1" AcceptedAnswerId="8754" CreationDate="2011-04-16T15:08:59.567" Score="1" ViewCount="117" Body="&lt;p&gt;What is a Kustaanheimo-Stiefel transformation? Which applications has it in physics? Can you point me to a reference, where this transformation is explained?&lt;/p&gt;&#xA;" OwnerUserId="1648" LastActivityDate="2011-04-16T21:53:35.233" Title="What is a Kustaanheimo-Stiefel transformation?" Tags="&lt;coordinate-systems&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8745" PostTypeId="1" CreationDate="2011-04-16T15:54:44.213" Score="4" ViewCount="115" Body="&lt;p&gt;Is it true that a spontaneously light-emitting atom changes its lifetime if it is put between two parallel plates that are so near that they attract each other through the Casimir effect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus: does the Casimir effect, and the changes in the vacuum state it induces, influence spontanenous emission?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, who measured this effect for the first time? Where can I read about it?&lt;/p&gt;&#xA;" OwnerUserId="3148" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-04-17T05:09:14.933" LastActivityDate="2011-04-17T05:09:14.933" Title="Does the Casimir effect allow to change the lifetime of a radiating atom?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-optics&gt;&lt;quantum-electrodynamics&gt;&lt;cavity-qed&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="8746" PostTypeId="2" ParentId="8745" CreationDate="2011-04-16T16:31:40.853" Score="5" ViewCount="" Body="&lt;p&gt;This is true. The simple explanation is this: For calculating the decay rate of an excited state, you use Fermi's Golden Rule, which involves the matrix element &#xA;$$|\langle f | V | i \rangle|^2$$&#xA;where $f$ and $i$ denote the final and initial state, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the final state contains the electron in its groundstate together with a photon created by this decay, the nature of your cavity determines what the matrix element will be:&#xA;For example, if your cavity forbids standing waves of the emission frequency, decay is suppressed. The study of these effects goes under the name of &lt;a href=&quot;http://en.wikipedia.org/wiki/Cavity_quantum_electrodynamics&quot; rel=&quot;nofollow&quot;&gt;Cavity QED&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would not say that this is &lt;em&gt;due to&lt;/em&gt; the Casimir effect. Rather, this effect and the Casimir effect are both due to the boundary conditions created by your plates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't exactly know who first studied this. I suggest consulting a review article such as &lt;a href=&quot;http://iopscience.iop.org/0034-4885/69/5/R02&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-04-16T16:31:40.853" />
  <row Id="8747" PostTypeId="2" ParentId="8727" CreationDate="2011-04-16T16:36:44.313" Score="3" ViewCount="" Body="&lt;p&gt;If you define valence electrons as those that belong to open shells, or as those who participate in chemical bonding, then no: The transition metal elements have open $d$-shells, and they play an important role in determining the properties of transition metal compounds.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-04-16T16:36:44.313" />
  <row Id="8748" PostTypeId="1" AcceptedAnswerId="8767" CreationDate="2011-04-16T17:06:45.230" Score="1" ViewCount="236" Body="&lt;p&gt;Why is it that photons are emitted in bundles? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My physics teacher's answer was &quot;it's complicated&quot;... &lt;/p&gt;&#xA;" OwnerUserId="2536" LastActivityDate="2011-06-20T03:27:58.717" Title="Nature of Photons" Tags="&lt;quantum-mechanics&gt;&lt;photons&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="8749" PostTypeId="1" AcceptedAnswerId="8753" CreationDate="2011-04-16T17:17:27.573" Score="0" ViewCount="106" Body="&lt;p&gt;&lt;a href=&quot;http://img696.imageshack.us/img696/9046/sphs.png&quot; rel=&quot;nofollow&quot;&gt;Figure&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dielectric shaped as on the figure has dielectric constant $\varepsilon=\varepsilon\left(r\right)$ and free charge density $\rho=\rho\left(r\right)$. What is the electric field and electrostatic potential as a function of $r$? The above is 3D object. We have  $R_{1}$, $R_{2}$ and $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really have no idea how to approach this kind problems. Does electric field in $r&amp;lt;R_{1}$ equal zero? Is the object a Gaussian surface? Does electric field in $R_{1} &amp;lt; r &amp;lt; R_{2}$ depend on dielectric constant?&lt;/p&gt;&#xA;" OwnerUserId="3138" LastActivityDate="2011-04-17T08:51:15.317" Title="Electric field of a charged spherical surface" Tags="&lt;homework&gt;&lt;electrostatics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8750" PostTypeId="2" ParentId="8748" CreationDate="2011-04-16T17:35:05.560" Score="-1" ViewCount="" Body="&lt;p&gt;An atom emits a photon when an electron moves from a higher energy orbital to a lower energy orbital. The energy difference between the two levels equals the energy contained within the &quot;bundle&quot; or &quot;packet&quot; of energy known as a photon. Most light sources have many atoms, any of which may emit a photon, thus resulting in many photons or &quot;bundles.&quot; &#xA;In defense of &quot;bundles of energy&quot; (since this is now at -1) read the top answer &lt;a href=&quot;http://physics.stackexchange.com/questions/11270/photon-absorption&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2843" LastEditorUserId="2843" LastEditorDisplayName="" LastEditDate="2011-06-20T03:27:58.717" LastActivityDate="2011-06-20T03:27:58.717" CommentCount="6" />
  <row Id="8751" PostTypeId="2" ParentId="8748" CreationDate="2011-04-16T17:45:34.863" Score="4" ViewCount="" Body="&lt;p&gt;In general photons are emitted one by one from individual atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect you were discussing laser photons, and also suspect that you still are in high school. Photons from lasers do come in bunches, and your teacher is correct that it is not simple to explain it. Let me try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you seen the domino set ups, where the tiles are finely balanced and then the first one is pushed and all the rest start falling in steps?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In lasers the atoms are artificially pushed up to an excited state, and when a photon of the right frequency passes, the atoms fall to the lower state all in step and thus appear as bunches.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-16T17:45:34.863" CommentCount="3" />
  <row Id="8752" PostTypeId="1" AcceptedAnswerId="8756" CreationDate="2011-04-16T17:50:08.460" Score="5" ViewCount="324" Body="&lt;p&gt;I'm familiar with sigma, and how its usually calculated and  used, but would like to know how it's applied to particle physics. I recall reading that the discovery of the Higgs would only be credible unless it meets a 5 sigma criterion. What are the particle physics factors/numbers that would go into a sigma formula for a credible Higgs discovery? &lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-04-16T19:37:59.643" Title="Standard Deviation in Particle Physics" Tags="&lt;particles&gt;&lt;statistics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="8753" PostTypeId="2" ParentId="8749" CreationDate="2011-04-16T18:02:37.377" Score="1" ViewCount="" Body="&lt;p&gt;You are looking for a solution of Gauss's law $$\nabla \cdot \mathbf{E}(r) =  \rho(r)/\epsilon_0\epsilon(r).$$&#xA;Because the system is static you can find a potential $\Phi(r)$ such that the electric field is given by $$\mathbf{E}(r) = - \nabla \Phi(r).$$  Thus, the equation becomes $$\Delta \Phi(r) = \frac{1}{r^2} \partial_r [ r^2 \partial_r \Phi(r)] = - \rho(r)/\epsilon_0\epsilon(r);$$&#xA;in the last step we used the isotropy of the system. Solving this ordinary differential equation, you obtain the electric potential $\Phi(r)$. In order to solve this equation, we use the substitution $\Phi(r) =f(r)/r$, after which the differential equation reads&#xA;$$f&amp;#39;&amp;#39;(r)=- r\rho(r)/\epsilon_0\epsilon(r).$$ Integrating twice yields&#xA;$$f(r) = -\int_0^r dr&amp;#39; \int_0^{r&amp;#39;} dr&amp;#39;&amp;#39; \frac{r&amp;#39;&amp;#39; \rho(r&amp;#39;&amp;#39;)}{\epsilon_0\epsilon(r&amp;#39;&amp;#39;)},$$&#xA;where we used the boundary conditions $f(0)=0$, $f&amp;#39;(0)=0$ (which requires that $\rho(r)$ goes to 0 for $r\to0$ sufficiently fast). In principal, you can add $c r$ to $f$ with $c$ an arbitrary constant in order to fulfill &quot;any&quot; boundary condition. Putting everything together, we have&#xA;$$\Phi(r) = -\int_0^r dr&amp;#39; \int_0^{r&amp;#39;} dr&amp;#39;&amp;#39; \frac{r&amp;#39;&amp;#39; \rho(r&amp;#39;&amp;#39;)}{r \epsilon_0\epsilon(r&amp;#39;&amp;#39;)}$$&#xA;and $$E_r(r)= -\partial_r \Phi(r) = \int_0^{r} dr&amp;#39; \frac{r&amp;#39; \rho(r&amp;#39;)}{r \epsilon_0\epsilon(r&amp;#39;)} -  \int_0^r dr&amp;#39; \int_0^{r&amp;#39;} dr&amp;#39;&amp;#39; \frac{r&amp;#39;&amp;#39; \rho(r&amp;#39;&amp;#39;)}{r^2 \epsilon_0\epsilon(r&amp;#39;&amp;#39;)}.$$ For more concrete results, you need to provide the explicit form of $\rho(r)$ and $\epsilon(r)$.&lt;/p&gt;&#xA;" OwnerUserId="2962" LastEditorUserId="2962" LastEditorDisplayName="" LastEditDate="2011-04-17T08:51:15.317" LastActivityDate="2011-04-17T08:51:15.317" CommentCount="2" />
  <row Id="8754" PostTypeId="2" ParentId="8744" CreationDate="2011-04-16T18:05:29.247" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/0803.4441&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0803.4441&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Kustaanheimo-Stiefel transform turns a gravitational two-body problem into a harmonic oscillator, by going to four dimensions. In addition to the mathematical-physics interest, the KS transform has proved very useful in N-body simulations, where it helps handle close encounters. Yet the formalism remains somewhat arcane, with the role of the extra dimension being especially mysterious. This paper shows how the basic transformation can be interpreted as a rotation in three dimensions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-16T18:05:29.247" />
  <row Id="8755" PostTypeId="1" AcceptedAnswerId="8764" CreationDate="2011-04-16T18:28:35.113" Score="8" ViewCount="308" Body="&lt;p&gt;Suppose I have an hourglass that takes 1 full hour on average to drain.  The grains of sand are, say, $1 \pm 0.1\ {\rm mm}$ in diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I replace this with very finely-grained sand $0.1 \pm 0.01\ {\rm mm}$ in diameter but keep the hourglass otherwise the same, how long a time will the &quot;hourglass&quot; now measure?  Does this depend on the size of the funnel, or should all one-hour hourglasses change in roughly the same way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus:  is the new hourglass more or less precise?  (precision defined at $\sigma_t /t$, with $t$ the time to drain)&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-04-16T18:38:38.827" LastActivityDate="2011-04-17T11:13:45.077" Title="How does the period of an hourglass depend on the grain size?" Tags="&lt;classical-mechanics&gt;&lt;scaling&gt;&lt;dimensional-analysis&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="8756" PostTypeId="2" ParentId="8752" CreationDate="2011-04-16T18:49:03.280" Score="5" ViewCount="" Body="&lt;p&gt;Errors in particle physics are of two kinds. Statistical, and systematic. Statistical is the usual standard deviation of gausian distributions, sqrt(n)/N for 1 sigma. It is the systematics that take a lot of effort, and often are not taken well into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Systematic errors come from &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) the background to the signal expected. The background is calculated theoretically and entered into a monte carlo program that simulates the production of the data. There are also errors in the  parameters the theory has used  that enter as systematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)Effects from the limitations of the measuring apparatus, which is also inserted into the monte carlo simulation of events&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) from the method of analysis, i.e. the cuts made in order to isolate the signal in the data and in the simulated events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Systematic errors are added linearly to the deviation from background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, the monte carlo simulation will have many more events than the data and so the statistical  error from the MC can be ignored. One should estimate the errors from uncertainty in theory and from defects of detectors by varying the parameters in the MC program to the 1 sigma level of the important parameters and cuts and observing the change in the distributions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there are so many parameters and cuts that this procedure is not adhered to strictly, as happened with the new 3 sigma announcement of the &quot;fifth force&quot; by CDF, where the errors are just statistical from the events.  That is why for discovery one asks 5 sigma. It is very hard to have an effect of 5 sigma due to the systematics enumerated above, whereas 3 sigma announcements have often disappeared . The ALEPH Higgs for example was a 3 sigma one that disappeared when the other 3  LEP experiments looked. I have known a 4 sigma resonance that was the effect of not estimating the systematics of the cuts. 5 sigma is for playing it safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should add here that the different systematics in different experiments are the main reason why at least two expensive experimental set ups are approved and built in the collider setups: independent confirmation in parallel .&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-04-16T19:36:27.700" LastActivityDate="2011-04-16T19:36:27.700" CommentCount="4" />
  <row Id="8757" PostTypeId="2" ParentId="8755" CreationDate="2011-04-16T19:28:34.030" Score="1" ViewCount="" Body="&lt;p&gt;As a first guess, I would think the strain rate under a fixed amount of shear stress for a unit cube of material should be expected to be proportional to the surface area of the grains. That would imply that the strain rate, and hence the flow rate for fixed geometry should be inversely proportional to grain size. So I think your fine grained hourglass would be much faster. I suspect in reality things would get more complicated then that. What shape are the grains? Spherical is probably faster than angular. What about grain on grain sliding coefficient of friction? Is there some stress level below which no flow/creep takes place?&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-04-16T19:28:34.030" />
  <row Id="8758" PostTypeId="2" ParentId="8752" CreationDate="2011-04-16T19:37:59.643" Score="-2" ViewCount="" Body="&lt;p&gt;Those 5 sigmas are much older than particle physics.&#xA;Gauß developed this calculus when he got a order from &#xA;Hannover Administration to check the precision of a &#xA;big trianglation reaching from Harz mountains to Neuwerk. &#xA;(One of the first triangulations by modern standards) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since then standard deviation was used in all fields of &#xA;precision measurement, about 200 Years now. Surveyors, &#xA;optics and machinery parts, were the pioneers in this field.  &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-16T19:37:59.643" CommentCount="3" />
  <row Id="8759" PostTypeId="2" ParentId="8755" CreationDate="2011-04-16T20:00:21.193" Score="1" ViewCount="" Body="&lt;p&gt;Flow of granular media is very complicated, depending on form of the &#xA;grains. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one big difference to real liquids: whereas liquids contain &#xA;molecules which can/do move by Brownian motion, granular solids &#xA;do not have such a &quot;lubrication&quot;. Any movement of a particle requests &#xA;some room to change place. Before flow can happen, some dilatation &#xA;is necessary. The amount of this is very different, small for spherical particles &#xA;and especially small for grain spectra where the voids of the bigger particles are filled &#xA;by smaller particles fitting into the voids. This dilatancy is basic reason for &#xA;such media to flow not at all or only after some &quot;initialisation&quot;. &#xA;I'd say, the only reliable source for an answer to this question is experience &#xA;of hourglass makers. An interesting thing were to know from where they get &#xA;their sand, is it natural, artificial?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit, from wikipedia &quot;hourglass&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While some hourglasses actually did&#xA;  use sand as the granular mixture to&#xA;  measure time, many did not use sand at&#xA;  all. The material used in most bulbs&#xA;  was a combination of “powdered marble,&#xA;  tin/lead oxides, and pulverized, burnt&#xA;  eggshell”. [4] Over time, different&#xA;  textures of granule matter were tested&#xA;  to see which gave the most constant&#xA;  flow within the bulbs. It was later&#xA;  discovered that for the perfect flow&#xA;  to be achieved the ratio of granule&#xA;  bead to the width of the bulb neck&#xA;  needed to be 1/12 or more but not&#xA;  greater than 1/2 the neck of the bulb.&#xA;  [7]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another Edit:&#xA;Using shredded marble or eggshell or tin/lead oxide has one thing in common:&#xA;all this substances are softer than glass. I think the idea is to avoid the hourglass &#xA;being scratched blind. In wiki they say at an other line that today glass beads &#xA;are used. One can produce glass beads (or plastics beads) with a very narrow &#xA;grain spectrum, whereas grinding and screening is rather inefficient. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-17T11:13:45.077" LastActivityDate="2011-04-17T11:13:45.077" />
  <row Id="8760" PostTypeId="2" ParentId="8744" CreationDate="2011-04-16T21:53:35.233" Score="0" ViewCount="" Body="&lt;p&gt;Concerning applications to quantum mechanics, the K-S transformation has been used to calculate the Green's function of the hydrogen atom via path integrals. See, for example, &lt;a href=&quot;http://link.aps.org/doi/10.1103/PhysRevLett.48.231&quot; rel=&quot;nofollow&quot;&gt;Ho &amp;amp; Inomata, Phys. Rev. Lett. 48, 231–234 (1982)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2011-04-16T21:53:35.233" />
  <row Id="8761" PostTypeId="2" ParentId="783" CreationDate="2011-04-16T22:29:23.623" Score="2" ViewCount="" Body="&lt;p&gt;Note that the Dr. Quantum video is from the pseudoscience film &quot;What the Bleep Do We Know?&quot;, which takes the following approach:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Use examples of quantum physics to show the viewer that the universe is far more weird and complex than our basic human perception/intuition suggests.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Attempt to convince the viewer that if quantum weirdness is real, then the weirdness of someone channeling a 35,000 year old warrior-spirit named Ramtha is also real.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Profit.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Despite its flaws, the Dr. Quantum video on its own isn't terrible.  But I think the source of scientific information should be taken into consideration when assessing its accuracy.  There's a similar but better video here: &lt;a href=&quot;http://www.youtube.com/watch?v=UMqtiFX_IQQ&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=UMqtiFX_IQQ&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3150" LastActivityDate="2011-04-16T22:29:23.623" />
  <row Id="8762" PostTypeId="1" AcceptedAnswerId="9738" CreationDate="2011-04-16T23:20:38.987" Score="6" ViewCount="145" Body="&lt;p&gt;Does there exist any good proposal on how the gauge coupling unification can be fixed in non-supersymmetric GUTs? If not, can we assert that non-supersymmetric GUTs have been experimentally ruled out? Of course, I'm referring to the category of GUTs not already ruled out by proton decay rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Somehow people take this issue as evidence for supersymmetry, but not as a death blow to non-supersymmetric GUTs. Why is it the case? Is there a simple fix?&lt;/p&gt;&#xA;" OwnerUserId="2643" LastEditorUserId="2643" LastEditorDisplayName="" LastEditDate="2011-04-17T04:18:51.630" LastActivityDate="2011-05-11T16:39:54.347" Title="Are non-supersymmetric GUTs ruled out due to lack of precise gauge coupling unification?" Tags="&lt;supersymmetry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8763" PostTypeId="2" ParentId="783" CreationDate="2011-04-16T23:31:06.903" Score="1" ViewCount="" Body="&lt;p&gt;I get this feeling that this is Jack Sarfatti as a digital avatar.  If you have ever encountered him he has all sorts of strange ideas about things.  Cedric’s comment about Area 51 sort of tipped me off, for Sarfatti has all sorts of UFO ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This little video is correct, but lapses into mysteriousness at the end.  The process of measurement of a system in a superposition is to replace that superposition with an entanglement.  We might think of it as a process where the phase associated with the “waviness” of the superposition of a system is transferred to a nonlocal property of this system with another.  Consider a two slit experiment where a photon wave function interacts with a screen.  The wave vector is of the form&#xA;$$&#xD;&#xA;|\psi\rangle~=~e^{ikx}|1\rangle~+~e^{ik’x}|2\rangle&#xD;&#xA;$$&#xA;as a superposition of states for the slits labeled $1$ and $2$.  The normalization is assumed.  The state vector is normalized as&#xA;$$&#xD;&#xA;\langle\psi|\psi\rangle~=~1~=~\langle 1|1\rangle~+~\langle 2|2\rangle~+~ e^{i(k’~+~k)x}\langle 1|2\rangle~+~ e^{-i(k’~+~k)x}\langle 2|1\rangle&#xD;&#xA;$$&#xA;The overlaps $\langle 1|2\rangle$ and $\langle 2|1\rangle$ are multiplied by the oscillatory terms which are the interference probabilities one measures on the photoplate. We now consider the classic situation where one tries to measure which slit the photon traverses.  We have a device with detects the photon at one of the slit openings.  We consider another superposed quantum state.  This is a spin space that is &#xA;$$&#xD;&#xA;|\phi\rangle~=~\frac{1}{\sqrt {2}}(|+\rangle~+~|-\rangle).&#xD;&#xA;$$&#xA;This photon quantum state becomes entangled with this spin state.  So we have&#xA;$$&#xD;&#xA;|\psi,\phi\rangle~=~e^{ikx}|1\rangle|+\rangle~+~e^{ik’x}|2\rangle|-\rangle&#xD;&#xA;$$&#xA;which means if the photon passes through slit number 1 the spin is + and if it passes through slit 2 the spin is in the – state.  Now consider the norm of this state vector&#xA;$$&#xD;&#xA;\langle\psi,\phi|\psi,\phi\rangle~=~\langle 1|1\rangle\langle +|+\rangle~+~\langle 2|2\rangle\langle-|-\rangle~+~ e^{i(k’~+~k)x}\langle 1|2\rangle\langle +|-\rangle~+~ e^{-i(k’~+~k)x}\langle 2|1\rangle\langle-|+\rangle.&#xD;&#xA;$$&#xA;The spin states $|+\rangle$ and $|-\rangle$ are orthogonal and thus $\langle +|-\rangle $ and $\langle-|+\rangle$ are zero.  This means the overlap or interference terms are removed.  In effect the superposition has been replaced by an entanglement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This analysis does not tell us which state is actually measured, but it does tell us how the interference term is lost due to the entanglement of the system we measure with an instrument quantum state.  So one does not need to invoke an outright collapse to illustrate how a superposition is lost.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How the actual state is obtained is some matter of debate.  We might think of there being some other system which now measures this spin state.  So with the $|\pm\rangle$ states we now entangle another system with two states.  Yet it is clear this does not help much, for we could do this inductively “forever” and get presumably no closer to finding out which state obtains.  However, maybe this third state could be a heftier spin, or an angular momentum in this case, say a spinning buckyball cooled to some low temperature.  The buckyball could enter into an entanglement, as various quantum properties of these have been observed.  What has this accomplished?  The path integral for the entire entangled system is now narrowed closer to a classical path.  We have a bit of quantum superposition properties here, but “barely.”  Now we need to measure the buckyball’s rotational state.  This carries us up to an even larger system and …, well we have the Schrodinger cat issue.  However, some sort of asymmetry enters into the picture with the buckyball which puts the buckyball in a .7 to .3 probability ratio of being either angular momentum configuration.  Further entangling this reduces the probability ratio further to .9 to .1 and the so forth.  Somehow entanglement phase is being transferred completely out of the picture or into the environment (or demolished) , which then gives this state reduction in a measurement.  From the perspective of a quantum path integral the set of paths are reduced to an ever narrow set of paths which corresponds to the outcome.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-04-16T23:31:06.903" />
  <row Id="8764" PostTypeId="2" ParentId="8755" CreationDate="2011-04-16T23:42:43.350" Score="7" ViewCount="" Body="&lt;p&gt;I have at least an answer :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does this depend on the size of the funnel ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, as it was expected (for a very large funnel, all sand falls in the same time, whereas for a very small funnel it doesn't fall at all. More precisely ... the complete answer is probably extremely complex, as one has to take into account the shape of the hourglass, the dynamics of the grains, and so on (see &lt;a href=&quot;http://arxiv.org/abs/0707.4550&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0707.4550&lt;/a&gt; for example). However we can have a rough idea with some dimensional analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider a cylinder of diameter $D$, with a circular hole punched on the bottom side with radius $a$. We fill the cylinder with a height $H$ of sand. If we look at the speed of sand grains going out the bucket, we observe (experimentally) that it does not depend of the height of sand $H$, if $H$ is big enough (compared to the diameter $D$ - because the constraint saturates). We are left with two parameters : the diameter of the grain $a$ and the gravity field $g$ that makes it fall, so the output speed $v$ has to be proportional to $\sqrt{g a}$. The flow rate is the speed times the section, thus it is $Q \propto v \, a^2$, so it is of order $Q \propto g^{1/2} a^{5/2}$ (this is the Beverloo law).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However we have to take into account the size of grains ; let us suppose they are spherical and let be $d$ their diameter. All grains that are less than half on the hole won't fall, so there is a ring-shaped exclusion region at the border of the hole. The effective diameter of the hole, were grains will actually fall, is thus reduced by $d/2 + d/2 = d$, and is thus $a-d$. We thus replace $a$ with $a-d$ in the Beverloo law and get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q \propto g^{1/2} (a-d)^{5/2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this becomes invalid when $H$ and $D$ are of the same order, but we can get an idea of the time to drain $T$ by saying that as the flow rate does not depend of $H$, $Q T = V = 4 \pi (D/2)^2 H = \pi D^2 H$ so, dropping the $\pi$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T \propto g^{1/2} \, \frac{(a-d)^{5/2}}{D^2 \, H}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thus for everything but the size of sand grains $d$ constant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{T(d_1)}{T(d_2)} = \left( \frac{a - d_1}{a - d_2} \right)^{5/2}$&lt;/p&gt;&#xA;" OwnerUserId="3151" LastEditorUserId="3151" LastEditorDisplayName="" LastEditDate="2011-04-17T00:58:22.117" LastActivityDate="2011-04-17T00:58:22.117" CommentCount="3" />
  <row Id="8765" PostTypeId="2" ParentId="8745" CreationDate="2011-04-16T23:51:26.247" Score="0" ViewCount="" Body="&lt;p&gt;Presence of something else next to an excited atom influences the lifetime of the excited state. Any such presence is described with some additional interaction energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of a cavity QED, you can get suppression of radiation rate: $e^{-\gamma_1 t}$ with smaller $\gamma_1$ due to suppression of the corresponding part of the electromagnetic spectrum of the whole system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of a neighboring similar atom in the ground state, you can obtain a decrease of the life-time due to additional independent channel of the excitation deactivation - a resonance transition, for example. So the resulting decay rate will be $e^{-(\gamma_1+\gamma_2) t}$.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-04-16T23:51:26.247" />
  <row Id="8766" PostTypeId="2" ParentId="2188" CreationDate="2011-04-16T23:58:01.373" Score="1" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;information is stored in capacitors&#xA;in the form of electrons. When a&#xA;capacitor is filled more than 50%&#xA;with electrons, it's considered to be&#xA;'on' (a bit with value '1').&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Incorrect.   Information is stored in capacitors in the form of electromagnetic energy.  It's also in the form of imbalance of electrons, not electrons themselves.  To &quot;charge&quot; a capacitor, we take some electrons out of one metal plate and deposit them on the other metal plate.  So, the number of electrons inside the capacitor never changes.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the information represented by a&#xA;certain current (or rather, a series&#xA;of on/off currents)  has mass,&#xA;because it consists of the electrons&#xA;that flow through the wire.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Incorrect, because the wires always contain the same number of electrons.  The information is stored by having those electrons moving or unmoving.  Analogy: a rubber drive belt is much like a current in an electric circuit, where the rubber is like the electrons.  The electrons are there inside the wires, even when they aren't moving.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a virtual object is tangible because&#xA;it exists in memory in the form of&#xA;electrons that are there (or not) in&#xA;a certain pattern.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Incorrect.  For example, in RAM memory, electrons behave analogous to beads on an Abacus.  To store ones and zeros we flip the beads left and right.  But we never add any beads to the abacus, or remove them.  Only the pattern is important, not the beads.  Digital information is like writing in sand, and we don't buy and sell sand, we buy and sell only the patterns.  In RAM memory the total number of electrons never changes.  But in each memory cell, in each flipflop, the electron flow is diverted to one of two possible paths to store one of two possible states: one or zero.&lt;/p&gt;&#xA;" OwnerUserId="3153" LastActivityDate="2011-04-16T23:58:01.373" CommentCount="2" />
  <row Id="8767" PostTypeId="2" ParentId="8748" CreationDate="2011-04-17T00:19:58.017" Score="1" ViewCount="" Body="&lt;p&gt;Photons aren't complicated,  Instead they're non-sensical!  This topic is really about &quot;Quantum Weirdness.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First let's get one thing straight: photons are not emitted in bundles.   Instead atoms emit photons one at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing here, but I think the teacher was trying to communicate the following:  EM wave-energy is emitted and detected as photons.   This could be translated as &quot;Light is emitted in bundles called Photons.&quot;    Each photon is something like a bundle of light.   So, the OP question is probably not about bundles of photons.  It's about photons themselves, where each photon is a &quot;bundle of electromagnetic energy.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But don't forget that this topic is non-sensical.  You cannot visualize photon behavior.  Nobody ever has.  Light is waves which are particles, and the infinitely small particles are waves at the same time.  It's like trying to visualize a square circle, or dark whiteness.  The experts get part-way there by using many different ideas.  Something here is very complicated: the ideas being used to explain photons.  Ideas such as instantaneous wave-function collapse, or multiple-universe splitting in the Everett-Wheeler Many Worlds theory.   Richard Feynman's QED assumes that each atom emits an infinite number of &quot;virtual&quot; photons which take all possible trajectories, then they combine together to form one photon which impacts an atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some good books:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the Quantum Leap, by Fred A. Wolf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shrodinger's Kittens, by John Gribben&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QED by Richard Feynman&lt;/p&gt;&#xA;" OwnerUserId="3153" LastActivityDate="2011-04-17T00:19:58.017" CommentCount="1" />
  <row Id="8768" PostTypeId="1" AcceptedAnswerId="11464" CreationDate="2011-04-17T00:37:53.383" Score="6" ViewCount="172" Body="&lt;p&gt;I am familiar with the idea of a BPS bound as in a lower limit on the mass of supermultiplets given by a certain function of the central charge and when I think of $\cal{N}=4$ SYM I see a complicated lagrangian in my mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somehow the above picture seems insufficient to understand the classification of what are called &quot;BPS sectors&quot; in $\cal{N}=4$ SYM. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know what is the meaning and the derivation of the following listing for $\cal{N}=4$ SYM ,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\frac{1}{2}$ BPS sector consists of multi-trace operators involving a single bosonic operator $Z$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\frac{1}{4}$ BPS sector consists of multi-trace operators involving two bosonic operator $Z, Y$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\frac{1}{8}$ BPS sector consists of multi-trace operators involving three bosonic operator $Z, Y, X$ and two fermionic operators $\lambda$, $\bar{\lambda}$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(everything above is apparently assumed to be in the the Lie algebra of $U(N)$) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also like to know what is the relation between the above classification and thinking in terms of short/semi-short/long multiplets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be happy to know of expository references for the above topics. I haven't been able to locate any on my own. &lt;/p&gt;&#xA;" OwnerUserId="1637" LastEditorUserId="1637" LastEditorDisplayName="" LastEditDate="2011-04-17T00:44:27.510" LastActivityDate="2011-06-22T22:57:09.953" Title="BPS sectors in $\cal{N}=4$ SYM" Tags="&lt;quantum-field-theory&gt;&lt;supersymmetry&gt;" AnswerCount="1" />
  <row Id="8769" PostTypeId="2" ParentId="8735" CreationDate="2011-04-17T01:24:19.880" Score="1" ViewCount="" Body="&lt;p&gt;I thought at first this question was more general involving the nature of the Virasoro algebra.  As a result the first two paragraphs are boiler plate discussions on that.  The actual question is addressed in the third paragraph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vector fields $T^m~=~z^{m+1}\partial z$ satisfy the Witt algebra or Virasoro algebra without central extension for each index value,&#xA;$$&#xD;&#xA;[T^m,~T^n]~=~(n~-~m) T^{m+n}.&#xD;&#xA;$$&#xA;The Virasoro algebra maybe extended by a one-dimensional center with the map&#xA;$$&#xD;&#xA;{m:Vir\oplus {\mathbb C}~\rightarrow~Vir,} &#xD;&#xA;$$&#xA;where this central extension is the kernel of the map.  The extended Virasoro group is then,&#xA;$$&#xD;&#xA;[T^m,~T^n]~=~(n~-~m)T^{m+n}~+~c(m,~n).&#xD;&#xA;$$&#xA;This is imposed since with the case $m~+~n~=~0$ then $[T^m,~T^{-m}]~=~-2mT^{2m}$ and there are two infinite sums with the same result, one for $m~\ge~0$ and $m~\le~0$.  To give a central extension for a specific index $m$ this is a c-number in the $\mathbb C$ in the domain of the above map, and $c(m,~n)~=~c(m)\delta_{m+n}$ The central extension may then be computed with the Jacobi identity&#xA;$$&#xD;&#xA;[T^k,~[T^m,~T^n]]~+~ [T^m,~[T^n,~T^k]]~+~ [T^n,~[T^k,~T^m]]~=~0.&#xD;&#xA;$$&#xA;which is equal to&#xA;$$&#xD;&#xA;(m~-~n)c(k)~+~(n~-~k)c(m)~+~(k~-~m)c(n)~=~0.&#xD;&#xA;$$&#xA;To get $c(m)$ we set $k~=~1$ and $n~=~-(m~=~1)$.  This gives us&#xA;$$&#xD;&#xA;c(m~+~1)~=~\frac{(m~+~2)c(m)~-~(2m~+~1)c(1)}{m~-~1}.&#xD;&#xA;$$&#xA;This is a nice recursion relationship and we get $c(2)~=~3c(1)~-~3c(1)$ $=~0$ and the general recursion&#xA;$$&#xD;&#xA;c(m)~=~c(3)m^3~-~c(1)m&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $T^m$ are expanded according to string modes&#xA;$$&#xD;&#xA;T^m~=~\frac{1}{2}\sum_{n=-\infty}^\infty\alpha_{m-n}\alpha_n.&#xD;&#xA;$$&#xA;Form here the coefficients $c(3)$ and $c(1)$ can be determined.  The $T^0,~T^{\pm 1}$ generate a closed $SL(2,{\mathbb R}$ subalgebra.  The $T^0~=~\frac{1}{2}\sum_{n=-\infty}^\infty\alpha_{-n}\alpha_n$ annihilates the vacuum state as does $L^{\pm}$.  For $m~=~2$ one can show by computing the vacuum expectation $\langle 0|T^{-2}T^2|0\rangle$&#xA;$$&#xD;&#xA;c(m)~=~\frac{D}{12}(m^3~-~m)&#xD;&#xA;$$&#xA;with some additional work, which is a bit lengthy, you can show $D~=~26$ and work up the bosonic string.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any field $\phi$ transforms under the conformal algebra as its commutator with a $T^m$ element of is $iz^{m+1}\partial_z\phi$.  The $SL(2,{\mathbb R})$ is the symmetries of conformal quantum mechanics, and the Witt algebra or Virasoro algebra is then its generalization as the set of diffeomorphisms of the circle.  Suppose a function or field that transforms under these diffeomorphisms requires this central extension.  For instance the Hamiltonian is $H~=~T^0$ and suppose the field is $\phi(\tau)~=~\phi(0)e^{iH\tau}$, then&#xA;$$&#xD;&#xA;\delta\phi~=~\delta\phi(0)e^{iT^0\tau}~+~i\phi(0)\epsilon\sum_m[T^m,~T^0]~+~O(\tau^2).&#xD;&#xA;$$&#xA;It is then clear that the central extension is $c(0)~=~0$.  It is not hard to see that a similar situation hold if the field is dependent on $T^{\pm}$.  However, if the field depends upon $L^m$, for $|m|~&amp;gt;~1$ then care needs to be taken.  The $L_m$ are the Fourier modes of the stress-energy tensor in the string $1~+~1$ dimensional spacetime.  So it is unclear whether a field is of this nature.  I would be tempted at this point to say no, except for the case of the $T^0$, which is the Hamiltonian.  However, in that case the central or anomaly term is zero.  The only reason one would need to be concerned with the central term is if $\phi$ is an element of the circle diffeomorphism, or is some function of it.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-04-17T01:30:14.087" LastActivityDate="2011-04-17T01:30:14.087" CommentCount="2" />
  <row Id="8770" PostTypeId="1" CreationDate="2011-04-17T01:25:30.857" Score="0" ViewCount="297" Body="&lt;p&gt;If you put a latex balloon in a vacuum, how much would it expand? And would it pop? Assume it doesn't leak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some numbers: Ambient pressure is 100 KPa, balloon is perfectly spherical with a diameter of 300 mm, deflated it has a diameter of 25mm, temperature is always at equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question components stated more formally:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What pressure does the balloon need to have been inflated to to reach the 300 mm radius?&lt;/li&gt;&#xA;&lt;li&gt;What is the relationship between the (gauge) pressure and volume? &lt;em&gt;I think this simply comes down to PV=nRT&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;If there is some sort of spring constant involved, what is this value for a typical latex balloon?&lt;/li&gt;&#xA;&lt;li&gt;What is the relationship between the pressure (or volume) and the tension on the material the balloon is made from?&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the limit the tension can reach before popping?&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Semi-related: how much would a balloon expand if you sealed it deflated and put it in a vacuum?&lt;/p&gt;&#xA;" OwnerUserId="3154" LastEditorUserId="3154" LastEditorDisplayName="" LastEditDate="2011-04-20T21:42:37.537" LastActivityDate="2011-04-20T21:42:37.537" Title="If you put a latex balloon in a vacuum, how much would it expand?" Tags="&lt;pressure&gt;&lt;inert-gases&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="8771" PostTypeId="2" ParentId="8748" CreationDate="2011-04-17T02:06:53.303" Score="-1" ViewCount="" Body="&lt;p&gt;Photons can be thought of as bundles of energy. There are various &quot;energy levels&quot; in an atom. The energy of an electron in an energy level is the same as the energy of the energy level. When an electron of an atom &quot;jumps&quot; from a higher energy level (say $E_1$) to a lower energy level (say $E_2$) a photon is emitted. Since the energy of those &quot;energy levels&quot; are fixed, a photon necessarily have a fixed energy whose frequency is given by the following equation. Here $\nu$ is the frequency of the emitted radiation, $h$ is the Planck's constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_1 - E_2 = h \nu$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or, $\nu = \frac{E_1 - E_2}{h}$&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-04-17T02:06:53.303" />
  <row Id="8772" PostTypeId="2" ParentId="8770" CreationDate="2011-04-17T02:15:55.950" Score="0" ViewCount="" Body="&lt;p&gt;it would expand untill it popped because the pressure from inside the balloon would be greater than the vacum since there is no air (pressure)&#xA;the only reason a balloon doesnt pop in a setting like earth is because the atmospheric pressure is pushing on the balloon with the same force as the helium is pushing outward on the balloon &lt;/p&gt;&#xA;" OwnerUserId="2530" LastActivityDate="2011-04-17T02:15:55.950" CommentCount="3" />
  <row Id="8773" PostTypeId="2" ParentId="6742" CreationDate="2011-04-17T03:09:13.283" Score="1" ViewCount="" Body="&lt;p&gt;No one mentioned &lt;a href=&quot;http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html&quot; rel=&quot;nofollow&quot;&gt;The Relativistic Rocket&lt;/a&gt;? Those are the special relativistic equations for acceleration.&lt;/p&gt;&#xA;" OwnerUserId="2314" LastActivityDate="2011-04-17T03:09:13.283" />
  <row Id="8774" PostTypeId="2" ParentId="8770" CreationDate="2011-04-17T04:39:30.593" Score="3" ViewCount="" Body="&lt;p&gt;It would depend on two things - how much air is in the balloon and the tensile strength of latex. To see why, I hope you'll find the following useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a balloon on earth near the surface, the pressure inside depends on how much air you blow into it. But in order for the balloon to be in static equilibirium, that internal pressure has to be matched by the (constant) atmospheric pressure that opposes outward expansion PLUS the elastic surface tension of the surface, which also tends to oppose outward expansion (wanting to minimize the surface area). $$P_i = P_0 + S$$ If you blow more air into it, $P_i$ increases, and since $P_0$ is constant at the same height, the surface tension has to increase for LHS and RHS to match. This will continue until the surface tension exceeds the tensile strength of the balloon, at which point the balloon pops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's say you have blown only enough air into the balloon to make it taut -- if you let this balloon go so that it floats up, the atmospheric pressure $P_0$ starts dropping with height, and so again, the surface tension has to keep rising so that the RHS matches the constant internal pressure $P_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two things could happen now. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Even before $P_0$ has dropped to zero (in the vacuum of outer space), the surface tension has had to become so high that it exceeds the tensile strength and makes the balloon pop. This is what happens to most helium balloons let go - they eventually pop even before reaching vacuum - because whatever material the surface is made of has a tensile strength too low to be able to sustain the minimum amount of internal pressure required to make the balloon taut once the external pressure has dropped enough. Note that this minimum amount of internal pressure could be made smaller by reducing the size of the balloon but you can't do this beyond the limit at which the balloon's mass overcompensates and makes it impossible for it to float in the first place!&lt;/li&gt;&#xA;&lt;li&gt;The other possibility is that the tensile strength is high enough that even after $P_0$ is zero, the surface tension required to match the internal pressure doesn't exceed the tensile strength and you'll have a nice happy balloon floating in the vacuum. This is what happens with space stations, for instance :)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-04-17T04:39:30.593" CommentCount="3" />
  <row Id="8775" PostTypeId="1" CreationDate="2011-04-17T05:49:54.427" Score="0" ViewCount="118" Body="&lt;p&gt;Everyone knows that mental agility goes down with age.  For ordinary physicists, as they age, they  compensate for it with increasing experience and background knowledge. Someone who only starts out when they are old have all the odds stacked out against them. They have neither sufficient experience nor mental agility. They often have less time as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given all this, are there examples of successful physicists who only started out when they were old?&lt;/p&gt;&#xA;" OwnerUserId="3156" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-04-17T06:10:54.353" LastActivityDate="2011-04-17T06:10:54.353" ClosedDate="2011-04-17T05:54:28.470" Title="A list of successful physicists who only started when they're old?" Tags="&lt;soft-question&gt;" CommentCount="7" />
  <row Id="8776" PostTypeId="1" CreationDate="2011-04-17T06:09:48.443" Score="11" ViewCount="219" Body="&lt;p&gt;Many years ago, a discrepancy was found between the experimentally measured value of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Anomalous_magnetic_dipole_moment&quot; rel=&quot;nofollow&quot;&gt;muon magnetic moment&lt;/a&gt;, and the theoretically calculated value. Shockingly, most physicists were blase about it. It was no big deal to them. They dismissed it as either an experimental error, or some mistake in the QCD calculations, even though error bars have been painstakingly computed for both of them, and the discrepency still survived up to a few sigmas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the current status of the anomalous muon magnetic moment?&lt;/p&gt;&#xA;" OwnerUserId="3157" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-04-17T12:40:15.400" LastActivityDate="2011-04-17T18:10:27.277" Title="What is the current status of the anomalous  muon magnetic moment?" Tags="&lt;particle-physics&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="8777" PostTypeId="1" AcceptedAnswerId="8781" CreationDate="2011-04-17T08:06:45.033" Score="2" ViewCount="92" Body="&lt;p&gt;I'm intrigued by this, and how it would work:- 3 sub-questions if I may:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QUESTION 1) Construction: &#xA;As I understand it's a flexible sphere constrained by a rigid edge&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. Do we simply glue 2 flat circular pieces of flexible material together at the edges and inflate?   &lt;a href=&quot;http://www.qsl.net/pe1rah/RAHdish.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.qsl.net/pe1rah/RAHdish.htm&lt;/a&gt;&#xA;Or, &#xA;b. do you have to cut out sections to make it parabolic, and the inflation simply gives it the rigidity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;example of latter: page 4&#xA;&lt;a href=&quot;http://www.google.com.hk/url?sa=t&amp;amp;source=web&amp;amp;cd=2&amp;amp;ved=0CB4QFjAB&amp;amp;url=http%3A%2F%2Fciteseerx.ist.psu.edu%2Fviewdoc%2Fdownload%3Fdoi%3D10.1.1.112.8157%26rep%3Drep1%26type%3Dpdf&amp;amp;ei=A5uqTdTqF4OgvQOq6rmfCg&amp;amp;usg=AFQjCNF0h6SMIPBsglPIBV8aKCTPlqpmaA&amp;amp;sig2=ryWREETqB2DR_oQMYMoSpQ&quot; rel=&quot;nofollow&quot;&gt;http://www.google.com.hk/url?sa=t&amp;amp;source=web&amp;amp;cd=2&amp;amp;ved=0CB4QFjAB&amp;amp;url=http%3A%2F%2Fciteseerx.ist.psu.edu%2Fviewdoc%2Fdownload%3Fdoi%3D10.1.1.112.8157%26rep%3Drep1%26type%3Dpdf&amp;amp;ei=A5uqTdTqF4OgvQOq6rmfCg&amp;amp;usg=AFQjCNF0h6SMIPBsglPIBV8aKCTPlqpmaA&amp;amp;sig2=ryWREETqB2DR_oQMYMoSpQ&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 2: Spherical Antenna: &lt;a href=&quot;http://www.gatr.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.gatr.com/&lt;/a&gt;&#xA;Would the design for this inflated dome with internal parabola be in effect 2 inflatable structures - 1 sphere (to give rigid circumference, and an internal parabola as in QUESTION 1, or could you do something like this with only 1 inflatable structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 3: Design&#xA;I'm assuming the parabola generated using Q1.a would be a factor of gas pressure + stretch of fabric given there is no slack in the material? Or are other factors involved?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions for design or theory papers would be most appreciated. I'm going to have a try at building one of these for portable wifi :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I just threw together a prototype from 2 plastic bags/drinking straw/tape - it's a bit rough to see a definitive answer, but looks parabolic.&lt;/p&gt;&#xA;" OwnerUserId="2951" LastEditorUserId="2951" LastEditorDisplayName="" LastEditDate="2011-04-17T09:08:13.567" LastActivityDate="2011-04-19T05:29:36.203" Title="How is an inflatable parabolic antenna created?" Tags="&lt;antennas&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="8778" PostTypeId="1" CreationDate="2011-04-17T09:00:51.500" Score="2" ViewCount="121" Body="&lt;p&gt;I was wondering about the relation between noise with distance, assuming a point source, using sound as the method for communication and air as the medium of communication. Obviously as the distance from the sound source increases, noise should increase- but what is the nature of this relationship? Is it linear or non-linear?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas? Is there any other way noise can be modeled (in the context of communication between individuals)? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3159" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-04-17T09:45:18.557" LastActivityDate="2011-04-18T02:30:48.287" Title="Modelling noise with distance" Tags="&lt;sound&gt;&lt;models&gt;" AnswerCount="1" CommentCount="13" FavoriteCount="0" />
  <row Id="8779" PostTypeId="2" ParentId="8776" CreationDate="2011-04-17T09:31:04.610" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Anomalous_magnetic_dipole_moment#Anomalous_magnetic_moment_of_the_muon&quot; rel=&quot;nofollow&quot;&gt;WP Anomalous_magnetic_moment_of_the_muon&lt;/a&gt; 3.4$\sigma$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_deviation&quot; rel=&quot;nofollow&quot;&gt;standard deviation&lt;/a&gt; (as of 2007) &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is a contribution of effects of&#xA;  quantum mechanics, expressed by&#xA;  Feynman diagrams with loops, to the&#xA;  magnetic moment of that particle.&#xA;   ...its measurement provides a &lt;strong&gt;precision&#xA;  test of the Standard Model.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.g-2.bnl.gov/&quot; rel=&quot;nofollow&quot;&gt;The experiment Home Page - E821 Muon (g-2)&lt;/a&gt; (as of 2004)&lt;br&gt;&#xA;&lt;a href=&quot;http://www.g-2.bnl.gov/physics/index.html&quot; rel=&quot;nofollow&quot;&gt;Overview of the experiment&lt;/a&gt;&lt;br&gt;&#xA;&lt;strong&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/hep-ph/pdf/0611/0611102v3.pdf&quot; rel=&quot;nofollow&quot;&gt;Improved predictions&lt;/a&gt; as of 2007 : 3.4$\sigma$ :&lt;/strong&gt;&lt;br&gt;&#xA;Abstract&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We update the Standard Model&#xA;  predictions of the anomalous magnetic&#xA;  moment of the muon..., incorporating&#xA;  the new $e_{+}e_{-}-&amp;gt; \pi\pi$ data&#xA;  obtained by CMD-2 and KLOE, as well as&#xA;  the corrected SND data, and other&#xA;  improvements. The prediction .. which&#xA;  corresponds to a 3.4 $sigma$ &#xA;  deviation from the measured value...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But the final &lt;em&gt;Note added in proof:&lt;/em&gt;(page 10) points to &lt;strong&gt;3.5$\sigma$ discrepancy&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.g-2.bnl.gov/future.html&quot; rel=&quot;nofollow&quot;&gt;Future Experiments&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will take this fact in consideration when I read that SM is OK.&lt;br&gt;&#xA;(there is a particle model by Douglas Pinnow (Our Resonant Universe) waiting to be read and commented. Have a glimpse of it &lt;a href=&quot;http://heldervelez1.tiddlyspace.com/#%5B%5BOur%20Resonant%20Universe%20by%20Douglas%20Pinnow%5D%5D%20GettingStarted&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;)   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT add&lt;/strong&gt;&lt;br&gt;&#xA;IMO, add: In the experiment the muons are obtained by p+p collisions; add: the proton &lt;strong&gt;'spin crisis'&lt;/strong&gt; antiquity; then the model defect hypothesis is enhanced. (both problems are linked, I think)  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-04-17T18:10:27.277" LastActivityDate="2011-04-17T18:10:27.277" CommentCount="5" />
  <row Id="8780" PostTypeId="2" ParentId="8776" CreationDate="2011-04-17T09:33:24.803" Score="6" ViewCount="" Body="&lt;p&gt;One can find a number of references by googling. Here is one:&#xA;&lt;a href=&quot;http://arxiv.org/PS_cache/hep-ph/pdf/0102/0102122v2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/hep-ph/pdf/0102/0102122v2.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discrepancy between measurements and standard model calculations for the muon anomalous magnetic moment  is of the order of &lt;a href=&quot;http://www.annualreviews.org/doi/abs/10.1146/annurev.nucl.53.041002.110618&quot; rel=&quot;nofollow&quot;&gt;2.6 sigma.&lt;/a&gt;  There are interpretations that this is a signal for new physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the &lt;a href=&quot;http://physics.stackexchange.com/questions/8752/standard-deviation-in-particle-physics&quot;&gt;discussion of standard deviations&lt;/a&gt; needed to declare a discovery in particle physics you will find that 5 sigma is really necessary. 2.5 sigma is barely at the level of &quot;interesting&quot; That is why there is not much discussion and you observe :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Shockingly, most physicists were blase about it. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It was no big deal to them except :interesting, more data needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were 4 sigma, they would be jumping up and down with excitement and speculating about the masses of new physics intermediate particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example is the accurate data from LEP that allowed to calculate the top mass, before it was observed at Fermi lab, from differences in expectations.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-17T09:33:24.803" CommentCount="5" />
  <row Id="8781" PostTypeId="2" ParentId="8777" CreationDate="2011-04-17T10:00:13.687" Score="2" ViewCount="" Body="&lt;p&gt;When You look with an &quot;math eye&quot; at the shape in Your first link,&#xA;You will see that the form is more a part of a sphere, as theory &#xA;predicts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In second link there is a picture of the parabolic &lt;strong&gt;mandrel&lt;/strong&gt; which &#xA;was used to &lt;strong&gt;form&lt;/strong&gt; the foil. The pressure will not deform the &#xA;foil much, just keep it inflated. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-17T10:00:13.687" CommentCount="1" />
  <row Id="8782" PostTypeId="1" AcceptedAnswerId="8783" CreationDate="2011-04-17T10:09:02.383" Score="3" ViewCount="211" Body="&lt;p&gt;I'm trying to determine the viscous dampening coefficient of a spring (c). Read about it on Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Damping&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two equations which I have are:&#xA;f=-cv and ma+cv = -kx&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the spring constant (k)=5, the mass is 50g and the initial amplitude of the spring is 10cm.&lt;/p&gt;&#xA;" OwnerUserId="3044" LastActivityDate="2011-06-27T17:26:37.157" Title="How to determine viscous dampening coefficient of spring?" Tags="&lt;harmonic-oscillator&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="8783" PostTypeId="2" ParentId="8782" CreationDate="2011-04-17T11:36:26.070" Score="3" ViewCount="" Body="&lt;p&gt;with the given $m$ and $k$ you indeed &lt;strong&gt;cannot&lt;/strong&gt; calculate the damping coefficient $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that you just have a &lt;strong&gt;model&lt;/strong&gt; where you put some constants in and you can only derive other constants which somehow depend on them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question concerning an actual &lt;strong&gt;measurement&lt;/strong&gt; was answered in &lt;a href=&quot;http://physics.stackexchange.com/questions/8495/investigating-the-dampening-of-a-spring/8499#8499&quot;&gt;Investigating the dampening of a spring&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Greets&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2011-04-17T15:42:40.723" LastActivityDate="2011-04-17T15:42:40.723" CommentCount="2" />
  <row Id="8784" PostTypeId="1" CreationDate="2011-04-17T13:40:28.747" Score="1" ViewCount="357" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/556/what-is-the-fallacy-in-this-infinite-motion-machine&quot;&gt;What is the fallacy in this infinite motion machine?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Most of the &quot;troll physics&quot; images I can figure out, but this one has me stumped. What is broken about the following machine's physics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/VbmLW.jpg&quot; alt=&quot;infinite electricity machine&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3162" LastActivityDate="2011-04-17T14:00:20.310" ClosedDate="2011-04-17T15:22:27.490" Title="&quot;Troll physics&quot;: Buoyancy for infinite power" Tags="&lt;untagged&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8785" PostTypeId="2" ParentId="8784" CreationDate="2011-04-17T14:00:20.310" Score="6" ViewCount="" Body="&lt;p&gt;The problem is in the seal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of work to move the seal against the water pressure is the same amount of energy that is gained by the balls when they are pushed up by the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if we remove the seal and we imagine a magic &quot;one-way pass-through&quot; wall, the ball would still need to displace the same volume of water as itself in order to get into the tank.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-04-17T14:00:20.310" />
  <row Id="8786" PostTypeId="1" CreationDate="2011-04-17T14:04:56.970" Score="0" ViewCount="109" Body="&lt;p&gt;Does angry birds have randomness or just chaos?&lt;/p&gt;&#xA;" OwnerUserId="3161" LastActivityDate="2011-04-17T14:04:56.970" ClosedDate="2011-04-17T15:21:45.150" Title="Does angry birds have randomness or just chaos" Tags="&lt;gravity&gt;&lt;mechanics&gt;" CommentCount="4" />
  <row Id="8787" PostTypeId="2" ParentId="7829" CreationDate="2011-04-17T14:38:16.537" Score="3" ViewCount="" Body="&lt;p&gt;To prevent particles from escaping, a few centimeters of concrete is enough.&#xA;To prevent gamma radiation from escaping, one has to define what levels one considers acceptable outside the enclosure.&#xA;To prevent the hot air emanating from press weenies and the anti-nuclear lobby from escaping, several lightyears' thickness of lead might not be enough.&lt;/p&gt;&#xA;" OwnerUserId="2625" LastActivityDate="2011-04-17T14:38:16.537" CommentCount="2" />
  <row Id="8788" PostTypeId="1" AcceptedAnswerId="8795" CreationDate="2011-04-17T15:15:50.287" Score="2" ViewCount="191" Body="&lt;p&gt;I'm using a game engine that has a ton of physics stuff built in. What I am currently trying to do is to simulation an explosion with F force in an x,y coordinate system and figure out how other items in the world are affected by it (they each have x,y as well).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ultimate function I call is applyForce and it takes an x,y vector in Newtons for the force and a point to act on (the other objects in this case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently I am doing the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Determining the distance of the explosion point for each of the objects' x,y coordinates&lt;/li&gt;&#xA;&lt;li&gt;If the distance is less than the force F (my total explosion force) then I proceed&lt;/li&gt;&#xA;&lt;li&gt;I subtract the distance from F and assign that back to F - so this effectively becomes the total force that can act on the object based on distance&lt;/li&gt;&#xA;&lt;li&gt;I find the angle of the explosion point from the object by atan2(x1-x2,y1-y2)*180/PI&lt;/li&gt;&#xA;&lt;li&gt;Find the resulting X force by F * cos(angle)&lt;/li&gt;&#xA;&lt;li&gt;Find the resulting Y force by F * sin(angle)&lt;/li&gt;&#xA;&lt;li&gt;Use these to assign the Force vector and apply it to the object&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In general I think this okay but I feel like I am missing something as right now the forces are still a bit sporadic. I think I am making the assumption that the signs of the force (- or +) will be handled by the Trig functions above but I have a feeling that is not the case. From my logs sometimes the angle is -167 and I'm wondering if that shouldn't just be based on a 360 degree system instead? I'm assuming if that's the case then I have to do some logic to assign the correct - or + signs to the forces so it moves in a direction away from the explosion point. Can anyone point me in the right direction? I am a developer so this could be completely off :)&lt;/p&gt;&#xA;" OwnerUserId="3163" LastActivityDate="2011-04-17T18:45:30.373" Title="Basic explosion physics - determining force" Tags="&lt;forces&gt;&lt;explosions&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="8789" PostTypeId="1" CreationDate="2011-04-17T16:33:07.237" Score="2" ViewCount="167" Body="&lt;p&gt;If a butterfly did not flap its wings some time ago, but instead decided to slide for that millisecond, can this cause a tornado on the other side of the earth if we just wait long enough? Does this perturbation die out, or average out in short time, or does it propagate and have grander and grander consequences?&lt;br&gt;&#xA;(Clearly there are cases where it propagates, ie if its wings were artificially connected to trigger some nuclear bomb, but is it always so that it propagates?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some clear-cut criteria to determine whether a perturbation propagates or dies out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the answer different if the world was purely classical versus QM?  &lt;/p&gt;&#xA;" OwnerUserId="1708" LastEditorUserId="1708" LastEditorDisplayName="" LastEditDate="2011-04-17T17:23:04.890" LastActivityDate="2011-04-17T17:34:28.347" Title="'A' butterfly effect" Tags="&lt;statistical-mechanics&gt;&lt;chaos-theory&gt;&lt;weather&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="8791" PostTypeId="2" ParentId="8789" CreationDate="2011-04-17T16:59:41.937" Score="2" ViewCount="" Body="&lt;p&gt;The weather is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Chaos_theory&quot; rel=&quot;nofollow&quot;&gt;chaotic system&lt;/a&gt; and has the characteristic of having &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/qj.49712252906/abstract&quot; rel=&quot;nofollow&quot;&gt;critical dependence on initial conditions&lt;/a&gt;. This in practice means that any forecast over 2 or 3 days starts to be flawed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chaotic systems are a feature of mathematics, and as such are present in classical physics:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://nylander.wordpress.com/2007/10/27/magnetic-pendulum-strange-attractor/&quot; rel=&quot;nofollow&quot;&gt;Magnetic pendulum&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.myphysicslab.com/dbl_pendulum.html&quot; rel=&quot;nofollow&quot;&gt;Double pendulum&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.seas.harvard.edu/softmat/downloads/2005-06.pdf&quot; rel=&quot;nofollow&quot;&gt;Dripping&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In practice, when people use the butterfly effect example, mean that the overall ignorance of the initial conditions determines the unpredictability of the larger system. In practice you should read that as: &lt;em&gt;in order to be able to predict the weather long term, we would need to know a lot of detail, up to the level of butterflies flapping their wings&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not the case of a single butterfly determining the weather.&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-04-17T16:59:41.937" />
  <row Id="8792" PostTypeId="2" ParentId="8789" CreationDate="2011-04-17T17:03:16.630" Score="7" ViewCount="" Body="&lt;p&gt;If you take a &lt;em&gt;well-behaved&lt;/em&gt; physical system and perturb it a &lt;em&gt;little bit&lt;/em&gt;, then you expect the total behavior or your system to be changed only a &lt;em&gt;little bit&lt;/em&gt;. You can quantify this by saying that if your initial perturbation is $\delta$, then the final perturbation can never exceed $\gamma \delta$ for some constant $\gamma$. In many cases, such perturbations manifest themselves as small oscillations around the unperturbed orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a &lt;strong&gt;chaotic&lt;/strong&gt; system, by definition, the deviation between the solutions for systems with minimally different initial conditions can become arbitrarily large. People first discovered this (at least a video on chaos theory I've seen in high school claimed so) when they tried to calculate the weather using some simple models. You quickly see that even very tiny changes in your initial conditions lead to &lt;em&gt;completely&lt;/em&gt; different results for your forecast. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as was pointed out in a comment, a &lt;em&gt;tsunami&lt;/em&gt; is not triggered by atmospheric disturbances, but by underwater earthquakes. Maybe you think of &quot;tornado&quot;. Here the answer would be yes and no. On the one hand, the very fact that there is such a thing as &quot;tornado season&quot; (i.e. right now in the US) and certain regions on earth with heightened frequency of tornadoes (tornado alley in the US) shows that it is not absolutely arbitrary when and where these things occur. The grand scheme, as you could call it, will not be affected by a butterfly's actions on the other side of the globe. &lt;strong&gt;But&lt;/strong&gt; if you set out to calculate then when and where up to the millimeter and the millisecond, you would have to take into account all the tiny atmospheric disturbances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a mathematical definition from the Wikipedia article on the Butterfly effect:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A dynamical system with evolution map ft displays sensitive dependence on initial conditions if points arbitrarily close together become separate with increasing t at an exponential rate. The definition is not topological, but essentially metrical.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-04-17T17:03:16.630" />
  <row Id="8793" PostTypeId="2" ParentId="8789" CreationDate="2011-04-17T17:04:11.857" Score="1" ViewCount="" Body="&lt;p&gt;You have to study a bit about &lt;a href=&quot;http://en.wikipedia.org/wiki/Deterministic_chaos&quot; rel=&quot;nofollow&quot;&gt;deterministic chaos.&lt;/a&gt;. The illustration in the article is what might look like a butterfly, and the popular butterfly narrative may have its origin to the similarity, but there is a wealth of clear mathematical derivations that lead to statements like  the effect of &quot; a butterfly waving its wings &quot; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is an illustration that when one tries to solve the known dynamical equations of physical systems, very small changes in the initial variables condition can have enormous consequences to the solution output. The system of solutions can be studied mainly through the use of computers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus your question does not have a meaning because you have taken seriously the butterfly waving its wings in Japan. It is the dynamical equations that might connect a small initial condition in Japan with a storm in London. No dynamical equations connect tsunamis with air disturbances.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-04-17T17:34:28.347" LastActivityDate="2011-04-17T17:34:28.347" CommentCount="2" />
  <row Id="8794" PostTypeId="2" ParentId="8789" CreationDate="2011-04-17T17:09:41.370" Score="2" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;The butterfly effect is to be taken metaphorically. The crux of it really is long-term unpredictability due to sensitivity to initial conditions. No serious meteorologist believes that the formation of a tornado could depend on whether a butterfly flaps its wings or not. Reliable short-term predictions of weather (over a few days) can be and are being made everyday without observing what butterflies do. But what the Butterfly Effect &lt;em&gt;does&lt;/em&gt; say is that small changes in the initial data that you provided for your weather forecasting model will lead to wildly different predictions for the long-term (a week or more from now).&lt;/li&gt;&#xA;&lt;li&gt;The effect is purely classical and would still be observed in a completely classical world.&lt;/li&gt;&#xA;&lt;li&gt;Tsunamis are caused by earthquakes on the ocean bed, which can´t have anything to do with local weather. &amp;lt;&amp;lt; &lt;em&gt;I notice this is irrelevant after your question edit&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-04-17T17:20:58.783" LastActivityDate="2011-04-17T17:20:58.783" CommentCount="2" />
  <row Id="8795" PostTypeId="2" ParentId="8788" CreationDate="2011-04-17T18:20:00.010" Score="6" ViewCount="" Body="&lt;p&gt;In the simplest approximation, an explosion is a shockwave moving out from some locus. The shockwave may be a compression front in a ambient medium, or may be a wave of gas propagating from the explosive into a vacuum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that's the first thing you need to tell us: in air, water, vacuum, or what?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the shockwave arrives at some material thing, it is the pressure exerted by the shockwave that transfers momentum (i.e. applies a force) to the target. The target object then accelrates as per Newton's law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec{F} = m\vec{a} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vector part of the above is the trigonometry that you show. I'm simply going to assume that you have your coordinate system squared away. However, we still haven't said &lt;em&gt;how much&lt;/em&gt; force. To a first approximation it goes by the shock pressure $P$ times the area $A$ the object presents to the shock wave. So that gets us to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec{a} = \frac{P A}{m} \hat{n} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the unit vector $\hat{n}$ is normal to the surface of the shockwave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're still not done because we don't know $P$. Again, we'll take the simplest approximation. Assume you know the shock pressure $P_0$ at the surface of the explosive (this is presumably something you can look up for chemical explosives). When the shockwave has total area $\mathcal{A}$, it's pressure is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P = P_0 \frac{\mathcal{A}_0}{\mathcal{A}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathcal{A_0}$ is the area over which $P_0$ applied. For small sources (like a bomb or stick of dynamite) this will give you a $1/r^2$ type dependence for the pressure. For long linear sources you get a $1/\rho$ dependence; and for large surface charges a pressure that is independent of distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That isn't a good approximation, because you can expect the shockwave to spread out of it's own accord over time, and I have not yet modeled that. The really, really simple thing to do here is to choose a maximum range $l = \tau v_s$ equal to some time interval times the speed of shock propagation, and apply a hard cutoff at that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know how much force as a function of distance from the explosive and geometry of the explosive, but to get to change in velocity we need to know for how long the force is applied. Obviously that will be related to the depth of the shockwave and to it's speed of propagation: $ \Delta t \approx v_s * d $. Unfortunately I'm hazy on how to estimate $d$. For a simple explosive I might go with half the &quot;size&quot; of the charge (radius, thickness, whatever). The speed of propagation depends on the medium. Again, for game purposes you could simple chose some number. Once we have $\Delta t$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta v = \Delta t * a . $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you say that your engine works on a applied impulse, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ I = \Delta t * P * A = \Delta t * \frac{P_0 \mathcal{A}_0}{\mathcal{A}} * A, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is actually pretty simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that should get you on your way.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-04-17T18:45:30.373" LastActivityDate="2011-04-17T18:45:30.373" />
  <row Id="8797" PostTypeId="1" CreationDate="2011-04-17T18:45:55.503" Score="1" ViewCount="129" Body="&lt;p&gt;Assume you have a rigid body falling into the ocean at terminal velocity. Also assume that the rigid body does not break on impact. How could you figure out how much kinetic energy would be lost in the resulting splash? I promise this isn't a homework question, but an engineering thought experiment. I began to reason that water is incompressible so the body has to displace the water. I know the volume of my object so I could say I know how much water gets displaced. I also have a figure that the splash (radial jet) leaves at 20-30 times the impact speed. From there I can calculate a KE, but I don't believe it because this doesn't account for the surface tension. Any guidance would really be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3166" LastActivityDate="2011-04-17T21:46:57.167" Title="Kinetic energy loss when a rigid body falls into water" Tags="&lt;fluid-dynamics&gt;&lt;energy-conservation&gt;&lt;collision&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="8798" PostTypeId="1" CreationDate="2011-04-17T19:21:48.150" Score="1" ViewCount="162" Body="&lt;p&gt;Recently I had a quiz in my physics class and I feel like the professor made a mistake on the solution for it. Yes, I already have the answer to this question. I am not trying to get people's answers for homework or anything. But I would like to hear people responses to this.&#xA;The question was:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which of the following conditions are necessary restrictions imposed on the solutions of the one dimensional Schrodinger equation? (Choose all the apply)&lt;br&gt;&#xA;1. The wave function must be continuous.&lt;br&gt;&#xA;2. The wave function needs to be normalized.&lt;br&gt;&#xA;3. The wave function must converge to zero as x tends to positive and negative infinity.&lt;br&gt;&#xA;4. The wave is zero at the boundary conditions  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the first three are correct because they are right out of the book. But my problem lies with the last one. So far we have only dealt with a particle in a rigid box and in the derivation, the wave function was zero at the boundary conditions. That's why I said 4 was correct too. This was before we studied finite potential wells where the wave function exponentially decays at the boundaries. So with knowing our limits to the material for the quiz, do you guys think it is fair that he take points off for saying the wave function zero at the boundary condition when it really is zero at the boundary with what we were studying?&lt;/p&gt;&#xA;" OwnerUserId="3137" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-17T22:27:00.230" LastActivityDate="2011-04-17T22:27:00.230" ClosedDate="2011-04-17T22:22:11.797" Title="Does the wave function need to be zero at the boundaries?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8799" PostTypeId="2" ParentId="8798" CreationDate="2011-04-17T19:42:55.683" Score="2" ViewCount="" Body="&lt;p&gt;I would say it's definitely fair. You're not supposed to just take (4) for granted in the rigid box case - you were probably expected to understand &lt;em&gt;why&lt;/em&gt; (4) holds in the rigid box case, and if you did, you would see why (4) doesn't have to hold in the general case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A particle inside a rigid box can never be found outside the box. So the wavefunction is zero everywhere outside the box, but non-zero generally inside the box. Now since the wavefunction is continuous everywhere, this necessarily means that it has to be zero at the boundary of the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that if (as in the case for a finite potential well) the wavefunction does not have to be zero outside the well (because of tunelling), then continuity does &lt;em&gt;not&lt;/em&gt; require it to be zero at the boundary, so clearly (4) is false in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, the thing that causes (4) to be true in the rigid box is the absence of a wavefunction outside the box. Since you wouldn't expect this to be true in general, (4) need not hold true in general.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-04-17T19:42:55.683" />
  <row Id="8800" PostTypeId="1" AcceptedAnswerId="8802" CreationDate="2011-04-17T19:58:16.687" Score="4" ViewCount="176" Body="&lt;p&gt;The book I am reading takes the unjustified step $$e^{-\frac{i}{\hbar}\vec{p}\cdot\vec{r}}f(\vec{p}) = f(\frac{\hbar}{i}\vec{\nabla})e^{-\frac{i}{\hbar}\vec{p}\cdot\vec{r}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and similarly, he uses elsewhere $$e^{-\frac{i}{\hbar}\vec{p}\cdot\vec{r}}g(\vec{r}) = g(i\hbar\vec{\nabla}_p)e^{-\frac{i}{\hbar}\vec{p}\cdot\vec{r}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a gut feeling this follows somehow from the commutation relation $[r^i,p_j] = i\hbar\delta^i_j$ but I cannot justify this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow, the book has not introduced the standard commutation relations, and is working only with the fact that $\phi(\vec{p},t)$ and $\psi(\vec{r},t)$ are Fourier transforms of each other, and $f(\vec{p})$ and $g(\vec{r})$ are analytic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone give me a hint?&lt;/p&gt;&#xA;" OwnerUserId="1671" LastEditorUserId="1671" LastEditorDisplayName="" LastEditDate="2011-04-17T21:57:22.263" LastActivityDate="2011-04-17T21:57:22.263" Title="Fourier transform and commutation of functions" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" />
  <row Id="8801" PostTypeId="1" CreationDate="2011-04-17T20:09:28.657" Score="3" ViewCount="203" Body="&lt;p&gt;I know that by Heisenberg's Uncertainty Principle that it is not possible to know the exact values of position and momentum of a particle simultaneously, but can we know the exact values of momentum and velocity of a particle simultaneously? I would think the answer would be no because even if we were 100% certain of the particle's position, we would be completely unsure of the particle's momentum, thus making us also completely unsure of the particle's velocity. Does anyone have any insight into this?&lt;/p&gt;&#xA;" OwnerUserId="3137" LastActivityDate="2011-04-18T13:08:25.153" Title="Is it possible to know the exact values of momentum and velocity of a particle simultaneously?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="3" />
  <row Id="8802" PostTypeId="2" ParentId="8800" CreationDate="2011-04-17T20:14:50.363" Score="5" ViewCount="" Body="&lt;p&gt;It follows from the rule&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial}{\partial x^i}  e^{a_j x^j}= a_i e^{a_j x^j},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which by repeated use leads to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(\frac{\partial}{\partial x^i})  e^{a_j x^j}= f(a_i)e^{a_j x^j},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f$ is a polynomial. More general classes of functions $f$ can be reached by taking appropriate limits.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-04-17T20:43:44.773" LastActivityDate="2011-04-17T20:43:44.773" CommentCount="5" />
  <row Id="8805" PostTypeId="2" ParentId="584" CreationDate="2011-04-17T20:32:35.453" Score="1" ViewCount="" Body="&lt;p&gt;I noticed that causal sets afford a definition of energy ratios and their quantum, thanks to the relative frequency ratios formed by re-convergent pathways.  It is then possible to construct an extendable 4-D lattice in conjunction with higher frequency sequences representing nuclei embedded in the lattice, such that time dilation is proportional to local energy density.  This is precisely the local condition that accumulates to General Relativity at the global scale.  See [http://vixra.org/pdf/1006.0070v1.pdf][1].&#xA;   The modeling produced above does not employ random sprinkling in order to make the 4-D lattice conform to Einstein's continuous manifold, but rather substitutes quantized mass-energy values for Einstein's tensor values, reconciling quantum theory with gravity via pure theoretical hypothesis.  The coherence of the overall scheme is the only justification for replacing the continuum version of General Relativity with the discrete version.&#xA;   The search for general conditions for faithful embedding of a causal set in a manifold becomes a moot point if the specific modeling above proves to be correct.  At the least, an example exists of a causal set with an inherent inflationary metric defined, which shows how non-uniform a discrete causal set manifold can be and still conform to Einstein's well-accepted scheme of variable space-time curvature.&lt;/p&gt;&#xA;" OwnerUserId="3167" LastActivityDate="2011-04-17T20:32:35.453" CommentCount="1" />
  <row Id="8807" PostTypeId="2" ParentId="8798" CreationDate="2011-04-17T20:37:30.010" Score="1" ViewCount="" Body="&lt;p&gt;Regarding (4), you are asking wether it was fair to have asked a question which employed stuff  you presumably hadn't covered. That is a question regarding academic and testing practice. However, to answer how you can know by just knowing the theory you knew:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the infinite potential well the hamiltonian has the potential term $$\hat{H}\psi(r) = -\frac{\hbar^2}{2m}\nabla^2 \psi(r) - V(r) \psi(r)$$ Because $V(x)=\infty$ at the boundary, the only way you can stop the hamiltonian from blowing up is by taking $\psi=0$. It becomes obvious hereon, that if the potential does not blow up at the boundary, there is no need for the wavefunction to be zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Regarding (3), although correct, is necessary but not sufficient. It lies in the requirement for square integrability (or belogning to $L_2$) $$\int|\psi(x,t)|^2 dx\lt \infty$$ Infact, a wavefunction that depends on one spatial coordinate $x$ must fall off at least as fast as $\frac{1}{\sqrt{x}}$. So just being zero at infinity is not enough.&lt;/p&gt;&#xA;" OwnerUserId="1671" LastEditorUserId="1671" LastEditorDisplayName="" LastEditDate="2011-04-17T21:53:52.330" LastActivityDate="2011-04-17T21:53:52.330" />
  <row Id="8808" PostTypeId="2" ParentId="8801" CreationDate="2011-04-17T21:20:04.600" Score="2" ViewCount="" Body="&lt;p&gt;If in your theory the momentum operator and velocity operator are proportional to each other, then yes. Knowing one's eigenvalue means knowing the other's. It is always the case with any function of a &quot;known&quot; operator.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-04-17T21:20:04.600" CommentCount="1" />
  <row Id="8809" PostTypeId="2" ParentId="8797" CreationDate="2011-04-17T21:46:57.167" Score="2" ViewCount="" Body="&lt;p&gt;The final velocity in the fluid is zero if there is no gravity so all kinetic energy is lost. If there is gravity, the body will move down with the velocity determined with balance of the gravity, Archimedes', and friction forces. It has nothing to do with the initial velocity.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-04-17T21:46:57.167" />
  <row Id="8810" PostTypeId="2" ParentId="8801" CreationDate="2011-04-18T00:03:15.173" Score="0" ViewCount="" Body="&lt;p&gt;For the example of the electron, the Dirac equation has eigenvalues for momentum with which we are all familiar. But the eigenvalues for velocity are rather strange. Because it is a relativistic particle they are not related to the momentum simply by $v = p/m$. Instead, the eigenvalues of velocity turn out to be $v = \pm c$. For example, from the web notes for the 130 quantum mechanics class at UC San Diego:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Dirac equation has some unexpected&#xA;  phenomena which we can derive.&#xA;  Velocity eigenvalues for electrons are&#xA;  always $\pm c$ along any direction.&#xA;  Thus the only values of velocity that&#xA;  we could measure are $\pm c$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://quantummechanics.ucsd.edu/ph130a/130_notes/node45.html&quot; rel=&quot;nofollow&quot;&gt;http://quantummechanics.ucsd.edu/ph130a/130_notes/node45.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a relation to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Zitterbewegung&quot; rel=&quot;nofollow&quot;&gt;zitterbewegung&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-04-18T00:03:15.173" CommentCount="1" />
  <row Id="8811" PostTypeId="2" ParentId="7140" CreationDate="2011-04-18T00:30:20.450" Score="1" ViewCount="" Body="&lt;p&gt;Almost all the various &quot;metals&quot; (like primitive tribes that count &quot;1, 2 many&quot;, the astronomers describe the elements as &quot;hydrogen, helium, metals&quot;)  in the earth come from supernovae explosions. So the uranium got there pretty much the same way that the silicon, iron, oxygen, potassium, etc.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-04-18T00:30:20.450" CommentCount="3" />
  <row Id="8812" PostTypeId="2" ParentId="8778" CreationDate="2011-04-18T00:45:49.727" Score="2" ViewCount="" Body="&lt;p&gt;I assume that what you're asking is about the signal to noise ratio. Clearly the noise will not depend on distance; the problem is that the signal drops below the noise floor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you're transmitting with frequency $\lambda$. As the distance increases the intensity drops but the relationship is not so simple as is suggested in the comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the case where sound propagates in three dimensions over a distance $r$ and $\lambda &amp;lt;&amp;lt; r$, the sound energy is spread over the surface of a sphere of radius $r$ and so decreases proportional to $r^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same situation, but $r &amp;lt;&amp;lt; \lambda$, the sound energy does not decrease. This is a near-field effect. The place in the real-world where I've noticed this effect is when walking near a jack-hammer. As you approach the jack-hammer from a great distance, the sound steadily increases. But when you get very close to it (and I've spent my share of time &quot;balling that jack&quot;), the high frequencies dominate. This is because the low frequencies can't increase substantially as one gets nearer to the source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of describing this effect is to note that at long distances, the energy of the sound is transmitted in a single direction, away from the source. But very close to the source of the sound, energy is transmitted both towards the source and away from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of sound waves produced from a small source but with a very long wave length is a room with a cylinder of compressed gas that steadily leaks. This produces an increase in pressure but the pressure is essentially the same throughout the room (despite the small size of the source of the pressure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, in situations where the sound spreads in less than 3 dimensions, the energy in the sound decreases at a slower rate. In 2 dimensions it generally (for $\lambda &amp;lt;&amp;lt; r$) goes down as $1/r$, and in 1 dimension the sound does not decrease (other than absorption which applies to all these cases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A classic example of sound restricted to 2 dimensions is the sound produced by a volcanic explosion, as heard many thousands of miles away.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-04-18T02:30:48.287" LastActivityDate="2011-04-18T02:30:48.287" CommentCount="3" />
  <row Id="8813" PostTypeId="1" CreationDate="2011-04-18T02:25:40.180" Score="2" ViewCount="105" Body="&lt;p&gt;In class we derived the 3D case, but there's a step I don't understand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ N = g \cdot {V \over (2 \pi \hbar)^3} \cdot \int\limits_{0}^{\infty}{1 \over{e^{\left( E_p \over{K_B T}\right)}-1}} d^3 p =  g \cdot {V \over (2 \pi \hbar)^3} \cdot 4 \pi \cdot \int\limits_{0}^{\infty}{p^2 \over{e^{\left( E_p \over{K_B T}\right)}-1}} dp $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... I feel like if I knew why that step made sense, I could figure out how to do the equivalent thing for the 2D case, but I'm stuck on that.&lt;/p&gt;&#xA;" OwnerUserId="3170" LastActivityDate="2011-04-18T02:25:40.180" Title="How do I derive the critical temperature for bose condensation in two dimensions?" Tags="&lt;thermodynamics&gt;&lt;homework&gt;&lt;statistical-mechanics&gt;&lt;mathematics&gt;&lt;bose-einstein-condensate&gt;" CommentCount="10" />
  <row Id="8814" PostTypeId="2" ParentId="6897" CreationDate="2011-04-18T03:08:05.937" Score="1" ViewCount="" Body="&lt;p&gt;Newton also assumed that the speed of gravity is infinite. This led him to further support his idea of an infinite universe. If he had somehow known that the &quot;c&quot; is finite, then he would have eventually reasoned that the universe can not be static. &lt;/p&gt;&#xA;" OwnerUserId="3171" LastActivityDate="2011-04-18T03:08:05.937" />
  <row Id="8815" PostTypeId="1" AcceptedAnswerId="9229" CreationDate="2011-04-18T03:17:10.013" Score="4" ViewCount="136" Body="&lt;p&gt;Could you suggest which metal/alloy shows the least activation &amp;amp; better long-term stability for space (LEO, inside Van allen belts, deep space) applications? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or stability of construction elements isn't an issue comparing to stability of electronics?&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-04-28T08:00:30.083" Title="The most stable metal for radioactive environment" Tags="&lt;space&gt;&lt;radiation&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8816" PostTypeId="1" CreationDate="2011-04-18T07:28:43.740" Score="5" ViewCount="148" Body="&lt;p&gt;Inspired by: &lt;a href=&quot;http://physics.stackexchange.com/questions/8011/angular-deficit&quot;&gt;Angular deficit&lt;/a&gt;&lt;br&gt;&#xA;The 2+1 spacetime is easier for me to visualize, so let's use that here. (so I guess the cosmic string is now just a 'point' in space, but a 'line' in spacetime)  Edward says it is possible to cut a wedge out of a flat spacetime and glue the edges together.  So in my mind this looks like a paper cone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble understanding why this is flat everywhere except at the 'tip' of the cone.  Imagine a triangle on the original paper, and now after piecing it together, wouldn't an observer think the lines are now curved?  And in piecing it together, wouldn't there now be another angle, so it is a 4 sided polygon, and the exterior angles won't correctly add up to 360 degrees anymore?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just very confused because Edward and Lubos say the spacetime is flat everywhere except at the center, so the Riemann curvature tensor is zero everywhere except at the center, but Lubos says a parallel transported vector around a path on this flat spacetime can change angle!?  Does this mean we can't describe the parallel transport of a vector with the local Riemann curvature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully I've said enough that someone knowledgeable can see what is confusing me and can help me understand.  If a clear question is needed then let it be this:&lt;br&gt;&#xA;&lt;strong&gt;How can we explicitly calculate the curvature, and the effect this has on angles of paths or vectors, in conical spacetime?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 'paste flat spacetime pieces together' process seems very fishy to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt;&lt;br&gt;&#xA;Okay, thanks to Ted and Edward I got most of it figured out (although my attempt couldn't say anything about the curvature 'spike' at the center)', but still can't figure out how to see the parallel transport of vector in a closed loop ala Lubos's comment.  It would be neat to see this last part worked out for an arbitrary loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular Ted's comment &lt;em&gt;&quot;that (in some appropriately-defined sense) the average curvature inside the triangle is nonzero. In this particular case, that average comes entirely from a curvature &quot;spike&quot; at the origin.&quot;&lt;/em&gt;  sounds like there may be an easy way to transfer the integral around a path to an integral over the area bounded by the path, ala Gauss-Bonnet but the integral I'm getting doesn't even look like a normal integral and I don't really understand what Gauss-Bonnet is saying physically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone work out this last little piece explicitly, and if you use something like Gauss-Bonnet maybe help explain what the math is telling us about the physics here?&lt;/p&gt;&#xA;" OwnerUserId="2655" LastEditorUserId="2655" LastEditorDisplayName="" LastEditDate="2011-04-21T15:01:11.123" LastActivityDate="2011-04-21T15:01:48.423" Title="Conical spacetime of cosmic string" Tags="&lt;general-relativity&gt;&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8817" PostTypeId="1" CreationDate="2011-04-18T08:16:43.710" Score="0" ViewCount="213" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/7112/why-do-people-still-talk-about-bohmian-mechanics-hidden-variables&quot;&gt;Why do people still talk about bohmian mechanics/hidden variables&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I've heard of De Broglie–Bohm theory a.k.a causal interpretation of quantum theory. The predictions match accurately with with the nondeterministic quantum theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a philosophy buff this one seems just like the classical universe. No funny new-age religion style gimmicks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why don't people use this interpretation instead of the Copenhagen &quot;magic&quot; interpretation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I'm not an expert. I possess only superficial knowledge of quantum physics.&lt;/p&gt;&#xA;" OwnerUserId="3172" LastActivityDate="2011-04-18T09:18:10.100" ClosedDate="2011-04-18T23:47:11.680" Title="What is wrong with the De Broglie–Bohm theory a.k.a &quot;Causal Interpretation&quot; of quantum theory?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="8818" PostTypeId="2" ParentId="8817" CreationDate="2011-04-18T08:35:27.050" Score="0" ViewCount="" Body="&lt;p&gt;First, it is a mistake to think that DBB theory is classical. Momenta and positions are still not simultaneously defined, Heiseinberg's uncertainty principle still holds. Second, DBB theory only clarifies the Copenhagen interpretation by providing it with an ontology. A couple of reasons why it is not liked: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It's thought to be cheap, i.e. the way positions are added to complete the theory.&lt;/li&gt;&#xA;&lt;li&gt;It's non-local, while most people argue that quantum mechanics is local.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/q/7112/369&quot;&gt;Here is one of the latest extensive threads&lt;/a&gt; about the subject which in the end got closed because it was too argumentative.&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2011-04-18T08:35:27.050" CommentCount="3" />
  <row Id="8819" PostTypeId="2" ParentId="8817" CreationDate="2011-04-18T09:18:10.100" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;So why don't people use this interpretation instead of the Copenhagen &quot;magic&quot; interpretation?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People don't use this interpretation, because it is not &lt;em&gt;useful&lt;/em&gt;.  On the experimental front it doesn't aid in prediction of experiments, and on the theoretical front it doesn't aid in advancing theoretical techniques or understanding of newly found phenomenology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fundamental assumption of Bohmian mechanics is that the unknown particle position when we do an experiment, has an ensemble probability density the same as that predicted by the wavefunction in the usual interpretation.  This is necessary for Bohmian mechanics to reproduce the Copenhagen approach.  So even in Bohmian mechanics, the results for experiments can only be given probabilistically, and the wavefunction encodes these probabilities.  So when calculating experimental predictions, adding additional pieces to the state function do not help, and just are not &lt;em&gt;useful&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if we wanted to calculate using Bohmian mechanics, even some simple things become quite needlessly complicated.  For example even if you somehow knew the particle positions and the wavefunction, you still wouldn't know how to update the positions unless you somehow also knew what coordinate system is defining the absolute simultaneity ... and of course this frame isn't measureable.  And even if you knew this as well somehow, things like spin are now not clearly defined.  Spin in Bohmian mechanics is purely contextual, which means to predict the outcome of an experiment (even if you knew all the things previously mentioned) you'd now need to know precise details of the entire measuring device and simulate that as well.  So it should be clear why people use the simpler calculational interpretations instead of Bohmian mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, it has not been useful in advancing our knowledge.  Bohmian mechanics comes from ordinary non-relativistic quantum mechanics.  A large portion of our advances in theoretical physics have come from understanding how to appropriately use and apply non-trivial symmetries.  For example Lorentz symmetry allows us to use certain representation for particles, and also allows many terms in a general Lagrangian to be thrown out.  Bohmian mechanics instead obscures this right from the start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of bringing us the spin-statistics theorem, or advancing to relativistic quantum mechanics and to quantum field theory, Bohmian mechanics confuses instead of clarifying.  To keep it up to date with current understanding, people would need to keep updating it and playing 'catch-up'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people find it interesting to try to answer 'is this possible?' type questions by trying to find if Bohmian mechanics can be adjusted or fixed to somehow create a Bohmian quantum field theory.  I have yet to see a convincing relativistic formulation.  It is a bit worrisome that there are also researchers (who publish at least, so its not complete fringe) that instead feel Bohmian mechanics gives &lt;em&gt;the&lt;/em&gt; correct picture.  At some point it often crosses into metaphysics.  So physicists also tend to avoid Bohmian mechanics, as dicussions of it can often become controversial but rarely ever &lt;em&gt;useful&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2652" LastActivityDate="2011-04-18T09:18:10.100" />
  <row Id="8820" PostTypeId="1" AcceptedAnswerId="8826" CreationDate="2011-04-18T09:43:38.653" Score="3" ViewCount="92" Body="&lt;p&gt;I googled the sentence in the title of this question and found the famous experiment by &#xA; Wu &lt;em&gt;et al&lt;/em&gt; demonstrating that electrons in weak decay are emitted ``in the direction of motion of a left-handed screw rotating with the nuclear spin''.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This show that: asymmetry in decay -&gt; parity violation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to prove the opposite? (parity violation -&gt; asymmetry ?)&lt;/p&gt;&#xA;" OwnerUserId="2399" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-18T09:49:35.870" LastActivityDate="2011-04-18T13:43:10.107" Title="Why does parity violation in weak decay imply decay asymmetry?" Tags="&lt;parity&gt;&lt;weak-force&gt;" AnswerCount="1" />
  <row Id="8821" PostTypeId="2" ParentId="8626" CreationDate="2011-04-18T11:06:14.713" Score="-4" ViewCount="" Body="&lt;p&gt;Dear Tobias, everything is mush simpler: the equations are constraints to possible system motion and they themselves provide conserved quantities. You may call these equation &quot;constraints&quot; or &quot;symmetry requirement&quot;, if you like but they are necessary and sufficient conditions for having conserved quantities (=solutions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider an ODE: $F(\dot{x},x,t) = 0$ with some initial data $x_0 (t_0)$. After solving this equation you obtain a solution that can be casted in an implicit form like this: $f(x(t), t, x_0)=0$. On the other hand, this implicit form of solution &lt;em&gt;is&lt;/em&gt; a conservation law: a combination of the problem variable and the time does not depend on time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I differentiate this implicit solution and obtain: $f&amp;#39;_x\cdot \dot{x} + \dot{f} = 0$. It is an expression of conserving something, isn't it? On the other hand, it should coincide with or be equivalent to the original equation $F=0$. Now you understand that equations are necessary and sufficient requirements to have &lt;strong&gt;conserved quantities&lt;/strong&gt;. The latter are to a great extent synonymous to &lt;strong&gt;solutions&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: There is no need in any symmetry to have conserved quantities. Only equations. Symmetries simplify the look of conserved quantities (solutions). Consider a 1D motion of a particle in a time-dependent external force. The energy and momentum are not conserved but there are still two conservation &quot;laws&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-04-18T13:14:57.410" LastActivityDate="2011-04-18T13:14:57.410" CommentCount="11" />
  <row Id="8823" PostTypeId="2" ParentId="8801" CreationDate="2011-04-18T13:08:25.153" Score="1" ViewCount="" Body="&lt;p&gt;It is quite common to discuss the two extremes of the uncertainty principle, sinusoid and delta function. One has a perfectly defined wavelength but no position, the other has a perfectly defined position but no wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, neither one of these shapes is terribly physical for a particle's position wavefunction. A true sinusoidal wavefunction would extend through all space, which is absurd for several reasons (including the presence of other matter). A true delta function would be equally likely to have any momentum, which would probably violate conservation of energy. So, these two extreme limits are mathematically interesting, but not physically relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the question &quot;Does the uncertainty principle put some bound on momentum and velocity being simultaneously well-defined?&quot;, the answer is no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the question &quot;Does the uncertainty principle forbid me from measuring any single variable with infinite precision?&quot;, the answer is no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the question &quot;Does &lt;strong&gt;anything&lt;/strong&gt; forbid me from measuring with infinite precision?&quot;, &lt;a href=&quot;http://physics.aps.org/synopsis-for/10.1103/PhysRevLett.106.110406&quot; rel=&quot;nofollow&quot;&gt;the answer is yes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, your question mentions 'exact values', which is a very interesting, thorny subject. (Is it ever possible to measure an exact value? How would we tell the difference?) Are you really curious about 'exact values'? Are you more curious about where the Heisenberg uncertainty principle does and does not apply? Or are you curious if there are other bounds on our ability to measure, in addition to the uncertainty principle?&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2011-04-18T13:08:25.153" CommentCount="1" />
  <row Id="8824" PostTypeId="2" ParentId="8735" CreationDate="2011-04-18T13:20:43.870" Score="2" ViewCount="" Body="&lt;p&gt;The main answer to the question is that the full generator&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L_n = L^{(\mathrm{m})}_n+L^{(\mathrm{g})}_n$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;in Bosonic string theory is a sum of a matter part with normal ordering constant $a=1$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L^{(\mathrm{m})}_n=\frac{1}{2} \eta_{\mu\nu}\sum_m:\alpha_{n-m}^{\mu}\alpha_m^{\nu}:-\hbar a\delta_n^0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and a ghost part&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ L^{(\mathrm{g})}_m=\sum_n(m-n):b_{m+n}c_{-n}:.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matter and ghost part satisfy (normal-order-shifted) Virosoro algebras with central charges $\pm 26$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ [L^{(\mathrm{m})}_m,L^{(\mathrm{m})}_n]=\hbar(m-n)L^{(\mathrm{m})}_{m+n}+\hbar^2 \left(\frac{Dm(m^2-1)}{12}+2ma \right) \delta_{m+n}^0, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ [L^{(\mathrm{g})}_m,L^{(\mathrm{g})}_n]=\hbar(m-n)L^{(\mathrm{g})}_{m+n}+\hbar^2 \frac{m(1-13m^2)}{6} \delta_{m+n}^0, $$&#xA;respectively, while the full generator $L_n$ satisfies a Witt algebra&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ [L_m,L_n]=\hbar(m-n)L_{m+n}&#xD;&#xA;+\hbar^2\frac{(D-26)m^3+(24a+2-D)m}{12}\delta_{m+n}^0 $$&#xA;$$=\hbar(m-n)L_{m+n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in the critical dimension $D=26$ and $a=1$. To summarize, while the individual matter and ghost parts, and more generally, various subsectors of the theory, can have separately non-vanishing Virasoro quantum anomalies, there is no net Virasoro quantum anomaly in the full construction. To return to the question, the full generators $L_n$ act/have a representation/interpretation as conformal transformations $\sim z^{n+1}\partial/\partial z$ on the world sheet. In precisely the critical dimension $D=26$, there is no Weyl/conformal anomaly.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-04-18T18:39:05.240" LastActivityDate="2011-04-18T18:39:05.240" />
  <row Id="8825" PostTypeId="2" ParentId="8816" CreationDate="2011-04-18T13:35:06.897" Score="5" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Imagine a triangle on the original&#xA;  paper, and now after piecing it&#xA;  together, wouldn't an observer think&#xA;  the lines are now curved?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is no: lines that were straight on the original paper are still straight once it's been made into a cone. One way to see why is to write things out explicitly in polar coordinates. Let $(r,\phi)$ be the polar coordinates of a point on the paper, before you've folded it up into a cone. Put the origin at the vertex of the cone. The missing wedge of the paper means that only angles from 0 to $2\pi-\alpha$ will exist on the paper, where $\alpha$ is the &quot;angle deficit.&quot; Once you've folded the paper up into a cone, points with $\phi=0$ will be identified with points with $\phi=2\pi-\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the flat paper, the distance between two nearby points is &#xA;$$&#xD;&#xA;ds^2=dr^2+r^2\,d\phi^2.&#xD;&#xA;$$&#xA;To understand why the cone has a flat geometry, the key observation is that &lt;em&gt;this relationship continues to hold after you've bent the paper into a cone.&lt;/em&gt; That may not be intuitively obvious, but you can convince yourself that it's true. One way, which is messy but completely rigorous, is to write down the mapping between $(r,\phi)$ on the original paper and coordinates in 3-D space on the cone, then apply the usual distance formula. A better way is to draw a picture of a tiny right triangle on the flat paper, with sides $dr, r\,d\phi,ds$, and then draw the same picture again on the cone-ified paper. You can convince yourself that the sides of the triangle don't change, and that it's still a right triangle, so you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the above equation holds for both the original paper and the cone is all you need to know to be sure that the geometry of the cone is flat, and that straight lines in one map onto straight lines on the other. For instance, suppose you've got a path joining two points $(r_1,\phi_1)$ and $(r_2,\phi_2)$. The length of that path is just the integral of $ds$ along that path. That integral will be exactly the same before and after you fold the paper into a cone. Therefore, the path that was shortest before cone-ifying will also be the shortest path afterwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, you only have curvature if you have to &quot;crumple&quot; or &quot;stretch&quot; the paper. When you form a cone, you don't have to do that, anywhere except at the vertex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to go further and actually calculate the curvature, you need the machinery of Riemannian geometry. If you think more like a physicist than like a mathematician, then I'd recommend learning that from a general relativity book. I like the one by Schutz and the one by Hartle for a first pass through the subject. All I'll say here is that there is a completely well-specified algorithm for translating the &quot;line element&quot; -- that is, the expression for $ds^2$ in terms of the coordinates -- into a mathematical object called the curvature tensor, which in two dimensions boils down to just a single number known as the curvature.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-18T13:35:06.897" CommentCount="8" />
  <row Id="8826" PostTypeId="2" ParentId="8820" CreationDate="2011-04-18T13:43:10.107" Score="2" ViewCount="" Body="&lt;p&gt;Dear Pie86, the emission of particles in a weak decay is a complicated reaction, and Gell-Mann's totalitarian principle applies: every process or effect that is not prohibited by a symmetry will occur at a nonzero probability. The asymmetry or the spin-momentum correlation for the electrons is such an effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, it is infinitely unlikely that the asymmetry will be exactly zero unless the symmetry is implied by a valid symmetry. Because parity is not a valid symmetry, there's no reason for the asymmetry - the correlation between the spin and direction of the electron, among other similar correlation coefficients - to be exactly zero, so it will probably not be exactly zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the particular case of the beta decay, one may calculate the corresponding probability amplitude and the asymmetry - or correlation coefficients - out of it. If the Lagrangian has the $(a+b\gamma_5)$ matrix to guarantee an asymmetry, the asymmetry of the particular decay - or the spin-momentum correlation coefficient - will be proportional to something like $ab$ or $(a^2+b^2)$ times something. More generally, it will be a function of $a,b$ that is manifestly nonzero if both $a,b$ are nonzero. And in weak interactions, they are nonzero; in fact, weak interactions are maximally parity violating so $a=\pm b$.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-18T13:43:10.107" />
  <row Id="8827" PostTypeId="1" CreationDate="2011-04-18T13:53:09.213" Score="5" ViewCount="193" Body="&lt;p&gt;One of the pertinent questions about many body systems that causes me much wonder is why the solar system is so stable for billions of years. I came across the idea of &quot;resonance&quot; and albeit an useful concept, it hardly explains the long stability of the solar system. Normally an N body problem with inverse square mutual interaction is an example of a chaotic system. Is there any real progress about this stability issue?&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-04-19T18:20:49.353" Title="Question on the stability of the solar system" Tags="&lt;many-body&gt;&lt;solar-system&gt;&lt;chaos-theory&gt;&lt;stability&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="8828" PostTypeId="1" CreationDate="2011-04-18T14:39:16.963" Score="6" ViewCount="113" Body="&lt;p&gt;When looking at the PDG,&#xA;there is a difference between the 'running' and the 'current' quark masses.&#xA;Does anyone know which is the difference between these two?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks very much in advance!&lt;/p&gt;&#xA;" OwnerUserId="3176" LastActivityDate="2011-04-25T06:56:44.683" Title="What is the difference between 'running' and 'current' quark mass?" Tags="&lt;particle-physics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8829" PostTypeId="2" ParentId="8827" CreationDate="2011-04-18T14:59:03.083" Score="-2" ViewCount="" Body="&lt;p&gt;The sun contains about 99.9% of the mass of the solar system, so the motion of a planet is approximately that of a two body problem with $m&amp;lt;&amp;lt;M$, which is quite stable.&lt;/p&gt;&#xA;" OwnerUserId="1708" LastActivityDate="2011-04-18T14:59:03.083" CommentCount="3" />
  <row Id="8830" PostTypeId="1" CreationDate="2011-04-18T15:47:32.223" Score="2" ViewCount="293" Body="&lt;p&gt;If I put a ping pong ball in a vacuum, would it pop?  If so, at what point would it happen?  Any standard table ping pong ball is acceptable.&lt;/p&gt;&#xA;" OwnerUserId="366" LastActivityDate="2011-04-22T08:22:13.470" Title="If I put a ping pong ball in a vacuum, would it pop?" Tags="&lt;pressure&gt;&lt;vacuum&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="1" />
  <row Id="8831" PostTypeId="2" ParentId="8830" CreationDate="2011-04-18T16:07:39.180" Score="3" ViewCount="" Body="&lt;p&gt;Probably not - but without knowing the details of the specific ping-pong ball it's hard to tell.&lt;br&gt;&#xA;There may be an international standard for ping-pong balls who would be happy to tell you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interesting part about the question is scaling laws. In a vacuum there is a pressure of one atmosphere (15psi or 10tons/m^2) acting outward on the inside of the ball.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BUT there aren't all that many square meters (or even square inches) on a ping-pong ball so the actual force is really quite low.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sphere is a good shape for resisting this force, the most likely failure is the glue seam that holds the two halves together.  And a ball has to survive the force of being hit pretty hard by a bat so the glue line is reasonably strong.&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-04-18T16:07:39.180" CommentCount="1" />
  <row Id="8834" PostTypeId="1" AcceptedAnswerId="8835" CreationDate="2011-04-18T18:49:45.710" Score="2" ViewCount="186" Body="&lt;p&gt;I am having a little bit of trouble with this following problem - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose a particle has initial velocity and is moving with a constant acceleration. After t seconds its velocity is v. What will its displacement be in 2t seconds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to solve this on my own with the help of these equations - &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;v = u + at ( a is acceleration, u is initial velocity)&lt;/li&gt;&#xA;&lt;li&gt;s = ut + 1/2 at^2 ( s is the displacement)&lt;/li&gt;&#xA;&lt;li&gt;v^2 = u^2 + 2as&lt;/li&gt;&#xA;&lt;li&gt;s = (u+v)*t/2&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I tried to replace all the acceleration parameter using equation 1(a = (v-u)/t) in all other equations. I've also tried some other ways but they didn't help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I couldn't figure out how to do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish I could have tagged it as &lt;code&gt;high-school physics&lt;/code&gt;, but couldn't find such tag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is appreciated!!!!!&lt;/p&gt;&#xA;" OwnerUserId="3178" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2011-04-18T19:19:21.443" LastActivityDate="2011-04-21T20:57:02.637" Title="How to calculate displacement of an object" Tags="&lt;homework&gt;&lt;kinematics&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="8835" PostTypeId="2" ParentId="8834" CreationDate="2011-04-18T19:30:56.220" Score="2" ViewCount="" Body="&lt;p&gt;You have actually solved the problem yourself already! The method you described, solving for the acceleration and plugging it into the formula for displacement, work just fine. &lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-04-18T19:30:56.220" />
  <row Id="8836" PostTypeId="2" ParentId="8834" CreationDate="2011-04-18T19:42:04.137" Score="2" ViewCount="" Body="&lt;p&gt;You have found $a = \dfrac{v-u}{t}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You now need to use $s = ut + \dfrac{at^2}{2}$ except that where $t$ appears, you need to replace it with $2t$, and remember that $(2t)^2 = 4 t^2$.  Only then should you replace $a$ with the expression you have already found.&lt;/p&gt;&#xA;" OwnerUserId="2714" LastActivityDate="2011-04-18T19:42:04.137" />
  <row Id="8837" PostTypeId="2" ParentId="8834" CreationDate="2011-04-18T20:22:39.817" Score="2" ViewCount="" Body="&lt;p&gt;The key to this is that you have two equations in five unknowns,&#xA;$$s_1=\frac{at^2}{2}+v_0t, \qquad s_2=2at^2+2v_0t.$$&#xA;You can solve these for any two variables, leaving three still unknown. You definitely want to know $s_2$, so that's one of them. The other variable you want to isolate is $t$, in an equation that doesn't have $s_2$ in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, solve the first equation for $t$, which is quadratic, choose the positive solution, on the assumptions that $v_0&amp;gt;0$, $t&amp;gt;0$, and $a&amp;gt;0$, which gives $t=\frac{-v_0+\sqrt{v_0^2+2as_1}}{a}$, in terms of $v_0$, $a$, and $s_1$. Substitute that into the equation for $s_2$, and do some tidying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: You can see in the result that, on the above assumptions, $s_2$ is between $2\!\!\times$ and $4\!\!\times s_1$, depending on whether the initial velocity or the acceleration dominates the calculation. You should also consider the case when the acceleration is in the opposite direction to the direction of the initial velocity.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-04-19T12:51:15.470" LastActivityDate="2011-04-19T12:51:15.470" />
  <row Id="8838" PostTypeId="2" ParentId="8830" CreationDate="2011-04-18T20:25:52.420" Score="4" ViewCount="" Body="&lt;p&gt;Ping-Pong balls have 40 mm diameter and a weight of 2.7 grams. &#xA;Material is still pure (not mostly) celluloid. &#xA;(And Yes celluloid is thermoplastic, which is as informative as cars having wheels)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;diameter 40 mm ; mass is 2.7 g &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Density of celluloid is ca. 1.2 g/cm³ , &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://resources.metapress.com/pdf-preview.axd?code=k189105l81787078&amp;amp;size=largest&quot; rel=&quot;nofollow&quot;&gt;tensile strength roughly 8 kg/mm²&lt;/a&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From mass and density we get a &lt;/p&gt;&#xA;&#xA;&lt;p&gt;volume of celluloid of 2.25 cm³ ; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The PingPongBalls surface is 50.26 cm², &lt;/p&gt;&#xA;&#xA;&lt;p&gt;from that we get a wall thickness of 0.45 mm &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross section of the annular ring is ball circumference X wall thickness,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;126 mm  X 0.45 mm = 56.7 mm² &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Together with the tensile strength this makes 56.7 mm² X 8 kg/mm² = 453 kg (wow) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the cross section of the ball: 12.5 cm² , &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus 12.5cm² times 1 kg/cm² (roughly atmospheric pressure)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;makes 12.5 kg the &quot;force&quot; which the annular ring has to withstand. &lt;strong&gt;Peanuts&lt;/strong&gt;, &#xA;if I didn't make some mistake. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-18T20:25:52.420" CommentCount="3" />
  <row Id="8839" PostTypeId="1" AcceptedAnswerId="8841" CreationDate="2011-04-18T20:56:12.970" Score="2" ViewCount="80" Body="&lt;p&gt;I'm currently going through statistical physics, especially on Fermi energy when I came across a term called &quot;quasi-continuum&quot;, what exactly is it?&lt;/p&gt;&#xA;" OwnerUserId="2057" LastActivityDate="2011-04-18T21:14:33.983" Title="What is a quasicontinuum?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" />
  <row Id="8841" PostTypeId="2" ParentId="8839" CreationDate="2011-04-18T21:14:33.983" Score="3" ViewCount="" Body="&lt;p&gt;&quot;Quasi-something&quot; is &quot;almost something&quot;. In particular, quasicontinuum is a discrete set that contains so many entries (typically energy eigenvalues) with so small spacings that it looks like a continuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the energy eigenvalues are&#xA;$$E_n = n\epsilon$$&#xA;where $n$ is integer and $\epsilon$ is $0.0001$ times some natural energy difference, then $E_n$ is almost continuous and the set $\{E_n|n\in Z\}$ is a quasi-continuum. In many physical questions, a quasi-continuum behaves just like a continuum.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-18T21:14:33.983" CommentCount="1" />
  <row Id="8842" PostTypeId="1" AcceptedAnswerId="8844" CreationDate="2011-04-18T21:15:03.187" Score="3" ViewCount="91" Body="&lt;p&gt;There is a way to photograph air in a room. It makes convection, breathing and movement visible. The result looks a bit like a soap bubble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is some kind of optical effect. No special gases or fogs or powders are used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's this special form of photography called?&lt;/p&gt;&#xA;" OwnerUserId="662" LastActivityDate="2011-04-19T14:25:30.710" Title="What's the name for taking pictures of air flow in a normal room?" Tags="&lt;optics&gt;&lt;air&gt;" AnswerCount="1" />
  <row Id="8843" PostTypeId="1" AcceptedAnswerId="8845" CreationDate="2011-04-18T21:18:00.880" Score="1" ViewCount="112" Body="&lt;p&gt;The Schrödinger Equation provides a Probability Density map of the atom. In light of that, are either of the following possible: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The orbital/electron cloud converges to a 2d surface without heat (absolute zero)?&lt;/li&gt;&#xA;&lt;li&gt;heat is responsible for the probability density variation from the above smooth surface?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I have taken two calculus based physics, and Modern Physics with the Schrödinger equation, Heisenberg Uncertainty Principle, Etc. &lt;/p&gt;&#xA;" OwnerUserId="2843" LastEditorUserId="661" LastEditorDisplayName="" LastEditDate="2011-04-18T22:02:28.517" LastActivityDate="2011-04-18T22:42:15.253" Title="Is it possible that Atomic Electron Probability Density is a result of Heat?" Tags="&lt;heat&gt;&lt;schrodinger-equation&gt;" AnswerCount="1" />
  <row Id="8844" PostTypeId="2" ParentId="8842" CreationDate="2011-04-18T21:29:07.573" Score="5" ViewCount="" Body="&lt;p&gt;You think of Schlieren method? &#xA;Google for Schlieren or Schlieren-photography.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT]: Wikipedia article &lt;a href=&quot;http://en.wikipedia.org/wiki/Schlieren_photography&quot; rel=&quot;nofollow&quot;&gt;Schlieren photography&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="662" LastEditorDisplayName="" LastEditDate="2011-04-19T14:25:30.710" LastActivityDate="2011-04-19T14:25:30.710" CommentCount="1" />
  <row Id="8845" PostTypeId="2" ParentId="8843" CreationDate="2011-04-18T21:39:43.567" Score="10" ViewCount="" Body="&lt;p&gt;1.) No. All the calculations one does in elementary quantum mechanics courses &lt;em&gt;are&lt;/em&gt; at zero temperature. If they were at a finite temperature, you could never reliably say what quantum mechanical state your system is in; it would always be in an ensemble of different states. Since the ground-state wavefunction and ground-state density is &lt;em&gt;not&lt;/em&gt; a 2d surface, you don't get one at $T = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) No. At zero temperature, the probability density of your electron is given by the ground state wavefunction:&#xA;$$\varrho(x) = \psi_0^*(x) \psi_0(x)$$&#xA;At &lt;em&gt;finite&lt;/em&gt; temperature, your system is best described by an ensemble of states. Basically, you get &#xA;$$\varrho(x) = \sum_i p_i \psi_i^*(x) \psi_i(x)$$&#xA;where $p_i$ is the ensemble-probability for your system to be in state $\psi_i(x)$. For a canonical ensemble, for example, you have $p_i \sim e^{-E_i/kT}$ if your $\psi_i(x)$ are the energy-eigenstates with eigenenergies $E_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true for any other expectation value:&#xA;$$\langle \hat A \rangle = \sum_i p_i \langle \psi_i | \hat A | \psi_i \rangle$$&#xA;Note the two different expectation value here: One is $\langle \psi_i | \hat A | \psi_i \rangle$, the &lt;em&gt;quantum mechanical&lt;/em&gt; expectation value of $\hat A$ when the system is in state $| \psi_i \rangle$. The sum over these, together with the $p_i$, then gives the &lt;em&gt;thermodynamic&lt;/em&gt; expectation value&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This framework is used everywhere in physics and has been proven to be mind-bogglingly exact.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditorDisplayName="" LastEditDate="2011-04-18T22:42:15.253" LastActivityDate="2011-04-18T22:42:15.253" CommentCount="1" />
  <row Id="8846" PostTypeId="1" AcceptedAnswerId="8851" CreationDate="2011-04-18T22:21:32.160" Score="2" ViewCount="208" Body="&lt;p&gt;Stimulated by the &lt;a href=&quot;http://physics.stackexchange.com/questions/8830/if-i-put-a-ping-pong-ball-in-a-vacuum-would-it-pop&quot;&gt;calculation showing that a ping-pong ball does not pop&lt;/a&gt; in vacuum, I'm driven to ask how deep in the ocean a ping-pong ball can be brought before it collapses due to pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has a practical application in that boats have been refloated by the expedient of pumping ping pong balls into their hull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Youtube ping pong ball boat refloating:&#xA;&lt;a href=&quot;http://www.youtube.com/watch?v=4MOJN07XRYw&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=4MOJN07XRYw&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-04-19T02:17:35.150" Title="How deep in the ocean can a ping-pong ball go before it collapses due to pressure?" Tags="&lt;pressure&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8847" PostTypeId="2" ParentId="8846" CreationDate="2011-04-18T22:39:50.463" Score="2" ViewCount="" Body="&lt;p&gt;Let me assume that the answer to the previous question is right that the tensile strength is 50 MPa. I think that Georg is right that it's not correct just to equate that to the pressure. (I thought it was right at first -- dimensional analysis is often your friend!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a very heuristic argument why not. Suppose that the ball is under high pressure from the outside. Then the downward force on the upper hemisphere (or the upward force on the downward hemisphere) is $\sim PA\sim P\pi r^2$, where $r$ is the ball's radius. The part of the ball around the equator has to withstand that force. If the tensile strength is $T$, then the maximum force it can withstand is something like $T$ times the area of the ball that you'd expose if you sliced the ball in half around the equator. That is, it's $T(2\pi r)t$ where $t$ is the thickness of the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the maximum pressure the ball can withstand is something like $P\sim Tt/r$. If $t/r\sim 0.01$, then $P\sim 5\times 10^5$ Pa. That's 5 atmospheres. One atmosphere is 10 meters of depth in water, so I estimate that it's of order 50 m.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-04-18T22:53:12.097" LastActivityDate="2011-04-18T22:53:12.097" />
  <row Id="8848" PostTypeId="1" AcceptedAnswerId="8849" CreationDate="2011-04-18T23:09:46.837" Score="3" ViewCount="138" Body="&lt;p&gt;Hopefully this is the correct forum for this. I felt that Physics Overflow may not be the correct place.  I had a student approach me ask me what kinds of mathematics goes into the study of wormholes. She specifically asked whether there is any amount of topology involved, but I'd appreciate it if I could give her a more full answer. So, topology answers are the best but other areas of math would be nice. I am a topology Ph.D. student but know almost zero physics, just in case my background helps with your answer. I didn't know which tag(s) to add.  Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3183" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-19T00:16:41.947" LastActivityDate="2011-04-19T01:30:08.127" Title="Math and Wormholes" Tags="&lt;general-relativity&gt;&lt;mathematical-physics&gt;&lt;wormholes&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="8849" PostTypeId="2" ParentId="8848" CreationDate="2011-04-19T01:30:08.127" Score="5" ViewCount="" Body="&lt;p&gt;The answer to your initial question is &quot;mostly differential geometry, with a little bit of topology&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the Kruskal diagram for the Schwarzschild spacetime:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://members.arstechnica.com/x/zeotherm/rindler1.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This represents the largest posible manifold covered by the coordinate system $(t,r,\theta,\phi)$ in which the Schwarzschild spacetime's metric is usually rendered (these coordinates only cover a patch of this spacetime).  The hyperbola labeled $r=0$ is a region in which the spacetime curvature is infinite.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diagonal directions on the diagram represent the paths traveled by light rays, while directions more vertical than this represents the possible paths of observers.  The diagonal region labeled II is therefore constrained to intersect the surface $r=0$ at some point--this is the interior of the black hole.  (note that all histories in region III have the r=0 surface in their past, and they all must exit the diagonal rays--this is the interior of the white hole).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not the case for regions I and IV, however.  Observers in these surfaces need never intersect the $r=0$ surface, either in their past or their future (see the line labeled $r=const.--this could represent a circular orbit happily orbiting in a circle forever).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, regions I and IV are similar enough that one could imagine at some point far to the right, a surface in region I is identified (in the sense of topology) with a surface in region IV.  Observers in this region would say that they are &quot;nearby,&quot; but travelling in this way would take a long time, as you would have to go far to the right, and then to the surface, and back around.  A faster route may be to travel through the point where region I and region IV intersect.  It should be clear that this is not a possible path for this diagram, however--you would have to travel at a steeper angle than $45^{\circ}$ in order to take this trip.  This wormhole is not transversible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out, however, that for perturbations of the Schwarzschild geomerty, however, you do get transversible wormholes, which you can picture by just having the two lines coming from the black hole intersect somewhere higher than the (T,X) origin, and the two lines coming from the white hole doing the symmetric thing.  Then, you are allowed to travel in either direction to get to the region where the two spheres are identified.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-04-19T01:30:08.127" CommentCount="1" />
  <row Id="8850" PostTypeId="1" CreationDate="2011-04-19T02:15:44.927" Score="4" ViewCount="193" Body="&lt;p&gt;On the &lt;a href=&quot;http://en.wikipedia.org/wiki/Alcubierre_drive&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article on the Alcubierre drive&lt;/a&gt;, it says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since the ship is not moving within this bubble, but carried along as the region itself moves, conventional relativistic effects such as time dilation do not apply in the way they would in the case of a ship moving at high velocity through flat spacetime relative to other objects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, this method of travel does not actually involve moving faster than light in a local sense, since a light beam within the bubble would still always move faster than the ship; it is only &quot;faster than light&quot; in the sense that, thanks to the contraction of the space in front of it, the ship could reach its destination faster than a light beam restricted to travelling outside the warp bubble.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm confused about the statement &lt;em&gt;&quot;conventional relativistic effects such as time dilation do not apply&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say Bob lives on Earth, and Jill lives on a planet in Andromeda, and we'll say for the sake of argument that they're stationary. If I were to travel from Bob to Jill using an Alcubierre drive such that the journey would take me, say, 1 week from my reference frame... how long would Jill have to wait from her reference frame? Do the time dilation effects cancel out altogether? Would she only wait 1 week?&lt;/p&gt;&#xA;" OwnerUserId="2348" LastActivityDate="2011-07-30T06:28:27.450" Title="Alcubierre Drive - Clarification on relativistic effects" Tags="&lt;general-relativity&gt;&lt;speed-of-light&gt;&lt;faster-than-light&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="8851" PostTypeId="2" ParentId="8846" CreationDate="2011-04-19T02:17:35.150" Score="8" ViewCount="" Body="&lt;p&gt;The collapse pressure of a ping-pong ball is probably limited by the eventual &lt;a href=&quot;http://en.wikipedia.org/wiki/Buckling&quot; rel=&quot;nofollow&quot;&gt;buckling&lt;/a&gt; of the wall, instead of a compressive failure of its walls. Assuming (unrealistically) that the ball is a perfect sphere, the critical external pressure &lt;a href=&quot;http://oai.dtic.mil/oai/oai?verb=getRecord&amp;amp;metadataPrefix=html&amp;amp;identifier=AD0422297&quot; rel=&quot;nofollow&quot;&gt;will be given by&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_e - P_i = \frac{2E\,\left(\frac{h}{R}\right)^2}{\sqrt{3(1-\nu^2)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $P_e$ is the external pressure, $P_i$ the internal pressure, $E$ the &lt;a href=&quot;http://en.wikipedia.org/wiki/Young%27s_modulus&quot; rel=&quot;nofollow&quot;&gt;Young's modulus&lt;/a&gt; of the material, $h$ the shell thickness, $R$ the sphere radius and $\nu$ the &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson%27s_ratio&quot; rel=&quot;nofollow&quot;&gt;Poisson's ratio&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ping_pong_ball#Ball&quot; rel=&quot;nofollow&quot;&gt;A ping-pong ball has&lt;/a&gt; a diameter of 40 millimeters, a weight of 2.7 grams and &lt;a href=&quot;http://www.britannica.com/EBchecked/topic/101626/celluloid&quot; rel=&quot;nofollow&quot;&gt;is made of celluloid&lt;/a&gt;. &lt;a href=&quot;http://www.efunda.com/materials/polymers/properties/polymer_datasheet.cfm?MajorID=cellulose&amp;amp;MinorID=6&quot; rel=&quot;nofollow&quot;&gt;As celluloid's&lt;/a&gt; density is $1.4\,\mathrm{g\ cm^{-3}}$, its wall thickness will be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h = \frac{2.7\,\mathrm{g}}{4\,\pi\,(2\,\mathrm{cm})^2\,1.4\,\mathrm{g\ cm^{-3}}} \approx 0.04\,cm$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Young's modulus for celluloid is approximately $1400\,\mathrm{MPa}$ and its Poisson's ratio can be estimated as 0.35. Replacing these values in the expression for the critical pressure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_e - P_i = \frac{2\cdot 1400\,\mathrm{MPa}\cdot 0.02^2}{\sqrt{3(1-0.35^2)}} = \frac{1.12\,\mathrm{MPa}}{1.62} \approx 0.7\,\mathrm{MPa}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a compressive strength of $50\,\mathrm{MPa}$), the ball will fail &quot;compressively&quot; at a pressure of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_e - P_i = \frac{50\,\mathrm{MPa}\cdot 0.04\,\mathrm{cm}\cdot 2\,\pi\cdot2\,\mathrm{cm}}{\pi(2\,\mathrm{cm})^2} = 2\,\mathrm{MPa}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the real failure mode will be buckling. If the internal pressure of the ball is not very different from the atmospheric pressure, the crush depth of a perfect ball will be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_c \approx \frac{0.7\,\mathrm{MPa}}{0.01\,\mathrm{MPa}\ \mathrm{m^{-1}}} = 70\,\mathrm{m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real crush depth will be between a half and a quarter of this value, matching &lt;a href=&quot;http://kwc.org/mythbusters/2004/11/mythbusters_ping_pong_balls_an.html&quot; rel=&quot;nofollow&quot;&gt;the experimental value of approximately 30 meters&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1656" LastActivityDate="2011-04-19T02:17:35.150" />
  <row Id="8852" PostTypeId="1" CreationDate="2011-04-19T03:26:19.117" Score="1" ViewCount="131" Body="&lt;p&gt;&lt;a href=&quot;http://www.getbuckyballs.com&quot; rel=&quot;nofollow&quot;&gt;Buckyballs&lt;/a&gt; are a desktoy that consist of 216 tiny spherical magnets. You can makes all kinds of interesting shapes and structures out of them (do a Google search).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to model them in a physics simulation, but I'm not sure how to do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, I'm thinking calculate the sum of the forces between each of the buckyball's poles and each of the other buckyballs' poles, and update its linear momentum. Then calculate the torque that each of those forces causes about the magnet's center, and use that to update angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I'm having is how to model the magnets touching. It's easy in real life to make a long chain or a hexagon. How should I model the magnets actually running into each other to mimic this behavior?&lt;/p&gt;&#xA;" OwnerUserId="3184" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-19T03:40:01.577" LastActivityDate="2011-04-19T08:35:30.833" Title="How can I model buckyballs sticking together? (Tiny spherical magnets)" Tags="&lt;electromagnetism&gt;&lt;simulation&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="8853" PostTypeId="2" ParentId="8850" CreationDate="2011-04-19T04:41:46.493" Score="2" ViewCount="" Body="&lt;p&gt;Relativistic effects happen when you travel near light speed through space-time. With the Alcubierre drive, you don't travel through space-time, but remain stationary and the space-time around you is warped in a way that brings you closer to your destination. So if it only took a week for you, it would only take a week for your observers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It'd be like if you put two objects on a sheet, and instead of moving them across the sheet, you scrunched up the part of the sheet between them--moving them closer together even though they both remained stationary relative to the sheet.&lt;/p&gt;&#xA;" OwnerUserId="417" LastActivityDate="2011-04-19T04:41:46.493" CommentCount="3" />
  <row Id="8854" PostTypeId="2" ParentId="8816" CreationDate="2011-04-19T04:57:57.703" Score="1" ViewCount="" Body="&lt;p&gt;Here is my attempt, hopefully people can learn from my mistakes or something (&lt;em&gt;please&lt;/em&gt; leave comments, so that I can learn more)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From suggestion, I'll focus on 2+0 space.  The parallel transport sounds like it can be gotten from the Christoffel symbols, and in turn the Riemann curvature tensor can be calculated from the Christoffel symbols. So I'll try to calculate those here (in gory detail to try to get a feel for the calculations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a flat piece of paper, cut a $\alpha$ degree wedge out of it and piece the edges together.  This is the &quot;cone space&quot; for this discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The line element can be written as:&lt;br&gt;&#xA;$$ds^2 = dr^2 + r^2 d\theta^2$$&#xA;for $0 \le r &amp;lt; \infty$ and $0 \le \theta \le 2\pi - \alpha$&lt;br&gt;&#xA;alternatively we can choose a scaled angle so that $k \phi=\theta, k=1 -\alpha/2\pi$ giving the line element in these coordinates as&lt;br&gt;&#xA;$$ds^2 = dr^2 + r^2 k^2 d\phi^2$$&#xA;for $0 \le r &amp;lt; \infty$ and $0 \le \phi \le 2\pi$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g_{00}=1,g_{11}=r^2k^2$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g_{mk,\ell} = &#xD;&#xA;\left\{&#xD;&#xA;     \begin{array}{ll}&#xD;&#xA;       2rk^2 &amp;amp;: m=k=1,\ell=0\\&#xD;&#xA;       0 &amp;amp; : \mathrm{otherwise}\\&#xD;&#xA;     \end{array}&#xD;&#xA;\right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Christoffel symbol is (defined to be?):&#xA;$$\Gamma^i_{k\ell}={1 \over 2} g^{im} (g_{mk,\ell} + g_{m\ell,k} - g_{k\ell,m})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only non-zero components are:&#xA;$$\Gamma^0_{11}= {1 \over 2} g^{00}(- g_{11,0}) = -rk^2$$&#xA;$$\Gamma^1_{10}=\Gamma^1_{01} = {1 \over 2} g^{11}g_{11,0} = {1 \over 2} \frac{1}{r^2k^2}2rk^2 = 1/r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Reimann curvature is (defined to be?):&#xA;$${R^\rho}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho_{\nu\sigma}&#xD;&#xA;    - \partial_\nu\Gamma^\rho_{\mu\sigma}&#xD;&#xA;    + \Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma}&#xD;&#xA;    - \Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the (0,0,?,?) components&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{split}&#xD;&#xA;{R^0}_{0\mu\nu} &amp;amp;= \partial_\mu\Gamma^0_{\nu0}&#xD;&#xA;    - \partial_\nu\Gamma^0_{\mu0}&#xD;&#xA;+ \Gamma^0_{\mu\lambda}\Gamma^\lambda_{\nu0}&#xD;&#xA;    - \Gamma^0_{\nu\lambda}\Gamma^\lambda_{\mu0} \\&#xD;&#xA;&amp;amp;= \Gamma^0_{\mu1}\Gamma^1_{\nu0}&#xD;&#xA;    - \Gamma^0_{\nu1}\Gamma^1_{\mu0} = 0&#xD;&#xA;\end{split}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the (0,1,?,?) components&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{split}&#xD;&#xA;{R^0}_{1\mu\nu} &amp;amp;= \partial_\mu\Gamma^0_{\nu1}&#xD;&#xA;    - \partial_\nu\Gamma^0_{\mu1}&#xD;&#xA;    + \Gamma^0_{\mu\lambda}\Gamma^\lambda_{\nu1}&#xD;&#xA;    - \Gamma^0_{\nu\lambda}\Gamma^\lambda_{\mu1} \\&#xD;&#xA; &amp;amp;= \partial_\mu\Gamma^0_{\nu1}&#xD;&#xA;    - \partial_\nu\Gamma^0_{\mu1}&#xD;&#xA;    + \Gamma^0_{\mu1}\Gamma^1_{\nu1}&#xD;&#xA;    - \Gamma^0_{\nu1}\Gamma^1_{\mu1} \\&#xD;&#xA;{R^0}_{100} &amp;amp;= {R^0}_{111} = 0 \\&#xD;&#xA;{R^0}_{101} &amp;amp;= \partial_0\Gamma^0_{11}&#xD;&#xA;    - \Gamma^0_{11}\Gamma^1_{01} \\&#xD;&#xA; &amp;amp;= -k^2 -(-k^2r)(1/r) = 0 \\&#xD;&#xA;{R^0}_{110} &amp;amp;= - \partial_0\Gamma^0_{11}&#xD;&#xA;    + \Gamma^0_{11}\Gamma^1_{01} \\&#xD;&#xA; &amp;amp;= -(-k^2) +(-k^2r)(1/r) = 0&#xD;&#xA;\end{split}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't feel like working out the (1,0,?,?) components.  So I'll skip the double check and just trust the antisymmetry to (0,1,?,?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the (1,1,?,?) components&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{split}&#xD;&#xA;{R^1}_{1\mu\nu} &amp;amp;= \partial_\mu\Gamma^1_{\nu1}&#xD;&#xA;    - \partial_\nu\Gamma^1_{\mu1}&#xD;&#xA;    + \Gamma^1_{\mu\lambda}\Gamma^\lambda_{\nu1}&#xD;&#xA;    - \Gamma^1_{\nu\lambda}\Gamma^\lambda_{\mu1} \\&#xD;&#xA; &amp;amp;= (\Gamma^1_{\mu0}\Gamma^0_{\nu1}&#xD;&#xA;    - \Gamma^1_{\nu0}\Gamma^0_{\mu1}) + (\Gamma^1_{\mu1}\Gamma^1_{\nu1}&#xD;&#xA;    - \Gamma^1_{\nu1}\Gamma^1_{\mu1}) = 0&#xD;&#xA;\end{split}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this coordinate system, all $dimension^4$ components of the Riemann tensor are zero. So this is indeed flat space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to look at parallel transport.  A vector is parallel transported if the covariant derivative is zero.&#xA;$${\lambda^a}_{;b} = \partial_b \lambda^a+\Gamma^a_{cb}\lambda^c = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an infinitesimal step $dx^b$, the new components $\tilde{\lambda}^a$ will be&#xA;$$\tilde{\lambda}^a = \lambda^a + \Gamma^a_{cb}\lambda^c dx^b = \lambda^c (\delta^a{}_c + \Gamma^a_{cb} \ dx^b)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So instead of a normal integral which is like a sum of infinitesimal pieces, we need something that does a series of infinitesimal multiplications along the path.  Apparently this is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Product_integral&quot; rel=&quot;nofollow&quot;&gt;product integral&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we have a path $s^b(t),0\le t\le 1$, then I think&#xA;$$\tilde{\lambda}^a = \lambda^c \prod_{t=0}^{t=1} \left(\delta^a{}_c + \Gamma^a_{cb} \ \frac{\partial s^b(t)}{\partial t} dt\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to be able to check out Lubos's previous comment that this is just the identity for closed paths unless it goes around the origin.  I'm not sure how to continue though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on Ted's comment that the angular deficit is somehow related to an average of the enclosed curvature, maybe there is some Gauss-Bonnet type way of doing this - relating a path integral to some integral over an area bounded by the path.&lt;/p&gt;&#xA;" OwnerUserId="2655" LastEditorUserId="2655" LastEditorDisplayName="" LastEditDate="2011-04-19T08:19:01.743" LastActivityDate="2011-04-19T08:19:01.743" CommentCount="2" />
  <row Id="8855" PostTypeId="2" ParentId="8777" CreationDate="2011-04-19T05:29:36.203" Score="0" ViewCount="" Body="&lt;p&gt;on further digging, i confirmed that the parabola in a sphere, holds its shape by continually topping up the pressure in the 'top' of the sphere to keep the parabolic shape bowed the right way.&lt;/p&gt;&#xA;" OwnerUserId="2951" LastActivityDate="2011-04-19T05:29:36.203" />
  <row Id="8856" PostTypeId="1" AcceptedAnswerId="8857" CreationDate="2011-04-19T06:21:43.570" Score="2" ViewCount="125" Body="&lt;p&gt;It has been more than a decade since I did all vector related math and physics so pardon me if my question does not make sense. I am reading some article that says it was inspired from Coulomb's law and gives the following expression to calculate the partial force that one particle is exerting on another at time $n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f^{i,j}_n = (c - |p^{i}_n - p^{j}_n|)\frac{p^{j}_n - p^{i}_n}{|p^{i}_n - p^{j}_n|}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c$ is the range of the particle over which it's force can spread and $p^{i}_n$ is the location of particle $i$ in the form of $(x,y)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I checked up the Wikipedia article on Coulomb's law but this formula does not make sense to me. What exactly is the formula trying to achieve? Can someone please help me understand this?&lt;/p&gt;&#xA;" OwnerUserId="3186" LastActivityDate="2011-04-19T11:17:33.610" Title="Need some help interpreting a formula inspired from Coulomb's law" Tags="&lt;electromagnetism&gt;" AnswerCount="2" />
  <row Id="8857" PostTypeId="2" ParentId="8856" CreationDate="2011-04-19T06:50:45.677" Score="7" ViewCount="" Body="&lt;p&gt;It's nothing you can rigorously derive from Coulomb's law, but the idea is probably the following: the rightmost factor&#xA;$$\frac{p^j - p^i}{|p^j - p^i|}$$ is just the unit vector pointing from particle $i$ to particle $j$, just as in Coulomb's law or Newton's law. Normally, for Coulomb's law, you'd have a factor of $q_1 q_2/4\pi\varepsilon_0 |p^j - p^i|^2$ in front of that, but here they replaced the $1/r^2-$ behaviour by a linearly decreasing factor&#xA;$$c - |p^j - p^i|,$$ which yields c when $p^i = p^j$, and vanishes as $|p^j - p^i| \rightarrow c.$ (Since you call c a range, they &lt;em&gt;might&lt;/em&gt; also mean that whenever $|p^j - p^i| &amp;gt; c$, the force should vanish, but it's not clear from the notation alone.)&lt;/p&gt;&#xA;" OwnerUserId="1471" LastEditorUserId="1471" LastEditorDisplayName="" LastEditDate="2011-04-19T11:17:33.610" LastActivityDate="2011-04-19T11:17:33.610" />
  <row Id="8858" PostTypeId="2" ParentId="8856" CreationDate="2011-04-19T06:53:30.603" Score="3" ViewCount="" Body="&lt;p&gt;It says that the force between two particles acts along the line joining them (with a magnitude that also depends on the distance between I and j) that last term is a unit vector pointing from i to j. &lt;/p&gt;&#xA;" OwnerUserId="2599" LastActivityDate="2011-04-19T06:53:30.603" />
  <row Id="8859" PostTypeId="2" ParentId="8642" CreationDate="2011-04-19T07:36:30.760" Score="4" ViewCount="" Body="&lt;p&gt;It means that the postulate is incorrect in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever the number of degrees of freedom is large, there is some &quot;extensivity&quot;. For example, a single string stores energy and entropy in many modes labeled by $n$, the Fourier mode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is surely incorrect that the entropy and other quantities is proportional to volume in general. In particular, quantum gravity obeys the holographic principle which implies that the entropy only scales as a surface area, not the volume. In that context, the disagreements are not just exceptions: they're generic and they show that &quot;modular nuclearity&quot; is much more wrong than the modest example of the Hagedorn temperature shows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many other wrong assumptions underlying AQFT, too. The closer one studies renormalized field theory; string theory; quantum gravity, the more visible the invalidity of the AQFT prejudices becomes.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-19T07:36:30.760" CommentCount="1" />
  <row Id="8860" PostTypeId="1" AcceptedAnswerId="8861" CreationDate="2011-04-19T07:41:00.193" Score="8" ViewCount="250" Body="&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/8626/if-all-conserved-quantities-of-a-system-are-known-can-they-be-explained-by-symme&quot;&gt;This question&lt;/a&gt; resulted, rather as by-product, the discussion on how to count degrees of freedom (DOF). I extend that question here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Are necessary&lt;sup&gt;1&lt;/sup&gt; derivatives such as velocities counted as individual DOFs or together with the respective coordinate&lt;sup&gt;2&lt;/sup&gt;?&lt;/li&gt;&#xA;&lt;li&gt;Should complex valued DOFs be counted twice as in &quot;two real-valued DOFs&quot; or once as in &quot;one complex-valued DOF&quot;? (I mean, when one does not want to specify this explicitly)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Please answer with some reference, unless it turns out this is actually a matter of taste rather than a strictly defined thing.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1)&lt;/sup&gt; I mean those a value of which is &lt;em&gt;required&lt;/em&gt; as an initial condition&lt;br&gt;&#xA;&lt;sup&gt;2)&lt;/sup&gt; I count fields in QFT as coordinates as well, while space-time-coordinates are parameters to me, if that matters. I know a field actually has $\infty$ (or rather, $2^\aleph$) DOFs itself, but let's say e.g. &quot;one $\mathbb R^3$ continous DOF&quot; in that case&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-04-19T08:03:58.067" LastActivityDate="2011-08-17T17:44:43.710" Title="What is the definition of how to count degrees of freedom?" Tags="&lt;definition&gt;&lt;degrees-of-freedom&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="8861" PostTypeId="2" ParentId="8860" CreationDate="2011-04-19T07:57:32.407" Score="5" ViewCount="" Body="&lt;p&gt;To name an simple example, a 1D simple gravity &lt;a href=&quot;http://en.wikipedia.org/wiki/Pendulum&quot; rel=&quot;nofollow&quot;&gt;pendulum&lt;/a&gt; with Lagrangian&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L(\theta,\dot{\theta}) = \frac{m}{2}\ell^2 \dot{\theta}^2 + mg\ell\cos(\theta)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;has &lt;em&gt;one&lt;/em&gt; d.o.f., $\theta$, although its solution $\theta=\theta(t)$ has &lt;em&gt;two&lt;/em&gt; integration constants. Here, $\theta$ is the angle of the pendulum; $\dot{\theta}$ is the (angular) velocity; and $p_{\theta}:=\frac{\partial L}{\partial\dot{\theta}}=m\ell^2\dot{\theta}$ is the (angular) momentum. Furthermore, the configuration space with coordinates $(\theta,\dot{\theta})$ and the phase space with coordinates $(\theta,p_{\theta})$ are both &lt;em&gt;two&lt;/em&gt; dimensional spaces. In other words, it takes &lt;em&gt;two&lt;/em&gt; coordinates to fully describe the instantaneous state of the pendulum at a given instant $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to answer the main question: no, the corresponding velocity (or momentum in the Hamiltonian formulation) is not counted as a separate d.o.f.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Landau &amp;amp; Lifshitz, &quot;&lt;a href=&quot;http://lccn.loc.gov/60011795&quot; rel=&quot;nofollow&quot;&gt;Mechanics&lt;/a&gt;&quot;: see &lt;a href=&quot;http://i.stack.imgur.com/jbecX.jpg&quot; rel=&quot;nofollow&quot;&gt;first page&lt;/a&gt; of Chapter 2; &lt;/li&gt;&#xA;&lt;li&gt;H. Goldstein, &quot;&lt;a href=&quot;http://lccn.loc.gov/79023456&quot; rel=&quot;nofollow&quot;&gt;Classical Mechanics&lt;/a&gt;&quot;: see pg.13 and first page of Chapter 8 in both 2nd and 3rd edition;&lt;/li&gt;&#xA;&lt;li&gt;J.V. Jose and E.J. Saletan, &quot;&lt;a href=&quot;http://lccn.loc.gov/97043733&quot; rel=&quot;nofollow&quot;&gt;Classical Dynamics: A Contemporary Approach&lt;/a&gt;&quot;: see p.18;&lt;/li&gt;&#xA;&lt;li&gt;or Wikipedia, either &lt;a href=&quot;http://en.wikipedia.org/wiki/Degrees_of_freedom_%28mechanics%29&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Degrees_of_freedom_%28physics_and_chemistry%29&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-08-17T17:44:43.710" LastActivityDate="2011-08-17T17:44:43.710" CommentCount="7" />
  <row Id="8862" PostTypeId="1" CreationDate="2011-04-19T08:13:00.323" Score="-6" ViewCount="242" Body="&lt;p&gt;This question raised since yesterday being full moon day :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a saying that&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;On Full moon day sea tide use to be more aggressive  because of the moon's gravity&lt;/li&gt;&#xA;&lt;li&gt;Neil Armstrong was not able to walk the way on moon as they way we use to walk on earth(where some video proof exist for his peculiar walk on moon) &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here is my doubt :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;If moon's gravity take effect on earth's sea tide why Armstrong was not able to walk free on moon?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="3187" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-04-19T16:41:05.913" LastActivityDate="2011-04-19T16:41:05.913" Title="Serious doubt about gravity of Moon" Tags="&lt;gravity&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="0" />
  <row Id="8863" PostTypeId="2" ParentId="8852" CreationDate="2011-04-19T08:35:30.833" Score="2" ViewCount="" Body="&lt;p&gt;In a first approximation, you can add an artificial potential which cases repulsive force when the spheres are overlapping -- this will however result in some oscillation artifacts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proper way to do it is to use rigid body dynamics with constraints, but this is a bit more complex thing; check out this page for more details: &lt;a href=&quot;http://www.cs.cmu.edu/~baraff/sigcourse/&quot; rel=&quot;nofollow&quot;&gt;http://www.cs.cmu.edu/~baraff/sigcourse/&lt;/a&gt;&#xA;To ease the thing up, I think you can resign from friction between spheres -- it would be negligible in comparison to the force with which magnetic moments will try to adjust.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2011-04-19T08:35:30.833" />
  <row Id="8865" PostTypeId="2" ParentId="8862" CreationDate="2011-04-19T09:17:52.747" Score="3" ViewCount="" Body="&lt;p&gt;Armstrong could not walk normally on the moon because the moon's gravity is much weaker than Earth's, and also because he was wearing an awkward, bulky, massive spacesuit.  However, although the moon's gravity is weaker, the moon still has gravity and that gravity is enough to influence the tides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take a stone and drop it on Earth, it accelerates towards the ground at about 10 m/s^2.  This means that it gets faster and faster as it falls, and that every second it picks up 10 m/s of speed.  So after 1 second it is going 10 m/s; after 2 seconds it is going 20 m/s; after 3 seconds it is going 30 m/s, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do the same on the moon, the rock falls in the same way, with the change that the rock now falls at 1.6 m/s^2 instead of 10 m/s^2.  It still gains the same amount of speed every second, but that speed is simply less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Armstrong was used to walking on earth, where he got pulled down strongly by gravity.  On the moon, he was pulled down by gravity, but more weakly, so his usual stride became a bouncing series of jumps.  He was still pulled down by gravity, though, and if you trace his trajectory on the moon, you get a parabola, just like you do on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity is caused by mass.  The entire mass of the Earth is pulling you down right now.  On the moon, the entire mass of the moon is pulling you down.  Since the moon is smaller and less dense, it has less mass, and so the gravitational force on you is smaller on the moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravity of an object isn't some holistic property of the entire object.  Gravity of the Earth isn't caused by Gaia.  Instead, every little bit of matter pulls on every other bit of matter by gravity, regardless of what celestial body that matter happens to be a part of.  In fact, the moon is pulling on you by gravity right now, but weakly so because it's far away.  The sun is pulling on you by gravity, but you don't notice because you're in free fall, and everything else around you is, too.  Even the different parts of your own body pull on each other by gravity.  (This pull is too weak to notice, but it's theoretically there.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the moon has gravity, and that gravity is enough to affect the tides on Earth.  However, its gravity is weaker than Earth's, so Armstrong's gait was unusual.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-04-19T09:17:52.747" />
  <row Id="8866" PostTypeId="2" ParentId="8626" CreationDate="2011-04-19T09:19:20.687" Score="2" ViewCount="" Body="&lt;p&gt;I'll start with some terminological subtleties, that have to be accounted for, when it comes to &quot;conserved quantities&quot; or &quot;integrals of motion&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all it is important to state on which variables the quantities can depend. &#xA;In the field of &lt;a href=&quot;http://en.wikipedia.org/wiki/Conserved_quantity&quot; rel=&quot;nofollow&quot;&gt;differential equations&lt;/a&gt;, Hamiltonian dynamics and &lt;a href=&quot;http://en.wikipedia.org/wiki/Constant_of_motion&quot; rel=&quot;nofollow&quot;&gt;dynamical systems&lt;/a&gt;, a conserved quantity by definition does not explicitly depends on time -- it is usually a function, defined on a space (a manifold), that fully describes the state of your system (uses all the degrees of freedom as you have defined them). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to be taken into account  -- It seems to be possible to take a function of a constant of motion, therefore arriving at another quantity that is constant, therefore producing any amount of constants of motion. So there is an implicit condition for functional independence between those $N$ constants of motion. Which can be formulated as the vanishing of the Poisson brackets between any pair of the quantities. Here is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Completely_integrable#Hamiltonian_systems_and_Liouville_integrability&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt; about it on Wikipedia, also I've asked &lt;a href=&quot;http://physics.stackexchange.com/questions/5600/what-does-it-mean-when-one-says-that-system-has-n-constants-of-motion&quot;&gt;a question&lt;/a&gt; about it some time ago...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think that your implicit statement: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If a system has N degrees of freedom (DOF) and therefore N conserved independent quantities&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is not totally correct, if one takes into account what was said about the notion of &quot;conserved quantity&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Concerning your question about finding the symmetry transformation, that corresponds to the quantity -- I've addressed the issue &lt;a href=&quot;http://physics.stackexchange.com/questions/5792/meaning-of-the-poisson-bracket-as-a-coordinate-transformation/6085#6085&quot;&gt;in this question&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Briefly -- given a &quot;generator&quot; $\delta G$ and some quantity $A$. The small transformation, generated by $\delta G$ is: $$A \to A+\delta A,\quad\quad \delta A = -\{\delta G, A\}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting $A = H$, and noting that, if $\delta G$ is a constant of motion, then $\{\delta G, H\} = 0$. One immediately arrives at the conclusion that the transformation, generated by $\delta G$ is the symmetry transformation. (Some examples are &lt;a href=&quot;http://physics.stackexchange.com/questions/5792/meaning-of-the-poisson-bracket-as-a-coordinate-transformation/6085#6085&quot;&gt;here&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-04-19T09:19:20.687" CommentCount="5" />
  <row Id="8867" PostTypeId="2" ParentId="8862" CreationDate="2011-04-19T09:22:32.740" Score="0" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When there is a full moon, the Moon, Earth and Sun are almost in a straight line, combining the gravitational effect of the Moon and Sun on the Earth's tides to make tides slightly higher.  There is a similar effect at a new moon.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The gravitational force on the surface of the Moon is smaller than on the Earth.  This (combined with wearing a spacesuit) affected how the Apollo astronauts had to walk. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2714" LastActivityDate="2011-04-19T09:22:32.740" />
  <row Id="8868" PostTypeId="1" AcceptedAnswerId="8885" CreationDate="2011-04-19T09:57:09.703" Score="3" ViewCount="158" Body="&lt;p&gt;It is easy to observe that on a windy day, the wind does not blow for several hours at constant speed, then gradually subside.  Instead, on a time scale of seconds or tens of seconds, there are stronger gusts of wind followed by lulls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably this is an effect of turbulence.  If so, is this turbulence due to the complicated geometry around me - buildings, trees, hills, etc.?  If we removed these features and had wind blowing over a flat ocean surface or a flat plain for hundreds of miles, would we still observe the wind to blow in the same sorts of gusts, or should I expect a more steady flow?&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-04-19T10:20:28.050" LastActivityDate="2011-04-19T23:18:10.133" Title="Does the wind gust over the ocean?" Tags="&lt;fluid-dynamics&gt;&lt;turbulence&gt;" AnswerCount="4" />
  <row Id="8869" PostTypeId="2" ParentId="8868" CreationDate="2011-04-19T10:45:23.670" Score="2" ViewCount="" Body="&lt;p&gt;Of course there ar gusts of wind on oceans, a lot of ships were lost by such. &#xA;But Your question as such cannot be answered this simple. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best thing to say, is that in general wind on sea is more steady. &#xA;The zone of &quot;trade winds&quot; have such steady conditions most of the year, &#xA;the spanish named them &quot;el golfo de las damas&quot;, &quot;Ladys sea&quot;. &#xA;But approaching cyclons, taifunes, hurricans or pamperos have big gusts, &#xA;those were feared by sailship crews a lot. (You need a certain time &#xA;to get sails down (or up) on a big ship.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowledge of this global winds (and dependence to season) was the big &#xA;thing for the captains of sailships. The fact that reliable trade was possible &#xA;despite weather proves that wind on sea is steadier globally, but not free from &#xA;gusts. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-19T10:45:23.670" CommentCount="3" />
  <row Id="8870" PostTypeId="1" CreationDate="2011-04-19T11:19:17.427" Score="2" ViewCount="168" Body="&lt;p&gt;I was watching a documentary last night on the first planet discovered outside our solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first one apparently is a giant gas planet that orbits very close to its sun over a very fast period (~1 week per orbit), and it's so hot all metal elements vaporise and form parts of it's atmosphere (so the current theory states anyway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of curiosity, when they estimate the surface heat of the planet (they give big numbers), did they take into account reflection of heat from it's metallic atmosphere, and internal reflection from heat bouncing off the surface and it's atmosphere (like global warming) when estimating surface temperature of the planet?&lt;/p&gt;&#xA;" OwnerUserId="2720" LastActivityDate="2011-04-19T14:43:19.247" Title="Heat reflection on distant planets" Tags="&lt;space&gt;&lt;heat&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8871" PostTypeId="2" ParentId="8870" CreationDate="2011-04-19T12:04:42.753" Score="3" ViewCount="" Body="&lt;p&gt;The temperature estimation procedure basically is to equate the star's incoming radiation to the planet's outgoing radiation. Factors like star type and distance, planetary albedo and greenhouse gases can be estimated. Here's a nice site with the necessary formulas, &lt;a href=&quot;http://www.dangermouse.net/gurps/science/temps.html&quot; rel=&quot;nofollow&quot;&gt;http://www.dangermouse.net/gurps/science/temps.html&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="1924" LastEditorDisplayName="" LastEditDate="2011-04-19T12:56:50.250" LastActivityDate="2011-04-19T12:56:50.250" />
  <row Id="8872" PostTypeId="2" ParentId="7129" CreationDate="2011-04-19T12:19:41.113" Score="1" ViewCount="" Body="&lt;p&gt;It has to be remembered there are two types of phonons.  Phonons with transverse modes are optical phonons.  These possess oscillatory motion of charged ions in a lattice, where that motion is perpendicular to the wave motion direction along the lattice.  These are phonons responsible for Raman scattering and related physics.  The other type of phonon has longitudinal modes which are parallel to the direction of motion.  These are acoustical phonons, which for large wavelength are sound waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two types of phonons are distinct in a number of ways.  Optical phonons occur with a lattice structure.  This tends to be because amorphous solids do not often have a conduction band of electrons in a medium of ions.  Ions in some array are necessary for the electric field oscillation in an optical phonon.  Disordered solids are most often not of this nature.  In fact it is one reason light can pass through them, such as glass, for a photon does not couple to a lattice of ions.  Acoustical phonons can exist in disordered solids, and these are responsible for sound waves.  Acoustical phonons of a very short wave length are attenuated in disordered solids.  Compare trying to drive a car down New York streets or any city laid out on a grid when compared to driving in Boston.  However, long wavelength acoustical phonons that are 10 or more times the average atom spacing can pass through a disordered solid.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-04-19T12:19:41.113" CommentCount="6" />
  <row Id="8873" PostTypeId="2" ParentId="8850" CreationDate="2011-04-19T13:05:38.033" Score="6" ViewCount="" Body="&lt;p&gt;Spacetime can dynamically evolve in a way which apparently violates special relativity.  A good example is how galaxies move out with a velocity v = Hd, the Hubble rule, where v = c = Hr_h at the de Sitter horizon (approximately) and the red shift is z = 1.  For z &gt; 1 galaxies are frame dragged outwards at a speed greater than light.  Similarly an observer entering a black hole passes through the horizon and proceeds inwards at v &gt; c by the frame dragging by radial Killing vectors.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Alcubierre warp drive is a little spacetime gadget which compresses distances between points of space in a region ahead of the direction of motion and correspondingly expands the distance between points in a leeward region.  If distances between points in a forwards region are compressed by a factor of 10 this serves as a “warp factor” which as I remember is $w~=~1~+~ln(c)$, so a compression of 10 is a warp factor 3.3.  The effect of this compression is to reduce the effective distance traveled on a frame which is commoved with the so called warp bubble.  This compression of space is given by $g_{tt}$ $=~1~-~vf(r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course as it turns out this requires exotic matter with $T^{00}~&amp;lt;~0$, which makes it problematic.  Universe is also a sort of warp drive, but this is not due to a violation of the weak energy condition $T^{00}~\ge~0$.  Inflationary pressure is due to positive energy.  The gravity field is due to the quantum vacuum, and this defines an effective stress-energy tensor $T^{ab}$ with components $T^{00}~=~const*\rho$, for $\rho$ energy density, and $T^{ij}~=~const*pu^iu^j$, for $i$ and $j$ running over spatial coordinates $u^i$ velocity and $p$ pressure density.  For the de Sitter spacetime the energy density and pressure satisfies a state $p~=~w*\rho$ where $w~=~-1$.  So the pressure in effect is what is stretching out space and frame dragging galaxies with it.  There is no need for a negative energy density or exotic matter.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Negative energy density or negative mass fields have serious pathologies.  Principally since they are due to quantum mechanics the negative eigen-energy states have no lower bound.  This then means the vacuum for these fields is unstable and would descend to ever lower energy levels and produce a vast amount of quanta or radiation.  I don’t believe this happens.  The Alcubierre warp drive then has a serious departure between local laws of physics and global ones, which is not apparent in the universe or de Sitter spacetime.  The Alcubierre warp drive is then important as a gadget, along with wormholes as related things, to understand how nature prevents closed timelike curves and related processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question was asked about the redshift factor and the cosmological horizon.  This requires a bit more than a comment post.  On a stationary coordinate region of the de Sitter spacetime $g_{tt}~=~1~-~\Lambda r^2/3$.  This metric term is zero for $r~=~\sqrt{3/\Lambda}$, which is the distance to the cosmological horizon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red shift factor can be considered as the expansion of a local volume of space, where photons that enter and leave this “box” can be thought of as a standing wave of photons.  The expansion factor is then given by the scale factor for the expansion of the box&#xA;$$&#xD;&#xA;z~=~\frac{a(t_0)}{a(t)}~-~1&#xD;&#xA;$$&#xA;The dynamics for the scale factor is given by the FLRW metric&#xA;$$&#xD;&#xA;\Big(\frac{\dot a}{a}\Big)^2~=~\frac{8\pi G\rho}{3}&#xD;&#xA;$$&#xA;for $k~=~0$.  The left hand side is the Hubble factor, which is constant in space but not time.  Writing the $\Lambda g_{ab}~=~8\pi GT_{ab}$ as a vacuum energy and $\rho~=~T_{00}$ we get &#xA;$$&#xD;&#xA;\Big(\frac{\dot a}{a}\Big)^2~=~H^2~=~\frac{\Lambda}{3}&#xD;&#xA;$$&#xA;the evolution of the scale factor with time is then &#xA;$$&#xD;&#xA;a(t)~=~\sqrt{3/\Lambda}e^{\sqrt{\Lambda /3}t}.&#xD;&#xA;$$&#xA;Hence the ratio is $a(t)/a(t_0)~=~ e^{\sqrt{\Lambda /3}(t-t_0)}$.The expansion is this exponential function, which is Taylor expanded to give to first order the ratio above&#xA;$$&#xD;&#xA;a(t)/a(t_0)~\simeq~1~+~H(t_0)(t_0-t)~=~1~+~H(t_0)(d-d_0)/c&#xD;&#xA;$$&#xA;which gives the Hubble rule.  $z~=~a(t)/a(t_0)~-~1$.  It is clear that from the general expression that $a(t)$ can grow to an arbitrarily large value, and so can $z$.  On the cosmological horizon for $d~-~d_0~=~r_h~=~\sqrt{3/\Lambda}$ we have $z~=~1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking beyond the cosmological horizon $r_h~\simeq~10^{10}$ly is similar to an observer in a black hole looking outside to the exterior world outside the black hole horizon.  People get confused into thinking the cosmological horizon is a black membrane similar to that on a black hole.  Anything which we do observe beyond the horizon we can never send a signal to, just as a person in a black hole can see the exterior world and can never send a message out.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-04-19T15:40:25.043" LastActivityDate="2011-04-19T15:40:25.043" CommentCount="1" />
  <row Id="8875" PostTypeId="1" CreationDate="2011-04-19T13:15:56.780" Score="3" ViewCount="242" Body="&lt;p&gt;I'm freshman of Math major, because we don't have any physics course. But I'm very interested in Physics. But I'm not very sure which is better or say, more suitable for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.Sears and Zemansky's University Physics (10th Edition)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.Berkeley Physics Course Series (Vol 1, 2, 3, 4, 5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because I have talked to some of my friends who are in the major of physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some said that Sears physics is more oriented to High school level, with lots of pics, examples, hence to recommand me the Berkeley series which is more 'academic'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some others said that the Berkeley Series is too beyond for me, because I'm in major of maths, with very limited foundation of physics.&lt;/p&gt;&#xA;" OwnerUserId="3189" LastEditorUserId="3189" LastEditorDisplayName="" LastEditDate="2011-04-19T15:48:21.240" LastActivityDate="2011-04-28T23:42:07.593" ClosedDate="2011-04-19T15:32:42.523" Title="Which is better intro physics for math major student ? Sears Physics and Berkeley Physics Series" Tags="&lt;soft-question&gt;&lt;books&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="8876" PostTypeId="2" ParentId="8862" CreationDate="2011-04-19T13:34:01.157" Score="7" ViewCount="" Body="&lt;p&gt;During full and new moons the gravitation of the moon and sun are combined creating what is called a spring tide (no relation to the season).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Completely unrelated is the issue of walking on the Moon.  The Moon's gravity is only 1/6th that of Earth, so the force you would normally use to, say, walk across the ground on Earth would be six times too much.  They found the best way to move on the Moon was sort of skipping or a long, jumping sort of run.&lt;/p&gt;&#xA;" OwnerUserId="3190" LastActivityDate="2011-04-19T13:34:01.157" CommentCount="1" />
  <row Id="8877" PostTypeId="2" ParentId="8870" CreationDate="2011-04-19T14:43:19.247" Score="2" ViewCount="" Body="&lt;p&gt;Tom,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if that planet was classified a &quot;gas giant&quot; then its atmosphere is not &#xA;&quot;metallic&quot; in the sense of reflection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1st be aware that astronomers call all elements heavier than helium &quot;metals&quot;. &#xA;(People with bad marks in chemistry become astronomers) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And 2nd this might result in a glowing atmosphere with some % of those &quot;metals&quot;. &#xA;Similar to our sun that &quot;metals&quot; will cause some Fraunhofer lines, thats all.&#xA;A reflection from &quot;surface&quot; (I assume You mean the surface of the solid core) &#xA;is absolutely impossible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More would need to read that paper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Last but not least astronomers are bad chemists, but when it comes to calculate &#xA;reflection or albedo or emission of celectial bodys, they know what to do :=)&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-19T14:43:19.247" />
  <row Id="8878" PostTypeId="2" ParentId="8875" CreationDate="2011-04-19T15:25:59.520" Score="2" ViewCount="" Body="&lt;p&gt;I'd suggest you instead consider Feynman Lectures on Physics. Also, note that Stanford have many freely-available lecture course videos online (their website, as well as their youtube channel.) All the best :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: In keeping with the comments posted to this answer, I felt I could add something by elaborating - To begin with, I agree with @Ted Bunn that the FLoP are best suited to give the reader a deeper insight, provided they have &quot;the basics&quot; covered beforehand. Whilst this may be true in general I feel that for graduate mathematicians, developing a new-found interest in physics, it is certainly worth consideration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, and to expand my answer, I'd like to suggests an alternative option: 'The Road to Reality' by Professor Sir Roger Penrose could well be exacly what you're looking for. For starters, he began his academic career as a mathematical physicist. Therefore, it is by his very nature that he presents concepts and also physical intuition from a mathematical standpoint. Secondly, Professor Penrose is undoubtably a deservingly reputable physicist and his text book has served many-a-PhD-student well in gaining a deeper understanding of physical concepts; together with maintaining a firm mathematical foundation. Indeed it is true to say that many (PhD students) see it as a physicists 'bible' for reviewing basic physics that they may have misunderstood or merely overlooked, during their undergraduate study.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="1696" LastEditorDisplayName="" LastEditDate="2011-04-28T23:42:07.593" LastActivityDate="2011-04-28T23:42:07.593" CommentCount="9" />
  <row Id="8879" PostTypeId="1" AcceptedAnswerId="8881" CreationDate="2011-04-19T16:16:43.167" Score="1" ViewCount="111" Body="&lt;p&gt;In many articles I read the term &quot;crossover phenomena&quot; and a lot of methodology discussed according to it, with little or no description about what is meant by it. Sometimes there is a connection to networks, matrices, and then to other structures of elements. Can someone describe the concept of the phenomena?&lt;/p&gt;&#xA;" OwnerUserId="813" LastActivityDate="2011-04-19T17:12:07.617" Title="what is meant by &quot;crossover phenomena&quot;?" Tags="&lt;theory&gt;&lt;definition&gt;&lt;graph-theory&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8880" PostTypeId="2" ParentId="8879" CreationDate="2011-04-19T16:35:41.270" Score="2" ViewCount="" Body="&lt;p&gt;It is fashionable jargon, suitable for Humpty-Dumpty: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When I use a word, it means just what&#xA;  I choose it to mean — neither more nor&#xA;  less.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/cond-mat/9810046&quot; rel=&quot;nofollow&quot;&gt;Gluzman and Yukalov&lt;/a&gt; describe it as being like a function in an interval where, although the behaviour of the function is qualitatively different in the vicinity of the different boundaries of the interval, the function is continuous and there is no identifiable phase transition point.&lt;/p&gt;&#xA;" OwnerUserId="2714" LastActivityDate="2011-04-19T16:35:41.270" />
  <row Id="8881" PostTypeId="2" ParentId="8879" CreationDate="2011-04-19T17:12:07.617" Score="2" ViewCount="" Body="&lt;p&gt;&quot;Crossover&quot; is a general term describing a situation when a system can go from being in one phase to being in another phase as a certain parameter is changed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best example of a &quot;crossover&quot; phenomenon I can think of is the BEC-BCS crossover which occurs in cold Fermi gases. There one can tune the scattering length $a$ for pairwise interactions to go from being positive to negative. For $a &amp;lt; 0$ bound states can form and we get a BCS state (Bardeen-Cooper-Schrieffer) due to formation of Cooper pairs. As $a$ is tuned to go from being negative to positive the system &quot;crosses over&quot; from being a BCS to a BEC (Bose-Einstein Condensate).&lt;/p&gt;&#xA;" LastActivityDate="2011-04-19T17:12:07.617" CommentCount="1" />
  <row Id="8882" PostTypeId="1" AcceptedAnswerId="8890" CreationDate="2011-04-19T18:08:03.940" Score="5" ViewCount="127" Body="&lt;p&gt;In Landau course, vol.1 Mechanics, one finds the statement: &quot;...the absoluteness of time necessarily implies that the ordinary law of composition of velocities is applicable to all phenomena.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see this implication clearly. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3064" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-04-20T03:13:19.933" LastActivityDate="2011-04-20T03:13:19.933" Title="Why absoluteness of time implies galilean transformations?" Tags="&lt;classical-mechanics&gt;&lt;special-relativity&gt;&lt;relativity&gt;&lt;relative-motion&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="8883" PostTypeId="2" ParentId="8868" CreationDate="2011-04-19T18:11:56.460" Score="1" ViewCount="" Body="&lt;p&gt;Strangely enough Nasa thinks that winds gust stronger &lt;a href=&quot;http://earthobservatory.nasa.gov/IOTD/view.php?id=8916&quot; rel=&quot;nofollow&quot;&gt;over the oceans&lt;/a&gt; than over land.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Wind tends to blow stronger over the ocean than over land. The ocean presents a smooth surface over which wind can glide without interruption, while hills, mountains, and forests tend to slow or channel wind over land.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This map is &lt;a href=&quot;http://www.worldchanging.com/archives/002770.html&quot; rel=&quot;nofollow&quot;&gt;interesting.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have watched gusts sweeping over calm seas , where one can predict the arrival of wind from the observed disturbance in the distance.  So both land and ocean have changing winds, but which is stronger in producing gusts, as defined rigorously, is not obvious.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-19T18:11:56.460" CommentCount="1" />
  <row Id="8884" PostTypeId="2" ParentId="8827" CreationDate="2011-04-19T18:20:49.353" Score="5" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://groups.csail.mit.edu/mac/users/wisdom/&quot; rel=&quot;nofollow&quot;&gt;Jack Wisdom&lt;/a&gt; at MIT has extensively studied the question of the stability of the solar system.  He has a list of papers with links to freely-readable PDF files on his website:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://groups.csail.mit.edu/mac/users/wisdom/&quot; rel=&quot;nofollow&quot;&gt;http://groups.csail.mit.edu/mac/users/wisdom/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A good starting point might be &quot;&lt;a href=&quot;http://groups.csail.mit.edu/mac/users/wisdom/measurements.pdf&quot; rel=&quot;nofollow&quot;&gt;Is the Solar System Stable? and Can We Use Chaos to Make Measurements?&lt;/a&gt;&quot; (PDF) (in &lt;em&gt;Chaos&lt;/em&gt;, proceedings of the ''Joint Soviet-American Chaos Conference'' held at Woods Hole, June, 1989).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The abstract begins:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This talk addresses two separate questions: &quot;Is the solar system stable?&quot; and &quot;Can we use chaos to make better measurements?&quot;  In the first part, a review is presented of the numerical experiments which indicate that the motion of Pluto, and indeed the whole solar system, is chaotic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-04-19T18:20:49.353" CommentCount="1" />
  <row Id="8885" PostTypeId="2" ParentId="8868" CreationDate="2011-04-19T18:37:54.470" Score="1" ViewCount="" Body="&lt;p&gt;I believe you're exactly right: it's the complexity of hills, buildings, trees, asphalt, water, etc that make surface winds complicated.  As you go higher in the atmosphere, these surface effects disappear and the winds become much more steady.  You can see this in the winds aloft forecasts issued by the FAA for use in aviation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://aviationweather.gov/adds/winds/&quot; rel=&quot;nofollow&quot;&gt;http://aviationweather.gov/adds/winds/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the upper left there is a drop-down box allowing you to select altitudes from the surface (SFC) to 48,000 feet (FL480).  As you go up in altitude, the relatively chaotic surface winds blend into a much smoother (and faster!) flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume the effect is similar at sea, given the simpler boundary conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I admit this does not directly address your question of, &quot;how variable are the winds at sea&quot;? I too am curious to hear an authoritive answer.)&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-04-19T18:46:12.950" LastActivityDate="2011-04-19T18:46:12.950" />
  <row Id="8886" PostTypeId="1" CreationDate="2011-04-19T18:57:58.367" Score="4" ViewCount="154" Body="&lt;p&gt;This is a &lt;em&gt;very&lt;/em&gt; basic question about the concepts of force and acceleration. Consider the following situations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) I sit on the floor of my room on earth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) I sit on the floor of a spaceship that accelerates with 1g (in a direction perpendicular to its &quot;floor&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both situations I feel the same (a force that presses me towards the floor). Both situations involve an acceleration of 1g. BUT: in situation (2) my velocity changes, while in (1) it does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: what is the essential difference between these situations?&lt;/p&gt;&#xA;" OwnerUserId="3193" LastActivityDate="2011-04-20T02:22:11.810" Title="Basic question about force and acceleration" Tags="&lt;newtonian-mechanics&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="8887" PostTypeId="2" ParentId="8886" CreationDate="2011-04-19T19:11:10.197" Score="3" ViewCount="" Body="&lt;p&gt;General relativity rests on the fact that there is no &lt;em&gt;essential difference&lt;/em&gt;. This means: It is absolutely impossible for you to design an experiment that distinguishes between (1) and (2). You can use this as the foundation of GR and do some cool &lt;em&gt;Gedankenexperimente&lt;/em&gt;, such as this: If in your spaceship you shine a laser parallel to the floor, you will, due to the acceleration, observe that it gets bent (very slightly). On earth, you will observe the same, which shows that photons are affected by gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, in your accelerated framework, light experiences a frequency shift due to the relativistic Doppler effect. On earth, you will observe the same, but this time it's because photons lose energy in earth's gravitational field and since $E = h\nu$, so $\nu$ gets smaller.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditorDisplayName="" LastEditDate="2011-04-19T20:38:46.660" LastActivityDate="2011-04-19T20:38:46.660" CommentCount="2" />
  <row Id="8888" PostTypeId="2" ParentId="8886" CreationDate="2011-04-19T19:16:43.347" Score="0" ViewCount="" Body="&lt;p&gt;From what I know the difference lays in the fact that a body in a spaceship will be in a homogenous force field, while the proximity of a mass (e.g. Earth) will cause a slight deformation of the body due to the changes of gravity force throught the volume of the body.&lt;/p&gt;&#xA;" OwnerUserId="2752" LastActivityDate="2011-04-19T19:16:43.347" />
  <row Id="8890" PostTypeId="2" ParentId="8882" CreationDate="2011-04-19T20:01:32.653" Score="5" ViewCount="" Body="&lt;p&gt;We have to be careful in stating exactly what we're going to allow ourselves to assume here. We need some sort of principle of relativity -- that the laws are the same for both observers. But we don't want to assume anything else a priori, right? For instance, we don't want to assume at first that rulers have the same length for both observers -- we need to prove that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's work in one dimension for simplicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that observer B is moving at constant velocity v relative to observer A. Suppose some object is moving along with some speed $u_B$ as measured by B. We want to show that the speed as measured by observer A is $u_A=u_B+v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the position of the object at two different times, separated by a small amount $dt$. Since time is absolute (all observers use the same $dt$), what we want to show is equivalent to&#xA;$$&#xD;&#xA;dx_A = dx_B+v\,dt&#xD;&#xA;$$&#xA;(multiplying the original equation through by $dt$). Here $dx_A$ means $x_A(t+dt)-x_A(t)$, that is, the change in the position of the object at the two times, as measured by A, and similarly for B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a useful fact: &lt;em&gt;If both observers measure the distance between two points at an instant of time $t$, they must get the same answer.&lt;/em&gt; The reason is symmetry. If the two disagreed, then one would have to get a bigger answer than the other. But for a measurement of this sort, there's nothing to break the symmetry between A and B -- that is, we can just change the sign of $v$, and consider B to be stationary and A to be moving, and that shouldn't affect the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that's enough to get us there. Suppose that observer B sets of a firecracker at his location at time $t$, and another at time $t+dt$. The two observers must agree on the distance from B to the object at the time the first firecracker went off, and they must agree on the distance from B to the observer at the time the second firecracker went off. The difference between these these two numbers is $dx_B$. But the difference between these two numbers is also $dx_A-v\,dt$, since observer A knows that observer B traveled a distance $v\,dt$ during that time interval. The conclusion follows.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-19T20:01:32.653" CommentCount="1" />
  <row Id="8891" PostTypeId="1" AcceptedAnswerId="8893" CreationDate="2011-04-19T18:33:00.410" Score="2" ViewCount="318" Body="&lt;p&gt;I know what centrifugal force is, but how does it work? Why are things forced to the outside?&lt;/p&gt;&#xA;" OwnerUserId="3196" LastEditorUserId="2911" LastEditorDisplayName="" LastEditDate="2011-04-19T20:24:36.510" LastActivityDate="2011-04-20T02:02:23.003" Title="How does centrifugal force work?" Tags="&lt;forces&gt;&lt;centrifugal-force&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" />
  <row Id="8892" PostTypeId="2" ParentId="8891" CreationDate="2011-04-19T18:39:28.387" Score="2" ViewCount="" Body="&lt;p&gt;Because things want to go in a straight line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine swinging an object around your head on a string. At any moment the object wants to go straight ahead (ie on a tangent) but it can't because of the string. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's as if the string was pulling it back in toward you - that's balanced by a force pushing it outward = centrifugal force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more technical explanation try physics.stackexchange.com, but expect an arguement about whether centrifugal force exists!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/l0uzq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" LastActivityDate="2011-04-19T18:55:06.360" CommentCount="2" />
  <row Id="8893" PostTypeId="2" ParentId="8891" CreationDate="2011-04-19T18:59:35.853" Score="6" ViewCount="" Body="&lt;p&gt;The real force at work is &lt;em&gt;centripetal&lt;/em&gt; force, or a force &lt;em&gt;pushing inwards&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you have a bucket on a string, and you swing that around in a circle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/X0bYe.png&quot; alt=&quot;Terrible drawing&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you swing the bucket, it travels in a circle. The red line shows the path the bucket takes. In order to make it swing like this, you have to apply a constant force on the rope -- this is the green arrow in the image. At any given moment in time, the bucket &lt;em&gt;wants&lt;/em&gt; to travel &lt;em&gt;straight&lt;/em&gt; -- the blue line in the diagram. By applying the &lt;em&gt;centripetal force&lt;/em&gt;, the &lt;em&gt;inward&lt;/em&gt; force, you change the motion from straight to the circular motion (the red line).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the contents of the bucket always want to go straight, and the force you apply always make them change direction, there &lt;em&gt;seems&lt;/em&gt; to be an &quot;outward&quot; or &quot;centrifugal&quot; force &quot;pushing&quot; the contents against the side of the bucket. But it's an an illusion -- it's really just the momentum of the bucket and it's contents.&lt;/p&gt;&#xA;" OwnerUserId="2098" LastEditorUserId="2098" LastEditorDisplayName="" LastEditDate="2011-04-19T22:43:57.953" LastActivityDate="2011-04-19T22:43:57.953" CommentCount="9" />
  <row Id="8894" PostTypeId="2" ParentId="8891" CreationDate="2011-04-19T20:35:43.407" Score="-1" ViewCount="" Body="&lt;p&gt;simply said centrifugal force is the force experienced by a stationary object placed in a rotating frame of reference. You might have noticed it when travelling in a bus that makes a sharp turn. If it turns to the left you are pushed to the right.&lt;/p&gt;&#xA;" OwnerUserId="3195" LastEditorUserId="3195" LastEditorDisplayName="" LastEditDate="2011-04-19T20:41:23.387" LastActivityDate="2011-04-19T20:41:23.387" />
  <row Id="8895" PostTypeId="1" CreationDate="2011-04-19T20:47:24.660" Score="7" ViewCount="400" Body="&lt;p&gt;This is not the same as: &lt;a href=&quot;http://physics.stackexchange.com/questions/4118/how-many-bytes-can-the-observable-universe-store&quot;&gt;How many bytes can the observable universe store?&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Bekenstein_bound&quot; rel=&quot;nofollow&quot;&gt;Bekenstein bound&lt;/a&gt; tells us how many bits of data can be stored in a space. Using this value, we can determine the number of unique states this space can be in.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine now, we are to simulate this space by enumerating each state along with which states it can transition to with a probability for each transition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much information is needed to encode the number of legal transitions and the probabilities? Does this question even make sense? If so, is there any indication that any of these probabilities are or are not &lt;a href=&quot;http://en.wikipedia.org/wiki/Computable_number&quot; rel=&quot;nofollow&quot;&gt;Computable numbers&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;br&gt;&#xA;Here's a thought experiment.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Select your piece of space and start recording all the different states you see. &lt;/li&gt;&#xA;&lt;li&gt;If the Bekenstein bound tells us we can store n bits in our space, wait until you see 2^n different states. Now we've seen all the states our space can be in (otherwise we can violate the Bekenstein bound).   &lt;/li&gt;&#xA;&lt;li&gt;For any state, record any other state that the space can legally transition to without violating any physical laws.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To simulate this portion of space, take it's state and transition it to a legal state. Repeat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have only used a finite number of bits and we have modeled a section of space.&lt;/p&gt;&#xA;" OwnerUserId="3194" LastEditorUserId="3194" LastEditorDisplayName="" LastEditDate="2011-05-26T22:00:47.367" LastActivityDate="2011-05-28T17:35:15.970" Title="How many bits are needed to simulate the universe?" Tags="&lt;cosmology&gt;&lt;quantum-information&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="4" />
  <row Id="8896" PostTypeId="2" ParentId="8868" CreationDate="2011-04-19T23:18:10.133" Score="1" ViewCount="" Body="&lt;p&gt;anecdotally from my sailing experience around a coast, and the conventional sailing knowledge:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A wind that has travelled over the sea to get there is more even that one that has come off off the land.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. The statement above: &quot;Wind tends to blow stronger over the ocean than over land&quot;&#xA; does not imply that it is more gusty&lt;/p&gt;&#xA;" OwnerUserId="3198" LastActivityDate="2011-04-19T23:18:10.133" />
  <row Id="8897" PostTypeId="2" ParentId="8074" CreationDate="2011-04-20T00:05:49.160" Score="8" ViewCount="" Body="&lt;p&gt;Here we would like to calculated analytically Lubos Motl's solution to the first order in the flatness parameter $f$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0&amp;lt;f:=1-\frac{b}{a}\approx\frac{a}{b}-1 \ll 1, $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a$ and $b$ are the equatorial and polar radius of the Earth, respectively, and $a&amp;gt;b$. (The $\approx$ symbol will from now on mean equality up to higher-order terms in $f$.) We assume that Earth is a massive oblate ellipsoid &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left(\frac{x}{a}\right)^2 + \left(\frac{y}{a}\right)^2 +\left(\frac{z}{b}\right)^2 \leq 1  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with uniform density $\rho$ and volume $V=\frac{4}{3}\pi a^2 b\approx\frac{4}{3}\pi b^3(1+2f)$. The eccentricity is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e=\sqrt{(2-f)f}\approx \sqrt{2f}\ll 1.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will assume that gravity is Newtonian. What we need to calculate is the quadrupole moment contribution to the gravitational potential&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ U_4 = \frac{x^i Q_{ij}x^j}{r^5}, $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For symmetry reasons, one of the principal directions of the quadrupole moment $Q_{ij}=Q_{ji}$ must be along the polar $z$-axis, and the two other principal directions must have same eigenvalues and be in the equatorial $xy$ plane. Thus $U_4$ must be of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ U_4 = \frac{Q_a s^2+Q_b c^2}{r^3}, $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $Q_a$ and $Q_b$ are the equatorial and polar eigenvalues, respectively; where we have introduced the short-hand notation $s\equiv\sin(\theta)$ and $c\equiv\cos(\theta)$; and where $\theta\in[0,\pi]$ is the polar angle. Since the quadrupole moment $Q_{ij}$ cannot contribute to Gauss' law, we must demand that the Laplacian $\nabla^2 U_4 = 0$ vanishes, which leads to that &lt;em&gt;the polar eigenvalue is minus two times the equatorial eigenvalue,&lt;/em&gt; $Q_b=-2Q_a$. In other words,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ U_4 \sim \frac{3c^2-1}{r^3}=\frac{2-3s^2}{r^3}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus it is enough to calculate the gravitational potential at a point $P=(0,0,r)$ on the polar $z$-axis, where we have azimuthal $U(1)$ symmetry $\varphi\to\varphi+\varphi_{0}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us calculate the contribution to the gravitational potential energy from a ring parallel to the $xy$ plane and with center $(0,0,z)$ on the $z$-axis. Because of the &#xA;$U(1)$ azimuthal symmetry, we may focus on a point on the ring with $y=0$ and $x\geq 0$,&#xA;and which lie in the $xz$ plane. Let this point have 2D polar coordinates $(R,\theta)$. &#xA;In other words, the point has 3D coordinates $(x,y,z)=(Rs,0,Rc)$. From the equation of an ellipse&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left(\frac{x}{a}\right)^2 +\left(\frac{z}{b}\right)^2 = 1,  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we obtain, after some elementary algebra,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ R \approx b (1 + fs^2) \quad \left\{\begin{array}{rcl}= b &amp;amp;\mathrm{for}&amp;amp; \theta=0, \\ \approx a &amp;amp;\mathrm{for}&amp;amp; \theta=\frac{\pi}{2} . \end{array}\right.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, the distance $\ell$ from the ring to the point $P$ is given by the cosine relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ell^2 = r^2 +R^2 - 2rRc \approx  r^2 +b^2(1 + 2fs^2) - 2rb(1 + fs^2)c$$&#xA;$$=(r^2 +b^2 -2rbc) +2fbs^2(b-rc).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;outer&quot; surface area of the ring is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ dA = 2\pi Rs \cdot R d\theta \approx 2\pi b^2(1 + 2fs^2) (-dc), $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;with infinitesimal &quot;radial&quot; thickness&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ dR \approx (1 + fs^2) db.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the volume of the ring is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d^2V = dA \ dR \approx 2\pi b^2db(1 + 3fs^2) (-dc). $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Eventually we want to integrate over the polar angle $\theta$ from $0$ to $\pi$. This corresponds to integrate over $c\equiv\cos(\theta)$ from $1$ to $-1$ in the negative direction. Therefore $dc&amp;lt;0$ is negative.) The contribution from the ring to the gravitational potential energy $d^2 U$ at the point $P$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ d^2 U = -G \frac{d^2 M}{\ell} =  - G\rho \frac{d^2 V}{\ell} &#xD;&#xA;\approx G\rho\frac{2\pi b^2db(1 + 3fs^2) dc}{\sqrt{(r^2 +b^2 -2rbc) +2fbs^2(b-rc)}} $$&#xA;$$\approx G\rho2\pi b^2db\ dc\left(\frac{1 + 3f(1-c^2)}{\sqrt{r^2 +b^2 -2rbc}}- \frac{1}{2}&#xD;&#xA;\frac{2fb(1-c^2)(b-rc)}{(r^2 +b^2 -2rbc)^{\frac{3}{2}}}\right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integration over cosine $c$ of the polar angle $\theta\in[0,\pi]$ to form a thin shell yields (with the help of the MAPLE program)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ dU = \int_{c=1}^{c=-1} d^2 U&#xD;&#xA;\approx -G\rho\frac{4\pi b^2db}{r}\left( (1+3f) -f( 1+ \frac{2b^2}{5r^2}) - \frac{4fb^2}{15r^2}\right) $$&#xA;$$ = -G\rho\frac{4\pi b^2db}{r}\left( (1+2f) - \frac{2fb^2}{3r^2}\right).  $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next we integrate over $b$ to get the polar potential of a massive ellipsoid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ U = \int^{b} dU \approx -G\rho\frac{4\pi b^3}{3r}\left( (1+2f) - \frac{2fb^2}{5r^2}\right) =U_1+U_{4b}.$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term $U_1$ is, of course, precisely the monopole potential&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U_1(r)=-\frac{GM}{r},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the second term $U_{4b}$ is the polar quadrupole  potential&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U_{4b}\approx-\frac{2fb^2}{5r^2}U_1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we know that the gravitational quadrupole potential in an arbitrary point (not necessarily on the $z$-axis) is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U_{4}(r)\approx-\frac{(2-3s^2)fR^2}{5r^2}U_1(r),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the full gravitational potential is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U(r) \approx U_1(r)+U_4(r) \approx -\frac{GM}{r}\left(1 - \frac{(2-3s^2)fR^2}{5r^2}\right). $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From now on, let us only consider points $P$ with $r=R\approx b (1 + fs^2)$ on the surface of the ellipsoid-shaped Earth. Then the monopole potential in $P$ becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ U_1(R) \approx -(1- fs^2)\frac{GM}{b}, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while the quadrupole potential in $P$ becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ U_4(R) \approx \frac{f(2-3s^2)}{5}\frac{GM}{b} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that the full gravitational potential in $P$ becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ U(R) \approx U_1(R) + U_4(R) \approx -(1-\frac{2f(1+s^2)}{5})\frac{GM}{b}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us use the North pole as a reference point, i.e., subtract the gravitational potential $U_b(b)$ at the North pole. Then the gravitational potential in $P$ becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ U(R)- U_b(b) \approx \frac{2fs^2}{5} \frac{GM}{b}\approx \frac{2}{5}g hs^2, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $h:=a-b=fa \approx fb &amp;gt;0$ is the difference in the equatorial and polar radii, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g= |\vec{\nabla} U(R)|  = \left. \frac{\partial U(r)}{\partial r}\right|_{r=R}+{\cal O}(f) = \frac{GM}{R^2}+{\cal O}(f).$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We should now add the centrifugal potential &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ V = -\frac{(\omega Rs)^2}{2}. $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the order in $f\ll 1$ that we are working, we see that the total potential $U(R)+V$ is constant (independent of the surface point $P$), if &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{2}{5}gh=\frac{(\omega R)^2}{2}. $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we find a factor $\frac{2}{5}$ in difference from Mark Eichenlaub's original monopole argument. &lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-04-22T19:07:58.300" LastActivityDate="2011-04-22T19:07:58.300" CommunityOwnedDate="2011-04-21T14:01:16.563" />
  <row Id="8898" PostTypeId="2" ParentId="8895" CreationDate="2011-04-20T00:31:46.387" Score="3" ViewCount="" Body="&lt;p&gt;There's a huge difference between the number of bits you can store in a given space and the number of bits you need to describe that space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a single atom of iron with its 26 electrons. For a complete description, you need the many-particle wavefunction $\psi(\vec{x}_1, \vec{x}_2, \vec{x}_3, \dots, \vec{x}_{26})$ (ignoring spin for the moment). Imagine you want to sample it in a given region of space with a very crude grid of 10 points for each direction, so you have $1000$ points in total. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means you need $1000^{26} = 10^{3\cdot 26} = 10^{78}$ numbers to store it. For decent precision, you want to use at least $16$ bits, so you end up with approximately $10^{79}$ to $10^{80}$ bits. This is more than (or of the same order as) there are atoms in the entire universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now taking it up from here, for a super-exact simulation of the universe you need the complete wavefunction of the entire universe, so replace the $26$ from the example above with something much higher, and of course you want it to be more precise, so replace the 1000 with something much higher, and then note that due to quantum field theory, the number of particles isn't even fixed, so a simple wavefunction isn't even enough... In a black body, for example, you can have an &lt;em&gt;infinite&lt;/em&gt; number of photons. Although the probability for this decays exponentially, you'd still have to include it in an exact simulation...&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-04-20T00:31:46.387" CommentCount="7" />
  <row Id="8899" PostTypeId="2" ParentId="8891" CreationDate="2011-04-20T00:54:41.940" Score="3" ViewCount="" Body="&lt;p&gt;Centrifugal force is a particular example of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Fictitious_force&quot; rel=&quot;nofollow&quot;&gt;fictitious force&lt;/a&gt;.  It is introduced so that Newton's second law holds in a rotating reference frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton's second law says&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = ma$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that whenever we find an object accelerating (speeding up, slowing down, turning, or some combination), we can look around and find a physical reason why this happens.  For example, a dropped stone accelerates towards the Earth, and this is due to Earth's gravity; if we drop the stone far from Earth, it won't fall.  Your car turns a corner.  This happens due to friction with the road.  If the road were perfectly slick, the car would simply slide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton's second law holds in an &lt;a href=&quot;http://en.wikipedia.org/wiki/Inertial_frame_of_reference&quot; rel=&quot;nofollow&quot;&gt;inertial reference frame&lt;/a&gt;.  It is simply a fact that such reference frames exist, and that they are all related to each other by moving past each other at constant velocities.  (This becomes more complicated in general relativity, but that is not a major concern in everyday situations.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, suppose you are in a train that begins accelerating forward (from the stationary track's point of view), and you are looking out the window at a ball sitting on the sidewalk nearby.  From your reference frame in the  train, the ball is accelerating backwards.  However, there is no obvious source of a force on the ball that would make it accelerate backwards.  This means that in an accelerating frame, Newton's second law doesn't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes we would still like to do physics in such an accelerating frame, so we simply invent a new force, called a fictitious force, and say that the ball has a fictitious force of just the right amount needed to give it the acceleration we observe.  Since the ball's acceleration is $a_b = -a_t$  with $a_t$ the acceleration of the train in an inertial frame, we need to introduce a fictitious force&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{fict} = -m a_t = m a_b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That way, Newton's laws still work and we can do physics as normal as long as the train's acceleration stays the same.  We could, for example, play billiards in the accelerating train, noticing the the balls have curved trajectories across the table, and these curved trajectories would be perfectly explained by a fictitious force $-ma_t$ acting on each ball (with $m$ changing for balls with different masses).  Keep in mind that the fictitious force points in the opposite direction of the train's acceleration.  If the train accelerates forward, the ball appears to accelerate backwards, so the fictitious force must point backwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another type of accelerating frame is a rotating reference frame, for example a carousel.  On the carousel, every part is accelerating towards the center (see &lt;a href=&quot;http://physics.stackexchange.com/questions/8891/how-does-centrifugal-force-work/8893#8893&quot;&gt;Josh's answer&lt;/a&gt;).  Therefore, to do physics in this frame, we must introduce a fictitious force&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = -m a_c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as before.  $a_c$ is the acceleration of the carousel at any point.  Because this acceleration points in towards the center of the carousel, the fictitious force points the opposite direction - out towards the edge.  This fictitious force is called the centrifugal force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Introducing the centrifugal force lets us do physics from the point of view of the rotating carousel, with the caveat that we can only handle statics this way.  If things are actually moving on the carousel, we need to include the Coriolis force, which pushes things sideways.  (See derivation &lt;a href=&quot;http://physics.stackexchange.com/questions/383/derivation-of-the-centrifugal-and-coriolis-force&quot;&gt; here &lt;/a&gt; or some discussion of it in my answers &lt;a href=&quot;http://physics.stackexchange.com/questions/7680/artificial-gravity-on-a-rotating-space-stations/7687#7687&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;http://physics.stackexchange.com/questions/1155/coriolis-force-and-newtons-third-law/1156#1156&quot;&gt;here&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for whether the centrifugal force is &quot;real&quot;, it depends on what that means.  In an inertial frame, each force can be traced back to some physical interaction like the exchange of a photon.  That's not true for the centrifugal force.  This is the essential difference people are referencing when they say the centrifugal force and other fictitious forces &quot;are not real&quot;.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-04-20T00:54:41.940" CommentCount="1" />
  <row Id="8900" PostTypeId="2" ParentId="8891" CreationDate="2011-04-20T02:02:23.003" Score="2" ViewCount="" Body="&lt;p&gt;Centrifugal force is not a real force in Newtonian mechanics. Whenever a frame of reference is accelerated (except the cases where the acceleration is due to gravity) w.r.t. an inertial frame of reference, an observer situating in that reference frame experiences a &quot;force&quot; in the opposite direction of the acceleration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an outside inertial observer this is nothing but the inertia of the objects situated in the accelerated frame. Since in Newtonian mechanics strictly the descriptions of only the inertial observers are legal hence the &quot;force&quot; experience by the objects in the accelerated frames is not a real force. It is called a pseudo force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rotating framed of reference is also an accelerated frame where the acceleration is directed towards the center. It is called the centripetal force. As always, as described above, a &quot;force&quot; will be experienced by the objects situated in that rotating frame opposite to the centripetal acceleration i.e. away from the center. This is called the centrifugal &quot;force&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously it is not a real force from the point of view of an outside inertial observer, who is the legal observer in the Newtonian mechanics. For this observer it is simply the inertia of the objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we call this centrifugal force as a pseudo force and not a real force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From General Relativistic considerations however &lt;em&gt;all&lt;/em&gt; observers are equivalent no matter how they are moving and all accelerations are equivalent to gravitational field in a short enough scaled. Therefore for an infinitesimally small rotating frame the centrifugal &quot;force&quot; can be thought of as a gravitational force! This directly follows from the equivalence principle.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-04-20T02:02:23.003" />
  <row Id="8901" PostTypeId="2" ParentId="8886" CreationDate="2011-04-20T02:22:11.810" Score="0" ViewCount="" Body="&lt;p&gt;It is called the principle of equivalence. It was discovered by Albert Einstein probably in 1907. This is the central principle of his theory of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply speaking, it says that gravity and accelerations are equivalent. Sitting on an &quot;accelerated frame&quot; (w.r.t. an out side observer) an observer can not determine whether he is accelerating or there is simply a gravitational field around him. This is due to the fact that inertial mass of an object is the same as its gravitational mass as is shown by the experiments till date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equivalence ensures that all freely falling objects will fall towards ground with the same acceleration. It also ensures that there could be a metric theory of gravity (not necessarily GR). There are many metric theory of gravity and Einstein's GR is the simplest possible and experimentally confirmed theory of gravity till today.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-04-20T02:22:11.810" />
  <row Id="8902" PostTypeId="2" ParentId="8882" CreationDate="2011-04-20T02:52:35.863" Score="0" ViewCount="" Body="&lt;p&gt;Q: absoluteness of time implies galilean transformations?&lt;br&gt;&#xA;A: No.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What implies galilean transformations (ordinary law of composition of velocities) is is the non-recognition of a speed limit of propagation on the physical environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The long text that follows is just to illustrate the above statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galileo could have known this? (lets change the course of past events with a fiction ;-)&lt;br&gt;&#xA;Perhaps he had noticed that a fast boat and a slow boat leaves waves that propagate without difference between them and concluded: The speeds are not additive.&#xA;As in his time all objects were slow he did not concluded that in sound and light the velocities are not additive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But assuming the contrary, let's see what he could have thought:&lt;br&gt;&#xA;A very fast plane sounds a tone and an observer, armed with an absolute ruler and a clock common to the pilot of the plane, with instant communications, measures a constant and finite speed for the sound and concludes: I can not add the speed of the plane to the speed of the sound, and observes that the sound must propagate in waves similar to the ones of the boats. These speed limits are characteristic of the propagation medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.speed-light.info/measure/speed_of_light_history.htm&quot; rel=&quot;nofollow&quot;&gt;Galileo had read&lt;/a&gt; the work of Robert Hooke '&lt;a href=&quot;http://en.wikipedia.org/wiki/Light#Wave_theory&quot; rel=&quot;nofollow&quot;&gt;the wave theory of light&lt;/a&gt;' (in 1660) and Ole Roemer in 1675 had informed that the speed of light is 200,000 km/s (only in 1728, 53 years later, James Bradley measured 301,000 km/s). Since I (Galileu) have no immediate vision I will have to measure the speed of sound with the aid of a flash of light and make the appropriate corrections to have a more accurate measure of the speed of sound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And after Galileo had known Einstein's work, which he aproved, he thought: it makes sense that the speed of light is also a constant feature of the medium. I measured its value in a closed circuit and Einstein confirmed to him: the 'c' is the average value of the round trip of a light ray, as it is written in his 1905 paper. To put an end to the confusion with the different observer referencials Galileo decided to change the perspective: I will adopt the uniform &lt;a href=&quot;http://arxiv.org/abs/physics/0205033&quot; rel=&quot;nofollow&quot;&gt;referential of the medium&lt;/a&gt; because it is &lt;strong&gt;common to all observers&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-04-20T02:52:35.863" />
  <row Id="8903" PostTypeId="1" CreationDate="2011-04-20T02:56:10.890" Score="11" ViewCount="665" Body="&lt;p&gt;What is Lagrangian mechanics, and what's the difference compared to Newtonian mechanics? I'm a mathematician/computer scientist, not a physicist, so I'm kind of looking for something like the explanation of the Lagrangian formulation of mechanics you'd give to someone who just finished a semester of college physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things I'm hoping to have explained to me:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What's the overall difference in layman's terms? From what I've read so far, it sounds like Newtonian mechanics takes a more local &quot;cause-and-effect&quot;/&quot;apply a force, get a reaction&quot; view, while Lagrangian mechanics takes a more global &quot;minimize this quantity&quot; view. Or, to put it more axiomatically, Newtonian mechanics starts with Newton's three laws of motion, while Lagrangian mechanics starts with the Principle of Least Action.&lt;/li&gt;&#xA;&lt;li&gt;How do the approaches differ mathematically/when you're trying to solve a problem? Kind of similar to above, I'm guessing that Newtonian solutions start with drawing a bunch of force vectors, while Lagrangian solutions start with defining some function (calculating the Lagrangian...?) you want to minimize, but I really have no idea.&lt;/li&gt;&#xA;&lt;li&gt;What are the pros/cons of each approach? What questions are more naturally solved in each? For example, I believe Fermat's Principle of Least Time is something that's very naturally explained in Lagrangian mechanics (&quot;minimize the time it takes to get between these two points&quot;), but more difficult to explain in Newtonian mechanics since it requires knowing your endpoint.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3199" LastActivityDate="2011-07-30T14:41:35.973" Title="Newtonian vs. Lagrangian Mechanics" Tags="&lt;mechanics&gt;&lt;newtonian-mechanics&gt;&lt;lagrangian-mechanics&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="4" />
  <row Id="8904" PostTypeId="1" AcceptedAnswerId="8911" CreationDate="2011-04-20T03:07:43.457" Score="1" ViewCount="210" Body="&lt;p&gt;The title says it all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example why is plutonium considered more dangerous than radioactive iodine?&lt;/p&gt;&#xA;" OwnerUserId="3201" LastActivityDate="2011-04-20T18:58:11.737" Title="Why do they consider radioactive matter with long half lives more dangerous than matter with a short half life?" Tags="&lt;radioactivity&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="1" />
  <row Id="8905" PostTypeId="2" ParentId="8904" CreationDate="2011-04-20T03:22:19.363" Score="0" ViewCount="" Body="&lt;p&gt;There are two aspects to be considered for radioactive material. Long half-lives like Pu will contaminate areas for decades or centuries but have relatively small decays/second per unit mass. Short half-lives  like Iodine disappear in months but have extremely high decays/second per unit mass. Of course, ingesting either can lead to serious illness or death.&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-04-20T03:22:19.363" />
  <row Id="8906" PostTypeId="1" CreationDate="2011-04-20T03:23:15.960" Score="3" ViewCount="261" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Superconducting_Super_Collider&quot; rel=&quot;nofollow&quot;&gt;Superconducting Super Collider&lt;/a&gt; was famously cancelled in 1993 after running enormously over budget.  According to the wikipedia page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;During the design and the first construction stage, a heated debate ensued about the high cost of the project. In 1987, Congress was told the project could be completed for &amp;#36;4.4 billion, and it gained the enthusiastic support of Speaker Jim Wright of nearby Fort Worth. By 1993, the cost projection exceeded 12 billion dollars. ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What went wrong? Why was it so much over budget?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can such a fate be avoided by future big physics projects in the US?  What did we learn?&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-04-20T03:34:33.457" LastActivityDate="2011-04-20T20:44:21.060" Title="The Superconducting Super Collider: what went wrong?" Tags="&lt;soft-question&gt;&lt;history&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="8907" PostTypeId="2" ParentId="8904" CreationDate="2011-04-20T03:26:45.053" Score="3" ViewCount="" Body="&lt;p&gt;Half  life is &lt;a href=&quot;http://en.wikipedia.org/wiki/Half_life&quot; rel=&quot;nofollow&quot;&gt;defined&lt;/a&gt;  as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the period of time it takes for a substance undergoing decay to decrease by half. The name was originally used to describe a characteristic of unstable atoms (radioactive decay), but it may apply to any quantity which follows a set-rate decay.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The shorter the life time the faster the material returns to normal levels of radioactivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iodine in eight days exudes  half the radiation it had the first day. In sixteen days it is 1/(2*2)=1/4 etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plutonium takes hundreds of years to get to halve its radiation level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition the danger of the body replacing its minerals with radioactive ones is a major factor of danger from long lived elements, like plutonium, through the food chain, from fallout and ground pollution. The shorter the lifetime , the better.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-20T03:26:45.053" />
  <row Id="8908" PostTypeId="2" ParentId="8903" CreationDate="2011-04-20T03:28:17.783" Score="7" ViewCount="" Body="&lt;p&gt;In Newtonian mechanics you have to use mainly rectangular co ordinate system and consider all the constraint forces. Lagrange's scheme avoids the considerations of the constraint forces deftly and you can use any set of &quot;generalized coordinates&quot; like angle, radial distance etc. consistent with the constraint relations. The number of those generalized coordinates are the same with the number of degrees of freedom of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all dynamical systems we arbitrarily choose some generalized co ordinates consistent with the constraints of the system. In Newtonian mechanics, the difference between the kinetic and potential energy of the system gives you the so called Lagrangian. Then we have n number of differential equations. $n$ is the number of degrees of freedom of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main advantage of Lagrangian mechanics is that we don't have to consider the forces of constraints and given the total kinetic and potential energies of the system we can choose some generalized coordinates and blindly calculate the equation of motions totally analytically unlike Newtonian case where one has to consider the constraints and the geometrical nature of the system.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="2146" LastEditorDisplayName="" LastEditDate="2011-04-20T16:16:02.233" LastActivityDate="2011-04-20T16:16:02.233" CommentCount="2" />
  <row Id="8909" PostTypeId="2" ParentId="8906" CreationDate="2011-04-20T03:32:03.337" Score="0" ViewCount="" Body="&lt;p&gt;There are two points of view. Because of cost overuns and fiscal issues, it was deemed too expensive to complete. The other side of the coin was that the administration didn't sufficiently value America's leadership in particle physics. &lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-04-20T03:32:03.337" CommentCount="3" />
  <row Id="8910" PostTypeId="2" ParentId="8904" CreationDate="2011-04-20T04:02:14.610" Score="6" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;why is plutonium considered more dangerous than radioactive iodine?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because the press have heard of Plutonium and Pu=atomic bombs=bad&lt;br&gt;&#xA;Plutonium's danger is over stated, it's insoluble so hard to get into the food chain and even if ingested is going to go straight through you. Pu is only a real concern if breathed into the lungs as a fine dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iodine is much more of a concern to human (and animal) health it is readily absorbed in the body and is sufficiently active to have serious radiological effects.  The only good thing about Iodine is that if it occurs in a reactor accident on the other side of the world and takes 8days to get to you - there is only half as much of it left.&lt;/p&gt;&#xA;" OwnerUserId="2525" LastEditorUserId="2525" LastEditorDisplayName="" LastEditDate="2011-04-20T15:49:36.783" LastActivityDate="2011-04-20T15:49:36.783" />
  <row Id="8911" PostTypeId="2" ParentId="8904" CreationDate="2011-04-20T08:14:23.237" Score="1" ViewCount="" Body="&lt;p&gt;Scientists consider radionuclides with short half lives to be a lot more dangerous because they are.&#xA;Press weenies and environmentalists have a political agenda and spread the lie that long half lives are bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranium and Plutonium are hardly a health hazard at all, at least in metalic form, and certainly not a radiation hazard unless you happen to get a critical mass together).&#xA;Of course in dust form, particularly as salts or oxides, they are highly toxic, but that's not a radiation hazard but a chemical one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short lived isotopes cause the high radiation environment you don't want to be exposed to for prolonged periods. Those environments of course also don't last very long as decay will cause those isotopes to disappear after short period (days to decades).&#xA;This is plainly visible in the area around Chernobyl today where background radiation (except in some hotspots) has fallen to a level where it's been safe for people to work right next to reactor 4 without special protective clothing for decades, where now the Ukraine is actively promoting tourism into the area which is a haven for wildlife.&#xA;It's also no surprise that both Hiroshima and Nagasaki are not wastelands but bustling cities where people live with no fear of radiation sickness and have since shortly after WW2.&#xA;This of course doesn't stroke with the alarmist message that &quot;the site will remain dangerously radioactive for a hundred thousand years&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The danger from short lived radioisotopes is twofold:&#xA;1) direct radiation exposure from decay is high and intense, which can cause radiation sickness, cancer, etc.&#xA;2) many are readily absorbed into the human body, and when decaying after being used as building blocks for cells (especially DNA) can lead to mutations, cancer, and other health problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While theoretically possible for some long lived radioisotopes as well, absorbing into your tissue some of those won't likely lead to decay related problems during your lifetime as there simply won't be enough decay during your lifetime for there to develop problems larger than your body can correct.&lt;/p&gt;&#xA;" OwnerUserId="2625" LastActivityDate="2011-04-20T08:14:23.237" />
  <row Id="8912" PostTypeId="2" ParentId="8906" CreationDate="2011-04-20T08:18:17.913" Score="0" ViewCount="" Body="&lt;p&gt;I cannot judge American practice, for Germany (and Europe) &#xA;2-fold over budget is not uncommon in &quot;state&quot; projects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about American colliders before SSC? &#xA;I assume those were all over budget. But that did not matter.&#xA;Wasn't the time of SSC refusal the big times of Newt Gingerich?&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-20T10:24:33.870" LastActivityDate="2011-04-20T10:24:33.870" CommentCount="3" />
  <row Id="8913" PostTypeId="2" ParentId="8906" CreationDate="2011-04-20T11:03:24.670" Score="2" ViewCount="" Body="&lt;p&gt;Several things went wrong. I was not part of the community until very late in the process, but my older colleagues report the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Project budget estimates are generally low-balled at some level.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It was &quot;designed&quot; (perhaps &quot;spec'ed out&quot; would be better) on the assumption that a break-through in super-conducting magnet design could be scheduled and made to happen by throwing money at it. This did not work out as well as they had hoped, so more and more money was thrown at the problem because the magnets were critical infrastructure. Of course, the LHC has benefited from the results.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The project budget was necessarily put into the hands of a few physicist-managers. All senior people who had run successful &quot;large&quot; projects in the past. But there is a difference between &quot;large&quot; and &lt;strong&gt;LARGE&lt;/strong&gt;. The total SSC budget was roughly ten times what they had handled in the past, and the procedures and heuristics they used we're not really good enough. Money was mis-spent. DOE has implemented a more strenuous oversight processes since then: lots of &quot;critical decisions&quot; to be met before you get the big bucks.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Even the parts of the project that weren't assuming a break-through suffered the usual budget creep.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A certain amount of effort went into selling it as a potential job creator to the legislators representing several potential sites. Once they settled on one, the losers obviously lost interest.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;&lt;a href=&quot;http://www.sciencemag.org/content/302/5642/40.full&quot; rel=&quot;nofollow&quot;&gt;Scientists Are Long Gone, But Bitter Memories Remain&lt;/a&gt;&quot;  (in &lt;em&gt;Science&lt;/em&gt; magazine, behind pay wall).  Like the title says; discusses the fate of the facility 10 years after cancellation.  No discussion of why it was cancelled.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="520" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-04-20T20:44:21.060" LastActivityDate="2011-04-20T20:44:21.060" CommunityOwnedDate="2011-04-20T11:03:24.670" />
  <row Id="8914" PostTypeId="1" AcceptedAnswerId="8916" CreationDate="2011-04-20T11:12:29.277" Score="3" ViewCount="167" Body="&lt;p&gt;The volume of a cylinder is given by the expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V=\pi r^2 h$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The uncertainties for $V$ and $h$ are as shown below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align}&#xD;&#xA;V&amp;amp;\pm 7\%\\&#xD;&#xA;h&amp;amp;\pm 3\%&#xD;&#xA;\end{align}$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the uncertainty in $r$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, the obvious answer would be $2\%$, from&#xA;$$\frac{dV}{V}=\frac{dh}{h}+2\frac{dr}{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, rearranging to $r^2=\frac{V}{\pi h}$ gives&#xA;$$2\frac{dr}{r}=\frac{dV}{V}+\frac{dh}{h}$$&#xA;which gives a different answer of $5\%$. Thus, by simply rearranging the formula, we get different values for uncertainty in $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&#xA;How do you explain this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(The mark scheme lists $5\%$ as the correct answer)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditorDisplayName="" LastEditDate="2011-04-20T11:32:59.727" LastActivityDate="2011-04-26T02:43:36.497" Title="The approximate uncertainty in $r$" Tags="&lt;homework&gt;" AnswerCount="5" />
  <row Id="8915" PostTypeId="1" CreationDate="2011-04-20T11:57:00.200" Score="0" ViewCount="95" Body="&lt;p&gt;Assuming any (optimal) body orientation/position relative to the direction of force and mechanism of death.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be interested in a rough estimate of threshold force and a suggested mechanism of death. A person experiencing a force just below this threshold should be able to survive for a few minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give more context the question derives from: 'How much centripetal acceleration is fatal?', or 'How fast and tight must a rollercoaster go to kill you?'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realise there may be some biological considerations for this question, but giving bounds on these seems within the remit of back-of-the-envelope physics.&lt;/p&gt;&#xA;" OwnerUserId="3207" LastActivityDate="2011-04-20T11:57:00.200" Title="What is the maximum non-fatal force withstandable by a human being for a short period?" Tags="&lt;centrifugal-force&gt;" CommentCount="4" />
  <row Id="8916" PostTypeId="2" ParentId="8914" CreationDate="2011-04-20T12:25:36.030" Score="7" ViewCount="" Body="&lt;p&gt;You're confusing independent and dependent variables.  When you propogate from uncertainties in the $x_{i}$ to some $f(x_{1},x_{2}...)$, the formula $\delta f(x_{1}...)=\sum \left|\frac{\partial f}{\partial x_{i}}\right|\delta x_{i}$ assumes that each of the $x_{i}$ is an independently measured variable and that $f$ is a dependent variable to be calculated from the $x_{i}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the example you give, you have two independent measurements of $V$ and $h$ and are expected to calculate the uncertainty in $r$.  Well, to use the above formula, you need to write $r$ as a dependent variable of $V$ and $h$.  Therefore, it's only correct to solve for $r$ first, and &lt;b&gt;then&lt;/b&gt; calculate the uncertainty.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-04-20T12:25:36.030" CommentCount="3" />
  <row Id="8917" PostTypeId="2" ParentId="8903" CreationDate="2011-04-20T12:44:38.947" Score="-3" ViewCount="" Body="&lt;p&gt;According to Newton: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r(t+dt)=r(t)+v(t)dt$    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{r}(t+dt)=\dot{r}(t)+\frac{F(t)}{m}dt$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, there is no freedom to choose the trajectory - it is determined with the instant values of force and velocity. &quot;Future&quot; is determined with &quot;present&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A particle never &quot;chooses&quot; the optimal trajectory to go from a known position in the past $r(t_1)$ to a known position in the future $r(t_2)$. The future data are not involved in the dynamics. But the least action &quot;principle&quot;, apart from good equations, is based on the future data $r(t_2)$ which is mathematically possible but physically meaningless. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no a &quot;least action principle&quot; proceeding only from the initial data. Instead, the Newton equations with the initial data suffice to solve physical problems ;-)&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-04-21T22:01:07.563" LastActivityDate="2011-04-21T22:01:07.563" CommentCount="9" />
  <row Id="8918" PostTypeId="1" AcceptedAnswerId="8920" CreationDate="2011-04-20T13:10:44.817" Score="3" ViewCount="123" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/3244/vortex-in-liquid-collects-particles-in-center&quot;&gt;Vortex in liquid collects particles in center&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Why when you mix a cup of tea with a teaspoon, tea-craps collect in the center of cup, instead of goes near the walls?&lt;/p&gt;&#xA;" OwnerUserId="3208" LastActivityDate="2011-04-21T15:50:52.900" ClosedDate="2011-04-21T21:44:47.450" Title="Mixing a cup of tea and break of centrifugal force" Tags="&lt;centrifugal-force&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="8919" PostTypeId="2" ParentId="8914" CreationDate="2011-04-20T13:32:48.873" Score="6" ViewCount="" Body="&lt;p&gt;Jerry Schirmer's right about why solving for $r$ first is the right procedure. One way to illustrate this is to notice that with the other procedure the uncertainty could go negative, which can't be right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the main thing I wanted to point out is that, if the measurements of $V$ and $h$ are independent, and if the &quot;errors&quot; mean standard deviations as usual, then the correct procedure is to add the errors in quadrature (i.e., to add the squares and take the square root):&#xA;$$&#xD;&#xA;{\delta r\over r}={1\over 2}\sqrt{\left(\delta V\over V\right)^2+\left(\delta h\over h\right)^2}.&#xD;&#xA;$$&#xA;See, e.g., any of the first few &lt;a href=&quot;http://www.google.com/search?q=propagation%20of%20errors&quot; rel=&quot;nofollow&quot;&gt;Google hits&lt;/a&gt; for &quot;propagation of errors.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-20T13:32:48.873" CommentCount="2" />
  <row Id="8920" PostTypeId="2" ParentId="8918" CreationDate="2011-04-20T13:34:43.103" Score="1" ViewCount="" Body="&lt;p&gt;A rotating fluid roughly observes a $v=r\omega$ law where $v$ is the velocity of a fluid element or anything moving with the fluid element, $r$ is the distance from the centre and $\omega$ is the roughly constant angular rotation rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if a large particle like a tea leaf residue is found in the fluid, it should rotate at a speed $v$, given a particular distance. However, such large particles will be more suspect to frictional forces by the viscous tea that reduce the velocity of the particle very quickly. So in order to satisfy $v=r\omega$, it has to collect at the centre of the cup once its velocity is completely reduced to zero.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There is yet another reason why such a particle should float to the centre, but I suspect it has a much smaller contribution. In the rotating frame, there is a centrifugal force acting outward on the large particle. Let's ignore Coriolis forces and look at the effect of only this centrifugal force. Now locally, in a small region around this particle, this is analogous to a gravitational field acting along the surface of the fluid outward to the edge of the cup (the Equivalence Principle). You'd expect the particle to move towards the edge then, but remember that this tea leaf particle is less dense than tea (it floats) and so it should feel a force opposite in direction to the imaginary gravitational field (the centrifugal force), just like a tea leaf particle in a stationary fluid should float upwards &lt;em&gt;against&lt;/em&gt; a real vertical gravitational field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sort of behaviour is counter-intuitive, but it is actually observed in centrifuges where the suspended particles are less dense than the fluid -- they collect in the center. If you want to check this for yourself without a centrifuge (or if you suspect that friction has a greater role in this), take a helium balloon into a train and you will observe that when the train brakes, while you move forward because of the non-inertial force, the balloon actually inclines backwards! (If the braking is long enough for the air to settle and there are no air currents in the cabin -- agreed, it's not the easiest thing to test)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See here for a more detailed explanation of the balloon example : &lt;a href=&quot;http://physics101.wordpress.com/2008/11/30/balloons-and-accelerations/&quot; rel=&quot;nofollow&quot;&gt;http://physics101.wordpress.com/2008/11/30/balloons-and-accelerations/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-04-20T13:50:41.850" LastActivityDate="2011-04-20T13:50:41.850" CommentCount="4" />
  <row Id="8921" PostTypeId="2" ParentId="8906" CreationDate="2011-04-20T14:14:16.237" Score="0" ViewCount="" Body="&lt;p&gt;The politics, not the cost overruns themselves, killed it:  If the cost killed it then it would have been a bureaucratic move.  What happened was that Congress publicly selected and sacrificed the SSC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was young, but the political story I recall has a few other elements: Congress wanted headlines about cutting spending (but did not want to really tackle controlling the budget).  The delegation from Texas was politically weak at this juncture (otherwise they could have defended the local SSC project).  All in all, many elected officials get more popular from cutting science spending than increasing science spending -- and the SSC was the largest undefended item. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This year (2011) the House is loudly cutting spending on public TV and public radio and other nickel-and-dime things to generate headlines (but does not want to really tackle controlling the budget).  They spent the last few years mocking various science projects (e.g. mocking &quot;volcano research&quot; right before a big eruption...).&lt;/p&gt;&#xA;" OwnerUserId="2633" LastActivityDate="2011-04-20T14:14:16.237" CommentCount="1" />
  <row Id="8922" PostTypeId="1" AcceptedAnswerId="8936" CreationDate="2011-04-20T15:48:19.593" Score="3" ViewCount="203" Body="&lt;p&gt;Astronomical telescopes are now mega projects and cost $1Bn and although they are pitched to solve the current interest of the day they are general purpose machines and with upgrades and new instruments have a life of perhaps 50years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that large accelerator projects are built to answer one question, to find one particle. But since the design must be based around the particle having a particular energy and the cost and timescale being so large - you have to be pretty damn sure that you expect the particle to exist and at the predicted energy.  It almost seems that if you had a good enough estimate to build the accelerator then you don't really need to!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is something like the LHC a one trick pony? You turn it on and confirm the Higgs or if not - build a bigger one?    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the LHC really a more general purpose experiment but the Higgs gets the press attention or is it just that the nature of discover in HEP is different and you need to build a single one shot experiment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: Please this wasn't intended as a flame war or an attack on the LHC (I was an astronomer) just an interesting discussion.&lt;/p&gt;&#xA;" OwnerUserId="2525" LastEditorUserId="2525" LastEditorDisplayName="" LastEditDate="2011-04-20T16:58:09.707" LastActivityDate="2011-04-20T20:41:30.090" Title="What does the LHC do afterwards?" Tags="&lt;soft-question&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="8923" PostTypeId="1" AcceptedAnswerId="8933" CreationDate="2011-04-20T16:18:37.177" Score="7" ViewCount="161" Body="&lt;p&gt;Sorry if this is a silly question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand correctly, for two atoms &quot;having the same number of protons&quot; is equivalent to &quot;being of the same element&quot;, while &quot;having the same number of protons and the same number of neutrons&quot; equates to &quot;being of the same isotope (of the same element)&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But does &quot;having the same number of neutrons&quot; in itself have some significance in physics? And what about &quot;having the same total number of protons and neutrons (but not necessarily with the same summands)&quot;?&lt;/p&gt;&#xA;" OwnerUserId="2651" LastActivityDate="2011-04-20T19:08:17.807" Title="Having the same number of neutrons" Tags="&lt;nuclear-physics&gt;" AnswerCount="2" />
  <row Id="8924" PostTypeId="2" ParentId="8923" CreationDate="2011-04-20T16:40:31.503" Score="5" ViewCount="" Body="&lt;p&gt;Two nuclides (a type of a nucleus with some values of $Z$ and $N$) that have the same number of neutrons $N$ have the same number of neutrons. The previous sentence is a tautology but I wrote it to show that the equality doesn't mean anything special, aside from the things that are obviously implied by it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two nuclides that have the same total number of nucleons - neutrons plus protons - have the same total number of nucleons - neutrons and protons. In the latter case, it also means that the two nuclides are approximately equally heavy and that there may be a beta-decay transformation from one to the other (beta decay doesn't emit any protons or neutrons, so it approximately conserves the mass of the nuclide).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases - equal $N$ or equal $A=Z+N$ - are less important than if two nuclides have the same $Z$. The number of protons $Z$ is by far the most important quantity that determines the character of the resulting material. It's because the electrons only feel the electrostatic force of the nucleus and it only depends on $Z$; it doesn't depend on $N$ or $A=Z+N$. A large chunk of matter has to neutralize the electric charge, so it has to assign the right number of electrons - namely $Z$ electrons for one nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, these $Z$ electrons get inevitably arranged into the shells of atomic physics and the properties of the material are therefore a quasi-periodic function of $Z$, the number of protons, as the periodic table makes very clear. That's why the chemical properties of e.g. iodium-127 and iodium-131 are almost identical while iodine-131 and xenon-131 are extremely different chemically, despite their having the same value of $A=N+Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the mass of the materials depends on $A=Z+N$ and not just $Z$. Also, some isotopes are radioactive while others are not. But radioactivity is not really &quot;chemistry&quot; or &quot;material science&quot;; radioactivity is a rare process that only influences the nuclei while chemistry and material science depends on the electron shells.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-20T16:40:31.503" />
  <row Id="8925" PostTypeId="2" ParentId="8922" CreationDate="2011-04-20T16:44:57.210" Score="-2" ViewCount="" Body="&lt;p&gt;When they've finished with the LHC, they'll sell it as scrap metal. You have to remember we're living in the age of recylcability and anything that can be reused will be so.&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-04-20T16:44:57.210" CommentCount="1" />
  <row Id="8926" PostTypeId="1" AcceptedAnswerId="8928" CreationDate="2011-04-20T16:47:24.113" Score="5" ViewCount="196" Body="&lt;p&gt;I see that it appears as a constant in the relation for the running of the strong coupling constant. What is its significance? Does it have to be established by experiment? Is it somehow a scale for quark confinement? If yes, how? I ask because I saw this in Perkins' &lt;em&gt;Particle Astrophysics&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;After kT fell below the strong quantum&#xA;  chromodynamics (QCD) scale parameter ∼ 200 MeV, the remaining quarks,&#xA;  antiquarks, and gluons would no longer exist as separate components of a&#xA;  plasma but as quark bound states, forming the lighter hadrons such as pions&#xA;  and nucleons.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-04-20T17:01:42.867" Title="What is the significance of the QCD scale parameter $\Lambda$?" Tags="&lt;particle-physics&gt;&lt;quantum-chromodynamics&gt;" AnswerCount="1" />
  <row Id="8927" PostTypeId="2" ParentId="8922" CreationDate="2011-04-20T16:57:03.183" Score="3" ViewCount="" Body="&lt;p&gt;The LHC has four main detectors and several fixed target 'sheds', the search for the Higgs is obviously the headline search but there are lots of bits of particle physics that will be refined and expanded upon by the knowledge learnt by work at the LHC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This CERN page goes into more detail:  &lt;a href=&quot;http://public.web.cern.ch/public/en/LHC/LHCExperiments-en.html&quot; rel=&quot;nofollow&quot;&gt;http://public.web.cern.ch/public/en/LHC/LHCExperiments-en.html&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once (if!) the Higgs is found, there will be much more work to properly characterise the situation, this isn't just a case of 'bird-spotting' where we are only tryin to tick off another particle!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as an aside some new astronomical telescopes are short term projects.  Herschel will likely have a productive lifespan of around 3 years only...&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-04-20T16:57:03.183" CommentCount="1" />
  <row Id="8928" PostTypeId="2" ParentId="8926" CreationDate="2011-04-20T17:01:42.867" Score="3" ViewCount="" Body="&lt;p&gt;Dear dbrane, $\Lambda_{\rm QCD}$ is the only dimensionful parameter of pure QCD (pure means without extra matter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is dimensionful and replaces the dimensionless parameter $g_{\rm QCD}$, the QCD coupling constant. The process in which a dimensionless constant such as $g$ is replaced by a dimensionful one such as $\Lambda$ is called the dimensional transmutation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dimensional_transmutation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dimensional_transmutation&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The constant $g$ isn't quite constant but it depends on the characteristic energy scale of the processes - essentially logarithmically. Morally speaking,&#xA;$$ \frac{1}{g^2(E)} =  \frac{1}{g^2(E_0)} + K \cdot \ln (E/E_0), $$&#xA;at least in the leading approximation. Because $g$ depends on the scale, it is pretty much true that every value of $g$ is realized for some value of the energy scale $E$. Instead of talking about the values of $g$ for many specific values of $E$, one may talk about the value of $E$ where $g$ gets as big as one or so, and this value of $E$ is known as $\Lambda_{\rm QCD}$ although one must be a bit more careful to define it so that it is 150 MeV and not twice as much, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it is the characteristic scale of confinement and all other typical processes of pure QCD - those that don't depend on the current quark masses etc. In most sentences about the QCD scale, including your quote, the detailed numerical constant is not too important and the sentences are valid as order-of-magnitude estimates. However, given a proper definition, the exact value of $\Lambda_{\rm QCD}$ may be experimentally determined. With this knowledge and given the known Lagrangian of QCD - and the methods to calculate its quantum effects - one may reconstruct the full function $g(E)$.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-20T17:01:42.867" />
  <row Id="8929" PostTypeId="2" ParentId="8922" CreationDate="2011-04-20T17:14:56.977" Score="0" ViewCount="" Body="&lt;p&gt;Real big parts of the forerunner of the LHC were used for LHC. &#xA;So, LHC might be part of the &quot;machine to come after it&quot; or &#xA;might be graded up for polarised beams, a very promising task &#xA;at the same energy levels. &#xA;DESY eg is used today as a luminous source for coherent vacuum UV &#xA;and Röntgen rays. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-20T20:41:30.090" LastActivityDate="2011-04-20T20:41:30.090" CommentCount="2" />
  <row Id="8930" PostTypeId="1" CreationDate="2011-04-20T17:17:15.853" Score="4" ViewCount="169" Body="&lt;p&gt;I was having a look at the original paper on supergravity by Ferrara, Freedman and van Nieuwenhuizen &lt;a href=&quot;http://www.adsabs.harvard.edu/abs/1976PhRvD..13.3214F&quot; rel=&quot;nofollow&quot;&gt;available here&lt;/a&gt;. The abstract has an interesting line saying that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Added note: This term has now been shown to vanish by a computer calculation, so that the action presented here does possess full local supersymmetry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But the paper was written in 1976! Do you have any info what kind of computer and computer algebra system did they use? Is it documented anywhere? &lt;/p&gt;&#xA;" OwnerUserId="1864" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-04-20T17:27:56.060" LastActivityDate="2011-04-23T04:42:08.093" Title="Supergravity calculation using computer algebra system in early days" Tags="&lt;history&gt;&lt;software&gt;&lt;supergravity&gt;&lt;computational-physics&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="8931" PostTypeId="2" ParentId="8904" CreationDate="2011-04-20T18:58:11.737" Score="0" ViewCount="" Body="&lt;p&gt;It might have to do with metrics. If Becquerels are used, 1 Becquerel equals one distintegration per second, then one becquerel of a long halflife substance means there are more radioactive atoms present (by the ratio of halflives). It all depends on how you normalize things. If you normalize by the number of radioactive atoms present, then the shorter halflife material will have many more disintegrations per second initially. If you normalize by instantaneous activity, then the shorter half-lived substance will always jave fewer distintegrations per second. As usual, some actual numerical understanding, as opposed to rhetorical understanding must be applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of contamination, contamination by short lived stuff, can probably be waited out. For instance if milk has too much I-131 (halflife 8days), and it were made into powdered milk or cheese, and left on the shelf for a few months it would then be harmless, whereas if a foodstuff were contaminated with Cesium-137 halflife 30years, it is impractical to wait for it to decay.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-04-20T18:58:11.737" />
  <row Id="8932" PostTypeId="1" AcceptedAnswerId="8935" CreationDate="2011-04-20T19:02:11.860" Score="8" ViewCount="257" Body="&lt;p&gt;If you draw a big triangle in earth 2D surface you will have an aproximated spherical triangle, this will be a non euclidean geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but from a 3D perspective, for example the same triangle from space, it could be depicted as euclidean in 3D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then why we talk about non euclidean instead of adding 1 dimension?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easier the non-euclidean approach? I don't see how!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if the question is perhaps naive, but this is a long time doubt for me, so I will be grateful for an answer, thanks &lt;/p&gt;&#xA;" OwnerUserId="1916" LastActivityDate="2011-04-21T13:45:38.310" Title="Can a non-euclidean space be descripted through an euclidean of higher dimension? so why use non-euclidean?" Tags="&lt;general-relativity&gt;&lt;spacetime&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="4" />
  <row Id="8933" PostTypeId="2" ParentId="8923" CreationDate="2011-04-20T19:08:17.807" Score="12" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Isotone&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Isotones&lt;/strong&gt;&lt;/a&gt; are nuclides having the same number of neutrons. &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Magic_number_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;Magic proton or neutron numbers&lt;/a&gt; give the nucleus greater stability. Magic 82-isotone nuclides for instance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/p05hZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Isobar_%28nuclide%29&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Isobars&lt;/strong&gt;&lt;/a&gt; are nuclides having the same mass number (i.e. sum of protons plus neutrons). &#xA;The number of protons in beta-plus (beta-minus) decay decreases (increases) by a unit and the number of neutrons increases (decreases) by a unit, so that an isobar standing to the left (right) of the original nucleus is formed.&#xA;It may be 1, 2 or 3 &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta-decay_stable_isobars#List_of_known_beta-decay_stable_isobars&quot; rel=&quot;nofollow&quot;&gt;beta-decay stable isobars&lt;/a&gt;.  Beta-decay energy of 154-isobar nuclides for instance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/QEUKY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isotones and isobars have great significance for studying  of nuclide stability.&lt;/p&gt;&#xA;" OwnerUserId="238" LastActivityDate="2011-04-20T19:08:17.807" CommentCount="1" />
  <row Id="8935" PostTypeId="2" ParentId="8932" CreationDate="2011-04-20T19:17:11.473" Score="12" ViewCount="" Body="&lt;p&gt;You have to add more than one dimension, in general. Mathematicians have studied in great detail the question of how many extra dimensions you need in order to embed a curved manifold in a flat one. One key result is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nash_embedding_theorem&quot; rel=&quot;nofollow&quot;&gt;Nash Embedding Theorem&lt;/a&gt;, which says that you can isometrically embed an $m$-dimensional Riemannian manifold in $n$-dimensional flat space, for some $n\le m(m+1)(3m+11)/2$. (Isometric embedding means embedding in a way that preserves lengths, etc., which is what seems to be relevant here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's 120 dimensions for 3-dimensional space!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This theorem only applies to Riemannian manifolds, not Lorentzian ones -- that is, it applies to space, not spacetime with its annoying minus sign in the metric. If it did apply to spacetime, so that we could apply the result with $m=4$, you'd need 230 dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, there's not a comparably clean result for Lorentzian spacetime. There are bunches of references &lt;a href=&quot;http://www.phy.olemiss.edu/~luca/Topics/e/embedding.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. But it certainly can't be any easier to embed spacetime than to embed space!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I think this illustrates why people don't like to think of general relativity in terms of embeddings in higher-dimensional space. It's going to be way, way harder than the standard approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, many people have a philosophical preference not to populate a theory with unobservable entities. If you don't need those extra dimensions, why postulate them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One addendum, after thinking about Marek's comment that one might expect to get by generically with only one (or maybe a few) extra dimensions, rather than the large number in the Nash theorem. I mentioned in the comments that my intuition was different, although I wasn't sure. I just want to expand on that a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can run into trouble when using two-dimensional intuition to make guesses about higher-dimensional manifolds. In two dimensions, the Riemann curvature tensor only has one component -- that is, the curvature is described by a single number at each point. I conjecture that that's the reason why it seems intuitive that you can embed 2 dimensions in 3: you just need to &quot;bend&quot; the surface one way at each point to account for the curvature. (Even so, it turns out that you can't embed 2 dimensions in 3, even in relatively simple cases.) But the number of components of the curvature tensor &lt;a href=&quot;http://mathworld.wolfram.com/RiemannTensor.html&quot; rel=&quot;nofollow&quot;&gt;grows rapidly&lt;/a&gt; with dimension. In 3 dimensions, there are 6 components, and in 4 there are 20. It's wildly implausible, to me, that you could &quot;usually&quot; account for all of those extra degrees of freedom with one or two extra dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A bit of a digression, just because I think it's cool: Another example of how 2-D intuition can be a bad guide to higher dimensions. The problem of topologically classifying 2-D manifolds was solved ages ago. One might have guessed that the problem of classifying 3-D manifolds would be similar, but it turns out to be vastly harder. Last time I checked, it was thought that this problem had been solved, but there was some doubt about whether the solution was correct. And in 4-D or more, the problem is apparently &lt;a href=&quot;http://en.wikipedia.org/wiki/Topological_manifold#Classification_of_manifolds&quot; rel=&quot;nofollow&quot;&gt;known to be undecidable&lt;/a&gt;!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more point: even if it's true that you can &quot;usually&quot; get by with fewer dimensions, I'm not sure how relevant that is. Any manifold is a possible solution to Einstein's equation (for some stress tensor), so if you try to recast the theory in terms of extra dimensions, you'll need enough extra dimensions to account for all possibilities, not just the simple ones.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-04-21T13:45:38.310" LastActivityDate="2011-04-21T13:45:38.310" CommentCount="6" />
  <row Id="8936" PostTypeId="2" ParentId="8922" CreationDate="2011-04-20T19:50:11.423" Score="11" ViewCount="" Body="&lt;p&gt;The LHC was envisioned as a &quot;discovery&quot; machine, a multipurpose one. The Higgs gets the press but the expectations is that new physics will become accessible with the higher energy available for center of mass collisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Z was discovered in the SPS the proton antiproton previous generation collider. &#xA;The previous machine in the same tunnel as the LHC,  LEP, an electron positron collider was needed to  establish with great accuracy the parameters of Z itself and  the standard model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general leptonic collisions probe elementary interactions with many less assumptions than proton proton or proton antiproton machines. This is because one is throwing balls of three quarks with their gluons at each other and measures the debris, in order to study the interactions. New physics, because of the high energy, &lt;em&gt;will&lt;/em&gt; appear, but will be  in a complexity unprecedented up to now. Hopefully the next generation will be a lepton machine that will allow to establish the appropriate models unequivocally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now on the question of one off detectors: CERN is practically using all the accelerators built up to now as increasing energy stages to feed the end machine, the LHC. Nothing is wasted. In addition a lot of &lt;a href=&quot;http://greybook.cern.ch/&quot; rel=&quot;nofollow&quot;&gt;experiments&lt;/a&gt; are approved and running in beam lines that are not in the mainstream but may prove valuable or have unexpected theoretical repercussions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus one expects that the LHC will open the window to the new physics that is tantalizing us,strings and unification of all forces at the moment, and the next generation machines will be leptonic ones to allow accurate measurements of parameters and decide between models.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-20T19:50:11.423" />
  <row Id="8937" PostTypeId="1" CreationDate="2011-04-20T20:15:40.653" Score="1" ViewCount="128" Body="&lt;p&gt;How is it that atoms with equal numbers of protons and electrons are described as &quot;electrically neutral&quot; when the proton is 1,800 times more massive than the electron?&lt;/p&gt;&#xA;" OwnerUserId="3210" LastActivityDate="2011-04-20T21:18:27.483" Title="Electrical neutrality of atoms" Tags="&lt;nuclear-physics&gt;&lt;atoms&gt;" AnswerCount="2" />
  <row Id="8938" PostTypeId="2" ParentId="8937" CreationDate="2011-04-20T20:21:19.277" Score="12" ViewCount="" Body="&lt;p&gt;Do not confuse mass with charge. Although the proton is more massive, the magnitude of its positive charge is equal to the magnitude of the electron's negative charge. Hence, neutral atoms! &lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-04-20T20:21:19.277" />
  <row Id="8939" PostTypeId="2" ParentId="8932" CreationDate="2011-04-20T20:21:30.320" Score="1" ViewCount="" Body="&lt;p&gt;I think there is some pedagogical value in the embedding approach you described. First courses on differential geometry often go through a detailed study of plane and space curves, and surfaces in $\mathbb{R}^3$ to build up intuition and then by observing that some geometric quantities are &lt;a href=&quot;http://en.wikipedia.org/wiki/Theorema_Egregium&quot; rel=&quot;nofollow&quot;&gt;intrinsic&lt;/a&gt; to the surface, start the study of intrinsic geometry of manifolds. Dirac wrote a &lt;a href=&quot;http://books.google.ca/books?id=Jtu-n8AzJhMC&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; on general relativity using this embedding approach. But in the end, as far as general relativity is concerned, this is only a pedagogical device since one has to abandon the extra dimensions as nonphysical. It is also true for intrinsic geometry of manifolds (as mathematical subject) since there can be countless different ways to choose the embedding.&lt;/p&gt;&#xA;" OwnerUserId="2299" LastEditorUserId="2299" LastEditorDisplayName="" LastEditDate="2011-04-20T23:43:58.773" LastActivityDate="2011-04-20T23:43:58.773" />
  <row Id="8940" PostTypeId="1" CreationDate="2011-04-20T14:20:50.253" Score="4" ViewCount="256" Body="&lt;p&gt;How does it work that gravitational forces can affect time and what usable applications could arise from this?&lt;/p&gt;&#xA;" OwnerUserId="934" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-04-21T07:56:31.267" LastActivityDate="2011-04-21T22:55:21.787" Title="How can gravitational forces influence time?" Tags="&lt;gravity&gt;&lt;time&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="8941" PostTypeId="2" ParentId="8940" CreationDate="2011-04-20T15:31:23.477" Score="0" ViewCount="" Body="&lt;p&gt;The easy answer is that in day to day life, it can't. And so not any really usable applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hard answer is that in special circumstances it can (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Special_relativity#Time_dilation_and_length_contraction&quot; rel=&quot;nofollow&quot;&gt;Special relativity&lt;/a&gt;), or more accurately the speed you travel at can affect how you experience time. To get a really good understanding of why would require a university degree in both maths and theoretical physics, which I don't have and neither would I have chance to explain in a couple of paragraphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put simply, the closer to the speed of light you travel, the slower you experience time, so that if you could theoretically reach the speed of light (which you can't without an infinite amount of energy, unless you have zero mass, which is a mind bender in its self) time would appear to stop (from your point of view at least).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider this (sorry its quoted directly from wikipedia but its illustrates it quite well): Since one can not travel faster than light, one might conclude that a human can never travel further from Earth than 40 light years, if the traveler is active between the age of 20 and 60. One would easily think that a traveller would never be able to reach more than the very few solar systems which exist within the limit of 20-40 light years from the earth. But that would be a mistaken conclusion. Because of time dilation, he can travel thousands of light years during his 40 active years. If the spaceship accelerates at a constant 1G, he will after a little less than a year (mathematically) reach almost the speed of light, but time dilation will increase his life span to thousands of years, seen from the reference system of the Solar System, but his subjective lifespan will not thereby change. If he returns to Earth he will land thousands of years into its future. Even if he should accelerate for a longer period, his speed will not be seen as higher than the speed of light by observers on Earth, and he will not measure his speed as being higher than the speed of light. This is because he will see a length contraction of the universe in his direction of travel. And during the journey, people on Earth will experience much more time than he does. So, although his (ordinary) speed cannot exceed c, his four-velocity (distance as seen by Earth divided by his proper (i.e. subjective) time) can be much greater than c. This is similar to the fact that a muon can travel much further than c times its half-life (when at rest), if it is traveling close to c.&lt;/p&gt;&#xA;" LastActivityDate="2011-04-20T15:31:23.477" CommentCount="3" />
  <row Id="8942" PostTypeId="1" CreationDate="2011-04-20T20:37:00.970" Score="11" ViewCount="225" Body="&lt;p&gt;I would like to support open-access journals by choosing to publish in journals which allow readers free online access.    Ideally I would also like to retain copyright instead of signing it over to the journal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some of the better open-access journals in physics?  I am particularly interested in journals focused on optics, but let's keep the question general.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-04-21T15:48:01.643" LastActivityDate="2011-04-22T10:06:40.103" CommunityOwnedDate="2011-04-22T22:24:35.010" Title="What are the best open-access journals in physics?" Tags="&lt;soft-question&gt;&lt;publishing&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="8943" PostTypeId="2" ParentId="8937" CreationDate="2011-04-20T20:53:01.530" Score="-3" ViewCount="" Body="&lt;p&gt;Due to different masses the charge distribution within an atom is a complicated function of distance. But outside the atom, at far distances the electric and magnetic fields become very small, nearly zero so at far distances the atom is like an empty space - nothing reveals its presence. If you do not touch it, you do not feel it.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-04-20T21:18:27.483" LastActivityDate="2011-04-20T21:18:27.483" CommentCount="2" />
  <row Id="8944" PostTypeId="1" AcceptedAnswerId="8949" CreationDate="2011-04-20T21:25:48.993" Score="2" ViewCount="189" Body="&lt;p&gt;Inspired by this: &lt;a href=&quot;http://physics.stackexchange.com/questions/8937/electrical-neutrality-of-atoms&quot;&gt;Electrical neutrality of atoms&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have a wavefunction of the 'reduced mass coordinate' for a hydrogen like atom made from an electron and a positron, what is the spatial charge distribution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we solve the hydrogen atom, we change into coordinates of the center of mass, and the separation distance with the reduced mass.  Here, the masses of the constituent particles are the same.  So the center of mass is equidistant from the positron and electron, and so discussing r and -r is just swapping the particles.  Since the probability distribution for all the energy levels of the hydrogen atom are symmetric to inversion (images can be seen here &lt;a href=&quot;http://panda.unm.edu/Courses/Finley/P262/Hydrogen/WaveFcns.html&quot; rel=&quot;nofollow&quot;&gt;http://panda.unm.edu/Courses/Finley/P262/Hydrogen/WaveFcns.html&lt;/a&gt; ), this seems to say no matter what energy level positronium is in, the charge distribution is neutral?  Since the energy level basis is complete, this seems to say we can't polarize a positronium atom without dissociating it!? This doesn't make sense to me, so I'm  probably making a big mistake here.&lt;/p&gt;&#xA;" OwnerUserId="2655" LastActivityDate="2011-04-21T11:33:27.883" Title="Charge distribution in positronium" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="8945" PostTypeId="2" ParentId="8944" CreationDate="2011-04-20T21:54:41.260" Score="-4" ViewCount="" Body="&lt;p&gt;Normally we speak of charge distribution in terms of atomic &lt;strong&gt;&lt;a href=&quot;http://vladimirkalitvianski.wordpress.com/2010/12/02/zoom-in-atom-or-unknown-physics-of-short-distances/&quot; rel=&quot;nofollow&quot;&gt;form-factors&lt;/a&gt;&lt;/strong&gt;. It means Born approximation which means in turn fast charged projectiles or photons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Para- and orto-positroniums have different magnetic fields due to different spin orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For slow charged projectiles its field strongly polarizes the atom and the charge distribution is not the same as in a &quot;free&quot; atom.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-04-21T11:33:27.883" LastActivityDate="2011-04-21T11:33:27.883" CommentCount="5" />
  <row Id="8946" PostTypeId="2" ParentId="8942" CreationDate="2011-04-20T22:59:59.707" Score="6" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://iopscience.iop.org/1367-2630&quot; rel=&quot;nofollow&quot; title=&quot;NJoP&quot;&gt;The New Journal of Physics&lt;/a&gt; has done a pretty good job attracting serious papers, and there are a number of quite good cold atoms/condensed matter/quantum optics papers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;APS is new to the open access game, but I've no doubt that &lt;a href=&quot;http://prx.aps.org/&quot; rel=&quot;nofollow&quot;&gt;PRX&lt;/a&gt; (which I think, as a community, we should all refer to as Phys. Rev. eXtreme...) will be a good resource when it debuts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: fixed so journal titles are links&lt;/p&gt;&#xA;" OwnerUserId="791" LastEditorUserId="791" LastEditorDisplayName="" LastEditDate="2011-04-20T23:32:48.147" LastActivityDate="2011-04-20T23:32:48.147" CommunityOwnedDate="2011-04-22T22:24:35.010" CommentCount="6" />
  <row Id="8947" PostTypeId="1" AcceptedAnswerId="8950" CreationDate="2011-04-20T23:13:47.383" Score="5" ViewCount="412" Body="&lt;p&gt;So I was cruising around at YouTube and saw this &lt;a href=&quot;http://www.youtube.com/watch?v=im1iNq02Kz0&amp;amp;feature=feedu&quot; rel=&quot;nofollow&quot;&gt;sweet vid&lt;/a&gt;, and as I was watching started to wonder: &quot;How is this possible?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a little bit of background, in case you decide to not watch the video, what is happening is an individual can hook out a 'pack' that sucks up water and then shoots the water downward:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/tMc9z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This water Jet Levitation (JetLev) causes the rider to be able to fly upward into the air and move about forward, backward, side-to-side and even underwater (you have to watch the vid for that).  So my question is:  How exactly is this possible?&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2011-04-22T01:03:32.393" Title="How does the 'water jet pack' work?" Tags="&lt;forces&gt;" AnswerCount="3" />
  <row Id="8948" PostTypeId="2" ParentId="8947" CreationDate="2011-04-20T23:30:11.650" Score="7" ViewCount="" Body="&lt;p&gt;As with any other rocket, ejecting the propellant out the nozzle generates an equal and opposite thrust. The difficulty here is that the propellant is in the body of water the device flys above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you read the &lt;a href=&quot;http://jetlev.com/Pages/faq.html&quot; rel=&quot;nofollow&quot;&gt;FAQ&lt;/a&gt; at the jetlev website you will see that power and pumping is provided by a separate floating unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The buoyancy of the boat unit supports the thrust of lifting the water to the backpack unit, and momentum transfer when the backpack unit turns it around supports the &quot;flyer&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They claim to have rolled it out as a commercial product last month (March 2011). Presumably you'll be able to rent one at club med or similar places.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-04-21T04:04:09.483" LastActivityDate="2011-04-21T04:04:09.483" CommentCount="6" />
  <row Id="8949" PostTypeId="2" ParentId="8944" CreationDate="2011-04-20T23:54:04.910" Score="5" ViewCount="" Body="&lt;p&gt;Okay we have the center of mass coordinate $r_{cm} = (r_e + r_p)/2$, and the reduced mass coordinate $r = r_e - r_p$.  So given the wavefunction $\psi(r_{cm},r)$ what you are asking is just a change of basis from $|r_{cm},r\rangle$ to $|r_e,r_p\rangle$.  So you just need to consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle r_e,r_p|r_{cm},r\rangle = \delta\left(r_{cm}-(r_e+r_p)/2\right) \ \delta\left(r-(r_e-r_p)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;To get more at what appears to be confusing you, let's focus on how positronium can be polarized.  The spherical harmonics will have a defined even or odd parity to inversion, and so &lt;em&gt;yes&lt;/em&gt; the square of the wavefunction (probability density) is invariant to inversion.  But this does not mean all &lt;em&gt;superpositions&lt;/em&gt; of these harmonics will have this property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider for instance just s + p_z orbital.  On one side the wavefunction amplitude will add constructively, while on the other it will add deconstructively.  The density is no longer symmetric to inversion.  This is the standard chemistry description of orbitals, so probably a good way to get follow up reading is searching for &lt;a href=&quot;http://en.wikipedia.org/wiki/Hybrid_orbital&quot; rel=&quot;nofollow&quot;&gt;hybrid orbitals&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a link I found with some images for you:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.uwosh.edu/faculty_staff/gutow/Orbitals/N/What_are_hybrid_orbitals.shtml&quot; rel=&quot;nofollow&quot;&gt;http://www.uwosh.edu/faculty_staff/gutow/Orbitals/N/What_are_hybrid_orbitals.shtml&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2652" LastEditorUserId="2652" LastEditorDisplayName="" LastEditDate="2011-04-21T09:36:27.563" LastActivityDate="2011-04-21T09:36:27.563" CommentCount="3" />
  <row Id="8950" PostTypeId="2" ParentId="8947" CreationDate="2011-04-21T00:02:33.433" Score="6" ViewCount="" Body="&lt;p&gt;Well I ought to be studying for a physics exam, but I'll consider answering this to be my studying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton's third law states that for every action, there is an equal an opposite reaction. In this case, the jetpack is ejecting water at high velocity toward the ground. This is generating a significant force downward. The resulting opposite force pushes upward, elevating the rider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual pump for the unit is located on a floating watercraft attached to but not elevated with the jetpack. This is how there is enough force for the unit to fly - otherwise, the mass of the pump would be so great that it would be difficult to maintain enough pressure to keep both the pump and rider aloft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know the specifics of how it orients itself, though I imagine that by making small variations in the water pressure and orientation of each nozzle, one can cause an imbalance in the forces acting on the rider, casing them to move one way or the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/mZmDz.png&quot; alt=&quot;force diagram&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="302" LastEditorUserId="302" LastEditorDisplayName="" LastEditDate="2011-04-21T00:38:37.913" LastActivityDate="2011-04-21T00:38:37.913" CommentCount="5" />
  <row Id="8951" PostTypeId="1" AcceptedAnswerId="8952" CreationDate="2011-04-21T00:19:38.143" Score="0" ViewCount="139" Body="&lt;p&gt;For those who don't know what a steadicam is, I'll explain it in two points to give a &lt;em&gt;basic&lt;/em&gt; idea. If you know what a steadicam is, skip to point 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) A &lt;a href=&quot;http://en.wikipedia.org/wiki/Steadicam&quot; rel=&quot;nofollow&quot;&gt;steadicam&lt;/a&gt; is a stabilizing mount for a motion picture camera, which mechanically isolates the operator's movement from the camera, allowing a very smooth shot even when the operator is moving quickly over an uneven surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) How does it work (image courtesy of &lt;a href=&quot;http://diycamera.com/stabiliser/index.html&quot; rel=&quot;nofollow&quot;&gt;wsclater&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/jMrJJ.jpg&quot; alt=&quot;Steadicam&quot;&gt;&#xA;Red dots = control masses, Blue dot = center of rotation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea behind a steadicam is that the center of the mass is taken &lt;strong&gt;out of the camera&lt;/strong&gt; itself. Therefore, when moving / panning the whole system (steadicam + camera), the motion of the camera itself is more fluid. This can be seen in action in &lt;a href=&quot;http://www.youtube.com/watch?v=R89VjDdvk2A&quot; rel=&quot;nofollow&quot;&gt;this amazing youtube example&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we determine the ideal distance of the center of rotation from the camera?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we determine the ideal distance of the center of mass from the center of rotation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the ideal length of the handle (the point where we hold the system), i.e. how far should we hold the system from its center of rotation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Disclaimer: This is neither a matter of profit, nor a matter of homework. I'm just trying to build some of those for my own personal use, especially to find out how it really works, as I'm kind of amazed from this.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2834" LastEditorUserId="2834" LastEditorDisplayName="" LastEditDate="2011-04-21T00:35:47.933" LastActivityDate="2011-05-21T15:37:56.907" Title="What are the ideal centers of mass and rotation for a steadicam?" Tags="&lt;rotational-dynamics&gt;&lt;stability&gt;&lt;mass-balance&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="8952" PostTypeId="2" ParentId="8951" CreationDate="2011-04-21T01:09:14.183" Score="0" ViewCount="" Body="&lt;p&gt;You need to know two things to design the system:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Where you want the systems's center of mass to be.&lt;/li&gt;&#xA;&lt;li&gt;How to calculate such a thing.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Having read the wikipedia and watch some video I'm guess the answer to (1) is &lt;em&gt;&quot;a short distance directly below the gimbal&quot;&lt;/em&gt; (for both high and low mode, though that means slightly different places). &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;How to calculate the CoM for a set of point masses. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Establish a coordinate system. Say $z$ increases up and starts at the camera platform, and $x$ increase to the front and starts at the location of the gimbal. This makes $y$ increase to the left if we want the usual convention, and we'll start it at the gimbal as well.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Find the (three dimensional!) location and mass of each part of the system (remember so far I'm only talking about little dense masses) and call these $(x,y,z)_i$ and $m_i$  where $i$ is just an index that runs over all you bits. Write&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$ x_{com} = \frac{\sum_i x_i m_i}{\sum_i m_i} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and an identical equation for $y$ and $z$. And now we know the location of the the Center of Mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For continuous objects, we could turn those sums into integrals, but you probably don't want to do that. Especially as you don't know the mass distribution of the camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead I will suggest finding the center of mass for each object (and this is why the wsclater link calls the red dots center of mass, they &lt;em&gt;are&lt;/em&gt; for each object), by the simple expedient of balancing or hanging them.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you hang an object from &lt;em&gt;any&lt;/em&gt; point, it's CoM will lie below the point you hang it from.&lt;/li&gt;&#xA;&lt;li&gt;If you balance an object on a thin edge, the objects CoM will lie above the edge.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Combine several such measurements and you can find the three diminsional position of the CoM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here's the trick. Once you know the CoM for an extended object, you can treat the whole thing as a point mass at that location (for the purposes of calculating the CoM of a system that includes the object).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;OK. Making progress, but were, still not done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know where you want the CoM, and need to find where to put the masses. Which is backwards from what we did above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to control &lt;em&gt;three&lt;/em&gt; values (the $x$, $y$, and $z$ positions of the CoM), and you have three equations. Which means that you can have only &lt;em&gt;three&lt;/em&gt; adjustable parameters in your system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side note: The value of $y$ is easy, if you position the camera so that it's CoM is centered. Then, because everything else is symmetric, it happens automatically. So I'm going to only worry about $x$ and $z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you do is fix all the position relationships in the system except for two. Maybe the height of the lower control mass and the horizontal position of the middle one. Those two you make variables. Now you write down the CoM equations above *with those two variables in there &lt;em&gt;and&lt;/em&gt; the desired values for the location of the CoM. The resulting system is solvable. Do the algebra and you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But leave some ability to fine tune things (say by putting the masses on threaded rods). This lets you only worry about getting close with the calculation. The pictured one appears to use a flexible support to provide the adjustability. That will work, but may require constant tinkering.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-04-21T01:14:45.910" LastActivityDate="2011-04-21T01:14:45.910" />
  <row Id="8953" PostTypeId="2" ParentId="8942" CreationDate="2011-04-21T01:13:55.850" Score="4" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.opticsinfobase.org/oe/home.cfm&quot; rel=&quot;nofollow&quot;&gt;Optics express&lt;/a&gt; is excellent. Fast turnaround, high impact factor, and decent peer review. &lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-04-21T04:36:23.320" LastActivityDate="2011-04-21T04:36:23.320" CommunityOwnedDate="2011-04-22T22:24:35.010" CommentCount="1" />
  <row Id="8954" PostTypeId="1" AcceptedAnswerId="8955" CreationDate="2011-04-21T03:43:34.797" Score="7" ViewCount="108" Body="&lt;p&gt;I know that it is somehow related to the parton distribution functions, allowing specific reactions with gluons instead of quarks and anti-quarks, but I would really appreciate more detailed answers !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="2022" LastActivityDate="2011-04-21T17:42:34.577" Title="What are the main differences between $p p$ and $p \bar p$ colliders" Tags="&lt;particle-physics&gt;&lt;experimental-physics&gt;&lt;large-hadron-collider&gt;&lt;tevatron&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="8955" PostTypeId="2" ParentId="8954" CreationDate="2011-04-21T04:09:59.163" Score="6" ViewCount="" Body="&lt;p&gt;The difference  in &lt;a href=&quot;http://pdg.lbl.gov/2010/reviews/rpp2010-rev-cross-section-plots.pdf&quot; rel=&quot;nofollow&quot;&gt;scattering cross&lt;/a&gt; sections is more evident the lower the energy of collisions. Fig 41.11. At the energies of TeV the probability of new physics observations is the same for both choices of collisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that at low energies the fact that the proton has three quarks and the anti proton three anti quarks predominates. Quark antiquark scattering at low energies has much higher cross section than quark quark due to the extra possibility of annihilation of the quarks. At low energy the gluon &quot;sea&quot; plays a small part. The higher the energy of interactions the higher the number of energetic gluons that scatter and finally at TEV energies that is what predominates and the two cross sections converge.      Thus for physics it makes no difference whether one uses as targets protons or antiproitons, as far as discovery potential goes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be some technical advantage in the construction, in that in principle the antiproton-proton beams can circulate in the same magnetic configuration as mirror images and make the magnet construction circuits simpler. I guess that the need for high luminosity made LHC a proton proton collider, since it is more difficult to store antiprotons. I would have to research this guess.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-21T04:09:59.163" CommentCount="3" />
  <row Id="8956" PostTypeId="2" ParentId="8895" CreationDate="2011-04-21T04:35:02.860" Score="2" ViewCount="" Body="&lt;p&gt;I will add to  @Lagerbaer's estimate that  the prevailing physics theories do not describe the universe as a LEGO constructable entity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you see a man figure sculpted by lego bricks, you can ask &quot; how many lego pieces went into simulating that man, because there is a finite size for a man and a finite size for the lego brick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though the Universe has a finite size, there are no finite lego bricks that can simulate it. You need some calculus similar to what mathematicians use when counting and manipulating infinities.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-21T04:35:02.860" CommentCount="2" />
  <row Id="8957" PostTypeId="2" ParentId="6157" CreationDate="2011-04-21T06:57:09.547" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Photonic Crystals:&#xA;Molding the Flow of Light&#xA;second edition&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ab-initio.mit.edu/book/&quot; rel=&quot;nofollow&quot;&gt;http://ab-initio.mit.edu/book/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1814" LastActivityDate="2011-04-21T06:57:09.547" CommunityOwnedDate="2011-04-21T06:57:09.547" />
  <row Id="8958" PostTypeId="2" ParentId="8706" CreationDate="2011-04-21T08:42:04.987" Score="1" ViewCount="" Body="&lt;p&gt;The Link Bjorn posted contains most of the information, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;currenty the AFS (active filling scheme) reads:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;50ns_480b+1small_424_12_468_36bpi15inj&lt;/p&gt;&#xA;&#xA;&lt;p&gt;50ns: bunch spacing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;480b: bunches per beam&lt;/p&gt;&#xA;&#xA;&lt;p&gt;424: colliding bunches in the interaction Points 1 (ATLAS) and 5 (CMS)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12: colliding bunches in the IP 2 (ALICE)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;468: colliding bunches in IP8 (LHCb)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;36bpi15inj: 36 bunches per injection and 15 injections per beam&lt;/p&gt;&#xA;" OwnerUserId="2151" LastActivityDate="2011-04-21T08:42:04.987" CommentCount="4" />
  <row Id="8959" PostTypeId="1" CreationDate="2011-04-21T09:32:17.553" Score="4" ViewCount="107" Body="&lt;p&gt;The Heisenberg Principle states that for each direction, $\Delta x\cdot \Delta p_x \ge \hbar , \Delta y\cdot \Delta p_y \ge \hbar$ and $\Delta z\cdot \Delta p_z \ge \hbar$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, can anything be said about $\Delta x\cdot \Delta p_y$? Can I measure the position in one direction and the momentum in an orthogonal direction in any required precision?&lt;/p&gt;&#xA;" OwnerUserId="2918" LastActivityDate="2011-04-21T09:50:34.240" Title="Uncertainly Principle in orthogonal directions" Tags="&lt;uncertainty-principle&gt;&lt;momentum&gt;&lt;planck&gt;" AnswerCount="1" />
  <row Id="8960" PostTypeId="2" ParentId="8959" CreationDate="2011-04-21T09:50:34.240" Score="6" ViewCount="" Body="&lt;p&gt;The uncertainty principle arises from non-commuting observables.  More information can be seen in a discussion and derivation at wikipedia: &lt;a href=&quot;http://en.wikipedia.org/wiki/Uncertainty_principle#Mathematical_derivations&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;  The result is the general uncertainty principle for two observables A,B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_A\sigma_B \ge \frac{1}{2} \left|\left\langle\left[{A},{B}\right]\right\rangle\right|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $p_y$ and $x$ commute, in principle you can specify both to arbitrary precision.&lt;/p&gt;&#xA;" OwnerUserId="2652" LastActivityDate="2011-04-21T09:50:34.240" CommentCount="2" />
  <row Id="8961" PostTypeId="2" ParentId="8940" CreationDate="2011-04-21T10:24:23.183" Score="8" ViewCount="" Body="&lt;p&gt;Gravity is currently understood to be curvature of spacetime, so mathematically it's unambiguous why gravity should affect time. But to leave it at that would be cheating, no? So I'll try to give an explanation of why you'd intuitively &lt;em&gt;expect&lt;/em&gt; gravity to affect time. This is called the Einstein Tower thought experiment. If you've seen the phenomenon of time dilation in special relativity before, you'll notice that the way time behaves depends a lot on how light behaves. It's not crucial you know this but the spirit of the following explanation is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume for a moment that gravity doesn't affect light in any way. Suppose you create a photon with energy $E$ on the ground and fire it up to the top of a tower where this photon is converted into an equivalent mass $m=E/c^2$. This mass will fall back to the ground and once it reaches the ground it will have gained extra energy $mgh$ because it fell through the gravitational field. If you started with energy $E$, you end up with energy $$E&amp;#39;=E+mgh=E(1+gh/c^2)$$  So every time you do this, you create energy! Surely something is wrong -- it must've been our assumption that gravity doesn't affect light. The most obvious fix then is to assume that the photon lost energy as it climbed up the gravitational field and that its energy at the top is $$E_{top}=E(1+gh/c^2)^{-1}$$ This is (a first approximation to) the phenomenon of gravitational red shift where photons lose energy as they climb out of a gravitational field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you remember that the photon frequency is related to its energy by $E=h\nu$, you'll see that the frequency (~$1/T$ which is the &quot;internal clock&quot; of a photon) obeys the same relation as above. This is indicative of gravity affecting time -- in fact it's immediately obvious that any clock based on the frequency of light will run at a slower rate higher up in the gravitational field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, the result is far more general, though you'll need the apparatus of general relativity to see this.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_time_dilation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Gravitational_time_dilation&lt;/a&gt; for more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's probably no practical application you can milk out of gravitational time dilation (yet) since the effect is weak. But you definitely need to &lt;em&gt;account&lt;/em&gt; for these effects in engineering projects such as &lt;a href=&quot;http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html&quot; rel=&quot;nofollow&quot;&gt;GPS satellites&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-04-21T22:55:21.787" LastActivityDate="2011-04-21T22:55:21.787" CommentCount="1" />
  <row Id="8963" PostTypeId="2" ParentId="8914" CreationDate="2011-04-21T11:14:07.333" Score="2" ViewCount="" Body="&lt;p&gt;This should be a comment to Jerry Schirmer's right answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solutions you give solve different problems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first calculation you measure V with its uncertainty $\Delta V$, you know it depends with r and h, you know the uncertainty you have in the measurement of h, $\Delta h$ and thus you infer the uncertainty you had associated with measuring r, $\Delta r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second calculation you want to measure r from V and h. Knowing the uncertainties in V and h you predict the uncertainty that you will have in r.&lt;/p&gt;&#xA;" OwnerUserId="3100" LastActivityDate="2011-04-21T11:14:07.333" />
  <row Id="8964" PostTypeId="2" ParentId="8930" CreationDate="2011-04-21T13:07:58.563" Score="1" ViewCount="" Body="&lt;p&gt;It sounds like it may have been an early version of SHEEP, or some extension thereof. SHEEP was 'officially' released in 1977, but its predecessor, ALAM, was developed by d'Inverno in 1969. It was used to automate some of the complicated algebra in early calculations of the Bondi mass. You can read a bit about the history here: &lt;a href=&quot;http://www.computeralgebra.nl/systemsoverview/special/tensoranalysis/sheep/shpdrv.ps.gz&quot; rel=&quot;nofollow&quot;&gt;notes on SHEEP&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1727" LastActivityDate="2011-04-21T13:07:58.563" />
  <row Id="8965" PostTypeId="2" ParentId="8940" CreationDate="2011-04-21T13:46:52.963" Score="1" ViewCount="" Body="&lt;p&gt;The Schwarzschild metric is&#xA;$$&#xD;&#xA;ds^2~=~(1~-~2GM/rc^2)(cdt)^2~-~(1~-~2GM/rc^2)^{-1}dr^2~-~r^2d\Omega^2&#xD;&#xA;$$&#xA;This has all the constants and the like here to illustrate how the time part of the metric has $O(c^2)$ larger dependency than the spatial part, so $|g_{00}~-~1|~&amp;gt;&amp;gt;$ $|g_{rr}~-~1|$.  So we may approximate this with&#xA;$$&#xD;&#xA;ds^2~\simeq~(1~-~2GM/rc^2)(cdt)^2~-~dr^2~-~r^2d\Omega^2&#xD;&#xA;$$&#xA;So we may then consider a stationary situation with $dr~=~0$ so the propertime is &#xA;$$&#xD;&#xA;d\tau~=~\sqrt{1~-~2GM/rc^2}dt~\simeq~(1~-~GM/rc^2)dt,&#xD;&#xA;$$&#xA;which for near Earth surface gravity the $1~-~GM/r~\simeq~1~+~gh/c^2$, $r~=~R~+~h$.  This recovers the dbrane result.  For a moving body $dr~=~vdt$ we may the compute a gamma factor&#xA;$$&#xD;&#xA;ds^2~\simeq~(1~-~2GM/rc^2~-~(v/c)^2)dt^2&#xD;&#xA;$$&#xA;which can be used to derive modified Lorentz transformations of spacetime.  These then may be used to compute time dilation results for satellites in Earth orbit.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-04-21T22:44:30.833" LastActivityDate="2011-04-21T22:44:30.833" CommentCount="1" />
  <row Id="8966" PostTypeId="2" ParentId="8940" CreationDate="2011-04-21T15:00:52.073" Score="1" ViewCount="" Body="&lt;p&gt;Since your question is from the How Things Work site I guess you're looking for some basic understanding rather than lots of equations. What follows is in that spirit ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question doesn't really have an answer, because it isn't meaningful, or at the least it's misleading, to say that gravitational forces affect time. Mainly that's because, assuming you believe in General Relativity, gravity isn't really a force at all. At least it's not a force in the same way that there's an electric force between two charges or a strong force between two subatomic particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General Relativity is based on the idea that mass (and energy and even pressure) cause spacetime to curve. The path that objects moving through spacetime follow is affected by this curvature so objects moving through curved space don't move in straight lines. If you've watched any popular science programmes you've probably seen the analogy of balls rolling on rubber sheets. The key point is that the freely moving object thinks it's moving in a straight line and it doesn't feel any gravitational force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose you leap off a high cliff. As long as nothing is in your way you'll move in a locally straight line and you won't feel any force at all (ignoring wind resistance). That is, you won't experience any gravitational force. The trouble is that the presence of the Earth curves spacetime in your vicinity, and as a result your locally straight line will hit the surface of the earth. Assuming you land with no injury, you'll now find you feel a force holding you to the ground that you (and Newton!) would describe as a gravitational force. But the apparent force is just because you and the Earth are trying to move along different locally straight lines. There isn't really any force there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my point is that it isn't meaningful to ask how a gravitational force affects time, because there isn't any such thing as a gravitational force. What does affect time, or more precisely spacetime, is the presence of matter, and that produces the appearance of a gravitation force. Your question should really be &quot;How can (the curvature of) time affect gravity?&quot; i.e. exactly the reverse of what you asked!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now you're going to ask &quot;How does matter affect time&quot;, and that's a good question that doesn't have an easy answer. The Einstein equation tells you how matter affects spacetime, and you can feed in numbers and get the right answers. What it doesn't tell you is why matter affects spacetime. We tend to leave the &quot;why&quot; questions to the philosophers :-)&lt;/p&gt;&#xA;" OwnerUserId="1325" LastActivityDate="2011-04-21T15:00:52.073" />
  <row Id="8967" PostTypeId="2" ParentId="8622" CreationDate="2011-04-21T15:43:02.787" Score="1" ViewCount="" Body="&lt;p&gt;Since&#xA;\begin{equation}&#xA;\langle \psi_1 \mid \psi_2 \rangle = \int d^3 x \ \psi_1^*(t,x) \psi_2(t,x),&#xA;\end{equation}&#xA;the quantity $\psi_1^* \psi_2$ is the spatial density of the transition amplitude from $\psi_2$ to $\psi_1$.  This amplitude density is in general complex, so it is not possible in general to interpret it as a probability density, but according to your equation it is conserved all the same.  The only exception is when $\mid \psi_1 \rangle = \mid \psi_2 \rangle = \mid \psi \rangle$, in which case the equation above becomes&#xA;\begin{equation}&#xA;1 = \int d^3 x \ \psi^*(t,x) \psi(t,x),&#xA;\end{equation}&#xA;which is why the non-negative quantity $\psi^* \psi$ can be interpreted as a probability density. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conservation of amplitude density implies the important fact that transition amplitudes between well-behaved solutions of the Schrodinger equation are time independent.  It follows from your equation that&#xA;\begin{equation}&#xA;\frac{\partial}{\partial t} \langle \psi_1 \mid \psi_2 \rangle = \int dx \ \frac{\partial}{\partial t} (\psi_1^* \psi_2) = \frac{i \hbar}{2 m} \int dx \ \nabla \cdot (\psi_1^* \nabla \psi_2 - \psi_2 \nabla \psi_1^*).&#xA;\end{equation}&#xA;If the wavefunctions $\psi_1$, $\psi_2$ are well-behaved at infinity, this final integral will vanish, so $\langle \psi_1 \mid \psi_2 \rangle = {\rm constant}$.  In hindsight, you could have inferred the necessity of the conservation of amplitude density from two facts: 1) transition amplitudes can be written as an integral of a local quantity, and 2) transition amplitudes are time-independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this help answer your question?&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2011-04-21T15:43:02.787" CommentCount="4" />
  <row Id="8968" PostTypeId="1" CreationDate="2011-04-21T15:47:00.053" Score="2" ViewCount="127" Body="&lt;p&gt;I have two air coils (assume they are simple, circular wire loops). They both have diameter &lt;strong&gt;d&lt;/strong&gt;. There is a distance &lt;strong&gt;D&lt;/strong&gt; between their centres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;D&lt;/strong&gt; is much greater than &lt;strong&gt;d&lt;/strong&gt; (more than 10x greater)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both coils are at different angles, &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;b&lt;/strong&gt;, relative to the line between their centers. There is an alternating current in one coil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some simple function which estimates the current induced in the second coil? I would be happy if the form of the function was:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k * f(D, a, b) (where k had to be measured)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: k * cos(a) * cos(b) * D^-2&lt;/p&gt;&#xA;" OwnerUserId="3220" LastEditorUserId="3220" LastEditorDisplayName="" LastEditDate="2011-04-21T16:38:26.357" LastActivityDate="2011-04-21T17:55:23.103" Title="How to estimate inductive coupling between distant air coils" Tags="&lt;electromagnetism&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="8969" PostTypeId="2" ParentId="8918" CreationDate="2011-04-21T15:50:52.900" Score="-1" ViewCount="" Body="&lt;p&gt;The previous life of this question gave an explanation based upon slowing convection, caused by the frictional slowing of the fluid, (slower at the edges), which seems a reasonable possibility. In that case fluid flow is needed because the surface of the fluid is changing because of decreasing angular velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if surface tension might be involved as well (or instead). Surface tension is modified by the curvature of the surface, and our surface does have curvature that varies as a function of radius. Could this be driving small hydrophobic particles towards the center? However, I think the phenomena is also seen in large ocean votices, with logs caught on the surface, and these objects would be too large to be affected by surface tension gradients.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-04-21T15:50:52.900" CommentCount="1" />
  <row Id="8970" PostTypeId="2" ParentId="1779" CreationDate="2011-04-21T16:36:00.817" Score="3" ViewCount="" Body="&lt;p&gt;Talks, talks talks. If you're near graduation, you need as much publicity as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Give as many talks as possible&lt;/strong&gt;. Give talks at conferences, especially. Ask your advisor to have his friends invite you to give talks at other universities. Consider even cold-calling other physics research programs asking if they're looking for a speaker. During your talks, mention that you're graduating soon, and looking for a postdoc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Attend as many talks as possible&lt;/strong&gt;. Attend any talks by visitors that sound even halfway interesting. Go to conferences, and attend every session that sounds even halfway interesting. Corner as many interesting speakers as you can, and ask them if they're hiring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Don't be shy&lt;/strong&gt;. Most hiring is 'right place, right time', not 'perfect fit'.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2011-04-21T16:36:00.817" />
  <row Id="8971" PostTypeId="2" ParentId="8968" CreationDate="2011-04-21T16:58:27.917" Score="1" ViewCount="" Body="&lt;p&gt;What You are looking for is named &quot;mutual inductance&quot;. &#xA;When You google for that You will get a lot of links,&#xA;this one looks very promising:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.google.de/url?sa=t&amp;amp;source=web&amp;amp;cd=3&amp;amp;ved=0CDsQFjAC&amp;amp;url=http%3A%2F%2Fweb.mit.edu%2F8.02t%2Fwww%2F802TEAL3D%2Fvisualizations%2Fcoursenotes%2Fmodules%2Fguide11.pdf&amp;amp;ei=P2CwTaKgMNHDswaSye3iCw&amp;amp;usg=AFQjCNGO_M43ZU9ftwT2gdOjEUSzeV5gBg&quot; rel=&quot;nofollow&quot;&gt;http://www.google.de/url?sa=t&amp;amp;source=web&amp;amp;cd=3&amp;amp;ved=0CDsQFjAC&amp;amp;url=http%3A%2F%2Fweb.mit.edu%2F8.02t%2Fwww%2F802TEAL3D%2Fvisualizations%2Fcoursenotes%2Fmodules%2Fguide11.pdf&amp;amp;ei=P2CwTaKgMNHDswaSye3iCw&amp;amp;usg=AFQjCNGO_M43ZU9ftwT2gdOjEUSzeV5gBg&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question of mutual rotation angle is very basic, it is just &#xA;the cos of the angle. (zero ° for collinear coil axes) &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-21T16:58:27.917" CommentCount="2" />
  <row Id="8972" PostTypeId="2" ParentId="8954" CreationDate="2011-04-21T17:42:34.577" Score="4" ViewCount="" Body="&lt;p&gt;I would add to @anna's answer that $p\bar{p}$ collider such as the Tevatron is CP symmetric. This was one of the arguments for continuing the Tevatron Experiments. To quote from the proposal&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Measurements that get a special&#xA;  advantage from the p-pbar environment.&lt;/strong&gt;&#xA;  The  primary example in this category&#xA;  is CP-violation, which strongly limits&#xA;  the range of allowed  models of new&#xA;  physics up to scales of several TeV.&#xA;  There are good a priori reasons to&#xA;  expect the  existence of some non-SM&#xA;  CP-violating processes, and finding&#xA;  them is of comparable  importance to&#xA;  addressing electroweak symmetry&#xA;  breaking. Precision measurements at&#xA;  the 1%  level or better are accessible&#xA;  at the Tevatron due to the&#xA;  CP-symmetric initial state (p-pbar),&#xA;  and  symmetry of the detectors that&#xA;  allow cancellation of systematics.&#xA;  Some of these measurements  already&#xA;  show tantalizing effects, like the&#xA;  recently published di-muon asymmetry&#xA;  result from the  DZero experiment,&#xA;  showing the first indication of a&#xA;  deviation from the Standard Model&#xA;  picture  of CP-violation. Other&#xA;  measurements are exploring a&#xA;  completely new field, as the recent&#xA;  CPVmeasurement with the D 0 mesons at&#xA;  CDF, yielding a substantial&#xA;  improvement in precision with  respect&#xA;  to previous B-factories data. This has&#xA;  provided a proof of feasibility of an&#xA;  exciting program of precision&#xA;  measurement with a unique possibility&#xA;  to find anomalous interactions in &#xA;  up-type quarks. A non CP-related&#xA;  example in this category is the&#xA;  forward-backward asymmetry  in top&#xA;  quark production. Current measurements&#xA;  by both CDF and DZero indicate an&#xA;  asymmetry  above the Standard Model&#xA;  prediction. If this persists with more&#xA;  data, it can be interpreted as  new&#xA;  dynamics. This is not an easy&#xA;  measurement to replicate in a&#xA;  proton-proton environment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.fnal.gov/directorate/Tevatron/Tevatron_whitepaper.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.fnal.gov/directorate/Tevatron/Tevatron_whitepaper.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2151" LastActivityDate="2011-04-21T17:42:34.577" CommentCount="1" />
  <row Id="8973" PostTypeId="2" ParentId="8968" CreationDate="2011-04-21T17:50:37.303" Score="4" ViewCount="" Body="&lt;p&gt;I think it's neater to express the various directions in terms of unit vectors. We can convert to angles at the end. So say that $\hat n_1$ and $\hat n_2$ are unit vectors perpendicular to the axes of the coils, and let $\hat r$ be a unit vector pointing from the center of coil 1 to the center of coil 2. Since the coils are separated by a distance that's large compared to their diameters, I'll assume that each one can be approximated as a magnetic dipole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnetic field produced by coil 1, at the location of coil 2, is &#xA;$$&#xD;&#xA;\vec B \propto {1\over D^3}[3(\hat n_1\cdot\hat r)\hat r-\hat n_1]&#xD;&#xA;$$&#xA;(This is a standard formula for a dipole field. See a textbook such as Griffiths. Since you graciously allowed an arbitrary constant that had to be measured, I'm writing proportionalities and ignoring boring constants).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mutual inductance is determined by the flux through coil 2 due to this field. Because of the large separation, we can assume the field is approximately constant at the coil's location, so this is just $\vec B\cdot\vec A$. So the flux is just proportional to the component of $\vec B$ that's normal to the second coil. That is,&#xA;$$&#xD;&#xA;\Phi\propto \vec B\cdot\hat n_2\propto {1\over D^3}[3(\hat n_1\cdot\hat r)(\hat n_2\cdot \hat r)-\hat n_1\cdot\hat n_2].&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the mutual inductance is proportional to the flux, this is the answer you're looking for. To express it in terms of the angles, I guess we have $\hat n_1\cdot\hat r=\cos a$ and $\hat n_2\cdot\hat r=\cos b$. The dot product $\hat n_1\cdot\hat n_2=\cos c$, where $c$ is the angle between the axes of the coils. &#xA;So I think the final answer is&#xA;$$&#xD;&#xA;k D^{-3}[3\cos a\cos b-\cos c].&#xD;&#xA;$$&#xA;If all of the various axes lie in a plane, then I guess $c=a-b$, in which case the angular dependence is something like $2\cos a\cos b-\sin a\sin b$, but if things can swivel around in all three dimensions, then $c$ is not uniquely determined by the other two.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-21T17:50:37.303" />
  <row Id="8974" PostTypeId="2" ParentId="8968" CreationDate="2011-04-21T17:55:23.103" Score="1" ViewCount="" Body="&lt;p&gt;I vaguely remember learning fifty or sixty years ago that induction is inversely proportional to the cube of the distance.  A glance at p263 of Jackson Classical Electrodynamics second edition confirms that the leading order of an infinite series for both coplanar and coaxial induction is 1/R^3.&lt;br&gt;&#xA;So your simple formula needs  D^-3, not D^-2.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can’t help you with the angle part.&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-04-21T17:55:23.103" />
  <row Id="8975" PostTypeId="1" AcceptedAnswerId="8977" CreationDate="2011-04-21T18:29:59.787" Score="3" ViewCount="276" Body="&lt;p&gt;Problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An object is undergoing simple harmonic motion with period 1.2s and amplitude 0.6m. At $t=0$, the object is at $x=0$. How far is the object from the equilibrium position when $t=0.480$s?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempt at solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used the displacement equation :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x(t)=A\cos(\omega t+\phi)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and also found out what the angular frequency $\omega$ is (5.2rad/s). Then I found that $\phi$ is 0. I plugged my results in the equation and when I looked at the solution they used the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x(t)=A\sin(\omega t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Im not quite sure how I was supposed to know to use this equation instead of the cosine equation that is written on my equation sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="1822" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-04-21T22:08:29.623" LastActivityDate="2011-05-14T13:22:44.907" Title="Simple harmonic motion problem" Tags="&lt;classical-mechanics&gt;&lt;homework&gt;" AnswerCount="2" />
  <row Id="8976" PostTypeId="1" CreationDate="2011-04-21T18:39:47.457" Score="5" ViewCount="304" Body="&lt;p&gt;what does string theory have to say about the trace anomaly in the expectation value of the stress energy tensor of massless quantum fields on a curved background and its interpretation as the emergence of an extra scalar degree of freedom in the semiclassical limit of GR + QFT?&#xA;Is it something that can be proved? that emerges naturally? Or that can be shown to be false?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question has actually two quite distinct parts: 1) the presence of the anomaly and its form and 2) the possibility of interpreting it as the emergence of an extra degree of freedom at the classical level with its own EOM and a specific stress energy tensor satisfying the constraint Trace (T_mu_nu) = some function of the curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot&lt;/p&gt;&#xA;" OwnerUserId="3229" LastEditorUserId="3229" LastEditorDisplayName="" LastEditDate="2011-04-24T05:44:25.813" LastActivityDate="2011-08-23T15:33:42.390" Title="String theory and trace anomaly in semiclassical gravity?" Tags="&lt;gravity&gt;&lt;string-theory&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" />
  <row Id="8977" PostTypeId="2" ParentId="8975" CreationDate="2011-04-21T18:43:33.990" Score="4" ViewCount="" Body="&lt;p&gt;The initial phase $\phi$ should be $-\pi/2$, or in other words, the solution should be $A\sin(\omega t)$ because $\cos(\theta-\pi/2)=\sin(\theta)$. How did you find that $\phi$ is zero? You're supposed to use the fact that $x$ is zero when $t$ is zero. Which is what let's you set $\phi $ to $-\pi/2$ (modulo $\pi$).&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-04-30T18:34:08.623" LastActivityDate="2011-04-30T18:34:08.623" />
  <row Id="8978" PostTypeId="2" ParentId="8622" CreationDate="2011-04-21T19:36:29.760" Score="1" ViewCount="" Body="&lt;p&gt;I'm not sure what a &quot;probability density for transition&quot; is - is it a probability density or a transition amplitude?  I think you mean transition amplitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you sure that $\rho = \psi_1^* \psi_2$ is a transition amplitude?  It looks more like simply an overlap.  How can it be a transition amplitude without any time evolution operator or Hamiltonian?  I believe that a transition amplitude should have an interaction which connects the two states appearing between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I assume that $\psi_1$ and $\psi_2$ separately satisfy Schrodinger's Equation, then the quantity $\psi_1^* \psi_2$ would be a conserved charge with the current calculated above. This &quot;conservation law&quot; is just a statement that the overlap between two states satisfying Schrodinger's Equation will be preserved as the states evolve in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I assume that $\psi_1$ and $\psi_2$ separately satisfy Schrodinger's Equation and the matrix element between the states is 1, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\int d^3 x \ \psi_1^*(x) \psi_2(x)&#xA;\end{equation}  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a transition amplitude, I think.  The probability of transition would then go like the modulus of the amplitude squared.  This is different from a probability density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you conflating two ideas?  Have I missed something?&lt;/p&gt;&#xA;" OwnerUserId="3230" LastActivityDate="2011-04-21T19:36:29.760" CommentCount="1" />
  <row Id="8979" PostTypeId="2" ParentId="8834" CreationDate="2011-04-21T20:57:02.637" Score="2" ViewCount="" Body="&lt;p&gt;What Mark Eichenlaub said. Or, by simply combining your first and second equations with the appropriate values: 2vt (if you're looking for an answer in terms of v and t).  Look at that!  The u cancelled!  That's kind of elegant.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-04-21T20:57:02.637" />
  <row Id="8980" PostTypeId="1" CreationDate="2011-04-21T23:01:05.610" Score="-2" ViewCount="105" Body="&lt;p&gt;A sphere contains all the mass on one side while the other side is empty of matter. Both sides are separated by a strong material, you name it, that will keep the sphere together. Given that the mass side is more massive than the sun or at least very massive, how would it behave in respect to gravity? Would it spin?&lt;/p&gt;&#xA;" OwnerUserId="3037" LastActivityDate="2011-04-22T01:03:59.683" Title="Half-massive sphere in space" Tags="&lt;astrophysics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="8981" PostTypeId="1" AcceptedAnswerId="8994" CreationDate="2011-04-21T23:25:46.717" Score="4" ViewCount="226" Body="&lt;p&gt;I have a two-part question:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;First and foremost: I have been going through the paper by Dijkgraaf and Witten &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.cmp/1104180750&quot; rel=&quot;nofollow&quot;&gt;&quot;Group Cohomology and Topological Field Theories&quot;&lt;/a&gt;. Here they give a general definition for the Chern-Simons action for a general $3$-manifold $M$. My question is if anyone knows of any follow-up to this, or notes about their paper?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To those who know the paper: They say that they have no problem defining the action modulo $1/n$ (for a bundle of order $n$) as $n\cdot S = \int_B Tr(F\wedge F)$ $(mod 1)$, but that this has an $n$-fold ambiguity consisting of the ability to add a multiple of $1/n$ to the action - What do they mean here? Also, later on they re-define the action as $S = 1/n\left(\int_B Tr(F\wedge F) - \langle \gamma^\ast(\omega),B\rangle\right)$ $(mod 1)$ - How does this get rid of the so-called ambiguity? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Basically my question is if anyone can further explain the info between equations 3.4 and 3.5 in their paper. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3233" LastEditorUserId="1469" LastEditorDisplayName="" LastEditDate="2011-04-22T11:52:27.470" LastActivityDate="2011-04-23T03:43:17.457" Title="Group Cohomology and Topological Field Theories" Tags="&lt;mathematical-physics&gt;&lt;topological-order&gt;&lt;topological-field-theory&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="8982" PostTypeId="2" ParentId="8589" CreationDate="2011-04-21T23:51:17.143" Score="0" ViewCount="" Body="&lt;p&gt;&quot;The 10.000 muons per minute per square meter are mainly above GeV energies whereas for atomic sizes one needs keV at best,&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps there is a certain depth in the ocean or underground where the surviving muons have been slowed to average keV energies?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, is there possibly a &quot;slow muon layer&quot; in the ocean where a tiny amount of natural deuterium and protium is being naturally muon-fused?  Would this layer be helium-enriched in a detectable way?&lt;/p&gt;&#xA;" OwnerUserId="3235" LastActivityDate="2011-04-21T23:51:17.143" CommentCount="1" />
  <row Id="8983" PostTypeId="1" AcceptedAnswerId="8990" CreationDate="2011-04-22T00:32:35.973" Score="2" ViewCount="206" Body="&lt;p&gt;The explanation I have heard of the difference between static and dynamic friction is that static friction is stronger because bonds form when one object is put on top of another object and these have to be overcome to get the movement started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a rotating tire, although the point on the ground will be stationary for an instant, it would seem that bonds wouldn't have time to form. So, why isn't the dynamic coefficient of friction used?&lt;/p&gt;&#xA;" OwnerUserId="119" LastActivityDate="2011-05-26T19:01:07.730" Title="Why does a rotating tire use the static, rather than the dynamic coefficient of friction?" Tags="&lt;friction&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="8984" PostTypeId="2" ParentId="8983" CreationDate="2011-04-22T00:48:46.350" Score="0" ViewCount="" Body="&lt;p&gt;You'd use the dynamic friction if you slam on your brakes and start skidding. Until then, your tire is rolling with the road and so the tire rubber isn't moving with respect to the road. Hence with tires rolling, you use static friction.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-04-22T00:48:46.350" CommentCount="3" />
  <row Id="8985" PostTypeId="1" CreationDate="2011-04-22T00:58:57.927" Score="4" ViewCount="65" Body="&lt;p&gt;Hey guys, I'm reading my physics book right now and I have a quick question about transitions in multielectron atom. I am posting a picture as well so you can see what I am talking about. How come the picture doesn't show a transition from the 5p state to the 5s state? I thought that as long as the different between the orbital quantum numbers was 1, then the emission could occur. In this case, 5s has l=0 and 5p has l=1, so the difference is one and I would think the transition would occur. Also in the picture, it doesn't show the transition of 4p to 4s even though the orbital quantum difference is also one. Is the picture wrong?&#xA;&lt;img src=&quot;http://i.stack.imgur.com/IgbO1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3137" LastActivityDate="2011-04-22T01:26:50.803" Title="Is a 5p to 5s transition possible in a multielectron atom?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" />
  <row Id="8986" PostTypeId="2" ParentId="8947" CreationDate="2011-04-22T01:03:32.393" Score="0" ViewCount="" Body="&lt;p&gt;To me, the &quot;paradox&quot; seems to be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An equal amount of water is goes up the yellow supply tube, and then down out the nozzles.&#xA;Equal amounts of water in opposite directions -- so it seems obvious that the net force is zero and it can't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once that central paradox is resolved, then little details like steering and underwater operation are easier to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 2 things that break the &quot;equal and opposite&quot; symmetry, each of them providing some lift:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;thrust from net change in velocity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personal watercraft such as the WaveRunner and the Jet Ski and this JetLev take water in through the intake and then out through the jet nozzle.&#xA;The water going out the nozzle is &lt;em&gt;much&lt;/em&gt; faster than the water coming in the intake,&#xA;so the drag from the intake is more than compensated for by the thrust from the jet nozzle.&#xA;The net difference in speed of the water&#xA;(from still lake water to relatively fast-moving exhaust water moving in one direction)&#xA;leads to a net acceleration of the stuff pushing against the water (in the opposite direction).&#xA;This effect would still happen if the pumping was done solely on the flying unit with a rigid pipe into the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Decoupled thrust on boat vs thrust on backpack&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the pump on the boat is off, the water on the lake &quot;comes up to&quot; the painted water line on the boat -- i.e., that's the point where the downward force on the boat (gravitational force on its own weight) balances the upward buoyancy force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pump on the boat accelerates approximately stationary lake water &lt;em&gt;upwards&lt;/em&gt;.&#xA;The reaction force pushes the boat &lt;em&gt;downwards&lt;/em&gt;, driving the painted water line underwater.&#xA;Eventually the boat settles to a (lower in the water) position where the downward force on the boat (gravitational force on its own weight, plus the reaction force from pushing water up the tube)  balances the upward buoyancy force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, the piping at the top of the takes the high-speed water coming up the tube and turns it around.&#xA;The change in direction of velocity leads to a net acceleration of that piping (and the backpack and pilot to whom it is firmly attached).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a sense, some of the force of gravity on the person is transferred to the boat, pushing the boat deeper into the water as if the person were standing on that boat, even though the person is not directly touching the boat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's vaguely like shooting a water cannon straight up, and balancing a ball centered over that cannon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This system also has similarities to the &lt;a href=&quot;http://en.wikipedia.org/wiki/launch_loop&quot; rel=&quot;nofollow&quot;&gt;&quot;Lofstrom loop&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ice analogy&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It helps me visualize the forces involved if I picture the water as chunks of ice that move around.&lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2011-04-22T01:03:32.393" />
  <row Id="8987" PostTypeId="2" ParentId="8980" CreationDate="2011-04-22T01:03:59.683" Score="1" ViewCount="" Body="&lt;p&gt;If the sphere is &lt;a href=&quot;http://en.wikipedia.org/wiki/Rigid_body&quot; rel=&quot;nofollow&quot;&gt;rigid&lt;/a&gt;, meaning it cannot change size or shape at all, then gravity will have no effect.  Specifically, the sphere will not rotate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distances between all parts of the sphere are fixed, so the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_potential&quot; rel=&quot;nofollow&quot;&gt;gravitational potential&lt;/a&gt;, which depends only on these distances, cannot change.  Therefore there is no way for the sphere to release any gravitational energy and so it cannot move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sphere is not rigid, it may be able to move some, and will oscillate in &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_mode&quot; rel=&quot;nofollow&quot;&gt;its normal modes of oscillation&lt;/a&gt;.  However, due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Momentum#Conservation_of_linear_momentum&quot; rel=&quot;nofollow&quot;&gt;conservation of linear momentum&lt;/a&gt;, the sphere can't go anywhere, and due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Angular_momentum#Conservation_of_angular_momentum&quot; rel=&quot;nofollow&quot;&gt;conservation angular momentum&lt;/a&gt;, it cannot rotate.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-04-22T01:03:59.683" CommentCount="2" />
  <row Id="8988" PostTypeId="2" ParentId="8985" CreationDate="2011-04-22T01:26:50.803" Score="4" ViewCount="" Body="&lt;p&gt;Yes, such transitions are possible.  The book just list the main emission lines below 1800 nm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, some missing lines in that image:&lt;br&gt;&#xA;2206.248 nm,  $2p^64p \rightarrow 2p^64s$&lt;br&gt;&#xA;5428.36 nm, $2p^65p \rightarrow  2p^65s$&lt;br&gt;&#xA;9093.03 nm, $2p^64p \rightarrow 2p^63d$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the book probably just left these longer wavelengths out to not make the image any more cluttered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can search for more lines here:&lt;br&gt;&#xA;&lt;a href=&quot;http://physics.nist.gov/PhysRefData/ASD/lines_form.html&quot; rel=&quot;nofollow&quot;&gt;http://physics.nist.gov/PhysRefData/ASD/lines_form.html&lt;/a&gt;&lt;br&gt;&#xA;Search for &quot;Na I&quot; otherwise you'll get lines for all the sodium ions as well.&lt;/p&gt;&#xA;" OwnerUserId="2652" LastActivityDate="2011-04-22T01:26:50.803" CommentCount="1" />
  <row Id="8990" PostTypeId="2" ParentId="8983" CreationDate="2011-04-22T03:12:39.153" Score="5" ViewCount="" Body="&lt;p&gt;A car drives at 20 m/s.  The circumference of the wheel is 2m, so the rotation rate is 10 Hz.  A reasonable percentage of the tire is in contact with the ground - maybe around 5%.  That would give a contact time of 5*10^-3 s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a pretty long time in molecular terms.  The distance between molecules divided by the speed of light is around 10^-18 s, so that's the fastest we can imagine some sort of bonding occurring.  That 16 orders of magnitude faster than the contact time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Real chemical reactions must be slower, but with 10^16 &quot;clicks&quot; of time for the chemistry to sort itself out, there should be plenty of time for the tire to stick.  &lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-04-22T03:12:39.153" CommentCount="4" />
  <row Id="8991" PostTypeId="1" AcceptedAnswerId="8992" CreationDate="2011-04-22T04:53:36.380" Score="0" ViewCount="40" Body="&lt;p&gt;So I am looking through my book and it says &quot;.... the order of the excited states is exactly the same order (3p-4s-3d-4p)&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now I am looking at a question in the book and it says &quot;Is 3d to 4s transition possible? Why or why not?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer to this question is: No it can't be because it doesn't abide by the selection rule because the difference in the orbital quantum numbers is 2 instead of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if this is the case, why does the above statement have the oder of (3p-4s-3d-4p) if we know there cannot be a transition between the 4s and 3d state like this order says?&lt;/p&gt;&#xA;" OwnerUserId="3137" LastActivityDate="2011-04-22T04:59:39.243" Title="Quick question on the ionization energy and the selection rule" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" />
  <row Id="8992" PostTypeId="2" ParentId="8991" CreationDate="2011-04-22T04:59:39.243" Score="4" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Order of excited states&lt;/em&gt; means ordered by their energy: 3p is lower than 4s is lower than 3d is lower than 4p. As you correctly point out, the transition from 4s to 3d is forbidden due to a selection rule concerning the angular momentum. So, 3d being higher in energy than 4s has nothing to with there being or not being a transition between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Nothing is ever &lt;em&gt;really&lt;/em&gt; forbidden. You might not have a dipole transition from 4s to 3d, but you might have an electric quadrupole or magnetic dipole or higher multipole transition. However, these have much lower intensity.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-04-22T04:59:39.243" CommentCount="1" />
  <row Id="8994" PostTypeId="2" ParentId="8981" CreationDate="2011-04-22T05:36:34.467" Score="3" ViewCount="" Body="&lt;p&gt;First, the full paper is here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://citeseer.ist.psu.edu/viewdoc/download;jsessionid=807BE383780883ACB4CAB8BD48E8C90B?doi=10.1.1.128.1806&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;http://citeseer.ist.psu.edu/viewdoc/download;jsessionid=807BE383780883ACB4CAB8BD48E8C90B?doi=10.1.1.128.1806&amp;amp;rep=rep1&amp;amp;type=pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second, the paper has 150 citations. See all this information at INSPIRE (updated SPIRES): &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://inspirebeta.net/record/278923?ln=en&quot; rel=&quot;nofollow&quot;&gt;http://inspirebeta.net/record/278923?ln=en&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Third, the text between 3.4 and 3.5 looks totally comprehensible. At that point, they are able to define $n\cdot S$ modulo 1, which is equivalent to defining the action $S$ modulo $1/n$. The goal is to define the action $S$ itself modulo 1; I suppose that their normalization of the path integral has to have $\exp(2\pi i S)$ with the atypical $2\pi$ factor. Yes, confirmed, it's equation 1.2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you shift the action by 1 - or $2\pi$ in the ordinary conventions - it doesn't change the integrand of the path integral; it doesn't change the physics. So quite generally, if one is able to say that the action $S$ is equal to $S_0+n$ (or $2\pi n$ normally) for some integer $n$, he knows everything about the physics of the action he needs; shifting it by an integer doesn't change anything. That's why, in fact, the action is often defined modulo 1 only (up to the addition of an integer multiple of 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's enough to know the &quot;fractional part&quot; of the action; the integer part is irrelevant. However, at the point of the equation 3.4, their uncertainty is larger than that: they only know the action modulo $1/n$. For example, if the action is $9.37$ modulo $1/2$, it means that the fractional part may be $0.37$ but it may also be $0.87$. These two values of $S$ would change the physics because the contribution of the configuration to the path integral changes the sign if one changes $S$ by $1/2$ (in normal conventions, by $\pi$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one only knows $S$ modulo $1/n$, and if he thinks it's $S_0$ - in this case, the $F\wedge F$ expression - it means that the real action is &#xA;$$ S = S_0 + K/n $$&#xA;and the integer $K$ has to be determined. Because the change of the action $S$ by an integer doesn't change physics, it doesn't matter if $K$ in the equation above is changed by a multiple of $n$. So the goal is to find the right $K$ to define the action - and $K$ is an unknown integer defined (or relevant) modulo $n$, i.e. up to the addition of an irrelevant and arbitrary multiple of $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some point, they find the right answer and it is&#xA;$$ K = -\langle \gamma^*(\omega),B\rangle $$&#xA;which removes the ambiguity of $S$ - the missing knowledge whether $S$ should be the original $S$ or higher or smaller by a particular multiple of $1/n$. If you don't understand the text above, then apologies, I have no way to find out why, so I can't give you a better answer unless you improve your question.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-22T05:42:45.977" LastActivityDate="2011-04-22T05:42:45.977" CommentCount="2" />
  <row Id="8995" PostTypeId="2" ParentId="8830" CreationDate="2011-04-22T08:22:13.470" Score="3" ViewCount="" Body="&lt;p&gt;The answer to your question is &quot;no&quot;.  A quick web search for &quot;ping-pong vacuum&quot; yields a wealth of experimental evidence, in the form of articles with photographs about cannons that operate by putting ping-pong balls in vacuum and selectively releasing the vacuum.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.google.co.jp/search?tbm=vid&amp;amp;q=ping-pong%20%22vacuum%20cannon%22&quot; rel=&quot;nofollow&quot;&gt;Here are some videos of demonstrations&lt;/a&gt;.  Note that the balls don't pop before being fired.&lt;/p&gt;&#xA;" OwnerUserId="761" LastActivityDate="2011-04-22T08:22:13.470" CommentCount="2" />
  <row Id="8996" PostTypeId="2" ParentId="8930" CreationDate="2011-04-22T09:30:08.130" Score="1" ViewCount="" Body="&lt;p&gt;(This is not really an answer, but here I have not yet enough reputation to post comments. If someone wants to move this to a comment, I won't object.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1976 is not a particularly early date for computer calculations: &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermi%E2%80%93Pasta%E2%80%93Ulam_problem&quot; rel=&quot;nofollow&quot;&gt;Fermi, Pasta, and Ulam&lt;/a&gt; used computer simulations in the early 50s for their 1955 paper.&lt;/p&gt;&#xA;" OwnerUserId="2651" LastActivityDate="2011-04-22T09:30:08.130" />
  <row Id="8997" PostTypeId="2" ParentId="8942" CreationDate="2011-04-22T10:06:40.103" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://arxiv.org/&quot; rel=&quot;nofollow&quot;&gt;arXiv&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though it is not a journal in the classical meaning (it is a repository of preprints), to some extent it serves as such. So when you care for the open-access, advisable it to upload your paper to arXiv and then send to a peer-review journal (open or not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sending a paper to arXiv give also other benefits - it is (almost) immediately accessible to the scientific community. So it both boosts the research process and prevents form being scooped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And out of my personal philosophy - publicly funded research should be released openly, not locked in private hands.)&lt;/p&gt;&#xA;" OwnerUserId="184" LastActivityDate="2011-04-22T10:06:40.103" CommunityOwnedDate="2011-04-22T22:24:35.010" CommentCount="3" />
  <row Id="8999" PostTypeId="1" CreationDate="2011-04-22T11:54:52.570" Score="8" ViewCount="375" Body="&lt;p&gt;Several years ago, &lt;a href=&quot;http://arxiv.org/abs/hep-th/0610241&quot; rel=&quot;nofollow&quot;&gt;noncommutative geometry was used to describe the standard model&lt;/a&gt;, somehow yielding a prediction of 170 GeV for the mass of the Higgs boson, a prediction which was &lt;a href=&quot;http://noncommutativegeometry.blogspot.com/2008/08/irony.html&quot; rel=&quot;nofollow&quot;&gt;falsified a few years later&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, &lt;a href=&quot;http://dorigo.wordpress.com/2008/03/29/115-gev-higgs-is-evidence-piling-up/&quot; rel=&quot;nofollow&quot;&gt;for a long time&lt;/a&gt; there have been indications of a Higgs boson at 115 GeV. As it happens, 115 GeV and 170 GeV form somewhat natural &lt;a href=&quot;http://motls.blogspot.com/2006/10/alain-connes-et-al-predictions-for.html&quot; rel=&quot;nofollow&quot;&gt;theoretical bounds on the mass of the Higgs&lt;/a&gt;: above 170 GeV, the theory would develop a &lt;a href=&quot;http://en.wikipedia.org/wiki/Landau_pole&quot; rel=&quot;nofollow&quot;&gt;Landau pole&lt;/a&gt;, and below 115 GeV, &lt;a href=&quot;http://motls.blogspot.com/2010/07/what-light-higgs-would-mean-for.html&quot; rel=&quot;nofollow&quot;&gt;the electroweak vacuum would become unstable&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm wondering whether the mechanism or the logic behind the original prediction can be reversed, so as to drive the Higgs mass to the lower bound, rather than to the upper bound, in an altered version of the noncommutative standard model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: I am nowhere near decoding how the prediction of 170 GeV was made, but simple algebra shows that it is amazingly easy to get very close to 115 GeV instead, by adjusting some of the penultimate quantities appearing in the calculation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://resonaances.blogspot.com/2011/04/higgs-at-atlas-maybe.html&quot; rel=&quot;nofollow&quot;&gt;A comment at Resonaances&lt;/a&gt; points out that $\sqrt{2} m_W$ is close to 115 GeV (it's a little over 113 GeV). In equation 5.15 of hep-th/0610241, we see this formula: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_H = \sqrt{2\lambda}\frac{2M}{g} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On page 36 (section 4.1), we read that $M = m_W$ and $g = \sqrt{4\pi\alpha}$. So for $m_H$ to be approximately 115 GeV, we need $\lambda = \pi\alpha$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In equation 5.10, we have that lambda-tilde is approximately $4/3 \pi\alpha_3$, and in remark 5.1 we read that &quot;the factor of 4/3 in (5.10) should be corrected to 1&quot;. So we have $\tilde\lambda = \pi\alpha_3$. Almost what we need - it's just a different lambda and a different alpha! :-) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I'll point out again that 115 GeV is a special value theoretically: it's in a narrow range of values for m_H for which, given the measured value of m_top, the vacuum of the minimal standard model is metastable (see figure 13, &lt;a href=&quot;http://arxiv.org/abs/0704.2232&quot; rel=&quot;nofollow&quot;&gt;arxiv:0704.2232&lt;/a&gt;). So I can't believe any of this is coincidence. &lt;/p&gt;&#xA;" OwnerUserId="1486" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-31T15:52:06.013" LastActivityDate="2011-08-31T15:52:06.013" Title="115 GeV, 170 GeV, and the noncommutative standard model" Tags="&lt;particle-physics&gt;&lt;standard-model&gt;&lt;higgs-boson&gt;" CommentCount="4" FavoriteCount="1" />
  <row Id="9000" PostTypeId="1" CreationDate="2011-04-22T12:28:42.823" Score="3" ViewCount="297" Body="&lt;p&gt;Why is mass the quadratic term in a Lagrangian?&lt;/p&gt;&#xA;" OwnerUserId="3240" LastActivityDate="2011-04-24T22:09:43.723" Title="Lagrangians and mass" Tags="&lt;quantum-field-theory&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="9001" PostTypeId="2" ParentId="9000" CreationDate="2011-04-22T13:19:11.280" Score="3" ViewCount="" Body="&lt;p&gt;Take the Lagrangian with only the kinetic and quadratic terms. Find the wave equation and propagator and you'll see that they correspond to those of a free relativistic particle with mass squared equal to the quadratic coefficient. So the interpretation is correct in free field theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding higher order terms doesn't change this interpretation, since the higher order terms correspond to interactions.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-04-22T13:25:13.080" LastActivityDate="2011-04-22T13:25:13.080" />
  <row Id="9002" PostTypeId="2" ParentId="9000" CreationDate="2011-04-22T13:20:47.207" Score="3" ViewCount="" Body="&lt;p&gt;I'm not sure if this is the sort of answer you want, but here goes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass, by definition, is the thing in $F=ma$. That is, it's the coefficient in front of the acceleration in the equation of motion. If you build a Lagrangian out of terms involving a coordinate $x$ and its derivative $\dot x$, and then work out the Euler-Lagrange equations, the only way to get a term of the form $m\ddot x$ at the end is to start with a term of the form ${1\over 2}m\dot x^2$ in the Lagrangian. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the classical-mechanics answer. If you were looking for a field-theory answer instead, the details are different, but the idea is the same. Think about how you expect mass to show up in the field equation, and work out what term in the original Lagrangian will give rise to the appropriate expression when you use the Lagrangian to derive the field equation.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-22T13:20:47.207" />
  <row Id="9004" PostTypeId="2" ParentId="9000" CreationDate="2011-04-22T14:24:29.157" Score="1" ViewCount="" Body="&lt;p&gt;For the free quantum field, the quadratic term determines that the energy-momentum 4-vector observable of a single-particle state in a single fourier mode would be on the forward mass-shell instead of on the forward light-cone. That is, $E^2+\mathbf{p}^2=m^2$ is a constant, which is algebraically of the same form as the relativistic mass-energy-momentum relationship. [Note, however, that fourier modes are strictly speaking an improper limit in quantum field theory.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again for the free quantum field, but in terms of QFT understood as a stochastic formalism for field observables instead of in terms of a particle interpretation, the two-point correlation function $C(x-y)=\left&amp;lt;0\right|\hat\phi(x)\hat\phi(y)\left|0\right&amp;gt;$ satisfies the classical Klein-Gordon field equation, where we can discuss energy-momentum density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More abstractly, one could take mass, or inertia, to denote a generalized &quot;resistance&quot; of a given simple evolution to a given deformation of the background or to the introduction of a new force law. In these terms, every constant term in a Lagrangian is part of a multi-dimensional description of what the response of an initial system would be to any given additional term in the Lagrangian. The mass term is singled out as the lowest order component of the inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There seem to me to be a surprising number of ways in which this Question can be taken. I decided only after a while that the Question is useful enough to obviate the lack of clarity. For me it's only because of the QFT tag that the Question becomes productive.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-04-22T14:24:29.157" />
  <row Id="9005" PostTypeId="1" AcceptedAnswerId="9007" CreationDate="2011-04-22T14:51:56.560" Score="0" ViewCount="186" Body="&lt;p&gt;Why does $E=mc^2$ give results in Joules ? Why didn't Einstein need to create a separate unit and how come using two standard units (speed of light in km.s and mass in grams) give another result in Joules ?&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-04-22T16:18:30.000" LastActivityDate="2011-04-22T16:18:30.000" Title="Why does $E=mc^2$ give results in Joules?" Tags="&lt;si-units&gt;&lt;unit-conversion&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="9006" PostTypeId="2" ParentId="9005" CreationDate="2011-04-22T15:18:13.800" Score="2" ViewCount="" Body="&lt;p&gt;Because of the definition of Joules in terms of the base kg/m/s units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using km and grams doesn't give Joules, the speed is squared so you would need to use units of mass that are 1000^2 smaller to cancel, ie speed in km/s and mass in milli-grams&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-04-22T15:18:13.800" />
  <row Id="9007" PostTypeId="2" ParentId="9005" CreationDate="2011-04-22T15:23:01.560" Score="5" ViewCount="" Body="&lt;p&gt;It's an energy.  The SI unit for energy is a Joule.  If the theory is going to predict an energy-mass correspondance, then it better give the energy in units of energy.  If the units of $mc^{2}$ didn't work out, for the equation to make any sense, you'd have to include some constant $\alpha$ so that $E=\alpha mc^{2}$ would have units of energy on both sides.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-04-22T15:23:01.560" />
  <row Id="9008" PostTypeId="1" CreationDate="2011-04-22T16:45:30.783" Score="0" ViewCount="65" Body="&lt;p&gt;A venturimeter is to be fitted to a 250mm dia pipe in which the max flow rate is 7200l/min and the pressure head is 6m of water.What is the min dia of throat so that there is no negative head in it?&lt;/p&gt;&#xA;" OwnerUserId="3242" LastActivityDate="2011-04-22T16:45:30.783" ClosedDate="2011-04-22T22:27:47.830" Title="venturimeter and head in it" Tags="&lt;fluid-dynamics&gt;" CommentCount="5" />
  <row Id="9009" PostTypeId="2" ParentId="8622" CreationDate="2011-04-22T17:49:33.037" Score="1" ViewCount="" Body="&lt;p&gt;Even the scalar product $\langle\psi_1|\psi_2\rangle$ is not a &lt;em&gt;transition&lt;/em&gt; probability but an amplitude to find the system in one state when it is know that it is in another. And the wave function product is just a part of this amplitude. It's like a single term of a scalar product of two vectors $(\vec{a}\vec{b})$: $a_i b_i$, $i$ = fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &quot;Transition&quot; implies evolution in time. It is not the case here. It is amplitude of &quot;presence&quot;. Although the wave functions may evolve in time (each is a non trivial superpositions of eigenstates with their &quot;energetic&quot; exponentials) the time evolution of each state is determined with the same Hamiltonian, so the product $\psi_1 (t)^*\psi_2 (t) = \psi_1 (0)^*\psi_2 (0)$ due to $U(t)^+U(t)=1$.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-04-22T22:38:58.280" LastActivityDate="2011-04-22T22:38:58.280" />
  <row Id="9010" PostTypeId="1" AcceptedAnswerId="9026" CreationDate="2011-04-22T18:20:34.513" Score="7" ViewCount="167" Body="&lt;p&gt;One possible outcome of the collision experiments at LHC is the discovery of new elementary particles with large mass. Is there a theoretical way to derive an upper bound on the mass of elementary particles? For example, we have the electron, the muon and the tau particle. Can we be sure that there are no heavier elementary fermions with charge $e$? (Maybe there are some symmetry arguments?)&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-04-23T14:38:06.507" Title="Is there a theoretical upper bound on the mass any new particles can have?" Tags="&lt;particle-physics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9011" PostTypeId="2" ParentId="8074" CreationDate="2011-04-22T18:52:51.800" Score="1" ViewCount="" Body="&lt;p&gt;Here I would like to numerically check the theoretical prediction of a factor $\frac{2}{5}$ in difference from Mark Eichenlaub's original monopole argument. In practice, this means calculating the difference in the gravitational potential between the North pole and the Equator, and divide by the corresponding difference in monopole potentials. For numerical reasons, it is better in practice to calculate the inverse (=reciproc) fraction, which then should be compared with $\frac{5}{2}$. Since my programming skills are limited, I have just written a slow MAPLE code to do the job.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; b:=100; f:=.10; a:=b*(1+f); V1:=evalf(4*Pi*a^2*b/3); &#xA; xa:=a; ya:= 0; za:=0; xb:=0; yb:=0; zb:=b; &#xA; U1a := evalf(V1/sqrt(xa^2 + ya^2 + za^2)); &#xA; U1b := evalf(V1/sqrt(xb^2 + yb^2 + zb^2));&#xA;&#xA; Ua:=0;Ub:=0;V:=0;&#xA; for x from -a-.5 by 1 to a+.5 do &#xA; for y from -a-.5 by 1 to a+.5 do &#xA; for z from -b-.5 by 1 to b+.5 do &#xA; if (x/a)^2 + (y/a)^2 + (z/b)^2 &amp;lt; 1 then &#xA; Ua:=Ua + 1/sqrt((x-xa)^2 + (y-ya)^2 + (z-za)^2); &#xA; Ub:=Ub + 1/sqrt((x-xb)^2 + (y-yb)^2 + (z-zb)^2); &#xA; V:=V+1;&#xA; end if;od;od;od;&#xA;&#xA; b;f;Ua;U1a;Ub;U1b;V;V1;Ub-Ua;U1b-U1a;(U1b-U1a)/(Ub-Ua);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The result with polar radius $b:=100$ was&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{array}{cc} &#xD;&#xA;\mathrm{Flatness}&amp;amp;\mathrm{Inverse \ fraction}  \\&#xD;&#xA;f &amp;amp; \frac{U_{1b}-U_{1a}}{U_b-U_a} \\ \\&#xD;&#xA;0.10 &amp;amp; 2.5549 \\&#xD;&#xA;0.05 &amp;amp; 2.5353 \\&#xD;&#xA;0.03 &amp;amp; 2.5149 \\&#xD;&#xA;0.02 &amp;amp; 2.5085 \\&#xD;&#xA;0.01 &amp;amp; 2.5088 \\&#xD;&#xA;\end{array}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the estimate does not improve from flatness $0.02$ to $0.01$ is a lattice artifact, because the lattice spacing is of same order as the difference in radius between the North pole and the Equator. &lt;/p&gt;&#xA;" OwnerUserId="2451" LastActivityDate="2011-04-22T18:52:51.800" />
  <row Id="9012" PostTypeId="1" CreationDate="2011-04-22T19:37:26.437" Score="2" ViewCount="167" Body="&lt;p&gt;The paradox I envision involves two objects that exist in very different gravitational potentials -- one very high (eg: close to a black hole) and one very low (eg: far away from a black hole).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, imagine these objects are actually little spaceships that are blasting their engines so that they remain a constant distance away from the black hole.  Also, imagine they are lined up with each other (the black hole, the near spaceship, and the far spaceship form a line).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be possible to lower a string from the further spaceship to the spaceship which is closer?  &lt;strong&gt;That is, will the distance between the two spaceships be measured to be a constant?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If it's not..&lt;/em&gt; please explain this!  It seems odd that the distance between the two ships wouldn't be constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If it is&lt;/em&gt;, then it seems like there would be a paradox.  First, let's label the spaceship nearer to the blackhole &quot;A&quot; and the one further &quot;B&quot;.  Next, let's say the string's length is exactly 1 light second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's start a light clock between &quot;A&quot; and &quot;B&quot; -- &quot;A&quot; will send a light pulse to &quot;B&quot;, and upon receiving this, &quot;B&quot; will send a light pulse to &quot;A&quot;, and so forth.  Since both &quot;A&quot; and &quot;B&quot; know that their distance is 1 light second apart, each time they receive the pulse, they will add 2 seconds to their clock.  While we're at it, they might as well send along their recorded time with the light pulse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paradox is that from the perspective of B viewing A, A should appear to be moving in &quot;slow motion&quot; -- everything will be redshifted, and everything will transpire slower.  From the perspective of A viewing B, B will be blueshifted and moving in &quot;fast motion&quot;.  Yet, this cannot be the case, because both will be receiving ticks at 2 second intervals, and each tick will represent 2 seconds of elapsed time on the other party.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what gives?  I'm assuming you cannot have a constant distance between two objects if their gravitational potentials are different.  But that just seems very strange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  It would be very helpful if the answers could give an example of what it would be like to be onboard A and onboard B.  If you're on A, and you send the pulse, how much time will you measure before you hear back from B?  If you're on B, and you send the pulse, how much time will you measure before you hear back from A?&lt;/p&gt;&#xA;" OwnerUserId="3243" LastEditorUserId="3243" LastEditorDisplayName="" LastEditDate="2011-04-22T21:13:23.670" LastActivityDate="2011-04-22T21:13:23.670" Title="General Relativity Paradox: Holding a string across a gravitational gradient" Tags="&lt;general-relativity&gt;&lt;gravity&gt;&lt;paradox&gt;" AnswerCount="2" />
  <row Id="9013" PostTypeId="2" ParentId="9012" CreationDate="2011-04-22T20:22:41.780" Score="5" ViewCount="" Body="&lt;p&gt;It certainly is possible for two observers to remain at rest near a black hole, as long as they are both outside the horizon. They can stretch a string from one to the other, and they can bounce light signals back and forth to each other as you describe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not true, however, that the elapsed time as measured by either observer between light signals will be 2 seconds (for a one-light-second string length). The reason is that, in curved spacetime, the very definitions of things like length, time, and speed only make sense (or to be precise only correspond to our special-relativity notions) &lt;em&gt;locally&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, even a rule like &quot;light travels at speed $c$&quot; only makes sense locally in general relativity. Here &quot;locally,&quot; to be precise, means &quot;in an infinitesimal region,&quot; although in practice you can stretch it to mean &quot;in a small region,&quot; where &quot;small&quot; is defined relative to the length scales associated with curvature of spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine stationing a bunch of observers all along the length of the string. If each observer uses his own rulers and clocks to measure the speed of a light pulse as it goes by, each will find that its speed is $c$ as expected. (These are &quot;local&quot; measurements.) But each observer's clocks are ticking at a different rate (as measured by a distant observer), so it's not the case that the elapsed time, as measured by A (or B) equals the sum of the elapsed times as measured by all of the other observers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over a long path (long enough to be &quot;not local&quot;), the speed of light is not necessarily $c$. In fact, the speed (of light or anything else) is not even well-defined in this situation! The reason is that in curved spacetime measurements of distances and times depend on the choice of coordinates one adopts. In flat spacetime, there are a bunch of natural coordinate systems to use, which we call &quot;inertial reference frames.&quot; With these coordinates, the special relativity rules apply. But when spacetime is curved, there are no inertial frames anymore (except locally). In fact, the nonexistence of inertial frames is one way do &lt;em&gt;define&lt;/em&gt; what we mean by &quot;curved spacetime.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-22T20:22:41.780" CommentCount="12" />
  <row Id="9014" PostTypeId="2" ParentId="9012" CreationDate="2011-04-22T20:25:45.280" Score="2" ViewCount="" Body="&lt;p&gt;Dear Sam, it's just not true that they will both receive ticks in 2-second intervals. I think that the assumption you used to derive this incorrect conclusion was &quot;flatness of the whole spacetime&quot;, which is nothing else than the denial of the basic point of the whole general theory of relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that you just assumed that the distance is $x=ct$ but this is only valid for a uniform motion in a flat spacetime or, equivalently, for infinitesimal pieces of the trajectory (because infinitesimal volumes of spacetime are flat at an arbitrarily good accuracy). But there's surely no law that would say that in a general curved spacetime, some integrated proper distance $s$ is equal to $ct$ where $t$ is some proper time measured along a different path in spacetime. Different paths in a general curved spacetime have different proper lengths or proper times - and the only case in which the simple linear relationship may hold is the case when the spacetime is linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proper times measured at places &quot;A&quot; and &quot;B&quot; in the static configuration that you defined are related exactly by the multiplicative factor - the ratio of the red shifts - and you seem to know very well what it is. So it's surprising that you also want to convince yourself that it is not the case even though you don't have any rational reason to think it is not the case.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-22T20:25:45.280" CommentCount="4" />
  <row Id="9015" PostTypeId="2" ParentId="3824" CreationDate="2011-04-22T20:33:14.920" Score="0" ViewCount="" Body="&lt;p&gt;You write &quot;rather than the electrical force between you and the earth&quot;. There is no electrical force between you and the earth because neither of you is charged! But you and the earth do both have mass so gravity acts but electricity doesn't. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;that the electrical force is so much stronger than gravitational force at atomic levels'&quot; is only true between two charged particles of say the mass of a proton and the charge of a proton.&lt;/p&gt;&#xA;" OwnerUserId="3244" LastActivityDate="2011-04-22T20:33:14.920" />
  <row Id="9016" PostTypeId="2" ParentId="6086" CreationDate="2011-04-22T22:05:37.333" Score="0" ViewCount="" Body="&lt;p&gt;It is impossible to have part of the rigid rod being under the horizon and part above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically on the horizon itself the speed of rod's movement becomes the speed of light. This means the linear contraction of length of the rod to zero for a stationary observer. Thus the rod crosses the horizon WHOLE AT ONCE and after this moment its speed becomes greater than speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this also means that the rod cannot be made of substance or carry any information since information cannot be transferred faster than light (in actual world the BH will evaporte earlier than any object can approach the horizon, to cross the horizon one would have to move towards it faster than light with only light rays in theory can reach the horizon exactly at the last moment of the BH's existence).&lt;/p&gt;&#xA;" OwnerUserId="1186" LastEditorUserId="1186" LastEditorDisplayName="" LastEditDate="2011-04-22T22:11:06.407" LastActivityDate="2011-04-22T22:11:06.407" CommentCount="18" />
  <row Id="9017" PostTypeId="1" AcceptedAnswerId="9060" CreationDate="2011-04-22T23:51:17.590" Score="2" ViewCount="170" Body="&lt;p&gt;If I put next to my cooler another similar cooler that produces similar white noise, will the overall noise level increase?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to point out that I am speaking about adding another independent, non-correlated white noise source, not just another pair of speakers that produce sound of the same phase and curve. It is evident for me that adding another speaker will increase amplitude twofold compared to one speaker, but this is not evident for coolers which produce non-correlated sound.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastEditorUserId="1186" LastEditorDisplayName="" LastEditDate="2011-04-23T10:37:33.580" LastActivityDate="2011-04-24T11:46:40.970" Title="Does a second similar source of white noise increase the overall noise level?" Tags="&lt;acoustics&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="9018" PostTypeId="2" ParentId="9017" CreationDate="2011-04-23T00:00:44.630" Score="0" ViewCount="" Body="&lt;p&gt;What you very probably have in mind is noise cancelling. If you have two sounds which are exactly the same, but one of them is inversed (phase and anti-phase) and you play them together, they cancel each other out. The problem is that having two exactly same sounds in real life is almost impossible, without using any microphones and/or digital processing that is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, to have two sounds with aforementioned qualities cancelled out, you'd need to listen to them together perfectly, that means they'd need to point to your ears in a way so that you listen to the phase without any offset to any of them. This is, again, in real life, impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your idea is based on something solid, but it can't work in your case, because (to sum it up):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the two sounds of the coolers are different&lt;/li&gt;&#xA;&lt;li&gt;the two sounds of the coolers are not in an inversed phase to each other&lt;/li&gt;&#xA;&lt;li&gt;the two sounds of the coolers aren't pointing perfectly to your ears :)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore, your two coolers cannot cancel each other out, nor can they together produce the same level of volume. The volume of course increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sengpielaudio.com/calculator-levelchange.htm&quot; rel=&quot;nofollow&quot;&gt;sengpielaudio.com&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;20 dB gain modification should give the ratio of 4 (four times) for sensed volume and loudness&lt;br&gt;&#xA;  20 dB gain modification gives the ratio of 10 for measured voltage and sound pressure and&lt;br&gt;&#xA;  20 dB gain modification gives the ratio of 100 for calculated sound power and acoustic intensity&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Doubling of the volume (loudness) should be felt by a level difference of 10 dB&lt;br&gt;&#xA;  Doubling the sound pressure (voltage) corresponds to a measured level change of 6 dB&lt;br&gt;&#xA;  Doubling of acoustic power (sound intensity) corresponds to a level change of 3 dB&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;3 dB = twice the power (Power respectively intensity - mostly calculated)&lt;br&gt;&#xA;  6 dB = twice the amplitude (Voltage respectively sound pressure - mostly measured)&lt;br&gt;&#xA;  10 dB = twice the perceived volume or twice as loud (Loudness nearly sensed - psychoacoustics)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2834" LastEditorUserId="2834" LastEditorDisplayName="" LastEditDate="2011-04-23T00:09:23.340" LastActivityDate="2011-04-23T00:09:23.340" />
  <row Id="9019" PostTypeId="2" ParentId="9000" CreationDate="2011-04-23T00:11:50.010" Score="2" ViewCount="" Body="&lt;p&gt;The mass is a quadratic term in a Lagrangian because it is the Lagrangian &lt;strong&gt;density&lt;/strong&gt;, the Lagrangian per unit of volume. The Lagrangian itself is obtained by integrating over the volume which contains the field. The &quot;classical&quot; relation is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H ~-~pv~=~-L$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These integrated quantities, the Hamiltonian $H$, the Lagrangian $L$ and the term $pv$ transform like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;   \begin{array}{lll}&#xD;&#xA;   H  &amp;amp;  \mbox{transforms as} &amp;amp; \gamma        \\&#xD;&#xA;   pv &amp;amp;  \mbox{transforms as} &amp;amp; \beta^2\gamma \\&#xD;&#xA;   L  &amp;amp;  \mbox{transforms as} &amp;amp; 1/\gamma      \\&#xD;&#xA;   \end{array}&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volume of a wave-functions transforms like $1/\gamma$ due to Lorentz contraction. So, the densities become higher by a factor $\gamma$, hence the Hamiltonian density ${\cal H}$, the Lagrangian density ${\cal L}$ and the density of the $pv$ term transform like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$   \begin{array}{lll}&#xD;&#xA;   \mathcal{H}                   &amp;amp;  \mbox{transforms as} &amp;amp; \gamma^2          \\&#xD;&#xA;   \mathcal{pv}  &amp;amp;  \mbox{transforms as} &amp;amp; \beta^2\gamma^2   \\&#xD;&#xA;   \mathcal{L}                  &amp;amp;  \mbox{transforms as} &amp;amp; &amp;quot;1&amp;quot; \\&#xD;&#xA;   \end{array}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding relation of the densities transforms like the basic energy/momentum relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;    {\cal H}\ -\ {\cal pv}\ =\ -{\cal L} \qquad \mbox{transforms as} \qquad E^2-p^2\ =\ m^2&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that the Lagrangian density is the same in all reference frames. It is a Lorentz scalar. This makes the Lagrangian density a fundamental quantity in quantum field theory. The Standard Model of physics is based on the Lagrangian density which in quantum physics is generally called just the Lagrangian, without the density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;equations of motion&quot; are based on the derivatives of the Lagrangian density which is a Lorentz scalar. This assures that the whole Standard Model of physics transforms in the right way, that is, the laws of physics are the same in every reference frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards, Hans&lt;/p&gt;&#xA;" OwnerUserId="3197" LastActivityDate="2011-04-23T00:11:50.010" CommentCount="2" />
  <row Id="9020" PostTypeId="2" ParentId="9017" CreationDate="2011-04-23T00:15:19.243" Score="3" ViewCount="" Body="&lt;p&gt;The white noise of both coolers will add up and appear to be one white noise with stronger intensity than the white noise of each individual cooler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that the white noise produced by each cooler can be seen as a sound wave that has a large amount of &quot;random walk&quot; behaviour to it.  The two similar white noise sources will both have a &quot;random walk&quot; behaviour but as they are just similar but otherwise not related to each other, they can both (at some points in time) cancel each other or (at some other points in time) amplify each other in some direction.  When they work against each other and when they &quot;cooperate&quot; is, well, random.  So the result when combining them is essentially... white noise.  However, as they sometimes &quot;cooperate&quot; and result in wave spikes in the same direction, you end up having bigger spikes and alterations back and forth in the combined white noise than in the individual white noises, so you end up with a more intense white noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(IIRC, the formal definition of white noise is such that if you add two white noise waves with the same frequency together, you end up with a result which mathematically &lt;em&gt;is&lt;/em&gt; white noise)&lt;/p&gt;&#xA;" OwnerUserId="3245" LastActivityDate="2011-04-23T00:15:19.243" CommentCount="7" />
  <row Id="9021" PostTypeId="2" ParentId="9017" CreationDate="2011-04-23T00:48:10.883" Score="2" ViewCount="" Body="&lt;p&gt;Yes. The noise from the two different sources should be uncorrelated, so they add in power. Adding a second identical noise source increases the power by a factor of 2. Since the power is proportional to the square of the amplitude, that means the amplitude increases by a factor of $\sqrt{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of decibels, this means that the power increases by 3 db, and the amplitude increases by 1.5 db. (Power is more commonly measured in db than amplitude, but there's no reason you can't use db for both power and amplitude.)&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-04-23T00:48:10.883" CommentCount="3" />
  <row Id="9022" PostTypeId="1" AcceptedAnswerId="9025" CreationDate="2011-04-23T01:19:04.443" Score="2" ViewCount="139" Body="&lt;p&gt;Assuming supersymmetry exists and a neutralino is stable, it's often seen as a leading dark matter candidate. What would be expected from the interaction of a neutralino and its anti-particle? Has there been any unambiguous detection of such an interaction? &lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-04-23T15:49:46.397" Title="Neutralino Dark Matter Detection" Tags="&lt;supersymmetry&gt;&lt;dark-matter&gt;" AnswerCount="2" />
  <row Id="9023" PostTypeId="2" ParentId="8981" CreationDate="2011-04-23T03:43:17.457" Score="0" ViewCount="" Body="&lt;p&gt;The integer part of this is a cocycle condition, which is a measure of the winding number for a gauge transformation.  The Chern-Simons (CS) theory is a $2~+~1$ dimensional quantum field theory for a non-dynamical gauge field$A_\mu$. The action for such a theory is&#xA;$$&#xD;&#xA;S_{CS}~=~\frac{k}{4\pi}\int A\wedge dA~+~\frac{2}{3}A\wedge A\wedge A&#xD;&#xA;$$&#xA;Where $A_\mu$ is a component of the one form ${\underline A}~=~A_\mu{\underline e}^\mu$ for a non-abelian gauge field transforming in the adjoint representation of the gauge group $U(N)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory to make sense must be well behaved under gauge transformations. While it is relatively easy to show invariance in the abelian case, the non-abelian case is a little more subtle. In this case&#xA;$$&#xD;&#xA;S_{CS}~\rightarrow~S_{CS}~+~2\pi kN&#xD;&#xA;$$&#xA;Where $N$ is a integer for the winding number of the gauge transformation performed. Quantization of the theory using Feynman’s path integral formalism requires that$e^{iS_{CS}}$ be gauge invariant. This leads to the condition that $k~\in~{\mathbb Z}$. The integer $k$ is the Chern-Simons level $A_\mu$. Typically every gauge group in the Chern-Simons theory has a level associated to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This form of the Chern-Simons theory is not supersymmetric. However it is possible to make the gauge field $A_\mu$ a component of an ${\cal N}~=~2$ vector multiplet. This necessarily introduces two scalar fields $A_\mu$ $F$, an auxiliary field, and a 2-component Dirac spinor $\psi$ to the theory in a superfield&#xA;$$&#xD;&#xA;\Psi~=~\psi~+~\theta \sigma^\mu A_\mu~+~H.C.~+~{\bar\theta}\theta F.&#xD;&#xA;$$&#xA;It is possible to extend this theory to admit the full ${\cal N}~=~8$ SUSY (16 supercharges).&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-04-23T03:43:17.457" CommentCount="2" />
  <row Id="9024" PostTypeId="2" ParentId="8930" CreationDate="2011-04-23T04:42:08.093" Score="1" ViewCount="" Body="&lt;p&gt;Van Nieuwenhuizen's PhD advisor, Matinus Veltman, was arguably the first person to develop a computer algebra system in the early 1960s, and the program was used in the proof of renormalization of gauge theories.&lt;/p&gt;&#xA;" OwnerUserId="2643" LastActivityDate="2011-04-23T04:42:08.093" />
  <row Id="9025" PostTypeId="2" ParentId="9022" CreationDate="2011-04-23T05:04:42.057" Score="1" ViewCount="" Body="&lt;p&gt;Dear Michael, neutralino carries no conserved charges that can take arbitrarily large values (and no spin) and it is identical with its antiparticle, much like in the case of photons. That still allows two neutralinos to annihilate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutralinos only carry a &quot;1&quot; charge under the $Z_2$ symmetry called R-parity. This means that the number of neutralinos must be conserved modulo 2. This is compatible with the annihilation of pairs of neutralinos. The annihilation of neutralino pairs was important when the Universe was young. With the right TeV-like mass and MSSM-like annihilation cross section, models may predict that the dark matter density has been diluted exactly to the observed value (plus minus a few orders of magnitude which is still a nontrivial agreement).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The products of this annihilation are most typically two photons (but two gluons or gamma-Z are also possible) and the resulting gamma rays are so rare in the present era that it's hard to observe them. However, people have still tried to argue that they should be observable, at least under various assumptions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/astro-ph/9712318&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/astro-ph/9712318&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As far as I know, none of those gamma rays has been observed as of today.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-23T05:27:01.873" LastActivityDate="2011-04-23T05:27:01.873" CommentCount="7" />
  <row Id="9026" PostTypeId="2" ParentId="9010" CreationDate="2011-04-23T05:08:58.957" Score="3" ViewCount="" Body="&lt;p&gt;This needs an input from a theoretician. From my experimentalist's view point of the matter, the answer is &quot;probably not&quot; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do have the standard model, and it is symmetry based and masses are limited within this symmetry. Nevertheless, there are tantalizing indications of physics beyond the standard model, from deviations of measurements and theoretical calculations in several quantities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Higher symmetries are envisioned, as supersymmetry, and indeed there, there would be an upper limit in the masses of all the particles, except that theoreticians will embed it into strings which are another ball park, where even a black hole can be considered an excitation on the string. So no upper bound calculable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question then will be answered by the Theory Of Everything, strings, according to string theorists. Or we will keep opening russian dolls to higher and higher symmetries and particles, as fans of the unpopular composite theories contend, until an alternate  TOE tells us whether we have reached a top mass.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-23T05:08:58.957" CommentCount="1" />
  <row Id="9028" PostTypeId="2" ParentId="7722" CreationDate="2011-04-23T08:36:28.237" Score="-1" ViewCount="" Body="&lt;p&gt;I think they add some vibration to the belt to help substance to stick off.&lt;/p&gt;&#xA;" OwnerUserId="3249" LastActivityDate="2011-04-23T08:36:28.237" CommentCount="1" />
  <row Id="9029" PostTypeId="1" CreationDate="2011-04-23T08:48:10.367" Score="1" ViewCount="126" Body="&lt;p&gt;How can I evaluate the band gap of my ZnO thin film. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thickness=d=80 nm&#xA;I’m aware of alpha=-1/d*log(T)&#xA;But I can’t  fit a line to my data.&#xA;And: How can I smoothen my data on about 346 nm?&#xA;Do you have any other idea for calculation of band gap from Transmittance and Absorbtion of light by thin films?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;h=4.13566733e-15;            %# eV.s &#xA;c=299792458;                 %# m/s &#xA;                             %# landa in (nm) &#xA;nu=c./(landa*1e-9);          %# 1/s &#xA;alpha=-1/(50e-7)*log(T50);   %# 1/cm &#xA;E=h*nu;                      %# eV &#xA;Y=alpha.^2; &#xA;plot(E,Y);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dl.dropbox.com/u/21031944/80.xls&quot; rel=&quot;nofollow&quot;&gt;Data: 80.txt &lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3208" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-23T15:12:13.040" LastActivityDate="2011-04-23T15:20:44.567" Title="Evaluation of band gap from transmittance" Tags="&lt;solid-state-physics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9030" PostTypeId="1" AcceptedAnswerId="9073" CreationDate="2011-04-23T11:27:47.610" Score="3" ViewCount="271" Body="&lt;p&gt;A few questions about the design of a common table fan, such as below.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Does the round disc (silver in the image) in front of the blades serve any functional purpose?&lt;/li&gt;&#xA;&lt;li&gt;Does the metal casing around the blades serve any functional purpose except protecting the blades?&lt;/li&gt;&#xA;&lt;li&gt;What exactly are the sources of noise in a well-designed fan? Does the metal casing in a well-designed fan have an impact on the noise level? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://img193.imageshack.us/img193/8081/60933751.gif&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3250" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-04-23T14:45:00.463" LastActivityDate="2011-04-25T02:02:10.460" Title="Design of a common table fan" Tags="&lt;sound&gt;&lt;vibration&gt;&lt;design&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="9031" PostTypeId="1" CreationDate="2011-04-23T14:29:47.477" Score="4" ViewCount="133" Body="&lt;p&gt;I have often heard statements like, Continuum limit of Lattice Yang-Mills doesnot exist in dimensions higher then four. Is there a neat way to see this? or could some one point me to some relevant literature where this has been shown.&#xA;thanks&lt;/p&gt;&#xA;" OwnerUserId="3251" LastActivityDate="2011-04-23T16:37:27.280" Title="Lattice Yang-Mills" Tags="&lt;yang-mills&gt;" AnswerCount="1" />
  <row Id="9032" PostTypeId="2" ParentId="9010" CreationDate="2011-04-23T14:38:06.507" Score="3" ViewCount="" Body="&lt;p&gt;A candidate for such an upper mass would be the Planck mass, at least from well accepted main stream physics. It is the mass where the Compton wavelength and Schwarzschild radius become equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Planck_mass&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Planck_mass&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards, Hans&lt;/p&gt;&#xA;" OwnerUserId="3197" LastActivityDate="2011-04-23T14:38:06.507" CommentCount="1" />
  <row Id="9033" PostTypeId="2" ParentId="9029" CreationDate="2011-04-23T15:20:44.567" Score="1" ViewCount="" Body="&lt;p&gt;I wonder what You are looking for? The center of the shoulder &#xA;in Your data is at 370 nm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 1 eV resembling 1,24 µm You get a bandgap of 3.35 eV. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is practically identical to 3.37 eV for ZnO in tables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do You think thin films have different values? Not at 80 nm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do You think that kink in Your data is an artifact? &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-23T15:20:44.567" CommentCount="0" />
  <row Id="9034" PostTypeId="2" ParentId="9022" CreationDate="2011-04-23T15:49:46.397" Score="0" ViewCount="" Body="&lt;p&gt;It could be that the neutralino is a Majorana fermion. The $Z_2$ structure could be a signature of a Majorana fermion.  The Majorana fermion is the charge conjugate fermion of the form $\psi_c~=~\gamma^2\psi^*$ with $\psi_c~=~\psi$.  This has some interesting statistics.  For fermions the exchange of a wave function introduces a negative sign.  The additional constraint that $\psi~=~\gamma^2\psi^*$, means that the $\gamma^2$ introduces a braid structure into the exchange statistics of Majorana fermions.  If this is the case the neutralino is its own anti-particle.  The annihilation of the neutralino particle could then be a type of decay process.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-04-23T15:49:46.397" CommentCount="1" />
  <row Id="9035" PostTypeId="2" ParentId="9031" CreationDate="2011-04-23T16:37:27.280" Score="8" ViewCount="" Body="&lt;p&gt;This may look like a deep question and in some sense, it is. But the explanation is completely trivial, doesn't really depend on the &quot;lattice&quot; at all, and boils down to dimensional analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The action of Yang-Mills theory&#xA;$$ S =\int d^D x \frac{1}{2 g^2} {\rm Tr} \, F_{\mu\nu} F^{\mu\nu} $$&#xA;Now, $F_{\mu\nu}\sim [D_\mu,D_\nu]$ has dimension of squared mass - because the (covariant) derivatives have the dimension of mass in these conventions. That's why $F^2$ has dimension $mass^4$. Because $d^D x$ has $mass^{-D}$, it's clear that the dimension of $g$ has to be&#xA;$$ mass^{(4-D)/2}$$&#xA;For $D=4$, the coupling constant is classically dimensionless. For $D&amp;gt;4$, the coupling constant has the dimension of a negative power of mass (positive power of length); for $D&amp;lt;4$, it's the other way around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strength of interactions - e.g. the probability of scattering - goes like a positive power of $g$. However, to get e.g. the probability I mentioned, we must make it dimensionless. Dimensional analysis tells us that the right &quot;dimensionless coupling&quot; is&#xA;$$g \cdot L^{(4-D)/2}$$&#xA;where $L$ is a characteristic distance scale of the experiment. For $D&amp;gt;4$, the interactions are therefore getting stronger - and infinitely strong in the $L\to 0$ limit. In particular, if $L=a$ is the lattice spacing which is send to zero, you will need to adjust the coupling constant at the lattice scale to be infinite if you want to get a finite $g \cdot L^{(4-D)/2}$ for a finite $L$. However, there's really no way to define a gauge theory with an infinite coupling, even at the level of a lattice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, for $D&amp;gt;4$, the gauge theory action above only works as an effective action for long enough distances $L$ where the &quot;dimensionless coupling&quot; is small enough. This is the only starting point where &quot;a theory&quot; exists and if you try to extrapolate it to much shorter distances, you will find out that the theory becomes ill-defined once the dimensionless coupling constant exceeds one or so - at the distance scale corresponding to a power of $g$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So no theory that could be identified with the gauge theory exists at supershort distances if $D&amp;gt;4$. Alternatively, you may naively define a lattice gauge theory for $D&amp;gt;4$ with a finite coupling constant. However, the scaling laws guarantee that at distances that are infinitely longer - as in the continuum limit - the interactions will vanish. At any rate, you can't get finite yet non-vanishing interactions at both distances $a$ that are sent to zero as well as finite distances $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case of $D&amp;lt;4$ is different because the interactions become stronger at longer distances. So one may start with a lattice with &quot;infinitesimal&quot; interactions and they go stronger at longer distances.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-23T16:37:27.280" CommentCount="2" />
  <row Id="9036" PostTypeId="2" ParentId="9030" CreationDate="2011-04-23T18:31:43.060" Score="1" ViewCount="" Body="&lt;p&gt;I won't address the details of the fluid dynamics that cause the cage to produce noise because it is well out of my area of expertise. But I have performed an experiment in my office. The small desk fan I bought was too noisy. Since it lives on the top of a tall bookcase and well out of reach, I removed the cage completely. It is a lot quieter now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that turbulence around the bars of the cage causes the noise. With the cage removed, the air stream throws further. There is still some noise, so I'm guessing that the fan's blades are not optimally shaped.&lt;/p&gt;&#xA;" OwnerUserId="1558" LastEditorUserId="1558" LastEditorDisplayName="" LastEditDate="2011-04-24T00:02:44.617" LastActivityDate="2011-04-24T00:02:44.617" CommentCount="1" />
  <row Id="9037" PostTypeId="2" ParentId="8975" CreationDate="2011-04-23T18:48:34.033" Score="1" ViewCount="" Body="&lt;p&gt;Also remember that the &quot;displacement equation&quot; you refer to could just as well be:&#xA;\begin{equation}&#xA;x(t) = A \sin{\left(\omega t + \phi\right)}&#xA;\end{equation}&#xA;The phase angle $\phi$ is a measure of when you decide to start your stopwatch, i.e. when you set $t=0$. For an easy way to remember this, just remember the graphs of $\sin{t}$ and $\cos{t}$ -- the sine function has the value $0$ at $t=0$, while the cosine function has a value $1$ at $t=0$. Thus, if you start your clock when the oscillating body is at the equilibrium position, you should use the sine, and if you start your clock when the oscillating body is at the maximum amplitude position, you should use the cosine.&lt;/p&gt;&#xA;" OwnerUserId="3252" LastActivityDate="2011-04-23T18:48:34.033" CommentCount="1" />
  <row Id="9039" PostTypeId="2" ParentId="728" CreationDate="2011-04-23T19:49:25.720" Score="3" ViewCount="" Body="&lt;p&gt;There are already many good answers and comments. Here is my answer, which borrows several observations from, in particular, Carl Brannen's answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the surface of the hill be described by a smooth function $x^3=h(x^1,x^2)$. Since we assume that the ball with radius $R$ continuous to touch the hill in precisely one point, the center of the ball will lie on a curve $x^3=f(x^1,x^2)$, where the distance, between the two graphs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{graph}(h)=\{(x^1,x^2,x^3)\in \mathbb{R}^3 \mid x^3=h(x^1,x^2)\},$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{graph}(f)=\{(x^1,x^2,x^3)\in \mathbb{R}^3 \mid x^3=f(x^1,x^2)\},$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is $R$ everywhere. (Formulated more precisely: The distance from one graph to an arbitrary point $P$ on the other graph is always $R$, independent of the point $P$.) We will next choose the center-of-mass (CM) coordinates $(x^1,x^2)$ as &lt;a href=&quot;http://en.wikipedia.org/wiki/Generalized_coordinates&quot; rel=&quot;nofollow&quot;&gt;generalized coordinates&lt;/a&gt; for the ball. The CM $x^3$-coordinate is then just $f(x^1,x^2)$, which determines the potential energy &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V(x^1,x^2)=mgf(x^1,x^2)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;of the ball. We are not able to describe the &lt;em&gt;orientation&lt;/em&gt; of the ball only with the help of the generalized coordinates $(x^1,x^2)$, but we don't care. Since we don't describe the orientation of the ball, we will also avoid non-holonomic constraints. Next note that friction and normal force do no work on the ball, and that the total mechanical energy $E=T+V$ of the ball is conserved. We have already enough information from the sliding vs. rolling conditions to fully describe the total kinetic energy $T$ in terms of the configuration space variables $(x^1,x^2,\dot{x}^1,\dot{x}^2)$. The total kinetic energy is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T = T_{CM}+T_{ROT} = \alpha T_{CM}, \qquad \alpha:=1+\frac{I}{mR^2},  $$&#xA;$$T_{CM}=\frac{m}{2}|\vec{v}|^{2}, \qquad \vec{v} = \left(\dot{x}^1,\dot{x}^2,\sum_{i=1}^2\frac{\partial f}{\partial x^i}\dot{x}^i\right), $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, as Carl Brannen observes, applies to the sliding case as well, if we artificially assign $I=0$ in that case. In particular, the total kinetic energy $T$ is the same up to a constant factor $\alpha$, and the question can be reformulated as asking if the path is independent of the moment of inertia $I$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, let us recall Jacobi's formulation of &lt;a href=&quot;http://en.wikipedia.org/wiki/Maupertuis%27_principle&quot; rel=&quot;nofollow&quot;&gt;Maupertuis principle&lt;/a&gt;. Ultimately, this approach does not solve the problem, but it does offer a conceptional understanding. The principle states that the ball chooses (among virtual same-energy-paths) a path that extremizes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{P_i}^{P_f} \sqrt{T} \ \mathrm{d}s,  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for fixed initial and final position, $P_i$ and $P_f$, respectively. Here the metric &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{d}s^2=\sum_{i,j=1}^2 g_{ij}\mathrm{d}x^i \mathrm{d}x^j$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;arises from the total kinetic energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T=\frac{1}{2}\sum_{i,j=1}^2 g_{ij}\dot{x}^i\dot{x}^j = \frac{1}{2}\left(\frac{ds}{dt}\right)^2, \qquad g_{ij}= \alpha m \left(\delta_{ij}+ \frac{\partial f}{\partial x^i}\frac{\partial f}{\partial x^j} \right).&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, an overall constant multiplicative factor $\alpha$ will clearly not affect the extremization process. Nevertheless, this does not solve the original problem, where we don't want to assume a specific (i.e., fixed) final position $P_f$. So Maupertuis principle does only cover the case, where the end points $P_f$ are assumed to be the same a priori. Then it states that the intermediate paths must also be the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, to solve the original question, it turns out to be better to just examine Lagrange equations directly,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}^i}\right) = \frac{\partial L}{\partial x^i}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total kinetic energy $T$ becomes independent of $\alpha$ under a scaling of time $t\to \sqrt{\alpha} t$. Also note that the Lagrange equations are invariant under the time scaling $t\to \sqrt{\alpha}t$. Thus the solution will be the same up to a rescaled time parameter and choice of initial conditions. This is also Carl Brannen's point. Interestingly, it is important that the initial velocity is assumed to be zero to achieve the same paths in the two situations, as the following simple example illustrates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: Let the hill be an inclined plane with inclination $45^{\circ}$, and let the $x^1$-axis be in the &quot;uphill direction&quot;. We take the function $f(x^1,x^2)=x^1$. Let the initial position be $(x^1_i,x^2_i)=(0,0)$, and the initial velocity be $(\dot{x}^1_i,\dot{x}^2_i)=(0,v_i)$ along the $x^2$-axis. Then the metric becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ g_{ij} = \alpha m \left[\begin{array}{cc} 2 &amp;amp; 0 \\ 0 &amp;amp; 1 \end{array} \right]. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations of motion are &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 2\alpha m  \ddot{x}^1 = -mg, \qquad \alpha m  \ddot{x}^2 =0. $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is a parabola&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x^1(t) = -\frac{g}{4\alpha} t^2, \qquad  x^2(t)=v_i t. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paths are only the same (i.e., independent of $\alpha$), if the initial velocity $v_i$ is zero.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-04-25T01:57:19.990" LastActivityDate="2011-04-25T01:57:19.990" />
  <row Id="9040" PostTypeId="1" AcceptedAnswerId="9042" CreationDate="2011-04-23T20:12:59.967" Score="3" ViewCount="182" Body="&lt;p&gt;Imposing SU(2) and U(1) local gauge invariance introduces 4 gauge bosons, two of which correspond to $W^{\pm}$ bosons. The other two gauge fields $W^{\mu}_3$ and $B^{\mu}$ however are said not to correspond to $Z$ and $\gamma$ bosons because of the incorrect chiral coupling of the $Z$ boson if this identification is made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;physical&quot; bosons are said to be Weinberg rotated by $\sin^2\theta_W=0.23$, where the angle is obtained from experiment. This gives the observed coupling of $Z$ to both left and right chiralities and also a different mass from the $W$ brothers, again observed experimentally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? Why do the physical $Z$ and $\gamma$ bosons have to be Weinberg rotated? (by Nature that is, not by phenomenologists) If there was no rotation, the three $W$ 's would have the same masses and would couple only to lefties. This doesn't immediately seem particularly disastrous to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see how &lt;a href=&quot;http://en.wikipedia.org/wiki/Totalitarian_principle&quot; rel=&quot;nofollow&quot;&gt;Gell-Mann's Totalitarian Principle&lt;/a&gt; can save the day here, because what I'm essentially asking is why our particle detectors observe the Weinberg rotated bosons and not the &quot;natural&quot; ones that come out of the Lagrangian? What is it that makes particular linear combinations of bosons the physical ones? And crucially, why aren't the $W^{\pm}$ bosons rotated in a similar manner? (Or would we not notice if they were?)&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-04-24T02:17:09.880" LastActivityDate="2011-04-24T02:17:09.880" Title="What's so unphysical about $W^{\mu}_3$ and $B^{\mu}$?" Tags="&lt;particle-physics&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" />
  <row Id="9041" PostTypeId="1" AcceptedAnswerId="9057" CreationDate="2011-04-23T20:30:40.853" Score="4" ViewCount="234" Body="&lt;p&gt;&quot;Field&quot; is a name for associating a value with each point in space. This value can be a scalar, vector or tensor etc. I read the wikipedia article and got that much, but then it goes it into more unfamiliar concepts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how to interpret a basic field. Lets say there is a field with momentum and energy. Does that mean, that any &quot;object&quot; which can interact with that field, can borrow that momentum and energy from that field and give back to it as well. Much like an electron in some (not its own) electromagnetic field, right? Now how do I extend this understanding to concepts like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$U_\lambda$ is the radiation energy density per unit wavelength of a thermodynamic equilibrium radiation field trapped in some cavity.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does that mean, that the &quot;numbers&quot; that make up the field in each point in space stay constant  with temperature or time (or both, I'm not sure.) And if there are object is that field which can interact with it, for example electrons, then they can take energy from that field or lose energy to that field. Then the author calls this field a &quot;photon gas&quot; without explanation. So does that mean a bag of photons in cavity is mathematically equivalent to specifying some numbers in space? Or something else. &lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="2599" LastEditorDisplayName="" LastEditDate="2011-04-23T20:43:21.433" LastActivityDate="2011-04-24T14:11:10.233" Title="How to think physically about basic &quot;fields&quot; " Tags="&lt;mathematical-physics&gt;&lt;thermodynamics&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="9042" PostTypeId="2" ParentId="9040" CreationDate="2011-04-23T20:44:31.433" Score="4" ViewCount="" Body="&lt;p&gt;If you break the $$SU(2)\times U(1)$$ symmetry spontaneously &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left|-ig\frac{\sigma}{2}W_\mu-ig&amp;#39;B_\mu\phi\right|^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and insert the vacuum expectation value of the scalar field you get terms&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{2}vgW_\mu^+W^{\mu-}+\frac{1}{8}v^2(W_\mu^3,B_\mu)\left( \begin{array}{cc}&#xD;&#xA;g^2 &amp;amp; -gg&amp;#39; \\&#xD;&#xA;-gg&amp;#39; &amp;amp; g&amp;#39;^2 &#xD;&#xA;\end{array} \right)\left( \begin{array}{c}&#xD;&#xA;W^{3\mu} \\&#xD;&#xA;B^\mu &#xD;&#xA;\end{array} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you diagonalize the mass matrix on the right to get the physical fields you get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{8}v^2[gW_\mu^3-g&amp;#39;B_\mu]^2+0\cdot[g&amp;#39;W_\mu^3+gB_\mu]^2=\frac{1}{2}m_ZZ_\mu^2+\frac{1}{2}m_AA_\mu^2$$, that is a massive Z boson and a massless photon&lt;/p&gt;&#xA;" OwnerUserId="2151" LastEditorUserId="2151" LastEditorDisplayName="" LastEditDate="2011-04-23T20:54:12.533" LastActivityDate="2011-04-23T20:54:12.533" CommentCount="3" />
  <row Id="9043" PostTypeId="2" ParentId="9041" CreationDate="2011-04-23T20:46:46.197" Score="5" ViewCount="" Body="&lt;p&gt;I'll start here -- &quot;So does that mean a bag of photons in cavity is mathematically equivalent to specifying some numbers in space?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It means much more than that. First, you define a field -- in the electromagnetic case, it's a set of vectors everywhere in space. Then you allow it to be dynamic, that is you write a Lagrangian for it that leads to classical equations of motion. Then you quantize it, which leads to (1) the idea of a vacuum, that is a state of the field that contains no excitations and (2) particles -- excitations of the field that can carry momentum and energy around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a photon gas is much more than just a simple set of numbers everywhere. The field is dynamic. It's &lt;em&gt;not&lt;/em&gt; in the vacuum state and so it contains particles. And finally, it is in a very precise kind of state - the photons are distributed in energy according to the Bose-Einstein distribution at a particular temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did it get into that equilibrium state? To answer that, you need to extend your field from the plain EM one to include coupling to other matter fields, say the electron. So the electron will have it's own kind of field and your combined Lagrangian will contain a term that couples the two fields together so that they can interact. This is how &quot;objects&quot; like electrons can borrow and lend energy to the EM field, through the coupling between the electron field (which by the way is a spinor field, and not a vector field) and the EM field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can have fields for all the other kinds of particles and have interaction terms for all the other known forces. So now you can have a cavity made of matter that interacts with the EM field. And through complicated dynamics that depend crucially on the Second Law of Thermodynamics, the photon gas can come into equilibrium with the matter cavity, all this mediated by the couplings in the Lagrangian.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-04-23T20:46:46.197" />
  <row Id="9044" PostTypeId="1" CreationDate="2011-04-23T20:54:36.297" Score="0" ViewCount="206" Body="&lt;p&gt;The question is very simple, but complicated to solve. I am looking for a &lt;em&gt;genera&lt;/em&gt; form solution to this problem. Thus there will be no numbers, just variables. What I mean by that is often in physics you have equations like $F = ma$. I don't have the numbers to the problem yet, because that will vary based on the spaceship, its engine, the distance, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newtonian physics works as an approximation, however its allows the spaceship to exceed $c$ during its travels. I would like to add relativity, with the frame of reference being the spaceship. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can already solve this problem in discrete time steps programatically, but it would be nice to have a general form solution that would be more accurate. (Because discrete time stepping is expensive computationally, and a general form solution would be much much faster.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my current algorithm the variables are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$M$ = base mass of spaceship in kkg (ie 1000kg)&lt;/li&gt;&#xA;&lt;li&gt;$C$ = cargo space of spaceship kkg &lt;/li&gt;&#xA;&lt;li&gt;$D$ = distance to be traveled in km&lt;/li&gt;&#xA;&lt;li&gt;$T$ = thrust of the engines in kN&lt;/li&gt;&#xA;&lt;li&gt;$F$ = fuel consumed per year in kkg/year &lt;/li&gt;&#xA;&lt;li&gt;$E$ = food consumed per year in kkg/year&lt;/li&gt;&#xA;&lt;li&gt;$A$ = maximum velocity you can possibly achieve {$A &amp;lt; c$} in m/s&lt;/li&gt;&#xA;&lt;li&gt;$B$ = maximum velocity you are willing to achieve {$B &amp;lt;= A$} in m/s&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some notes. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cargo space can be used to be store fuel and/or food.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The way I solved this problem before, was calculated fuel costs first, and then decided if there was enough food afterwards. Based on $C_{remaining} &amp;gt; t_{travel} * E$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Loaded cargo adds to $M$ so $M_{total} = M + C_{fuel} + C_{food}$&lt;/li&gt;&#xA;&lt;li&gt;As you fly the mass of the fuel is consumed, reducing the weight of the spaceship, allowing you accelerate faster.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Engine efficiency is not part of the equation. The engine produces $T$ thrust and consumes $F$ fuel per year. (unless I am wrong, I shouldn't have to worry about it, if it does matter, lets say 100%; its ideal.)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;The ship needs to spend fuel accelerating (to the halfway point or until B it achieved), and then spend fuel decelerating on the other end. the $t_{accelerating} \neq t_{decelerating}$ due to the change in mass during voyage. This also means less fuel will be consumed to decelerate as well.&lt;/li&gt;&#xA;&lt;li&gt;As length inversely dilates with velocity, the length of the trip shrinks as the spaceship nears $c$ and expands as you decelerate as you near the destination.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What I am looking for is a function that determines if the one-way trip is possible, how long it will take, and how much fuel is consumed in the process, in the form of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(M, C, D, T, F, E, A, B) -&amp;gt; (isPossible, t_{total}, C_{fuel-consumed})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that with B there will be conditionals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can tell, there are 3 rate of change problems (more?)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mass increases as velocity increases $M_{relative} = M_{current} * \gamma$&lt;/li&gt;&#xA;&lt;li&gt;Mass decreases as fuel is expended $M_{current} = M + C_{fuel} - C_{fuel-consumed}$&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$C_{fuel-consumed} = F * t_{running-engines}$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Distance decreases as velocity increases $D_{relative} = D_{base} / \gamma$&lt;/li&gt;&#xA;&lt;li&gt;Am I missing something?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some references:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Speed_of_light#Upper_limit_on_speeds&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Speed_of_light#Upper_limit_on_speeds&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;$\gamma = \dfrac{1}{\sqrt{1-v^2/c^2}}$&lt;/li&gt;&#xA;&lt;li&gt;$a = \dfrac{T}{M_{relative}}$ Note: $a$ will change as $M_{relative}$ changes, but $T$ remains constant&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3254" LastEditorUserId="3254" LastEditorDisplayName="" LastEditDate="2011-04-24T05:58:22.350" LastActivityDate="2011-04-24T05:58:22.350" ClosedDate="2011-04-24T00:41:57.533" Title="A Spaceship travels from Star A to B, but food/fuel and relativity!" Tags="&lt;special-relativity&gt;" CommentCount="14" />
  <row Id="9045" PostTypeId="1" AcceptedAnswerId="9046" CreationDate="2011-04-23T23:17:46.820" Score="2" ViewCount="138" Body="&lt;p&gt;I watched on TV as they where showing gold bars stored in bank vaults and I noticed that they always stack them with the narrow side down and the wide side up. Like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Mvllb.jpg&quot; alt=&quot;Gold Bar Stacking&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there has to be a mechanical reason why is that. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-04-24T00:23:44.077" ClosedDate="2011-04-24T00:25:30.363" Title="Why do they store gold bars with the narrow side down?" Tags="&lt;stability&gt;&lt;stacking&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9046" PostTypeId="2" ParentId="9045" CreationDate="2011-04-24T00:14:32.177" Score="4" ViewCount="" Body="&lt;p&gt;They do not always seem to be that shape.  For some cuboids which are not designed to be moved regularly, see this &lt;a href=&quot;http://www.britannica.com/EBchecked/media/119692/Gold-bars-stored-at-the-US-Bullion-Depository-Fort-Knox&quot; rel=&quot;nofollow&quot;&gt;picture from Fort Knox&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for bullion bars in the world market (about 10.9-13.4 kg of gold) which have the standard sort of trapezium cross-section, the top when casting is the wider area to get them out of the mould, so the wider area is defined as the &quot;top&quot; in the &lt;a href=&quot;http://www.lbma.org.uk/pages/index.cfm?page_id=27&amp;amp;title=specifications&quot; rel=&quot;nofollow&quot;&gt;international specification&lt;/a&gt;) and the fineness, hallmarks and serial numbers usually stamped on top (&lt;a href=&quot;http://gold.bullionvault.com/How/GoldBullion&quot; rel=&quot;nofollow&quot;&gt;as illustrated here&lt;/a&gt;).  Stacking them top face up makes these easier to read for the bars on top, as well as easier to move.  Pure gold bars may &lt;a href=&quot;http://www.usmint.gov/kids/coinNews/mintFacilities/ftk/#gold&quot; rel=&quot;nofollow&quot;&gt;dent if dropped&lt;/a&gt; because of gold's softness, requiring recasting, so some care is needed. &lt;/p&gt;&#xA;" OwnerUserId="2714" LastActivityDate="2011-04-24T00:14:32.177" />
  <row Id="9047" PostTypeId="2" ParentId="9045" CreationDate="2011-04-24T00:23:44.077" Score="0" ViewCount="" Body="&lt;p&gt;The primary reason I have usually heard is that this makes it easier to lift them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pure gold is quite smooth, and very dense, and would be near impossible to life from the other side.&lt;/p&gt;&#xA;" OwnerUserId="3258" LastActivityDate="2011-04-24T00:23:44.077" />
  <row Id="9048" PostTypeId="2" ParentId="9030" CreationDate="2011-04-24T04:18:47.517" Score="0" ViewCount="" Body="&lt;p&gt;A related question. Compare a table fan's blades versus a wind turbines blades. It is obvious that the former covers more than half the area of the disk prescribed by its circumference, but the wind turbines swept area is mostly empty air. I've heard the power of a wind turbine depends upon the swept area, which implies that the blades already cover enough area, so I would surmise the table fan is probably seriously inoptimal. Probablt the blades are as wide as they are because of pschological marketting, rather than aerodynamics. Does anyone know if this is in fact the case?&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-04-24T04:18:47.517" CommentCount="2" />
  <row Id="9049" PostTypeId="1" CreationDate="2011-04-24T05:29:14.143" Score="9" ViewCount="719" Body="&lt;p&gt;Why doesn't the Moon, or for that matter anything rotating another larger body, ever fall into the larger body?&lt;/p&gt;&#xA;" OwnerUserId="2530" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-24T12:38:48.720" LastActivityDate="2011-06-27T05:53:27.513" Title="Why doesn't the Moon fall upon Earth?" Tags="&lt;gravity&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="2" />
  <row Id="9050" PostTypeId="2" ParentId="9049" CreationDate="2011-04-24T05:49:24.757" Score="-5" ViewCount="" Body="&lt;p&gt;The reason the smaller body doesn't fall into the bigger body is because of einstiens third law an object will not change it's momentum unless an outside force acts upon it in the case of the earth and the moon the moon is at a specific velocity and the earth is in this case stationary there is also the centripetal force in which the grAvity of earth is pulling the moon toward it therefore there is a complete equilibrium betweeen the two bodies the only way that the moon would fall into earth would be if something in the entire system changed for example the earth in greasing in size &lt;/p&gt;&#xA;" OwnerUserId="3260" LastActivityDate="2011-04-24T05:49:24.757" CommentCount="3" />
  <row Id="9051" PostTypeId="2" ParentId="9049" CreationDate="2011-04-24T05:52:42.613" Score="9" ViewCount="" Body="&lt;p&gt;Moon is continuously falling towards earth but missing all the time! Same with other planets too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, in an inverse square central force field one can calculate the trajectory of a particle and verify that the trajectory is either a parabola or ellipse or hyperbola (conic sections) depending upon the initial position and initial momentum of the particle. For a two body system with certain initial conditions, it is a stable elliptical orbit. In case of the sun and the earth it is an ellipse (ignoring gravitation of other objects and also ignoring the relativistic precision of orbit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.teachersdomain.org/resource/ess05.sci.ess.eiu.moonorbit/&quot; rel=&quot;nofollow&quot;&gt;This page has a nice video.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-04-24T09:55:36.823" LastActivityDate="2011-04-24T09:55:36.823" />
  <row Id="9052" PostTypeId="1" AcceptedAnswerId="9071" CreationDate="2011-04-24T06:39:04.967" Score="8" ViewCount="320" Body="&lt;p&gt;One of the things I never understood but was too afraid to ask is this: how should I think of things like kg/&lt;strong&gt;s^2&lt;/strong&gt;. What exactly is a square second? Square foot makes sense to me because I can see it, but square second? I've always assumed it's just one of those things you're supposed to deal with abstractly and not try to get an intuition for, the same way you deal with 4-dimensional space. Should I just keep doing that or is there some bit of wisdom I've been deprived of all these years?&lt;/p&gt;&#xA;" OwnerUserId="1827" LastActivityDate="2011-05-02T00:38:43.843" Title="Understanding counterintuitive units like s^2" Tags="&lt;dimensionality&gt;&lt;units&gt;&lt;dimensional-analysis&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="9053" PostTypeId="1" CreationDate="2011-04-24T06:46:44.690" Score="2" ViewCount="205" Body="&lt;p&gt;If one has a deterministic model of physics, why is causality so important?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's work in a fixed frame. Suppose that event A in the future causes event B, which happens before event A. Now, given the conditions at some time $t_0$ before both events (or at the time of event B), one can predict what will happen in the future, hence one already knows that event A will happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, there's no event that happens after B that causes B and that we didn't already know about at the time of event B. Rather, we could say that because event A is caused by the conditions at time $t_0$, event B is in essence caused by the conditions at time $t_0$ and not by event A (in the sense that it can be explained completely in terms of what happens at time $t_0$). &lt;strong&gt;Therefore, if we have a theory that says that event A can cause event B, we could re-explain the exact same theory by considering the mechanism that allows A to cause B as simply part of the laws which allow us to determine all future events from the initial conditions at time $t_0$.&lt;/strong&gt; Maybe, there's some mathematical aspect of those laws, some term in the equations, which we would philosophically like to think of as 'event A causing event B.' But physically, causality is already in this sense satisfied under a deterministic model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came to wonder about this as a result of classical electrodynamics (without any other forces or gravity). If we know the electromagnetic field at a given point in time in addition to its time derivative, as well as the position and velocity of all charge and mass, we can use Maxwell's equations and the Lorentz force law to find out what will happen at all future events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The implication of this is that in special relativity, we shouldn't really have to worry so much about information traveling faster than the speed of light (something we impose on a system because otherwise it would violate the law of causality after applying the Lorentz transformations). It does so happen that the current classical special theory of relativity works nicely mathematically and also satisfies that principle, so we have no reason to reject it, but I don't see why it's so important for information not to travel faster than the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One further idea: We are assuming that we can choose some time $t_0$ before all events caused by A, but what if A causes events arbitrarily far into the past. One idea is that we could in some sense consider the limit of the state of the universe as $t$ approaches $-\infty$ as our initial conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe the independence of causality and determinism could make sense in a universe with 'divine intervention' of sorts, where we could suddenly decide to intervene at any point and break the deterministic model. For example, suppose we could suddenly give an electron in the universe a mysterious push, or we could even magically make an electron appear or disappear. Then if the laws of physics allowed that divine push to influence what happened in the past, that would violate causality. But that situation would not be deterministic.&lt;/p&gt;&#xA;" OwnerUserId="2591" LastActivityDate="2011-04-24T14:56:59.880" Title="Causality and Determinism" Tags="&lt;causality&gt;" AnswerCount="4" />
  <row Id="9054" PostTypeId="2" ParentId="9049" CreationDate="2011-04-24T06:48:53.800" Score="23" ViewCount="" Body="&lt;p&gt;The moon does not fall to Earth because it is in an &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbit&quot; rel=&quot;nofollow&quot;&gt;orbit&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the most difficult things to learn about physics is the concept of force.  Just because there is a force on something does not mean it will be moving in the direction of the force.  Instead, the force influences the motion to be a bit more in the direction of the force than it was before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you roll a bowling ball straight down a lane, then run up beside it and kick it towards the gutter, you apply a force towards the gutter, but the ball doesn't go straight into the gutter. Instead it keeps going down the lane, but picks up a little bit of diagonal motion as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you're standing at the edge of a cliff 100m tall. If you drop a rock off, it will fall straight down because it had no velocity to begin with, so the only velocity it picks up is downward from the downward force.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you throw the rock out horizontally, it will still fall, but it will keep moving out horizontally as it does so, and falls at an angle. (The angle isn't constant - the shape is a curve called a parabola, but that's relatively unimportant here.) The the force is straight down, but that force doesn't stop the rock from moving horizontally. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you throw the rock harder, it goes further, and falls at a shallower angle. The force on it from gravity is the same, but the original velocity was much bigger and so the deflection is less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine throwing the rock so hard it travels one kilometer horizontally before it hits the ground. If you do that, something slightly new happens. The rock still falls, but it has to fall more than just 100m before it hits the ground. The reason is that the Earth is curved, and so as the rock traveled out that kilometer, the Earth was actually curving away underneath of it. In one kilometer, it turns out the Earth curves away by about 10 centimeters - a small difference, but a real one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you throw the rock even harder than that, the curving away of the Earth underneath becomes more significant. If you could throw the rock 10 kilometers, the Earth would now curve away by 10 meters, and for a 100 km throw the Earth curves away by an entire kilometer. Now the stone has to fall a very long way down compared to the 100m cliff it was dropped from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out the following drawing. It was made by Isaac Newton, the first person to understand orbits. IMHO it is one of the greatest diagrams ever made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/tuwXa.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What it shows is that if you could throw the rock hard enough, the Earth would curve away from underneath the rock so much that the rock actually never gets any closer to the ground. It goes all the way around in the circle and might hit you in the back of the head!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an orbit. It's what satellites and the moon are doing. We can't actually do it here close to the surface of the Earth due to wind resistance, but on the surface of the moon, where there's no atmosphere, you could indeed have a very low orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the mechanism by which things &quot;stay up&quot; in space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity gets weaker as you go further out. The Earth's gravity is much weaker at the moon than at a low-earth orbit satellite.  Because gravity is so much weaker at the moon, the moon orbits much more slowly than the International Space Station, for example. The moon takes one month to go around. The ISS takes a few hours. An interesting consequence is that if you go out just the right amount in between, about six Earth radii, you reach a point where gravity is weakened enough that an orbit around the Earth takes 24 hours. There, you could have a &quot;geosynchronous orbit&quot;, a satellite that orbits so that it stays above the same spot on Earth's equator as Earth spins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although gravity gets weaker as you go further out, there is no cut-off distance. In theory, gravity extends forever. However, if you went towards the sun, eventually the sun's gravity would be stronger than the Earth's, and then you wouldn't fall back to Earth any more, even lacking the speed to orbit. That would happen if you went about .1% of the distance to the sun, or about 250,000 km, or 40 Earth radii. (This is actually less than the distance to the moon, but the moon doesn't fall into the Sun because it's orbiting the sun, just like the Earth itself is.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the moon &quot;falls&quot; toward Earth due to gravity, but doesn't get any closer to Earth because its motion is an orbit, and the dynamics of the orbit are determined by the strength of gravity at that distance and by Newton's laws of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;note: adapted from an answer I wrote to &lt;a href=&quot;http://www.quora.com/How-close-does-an-object-have-to-be-to-earth-to-be-pulled-by-gravity&quot; rel=&quot;nofollow&quot;&gt;a similar question&lt;/a&gt; on quora&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-06-27T05:53:27.513" LastActivityDate="2011-06-27T05:53:27.513" />
  <row Id="9055" PostTypeId="2" ParentId="9052" CreationDate="2011-04-24T06:56:13.790" Score="9" ViewCount="" Body="&lt;p&gt;There is no reason why you should be &quot;imagining&quot; a squared second. Most quantities in physics don't have any canonical &quot;geometric&quot; visualization and there is no reason why they should have. What matters is that you should be able to calculate with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the gravitational acceleration on Earth is $9.81\,\,{\rm m/s}^2$. This simply means that the acceleration is&#xA;$$ g = \frac{9.81 \,\,{\rm m/s}}{{\rm s}}$$&#xA;Every second, the velocity increases by $9.81 \,\,{\rm m/s}$ in the downward direction. The acceleration is 9.81 meters per second per second. If you divide the unit ${\rm m/s}$ by another ${\rm s}$, you get ${\rm m/s/s}$ which is the same thing as ${\rm m/s}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A squared second would still be very simple to imagine: just imagine a square in a fictitious spacetime with two time coordinates whose side is one second. There's no problem with the fact that this spacetime is not real: you are just trying to imagine something that shouldn't be imagined, so it's not surprising that the imaginations are unphysical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exist much more &quot;bizarre&quot; units for seemingly simple quantities. For example, the unit of electric charge in the CGSE system is one statcoulomb &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Statcoulomb&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Statcoulomb&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is just a different way of saying&#xA;$$ 1 {\rm g}^{1/2} {\rm cm}^{3/2} {\rm s}^{-1} $$&#xA;which contains fractional powers. You can't imagine any shapes whose &quot;volumes&quot; are fractional powers of the sides. Still, there is no difficulty with calculating with these units. There are lots of formulae in physics which are &quot;nonlinear&quot; - in which one quantity has to be inverted, squared, cubed, or exponentiated to another (possibly fractional) power - to obtain another quantity. The units must also be exponentiated accordingly.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-24T06:56:13.790" />
  <row Id="9056" PostTypeId="2" ParentId="9053" CreationDate="2011-04-24T07:06:13.427" Score="2" ViewCount="" Body="&lt;p&gt;Causality means that for any two events A,B, there has to exist an ordering that says whether A can influence B or B can influence A, and in the normal examples with &quot;time&quot;, the ordering is the condition&#xA;$$ t(A) &amp;lt; t(B). $$&#xA;If the condition above is satisfied (i.e. if A precedes B), then A may influence B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An ordering - a transitive relation - has to exist in order to avoid logical contradictions. If the relationship were not transitive, for example, it would be possible to find triplets of events A,B,C such that A influences B, B influences C, C influences A. That would be a &quot;closed time-like curve&quot; and it would lead to logical inconsistencies because in general, there would be no way to choose the outcomes of the events A,B,C so that all the three implications are preserved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those contradictions are avoided in any causal theory because the outcome of event A (in a causal and deterministic theory, to be specific) is never calculated from conditions at event B if A is the cause of B (if A precedes B). It's the other way around. Causality makes it clear which data are &quot;inputs&quot; and which data are their &quot;outputs&quot;, so because of this orientation, there can't be any contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a geometric setup, the comparison of a coordinate associated with the events is the only way how to produce ordering as a relationship. We call this coordinate &quot;time&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In special (and similarly general) relativity, the condition for A to be able to influence B becomes sharper - $t(A)&amp;lt;t(B)$ has to hold in all reference frames which means that B has to belong to the future light cone of A.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-24T07:06:13.427" />
  <row Id="9057" PostTypeId="2" ParentId="9041" CreationDate="2011-04-24T07:19:29.407" Score="2" ViewCount="" Body="&lt;p&gt;Think about this: a function that maps points on a 2D space to numbers can describe the shape of terrain, but I wouldn't say that it &lt;em&gt;is&lt;/em&gt; the terrain. In the same way, a mapping of points to objects (scalars, vectors, tensors, etc.) is the mathematical description of a field, but if you think of the field as &lt;em&gt;just&lt;/em&gt; the mapping, you're missing out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fields can have various physical properties. For example, just as a particle can have a certain amount of energy, so can an electromagnetic field. The difference is, since the field is spread throughout space, so is the energy; therefore, it makes more sense to talk about the &lt;em&gt;density&lt;/em&gt; of energy rather than the amount. Same applies for momentum, or any other physical quantity carried by the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as the field could be described by a mapping of points to vectors, so the energy density can be described by a mapping of points to numbers. Given the vector value of the field at any point, you can calculate the numeric value of the energy density at that point. But remember that these numbers (i.e. the mapping) are just a mathematical description of the energy density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you may notice that the mapping that describes the energy density ($u(x)$) satisfies the naive definition of a mathematical description of a field: it associates a number with each point in space. But physicists wouldn't normally call that mapping a &quot;field,&quot; because in a sense, it's not really independent. Mathematically, you can calculate $u(x)$ from $A(x)$; physically, the energy &quot;field&quot; is completely determined by the EM field. In physics, we tend to reserve the term &quot;field&quot; to talk about something that can't be obtained by a simple calculation from some other field.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does that mean, that the &quot;numbers&quot; that make up the field in each point in space stay constant with temperature or time (or both, I'm not sure.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how you got that from the quote you listed... no, the numbers that make up the mathematical description of the EM field do not stay constant with either time or temperature. In fact, one of the things that characterizes a physical field is that it has dynamics - mathematically, this means that the numbers (or whatever) making up the field change with respect to time and space, but in a predictable manner which can be described with differential equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are things you can calculate from a field that &lt;em&gt;do&lt;/em&gt; stay constant. For instance, you can calculate the total energy stored in the field by calculating the energy density and then integrating it over the volume of the field. You could also calculate the temperature of the field, by some mathematical procedure. In many cases, these quantities are more closely based on the manner in which the field changes than the actual values that describe it. (In fact, in some sense, it turns out that you can describe a field by the way that the numbers change, just as well as you can with the numbers themselves. Read up on the Fourier transform and momentum space if you are interested.)&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-04-24T07:19:29.407" />
  <row Id="9058" PostTypeId="2" ParentId="9053" CreationDate="2011-04-24T09:07:22.143" Score="0" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;we shouldn't really have to worry so much about information traveling faster than the speed of light (something we impose on a system...  I don't see why it's so important for information not to travel faster than the speed of light.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fact that nothing can move with a speed greater than the speed of light is not an &quot;assumption&quot; that ensures causality. Rather, it is a consequence of the special theory of relativity as follows: the energy of a massive body of mass $m$ moving with a speed $v$ is $\gamma m c^2$, where $\gamma$ is the Lorentz factor. Thus, for a body to be accelerated to the speed $c$ you will require an infinite amount of energy! This makes $c$ a natural upper bound to the velocity a massive body can attain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Causality is a logical principle that is appealed to when you want to explain why $c$ is the upper bound for velocities using the relativity of simultaneity. Like Lubos said, you run into all sorts of contradictions if superluminal velocities are possible. I don't see how causality has anything to do with determinism, though.&lt;/p&gt;&#xA;" OwnerUserId="3252" LastActivityDate="2011-04-24T09:07:22.143" CommentCount="5" />
  <row Id="9059" PostTypeId="1" CreationDate="2011-04-24T11:43:58.163" Score="3" ViewCount="335" Body="&lt;p&gt;Two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Assuming you dive head first or fall straight with your legs first, what is the maximal height you can jump into water from and not get hurt?&#xA;&lt;br&gt;In other words, an H meter fall into water is equivalent to how many meters concrete-pavement fall, force wise? (I'm assuming the damage caused will be mainly due to amount of force and not the duration)&lt;/li&gt;&#xA;&lt;li&gt;Assume you jump head first and hold a sharp and strong long object that cuts the water before you arrive, will that make the entrance to the water more smooth and protect you?&lt;img src=&quot;http://i.stack.imgur.com/RGzj6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="72" LastActivityDate="2011-04-25T20:26:18.083" Title="jumping into water" Tags="&lt;gravity&gt;&lt;newtonian-mechanics&gt;&lt;water&gt;&lt;free-fall&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="9060" PostTypeId="2" ParentId="9017" CreationDate="2011-04-24T11:46:40.970" Score="2" ViewCount="" Body="&lt;p&gt;I assume you're talking about &quot;coolers&quot; as in CPU-coolers or other cooling systems in computers. The noise-spectrum from such a fan is not white strictly speaking, you can see it measured &lt;a href=&quot;http://en.wikibooks.org/wiki/Engineering_Acoustics/Noise_from_cooling_fans&quot; rel=&quot;nofollow&quot;&gt;on this page&lt;/a&gt;. The spectrum is fairly level if you look at small parts though and ignore the tonal components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, by your definition, two uncorrelated white noise sources add without interference. The &quot;peaks and troughs&quot; (sound pressures) don't add up coherently (in ideal sources of course :), but they don't cancel each other perfectly either. Therefore the sound &lt;em&gt;power&lt;/em&gt; is doubled. By definition the RMS-amplitude (proportional to the sound &lt;em&gt;pressure&lt;/em&gt;) is then multiplied by $\sqrt 2$. The sound &lt;em&gt;intensity&lt;/em&gt; is proportional to the sound &lt;em&gt;power&lt;/em&gt; and hence is doubled as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I actually simulated this in Matlab, because the terminology with sound power, sound energy, sound levels, sound pressures etc is daunting, misleading and confusing.. I'm specifically avoiding talking about decibels.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&#xA;&gt;&gt; w1 = wgn(8192, 1, 10, 'real');  % generates real white noise, 8192 samples&#xA;&gt;&gt; w2 = wgn(8192, 1, 10, 'real');&#xA;&gt;&gt; mean(([w1 w2]).^2)              % show mean powers of both noise signals&#xA;&#xA;ans =&#xA;&#xA;   10.0410   10.0832&#xA;&#xA;&gt;&gt; mean(([(w1+w1) (w1+w2)]).^2)    % add one of them to itself, add both&#xA;&#xA;ans =&#xA;&#xA;   40.1640   20.6295&#xA;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So yes you do get a doubling of power when adding uncorrelated white-noise sources.. and when &lt;em&gt;perfectly&lt;/em&gt; adding a single source, you quadruple the power. By definition the RMS amplitude is the root of the above.&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-04-24T11:46:40.970" CommentCount="2" />
  <row Id="9061" PostTypeId="2" ParentId="9049" CreationDate="2011-04-24T12:05:11.057" Score="4" ViewCount="" Body="&lt;p&gt;Another way to look at it: In the frame of reference of earth, the moon has angular momentum. Angular momentum is preserved if no torque is applied ($\tau=dL/dt$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravity forces between earth and the moon are in the direction of the center of mass, so they produce no torque ($\tau=mv\times R$), so the angular momentum ($L$) cannot change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity is perpendicular to the moon's speed, so it changes the direction and not the magnitude of the speed itself. $L=mv\times R$ and if $L$, $m$ and $v$ are constant, $R$ must remain constant as well, so the radius doesn't change.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-04-24T12:11:54.260" LastActivityDate="2011-04-24T12:11:54.260" CommentCount="1" />
  <row Id="9062" PostTypeId="2" ParentId="9059" CreationDate="2011-04-24T12:27:41.193" Score="3" ViewCount="" Body="&lt;p&gt;From searching I find that the survival rate from a fall depends on the deceleration rate that the body undergoes on impact, and that depends on the surface of the fall. Water is less punishing than land.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a quote, which I will not source, but can be found on the net.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Stone states that jumping from 150 feet (46 metres) or higher on land, and 250 feet (76 metres) or more on water, is 95% to 98% fatal. 150 feet/46 metres, equates to roughly 10 to 15 stories in a building, depending on the height of one story. 250 feet is the height of the Golden Gate Bridge in San Francisco.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most publications are behind pay walls and have to do with statistics of suicide survival, not a pleasant topic.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-24T12:27:41.193" CommentCount="1" />
  <row Id="9063" PostTypeId="2" ParentId="9049" CreationDate="2011-04-24T12:50:21.593" Score="1" ViewCount="" Body="&lt;p&gt;The Moon does not fall towards Earth right now because Earth rotates itself. The energy from the Earth's own rotation around its axis is gradually tranferred into energy of the Moon's orbital motion. That's why the Earth's rotating speed decreases but the distance to the Moon increases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process will continue until Earth's proper rotation will slow to the point where it has the same angular velocity as the Moon's orbital motion. From that moment on the Moon will start to gradually approach Earth.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastEditorUserId="1186" LastEditorDisplayName="" LastEditDate="2011-04-24T14:11:02.333" LastActivityDate="2011-04-24T14:11:02.333" CommentCount="8" />
  <row Id="9064" PostTypeId="2" ParentId="9053" CreationDate="2011-04-24T13:58:44.097" Score="-2" ViewCount="" Body="&lt;p&gt;Good question. Yes, it is possible to travel faster than light (and faster than infinite speed, i.e backwards in time) in a relativistic theory that is symmetric against the direction of time, i.e. reversible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that classical physics is not time-reversible due to the second law of thermodynamics. The second law of thermodynamic in turn originates in quantum theory, namely in the postulate of the wave function collapse. The collapse cannot be reversed, so the casualty means that the result of a completed measurement cannot be affected by any future measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we do not make any measurements, the system undergoes unitary, time-reversible evolution. This means that faster-than-light travel is actually possible in quantum mechanics in well isolated quantum system. And this is supported by experimental evidence (for example, photons can travel faster than light due to uncertainty principle). One just cannot transfer classical information this way because such transfer unavoidably includes measurements.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastActivityDate="2011-04-24T13:58:44.097" CommentCount="2" />
  <row Id="9065" PostTypeId="2" ParentId="9041" CreationDate="2011-04-24T14:11:10.233" Score="1" ViewCount="" Body="&lt;p&gt;dbrane and DZ have given Useful conventional Answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dbrane's characterization of the vacuum as &quot;contains no excitations&quot;, however, is not very helpful because there is no clear Correspondence with the classical idea of no excitations, the everywhere zero classical field. Local measurements of the vacuum in general give non-zero results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IMO (not a conventional Answer, so Useful only with care for the purposes of exams, etc.) it's more helpful to think of particles as &lt;em&gt;modulations&lt;/em&gt; of the vacuum, which is abstractly constructed as a Poincaré invariant state over an algebra of observables. An alternative perspective is that a free field can be constructed from its Wightman functions, which are essentially &lt;em&gt;correlations&lt;/em&gt; between the measured local values of the field when the localizations &lt;em&gt;are at space-like separation&lt;/em&gt; [it's more usual to construct the Wightman functions from the quantum field, $W(x_1,x_2,...x_n)=\left&amp;lt;0\right|\hat\phi(x_1)\hat\phi(x_2)...\hat\phi(x_n)\left|0\right&amp;gt;$, but the other way round works too; note that, as generalized correlations, the Wightman functions are functions on the (symmetrized) space $M\oplus (M\times M)\oplus(M\times M\times M)\oplus...$, not on Minkowski space $M$ simpliciter]. The Wightman functions are &lt;strong&gt;&lt;em&gt;not&lt;/em&gt;&lt;/strong&gt; zero for the vacuum state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-vacuum states have Wightman functions that are different from the vacuum state. In particular, they are not Poincaré invariant (which is why we can say that they &quot;carry momentum and energy around&quot;). Nonetheless, they are systematic &lt;em&gt;deformations&lt;/em&gt; of the vacuum state, and hence of the Wightman functions, constructed by the action of the quantum field on the vacuum vector, which I choose to call &lt;em&gt;modulations&lt;/em&gt; because I wish to emphasize the signal processing aspect of quantum field theory. [This doesn't make noncommutativity of measurements go away, but the relationship between quantum fields and classical signal processing is substantially different from the relationship between quantum mechanics and classical particle physics.] Although there is a continuum of possible modulations, there is also a discrete structure of states that can be constructed by the action of $1, 2, ...$ field operators on the vacuum vector, which we call the number of particles in the state. Unless there are superselection rules in place, we can construct (1) linear compositions of vectors and (2) linear compositions of states (superpositions and mixtures, respectively).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that by the vacuum state I mean a linear map from the space of operators $\omega_0:\mathcal{A}\rightarrow\mathbb{C};\omega_0(\hat A)=\left&amp;lt;0\right|\hat A\left|0\right&amp;gt;$, where I've used the vacuum &lt;em&gt;vector&lt;/em&gt; $\left|0\right&amp;gt;$ to construct the vacuum &lt;em&gt;state&lt;/em&gt; $\omega_0$. This is now a fairly universal distinction in mathematical physics, but not, I think, in Physics generally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to your Question's more specific points. Interacting quantum fields defined in terms of deformed Lagrangians and Hamiltonians are only as well-defined as the degree of your acceptance of the mathematics of renormalization. From moment to moment, unitary evolution of a state preserves the Hilbert space norm (by definition), but may or may not conserve the number of particles in the state, nor even the energy and momentum if we construct an ad-hoc Hamiltonian (a thermal field in a box &lt;em&gt;is&lt;/em&gt; an ad-hoc system, insofar as the Hamiltonian is not translation or boost invariant). Unitary evolutions are generalizations of sinusoidal motion to higher dimensional spaces, so insofar as we think of sinusoidal motion as borrowing potential energy to create kinetic energy, sure, it's &quot;borrowing&quot;, but there are other, and I think better ways to think about such models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a field perspective that starts from the vacuum, a thermal state is a thermodynamic limit of mixtures of $0,1,2,...,n$ particle states (thermodynamic in the sense that $n\rightarrow\infty$), with weight for different states that is determined by the energy. The energy and the thermal state constructed using it are invariant under translations and under rotations, but not under boosts. Thermal states are special because the limit is not in the Hilbert space of bounded states. I regret that I don't have the time (I'm not sure if it's an hour, a week, or a Ph.D. thesis) to construct a field theoretic analysis of your thermal field in a box example, I'll have to leave it to you, expanding upon the principled approach I've laid out above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yayu, I've sent you to my papers before, so I won't do so again. I've riffed on your interestingly asked Question more than I've Answered it; I think of it as more up to you to make it Useful rather than Useful in itself. Best wishes.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-04-24T14:11:10.233" />
  <row Id="9066" PostTypeId="2" ParentId="9053" CreationDate="2011-04-24T14:56:59.880" Score="1" ViewCount="" Body="&lt;p&gt;Lubos has already provided the correct answer. However let me add something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to discard the idea of &quot;causality&quot;. At the same time you are talking about &quot;past&quot; and &quot;future&quot;, a time $t_0$ which is earlier than A and B etc. etc. which is funny since the very concept of past and future depends crucially on the concept of causality!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the causality which enables you to decide which event has taken place earlier and which latter. So all the logic of your arguments falls flatly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As already pointed out that any kind of &quot;retro-causality&quot; will bring inconsistency in the logical structure itself. In special relativity the concept of causality is even more forceful. Without that it can't work.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-04-24T14:56:59.880" CommentCount="2" />
  <row Id="9067" PostTypeId="2" ParentId="9059" CreationDate="2011-04-24T15:20:38.363" Score="2" ViewCount="" Body="&lt;p&gt;It obviously depends quite strongly on how hydrodynamic your entry is. Try doing a belly flop from 2 meters: ouch! Wheras competition divers routinely dive from 10M. Having grown up in a pool, and jumping/diving off the ten foot board I am amazed at current law, which makes it illegal to lave a 3foot dive without the water being something like 17feet deep. Ours was a mere 9feet, and thousands jumped dived in over a period of decades with noone getting smashed on the bottom. So I suspect if you get the hydrodynamics just right, you can end up pretty deep/fast. Thats the position you need to hit in, but its probably well out on the tail of the distribution.&#xA;   I believe I had heard about a case of someone falling into the ocean from 20,000feet and surviving (must have been a parachute failure or such), and it may have been only an urban legend, but I have no reason to doubt that a lucky landing at terminal velocity is survivable.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-04-24T15:20:38.363" CommentCount="3" />
  <row Id="9068" PostTypeId="2" ParentId="9052" CreationDate="2011-04-24T16:05:50.603" Score="13" ViewCount="" Body="&lt;p&gt;Lubos Motl's answer is completely right, but I'll add my perspective anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For many compound units, you shouldn't try to &quot;visualize&quot; the meaning of the unit, but you should think of it as reminding you about relationships between that quantity and others. Why are the units of Newton's constant $G$ ${\rm N\ m^2/kg^2}$? It's because $G$'s &quot;purpose in life&quot; is to be multiplied by a couple of masses and divided by a squared distance, leaving you with a force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, this sometimes means that, at least when you're new to a quantity, it's often nice &lt;em&gt;not&lt;/em&gt; to reduce its units to the simplest form. Lubos's example is probably the best one here: The meaning of ${\rm m/s^2}$ is obscure to some beginning physics students, whereas $\rm (m/s)/s$ or &#xA;$${\rm m/s}\over {\rm s}$$&#xA;is a clearer reminder of the meaning. (If you write it in the first form, please use the parentheses. Older books used to write the potentially ambiguous m/s/s.) Similarly, I wrote the units of $G$ in the form I did because that's the way it's easy to remember. It's equivalent to $\rm m^3/(kg\ s^2)$, but the &quot;meaning&quot; of this is harder to see at a glance.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-24T16:05:50.603" />
  <row Id="9069" PostTypeId="1" CreationDate="2011-04-24T20:44:32.027" Score="3" ViewCount="166" Body="&lt;p&gt;I have some questions about the statistics of Bell Tests. For convenience I'm going to refer to the test described in this paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/quant-ph/0205171&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/quant-ph/0205171&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They show a coincidence rate of 300 counts with polarizers aligned. I'm wondering how many counts there would be if you took out both polarizers altogether, and how many of those would be coincidence counts. I'm interested in the ideal case of 100% detector efficiency and 100% downcoversion efficiency, but I also wonder how close the ideal case can be approached in practise.&lt;/p&gt;&#xA;" OwnerUserId="3253" LastEditorUserId="2936" LastEditorDisplayName="" LastEditDate="2011-04-24T22:06:08.243" LastActivityDate="2011-05-24T23:15:00.130" Title="Bell Test statistics" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="9070" PostTypeId="2" ParentId="9000" CreationDate="2011-04-24T22:09:43.723" Score="2" ViewCount="" Body="&lt;p&gt;For a relativistic particle (free or in an external filed), the Lagrangian contains mass differently, not in a &quot;quadratic&quot; way.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-04-24T22:09:43.723" />
  <row Id="9071" PostTypeId="2" ParentId="9052" CreationDate="2011-04-25T00:53:46.713" Score="2" ViewCount="" Body="&lt;p&gt;Every now and then I run into this question from students. I approach it as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In computing how much paint you will need to cover a wall, it's natural to think in terms of square meters $m^2$. Students seem to accept this intuitively. The problem arises when having to think about things like time squared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) When you use units of square meters you must recognize that this is already a derived unit. You don't use a square meter to measure square meters. You use a meter stick, and then you use some math. So you already lost your virginity in square meters, square seconds is just another step on the path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) When you go down to buy a car (in the US), one of the attributes they will sell you the car on, in terms of measuring its acceleration, is how many seconds it takes to reach a speed, typically 60 mph. So a car might take 10 seconds to reach 60 miles per hour. Assuming the acceleration is constant, this is 6 miles per hour each second. We write this as&#xA;$$\frac{\textrm{60 miles per hour}}{\textrm{10 seconds}} = &#xD;&#xA;\frac{\textrm{6 miles per hour}}{\textrm{second}}.$$&#xA;And this can be rewritten as:&#xA;$$\frac{\textrm{6 miles /hour}}{\textrm{second}}&#xD;&#xA;= \frac{\textrm{6 miles /hour}}{\textrm{second/1}}$$&#xA;$$= \frac{\textrm{6 miles}}{\textrm{hour}} \frac{\textrm{1}}{\textrm{second}}$$&#xA;$$= \frac{\textrm{6 miles}}{\textrm{hour-second}}$$&#xA;and so we are left with the unit &quot;hour-second&quot; which indeed is time squared, and is easily seen to be equal to minute$^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, we must always remember that the units we use are there only to allow us to make calculations. We defined them, we use them, nature abides by them, but She does not contain them.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-05-02T00:38:43.843" LastActivityDate="2011-05-02T00:38:43.843" />
  <row Id="9072" PostTypeId="1" CreationDate="2011-04-25T01:55:10.597" Score="4" ViewCount="143" Body="&lt;p&gt;I know that AM radio station signals fluctuate in quality with time of day because of changes in the ionosphere. How do these changes trend with time of day?&lt;/p&gt;&#xA;" OwnerUserId="3135" LastEditorUserId="3135" LastEditorDisplayName="" LastEditDate="2011-04-25T02:45:42.707" LastActivityDate="2011-04-25T10:13:17.307" Title="How do AM radio signals change with time of day?" Tags="&lt;weather&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9073" PostTypeId="2" ParentId="9030" CreationDate="2011-04-25T02:02:10.460" Score="0" ViewCount="" Body="&lt;p&gt;To the last comment - I disagree.  Wind turbines operate in a fairly different flow regime, and compare the table fan to the prop of a ship, they are about the same shape.  The key difference is moving fluid volume.  The larger blades push against the fluid more strongly, which is desirable in the fan case and in the ship case.  Wind turbines are different as the builder cares nothing about the exhaust wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway...&#xA;1. I don't think it's 'functional' in any sense other than the fact that the motor is behind it.&#xA;2. I don't believe so and I agree with other commentators that the fan would be more simple and less noisy without them.  I'll withhold comment on the obvious safety considerations.&#xA;3. Everything dealing with pumping or pushing a fluid makes noise, and it is quite considerable.  You could remove the blades and listen to the electric motor itself and you would find it constitutes only a small fraction of the noise.  Noise minimization for fans is a major engineering topic and it's more or less impossible to reduce it.  Obviously, the faster you get the bigger of a problem it is.  I believe we're talking about turbulent flow in general, I don't think many pumps are actually laminar.  The metal bars will have an influence, but I could not say how much.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-04-25T02:02:10.460" />
  <row Id="9074" PostTypeId="2" ParentId="9072" CreationDate="2011-04-25T03:33:31.423" Score="4" ViewCount="" Body="&lt;p&gt;I'll give this one a shot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ionosphere is a region of the upper atmosphere, where the atoms and molecules are ionized by photons coming primarily from the sun, but also from cosmic rays. During the day, the ionosphere is said to be &quot;charged&quot; - there are loads of free electrons bumping into each other. This charged state of the ionosphere is what causes radio waves to lose strength - a phenomenon called ionospheric absorption. At night, due to the absence of ionizing sunlight, the ions recombine and this charge is drained due to absence of a source of ionizing radiation. A small part, however, remains due to cosmic rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A consequence of all this is that during the day, AM radio signals can't really travel by &lt;em&gt;skywave&lt;/em&gt;, and so travel by &lt;em&gt;groundwave&lt;/em&gt;, diffracting around the curve of the earth. Since the ground is not a perfect electrical conductor, ground waves lose strength as they follow the earth’s surface, although this loss of signal strength is low for lower frequencies. The conductivity of the surface affects propagation of groundwaves - that means if the groundwave is travelling over water, it will travel further for the same loss in signal strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, after sunset, the the AM radio signals can bounce off the bottom of the ionosphere and head back towards the earth as if they were reflected by a mirror. These 'hops' are quite large. The surface of the earth and the ionosphere form a kind of waveguide in which the AM radio signals can travel. The range using skywave is often &lt;strong&gt;much&lt;/strong&gt; larger than it is using groundwave, which is why AM radio stations can be heard much farther from their point of origin during the night than during the day. Since skywave propagation depends on the state of the ionosphere, it should be sensitive to ionospheric disturbances caused by solar flares or geomagnetic storms. &lt;/p&gt;&#xA;" OwnerUserId="3252" LastEditorUserId="3252" LastEditorDisplayName="" LastEditDate="2011-04-25T03:52:19.760" LastActivityDate="2011-04-25T03:52:19.760" CommentCount="1" />
  <row Id="9075" PostTypeId="2" ParentId="6446" CreationDate="2011-04-25T04:08:45.763" Score="0" ViewCount="" Body="&lt;p&gt;Brian Greene author of 2 books on Superstrings discovered something about black holes evaporating.  So did Witten at the same time but using a different approach.  They agreed to submit their discovery to arXiv.com at the same time and did.  arXiv is not peer reviewed.  So, are their discoveries without merit?  Recently arXiv.com requires an endorser supposedly to rule out &quot;junk&quot; math-science. However I noticed that a number of endorsers have cited works of El Naschie who seems to be catching a lot of flack from 1000's of people.  I am not in a position to decide for myself whether his works are those of a crackpot or numerologist.  What he seems to say seem to be interesting and maybe worthwhile.  I'll just wait until the dust settles before deciding what to make of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SiG&lt;/p&gt;&#xA;" OwnerUserId="3273" LastActivityDate="2011-04-25T04:08:45.763" />
  <row Id="9076" PostTypeId="1" CreationDate="2011-04-25T05:58:04.840" Score="10" ViewCount="533" Body="&lt;p&gt;I was confused when I was listening to a Quantum Mechanics lecture online. Are space and time assumed to be continuous or discrete in Quantum Mechanics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see the question is vague, but this is so since I am confused.&lt;/p&gt;&#xA;" OwnerUserId="403" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-04-25T13:45:11.300" LastActivityDate="2011-04-25T19:09:59.250" Title="Does Quantum Mechanics assume space and time are continuous?" Tags="&lt;quantum-mechanics&gt;&lt;spacetime&gt;&lt;quantization&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="9077" PostTypeId="2" ParentId="9076" CreationDate="2011-04-25T06:03:03.073" Score="8" ViewCount="" Body="&lt;p&gt;Space and time are continuous, in quantum mechanics or otherwise. In particular, whenever our theories of any kind talk about time, it is always a real continuous parameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, spatial positions of particles in ordinary quantum mechanics are operators $\hat x$ whose eigenvalues are continuous, too. This fact is related to the continuity of time by the Lorentz symmetry required by the special theory of relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The word &quot;quantum&quot; in &quot;quantum mechanics&quot; is often misunderstood. It refers to the fact that some quantities - such as the angular momentum and energy of bound states (pairs of particles orbiting each other, for example) - may be quantized i.e. take discrete values. However, it's not true that &lt;em&gt;all&lt;/em&gt; quantities have to be discrete. Positions and momenta of particles are continuous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Momenta may become discrete, much like energy, but only if particles are propagating in a finite region of space.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many people have invested thousands of manhours to the research of the possibility that space or even time are discrete. However, the results of their research have only reinforced the conclusion that such theories are inevitably inconsistent.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-25T06:03:03.073" CommentCount="13" />
  <row Id="9079" PostTypeId="2" ParentId="9076" CreationDate="2011-04-25T06:22:19.073" Score="2" ViewCount="" Body="&lt;p&gt;No, space and time are continuous in ordinary quantum mechanics (not QFT or quantum gravity or string theory or whatever). Keep in mind, though, that this does not constitute a prediction that spacetime is &lt;em&gt;actually&lt;/em&gt; continuous. Basic quantum mechanics has nothing to say on that matter. It's simply that the theory as it is normally taught and used is constructed on a continuous manifold.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-25T18:02:49.057" LastActivityDate="2011-04-25T18:02:49.057" CommentCount="4" />
  <row Id="9080" PostTypeId="2" ParentId="8828" CreationDate="2011-04-25T06:56:44.683" Score="1" ViewCount="" Body="&lt;p&gt;It's possible I have the terminology wrong, because this isn't my field, but as far as I can tell &quot;current quark&quot; masses &lt;strong&gt;are&lt;/strong&gt; &quot;running&quot; masses. The distinction isn't between &quot;current&quot; masses and &quot;running&quot; masses, it's between &quot;current quark&quot; masses and &quot;constituent quark&quot; masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's back up and start from some basics. The reason quark masses are hard to define compared to, for example, lepton masses is that free quarks don't exist in the IR limit. QFT says that the &quot;bare&quot; masses in the Lagrangian are infinite for all particles, but divergent loop contributions to the propagator cancel them out to give finite &quot;dressed&quot; masses that you actually measure. So in the case of an electron, for example, you plug in the experimentally measured electron mass as an input parameter to your theory, and QFT tells you that, well, the bare electron mass must really be infinite, but there's a nice well-defined way that the mass &quot;runs&quot; from infinity at very small length scales, to a constant at very large length scales (&quot;IR fixed point&quot;). So when someone just says &quot;electron mass&quot; they're talking about this IR-limit value, which is the same as the experimentally measured value (although I remember hearing that in some high-energy collisions of electrons they could actually measure an electron mass significantly different from the IR-limit value).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For quarks masses the same kind of &quot;running&quot; happens, but instead of converging to a constant, they diverge at the energy scale called $\Lambda_\text{QCD}$. So if you ask the question &quot;what should the quark masses be for me, a macroscopic observer?&quot;, the answer is undefined, because they already became infinite at a much smaller length scale. This makes perfect sense because quarks are confined into hadrons and can't be observed macroscopically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The masses that PDG gives, then, are the values of the &quot;running&quot; masses at some energy scale greater than (length scale smaller than) $\Lambda_\text{QCD}$, defined in some specific renormalization scheme. Specifically, it says they're the values in the $\overline{MS}$ scheme at renormalization scale $\mu = 2\, \text{GeV}$. Without this information the mass numbers are pretty much meaningless, because at a renormalization scale closer to $\Lambda_\text{QCD}$ the masses will of course be much bigger. The way you can convert between different renormalization scales is to use the renormalization group equation, which can be calculated in perturbation theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that's what &quot;running&quot;/&quot;current quark&quot; masses are. It's a pretty subtle concept but it's rigorously defined because the $\overline{MS}$ renormalization scheme is. If somebody else calculates the mass at a different scale or with a different renormalization scheme, they'll get a different number, but you can compare them by using the renormalization group equation (as long as all the scales are far to the UV side of $\Lambda_\text{QCD}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for &quot;constituent quark&quot; masses, I don't really know how those are defined, so I'll just quote this paragraph from the PDG explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The quark masses for light quarks discussed so far are often&#xA;referred to as current quark masses. Nonrelativistic quark models use constituent quark masses, which are of order 350 MeV&#xA;for the u and d quarks. Constituent quark masses model the&#xA;effects of dynamical chiral symmetry breaking, and are not related to the quark mass parameters $m_k$ of the QCD Lagrangian&#xA;Eq. (1). Constituent masses are only defined in the context of&#xA;a particular hadronic model.&quot;&#xA;&amp;mdash;&quot;Note on Quark Masses&quot;: &lt;a href=&quot;http://pdg.lbl.gov/2010/reviews/rpp2010-rev-quark-masses.pdf&quot; rel=&quot;nofollow&quot;&gt;http://pdg.lbl.gov/2010/reviews/rpp2010-rev-quark-masses.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that about sums it up. The &quot;constituent quark&quot; masses (which are the only other finite quark masses I've heard of) are not directly related to the mass parameters in the QCD Lagrangian, but are supposed to represent all this nonperturbative stuff that happens as you cross $\Lambda_\text{QCD}$ and quarks become very significantly &quot;dressed&quot; and confined into hadrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://en.wikipedia.org/wiki/Current_quark&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Current_quark&lt;/a&gt; , &lt;a href=&quot;http://en.wikipedia.org/wiki/Constituent_quark&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Constituent_quark&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-04-25T06:56:44.683" />
  <row Id="9081" PostTypeId="1" CreationDate="2011-04-25T08:22:59.673" Score="0" ViewCount="61" Body="&lt;p&gt;We know that solar panels don't work at their best under cloudy skies. Wind turbines also don't turn when the wind doesn't blow. I like to imagine a 100% renewable energy portfolio (solar PV/thermal, wind, ocean wave, etc.) for everybody. How do we deal with intermittence of such sources without resorting to battery banks (toxic or otherwise), and still enjoy continuous electricity supply when the sun doesn't shine or the wind doesn't blow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A crazy idea about solar PV and thermal. Given that in one full day, the sun bombards the earth's surface with enough energy that we need for one year, how about having the daytime half of the world power itself &lt;em&gt;and&lt;/em&gt; the nighttime half?&lt;/p&gt;&#xA;" OwnerUserId="293" LastActivityDate="2011-04-25T08:22:59.673" ClosedDate="2011-04-25T08:36:38.220" Title="How do we deal with the intermittence of solar and wind energy?" Tags="&lt;electricity&gt;&lt;energy&gt;&lt;electrical-engineering&gt;" CommentCount="4" />
  <row Id="9082" PostTypeId="2" ParentId="9076" CreationDate="2011-04-25T09:16:04.857" Score="6" ViewCount="" Body="&lt;p&gt;All verified quantum mechanical theories till date assumes space and time are continuous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in several approaches of quantum gravity theories (like LQG) spacetime may assume discrete structure. On the other hand in string theory spacetime is assumed continuous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are controversies and animosity between different camps of quantum gravity research. All these approaches have the shortcoming till now in that they have failed to produce any exclusive verifiable predictions (means not something like S.T. predicts Equivalence principle etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fair and honest answer is no body knows for sure whether spacetime is ultimately continuous or discrete. Only time will tell.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-04-25T09:36:50.267" CommentCount="5" />
  <row Id="9084" PostTypeId="1" CreationDate="2011-04-25T10:56:55.910" Score="2" ViewCount="593" Body="&lt;p&gt;Let us suppose that we have a known electromagnetic wave-train of finite size propagating in a certain direction. There is a probe charge on its way. This EMW is an external field for the charge. The EMW has a certain energy-momentum (integral over the whole space). After acting on the probe charge the wave continues its way away. In the end we have the energy of the initial wave (displaced somewhere father), the kinetic energy of the charge (hopefully it starts moving), and the energy of the radiated EMF propagating in other directions. Thus the total energy may become different from the initial one. How to show that the total energy is conserved in this case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not a Compton scattering. Just a regular electrodynamics problem. How EM energy can change appropriately? Via destructive interference? How to show it if the incident field is a known function of space-time? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I can emit a half-period long wave from a radio-transmitter:$E(t)=E_0 sin(\Omega t), 0 &amp;lt; t &amp;lt; \pi/\Omega $. Then the final charge velocity will be clearly different from zero:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ma=F(t), v(t&amp;gt;\pi/\Omega)=\int_{0}^{t}F(t&amp;#39;)dt&amp;#39;=\frac{2qE_0}{m\Omega}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, the charge itself radiates some new wave during acceleration period. The radiated energy is only a small fraction of $\frac{mv^2}{2}$. What can guarantee that the total energy remains the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2&lt;/strong&gt;: OK, let us simplify the task. I wonder if there is a simplest problem in CED where the total energy with a radiating charge is conserved explicitly?&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-05-21T17:15:09.163" LastActivityDate="2011-05-21T17:15:09.163" Title="Energy conservation in Electrodynamics" Tags="&lt;electrodynamics&gt;&lt;conservation-laws&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="2" />
  <row Id="9085" PostTypeId="2" ParentId="9084" CreationDate="2011-04-25T11:22:17.600" Score="2" ViewCount="" Body="&lt;p&gt;In order for energy to be conserved in a model, one must use a translation invariant formalism, then, with other assumptions about the space of models, one can prove Noether's theorem. As soon as one says that some part of the interaction is &quot;external&quot;, one is working in a formalism in which energy will not be conserved. Making the EMW &lt;em&gt;internal&lt;/em&gt; to the system is your first step.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-04-25T11:22:17.600" CommentCount="4" />
  <row Id="9086" PostTypeId="2" ParentId="9084" CreationDate="2011-04-25T12:15:47.293" Score="5" ViewCount="" Body="&lt;p&gt;The proof that classical electrodynamics conserves energy is found in all sorts of textbooks. I'd start with Griffiths's &lt;em&gt;Introduction to Electrodynamics&lt;/em&gt;, and go on to Jackson's &lt;em&gt;Classical Electrodynamics&lt;/em&gt; if you want more.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-25T12:15:47.293" CommentCount="2" />
  <row Id="9087" PostTypeId="2" ParentId="9084" CreationDate="2011-04-25T14:27:43.947" Score="10" ViewCount="" Body="&lt;p&gt;The stress energy tensor $T^{\mu\nu}$ contains all the energy/momentum components of the elctromagnetic field and the conservation of these components is expressed by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\partial_{\nu}T^{\mu \nu} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which states that the &lt;em&gt;change in time&lt;/em&gt; of energy/momentum is zero. If the above is non-zero then electromagnetic field energy/momentum is transferred to charged matter and in this case the conservation law becomes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\partial_{\nu}T^{\mu \nu} + \eta^{\mu \rho} \, f_{\rho} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $f$ is the force density four vector acting on the charge matter. If we talk specifically about the energy, as in your case, then $f_o$ is given by $\vec{J}\cdot\vec{E}~$ representing charged matter moving up or down a potential field which causes a change in time  of the potential energy of the charged matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards, Hans  &lt;/p&gt;&#xA;" OwnerUserId="3197" LastActivityDate="2011-04-25T14:27:43.947" CommentCount="5" />
  <row Id="9088" PostTypeId="1" AcceptedAnswerId="9093" CreationDate="2011-04-25T15:01:16.540" Score="2" ViewCount="161" Body="&lt;p&gt;Perhaps this should be a chemistry question, but it seems to have physics attributes. There's a perennial ozone &quot;hole&quot; around the south pole created by destruction from Cl based chemicals like CFCs. Given that most CFC emissions originated in the northern hemisphere, why isn't there a significantly similar effect around the north pole?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-05-03T16:51:40.077" Title="Ozone Hole in the South Pole" Tags="&lt;air&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="9089" PostTypeId="1" AcceptedAnswerId="9095" CreationDate="2011-04-25T15:09:14.677" Score="5" ViewCount="236" Body="&lt;p&gt;I am trying to teach myself some basic physics, here is something I do not really understand about the definition of work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when moving from $a$ to $b$ (in one dimension), the work done is defined to be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ W = \int_{a}^{b} F dx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, say, I lift something up from the floor to a table covering a distance $h$. The work done is supposed to be $mgh$. This result is obtained by setting $F = mg$ in the equation above, which is the force exerted by gravity on the object. However, gravity is not the only force acting upon the object. What about the force used by me to lift the object, why does it not enter the equation?&lt;/p&gt;&#xA;" OwnerUserId="3275" LastActivityDate="2011-04-25T16:50:07.897" Title="What exactly is &quot;F&quot; in $W = \int_{a}^{b} F dx$?" Tags="&lt;classical-mechanics&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" />
  <row Id="9090" PostTypeId="2" ParentId="9089" CreationDate="2011-04-25T15:20:16.733" Score="2" ViewCount="" Body="&lt;p&gt;The work by your lifting is equal and opposite the gravitational mgh work via Newton's Third Law.&lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="1924" LastEditorDisplayName="" LastEditDate="2011-04-25T15:28:49.630" LastActivityDate="2011-04-25T15:28:49.630" CommentCount="1" />
  <row Id="9091" PostTypeId="2" ParentId="9089" CreationDate="2011-04-25T15:35:37.990" Score="1" ViewCount="" Body="&lt;p&gt;The force that you use is not included because the work is defined with the idea in mind to account for the work that &lt;em&gt;you&lt;/em&gt; must put into a system to get it from $a$ to $b$. In the case of the table of height $h$, it is $mgh$, as you correctly point out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you do not work against yourself, you don't include your own force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other point is that work done on a system is (at least in so-called conservative potentials) not lost, but stored in the object: If you lift your object to the table, you have done work $mgh$ and this is stored as &lt;em&gt;potential energy&lt;/em&gt; in the object. If you let the object drop from the table, this energy gets converted into kinetic energy. At the bottom, you'll have $\frac{1}{2}mv^2 = mgh$, so you can calculate the velocity with which the object will hit the ground. The force that you used to lift the object does not contribute to this potential energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, in addition to gravity, a friend of yours would also pull down on the object with a constant force, then you'd have to add the force &lt;em&gt;he&lt;/em&gt; uses to the gravitational force in order to figure out your work.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-04-25T15:35:37.990" />
  <row Id="9092" PostTypeId="2" ParentId="9089" CreationDate="2011-04-25T15:44:13.820" Score="3" ViewCount="" Body="&lt;p&gt;I'm going to bet that you're being asked what the work done by nonconservative forces is.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Properly, you can imagine a conservative force being a force that satisfies $\nabla \times F =0$.  If this is true, then the net work done by this force around any closed curve is zero.  It turns out that there is also a central identity of vector calculus that tells you that if $\vec F$ is conservative, then for some scalar field $V(x)$, you have $\vec F=-\vec \nabla V$.  Then, you start with the work energy theorem, and split your forces into conservative and nonconservative forces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}\Delta KE &amp;amp;= \int \vec F \cdot d\vec x\\ &amp;amp;=\int (\vec F_{con} + \vec F_{noncon})\cdot d\vec x\\ &amp;amp;= -\int (\vec \nabla V)\cdot d\vec x + W_{noncon} \\ \Delta KE + V_{f} - V_{i} &amp;amp;= W_{noncon} \\ \Delta KE + \Delta PE &amp;amp;= W_{noncon} \end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where we defined $\Delta PE$ as the total change in $V(x)$ over our path, and we used the fundamental theorem of calculus to do the integral defining the work done by conservative forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the change in mechanical energy of the system is given by the work done by nonconservative forces.  Since your system gains $mgh$ potential energy, it was necessary that external forces did $mgh$ of work.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you are right in saying that the &lt;i&gt;net&lt;/i&gt; work on the system is zero--the change in kinetic energy of the system is zero, so the work-energy theorem says that the net work done must also be zero.  The work done by the gravitational force cancels out the work done in lifting the book.  It's just that we rarely treat problems this way, preferring to account for conservative forces using the concept of potential energy.&lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2011-04-25T16:04:54.310" LastActivityDate="2011-04-25T16:04:54.310" />
  <row Id="9093" PostTypeId="2" ParentId="9088" CreationDate="2011-04-25T15:54:10.820" Score="5" ViewCount="" Body="&lt;p&gt;From the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ozone_depletion&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;:(My emphasis)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some people thought that the ozone hole should be above the sources of CFCs. However, CFCs are &lt;strong&gt;well mixed globally&lt;/strong&gt; in the troposphere and the stratosphere. The reason for occurrence of the ozone hole above Antarctica is &lt;strong&gt;not because there are more CFCs concentrated but because the low temperatures help form polar stratospheric clouds&lt;/strong&gt;.[82] In fact, there are findings of significant and localized &quot;ozone holes&quot; above other parts of the earth.[83]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The chemical processes are complicated and involve UV-dissociation of the CFCs to release chlorine which then destroys the ozone. However:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Cl-catalyzed ozone depletion can take place in the gas phase, but it is dramatically enhanced in the presence of polar stratospheric clouds (PSCs).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first quote states that the conditions over the Antarctic are more favorable for the chemical process of ozone depletion to occur.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-04-25T15:54:10.820" CommentCount="4" />
  <row Id="9094" PostTypeId="2" ParentId="2542" CreationDate="2011-04-25T16:08:39.953" Score="1" ViewCount="" Body="&lt;p&gt;Leonidas is completely wrong! The color of light produced from a light emitting diode is manly dependent of the band gap of the semiconductor material that is used to produce the LED.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&#xA;• Infrared: GaAs or AlGaAs&#xA;• Red: GaInAlP or GaAlP&#xA;• Yellow/Orange: GaInAlP or GaAlP.&#xA;• Green: GaAlP or InGaN&#xA;• Blue: GaN or InGaN.&#xA;• Ultraviolet: GaN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may note that some of the materials are the same for different colors of light. This is due to the change in band gap that can be produced by changing the doping concentration and doping materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some LEDs have a layer of fluorescent material though. To shift the color slightly in an easy way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to answer the original question:&#xA;Yes, as BarsMonster pointed out, there are Ultraviolet LEDs, but the selection is not that big.&lt;/p&gt;&#xA;" OwnerUserId="3276" LastActivityDate="2011-04-25T16:08:39.953" CommentCount="2" />
  <row Id="9095" PostTypeId="2" ParentId="9089" CreationDate="2011-04-25T16:40:00.650" Score="9" ViewCount="" Body="&lt;p&gt;What I've been telling the students in the intro mechanics class I TA is that each force corresponds to some amount of work. The individual contributions of work add up to the net (or total) work, just as the individual forces add up to the net force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for instance, when you lift a box up from the floor to a table, there are two forces acting on that box: the force $\vec{F}_\text{lift}$ you apply to lift it, and the force of gravity, $\vec{F}_g = -mg\hat{y}$. You can calculate the amount of work done by the lifting force as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W_\text{lift} = \int_0^h \vec{F}_\text{lift}\cdot\mathrm{d}\vec{x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you can calculate the amount of work done by the gravitational force as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W_g = \int_0^h \vec{F}_g\cdot\mathrm{d}\vec{x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a typical physics-problem scenario, you would be lifting the box at constant velocity (except for the brief moments when you start and stop its motion, which I will ignore for now). In this case, you could draw a free-body diagram and use Newton's second law $\vec{F} = m\vec{a}$ to discover that your lifting force has to be equal in magnitude and opposite in direction to the gravitational force:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{F}_\text{lift} = mg\hat{y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Armed with this result, you can calculate the work done by each individual force (try it!): $W_\text{lift} = mgh$ and $W_g = -mgh$. The net work done is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can just as well calculate the net work from the net force,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W_\text{net} = \int_0^h \vec{F}_\text{net}\cdot\mathrm{d}\vec{x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the net force is $\vec{F}_\text{lift} + \vec{F}_g$, which is zero (because the box is moving at constant velocity, remember $\vec{F}_\text{net} = m\vec{a}$). So, again, the net work is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;But wait,&quot; you say, &quot;I thought the work done was $mgh$!&quot; Well, it is, if you only consider the work &lt;em&gt;you&lt;/em&gt; did to lift the box. This is what you would measure if you used a crane to lift the box, for example: the amount of work that needs to be done by the crane is $mgh$. However, gravity did the opposite amount of work, which canceled out the work you did (except for those brief moments at the beginning and the end, but those two contributions cancel each other out anyway). So the total work done by &lt;em&gt;all&lt;/em&gt; forces involved is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might seem a little confusing - perhaps you're wondering, if the total work done is zero, how did the gravitational potential energy increase from $0$ to $mgh$? The trick there is that the &quot;work done by gravity&quot; $W_g$ is actually the gravitational potential energy under a different name. For conservative forces, we define $\Delta U_i = -W_i$: the change in a certain kind of potential energy is the negative of the work done by the force corresponding to that potential energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the gravitational force does work $-mgh$, so the change in gravitational potential energy should be $mgh$. But once you start calling it a change in gravitational potential energy, you can no longer call it a contribution to the work - and that leaves you with just the $W_\text{lift} = mgh$ contribution. So in that case, if you label the &quot;action&quot; of gravity as a change in potential energy rather than a work, the work done is $mgh$.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-25T16:50:07.897" LastActivityDate="2011-04-25T16:50:07.897" CommentCount="4" />
  <row Id="9096" PostTypeId="1" CreationDate="2011-04-25T17:24:11.970" Score="2" ViewCount="143" Body="&lt;p&gt;Rocket engine, perpendicular to the end of stick, located in cosmos, without atmosphere and gravitation. As in picture below:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/DM0AM.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In what direction will the stick move? In &lt;em&gt;V&lt;/em&gt; or another direction?&lt;/p&gt;&#xA;" OwnerUserId="3277" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-04-25T18:35:20.290" LastActivityDate="2011-04-27T02:15:33.633" Title="Rocket engine, perpendicular to the end of stick, in space" Tags="&lt;forces&gt;&lt;newtonian-mechanics&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="9097" PostTypeId="2" ParentId="9096" CreationDate="2011-04-25T17:29:14.113" Score="0" ViewCount="" Body="&lt;p&gt;Without atmosphere you have the simple case of no movement due to no atmosphere for the jet action to work upon, and the not so simple case of the stick rotating in a plane perpendicular to &lt;em&gt;v&lt;/em&gt; due to conservation of momentum when its turbine blades rotate.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-04-25T17:29:14.113" CommentCount="3" />
  <row Id="9098" PostTypeId="1" AcceptedAnswerId="9100" CreationDate="2011-04-25T18:00:43.047" Score="3" ViewCount="164" Body="&lt;p&gt;I know outside a nucleus, neutrons are unstable and they have half life of about 15 minutes. But when they are together with protons inside the nucleus, they are stable. How does that happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got this from wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When bound inside of a nucleus, the instability of a single neutron to beta decay is balanced against the instability that would be acquired by the nucleus as a whole if an additional proton were to participate in repulsive interactions with the other protons that are already present in the nucleus. As such, although free neutrons are unstable, bound neutrons are not necessarily so. The same reasoning explains why protons, which are stable in empty space, may transform into neutrons when bound inside of a nucleus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I don't think I get what that really means. What happens inside the nucleus that makes neutrons stable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it the same thing that happens inside a neutron star's core? Because, neutrons seem to be stable in there too.&lt;/p&gt;&#xA;" OwnerUserId="1927" LastActivityDate="2011-04-25T18:58:47.237" Title="How come neutrons in a nucleus don't decay?" Tags="&lt;particle-physics&gt;&lt;neutron-stars&gt;&lt;weak-force&gt;&lt;neutron&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="9099" PostTypeId="1" AcceptedAnswerId="9103" CreationDate="2011-04-25T18:08:15.583" Score="0" ViewCount="100" Body="&lt;p&gt;Does anyone knows if the chiral ring of the Kazama-Suzuki models have been found? If yes, please send me the reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1531" LastActivityDate="2011-04-25T18:32:26.093" Title="The chiral ring of N=2 WZW models" Tags="&lt;string-theory&gt;" AnswerCount="1" />
  <row Id="9100" PostTypeId="2" ParentId="9098" CreationDate="2011-04-25T18:13:14.247" Score="6" ViewCount="" Body="&lt;p&gt;Spontaneous processes such as neutron decay require that the final state is lower in energy than the initial state. In (stable) nuclei, this is not the case, because the energy you gain from the neutron decay is lower than the energy it costs you to have an additional proton in the core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For neutron decay in the nuclei to be energetically favorable, the energy gained by the decay must be larger than the energy cost of adding that proton. This generally happens in neutron-rich isotopes:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/tt0Vd.png&quot; alt=&quot;Radioactive decay of isotopes.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example is the $\beta^-$-decay of Cesium:&#xA;$$\phantom{Cs}^{137}_{55} Cs \rightarrow \phantom{Ba}^{137}_{56}Ba + e^- + \bar{\nu}_e$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a first impression of the energies involved, you can consult the semi-empirical &lt;a href=&quot;http://en.wikipedia.org/wiki/Bethe-Weizs%C3%A4cker_formula&quot; rel=&quot;nofollow&quot;&gt;Bethe-Weizsäcker formula&lt;/a&gt; which lets you plug in the number of protons and neutrons and tells you the binding energy of the nucleus. By comparing the energies of two nuclei related via the $\beta^-$-decay you can tell whether or not this process should be possible.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditorDisplayName="" LastEditDate="2011-04-25T18:58:47.237" LastActivityDate="2011-04-25T18:58:47.237" CommentCount="1" />
  <row Id="9101" PostTypeId="2" ParentId="9098" CreationDate="2011-04-25T18:16:19.397" Score="1" ViewCount="" Body="&lt;p&gt;The Pauli Exclusion Principle states that no two identical fermions (neutrons and protons are fermions - they have half-integer spins and obey Fermi-Dirac statistics) can occupy the same quantum state at the same time. If the neutron were to $\beta$-decay as:&#xA;\begin{equation}&#xA;n \longrightarrow p + e^- + \bar{\nu_e}&#xA;\end{equation}&#xA;then this freshly minted proton will try to occupy the quantum state with the lowest possible energy. However, since there are already loads of protons in the nucleus, this 'new' proton can't do that, and so will be forced to occupy a state with higher energy. In order to get to that state, it must absorb some energy. This is why neutrons don't usually $\beta$-decay inside the nucleus. Do remember that $\beta$-decay of neutrons inside the nucleus isn't unheard of - just uncommon.&lt;/p&gt;&#xA;" OwnerUserId="3252" LastActivityDate="2011-04-25T18:16:19.397" CommentCount="2" />
  <row Id="9102" PostTypeId="2" ParentId="9096" CreationDate="2011-04-25T18:31:47.403" Score="2" ViewCount="" Body="&lt;p&gt;The obvious answer is: &quot;In a different direction.&quot;  Or, more correctly, the direction of vector V will change continuously.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The engine will force the stick to rotate, continuously gaining angular velocity.  In addition, assuming the stick begins at rest, there will be a linear component to the AVERAGE OVERALL velocity of the center of mass of the contraption.  But it will not  be in the direction of vector 'V' as you have it drawn.  Somewhere slightly clockwise of 'V' for most reasonable choices of (impulse/mass) and (impulse/rotational_inertia)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming duration of thrust is long enough, vector 'V' will eventually point in EVERY direction in the 2-D plane of the page in which it is drawn, multiple times.  Any more significant detail than that requires more information on the mass, center of mass, moment of internia, initial angular velocity, thrust magnitude, and thrust duration.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-04-25T20:31:48.660" LastActivityDate="2011-04-25T20:31:48.660" />
  <row Id="9103" PostTypeId="2" ParentId="9099" CreationDate="2011-04-25T18:32:26.093" Score="1" ViewCount="" Body="&lt;p&gt;See Lerche, Vafa, Warner 1989, Nuclear Physics B,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.mth.kcl.ac.uk/~anderl/lgcft/1989%20--%20Lerche,%20Vafa,%20Warner%20--%20Chiral%20Rings%20in%20N=2%20Superconformal%20Theories.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.mth.kcl.ac.uk/~anderl/lgcft/1989%20--%20Lerche,%20Vafa,%20Warner%20--%20Chiral%20Rings%20in%20N=2%20Superconformal%20Theories.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-25T18:32:26.093" CommentCount="1" />
  <row Id="9105" PostTypeId="2" ParentId="9076" CreationDate="2011-04-25T19:09:59.250" Score="1" ViewCount="" Body="&lt;p&gt;There is fundamental problem with the meaning of the notion of &quot;discrete time and space for QM&quot;. In QM if You need to look at some very small structure, You have to use very large energy. So answer to Your question may have unintended effects included. I think about situation when someone asks about how it looks like space and time form very small time intervals and small spatial structures but do not think that it implies very large energies at all! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone do not think about it it may lead to false conclusions, that &quot;normal space and time&quot; may be discrete all the time, whilst this may be not the case. At normal energies we probably may safely say that space and time are continuous enough to do not worry about any discretization because of various symmetries which are used in quantum mechanics during description of interaction with matter, about which Lubos mentioned above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the other hand if discretization of space and time will be required in order to describe effects of such interaction for very large energies - it may be not satisfactory answer to Your question, because it may be an effect ( for example unification of QGravity and QMechanics) which may appear only during early stages of Big_Bang and then do not lead to any results which may be valid now -  where gravity and quantum world are rather weekly coupled. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it may important philosophical and historical meaning but may not have any any practical meaning in a sense that there may be no way to see any effects of this discretization here and now - then QM may not need any &quot;correction from discrete time and space&quot;. &lt;/p&gt;&#xA;" OwnerUserId="1012" LastActivityDate="2011-04-25T19:09:59.250" />
  <row Id="9106" PostTypeId="1" AcceptedAnswerId="9110" CreationDate="2011-04-25T19:27:41.723" Score="8" ViewCount="217" Body="&lt;p&gt;What does it mean when we say that the CMBR is mostly gaussian? What are non-gaussianities in CMBR? How does evaluation of 3-point correlation functions of the inflaton field tells us that there is non-Gaussianity?&lt;/p&gt;&#xA;" OwnerUserId="2973" LastActivityDate="2011-04-27T19:16:18.920" Title="What exactly is meant by the &quot;Gaussianity&quot; of CMBR?" Tags="&lt;cosmology&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="9107" PostTypeId="2" ParentId="9096" CreationDate="2011-04-25T19:32:27.007" Score="4" ViewCount="" Body="&lt;h3&gt;Assumptions&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The rocket is &quot;magic&quot; - it has no weight and the fuel it burns doesn't decrease the mass of the system.  This can also be thought of as working in the limit where the time of burning a more realistic rocket is small enough that the mass of fuel burned is negligible compared to the mass of the stick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force supplied by the rocket is constant and always perpendicular to the orientation of the stick.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Hueristic&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Initially, the stick will move in the direction of $\vec{V}$, but it will also start rotating, and as it does, its path will curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stick spins faster and faster.  Call a rotation of the stick the time during which $\theta$, the angle the stick makes with horizontal, starts at a multiple of $2\pi$ and increases to the next multiple of $2\pi$.  In each rotation the stick spins a full circle, and the rocket points in all directions once.  Thus, the forces at different times in the same rotation mostly cancel, and the stick doesn't simply blast away.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the stick spins a bit more slowly at the beginning of each rotation than at the end.  At the beginning, the thrust is pointing up and to the right, so there is a bit more momentum transfer in those directions.  The net result is that the stick picks up a little bit of momentum up and to the right with each spin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the rocket spins faster and faster, the percentage change in its spin rate during each spin goes down, and the time-averaged acceleration of the center of mass decreases.  &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Calculation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The stick is subjected to a constant torque $Fl$, with $F$ the force exerted on the stick by the rocket and $l$ the half-length of the stick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moment of inertia of the stick about its center is $ml^2/3$, with $m$ the mass of the stick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angular acceleration of the stick is the torque divided by the moment of inertia, $Fl/(ml^2/3) = 3F/ml$.  Integrating twice with respect to time, we get the angle the stick makes with horizontal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta(t) = \frac{3F}{2ml}t^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we choose units correctly we can set $3F/2ml = F = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motion of the center of mass comes from $F = ma$.   At any given moment, the force on the stick has components&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_x = \sin\theta = \sin(t^2)$$&#xA;$$F_y = \cos\theta = \cos(t^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating these gives the motion of the center of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integrals for velocities are called &lt;a href=&quot;http://en.wikipedia.org/wiki/Fresnel_integral&quot; rel=&quot;nofollow&quot;&gt;Fresnel Integrals&lt;/a&gt;.  They are mostly used in optics.  The intriguing result is that a plot of the velocity as a function of time is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler_spiral&quot; rel=&quot;nofollow&quot;&gt;Cornu Spiral or Euler Spiral&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/3zMY5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture is copied from Wikipedia.  It's a little bit backwards because $x$ is usually taken to be the integral of $\cos t^2$, whereas we've taken it as the integral of $\sin t^2$.  If you imagine rotating the picture 90 degrees clockwise, then mirror-flipping left and right, you get something that matches up with the original drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left half of the (transformed) picture shows the velocity integrating backwards into the past.  The right half shows it integrating into the future.  The velocity is asymptotic to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_x = v_y = \frac{\sqrt{\pi/2}}{2} \approx .627$$&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-04-25T19:46:33.650" LastActivityDate="2011-04-25T19:46:33.650" CommentCount="11" />
  <row Id="9108" PostTypeId="1" AcceptedAnswerId="9119" CreationDate="2011-04-25T19:53:13.750" Score="3" ViewCount="117" Body="&lt;p&gt;What happens if different size atoms meet?  We've just created anti-helium, I think.  What if one atom of anti-helium collided with one atom of iron.  Would some of the iron be left over as a new element?&lt;/p&gt;&#xA;" OwnerUserId="3282" LastActivityDate="2011-04-25T21:41:17.067" Title="Matter-antimatter annihilation" Tags="&lt;nuclear-physics&gt;" AnswerCount="1" />
  <row Id="9109" PostTypeId="1" CreationDate="2011-04-25T20:00:25.563" Score="4" ViewCount="364" Body="&lt;p&gt;The Schrodinger equation in non-relativistic quantum mechanics yields the time-evolution of the so-called wavefunction corresponding to the system concerned under the action of the associated Hamiltonian. And this wavefunction is, in general, complex, and its modulus squared yields the probabilities observed experimentally. Though, perhaps, this question has been asked many times, I am wondering if there is a direct physical interpretation - something that physically corresponds to - the wavefunction. Or is it just an intermediate calculational tool to arrive at the appropriate predictions for experimental outcomes, and nothing more? Of course, things like superposition and interference effects follow from the complex nature of the probability amplitude. So there must be something physical about it. What is it? Or are we not &lt;em&gt;supposed&lt;/em&gt; to ask that question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is is because the probability amplitude is complex that we have difficulty in relating it to something physical? Can we do quantum mechanics without complex numbers?&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2011-04-26T16:22:11.697" Title="Is there a direct physical interpretation for the quantum mechanical wavefunction?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="3" />
  <row Id="9110" PostTypeId="2" ParentId="9106" CreationDate="2011-04-25T20:10:18.770" Score="3" ViewCount="" Body="&lt;p&gt;The physical meaning of gaussianity is that the CMBR is as random as possible, with no correlation at all between the independent fourier components of the fluctuation field. Each fourier mode of the inflationary perturbations that gave rise to the CMBR anisotropy evolved as independent uncoupled harmonic oscillators (albeit with a time dependent frequency).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gaussianity of any cosmological variable $g(\textbf{x})$ (say, the scalar field perturbation $\delta\phi(\textbf{x})$ or the density $\rho(\textbf{x})$) is defined by the structure of the two-point correlator. If it has the form $$\langle g_{\textbf{k}}g_{\textbf{k&amp;#39;}}\rangle=(2\pi)^3\delta^3_{\textbf{k}+\textbf{k&amp;#39;}}P_g(k)$$ then &#xA;$g(\textbf{x})$ is said to be Gaussian, where we've also assumed isotropy so that the power spectrum depends only on the magnitude $k$. (Homogenity means that $\langle g_{\textbf{k}}\rangle=0$). The 3-point-correlator (and all odd point correlators) vanish. This is the minimal possible correlation that you can have in a stochastic field. You're allowed to have any form for the power spectrum $P_g(k)$, although inflation predicts a nearly scale invariant spectrum, with it depending very weakly on $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're wondering where the &quot;gaussianity&quot; actually comes in, it can be shown that the probability distribution for $g(\textbf{x})$ at any given point actually follows an exact gaussian: $$\textbf{P}(g)=\frac{1}{\sqrt{2\pi\sigma}}\exp{(-\frac{g^2}{2\sigma^2})}$$ So, for things like the scalar field, density and curvature, zero-deviation is most likely and higher fluctuations are exponentially suppressed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can start introducing additional correlations through the 3-point-correlator by making it non-zero and then you will have destroyed the gaussian structure of $\textbf{P}(g)$ above, hence it's called non-gaussianity. Such deviations are predicted by some models of inflation, but none have been observed so far. Tighter constraints on cosmological parameters from experiments like Planck should start coming in soon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chapter 6 of Lyth and Liddle's &lt;em&gt;Primordial Density Perturbation&lt;/em&gt; is a good reference for this.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-04-26T01:01:18.497" LastActivityDate="2011-04-26T01:01:18.497" CommentCount="8" />
  <row Id="9111" PostTypeId="1" CreationDate="2011-04-25T20:14:41.130" Score="12" ViewCount="240" Body="&lt;p&gt;Quantum fields are presented as operator-valued &lt;em&gt;distributions&lt;/em&gt;, so that the &lt;em&gt;operators&lt;/em&gt; in the theory are linear functionals of some test function space. This works well for free fields, giving us a particular form for VEVs, in which the 2-point connected correlation function determines the whole theory, and a trivial S-matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For interacting fields, however, the VEVs and the statistics of S-matrix observables that are predicted by the theory depend not only on the test functions (generally given for the S-matrix in terms of pure, improper wave-number modes of the free field at $t=\pm\infty$) but also on the &quot;energy scale of measurements&quot;. One obtains a different functional form at different energy scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy scales at which one is measuring are already encoded in the wave-numbers associated with the test functions, so it seems that the test functions play extra duty, to determine the functional form of the VEVs (at least in their S-matrix form) both by linearly smearing the quantum field and, more indirectly, by determining the renormalization scale. As a consequence, the VEVs, the Wightman functions, seem clearly to be nonlinear functionals of the test functions for interacting QFTs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've worried at this for a while. The argument as put brings into question the Wightman axioms' insistence that a quantum field must be a linear map from the test function space to the space of operators, although I suppose the effect would be only logarithmic in some measure of the frequencies of the test functions. The Haag-Kastler additivity axiom also seems problematic. I'm not aware of weakening this particular aspect of the Wightman or Haag-Kastler axioms so as to be able to construct models of the axioms that parallel empirically successful QFT having been discussed in the literature, but have they?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Note that dropping the linearity of the map from test functions to operators in the algebra of observables does not affect the linearity of the algebra. There's still a perfectly good probability interpretation of a state over a *-algebra and a perfectly good Hilbert space, but the relationship of measurements to space-time is somewhat modified.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I'll &lt;em&gt;attempt&lt;/em&gt; to make the Question slightly more accessible. It turns out that what follows makes it significantly longer. The Wightman functions (effectively a different name for Vacuum Expectation Values (VEVs)) allow the reconstruction of a Wightman (quantum) field. That is, given the vacuum state over the algebra one can construct the Wightman functions, $$W(x_1,x_2,...x_n)=\left&amp;lt;0\right|\hat\phi(x_1)\hat\phi(x_2)...\hat\phi(x_n)\left|0\right&amp;gt;,$$ but also, given the Wightman functions one can construct the algebra. Wightman functions satisfy somewhat arcane relationships, so that just any function $VV(x_1,x_2,...,x_n)$ is almost certainly not a Wightman function and cannot be used to reconstruct a Wightman field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The object $\hat\phi(x)$ is not an operator in the algebra of observables, it is an &lt;em&gt;operator-valued distribution&lt;/em&gt;, so that if we have a well-behaved &quot;test function&quot; $f(x)$, where &quot;well-behaved&quot; most often means that it is smooth (infinitely differentiable) and has a smooth fourier transform, then $\hat\phi_f=\int\hat\phi(x)f(x)\mathrm{d}^4x$ &lt;em&gt;is&lt;/em&gt; an operator. This is often called &quot;smearing&quot; with the test function $f(x)$. The test function $f(x)$ effectively tells us what relative amplification we apply to the field at each point in space-time for a given measurement, and its fourier transform $\tilde f(k)$ tells us what relative amplification we apply to the field at each point in momentum space. In signal analysis the same object is called a &quot;window function&quot;, which can be looked up on Wikipedia. One can talk about the bandwidth of a measurement, or say that a signal transformation is high-pass or low-pass. If it were not for non-commutativity of measurement operators at time-like separation, quantum field theory would be &lt;em&gt;exactly&lt;/em&gt; what one would want for a mathematical formalism for modeling stochastic signal processing. If one only talks about measurements that are space-like separated &amp;mdash;admittedly a tight constraint&amp;mdash;, QFT &lt;em&gt;is&lt;/em&gt; a stochastic signal processing formalism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast to $\hat\phi(x)$, $\hat\phi_f$ is not a singular mathematical object; the difference is comparable to the difference between a smooth function and the Dirac delta function. In terms of these operators, the Wightman functions become $$W(f_1,f_2,...,f_n)=\left&amp;lt;0\right|\hat\phi_{f_1}\hat\phi_{f_2}...\hat\phi_{f_n}\left|0\right&amp;gt;,$$&#xA;which is linear in all the test functions.&#xA;When we move to the practical formalisms of interacting QFT, however, the VEVs predicted by the bare theory are infinite. Renormalization fixes that in a Lorentz invariant way, but it leaves the VEVs a function of the energy scale $\mu$ of the measurements that are involved in an experiment. Thus, one finds that the Wightman functions are now a function of $\mu$ as well as of the points $x_i$, $W_\mu(x_1,x_2,...x_n)$, or, in terms of operators, $W_\mu(f_1,f_2,...f_n)$. [It's perhaps worth noting that we don't know whether these $W_\mu(...)$ satisfy the relationships required for them to be Wightman functions, so that we could reconstruct a Wightman field, but I've never seen a discussion of perturbative QFT  conducted in such terms.] The Wightman functions in terms of positions do not encode energy scales, but they are &lt;em&gt;not&lt;/em&gt; observables, they are only a template for constructing observables. The fourier transforms of the test functions, $\tilde f_i(k)$, however, already determine various energy scales at which the measurements made in an experiment operate, potentially in &lt;em&gt;much&lt;/em&gt; more detail than a single number $\mu$. We could just say that the energy scale of measurements $\mu$ is separate from the energy scales determined by the test functions $f_i$, but it seems more natural &lt;em&gt;to me&lt;/em&gt; to say that $\mu$ is a functional of the test functions $f_i$, in which case we can write $W&amp;#39;(f_1,f_2,...f_n)=W_\mu(f_1,f_2,...f_n)$, where $W&amp;#39;$ is now a non-linear functional of the $f_i$. I suppose we expect $W&amp;#39;$ to be a symmetric function of each of the $f_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more-or-less trivial example, suppose we have two ordinary quantized Klein-Gordon fields $\hat\phi_f$ and $\hat\xi_f$, not necessarily having the same mass spectrum, then we could construct a quantum field $\hat\Phi_f=\hat\phi_f+\hat\xi_{f+\epsilon f^2}$ (where the square $f^2$ is $[f^2](x)=[f(x)]^2$, so that $\hat\Phi_f$ satisfies microcausality), so that $\hat\Phi:\mathcal{S}\rightarrow\mathcal{A};\hat\Phi:f\mapsto\hat\Phi_f$ is a nonlinear map from the test function space into the algebra of observables. This is still a Gaussian field, so not interacting. Already there start to be significant complications. One of the most fundamental Wightman axioms is the restriction to positive spectrum, which in terms of test functions is a projection to forward light-cone components of the test functions. Only components of $\tilde f(k)$ for which $k_\alpha$ is in the forward light-cone have any effect on measurement results. In the above, however, forward light-cone components of $f^2(x)$ contribute, which in fourier space, where $f^2(x)$ is a convolution, means that negative frequency components of $f(x)$ will contribute. Although this means that this kind of nonlinearity apparently does not satisfy positivity of the energy, it's not clear that this means such a system is not &lt;strong&gt;&lt;em&gt;stable&lt;/em&gt;&lt;/strong&gt;, which is the (non-axiomatic) reason why we insist on positivity of the energy as an axiom. Which, whoops, brings into question positivity of the energy, one of the &lt;em&gt;big&lt;/em&gt; axioms, which I asked about &lt;a href=&quot;http://physics.stackexchange.com/questions/5812/there-seems-to-be-no-definition-of-stability-in-axiomatic-qft-is-there-and-i&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, of course, I'm trying out doing research here as if it was a virtual blackboard. Many of the ideas above ought to be rubbed out, different ideas tried and equally or more vigorously discarded. I've got lots of pieces of paper that have discarded ideas on them that you don't want to see, because I hope you have your own discarded ideas, as anyone who doesn't get too fixed in their ways should, most of which I probably don't want to see (not that I'm not a little fixed in my ways, but I've discarded &lt;em&gt;some&lt;/em&gt; ideas). The trouble is that I can't see how we can go forward from me just typing away, I think the feedback loop may be too long for this to work. Still, the alternative is to write a paper and get feedback six months later. This introduces Questions that I may try to put on meta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, making a Question Favorite counts for no Reps, but it's much appreciated. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-04-26T13:45:45.683" LastActivityDate="2011-06-27T18:07:33.483" Title="Does renormalization make quantum fields into (slightly) nonlinear functionals of test functions?" Tags="&lt;quantum-field-theory&gt;&lt;mathematical-physics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="9" />
  <row Id="9112" PostTypeId="2" ParentId="9059" CreationDate="2011-04-25T20:15:55.313" Score="2" ViewCount="" Body="&lt;p&gt;Answering your questions in reverse order:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, a long pointy object (like your arms over your head, in a dive, or your pointed toes in a feet-first entry) will make a big difference.  Remember the tongue-in-cheek adage, &quot;it's not the fall that kills you; it's the sudden stop?&quot;  That is exactly what differentiates a fall onto concrete from a fall into water: how sudden is the stop.  And making that stop LESS sudden (decreasing the magnitude of deceleration during the stop) is exactly how airbags save your life in a car crash.  One can decrease the magnitude of deceleration by reducing the ratio --&gt; (delta_V / delta_t).  Since there is roughly a linear relationship between time and distance traveled during the instant of impact, you can achieve the same effect by reducing the ratio --&gt; (delta_V / delta_s) where 's' = distance traveled during the deceleration event.  The easiest way to do this is to lengthen 's'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing to remember about the water fall statistics is that a large number of them are likely &quot;unpracticed.&quot;  These are not olympic divers working up to 250 feet.   A large proportion of them are unconditioned people forced into a water &quot;escape;&quot; or, worse, are people TRYING to die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you are doing the right thing, and optimizing your form for water entry, you will simultaneously be minimizing your wind resistance during the fall:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) A fall from 30 feet will result in a velocity of roughly 44 ft/s = 30 mph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) A fall from 100 feet will result in a velocity of roughly 80 ft/s = 54 mph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) A fall from 150 feet will result in a velocity of roughly 97 ft/s = 66 mph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.) A fall from 250 feet will result in a velocity of roughly 125 ft/s = 85 mph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first case is a tower jump I did for the Navy, and is trival for anyone who is HWP and doesn't belly flop.  The second is an approximation of a leap from a carrier deck, which the tower jump was supposed to teach you how to survive (be able to swim after the fall).  The third is only 20% faster entry speed (and force) and should be survivable by anyone in good shape and able to execute good form (pointed toe entry, knees locked, head up, arms straight up).  The La Quebrada cliff divers routinely dive from 125 feet as a tourist attraction.  If forced to choose, I'd pick a feet-first entry at 150 feet over a dive at 125.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the interesting part is the stretch from 150 to 250 feet.  My guess is that the limit for someone voluntarily performing repeated water dives/jumps from a height of 'x' will show 'x' to be somewhere around 225 feet +/- 25 feet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:  There are documented cases of people surviving falls from thousands of feet (failed parachute) onto LAND.  These freaky cases of surviving terminal velocity falls do not answer the question practically; but they are there.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-04-25T20:26:18.083" LastActivityDate="2011-04-25T20:26:18.083" />
  <row Id="9113" PostTypeId="1" CreationDate="2011-04-25T20:28:51.507" Score="3" ViewCount="214" Body="&lt;p&gt;Source:&#xA;&lt;a href=&quot;http://www.engineeringtoolbox.com/sound-power-intensity-pressure-d_57.html&quot; rel=&quot;nofollow&quot;&gt;http://www.engineeringtoolbox.com/sound-power-intensity-pressure-d_57.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both sound intesity and pressure level are measured in dB. Given a specific sound, are these two dB values the same?&lt;/p&gt;&#xA;" OwnerUserId="3283" LastActivityDate="2011-04-25T23:28:39.243" Title="How sound intensity (dB) and sound pressure level (dB) are related?" Tags="&lt;sound&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="9114" PostTypeId="2" ParentId="9109" CreationDate="2011-04-25T20:43:29.887" Score="2" ViewCount="" Body="&lt;p&gt;The problem is not so much that you're not supposed to ask the Question, it's more that if you ask the Question you may be swamped by many different Answers, which will have relationships between them that you may well not be able to understand unless you have already read widely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A moderately standard Answer is that the Born interpretation of the wave function gets you a long way. You can model the statistics of experimental raw data quite nicely by the probability measures that come out of the mathematics of quantum mechanics, if you get the right model for the experimental apparatus. &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; One simple-minded way to say what QM predicts when measurements don't commute (that is &lt;em&gt;not&lt;/em&gt; very standard) is that the probabilities come out negative, and you can't do an experiment that gets statistics that match those probabilities, of which we can say that those measurements are incompatible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An off-beat justification for complex numbers &amp;mdash;IMO, definitely not standard, and there are certainly other attempts at this&amp;mdash; is Leon Cohen's paper &quot;Rules of Probability in Quantum Mechanics&quot;, Foundations of Physics 18, 983(1988) (which ties probabilities to complex structure by showing that the introduction of a characteristic function approach makes a complex structure natural &amp;mdash; though this should make you worry about circularity), which sadly is only available behind a paywall, at &lt;a href=&quot;http://www.springerlink.com/content/x38rw11764812349/&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/x38rw11764812349/&lt;/a&gt;, being too early for an arXiv preprint version to exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: But algebraic completeness is a very good reason, which has the advantage that it trips off the tongue nicely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT(2): The question is perhaps whether there is a &lt;strong&gt;&lt;em&gt;natural&lt;/em&gt;&lt;/strong&gt; complex structure. The only possible candidate, as far as I've ever seen, is the Hodge dual, in tensor form ${\epsilon^{\alpha\beta}}_{\mu\nu}$, in the exterior calculus $\star$, but so far I've not liked anything I've seen or that I've tried to construct that uses this structure. Frankly, it's not often easy to take seriously approaches that take the Hodge dual with ontological seriousness. The usual approach effectively introduces a complex structure $i$ as the imaginary that is used whenever one constructs a fourier transform, which &lt;em&gt;is&lt;/em&gt; a quite natural introduction, but is not for any &lt;em&gt;other&lt;/em&gt; reason a natural structure.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-04-25T23:06:23.370" LastActivityDate="2011-04-25T23:06:23.370" CommentCount="3" />
  <row Id="9115" PostTypeId="2" ParentId="9049" CreationDate="2011-04-25T20:52:41.970" Score="3" ViewCount="" Body="&lt;p&gt;The truth is that the moon IS constantly trying to fall upon the earth, due to the force of gravity; but it is constantly missing, due to its tangential velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand this, think of whirling a rock, tied to the end of a string, around and around, with your hand just above your head.  As the rock travels in circles it is constantly being pulled toward you by the  force on the string (which is like Earth's pull of gravity on the moon).  Why doesn't the rock come bonk you on the head, if you are constantly pulling it toward your head?  The answer is that the rock is always trying to change its velocity vector to come do just that; but the change is only enough to just keep it in a circular path, like the pull on the Moon is just enough to keep it in a circular orbit around Earth.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-04-25T20:52:41.970" CommentCount="1" />
  <row Id="9116" PostTypeId="2" ParentId="9113" CreationDate="2011-04-25T20:59:22.300" Score="1" ViewCount="" Body="&lt;p&gt;I think so, yes. They are both measures of the loudness of what we hear. If $L_p$ is the sound pressure level and $L_i$ is the sound intensity, then:&#xA;\begin{equation}&#xA;L_p =20 \log_{10}\left(\frac{p_{\mathrm{rms}}}{p_{\mathrm{ref}}}\right)\mbox{ dB} = 10 \log_{10}\left(\frac{{p_{\mathrm{{rms}}}}^2}{{p_{\mathrm{ref}}}^2}\right)\mbox{ dB}&#xA;\end{equation}&#xA;The intensity of the sound wave, however, is proportional to the square of the sound pressure, or $I \propto p_{\mathrm{rms}}^2$. This gives:&#xA;\begin{equation}&#xA;L_p = 10 \log_{10}\left(\frac{{p_{\mathrm{{rms}}}}^2}{{p_{\mathrm{ref}}}^2}\right)\mbox{ dB} = 10 \log_{10}\left(\frac{{I_{\mathrm{{rms}}}}}{{I_{\mathrm{ref}}}}\right)\mbox{ dB} = L_i&#xA;\end{equation}&lt;/p&gt;&#xA;" OwnerUserId="3252" LastActivityDate="2011-04-25T20:59:22.300" />
  <row Id="9117" PostTypeId="2" ParentId="9109" CreationDate="2011-04-25T21:13:04.667" Score="7" ViewCount="" Body="&lt;p&gt;I'll leave an answer for your last question on whether complex numbers are necessary for QM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scott Aaronson has a nice lecture here &lt;a href=&quot;http://www.scottaaronson.com/democritus/lec9.html&quot; rel=&quot;nofollow&quot;&gt;http://www.scottaaronson.com/democritus/lec9.html&lt;/a&gt; , scroll down to the section on Real vs. Complex numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My favorite argument there is the first one -- that if you have a linear operator $U$, then you would want to have operators like $V$ where $V^2=U$, simply because you expect continuity; i.e. if you're allowed to do one full transformation, you should be able to do &quot;half&quot; of it too. (If waiting for one second is allowed, then waiting for half a second should also be allowed). In order to have square roots of operators in general, you'll need to allow operator matrices with complex elements. And once you allow that, the state vectors that they act on will also need to be complex in general. And so your wavefunction will also need to be complex.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-04-25T21:13:04.667" CommentCount="3" />
  <row Id="9118" PostTypeId="2" ParentId="9113" CreationDate="2011-04-25T21:20:51.177" Score="2" ViewCount="" Body="&lt;p&gt;The first thing to understand is that dB is a logarithm of a ratio.  More specifically, a logarithm of a ratio of power/power (which is why dB has no units).  A Bel is the base-10 logarithm of a power ratio.  A decibel slices each Bel into ten parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power is power.  It does not matter whether it is sound or electricity or whatever.  And dB always refers to a ratio of power. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most sound references, &quot;dB&quot; means &quot;dB in reference to (as a ratio over) 'X'.&quot;  Where 'X' is a reference level.  I think the reference used is the sound pressure which is the threshold of human hearing.  So a sound with 10 times the POWER of that reference would be 10 dB.  A sound with POWER 100 times the reference would be 20 dB. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your question is:  It can be.  Sound intensity (if expressed as a pressure) and Sound Pressure Level (SPL) are the same.  But neither is power.  Sound power is proportional to sound pressure squared.  Thus, if you increase sound PRESSURE by a factor of 10, you have increased POWER by a factor of 10^2 = 100.  These two actions are synonomous, and result in a 20dB increase.  Note that I did not explicitly state what increased by 20dB in that last sentence...  because it is ALWAYS power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you need to watch your units, and make sure the other guy is watching his.  If he doubles his POWER level, that is a 3 dB increase.  If he doubles his sound PRESSURE, that is a 6dB increase. If he doubles his sound INTENSITY, you need to nail down whether that is in units of pressure or power.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that helped.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-04-25T23:28:39.243" LastActivityDate="2011-04-25T23:28:39.243" />
  <row Id="9119" PostTypeId="2" ParentId="9108" CreationDate="2011-04-25T21:41:17.067" Score="2" ViewCount="" Body="&lt;p&gt;Yes, some of the nucleons (protons and neutrons) from the iron would remain because of baryon number conservation. If you start with 4 antibaryons and 56 baryons, you're most likely going to end up with 52 baryons. (Of course, you could end up with 2 antibaryons and 54 baryons, or whatever, but it's impossible to end up with 0 baryons.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you can't just say &quot;oh, I started with 26 protons and 2 antiprotons, and 30 neutrons and 2 antineutrons, so I must end up with 24 protons and 28 neutrons, or chromium-52&quot;. Although it's theoretically possible to end up with a chromium-52 nucleus and a bunch of gamma rays, that must be very unlikely because there are many other allowed possibilities for the products of the reaction, and the final state with a 52Cr nucleus and gamma rays corresponds to only a small fraction of the parameter space of possible products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More likely, the 56Fe nucleus is going to explode into a bunch of smaller fragments because so much energy is released inside it. Most of these fragments will probably be smaller nuclei, but it's possible to get other kinds of particles too: pions, muons... anything there's enough energy to create can be created (as long as the whole process doesn't violate any conservation laws).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that the numbers of protons and neutrons don't have to be separately conserved. It's completely possible to produce some smaller nuclei which are unstable and subsequently beta-decay, changing protons into neutrons or vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically the answer is &quot;yes, some of the baryons would be left over as new elements, but there's no way to tell which ones and it will be a messy, complicated process&quot;.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-04-25T21:41:17.067" />
  <row Id="9120" PostTypeId="2" ParentId="9109" CreationDate="2011-04-25T22:00:31.437" Score="4" ViewCount="" Body="&lt;p&gt;On your question: &lt;em&gt;Can we do quantum mechanics without complex numbers?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes. One can in general replace any complex number by a 2x2 real valued matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a+ib ~=~\left(\begin{array}{rr} ~~a &amp;amp; -b \\ b &amp;amp; a \end{array}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other examples are the complex Pauli matrices and the quaternions which can be both replaced by 4x4 real valued matrices. There isn't anything magical or special in the use of complex values in physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards, Hans&lt;/p&gt;&#xA;" OwnerUserId="3197" LastActivityDate="2011-04-25T22:00:31.437" CommentCount="6" />
  <row Id="9121" PostTypeId="2" ParentId="8914" CreationDate="2011-04-25T22:09:05.243" Score="1" ViewCount="" Body="&lt;p&gt;Your first formula is correct, but it looks like you substituted in the incorrectly signed value for (dh/h).  Your second formula is missing a sign change for the (dh/h) term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first equation, substituting, to get the MAXIMUM (dr/r) ---&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;+7% = (-3%) + (2 x (+5%))&#xA; -OR-&#xA;-7% = (+3%) + (2 x (-5%))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, your text is correct: +/-5% is the (approximate) uncertainty for r.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you solve this problem algebraically, you will find that the (corrected) differential equations you constructed gives a very close but not exact answer.  The correct uncertainty for r is something like + SQRT(1.07/0.97) and - SQRT(0.93/1.03) = +5.0282% / -4.9783%.  This is because the differential eqns are valid only for infinitesimal deltas; but we are using deltas on the order of +/-5% (hardly infinitesimal).&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-04-25T22:09:05.243" />
  <row Id="9122" PostTypeId="1" CreationDate="2011-04-25T22:31:34.363" Score="12" ViewCount="327" Body="&lt;p&gt;I know one is a partial derivative and the other is a total derivative. But physically I cannot differentiate between them. I have a clue that my doubt really might be understanding the difference between explicit and implicit time dependance. &lt;/p&gt;&#xA;" OwnerUserId="3239" LastEditorUserId="3239" LastEditorDisplayName="" LastEditDate="2011-04-25T22:36:41.057" LastActivityDate="2011-04-26T10:12:59.923" Title="What is the differenct between implicit and explicit time dependance e.g: $\frac{\partial \rho}{\partial t}$ and $\frac{d \rho} {dt}$" Tags="&lt;mathematical-physics&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="9123" PostTypeId="2" ParentId="8676" CreationDate="2011-04-25T23:02:32.903" Score="3" ViewCount="" Body="&lt;p&gt;Another way to look at this is to solve for what weight/area you are allowed to use.  Assume a spherical craft.  Surface area goes up in proportion to r^2.  Volume (which will generate your lift) is proportional to r^3.  If you do the math, you find that you are allowed to use (r x 0.4)kg/m^2, or about (r x 0.88) lbs/m^2.  So the problem gets easier to solve as the craft gets bigger.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force experienced by a totally evacuated body at sea level, would be about 14.7 psi, which is about 11.4 ~TONS~ per m^2.  (Don't you just love mixing metric and English units?  I do...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At r = 1m, you can only use 0.88 lbs of material for skin and structure to hold back that massive force (good luck with that).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At r = 10m, you get 8.8 lbs/m^2.  That equates to a (1/17) inch thick aluminum skin and no internal supporting structure. ---&gt; Still a no-go.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At r = 100m, you are allowed 88 lbs/m^2. ... O.K., now we are talking. 1/8 inch aluminum plate weighs about 19lb./m^2.  That leaves 69 lbs./m^2 for ridigization of the skin panels and internal structure.  Something that approximated a monocoque structure (like an egg shell: able to take tremendous pressure with no deep internal bracing) might work.  As an engineer, I'd be willing to take on that challenge; but... that is a H-U-G-E craft!  OVER two football field lengths in diameter, and weighing in at about 5 million kg before evacuation!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe someone would be able to do it for r = 50m, which would be a budget of 44lbs./m^3.  My guess is that technical feasibility lies somewhere in the design area of 50m &amp;lt; r &amp;lt; 100m.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-04-25T23:22:52.053" LastActivityDate="2011-04-25T23:22:52.053" />
  <row Id="9124" PostTypeId="2" ParentId="9122" CreationDate="2011-04-26T00:28:58.603" Score="15" ViewCount="" Body="&lt;p&gt;You are essentially asking about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Material_derivative&quot; rel=&quot;nofollow&quot;&gt;material derivative&lt;/a&gt; when discussing a total derivative with respect to time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you are looking at the velocity of the air in your room. There is a different velocity everywhere, and it changes with time, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = v(x,y,z,t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you take a derivative like &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial v}{\partial t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you are saying &quot;I will keep sampling the wind velocity at the exact same point in my room, and find how quickly that velocity changes.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, on the other hand, you take&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\textrm{d}v}{\textrm{d}t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you are now saying, &quot;keep following one particular little bit of air, and see how quickly its velocity changes (i.e. find its acceleration).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(note: Marek has made a nice clarification about the difference between these two uses of $t$ in the comments to this answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are related by the chain rule&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\textrm{d}v}{\textrm{d}t} = \frac{\partial v}{\partial t} + \frac{\partial v}{\partial x}\frac{\textrm{d}x}{\textrm{d}t} + \frac{\partial v}{\partial y}\frac{ \textrm{d}y}{\textrm{d}t} + \frac{\partial v}{\partial z}\frac{\textrm{d}z}{\textrm{d}t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This says that the if you look at one particular little air particle, its velocity is changing partially because the entire velocity field is changing. But even if the entire velocity field weren't changing, the particle's velocity would still change because it moves to a new spot, and the velocity is different at that spot, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As another example, say there is an ant crawling over a hill. It has a height that is a function of two-dimensional position&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h = h(x,y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at $\partial h/\partial x$, we're looking at the slope in the x-direction. You find it by moving a little bit in the x-direction while keeping y the same, finding the change in z, and dividing by how far you moved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, since we're tracking the ant, we might want to know how much its height changes when it moves a little bit in the x-direction. But the ant is traveling along its own convoluted path, and when it moves on the x-direction, it winds up changing its y-coordinate as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total change in the ant's height is the change in its height due to moving in the x-direction plus the change due to moving in the y-direction. The distance the ant moves in the y-direction in turn depends on the x-direction movement. So now we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\textrm{d}h}{\textrm{d}x} = \frac{\partial h}{\partial x} + \frac{\partial h}{\partial y}\frac{\textrm{d}y}{\textrm{d}x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the right hand side of that equation, the first term corresponds to the change in height due to moving in the x-direction. The second term is the change in height due to moving in the y-direction. The first part of that, $\partial h/\partial y$ is the change in height due to changing y, while the second part, $\textrm{d}y/\textrm{d}x$ describes how much y itself actually changes as you change x, and depends on the particulars of the ant's movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit&lt;/b&gt;&#xA;I now see that you're specifically concerned with the quantum mechanics equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\textrm{d}}{\textrm{d}t}\langle A \rangle = -\frac{\imath}{\hbar}\langle[A,H]\rangle + \langle \partial A/\partial t \rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $\langle \partial A/\partial t\rangle$ is the expectation value of the partial derivative of the operator $A$ with respect to time.  For example, if $A$ is the Hamiltonian for a particle in a time-dependent electric field, that operator would contain time explicitly.  We begin by formally differentiating the operator itself, then taking the expectation value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand $\langle A \rangle$ is simply a real-valued function of time (if $A$ is Hermitian), so $\textrm{d} \langle A \rangle / \textrm{d} t$ is the usual derivative of a real function of a single variable.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-04-26T07:56:05.693" LastActivityDate="2011-04-26T07:56:05.693" CommentCount="2" />
  <row Id="9125" PostTypeId="2" ParentId="9122" CreationDate="2011-04-26T00:31:35.623" Score="3" ViewCount="" Body="&lt;p&gt;Maybe an intuitive answer is best given in terms of classical physics. Suppose you are looking at the movement of a classical particle. The relevant variables here are position and momentum. If you solve the motion of your system, you are presented with functions $x(t)$ and $p(t)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there are a lot of derived quantities you can build from these trajectories. For example, angular momentum $\vec{L} = \vec{x} \times \vec{p}$. Since $x$ and $p$ depend on the time, $L$ also depends on time, &lt;strong&gt;but&lt;/strong&gt; in this case it does so only because $x$ and $p$ depend on time. You have basically a function $L = L(x,p)$ which then becomes $L(x(t), p(t))$. This is because in the &lt;em&gt;definition&lt;/em&gt; of $L$, the time does not play a role. Therefore, we say that this quantity has only an &lt;em&gt;implicit&lt;/em&gt; time dependence. In particular, &#xA;$\frac{\partial L}{\partial t} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, your derived quantity $f$ is defined for some reason such that the time occurs explicitly in the definition, then $\frac{\partial f}{\partial t} \not= 0$. For example, you might want to add a time-dependent phase factor to your quantitiy, e.g.&#xA;$f = \vec{x} \cdot \vec{p} \cdot e^{i\omega t}$.&#xA;Then we have $f = f(x,p,t) = f(x(t), p(t), t)$, and now $\frac{\partial f}{\partial t}$ is not zero.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-04-26T00:31:35.623" />
  <row Id="9126" PostTypeId="2" ParentId="9106" CreationDate="2011-04-26T01:04:38.853" Score="8" ViewCount="" Body="&lt;p&gt;(&lt;strong&gt;Updated&lt;/strong&gt; on 27 April to expand on and justify some points in response to comments.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@dbrane's answer conveys a lot of the flavor of the subject correctly, but I'm sorry to say that it's not right in detail. Because there's plenty of room for confusion here, I'm going to spend quite a bit of time on the mathematics before saying anything about the physics or cosmology. Sorry about that. Feel free to skip to the bottom if you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mathematics.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of Gaussian in this context actually goes like this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A function $g$ is Gaussian if the&#xA;  $n$-tuple $(g({\bf x}_1),g({\bf x}_2),\ldots,g({\bf x}_n))$ is drawn&#xA;  from an $n$-dimensional Gaussian&#xA;  distribution for all $n$ and for all&#xA;  choices of ${\bf x}_i$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is, if you look at any one point, that the value at that point is drawn from a Gaussian distribution. If you look at any two points, the values are drawn from a bivariate Gaussian distribution (possibly with correlations), etc. The above is my phrasing, but it's equivalent to things you'll find elsewhere, e.g., &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_process&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That definition is kind of a mouthful, but there's not much I can do about it: it's the way it is!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before digging into the subject in more detail, it's worth defining one additional term, &lt;em&gt;statistically homogeneous&lt;/em&gt; random fields. This just means that the statistical properties are unchanged if you slide all points over by any given amount. For instance, the average value $\langle g({\bf x})\rangle$ is independent of ${\bf x}$ (and indeed so is $\langle g({\bf x})^k\rangle$ for any $k$). Moreover, $\langle g({\bf x}_1)g({\bf x}_2)\rangle$ depends only on the difference ${\bf x}_1-{\bf x}_2$, and so forth. We often assume that the fields in cosmology have this property, because we think that, on average, the Universe is the same everywhere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, we also often assume that the statistical properties are invariant under rotation -- e.g., two-point correlations depend only on the magnitude of ${\bf x}_1-{\bf x}_2$ and not on its direction. The latter property is called &lt;em&gt;statistical isotropy&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the next little while, I'm going to focus on mathematical points that don't really depend on how many spatial dimensions we have, so for simplicity I'll work only in one dimension. In that case, of course, there's no such thing as isotropy. (By the way, I'm also going to start abbreviating &quot;statistical homogeneity&quot; as just &quot;homogeneity.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before proceeding further, let me note explicitly that Gaussianity and statistical homogeneity are independent properties: neither one implies the other.  For example, suppose that $g$ is Gaussian and homogeneous, and define another random field $h$ by setting $h(x)=[g(x)]^2$. Then $h$ is homogeneous but not Gaussian. (The pdf at any point is not Gaussian, for instance.) On the other hand, if we define $h(x)=xg(x)$, then $h$ is Gaussian but not homogeneous. (For instance, the variance $\langle h(x)^2\rangle$ depends on $x$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nice property that dbrane mentions, namely that the &lt;em&gt;Fourier transform&lt;/em&gt; of $g$ has no correlations,&#xA;$$&#xD;&#xA;\langle \tilde g({k_1})\tilde g({k}_2)\rangle=P_g({k}_1)\delta({k}_1+{k}_2),&#xD;&#xA;$$&#xA;is actually a consequence of &lt;em&gt;homogeneity&lt;/em&gt;, not &lt;em&gt;Gaussianity&lt;/em&gt;. (This property means that each Fourier mode $\tilde g({k})$ is uncorrelated with the others.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one's actually not hard to prove. The key is to think about the correlations between two points $x$ and $x+y$: $\langle g(x)g(x+y)\rangle$. Writing both of the $g$'s as Fourier integrals, we have&#xA;$$&#xD;&#xA;\langle g(x)g(x+y)\rangle=&#xD;&#xA;\int\int\langle\tilde g(k_1)\tilde g(k_2)\rangle e^{i(k_1+k_2)x}e^{ik_2y}\,dk_1\,dk_2.&#xD;&#xA;$$&#xA;Let me call the quantity on the left $G(x,y)$. This equation shows that the Fourier transform of $G$ with respect to both variables is given by&#xA;$$&#xD;&#xA;\tilde G(k_1+k_2,k_2)=\langle \tilde g(k_1)\tilde g(k_2)\rangle.&#xD;&#xA;$$&#xA;Homogeneity implies that $G$ depends only on $y$, not $x$, so $\tilde G(q,q&amp;#39;)=0$ for all $q\ne 0$. Therefore $\langle\tilde g(k_1)\tilde g(k_2)\rangle$ is nonzero only for $k_1+k_2=0$. That proves the desired result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're lucky enough to have &lt;em&gt;both&lt;/em&gt; homogeneity and Gaussianity, then the &quot;power spectrum&quot; $P_g$ is a complete description of the statistical properties of $g$. But you have to &lt;em&gt;assume&lt;/em&gt; Gaussianity in that argument; it doesn't follow from the assumption of uncorrelated Fourier modes. The simplest cosmological models predict both homogeneity and Gaussianity, so people often forget which useful properties follow from one as opposed to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Digression: Other definitions found in the literature.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the comments, a couple of people have pointed out quotes from very well-respected cosmologists (Liddle &amp;amp; Lyth, Komatsu) contradicting what I said and &lt;em&gt;defining&lt;/em&gt; Gaussianity as uncorrelated Fourier modes. These folks are great cosmologists, but they're just wrong on this. I guess I can't expect you to take my word for it, so let me try to convince you in a couple of ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let me point out that Gaussian random processes and Gaussian random fields exist in many areas of applied mathematics, and the definition I gave is utterly standard everywhere else. (In case you're wondering, one tends to say &quot;process&quot; in one dimension, especially when you want to think of that one dimension as time, and &quot;field&quot; in 2 or more dimensions.) &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_process&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; confirms this, as do plenty of &lt;a href=&quot;http://www.ifp.illinois.edu/~vvv/ece359/handouts/h13.pdf&quot; rel=&quot;nofollow&quot;&gt;other&lt;/a&gt; &lt;a href=&quot;http://publications.nr.no/917_Rapport.pdf&quot; rel=&quot;nofollow&quot;&gt;sources&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, sometimes people in one discipline use a term differently from people in all other disciplines. Maybe cosmologists are doing that here? That turns out not to be the case. For instance, &lt;a href=&quot;http://adsabs.harvard.edu/full/1986ApJ...304...15B7&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; is one of the ones that started the whole idea of considering Gaussianity in cosmology, and it uses the standard definition. Perhaps even more convincing is the fact that the main form of non-Gaussianity that cosmologists try to test these days is called &lt;a href=&quot;http://arxiv.org/pdf/1004.0818v1&quot; rel=&quot;nofollow&quot;&gt;local non-Gaussianity&lt;/a&gt;, which is defined as&#xA;$$&#xD;&#xA;\Phi({\bf x})=\phi({\bf x})+f_{NL}(\phi({\bf x})-\langle\phi\rangle)^2.&#xD;&#xA;$$&#xA;Here $f_{NL}$ is a constant, $\phi$ is a homogeneous Gaussian random field, and $\Phi$ is the physical field to be measured. The field $\Phi$ has uncorrelated Fourier modes (by the proof I gave above). So if &quot;uncorrelated Fourier modes&quot; were the definition of Gaussianity, then this field would be Gaussian. But if you talk to cosmologists working in the field, you'll find that every single one knows that this field is non-Gaussian -- in fact, it's regarded as the very archetype of non-Gaussianity. (Here's &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v100/i18/e181301&quot; rel=&quot;nofollow&quot;&gt;one example&lt;/a&gt; out of many I could have chosen.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the error these folks have made is quite common. I think that there are several reasons.  The main one is that the hypothesis of uncorrelated Fourier modes does imply approximate Gaussianity by a sort of central limit theorem argument, &lt;em&gt;if&lt;/em&gt; some additional assumptions hold. Essentially, whatever quantities you're calculating have to have comparably-big contributions from many different Fourier modes, so that non-Gaussianities get &quot;washed out.&quot; Back in the old days, people were willing to make that assumption, but when you're actually trying to measure non-Gaussianity, this is precisely the level of approximation you're &lt;em&gt;not&lt;/em&gt; allowed to make.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason is that everyone learns the &quot;standard&quot; (homogeneous, isotropic, Gaussian) picture of cosmological perturbations really well, then they forget which properties of the standard picture depend on which of the hypotheses. And of course once something is incorrectly stated somewhere, it propagates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Physics.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the physics is concerned, the main point is that some standard theoretical models, such as inflation, predict that the CMB (among other observables) should be Gaussian, as well as homogeneous and isotropic. If this is true, then the &lt;em&gt;power spectrum&lt;/em&gt;, which I called $P_g(k)$ above, is a complete description of the statistical properties of the field. To define a Gaussian distribution, you just need to specify its first two moments, i.e., the mean and the variances &amp;amp; covariances. The mean is zero (mostly because we define it that way -- the observables are fluctuations about the mean). With these assumptions, there are no correlations in Fourier space, so all that leaves is the variances of the Fourier modes. That's what the power spectrum is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some slightly more exotic cosmological models (fancier versions of inflation as well as non-inflationary models) predict non-Gaussianity. One form this might take is nonzero 3rd-order correlations: things of the form $\langle g({\bf x}_1)g({\bf x}_2)g({\bf x}_3)\rangle$ should be zero in a Gaussian theory but not in a non-Gaussian one. That's the reason people look for 3-point correlations. If you look for them in position space, you call the thing you're trying to measure the &quot;3-point correlation function.&quot; If you try to measure it in Fourier space, you call it the &quot;bispectrum.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even if there are no 3-point correlations, things might still be non-Gaussian. Gaussianity is a very specific set of mathematical properties, and there are infinitely many ways to be non-Gaussian. As a result, there are tons of proposed tests for non-Gaussianity, all looking at completely different things. For instance, a form of non-Gaussianity that was symmetric under a sign flip (the statistical properties of $-g$ are the same as those of $g$) would have no 3-point correlations, but it could still be non-Gaussian. You could search for that by looking at 4th-order correlations, although at some point thinking of things in terms of these correlations is not the best way to do it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, there are a bewildering variety of completely different tests people use. To cite one more-or-less randomly chosen example, people calculate the &lt;a href=&quot;http://en.wikipedia.org/wiki/Minkowski_functional&quot; rel=&quot;nofollow&quot;&gt;Minkowski functionals&lt;/a&gt; of excursion sets in CMB maps. The connection between Minkowski functionals and non-Gaussianity is not obvious, but the point is just that in a Gaussian process you can predict various properties of the functionals, and if the observations give different answers, that's non-Gaussianity.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-04-27T19:16:18.920" LastActivityDate="2011-04-27T19:16:18.920" CommentCount="10" />
  <row Id="9127" PostTypeId="2" ParentId="8914" CreationDate="2011-04-26T02:43:36.497" Score="0" ViewCount="" Body="&lt;p&gt;I will use two ways to derive the uncertainity on r&lt;br&gt;&#xA;1 - very simple method: the 'interval arithmetic' present in the Euler math package (free)&lt;br&gt;&#xA;2 - derive the formula of dr(V,h, dV,dh)&lt;br&gt;&#xA;Both answers : 5%  (for any value V or h that I choose for V and h)   &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;// 'using interval arithmetic'    &#xA;&amp;gt;dV=0.07 dh=0.03         // deltas  &#xA;&amp;gt;V=1000 h=40             //arbitrary choosen values  &#xA;&amp;gt;r0=sqrt(1/pi*V/h)       //central value for r    &#xA; 2.82094791774  &#xA;&amp;gt;r=sqrt(1/pi*((V±V*dV)/(h±h*dh))) //interval values for r   &#xA; ~2.6,3~                          //    from 2.6 to 3  &#xA;&amp;gt;(r-r0)  &#xA; ~-0.14,0.14~  &#xA;&amp;gt;(r-r0)/r0                           &#xA; **~-0.05,0.05~                     // 5% for dr/r0 for any V, h**  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;//calc dr(V,h, dV,dh) (see&lt;a href=&quot;http://en.wikipedia.org/wiki/Propagation_of_uncertainty#Example_application%3a_Resistance_measurement&quot; rel=&quot;nofollow&quot;&gt;Resistance_measurement Example&lt;/a&gt; and adapting the derivatives)       &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;dr=1/2/sqrt(pi)*sqrt(dV^2/V/h+dh^2*V/h^3)&lt;br&gt;&#xA;   0.00106245302667&lt;br&gt;&#xA;  1-(r-dr)/r0&lt;br&gt;&#xA;   &lt;strong&gt;~-0.05,0.05~                     // the result is again 5%&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-04-26T02:43:36.497" />
  <row Id="9128" PostTypeId="2" ParentId="7962" CreationDate="2011-04-26T04:13:31.580" Score="2" ViewCount="" Body="&lt;p&gt;I am stunned -- STUNNED! -- that no one mentioned the difference in voltage between &quot;neutral&quot; and &quot;safety ground&quot;.&#xA;Anyone with a multimeter can jam the probes into the &quot;neutral&quot; and &quot;ground&quot; slots of a standard 3-pin electrical outlet and measure for himself the difference in voltage between those two points.&#xA;That difference is &lt;em&gt;not&lt;/em&gt; zero -- even when both lines are properly and solidly connected to &quot;earth ground&quot; at a single point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This situation involves small resistances that normally can be neglected, but not in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a situation with &quot;no voltage difference&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;   ________________ 30 amps -&amp;gt; ________________&#xA;  |                                            |&#xA;  +                                            |&#xA; 220V                                        Load&#xA;  -                                            |&#xA;  |__________________________&amp;lt;- 30 amps________|&#xA;  |                       |&#xA;  |______(YOU ARE HERE)___|&#xA;  |&#xA;  |&#xA;Ground&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;The wire you touched was not only at 0&#xA;  volts, but also grounded&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. One end of the wire connected to the ground post is grounded at 0 volts, but the other end that you are holding is not exactly 0 volts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&#xA;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I don't get how this can work.&#xA;  If the wire is at 0 volts and you are at 0 volts,&#xA;  then there is no potential difference and hence I'd expect no current.&#xA;  Is this physics correct?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The wire is not at 0 volts, and therefore there &lt;em&gt;is&lt;/em&gt; current through your body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's redraw this diagram, including a few resistances that are &lt;em&gt;normally&lt;/em&gt; negligible:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;   ________________ 30 amps -&amp;gt; ________________&#xA;  |                                            |&#xA;  +                                            |&#xA; 220V                                        Load: R1&#xA;  -                                            |&#xA;  |_r2__________r6_______r4____&amp;lt;- 30 amps______|&#xA;  |                         |&#xA;  |r3_____(YOU ARE HERE)_r5_|&#xA;  |&#xA;  |&#xA;Ground&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When 2 physical wires touch each other, there is a contact resistance.&#xA;(Rule of thumb: it's typically around 0.1 Ohm). That's r2, r3, r4, r5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have current flowing from one end of a long wire to the other end, the resistance (r6) is often very negligible, but not in this case.&#xA;A copper-conductor 16 AWG extension cord has a resistance of about 4 milli-Ohm per foot.&#xA;So the &quot;neutral&quot; conductor of a 20 foot long cord has about 0.08 Ohm of resistance from the copper (not including the contact resistance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though &lt;em&gt;most&lt;/em&gt; of the 30 A bypasses the human and flows in the copper wires,&#xA;a small amount of that 30 A goes through the human.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The voltage across the human is approximately 30 A * ( 0.1 Ohm + 0.08 Ohm + 0.1 Ohm) =~= 8 V.&#xA;Out of 220 VAC, a mere 8 VAC is often negligible -- but not in this case.&#xA;If some human has a resistance of 100 Ohms (it's typically an order of magnitude more),&#xA;The current through that human would be approximately 8 V / 100 Ohm =~= 80 mA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is more than the 60 mA AC which can cause death from ventricular fibrillation.&lt;/p&gt;&#xA;" OwnerUserId="965" LastActivityDate="2011-04-26T04:13:31.580" CommentCount="2" />
  <row Id="9129" PostTypeId="1" AcceptedAnswerId="9130" CreationDate="2011-04-26T06:01:53.230" Score="7" ViewCount="256" Body="&lt;p&gt;What force do the arms have to generate to do a pushup? Let us look a this simplified model:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ueWHS.jpg&quot; alt=&quot;pushup physics&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The body can be represented by the green plank of mass B. Its angle to the ground is $\theta$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This question was asked on &lt;a href=&quot;http://fitness.stackexchange.com/questions/1717/what-is-the-effective-weight-when-doing-push-ups&quot;&gt;fitness.stackexchange.com&lt;/a&gt;, although not phrased with physics in mind.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3287" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-04-26T06:08:12.013" LastActivityDate="2011-04-26T21:45:15.463" Title="What is the force on the arms in a pushup?" Tags="&lt;forces&gt;&lt;newtonian-mechanics&gt;&lt;statics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="9130" PostTypeId="2" ParentId="9129" CreationDate="2011-04-26T06:35:12.277" Score="4" ViewCount="" Body="&lt;p&gt;A push up is a form of &lt;a href=&quot;http://en.wikipedia.org/wiki/Lever&quot; rel=&quot;nofollow&quot;&gt;lever&lt;/a&gt;.  The athlete must exert roughly half her body weight (under some assumptions I'll clarify at the end of the post.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can solve this problem using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Virtual_work&quot; rel=&quot;nofollow&quot;&gt;principle of virtual work&lt;/a&gt;.  Assume the athlete raises her body through a small angle $\textrm{d}\theta$.  Then her center of mass rises by $l \cos\theta \ \textrm{d}\theta$, with $l$ the distance from her feet to her center of mass.  The work done is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\textrm{d}W = mgl\cos\theta \ \textrm{d}\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This work is equal to the force multiplied by the distance over which the force is exerted.  If we call the distance from her feet to her shoulders $L$, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$mgl\cos\theta \ \textrm{d}\theta = F L \cos\theta \ \textrm{d}\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = \frac{mgl}{L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your center of mass is roughly half way up your body, so a push up requires you to lift about half your weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer assumes the pushup is infinitely slow (i.e. no acceleration).  This is actually not so bad an approximation as it sounds, but people doing fast push ups will probably exert a higher force at the bottom of the push up, then let gravity do negative work to slow them down as they near the top.  This is the basis for &quot;clapping push ups&quot; (which I am too weak to perform).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have only calculated the component of force in the direction of motion, so, we are assuming the athlete pushes directly in the direction of motion of her shoulders.  This means she doesn't push along the red line in your picture, but diagonally ahead of it.  In reality, she might push more straight down, increasing the force and decreasing the angle through with her elbows flex.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force may actually change throughout the push up due to this effect.  If you lean into the arm of a couch and do a push up off that, you'll find it's much easier at this high angle.  The reason is you aren't pushing straight down any more, so it's a little bit like going up a &lt;a href=&quot;http://en.wikipedia.org/wiki/Inclined_plane&quot; rel=&quot;nofollow&quot;&gt;ramp&lt;/a&gt; (in that the distance over which the force is exerted is lengthened to accomplish the same amount of work).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also assumed the athlete's center of mass doesn't move, which means I'm pretty much thinking of her arms as massless, and the rest of her is a rigid board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I'm assuming her hands and feet don't slip, the floor is too big to move, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An empirical way to determine the force would simply be to put scales underneath the hands, but the reading could be a little off depending on whether the scales are constructed to measure total force on them, or only the vertical component of the force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, an actual human is not simply one force pushing in one direction.  There are muscles, bones, ligaments, etc.  All sorts of different forces are going on in different places.  This calculation gives the force that the arms exert on the main body.  The force that the triceps exert on the elbow, for instance could be much higher.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-04-26T06:35:12.277" CommentCount="1" />
  <row Id="9131" PostTypeId="1" CreationDate="2011-04-26T09:00:59.657" Score="5" ViewCount="108" Body="&lt;p&gt;If I have two collinear beams with opposite velocities the cross section is defined by the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dN}{dtdV}=\sigma \rho_1 \rho_2 |\underline{v}_1-\underline{v}_2|$&#xA;where $\sigma$ is the cross section, $\underline{v}_1$,  $\underline{v}_2$ the velocities of the beams and $\rho_1$, $\rho_2$ their densities.&#xA;I read that if I have two non-collinear beams I must substitute the factor $|\underline{v}_1-\underline{v}_2|$ with $\sqrt{|\underline{v}_1-\underline{v}_2|^2-\frac{|\underline{v}_1 \wedge\underline{v}_2|^2}{c^2}}$. Is that true? How can I derive this?&lt;/p&gt;&#xA;" OwnerUserId="592" LastEditorUserId="592" LastEditorDisplayName="" LastEditDate="2011-04-26T09:55:33.497" LastActivityDate="2011-08-06T11:39:53.457" Title="Cross section fon non-collinear beams" Tags="&lt;cross-section&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="9132" PostTypeId="1" CreationDate="2011-04-26T09:38:36.147" Score="0" ViewCount="91" Body="&lt;p&gt;I've got loose tea that I pour to the bottom of a glass, then apply water, stir and drink. I've noticed that when the water in the water boiler has gone cold a larger portion of the tea remains afloat. This is making the drinking experience far less enjoyable as I don't like to chew my tea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why would it matter if the water is hot or cold, wouldn't the tea soak up the water and lose buoyancy either way? Is it all in my head? Can I do something about it besides adding warm water?&lt;/p&gt;&#xA;" OwnerUserId="863" LastActivityDate="2011-04-26T09:38:36.147" ClosedDate="2011-04-26T17:40:36.630" Title="Why is it that when I mix tea with hot water there appear to be far less floaters on the surface than if you mix with regular temperature water?" Tags="&lt;water&gt;&lt;chemical-compounds&gt;" CommentCount="7" />
  <row Id="9133" PostTypeId="1" AcceptedAnswerId="9138" CreationDate="2011-04-26T09:51:34.617" Score="9" ViewCount="418" Body="&lt;p&gt;If you had a glass ball filled with water, completely sealed and containing a fish, could the fish move the ball?&lt;/p&gt;&#xA;" OwnerUserId="3289" LastEditorUserId="3289" LastEditorDisplayName="" LastEditDate="2011-04-26T10:55:34.223" LastActivityDate="2011-04-26T20:55:02.273" Title="Could a fish in a sealed ball, move the ball?" Tags="&lt;fluid-dynamics&gt;&lt;motion&gt;&lt;momentum&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="9134" PostTypeId="1" AcceptedAnswerId="9139" CreationDate="2011-04-26T10:07:25.767" Score="3" ViewCount="178" Body="&lt;p&gt;I'm wondering how would you move without gravity? You would still have air pressure at 1Atm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would you &quot;swim&quot; in the air or would you have do something else?&lt;/p&gt;&#xA;" OwnerUserId="3290" LastActivityDate="2011-04-26T14:16:44.783" Title="How would you move without gravity?" Tags="&lt;gravity&gt;&lt;space&gt;&lt;air&gt;&lt;motion&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="9135" PostTypeId="2" ParentId="9122" CreationDate="2011-04-26T10:12:59.923" Score="6" ViewCount="" Body="&lt;p&gt;I usually explain it this way:&#xA;$$\rho = \rho(t,x(t),p(t))$$&#xA;$$\frac{\partial\rho}{\partial t} = \lim_{\Delta t \to 0} \frac{\rho(t+\Delta t,x(t),p(t))-\rho(t,x(t),p(t))}{\Delta t}$$&#xA;$$\frac{d\rho}{d t} = \lim_{\Delta t \to 0} \frac{\rho(t+\Delta t,x(t+\Delta t),p(t+\Delta t))-\rho(t,x(t),p(t))}{\Delta t}$$&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-04-26T10:12:59.923" CommentCount="1" />
  <row Id="9136" PostTypeId="1" AcceptedAnswerId="9137" CreationDate="2011-04-26T11:20:07.640" Score="2" ViewCount="148" Body="&lt;p&gt;We know that metal halide lamps and other HID lamps (like xenon/mercury arc ones) require 2-30kV+ spark to initiate discharge. Handling these voltages is not easy, and I am thinking if it's possible to initiate discharge by applying radiofrequency on the electrodes (for example 13.56 MHz) for a short time, just like in induction lamps or plasma etching chambers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does required power depend on gas pressure? Do I just need higher amplitude to excite high-pressure lamp or it's just impossible? &lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-04-28T14:15:27.667" Title="RF ignition in metal halide lamps" Tags="&lt;hid&gt;&lt;rf&gt;" AnswerCount="1" />
  <row Id="9137" PostTypeId="2" ParentId="9136" CreationDate="2011-04-26T11:51:18.587" Score="3" ViewCount="" Body="&lt;p&gt;The ignition of such discharge lamps is done by &quot;high&quot; frequency &#xA;since long. But using 13.565 MHz for that is nonsense, because You &#xA;invite a lot of short-wave problems without any profit. &#xA;The ignition works at frequencies from some kHz to some dozen kHz. &#xA;Gas pressure: only Xenon high pressure lamps have really high (some bars) &#xA;pressure before start. All other lamps have not more than some &#xA;millibars or less. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And: Power/pressure is a question of someone lacking simplest gas &#xA;discharge basics. To ignite You need a voltage, and after that &#xA;some current, µA or in extreme cases some milliamperes are sufficient.&#xA;The main voltage source takes over and drives the lamp warm &#xA;within some seconds or minutes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally: Why do You try to reinvent wheels? The mere fact that &#xA;millions of discharge lamps are around in the world says that ignition &#xA;has been solved in some way. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-26T11:51:18.587" CommentCount="12" />
  <row Id="9138" PostTypeId="2" ParentId="9133" CreationDate="2011-04-26T11:53:22.123" Score="5" ViewCount="" Body="&lt;p&gt;Yes, with gravity and a generous definition of &quot;moving&quot;.. it would be the same principle as the toys where you can control a sphere using a radio control (&lt;a href=&quot;http://www.ohgizmo.com/2011/01/10/ces-2011-hands-on-with-sphero/&quot; rel=&quot;nofollow&quot;&gt;or using your iphone)&lt;/a&gt;. The fish swims along the edge and gravity pulls it back down, which starts a rotation of the water and by friction to the sphere starts the rolling motion of the sphere on the ground or other surface. Obviously the water/sphere friction will probably be miniscule, but at least it is possible in theory :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A follow-up question would of course be if it's possible to move a hermetically sealed sphere freefloating in vacuum and without gravity or any other appreciable fields intersecting it. If you solve this, I'm pretty sure NASA will want to talk to you (or the fish)!&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-04-26T11:53:22.123" />
  <row Id="9139" PostTypeId="2" ParentId="9134" CreationDate="2011-04-26T11:54:42.543" Score="2" ViewCount="" Body="&lt;p&gt;The answer is: with great difficulty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a space ship there should be hand holds and footholds and a grid of ropes crisscrossing and you would move by pulling and pushing yourself: muscle power. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air is too thin to get a reactive force to propel yourself effectively by swimming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once a motion is started, stopping will have to be taken into account too.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-26T11:54:42.543" CommentCount="5" />
  <row Id="9140" PostTypeId="2" ParentId="9134" CreationDate="2011-04-26T12:04:13.837" Score="1" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/kN5yy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well we could use arrangements springs and nets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&#xA;Suppose you want to get from A to B. There is spring at A (launcher spring) and there is another spring at C(deflector spring). There is a net at B to catch you. And the big black rectangle is a house. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You firmly push your head against spring at A and let go. Once you reach spring C it will deflect you and at B the net will catch you. Please wear helmet! &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2011-04-26T12:04:13.837" />
  <row Id="9141" PostTypeId="1" CreationDate="2011-04-26T13:20:28.497" Score="0" ViewCount="104" Body="&lt;p&gt;A supernova's Ni/Fe core remnant with mass exceeding about three solar masses would result in a black hole. Although the core's temperature was in the ~1E10K range, the black hole temperature based on predicted Hawking radiation would be very near absolute zero, well below CMB temperature. It seems implausible that a very high temperature supernova remnant can instantly evolve to a near-zero K temperature black hole. How is this consistent with conservation of energy?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-04-26T15:41:41.097" Title="Black Holes and Energy Conservation" Tags="&lt;black.holes&gt;&lt;energy.conservation&gt;&lt;supernova&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="9142" PostTypeId="2" ParentId="9134" CreationDate="2011-04-26T14:16:44.783" Score="0" ViewCount="" Body="&lt;p&gt;I would grow long nails on my hands and feet, and then run like a beast with the nails firmly in the soil at all times :) If I accidentally leave the surface, I would huff and puff in the reverse direction until I reach the surface again.&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-04-26T14:16:44.783" CommentCount="1" />
  <row Id="9143" PostTypeId="2" ParentId="9141" CreationDate="2011-04-26T14:26:36.820" Score="2" ViewCount="" Body="&lt;p&gt;When the matter of the Ni/Fe core collapses, it releases huge amounts of energy in the form of radiation as well as high energy jets (ejected from the rotational poles of the black hole), so a lot of the mass + energy the black hole is trying to engulf (and keep for itself) is actually escaping before being trapped in the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, do not be confused by the fact that the celestial body goes from ca 1E10K range to near 0K in it's black body radiation.  Sure, in &lt;em&gt;that&lt;/em&gt; respect it becomes less energetic (by far) but it will retain a lot of mass (which is a lot of energy, right there) and it will also be rotating (I won't speculate how much that energy, the one stored in the celestial body's angular momentum changes when it goes from Ni/Fe core to black hole though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then of course, if Hawkings radiation really do exist and enough time elapses, all the energy trapped in the black hole will eventually seep out of it as it evaporates, but yes, it would take absurd amount of time (even by cosmological standards) for that to happen.&lt;/p&gt;&#xA;" OwnerUserId="3245" LastActivityDate="2011-04-26T14:26:36.820" CommentCount="3" />
  <row Id="9144" PostTypeId="2" ParentId="9109" CreationDate="2011-04-26T15:13:16.920" Score="1" ViewCount="" Body="&lt;p&gt;The physical interpretation of a wave-function is correctly given in almost all text-books. Its being &quot;unusual&quot; is due to too simplified teaching the classical mechanics. For example, ask yourself what the Moon position is. It is an average of many data points. The CM certainty is obtained as a result of averaging many data points. &quot;Many data points&quot; is an intrinsic thing for physical phenomena. Indeed, can you convince anybody in anything if you bring only one point on your photo-film? In QM the position is not a function of time anymore but an operator with different eigenvalues. Ensemble of these eigenvalues describes a state. One point does not describe a state, unfortunately. A photo of the Moon is different from a photo of Mars in details that are different points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So arrays of data are not unusual to physics. They are necessary and are implied in our notions of space, time, reference systems, etc. These arrays obey their own laws. These laws are sometimes wave laws. So the wave function is a representation of the data describing a given physical system if it is observed &quot;many times&quot;. Without averaging it is more detailed than after averaging. To cope with complex waves, think of the light description in terms of complex amplitudes and of the way to get a real-valued intensity.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-04-26T16:22:11.697" LastActivityDate="2011-04-26T16:22:11.697" />
  <row Id="9145" PostTypeId="2" ParentId="9141" CreationDate="2011-04-26T15:41:41.097" Score="0" ViewCount="" Body="&lt;p&gt;I had to think about this a while before I came up with what I think is a good intuitive picture. Imagine a neutron star emitting radiation with whatever its characteristic energy distribution is. Now, slowly dial up the gravitational constant and watch what happens. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that, when it gets high enough, the neutron star will collapse into a black hole. But, what happens in the intermediate time? Well, as G gets larger, the neutron star's radiation will be red-shifted to lower and lower energies as more of the energy is needed to escape the increasing gravitational well. Whatever the lower limit of the detector is, more and more radiation will fall below it, gradually making the star disappear. When the star becomes a black hole, then, we see no radiation coming from it at all (putting aside external sources) except for Hawking radiation, which is of extremely low energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't say how well this picture translates to actual black hole creation, but it gives an intuitive picture of how the core temperature appears to drop, from an outside perspective, without an instantaneous transition.&lt;/p&gt;&#xA;" OwnerUserId="1196" LastActivityDate="2011-04-26T15:41:41.097" />
  <row Id="9146" PostTypeId="2" ParentId="9109" CreationDate="2011-04-26T15:54:24.117" Score="2" ViewCount="" Body="&lt;p&gt;Can we do quantum mechanics without complex numbers? Yes.&lt;br&gt;&#xA;Use Geometrical Algebra (GA) as a simpler framework to express physics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://geocalc.clas.asu.edu/pdf/OerstedMedalLecture.pdf&quot; rel=&quot;nofollow&quot;&gt;Oersted Medal Lecture 2002: Reforming the Mathematical Language of Physics&lt;/a&gt;&lt;br&gt;&#xA;Geometry Algebra (GA) encompasses in a single framework for all this:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Synthetic Geometry,&lt;/li&gt;&#xA;&lt;li&gt;Coordinate Geometry,  &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Complex Variables&lt;/strong&gt;,  &lt;/li&gt;&#xA;&lt;li&gt;Quaternions,  &lt;/li&gt;&#xA;&lt;li&gt;Vector Analysis,  &lt;/li&gt;&#xA;&lt;li&gt;Matrix Algebra,  &lt;/li&gt;&#xA;&lt;li&gt;Spinors,  &lt;/li&gt;&#xA;&lt;li&gt;Tensors,  &lt;/li&gt;&#xA;&lt;li&gt;Differential forms.   &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is one language for all physics.&lt;br&gt;&#xA;Probably Schrödinger, Dirac, Pauli, etc ... would have used GA if it existed at the time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;GA Reduces “grad, div, curl and all that” to a single vector&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;  &#xA;  &lt;p&gt;derivative that, among other things,&#xA;  combines the standard set of four&#xA;  Maxwell equations into a single&#xA;  equation and provides new methods to&#xA;  solve it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using Geometric Algebra an intuitive view is around the corner (geometry notions fit better in my head). &#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/8062/about-the-complex-nature-of-the-wave-function/8085#8085&quot;&gt;In this PSE I link a list of resources of GA.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-04-26T15:54:24.117" />
  <row Id="9147" PostTypeId="2" ParentId="9133" CreationDate="2011-04-26T16:09:10.743" Score="0" ViewCount="" Body="&lt;p&gt;Yes, the fish can move the sealed ball, but the centre centre of mass of the fish and the ball will not move.&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-04-26T16:09:10.743" CommentCount="1" />
  <row Id="9149" PostTypeId="2" ParentId="9111" CreationDate="2011-04-26T19:07:27.287" Score="3" ViewCount="" Body="&lt;p&gt;G. &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540601422&quot; rel=&quot;nofollow&quot;&gt;Scharf&lt;/a&gt;, instead of renormalizations, splits arguments in the distributions. I think it is equivalent to replacing one of distributions with a regular function.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-04-26T19:07:27.287" CommentCount="1" />
  <row Id="9150" PostTypeId="1" CreationDate="2011-04-26T19:07:49.813" Score="5" ViewCount="298" Body="&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/9133/could-a-fish-in-a-sealed-ball-move-the-ball&quot;&gt;This question&lt;/a&gt; led me to wonder whether swimming would be the same experience for a fish in a full, sealed ball as it is normally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the fish is about 7cm from the walls of the tank, a pressure wave can propagate from the fish to the wall and back in .0001 seconds, while the time scale on which a fish wiggles is tenths of a second.  So unlike the open ocean, the water surrounding the fish can all communicate with itself on the time scale that the fish wiggles, and unlike a normal fish tank, the water has nowhere to go and so can't change its shape.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the fish notice any hydrodynamic effects in a full, sealed tank compared to normal swimming?&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-06-27T17:26:57.200" Title="Would a fish in a sealed ball swim normally?" Tags="&lt;fluid-dynamics&gt;&lt;propulsion&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="9151" PostTypeId="1" AcceptedAnswerId="9178" CreationDate="2011-04-26T19:46:37.237" Score="2" ViewCount="116" Body="&lt;p&gt;Suppose I have a compactification of string theory to four Minkowski dimensions of the form $M^{1,3}\times X$, where the internal CFT on $X$ is a $c=9$, $(2,2)$ SCFT. For example, let the SCFT on $X$ be described by the IR fixed point of an LG model with the superpotential $W(\Phi)=\Phi^p$. Can someone explain to me (hopefully in details) how can we extract the space-time physics in $M^{1,3}$ from the data of the internal CFT. What is the relation between the world-sheet superpotential $W(\Phi)=\Phi^p$ and the space-time superpotential? What is the space-time interpretation of the chiral ring of $W(\Phi)$? How does the non-perturbative corrections to the space-time superpotential are related to the correction to $W(\Phi)$. I know this is a basic question in string theory and it should be easy to find an answer in Polchinski but I hope someone will clarify things further. &lt;/p&gt;&#xA;" OwnerUserId="1531" LastActivityDate="2011-04-27T05:00:51.697" Title="The 4D Physics of a Superstring Background" Tags="&lt;string-theory&gt;" AnswerCount="1" />
  <row Id="9152" PostTypeId="1" CreationDate="2011-04-26T19:57:54.867" Score="2" ViewCount="146" Body="&lt;p&gt;1.) Does every particle that has rest mass also have an anti-particle with which it would annihilate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) Does annihilation only occur between like particles?  For example what happens if a antineutron (anti u, anti d, anti d) collides with a proton (uud)?  What happens if a positron collides with a proton?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) Since the Tevatron accelerates antiprotons is this more difficult to handle and dump?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.) I've read about WIMP annihilation detection.  Why would one assume there is any different proportion of WIMPS to anti-WIMPS than as is for non-WIMP matter (where there is far more matter than anti-matter).&lt;/p&gt;&#xA;" OwnerUserId="3282" LastEditorUserId="3282" LastEditorDisplayName="" LastEditDate="2011-04-26T20:35:01.263" LastActivityDate="2011-04-27T18:59:02.390" Title="More on matter and anti-matter" Tags="&lt;particle-physics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9153" PostTypeId="2" ParentId="9133" CreationDate="2011-04-26T20:55:02.273" Score="2" ViewCount="" Body="&lt;p&gt;Yes, the fish could (theoretically at least) move the ball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick way to visualize why this is so is to imagine yourself sealed in a 7-foot diameter ball of air.  You could easily move the ball by walking in the direction you wanted it to go.  Of course you would have a much easier time of it, since your density is much higher relative to the surrounding fluid (air).  The fish's density is much closer to the surrounding fluid (water), so it would have a harder time of it.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-04-26T20:55:02.273" />
  <row Id="9154" PostTypeId="1" AcceptedAnswerId="13219" CreationDate="2011-04-26T20:55:27.200" Score="3" ViewCount="167" Body="&lt;p&gt;What subject (suggest book titles, etc.) should I study to get a clear grasping of hypersurfaces, 2-surfaces, and integration on them, mostly in special relativity (I'm not messing with general relativity yet).&lt;/p&gt;&#xA;" OwnerUserId="3064" LastEditorUserId="3064" LastEditorDisplayName="" LastEditDate="2011-08-05T20:33:28.857" LastActivityDate="2011-08-05T21:56:23.590" Title="Relativistic space-time geometry" Tags="&lt;special-relativity&gt;&lt;differential-geometry&gt;&lt;geometry&gt;&lt;minkowski-space&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="9155" PostTypeId="2" ParentId="9150" CreationDate="2011-04-26T21:04:46.947" Score="2" ViewCount="" Body="&lt;p&gt;Yes, the fish will swim normally.  The fish does not make itself move by radiating pressure waves to infinity, it creates and sheds small local vortices with its fins which move at a much lower speed. edit&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-28T09:52:27.320" LastActivityDate="2011-04-28T09:52:27.320" CommentCount="4" />
  <row Id="9156" PostTypeId="1" CreationDate="2011-04-26T21:24:39.373" Score="7" ViewCount="201" Body="&lt;p&gt;Consider the normalized state, written in some orthonormal basis as:&#xA;$$\psi = A |0\rangle + B |1\rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's define a &quot;purity operator&quot; for a basis as any operator whose &lt;em&gt;expectation value&lt;/em&gt; gives 1 for a pure state in this basis, and 0 for the most mixed state in this basis. Inbetween states should give between 0 and 1, although the specific value doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible example (please note my question is on the general case though, this specific example is just to aid discussion), is &#xA;$$\langle \mathcal{O} \rangle = 1 - 4 \frac{|A||B|}{|A|+|B|}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What mathematically prevents such measurements in quantum mechanics?&lt;/p&gt;&#xA;" OwnerUserId="3299" LastActivityDate="2011-04-27T14:40:35.170" Title="What are constraints on a &quot;purity&quot; operator in quantum mechanics?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="9157" PostTypeId="2" ParentId="9129" CreationDate="2011-04-26T21:26:17.030" Score="0" ViewCount="" Body="&lt;p&gt;As an independent check on Mark's post, I would solve this as a summation of moments problem, with the exerciser's toes as the center point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming you are more interested in how hard the exercise is, and not in how much weight the floor is supporting.  In performing a push-up, most of your arm weight does not count, since the lower arms do not go up and down, and the center of mass of the upper arms only moves half as far as your chest.  Ballpark this as a 10% credit against total body weight.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leaves 90% of your body weight, with a center of mass at approximately 4.5/9 body lengths from your toes.  The moment generated is 90% x ((4.5/9) body lengths) = 45 (% x BL). At the lowest point in the pushup, your chest muscles are generating enough force to counteract this moment, on a lever arm of about 8/9 body lengths.  the force is 45 (% x BL) / ((8/9) BL) = about 51% of your body weight.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-04-26T21:45:15.463" LastActivityDate="2011-04-26T21:45:15.463" />
  <row Id="9158" PostTypeId="1" AcceptedAnswerId="9175" CreationDate="2011-04-26T21:32:39.227" Score="3" ViewCount="132" Body="&lt;p&gt;What is an adequate way to understand this simultaneously. One has the underlying assumption that matter is saturated and has the merit of being able to come up with an accurate formula for the Binding Energy (the SEMF/Bethe-Weiszacker) and the other can explain the magic numbers by building up energy levels in a shell structure similar to atomic orbitals. However, the underlying assumptions appear to be completely contradictory. I am looking for a better perspective on how to understand these two simultaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Apologies for the cheesy title.&lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="2599" LastEditorDisplayName="" LastEditDate="2011-05-21T20:03:02.513" LastActivityDate="2011-05-21T20:03:02.513" Title="$\frac{1}{\sqrt{2}}$|Independent particle Model $\rangle$ + |Strong Interaction Model$\rangle$?" Tags="&lt;nuclear-physics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="9159" PostTypeId="1" CreationDate="2011-04-26T21:47:54.977" Score="6" ViewCount="202" Body="&lt;p&gt;Consider an animal like a horse.  Now scale its neck longer and longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can a giraffe, or even worse a huge dinosaur, raise its neck without the tendons snapping?  The dinosaur case in particular seems ridiculous.  Is there a &quot;physics trick&quot; the animals use to make this more manageable?  Or does the tendon tension not scale as badly as my intuitiion is claiming?&lt;/p&gt;&#xA;" OwnerUserId="3299" LastActivityDate="2011-05-10T03:51:28.307" Title="Physics of scaling up an animal: the neck" Tags="&lt;classical-mechanics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="9160" PostTypeId="2" ParentId="9156" CreationDate="2011-04-26T22:07:33.233" Score="3" ViewCount="" Body="&lt;p&gt;From the example you've given, it's clear that you're using the wrong terms to describe what you want. Purity and mixedness apply to density operators and not to state vectors -- if your system is described by a state vector $|\psi\rangle$, it is already pure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you seem to want to know is whether there is an operator that, in a particular basis, has an expectation value that lets you know to what degree the state is in a superposition of basis states. I think the uncertainty operator might work for you here. If $S$ is the operator whose eigenstates are $|0\rangle$ and $|1\rangle$, then the operator $(S-\langle S\rangle)^2$ has zero expectation value for the basis states and a non-zero value for all superpositions. You could scale this in some way to give you a value between 0 and 1.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-04-26T22:07:33.233" />
  <row Id="9161" PostTypeId="1" CreationDate="2011-04-26T22:58:07.690" Score="8" ViewCount="194" Body="&lt;p&gt;In physics and engineering sources, calculus-based formalisms - whether differential forms on a manifold, or &quot;differentials&quot; of functions of several variables - are presented as a way of modeling and reasoning about thermodynamic systems. However, I've found little to no mathematical background given to justify these formal manipulations when there are phase changes or other discontinuities. In such regions, one would expect the theory to break down since partial derivatives and differential forms are not defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, if you &quot;shut up and calculate&quot;, everything seems to work out fine. Why is this the case, and what is the proper mathematical state space and framework for systems with phase transitions or other non-smooth properties? (perhaps some sort of &quot;weakly-differentiable&quot; manifold?)&lt;/p&gt;&#xA;" OwnerUserId="821" LastActivityDate="2011-05-28T15:07:32.550" Title="Discontinuities and nondifferentiability in thermodynamics" Tags="&lt;thermodynamics&gt;&lt;differential-geometry&gt;&lt;phase-transition&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9162" PostTypeId="2" ParentId="9156" CreationDate="2011-04-26T23:06:48.573" Score="4" ViewCount="" Body="&lt;p&gt;The question (v1) abuses language slightly no matter how one understands it. Here I sketch three interpretations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Interpretation in terms of a density matrix $\rho$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the operator ${\cal O}$ is diagonalizable. Since the eigenstates are pure states, the eigenvalues must be $1$. In other words, ${\cal O}$ is the identity operator.&#xA;Then $\langle {\cal O} \rangle = \mathrm{tr}{\cal O}\rho=\mathrm{tr}\rho=1$ for any mixed or pure density matrix $\rho$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Interpretation in terms of an operator ${\cal O}$ that satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle 0 |{\cal O}| 0\rangle =1,$$&#xA;$$\langle 1 |{\cal O}| 1\rangle =1,$$&#xA;$$\langle \psi |{\cal O}| \psi\rangle =0 \qquad \mathrm{for~~all} \qquad |A|=|B|=\frac{1}{\sqrt{2}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\psi\rangle = A| 0\rangle+B| 1\rangle, \qquad 1=||\psi||^2 = \langle \psi|\psi\rangle =|A|^2+|B|^2 .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|A|=|B|=\frac{1}{\sqrt{2}}$ is &lt;em&gt;not&lt;/em&gt; what is traditionally meant by being &quot;the most mixed state&quot;. Note that there are not just one of these states, but infinitely many pairs of complex numbers $(A,B)$ that satisfy $|A|=|B|=\frac{1}{\sqrt{2}}$. Even after removing an overall phase, there is a relative phase left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume ${\cal O}$ is a Hermitian operator. Then ${\cal O}$ has a matrix of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\cal O} = \left[\begin{array}{cc} 1 &amp;amp; c \\ c^* &amp;amp; 1 \end{array}\right]. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0=\langle \psi |{\cal O}| \psi\rangle = |A|^2+|B|^2+A^*cB+B^*c^*A = 1+2\mathrm{Re}(A^*cB).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is impossible for general complex phases of $A$ and $B$ with $|A|=|B|=\frac{1}{\sqrt{2}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Interpretation in terms of an operator ${\cal O}$ that satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle 0 |{\cal O}| 0\rangle =1,$$&#xA;$$\langle 1 |{\cal O}| 1\rangle =1,$$&#xA;$$\langle \psi |{\cal O}| \psi\rangle =0 \qquad \mathrm{for~at~least~one} \qquad |A|=|B|=\frac{1}{\sqrt{2}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The characteristic polynomial $p_{\cal O}(\lambda)$ for the matrix ${\cal O}$ reads&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_{\cal O}(\lambda)=\det({\cal O}-\lambda {\bf 1})= (\lambda-1)^2-|c|^2= (\lambda-1-|c|) (\lambda-1+|c|).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the two eigenvalues of ${\cal O}$ are $\lambda=1\pm|c|$. Since the expectation values for ${\cal O}$ should be between $0$ and $1$, we must demand that $c=0$. In other words, ${\cal O}$ is the identity operator, which leads to the following inconsistency &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0=\langle \psi |{\cal O}| \psi\rangle =\langle \psi |\psi\rangle =1.$$&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-04-27T14:40:35.170" LastActivityDate="2011-04-27T14:40:35.170" CommentCount="3" />
  <row Id="9163" PostTypeId="1" CreationDate="2011-04-26T23:14:40.270" Score="5" ViewCount="177" Body="&lt;p&gt;I'm trying to imagine the geometry &quot;operations&quot; here:&lt;br&gt;&#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/8011/angular-deficit&quot;&gt;Angular deficit&lt;/a&gt;&lt;br&gt;&#xA;and&lt;br&gt;&#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/8816/conical-spacetime-of-cosmic-string&quot;&gt;Conical spacetime of cosmic string&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If we sew flat spacetime pieces together, what is the requirement for the sewing to not create curvature at that seam?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly there has to be some condition, because in the examples above, sewing together the wedge &quot;sides&quot; doesn't produce cuvature, but somehow at the corner of the removed wedge, curvature appears.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, more concretely, if we remove this chunk of flat spacetime:&#xA;$$(x&amp;gt;0) \mathrm{\ and \ } (0&amp;lt;y&amp;lt;x)$$&#xA;and now sew it together, somehow this creates flat spacetime everywhere &lt;em&gt;except&lt;/em&gt; at $x=y=0$.  Is there some intuitive physical meaning to this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the issue is the sharp angle there, then what if we remove instead a nice smooth parabola:&#xA;$$(x&amp;gt;0) \mathrm{\ and \ } (-\sqrt{x}&amp;lt;y&amp;lt;\sqrt{x})$$&#xA;or maybe a wedge that is completely smooth to zero angle&#xA;$$[x&amp;gt;0] \mathrm{\ and \ } [0&amp;lt;y&amp;lt;\exp(-1/x)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these completely flat everywhere (no string defect) AND have angular deficit?&lt;/p&gt;&#xA;" OwnerUserId="2702" LastEditorUserId="2702" LastEditorDisplayName="" LastEditDate="2011-04-27T00:13:51.473" LastActivityDate="2011-04-27T16:14:46.193" Title="Sewing together flat spacetime pieces = flat spacetime?" Tags="&lt;differential-geometry&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="9165" PostTypeId="1" CreationDate="2011-04-26T23:50:25.467" Score="3" ViewCount="366" Body="&lt;p&gt;I'm a bachelor student majoring in math, and pretty interested in physics. I would like a book to study for classical mechanics, that will prepare me to work through &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201657023&quot; rel=&quot;nofollow&quot;&gt;Goldstein's Classical Mechanics&lt;/a&gt;. What books would be good for a beginner?&lt;/p&gt;&#xA;" OwnerUserId="3189" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-12T00:20:06.417" LastActivityDate="2011-08-12T00:20:06.417" CommunityOwnedDate="2011-08-12T00:20:14.823" Title="Which Mechanics book is the best for beginner in math major?" Tags="&lt;soft-question&gt;&lt;books&gt;" AnswerCount="6" CommentCount="6" FavoriteCount="1" />
  <row Id="9166" PostTypeId="2" ParentId="8381" CreationDate="2011-04-27T00:27:29.570" Score="2" ViewCount="" Body="&lt;p&gt;I will now answer yes to my own question - there is a relationship - though I can't prove it yet. I was thinking about the primordial example of AdS/CFT - N=4 Yang-Mills - and I was wondering how to interpolate between flat space and AdS space (&lt;a href=&quot;http://ocw.mit.edu/courses/physics/8-821-string-theory-fall-2008/lecture-notes/lecture03.pdf&quot; rel=&quot;nofollow&quot;&gt;exposition here&lt;/a&gt;) in the string picture; that is, between the situation in which there is no geometric back-reaction due to the stack of branes, and the situation in which the stack of branes forms a black hole. AdS space shows up as the geometry around the branes in the latter case, the geometry at the bottom of the black hole's &quot;throat&quot;. The throat is infinitely deep, so the space at the bottom is decoupled from the space outside the black hole, and one is left with a pure AdS geometry. Yet this lies on a continuum with the flat space case; how can this be? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It suddenly occurred to me that the AdS geometry is already present in the flat space geometry - it is the &quot;geometry&quot; created by RG flow in the worldvolume theory! That's the emergent geometry showing up in MERA, including its &lt;a href=&quot;http://arxiv.org/abs/1102.5524&quot; rel=&quot;nofollow&quot;&gt;continuum version&lt;/a&gt;; this must be &lt;a href=&quot;http://arxiv.org/abs/1010.1264&quot; rel=&quot;nofollow&quot;&gt;&quot;holographic renormalization&quot;&lt;/a&gt; by another name. In the paper I cited, on constructing a Vasiliev dual to a free field theory, they explicitly write down equations of motion on the RG geometry. However you approach it, this must be how you interpolate smoothly between flat space and AdS space - the AdS space goes from being the RG geometry of the worldvolume theory, to the physical geometry, while the physical space spanned by the branes recedes to the boundary of AdS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the conceptual details remain to be worked out, of course. :-)  But I imagine that will be done soon. Perhaps the real open question is still the relationship between string theory and Vasiliev theory. &lt;/p&gt;&#xA;" OwnerUserId="1486" LastEditorUserId="1486" LastEditorDisplayName="" LastEditDate="2011-04-27T01:24:27.033" LastActivityDate="2011-04-27T01:24:27.033" CommentCount="2" />
  <row Id="9167" PostTypeId="2" ParentId="9163" CreationDate="2011-04-27T01:52:49.550" Score="4" ViewCount="" Body="&lt;p&gt;The mathematics actually tells you that your second and third constructions are physically impossible!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I shall assume here the &quot;lack of string defects&quot; is equivalent to your space-time being globally a $C^2$-Riemannian manifold. As David mentioned in the comments, for any null-homotopic loop in such a manifold (null-homotopic just means that the loop can be continuously shrunken to a point) there can be no angular defect. (In two dimensions this is a Corollary of the Gauss-Bonnet theorem.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the objects that you want to construct by removing a wedge and sewing up will obviously be simply connected, so every loop is null homotopic. So if you were actually able to do the sewing without creating a lack of smoothness (a string defect), the above will imply that there cannot be an angular defect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, by construction, if this object were to exist, it is easy to argue, as in the cone case, that there must be an angular defect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So by eliminating all the impossibilities, the only option left is that &lt;strong&gt;whatever the shape you cut out of the plane, you can not &quot;sew up the new edges&quot; in a way without introducing a singularity&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="829" LastEditorUserId="829" LastEditorDisplayName="" LastEditDate="2011-04-27T16:14:46.193" LastActivityDate="2011-04-27T16:14:46.193" CommentCount="3" />
  <row Id="9168" PostTypeId="2" ParentId="9096" CreationDate="2011-04-27T02:15:33.633" Score="0" ViewCount="" Body="&lt;p&gt;If the force of the engine is constant and equal to $F$, while the length of the stick is $L$, and it's mass moment of inertia $I=m\,L^2/12$ then there is a constant torque applied resulting in a constant rotational acceleration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ddot{\theta} = \mbox{-}\dfrac{F\,L}{2 I} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\theta$ is the orientation of the stick relative some arbitrary fixed axis. Positive $theta$ is counter-clockwise from horizontal in the picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \theta(t) = \theta_0 + \omega_0\,t-\dfrac{F\,L}{4\,I}\,t^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while the center of gravity is accelerating linear in whatever direction the engine in point at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec{a}_G = \begin{pmatrix} \mbox{-}\frac{F}{m}\sin\theta &amp;amp; \frac{F}{m}\cos\theta \end{pmatrix} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integration of the acceleration vector over time escapes me, but it does look like it is related to &lt;em&gt;Fresnel&lt;/em&gt; integrals.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-04-27T02:15:33.633" CommentCount="0" />
  <row Id="9169" PostTypeId="1" CreationDate="2011-04-27T02:27:48.687" Score="1" ViewCount="250" Body="&lt;p&gt;I am having this very basic problem. In e.g Maldacena's AdS/CFT review (0309246) (page 5), he has defined operators as $\mathcal{O}=N\,{\rm tr}[f(M)]$ for some matrices $M$ and got the connected correlators as $\langle\mathcal{O}\mathcal{O}\rangle_c\sim N^2$ and $\langle\mathcal{O}\mathcal{O}\mathcal{O}\rangle_c\sim N^2$. Whereas in the MAGOO review (9905111) (page 14-15 e.g) we see some operators defined as $G$ which is added to the action as $S\rightarrow S+N\sum g_jG_j$ which gives the correlators as $\langle\prod_j G_j\rangle\sim N^{2-n}$. So that three points and higher vanish at large N. I thought this one is generic. If so I wanted to understand how is the first conventions/case are different and their difference (has it something to do with connected/disconnected diagram? etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related to that I wanted to know about large N factorization. For a particular model if I calculate disconnected  diagrams, I see that at large N I indeed get this factorization. I learned that its like WKB in $\hbar\sim 1/N$. But I wanted to feel it more generally in possibly a mathematical+insightful way if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: If it has a long answer please at least let me know which way to think or any references where I can get some ideas. Thanks again.&lt;/p&gt;&#xA;" OwnerUserId="1349" LastEditorUserId="1349" LastEditorDisplayName="" LastEditDate="2011-04-27T21:02:27.060" LastActivityDate="2011-04-27T21:02:27.060" Title="Correlators at large N and large N factorization" Tags="&lt;quantum-field-theory&gt;&lt;ads-cft&gt;" FavoriteCount="1" />
  <row Id="9170" PostTypeId="1" CreationDate="2011-04-27T02:36:06.150" Score="2" ViewCount="124" Body="&lt;p&gt;I don't know about all the details of Bell tests using methods like parametric down conversion, but at least in some versions of the EPR paradox you get two photons moving apart in opposite directions. I wonder if you can look for detection coincidences by using photographic plates instead of coincidence counters? The idea would be that if you had two photographic plates on opposite sides of the source, you would get some instances where you would have a perfect matchup of reduced silver crystals, or dots. And that if you inserted crossed polarizers in front of the plates, the coincidences would disappear. I wonder if this analysis is correct, and if so, whether such experiments have been done or proposed? &lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-05-28T20:51:12.210" Title="Bell Tests using position measurement" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="9171" PostTypeId="2" ParentId="9165" CreationDate="2011-04-27T02:42:34.737" Score="1" ViewCount="" Body="&lt;p&gt;Goldstein is good. However, since you are anyways a math major, you can easily start off with Jose and Saletan or Arnold's \emph{Mathematical Methods of Classical Mechanics}. There are other books like Hand &amp;amp; Finch and definitely the classics by Landau.&lt;/p&gt;&#xA;" OwnerUserId="2973" LastActivityDate="2011-04-27T02:42:34.737" CommunityOwnedDate="2011-08-12T00:20:14.823" CommentCount="4" />
  <row Id="9173" PostTypeId="2" ParentId="9165" CreationDate="2011-04-27T02:45:29.087" Score="3" ViewCount="" Body="&lt;p&gt;Start here: &lt;a href=&quot;http://www.math.uga.edu/~shifrin/Spivak_physics.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.math.uga.edu/~shifrin/Spivak_physics.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2599" LastActivityDate="2011-04-27T02:45:29.087" CommunityOwnedDate="2011-08-12T00:20:14.823" CommentCount="2" />
  <row Id="9174" PostTypeId="2" ParentId="9163" CreationDate="2011-04-27T03:32:37.670" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;what is the requirement for the sewing to not create curvature at that seam?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Ted's answer to my question, he wrote:&lt;br&gt;&#xA;&lt;em&gt;&quot;In general, you only have curvature if you have to &quot;crumple&quot; or &quot;stretch&quot; the paper. When you form a cone, you don't have to do that, anywhere except at the vertex.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand why the vertex counts as crumpling or stretching, but the comment makes intuitive sense.  So one condition is that you need to sew the pieces together without stretching.  If sewing is mapping a point on one edge to a point on the other edge, then once one pair of points are equated there will be curvature if the other points aren't equated with equal path length along the sewing seam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to go further in general, so I'll simplify the problem some.  Let's cut the following out of flat space (in polar coordinates):&#xA;$0&amp;lt;r&amp;lt;\infty, -f(r)&amp;lt;\theta&amp;lt;f(r)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sewing it up we can change coordinates&lt;br&gt;&#xA;$k \phi=\theta, k=\frac{2\pi - 2f(r)}{2\pi} = 1- f(r)/\pi$&lt;br&gt;&#xA;giving the line element in these coordinates as&lt;br&gt;&#xA;$$d\theta = \phi \frac{\partial k}{\partial r} dr + k\ d\phi = \phi F(r) dr + k\ d\phi$$&#xA;$$ds^2 = dr^2 + r^2 (\phi^2 F^2\ dr^2 + 2\phi F k\ dr\ d\phi + k^2\ d\phi^2)$$&#xA;and the coordinates go from $0 \le r &amp;lt; \infty$ and $0 \le \phi \le 2\pi$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh god this is going to be messy.  But you can in principle get the Riemann curvature and a condition on f(r) for the curvature to be zero.  (Does anyone know how to make Mathematica or something calculate the answer here?)&lt;/p&gt;&#xA;" OwnerUserId="2655" LastEditorUserId="2655" LastEditorDisplayName="" LastEditDate="2011-04-27T03:38:48.740" LastActivityDate="2011-04-27T03:38:48.740" CommentCount="1" />
  <row Id="9175" PostTypeId="2" ParentId="9158" CreationDate="2011-04-27T03:36:20.787" Score="1" ViewCount="" Body="&lt;p&gt;Here comes an experimentalist point of view.These models are approximations of the real theoretical description of matter in aggregate,which at the moment is not a simple one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To truly describe nuclear forces in the language of QCD , our current knowledge of the theory of strong interactions, is a formidable many body task.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the calculational methods that exist now you would need thousands of Feynman diagrams calculated for each specific question. The models, fitted to real data, allow one to describe the behavior of nuclei and have some predictive power  over it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Qualitatively one can think of the drop model as an approximation to the continuity of force interchanges with gluon exchanges between the quarks of the nuclei within the nuclear bag. At high energies one talks of quark-gluon plasma. This is the liquid phase analogy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shell model works as a total approximation to a potential well, similar to the way the Bohr model worked for describing the atom. There is a collective potential well there, created by all the gluon exchanges, and it is  approximated in this simplified form using data to be able to predict further data. It works.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-27T03:36:20.787" CommentCount="1" />
  <row Id="9177" PostTypeId="2" ParentId="9152" CreationDate="2011-04-27T04:02:22.273" Score="1" ViewCount="" Body="&lt;p&gt;For 1) Yes for particles in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_Model&quot; rel=&quot;nofollow&quot;&gt;standard model&lt;/a&gt; scheme. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 2) You have to figure it out from the SM scheme, linked above and the energies involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 3) It is more difficult to create and handle an anti proton beam as well as to build up a sufficient number of antiprotons in the beam. Dumping is the same for both beams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 4)It depends on your model, and how one assumes the WIMPs appear during the cooling after the big bang. It is reasonable to assume that the same CP violation that created the observed asymmetry between matter and antimatter holds for WIMPS too. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-27T04:02:22.273" />
  <row Id="9178" PostTypeId="2" ParentId="9151" CreationDate="2011-04-27T05:00:51.697" Score="1" ViewCount="" Body="&lt;p&gt;There is no simple relationship between the worldsheet objects and spacetime objects (such as superpotentials) in this curved case (and many other general cases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LG-like models correspond to Calabi-Yau compactifications whose typical radii are comparable to the string length so it is very hard to see the spacetime geometry - and moreover, the mirror symmetric geometry is as easy or as hard to see as the &quot;original&quot; one, whichever is which.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, much of the world sheet objects are auxiliary and don't survive into spacetime. In particular, the whole world sheet supersymmetry is an auxiliary symmetry - a gauge symmetry - that doesn't really survive in spacetime. Its existence has consequences in spacettime (nonrenormalization theorems etc.) but its detailed building blocks don't have any counterparts because they're partly &quot;unphysical states&quot; from the spacetime viewpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the way to proceed when identifying the topology of Gepner-like models is to find the Hodge numbers and primary operators, choose a corresponding topology of a Calabi-Yau space that matches the constraints, and verify that this candidate is indeed equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that there can't be any coordinate-by-coordinate map between the world sheet pieces and spacetime physics, note that the central charges of the world sheet CFT are fractional. The relationship between Gepner models and particular Calabi-Yaus is a form of duality, and as any duality, it must be nontrivial to be seen, otherwise it wouldn't be interesting and it wouldn't really be a &quot;duality&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-27T05:00:51.697" />
  <row Id="9181" PostTypeId="2" ParentId="9165" CreationDate="2011-04-27T09:25:50.417" Score="1" ViewCount="" Body="&lt;p&gt;You might be interested in Eugene Hecht's book: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0534339859&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Physics-Calculus-Eugene-Hecht/dp/0534339859&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I like about it is that both fundamental principles (notably Noether's principle) and basic but important techniques (like dealing with significant digits) are consistently mentioned and enforced throughout the book. It also contains a lot of examples and exercices of varying levels of difficulty.&lt;/p&gt;&#xA;" OwnerUserId="1725" LastActivityDate="2011-04-27T09:25:50.417" CommunityOwnedDate="2011-08-12T00:20:14.823" CommentCount="1" />
  <row Id="9182" PostTypeId="1" CreationDate="2011-04-27T09:41:29.677" Score="0" ViewCount="115" Body="&lt;p&gt;I have a vehicle traveling at initial speed 100m/s.&#xA;Expected destination speed is also 100m/s.&#xA;Maximum deceleration is 3.&#xA;Maximum acceleration is 10.&#xA;Distance to the target is 2000m.&#xA;The vehicle should reach the target in 40s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vehicle should follow a linear path, decelerate(a1) to a certain speed(v1) and accelerate back(a2) to reach the target precisely on 40s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: v1 &gt; 0, a1 &amp;lt; 3, a2 &amp;lt; 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we find a1, a2 and v1 ??&lt;/p&gt;&#xA;" OwnerUserId="3306" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-04-27T09:55:51.763" LastActivityDate="2011-04-27T20:35:59.720" Title="Deccelerate and Accelerate to achieve timed journey" Tags="&lt;homework&gt;&lt;kinematics&gt;&lt;motion&gt;&lt;velocity&gt;&lt;acceleration&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="9183" PostTypeId="1" AcceptedAnswerId="9184" CreationDate="2011-04-27T09:58:01.770" Score="3" ViewCount="184" Body="&lt;p&gt;Once a quantum partition function is in path integral form, does it contain any operators?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. The quantum partition function is $Z=tr(e^{-\beta H})$ where H is an operator, the Hamiltonian of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if I put this into the path integral formalism so that we have something like $Z= \int D(\bar{\gamma},\gamma) e^{-\int_0^\beta d\tau H(\bar{\gamma},\gamma)}$, is the $H(\bar{\gamma},\gamma)$ an operator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="873" LastActivityDate="2011-09-04T23:18:59.100" Title="Once a quantum partition function is in path integral form, does it contain any operators?" Tags="&lt;quantum-mechanics&gt;&lt;statistical-mechanics&gt;&lt;path-integral&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" />
  <row Id="9184" PostTypeId="2" ParentId="9183" CreationDate="2011-04-27T10:07:35.000" Score="5" ViewCount="" Body="&lt;p&gt;Nope, Feynman's path integral formulation of quantum mechanics is a method to directly calculate the complex probability amplitudes and all objects that appear in its formalism - not counting proofs of equivalence with other approaches to quantum mechanics - are $c$-numbers representing classical observables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the exponent in the path integral - which should be $iS$ ($i$ times the action i.e. $i$ times the integrated Lagrangian), not the Hamiltonian - is a $c$-number-valued function of the &quot;classical observables&quot;, the same function that is relevant for the classical (non-quantum) theory. So the path integral is an infinite-dimensional integral over otherwise &quot;ordinary classical variables&quot; that produces some probability amplitudes - the same ones that may be (but don't have to be) obtained from the operator formalism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Uncertainty principle in path integral&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, some sorts of the path integral include integration over both positions and momenta, $\int Dx(t)\, Dp(t)$. How it is possible that both of them are treated &quot;classically&quot; as $c$-numbers? Doesn't it violate the uncertainty principle? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that it doesn't violate the uncertainty principle. One may still deduce that $xp-px=i\hbar$ from the path integral as long as she is careful about putting the right values of the time $t$ (the argument). The quantities $x(t)p(t-\epsilon)$ and $p(t)x(t-\epsilon)$ differ. A necessary condition for this difference to exist is the fact that &quot;most&quot; trajectories contributing to the path integral are discontinuous.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-04-27T12:10:05.847" LastActivityDate="2011-04-27T12:10:05.847" CommentCount="3" />
  <row Id="9186" PostTypeId="2" ParentId="9182" CreationDate="2011-04-27T12:11:33.023" Score="1" ViewCount="" Body="&lt;p&gt;This problem looks likely to be badly posed: either there are a large number of solutions or there are none.  Only in the most artificial cases will there be a single solution at an extreme: either do not decelerate or accelerate at all so you go as far as possible in the time, or decelerate and then accelerate as hard as possible all the time to go as short a distance as possible in the time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take the hard deceleration and acceleration possibility.  Since the final speed is the same as the initial speed, the changeover time must be at about time 30.769 seconds.  This means $v_1$ is about 7.692 m/s and the distance travelled to the changeover point is about 1656.805 metres; so at 40 seconds the speed is back to 100 m/s and the further distance travelled is about 497.041 metres, making a total of well over 2000 metres.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Too far.&lt;/p&gt;&#xA;" OwnerUserId="2714" LastActivityDate="2011-04-27T12:11:33.023" />
  <row Id="9187" PostTypeId="1" CreationDate="2011-04-27T12:17:16.287" Score="-1" ViewCount="174" Body="&lt;p&gt;Currently I am planning to get masters degree. So I am thinking about a subject in which I have to get masters degree. Following are my questions to leading physicists..&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Which technology is the future of telecommunication? Currently electromagnetic waves only rules the world for telecommunication. But now a days there is very vast amount of research going on in quantum mechanics and particle physics. According my knowledge the quantum entanglement and delocalization is the base for teleportation and future of communication. I want from the experts to suggest for choosing the topic for masters degree. Is the quantum communication is possible in real world one day? Can it replace our way of telecommunication through electromagnetic waves?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If i will do masters in electromagnetism, will i stay behind in time?&lt;/p&gt;&#xA;" OwnerUserId="3309" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-27T12:22:36.003" LastActivityDate="2011-04-27T12:30:04.443" Title="Can quantum communication really replace electromagnetic waves for telecommunication medium in future?" Tags="&lt;quantum-mechanics&gt;&lt;electromagnetism&gt;&lt;teleportation&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="9188" PostTypeId="2" ParentId="9187" CreationDate="2011-04-27T12:30:04.443" Score="5" ViewCount="" Body="&lt;p&gt;Quantum communication &lt;em&gt;uses&lt;/em&gt; electromagnetism. It's not a magic technology which communicates information from point A to point B with no physical objects carrying it. Quantum communication is carried by qubits, and you need to get qubits from one point to another; right now, it looks like the most practical way to do that is by sending photons through fiber optic cables (or maybe sending photons directly through the air or through space). Photons are the quantum version of electromagnetic waves. So studying electromagnetism will be useful to you.&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2011-04-27T12:30:04.443" />
  <row Id="9189" PostTypeId="1" CreationDate="2011-04-27T12:37:58.320" Score="1" ViewCount="137" Body="&lt;p&gt;I often see the expression $W = V \Delta P$ for the work of a constant volume compression where there are a fixed number of moles and the compression is caused by heating.  Is this the work equation for a constant volume, isothermal process where the pressure is increased by adding moles of a gas?&lt;/p&gt;&#xA;" OwnerUserId="3311" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-04-27T15:25:07.657" LastActivityDate="2011-04-28T22:54:18.277" Title="Work for constant volume compression by adding moles of gas" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9192" PostTypeId="2" ParentId="9182" CreationDate="2011-04-27T14:43:31.610" Score="0" ViewCount="" Body="&lt;p&gt;In general, Henry's answer is correct - this task has no solution to satisfy all the conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The normal set of equations would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_1=100, v_3=100, t=40, s=2000, a1&amp;lt;3, a2&amp;lt;10$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s_1 = v_1 t_1$  - first idle part at max speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s_2 = v_1 t_2 - a_1 t_2^2/2$  - deceleration part&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s_3 = v_2 t_3 + a_2 t_3^2/2$ - acceleration part&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_2 = v_1 - a_1*t_2$ - the minimum speed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_2 = v_3 - a_2*t_3$ - we must accelerate to end speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s_1 + s_2 + s_3 = s$ - total distance = 2000m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t_1 + t_2 + t_3 = t$ - total time = 40s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t_1 &amp;gt;= 0, t_2 &amp;gt;= 0, t_3 &amp;gt;= 0, s_1 &amp;gt;= 0, s_2 &amp;gt;= 0, s_3 &amp;gt;= 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_1 &amp;gt;= 0, a_2 &amp;gt;= 0, a_1 &amp;lt;= 3, a_2 &amp;lt;= 10, v_2&amp;gt;0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want, try to solve it fully. You'll eventually arrive at a quadric equation with negative delta. If you want a simplified solution, assume $a_1 = 3, a_2 = 10, t_0=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $v_2$ equations give $t_3= 0.3 t_2$;  $t_2 + t_3 = 40$. You can calculate the results of this time split yourself.&lt;/p&gt;&#xA;" OwnerUserId="2754" LastActivityDate="2011-04-27T14:43:31.610" />
  <row Id="9193" PostTypeId="2" ParentId="9161" CreationDate="2011-04-27T15:49:09.757" Score="1" ViewCount="" Body="&lt;p&gt;Far from a phase transition, the thermodynamics parameters are well defined things and are sufficient to describe thermodynamical processes. Close to a phase transition, these parameters may become highly fluctuating, and indicating only an average value is a too poor description. It is not only mathematical &quot;discontinuity&quot; but also physical &quot;uncertainty&quot; that is implicated. Here one needs a more sophisticated picture.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-04-27T15:49:09.757" CommentCount="3" />
  <row Id="9194" PostTypeId="1" AcceptedAnswerId="9199" CreationDate="2011-04-27T16:08:10.177" Score="3" ViewCount="320" Body="&lt;p&gt;I understand the mathematics of commutation relations and anti-commutation relations, but what does it mean for something to commute with another in quantum mechanics? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. an operator with the hamiltonian&lt;/p&gt;&#xA;" OwnerUserId="1123" LastActivityDate="2011-04-28T07:45:55.907" Title="What is the meaning of Commutation?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="9195" PostTypeId="2" ParentId="9194" CreationDate="2011-04-27T16:28:46.473" Score="0" ViewCount="" Body="&lt;p&gt;It means you can (in principle) measure both quantities to arbitrary precision at the same time. If they didnt commute then this would be impossible by the uncertainty principle.&lt;/p&gt;&#xA;" OwnerUserId="3314" LastActivityDate="2011-04-27T16:28:46.473" CommentCount="1" />
  <row Id="9196" PostTypeId="2" ParentId="9194" CreationDate="2011-04-27T16:31:37.330" Score="-2" ViewCount="" Body="&lt;p&gt;You may consider commutativity of &lt;em&gt;different&lt;/em&gt; variables as physical independence, something like separated independent variables:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial}{\partial x} \frac{\partial}{\partial y} = \frac{\partial}{\partial y} \frac{\partial}{\partial x}$.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-04-27T21:06:27.743" LastActivityDate="2011-04-27T21:06:27.743" />
  <row Id="9197" PostTypeId="2" ParentId="9194" CreationDate="2011-04-27T16:46:28.180" Score="0" ViewCount="" Body="&lt;p&gt;Physically it means, it does not matter in which temporal order you measure the two commuting observables.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-04-27T16:46:28.180" />
  <row Id="9198" PostTypeId="2" ParentId="9165" CreationDate="2011-04-27T16:58:22.917" Score="1" ViewCount="" Body="&lt;p&gt;Goldstein's Classical Mechanics is a masterpiece. That's why it's still held in such high regard as one of the best mechanics books available 60 years later after after 3 revisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest that you should stick with this book and work through it sloooooooooooooooooooooooooooooooowly. If that is too difficult then sloooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty thick, to be honest, which is why I had no problem over the embarrassement of taking a week trying to understand a page of the first chapter sometimes. Others would take a month, and others I still don't understand fully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think any book comes anywere close to explaining mechanics concisely and accurately in the first chapter, where others take 300 pages and yet still gloss over the details.&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-04-27T16:58:22.917" CommunityOwnedDate="2011-08-12T00:20:14.823" CommentCount="4" />
  <row Id="9199" PostTypeId="2" ParentId="9194" CreationDate="2011-04-27T17:17:50.457" Score="10" ViewCount="" Body="&lt;p&gt;Let us first restate the mathematical statement that two operators $\hat A$ and $\hat B$ commute with each other. It means that&#xA;$$\hat A \hat B - \hat B \hat A = 0,$$&#xA;which you can rearrange to &#xA;$$\hat A \hat B = \hat B \hat A.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you recall that operators &lt;em&gt;act&lt;/em&gt; on quantum mechanical states and give you a new state in return, then this means that with $\hat A$ and $\hat B$ commuting, the state you obtain from letting first $\hat A$ and then $\hat B$ act on some initial state is &lt;em&gt;the same&lt;/em&gt; as if you let first $\hat B$ and then $\hat A$ act on that state:&#xA;$$\hat A \hat B | \psi \rangle = \hat B \hat A | \psi \rangle.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is &lt;strong&gt;not&lt;/strong&gt; a trivial statement. Many operations, such as rotations around different axes, do &lt;em&gt;not&lt;/em&gt; commute and hence the end-result depends on how you have ordered the operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what are the important implications? Recall that when you perform a quantum mechanical measurement, you will always measure an &lt;em&gt;eigenvalue&lt;/em&gt; of your operator, and after the measurement your state is left in the corresponding eigenstate. The eigenstates to the operator are precisely those states for which there is no uncertainty in the measurement: You will always measure the eigenvalue, with probability $1$.&#xA;An example are the energy-eigenstates. If you are in a state $|n\rangle$ with eigenenergy $E_n$, you know that $H|n\rangle = E_n |n \rangle$ and you will &lt;em&gt;always&lt;/em&gt; measure this energy $E_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what if we want to measure two different observables, $\hat A$ and $\hat B$? If we first measure $\hat A$, we know that the system is left in an eigenstate of $\hat A$. This might alter the measurement outcome of $\hat B$, so, in general, the order of your measurements is important. Not so with commuting variables! It is shown in every textbook that if $\hat A$ and $\hat B$ commute, then you can come up with a set of &lt;em&gt;basis states&lt;/em&gt; $| a_n b_n\rangle$ that are eigenstates of &lt;em&gt;both&lt;/em&gt; $\hat A$ and $\hat B$. If that is the case, then any state can be written as a linear combination of the form&#xA;$$| \Psi \rangle = \sum_n \alpha_n | a_n b_n \rangle$$&#xA;where $|a_n b_n\rangle$ has $\hat A$-eigenvalue $a_n$ and $\hat B$-eigenvalue $b_n$.&#xA;Now if you measure $\hat A$, you will get result $a_n$ with probability $|\alpha_n|^2$ (assuming no degeneracy; if eigenvalues are degenerate, the argument still remains true but just gets a bit cumbersome to write down). What if we measure $\hat B$ first? Then we get result $b_n$ with probability $|\alpha_n|^2$ and the system is left in the corresponding eigenstate $|a_n b_n \rangle$. If we &lt;em&gt;know&lt;/em&gt; measure $\hat A$, we will always get result $a_n$. The overall probability of getting result $a_n$, therefore, is again $|\alpha_n|^2$. So it didn't matter that we measure $\hat B$ before, it did not change the outcome of the measurement for $\hat A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; Now let me expand even a bit more. So far, we have talked about &lt;em&gt;some&lt;/em&gt; operators $\hat A$ and $\hat B$. We now ask: What does it mean when some observable $\hat A$ commutes with the &lt;em&gt;Hamiltonian&lt;/em&gt; $H$? First, we get all the result from above: There is a simultaneous eigenbasis of the energy-eigenstates and the eigenstates of $\hat A$. This can yield a tremendous simplification of the task of diagonalizing $H$. For example, the Hamiltonian of the hydrogen atom commutes with $\hat L$, the angular momentum operator, and with $\hat L_z$, its $z$-component. This tells you that you can classify the eigenstates by an angular- and magnetic quantum number $l$ and $m$, and you can diagonalize $H$ for each set of $l$ and $m$ independently. There are more examples of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another consequence is that of time dependence. If your observable $\hat A$ has no explicit time dependency introduced in its definition, then if $\hat A$ commutes with $\hat H$, you immediately know that $\hat A$ is a constant of motion. This is due to the &lt;strong&gt;Ehrenfest Theorem&lt;/strong&gt;&#xA;$$\frac{d}{dt} \langle \hat A \rangle = \frac{-i}{\hbar} \langle [\hat A, \hat H] \rangle &#xD;&#xA;+ \underbrace{\langle \frac{\partial \hat A}{\partial t} \rangle}_{=0\,\text{by assumption}}$$&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditorDisplayName="" LastEditDate="2011-04-27T23:13:16.540" LastActivityDate="2011-04-27T23:13:16.540" CommentCount="1" />
  <row Id="9200" PostTypeId="2" ParentId="9182" CreationDate="2011-04-27T18:27:31.117" Score="0" ViewCount="" Body="&lt;p&gt;The units of a1 and a2 are m/(s^2)?  Gravities?  I will assume the former.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you define: t1 = time of acceleration a1 (with 0&gt;a1&gt;-3); t2 = time of acceleration a2 (t2 = 0 at start of acceleration a2); and restrict the solution space to t1+t2=40 (no dead time between a1 and a2). Then you can get a quadratic to describe motion from start to intermediate point, and another quadratic to describe motion from the intermediate point to 2000m.  The intermediate point is where you transition from a1 to a2.  That is 3 equations and 4 unknowns.  When I do the substitution (no guarantee I got it right) I get a quadratic in t1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a(t1^2) + b(t1) + c = 0; where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b = -80&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c = (4000+1600(a2))/(a2-a1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a solution space which gets you to 2000m at t1+t2 = 40 seconds.  The next thing you need to do is ensure you get to 2000m with a velocity of 100m/s.  Since there is not dead time, and since v(t=0s) = v(t=40s) = 100m/s; then (t1)(a1) + (t2)(a2) = 0.  This eqn yields a1 = -(a2)(t2)/(t1) = -(a2)(40-t1)/(t1).  substitute that back into the quadratic.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-04-27T20:35:59.720" LastActivityDate="2011-04-27T20:35:59.720" />
  <row Id="9201" PostTypeId="1" CreationDate="2011-04-27T18:28:22.520" Score="1" ViewCount="166" Body="&lt;p&gt;I have recently come across some cosmological assertions (based on empirical data) about the universe being self contained in the sense that it is entirely capable of coming into existence  from a zero-energy initial state . This is based on the observation that at grand scale the positive and negative gravity/energy etc. cancel out each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do the terms positive and negative actually mean in this context ?&lt;/p&gt;&#xA;" OwnerUserId="152" LastActivityDate="2011-06-18T07:00:38.543" Title="What is meant by positive and negative gravity/energy/spactimecurvature?" Tags="&lt;gravity&gt;&lt;cosmology&gt;&lt;astronomy&gt;&lt;cosmological-constant&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9202" PostTypeId="2" ParentId="9152" CreationDate="2011-04-27T18:59:02.390" Score="0" ViewCount="" Body="&lt;p&gt;1) Yes - for charged  particles and some neutral particles. But there are so called Majorana neutral particles that has no antiparticles (neutral pion or eta-meson, for example). In other words Majorana particle is its own antiparticle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) At low energies proton annihilates with antineutron like proton with antiproton - mainly into neutral and charged pions (charge should be conserved). &#xA;Positron is not annihilate with a proton. They have the same charge.&lt;/p&gt;&#xA;" OwnerUserId="238" LastActivityDate="2011-04-27T18:59:02.390" />
  <row Id="9203" PostTypeId="1" AcceptedAnswerId="9215" CreationDate="2011-04-27T19:24:35.903" Score="5" ViewCount="259" Body="&lt;p&gt;Title says it all - recently I encountered a strange homework exercise on de Broglie dual theory with an electron wavelength of few millimeters - which implies the velocity lower than 1 m/s. I personally considered the problem very artificial and unrealistic, but it made me wonder - what is the smallest electron velocity recorded in experiments? What does the theory on the other hand say?&lt;/p&gt;&#xA;" OwnerUserId="666" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-27T20:39:42.370" LastActivityDate="2011-04-28T12:11:58.273" Title="How low can an electron go?" Tags="&lt;quantum-mechanics&gt;&lt;velocity&gt;&lt;electron&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="3" />
  <row Id="9204" PostTypeId="1" CreationDate="2011-04-27T19:56:54.390" Score="0" ViewCount="251" Body="&lt;p&gt;Given I have two identical balloons on earth, how will the buoyancy compare between the one filled with helium and another filled with hydrogen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the ratio of buoyancy given two different substances and identical balloons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also interested in the equations relating to this problem. &lt;/p&gt;&#xA;" OwnerUserId="3283" LastActivityDate="2011-04-27T21:07:15.450" Title="Buoyancy: helium vs hydrogen balloons" Tags="&lt;buoyancy&gt;" AnswerCount="3" />
  <row Id="9205" PostTypeId="1" CreationDate="2011-04-27T20:33:04.587" Score="0" ViewCount="146" Body="&lt;p&gt;I'm trying to design a quasi-simple vertical axis wind turbine, and a coworker came up with &lt;a href=&quot;https://docs.google.com/drawings/d/1B-NzCTWUyBIhNcv9b46hKvyk6R-l4NXiVzGHsb01fTI/edit?hl=en&amp;amp;authkey=CNeWgdwL&quot; rel=&quot;nofollow&quot;&gt;this design&lt;/a&gt; to focus the wind as it reaches the turbine in a wind tunnel.  He says that there's going to be turbulence caused by some of the design factors, but I'm not quite sure why and he doesn't know why, either.  Why, and where, would this design cause turbulence, and what improvements could be made to improve it?&lt;/p&gt;&#xA;" OwnerUserId="1699" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-26T13:14:28.833" LastActivityDate="2011-07-26T13:14:28.833" Title="What is it that makes this structure cause turbulence?" Tags="&lt;fluid-dynamics&gt;&lt;air&gt;&lt;renewable-energy&gt;&lt;turbulence&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9206" PostTypeId="2" ParentId="9204" CreationDate="2011-04-27T20:37:52.993" Score="4" ViewCount="" Body="&lt;p&gt;Assuming they are filled to the same volume, and the air surrounding them is of the same density, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Buoyancy&quot; rel=&quot;nofollow&quot;&gt;buoyant force&lt;/a&gt; acting on the balloons will be the same. Buoyant force is simply equal to the weight of the amount of surrounding fluid that would occupy the space filled by the balloon, if the balloon were not there. It has nothing to do with the contents or nature of the balloon itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the &lt;em&gt;net&lt;/em&gt; force on the balloon filled with hydrogen would be greater, because helium is more dense than hydrogen. Given that the balloons are filled to the same volume (and the pressure and temperature are the same), the one with helium in it will be heavier; the weight of the helium can cancel out more of the buoyant force than the weight of the hydrogen can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, the buoyant force is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_b = \rho_\text{fluid}gV$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\rho_\text{fluid}$ is the density of the surrounding fluid (air in this case), and $V$ is the volume of the balloon. The gravitational force on the balloon itself is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_g = -mg - \rho_\text{contents}gV$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ is the mass of the balloon itself (not counting the gas inside), and $\rho_\text{contents}$ is the density of the gas filling the balloon. The net force can be calculated as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_\text{net} = (\rho_\text{fluid} - \rho_\text{contents})gV - mg$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and in this equation, only $\rho_\text{contents}$ depends on the gas filling the balloon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've written a &lt;a href=&quot;http://www.ellipsix.net/blog/2009/11/buoyancy-part-1.html&quot; rel=&quot;nofollow&quot;&gt;blog post&lt;/a&gt; about buoyancy in the context of an old Mythbusters episode; some of the information in there might be of interest to you.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-04-27T20:37:52.993" />
  <row Id="9207" PostTypeId="2" ParentId="9204" CreationDate="2011-04-27T20:39:58.367" Score="5" ViewCount="" Body="&lt;p&gt;Balloons are buoyant because the air pushes on them.  The air doesn't know what's in the balloon, though.  It pushes on everything the same, so the buoyant force is the same on all balloons of the same size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the &quot;balloon&quot; is just a lump of air with an imaginary boundary, then the lump won't go anywhere because the air isn't moving on average.  So the buoyant force must exactly cancel the gravitational force (the weight).  Since the buoyant force  is the same on everything, the buoyant force on a balloon is equal to the weight of the air it displaces.  In symbols this is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{buoyant} = \rho g V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\rho$ is the density of air, $g$ is gravitational acceleration, and $V$ is the balloon's volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen and helium have less weight than a similar volume of air at the same pressure.  That means the buoyant force on them, which is just enough to hold up air, is more than enough to hold up the balloons, and they have to be tethered down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming they have the same pressure and volume, a hydrogen balloon has less weight than a helium balloon.  Things like pressure and volume are roughly decided on a per-molecule basis, at least in gases at low pressure, so at the same pressure and volume hydrogen and helium will have the same number of molecules.  Hydrogen is lighter per molecule, so the hydrogen weighs less, and less of the buoyant force is canceled out.  That means the net force on a hydrogen balloon is greater.  The difference in the net force is small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen is $H_2$, which has atomic mass 2, while air is mostly $N_2$, which has atomic mass 28, so the hydrogen balloon has a net force of about (28-2)/28  = .93 the weight of the air it displaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Helium is mostly helium-4, so the net force on a helium balloon is about (28-4)/28 = .86 the  weight of the air is displaces.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So net force on the hydrogen balloon is in the neighborhood of 10% more.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-04-27T20:45:34.680" LastActivityDate="2011-04-27T20:45:34.680" />
  <row Id="9208" PostTypeId="2" ParentId="9165" CreationDate="2011-04-27T20:49:01.643" Score="0" ViewCount="" Body="&lt;p&gt;I think Goldstein is a great book which should be supplemented by the internet, rather than a mass of books to clarify every detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's the MIT OpenCourseWare program for classical mechanics: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/index.htm&quot; rel=&quot;nofollow&quot;&gt;http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/index.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you'll find the videos by prof Lewin, lectures notes, exam questions, study groups etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also the 10 classical mechanics videos by Prof Susskind of Stanford, but on a level around that of Goldstein:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=pyX8kQ-JzHI&amp;amp;p=189C0DCE90CB6D81&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=pyX8kQ-JzHI&amp;amp;p=189C0DCE90CB6D81&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1345" LastActivityDate="2011-04-27T20:49:01.643" CommunityOwnedDate="2011-08-12T00:20:14.823" />
  <row Id="9209" PostTypeId="2" ParentId="9204" CreationDate="2011-04-27T20:53:28.720" Score="1" ViewCount="" Body="&lt;p&gt;To add just a little more to David's excellent post: the densities of interest to you to use in his last equation are...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;air (fluid): 1.2 kg/(m^3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;helium (contents): 0.0899 kg/(m^3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hydrogen (contents): 0.1786 kg/(m^3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Helium has an atomic mass of about four times that of hydrogen.  But it does not form a diatomic molecule like hydrgen (H2) does.  So it's density is only twice that of hydrogen, rather than four times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your question is that, at STP, the CONTENTS of a hydrogen filled ballon will give a net bouyant force of about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(9.8)(1.2kg - 0.0899kg)/(m^3) = 10.0N/(m^3) ---&gt; about 2.44 lb./(m^3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CONTENTS of a helium filled ballon will give a net bouyant force of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(9.8)(1.2kg - 0.1786kg)/(m^3) = 10.0N/(m^3) ---&gt; about 2.25 lb./(m^3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydrogen filled ballon gives about 9% more net bouyant force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that temperature and pressure both dramatically affect actual bouyant force.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-04-27T21:07:15.450" LastActivityDate="2011-04-27T21:07:15.450" CommentCount="1" />
  <row Id="9210" PostTypeId="2" ParentId="9203" CreationDate="2011-04-27T21:00:36.623" Score="-2" ViewCount="" Body="&lt;p&gt;The minimum electron velocity depends on the external potential and can be as low as zero m/s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider, for example an electron bouncing on a horizontal mirror in the gravitational field. Such a system has a minimal energy $E_0$ and therefore a minimal non-zero velocity.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-04-27T21:00:36.623" />
  <row Id="9212" PostTypeId="1" AcceptedAnswerId="9216" CreationDate="2011-04-27T21:51:18.200" Score="5" ViewCount="200" Body="&lt;p&gt;When a charge is accelerated, it radiates and loses kinetic energy. This can be modeled by having another force act on the charge, which is proportional to the derivative of the acceleration. So if this force is included the equations of motion, we end up with 3rd order equations. Is this allowed? Why don't we need to fix another constant to determine the motion? Is it because we usually neglect this force because of its small magnitude?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the electromagnetic Lagrangian only contains up to first derivatives (of the 4-potential in this case). How can this result in 3rd order equations of motion?&lt;/p&gt;&#xA;" OwnerUserId="1635" LastActivityDate="2011-04-28T21:42:30.913" Title="Radiation Resistance" Tags="&lt;electromagnetism&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" />
  <row Id="9213" PostTypeId="1" AcceptedAnswerId="9218" CreationDate="2011-04-27T22:09:11.140" Score="3" ViewCount="159" Body="&lt;p&gt;I get &lt;a href=&quot;http://physics.stackexchange.com/questions/3489/what-is-quantum-entanglement&quot;&gt;quantum entanglement&lt;/a&gt; but I don't quite get how one would go about generating two complementary particles that are entangled (a photon and its entangled sibling, an electron and its entangled sibling, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain this referencing how this could be achieved in a lab?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3316" LastActivityDate="2011-04-28T01:21:46.193" Title="Quantum Entanglement: how to generate 2 entangled particles?" Tags="&lt;quantum-mechanics&gt;&lt;experimental-physics&gt;&lt;entanglement&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9214" PostTypeId="2" ParentId="9212" CreationDate="2011-04-27T22:55:49.427" Score="-4" ViewCount="" Body="&lt;p&gt;Yes, they often say that an electron acts on itself. This model has two flaws: large inertial term $\delta m \ddot{x}$ which is usually discarded from the equations, and a third time derivative $\dddot{x}$ leading to self-accelerating (runaway) solutions. In other words, such an idea just does not work. It is a complete failure whatever they say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, one can introduce &lt;em&gt;potential forces&lt;/em&gt; to take into account radiative resistance. For that one has to proceed from a different electron model. This model is just outlined but not yet developed properly and is a not-yet mainstream direction (it's my personal model).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: If one admits an idea of self-action, the additional force term is the inertial one. It is very large so it does not agree with experiments. The third derivative term is &quot;derived&quot; under condition that the inertial term is discarded, so it is not a consequence of self-action in a true sense. It is a consequence of deliberately modifying the equations. An artificial intervention modifies the original idea and the corresponding results. So the third derivative is not contained in the original Lagrangian.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-04-28T21:42:30.913" LastActivityDate="2011-04-28T21:42:30.913" CommentCount="2" />
  <row Id="9215" PostTypeId="2" ParentId="9203" CreationDate="2011-04-27T23:03:01.223" Score="4" ViewCount="" Body="&lt;p&gt;If you define the speed of your electron in a de Broglie way and just set $v = \frac{h}{m\cdot \lambda}$, you can rewrite that expression to involve the wavenumber $k$, i.e. &#xA;$v = \frac{\hbar k}{m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k$ is constrained by the boundary conditions of the electron. In a large solid with periodic boundary conditions, $k$ can be arbitrarily small, so the particular electron would be arbitrarily slow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you don't see these electrons, as, due to the Pauli principle, when you start filling up your solid with electrons, you have to go to ever larger values of $k$. The largest of them is called the Fermi wavenumber $k_F$, which then defines you a Fermi velocity $v_F$ which can be quite large. In experiments, you typically interact with electrons at, or close to, the Fermi level, since these are the ones with the highest energy and hence the ones you can remove the easiest. Therefore, you will not really &lt;em&gt;see&lt;/em&gt; the slow electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, on the other hand, you don't consider electrons, which are Fermions and thus subject to the Pauli exclusion principle, but &lt;em&gt;Bosons&lt;/em&gt; (such as ${}^{85}Rb$ atoms), you don't have to put them all in different quantum states. You are then free to put them &lt;em&gt;all&lt;/em&gt; into the $k = 0$ state. Indeed, this happens at &lt;em&gt;very&lt;/em&gt; low temperatures. Such a state is called a Bose-Einstein condensate. In time-of-flight experiments, it is then indeed observed that they all (or, at finite temperature, at least &lt;em&gt;most&lt;/em&gt; of them) have essentially a velocity of $0$.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditorDisplayName="" LastEditDate="2011-04-28T04:20:21.650" LastActivityDate="2011-04-28T04:20:21.650" CommentCount="2" />
  <row Id="9216" PostTypeId="2" ParentId="9212" CreationDate="2011-04-27T23:33:56.513" Score="7" ViewCount="" Body="&lt;p&gt;This is a hard problem! I don't think that a completely satisfactory theory of classical electrodynamics, with point charges, including radiation reaction, is known. In fact, if I had to conjecture, I'd bet that there is no such theory -- that is, classical theory simply doesn't play nicely with pointlike charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing to point out is that if we don't have truly pointlike charges, there's no problem. If you're willing to model all of your charges as spheres of some small but nonzero radius, or anything like that, then the whole theory is perfectly self-consistent. The radiation reaction force on one of these spheres is just caused by the force on each little element of the sphere, due to all the other elements. The whole theory, consisting of just Maxwell's equations, the Lorentz force law, and some assumption about the additional stresses required to hold the sphere together against its own repulsion, is self-consistent, conserves energy-momentum, and generally behaves the way you want it to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The details of this theory depend on precisely what you assume about the arrangement of charge on the sphere, how it responds to stresses (e.g., does it always remain spherical in its own instantaneous rest frame?), etc. But in cases where the radius of the sphere is small compared to other length scales in the problem, you'd expect those details not to matter much, and indeed that is the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In these models, the equation of motion for one of these not-quite-point charges turns out to be an integro-differential equation, in which the acceleration of the charge at any time $t$ is determined by an integral over the charge's position for times $t&amp;#39;$ in the range $t-2a/c&amp;lt;t&amp;#39;&amp;lt;t$, where $a$ is the radius of the sphere. This makes sense: the force on each point on the surface of the sphere depends on what the other points were doing when they cross the given point's past light cone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Griffiths's textbook has a nice &quot;toy model&quot; discussion of this, which is good for building intuition. For all the details, the best places to go are a &lt;a href=&quot;http://books.google.com/books?id=lJrPrP6L6tQC&amp;amp;printsec=frontcover&amp;amp;dq=rohrlich%20classical%20charged%20particles&amp;amp;source=bl&amp;amp;ots=jlIgWKN3mY&amp;amp;sig=WEfDzZSGNCJKhWaVmN4Xs7aOg8o&amp;amp;hl=en&amp;amp;ei=OaS4Td3AGNC9tgfM2eXeBA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBwQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; and a bunch of articles (e.g., &lt;a href=&quot;http://www.sciencedirect.com/science?_ob=ArticleURL&amp;amp;_udi=B6TVM-4349XVR-2&amp;amp;_user=2065684&amp;amp;_coverDate=05/21/2001&amp;amp;_rdoc=1&amp;amp;_fmt=high&amp;amp;_orig=gateway&amp;amp;_origin=gateway&amp;amp;_sort=d&amp;amp;_docanchor=&amp;amp;view=c&amp;amp;_searchStrId=1733243793&amp;amp;_rerunOrigin=scholar.google&amp;amp;_acct=C000056005&amp;amp;_version=1&amp;amp;_urlVersion=0&amp;amp;_userid=2065684&amp;amp;md5=72b110dc2e4ab1b136405298aa9cc6dd&amp;amp;searchtype=a&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;) by F. Rohrlich.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you've got a good theory for spheres of radius $a$, the natural thing to do is to take the limit as $a\to 0$. When you do, you get a third-order differential equation of motion for the (now pointlike) charge. This same third-order equation can be derived in a bunch of different ways, going all the way back to the early 20th century. It's often called the Lorentz-Dirac equation, although Abraham's name gets thrown in too sometimes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the equation is third-order is actually not as bad as it seems. &lt;a href=&quot;http://www.springerlink.com/content/09gab1qnghhl5463/&quot; rel=&quot;nofollow&quot;&gt;Kijowski&lt;/a&gt; makes a nice argument that you should have expected a third-order equation all along! To formulate this setup as an initial-value problem, you need to specify the position and velocity of the charge, &lt;em&gt;and&lt;/em&gt; the electromagnetic field. The latter actually determines the charge's acceleration at that moment. To be specific, the specific way the field diverges as you approach the charge depends on the acceleration. Since the acceleration, as well as the position and velocity, are determined by the initial conditions, it makes sense that the equation of motion is third order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even if you're OK with the fact that the equation of motion is third-order, the Lorentz-Dirac equation is pathological. It contains &quot;runaway&quot; solutions, in which the charge's acceleration grows exponentially forever after any external force is removed. You can artificially remove these runaway solutions, but the solutions left behind suffer from a problem of &quot;preacceleration&quot;: the motion of the charge at time $t$ depends on what external forces will be applied at all &lt;em&gt;later&lt;/em&gt; times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there's a finite-sized-particle theory that works fine, but it goes bad as the size goes to zero. In fact, it actually goes bad even earlier: typically, the finite-sized theory starts to have pathological solutions when the size of the particle gets to be less than the classical particle radius $q^2/(4\pi\epsilon_0mc^2)$. I strongly suspect (but I can't prove) that what this is telling us is that classical electrodynamics simply doesn't work with particles smaller than this size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a surprising situation, but it doesn't contradict anything fundamental that I can think of. In particular, we have no right to expect there to be a &quot;correct&quot; theory of classical point particles, if &quot;correct&quot; is taken to mean &quot;in agreement with physical reality.&quot; After all, physical reality is quantum!&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-27T23:33:56.513" CommentCount="8" />
  <row Id="9217" PostTypeId="2" ParentId="9203" CreationDate="2011-04-28T00:19:49.357" Score="2" ViewCount="" Body="&lt;p&gt;Smallest recorded in experiments?  The experiment that comes to my mind is the work done in the Gabrielse lab at Harvard, where they've trapped a single in a Penning trap, and cooled it down to less than a tenth of a Kelvin (which pretty much puts the electron in the ground state of the trap).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electron has fast cyclotron motion, but I'd estimate that in one direction it should have a velocity around 1000 m/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a bare electron with a velocity of 1 m/s would require cooling the thing down below a microkelvin (if I'm doing my arithmetic right).  As far as I know, that's never been done experimentally.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-04-28T00:19:49.357" />
  <row Id="9218" PostTypeId="2" ParentId="9213" CreationDate="2011-04-28T01:21:46.193" Score="4" ViewCount="" Body="&lt;p&gt;For photons, the process that immediately comes to mind is &lt;a href=&quot;http://en.wikipedia.org/wiki/Spontaneous_parametric_down-conversion&quot; rel=&quot;nofollow&quot;&gt;Spontaneous Parametric Down Conversion&lt;/a&gt; where a single photon is split into two by a nonlinear crystal. I've never actually bothered to look up methods for other qubit systems, but it looks like a search for &quot;generating entangled electrons&quot; without quotes in Google Scholar turns up quite a few results.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-04-28T01:21:46.193" />
  <row Id="9219" PostTypeId="1" CreationDate="2011-04-28T02:38:58.200" Score="0" ViewCount="348" Body="&lt;p&gt;Can spin of a particle or a group of particles become infinity?Explain plz.Is there any representation for spins like dot(for s=0) and arrow(for s=1)?If so what for s= infinity?&lt;/p&gt;&#xA;" OwnerUserId="3321" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-04-28T04:52:11.110" LastActivityDate="2011-05-08T09:29:20.403" Title="Can spin be infinite?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="9220" PostTypeId="2" ParentId="9205" CreationDate="2011-04-28T03:58:00.133" Score="1" ViewCount="" Body="&lt;p&gt;The diagram is not labeled, but if the green lines are rigid walls I can see that the wind is given angular momentum by the shape. Angular momentum is conserved and will create turbulence in a fluid.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-28T03:58:00.133" />
  <row Id="9221" PostTypeId="1" AcceptedAnswerId="9224" CreationDate="2011-04-28T04:57:48.293" Score="0" ViewCount="147" Body="&lt;p&gt;Imagine there are two cars travelling &quot;straight&quot; at the speed of light*, $A$, and $B$. $B$ is following directly behind $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suddenly, $B$ switches on its headlights. Will $A$ be able to see this light?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My answer is no, since $A_v = B_v = c$ (the light will always stay stationary relative to $B$. This will probably lead to it gathering up, and intensifying.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;*I realize this is impossible, but it's a question my Grade 9 [Honours] teacher asked, so we don't need to get into Relativity, $m = \frac{m_0}{\sqrt{1 – (v / c)^2}}$, &lt;em&gt;cough cough&lt;/em&gt;. (I think.)&lt;/p&gt;&#xA;" OwnerUserId="125" LastActivityDate="2011-04-28T18:54:07.200" Title="Seeing light travelling at the speed of light" Tags="&lt;speed-of-light&gt;&lt;velocity&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="9222" PostTypeId="1" AcceptedAnswerId="9226" CreationDate="2011-04-28T05:16:44.887" Score="0" ViewCount="135" Body="&lt;p&gt;I believe the quarks would flip the proton's internal charge, reverse the spin of the proton, transform into another flavor, and then become unconfined and free.&lt;/p&gt;&#xA;" OwnerUserId="3322" LastActivityDate="2011-04-28T05:37:16.877" Title="What happens if up quarks are replaced by down quarks and down quarks are replaced by up quarks?" Tags="&lt;quantum-mechanics&gt;&lt;particle-physics&gt;&lt;spin&gt;&lt;proton&gt;&lt;quark-model&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="9223" PostTypeId="2" ParentId="7129" CreationDate="2011-04-28T05:20:55.087" Score="1" ViewCount="" Body="&lt;p&gt;Consider a solid rod made of a glassy substance, and model it as a set of atoms in random locations, held together by randomly oriented harmonic springs (enough so that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Structural_rigidity&quot; rel=&quot;nofollow&quot;&gt;graph is rigid&lt;/a&gt;). The Hamiltonian of this system is in principle diagonalizable, and since all the springs are harmonic, the potential energy is quadratic in all the atomic coordinates, so the whole system is equivalent to a set of &lt;strong&gt;uncoupled&lt;/strong&gt; harmonic oscillators, which are of course the normal modes. The lowest frequency mode will be rod flexing back and forth with two nodes, the second will be the second overtone with three nodes, and so on, but importantly this set of normal modes goes all the way up to modes with a number of nodes on the order of the number of atoms in the rod. None of these normal modes has a &lt;strong&gt;precisely&lt;/strong&gt; defined wavevector, because of the lack of periodicity in the glass, but they do have approximate wavevectors, and of course they have precise frequencies corresponding to the eigenvalues of the Hamiltonian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One weird thing about this that seems completely different from the usual treatment of phonons in crystals is that all these modes are standing waves - they don't propagate in a particular direction, and their approximate wavevectors are only defined up to sign. This is actually the case for a &lt;strong&gt;finite-size&lt;/strong&gt; crystal as well - no propagating waves can actually be eigenstates, and instead the boundary conditions at the ends of the crystal cause them to mix into standing waves that are the exact physical eigenstates. The only reason we introduce periodic boundary conditions and talk about the propagating waves in crystals is that it's so much more convenient. Of course, if you create a wave &lt;strong&gt;packet&lt;/strong&gt; at one end of the material, either crystal or glass, you can always express that in whatever basis of eigenstates you want, and as they evolve the packet will end up moving through the crystal and spreading out according to some dispersion relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how you would actually calculate that dispersion relation for a glass (other than brute-force computation), but it's possible in principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same considerations also apply to quasicrystals, but with the interesting addition that there are now diffusively propagating modes called &lt;em&gt;phasons&lt;/em&gt; with long relaxation times. &lt;a href=&quot;http://www.physics.princeton.edu/~steinh/PhoPhaDef.pdf&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phonons in a gas is a really weird thing to think about because in an ideal gas, the particles are assumed to be non-interacting, and they have to be in a thermal distribution of single-particle quantum states for it to be a gas (rather than a Bose-Einstein condensate or something). If the gas particles are delocalized, and don't interact with each other, then what the heck is a phonon? Yet sound waves in a gas obviously exist, so the question remains whether they're quantized or not. I can't answer this part of the question.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-04-28T05:20:55.087" />
  <row Id="9224" PostTypeId="2" ParentId="9221" CreationDate="2011-04-28T05:33:58.847" Score="4" ViewCount="" Body="&lt;p&gt;I can think of three ways to answer this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It can't happen.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It really can't happen.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;See #1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Okay, that's probably enough ;-) Since you say we don't need to consider special relativity, suppose that the universe actually obeys &lt;a href=&quot;http://en.wikipedia.org/wiki/Galilean_relativity&quot; rel=&quot;nofollow&quot;&gt;Galilean relativity&lt;/a&gt;. That's the technical term for the intuitive way to think about motion, where velocities are measured with respect to some absolute rest frame, and there's nothing special about the speed of light or any other speed. If that were the case, then yes, the light beam would never catch up to car A. The energy contained in the light would presumably pile up in the headlight where it was emitted at first, but afterwards perhaps it would spread out sideways, or would be reabsorbed by the headlight as heat. We don't really have a good answer, because that's not the way the universe works - in fact, there's a lot of physics, both experimental and theoretical, that has been done to &lt;em&gt;prove&lt;/em&gt; that it can't work that way. No matter how you try to resolve the problem, at some point you will run into a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best thing you could probably do would be to draw a parallel to some sort of wave that travels with respect to some fixed reference frame, at a speed much less than that of light. Sound, for instance. Sound waves travel with a certain speed with respect to the air, which defines a single absolute reference frame, and their speed is much less than that of light, so there are no special relativistic effects to worry about. Your headlight scenario would then be roughly equivalent to an airplane traveling at the speed of sound. What happens in that case is that the airplane creates a &lt;a href=&quot;http://en.wikipedia.org/wiki/Sonic_boom&quot; rel=&quot;nofollow&quot;&gt;sonic boom&lt;/a&gt;, a shock wave which results from the energy in the emitted sound waves piling up at the airplane and eventually being forced to spread out sideways. So one might guess that in your hypothetical situation, the headlights of car B would create a light shock wave that would spread out perpendicular to the direction of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This actually can happen in certain physical situations, namely when something is traveling through a transparent material that slows down the speed of light. This means that light itself travels at a slower speed, but not that the &quot;universal speed limit&quot; is any different. The effect is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Cherenkov_radiation&quot; rel=&quot;nofollow&quot;&gt;Cherenkov radiation&lt;/a&gt; and it does indeed work out much like a sonic boom would.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-04-28T05:33:58.847" CommentCount="3" />
  <row Id="9225" PostTypeId="2" ParentId="9221" CreationDate="2011-04-28T05:35:33.867" Score="3" ViewCount="" Body="&lt;p&gt;The question is, as you stated, impossible.  However, something roughly like it occurs when the cars are driving at sub-light speeds, but going faster than the group velocity of light in a medium with a refractive index.  It's called &lt;a href=&quot;http://en.wikipedia.org/wiki/Cherenkov_radiation&quot; rel=&quot;nofollow&quot;&gt;Cerenkov radiation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the cars are moving at the speed of light, they must be massless, and also from our perspective they don't experience any proper time.  They aren't cars and can't turn on their lights - the problem simply doesn't make sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the cars are moving very close to the speed of light in our reference frame, (perhaps $(1 - 10^{-10})c$), then by the principle of relativity in their own reference frame everything is business as usual and the lead car will see the headlights turn on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another interpretation might be &quot;what if one photon split into two photons?&quot;  This could conserve energy and momentum if and only if the two photons continued traveling next to each other in the same direction.  If you have photon A in front and photon B in back, and photon B split into two lower-energy photons, then no, neither of them would catch up to photon A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(note: this answer is pure special relativity.  I'm not actually sure whether one photon can simply turn into two photons.  Maybe someone could fill me in on this.)&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-04-28T08:56:39.780" LastActivityDate="2011-04-28T08:56:39.780" CommentCount="3" />
  <row Id="9226" PostTypeId="2" ParentId="9222" CreationDate="2011-04-28T05:35:57.923" Score="6" ViewCount="" Body="&lt;p&gt;The proton ($uud$) turns into a neutron ($ddu$). Up and down quarks don't have equal charges; the up is $+\frac{2}{3}e$ and the down is $-\frac{1}{3}e$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, such an operation has a name - &lt;a href=&quot;http://en.wikipedia.org/wiki/Isospin&quot; rel=&quot;nofollow&quot;&gt;isospin symmetry transformation&lt;/a&gt; - corresponding to an approximate SU(2) symmetry that makes the proton and neutron have almost similar masses.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-04-28T05:35:57.923" />
  <row Id="9227" PostTypeId="2" ParentId="9222" CreationDate="2011-04-28T05:37:16.877" Score="1" ViewCount="" Body="&lt;p&gt;Um... no, there's no reason to think any of that stuff would happen. The transformation of up quarks into down quarks and vice versa is called an &lt;a href=&quot;http://en.wikipedia.org/wiki/Isospin&quot; rel=&quot;nofollow&quot;&gt;isospin&lt;/a&gt; transformation, and it's been quite well studied. For example, the proton and neutron are members of an isospin doublet, which means that making this transformation would convert each one into the other.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-04-28T05:37:16.877" CommentCount="1" />
  <row Id="9229" PostTypeId="2" ParentId="8815" CreationDate="2011-04-28T08:00:30.083" Score="0" ViewCount="" Body="&lt;p&gt;No answer yet, have to back to goodle.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-04-28T08:00:30.083" />
  <row Id="9230" PostTypeId="2" ParentId="9189" CreationDate="2011-04-28T08:09:09.947" Score="2" ViewCount="" Body="&lt;p&gt;In classical gas dynamics, in order to have work, you need a volume change $dV$. If volume $V$ is constant, no work is performed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The traditional picture is to think of a piston closing a cylinder containing the gas. The gas inside the cylinder will be characterized by pressure $p$, temperature $T$ and volume $V$. To actually perform work, the piston needs to be moved, which implies a change of volume $dV=S\, dx$, where $S$ is the moving surface of the piston and $dx$ the distance it moved. Work is then performed by the force pushing the piston, related to pressure by $F=p\cdot S$, and during a slight motion $dx$ of the piston, work performed is &#xA;$dW=F\cdot dx=p\,S\,dx=p\, dV$. Adding work on small motions yields an integral, $W=\int p\, dV$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If volume doesn't change, no work is performed. Energy may still be transfered into or out of the &quot;cylinder&quot; (whatever the gas' container is), but in a disorderly, non-directional fashion, that is as Heat, not Work.&lt;/p&gt;&#xA;" OwnerUserId="1725" LastEditorUserId="1725" LastEditorDisplayName="" LastEditDate="2011-04-28T22:54:18.277" LastActivityDate="2011-04-28T22:54:18.277" CommentCount="2" />
  <row Id="9231" PostTypeId="1" AcceptedAnswerId="9233" CreationDate="2011-04-28T08:56:52.420" Score="4" ViewCount="150" Body="&lt;p&gt;I want to solve the transient heat equation on a ball. The boundary condition is the same over the hole outer surface. So this should reduce to a one-dimensional problem in radial direction. However I cannot use the one-dim heat equation, since the surface through which the heat flows goes quadratic with the radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the following approach correct: Take the heat equation, transform it into sperical coordinates and eliminate the derivatives in angular directions. I think the last part is justified, because of the rotational symmetry of the boundary condition. So I end up with this equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\frac{\partial T}{\partial t} = \alpha \frac{1}{r^2}\frac{\partial}{\partial r} \left( r^2\frac{\partial T}{\partial r}\right)&#xD;&#xA;$$&#xA;with thermal diffusivity $\alpha$. Is this the correct reformulation of the heat equation on a ball to a one-dimensional problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: When I solve this with the finite element method, will there be a problem in the centre of the ball? The $1/r^2$ factor might cause a problem.&lt;/p&gt;&#xA;" OwnerUserId="2480" LastEditorUserId="2480" LastEditorDisplayName="" LastEditDate="2011-04-28T09:04:13.397" LastActivityDate="2011-04-28T11:48:44.557" Title="Heat equation on ball - one-dimensional description" Tags="&lt;heat&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9233" PostTypeId="2" ParentId="9231" CreationDate="2011-04-28T09:15:18.697" Score="2" ViewCount="" Body="&lt;p&gt;That sounds about right to me. Although I've never done this for the heat equation specifically, that's much the same approach we use to solve the Schrödinger equation in spherical coordinates, e.g. for the hydrogen atom. Your full solution will be products of the radial functions $T(r,t)$ with appropriate spherical harmonics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When determining the radial part of the solution, assuming this goes like the Schrödinger equation, one of the boundary conditions you'll need to use is consistency at the origin: basically you need your solution for $T(r)$ to be finite and differentiable at $r=0$. I can come back and expand on this later, or perhaps someone else will give you a more detailed explanation before I get to it.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-04-28T09:15:18.697" CommentCount="4" />
  <row Id="9234" PostTypeId="1" AcceptedAnswerId="9236" CreationDate="2011-04-28T10:46:00.693" Score="2" ViewCount="175" Body="&lt;p&gt;I don't know much about string theory, apart from it being a theory of everything which brings QM, QED and nuclear forces and gravity under one single roof. I am curious to know from a mathematical view point, how the quantization of space-time is achieved in super string theory.&lt;/p&gt;&#xA;" OwnerUserId="540" LastActivityDate="2011-04-28T11:31:10.300" Title="What is the mathematical nature of space time quantization in string theory/super string theory?" Tags="&lt;gravity&gt;&lt;string-theory&gt;&lt;soft-question&gt;&lt;mathematical-physics&gt;&lt;mathematics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9235" PostTypeId="2" ParentId="9203" CreationDate="2011-04-28T10:48:49.593" Score="1" ViewCount="" Body="&lt;p&gt;If this surprises you it means you need to understand the role of &lt;strong&gt;Special Relativity!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wavelength of a free electron is infinite in its rest frame. Furthermore, the direction of the wavefront of an electron depends on the reference frame. Whatever direction you choose, there is a reference frame in which the wavefront of the electron has that particular direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Believe me, it all makes sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The wavefront is about the phase, not the magnitude. So the density of the electron is the same at all places of the wavefront. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Think of the phase as a local clock. In the rest frame all clocks show the same time. If you choose a different reference frame then some clocks in a certain direction show a later time while the clocks in the opposite direction show an earlier time. This direction is the same direction as that of the wavefront.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are images in my book which illustrate this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics-quest.org/Book_Chapter_Klein_Gordon.pdf&quot; rel=&quot;nofollow&quot;&gt;http://physics-quest.org/Book_Chapter_Klein_Gordon.pdf&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look for instance at figure 9.2 and 9.3 to start with. For the rotation of the wavefront you can look at 9.4 and 9.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards, Hans&lt;/p&gt;&#xA;" OwnerUserId="3197" LastEditorUserId="3197" LastEditorDisplayName="" LastEditDate="2011-04-28T10:56:14.843" LastActivityDate="2011-04-28T10:56:14.843" />
  <row Id="9236" PostTypeId="2" ParentId="9234" CreationDate="2011-04-28T11:31:10.300" Score="3" ViewCount="" Body="&lt;p&gt;Dear Rajesh, the goal as well as main achievement of string theory is not to &quot;achieve spacetime quantization&quot; whatever this phrase is supposed to mean but to allow calculations about spacetime that are compatible with the postulates of quantum mechanics. It's not the same thing. The statement that &quot;spacetime is quantized&quot;, whatever it means, is just a working hypothesis, not a holy grail that should be &quot;achieved&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, what is needed is to have a theory that is a quantum theory as a whole, i.e. one that agrees with the uncertainty principle, probabilistic character of predictions, observables' being expressed by linear operators on the Hilbert space, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the spacetime is an approximate concept in string theory. At distances much longer than the characteristic distance scale of string theory, the string scale (or the related gravitational Planck scale), classical general relativity is a good approximation - and some of its aspects remain exact at all distance scales. At distances comparable to the string scale (or Planck scale), entirely new set of physical phenomena take over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not true that the only difference of these new phenomena from classical general relativity is that &quot;spacetime is quantized&quot;. And in fact, it is not true in any sense and it cannot be true in a consistent theory of quantum gravity that spacetime becomes discrete. And in fact, geometric quantities - while remaining continuous whenever it makes sense to use them - may be shown in string theory to be invalid variables to describe physics at very short distances. To say the least, they're incomplete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if your question is How string theory confirms the prejudices and beliefs that the spacetime quantities survive as good variables up to arbitrarily short distance scales and/or that they become discrete, the answer is that string theory is a way to prove that both of these prejudices are dead wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your question is which quantum phenomena affecting spacetime are predicted by string theory, it is a valid question but it is too broad and a proper answer would require to review all of string theory - because all important insights of string theory, in some sense, show the consequences of the co-existence of quantum mechanics with a dynamical spacetime.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-28T11:31:10.300" CommentCount="0" />
  <row Id="9237" PostTypeId="2" ParentId="9231" CreationDate="2011-04-28T11:48:44.557" Score="2" ViewCount="" Body="&lt;p&gt;your calculation is fine but there is &lt;strong&gt;no need&lt;/strong&gt; to employ &lt;strong&gt;numerical methods&lt;/strong&gt; for the given equation at hand. The equation can be solved using &lt;a href=&quot;http://en.wikipedia.org/wiki/Separation_of_variables&quot; rel=&quot;nofollow&quot;&gt;separation of variables&lt;/a&gt; with a physical ansatz for the radial part. Of course you need not restrict your problem to a radial symmetry of the solution, the full Laplace will lead to &lt;a href=&quot;http://en.wikipedia.org/wiki/Spherical_harmonics&quot; rel=&quot;nofollow&quot;&gt;spherical harmonics&lt;/a&gt; for the angular parts. The following derivation can directly be applied in this case.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Solution using separation of variables&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Taking $$T(r,t) = f(t)\cdot B(r)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;results in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f&amp;#39;(t) B(r) = \alpha f(t)\frac{1}{r^2}\partial_r \left(r^2 B&amp;#39;(r) \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we see that we can re-write this as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{f&amp;#39;(t)}{f(t)} = \alpha \frac{1}{r^2}\frac{\partial_r \left(r^2 B&amp;#39;(r) \right)}{B(r)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and see that both sides are &lt;strong&gt;independent&lt;/strong&gt; of each other because of the different dependencies, hence they must be a &lt;strong&gt;constant&lt;/strong&gt;, say $-\lambda$, for the left hand side $$\frac{f&amp;#39;(t)}{f(t)} = -\lambda$$ for which we find the solution &#xA;$$f(t) = f_0 e^{-\lambda t}\ .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda$ can be seen as a &lt;strong&gt;dissipation constant&lt;/strong&gt; defining the loss of heat per time unit. I found a nice illustration of this process on &lt;a href=&quot;http://www.youtube.com/watch?v=ohAAS7uI8no&quot; rel=&quot;nofollow&quot;&gt;youtube&lt;/a&gt; or on &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_equation&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; for another geometry: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/xKIKG.gif&quot; alt=&quot;wiki heat dissipation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to the radial part, &#xA;$$\alpha \frac{1}{r^2}\partial_r \left(r^2 B&amp;#39;(r) \right) = -\lambda B(r)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for which we find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(r) = c_1 \frac{e^{-\mathrm{i}\kappa r}}{r} + c_2\frac{e^{\mathrm{i}\kappa r}}{r} $$&#xA;with $$\kappa =  \sqrt{\lambda/ \alpha}\ .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q&lt;/strong&gt;: Can you tell me what kind of solutions we just arrived at?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Application to a spherical body&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now this was the &lt;strong&gt;general solution&lt;/strong&gt; to the problem but we have to apply it to your situation. First of all, wee need some &lt;strong&gt;initial&lt;/strong&gt; thermal distribution: $$T_0(r) = T(r,0)$$ to determine the dynamics from this point. Second of all, we need to know what it means to have some &lt;strong&gt;sphere&lt;/strong&gt; in our system. We don't have a choice but to assign some distribution to $\alpha$, determining different heat coefficients in different regions, e.g.&#xA;$$\alpha = \begin{cases}&#xD;&#xA;\alpha_{i} &amp;amp; r\leq R\\&#xD;&#xA;\alpha_{0} &amp;amp; r&amp;gt;R\end{cases} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing left to do is to &lt;strong&gt;determine the constants&lt;/strong&gt; $\lambda$ and $c_i$ from the continuity of $T$ at $r=R$ and meaningful values at $r = \infty$ ($T\rightarrow 0$) and $r = 0$ ($T$ finite). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will leave this step to you.&lt;br&gt;&#xA;Having the solution at hand you can compare it to your simulation results and further extend it, e.g. to a system of shells or special boundary conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2011-04-28T11:48:44.557" CommentCount="1" />
  <row Id="9238" PostTypeId="2" ParentId="9203" CreationDate="2011-04-28T11:58:03.153" Score="-1" ViewCount="" Body="&lt;p&gt;I turns out that when you find the expectation value of free electrons speed that tend to always get c, when interpreted by the dirac equation!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This phenomenon is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Zitterbewegung&quot; rel=&quot;nofollow&quot;&gt;zitterbewegung&lt;/a&gt;, The idea being that you have an electron moving at c, but around a path with a low drift velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That probably didn't help but I think it is nice.&lt;/p&gt;&#xA;" OwnerUserId="1123" LastEditorUserId="1123" LastEditorDisplayName="" LastEditDate="2011-04-28T12:11:58.273" LastActivityDate="2011-04-28T12:11:58.273" CommentCount="1" />
  <row Id="9239" PostTypeId="1" CreationDate="2011-04-28T14:27:32.303" Score="2" ViewCount="185" Body="&lt;p&gt;how can we find out the effective mass of a hole,since a hole in the valence band is just an absence of electron?&lt;/p&gt;&#xA;" OwnerUserId="3321" LastActivityDate="2011-04-29T17:28:26.843" Title="how to find the effective mass of a hole" Tags="&lt;solid-state-physics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9240" PostTypeId="1" AcceptedAnswerId="9245" CreationDate="2011-04-28T16:57:18.330" Score="4" ViewCount="187" Body="&lt;p&gt;I'm following my first course in field theory and the professor began, like many books do, by introducing the scalar field. However, I am a bit hesitant about the physical idea of fields. My question is: what is the physical meaning of the fields? Why they are introduced?&#xA;I read the introduction of the books of Peskin and Weinberg but I'm not satisfied.&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2011-04-28T18:50:02.670" Title="What are fields?" Tags="&lt;quantum-field-theory&gt;&lt;field-theory&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="9241" PostTypeId="2" ParentId="9239" CreationDate="2011-04-28T17:18:17.040" Score="1" ViewCount="" Body="&lt;p&gt;Actually the concept of hole is an idealization because is easier to describe how electrons behave in a band describing the behavior of the holes rather than that of all the electrons that compose it. So the hole is our own definition and is defined as the absence of an electron and all its properties are obtained according to this definition.&#xA;in particular we find that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ ( \frac{1}{m})_{ij}^{(h)}=-( \frac{1}{m})_{ij}^{(e)} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m_{ij}$ is the effective mass tensor.&lt;/p&gt;&#xA;" OwnerUserId="592" LastEditorUserId="592" LastEditorDisplayName="" LastEditDate="2011-04-28T19:05:59.167" LastActivityDate="2011-04-28T19:05:59.167" />
  <row Id="9242" PostTypeId="1" CreationDate="2011-04-28T17:25:17.967" Score="2" ViewCount="100" Body="&lt;ul&gt;&#xA;&lt;li&gt;For a general ${\cal N}$ the R-symmetry group is $U({\cal N})$ but for the ${\cal N}=2$ case why is it $SU(2)$ ? I guess it is again different for ${\cal N}=4$. How does one understand this?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One denotes the generators of the R-symmetry group by $B^i$ and let $\alpha$ be the spinor index and let $a,b$ be the R-charge indices in a ${\cal N}=2$ theory. Then one writes the defining equations as,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[Q_{\alpha a}, B^i] = -\frac{1}{2} (\tau ^i)_a ^b Q_{\alpha b}$ and&#xA;$[\bar{Q}_{\dot{\alpha}}^a, B^i] = \frac{1}{2}  \bar{Q}_{\dot{\alpha}}^b  (\tau ^i)_b ^a$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;But it is not clear to me as to why over and above the aforementioned equations there should be yet another charge $R$ (called the $U(1)_R$ charge) with the defining equations,&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$[Q_{\alpha a},R] = Q_{\alpha a}$ and&#xA;$[\bar{Q}_{\dot{\alpha} a}, R] = - \bar{Q}_{\dot{\alpha} a}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know why the above charge should exist in ${\cal N}=2$ supersymetry separate from the R-symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an analogue for this $R$ for other values of ${\cal N}$ ?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Typically this R-charge as defined above is anomalous.&#xA;How does one see that?&#xA;Also what is the analogous statement for R-symmetry? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1637" LastActivityDate="2011-04-28T17:25:17.967" Title="Definition and difference between the R-symmetry and the $U(1)_R$ internal symmetry" Tags="&lt;quantum-field-theory&gt;&lt;supersymmetry&gt;" FavoriteCount="1" />
  <row Id="9243" PostTypeId="2" ParentId="9240" CreationDate="2011-04-28T17:34:34.263" Score="4" ViewCount="" Body="&lt;p&gt;Fields are important in both classical and quantum physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first it was thought they are just some imaginary mapping of the forces of gravitation or electric and magnetic forces. Gradually it became clear one can not keep it away from physics. They are more than imagination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? It is because electromagnetic effects take some finite nonzero time to reach from one point to other. A moving charge takes a nonzero time to influence another charge or a magnetic needle. If we want to preserve the laws of conservation of momentum and energy during that nonzero time we must have to take the concept of field as real and it should be thought of as the seat of energy and momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first people tried to understand field in terms of classical mechanics. It was believed at first that there must be some mechanical medium whose stress and strain must be the manifestation of field. The medium was named as aether. But all such efforts failed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It became clear that field is a fundamental entity of nature just as matter is. Or maybe matter is itself a kind of concentrated field!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum theory has given rise to quantum field theories and the classical fields have been quantized in QFT. It is now understood that fields can be thought of as consist of field quantas or particles with integer spin (bosons) and the so called matters are particles of half interger spin (fermions).&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-04-28T17:34:34.263" />
  <row Id="9244" PostTypeId="1" CreationDate="2011-04-28T17:44:50.520" Score="2" ViewCount="162" Body="&lt;p&gt;Ok, so bearing in mind I only have a brief understanding of quantum mechanics (no formal education, only from reading about concepts in books), so I could be way off here, I have a question regarding probability wave functions and light waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my understanding, every particle of matter has a probability wave which fills the entire universe, that's to say the only reason a particle in my fingernail is in my fingernail is because that's where the highest probability of it being is (with the chance that it's anywhere else in the universe being infinitesimally small), so my question is what is a light wave? Also from my understanding, a light wave is simply a probability wave for the location of a photon, and only upon observation does that wave collapse and does the photon 'take on' a definite position in space, so if I compare a particle of matter to a particle of energy (a photon in this case), how come the particle of matter has a probability wave that fills the entire universe, and the photon have only a small defined wave which virtually makes up a straight line in space? Is the light wave just an area (straight line?) of space where the probability is magnitudes higher than everywhere else and the photon's probability wave is still permeating the rest of the universe, or is the photon's location DEFINITELY somewhere in the area of space defined by the light wave?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, could a photon that's part of a laser beam ever interact with a photon from a laser beam travelling parallel to the first beam? (Disregarding any other physical theories that could cause the two beams to converge through other means, I'm wanting to stay on the topic of Schroedinger probability waves)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I might well be completely misunderstanding/mixing concepts here, so that might be the reason for my confusion, if so, please enlighten me :) &lt;/p&gt;&#xA;" OwnerUserId="3331" LastActivityDate="2011-04-29T05:07:45.983" Title="Light waves and Schrödinger probability waves" Tags="&lt;quantum-mechanics&gt;&lt;light&gt;&lt;waves&gt;&lt;photons&gt;&lt;schrodinger-equation&gt;" AnswerCount="1" />
  <row Id="9245" PostTypeId="2" ParentId="9240" CreationDate="2011-04-28T18:28:06.427" Score="2" ViewCount="" Body="&lt;p&gt;Probably the most fundamental and simple idea of the field arises from heat equation. You have a heat source and heat diffusion through media. It is described by field of temperature. It is the simplest scalar field I can imagine. But it has nontrivial equation of motion -  has it? From that simplest cases more complicated arises: force fields introduced without any complicated mathematics by well known genius Michael Faraday, who just draw lines of equal potential from one charge to another in order to understand how electrostatic and magnetic forces works. Of course You may imagine even more complicated fields - general tensor or spinor ones. &#xA;Strict definition of fields is of course a function which for given point on the manifold assign tensor or spinor to it. Physical fields often has strong continuity and differential conditions on them and obeys complicated differential or integral equations, but the most fundamental idea is still the same: lines which was drawn by Faraday, temperature in continuous media, velocity field of the fluid passing through water canal. Try to understand that simple ideas and return to &quot;quantum-compacted-26-dimensional-theories of everything&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1012" LastActivityDate="2011-04-28T18:28:06.427" />
  <row Id="9246" PostTypeId="1" AcceptedAnswerId="9276" CreationDate="2011-04-28T18:43:49.570" Score="3" ViewCount="129" Body="&lt;p&gt;So basically, I was trying to find a good answer to the question of how light interacts matter. Namely on the quantum level what causes matter to appear transparent, reflective, opaque, etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came upon the answer to transparency here &lt;a href=&quot;http://www.sixtysymbols.com/videos/energygap.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.sixtysymbols.com/videos/energygap.htm&lt;/a&gt; but I feel that the concept isn't fully explored here and that something is missing (particularly in the coloured light talk). The jist is that light passes through some matter because the energy gap between the matter's electron's low and excited state exceeds the energy of the photon, and thus the light is not absorbed and the photon passes through the matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone provide more detailed information about transparency, and information about reflection etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I guess for the purposes of forming this into a question I'll simply ask why some materials reflect light and look like mirrors while other materials reflect light with less precision? (thought I get the feeling it has similarly to do with basic mechanics and the regularity of the electron distribution)&lt;/p&gt;&#xA;" OwnerUserId="2806" LastActivityDate="2011-04-29T14:38:57.023" Title="The Interactions of Light and Matter" Tags="&lt;light&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="9247" PostTypeId="1" CreationDate="2011-04-28T18:48:57.210" Score="0" ViewCount="277" Body="&lt;p&gt;How can I calculate the current (instantaneous) mpg of my car if I know the speed and acceleration of the car? From reading various answers for the &quot;car going level or up/down hill&quot; question asked earlier by someone, I understand there are 4-5 factors to consider:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Rolling resistance (how do I get or calculate this value for my make/model?)  &lt;/li&gt;&#xA;&lt;li&gt;Engine resistance (how do I get or calculate this value for my make/model?)  &lt;/li&gt;&#xA;&lt;li&gt;Air drag coefficient (assume I know this)  &lt;/li&gt;&#xA;&lt;li&gt;Mass of car (assume I know this)  &lt;/li&gt;&#xA;&lt;li&gt;&quot;Power curve&quot; of car? What is power curve? Do I  need to know this? How do I get or calculate this?  &lt;/li&gt;&#xA;&lt;li&gt;Slope of road (assume level for simplification)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3333" LastActivityDate="2011-08-26T00:34:49.020" Title="How to calculate fuel consumption of car (mpg) from speed and accleration knowing mass, drag coeff and rolling resistance?" Tags="&lt;energy&gt;&lt;acceleration&gt;&lt;speed&gt;&lt;power&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="9248" PostTypeId="2" ParentId="9240" CreationDate="2011-04-28T18:50:02.670" Score="-2" ViewCount="" Body="&lt;p&gt;Quantum fields are somewhat opposite to classical fields. If a classical field is the whole ensemble of the Faraday lines, a quantum filed is only one line ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a classical point-like source of light creates a spherical wave visible at any angle, a quantum source creates only one photon visible with only one photo-detector. Quantum waves are amplitudes of probability rather than matter waves. Classical fields are an inclusive quantum picture.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-04-28T18:50:02.670" />
  <row Id="9249" PostTypeId="2" ParentId="9221" CreationDate="2011-04-28T18:54:07.200" Score="0" ViewCount="" Body="&lt;p&gt;I think Asimov's answer to the age-old question &quot;What happens if an unstoppable force hits an unmovable object&quot; (in his book &lt;em&gt;Please Explain&lt;/em&gt;) is applicable here: If you are to play the scientific game you have to agree terms that make sense. So no good science is going to come out of saying &quot;going at the speed of light is impossible, but what if ...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this is different than the line of thought &quot;What if the sum of angles in a triangle add to more than 180 degrees?&quot; This is because mathematics is axiomatic (i.e. with different axioms you can arrive at different, but equally valid results) whereas science (e.g. physics) is (mostly) not: We &lt;em&gt;know&lt;/em&gt; that objects with nonzero mass cannot reach the speed of light. Note that this is different than knowing &lt;em&gt;why&lt;/em&gt; this is true.&lt;/p&gt;&#xA;" OwnerUserId="852" LastActivityDate="2011-04-28T18:54:07.200" CommentCount="1" />
  <row Id="9250" PostTypeId="2" ParentId="8626" CreationDate="2011-04-28T19:36:35.290" Score="2" ViewCount="" Body="&lt;p&gt;There are already several good answers. However, the off-shell aspect related to Noether Theorem has not been addressed so far. (The words &lt;em&gt;on-shell&lt;/em&gt; and &lt;em&gt;off-shell&lt;/em&gt; refer to whether the equations of motion (e.o.m.) are satisfied or not.) Let me rephrase the problem as follows. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider a (not necessarily isolated) Hamiltonian system with $N$ degrees of freedom (d.o.f.). The phase space has $2N$ coordinates, which we denote $(z^1, \ldots, z^{2N})$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(We shall have nothing to say about the corresponding Lagrangian problem.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Symplectic structure.  Usually, we work in Darboux coordinates $(q^1, \ldots, q^N; p_1, \ldots, p_N)$, with the canonical symplectic potential one-form  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vartheta=\sum_{i=1}^N p_i dq^i.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it turns out to be more efficient in later calculations, if we instead from the beginning consider general coordinates $(z^1, \ldots, z^{2N})$ and a general (globally defined) symplectic potential one-form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vartheta=\sum_{I=1}^{2N} \vartheta_I(z;t) dz^I,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with non-degenerate (=invertible)  symplectic two-form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega = \frac{1}{2}\sum_{I,J=1}^{2N} \omega_{IJ} \ dz^I \wedge dz^J = d\vartheta,\qquad\omega_{IJ} =\partial_{[I}\vartheta_{J]}=\partial_{I}\vartheta_{J}-\partial_{J}\vartheta_{I}. $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding Poisson bracket is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\{f,g\} = \sum_{I,J=1}^{2N} (\partial_I f) \omega^{IJ} (\partial_J g), \qquad \sum_{J=1}^{2N} \omega_{IJ}\omega^{JK}= \delta_I^K.  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Action. The Hamiltonian action $S$ reads&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S[z]= \int dt\ L_H(z^1, \ldots, z^{2N};\dot{z}^1, \ldots, \dot{z}^{2N};t),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ L_H(z;\dot{z};t)= \sum_{I=1}^{2N} \vartheta_I(z;t)  \dot{z}^I- H(z;t)  $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the Hamiltonian Lagrangian. By infinitesimal variation &#xA;$$\delta S = \int dt\sum_{I=1}^{2N}\delta z^I \left( \sum_{J=1}^{2N}\omega_{IJ} \dot{z}^J-\partial_I H - \partial_0\vartheta_I\right)+ \int dt \frac{d}{dt}\sum_{I=1}^{2N}\vartheta_I \delta z^I, \qquad \partial_0 \equiv\frac{\partial }{\partial t},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of the action $S$, we find the Hamilton e.o.m. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  \dot{z}^I \approx \sum_{J=1}^{2N}\omega^{IJ}\left(\partial_J H + \partial_0\vartheta_J\right) = \{z^I,H\} + \sum_{J=1}^{2N}\omega^{IJ}\partial_0\vartheta_J. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(We will use the $\approx$ sign to stress that an equation is an on-shell equation.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Constants of motion. The solution &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z^I = Z^I(a^1, \ldots, a^{2N};t)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;to the first-order Hamilton e.o.m. depends on $2N$ constants of integration $(a^1, \ldots, a^{2N})$. Assuming appropriate regularity conditions, it is in principle possible to invert locally this relation such that the constants of integration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a^I=A^I(z^1, \ldots, z^{2N};t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are expressed in terms of the $(z^1, \ldots, z^{2N})$ variables and time $t$. These functions $A^I$ are $2N$ constants of motion (c.o.m.), i.e., constant in time $\frac{dA^I}{dt}\approx0$. Any function $B(A^1, \ldots, A^{2N})$ of the $A$'s, but without explicit time dependence, will again be a c.o.m. In particular, we may express the initial values $(z^1_0, \ldots, z^{2N}_0)$ at time $t=0$ as functions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z^J_0(z;t)=Z^J(A^1(z;t), \ldots, A^{2N}(z;t); t=0)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of the $A$'s, so that $Z^J_0$ become c.o.m.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now, let &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$b^I=B^I(z^1, \ldots, z^{2N};t)$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;be $2N$ independent c.o.m., which we have argued above must exist. The question is if there exist $2N$ off-shell symmetries of the action $S$, such that the corresponding Noether currents are on-shell c.o.m.?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Remark. It should be stressed that an &lt;em&gt;on-shell symmetry&lt;/em&gt; is a vacuous notion, because if we vary the action $\delta S$ and apply e.o.m., then $\delta S\approx 0$ vanishes by definition (modulo boundary terms), independent of what the variation $\delta$ consists of. For this reason we often just shorten &lt;em&gt;off-shell symmetry&lt;/em&gt; into &lt;em&gt;symmetry&lt;/em&gt;. On the other hand, when speaking of c.o.m., we always assume e.o.m. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Change of coordinates. Since the action $S$ is invariant under change of coordinates, we may simply change coordinates $z\to b = B(z;t)$ to the $2N$ c.o.m., and use the $b$'s as coordinates (which we will just call $z$ from now on). Then the e.o.m. in these coordinates are just&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dz^I}{dt}\approx0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so we conclude that in these coordinates, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \partial_J H + \partial_0 \vartheta_J=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as an off-shell equation. [An aside: This implies that the symplectic matrix $\omega_{IJ}$ does not depend explicitly on time, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\partial_0\omega_{IJ} =\partial_0\partial_{[I}\vartheta_{J]}=\partial_{[I} \partial_0\vartheta_{J]}=-\partial_{[I}\partial_{J]} H=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the Poisson matrix $\{z^I,z^J\}=\omega^{IJ}$ does not depend explicitly on time. &#xA;By Darboux Theorem, we may locally find Darboux coordinates $(q^1, \ldots, q^N; p_1, \ldots, p_N)$, which are also c.o.m.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Variation. We now perform an infinitesimal variation  $\delta= \varepsilon\{z^{I_0}, \cdot \}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta z^J = \varepsilon\{z^{I_0}, z^J\}=\varepsilon \omega^{I_0 J},$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;with Hamiltonian generator $z^{I_0}$, where $I_0\in\{1, \ldots, 2N\}$. It is straightforward to check that the infinitesimal variation $\delta= \varepsilon\{z^{I_0}, \cdot \}$ is an off-shell symmetry of the action (modulo boundary terms)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta S = \varepsilon\int dt \frac{d f^0}{dt},  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f^0 = z^{I_0}+ \sum_{J=1}^{2N}\omega^{I_0J}\vartheta_J.$$ The bare Noether current is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$j^0 = \sum_{J=1}^{2N}\frac{\partial L_H}{\partial \dot{z}^J} \omega^{I_0 J}=\sum_{J=1}^{2N}\omega^{I_0J}\vartheta_J,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that the full Noether current&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ J^0=j^0-f^0=-z^{I_0} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;becomes just (minus) the Hamiltonian generator $z^{I_0}$, which is conserved on-shell $\frac{dJ^0}{dt}\approx 0$ by definition.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So the answer is yes in the Hamiltonian case.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-04-28T21:06:26.180" LastActivityDate="2011-04-28T21:06:26.180" />
  <row Id="9251" PostTypeId="2" ParentId="9170" CreationDate="2011-04-28T20:43:21.993" Score="1" ViewCount="" Body="&lt;p&gt;Yes, something like that is possible, and they've done it in Anton Zeilinger's lab.  It's not a Bell's inequality test, and to get it to work it's more complicated than what you're describing, but: they do see spatial (i.e. position) correlations/interference disappear or not based on measurement of one of the entangled particles.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's described in a nice review paper (which has a bunch of other nice EPR/Bell experiments):&#xA;Rev. Mod. Phys. 71 S288 (1999)&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-04-28T20:43:21.993" CommentCount="1" />
  <row Id="9252" PostTypeId="1" CreationDate="2011-04-28T20:13:39.223" Score="9" ViewCount="299" Body="&lt;p&gt;In my poor understanding of quantum physics, quantum entanglement means that certain properties of one of two 'entangled' quantum particles can lead to change over infinitely large distances when the other particles' properties are changed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disregarding this already mind-boggling event taking place over say 10 meters distance; how have physicists been able to demonstrate, beyond reasonable doubt, that this can take place over infinitely large distances?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance: have they done some of these tests between &lt;a href=&quot;http://en.wikipedia.org/wiki/ISS&quot; rel=&quot;nofollow&quot;&gt;ISS&lt;/a&gt; and Earth perhaps?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can they be so sure?&lt;/p&gt;&#xA;" OwnerUserId="2808" LastEditorUserId="863" LastEditorDisplayName="" LastEditDate="2011-04-29T02:26:31.883" LastActivityDate="2011-04-29T02:26:31.883" Title="Has quantum entanglement been demonstrated to be able to take place over infinite distances?" Tags="&lt;quantum-mechanics&gt;&lt;entanglement&gt;&lt;experimental-evidence&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="9253" PostTypeId="2" ParentId="9252" CreationDate="2011-04-28T21:01:32.983" Score="14" ViewCount="" Body="&lt;p&gt;First of all, it is very important to note that quantum entanglement is &lt;em&gt;not&lt;/em&gt; a &lt;em&gt;spooky action-at-a-distance&lt;/em&gt; as Einstein once called it! It is a strong correlation of measurements that is stronger than any classical correlation could ever be. This has been experimentally verified by the so-called &lt;em&gt;violation of Bell's inequalities&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, quantum entanglement has been successfully demonstrated for distance of some 100 km, as the Wikipedia article on quantum teleportation states (references in Journales such as Nature are given there)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Zeilinger’s experiments on the distribution of entanglement over large distances began with both free-space and fiber-based quantum communication and teleportation between laboratories located on the different sides of the river Danube. This was then extended to larger distances across the city of Vienna and most recently over &lt;strong&gt;144 km&lt;/strong&gt; between two Canary Islands, resulting in a successful demonstration that quantum communication with satellites is feasible. His dream was to bounce entangled light off of satellites in orbit.[1] This was achieved during an experiment at the Italian Matera Laser Ranging Observatory.[4]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Admittedly, that is not a demonstration &lt;em&gt;at infinite distance&lt;/em&gt;, but then, &lt;em&gt;nothing&lt;/em&gt; has ever been demonstrated at that length scale and we would never be able to do and to demand such a demonstration is unreasonable. But please not the different scales involved here! Usually, we think of quantum mechanics to govern the microscopic world, involving length scales of under a micrometer, i.e. $10^{-6} m$. Achieving quantum effects on a length scale of $100 km = 10^5 m$ means you span 11 orders of magnitude. You can jokingly call this &quot;infinite for all practical purposes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The misunderstanding is that nothing changes &lt;em&gt;in the moment of your measurement&lt;/em&gt;. There is a classical analogue that is wrong on the details but emphasizes the main idea: Suppose you take two playing cards, one is a king and the other a queen. You shuffle them and let your friend pick one at random while you keep the other one. Now, you have the king with probability $1/2$ and the queen also with probability 1/2, but you don't know which one you have without looking. You also don't know which one your friend has. But as soon as you look at &lt;em&gt;your&lt;/em&gt; card, you immediately know which one your friend has: If you got the king, he must have got the queen. There is no spooky action at a distance that somehow changes what the card of your friend looks like!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum mechanics, it is admittedly a bit more complex, because not only would you not know what card you have without looking, it isn't even &lt;em&gt;determined&lt;/em&gt; until you look. It is, however, possible to go through the math and the mechanisms and come up with a satisfactory explanation that does not involve faster-than-light communication and processes.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditorDisplayName="" LastEditDate="2011-04-28T21:49:24.990" LastActivityDate="2011-04-28T21:49:24.990" CommentCount="4" />
  <row Id="9254" PostTypeId="2" ParentId="9252" CreationDate="2011-04-28T21:01:38.857" Score="8" ViewCount="" Body="&lt;p&gt;Since my Alma Mater was involved I can point to &lt;a href=&quot;http://www.esa.int/esaCP/SEMXM7Q08ZE_index_0.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;A team of European scientists has&#xA;  proved within an ESA study that the&#xA;  weird quantum effect called&#xA;  'entanglement' remains intact over a&#xA;  distance of 144 kilometres.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ziXSf.jpg&quot; alt=&quot;enter image description here&quot;&gt; (&lt;a href=&quot;http://www.qubitapplications.com/Partners/austria/oeaw.asp&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In September 2005, &lt;strong&gt;the European team&#xA;  aimed ESA's one-metre telescope on the&#xA;  Canary Island of Tenerife toward the&#xA;  Roque de los Muchachos Observatory on&#xA;  the neighbouring island of La Palma,&#xA;  144 kilometres away.&lt;/strong&gt; On La Palma, a&#xA;  specially built quantum optical&#xA;  terminal generated entangled photon&#xA;  pairs, using the SPDC process, and&#xA;  then sent one photon towards Tenerife,&#xA;  whilst keeping the other for&#xA;  comparison.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There is an effort for &lt;a href=&quot;http://www.quantum.at/quest&quot; rel=&quot;nofollow&quot;&gt;Quantum Entanglement in Space Experiments&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Experimentally is has not been proven to work over an infinite distance, since that would be impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To better understand &lt;em&gt;Quantum Entanglement&lt;/em&gt; (and why distance is not a problem) I suggest starting with the &lt;a href=&quot;http://en.wikipedia.org/wiki/EPR_paradox&quot; rel=&quot;nofollow&quot;&gt;EPR Paradox&lt;/a&gt;. &lt;/p&gt;&#xA;" LastActivityDate="2011-04-28T21:01:38.857" />
  <row Id="9255" PostTypeId="1" AcceptedAnswerId="14213" CreationDate="2011-04-28T21:53:15.643" Score="4" ViewCount="358" Body="&lt;p&gt;We know that a point charge $q$ located at the origin $r=0$ produces a potential $\sim \frac{q}{r}$, and this is consistent with the fact that the Laplacian of $\frac{q}{r}$ is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla^2\frac{q}{r}=-4\pi q\cdot \delta^3(\vec{r}).$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what is the Laplacian of $\frac{1}{r^2}$ (at the origin!)? Is there a charge distribution that would cause this potential? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3064" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-04-29T20:25:43.357" LastActivityDate="2011-09-01T03:17:02.607" Title="Laplacian of $1/r^2$ (context: electromagnetism and poisson equation)" Tags="&lt;electromagnetism&gt;&lt;mathematical-physics&gt;&lt;electrostatics&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="9256" PostTypeId="2" ParentId="9255" CreationDate="2011-04-28T22:14:44.130" Score="2" ViewCount="" Body="&lt;p&gt;Let us see: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta \frac{q}{r^2} = \frac{1}{r^2}\frac{\partial}{\partial r} \left( r^2\frac{\partial} {\partial r}\frac{q}{r^2}\right)=\frac{2q}{r^4}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the charge density $\rho(r)$ is proportional to $r^{-4}$. Such a charge density is only possible to create at a macroscopic scale where the charge may be considered continuous (charged dielectric spherical layers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; If the potential is spherically symmetric, it is not a dipole field but a monopole one with the charge continuously distributed along $r$!&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-04-30T12:11:13.763" LastActivityDate="2011-04-30T12:11:13.763" CommentCount="11" />
  <row Id="9257" PostTypeId="2" ParentId="9255" CreationDate="2011-04-28T22:43:25.307" Score="0" ViewCount="" Body="&lt;p&gt;The operator to go from one potential to the other is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-\frac{\partial }{\partial r}\left\{\frac{q}{4\pi r}\right\} ~~=~~ \frac{q}{4\pi r^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and therefor the source in the center which is given by the Laplacian of the new potential is obtained by using the same operator on the source of the original potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-\frac{\partial }{\partial r}\Big\{\nabla^2\frac{q}{4\pi r} \Big\} ~~=~~ -\frac{\partial }{\partial r}\Big\{q\,\delta(r)\Big\}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the source charge would be the radial derivative of the delta function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards, Hans&lt;/p&gt;&#xA;" OwnerUserId="3197" LastEditorUserId="3197" LastEditorDisplayName="" LastEditDate="2011-04-28T22:57:53.360" LastActivityDate="2011-04-28T22:57:53.360" CommentCount="3" />
  <row Id="9258" PostTypeId="1" AcceptedAnswerId="9264" CreationDate="2011-04-29T02:04:11.037" Score="1" ViewCount="100" Body="&lt;p&gt;In the MSUGRA breaking scenario, the stop particle typically appears at energies reachable at the LHC. Other sfermions, notably the partners of up, down, strange and charm are assumed to be degenerate in mass, and also heavier than the stop. Something similar holds for the stau.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the third generation different in MSUGRA (not degenerate as the first two), and why is the mass hierarchy inverted wrt. the Standard Model sector (3rd generation sparticles lighter)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I guess these features are not neccessarily specific to MSUGRA, but might apply to more general models as well.)&lt;/p&gt;&#xA;" OwnerUserId="825" LastActivityDate="2011-04-29T06:04:38.157" Title="Why are the third generation superpartners lighter than the other sfermions in MSUGRA" Tags="&lt;particle-physics&gt;&lt;supersymmetry&gt;" AnswerCount="1" />
  <row Id="9259" PostTypeId="1" CreationDate="2011-04-29T02:23:28.390" Score="3" ViewCount="149" Body="&lt;p&gt;I know that if one astronaut falls into a black hole, then a distant observer will see him take an infinite amount of time to reach the event horizon (provided the observer can see light of arbitrarily large wavelengths).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the falling astronaut will only take a finite amount of time to reach the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: What will the falling astronaut see if he &quot;looks backwards&quot; while falling.&#xA;Will he see the distant observer growing old and all the stars dying by the time he reaches the horizon ?&lt;/p&gt;&#xA;" OwnerUserId="3335" LastActivityDate="2011-04-29T08:10:05.853" Title="Time dilation when falling into black hole" Tags="&lt;general-relativity&gt;&lt;black-holes&gt;" AnswerCount="1" />
  <row Id="9260" PostTypeId="2" ParentId="9244" CreationDate="2011-04-29T05:07:45.983" Score="1" ViewCount="" Body="&lt;p&gt;Let's take it a step at a time:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From my understanding, every particle of matter has a probability wave which fills the entire universe, that's to say the only reason a particle in my fingernail is in my fingernail is because that's where the highest probability of it being is (with the chance that it's anywhere else in the universe being infinitesimally small),&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;true&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;so my question is what is a light wave? Also from my understanding, a light wave is simply a probability wave for the location of a photon, and only upon observation does that wave collapse and does the photon 'take on' a definite position in space,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A light wave is also a particle, called the photon. Depending on the observation it can display its particle nature or its wave nature.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;so if I compare a particle of matter to a particle of energy (a photon in this case), &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a photon is a particle too, not different than other particles&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;how come the particle of matter has a probability wave that fills the entire universe, and the photon have only a small defined wave which virtually makes up a straight line in space? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A photon has the same  behavior as any other particle, depending on the experiment: it either displays its wave nature, or its particle nature depending on the observation: through two slits shows interference,  therefore wave properties, absorbed in an atom, particle. Both particles   and photons behave the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the light wave just an area (straight line?) of space where the probability is magnitudes higher than everywhere else and the photon's probability wave is still permeating the rest of the universe, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a volume , and a very  small one, like all particles&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;or is the photon's location DEFINITELY somewhere in the area of space defined by the light wave?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no definite in quantum mechanics. Only after the experiment the values are known. Only probabilities can be calculated for individual particles, photons or not. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For example, could a photon that's part of a laser beam ever interact with a photon from a laser beam travelling parallel to the first beam?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here we are entering deeper into quantum theory, into what is called &quot;second quantization&quot;.&#xA;Yes, two photons parallel to each other in their probability paths can interact with tiny tiny probabilities by exchanging quanta of energy between them, these are carried by virtual particles covering the whole spectrum that is permitted by quantum number conservations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Digressing into lasers:&#xA;Lasers are another story, and are the result of the possibility that atomic physics allows to have coherent waves of photons. Coherent means &quot;in step&quot; in time and space. Atomic physics because the transitions between energy states in atoms happen with photon exchanges and interactions, and one can easily make a coherent beam of photons, all in phase with each other, as is the laser beam. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A way for a layman to understand the possibilities of the effects of coherence is the famous &quot;soldiers crossing a bridge&quot;. In olden times, bridges were mainly stable because the arches held them up statically. The connecting mortar did not have much structural strength. Soldiers when crossing a bridge had to break step, otherwise the  hitting of boots in phase could destroy the bridge by building up coherently  large forces that the arch could not hold.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-29T05:07:45.983" CommentCount="3" />
  <row Id="9261" PostTypeId="1" CreationDate="2011-04-29T05:12:33.277" Score="0" ViewCount="97" Body="&lt;p&gt;So it has been asserted that a Stirling engine with proper regeneration can be made reversible. It will consist of two isothermal quasi-static processes connected by two constant-volume processes. The working gas is supposed to be heated/cooled through infinitesimal temperature differences (thus preventing the creation of entropy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know an explanation on the working gas properties during each step of the cycle. What are the state conditions for the gas at each of the points in the cycle (including inside the regenerator)? I've been looking for a the past week and found nothing.&lt;/p&gt;&#xA;" OwnerUserId="3337" LastActivityDate="2011-04-29T11:10:42.867" Title="Reversible Stirling Engine" Tags="&lt;thermodynamics&gt;" AnswerCount="1" />
  <row Id="9262" PostTypeId="1" AcceptedAnswerId="9288" CreationDate="2011-04-29T05:30:34.260" Score="2" ViewCount="188" Body="&lt;p&gt;Apologies if this question is a bit too chemistry-flavoured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In electron paramagnetic resonance spectroscopy, there's a practically ubiquitous convention of plotting the first derivative of the absorption with respect to field, rather than the simple absorption. Why is this?&lt;/p&gt;&#xA;" OwnerUserId="1876" LastActivityDate="2011-06-14T13:52:24.460" Title="Why do we take the first derivative of EPR/ESR spectra?" Tags="&lt;spectroscopy&gt;&lt;spin&gt;&lt;resonance&gt;" AnswerCount="4" />
  <row Id="9263" PostTypeId="1" AcceptedAnswerId="9605" CreationDate="2011-04-29T05:36:09.900" Score="2" ViewCount="192" Body="&lt;p&gt;Do surfaces that reflect visible light efficiently also reflect UV light?  If not, are there surfaces that do?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I have large array of UV LEDs that I need to make larger and more diffuse, so I'm considering reflecting it off nearby surfaces and into some diffuse material.)&lt;/p&gt;&#xA;" OwnerUserId="3338" LastActivityDate="2011-05-08T20:02:58.703" Title="UV reflective surfaces" Tags="&lt;light&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="9264" PostTypeId="2" ParentId="9258" CreationDate="2011-04-29T06:04:38.157" Score="1" ViewCount="" Body="&lt;p&gt;In short, the answer goes like this:  All fermion masses are assumed to unify at some high scale (e.g.~$10^{16} GeV$) in MSUGRA.  So, the mass differences between them at low energies are due to the running of the masses from that high scale down to the observed scale (e.g. ~1TeV at the LHC).  The $\beta$ function for the stop mass has a positive contribution due to the top Yukawa coupling,$y_t$, which is large due to the fact that the top mass is not small compared to the Higgs vacuum expectation value,$v$, ($m_t = y_tv$).  This implies that the stop mass drops more rapidly as the renormalization scale is lowered than the 1st and 2nd generation squarks, which have negligible Yukawa couplings.  This leads to a mass spectrum where the stop is light and the first two generation squarks are nearly degenerate (ditto for the stau).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Supersymmetry Primer by Martin has a good discussion of this (http://arxiv.org/abs/hep-ph/9709356), particularly on pg. 46 (where you can find the $\beta$ functions) and 75 (which discusses the squark and slepton spectrum).  &lt;/p&gt;&#xA;" OwnerUserId="3339" LastActivityDate="2011-04-29T06:04:38.157" />
  <row Id="9265" PostTypeId="1" AcceptedAnswerId="9266" CreationDate="2011-04-29T06:42:39.567" Score="0" ViewCount="109" Body="&lt;p&gt;As described in the wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Wave_function_collapse#History_and_context&quot; rel=&quot;nofollow&quot;&gt;wave function collapse&lt;/a&gt;, the mathematical formulation of quantum mechanics postulates that wave functions change according to two processes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;When not being observed, a wave function will evolve (deterministically) according to some relevant differential equation, e.g., Schrödinger's equation. &lt;/li&gt;&#xA;&lt;li&gt;Upon measurement, a wave function will collapse (probabilistically) to one of its component eigenfunctions. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What I'd like to understand is how quickly after a measurement a particle's wave function will wander away, according to (1), from whatever eigenfunction it collapsed to.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a concrete example (I hope!), consider &lt;a href=&quot;http://en.wikipedia.org/wiki/Stern%E2%80%93Gerlach_experiment#Sequential_experiments&quot; rel=&quot;nofollow&quot;&gt;two Stern-Gerlach experiments set up in sequence&lt;/a&gt;, both arranged to measure the z component of spin, in which we feed to the second experiment only the spin up half of the beam from the first experiment. When one learns about this in an introductory QM course, it is said that the observed output from the second experiment will still be all spin up. But is this still true if the experiments are spaced really far apart? How will the time evolution in (1) affect the outcome then? (Or is there something I have missed that prevents time evolution from affecting the intermediate beam?) If the observed outcome from the second experiment does depend on the distance between the experiments, then it seems this would give a way to measure (at least in principle) how rapidly a pure eigenstate can degenerate (by measuring, for example, the distance between the two experiments needed to get an output beam from the second experiment which is 75% up, 25% down).   &lt;/p&gt;&#xA;" OwnerUserId="2966" LastActivityDate="2011-04-29T06:54:13.640" Title="Time evolution of quantum state after an observation" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" />
  <row Id="9266" PostTypeId="2" ParentId="9265" CreationDate="2011-04-29T06:54:13.640" Score="5" ViewCount="" Body="&lt;p&gt;It depends on the Hamiltonian, i.e., the interactions the beam is subject to after the first measurement. If the beam is allowed to propagate freely, then a pure state, say $|0\rangle$, will remain in that state always because the Hamiltonian only contains a momentum operator that will not affect the spin component of the state vector. If on the other hand, after the first measurement, there are other interactions like a magnetic field, then it can lead to a non-trivial Schrodinger evolution of the pure state (spin part) which will change the outcome at the second measurement.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-04-29T06:54:13.640" CommentCount="1" />
  <row Id="9267" PostTypeId="2" ParentId="9259" CreationDate="2011-04-29T07:48:36.753" Score="2" ViewCount="" Body="&lt;p&gt;No. He would see that only if he stopped falling close to and above the horizon, by thrusting or going into orbit, say. (No stable orbits exist too close.) An observer falling straight in won't see time speed up above him.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A way to visualize this is with the river model of black holes (googleable). An observer who flows with the river, by falling straight inward toward a center of gravity, doesn't notice time speed up above him. To notice that the astronaut must &quot;fight the current&quot; somehow. By fighting the current the astronaut can receive information that is &quot;flowing with the river&quot; faster than normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are able to do that just sitting in our chairs. It just requires fine instruments to measure. If we jump off a cliff (or even a curb) we can no longer notice it.&lt;/p&gt;&#xA;" OwnerUserId="2314" LastEditorUserId="2314" LastEditorDisplayName="" LastEditDate="2011-04-29T08:10:05.853" LastActivityDate="2011-04-29T08:10:05.853" />
  <row Id="9268" PostTypeId="2" ParentId="331" CreationDate="2011-04-29T08:28:14.847" Score="-1" ViewCount="" Body="&lt;p&gt;usually, when the top is spinned, the base of the cone generated by the axis changes its radius periodically. Thus the angular frequency coming to assist is little doubtful. Rather an average angular velocity or so shall be the appropriate term.&lt;/p&gt;&#xA;" OwnerUserId="3340" LastActivityDate="2011-04-29T08:28:14.847" CommentCount="1" />
  <row Id="9271" PostTypeId="2" ParentId="9262" CreationDate="2011-04-29T09:14:56.740" Score="1" ViewCount="" Body="&lt;p&gt;As far as my (limited) knowledge on microwaves reaches, &#xA;You would have a fixed frequency transmitter, a wave duct, &#xA;then the sample in a magnetic field. This field is sweeped, &#xA;rather than frequency. One of the typical arrangements of &#xA;the sample to get sensitive measurement, is to use it as &#xA;the active incredient of a &quot;gyrator&quot;, or ciculator, or something similar. &#xA;All that may be just &quot;traditional&quot;, because technical &#xA;possibilities were that way when this spectroscopy was first &#xA;developed, maybe the situation is still the same. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-29T11:13:53.127" LastActivityDate="2011-04-29T11:13:53.127" />
  <row Id="9272" PostTypeId="2" ParentId="9255" CreationDate="2011-04-29T10:22:15.000" Score="3" ViewCount="" Body="&lt;p&gt;Vladimir's answer is off by factor of 2. The laplacian is $\nabla^2(\frac{1}{r^2}) = \frac{4}{r^4}$ A potential that falls of as $\frac{1}{r^2}$ is a dipole (In general, if it falls off as $r^{-n}$ its an ($2^{n-1}$)-pole, e.g $\frac{1}{r^3}$ bheaviour is quadrupole, etc). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a dirac delta? To find out, check : $$\int_{\mbox{All space}}\nabla^2(\frac{1}{r^2})d^3r  $$ $$=16\pi\int_0^\infty \frac{1}{r^2}dr\neq 1$$ Yup, integral diverges, so it is &lt;strong&gt;NOT&lt;/strong&gt; a delta function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think your confusion is regarding the nature of a delta function. If something blows up at the origin it does not mean it is necessarily a delta function. &lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="3239" LastEditorDisplayName="" LastEditDate="2011-04-29T18:06:59.007" LastActivityDate="2011-04-29T18:06:59.007" CommentCount="2" />
  <row Id="9273" PostTypeId="2" ParentId="9261" CreationDate="2011-04-29T11:10:42.867" Score="1" ViewCount="" Body="&lt;p&gt;The stirling engine is reversible in a common meaning of the word, that means - apply temperature difference, get torque; apply torque, get temperature difference. Of course it won't be fully reversible in thermodynamic sense, due to its own lossy character of operation (like loose displacer that allows heat transfer from hot to cold regenerator without producing usable work.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stirling engine and stirling cryocooler are essentially the same device, they work on exactly the same principle - take unpowered stirling engine (with no heat/cold sources) and start rotating the axis and you will cause the two plates to obtain temperature difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course in practical uses there are serious technological differences that may make the operation impossible, but the principles are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The four phases of stirling engine in &quot;cryocooler&quot; mode? &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Compress the air on the &quot;hot&quot; side by artificially pushing the work piston, providing external work. Compression heats the air. Hot air passes heat to the regenerator, losing some temperature.&lt;/li&gt;&#xA;&lt;li&gt;Displace the air towards the cold side, using the displacer.&lt;/li&gt;&#xA;&lt;li&gt;Decompress the air by pulling the work piston. Some work is provided by air pressure, more, external work is provided. Air is cooled and extracts the temperature from the regenerator, making it cooler.&lt;/li&gt;&#xA;&lt;li&gt;Displace decompressed air back to the hot side using the displacer.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;engine:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;hot side regenerator transfer heat into gas. Gas heats, expands, pressure rises, this pushes work piston providing work, reducing pressure.&lt;/li&gt;&#xA;&lt;li&gt;gas is displaced to cold side regenerator.&lt;/li&gt;&#xA;&lt;li&gt;gas transfers heat into the cold side regenerator. As its temperature drops, pressure and volume drops, pulling the work piston in.&lt;/li&gt;&#xA;&lt;li&gt;Displacer moves cooled gas back to the hot side.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2754" LastActivityDate="2011-04-29T11:10:42.867" />
  <row Id="9274" PostTypeId="1" AcceptedAnswerId="9278" CreationDate="2011-04-29T13:10:10.420" Score="3" ViewCount="110" Body="&lt;p&gt;I was wondering the other day an X-ray detector (like the ones used at airports) can detect gamma-rays lets say from a sample of uranium. I know its all electro-magnetic waves but I'm really unsure about how it works in practical terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now out of curiousity, if you got an beta-radiator, might that go unnoticed?&lt;/p&gt;&#xA;" OwnerUserId="3341" LastEditorUserId="3341" LastEditorDisplayName="" LastEditDate="2011-04-29T15:40:50.457" LastActivityDate="2011-04-29T15:40:50.457" Title="Can one see radioactive substances with an X-ray detector?" Tags="&lt;electromagnetic-radiation&gt;&lt;radioactivity&gt;&lt;x-ray&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9275" PostTypeId="2" ParentId="9274" CreationDate="2011-04-29T14:31:46.120" Score="3" ViewCount="" Body="&lt;p&gt;In wikipedia there is an extensive article about &lt;a href=&quot;http://en.wikipedia.org/wiki/X-ray_detector#Detectors&quot; rel=&quot;nofollow&quot;&gt;x-ray detectors&lt;/a&gt; and one on &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma-ray_detectors&quot; rel=&quot;nofollow&quot;&gt;gamma ray&lt;/a&gt; detectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Detectors are calibrated to work with specific frequencies and though there will be some effect  from a gamma ray source on an x ray detector, and vice versa, to make sense of it one would need the proper calibration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you are asking whether when luggage is radiated by x rays and a &quot;picture&quot; taken if gamma radiation will show up, it is  improbable mainly due to the distances involved: radiation from a source falls as 1/r^2 whereas the xray detector uses coherent/and or parallel x-rays. It would need a very large source to show up on the picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They must have special Geiger counter detectors to catch illegal radioactive material being passed.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-29T14:31:46.120" CommentCount="2" />
  <row Id="9276" PostTypeId="2" ParentId="9246" CreationDate="2011-04-29T14:38:57.023" Score="3" ViewCount="" Body="&lt;p&gt;The optical properties of matter are a highly non-trivial matter. Let me try to explain some of the basic mechanisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you know, light can be described either as an electromagnetic wave (appropriate at a macroscopic level) or as consisting of photons of certain energy and frequency (more appropriate at a microscopic level).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phenomena of absorption, transmission, reflection etc. often require to use both concepts, depending on the material. Let us therefore look at different classes of materials:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Gases:&lt;/strong&gt; Let us look at a gas made of a bunch of atoms or molecules. Quantum mechanics tells you that in these, the electrons occupy discrete orbitals (atomic orbitals or molecular orbitals). A photon of the right energy can excite electrons from lower to higher orbitals. If this happens, the photon is destroyed (absorbed). Consequently, it is missing in the transmitted beam. Since this happens only at discrete energies, you see this as the black spots in the spectrum when you use white light (i.e. light that contains photons of all visible light):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/6WmE7.png&quot; alt=&quot;Spectral Light&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above image shows first the continuous spectrum of white light, then the emission lines of some element and then the spectrum you see when looking at white light after it has passed through that gas. This information can be used to identify unknown elements and is, for example, used in astronomy to determine what stars are made of.&#xA;In conclusion, the relevant process here is the discrete energy spectrum of gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solids:&lt;/strong&gt; In solids, the discrete energy levels of the atoms &quot;interact&quot; with each other and form continuous bands. These continuous bands are described by the &lt;em&gt;density of states&lt;/em&gt; $D(E)$ which tells you how many states there are in a small interval around energy $E$. You then fill your system up with electrons, beginning at the lowest possible energy and up to some maximum energy which is called the Fermi energy $E_F$. If you have an energy $E_1 &amp;lt; E_F$ and an energy $E_2 &amp;gt; E_F$, and if $D(E_1) \not= 0$ and $D(E_2) \not= 0$, a photon can in principle be absorbed if it has energy $E_2 - E_1$. In a semiconductor, you have a gap in the band: All bands below the Fermi energy are filled, and all bands above the Fermi energy are empty, and the last occupied and the first unoccupied states are separated by a gap $E_g$.&#xA;&lt;img src=&quot;http://i.stack.imgur.com/uCaTU.png&quot; alt=&quot;Band diagram of a semiconductor&quot;&gt;&#xA;Hence, if an incident photon has energy lower than this gap, it will not be absorbed.&#xA;However, there are also a lot of other effects at work. On a macroscopic level, light interacts with a dielectric medium as characterized by the dielectric function $\varepsilon(\omega)$ via&#xA;$$D(\omega) = \varepsilon(\omega) \cdot E(\omega)$$ where $D$ is the dielectric displacement and $E$ is the electric field. The function can, in general, by complex. This gives rise to a complex index of refraction, where the imaginary part is the absorption coefficient. The reflection coefficient can be calculated from the real and the imaginary part. &#xA;A typical dielectric function looks like the picture below. Peeks in the imaginary part relate to absorption. So, if there are no such peaks in the visible spectrum, no absorption takes place and the material becomes transparent.&#xA;&lt;img src=&quot;http://i.stack.imgur.com/5PNYp.png&quot; alt=&quot;dielectric function&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strong reflectivity of metals is best explained by the free electron model. Since electrons in a model are relatively free to move, they can take on the oscillation of incident light waves and re-emit light of the same frequency for a wide range of frequencies.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-04-29T14:38:57.023" />
  <row Id="9277" PostTypeId="1" CreationDate="2011-04-29T14:50:37.727" Score="-1" ViewCount="253" Body="&lt;p&gt;I have a conceptual difficulty in understanding the electron spin. On the one hand, it is an experimental, observable feature of electrons. The problem is in understanding to what it belongs - to a bare electron or to an electron already coupled to the electromagnetic field. I think it is the latter. In other words, the electron spin characterizes a complicated object (a real system) rather than a &quot;free&quot; (non interacting) electron. It is like the total spin of a compound system. On the other hand, in theory we ascribe it to a &quot;bare&quot; electron solely. Do you think we have a conceptual problem here?&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-04-29T17:25:17.300" ClosedDate="2011-04-29T19:57:46.870" Title="Spin of an electron" Tags="&lt;quantum-electrodynamics&gt;&lt;spin&gt;&lt;electron&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="9278" PostTypeId="2" ParentId="9274" CreationDate="2011-04-29T15:24:40.883" Score="1" ViewCount="" Body="&lt;p&gt;Short answer to the title question: &lt;strong&gt;yes&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are both species of ionizing photons, and you use the same set of mechanism to detect either band. Typically hodoscopes (perhaps based on fiber and high gain photo-diodes), multi-channel plates, or silicon strip detectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, I suspect that &lt;a href=&quot;http://physics.stackexchange.com/questions/9274/can-one-see-gamma-rays-in-an-x-ray-detector/9275#9275&quot;&gt;anna is right&lt;/a&gt; when she says that the machines used to scan luggage at the airport are not designed, or tuned to detect low levels of radioactivity (say kBq) in the baggage. The geometry that is most favorable for fast planar scanning will not be good at imaging a point source, and the designed flux is probably pretty high, meaning that weak sources only tweak the results by a little bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen colloquia from people involved in designing large scale scanning machines for (shipping containers), and they but a great deal of thought and effort into the matter. Hand luggage is surely easier, but still non-trivial.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-04-29T15:24:40.883" CommentCount="2" />
  <row Id="9279" PostTypeId="2" ParentId="9262" CreationDate="2011-04-29T15:41:16.153" Score="1" ViewCount="" Body="&lt;p&gt;The first derivative exploits minor differences in the absorption curve that when taken  yields a chracateristic curve for a particular species of analysis.  The subsequent EPR spectra can be analyzed and can  provide information about material and the properties based on the hyper fine splitting and other details. &lt;/p&gt;&#xA;" OwnerUserId="3342" LastActivityDate="2011-04-29T15:41:16.153" />
  <row Id="9281" PostTypeId="2" ParentId="9277" CreationDate="2011-04-29T17:25:17.300" Score="2" ViewCount="" Body="&lt;p&gt;There is no conceptual problem. The bare electron field is a Dirac, spin-1/2 field, and so is the renormalized field. A bare electron carries $j=1/2$, and so does the physical, renormalized electron. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, because spin is a discrete quantum number, a continuous process such as renormalization (a process that depends either on a continuous coupling $e$ or the mass scale $\mu$) couldn't change the value of such a quantum number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the equal value of both spins is not only free of contradictions; it is inevitable.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-29T17:25:17.300" CommentCount="1" />
  <row Id="9282" PostTypeId="2" ParentId="9239" CreationDate="2011-04-29T17:28:26.843" Score="0" ViewCount="" Body="&lt;p&gt;I only know the basics.. I won't care to explain but look up the Tight Binding Model. Kittel explains it in this book. That should give you the basic explanation.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2011-04-29T17:28:26.843" CommentCount="1" />
  <row Id="9283" PostTypeId="2" ParentId="8426" CreationDate="2011-04-29T21:13:45.803" Score="0" ViewCount="" Body="&lt;p&gt;No. The Reeh-Schlieder theorem is a consequence of the Wightman axioms and of the Haag-Kastler axioms, both of which assume a Minkowski space, complete with the trivial metric structure, underlies everything. Variations of these axiomatic systems either allow or do not allow an analogue of the Reeh-Schlieder theorem to be derived. The idea that we might build space-time from quantum entanglement, while conceivable, presumably introduces either an axiomatic or some less well-defined mathematical structure, which will then either allow or not allow a Reeh-Schlieder-like theorem to be derived. Some thought this afternoon has suggested to me no way in which the Reeh-Schlieder theorem might of itself suggest a particular underlying mathematical structure. I take the lack of other Answers to suggest that no-one else immediately thought of such a thing either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally take the Reeh-Schlieder theorem to say that there is a sense in which &lt;em&gt;modulations&lt;/em&gt; of vacuum fluctuations are nonlocal, in a way that is weakly analogous to the nonlocality of analytic complex functions, for which the equivalent to the Reeh-Schlieder theorem is the possibility of analytic continuation. Analyticity, in a relatively sophisticated sense, is fundamental to the proof of the Reeh-Schlieder theorem. This is talk rather than mathematics, but to me it suggests the negative reply that I began with.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-04-30T11:13:18.477" LastActivityDate="2011-04-30T11:13:18.477" CommentCount="6" />
  <row Id="9284" PostTypeId="2" ParentId="9255" CreationDate="2011-04-29T21:25:17.917" Score="4" ViewCount="" Body="&lt;p&gt;We put the permittivity $\varepsilon=1$ to one from now on. Let us first rephrase the question a bit. Instead of starting from the potential &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Phi=\frac{1}{r} \qquad \mathrm{and} \qquad \Phi=\frac{1}{2r^2}, \qquad r\neq 0, $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;respectively, let us assume that the electric field has be given as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{E}=\frac{\vec{r}}{r^3} \qquad \mathrm{and} \qquad\vec{E}= \frac{\vec{r}}{r^4},\qquad r\neq 0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;respectively, and we want to know the charge density $\rho=\vec{\nabla} \cdot \vec{E}$, in particular, at the origin $r=0$. For $r\neq 0$, the charge density is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho=0 \qquad \mathrm{and} \qquad \rho=-\frac{1}{r^4},\qquad r\neq 0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;respectively. This rephrasing is just so we only have to differentiate one time instead of two times, but the argument is in principle the same, cf. footnote $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A minor issue is that the electric field $\vec{E}$ has not been specified at $r=0$. A way to make sense of this is to use distributions and test functions $f\in C^{\infty}_c(\mathbb{R}^3)$, i.e., infinitely often differentiable functions $f$ with compact support. We now declare that a smeared electric field $E^i$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E^i[f] = \int_{\mathbb{R}^3} d^3r \  E^i(\vec{r})f(\vec{r}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Lebesgue majorant of the integrand is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{|f(\vec{r})|}{r^2} \qquad \mathrm{and} \qquad \frac{|f(\vec{r})|}{r^3},\qquad r\neq 0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;respectively. Only the first case is Lebesgue integrable in $\mathbb{R}^3$. This is the heart of the problem. In the second case, even after smearing with a test function $f$, the electric field $E^i$ does not make sense as a distribution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In distribution theory, the derivative of a distribution is always defined by applying the derivative to the test function with a minus sign$^{1}$. If $E^i$ is a distribution, we can carry out the next step to define the derivative of $E^i$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\vec{\nabla} \cdot \vec{E}[f] =   \int_{\mathbb{R}^3} d^3r \ \vec{E}(\vec{r})\cdot\vec{\nabla} f(\vec{r}) =\lim_{\varepsilon\to 0} \int_{\{r\geq\varepsilon\}} d^3r \ \vec{E}(\vec{r})\cdot\vec{\nabla} f(\vec{r}) $$&#xA;$$ =\lim_{\varepsilon\to 0} \int_{\{r\geq\varepsilon\}} d^3r \left[ \vec{\nabla} \cdot\left( \vec{E}(\vec{r})f(\vec{r})\right)-f(\vec{r}) \underbrace{\vec{\nabla} \cdot \vec{E}(\vec{r})}_{=0}\right] $$&#xA;$$ =-\lim_{\varepsilon\to 0} \int_{\{r=\varepsilon\}} d^2\vec{A} \cdot \vec{E}(\vec{r})f(\vec{r})=-\lim_{\varepsilon\to 0} \int_{\{r=\varepsilon\}} \frac{d^2A}{\varepsilon^2}f(\vec{r}) =-4\pi f(0), $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where we performed the well-known manipulations in the first case to show that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec{\nabla} \cdot\frac{\vec{r}}{r^3}  = 4\pi\delta^3(\vec{r}). $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, so much for distribution theory and mathematical idealization. In reality, the $\frac{-1}{r^4}$ charge  density, $r\neq 0$, would break down as we approach the singularity $r\to 0$, so that we never get to ask: &lt;em&gt;What sits at $r=0$?&lt;/em&gt; This leads to the idea of regularization&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Phi_{\varepsilon}=\frac{1}{\sqrt{r^2+\varepsilon}} \qquad \mathrm{and} \qquad \Phi_{\varepsilon}=\frac{1}{2(r^2+\varepsilon)}, \qquad \varepsilon&amp;gt;0,$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;respectively. The regularized charge density $\rho_{\varepsilon}=-\vec{\nabla}^2\Phi_{\varepsilon}\in C^{\infty}(\mathbb{R}^3)$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho_{\varepsilon}=\frac{3\varepsilon}{(r^2+\varepsilon)^{\frac{5}{2}}} \qquad \mathrm{and} \qquad \rho_{\varepsilon}=\frac{3\varepsilon-r^2}{(r^2+\varepsilon)^{3}}, \qquad \varepsilon&amp;gt;0,$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;respectively. We can now smear with a test function $f$. It is straightforward to check that the first $\rho_{\varepsilon}$ becomes $4\pi\delta^3(\vec{r})$ as $\varepsilon\to0$, while the second $\rho_{\varepsilon}$ does not make sense as a distribution when $\varepsilon\to0$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^{1}$ In the second case, we could in principle define the smeared potential&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  \Phi[f] = \int_{\mathbb{R}^3} d^3r \ \Phi(\vec{r})f(\vec{r}),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because $\Phi=\frac{1}{2r^2}$, $r\neq 0$, is locally Lebesgue integrable in $\mathbb{R}^3$, and then define the electric field as a distribution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec{E}[f] =   \Phi[\vec{\nabla} f] = \int_{\mathbb{R}^3} d^3r \ \Phi(\vec{r})\vec{\nabla}f(\vec{r}),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\vec{\nabla} \cdot \vec{E}[f] =   \int_{\mathbb{R}^3} d^3r \ \Phi(\vec{r})\vec{\nabla}^2  f(\vec{r}) = \lim_{\varepsilon\to 0} \int_{\{r\geq\varepsilon\}} d^3r \ \Phi(\vec{r})\vec{\nabla}^2  f(\vec{r}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this mathematical construct is not as useful in practice as one might naively have hoped for. For instance, if we try to integrate by part, we essentially get back to the problem that the electric field $\vec{E}= \frac{\vec{r}}{r^4}$, $r\neq 0$, is not locally Lebesgue integrable in $\mathbb{R}^3$.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-04-30T13:15:09.657" LastActivityDate="2011-04-30T13:15:09.657" CommentCount="2" />
  <row Id="9286" PostTypeId="2" ParentId="9263" CreationDate="2011-04-30T02:08:01.103" Score="1" ViewCount="" Body="&lt;p&gt;In physics, the reflectivity of a surface is described by &lt;a href=&quot;http://en.wikipedia.org/wiki/Fresnel_equations&quot; rel=&quot;nofollow&quot;&gt;Fresnel Equations&lt;/a&gt;. One of the parameters of this equations is the refractive index which in general depends on the wavelength of light used. So surfaces that reflect visible light need not be good reflectors in another wavelength. A very good example of this is the phosphorescent coating that you find inside fluorescent lights. These coatings reflect visible light very efficiently (hence appear white to our eyes). But they absorb UV and re-emit it as visible light (this is essentially what happens when you switch the light on).&lt;/p&gt;&#xA;" OwnerUserId="966" LastActivityDate="2011-04-30T02:08:01.103" CommentCount="3" />
  <row Id="9287" PostTypeId="1" AcceptedAnswerId="9295" CreationDate="2011-04-30T02:48:05.440" Score="3" ViewCount="200" Body="&lt;p&gt;My question essentially revolves around multi-electron atoms and spectroscopic terms.  I understand the idea that the total wavefunction for Fermions should be antisymmetric.  Consider as an example, the $2p^2$ electrons in a partially filled p shell; that is, the outer shell of Carbon.  The two electrons both have $l=1$, and hence total orbital angular momentum takes the values:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L = L1+L2, L1+(L2-1),...,|L1-L2| = 0,1,2$&#xA;and &#xA;$S = 0,1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can sort of intuitively see that $L=2$ must refer to a symmetric spatial wavefunction and hence an antisymmetric spin wavefunction.  I can handwave and say that for $L=2$, we must have $m_{l1}=m_{l2}=\pm1$ and hence they must have opposing spin to satisfy PEP which gives S=0 - but I'm not sure how to express that in terms of an actual wavefunction and it seems to be a bit of a circular argument.  However, I don't see why $L=1$ must have $S=1$ (a triplet) and $L=0$, $S=0$ (another singlet). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone shed some light on this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3346" LastActivityDate="2011-05-02T03:37:28.467" Title="Relation between total orbital angular momentum and symmetry of the wavefunction" Tags="&lt;quantum-mechanics&gt;&lt;angular-momentum&gt;&lt;spectroscopy&gt;&lt;symmetry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="9288" PostTypeId="2" ParentId="9262" CreationDate="2011-04-30T02:55:38.933" Score="1" ViewCount="" Body="&lt;p&gt;Sorry I'm answering my own question, but a major figure in biological EPR popped in and informed me that the reason EPR spectra are taken as first derivatives is largely historical - old EPR spectrometers had large linearly field-dependent baseline shifts which could be eliminated easily by taking the first derivative, which changed the constantly increasing shift into a simple vertical offset on the plot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This said, Sam's answer hints at a secondary benefit of taking the first derivative, which is in amplifying minor asymmetries in the peaks.&lt;/p&gt;&#xA;" OwnerUserId="1876" LastActivityDate="2011-04-30T02:55:38.933" />
  <row Id="9289" PostTypeId="1" AcceptedAnswerId="9293" CreationDate="2011-04-30T06:53:10.370" Score="0" ViewCount="111" Body="&lt;p&gt;Does Coulomb's law applicable for Plasma substance?&lt;/p&gt;&#xA;" OwnerUserId="3347" LastEditorUserId="3347" LastEditorDisplayName="" LastEditDate="2011-04-30T08:22:46.810" LastActivityDate="2011-04-30T09:26:48.383" Title="Coulomb's law and Plasma" Tags="&lt;plasma-physics&gt;&lt;coulombs-law&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9290" PostTypeId="1" AcceptedAnswerId="9292" CreationDate="2011-04-30T08:17:18.077" Score="5" ViewCount="376" Body="&lt;p&gt;I once saw on TV that the moon is slowly drifting away from the earth, something like an inch a year. In relation to that the day on earth what also increase in time.&#xA;I wonder why is that?&lt;/p&gt;&#xA;" OwnerUserId="3341" LastEditorUserId="3341" LastEditorDisplayName="" LastEditDate="2011-05-03T10:10:40.877" LastActivityDate="2011-05-03T11:00:34.750" Title="Why does the moon drift away from earth?" Tags="&lt;gravity&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" />
  <row Id="9291" PostTypeId="2" ParentId="9290" CreationDate="2011-04-30T09:08:19.953" Score="1" ViewCount="" Body="&lt;p&gt;Schau mal unter &quot;Gezeitenkraft&quot; nach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google translation: Look for the &quot;tidal power&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-04-30T09:27:56.280" LastActivityDate="2011-04-30T09:27:56.280" CommentCount="3" />
  <row Id="9292" PostTypeId="2" ParentId="9290" CreationDate="2011-04-30T09:19:19.207" Score="5" ViewCount="" Body="&lt;p&gt;This &lt;a href=&quot;http://en.wikipedia.org/wiki/Moon&quot; rel=&quot;nofollow&quot;&gt;says it concisely&lt;/a&gt;, when describing the effect of tides:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Gravitational coupling between the Moon and the tidal bulge nearest the Moon acts as a torque on the Earth's rotation, draining angular momentum and rotational kinetic energy from the Earth's spin. In turn, angular momentum is added to the Moon's orbit, accelerating it, which lifts the Moon into a higher orbit with a longer period. As a result, the distance between the Earth and Moon is increasing, and the Earth's spin slowing down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In fewer words: it is the tides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I am copying from a comment:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To show the right sign, one must show that the orbital angular momentum of the Moon actually increases with the radius - despite the decreasing velocity as the function of the radius &#xA;  For a $1/r$ potential, $mv^2\propto m/r$ says $v\propto 1/\sqrt{r}$, so the angular momentum $L=rp=mrv=mr/\sqrt{r}\propto \sqrt{r}$ which increases with $r$. – Luboš Motl &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In addition I found this &lt;a href=&quot;http://books.google.gr/books?id=dw2GadaPkYcC&amp;amp;pg=PA826&amp;amp;lpg=PA826&amp;amp;dq=LOD%20and%20sun%20tides&amp;amp;source=bl&amp;amp;ots=TrE7BS2bZi&amp;amp;sig=KpuPYsGhQtqJzlQfRuRxC54w-jE&amp;amp;hl=el&amp;amp;ei=oli9TYO6Es-0hAfb6YTQBQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBcQ6AEwAA#v=onepage&amp;amp;q=LOD%20and%20sun%20tides&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;better link&lt;/a&gt; by googling.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-05-03T11:00:34.750" LastActivityDate="2011-05-03T11:00:34.750" CommentCount="11" />
  <row Id="9293" PostTypeId="2" ParentId="9289" CreationDate="2011-04-30T09:26:48.383" Score="2" ViewCount="" Body="&lt;p&gt;You will be enlightened by this &lt;a href=&quot;http://en.wikipedia.org/wiki/Plasma_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;article in wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Except near the electrodes, where there are sheaths containing very few electrons, the ionized gas contains ions and electrons in about equal numbers so that the resultant space charge is very small. We shall use the name plasma to describe this region containing balanced charges of ions and electrons.[4]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So plasma is mainly neutral, viewed from afar. Coulombs law applies within the plasma for each individual ion and electron, in a many body statistical conglomerate, a different phase of matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus it will depend on the problem you want to solve whether coulomb's law should be used explicitly, for example near electrodes, or conducting surfaces. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-04-30T09:26:48.383" />
  <row Id="9294" PostTypeId="2" ParentId="9287" CreationDate="2011-04-30T09:42:00.313" Score="4" ViewCount="" Body="&lt;p&gt;Dear Josh, the wave functions are only perfectly symmetric and/or perfectly antisymmetric - each of the factors is - in the case of two particles. As Fabian correctly says, for more than 2 particles, the wave function isn't perfectly symmetric and isn't perfectly antisymmetric with respect to particular transpositions of the two particles. The general character of the wave function's behavior is given by a Young diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why what you want to be proved can only be proved for two particles. In the text below, I will therefore assume that there are just two electrons. What you want to be proved is that if the total $L=L_1+L_2$ of the two electrons is even, the wave function is even under the exchange of the two electrons, and it's odd if $L$ is odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not hard to see. By rotational symmetry, the states with a given value of $L$ form a multiplet with $2L+1$ components because $L_z$ may go from $-L$ to $+L$ with the spacing one. Without the loss of generality, you may focus on the wave function with the maximum value of $L_3$, namely $L_z=L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get this maximum value, you need $L_z=L_{z,1}+L_{z,2}$ to be composed of two equally maximal, equally oriented terms. Because $L_{z,1}$ and $L_{z,2}$ also go between $-L_1$ and $+L_1$, or similarly between $-L_2$ and $+L_2$, it's clear that the only way to get $L_z=L$ is to have $L_{z,1}=L_1$ and $L_{z,2}=L_2$: both $z$-components must be maximum, too. But if it is so, then the wave function is simply&#xA;$$ \psi_{L_3=L} = \psi_{L_{z,1}=L_1} \otimes \psi_{L_{z,2}=L_2} $$&#xA;For $L_1=L_2$, you can easily see that this wave function has to be symmetric under the permutation of $1$ and $2$. After all, it is the tensor product of the two equal pieces. If you antisymmetrized it, you would get zero. And in fact, the total wave function cannot have any complete symmetry or complete antisymmetry under the exchange of the particles $1,2$ if $L_1\neq L_2$. It's because $L_1^2$ acting on the total wave function gives you $L_1(L_1+1)$ times the wave function, while $L_2^2$ acting on the total wave function gives you $L_2(L_2+1)$. Because the two eigenvalues are not equal for $L_1\neq L_2$, the total wave function can't be symmetric under the exchange of $1,2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, the question about the symmetry of the orbital wave function only has a sharp answer is there are two particles and if they have the same $L_1=L_2$ - and in that case, the function is symmetric under the permutation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, one can prove that the exchange of the two particles with the same half-integer spins $S_1=S_2$ produces a minus sign - assuming that $S_1=S_2$ differs from an integer by $1/2$. To do such things, it's useful to imagine that the components of the multiplet with a given $J$ are organized into a symmetric spintensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the $2J+1$ components of the multiplet with the angular momentum $J$ may be expressed as a completely symmetric &quot;tensor&quot; with 2-valued spinor indices,&#xA;$$ T_{abc\dots z} $$&#xA;Each index is either $0$ or $1$. The number of ones goes from $0$ to the number of indices $N$ - so there are $N+1$ components of this tensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because each index brings $1/2$ to the total angular momentum, it's clear that the number of indices is $N=2J$. Indeed, then we have $N+1=2J+1$ components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there were pairs of indices that are antisymmetric with respect to the exchange of the two indices, one could factorize $\epsilon_{ab}$, a totally antisymmetric object. So only the total symmetry is relevant for our simplest case. Now, the spintensor for an angular momentum $J=J_1+J_2$ object may be written as the symmetrization of&#xA;$$ T_{abc\dots z} = T^{1}_{(abc\dots m} T^{2}_{nop\dots z)} $$&#xA;where the parentheses represent the complete symmetrization - because the multiplet is represented by the totally symmetric tensor, as we said. The tensors $T^1$ and $T^2$ correspond to the $J_1$ and $J_2$ pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the total angular momentum $J$ differs by an odd number from its maximum value $J_1+J_2$, like in the case of the singlet where $J_1=J_2=1/2$ but $J=0$, then we must factorize those epsilons.&#xA;$$ T_{abc\dots z, \,\,{\rm missing\,\,}{mn}} = T^{1}_{(abc\dots l} T^{2}_{op\dots z)} \epsilon^{mn} $$&#xA;The epsilon was added to the right hand side to reduce the number of indices by two - i.e. the total spin by one. Because the epsilon is antisymmetric, it changes the symmetry of the whole $T$ under the exchange of the two groups of indices. Each time you reduce the total $J$ by one, the symmetry property changes from symmetry to antisymmetry or vice versa. So the sign obtained from the permutation is given, for $J_1=J_2$, by $(-1)^{J-J_1-J_2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also useful to know that the orbital wave functions of a single particle that can be expressed as spherical harmonics $Y_{lm}$ pick the factor of $(-1)^{l}$ under parity i.e. the factor of $(-1)^{l+m}$ under $\theta\to\pi-\theta$.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-30T09:42:00.313" CommentCount="1" />
  <row Id="9295" PostTypeId="2" ParentId="9287" CreationDate="2011-04-30T12:30:25.970" Score="2" ViewCount="" Body="&lt;p&gt;There are several ways of knowing what states should be there. For simple cases such as this, the easiest way is just by counting all the possibilities or micro-states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you have 2 &quot;equivalent&quot; electrons in $p^2$ (equivalent meaning that they share quantum numbers $n$ and $l$, related to the energy of the system) there are &#xA;$$\left( \begin{array}{l} 6 \\ 2 \end{array} \right) = 6 \cdot 5/2 = 15$$&#xA;micro-states (possible ways of assigning $n$, $l$, $m_l$ and $m_s$ to the outer (valence) electrons. Know you have to count all the possible (allowed by Pauli's principle) different arrangements of $m_{l,1}$, $m_{l,2}$, $m_{s,1}$, $m_{s,2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find them explicitly in any Physical Chemistry book (e.g. McQuarrie and Simon). Writing $m_{l,1}, m_{l,2}$ and $m_s$ omitted ($m_s = 1/2$) or with an over bar (for spin $m_s = -1/2$), the microstates are:&#xA;$(1,\bar{1}), (1,0),(1,\bar{0}), (1,-1),(1,-\bar{1})$, $(0,1),(0,\bar{1}), (0,\bar{0}), (0,-1),(0,-\bar{1})$, $(-1,1),(-1,\bar{1}), (-1,0),(-1,\bar{0}), (-1,-\bar{1})$&#xA;Now you have to group these states by characterising $L$ and $S$ (since, barring spin-orbit coupling, for a given $L$ and $S$ the $M_L$ and $M_S$ states form a manifold of degenerate states).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several ways of doing so, but I'll be sketchy to avoid lengthiness. For instance you can first think in those cases with $m_{s,1} \neq m_{s,2}$ where $m_{l,1}$ can be equal to $m_{l,2}$. Then $M_S = m_{s,1} + m_{s,2} = 0$ ($S$ can still be 1 -triplet or 0 - singlet). $M_L = m_{l,1} + m_{l,2} = 2,1,0$. So you can have states with $L = 2,1,0$. The states with $L = 2$ must be $S = 0$ since, as you realised, $m_{s,1}$ cannot be equal to $m_{s,2}$. Thus you identify 5 micro-states ($M_L = +L, \ldots, 0, \ldots -L$) corresponding to a single level $^1$D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the remaining 10 states you can clearly see from the listing of micro-states that your $L=1$ level is a triplet (you can find microstates with $M_S = 1$ and $M_L = 1$ so the remaining $M_S = 0,-1$ and $M_S = 0, -1$ have to be there too). For a $^3$P level there are $3\times 3 = 9$ microstates. Finally, the remaining micro-state must correspond to a $^1$S state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your question of the wave function. Again, think only on your last 2 electrons (it is not difficult to &quot;enlarge&quot; your Slater determinant with the other electrons). Each of your previous micro-states would correspond to a single Slater determinant. For example, for $(1,\bar{0})$ you would have the wave function&#xA;$$ \frac{1}{\sqrt{2}} \left| \begin{array}{cc} 2p_1(1)\alpha(1) &amp;amp; 2p_0(1)\beta(1) \\&#xD;&#xA;2p_1(2)\alpha(2) &amp;amp; 2p_0(2)\beta(2) \end{array} \right|$$&#xA;where $2p_{m}$ is the wave function in atomic orbital $2p_{m}$ and $\alpha$ and $\beta$ are the spin functions. Some states belonging to the levels with well-defined $L$ and $S$ correspond directly to the micro-states. For instance $(1,\bar{1})$ is exactly $|L=2,S=0,M_L=2,M_S=0\rangle$. However, most states must be written as linear combinations of Slater determinants. For instance, to the state $|L=2,M_L=1,S=0,M_S=0\rangle$ corresponds the wave function&#xA;$$\frac{1}{2} \left| \begin{array}{cc} 2p_1(1)\alpha(1) &amp;amp; 2p_0(1)\beta(1) \\&#xD;&#xA;2p_1(2)\alpha(2) &amp;amp; 2p_0(2)\beta(2) \end{array} \right| + \frac{1}{2} \left| \begin{array}{cc} 2p_0(1)\alpha(1) &amp;amp; 2p_1(1)\beta(1) \\&#xD;&#xA;2p_0(2)\alpha(2) &amp;amp; 2p_1(2)\beta(2) \end{array} \right|$$&#xA;whereas for the state $|L=2,M_L=0,S=0,M_S=0\rangle$ the wave function would be&#xA;$$\frac{1}{2\sqrt{3}} \left| \begin{array}{cc} 2p_1(1)\alpha(1) &amp;amp; 2p_{-1}(1)\beta(1) \\&#xD;&#xA;2p_1(2)\alpha(2) &amp;amp; 2p_{-1}(2)\beta(2) \end{array} \right| + \frac{1}{\sqrt{3}} \left| \begin{array}{cc} 2p_0(1)\alpha(1) &amp;amp; 2p_0(1)\beta(1) \\&#xD;&#xA;2p_0(2)\alpha(2) &amp;amp; 2p_0(2)\beta(2) \end{array} \right| + \frac{1}{2\sqrt{3}} \left| \begin{array}{cc} 2p_{-1}(1)\alpha(1) &amp;amp; 2p_{1}(1)\beta(1) \\&#xD;&#xA;2p_{-1}(2)\alpha(2) &amp;amp; 2p_{1}(2)\beta(2) \end{array} \right|$$&lt;/p&gt;&#xA;" OwnerUserId="3100" LastEditorUserId="3100" LastEditorDisplayName="" LastEditDate="2011-05-02T03:37:28.467" LastActivityDate="2011-05-02T03:37:28.467" CommentCount="3" />
  <row Id="9296" PostTypeId="1" CreationDate="2011-04-30T12:58:00.463" Score="4" ViewCount="204" Body="&lt;p&gt;We are all told at school water vapour and carbon dioxide are the top two greenhouse gases, and that they absorb thermal infra red radiation, trap heat and warm up the Earth. My question is how do they do that? Why can't Oxygen or Nitrogen or any other gas not absorb infra red radiation as well as water vapour or CO2?&lt;/p&gt;&#xA;" OwnerUserId="2057" LastActivityDate="2011-04-30T14:18:30.843" Title="How does carbon dioxide or water vapour absorb thermal infra red radiation from the sun?" Tags="&lt;atmosphere&gt;" AnswerCount="2" />
  <row Id="9297" PostTypeId="2" ParentId="9296" CreationDate="2011-04-30T14:16:24.667" Score="3" ViewCount="" Body="&lt;p&gt;The structure of water or CO2 or methane enables vibrations &#xA;which are accompanied by variations in dipole moments. &#xA;This is reason for absorption of IR radiation at &#xA;certain wavelenghts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen and Nitrogen and similar symmetric diatomic &#xA;molecules do not absorb IR at all, they lack dipole &#xA;moments in any state of vibrational deformation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &quot;trapping&quot; is not due to water or CO2, this is &#xA;done by the atmosphere as en entity, the absorbed &#xA;IR (coming from 6000 K sun) is  dissipated to the &#xA;entire atmospheric gases and they do not radiate &#xA;back either because they can't (Nitrogen , oxygen) &#xA;or because the temperature is too low. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-04-30T14:16:24.667" CommentCount="4" />
  <row Id="9298" PostTypeId="2" ParentId="9296" CreationDate="2011-04-30T14:18:30.843" Score="0" ViewCount="" Body="&lt;p&gt;As you can see on these absorption spectra for &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C7732185&amp;amp;Units=SI&amp;amp;Type=IR-SPEC&amp;amp;Index=1#IR-SPEC&quot; rel=&quot;nofollow&quot;&gt;$\textrm{H}_2\textrm{O}$&lt;/a&gt; and &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C124389&amp;amp;Units=SI&amp;amp;Type=IR-SPEC&amp;amp;Index=1#IR-SPEC&quot; rel=&quot;nofollow&quot;&gt;$\textrm{C}\textrm{O}_2$&lt;/a&gt;, both molecules have moderate to strong absorbtion in the mid-IR wavelengths, with the absorption of $\textrm{C}\textrm{O}_2$ extending out into the longer wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other molecules common in the atmosphere don't have such strong absorption at the wavelengths given off by thermal radiation. If you are asking why that is, I'm afraid I can't give you a very detailed answer, except to say that the absorption spectra of molecules (and atoms) is governed by quantum mechanics. Maybe somebody else can explain how they would be calculated from principles, but that is beyond my education.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-04-30T14:18:30.843" />
  <row Id="9299" PostTypeId="2" ParentId="8134" CreationDate="2011-04-30T14:57:11.593" Score="2" ViewCount="" Body="&lt;p&gt;For modeling of physical (and chemical) systems on quantum computer even 25-30 qubits would be already quite nice, see Lanyon, et al, “Towards Quantum Chemistry on a Quantum Computer”, Nature Chemistry 2, 106 - 111 (2009) (see also &lt;a href=&quot;http://arxiv.org/abs/0905.0887&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0905.0887&lt;/a&gt; )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really, quant-ph section in arXiv.org is standard place for papers about quantum computers, the paper from PRL you mentioned also may be found there (but seems I may post only one link).&lt;/p&gt;&#xA;" OwnerUserId="3351" LastActivityDate="2011-04-30T14:57:11.593" CommentCount="2" />
  <row Id="9300" PostTypeId="1" AcceptedAnswerId="9311" CreationDate="2011-04-30T15:03:37.047" Score="14" ViewCount="472" Body="&lt;p&gt;The Ising model is a well-known and well-studied model of magnetism. Ising solved the model in one dimension in 1925. In 1944, Onsager obtained the exact free energy of the two-dimensional (2D) model in zero field and, in 1952, Yang presented a computation of the spontaneous magnetization. But, the three-dimensional (3D) model has withstood challenges and remains, to this date, an outstanding unsolved problem.&lt;/p&gt;&#xA;" OwnerUserId="47" LastEditorUserId="47" LastEditorDisplayName="" LastEditDate="2011-04-30T17:15:51.550" LastActivityDate="2011-04-30T18:29:46.760" Title="Why is it hard to solve the Ising-model in 3D?" Tags="&lt;ising-model&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="9301" PostTypeId="2" ParentId="9290" CreationDate="2011-04-30T15:06:49.440" Score="-4" ViewCount="" Body="&lt;p&gt;The tidal effect was supposed to deccelerate the Earth rotation and the lost angular momentum should be transferred to the receding Moon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But facts go on the contrary: &lt;strong&gt;LOD - Length Of Day is decreasing&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moon-Earth distance displays an increase of 3.8 cm/yr of the semimajor axis of 384,399 km. ([Williams J.G. et al, 2008])&#xA;There is an &lt;strong&gt;increase of orbital radius at a ratio of $2H_{0}$ as modeled &lt;a href=&quot;http://arxiv.org/PS_cache/astro-ph/pdf/0208/0208365v1.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/strong&gt; (eqs 35 and 36).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a surprise to many.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT add (to address some concerns evident in the comments)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement 1: &lt;strong&gt;Established theory:&lt;/strong&gt; Any transfer of momentum is delayed only by the speed of gravity (the usual light 'c' speed). It happens in a gravitationally bound system, like the Moon-Earth, that the angular momentum is conserved.&lt;br&gt;&#xA;Statement 2: &lt;strong&gt;It is a widespread beleif&lt;/strong&gt; (a wish is not theory, Ok?): the Earth slows down the rotation due to tidal effects. The paper with the computations is ..., please?&lt;br&gt;&#xA;Statement 3: &lt;strong&gt;Facts:&lt;/strong&gt; We are recording the most precise direct measurements ever made:&lt;br&gt;&#xA; - The Moon is receding.&lt;br&gt;&#xA; - The LOD ( Length of Day ) is shorter and shorter, implying that the the Earth's angular momentum is increasing!&lt;br&gt;&#xA;&lt;strong&gt;A choice is mandatory because it exists an internal contradition between the Statements 1, 2 and 3.&lt;/strong&gt;&lt;br&gt;&#xA;My choice is that the tidal effect must be of a low order.&lt;br&gt;&#xA;The question remains: Why is the Moon receding?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All previous studies about the past, based in proxies, shall be ignored because the inconsistency exist now, and we must seek a credible answer to explain the present time.&lt;br&gt;&#xA;Those studies deserve a few words:&lt;br&gt;&#xA;Eclipses - The lack of correct time keeping in the past, and imprecisions in the recorded time and location of eclipses can put to trash those studies.&lt;br&gt;&#xA;Tides - the tide amplitude can be affected by several localized factors as coastal line configuration, how deep is the ocean here and there, etc,... Today, as in the past, the tide values are not uniform along the globe because the Earth is a dynamical system.  It is not enough to line up some values.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder why persists the spreading of the misconception about the LOD?&lt;br&gt;&#xA;(IMO, the unpleasant answer to this is out of the scope of PSE and I will keep it to myself)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Ql7BA.png&quot; alt=&quot;LOD - from maia.usno.navy.mil/lplot1.png &quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-04-30T23:34:20.813" LastActivityDate="2011-04-30T23:34:20.813" CommentCount="15" />
  <row Id="9302" PostTypeId="1" AcceptedAnswerId="9309" CreationDate="2011-04-30T15:49:45.440" Score="2" ViewCount="162" Body="&lt;p&gt;I think I saw in a video that if dark matter wasn't repulsive to dark matter, it would have formed dense massive objects or even black holes which we should have detected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, could dark matter be repulsive to dark matter? If so, what are the reasons? Could it be like the opposite pole of gravity that attracts ordinary matter and repulses dark matter?&lt;/p&gt;&#xA;" OwnerUserId="1927" LastActivityDate="2011-04-30T18:12:24.280" Title="Is dark matter repulsive to dark matter? Why?" Tags="&lt;gravity&gt;&lt;cosmology&gt;&lt;dark-matter&gt;&lt;general-relativity&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="9303" PostTypeId="1" CreationDate="2011-04-30T16:26:44.020" Score="2" ViewCount="133" Body="&lt;p&gt;In the Karlsruhe physics course one defines the term &quot;substance-like&quot; quantity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let my cite the definition from a paper by &lt;a href=&quot;https://oebv.at/sixcms/media.php/229/energy_forms_and_carriers.pdf&quot; rel=&quot;nofollow&quot;&gt;Falk, Herrmann and Schmid&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;There is a class of physical quantities whose characteristics are especially easy to visualize: those &lt;strong&gt;extensive&lt;/strong&gt; physical quantities &lt;strong&gt;to which a density can be assigned&lt;/strong&gt;. These include electric charge, mass, amount of substance (number of particles), and others. Because of the fundamental role these quantities play throughout science and because such quantities can be distributed in and flow through space, we give them a designation of their own: &lt;strong&gt;substance-like&lt;/strong&gt;.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there examples of extensive quantities, which are not substance-like? I think volume  is one example, since it seems to make no sense to assign a density to it, are there others?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the authors write that a quantity can only be conserved if it is substance-like, let my cite this from an other publication:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F. Herrmann, &lt;a href=&quot;http://www.physikdidaktik.uni-karlsruhe.de/kpk/english/KPK_Teacher.pdf&quot; rel=&quot;nofollow&quot;&gt;writes&lt;/a&gt;: &quot;It is important to make clear that the question of conservation or non-conservation only makes sense with substance-like quantities. Only in the context of substance-like quantities does it make sense to ask the question of whether they are conserved or not. The question makes no sense in the case of non-substance-like quantities such as field strength or temperature.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my second question is: &lt;strong&gt;Why&lt;/strong&gt; has a conserved quantity to be substance like? It would be great if one could give me a detailed explanation (or a counterexample if he thinks the statement is wrong).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there resources where the ideas cited above are introduced with some higher degree of detail and rigour?&lt;/p&gt;&#xA;" OwnerUserId="2740" LastEditorUserId="2740" LastEditorDisplayName="" LastEditDate="2011-04-30T16:40:03.927" LastActivityDate="2011-05-30T17:52:02.683" Title="Substance like quanties and conserved quantities, Karlsruhe physics course" Tags="&lt;education&gt;&lt;conservation-laws&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="9304" PostTypeId="2" ParentId="9300" CreationDate="2011-04-30T17:12:46.553" Score="0" ViewCount="" Body="&lt;p&gt;Two-dimensional theories simply have much more mathematical structure that makes many such models mathematically solvable - integrable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, in the long-distance limit, one obtains a scale-invariant theory that is typically conformally invariant as well. Two-dimensional conformal symmetry is - unlikely any higher-dimensional symmetry - infinite-dimensional. This fact plays a very important role in string theory which has 2-dimensional world sheets. In some sense, the 2D systems lead to &quot;infinitely many conserved quantities&quot; which often makes their physics solvable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perturbative string theory reflects much of the special mathematics that makes problems tractable in two dimensions. Needless to say, 1D systems may be as solvable as 2D systems or more so. Integrable - analytically solvable - systems have also included spin chains. All these things are parts of string theory in one way or another. On the other hand, problems in three or more dimensions are qualitatively harder and most of the questions about the Ising model in 3D and similar models are not analytically solvable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question is whether one may understand a model qualitatively. Of course, an analytic understanding gives one a superior tool to answer this question as well. When it doesn't, it's still possible to gain some qualitative understanding - numerically or by various approximation schemes - and the fact that as of 2011, it hasn't been done, is just a historical accident that is more likely because it's a difficult problem.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-30T17:12:46.553" CommentCount="7" />
  <row Id="9305" PostTypeId="2" ParentId="9302" CreationDate="2011-04-30T17:18:09.297" Score="2" ViewCount="" Body="&lt;p&gt;Dark matter surely has to carry a positive mass, and by the equivalence principle, all positive masses have to exert attractive gravity on other masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, from the viewpoint of phenomenological cosmology, we obviously want dark matter to attract itself. It has to attract visible matter because this is why dark matter was introduced in the first place: it helps to keep the stars in a galaxy even though they're orbiting more quickly than one would expect from the distribution of visible mass in the galaxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this reason, the force between dark matter and ordinary is surely attractive. The force between dark matter and dark matter has to be attractive, too. In fact, dark matter has played the dominant role in the structure formation - the creation of the initial non-uniformities that ultimately became galaxies, clusters of galaxies, and so on. The dark matter halos are larger than the visible parts of the galaxies: the visible stars arose as &quot;cherries on the pie&quot; near the centers of the dark matter halos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no doubt that the gravitational force between any pair of particle-like entities is attractive. This is linked to the positive mass i.e. positive energy - which is needed for stability of the vacuum (if there existed negative-energy states, the vacuum would decay into them spontaneously which would be catastrophic and is not happening) - and the basic properties of general relativity. In particular, there's a lot of confusion among the laymen whether antimatter has an attractive gravity. Yes, of course, the matter-antimatter and antimatter-antimatter gravitational forces are known to be attractive, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The non-gravitational forces between dark matter are almost certainly short-range forces. In particular, dark matter doesn't interact with electromagnetism, the only long-range non-gravitational force (mediated by a massless photon) we know - that's why it's dark (emits no light).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only repulsive force that arises in similar cosmological discussions is one due to dark energy - or the cosmological constant, to be more specific. Dark energy is something very different than dark matter. This force makes the expansion of the Universe accelerate and it is due to the negative pressure of dark energy which may be argued to cause this &quot;repulsive gravity&quot;. However, dark energy is not composed of any particles. It's just a number uniformly attached to every volume of space.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-30T17:18:09.297" CommentCount="1" />
  <row Id="9306" PostTypeId="2" ParentId="9303" CreationDate="2011-04-30T17:30:24.563" Score="1" ViewCount="" Body="&lt;p&gt;Concerning your first question, whether extensive quantities may fail to be substance-like, the answer is No. The very citation you quoted says that &quot;substance-like&quot; is a synonym that the authors chose for an &quot;extensive&quot; quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning your second question, yes, conserved quantities have to be substance-like i.e. extensive. It's because our world may be approximated by a field theory governed by the action which is an integral of the Lagrangian density over spacetime.&#xA;$$ S = \int d^d x \, {\mathcal L} $$&#xA;Conserved quantities are associated with symmetries and Noether's procedure applied to all theories described by such an action leads to a formula for the conserved quantity that is an integral over space, e.g.&#xA;$$ J = \int d^{d-1} x\, j_0  $$&#xA;That means that $J$ is extensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if one could show that despite all the positive experiments of the last centuries, physics does &lt;em&gt;not&lt;/em&gt; take place in space or it is not local, the argument above could break down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also want to mention that the substance-like character of the conserved quantities is kind of obvious. It just says that the conserved things must be &quot;somewhere&quot; in space. To mention a particular example of quantities that can't be conserved, it's the &quot;intensive&quot; quantities such as temperature and pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, they can't be conserved because they're not even well-defined in a generic situation - different parts of the physical system have different pressures. If two similar objects have different temperatures, the &quot;natural&quot; other temperatures we can construct out of these two is (approximately) the average, rather than the sum, of the original two temperatures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that addition - something we do with conserved quantities - isn't appropriate for intensive quantities. Instead, averaging is.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-04-30T17:30:24.563" CommentCount="2" />
  <row Id="9307" PostTypeId="1" AcceptedAnswerId="9433" CreationDate="2011-04-30T17:45:13.670" Score="2" ViewCount="135" Body="&lt;p&gt;Background:&#xA;I'm on the pledge drive committee for a non-profit group, and there is an issue that is always contentious within this group; thankfully the group is amicable, although competitive, about the issue.  So, we want to develop a computer-based equal-arm, weighing scale (e.g., the kind Lady Justice holds).  During our pledge event, individuals will be able to donate to one side of the issue or the other and see the scale change accordingly.  None of us on the pledge committee are well-versed in physics, so we aren't familiar with, and haven't been able to find, a formula for coding the angle of the scale in relation to the current donations.  As a non-physicist, this seems like an equation that would have been identified, but we just don't know where to find it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a standard formula for the angle of the beam, in relation to the fulcrum, based on the relative 'weights' on each side?  If so, what is it or where can we learn about it?  If not, are there formulas that we could use as a basis for our simulation?&lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-04-30T18:26:11.580" LastActivityDate="2011-05-04T00:12:06.557" Title="Formula for implementing a simulated weighing scale" Tags="&lt;gravity&gt;&lt;mechanics&gt;&lt;statics&gt;&lt;formula&gt;" AnswerCount="3" />
  <row Id="9309" PostTypeId="2" ParentId="9302" CreationDate="2011-04-30T18:12:24.280" Score="5" ViewCount="" Body="&lt;p&gt;Lubos Motl's answer is exactly right. Dark matter has &quot;ordinary&quot; gravitational properties: it attracts other matter, and it attracts itself (i.e., each dark matter particle attracts each other one, as you'd expect).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's true that dark matter doesn't seem to have collapsed into very dense structures -- that is, things like stars and planets. Dark matter does cluster, collapsing gravitationally into clumps, but those clumps are much larger and more diffuse than the clumps of ordinary matter we're so familiar with. Why not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer seems to be that dark matter has few ways to dissipate energy. Imagine that you have a diffuse cloud of stuff that starts to collapse under its own weight. If there's no way for it to dissipate its energy, it can't form a stable, dense structure. All the particles will fall in towards the center, but then they'll have so much kinetic energy that they'll pop right back out again. In order to collapse to a dense structure, things need the ability to &quot;cool.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ordinary atomic matter has various ways of dissipating energy and cooling, such as emitting radiation, which allow it to collapse and not rebound. As far as we can tell, dark matter is weakly interacting: it doesn't emit or absorb radiation, and collisions between dark matter particles are rare. Since it's hard for it to cool, it doesn't form these structures.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-04-30T18:12:24.280" CommentCount="4" />
  <row Id="9310" PostTypeId="2" ParentId="9307" CreationDate="2011-04-30T18:25:47.477" Score="2" ViewCount="" Body="&lt;p&gt;To a first approximation, the scale is either balanced or not balanced, and remains so at all angles, so there is no formula that does what you're looking for.  A simple model of a scale says that the angle should be either&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;whatever angle you want if the weights on both sides are equal&lt;/li&gt;&#xA;&lt;li&gt;tipped all the way over to the heavier side if the weights are unequal&lt;/li&gt;&#xA;&lt;/ol&gt;   &#xA;&#xA;&lt;p&gt;This assumes that the pans are equally-spaced from the fulcrum and the fulcrum is an idealized point and the scale is rigid and the masses are the same shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can try this for yourself easily with an experiment I did just now.  Balance a spoon on your finger.  Put a penny in the bowl of the spoon.  The spoon will tip all the way over, the penny will fall off, and the spoon falls off your finger.  (It did for me at least - your results may vary depending on how ideal spoons and fingers are where you live.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to modify this using a physics concept would be to move the fulcrum towards the heavier side.  A scale is balanced when&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L_1w_1 = L_2w_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $L_1$ and $L_2$ the length of the sides of the scale from the fulcrum and $w_1$ and $w_2$ the two weights.  For example, if you put a 2 kg weight 10cm from the fulcrum to the right, it can be balanced by a 1 kg weight on the left if you put it 20cm away.  This is the same principle a parent can use to play see-saw with their child.  The physics term is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Lever&quot; rel=&quot;nofollow&quot;&gt;lever&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to see where this equation comes from is by balancing the &lt;a href=&quot;http://en.wikipedia.org/wiki/Torque&quot; rel=&quot;nofollow&quot;&gt;torques&lt;/a&gt; exerted by the weights. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another is to consider energy.   The energy of the weights comes from the formula &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U = w h$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U$ is the potential energy, $w$ is the weight, and $h$ is the height above ground.  A basic principle of statics is that the scale will be balanced whenever this &lt;a href=&quot;http://en.wikipedia.org/wiki/Potential_energy&quot; rel=&quot;nofollow&quot;&gt;potential energy&lt;/a&gt; stays the same as you change the angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose one arm is twice as long as the other.  Then when you tilt the scale a little, that side will drop twice as far, so $h$ changes by twice as much for the long side as for the short side.  That means that to have the scale balanced $w$ must be half as much for the long side.  That way the change in $h$ is twice as much but $w$ is half at much, so when you multiply them to get the energy, you find the same thing for both sides.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if with the 2kg and 1kg weights described above, the 1kg weight is on the long side.  If it falls 1cm, then 2kg weight will  rise 1/2 cm.  Multiplying the change in height by the weight gives the same for both.  (The weight is technically gravitational acceleration times the mass, so the weights are 10 Newtons and 20 Newtons because gravitational acceleration is 10 m/s^2.) This is true regardless of the angle, so the scale is either balanced or unbalanced at all angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To read some more, the scale is a simple example of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Potential_energy&quot; rel=&quot;nofollow&quot;&gt;lever&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-04-30T18:25:47.477" CommentCount="2" />
  <row Id="9311" PostTypeId="2" ParentId="9300" CreationDate="2011-04-30T18:29:46.760" Score="6" ViewCount="" Body="&lt;p&gt;There is a result I only heard about recently: it has been proven that computing partition functions for the Ising-model in dimensions &gt; 2 is NP-complete. (The paper can be found at &lt;a href=&quot;http://www.cs.brown.edu/people/sorin/pdfs/Ising-paper.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.cs.brown.edu/people/sorin/pdfs/Ising-paper.pdf&lt;/a&gt;; a more readable one is here &lt;a href=&quot;http://www.siam.org/pdf/news/654.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.siam.org/pdf/news/654.pdf&lt;/a&gt; - both can be found on the Wikipedia on the Ising model). I'm far from an expert on this, but the main idea is that a certain NP-complete graph theory problem on finding maximal sets of edges can be mapped to ground states of Ising-3D. Roughly, this means that you can't find ground states in polynomial time, and as most physicists know, if the difficulty of your problem scales exponentially, solving something exactly for large systems quickly becomes impossible.&lt;/p&gt;&#xA;" OwnerUserId="1471" LastActivityDate="2011-04-30T18:29:46.760" CommentCount="5" />
  <row Id="9312" PostTypeId="2" ParentId="9307" CreationDate="2011-04-30T19:02:06.307" Score="2" ViewCount="" Body="&lt;p&gt;O sancta simplicitas! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A balance is a pendulum, but a very short one! The function is given by having the central bearing a tiny bit higher than the bearings for the scales. This means, they are not on a staight line, but making a flat triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one knows the center of gravity of the beam (inluding the pointer!) and the distance of that center from the central bearing, one can calculate the angle of the beam. &#xA;(elementary vector calculation) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a computer-simulated balance all physics &quot;reality&quot; is not necessary, just programm a angle/donation ratio. In a luxury version You can add some swing motion (some seconds swing time)  with undercritical damping, this makes the motion of the balance look &quot;real&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-04-30T19:27:24.933" LastActivityDate="2011-04-30T19:27:24.933" />
  <row Id="9313" PostTypeId="1" CreationDate="2011-04-30T21:29:23.780" Score="1" ViewCount="85" Body="&lt;p&gt;I've been doing an experiment using liquid nitrogen, and am aware that pouring it into different containers gives a chance for other gases to liquify and contaminate it; what's a good estimate for the amount of other gases in the flask in such a situation?&lt;/p&gt;&#xA;" OwnerUserId="3356" LastActivityDate="2011-04-30T21:29:23.780" Title="How 'pure' is liquid nitrogen?" Tags="&lt;experimental-physics&gt;&lt;phase-transition&gt;&lt;inert-gases&gt;" />
  <row Id="9314" PostTypeId="1" AcceptedAnswerId="9324" CreationDate="2011-04-30T21:55:24.230" Score="2" ViewCount="437" Body="&lt;p&gt;Why is the speed of light defined as 299792458 m/s? Why did they choose that number and no other number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or phrased differently:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is a metre 1/299792458 of the distance light travels in a sec in a vacuum?&lt;/p&gt;&#xA;" OwnerUserId="3290" LastEditorUserId="3290" LastEditorDisplayName="" LastEditDate="2011-04-30T22:01:31.153" LastActivityDate="2011-05-03T04:11:32.527" Title="Why is the speed of light defined as 299792458 m/s?" Tags="&lt;speed-of-light&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="9315" PostTypeId="2" ParentId="9314" CreationDate="2011-04-30T22:33:31.420" Score="10" ViewCount="" Body="&lt;p&gt;The number is arbitrary.  We could have chosen it to be 1 m/s, but that would have been a little awkward, as we'd be going around telling people to go left at the next corner .00000084 meters down the road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It used to be that the meter was defined in terms of two scratches on a stick.  Therefore, we could measure the speed of light experimentally.  Before redefining things t&lt;a href=&quot;http://en.wikipedia.org/wiki/Speed_of_light#Increased_accuracy_of_c_and_redefinition_of_the_metre&quot; rel=&quot;nofollow&quot;&gt;hey measured as accurately as they could&lt;/a&gt; and then used the result as the new official standard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Defining the speed of light means that the meter is no longer defined in terms of scratches on a stick.  If it were, we'd have the second, the meter, and the speed of  light all defined independently, while really they are dependent.  That means that future, more accurate measurements could show that our system was inconsistent.  As a result, the meter is now simply the distance that light moves in 1/299792458 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the main motivation behind this is that speed of light is always &lt;a href=&quot;http://physics.stackexchange.com/questions/2230/why-and-how-is-the-speed-of-light-constant&quot;&gt;the same everywhere&lt;/a&gt; (for a local measurement) and is &lt;a href=&quot;http://en.wikipedia.org/wiki/Speed_of_light#Measurement&quot; rel=&quot;nofollow&quot;&gt;fairly easy to measure&lt;/a&gt;.  Before the redefinition, the accuracy in the measurement of the speed of light was getting to be so good that &lt;a href=&quot;http://en.wikipedia.org/wiki/Speed_of_light#Increased_accuracy_of_c_and_redefinition_of_the_metre&quot; rel=&quot;nofollow&quot;&gt;it was limited by the accuracy with which we could measure a meter&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many precise distance measurements actually are made using light rather than meter sticks.  See, for example, &lt;a href=&quot;http://en.wikipedia.org/wiki/LIDAR&quot; rel=&quot;nofollow&quot;&gt;LIDAR&lt;/a&gt;, in which light is bounced off a source, and you see how long it takes to come back.   Precise distance measurements (or measurements of change in a distance) can also be made with light using interferometry.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/LIGO&quot; rel=&quot;nofollow&quot;&gt;LIGO&lt;/a&gt; for example.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-04-30T22:33:31.420" />
  <row Id="9316" PostTypeId="2" ParentId="8426" CreationDate="2011-04-30T22:38:56.860" Score="1" ViewCount="" Body="&lt;p&gt;No, as Peter said, you are jumping the gun: A profound precondition of Reeh-Schlieder theorems is that spacetime is a smooth Lorentzian manifold. You can find a lot of information on the nLab:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ncatlab.org/nlab/show/Reeh-Schlieder+theorem&quot; rel=&quot;nofollow&quot;&gt;Reeh-Schlieder theorem (nLab)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most pedagogical exposition I know is the paper by Summers referenced there:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Stephen J. Summers: &quot;Yet More Ado About Nothing: The Remarkable Relativistic Vacuum State&quot; (&lt;a href=&quot;http://de.arxiv.org/abs/0802.1854v2&quot; rel=&quot;nofollow&quot;&gt;arXiv&lt;/a&gt;.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There is also a proof on that page for the vacuum representation in Minkowski spacetime based on the Haag-Kastler axioms, which shows that the proof is basically mathematical gynmnastics starting with the axioms, that invokes some heavy machinery like the SNAG-theorem, the edge of the wedge theorem and a Lesbesgue dominated convergence theorem for spectral integrals (as you can see: all tools from calculus). The proof is mainly there to illustrate that the axiom of locality is not needed in the first part of the proof, as claimed by Halvorson in the paper referenced on the page. So, this version of the Reeh-Schlieder follows directly from the axioms :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In curved spacetime you have to replace the axioms that use the Poincarè group with versions that make use of the local structure only, which can be done, but again this relies heavily on the structure of a smooth manifold. (BTW: QFT on curved spacetimes has produced a lot of results that are very important for quantum gravity research, beside beeing useful by providing a different context for QFT in flat spacetimes.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If and how any of this will be useful for future theories time will tell, but I think that understanding the properties of the vacuum state in AQFT as explained by Summers in his paper should be very useful for anyone working in QFT or quantum gravity. For example this theorem is the clearest explanation I know of, of how and why Einstein causality and quantum entanglement are complementary concepts. Locality in QFT is one of the most profound, difficult and misunderstood concepts...&lt;/p&gt;&#xA;" OwnerUserId="3357" LastActivityDate="2011-04-30T22:38:56.860" CommentCount="2" />
  <row Id="9317" PostTypeId="1" CreationDate="2011-04-30T22:58:10.657" Score="1" ViewCount="115" Body="&lt;p&gt;With Geostationary orbit your limited to the equator and ~36000km. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be possible to orbit the sun at the same speed as Earth, and then in basically the same place so that the distance to the Sun is slightly less or more than Earth, and the inclination is slightly less or more, in order to achieve a geostationary effect over any point at any height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wouldn't the Earth's gravity interfere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The reason I ask is, would this be a way of creating a space elevator?)&lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2011-05-01T00:46:54.490" Title="Is it possible to orbit the sun next to Earth?" Tags="&lt;astrophysics&gt;&lt;orbital-motion&gt;" AnswerCount="2" />
  <row Id="9318" PostTypeId="2" ParentId="9317" CreationDate="2011-04-30T23:03:17.893" Score="7" ViewCount="" Body="&lt;p&gt;It took me several tries to properly understand the question.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You may be thinking about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrangian_point&quot; rel=&quot;nofollow&quot;&gt;Lagrange points&lt;/a&gt;. There are a set of equilibrium points associated with every pair of bodies in simple orbits around each other where third (light) bodies may be placed. The ones you are looking for are L1 and L2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are however,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Not stable, which means you would need active station-keeping. (L4 and L5 &lt;em&gt;are&lt;/em&gt; stable, but they make point (3) below much &lt;em&gt;much&lt;/em&gt; worse.)&lt;/li&gt;&#xA;&lt;li&gt;Not usable for building a space elevator in the Earth--Sun system, because the moon will crash into the cable from time to time.&lt;/li&gt;&#xA;&lt;li&gt;Much further away than a geo-stationary orbit.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you do not mean the Lagrange points, then the short answer is &lt;strong&gt;no&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every thing in orbit around the Earth is also in orbit round the sun. (Indeed, the gravitational force exerted by the Earth on the Moon is considerably less than that exerted by the sun.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the fact that an object orbits the sun does not prevent it from being acted upon gravitationally by the Earth, so it must also be in orbit around the Earth if the situation is to be stable.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-01T00:46:54.490" LastActivityDate="2011-05-01T00:46:54.490" CommentCount="6" />
  <row Id="9319" PostTypeId="1" AcceptedAnswerId="9329" CreationDate="2011-04-30T23:32:31.410" Score="6" ViewCount="472" Body="&lt;p&gt;How many $fb^{-1}$ of integrated luminosity at the 7 Tev LHC do physicists expect are needed, to make a $5\sigma$ discovery of the most likey 115 Gev Higgs, if it exists?&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="1345" LastEditorDisplayName="" LastEditDate="2011-05-01T11:35:53.497" LastActivityDate="2011-05-01T11:35:53.497" Title="How many $fb^{-1}$ for the most likely $5\sigma$ 115 Gev Higgs at the 7 Tev LHC?" Tags="&lt;large-hadron-collider&gt;&lt;higgs-boson&gt;" AnswerCount="1" />
  <row Id="9320" PostTypeId="2" ParentId="9317" CreationDate="2011-05-01T00:39:00.283" Score="1" ViewCount="" Body="&lt;p&gt;Satellites orbiting the Earth, including geostationary satellites, already are orbiting the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To build a geostationary satellite, you need to be about six Earth radii from the Earth.  A space elevator built there would feel some &lt;a href=&quot;http://en.wikipedia.org/wiki/Tidal_force&quot; rel=&quot;nofollow&quot;&gt;tidal effects&lt;/a&gt; from the sun.  Other than those tidal effects, the sun's gravity is already accounted for when we talk about satellites orbiting Earth.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-05-01T00:39:00.283" />
  <row Id="9321" PostTypeId="1" CreationDate="2011-05-01T03:54:40.497" Score="0" ViewCount="122" Body="&lt;p&gt;The principle of equivalence - that, locally, you can't distinguish between a uniform gravitational field and a noninertial frame accelerating in the sense opposite to the gravitational field - is dependent on the equality of gravitational and inertial mass. Is there any deeper reason for why this equality of &quot;charge corresponding to gravitation&quot; (that is, the gravitational mass) and the inertial mass (that, in newtonian mechanics, enters the equation F=ma) should hold? While it has been observed to be true to a very high precision, is there any theoretical backing or justification for this? You could, for example (i wonder what physics would look like then, though), have the &quot;charge corresponding to electromagnetic theory&quot; equal to the the inertial mass, but that isn't seen to be the case.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2011-05-01T05:08:42.197" Title="Is there a fundamental reason why gravitational mass is the same as inertial mass?" Tags="&lt;general-relativity&gt;&lt;gravity&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="9322" PostTypeId="2" ParentId="9321" CreationDate="2011-05-01T04:40:47.767" Score="2" ViewCount="" Body="&lt;p&gt;This is a tricky question.  Fundamentally, this is the motivation of general relativity (and all metric theories of gravity)--if all masses interact with a gravitational field in the same way, then, in a sense, the motion of a particular mass is determined by the local gravitational field, independently of the mass.  This then leads you into explaining the gravitational &quot;force&quot; as an emergent property of the local spacetime curvature.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then, what came first?  The explanation of gravity as curvature, or the equivalence of gravitational and inertial mass?  In a way, they are just dual pictures of the same thing.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-05-01T04:40:47.767" />
  <row Id="9323" PostTypeId="2" ParentId="9321" CreationDate="2011-05-01T05:08:42.197" Score="0" ViewCount="" Body="&lt;p&gt;The fact that the value of inertial mass and gravitational mass of a particle are same is the result of the &quot;equivalence principle&quot; which is an essential principle for any metric theory of gravity like general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were different, then the whole edifice of GR will fall down like a house of cards. Fortunately all experiments up to now have confirmed that they are same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the only deeper significance of this fact can be attributed to the fact that gravity is not a &quot;force&quot; like other forces but a consequence of the fact that geometry of the spacetime is not static and unchanging but it is a dynamic arena, responding to the presence of mass/energy.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-05-01T05:08:42.197" />
  <row Id="9324" PostTypeId="2" ParentId="9314" CreationDate="2011-05-01T05:09:44.210" Score="11" ViewCount="" Body="&lt;p&gt;The speed of light is 299 792 458 m/s because people used to define one meter as 1/40,000,000 of the Earth's meridian - so that the circumference of the Earth was 40,000 kilometers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, they used to define one second as 1/86,400 of a solar day so that the day may be divided to 24 hours each containing 60 minutes per 60 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our Universe, it happens to be the case that light is moving by speed so that in 1 second defined above, it moves approximately by 299,792,458 meters defined above. In other words, during one solar day, light changes its position by&#xA;$$ \delta x = 86,400 \times 299,792,458 / 40,000,000\,\,{\rm circumferences\,\,of\,\,the\,\,Earth}.$$&#xA;The number above is approximately 647,552. Try it. Instruct light to orbit along the surface of the Earth and you will find out that in between two noons, it will complete 647,552 orbits. Why it is exactly this number? Well, it was because how the Earth was created and evolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it didn't hit a big rock called Megapluto about 4,701,234,567.31415926 years ago, it would have been a few percent larger and it would be rotating with a frequency smaller by 1.734546346 percent, so 647,552 would be replaced by 648,243.25246 - but because we hit Megapluto, the ratio eventually became what I said. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There were about a million of similarly important big events that I skip, too.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth's size and speed of spinning were OK for a while but they're not really regular or accurate so people ultimately switched to wavelengths and durations of some electromagnetic waves emitted by atoms. Spectroscopy remains the most accurate way in which we can measure time and distances. They chose the new meter and the new second as a multiple of the wavelength or periodicity of the photons emitted by various atoms - so that the new meter and the new second agreed with the old ones - those defined from the circumference of the Earth and from the solar day - within the available accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some time, people would use two independent electromagnetic waves to define 1 meter and 1 second. In those units, they could have measured the speed of light and find out that it was 299,792,458 plus minus 1.2 meters per second or so. (The accuracy was not that great for years, but the error of 1.2 meters was the final accuracy achieved in the early 1980s.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the speed of light is so fundamental - adult physicists use units in which $c=1$, anyway - physicists decided in the 1980s to redefine the units so that both 1 meter and 1 second use the same type of electromagnetic wave to be defined. 1 meter was defined as 1/299,792,458 of light seconds which, once again, agreed with the previous definition based on two different electromagnetic waves within the accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantage is that the speed of light is known to be accurately, by definition, today. Up to the inconvenient numerical factor of 299,792,458 - which is otherwise convenient to communicate with ordinary and not so ordinary people, especially those who have been trained to use meters and seconds based on the solar day and meridian - it is as robust as the $c=1$ units.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-01T13:17:25.253" LastActivityDate="2011-05-01T13:17:25.253" />
  <row Id="9325" PostTypeId="1" AcceptedAnswerId="9330" CreationDate="2011-05-01T05:13:53.503" Score="9" ViewCount="290" Body="&lt;p&gt;Mach's principle says that it is impossible to tell if something is accelerating unless there is something else in the universe to compare that motion to, which seems reasonable. However, if you had one detector in the universe, you seem to be able to tell if it is accelerating because an accelerating detector would record radiation where a non-accelerating detector would not, due to the Unruh effect. So, my question is, does the Unruh effect provide a way to tell if something is accelerating, even if it is the only thing in otherwise &quot;empty&quot; space, thereby violating Mach's principle? (At least Mach's principle in its form stated above.)&lt;/p&gt;&#xA;" OwnerUserId="1040" LastActivityDate="2011-08-13T00:02:30.903" Title="Does the Unruh effect violate Mach's principle?" Tags="&lt;quantum-field-theory&gt;&lt;general-relativity&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="4" />
  <row Id="9326" PostTypeId="1" CreationDate="2011-05-01T00:16:43.850" Score="2" ViewCount="176" Body="&lt;p&gt;Ok, I am stumped by this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the system in the diagram below, block M (15.7 kg) is initially moving to the left. A force F , with magnitude 60.5 N, acts on it directed at an angle of 35.0 degrees above the horizontal as shown. The mass m is 8.2 kg. There is no friction and the pulley and string are massless. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/r3W4t.jpg&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the normal force (N) on M ?&lt;/li&gt;&#xA;&lt;li&gt;What is the tension (N) in the string?&lt;/li&gt;&#xA;&lt;li&gt;What is the acceleration (m/s2) of M (positive is to the right)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I tried and got the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Got the X component of the force F: F * Cos (35) = 49.55869868 N&lt;br&gt;&#xA;Got the Y component of the force F: F * Sin (35) = 34.7013744 N&lt;br&gt;&#xA;Got the force of m on M: 8.2 * 9.8 = 80.36 N (which is what I assumed the tension on the string was)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the acceleration of M caused by m = -5.118471338&lt;br&gt;&#xA;Then I got the acceleration of M cause by F = 3.156605011  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting it all together I got the following answers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;129 N&lt;/li&gt;&#xA;&lt;li&gt;80.4 N&lt;/li&gt;&#xA;&lt;li&gt;-1.96 m/s^2&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I (obviously) got all three wrong.  The correct answers are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;119 N&lt;/li&gt;&#xA;&lt;li&gt;69.8 N&lt;/li&gt;&#xA;&lt;li&gt;-1.29 m/s^2&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, I'm stuck.  There is something missing that is causing my string tension to be off.  If I would have gotten the right string tension, all the other answers would have been correct.  I know that is where the problem is, so if someone could point out my mistake, that would be wonderful!&lt;/p&gt;&#xA;" LastActivityDate="2011-05-01T05:17:32.260" ClosedDate="2011-05-01T14:04:13.107" Title="Physics: Tension on a string from a system" Tags="&lt;homework&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="9327" PostTypeId="2" ParentId="9326" CreationDate="2011-05-01T00:37:55.697" Score="0" ViewCount="" Body="&lt;p&gt;If m is accelerating, the tension in the string is not $mg$, but $m(g-a)$ where $a$ is the acceleration.  If there were no string, there would be no tension, and the mass would be accelerating at $g=9.8 m/sec^2$.  So you need to calculate the total acceleration on M, which will also be the acceleration on m, then reduce the string tension appropriately.&lt;/p&gt;&#xA;" OwnerUserId="3257" LastActivityDate="2011-05-01T00:37:55.697" />
  <row Id="9328" PostTypeId="2" ParentId="9326" CreationDate="2011-05-01T01:10:51.667" Score="1" ViewCount="" Body="&lt;p&gt;For 1, you should have an normal force of $Mg −F \sin \theta $ which being positive means that the large mass does not move vertically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 3, you know the net accelerating force (upwards for the small mass or to the right for the large mass) on the two masses combined is $F \cos \theta - mg$, and you know the combined mass $m+M$ so you can calculate the acceleration $a$ in the same directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 2, your answer to 3 has shown that the small mass is accelerating downwards by less than $g$ (or upwards by more than $-g$), so there is an upward force on the small mass caused by tension in the rope, and this is (watch out for signs) $m(a-(-g))$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you get the arithmetic correct this should give you the right answers.&lt;/p&gt;&#xA;" OwnerUserId="2714" LastActivityDate="2011-05-01T01:10:51.667" />
  <row Id="9329" PostTypeId="2" ParentId="9319" CreationDate="2011-05-01T05:23:13.000" Score="6" ViewCount="" Body="&lt;p&gt;Dear John, your question is answered by this graph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/RmFHL.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was taken from &lt;a href=&quot;http://www.science20.com/quantum_diaries_survivor/lhc_will_run_7_tev_2011_and_2012-75788&quot; rel=&quot;nofollow&quot;&gt;Tommaso Dorigo's blog&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 115 GeV Higgs is the very left beginning of all the lines in the graph. The full lines are the 2x 3.5 TeV beams; the dotted lines describe the 2x 4 TeV beams that were planned as a small upgrade of the energy for a while but the plans were abolished and 7 TeV was kept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see that with 5 inverse femtobarns at 7 TeV, you only reach about 2.7 sigma (the left beginning of the full blue line). Because the significance in standard deviations only goes as $\sqrt{N}$ where $N$ is the number of collisions, to achieve 5 sigma, you need $(5/2.7)^2=3.43$ times higher amount of data than 5/fb which is about 17 inverse femtobarns. Because of statistical flukes, the 5-sigma discovery could of course occur earlier or later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 115 GeV Higgs is the most challenging one to be seen by the LHC. But the run to be closed by the end of 2012 - before the 1-year upgrade - was ultimately chosen in such a way that the LHC should safely accumulate even those 17 inverse femtobarns by the end of 2012 - and maybe even sooner - so that it will discover the Higgs boson even at 115 GeV. One assumes that the additional improvements to the luminosity will materialize. But even if they used the bunch schemes they use now, without any further improvements, they should get to 10/fb or so by the end of 2012 - enough for a 4-sigma &quot;discovery&quot; or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, that assumes that the cross sections for the Higgs production are given by the Standard Model. New physics such as supersymmetry may make the Higgs both easier and harder to be detected - usually scenarios are studied where it is easier because the production rate is enhanced. Also, if there's more new physics than just the Standard Model Higgs, the LHC could find something more interesting than the Higgs.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-01T05:28:53.247" LastActivityDate="2011-05-01T05:28:53.247" />
  <row Id="9330" PostTypeId="2" ParentId="9325" CreationDate="2011-05-01T07:31:21.627" Score="9" ViewCount="" Body="&lt;p&gt;Yes, the Unruh effect violates Mach's principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In effective quantum field theory, the Unruh effect arises because the ground state of the Hamiltonian $H_a$ naturally associated with an accelerating observer is different than the ground state of the Hamiltonian $H_s$ generating time translations in a &quot;static&quot; frame. The static and accelerating frames have different time coordinates $t$, and because the Hamiltonian generates infinitesimal changes of $t$, it is different. And different operators have different eigenstates, including the lowest-eigenvalue one (the ground state).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One vacuum may be viewed as a squeezed/coherent state built upon the other, which may be interpreted as particle production, and creation operators have to be accordingly mixed with the annihilation operators according to the other frame - by the so-called Bogoliubov transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although Mach's principle has never been properly defined or turned into a realistic theory, it seems pretty clear that according to all of its interpretations, it postulated that it was impossible to have two different notions of a &quot;vacuum&quot; depending on the acceleration of the observer. However, that's exactly the case because of the Unruh effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An acceleration-dependent definition of the ground state (the vacuum), as realized in the derivation of the Unruh effect, exactly means that the acceleration may be distinguished from no acceleration even in the vacuum - which is exactly what Mach's principle would like to prohibit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Unruh effect therefore violates Mach's principle. More precisely, it refutes it because Mach's principle is wrong and the Unruh effect is real. However, one doesn't really have to go to the quantum theory to see that Mach's principle has been showed incorrect by subsequent developments. The existence of a dynamical metric tensor - even in the classical, non-quantum theory - is enough. The gravitational waves are the simplest classical entities that show that the gravitational field is real even in the vacuum, and it allows the particles moving through it to distinguish free fall from accelerating motion.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-01T07:31:21.627" CommentCount="2" />
  <row Id="9331" PostTypeId="1" AcceptedAnswerId="9436" CreationDate="2011-05-01T07:56:38.977" Score="0" ViewCount="295" Body="&lt;p&gt;I understand that perpetual motion is impossible, but I became confused when a friend of mine told me that the orbits of the planets is an example of perpetual motion. Will earth's orbit ever end, and whats wrong with his statement. I'm asking for a simple answer to 'disprove' my friends 'theory'.&lt;/p&gt;&#xA;" OwnerUserId="3361" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-05-01T08:42:25.603" LastActivityDate="2011-05-04T01:06:28.913" Title="Are orbiting planets an  example of perpetual motion?" Tags="&lt;orbital-motion&gt;&lt;perpetual-motion&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="0" />
  <row Id="9332" PostTypeId="1" CreationDate="2011-05-01T08:19:13.583" Score="4" ViewCount="104" Body="&lt;p&gt;You always hear theoreticians proudly proclaim the AdS/CFT correspondence implies time evolution for evaporating black holes is unitary. But if you examine the argument carefully, you find AdS black holes can't completely radiate away because the outgoing Hawking radiation always fall back into the black hole in an asymptotically anti de Sitter geometry. To get a black hole which completely evaporates away, we need a zero or positive cosmological constant.&lt;/p&gt;&#xA;" OwnerUserId="3034" LastActivityDate="2011-05-01T10:51:24.760" Title="Does the AdS/CFT correspondence for thermal states really imply time evolution for evaporating black holes is unitary?" Tags="&lt;quantum-gravity&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="9333" PostTypeId="1" AcceptedAnswerId="9336" CreationDate="2011-05-01T08:26:11.273" Score="2" ViewCount="247" Body="&lt;p&gt;Have ever noticed? When water is about to boil, no matters the kettle, there is some &lt;strong&gt;sound&lt;/strong&gt; I have no idea where it comes from, sometimes long before it boils.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you any explanation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isaac&lt;/p&gt;&#xA;" OwnerUserId="2499" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-05-01T10:12:46.730" LastActivityDate="2011-05-01T22:01:23.747" Title="When water is about to boil" Tags="&lt;fluid-dynamics&gt;&lt;sound&gt;" AnswerCount="3" />
  <row Id="9334" PostTypeId="2" ParentId="9333" CreationDate="2011-05-01T08:35:00.603" Score="2" ViewCount="" Body="&lt;p&gt;I don't know how good &lt;a href=&quot;http://www.worsleyschool.net/science/files/boilnoise/page.html&quot; rel=&quot;nofollow&quot;&gt;this explanation&lt;/a&gt; is, but it is certainly plausible. The sound must be from bubble formation/popping. It seems the difference between the noise in the beginning and when the water is really boiling well is because in the initial phase, bubbles don't reach the surface.&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2011-05-01T08:35:00.603" />
  <row Id="9335" PostTypeId="2" ParentId="9331" CreationDate="2011-05-01T08:39:49.867" Score="3" ViewCount="" Body="&lt;p&gt;Actually, what is impossible is a perpetuum mobile which is a machine or system that breaks the first or second law of thermodynamics. Planetary motions break none of those laws, therefore there is no contradiction with their &quot;perpetual motion&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question of the end of Earth's orbit is a different one though. Certainly, the Earth's orbit will end, most likely because the Sun will engulf the Earth as it grows to a red giant. If it doesn't end sooner. &lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2011-05-01T08:39:49.867" />
  <row Id="9336" PostTypeId="2" ParentId="9333" CreationDate="2011-05-01T08:44:12.960" Score="3" ViewCount="" Body="&lt;p&gt;The water near the heating element turns into water vapor. This vapor then rises up to the surface but as it meets colder water upwards it turns back into water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the water/steam transition is not smooth (e.g. the volume changes rather rapidly during phase change), the constant transition between vapor and liquid states produces noise. As the water gets to +100C (boiling point), the vapor bubble will not turn back into liquid before reaching the surface thus making less noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more on this &lt;a href=&quot;http://wiki.answers.com/Q/Why_does_a_kettle_make_noise_before_boiling&quot; rel=&quot;nofollow&quot;&gt;click this.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-05-01T08:56:40.030" LastActivityDate="2011-05-01T08:56:40.030" CommentCount="1" />
  <row Id="9337" PostTypeId="1" CreationDate="2011-05-01T08:44:31.200" Score="1" ViewCount="245" Body="&lt;p&gt;What is the smoking gun signature of string theory? Suppose we have a complete and consistent model of quantum gravity with a zero or negative cosmological constant, but all we are given is its complete S-matrix. Using only this information, how do we figure out whether or not this model corresponds to a string theory compactification?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the cosmological constant is positive, what can we use in place of the S-matrix?&lt;/p&gt;&#xA;" OwnerUserId="3034" LastActivityDate="2011-05-31T16:52:32.250" Title="What is the smoking gun signature of string theory?" Tags="&lt;string-theory&gt;&lt;quantum-gravity&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="9338" PostTypeId="2" ParentId="9332" CreationDate="2011-05-01T10:51:24.760" Score="2" ViewCount="" Body="&lt;p&gt;For the information loss paradox, it's completely irrelevant that the radiation may get reflected from the AdS boundary. What is important is that the information is not being lost while the radiation is leaving the region around the black hole horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the causal arguments based on classical general relativity were valid, the information about the state would already disappear when the radiation leaves the black hole. By causality, one can't imprint the information from the black hole interior to the exterior region (i.e. to the radiation). This failure to preserve the information would violate unitarity already at the moment when the radiation is emitted. However, unitarity holds in AdS/CFT at all times which proves that the classical argument based on causality is circumvented in quantum gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much more generally, it is not true that the negative cosmological constant makes any difference to the paradox. In particular, it is not true that all the radiation has to return to the same black hole. If the AdS curvature radius is large enough, the AdS space may host millions of other black holes and/or stars aside from the black hole whose information we analyze (it's a large universe just like ours!), and the radiation may get reflected to those other objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even more generally, physics in an AdS space with a low enough curvature is clearly locally indistinguishable from physics in flat space. Assuming nonlocalities at most at the distance scale of the black hole radius, it's clear that if the AdS radius is much longer than the black hole radius, physics can't be qualitatively affected by the nonzero AdS curvature.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-01T10:51:24.760" />
  <row Id="9339" PostTypeId="2" ParentId="9325" CreationDate="2011-05-01T11:50:23.237" Score="6" ViewCount="" Body="&lt;p&gt;Luboš's Answer is perfectly fine, +1, but whether the Unruh effect violates Mach's Principle depends on whether one takes the vacuum state to be something or nothing. One can consider measurement of one's acceleration by taking advantage of the Unruh effect to be possible just because it is relative to the vacuum state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vacuum state is a somewhat awkward object for 19th Century Physics, in that it does not allow us to measure velocity relative to it, but that it &lt;em&gt;does&lt;/em&gt; allow measurement of acceleration relative to it can be said to make it something. The something that it is has a Poincar&amp;eacute; invariant description, not a Galilean invariant description. This can be regarded as just what one has to do if one wants to save Mach's Principle. One doesn't have to save Mach's Principle, one can just let it go, but one can save it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vacuum state can easily be taken to be something rather than nothing if one takes the view that there &lt;em&gt;are&lt;/em&gt; quantum fluctuations, which are distinct from thermal fluctuations because of their different symmetry properties. It is possible to do a lot of Physics while maintaining that there are no quantum fluctuations, essentially considering the vacuum to be nothing, but it is not essential that we take this point of view, we can use either or both points of view (we can choose to look at a table with one eye or with both, right?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one moves to General Relativity, which requires a manifestly generally covariant description, we do not have an adequate theory of the vacuum state, so we have yet to see the fate of Mach's Principle in that conceptual background.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-05-01T11:50:23.237" CommentCount="7" />
  <row Id="9340" PostTypeId="1" CreationDate="2011-05-01T15:25:56.557" Score="5" ViewCount="162" Body="&lt;p&gt;Can we implement a scale dependent cutoff &amp;Lambda; to string theory? Can we perform a renormalization group analysis of string theory consistently?&lt;/p&gt;&#xA;" OwnerUserId="3363" LastActivityDate="2011-05-01T16:48:42.870" Title="Does the renormalization group apply to string theory?" Tags="&lt;string-theory&gt;&lt;renormalization&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="9341" PostTypeId="1" CreationDate="2011-05-01T15:52:42.387" Score="3" ViewCount="351" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Ultimately, the factor limiting the&#xA;  maximum speed of a rocket is:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;the amount of fuel it carries&lt;/li&gt;&#xA;  &lt;li&gt;the speed of ejection of the gases&lt;/li&gt;&#xA;  &lt;li&gt;the mass of the rocket&lt;/li&gt;&#xA;  &lt;li&gt;the length of the rocket&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This was a multiple-choice question in a test I've recently taken. The answer was (1), however, is this disputable, for if we assume that this rocket can potentially achieve relativistic speeds, what implications would this present to the limiting factor on maximum speed?&lt;/p&gt;&#xA;" OwnerUserId="3365" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-05-01T17:13:24.690" LastActivityDate="2011-06-07T12:03:41.450" Title="Maximum speed of a rocket with a potential of relativistic speeds" Tags="&lt;classical-mechanics&gt;&lt;special-relativity&gt;&lt;speed-of-light&gt;&lt;speed&gt;&lt;rocket&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="1" />
  <row Id="9342" PostTypeId="2" ParentId="9337" CreationDate="2011-05-01T15:54:15.517" Score="1" ViewCount="" Body="&lt;p&gt;I ( an experimentalist) will wade  in where angels ( theorists)  fear to tread :).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exists this &lt;a href=&quot;http://cdsweb.cern.ch/record/1129810/files/jinst8_08_s08004.pdf&quot; rel=&quot;nofollow&quot;&gt;great in length report by CMS&lt;/a&gt;  where  you will see in the introduction part of the motivation for the experiment as :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Various alternatives to the Standard Model invoke new symmetries,&#xA;  new forces or constituents. Furthermore, there are high hopes for discoveries that could pave the way toward a unified theory. These discoveries could take the form of supersymmetry or extra dimensions, the latter often requiring modification of gravity at the TeV scale.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So experimentalist have been lured to look for extra dimensions, one of the possible marks of string theory, as &quot;predicted&quot; by various phenomenological models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If nothing new/unexpected  like this is found, it will not preclude that string theories are waiting for us as the theory of everything in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a signal of extra dimensions is found in the data, then string theorists will be singing a different, not so cautious song.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-01T15:54:15.517" />
  <row Id="9343" PostTypeId="2" ParentId="9341" CreationDate="2011-05-01T16:08:09.520" Score="1" ViewCount="" Body="&lt;p&gt;I am not sure whether I have understood the question clearly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Firstly&lt;/strong&gt; you need a HUGE fuel resource in order to attain anything even remotely comparable to the speed of light which is practically almost an impossibility for a practical size rocket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may calculate the velocity of the rocket assuming that the gas is ejected at a constant rate and at constant velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider the rate of ejection is constant = $\alpha$ at constant velocity then we have $(m_0 - \alpha \times t) \times \frac {dv}{dt} - \alpha{v_0} = - (m_0 - \alpha{t})g$&#xA;Solving it we find,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v = -gt + v_0ln(\frac{m_0}{m_0 - \alpha t})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a general case where rate of ejection is not constant we have to know the exact dependence of mass of gas with time and solve the equation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m\frac{dv}{dt} + v_0 \frac{dm}{dt} = F$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second&lt;/strong&gt;, as usual in S.R., as the rocket approaches the speed of light w.r.t. a stationary inertial observer, it needs more and more energy to accelerate further. It requires an infinite amount of energy to attain the exact speed of light which is the limiting speed for any physical object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E = \frac {E_0}{\sqrt {1- v^2/c^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_0$ is the rest mass energy of the rocket and $E$ is the required energy to attain the speed $v$, $c$ is the speed of light. It is clear that as $v$ approaches $c$, $E$ approaches $\infty$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, ultimately it is the law of nature which will limit the maximum speed attainable by the rocket (or any other physical object).&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-05-01T16:43:52.577" LastActivityDate="2011-05-01T16:43:52.577" CommentCount="2" />
  <row Id="9344" PostTypeId="1" CreationDate="2011-05-01T16:22:33.490" Score="4" ViewCount="126" Body="&lt;p&gt;To completely localize a string within any bounded region of space, no matter how large, requires an infinite energy. Does this mean strings modes are inherently nonlocal?&lt;/p&gt;&#xA;" OwnerUserId="3052" LastActivityDate="2011-05-31T17:52:34.390" Title="If strings are infinitely extended, are they nonlocal?" Tags="&lt;string-theory&gt;" AnswerCount="1" />
  <row Id="9345" PostTypeId="1" CreationDate="2011-05-01T16:30:14.910" Score="1" ViewCount="142" Body="&lt;p&gt;I think one catch in Twin Paradox was about the big acceleration that can turn back the traveling twin from light speed outward bound, to become light speed inward bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if there is strictly no acceleration?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Peter is on a space ship, traveling 99% of light speed.  He is exactly 20 years ago.  &lt;/li&gt;&#xA;&lt;li&gt;Michael is on Earth (&lt;strong&gt;or a planet similar to Earth, but with a radius so small that any centripetal acceleration is negligible&lt;/strong&gt;... or consider him standing just on a piece of concrete in space with oxygen supply)&lt;/li&gt;&#xA;&lt;li&gt;Michael is also exactly 20 years old.&lt;/li&gt;&#xA;&lt;li&gt;According to time dilation, Peter's clock in the spaceship is slowing than Michael's clock.&lt;/li&gt;&#xA;&lt;li&gt;According to time dilation, Michael's clock on Earth is slower than Peter's clock.  (since motion is relative, if we consider Peter to be stationary, and Michael is traveling)&lt;/li&gt;&#xA;&lt;li&gt;Peter's spaceship is traveling towards Michael.&lt;/li&gt;&#xA;&lt;li&gt;After 30 years on Earth, Peter's spaceship went past Michael's face, so Peter and Michael is 1 cm apart, face to face and eye to eye.&lt;/li&gt;&#xA;&lt;li&gt;Now, would Peter see Michael quite older than him, and also, Michael sees Peter quite older than him?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2701" LastEditorUserId="2701" LastEditorDisplayName="" LastEditDate="2011-05-01T23:46:05.677" LastActivityDate="2011-05-01T23:46:05.677" Title="How will the Twin Paradox become, for Time Dilation, if no acceleration was ever involved?" Tags="&lt;general-relativity&gt;&lt;special-relativity&gt;&lt;relativity&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="9346" PostTypeId="2" ParentId="9340" CreationDate="2011-05-01T16:48:42.870" Score="4" ViewCount="" Body="&lt;p&gt;Good question. The short answer is no, cutoff scales have no relevance to string theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cutoff scales are given by maximum or minimum energies or distances where the given theory may be applied. This concept is only useful because in quantum field theory, such cutoffs are natural regulators to get rid of short-distance divergences. These divergences reflect the fact that field theories should only be thought of as approximate theories that are only valid at distances longer than the cutoff distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of renormalizable quantum field theories, the cutoff distance may be made arbitrarily short in principle; in the case of non-renormalizable quantum field theories, it may not; in those cases, the theory is literally ill-behaved at short enough distances. However, in both cases, some cutoff is useful because only if this characteristic scale is finite, one may assume that the fundamental parameters of the Lagrangian are finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;String theory doesn't suffer from any short-distance problems. It's an almost defining property of string theory, one that can really be demonstrated even in its simplest &quot;toy models&quot; such as bosonic string theory. That means that it is incorrect to think of string theory as an approximate theory that only works at long enough distances. Quite on the contrary, the very purpose of string theory - one of its main advantages over quantum field theory - is that it is valid up to arbitrarily short distances, despite the fact that the gravitational interaction is one of the major predictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the integrals in string theory that replace the divergent integrals in quantum field theory are convergent; there are no short-distance divergences we get in string theory at all. For this reason, there is no reason to cutoff the integrals i.e. to introduce cutoffs in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The characteristic scale of perturbative string theory is the string scale - and at the non-perturbative level, various Planck scales may be more universal typical distance scales - but that doesn't mean that string theory breaks down at the string scale. Quite on the contrary, the bulk of computations in string theory is exactly about the exact physics that takes place at the string scale energies or slightly larger ones. To describe (much) lower energies than the string scale, an effective quantum field theory would be enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So string theory incorporates all the effects from particles that are 2 times or 10 times heavier than the string scale automatically. It's how its formalism works. There is no reason to have cutoffs. They exist in effective field theories - which are assumed to be long-distance approximations of whatever more detailed theory that is relevant at all scales. String theory is such a theory that is relevant at all scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, descriptions of stringy physics in terms of effective field theories are omnipresent. Those universally depend on a cutoff, much like effective field theories in any other context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;String field theory: level truncation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just an addition. There is an approach to string theory called string field theory - which leads to the same results as more standard calculations in string theory but that is formulated to be as similar to quantum field theory as possible. In particular, string field theory is a quantum field theory with infinitely many fields that are associated with the individual string vibration patterns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In string field theory, people often do numerical calculations - e.g. solutions with D-branes and tachyon condensation - and they often do so numerically (although analytic solutions became common after the groundbreaking advances by Martin Schnabl and others a few years ago). When numerical calculations are being done, one can't really deal with infinitely many fields. In that case, people only deal with fields corresponding to &quot;light enough particles&quot;, e.g. those that are lighter than $\sqrt{13} m_{s}$. This scheme looks like a refined cutoff - and it is known as the level truncation scheme. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the reason why the levels are truncated is somewhat different from effective field theory. In effective field theory, a cutoff is needed for finite results. In string field theory, level truncation is just a way to make numerical calculations more tractable; however, the exact results that include the whole infinite tower of states are still totally finite.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-01T16:48:42.870" CommentCount="1" />
  <row Id="9347" PostTypeId="2" ParentId="9344" CreationDate="2011-05-01T17:02:20.677" Score="1" ViewCount="" Body="&lt;p&gt;Yes (formally) and no (physically). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The full extent of a string with finite energy is infinitely large; the expectation value of $\int d\sigma (x^i(\sigma))^2$ is proportional to $\sum 1/n$ and is logarithmically divergent. However, the contributions that make it infinitely large arise from very high-energy vibrations of the string, i.e. high values of the Fourier mode label $n$. And these vibrations are so fast that they average out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if one will study the &quot;size&quot; of a string with any other finite-energy probe, she will inevitably find out that the string is effectively finite. As a result, the scattering amplitudes in perturbative string theory actually strictly obey some inequalities that one may derive to hold in any local theory. In some sense, string theory saturates them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The divergent value of the string's radius is also a perturbative artifact. It's clear that when the string coupling constant is nonzero, the highly excited modes become unstable and introduce an effective cutoff on the world sheet, anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;String theory is local in some other senses, too. Physics is local on the world sheet. So if one acknowledges that a string is an extended object, physical phenomena become completely local in spacetime once again. Because of the arguments two paragraphs above, some locality is preserved even if we associate every string with its center of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same non-locality discussion applies not only to strings but also to arbitrary bound states in BFSS-like matrix models of string/M-theory. In those cases, one may also see that the size of the bound states - the whole &quot;clouds&quot; - is infinitely large in the relevant large $N$ limit. However, these large clouds will also penetrate through one another because the effect of very distant loci of the clouds (from the center) is associated with very fast vibrations that get rapidly averaged out.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-01T17:02:20.677" />
  <row Id="9348" PostTypeId="2" ParentId="9333" CreationDate="2011-05-01T17:27:11.797" Score="1" ViewCount="" Body="&lt;p&gt;As Rasolnikov said, this is not when water &quot;is about to boil&quot; ! &#xA;The sound occures when the water still is cold, after the &#xA;electric heater has reached its full temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The noise is produced by the collapse (implosion) of the &#xA;steam bubbles rising from the bottom and reaching the water &#xA;still far from boiling point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this would produce some noise (hiss), not a musical sound. &#xA;For the sound, one needs a mechanism to synchronize this bubble &#xA;production, which I do not know. &#xA;The processes involved in bubble versus film evaporation &#xA;are very turbulent and fast. The strong drop in heat transfer &#xA;involved in that domain might play a role. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am quite shure that this question was here already, but I don't find it. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-05-01T22:01:23.747" LastActivityDate="2011-05-01T22:01:23.747" />
  <row Id="9349" PostTypeId="1" AcceptedAnswerId="9350" CreationDate="2011-05-01T17:28:59.260" Score="1" ViewCount="135" Body="&lt;p&gt;I'm having trouble doing it. I know so far that if we have two Hermitian operators $A$ and $B$ that do not commute, and suppose we wish to find the quantum mechanical Hermitian operator for the product $AB$, then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{AB+BA}{2}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if I have to find an operator equivalent for the radial component of momentum, I am puzzled. It does not come out to be simply &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\vec{p}\cdot\frac{\vec{r}}{r}+\frac{\vec{r}}{r}\cdot\vec{p}}{2},$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\vec{r}$ and $\vec{p}$ are the position and the momentum operator, respectively. Where am I wrong in understanding this?&lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-02T05:17:19.980" LastActivityDate="2011-05-07T06:03:17.697" Title="How to construct the radial component of the momentum operator?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="9350" PostTypeId="2" ParentId="9349" CreationDate="2011-05-01T17:46:12.210" Score="2" ViewCount="" Body="&lt;p&gt;You would have to use the fact that the momentum operator in position space is $\vec{p} = -i\hbar\vec{\nabla}$ and use the definition of the gradient operator in spherical coordinates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{\nabla} = \hat{r}\frac{\partial}{\partial r} + \hat{\theta}\frac{1}{r}\frac{\partial}{\partial\theta} + \hat{\phi}\frac{1}{r\sin\theta}\frac{\partial}{\partial\phi}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the radial component of momentum is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_r = -i\hbar\hat{r}\frac{\partial}{\partial r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However: after a bit of investigation prompted by the comments, I found that in practice this is not used very much. It's more useful to have an operator $p_r&amp;#39;$ that satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{\hbar^2}{2m}\nabla^2 R(r) = \frac{p_r&amp;#39;^2}{2m} R(r)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This lets you write the radial component of the time-independent Schrödinger equation as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\biggl(\frac{p_r&amp;#39;^2}{2m} + V(r)\biggr)R(r) = E R(r)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The action of the radial component of the Laplacian in 3D is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla^2 R(r) = \frac{1}{r^2}\frac{\partial}{\partial r}\biggl(r^2\frac{\partial R(r)}{\partial r}\biggr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if you solve for the operator $p&amp;#39;_r$ that satisfies the definition above, you wind up with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p&amp;#39;_r = -i\hbar\biggl(\frac{\partial}{\partial r} + \frac{1}{r}\biggr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is called the &quot;radial momentum operator.&quot; Strictly speaking, it is different from the &quot;radial component of the momentum operator,&quot; which is, by definition, $p_r$ as I wrote it above, although I wouldn't be surprised to find people mixing up the terminology relatively often.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-01T18:17:02.750" LastActivityDate="2011-05-01T18:17:02.750" CommentCount="4" />
  <row Id="9351" PostTypeId="2" ParentId="9345" CreationDate="2011-05-01T17:52:27.707" Score="7" ViewCount="" Body="&lt;p&gt;You say that both twins are &quot;exactly 20 years old.&quot; I assume you mean that they are both 20 years old &lt;em&gt;at the same time&lt;/em&gt;. But part of the point of special relativity is that a phrase like &quot;at the same time&quot; means different things in different reference frames. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be specific, suppose that these two moments (Peter's birthday party and Michael's birthday party) are simultaneous in a reference frame in which the Earth is at rest. Then, performing the entire analysis in that same frame, we would say the following: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Peter's clock ticks slower. &lt;/li&gt;&#xA;&lt;li&gt;Therefore, at the moment the two pass each other, Peter is younger than Michael.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now let's look at things as measured in Peter's reference frame. In his frame, Michael's clock ticks slower. Therefore, from the time of his 20th birthday until the time the two of them meet, the amount of time as measured by Michael's clock is less than the amount of time as measured by Peter's clock. If we then concluded that Michael would be younger than Peter, we would indeed have a paradox. But that conclusion doesn't follow, because, in this reference frame, &lt;em&gt;the two of them didn't start out the same age.&lt;/em&gt; To be specific, the event &quot;Michael's 20th birthday&quot; and the event &quot;Peter's 20th birthday&quot; were not simultaneous. Michael's birthday happened earlier. So in this reference frame, Michael started out older, and even though his clock ticked slower, he was still older when the two met.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of that is the way things play out if the two birthdays were simultaneous in Michael's reference frame. If on the other hand the two events were simultaneous in Peter's reference frame, then you can just switch the names &quot;Michael&quot; and &quot;Peter&quot; throughout, and everything will work the same way.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-05-01T18:22:59.267" LastActivityDate="2011-05-01T18:22:59.267" CommentCount="7" />
  <row Id="9352" PostTypeId="1" CreationDate="2011-05-01T18:20:12.933" Score="1" ViewCount="94" Body="&lt;p&gt;&lt;a href=&quot;http://www.reformation.org/tesla-wireless-transmission.html&quot; rel=&quot;nofollow&quot;&gt;This description&lt;/a&gt; of Tesla's &quot;magnifying transmitter&quot;, which supposedly used electrical resonance to transmit energy (similar to &lt;a href=&quot;http://en.wikipedia.org/wiki/Resonant_inductive_coupling&quot; rel=&quot;nofollow&quot;&gt;resonant inductive coupling&lt;/a&gt;?) states that the coils (or at least part of them) are submerged in liquid air, which &quot;causes an extraordinary magnification of oscillation in the resonating circuits&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What effect would submerging a coil in liquid air actually have?  Does it just decrease resistance of the coils?  I know with resonant inductive coupling, efficiency is improved with higher &lt;a href=&quot;http://en.wikipedia.org/wiki/Q_factor#RLC_circuits&quot; rel=&quot;nofollow&quot;&gt;Q factor of the RLC circuit&lt;/a&gt;, and Q is improved by reducing the resistive component, so would this make some sense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it have any other effects?  Electrical conductivity, magnetic permeability, etc?&lt;/p&gt;&#xA;" OwnerUserId="176" LastEditorUserId="176" LastEditorDisplayName="" LastEditDate="2011-05-01T18:28:30.573" LastActivityDate="2011-07-31T02:22:49.950" Title="What effect would liquid air have on a resonant coil?" Tags="&lt;electromagnetism&gt;&lt;temperature&gt;&lt;resonance&gt;" AnswerCount="2" />
  <row Id="9353" PostTypeId="2" ParentId="9349" CreationDate="2011-05-01T18:45:25.027" Score="1" ViewCount="" Body="&lt;p&gt;I was able to figure it out, so here goes the clarification for the record. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classically $$p_r=\hat{D_r} = \hat{r}\cdot\hat{p} = \frac{\hbar}{i}\frac{\partial}{\partial r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However $\hat{D_r}$ is not hermitian. Consider the adjoint $$\hat{D_r^+}= \hat{p}\cdot\hat{r} =\frac{\hbar}{i}\frac{\partial}{\partial r}+\frac{2}{r} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know from linear algebra how to construct a hermitian operator from an operator and its adjoint:  $$\hat{p_r} = \frac{\hat{D_r^+}+\hat{D_r}}{2}=\frac{\hbar}{i}\frac{\partial}{\partial r}+\frac{1}{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And btw, for those who followed my initial question, don't do the following calculational mistake that I committed: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_r = \frac{1}{2}\left(\frac{1}{r}(\vec{r}\cdot\vec{p})+(\vec{p}\cdot\vec{r})\frac{1}{r}\right)\neq \frac{1}{2}\frac{1}{r}(\vec{r}\cdot\vec{p}+\vec{p}\cdot\vec{r})$$&lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="2599" LastEditorDisplayName="" LastEditDate="2011-05-07T06:03:17.697" LastActivityDate="2011-05-07T06:03:17.697" CommentCount="3" />
  <row Id="9354" PostTypeId="1" AcceptedAnswerId="9358" CreationDate="2011-05-01T19:10:21.700" Score="1" ViewCount="180" Body="&lt;p&gt;I just have a quick question about time dilation/proper time because my physics book makes it a little confusing. Let's say we have an observer on Earth, and then an observer on a space ship. The space ship leaves Earth, flies to the Moon, and then returns to Earth. Who is the person measuring the proper time and why? I know that a clock &quot;runs slower&quot; when it is in motion because it is in frame S' which is the rest frame of the clock, but doesn't the observer on Earth also have a clock that is in it's rest frame?&lt;/p&gt;&#xA;" OwnerUserId="3137" LastActivityDate="2011-05-01T20:26:12.977" Title="Question about Time Dilation" Tags="&lt;relativity&gt;" AnswerCount="1" />
  <row Id="9355" PostTypeId="1" AcceptedAnswerId="9380" CreationDate="2011-05-01T19:23:02.993" Score="5" ViewCount="191" Body="&lt;p&gt;My question is inspired by the analogy of the Berry phase in the spin coherent state representation of a rotator and the Aharonov-Bohm phase of a magnetic monopole (see e.g., Section 1.8.3 in &lt;a href=&quot;http://www.physics.ubc.ca/~berciu/TEACHING/PHYS503/section1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.physics.ubc.ca/~berciu/TEACHING/PHYS503/section1.pdf&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a physical system invariant under a (continuous) group of transformations G. The aim is to define the corresponding quantum mechanical Hilbert space H ``in the richest possbile way'', that is, a representation which has as many state vectors as group elements. In case of  G=SO(3) it is the 2-sphere of spin coherent states. The Aharonov-Bohm analogy says that H is a manifold with a connection corresponding to vector potential of some specific kind (in the example of spin it is the vector potential which gives the magnetic field of the monopole). Once we know that connection, can we classify &#xA;the possible global topologies of representations of G as the angular momentum representations are classified by requiring the uniqueness of the Berry phase of a closed loop?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can such a manifold be constructed for any G (from a sufficiently interesting class of groups)? My feeling is that this should correspond to some theorem in topolgy/representation theory. I'm very ignorant in these fields of math, and would appreciate a hint on where to look further.&lt;/p&gt;&#xA;" OwnerUserId="1739" LastActivityDate="2011-05-20T09:39:58.530" Title="Generalization of spin coherent states for an arbitrary group?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" />
  <row Id="9356" PostTypeId="2" ParentId="9352" CreationDate="2011-05-01T19:23:24.880" Score="0" ViewCount="" Body="&lt;p&gt;Liquid oxygen is paramagnetic. A Tesla coil is basically an air-core, high-frequency transformer. Immersing the coil in liquid air would be like putting a core in the transformer with a permeability higher than non-liquified air.&lt;/p&gt;&#xA;" OwnerUserId="3368" LastActivityDate="2011-05-01T19:23:24.880" CommentCount="1" />
  <row Id="9357" PostTypeId="2" ParentId="9341" CreationDate="2011-05-01T19:28:16.310" Score="3" ViewCount="" Body="&lt;p&gt;Let us take the rocket to start with an initial mass of $M_{0}$ and to eject mass at a rate $\frac{dm}{dt}$, so after a time $t$, it has ejected a total mass $m$.  Furthermore, let the ejection speed be given by $v_{e}$ for simplicity.  Then, using the formula for relativistic momentum, we have the momentum gained by the rocket equalling the momentum of the expelled rocket fuel (we also assume the rocket starts from rest):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}&#xD;&#xA;\frac{dm}{dt}v_{e} &amp;amp;=\frac{d}{dt}\left(\frac{mv}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\right)\\&#xD;&#xA;m\,v_{e} &amp;amp;=\frac{(M_{0}-m)v}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}\\&#xD;&#xA;\frac{m^{2}v_{e}^{2}}{(M_{0}-m)^{2}}&amp;amp;= \frac{v^{2}}{1-\left(\frac{v}{c}\right)^{2}}&#xD;&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some algebra, you get the final result,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v=\frac{v_{e}}{\sqrt{\left(\frac{M_{0}-m}{m}\right)^2+\left(\frac{v_{e}}{c}\right)^{2}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final speed of the rocket will then be given as a function of the mass $M_{0}-m$ of the rocket itself, the mass $m$ of the fuel expelled, and the velocity $v_{e}$ at which the fuel was expelled.  Therefore, the correct answer is definitely everything but the length of the rocket.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-05-01T19:28:16.310" CommentCount="5" />
  <row Id="9358" PostTypeId="2" ParentId="9354" CreationDate="2011-05-01T19:31:31.877" Score="4" ViewCount="" Body="&lt;p&gt;Each observer has their own proper time measured by the clock in their rest frame. However, one man's proper time is not another man's proper time. Time dilation means that each observer will see the other observer's clock running slower (compared to their own proper time measuring clock). But everything is perfectly symmetric from either observer's point of view as long as the relative motion is uniform. You measure your clock ticking at the &quot;normal rate&quot; (your proper time) and you see the other person's clock ticking at a slower rate. Similarly the other person measures their clock ticking at the &quot;normal rate&quot; (their proper time) and they see your clock ticking at a slower rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all well and nice, but it gets interesting when the two compare their clocks after one of them does a round trip. This means that one of them necessarily had to accelerate and decelerate and was not in uniform motion (technically, was not on a geodesic). Now you have an opportunity to actually compare those two clocks and you'll always find that the person in uniform motion (in this case, the observer at rest on Earth) was the one whose clock has ticked the most, and hence aged the most. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way to understand this is to realize that the length of paths in spacetime is measured by the total proper time along that path (measured by that path-traveller's clock in their rest frame). One can show that the paths of uniform motion (geodesics) have that length maximized, so any path that deviates from a geodesic (because of accelerations), will necessarily measure a shorter total proper time after a round trip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT AFTER FIRST COMMENT:&#xA;Time dilation isn't the appropriate effect to consider in this particular problem -- length contraction is. In Nick's frame, a length contracted ship passes by at speed $v$. In Molly's frame, a point-object (heh) Nick passes by an uncontracted ship at speed $v$. Clearly, this should happen quicker in Nick's frame because of the length contraction. Thinking in terms of time dilation simply doesn't help here. Think from the point of view of each observer and it will be quickly obvious which effect to use.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-05-01T20:26:12.977" LastActivityDate="2011-05-01T20:26:12.977" CommentCount="2" />
  <row Id="9359" PostTypeId="1" CreationDate="2011-05-01T23:59:38.123" Score="5" ViewCount="153" Body="&lt;p&gt;This question is slightly related to this one &lt;a href=&quot;http://physics.stackexchange.com/questions/2696/do-all-massive-bodies-emit-hawking-radiation&quot;&gt;Do all massive bodies emit Hawking radiation?&lt;/a&gt;, which I think was poorly posed and so didn't get very useful answers. There are several questions in this post so I hope people who answer will follow the question marks and give their opinion on each one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at flat spacetime first. Inertial observers detect no particles in empty space. Non-inertial observers do detect particles because there is a non-trivial Bogolyubov transformation that mixes positive and negative frequency modes of the matter field. And finally, if the non-inertial observers see a horizon as in the Unruh case, the spectrum of detected particles is precisely thermal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now look at a black hole spacetime. Free falling observers will not detect any particles but distant/stationary observers will detect a &lt;em&gt;flux&lt;/em&gt; of radiation from the black hole outward to infinity. &lt;strong&gt;Q1:&lt;/strong&gt; Why is there a flux in this case and none in the Unruh case? Shouldn't the two situations be nearly identical for the case of a very large black hole or very near to the horizon? How would you physically distinguish between the two anyway? (Maybe this is equivalent to asking why the black hole should lose mass and evaporate)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q2:&lt;/strong&gt; Is it natural to expect that an observer who is not in free fall but is not stationary either (and is on some weird trajectory) will detect particles but that they will not be in a thermal state?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For massive compact objects without a horizon, free fallers should see no particles but other non-inertial observers should detect particles because of the Bogolyubov transformation as in the Unruh case. This probably depends a lot on the answer to Q1 but &lt;strong&gt;Q3:&lt;/strong&gt; suppose you bring such an object arbitrarily close to the point of forming a black hole with a horizon (by adding mass) -- what would a stationary observer see? Would radiation and evaporation be observed only once the horizon is formed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q4:&lt;/strong&gt; In the Unruh case, the energy from the radiation is accounted for by the agency that accelerates the detector. In the black hole case, it is accounted for by the evaporation of the black hole. How is it accounted for in the case of a stationary detector outside a massive compact object without a horizon (which doesn't evaporate)?&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-05-02T09:05:31.790" Title="Why is there a flux of radiation in the Hawking effect but not in the Unruh effect? (and other questions)" Tags="&lt;quantum-field-theory&gt;&lt;general-relativity&gt;&lt;black-holes&gt;&lt;hawking-radiation&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="9360" PostTypeId="2" ParentId="4384" CreationDate="2011-05-02T02:15:19.023" Score="1" ViewCount="" Body="&lt;p&gt;It isn't &quot;physical&quot; intuition per-se, but I find the convex-analysis interpretation of the Legendre transform to be the most enlightening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A convex set is uniquely determined by it's supporting hyperplanes, and the Legendre transform is an encoding of the convex hull of a function's epigraph in terms of it's supporting hyperplanes. If the function is convex and differentiable, then the supporting hyperplanes correspond to the derivative  at each point, so the Legendre transform is a reencoding of a function's information in terms of it's derivative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a few links that illustrate the concept:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://jmanton.wordpress.com/2010/11/21/introduction-to-the-legendre-transform/&quot; rel=&quot;nofollow&quot;&gt;http://jmanton.wordpress.com/2010/11/21/introduction-to-the-legendre-transform/&lt;/a&gt; (great in-depth explanation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mia.uni-saarland.de/Teaching/NAIA07/naia07_h3_slides.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.mia.uni-saarland.de/Teaching/NAIA07/naia07_h3_slides.pdf&lt;/a&gt; (compuational perspective)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://maze5.net/?page_id=733&quot; rel=&quot;nofollow&quot;&gt;http://maze5.net/?page_id=733&lt;/a&gt; (graphical/visual explanation on my website. Not that advanced, but but with a lot of pictures)&lt;/p&gt;&#xA;" OwnerUserId="821" LastActivityDate="2011-05-02T02:15:19.023" />
  <row Id="9361" PostTypeId="2" ParentId="9359" CreationDate="2011-05-02T05:16:04.630" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Q1&lt;/strong&gt; - The black hole spacetime sees actual particles in the &quot;static&quot; frame because the diminishing curvature across the black hole spacetime continuously connects the static frame at infinity - which is really static and corresponds to freely falling, static observers - with the static observers who otherwise live in a strong gravitational field near the black hole event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the Hamiltonian corresponding to the static time in the black hole case looks like the &quot;static frame&quot; of the Unruh case at infinity while it looks like the &quot;accelerating frame&quot; of the Unruh case near the event horizon. However, the natural Hamiltonian near the event horizon is linked to the freely falling frame - because the particles can continue to fall after they cross the event horizon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that's the frame in which there are no particles near the event horizon. Because the static frame is accelerating relatively to it, it will see a particle flux near the event horizon. And because the static coordinates connect this place with infinity, the same particle flux must be seen at infinity, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Unruh case, you can't turn the radiation to real particles in a flat space because every frame is either inertial or accelerating and we know very well that only the inertial frame has a flat spacetime and it sees no particles. Note that Unruh's argument is only valid locally - you may show that if one observer sees nothing, a &lt;em&gt;nearby&lt;/em&gt; accelerating observer does see particles. You need to propagate the particles through a general curved spacetime to figure out how these particles will manifest themselves at a completely different location - and that's where the nontrivial curvature of the black hole spacetime kicks in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, as Anna suggested, it may be more pedagogic to link the creation of Hawking particles to the split virtual pairs near the event horizon. Or to the quantum tunneling from the black hole - the uncertainty principle prevents particles from being completely safely hidden inside the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q2&lt;/strong&gt; - An observer in a general state of acceleration will detect a radiation that will generalize the radiation or no radiation seen in the two frames and it will be approximately thermal, too. There is always a frame in which the radiation may be argued to vanish - and all other frames in the same place will see Unruh-like radiation corresponding to their acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q3&lt;/strong&gt; - Before the horizon gets formed, the situation is not translationally symmetric in time. So the nice Hamiltonian associated with the Killing vector field - showing that the spacetime is static - doesn't exist yet. At that moment, it's not possible to say whether some particles are &quot;real&quot;. However, quite generally, freely falling observers define the frames in which there are no particles produced by the future black hole (there can be other &quot;real&quot; particles participating in the collapse, of course). That's not surprising because these freely falling particles are analogous to those that cross the event horizon of the completed black hole. All observers that accelerate relatively to them - in the same place - will see an Unruh-like radiation. It's important to realize that it's hard to detect the &quot;exact radiation&quot; during the formative moments of the black hole. The reason is that the creation of the black hole approximately takes time $t = R/c$ or so, and because the typical energy of the Hawking quanta is just $E = \hbar / t$ or so, we're near the saturation point of the uncertainty principle, so we can't measure the energy of the Hawking quanta too accurately. And the density of the Hawking particles is of order one per this $R^3\times t$ region of spacetime, too. Also note that once it's decided that the black hole is going to form, you can't slow the process down. A much longer time than the time of the black hole birth is needed to be sure about the precise spectrum of the Hawking radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q4&lt;/strong&gt; - The static observer sitting on a heavy star doesn't see any radiation! It's because there is no event horizon. So the accounting is zero equals zero. There is not even an outgoing flux equal to an incoming flux - after all, one couldn't guarantee any permanent incoming flux from an empty space. The only sensible frame - relatively to which the actual state of the empty space is approximately the ground state - is the static frame. And there are no particles. The presence of the rock solid star's surface guarantees that the freely falling frame is no good to define the vacuum state.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-02T09:05:31.790" LastActivityDate="2011-05-02T09:05:31.790" />
  <row Id="9363" PostTypeId="1" CreationDate="2011-05-02T09:00:10.593" Score="-3" ViewCount="111" Body="&lt;p&gt;kindly, provide answers as soon as possible. &lt;/p&gt;&#xA;" OwnerUserId="3376" LastActivityDate="2011-05-02T11:43:56.287" ClosedDate="2011-05-02T15:44:18.187" Title="what is the maxwell's equation for electromagnetism?" Tags="&lt;homework&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9364" PostTypeId="1" CreationDate="2011-05-02T09:11:16.757" Score="0" ViewCount="305" Body="&lt;p&gt;This will perhaps look like a very basic and trivial question. But I find it confusing. As an experience, when you are travelling in non-AC car in summer, have people felt if putting on or putting off the windowpane affect how much heat it feels inside to passengers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On one hand, I feel that heat can still come inside thru glass particles (This makes me ask if heat needs media to travel?) and so putting on windowpane should not reduce heat felt inside. But psychologically it feels like by putting on the pane, you are blocking heat. Also the windowpane will at least block the heated air from coming inside. But does this heated air really brings much heat apart from the heat coming through anyway or it actually in practical acts more like a ventillator thus reducing heat felt inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that this question may appear strange, but it pops up to my head everytime I travel in non-AC car in summer. I have tried to articulate it as well as I could.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do insulator windowpane glasses reduce heat felt inside? because they will transmit lesser heat inside. How do black shade glasses work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="1093" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-05-02T11:11:08.417" LastActivityDate="2011-07-01T20:13:31.270" Title="Does opening or closing the window in a non-AC car in the summer affect how much heat is felt inside?" Tags="&lt;thermodynamics&gt;&lt;light&gt;&lt;energy&gt;&lt;heat&gt;&lt;weather&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="9365" PostTypeId="2" ParentId="9088" CreationDate="2011-05-02T09:27:04.597" Score="-2" ViewCount="" Body="&lt;p&gt;this is because the gases that are heavier than air, sink to the bottom of the globe thanks to the centrifugal effect of the earths rotation.&lt;/p&gt;&#xA;" OwnerUserId="3377" LastActivityDate="2011-05-02T09:27:04.597" CommentCount="1" />
  <row Id="9366" PostTypeId="2" ParentId="9364" CreationDate="2011-05-02T10:29:15.800" Score="0" ViewCount="" Body="&lt;p&gt;Keeping your windows closed increases the temperature inside the car because of a greenhouse-like effect. The inside of the car continually receives radiation from the sun, re-emits it to warm the air inside the car but doesn't allow the heated air to leave the car by convection. Hence the car will be considerably hotter inside if the windows are kept closed. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Greenhouse_effect#Real_greenhouses&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Greenhouse_effect#Real_greenhouses&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black shade glasses simply reduce the amount of radiation getting in thus making things slightly cooler.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-05-02T10:29:15.800" />
  <row Id="9367" PostTypeId="2" ParentId="9363" CreationDate="2011-05-02T11:43:56.287" Score="-1" ViewCount="" Body="&lt;p&gt;Normally Maxwell equations show how electric and magnetic forces are created for a probe charge with given charges and currents.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-05-02T11:43:56.287" />
  <row Id="9368" PostTypeId="1" CreationDate="2011-05-02T12:20:11.120" Score="0" ViewCount="130" Body="&lt;p&gt;How is this Theorem useful, and how can it be used to advance mathematics, or science?&lt;/p&gt;&#xA;" OwnerUserId="2881" LastActivityDate="2011-05-02T12:20:11.120" ClosedDate="2011-05-02T15:46:23.927" Title="Fermat's Last Theorem" Tags="&lt;mathematics&gt;" CommentCount="7" />
  <row Id="9369" PostTypeId="1" AcceptedAnswerId="9370" CreationDate="2011-05-02T13:10:16.850" Score="3" ViewCount="299" Body="&lt;p&gt;I'm a non-physicist with a basic high-school understanding of physics. I've always wondered what it is that makes things &quot;solid&quot;. Why do molecules rebound from each other? There's just a bunch of tiny atoms there with (comparatively) large spaces between them. Why don't they just &quot;slide&quot; around or between each other?&lt;/p&gt;&#xA;" OwnerUserId="3381" LastActivityDate="2011-05-03T01:30:57.497" Title="What makes matter &quot;solid&quot;?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" />
  <row Id="9370" PostTypeId="2" ParentId="9369" CreationDate="2011-05-02T13:45:47.650" Score="7" ViewCount="" Body="&lt;p&gt;What makes it solid is a combination of the uncertainty principle and Pauli's exclusion principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the uncertainty principle, electrons can't have a well-defined position if they have a sufficiently well-defined momentum (mass times velocity). For the energy of electrons to be low enough, the momentum also has to be low enough. That also means that the uncertainty of the momentum has to be low enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The uncertainty principle then implies that the uncertainty of the position of the electron has to be large enough, at least 0.1 nanometers or so - the atomic radius - for the kinetic energy to be smaller than a few electronvolts, a decent amount of energy that is comparable to the potential energy of electrons near protons if the electron clouds is similarly large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pauli's exclusion principle then guarantees that in each volume comparable to the volume of an atom, there can be at most 1 (or 2) electrons. That's why matter is impenetrable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, to derive the actual maximum density of electrons, one also needs to know the strength of the electrostatic attraction between the electrons and the protons that neutralize the charge. The Bohr radius goes like $1/\alpha$ - the inverse fine-structure constant - so if the electric force were stronger, matter could actually become denser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was an explanation why matter based on nuclei and electrons can't really significantly exceed the density of ordinary materials. Still, there are different phases. In gases, the molecules are separated by big gaps - so most of the space is empty and the exclusion principle is not too important. For liquids, the distance between the molecules is near the saturation point - like dense gases - but they still don't keep the shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Glass is an example of a liquid, in some sense, that is however behaving almost as solids. The canonical solids prefer to be crystals - like metals or diamond. In that case, it's energetically favored for the atoms or molecules to be organized to regular cubic or hexagonal or similar lattices. They like to keep this shape because it saves energy. It's still true that such solids are impenetrable because of the explanations at the top.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-02T13:45:47.650" CommentCount="1" />
  <row Id="9371" PostTypeId="1" AcceptedAnswerId="9372" CreationDate="2011-05-02T14:02:10.420" Score="2" ViewCount="466" Body="&lt;p&gt;Confinement of antihydrogen might help provide a future answer. &lt;a href=&quot;http://arxiv.org/abs/1104.4982&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1104.4982&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-09-01T12:47:44.427" Title="Why would Antimatter behave differently via Gravity?" Tags="&lt;gravity&gt;&lt;antimatter&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="9372" PostTypeId="2" ParentId="9371" CreationDate="2011-05-02T14:25:24.230" Score="0" ViewCount="" Body="&lt;p&gt;You're of course right. Experimentally established, known laws of physics - especially the equivalence principle - are enough to be certain that antimatter has the same gravitational properties - including universal attraction - as ordinary matter.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2010/09/can-antimatters-gravity-be-repulsive.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/09/can-antimatters-gravity-be-repulsive.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The justification of further experiments by &quot;tests of antimatter's gravity&quot; is partly based on ignorance and partly on deliberate deception to get funding.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-02T14:25:24.230" CommentCount="7" />
  <row Id="9373" PostTypeId="1" AcceptedAnswerId="9374" CreationDate="2011-05-02T14:28:12.570" Score="5" ViewCount="205" Body="&lt;p&gt;When we introduce minimal coupling for the Dirac spinor field, we introduce terms into the Lagrangian, by the substitution $i\frac{\partial}{\partial x^\mu}\mapsto i\frac{\partial}{\partial x^\mu}+eA_\mu$, so that the Lagrangian is invariant under arbitrary changes of phase at every point of space-time, $$\psi(x)\mapsto e^{-ieG(x)}\psi(x),\qquad A_\mu\mapsto A_\mu+\partial_\mu G(x),$$ where $G(x)$ is a scalar function.&#xA;We don't make this substitution, however, in the anti-commutator $$\left\{\psi_\xi(x),\overline{\psi_{\xi&amp;#39;}(x&amp;#39;)}\right\}=\left(i\gamma^\mu\frac{\partial}{\partial x^\mu}+m\right)_{\xi\xi&amp;#39;}i\Delta(x-x&amp;#39;),$$&#xA;which, &lt;em&gt;it seems&lt;/em&gt;, is therefore not invariant under the transformation $\psi(x)\mapsto e^{-ieG(x)}\psi(x)$,&#xA;$\psi(x&amp;#39;)\mapsto e^{-ieG(x&amp;#39;)}\psi(x&amp;#39;)$.&#xA;Presumably the path-integral formalism doesn't care about this, but does this have consequences in subsequent derivations in the canonical approach?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect that in fussing about this I'm missing something very obvious. Am I? What is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: So it appears from the Answers, for which Thanks, that the anti-commutator given is valid for both space-like and time-like separation for the free Dirac field, and it's valid for space-like separation in QED, but it's nae valid for time-like separation in QED. Put somewhat loosely, this is OK in the canonical formalism because this anti-commutator is valid for the phase space. I'd rather like to know what a valid expression for the anti-commutator is at time-like separation in QED, but since that would, in a sense, be a solution of the theory I guess I'll have to whistle for it. I'll be grateful for Comments if this is an obviously obtuse reading of the Answers, though I ask that you consider that I would like a manifestly covariant version (so to speak, perhaps obscurely) of the canonical formalism before leaping in.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-05-02T23:01:06.737" LastActivityDate="2011-05-02T23:01:06.737" Title="Why is the spinor field anti-commutator not made gauge invariant?" Tags="&lt;quantum-field-theory&gt;&lt;quantum-electrodynamics&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" />
  <row Id="9374" PostTypeId="2" ParentId="9373" CreationDate="2011-05-02T15:04:29.747" Score="2" ViewCount="" Body="&lt;p&gt;Because $\left(i\gamma^\mu\frac{\partial}{\partial x^\mu}+m\right)_{\xi\xi&amp;#39;}i\Delta(x-x&amp;#39;)$ is a symbolic expression for a given analytical right-hand side. It is written so for convenience (not yet calculated) but it is a specific expression like $\delta(x-x&amp;#39;)$ or $\delta(x-x&amp;#39;)&amp;#39;$. It should not acquire any &quot;gauge extension&quot; by definition. This expression does not contain a &quot;particle momentum&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-05-02T15:12:42.893" LastActivityDate="2011-05-02T15:12:42.893" CommentCount="10" />
  <row Id="9375" PostTypeId="1" CreationDate="2011-05-02T15:18:02.397" Score="0" ViewCount="194" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/9371/why-would-antimatter-behave-differently-via-gravity&quot;&gt;Why would Antimatter behave differently via Gravity?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Regarding the following statement in &lt;a href=&quot;http://www.technologyreview.com/blog/arxiv/26709/&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Most important of these is whether ordinary gravity attracts or repels antimatter. In other words, does antihydrogen fall up or down?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is this a seriously considered hypothesis? What would be the consequences on general relativity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is seriously studied, can you point to some not-too-cryptic studies on the (anti ;-)matter?&lt;/p&gt;&#xA;" OwnerUserId="66" LastActivityDate="2011-05-02T18:24:16.307" ClosedDate="2011-05-04T17:23:27.047" Title="Anti-matter repelled by gravity - is it a serious hypothesis?" Tags="&lt;general-relativity&gt;&lt;antimatter&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="9376" PostTypeId="2" ParentId="9375" CreationDate="2011-05-02T15:30:11.930" Score="1" ViewCount="" Body="&lt;p&gt;I can't guarantee the authenticity of the article. But it seems to me quite bizarre since I fail to see how something (even antimatter) can behave differently than matter in a gravitational field without violating the equivalence principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A positron for example is a hole in the Dirac sea and it has the same mass as of the electron and behaves exactly similar to an electron in a gravitational field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only repulsive gravity that exists in GR is the cosmological constant in the Einstein's equations which is in no way related to antimatter.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-05-02T15:30:11.930" CommentCount="1" />
  <row Id="9377" PostTypeId="2" ParentId="9375" CreationDate="2011-05-02T15:56:27.260" Score="3" ViewCount="" Body="&lt;p&gt;If antimatter anti-gravitates, it would be possible to build a perpetual motion machine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Build some sort of collider that produces antimatter as an end product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) allow antimatter to &quot;fall&quot; up to some height.  It will gain kinetic energy as it falls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) at top of &quot;fall&quot;, have antimatter annihlate with matter, producing light that you send through some focusing insturment &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) send photons back down to original collider (they even blueshift as they go back down!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) use photons to power original collider, and siphon off surplus energy &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-05-02T15:56:27.260" CommentCount="8" />
  <row Id="9378" PostTypeId="2" ParentId="9364" CreationDate="2011-05-02T16:08:47.180" Score="1" ViewCount="" Body="&lt;p&gt;A couple of different effects here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A glass windowed box will heat up to considerably more than the outside temperature. Radiation from the sun, as visible and near infrared light, passes easily through the glass - hits the black interior surfaces and heats them up, this in turn heats the air and the glass stops the warm air escaping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Opening a window will let the hot air escape and the inside cool to approximately the outside air temperature - it can't cool below this without some power source (eg AC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting your hand out of a moving car window will strongly cool your hand. Water from your hand evaporates, cooling it, the wind blows this warm wet air away quickly allowing more fresh moisture to leave your hand. This is the wind-chill which although it can only cool your body to the air temperature makes you feel very cold as your body tries to maintain it's normal temperature.&lt;/p&gt;&#xA;" OwnerUserId="2525" LastEditorUserId="2525" LastEditorDisplayName="" LastEditDate="2011-05-02T16:52:54.147" LastActivityDate="2011-05-02T16:52:54.147" CommentCount="5" />
  <row Id="9379" PostTypeId="2" ParentId="9290" CreationDate="2011-05-02T17:16:11.003" Score="-1" ViewCount="" Body="&lt;p&gt;from &lt;a href=&quot;http://arxiv.org/abs/0907.2469&quot; rel=&quot;nofollow&quot;&gt;Astrometric Solar-System Anomalies&lt;/a&gt; of Anderson and Nieto, 2009, page 9,&lt;br&gt;&#xA;&lt;strong&gt;Increase in the eccentricity of the Moon's orbit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While the mean motion and semi-major&#xA;  axis rates of the lunar orbit are&#xA;  consistent with physical models for&#xA;  dissipation in Earth and Moon, LLR&#xA;  orbital solutions consistently reveal&#xA;  an anomalous secular eccentricity&#xA;  variation. After accounting for tides&#xA;  on the Earth that produce an&#xA;  eccentricity change of $1.3*10^{-11} yr^{-1}$ and tides on the Moon that&#xA;  produce a change of $-0.6*10^{-11} yr^{-1}$, &lt;strong&gt;there is an anomalous rate&lt;/strong&gt;&#xA;  of $(0.9\pm0.3)*10^{-11} yr^{-1}$ ,&#xA;  &lt;strong&gt;equivalent to an extra 3.5 mm&#xA;  $yr^{-1}$&lt;/strong&gt; in perigee and apogee&#xA;  distance (Williams \&amp;amp; Boggs 2009).&#xA;  &lt;strong&gt;This anomalous eccentricity rate is&#xA;  not understood and it presents a&#xA;  problem&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1102/1102.0212v6.pdf&quot; rel=&quot;nofollow&quot;&gt;On the anomalous secular increase of the eccentricity of the orbit of the Moon&lt;/a&gt;, by L. Iorio, 2011,  explores several alternatives to explain the problem. All of them were inviable, concluding:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thus, the issue of ﬁnding a satisfactorily explanation&#xA;  for the anomalous behavior of the Moon’s eccentricity remains open.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;also, from Iorio slides &lt;a href=&quot;http://moriond.in2p3.fr/J11/transparents/iorio.pdf&quot; rel=&quot;nofollow&quot;&gt;ON THE ANOMALOUS INCREASE OF THE ECCENTRICITY OF THE LUNAR ORBIT: SEARCH FOR POSSIBLE EXPLANATIONS&lt;/a&gt; , 2011, he try to offer: A VIABLE, &lt;strong&gt;empirical&lt;/strong&gt; EXPLANATION&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let us assume that there is a small radial extra-acceleration of the&#xA;  form...  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This procedure is called 'data fit' and obviously it is not an &lt;em&gt;EXPLANATION&lt;/em&gt; at all.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I based my other answer on the decreasing LOD FACT, and all other answers are saying the contrary.&lt;br&gt;&#xA;With this anser I make notice to the fact that an anomaly is present and that the present physics is unable to model it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again I point to a &lt;a href=&quot;http://arxiv.org/PS_cache/astro-ph/pdf/0208/0208365v1.pdf&quot; rel=&quot;nofollow&quot;&gt;MODEL&lt;/a&gt; where the reported anomaly is not present because data is along with theory (eq 35 and 36). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm expecting a reception to this answer in line with the reception to the other answer.&lt;br&gt;&#xA;May be the case that some doubt in your minds can rise the need to read 'out-of-the-box' and try to see if the Sun-Earth anomaly (AU increase) or the Pioneer anomaly can be explained by this model.&lt;br&gt;&#xA;When we have the chance to make measures with more precision we have to report an anomaly.&lt;br&gt;&#xA;A bunch of anomalies is a call for a new model.&lt;br&gt;&#xA;I'm expecting that someday, somebody, will read that paper. It is a cosmological model that do not need DM, DE, Cosmological Constant, it goes well with GR, and agrees with data both at local scale and cosmological scale.   &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-05-02T17:16:11.003" />
  <row Id="9380" PostTypeId="2" ParentId="9355" CreationDate="2011-05-02T17:29:00.413" Score="2" ViewCount="" Body="&lt;p&gt;There are essentially three ways to generalize the notion of coherent state, based on extensions of the concept of coherent states for the case of the harmonic oscillator.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The first is to note that in the case of the harmonic oscillator, the coherent states can be obtained by acting on the vacuum state with translations in position and momentum space. This is the approach that &lt;a href=&quot;http://arxiv.org/PS_cache/math-ph/pdf/0203/0203002v1.pdf&quot; rel=&quot;nofollow&quot;&gt;Perelomov&lt;/a&gt; generalizes by replacing the action with arbitrary Lie groups.  &lt;/li&gt;&#xA;&lt;li&gt;The second approach is to note that coherent states of the harmonic oscillator are saturating the Heisenberg inequalities, i.e. they have minimal uncertainty in both momentum and position. This can be generalized as well to give other kinds of coherent states which are not necessarily equivalent to Perelomov generalized coherent states.&lt;/li&gt;&#xA;&lt;li&gt;Then there is a third approach, that generalizes the annihilation/creation operators for any Hamiltonian with discrete spectrum and builds an overcomplete set of states out of the eigenstates of the annihilation operator. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All three methods are mentioned in &lt;em&gt;J.R. Klauder and B. Skagerstam, &quot;Coherent States&quot;, World Scientific Publishing Co. Pte. Ltd., Singapore, 1985&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2011-05-20T09:39:58.530" LastActivityDate="2011-05-20T09:39:58.530" CommentCount="5" />
  <row Id="9381" PostTypeId="2" ParentId="9373" CreationDate="2011-05-02T17:45:59.907" Score="6" ViewCount="" Body="&lt;p&gt;Dear Peter, there is no freedom to &quot;make&quot; the (anti)commutator of your fields whatever you want, so there is no freedom to &quot;make&quot; it gauge-invariant, either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The canonical momentum is obtained as the derivative of the Lagrangian density with respect to the time-derivative of the canonical coordinate, and the (anti)commutator of these two is just the Kronecker delta (multiplied by the delta-function). The remaining (anti)commutators are defined to vanish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's clear that the cubic coupling of the Dirac field to the electromagnetic field, $\bar\psi A_\mu \gamma^\mu \psi$, doesn't contain any time derivatives, so it also doesn't modify the canonical momenta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that the anticommutators of $\psi$ and $\bar \psi$ - which are just a Kronecker delta times a delta function (arising from the $\bar\psi p_\mu \gamma^\mu \psi$ kinetic term in the Lagrangian) - are completely unaffected by this coupling to the electromagnetic field - which is needed for gauge invariance of the action. And indeed,&#xA;$$   \{ \psi_a(x), \psi^\dagger_b(x&amp;#39;) \} = \delta_{ab} \delta^{(d-1)} (x-x&amp;#39;).$$&#xA;The equation above transforms trivially under the gauge transformation. The left-hand side gets multiplied by &#xA;$$ e^{-ieG(x)+ieG(x&amp;#39;)} $$&#xA;and you may add the same factor to the right hand side as well because the phase above is equal to one for $x=x&amp;#39;$, and it's the only point where it matters because of the $\delta(x-x&amp;#39;)$ factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some reasons, you wrote the anticommutator as the Dirac differential operator acting on the Dirac propagator (although I am confused about your method to deal with the Dirac spinor indices - it's probably wrong), but the propagator is nothing else than the inverse Dirac differential operator (acting on a delta function), so the Dirac operators cancel and you're left with the delta function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The previous sentence is meant to be an explicit elegant proof, but even if you decided to write the position representation of the propagator in terms of Bessel's functions, you should be able to see that the right hand side of your complicated form of the anticommutator will be just a delta function. Note that you only want to evaluate the commutator at equal times, $t=t&amp;#39;$, and as you have already established - as far as your comments under Vladimir's answer go - the function vanishes for spacelike separations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You only need to be careful about the behavior of the right hand side as a distribution for very small spatial separations of $x$ and $x&amp;#39;$ and for $t=t&amp;#39;$. That's why the non-vanishing of the function for timelike separations would be irrelevant for the anticommutator. However, despite the irrelevance, it's still true that when you act on the Dirac propagator with the Dirac differential operator, you get a delta-function that vanishes everywhere in the spacetime except for $x=x&amp;#39;$ - localized in all spacetime dimensions. That's the function whose spacetime Fourier transform is proportional to $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But don't get me wrong: the anticommutator does &lt;em&gt;not&lt;/em&gt; vanish for time-like separations.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-02T19:26:05.483" LastActivityDate="2011-05-02T19:26:05.483" CommentCount="4" />
  <row Id="9382" PostTypeId="2" ParentId="9375" CreationDate="2011-05-02T18:24:16.307" Score="0" ViewCount="" Body="&lt;p&gt;The only problem with this is that antimatter has the same mass as its matter counterpart. So this means that is effected the same way by gravity as normal matter.&lt;/p&gt;&#xA;" OwnerUserId="3290" LastActivityDate="2011-05-02T18:24:16.307" CommentCount="1" />
  <row Id="9384" PostTypeId="2" ParentId="9364" CreationDate="2011-05-02T19:00:11.977" Score="0" ViewCount="" Body="&lt;p&gt;If the car is in the sun closed windows will trap heat due to two mechanisms. As dbrane says the greenhouse effect is named after onr of these this effects, the radiation comes in at shortwave lengths, and the reradiated thermal energy is partially blocked by the glass. But any dark object in direct sunlight will get surprisingly hot, 150-200 F is not uncommon, so leaving the windows open a crack allows some airflow, bringing the inside temperature closer to ambient. Also in dry heat, opening your windows for the first couple of minutes of driving to drive out the superheated air is more efficient then immediately turning on the AC.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-05-02T19:00:11.977" />
  <row Id="9385" PostTypeId="2" ParentId="9314" CreationDate="2011-05-02T20:19:33.973" Score="3" ViewCount="" Body="&lt;p&gt;It's a rather nice historical full circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The meter was originally defined as 1/10,000,000 the distance from the equator to the north pole - through Paris - as Luboš explained above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was dreamed up by some guys in the French revolution with the idea that units of physics should depend only on the universe (or at least Paris - which is pretty much the same thing) and not on historic details like the length of a king's arm or foot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also meant that you could independently arrive at your own impartial measurement of the unit - there was no master metre in a vault someone. Thus making all men, and their measurements, equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like lots of the French revolution this didn't exactly work out exactly as planned - the survey was off - and soon the demands of industry meant that the only sufficently accurate standard was a master metre in a vault somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By redefining the metre in terms of the speed of light - which is universal (even outside Paris) we are going back to that original noble principle.&lt;/p&gt;&#xA;" OwnerUserId="2525" LastEditorUserId="2525" LastEditorDisplayName="" LastEditDate="2011-05-03T04:11:32.527" LastActivityDate="2011-05-03T04:11:32.527" />
  <row Id="9386" PostTypeId="1" AcceptedAnswerId="10851" CreationDate="2011-05-02T21:00:19.540" Score="4" ViewCount="247" Body="&lt;p&gt;I'm trying to get a more intuitive understanding of &lt;a href=&quot;http://en.wikipedia.org/wiki/Resonant_inductive_coupling&quot; rel=&quot;nofollow&quot;&gt;resonant inductive coupling&lt;/a&gt;.  It's supposed be a more efficient way to transfer electrical energy wirelessly, because the coils are only coupled by near fields, and don't waste energy as far field radiation.  The efficiency is supposed to increase as Q increases (the resistive component decreases).&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/1zhUl.png&quot; alt=&quot;LC circuit schematic&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, an LC circuit has a Q of ∞.  If you charge up the capacitor in an LC circuit, and then leave it floating in space away from anything else, the energy will cycle back and forth between the L and C.  The voltage, current, electric field and magnetic field will all cycle sinusoidally (right?)  With ideal components, and no resistive component, this will continue cycling back and forth forever.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Real LC circuits have some resistance, which wastes some of the energy as thermal radiation, and the cycling eventually dies.  I think they also have some other non-idealities that allow energy to escape as far field electromagnetic radiation, correct?  What are these non-idealities?  Are they independent of the resistive component?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If another identical LC circuit is brought near this oscillating circuit, will energy transfer from one circuit to the other due to the changing magnetic fields?  Will this happen even with ideal components that don't have far field radiation?  Is there no energy transfer if the coils are perpendicular and the axis of one is coplanar with the other, like this?:  | —&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the energy distribute itself evenly among the circuits, so they reach an equilibrium where each is circulating half of the original energy?  Or will the energy oscillate back and forth between the two circuits, with one alternately being depleted or full?&lt;/p&gt;&#xA;" OwnerUserId="176" LastEditorUserId="176" LastEditorDisplayName="" LastEditDate="2011-05-03T03:00:10.243" LastActivityDate="2011-06-08T15:21:01.070" Title="Lots of little questions about radiation of LC circuits" Tags="&lt;electromagnetism&gt;&lt;electromagnetic-radiation&gt;&lt;resonance&gt;&lt;electric-circuits&gt;&lt;electrical-engineering&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="9387" PostTypeId="1" AcceptedAnswerId="9395" CreationDate="2011-05-02T21:45:46.963" Score="-1" ViewCount="146" Body="&lt;p&gt;I don't understand how a vacuum, the absence of matter, can hold energy.  How can it hold energy when Einstein proved that matter is energy?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And a second related question; how does the energy in a vacuum allow the Universe to produce itself?  And how does that account for why the Universe is expanding?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw Krauss' lecture on YouTube called &quot;Universe from Nothing&quot; and only understood some of it, even though he was fantastic at putting things in layman's terms and making these concepts easy to visualize.  However, he had a bunch of &quot;asides&quot;, so it was a bit confusing.  He mentioned that the curvature of the Universe was flat, but then also went on to say how that was the worst possible scenario.  Why is a flat Universe such a bad thing?&lt;/p&gt;&#xA;" OwnerUserId="3385" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-05-03T04:39:56.610" LastActivityDate="2011-05-03T04:39:56.610" Title="Lawrence Krauss' Theory of Origin" Tags="&lt;cosmology&gt;&lt;theory&gt;&lt;vacuum&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9388" PostTypeId="2" ParentId="6027" CreationDate="2011-05-02T21:51:09.953" Score="1" ViewCount="" Body="&lt;p&gt;All the basic kinematic equation you've learned come from one basic equation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\textrm{d}^2\vec{x}}{\textrm{d}t^2} = \vec{g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_vector&quot; rel=&quot;nofollow&quot;&gt;vector&lt;/a&gt;-valued &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_equation#Nomenclature&quot; rel=&quot;nofollow&quot;&gt;second order&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_differential_equation&quot; rel=&quot;nofollow&quot;&gt;linear&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_equation&quot; rel=&quot;nofollow&quot;&gt;differential equation&lt;/a&gt;.  $ \vec{x}$ is the position of the projectile.  $m$ is its mass.  $\vec{g}$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_gravity&quot; rel=&quot;nofollow&quot;&gt;acceleration due to gravity&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation can be read in simplified form as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\vec{F}}{m} = \vec{a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion#Newton.27s_second_law&quot; rel=&quot;nofollow&quot;&gt;Newton's Second Law&lt;/a&gt;.  $\vec{F}$ is the force on an object, which for a projectile near Earth's surface is $m\vec{g}$, ignoring all effects besides simple gravity (air resistance is the big one).  $\vec{a}$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Acceleration&quot; rel=&quot;nofollow&quot;&gt;acceleration&lt;/a&gt;, which by definition is $\frac{\textrm{d}^2\vec{x}}{\textrm{d}t^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinematic equations that tell you, for example, the distance traveled by a projectile or its speed as a function of height can all be derived from this differential equation using calculus and algebra.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The differential equation itself is a postulate of Newtonian mechanics.  In this sense it doesn't come from anywhere; it is simply assumed, along with the assumption that the force of gravity is constant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, with more advanced physics it is possible to explain Newtonian mechanics in terms of other theories.  This simplest example is that the assumption that the force from gravity is $m\vec{g}$ can be understood in terms of &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation&quot; rel=&quot;nofollow&quot;&gt;Newton's theory of gravity&lt;/a&gt;.  What we see is that the simple gravity law is only an approximation, but a good one.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, as we bring in more and more considerations, we tend to find that almost everything we've done in the beginning is only an approximation.  The kinematic equations aren't fundamental laws of nature.  They're mathematical consequences of a simple differential equation that is roughly true for everyday things.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-05-02T21:51:09.953" />
  <row Id="9389" PostTypeId="1" CreationDate="2011-05-02T22:50:37.927" Score="2" ViewCount="220" Body="&lt;p&gt;I need a good book about elementary particles. I am a high school student and don't want anything to technical. I read a brief history of time and the universe in a nutshell but i want something that centers more around elementary particles.&lt;/p&gt;&#xA;" OwnerUserId="3387" LastEditorUserId="2599" LastEditorDisplayName="" LastEditDate="2011-05-03T08:13:36.437" LastActivityDate="2011-05-04T17:37:06.540" Title="Good book about elementary particles for high school students?" Tags="&lt;books&gt;&lt;particles&gt;" AnswerCount="5" />
  <row Id="9390" PostTypeId="1" CreationDate="2011-05-03T00:22:30.910" Score="0" ViewCount="146" Body="&lt;p&gt;i was a little confused about the casimir effect. my understanding is with the 2 plate experiment there was a force pushing the plates together because there were less virtual particles in between the plates than outside them. Why are there less particles in between the plates?&lt;/p&gt;&#xA;" OwnerUserId="3387" LastActivityDate="2011-05-03T13:42:24.783" Title="Casimir effect plate experiment" Tags="&lt;casimir-effect&gt;" AnswerCount="2" />
  <row Id="9391" PostTypeId="1" CreationDate="2011-05-03T00:42:25.463" Score="0" ViewCount="268" Body="&lt;p&gt;I know that a polytropic process equation is often the result of an empirical fit of a P-V curve that allows one to derive some nice analytical equations for the process.  Is there actually a derivation for the polytropic process equation?&lt;/p&gt;&#xA;" OwnerUserId="3311" LastActivityDate="2011-05-03T13:12:46.773" Title="Derivation of polytropic process equation" Tags="&lt;thermodynamics&gt;" AnswerCount="2" />
  <row Id="9392" PostTypeId="1" CreationDate="2011-05-03T01:15:09.807" Score="4" ViewCount="190" Body="&lt;p&gt;Whether gravity waves are real or just a coordinate freedom was argued in the early days of GR.  Eventually the conclusion was that they were real.  And if they are 'real' then I'm curious if...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Has it ever been shown that in General Relativity, two (or more) colliding gravitational waves can lead to a black hole?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2655" LastActivityDate="2011-05-03T07:49:08.133" Title="Can colliding gravity waves create a black hole?" Tags="&lt;general-relativity&gt;&lt;black-holes&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9393" PostTypeId="2" ParentId="9369" CreationDate="2011-05-03T01:30:57.497" Score="0" ViewCount="" Body="&lt;p&gt;I was told in College that is was simple electrostatics. The electrons never come close enough to make the Pauli Exclusion, or uncertainty principle effective. It is simply that atoms repel each other at large distances due to the negative charge of the outer shell electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just wanted to throw that out there.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-05-03T01:30:57.497" CommentCount="1" />
  <row Id="9394" PostTypeId="2" ParentId="9386" CreationDate="2011-05-03T02:56:24.690" Score="0" ViewCount="" Body="&lt;p&gt;You have asked many questions. Let me answer them one by one.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If ideally there is no resistance then the only loss will be core loss. Whatever (except free space) is within the inductor undergoes hysteresis and because of that some energy will be lost. Also there will be some loss in the core due to eddy currents in the core.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Stray magnetic field loss. Some  energy will be lost due to whatever stray magnetic field present there. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If another LC circuit is brought near the original circuit then it will act as a transformer and some energy will transferred from one inductor to another by transformer action. The usual losses like core loss, hysteresis and stray magnetic field losses (except copper loss) will be there.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the inductor of the second circuit is kept at perpendicular position to the first one then if the magnetic circuit between the two inductor closes itself through the coils then energy will be transferred as usual.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-05-03T04:02:22.287" LastActivityDate="2011-05-03T04:02:22.287" CommentCount="9" />
  <row Id="9395" PostTypeId="2" ParentId="9387" CreationDate="2011-05-03T04:33:12.783" Score="2" ViewCount="" Body="&lt;p&gt;First of all, Lawrence Krauss has nothing to do with the discovery of dark energy, cosmological constant, or &lt;em&gt;any&lt;/em&gt; important discovery, for that matter. This doesn't prove that his ambiguous formulations to suggest otherwise were deliberately written as ambiguous - but it doesn't disprove it, either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cosmological constant was introduced by Einstein right after 1916 in an incorrect attempt to make the Universe static and it was observed - at a lower value - in the 1990s experimentally by experimenters which couldn't include Lawrence Krauss. A rather detailed history, explanations, and names of the discoverers of these advances may be found in the very book by Brian Greene, &lt;em&gt;The Hidden Reality&lt;/em&gt;, that my Dopplegänger Sheldon Cooper downplayed in the video offered by Raskolonikov, and that I am just translating, in a parallel universe, to my native tongue. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, Einstein's $E=mc^2$ is not an equivalence of energy and &lt;em&gt;matter&lt;/em&gt;. It is the equivalence of energy and &lt;em&gt;mass&lt;/em&gt; (i.e. the number of kilograms). So a unit volume of the empty space carries some &lt;em&gt;mass&lt;/em&gt; equivalent to the energy - it's a mass of a few protons per cubic meter. But the $E=mc^2$ equation does &lt;em&gt;not&lt;/em&gt; imply, in any sense, that the mass equivalent to the energy has to take the form of localized particles. It may be dispersed, much like the cosmological constant - whose generalized form is also called dark energy. The main reason why the vacuum contains &lt;em&gt;mass&lt;/em&gt; is that this mass contributes to the curvature of spacetime - the gravitational field of mass - and be sure that dark energy does. That is why it was introduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark energy, unlike mass, carries a negative pressure, and it's the real source of the accelerating expansion it induces. Ordinary matter has attractive gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, physics can't label things by moral evaluations such as &quot;worst possible&quot; etc. Krauss probably meant that if the spatial slices of our spacetime are exactly flat - which is the boundary scenario in between the positive curvature and negative curvature - then we have no chance to ever experimentally find out what the magnitude of the curvature is because the sign may always be 0 or $\pm \epsilon$ where $\epsilon$ is sufficiently small relatively to the experimental resolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some sense, the exact flat slices may also be viewed as an example of fine-tuning - something would be zero without any good reason. Some advanced cosmological considerations could imply that the spatial curvature is positive (we can swim around the Universe in principle, after some time), or negative (curved like the Lobachevsky plane, a saddle if you wish).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-03T04:38:31.533" LastActivityDate="2011-05-03T04:38:31.533" />
  <row Id="9396" PostTypeId="2" ParentId="9389" CreationDate="2011-05-03T04:41:17.740" Score="1" ViewCount="" Body="&lt;p&gt;You should definitely pre-order Lisa Randall's &lt;em&gt;Knocking on Heaven's door&lt;/em&gt;, to be released in September,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/006172372X&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Knocking-Heavens-Door-Scientific-Illuminate/dp/006172372X&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is detailed, excellent, and addressed to the laymen such as bright high school students (no equations). Randall is a top particle physicist - and some advanced enough stuff about particle physics, including the LHC collider, is the bulk of the book. Try also &lt;em&gt;Warped Passages&lt;/em&gt; by the same author which is more theoretical and less experimental:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B003A02WS0&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Warped-Passages-Unraveling-Mysteries-Dimensions/dp/B003A02WS0/ref=sr_1_1&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-03T06:07:16.263" LastActivityDate="2011-05-03T06:07:16.263" CommentCount="2" />
  <row Id="9397" PostTypeId="2" ParentId="9390" CreationDate="2011-05-03T04:50:53.727" Score="4" ViewCount="" Body="&lt;p&gt;The objects whose number is lower in between the plates are not really particles &lt;em&gt;per se&lt;/em&gt; but the different modes - different possible values of the wavelength or frequency, in particular - in which the particles may be created.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the distance of the parallel plates is $L$, then the electric field has to vanish at the boundary between the vacuum and the metals - at the plates - and this implies that the electromagnetic waves may be decomposed to standing waves. The wavelength of such photons in the $x$ direction (the direction in which the plates are separated) has to be an integer multiple of $2L$. So only frequencies &#xA;$$f_n = n\times c / 2L$$&#xA;are possible ($c$ is the speed of light). That's needed, once again, for the electric field to vanish at the plates. Note that I am talking about the transverse electric field and $-\nabla \Phi$ has to vanish in the directions $y,z$ because the potential $\Phi$ is constant throughout the metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there were no plates, the frequency $f$ could be any real number - which is equivalent to the $L\to\infty$ limit of the formula above. In this sense, the plates restrict the allowed frequency - they reduce the number of possible values of the frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For the sake of simplicity, I was assuming that the photons don't have any motion in the directions $y,z$ along the plates - in general, they do have this motion. This fact would be correctly and easily accounted for by switching from the wavelength to the wavenumbers $k$.)&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-03T04:50:53.727" CommentCount="2" />
  <row Id="9398" PostTypeId="2" ParentId="9391" CreationDate="2011-05-03T04:58:24.033" Score="1" ViewCount="" Body="&lt;p&gt;Indeed, the general polytropic equation $pV^n=C$ is meant as a fit to describe a wide variety of situations - but the general situations don't have to exactly obey any polytropic equation with any $n$, anyway (because the right $n$ depends on $p$ and/or $V$), so there is nothing to derive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A major, general enough exception is the adiabatic equation $pV^\gamma = C$. It only differs by using the name $\gamma$ for $n$ where $\gamma = C_P/C_V$, the ratio of heat capacities at constant pressure and volume, respectively. The adiabatic process is one in which no heat is transferred between the gas and the environment, so all the work being used or consumed is used to change the internal thermal energy of the gas. Nevertheless, $\gamma$ can have many different values such as $5/3$ etc. Microscopically, the value is linked to a function of the degrees of freedom of a single (average) molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, the derivation of the adiabatic equation - based on $\alpha p \,dV = dU $ - is here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Adiabatic_process#Derivation_of_continuous_formula&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Adiabatic_process#Derivation_of_continuous_formula&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-03T04:58:24.033" />
  <row Id="9399" PostTypeId="2" ParentId="9392" CreationDate="2011-05-03T05:02:28.353" Score="2" ViewCount="" Body="&lt;p&gt;It was not showed experimentally - the only experimental evidence of gravitational waves is that some binary pulsars change their orbiting frequency exactly as expected from their losing energy by gravitational waves of the GR-predicted intensity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is surely established theoretically. The answer is Yes, gravitational waves that carry enough energy and squeeze it to a small enough volume - essentially the volume of the Schwarzschild black hole whose radius $R = 2GM/c^2$ is calculated from the mass $M=E/c^2$ where $E$ is the energy carried by the waves (the numerical coefficient $2$ is not necessarily accurate) - inevitably collapse into a black hole. In fact, a collision of two particles with high enough energies is enough, too. The conjecture due to Thorne that the squeezing of the energy is enough is known as the hoop conjecture.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hoop_conjecture&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hoop_conjecture&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This fact can be numerically calculated by simulating general relativity on a computer. The process in which high-energy collisions of particles - not only gravitons - lead to the birth of black holes is also the dominant or universal scattering effect at huge, trans-Planckian energies (when the energy of a particle is comparable to the lightest black hole or higher).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-03T07:49:08.133" LastActivityDate="2011-05-03T07:49:08.133" CommentCount="5" />
  <row Id="9400" PostTypeId="1" CreationDate="2011-05-03T05:46:15.837" Score="2" ViewCount="214" Body="&lt;p&gt;Hey everyone, I'm currently taking an Eletricity and Magnetism course and I'm having dificulty answering this problem. I've gone to my tutor, but sadly, he hasn't taken E+M for a while now and couldn't recall off the top of his head how to go about solving this. We went through my book together, but couldn't find any examples pertaining to this type of problem. (We did find examples of induced EMF's obviously, but making the leap to a spinning rod seems to be too great a task)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways, here is the problem I'm working on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A metal rod is spinning on an axis that is located in the middle of the rod, and perpendicular to the rod's length. If the rod is in an external magnetic field that is parallel to the axis the rod is spinning on, how would I go about finding the induced emf?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rod's thickness is negligible but has length L. It's spinning at 1500 rev/s and the magnetic field is 7.5T&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any pointers would be appreciated!&lt;/p&gt;&#xA;" OwnerUserId="3391" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-05-03T06:00:55.697" LastActivityDate="2011-05-04T00:05:24.813" Title="Induced EMF of a spinning metal rod" Tags="&lt;electromagnetism&gt;&lt;homework&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="9401" PostTypeId="2" ParentId="9389" CreationDate="2011-05-03T05:47:49.627" Score="2" ViewCount="" Body="&lt;p&gt;Martinus Veltman &lt;a href=&quot;http://books.google.com/books?id=CNCHDIobj0IC&amp;amp;printsec=frontcover&amp;amp;dq=facts+and+mysteries+elementary+particle&amp;amp;hl=en&amp;amp;ei=pqq_TdGcHojYgQfQp7j2BQ&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=book-preview-link&amp;amp;resnum=1&amp;amp;ved=0CD4QuwUwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Facts and Mysteries in Elementary particle physics&lt;/a&gt; He introduces almost all the concepts he talks about, provides interesting biographies and stories, and keeps the material at a level perfect for layman and high school students. And he is a Nobel Laureate.&lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="2599" LastEditorDisplayName="" LastEditDate="2011-05-03T07:17:48.693" LastActivityDate="2011-05-03T07:17:48.693" />
  <row Id="9402" PostTypeId="2" ParentId="9389" CreationDate="2011-05-03T06:26:07.473" Score="0" ViewCount="" Body="&lt;p&gt;Robert Oerter's &quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0452287863&quot; rel=&quot;nofollow&quot;&gt;The Theory of Almost Everything&lt;/a&gt;&quot; is a very readable book that focuses on the development of the Standard Model. It's a gentle introduction to some crucial ideas in particle physics (that aren't string theory...) and also gives a bit of historical context for how the field developed.&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2011-05-03T06:26:07.473" />
  <row Id="9403" PostTypeId="2" ParentId="9400" CreationDate="2011-05-03T06:53:47.237" Score="2" ViewCount="" Body="&lt;p&gt;It is zero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_induction&quot; rel=&quot;nofollow&quot;&gt;Faraday's law&lt;/a&gt; states that the induced EMF around the boundary of a surface is proportional to the time derivative of the flux through a that surface.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, when the magnetic field is constant in both space and time there is no electromagnetic induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There would be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hall_effect&quot; rel=&quot;nofollow&quot;&gt;Hall effect&lt;/a&gt;.  Its magnitude would depend on the material, especially the density of charge carriers.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-05-03T06:59:23.987" LastActivityDate="2011-05-03T06:59:23.987" CommentCount="11" />
  <row Id="9404" PostTypeId="2" ParentId="9392" CreationDate="2011-05-03T07:20:03.773" Score="2" ViewCount="" Body="&lt;p&gt;A couple of links, they could be useful:    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On critical collapse of gravitational waves.&#xA;Evgeny Sorkin. &lt;a href=&quot;http://arxiv.org/abs/arXiv:1008.3319&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/arXiv:1008.3319&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultra Relativistic Particle Collisions&#xA;Matthew W. Choptuik, Frans Pretorius&#xA;&lt;a href=&quot;http://arxiv.org/abs/arXiv:0908.1780&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/arXiv:0908.1780&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3229" LastActivityDate="2011-05-03T07:20:03.773" />
  <row Id="9405" PostTypeId="2" ParentId="9331" CreationDate="2011-05-03T09:00:49.460" Score="0" ViewCount="" Body="&lt;p&gt;Strictly speaking, planetary orbits are not perpetual motion.  As the planet (and their star) rotate around their common center of gravity, they emit gravitational waves.  Those gravitational waves drains the planet/star orbital system on energy so the planet eventually get's closer and closer to its star.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this orbital decay due to gravitational wave emission is ridicoulusly small so it has only been measured for extreme systems like binary neutron stars (which are heavy and may orbit each other within minutes, seconds or just a fraction of a second).  Our planet Earth is subject to gravitational wave emission as well, but the orbital decay is so small that it in practice won't affect earth within the Suns lifetime, instead of Earth spiralling into and being devoured by our Sun, it will rather be our Sun (turning into a red giant) which extends beyond the current orbit of Earth and thus devours it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even by cosmological timescales, planetary motion around a star will go on for a long time, e.g for the lifetime of a sun-like star and beyond.&lt;/p&gt;&#xA;" OwnerUserId="3245" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-05-03T09:38:40.700" LastActivityDate="2011-05-03T09:38:40.700" />
  <row Id="9406" PostTypeId="2" ParentId="9400" CreationDate="2011-05-03T10:49:09.353" Score="0" ViewCount="" Body="&lt;p&gt;You are asking about a &lt;a href=&quot;http://en.wikipedia.org/wiki/Homopolar_generators&quot; rel=&quot;nofollow&quot;&gt;homopolar generator&lt;/a&gt;. The excellent UMD physics demo page shows an &lt;a href=&quot;http://www.physics.umd.edu/lecdem/services/demos/demosk1/k1-41.htm&quot; rel=&quot;nofollow&quot;&gt;experimental demonstration&lt;/a&gt; and includes &lt;a href=&quot;http://dx.doi.org/10.1119/1.1987136&quot; rel=&quot;nofollow&quot;&gt;references&lt;/a&gt;. I'll try to find a free version of that article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &lt;a href=&quot;http://farside.ph.utexas.edu/teaching/plasma/lectures/node71.html&quot; rel=&quot;nofollow&quot;&gt;Fitzpatrick&lt;/a&gt; discusses homopolar generators in his online plasma physics textbook.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="2359" LastEditorDisplayName="" LastEditDate="2011-05-03T10:56:40.220" LastActivityDate="2011-05-03T10:56:40.220" CommentCount="2" />
  <row Id="9407" PostTypeId="1" AcceptedAnswerId="9408" CreationDate="2011-05-03T11:36:24.163" Score="1" ViewCount="84" Body="&lt;p&gt;Why do bar magnets have the least attraction in its center?&lt;/p&gt;&#xA;" OwnerUserId="3097" LastActivityDate="2011-05-03T12:29:34.757" Title="Why do bar magnets have least attraction in its center?" Tags="&lt;electromagnetism&gt;" AnswerCount="1" />
  <row Id="9408" PostTypeId="2" ParentId="9407" CreationDate="2011-05-03T12:29:34.757" Score="3" ViewCount="" Body="&lt;p&gt;Their field lines look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/bTeXS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see that the field lines near the center are parallel to the length of the bar. Moreover, the density of the field lines on the picture is maximized at the &quot;poles&quot; of the bar magnet and minimized at the center which means that the magnitude of the magnetic field is smallest at the center, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not hard to see why the field lines look approximately as those on the picture above. For an infinitely long magnet, the magnetic field would be fully confined to the cylinder - think of an infinitely long solenoid which essentially has the same magnetic field as a bar magnet. For a bar magnet of a finite length, the field lines ultimately split behind the poles but they're still comparably strong over there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the field outside the bar, and far from the poles, is only nonzero because of some subleading effects; it would vanish for an infinitely long solenoid - that's why the field is so small over there.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-03T12:29:34.757" />
  <row Id="9409" PostTypeId="2" ParentId="9389" CreationDate="2011-05-03T12:49:48.927" Score="1" ViewCount="" Body="&lt;p&gt;Bruce Schumm's &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/080187971X&quot; rel=&quot;nofollow&quot;&gt;Deep down things: the breathtaking beauty of particle physics&lt;/a&gt;&lt;/em&gt; is also quite readable and not too technical.  (See the Amazon reviews)  I found elementary descriptions of esoteric concepts like gauge theory, renormalization, the need for Lie algebras, the evolution of various theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally speaking, it's best to browse multiple books because they tend to complete each other.&lt;/p&gt;&#xA;" OwnerUserId="1468" LastActivityDate="2011-05-03T12:49:48.927" />
  <row Id="9410" PostTypeId="2" ParentId="9390" CreationDate="2011-05-03T13:00:26.177" Score="0" ViewCount="" Body="&lt;p&gt;The Casimir effect (plate attraction) is better understood as a Van der Waals force due to mutual polarization of neutral but extended quantum mechanical systems. Consider (find out about) two distant neutral atoms, for example. The case of plates is physically not too different from that. It is funny, but this attraction can be attributed to the virtual particles &lt;em&gt;between&lt;/em&gt; atoms, contrary to the OP statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force is a (minus) gradient of the electric potential of interaction. Calculation of possible modes inside plates helps calculate this potential energy.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-05-03T13:42:24.783" LastActivityDate="2011-05-03T13:42:24.783" />
  <row Id="9411" PostTypeId="1" CreationDate="2011-05-03T13:07:23.160" Score="3" ViewCount="103" Body="&lt;p&gt;Mixing of two different fluids is associated with an increase of entropy. Conversely, separation of two gases must be associated with a decrease of the entropy of the two fluids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a minimum requirement imposed by this thermodynamic constraint on the minimum energy needed to separate two fluids? &lt;/p&gt;&#xA;" OwnerUserId="1559" LastActivityDate="2011-07-14T14:19:06.340" Title="Energy of unmixing" Tags="&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="9412" PostTypeId="2" ParentId="9391" CreationDate="2011-05-03T13:12:46.773" Score="0" ViewCount="" Body="&lt;p&gt;In addition to Lubos’ answer, it can be said that the polytropic exponent n is a representation of our ignorance in the actual transformation undergone by a gas. There are basically two limiting cases : adiabatic and isothermal. The later is characterized by PV=Cst, ie n=1, while the former has n= gamma. A real process is never perfectly isothermal or adiabatic, and this results in a wide range of values possible for n.&lt;/p&gt;&#xA;" OwnerUserId="1559" LastActivityDate="2011-05-03T13:12:46.773" />
  <row Id="9413" PostTypeId="2" ParentId="208" CreationDate="2011-05-03T14:17:15.340" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Definitely Yes.&lt;br&gt;&#xA;The Earth is moving away from the Sun at a rate ($0.57H_{0}$) that can not be explained by any of the current &lt;em&gt;official&lt;/em&gt; models.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once again I point to a &lt;strong&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/astro-ph/pdf/0208/0208365v1.pdf&quot; rel=&quot;nofollow&quot;&gt;MODEL&lt;/a&gt;&lt;/strong&gt; in which the data is consistent with the theory. This model is public since 2002 (and my personal knowledge since 1982) in anticipation of the factual finding that the Earth was moving away from the Sun.&lt;br&gt;&#xA;It is very problematic to place the origin of life in a cold world. With this model the life has started on an Earth full of energy.&lt;br&gt;&#xA;I think that someday this model will be studied with due attention.&#xA;I am not the author of the model. I'm just a messenger (preaching in the desert?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by G. A. Krasinsky and V. A. Brumberg, 2004&lt;br&gt;&#xA;&lt;a href=&quot;http://www.springerlink.com/content/g5051650115444k9/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Secular Increase of astronomical unit from analysis of the major planet motions, and Its Interpretation&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;$\frac{dAU}{dt}=15\pm4\: m/cy$&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;at present there is no satisfactory&#xA;  explanation of the detected secular&#xA;  increase of AU&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;WeiJia Zhang, ZhengBin Li and Yang Lei, 2010&lt;br&gt;&#xA;&lt;a href=&quot;http://www.springerlink.com/content/k130348656287201/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Experimental measurement of growth patterns on fossil corals: Secular variation in short distances ancient Earth-Sun&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;both the modern and ancient leaving&#xA;  rates could be measured with high&#xA;  precision, &lt;strong&gt;and it was found that the&#xA;  Earth has been leaving the Sun over&#xA;  the past 0.53 billion years&lt;/strong&gt;. The&#xA;  Earth’s semi-major axis was 146&#xA;  million kilometers at the beginning of&#xA;  the Phanerozoic Eon, equating to 97.6%&#xA;  of its current value. Measured modern&#xA;  leaving rates are 5–14 m/cy, whereas&#xA;  the ancient rates were much higher.&#xA;  &lt;strong&gt;Experimental results indicate a special expansion with an average&#xA;  expansion coefficient of $0.57H_{0}$&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-05-03T14:17:15.340" />
  <row Id="9414" PostTypeId="2" ParentId="9088" CreationDate="2011-05-03T15:29:31.927" Score="0" ViewCount="" Body="&lt;p&gt;Sometimes the world of physics is influenced by the '&lt;a href=&quot;http://en.wikipedia.org/wiki/Realpolitik&quot; rel=&quot;nofollow&quot;&gt;real politik&lt;/a&gt;' and I bold read in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ozone_depletion#Public_policy&quot; rel=&quot;nofollow&quot;&gt;same WP-Ozone link:&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;A critical DuPont manufacturing patent&#xA;  for Freon was set to expire in 1979&lt;/strong&gt;&#xA;  ...&#xA;  The United States banned the use of&#xA;  CFCs in aerosol cans in 1978...&lt;br&gt;&#xA;  Dr. Mostafa Tolba, former head of the UN Environment Programme, who was quoted in the 30 June 1990 edition of The New Scientist, '...the &lt;strong&gt;chemical industry supported the Montreal Protocol&lt;/strong&gt; in 1987 because it set up a worldwide schedule for phasing out CFCs, &lt;strong&gt;which [were] no longer protected by patents. This provided companies with an equal opportunity to market new, more profitable compounds.&lt;/strong&gt;'&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Simply put: without regulation = small business due to the low manufacturing price; with regulation = new protected compounds and big business, as usual;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The subject was born, grew up in a short time and bore fruits (a regulation and a Nobel). It is now almost dead and buried.&lt;br&gt;&#xA;Now back to physics speculation:&lt;br&gt;&#xA;I wonder: Would have been ruled out other possible explanations for the phenomenon?&#xA;For example: variation of the Earth magnetic axis, some characteristic of the solar cycle, cosmic rays, a mix of causes, etc...&lt;br&gt;&#xA;As protons and electrons are differently attracted to North and South magnetic Poles a distinct pattern could arise and explain your question in a different way.   &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-05-03T16:51:40.077" LastActivityDate="2011-05-03T16:51:40.077" />
  <row Id="9415" PostTypeId="1" AcceptedAnswerId="9418" CreationDate="2011-05-03T16:32:10.207" Score="6" ViewCount="356" Body="&lt;p&gt;We all learn in grade school that electrons are negatively-charged particles that inhabit the space around the nucleus of an atom, that protons are positively-charged and are embedded within the nucleus along with neutrons, which have no charge. I have read a little about electron orbitals and some of the quantum mechanics behind why electrons only occupy certain energy levels. However...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the electromagnetic force work in maintaining the positions of the electrons? Since positive and negative charges attract each other, why is it that the electrons don't collide with the protons in the nucleus? Are there ever instances where electrons and protons &lt;em&gt;do&lt;/em&gt; collide, and, if so, what occurs?&lt;/p&gt;&#xA;" OwnerUserId="3395" LastActivityDate="2011-05-05T02:14:42.123" Title="Why do electrons occupy the space around nuclei, and not collide with them?" Tags="&lt;quantum-mechanics&gt;&lt;electromagnetism&gt;&lt;nuclear-physics&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="9416" PostTypeId="2" ParentId="9415" CreationDate="2011-05-03T16:40:57.137" Score="3" ViewCount="" Body="&lt;p&gt;This was the basic reason for the invention of quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple mechanics with electromagnetism do not work in atomic dimensions, particularly with the charged electrons. Classical electromagnetism would have the electrons radiate energy away  because of the continuous acceleration of a circular path and finally fall in the nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is : because in the microscopic world nature  follows quantum mechanics equations and not classical mechanics equations. Quantum mechanics equations  include electromagnetic fields, and their solutions are stable and allow for the existence of atoms, which is what we experimentally observed to start with.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-05-03T17:49:50.690" LastActivityDate="2011-05-03T17:49:50.690" CommentCount="2" />
  <row Id="9417" PostTypeId="2" ParentId="9415" CreationDate="2011-05-03T16:43:38.480" Score="0" ViewCount="" Body="&lt;p&gt;It is due to the uncertainty principle of quantum mechanics. $\delta x \delta p_x &amp;gt; h$&#xA;Here $h$ is the Planck constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an electron comes too close to the nucleus then its position is determined within a very narrow margin of error. According to the uncertainty principle, the uncertainty in its momentum should be too large then. Accordingly it can't stay close for too long in the vicinity of the nucleus. Instead it is smeared out around the nucleus with some probability distribution.&lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2011-05-03T16:43:38.480" CommentCount="9" />
  <row Id="9418" PostTypeId="2" ParentId="9415" CreationDate="2011-05-03T16:50:25.570" Score="5" ViewCount="" Body="&lt;p&gt;In fact the electrons (at least those in s-shells) &lt;em&gt;do&lt;/em&gt; spend some non-trivial time inside the nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason they spend a lot of time outside the nucleus is essentially quantum mechanical. To use too simple an explanation their momentum is restricted to a range consistent with begin captured (not free to fly away), and as such there is a necessary uncertainty in their position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of physics arising because they spend &lt;em&gt;some&lt;/em&gt; time in the nucleus is so called &lt;em&gt;&quot;beta capture&quot;&lt;/em&gt; radioactive decay in which&#xA;$$ e + p \to n + \nu $$&#xA;occurs within the nucleus. The reason this does not happen in most nuclei is &lt;em&gt;also&lt;/em&gt; quantum mechanical and is related to energy levels and Fermi-exclusion.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To expand on this picture a little bit, let's appeal to de Broglie and Bohr. Bohr's picture of the electron orbits being restricted to a set of finite energies $E_n \propto 1/n^2$ and frequencies can be given a reasonably natural explanation in terms of de Broglie's picture of all matter as being composed of waves of frequency $f = E/h$ by requiring that a integer number of waves fit into the circular orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to a picture of the atom in which all the electrons occupy neat circular orbits far away from the nucleus, and provides one explanation of why the electrons don't just fall into the nucleus under the electrostatic attraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's not the whole story for a number of reasons; for our purposes the most important one is that Bohr's model predicts a minimum angular momentum for the electrons of $\hbar$ when the experimental value is 0.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Pushing on, can solve the three dimensional Schrödinger equation in three dimesions for Hydrogen-like atoms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left( i\hbar\frac{\partial}{\partial t} - \hat{H} \right) \Psi = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for electrons in a $1/r^2$ electrostatic potential to determine the wavefunction $\Psi$. The wave function is related to the probability $P(\vec{x})$ of finding an electron at a point $\vec{x}$ in space by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P(\vec{x}) = \left| \Psi(\vec{x}) \right|^2 = \Psi^{*}(\vec{x}) \Psi(\vec{x}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $^{*}$ means the complex conjugate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solutions are usually written in the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Psi(\vec{x}) = Y^m_l(\theta,\phi) L^{2l+1}_{n-l-1}(r) e^{-r/2} * \text{normalizing factors} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the $Y$'s are the spherical harmonics and the $L$'s are the generalized Laguerre polynomials. But we don't care for the details. Suffice it to say that that these solutions represent a probability density for the electrons that is smeared out over a wide area near around the nucleus. Also of note, for $l=0$ states (also known as s orbitals) there is a non-zero probability at the center, which is to say &lt;em&gt;in&lt;/em&gt; the nucleus (this fact arises because these orbital have zero angular momentum, which you might recall was not a feature of the Bohr atom).&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-05T02:14:42.123" LastActivityDate="2011-05-05T02:14:42.123" CommentCount="6" />
  <row Id="9419" PostTypeId="1" AcceptedAnswerId="9420" CreationDate="2011-05-03T17:11:40.820" Score="2" ViewCount="202" Body="&lt;p&gt;How can the universe become infinite in spatial extent if it started as a singularity, wouldnt it take infinite time to expand into an infinite universe?&lt;/p&gt;&#xA;" OwnerUserId="3396" LastActivityDate="2011-05-03T17:22:01.827" Title="How can something finite become infinite?" Tags="&lt;cosmology&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="9420" PostTypeId="2" ParentId="9419" CreationDate="2011-05-03T17:22:01.827" Score="3" ViewCount="" Body="&lt;p&gt;If the Universe is spatially infinite, it always had to be spatially infinite, even though the distances were shortened by an arbitrary factor right after the Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a spatially infinite Universe, one has to be careful that the singularity doesn't necessarily mean a single point in space. It is a place - the whole Universe - where quantities such as the density of matter diverge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general relativity, people use the so-called Penrose (causal) diagrams of spacetime in which the light rays always propagate along diagonal lines tilted by 45 degrees. If you draw the Penrose diagram for an old-fashioned Big Bang cosmology, the Big Bang itself is a horizontal line - suggesting that the Big Bang was a &quot;whole space worth of points&quot; and not just a point. This is true whether or not the space is spatially infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the popular level - and slightly beyond - these issues are nicely explained in Brian Greene's new book, &lt;em&gt;The Hidden Reality&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-03T17:22:01.827" />
  <row Id="9421" PostTypeId="1" AcceptedAnswerId="9501" CreationDate="2011-05-03T19:52:36.607" Score="7" ViewCount="271" Body="&lt;p&gt;Recently i found out this &lt;a href=&quot;http://news.stanford.edu/news/2010/august/sun-082310.html&quot; rel=&quot;nofollow&quot;&gt;uber strange article&lt;/a&gt; about nuclear decay rates being somehow showing seasonal variations with a high correlation with sun activity. Two very precise questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;strong&gt;has this been experimentally confirmed/disproved?&lt;/strong&gt; an experiment using neutrinos from a fission reactor would be awesome, although probably a couple orders of magnitude below the required luminosity (at least to be comparable with solar sources)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;strong&gt;could the standard model possibly allow neutrinos to modulate decay rates in this way? or do we need new physics?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;link to the &lt;a href=&quot;http://arxiv.org/abs/1007.0924&quot; rel=&quot;nofollow&quot;&gt;public version of the paper&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; brief explanation why i tag this question as &lt;code&gt;cavity-qed&lt;/code&gt;; because &lt;em&gt;it is the only other known mechanism we are aware that you can use to shift decay rates of energy levels&lt;/em&gt;, it might be interesting to see if there are deeper relationships between both mechanisms involved&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-05-05T08:37:52.333" LastActivityDate="2011-05-07T18:35:48.977" Title="strange modulation of radiactive decay rates with solar activity" Tags="&lt;standard-model&gt;&lt;radioactivity&gt;&lt;neutrino&gt;&lt;cavity-qed&gt;&lt;cmb&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" />
  <row Id="9422" PostTypeId="1" CreationDate="2011-05-03T20:24:00.383" Score="2" ViewCount="137" Body="&lt;p&gt;I'll keep it simple. How does inflation drive Ω close to 1?&lt;/p&gt;&#xA;" OwnerUserId="3398" LastActivityDate="2011-05-05T14:50:06.370" Title="How does inflation drive Ω close to 1?" Tags="&lt;astrophysics&gt;&lt;cosmology&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="9423" PostTypeId="1" AcceptedAnswerId="9427" CreationDate="2011-05-03T20:29:23.707" Score="-2" ViewCount="86" Body="&lt;p&gt;What kind of object is (656) Beagle [1908BU]?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know it's a minor planet, but that includes a large array of different stuff. Specifically, I am looking at the general chemistry/geology of the object.&lt;/p&gt;&#xA;" OwnerUserId="3399" LastActivityDate="2011-05-03T21:36:09.537" ClosedDate="2011-05-03T21:48:05.757" Title="What is 656 Beagle?" Tags="&lt;astrophysics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9425" PostTypeId="5" CreationDate="2011-05-03T21:11:54.057" Score="0" ViewCount="" Body="&lt;p&gt;Cavity quantum electrodynamics (cavity QED) is the study of the interaction between light confined in a reflective cavity and atoms or other particles, under conditions where the quantum nature of light photons is significant. &lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-05-03T21:48:42.450" LastActivityDate="2011-05-03T21:48:42.450" />
  <row Id="9426" PostTypeId="4" CreationDate="2011-05-03T21:11:54.057" Score="0" ViewCount="" Body="Cavity QED is the study of quantum effects on small cavities, manifested in variations in the lifetimes of unstable energy levels" OwnerUserId="-1" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-05-03T21:48:39.627" LastActivityDate="2011-05-03T21:48:39.627" />
  <row Id="9427" PostTypeId="2" ParentId="9423" CreationDate="2011-05-03T21:19:43.930" Score="0" ViewCount="" Body="&lt;p&gt;656 Beagle is a small (~53 km diameter) asteroid with a 3.15 AU semimajor axis. It's luminosity is very faint, with H = 9.92. Nothing seems to be known about its geologic basis, but it certainly does not possess a measurable atmosphere. See &lt;a href=&quot;http://occsec.wellington.net.nz/planet/2008/updates/080912_656_19331_u.htm&quot; rel=&quot;nofollow&quot;&gt;http://occsec.wellington.net.nz/planet/2008/updates/080912_656_19331_u.htm&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="1924" LastEditorDisplayName="" LastEditDate="2011-05-03T21:36:09.537" LastActivityDate="2011-05-03T21:36:09.537" />
  <row Id="9428" PostTypeId="2" ParentId="9400" CreationDate="2011-05-03T21:25:07.053" Score="1" ViewCount="" Body="&lt;p&gt;For a wire moving in a magnetic field,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;emf = rate of cutting of magnetic flux&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is one form of Faraday's Law. This is straight forward to use when all parts of the wire are moving at the same velocity. If not, you can divide the wire up into small parts and imagine each small element having a unique velocity, with an emf across it, and the sum across each element giving the total emf across the ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One half of the rod sweeps out an area $\pi r^2$ in a time $1/f$ giving a rate of cutting of magnetic flux and hence emf between the center and r as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$emf = Bf\pi r^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more sophisticated approach is to use the Lorentz force law which needs a deeper understanding of what's going on inside the rod.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = e( E + v \times B)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the rod moves through the magnetic field, it carries with it electrons and metal ions, the former being of a much smaller mass so their motion is affected much more by electric and magnetic fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example, let $B$ point into the screen, the rod rotating clockwise about $r = 0$ in the plane of the screen. From the definition of the cross product $\times$, the magnetic force has a magnitude $evB$, and points towards r = 0 along the rod for an electron. The electrons being forced in this direction disrupts the neutrality of the rod, so creating an internal E which prevents further electron movement. So we can say that since $F = 0$ then $E = -vB$ and we've now got an internal electric field $E$ which can do work, unlike the static magnetic field B. To calculate the emf, we need the work done in moving a charge of 1 Coulomb between two points&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$emf = -\int_{0}^{r}Edr = \int_{0}^{r}vBdr$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the speed of an electron at $r$ is $2 \pi fr$, finally&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$emf = Bf\pi r^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which gives the same result as before. As expected, there is an emf between the center and each end, but not between the ends.&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="1345" LastEditorDisplayName="" LastEditDate="2011-05-04T00:05:24.813" LastActivityDate="2011-05-04T00:05:24.813" />
  <row Id="9429" PostTypeId="1" AcceptedAnswerId="9450" CreationDate="2011-05-03T22:19:38.390" Score="3" ViewCount="156" Body="&lt;p&gt;Do the apperance in the atomic nucleus of virtual matter-antimatter particle pairs play a role in the random nature of radioactive decay?&lt;/p&gt;&#xA;" OwnerUserId="2942" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-03T17:11:49.483" LastActivityDate="2011-06-03T17:11:49.483" Title="The vacuum as trigger" Tags="&lt;quantum-field-theory&gt;&lt;nuclear-physics&gt;&lt;radioactivity&gt;&lt;vacuum&gt;" AnswerCount="1" />
  <row Id="9430" PostTypeId="1" CreationDate="2011-05-03T21:25:19.347" Score="1" ViewCount="184" Body="&lt;p&gt;Interstellar space propulsion...if a spaceship were to get beyond our Sun's gravitational pull and since there is no atmosphere/wind/friction in space...does that mean, if an engine was constantly thrusting, that the spaceship would continue to accelerate to the speed of light, as long as the engine was propulsing forward?  In other words, does speed constantly increase in a vacuum? or are the two ideas (Space/Vacuum) seperate ideas?  Thanks!&lt;/p&gt;&#xA;" LastActivityDate="2011-05-06T07:29:38.783" Title="Interstellar space propulsion...?" Tags="&lt;space&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="9431" PostTypeId="2" ParentId="9430" CreationDate="2011-05-03T21:43:40.333" Score="4" ViewCount="" Body="&lt;p&gt;The answer is no, according to &lt;a href=&quot;http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html&quot; rel=&quot;nofollow&quot;&gt;Einstein's Theory of Relativity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As an object approaches the speed of light, more and more energy is needed to accelerate it further.  To reach the speed of light an infinite amount of energy would be required. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, there is wind/friction in space. There is no absolute vacuum, there is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Interstellar_medium&quot; rel=&quot;nofollow&quot;&gt;interstellar medium&lt;/a&gt;. Indeed, relativistic rockets would have to account for the medium, and perhaps even use it as a source of fuel, see: &lt;a href=&quot;http://en.wikipedia.org/wiki/Bussard_ramjet&quot; rel=&quot;nofollow&quot;&gt;Bussard Ramjet&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2872" LastActivityDate="2011-05-03T21:50:07.933" CommentCount="6" />
  <row Id="9432" PostTypeId="2" ParentId="9421" CreationDate="2011-05-03T23:37:17.413" Score="-3" ViewCount="" Body="&lt;p&gt;I follow the work of Cahill (out of mainstream) about the The &lt;strong&gt;anisotropy of the speed of light&lt;/strong&gt;.&#xA;Although I think that we have no chance of measure this directlly I think that this effect can be an indirect manifestation of it. &lt;strong&gt;The path of Earth thru CMB referential is cyclic (annual)&lt;/strong&gt;. May be he is not far from a good viewpoint when he tried to model the &lt;strong&gt;flyby anomaly&lt;/strong&gt; (&lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0804/0804.0039v4.pdf&quot; rel=&quot;nofollow&quot;&gt;Resolving Spacecraft Earth-Flyby Anomalies with Measured Light Speed Anisotropy&lt;/a&gt; - Cahill, Progress in physics 3, 9, (2008)) (*)&lt;br&gt;&#xA;Of course I'm not absolutelly sure that this is the correct answer that, in the future, will be adopted. Anyway I keep myself vigilant with crazy ideas.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT add :&lt;br&gt;&#xA;The answer to this question may turn interesting:&lt;br&gt;&#xA;&lt;strong&gt;The seasonal data as measured in this paper are better aligned with the direction of the dipole of the CMB (anisotropy of the speed of light) or the major axis of the trajectory of the Earth around the Sun (neutrinos) ?&lt;/strong&gt;&lt;br&gt;&#xA;I didnt found a graph of a seasonal variation of the projection (dot product) of the Earth's speed vector with the CMB dipole direction vector to compare with the graphs of this paper. I hope that someone can help with this.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also from &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_microwave_background_radiation#CMBR_dipole_anisotropy&quot; rel=&quot;nofollow&quot;&gt;WP-CMBR_dipole_anisotropy&lt;/a&gt;:&lt;br&gt;&#xA;  The dipole anisotropy and others due to Earth's annual motion relative to the Sun...&#xA;  moving at 627±22 km/s relative to the reference frame of the CMB (also called the CMB rest frame, or the frame of reference in which there is no motion through the CMB) in the direction of galactic longitude l = 276±3°, b = 30±3°.&lt;br&gt;&#xA;  Some observers have pointed out that the&#xA;  anisotropies in the WMAP data did not&#xA;  appear to be consistent with the big&#xA;  bang picture. In particular, &lt;strong&gt;the&#xA;  quadrupole and octupole (l=3) modes&#xA;  appear to have an unexplained&#xA;  alignment with each other &lt;em&gt;and with&#xA;  the ecliptic plane&lt;/em&gt;&lt;/strong&gt;, an alignment&#xA;  sometimes referred to as &lt;strong&gt;&lt;a href=&quot;http://blog.lib.umn.edu/mill1974/EGAD/041107.html&quot; rel=&quot;nofollow&quot;&gt;the axis of evil&lt;/a&gt;&lt;/strong&gt;. A number of groups have&#xA;  suggested that &lt;strong&gt;this could be the&#xA;  signature of new physics&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I'm reading ok &lt;strong&gt;the ecliptic plane&lt;/strong&gt; reads: the trajectory of Earth thru space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a comment: I've in my mind some form of &lt;em&gt;why&lt;/em&gt; the &lt;strong&gt;anisotropy of the speed of light&lt;/strong&gt; is important at nuclear scale but that is more than we need to discuss here.&lt;br&gt;&#xA;The neutrino was the first darkness to enter in the menu and it will be the last to evade. Of course that nobody will take this statement seriously. I bet 100 % that the correlation to the CMB dipole direction is much more interesting than the neutrino link.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) I found a wild comment in the net that someone bet, and won, that the flyby RosetaIII (Nov/13/2009) will have a null result just because of the calendar date. Coincidence?&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-05-04T16:01:12.873" LastActivityDate="2011-05-04T16:01:12.873" CommentCount="13" />
  <row Id="9433" PostTypeId="2" ParentId="9307" CreationDate="2011-05-03T23:38:14.120" Score="0" ViewCount="" Body="&lt;p&gt;Mark is right: If your fulcrum lies on the straight line between the hanging points for weight 1 (w1) and weight 2 (w2); then the angle can be anything if w1=w2, and will be straight up and down if w1 does not equal w2 (the heavier weight being at the bottom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Georg is correct, that the way to change this is to offset the fulcrum (make is higher than the straight line between w1 and w2 (or, equivalently, make it higher than the hanging points for the weighing baskets for w1 and w2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to get a simple closed form which described the balance angle as a function of w1/w2, but could not in the short time I have to devote to this problem.  (Maybe someone can pick up where I left off).  What I do have is an equation which describes that relationship, as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let B be the half-length of a balance beam between w1 (on the left) and w2 (on the right).  Let H be the height of the fulcrum point (F) above this straight line.  (So, lines drawn between points w1, w2 and F make a short, wide triangle.) Let L = the slant distance from F to w1 (and also from F to W2).  L = SQRT(B^2 + H^2).  If you draw a line through F and parallel to line w1w2 (the base of the triangle) then lines w1F and w2F both make a small angle 'A' (radians) with it.  Use the convention that if w1 is heavier, it will cause a positive rotation of the balance beam by an angle of 'D' (radians).  By writing and solving a moments eqn around point F, you get the eqn:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;w1/w2 = (cos(D-A)) / (cos(D+A))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D must (and in the real world will) be limited to the range:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-((Pi/2)-A) radians&amp;lt; D &amp;lt; ((Pi/2)-A) radians.&lt;/p&gt;&#xA;" OwnerUserId="3400" LastEditorUserId="3400" LastEditorDisplayName="" LastEditDate="2011-05-04T00:12:06.557" LastActivityDate="2011-05-04T00:12:06.557" />
  <row Id="9434" PostTypeId="2" ParentId="9341" CreationDate="2011-05-04T00:48:22.937" Score="0" ViewCount="" Body="&lt;p&gt;Both answers 1 and 2 are the correct answers.  It appears your teacher made some mental leaps (and expected you to come along) to narrow it down to answer #1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring gravitational and relativistic considerations (and for practical rockets in deep space, this is reasonable), and assuming an ideal rocket engine, the final rocket velocity can be expressed as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pv = Ev x ln((Pm+Em)/Pm)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pv = Empty rocket (Payload) Velocity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pm = Empty rocket (Payload) mass &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ev = Exhaust velocity of rocket engine gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Em = Total Ejected propellant (fuel + oxidizer) mass&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;ln(x)&quot; means &quot;take the natural logarithm of x.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I saw something approximating this form in sb1's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the eqn, you can see that final payload velocity is linearly proportional to exhaust velocity, and very non-linearly proportional to Em/Pm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first mental leap you teacher wanted was for you to realize that there are practical limitations on Ev.  If you want to increase the speed of a modern rocket that is pushing the envelope of state of the art science, by 45%, you can't just go out and grab a propellent/rocket combo that increases Ev by 45%.  We don't have that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second mental leap you teacher wanted was for you to realize that (s)he meant &quot;amount of fuel&quot; as a percentage of total rocket mass.  As you increase this percentage you have more leverage to increase final payload velocity.&lt;/p&gt;&#xA;" OwnerUserId="3400" LastActivityDate="2011-05-04T00:48:22.937" CommentCount="3" />
  <row Id="9435" PostTypeId="1" AcceptedAnswerId="9443" CreationDate="2011-05-04T01:05:48.777" Score="1" ViewCount="171" Body="&lt;p&gt;I recently found a &lt;a href=&quot;http://en.wikipedia.org/wiki/Half_dollar_%28United_States_coin%29&quot; rel=&quot;nofollow&quot;&gt;half dollar coin&lt;/a&gt; in my basement.  It's strange but I think when you flip it, that heads is a big favorite to land face up!  I really could have sworn I heard that somewhere before; am I wrong or is the half dollar weighted for heads?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's funny, I actually said to my friend yesterday, &quot;Here look at this coin, I think it's weighted for heads, like 3 out of 4 times it lands heads.&quot;  Well I flipped it 4 times and that's exactly what happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion it's because the coin is large and the face seems to be pretty heavy; I'm curious to know if anyone can derive the exact chance of a half dollar landing heads, or to convince me it's not weighted.&lt;/p&gt;&#xA;" OwnerUserId="442" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-04T03:16:33.230" LastActivityDate="2011-05-04T03:16:33.230" Title="What is the probability that a half dollar coin lands on heads?" Tags="&lt;newtonian-mechanics&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="9436" PostTypeId="2" ParentId="9331" CreationDate="2011-05-04T01:06:28.913" Score="0" ViewCount="" Body="&lt;p&gt;Perpetual motion is a system that produces work (energy) while maintaining its state (which implies it can forever produce energy without changing state).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbiting planets are not a perpetual motion system.  NOTHING we know of or can explain is a perpetual motion system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbiting planets APPEAR to be a perpetual motion system, because the length of time needed for an OBSERVABLE (measurable) state change to occur is enormous (many, many, many, many lifetimes).&lt;/p&gt;&#xA;" OwnerUserId="3400" LastActivityDate="2011-05-04T01:06:28.913" />
  <row Id="9437" PostTypeId="2" ParentId="9435" CreationDate="2011-05-04T01:21:08.147" Score="2" ViewCount="" Body="&lt;p&gt;Measuring the probability of heads and it being weighted are mutually exclusive up to a point. To measure the probability it needs to be flipped statistically, to measure if it's weighted then need find its centre of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are at least two types of weighted that needs to be considered:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physically : you can measure if one side is heavier than other. But this does not neccerily imply that you would get one side coming up more than other. (This is the physics forum answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically : Toss the coin and see what happens statiscally, the coin might be weighted (physically or not) but if one side turns up more than other then yes it mathematically weighted :&#xA; &lt;a href=&quot;http://www.vosesoftware.com/ModelRiskHelp/Analysing_and_using_data/Bayesian/Bayesian_analysis_example_identifying_a_weighted_coin.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.vosesoftware.com/ModelRiskHelp/Analysing_and_using_data/Bayesian/Bayesian_analysis_example_identifying_a_weighted_coin.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Need to correlate the both to see at what point one type of weighted can be inferred from the other.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-05-04T01:45:06.340" LastActivityDate="2011-05-04T01:45:06.340" />
  <row Id="9438" PostTypeId="2" ParentId="9435" CreationDate="2011-05-04T01:28:28.910" Score="0" ViewCount="" Body="&lt;p&gt;To determine EXPERIMENTALLY, by flipping, that the coin was or was not weighted would take a very tightly contolled experiment with many repititions.  Certainly 4 flips is not statistically significant; and flipping by hand is not random.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally knew a guy who was prcticed at making coin tosses come out per his wish.  He was not perfect, but very good at it.  You may inadvertently be doing the same thing.  Of course, you can't tell after 4 flips.  Try 10,000.&lt;/p&gt;&#xA;" OwnerUserId="3400" LastActivityDate="2011-05-04T01:28:28.910" CommentCount="1" />
  <row Id="9439" PostTypeId="1" AcceptedAnswerId="9444" CreationDate="2011-05-04T01:51:34.693" Score="0" ViewCount="151" Body="&lt;p&gt;In a test I took recently in my beginning Physics class there was a question that really irked me: &quot;A man is swimming upstream at 10 mph.  The water is flowing downstream at 5 mph.  How fast is the man moving?&quot;  Well, you can actually take this question two ways.  Firstly, &quot;a man is swimming upstream at 10 mph&quot; could mean exactly what it sounds like, and therefore the 5 mph down-streaming water is irrelevant and the man was actually using enough force to swim 15 mph in stationary water.  That means he was moving at a speed of 10 mph, like it stated.  Then, it could be taken to mean &quot;a man is swimming upstream at 10 mph&quot;, meaning that that is the amount of force he is using.  The man is using enough force to swim 10 mph in stationary water.  Then, if the water is moving downstream at 5 mph, then he is moving at a speed of 5 mph.  I chose the first answer, but I wanted to figure if I had gotten the question right.  Is there a way to tell, sort of an order of operations for Physics?&lt;/p&gt;&#xA;" OwnerUserId="3385" LastEditorUserId="3385" LastEditorDisplayName="" LastEditDate="2011-05-04T02:05:27.297" LastActivityDate="2011-05-05T07:08:06.363" Title="Relativity Question" Tags="&lt;classical-mechanics&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="9440" PostTypeId="2" ParentId="9439" CreationDate="2011-05-04T01:56:41.800" Score="3" ViewCount="" Body="&lt;p&gt;Nope. The question is ambiguous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, if there is a way to resolve this ambiguity, it's a question of language interpretation. The physics only really starts once you've figured out what the question means. (To be fair, it's often the case that most of the difficulty in doing a physics problem is just translating it from English into math :-P)&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-05-04T01:56:41.800" CommentCount="1" />
  <row Id="9441" PostTypeId="1" AcceptedAnswerId="9446" CreationDate="2011-05-04T02:16:31.533" Score="2" ViewCount="119" Body="&lt;p&gt;In the question &lt;a href=&quot;http://physics.stackexchange.com/questions/2901/is-there-a-black-hole-in-the-centre-of-the-milky-way&quot;&gt;Is there a black hole in the centre of the Milky Way?&lt;/a&gt; the answer by Motl &lt;em&gt;seems&lt;/em&gt; to all but say the existence of that black hole is a fact (see also &lt;a href=&quot;http://physics.stackexchange.com/questions/3349/evidence-for-black-hole-event-horizons&quot;&gt;Evidence for black hole event horizons&lt;/a&gt;), whereas the answer by Bunn says &quot;There's a strong consensus among astrophysicists&quot;. According to the &lt;a href=&quot;http://chandra.harvard.edu/resources/faq/black_hole/bhole-10.html&quot; rel=&quot;nofollow&quot;&gt;Chandra X-ray Observatory FAQ&lt;/a&gt; the existence of that black hole is conditioned upon the validity of general relativity (GR); the answer there concludes &quot;So, unless Einstein's theory of gravity breaks down, our galactic center must contain a black hole.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Does humankind's discovery of a black hole in the center of the Milky Way depend on the validity of GR? (If no, can you say why the Chandra X-ray Observatory FAQ contradicts that?) If yes:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do &lt;em&gt;all&lt;/em&gt; of our discoveries of black holes in nature depend on the validity of GR? (If no, can you please elaborate on how one was discovered in a way that didn't depend on GR?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Before answering please note&lt;/strong&gt;: I'm not challenging GR in any way whatsoever here. I've tried to phrase the main questions so they can have unambiguous &quot;yes&quot; or &quot;no&quot; answers. As far as I can tell my questions aren't clearly answered elsewhere on this site. I realize that it would be difficult to discover a black hole without in any way depending on a theory of gravity, but I'm asking about a &lt;em&gt;particular&lt;/em&gt; theory of gravity here, a theory that predicts black holes.&lt;/p&gt;&#xA;" OwnerUserId="2314" LastEditorUserId="2314" LastEditorDisplayName="" LastEditDate="2011-05-04T02:32:15.927" LastActivityDate="2011-05-04T05:21:52.897" Title="Do all of our discoveries of black holes in nature depend on the validity of GR?" Tags="&lt;general-relativity&gt;&lt;black-holes&gt;" AnswerCount="2" />
  <row Id="9442" PostTypeId="2" ParentId="9430" CreationDate="2011-05-04T02:19:40.270" Score="1" ViewCount="" Body="&lt;p&gt;You asked three questions...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) ... does that mean, if an engine was constantly thrusting, that the spaceship would continue to accelerate to the speed of light, as long as the engine was propulsing forward? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No. For reasons ghoppe gave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) ... In other words, does speed constantly increase in a vacuum? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, speed would increase continuously, for the example you gave, asymtotically approaching (but never reaching) the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) ... or are the two ideas (Space/Vacuum) seperate ideas? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No.  A vacuum is an excellent and practical description of what most space is. &lt;/p&gt;&#xA;" OwnerUserId="3400" LastActivityDate="2011-05-04T02:19:40.270" CommentCount="1" />
  <row Id="9443" PostTypeId="2" ParentId="9435" CreationDate="2011-05-04T02:20:26.573" Score="8" ViewCount="" Body="&lt;p&gt;The flipping of a half dollar has been investigated (theoretically and experimentally) by the great Persi Diaconis along with Susan Holmes, and Richard Montgomery: see &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www-stat.stanford.edu/~cgates/PERSI/papers/headswithJ.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www-stat.stanford.edu/~cgates/PERSI/papers/headswithJ.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is rather impressive.  They conclude that &quot;vigorously-ﬂipped coins are biased to come up the same way they started&quot; in that a coin has about a 51% chance of landing on the side that started up when the coin was flipped.  They call this a &quot;dynamical bias.&quot; It has nothing to do with the asymmetry of the coin (which they take into account - they even give the principle moments of inertia of an American half dollar).  If the initial condition (heads up or tails up) is chosen randomly (and fairly!), then according to Diaconis et al. the result is essentially a fair coin toss. This of course raises the question of how this initial state should be generated in practice.  But in any case this dynamical bias, in practice, favors heads and tails equally.&lt;/p&gt;&#xA;" OwnerUserId="266" LastActivityDate="2011-05-04T02:20:26.573" CommentCount="2" />
  <row Id="9444" PostTypeId="2" ParentId="9439" CreationDate="2011-05-04T02:25:53.180" Score="1" ViewCount="" Body="&lt;p&gt;David answer is ok.&#xA;A man is swimming upstream at 10 mph  -- is undefined : &lt;strong&gt;in relation to&lt;/strong&gt; the ground or in relation to the water?,    &lt;em&gt;probably&lt;/em&gt; in relation to the ground  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;water is flowing downstream at 5 mph. -- &lt;em&gt;probably&lt;/em&gt; in relation to the ground  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How fast is the man moving?  --   is undefined : &lt;strong&gt;in relation to&lt;/strong&gt; the ground or in relation to the water?   &lt;em&gt;probably&lt;/em&gt; in relation to the ground  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;motion begs &lt;strong&gt;in relation to&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-05-04T02:25:53.180" CommentCount="1" />
  <row Id="9445" PostTypeId="2" ParentId="7644" CreationDate="2011-05-04T04:23:14.203" Score="0" ViewCount="" Body="&lt;p&gt;go to this chart&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nndc.bnl.gov/chart/reColor.jsp?newColor=235ufy&quot; rel=&quot;nofollow&quot;&gt;http://www.nndc.bnl.gov/chart/reColor.jsp?newColor=235ufy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in little box on right marked &quot;nucleus&quot; type 134I and click GO&lt;/p&gt;&#xA;&#xA;&lt;p&gt;click at top the box 235U FY&lt;/p&gt;&#xA;&#xA;&lt;p&gt;click just below on &quot;Zoom = 1 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you're there you'll see colored squares,  one in center says 134I&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that means iodine 134 .  the other squares in the row are other isotopes of iodine, the row above is xenon and one below is tellurium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the color gives a clue as to likelihood a uranium atom will split into a I134 atom. red is fairly likely.&lt;br&gt;&#xA; Number at bottom IS that probability,  0.0036 means that you'll get 3.6 I134's for every thousand fissions of U235. That's what the 235U FY button meant,  makes chart show fission yield.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the B- 100%. That means 100% of I134 decays by Beta which is electron emission.&#xA;Beta decay moves the isotope diagonally up one and left one, into 134Xe's box.&#xA;Note they also give the halflife 52.5 Minutes.&#xA;That means : however much I134 you have right now,  in 52.5 minutes half of it will have turned into Xe134, moving up one and leaft one..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that mean you lost half your i134?&lt;br&gt;&#xA;Well sorta,  half of the atoms you had indeed morphed into Xenon. &#xA;BUT look one square right and one down at13Te. It is a almost 20X more likely fission product -- 62.2 atoms of Te134 per thousand fissions compared to 3.6 of I134.&#xA;And it decays by beta so it morphs into I134.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your I134 box will simultaneously empty into the Xe134 box and  fill from the Te134 box.&#xA;Since fission made more Te atoms than I,  the inflow from Te will exceed outflow to Xe until you run low on Te. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes,  I134 comes from fission  but by two paths -  directly and via Te134. &#xA;The Te path looks to make most of the I134,  judging by fission yields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;did i get it across?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i dont see any precursors with long halflives.&#xA;You'll enjoy this site too...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.periodictable.com/Isotopes/053.134/index.html&quot; rel=&quot;nofollow&quot;&gt;http://www.periodictable.com/Isotopes/053.134/index.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;old jim&lt;/p&gt;&#xA;" OwnerUserId="3401" LastActivityDate="2011-05-04T04:23:14.203" />
  <row Id="9446" PostTypeId="2" ParentId="9441" CreationDate="2011-05-04T04:37:19.627" Score="1" ViewCount="" Body="&lt;p&gt;I would rephrase your question as : what is the experimental signature of a black hole. If there exists a definite experimental signature of a celestial body that defines a black hole, your question is answered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0701228&quot; rel=&quot;nofollow&quot;&gt;following&lt;/a&gt; paper that clarifies the issue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Classical black holes are solutions of the field equations of General Relativity. Many astronomical observations suggest that black holes really exist in nature. However, an unambiguous proof for their existence is still lacking. Neither event horizon nor intrinsic curvature singularity have been observed by means of astronomical techniques. This paper introduces to particular features of black holes. Then, we give a synopsis on current astronomical techniques to detect black holes. Further methods are outlined that will become important in the near future. For the first time, the zoo of black hole detection techniques is completely presented and classified into kinematical, spectro-relativistic, accretive, eruptive, obscurative, aberrative, temporal, and gravitational-wave induced verification methods. Principal and technical obstacles avoid undoubtfully proving black hole existence. We critically discuss alternatives to the black hole. However, classical rotating Kerr black holes are still the best theoretical model to explain astronomical observations. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From this it is evident that in contrast to most physics subjects where first there is experimental evidence needing explanation and then theory arrives, black holes are an &quot;artefact&quot; of general relativity theory, i.e. the concept carries all the baggage of GR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, what is called a &quot;black hole&quot; in GR has some experimental signatures which any other competing gravitational theory would have to account for. It might not be in the enticing format of a &quot;black hole&quot;, but some data are there. These at the moment are consistent with the theoretical black hole expected from GR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in a sense your question has as answer : &quot;yes the definition of a black hole is within the terminology introduced by general relativity&quot; and may not be defined outside it;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but also&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;no the experimental signatures do not depend on general relativity in order to exist, just their interpretation and attribution as a black hole&quot; may be in question,  if an alternative theory to GR succeeds to describe reality.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-04T04:37:19.627" CommentCount="1" />
  <row Id="9447" PostTypeId="2" ParentId="9441" CreationDate="2011-05-04T05:21:52.897" Score="2" ViewCount="" Body="&lt;p&gt;The questions are ambiguous but let me answer:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Yes&lt;/li&gt;&#xA;&lt;li&gt;Yes&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary:&lt;/strong&gt; it's completely irrelevant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could also claim that the answers to your questions are ambiguous because you haven't specified what part of GR the black hole discoveries should be dependent upon. Of course that they're dependent on GR at least to some extent because GR is our theory of gravity and gravity is essential both for the existence of black holes and their observation. One needs to use &lt;em&gt;some&lt;/em&gt; knowledge about gravity and GR - with some possibly adjustable features - is the best theory we can use. One cannot interpret &lt;em&gt;any&lt;/em&gt; observations without &lt;em&gt;any&lt;/em&gt; theory. Ever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you ask whether the evidence for e.g. the black hole at the galactic center depends on the special features of GR that are not present in Newton's theory, well, that would be a more specific question but the answer would still be ambiguous. Of course that the existence of black holes depends on the difference of Einstein's gravity and Newton's gravity. In Newton's mechanics, there is no speed limit, so one can &lt;em&gt;never&lt;/em&gt; create an object that would prevent light from escaping even in principle. One needs a relativistic theory of gravity to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone might claim that this observation weakens some claims but it doesn't weaken anything at all because Newton's theory of gravity is perfectly ruled out (in the realm of high velocities or strong gravitational fields). Of course that only a relativistic theory is a viable candidate for a theory of gravity. So we are allowed to assume a relativistic theory of gravity and it doesn't weaken the conclusions, not even infinitesimally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it is fair to say that the main empirical evidence in favor of the galactic black hole has really nothing to do with any subtle features of general relativity. As discussed in the previous question, the main evidence is that energy is apparently being lost from our sight. It is not being returned by a corresponding thermal radiation at some point, and so on. More generally, the object is much less luminous than any object with an imaginable surface would be. This is not an argument depending on some subtle third-order corrections in GR; it is a very old-fashioned, down-to-earth, empirical argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two points you're not saying explicitly but you're saying them in between the lines as far as I can say - and they're the only reason I can imagine why someone would post this question in this form. The points of yours are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The evidence for the existence of black holes depends on details of GR - that's what you're asking about.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The dependence on general relativity influences the robustness of the evidence.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The first point is strictly speaking tautologically right (we can't describe gravitational phenomena without any theory) but it is morally wrong because the main empirical evidence is of empirical character, indeed. But it's really the second assumption that is totally invalid and the very source of your problems. General relativity itself - or its key portions for this discussion - is even more well-established than the existence of black holes, so if the arguments that black holes exist depend on GR, it is not a problem at all.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-04T05:21:52.897" CommentCount="1" />
  <row Id="9448" PostTypeId="2" ParentId="2824" CreationDate="2011-05-04T05:52:51.353" Score="1" ViewCount="" Body="&lt;p&gt;So I've been feeling the same thing with my girlfriend.  I believe it has more to do with what Peer is talking about.  Both my girlfriend and I are on our laptops, both plugged into power sources.  She has a dell, I have a mac.  At first I thought I could be that our comps are spinning at different rates, until I read this blog and experimented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I touched her arm while we both had our comps on our laps and plugged into power source.&#xA; -vibration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took my comp off my lap while she still had hers on her lap, then touched her arm.&#xA; -no vibration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I put my comp back on my lap and unplugged it from power source, she had hers on her lap still plugged into power source and I touched her arm.&#xA; -no vibration&lt;/p&gt;&#xA;" OwnerUserId="3402" LastActivityDate="2011-05-04T05:52:51.353" />
  <row Id="9449" PostTypeId="1" CreationDate="2011-05-04T06:08:42.967" Score="2" ViewCount="494" Body="&lt;p&gt;This is a rather broad question. But does anyone know of good video lectures for grad level classical electrodynamics? &lt;/p&gt;&#xA;" OwnerUserId="1548" LastActivityDate="2011-05-07T20:53:34.947" Title="Graduate level Classical Electrodynamics" Tags="&lt;classical-electrodynamics&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="9450" PostTypeId="2" ParentId="9429" CreationDate="2011-05-04T06:10:41.203" Score="2" ViewCount="" Body="&lt;p&gt;The random character of the process itself doesn't depend on any specific virtual particles - instead, the random character of all microscopic processes in the world is a basic consequence of quantum mechanics. The existence of virtual particles is a consequence of quantum mechanics, too. Obviously, virtual particle pairs are important - in a particular computational scheme - to calculate the actual decay rate of a particular unstable object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the virtual pairs are not the &lt;em&gt;cause&lt;/em&gt; of the randomness itself - they're a consequence of the randomness - but in any calculation, the virtual pairs are a part of the cause of the actual numerical value of the decay rate.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-04T06:10:41.203" CommentCount="5" />
  <row Id="9451" PostTypeId="2" ParentId="9449" CreationDate="2011-05-04T08:11:09.573" Score="1" ViewCount="" Body="&lt;p&gt;You may try this sequence of 36 MIT lectures by Walter Lewin:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/&quot; rel=&quot;nofollow&quot;&gt;http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would guess that at the end, they're undergraduate lectures - but if it is true, I would find it natural because there is nothing such as &quot;universal graduate classical electromagnetism&quot;. The general shared material of classical electromagnetism belongs to the undergraduate curriculum. Graduate students may also learn classical electromagnetism but it must not be the basic or universal material but rather some &quot;twist&quot; that is appropriate for a narrower physics discipline that is still actively evolving. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classical electromagnetism was researched mainly in the 19th century and as a general research subject, it's not really alive today. There are related active fields of research but they're not &quot;just classical electromagnetism&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-04T08:17:29.617" LastActivityDate="2011-05-04T08:17:29.617" CommentCount="9" />
  <row Id="9452" PostTypeId="1" CreationDate="2011-05-04T08:34:28.763" Score="1" ViewCount="111" Body="&lt;p&gt;For example how radio signals of a base transceiver station (BTS) penetrate through buildings?&lt;/p&gt;&#xA;" OwnerUserId="3403" LastActivityDate="2011-05-04T08:55:40.563" Title="How radio waves penetrate through buildings?" Tags="&lt;electromagnetism&gt;&lt;waves&gt;&lt;radio&gt;" AnswerCount="1" />
  <row Id="9453" PostTypeId="2" ParentId="9452" CreationDate="2011-05-04T08:55:40.563" Score="6" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Radio_waves&quot; rel=&quot;nofollow&quot;&gt;Radio waves&lt;/a&gt; are large wavelength waves, and non metal walls are transparent to the radiation at those wavelengths, depending on the thickness of the walls,  because there are no energy level &quot;receptors&quot; to absorb them in bulk by excitation of electronic orbits. The wavelengths start from a centimetre up to kilometre, which is the width of the cycle of the varying electric and magnetic fields.Dissipation  happens  as the depth/thickness grows,  depending on the wavelength because all materials display bulk properties of conductivity at some level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metal has organized large distance electronic behaviour and can completely absorb radio waves, again depending on the wavelength and the dimensions .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a related discussion here: &lt;a href=&quot;http://physics.stackexchange.com/questions/7437/why-glass-is-transparent&quot;&gt;Why glass is transparent?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-04T08:55:40.563" />
  <row Id="9454" PostTypeId="2" ParentId="2824" CreationDate="2011-05-04T11:40:36.527" Score="2" ViewCount="" Body="&lt;p&gt;Its due to residual transductance of the live AC current into the shielding of the device. Get an earthed plug and it will disappear. I've experienced the issue in dozens of shielded but unearthed electrical appliances, not just laptops. It feels exactly like a vibration (when you move the finger over the surface) but it's not, my guess is that the mechanoreceptive nerve-endings in the skin react to the small current. &lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-05-04T11:40:36.527" />
  <row Id="9456" PostTypeId="1" AcceptedAnswerId="9470" CreationDate="2011-05-04T15:05:24.090" Score="3" ViewCount="129" Body="&lt;p&gt;Would a submarine float in the atmosphere of Jupiter, or would it get crushed?&lt;/p&gt;&#xA;" OwnerUserId="3404" LastActivityDate="2011-05-04T21:21:36.090" Title="Would a submarine float in the atmosphere of Jupiter?" Tags="&lt;astrophysics&gt;&lt;buoyancy&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="9457" PostTypeId="1" AcceptedAnswerId="9460" CreationDate="2011-05-04T16:01:39.187" Score="0" ViewCount="127" Body="&lt;p&gt;Electromagnetic energy can be related to it's frequency via E = h x nu. Is there a comparable relationship between gravitational wave energy and frequency?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-05-04T19:45:55.853" Title="Gravitational wave energy" Tags="&lt;general.relativity&gt;&lt;electromagetism&gt;" AnswerCount="1" />
  <row Id="9459" PostTypeId="1" AcceptedAnswerId="9823" CreationDate="2011-05-04T16:16:06.907" Score="1" ViewCount="308" Body="&lt;p&gt;This is a theoretical question for which i would like to know an answer with an example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know if its possible to create a setup where the electric field at a point $P$ is $n^{th}$ derivative (for some number $n$) of the electric (or magnetic) field at some other point $Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumption : Continuous electric charge distributions exist, which means there is no need to base everything on an electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: derivative is with respect to space or time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT : I am adding this in view of comments given below.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that the derivative is with respect to time. My question is some what similar to this situation here, assuming an inductor connected to an ideal current source (AC), then the potential drop across the inductor is proportional to the derivative of the current through it.The only difference here is i need it in terms of fields (assuming any type of charge distributions (except distributions(math))) rather than in a circuit with lumped elements.  &lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="540" LastEditorDisplayName="" LastEditDate="2011-05-12T08:55:05.707" LastActivityDate="2011-05-15T11:54:03.120" Title="Electric field at a point being an $n^{th}$ derivative of electric (or magnetic) field at some other point" Tags="&lt;electromagnetism&gt;&lt;mathematical-physics&gt;&lt;theory&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="2" />
  <row Id="9460" PostTypeId="2" ParentId="9457" CreationDate="2011-05-04T16:17:43.940" Score="1" ViewCount="" Body="&lt;p&gt;Yes, it is $E=h\nu$, too. This relation is totally universal across the world of quantum mechanical theories. It holds for photons, gravitons, electrons, Higgs bosons, or any other particle. It's because of the very general Schrödinger equation that says &#xA;$$ i\hbar \frac{d}{dt} |\psi\rangle = H | \psi \rangle $$&#xA;The operator on the left hand side adds $\hbar\omega = h\nu$, the operator (Hamiltonian) on the right hand side adds $E$ if the state $|\psi\rangle$ is an energy eigenstate.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-04T16:17:43.940" CommentCount="4" />
  <row Id="9461" PostTypeId="2" ParentId="9389" CreationDate="2011-05-04T17:37:06.540" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201408260&quot; rel=&quot;nofollow&quot; title=&quot;The Particle Garden&quot;&gt;The Particle Garden&lt;/a&gt; is a good read, and accessible to high school students, providing interesting historical accounts.&lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2011-05-04T17:37:06.540" />
  <row Id="9462" PostTypeId="1" CreationDate="2011-05-04T18:52:07.477" Score="0" ViewCount="239" Body="&lt;p&gt;Is it because of the upward force that stops the object? for example-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/80t15.png&quot; alt=&quot;http://i.stack.imgur.com/gqXyY.png&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3408" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-05-04T20:00:07.563" LastActivityDate="2011-05-05T09:57:54.587" Title="Why is heavier object more reluctant to get falling down?" Tags="&lt;newtonian-gravity&gt;&lt;free-fall&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="9463" PostTypeId="2" ParentId="9459" CreationDate="2011-05-04T19:04:49.567" Score="0" ViewCount="" Body="&lt;p&gt;For a harmonic field like $E\cdot cos(\omega t - \vec{k}\vec{r})$ is is possible since the second time and space derivatives are proportional to each other. For exponentially depending field (decreasing or increasing) it is also possible. No for any other case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is not calculated as an $n^{th}$ derivative, of course.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-05-10T10:39:39.873" LastActivityDate="2011-05-10T10:39:39.873" />
  <row Id="9464" PostTypeId="2" ParentId="9462" CreationDate="2011-05-04T19:21:39.983" Score="6" ViewCount="" Body="&lt;p&gt;OK, based on the comments I interpret the question as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Universal gravitation tells us that the gravitation force on a heavy object is larger than that on a light object, so why doesn't the heavy one fall faster?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Start with Newton's gravitation (as simplified for objects in the neighborhood of the Earth's surface):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F_g = mg $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer arises from another of Newton's observations: the second law&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F = ma $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;speed&lt;/em&gt; at which something falls is a colloquial term. More rigorously we talk about the &lt;em&gt;acceleration&lt;/em&gt; which is the $a$ above. Setting the forces above equal, it becomes obvious that the acceleration due to gravity is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a_g = \frac{F_g}{m} = \frac{mg}{m} = g $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the $m$'s cancel out. This is because mass &lt;em&gt;is&lt;/em&gt; coefficient that connects force and acceleration &lt;em&gt;and&lt;/em&gt; the &quot;charge&quot; equivalent for the gravitational force.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-05-04T19:21:39.983" />
  <row Id="9466" PostTypeId="1" CreationDate="2011-05-04T19:53:57.057" Score="0" ViewCount="132" Body="&lt;p&gt;&lt;strong&gt;TRIAL 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. )&#xA;Bag A + 6 stones&#xA;Mass (kg) - 0.037 kg&#xA;Mass of weight hanger + slotted weights - 210 g/ 0.21kg&#xA;Tension 1 (N) - ??&#xA;Acceleration (m/s2) - ??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. )&#xA;Bag B + 4 stones&#xA;Mass (kg) - 0.08&#xA;Mass of weight hanger + slotted weights - 210 g/ 0.21kg&#xA;Tension 1 (N) - ??&#xA;Acceleration (m/s2) - ??&#xA;Tension 2 (N) - ??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. )&#xA;Bag C + 2 stones&#xA;Mass (kg) - 0.074&#xA;Mass of weight hanger + slotted weights - 210 g/ 0.21kg&#xA;Tension 1 (N) - ??&#xA;Acceleration (m/s2) - ??&#xA;Tension 2 (N) - ??&#xA;Tension 3 (N) - ??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;======================&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;if you please write it in details so I can do trial 2 and 3 on my own then&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;======================&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRIAL 2&lt;/strong&gt; - Mass of weight hanger + slotted weights = 0.26kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bag A - 6 stones - 0.058g; Tension 1 = ?, acceleration = ?&#xA;Bag B - 4 stones - 0.090g; Tension 1 = ?, acceleration = ?, Tension 2 = ?&#xA;Bag C - 2 stones - 0.081g; Tension 1 = ?, acceleration = ?, Tension 2 = ?, Tension 3 = ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;======================&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TRIAL 3&lt;/strong&gt; - Mass of weight hanger + slotted weights = 0.36kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bag A - 6 stones - 0.078g; Tension 1 = ?, acceleration = ?&#xA;Bag B - 4 stones - 0.159g; Tension 1 = ?, acceleration = ?, Tension 2 = ?&#xA;Bag C - 2 stones - 0.091g; Tension 1 = ?, acceleration = ?, Tension 2 = ?, Tension 3 = ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;======================&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;FORMULAS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F=ma&#xA;T1 - Tabc = (Mabc)a&#xA;T2= ?&#xA;T3= ?&#xA;T2- Fbc = (Mbc)a&#xA;T3 - Fc = (Mc)a&lt;/p&gt;&#xA;" OwnerUserId="3409" LastActivityDate="2011-05-04T19:53:57.057" ClosedDate="2011-05-04T20:42:28.367" Title="physics newtons third law experiment with pulley" Tags="&lt;experimental-physics&gt;&lt;newton&gt;" CommentCount="1" />
  <row Id="9467" PostTypeId="2" ParentId="9462" CreationDate="2011-05-04T19:54:35.557" Score="4" ViewCount="" Body="&lt;p&gt;The big object has more &lt;a href=&quot;http://en.wikipedia.org/wiki/Inertia&quot; rel=&quot;nofollow&quot;&gt;inertia.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a big object and a small object, the gravitational force on the big object is greater.  Why, then, doesn't it fall faster?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that the big object needs more force to accelerate the same way.  This is actually quite obvious if you view it in a different context.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose you have a compact car and a school bus.  Which one needs a bigger engine?  The school bus, of course. You need more force to accelerate a bigger thing.  This is summarized by saying the school bus has more inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we drop the small car and the school bus off a cliff.  Which one falls faster?  There is more gravitational force on the school bus, but you also need more force to get the school bus going, so it is not obvious which of these effects will win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that the extra force on the school bus is just enough to cancel its extra inertia.  The bus and car fall at the same rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Newtonian mechanics, this is a coincidence.  It was later explained by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalence_principle&quot; rel=&quot;nofollow&quot;&gt;equivalence principle&lt;/a&gt;, a feature of general relativity.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-05-04T19:54:35.557" />
  <row Id="9468" PostTypeId="1" AcceptedAnswerId="9472" CreationDate="2011-05-04T20:56:22.597" Score="3" ViewCount="185" Body="&lt;p&gt;I know: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Quantum Mechanics (Griffiths Level, currently doing Sakurai Level)&lt;/li&gt;&#xA;&lt;li&gt;Mechanics (Newtonian+ Lagrangian/Hamiltonian but at level lower than Goldstein/Landau)&lt;/li&gt;&#xA;&lt;li&gt;Classical Electrodynamics (Griffiths + electro/magnetostatics from Jackson)&lt;/li&gt;&#xA;&lt;li&gt;Statistical Physics (Pathria)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I know very little relativity from undergrad mechanics class. Nothing about General Relativity, nothing about QFT, etc. I wish to reach string theory in a proper way without leaving any gaping holes in my education. What subjects should be studied in what order?&lt;/p&gt;&#xA;" OwnerUserId="2599" LastActivityDate="2011-05-04T22:09:34.110" Title="In what order should the subjects be studied in order to get to String Theory" Tags="&lt;reference-request&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="9469" PostTypeId="1" AcceptedAnswerId="9471" CreationDate="2011-05-04T21:06:32.290" Score="2" ViewCount="238" Body="&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=yw075uSG534&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=yw075uSG534&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;&lt;a href=&quot;http://tvnoviny.sk/sekcia/spravy/technologie/v-lacnove-sa-predmety-kotulaju-do-kopca.html&quot; rel=&quot;nofollow&quot;&gt;http://tvnoviny.sk/sekcia/spravy/technologie/v-lacnove-sa-predmety-kotulaju-do-kopca.html&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;Can someone explain this to me? Is it a &quot;hoax&quot;?&lt;/p&gt;&#xA;" OwnerUserId="3410" LastActivityDate="2011-05-04T21:45:09.743" Title="gravity ashamed" Tags="&lt;gravity&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="9470" PostTypeId="2" ParentId="9456" CreationDate="2011-05-04T21:11:57.297" Score="3" ViewCount="" Body="&lt;p&gt;As the comments above indicate, factors like density, pressure and temperature are important for a Jupiter submariner. Of course nobody yet has the exact details of Jupiter's interior structure, but there's a diagram in &lt;a href=&quot;http://lasp.colorado.edu/education/outerplanets/giantplanets_interiors.php&quot; rel=&quot;nofollow&quot;&gt;http://lasp.colorado.edu/education/outerplanets/giantplanets_interiors.php&lt;/a&gt; that indicates the following:                                                                       At the intersection of Jupiter's liquid hydrogen/metallic hydrogen layer, the density is about 1 g/cm^3, which would permit the submarine to float. However, the temperature there is ~5000K, and the pressure is ~2E6 bars, so it can't be done with normal equipment.  &lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="1924" LastEditorDisplayName="" LastEditDate="2011-05-04T21:21:36.090" LastActivityDate="2011-05-04T21:21:36.090" />
  <row Id="9471" PostTypeId="2" ParentId="9469" CreationDate="2011-05-04T21:27:29.237" Score="8" ViewCount="" Body="&lt;p&gt;Dear LanceBaynes, maybe you shouldn't expect that all users speak almost perfect Slovak on this server - even though you may be right with some of us. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This video&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=3JkjeWKQnz0&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=3JkjeWKQnz0&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;shows that the cars are moving exactly as the altimeters and level indicate. There's nothing anomalous about that place. Markíza TV has already aired an official spot explaining that this story is just a hoax. See also TV Noviny.sk,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://tvnoviny.sk/sekcia/spravy/zaujimavosti/zahada-kopca-pri-lacnove-kde-voda-tecie-hore-brehom-je-odhalena.html&quot; rel=&quot;nofollow&quot;&gt;http://tvnoviny.sk/sekcia/spravy/zaujimavosti/zahada-kopca-pri-lacnove-kde-voda-tecie-hore-brehom-je-odhalena.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When the cars move spontaneously, they reduce their altitude by 150 meters or so. And I think that this stunt has been copied from some foreign places where people have equally nonsensically claimed that cars are moving uphill. For some list of &quot;gravity hills&quot; (with the same claimed property) in the world, see&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://paranormal.about.com/library/weekly/aa120301a.htm&quot; rel=&quot;nofollow&quot;&gt;http://paranormal.about.com/library/weekly/aa120301a.htm&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravity_hill&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Gravity_hill&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia link explains that all &quot;gravity hills&quot; are optical illusions. In general, some of these &quot;mystery spots&quot; may even be cleverly engineered tourist attractions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.sandlotscience.com/MysterySpots/Mystery_Spots_1.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.sandlotscience.com/MysterySpots/Mystery_Spots_1.htm&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://en.wikipedia.org/wiki/Mystery_Spot&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mystery_Spot&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Optical illusions are often not accidental - they're deliberately fabricated by houses built with wrong inclinations, and so on. Especially in the case of the most famous mystery spot near Santa Cruz in California (where I've spent half a year) opened around 1940, see the Wikipedia link above, this has been designed as a business project. You can be pretty much sure that everyone involved in the promotion of that place in Slovakia is either a naive simpleton or a participant in the hoax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite generally, water is not magnetic so it couldn't be affected, and the non-verticality of the gravitational field cannot exceed a certain tiny angle anywhere on the globe. Even near Mount Everest, which is huge relatively to anything you find in Slovakia, the angle between the direction of the gravitational field and the direction towards the center of the Earth is much smaller than a degree - while the hoax in Lačnov claims that the angular anomaly is of order dozens of degrees which is clearly nonsense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claims that the surprising effects are caused by &quot;magnetism&quot; are almost on par with the claims that they're caused by &quot;paranormal phenomena&quot;: both of these claims would manifestly contradict even elementary basic-school science.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-04T21:45:09.743" LastActivityDate="2011-05-04T21:45:09.743" />
  <row Id="9472" PostTypeId="2" ParentId="9468" CreationDate="2011-05-04T22:04:33.877" Score="2" ViewCount="" Body="&lt;p&gt;You can try reading Zwiebach 'A first course in String theory' which is roughly at your level right now.  Its very handwavy, but well thats the best you can hope for at this level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, to really learn the subject you will absolutely need &#xA;1) Grad level GR&#xA;2) Quantum II, + 2 semester long courses in QFT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then you can start thinking about it.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally I find the above level a little loose, so to make it more comprehensive and less opaque i'd recommend in addition to the above, to have some experience in Semiclassical gravity (Wald or Birrel and Davies), Conformal field theory (Di Francesco) and Supersymmetry (Weinberg or Wess and Bagger)&lt;/p&gt;&#xA;" OwnerUserId="1350" LastActivityDate="2011-05-04T22:04:33.877" />
  <row Id="9473" PostTypeId="2" ParentId="9468" CreationDate="2011-05-04T22:09:34.110" Score="0" ViewCount="" Body="&lt;p&gt;I would say Special Relativity, then General Relativity, and finally QFT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Books:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Special Relativity -- I would recommend Wheeler &amp;amp; Taylor and Woodhouse&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General Relativity -- Woodhouse&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QFT -- Zee and Aitchison &amp;amp; Hey&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also an undergrad physics book [junior level] on String Theory by Zwieback that's highly recommended by many.&lt;/p&gt;&#xA;" OwnerUserId="3270" LastActivityDate="2011-05-04T22:09:34.110" CommentCount="2" />
  <row Id="9474" PostTypeId="1" AcceptedAnswerId="9475" CreationDate="2011-05-05T04:09:11.507" Score="2" ViewCount="120" Body="&lt;p&gt;Ever since Einstein published his GR theory in 1916, there have been numerous experimental tests to confirm its correctness, and it has passed with flying colors. NASA and Stanford have just announced that their Gravity Probe B activity has confirmed GR's predicted geodetic and frame-dragging effects. Are there any other facets of GR that need experimental verification?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-05-05T06:03:45.247" Title="Testing General Relativity" Tags="&lt;general.relativity&gt;" AnswerCount="1" />
  <row Id="9475" PostTypeId="2" ParentId="9474" CreationDate="2011-05-05T04:24:45.460" Score="1" ViewCount="" Body="&lt;p&gt;Sure there are. The theory has been tested within only a teeny tiny part of the range of its predictions. For example it predicts gravitational redshift in the range of 0% (no redshift) to 100% (black hole), but experiments to date have shown a maximum gravitational redshift less than 0.01%. It matters less how many tests of GR are done than how extensively those tests cover the range of what GR predicts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While we have little experimental data to definitively show that GR is the correct theory of gravity, we do know that it leads to major problems for physics, like its breakdown at gravitational singularities, its incompatibility with quantum mechanics, and the black hole information loss paradox. A competing theory of gravity that is confirmed by all experimental tests of GR to date need not have any of those problems, indicating that a lot more testing of GR is warranted.&lt;/p&gt;&#xA;" OwnerUserId="2314" LastEditorUserId="2314" LastEditorDisplayName="" LastEditDate="2011-05-05T06:03:45.247" LastActivityDate="2011-05-05T06:03:45.247" CommentCount="10" />
  <row Id="9476" PostTypeId="1" AcceptedAnswerId="9740" CreationDate="2011-05-05T05:59:57.483" Score="1" ViewCount="139" Body="&lt;p&gt;Can the equivalence principle be tested to high precision in a human-sized lab falling through the horizon of a black hole, in principle? By &quot;human-sized lab&quot; I mean a lab the size of the International Space Station, say, with typically-sized lab equipment. I also mean the absolute kind of (event) horizon, not the apparent kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The #1 answer at &lt;a href=&quot;http://physics.stackexchange.com/questions/461/why-is-it-thought-that-normal-physics-doesnt-exist-inside-the-event-horizon-of-a&quot;&gt;this question&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For large enough black holes, space is still weakly curved at the event horizon, so of course we should expect that normal physics still exists there. An infalling observer wouldn't experience anything out of the ordinary when crossing an event horizon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This suggests that the answer to my question here is &quot;yes&quot;. However, when I asked a &lt;a href=&quot;http://physics.stackexchange.com/questions/6086/how-does-this-thought-experiment-not-rule-out-black-holes&quot;&gt;question involving such a test&lt;/a&gt;, the highest-rated answer with 23 votes clearly says &quot;no&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The equivalence principle only allows you to transform to an inertial frame locally. This means that if your spacetime is curved, then the falling observer can only choose Minkowski coordinates for an infinitesimal region around her.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've seen this contradiction many times elsewhere. Certainly the EP has actually been tested to high precision in larger-than-infinitesimal labs. Yet once one talks about testing the EP in a lab falling through the horizon of a black hole it seem respondents &lt;em&gt;in that case&lt;/em&gt; think the lab must be infinitesimal in size, so that the EP can't be tested in any practical sense, even though &quot;An infalling observer wouldn't experience anything out of the ordinary when crossing an event horizon&quot; if the black hole was sufficiently large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the EP is strictly true for only a point in spacetime. But that hasn't prevented it from being validly tested to high precision in a larger lab, where the tidal force in the lab was weak enough that it didn't change the result at that precision. I'm not asking about testing the EP to &lt;em&gt;infinite&lt;/em&gt; precision, nor am I asking about a test involving a black hole of &lt;em&gt;any&lt;/em&gt; size, like one where the lab would be spaghettified by the tidal force before reaching the horizon. That's why I qualified the question with &quot;in principle&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that I'm talking about a test being done as the lab equipment (temporarily) straddles the horizon, not a test limited to one side of the horizon or the other within the lab. A test using equipment straddling the horizon wouldn't be allowed according to the other answer with 23 votes, which says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The infinitesimal flat patch in which you're allowed to play with the EP does not include ... anything beyond the horizon ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't see how a test of the EP would be invalidated due to the equipment straddling the horizon.&lt;/p&gt;&#xA;" OwnerUserId="2314" LastEditorUserId="2314" LastEditorDisplayName="" LastEditDate="2011-05-05T06:28:52.933" LastActivityDate="2011-05-11T17:56:49.407" Title="Can the equivalence principle be tested to high precision in a human-sized lab falling through the horizon of a black hole, in principle?" Tags="&lt;general-relativity&gt;&lt;black-holes&gt;&lt;equivalence-principle&gt;" AnswerCount="2" CommentCount="10" FavoriteCount="1" />
  <row Id="9477" PostTypeId="2" ParentId="7903" CreationDate="2011-05-05T06:21:05.170" Score="2" ViewCount="" Body="&lt;p&gt;The field lines of earth's magnetic field go from south pole to north pole. So they are parallel to the surface at the equator, and perpendicular at the (magnetic) poles (going vertically into the ground).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what happens to a moving charge in a magnetic field? A particle moving parallel to the field lines will not feel any force, however when it's moving perpendicularly to the field, it will be deflected according to the right hand rule. Note that the direction of $\vec v$ will change, but not the magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The source of this behavior comes from the cross product in the formulation of Lorentz' force: $F_L = q \, (\vec v \times \vec B)$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say your thumb points into the direction of the field line, your index finger in the direction of the particle movement. Then it will be deflected in the direction of your middle finger. Now turn your hand such that your index finger points into the &lt;em&gt;new&lt;/em&gt; direction. You will see that the particle will describe a &lt;strong&gt;circle around the field line&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An incoming particle, say an electron from the ion wind of the sun, will in general have both velocity components perpendicular and parallel to the field. It will be moved away while describing the circle, and thus fly on a &lt;strong&gt;corkscrew-like path&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, this is also the source of the northern lights (aurora borealis):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Near the equator, the component of $\vec v$ pointing to the surface is perpendicular to the field lines, so that component is &quot;trapped&quot; in the circular motion. The other component parallel to the field lines leads the particle to the poles.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Near the poles, the component of $\vec v$ pointing to the surface is parallel to the field lines, so the particles can descend to the ground while describing circles.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Because of the circular motion, the particles a) need much longer to reach the ground, and b) loose energy due to syncroton radiation. In this way, earth's magnetic field protects us from certain types of cosmic radiation!&lt;/p&gt;&#xA;" OwnerUserId="825" LastActivityDate="2011-05-05T06:21:05.170" CommentCount="3" />
  <row Id="9478" PostTypeId="2" ParentId="9219" CreationDate="2011-05-05T06:34:45.487" Score="2" ViewCount="" Body="&lt;p&gt;I am not aware of any theory involving a notion of infinite spin. Generally, spin is a quantum number that takes (typically) small integer or half integer values. In principle, you can have a system with as high a spin as you would like, but that's not infinite. So I'd say the answer is no.&lt;/p&gt;&#xA;" OwnerUserId="825" LastActivityDate="2011-05-05T06:34:45.487" CommentCount="4" />
  <row Id="9479" PostTypeId="2" ParentId="9439" CreationDate="2011-05-05T07:08:06.363" Score="0" ViewCount="" Body="&lt;p&gt;If this is only a simple High School Test, then they probably wouldn't be going in so deep as to whether the man was moving in relation as to what. All they would want is probably the man's moving speed after being slowed down by the flow of the stream&lt;/p&gt;&#xA;" OwnerUserId="2881" LastActivityDate="2011-05-05T07:08:06.363" CommentCount="1" />
  <row Id="9480" PostTypeId="2" ParentId="9462" CreationDate="2011-05-05T09:51:38.590" Score="0" ViewCount="" Body="&lt;p&gt;I'will avoid the use of the concept of Inertia because quoting &lt;a href=&quot;http://en.wikipedia.org/wiki/Inertia&quot; rel=&quot;nofollow&quot;&gt;WP-Inertia&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is no single accepted theory that explains the source of Inertia.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'will avoid the use of equations because they describe and do not explain.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will answer to the question:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you have an object with more mass (heavier) than other object&#xA;  the gravitational force on the heavier object is greater. Why, then,&#xA;  doesn't it fall faster?&#xA;  The concept of &lt;a href=&quot;http://en.wikipedia.org/wiki/Mass&quot; rel=&quot;nofollow&quot;&gt;mass&lt;/a&gt; can have different meanings that, only in pratical terms, are the measure of the same property. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;gravitational passive mass - how a&#xA;body reacts under the effect of&#xA;gravity&lt;/li&gt;&#xA;&lt;li&gt;gravitational active  mass - how a body contributes to the effect of gravity (as a source of it)  &lt;/li&gt;&#xA;&lt;li&gt;inertial mass - how a body reacts under the influence of a generic force  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The answer begs to use the 'gravitational passive mass' concept.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To simplify the anser I assume that they are made of the same substance and one has the double of mass of the other.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: &lt;strong&gt;Why an object have more mass than the other?&lt;/strong&gt;&lt;br&gt;&#xA;Answer: Because it has the double of &lt;strong&gt;particles&lt;/strong&gt; of the other.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this simplification we see that it is valid &lt;strong&gt;Mass = Number of Particles&lt;/strong&gt;.&lt;br&gt;&#xA;(In this problem the mass of each particle  is 1 unit of mass) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further simplification: One body has two particles and the other only one.  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         'Body 1'          'Body 2'&#xA;Two particles OO  One Particle  O   &#xA;Gravit Force  ||                |&#xA;Ground        ||                | &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(Where the downward '|' represent the value and direction of the gravitational force)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Each particle is downward atracted with the &lt;em&gt;same value&lt;/em&gt;&lt;/strong&gt;. This is the answer to question as I pose it.  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;We can not transform || into this |    because the force act in parallel &#xA;                                  |        on 'each particle'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To stress the idea I will transform again the body into other equivalent system (same mass) where each particle is separated from the others:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                 'Body 1'            'Body 2'&#xA;Two particles      O  O      One Particle  O   &#xA;Gravit Force       |  |                    |&#xA;Ground             |  |                    | &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And now you can 'see' that there is no chance to add the forces. In other words it is not additive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally : the answer to the original question is a clear NO : The heavier object is NOT more reluctant to get falling down.  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-05-05T09:57:54.587" LastActivityDate="2011-05-05T09:57:54.587" />
  <row Id="9481" PostTypeId="1" CreationDate="2011-05-05T10:55:02.377" Score="0" ViewCount="97" Body="&lt;p&gt;Is the distance between the Sun and Venus increasing?&lt;br&gt;&#xA;What is the measurement?&lt;br&gt;&#xA;May I request for a formula?&lt;/p&gt;&#xA;" OwnerUserId="3412" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-05-05T11:41:27.020" LastActivityDate="2011-05-05T12:50:50.463" Title="Distance increase between Sun and Venus" Tags="&lt;gravity&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9482" PostTypeId="1" AcceptedAnswerId="9534" CreationDate="2011-05-05T11:42:45.200" Score="14" ViewCount="399" Body="&lt;p&gt;Why is the angle of a wake of a duck constant? And why are some conditions on the water depth necessary?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that this question turns up in google searches, but I did not see a good discussion. I will be quite happy with a link.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited to add:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone tell me how the two up-voted answers are related?&lt;/p&gt;&#xA;" OwnerUserId="3415" LastEditorUserId="3415" LastEditorDisplayName="" LastEditDate="2011-05-06T14:13:59.200" LastActivityDate="2011-05-07T21:29:39.870" Title="Why is the angle of the wake of a duck constant?" Tags="&lt;waves&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="9483" PostTypeId="2" ParentId="9482" CreationDate="2011-05-05T11:58:04.203" Score="5" ViewCount="" Body="&lt;p&gt;In particle physics, we usually solve a similar problem with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cerenkov_radiation&quot; rel=&quot;nofollow&quot;&gt;Čerenkov radiation&lt;/a&gt; whose angle is given by &#xA;$$ \cos\theta = \frac{v_{\rm radiation}}{v_{\rm particle}} $$&#xA;Note that to get the radiation bounded by a specific angle, the particle has to be faster than the radiation in the given environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/kk472.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for the duck - assuming it is moving in a deep water - the velocity of the &quot;radiation&quot; depends on the frequency. The speed of the duck's paddling is also variable. However, these two uncertainties combine in such a way that the angle is universally 39 degrees or so, independently of the duck's velocity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, the numerical calculation of the precise numerical value of the angle is clearly difficult and depends on real-world factors. The constancy of the angle may be argued to hold by dimensional analysis and/or scaling laws. &lt;a href=&quot;http://en.wikipedia.org/wiki/Dispersion_%28water_waves%29&quot; rel=&quot;nofollow&quot;&gt;The deep water waves&lt;/a&gt; have phase velocity proportional to the frequency,&#xA;$$  c_{\rm phase} \sim f $$&#xA;This was previously denoted $v_{\rm radiation}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The velocity of the duck is also proportional to the frequency of its paddling; each cycle  gets you that far (by a fixed distance $\delta x$), so $v_{\rm particle=duck} = \delta x\cdot f$. So if you compute the ratio of the velocities, as in the Čerenkov formula at the top, you get a constant angle because the frequency $f$ cancels.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-05T12:13:11.497" LastActivityDate="2011-05-05T12:13:11.497" CommentCount="9" />
  <row Id="9484" PostTypeId="1" AcceptedAnswerId="9493" CreationDate="2011-05-05T12:06:07.900" Score="2" ViewCount="199" Body="&lt;p&gt;I have learned that the classical measurement of the speed of light with a rotating mirror does not work with a laser (as opposed to, say, a mercury-vapor lamp).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you tell me if and how coherency of light is necessary/contraindicated for the success of this experiment?&lt;/p&gt;&#xA;" OwnerUserId="3415" LastEditorUserId="3415" LastEditorDisplayName="" LastEditDate="2011-05-05T13:21:53.173" LastActivityDate="2011-05-05T15:33:21.747" Title="Can you use a laser to measure the speed of light with a rotating mirror?" Tags="&lt;speed-of-light&gt;&lt;measurement&gt;&lt;laser&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="9485" PostTypeId="2" ParentId="9481" CreationDate="2011-05-05T12:15:05.787" Score="1" ViewCount="" Body="&lt;p&gt;As you are new here, welcome. It would be good if you could give a justification  to why you are asking this question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A relevant thread to this  is &lt;a href=&quot;http://physics.stackexchange.com/questions/9290/why-does-the-moon-drift-away-from-earth&quot;&gt;Why does the moon drift away from earth?&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will find there the simple mechanics formulae of why the moon, because of the earth tides it induces, drifts away slowly from the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Venus does induce tides on the sun, but the sun also induces tides on Venus so one has a more complicated problem. I expect though that the effect would be the same over centuries, that the average distance  between Venus and Sun would grow.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-05T12:15:05.787" />
  <row Id="9486" PostTypeId="2" ParentId="9481" CreationDate="2011-05-05T12:50:50.463" Score="-1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Most Probably Yes.&lt;/strong&gt;&lt;br&gt;&#xA;The Earth is moving away from the Sun (&lt;a href=&quot;http://physics.stackexchange.com/questions/208/is-the-distance-between-the-sun-and-the-earth-increasing/&quot;&gt;see PSE-Question&lt;/a&gt;) at a rate $0.57H_{0}$, the most recent reference AFAIK, that can not be explained by any of the current official models, one can expect that the same happens with the Earth-Venus distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 'simple' triangulation when the 'transit of Venus' (in the line of sight Earth-Sun)  happens will reveal if the answer is yes or no. The natural answer is yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;by G. A. Krasinsky and V. A. Brumberg, 2004&lt;br&gt;&#xA;&lt;a href=&quot;http://www.springerlink.com/content/g5051650115444k9/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Secular Increase of astronomical unit from analysis of the major planet motions, and Its Interpretation&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;$\frac{dAU}{dt}=15\pm4\: m/cy$&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;at present there is no satisfactory&#xA;  explanation of the detected secular&#xA;  increase of AU&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;WeiJia Zhang, ZhengBin Li and Yang Lei, 2010&lt;br&gt;&#xA;&lt;a href=&quot;http://www.springerlink.com/content/k130348656287201/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Experimental measurement of growth patterns on fossil corals: Secular variation in short distances ancient Earth-Sun&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;both the modern and ancient leaving&#xA;  rates could be measured with high&#xA;  precision, &lt;strong&gt;and it was found that the&#xA;  Earth has been leaving the Sun over&#xA;  the past 0.53 billion years&lt;/strong&gt;. The&#xA;  Earth’s semi-major axis was 146&#xA;  million kilometers at the beginning of&#xA;  the Phanerozoic Eon, equating to 97.6%&#xA;  of its current value. Measured modern&#xA;  leaving rates are 5–14 m/cy, whereas&#xA;  the ancient rates were much higher.&#xA;  &lt;strong&gt;Experimental results indicate a special expansion with an average&#xA;  expansion coefficient of $0.57H_{0}$&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-05-05T12:50:50.463" />
  <row Id="9487" PostTypeId="2" ParentId="9484" CreationDate="2011-05-05T12:54:23.363" Score="2" ViewCount="" Body="&lt;p&gt;Did the Wikipedia article specifically mention using a rotating mirror with laser light? When I was in undergrad we used a pulsed laser to measure the speed of light, but not with a rotating mirror. We split the beam and directed one one half at a photodiode. The other half was sent across a long room a few times with mirrors before hitting the photodiode. We used the timing measurement from the photodiode and manually measured the path length. &lt;/p&gt;&#xA;" OwnerUserId="3418" LastActivityDate="2011-05-05T12:54:23.363" CommentCount="1" />
  <row Id="9488" PostTypeId="2" ParentId="9476" CreationDate="2011-05-05T13:02:21.040" Score="4" ViewCount="" Body="&lt;p&gt;The answer to the question posed in your title is &lt;strong&gt;no&lt;/strong&gt;, but not for the reason which you may think is the case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the &lt;em&gt;Einstein equivalence principle&lt;/em&gt; is &lt;strong&gt;not&lt;/strong&gt; a precise principle, so the statement &quot;testing the equivalence principle to high precision&quot; is in itself meaningless. The principle is usually stated informally as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The outcome of any local non-gravitational experiment in a freely falling laboratory is independent of the velocity of the laboratory and its location in spacetime.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with the caveat in the definition of &quot;local&quot; and &quot;independent&quot;. Indepdence is generally taken to mean that indistinguishable to the precision allowed by your experiment. The definition of local, besides the obvious criteria of &lt;em&gt;non-interaction&lt;/em&gt; with objects outside the laboratory except through gravitational coupling, contains also the requirement that throughout the space-time region occupied by the laboratory during the extent of the experiment, the gravitational field changes negligibly (where negligible is defined, again, as below experimental precision). So in particular you can not have large tidal forces. To achieve this you need the total curvature contained in the space-time region occupied by the laboratory during the experiment (obtained by something like integrating the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kretschmann_scalar&quot; rel=&quot;nofollow&quot;&gt;Kretschmann scalar&lt;/a&gt; over the space-time domain occupied by the laboratory) to be small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I getting at? The equivalence principle only requires that at any space-time location, ignoring quantum limits, you can build a sufficiently small laboratory that runs for a sufficiently short (proper) time, so that any (non-gravitational) experiments performed within it is indistinguishable from one performed in ideal Minkowski space. So the interesting question to ask is: given that your laboratory is roughly the size of the international space station (roughly $10^3 m^3$), and you run your experiment for a fixed lenghth of time (say 1 day in proper time), and your laboratory is straddling the apparent horizon of a black hole, what is the minimum mass of the black hole such that you won't be able to tell the difference, to say, one part in $10^{10}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rough estimate for this can be done with some dimensional analysis. As is well known, the Kretschmann scalar is rougly on the order of $10r_S^{-4}$ for a black hole at the apparent horizon, where $r_S$ is the Schwarzschild radius, given by the black hole mass $M$ via the formula $2GM/c^2 = r_S$. Integrating the Kretschmann scalar over a space-time region will then be a dimensionless quantity, which we would to be small $\sim 10^{-10}$. The space-time volume occupied by our laboratory is $cT V$ where $V$ is the spatial volume of the lab and $T$ is the amount of time in which to run our experiment. In our assumptions above, we have that $cTV \sim 3 \times 10^{8} \frac{m}{s} \times 9\times 10^4 s \times 10^3 m^3 \sim 2.5 \times 10^{15} m^4$. So we need &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ r_S^4 \sim 2.5\times 10^{15} m^4 \times 10^{10} \times 10 = 2.5 \times 10^{26} m^4 \implies r_S \sim 4 \times 10^6 m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives a mass of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M = \frac{c^2 r_S}{2G} \sim 9 \times 10^{16} \frac{m^2}{s^2} \times 2\times 10^6 m \times \frac{1}{7} \times 10^{11} \frac{kg \cdot s^2}{m^3} \sim 2\times 10^{33} kg $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is about 1000 solar masses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarise, if you drive the international space station to the vicinity of a 500 solar mass black hole, and you have some experimental apparatus with sensitivity 1 part in $10^10$, and you accidentally crossed the apparent horizon without looking out the window, then you won't notice until about a day later, which is, well, rather late. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let me make a note here that it is rather impossible for a free falling laboratory to straddle the apparent/event horizons for any length of time. The event horizon is null, and for Schwarzschild black holes (as well as black holes in certain other nice matter models) the apparent horizon is achronal. So for your laboratory to remain straddling the horizon, an upperbound of the time you can do so is roughly $T = L/c$, where $L$ is the linear size of your laboratory. For something of the size of the ISS, that makes $T$ on the order of $10^{-6}$ seconds in proper time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you were to perform an experiment only when you are straddling the horizon, you gain a factor of $10^{10}$ compared to the previous computation, and the minimal mass of the black hole such that you won't be able to tell the difference is on the order of the mass of the Earth. &lt;/p&gt;&#xA;" OwnerUserId="829" LastEditorUserId="829" LastEditorDisplayName="" LastEditDate="2011-05-05T13:13:58.737" LastActivityDate="2011-05-05T13:13:58.737" CommentCount="6" />
  <row Id="9489" PostTypeId="2" ParentId="9482" CreationDate="2011-05-05T13:41:58.810" Score="-2" ViewCount="" Body="&lt;p&gt;EDIT&lt;br&gt;&#xA;This answer is not correct and I assume my mistake. Skip to other answer, pls.&lt;br&gt;&#xA;I should have been more cautious before writing, find some references, think more, wait and see...&#xA;Its a learning process. 'Errare humanum est' and so, I am very human.&lt;br&gt;&#xA;END_EDIT  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;;-) - the little ducks follow the parents and go at the same speed. Older ducks dont get much pleasure making races into water. Then, I suspect that we can not see a great dispersion of the surface speed of ducks if they always paddle at the same rithm.&lt;br&gt;&#xA;impulse/the resistance of water = constant = duck foot area/volume of water displaced&lt;br&gt;&#xA;The caracteristic of the medium 'dispersion relation' determines the value of 'c'. Any perturbation (deep waters) propagate always at 'c'. &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-05-07T21:29:39.870" LastActivityDate="2011-05-07T21:29:39.870" />
  <row Id="9490" PostTypeId="1" AcceptedAnswerId="9499" CreationDate="2011-05-05T14:08:17.467" Score="2" ViewCount="92" Body="&lt;p&gt;I am reading a physiology book chapter (&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387983813&quot; rel=&quot;nofollow&quot;&gt;Mathematical Physiology, by Keener&lt;/a&gt; --Respiration chapter) about the gas exchange between capillaries and alveoli. It seems that this gas exchange can be modeled after some simple physical relationships. Since I am not usually studying physics, I do not fully understand some concepts that the author uses to derive a conservation law and I would appreciate any help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do admit, however, that this question may be off-topic since it could be a mathematical lapsus rather than a physics question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Preliminaries&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the author indicates that if a gas with partial pressure $P_g$ is in contact with a liquid, the steady-state concentration $U$ of gas is given by:&#xA;$$&#xD;&#xA;U = \sigma P_s&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\sigma$ is the solubility of the gas in the liquid. I assume that this is a particular version of Henry's Law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, it explains if there is a difference between the partial pressure of the gas ($P_g$) and the partial pressure on the fluid ($\frac{U}{\sigma}$), then there should be some flux between the gas and the fluid and the simplest model would be to assume that this flux is linearly proportional to the pressure difference:&#xA;$$&#xD;&#xA;q = D_s \left(P_g - \frac{U}{\sigma}\right)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author then considers a segment of a capillary (a cylindrical tube) of length $L$, constant cross-sectional area $A$ and perimeter $p$, that is in contact with a gas with partial pressure $P_g$. The fluid moves through the tube with a velocity $v(x)$. Finally, they say that since mass is conserved:&#xA;$$&#xD;&#xA;\frac{d}{dt} \left( A \int_{0}^{L} U(x,t) dt \right) = v(0)AU(0,t) - v(L)AU(L,t) + p \int_{0}^{L} q(x,t) dx&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is the relationship below derived? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that $A \int_{0}^{L} U(x,t) dt$ is in fact the total amount of the dissolved gas in the tube at a given time. I also understand how $\int_{0}^{L} q(x,t) dx$ represents the total flux of gas across the whole capillary wall. However, I fail to see what the two first elements of the right-hand represent and why is the left-side derived with respect to $t$.&lt;/p&gt;&#xA;" OwnerUserId="3413" LastEditorUserId="3413" LastEditorDisplayName="" LastEditDate="2011-05-05T14:41:12.683" LastActivityDate="2011-05-05T18:39:28.583" Title="Diffusion across an interface and conservation of mass" Tags="&lt;mathematics&gt;&lt;diffusion&gt;" AnswerCount="2" />
  <row Id="9491" PostTypeId="2" ParentId="9422" CreationDate="2011-05-05T14:50:06.370" Score="1" ViewCount="" Body="&lt;p&gt;There are a few different perspectives from which one can answer this question. Here's one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|\Omega-1|$ is a measure of the curvature of the Universe. To be specific, it's essentially the ratio of the Hubble length $1/H$ to the curvature radius $R_c$:&#xA;$$&#xD;&#xA;|\Omega-1|={H^{-1}\over R_c}.&#xD;&#xA;$$&#xA;(To be honest, there may be constants of order 1 in there.) &#xA;The Hubble length is the length scale set by the expansion rate of the Universe, and the curvature radius is the length scale on which the effects of curvature become noticeable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the Universe expands, the curvature radius expands with it. A large curvature radius means a Universe that's close to flat: curvature is hard to notice until you go out to large scales. During inflation, the Universe expands rapidly, so $R_c$ grows to be very large. On the other hand, the Hubble length remains roughly constant, so $|\Omega-1|$ plummets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More geometrically, what we're saying here is that the length scale of curvature (the size of any &quot;wrinkles&quot; in space, if you like) grows, but the natural length scale to use for comparison doesn't. So curvature becomes less and less noticeable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During &quot;normal&quot; expansion, both length scales grow, and in fact when the expansion is decelerating, $H^{-1}$ grows faster than $R_c$, so curvature becomes more noticeable -- that is, $|\Omega-1|$ grows.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-05-05T14:50:06.370" />
  <row Id="9492" PostTypeId="2" ParentId="2065" CreationDate="2011-05-05T15:16:54.450" Score="1" ViewCount="" Body="&lt;p&gt;WiFi signals are just microwaves, so they are affected by the same things. If something would prevent a microwave oven from working, there is a good chance it will also block a wireless signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of a wooden bed, humidity in the wood could reduce the signal, although not by much. Even if wood would absorb microwaves, a bed is typically not a massive slab of it, but has openings between the slats etc.. You would more likely get an effect from any springs in the matress. Other things that commonly weaken the signal are steel in reenforced concrete walls, and anti-glare coatings on windows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be a great case for an experiment btw..&lt;/p&gt;&#xA;" OwnerUserId="825" LastActivityDate="2011-05-05T15:16:54.450" />
  <row Id="9493" PostTypeId="2" ParentId="9484" CreationDate="2011-05-05T15:33:21.747" Score="3" ViewCount="" Body="&lt;p&gt;Coherence of light doesn't harm this experiment because this experiment is based on geometric optics. However, laser is actually an immensely good source of light which has a very well-defined direction of the light rays. That's why it's much better to use a laser as the light source than the candles or whatever Foucault used to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see a web page of people who have repeated the Foucault rotating mirror measurement of the speed of light using a He-Ne laser, see e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.pas.rochester.edu/~pavone/particle-www/teachers/demonstrations/FoucaultDemonstration.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.pas.rochester.edu/~pavone/particle-www/teachers/demonstrations/FoucaultDemonstration.htm&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-05T15:33:21.747" CommentCount="1" />
  <row Id="9494" PostTypeId="1" AcceptedAnswerId="9503" CreationDate="2011-05-05T16:26:55.717" Score="0" ViewCount="120" Body="&lt;p&gt;My question is mainly an engineering question. Assume I have a turbine in which I adiabatically expand compressed air. The air cools down and does work to its surroundings, which is captured by the blades of the turbine and then is transformed into rotational energy/electricity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When instead I let the compressed air flow through a nozzle into open space (no vacuum) and it expands adiabatically, it also cools down. But where is the internal energy of the air going? It does work to its surroundings, but where does this work end up? Does the surrounding heat up? Does it produce wind, i.e. kinetic energy in the surrounding air?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feel free to replace adiabatic with polytropic. Whenever the gas cools down, I ask: Where does the energy end up, when I do not capture it with a turbine?&lt;/p&gt;&#xA;" OwnerUserId="2480" LastActivityDate="2011-05-05T22:41:17.803" Title="free adiabatic expansion" Tags="&lt;thermodynamics&gt;" AnswerCount="1" />
  <row Id="9495" PostTypeId="1" AcceptedAnswerId="9496" CreationDate="2011-05-05T16:44:27.227" Score="4" ViewCount="237" Body="&lt;p&gt;for example we know that a the Dielectric relaxation time is about 10^-14 sec. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you want a motivation- i would like to think hypothetically of very accurate switching or clocking...&lt;/p&gt;&#xA;" OwnerUserId="3421" LastActivityDate="2011-05-06T16:22:02.520" Title="what is the fastest process in nature?" Tags="&lt;particle-physics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="9496" PostTypeId="2" ParentId="9495" CreationDate="2011-05-05T17:05:20.377" Score="2" ViewCount="" Body="&lt;p&gt;The fastest fully localized process is the evaporation of the smallest black hole worth the time - it takes one &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck_time&quot; rel=&quot;nofollow&quot;&gt;Planck time&lt;/a&gt; or so, $10^{-43}$ seconds. There are other characteristic processes in quantum gravity that take a Planck time - the shortest time scale for which the usual spacetime geometry works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you allow changes in collective properties of large objects, there are much shorter times. For example, the mass of the visible Universe is $3\times 10^{52}$ kilograms or so. The corresponding energy, via $E=mc^2$, is $10^{61}$ Joules. The periodicity of the corresponding quantum wave, via $E=\hbar\omega$, is about $10^{-95}$ seconds. So the wave function of the Universe periodically changes its phase $10^{95}$ times every second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess that you mean local processes, and moreover some processes accessible experimentally. That's a question with no permanent answer. The whole field of particle physics may be classified according to the time scale we can resolve. The time scale you mention is that of atomic physics; the time scales in nuclear physics are up to 10 orders of magnitude shorter, $10^{-24}$ seconds or so. The LHC is probing time scales that are 4 orders of magnitude shorter than that. In principle, this progress could continue. The &quot;higher energy&quot; we have, the shorter times (and distances) we may resolve.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-05T17:05:20.377" CommentCount="5" />
  <row Id="9497" PostTypeId="2" ParentId="9490" CreationDate="2011-05-05T18:03:52.313" Score="1" ViewCount="" Body="&lt;p&gt;It's just the usual conservation law. When something is conserved then its amount in some observed region can't disappear but must be flowing in or out. So this means that time derivative of the integral of the quantity in a given volume (the LHS) must be equal to total flux associated with the quantity through a surface of that volume (the RHS).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case surface (and flux through it) splits into horizontal and vertical parts. Horizontal part is just diffusion across the wall due to pressure difference while vertical part is a fluid's flow through the tube. The amount that flows in depends on the concentration at a given location $U(0,t)$, the speed $v(0)$ at that location and the cross-section $A$. Similarly for the outflow (but note the minus sign).&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-05-05T18:03:52.313" CommentCount="1" />
  <row Id="9498" PostTypeId="2" ParentId="9482" CreationDate="2011-05-05T18:34:17.150" Score="7" ViewCount="" Body="&lt;p&gt;The constant wake angle is a well known phenomenon for boats and ships and is known as the Kelvin wake.  As others have alluded to, it is the result of the unique relation between group and phase velocity for (deep water) gravity waves.  The best detailed explanation I have seen in in the textbook &quot;The Physics of Waves&quot; by Howard Georgi, which is available online for free here: &lt;a href=&quot;http://www.people.fas.harvard.edu/~hgeorgi/new.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.people.fas.harvard.edu/~hgeorgi/new.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-05-05T18:34:17.150" />
  <row Id="9499" PostTypeId="2" ParentId="9490" CreationDate="2011-05-05T18:39:28.583" Score="1" ViewCount="" Body="&lt;p&gt;Think about it physically. How much gas is in the tube? There are three sources:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The gas coming in at $x=0$ which was already in solution.&lt;/li&gt;&#xA;&lt;li&gt;The gas being carried away by the fluid at the other end, $x=L$&lt;/li&gt;&#xA;&lt;li&gt;Diffusion across the walls of the tube.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These three terms are represented in the equation $$&#xD;&#xA;\frac{d}{dt} \left( A \int_{0}^{L} U(x,t) dt \right) = v(0)AU(0,t) - v(L)AU(L,t) + p \int_{0}^{L} q(x,t) dx&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term is the velocity of the liquid, $v$, times the cross section, $A$, giving the total volume of liquid coming in at $x=0$. We multiply this by the concentration of gas in the liquid at $x=0$ to get the velocity that gas enters at $x=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second term goes in much the same way, but we have a negative sign because it is the gas which is &lt;em&gt;leaving&lt;/em&gt; at the other end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say you understand that the third term as diffusion through the walls, which is correct.&lt;/p&gt;&#xA;" OwnerUserId="1655" LastActivityDate="2011-05-05T18:39:28.583" CommentCount="1" />
  <row Id="9500" PostTypeId="1" AcceptedAnswerId="9502" CreationDate="2011-05-05T18:55:16.047" Score="1" ViewCount="103" Body="&lt;p&gt;A ball is struck such that its max height is equal to its range. determine the angle its struck at. Air resistance is ignored and the ground is horizontal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my reasoning is that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using vectors&#xA;Xsin(angle) = Xcos(angle)&#xA;so tan(angle) = x/x&#xA;so inverse tan of 1 is 45degrees&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this be the right approach to the question&#xA;Id ask my lecturer but im at home &#xA;Cheers&lt;/p&gt;&#xA;" OwnerUserId="3422" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-05T19:26:15.973" LastActivityDate="2011-05-05T20:51:36.810" Title="Projectile motion question" Tags="&lt;homework&gt;&lt;newtonian-mechanics&gt;&lt;projectile&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9501" PostTypeId="2" ParentId="9421" CreationDate="2011-05-05T19:32:52.807" Score="4" ViewCount="" Body="&lt;p&gt;Half life variations have been suspected for decades, and almost all (maybe all…but I do not pretend to have a comprehensive knowledge) have been shown to be caused by limits in experimental design.  This latest set from this (now expanded) group (they did another paper on this topic a couple of years ago &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0808/0808.3283v1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/arxiv/pdf/0808/0808.3283v1.pdf&lt;/a&gt;) is interesting.  Any periodic variation in decay rates is of great importance regardless of whether it correlates to some astronomical phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess that is a good setup for a “I’m not really going to answer your question, but” type answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2008 paper was followed by a set of measurements from the Cassini probe (http://arxiv.org/PS_cache/arxiv/pdf/0809/0809.4248v1.pdf).  The distance between that probe and the sun was such that any sun effect should have been magnified.  None was seen.  This may disprove the solar effect, or it could say something about the isotopes used.  The Cassini isotopes were different than the Purdue group’s isotopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The isotopes they used in the paper you cite were interesting:   Mn 56: 2.6 hr, Si 32: 101 yr, Ra 226: 1,600 yr.  From a quick scan of the plots it looks like the longer the half-life the greater the adherence to the annual periodicity.  I also notice that the bulk of the data was accumulated before 1990, and the data after 1996 does not show the same adherence to the annual periodicity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m trying not to discount these decay-rate observations, and there are good reasons to not discard them out-of-hand.  Most observations of constant decay rates have been made with rapidly decaying isotopes over a short time frame.  The experiments are just much easier to set up.  Very long term experiments are relatively rare, and could illuminate previously unobserved phenomena.  If the periodicity is on the order of a year it should only take a couple of years to confirm the data using available methodologies.    &lt;/p&gt;&#xA;" OwnerUserId="2708" LastActivityDate="2011-05-05T19:32:52.807" CommentCount="2" />
  <row Id="9502" PostTypeId="2" ParentId="9500" CreationDate="2011-05-05T19:57:26.383" Score="2" ViewCount="" Body="&lt;p&gt;Range $  = v^2 \sin(2\theta) / g = (v \sin \theta)^2 / 2g = $ Max height &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4 \sin\theta  \cos\theta = \sin\theta  \sin\theta $ because $\sin2\theta = 2 \sin\theta \cos\theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 4 \cos \theta = \sin\theta $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \tan\theta = 4$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta = \tan^{-1} 4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem is $X\cos\theta = X\sin\theta$ equates horizontal and vertical components of $X$ (whatever $X$ is), which isn't same as equating range of projectile to maximum possible height when thrown at certain angle. &lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2011-05-05T20:51:36.810" LastActivityDate="2011-05-05T20:51:36.810" CommentCount="2" />
  <row Id="9503" PostTypeId="2" ParentId="9494" CreationDate="2011-05-05T20:09:54.617" Score="0" ViewCount="" Body="&lt;p&gt;Adiabatic expansion in a nozzle results in internal energy being converted into kinetic energy. A generic steady-state energy balance is :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta (H + \frac{1}{2v^2} + zg) = Q + W$$&lt;/p&gt;&#xA;" OwnerUserId="1559" LastEditorUserId="869" LastEditorDisplayName="" LastEditDate="2011-05-05T22:41:17.803" LastActivityDate="2011-05-05T22:41:17.803" CommentCount="2" />
  <row Id="9504" PostTypeId="1" CreationDate="2011-05-05T20:23:28.650" Score="0" ViewCount="249" Body="&lt;p&gt;I am trying to understand Space-Charge-Limited-Current (SCLC) and the resulting Mott-Gurney Law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main problem is conceptual; I don't get why SCLC doesn't produce Ohm's law (j $\propto$ V). &lt;/p&gt;&#xA;" OwnerUserId="1939" LastEditorUserId="869" LastEditorDisplayName="" LastEditDate="2011-05-06T00:30:31.593" LastActivityDate="2011-06-05T05:54:53.227" Title="Ohm's Law and Space-Charge Limited Current" Tags="&lt;semiconductor-physics&gt;&lt;electric-circuits&gt;" AnswerCount="1" />
  <row Id="9505" PostTypeId="1" CreationDate="2011-05-05T20:24:09.103" Score="4" ViewCount="84" Body="&lt;p&gt;Have the correlation functions of the XY spin chain model,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;H=-\sum_l (J_x \sigma_l^x \sigma_{l+1}^x+J_x \sigma_l^y \sigma_{l+1}^y)-B\sum_l \sigma_l^z&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;been calculated using a functional partition function with source terms? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By functional partition function I mean the partition function plus source terms (or generating functional) which allow correlation functions such as $\langle \sigma_l^x \sigma_{l+1}^x \rangle$ to be calculated analytically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;Z[\bar{\omega},\omega] = Tr \left[Exp \left[-\int_0^\beta d\tau H(a^{\dagger},a) - \int_0^\beta d\tau \sum_l (\bar{\omega} a + a^{\dagger} \omega)  \right] \right]&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $H(a^{\dagger},a)$ is the XY Hamiltonian expressed in terms of spinless fermions (i.e. after a Jordan-Wigner transformation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Anyone? An Ising chain would be fine too!&lt;/p&gt;&#xA;" OwnerUserId="873" LastEditorUserId="873" LastEditorDisplayName="" LastEditDate="2011-05-06T22:01:07.323" LastActivityDate="2011-05-06T22:01:07.323" Title="Have the correlation functions of the XY spin chain model been calculated using a functional partition function with source terms?" Tags="&lt;path-integral&gt;&lt;spin-chains&gt;&lt;correlation-function&gt;" FavoriteCount="2" />
  <row Id="9506" PostTypeId="1" CreationDate="2011-05-05T20:35:16.893" Score="2" ViewCount="88" Body="&lt;p&gt;I am trying to &quot;decode&quot;/derive an expression for the macroscopic rate constant for the tunneling of protons through a potential energy barrier that I read in a journal article:&#xA;$$&#xD;&#xA;k_{\rm tun}(T)=(2\pi\hbar)^{-1}\int_0^{V_{\rm max}} Q(V,T) P_{\rm tun}(V)\ dV.&#xD;&#xA;$$&#xA;So basically: the authors say work out the probability of tunneling at each point up the potential energy barrier (from 0 to $V_{\rm max}$), multiply this by the Boltzmann factor ($Q(V,T)$), integrating over all energies (i.e. we are taking a Boltzmann average )and then convert from energy in J to rate by multiplying by $(2\pi\hbar)^{-1}$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors use $(2\pi\hbar)^{-1}$ to convert from an energy in J to a rate in per second. This implies that the relationship between energy and frequency being used is: &#xA;$$&#xD;&#xA;E=\hbar\omega &#xD;&#xA;$$&#xA;I thought this just applied to electromagnetic radiation and free particles not in a potential. Is it OK to use this relation for protons tunneling in a potential? Or should you use some info about the shape of the potential?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N26&lt;/p&gt;&#xA;" OwnerUserId="3424" LastEditorUserId="3424" LastEditorDisplayName="" LastEditDate="2011-05-05T21:37:36.957" LastActivityDate="2011-05-06T04:47:15.757" Title="Tunneling Rate Constant" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="9507" PostTypeId="1" AcceptedAnswerId="9508" CreationDate="2011-05-05T23:53:08.410" Score="2" ViewCount="107" Body="&lt;p&gt;I'm working in a book on relativity. The author states that if $u$ and $u&amp;#39;$ are a velocity referred to two inertial frames with relative velocity $v$ confined to the $x$ axis, then the quantities $l$, $m$, $n$ defined by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;&#xD;&#xA;(l, m, n) = \frac{1}{|u|}(u_x, u_y, u_z) &#xD;&#xA;&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;(l&amp;#39;, m&amp;#39;, n&amp;#39;) = \frac{1}{|u&amp;#39;|}(u&amp;#39;_x, u&amp;#39;_y, u&amp;#39;_z) &#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are related by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;&#xD;&#xA;(l&amp;#39;, m&amp;#39;, n&amp;#39;) = \frac{1}{D}(l - \frac{v}{u}, m\gamma ^{-1}, n\gamma ^{-1})&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that this can be considered a relativistic aberration formula.&#xA;The author gives the following definition for $D$, copied verbatim. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;&#xD;&#xA;D = \frac{u&amp;#39;}{u}\left( 1 - \frac{u_xv}{c^2}\right) = \left[1 - 2l\frac{v}{u} + \frac{v^2}{u^2} - (1 - l^2)\frac{v^2}{c^2} \right]^{\frac{1}{2}}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why the author finessess the expression into the third/final form. I was able to get it but it seems like a pain. Why is that better than the second expression? It seems more difficult to calculate and not at all clear or meaningful.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Also, in case it's not clear, $\gamma =1/ \sqrt{1 - \frac{v^2}{c^2}}$ and $|u| = |(u_x, u_y, u_z)| = \sqrt{u_x^2 + u_y^2 + u_z^2}$ &lt;/p&gt;&#xA;" OwnerUserId="2953" LastEditorUserId="2953" LastEditorDisplayName="" LastEditDate="2011-05-06T22:10:24.177" LastActivityDate="2011-05-06T22:10:24.177" Title="Physical interpretation of equation for relativistic aberration" Tags="&lt;special-relativity&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9508" PostTypeId="2" ParentId="9507" CreationDate="2011-05-06T04:44:04.230" Score="1" ViewCount="" Body="&lt;p&gt;The last form avoids any usage of $u&amp;#39;$ as well as $u_x$ which may be considered derived or non-covariant quantities, respectively, so the last form may be superior in many contexts. Moreover, it makes it much easier to imagine how far the result is from one - because it resembles a Taylor expansion of a sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even if one didn't write the comments above, why would it be a problem to write a result in yet another mathematically equivalent way?&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-06T04:44:04.230" CommentCount="1" />
  <row Id="9509" PostTypeId="2" ParentId="9506" CreationDate="2011-05-06T04:47:15.757" Score="1" ViewCount="" Body="&lt;p&gt;$E=\hbar\omega$ is a totally universal formula that holds for all particles and everywhere in quantum mechanics. Schrödinger's equation guarantees that. The same question was being answered yesterday:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/9457/gravitational-wave-energy/9460&quot;&gt;Gravitational wave energy&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-06T04:47:15.757" />
  <row Id="9510" PostTypeId="2" ParentId="9504" CreationDate="2011-05-06T05:01:28.353" Score="1" ViewCount="" Body="&lt;p&gt;Semiconductors are complicated, nonlinear materials so there's no reason to assume a particular simple relationship between any two inequivalent quantities. They're not just conductors with resistance - the non-Ohmic relationship between current and voltage is really what they're famous for and good for. I think that a better approach to similar questions is to look at a derivation of the Mott-Gurney law (yes, Mott was my uncle) such as &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://my.ece.ucsb.edu/mhwong/documents/thesis_app2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://my.ece.ucsb.edu/mhwong/documents/thesis_app2.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and ask about particular assumptions of the derivation rather than directly about the final result. One can't see what the result is or is not without actually deriving it.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-06T05:01:28.353" />
  <row Id="9511" PostTypeId="1" CreationDate="2011-05-06T05:11:09.663" Score="3" ViewCount="165" Body="&lt;p&gt;Why only big rocks (planets) have satellites and not small ones? Why cosmic dust doesn't orbit rocks that are many times heavier than the dust grains? If dust is still too heavy then what about molecules, atoms, or any particle for that matter? The mass difference should be millions of millions times, isn't it enough for orbiting? Moon is 1% of Earth mass, yet we don't see 1kg rocks orbiting 100kg ones.&lt;/p&gt;&#xA;" OwnerUserId="213" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-06T18:24:15.447" LastActivityDate="2011-05-06T18:24:15.447" Title="Why all rocks are not orbiting bigger rocks in space?" Tags="&lt;gravity&gt;&lt;orbital-motion&gt;&lt;solar-system&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="9512" PostTypeId="1" CreationDate="2011-05-06T05:12:40.470" Score="1" ViewCount="84" Body="&lt;p&gt;This was just a random thought that crossed my mind, and I'm sure there is a simple physics explanation. Say we have a leaky faucet 20 feet above the ground that lets out the exact same amount of water every few seconds or so. Why is it that the splash pattern for every drop is never the same as any other drop? Now it's not like I've sat there and observed every splash, but sometimes its obvious, like a certain splash's water reached a farther distance than another splash. If we are letting the same amount of water come out everytime, and nothing else changes, why do the splashes vary? (pretend the water disappears off the ground after every splash, that way every splash has the same environment.)&lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2011-05-06T05:45:45.783" Title="Water dripping on the floor" Tags="&lt;gravity&gt;&lt;water&gt;" AnswerCount="2" />
  <row Id="9513" PostTypeId="2" ParentId="9512" CreationDate="2011-05-06T05:24:53.067" Score="2" ViewCount="" Body="&lt;p&gt;A dripping faucet is a classical setup in chaos experiments. &#xA;The drops do not fall at equidistant time intervals and they &#xA;are not identical in mass. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-05-06T05:24:53.067" CommentCount="2" />
  <row Id="9514" PostTypeId="2" ParentId="9511" CreationDate="2011-05-06T05:42:32.060" Score="3" ViewCount="" Body="&lt;p&gt;The mass difference has practically nothing to do with whether two things can orbit each other. You can have two objects of the same mass in orbit (e.g. binary star systems), or you can have two objects of wildly different masses in orbit (the Earth and a piece of space junk), or anything in between. There are a lot of factors involved, for example:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A random dust grain probably won't be going at the right velocity to be captured into orbit around a bigger rock (like an asteroid).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When dust grains do orbit asteroids, there will probably be a lot of them, so they're going to collide with each other and lose energy, so they fall out of orbit rather quickly.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Planets tend to be pretty isolated, so things can orbit them without getting disturbed by other planets. In contrast, smaller rocks like asteroids are often found in large groups, so a dust grain couldn't orbit just one - it would also be affected significantly by the gravity of other asteroids in the vicinity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-05-06T05:42:32.060" CommentCount="4" />
  <row Id="9515" PostTypeId="2" ParentId="9512" CreationDate="2011-05-06T05:45:45.783" Score="4" ViewCount="" Body="&lt;p&gt;In order for the splashes to actually be the same, each drop would have to be &lt;em&gt;exactly&lt;/em&gt; identical, down to the exact positions and velocities of each molecule in the drop. It's basically impossible that two drops would ever be the same to that level of detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that this is a chaotic system, as Georg said, means that even tiny differences in the state of each drop can have a significant effect on the splash pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the same would have to apply to the environment: it would have to be identical, down to the molecular structure of the floor, for every drop. That would never happen in real life.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-05-06T05:45:45.783" />
  <row Id="9516" PostTypeId="1" CreationDate="2011-05-06T06:44:06.227" Score="4" ViewCount="163" Body="&lt;p&gt;Here's a puzzle I have been pondering over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have two extremal black holes with the same charge, the electrostatic repulsion between them ought to cancel the gravitational attraction between them. Without any net attraction or repulsion between them, how close can we bring these two black holes to each other without merging? A peculiarity of the metric seems to suggest the event horizon is always infinitely far away for extremal black holes $\int_R^r dr&amp;#39; \frac{1}{r&amp;#39;-R} = \infty$. Does this give enough elbow room for both black holes to get arbitrarily close to each other without merging?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="3430" LastActivityDate="2011-05-11T07:23:51.213" Title="How close can two extremal black holes with the same charge can get?" Tags="&lt;general-relativity&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="9517" PostTypeId="2" ParentId="9430" CreationDate="2011-05-06T07:29:38.783" Score="1" ViewCount="" Body="&lt;p&gt;Assuming that our perceptions of Relativity hold up in interstellar space...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ship would continue to accelerate as long as fuel held out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speed would continue its asymptotic clime towards C... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;until drag was sufficient to end that climb. Drag would also increase due to interstellar medium collisions as speed increased. But note also: a sufficiently high thrust will not be exceeded by drag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that as speed increases, and drag increases, so does radiation. The interstellar medium results in Alpha &amp;amp; Beta radiation increases, and high energy photon releases from same producing X-ray and even gamma radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that several science fiction works have relativity have a cutoff point. In a few more, Relativity is ignored entirely; the most egregious offender is E.E. &quot;Doc&quot; Smith's &lt;em&gt;Lensmen&lt;/em&gt; series. John Ringo's Looking Glass series has heliopause boundary changes to the nature of physics, including relativity&lt;/p&gt;&#xA;" OwnerUserId="3200" LastActivityDate="2011-05-06T07:29:38.783" CommentCount="2" />
  <row Id="9518" PostTypeId="2" ParentId="9511" CreationDate="2011-05-06T11:32:11.143" Score="2" ViewCount="" Body="&lt;pre&gt;&lt;code&gt;An intriguing picture of small solar systems comes to mind.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Nevertheless,  Luboš is correct that for any recognizable system the velocities of capture would be so low as to make the question moot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rough example , equating centripetal gravitational with centrifugal forces, a particle orbiting a 100 kg mass 100 meters away from it would need a velocity order of magnitude of ten microns per second, something that would hardly be observable: the orbit of 628 meters would take 1.7*10^4 hours to complete.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, in real life in our space around the sun it is a many body problem, as David suggests above, so statistical and chaotic behaviours will enter the problem and destroy any small scale regularity. Only in deep outer space outside the sun's field such an orbit might be undisturbed and observed by a patient observer :).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for such small velocities is the weakness of the gravitational constant G. The velocity of a stable orbit is proportional to the square root of G. The mass of the orbiting particle does not matter as it is eliminated in equating  centripetal and centrifugal forces.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-05-06T11:41:56.627" LastActivityDate="2011-05-06T11:41:56.627" />
  <row Id="9519" PostTypeId="1" CreationDate="2011-05-06T12:17:50.600" Score="1" ViewCount="96" Body="&lt;p&gt;Metallic strings exist in different kinds but I would like to measure for a metallic string:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the section/geometry along its length, to a precision of 1/100 mm&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;its elongation when a tension is exerted on it on a measuring system I would construct to apply forces of the order of 1 to 100 daN. I have no clear idea of the order of magnitude of the elongation, so it might be too small to measure with reasonable equipement.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Any ideas, suggestions, theoretical and practical remarks welcome.&lt;/p&gt;&#xA;" OwnerUserId="2706" LastActivityDate="2011-05-11T06:10:09.910" Title="How can I measure the physical properties of a metallic string?" Tags="&lt;experimental-physics&gt;&lt;measurement&gt;&lt;applied-physics&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="9520" PostTypeId="1" AcceptedAnswerId="9525" CreationDate="2011-05-06T13:44:11.857" Score="1" ViewCount="63" Body="&lt;p&gt;This question is an attempt to complement &lt;a href=&quot;http://physics.stackexchange.com/questions/9421/strange-modulation-of-radiactive-decay-rates-with-solar-activity&quot;&gt;this other question&lt;/a&gt; about fluctuations in radiactive decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is completely experimental though: in general, suppose i have certain sample of a radiactive substance. My goal is to gather experimental data about fluctuations in net momenta of all decay products. I certainly would have to set the sample temperature near to zero Kelvin in order to reduce or eliminate thermal contributions to net momenta. However, what is the experimental setup one would have to have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should one only consider individual events where all decay products have been detected and their individual momenta obtained?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you make sure that a set of decay product events are associated to the same event? time correlation? does this mean that the sample size needs to be small enough so overlapping decay events are percentually few and can be filtered out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what other considerations are required?&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-05-06T15:52:25.603" Title="experimental setup to measure Center-Of-Momenta of products from spontaneous radiactive decay" Tags="&lt;particle-physics&gt;&lt;experimental-physics&gt;&lt;radioactivity&gt;&lt;particle-detectors&gt;" AnswerCount="1" />
  <row Id="9521" PostTypeId="2" ParentId="9519" CreationDate="2011-05-06T13:56:27.880" Score="2" ViewCount="" Body="&lt;p&gt;In a physics laboratory session at university we measured the extension of copper wire with different weights attached.  To do so requires two wires hung vertically, one has the weights attached and is stretched, the other acts a sort of control.  Between them a device is attached, this contains an adjustable spirit level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before the weight is applied the spirit level is set to horizontal, after the weights have stretched one string the device can be re-levelled and this 'correction angle' implies a certain extension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To predict the sort of stretch you might achieve I would suggest looking up Youngs Modulus: &lt;a href=&quot;http://en.wikipedia.org/wiki/Young%27s_modulus&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Young%27s_modulus&lt;/a&gt;.  (For metal wires and 'safe' loads don't expect very much!).&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-05-06T13:56:27.880" />
  <row Id="9522" PostTypeId="2" ParentId="9511" CreationDate="2011-05-06T14:57:27.207" Score="3" ViewCount="" Body="&lt;p&gt;I want to expand a bit on David Zaslavsky's point #1: &quot;A random dust grain probably won't be going at the right velocity to be captured into orbit around a bigger rock (like an asteroid).&quot; Without the assistance of a third body, it's actually &lt;em&gt;impossible&lt;/em&gt; for an object to be gravitationally &quot;captured&quot; in an orbit by another: if it wasn't already in orbit before, there's no way for it to start orbiting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is just conservation of energy. If the object wasn't already in an orbit, then it was moving too fast, given its distance, to be in an orbit. (To put it the other way around, it was too far away, given its speed.) Even if the little guy was moving in just the right direction to come close to the big guy, it won't be captured: it will move in on a hyperbolic path, approach the other object once, and then fly away again, simply because it has too much energy to be captured. The only way to produce an orbit is for a third body to interact with the system and siphon off some energy at the right time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That can happen, but unless the density of things flying around is high, it's rare. And if the density is high, then subsequent collisions, which will disrupt the orbit, will also be common. &lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-05-06T14:57:27.207" CommentCount="1" />
  <row Id="9523" PostTypeId="1" AcceptedAnswerId="13294" CreationDate="2011-05-06T14:59:40.443" Score="6" ViewCount="206" Body="&lt;p&gt;Howdy, I'm building a simulation for looking at the light field underwater. In order to verify my simulation, I'm looking for some data showing the far-field intensity that comes from single scattering from many small particles in suspension. I suspect Mie theory plays a part here, but I'm having a hard time finding some results, rather than doing all the derivations myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, I want to know the power distribution on a plane after a beam of light has been scattered by a bunch of small particles through a volume. I know Oregon Medical has a nice online simulation that produces scattering phase functions (http://omlc.ogi.edu/calc/mie_calc.html), but that doesn't give me the power on a plane - only the scattering profile from individual particles. I'm fine with only a single scattering result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to do initial verification using a fixed particle size. Having a hard time finding a reference with this data. Help? &lt;/p&gt;&#xA;" OwnerUserId="2735" LastEditorUserId="2735" LastEditorDisplayName="" LastEditDate="2011-05-24T18:55:12.377" LastActivityDate="2011-08-10T19:33:25.390" Title="Far-field intensity from scattering of small particles" Tags="&lt;optics&gt;&lt;laser&gt;&lt;scattering&gt;" AnswerCount="5" FavoriteCount="2" />
  <row Id="9524" PostTypeId="1" AcceptedAnswerId="9540" CreationDate="2011-05-06T15:07:43.283" Score="3" ViewCount="307" Body="&lt;p&gt;I learned programming as a child, for fun. Now I am working as a programmer, even though I got a business major degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if there are career paths for doing physics &lt;em&gt;other&lt;/em&gt; than becoming a researcher in a University or a professor, that are not &lt;em&gt;all&lt;/em&gt; about credentials?&lt;/p&gt;&#xA;" OwnerUserId="602" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-06T20:17:27.203" LastActivityDate="2011-05-07T20:37:47.307" Title="A professional physics career without a degree?" Tags="&lt;soft-question&gt;&lt;physics-careers&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="9525" PostTypeId="2" ParentId="9520" CreationDate="2011-05-06T15:35:56.377" Score="1" ViewCount="" Body="&lt;p&gt;That's a hard experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remnant nucleus generally has a very low (non-relativistic) velocity{*}, making it difficult to detect and characterize. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Should one only consider individual events where all decay products have been detected and their individual momenta obtained?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the formal definition of an &quot;exclusive&quot; event, but experimentally we (that's the nuclear and particle physics &quot;we&quot;) almost always relax it with respect to the remnant nucleus.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do you make sure that a set of decay product events are associated to the same event? time correlation? does this mean that the sample size needs to be small enough so overlapping decay events are percentually few and can be filtered out?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where possible you set your event rate slow compared to the DAQ latch and report rate. So almost all reported DAQ events are associated with only one physics event. There is a bit of art and science in this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I was really going to do it, I'd consider measuring the transverse direction only at a radioactive beam accelerator{**}. That won't make it easy, mind you, just less horribly difficult. That simplifies your life in the sense that the remnant is now a ionizing particle with a macroscopic track length, so you can ID it, but of course you no longer have good information on exactly where the decay occurs and may not be able to extract a good value for the longitudinal component of the momentum due to the decay {+}.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;{*} Momentum of a few MeV, and mass of multiple to hundreds of GeV. You might ask the AMO people about this, they are more used to working with just barely ionizing energies than particle physicists and may have an answer. Perhaps it is a good application for a multi-channel plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{**} They do exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{+} Lost in the noise of the much larger component due to the beam momentum.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-06T15:52:25.603" LastActivityDate="2011-05-06T15:52:25.603" />
  <row Id="9526" PostTypeId="1" CreationDate="2011-05-06T15:44:46.433" Score="3" ViewCount="171" Body="&lt;p&gt;You buy one of those remote control toy helicopters. You bring it into an elevator. The elevator goes up. Does the helicopter hit the floor or does the floor of the elevator push the air up into the bottom of the helicopter so that it maintains altitude relative to the elevator? The same question can be reversed. If the elevator goes down will it hit the cieling? If the elevator is air-tight, would it make a difference?&lt;/p&gt;&#xA;" OwnerUserId="3435" LastActivityDate="2011-05-06T15:53:28.793" Title="Helicopter in an Elevator" Tags="&lt;forces&gt;&lt;aerodynamics&gt;" AnswerCount="2" />
  <row Id="9527" PostTypeId="2" ParentId="9526" CreationDate="2011-05-06T15:49:18.770" Score="6" ViewCount="" Body="&lt;p&gt;The air in an elevator does tend to move with the elevator, because it has relatively little inertia. However, thinking about the problem in these terms seems, to me, misleading. The simplest way to think about this is to consider the acceleration of the elevator as and addition to the normal acceleration due to gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this light, it would be as if the helicopter were momentarily heavier wen the elevator accelerated upwards, and momentarily lighter when it accelerates downwards. This would inevitably cause changes in the height of the helicopter above the floor of the elevator, but I expect that most real-world elevators would not accelerate fast enough nor long enough for the helicopter to be smashed to the floor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, toy helicopters are not all alike, so your mileage may vary!&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-05-06T15:49:18.770" />
  <row Id="9528" PostTypeId="2" ParentId="9526" CreationDate="2011-05-06T15:53:28.793" Score="0" ViewCount="" Body="&lt;p&gt;Assuming an initially vertically stationary hover I would say that the floor closes in on the helicopter to some small distance and the helicopter then moves up with the floor to some degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reasoning is that the helicopter is creating a downwards force via its rotors but this is poorly coupled to the elevator floor and so is essentially independent of its movement.  Until the floor closes in on some length where the downward force couples with the floor of the elevator and allows the helicopter to move upwards at some velocity.  This velocity is very hard to estimate without a hatful of assumptions.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-05-06T15:53:28.793" />
  <row Id="9529" PostTypeId="1" AcceptedAnswerId="9838" CreationDate="2011-05-06T16:04:05.567" Score="3" ViewCount="270" Body="&lt;p&gt;Over the past hour or so I've been following one of my standard physics-based, wanders-through-the-internet. Specifically, I began by reviewing some details of dark energy theory but soon found myself pondering a totally unassociated topic..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of today's tweets by #arxivblog concerns a speculative (IMO) paper about the possibility that &quot;Mini Black Holes Could Form Gravitational Atoms&quot;. I read the article on it (&lt;a href=&quot;http://www.technologyreview.com/blog/arxiv/26726/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) and a few of the comments and soon became perplexed by how it contradicted something I already knew. Namely, small black holes don't last very long so how-on-earth (or in space) could one &quot;of about 10 to 1000 tonnes&quot; exist long enough to capture a passing particle into a quasi-stable orbit? (I was recalling something I'd read about possible micro black holes, being produced at the LHC, and subsequently dissapearing in a tiny fraction of a second, due to Hawking radiation.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;~&lt;/strong&gt; My first question is therefore do these micro black holes decay in a &lt;em&gt;particle-like&lt;/em&gt; fashion or does Hawking radiation theory have significant consequences on the possible decay channels?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little more digging and I soon found that the lifetime of a black hole is $ t_{l}=M^{3}/3K $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $ K=h .c^{4}/30720.\pi^{2}.G^{2}=3.98\times10^{15} kg^{3}/s^{-1} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Giving, for a mini black hole of mass, $M=100\times10^{3}kg$, $ t_{l}=0.084s $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which, I'm sure you'll agree, is clearly too short to allow anything that could meaningfully be described as a gravitational atom. (Incidentally, my calculation for the lifetime of an LHC black hole gave something of order $10^{-94}s$!!) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just then, the resolution to this apparent contradiction dawned on me - &lt;em&gt;the above formula is derived from the radiated energy given off by a black hole via Hawking Radiation. Conversely, the paper in question is written from an initial, unstated, assumption that Hawking Radiation does not exist.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;~&lt;/strong&gt; This led me wonder, to what extent is Hawking radiation accepted amongst professional theorists? (I'm aware that it has not yet been directly prooven, so consideration of the implications of it's non-existence must surely be worthwhile, and interesting, IMO.) &#xA;I realise this question is somewhat vague but I am hoping for some elucidation on how firm-a-foundation Hawking first derived this phenomena and, given it is based on QFT &lt;em&gt;bolted-onto&lt;/em&gt; a curved spacetime, has modern developments (in for example, String theory) corroborated the possibility of energy loss over the event horizon in this way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;~&lt;/strong&gt; A quick glance at the wiki article on primordial black holes (&lt;a href=&quot;http://en.wikipedia.org/wiki/Primordial_black_hole#String_theory&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) informed me that the Fermi Gamma-ray Space Telescope (GLAST experiment), launched in 2008, is hoping to find evidence of primordial black holes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;If they observe specific small interference patterns within gamma-ray bursts (GRB), it could be the first indirect evidence for primordial black holes and string theory.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reference to string theory appears only to concern the longer predictions of the primordial back hole's lifetimes; based on the extra, 'rolled-up' spatial dimensions posited by some/all(?) string theories (AFAIK due to gravity being able to propagate in these extra-dimensions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;~&lt;/strong&gt; Finally, therefore, if GLAST does/has find/found certain characterists in GRB data, how strongly does it (will it) rate as evidence for or against: the existence of primordial black holes, the existence of Hawking radiation and indirectly as evidence for String theory? I have had a look at papers relating to recent GRB data from the LAT experiment (&lt;a href=&quot;http://arxiv.org/find/astro-ph/1/ANDNOT+OR+abs%3a+AND+Fermi+gamma+ti%3a+AND+Fermi+gamma+au%3a+OR+Abdo+Collaboration/0/1/0/past/0/1&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for eg) but, as a non-specialist, it is very unclear to me whether the data has any implications to the questions posed above.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="1696" LastEditorDisplayName="" LastEditDate="2011-05-13T11:18:59.410" LastActivityDate="2011-05-14T04:39:08.377" Title="On black holes, Hawking radiation and gravitational atoms" Tags="&lt;string-theory&gt;&lt;black-holes&gt;&lt;hawking-radiation&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="9530" PostTypeId="2" ParentId="9524" CreationDate="2011-05-06T16:14:41.107" Score="0" ViewCount="" Body="&lt;p&gt;No, you haven't got a chance unless you want to be a gofor in an R&amp;amp;D department of an engineering company. Physicists become programmers because physics jobs are hard to come by and difficult to do compared to programming, not the other way round.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking about it further, yes, you can forge a professional physics career without a degree, as long as you target a company where the applicants don't have a degree. I.e, do voluntary work where you don't expect to get paid much, or where the work conditions are poor, so that no professional physicist would want to work for that company.&lt;/p&gt;&#xA;" OwnerUserId="2146" LastEditorUserId="2146" LastEditorDisplayName="" LastEditDate="2011-05-07T20:37:47.307" LastActivityDate="2011-05-07T20:37:47.307" CommentCount="4" />
  <row Id="9531" PostTypeId="2" ParentId="9495" CreationDate="2011-05-06T16:22:02.520" Score="0" ViewCount="" Body="&lt;p&gt;a photon traveling a planck length.&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-05-06T16:22:02.520" CommentCount="2" />
  <row Id="9532" PostTypeId="1" CreationDate="2011-05-06T16:26:14.743" Score="1" ViewCount="122" Body="&lt;p&gt;Will coffee in two differant size styrofoam cups in the same room conditions cool at the same rate? One cup is 8 oz. and the other would be 12 oz.  &lt;/p&gt;&#xA;" OwnerUserId="3436" LastActivityDate="2011-05-10T18:19:18.093" Title="Coffee cooling in a cup" Tags="&lt;thermodynamics&gt;&lt;fluid-dynamics&gt;" AnswerCount="2" />
  <row Id="9534" PostTypeId="2" ParentId="9482" CreationDate="2011-05-06T16:46:15.343" Score="5" ViewCount="" Body="&lt;p&gt;The ideal Kelvin boat wake ignores surface tension, and it assumes deep water waves with an (in general) broad spectrum of frequencies $\omega$ with dispersion relation $\omega^2=gk$, where $g\approx 9.8 \frac{m}{s^2}$. The ideal Kelvin wake furthermore assumes that the ship sails with a constant velocity, and that the wave amplitudes of the partial waves are so small that they obey a linear superposition principle. The Kelvin wake does not describe the narrow turbulent band behind a ship, nor shock waves. The Kelvin wake consists of two types of waves: &lt;em&gt;transverse&lt;/em&gt; and &lt;em&gt;divergent&lt;/em&gt; waves. There are two characteristic angles &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha\approx 19^{\circ} \qquad \mathrm{and} \qquad \beta\approx 35^{\circ},$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;corresponding to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan(\alpha)= \frac{1}{2\sqrt{2}} \qquad \mathrm{and} \qquad \tan(\beta) = \frac{1}{\sqrt{2}},$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or equivalently,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin(\alpha)= \frac{1}{3} \qquad \mathrm{and} \qquad \sin(\beta) = \frac{1}{\sqrt{3}}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In polar coordinates $(r,\theta)$ of a co-moving coordinate system, where the position of the boat is at the origin, the transverse waves are in the region $|\theta|\leq \beta$, and divergent waves are in the region $\alpha\leq |\theta|\leq \beta$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angles $\alpha$ and $\beta$ are constant in at least two ways: Firstly, they don't depend on the distance $r$ to the ship. This is because the speed of each partial wave (with frequency $\omega$) is independent of the position $(x,y)$. Secondly, $\alpha$ and $\beta$ are, evidently, universal angles, independent of, for instance, $g$. This is explained in the references below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.wikiwaves.org/images/a/a1/Wake.avon.gorge.arp.750pix.jpg&quot; alt=&quot;Image taken from Ref. 3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Howard Georgi, &lt;a href=&quot;http://www.people.fas.harvard.edu/~hgeorgi/new.htm&quot; rel=&quot;nofollow&quot;&gt;&quot;The Physics of Waves&quot;&lt;/a&gt;, Chapter 14. (Hat tip:user1631.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) MIT on-line open course ware, mechanical engineering, wave propagation, lecture notes, fall 2006, &lt;a href=&quot;http://ocw.mit.edu/courses/mechanical-engineering/2-062j-wave-propagation-fall-2006/lecture-notes/&quot; rel=&quot;nofollow&quot;&gt;Chapter 4.7&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) &lt;a href=&quot;http://www.wikiwaves.org/index.php/Ship_Kelvin_Wake&quot; rel=&quot;nofollow&quot;&gt;Wikiwaves&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-06T19:59:35.477" LastActivityDate="2011-05-06T19:59:35.477" />
  <row Id="9535" PostTypeId="2" ParentId="9532" CreationDate="2011-05-06T16:56:30.560" Score="4" ViewCount="" Body="&lt;p&gt;Because styrofoam is a rather good isolator, cooling is dertermined &#xA;by surface area predominantly. From surface area vaporisation is &#xA;possible, which is a very effective way of heat transfer &#xA;(Water has a very high heat of vaporisation). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If both cups have the same cross section, the 12 oz will last longer. &#xA;If the 12 oz cup has 3/2 the surface of the 8 oz cup, both will cool &#xA;at &lt;strong&gt;about&lt;/strong&gt; the same speed.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-05-06T17:29:58.753" LastActivityDate="2011-05-06T17:29:58.753" />
  <row Id="9536" PostTypeId="2" ParentId="9516" CreationDate="2011-05-06T19:09:45.377" Score="4" ViewCount="" Body="&lt;p&gt;The papers you probably should look at are &lt;a href=&quot;http://dx.doi.org/10.1007%2FBF01645696&quot; rel=&quot;nofollow&quot;&gt;the one by Hartle and Hawking&lt;/a&gt; establishing existence of the Majumdar-Papatetrou multi-black-hole solutions, and the &lt;a href=&quot;http://www.jstor.org/stable/2397048&quot; rel=&quot;nofollow&quot;&gt;one by Gibbons&lt;/a&gt; establishing their uniqueness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at Hartle and Hawking, if you fix a background Euclidean metric and &quot;place&quot; the black holes, you can place them arbitrarily close to each other as measured on the background Euclidean metric. However, this tells you pretty much nothing about any physical intuition. The problem is mainly that for a Lorentzian geometry, while the &quot;proper time&quot; for any two time-like related points are well-defined (modulo causality assumptions), the &quot;distance&quot; for any two space-like related points is not. In Riemannian geometry you can minimize over all curves connecting the two points. But in the Lorentzian case by drawing two &quot;almost null&quot; paths from the two points that intersect in the future, you can construct a curve in space-time connecting the two points with &quot;length&quot; integral as close to zero as you want. But this has nothing to do with the black hole geometry what so ever, but just a fact about reasonably nice Lorentzian manifolds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to consider distance between two objects is to compute them relative to some preferred space-like foliation. Here you run into a similar problem as before: in general you can find a foliation that is sufficiently clost to being null that the Riemannian distance along a leave of the foliation is as small as you want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you are down to hoping for the existence of a preferred foliation, perhaps one defined by the static time translation. Unfortunately, as you noted already in your question, the constant $t$ surfaces here do not actually intersect the event horizon (or, as you say it, it intersects the event horizon at &quot;infinite distance&quot;). So measured in this sense the two black holes will necessarily be &quot;infinitely far away&quot; from each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the main difficulty, in fact, is for &lt;em&gt;you&lt;/em&gt; to provide a definition of how you want to measure the distance between two points in Lorentzian geometry. &lt;/p&gt;&#xA;" OwnerUserId="829" LastActivityDate="2011-05-06T19:09:45.377" />
  <row Id="9537" PostTypeId="1" AcceptedAnswerId="9676" CreationDate="2011-05-06T21:18:10.277" Score="1" ViewCount="126" Body="&lt;p&gt;I work with a lot of Boy Scouts who want to complete the &lt;a href=&quot;http://meritbadge.org/wiki/index.php/Space_Exploration&quot; rel=&quot;nofollow&quot;&gt;Space Exploration merit badge&lt;/a&gt;.  It's a great program, and one of the coolest requirements is building, launching, and recovering a model rocket.  Unfortunately, my Scouts are in the city and finding a location to launch rockets is ... well ... tricky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the past, we'd drive out to the countryside and use either a park or a friendly farm to launch the rockets off.  It never used to be a big deal ... but someone got it into their head recently that these rockets pose a significant fire hazard (despite over a decade of fun without a single problem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest argument I've heard against the rockets is the nature of the engine.  Basically, model rocket engines are black powder-based.  A small electric igniter lights the engine and it produces a lot of heat and smoke to propel the rocket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parks and farms that have since banned us from launching rockets finally conceded that, if I could find a way to launch rockets &lt;em&gt;without&lt;/em&gt; burning anything, they'd let me bring the Scouts back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every reference I can find to &quot;cold rocket engines&quot; seems to refer to older 1970s freon-based engines.  I don't want to go that route, but I'm looking for some ideas for cold or low-temperature rocket engines.  Even if it's something I'd have to build myself ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas or suggestions?&lt;/p&gt;&#xA;" OwnerUserId="2197" LastEditorUserId="2197" LastEditorDisplayName="" LastEditDate="2011-05-06T21:53:29.937" LastActivityDate="2011-05-11T00:39:14.560" Title="Low-temperature hobby rocketry" Tags="&lt;rocket&gt;" AnswerCount="3" CommentCount="8" />
  <row Id="9538" PostTypeId="2" ParentId="9524" CreationDate="2011-05-06T21:36:50.557" Score="0" ViewCount="" Body="&lt;p&gt;Are you asking about never getting a formal education and never getting an academic job but still doing physics? For example you have a rich aunt, you don't have to work so you can write papers from the library in your mansion, where you learnt physics? I would say that would be hard because you need the communication with other physicists. So if you go regularly to the near by university to attend seminars and discuss and collaborate then why not.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2011-05-06T21:36:50.557" CommentCount="3" />
  <row Id="9539" PostTypeId="2" ParentId="8390" CreationDate="2011-05-06T21:45:29.570" Score="0" ViewCount="" Body="&lt;p&gt;I found a &lt;a href=&quot;http://www.greenliter.com/gas-savings/the-wave-disk-generator-a-new-fuel-efficient-engine/&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; that explains it nicely.&lt;/p&gt;&#xA;" OwnerUserId="2843" LastActivityDate="2011-05-06T21:45:29.570" />
  <row Id="9540" PostTypeId="2" ParentId="9524" CreationDate="2011-05-06T21:53:04.557" Score="4" ViewCount="" Body="&lt;p&gt;Although I haven't researched it, you can learn some physics and try to get into computational physics. There is &lt;em&gt;some&lt;/em&gt; industry for it out there (e.g the materials sector, though I would doubt they hire anything but Phds). However, if you're a really great programmer and bring in your ideas and experience creating software, with the additional knowledge of being able to conduct simulations of physical systems and conduct solid quantitative analysis, then why not. There is also a market for physics specific software. For example, you can learn electrodynamics and create a (hopefully open source or atleast free ;)) counterpart to &lt;a href=&quot;http://simion.com/&quot; rel=&quot;nofollow&quot;&gt;Simion&lt;/a&gt;. If developing independent software alternative is too much, you can contribute by creating physics modules, writing patches etc. to products like &lt;a href=&quot;http://www.sagemath.org/&quot; rel=&quot;nofollow&quot;&gt;Sage&lt;/a&gt;. There are more possible places where you can develop, off the top of my mind the &lt;a href=&quot;http://root.cern.ch/drupal/content/root-development-team&quot; rel=&quot;nofollow&quot;&gt;ROOT&lt;/a&gt; develeopment team at CERN has two non-physicists working for them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best strategy I would recommend is to start learning basic physics, and simultaneously research what is going on at the interface of physics and computation. One guide would be to look at the conferences and seminars that are held on the subject and find out what currently engages physicists. For example, have a look at: &lt;a href=&quot;http://www.pc2010.uac.pt/index.html&quot; rel=&quot;nofollow&quot;&gt;Physics and Computation 2010&lt;/a&gt; and &lt;a href=&quot;http://www.ccp2010.no/programme.php&quot; rel=&quot;nofollow&quot;&gt;Conference of Computational Physics&lt;/a&gt;. Look at the titles of talks and submitted papers, the workshops, tutorials etc. Find more on the web (keyword search on the arxiv) etc, and you will get a rough idea of the status of the field. Start working on something, make some contributions to open source initiatives or get your own results, basically get some credentials so that employers might pay attention to you and you might find yourself working along with Phds.&lt;/p&gt;&#xA;" OwnerUserId="1671" LastEditorUserId="1671" LastEditorDisplayName="" LastEditDate="2011-05-06T22:02:57.197" LastActivityDate="2011-05-06T22:02:57.197" CommentCount="4" />
  <row Id="9541" PostTypeId="1" AcceptedAnswerId="9831" CreationDate="2011-05-06T21:58:54.867" Score="2" ViewCount="112" Body="&lt;p&gt;I'm a non-engineer interested in the recent GP-B mission results: &lt;a href=&quot;http://www.engadget.com/2011/05/06/nasa-concludes-gravity-probe-b-space-time-experiment-proves-e/#disqus_thread&quot; rel=&quot;nofollow&quot;&gt;http://www.engadget.com/2011/05/06/nasa-concludes-gravity-probe-b-space-time-experiment-proves-e/#disqus_thread&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it correct that this means that both the rotation of the earth and the size affect time?  Would a result be that a person living on an earth-sized planet spinning faster make that person's perception of time relative to an earth-bound person different, and if so, in what way?  Also, how massive must an object be and and how fast must an object rotate to have noticable effect?  &lt;/p&gt;&#xA;" OwnerUserId="3439" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-13T16:57:01.707" LastActivityDate="2011-06-12T22:57:33.390" Title="Does the recent Gravity Probe - B mission mean both the mass of an object and the spin of an object affect time?" Tags="&lt;general-relativity&gt;&lt;experimental-physics&gt;&lt;spacetime&gt;" AnswerCount="1" />
  <row Id="9542" PostTypeId="2" ParentId="9537" CreationDate="2011-05-06T22:04:23.023" Score="1" ViewCount="" Body="&lt;p&gt;I found &lt;a href=&quot;http://www.youtube.com/watch?v=CAAhQ6xcAnw&quot; rel=&quot;nofollow&quot;&gt;this water rocket video&lt;/a&gt;, which appears to attain sufficient altitude. The group that makes them has a nice website and &lt;a href=&quot;http://www.aircommandrockets.com/&quot; rel=&quot;nofollow&quot;&gt;tutorial&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps launching black powder based rockets from a boat might be an alternative solution. &lt;/p&gt;&#xA;" OwnerUserId="2843" LastEditorUserId="2843" LastEditorDisplayName="" LastEditDate="2011-05-07T01:53:00.613" LastActivityDate="2011-05-07T01:53:00.613" CommentCount="2" />
  <row Id="9543" PostTypeId="1" AcceptedAnswerId="9580" CreationDate="2011-05-06T22:22:13.140" Score="3" ViewCount="276" Body="&lt;p&gt;Does the nature of the assortment of virtual particles depend upon the warping of spacetime in a direct manner ?&lt;/p&gt;&#xA;" OwnerUserId="2942" LastEditorUserId="2942" LastEditorDisplayName="" LastEditDate="2011-05-06T22:27:15.570" LastActivityDate="2011-05-09T03:42:33.157" Title="Effect of spacetime curvature on the sea of virtual particles" Tags="&lt;vacuum&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9544" PostTypeId="1" CreationDate="2011-05-07T00:48:22.237" Score="3" ViewCount="147" Body="&lt;p&gt;Can someone suggest good books/textbooks/treatises/etc on elasticity? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3064" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-08-05T20:21:07.477" LastActivityDate="2011-08-05T20:21:07.477" CommunityOwnedDate="2011-05-07T03:11:29.283" Title="Good books on elasticity" Tags="&lt;education&gt;&lt;books&gt;&lt;reference-request&gt;&lt;big-list&gt;&lt;elasticity&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9545" PostTypeId="2" ParentId="2824" CreationDate="2011-05-07T00:59:07.020" Score="0" ViewCount="" Body="&lt;p&gt;My boyfriend noticed the same thing when I was on his laptop. The sensation stopped when I put it down. We thought it was the laptop too but recently found it happens when I have the electric blanket on, or when using our phones on the charger. I'm just wondering if it's dangerous - health wise and to the safety of our house &lt;/p&gt;&#xA;" OwnerUserId="3440" LastActivityDate="2011-05-07T00:59:07.020" />
  <row Id="9546" PostTypeId="1" AcceptedAnswerId="9550" CreationDate="2011-05-07T01:05:50.637" Score="2" ViewCount="220" Body="&lt;p&gt;I'm currently reading the Seiberg-Witten paper on $N=2$ supersymmetric Yang Mills pure gauge theory (i.e. no hypermultiplets). I have the following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one understand that the metric on the moduli space of the full quantum theory is the same as the metric obtained from the Kahler potential for the scalar field (or in general the $N=1$ chiral superfield) in the low-energy effective theory? On the face of it, the two things seem quite different - while the moduli space is the space of all gauge-inequivalent vacua in the full theory, the Kahler metric is derived from the Kahler potential in the low energy theory. &lt;/p&gt;&#xA;" OwnerUserId="3441" LastActivityDate="2011-05-07T15:44:20.273" Title="Seiberg Witten theory" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;supersymmetry&gt;" AnswerCount="1" />
  <row Id="9547" PostTypeId="2" ParentId="9516" CreationDate="2011-05-07T01:55:29.097" Score="-1" ViewCount="" Body="&lt;p&gt;It is not possible in any trivial manner to consider the interaction between two black holes.  This is not entirely tractable.  However, we can examine what a single charge experiences.  The Reissner-Nordstrom (RN) metric is&#xA;$$&#xD;&#xA;ds^2~=~-\alpha dt^2~+~\alpha^{-1}dr^2~+~r^2d\Omega^2,~\alpha~=~{{(r~-~r_+)(r~-~r_-)}\over{r^2}},&#xD;&#xA;$$&#xA;for $r_\pm~\simeq~m~\pm~\sqrt{m^2~-~q^2}$ are the horizons at coordinate singularities. The vector potentials $A_t~=~\Phi_tdt$ $={q_e\over r}dt$, $A_\phi~=~\Phi_\phi d\phi$ $=~q_mcos\theta d\phi$ correspond to the field strength two-form is &#xA;$$&#xD;&#xA;F~=~{q_e\over{r^2}}dt\wedge dr~+~q_m sin\theta d\theta\wedge d\phi,&#xD;&#xA;$$&#xA;for the electric and magnetic charges $q_e,~q_m$ and $q~=~\sqrt{q_e^2~+~q_m^2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classical connection coefficients of interest are for $g_{tt}~=~\alpha$ &#xA;$$&#xD;&#xA;{\Gamma^t}_{rt}~=~{1\over 2}\alpha^{-1}\partial_r\alpha,~ {\Gamma^r}_{tt}~=~-{1\over 2}\alpha^{-1}\partial_r\alpha,~{\Gamma^r}_{rt}~=~{1\over 2}\alpha\partial_r\alpha&#xD;&#xA;$$&#xA;$$&#xD;&#xA;{\Gamma^\phi}_{\theta\phi}~=~cot\theta,~{\Gamma^\theta}_{\phi\phi}~=~-sin\theta cos\theta,&#xD;&#xA;$$&#xA;which give the electromagnetic field components from $F~=~dA~+~\Gamma\wedge A$ as&#xA;$$&#xD;&#xA;F_{rt}~=~{q_e\over{r^2}}~-~{q_e\over {2r}}\alpha^{-1}{{\partial\alpha}\over{\partial t}},~F_{\phi\theta}~=~~q_m sin\theta.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extremal black hole has $q_e~=~m$, or $r_+~=~r_-$.  In working the geodesic equation let me consider the far field situation.  In this way I have $U^t~\simeq~1$ and $r~&amp;gt;&amp;gt;~m$.  The geodesic equation then gives the gravitational force&#xA;$$&#xD;&#xA;\frac{d^2r}{dt^2}~\simeq~{\Gamma^r}_{tt}~\simeq~-\frac{m}{r^2}&#xD;&#xA;$$&#xA;The force due to the charge $q_e~=~m$ is then equal to this if $\partial\alpha/\partial t~=~0$.  As a result the charge will not experience any net force.  So a single charge far removed from the extremal black hole experiences no acceleration in a Newtonian sense.  Since we are considering this force far from the black hole it is not hard to see that an identical black hole can be treated as a mass $m$ a charge distribution that sums to $q_e$.  There will then experience no net attraction between the two black holes.  So within this approximation where $r~&amp;gt;&amp;gt;~m$ it is clear that the two extermal black holes can exist without any net attraction.  Departures of course set in with close proximity, and to study the interaction of two extremal black holes in such a configuration would likely requires numerical methods.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-05-07T01:55:29.097" />
  <row Id="9548" PostTypeId="2" ParentId="9544" CreationDate="2011-05-07T04:40:15.523" Score="3" ViewCount="" Body="&lt;p&gt;For me the best is &quot;Landau, Lifschitz: Vol. 7&quot;...&lt;/p&gt;&#xA;" OwnerUserId="2962" LastActivityDate="2011-05-07T04:40:15.523" CommunityOwnedDate="2011-05-07T04:40:15.523" />
  <row Id="9549" PostTypeId="2" ParentId="9544" CreationDate="2011-05-07T05:04:51.887" Score="1" ViewCount="" Body="&lt;p&gt;You could try chapters 10 and 11 of Thorne and Blandford, online here: &lt;a href=&quot;http://www.pma.caltech.edu/Courses/ph136/yr2008/&quot; rel=&quot;nofollow&quot;&gt;http://www.pma.caltech.edu/Courses/ph136/yr2008/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-05-07T05:04:51.887" CommunityOwnedDate="2011-05-07T05:04:51.887" />
  <row Id="9550" PostTypeId="2" ParentId="9546" CreationDate="2011-05-07T06:47:25.230" Score="2" ViewCount="" Body="&lt;p&gt;Dear Onkar, they're the same thing. The (massless or at least light) scalar fields are what &lt;em&gt;parameterizes&lt;/em&gt; the moduli spaces - in any theory - and the metric on the moduli space (which is a mathematical concept that doesn't &quot;a priori&quot; exist in physics) is &lt;em&gt;defined&lt;/em&gt; from the (ultimate low energy) kinetic terms of these scalar fields. In a supersymmetric theory, these kinetic terms &#xA;$$\frac{1}{2}g_{ij}(\phi_a)  \partial_\mu \phi^i \partial^\mu \phi^j$$&#xA;are determined from the Kähler potential, $g_{i\bar j}\sim\partial_i \partial_{\bar j} K$ because of the basic supersymmetric calculus. There is a lot of nontrivial maths here but the particular statement you're quoting is a tautology.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-07T15:44:20.273" LastActivityDate="2011-05-07T15:44:20.273" CommentCount="2" />
  <row Id="9551" PostTypeId="1" CreationDate="2011-05-07T08:14:49.100" Score="2" ViewCount="228" Body="&lt;p&gt;I'm currently learning some basic functional analysis. Yesterday I arrived at the &#xA;&quot;spectral theorem of self-adjoint operators&quot;. I've heard that this theorem has lots of applications in Quantum Mechanics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let me first state the formulation of the theorem that I'm using:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let H be a Hilbert space.&#xA;There's a 1-to-1-correspondence between self-adjoint operators A on H &#xA;and spectral measures $P^{A}$ given by &#xA;$A=\int_{R} \lambda dP^{A}$. ($\lambda$ denotes a constant, R denotes the real numbers)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A corollary is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let g:R-&gt;R be a function. (Again: R denotes the set of real numbers)&#xA;Then: $g(A):=\int_{R} g(\lambda) dP^{g(A)}$  $P^{g(A)}(\Delta) = P^{A}(g^{-1}(\Delta))$&#xA;where $\Delta$ denotes a set in the $\sigma$-algebra of R. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay. Now this is the theorem. &#xA;First I don't really the application of the corrolary in Quantum mechanics.&#xA;I've heard that suppose you're given an operator A this means that it's easy for you&#xA;to define operators like exp(A), especially on infinite dimensional Hilbert spaces. &#xA;This indeed could be useful in quantum mechanics. Especially when thinking about the &quot;time-evlution operator&quot; of a system. However then I say: Why do you make things so complicated? &#xA;Suppose you want to calculte exp(A). Why don't you define exp(A):=1+A+1/2 A^2 + ... &#xA;and require convergence with respect to the operator norm. &#xA;An example: Consider the vectorspace spanned by the monomials 1,x,x^2,... and let A=d/dx&#xA;Then you can perfectly define exp(d/dx):=1+ d/dx + 1/2 d^2/dx^2 + ...  and require convergence with respect to the operator norm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to that I've heard that the spectral theorem gives a full description of all &#xA;self-adjoint operators. Now why is that the case? I mean okay..there's a one to one correspondence between self-adjoint operators and spectral measures...but why does this give me any information about &quot;the inner structure of the operator&quot;? (And why is there this $\lambda$ in the integral? Looks somehow like an eigenvalue of A? But I'm just guessing) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd me more than happy, if you could provide me with some intuition and ideas &#xA;of how the theorem can be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="3445" LastActivityDate="2011-05-09T08:26:50.147" Title="Applications of the Spectral Theorem to Quantum Mechanics" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" />
  <row Id="9552" PostTypeId="2" ParentId="8149" CreationDate="2011-05-07T10:14:05.127" Score="1" ViewCount="" Body="&lt;p&gt;Here is an answer in the first case (unbounded domain). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As already noted, the difference between the positions of the two particles performs a Brownian motion starting from point $(d,0,0)$ with diffusion coefficient $D=D_1+D_2$ and one is interested in the time $\tau$ of the first hitting of the ball $B(r)$ around zero with radius $r=r_1+r_2$ when $d&amp;gt;r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It happens that in three or more dimensions, the Brownian motion is &lt;strong&gt;transient&lt;/strong&gt;. This means that the particle goes to infinity almost surely, in the sense that for any radius $R$, there will exist an almost surely finite time $L_R$ such that after time $L_R$, the particle is not in $B(R)$. The same holds if one replaces $B(R)$ by any bounded domain: after a random time, finite with full probability, the particle wanders out of the domain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another consequence is that, starting from a point not in $B(r)$, for example from the point $(d,0,0)$, the particle has a nonzero probability to never hit $B(r)$. Thus, it happens with positive probability that the two particles the OP is considering never collide in $\mathbb{R}^3$ (and of course this probability depends on $d$, $r_1$ and $r_2$, but for any $d&amp;gt;r_1+r_2$ it is positive). In particular:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The collision time is infinite with nonzero probability. All its (positive) moments are infinite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two more remarks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First, in one dimension&lt;/strong&gt; the collision time $\tau$ would be almost surely finite, with the well known density&#xA;$$&#xD;&#xA;P(\tau\in\mathrm{d}t)=\frac{a}{\sqrt{2\pi}}\mathrm{e}^{-a^2/2t}\frac{\mathrm{d}t}{t^{3/2}},\qquad&#xD;&#xA;a=d-r_1-r_2.&#xD;&#xA;$$&#xA;Note that, since the density above is equivalent to a multiple of $t^{-3/2}$ when $t\to+\infty$, the mean of $\tau$ is infinite. More precisely, for positive $\gamma$, $\tau^\gamma$ is integrable if $\gamma&amp;lt;1/2$ but not if $\gamma\ge1/2$.&#xA;Beware that we followed the mathematicians' convention for Brownian motion, that is, the formula above applies to a one dimensional Brownian motion $(B_t)$ with transition probabilities&#xA;$$&#xD;&#xA;P(B_t\in\mathrm{d}x\vert B_0=y)=\frac1{\sqrt{2\pi t}}\mathrm{e}^{-(x-y)^2/2t}\mathrm{d}x.&#xD;&#xA;$$&#xA;By scaling, it should be straightforward to deduce the distribution of $\tau$ for a one dimensional Brownian motion with diffusion constant $D$ since $\tau$ for a diffusion constant $D$ is distributed like $\tau^{(1)}/D$ where $\tau^{(1)}$ corresponds to the first hitting time for a diffusion constant $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second, in case 2 (bounded domain),&lt;/strong&gt; one can expect the collision time to be almost surely finite and integrable (and even exponentially integrable) but to make it possible to answer the question, one should explain what happens to the particles when one of them hits the boundary of the domain before their collision time.&lt;/p&gt;&#xA;" OwnerUserId="3446" LastEditorUserId="3446" LastEditorDisplayName="" LastEditDate="2011-05-07T18:01:48.443" LastActivityDate="2011-05-07T18:01:48.443" />
  <row Id="9554" PostTypeId="2" ParentId="9219" CreationDate="2011-05-07T10:42:34.900" Score="0" ViewCount="" Body="&lt;p&gt;The three-dimensional spin group is the compact group $SU(2)$, which has only finite-dimensional representations. Hence, the spin is always a non-negative integer or half-integer.&lt;/p&gt;&#xA;" OwnerUserId="3131" LastActivityDate="2011-05-07T10:42:34.900" CommentCount="5" />
  <row Id="9555" PostTypeId="1" AcceptedAnswerId="9558" CreationDate="2011-05-07T14:19:52.637" Score="1" ViewCount="140" Body="&lt;p&gt;I have got some mathematical difficulties in the following exercise :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the potential of the polarized sphere along the &lt;em&gt;z&lt;/em&gt;-axis. There are no free charges.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For this, we need to solve Laplace's equation, by using the method of separation of variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi (r, \theta, \phi) = R(r) \Theta(\theta) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We obtain a partial differential equation on $r$ and $\theta$ :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{R} \frac{\partial}{\partial r} r^2 \frac{\partial}{\partial r} R = l(l+1) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{\Theta} \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \sin \theta \frac{\partial}{\partial \theta} \Theta = - l(l+1) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is : &lt;strong&gt;&quot;Why is $l$ to be a positive integer?&quot;&lt;/strong&gt; The solution I got in some course takes it for granted, but I do not know why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isaac&lt;/p&gt;&#xA;" OwnerUserId="2499" LastActivityDate="2011-05-07T16:07:27.827" Title="Laplace's equation" Tags="&lt;electromagnetism&gt;&lt;polarization&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="9556" PostTypeId="2" ParentId="9555" CreationDate="2011-05-07T15:04:00.480" Score="0" ViewCount="" Body="&lt;p&gt;It need not be positive but can also be zero. So let's suppose you're asking why it has to be non-negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to this question is that Laplace's operator $\Delta$ splits into radial and angular parts (these are precisely the operators that can be found in your equations, if you discard the $\phi$ dependence). The angular part is then simply $L^2$ where $\mathbf L$ is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Angular_momentum_operator&quot; rel=&quot;nofollow&quot;&gt;angular momentum&lt;/a&gt; operator. Now because $L^2$ is a non-negative operator (convince yourself that a square of an operator must be non-negative) it must have only non-negative eigenvalues.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-05-07T15:04:00.480" />
  <row Id="9557" PostTypeId="1" AcceptedAnswerId="9559" CreationDate="2011-05-07T15:59:34.923" Score="4" ViewCount="207" Body="&lt;p&gt;My book on quantum mechanics suggests a derivation of the wave equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(\Delta - \frac{1}{c^2} \frac{\delta^2}{\delta t^2} \right) \psi(\bar{r},t) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from the photon energy-impulse relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E^2 = c^2p^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the substitutions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bar{p} \to -i\hbar \nabla$ and $E \to i \hbar \frac{\delta}{\delta t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose the $\hbar$s are there for the non-photon generalisation, but is there a reason for the minus sign in the $p$ substitution? Removing it would yield the same result.&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2011-05-08T15:43:18.910" Title="Why is there a minus sign in this wave equation derivation?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" />
  <row Id="9558" PostTypeId="2" ParentId="9555" CreationDate="2011-05-07T16:01:06.553" Score="3" ViewCount="" Body="&lt;p&gt;Dear Isaac, with all my respect, I don't think that Marek is answering your question. You're primarily asking why it's integer, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, $l$ can be both positive or negative integer. If $l=-10$, for example, $l(l+1)=90$ is clearly the same number as $l(l+1)$ for $l=9$. If you replace $l$ by $-1-l$, you get the same value of $l(l+1)$. The value $l=-1$ is equivalent to $l=0$, $l=-2$ is equivalent to $l=+1$, and so on. These eigenvalues of $l(l+1)$ are equal so the corresponding solutions shouldn't be double-counted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, why $l$ is an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Confusingly, it's because how the Legendre polynomials work:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Legendre_polynomials&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Legendre_polynomials&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It just happens that for integer values of $l$, the function $\Theta$ is non-singular for $\sin\theta=\pm 1$. Surprisingly, the same condition - integrality of $l$ - guarantees that the function $\Theta$ reduces to a polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, this doesn't explain anything because I would have to use some maths to prove the propositions in the previous paragraph. There is a very transparent, more physical explanation why it's so. The operator acting on your $\Theta$ function - on the left-hand side - is $-\vec L\cdot \vec L$, usually denoted as $L^2$, and this operator has eigenvalues $l(l+1)$ for integer values of $l$. Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's because one may write &#xA;$$ L^2 = L_- L_+ + L_z^2 + L_z $$&#xA;where $L_{\pm} = L_x\pm i L_y$. The last term, $L_z$, has to be added because $L_+ L_-$ was only written in one order and $L_x,L_y$ (or, equivalently, $L_+,L_-$) don't commute with each other - their commutator is $i L_z$ (in $\hbar=1$ units). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $L\pm$ commute with $L^2$, so the state $L^+ |\psi\rangle$ has the same eigenvalue of $L^2$ as $|\psi\rangle$ itself. However, the eigenvalue of $L_z$ of the former state - with the action of $L_+$ - is greater by one than the $L_z$ eigenvalue of $|\psi\rangle$ itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Act with $L^+$ as many times as you need to get zero. You eventually have to get zero because the eigenvalue of $L_z^2$ can't be greater than the eigenvalue of $L^2$. When you get to the state with the maximum eigenvalue of $L_z$ which is non-vanishing, you will have $L_+|\psi\rangle = 0$, so the first term in the formula for $L^2$ above will vanish. The remaining terms will give you $l(l+1)$ where $l$ is the maximum value of $L_z$ that you can get from the original state by actions of $L_+$. That's called the orbital angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a similar way, I can act with $L_-$ to get to the minimum value of $L_z$. This minimum value of $L_z$ has to be the minus the maximum value, by rotational symmetry or by the $l\to -1-l$ symmetry above. So the difference between the maximum and minimum values of $L_z$ has to be integer - because $L_\pm$ was changing $L_z$ by steps equal to one. It follows that $l$ has to be either half-integer or integer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, only integer values are possible for orbital angular momentum because $L_z$ has to be integer because the $\phi$-dependent wave function $\exp(im\phi)$ is only single-valued on the circle for integer values of $m$. That's why for orbital wave functions all eigenvalues $L_z$ are integer, and because $l$ is defined as the maximum $L_z$ in the &quot;multiplet&quot; of states that can be obtained from each other by the action of $L_\pm$, it follows that $l$ is integer, too. The formula for $L^2$ above guarantees that its eigenvalue is $l(l+1)$ for this integer $l$.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-07T16:07:27.827" LastActivityDate="2011-05-07T16:07:27.827" CommentCount="1" />
  <row Id="9559" PostTypeId="2" ParentId="9557" CreationDate="2011-05-07T16:38:39.013" Score="8" ViewCount="" Body="&lt;p&gt;The relative sign is &lt;em&gt;not&lt;/em&gt; just a convention. Once you decide that $E$ is represented by $i\hbar \partial/\partial t$, there must be a minus sign in the formula for $p$, namely $p=-i\hbar \partial / \partial x$. Or vice versa. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, there has to be $i$ or $-i$ in all the formulae because $\partial/\partial x$ is an anti-Hermitian operator (because of the minus sign in the integration by parts) and we need Hermitian operators (which has real, measured eigenvalues) for the energy, momentum, and others. What about the signs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only sign convention that was chosen in the early days of quantum mechanics was one for the energy; indeed, one could have replaced $i$ by $-i$ in that equation because $i$ and $-i$ play the same algebraic role: exchanging $i$ and $-i$ is an &quot;outer automorphism&quot; of complex numbers. But once this sign is fixed, all other signs are fixed, too. That includes minus signs in momentum, angular momentum, gauge transformations, Schrödinger's picture, Heisenberg's picture, Feynman's path integral, and any other formula of quantum mechanics. There is only one way to define quantum mechanics given a classical limit (with its sign conventions) we need to get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relative minus sign in $p,E$ may become invisible if you only act with second derivatives - squared momentum, squared energy - but it is visible if you act with first powers of the operators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;De Broglie's wave - or a wave associated with a particle - is proportional to &#xA;$$ \exp \left[ \frac{i}{\hbar} (\vec p\cdot \vec x-Et)\right] $$&#xA;Note that if you differentiate with respect to $x$ and $t$, and multiply the result by $i\hbar$, you get $-p$ and $E$, respectively. (Omit the vector signs if you want just one-dimensional space with one $x$ and one $p$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relative sign between $\vec p\cdot \vec x$ and $Et$ in de Broglie's formula above is physically necessary because only $Et - \vec p \cdot \vec x$ is the correct Lorentzian inner product of the vectors $(E,\vec p)$ and $(t,\vec x)$: the relative minus sign comes from the opposite signs of space and time in the signature of spacetime. Note that the doubly relative sign in $(E,\vec p)$ and  $(t,\vec x)$ can't be flipped because in relativity, $\vec pc^2/E$ is the velocity $\vec v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument above is for a relativistic interpretation of $E$. However, the non-relativistic kinetic energy is defined just as a shifted relativistic energy,&#xA;$$ E_{\rm nonrel} = E_{\rm rel} - mc^2 $$&#xA;so the coefficient (and sign) in front of $E$ remains unchanged and $\vec p$ is totally unchanged. One may also design related arguments analyzing where the de Broglie wave is moving. For it to be moving in the right direction, there has to be a minus sign in $Et-px$. It's related to the fact that the shape of objects moving by velocity $v$ only depends on $x-vt$ because $x=vt$ (without a minus sign, which becomes a minus sign if you move both terms to the same side) is the equation for their center of mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hypersurfaces of constant phase of the de Broglie wave are orthogonal to the world lines of the particles which dictates the sign. After all, the waves' maxima and minima should be moving in the same direction as the particles themselves.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-07T16:49:55.853" LastActivityDate="2011-05-07T16:49:55.853" />
  <row Id="9560" PostTypeId="1" AcceptedAnswerId="9671" CreationDate="2011-05-07T16:46:56.003" Score="1" ViewCount="265" Body="&lt;p&gt;This is not mainstream physics. There are some people who (without having a stated theory that I know of) insist that Gravity, Electricity, and Magnetism are related. Some point to symmetry in Maxwell's Equations as a potential indication of this (intuitive) connection. Is it possible that there are &quot;Maxwell-like&quot; equations that relate Electricity and Magnetism directly to Gravity? All forms of electromagnetic energy do interact with gravity. Are there any physicists who are working on this sort of theory?&lt;/p&gt;&#xA;" OwnerUserId="2843" LastEditorUserId="1782" LastEditorDisplayName="" LastEditDate="2011-05-09T18:50:04.420" LastActivityDate="2011-07-01T02:05:32.150" Title="Theories that Relate Gravity, Electricity, and Magnetism" Tags="&lt;gravity&gt;&lt;electromagnetism&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="9561" PostTypeId="2" ParentId="7700" CreationDate="2011-05-07T18:26:59.077" Score="1" ViewCount="" Body="&lt;p&gt;I'm from the old school of physics (1970).  This is what I remember:  ALL laws of nature must be so constituted that it is transformed into a law of exactly the same form when, instead of the space-time variables x, y, z, t of the original co-ordinate system K, we introduce new space-time variables x', y', z', t' of a co-ordinate system K'.  OR - another way of saying this is: General laws of nature are co-variant with respect to Lorentz transformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is called the Heuristic value of Relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me Co-variant means that when a change is measured, the Formula to the natural law stays the same, but time and position of the object changes.  Therefore, the natural law constituted works together with Lorentz Transformations.&lt;/p&gt;&#xA;" OwnerUserId="3451" LastActivityDate="2011-05-07T18:26:59.077" />
  <row Id="9562" PostTypeId="2" ParentId="9421" CreationDate="2011-05-07T18:35:48.977" Score="0" ViewCount="" Body="&lt;p&gt;A quite detailed article (in german though) discussing this &quot;effect&quot; can be found here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.wissenschaft-online.de/artikel/1047442&quot; rel=&quot;nofollow&quot;&gt;Radioaktivität: Wer hat an der Uhr gedreht?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At the end of the article there is a list of interesting related papers and literature.  According to this article first experiments trying to use artificially produced neutrinos to influence the decay rates of radioactive materials came out with negative results. So if the sun should influence the decay rate, it doesn't seem to be via neutrinos, but must be by some mysterious unknown force or particle. However most physicist seem to believe more in a systematic error produced by the used measurement equipment as explanation for this &quot;effect&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1648" LastActivityDate="2011-05-07T18:35:48.977" />
  <row Id="9563" PostTypeId="2" ParentId="9219" CreationDate="2011-05-07T18:54:26.260" Score="2" ViewCount="" Body="&lt;p&gt;Compact group $SU(2)$ really has only finite-dimensional unitary irreducible representations, but&#xA;formally it is not enough to close the question, because there are unitary irreducible infinite-dimensional representations of spin group $SL(2,C)$ of four-dimensional relativistic Lorentz group and they were used in some models, below is a cite from: N.N. Bogolubov, A. A. Logunov, A.I. Oksak, I. Todorov, General principles of quantum field theory, Springer, 1989. Appendix I for chapter 9&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The concept of an infinite-component&#xA;  field (ICF for short) is the result of&#xA;  abandoning the &quot;technical&quot; requirement&#xA;  that the representations of the&#xA;  Lorentz group according to which the&#xA;  fields transform (say, in the Wightman&#xA;  formalism) be finite-dimensional. This&#xA;  idea turned up at the earliest stages&#xA;  of quantum field theory: in 1932,&#xA;  Majorana gave an example of an&#xA;  infinite-dimensional wave equation &#xA;  $(i \Gamma^\mu \partial_\mu – M) \psi(x) = 0$ &#xA;  without negative-energy solutions&#xA;  of non-negative square mass, that is,&#xA;  without &quot;antiparticles&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;…&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Running ahead (see&#xA;  §1.3), it should be noted, however,&#xA;  that the description of composite&#xA;  systems by means of ICF's has met with&#xA;  difficulties which, it would seem,&#xA;  require a weakening of the postulate&#xA;  of (strict) locality.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3351" LastEditorUserId="3351" LastEditorDisplayName="" LastEditDate="2011-05-08T09:29:20.403" LastActivityDate="2011-05-08T09:29:20.403" CommentCount="9" />
  <row Id="9564" PostTypeId="2" ParentId="9551" CreationDate="2011-05-07T19:56:01.153" Score="2" ViewCount="" Body="&lt;p&gt;Dear Constantin, the $A=\int_R \lambda\,dP^A$ is just a continuous version of spectral decomposition. Here $dP^A$ is a differential version of the projection operators that define the Hermitian operator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a discrete spectrum, the corresponding equation would be&#xA;$$ A = \sum_{i} \lambda_i P_{\lambda_{i}} $$&#xA;where the sum goes over the eigenvalues $\lambda_i$ and $P_{\lambda_i}$ are the projection operators on the subspace of the Hilbert space that contains the eigenvectors with the eigenvalue $\lambda_i$. Indeed, $\lambda$ is always meant to be a possible eigenvalue of the operator. And indeed, a Hermitian operator is fully determined by its spectrum and the corresponding eigenvectors (and multiplicities) for each eigenvalue, which is why the formula above is an equivalent way to rewrite a Hermitian operator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the spectrum of $A$ is continuous, the summation over $i$ has to be replaced by an integral, and the corresponding differential $d$ is added in front of $dP_{\lambda}$: it's really the differential of the projection operator on the space of eigenstates with eigenvalues in the interval $[-\infty, \lambda]$; $dP_\lambda = dP_\lambda / d\lambda \cdot d\lambda$, if you wish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's really morally the same thing as in the case of the discrete spectrum (which produces delta functions in $dP_\lambda / d\lambda$ if we adopt this terminology). Also, some of your additional proofs involving $\Delta$ are just trivial substitutions under the integral sign. One would need to know lots of details of your mathematical axioms - lots of the particular &quot;math culture&quot; you're coming from - to figure out what could exactly be difficult for a mathematician about the substitutions under the integral sign. There are no difficulties from a physicist's perspective - it's high school maths.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Spectral_theorem#Hermitian_matrices&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Spectral_theorem#Hermitian_matrices&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mathematicians may worry about boundedness and well-definedness of all these things for most of their careers but these things are totally vacuous from the viewpoint of physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a physicist finds out that the answer to a physics question requires him to calculate $f(A)$, a function of an operator - an observable - he just has to calculate it whether or not it looks hard or well-defined. In particular, the Taylor expansion for functions such as the exponential is always assumed to hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You discuss the function $g(A)$ of the operator as an example. The procedures you outline physically mean that one diagonalizes $A$ - which brings the projection operators to a simple form (only one number $1$ on the diagonal) - and then he simply applies the function $g$ to the eigenvalues. In other words,&#xA;$$ A = U D U^{-1} \quad \Rightarrow \quad g(A) = U g(D) U^{-1} $$&#xA;where $g(D)$ is simply a diagonal matrix with entries $g(D_{ii})$ on the diagonal. The formula above works because $U^{-1} U$ cancel everywhere in the middle if we write $g(A)$ e.g. as a Taylor expansion - and by generalization, we just declare the formula above to be right even if the Taylor expansion is not appropriate for a mathematician.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Taylor expansion for the exponential is always OK from a physicist's viewpoint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these objects - Hilbert spaces, operators, their functions (especially exponentials), spectra, eigenvalues - and all these operations - exponentiation, search for projection operators etc. - are important in physics, indeed. And that's what mathematicians often declare when they want their students to listen. But it's equally true that all the points that mathematicians actually focus on most of their lives are totally uninteresting from a scientific viewpoint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why the comments that the &quot;material is important in physics&quot; is morally wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematicians shouldn't try to support the attractiveness of their teachings by physical applications - especially because their real goal (and the real goal of pure maths) is to make them as independent of natural science as possible. One can't have it both ways. Doing physics or science means that one is allowed to &quot;prove&quot; all these things - such as $g(A)=g(A)$ which is really the content of the &quot;difficult&quot; proof you sketched) - much more elegantly and arguably naively than in maths. One is only worried about the lack of rigor if he can actually find a contradiction with the experiment or other calculations. If it doesn't exist, the science is just perfectly OK.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, mathematicians are usually looking for problems even if no problems from a scientist's viewpoint exist. This implies a totally different set of priorities and it is unlikely that a student is going to be excited about both. One either prefers to measure the truth according to nitpicking based on predetermined sets of axioms - which is the mathematician's viewpoint - or one is ready to adjust his methods, axioms, and precise definitions of objects (and invent or learn totally new branches of physics) as needed to agree with the empirical data and other accurate calculations - which is the physicist's viewpoint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They're different attitudes and this is why I think that your question should have been posted on a math forum because it's not really following physicist's way of thinking and it is not really motivated by the desire to understand Nature.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-07T20:11:21.583" LastActivityDate="2011-05-07T20:11:21.583" />
  <row Id="9566" PostTypeId="2" ParentId="6464" CreationDate="2011-05-07T20:13:19.420" Score="0" ViewCount="" Body="&lt;p&gt;Let me throw my two cents in.  Space is nothing more than Distance.&lt;/p&gt;&#xA;" OwnerUserId="3451" LastActivityDate="2011-05-07T20:13:19.420" />
  <row Id="9567" PostTypeId="2" ParentId="9449" CreationDate="2011-05-07T20:33:01.027" Score="1" ViewCount="" Body="&lt;p&gt;My understanding of Graduate Level is an overlap between JD Jackson's Classical electrodynamics, Landau's Electrodynamics of continuous media and Landau's Classical Theory of fields. Unfortunately, there isnt much video material out there, which is justifiable because there is no great pedagogical need here. If you understand Griffith's level electrodynamics the Jackson's book is an advanced methods to solve sophisticated problems book. Which is best learnt by doing problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ALthough I am not greatly impressed, this is a set of video lectures that treats Landau and Jackson as textbooks. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://vubeam.pa.msu.edu/lectures/phy962/962d/electrodynamics/&quot; rel=&quot;nofollow&quot;&gt;http://vubeam.pa.msu.edu/lectures/phy962/962d/electrodynamics/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It might be worthwhile to have a look at Leonard Susskind's lecture on classical electrodynamics and classical theory of fields in the special relativity module.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.cosmolearning.com/video-lectures/electrodynamics/&quot; rel=&quot;nofollow&quot;&gt;http://www.cosmolearning.com/video-lectures/electrodynamics/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you're looking for companion notes, then these lecture slides would help you a lot more specifically with understanding the material presented in Jackson, I found it really helpful.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.gmu.edu/~joe/PHYS685/&quot; rel=&quot;nofollow&quot;&gt;http://physics.gmu.edu/~joe/PHYS685/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1671" LastEditorUserId="1671" LastEditorDisplayName="" LastEditDate="2011-05-07T20:53:34.947" LastActivityDate="2011-05-07T20:53:34.947" />
  <row Id="9568" PostTypeId="2" ParentId="9557" CreationDate="2011-05-07T22:16:30.930" Score="-1" ViewCount="" Body="&lt;p&gt;Yes, if the time dependence is written as $e^{-i\omega t}$, then the other signs get certain. It's a convention in QM. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In plasma physics they use sometimes the opposite convention for the time Fourier transform. In the opposite convention the imaginary part of $\omega$  = $\omega&amp;#39; + i\omega&amp;#39;&amp;#39;$ describes the decrement of wave fading out, if positive.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-05-08T15:43:18.910" LastActivityDate="2011-05-08T15:43:18.910" />
  <row Id="9569" PostTypeId="1" AcceptedAnswerId="13352" CreationDate="2011-05-07T23:06:44.330" Score="1" ViewCount="112" Body="&lt;p&gt;I'd like to have access to the datafile of the electrical output of the RTG (Radioisotope Thermoelectric Generator) of  the Pioneer's and Cassini missions, as can be seen in the figures 1 of the papers of &lt;a href=&quot;http://arxiv.org/PS_cache/gr-qc/pdf/0107/0107022v3.pdf&quot; rel=&quot;nofollow&quot;&gt;Anderson,2002&lt;/a&gt; and &lt;a href=&quot;http://arxiv.org/abs/0809.4248&quot; rel=&quot;nofollow&quot;&gt;Peter S. Cooper, publ 2009&lt;/a&gt;.&lt;br&gt;&#xA;I know that I can grab the aprox values  directly from the published graphs with the help of software, but accessing the original data is better.&#xA;I g-searched but I did not found them.&#xA;Can someone give advice, pls?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is specific but the answers can say more about the difficulties and solutions of accessing scientific data.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-08-10T15:43:21.527" Title="Cassini 's and Pioneers RTG power output datafile" Tags="&lt;data&gt;" AnswerCount="1" />
  <row Id="9570" PostTypeId="1" CreationDate="2011-05-07T23:28:41.600" Score="2" ViewCount="202" Body="&lt;p&gt;Does anyone know how the Breit-Rabi formula (energy dependence of the hyper-fine splitting in presence of any external magnetic field for alkali atoms when $F=I \pm 1/2$) is derived, or where to find a reference for it? It doesn't seem to be in the original paper from 1931, or in the textbooks I've looked at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The low- and high-field solutions I understand, but in which step in the derivation is $F=I \pm 1/2$ used, and how does this lead to the exact solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="2445" LastEditorUserId="2445" LastEditorDisplayName="" LastEditDate="2011-05-07T23:41:15.867" LastActivityDate="2011-07-04T17:45:25.377" Title="How is the Breit-Rabi formula derived?" Tags="&lt;quantum-mechanics&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="9571" PostTypeId="1" AcceptedAnswerId="9572" CreationDate="2011-05-07T23:53:49.357" Score="1" ViewCount="112" Body="&lt;p&gt;One of my more recent hobbies is making &quot;swords&quot; out of PVC conduit, foam, and duct tape. As all the weight is forward of the cross guards, I needed to put some amount of some substance inside the hilt to counter-balance it. Curious as to how much counter-weight I needed, I decided to see if I could develop a continuous version of the typical moment formula (moment = force * distance, or $M = F*D$) as there was no way the counter-weight would be close enough to a point mass to use the usual formula. Here's how I set about it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m$ = mass&lt;br&gt;&#xA;$l$ = length&lt;br&gt;&#xA;$d$ = fraction of whole&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I have a rod where $m = 80 kg$ and $l = 16 ft$ and cut it into 8 pieces, $d = .125$ and the mass of each piece will be $m*d = 10kg$ and the length of each piece will be $l*d=2ft$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, treating each piece as a point mass centered at the point closest to the pivot, adding up the moments gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum F*D$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\displaystyle \lim_{d \to 0} \; \sum_{k = 0}^{\frac{1}{d}} mld^2k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\displaystyle \lim_{d \to 0} \;\; mld^2\sum_{k = 0}^{\frac{1}{d}}k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\displaystyle \lim_{d \to 0} \;\; mld^2 \left( \frac{ \left( \frac{1}{d} \right) \left( \frac{1}{d}+1 \right) }{2} \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\displaystyle \lim_{d \to 0} \; \frac{ml}{2} + \frac{mld}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\displaystyle \frac{ml}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I've noticed with this formula is that it means that the further half of an object has the exact same moment as the nearer half (a rod twice as long has twice the moment, not four times as it should be). This is clearly wrong, but I don't know where I made an invalid assumption. Can someone else see it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, I had a brain flash while writing this and realized the simplest (and correct) way to derive it is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \int_{a}^{b} (F*D)\;dD = \frac{F}{2}(b^2-a^2)$&lt;/p&gt;&#xA;" OwnerUserId="3454" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-11T20:02:48.703" LastActivityDate="2011-05-11T20:02:48.703" Title="How should I REALLY have done this continuous moment formula derivation?" Tags="&lt;gravity&gt;&lt;classical-mechanics&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="9572" PostTypeId="2" ParentId="9571" CreationDate="2011-05-08T00:58:00.370" Score="2" ViewCount="" Body="&lt;p&gt;I'm not sure I understand what you're really trying to calculate, or what you're doing to try to calculate it. What are $F$ and $D$, for instance? And what is $\Delta x$? That variable is normally used for an increment length, but you seem to be using it as an index that labels the various pieces of the rod.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I can gather, your example calculation seems to be calculating the first moment of mass of the rod around one end. The first mass moment in one dimension can be defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_x(c) = \int (x - c)\ \mathrm{d}m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x$ is a coordinate that measures along the length of the rod, and $c$ is the position of the end in this coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use this formula for the half of the rod that is closest to the chosen end, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_x(0) = \int_0^{l/2} x\frac{m}{l}\mathrm{d}x = \frac{ml}{8}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and for the other half,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_x(0) = \int_{l/2}^{l} x\frac{m}{l}\mathrm{d}x = \frac{3ml}{8}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the moment of the further half is indeed more than that of the closer half. In your calculation, I don't see any place where you distinguish between the two halves of the rod; all you're doing is calculating the moment of the whole thing. I'm not sure how you're concluding that your formula implies that both halves have the same moment. It doesn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're looking into minimizing the resistance of your swords to swinging, what you should be looking at is the &lt;em&gt;second&lt;/em&gt; moment of mass, also known as the moment of inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_{xx}(c) = \int (x - c)^2\ \mathrm{d}m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the quantity that measures the resistance of an object to rotation around the point $c$. So to make the sword as easy to swing as possible, you want the moment of inertia around the hilt to be as small as possible. (At least, I'm guessing so, but I don't know anything about wielding a sword specifically)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out (I'll omit the mathematical detail, unless you're interested) that the way to do that involves distributing the weight so that the center of mass is inside the hilt of the sword. To put it another way, the moment of inertia around a point $c$ for a given mass and length is smallest when the center of mass is at the point $c$. Mathematically, the way to figure out the center of mass is to divide the first moment of mass by the total mass,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_\text{CM} = \frac{\int x\ \mathrm{d}m}{\int\mathrm{d}m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, in practice, you wouldn't bother with all this mass. Just move the counterweight such that the sword is balanced around the point where you're going to hold it.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-05-08T00:58:00.370" CommentCount="3" />
  <row Id="9574" PostTypeId="1" AcceptedAnswerId="9602" CreationDate="2011-05-08T04:02:00.687" Score="2" ViewCount="315" Body="&lt;p&gt;I have a quick question that just came up in my research and I could not find an answer anywhere so I thought I'd try here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one of the definitions of the Dirac Delta is the limit of the Lorentzian function's with going to zero. See here &lt;a href=&quot;http://hitoshi.berkeley.edu/221a/delta.pdf&quot; rel=&quot;nofollow&quot;&gt;http://hitoshi.berkeley.edu/221a/delta.pdf&lt;/a&gt; for the expression on the first page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, can I define the Dirac Delta just as well with this&#xA;$\delta(t) = \lim_{\epsilon\rightarrow 0} \frac{1}{\pi}\frac{\epsilon^2}{\epsilon^2+t^2}$ where I have included an extra $\epsilon$ in the numerator. My hunch is that this is no problem since the limiting behavior looks the same to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1939" LastActivityDate="2011-05-09T05:32:09.883" Title="Limit of Lorentzian is Dirac Delta" Tags="&lt;mathematical-physics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="9575" PostTypeId="2" ParentId="9574" CreationDate="2011-05-08T04:27:27.440" Score="3" ViewCount="" Body="&lt;p&gt;It looks like a delta-function. However, because $\epsilon / (\epsilon^2+t^2)$ - you should omit one $\epsilon$ in the numerator, to get the right integral equal to one, by the way - decreases too slowly as $|t|\to\infty$, as $1/t^2$, it will only work as the Dirac delta distribution for test functions that decrease at infinity or at least increase slower than as $O(t)$. If the test function is $t^2$, for example, the integral&#xA;$$ \int_{-\infty}^\infty dt\,t^2\,\delta(t) $$&#xA;should yield 0 because $t^2=0$ for $t=0$. However, with your definition of the delta function, you will get a divergent answer because the infinite-range integral ultimately beats any $\epsilon$. For this reason, one usually wants approximations of delta functions that decrease faster at $|t|\to\infty$ than the Lorentzian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, if you include one extra $\epsilon$, you get $\epsilon\cdot \delta(t) = 0$ regardless of details about the $|t|\to\infty$ behavior.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-09T04:53:03.850" LastActivityDate="2011-05-09T04:53:03.850" CommentCount="3" />
  <row Id="9576" PostTypeId="2" ParentId="2759" CreationDate="2011-05-08T05:00:22.140" Score="2" ViewCount="" Body="&lt;p&gt;I know this question is a little old, but so far there is no correct answer posted. It happens that I used to argue this question with Civil Engineering students years ago. They had trouble with it because they learned about &quot;bending moment&quot;, which is basically the same thing as curvature only no one told them that. The curvature, or the bending moment if you will, goes to zero at the free end and increases uniformly (linearly) up to the point of constraint. You can tell that this makes sense because if the bar is over-stressed, that's where it will break. The question asks us to plot the graph of the bar, which you basically get by integrating twice. (In the limit of small displacements, the curvature of course is equal to the second derivative of displacement.) If you differentiate twice in the other direction, you get the distribution of forces. You can see that the step function at the point of constraint gives you a couple when you differentiate twice.&lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-05-08T05:00:22.140" CommentCount="2" />
  <row Id="9577" PostTypeId="1" AcceptedAnswerId="9741" CreationDate="2011-05-08T05:55:58.007" Score="10" ViewCount="284" Body="&lt;p&gt;Maxwell's equations are nearly symmetric between $E$ and $B$.  If we add magnetic monopoles, or of course if we restrict ourselves to the sourceless case, then this symmetry is exact.  This is not just a discrete symmetry of exchange.  A &quot;duality rotation&quot; of $(E, B) \rightarrow (E \cos \theta - B \sin \theta, B \cos \theta + E \sin \theta)$ preserves the form of the equations.  (The charges and currents, if any, would have to similarly mix as well.  The upshot is that if all particles have the same ratio between their electric and magnetic charges, it's indistinguishable from just having electric charges.  See e.g. Jackson 6.12.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows up quite nicely in the Hamiltonian formulation as the invariance of the free Hamiltonian density $\mathcal{H} = E^2 + B^2$.  The free Lagrangian density, however, is the Lorentz invariant scalar $\mathcal{L} = E^2 - B^2$, which does not appear to have the full continuous symmetry, though a quarter rotation leads to $\mathcal{L} \rightarrow - \mathcal{L}$, leaving the equations of motion unchanged.  Is there an alternate free Lagrangian density that makes the full continuous symmetry manifest?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The normal Lagrangian expression is dependent on a gauge field A, whose existence depends on the non-existence of monopoles.  &lt;a href=&quot;http://physics.stackexchange.com/questions/9577/making-symmetry-between-e-and-b-fields-manifest-in-lagrangian/9589#9589&quot;&gt;As Peter Morgan points out&lt;/a&gt;, this obviously breaks the symmetry and explains why this Lagrangian needn't have this symmetry.  However, that leaves the following questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is there another Lagrangian giving the same equations of motion that does have this symmetry?  If yes, can it be demonstrated?  If not, is there a proof?&lt;/li&gt;&#xA;&lt;li&gt;Even if the answer to 1 is no, is there a family of Lagrangians related by this symmetry that each give the same equations of motion?  If yes, can it be demonstrated?  If not, is there a proof?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1233" LastEditorUserId="1233" LastEditorDisplayName="" LastEditDate="2011-05-10T02:17:08.503" LastActivityDate="2011-05-12T13:24:01.557" Title="Making symmetry between E and B fields manifest in Lagrangian" Tags="&lt;electromagnetism&gt;&lt;lagrangian-mechanics&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="3" />
  <row Id="9578" PostTypeId="1" CreationDate="2011-05-08T07:08:02.727" Score="0" ViewCount="124" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/2175/is-it-possible-for-information-to-be-transmitted-faster-than-light&quot;&gt;Is it possible for information to be transmitted faster than light?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;So this isn't a very official or serious source of information I'm aware.. but this picture makes me curious. Can someone explain why or why not this is correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cache.ohinternet.com/images/7/7f/Troll_physics1.png&quot; rel=&quot;nofollow&quot;&gt;http://cache.ohinternet.com/images/7/7f/Troll_physics1.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arguments is that if you have a stick a light year long and you move it forward as if to poke someone, they will feel it immediately before light sent at the same time reaches them. (I think) &lt;/p&gt;&#xA;" OwnerUserId="2536" LastActivityDate="2011-05-08T07:08:02.727" ClosedDate="2011-05-08T07:14:10.543" Title="Stick a Lightyear Long, Travels faster than light?" Tags="&lt;speed-of-light&gt;" CommentCount="1" />
  <row Id="9579" PostTypeId="2" ParentId="9577" CreationDate="2011-05-08T08:32:24.953" Score="2" ViewCount="" Body="&lt;p&gt;No, the Lagrangian of electrodynamics indeed fails to be invariant under the particular continuous $SO(2)$ symmetry that you described. You can't change this fact by any field redefinition etc. because physically harmless field redefinitions don't change the action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the Lagrangian is invariant under&#xA;$$ (E,B) \to  (E\cosh \eta + B \sinh \eta, B\cosh \eta + E \sinh \eta) $$&#xA;which corresponds to changing the metric on the $(E,B)$ space from the $(++)$ Euclidean signature to the $(+-)$ Lorentzian signature (like in special relativity). On the other hand, the energy density (with the positively definite quadratic invariant) obviously fails to be invariant under this modified transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it's the latter, indefinite &quot;Lorentzian&quot; transformation that is closer to the actual continuous symmetries that may exist between electricity and magnetism. In $N=4$ gauge theory or type IIB string theory, it's easy to see that the most natural continuous symmetry is $SL(2,R)$ which is, for two possible &quot;types of charges&quot;, equivalent to $Sp(2,R)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantization of electric charges breaks it to $SL(2,Z)$ which is the actual symmetry group - S-duality - of $N=4$ gauge theory or type IIB string theory. It is isomorphic neither to $SO(2)$ nor to $SO(1,1)$ because its bilinear invariant is antisymmetric, not symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This $SL(2,Z)$ group contains the &quot;quarter rotation&quot; you mentioned but it doesn't contain the full continuous group. Even the $SL(2,R)$ doesn't. At the level of the Lagrangian or quantum theory, one may see why the continuous group mixing $E,B$ is morally wrong. The action is not invariant, as you noticed. Also, the Dirac quantization rule leads one to consider the product of the elementary electric charge and elementary magnetic charge, which should be a multiple of $2\pi$, and this &quot;symplectic&quot; structure has to be preserved as well.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-08T08:32:24.953" CommentCount="2" />
  <row Id="9580" PostTypeId="2" ParentId="9543" CreationDate="2011-05-08T08:47:03.910" Score="1" ViewCount="" Body="&lt;p&gt;I will undertake a hand waving answer and maybe since the question will come to the top somebody knowledgeable will give a full answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Virtual particles are the province of quantum mechanics and led to the development of quantum field theory with creation and annihilation operators. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This led to the realization that the vacuum, in the QFT description is not &quot;empty&quot; but is like a sea where virtual particles continuously create and annihilate with no loss of energy.&#xA;The vacuum is a &quot;ground state&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly enough this field theoretic description with creation and annihilation operators does not correspond one to one with particle physics. Back in 1963 I sat through a field theoretical course for nuclear physics where creation and annihilation operators acted on nuclear levels. But I digress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the curvature of space is a unique proposal of General Relativity. General relativity has not been quantized in an irrefutable manner. String theorists believe that they have managed to do that, but I leave it to them to describe what a sea of virtual particles in a string universe is like. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to naively say: suppose gravity is quantized in the classic QFT manner and gravitons exist in the vacuum sea too, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the answer would be: the higher the curvature the more the distribution of the particle antiparticle sea would be weighted statistically towards heavier pairs, due to energy considerations with respect to flat space.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-08T08:47:03.910" CommentCount="4" />
  <row Id="9581" PostTypeId="2" ParentId="9577" CreationDate="2011-05-08T09:26:34.540" Score="2" ViewCount="" Body="&lt;p&gt;a somehow hidden assumption of your question is that you consider a theory in which the &lt;strong&gt;material response&lt;/strong&gt; is the same to a magnetic as well as to an electric field. We have to elaborate on this a little now to answer your question from an electrodynamic point of view in addition to the nice general &lt;a href=&quot;http://physics.stackexchange.com/questions/9577/making-symmetry-between-e-and-b-fields-manifest-in-lagrangian/9579#9579&quot;&gt;answer given by Luboš&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose you know the &lt;strong&gt;linear constitutive relations&lt;/strong&gt; $$\mathbf{D} = \epsilon \mathbf{E}\quad \text{and}\quad \mathbf{B}=\mu \mathbf{H}\ .$$&#xA;These link the electric and magnetic field to their &quot;dual&quot; counterparts using the permittivity $\epsilon = \epsilon_0 \epsilon_r$ and permeability $\mu = \mu_0 \mu_r$ assumed to be not depending on the fields itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case you will have a Lagrangian of the form $$\mathcal{L} \propto F_{\mu\nu}M^{\mu\nu}$$&#xA;where now the tensor $M$ is a quantity that is linearly linked to $\epsilon$ and $\mu$ and $F$ via $M^{\mu\nu} = \epsilon^{\mu\nu\alpha\beta}F_{\alpha\beta}$. Generally, $M$ will &lt;strong&gt;depend&lt;/strong&gt; on $F$ in some arbitrary complex way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you assume linear materials with the same response to electric and magnetic excitation, $$\epsilon_r \equiv \mu_r\ ,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you find that you can write $$\mathcal{L} \propto F\wedge\star F$$ as in the case of vacuum where $\epsilon_r=\mu_r = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These materials are called &lt;strong&gt;impedance matched&lt;/strong&gt; and only a special subclass of all materials as describes.&#xA;The symmetry you refer to can then be used to interprete $\epsilon$ as a &lt;strong&gt;metric&lt;/strong&gt; which is done in a field called &lt;strong&gt;transformation optics&lt;/strong&gt;, see eg. &lt;a href=&quot;http://arxiv.org/abs/0805.4778&quot; rel=&quot;nofollow&quot;&gt;Leonhardt, Philbin: Transformation Optics and the Geometry of Light&lt;/a&gt; where all the concepts are discussed in greater detail along with spectacular applications as cloaking:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/PnhxA.jpg&quot; alt=&quot;cloaking&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to sum up, in general (macroscopic) electrodynamics, the &lt;strong&gt;symmetry&lt;/strong&gt; between electric and magnetic fields &lt;strong&gt;breaks down&lt;/strong&gt;. &#xA;In e.g. optical frequencies, materials tend to respond only in an electric way, $$\mu^{\text{opt}}_r = \mu_0 \quad ,\ \epsilon^{\text{opt}}_r \text{ arbitrary}\ .$$&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2011-05-08T09:26:34.540" CommentCount="2" />
  <row Id="9582" PostTypeId="1" CreationDate="2011-05-08T10:32:30.270" Score="2" ViewCount="151" Body="&lt;p&gt;In standard college physics text books, high-school books and popular level physics books, the collapse of the Tacoma Narrows Bridge is often taken as an example of &lt;em&gt;resonance&lt;/em&gt;. However a more detailed analysis shows that the collapse was not due to simple resonance (see the article by &lt;a href=&quot;http://www.ketchum.org/billah/Billah-Scanlan.pdf&quot; rel=&quot;nofollow&quot;&gt;Billah and Scanlan&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Tacoma_Narrows_Bridge_%281940%29#Resonance_hypothesis&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is, whether there are further examples from mechanics, which are often incorrectly or oversimplified declared as examples of resonance phenomena (for example in physics text books)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;I should add, that the definition usually given in textbooks is the following: &#xA;Consider a mechanical system which can be described by a differential equation like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d^2}{d t^2}x(t) + 2 \rho \frac{d}{d t} x(t) + \omega_0^2 x(t) = f(t) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f(t) = f_0 \cos(\omega t)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then for example the amplitude or the energy of the system as function of $\omega$ has a maximum near $\omega_0$. In this case one speaks of &lt;em&gt;resonance&lt;/em&gt; of the system.&lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorUserId="667" LastEditorDisplayName="" LastEditDate="2011-05-08T13:07:00.133" LastActivityDate="2011-05-08T13:07:00.133" Title="Phenomena which are incorrectly declared as resonance phenomena?" Tags="&lt;classical-mechanics&gt;&lt;education&gt;&lt;applied-physics&gt;&lt;vibration&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="9583" PostTypeId="2" ParentId="9131" CreationDate="2011-05-08T10:41:48.063" Score="1" ViewCount="" Body="&lt;p&gt;I try to give my answer in case anyone is interested:&#xA;Weinberg in his book &quot;Quantum theory of fields&quot; Vol 1 in section 3.4 defines the flow factor $v_{\alpha}$ in the formula of the cross section so that it coincides with the velocity of particle 1 in the reference frame in which particle 2 is at rest, in such a way that the cross section remains Loretz invariant in different reference frame. In other word:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_{\alpha}= \frac{ ({(p_1 \cdot p_2)^2-(m_1m_2)^2})^{(1/2)} } {E_1E_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you try to simplify this expression using the equations $\underline{v}=\frac{\underline{p}}{E}$ you find the factor of my question above.&#xA;To summarize this factor is a definition to make the cross section Lorentz invariant in any reference frame.&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2011-05-08T10:41:48.063" />
  <row Id="9584" PostTypeId="1" AcceptedAnswerId="9595" CreationDate="2011-05-08T11:40:16.227" Score="1" ViewCount="86" Body="&lt;p&gt;Assume that by some means, the moon could be given an atmosphere, of the same density and pressure at the surface as the earth's. Obviously in a stable atmosphere there are temperature variations from pole to equator, day to night, and a lot of interesting dynamics follow from that, but for argument's sake assume an average of something like 0 to 5 C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How long would it last?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly there's no air there now (unless you count traces of potassium, or whatever it has), so such at artificial atmosphere would not be permanent, but presumably the moon could hang onto some air, for some time. With the moon's lower gravity, the density or pressure vs altitude curve will be different, and I'd guess that a significantly larger fraction of the mass would be at higher altitudes than on earth, and there's no protection from the solar wind. Someone with the right equations could probably work out how long it would take the pressure at the surface to drop by half.&lt;/p&gt;&#xA;" OwnerUserId="717" LastActivityDate="2011-05-10T18:53:48.650" Title="stability of hypothetical lunar atmosphere" Tags="&lt;astrophysics&gt;&lt;atmosphere&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9585" PostTypeId="2" ParentId="9582" CreationDate="2011-05-08T11:41:20.697" Score="3" ViewCount="" Body="&lt;p&gt;looking at the link you gave, I think it is just a problem of &lt;strong&gt;name convention&lt;/strong&gt; and has nothing to do with oversimplifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was indeed ruled out that the reason of the failure was not due to a resonance phenomenon called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Von_K%C3%A1rm%C3%A1n_vortex_street&quot; rel=&quot;nofollow&quot;&gt;Kármán vortex street&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/iwIKv.gif&quot; alt=&quot;Karman vortex street&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a phenomenon arising in fluiddynamics with a certain resonance frequency, called the &lt;strong&gt;Strouhal frequency&lt;/strong&gt;. However, if you look a little further you find that the cause for failure was due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Aeroelasticity#Flutter&quot; rel=&quot;nofollow&quot;&gt;aeroelastic flutter&lt;/a&gt;. According to the wikipedia article &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Flutter is a self-feeding and potentially destructive vibration where aerodynamic forces on an object couple with a structure's natural &lt;strong&gt;mode of vibration&lt;/strong&gt; to produce rapid &lt;strong&gt;periodic motion&lt;/strong&gt;. Flutter can occur in any object within a strong fluid flow, under the conditions that a positive &lt;strong&gt;feedback&lt;/strong&gt; occurs between the structure's natural vibration and the aerodynamic forces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, it is a kind of periodic &lt;strong&gt;process&lt;/strong&gt;, driven by an &lt;strong&gt;excitation&lt;/strong&gt; (here aerodynamic force) causing an energy flow into an &lt;strong&gt;eigenmode&lt;/strong&gt; of the system possibly leading to destruction.&lt;br&gt;&#xA;I think there is &lt;strong&gt;no problem&lt;/strong&gt; calling this a resonance unless you give a very strict definition of such a process.&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2011-05-08T11:41:20.697" CommentCount="2" />
  <row Id="9586" PostTypeId="1" AcceptedAnswerId="9590" CreationDate="2011-05-08T11:44:48.150" Score="0" ViewCount="73" Body="&lt;p&gt;Let's say an array of rays of light is given.&lt;br&gt;&#xA;Each ray has a specific wavelength (in the range of visible light).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ray1 = 470 nm&#xA;ray2 = 640 nm&#xA;ray3 = 510 nm&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Can this rays be merged into a single ray of &lt;code&gt;white light&lt;/code&gt; which preserve the information of it's components?&lt;/p&gt;&#xA;" OwnerUserId="3459" LastActivityDate="2011-05-08T13:23:50.670" Title="Merge different wavelength rays" Tags="&lt;optics&gt;&lt;light&gt;" AnswerCount="1" />
  <row Id="9587" PostTypeId="2" ParentId="9582" CreationDate="2011-05-08T12:30:39.793" Score="-1" ViewCount="" Body="&lt;p&gt;The main problem is the understanding of &quot;resonance&quot;. Resonance is &#xA;eg the wineglass smashed by Caruso singing (a fairy tale).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tacoma bridge is an example of an oscillator, who worked up to &#xA;destuctive amplitudes. Any oscillator resonates at its natural &#xA;frequency, this is often called resonance, but that is not &#xA;really correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aeroelastic flutter is the name of winddriven oscillators &#xA;on basis of elastic masts or towers or chimneys. This phenomenon &#xA;can be driven by a Karman street mechanism! This is no contradiction. &#xA;The Karman air flow is rather adoptable and will synchronize &#xA;easily to the natural vibration of the elastic object. &#xA;Have You ever seen some spiral ridge on slim steel chimneys in &#xA;chemical plants or on car antennas? This is to avoid flutter &#xA;driven by Karman vortices. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This difference between resnoance and exitation as a oscillator &#xA;is quite common among civil engineers, thes are too much trained &#xA;for &lt;strong&gt;statics&lt;/strong&gt;, dynamics is terra incognita for them. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-05-08T12:30:39.793" CommentCount="5" />
  <row Id="9588" PostTypeId="1" AcceptedAnswerId="9596" CreationDate="2011-05-08T12:43:42.680" Score="1" ViewCount="100" Body="&lt;p&gt;We seem to know that black hole energy, and apparently event horizon geometry and topology, are functions of Mass, Charge and Angular Momentum.  Area is a function of geometry and topology as well, which apparently controls entropy.  Strominger, et al, can show that the apparent agreement of string states to event horizon in extremal cases.  What is the study of diffeomorphisms from extremal black holes solutions to schwarzschild solutions called?  What are routes to show string states describe schwarzschild black holes as charge and angular momentum are taking to limit zero?&lt;/p&gt;&#xA;" OwnerUserId="3460" LastActivityDate="2011-05-08T13:54:55.903" Title="Black Holes and String states" Tags="&lt;string-theory&gt;&lt;astrophysics&gt;" AnswerCount="1" />
  <row Id="9589" PostTypeId="2" ParentId="9577" CreationDate="2011-05-08T13:13:15.397" Score="1" ViewCount="" Body="&lt;p&gt;An alternative way to look at this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hamiltonian and the associated Lagrangian formulation that you give presupposes that the field is the electromagnetic potential, with the electric field as the momentum, which is only possible if we do not have monopoles in the mix. The electromagnetic field has to satisfy &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%27s_law_for_magnetism&quot; rel=&quot;nofollow&quot;&gt;Gauss's law for magnetism&lt;/a&gt; and &lt;a href=&quot;http://electron9.phys.utk.edu/phys136d/modules/m8/inductance.htm&quot; rel=&quot;nofollow&quot;&gt;Faraday's law of induction&lt;/a&gt;(chosen from the first page of a Google search page because I didn't like the Wikipedia page much) in order for the &lt;em&gt;six&lt;/em&gt; components of the electromagnetic field to be expressible as the exterior derivative of the &lt;em&gt;four&lt;/em&gt; components of the electromagnetic potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without the reduction of the EM field to the EM potential because of the no monopole constraint, one way to proceed to a Hamiltonian or Lagrangian formalism would be to introduce the time-like derivatives of the electromagnetic field as the momentum components, which, however, would change the EM field equations into second order equations that are not the Maxwell equations. I'm sure this must have been done somewhere in the literature, and various possible constrained dynamics will have been considered, because people have investigated the possibility of monopoles, but I regret that I don't know a good reference. The people who have investigated the possibility of monopoles have apparently not so far managed to make the phenomenology of such a dynamics work out nicely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we deal with the source-free Maxwell's equations, we could equally well express the electromagnetic field in terms of a potential as $\mathbf{B}(x)=\nabla\phi&amp;#39;(x)$ and $\mathbf{E}=\nabla\times\mathbf{A}&amp;#39;(x)$ instead of the usual prescription, or indeed in terms of any linear transform, $\mathbf{B}(x)=\cos(\alpha)\nabla\phi_\alpha(x)+\sin(\alpha)\nabla\times\mathbf{A}_\alpha(x)$ and $\mathbf{E}(x)=-\sin(\alpha)\nabla\phi_\alpha(x)+\cos(\alpha)\nabla\times\mathbf{A}_\alpha(x)$, although we have to keep sight of the fact that this breaks down as soon as we introduce electric charges or monopoles. I suggest you try seeing, for the free field case, what the effect is of changing the definition of the electromagnetic field in terms of the electromagnetic potential as you transform the electromagnetic field.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-05-08T13:13:15.397" />
  <row Id="9590" PostTypeId="2" ParentId="9586" CreationDate="2011-05-08T13:23:50.670" Score="2" ViewCount="" Body="&lt;p&gt;Short answer: No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer: What is the 'information' you wish to preserve? The phase? In that case, if you manage to make the rays collinear, you will have three monochromatic rays that each have a separate phase, so yes, the information will be preserved, but it will not be white light - it will be three monochromatic rays of 470, 510, and 640 nm. What is generally referred to as 'white light' has a broad spectrum and is not monochromatic (or trichromatic) at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's more, when you are dealing with real 'rays', they will never be perfectly collinear, so they will separate after some distance. If they are travelling through a medium, like air, they will also have slightly different indices of refraction, so their relative phases will not be preserved either.&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2011-05-08T13:23:50.670" CommentCount="2" />
  <row Id="9591" PostTypeId="1" AcceptedAnswerId="9593" CreationDate="2011-05-08T13:29:57.780" Score="2" ViewCount="108" Body="&lt;p&gt;I'm looking for an originally German introduction to quantum mechanics. Is there such a canonical book used in German QM undergraduate courses?&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2011-05-08T14:07:40.443" Title="German book on introductory quantum mechanics" Tags="&lt;quantum-mechanics&gt;&lt;education&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9592" PostTypeId="2" ParentId="9591" CreationDate="2011-05-08T13:35:43.423" Score="2" ViewCount="" Body="&lt;p&gt;Walter Greiner's first part on Quantum Mechanics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540674586&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.de/Quantummechanics-Introduction-Walter-Greiner/dp/3540674586&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1671" LastActivityDate="2011-05-08T13:35:43.423" CommentCount="0" />
  <row Id="9593" PostTypeId="2" ParentId="9591" CreationDate="2011-05-08T13:37:47.747" Score="3" ViewCount="" Body="&lt;p&gt;I think a very good orignally german QM book is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540428887&quot; rel=&quot;nofollow&quot;&gt;Straumann: &quot;Quantenmechanik: Ein Grundkurs über nichtrelativistische Quantentheorie&quot; &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there is also a second volume:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540229515&quot; rel=&quot;nofollow&quot;&gt;Straumann: &quot;Relativistische Quantenfeldtheorie&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another good book is from &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3923944403&quot; rel=&quot;nofollow&quot;&gt;G. Grawert: Quantenmechanik&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also have a look at &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3211825355&quot; rel=&quot;nofollow&quot;&gt;Thirrings&lt;/a&gt; books about QM (it is mathematically more advanced).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think Straumanns book is not &quot;canonical&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often used by students are the books by &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540688684&quot; rel=&quot;nofollow&quot;&gt;Nolting&lt;/a&gt; and Greiner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the experimental point of view you may have a look at &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3642039103&quot; rel=&quot;nofollow&quot;&gt;Demtröders&lt;/a&gt; book.&lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorUserId="667" LastEditorDisplayName="" LastEditDate="2011-05-08T14:07:40.443" LastActivityDate="2011-05-08T14:07:40.443" />
  <row Id="9594" PostTypeId="1" CreationDate="2011-05-08T13:43:35.470" Score="0" ViewCount="89" Body="&lt;p&gt;The time has come for me to look for a thesis. I would like to do a thesis in theoretical physics concerning the structure of matter or particle physics. What, in your opinion, some interesting proposals on this topics where you can make something original?&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2011-05-08T13:43:35.470" ClosedDate="2011-05-08T15:33:45.407" Title="proposals for a thesis" Tags="&lt;soft-question&gt;" CommentCount="4" />
  <row Id="9595" PostTypeId="2" ParentId="9584" CreationDate="2011-05-08T13:49:26.307" Score="2" ViewCount="" Body="&lt;p&gt;Not sure about the solar wind but when it comes to Jeans escape: The escape velocity from the Moon is 2.4 km/s:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=escape+velocity+from+moon&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=escape+velocity+from+moon&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not much higher than the velocity of the molecules. The mass of Nitrogen molecule, the prevailing one, is 28 u, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=mass+of+nitrogen+molecule&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=mass+of+nitrogen+molecule&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is $4.65 \times 10^{-26}$ kg. In the $\exp(-mv^2/2kT)$ Maxwell factor for $T=275$ Kelvin, we may see that &#xA;$$\langle v^2\rangle = kT/m = 1.38 \times 10^{-23} \times 275 / 4.65 \times 10^{-26} =81,600$$&#xA;squared meters over squared seconds. We need $v^2=2,400^2 = 5,760,000$ meters per second which is 70 times more than $81,600$, so the relevant exponential is $\exp(-35)$ (don't forget the factor of $1/2$) which is $6\times 10^{-16}$ or so. Only this fraction of molecules going up escape. It takes about 100 seconds for a molecule to try another value of its velocity, after a collision and a return from the upper atmosphere, so one needs something like $10^{16}\times 100 = 10^{18}$ seconds for the Nitrogen molecules to reach the escape velocity. That's something like 30 billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitrogen in the lunar atmosphere (and heavier molecules) would be safe against Jeans escape, I think. Of course, it would be much easier for Hydrogen to escape. 28 drops to 2, 35 drops to something like 2 as well, so much of the Hydrogen escape during its 10th attempt or so, within minutes. Modifying the prefactors in the exponent makes a huge impact on the result.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-08T13:49:26.307" CommentCount="4" />
  <row Id="9596" PostTypeId="2" ParentId="9588" CreationDate="2011-05-08T13:54:55.903" Score="0" ViewCount="" Body="&lt;p&gt;The generalization of the Strominger-Vafa calculation of the extremal black hole entropy to the case of neutral black holes is not just about a &quot;diffeomorphism&quot; (which is a word representing just a trivial coordinate transformation in the context of GR). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a totally different problem - something that you were implicitly aware of at the beginning of your question, before you contradicted yourself. So the whole calculation has to be done from scratch while one cannot rely on supersymmetry because non-extremal black holes such as Schwarzschild cannot be supersymmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite this complication, the correct entropy of various non-supersymmetric black holes has been calculated from string theory, too.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-08T13:54:55.903" CommentCount="1" />
  <row Id="9597" PostTypeId="2" ParentId="9084" CreationDate="2011-05-08T13:55:31.037" Score="0" ViewCount="" Body="&lt;p&gt;I can't help thinking you are needlessly complicating the physics by focussing on the transient case. In the steady state it is much easier to do the energy balancing. Of course there are several possible versions of this problem in the &quot;steady state&quot;: you could have a free charge which is allowed to pick up speed in the direction of the propagating wave (via v X B forces), a free charge which is constrained to move only perpendicular to the propagating wave motion, or a charge on a spring which may oscillate in resonance with the excitation. I am most familiar with the third version: the driven oscillator. In the steady state, there is energy which is scattered everywhere, and there is a &quot;shadow zone&quot; behind the oscillator, roughly paraboloid in shape, where the secondary radiation interferes destructively with the incident wave. The amount of destructive interference is linear in the oscillator strength, and the spherically scattered radiation is of course quadratic. The balance occurs when the linear term is equal to the quadratic term.&lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-05-08T13:55:31.037" />
  <row Id="9598" PostTypeId="1" CreationDate="2011-05-08T16:41:07.650" Score="0" ViewCount="246" Body="&lt;p&gt;I must apologize, I was a little bit excited when I began understanding some of this, I can not say I can compete with professionals, and words are still difficult concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In (&lt;a href=&quot;http://arxiv.org/abs/gr-qc/9409013&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;) S.H. et al. it is discussed that in extremal cases of charged black holes in a euclidean space, there is a topological change as compared to the non extremal case.  This is viewed as a discontinuity by (&lt;a href=&quot;http://arxiv.org/abs/0901.0931&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;) who attempts a similar calculation in AdS.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The notion of charge being associated with topology comes from some versions (&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0071498702&quot; rel=&quot;nofollow&quot;&gt;3&lt;/a&gt;,&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521831431&quot; rel=&quot;nofollow&quot;&gt;4&lt;/a&gt;) of the calculation performed in (&lt;a href=&quot;http://arxiv.org/abs/hep-th/9601029&quot; rel=&quot;nofollow&quot;&gt;5&lt;/a&gt;). Where the &quot;charges are Q1 and Q5 are generated by wrapping a number Q1 of D1-branes around the circle $x^5$ and a number Q5 of D5-branes around the five circles...&quot; and N is the momentum around the circle $x^5$ (&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521831431&quot; rel=&quot;nofollow&quot;&gt;4&lt;/a&gt;). Since the entropy calculation relies on those charges, it should follow that the number of string states is also dependent on the topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in (&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0071498702&quot; rel=&quot;nofollow&quot;&gt;3&lt;/a&gt;, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521831431&quot; rel=&quot;nofollow&quot;&gt;4&lt;/a&gt;, &lt;a href=&quot;http://arxiv.org/abs/hep-th/9601029&quot; rel=&quot;nofollow&quot;&gt;5&lt;/a&gt;), the calculation is done on a supersymmetric extremal black hole.  In (&lt;a href=&quot;http://arxiv.org/abs/gr-qc/9409013&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;) the calculation is not on a supersymmetric extremal black hole, but there is difference in topology at the extremal limit, which is implied in (&lt;a href=&quot;http://arxiv.org/abs/0901.0931&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;) to be discontinuous.  In both case the change in topology is related to their being two event horizons that do connect to each other, implying a toroidal shape.  In (&lt;a href=&quot;http://arxiv.org/abs/gr-qc/9409013&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;) the second horizon extends to infinity, causing the change in topology, in (&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0071498702&quot; rel=&quot;nofollow&quot;&gt;3&lt;/a&gt;), the two horizons are described as meeting at the same radius.  This suggests some sort of degeneration of the torus, implying a change in topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial query is whether we should think of these changes in topology as discontinuous (assuming that what I described isn't horribly flawed), or possibly a smooth transitions due to quantum effects, such as uncertainty and tunneling, where the change only occur discontinuously at a microscopic level so that it appears smooth at a macroscopic level?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part of the question is that charges are described as being dependent on topology, and they are also related to symmetries.  So it seems natural to think that symmetries are dependent on topology.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although we live in a relativistic world, in our every day life, we still think in non-relativistic terms, and most of our theories have non-relativistic roots, so is it natural to think of there being some sort of low energy topological state (I don't know, something like Ball-3 x Real).  That low energy state would apparently have certain symmetries that are dependent on that topology and would make it distinct from other possible topologies.  Is that possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/9409013&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;- &lt;a href=&quot;http://arxiv.org/abs/gr-qc/9409013&quot; rel=&quot;nofollow&quot;&gt;Entropy, Area and Black Holes&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/0901.0931&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;- &lt;a href=&quot;http://arxiv.org/abs/0901.0931&quot; rel=&quot;nofollow&quot;&gt;Extremal Limits and Black Hole Entropy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0071498702&quot; rel=&quot;nofollow&quot;&gt;3&lt;/a&gt;- &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0071498702&quot; rel=&quot;nofollow&quot;&gt;String Theory Demystified&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521831431&quot; rel=&quot;nofollow&quot;&gt;4&lt;/a&gt;- &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521831431&quot; rel=&quot;nofollow&quot;&gt;A First Course in String Theory&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/9601029&quot; rel=&quot;nofollow&quot;&gt;5&lt;/a&gt;- &lt;a href=&quot;http://arxiv.org/abs/hep-th/9601029&quot; rel=&quot;nofollow&quot;&gt;Microscopic Origin of Bekenstein-Hawking Entropy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further reference (pg 355 in &lt;a href=&quot;http://books.google.com/books?id=LFakv848R2oC&amp;amp;lpg=PA355&amp;amp;ots=rNq7UjGfvC&amp;amp;dq=loop%20at%20each%20point%20of%20sphere&amp;amp;pg=PA355#v=onepage&amp;amp;q=loop%20at%20each%20point%20of%20sphere&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Group Theory: An intuitive approach&lt;/a&gt; by R Mirman):&lt;/p&gt;&#xA;" OwnerUserId="3460" LastEditorUserId="3460" LastEditorDisplayName="" LastEditDate="2011-05-17T23:25:10.263" LastActivityDate="2011-09-03T23:31:52.930" Title="Charges and Topology" Tags="&lt;string-theory&gt;&lt;black-holes&gt;&lt;ads-cft&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="9599" PostTypeId="2" ParentId="9459" CreationDate="2011-05-08T17:05:44.810" Score="0" ViewCount="" Body="&lt;p&gt;The electric field of a dipole is the derivatve of the monopole field: to put it more accurately, if you create a dipole with a z orientation, the field is obviously d/dz of the field of a monopole. So if you have a given charge distribution, you just replace all the charges with dipoles and the resulting field is the derivative (with respect to z) of the original field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think this is what you were asking though.&lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-05-08T17:05:44.810" />
  <row Id="9600" PostTypeId="2" ParentId="647" CreationDate="2011-05-08T17:24:36.067" Score="0" ViewCount="" Body="&lt;p&gt;There is only a paradox if you assume an infinitely rigid spring AND and infinitely elastic collision. For a collision that occurs on a timescale much smaller than the natural oscillation period of the molecule, the incident atom stops dead. For a collision that is spread out over several oscillations of the molecule, the atom recoils as though it had struck a target of mass 2m. The problem is solved by trivial methods so long as you can assume that the ratio between the oscillation period and the contact time is very large or very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, this isn't how atoms and molecules work at all. It's really a billiard ball and spring problem, and we've just used the &quot;atom/molecule&quot; terminology as a convenient shorthand.)&lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-05-08T17:24:36.067" CommentCount="1" />
  <row Id="9601" PostTypeId="1" AcceptedAnswerId="9644" CreationDate="2011-05-08T17:38:26.347" Score="4" ViewCount="220" Body="&lt;p&gt;Let $\psi (x,y,z)$ be a scalar field. I found the following statement in Morse &amp;amp; Feshbach Methods of Theoretical Physics:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The limiting value of the difference between $\psi$ at a point and the average value of $\psi$ at neighboring points is $-\frac{1}{6}(dxdydz)^2\nabla ^2\psi$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By taking a small sphere centered at the point of radius $r$, I was able to show that the difference between $\psi$ at a point and the average value of $\psi$ at neighboring points equals $-\frac{1}{6}r^2\nabla ^2\psi$. But I have not been able to get the expression given in the book. Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: To obtain $-\frac{1}{6}r^2\nabla ^2\psi$ I first expanded $\psi$ in a Taylor series up to quadratic terms about the point in question. Then I integrated $\psi$ (rather, its Taylor expansion up to quadratic terms) over the surface of a small sphere of radius $r$ centered at the given point and divided this by the surface are of the sphere (this gives the average value of $\psi$ in the sphere). Subtracting this average from the value of $\psi$ at the center of the sphere gives the result $-\frac{1}{6}r^2\nabla ^2\psi$.&lt;/p&gt;&#xA;" OwnerUserId="3064" LastEditorUserId="394" LastEditorDisplayName="" LastEditDate="2011-05-10T01:07:03.427" LastActivityDate="2011-05-10T01:07:03.427" Title="$\nabla ^2\psi$ equals $\psi -$ average value of $\psi$ at neighboring points" Tags="&lt;mathematical-physics&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="1" />
  <row Id="9602" PostTypeId="2" ParentId="9574" CreationDate="2011-05-08T18:00:43.863" Score="3" ViewCount="" Body="&lt;p&gt;The answer is no, the generalized function (=&lt;a href=&quot;http://en.wikipedia.org/wiki/Distribution_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;distribution&lt;/a&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lim_{\epsilon\rightarrow 0} \frac{\epsilon^2}{\epsilon^2+t^2}=0 &#xD;&#xA;\qquad\mathrm{a.e.}&#xD;&#xA;$$ &#xA;is &lt;a href=&quot;http://en.wikipedia.org/wiki/Almost_everywhere&quot; rel=&quot;nofollow&quot;&gt;almost everywhere&lt;/a&gt; (a.e.) equal to the ordinary zero function $0:\mathbb{R}\to\mathbb{R}$ that sends $t\mapsto 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof. Consider a test function $f\in C^{\infty}_c(\mathbb{R})$, i.e., an infinitely often differentiable function $f$ with compact support. Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{\mathbb{R}} dt \ f(t)\frac{\epsilon^2}{\epsilon^2+t^2}  ~\stackrel{t=\epsilon x}{=}~\epsilon \cdot \int_{\mathbb{R}} dx \ f(\epsilon x)\cdot\frac{1}{1+x^2} $$&#xA;$$ \longrightarrow 0\cdot f(0)\cdot\int_{\mathbb{R}} dx \ \frac{1}{1+x^2} = 0 \cdot f(0)\cdot\pi =0 \qquad \mathrm{for} \qquad \epsilon \to 0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because of, e.g., &lt;a href=&quot;http://en.wikipedia.org/wiki/Dominated_convergence_theorem&quot; rel=&quot;nofollow&quot;&gt;Lebesgue's dominated convergence theorem&lt;/a&gt;. $\Box$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distribution only becomes $\pi\delta(t)$, if we remove one factor of $\epsilon$. Here $\delta(t)$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_delta_function&quot; rel=&quot;nofollow&quot;&gt;Dirac delta distribution&lt;/a&gt; (often called the Dirac delta function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of using distribution theory, we may simply interpret the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lim_{\epsilon\rightarrow 0} \frac{\epsilon^2}{\epsilon^2+t^2}~=~\delta_{t,0}~=~\left\{\begin{array}{rcl} 1 &amp;amp;\mathrm{for}&amp;amp; t=0 \\  0 &amp;amp;\mathrm{for}&amp;amp; t\in\mathbb{R}\backslash\{0\} \end{array}\right. $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;as a $t$-pointwise limit. Here $\delta_{t,0}$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kronecker_delta&quot; rel=&quot;nofollow&quot;&gt;Kronecker delta function&lt;/a&gt;, which should &lt;em&gt;not&lt;/em&gt; be confused with the Dirac delta distribution. The former is an ordinary function, while the latter is not. The last formula has the added benefit, that it is true both in a $t$-pointwise sense and in a distribution sense, since the Kronecker delta function is zero almost everywhere (with respect to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lebesgue_measure&quot; rel=&quot;nofollow&quot;&gt;Lebesgue measure&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-09T05:32:09.883" LastActivityDate="2011-05-09T05:32:09.883" CommentCount="3" />
  <row Id="9603" PostTypeId="2" ParentId="9263" CreationDate="2011-05-08T18:09:13.093" Score="-1" ViewCount="" Body="&lt;p&gt;Most of materials that highly reflect the visible spectrum of light are highly absorbing materials for UV- spectrum, like noble metals(Au, Ag and Cu), and also alkali metals.  To have a mirror that reflect light in a long spectrum we can use multilayer mirrors that have constructed from complex of several different layers with thickness of several nanometers,(coated on a substrate) that known as VUV-UV Mirrors .&#xA;You can search for “multilayer uv mirrors:pdf” in net to find something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.optica.tn.tudelft.nl/publications/docs/2000%20pdf%20filed/2000.006.pdf&quot; rel=&quot;nofollow&quot;&gt;Design of multilayer extreme-ultraviolet mirrors for enhanced reflectivity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever the range of spectrum required for reflection increases, the number of layers increase.&#xA;&lt;img src=&quot;http://i.stack.imgur.com/VZ8l9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3208" LastEditorUserId="3208" LastEditorDisplayName="" LastEditDate="2011-05-08T18:38:12.390" LastActivityDate="2011-05-08T18:38:12.390" />
  <row Id="9605" PostTypeId="2" ParentId="9263" CreationDate="2011-05-08T20:02:58.703" Score="1" ViewCount="" Body="&lt;p&gt;Aluminium is about the best coating for mirrors below 400nm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most commercial small mirrors for makeup etc will be aluminium behind glass. Although the glass isn't very good in the UV it's thin enough not to have too much absorption&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-05-08T20:02:58.703" CommentCount="1" />
  <row Id="9606" PostTypeId="1" AcceptedAnswerId="9722" CreationDate="2011-05-08T23:27:29.913" Score="2" ViewCount="389" Body="&lt;p&gt;Could anyone suggest where can I find absorption spectra &amp;amp; refractive index of Glycerol?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am specifically interested in UV range, 200-400nm, everything I was able to find out was for standard conditions only...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any software which can get these parameters through simulation?&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-05-11T15:17:11.983" Title="Glycerol: refractive index &amp; absorption spectra in 0.2-0.4um range" Tags="&lt;glycerol&gt;&lt;spectra&gt;&lt;refractive&gt;&lt;index&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="9609" PostTypeId="1" AcceptedAnswerId="9627" CreationDate="2011-05-09T02:49:44.727" Score="4" ViewCount="158" Body="&lt;p&gt;I've read that neutrinos in supernova can be affected by &quot;neutrino refraction.&quot; Is this analagous to the refraction of light, and if so, is the speed of these neutrinos similarly reduced from their near c speeds via this index of refraction? &lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="1924" LastEditorDisplayName="" LastEditDate="2011-05-09T03:10:10.177" LastActivityDate="2011-05-09T10:01:26.410" Title="Neutrino Speed in Supernova" Tags="&lt;neutrino&gt;&lt;supernova&gt;" AnswerCount="1" />
  <row Id="9610" PostTypeId="2" ParentId="9577" CreationDate="2011-05-09T03:04:36.230" Score="1" ViewCount="" Body="&lt;p&gt;But the electric and magnetic field are fundamentally different so handling them as symmetric just obscures their elementarry role in the Pointcaré algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fields tensor expressed in Lorentz group generator form with $\mathsf{J}$ and $\mathsf{K}$ as the generators of rotations and boosts for (axial-)vector fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;F^\mu_{~\nu} ~~=~~ \mathsf{B}\cdot\mathsf{J} ~+~ \mathsf{E}\cdot\mathsf{K} ~~=~~&#xD;&#xA;\left(&#xD;&#xA;\begin{array}{rrrr}&#xD;&#xA; \  0\ \      &amp;amp;  ~~\mathsf{E}_x &amp;amp; ~~\mathsf{E}_y &amp;amp; ~~\mathsf{E}_z \ \\&#xD;&#xA; \mathsf{E}_x &amp;amp; \          0\ \ &amp;amp; ~~\mathsf{B}_z &amp;amp; - \mathsf{B}_y \ \\&#xD;&#xA; \mathsf{E}_y &amp;amp; - \mathsf{B}_z  &amp;amp; \         0\ \ &amp;amp; ~~\mathsf{B}_x \ \\&#xD;&#xA; \mathsf{E}_z &amp;amp; ~~\mathsf{B}_y  &amp;amp; - \mathsf{B}_x &amp;amp; \          0\ \ \&#xD;&#xA;\end{array}&#xD;&#xA;    \right)&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you see the fundamental difference: The magnetic field rotates while the electric field boosts. If you simply look at the signs then you'll recognize the $\mathsf{J}$ and $\mathsf{K}$ in the field tensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The field tensor is the boost/rotate generator and so it defines the changes in (proper) time of vector currents and axial currents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;\begin{aligned}&#xD;&#xA;&amp;amp;\frac{\partial}{\partial \tau}\Big(\bar{\psi}\gamma^\mu\psi\Big) &amp;amp;=~~&#xD;&#xA;&amp;amp;\tfrac{q}{m}\,F^\mu_{~\nu}\Big(\bar{\psi}\gamma^\nu\psi\Big)&#xD;&#xA;\\ \\&#xD;&#xA;&amp;amp;\frac{\partial}{\partial \tau}\Big(\bar{\psi}\gamma^5\gamma^\mu\psi\Big) &amp;amp;=~~&#xD;&#xA;&amp;amp;\tfrac{q}{m}\,F^\mu_{~\nu}\Big(\bar{\psi}\gamma^5\gamma^\nu\psi\Big)&#xD;&#xA;\end{aligned}&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are just the QFT equivalents of the classical expressions. See for instance Jackson section 11.11. where the axial expression is just the first term in the Thomas-Bargmann-Michel-Telegdi equation. The equivalent Lorentz generator ${\cal F}^\mu_{~\nu}$ for chiral spinors is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;\mathcal{F}^\mu_{~\nu} ~~=~~ \mathsf{B}\cdot J ~+~ \mathsf{E}\cdot K ~~=~~&#xD;&#xA;-\tfrac12\left( \begin{array}{c}&#xD;&#xA;(i\mathsf{B}^i+\mathsf{E}^i)\sigma_i ~~~~~~0~~~~~~&#xD;&#xA;\\ \\&#xD;&#xA; ~~~~~~0~~~~~~ (i\mathsf{B}^i-\mathsf{E}^i)\sigma_i  \\&#xD;&#xA;\end{array} \right)&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These act on the spinors and, as it should be, the result for the vector and axial current is just the same: Let $\psi_{\,\tau}$ represent the chiral spinor field $\psi_o$ after the corresponding operator has acted on it for a total (proper) time $\tau$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;\psi_{\,\tau} ~~=~~ \exp\left(\tau\tfrac{q}{m}\,{\cal F}^\mu_{~\nu}\right)\psi_o&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can write.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;\begin{aligned}&#xD;&#xA;&amp;amp;\exp\left(\tau\tfrac{q}{m}\,F^\mu_{~\nu}\right)~\bar{\psi}_o\gamma^\nu\psi_o &amp;amp;=~~&#xD;&#xA;&amp;amp;\bar{\psi}_{\,\tau}\gamma^\nu\psi_{\,\tau}&#xD;&#xA;\end{aligned}&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that operating with the Field tensor on the vector current is just the same as operating with the equivalent spinor boost/rotate operator on the spinor field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards, Hans&lt;/p&gt;&#xA;" OwnerUserId="3197" LastActivityDate="2011-05-09T03:04:36.230" />
  <row Id="9611" PostTypeId="1" CreationDate="2011-05-09T03:23:44.137" Score="0" ViewCount="56" Body="&lt;p&gt;I live in a location that I cannot evacuate from tsunami (26 mile peninsula).  I figure I can Anchor a vehicle or go up.  Is going up in a gas filled balloon an option (2 people)?  &lt;/p&gt;&#xA;" OwnerUserId="3464" LastActivityDate="2011-05-09T03:23:44.137" ClosedDate="2011-05-09T12:45:55.327" Title="Tsunami evacuation baloon" Tags="&lt;tsunami&gt;" CommentCount="2" />
  <row Id="9612" PostTypeId="1" AcceptedAnswerId="9639" CreationDate="2011-05-09T03:34:16.087" Score="8" ViewCount="192" Body="&lt;p&gt;The radar image of the midwest provided by &lt;a href=&quot;http://www.wunderground.com/cgi-bin/findweather/getForecast?brand=wxmap&amp;amp;query=40.10669,-88.21381&amp;amp;lat=40.10669&amp;amp;lon=-88.21381&amp;amp;zoom=6&amp;amp;type=ter&amp;amp;units=english&amp;amp;wxsn=0&amp;amp;rad=1&amp;amp;rad.num=1&amp;amp;rad.spd=25&amp;amp;rad.opa=70&amp;amp;rad.stm=1&amp;amp;rad.type=N0R&amp;amp;rad.type2=&amp;amp;rad.smo=1&amp;amp;rad.mrg=0&amp;amp;sat=0&amp;amp;cams=0&amp;amp;svr=0&amp;amp;tor=0&quot; rel=&quot;nofollow&quot;&gt;Weatherunderground&lt;/a&gt; at 10:30 PM Central time, May 8 2011 has odd patterns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these patterns real? Perhaps caused by large scale convection over cities? Or are they artifacts of radar placement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the image that I am referring to, where green indicates light and yellow moderate rain:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/xlvW8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2809" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-09T19:59:28.670" LastActivityDate="2011-05-09T19:59:28.670" Title="Is this a weather phenomenon or an instrumental artifact?" Tags="&lt;atmosphere&gt;&lt;weather&gt;&lt;convection&gt;&lt;radar&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="9614" PostTypeId="2" ParentId="9543" CreationDate="2011-05-09T03:42:33.157" Score="1" ViewCount="" Body="&lt;p&gt;Unruh effect, coordinate dependence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Creation and annihilation operators are coordinate-dependent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ground state in one is a squeezed state in another.&lt;/p&gt;&#xA;" OwnerUserId="3466" LastActivityDate="2011-05-09T03:42:33.157" />
  <row Id="9615" PostTypeId="1" AcceptedAnswerId="9617" CreationDate="2011-05-09T05:30:48.823" Score="6" ViewCount="243" Body="&lt;p&gt;I was reading some information about the 2003 power blackout in the Northeastern US. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Beginning early in the afternoon of August 14, 2003 big transmission lines began to fail in First Energy's operating area, several because the utility had not kept up with tree-trimming, so that &lt;strong&gt;as heavily loaded lines heated up, they sagged into brush and shorted out&lt;/strong&gt;. As one went down, the next would become too loaded, sag still more, and short, and so on. All that, the result of a serious infringement of operating standards and no small matter in its own right,&#xA;  would have remained a local problem if First Energy and the midwestern power regulator had quickly recognized what was going on and had promptly cut service to enough customers to keep the whole system from getting overloaded. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;From &quot;&lt;a href=&quot;http://ic.galegroup.com:80/ic/ovic/ViewpointsDetailsPage%20/MLA7CitationWindow?displayGroupName=Viewpoints&amp;amp;prodId=&amp;amp;action=2&amp;amp;catId=&amp;amp;%20documentId=GALE%7CEJ3010502219&amp;amp;userGroupName=s8405248&amp;amp;%20jsid=67473c81a334b747667195a4e010fca&quot; rel=&quot;nofollow&quot;&gt;A Coal Power Ban Is a Necessity.&lt;/a&gt;&quot; Coal. Ed. Michael Logan. 5 Aug. 2010&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about overheating causes power lines to sag? I understand that increasing the load on the line will increase the line's temperature due to the resistance in the cables. Why does this increased heat cause the lines to sag though?&lt;/p&gt;&#xA;" OwnerUserId="302" LastActivityDate="2011-05-09T18:09:10.197" Title="Why do power lines sag when they are heated up?" Tags="&lt;electricity&gt;&lt;heat&gt;&lt;power&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="9616" PostTypeId="1" CreationDate="2011-05-09T05:44:47.840" Score="5" ViewCount="425" Body="&lt;p&gt;After watching some ants in my garden today, and then looking at &lt;a href=&quot;http://learn.genetics.utah.edu/content/begin/cells/scale/&quot; rel=&quot;nofollow&quot;&gt;this very illuminating demonstration&lt;/a&gt;, I got to wondering, about what they would see.  Not specifically ants (I understand their eyesight is quite poor), but similarly small, or even smaller creatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess I'm asking more about the nature of light and how photons are reflected off very small surfaces.  Would a very small creature, like say, an ant, with vision, be able to see something as small as a single e. coli bacterium? or a virus? Would their world 'look' the same as ours or does the viewers relative size have a bearing on the quality of their perception?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And additionally beyond the realm of reality, if I could shrink myself down to the size of a bacterium, could I see atoms?&lt;/p&gt;&#xA;" OwnerUserId="3190" LastActivityDate="2011-05-10T18:05:32.770" Title="What do ants see?" Tags="&lt;light&gt;&lt;photons&gt;&lt;microscopy&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="9617" PostTypeId="2" ParentId="9615" CreationDate="2011-05-09T05:52:56.920" Score="12" ViewCount="" Body="&lt;p&gt;The conductor material (steel, aluminum) expands when heated. When the temperature increases, the length of the power line between two towers increases due to thermal expansion, and the line sags because of the increased slack.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditorDisplayName="" LastEditDate="2011-05-09T18:09:10.197" LastActivityDate="2011-05-09T18:09:10.197" CommentCount="0" />
  <row Id="9618" PostTypeId="2" ParentId="9615" CreationDate="2011-05-09T05:53:08.680" Score="7" ViewCount="" Body="&lt;p&gt;There is &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_expansion&quot; rel=&quot;nofollow&quot;&gt;thermal expansion&lt;/a&gt; of metals, and in a power line the length will become greater than the installation one. Gravity takes up the slack and lowers the lowest point of the line. The shape taken by suspended strings/cables  with only their own weight is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Catenary&quot; rel=&quot;nofollow&quot;&gt;catenary&lt;/a&gt;, and the length depends on the constants of the solution ( &lt;em&gt;alternative analysis&lt;/em&gt; in the catenary link). &lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-09T05:53:08.680" />
  <row Id="9619" PostTypeId="2" ParentId="9616" CreationDate="2011-05-09T06:40:24.473" Score="6" ViewCount="" Body="&lt;p&gt;Ants only have low-resolution eyes, aside from three ocelli - simple eyes - that only detect an overall light level and polarization, see&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ant#Morphology&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Ant#Morphology&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Their ability to see details - small objects and their features - is much worse than for vertebrates like us. To suggest that animals - especially as primitive animals as ants - could see bacteria is preposterous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wavelength of visible light is about half a micron - which is also the size of many bacteria. So you can't see anything inside bacteria with the visible light, not even with cutting-edge technology. To see more detailed objects, you have to switch to X-rays or electrons and create better microscopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's even more unrealistic to propose that one - or even an ant - could see an atom (which is 10,000 times smaller than a bacterium) through visible light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't just scale things up and down. The world is not invariant under scale transformations, we say. Different length scales see different kinds of physical phenomena and different physical objects. The atom of a given kind has always the same size and you can't scale it up. Moreover, you didn't even do the scaling properly because you didn't scale the wavelength of the light. Also, vision with detailed resolution requires some &quot;large enough circuits&quot; to deal with the information etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, this holds even for accelerators. The LHC is our best &quot;microscope&quot; that can see distances shorter than $10^{-19}$ meters - but to do so, it requires tunnels with the best magnets that are 27 kilometers long. Objects as small as ants can't see with this good resolution, and even if they could, they couldn't deal with the huge amount of information that their eyes would be giving to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Large enough animals - e.g. mammals - see the world much like we do. There are well-known differences between the colors that different mammals are sensitive to. Dogs are, for instance, partly color-blind, relatively to what we can do.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-09T09:41:34.713" LastActivityDate="2011-05-09T09:41:34.713" CommentCount="13" />
  <row Id="9621" PostTypeId="1" AcceptedAnswerId="9622" CreationDate="2011-05-09T06:51:33.453" Score="1" ViewCount="433" Body="&lt;p&gt;In other words: which physics experiment requires to know Pi with the highest precision?&lt;/p&gt;&#xA;" OwnerUserId="3469" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-05-09T16:44:36.883" LastActivityDate="2011-05-10T12:59:52.237" Title="How many digits of Pi are required in physics?" Tags="&lt;mathematical-physics&gt;&lt;experimental-physics&gt;&lt;mathematics&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="9622" PostTypeId="2" ParentId="9621" CreationDate="2011-05-09T07:42:57.530" Score="4" ViewCount="" Body="&lt;p&gt;Pi is very far from being the only number we need in physics. Typical theoretical predictions depend on many other measured and calculated (or both) numbers besides pi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, it is true that one needs to substitute the right value of pi to get the right predictions. Therefore, the right answer to your question is the most accurately experimental verified theoretical prediction we have in physics as of today, namely the anomalous magnetic dipole moment of the electron&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Anomalous_magnetic_dipole_moment&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Anomalous_magnetic_dipole_moment&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some natural units, the magnetic moment of the electron is expressed as a g-factor which is somewhat higher than two. Experimentally,&#xA;$$\frac g2 = 1.00115965218111 \pm 0.00000000000074$$&#xA;Theoretically, $g/2$ may be written as&#xA;$$\frac g2 = 1+\frac{\alpha}{2\pi} + \dots$$&#xA;where the $\alpha/2\pi$ first subleading term was obtained by Schwinger in 1948 and many other, smaller terms are known today. The theoretical prediction agrees with the experimental measurement within the tiny error margin; the theoretical uncertainty contains the effect of new species of virtual particles with the masses and couplings that have not yet been ruled out. This requires, among many and many other things, to substitute the correct value of $\pi$ in Schwinger's leading correction $\alpha/2\pi$. You need to know 9-10 decimal points of $\pi$ to make this correction right within the experimental error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in practice, $\pi\approx 3.141592654$ would be OK everywhere in the part of physics that is testable. However, theoretical physicists of course often need to make calculations more accurately if not analytically, to figure out what's really happening with the formulae.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-09T07:42:57.530" CommentCount="3" />
  <row Id="9623" PostTypeId="2" ParentId="9621" CreationDate="2011-05-09T08:09:50.260" Score="2" ViewCount="" Body="&lt;p&gt;In principle you will &lt;strong&gt;never reach the accuracy&lt;/strong&gt; of the numerical value of $\pi$ in an experiment, it is much &lt;strong&gt;more important as an analytical tool&lt;/strong&gt;. The part of the question concerning the most accurate experiment was given by &lt;a href=&quot;http://physics.stackexchange.com/questions/9621/how-many-digits-of-pi-are-required-in-physics/9622#9622&quot;&gt;Luboš&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, just to give a little thing about the importance of $\pi$ as the area of a unit circle and also the ratio of a circle's circumference to diameter,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ToHgW.gif&quot; alt=&quot;wiki-pi&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(from the corresponding &lt;a href=&quot;http://en.wikipedia.org/wiki/Pi&quot; rel=&quot;nofollow&quot;&gt;wiki-article&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One very famous examples are &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;Fourier transforms&lt;/a&gt; where you ask a periodic function to be represented by $\sin$ and $\cos$-terms,&#xA;$$\mathcal{F}(f)(\xi)=\int_{-\infty}^{\infty}f(x)e^{2\pi\mathrm{i} \xi x}dx\ \mathrm{with}\  e^{\mathrm{i}\varphi}=\cos(\varphi) + \mathrm{i}\sin(\varphi)\ .$$&#xA;This is linked to $\pi$ because of the periodicity of the circle, $$r_1(\varphi) = r_2(\varphi + 2\pi)\ \mathrm{with}\ r_i\in \mathrm{circle}\ .$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;strong&gt;much more&lt;/strong&gt; things to add as the underlying &lt;strong&gt;group of rotations&lt;/strong&gt; is $U(1)$ which is in turn the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauge_theory&quot; rel=&quot;nofollow&quot;&gt;gauge group&lt;/a&gt; for &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_electrodynamics&quot; rel=&quot;nofollow&quot;&gt;(quantum)electrodynamics&lt;/a&gt; explaining why you will almost certainly find the most prominent examples of results linked to the value of $\pi$ in this field.&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-05-10T12:59:52.237" LastActivityDate="2011-05-10T12:59:52.237" CommentCount="2" />
  <row Id="9624" PostTypeId="2" ParentId="9551" CreationDate="2011-05-09T08:26:50.147" Score="1" ViewCount="" Body="&lt;p&gt;Meta:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that a lot of quantum mechanics can be taught and understood without much knowledge of the mathematical foundations, and usually it is. Since QM is a mandatory class at many faculties that future experimental physicists have to attend, too, this also makes sense. But for future theoretical and mathematical physicists, it may pay off to learn a little bit about the math, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little anecdote: John von Neumann once said to Werner Heisenberg that mathematicians should be grateful for QM, because it lead to the invention of a lot of beautiful mathematics, but that mathematicians payed back by clarifying e.g. the difference of selfadjoint and symmetic operators. Heisenberg asked back: &quot;What is the difference?&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose you want to calculte exp(A). Why don't you define exp(A):=1+A+1/2 A^2 + ... and require convergence with respect to the operator norm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's correct. The benefit of the spectral theorem is that you can define f(A) for any selfadjoint (or more generally, normal) operator for any bounded Borel function. This comes in handy in many proofs in operator theory.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In addition to that I've heard that the spectral theorem gives a full description of all self-adjoint operators. Now why is that the case? I mean okay..there's a one to one correspondence between self-adjoint operators and spectral measures..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's correct, too. Spectral measures are much much simpler objects than selfadoint operators, that's why. Futhermore, you can use the spectral theorem to proove that every selfadjoint operator is unitarily equivalent to a multiplication operator (multiply f(x) by x). From an abstract viewpoint, this is a very satisfactory characterization. It does not help much for concrete calcuations in QM, though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW: On a more advanced level, you'll need to understand the spectral theorem to understand what a mass gap is in &lt;a href=&quot;http://www.claymath.org/millennium/Yang-Mills_Theory/&quot; rel=&quot;nofollow&quot;&gt;Yang-Mills Theory&lt;/a&gt; (millenium problem). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: In QFT in Minkowski-Spacetime, one usually assumes that there is a continuous representation of the Poincaré group, especially of the commutative subgroup of translations, on the Hilbert space that contains all physical states. The operators that form the representation have a common spectral measure, this is an application of the SNAG-theorem. The support of this spectral measure is bounded away from zero, that's the definition of the mass gap.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-05-09T08:26:50.147" />
  <row Id="9625" PostTypeId="1" AcceptedAnswerId="9626" CreationDate="2011-05-09T09:02:33.640" Score="2" ViewCount="104" Body="&lt;p&gt;Quite new to all this - simple question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A ball starting from rest rolls down a hill and reaches 10m/s at the bottom.  If the same ball starts rolling down the hill from an initial velocity of 5m/s, will it's velocity at the bottom be:&#xA;(A) Less than 15m/s;&#xA;(B) Equal to 15m/s;&#xA;(C) Greater than 15m/s.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer I have marked in the book this question comes from says (A).  However, I am unable to reason to that conclusion from the way the question is stated.  Here is my reasoning, please point out where it is wrong.  If the same ball (the same mass) traveses the same distance down this same hill (the same change in height), the gravitational potential energy must be the same at the point on top of the hill, regardless of its initial velocity.  Therefore, on rolling down the hill, by the time it reaches the bottom, this potential energy would have transferred to kinetic energy - indicating (B)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tim.&lt;/p&gt;&#xA;" OwnerUserId="3350" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-09T09:05:51.063" LastActivityDate="2011-05-10T17:14:15.383" Title="Mechanical energy problem - please help" Tags="&lt;homework&gt;&lt;newtonian-mechanics&gt;&lt;energy-conservation&gt;" AnswerCount="2" />
  <row Id="9626" PostTypeId="2" ParentId="9625" CreationDate="2011-05-09T09:30:23.247" Score="2" ViewCount="" Body="&lt;p&gt;Dear nulliusinverba, this server usually tries to avoid solving homework problems but let me give you a hint where your reasoning goes awry. You have assumed that the kinetic energy is proportional to $v$, haven't you? That's why you could think that by adding a fixed amount of kinetic energy, you always increase the velocity $v$ by the same amount, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you really think that the kinetic energy is proportional to $v$? And if it happens not to be, don't you think that the velocity change obtained by the same increase of the kinetic energy could depend on the initial velocity? In fact, shouldn't you be able to calculate the final velocity exactly with these hints?&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-09T09:30:23.247" CommentCount="4" />
  <row Id="9627" PostTypeId="2" ParentId="9609" CreationDate="2011-05-09T10:01:26.410" Score="2" ViewCount="" Body="&lt;p&gt;First, &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutrino#Analogies_to_the_index_of_refraction_and_the_MSW_effect&quot; rel=&quot;nofollow&quot;&gt;check this reference on Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it is generally true that the &quot;speed&quot; (or, more accurately, the dispersion relation) of &lt;strong&gt;any&lt;/strong&gt; particle is affected by a medium, where it travels. Well, of course, if the particle interacts with the medium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For neutrinos the &quot;slowing down&quot; itself is absolutely negligible even in very dense media. What is important is that interaction of the electron neutrino with ordinary matter is much stronger, so it affects the patterns of neutrino oscillations -- the effect is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/MSW_effect&quot; rel=&quot;nofollow&quot;&gt;MSW effect&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, this is not particularly related to supernovae. The idea behind interest in supernovae is that during an explosion there are a lot of neutrinos so one must account for the &quot;neutrino matter&quot; and its effect on oscillations as well. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-05-09T10:01:26.410" CommentCount="1" />
  <row Id="9628" PostTypeId="1" AcceptedAnswerId="9629" CreationDate="2011-05-09T10:15:21.570" Score="1" ViewCount="215" Body="&lt;p&gt;Recently I've had several non-physicist friends ask me, independently of each other, about an experiment where two collinear laser beams destructively interfere along a certain length. Everybody wants to know &quot;where does the energy go?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answering that question is not the problem, but I would be more convincing if I knew what experiment they were talking about! None of them can recall where they read it, but it seems to have made the rounds of popular science websites sometime in the last year. I am especially surprised that this has been published &lt;em&gt;now&lt;/em&gt;, since similar experiments have been done for decades, so I'm guessing there must be some other twist to the experiment that didn't register in my friends' memories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody point me towards a published paper or a popular science article?&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2011-06-21T15:57:31.673" Title="Recently publicized experiment on destructive interference between two laser beams" Tags="&lt;optics&gt;&lt;waves&gt;&lt;laser&gt;&lt;interference&gt;&lt;literature&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="9629" PostTypeId="2" ParentId="9628" CreationDate="2011-05-09T10:33:31.787" Score="3" ViewCount="" Body="&lt;p&gt;If we have the same popular press in mind, the official name of this device - informally known as anti-laser - is a &quot;coherent perfect absorber&quot;. See&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Coherent_perfect_absorber&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Coherent_perfect_absorber&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It was proposed in early 2010 by A. Douglas Stone and collaborators:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1003.4968&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1003.4968&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://www.pqeconference.com/pqe2011/abstractd/170.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.pqeconference.com/pqe2011/abstractd/170.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It was already observed in late 2010, see e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.opticsinfobase.org/abstract.cfm?URI=FiO-2010-PDPC11&quot; rel=&quot;nofollow&quot;&gt;http://www.opticsinfobase.org/abstract.cfm?URI=FiO-2010-PDPC11&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The anti-laser is just a time-reversal of a laser, so whatever problem with energy conservation you could see exists for both devices. Obviously, there is no problem - the energy of light is being taken from, or deposited to, the atomic excitation energies in the cavity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are roughly 10 papers about this subject only&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://scholar.google.com/scholar?q=%22coherent+perfect+absorber%22&quot; rel=&quot;nofollow&quot;&gt;http://scholar.google.com/scholar?q=%22coherent+perfect+absorber%22&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so one shouldn't buy some popular media's hype suggesting that it's one of the hottest things in physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your question were not about anti-laser but about ordinary interference between two laser beams, well, I would leave it as no comment. Of course that the energy is conserved at the end. One may get destructive interference in a big region but this is always compensated by the existence of constructive interference in other regions - e.g. on the boundary of the beams. Maxwell's equations may be showed to imply energy conservation so the total energy can never get lost.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-09T10:33:31.787" CommentCount="1" />
  <row Id="9630" PostTypeId="2" ParentId="9628" CreationDate="2011-05-09T10:43:55.453" Score="0" ViewCount="" Body="&lt;p&gt;The electromagnetic wave energy is an integral notion. The energy space density may depend on particular place but it is not the total energy. As Lubosh says, according to the Maxwell's equations the total energy is conserved in absence of charges. If it is not conserved, then the filed is not governed with the Maxwell's equations.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-05-09T10:43:55.453" />
  <row Id="9631" PostTypeId="2" ParentId="9621" CreationDate="2011-05-09T12:11:40.173" Score="5" ViewCount="" Body="&lt;p&gt;39 digits of pi is enough to calculate circumference of visible universe to a margin of error equal to the width of a proton ( unverified computation, according to &lt;a href=&quot;http://www.guardian.co.uk/science/blog/2011/mar/14/pi-day&quot; rel=&quot;nofollow&quot;&gt;http://www.guardian.co.uk/science/blog/2011/mar/14/pi-day&lt;/a&gt; CordwainerBird ) .&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-05-09T12:11:40.173" CommentCount="1" />
  <row Id="9632" PostTypeId="2" ParentId="9616" CreationDate="2011-05-09T12:29:49.530" Score="3" ViewCount="" Body="&lt;p&gt;The ant world is ordered far more by chemical reception and pheromones than by vision.  Ants produce an array of such chemicals which act as signals.  They also sense other chemicals in their environment, and as what might be called a “super organism” they have some collective map, a chemical map, of the terrain they inhabit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ants have compound eyes, and they are pretty small.  For the most part their purpose is to sense sudden changes in light levels.  An ant which perceives such then gets a signal that some predator might be present and so getting out of there is in order.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some species of baceteria have opsin molecules that are photoactive.  So the reception of photons can result in changes in molecular pathway activity.  The rhodopsin molecule in our eyes or retina has two conformal states for reception and nonreception of a photon.  The energy of the photon changes the shape of the molecule and this then acts to initiate a GTP molecular pathway which is amplified ultimately into a neural action potential.  Rhodopsin is one form of opsin molecules, which in their general classification overlap with photosynthetic molecules in some bacteria as well.  However, the bacilli do not form any sort of image of anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for a bacillus to “see” an atom they would need to detect gamma rays.  Gamma rays are largely outside the EM spectrum available to biological systems.  In fact they are lethal. &lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-05-09T12:29:49.530" CommentCount="2" />
  <row Id="9633" PostTypeId="2" ParentId="9628" CreationDate="2011-05-09T13:55:29.950" Score="-1" ViewCount="" Body="&lt;p&gt;EDIT add&lt;br&gt;&#xA;The papers are correctly posted in Lubos answer and, if this is all you need, than you can skip this answer&lt;br&gt;&#xA;EDIT add end  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two coherent light beams are counter propagating in perfect phase opposition the Poynting vectors are of equal magnitude and directly opposed. The total field is null everywhere those conditions apply and energy is null. This comply with Maxwell equations and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Interference_%28wave_propagation%29&quot; rel=&quot;nofollow&quot;&gt;Superposition Principle&lt;/a&gt;.  The issue that Energy+Energy=0 derives from Superposition Principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that the &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nonradiation_condition&quot; rel=&quot;nofollow&quot;&gt;Nonradiation_condition&lt;/a&gt;&lt;/strong&gt; uses the same superposition principle to cancel the energy in a region outside a finite one. Paul Ehrenfest, Goedeck, Haus, and Douglas Pinnow.&#xA;And goes like this (from the same WP-link):&lt;br&gt;&#xA;&quot;a distribution of accelerated charges &lt;em&gt;will radiate if and only if&lt;/em&gt; it has Fourier components synchronous with waves traveling at the speed of light.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that the charge do not have a sensor to see if the conditions of radiation are met or not met. The charge always radiate electrostatic field and due to the relativity of motion the radiated pattern can be constructive or not. When it is desctructive (as in the sense of the nonradiation condition) we measure and obtain 0 energy outside the region but the particles always radiate energy at the same rate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motion of the particle induce the waves in the radiated electrostatic field.&lt;br&gt;&#xA;And when it cancels: energy is not there. 'Cancel', 'Vanish', 'Destroyed'.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Nonradiation condition, and also the anti-laser experiment work the same way in the vacuum where there are no absorbers, no heat, no particles, ...    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm clearly stating that the only radiation of any charge is originated only on its continuos electrostatic field.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a magical thinking when we say that the particle radiates if it is in motion or other particles in the vicinity are in motion. To be physical It can not be like this, simply.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explore visually this &lt;a href=&quot;http://www-xfel.spring8.or.jp&quot; rel=&quot;nofollow&quot;&gt;Radiation2D.exe&lt;/a&gt; and its formulation &lt;a href=&quot;http://accelconf.web.cern.ch/accelconf/e02/PAPERS/WEPRI038.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;NEW MATHEMATICAL METHOD FOR RADIATION FIELD OF MOVING CHARGE&quot;&lt;/a&gt; by T. Shintake.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how we like to complicate the radiation issue see : &#xA;&lt;a href=&quot;http://physics-quest.org/Magnetism_from_ElectroStatics_and_SR.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;The simplest, and the full derivation of Magnetism as a Relativistic side efect of ElectroStatics&quot;&lt;/a&gt; by Hans de Vries;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I've explored this issue &lt;a href=&quot;http://physics.stackexchange.com/questions/5743/anti-laser-how-sure-we-are-that-energy-is-transported&quot;&gt;PSE-here&lt;/a&gt;, but here I've added the nonradiation condition  argument)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT add: time-reversal and information loss&lt;br&gt;&#xA; The experiment is not time-simetric because after the cancelation any information that could be equally encoded in both of the beams are definitelly lost.&lt;br&gt;&#xA;EDIT ad end  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-05-09T16:13:05.587" LastActivityDate="2011-05-09T16:13:05.587" CommentCount="3" />
  <row Id="9635" PostTypeId="1" CreationDate="2011-05-09T14:22:54.347" Score="8" ViewCount="290" Body="&lt;p&gt;The 115 GeV ATLAS Higgs with enhanced diphoton decays has &lt;a href=&quot;http://motls.blogspot.com/2011/05/atlas-at-94pb-diphoton-hysteria-goes.html&quot; rel=&quot;nofollow&quot;&gt;gone away&lt;/a&gt; but there are several other recent tantalizing hints relevant for particle physics, namely&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://motls.blogspot.com/2011/05/cogent-sees-seasons-and-maybe-dark.html&quot; rel=&quot;nofollow&quot;&gt;CoGeNT's 7-8 GeV dark matter particle&lt;/a&gt; that seems to support DAMA's and other signals&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://motls.blogspot.com/2011/04/d0-3-sigma-evidence-for-325-gev-top.html&quot; rel=&quot;nofollow&quot;&gt;D0's fourth-generation top prime quark&lt;/a&gt; that may be there at 325 GeV if a 3-sigma signal is right&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://motls.blogspot.com/2011/04/fermilab-cdf-new-force-press-conference.html&quot; rel=&quot;nofollow&quot;&gt;CDF's &quot;new&quot; force Z prime boson&lt;/a&gt; at 144 GeV which could be there if another 3-sigma bump is real&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://motls.blogspot.com/2011/04/fermilab-cdf-new-force-press-conference.html&quot; rel=&quot;nofollow&quot;&gt;Tevatron's top-antitop asymmetry&lt;/a&gt; above 400 GeV which seems very large&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://motls.blogspot.com/2011/01/cdf-sees-another-top-quark-related.html&quot; rel=&quot;nofollow&quot;&gt;A related CDF top-quark anomaly&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and maybe others I missed. My question is simple:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there some sensible theoretical basis (e.g. paper) that would simultaneously explain at least two of the observations above if they were real?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-20T22:45:41.660" Title="Correlation between outstanding hints in experimental particle physics" Tags="&lt;particle-physics&gt;&lt;large-hadron-collider&gt;&lt;dark-matter&gt;&lt;tevatron&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="4" />
  <row Id="9636" PostTypeId="1" AcceptedAnswerId="9638" CreationDate="2011-05-09T15:17:23.577" Score="5" ViewCount="258" Body="&lt;p&gt;This posting is regarding the recent confirmation of the DAMA results that might be due to underlying differences in proton and neutron cross section with the dark matter particles, which reflect on the differences between Xenon and Germanium detectors&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in the early days of particle physics, neutrinos were discovered as missing momentum in decay events&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my question is the following: &lt;strong&gt;How did the current accelerators missed a dark matter product of 7GeV?&lt;/strong&gt; possible answers i think:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) does by any chance theory predict that these relatively low energy particles will not be created with normal particle collisions in our current multi-TeV range?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) does the current complexity of analysis of TeV-magnitude collision products just cannot detect such missing momenta signals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) there are no dark matter particles in such low energy range, otherwise we would have already spotted them in particle accelerators&lt;/p&gt;&#xA;&#xA;&lt;p&gt;references:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2011/05/cogent-sees-seasons-and-maybe-dark.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2011/05/cogent-sees-seasons-and-maybe-dark.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://sixwoffers.blogspot.com/2011/05/second-experiment-hints-at-seasonal.html&quot; rel=&quot;nofollow&quot;&gt;http://sixwoffers.blogspot.com/2011/05/second-experiment-hints-at-seasonal.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-05-09T21:08:35.573" LastActivityDate="2011-05-09T21:08:35.573" Title="7 GeV dark matter particle: how particle accelerators missed it?" Tags="&lt;particle-physics&gt;&lt;dark-matter&gt;&lt;particle-detectors&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="9638" PostTypeId="2" ParentId="9636" CreationDate="2011-05-09T15:57:15.753" Score="3" ViewCount="" Body="&lt;p&gt;If dark matter interacts only gravitationally, then the cross section for producing it in the e+e- machines is inherently too low to be detected. I am discussing e+e- machines because those are the ones that can give a closed enough system to be able to detect missing mass and energy cleanly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://pdg.lbl.gov/2010/reviews/rpp2010-rev-cross-section-plots.pdf&quot; rel=&quot;nofollow&quot;&gt;cross section&lt;/a&gt; at the Y ( about 10GeV in mass)is something like 10^-2 millibarn. Now the coupling constant in front of the calculations (squared) is the electromagnetic one, which is orders of magnitude larger than the gravitational one. This will affect  to practically zero both the magnitude and the width of any reaction producing the hypothetical 7 Gev particle, either in some pair production, or associated production. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was some talk of finding more positrons than electrons associated with the measurements reported. In that case there exists a coupling between electromagnetic fields and these proposed particles, but a specific model would be needed to say at what level the production of these would be excluded by the existing world data from e+e- machines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are limits given assuming super symmetry is the valid theory. See &lt;a href=&quot;http://cdsweb.cern.ch/record/601840/files/thesis-2003-003.pdf&quot; rel=&quot;nofollow&quot;&gt;this ALEPH&lt;/a&gt; thesis which gives limits over 40 GeV .&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-05-09T16:17:32.587" LastActivityDate="2011-05-09T16:17:32.587" CommentCount="6" />
  <row Id="9639" PostTypeId="2" ParentId="9612" CreationDate="2011-05-09T16:18:21.683" Score="2" ViewCount="" Body="&lt;p&gt;No proposed urban rainfall effect is dominant enough to cause that picture. But limited radar range, with radars being located in larger cities would easily explain it. The urban rainfall enhancement effect would have to be pretty extreme for it to be otherwise.&lt;/p&gt;&#xA;" OwnerUserId="377" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-09T19:53:01.883" LastActivityDate="2011-05-09T19:53:01.883" CommentCount="1" />
  <row Id="9640" PostTypeId="2" ParentId="9621" CreationDate="2011-05-09T16:31:31.763" Score="4" ViewCount="" Body="&lt;p&gt;16 digits, for converting frequencies from Hz to angular frequency.  Frequencies can now be measured with a precision approaching 1 part in 10^16, so dealing with those numbers would require knowing Pi to 16 digits or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this really &quot;which physics experiment REQUIRES to know Pi with the highest precision&quot;?  A purist (or theorist) might disagree that this example doesn't count, and I'd agree that there's nothing fundamental here, just conversion from one convention to another.  But from a practical point of view, it's something people do in physics and it requires knowing Pi to 16 digits.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-05-09T16:31:31.763" />
  <row Id="9641" PostTypeId="1" AcceptedAnswerId="9642" CreationDate="2011-05-09T17:23:58.287" Score="1" ViewCount="155" Body="&lt;p&gt;What is the physical meaning of the $c$-numbers $Q, P\in \mathbb{R}$ in the exponent of the&#xA;Weyl system  $\exp\left[\frac{i}{\hbar} Q \hat{p}\right]$ and $\exp\left[\frac{i}{\hbar}P\hat{q}\right]$? Here $\hat{q},\hat{p}$ are operators of position and momentum with canonical commutation relation $[\hat{q},\hat{p}]=i\hbar$.&lt;/p&gt;&#xA;" OwnerUserId="1901" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-09T18:50:54.683" LastActivityDate="2011-05-09T18:50:54.683" Title="Weyl exponential form of the Canonical Commutation Relations" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" />
  <row Id="9642" PostTypeId="2" ParentId="9641" CreationDate="2011-05-09T18:43:09.803" Score="0" ViewCount="" Body="&lt;p&gt;There's a considerable danger of oversimplification here, but I find it helpful to think of the construction $e^{i\lambda \hat q}$ as an operator that generates &lt;a href=&quot;http://en.wikipedia.org/wiki/Characteristic_function_%28probability_theory%29&quot; rel=&quot;nofollow&quot;&gt;the characteristic function&lt;/a&gt; of a probability distribution in a particular state. Being a little free with mathematical details, we can write the probability density of observing the value $q$ in a vector state $\left|\psi\right&amp;gt;$ as $\mathsf{Pr}(q)=\left&amp;lt;\psi\right|\delta(q-\hat q)\left|\psi\right&amp;gt;$, for which the characteristic function is the fourier transform&#xA;$$C(\lambda)=\int\left&amp;lt;\psi\right|\delta(q-\hat q)\left|\psi\right&amp;gt;e^{i\lambda q}\mathrm{d}q=&#xD;&#xA;      \left&amp;lt;\psi\right|e^{i\lambda \hat q}\left|\psi\right&amp;gt;.$$&#xA;We can inverse fourier transform this back to a probability density. What is commonly labeled $Q$, as you have here, is more helpfully labeled in a way that emphasizes its difference from the position operator and the values that position measurements may take. Mathematically, $\lambda$ is a linear dual of the position. It's very clear what a fourier transform of a probability density is algebraically, but perhaps not so clear what its &lt;em&gt;physical&lt;/em&gt; meaning is. It's perhaps best to think of it as a formal device that uses $\lambda$ for keeping track of all the moments of the probabilities of different measurement results, which allows us to say that $\lambda^n$ is associated with the $n$-th moment. [You could do a lot worse, if you want to understand generating functions, though this is slightly idiosyncratic suggestion on my part, to read John Baez's recent series on Network Theory with an open mind; a Google search for &lt;a href=&quot;http://www.google.com/search?q=%22network%20theory%20%28part%22%20baez&quot; rel=&quot;nofollow&quot;&gt;&quot;network theory (part&quot; baez&lt;/a&gt; finds them. It may well twist your head a little, however, so if you want quick understanding this may not be for you.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything is essentially the same for the momentum operator taken alone, but if we introduce both $\hat q$ and $\hat p$, so that we consider an object such as&#xA;$$\tilde W(\lambda,\mu)=\int\left&amp;lt;\psi\right|\delta(q-\hat q)\delta(p-\hat p)\left|\psi\right&amp;gt;&#xD;&#xA;        e^{i\lambda q+i\mu p}\mathrm{d}q\mathrm{d}p=&#xD;&#xA;      \left&amp;lt;\psi\right|e^{i\lambda \hat q+i\mu\hat p}\left|\psi\right&amp;gt;,$$&#xA;and then inverse fourier transform this object, we obtain negative &quot;probability densities&quot; for some values of $p$ and $q$. This is the Wigner function, about which &lt;a href=&quot;http://en.wikipedia.org/wiki/Wigner_function&quot; rel=&quot;nofollow&quot;&gt;much has been written&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-05-09T18:43:09.803" CommentCount="1" />
  <row Id="9643" PostTypeId="1" CreationDate="2011-05-09T19:21:38.147" Score="4" ViewCount="196" Body="&lt;p&gt;How is the shape of the blades of an air fan determined? Trial and error, or is there a theory behind it? What are they trying to maximize, volume of air dislocated per rotation?&lt;/p&gt;&#xA;" OwnerUserId="668" LastActivityDate="2011-05-10T15:33:14.597" Title="Shape of Fan Blades" Tags="&lt;mechanics&gt;&lt;engineering&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="9644" PostTypeId="2" ParentId="9601" CreationDate="2011-05-09T19:54:56.757" Score="2" ViewCount="" Body="&lt;p&gt;I think this is a typo in Morse &amp;amp; Feshbach Methods of Theoretical Physics. The correct expression is $-\frac{1}{6}r^2\nabla ^2\psi$ or $-\frac{1}{6}(dx^2+dy^2+dz^2)\nabla^2 \psi$.&lt;/p&gt;&#xA;" OwnerUserId="3064" LastActivityDate="2011-05-09T19:54:56.757" />
  <row Id="9645" PostTypeId="1" AcceptedAnswerId="9654" CreationDate="2011-05-09T22:34:13.563" Score="1" ViewCount="83" Body="&lt;p&gt;it is commonly argued that the equivalence principle implies that everything must fall &lt;em&gt;locally&lt;/em&gt; in the same direction, because any local variation of accelerations in a small enough neighbourhood is equivalent to an inertial force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the purpose of this question is to understand, what are the bounds for a frame to be considered local, when the space might be compactified, and some of the macroscopic asymmetries of fields we see are explainable by symmetries in the inner dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, consider the following argument: &lt;em&gt;antimatter cannot fall away from matter, otherwise a local frame falling toward matter will not see local antimatter falling inertially&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;don't get me wrong, there might be good reasons why in the particular case of anti-matter, it will fall exactly like regular matter (some of those might be found on &lt;a href=&quot;http://physics.stackexchange.com/questions/9375/anti-matter-repelled-by-gravity-is-it-a-serious-hypothesis/9377#9377&quot;&gt;this answer&lt;/a&gt;), however, i'm concerned about this particular argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a compactified space, what we mean with &lt;em&gt;locally&lt;/em&gt; in the context of macroscopic observations needs to be refined; for instance consider the scenario where antimatter (or for the purpose of avoiding controversy, lets call it $matter \dagger$ ) falls gravitationally away from regular matter; it seems to me that the principle of equivalence still holds if what we mean as a local neighbourhood means actually:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lbrace$ usual local neighbourhood in macroscopic spacetime $\rbrace$ $\times$ $\lbrace$ the subset of the compactification space where regular matter lives $\rbrace$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in this scenario, $matter \dagger$ falls inertially in its own neighbourhood given by removing the regular matter sector from the local neighbourhood in the whole space (macroscopic plus compactified)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I even made a ugly picture of the so called scenario:&lt;img src=&quot;http://i.stack.imgur.com/Flz1u.jpg&quot; alt=&quot;two particles in the same macroscopic local neighbourhood see widely different curvatures, which macroscopically seems to contradict EP&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 1: &lt;strong&gt;Are there known reasons why such reinterpretations of the equivalence principle, taking into account the additional dimensions, are flawed?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 2: &lt;strong&gt;What does the original Kaluza-Klein theory has to say about how and under what forces the electric negative and positive charges move?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry for the rather crude arguments, hope the intent is rather clear though&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-05-09T22:46:39.710" LastActivityDate="2011-05-10T04:24:25.240" Title="equivalence principle and nontrivial compactifications" Tags="&lt;equivalence-principle&gt;&lt;kaluza-klein&gt;&lt;compactification&gt;&lt;inertial-frames&gt;" AnswerCount="1" />
  <row Id="9646" PostTypeId="2" ParentId="9601" CreationDate="2011-05-09T22:54:14.863" Score="2" ViewCount="" Body="&lt;p&gt;Formulated this way, $-\frac16\,r^2\,\nabla^2\psi$ is indeed the correct result. While it is certainly more elegant to express such things by means of differential forms rather than plain functions with finite $r$, the result as given in the book is not a meaningful differential form at all, since&#xA;$$&#xD;&#xA;(\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z)^2&#xD;&#xA;= \mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z\wedge\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z&#xD;&#xA;$$&#xA;$$&#xD;&#xA;= (-1)(-1)\mathrm{d}x\wedge\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z\wedge\mathrm{d}y\wedge\mathrm{d}z = 0,&#xD;&#xA;$$&#xA;due to $\mathrm{d}x_i\wedge\mathrm{d}x_j=-\mathrm{d}x_j\wedge\mathrm{d}x_i$. The correct way might be something like&#xA;$$&#xD;&#xA;-\frac16\,(\mathrm{d}y\mathrm{d}z+\mathrm{d}z\mathrm{d}x+\mathrm{d}x\mathrm{d}y)\,\nabla^2\psi&#xD;&#xA;$$&#xA;but I'm not sure if this would much more meaningful. The plain finite-small-$r$ expression is probably not the worst after all.&lt;/p&gt;&#xA;" OwnerUserId="3475" LastActivityDate="2011-05-09T22:54:14.863" />
  <row Id="9647" PostTypeId="1" CreationDate="2011-05-09T23:04:32.870" Score="0" ViewCount="297" Body="&lt;p&gt;Are there any reasons to assume that organic processes cannot arise in relatively flat spacetime environments?.. An argument for the existence of a critial value or window ..  &lt;/p&gt;&#xA;" OwnerUserId="2942" LastEditorUserId="2942" LastEditorDisplayName="" LastEditDate="2011-05-10T12:43:56.260" LastActivityDate="2011-08-10T14:21:57.843" Title="Space-time hospitality" Tags="&lt;spacetime&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="9648" PostTypeId="2" ParentId="9159" CreationDate="2011-05-10T01:06:11.867" Score="1" ViewCount="" Body="&lt;p&gt;Let's scale up an animal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If length is $L$, and we don't change the proportions, then the mass of the head+neck will scale as $L^3$.  If the neck &amp;amp; head are being held horizontally, the lever arm scales as $L$, so the torque at the base of the neck scales as $L^4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The width of the neck scales as $L$, so the force on the tendons/muscles scales as $L^3$.  The yield strength of a tendon/muscle scales with its cross-sectional area: $L^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force is going as $L^3$, but the yield strength is going as $L^2$.  My math agrees with your intuition: as you make an animal bigger, eventually the neck won't be able to handle the stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why the giraffe can get away with long necks is twofold, I'd guess:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) They have a proportionally thicker neck.  Look at a picture of giraffe.  Note that the neck gets extremely thick toward the base along the front-to-back axis (which is the the axis along which they lower their head), while it's slender side-to-side.  Of course, this you can only change proportions so far: eventually the animal will be all neck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) We've got a lot of overhead with our short human necks.  Ferinstance, I can support weights much heavier than my head with my arms, despite my arms being much longer AND a bit thinner than my neck.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-05-10T01:06:11.867" />
  <row Id="9649" PostTypeId="2" ParentId="9643" CreationDate="2011-05-10T02:12:06.167" Score="1" ViewCount="" Body="&lt;p&gt;I would guess the Lift/Drag ratio of the blades is important. Lift = wind thrust (and speed), and Drag = friction and power loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the part of the blade moves slower the closer to the center, the angle of attack needs to increase to provide the same wind speed. Keeping an even velocity profile is probably important in order to minimize losses due to turbulence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is probably a lot to be done with the wing-tips that is &lt;em&gt;not&lt;/em&gt; done due to aesthetic reasons. pffft.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-05-10T02:12:06.167" />
  <row Id="9650" PostTypeId="1" AcceptedAnswerId="9656" CreationDate="2011-05-10T02:14:55.103" Score="3" ViewCount="103" Body="&lt;p&gt;There has been recent activity by astrophysicists to determine whether a fourth flavor of neutrino, a sterile neutrino, exists. It would likely be more massive than electron, muon or tau neutrinos. However, it wouldn't be affected by the weak force, only by gravity. It would therefore have similar characteristics to dark matter. The LHC is looking for evidence of supersymmetry, and the existence of the stable neutralino, another dark matter candidate. Can the LHC be of any assistance in helping determine the existence of the hypothetical sterile neutrino? &lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-10T02:34:45.813" LastActivityDate="2011-05-10T06:18:32.380" Title="Sterile Neutrinos as Dark Matter" Tags="&lt;supersymmetry&gt;&lt;dark-matter&gt;&lt;neutrino&gt;&lt;weak-force&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9652" PostTypeId="1" CreationDate="2011-05-10T02:45:16.317" Score="4" ViewCount="92" Body="&lt;p&gt;Simulations of the Bak-Sneppen Model of Species Evolution (introduced in &lt;a href=&quot;http://link.aps.org/doi/10.1103/PhysRevLett.71.4083&quot; rel=&quot;nofollow&quot;&gt;http://link.aps.org/doi/10.1103/PhysRevLett.71.4083&lt;/a&gt;) show that it exhibits Self-Organized Criticality where after a transient only mutation through barriers of height $B_i &amp;lt; B_C$ where $B_C = 0.67 \pm 0.01$ occur spontaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can tell two studies attempt to show the SOC property of the Bak-Sneppen Model in the Mean Field approximation:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Flyvbjerg et al. (1993, &lt;a href=&quot;http://link.aps.org/doi/10.1103/PhysRevLett.71.4087&quot; rel=&quot;nofollow&quot;&gt;http://link.aps.org/doi/10.1103/PhysRevLett.71.4087&lt;/a&gt;) formulate the evolution equation for the probability distribution of the barrier values $P(B, t)$ but the argument for the existence of a global attractor is given very tersely as:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Our mean field dynamics is an approximation to the master equation for the Markov process of the random neighbor model, both having one unique attractive fixed point. (Page 4088, last paragraph left column)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In de Boer et al. (1994, &lt;a href=&quot;http://link.aps.org/doi/10.1103/PhysRevLett.73.906&quot; rel=&quot;nofollow&quot;&gt;http://link.aps.org/doi/10.1103/PhysRevLett.73.906&lt;/a&gt;) the Markov Process hinted at in the previous publication is studied by deriving the Master equation $P_{n,\lambda}(t)$ where $n$ is the number of sites with a barrier value less than $\lambda$. However as $\lambda \rightarrow \frac{1}{2}$, $P_{n,\lambda} \rightarrow 0$, which I interpret as meaning that there exists a $B_C \ge \frac{1}{2}$. But I do not see how this proves that the critical state is an attractor.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Therefore what is the argument for the critical state being a global attractor in the Mean Field Approximation?&lt;/p&gt;&#xA;" OwnerUserId="94" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-05-10T02:55:55.270" LastActivityDate="2011-09-04T16:36:40.173" Title="Why is the critical state in the Mean Field Bak-Sneppen Model a global attractor?" Tags="&lt;non-equilibrium&gt;&lt;non-linear-dynamics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9653" PostTypeId="2" ParentId="9159" CreationDate="2011-05-10T03:51:28.307" Score="0" ViewCount="" Body="&lt;p&gt;Why not? Even though, as Anonymous Coward pointed out very well, the force scales as a cube while strength goes as a square, this only implies that infinitely long necks aren't possible. Obviously, lots of coefficients are being dropped in this rough sketch - stuff that pertains to the placement of tendons and arrangement of muscles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Giraffes &lt;em&gt;do&lt;/em&gt; exist; this guarantees that necks can be at least 7 or 8 feet once the optimal neck-construction is done. Beyond that, it could be that the optimal neck-construction lets necks get to 10 or 20 or 40 feet. Impossible to say. All your intuition really tells you - assuming you don't know a whole lot about the architecture of giraffe necks - is that the neck can't be infinitely long.&lt;/p&gt;&#xA;" OwnerUserId="1655" LastActivityDate="2011-05-10T03:51:28.307" />
  <row Id="9654" PostTypeId="2" ParentId="9645" CreationDate="2011-05-10T04:24:25.240" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Question 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the error in your scenario is that matter and antimatter live at &quot;different places&quot; of extra dimensions which they surely can't. By definition, antimatter is being created by the Hermitian conjugates of the operators that create matter. So if a matter species is associated with a particular function of the extra dimensions, $f(x_{\rm hidden})$, then indeed, the corresponding antiparticle species is associated with the function $f(x_{\rm hidden})^*$. The antiparticle has the same probability density to occupy a given point in the extra dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position in the extra dimensions - the extra profile - is a part of the definition of a particle species. In fact, electron and muon (and similarly pairs of particles of different generations) may be created by the same higher-dimensional field but with a different position (or wave function) in extra dimensions. A different position, like in your scenario, means different species with different properties (and masses) - the species are surely not &quot;matter and antimatter&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, you only need the ordinary Kaluza-Klein theory (i.e. higher-dimensional general relativity with an extra dimension) - and not the full string theory - to make the arguments above as well as to make your original proof that the antimatter is being attracted to the matter just like other matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Additional comment&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could still create some extra matter species that live elsewhere (without calling it antimatter which is surely a wrong label) and whose mass is negative - which is a different way of saying that they have a repulsive gravity. In that case, you live at risk that your vacuum will be unstable because pairs of matter and mirror matter may be spontaneously created out of the vacuum without spoiling the energy/mass conservation law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite this pathology, similar scenarios have been proposed to explain why the cosmological constant was small or zero. Most physicists think that the instability kills the model.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-10T04:24:25.240" CommentCount="2" />
  <row Id="9655" PostTypeId="1" CreationDate="2011-05-10T05:07:40.087" Score="1" ViewCount="107" Body="&lt;p&gt;How do you obtain the spectral function for advection equation in one dimension?&#xA;2.How about 2D?&#xA;thanks a lot&lt;/p&gt;&#xA;" OwnerUserId="3477" LastActivityDate="2011-05-10T05:07:40.087" ClosedDate="2011-05-29T02:20:32.017" Title="spectral function" Tags="&lt;mathematical-physics&gt;" CommentCount="3" FavoriteCount="1" />
  <row Id="9656" PostTypeId="2" ParentId="9650" CreationDate="2011-05-10T05:32:42.840" Score="3" ViewCount="" Body="&lt;p&gt;The LHC is a discovery machine. It was designed to maximize the probability of seeing new physics outside the Standard Model, but it is a poor experiment for discriminating between two or three states that the only signature they have is missing energy and small to zero missing mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that the strong interactions initiating the scattering are a many body problem, 3 quarks on three quarks and a great number of gluons on gluons, thus  the only &quot;closed&quot; system is a subset  defined by a high transverse momentum transfer, and this is contaminated by debris that is not really relevant to the deep scattering. The use of Monte Carlo simulations is imperative, and MC means that a model of the interactions is assumed in the creation of the sample. Not exactly circular, but not helping in fine details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the LHC finds  photino or neutralino candidates  a leptonic   collider would be needed to be able to discriminate theories like the one proposing a fourth neutrino. IMO of course&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-05-10T06:18:32.380" LastActivityDate="2011-05-10T06:18:32.380" />
  <row Id="9657" PostTypeId="1" CreationDate="2011-05-10T06:05:02.280" Score="1" ViewCount="134" Body="&lt;p&gt;I recently saw cm^3/g as a unit for amount adsorbed.  Usually, you see either kg adsorbate/kg adsorbent or mole adsorbate/kg adsorbent.  Does anyone know the meaning of this unit?&lt;/p&gt;&#xA;" OwnerUserId="3311" LastActivityDate="2011-05-10T09:12:42.190" Title="cm^3/g as a unit of adsorption" Tags="&lt;thermodynamics&gt;" AnswerCount="3" />
  <row Id="9658" PostTypeId="2" ParentId="9657" CreationDate="2011-05-10T06:40:10.390" Score="1" ViewCount="" Body="&lt;p&gt;I've only ever seen those units described as &quot;specific surface area&quot;. It does have implications for adsorption, but I don't think it is the be all and end all.&lt;/p&gt;&#xA;" OwnerUserId="3478" LastActivityDate="2011-05-10T06:40:10.390" />
  <row Id="9659" PostTypeId="2" ParentId="9657" CreationDate="2011-05-10T07:12:13.423" Score="2" ViewCount="" Body="&lt;p&gt;Indeed, when you just want to measure how much material has been adsorbed, it's more natural to use the same units for both materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that 1 cubic centimeter per gram is not a unit of &quot;adsorption&quot; per se - adsorption is a process, not a quantity, after all - but it's a unit of &quot;specific pore volume&quot;. You take one gram of a material and measure the volume of the pores inside this material in cubic centimeters - and you get &quot;specific pore volume&quot; in cubic centimeters per gram. Note that pores - holes - are empty so they may only be measured by a unit of volume, and not a unit of mass or moles. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas adsorption is a major method to measure &quot;specific pore volume&quot;. In some sense, the &quot;specific pore volume&quot; also does quantify adsorption because it tells you at how many microscopic places the gas may get adsorbed inside a material. See some papers about the two concepts to get a flavor:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://scholar.google.com/scholar?q=%22specific+pore+volume%22+adsorption&quot; rel=&quot;nofollow&quot;&gt;http://scholar.google.com/scholar?q=%22specific+pore+volume%22+adsorption&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-10T07:12:13.423" />
  <row Id="9660" PostTypeId="2" ParentId="9657" CreationDate="2011-05-10T09:12:42.190" Score="1" ViewCount="" Body="&lt;p&gt;When determining surface areas, &quot;permanent&quot; gases are the prefered test gases. &#xA;You handle those gases in apparatus which is calibrated in volume units, and &#xA;record pressure. So, experimental results are in gas volume/mass of adsorbent.&#xA;Maybe this unit was left as it was? &#xA;In the &quot;gas business&quot; the base unit long was (still is?)  &quot;norm Volume&quot; &#xA;ie. volume at 0°C and 1 atmosphere. In case that literature was from this &#xA;area, the gas volume as unit is quite &quot;natural&quot;. &#xA;If we had the citation at least, one could derive more, evtl. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-05-10T09:12:42.190" />
  <row Id="9661" PostTypeId="1" AcceptedAnswerId="9662" CreationDate="2011-05-10T09:56:01.627" Score="8" ViewCount="192" Body="&lt;p&gt;Do the protons inside the nucleus repel each other by the electrostatic force? If they do, why doesn't the repulsion drive the protons apart so that the nuclei get disintegrated?&lt;/p&gt;&#xA;" OwnerUserId="3347" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-10T10:09:49.720" LastActivityDate="2011-08-01T17:31:47.617" Title="Protons' repulsion within a nucleus" Tags="&lt;nuclear-physics&gt;&lt;proton&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="9662" PostTypeId="2" ParentId="9661" CreationDate="2011-05-10T10:08:02.717" Score="9" ViewCount="" Body="&lt;p&gt;There is an electrostatic repulsion between the protons in the nucleus. However, there is also an attraction due to another kind of force besides electromagnetism, namely the so-called &quot;strong nuclear interaction&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strong nuclear interaction ultimately boils down to the forces between the &quot;colorful&quot; quarks inside the protons - and neutrons. It is mediated by gluons, much like electromagnetism is mediated by photons, described by Quantum Chromodynamics (QCD), much like electromagnetism is described by Quantum Electrodynamics (QED), and it acts (almost) equally on protons and neutrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The attractive strong nuclear interaction inside the nuclei is 1-2 orders of magnitude stronger than the repulsive electrostatic interaction which is what keeps the nuclei together despite the repulsive electrostatic force.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-10T10:08:02.717" CommentCount="1" />
  <row Id="9663" PostTypeId="1" AcceptedAnswerId="9668" CreationDate="2011-05-10T10:58:18.000" Score="6" ViewCount="146" Body="&lt;p&gt;From my recent experience teaching high school students I've found that they are taught that the strong force between nucleons is mediated by virtual-pion exchange, whereas between quarks it's gluons. They are not, however, taught anything about colour or quark-confinement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a more sophisticated level of physics, is it just that the maths works equally well for either type of boson, or is one (type of boson) in fact more correct than the other?&lt;/p&gt;&#xA;" OwnerUserId="1696" LastActivityDate="2011-05-10T12:55:15.143" Title="Is it pions or gluons that mediate the strong force between nucleons?" Tags="&lt;particle-physics&gt;&lt;quantum-chromodynamics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="3" />
  <row Id="9664" PostTypeId="2" ParentId="7506" CreationDate="2011-05-10T11:35:42.743" Score="0" ViewCount="" Body="&lt;p&gt;After reading relevant papers and comments I have impression that it is sometimes using the same terms by different people for different things. In his paper Bell described concrete values, particular physical experiments, but not some abstract theoretical formula and he discussed that the results of the experiments may not be explained by classical correlations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joy Christian wrote: “Central to Bell’s theorem [1] is the claim that no local and realistic model can reproduce the correlations observed in the EPR-Bohm experiments.” &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I doubt, Bell achievement could be formulated in such a simple way: let us recall discussions that even the usual quantum mechanics without reduction is local theory (from some works of Deutsch et al or maybe even directly from definitions). It is possible to consider  standard expression for correlations in quantum mechanics as a formal postulate and it is precisely reproduced in all experiments. From mathematical point of view this expression might be equivalent with formula of Christian (if later is correct, after all). The problem is not to write an abstract expression for correlation, but to explain, how it appears in real experiment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The subtlety is that reduction or some other way of transition from quantum to classical world used in description of experiments produce problems and misconceptions with interpretation of non-local effects and it is just a reason of late understanding of significance of Bell work and subject of many discussions on foundations of quantum mechanics. &lt;/p&gt;&#xA;" OwnerUserId="3351" LastActivityDate="2011-05-10T11:35:42.743" />
  <row Id="9665" PostTypeId="1" AcceptedAnswerId="9666" CreationDate="2011-05-10T12:06:23.583" Score="1" ViewCount="122" Body="&lt;p&gt;If I have a container that is full of water and I attempt to freeze the water by freezing the container, what would happen if the container is strong enough to prevent the water expansion? Could the water be in liquid state even though it is at or below freezing point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is probably a daft question but its something that's been bugging me.&lt;/p&gt;&#xA;" OwnerUserId="3479" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-10T14:21:12.760" LastActivityDate="2011-05-10T14:21:12.760" Title="What happens if you try to freeze water in an water tight container" Tags="&lt;water&gt;&lt;phase-transition&gt;&lt;states-of-matter&gt;" AnswerCount="2" />
  <row Id="9666" PostTypeId="2" ParentId="9665" CreationDate="2011-05-10T12:40:44.840" Score="7" ViewCount="" Body="&lt;p&gt;Yes, of course, the freezing point will decrease by the pressure &#xA;developed, while part of the water freezes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But do not underestimate the pressures! In such an experiment &#xA;easily some thousand bares may be developed. (Depends on the &#xA;rigidity of the vessel and the volume of water) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a video showing how freezing water cracks a cast iron sphere:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.physik.uni-wuerzburg.de/video/waermelehre/grundlagen/filme/g07ausdehnfluessig.mpg&quot; rel=&quot;nofollow&quot;&gt;http://www.physik.uni-wuerzburg.de/video/waermelehre/grundlagen/filme/g07ausdehnfluessig.mpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The bath is ice/salt at about -10°C)&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-05-10T13:04:26.100" LastActivityDate="2011-05-10T13:04:26.100" CommentCount="3" />
  <row Id="9667" PostTypeId="2" ParentId="9665" CreationDate="2011-05-10T12:44:11.190" Score="2" ViewCount="" Body="&lt;p&gt;I think what you are really getting at is the nature of phase changes in respect of changes in pressure, volume and temperature.  This 3D diagram explains a lot though it can a while to intepret it:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/File:PVT_3D_diagram.png&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/File:PVT_3D_diagram.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the associated article should explain in more detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  Here is a simpler 2D version:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/File:Phase-diag2.svg&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/File:Phase-diag2.svg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-05-10T12:44:11.190" />
  <row Id="9668" PostTypeId="2" ParentId="9663" CreationDate="2011-05-10T12:55:15.143" Score="8" ViewCount="" Body="&lt;p&gt;Dear qftme, I agree that your question deserves a more expansive answer. The answer, &quot;pions&quot; or &quot;gluons&quot;, depends on the accuracy with which you want to describe the strong force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Historically, people didn't know about quarks and gluons in the 1930s when they began to study the forces in the nuclei for the first time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1935, Hideki Yukawa made the most important early contribution of Japanese science to physics when he proposed that there may be short-range forces otherwise analogous to long-range electromagnetism whose potential is&#xA;$$V(r) = K\frac{e^{-\mu r}}{r} $$&#xA;The Fourier transform of this potential is simply $1/(p^2+\mu^2)$ which is natural - an inverted propagator of a massless particle. (The exponential was added relatively to the Coulomb potential; and in the Fourier transform, it's equivalent to the addition of $\mu^2$ in the denominator.) The Yukawa particle (a spinless boson) was mediating a force between particles that was only significantly nonzero for short enough distances. The description agreed with the application to protons, neutrons, and the forces among them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the mediator of the strong force was thought to be a pion and the model worked pretty well. (In the 1930s, people were also confusing muons and pions in the cosmic rays, using names that sound bizarre to the contemporary physicists' ears - such as a mesotron, a hybrid of pion and muon, but that's another story.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pion model was viable even when the nuclear interactions were understood much more quantitatively in the 1960s. The pions are &quot;pseudo-Goldstone bosons&quot;. They're spinless (nearly) massless bosons whose existence is guaranteed by the existence of a broken symmetry - in this case, it was the $SU(3)$ symmetry rotating the three flavors we currently know as flavors of the $u,d,s$ light quarks. The symmetry is approximate which is why the pseudo-Goldstone bosons, the pions (and kaons), are not exactly massless. But they're still significantly lighter than the protons and neutrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the theory with the fundamental pion fields is not renormalizable - it boils down to the Lagrangian's being highly nonlinear and complicated. It inevitably produces absurd predictions at short enough distances or high enough energies - distances that are shorter than the proton radius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better theory was needed. Finally, it was found in Quantum Chromodynamics that explains all protons, neutrons, and even pions and kaons (and hundreds of others) as bound states of quarks (and gluons and antiquarks). In that theory, all the hadrons are described as complicated composite particles and all the forces ultimately boil down to the QCD Lagrangian where the force is due to the gluons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So whenever you study the physics at high enough energy or resolution so that you see &quot;inside&quot; the protons and you see the quarks, you must obviously use gluons as the messengers. Pions as messengers are only good in approximate theories in which the energies are much smaller than the proton mass. This condition also pretty much means that the velocities of the hadrons have to be much smaller than the speed of light.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-10T12:55:15.143" CommentCount="8" />
  <row Id="9671" PostTypeId="2" ParentId="9560" CreationDate="2011-05-10T14:09:04.893" Score="0" ViewCount="" Body="&lt;p&gt;I know two (non-mainstream) distinct ways.&lt;br&gt;&#xA;&lt;strong&gt;Because this simple Question atracted downvotes I will have to say something in defense:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A particular physics viewpoint that &lt;em&gt;'is not mainstream'&lt;/em&gt; do not cast in itself a ban, at least it should not be so. To a physicist minded person there are no taboos. To be &lt;em&gt;mainstream&lt;/em&gt; means that, in the present, the main focus of the mindforce is concentrated in the exploration of certain &lt;em&gt;labyrinths&lt;/em&gt; that once were hidden behind a closed door. Other doors were closed without further exploration and a  sign was placed at the entrance: &lt;em&gt;Dead End&lt;/em&gt;. Some solitary minds pursue the difficult job of digging other labyrints. If they succeed they will be our heros. The only quality that is required is honesty, IMO, as is posted at the entrance of this site.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Polarizable_vacuum&quot; rel=&quot;nofollow&quot;&gt;Polarizable-Vacuum&lt;/a&gt; approach:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/gr-qc/9909037&quot; rel=&quot;nofollow&quot;&gt;Polarizable-Vacuum (PV) representation of general relativity&lt;/a&gt;  by Puthoff, 1999  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Having shown by specific calculation&#xA;  that the PV approach to the three&#xA;  classical tests of GR reproduces the&#xA;  traditional GR results  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;from the Discussion section:    &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;alternatively, one can define space as&#xA;  Euclidean and time as the same&#xA;  everywhere, and discover (from exactly&#xA;  the same measurements) how the&#xA;  velocity of light, and natural clocks,&#xA;  rods, and particle inertias 'really'&#xA;  behave in the neighborhood of large&#xA;  masses. There is just as much (or as&#xA;  little) content for the word 'really'&#xA;  in the one approach as in the other;&#xA;  provided that each is self-consistent,&#xA;  the ultimate appeal is only to&#xA;  convenience and fruitfulness, and even&#xA;  'convenience' may be largely a matter&#xA;  of  personal taste...&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Linked with this paper &lt;a href=&quot;http://arxiv.org/abs/0902.1305&quot; rel=&quot;nofollow&quot;&gt;Polarizable vacuum analysis of electric and magnetic fields&lt;/a&gt; by Xing-Hao Ye, 2009    &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The electric and magnetic ﬁelds are&#xA;  investigated on the basis of quantum&#xA;  vacuum. The analysis of the&#xA;  electromagnetic energy and force&#xA;  indicates that an electric ﬁeld is a&#xA;  polarized distribution of the vacuum&#xA;  virtual dipoles, and that a magnetic&#xA;  ﬁeld in vacuum is a rearrangement of&#xA;  the vacuum polarization. It means that&#xA;  an electromagnetic wave is a&#xA;  successional changing of the vacuum&#xA;  polarization in space. Also, it is&#xA;  found that the average half length of&#xA;  the virtual dipoles around an&#xA;  elementary charge is $a = 2.8 &amp;#215;&#xD;&#xA;&amp;gt; 10^{−15}m$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The other approach that I'm aware is explored in &lt;a href=&quot;http://www.authorhouse.com/BookStore/BookDetail.aspx?Book=233588&quot; rel=&quot;nofollow&quot;&gt;Douglas Pinnow book 'Our Resonant Universe'&lt;/a&gt;&lt;/strong&gt;  it follows after the steps of Kaluza-Klein, Goedecke and Haus (and Maxwell ;-) and ...&lt;br&gt;&#xA;It is a monography of a &lt;strong&gt;model of particles, based only on Electromagnetism&lt;/strong&gt; (EM), that has only one parameter (electron mass) and derives the particle properties to within 1% of their values (barion masses bellow 0.1%) and does not suffer of the barionic spin crisis.&#xA;The model uses only three building blocks: Electron, Pion and Muon.&lt;br&gt;&#xA;The concepts of &lt;strong&gt;Mass, Charge and Gravitational Force became crystal clear&lt;/strong&gt;, IMO.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can have a glimpse of the model and explore further starting from &lt;a href=&quot;http://heldervelez1.tiddlyspace.com/#%5B%5BOur%20Resonant%20Universe%20by%20Douglas%20Pinnow%5D%5D%20GettingStarted&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I'm used to see two arguments against the possibility of unification of EM and gravity:&lt;/strong&gt;&lt;br&gt;&#xA;The different scale of the forces and the fact that gravity is only attractive.&lt;br&gt;&#xA;IMO both of them can be easily dismissed.  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-05-10T14:09:04.893" CommentCount="2" />
  <row Id="9672" PostTypeId="2" ParentId="9643" CreationDate="2011-05-10T15:33:14.597" Score="1" ViewCount="" Body="&lt;p&gt;Mostly trial and error - but in a computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A combination of &lt;a href=&quot;http://en.wikipedia.org/wiki/Computational_fluid_dynamics&quot; rel=&quot;nofollow&quot;&gt;Computational Fluid Dynamics&lt;/a&gt; (modelling how the air flows over the blade) and &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_element_method&quot; rel=&quot;nofollow&quot;&gt;Finite Element Modelling&lt;/a&gt; (how the stresses in the metal behave)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these are complex areas - and when they come to together you need a lot of expensive computers and some even more expensive engineers.&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-05-10T15:33:14.597" />
  <row Id="9673" PostTypeId="2" ParentId="9647" CreationDate="2011-05-10T16:05:11.420" Score="0" ViewCount="" Body="&lt;p&gt;First, let's assume that the heavier elements already exit but then find themselves in a relatively flat area of space (ie low &lt;em&gt;g&lt;/em&gt;). In order for life to begin, and to continue (initially at least), one requires an environment containing a complex mixture of many compounds, in a warm liquid state. To expand on why it would require each of these conditions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;~&lt;/em&gt; complex mixture of compounds&lt;/strong&gt; would be required as, together, they have the ability to form amino acids, which in turn can form the building blocks of all 'living' things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;~&lt;/em&gt; warm liquid state&lt;/strong&gt; would be required to ensure that the compounds were close enough to eachother for them to frequently interact (&lt;strong&gt;liquid&lt;/strong&gt;) and moving around eachother enough (&lt;strong&gt;warm&lt;/strong&gt;) to overcome the extremely narrow possibility of their interacts leading to creation of something 'living'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How exactly these conditions could be met without a planet, for the liquid to be attracted to, and a star (or geothermal energy as in Earth's case,) to provide the required heat is difficult to imagine. Not to mention, as @Georg pointed out, older stars are needed to make the heavier elements (fusion) and distribute them (supernovae,) in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, and to get back to physics, it seems gravity is essential at every stage. Quantifying this conclusion would involve many assumptions and a great deal of calculation. With regard to star formation however, I believe it has already been done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;~ The steps I have described are (loosely-) based on the first episode of Sir David Attenborough's recent TV series; entitled &lt;a href=&quot;http://en.wikipedia.org/wiki/First_Life_%28TV_series%29&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;First Life&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;~ To me, his arguments seem plausible and firmly based on scientific research. As such, I am inclined to believe this process as being the most likely 'origin of life'. (Of course, not everyone will agree on either the details or the principle as a hole.)&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="1696" LastEditorDisplayName="" LastEditDate="2011-05-12T13:50:26.283" LastActivityDate="2011-05-12T13:50:26.283" />
  <row Id="9674" PostTypeId="2" ParentId="9647" CreationDate="2011-05-10T16:11:13.240" Score="0" ViewCount="" Body="&lt;p&gt;Freeman Dyson gave a lecture on this topic (sort of) at Berkeley about a decade ago.  Here is the transcript:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;&lt;a href=&quot;http://www.ocf.berkeley.edu/~fricke/dyson.html&quot; rel=&quot;nofollow&quot;&gt;Gravity is Cool, or, Why our Universe is Hospitable to Life&lt;/a&gt;&quot;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-05-10T16:11:13.240" CommentCount="1" />
  <row Id="9675" PostTypeId="1" AcceptedAnswerId="9679" CreationDate="2011-05-10T16:55:31.750" Score="2" ViewCount="146" Body="&lt;p&gt;For the Lagrangian $$L = \frac{1}{2}mu^2 - q(\phi - \frac{\vec{A}}{c}\cdot \vec{u})$$ of a non-relativistic point particle in an electromagnetic potential, what gauge is used for the electromagnetic potential $\vec{A}$ and $\phi$?&lt;/p&gt;&#xA;" OwnerUserId="2146" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-10T17:49:05.450" LastActivityDate="2011-05-10T19:38:53.203" Title="What gauge is used in the Lagrangian for a non-relativistic point particle in an electromagnetic potential" Tags="&lt;electrodynamics&gt;&lt;lagrangian-mechanics&gt;" AnswerCount="1" />
  <row Id="9676" PostTypeId="2" ParentId="9537" CreationDate="2011-05-10T17:01:33.213" Score="0" ViewCount="" Body="&lt;p&gt;EAMann,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is a physics site, and you are looking for a physics answer; but I noticed that you are in Portland, Oregon.  It seems amazing to me that you can not find a site reasonably close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If nothing is closer, it seems to me that one of your beautiful beaches would make a nice launch site.  That is (IIRC) more than an hour away, but you could combine it with a camping trip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My high school biology professor used to take us out to the Gingko petrified forest in eastern WA.  Along the way we would stop and jump down the sand dunes on the WA side of the Columbia River, not too far (a few hours) from where you are. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a lot of desert in eastern WA and eastern OR.  Can't imagine not being able to find a launch site out there.  The other thing I would do is network through your own organization, BSA, to find out where other nearby scout troops are going.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-05-10T17:01:33.213" CommentCount="1" />
  <row Id="9677" PostTypeId="2" ParentId="9625" CreationDate="2011-05-10T17:14:15.383" Score="0" ViewCount="" Body="&lt;p&gt;You reasoned correctly that in both cases the ball converts the same amount of potential energy into kinetic energy (at least it appears that you came to this conclusion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem you are having is that you need to realize that kinetic energy is proportional to v^2.  So, it takes NINE times as much energy to get from 0 m/s to 15 m/s as it does to get from 0 to 5 m/s.  If the potential energy available is just enough to go from 0 to 10 m/s, it is then obvious that it is NOT going to be enough to go from 5 to 15 m/s.  Answer A.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-05-10T17:14:15.383" />
  <row Id="9678" PostTypeId="2" ParentId="9616" CreationDate="2011-05-10T17:36:47.747" Score="1" ViewCount="" Body="&lt;p&gt;As far as the function of light goes: Yes you can scale down (to a point).  I have worked on an ASIC (Application Specific Integrated Circuit) which used an 8 micron process (Cro-magnon by today's standards).  I could not see the detail of these circuits in the finished product (far too small) BUT they were made basically (I am greatly simplifying) with itsy-bitsy photographic images produced by light (beyond the range of visible light).  To say it another way: the resolvable detail available from light is far, FAR finer than the un-aided human eye can see.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Biologists claim that eagles can see about 10x sharper detail than a human (and an eagle's eye is noticably smaller than a human eye).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't know, is where the physical size of the eye limits detail.  I see no reason why it couldn't scale down... WAY down.  But I am not a biologist and (thankfully) not an ant.  It would be interesting to find out where the smallest eyeballs end and where other seeing apparati take over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, your example of getting down to where you could see bacteria presents an interesting break point: There is theoretically no problem seeing a bacterium (size about 1000 nm) at the lower end of UV (upper end of human-visible) light, wavelength about 400 nm.  But detail would obviously be a bit hazy.  The bacterium would appear as a blurry blob, and no glasses would help.  The theoretical limit for modern optical microscopes to resolve detail is 200 nm (using 550 nm &quot;green&quot; light).&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-05-10T18:05:32.770" LastActivityDate="2011-05-10T18:05:32.770" />
  <row Id="9679" PostTypeId="2" ParentId="9675" CreationDate="2011-05-10T17:39:57.630" Score="8" ViewCount="" Body="&lt;p&gt;There is no specific gauge being used in this electromagnetic Langrangian interacting with classical particles; in other words, the Lagrangian you wrote is still gauge-invariant so you may use any gauge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that it's gauge invariant, perform a gauge transformation&#xA;$$ \phi \to \phi + \frac{\partial}{\partial t} \lambda,\quad \vec A\to \vec A + c \nabla \lambda $$&#xA;where $\lambda = \lambda(\vec x,t) = \lambda(\vec x(t),t)$. With this change of the potentials, your Lagrangian will change to&#xA;$$ L\to L - q \frac{\partial}{\partial t} \lambda - q\vec u \cdot\nabla \lambda $$&#xA;The final $L$ may be rewritten if one knows that $\vec u = \partial \vec x / \partial t$:&#xA;$$ L\to L - q \frac{\partial}{\partial t} \lambda - \frac{q\partial \vec x}{\partial t} \cdot\nabla \lambda $$&#xA;But via something like the chain rule, that's nothing else than the total time derivative of $\lambda$:&#xA;$$ L \to L - q \frac{d}{dt}\lambda $$&#xA;which is seen because $\lambda$ is a function of $\vec x, t$ but we must use the right value of the argument $\vec x$, and because $\vec x$ itself depends on $t$, the total time derivative is given by the partial time derivative plus the extra spatial term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the action is &#xA;$$ S = \int dt\, L $$&#xA;and if you add a &quot;total time derivative&quot; to the Lagrangian, just like we did above, this extra term may be integrated and gives you the difference of $q\lambda$ at $t=+\infty$ and $t=-\infty$. Assuming that at both of these extreme moments, the particle had or will have positions $\vec x$ that are far from the spacetime region where the gauge transformation was nontrivial i.e. where $\lambda$ was nonzero, the integral of the total derivative vanishes. So we have &#xA;$$ S \to S,$$&#xA;the action is gauge-invariant. It's common in physics that the argument why the boundary things - and total derivatives - don't matter is simplified. There are other reasons to ignore the total derivatives &quot;immediately&quot; and &quot;generally&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Generalizations: quantum physics and QFT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum physics, particles are described as fields, so one would couple the particles to the electromagnetic field a bit differently, e.g. via $\bar\psi \gamma^\mu A_\mu \psi$, and the phase of the field $\psi$ would have to be changed by something like $\exp(i\lambda)$ to make the gauge invariance work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it would still be true that the Lagrangian as naturally written is gauge-invariant. If one couldn't write basic Lagrangians (e.g. for a single charged particle) that are gauge-invariant, gauge invariance wouldn't exist (it wouldn't really be possible in an interesting theory) and we wouldn't ever be talking about it. If you want the Lagrangian to pick a specific gauge, you have to include some extra gauge-fixing terms to the Lagrangian such as $K&amp;#39;(\partial_\mu A^\mu)^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it would be silly to do such a thing without a good reason: in most cases, gauge invariance is a damn useful symmetry (more precisely: a redundancy) that allows you to simplify some calculations. In quantum field theory, a good reason to gauge-fix the Lagrangian is the desire to have a well-defined photon propagator. The propagator is the inverse of the differential operator in the equations of motion; without gauge-fixing, the operator has a reduced rank and its inverse would be singular.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-10T19:38:53.203" LastActivityDate="2011-05-10T19:38:53.203" CommentCount="2" />
  <row Id="9680" PostTypeId="2" ParentId="9532" CreationDate="2011-05-10T18:19:18.093" Score="-1" ViewCount="" Body="&lt;p&gt;If we may assume that the cups have the same form factor, or that the cups are identical and that the 12 oz just fills it up farther; in either case the 12 oz. will stay warmer longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Agreed with Georg: that the dominant heat loss mechanism in this case will be evaporative cooling&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-05-10T18:19:18.093" />
  <row Id="9681" PostTypeId="2" ParentId="9519" CreationDate="2011-05-10T18:47:04.153" Score="1" ViewCount="" Body="&lt;p&gt;Ah.  Hobby engineering.  You have strayed into an area of intense interest.  Here is a simple Rube Goldberg device which may give you other ideas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;build a short teeter-totter (&quot;TT&quot;) with a laser pointer clamped to it, so that the laser points at a wall.  Make sure the TT is fairly level, the fulcrum of the TT is firmly supported and that it is fairly balanced (so as to not add weight to your string).  Attach one end of the TT near the free end of the string.  Turn on the laser, make a pencil mark on the wall where it hits, load the string, and make your second mark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call the length from the TT fulcrum to the string = TS.&#xA;Call the length from the string to the wall = SW.&#xA;Call the length from your first to second wall mark = WW&#xA;Call the length of stretch of the string = SS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then: SS/TS = WW/(TS+SW) ---&gt;  SS = (WW)(TS/(TS+SW))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you build a TT with a 2 inch half-arm (TS), and your wall is 9 feet away, you will get a movement magnification of 110/2 = 55.  This might not be enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another idea would be to mount a small mirror on the TT and hit it with a fixed laser, so that it bounces back to the wall. This would double your sensitivity (movement magnification of 110).  If you made TT for the mirror = 1 inch, you could get approximately double that (218).  If you wanted to add another mirror at the target spot on the wall and make your pencil marks on the opposite wall... well... you do the math.  :-)  As long as you keep the angles small (less than 5 degrees) you can dispense with any trig and just use simple geometry.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-05-10T18:54:49.793" LastActivityDate="2011-05-10T18:54:49.793" />
  <row Id="9683" PostTypeId="1" AcceptedAnswerId="9694" CreationDate="2011-05-10T19:09:50.960" Score="1" ViewCount="148" Body="&lt;p&gt;In a nutshell, Noether's theorem states that for every continuous symmetry a corresponding conserved quantity exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the Hamiltonian equations of motion (let's talk about a classical system here) are invarinat under addition of any constant to the Hamiltonian function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H\rightarrow H+ \text{const.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a corresponding conserved quantity?&lt;/p&gt;&#xA;" OwnerUserId="2726" LastActivityDate="2011-05-11T05:44:43.233" Title="Noether's theorem and &quot;translations&quot; of the Hamiltonian function" Tags="&lt;symmetry&gt;&lt;hamiltonian-mechanics&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="9684" PostTypeId="2" ParentId="9683" CreationDate="2011-05-10T19:21:06.213" Score="3" ViewCount="" Body="&lt;p&gt;Noether's theorem is that for every (differentiable) symmetry of the action (integral of the Lagrangian) there is a conserved quantity.  It does not directly say anything about symmetries of the equations of motion, nor about the Hamiltonian.&lt;/p&gt;&#xA;" OwnerUserId="1233" LastActivityDate="2011-05-10T19:21:06.213" />
  <row Id="9685" PostTypeId="2" ParentId="9683" CreationDate="2011-05-10T19:21:13.037" Score="4" ViewCount="" Body="&lt;p&gt;Nope, this operation is not a symmetry in the physics sense. A symmetry transformation is a transformation that changes or mixes the values of the basic degrees of freedom such as positions and momenta $x(t)$, $p(t)$ in mechanics or the values of fields such as $\vec E(\vec x,t)$ and $\vec B(\vec x,t)$ in electromagnetism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hamiltonian is not an independent variable or a basic degree of freedom; it is a function of them. You're not changing the values of any quantities that evolve with time; instead, you're changing some formulae for derived and in principle unnecessary auxiliary objects in the theory (in this case the Hamiltonian), claiming that other formulae are preserved. This is not a symmetry so there is no conservation law associated with this operation. You're not &quot;rotating&quot; real objects which is what symmetry transformations should do: you're just redefining auxiliary, derived objects on the paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, the operation you mention fails to be &quot;harmless&quot; in general relativity because the energy is a source of gravitational field - curvature of space - so if you move it by a constant, you &lt;em&gt;do&lt;/em&gt; change physics.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-10T19:21:13.037" CommentCount="3" />
  <row Id="9686" PostTypeId="1" AcceptedAnswerId="9719" CreationDate="2011-05-10T19:21:33.707" Score="6" ViewCount="338" Body="&lt;p&gt;The action&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=\int L \;\mathrm{d}t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is an important physical quantity. But can it be understood more intuitively? The Hamiltonian corresponds to the energy, whereas the action has dimension of energy × time, the same as angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard the action being described as a measure of &lt;em&gt;change&lt;/em&gt;, although I don't know how this description can be justified.&lt;/p&gt;&#xA;" OwnerUserId="2726" LastActivityDate="2011-05-12T10:01:27.903" Title="The meaning of action" Tags="&lt;lagrangian-mechanics&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" />
  <row Id="9687" PostTypeId="2" ParentId="9686" CreationDate="2011-05-10T20:40:46.870" Score="3" ViewCount="" Body="&lt;p&gt;Here comes an attempt at an intuitive analogy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must have heard the classic &quot;the shortest distance between two points is a straight line&quot; , in euclidean geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The action integral is analogous  to distance in the generalized coordinates of the system under study. Its minimization gives the shortest path in configuration space going from time t1 to time t2  and is called the principle of least action. Applying it leads directly to Lagrange's equations, from which the equations of motion of the system are derived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chapter 2 of H. Goldstein's &quot;classical mechanics&quot; (available on the web)  is a good introduction to this and to variational principles.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-10T20:40:46.870" />
  <row Id="9688" PostTypeId="2" ParentId="9686" CreationDate="2011-05-10T21:05:51.257" Score="-3" ViewCount="" Body="&lt;p&gt;Action is a function of the upper limit of integration (if $L(t)$ is specified), so it grows from zero when $t_2$ goes farther and farther from $t_1$. In this sense, action is a &quot;measure of change&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But normally we do not calculate the action values with &lt;strong&gt;solutions&lt;/strong&gt; inserted in $L$. Action is a functional of &lt;strong&gt;possible trajectories&lt;/strong&gt; before varying $q(t)$ and $\dot{q} (t)$. Before varying, $q(t)$ and $\dot{q} (t)$ are arbitrary, not specific solutions of equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An intuitive understanding of action might be the following: on real trajectories its growth rate might be minimal but it is not the case, unfortunately. Instead, we have the least action requirement: $\delta S = 0$ between two fixed points.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-05-10T22:23:49.927" LastActivityDate="2011-05-10T22:23:49.927" CommentCount="1" />
  <row Id="9689" PostTypeId="1" CreationDate="2011-05-10T21:15:07.270" Score="2" ViewCount="113" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/8659/invariant-spacetime-distance-circular-motion&quot;&gt;Invariant spacetime - distance - Circular Motion&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;This is a question that I thought up a few years ago when I was taking mechanics. I asked the professor but didn't really get a straight answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a record spinning at relativistic speeds. In the lab frame, the circumference of the record should decrease according to the Lorentz contraction. However the radius of the record should remain fixed since it is orthogonal to the direction of motion. So does the shape of the record appear distorted since the circumference is smaller but the radius is the same? What would it look like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a feeling the answer is obvious and/or well known since it would seem to be similar to the situation of relativistic particles traveling around a circular accelerator but I can't think of how it can be solved.&lt;/p&gt;&#xA;" OwnerUserId="1939" LastActivityDate="2011-05-11T04:52:23.297" ClosedDate="2011-05-11T05:10:38.927" Title="Paradox of the Relativistic Record Player" Tags="&lt;special-relativity&gt;&lt;paradox&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="9690" PostTypeId="1" CreationDate="2011-05-10T22:16:36.330" Score="1" ViewCount="75" Body="&lt;p&gt;I've got a flywheel with angular momentum $\vec L$. Now I'd like to change the orientation of $\vec L$ by an angle $φ$ (in degrees). How do I calculate the angular momentum $\vec M$ for that change?&lt;/p&gt;&#xA;" OwnerUserId="3483" LastEditorUserId="3483" LastEditorDisplayName="" LastEditDate="2011-05-11T09:19:14.797" LastActivityDate="2011-05-11T09:19:14.797" Title="resistance of a flywheel to change of angular momentum" Tags="&lt;newtonian-mechanics&gt;&lt;angular-momentum&gt;&lt;gyroscopes&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9691" PostTypeId="2" ParentId="9537" CreationDate="2011-05-11T00:39:14.560" Score="0" ViewCount="" Body="&lt;p&gt;This doesn't answer the question, but it does solve the problem. Do it on the 4th of July along side your neighbors who (I assume) will also be launching bottle rockets etc. Here in South Carolina it sounds like a war zone on the 4th from about 8-12pm. It that doesn't work, maybe ask to do it before/after a fireworks display?&lt;/p&gt;&#xA;" OwnerUserId="2843" LastActivityDate="2011-05-11T00:39:14.560" />
  <row Id="9692" PostTypeId="2" ParentId="9690" CreationDate="2011-05-11T02:11:09.740" Score="1" ViewCount="" Body="&lt;p&gt;Rate of change of angular momentum is equal to sum of applied moments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec{M} = \dfrac{\rm{d}\vec{L}}{\rm{d}t} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To impose a diffrence in angular momentum, you have to apply an moment action&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int\vec{M}\,{\rm d}t={\rm Rot}(\hat{k},\varphi)\,\vec{L}-\vec{L} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if the moment is constant, it can be taken outside the integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;=== Euler Equations ===&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can use the equations of motion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec{M}=I\,\vec{\alpha}+\vec{\omega}\times I\,\vec{\omega} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;expressed in an inertial frame, and use $\vec{L}=I\,\vec{\omega}$ at any instant.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-05-11T02:11:09.740" />
  <row Id="9693" PostTypeId="2" ParentId="9689" CreationDate="2011-05-11T04:52:23.297" Score="0" ViewCount="" Body="&lt;p&gt;As MBN pointed out, this is called the Ehrenfest paradox.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Ehrenfest_paradox&quot; rel=&quot;nofollow&quot;&gt;Wikipedia summarizes&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;The modern resolution can be briefly summarized as follows:&#xA;Small distances measured by disk-riding observers are described by the Langevin-Landau-Lifschitz metric, which is indeed well approximated (for small angular velocity) by the geometry of the hyperbolic plane, just as Kaluza had claimed.&#xA;For physically reasonable materials, during the spin-up phase a real disk expands radially due to centrifugal forces; relativistic corrections partially counteract (but do not cancel) this Newtonian effect. After a steady-state rotation is achieved and the disk has been allowed to relax, the geometry &quot;in the small&quot; is approximately given by the Langevin-Landau-Lifschitz metric.&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't have anything to add - I think Wikipedia is good for this one.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-05-11T04:52:23.297" />
  <row Id="9694" PostTypeId="2" ParentId="9683" CreationDate="2011-05-11T05:44:43.233" Score="4" ViewCount="" Body="&lt;p&gt;What you're describing is not a symmetry in the sense of Noether's theorem of the (classical i.e. non relativistic and non quantum) system, but it is a symmetry of our description of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good example is provided when the Hamiltonian can be written as $H=T+V$ where $T$ is a kinetic energy and $V$ is a potential energy. Modifying the Hamiltonian by adding a constant to the potential energy does not change the behavior of the system in either the classical case or in the non relativistic quantum case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sort of symmetry is called a &quot;gauge symmetry&quot; and is discussed, in the quantum mechanics case, at length in Sakurai's quantum textbook &quot;Modern Quantum Mechanics&quot;:&lt;br&gt;&#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0805382917&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Modern-Quantum-Mechanics-2nd-Sakurai/dp/0805382917&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See section (2.6) page 123, &quot;Potentials and Gauge Transformations&quot;, which describes the difference between the classical case, where changing the potential has no physical significance, and the quantum case, where changing the potential changes the phase, but doesn't result in physically observable consequences.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-05-11T05:44:43.233" CommentCount="1" />
  <row Id="9695" PostTypeId="2" ParentId="9519" CreationDate="2011-05-11T06:10:09.910" Score="1" ViewCount="" Body="&lt;p&gt;To accurately measure the cross section, assuming you know the material's density, just measure the length, mass and use&#xA;$$\sigma = \frac{M}{\rho\;L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're interested in the answer (rather than having to measure it), and you know the material's Young's modulus, it's easy to calculate the extension. But I'll assume you don't know this (which strength of a material is a lot harder to predict than its density).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're having trouble measuring the extension due to its small amount, then use the following fixture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/GXEjk.jpg&quot; alt=&quot;Leverage on string extension&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to use leverage so you can measure a smaller change in string length. There are two strings, one with no weight applied, the other stretches slightly. The result is that the blue stick moves quite a lot. (Probably it would be better to have the blue stick set up so that it has equal lengths on either sides of where you attach the strings. I'd use balsa for the stick.)&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-05-11T06:10:09.910" />
  <row Id="9696" PostTypeId="1" CreationDate="2011-05-11T06:33:04.190" Score="3" ViewCount="83" Body="&lt;p&gt;Feynman diagrams are ordinarily the usual method of perturbative analysis for weakly interacting quantum field theories. However, over a de Sitter background, the total number of particles isn't even conserved in the free field theory. So, how does one apply perturbation theory to the Bunch-Davies state? Feynman diagrams presuppose particle number is conserved in the unperturbed model. This would definitely help us to compute the nongaussian correlations in the cosmic microwave background radiation in inflationary models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the closed time path formalism absolutely necessary?&lt;/p&gt;&#xA;" OwnerUserId="3491" LastEditorUserId="3491" LastEditorDisplayName="" LastEditDate="2011-05-11T06:51:55.280" LastActivityDate="2011-06-10T08:56:36.063" Title="How is perturbation theory applied to the Bunch-Davies state for an interacting quantum field theory?" Tags="&lt;quantum-field-theory&gt;&lt;cosmological-inflation&gt;&lt;de-sitter-spacetime&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="9697" PostTypeId="2" ParentId="9606" CreationDate="2011-05-11T06:38:39.007" Score="1" ViewCount="" Body="&lt;p&gt;This can probably be achieved with time dependent density functional theory, however it's nontrivial. Regrettably, I have no real experience in this area, however TD-DFT is a good place to start if you're interested in calculating UV/Vis and (hyper)polarisability properties almost from first principles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good electronic structure codes should be capable of directly outputting UV/Vis spectra, whilst the refractive index is related to the &lt;a href=&quot;http://prb.aps.org/abstract/PRB/v38/i8/p5694_1&quot; rel=&quot;nofollow&quot;&gt;unit cell of the material and the frequency dependent polarisability tensor&lt;/a&gt;(1). Accordingly, you'll only be able to get hard results for crystalline glycerol, rather than liquid or glassy phases, however the index may not change much. Because the index is frequency dependent, you may have to do calculations over a subdivided range of frequencies, which may be quite time consuming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to look at software packages such as GAUSSIAN, ADF, GAMESS/Firefly, ORCA, MPQC, NWChem, etc., which range in price between free and very expensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if you have access to the right facilities, you could always make empirical measurements in about 5 minutes, but where's the fun in that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt;: regarding Georg's statement answer, glycerol &lt;a href=&quot;http://www.azom.com/article.aspx?ArticleID=4183#_Spectral_Range_of&quot; rel=&quot;nofollow&quot;&gt;has a small absorption in the range of interest&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Tossell, J. A. and Lazzeretti, Paolo; &lt;em&gt;Ab initio&lt;/em&gt; calculation of the refractive indices and related properties of $CaCO_{3}$; Phys. Rev. B; 38, 8, &lt;strong&gt;1988&lt;/strong&gt;, pp. 5694-5698&lt;/p&gt;&#xA;" OwnerUserId="1876" LastEditorUserId="1876" LastEditorDisplayName="" LastEditDate="2011-05-11T11:13:51.290" LastActivityDate="2011-05-11T11:13:51.290" CommentCount="1" />
  <row Id="9698" PostTypeId="1" CreationDate="2011-05-11T06:38:53.380" Score="1" ViewCount="117" Body="&lt;p&gt;I'm currently resident in Hong Kong a country which appars to import heavily from Japan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last saturday 7th May I went to a restaurant:&#xA;&lt;a href=&quot;http://www.openrice.com/english/restaurant/sr2.htm?shopid=39760&quot; rel=&quot;nofollow&quot;&gt;http://www.openrice.com/english/restaurant/sr2.htm?shopid=39760&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which I later discovered is themed after a popular Japanese coffee brand. I had one of those Fried Pork Chops in breadcrumbs which was served on a bed of white &quot;pearl&quot; rice with Japanese (style?) curry sauce&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It occured to me that the rice may have been imported from Japan.&#xA;I was interested in trying to identify the lifetime cancer risk from such a potential exposure. I have read that ingested isotopes are worse than external contaminations which can be washed away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware that imports have been screened by the local goverment but from what I saw on the television it looked like they were just using dosimeters on the outside of cargo containers rather than in depth sampling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also wonder what types of isotopes would have been released in the fallout from Fukushima and in what ratios. I'm aware of Iodine 131, Caesium 134 and 147 and Strontium 90 but I've not seen any data on the ratios of what was released.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read that there are many Rice farms around Fukushima in Northern Japan and these provide approvimately 25% of the nations annual production.&lt;/p&gt;&#xA;" OwnerUserId="3492" LastActivityDate="2011-05-11T13:47:12.497" Title="What is the risk from radiation on imported food from Japan" Tags="&lt;radiation&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="9699" PostTypeId="2" ParentId="9698" CreationDate="2011-05-11T06:58:09.820" Score="4" ViewCount="" Body="&lt;p&gt;Indeed, it's likely that the rice came from Japan, and if it did, it's pretty likely it came from the Fukushima region which is famous across Japan and beyond for its rice - and other products. However, it could have been harvested before the tsunami. But as discussed here,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/8264/are-these-radioactive-particle-matter-and-air-emmissions-dangerous-2000km-from-f&quot;&gt;Are these radioactive particle matter and air emmissions dangerous, 2000KM from Fukushima?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the measurements indicate that the contamination of all these products has been minimal even when it was found, and you would need to eat many kilograms of this rice - or something else - to approach the radioactive dose that you obtain from one healthy banana (produced outside Japan).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, I can't guarantee that one particular package of rice couldn't contain an amount of radioactivity that is higher by three orders of magnitude - but it's just unlikely. Don't get me wrong: if I were you, I could react in the same sensitive way. But from my perspective right now, you will shorten your life much more visibly by the worries than by the actual radioactivity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radioactive content in that food is almost certainly pretty much indistinguishable from the natural levels. No one could have harvested rice and other plants in the truly contaminated areas around the power plant after the nuclear hassle (and only then it mattered) - because they had been evacuated. And the contamination in the more distant places is just small and when one computes what can get inside rice and survive for long enough time, it's negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, when you eat radioactive isotopes, it's actually not the most permanent source of radioactivity and cancer risk. This stuff mostly gets out of your body within two months or so. Breathing plutonium is more risky. See this summary of the health impact of the basic radioactive isotopes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2011/03/radioactivity-sieverts-and-other-units.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2011/03/radioactivity-sieverts-and-other-units.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, there is no reason to think that by eating the rice, you have been overexposed to these isotopes.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-11T06:58:09.820" CommentCount="1" />
  <row Id="9700" PostTypeId="2" ParentId="9696" CreationDate="2011-05-11T07:13:55.963" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://cosmos.asu.edu/publications/papers/QuantumFieldTheoryInDeSitterSpaceRenormalization%2030.pdf&quot; rel=&quot;nofollow&quot;&gt;The Bunch-Davies (BD) vacuum&lt;/a&gt; is a &quot;pure state&quot; when it comes to all the creation and annihilation operators in the patch where the BD vacuum is relevant. Inside this patch, we can say what operators annihilate the BD state which uniquely specifies it (well, for massive fields, one has the $\alpha$-vacua as generalizations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there's also a thermal radiation coming from the boundary of this patch. This is the reason for the non-conservation of the particle number. Conceptually, this non-conservation is equivalent to the ordinary thermal ensemble in the Minkowski space - which is a &quot;mixed state&quot;. So the two-point functions are essentially thermal two-point functions, and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The correlation functions are computed via Feynman rules that are analogous to those in the Minkowskian thermal state. We don't compute scattering amplitudes because there is no scattering S-matrix in the de Sitter space. That's because an observer's causal diamond &lt;a href=&quot;http://arxiv.org/abs/hep-th/0412197&quot; rel=&quot;nofollow&quot;&gt;misses almost all of the asymptotic regions&lt;/a&gt; where global in- and out-states may be defined.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, it is not true that the Feynman diagrams for thermal field theory require the particle numbers to be well-defined (thermal states are always mixtures of different particle number states) or conserved (interactions always change the number of particles, and interactions had to be incorporated to get the right thermal state, anyway). The number of particles is &lt;em&gt;not&lt;/em&gt; a good quantum number in violently cosmological backgrounds and one will find nothing interesting if he or she focuses on this &lt;em&gt;bad&lt;/em&gt; quantum number, assuming that it carries some wisdom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the level of field theory in a curved spacetime, the right rules are known and analogous to thermal QFT; correlation functions and not (unmeasurable) amplitudes may be calculated. When gravity is made dynamical, the calculations only work at the semiclassical level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Beyond semiclassical QFT in curved spaces&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are subtle things and no one knows the totally rigorous way to describe the de Sitter space. In particular, we don't know whether the relevant Hilbert space is finite-dimensional (as suggested by the finite de Sitter entropy) or infinite-dimensional; whether the degrees of freedom behind the cosmic horizon are &quot;independent&quot; or just reparametrizations of the degrees of freedom inside the visible universe, because of the complementarity principle; whether there can be a fully predictive framework or not, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, at the same moment, one should realize that all these uncertainties are &quot;very tiny&quot; when the de Sitter radius is large. The thermal wavelength is comparable to the de Sitter radius and the expected number of thermal particles within the whole visible part of the de Sitter space is of order one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we're really uncertain about roughly one photon and one graviton that may or may not live somewhere in the whole visible Universe and whose energy is so tiny that the wavelength is comparable to the de Sitter radius, too. In any practical context, even during inflation, this is a very small effect.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-11T08:12:38.027" LastActivityDate="2011-05-11T08:12:38.027" />
  <row Id="9701" PostTypeId="2" ParentId="9516" CreationDate="2011-05-11T07:23:51.213" Score="1" ViewCount="" Body="&lt;p&gt;There's an exact solution. The Majumdar-Papapetrou solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ds^2 = - \left( 1 + \frac{m_1}{\left| \vec{x} - \vec{x}_1 \right|} + \frac{m_2}{\left|  \vec{x} - \vec{x}_2 \right|} \right)^{-2} dt^2 + \left( 1 + \frac{m_1}{\left| \vec{x} - \vec{x}_1 \right|} + \frac{m_2}{\left|  \vec{x} - \vec{x}_2 \right|} \right)^2 d\vec{x}\cdot d\vec{x}$$&#xA;$$A_t = \pm \left( 1 + \frac{m_1}{\left| \vec{x} - \vec{x}_1 \right|} + \frac{m_2}{\left|  \vec{x} - \vec{x}_2 \right|} \right)^{-1},\, \vec{A}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left| \vec{x}_2 - \vec{x}_1 \right|$ can be arbitrarily small, but the distance between both event horizons will always be infinite. The areas of the event horizons will always remain $4\pi m_1^2$ and $4\pi m_2^2$ respectively, and the geometry will always exactly be that of a sphere.&lt;/p&gt;&#xA;" OwnerUserId="3493" LastActivityDate="2011-05-11T07:23:51.213" />
  <row Id="9702" PostTypeId="1" CreationDate="2011-05-11T07:27:16.997" Score="0" ViewCount="125" Body="&lt;p&gt;Here is a description of the principle of the operation of a new lightning device:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;During a storm the ambient electric field may rise to&#xA;between 10 to 20 kV/m. As soon as the field exceeds&#xA;a thresold representing the minimum risk of a lightning&#xA;strike, the OPR lightning terminal is activated. &lt;strong&gt;It draw&#xA;its energy from the ambient electric field the energy&#xA;required to generate high voltage pulses, creating&#xA;and propagating an upward leader.&lt;/strong&gt; No other power&#xA;sources are required, and no radioactive components&#xA;are used.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It remains unclear to me the physics of the drawing the energy from the electric field. Obviously, you can not draw the energy from the electrostatic field, no matter how strong it is. So i suppose that the electric field must oscillate more or less. The question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is a minimal frequency of the oscillations so that such device can work properly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://2.imimg.com/data2/LX/XS/MY-/la_opr-ese-terminal.pdf&quot; rel=&quot;nofollow&quot;&gt;A full description of the device&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="552" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-05-11T08:37:44.983" LastActivityDate="2011-05-12T04:13:32.707" Title="A lightning protection device physics" Tags="&lt;electromagnetism&gt;&lt;geophysics&gt;&lt;electrical-engineering&gt;&lt;lightning&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="9703" PostTypeId="2" ParentId="9696" CreationDate="2011-05-11T07:31:02.860" Score="-2" ViewCount="" Body="&lt;p&gt;You're not answering my question at all. This is a question about quantum field theory in curved spacetime, not quantum gravity. This is also a question about perturbation theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not asking why we have a Bunch-Davies state. I'm asking how to compute it using perturbation theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also not asking about the current accelerating universe, but the inflationary epoch, when the de Sitter radius was tiny.&lt;/p&gt;&#xA;" OwnerUserId="3494" LastActivityDate="2011-05-11T07:31:02.860" CommentCount="2" />
  <row Id="9704" PostTypeId="1" CreationDate="2011-05-11T07:48:15.683" Score="1" ViewCount="109" Body="&lt;p&gt;Ref: &lt;a href=&quot;http://en.wikipedia.org/wiki/Yang%E2%80%93Mills_existence_and_mass_gap&quot; rel=&quot;nofollow&quot;&gt;Yang–Mills existence and mass gap&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone please explain why the Yang–Mills Existence and Mass Gap problem is so important / fundamental to contemporary mathematics (and, presumably, theoretical / mathematical physics)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible, please include the following in your answer / response:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Why the problem is so fundamental / important - for both mathematics and theoretical / mathematical physics;&#xA;(2) How the solution to the problem would impact research in the above two fields;&#xA;(3) (If applicable) What progress has been made so far in resolving it; (references to papers, books, other resources would be helpful).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm an upper-level undergraduate in theoretical physics, and so am seeking an answer / response at such level.&lt;/p&gt;&#xA;" OwnerUserId="1760" LastActivityDate="2011-05-11T08:36:11.913" Title="Why is the Yang–Mills existence and mass gap problem so fundamental?" Tags="&lt;mathematical-physics&gt;&lt;mathematics&gt;" AnswerCount="1" />
  <row Id="9705" PostTypeId="1" AcceptedAnswerId="9713" CreationDate="2011-05-11T07:55:58.093" Score="1" ViewCount="153" Body="&lt;p&gt;In the following excerpt from S. Gasiorowicz's &lt;em&gt;Quantum Physics&lt;/em&gt;, he derives an expression for the average momentum of a free particle. $\psi(x,t)$ is the wave function of a free particle, $\psi^*$ denotes its complex conjugate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We try the following: since classically,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$ p = mv = m\frac{dx}{dt} $$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;we shall write&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$ &amp;lt;p&amp;gt; = m\frac{d}{dt}&amp;lt;x&amp;gt; = m\frac{d}{dt}\int{dx \psi^*(x,t) x \psi(x,t)} $$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This yields&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$ &amp;lt;p&amp;gt; = m\int_{-\infty}^\infty{dx\left( \frac{\partial\psi^*}{\partial t} x \psi + \psi^* x \frac{\partial\psi}{\partial t} \right)} $$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Note that there is no $dx/dt$ under the integral sign. The only quantity that varies with time is $\psi(x,t)$, and it is this variation that gives rise to a change in $x$ with time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I seem to have trouble understanding the difference between the position $x$ and the average position $&amp;lt;x&amp;gt;$. Why can it be assumed that $\frac{dx}{dt}=0$? What &lt;em&gt;is&lt;/em&gt; x?&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2011-05-12T08:30:18.697" Title="Why is $\frac{dx}{dt}=0$ in this average momentum calculation?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" />
  <row Id="9706" PostTypeId="1" AcceptedAnswerId="9709" CreationDate="2011-05-11T08:06:23.997" Score="1" ViewCount="181" Body="&lt;p&gt;At low energy, 511 keV, the value of the fine structure constant is 1/137.03599...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At Planck energy $\sqrt{\frac{\hbar c^5}{G}}$, or  1.956 $\times$ 10&lt;sup&gt;9&lt;/sup&gt; Joule, or  1.22 $\times$ 10&lt;sup&gt;28&lt;/sup&gt; eV, it has a value around 1/100 .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know a more precise value, maybe with error bars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarification: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to take the famous &quot;semi-official&quot; plot for the standard model: go to page &lt;a href=&quot;http://www-ekp.physik.uni-karlsruhe.de/~deboer/&quot; rel=&quot;nofollow&quot;&gt;http://www-ekp.physik.uni-karlsruhe.de/~deboer/&lt;/a&gt; then click on &quot;our research&quot; then go towards the end of the page:  the graph is the link &quot;Unification plot.ps (update from Phys. Lett. B260 (1991) 447)&quot;. Now take the left graph (not the right one for supersymmetry). The upper curve shows the fine structure constant, but scaled with some funny numbers suggested by GUTs. The precise question is: what value for the fine structure constant would this normal QED/standard model graph show at Planck energy if these funny GUT numbers would not be in the graph? My estimate is about 1/100, but I would be interested in a more precise value. It is clear that this question &lt;em&gt;assumes&lt;/em&gt;, like the graph itself, that both QED and the standard model are correct all the way up to Planck energy.&lt;/p&gt;&#xA;" OwnerUserId="3495" LastEditorUserId="3495" LastEditorDisplayName="" LastEditDate="2011-05-13T07:21:22.453" LastActivityDate="2011-05-13T07:21:22.453" Title="What is the value of the fine structure constant at Planck energy?" Tags="&lt;quantum-field-theory&gt;&lt;quantum-electrodynamics&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="9707" PostTypeId="2" ParentId="9705" CreationDate="2011-05-11T08:17:07.320" Score="3" ViewCount="" Body="&lt;p&gt;This $x$ is a position in the reference frame's coordinate system, which is just static by design. You can imagine it as a ruler, with a probability cloud in a foreground; the ruler stays on its place while the cloud moves and deforms changing its mean position.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2011-05-11T08:17:07.320" />
  <row Id="9708" PostTypeId="1" CreationDate="2011-05-11T08:20:26.863" Score="1" ViewCount="342" Body="&lt;p&gt;I wanted to know that if FIRE is Matter or Energy. I know that both are inter-convertible. but that doesn't mean there is interconversion taking place just like nuclear fusion or fission. If it is matter, then its has to occupy space and have mass. I really doubt that fire has mass. If it is energy, then i don't get it. Please enlighten me from my ignorance.&lt;/p&gt;&#xA;" OwnerUserId="3496" LastEditorUserId="987" LastEditorDisplayName="" LastEditDate="2011-06-18T21:00:14.040" LastActivityDate="2011-06-19T01:34:26.823" Title="Is fire matter or energy?" Tags="&lt;energy&gt;&lt;states-of-matter&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="9709" PostTypeId="2" ParentId="9706" CreationDate="2011-05-11T08:20:50.903" Score="6" ViewCount="" Body="&lt;p&gt;Dear Claude, you are extrapolating electromagnetism way too high. You're going from low energies to the Planck scale, assuming that nothing qualitatively changes, but this assumption is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fine-structure constant is essentially constant below the mass of the electron - the lightest charged particle - which is 511,000 eV or so. You are extrapolating the running of the electromagnetic fine-structure constant $\alpha = 1/137.03599$ all the way up to the Planck scale, about 10,000,000,000,000,000,000,000,000,000 eV. I chose to avoid the scientific notation to make it more explicit how far you have extrapolated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, at the electroweak scale, about 247,000,000,000 eV, which is much lower than the Planck scale, the electromagnetic force is no longer the right description. The weak nuclear force gets as strong and important as electromagnetism and in fact, they start to mix in nontrivial ways. The right theory doesn't use $U(1)_{\rm electromagnetism}$ but $SU(2)_{\rm weak}\times U(1)_{\rm hypercharge}$. Note that the electromagnetic $U(1)$ is not the same thing as the hypercharge $U(1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So instead of the fine-structure for the electromagnetic $U(1)$, one must express physics in terms of the fine-structure constants for the new $SU(2)$ and $U(1)$ electroweak gauge groups. These fine-structure constants are not as tiny as the electromagnetic one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hypercharge $U(1)$ fine-structure constant gets stronger as the energy grows - much like the electromagnetic one would - while the $SU(2)$ fine-structure constant gets weaker as the energy grows (however, it would be also getting stronger, like electromagnetism, if we included new $SU(2)$-charged particles such as superpartners).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same time, the $SU(3)$ QCD fine-structure constant is getting weaker as we raise the energy. Ultimately, all the three fine-structure constants, when properly normalized, become close near $10^{16}$ GeV which is the GUT scale - they exactly cross if one includes the superpartners of the known particles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The common value of all these three fine-structure constants at the GUT scale is something like $1/24$ or $1/25$ - in fact, there exist somewhat preliminary arguments based on F-theory (in string theory) and similar frameworks that suggest that the number could be exactly $1/24$ or $1/25$ when certain conventions are carefully followed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your question was not one about the real world, but one about a fictitious world that only contains QED up to the Planck scale, then indeed, the fine-structure constant would increase just &quot;somewhat&quot;, perhaps to $1/100$ or so; indeed, if the world were pure QED, the Landau pole with $\alpha=\infty$ would occur at much higher energies than the Planck scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number can be calculated as a function of $\alpha_{E=0}$. However, it is meaningless to assign error margins to a theory with adjustable parameters that doesn't describe the real world - and QED above the electroweak scale doesn't. A theory that doesn't describe the real world is always in error, even when you're within any error margins, and you can't just &quot;match&quot; it to the real world because different ways of &quot;matching&quot; two different theories would yield different results.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-11T08:20:50.903" CommentCount="6" />
  <row Id="9710" PostTypeId="2" ParentId="9708" CreationDate="2011-05-11T08:27:27.177" Score="7" ViewCount="" Body="&lt;p&gt;Fire is neither.  Fire is a &lt;em&gt;process&lt;/em&gt; involving both.  Fire is the energetic combination of various substances with oxygen to release light and heat.  In a gas fire, such as might be found on a stove or in a heater, a light hydrocarbon such propane is broken down into components of hydrogen and carbon which unite with oxygen from the atmosphere to form water and carbon dioxide.  In an ordinary wood fire, heat causes the wood to break down and release hydrocarbons which then burn as propane would.&lt;/p&gt;&#xA;" OwnerUserId="1233" LastActivityDate="2011-05-11T08:27:27.177" CommentCount="3" />
  <row Id="9712" PostTypeId="2" ParentId="9708" CreationDate="2011-05-11T08:35:12.493" Score="1" ViewCount="" Body="&lt;p&gt;welcome to the site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your profile you do not state your background , but from your question I assume that physics is not your strong point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this &lt;a href=&quot;http://en.wikipedia.org/wiki/Fire&quot; rel=&quot;nofollow&quot;&gt;Wiki article&lt;/a&gt; fire is described in physics terms, so I will not repeat it here.&#xA;Wnoise is correct that fire is a process, so your question is more appropriate to &quot;flames&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the article you can see that flames are gases in various temperatures while molecular interactions   increase the kinetic energy of the molecules and are also releasing radiation which goes into heat and light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A flame has both matter and energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matter is  the molecules of the gas, Oxygen, Carbon, which combine and release radiation by combining,  which radiation heats the whole gas/air mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy in the flame is in the form of heat, i.e. kinetic energy of the molecules,plus in the form of electromagnetic radiation, photons, both infrared and visible. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-05-11T12:04:35.693" LastActivityDate="2011-05-11T12:04:35.693" CommentCount="1" />
  <row Id="9713" PostTypeId="2" ParentId="9705" CreationDate="2011-05-11T08:35:57.640" Score="4" ViewCount="" Body="&lt;p&gt;The confusion seems to stem from a) not understanding what kind of objects you are dealing with and b) usual custom of not writing (all) arguments of functions when they are understood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify a) note that the position operator $\hat x$ &lt;strong&gt;does not&lt;/strong&gt; depend on time, and so also its kernel $\left&amp;lt; x \right | \hat{x} \left | x&amp;#39; \right&amp;gt; = x\delta(x-x&amp;#39;)$ with respect to &quot;position vectors&quot; $\left | x \right &amp;gt;$  also doesn't depend on time. This $x$ is the one that is present in your integral. So in particular ${{\rm d} x \over {\rm d} t} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the average of the operator $\hat A$ in the state $\psi$ (which depends on time) obviously depends on the state $\psi$: $\left&amp;lt; \hat {A} \right&amp;gt; := \left&amp;lt; \psi \right | \hat {A} \left | \psi \right&amp;gt;$ and so if you perform averages on a family of vectors $\psi(t)$ so also the average will depend on time. In your case, this should be written $\left&amp;lt; \hat{x} \right &amp;gt; (t)$ to make it obvious that one is dealing with a function of time. But this dependence is usually understood and omitted.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-11T10:59:24.950" LastActivityDate="2011-05-11T10:59:24.950" CommentCount="1" />
  <row Id="9714" PostTypeId="2" ParentId="9704" CreationDate="2011-05-11T08:36:11.913" Score="1" ViewCount="" Body="&lt;p&gt;Dear Anonymous, gauge theories such as QCD are among the easiest theories to be formulated yet richest theories when it comes to the phenomena they cover, many of which are important in the real world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The existence of the mass gap - the absence of arbitrarily small positive values in the mass spectrum - is a simple property of QCD that holds but that hasn't been rigorously demonstrated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To demonstrate it and win the $1 million from the Clay Institute, she has to define the quantum field theory at a rigorous mathematical level and master much of its physics in an equally rigorous way. So it's a good, simple enough to be formulated, mathematical problem whose solution would bring mathematicians' mastery to a higher level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same moment, the paper that would win the $1 million award would almost certainly not be very important for physicists. Physicists have found lots of complementary ways and insights that made them sure that the mass gap exists. Harboring doubts about the mass gap or trying to &quot;totally&quot; eliminate these doubts is simply not what theoretical physicists in this discipline spend most of their man-hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The evidence that the mass gap is real comes from renormalization group calculations of the strength of various interactions; simulations in lattice QCD; and, among other approaches, the most modern tools are based on the holographic AdS/CFT correspondence. These physical insights are arguably much more important and valuable than whatever could be included in the hypothetical math paper that proves the existence of the mass gap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would summarize this answer by saying that despite the positive hype I started with, and despite the prize that has been offered, a fully mathematical proof of the mass gap is not one of the most important problems in physics - and maths. When it comes to maths, I personally view it as much less profound than e.g. the Riemann Hypothesis. When it comes to physics, I could enumerate hundreds of problems that are more important than a rigorous proof of the mass gap.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-11T08:36:11.913" CommentCount="2" />
  <row Id="9715" PostTypeId="1" AcceptedAnswerId="10980" CreationDate="2011-05-11T08:40:40.643" Score="5" ViewCount="157" Body="&lt;p&gt;Consider a collimated light beam in a vacuum (I am being unspecific about the frequency, anything from radio to gamma). If the beam power/cross section was increased &lt;em&gt;indefinitely&lt;/em&gt; would new phenomena occur at certain energy densities?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I gather that pair-production couldn't occur due to momentum considerations but would there be some interaction with the quantum vacuum? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At sufficiently high power is there a self-gravitation effect of the beam, a point touched on by a &lt;a href=&quot;http://physics.stackexchange.com/questions/6197/do-two-beams-of-light-attract-each-other-in-general-theory-of-relativity&quot;&gt;previous question&lt;/a&gt;? What if the beam was reflected back on itself by a mirror?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: the motivation for this question relates to theoretical limits for beam weapons.&lt;/p&gt;&#xA;" OwnerUserId="1394" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-05-11T10:44:06.527" LastActivityDate="2011-07-27T09:54:35.283" Title="Very High Power Light Beams" Tags="&lt;electromagnetic-radiation&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="9716" PostTypeId="1" AcceptedAnswerId="9726" CreationDate="2011-05-11T08:47:27.303" Score="5" ViewCount="141" Body="&lt;p&gt;In that &lt;a href=&quot;http://physics.stackexchange.com/questions/3228/is-there-any-thing-other-than-time-that-triggers-a-radioactive-atom-to-decay&quot;&gt;question&lt;/a&gt; and its answers it was mentioned that you could trigger radioactive decay by bombarding atoms with gamma rays of the right energy level (there may be other solutions I do not know about, but of course if you bombard with neutrons you can trigger nuclear reactions)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am mainly interested by beta-decay. Is it possible to &lt;em&gt;decrease&lt;/em&gt; the probability of the beta decay of some radioactive material by a physical treatment? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the rate fairly independant of temperature and external magnetic field for instance?&lt;/p&gt;&#xA;" OwnerUserId="3497" LastActivityDate="2011-05-13T17:16:06.677" Title="Is there a way to decrease the rate of nuclear Beta decay?" Tags="&lt;experimental-physics&gt;&lt;radioactivity&gt;&lt;radiation&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="9717" PostTypeId="2" ParentId="9702" CreationDate="2011-05-11T09:09:52.073" Score="1" ViewCount="" Body="&lt;p&gt;There is not only an electric field in air, there is a small &#xA;current as well. So called air ions (rather big/slow aggregates &#xA;of several molecules of nitrogen/oxygen) stream along the field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This small current can be collected by a antenna-like device,&#xA;and stored in a capacitor. This is a question of the most &#xA;sophisticated  insulation properties of this antenna/capacitor and all &#xA;the included electronics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that the apparatus contains a vacuum capacitor of, &#xA;maybe some nanofarad, and a electronic circuit to measure &#xA;the voltage, and on exceeding the threshold, makes some &#xA;high voltage impulse to the tip to create a streamer. &#xA;What I'd like to know is, how do they protect the circuit &#xA;from the lightning which then flashes along that streamer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radio hams are well aware of such effects, and connect some &#xA;overvoltage protection to their antannas, to avoid shocks/damage &#xA;under fine weather conditions. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which is a minimal frequency of the&#xA;  oscillations so that such device can&#xA;  work properly?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This part of Your question I do not understand. D You have any &#xA;information on the interior of the device? What &quot;frequency&quot; do &#xA;You refer to?&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-05-11T17:39:17.073" LastActivityDate="2011-05-11T17:39:17.073" CommentCount="7" />
  <row Id="9718" PostTypeId="1" CreationDate="2011-05-11T09:11:29.530" Score="2" ViewCount="134" Body="&lt;p&gt;I've noticed that for many solids, specific heat is around 1 Kj/Kg*K is there a theoretical explanation for this?  What common solids have the highest heat capacity per mass and per volume?&lt;/p&gt;&#xA;" OwnerUserId="3311" LastEditorUserId="3311" LastEditorDisplayName="" LastEditDate="2011-05-11T12:21:14.440" LastActivityDate="2011-05-13T06:45:59.337" Title="Solids with very large specific heats" Tags="&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="9719" PostTypeId="2" ParentId="9686" CreationDate="2011-05-11T09:20:43.277" Score="7" ViewCount="" Body="&lt;p&gt;The action $S$ is not a well-known object for the laymen; however, when one seriously works as a physicist, it becomes as important and natural as the energy $H$. So the action is probably unintuitive for the inexperienced users - and there's no reason to hide it - but it is important for professional physicists, especially in particle and theoretical physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The OP's statement that the Hamiltonian corresponds to energy is a vacuous tautology because the Hamiltonian is a technical &lt;em&gt;synonym&lt;/em&gt; for energy. In the same way, one might say that the action intuitively corresponds to Wirkung (a German name) because it's the same thing, too. Because it now has two names, it becomes more natural :-) and the OP could also blame the energy for having &quot;unnatural&quot; units of action per unit time. In other words, the question assumes that energy (and its unit) is more fundamental and intuitive than the action (and its unit) - so it shouldn't be surprising that using his assumptions, the OP may also &quot;deduce&quot; the conclusion that the energy is more fundamental and intuitive than the action. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is the assumption = conclusion right? Well, energy is intuitive because it's conserved, and the action is intuitive because it's minimized - so there is no qualitative difference in their importance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the only difference is that non-physicists don't learn to use the action at all. The energy may be imagined as &quot;potatoes&quot; which every can do; the action is an abstract score on the history that is only useful once we start to derive differential equations out of it - which almost no layman can imagine. If the laymen's experience with a concept measures whether something is &quot;intuitive&quot;, then the action simply is less intuitive and there is no reason to pretend otherwise. However, physicists learn that it's in some sense more fundamental than the energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, the Hamiltonian is the key formula defining time evolution in the Hamiltonian picture while the action is the key formula to define the evolution in the nicer, covariant, &quot;spacetime&quot; picture, which is why HEP physicists use it all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What the action is in general&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, the main raison d'etre for the action is the principle of least action, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Principle_of_least_action&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Principle_of_least_action&lt;/a&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is what everyone should learn if he wants to know anything about the action itself. Historically, this principle - and the concept of action - generalized various rules for the light rays that minimize time to get somewhere, and so on. It makes no sense to learn about a quantity without learning about the defining &quot;application&quot; that makes it important in physics. Energy is defined so that it's conserved whenever the laws of Nature are time-translational symmetric; and action is defined as whatever is minimized by the history that the system ultimately takes to obey the same laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy is a property of a system at a fixed moment of time - and because it's usually conserved, it has the same values at all moments. On the other hand, the action is not associated with the state of a physical object; it is associated with a history.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one point I need to re-emphasize. For particular systems, there may exist particular &quot;defining&quot; formulae for the Hamiltonian or the action, such as $E=mv^2/2$ or $S = \int dt(mv^2/2-kx^2/2)$. However, they're not the most universal and valid &lt;em&gt;definitions&lt;/em&gt; of the concepts. These formulae don't explain why they were chosen in this particular way, what they're good for, and how to generalize them in other systems. And one shouldn't be surprised that one may derive the right equations of motion out of these formulae for $H$ or $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, the energy is universally &lt;em&gt;defined&lt;/em&gt; in such a way that it is conserved as a result of the time-translational symmetry; and the action is &lt;em&gt;defined&lt;/em&gt; in such a way that the condition $\delta S = 0$ (stationarity of the action) is equivalent to the equations of motion. These are the general conditions that define the concepts in general and that make them important; particular formulae for the energy or action are just particular applications of the general rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the text above, I was talking about classical i.e. non-quantum physics. In quantum physics, the action doesn't pick the only allowed history; instead, one calculates the probability amplitudes as sums over all histories weighted by $\exp(iS/\hbar)$ which may be easily seen to reduce to the classical predictions in the classical limit. A stationary action of a history means that the nearby histories have a similar phase and they constructively interfere with each other, making the classically allowed history more important than others.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-11T09:29:13.563" LastActivityDate="2011-05-11T09:29:13.563" />
  <row Id="9720" PostTypeId="1" AcceptedAnswerId="9721" CreationDate="2011-05-11T09:34:34.640" Score="4" ViewCount="248" Body="&lt;p&gt;On a quantum scale the smallest unit is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck_scale&quot; rel=&quot;nofollow&quot;&gt;Planck scale&lt;/a&gt;, which is a discrete measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There several question that come to mind:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Does that mean that particles can only live in a discrete grid-like structure,  i.e. have to &quot;magically&quot; jump from one pocket to the next? But where are they in between? Does that even give rise to the old paradox that movement as such is impossible (e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Zeno%27s_paradoxes&quot; rel=&quot;nofollow&quot;&gt;Zeno's paradox&lt;/a&gt;)?&lt;/li&gt;&#xA;&lt;li&gt;Does the same hold true  for time (i.e. that it is discrete) - with all the ensuing paradoxes?&lt;/li&gt;&#xA;&lt;li&gt;Mathematically does it mean that you have to use difference equations instead of differential equations? (And sums instead of integrals?)&lt;/li&gt;&#xA;&lt;li&gt;From the point of view of the space metric do you have to use a discrete metric (e.g. the &lt;a href=&quot;http://en.wikipedia.org/wiki/Manhattan_metric&quot; rel=&quot;nofollow&quot;&gt;Manhatten metric&lt;/a&gt;) instead of good old Pythagoras?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thank you for giving me some answers and/or references where I can turn to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I just saw this call for papers - it seems to be quite a topic after all:&#xA;&lt;a href=&quot;http://www.fqxi.org/community/essay&quot; rel=&quot;nofollow&quot;&gt;http://www.fqxi.org/community/essay&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2011-05-25T14:47:30.813" LastActivityDate="2011-05-25T14:47:30.813" Title="Is reality discrete at the quantum level? (...and what does it imply not only mathematically?)" Tags="&lt;quantum-mechanics&gt;&lt;spacetime&gt;&lt;space&gt;&lt;time&gt;&lt;planck&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="4" />
  <row Id="9721" PostTypeId="2" ParentId="9720" CreationDate="2011-05-11T09:45:39.347" Score="5" ViewCount="" Body="&lt;p&gt;The answer to all questions is No. In fact, even the right reaction to the first sentence - that the Planck scale is a &quot;discrete measure&quot; - is No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Planck length is a particular value of distance which is as important as $2\pi$ times the distance or any other multiple. The fact that we can speak about the Planck scale doesn't mean that the distance becomes discrete in any way. We may also talk about the radius of the Earth which doesn't mean that all distances have to be its multiples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum gravity, geometry with the usual rules doesn't work if the (proper) distances are thought of as being shorter than the Planck scale. But this invalidity of classical geometry doesn't mean that anything about the geometry has to become discrete (although it's a favorite meme promoted by popular books). There are lots of other effects that make the sharp, point-based geometry we know invalid - and indeed, we know that in the real world, the geometry collapses near the Planck scale because of other reasons than discreteness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum mechanics got its name because according to its rules, &lt;em&gt;some&lt;/em&gt; quantities such as energy of bound states or the angular momentum can only take &quot;quantized&quot; or discrete values (eigenvalues). But despite the name, that doesn't mean that &lt;em&gt;all&lt;/em&gt; observables in quantum mechanics have to possess a discrete spectrum. Do positions or distances possess a discrete spectrum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proposition that distances or durations become discrete near the Planck scale is a scientific hypothesis and it is one that may be - and, in fact, has been - experimentally falsified. For example, these discrete theories inevitably predict that the time needed for photons to get from very distant places of the Universe to the Earth will measurably depend on the photons' energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Fermi satellite has showed that the delay is zero within dozens of milliseconds&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2009/08/fermi-kills-all-lorentz-violating.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2009/08/fermi-kills-all-lorentz-violating.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which proves that the violations of the Lorentz symmetry (special relativity) of the magnitude that one would inevitably get from the violations of the continuity of spacetime have to be much smaller than what a generic discrete theory predicts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the argument used by the Fermi satellite only employs the most straightforward way to impose upper bounds on the Lorentz violation. Using the so-called birefringence, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1102.2784&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1102.2784&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;one may improve the bounds by 14 orders of magnitude! This safely kills any imaginable theory that violates the Lorentz symmetry - or even continuity of the spacetime - at the Planck scale. In some sense, the birefringence method applied to gamma ray bursts allows one to &quot;see&quot; the continuity of spacetime at distances that are 14 orders of magnitude shorter than the Planck length. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't mean that all physics at those &quot;distances&quot; works just like in large flat space. It doesn't. But it surely does mean that some physics - such as the existence of photons with arbitrarily short wavelengths - has to work just like it does at long distances. And it safely rules out all hypotheses that the spacetime may be built out of discrete, LEGO-like or any qualitatively similar building blocks.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-11T10:35:05.883" LastActivityDate="2011-05-11T10:35:05.883" CommentCount="4" />
  <row Id="9722" PostTypeId="2" ParentId="9606" CreationDate="2011-05-11T10:28:08.637" Score="1" ViewCount="" Body="&lt;p&gt;1st, UV-spectrum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is none! To have some absorption above 200 nm You need some &#xA;conjugated double bonds. The standard solvent for UV-spectroscopy is &#xA;ethanol for that reasons, if one needed a much more polar solvent&#xA;one would use water or in extreme cases (of polarity) glycerol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd, Refractive index. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a real problem, I do not have this values. Maybe &#xA;You can use this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cauchy%27s_equation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Cauchy%27s_equation&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;to extrapolate values from visible domain to the UV. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/UV-Vis/spectrum.htm&quot; rel=&quot;nofollow&quot;&gt;http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/UV-Vis/spectrum.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is some link which might help You. Just google for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot; UV/Vis spectroscopy  &quot; or &quot; UV/Vis Spectroskopie&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the latter gives a lot of scripts from German universities, &#xA;the tables included and the formulas You should understand &#xA;without speaking German.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-05-11T15:17:11.983" LastActivityDate="2011-05-11T15:17:11.983" CommentCount="1" />
  <row Id="9723" PostTypeId="2" ParentId="9718" CreationDate="2011-05-11T10:38:58.453" Score="3" ViewCount="" Body="&lt;p&gt;Looking at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_capacity#Table_of_specific_heat_capacities&quot; rel=&quot;nofollow&quot;&gt;table&lt;/a&gt; of specific heat capacities on Wikipedia, there is quite a wide range of heat capacities for the solids listed and many are much less that $1~\mathrm{kJ}~\mathrm{kg}^{-1}~\mathrm{K}^{-1}$. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Dulong%E2%80%93Petit_law&quot; rel=&quot;nofollow&quot;&gt;Dulong-Petit law&lt;/a&gt; provides an approximate limit of $25~\mathrm{J}~\mathrm{mol}^{-1}~\mathrm{K}^{-1}$ for the &lt;em&gt;molar&lt;/em&gt; heat capacity of atomic crystals (and as you can see, predicts the molar heat capacities of many metals pretty well), but makes no statements about mass heat capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as solids with very high specific heat capacities go, try long chain hydrocarbons (paraffin and polyethylene are listed) or water ice, as well as anything that has a lot of water in it. For volumetric heat capacities, metals become competitive because of their typical high densities.&lt;/p&gt;&#xA;" OwnerUserId="1876" LastEditorUserId="1876" LastEditorDisplayName="" LastEditDate="2011-05-13T06:45:59.337" LastActivityDate="2011-05-13T06:45:59.337" CommentCount="1" />
  <row Id="9724" PostTypeId="2" ParentId="9715" CreationDate="2011-05-11T10:45:27.587" Score="0" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What if the beam was reflected back on&#xA;  itself by a mirror?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then the defense against this beam would be easy: &#xA;just coat the target with such a mirror surface. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-05-11T10:45:27.587" CommentCount="1" />
  <row Id="9725" PostTypeId="2" ParentId="9702" CreationDate="2011-05-11T11:10:57.197" Score="1" ViewCount="" Body="&lt;p&gt;The quotes you give do not really explain much. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a conductor one can short the static field of the atmosphere, one does not need oscillations for that. I think it was Benjamin Franklin with &lt;a href=&quot;http://en.wikipedia.org/wiki/Lightning&quot; rel=&quot;nofollow&quot;&gt; &lt;/a&gt; reaching the clouds who discovered that first hand?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lightening strikes when naturally induced upward leaders reach and meet  the charged cloud downward leaders creating a path of ions for the energy to go through. This happens when the fields are strong to strip neutral molecules from their electrons and create an ionisation path. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;, correction:&#xA;Just saw the link you provided. It seems that it works in channelling the lightning to the rod instead of playing russian roulette on where it will hit. It is the logic of simple lightning rods, so that the energy is sent to the ground without damaging the buildings. The proposed design makes sure that this will happen instead of leaving it to the probable fields that would develop naturally with the simple lightning rods.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-05-11T11:33:36.777" LastActivityDate="2011-05-11T11:33:36.777" CommentCount="2" />
  <row Id="9726" PostTypeId="2" ParentId="9716" CreationDate="2011-05-11T12:45:16.387" Score="5" ViewCount="" Body="&lt;p&gt;Yes. Have a &lt;a href=&quot;http://byzipp.com/animaProof.htm&quot; rel=&quot;nofollow&quot;&gt;look&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;“The ‘Reifenschweiler effect’ is the observation that the beta-decay of tritium half-life 12.5 years is delayed reversibly by about 25-30% when the isotope is absorbed in 15 nm titanium-clusters in a temperature window in between 160-275 C. Remarkably at 360 C the original radioactivity reappears. The effect is absent in bulk metal. Discovered around 1960/1962 at Philips Research Eindhoven, The Netherlands Reifenschweiler extensively discussed his observation with o.a Casimir (the director of research at the time), Kistemaker (ultracentrifuge expert), and although no satisfactory explanation was found, R. was allowed to publish it. At the time a unique example as to how an electronic environment might affect nuclear phenomena.” &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-11T12:45:16.387" CommentCount="5" />
  <row Id="9727" PostTypeId="1" AcceptedAnswerId="12469" CreationDate="2011-05-11T13:30:02.913" Score="17" ViewCount="448" Body="&lt;p&gt;Pioneer &lt;a href=&quot;http://en.wikipedia.org/wiki/Pioneer_10&quot; rel=&quot;nofollow&quot;&gt;10&lt;/a&gt; &amp;amp; &lt;a href=&quot;http://en.wikipedia.org/wiki/Pioneer_11&quot; rel=&quot;nofollow&quot;&gt;11&lt;/a&gt; are robotic space probes launched by the NASA in the early 1970's. After leaving our solar system, an &lt;strong&gt;unusual deceleration&lt;/strong&gt; of both spacecrafts has been measured to be approximately $$\ddot{r}_p = -(8.74&amp;#177;1.33)&amp;#215;10^{−10} \frac{m}{s^2}$$ with respect to our solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pioneer_anomaly#Proposed_explanations&quot; rel=&quot;nofollow&quot;&gt;Several attempts&lt;/a&gt; were made to &lt;strong&gt;explain this tiny effect&lt;/strong&gt;, called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pioneer_anomaly&quot; rel=&quot;nofollow&quot;&gt;Pioneer anomaly&lt;/a&gt;, but none was fully accepted in the scientific community so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two months ago, &lt;a href=&quot;http://arxiv.org/abs/1103.5222v1&quot; rel=&quot;nofollow&quot;&gt;Frederico Francisco et al&lt;/a&gt; have proposed another solution to the problem. They assume, roughly speaking, that the &lt;strong&gt;thermal radiation&lt;/strong&gt; of the spacecraft caused by plutonium on board along with the actual structure of the probes is responsible for this mystery of modern physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here an image of Pioneer 10 taken from Wikipedia along with a sketch of the &lt;strong&gt;radiation model&lt;/strong&gt; employed in the paper:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/Ukwcx.png&quot; alt=&quot;Pioneer 10 and heat model&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence my question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h3&gt;Is the Pioneer anomaly finally explained?&lt;/h3&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sincerely&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2011-08-20T12:24:38.130" Title="The Pioneer anomaly finally explained?" Tags="&lt;general-relativity&gt;&lt;astrophysics&gt;&lt;cosmology&gt;&lt;thermal-radiation&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="5" />
  <row Id="9728" PostTypeId="2" ParentId="9698" CreationDate="2011-05-11T13:47:12.497" Score="2" ViewCount="" Body="&lt;p&gt;To address the component about in-depth searches of containers - Iodine 131, Caesium 134 and 147 and Strontium 90 decays involve high energy gamma rays, which readily penetrate cargo containers, so scanning the surface of the shipping container is sufficient to detect these isotopes with the appropriate equipment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The science correspondent at National Public Radio in the US did a very nice piece in the aftermath of the disaster (&lt;a href=&quot;http://www.npr.org/blogs/health/2011/03/22/134746912/radioactive-milk-only-a-danger-after-58-000-glasses,&quot; rel=&quot;nofollow&quot; title=&quot;NPR story about the impact of Fukushima fallout on food&quot;&gt;link to NPR story about the impact of Fukushima fallout on food&lt;/a&gt;) where he interviewed a health physicist studying the fallout situation.  The health physicist had demonstrated that you would need to drink 58,000 glasses of milk contaminated at the highest levels found by the date of the article to raise your lifetime cancer risk by 4%. As Luboš noted above, the farmlands have been evacuated.  So the evacuation and halt in production combined with the scanning of imports at the port of arrival and the natural boost in isotopic reduction through decay means the likelihood of ingesting any amount of harmful isotope is small, and very likely nowhere near enough to cause serious health problems.&lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2011-05-11T13:47:12.497" />
  <row Id="9729" PostTypeId="1" CreationDate="2011-05-11T14:19:45.717" Score="1" ViewCount="88" Body="&lt;p&gt;consider object A with mass $m_{A}$ and positional vector $\overrightarrow{r_{A}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;object B with mass $m_{B}$ and positional vector $\overrightarrow{r_{B}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;object C with mass $m_{C}$ and positional vector $\overrightarrow{r_{C}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since reference frame is inertial (assumed) so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_{A}\frac{d^{2}}{dt^{2}}\overrightarrow{r_{A}}=\overrightarrow{F}{}_{AB}+\overrightarrow{F}{}_{AC}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow$$m_{A}\frac{d^{2}}{dt^{2}}\overrightarrow{r_{A}}=G\frac{m_{A}m_{B}}{\left|\overrightarrow{r_{B}}-\overrightarrow{r_{A}}\right|^{3}}(\overrightarrow{r_{B}}-\overrightarrow{r_{A}})+G\frac{m_{A}m_{C}}{\left|\overrightarrow{r_{C}}-\overrightarrow{r_{A}}\right|^{3}}(\overrightarrow{r_{C}}-\overrightarrow{r_{A}})$ ....[1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for potential energy equation is to be integrated with a positional&#xA;vector and i am trying&#xA;hard to figure out what that vector could be, but still no success. please&#xA;help&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Potential energy of a object due to another object&#xA;$m_{A}\frac{d^{2}}{dt^{2}}\overrightarrow{r_{A}}=\overrightarrow{F_{A}}=G\frac{m_{A}m_{B}}{\left|\overrightarrow{r_{A}}-\overrightarrow{r_{B}}\right|^{3}}(\overrightarrow{r_{B}}-\overrightarrow{r_{A}})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;assuming $\overrightarrow{r_{0}}=\overrightarrow{r_{B}}-\overrightarrow{r_{A}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow m_{A}\frac{d^{2}}{dt^{2}}\overrightarrow{r_{0}}=-(1+\frac{m_{A}}{m_{B}})(G\frac{m_{A}m_{B}}{r_{0}^{2}}\hat{r_{0}})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow\intop_{\overrightarrow{r_{0i}}}^{\overrightarrow{r_{0f}}}\left(m_{A}\frac{d^{2}}{dt^{2}}\overrightarrow{r_{0}}\right).d\overrightarrow{r_{0}}=-\intop_{\overrightarrow{r_{0i}}}^{\overrightarrow{r_{0f}}}\left((1+\frac{m_{A}}{m_{B}})(G\frac{m_{A}m_{B}}{r_{0}^{2}}\hat{r_{0}})\right).d\overrightarrow{r_{0}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow\left.\frac{1}{2}m_{A}v_{0}^{2}\right|_{v_{0}(\overrightarrow{r_{0i}})}^{v_{0}(\overrightarrow{r_{0f}})}=\left.G\frac{m_{A}(m_{A}+m_{B})}{r_{0}}\right|_{\overrightarrow{r_{0i}}}^{\overrightarrow{r_{0f}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So What positional vector should be integrated with equ [1] ?&#xA;How it is decided ? (i mean what kind of property(s) that vector should possess ?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_{\overrightarrow{r_{i}}}^{\overrightarrow{r_{f}}}\left(m_{A}\frac{d^{2}}{dt^{2}}\overrightarrow{r_{A}}\right).d\overrightarrow{r}=\int_{\overrightarrow{r_{i}}}^{\overrightarrow{r_{f}}}\left(G\frac{m_{A}m_{B}}{\left|\overrightarrow{r_{B}}-\overrightarrow{r_{A}}\right|^{3}}(\overrightarrow{r_{B}}-\overrightarrow{r_{A}})+G\frac{m_{A}m_{C}}{\left|\overrightarrow{r_{C}}-\overrightarrow{r_{A}}\right|^{3}}(\overrightarrow{r_{C}}-\overrightarrow{r_{A}})\right).d\overrightarrow{r}$&#xA;.....[2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;questions &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) what is $\overrightarrow{r}$ in terms of $\overrightarrow{r_{A}}$,$\overrightarrow{r_{B}}$or&#xA;$\overrightarrow{r_{C}}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) solve equ [2].&lt;/p&gt;&#xA;" OwnerUserId="3462" LastEditorUserId="3462" LastEditorDisplayName="" LastEditDate="2011-05-11T16:37:01.337" LastActivityDate="2011-09-03T23:44:55.793" Title="potential energy of an object due to other two objects" Tags="&lt;mathematical-physics&gt;&lt;energy&gt;&lt;energy-conservation&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="9730" PostTypeId="1" AcceptedAnswerId="9774" CreationDate="2011-05-11T14:46:53.030" Score="7" ViewCount="159" Body="&lt;p&gt;Before I start defining the situation and asking a question, I'd like to make a few things clear:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;this is &lt;strong&gt;not&lt;/strong&gt; a homework, merely a matter of personal interest and enthusiasm&lt;/li&gt;&#xA;&lt;li&gt;I am neither a physicist, nor a native English speaker, therefore I encourage anyone with proper knowledge to go ahead and edit anything in my post to correct/improve any terminology, thread title including&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'd like to start with an image I created for this purpose, I hope I didn't make any huge mistakes in it. So, here's the situation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Ommgm.png&quot; alt=&quot;The situation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a biker weighing &lt;em&gt;m1&lt;/em&gt; riding a bicycle weighing &lt;em&gt;m2&lt;/em&gt;, therefore the total weight &lt;em&gt;m = m1 + m2&lt;/em&gt;.&lt;br&gt;&#xA;The biker is driving at a constant speed of &lt;em&gt;v&lt;/em&gt; and is about to jump down from a higher ground to a lower ground, while the total height he's jumping, the height difference of those two grounds is &lt;em&gt;h&lt;/em&gt;.&lt;br&gt;&#xA;According to how fast he's going and how high he's jumping from, he'll jump a distance &lt;em&gt;d&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The question:&lt;/strong&gt; What's the biker's weight at the moment of landing with which he's &quot;pushing&quot; the bicycle down? As I'm not a physicist and I'm not sure what's the proper term, I'll try to form it like this:&lt;br&gt;&#xA;I know what the maximum weight the bicycle can hold is according to the manufacturer and I need to calculate whether I will exceed this limit when performing such a jump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Things to consider:&lt;/strong&gt; I believe there are many factors which can't be counted with properly, therefore&lt;br&gt;&#xA;I &lt;em&gt;think&lt;/em&gt;, and now correct me if I'm wrong:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;that we can assume that the forward speed when landing will be the same as the forward speed before jumping. Of course it won't, but I believe this speed will not drop so much, so possibly we can ignore it&lt;/li&gt;&#xA;&lt;li&gt;I believe there's a huge difference between locking one's joints and dropping with a rigid body versus using the proper technique, keeping joints free and flexing one's arms and legs to absorb the mass of landing. I don't know how it'd be possible to include this factor to the calculation, I'll let people decide.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Feel free to propose anything I forgot to include in my question, and again, if anyone who can edit posts can improve anything I wrote, please, don't hesitate and do so.&lt;/p&gt;&#xA;" OwnerUserId="2834" LastEditorUserId="2834" LastEditorDisplayName="" LastEditDate="2011-05-11T14:52:06.540" LastActivityDate="2011-05-12T20:23:18.203" Title="What's my weight when jumping on a bicycle from a higher to a lower ground?" Tags="&lt;kinematics&gt;" AnswerCount="3" CommentCount="7" />
  <row Id="9731" PostTypeId="1" AcceptedAnswerId="9735" CreationDate="2011-05-11T14:54:57.540" Score="1" ViewCount="324" Body="&lt;p&gt;I used to read the term &quot;pure energy&quot; in the context of matter antimatter annihilation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are them photons? is it other form of heat, are particles with mass ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does it refer to? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="1916" LastActivityDate="2011-06-07T12:08:07.000" Title="What is pure energy in matter antimatter annihilation?" Tags="&lt;energy&gt;&lt;antimatter&gt;" AnswerCount="4" />
  <row Id="9732" PostTypeId="1" CreationDate="2011-05-11T14:55:00.540" Score="5" ViewCount="157" Body="&lt;p&gt;From the Penrose diagram of de Sitter space, we see it has a future and past conformal boundary, and they are both spacelike. So, does de Sitter space admit an asymptotic S-matrix? Sure, in the usual coordinates which only cover half of the full space, we don't have an S-matrix because it's not causally complete, but in a coordinate system which covers the entire space, why not? Just because we don't have a globally timelike Killing vector field doesn't mean we can't have an S-matrix.&lt;/p&gt;&#xA;" OwnerUserId="3500" LastActivityDate="2011-05-11T18:08:37.973" Title="Does de Sitter space admit an asymptotic S-matrix?" Tags="&lt;quantum-field-theory&gt;&lt;de-sitter-spacetime&gt;" AnswerCount="1" />
  <row Id="9733" PostTypeId="2" ParentId="9730" CreationDate="2011-05-11T15:00:02.297" Score="2" ViewCount="" Body="&lt;p&gt;The most crucial factor here is of course the nature of the solid you land on. If you're landing on a mattress, your bike can handle a much larger height of fall than if you're landing on concrete (much much larger height if it's water). This is because the average force your bike endures during contact is inversely proportional to the time of contact, i.e, concrete will reduce your speed from $v$ to zero in milliseconds, whereas the mattress will take slightly longer $$F\approx m \frac{v}{t} = m\frac{\sqrt{2gh}}{t}$$ Dividing this by $g$ will give you the effective mass the bike feels during contact. But the formula is only useful if you know what the contact time $t$ is and I'm not sure if that's easy to determine for different materials.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-05-11T15:00:02.297" CommentCount="7" />
  <row Id="9734" PostTypeId="2" ParentId="9731" CreationDate="2011-05-11T15:03:42.370" Score="0" ViewCount="" Body="&lt;p&gt;In particle antiparticle annihilation the end products come from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_model&quot; rel=&quot;nofollow&quot;&gt;table of elementary&lt;/a&gt; particles  created so  that  quantum numbers are conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;photons, two to conserve momentum in the cmsystem, pions and other mesons etc.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-11T15:03:42.370" />
  <row Id="9735" PostTypeId="2" ParentId="9731" CreationDate="2011-05-11T15:05:56.127" Score="6" ViewCount="" Body="&lt;p&gt;If I ruled the world, I would ban the phrase &quot;pure energy&quot; in contexts like this. There's no such thing as pure energy!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When particles and antiparticles annihilate, the resulting energy can take many different forms -- one of the basic principles of quantum physics is that any process that's not forbidden (say, because of violation of some sort of conservation law) has some probability of happening. So when a proton and an antiproton annihilate, they can produce photons, or pairs of other particles and antiparticles, such as a neutrino-antineutrino pair, or a positron-electron pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although all sorts of things are possible, by far the most common product of matter-antimatter annihilation is photons, especially if the collision occurs at low energy. One reason is that lower-mass particles are easier to create than high-mass particles, and nothing has less mass than a photon. (Other particles, particularly neutrinos, have so little mass that they are &quot;effectively massless,&quot; but neutrinos are weakly interacting particles, which means that the probability of producing them is lower.)&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-05-11T15:05:56.127" CommentCount="5" />
  <row Id="9736" PostTypeId="2" ParentId="9731" CreationDate="2011-05-11T15:10:11.797" Score="1" ViewCount="" Body="&lt;p&gt;At low energies of matter-antimatter collision, the end products can only be massless particles like photons -- hence annihilation is conventionally associated with the production of radiation/light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At higher energies, the collision can have enough energy to generate pairs of massive particles like mesons and W bosons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LEP was an electron-positron collider and it has the record for producing the most massive particles from such an annihilation process (W+ and W- bosons).&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-05-11T15:10:11.797" CommentCount="1" />
  <row Id="9737" PostTypeId="2" ParentId="9730" CreationDate="2011-05-11T15:56:44.383" Score="2" ViewCount="" Body="&lt;p&gt;Well, this will only be a partial answer, because why the physics is simple, this is actually a relatively hard question to answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As has been alluded to, the force will depend heavily on the amount of time over which the impact occurs. This, in turn, depends on the rigidity of the objects involved. Let me give some background here. (I'm going to be very hand-wavey here, both because I'm trying to keep the concepts clear and because I'm short on time.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizontal motion of the bike is unimportant here, so lets imagine the bike and rider being dropped vertically from the same vertical height that the bike reaches during its jump. The instant before the bike makes contact with the ground, it is moving at some speed in the vertical direction $v_i$. Thus, it has momentum $p_i=mv_i$ and some kinetic energy $K=\frac{1}{2}mv_i^2$. During the collision, the vertical velocity of the bike is reduced to zero, so there must be some impulse exerted which equals the momentum of the bike. Impulse is force times time, and the force being exerted here is going to be varying strongly with time, so we're going to need an integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ p_i = \int_0^{t_c} F(\tau) d\tau $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $t_c$ is the time it takes for the collision to happen, and $\tau$ is just a dummy variable of integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinetic energy of the bike will also be reduced to zero, so the work done on the ground by the bike must equal the kinetic energy. Again we have an integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ K=\int_0^{x_f} F(\xi) d\xi $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x_f$ is the equilibrium deformation of the ground/bike/shock absorbers etc. and $\xi$ is again a dummy integration variable. Notice that this requires the collision force to come in as a function displacement, while the momentum relation requires it as a function of time. We can get one from the other by assuming that the whole system can be treated as a spring with effective spring constant $k_{eff}$, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F=k_{eff}x(t) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which means we ultimately need to find the deformation of the system as a function of time to give us both functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it would be quite reasonable here to make some approximations, and assume that the force is a constant equal to the average value of the time dependent force, but we are still left with this unknown $k_{eff}$ which will really be needed to solve our system of equations. In practice this parameter is very hard to know, or even guess at. It will also depend on the bike frame and, worst of all for those of us who like nice rigid mathematical certainty, will depend on how the rider uses his limbs to help cushion the impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very rough estimate (technique suggested by @dmckee) would be to just pick some reasonable value for $t_c$ and use that to calculate the average force, then double it to estimate the peak force. Using this technique, and estimating a quarter second for the collision ($t_c=0.25 \textrm{sec}$) we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{eqnarray}&#xD;&#xA;mv_i &amp;amp;= &amp;amp;\overline{F}t_c \\&#xD;&#xA;\frac{mv_i}{t_c} &amp;amp;= &amp;amp;\overline{F} \\&#xD;&#xA;\frac{mv_i}{0.25 \textrm{sec}} &amp;amp;= &amp;amp;\overline{F}&#xD;&#xA;\end{eqnarray} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\overline{F}$ is the average force. This is a very rough estimate, and will really only get you the right order of magnitude, but without a &lt;strong&gt;lot&lt;/strong&gt; of measurements to determine $k_{eff}$ it isn't really possible to do much better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I realized that I never explained how to get $v_i$. @dbrane included this in his answer, but I will plug the result into my equations for completeness. Assuming the bike drops from a height of $h$, whether this is the apex of a jump, or simply the height from which the pike is dropped, the velocity at the ground will be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v_i=\sqrt{2gh} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $g=9.8 \frac{m}{s^2}$ is the acceleration due to gravity. Plugging this in to the result above, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray}&#xD;&#xA;\frac{mv_i}{0.25 \textrm{sec}} &amp;amp;= &amp;amp;\overline{F}\\&#xD;&#xA;\frac{m\sqrt{2gh}}{0.25 \textrm{sec}} &amp;amp;= &amp;amp;\overline{F}\\&#xD;&#xA;\end{eqnarray} $$&lt;/p&gt;&#xA;" OwnerUserId="869" LastEditorUserId="869" LastEditorDisplayName="" LastEditDate="2011-05-11T16:03:00.777" LastActivityDate="2011-05-11T16:03:00.777" CommentCount="3" />
  <row Id="9738" PostTypeId="2" ParentId="8762" CreationDate="2011-05-11T16:39:54.347" Score="3" ViewCount="" Body="&lt;p&gt;There is the possibility to have particles at an intermediate scale which correct this dilemma, but this is not very appealing since the new scale is essentially a free parameter which takes away the cool GUT prediction for the strong coupling. Another alternative to supersymmetry is having other additional light particles at the TeV scale in the correct amount (for example four weak scalar doublets or something like that has been discussed before) which can very much improve the precision of unification in nonsupersymmetric models, and since there is more or less only a discrete parameter (the number of light new fields) it kind of is predictive and falsifiable, since those fields should in principle be observable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose the main reason why many people are scared away from such ideas is that one explicitely introduces new physics at the gut scale without solving the higgs hierarchy  problem. That means that at the GUT scale, you really do have one parameter in your theory that has to be tuned to many digits precision in order to get a small electroweak scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&lt;/p&gt;&#xA;" OwnerUserId="3502" LastActivityDate="2011-05-11T16:39:54.347" />
  <row Id="9739" PostTypeId="2" ParentId="9732" CreationDate="2011-05-11T17:24:41.787" Score="3" ViewCount="" Body="&lt;p&gt;Dear inflation, the answer is No, one can't define an asymptotic S-matrix in de Sitter space. See e.g. section V.A. in this paper by Bousso:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0412197&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0412197&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The problem is that the Penrose diagram of a de Sitter space starts and ends with a horizontal line. So at the beginning, the observer is forbidden - by causality - to set up the initial state on the initial horizontal line (because he can only affect its future light cone, a small triangle); and at the end, for a mirror reason, he can't measure what happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/GkBqC.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at most, if such an S-matrix existed, it would dictate the evolution of states that no one can prepare to states that no one can observe, as Bousso says. This S-matrix would not be an observable, it would be a &quot;meta-observable&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are good reasons to think that this is not just an aesthetic conflict with the &quot;testability&quot; criteria in science. It's usually the case in physics that if one can prove that something can't be measured, not even in principle, it doesn't make sense. Attempts to measure/define the values of $x,p$ with a high accuracy that violates the uncertainty principle is a canonical example. We know that it's not just because of our technological limitations; accurate values of $x,p$ simply can't exist at the same moment. The operators don't commute with one another, after all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the hypothetical de Sitter S-matrix &quot;meta-observable&quot;, the experience with the uncertainty principle suggests that e.g. the &quot;arbitrary&quot; initial state indeed can't be prepared - surely not by an observer, as argued above, but not even by God or Nature - because the degrees of freedom that belong to vastly separated regions on the infinite initial line are not really independent of each other. This notion is known as the complementarity - and while it's usually talked about in the case of black holes, de Sitter space offers a straightforward generalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;De Sitter space may be thought of as a black hole but the &quot;inner visible part&quot; of the Universe should be identified with the black hole &lt;em&gt;exterior&lt;/em&gt; while the black hole &lt;em&gt;interior&lt;/em&gt; is the infinite region behind the cosmic horizon of the de Sitter space. Much like the microscopic information about the black hole interior is ultimately imprinted in the Hawking radiation, one should believe that all the information about physics beyond the cosmic horizon is imprinted to detailed properties of particles in the visible patch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For similar reasons, the fact that the de Sitter horizon has a finite entropy - because the cosmic horizon has a finite area - should be taken pretty seriously and it arguably applies to the whole space and not just the visible patch. Because of the finiteness of the entropy, it seems that one can't define &lt;em&gt;any&lt;/em&gt; observable that is arbitrarily accurate and that depends on continuous parameters such as momenta, not even in principle.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-11T18:08:37.973" LastActivityDate="2011-05-11T18:08:37.973" CommentCount="3" />
  <row Id="9740" PostTypeId="2" ParentId="9476" CreationDate="2011-05-11T17:56:49.407" Score="-1" ViewCount="" Body="&lt;p&gt;The answer is yes. To Willie Wong's &quot;no&quot; answer I commented:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yes there is &quot;inherent fuzziness&quot; to use your words. But that doesn't mean the EP can't be tested to high precision in a human-sized lab in principle; those tests can be easily found with a simple web search. I've read your post and comments a few times; what I see is that you're just saying the black hole can't be too small or else it'll invalidate the result at some precision. Yes that's true. But it doesn't prevent a high-precision test of the EP in principle, like for a larger black hole.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The EP has indeed been tested to high precision in labs on Earth. &lt;em&gt;In principle&lt;/em&gt; could one such lab fall through the horizon of a black hole during the test, without the experimenters (or anyone else on Earth) noticing? Yes; the tidal force in the lab needn't noticably change in that event, when the black hole is sufficiently massive. There needn't been anything out of the ordinary noticed at all--many relativity texts note that. &quot;Nothing special need happen there&quot;, they say. The tidal force is the only reason there is any &quot;inherent fuzziness&quot; when testing the EP. In the question I included &quot;in principle&quot; meaning &quot;in some cases according to the theory&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2314" LastActivityDate="2011-05-11T17:56:49.407" CommentCount="18" />
  <row Id="9741" PostTypeId="2" ParentId="9577" CreationDate="2011-05-11T19:00:26.330" Score="4" ViewCount="" Body="&lt;p&gt;Let us rephrase the question as follows.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a Lagrangian density ${\cal L}$ that (1) implements the global $SO(2)$ symmetry manifestly, and (2) whose Euler-Lagrange equations yield Maxwell's equations in vacuum where there are no sources? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is Yes. Below we will show this. Let the speed of light be $c=1$ from now on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Field variables. The model has $2\times 3=6$ gauge potential fields ${\cal A}^a_i(\vec{x},t)$. Here $i=1,2,3$ are three spatial directions, and $a=1,2$ is an internal $SO(2)$ index. The gauge potential transforms&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\cal A}^a_i\to \sum_{b=1}^2M^a{}_b {\cal A}^b_i $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in the 2-dimensional fundamental representation of $SO(2)$, where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M^a{}_b =\left[\begin{array}{cc} \cos(\theta) &amp;amp; \sin(\theta) \\ -\sin(\theta) &amp;amp;\cos(\theta)\end{array}\right] \in SO(2), \qquad \sum_{b,c=1}^2 (M^{t})_a{}^b g_{bc}  M^c{}_d = g_{ad}, \qquad  g_{ab}\equiv \delta_{ab}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnetic field $\vec{B}$ and electric field $\vec{E}$ are given by the curl of the gauge potential $\vec{\cal A}^a$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{\cal B}^a := \vec{\nabla} \times \vec{\cal A}^a, \qquad a=1,2, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec{B}\equiv\vec{\cal B}^1 \qquad \mathrm{and} \qquad \vec{E}\equiv \vec{\cal B}^2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to check that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\cal B}_i^a\to \sum_{b=1}^2 M^a{}_b {\cal B}_i^b$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;implements the sought-for $SO(2)$ transformation on the magnetic and electric fields $\vec{B}$ and $\vec{E}$. The two scalar-valued Maxwell equations &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{\nabla}\cdot\vec{B}=0 \qquad \mathrm{and} \qquad \vec{\nabla}\cdot\vec{E}=0 $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;are identically satisfied, because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{\nabla}\cdot\vec{\cal B}^a=\vec{\nabla}\cdot(\vec{\nabla}\times\vec{\cal A}^a)=0, \qquad a=1,2.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Lagrangian density. The Lagrangian density ${\cal L}$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;{\cal L} = \frac{1}{2}\sum_{a,b=1}^2 \vec{\cal B}^a \cdot\left( \epsilon_{ab} \frac{\partial \vec{\cal A}^b }{\partial t} - g_{ab} \vec{\cal B}^b\right), &#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\epsilon_{ab} = - \epsilon_{ba}$ is the Levi-Civita tensor in 2 dimensions (with $\epsilon_{12}  = 1$). It is easy to check that the Lagrangian density ${\cal L}$ is manifestly invariant under global $SO(2)$ transformations, because both $\epsilon_{ab}$ and $g_{ab}\equiv \delta_{ab}$ are invariant tensors for $SO(2)$. The action is by definition&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  S[{\cal A}]=\int d^4x \ {\cal L}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extremizing the action with respect to $2\times 3=6$ gauge potential fields ${\cal A}^a_i$ produces $6$ Euler-Lagrange equations. In detail, an arbitrary infinitesimal variation ${\cal A}^a_i\to {\cal A}^a_i+\delta{\cal A}^a_i$ induces a change $\delta{\cal L}$ in the Lagrangian density&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\delta{\cal L} ~\sim~ \sum_{a,b=1}^2\delta \vec{\cal B}^a \cdot\left( \epsilon_{ab}  \frac{\partial \vec{\cal A}^b }{\partial t} - g_{ab} \vec{\cal B}^b\right)&#xD;&#xA; ~\sim~  \sum_{a,b=1}^2\delta \vec{\cal A}^a\cdot \left( \epsilon_{ab} \frac{\partial \vec{\cal B}^b }{\partial t} - g_{ab} \vec{\nabla} \times \vec{\cal B}^b\right),&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the &quot;$\sim$&quot; sign means equality modulo divergence terms. The variation $\delta S=0$ of the action vanishes iff the last parenthesis is zero. This yield precisely the two vector-valued Maxwell equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  \frac{\partial \vec{B}}{\partial t} + \vec{\nabla} \times \vec{E}=\vec{0} \qquad \mathrm{and} \qquad \frac{\partial \vec{E}}{\partial t} = \vec{\nabla} \times \vec{B}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) S. Deser and C. Teitelboim, &quot;Duality Transformations Of Abelian And Nonabelian Gauge Fields&quot;, Phys.Rev.D 13 (1976) 1592.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) C. Bunster and M. Henneaux, &quot;Can (Electric-Magnetic) Duality Be Gauged?&quot;, Phys.Rev.D83 (2011) 045031, &lt;a href=&quot;http://arxiv.org/abs/1011.5889&quot; rel=&quot;nofollow&quot;&gt;arXiv:1011.5889.&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) S. Deser, &quot;No local Maxwell duality invariance&quot;, Class.Quant.Grav.28 (2011) 085009, &lt;a href=&quot;http://arxiv.org/abs/1012.5109&quot; rel=&quot;nofollow&quot;&gt;arXiv:1012.5109.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-12T13:24:01.557" LastActivityDate="2011-05-12T13:24:01.557" CommentCount="1" />
  <row Id="9742" PostTypeId="2" ParentId="9716" CreationDate="2011-05-11T19:00:44.100" Score="3" ViewCount="" Body="&lt;p&gt;It is possible  &lt;a href=&quot;http://www.springerlink.com/content/g482g72871006420/&quot; rel=&quot;nofollow&quot;&gt;to increase &lt;/a&gt; the probability of the beta decay &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For the $\beta^+$-decay of $^{22}\text{Na}$ in the metallic environment Pd cooled to T=12K the $^{22}\text{Na}$ half-life was observed to be shorter by $1.2\pm 0.2$%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="238" LastActivityDate="2011-05-11T19:00:44.100" />
  <row Id="9743" PostTypeId="2" ParentId="9708" CreationDate="2011-05-11T20:41:53.310" Score="2" ViewCount="" Body="&lt;p&gt;I suggest that looking for an explanation for 'fire' may be the wrong approach.  Science is a process where we try to explain what we perceive.  In doing so, we may have to sacrifice (or at least temporarily suspend) a common perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My presumption is that you are probably mostly interested in the visual manifestation of fire - the dancing flames.  Explaining the existence of light and its color requires only correctly identifying the sources that emit the light (and showing how they got there).  Those sources are known to be from excited electron and molecule de-excitement, plus some blackbody radiation from the larger particles, but this depends on the type of reaction.  An ordinary camp fire or a candle, to the extent of my understanding, is a mix of the different sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The location of the flame above the reaction is due to convection, powered by the lower density in the hot gas (pushing it up).  You also know that the flame follows the wind, which further substantiates the fact that the observed color is, in fact, afterglow kind of effects that come a short time after the main reaction.  The dancing of the flame is the same type of movement you will see in any gas with different temperatures or subject to a heat source, which is a great breeding ground for turbulence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fire can not be contained and kept indefinitely.  It is only there as long as the reactants are present.  I mean, you have to have both the &quot;fuel&quot; and oxygen for the case of common fires.  One claim that could be made is that fire is a gas that is actively emitting those colors.  This may fit some people's definition for fire, but not others.  It should again be stressed that a gas meeting this criteria must be actively loosing energy, meaning that it can not exist without constant energy input.  Others may focus on the reaction itself, saying that fire is defined by the reaction, so your car uses fire.  A favorite term of mine regarding nuclear power plants is the &quot;nuclear fire&quot;, indicating that self-sustaining reactions in general are fire-like to some people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how I slice it: I feel comfortable saying that fire is matter emitting energy in a self-sustaining reaction, or the energy that comes from such reactions, or the reaction itself.  I see that giving answers of &quot;matter releasing energy&quot;, &quot;an energy flow&quot;, and &quot;a process&quot;.  I hope this helps clarify why physicists will often answer &quot;neither&quot; to the question.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-05-11T20:41:53.310" />
  <row Id="9745" PostTypeId="1" AcceptedAnswerId="9749" CreationDate="2011-05-12T00:48:31.063" Score="1" ViewCount="120" Body="&lt;p&gt;Both beam energy and luminosity are important for succesful particle accelerator experiments. The LHC's nominal design is for 7 TeV beams and 1E34 cm^-2 sec^-1 luminosity, while the SSC's nominal design was for 20 TeV beams and 1E33 cm^-2 sec^-1 luminosity. The LHC has 10x greater luminosity while the SSC would have had almost 3x greater energy. If the SSC was alive and both machines were operating at nominal levels, which would have the greater advantage for discovery of: i. the Higgs boson, ii. supersymmetry, iii. extra dimensions? I'm asking on the assumption that all three of these exist.&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-05-12T04:04:21.567" Title="Particle Accelerator Energy and Luminosity" Tags="&lt;energy&gt;&lt;large-hadron-collider&gt;&lt;ssc&gt;&lt;luminosity&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9746" PostTypeId="2" ParentId="9686" CreationDate="2011-05-12T01:29:44.020" Score="1" ViewCount="" Body="&lt;p&gt;As a layman, I can't offer too much, but I can offer how I think about action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=\int L \;\mathrm{d}t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key distinction in classical mechanics between the Hamiltonian and the Lagrangian is that Hamiltonian (H) is the sum of kinetic energy (T) and potential energy (V), where as the Lagrangian (L) is the difference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H=T+V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L=T-V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of simple harmonic motion (a mass, spring system for instance) we can use the following equations for the different types of energy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T=mv^2/2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V=kx^2/2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would refer you to most elementary texts, but the solution for position as a function of time for the simple harmonic system described is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=A \cos({t\sqrt{k/m} + \phi})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and for velocity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v=-A(\sqrt{k/m}) \sin({t\sqrt{k/m} + \phi})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting $A = 2$, and $m=k=1$ we can plot the following graph over time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/TpnTH.jpg&quot; alt=&quot;Hamiltonian vs Lagrangian&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this graph, the highlighted area, the area under the curve plotted by the lagrangian, is the action.  The area under the curve depends on the interval of interest (however, for integer multiples of the wave period, and for infinite time, the action would equal zero [see note]).  In any case, this should give an intuitive notion for action, and for the concept of a functional...the area under the curve for some interval is a number determined by the integral of a function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=\int L \;\mathrm{d}t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to the least action principle (as discussed by others), we set: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta S = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is done by considering the effects of a infinitesimal change in position $\epsilon$ so that $x$ becomes $x+\epsilon$ (I will refer you to chapter 2 of Quantum Field Theory Demystified for a good discussion on this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the objective is that the equation that one desires (newton's equation of motion, or what I like to think of as a force equation) is one that satisifies the $\delta S = 0$ constraint, which in this example, has the effect of keeping the value of the action (the area under the curve) constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that helps, I will edit for more content if needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Note: I would add that if one is clever, they can make the half cycle infinitely long, which would make the action non-zero in the case of a SHO]&lt;/p&gt;&#xA;" OwnerUserId="3460" LastEditorUserId="3460" LastEditorDisplayName="" LastEditDate="2011-05-12T10:01:27.903" LastActivityDate="2011-05-12T10:01:27.903" CommentCount="2" />
  <row Id="9747" PostTypeId="2" ParentId="8453" CreationDate="2011-05-12T03:30:00.807" Score="0" ViewCount="" Body="&lt;p&gt;You can have instances where there is no local extremum of the action--for instance, take the lagrangian $L=m\left(\dot x ^{2}+\dot y^{2}\right)$ over the space defined by a crescent embedded in $\mathbb{R}^2$--then, even though the tips of the crescent are both perfectly good starting and ending points in your domain, there is no extremal path connecting them--it would have to be the straight line that leaves the domain of your configuration space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is admittedly a contrived example.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-05-12T03:30:00.807" CommentCount="2" />
  <row Id="9748" PostTypeId="1" CreationDate="2011-05-12T03:55:49.167" Score="3" ViewCount="74" Body="&lt;p&gt;I am just reading Ryu, Moore and Ludwig's paper on classifications of topological insulators and quantum anomaly(arXiv:1010.0936v1). They are trying to relate the quantum anomaly as a signal of the presence of a topological phase in higher dimension. However, the descent equation they used in their paper has the sign different from the usual descent equations. I am wondering why do they need to change the sign?&lt;/p&gt;&#xA;" OwnerUserId="2397" LastActivityDate="2011-05-12T03:55:49.167" Title="Descent equation and anomaly polynomial" Tags="&lt;quantum-field-theory&gt;&lt;condensed-matter&gt;" />
  <row Id="9749" PostTypeId="2" ParentId="9745" CreationDate="2011-05-12T04:04:21.567" Score="1" ViewCount="" Body="&lt;p&gt;Dear Michael, with the energies and luminosities you described (which are not necessarily those that may be quickly achieved by either machine, and the LHC is still not there), the SSC would be a more potent machine to discover all the things you mentioned, especially the physics associated with the very heavy scales - a few TeV (which is surely extra dimensions, if they exist, and maybe even SUSY) - that the LHC may be unable to reach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The new particles near the upper bound of a collider's energy reach become extremely rare - so the LHC is producing e.g. a top quark many times a second while it used to be produced once a month at a lower energy but comparable luminosity by the Tevatron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even if the luminosity of a higher-energy machine is smaller, it may still produce many more particles even if they're lighter because a more energetic collision contains more particles in the final state. So for example, despite the fact that the LHC has only collided 0.3/fb in each detector, it has beaten the Tevatron with its 10/fb in pretty much everything - and in some &quot;disciplines&quot;, the LHC advantage has become overwhelming. Of course, examples of the latter include supersymmetry - the exclusion limits by the LHC are far more far-reaching than those by the Tevatron and the Tevatron couldn't even match the current LHC results even if it were running for another decade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There remain some special features, such as the top-antitop asymmetry, that are harder to be seen on the LHC than on the Tevatron because the latter is a proton-antiproton collider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite the advantage of the SSC, it would be a kind of a brute-force, expensive machine. The LHC is doing many things in a smarter, cheaper, and technologically more advanced way. The names may be confusing. The SSC was &quot;super&quot;, like in &quot;superconducting&quot;. Of course, the magnets in the LHC are also superconducting and technologically more advanced, in fact, than the SSC plans.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-12T04:04:21.567" CommentCount="1" />
  <row Id="9750" PostTypeId="2" ParentId="9702" CreationDate="2011-05-12T04:07:11.660" Score="1" ViewCount="" Body="&lt;p&gt;Whether the mentioned device is effective or not I can not guarantee (Although I don't see any problem in the working of the device and its basic principle seems to be more or less same as an ordinary lightning arrestor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me slightly elaborate the way a lightning arrestor works in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lightning arrestor generally consists of a metallic rod with a sharp pointed terminal, called prong, at the upper end and it is fixed at the highest point of a building. It is grounded by a good conductor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever a charged cloud comes above the device it induces opposite charges in the upper portion of the lightning arrestor. Since the upper terminal is pointed the charge density is maximum there and a lot of charge leakage takes place from the prong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those charges move upward by Coulomb attraction and neutralize the cloud as much as possible so that the chance of a lightning strike becomes less. However for a heavily charged cloud the discharge phenomenon is so fast that before it is neutralized fully by the upward moving leakage charges of opposite sign, a lightning strike takes place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the L.A. is installed at a high enough place, the chances of catching a lightning strike by it is high, since it provides a path of very low resistance to the ground. In other words, it provides a safe passage to the very high electrical discharge thereby saving the building. Normally it safeguards an area which lies within an imaginary inverted cone where the tip of the cone is on the point of the prong. Apology for the bad drawing by me on paint brush :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/YbSm4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="1355" LastEditorDisplayName="" LastEditDate="2011-05-12T04:13:32.707" LastActivityDate="2011-05-12T04:13:32.707" CommentCount="1" />
  <row Id="9751" PostTypeId="1" AcceptedAnswerId="9752" CreationDate="2011-05-12T04:59:50.357" Score="4" ViewCount="173" Body="&lt;p&gt;Let us say a block of mass is placed on the surface of earth. Then while drawing the forces on that body, we say:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Force F = mg acting towards the center of earth&lt;/li&gt;&#xA;&lt;li&gt;Normal reaction N offered by the surface of earth&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;if no other forces are acting. We say, F=N. But what about the centrifugal force $m\omega^2R$ . Why don't we ever bring that into picture? What am I missing?&lt;/p&gt;&#xA;" OwnerUserId="447" LastEditorUserId="447" LastEditorDisplayName="" LastEditDate="2011-05-12T05:05:35.097" LastActivityDate="2011-05-12T14:57:20.387" Title="Why don't we consider centrifugal force on a mass placed on earth?" Tags="&lt;classical-mechanics&gt;&lt;forces&gt;&lt;centrifugal-force&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="9752" PostTypeId="2" ParentId="9751" CreationDate="2011-05-12T05:06:14.923" Score="6" ViewCount="" Body="&lt;p&gt;Because it's effect is smaller than the variation in g due to earth's bulge (caused by the same centrifugal force) or the local geology - when you use 9.8m/s^2 that's just an approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect of the bulge and centrifugal force mean that 'g' at the equator is about 0.5% lower than 'g' at the poles&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: velocity at equator   40,000 km / 24 h = 1666.7 km/h = 0.463 km/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'centrifugal g' =  (0.463 km/s)^2 )/ 6375 km = 0.03 m/s^2  or 0.3% of 'g'  &lt;/p&gt;&#xA;" OwnerUserId="2525" LastEditorUserId="2525" LastEditorDisplayName="" LastEditDate="2011-05-12T05:15:37.340" LastActivityDate="2011-05-12T05:15:37.340" CommentCount="8" />
  <row Id="9753" PostTypeId="2" ParentId="3518" CreationDate="2011-05-12T06:37:03.297" Score="2" ViewCount="" Body="&lt;p&gt;This is a question that has been bothering me for a while, and is not definitively cleared up by the references found in the links provided by Johanes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously at the expansion following the Big Bang everything is the stress energy tensor and  is expanding as the universe expands. It is when the narrative of the appearance of particles comes about that extra hypothesis have to be taken into account. This is that once QCD,  maxwells equations and the weak lagrangian are in effect, and particles are generated from the primordial soup,  the mass systems formed keep their cohesiveness like corks floating on a river due to the strength of the interactions. Some references say that in the ADS calculations there is a small expansion for atoms, but nothing like the cosmological size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, until we have a theory of everything, and maybe strings et al  will provide us with  one, one cannot be dogmatic about this. After all, the wavelengths are expanding, which is why we know of the expanding universe, and electromagnetic waves are also matter and also governed by Maxwell's equations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment my answer is, probably you will have the same density until you hit the particle  formation time, with a question mark.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-12T06:37:03.297" />
  <row Id="9754" PostTypeId="1" AcceptedAnswerId="9780" CreationDate="2011-05-12T08:33:54.673" Score="1" ViewCount="183" Body="&lt;p&gt;What are the units of the damping constant from the following equation by dimensional analysis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta = \frac{c}{2\sqrt{mk}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming the units have to be s^-1, as the damping constant is present in the exponential equation which plots damping of y=Ae^kt (which plots amplitude vs time). Is that a correct assumption?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If somebody could do a quick dimensional analysis to confirm it would be great.&lt;/p&gt;&#xA;" OwnerUserId="3511" LastActivityDate="2011-05-16T18:25:18.743" Title="Dimensonal analysis of damping constant?" Tags="&lt;dimensional-analysis&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="9756" PostTypeId="1" CreationDate="2011-05-12T09:26:55.700" Score="-1" ViewCount="144" Body="&lt;p&gt;Where the gravitational pull of earth exist up to?  What distance from earth it will be zero?&#xA;How do the skydivers fly at a same altitude? Won't they feel gravitational pull? &lt;/p&gt;&#xA;" OwnerUserId="3512" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-05-12T10:30:28.513" LastActivityDate="2011-05-12T10:30:28.513" Title="What is the extent of Earth's gravitational pull?" Tags="&lt;gravity&gt;&lt;astrophysics&gt;&lt;forces&gt;&lt;newtonian-gravity&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9757" PostTypeId="2" ParentId="9756" CreationDate="2011-05-12T09:43:32.917" Score="2" ViewCount="" Body="&lt;p&gt;The gravitational pull of the Earth is never zero; the force (and therefore the acceleration) decreases as you go further from the Earth like $1/r^2$ where $r$ is the distance from the center of the Earth. At altitudes equal to the radius of the Earth, $6378$ kilometers or so, the force drops to $1/4$ to what it is on the surface but it is not zero. It is not zero even at 400,000 kilometers from the Earth - which is why the Moon is orbiting the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't feel any Earth's gravitational pull on the International Space Station or on the Moon or on any orbit because the attractive gravitational acceleration is exactly compensated by the fictitious centrifugal force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Skydivers may fly at the same altitude as other skydivers but all of them are attracted by the Earth and all of them fall down. Because of the air resistance, the skydivers velocity doesn't increase arbitrarily high. Instead, it ultimately converges to 200 km/h or so which is approximately 55 m/s. (Of course, it's the speed before they open the parachutes - if this were the speed after they open it, that would be pretty dangerous.) In the vacuum, you would reach this speed in 5.5 seconds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So approximately after 6 seconds, the skydivers no longer accelerate. All of them are falling at the same speed. By changing the shape of your body, you may influence the air resistance which modifies your velocity - both speed and the direction. I've tried it, too. It's quite a wind over there.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-12T09:43:32.917" />
  <row Id="9758" PostTypeId="2" ParentId="9756" CreationDate="2011-05-12T09:47:16.130" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Where the gravitational pull of earth&#xA;  exists upto?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This question makes no sense to me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;what distance from earth it will be&#xA;  zero?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Never! The pull will be ever smaller with increasing distance, but never zero.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do the skydivers fly at a same&#xA;  altitude?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They don't! What makes You think that?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Won't they feel gravitational pull?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They do!&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-05-12T09:47:16.130" />
  <row Id="9759" PostTypeId="2" ParentId="3228" CreationDate="2011-05-12T11:10:50.927" Score="-2" ViewCount="" Body="&lt;p&gt;There are more reasons then simple random time as we can see in this image.&lt;br&gt;&#xA; (M. Yamamoto et. al. Journal of Environmental Radioactivity, 2006, 86, 110-131) &lt;img src=&quot;http://i.stack.imgur.com/rBAtd.png&quot; alt=&quot;Graph of lead-210 deposited per month in Japan&quot;&gt; (from &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aLead210inairatjapan.png&quot; rel=&quot;nofollow&quot;&gt;WP&lt;/a&gt; )&lt;br&gt;&#xA;EDIT add 1&lt;br&gt;&#xA;This data is based in sediments. As sediments trace the environment conditions, they act as a proxy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next image is a variation of The decay rate of the radioactive isotope 32Si, and is found in&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/CijNb.jpg&quot; alt=&quot;from (Jenkins et al. 2008)&quot;&gt; &quot;“Evidence for Correlations Between Nuclear Decay Rates and Earth-Sun Distance”  by Jenkins et al. 2008.&lt;br&gt;&#xA;Later the distance from Sun connection was dismissed, but data is still valid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar data about neutrino and Wimp seasonal variarance can be found.&lt;br&gt;&#xA;EDIT add 1 end&lt;br&gt;&#xA;EDIT add 2&lt;br&gt;&#xA;a related paper, in the opposite direction can be found here&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/0810.3265&quot; rel=&quot;nofollow&quot;&gt;&quot;Evidence against correlations between nuclear decay rates and Earth-Sun distance&quot;&lt;/a&gt;&lt;br&gt;&#xA;&quot;We have reexamined our previously published data .. We find no evidence for such correlations&quot;&lt;br&gt;&#xA;They used ratios and this null result is expectable if both the samples and the 'presumed' reference are affected by the same nuclear processes. IMO, in this situation a counting procedure is better than to use ratios.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the Jenkins proposal were correct,&#xA;  it is very unlikely that the alpha,&#xA;  beta-minus, beta-plus, and&#xA;  electron-capture decays of all&#xA;  radioactive isotopes would be affected&#xA;  in quantitatively the same way. Thus&#xA;  the ratios of counts observed from two&#xA;  different isotopes would also be&#xA;  expected to show annual variations.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In order to minimize the influence of&#xA;  any changes in detector and/or&#xA;  electronics performance, we analyzed&#xA;  ratios of gamma-ray peak areas from&#xA;  the isotope of interest and those from&#xA;  a reference isotope whose half life&#xA;  was well known.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT add 2 end&lt;br&gt;&#xA;&lt;strong&gt;There are seasonal variations (diurnal and annual) in radioactive processes:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Radon, Lead, etc ...  &lt;/li&gt;&#xA;&lt;li&gt;Neutron production on reactors &#xA;(at lab   and in space missions&#xA;RTGs)&lt;/li&gt;&#xA;&lt;li&gt;Neutrino&lt;/li&gt;&#xA;&lt;li&gt;DM - WIMPs &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I dont know what the experts say about the actual explanations. I think that they dont know the whys. &#xA;I'm in the chase of data labeled with the timedate and geographical local of the 'crime'.&lt;br&gt;&#xA;&lt;strong&gt;Can someone help, pls?&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are the atoms of the same isotope equal?&lt;/strong&gt;&lt;br&gt;&#xA;I can not agree with the most voted answer, by Joe:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Actually, all the atoms are&#xA;  identical. The time at which it is&#xA;  observed to decay is not an intrinsic&#xA;  property of a given atom, but rather&#xA;  an effect of quantum mechanics.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since when quantum mechanics has effects? &lt;strong&gt;QM does not produce any effect&lt;/strong&gt;, QM describes what we see at a statistical level.&lt;br&gt;&#xA;We see that at a particular time moment, a particular atom decayed, and not the other. It  has to be an intrinsic property of that particular atom that made it decay at that precise moment.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know of any experiment that tried to measure how much equal or distinct can be a similar group of atoms. The community has  the hope that they are identical. &#xA;I'm skeptical about this issue and I take nothing from granted, that I can say &lt;strong&gt;'They are different one from the others'&lt;/strong&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-05-12T23:16:31.733" LastActivityDate="2011-05-12T23:16:31.733" CommentCount="5" />
  <row Id="9760" PostTypeId="2" ParentId="7872" CreationDate="2011-05-12T14:46:37.830" Score="9" ViewCount="" Body="&lt;p&gt;It's difficult to say how close we are to &quot;resolving high temperature superconductivity&quot;, as the answer depends very much on your definition of &quot;resolved&quot;. For example, have we mapped the phase diagrams? Yes. Do we understand the relevant experimental facts? Yes and no. Is there a complete theory of HTSC that predicts how to create a high temperature superconductors from first principles?  Not yet.  There are a lot of scientists working on these questions, and a lot of very different answers, so it will take some time to come to a consensus, even if the true answer is found today, so you may wish to refine your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, the largest &quot;recent&quot; discovery (Jan. 2008) in high temperature superconductivity is the discovery of iron based superconductors.  This is particularly important as iron is nominally ferromagnetic, and magnetism and superconductivity are not natural bedfellows. The Tc of these iron compounds is not close to that of cuprates (highest observed is 55K for SmFeAsO compound, whereas ~150K for Hg-1223 cuprate), but there are many similarities between the cuprates and iron based superconductors which have investigators hoping to find a common mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are quite seriously hundreds of thousands of research works on high temperature superconductors, but facts which are pretty well established by experiment are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) In cuprates, all static magnetism must be fully suppressed for superconductivity to appear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The symmetry of the superconducting order parameter for cuprates is d-wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) In iron-based superconductors, static magnetism must be reduced for superconductivity to appear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) There is a magnetic resonance appearing in unconventional superconductors which scales in energy with the superconducting critical temperature, and this is thought to be a signature of a magnetic pairing mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And some main outstanding questions remain: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) what is the precise mechanism of electron pairing in HTSC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) how can we &lt;em&gt;predict&lt;/em&gt; superconductivity from first principles - ie, can we create an algorithm which tells us how to mix different elements together to make an HTSC, and what the Tc will be.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) is there a common mechanism for high temperature superconductivity, or are there different mechanisms between the different families.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) is room temperature superconductivity possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For technical applications, creating robust wires cheaply and creating materials with higher critical fields is of paramount importance, but that's a topic for someone with more applied knowledge than I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, as noted in the comments, to understand the open questions in HTSC, one must invest some substantial time investigating the literature.  The most bleeding edge of the research is posted on pre-print servers: &lt;a href=&quot;http://arxiv.org/list/cond-mat.supr-con/recent&quot; rel=&quot;nofollow&quot;&gt;ArXiv of Superconductivity Research&lt;/a&gt;, while most journals covering physics provide some sort of superconductivity coverage. If your library doesn't have access to those journals, the &lt;a href=&quot;http://iopscience.iop.org/1367-2630&quot; rel=&quot;nofollow&quot;&gt;New Journal of Physics&lt;/a&gt; and &lt;a href=&quot;http://www.virtualjournals.org/super/&quot; rel=&quot;nofollow&quot;&gt;The Virtual Journal of Superconductivity&lt;/a&gt; are open access. Excellent review articles are available in Nature magazine and Reviews of Modern Physics.  &lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2011-05-12T14:46:37.830" CommentCount="1" />
  <row Id="9761" PostTypeId="2" ParentId="9751" CreationDate="2011-05-12T14:57:20.387" Score="2" ViewCount="" Body="&lt;p&gt;I think that all the right physics is contained in Martin Beckett's answer and the comments on it, but I'd like to restate it in a way that may bring out what I think the key point is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, when we do experiments in a lab near Earth's surface, we use a value of $g$ that's been determined empirically at that location. For instance, we might determine it by dropping something in vacuum and measuring its acceleration with respect to our lab. That value of $g$ already includes the centrifugal contribution, so we don't need to (indeed we must not!) include it separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We often tell introductory physics students that $g$ is the &quot;acceleration due to gravity,&quot; but strictly speaking we're telling a small lie when we do this: $g$ is really the acceleration due to gravity &lt;em&gt;and inertial forces&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, that lie is only a lie in the context of Newtonian mechanics: when we get to general relativity the distinction between gravity and inertial forces goes away anyway! The acceleration of a falling object in general relativity is most naturally thought of as being &lt;em&gt;all&lt;/em&gt; inertial force: the falling object is moving along a geodesic, and the reason we see it as accelerating is that our lab is not in an inertial reference frame.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-05-12T14:57:20.387" />
  <row Id="9762" PostTypeId="2" ParentId="8356" CreationDate="2011-05-12T15:16:09.650" Score="1" ViewCount="" Body="&lt;p&gt;I dont know if this is an answer to the Question, anyway, from&lt;br&gt;&#xA; &lt;a href=&quot;http://arxiv.org/abs/1011.6451&quot; rel=&quot;nofollow&quot;&gt;Informational derivation of Quantum Theory&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Quantum theory can be derived from&#xA;  purely informational principles.&lt;/strong&gt; Five&#xA;  elementary axioms-causality, perfect&#xA;  distinguishability, ideal compression,&#xA;  local distinguishability, and pure&#xA;  conditioning-define a broad class of&#xA;  theories of information-processing&#xA;  that can be regarded as a standard.&#xA;  One postulate-purification-singles out&#xA;  quantum theory within this class. The&#xA;  main structures of quantum theory,&#xA;  such as the representation of mixed&#xA;  states as convex combinations of&#xA;  perfectly distinguishable pure states,&#xA;  are derived directly from the&#xA;  principles &lt;strong&gt;without using the Hilbert&#xA;  space framework.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Complex numbers, tensors, matrix and vector algebra, etc, can be substituded by a simpler analog in  Geometric Algebra framework.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the problem: can a different set of principles achieve the same as QM? In principle yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QM, has any other theory, is a framework of principles and conclusions based on that. We validade the conclusions but we can not prove that, in some future, other theory using a diferent set of principles can not arrive to the same conclusions. The history has proved that all theories are temporary.  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-05-12T15:16:09.650" />
  <row Id="9763" PostTypeId="2" ParentId="8356" CreationDate="2011-05-12T15:22:01.600" Score="0" ViewCount="" Body="&lt;p&gt;No, in fact our current postulates allow certain ambiguity in the description of the same physical system (or at least of the possible set of measurements we can extract from them), suggesting that there might be a more concise underlying theory (or set of postulates) that groups such descriptions under a equivalence class&lt;/p&gt;&#xA;&#xA;&lt;p&gt;look at &lt;a href=&quot;http://physics.stackexchange.com/questions/8049/are-these-two-quantum-systems-distinguishable&quot;&gt;the answer to this question&lt;/a&gt; for an example.&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-05-12T15:22:01.600" />
  <row Id="9764" PostTypeId="2" ParentId="6768" CreationDate="2011-05-12T15:58:27.557" Score="1" ViewCount="" Body="&lt;p&gt;In a nuclear power plant, it is designed so that, if the power fails, the rods will slip into place (and in fact, the problem in Chernobyl was that they didn't). The control rods DID slip into place - but the amount of damage caused to the plant was such that other cooling systems began to fail. The problem was not an ongoing chain reaction, but rather the inability to keep cool the reactors. As a result, water began to evaporate off, which meant even more radioactivity was given off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, the control rods weren't the issue. Nuclear power plants have such complex safety structures, and quite right. Just turns out that Fukushima Daiichi wasn't prepared for this disaster. But then, what can you do? It's impossible to prevent all outcomes.&lt;/p&gt;&#xA;" OwnerUserId="3513" LastActivityDate="2011-05-12T15:58:27.557" />
  <row Id="9765" PostTypeId="1" AcceptedAnswerId="9767" CreationDate="2011-05-12T17:37:52.150" Score="0" ViewCount="100" Body="&lt;p&gt;I just recently finished a test on Physics, and when receiving my scores back I missed a question.  Here's what it was if I can remember it correctly:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Three resistors (10.0, 15.0, and 6.00 ohms) are parallel and the circuit is powered by 12.0V.  How much current (in amps) is the 6.00 ohm resistor using?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;|-----------------|----------------|---------------|&#xA;|                 |                |               |&#xA;--(12.0 V)        &amp;lt;(10.0 ohms)     &amp;lt;(15.0 ohms)    &amp;lt;(6.00 ohms)&#xA;-                 &amp;gt;                &amp;gt;               &amp;gt;&#xA;|                 |                |               |&#xA;|                 |                |               |&#xA;----------------------------------------------------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So, I assumed that since you can combine parallel resistors into one equivilent resistor, they use the same amount of current.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I just found the current used (&lt;strong&gt;4.00 A&lt;/strong&gt;), and assumed that the 6.00 ohms resistor used the same current. The teacher, however, stated that the correct answer was &lt;strong&gt;2.00 A&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right or wrong?  Why?&lt;/p&gt;&#xA;" OwnerUserId="3514" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-12T17:53:57.733" LastActivityDate="2011-05-12T22:55:56.800" Title="Calculating current used by individual resistor?" Tags="&lt;homework&gt;&lt;electricity&gt;&lt;electric-circuits&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="9766" PostTypeId="2" ParentId="9765" CreationDate="2011-05-12T17:46:13.200" Score="1" ViewCount="" Body="&lt;p&gt;Only if the resistors are the same, they carry the same amount of current. Otherwise, the current is distributed according to the ratio of their resistances.&lt;/p&gt;&#xA;" OwnerUserId="2962" LastActivityDate="2011-05-12T17:46:13.200" />
  <row Id="9767" PostTypeId="2" ParentId="9765" CreationDate="2011-05-12T17:47:31.103" Score="1" ViewCount="" Body="&lt;p&gt;Think about this: the current through a certain circuit element represents the amount of charge that flows through that element per second. In a case like this, it's electrons that carry the current, so the amount of current is related to the amount of electrons that flow per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you found that a total of $4\ \mathrm{A}$ of current flow through the circuit. Suppose that corresponds to a total of $N$ electrons per second. Are all $N$ of those electrons going to go through the $6\ \Omega$ resistor?&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-05-12T17:47:31.103" CommentCount="7" />
  <row Id="9768" PostTypeId="2" ParentId="9765" CreationDate="2011-05-12T18:03:57.373" Score="0" ViewCount="" Body="&lt;p&gt;The answer is 2.0 Amps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a parallel circuit.  Each resistor sees exactly the same voltage (assuming negligible wire losses).  The current through each resistor is 12V/(R Ohms) = X Amps, where X = 12/R.  The currents through the three resistors (starting from the left) are 1.2A, 0.8A and 2.0A.  TOTAL current is 0.8+1.2+2 = 4.0 Amps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to combine the parallel resistances into one equivalent resistance, you could use the formula Reqv = 1/((1/R1)+(1/R2)+(1/R3)).  But in this case there is an easier way:  Reqv = 12V/4A = 3 Ohms.  We can check this using the first equation: Reqv = 1/(0.1 + 0.066667 + 0.166667) = 1/(0.33333) = 3.00 Ohms.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-05-12T18:11:07.123" LastActivityDate="2011-05-12T18:11:07.123" CommentCount="0" />
  <row Id="9769" PostTypeId="2" ParentId="9718" CreationDate="2011-05-12T18:21:50.883" Score="0" ViewCount="" Body="&lt;p&gt;I know of nothing which has a higher specific heat than water (though I won't be dogmatic about water being the highest).  For a high specific heat solid, look for something like ice or another substance which is solid and has high water content at the temperature of interest.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-05-12T18:21:50.883" CommentCount="1" />
  <row Id="9770" PostTypeId="2" ParentId="7068" CreationDate="2011-05-12T18:48:09.860" Score="2" ViewCount="" Body="&lt;p&gt;There are a lot of works contained in proceedings related to the IAEA, but one very interesting publication is a study of deposition of &lt;em&gt;long lived&lt;/em&gt; isotopes as monitored at Tsukuba since 1956.  The period of study includes all of the nuclear weapons testing period, and also capturing the Chernobyl accident.  The report cited the total deposited activity (in Bq/m^2) for the year, and essentially shows that the deposition at it's highest for Cs-137 is ~2x10^3 Bq/m^2 for the year, in the early 60's (Bq=Becquerel=1 decay/second). After the testing stopped, this amount gradually decreased, spiking high for the year 1986 (Chernobyl), and then decreasing again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could make the crude approximation that the activity is uniformly dispersed in the atmosphere due to the prevailing winds causing dilution and mixing, so that sites world wide would note a similar deposition of radionuclides over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one makes a further rough approximation that the Fukushima releases are about a tenth that of the Chernobyl releases, which is a number cited in the news media, one could expect about one tenth of 100 Bq/m^2 for long-lived isotopes, or 10 Bq/m^2 for the year.  This may be incorrect, as the values cited for release of Cs-137 are pretty disparate in published reports, with some citing significantly more release of Cs-137 (up to and exceeding the Chernobyl levels).  It's probably still a little too soon to tell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Citation: Hirose, et. al: Analysis of the 50-year records of the atmospheric deposition of long-lived radionuclides in Japan, Applied Radiation and Isotopes, Volume 66, Issue 11, November 2008, Pages 1675-1678&lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2011-05-12T18:48:09.860" />
  <row Id="9772" PostTypeId="1" AcceptedAnswerId="11026" CreationDate="2011-05-12T19:16:45.533" Score="2" ViewCount="99" Body="&lt;p&gt;I need to process some images taken from a telescope to determine the intensity of an astroid. &#xA;This way I can determine the rotation period of this asteroid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pictures were taken the usual way (I guess, I'm not an astronomer, this is a school project), setting the telescope to track a nearby star. I would need to do dark frame and flat field correction on a lot of images (so batch processing would be great). Since the asteroid is moving in relation to the star, there would also have to be functionality to bring this into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a piece of software (&lt;strong&gt;for mac&lt;/strong&gt;) around that can do this? I know Iris exist for windows, but I don't have a PC with windows available, and learning a new piece of software on a computer that isn't mine (eg in the library) will only lead to frustration.&lt;/p&gt;&#xA;" OwnerUserId="3515" LastActivityDate="2011-06-12T15:18:26.587" Title="IRIS alternative on mac?" Tags="&lt;astronomy&gt;&lt;software&gt;&lt;astrometrics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="9773" PostTypeId="1" AcceptedAnswerId="9777" CreationDate="2011-05-12T19:58:20.007" Score="3" ViewCount="113" Body="&lt;p&gt;Why are these states called as such, and how do they differ? I vaguely understand that when E &gt; 0 you obtain a scattering state, but when E &amp;lt; 0 you have a bound state.&lt;/p&gt;&#xA;" OwnerUserId="202" LastActivityDate="2011-05-12T21:20:12.393" Title="Scattering vs bound states" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" />
  <row Id="9774" PostTypeId="2" ParentId="9730" CreationDate="2011-05-12T20:18:05.303" Score="2" ViewCount="" Body="&lt;p&gt;You are trying to figure out the force your bicycle can withstand, although you have posed the question so as to get that answer back as a &quot;mass multiplier.&quot;  The answer to your question really varies a ~LOT~ with technique.  The best we can do here is give boundaries to the possible set of answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming anything approaching good technique, the bike will land and start a rebound without bearing any appreciable weight from the rider.  So mass of the bike ('m2') may be ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forward velocity 'v' and distance 'd' both have nothing to do with the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does matter is h, m1, hardness of the ground, and technique (to include rider's physical fitness and anthropomorphic measurments).  Technique is the ability to decelerate the downward velocity of m1 to 0, while exerting the lowest peak downward force on the bike.  Once you figure out what this peak force is, you have &quot;effective&quot; weight at landing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume hard landing surface, so we can remove that from the equation.  Assume a simultaneous two-wheel landing (I am not saying this is best, just makes calculation a bit easier).  Assume that the tires plus your body mechanics give you 0.3m (just over 1 foot)of deceleration distance for m1.  You did not give a weight, so to make math (and subsequent scaling) easy, I will assume 100 kg. y = 0.3m is height of m1 at touchdown and y=0 is height of m1 when your face and/or crotch smash into the bike frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F=ma.  Or, more completely in this case: (The sum of forces) = ma.  In the space of 0.3 m, m1 must come to rest (vertically).  To do that you need to generate an 'a' which will result in dh/dt = 0.  This in turn will require a net force upward.  We need to find dh/dt at (wheels) touchdown; then calculate a; then derive a net force to accomplish that.  Three problems.  here we go...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) dh/dt at contact was given by dbrane (sign correction add for my frame of reference) --&gt; -SQRT(2gh).  So if h = 2m, then dh/dt = about -6.3 m/s at wheel contact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) To bring dh/dt to 0 in the space of 0.3 m, we need to use the same form of the equation used in step one, but this time: -(dh/dt) = SQRT(2a(-dy)).  For our 2m jump, this is -(-6.3 m/s) = 6.3 m/s = SQRT(0.6a).  'a' = about 66 m/(s^2), which is equivalent to about 6.7 gravities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) We are back to F=ma, with a twist.  Define the amount of force exerted by the rider = Fr.  Define the amount of force exerted on the rider by gravity = Fg = (100Kg)(-9.8m/(s^2)) = -980N.  Then (Fr + Fg) = (100kg)(66m/(s^2)) = 6600N.  Fr = 6600N + 980N = 7580N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force exerted (assuming the rider can apply it smoothly) is about 1700 pounds force.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By doing a one wheel landing, you could simultaneously shorten the effective drop and lengthen the subsequent deceleration distance; both of which would significantly lower the required deceleration forces.  My advice:  Limit your jumps to under 2m vertical drop.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-05-12T20:23:18.203" LastActivityDate="2011-05-12T20:23:18.203" CommentCount="1" />
  <row Id="9775" PostTypeId="1" AcceptedAnswerId="9798" CreationDate="2011-05-12T20:44:11.807" Score="2" ViewCount="75" Body="&lt;p&gt;If an external EM field (a laser, for example) act on a dielectric (a glass, for example) what will be the effect of this field on the dielectric constant and on the refractive index of the material? How, mathematically speaking, could I explain what is happen? &lt;/p&gt;&#xA;" OwnerUserId="1618" LastEditorUserId="61" LastEditorDisplayName="" LastEditDate="2011-05-13T07:21:03.813" LastActivityDate="2011-05-13T11:33:19.313" Title="Effect of a external EM field on a dielectric" Tags="&lt;electromagnetism&gt;&lt;condensed-matter&gt;&lt;non-linear-optics&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="9776" PostTypeId="1" AcceptedAnswerId="9834" CreationDate="2011-05-12T21:02:32.517" Score="3" ViewCount="209" Body="&lt;p&gt;I am looking for interesting ways to introduce the Doppler effect to students. I want some situations in nature or every day life, where a student is possibly surprised and may ask &quot;how could it be&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The common example for Doppler effect of sound waves is the siren of a moving emergency vehicle. Clearly the siren example is one way to introduce it starting from a phenomenon. However are there other interesting, perhaps astonishing phenomena in nature or in every day life which can explained by Doppler effect of sound waves?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this question I am only interested in Doppler effect of sound waves, not that one of light.&lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-13T23:47:20.500" LastActivityDate="2011-07-15T17:24:34.830" CommunityOwnedDate="2011-05-14T05:44:18.433" Title="Doppler effect of sound waves" Tags="&lt;education&gt;&lt;sound&gt;&lt;acoustics&gt;&lt;doppler-effect&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="9777" PostTypeId="2" ParentId="9773" CreationDate="2011-05-12T21:04:14.637" Score="6" ViewCount="" Body="&lt;p&gt;These terms apply when you're solving the Schrodinger equation with a potential that goes to zero at large distances. In this situation, the solutions with $E&amp;lt;0$ have the property that $\psi$ dies away to zero for large distance. So the particle is, with high probability, guaranteed to be in a confined region (not at large distance). So those are bound states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solutions with $E&amp;gt;0$, on the other hand, do not die away to zero at large distances -- instead, they go like $e^{ikr}$ where $k=\sqrt{2mE}/\hbar$. So these solutions represent particles that have high probability to be arbitrarily far away. Physically, they are useful when describing particles that start far away, approach the scattering center, and end up far away again. Hence the name &quot;scattering states.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-05-12T21:04:14.637" />
  <row Id="9778" PostTypeId="2" ParentId="9773" CreationDate="2011-05-12T21:20:12.393" Score="3" ViewCount="" Body="&lt;p&gt;Let me explain with a simple example. Consider a particle in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_potential_well&quot; rel=&quot;nofollow&quot; title=&quot;finite potential well&quot;&gt;finite potential well&lt;/a&gt;. There will be two cases: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) $E&amp;lt;V$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii) $ E&amp;gt;V $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the energy of the particle its smaller than the magnitude of the potential, the particle will be confined in the box forever, that is, the particle its bounded to the thing that is generating the potential. In this case where the particle its confined to a finite space, in a bound state, the energy will be quantized, that is, only multiples of a certain quantity of energy will be allowed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if the energy of the particle it's greater than the intensity of the potential the still will &quot;feel&quot; the &quot;hole&quot; below it, a portion of the &quot;particle&quot; will be reflected, and will go back, and the other portion will cross the well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy did not need to be necessarily smaller than zero to a bound state occur, as a matter of fact it only need to be smaller than the intensity of the potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the information described above can be obtained by solving the Schrödinger Equation to the potential in question, as was done in the link in the top of the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very good introductory book in this subject its &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0131244051&quot; rel=&quot;nofollow&quot;&gt;&quot;Introduction to Quantum Mechanics&quot;&lt;/a&gt; by David J. Griffiths. Read it, it is very nice! &lt;/p&gt;&#xA;" OwnerUserId="1618" LastActivityDate="2011-05-12T21:20:12.393" CommentCount="1" />
  <row Id="9779" PostTypeId="2" ParentId="9765" CreationDate="2011-05-12T22:55:56.800" Score="0" ViewCount="" Body="&lt;p&gt;I would suggest investigating Kirchoff's Laws of 'Current' and 'Voltage'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find the following website very helpful in teaching some circuit theory and this problem is covered as well,&#xA;&lt;a href=&quot;http://www.utwired.engr.utexas.edu/rgd1/lesson03.cfm&quot; rel=&quot;nofollow&quot;&gt;http://www.utwired.engr.utexas.edu/rgd1/lesson03.cfm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-05-12T22:55:56.800" />
  <row Id="9780" PostTypeId="2" ParentId="9754" CreationDate="2011-05-12T22:58:43.870" Score="0" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm assuming the units have to be&#xA;  s^-1, as the damping constant is&#xA;  present in the exponential equation&#xA;  which plots damping of y=Ae^kt (which&#xA;  plots amplitude vs time). Is that a&#xA;  correct assumption?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, the damping ratio $\zeta$ is dimensionless:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[c] = \frac{[F]}{\left[\frac{dx}{dt}\right]} = \frac{\mathrm{N}}{\mathrm{m}\cdot\mathrm{s}^{-1}} = \frac{\mathrm{kg}\cdot\mathrm{m}\cdot\mathrm{s}^{-2}}{\mathrm{m}\cdot\mathrm{s}^{-1}} = \mathrm{kg}\cdot\mathrm{s}^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\zeta] = \frac{[c]}{\sqrt{[m][k]}} = \frac{\mathrm{kg}\cdot\mathrm{s}^{-1}}{\sqrt{\mathrm{kg}\cdot\mathrm{N}\cdot\mathrm{m}^{-1}}} = \frac{\mathrm{kg}\cdot\mathrm{s}^{-1}}{\sqrt{\mathrm{kg}^2\cdot\mathrm{s}^{-2}}} = \frac{\mathrm{kg}\cdot\mathrm{s}^{-1}}{\mathrm{kg}\cdot\mathrm{s}^{-1}} = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Damped_harmonic_oscillator#Damped_harmonic_oscillator&quot; rel=&quot;nofollow&quot;&gt;The solution of the damped harmonic oscillator differential equation (when underdamped) is&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x(t) = A e^{-\zeta \omega_0 t} \ \sin \left( \sqrt{1-\zeta^2} \ \omega_0 t + \varphi \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the exponent is dimensionless (as it must be):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\zeta \omega_0 t] = 1\cdot\mathrm{s}^{-1}\cdot\mathrm{s} = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dimensionless and dimensionful parameters&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The differential equation for a damped harmonic oscillator is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can reduce the number of parameters to 2 just by dividing by $m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d^2x}{dt^2} + \frac{c}{m}\frac{dx}{dt} + \frac{k}{m}x = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can transform the two remaining parameters to get a dimensionless one, controlling the shape of the solution, and a dimensionful one, setting the timescale. One way of doing that is to define&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega_0 = \sqrt{\frac{k}{m}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\zeta = \frac{\frac{c}{m}}{\omega_0} = \frac{c\sqrt{m}}{m\sqrt{k}} = \frac{c}{\sqrt{k\,m}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that the differential equation takes the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d^2x}{dt^2} + \zeta\omega_0\frac{dx}{dt} + \omega_0^2x = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason to choose $\omega_0$ as the dimensionful parameter is physical: when the system is &lt;a href=&quot;http://en.wikipedia.org/wiki/Damping#Under-damping_.280_.E2.89.A4_.CE.B6_.3C_1.29&quot; rel=&quot;nofollow&quot;&gt;underdamped&lt;/a&gt;, $\omega_0$ is the angular frequency of oscillation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Damping&quot; rel=&quot;nofollow&quot;&gt;More information about this differential equation and its physical interpretation can be seen in Wikipedia.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1656" LastEditorUserId="1656" LastEditorDisplayName="" LastEditDate="2011-05-14T05:52:48.790" LastActivityDate="2011-05-14T05:52:48.790" CommentCount="6" />
  <row Id="9781" PostTypeId="1" AcceptedAnswerId="9809" CreationDate="2011-05-13T01:26:10.157" Score="2" ViewCount="210" Body="&lt;p&gt;I'm trying to determine the degeneracy of states given by $g(\epsilon)=g_{0} \epsilon$ for a system that is trapped in a quite specific potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In two dimensions, the particle has a potential as in the infinite square well, while on the third dimension, the potential behaves as if a harmonic oscillator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy for a single particle is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\epsilon = \frac{h^{2}}{8mL^{2}}(n_{x}^{2} + n_{y}^{2}) + hf \biggl(n_{z} - \frac{1}{2}\biggr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having some difficulty with this problem because of the broken symmetry. Is there any analytic way to determine the degeneracy as a function of energy?&lt;/p&gt;&#xA;" OwnerUserId="3518" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-13T18:43:24.600" LastActivityDate="2011-05-15T13:00:56.050" Title="Degeneracy of states  in mixed infinite square well, harmonic oscillator" Tags="&lt;quantum-mechanics&gt;&lt;energy&gt;&lt;harmonic-oscillator&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="9783" PostTypeId="2" ParentId="9781" CreationDate="2011-05-13T04:11:30.620" Score="2" ViewCount="" Body="&lt;p&gt;This seems to scream out for a &lt;a href=&quot;http://en.wikipedia.org/wiki/Separation_of_variables&quot; rel=&quot;nofollow&quot;&gt;Separation&lt;/a&gt; of &lt;a href=&quot;http://mathworld.wolfram.com/SeparationofVariables.html&quot; rel=&quot;nofollow&quot;&gt;Variables&lt;/a&gt; approach. In addition to those links you can find it in any book on math methods in physics or any reasonably advanced book on differential equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short--short version is you will write you solution in parts that depend only on the independent bits&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Psi(x,y,z) = W(x,y) Z(z) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and after applying the partial derivatives you will find that you have two much simpler equations to work with.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-05-13T04:11:30.620" />
  <row Id="9784" PostTypeId="1" AcceptedAnswerId="9785" CreationDate="2011-05-13T04:35:50.097" Score="-1" ViewCount="118" Body="&lt;p&gt;How do we normalize a wavefunction that's a linear combination of sines and cosines (or of $Ae^{ikx}+Be^{-ikx}$ -- they're the same, right)? One you square it, wouldn't the integrand be oscillating through all space, and thus infinite?&lt;/p&gt;&#xA;" OwnerUserId="202" LastEditorUserId="202" LastEditorDisplayName="" LastEditDate="2011-05-13T04:47:16.897" LastActivityDate="2011-05-13T05:16:40.950" Title="Wavefunction normalization" Tags="&lt;quantum-mechanics&gt;&lt;wavefunction&gt;&lt;normalization&gt;" AnswerCount="1" CommentCount="11" />
  <row Id="9785" PostTypeId="2" ParentId="9784" CreationDate="2011-05-13T05:16:40.950" Score="6" ViewCount="" Body="&lt;p&gt;You don't. Sinusoidal wavefunctions like these, a.k.a. &lt;em&gt;plane waves&lt;/em&gt;, are non-normalizable, because the integral which defines the norm does not converge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle\psi|\psi\rangle = \int_{-\infty}^{\infty} |A e^{ikx} + B e^{-ikx}|^2\mathrm{d}x = \infty$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;EDIT&lt;/em&gt;: just thought I should mention that $\ldots = \infty$ doesn't mean the integral literally equals infinity, it's a notation for &quot;does not converge&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People tend to talk about these plane waves because it's easy to figure out how they behave, but in reality, a wavefunction is never just a pure plane wave. It's always some linear combination of plane waves with different frequencies/wavelengths, such that the overall wavefunction goes to zero at large distances quickly enough to make the integral converge. For example, you might have a plane wave confined within a potential well, so that the wavefunction is zero everywhere outside the well, or you could have a Gaussian wavepacket.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-05-13T05:16:40.950" />
  <row Id="9786" PostTypeId="1" AcceptedAnswerId="11597" CreationDate="2011-05-13T05:48:21.880" Score="2" ViewCount="98" Body="&lt;p&gt;Does any known substance have a heat capacity at constant volume ($C_V$) &lt;strong&gt;per mole of atoms&lt;/strong&gt; greater than $3k_B$ ~ 24.94 J/(mol K)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to count, the substance must actually be made of atoms, that is, ordinary nuclei and electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, what are the extra degrees of freedom responsible for this unusually large heat capacity?&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-05-13T09:40:17.127" LastActivityDate="2011-06-26T17:45:38.500" Title="Violations of Dulong-Petit rule as an upper limit to heat capacity" Tags="&lt;thermodynamics&gt;&lt;degrees-of-freedom&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="9787" PostTypeId="1" CreationDate="2011-05-13T05:52:46.807" Score="3" ViewCount="83" Body="&lt;p&gt;Logical positivism is often invoked to explain why external observers can't talk about black hole interiors, or why we can't talk about what happens outside our causal patch in inflationary models. But just exactly how justified is this questionable philosophy of logical positivism? According to another philosophical tradition, scientific realism, it is meaningful to ask about what happens inside a black hole, or beyond our causal patch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just because we can't measure something in principle doesn't mean it doesn't exist!&lt;/p&gt;&#xA;" OwnerUserId="3500" LastActivityDate="2011-05-13T06:12:12.390" Title="Please justify invoking logical positivism to causal patches and black hole interiors in quantum gravity!" Tags="&lt;black-holes&gt;&lt;quantum-gravity&gt;&lt;de-sitter-spacetime&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9788" PostTypeId="1" CreationDate="2011-05-13T06:10:13.937" Score="0" ViewCount="182" Body="&lt;p&gt;Gravitons are string modes which can have arbitrarily small energies. Just make the wavelength arbitrarily large. String theory has no mass gap, but why? A classical string can only have positive energies. The string tension makes a positive contribution. The string kinetic energy also makes a positive contribution. A classical string can have an arbitrarily small mass because it can be arbitrarily small in length. But quantum strings can't shrink beyond the string scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lack of a mass gap can be traced to a shift factor to the string energy in the Virasoro constraint. For a single string, this shift is unconstrained and arbitrary. Without interactions, this is none other than the &quot;string chemical potential&quot;. But we can create/annihilate strings after second quantizing them. This shift becomes physical. Modular invariance of worldsheets of genus one fixes the shift, and it turns out to be negative. String theorists explain this as the Casimir effect on the worldsheet, but this is just a name, and not an explanation. Why is this Casimir shift negative? Only a negative value can cancel the mass gap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't say &amp;zeta;-function regularization, because it's purely formal and doesn't explain a thing.&lt;/p&gt;&#xA;" OwnerUserId="3522" LastActivityDate="2011-05-13T06:53:50.247" Title="Why doesn't string theory have a mass gap?" Tags="&lt;string-theory&gt;" AnswerCount="1" FavoriteCount="0" />
  <row Id="9789" PostTypeId="2" ParentId="9787" CreationDate="2011-05-13T06:12:12.390" Score="2" ViewCount="" Body="&lt;p&gt;That's a good question. These issues have an important philosophical core. But science is ultimately not about philosophy. The robust propositions made by science are about calculations and proofs obtained from empirically validated theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is appropriate to view your question as a scientific one. Then there may actually be a pretty good reason, besides the philosophical stance you mention, why exterior observers shouldn't talk about the detailed microscopic events in the black black hole interior. If the black hole complementarity is right, then&#xA;$$ [\phi(x,t)_{\rm out},\phi(x&amp;#39;,t&amp;#39;)_{\rm in}] \neq 0.$$&#xA;If you take two points $x,t$ and $x&amp;#39;,t&amp;#39;$ outside and inside a black hole that are space-like separated, according to classical geometry, the corresponding fields in effective field theory should commute with each other. But it's very likely that in the exact quantum gravity, they don't. The non-vanishing of the commutator seems to be necessary to get rid of the strict locality and strict causality. The strict locality and strict causality are bad because they are the key assumptions that allow one to prove the demonstrably wrong proposition that the information is being lost during black hole evaporation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hypothetical - and likely - situation is called &lt;em&gt;black hole complementarity&lt;/em&gt;. If that's so, the fields in the interior are complicated functions of the fields and their derivatives outside the black hole. So observables inside the black hole are actually encoded in some subtle correlations of physical features of the region outside the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much like in the uncertainty principle you can't talk about the exact values of $x,p$ at the same moment because they don't commute with each other, you can't talk about physical observations inside and outside at the same moment because they don't commute with each other. So the question &quot;what actually happened inside the black hole&quot; may really be ill-defined for the observers outside. Not only they have no way to find out; but even in principle, the answer may fail to exist. The infalling observers do see some outcomes; but their perspective may be fundamentally incompatible with the observers who stay outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expanding cosmological case is probably analogous, with the region behind the cosmic horizon following the same rules as the black hole interior. None of the propositions above has been rigorously proved. After all, our functioning theory of quantum gravity isn't really a local field theory with negligible refinements; it is fundamentally formulated in a nonlocal way which becomes particular manifest in extreme environments such as those with the event horizons. However, if one formulates questions about the &quot;interior&quot;, it's clear that we're using concepts of some effective field theory - the most general description of quantum gravity may refuse to separate the spacetime (and its degrees of freedom) into such simple regions, and when we do talk about the effective fields, it seems inevitable that the rules have to be modified relatively to the flat space if there are event horizons.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-13T06:12:12.390" CommentCount="1" />
  <row Id="9790" PostTypeId="2" ParentId="9788" CreationDate="2011-05-13T06:31:04.427" Score="5" ViewCount="" Body="&lt;p&gt;Dear Iskandar, as we're getting used to, a majority of your question is not a question; it is a sequence of fundamentally incorrect propositions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Greek-letter-you-have-prohibited-here-so-I-will-not-write-it function regularization actually explains &lt;em&gt;everything&lt;/em&gt; about the issues you're asking. It is a canonical example of an actual calculation that may produce correct values of the string energy - which gets translated to the squared mass of the corresponding particle in spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this regularization or any other calculation that is equally valid, one may see that the ground state of the string has a negative energy so the corresponding particle has a negative squared mass. In particular, in bosonic string theory, the ground state of a single string is a tachyon and its mass is&#xA;$$ m^2 = -\frac{1}{\alpha&amp;#39;} $$&#xA;or its multiples, depending on whether the string is open or closed, and so on. The negative value - the existence of a tachyon - implies that the spacetime of bosonic string theory is unstable and the theory is kind of sick and may only be used as a toy model (at least for closed bosonic strings).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the ground state world sheet energy - i.e. the squared spacetime mass - has to have this negative value can be showed by various controlled calculations of the single string or, if one decides to deny that the right value may be calculated already for a single string (which it can), the correct shift in the energy is certainly needed for consistent interactions at the one-loop level and beyond. Needless to say, the latter calculations are more demanding than the regularization of the single-string ground state energy but ultimately, the complicated calculations of interactions justify the correct regularizations of the single string, too. (The critical dimension, related to the ground state energy, was first encountered while people calculated one-loop diagrams.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the superstring case, the tachyons are projected out of the spectrum but there are always massless particles such as the graviton and indeed, the vanishing of their mass is linked to the existence of a negative-energy state on the world sheet that was removed. At least, this is the case in the RNS formulation of the superstring. In the Green-Schwarz formalism with world sheet fermionic fields that are spacetime spinors, there are no tachyons to start with and the vanishing of the ground state energy directly comes from the cancellation between fermions' and bosons' zero-point energies; you may prefer this formalism if you manage to learn it but the resulting spectrum and interactions are totally equivalent to the RNS formalism. In the Green-Schwarz formalism, the dominant negative contribution is one from the zero-point energy of a fermionic harmonic oscillator that is $-\hbar\omega/2$ much like it's $+\hbar\omega/2$ for the bosonic harmonic oscillators, and those cancel. (There may be other formalisms to formulate a CFT for a given string background, too, and the equivalences may be proved rigorously.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of the ground state energy - the tachyon's squared mass - is negative because we can actually calculate not just the sign but its exact value and this value is negative. The forbidden-Greek-letter function regularization is a transparent way to do it - one that many people, including myself, view as &quot;totally intuitive&quot;. But at the end, one has to apply &lt;em&gt;a&lt;/em&gt; method to calculate it. One can't scream the right result without doing any calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should look at the question in the opposite way. Other people may ask you why you think that the ground state energy should be non-negative. And you would answer with some arguments about the positivity of the kinetic and potential energy of a string but these arguments are ultimately wrong when applied to a string which has infinitely many modes. A consistent theory simply can work only when the ground state energy is shifted to the actual level required by consistency - or the world sheet conformal and modular invariance - and the right ground state value is negative. You may imagine that you treat the whole ground state energy as an unknown; you add a possible shift. And at the end, or earlier, you will find out that only for one value of the shift, one may get a consistent theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your last sentence about &quot;formal&quot; arguments betrays your complete misunderstanding of physics as a discipline. Solid answers in physics are all about - and have always been all about, at least from Newton's era - careful mathematical arguments and calculations. To prohibit the people who are supposed to answer your question - or yourself - from using maths is equivalent to make it totally impossible to do any serious physics whatsoever. You can't understand arguments of string theory with no knowledge of maths, or even by denial of the importance and total legitimity of maths. One may cook a soup but he can't do string theory with this lack of pre-requisites. It's simply not possible. String theory - and theoretical physics - is all about maths. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in your language, the &lt;em&gt;whole discipline&lt;/em&gt; is formal in this sense, and it is how it should be and how it must be. &quot;Informal intuition&quot; may arise by experience but it's clear that one has to work a lot with the maths before he begins to view insights about these theories at the intuitive level. That's simply because for millions of years, our ancestors haven't been trained by evolution to understand the quantum or microscopic world. So it shouldn't be surprising that we encounter lots of things that are not &quot;hard-wired&quot; in our brains and where we really have to rely on very careful, &quot;formal&quot; arguments and calculations rather than primitive emotions. There is no hormone and nothing else in the generic mammals' emotions that can make them instinctively understand why $1+2+3+4+5+\dots = -1/12$ whenever the sum of integers appears in an important calculation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only careful maths may shed light on the identity. One may always &quot;pre-regulate&quot; the theory so that the sum of integers appears in a regulated form, with extra terms that ultimately make it finite, and the right shift is added by hand, and is later adjusted to get an interesting theory. So the $1+2+3+4+5+\dots$ formula may be viewed as a mere &quot;sketch&quot; that misses lots of &quot;bureaucratic&quot; additions, regulators, shifts, and so on. But the ultimate result, in this case and in many others, is completely identical to the result of a calculation that adds no bureaucratic baggage of this kind and that simply sets $1+2+3+\dots$ equal to $-1/12$ whenever it appears. These are aesthetic differences in ways how physicists may calculate various things - they may differ in the way how they think about a problem - but at the end, competent physicists have to agree about the values of all predicted quantities that may be measured.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-13T06:53:50.247" LastActivityDate="2011-05-13T06:53:50.247" CommentCount="5" />
  <row Id="9791" PostTypeId="1" AcceptedAnswerId="9803" CreationDate="2011-05-13T06:39:40.990" Score="-1" ViewCount="109" Body="&lt;p&gt;&lt;em&gt;What is a distance an object went through at the speed of 14hm/min for 6 hours?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having problems with conversions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I tried passing 14hm/min to m/s. I did it alright. But now I am unsure. Do  I multiply it by 6 hours and I'm done?&lt;/p&gt;&#xA;" OwnerUserId="2950" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-13T15:31:19.013" LastActivityDate="2011-05-13T19:45:31.710" Title="Help getting the distance an object has moved" Tags="&lt;homework&gt;&lt;kinematics&gt;&lt;si-units&gt;&lt;unit-conversion&gt;&lt;units&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="9792" PostTypeId="1" AcceptedAnswerId="9794" CreationDate="2011-05-13T06:54:07.567" Score="3" ViewCount="844" Body="&lt;p&gt;After hearing so much about PLANCK, people these days are still constraining models with WMAP results. When will PLANCK finally release its much more superior data?&lt;/p&gt;&#xA;" OwnerUserId="2643" LastActivityDate="2011-05-13T08:28:47.897" Title="When will PLANCK release all-sky CMB anisotropy measurements?" Tags="&lt;planck&gt;&lt;cmb&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="9793" PostTypeId="2" ParentId="9716" CreationDate="2011-05-13T07:08:11.860" Score="1" ViewCount="" Body="&lt;p&gt;A special kind of decay, electron capture $\beta^+$ decay, by definition depends on the electron density of the environment, which may change with pressure, chemical composition etc. There's a Wikipedia article: &lt;a href=&quot;http://en.wikipedia.org/wiki/Change_of_decay_rate&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Change_of_decay_rate&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2643" LastEditorUserId="2643" LastEditorDisplayName="" LastEditDate="2011-05-13T17:16:06.677" LastActivityDate="2011-05-13T17:16:06.677" CommentCount="1" />
  <row Id="9794" PostTypeId="2" ParentId="9792" CreationDate="2011-05-13T07:28:04.107" Score="3" ViewCount="" Body="&lt;p&gt;Dear Felix, by the middle of 2010, Planck had already covered 100% of the skies for the first time. The first version of the final results are scheduled to be released in January 2013 (previously by the end of 2012), see&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Planck_(spacecraft)#Results&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Planck_(spacecraft)#Results&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, during a January 2011 Planck conference in Paris, they have released &quot;early results&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1101.2022&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1101.2022&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's mostly about the health status - the spacecraft seems to work astoundingly well so far. The package is available at&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.rssd.esa.int/index.php?project=Planck&quot; rel=&quot;nofollow&quot;&gt;http://www.rssd.esa.int/index.php?project=Planck&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://www.sciops.esa.int/index.php?project=PLANCK&amp;amp;page=Planck_Published_Papers&quot; rel=&quot;nofollow&quot;&gt;http://www.sciops.esa.int/index.php?project=PLANCK&amp;amp;page=Planck_Published_Papers&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second link contains 25 papers on the arXiv. Again, the maps and timelines will be released in January 2013 according to the current plan - see the conclusions of the 1101.2022 paper above.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-13T08:28:47.897" LastActivityDate="2011-05-13T08:28:47.897" CommentCount="1" />
  <row Id="9795" PostTypeId="1" CreationDate="2011-05-13T08:36:26.417" Score="0" ViewCount="128" Body="&lt;p&gt;A tricky question. Here is the famous graph of the running of the three coupling constants in the standard model: &lt;a href=&quot;http://www-ekp.physik.uni-karlsruhe.de/~deboer/html/Forschung/unification_eng.eps&quot; rel=&quot;nofollow&quot;&gt;http://www-ekp.physik.uni-karlsruhe.de/~deboer/html/Forschung/unification_eng.eps&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph shows, in its top curve, the running of the coupling constant alpha_1. This is the coupling of the weak hypercharge coupling constant for the weak hypercharge group U(1)_Y, which is one of the three gauge groups of the standard model of particle physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is a tricky detail. In that curve, alpha_1 is multiplied with 5/3. This factor 5/3 comes from the assumption that GUTs are valid. The factor ensures that the various group traces of U(1)_Y, SU(2) and SU(3) are normalized in the correct way when they form the SU(5), SO(10) or any other grand unification gauge group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case that grand unification is wrong, the factor 5/3 cannot be deduced. Which factor would be natural in this case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarification added, after remarks by Lubos Motl: it is assumed in the question that the usual definition of the weak hypercharge is used, Y_W=2(Q-T_3), in which left-handed quarks have hypercharge 1/3.&lt;/p&gt;&#xA;" OwnerUserId="3495" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-05-15T08:39:15.383" LastActivityDate="2011-05-15T08:39:15.383" Title="What is the most natural definition of the weak hypercharge coupling constant if grand unification is wrong?" Tags="&lt;quantum-field-theory&gt;" AnswerCount="1" />
  <row Id="9796" PostTypeId="2" ParentId="9795" CreationDate="2011-05-13T08:57:56.350" Score="1" ViewCount="" Body="&lt;p&gt;There is of course no unique answer because if the normal GUT embedding and theory is wrong, there may still be another valid theory that naturally involves the same or (almost) any other rescaling of the hypercharge fine-structure constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you say that GUT is &lt;em&gt;wrong&lt;/em&gt;, it is far from specifying what is actually &lt;em&gt;right&lt;/em&gt;. When it comes to coupling constants and other things, GUT is more predictive than theories without any unification - that's a characteristic feature of unifying theories. So if you say that GUT is &lt;em&gt;wrong&lt;/em&gt;, you just reduce the ability to predict and explain patterns in Nature, so you can't expect that by this negative assumption, you will obtain positive predictions. One shouldn't expect that by throwing out theories that are meant to explain something, one automatically finds different answers to the questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you assume a normalization with no relationship between any groups where the hypercharge $U(1)$ is normalized just like any other $U(1)$ would be, and your goal is to make the formulae as simple as possible on the paper (which is not really a physical criterion), then $5/3$ is replaced by $1$. You just omit the $5/3$ factor. But this is a kind of vacuous statement because one may only compare the fine-structure constants of the different group factors if there is &lt;em&gt;some&lt;/em&gt; relationship between them which is either grand unification or plays the same role.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more preemptive comment: at low energies, it is not true that the hypercharge fine-structure constant renormalized by another simple factor such as $5/3$ yields the electromagnetic fine-structure constant. At the GUT scale, similar relations exist but the electromagnetic fine-structure constant is not well-defined there.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-13T09:26:23.567" LastActivityDate="2011-05-13T09:26:23.567" CommentCount="10" />
  <row Id="9797" PostTypeId="2" ParentId="9791" CreationDate="2011-05-13T09:21:04.803" Score="1" ViewCount="" Body="&lt;p&gt;The answer to this question is yes, but... Remember that when we multiple quantities together it is important that they are all in standard (&lt;a href=&quot;http://en.wikipedia.org/wiki/International_System_of_Units&quot; rel=&quot;nofollow&quot;&gt;SI&lt;/a&gt;) units. This will make it easier to keep track of what units your answer will be in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is one that concerns speed ($s$), distance ($d$) and time ($t$). The equation relating them is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ s=d/t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging this to get the distance as the subject (as per your question) gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ d=st$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like I said- once you've converted all the values to SI units, it's merely a case of putting the values into your calculator and seeing what you get.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastActivityDate="2011-05-13T09:21:04.803" />
  <row Id="9798" PostTypeId="2" ParentId="9775" CreationDate="2011-05-13T11:33:19.313" Score="2" ViewCount="" Body="&lt;p&gt;In linear electrodynamics (i.e. low intensities), the dielectric constant and refractive index remain unchanged. A different thing is &lt;a href=&quot;http://en.wikipedia.org/wiki/Nonlinear_optics&quot; rel=&quot;nofollow&quot;&gt;nonlinear optics&lt;/a&gt; (applying to ED in general as well), for more theory see e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Nonlinear_optics#Theory&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. But this happens only for materials with strong non-linear parameters (usually $\chi^{(2)}$ or $\chi^{(3)}$) and for rather high field intensities. This usually yields effect such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Second_harmonic_generation&quot; rel=&quot;nofollow&quot;&gt;second harmonic generation&lt;/a&gt;, but there is also the effect of &lt;a href=&quot;http://en.wikipedia.org/wiki/Self-focusing&quot; rel=&quot;nofollow&quot;&gt;self-focusing&lt;/a&gt; among others, which stems from the dependence of the refractive index on the field amplitude.&lt;/p&gt;&#xA;" OwnerUserId="97" LastActivityDate="2011-05-13T11:33:19.313" CommentCount="1" />
  <row Id="9799" PostTypeId="1" AcceptedAnswerId="9828" CreationDate="2011-05-13T11:52:30.840" Score="0" ViewCount="110" Body="&lt;p&gt;Can phone signals penetrate glass, so can I expect phone signals to come if I am sitting inside a closed AC car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="1093" LastActivityDate="2011-05-13T20:18:51.393" Title="Do and can phone signals come inside AC car which is glass-packed" Tags="&lt;electromagnetism&gt;&lt;optics&gt;&lt;light&gt;&lt;electromagnetic-radiation&gt;&lt;radiation&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="9800" PostTypeId="1" AcceptedAnswerId="9804" CreationDate="2011-05-13T12:10:36.333" Score="1" ViewCount="212" Body="&lt;p&gt;If I see the steam coming out of cooling towers at an altitude of 200 meters, I cannot help but think that this energy is wasted. My question is: Why isn't this steam cooled enough to become water and used to create hydroelectric power due to the drop from 200 meters?&lt;/p&gt;&#xA;" OwnerUserId="904" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-27T09:49:56.957" LastActivityDate="2011-07-27T09:49:56.957" Title="Why is the steam from cooling towers not used?" Tags="&lt;electricity&gt;&lt;energy&gt;" AnswerCount="2" />
  <row Id="9801" PostTypeId="1" CreationDate="2011-05-13T12:19:19.383" Score="1" ViewCount="141" Body="&lt;p&gt;I have a buoy that can hold 2943.02 cubic millimeters of non-pressurized air, I need and equation to work out the maximum weight that the buoy can float.&lt;/p&gt;&#xA;" OwnerUserId="3523" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-13T20:07:33.393" LastActivityDate="2011-05-13T20:07:33.393" ClosedDate="2011-05-13T20:51:20.690" Title="Maxiumum weight a buoy can float" Tags="&lt;homework&gt;&lt;water&gt;&lt;pressure&gt;&lt;air&gt;" AnswerCount="3" CommentCount="10" FavoriteCount="0" />
  <row Id="9802" PostTypeId="2" ParentId="9772" CreationDate="2011-05-13T12:32:54.800" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://iris.en.softonic.com/mac&quot; rel=&quot;nofollow&quot;&gt;http://iris.en.softonic.com/mac&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3523" LastActivityDate="2011-05-13T12:32:54.800" CommentCount="1" />
  <row Id="9803" PostTypeId="2" ParentId="9791" CreationDate="2011-05-13T12:37:28.010" Score="2" ViewCount="" Body="&lt;p&gt;You can leave the 14hm/min just as it is, if you like. Convert the 6 hours to how many minutes there are in 6 hours, 6 hours times 60 minutes per hour = 360 minutes. The comments on your Question are beside the point, hm could be half-meters, hectometers, whatever, ... &amp;mdash; if one travels 14 &amp;ldquo;&lt;em&gt;hm&lt;/em&gt;&amp;rdquo;s in 1 minute, then one travels 360 times 14 &amp;ldquo;&lt;em&gt;hm&lt;/em&gt;&amp;rdquo;s in 360 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also convert the 14 hm/min into hm/hour, $60\times 14$ hm/hour, then multiply that by the number of hours. It's possible to keep track of the units quite nicely by arranging everything on two lines,&#xA;$$\frac{14\ {\rm hm}}{{\rm minute}}\times\frac{60\ {\rm minute}}{{\rm hour}}\times 6\ {\rm hour} = 14\times 60\times 6 {\rm hm}.$$&#xA;The different units all cancel, leaving behind the conversion factors. If you do this methodically every time then this kind of conversion, and harder ones, will become second nature.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-05-13T14:11:18.887" LastActivityDate="2011-05-13T14:11:18.887" CommentCount="5" />
  <row Id="9804" PostTypeId="2" ParentId="9800" CreationDate="2011-05-13T12:40:47.220" Score="3" ViewCount="" Body="&lt;p&gt;Let us see -- how much water per second is getting at the height?&#xA;Let's estimate it as 1 ton of water per second. &#xA;Then the energy you can extract form the water is:&lt;br&gt;&#xA;$$ 1 [\mbox{ton/s}] * 200 [meters] * g \simeq 20000000 J/s = 2 \mbox{megaWatts}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compared to hundreds or even thousands of megawatts produced and the power plants -- doesn't worth the fuss. And I'm pretty sure that it is a huge overestimation anyway. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-05-13T12:40:47.220" CommentCount="6" />
  <row Id="9805" PostTypeId="1" CreationDate="2011-05-13T13:01:48.217" Score="6" ViewCount="409" Body="&lt;p&gt;Why don't spinning tops fall over? (The young scientist version)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My nine year old son asked me this very question when playing with his &quot;Battle Strikers&quot; set. Having studied Physics myself, I am very keen to encourage him to take an interest in Science and I am delighted when he asks me such questions. In this case, however, I'm stumped. How do I explain why a spinning top doesn't fall down without going into the mathematics of angular momentum? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought the following post would help, but the answers would just bamboozle him I'm afraid. &#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/271/why-dont-spinning-tops-fall-over&quot;&gt;Why don&amp;#39;t spinning tops fall over?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how &lt;em&gt;do&lt;/em&gt; you explain spinning tops to a nine year old? &lt;/p&gt;&#xA;" OwnerUserId="3525" LastActivityDate="2011-06-05T16:00:29.867" Title="How do you explain spinning tops to a nine year old?" Tags="&lt;mechanics&gt;&lt;angular-momentum&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" />
  <row Id="9806" PostTypeId="2" ParentId="9805" CreationDate="2011-05-13T13:12:33.783" Score="9" ViewCount="" Body="&lt;p&gt;Well, the angular momentum conservation is still the essence although it may be formulated in a different language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top is spinning around a vertical axis and the spinning around this axis can't disappear. if the top decided to fall, the spinning would either disappear or would be replaced by a totally different spinning around a horizontal axis, and Nature doesn't allow such a change of the amount of spinning to occur quickly. One has to have a torque to change the amount of spinning, some force attempting to change the rotation, but the torque acting on the bottom tip of the top is so small that with a fast enough initial spinning, it takes a lot of time to change the spin substantially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, energy conservation guarantees that if there's no friction, the top can't ever fall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/eviqf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More practically, I would probably take a wheel from a bicycle, made the kid hold it, rotate it quickly, and then make him or her feel the forces when he tries to change the direction of the wheel. This is a pretty nice yet simple toy in various science museums, including &quot;Techmania&quot; we have here in Pilsen. See also &lt;a href=&quot;http://sirius.ucsc.edu/demoweb/cgi-bin/?mechan-rotate-rot_stool&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt; which contains the picture above as well as some other insightful games and experiments relevant for the angular momentum.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-14T15:17:22.793" LastActivityDate="2011-05-14T15:17:22.793" CommentCount="3" />
  <row Id="9808" PostTypeId="2" ParentId="9799" CreationDate="2011-05-13T14:22:49.590" Score="2" ViewCount="" Body="&lt;p&gt;This seems like a situation tailor-made for an experiment! Get in your car, close the windows, and call someone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But anyway, the answer is yes, these signals do pass through glass, as well as other solid materials: you can get mobile phone calls in your home, for instance, even if you're not near a window.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-05-13T14:22:49.590" CommentCount="1" />
  <row Id="9809" PostTypeId="2" ParentId="9781" CreationDate="2011-05-13T14:44:14.760" Score="1" ViewCount="" Body="&lt;p&gt;The Hamiltonian is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H=-\frac{\hbar^2}{2m}\vec{\nabla}^2 + V(\vec{r}) = \sum_{j=1}^3 H_{x^j}, $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;with potential &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ V(\vec{r}) = V_x(x) + V_y(y) + V_z(z), $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ V_x(x)= \left\{\begin{array}{ccc} 0 &amp;amp;\mathrm{for}&amp;amp; |x|&amp;lt;\frac{L}{2} \\&#xD;&#xA;\infty &amp;amp;\mathrm{for}&amp;amp; |x|\geq \frac{L}{2}\end{array}\right\}, $$&#xA;$$ V_y(y)= \left\{\begin{array}{ccc} 0 &amp;amp;\mathrm{for}&amp;amp;  |y|&amp;lt;\frac{L}{2} \\&#xD;&#xA;\infty &amp;amp;\mathrm{for}&amp;amp;  |y|\geq \frac{L}{2}\end{array}\right\}, $$ &#xA;$$ V_z(z)= \frac{m\omega^2}{2}z^2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_{x^j}=-\frac{\hbar^2}{2m}\partial^2_{x^j} + V_{x^j}(x^j) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;only depends on one coordinate $x^j$, where $j=1,2,3$. Thus we can use separation of variables, as suggested by dmckee, and first solve the time-independent &lt;a href=&quot;http://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation&quot; rel=&quot;nofollow&quot;&gt;Schroedinger equation&lt;/a&gt; for &lt;a href=&quot;http://en.wikipedia.org/wiki/Particle_in_a_box&quot; rel=&quot;nofollow&quot;&gt;three&lt;/a&gt; well-known one-dimensional &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator&quot; rel=&quot;nofollow&quot;&gt;problems.&lt;/a&gt; It is not hard to see that the 3D energy-levels $E_{\vec{n}}$ becomes a sum of the 1D energy-levels,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_{\vec{n}} = E_{xy}(n^2_x+n^2_y)+E_z(n_z+\frac{1}{2}),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the states are labeled by three integers $\vec{n}=(n_x,n_y,n_z)$ with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_x\geq 1,\qquad  n_y\geq 1,\qquad \mathrm{and}\qquad n_z\geq 0, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E_{xy}:= \frac{1}{8m} \left(\frac{h}{L}\right)^2 \qquad \mathrm{and}\qquad E_z :=\hbar\omega=hf$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are two constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there is one state per unit-volume in $\vec{n}$ space.&#xA;For large enough energy $E$, where $E\gg E_{xy}$ and $E\gg E_z$, the zero-point energy $\frac{1}{2}E_z$ and some other discrete features become irrelevant, and we can approximate the total number $N(E_{\vec{n}}\leq E)$ of states $\vec{n}$ with energy $E_{\vec{n}}\leq E$ by the volume of (a quarter of) a paraboloid with finite height in the following way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ N(E_{\vec{n}}\leq E) \approx\mathrm{Volume}\left\{(n_x,n_y,n_z) \in \mathbb{R}^3_+\mid E_{xy}(n^2_x+n^2_y)+E_z n_z \leq E \right\} $$&#xA;$$= \int_0^{\frac{E}{E_z}} dn_z \ \mathrm{Area}\left\{(n_x,n_y) \in \mathbb{R}^2_+\mid n^2_x+n^2_y\leq \frac{E - E_z n_z}{E_{xy}}\right\}$$&#xA;$$= \int_0^{\frac{E}{E_z}} dn_z \ \frac{\pi}{4} \frac{E - E_z n_z}{E_{xy}} = \frac{\pi}{4E_{xy}} \left[E n_z - E_z \frac{n_z^2}{2} \right]^{n_z=\frac{E}{E_z}}_{n_z=0} =  \frac{\pi}{8} \frac{E^2}{E_{xy}E_z},&#xD;&#xA;$$&#xA;where we used that the area of a quarter disc with radius $r$ is $\frac{1}{4}\pi r^2$.&#xA;Thus the density of states $g(E)$ at energy $E$ is then approximately&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(E) = \frac{dN(E_{\vec{n}}\leq E)}{dE} \approx \frac{\pi}{4} \frac{E}{E_{xy}E_z}=g_0 E, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ g_0 = \frac{\pi}{4E_{xy}E_z} =\frac{2\pi m L^2}{h^3f} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a constant.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-15T13:00:56.050" LastActivityDate="2011-05-15T13:00:56.050" />
  <row Id="9810" PostTypeId="1" CreationDate="2011-05-13T12:15:50.103" Score="0" ViewCount="125" Body="&lt;p&gt;I have a buoy that can hold 2943.02 cubic millimeters of non pressurized air. I need to know what the maximum weight this buoy can float on water.&lt;/p&gt;&#xA;" OwnerUserId="3523" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2011-05-13T15:24:52.150" LastActivityDate="2011-05-13T15:24:52.150" Title="Maximum weight a buoy can float" Tags="&lt;homework&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" />
  <row Id="9811" PostTypeId="2" ParentId="9801" CreationDate="2011-05-13T13:10:27.840" Score="1" ViewCount="" Body="&lt;p&gt;The buoyancy is the weight of water displaced.  Usually for problems like this we assume 1 g/cm^3 (which is fresh water at 4C).  That is the weight it can support, but you have to count its own weight as part of what is supported.&lt;/p&gt;&#xA;" OwnerUserId="3257" LastActivityDate="2011-05-13T13:10:27.840" />
  <row Id="9812" PostTypeId="2" ParentId="9801" CreationDate="2011-05-13T13:29:52.457" Score="0" ViewCount="" Body="&lt;p&gt;Seems like a very tiny buoy!  Also, the amount of air is given to surprising precision.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a first (and excellent) approximation, you can assume that air is weightless.  Water weighs roughly $1$ gram per cc, you have about $3$ cc of air, so the air can support about $3$ grams. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not an awful lot, compared to your estimated weight of the buoy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important consideration would be the density of the material that the buoy is made of.  If it is substantially greater than the density of water, the buoy will sink and support nothing.  But if the buoy is bulky and made of material of low density, the solid material of the buoy could contribute a lot more to the buoyancy than the air.  Imagine for example a buoy made of very lightweight wood with a tiny $3$ cc bubble of air in the middle.  This bubble will contribute virtually nothing to the buoyancy.  The &quot;solid&quot; material of the buoy will make a much greater contribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please do check your units.  As mentioned above, $2943.02$ cubic millimetres is really really tiny.  Do you mean $2943.02$ cc?  That would be more realistic, though still smallish.  And the density of the &quot;solid&quot; part of the buoy is relevant.&lt;/p&gt;&#xA;" LastActivityDate="2011-05-13T14:21:43.927" />
  <row Id="9813" PostTypeId="2" ParentId="9800" CreationDate="2011-05-13T16:16:34.797" Score="4" ViewCount="" Body="&lt;p&gt;There are a number of issues with recovering the energy (mechanical or thermal) from the stack exhaust of a cooling tower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the temperature and energy (enthalpy) of the exhaust is too low to effectively recover. The exhaust is full saturated or over saturated air typically at a temperature of 110 F, not above 120 F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the amount of water discussed above is way too high. The evaporation rate for a cooling tower can be estimated at 0.8% of the circulating water rate per 10 F of range. For a typical power plant this will max out at 2.0% of the cooling towers flow rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A typical 2X1 combined cycle plant has a 125 to 150 MW steam turbine. This equates to about 1,300 gallons per minute of evaporation (177 lbs/sec). Assuming that you could collect all of the water and channel the flow to be able to generate hydroelectric power from a turbine is 10% of the value discussed above, or 20 kw (assuming a 100% efficient cycle, which goes against entropy and the 2nd law).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cooling towers have been around and in use for a long time and in a lot of applications beyond power generation (petro-chem, refineries, building HVAC, etc.) and there has yet to be a cost effective way to harness the waste heat. Some btu's are just not worth trying to collect and are better sent to the universe to increase universal entropy.&lt;/p&gt;&#xA;" OwnerUserId="3527" LastActivityDate="2011-05-13T16:16:34.797" CommentCount="4" />
  <row Id="9814" PostTypeId="1" AcceptedAnswerId="9816" CreationDate="2011-05-13T16:36:46.480" Score="5" ViewCount="226" Body="&lt;p&gt;I found a pair of polarizing &quot;3D glasses&quot; lying around, and tried to look at myself in the mirror while wearing them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my utter confusion, when closing the left eye and only looking through the right eye, I could not see the right eye in the mirror.  The light could not pass through the same polarized lens twice.  (I could, however, see the closed left eye clearly.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would expect the opposite to be true, as light going out the right lens with polarization X and coming back in with the same polarization X should pass through unaffected.&lt;/p&gt;&#xA;" OwnerUserId="3528" LastActivityDate="2011-05-13T17:25:03.037" Title="Why can't my eye see itself in the mirror through polarizing 3D-glasses?" Tags="&lt;polarization&gt;&lt;mirror&gt;" AnswerCount="2" />
  <row Id="9815" PostTypeId="1" CreationDate="2011-05-13T16:40:31.090" Score="5" ViewCount="152" Body="&lt;p&gt;I'm trying to understand boundary layer theory in fluids. All I've found are dimensional arguments,  order of magnitude arguments, etc... What I'm looking for is more mathematically sound arguments. Not rigorous as in keeping track of all epsilons and deltas, but more rigorous than an heuristic argument. Hope you understand what I mean. Some free resources available on the web would be preferred, but if you can suggest book titles that's also helpful. Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The applications I have in mind are the calculation of damping in surface waves in basins of various shapes (circular, rectangular, etc).&lt;/p&gt;&#xA;" OwnerUserId="3064" LastEditorUserId="3064" LastEditorDisplayName="" LastEditDate="2011-05-18T13:33:07.787" LastActivityDate="2011-05-18T13:33:07.787" Title="Boundary layer theory in fluids learning resources" Tags="&lt;classical-mechanics&gt;&lt;mathematical-physics&gt;&lt;fluid-dynamics&gt;&lt;books&gt;&lt;reference-request&gt;" CommentCount="9" />
  <row Id="9816" PostTypeId="2" ParentId="9814" CreationDate="2011-05-13T16:49:46.263" Score="5" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Polarized_3D_glasses&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Polarized_3D_glasses&lt;/a&gt;. Most likely you have a pair of circularly polarized glasses. The mirror reverses the circular polarization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &lt;a href=&quot;http://en.wikipedia.org/wiki/Circular_polarization&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Circular_polarization&lt;/a&gt; does it better than I would be likely to achieve in less than an hour or two. Or Hyperphysics, &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/polclas.html&quot; rel=&quot;nofollow&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/polclas.html&lt;/a&gt;, or &lt;a href=&quot;http://www.google.com/search?q=circular%20polarization&quot; rel=&quot;nofollow&quot;&gt;Google&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-05-13T17:04:56.483" LastActivityDate="2011-05-13T17:04:56.483" CommentCount="1" />
  <row Id="9817" PostTypeId="2" ParentId="9814" CreationDate="2011-05-13T17:25:03.037" Score="4" ViewCount="" Body="&lt;p&gt;Cinema 3D glasses (at least those made by Read-D) are circular polarized. This has the advantage that the polarized light reflected form the screen doesn't depend on the angle between you head and the screen and so you can move your head around while watching. But when you look in a mirror the rotation direction is reversed on reflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shutter glasses used with most home 3D LCD-TV have a linear polarizer and when turned off you can see through them in a mirror.&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-05-13T17:25:03.037" />
  <row Id="9818" PostTypeId="1" CreationDate="2011-05-13T17:26:04.220" Score="-3" ViewCount="200" Body="&lt;p&gt;A black hole exists. Eventually, it will completely evaporate away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alice falls into the hole. Imagine you are Alice. According to logical positivism, the interior of the black hole exists. But you will die at the singularity. You can't measure anything after you're dead, so any event outside after the black hole has vanished doesn't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob stays outside the hole. Imagine you are Bob. The interior doesn't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enter Quentin. Right now, he holds a qubit in the state ${1\over\sqrt 2} (|0\rangle + |1\rangle)$. He will measure the value of the qubit shortly in the future. If 0, he will jump into the hole. If 1, he will stay out forever. Imagine you are Quentin. Right now, does the interior exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Curiouser and curiouser.&lt;/p&gt;&#xA;" OwnerUserId="3500" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-05-13T17:33:38.697" LastActivityDate="2011-08-13T00:18:04.777" Title="Logical positivism and black hole interiors" Tags="&lt;quantum-gravity&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="9819" PostTypeId="1" CreationDate="2011-05-13T17:44:34.233" Score="-4" ViewCount="93" Body="&lt;p&gt;The great physicist Raphael Bousso predicted time will end in &lt;a href=&quot;http://arxiv.org/abs/1009.4698&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;. We can't measure anything after our death in principle. So, does time end when we die?&lt;/p&gt;&#xA;" OwnerUserId="3500" LastActivityDate="2011-05-13T18:38:16.457" ClosedDate="2011-05-13T18:14:20.877" Title="Do events exist after our death if we can't measure them?" Tags="&lt;quantum-gravity&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9820" PostTypeId="2" ParentId="9819" CreationDate="2011-05-13T17:59:48.193" Score="2" ViewCount="" Body="&lt;p&gt;Uhm, definitely not. People are a part of time and space, they don't create it by being &quot;over&quot; it. If you die, there's absolutely no reason why time or space should cease to exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that you can't measure something doesn't mean it doesn't exist. That would be like saying that hundreds of years ago, when people couldn't measure temperature, the temperature didn't exist. That wouldn't work.&lt;/p&gt;&#xA;" OwnerUserId="2834" LastActivityDate="2011-05-13T17:59:48.193" />
  <row Id="9821" PostTypeId="2" ParentId="9818" CreationDate="2011-05-13T18:21:00.250" Score="0" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Alice falls into the hole. Imagine&#xA;  you are Alice. According to logical&#xA;  positivism, the interior of the black&#xA;  hole exists.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. Logical positivism requires you to look at all possible observers (or rather all possible experiments) to determine if something exists. So statements like &quot;for observer A, logical positivism implies...&quot; are meaningless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that understanding, black hole interiors definitely do exist. Plenty of experiments can deduce the effect of the interior like measuring tidal forces, or detecting Hawking radiation, or hitting a singularity or sending a probe towards the hole and seeing it get more and more redshifted.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-05-13T18:21:00.250" />
  <row Id="9822" PostTypeId="1" CreationDate="2011-05-13T18:24:24.983" Score="2" ViewCount="107" Body="&lt;p&gt;Jupiter has about twice the density  of Saturn (1.33 versus 0.69 g/cm^3) because it apparently has a higher mass percentage of rocky core and of metallic hydrogen in its interior. Available density references of this hydrogen state differ widely from 0.3 - 2.8 g/cm^3. Is there a theoretical density of metallic hydrogen, and would its density remain relatively constant given the varyng high pressures and temperatures found in these planetary interiors?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-05-16T08:19:58.087" Title="Metallic Hydrogen and Gaseous Planets" Tags="&lt;hydrogen&gt;&lt;planets&gt;" AnswerCount="1" />
  <row Id="9823" PostTypeId="2" ParentId="9459" CreationDate="2011-05-13T18:35:11.263" Score="0" ViewCount="" Body="&lt;p&gt;Here we will consider a single time derivative for simplicity. For technical reasons we will also assume that the band of frequencies $\omega$ for the time-varying electric fields $E_1(t)$ and $E_2(t)$, i.e, the support of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;Fourier transforms&lt;/a&gt; $\hat{E}_1(\omega)$ and $\hat{E}_2(\omega)$, are restricted within certain limits $\omega_1 \ll \omega \ll \omega_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is now to use a power supply with a time-varying voltage $V(t)$ and two &lt;a href=&quot;http://en.wikipedia.org/wiki/RC_circuit&quot; rel=&quot;nofollow&quot;&gt;RC circuits,&lt;/a&gt; where each capacitor is a &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html&quot; rel=&quot;nofollow&quot;&gt;parallel-plate capacitor&lt;/a&gt; with electric field $E_k=\frac{V_k}{d_k}$, where $d_k$ is the gab distance of the plates, $k=1,2$.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     _________________________&#xA;     |           |           |&#xA;     &amp;lt;           |           &amp;lt;&#xA;     &amp;gt;  R_1      |           &amp;gt;  R_2&#xA;     &amp;lt;          --- V(t)     &amp;lt;    &#xA;     |           -           |&#xA;V_1 === C_1      |      V_2 === C_2&#xA;     |           |           |&#xA;     |___________|___________|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It follows from, e.g., Kirchhoff's and Ohm's laws that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{V-V_k}{R_k}= I_k = \frac{dQ_k}{dt}= C_k \frac{dV_k}{dt}, \qquad k=1,2,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ V(t) = \left(1 + R_k C_k  \frac{d}{dt}\right) V_k(t), \qquad k=1,2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fourier transformation yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \hat{V}(\omega) = \int dt \ e^{-i\omega t} V(t) = \left(1 + R_k C_k  i\omega \right)  \hat{V}_k(\omega), \qquad k=1,2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now adjust the four components $R_k$ and $C_k$, $k=1,2$, such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega_k=\frac{1}{ R_k C_k}, \qquad k=1,2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{V}_2(\omega) \approx  \hat{V}(\omega) \approx  i\frac{\omega}{\omega_1} \hat{V}_1(\omega)  \qquad\mathrm{for}  \qquad \omega_1\ll \omega \ll \omega_2. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By inverse Fourier transformation, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_2(t) \approx V(t) \approx \frac{1}{\omega_1} \frac{dV_1(t)}{dt},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and hence, we achieve:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The electric field $E_2(t)=\frac{V_2(t)}{d_2}$ in parallel-plate capacitor 2 is proportional to the time derivative of the electric field $E_1(t)=\frac{V_1(t)}{d_1}$ in &#xA;  parallel-plate capacitor 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It should be emphasized that $V(t)\approx V_2(t)$ is the input, so the circuit is actually an integrator rather than a differentiator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, let us mention that the $n$'th (anti)derivative for instance can be build by appropriately connecting $n$ of such modules, perhaps with $n-1$ amplifiers in-between.    &lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-15T11:54:03.120" LastActivityDate="2011-05-15T11:54:03.120" />
  <row Id="9824" PostTypeId="2" ParentId="9819" CreationDate="2011-05-13T18:38:16.457" Score="1" ViewCount="" Body="&lt;p&gt;The question, on a first read, seems to refer to a single human life, but the paper linked to is about the end of the universe itself.  This is an important distinction, as is clear from the last answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum physics is often misunderstood, and in particular, the concepts of entanglement and observation.  In fact, the essence of two entangled particles is that the particles have &quot;observed&quot; each other.  Like any physics worth its salt, systems can be drawn around any given conglomeration of particles.  I am a layman myself, but here is a good video to help allay some common misconceptions.&#xA;&lt;a href=&quot;http://www.youtube.com/watch?v=dEaecUuEqfc&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=dEaecUuEqfc&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it is common to look at our own personal being as an observer of a quantum system, which is technically true.  But it's also true that the universe is filled with infinite definable &quot;observers&quot;, including all other people in the world, as well as subsystems of them and you.  Although not a formal physics idea, I do hear informally technical people express the view that there is nothing &quot;special&quot; about our own perception of the universe.  Other animals most certainly experience cognizance similar to what we do and future or unknown being could experience the universe with much greater cognizance.  The typical thinking on death is mostly concerned with the loss of memory continuity, but the concept of the &quot;same&quot; you existing from one second to the next is not established in the first place.  In a sense you may be dying and giving birth to a different being with every shift of your mental state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you want to talk about the subject matter of the paper, where time literally comes to an end in our universe - that is a clear possibility.  I believe it would be correct to say that time in the proper sense of time in our universe ends.  Granted, if our universe is part of a multiverse, those others may have something like time or similar to it.  Such different timelines may or may not follow in some causality relative to each other.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-05-13T18:38:16.457" />
  <row Id="9825" PostTypeId="2" ParentId="9801" CreationDate="2011-05-13T19:04:04.427" Score="0" ViewCount="" Body="&lt;p&gt;2943 mm^3 equals 2.943 cc.  This is the equivalent force of submerging a 0.7 inch diameter ball in water.  It will give about 0.002943 kg * (9.8N/kg) of bouyant force in fresh water.  You say it weighs 800 grams = 0.800 kg * (9.8N/kg).  0.800 kg is a lot more than 0.002943 kg.  Unless the enclosure is built of something less dense than water, this bouy is going to drop like a rock without any load at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you meant 2943 cc?  That is a volume of just under 3 liters, and equates to a sphere of exactly 7 inches diameter.  In that case, assuming negligible added volume of the construction material, then your net bouyant force in fresh water will be about (2943 grams - 800 grams)&lt;em&gt;(1kg/1000g)&lt;/em&gt;(9.8N/kg) = 2.143 kg * (9.8N/kg) = 21.00N = 4.71 lbs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explictly answer your exact question; the equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Net_Bouyant_Force = ((Total_Volume_of_bouy) * (Density_of_displaced_liquid)) - Weight_of_bouy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure you keep your units straight.  Density is often given as mass/volume.  In the above equation you want to convert to weight/volume.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-05-13T19:21:44.737" LastActivityDate="2011-05-13T19:21:44.737" />
  <row Id="9826" PostTypeId="2" ParentId="9791" CreationDate="2011-05-13T19:45:31.710" Score="0" ViewCount="" Body="&lt;p&gt;Peter Morgan's answer is the most straightforward.  It doesn't matter what an 'hm' is.  If you go 14 of them in one minute, then you go:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(14hm/min) * (60 min/hr) * (6 hr) = (14 * 360) hm = 5040 hm in 6 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The technique you need to master, in unit conversion, is unit cancellation.  Notice, in the equation above, that 'min' in the numerator of the first term cancels out 'min' in the denominator in the second.  Likewise for 'hr' in the second and third terms.  All that is left is 'hm' and a little math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you want to convert 'hm' to meters, you'll need to tell us what an 'hm' is.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-05-13T19:45:31.710" />
  <row Id="9827" PostTypeId="2" ParentId="9805" CreationDate="2011-05-13T19:54:33.033" Score="2" ViewCount="" Body="&lt;p&gt;Your 9-year old son?  OK, I'll give it a shot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Warning: I'm gonna play fast and loose with my terminology, and use the word &quot;force&quot; when  &quot;impulse&quot; or &quot;torque&quot; or something else might be more appropriate.  I'm trying to figure out how to relate things to a 9-year old, and I think things can be conveyed more simply by describing everything in terms of force (which is something he can feel).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you want to get him familiar with the idea of linear momentum.  If you have a dolly with rotatable wheels, sit on it and have him push you around.&#xA;Point out that it takes force to get it moving and to stop it moving.  Then, have him push you around in a circle (or an arc), so he can understand that it takes force to change the direction something is moving in.  If you don't have a dolly with rotatable wheels, you can probably improvise something along these lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you've built up that framework, you can show him the same principles apply to things spinning.  Dr. Motl's suggestion of a bicycle wheel is a good one.  First, note that it takes force to get the wheel spinning.  Then note that it takes force to stop it from spinning.  Then note that it takes force to change the direction of spinning (i.e. the direction of the angular momentum vector).  A bicycle wheel makes an excellent gyroscope, but may be hazardous for a 9-year old, so watch out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can try to explain the gyroscope using these two principles that the kid understands.  Stuff falls over because the earth's gravity acts on it with a force to pull it down.  For an object that's not spinning, all it has is the linear momentum resisting gravity.  When it's spinning, you have the linear momentum AND the angular momentum resisting gravity.  And that's why it takes longer to fall over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this is not a complete description by any means, and misses a lot of the extremely important details (like precession), but c'est la vie.  It's not as if you're gonna have the kid understanding all the details (like nutation) without going through the math.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-05-13T19:54:33.033" />
  <row Id="9828" PostTypeId="2" ParentId="9799" CreationDate="2011-05-13T20:18:51.393" Score="1" ViewCount="" Body="&lt;p&gt;I assume you mean an Air Conditioned (AC) car with the windows rolled up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, your cell phone will work wonderfully inside your car.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cell phone signals are electromagnetic (E-M) waves, just like light; but the cell phone E-M waves are longer than the E-M waves which make up visible light.  The reason we see differnt colors is that objects reflect and absorb light based on the material's interaction with different wavelengths of light.  For instance, you see a blue object, because it absorbs all other colors, but reflects blue.  Glass is transparent because is doesn't reflect or absorb any visible color.  You can think of E-M waves outside the visible spectrum as 'colors' of light which we can't see.  It just so happens that glass doesn't significantly reflect or absorb the 'color' of E-M waves that cell phones use (neither does wood, brick, drywall, and most other non-metallic substances). &lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-05-13T20:18:51.393" CommentCount="1" />
  <row Id="9829" PostTypeId="2" ParentId="9822" CreationDate="2011-05-13T20:24:47.453" Score="1" ViewCount="" Body="&lt;p&gt;There will be a density gradient of the hydrogen for the planet, but you can probably quote an average density for the planet if you need to estimate something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was work done by Ashcroft in 1968 (&lt;a href=&quot;http://link.aps.org/doi/10.1103/PhysRevLett.21.1748&quot; rel=&quot;nofollow&quot;&gt;N. Ashcroft, PRL 21, 1748, 1968&lt;/a&gt;) which addresses the topic of metallic hydrogen and density thereof, working to the conclusion that metallic hydrogen may be a high temperature superconductor. This was actually cited as a possible justification for the large magnitude of the magnetic field of the planet Jupiter.  I don't think that answers your question, but I think it's pretty neat. :)&lt;/p&gt;&#xA;" OwnerUserId="2433" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-05-16T08:19:58.087" LastActivityDate="2011-05-16T08:19:58.087" CommentCount="5" />
  <row Id="9830" PostTypeId="1" AcceptedAnswerId="10264" CreationDate="2011-05-13T20:57:24.277" Score="7" ViewCount="249" Body="&lt;p&gt;Consider a conducting wire bent in a circle (alternatively, a perfectly smooth metal ring) with a positive (or negative) electric charge on it.  Technically, this shape constitutes a torus.  Assume that the radius of the wire is much smaller than the radius of the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Find the tension in the wire due to the repulsive electrostatic force&lt;/strong&gt; (in Newtons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this problem, here are example physical values, (may or may not be necessary to solve the problem) charge, $Q = 1 C$, circle radius, $R = 1 m$, wire radius, $r = 1 mm$.  Use the most straightforward notations possible for Coulomb's law, $F=\frac{k q_1 q_2}{r^2}$.  This is a question about the mathematics of a set of inverse-square problems, not about theoretical E&amp;amp;M.  Other notations that will be of useful to this discussion:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&lt;li&gt; $\lambda = \frac{Q }{ 2 \pi R } $ = linear charge density&#xA;&lt;li&gt; $\sigma = \frac{\lambda }{ 2 \pi r }$ = surface charge density&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What happens to the tension in the wire as the wire radius goes to zero&lt;/strong&gt; with a constant circle radius, assuming:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&lt;li&gt;A constant charge on the entire thing (also constant linear charge density)&#xA;&lt;li&gt;A constant surface charge density&#xA;&lt;li&gt;A constant voltage&#xA;&lt;li&gt;If it really bothers you, consider the tensile stress for the above as opposed to the wire tension, although my question is for the latter&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The following are some helping equations.  Any of them could be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electric field from an infinite straight line charge&#xA;$$F(r) = \frac{ 2 \lambda k }{ r }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electric field face-on for a ring of charge, R is radius of ring and d is distance away&#xA;$$F(d) = \frac{2 \pi \lambda R k d }{ (d^2+r^2)^{3/2} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electric potential in vicinity of ring of charge, $\rho$ is distance from z-axis with torus in xy-plane, centered at origin.  Electric field is simple knowing this.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&lt;li&gt;$l = \sqrt{ (\rho+R)^2 + z^2 }$ = distance to furthest point on ring&#xA;&lt;li&gt;$K(x) =$ Elliptic integral of the 2nd kind&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$E(\rho,z) = \frac{ 4 R k \lambda }{l} K\left( \frac{ 2 \sqrt{ R \rho } }{  l } \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I think that a valid approach would be to find the net effective field that a differential part of the ring experiences.  &lt;em&gt;If&lt;/em&gt; you found this, I believe the tension could be written as follows, with T being tension, R the circle radius, and F the electric field.&#xA;$$T = \lambda R F$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have my own thoughts on the answer (and more equations than you would ever want), and I'll return to offer them later, but I certainly don't have a complete picture, and I want to see what approaches people here use.  Maybe this could be answered very simply and maybe some people will know intuitively which cases limit to infinity.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-06-02T14:46:27.173" LastActivityDate="2011-06-02T14:57:26.040" Title="Tension in a curved charged wire (electrostatic force) - does wire thickness matter?" Tags="&lt;forces&gt;&lt;electrostatics&gt;&lt;singularity&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="9831" PostTypeId="2" ParentId="9541" CreationDate="2011-05-13T22:28:48.250" Score="0" ViewCount="" Body="&lt;p&gt;the &lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein_field_equations&quot; rel=&quot;nofollow&quot;&gt;Einstein field equations&lt;/a&gt;, describing the gravitational field are given by&#xA;$$R_{\mu\nu}-\frac12 g_{\mu\nu}R = -\frac{8\pi G}{c^4}T_{\mu\nu}\ .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They relate in a complicated manner the &lt;strong&gt;gravitational field&lt;/strong&gt; that can be seen as the &lt;strong&gt;metric&lt;/strong&gt; $g_{\mu\nu}$ itself to the &lt;strong&gt;stress-energy tensor&lt;/strong&gt; $T_{\mu\nu}$ that might also depend on $g$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Newtonian gravitation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If now velocities are small compared to the speed of light $c$, the stress-energy tensor approximately only consists of its time-component, $$T_{tt} \approx \rho c^2$$ and the metric is flat with the exception of $$g_{tt} \approx 1 + \frac{2 U}{c^2}$$ where we can find, directly from the Einstein equations, that the Newtonian $$\Delta U = 4\pi G \rho$$ holds. This is a static approach and we see that there is no dependence on any flux term $j_i \propto T_{ti}$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Velocity matters: rotating disc of dust&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Considering the &lt;strong&gt;whole theory&lt;/strong&gt;, we find that of course the metric will &lt;strong&gt;depend&lt;/strong&gt; on contributions of all components of $T$ but only have a meaningful effect if associated &lt;strong&gt;characteristic velocities&lt;/strong&gt; are approaching the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A prominent analytic solution is that of a &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v75/i17/p3046_1&quot; rel=&quot;nofollow&quot;&gt;rigidly rotating disc of dust&lt;/a&gt;. Taking this solution, you can get an idea of how relativistic effects are important for the theory calculating the multipole moments $Q_n$ with respect to some relativistic parameter $\mu$  (corresponding also to the angular frequency $\Omega$ of the disc).&lt;br&gt;&#xA;In the following picture you can see that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kerr_metric&quot; rel=&quot;nofollow&quot;&gt;Kerr-spacetime&lt;/a&gt; is approached (from above!) for all moments $Q_n(\mu)$ for $\mu \rightarrow \infty$. This means that there is some $\mu$ where the effects of &lt;strong&gt;rotation dominate&lt;/strong&gt; those of the mass itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/lGvFD.png&quot; alt=&quot;Quadrupole Conjecture Disc&quot;&gt;&#xA;(Picture taken from &lt;a href=&quot;http://arxiv.org/abs/0902.1859&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to &lt;strong&gt;conclude&lt;/strong&gt;, there will only be some measurable time change for a person living on a rotating planet if it is &lt;strong&gt;extremely fastly rotating&lt;/strong&gt;. It is hence needless to say that this person would have some &lt;strong&gt;other difficulties&lt;/strong&gt; than to measure this deviation from an almost flat metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2011-05-13T22:28:48.250" />
  <row Id="9832" PostTypeId="2" ParentId="9805" CreationDate="2011-05-13T23:54:12.403" Score="1" ViewCount="" Body="&lt;p&gt;Momentum as an english term is somewhat abstract, imho. Personally I consider the Russian equivalent  &quot;момент количества движения&quot; (roughly translated as the moment of the &lt;em&gt;amount&lt;/em&gt; of motion) to be much more illuminating to a layman. Disregarding the term moment you can describe angular momentum (and linear momentum) as an &lt;em&gt;amount of motion&lt;/em&gt;, since quantities of things is something a 9 year old should be fairly familiar with, I propose pivoting explanations around this interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am of the opinion that self-discovery is always better than lecture or explanation, so I try to facilitate the child coming to the conclusion (through leading questions and hints) rather than just telling them the answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 1&lt;/strong&gt;&#xA;Give the child a top and ask them to come up with some way to &lt;em&gt;give a number&lt;/em&gt; to describe the spinning. Presumably they will come up with something akin to angular velocity, the number of rotations in some time, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2&lt;/strong&gt; Proceed to ask the child to assign a &lt;em&gt;direction&lt;/em&gt; to the spinning. Presumably their intuition will jump to clockwise and counterclockwise. Lead them into the right direction by challenging them to more &lt;em&gt;precisely&lt;/em&gt; state the direction of the spinning. The answer you are looking for is &quot;clockwise around an axis of rotation that is pointing up&quot;. Leading questions like &quot;what is the top spinning about?&quot; &quot;Which way is up for the top&quot; can lead to this kind of understanding&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3&lt;/strong&gt;&#xA;Tell the child that the direction and the number they came up with in the previous parts (disregarding mass for the moment, since it is fairly just a scaling factor) can be thought of as an &quot;amount&quot; of spinning, the same way that saying &quot;there are 5 apples&quot; is an amount of objects. The analogy being that you have a number and a qualifier, &lt;em&gt;apples&lt;/em&gt; corresponds to a direction, &lt;em&gt;5&lt;/em&gt; corresponds to the angular velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4&lt;/strong&gt;&#xA;Ask the child if &lt;em&gt;5 apples&lt;/em&gt; can become &lt;em&gt;5 oranges&lt;/em&gt; magically. Use this line of reasoning (and the analogy) to explain that &lt;em&gt;5 clockwise around that axis pointing up&lt;/em&gt;, can't magically become &lt;em&gt;5 clockwise around the axis pointing sideways&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5&lt;/strong&gt;&#xA;Asking the child if they can identify if anything is actually touching the spinning top. Conclude by asking them the leading question &quot;If nothing is touching or bothering the top, why would the amount of spinning change?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a theoretical discussion it is good to have a physical demonstration - Lubos's suggestion of using a bycicle wheel is perhaps the easiest way to demonstrate the fact that &quot;things want to keep spinning the same direction* in a way that the child can feel. &lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2011-05-13T23:54:12.403" CommentCount="2" />
  <row Id="9833" PostTypeId="2" ParentId="9776" CreationDate="2011-05-14T01:06:22.920" Score="1" ViewCount="" Body="&lt;p&gt;As a piano player, I sometimes shake my head back and forth to give myself the sensation of hearing vibrato.&lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-05-14T01:06:22.920" CommunityOwnedDate="2011-05-14T05:44:18.433" />
  <row Id="9834" PostTypeId="2" ParentId="9776" CreationDate="2011-05-14T02:38:30.120" Score="2" ViewCount="" Body="&lt;p&gt;There is &lt;a href=&quot;http://en.wikipedia.org/wiki/Doppler_effect#Application&quot; rel=&quot;nofollow&quot;&gt;a list on Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radar guns use an optical Doppler effect to measure speed.  Their acoustic equivalent is used in medicine, where it's called an ultrasound and used to measure blood flow or other sorts of motion in the body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Animals that use &lt;a href=&quot;http://en.wikipedia.org/wiki/Animal_echolocation&quot; rel=&quot;nofollow&quot;&gt;echolocation&lt;/a&gt; can use the Doppler shift to gain information about the motion of their surroundings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Sonic_boom&quot; rel=&quot;nofollow&quot;&gt;sonic boom&lt;/a&gt; occurs when the Doppler shift shifts a frequency to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess one can continue to concoct scenarios.  I wonder whether, when you drop a cat off a cliff, you can hear the pitch of its screaming drop as  it accelerates.  (It's not all that cruel - cats can usually survive a fall at terminal velocity.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could use it to determine which way a whale is swimming if you have two boats, both listening to the same whalesong.  You could even use the Doppler shift to gain information about the position of the whale because both the whale's position and its velocity contribute to the observed Doppler shift at any given place.  (I don't have any information about this actually being done, but it might be an interesting problem to work out the locus of possible whale locations for an observed Doppler shift.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was also curious about whether the Doppler effect gives us information on the motion of the crust that moves during an earthquake.  I found &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1111/j.1365-246X.1981.tb02664.x/abstract&quot; rel=&quot;nofollow&quot;&gt;this reference&lt;/a&gt; which suggests it does.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-05-14T05:54:11.987" LastActivityDate="2011-05-14T05:54:11.987" CommunityOwnedDate="2011-05-14T05:44:18.433" />
  <row Id="9835" PostTypeId="1" CreationDate="2011-05-14T02:42:27.923" Score="1" ViewCount="177" Body="&lt;p&gt;Does anyone have any references that discuss gravity in three-dimensions? I'm trying to make my way through some papers by Witten relating $SL(2,\mathbb{C})$ Chern-Simons theory and gravity in three spacetime dimensions and would like some explanation about the latter of the two topics. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3233" LastActivityDate="2011-05-14T02:42:27.923" Title="Three-Dimensional Gravity" Tags="&lt;gravity&gt;&lt;mathematical-physics&gt;&lt;quantum-gravity&gt;" CommentCount="2" />
  <row Id="9836" PostTypeId="1" AcceptedAnswerId="9924" CreationDate="2011-05-14T02:55:10.850" Score="2" ViewCount="197" Body="&lt;p&gt;Inspired by &lt;a href=&quot;http://physics.stackexchange.com/questions/9776/doppler-effect-of-sound-waves/9834#9834&quot;&gt;this question&lt;/a&gt; on the Doppler shift, suppose there is buoy somewhere on the surface of the ocean emitting a pure frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You get to place some boats wherever you want on the surface of the ocean, moving whatever direction you want with whatever speed you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boats listen to the pure frequency, which will in general be Doppler shifted due to the relative motion between buoy and boat.  If you do not know ahead of time the frequency the buoy is emitting, is it possible to deduce the buoy's location simply by observing the frequency measured by the boats?  What is the minimum number of boats necessary, how should they be positioned, and what velocities should they have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a toy problem with the mathematics of the Doppler effect, so let's leave out attenuation of sound over distance and assume the surface of the ocean is a plane.  Also, if the boats could continuously monitor the Doppler shift, they could collect extra information based on their own changes in position and velocity, so imagine locating the buoy based solely on a single frequency measurement from each boat.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-05-16T20:15:44.013" Title="How many boats does it take to find an acoustic buoy by Doppler shift?" Tags="&lt;acoustics&gt;&lt;doppler-effect&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="9837" PostTypeId="1" AcceptedAnswerId="9841" CreationDate="2011-05-14T03:43:29.723" Score="0" ViewCount="122" Body="&lt;p&gt;This problem is part of one of the free response questions on the 2011 AP Physics C Mechanics exam [&lt;a href=&quot;http://apcentral.collegeboard.com/apc/public/repository/ap11_frq_physics_cm.pdf&quot; rel=&quot;nofollow&quot;&gt;PDF of free-response questions&lt;/a&gt;] that was administered last week.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry the problem statement is so long; I wanted to make sure I'm not inadvertently changing the meaning by shortening it.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A projectile with mass $m$ and velocity $v_x$ is fired horizontally into a block of wood that is clamped to a tabletop so that it cannot move. The projectile travels a distance $d$ into the block before it stops.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(d) Derive an expression for the average force $F_b$ exerted on the projectile as it comes to rest in the block.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now a new projectile and block are used, identical to the first ones, but the block is &lt;em&gt;not&lt;/em&gt; clamped to the table. The projectile is again fired into the block of wood and travels a new distance $d_n$ into the block while the block slides across the table a short distance $D$. Assume the following: the projectile enters the block with speed $v_x$, the average force $F_b$ between the projectile and the block has the same value as determined earlier, the average force of friction between the table and the block is $f_T$, and the collision is instantaneous so the frictional force is negligible during the collision.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(e) Derive an expression for $d_n$ in terms of $d$, $D$, $f_T$, and $F_b$, as appropriate.&lt;br&gt;&#xA;  (f) Derive an expression for $d_n$ in terms of $d$, the mass $m$ of the projectile, and the mass $M$ of the block.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For part (d), I simply used energy to find $F_b$ using that $E = \frac12 m v_x^2 = F_b d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For part (e), I used energy again but with another term for friction: $\frac12 m v_x^2 = F_b d_n + f_T D$, giving $$d_n = \frac{ m v_x^2 - 2 f_T D }{ 2 F_b } = \frac{ F_b d - f_T D }{ F_b }.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure I'm right up to this point, but part (f) is quite confusing – it seems to me that there's absolutely no way that $d_n$ can be expressed only in terms of those three quantities because it needs to depend on the frictional force between the block and the table. Can anyone explain if I'm missing something? If I'm not missing anything, what is the most likely original intention of the question-writers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3532" LastActivityDate="2011-05-14T22:56:07.513" Title="Distance traveled by a projectile into a block" Tags="&lt;mechanics&gt;&lt;projectile&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9838" PostTypeId="2" ParentId="9529" CreationDate="2011-05-14T04:39:08.377" Score="3" ViewCount="" Body="&lt;p&gt;Hawking radiation is a very robust prediction. It comes simply from applying quantum field theory in the curved space-time near the event horizon. It's also part of the synthesis called &quot;black hole thermodynamics&quot;, for which string theory provides an explanation in terms of the statistical mechanics of microstates. In the S-matrix of quantum gravity, if black holes didn't evaporate, they'd show up as asymptotic states, but they don't. (There are eternal black holes in anti de Sitter space, but they still evaporate, they just don't get to evaporate completely; the particles produced by the evaporation can't escape to infinity because of the peculiarities of AdS geometry, and fall in again.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So denying the existence of Hawking radiation would screw up many other things. You could say that Hawking radiation is real but that it falls back in, like in AdS space, but there's no reason for it to do so. The paper featured at arxivblog is a &quot;what if&quot; paper which ignores all these problems and proceeds to calculate some of the consequences. You could compare it to an engineering study of one of M.C. Escher's impossible structures: if you ignore the contradictions in its design, maybe you can calculate some of its properties, but it only has recreational value to do so. We don't quite &lt;em&gt;know&lt;/em&gt; that a nonevaporating quantum black hole is logically impossible, in the way that we know the impossible staircase is impossible, but in the future a genuine proof may be available. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But empirical confirmation of black hole evaporation is rather unlikely. If we could produce mini black holes in colliders, then we'd see it, but those models aren't especially favored; they are a &quot;what if&quot; of a different sort, one in which there is at least a consistent fundamental picture behind the hypothesis (particular braneworld models), but it's just one of many possibilities about what happens at the next frontier of physics and those models are not significantly favored. (These models are also the ones which predict a detectable signature in GRB data.) If we could send a probe to the edge of an astrophysical black hole, maybe the radiation could be detected, but that is a job for interstellar civilizations, if they exist. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe you could find indirect evidence for Hawking evaporation of primordial black holes in the cosmic microwave background. But I don't know how likely that is - again, it would be highly model-dependent. &lt;/p&gt;&#xA;" OwnerUserId="1486" LastActivityDate="2011-05-14T04:39:08.377" />
  <row Id="9839" PostTypeId="2" ParentId="9836" CreationDate="2011-05-14T05:21:21.960" Score="1" ViewCount="" Body="&lt;p&gt;Really idealized approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the buoy is on the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Three boats.&lt;/strong&gt; They move at the same non-zero velocity much less than the speed of sound (but enough for the instruments to get read a change) perpendicular to the distance between them. Their separation is small compared to the length scales of the problem, but nearly twice the &quot;spotting something on the surface&quot; distance. At a given time they take a reading. Each boat gets a angle between it's velocity and the buoy, leading to a left-right ambiguity for each boat as well as the range ambiguity. Further, while the pairs should be able to resolve the range issue, there remains the possibly of several solution ambiguity for each pair. The third boat should resolve them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pathological case:&lt;/strong&gt; the buoy lies on the line between the boats when they take their reading. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution to the above:&lt;/strong&gt; Turn the formation through a significant angle and try again. Or just wait a while. Or if we insist on only one reading, have the center boat lead or trail the others by a significant distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problems with this:&lt;/strong&gt; The big one seems to be precision. Each boat does &lt;em&gt;not&lt;/em&gt; get a perfectly defined opening angle. They get a rough value. So in principle resolving the ambiguities could depend on the spacing of the boats and the relative location of the buoy. BUut I'm not going to do any math tonight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To make it harder:&lt;/strong&gt; Allow that the buoy might be sunken. Now the pairwise ambiguities are the intersections of cones instead of discrete points.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-15T20:42:56.960" LastActivityDate="2011-05-15T20:42:56.960" CommentCount="5" />
  <row Id="9840" PostTypeId="1" AcceptedAnswerId="9902" CreationDate="2011-05-14T05:39:46.840" Score="3" ViewCount="202" Body="&lt;p&gt;What specific behaviour confirmed the existence of the W and Z bosons at the UA1 and UA2 experiments?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1876" LastActivityDate="2011-05-16T09:17:54.630" Title="Detection of W and Z bosons" Tags="&lt;particle-physics&gt;&lt;standard-model&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9841" PostTypeId="2" ParentId="9837" CreationDate="2011-05-14T05:49:36.387" Score="1" ViewCount="" Body="&lt;p&gt;You certainly have a right to be upset about a poorly-written question.  The question talks about &quot;average force&quot; without defining what that means.  Usually, I would interpret &quot;average force&quot; to mean a time-average, but it looks like the person who wrote the problem intended it to mean an average over space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An average over space is an ill-defined concept in this case.  When the bullet enters the block and the block starts sliding backwards, are we averaging over the relative motion of the bullet through the block, or over the motion of the bullet relative to the table top?  Again, from context I'll infer it's an average over the path of the bullet relative to the block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, you can solve the problem by considering where the energy goes.  In the first case with the clamped block, all the energy goes into driving the bullet into the block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second case, some of the energy goes into stopping the bullet during the collision, just as before, but some of the energy goes into kinetic energy of the block+bullet system after the collision (and is then dissipated by friction with the table).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From $KE = p^2/2m$ and conservation of momentum, we see that the fraction of the bullet's original kinetic energy that goes into the block+bullet's kinetic energy immediately after impact is $m/(M+m)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the fraction of the kinetic energy that does not go into stopping the bullet, so the bullet's penetration depth goes down by a factor of one minus that, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_n = d \left(1 - \frac{m}{M+m}\right)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;edit&lt;/b&gt;: I simplified this a bit after re-reading the part about how we're supposed to consider the collision and the slowing of the block/bullet system due to friction as separate processes.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-05-14T07:33:50.777" LastActivityDate="2011-05-14T07:33:50.777" CommentCount="3" />
  <row Id="9842" PostTypeId="1" AcceptedAnswerId="9843" CreationDate="2011-05-14T07:37:02.510" Score="4" ViewCount="197" Body="&lt;p&gt;If, say, a particle with energy $E&amp;lt;V_0$, approaches a finite potential barrier with height $V_0$, and happens to tunnel through, where would the particle be during the time period when it is to the left of the potential barrier and to the right of the potential barrier? Surely there must be a finite amount of time for it to travel through to the other side, unless it simply teleports there? If it travels through with energy less than $V_0$, however, doesn't that mean it cannot enter in the region of the potential barrier?&lt;/p&gt;&#xA;" OwnerUserId="202" LastActivityDate="2011-05-30T15:48:10.427" Title="Where is the particle during a tunneling event?" Tags="&lt;quantum-mechanics&gt;&lt;tunneling&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="9843" PostTypeId="2" ParentId="9842" CreationDate="2011-05-14T08:37:37.443" Score="7" ViewCount="" Body="&lt;p&gt;Isn't the whole point here that one cannot say where the particle IS exactly? One can only calculate the probabilities of it being at one place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tunneling means the probability of it being inside the barrier isn't zero (since we want the probability distrubition to be continuous). There is always penetration of the wave function into the barrier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;IMHO tunneling means the penetration goes deep enough to actually reach the other side, so the wave function of the particle is propagated further on that side too, meaning there's a chance the particle went through. During the passing the particle has had a chance of being inside the barrier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if it's correct to say that when the particle has passed, it has been inside the barrier, but that just because the notion of the particle actually &lt;em&gt;being&lt;/em&gt; somewhere is somewhat wrong. &lt;/p&gt;&#xA;" OwnerUserId="3515" LastActivityDate="2011-05-14T08:37:37.443" />
  <row Id="9844" PostTypeId="2" ParentId="9842" CreationDate="2011-05-14T08:39:00.840" Score="4" ViewCount="" Body="&lt;p&gt;As Marek sais, the particle may be found &quot;inside&quot; the barrier, if you like. It means you can really find it there as well as outside. But in QM a particle is a wave and is &quot;created&quot; by the whole volume involved. It is not permanently &quot;localized&quot; or &quot;concentrated&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-05-14T08:55:01.523" LastActivityDate="2011-05-14T08:55:01.523" />
  <row Id="9846" PostTypeId="1" AcceptedAnswerId="9854" CreationDate="2011-05-14T11:02:38.527" Score="0" ViewCount="140" Body="&lt;p&gt;Two point charges, -2.5 micro coulombs and 6 micro coulombs, are separated by a distance of 1m (with the -2.5 charge on the left and 6 on the right).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the point where the electric field is zero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems exceptionally easy, but I can't figure it out. I can calculate the answer if both charges are negative/positive easily, but the fact that they're different is confusing me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3044" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-14T11:12:44.940" LastActivityDate="2011-05-16T19:31:15.707" Title="Determine the point at which the electric field is equal to zero?" Tags="&lt;homework&gt;&lt;electricity&gt;&lt;electrostatics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="9847" PostTypeId="1" AcceptedAnswerId="9851" CreationDate="2011-05-14T11:23:30.130" Score="0" ViewCount="94" Body="&lt;p&gt;this is my first question on PhysicsSE (I'm already an user of MathSE). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a mathematics students trying to understand Faraday's law, that is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\varepsilon= -\frac{d \Phi_B}{dt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\varepsilon$ means electromotive force and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Phi_B=\iint \mathbf{B}\cdot d\mathbf{S}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;means flux of magnetic field. As my textbook points out, there is an interpretation problem here: if the change in magnetic flux is due to movement of the conductor, then free charges in it are subject to Lorentz force, which then causes a current. On the contrary, if the conductor holds steady in a changing magnetic field, the induced current must be explained in terms of an electric field $\mathbf{E}$, described by the equations &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{cases} \nabla \cdot \mathbf{E}=\frac{\rho}{\varepsilon_0} \\ \nabla \times \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t}\end{cases}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt; Do those equations hold if we have a moving conductor in a stationary magnetic field? I guess not: this would mean $\mathbf{E}=\mathbf{0}$. How to solve this?&lt;/p&gt;&#xA;" OwnerUserId="3255" LastActivityDate="2011-05-14T12:21:25.137" Title="Moving conductors in magnetic fields: is there electric field or not?" Tags="&lt;electromagnetism&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9848" PostTypeId="1" AcceptedAnswerId="9855" CreationDate="2011-05-14T11:24:17.220" Score="2" ViewCount="86" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bessel_function&quot; rel=&quot;nofollow&quot;&gt;Hankel functions&lt;/a&gt; are solutions to the scalar Helmholtz-equation $$\Delta\psi + k_e^2\psi = 0$$ in cylindrical and spherical geometry (with respect to a separated angular dependence). Thus, they are very important describing &lt;strong&gt;spherical&lt;/strong&gt; and &lt;strong&gt;cylindrical&lt;/strong&gt; waves. Here is an example of such a propagation in the spherical case taken from &lt;a href=&quot;http://iem.at/Members/zotter/spherical_soundfield/gif_anims/RadialPropagationOutgoing.html&quot; rel=&quot;nofollow&quot;&gt;Franz Zotter&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/GSClW.gif&quot; alt=&quot;animation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am searching for a reference that states the &lt;strong&gt;phase accumulation&lt;/strong&gt; of &lt;a href=&quot;http://en.wikipedia.org/wiki/Bessel_function#Hankel_functions%3a_H.CE.B1.281.29.2C_H.CE.B1.282.29&quot; rel=&quot;nofollow&quot;&gt;Hankel waves&lt;/a&gt; of the form $$F_H^{\mathrm{out/in}}(\mathbf{r}) = H_m^{1/2}(k\rho)\ .$$&#xA;Assumed is stationarity with an $e^{-\mathrm{i}\omega t}$ time dependence fixing the meaning of the two different Hankel-waves as outgoing/incoming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;strong&gt;plane waves&lt;/strong&gt; one finds that the accumulated phase of a wave in $x$-direction, $$F_p=e^{\mathrm{i}kx}$$ is simply related to its argument, $$\phi_{\mathrm{acc}}(x_1,x_2)=\mathrm{Arg}(F_p(x_2))-\mathrm{Arg}(F_p(x_1)) = k(x_2 - x_1)$$ and it is &lt;strong&gt;natural&lt;/strong&gt; to just use this formula in the Hankel-case, e.g. $$\phi_{\mathrm{acc}}(\rho_1,\rho_2)=\mathrm{Arg}(F_H^{\mathrm{out/in}}(\rho_2))-\mathrm{Arg}(F_H^{\mathrm{out/in}}(\rho_1))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I was not able to find a &lt;strong&gt;suitable reference&lt;/strong&gt;. Hence my question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h3&gt;Is there a reference defining the phase accumulation of Hankel waves?&lt;/h3&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thank you&lt;/strong&gt; in advance.&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2011-05-14T17:35:42.777" Title="Phase Accumulation of Hankel-waves upon propagation" Tags="&lt;mathematical-physics&gt;&lt;electrodynamics&gt;" AnswerCount="2" />
  <row Id="9849" PostTypeId="2" ParentId="9842" CreationDate="2011-05-14T11:28:33.547" Score="-2" ViewCount="" Body="&lt;p&gt;The question &quot;how long does it take for the particle to tunnel?&quot; seems to be  more interesting than &quot;where is the particle during a tunneling&quot;. Quantum mechanics predicts that for an opaque barrier the delay time is independent on the length of the barrier (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Hartman_effect&quot; rel=&quot;nofollow&quot;&gt;Hartman effect&lt;/a&gt;) with the obvious problems when the barrier becomes very long.&lt;/p&gt;&#xA;" OwnerUserId="2962" LastEditorUserId="2962" LastEditorDisplayName="" LastEditDate="2011-05-30T15:48:10.427" LastActivityDate="2011-05-30T15:48:10.427" />
  <row Id="9850" PostTypeId="1" CreationDate="2011-05-14T12:14:02.110" Score="6" ViewCount="207" Body="&lt;p&gt;Ref: &lt;a href=&quot;http://arxiv.org/abs/1104.5675&quot; rel=&quot;nofollow&quot;&gt;CCC-predicted low-variance circles in CMB sky and LCDM&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To all cosmology / theoretical physics / related or similar researchers and academics,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I humbly solicit your view/s on the above-mentioned paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please feel free to provide your honest and frank opinion(s) on the paper's contents, it's validity, and potential future prospects for doing research in the area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Warmest gratitude in advance.&lt;/p&gt;&#xA;" OwnerUserId="1760" LastActivityDate="2011-05-17T12:05:41.507" ClosedDate="2011-05-19T18:06:00.830" Title="Feedback on the paper, 'CCC-predicted low-variance circles in CMB sky and LCDM' by V. G. Gurzadyan and R. Penrose" Tags="&lt;soft-question&gt;&lt;cosmology&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="9851" PostTypeId="2" ParentId="9847" CreationDate="2011-05-14T12:21:25.137" Score="2" ViewCount="" Body="&lt;p&gt;Dear dissonance, as discussed e.g. in these two questions&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/6457/calculate-the-electric-field-of-a-moving-infinite-magnet-without-boosting&quot;&gt;Calculate the electric field of a moving infinite magnet, without boosting&lt;/a&gt;&lt;br&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/6479/whats-a-good-reference-for-the-electrodynamics-of-moving-media&quot;&gt;What&amp;#39;s a good reference for the electrodynamics of moving media?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the laws of electrodynamics in moving materials may be a bit subtle (but they may be determined). However, whenever the motion is uniform, it's straightforward to transform Maxwell's equations back to the rest frame of the moving objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, a conductor will naturally have $\vec E =0$ in its rest frame while $\vec B$ is arbitrary. However, these two propositions get modified in a frame that is moving because the values of $\vec E,\vec B$ have to be transformed and mixed into one another if one changes the inertial frame. See&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity#Joules-Bernoulli_equation_for_fields_and_forces&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity#Joules-Bernoulli_equation_for_fields_and_forces&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Approximately, neglecting the terms of order $(v/c)^2$, we have&#xA;$$ \vec E&amp;#39; = \vec E + \vec v \times \vec B$$&#xA;$$ \vec B&amp;#39; = \vec B - \frac{\vec v}{c^2} \times \vec E  $$&#xA;Note that even if $\vec E=0$, it doesn't mean that the value $\vec E&amp;#39;$ in the moving frame is zero. Instead, it will be approximately $\vec v \times \vec B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of Faraday's effect, this electric field will have all the usual consequences and it will move the electrons just like you expect, proving that changing magnetic fields do have an impact on the flows of electrons, whether you view the situation from the viewpoint of the wires or the magnets.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-14T12:21:25.137" />
  <row Id="9852" PostTypeId="2" ParentId="9850" CreationDate="2011-05-14T12:45:48.677" Score="7" ViewCount="" Body="&lt;p&gt;Especially one of the authors is famous but the research of these CMB circles produced before the Big Bang resembles the research of crop circles produced by extraterrestrial aliens. In reality, the actual number of such &quot;strong enough circles&quot; is exactly what the statistical predictions based on standard cosmological models predict. It has to be so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors have posted several papers with almost the same content and they're slightly changing the message. For example, in the newest paper you mentioned, they try to place their &quot;new effects&quot; in the large-$L$ region (low angular separation or apparent radii of the circles). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the fundamental flawed content remains unchanged. It's a simple mathematical fact that the excess of arbitrary spherically symmetric &quot;special patterns&quot; that can be described by a quadratic score &#xA;$$\int d \theta_1 d\phi_1\, d\theta_2 d\phi_2 K(\theta_1,\phi_1,\theta_2,\phi_2) \Delta T(\theta_1,\phi_1)\Delta T(\theta_2,\phi_2)$$ &#xA;and the Penrose-Gurzadyan &quot;concentric circles&quot; surely belong to this category (they haven't studied any non-Gaussianities, and even if they had, none of them are detectable in the measured WMAP spectrum as of today, anyway) - is inevitably included in the power spectrum as a function of $L$. For a simple proof, see&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2010/12/why-penrose-and-gurzadyan-cannot-see.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/12/why-penrose-and-gurzadyan-cannot-see.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The kernel of their quadratic score $K(\hat n_1,\hat n_2)$ may be expressed as an operator acting on functions of $S^2$, the celestial two-sphere, and if this operator (e.g. if their rule to detect circles) is spherically symmetric, it has to be a function of $L^2$ only which means that this whole operator may be fully described by the eigenvalues it has when acting on $L^2$ eigenstates. In other words, it is fully encoded in the WMAP spectrum as a function of $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Z4xtG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The WMAP spectral curve knows &lt;strong&gt;everything&lt;/strong&gt; about the abundance of arbitrary patterns as long as this abundance may be described by a spherically symmetric score that is bilinear in $\Delta T$.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one doesn't have to look at the particular data to be sure that the authors are misleading themselves. What they write about the excess, relatively to the predictions of the famous WMAP curve included on the picture above, is as impossible as $2+2=5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before I admitted to myself that such famous authors (or one of them) could be ignorant about basic methods linked to the spherical harmonics, I wrote two texts speculating that they rediscovered something that is actually valid. One of them was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2010/11/what-penrose-and-gurzadyan-have.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/11/what-penrose-and-gurzadyan-have.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I no longer think that the article above is relevant. Their methodology is way too rough and unrefined for them to rediscover the things I thought that they rediscovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as one defines the &quot;score&quot; quantifying the number of concentric circles with various properties in terms of squared temperature differences in various pairs of the skies, this &quot;score&quot; is fully included in the WMAP spectrum that depends on $L$, and because this spectrum beautifully agrees with the predictions of the LCDM standard cosmological model, there's no room for some &quot;improvements&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Independently of this basic mistake, I am surprised how incredibly superficial analysis they want to be making. They just want to answer the question &quot;too many concentric circles Yes/No&quot;, and moreover, they give a wrong &quot;Yes&quot; answer to this question (by cherry-picking and irrationally overemphasizing some vaguely defined circles they like, without doing a proper statistical scrutiny of well-defined properties of the WMAP maps). Meanwhile, proper cosmology is predicting and testing whole detailed functions of $L$ with many bumps etc. - something that they're extremely far from addressing. The idea that they could compete with the standard cosmological model, without saying iota of valid things about the WMAP spectral curve, is preposterous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the mechanism of the proposed cyclic cosmology couldn't work to produce the alleged patterns. Incredibly enough, the main mistake that Penrose et al. do is a wrong Penrose diagram (note that the same name appears twice) for a Big Bang Cosmology. There can't be any &quot;non-inflating&quot; cosmology that has this big volume &quot;before the Big Bang&quot;. If one wants to extend the diagram in this way, it's inevitable to include some very fast expansion of space that &lt;em&gt;is&lt;/em&gt; inflation - at least when it comes to its consequences for the spacetime geometry. One should then ask what this inflation-like evolution of the geometry was caused by. And all known legitimate solutions are &quot;morally&quot; versions of inflation; it has to be so. I discussed these matters here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2010/11/penroses-ccc-cosmology-is-either.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/11/penroses-ccc-cosmology-is-either.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So in my opinion, the authors have to get a failing grade in modern cosmology. Moreover, I am somewhat afraid that especially Penrose has to have understood the spherical harmonics calculus by now and he's defending the indefensible partly in order to promote a popular book he no longer wants to scratch (even though an honest scientist definitely should) - and this book is based on similarly incorrect statements.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-16T06:20:29.873" LastActivityDate="2011-05-16T06:20:29.873" CommentCount="7" />
  <row Id="9854" PostTypeId="2" ParentId="9846" CreationDate="2011-05-14T13:22:53.470" Score="1" ViewCount="" Body="&lt;p&gt;Because of symmetry, the point at which the forces cancel must be on the same line as the two charges -- it's effectively a 1-dimensional problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a small positive test charge were &lt;em&gt;between&lt;/em&gt; the two charges, it would be &quot;pushed&quot; or &quot;pulled&quot; in the same direction by both charges, so it cannot be there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a small positive test charge were to the right of the two charges, it would always be &quot;pushed&quot; harder to the right by the bigger charge than it would be &quot;pulled&quot; to the left by the smaller charge, so it cannot be there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small positive charge that is placed to the left will be pulled to the right by the nearby negative charge and pushed to the left by more distant, but larger positive charge, so the balance point must be somewhere here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, taking the force that pushes or pulls a positive test charge $q$ at a distance $x$ to the right of the negative charge to be positive, there is a force proportional to $2.5q/x^2$ due to the nearby negative charge and a force proportional to $-6q/(1+x)^2$ due to the more distant positive charge. Adding the two together and asking the sum to be zero gives you an equation that you can convert into a quadratic equation. You only want the positive solution, right? Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the key to getting this kind of thing right? Decide what direction you're going to call positive and stick to it ruthlessly. Introducing a test charge and choosing it to be positive helps here by making the question less abstract.  Draw a rough diagram to help you keep which is which straight in your head. I had to be careful to make sure that choosing a positive charge instead of a negative charge doesn't make a difference (check why not, why it helped me draw a diagram, and what the diagram would look like if I were to use a negative charge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This homework question is too easy for Physics SE. Ask a different kind of Question next time. It would have been better if you had set out carefully and in more detail in your Question why the approach you took didn't work out. There's a good chance that if you had set it out carefully you would have seen why you had a problem with the calculation and the answers you were getting.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-05-14T13:22:53.470" />
  <row Id="9855" PostTypeId="2" ParentId="9848" CreationDate="2011-05-14T13:25:59.667" Score="1" ViewCount="" Body="&lt;p&gt;A closed form solution (in terms of more elementary function than Hankel functions) does not exist. However, typically one is only interested in the regime where $k\rho \gg1$, i.e., the asymptotic region far away from the source. There one can use the asymptotic form of the Hankel functions&#xA;$$H^{1/2}_m (x) \sim \sqrt{\frac2{\pi x}} e^{\pm (ix -i \frac\pi2 m - i\frac\pi4)}.$$&#xA;Thus, the accumulated phase is given by&#xA;$$\phi_\text{acc}(\rho_1,\rho_2) = \pm k(\rho_2 - \rho_1),$$&#xA;i.e., the same as for a plane wave.&lt;/p&gt;&#xA;" OwnerUserId="2962" LastActivityDate="2011-05-14T13:25:59.667" CommentCount="4" />
  <row Id="9856" PostTypeId="2" ParentId="9850" CreationDate="2011-05-14T15:38:15.607" Score="2" ViewCount="" Body="&lt;p&gt;It's very easy to go wrong in analyzing microwave background maps for this sort of signal. In particular, unless you're quite used to doing this sort of analysis, it's easy to make mistakes in calculating the probability of getting a false positive -- that is, of seeing patterns like the ones you're looking for, even if the standard model is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't reproduced the calculations myself (although I've done a number of very similar analyses in the past), but I have read papers by &lt;a href=&quot;http://arxiv.org/abs/1012.1305&quot; rel=&quot;nofollow&quot;&gt;various&lt;/a&gt; &lt;a href=&quot;http://arxiv.org/abs/1012.1656&quot; rel=&quot;nofollow&quot;&gt;other&lt;/a&gt; &lt;a href=&quot;http://arxiv.org/abs/1105.1081&quot; rel=&quot;nofollow&quot;&gt;people&lt;/a&gt; arguing that the original result was incorrect. All of these papers present very clear and cogent explanations of the correct way to do this analysis and point out errors in the original analysis, which would have precisely the effect of inflating the statistical significance (by reducing the computed false-positive probability). Moreover, the rebuttals are written by people with extensive experience doing this sort of analysis, while the original paper was written by people with no such experience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I conclude that there's no convincing evidence in support of the Gurzadyan-Penrose model. As far as I can tell, there's essentially complete consensus among experts in the field that this is the correct conclusion.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-05-14T15:38:15.607" />
  <row Id="9857" PostTypeId="1" AcceptedAnswerId="9861" CreationDate="2011-05-14T15:43:54.227" Score="5" ViewCount="215" Body="&lt;p&gt;My question is not about (pseudo) philosophical debate; it concerns mathematical operations and experimental facts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is short: &lt;strong&gt;&quot;What is an observer?&quot;&lt;/strong&gt; What are the conditions required to be qualified of &lt;strong&gt;observer&lt;/strong&gt;, both mathematically and experimentally?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In advance, thanks a million,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isaac&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit] in QM&lt;/p&gt;&#xA;" OwnerUserId="2499" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-16T18:50:35.563" LastActivityDate="2011-07-16T18:50:35.563" Title="nature of an observer" Tags="&lt;quantum-mechanics&gt;&lt;observables&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="9860" PostTypeId="2" ParentId="9848" CreationDate="2011-05-14T17:35:42.777" Score="0" ViewCount="" Body="&lt;p&gt;I am not sure how you get outwardly travelling waves from Hanken functions alone without also using Bessel functions. My understanding is that by analogy with plane waves, you have the stading wave solutions eg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sin(kx)*cos(wt)  and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cos(kx)*sin(wt)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the travelling wave comes from the sum of these two. Similarly in cylindrical geometry you have the standing waves&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bess(kr)*cos(wt) and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hank(kr)*sin(wt)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the travelling waves shown in the applets are really the sum of these. I'm not sure if this point is relative to the question being asked, but this is how I understand the pictures.&lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-05-14T17:35:42.777" CommentCount="2" />
  <row Id="9861" PostTypeId="2" ParentId="9857" CreationDate="2011-05-14T17:51:56.237" Score="7" ViewCount="" Body="&lt;p&gt;Are we talking quantum mechanics?  Then I'd say that a &quot;measurement&quot; is any operation that entangles orthogonal states of the system under consideration with orthogonal states of the environment.  &quot;Measurement&quot; is the important thing in most formulations of QM.  Colloquially speaking, an observer is something that performs measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only other place in physics I can think of where &quot;observer&quot; shows up is in the oft-used phrase &quot;This is obvious to the casual observer&quot;.  This is just shorthand for &quot;I can't be bothered to write out the mathematical proof&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-05-14T17:51:56.237" CommentCount="4" />
  <row Id="9862" PostTypeId="1" CreationDate="2011-05-14T18:41:47.813" Score="0" ViewCount="86" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/8701/is-microcausality-necessary-for-no-signaling&quot;&gt;Is microcausality &lt;em&gt;necessary&lt;/em&gt; for no-signaling?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;If we try to measure two non-commuting observables, only the earlier measurement would be valid. The later measurement would be messed up by the first. What if both measurements are space-like separated? If space-like observables always commute, simultaneous measurements are possible. But what if we have space-like observables which do not commute? Is it possible to construct a theory where this is possible but no information can be signaled faster than light?&lt;/p&gt;&#xA;" OwnerUserId="3538" LastActivityDate="2011-05-14T18:41:47.813" ClosedDate="2011-05-15T01:47:49.107" Title="Can space-like fields not commute and still forbid faster-than-light signalling?" Tags="&lt;quantum-mechanics&gt;&lt;relativity&gt;" CommentCount="2" />
  <row Id="9863" PostTypeId="1" CreationDate="2011-05-14T21:17:38.880" Score="-1" ViewCount="138" Body="&lt;p&gt;I've been having difficulty creating a mathematical model of a pendulum in my code. While there is plenty of info on general equations describing a the motion of a pendulum, I seem to be having difficulty translating those into code, or more specifically cartesian coordinates. These difficulties no doubt stem from my poor math skills, thus, I am looking for help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial thought before even researching the matter was to simply use two independent Sine Wave oscillators. I used a basic equation &lt;code&gt;y(t) = A*sin(ω*t+φ)&lt;/code&gt;. I think that its probably worth mentioning here that while I &lt;em&gt;understand what that equation does&lt;/em&gt;. I'm not 100% clear one why it behaves periodically. Of course have dealt with trigonometric ratios before, but I don't feel like I have a grasp on the actual concepts behind it. I don't fully appreciate all of its consequences I guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways, I initially figured I could determine the x coordinate and the y coordinate using 2 separate sine wave oscillators, while having the period of the y oscillator be half of the value of the x oscillator. This however (possibly obvious to you...) resulted in something closely resembling lissajous curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I've been reading explanations on the movement of a pendulum, yet while reading I get lost at two points regardless of the source I am reading from. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, everyone refers to a differential equation that I don't understand, probably due to never taking calculus. Here is a link to them all...&#xA;&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/pend.html#c3&quot; rel=&quot;nofollow&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/pend.html#c3&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And secondly, how can I derive two separate coordinates from one of these equations?&lt;/p&gt;&#xA;" OwnerUserId="3539" LastActivityDate="2011-05-22T23:32:32.180" Title="Modeling and Simulating Pendulum Motion" Tags="&lt;gravity&gt;&lt;classical-mechanics&gt;&lt;angular-momentum&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9864" PostTypeId="1" AcceptedAnswerId="9873" CreationDate="2011-05-14T21:25:00.330" Score="8" ViewCount="306" Body="&lt;p&gt;In classical mechanics the angular momentum is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{L} = \mathbf{r} \times \mathbf{p}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In relativistic mechanics we have the 4-vectors $x^{\mu}$ and $p^{\mu}$, but the cross product in only defined for 3 dimensions. So how to define angular momentum in special/general relativity in terms of 4-vectors?&lt;/p&gt;&#xA;" OwnerUserId="1648" LastActivityDate="2011-05-15T06:56:59.533" Title="How to define angular momentum in special/general relativity?" Tags="&lt;general-relativity&gt;&lt;special-relativity&gt;&lt;angular-momentum&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" />
  <row Id="9865" PostTypeId="2" ParentId="9864" CreationDate="2011-05-14T21:44:22.357" Score="-2" ViewCount="" Body="&lt;p&gt;In relativistic mechanics the forth (or zeroth) components are not independent so everything can be cast and is cast in a 3D form.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-05-14T21:44:22.357" />
  <row Id="9866" PostTypeId="2" ParentId="9863" CreationDate="2011-05-14T22:33:48.437" Score="2" ViewCount="" Body="&lt;p&gt;Forget about the motion for a moment. First just consider a pendulum of length $\ell$ at an angle $\vartheta$ from rest. If $(0,0)=0\mathbf{e}_x+0\mathbf{e}_y$ is where the pendulum is attached, then the end of it is at&#xA;$$&#xD;&#xA;\bigl(\ell\cdot\sin(\vartheta), -\ell\cdot\cos(\vartheta)\bigr)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, now let's describe the motion in terms of these variables. $\ell$ would usually be constant, so we don't need to care about this. It's $\vartheta$ that varies in time, so we get&#xA;$$&#xD;&#xA;y(t) = -\ell\cdot\cos(\vartheta(t)),&#xD;&#xA;$$&#xA;$$&#xD;&#xA;x(t) = \ell\cdot\sin(\vartheta(t)),&#xD;&#xA;$$&#xA;Unfortunatly, we don't know $\vartheta(t)$ yet.&#xA;We can either&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Assume that it does oscillate harmonically, that is, $\vartheta(t)=A\cdot\sin(\omega t)$ for some $A$ and $\omega$, which is often a good approximation.&lt;/li&gt;&#xA;&lt;li&gt;Decide we can do better, which is necessary if you consider big angles like $30^\circ$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Lets stick to small angles here. $\vartheta(t)=A\cdot\sin(\omega t)$ with $A\ll1$. We can then try to further simplify the cartesian coordinate representations: in&#xA;$$&#xD;&#xA;x(t) = \ell\cdot\sin(\vartheta(t)) = \ell\cdot\sin\bigl(A\cdot\sin(\omega t)\bigr),&#xD;&#xA;$$&#xA;we can use that $A\cdot\sin(\omega t)\ll 1$ and $\sin(x)\approx x$ for $x\ll 1$, therefore&#xA;$$&#xD;&#xA;x(t) \approx \ell\cdot A\cdot\sin(\omega t).&#xD;&#xA;$$&#xA;In&#xA;$$&#xD;&#xA;y(t) = -\ell\cdot\cos(\vartheta(t)) = -\ell\cdot\cos\bigl(A\cdot\sin(\omega t)\bigr),&#xD;&#xA;$$&#xA;we can use that $\cos(x)\approx 1-\tfrac{x^2}2$ for $x\ll 1$, therefore&#xA;$$&#xD;&#xA;y(t) \approx \ell\cdot \bigl(\tfrac{A^2}2\cdot\sin^2(\omega t)-1\bigr)&#xD;&#xA;$$&#xA;which, due to $\sin^2(x)=\tfrac12-\tfrac12\cos(2x)$, can be further simplyfied to&#xA;$$&#xD;&#xA;y(t) \approx \tfrac{A^2}{4}\ell\cdot\cos(2\omega t) - \tfrac\ell2.&#xD;&#xA;$$&#xA;As you see, two harmonic oscillators is in fact not such a bad assumption, and the important thing is that $y$ must oscillate at twice the frequency of $x$.&#xA;&lt;hr&gt;&#xA;Sorry, I forgot about this question...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I say $\sin(x)$ is approximately $x$, I mean just what I say: if you look at the sine function, and &quot;zoom in&quot; very close to 0 (that is, $x\ll1$, for instance $x=.001$), then this function looks very much like the simple $f(x)=x$.&#xA;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=plot+sin%28x%29+from+-.01+to+.01&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=plot+sin%28x%29+from+-.01+to+.01&lt;/a&gt;&#xA;It's just when you look &lt;em&gt;very&lt;/em&gt; carefully that you notice it's in fact bent a little bit.&lt;/p&gt;&#xA;" OwnerUserId="3540" LastEditorUserId="3540" LastEditorDisplayName="" LastEditDate="2011-05-22T23:32:32.180" LastActivityDate="2011-05-22T23:32:32.180" CommentCount="2" />
  <row Id="9868" PostTypeId="2" ParentId="9857" CreationDate="2011-05-14T23:09:23.653" Score="1" ViewCount="" Body="&lt;p&gt;J von Neumann in Mathematical foundations of quantum mechanics (eg Princeton 1955, 1996),&#xA;Chapter IV.1 explained idea of &quot;observer&quot; and I think it is appropriate introduction (and not only for the history of the problem). For more recent discussion and more references it is possible to see, eg: M. Schlosshauer, Decoherence, the measurement problem, and interpretations of quantum mechanics,  Rev. Mod. Phys. (2005), etc, etc.&lt;/p&gt;&#xA;" OwnerUserId="3351" LastActivityDate="2011-05-14T23:09:23.653" />
  <row Id="9869" PostTypeId="1" AcceptedAnswerId="9872" CreationDate="2011-05-15T02:47:58.520" Score="0" ViewCount="180" Body="&lt;p&gt;Let Suppose a bullet of 2 gram is falling with the speed of 200km/h.&#xA;Now how can we know the time after which it gains it terminal velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also when it gain its terminal velocity the force of friction and gravity will becomes equal and it stop accelarating do at this point the force of friction acting on it becomes constant or it go on increasing and also decrease the velocity of object.   &lt;/p&gt;&#xA;" OwnerUserId="2671" LastActivityDate="2011-05-15T17:32:31.080" Title="Terminal Velocity" Tags="&lt;gravity&gt;&lt;velocity&gt;&lt;friction&gt;" AnswerCount="3" />
  <row Id="9870" PostTypeId="2" ParentId="9869" CreationDate="2011-05-15T03:18:50.930" Score="1" ViewCount="" Body="&lt;p&gt;If you assume that the atmosphere is constant (i.e. ignore the fact that density and pressure increase as you go downwards), then you could linearize the differential equation for the velocity versus time. [I.e. you expand the drag as a function of velocity about the terminal velocity, and keep the first order correction to drag]. If you were to do so, you will find that it exponentially approaches terminal velocity with some characteristic time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the more realistic case where density (and hence drag) is a function of altitude, you'll have to solve the 2nd order diffeq numerically.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-05-15T03:18:50.930" />
  <row Id="9871" PostTypeId="2" ParentId="9869" CreationDate="2011-05-15T04:11:58.767" Score="0" ViewCount="" Body="&lt;p&gt;Here is a &lt;a href=&quot;http://math.kennesaw.edu/~sellerme/sfehtml/classes/math2202/fallingbodies.pdf&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d/dt(v)=g-pv&lt;/p&gt;&#xA;&#xA;&lt;p&gt;plug in the values for g and p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the earths surface g=-9.8 m/s^2, and p depends on altitude/air density. &#xA;d/dt(v)=-9.8-p*v&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d/dt(v)=p*(v+(-9.8/p))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/(v+(-9.8/p),dv=p,dt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;integrate both sides to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ln(v+(-9.8/p)=p*t+C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;raising both sides as exponents of e gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v+(-9.8/p)=e^(p*t+C)=C*e^(p*t+C)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v(t)=C*e^(p*t)-(-9.8/p)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if p is known, and the velocity at time=zero is known, then the following will yield the value of C;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v(0)=C*1-(-9.8/p)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You said (I think) that v(0)=-200km/hr=-55.56m/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so: -55.56m/s=C*1-(9.8/p)        solve for C and plugin C into: v(t)=C*e^(p*t)-(-9.8/p)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C=-55.56+(9.8/p)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v(t)=(-55.56+(9.8/p))*e^(p*t)-(-9.8/p)     this equation is (after p is plugged in) complete&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tv=g/p&lt;/p&gt;&#xA;&#xA;&lt;p&gt;set the v(t) equal to tv:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g/p=(-55.56+(9.8/p))*e^(p*t)-(-9.8/p)     solve this equation for t at which tv is attained&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tv=terminal velocity&#xA;v=velocity&#xA;v(t)=velocity as a function of time&#xA;g=acceleration due to gravity&#xA;p=drag coefficient&#xA;t=time&#xA;C=constant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Termial velocity is a constant unless the altitude changes enough to change the drag coefficient p. If the altitude decreases, the drag coefficient (air resistance) will likely increase and thereby slow decread the tv since tv=t/p: as p increases tv decreases. &lt;/p&gt;&#xA;" OwnerUserId="2843" LastEditorUserId="2843" LastEditorDisplayName="" LastEditDate="2011-05-15T17:32:31.080" LastActivityDate="2011-05-15T17:32:31.080" CommentCount="20" />
  <row Id="9872" PostTypeId="2" ParentId="9869" CreationDate="2011-05-15T04:48:41.457" Score="3" ViewCount="" Body="&lt;p&gt;It's not actually friction, it's the &lt;em&gt;drag force&lt;/em&gt;, also called air resistance. Technically, they're different (even though people do sometimes say &quot;air friction&quot;). The drag force depends on the object's speed, shape, and size, as well as the density of the air, but in the ideal case you can assume that the object's shape and size and the density are constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, when an object is falling at terminal velocity, the forces on it balance out, which means it has zero acceleration - in other words, its velocity doesn't change. And if the velocity doesn't change, then the drag force, which depends on velocity, won't change either. So the object is &quot;stuck&quot; in a steady state in which the drag force on it will remain constant as it falls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, though, an object never actually reaches terminal velocity unless it starts at terminal velocity. As it gets closer and closer to terminal velocity, the drag force on it gets closer and closer to exactly balancing out the gravitational force, which means that the net force on it, and thus its acceleration, becomes less and less. Basically, as the object gets closer to terminal velocity, the rate at which it continues to approach terminal velocity gets slower. It'll get continually closer to terminal velocity as time goes on, but it never quite reaches it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last year I wrote a &lt;a href=&quot;http://www.ellipsix.net/blog/2010/04/calculating-terminal-speed.html&quot; rel=&quot;nofollow&quot;&gt;blog post&lt;/a&gt; about this issue which discusses it using the actual math. It might be of interest to you.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-05-15T04:48:41.457" CommentCount="15" />
  <row Id="9873" PostTypeId="2" ParentId="9864" CreationDate="2011-05-15T05:26:56.933" Score="9" ViewCount="" Body="&lt;p&gt;Dear asmaier, you shouldn't view $\vec L = \vec x \times \vec p$ as a primary &quot;definition&quot; of the quantity but rather as a nontrivial result of a calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angular momentum is &lt;em&gt;defined&lt;/em&gt; as the quantity that is conserved because of the rotational symmetry - and this definition is completely general, whether the physical laws are quantum, relativistic, both, or nothing, and whether or not they're mechanics or field theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To derive a conserved charge, one may follow the Noether's procedure that holds for any pairs of a symmetry and a conservation law:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Noether_charge&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Noether_charge&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In particular, the angular momentum has no problem to be evaluated in relativity - when the background is rotationally symmetric. The fact that you write $\vec L$ as a vector is just a bookkeeping device to remember the three components. More naturally, even outside relativity, you should imagine&#xA;$$ L_{ij} = x_i p_j - x_j p_i $$&#xA;i.e. $L_{ij}$ is an antisymmetric tensor with two indices. Such a tensor, or 2-form, may be mapped to a 3-vector via $L_{ij} = \epsilon_{ijk} L_k$ but it doesn't have to be. And in relativity, it shouldn't. So in relativity, one may derive the angular momentum $L_{\mu\nu}$ which contains the 3 usual components $yz,zx,xy$ (known as $x,y,z$ components of $\vec L$) as well as 3 extra components $tx,ty,tz$ associated with the Lorentz boosts that know something about the conservation of the velocity of the center-of-mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, the general $x\times p$ Ansatz doesn't get any additional &quot;gamma&quot; or other corrections at high velocities. It's because you may imagine that it's the generator of rotations, and rotations are translations (generated by $\vec p$) that linearly depend on the position $x$. So the formula remains essentially unchanged. In typical curved backgrounds which still preserve the angular momentum, the other non-spatial components of the relativistic angular momentum tensor are usually not preserved because the background can't be Lorentz-boost-symmetric at the same moment.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-15T06:56:59.533" LastActivityDate="2011-05-15T06:56:59.533" CommentCount="2" />
  <row Id="9874" PostTypeId="1" CreationDate="2011-05-15T09:24:37.793" Score="1" ViewCount="121" Body="&lt;p&gt;Does anyone know about good video documentaries about physics, possibly on youtube?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen a couple of them about string theory, but I could not find many others.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="3545" LastActivityDate="2011-05-15T13:16:02.087" CommunityOwnedDate="2011-05-15T15:23:05.480" Title="video documentaries about physics?" Tags="&lt;education&gt;&lt;history&gt;" AnswerCount="3" />
  <row Id="9875" PostTypeId="2" ParentId="9874" CreationDate="2011-05-15T10:38:17.620" Score="2" ViewCount="" Body="&lt;p&gt;They are probably not on Youtube, but many of BBC's series are pretty good documentaries, although, obviously, quite &quot;popularized&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, their 2007 series &quot;Visions of the future&quot; was a 3-part series about &quot;The Intelligence Revolution&quot;, &quot;The Biotech revolution&quot; and &quot;The Quantum revolution&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their Horizon-series also encompasses many popular physics-subjects, perhaps these are what you've seen about string-theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't remember the name, but one historically interesting documentary was one about how temperature and heat was discovered and how the quest for cooling to yet colder temperatures proceeded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Youtube, the Feynman lectures stand out. Just search for &quot;Feynman lectures&quot; on Youtube, there are many. &lt;a href=&quot;http://www.youtube.com/watch?v=LPDP_8X5Hug&quot; rel=&quot;nofollow&quot;&gt;This is the first of a series of lectures of QED, Quantum Electrodynamics&lt;/a&gt; of which he was a pioneer. This lecture is an absolute must-view!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/user/carlosjerez23&quot; rel=&quot;nofollow&quot;&gt;This youtube-user&lt;/a&gt; has posted a lot of interviews and talks.&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-05-15T10:38:17.620" CommunityOwnedDate="2011-05-15T15:23:05.480" CommentCount="1" />
  <row Id="9876" PostTypeId="1" AcceptedAnswerId="9880" CreationDate="2011-05-15T11:13:03.210" Score="5" ViewCount="223" Body="&lt;p&gt;I imagine a point-like particle can only experience the local properties of spacetime. But locally there is no curvature and no gravity, as it is often stated that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Locally, as expressed in the equivalence principle, spacetime is Minkowskian, and the laws of physics exhibit local Lorentz invariance.(&lt;a href=&quot;http://en.wikipedia.org/wiki/General_relativity#Definition_and_basic_properties&quot; rel=&quot;nofollow&quot;&gt;Wikipedia:General relativity&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if this is the case how does a point-like particle know, which direction it should follow in a gravitational field? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this be a seen as a hint that point-like particles cannot be used as underlying concept for a field theory like general relativity?&lt;/p&gt;&#xA;" OwnerUserId="1648" LastActivityDate="2011-05-15T14:59:36.380" Title="How can a point-like particle &quot;feel&quot; gravity, if locally the curvature of spacetime is always flat?" Tags="&lt;general-relativity&gt;&lt;locality&gt;&lt;curvature&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="9877" PostTypeId="2" ParentId="9874" CreationDate="2011-05-15T11:56:44.973" Score="3" ViewCount="" Body="&lt;p&gt;Feynman's &lt;a href=&quot;http://research.microsoft.com/apps/tools/tuva/index.html&quot; rel=&quot;nofollow&quot;&gt;Messenger Lectures&lt;/a&gt; are fun.  They require some special Microsoft video software, though, and only work on Bill Gates-certified browsers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are not really documentaries.  IMHO the physics documentaries I've seen (things like PBS's NOVA) are entertainment more than education.  That's not to say they don't have interesting historical value (as opposed to physics pedagogical value), so whether they're worth watching depends on what you want from them.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-05-15T12:03:51.770" LastActivityDate="2011-05-15T12:03:51.770" CommunityOwnedDate="2011-05-15T15:23:05.480" />
  <row Id="9878" PostTypeId="2" ParentId="9876" CreationDate="2011-05-15T12:43:27.463" Score="4" ViewCount="" Body="&lt;p&gt;&quot;How can a point-like particle “feel” gravity&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can't. That's in fact true not only for perfect point like particles, but also for sufficiently small objects, like... Can &lt;em&gt;you&lt;/em&gt; feel gravity? I don't think so. What you can feel is the ground pushing against your feet, but there is no way of knowing whether this is because of gravity or because the earth accellerates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that an object does not need to &lt;em&gt;know&lt;/em&gt; about gravity in order to obey it. It follows some path which is indeed locally indistinguishable from a zero-curvature path, but if you integrate over time the locally neglectable $\mathcal{O}(\Delta x^2)$ components will sum up to something finite, thus yielding a nonzero-curvature path.&lt;/p&gt;&#xA;" OwnerUserId="3540" LastActivityDate="2011-05-15T12:43:27.463" />
  <row Id="9879" PostTypeId="2" ParentId="9874" CreationDate="2011-05-15T13:16:02.087" Score="2" ViewCount="" Body="&lt;p&gt;You could try Shankar's two lecture courses at Yale, &lt;a href=&quot;http://oyc.yale.edu/physics&quot; rel=&quot;nofollow&quot;&gt;http://oyc.yale.edu/physics&lt;/a&gt;. He's pretty highly rated by students at Yale, and there was positive feedback from someone I recommended this to on PhysicsForums. Yale open courses are one of the offerings mentioned on Wikipedia at &lt;a href=&quot;http://en.wikipedia.org/wiki/OpenCourseWare&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/OpenCourseWare&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously these are not documentaries as such. They will require more engagement to get something from them. This is not intended to be an Answer in the &lt;em&gt;precise&lt;/em&gt; terms of your Question.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-05-15T13:16:02.087" CommunityOwnedDate="2011-05-15T15:23:05.480" />
  <row Id="9880" PostTypeId="2" ParentId="9876" CreationDate="2011-05-15T14:59:36.380" Score="3" ViewCount="" Body="&lt;p&gt;Different regions of a general spacetime that are Minkowskian to $O(Delta x^2)$ can have light cones which have null rays that point in different directions.  A particle in this spacetime moves from one such region to another by connection coefficients, sometimes called the Christoffel symbols, which patch together these different locally flat regions.  This patching together is what defines the geodesic equation, which in effect cancels out any measurable effect of passing from one region of flatness to another.  In that way a particle falling in a gravity field is equivalent to a local frame that is globally flat.  This is the equivalence principle.  One can patch up a spacetime into locally flat spacetime regions any possible way, which is a matter of choosing a coordinate system to work in.  This results in different connection terms which define the motion through this different coordinate choice.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An extended body “feels” gravity because different points, or small masses which compose it at various points, will tend to move on different geodesics.  This is actually more physical.  The deviation between two geodesics is a measure of curvature&#xA;$$&#xD;&#xA;\frac{dU^a}{ds}~=~R^a_{bcd}V^bU^cV^d,&#xD;&#xA;$$&#xA;where we think of $V^b$ as tangent vectors to two geodesics and $U^a$ as a vector between them at each point.  This will cause a body do distend and is what causes tides.  &lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-05-15T14:59:36.380" CommentCount="1" />
  <row Id="9881" PostTypeId="1" CreationDate="2011-05-15T15:28:24.833" Score="3" ViewCount="135" Body="&lt;p&gt;The geodesic equation (let's suppose that we're talking about massive particles, so I'll parameterize the path by proper time $\tau$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d^2 x^\mu}{d \tau^2} + \Gamma^\mu_{\rho \sigma}\frac{d x^\rho}{d \tau} \frac{d x^\sigma}{d \tau}=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is invariant under $\tau \rightarrow -\tau$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, falling particles clearly have a direction, they always fall in instead of out. Formally, how does a falling particle 'know' which way to move in $\tau$?&lt;/p&gt;&#xA;" OwnerUserId="3546" LastActivityDate="2011-05-15T18:39:22.667" Title="In GR, how do particles know how to fall in instead of out of a gravitational well?" Tags="&lt;general-relativity&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="9882" PostTypeId="2" ParentId="9881" CreationDate="2011-05-15T15:49:04.393" Score="2" ViewCount="" Body="&lt;p&gt;A geodesic parallel-transports its own tangent vector, so if you &quot;start&quot; the particle pointing into the local (time-like) future it will follow the geodesic in that direction by definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, how does a particle &quot;know&quot; how to move forwards in time in plain Minkowski space for that matter? Microscopically it doesn't, this is also put in by definition as for example you can consider antiparticles to be particles travelling backwards through time. Then you have to decide which sort to call particles and which to call antiparticles..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tau is just enumerating the points on a geodesic path in GR. You still have both forward and backward-tilted lightcones along the path, so somewhere you have to make a choice (which will be conserved by the geodesic as I wrote above).&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-05-15T15:49:04.393" />
  <row Id="9883" PostTypeId="2" ParentId="9881" CreationDate="2011-05-15T15:59:26.350" Score="6" ViewCount="" Body="&lt;p&gt;If you take any solution to the geodesic equation, the time-reversal of that will also be a solution. If one describes a rock falling down in the Earth's gravitational field, the other will describe a rock that was tossed up at some point in the past. The situation is just like Newtonian gravity in that respect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, $\tau$ is just a parameter used to label points on the path. Physically, we have a clear understanding that one direction of $\tau$ is &quot;toward the future&quot; and one is &quot;toward the past,&quot; but the fundamental equations of (non-statistical) physics don't distinguish between future and past. That's true for general relativity, Newtonian mechanics, electromagnetism, etc. &lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-05-15T15:59:26.350" CommentCount="6" />
  <row Id="9884" PostTypeId="1" CreationDate="2011-05-15T17:13:39.863" Score="0" ViewCount="91" Body="&lt;p&gt;The theory of relativity makes very precise predictions about how an object's motion through space-time affects the passage of time for both the object and observers in other frames of reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious, however, whether this motion is equivalent to changing / fluctuating gravitational fields? What if instead of an object moving through space/time we instead had the same object exposed to shifting gravitational fields, similar to what it would experience if it were actually moving relative to other masses? Can it be shown that the same set of effects would still exist from the object's frame of reference (eg time dialation, length-contraction, etc)?&lt;/p&gt;&#xA;" OwnerUserId="3548" LastActivityDate="2011-05-15T18:41:57.213" Title="Is relativistic motion equivalent to fluctuating gravitational fields?" Tags="&lt;general-relativity&gt;&lt;gravity&gt;&lt;spacetime&gt;&lt;time&gt;&lt;equivalence-principle&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="9885" PostTypeId="2" ParentId="9884" CreationDate="2011-05-15T17:34:57.683" Score="1" ViewCount="" Body="&lt;p&gt;Absolute motion is a very, very tricky concept in General Relativity.  You can always change coordinates to a coordinate system where an arbitrary path is 'stationary'${}^{1}$, thereby embedding the dynamics into the spacetime, and removing them from the matter of interest.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, since measurable quantities are usually either scalars formed by contracting $4-D$ quantities, or coordinate invariant integrals, the machinery of differential geometry will guarantee that all of the same effects will happen in any coordinate system.  What you do, then, is choose a set of coordinates to make the calculation that you're doing look simpler.  So, for instance, if we're studying the expansion of a flat universe, we can look at it in the so-called &quot;comoving coordinates&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ds^{2} = -dt^{2} + a(t)^{2}d^{3}{\vec x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $d^{3}{\vec x}$ is the three-metric of flat $3-D$ space.  Here, you see the explicit expansion and contraction of the $3-D$ space by multiplying by the factor of $a(t)$, which governs the expansion and contraction of the space.  Also, it is readily apparent in this system that every person in the universe can measure their clocks in the same way--apparently an expanding spatially flat universe has a universal time coordinate.  It looks like space is expanding.  But, you can also look at this spacetime using a different radial coordinate: $R=a(t)r$.  If you do this, then the metric becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ds^{2} = -\left(1-\left(\frac{\dot a}{a}\right)^{2}r^{2}\right)dt^{2} - 2\,dt\,dR \left(\frac{\dot a}{a}\right)r + d^{3}{\vec x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it looks like space isn't expanding at all--there is no time dependence in the spacelike 3-metric!  But what have we changed?  Well, we've introduced terms that look like a motion into our metric--the factor $\left(1-\left(\frac{\dot a}{a}\right)^{2}r^{2}\right)dt^{2}$ is associated with a time dialation effect, while the factor $2\,dt\,dR \left(\frac{\dot a}{a}\right)r$ will give you a doppler effect--in this coordiante system, we can describe the distant galaxies as moving away from us on top of a fixed $3-D$ metric.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which picture is right?  Both are.  Any quantities that we calculate will give us the exact same answers, thanks to the rules of differential geometry (which I won't get into here).  All we're doing is using different variables to describe the effects--and depending on which variables you use, different effect become more obvious. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{1}$The easiest way to do this is by finding the path of the particle of interest, and finding a set of &quot;parallel&quot; curves, and then choosing their proper time as your time coordinate, but there are other ways, certainly.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2011-05-15T18:41:57.213" LastActivityDate="2011-05-15T18:41:57.213" CommentCount="1" />
  <row Id="9886" PostTypeId="2" ParentId="9881" CreationDate="2011-05-15T18:39:22.667" Score="0" ViewCount="" Body="&lt;p&gt;What you have noticed implies that both solutions (in and out) are acceptable physical solutions (that is, observing them would not be in violation of known principles). But which solution the particle happens to follow depends on the initial conditions of the motion.&#xA;I would add that this is not restricted to GR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sensing the direction of the time is an &quot;emerging&quot; feature (we have found out that entropy increases, so we can &quot;order&quot; two snapshots taken at different times: the egg was whole, then it was broken and not viceversa). &lt;/p&gt;&#xA;" OwnerUserId="525" LastActivityDate="2011-05-15T18:39:22.667" />
  <row Id="9887" PostTypeId="1" CreationDate="2011-05-15T19:28:13.053" Score="10" ViewCount="381" Body="&lt;p&gt;My camera, which is powered by two AA batteries in series, would not power on. I removed the batteries, exchanged their locations, and the device worked again - for another 15 minutes or so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature of the batteries did not change, and only about few minutes passed before I discovered this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a physical explanation for this phenomenon?&lt;/p&gt;&#xA;" OwnerUserId="2809" LastEditorUserId="113" LastEditorDisplayName="" LastEditDate="2011-05-15T23:27:38.993" LastActivityDate="2011-05-15T23:27:38.993" Title="Why do 'dead' batteries work again after exchanging the places of the batteries in an electronic device?" Tags="&lt;electricity&gt;&lt;electric-circuits&gt;&lt;laws-of-physics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="9888" PostTypeId="1" CreationDate="2011-05-15T19:41:40.087" Score="3" ViewCount="220" Body="&lt;p&gt;If the friction from brakes, wind resistance and all such factors remain constat, which will stop first? A heavier car or a lighter car? How will the momentum of the car and graviational pull on a heavier object influence the stopping of the car?&lt;/p&gt;&#xA;" OwnerUserId="3549" LastActivityDate="2011-05-16T18:46:54.060" Title="Which will stop first a heavier car or a lighter car?" Tags="&lt;forces&gt;&lt;momentum&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="9889" PostTypeId="1" CreationDate="2011-05-15T19:55:22.433" Score="2" ViewCount="150" Body="&lt;p&gt;My understanding is that dark matter cannot be (or is at least highly unlikely to be) an exotic form of any known particle. On the other hand, articles about particle accelerators seem to say that the Higgs is the last piece missing in the Standard Model jigsaw puzzle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If&lt;/em&gt; dark matter is determined to be some form of new particle, what are the certain implications? Might such a discovery &quot;stand to the side&quot; of the Standard Model or would it certainly change the foundations? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Forgive my extremely lay understanding and vocabulary -- please feel free to correct my mistakes.)&lt;/p&gt;&#xA;" OwnerUserId="3550" LastActivityDate="2011-05-16T05:02:10.470" Title="If dark matter is a new type of particle, what does that imply?" Tags="&lt;cosmology&gt;&lt;standard-model&gt;&lt;dark-matter&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9890" PostTypeId="2" ParentId="9887" CreationDate="2011-05-15T20:01:33.250" Score="1" ViewCount="" Body="&lt;p&gt;The only &quot;reason&quot; I can make up, is the assumption that there was some &#xA;resistance caused by corrosion at one or more of the four contacts &#xA;between batteries and some springs, which You removed when You &#xA;changed places of the batteries. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-05-15T20:01:33.250" CommentCount="2" />
  <row Id="9891" PostTypeId="2" ParentId="9887" CreationDate="2011-05-15T20:51:32.337" Score="1" ViewCount="" Body="&lt;p&gt;Sometimes dead batteries will return to life if the are shaken. This only works for batteries that have liquid energy storage chambers. I have never heard of it working with AA batteries before, but I haven't heard of a lot of things. &lt;/p&gt;&#xA;" OwnerUserId="2843" LastActivityDate="2011-05-15T20:51:32.337" CommentCount="3" />
  <row Id="9892" PostTypeId="2" ParentId="9888" CreationDate="2011-05-15T20:57:00.977" Score="6" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Physics Land:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physics 101 answer to there kinds of questions goes like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the cars will lock up their wheels so that I can apply a simple analysis of sliding friction. The frictional force $F_f$ is dependent &lt;em&gt;only&lt;/em&gt; on the normal force $N$ between the car and the surface and on the coefficient of friction $\mu$ between the tires and the road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F_f = \mu N = \mu M g $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $M$ is the mass of the vehicle and $g$ is the acceleration due to gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this we can compute the acceleration due to friction as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a_f = \frac{F_f}{M} = \mu g $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the stopping distance as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ d = \frac{v^2}{2(a_f)} = \frac{v^2}{2 \mu g} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $v$ is the speed of the car at the moment the brakes are applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will notice that this does &lt;em&gt;not&lt;/em&gt; depend on the mass of the car. So &lt;strong&gt;both cars stop in the same distance&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Welcome to the real world:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The first expression above: $F_f = \mu N$ is an approximation that only applies when there are negligible deformations of either body. This is simply not true in the case of a car riding on inflated rubber tires in contact with a rough road surface. In the real world there are complicated interactions between the wheel and the tire and the road. The exact effects of this depend on the nature of the tire and the nature of the road. We are not in a position to say much.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The above analysis relies on both cars being able to lock up their wheels. In principle either car might fail to have the braking power to do so.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The above analysis relies on both cars skidding, but almost all modern cars have a sophisticated anti-lock braking system to prevent this and get a little more out of the brakes (skidding is slightly suboptimal, but better than a bad driver). Even without ABS sufficiently practiced and skilled drivers can sometimes outperform skidding stops by pumping the brakes and riding the edge of lock-up (this isn't easy, but we used to learn and practice it). The braking performance in a mixed skidding/rolling scenario is much harder to analyze.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So real world answer: &lt;strong&gt;beats the heck out of me.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Side note:&lt;/em&gt; the conventional wisdom has it that big rigs brake faster and better when fairly heavily loaded than either running empty &lt;em&gt;or&lt;/em&gt; loaded to the gills.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-16T14:45:08.697" LastActivityDate="2011-05-16T14:45:08.697" CommentCount="7" />
  <row Id="9893" PostTypeId="2" ParentId="9888" CreationDate="2011-05-15T20:58:04.440" Score="-1" ViewCount="" Body="&lt;p&gt;The lighter car will stop sooner. A heavier car has more momentum or kinetic energy than a lighter car. Friction reduces the momentum of an object. Assuming friction is the same for both cars means the momentum will decrease at the same rate for both. Since an object is at rest when its momentum equals zero, Whichever car starts out with more momentum will roll further. So far I have assumed a flat surface/no acceleration due to gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a slope, then gravity will act more strongly on the heavier vehicle. If the slope is uphill, then the heavier vehicle will require more momentum/kinetic energy to attain the same altitude. Going uphill the angle of the slope will determine which car goes further. With a downward slope, the larger vehicle will gain more kinetic energy/momentum over the same drop and certainly go further. &lt;/p&gt;&#xA;" OwnerUserId="2843" LastEditorUserId="2843" LastEditorDisplayName="" LastEditDate="2011-05-15T21:14:48.153" LastActivityDate="2011-05-15T21:14:48.153" CommentCount="2" />
  <row Id="9894" PostTypeId="1" AcceptedAnswerId="9903" CreationDate="2011-05-15T21:02:37.977" Score="2" ViewCount="184" Body="&lt;p&gt;Why is there an absolute entropy?  Given any non-discrete probability distribution, we don't really have an absolute entropy because the entropy depends on the parametrization of the distribution (e.g. Beta vs. Beta-prime) which was arbitrarily chosen.  Another way to put it is that instead of entropies we only have &lt;a href=&quot;http://en.wikipedia.org/wiki/Kullback%E2%80%93Leibler_divergence&quot; rel=&quot;nofollow&quot;&gt;Kullback–Leibler (KL) divergences&lt;/a&gt; aka relative entropies.  So, why isn't there a physics analogue to KL-divergence?  Just as we have relativistic velocity, which has some properties, why don't we also have relative entropies, which have some properties?  Instead of saying the absolute entropy of the universe increases, why don't we say that the relative entropy given our prior belief of the universe increases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, what is the relation between entropy and number of microstates when the physical system is continuous, and how do we &quot;count microstates&quot;?&lt;/p&gt;&#xA;" OwnerUserId="569" LastEditorUserId="569" LastEditorDisplayName="" LastEditDate="2011-05-16T02:18:52.550" LastActivityDate="2011-05-16T06:58:55.020" Title="Why is there an absolute entropy?" Tags="&lt;entropy&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="9895" PostTypeId="2" ParentId="9889" CreationDate="2011-05-15T21:04:38.380" Score="9" ViewCount="" Body="&lt;p&gt;The conventional wisdom about dark matter is that it is likely to be a new kind of particle that is not part of the standard model. Basically, the reason for this is that most of the stable standard model particles interact electromagnetically (and so wouldn't be &quot;dark&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exception is neutrinos, and for a long time neutrino dark matter was considered to be a viable possibility, but it doesn't seem to work for several reasons. The main one is that neutrino dark matter would be &quot;hot&quot; (meaning that the particles would have had relativistic speeds in the not-too-distant past), whereas the way dark matter is observed to cluster gravitationally only seems to work if the dark matter is &quot;cold.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For quite a while, people tried to come up with models in which the dark matter was ordinary matter (made of atoms), but that also doesn't seem to work for several reasons. In a Universe with a high enough density of atomic matter, the abundances of light elements produced in the early Universe would be quite different from what is seen. There's no good way to make this sort of matter &quot;dark enough&quot;: even if you make it cold and neutral,\ it still interacts with radiation too strongly to remain hidden. Also, a Universe made of only atomic matter predicts a spectrum of fluctuations in the microwave background, and a bunch of other cosmological observables, that differ by a huge amount from what we observe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the leading theory is that the dark matter is a different sort of stable, neutral, weakly-interacting particle. Such a particle would necessarily be &quot;beyond the standard model.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the least exotic possibility is that it's a supersymmetric particle. If supersymmetry is right, there are a lot of new particles out there waiting to be discovered. Most of them are unstable, but the lightest one is stable and would make an excellent dark matter candidate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If supersymmetry is right, there's a good chance the LHC will detect it, so we may actually know the answer to this in the not too distant future. Then again, we may not.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-05-15T21:04:38.380" CommentCount="5" />
  <row Id="9896" PostTypeId="1" AcceptedAnswerId="9900" CreationDate="2011-05-16T00:37:17.750" Score="1" ViewCount="209" Body="&lt;p&gt;I am working on a game involving flying and steering a paper airplane for WP7. I want the plane to fly just like how normal paper airplanes fly (see this game for an example &lt;a href=&quot;http://armorgames.com/play/7598/flight&quot; rel=&quot;nofollow&quot;&gt;http://armorgames.com/play/7598/flight&lt;/a&gt;) but I can't seem to find an equation for how paper airplanes fly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone have any experience with this? In my game now, it just follows the usual motion for an object in a vacuum, which makes for some flight, but it doesn't feel perfect, and traveling at a slight downward angle makes you lose speed, which isn't right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thoughts?&lt;/p&gt;&#xA;" OwnerUserId="3552" LastActivityDate="2011-05-16T02:17:28.877" Title="Paper airplane physics" Tags="&lt;airplanes&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="9897" PostTypeId="1" CreationDate="2011-05-16T01:19:35.363" Score="1" ViewCount="134" Body="&lt;p&gt;Euler's laws of motion for a distributed mass are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = \frac{d}{dt} MV_{cm},\ N = \frac{d}{dt} L$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F$ are the sum of the external forces, $M$ the total mass, $V_{cm}$ the velocity of the centre of mass.  $N$ are the sum of the moments of the external forces about some given point, L the total angular momentum about the same point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a gyroscope is supported at its base with its axis horizontal, it precesses at a constant angular velocity. Using the above equations, how does one show this?&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-05-16T20:09:19.743" Title="How are Euler's laws of motion applied to gyroscopes?" Tags="&lt;classical-mechanics&gt;&lt;gyroscopes&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="9898" PostTypeId="1" AcceptedAnswerId="9907" CreationDate="2011-05-16T01:29:43.340" Score="2" ViewCount="140" Body="&lt;p&gt;Clouds are made up of tiny water or ice droplets, depending on temperature. This implies that cloud density is greater than that of dry air. Why don't clouds sink through their surrounding atmosphere rather than float by in a variety of formations? &lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-05-16T18:32:42.093" Title="The Density of Clouds" Tags="&lt;atmosphere&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="9899" PostTypeId="1" CreationDate="2011-05-16T02:09:39.130" Score="3" ViewCount="154" Body="&lt;p&gt;Matt in &lt;a href=&quot;http://photo.stackexchange.com/questions/9597/what-does-a-hexagonal-sun-tell-us-about-the-camera-lens-sensor/9601#9601&quot;&gt;his answer&lt;/a&gt; on &lt;a href=&quot;http://photo.stackexchange.com/q/9597/64&quot;&gt;What does a hexagonal sun tell us about the camera lens/sensor?&lt;/a&gt; mentions&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Incidentally the number of (distinct)&#xA;  points to the star is equal to double&#xA;  the total number of unique&#xA;  orientations* in the sides of the&#xA;  aperture shape i.e. three blades would&#xA;  be six points, six blades would also&#xA;  be six points as a hexagon has only&#xA;  three unique orientations in its&#xA;  sides.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;* a hexagonal aperture has six sides but only three unique orientations as&#xA;  there are three pairs of parallel&#xA;  sides.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think It makes sense to have three pointed star in case of three blades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how does a six blade aperture also result in a three pointed star?&lt;/p&gt;&#xA;" OwnerUserId="230" LastActivityDate="2011-05-16T14:56:27.997" Title="How does Fraunhofer diffraction depend on the orientation of the sides of a lens?" Tags="&lt;optics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="9900" PostTypeId="2" ParentId="9896" CreationDate="2011-05-16T02:17:28.877" Score="2" ViewCount="" Body="&lt;p&gt;There is no simple equation for how a paper airplane flies like there is for a simple projectile because the airplane can interact with the air in complicated ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physics of a paper airplane is described by &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion&quot; rel=&quot;nofollow&quot;&gt;Newton's laws of motion&lt;/a&gt;.  These laws apply to both the airplane and the air it travels through.  The plane is acted on by a &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_gravity&quot; rel=&quot;nofollow&quot;&gt;constant gravitational force&lt;/a&gt; and by &lt;a href=&quot;http://en.wikipedia.org/wiki/Contact_force&quot; rel=&quot;nofollow&quot;&gt;contact forces&lt;/a&gt; with the air, especially &lt;a href=&quot;http://en.wikipedia.org/wiki/Drag_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;drag&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Lift_%28force%29&quot; rel=&quot;nofollow&quot;&gt;lift&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nature of the force between the air and the plane can be quite complicated, and requires an extremely detailed analysis for accurate simulation.  For example, by constructing the plane slightly differently, you can make it fly faster, slower, further, curve left or right, or bob up and down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic physical ideas are those of &lt;a href=&quot;http://en.wikipedia.org/wiki/Fluid_dynamics&quot; rel=&quot;nofollow&quot;&gt;fluid dynamics&lt;/a&gt; and the basic equation involved is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations&quot; rel=&quot;nofollow&quot;&gt;Navier-Stokes equation&lt;/a&gt;.  Modeling something like an airplane accurately is mostly the domain of expertise of aeronautical engineers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make a simple model for a game, you might want to start with a simple constant gravity force, a drag force proportional to the square of  the velocity, and a lift force also proportional to the square of velocity (which comes from &lt;a href=&quot;http://en.wikipedia.org/wiki/Lift_%28force%29#Lift_coefficient&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;), and then play around with the parameters until you find something pleasing to your eye.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-05-16T02:17:28.877" CommentCount="2" />
  <row Id="9901" PostTypeId="2" ParentId="9889" CreationDate="2011-05-16T05:02:10.470" Score="0" ViewCount="" Body="&lt;p&gt;Let me address this part of the question, as the physics part is covered  by Ted Bunn.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If dark matter is determined to be some form of new particle, what are the certain implications? Might such a discovery &quot;stand to the side&quot; of the Standard Model or would it certainly change the foundations? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Up to now, progress in understanding particle physics has been happening not by completely discarding previous theories, but by assimilating them. The reason is that previous theories, as the standard model will become too in the future, were based on solid data. Being based on data means that they are just a shorthand of describing them. The old data still exists when new data appear, as the example of dark matter , so what will happen is that the standard model will be incorporated in any new theory describing matter in the microcosm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Already string theories include the standard model within their structure so no &quot;stand aside&quot; will happen if they are found sufficient to describe all new data. String theories have a plethora of matter forms that could well explain/describe dark matter.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-16T05:02:10.470" />
  <row Id="9902" PostTypeId="2" ParentId="9840" CreationDate="2011-05-16T05:51:18.603" Score="2" ViewCount="" Body="&lt;p&gt;Since Lubos did not make an answer of his comment I would suggest reading the &quot;discovery&quot; paragraph in the &lt;a href=&quot;http://en.wikipedia.org/wiki/W_boson#W_bosons&quot; rel=&quot;nofollow&quot;&gt;Wiki article.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;specific behaviour&quot;  are the characteristic decay modes of each particle, that clustered at the masses finally assigned, as the reference in the comment above shows.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need a library to go to the original  discovery references, but the decay modes are in the &lt;a href=&quot;http://cdsweb.cern.ch/record/180574/files/198710042.pdf&quot; rel=&quot;nofollow&quot;&gt;article in this link&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: from the last link:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/LsDL5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-05-16T09:17:54.630" LastActivityDate="2011-05-16T09:17:54.630" CommentCount="2" />
  <row Id="9903" PostTypeId="2" ParentId="9894" CreationDate="2011-05-16T06:58:55.020" Score="1" ViewCount="" Body="&lt;p&gt;The measure isn't arbitrary. In classical mechanics, the symplectic structure of phase space defines the Liouville measure. In quantum mechanics, the Hilbert space norm plays this role.&lt;/p&gt;&#xA;" OwnerUserId="3554" LastActivityDate="2011-05-16T06:58:55.020" CommentCount="1" />
  <row Id="9904" PostTypeId="1" CreationDate="2011-05-16T07:54:06.000" Score="-2" ViewCount="190" Body="&lt;p&gt;According to black hole complementarity, if there is a black hole and Alice falls into it carrying a qubit, but Bob stays out, then Alice can measure the qubit inside the black hole, and confirm it exists inside, but Bob can observe the Hawking radiation coming out, collect all the data, feed them into a giant quantum computer, and extract the qubit from the outgoing Hawking radiation. Either the no xerox theorem is violated, or Alice' and Bob's observations are incommensurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In plain old quantum mechanics, it is possible to set up an experiment where we can choose between measuring the position, or the momentum of a particle, but once that choice is made, all observers will agree upon whether it's the position or the momentum which is being measured. But not for black holes? Is the qubit inside, or outside?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bob has a problem. The qubit is scrambled and deeply encoded within the Hawking radiation. After collecting all the relevant Hawking radiation, would Alice have hit the singularity by then? If so, Alice measured first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things are actually much worse. According to complementarity, encoded within the Hawking radiation is also an extremely scrambled version of Alice as she would look like after drinking too many &quot;Drink me&quot;s, eating too many &quot;Eat me&quot;s, and nibbling way too many mushrooms. To make matters even worse, the scrambling scrambles the scrambled Alice with the scrambled qubit, and to extract the qubit, his quantum computer would also have to unscramble Alice from the qubit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PS&lt;/strong&gt;: If Alice is encoded holographically within the Hawking radiation, does that mean Bob can reconstruct what Alice did in the hole by observing all the Hawking radiation and feeding them into his quantum computer? And if he can reconstruct what Alice did, in what sense can't she communicate with Bob?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to simultaneously measure observables which don't commute with each other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does Alice commute with Bob? They can't compare notes, that's for sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Niels Bohr, something is objective only if an observer can measure it and unambiguously communicate the results in public using common language. If Alice can't communicate anything with the outside world, is anything she observes objective? Is she even qualified to be considered an observer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say Humpty-Dumpty had a great fall to the black hole. According to Bob, as he approaches the hot stretched horizon, he becomes heated and scrambled, until we get scrambled eggs. The scrambled eggs evaporate as Hawking radiation. What Bob wishes to do is put Humpty-Dumpty back together again. To do that, he has at his disposal the entire royal team consisting of all the king's horses, and all the king's men. They engineer a device which can detect and measure every single quanta of Hawking radiation from the beginning right until the very end. Not only that, the measuring devices can preserve the quantum coherence of all the Hawking radiation quanta. All this data is transmitted along a coherence preserving perfect quantum channel to a quantum computer far away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to reconstruct Humpty-Dumpty inside the quantum computer? Would this violate the second law of thermodynamics? According to Szilard, Landauer and Bennett, no, because we can convert entropy into information. Bob's quantum computer has enough memory to store all the information of the black hole. Not only that, it's much larger than the black hole in size so it doesn't collapse into a black hole of its own. It preserves quantum coherence. In fact, it's so large, and we are so patient to wait for an astronomically long time that all the quantum computer processes are low energy, and never remotely approach Planck energies at all. All the redshifted Hawking radiation are also low energy. Surely the universe is Turing complete enough for Bob to construct and program such a quantum computer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can Bob reconstruct Humpty-Dumpty in principle? What about in practice? If something can be done in principle, but not in practice, is it still possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Alice inside the hole, Humpty-Dumpty said a word inside the hole. This word is holographically encrypted within the Hawking radiation, and is all there outside in encrypted form for Bob to read. The information is all out there, and if Bob takes the detected encrypted pattern as the word itself, he gets the word. But to find out what that word means, he needs a &quot;common language&quot;, and he can use the quantum computer as the translator. If translated, did Humpty-Dumpty send a word faster-than-light? If untranslated, did he still send a word faster-than-light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside the hole, Alice measures the qubit (Tweedle) along a basis consisting of the states &quot;DEE&quot; and &quot;DUM&quot;. Outside, Bob detects all the Hawking radiation and feeds them into his quantum computer. He runs a program to extract Tweedle. But Tweedle is scrambled with Alice and her measuring apparatus, and in fact, just about almost everything else inside the hole. That's the nature of holographic encryption. According to Bohr, we have to include the measuring apparatus in the description. To unscramble Tweedle requires the quantum computer to extract a simulated Alice and simulated apparatus as well, and store them in the memory needed according to Landauer and Bennett. We can't erase the information! But the holographic Alice has already measured the holographic Tweedle using the holographic apparatus, and now, they are all entangled with each other. From the perspective of these holographic entities, this entanglement has decohered. To extract the pure qubit, Bob has to run another problem implementing the unitary transformation reversing the measurement and run them on the simulated Alice-apparatus-qubit system. This recombines all the decohered many worlds, recohering them. After extracting Tweedle, he measures it along an incompatible basis. But the simulated Alice now no longer has any memory of the measurement after the time-reversed transformation. The simulated Alice can remember the future, but not the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about it this way; the Hawking radiation is what we get after evolving time forward for the system consisting of the exterior and the stretched horizon. To reconstruct Alice requires the quantum computer to reverse time in the simulation right back to the moment when Alice and the qubit fall into the hole, then run the simulation forward in time again from the perspective of Alice. To simulate this time evolution, the quantum computer also has to include a significant part of the state outside the hole to take into account infalling matter, but this can be done in principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was a time when a simulated Alice knew whether the simulated Tweedle was DEE or DUM, but Bob had no clue whatsoever about Tweedle, and can't without decohering the simulation along the DEE/DUM basis. After reversing time in the simulation, the simulated Alice had her memory of the value of Tweedle wiped clean, but now, Bob can measure the simulated qubit along the incompatible basis. There was never any time when both the simulated Alice, and Bob knew what the value of Tweedle was along incompatible bases simultaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was Alice teleported from the interior into the simulation faster than light? The simulated Alice would have no clue she is inside a simulation, and would think she exists in an objective black hole interior. The faster-than-light interpretation only happens if we mix up incompatible frames. From Bob's frame consistently, it's the holographic Alice which was teleported, and this never happens faster than light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do Alice and Tweedledee/Tweedledum really exist, or are they all a simulated dream inside the quantum computer? Remember that for Bob, the interior doesn't exist; only the stretched horizon does. Are the holographic Alice and Tweedledee/Tweedledum distinct as objects from the simulated Alice and Tweedledee/Tweedledum? Which of them is the real Alice? If you say neither of them are the real Alice, by the same reasoning, Bob doesn't have the real qubit either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, as Alice and Tweedle are quantum systems, they obey the no Xerox principle. So, the holographic Alice and the simulated Alice can' both exist at the same time, although they can exist at different times. What no Xerox is telling us is from a quantum information theoretic point of view, both the holographic and simulated Alice are one and the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Bob, the disembodied information encrypted within the stretched horizon and the Hawking radiation hallucinated itself as Alice, and this illusion hallucinated a nonexistent black hole interior. This disembodied delocalized illusion is transfered and reconstructed as a simulation inside a quantum computer, all for the sake of getting the qubit, but only after the original disembodied illusion had dematerialized to satisfy the no Xerox principle. The simulated Alice had her memory erased, all for the sake of getting the qubit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things get even more exciting. After measuring the value of the simulated Tweedle along an incompatible basis, Bob has to decide what to do with the simulation running on his quantum computer. He could decide his task is done, and end the simulation, effectively killing the simulated Alice. But what if he decided to keep on running the simulation, this time with the simulated time running forward? The simulated Alice, who still thinks she's inside an objective black hole, would notice a violation of the laws of physics leading to an unexpected change of the qubit, projecting it into one of Bob's basis states.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There is an entirely incommensurable point of view to this story. According to Alice, she is objectively real, and is inside an objectively real black hole. The part of spacetime outside the black hole which exists before the black hole finally evaporates away is also real. More precisely, those events in spacetime for which there exist future directed trajectories which can end inside the black hole. So, Bob and his instruments measuring the Hawking radiation are real before the final evaporation. Their reality can be confirmed by messengers (maybe photons, or something more elaborate) from Bob falling into the black hole to be picked up by Alice. But time would end at the singularity, and with it, the exterior of the black hole too, because messengers can't reach beyond the singularity. Actually, there is no beyond the singularity. But Alice can imagine an extrapolation of Bob and his quantum computer into the future beyond that, but alas, not simulate them because the computational resources inside the hole are too limited. Maybe she does this because she's a cosmologist, or a philosopher ;) She knows from the messengers that the real Bob before the final evaporation had built a quantum computer and detectors collecting all the Hawking radiation with the express purpose of reconstructing her and Tweedle. She also knows that if time didn't end at the singularity, this Bob would actually have the resources to reconstruct her, and will, but she also knows time will end, so that won't really happen in reality. But she's free to imagine what would happen if time didn't end for Bob. The imaginary future Bob will be under the delusion he's the one who is real, and so is his quantum computer, but Alice, who is very real, is but a simulation running on his quantum computer. Well, Alice has already measured Tweedle, and observes it's in a definite state, either DEE or DUM. So, in her imagination, if she were in a simulation, which she's not, this would be the time before the imaginary future Bob made the measurement, when the simulation is being run backward... Ah, what about earlier, before she measured Tweedle? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the imaginary Bob decides never to run the simulation forward in time all the way to the singularity, then she won't experience the singularity if her imaginary scenario were actually true, and that would be a prediction distinguishing whether she was real, or simulated, but that's in the future and she hasn't experienced the future yet...&lt;/p&gt;&#xA;" OwnerUserId="3500" LastEditorUserId="3500" LastEditorDisplayName="" LastEditDate="2011-05-23T15:58:36.463" LastActivityDate="2011-05-23T15:58:36.463" Title="Incommensurability between different observers describing the same universe?" Tags="&lt;black-holes&gt;&lt;quantum-gravity&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="9906" PostTypeId="2" ParentId="9904" CreationDate="2011-05-16T08:14:19.837" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But not for black holes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not for black holes because for black holes, the infalling and exterior observers are spatially separated, so they can't communicate about what they should be measuring. Moreover, whatever A and B measure are observables that don't strictly commute with each other.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the qubit inside, or outside?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Both. This answer &quot;both&quot; is the very content of the &lt;a href=&quot;http://arxiv.org/abs/hep-th/9306069&quot; rel=&quot;nofollow&quot;&gt;black hole complementarity principle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bob has a problem. The qubit is scrambled and deeply encoded within the Hawking radiation. After collecting all the relevant Hawking radiation, would Alice have hit the singularity by then? If so, Alice measured first.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a simple look at the causal diagram of an evaporating neutral black hole shows,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/RWl5I.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the black hole singularity is spacelike-separated from the exterior Minkowski spacetime - even the portion of the Minkowski spacetime &quot;after&quot; the black hole evaporated - so one cannot say that Alice measured the qubit first. To say that Alice did something near the singularity &lt;em&gt;before&lt;/em&gt; Bob detected the last quanta of the Hawking radiation, the detection of the last quanta by Bob would have to belong to the future light cone of the singularity and it obviously doesn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the two measurements are spacelike-separated, there can't be a classical causal relationship between them, in either direction, and whether $t_A &amp;lt; t_B$ or $t_A &amp;gt; t_B$ depends on the coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Things are actually much worse.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Things cannot be &quot;worse&quot; before they are &quot;bad&quot;, and you haven't offered a single glimpse of any inconsistency of the black hole complementarity principle. There is no inconsistency exactly because of the basic point that you are fruitlessly attempting to deny - namely that the observations of Alice and Bob are complementary. They will never be compared with each other in the future because Alice is going to be destroyed at the singularity, without any hope of informing the external post-black-hole world about her perceptions. That's why there's no contradiction in assuming that Alice and Bob did measure observables that didn't commute with one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;P.S.&lt;/strong&gt; No, Bob can't reconstruct what Alice has actually measured inside the black hole because the corresponding observables don't commute with those that Bob inevitably measures when he wants to have any local device that is located outside the black hole event horizon. Alice's degrees of freedom - those from the black hole interior - are holographically encoded in the degrees of freedom outside the black hole. But that doesn't mean that they're actually being measured by Bob. At least if Bob is doing some doable low-energy local experiments outside the black hole, he can't measure the natural low-energy degrees of freedom relevant for Alice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, if Bob could make a sophisticated correlated measurement of &lt;em&gt;everything&lt;/em&gt;, such a measurement could perhaps be made equivalent to what Alice can measure. In that case, he would get the same results as Alice has. However, this could only be done after all the Hawking radiation has been emitted.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-16T11:21:00.900" LastActivityDate="2011-05-16T11:21:00.900" CommentCount="5" />
  <row Id="9907" PostTypeId="2" ParentId="9898" CreationDate="2011-05-16T13:02:48.737" Score="2" ViewCount="" Body="&lt;p&gt;The short answer is, it's a colloidal suspension, effectively. See &lt;a href=&quot;http://www.scientificamerican.com/article.cfm?id=why-do-clouds-float-when&quot; rel=&quot;nofollow&quot;&gt;http://www.scientificamerican.com/article.cfm?id=why-do-clouds-float-when&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1196" LastActivityDate="2011-05-16T13:02:48.737" CommentCount="1" />
  <row Id="9908" PostTypeId="2" ParentId="1111" CreationDate="2011-05-16T14:01:12.447" Score="-3" ViewCount="" Body="&lt;p&gt;So this is how I see it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...particle spins,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if momentum direction = spin direction, helicity is positive,&#xA;if momentum direction &amp;lt;&gt; spin direction, helicity is negative...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;chirality is complementary to helicity in that...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if the elements (momentum and spin) are reversed,&#xA;the same outcome is observed, but is opposite in direction.&lt;/p&gt;&#xA;" OwnerUserId="3556" LastActivityDate="2011-05-16T14:01:12.447" />
  <row Id="9909" PostTypeId="2" ParentId="9899" CreationDate="2011-05-16T14:55:36.703" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As has been stated in the answer you linked to, the propagation of light from the pupil of an imaging system to the image plane can be modeled by a very close approximation called &lt;a href=&quot;http://en.wikipedia.org/wiki/Fraunhofer_diffraction&quot; rel=&quot;nofollow&quot;&gt;Fraunhofer Diffraction&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ U(x,y) \propto \int\int u(\xi, \eta) e^{-i\frac{2\pi}{\lambda z}(x\xi+y\eta)} d\xi d\eta $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\xi$ and $\eta$ are the x- and y- coordinates in the pupil plane, $u(\xi, \eta)$ is the optical field in the pupil plane, and $U&amp;#39;$ is the optical field in the image plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the substitutions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{eqnarray}&#xD;&#xA;\mathcal{f}_x &amp;amp;= &amp;amp;\frac{x}{\lambda z} \\&#xD;&#xA;\mathcal{f}_y &amp;amp;= &amp;amp;\frac{y}{\lambda z} \\&#xD;&#xA;\end{eqnarray} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fraunhoffer diffraction is simply a Fourier transform (FT). Thus it is easier to write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ U(x,y) \propto \mathcal{F}\left[ u(\xi,\eta) \right] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathcal{F}$ will represent the FT, including the necessary scale factors and substitutions mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A property of the FT important for this discussion is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Convolution_theorem&quot; rel=&quot;nofollow&quot;&gt;Convolution Theorem&lt;/a&gt;. In terms relevant to this answer, this theorem states that the multiplication of two patterns in the pupil of the lens is equivalent to the convolution of the FT of those patterns in the image plane. We can write this as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathcal{F}\left[ u(\xi,\eta) \right] \otimes \mathcal{F}\left[ u(\xi,\eta) \right] \propto \mathcal{F}\left[ u(\xi,\eta) \times p(\xi,\eta) \right] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ U(x, y) \otimes P(x,y) \propto \mathcal{F}\left[ u(\xi,\eta) \times p(\xi,\eta) \right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $P(x,y)$ is the FT of the pupil function $p(\xi,\eta)$, and $\otimes$ is the convolution operator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how does this apply in the case of a camera lens? Well, in the pupil of a camera lens, there is some optical field -- i.e. some pattern of light on its way to the image plane. Keep in mind that it is not yet a clearly focused image, but an optical field which will &lt;em&gt;become&lt;/em&gt; an image by the time it arrives at the image sensor. Also in this plane is the aperture stop, which blocks some of the light. This aperture stop can be considered as a function which is multiplied by the light field on its way through the pupil. It multiplies the incoming filed by 1 where there aperture is clear, and by 0 where light is blocked. In other words:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ p = \left\{&#xD;&#xA;\begin{array}{rl}&#xD;&#xA;1 &amp;amp; \text{where the aperture is clear} \\&#xD;&#xA;0 &amp;amp; \text{where the aperture blocks light}&#xD;&#xA;\end{array} \right. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, based on the background given above, we know that the image we see will be the convolution of the ideal image $U$ with the FT of the pupil function, $P$. This will make bright light sources in the image take on the shape of the FT of the pupil function, which is why $P$ is typically called the &lt;em&gt;point spread function&lt;/em&gt; (PSF) of the camera lens*.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what does $P$ look like? The exact computation of the necessary FT is messy and complicated, but there is a basic rule of thumb which provides a good qualitative description of the FT of some basic shapes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sharp edges in the input to the FT are represented by bright streaks in the output, with the orientation of the streak being perpendicular to the edge which produced it.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, a triangular aperture has 3 sharp edges, and therefore the resulting PSF has 3 streaks, which produces a 6-pointed star. In the images below you can see an example in which I have produced a triangular shape to represent the lens aperture, and the resulting PSF:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Chha3.png&quot; alt=&quot;aperture shape&quot;&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/Bvz2r.png&quot; alt=&quot;PSF&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the (more realistic) case of a hexagonal aperture, there will technically be 6 streaks, but because the opposite sides of a hexagon are parallel to each other, the streaks in the PSF will be in parallel pairs, so they will overlap and you will only see 3 distinct streaks in the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*This is only true in the case where the lens is optically perfect. In reality, aberrations in the optics also contribute to the PSF, but a description of that is beyond the scope of this answer.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-05-16T14:55:36.703" />
  <row Id="9910" PostTypeId="2" ParentId="9899" CreationDate="2011-05-16T14:56:27.997" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fraunhofer_diffraction&quot; rel=&quot;nofollow&quot;&gt;Fraunhofer diffraction&lt;/a&gt; is virtually a two-dimensional Fourier transform. So we just have to attempt to make a Fourier transform of an appropriate function and see how many rays the resulting &quot;star&quot; look like.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Hexagonal aperture. I take the following function:&#xA;$$ \begin{split}&#xD;&#xA;h(x,y) = \left[\theta(2y+\sqrt{3})-\theta(2y-\sqrt{3})\right]\cdot&#xD;&#xA;\left[\theta(y-\sqrt{3}(x-1))-\theta(y-\sqrt{3}(x+1))\right]\cdot\\&#xD;&#xA;\left[\theta(y-\sqrt{3}(-x-1))-\theta(y-\sqrt{3}(1-x))\right]&#xD;&#xA;\end{split}$$&#xA;With $\theta(x)$ -- is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Heaviside_step_function&quot; rel=&quot;nofollow&quot;&gt;Heaviside step function&lt;/a&gt;.&lt;br&gt;&#xA;While it is in principle possible to calculate the Fourier transform of this by hand, I've just fed it to CAS and I've got something like this:&#xA;$$\frac{2\sqrt{3}\omega_x\left(\cos\frac{\omega_x}{2}\cos\frac{\sqrt{3}\omega_y}{2}&#xD;&#xA;-\cos\omega_x\right)-6\omega_y\sin\frac{\omega_x}{2}\sin\frac{\sqrt{3}\omega_y}{2}}&#xD;&#xA;{\pi \omega_x(\omega_x^2-3\omega_y^2)}$$&#xA;...plus some singular terms that I've dropped.&lt;br&gt;&#xA;Plotting square of the result (from -100 to 100):&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/n9oxd.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Triangular aperture. I take the following function:&#xA;$$ t(x,y) =\theta(2y+\sqrt{3})\theta(\sqrt{3}(x+1)-y)\theta(-\sqrt{3}(x-1)-y)$$&#xA;After Fourier transform, taking square and plotting:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/Mti5W.gif&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-05-16T14:56:27.997" CommentCount="2" />
  <row Id="9911" PostTypeId="1" AcceptedAnswerId="9913" CreationDate="2011-05-16T16:16:41.647" Score="16" ViewCount="454" Body="&lt;p&gt;I have a paper which lies on a flat surface. The paper is fixed on one side and the opposite side can slide in the direction of the opposing side. As side end slides toward the other, a &quot;bump&quot; forms. I want to know what the solution is to the shape of this bump. There must be some standard solution for this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT&lt;/em&gt;: Luboš Motl did provide a really nice answer for the case when the paper is assumed to lie flat against the surface at the ends (also with the assumption that the bump is small).&#xA;I also found this &lt;a href=&quot;http://sci-toys.com/bent_paper_problem.pdf&quot; rel=&quot;nofollow&quot;&gt;interesting paper&lt;/a&gt; on a similar topic.&lt;/p&gt;&#xA;" OwnerUserId="3558" LastEditorUserId="3558" LastEditorDisplayName="" LastEditDate="2011-05-17T17:30:35.467" LastActivityDate="2011-05-17T19:08:17.067" Title="Equations for bent paper" Tags="&lt;classical-mechanics&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="9912" PostTypeId="2" ParentId="9911" CreationDate="2011-05-16T16:30:31.480" Score="2" ViewCount="" Body="&lt;p&gt;I'm quite sure that the idealized shape is one cycle of a cosine. This is due to the differential equations governing the buckling of thin structures (beams).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there has been many, many years since I studied this kind of stuff, so I don't have reference at hand.&lt;/p&gt;&#xA;" OwnerUserId="3542" LastEditorUserId="3542" LastEditorDisplayName="" LastEditDate="2011-05-16T16:39:03.363" LastActivityDate="2011-05-16T16:39:03.363" CommentCount="3" />
  <row Id="9913" PostTypeId="2" ParentId="9911" CreationDate="2011-05-16T16:38:39.943" Score="9" ViewCount="" Body="&lt;p&gt;A very natural and interesting question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the paper stays nearly flat, we may use the linearized approximation. I will neglect the gravitational potential energy because it's probably much smaller than the bending energy (note that the shape of the bent paper is almost the same when the desk is vertical as when it is horizontal). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper doesn't want to bend much, so the energy contains a term that punishes the second derivatives of $y$ (the curvature)&#xA;$$E = K \int_0^L {\rm d}x\,(y&amp;#39;&amp;#39;)^2$$&#xA;where $y&amp;#39;\equiv dy/dx$, and so on. We want to keep the length of the paper fixed - it's prescribed by the way how you attach it at the end points. In the linear approximation, the length is a linear function of &#xA;$$D = \int_0^L {\rm  d} x\, (y&amp;#39;)^2 $$&#xA;which may be seen, if you need it, by a Taylor expansion of the exact expression for the length of the graph, $\int (1+y^{\prime 2})^{1/2}{\rm d}x$. Add it with a Lagrange multiplier, requiring&#xA;$$\delta (E+\lambda D) = 0.$$&#xA;I think that the resulting equations say&#xA;$$ y&amp;#39;&amp;#39;&amp;#39;&amp;#39; = \frac{\lambda}{K} y&amp;#39;&amp;#39;.$$&#xA;Note that in the Euler-Lagrange equations, all the primes &quot;clump&quot;. If you write $Y=y&amp;#39;&amp;#39;$, it says that the second derivative of $Y$ is proportional to $Y$ itself. The physically relevant solution is&#xA;$$ Y = Y_0 \cos (2\pi x n / L) $$&#xA;where only $n=1$ is possible if there is a table underneath the paper. Consequently,&#xA;$$ y = y_0 [1-\cos (2\pi x / L)],$$&#xA;too. I added the term $1$ as an integration constant (while the other is zero) to guarantee that $y=0$ and $y&amp;#39;=0$ at the boundaries. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;So what you get in practice is one wave of a cosine, from one minimum to the next one.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A more detailed discussion would be needed to eliminate the other cosine-like function with the same frequency, the sine, as well as linear functions, and to discuss whether the exponentially increasing/decreasing solutions may ever be relevant. Well, it wouldn't be too difficult. A more general solution for the same (cosine-like) sign of $\lambda$ would be&#xA;$$ y = y_0 [1-\cos (\phi_0+ 2\pi x N / L)] + Ax^2+B,$$&#xA;and the four conditions $y=0$, $y&amp;#39;=0$ at $x=0,L$ as well as the fixed length of the paper would imply $N=1$, $\phi_0=0$, $A=0$, $B=0$, as well as the right normalization $y_0$. As suggested above, the conditions would have other solutions where $N=2,3,4\dots$ but these solutions wouldn't satisfy $y\geq 0$ for $0\leq x \leq L$, so the paper couldn't arrange itself above the table. If there were a hole in the table, these higher harmonics (several periods of the wave containing) solutions would be possible - but I guess that they would be unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure whether I would be able to solve it analytically if the angle $y&amp;#39;$ were not infinitesimal. It seems clear that the exact solution for significant angles is not just a cosine: the bent paper tends to resemble a balloon - for which $y(x)$ is not even single-valued - when there's too much redundant paper in the middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you sharply bend the paper at $x=0,L$ so that $y&amp;#39;$ may be arbitrary at these two points, the paper in between will be bent as an arc of a circle - see another answer below - and this result is accurate even if $y&amp;#39;$ is of order one. Note that in the linearized approximation, the arc gives $y$ being a quadratic function of $x$ so that $y&amp;#39;&amp;#39;=0$ still solves the fourth-order equation above.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-17T07:03:35.787" LastActivityDate="2011-05-17T07:03:35.787" CommentCount="11" />
  <row Id="9914" PostTypeId="2" ParentId="9904" CreationDate="2011-05-16T17:00:31.213" Score="-1" ViewCount="" Body="&lt;p&gt;The question posed here is not entirely vacuous.  The hole in the BH complementarity is with black holes that are composed of a few Planck units of mass, say less than $10^3M_p$.  For black holes of this scale the event horizon is probably not a well defined null surface, but is quantum mechanically blurred.  The radius of the event horizon classically determines the mass of the BH $R~=~2M$, but quantum mechanically there is an uncertainty relationship which will adulterate this.  The uncertainty in the $\Delta E~\simeq~\hbar/\Delta t$ does have a funny situation, where if the energy is $\Delta E$ is larger than the Planck mass there is the occurrence of an event horizon.  The horizon has a radius $R~\simeq~2G\Delta E/c^4$, which is the uncertainty in the radial position $R~=~\Delta r$ associated with the energy fluctuation.  Putting this together with the Planckian uncertainty in the Einstein box we then have&#xA;$$&#xD;&#xA;\Delta r\Delta t~\simeq~\frac{2G\hbar}{c^4}~=~{\ell}^2_{Planck}/c.&#xD;&#xA;$$&#xA;This connects with noncommutative coordinates in quantum gravity.  This means exterior and interior states are not sharply distinguished from each other.  An exterior observer who measures states of the black hole has a non-zero probability of measuring a state that would be associated with the classical black hole interior.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black hole is a sort of processor, and it processes exterior states into interior states and then into exterior states as Hawking radiation.  Of course this statement betrays one problem with understanding this, for it is tensed.  The English language tends to be very time ordered, and yet this is a quantum process that nonlocally connects the interior of the BH that is causally disconnected from the exterior.  The exterior states have an S-matrix description which extends from $\infty$ to $2M$, or more properly for a stationary observer to $\lim_{r\rightarrow 2M}~r^*~\rightarrow~-\infty$ for&#xA;$$&#xD;&#xA;r^*~=~r~+~2Mlog\Big(\frac{r~-~2M}{2M}\Big)&#xD;&#xA;$$&#xA;This choice of coordinates extends the S-matrix to a domain $-\infty,~\infty$.  The interior observer observes quantum states on a different domain $-\infty,~0$.  The two sets are inequivalent, and so the black hole entropy which connects interior microstates with exterior QM is not complete.  The classical black hole radius = mass relationship is used to get the area-entropy relationship, which is an imposed rule and in some ways ad hoc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To return to the question at hand, even with a quantum mechanical black hole if an observer in the exterior measures quantum states associated with the BH the observer might have a nonzero probability of measuring an interior state, but that state is not equivalent to a measurement of an exterior state.  Further, this observer has no way of determining whether that state pertains to the exterior or the interior.  What is not well understood is how the BH complementarity operates in this case.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-05-16T17:00:31.213" />
  <row Id="9916" PostTypeId="1" CreationDate="2011-05-16T17:35:57.923" Score="0" ViewCount="140" Body="&lt;p&gt;Related to this question&#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/35/where-do-magnets-get-the-energy-to-repel&quot;&gt;Where do magnets get the energy to repel?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have a magnet repelling another, eg one in my hand, the other being pushed along the desk, how do the each of the magnet's fields actually &quot;push&quot; against each other? What translates the magnetic field to kinetic energy?&lt;/p&gt;&#xA;" OwnerUserId="3560" LastActivityDate="2011-05-16T18:09:48.457" Title="How is a magnetic field translated into physical force?" Tags="&lt;electromagnetism&gt;&lt;forces&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="9917" PostTypeId="2" ParentId="9911" CreationDate="2011-05-16T17:43:59.353" Score="2" ViewCount="" Body="&lt;p&gt;You gave some boundary contraints, but not enough to solve this problem without some assumptions on the part of the answerer.  The missing piece is:  For each end; is the end... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) constrained to lie flat against the table, or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) allowed to pivot up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer is 1 for either or both ends, the answer is non-trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer is 2 for BOTH ends, and the weight of the paper is non-negligible, then, again the answer is non-trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer is 2 for BOTH ends -AND- the weight of the paper is negligible, then the paper will bend in the arc of an almost perfect circle.  With the distace between two ends = paper height, the circle radius is infinity.  With the distance between the ends = 0, the radius is paper height divided by (2*Pi); and the arc is the complete circle.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT, based on fearlesscoder's, Lubos' and Ysap's comments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the paper is constrained to lie in the plain of the table at the ends (dy/dx = 0), then the answer could indeed be (or approximate) a complete cycle of a cosine wave (with an offset in the vertical axis); as long as the ends are constrained to not approach closer than L/8 to each other, where L is the length of the sheet.  Not sure what to do as the ends of the paper get really close to one another: the cosine thingy breaks down in that region.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would not, in any case (where the ends are constrained to lie flat), resemble a circle, nor a segment of a circle, nor a quadratic nor a Gaussian curve (which is the first thing that popped to mind when I realized exactly what boundary conditions the OP meant).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would make the assumption that the resistance of the sheet to bending is proportional to the radius of curvature (((d^2)x)/(dy^2) = ((d^2)y)/(dx^2)).  I would make the peak of the bump at x=0.  I would have to put in a lot of thought getting to the point where I can find out if y=cos(x) is a sol'n to that problem. &lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-05-17T19:08:17.067" LastActivityDate="2011-05-17T19:08:17.067" CommentCount="5" />
  <row Id="9918" PostTypeId="2" ParentId="9916" CreationDate="2011-05-16T17:53:13.790" Score="3" ViewCount="" Body="&lt;p&gt;You are confusing force with energy.  Energy is force x distance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are adding energy to a system when you slide one magnet (the one in your hand) along the table, forcing it closer to the other magnet.  By doing this you are forcing the system (the two magnets) into a higher energy state.  Every system (at least everyone I can think of) seeks a lower energy state.  At some point in your slide, the forces build up until the magnet not in your hand breaks free of static friction and slides to a lower energy state (away from the magnet in your hand.  The energy expended to move that magnet is the energy YOU stored in the system by forcing the one magnet near to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to think of it is in terms of potential energy.  The erarth's gravitational field is not energy, BUT you can store potenial energy in a system by raising a weight to ahigher elevation within that force (gravitational) field.  The energy is then reclaimed from the system when the weight drops.  With magnets the story is the same, except that the potential energy is due to position in a magnetic field rather than in a gravitational one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could go into detail about how dipoles in a ferromagnetic material add up to create a magnetic field, or how you can create a magnetic field with no magnetic magnet at all; but I don't think that is what you are asking.  Let me know if i missed the mark.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-05-16T18:09:48.457" LastActivityDate="2011-05-16T18:09:48.457" CommentCount="6" />
  <row Id="9919" PostTypeId="2" ParentId="9897" CreationDate="2011-05-16T18:11:40.983" Score="0" ViewCount="" Body="&lt;p&gt;You have to expand out $L=I\,\omega$ and notice how both terms vary with time. In many books they expand upon this to come to the equation of $\rm{d}L/\rm{d}t = I\,\dot{\omega} + \omega\times I\,\omega$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is a solution with $\omega&amp;gt;0$ and $\dot\omega=0$ for a constant $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. Is this related to a homework problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;Do a &lt;a href=&quot;http://en.wikipedia.org/wiki/Free_body_diagram&quot; rel=&quot;nofollow&quot;&gt;Free Body Diagram&lt;/a&gt; to balance gravity with the reaction force to notice there is a net moment of the center of gravity $N&amp;gt;0$. So the above has a stead state solution of $N=\omega\times I\,\omega$ (with $\dot\omega=0$, $L=I\,\omega$). If the angular momentum $L$ is not aligned with the rotation axis $\omega$ then there exists a non-zero vector $\omega&amp;gt;0$ to satisfy the above equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt;&#xA;if $\vec\omega = [\cos\varphi\,\dot\theta,\sin\varphi\,\dot\theta,\Omega]$ where $\Omega$ is the precession speed, $\varphi$ the orientation/precession angle and $\dot\theta$ the spin rate, then the angular momentum in an axis alinged with the object but not spining is $L_\rm{body}=[I_{xx}\,\dot\theta,0,\Omega\,(I_{zz}+m\,L^2)]$. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ N = [\dot\theta,0,\Omega] \times [I_{xx}\,\dot\theta,0,\Omega\,(I_{zz}+m\,L^2)] = [0,\Omega\dot\theta (I_{xx}-I_{zz}-m\,L^2),0]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;all expressed in this intermediate coordinate frame. So a torque component about the $y$-axis supports the gyroscope precession.&lt;/p&gt;&#xA;" OwnerUserId="392" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2011-05-16T19:40:57.913" LastActivityDate="2011-05-16T19:40:57.913" CommentCount="1" />
  <row Id="9920" PostTypeId="2" ParentId="9898" CreationDate="2011-05-16T18:21:34.033" Score="0" ViewCount="" Body="&lt;p&gt;Clouds don't start as clouds.  Most of them start out as humid air, typically from evaporation from a large body of water (say, the Pacific or Atlantic ocean, and most of that between latitudes 30 degrees north and south).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now humid air is LIGHTER (less dense) than normal air.  This is because gases like to run the same number of moles (number of molecules) per cubic meter, and an H2O molecule weighs less (much less) than an N2, O2 or CO2 molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the humid air rises until is cools at higher elevations, forming clouds.  Even at this point, where the microdroplets are denser than the surrounding air, they are not so heavy that they can force their way down.  They are trapped, sort of like grains of sand that get picked up in a sandstorm.  At some point, though, the micro-drops collect into not-so-micro drops and start falling.  As they fall they merge with other moisture and micro-drops to form rain or other precipitation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand how these micro-droplets or micro-crystals can stay up so long, think about hail storms.  We know now that larger hail sizes can be attributed to the initial tiny hail dropping for a distance and then being blown back up by thunderstorm updrafts.  The tiny hail balls go up and down, up and down; and get coated with a new layer of ice each time, until they are finally heavy enough to overcome the updrafts and fall to the earth.  At his point they may be golf ball sized or even larger.  Think of it: billions of balls of hail, each the weight of a snowball, being suspended in space by atmospheric conditions.  After pondering that, it is not hard to believe that micro-droplets can stay up there.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-05-16T18:32:42.093" LastActivityDate="2011-05-16T18:32:42.093" />
  <row Id="9921" PostTypeId="1" CreationDate="2011-05-16T18:25:01.407" Score="1" ViewCount="141" Body="&lt;p&gt;Voyager I, as an example, taking account gravity&lt;br&gt;&#xA;and setting aside effects of speed as cause of time dilation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is very far away from earth and sun, &#xA;so then there must be a difference in the spacetime curvature there &#xA;in the ship compared with here in earth,&#xA;It means a detectable difference between our local clocks and its onboard clocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a signal transmision was designed to be at 1 byte for second &#xA;at local Voyager clock&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should we receive it at higher and higher rates ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;( because of solar system gravity decrease as it(Voyager) moves away and it will keep decreasing while it doesn't reach a middle point between another massive object)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or the middle Voyager-Earth light path would compensate the effect, making the high rates generation from low curvature zones being delayed enough for us to receive it at same rate that it was generated? &lt;/p&gt;&#xA;" OwnerUserId="1916" LastEditorUserId="1916" LastEditorDisplayName="" LastEditDate="2011-05-16T18:34:18.680" LastActivityDate="2011-05-16T19:10:10.363" Title="Voyager local time dilation (caused by gravity)" Tags="&lt;general-relativity&gt;&lt;gravity&gt;&lt;spacetime&gt;" AnswerCount="1" />
  <row Id="9922" PostTypeId="2" ParentId="9754" CreationDate="2011-05-16T18:25:18.743" Score="0" ViewCount="" Body="&lt;p&gt;The above is unit less. How? well damping is always force/speed thus $c=[\rm{N}\,\rm{s}\,\rm{m^{-1}}]$, and stiffness is force/distance $k=[\rm{N}\,\rm{m^{-1}}]$, and of course a newton is $[\rm{N}]=[\rm{kg}\,\rm{m}\,\rm{s^{-2}}]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combine them all to make&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dfrac{\mathrm{N\,/(m/s)}}{\sqrt{{\rm kg\, N/m}}}=\sqrt{\dfrac{{\rm N}\,{\rm s}^{2}}{{\rm kg\, m}}}=\sqrt{\dfrac{{\rm kg\, (m/s^{2})\, s^{2}}}{{\rm kg\, m}}}=1 $$&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-05-16T18:25:18.743" />
  <row Id="9923" PostTypeId="2" ParentId="9888" CreationDate="2011-05-16T18:46:54.060" Score="0" ViewCount="" Body="&lt;p&gt;dmckee has answered what I believe to be the intent of the question; but Joe has correctly answered the absolute letter of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Level ground seems to be implicitly assumed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If  all other things are EQUAL (e.g. coefficients of friction), except where they need to be PROPORTIONAL to be equal (e.g. wind resistance), then the cars will stop in the same distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the cars both ride the the same coefficients of friction and have the same wind resistance (assumed to be a retarding force, not a tail wind) the lighter car will stop first. &lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-05-16T18:46:54.060" />
  <row Id="9924" PostTypeId="2" ParentId="9836" CreationDate="2011-05-16T19:02:32.020" Score="3" ViewCount="" Body="&lt;p&gt;Lemme make sure I have the assumptions correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The buoy is at rest in a known inertial frame in a known plane; the boats are in the same plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Each boat may make a &quot;single measurement&quot; of the frequency.  I will assume this means that the time of measurement is long enough that the boat can resolve the frequency to arbitrary precision, but the boat's velocity is constant over the measurement time and the change in the boat's position (relative to the buoy) is negligible.  I will assume the angle of origin of the signal cannot be determined, only the frequency.  The velocity and positions of the boats are known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A three-boat solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First boat: at rest in the inertial frame.  Measures the frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second boat: nonzero velocity in the inertial frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the Doppler shift measured by the second boat, you can determine the absolute value of the angle of the buoy with respect to the velocity vector of the second boat (but can't distinguish &quot;right&quot; from &quot;left&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third boat: nonzero velocity, nonparallel to the second boat.  Different location than the second boat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the Doppler shift measured by the third boat, you can determine the absolute value of the angle of the buoy with respect to the velocity vector of the third boat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Draw lines from the 2nd and 3rd boats along all the two possible angles from each.  Where they intersect is the location of the buoy.  That's the &quot;straightforward&quot; solution.  Is it possible to do it with two boats if you get tricky?&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-05-16T19:02:32.020" CommentCount="7" />
  <row Id="9925" PostTypeId="2" ParentId="9921" CreationDate="2011-05-16T19:10:10.363" Score="5" ViewCount="" Body="&lt;p&gt;There is in principle a &quot;gravitational blueshift&quot; for signals traveling from Voyager to us. The data rate we receive will be higher than the data rate transmitted by a factor $(1+\Delta\Phi/c^2)$, where $\Delta\Phi$ is the difference in Newtonian gravitational potential between the locations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, this formula is only valid in weak gravitational fields, where it makes sense to talk in terms of Newtonian gravitational potentials.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm not mistaken, $\Delta\Phi/c^2\approx 10^{-8}$ for this sort of system, so the shift is quite small in practice. In particular, it's much smaller than than the ordinary Doppler effect due to the fact that both Voyager and Earth are moving. A motion at a speed of just 3 m/s would cause a Doppler shift as large as this gravitational shift, and both bodies are moving much faster than that.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-05-16T19:10:10.363" CommentCount="6" />
  <row Id="9926" PostTypeId="2" ParentId="9846" CreationDate="2011-05-16T19:31:15.707" Score="0" ViewCount="" Body="&lt;p&gt;Following Peter Morgan's eqn, I worked out the answer to be ABOUT (I did not do this on a spreadsheet) (13/7) meters on a straight line away from the smaller charge and opposite the larger charge.  That is, the smaller charge is between the zero point and on a straight line from the zero point and the larger charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to visualize what is happening: Think of the earth have a gravity field of -2.5/(x^2), where x is your height above earth, measured is AU's (distance of Earth to Sun).  So, the earth is going to repel you, and you are floating out in space.  Now assume that the Sun is pulling you toward it with a force of 6/(z^2).  If you fall off the Earth in any other direction other than straight away from the Sun, you will be pulled into the Sun.  But if you are lucky enough to fall off the Earth directly away from the Sun, you will reach a point where the Earth's repulsion of you is exactly countered by the Sun's pull.  It would be a Lagrangian point of sorts (the unstable sort, actually).  That point would be a little less than (13/7) AUs away from the Earth, and a little less than (20/7) AUs away from the Sun.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-05-16T19:31:15.707" CommentCount="0" />
  <row Id="9927" PostTypeId="2" ParentId="9897" CreationDate="2011-05-16T20:09:19.743" Score="1" ViewCount="" Body="&lt;p&gt;Start off ignoring gravity.  The spin axis is horizontal?  Well then, you have an L vector.  Very simple, nothing happens.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now turn on gravity.  This pulls down on the centre of mass, which is elsewhere from the pivot.  That gives you an N vector - cross product of force with location relative to the pivot.  This is perpendicular to the axis - so perpendicular to L.   This is the change in L, according to N=dL/dt.   In an arbitrarily small change in time, dt, we find dL will not change the length of L but will change its directions.  Repeat indefinitely for every dt in a finite time interval t_1 to t_2.  &lt;/p&gt;&#xA;" OwnerUserId="353" LastActivityDate="2011-05-16T20:09:19.743" CommentCount="1" />
  <row Id="9928" PostTypeId="2" ParentId="9836" CreationDate="2011-05-16T20:15:44.013" Score="1" ViewCount="" Body="&lt;p&gt;Three boats is the minimum.  And I don't think even that is enough in the genral case, if one of them is not moving (i.e., dedicated to directly solving bouy frequency).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using just two boats with doppler lines will result in two lines of position for each boat.  The intersection of these lines from two boats will result in 4 possible positions.  You need a third boat to resolve the ambiguity.  Even using a third boat, there are special cases where its LOPs would intersect 2 of the intersections of LOPs from the first two boats; but it would not be likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would use 3 boats, which have traveled some distance along rays separated by 120 degrees and radiating from the same point.  Doppler shifts could be used to qualitatively isolte the bouy to the 1/3 of the ocean directly away from the boat with lowest received frequency.  The actual bouy frequency could also be bracketed to a very narrow frequency.  From there it is fairly simple to solve for the actual frequency.  Proceed from there to draw the LOPs from each boat.  The two boats whose paths bound the wedge of water containing the bouy would have LOPs defining 1 possible location for the bouy.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-05-16T20:15:44.013" CommentCount="2" />
  <row Id="9929" PostTypeId="5" CreationDate="2011-05-16T21:43:57.893" Score="0" ViewCount="" Body="&lt;p&gt;Particle physics is distinct from &lt;a href=&quot;http://physics.stackexchange.com/questions/tagged/nuclear-physics&quot;&gt;nuclear physics&lt;/a&gt; in that it generally involves energies higher than those that allow nucleon and especially nuclei to remain bound states, though there is a rich transition region (sometimes denoted &quot;non-perturbative&quot;) investigated by physicists from both disciplines.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Experiment&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The most visible experimental effort in particle physics in the early 2010s is the recently started Large Hadron Collider (LHC) at CERN. However the discipline is much larger than that encompassing &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/tagged/high-energy-physics&quot;&gt;high energy&lt;/a&gt; collider work with the LHC, the Tevatron at Fermilab, and RHIC at BNL&lt;/li&gt;&#xA;&lt;li&gt;fixed target efforts at many sites&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/tagged/neutrino&quot;&gt;neutrino&lt;/a&gt; studies based on beams, cosmic rays, solar neutrinos, reactor anti-neutrinos, and radioactive decay &lt;/li&gt;&#xA;&lt;li&gt;ultra-high energy cosmic ray shower telescopes&lt;/li&gt;&#xA;&lt;li&gt;direct dark matter detection&lt;/li&gt;&#xA;&lt;li&gt;proton decay monitoring at essentially every large, low back-ground detector used for other purposes&lt;/li&gt;&#xA;&lt;li&gt;a wide variety of astrophysical instruments probing the Cosmic Microwave Background, baryon acoustic resonances, active-galactic nuclei, and the large scale structure of the universe&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Experimental particle physics tends to require large, expensive equipment and many experimenters, and as such is mostly carried out by government funded scientists at a few dozen major facilities world wide.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Theory&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The core theory of particle physics currently is the so-called &lt;a href=&quot;http://physics.stackexchange.com/questions/tagged/standard-model&quot;&gt;Standard Model&lt;/a&gt; which embodies the theories of the strong and electro-weak forces with six quarks coming in three colors ($u$, $d$, $c$, $s$, $t$, $b$), three charged leptons ($e$, $\mu$, $\tau$), three un-charged leptons called neutrinos, and four gauge bosons (the eight color states of the gluon mediating the strong force, the photon ($\gamma$) and weak gauge bosons ($W^{\pm}$ and $Z$) mediating the electo-weak interaction, and the &lt;em&gt;assumed&lt;/em&gt; Higgs boson). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard model is known to be wrong about the neutrino masses (assumed to be zero, but experimentally shown to mix implying non-zero mass).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many candidates for beyond the standard model theories exists in several large classes with names like super-symmetry, string theories, and technicolor. Some of these can be shown to be equivalent to other under certain conditions.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-31T15:54:03.090" LastActivityDate="2011-05-31T15:54:03.090" />
  <row Id="9930" PostTypeId="4" CreationDate="2011-05-16T21:43:57.893" Score="0" ViewCount="" Body="Particle physics is the study of the fundamental forces of nature as they are embodied in the interactions of elementary and composite particles at high energies and short time and distance scales." OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-16T21:43:57.893" LastActivityDate="2011-05-16T21:43:57.893" />
  <row Id="9931" PostTypeId="5" CreationDate="2011-05-16T21:55:59.397" Score="0" ViewCount="" Body="&lt;p&gt;Nuclear physics is the study of the composition, behavior and interaction of atomic nuclei and their constituent parts. It differs from &lt;a href=&quot;http://physics.stackexchange.com/questions/tagged/particle-physics&quot;&gt;particle physics&lt;/a&gt; in that it spans a lower energy range where nucleons and even nuclei are stable and interactions can generally be described in terms of nucleon and meson degrees of freedom instead of quark and gluon degrees of freedom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effective theory known as Quantum Hadron Dynamics (QHD) is useful for some problems in nuclear despite being known to be incomplete. Nuclear physicists also study the transition region where QCD corrections become important but the theory is not fully perturbative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nuclear physics provides the basic tools for &lt;a href=&quot;http://physics.stackexchange.com/questions/tagged/nuclear-engineering&quot;&gt;nuclear power engineering&lt;/a&gt; and nuclear weaponry, though most nuclear physicists are not-involved in either of there fields.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-31T15:55:17.317" LastActivityDate="2011-05-31T15:55:17.317" />
  <row Id="9932" PostTypeId="4" CreationDate="2011-05-16T21:55:59.397" Score="0" ViewCount="" Body="Nuclear physics is the study of the composition, behavior and interaction of atomic nuclei and their constituent parts." OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-16T21:55:59.397" LastActivityDate="2011-05-16T21:55:59.397" />
  <row Id="9933" PostTypeId="5" CreationDate="2011-05-16T22:17:38.787" Score="0" ViewCount="" Body="&lt;p&gt;Newtonian mechanics is a branch of physics developed from the relationship between  force, mass and acceleration. The concepts of energy, momentum, and angular momentum are developed as consequences of Newton's laws and provide an extensive toolkit for describing the universe.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Newton's laws&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Issac Newton postulated three laws in verbal language, but they are representable in two equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec{F} = m \vec{a} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec{F_{1,2}} = -\vec{F_{2,1}} .$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Important concepts&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Starting from the force concept one can develop additional tools&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Energy found in terms of force applied of a distance $E = \int_S \vec{F} \cdot d\vec{s}$&lt;/li&gt;&#xA;&lt;li&gt;Momentum found in terms of force applied over a time $\vec{p} = \int_{a}^{b} \vec{F} dt$&lt;/li&gt;&#xA;&lt;li&gt;Angular momentum $\vec{L} = \vec{r} \times \vec{p}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;all of which obey conservation laws and provide powerful tools for understanding the universe.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Alternate formulations&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Newtonian mechanics turns out to be only one of three equally valid ways describing mechanics. The other are called &lt;a href=&quot;http://physics.stackexchange.com/questions/tagged/lagrangian-mechanics&quot;&gt;Lagrangian&lt;/a&gt; and &lt;a href=&quot;http://physics.stackexchange.com/questions/tagged/hamiltonian-mechanics&quot;&gt;Hamiltonian mechanics&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-17T18:22:56.470" LastActivityDate="2011-05-17T18:22:56.470" />
  <row Id="9934" PostTypeId="4" CreationDate="2011-05-16T22:17:38.787" Score="0" ViewCount="" Body="Newtonian mechanics is a branch of physics based on the ideas of force, mass and acceleration. The concepts of energy, momentum, and angular momentum are developed as consequences of Newton's laws and provide an extensive toolkit for describing the universe." OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-16T22:17:38.787" LastActivityDate="2011-05-16T22:17:38.787" />
  <row Id="9935" PostTypeId="1" CreationDate="2011-05-17T01:27:35.750" Score="6" ViewCount="387" Body="&lt;p&gt;I apologize if this is off topic since it does not relate directly to the study of physics, but I could not think of a better place to ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a high school student graduating in a few days and have recently developed an interest in quantum mechanics. I began to study physics about one and a half years ago and calculus two years ago, and I feel confident about my abilities in integral and vector calculus (at a typical college Calc I–III level), differential equations, mathematical problem solving, and some set theory. I have been introduced to quantum behavior, particle physics, and nuclear physics, but none of these were delivered with a heavy focus on mathematics. Quantum physics, however, caught my attention as a future field of study due to its bizarre nature and inquisitive mathematical models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without prior knowledge of more difficult topics, I am concerned that attempting to study quantum mechanics will only provide entertainment rather than benefit my college career. I have selected a few books (Griffiths, Sakurai, Feynman, etc.) of which I will choose one to read over the summer of this year and onto college if I determine to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I follow these plans without first strengthening my knowledge in other fields of physics and mathematics, as I dive directly into the topic? Or should I wait until I am more experienced in other fields before discouraging myself with the difficulty of quantum mechanics? If so, what areas of physics or mathematics would you recommend as prerequisites to the topic?&lt;/p&gt;&#xA;" OwnerUserId="69" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-17T02:18:38.323" LastActivityDate="2011-05-30T06:00:44.957" CommunityOwnedDate="2011-05-17T01:34:58.567" Title="Should I begin to study quantum mechanics or wait until I have a stronger base on easier topics?" Tags="&lt;quantum-mechanics&gt;&lt;soft-question&gt;&lt;education&gt;" AnswerCount="8" CommentCount="5" FavoriteCount="1" />
  <row Id="9936" PostTypeId="2" ParentId="9935" CreationDate="2011-05-17T01:37:04.913" Score="3" ViewCount="" Body="&lt;p&gt;All of those references will assume that you know certain things about mechanics and E&amp;amp;M as well as a certain amount of math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is that you could bump into a line like&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It clearly follow that...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and something that seems totally opaque to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are willing to risk this and able to take hours or days off while you go figure our why that should be &lt;em&gt;&quot;clear&quot;&lt;/em&gt;, then there is nothing stopping you from diving right it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you prefer things presented in a more considered and orderly manner you might want to start at the beginning.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-05-17T01:37:04.913" CommunityOwnedDate="2011-05-17T01:37:04.913" CommentCount="2" />
  <row Id="9937" PostTypeId="2" ParentId="9935" CreationDate="2011-05-17T02:03:22.307" Score="7" ViewCount="" Body="&lt;p&gt;I have had to take two or three swipes at every technical topic I learned. Yes, more electromagnetism, vibrations and waves, and theoretical mechanics, along with linear algebra an Fourier analysis will make it easier to learn quantum mechanics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you study those things first, though, you will still probably need to study quantum mechanics more than once. I'd say take a shot at quantum mechanics now, then study some more and come back to learn it more deeply. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out Thomas Jordan's &quot;Quantum mechanics in simple matrix form&quot; for a simplified but technical introduction to quantum mechanics before trying Griffiths. &lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-05-17T02:03:22.307" CommunityOwnedDate="2011-05-17T02:03:22.307" CommentCount="3" />
  <row Id="9938" PostTypeId="5" CreationDate="2011-05-17T03:39:34.693" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-05-17T03:39:34.693" LastActivityDate="2011-05-17T03:39:34.693" />
  <row Id="9939" PostTypeId="4" CreationDate="2011-05-17T03:39:34.693" Score="0" ViewCount="" Body="Optics is the study of light, and its interaction with matter. It includes topics such as imaging systems, fiber optics, lasers, quantum optics, and more." OwnerUserId="-1" LastEditorUserId="869" LastEditorDisplayName="" LastEditDate="2011-05-17T03:40:54.100" LastActivityDate="2011-05-17T03:40:54.100" />
  <row Id="9940" PostTypeId="1" AcceptedAnswerId="9944" CreationDate="2011-05-17T04:20:21.333" Score="1" ViewCount="146" Body="&lt;p&gt;I'm working through some exam problems, and I came across this one - the solution of which baffles me considerably.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A two-dimensional jet emerges from a narrow slit in a wall into fluid which is at rest. If the jet is thin, so that velocity $\vec u = (u, v)$ varies much more rapidly across the jet than along it, the fluid equation becomes:&lt;br /&gt;&#xA;  $u\frac{\partial u}{\partial x} +v\frac{\partial u}{\partial y} = \nu \frac{\partial^2 u}{\partial y^2}$&lt;br /&gt;&#xA;  where constant $\nu$ is the viscosity coefficient.  The bounday conditions are that the velocity and its derivatives tend to zero as we leave the jet (that is as $|y|\rightarrow\infty$) and $\frac{\partial u}{\partial y}$ at $y = 0$, as the motion is symmetrical about the x-axis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first question involves integrating across the jet to show that $\int u^2 dy$ is $x$ independent.  So you end up with three integrals (subscripts denoting derivatives),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $\int u u_x dy$ &lt;br/&gt;&#xA;2) $\int v u_y dy$&lt;br/&gt;&#xA;3) $\nu\int u_{yy} dy$&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somehow, for 1), you can write $\int u u_x dy = \frac{1}{2}\partial_x\int u^2 dy$ - is this an identity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly for 2) the solution states that: $\int v u_y dy = -\int v_y u dy$ from which you use the incompressibility condition to give $\int u_x u dy = \frac{1}{2}\partial_x\int u^2 dy$.  Where does $vu_y=-v_yu$ come from, and again, is there an identity used in the final step?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And thirdly, this is a smaller issue but again, one that confused me a bit, the second question supposes that the streamfunction is self-similar and takes the form: $\psi = x^a f(\eta)$, $\eta = yx^b$.  Which is fine, sub in for $u^2 = \psi_y^2$ and equate the power of the factor of $x$ that comes out to be zero.  However, in the solution, it comes out as $2a = -b$ and I don't see how that works.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you take: $\eta = yx^b$, differentiate to get $\frac{d}{d\eta}\eta = \frac{dy}{d\eta}x^b$ so that $dy = d\eta x^{-b}$.  When that's substituted into the integral, the factor that comes out is $x^{2a-b}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have noticed some errors in the solutions, so just checking it's me, not them!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure this is a simple problem once you see the trick, thanks very much!&lt;/p&gt;&#xA;" OwnerUserId="3346" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-17T17:56:09.433" LastActivityDate="2011-05-17T19:00:14.443" Title="Fluid dynamics, flow of a 2D jet from a narrow slit" Tags="&lt;fluid-dynamics&gt;" AnswerCount="1" />
  <row Id="9941" PostTypeId="2" ParentId="9935" CreationDate="2011-05-17T04:48:23.210" Score="4" ViewCount="" Body="&lt;p&gt;Certainly with a basic knowledge of calculus you should be able to grasp some basic ideas of quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best book I've ever read on quantum mechanic is Shankar, the first chapter is absolutely solid maths, but once you get over it, it's all very straightforward.  The maths itself is very well explained, and I would imagine a first year undergrad wouldn't have much of a problem.  Buy yourself a copy of the Feynman Lectures anyway, they're invaluable for anyone interested in university level physics.  Be wary of the high-ranked books on Amazon, many of them are excellent texts, but only much use at graduate level (Sakurai, for instance).  Most people recommend reading a &quot;soft&quot; textbook like Griffiths before tackling the classics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not entirely necessary to know other fields of physics before you dive into quantum mechanics, but as with everything, knowing the bigger picture helps a lot.  Certainly when learning quantum mechanics it is invaluable to be able to compare its predictions with those of classical physics.  You needn't know much electromagnetism besides basic electrostatics, pretty much all I've ever used as an undergrad in QM was the Coulomb force and dipoles (when dealing with spin-orbit and the Zeeman effect).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other answer about multiple coats is pretty much spot on.  Most of the QM I've been taught started out with a statement or observation and the actual explanation came a year later.  QM can very quickly get very complicated and most of the major results are very subtle, the Exclusion principle, for instance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the only way to find out would be to dive in and go for it.  If you find it's not for you then you haven't lost anything.&lt;/p&gt;&#xA;" OwnerUserId="3346" LastActivityDate="2011-05-17T04:48:23.210" CommunityOwnedDate="2011-05-17T04:48:23.210" CommentCount="1" />
  <row Id="9944" PostTypeId="2" ParentId="9940" CreationDate="2011-05-17T10:09:07.740" Score="3" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is $\int u u_x dy = \frac{1}{2}\int \partial_x (u^2) dy = \frac{1}{2}\partial_x\int u^2 dy$ an identity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, the order of differentiation in the $x$-direction and integration in the $y$-direction can be &lt;a href=&quot;http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign&quot; rel=&quot;nofollow&quot;&gt;exchanged&lt;/a&gt; under some mild technical assumptions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where does $vu_y= -v_y u$ come from?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nowhere, it is not true. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there an identity to use in the final step?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, you need the given information that the boundary terms at $y=\pm\infty$ vanish, so that you can integrate by part, e.g.,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int (v u_y + v_y u ) dy = \int \partial_y(v u) dy = \left[ v u\right]^{y=\infty}_{y=-\infty}=0,$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int u_{yy} dy  = \int \partial_y(u_y) dy = \left[u_y\right]^{y=\infty}_{y=-\infty}=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Piecing together the various parts yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \partial_x\int u^2 dy = 2 \int uu_x   dy= \int u(u_x -v_y)  dy = \int (u u_x + v u_y)dy=\nu\int u_{yy} dy  = 0. $$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The second question supposes that the streamfunction is self-similar and takes the form: $\psi = x^a f(\eta)$, $\eta = yx^b$.  Which is fine, sub in for $u^2 = \psi_y^2$ and equate the power of the factor of $x$ that comes out to be zero. However, in the solution, it comes out as $2a=−b$ and I don't see how that works.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are almost there. Just be careful with the algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi = x^af(yx^b),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ u = \psi_y = x^{a+b}f^{\prime}(yx^b),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \underbrace{\int u^2 dy}_{\mathrm{independent~of~} x} = x^{2a+b} \int f&amp;#39;(yx^b)^2 d(yx^b) =  x^{2a+b} \underbrace{\int f^{\prime}(\eta)^2 d\eta}_{\mathrm{independent~of~} x}. $$&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-17T19:00:14.443" LastActivityDate="2011-05-17T19:00:14.443" CommentCount="1" />
  <row Id="9945" PostTypeId="1" AcceptedAnswerId="9951" CreationDate="2011-05-17T10:17:05.993" Score="3" ViewCount="212" Body="&lt;p&gt;I believe the answer to be yes, but I realize that sometimes physicists place additional constraints that might not be obvious.  If superalgebras are clifford algebras, why make a literary distinction?&lt;/p&gt;&#xA;" OwnerUserId="3460" LastActivityDate="2011-05-17T16:19:27.237" Title="Are all superalgebra's clifford algebra's" Tags="&lt;mathematical-physics&gt;" AnswerCount="3" />
  <row Id="9946" PostTypeId="2" ParentId="9945" CreationDate="2011-05-17T10:28:51.663" Score="1" ViewCount="" Body="&lt;p&gt;On a mathematics stack exchange, you could succeed with such an &quot;isomorphism&quot;. But in physics, superalgebras are interpreted totally differently than Clifford algebras. So not only you're forbidden to identify the two terms universally; but in reality, there is not a single example in physics that could be called both a Clifford algebra and a superalgebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clifford algebras are reserved for algebras of matrices that have explicit Grassmann-even i.e. bosonic entries - especially the algebra of Dirac matrices&#xA;$$ \gamma_i \gamma_j + \gamma_j \gamma_i = 2 \delta_{ij} $$&#xA;while superalgebras are reserved for algebras with Grassmann-odd generators that can't be represented by Grassmann-even matrices in any natural way, e.g.&#xA;$$ \{Q_a,\bar Q_{\bar b}\} := Q_a Q_{\bar b} + Q_{\bar b} Q_a = 2 p_{a\bar b}. $$&#xA;So despite the mathematical similarities in the anticommutators, they're just totally different things. The difference between Grassmann-even (bosonic) and Grassmann-odd (fermionic) objects in physics is viewed as a very important one; one can't say that the bosonic and fermionic objects are the same thing. Whether they change their sign or not after a 360-degree rotation decides about their statistics, too - by the spin-statistics theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the $\{A,B\}$ brackets for the anticommutator are only &quot;natural&quot; in the case of the superalgebras. For example, for fermionic fields, it's the anticommutator that should vanish at spacelike separations. In the case of the Clifford algebras, the objects such as $\gamma_i$ are not generators of any physically natural groups, especially not fermionic ones, so one shouldn't talk about their anticommutators. In the Dirac matrix case, the anticommutators are sometimes used but they're just a bookkeeping device. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, in the superalgebra case, they're real replacements for commutators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, there may be mathematical analogies in the formulae defining the algebras but these similarities ignore the physical meaning that is very different.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-17T10:35:40.113" LastActivityDate="2011-05-17T10:35:40.113" CommentCount="0" />
  <row Id="9947" PostTypeId="1" AcceptedAnswerId="9948" CreationDate="2011-05-17T10:33:56.293" Score="0" ViewCount="171" Body="&lt;p&gt;If I had a mass of 100kg accelerating due to gravity, using F=ma:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F = 100kg * 9.8m/s/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F = 980 N...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I increased the mass to 200kg, the force would be 1960 N:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F = 200kg * 9.8m/s/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F = 1960 N&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, finally getting to my question: Does this increase in force (which is supposed to be a push/pull) mean that the object would fall faster when it weighs more?&lt;/p&gt;&#xA;" OwnerUserId="3569" LastActivityDate="2011-05-17T16:49:12.460" Title="Does an Increase of Force affect the Speed of an object if the acceleration stays constant?" Tags="&lt;forces&gt;&lt;mass&gt;&lt;acceleration&gt;&lt;speed&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="9948" PostTypeId="2" ParentId="9947" CreationDate="2011-05-17T11:06:12.250" Score="1" ViewCount="" Body="&lt;p&gt;No, the heavier object does not fall faster.  Instead, they heavy and light object fall at the same acceleration (and hence the same speed if they are both simply dropped).  This is an example of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalence_principle&quot; rel=&quot;nofollow&quot;&gt;equivalence principle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more massive object has more gravitational force on it, but it also has more inertia.  Specifically, because the object is twice as massive, it has twice the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mass#Inertial_mass&quot; rel=&quot;nofollow&quot;&gt;inertial mass&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force on it is doubled, so the acceleration stays the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = ma$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we see that when $F$ and $m$ are both multiplied by 2, $a$ stays the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check these questions for more:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/5973/free-falling-of-object-with-no-air-resistance&quot;&gt;Free falling of object with no air resistance&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/9462/why-is-heavier-object-more-reluctant-to-get-falling-down&quot;&gt;Why is heavier object more reluctant to get falling down?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/6269/projectile-motion-without-air-resistance&quot;&gt;Projectile motion without air resistance&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-05-17T11:06:12.250" CommentCount="3" />
  <row Id="9949" PostTypeId="2" ParentId="9850" CreationDate="2011-05-17T12:05:41.507" Score="0" ViewCount="" Body="&lt;p&gt;Since the first paper that I associate the circles with the &lt;strong&gt;voids&lt;/strong&gt;.&lt;br&gt;&#xA;in the paper, about Fig 4  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The distortions could be the result of&#xA;  ... or more likely, ...Indeed, the&#xA;  presence of giant &lt;strong&gt;voids&lt;/strong&gt; could&#xA;  particularly influence such images&#xA;  Gurzadyan V.G., Kocharyan A.A. (2009) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;concentric features&lt;/strong&gt; reminds me a X-ray crystal diffraction pattern. &lt;strong&gt;It can trace a spatial volumetric regularity&lt;/strong&gt; present in the creation of the photons of the CMB.&lt;br&gt;&#xA;goggle for images of &quot;X-ray crystal&quot; and compare with Figs 4 and 5.&lt;br&gt;&#xA;I think that this viewpoint is not allowed by the current mainstream cosmological model.&lt;br&gt;&#xA;If you want to understand what it can trace, or not, one must ignore any preconceived model and start with the data analisys and discover if it was properly done, and then answer to the question: can it trace a volumetric regularity ?&lt;br&gt;&#xA;If the answer is 'yes' or a 'may be' I can assure that a cosmological model already exists.&lt;br&gt;&#xA;(not the CCC one)  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-05-17T12:05:41.507" />
  <row Id="9950" PostTypeId="2" ParentId="9945" CreationDate="2011-05-17T12:35:31.277" Score="2" ViewCount="" Body="&lt;p&gt;A Clifford algebra is a geometrically rich algebraic structure. In the case of the Dirac algebra, which is a Clifford algebra over the complex field, we introduce a map from the tangent space of Minkowski space at a point into a superalgebra. Given a vector $V\in\mathcal{V}$, we can introduce a map into an algebra, $\gamma:\mathcal{V}\rightarrow\mathcal{A};V\mapsto\gamma(V)$. The algebraic structure can be presented in a number of ways, but informally they amount to working systematically with the linear dependence $\gamma(U)\gamma(V)+\gamma(V)\gamma(U)=2(U,V)$. In the Dirac algebra case, the inner product $(U,V)$ in component terms is obviously $U_\mu g^{\mu\nu}V_\nu$. The complex structure makes no appearance in the linear dependence that I have used to informally present the algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a number of reasons, I haven't worked with superalgebras in the terms described by Luboš, so I'm scandalously uncertain of his notation, but it looks as if the two distinct conjugations in his Answer, of the $Q$'s and of the $Q$-indices, signify that the inner product in the &amp;ldquo;superalgebra in Physics&amp;rdquo; case is sesquilinear (which seems to be complex anti-linear in the second component, though that's immaterial except as a convention). I presume that the inner product presented by Luboš as $p_{a\bar{b}}$ in general has no connection to the space-time metric. There is a level of abstraction at which the sesquilinear structure makes no difference, however it's enough that as categories the two constructions &lt;em&gt;are&lt;/em&gt; different in detail because the complex structure is now nontrivially involved in the construction of the algebra, we're not just working with the complexification of a Clifford algebra over the reals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made quite a few starts at writing interpretational remarks, but since they were essentially responses to aspects of Luboš's Answer, not to your Question, I've decided to leave you to draw your own conclusions.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-05-17T12:35:31.277" CommentCount="1" />
  <row Id="9951" PostTypeId="2" ParentId="9945" CreationDate="2011-05-17T13:09:24.017" Score="2" ViewCount="" Body="&lt;p&gt;The answer is No, as Lubos Motl has already pointed out. Here I would like to make a couple of general remarks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) On one hand, the notion of &lt;a href=&quot;http://en.wikipedia.org/wiki/Superalgebra&quot; rel=&quot;nofollow&quot;&gt;superalgebras&lt;/a&gt; is a huge topic, which includes, e.g., associative superalgebras and &lt;a href=&quot;http://en.wikipedia.org/wiki/Lie_superalgebra&quot; rel=&quot;nofollow&quot;&gt;Lie superalgebras&lt;/a&gt;. Important examples of Lie superalgebras are &lt;a href=&quot;http://en.wikipedia.org/wiki/Super-Poincar%C3%A9_algebra&quot; rel=&quot;nofollow&quot;&gt;super-Poincaré algebras&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) On the other hand, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Clifford_algebra&quot; rel=&quot;nofollow&quot;&gt;Clifford algebra&lt;/a&gt; $\mathrm{Cl}(V,g)$ over a vector space $V$ satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$vw+wv=2g(v,w){\bf 1}, \qquad v,w\in V.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In physics applications, the vector space $V$ is often a vector space spanned by a basis of gamma matrices,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\gamma^{\mu}\gamma^{\nu}+\gamma^{\nu}\gamma^{\mu}=2g^{\mu\nu}{\bf 1}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more mathematical applications, the vector space $V$ sometimes comes with an odd grading, so that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Commutator#Anticommutator&quot; rel=&quot;nofollow&quot;&gt;anticommutator&lt;/a&gt; is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Lie_superalgebra#Examples&quot; rel=&quot;nofollow&quot;&gt;supercommutator&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[v,w]=2g(v,w){\bf 1},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be viewed as a super &lt;a href=&quot;http://planetmath.org/encyclopedia/HeisenbergAlgebra.html&quot; rel=&quot;nofollow&quot;&gt;Heisenberg algebra&lt;/a&gt; with odd grading, and which is hence an example of a superalgebra.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, the vector space $V$ could be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Super_vector_space&quot; rel=&quot;nofollow&quot;&gt;super vector space&lt;/a&gt; $V=V_0\oplus V_1$ with both an even and an odd sector, leading to a notion of super Clifford algebras. &lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-17T16:19:27.237" LastActivityDate="2011-05-17T16:19:27.237" CommentCount="3" />
  <row Id="9952" PostTypeId="2" ParentId="9935" CreationDate="2011-05-17T13:17:40.253" Score="2" ViewCount="" Body="&lt;p&gt;When I was a high school student (and probably not as knowledgeable as you) I remember studying &quot;A Quantum Mechanics Primer&quot; by Gillespie. It is all about grasping what do the postulates imply, working with wave functions. Probably there are better ways of introducing the material right now, and more physically motivated, but I remember fully enjoying (and studying in every detail) that little book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, from a very different point of view (working with discrete spaces, e.g. state vectors, so that all the math is algebra), &quot;The Strange World of Quantum Mechanics&quot; by Styer is also a quite nice little book. This is the preferred type of presentation today, as many introductory books (and lectures) on Quantum Mechanics work with discrete 2-state spaces (e.g. spin or polarization).&lt;/p&gt;&#xA;" OwnerUserId="3572" LastEditorUserId="3572" LastEditorDisplayName="" LastEditDate="2011-05-17T19:18:38.073" LastActivityDate="2011-05-17T19:18:38.073" CommunityOwnedDate="2011-05-17T13:17:40.253" />
  <row Id="9953" PostTypeId="1" CreationDate="2011-05-17T14:45:32.053" Score="-3" ViewCount="96" Body="&lt;p&gt;Thinking in the posibility of extraterrestrial life, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that gravity could be a fundamental factor for evolution,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean, in darkest zones of universe,&#xA;where is dust, little mass accumulation, and perhaps low bright, low energy stars,&#xA;there must be low gravity, and finally, would be high clock rates..(relative to earth)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the bigger mass (perhaps brighter stars)&#xA;it is subject to more intense gravity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally we could be (relative to other zones of universe) in a high clock rate zone&#xA;or a in a very low, and that relative difference of clocks could be a clue for our search&#xA;of intelligent life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the search of low gravity zones, could be the search of (relative to us) faster rates of evolution, to exaggerate: a millon years in what it takes a year here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Life could need Carbon, Life could need star light, but I am deeply sure that first of all, Life need time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it be a possible guide to search intelligence, stop searching earth like with sun like stars, and start looking at the dark (or whatever that indicate less gravity) ?&lt;/p&gt;&#xA;" OwnerUserId="1916" LastActivityDate="2011-05-17T16:45:08.737" ClosedDate="2011-05-17T17:55:19.170" Title="Gravitational time dilation and intelligent life" Tags="&lt;gravity&gt;&lt;soft-question&gt;&lt;astrophysics&gt;&lt;spacetime&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="9954" PostTypeId="1" AcceptedAnswerId="9959" CreationDate="2011-05-17T16:24:19.043" Score="0" ViewCount="123" Body="&lt;p&gt;This is a theory vs experiment problem for me I think in that - as usual - I think I know the theory, but when it comes time to run an experiment, I can't actually extract squat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, here is my light source: &lt;a href=&quot;http://www.thorlabs.com/NewGroupPage9.cfm?ObjectGroup_ID=2692&quot; rel=&quot;nofollow&quot;&gt;http://www.thorlabs.com/NewGroupPage9.cfm?ObjectGroup_ID=2692&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using the Cold White LED (MCWHL2).  If you scroll down a bit, you'll see the LED normalized intensity vs. wavelength profiles for all the LEDs.  The black line on this graph is my LED.  As you can see, it meanders instead of being a simple Guassian-like thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I measure the intensity of this light by using this power meter:  Scientech Astral AD30 meter - which displays power and energy - with an AC2501 detector.  Specs are here: &lt;a href=&quot;http://www.scientech-inc.com/LPM/Tables/AstralCalorimeter.jpg&quot; rel=&quot;nofollow&quot;&gt;http://www.scientech-inc.com/LPM/Tables/AstralCalorimeter.jpg&lt;/a&gt;.  So I'm calculating total intensity by dividing by the area of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here's my question: say I get a reading of $ 10\frac{mW}{cm^2} $. If I want to know what is the intensity of the output only over the range 400nm to 455 nm, based on the Intensity vs Wavelength graphs and the characteristics of the meter and detector, what do I do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, Planck's equation for a black body was where I first looked.  My procedure was to find the total intensity for the light temperature (6500 K) and integrate from 0 to infinity.  I then numerically integrated from 440 nm to 455 nm and divided the two.  Multiplying this by the output on the detector gave a ridiculous solution.  Given the intensity profile - not really black bodyish - the ridiculous results make sense.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was at this point that I started searching and found nothing but Planck stuff.  Thus, I humbly come before all of you.  Please help!  There must be some sort of standard procedure  for those without specialized light analyzing equipment!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&#xA;Sam&lt;/p&gt;&#xA;" OwnerUserId="1037" LastActivityDate="2011-05-17T18:56:14.707" Title="How to get Intensity over a range of wavelengths from a Total Intensity" Tags="&lt;optics&gt;&lt;light&gt;" AnswerCount="4" />
  <row Id="9955" PostTypeId="2" ParentId="9953" CreationDate="2011-05-17T16:45:08.737" Score="0" ViewCount="" Body="&lt;p&gt;Although Dmckee has already answered your question (allowing for a bit of leeway for you to figure that answer out yourself), the fact of the matter is that in order for the time differential to be a million times faster, you'd basically have to have one planet sitting on the surface of a black hole, and the other planet sitting out in a void between galaxies. And even then, I don't think you'd get up to a million times - probably more like a hundred times or &lt;em&gt;maybe&lt;/em&gt; a thousand times, but I'm just pulling numbers out of my butt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also worth noting that the tidal effects of a black hole will rip the planet apart (and every other bit of matter, right down to the atomic level) before it gets anywhere near it. So &quot;evolution happening a million times faster&quot; is totally out of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to be creative about crazy methods of civilisations being far older than ours, this isn't the way. A much better way would be if purely by chance, intelligent life had evolved on a planet in only 3 billion years, instead of 4 billion. Even that's pretty crazy, since life on our own planet was basically dominated by nothing more complicated than single-celled bacteria and other sorts of goo for about 3 billion years. But replace &quot;dinosaurs&quot; with &quot;intelligent civilisation&quot; and you have a decent timeline that's millions of years ahead of us, and just about as likely to happen, IMHO.&lt;/p&gt;&#xA;" OwnerUserId="529" LastActivityDate="2011-05-17T16:45:08.737" CommentCount="1" />
  <row Id="9956" PostTypeId="2" ParentId="9947" CreationDate="2011-05-17T16:49:12.460" Score="1" ViewCount="" Body="&lt;p&gt;Agreed that the answer to the question is, &quot;No.&quot;   Acceleration remains constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to think of it is this:  The first 980N is accelerating the first 100kg at 9.8m/(s^2).  The second 980N is accelerating the second 100kg at 9.8m/(s^2).  Both masses fall at the same velocity and acceleration (neglecting wind resistance, etc.).  So it should be easy to see how you could join these to 100kg masses into one 200kg mass, with 1960N pulling it down, without changing the speed or acceleration of either mass. &lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-05-17T16:49:12.460" />
  <row Id="9957" PostTypeId="2" ParentId="9954" CreationDate="2011-05-17T17:16:53.820" Score="2" ViewCount="" Body="&lt;p&gt;You are on the right lines with integrating, the problem lies in the fact that the intensity profile for your LED isn't particularly uniform.  Clearly looking to the Planck function is going to give you problems because your LED does not emit like a blackbody.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solution would be to produce an equation that roughly replicates what your LED does.  Looking at the graph, you could do a sum of 3 or perhaps 4 Gaussian curves peaking at 450/530/560 ish (eyeballing).  The curve would be something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = a(\exp(f_1(x))+0.4\exp(f_2(x))+0.3\exp(f_3(x))+ \cdots )$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f_n$ are the parameters for the Gaussian distributions for each peak.  I'm using the fact that the 450nm peak is full intensity and the other peaks are scaled relatively by 0.4, 0.3 and so on.  It's not too difficult to produce a peak with the correct x-shift and width.  Playing around on Wolfram Alpha would do, or if you have access to Matlab/Mathematica then you could of course use them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can numerically integrate it for the area which will be a function of the scaling parameter, $a$.  Find the value of $a$ that gives you an area of 10 and that's the equation you use.  Then you can integrate between the wavelength values you're interested in.  This is all without units, of course, but as it's all relative, it doesn't make a difference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like a little bit of a bodge, but unless you have the profile in equation form or in a table, I don't see why this wouldn't work.&lt;/p&gt;&#xA;" OwnerUserId="3346" LastActivityDate="2011-05-17T17:16:53.820" />
  <row Id="9958" PostTypeId="2" ParentId="9954" CreationDate="2011-05-17T17:21:47.813" Score="1" ViewCount="" Body="&lt;p&gt;The white led is not a black-body source so you should just ignore the &quot;Planck stuff&quot; and use the specified profile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, your intensity meter is (hopefully) calibrated so as long as your input to the calorimeter is within its range (266-1200 nm) you don't need to consider its spectral response either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just take the given spectral profile of the white LED and integrate the portion from 400 to 455 nm and divide by the complete integral. Multiply this ratio with your total intensity measurement. You can trace the curve in a suitable graphics program or something, or just approximate it as a trapezoidal shape for example by measuring some points with a ruler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A spectroradiometer, as you imply, is a better tool for this but in your case you are saved by having a pre-measured spectral profile.&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-05-17T17:21:47.813" />
  <row Id="9959" PostTypeId="2" ParentId="9954" CreationDate="2011-05-17T17:24:11.977" Score="2" ViewCount="" Body="&lt;p&gt;There is an Excel spreadsheet you can download on the page you linked. It has the numerical data that was used to produce that plot. Just download that spreadsheet, and integrate numerically. If you want to be really really insanely cautious, you could even interpolate the data and numerically integrate that.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-05-17T17:24:11.977" CommentCount="4" />
  <row Id="9960" PostTypeId="1" AcceptedAnswerId="9963" CreationDate="2011-05-17T17:26:04.957" Score="0" ViewCount="79" Body="&lt;p&gt;are grassmaniann numbers a concept of graded lie algebras or is something specific to superalgebras? what are they (i.e: how are they defined, important properties, etc.)? is there a reasonable introduction to them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i think that what makes me wonder a little is , since there does not seem to be a sensible constructivist approach to these entities (other than accepting them as the entities that satisfy the required properties) is that nothing stops someone from going into 'constructing' meta-superalgebras by defining 'numbers' $\kappa_{i}$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\kappa_{i} \kappa_{j} = \theta_{k} (odd)$$&#xA;$$\kappa_{i} \kappa_{j}\kappa_{m} \kappa_{n} = \theta_{p} (even)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so i define such numbers as 'square-roots' of grassmannian A-numbers. It seems nothing stops this process ad infinitum. Maybe there is some property i'm missing that will allow the algebra to be closed but i dont know what that could be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;btw, i think this is a great reference question regarding this topic: &lt;a href=&quot;http://physics.stackexchange.com/questions/5005/velvet-way-to-grassmann-numbers&quot;&gt;&amp;quot;Velvet way&amp;quot; to Grassmann numbers&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-05-17T19:47:50.957" LastActivityDate="2011-05-17T20:00:20.857" Title="what are grassmanian (even/odd) numbers used in superalgebras?" Tags="&lt;mathematical-physics&gt;&lt;grassmann-numbers&gt;&lt;superalgebra&gt;" AnswerCount="1" />
  <row Id="9962" PostTypeId="2" ParentId="9935" CreationDate="2011-05-17T17:57:25.030" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Go for it.&lt;/strong&gt;  &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0131118927&quot; rel=&quot;nofollow&quot;&gt;Griffiths&lt;/a&gt; would make fine summer-time reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you might consider making the Feynman lectures your summer reading.  They provide a great deal of insight but are better to read before (or at the same time) you take the intro courses.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-05-17T17:57:25.030" CommunityOwnedDate="2011-05-17T17:57:25.030" />
  <row Id="9963" PostTypeId="2" ParentId="9960" CreationDate="2011-05-17T18:12:23.483" Score="1" ViewCount="" Body="&lt;p&gt;Here I will just make a couple of general remarks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;a href=&quot;http://en.wikipedia.org/wiki/Graded_algebra&quot; rel=&quot;nofollow&quot;&gt;Graded&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Graded_Lie_algebra&quot; rel=&quot;nofollow&quot;&gt;algebras&lt;/a&gt; usually refer to $\mathbb{Z}$- or $\mathbb{N}$-graded algebras, while &lt;a href=&quot;http://en.wikipedia.org/wiki/Superalgebra&quot; rel=&quot;nofollow&quot;&gt;superalgebras&lt;/a&gt; are $\mathbb{Z}_2$-graded algebras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;a href=&quot;http://en.wikipedia.org/wiki/Grassmann_number&quot; rel=&quot;nofollow&quot;&gt;Grassmann numbers&lt;/a&gt; are oddly graded &lt;a href=&quot;http://planetmath.org/encyclopedia/Supernumber.html&quot; rel=&quot;nofollow&quot;&gt;supernumbers.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please click on the links for further information, important properties and references. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;References: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Bryce De Witt, &quot;Supermanifolds&quot;, Cambridge Univ. Press, 1992.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Deligne, Pierre and John W. Morgan, &quot;Notes on Supersymmetry (following Joseph Bernstein)&quot;. Quantum Fields and Strings: A Course for Mathematicians (1999). American Mathematical Society. pp. 41–97. ISBN 0-8218-2012-5.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Concerning v3 of the question. The $\kappa_i$'s correspond to a $\mathbb{Z}_4$ grading, and there are indeed research works in that direction. However, many properties of numbers and supernumbers do not generalize easily to $\mathbb{Z}_n$-grading with $n&gt;$2. For instance, I think that already Berezin showed that it is not possible to define a useful notion of (Berezin) determinant of matrices with $\mathbb{Z}_n$-graded entries if $n&gt;$2. &lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-17T20:00:20.857" LastActivityDate="2011-05-17T20:00:20.857" CommentCount="1" />
  <row Id="9964" PostTypeId="2" ParentId="9954" CreationDate="2011-05-17T18:28:19.843" Score="0" ViewCount="" Body="&lt;p&gt;To get fairly close, numerical integration appears to be the easiest solution.  If you can give up a little accuracy, simply blow up the section from 400nm-700nm, overlay a grid, and enter height of the curve, from 420nm to 670nm, in an Excel spreadsheet for each 5nm slice.  Then sum up the slices:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) w(0) = 420; I(0) = 0; Sum_Intensity(0) = 0; Sum_Target(0) = 0;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) For N = 1 to 50;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) w(N) = (420 + 5N); I(N) = &amp;lt; pick curve height off chart for wavelength = w(N) nm &gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.) Sum_Intensity(N) = Sum_Intensity(N-1) + 5 x ((I(N-1)+I(N))/2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5.) Sum_Target(N) = Sum_Target(N-1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6.) If 399 &amp;lt; W(N) &amp;lt; 456, Then Sum_Target(N) = Sum_Target(N-1) + 5 x ((I(N-1)+I(N))/2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7.) Next N;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8.) 400-455nm power, as a percentage of total power = Sum_Target(50) / Sum_Intensity(50);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Step 8 assumes intensities were measured in power or power/area units (e.g. mW/cm^2).  Otherwise you need to square the intensity raio to get to the power ratio.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might look a little daunting at first, but should be easy to implement as a 52 row by 5 column-wide block in an Excel SS.  Maybe take you two hours of blowing up the chart and picking off points and entering them into the SS.  I would guess your accuracy to be better than +/- 3%.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give you a check: my calibrated eyeball says you should expect Sum_Target(50) and Sum_Intensity(50) (which are the numbers you are interested in) to be ~ON THE ORDER OF~ 65 and 10, respectively.  IF those numbers were actually 65 and 10, then your ratio of 400-455 power to total power would be 10/65 = 15.4%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, and this all assumes that your receiver's response is fairly linear across the spectrum of 430-670nm.  It appears that the manufacturer claims 1%, so you should be good to go.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-05-17T18:56:14.707" LastActivityDate="2011-05-17T18:56:14.707" />
  <row Id="9965" PostTypeId="2" ParentId="9935" CreationDate="2011-05-17T18:44:44.840" Score="1" ViewCount="" Body="&lt;p&gt;I think that if you are to really develop a sense for physics, you must spend time cultivating a physics intuition in addition to mathematical prowess, so if you're motivated enough to take a crack at Griffiths, take time to investigate other areas, like E&amp;amp;M and Statistical Physics, and try to really relate the ideas you find there to things that you already understand.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, spend some time with a computer language, see if you can create simulations to help your understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have iTunes, there are a number of lecture videos at iTunesU that discuss quantum, and also the &lt;a href=&quot;http://ocw.mit.edu/courses/#physics&quot; rel=&quot;nofollow&quot;&gt;MIT Open Course Ware Site:Physics&lt;/a&gt; has a number of very courses with lecture notes and some with full video sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck.&lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2011-05-17T18:44:44.840" CommunityOwnedDate="2011-05-17T18:44:44.840" />
  <row Id="9966" PostTypeId="1" CreationDate="2011-05-17T19:31:21.823" Score="1" ViewCount="77" Body="&lt;p&gt;I have a kayak and a bike. I routinely put them on top of my car and drive 60-70 mph for hundreds of miles. I am curious how much this affects my gas mileage. &lt;/p&gt;&#xA;" OwnerUserId="2843" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-17T19:54:02.137" LastActivityDate="2011-05-17T20:08:07.960" Title="How to calculate the effect of roof items on gas mileage?" Tags="&lt;fluid-dynamics&gt;&lt;aerodynamics&gt;" AnswerCount="2" />
  <row Id="9967" PostTypeId="2" ParentId="9966" CreationDate="2011-05-17T19:53:23.090" Score="4" ViewCount="" Body="&lt;p&gt;The best way: just do it. Go out and drive on the highway for some distance with the kayak and bike on top of your car, and measure the amount of fuel used, then do the same trip without the items on top of your car and measure the amount of fuel used in that case. The trip would have to be long enough that you can get a fairly accurate measurement of how much fuel you used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you could do is fill up your fuel tank immediately before setting out each time, then fill it up again when you get back, and the amount of gas you need to buy will tell you how much you used. Just make sure to use the same gas station, and preferably the same pump, for every measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to calculate it theoretically, that's a whole different story. Presumably the kayak and bike on top of your car increase the drag force and thus reduce its fuel efficiency, but as far as I know there's no way to do a calculation without a complex aerodynamic simulation.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-17T20:08:07.960" LastActivityDate="2011-05-17T20:08:07.960" CommentCount="3" />
  <row Id="9968" PostTypeId="2" ParentId="9966" CreationDate="2011-05-17T19:59:24.037" Score="1" ViewCount="" Body="&lt;p&gt;There are two things that get affected by the roof items. One is the frontal area of the car, and the second is the coefficient of drag for the car. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no rule of thumb, but in order to estimate the combined effect, then drive up to 60mph, put the car in neutral and measure how long it takes to coast down to like 50mph. Then try again with stuff on your roof. If time decreases $X$% then you can estimate that the rolling resistance increases by $X$% also, and hence your consumption goes up. Note that it makes sense to use a very flat road, and use measurements in both directions in negate the wind effects. If you average 3-5 measurements per direction you will get a good idea of the effect of the roof stuff.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-05-17T19:59:24.037" CommentCount="1" />
  <row Id="9969" PostTypeId="1" AcceptedAnswerId="9972" CreationDate="2011-05-17T20:34:46.033" Score="2" ViewCount="280" Body="&lt;p&gt;I learned that a moving charge creates a magnetic field perpendicular to its direction of motion. I also learned that charged particles like electrons have spin and they also create a magnetic field because of their magnetic dipole moment. I don't understand what magnetic dipole moment is, and I can't find a decent explanation for it by using Google.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, Can someone please explain magnetic dipole moment arising from spin in an effective way? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, is my main question; Does a magnetic field arise from a moving charge or from its spin, or both?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, I am very confused. I think I am mixing together the classical electrodynamics description of magnetic field, (moving charge), with quantum mechanics description of magnetic field, (magnetic dipole moment due to spin?)&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2011-07-24T06:35:31.063" Title="Does a magnetic field arise from a moving charge or from its spin, or both?" Tags="&lt;quantum-mechanics&gt;&lt;electromagnetism&gt;&lt;spin&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" />
  <row Id="9971" PostTypeId="2" ParentId="237" CreationDate="2011-05-17T21:28:15.590" Score="1" ViewCount="" Body="&lt;p&gt;Since the object is basketball size, it would displace a significant volume and weight of water when submerged. Weigh a container with some water and take a reading. Then attach a wire to the object, suspend it in the container of water till it's submerged without touching the container, and take a second reading. The difference between the two readings is the water weight that is displaced by the object. Volume (m^3) = weight difference (kg) / 1000 (kg/m^3).&lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="1924" LastEditorDisplayName="" LastEditDate="2011-05-18T01:04:25.920" LastActivityDate="2011-05-18T01:04:25.920" />
  <row Id="9972" PostTypeId="2" ParentId="9969" CreationDate="2011-05-17T21:28:32.993" Score="3" ViewCount="" Body="&lt;p&gt;The way you've asked your question (especially part 3), it sounds like you're trying to understand how magnetic dipole moments and moving charges are related to each other. But they're not. Moving charges and magnetic dipole moments don't &lt;em&gt;describe&lt;/em&gt; a magnetic field, they &lt;em&gt;produce&lt;/em&gt; a magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Off the top of my head, I can actually think of three ways that magnetic fields are produced:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Moving electric charge&lt;/strong&gt;: any time you have an electrically charged object in motion, it will produce a magnetic field. Electric currents fall into this category. This is usually the first way that physics students learn to generate a magnetic field; it's described by Ampere's law (one of Maxwell's equations).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Changing electric field&lt;/strong&gt;: any time an electric field changes in time, it will produce a magnetic field, even if there is no current around. This is usually the second way that physics students learn that a magnetic field can be generated. It's also described by Ampere's law (technically the &quot;Maxwell correction&quot; term).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Static magnetic multipoles&lt;/strong&gt;: this one is a little more complicated because it's not described by any of Maxwell's equations, at least not directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me start with an analogy. Hopefully you know that a charged object produces an electric field. But you don't &lt;em&gt;have&lt;/em&gt; to have a net charge to produce a field. If you take a positive charge and a negative charge of equal magnitude and put them very close to each other, you'll still get an electric field, because the field from the positive charge and the field from the negative charge don't &lt;em&gt;exactly&lt;/em&gt; cancel each other out. This is an example of an electric dipole. You can think of this as a &quot;secondary source&quot; of the field, which depends not on the total amount of charge, but on how the charge is distributed within the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, when the total amount of charge is nonzero, the distribution of the charge has a small enough effect that we don't have to care about it, but when there is no net charge, the way the charge is distributed becomes important. Obviously, in order to do physics we need to have a physical quantity that describes the distribution. This is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electric_dipole_moment&quot; rel=&quot;nofollow&quot;&gt;electric dipole moment&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, we can measure the electric dipole moment of an object and use it to do useful calculations even if we don't know anything about the charge distribution - or even if there may not be a charge distribution at all. In other words, one could imagine that there might be some unknown physical mechanism, completely separate from electric charge, that causes some object to have an electric dipole moment. So it makes sense to define an &quot;electric dipole&quot; as &quot;something that has a nonzero electric dipole moment,&quot; whether or not that thing has a charge distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thing applies to magnetic dipoles and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_dipole_moment&quot; rel=&quot;nofollow&quot;&gt;magnetic dipole moment&lt;/a&gt;. It works just like the electric dipole moment, except with the magnetic field and &quot;magnetic charge&quot; instead of electric field and electric charge. The thing is, as far as we know, there are no magnetically charged objects (the so-called &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_monopole&quot; rel=&quot;nofollow&quot;&gt;magnetic monopoles&lt;/a&gt;&quot;). So the magnetic dipole moment never gets masked by magnetic charge, the way the electric dipole moment usually does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with the electric dipole, a magnetic dipole of any sort will generate a magnetic field. One kind of magnetic dipole is a small loop of current. If the current is made of physical charges moving around in a circle, then it will have some angular momentum. So once it was discovered that the electron has intrinsic angular momentum (spin), physicists naturally wondered whether that angular momentum was due to constituent particles moving in circles inside the electron. One way to test this theory would be to measure the magnetic dipole moment of the electron and calculate whether it corresponds to the prediction of the current-loop model. As it turns out, it doesn't. So evidently something else is going on; the magnetic dipole moment of the electron is not just produced by classical charges moving in a circle. It's something intrinsic to the electron. (Quantum electrodynamics correctly predicts the exact value of the electron's magnetic dipole moment, but it doesn't offer a simple physical picture.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-05-17T21:28:32.993" CommentCount="3" />
  <row Id="9973" PostTypeId="2" ParentId="5296" CreationDate="2011-05-17T21:28:56.517" Score="0" ViewCount="" Body="&lt;p&gt;Actually, the answer is far worse than that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe was expanding more &lt;em&gt;slowly&lt;/em&gt; in the past than it is today. Of course, the initial reaction to this discovery was &quot;No, that can't be right. There must be something wrong with the way we're measuring it&quot;. But after some oh, 20 years of observations with different equipment and methods, we're coming up with the same result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which, to anyone with a basic knowledge of Newtonian physics and an inkling of the mass of the universe, just trying to &lt;em&gt;contemplate&lt;/em&gt; the amount of force required to make that happen could make your brain explode. Even astronomy has not before come up with numbers quite astronomical enough to describe the amount of energy required to make this happen. And to top it all off, we have as of yet no way of detecting the nature of this energy, or any of its properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that whatever this energy is, it makes up an astounding 73% of the universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately for astrophysicists, this utterly insane phenomenon is really real. More information: &lt;a href=&quot;http://en.wikipedia.org/wiki/Dark_energy&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dark_energy&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="529" LastActivityDate="2011-05-17T21:28:56.517" />
  <row Id="9975" PostTypeId="1" AcceptedAnswerId="9985" CreationDate="2011-05-18T01:05:40.697" Score="1" ViewCount="67" Body="&lt;p&gt;Wien's fifth power law says that emissive&#xA;power is proportional to&#xA;temperature to the fifth power.&#xA; But the Stefan–Boltzmann&#xA;law says emissive power is&#xA;proportional to temperature to the fourth power.&#xA;How can both of these be true?&lt;/p&gt;&#xA;" OwnerUserId="3576" LastActivityDate="2011-05-18T02:43:10.427" Title="Wein's fifth power Law and Stephan Boltzmann's fourth power laws of emissive power" Tags="&lt;energy&gt;&lt;radiation&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="9976" PostTypeId="5" CreationDate="2011-05-18T01:21:35.457" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-18T01:21:35.457" LastActivityDate="2011-05-18T01:21:35.457" />
  <row Id="9977" PostTypeId="4" CreationDate="2011-05-18T01:21:35.457" Score="0" ViewCount="" Body="A theory that describes how matter produces and responds to the geometry of space and time. It was first published by Einstein in 1915 and is currently used to study the structure and evolution of the universe, as well as having practical applications like GPS." OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-18T01:21:35.457" LastActivityDate="2011-05-18T01:21:35.457" />
  <row Id="9978" PostTypeId="5" CreationDate="2011-05-18T01:23:38.507" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-18T01:23:38.507" LastActivityDate="2011-05-18T01:23:38.507" />
  <row Id="9979" PostTypeId="4" CreationDate="2011-05-18T01:23:38.507" Score="0" ViewCount="" Body="Applies to questions of primarily educational value - not only questions that arise from actual homework assignments, but any question where it is preferable to guide the asker to the answer rather than giving it away outright." OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-18T01:23:38.507" LastActivityDate="2011-05-18T01:23:38.507" />
  <row Id="9980" PostTypeId="5" CreationDate="2011-05-18T01:25:06.117" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-18T01:25:06.117" LastActivityDate="2011-05-18T01:25:06.117" />
  <row Id="9981" PostTypeId="4" CreationDate="2011-05-18T01:25:06.117" Score="0" ViewCount="" Body="A branch of physics that studies heat and entropy of macroscopic objects." OwnerUserId="-1" LastEditorUserId="1501" LastEditorDisplayName="" LastEditDate="2011-09-02T14:03:24.340" LastActivityDate="2011-09-02T14:03:24.340" />
  <row Id="9982" PostTypeId="1" CreationDate="2011-05-18T01:47:31.293" Score="3" ViewCount="182" Body="&lt;p&gt;The field equations of the Brans-Dicke gravity are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Box\phi = \frac{8\pi}{3+2\omega}T$$&#xA;$$G_{ab} = \frac{8\pi}{\phi}T_{ab}+\frac{\omega}{\phi^2}&#xD;&#xA;(\partial_a\phi\partial_b\phi-\frac{1}{2}g_{ab}\partial_c\phi\partial^c\phi)&#xD;&#xA;+\frac{1}{\phi}(\nabla_a\nabla_b\phi-g_{ab}\Box\phi)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general relativity, the singularity in the Schwarzchild black hole seems to be considered either a 'delta function' like mass source term, or the &quot;edge&quot;/boundary condition of a manifold.  Either way, the equations can't tell us how the &quot;infinite curvature&quot;/singularity evolves and therefore we are forced in some sense to &quot;impose it&quot; as a boundary condition either way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in Brans-Dicke gravity the effective gravitational coupling constant appears to be sourced by $T$.  So it appears in a true vacuum $T \rightarrow 0$, $\phi \rightarrow 0$, and therefore no Schwarzchild solution is possible &lt;em&gt;regardless&lt;/em&gt; of the value of the parameter $\omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I've heard that Brans-Dicke is equivalent to General Relativity at least in some limit, so what is going on here?  Brans-Dicke seems to really care precisely how we deal with singularity, while General Relativity doesn't care what is behind the event horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1) Can Brans-Dicke gravity support black hole vacuum + curvature singularity solutions?&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;2) Does it force us to interpret the singularity a certain way, and if so does the Brans-Dicke $\longrightarrow$ General Relativity limit tell us the singularity should be interpretted as non-vacuum to make sense?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="3577" LastActivityDate="2011-07-29T15:09:02.160" Title="Brans-Dicke gravity and Black-Hole &quot;vacuum&quot;" Tags="&lt;general-relativity&gt;&lt;quantum-gravity&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="9983" PostTypeId="1" CreationDate="2011-05-18T02:10:30.927" Score="3" ViewCount="133" Body="&lt;p&gt;Let's say we have a spherically symmetric fluid:&#xA;$$&#xD;&#xA;T^{\alpha \beta} = \begin{bmatrix}&#xD;&#xA;  \rho &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA;     0 &amp;amp; p &amp;amp; 0 &amp;amp; 0 \\&#xD;&#xA;     0 &amp;amp; 0 &amp;amp; p &amp;amp; 0 \\&#xD;&#xA;     0 &amp;amp; 0 &amp;amp; 0 &amp;amp; p    &#xD;&#xA;\end{bmatrix}&#xD;&#xA;$$&#xA;where $\rho$ and $p$ are functions of the radial coordinate r.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $p=-\rho/3$ this would be traceless, and possibly compatible with an electro-vacuum solution (Are there other necessary conditions for compatibility besides the stress-energy tensor being traceless?). While it would probably be unstable, &lt;strong&gt;is there a singularity free electrovacuum solution of GR which is stationary?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that this has enough symmetry that someone skilled with differential-geometry and GR could actually probably just solve this themselves.  I'm interested in not just the answer, but also in seeing the process of how to solve this in GR.  So please include any work in your solution, thank you!&lt;/p&gt;&#xA;" OwnerUserId="3577" LastActivityDate="2011-08-09T04:21:20.087" Title="Singularity-free stationary electro-vacuum solution" Tags="&lt;general-relativity&gt;&lt;electromagnetism&gt;&lt;differential-geometry&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="9984" PostTypeId="2" ParentId="9935" CreationDate="2011-05-18T02:12:28.497" Score="0" ViewCount="" Body="&lt;p&gt;As others have stated or suggested, there is no right answer, although I get the notion that some feel there is a proper order to these things, and being a layman I am not qualified to judge what is proper when it comes to physics education.  However, I will offer you this nugget:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Understand the difference between an&#xA;equation and a solution.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some may laugh at this statement, or even shake their head in empathetic embarrassment, but I will tell you that one of the hardest concepts people have grasping (even extraordinarily intelligent people) when starting in QM is what is meant when one says Schrodinger wave equation and what one means when they say wave function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$i\hbar\frac{\partial}{\partial t} \Psi(\mathbf{r},\,t) \&#xA; = -\frac{\hbar^2}{2m}\nabla^2\Psi(\mathbf{r},\,t) + V(\mathbf{r})\Psi(\mathbf{r},\,t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E\Psi(\mathbf{r}) \&#xA; = -\frac{\hbar^2}{2m}\nabla^2\Psi(\mathbf{r}) + V(\mathbf{r})\Psi(\mathbf{r})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above are time dependent and time independent Schrodinger equations.  Most people would not be able to distinguish which component of the above expressions is the wave function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first case its (the time dependent case):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Psi(\mathbf{r},\,t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second its (the time independent case):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Psi(\mathbf{r})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without this distinction it is very hard to understand statements like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;i&gt;Solutions to the wave equation are wave functions which represent the probability amplitude of a particle being found at position r and time t.&lt;/i&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;i&gt;The time independent equation describes the standing wave solutions, or the energy eigenstates (states with definite energy), of the time dependent equation.&lt;/i&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest way I have in conveying understanding, without being overly pedantic, is to explain to a person that functions are simply all the possible lines one can draw on a paper that do not cross themselves (like in a loop) and do not make abrupt changes, jumps or skips.  The wave functions are those sets of lines that also satisfy the constraints place on them by the wave equation's operators; e.g. the linear differential operators like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$i\hbar\frac{\partial}{\partial t}$$ and, $$-\frac{\hbar^2}{2m}\nabla^2\ + V(\mathbf{r})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think understanding these distinctions transcend most physical subjects, even if the specific equation is not related to QM.  So I think it is something that one should feel comfortable understanding, and if one uses QM as the context I think it is o.k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to add, I am not trying to pick on the lack of understanding of these things by my fellow laymen, I just think that it is something that represents the first initial barrier for most people, and thus is a source of discouragement.  So if anything in the above can help aid people, I think it is a good thing to provide as input.&lt;/p&gt;&#xA;" OwnerUserId="3460" LastActivityDate="2011-05-18T02:12:28.497" CommunityOwnedDate="2011-05-18T02:12:28.497" />
  <row Id="9985" PostTypeId="2" ParentId="9975" CreationDate="2011-05-18T02:13:40.533" Score="2" ViewCount="" Body="&lt;p&gt;Wien's fifth power law applies to the maximum height of the emissive power density. The Stefan–Boltzmann law applies to the total emissive power (the integration of the emissive power density). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Wien's displacement law (derived &lt;a href=&quot;http://en.wikipedia.org/wiki/Wein%27s_displacement_law&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;), $\lambda_\max T = b$ is constant. So, from Planck's law (e.g. see &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck%27s_law&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I&amp;#39;(\lambda_\max, T) \propto \frac{\lambda_\max^{-5}}{e^{hc/(\lambda_\max k T)} -1} = \frac{T^5/b^5}{e^{hc/(k b)} -1} \propto T^5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and integrating Plank's law (e.g.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; except I keep in terms of wavelength for slightly longer) we get, by changing variables to $x=hc/(\lambda kT)$ so that $\lambda^{-5}d\lambda \propto -T^4x^3dx$ (the negative is absorbed by changing the integration interval limits):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_0^\infty I&amp;#39;(\lambda,T)d\lambda \propto \int_0^\infty \frac{T^4 x^3dx}{e^{x}-1} \propto T^4$.&lt;/p&gt;&#xA;" OwnerUserId="3325" LastEditorUserId="3325" LastEditorDisplayName="" LastEditDate="2011-05-18T02:43:10.427" LastActivityDate="2011-05-18T02:43:10.427" />
  <row Id="9986" PostTypeId="2" ParentId="9969" CreationDate="2011-05-18T03:48:20.483" Score="2" ViewCount="" Body="&lt;p&gt;In the end it is all moving charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that this is also the case for the magnetic field from the spin you need to perform the so-called &lt;em&gt;Gordon decomposition&lt;/em&gt; on the charge-current density of the electron as defined by Dirac's equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;see for instance 18.2 in this chapter of my book:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics-quest.org/Book_Chapter_Gordon_Decomposition.pdf&quot; rel=&quot;nofollow&quot;&gt;http://physics-quest.org/Book_Chapter_Gordon_Decomposition.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gordon decomposition of the charge-current density of the electron shows an additional part due to the magnetic moment of the electron. This additional part is equivalent to the classical &lt;em&gt;&quot;current from magnetization&quot;&lt;/em&gt; given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$j=\nabla\times M$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where M is the magnetization of the medium. Figure 18.1 of the link above shows&#xA;how this is just the familiar Stokes theorem in action. The current $j$ is an effective current caused by the inherent magnetization of the electron field. This is the current which is the source of the magnetic field due to the magnetic moment of the electron. For Stokes theorem see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.umn.edu/~nykamp/m2374/readings/stokesidea/&quot; rel=&quot;nofollow&quot;&gt;http://www.math.umn.edu/~nykamp/m2374/readings/stokesidea/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards, Hans&lt;/p&gt;&#xA;" OwnerUserId="3197" LastEditorUserId="3197" LastEditorDisplayName="" LastEditDate="2011-05-18T14:05:38.360" LastActivityDate="2011-05-18T14:05:38.360" CommentCount="3" />
  <row Id="9987" PostTypeId="2" ParentId="9983" CreationDate="2011-05-18T04:08:57.503" Score="-3" ViewCount="" Body="&lt;p&gt;I don't think extremal black holes have singularities.&lt;/p&gt;&#xA;" OwnerUserId="3578" LastActivityDate="2011-05-18T04:08:57.503" CommentCount="1" />
  <row Id="9992" PostTypeId="1" CreationDate="2011-05-18T07:27:35.693" Score="2" ViewCount="202" Body="&lt;p&gt;Are there any materials exploring intermediate levels (of any sub-subject) which give plain English descriptions along side formulas/equations/notation?&lt;/p&gt;&#xA;" OwnerUserId="2309" LastEditorUserId="2309" LastEditorDisplayName="" LastEditDate="2011-05-20T17:23:33.547" LastActivityDate="2011-05-21T13:05:41.893" Title="Any resources to learn how to interpret formulae/diagrams?" Tags="&lt;quantum-mechanics&gt;&lt;soft-question&gt;&lt;formula&gt;&lt;publications&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="4" />
  <row Id="9994" PostTypeId="1" CreationDate="2011-05-18T07:52:37.450" Score="1" ViewCount="81" Body="&lt;p&gt;A 6500-kg helicopter accelerates upward at 0.60 m/s² while lifting a 1200-kg car. &#xA; (a) What is the lift force exerted by the air on the rotors?&lt;br&gt;&#xA;(b) What is the tension in the cable (ignore its mass) that connects the car to the helicopter?&lt;/p&gt;&#xA;" OwnerUserId="3580" LastEditorUserId="2843" LastEditorDisplayName="" LastEditDate="2011-05-18T09:37:29.650" LastActivityDate="2011-05-18T09:37:29.650" ClosedDate="2011-05-18T09:37:39.543" Title="Acceleration Force" Tags="&lt;homework&gt;&lt;forces&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="9995" PostTypeId="1" AcceptedAnswerId="10423" CreationDate="2011-05-18T08:20:50.640" Score="0" ViewCount="118" Body="&lt;p&gt;I want to compute the viscosity of argon at different temperatures. What is &lt;a href=&quot;http://en.wikipedia.org/wiki/Viscosity#Gases&quot; rel=&quot;nofollow&quot;&gt;Sutherland's constant&lt;/a&gt; of argon?&lt;/p&gt;&#xA;" OwnerUserId="2480" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-18T09:37:05.407" LastActivityDate="2011-05-27T09:54:13.110" Title="Sutherland's constant of argon" Tags="&lt;fluid-dynamics&gt;" AnswerCount="1" />
  <row Id="9997" PostTypeId="2" ParentId="9994" CreationDate="2011-05-18T08:25:29.540" Score="1" ViewCount="" Body="&lt;p&gt;Force = (Mass)*(Acceleration)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the above equation to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) find the force required to accelerate the helicopter and car 0.60 m/s² (NOT upward)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) find the force required to counteract the acceleration due to gravity (9.8m/s²) for both the car and the heicopter &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) adding 1. and 2. together will equal the lift force exerted by the air on the rotors&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.) do the same thing as in 1-3 except for only the car, which will yield the tension in the cable.&lt;/p&gt;&#xA;" OwnerUserId="2843" LastActivityDate="2011-05-18T08:25:29.540" />
  <row Id="9998" PostTypeId="1" CreationDate="2011-05-18T08:29:09.667" Score="8" ViewCount="203" Body="&lt;p&gt;Is there any interacting quantum field theory of massless fields with helicity $\pm 1$ which can be expressed entirely locally in terms of the field strength F&lt;sub&gt;&amp;mu;&amp;nu;&lt;/sub&gt; with no reference to vector potentials at all? Clearly, quantum electrodynamics doesn't fall into this category. The Aharonov-Bohm effect is the reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classically, it's easy to come up with many such theories, but unfortunately, they happen to be nonrenormalizable. But do asymptotically safe models exist?&lt;/p&gt;&#xA;" OwnerUserId="3583" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-05-18T17:36:52.567" LastActivityDate="2011-08-10T07:45:38.520" Title="Is there any interacting quantum field theory of massless spin-1 fields expressed locally entirely in terms of F, with no vector potential?" Tags="&lt;quantum-field-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="10001" PostTypeId="2" ParentId="9992" CreationDate="2011-05-18T10:19:03.330" Score="2" ViewCount="" Body="&lt;p&gt;This is series from Stanford university by Leonard Susskind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=hbmf0bB38h0&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=hbmf0bB38h0&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which introduces and uses tensor notation.  This is a fairly high level course with a good introduction (the first 2 or 3 videos!) to various mathematical descriptions.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-05-18T10:19:03.330" CommentCount="1" />
  <row Id="10002" PostTypeId="1" CreationDate="2011-05-18T10:43:28.100" Score="0" ViewCount="271" Body="&lt;p&gt;Could anyone help me to understand the concept involved?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Double slit experiment can be easily understood by Wave nature of light, but while explaining it with photons , it required a lot of mathematics(Englert–Greenberger duality relation). Can anybody tell me what is the physical interpretation of theory rather than using probability functions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="3587" LastActivityDate="2011-07-21T04:24:33.843" Title="Quantum Mechanics" Tags="&lt;double-slit-experiment&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="10003" PostTypeId="1" AcceptedAnswerId="10005" CreationDate="2011-05-18T10:48:27.640" Score="2" ViewCount="98" Body="&lt;p&gt;As you probably know, the two next Olympic games will take place in London and Rio. Randall, author of xkcd, did an interesting observation (that I had never thought of) in &lt;a href=&quot;http://xkcd.com/852/&quot; rel=&quot;nofollow&quot;&gt;http://xkcd.com/852/&lt;/a&gt; , arguing that the difference in latitude may influence comparisons between different world records in pole vaulting. As I started considering other results that might be affected by a small change in the local g, I could not find an interesting model for swimming. Does anyone know a model to determinate if the swimming results would be better or worst in a gravity-increased scenario?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e., would it be easier to swim on Jupiter or on the moon?&lt;/p&gt;&#xA;" OwnerUserId="3586" LastActivityDate="2011-05-18T14:22:09.657" Title="Olympic games and the local g" Tags="&lt;gravity&gt;&lt;hydrodynamics&gt;&lt;models&gt;" AnswerCount="1" />
  <row Id="10004" PostTypeId="1" CreationDate="2011-05-18T11:08:13.830" Score="3" ViewCount="165" Body="&lt;p&gt;I know how to calculate the expectation of &amp;lt; $\Psi$|A|$\Psi$ &gt;&#xA;where the operator A is the eigenfunction of energy, momentum or position, but I'm not sure how to perform this for a pure frequency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, what is the expectation of frequency? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, is there an expectation, and is it solved using a fourier transform from position space into frequency space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&#xA;In response to the posts by lurscher and David Zaslavsky below, I think both may be right. Frequency f is a parameter and can be considered an operator. &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) For the plane wave there exists temporal and spatial frequencies which act as parameters. $|\textbf{u}_k\rangle = e^{-i k x }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) there is also an operator derived from the Hamiltonian &#xA;  E&#xA;  =&#xA;  h&#xA;  f&#xA; where &lt;br/&gt;  $\hat{H} =-  \frac{\hbar  }{i}\frac{\partial }{\partial t}$&lt;br/&gt;&#xA;Rearranging and inverting $i$ &lt;br/&gt;&#xA;$\hat{f} = \frac{H}{\hbar } = i \frac{\partial }{\partial t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively this appears like an operator, it has the right units and it transforms into the frequency domain, however, it still requires that  $\hat{f}$ be shown to be Hermitian and is an operator that is its own adjoint or satisfies a Poisson algebra. Also intuitively, we can observe a frequency, by taking a measurement. &lt;/p&gt;&#xA;" OwnerUserId="3588" LastEditorUserId="3588" LastEditorDisplayName="" LastEditDate="2011-05-20T02:32:40.427" LastActivityDate="2011-06-17T10:51:22.930" Title="How to calculate the quantum expectation of frequency of a particle?" Tags="&lt;quantum-mechanics&gt;&lt;mathematical-physics&gt;" AnswerCount="2" />
  <row Id="10005" PostTypeId="2" ParentId="10003" CreationDate="2011-05-18T11:35:07.903" Score="4" ViewCount="" Body="&lt;p&gt;You can estimate that someone swimming in water has a Reynolds number of about $10^6 - 10^7$; what counts is that this number is $\gg 1.$ In that case, you're dealing with the drag equation: &lt;a href=&quot;http://en.wikipedia.org/wiki/Drag_(physics)&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Drag_(physics)&lt;/a&gt;. If we assume that our swimmer has the same power $P$ on the moon and on Jupiter, his velocity $v$ scales as&#xA;$$v \propto \left(\frac{P}{\rho A}\right)^{1/3}$$&#xA;(modulo a dimensionless constant), where $\rho$ is the density of water and $A$ the size of Michael Phelps' torso. In particular, this doesn't depend on gravity, so it should be the same everywhere.*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*This is somewhat of a simplification of the very complicated physics behind the problem, but I hope that it's easy to understand.&#xA;[edit: fixed the HTML, removed the low-Reynolds thing.]&lt;/p&gt;&#xA;" OwnerUserId="1471" LastEditorUserId="1471" LastEditorDisplayName="" LastEditDate="2011-05-18T14:22:09.657" LastActivityDate="2011-05-18T14:22:09.657" CommentCount="3" />
  <row Id="10006" PostTypeId="1" CreationDate="2011-05-18T11:41:11.723" Score="0" ViewCount="95" Body="&lt;p&gt;An astronaut is conducting repairs on a satellite 50.0m from a space shuttle. Her mass is 120kg (including space suit) and she is using a 0.5kg spanner. She realises that she only has 5 minutes oxygen supply remaining. With what velocity must the spanner be thrown for her to return to the shuttle safely.&lt;/p&gt;&#xA;" OwnerUserId="3589" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-05-18T11:56:08.877" LastActivityDate="2011-05-18T14:47:02.447" ClosedDate="2011-05-18T19:21:09.710" Title="physics- momentum ( a space question)" Tags="&lt;homework&gt;&lt;newtonian-mechanics&gt;&lt;momentum&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10007" PostTypeId="2" ParentId="10004" CreationDate="2011-05-18T12:17:42.783" Score="1" ViewCount="" Body="&lt;p&gt;the expectation for &lt;strong&gt;time frequency&lt;/strong&gt; is obtained replacing A with the Hamiltonian operator, for &lt;strong&gt;space frequency&lt;/strong&gt; is obtained replacing A with the momentum operator&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; from your comments i think i have a better idea of the source of your confusion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;first, frequency is not an operator; it is a parameter (a c-number); for instance, plane waves $ | \textbf{u}_k \rangle = e^{-i k x } $ have a single parameter $k$ called &lt;em&gt;spatial-frequency&lt;/em&gt; (the x is not really a parameter of the state $ | \textbf{u}_k \rangle$ is just an artifact that appears only when you write the coefficients of this state in the basis of eigenstates of the position operator )  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;De Broglie had the insight of associating classical particle quantities like energy and momentum to &lt;em&gt;temporal and spatial frequencies of plane waves&lt;/em&gt;. This means that  $ | \textbf{u}_k \rangle$ represents a physical state of momentum $p = \hbar k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you consider the algebraic entity $ | \textbf{u}_k \rangle \langle \textbf{u}_k | $ and you accept the bra-ket convention, if you multiply this operator with an eigenfunction $ | \textbf{u}_j \rangle$ you will basically have zero if $k \neq j$ or $ | \textbf{u}_j \rangle$ if $k=j$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then, a correct way to write the momentum operator (which is the same for any operator, the only thing that will change is the form of the basis function) is like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\textbf{P} = \sum_{k} { \hbar k | \textbf{u}_k \rangle \langle \textbf{u}_k |}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ANY&lt;/strong&gt; state vector can be written in the following form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ | \Psi \rangle = \sum_{j} { \psi_{j} | \textbf{u}_k \rangle }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I leave you as an exercise to write now the expression $\langle \Psi | \textbf{P} | \Psi \rangle$ by replacing each term with the above sums&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;hint 1:&lt;/em&gt; what you want is to obtain an expression that looks like an average after all! but you need to identify the weight in the average&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;hint 2:&lt;/em&gt; the fact that $\textbf{P}$ is hermitian is equivalent to saying that 1) the scalar products $\langle \textbf{u}_j | \textbf{u}_k \rangle$ will be always zero if $j \neq k$ and 2) that the eigenvalues $\hbar k$ will always be real&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-05-19T05:40:47.387" LastActivityDate="2011-05-19T05:40:47.387" CommentCount="5" />
  <row Id="10008" PostTypeId="1" AcceptedAnswerId="10010" CreationDate="2011-05-18T14:02:32.533" Score="2" ViewCount="119" Body="&lt;p&gt;I'm myopic.&#xA;It's a fact.&#xA;I understand exactly how it works because Internet told me light rays encounter themeselves too soon in my eyes... that is why I can't see far objects even if I see near objects very good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yesterday, I was looking myself in a mirror (it seems that I'm a very narcissistic person :) ) and I discovered that I can't see far objects in a mirror exactly like if I was looking to them directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question : light rays bouncing on the mirror to my eyes have the exact same length (I guess it works like that but I suppose I'm wrong), the length mirror -&gt; my eyes. &#xA;So why can't I see far objects good when I look at them using a mirror ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="3592" LastActivityDate="2011-05-18T14:33:43.333" Title="Why can't I see far when I look in a mirror?" Tags="&lt;optics&gt;&lt;mirror&gt;&lt;eye&gt;" AnswerCount="1" />
  <row Id="10009" PostTypeId="2" ParentId="10006" CreationDate="2011-05-18T14:07:34.790" Score="2" ViewCount="" Body="&lt;p&gt;To start off, I assume that the throw is instantaneous, and that it's ok to arrive in exactly 5 minutes (no time for the airlock!).  That means that the required velocity is quite easily:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{50 m}{5 minutes \frac{ 60 seconds }{ minute} } = 0.166 \frac{m}{s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the throw, we have $(Mv)_1=(Mv)_2$, and the ratios of the masses are know to be $M_1/M_2=120/0.5=240$, so from that equation the ratios of velocities has to be reverse of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_2=v_1 \frac{M_1}{M_2}=0.166 \frac{m}{s} 240 = 40 \frac{m}{s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I take it that this would not be feasible and that she would need to think of another plan.  Granted, I think that even if you could do it, aiming would be difficult since 50.0 m is a long way and I don't know how much practice she has in space sports.  But then again, maybe this is just unlikely enough to make for a good movie plot.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-05-18T14:47:02.447" LastActivityDate="2011-05-18T14:47:02.447" CommentCount="4" />
  <row Id="10010" PostTypeId="2" ParentId="10008" CreationDate="2011-05-18T14:23:40.860" Score="4" ViewCount="" Body="&lt;p&gt;Dear Rootosaurus, when you're looking at an image of a chair behind you in a flat mirror, then you're observing the so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Virtual_image&quot; rel=&quot;nofollow&quot;&gt;virtual image&lt;/a&gt; of the chair. If the mirror's surface is located in the $x=0$ plane and the coordinate of the real chair is at $(x,y,z)$, then the virtual image of the chair is at $(-x,y,z)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the light rays coming from the real chair that are reflected by the mirror and that ultimately end up in your eyes have &lt;em&gt;exactly&lt;/em&gt; the same directions as the light rays of a hypothetical chair that would be &lt;em&gt;actually&lt;/em&gt; located at the point $(-x,y,z)$ behind the mirror. So geometrically, you can't really distinguish a reflection of a chair that seems to be located at $(-x,y,z)$ from a real chair that is located at $(-x,y,z)$ and that you're observing through a window without a mirror. The geometry of the light rays is identical. That's why the concept of images is so useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, myopia means that one has some trouble to observe distant objects. Distant objects - imagine a distant point - have the property that they emit light rays that are nearly parallel. The further an object is, the more parallel its light rays look when they arrive to your eye. However, the lenses in your eyes have to convert these parallel mirrors into converging mirrors - so that all the light rays coming from the distant star end up at one point of the retina.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Myopic eyes are good in converting &quot;steeply divergent&quot; light rays from nearby objects to converging ones, but they're doing too much of a good thing. When you get too parallel light rays, myopic eyes make them converge too much, too early - the intersection will be inside the liquid in your eye. Hyperopia is the opposite disorder in which eyes make the light converge less than is needed. But what's relevant for your question is that the virtual image of the chair at $(-x,y,z)$ &quot;emits&quot; the same excessively parallel rays as a real chair at the same point, so a myopic eye will have the same trouble seeing it. After all, it shouldn't be paradoxical: the total distance that the light rays have traveled includes the distance of you from the mirror as well as the distance of the mirror from the real chair - because the colorful photons ultimately came from the chair and they were just reflected, not created, at the plane of the mirror.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-18T14:33:43.333" LastActivityDate="2011-05-18T14:33:43.333" CommentCount="2" />
  <row Id="10011" PostTypeId="1" CreationDate="2011-05-18T14:37:31.437" Score="0" ViewCount="147" Body="&lt;p&gt;I recently watched Sydney Coleman's '&lt;a href=&quot;http://media.physics.harvard.edu/video/index.php?id=SidneyColeman_QMIYF.flv&quot; rel=&quot;nofollow&quot;&gt;Quantum Mechanics In Your Face&lt;/a&gt;,' and very much enjoyed it. (Highly recommended.) In it, Professor Coleman quotes a very clever comment by a philosopher called Ludwig Wittgenstein. Wittgenstein found it strange that people used to believe in geocentrism (Sun goes around the Earth.) When asked why he felt it strange, he responded: &quot;&lt;em&gt;well, how would it have have looked if it were the other way around.&lt;/em&gt;&quot; Of course it would look exactly the same. Personally, I loved Wittgenstein's response and thought it showed impressive insight although, clearly, things are always simpler in hindsight than in foresight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This got me thinking. There must be hundreds of equally elegant / beautiful quotes of famous physicists out there. I'm sure almost everyone has a favourite. So, the question, quite simply, is this: &lt;strong&gt;what is your favourite famous-physicist quote, and why??&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB/ For another couple of very interesting &lt;em&gt;list&lt;/em&gt;-type questions see: &lt;a href=&quot;http://physics.stackexchange.com/questions/1019/common-false-beliefs-in-physics&quot;&gt;Common false beliefs in Physics&lt;/a&gt; and/or &lt;a href=&quot;http://physics.stackexchange.com/questions/1453/great-unsolved-physics-problems&quot;&gt;Great unsolved physics problems&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="1696" LastEditorDisplayName="" LastEditDate="2011-05-18T14:50:37.733" LastActivityDate="2011-05-18T15:43:25.973" ClosedDate="2011-05-18T15:58:53.977" Title="Famous physicists' quotations" Tags="&lt;soft-question&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="10012" PostTypeId="1" CreationDate="2011-05-18T14:51:14.817" Score="3" ViewCount="91" Body="&lt;p&gt;The cosmological constant is 10&lt;sup&gt;-120&lt;/sup&gt; times its natural value, but it is yet nonzero. Even TeV-scale supersymmetry breaking can't save it. The renormalization group would seem to imply it ought to be at the Planck scale, assuming naturalness, and no fine-tuning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The anthropic principle explains the size of the cosmological constant nicely, but are there any other explanations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dynamical mechanisms run into problems with the renormalization group. The mechanism has to act in the infrared, but any ultraviolet definition will be modified by the renormalization group.&lt;/p&gt;&#xA;" OwnerUserId="3593" LastActivityDate="2011-05-18T16:22:31.960" Title="What are the viable non-anthropic explanations for the tininess of the cosmological constant?" Tags="&lt;cosmology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10013" PostTypeId="2" ParentId="10011" CreationDate="2011-05-18T15:34:23.937" Score="-2" ViewCount="" Body="&lt;p&gt;The most important (for a forum like this) quote from &#xA;Wittgenstein is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alles, was gefragt (gesagt) werden kann, kann klar gefragt (gesagt) werden!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;:=)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS Look for &lt;strong&gt;Lichtenberg&lt;/strong&gt;, he is more famous for &#xA;his aphorisms than for his figures today. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-05-18T15:34:23.937" />
  <row Id="10014" PostTypeId="2" ParentId="10011" CreationDate="2011-05-18T15:43:25.973" Score="2" ViewCount="" Body="&lt;p&gt;I'm not sure if I may post my top 5 quotes, but surely my number one would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Prediction is very difficult, especially about the future”. (Niels Bohr)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I was allowed to add the other four, even though they are not all that insightful, I would say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Physics is like sex. Sure, it may give some practical results, but that's not why we do it”. (Richard Feynman)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“For a successful technology, reality must take precedence over public relations, for Nature cannot be fooled.&quot; (Richard Feynman)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“The scientist does not study nature because it is useful to do so. He studies it because he takes pleasure in it, and he takes pleasure in it because it is beautiful. If nature were not beautiful it would not be worth knowing, and life would not be worth living”. (Henri Poincaré)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“It is not even wrong&quot;. (Wolfang Pauli, about a paper of a young physicist whose theory could not be falsified).&lt;/p&gt;&#xA;" OwnerUserId="3586" LastActivityDate="2011-05-18T15:43:25.973" />
  <row Id="10015" PostTypeId="1" AcceptedAnswerId="10017" CreationDate="2011-05-18T16:21:32.213" Score="3" ViewCount="81" Body="&lt;p&gt;Looking at pictures like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/DzlrB.jpg&quot; alt=&quot;enter image description here&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I bet it is a picture of a giant cloud of dust, bigger than a galaxy and made of asteroids and planets (instead of stars).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But probably I'm wrong. What is that?&lt;/p&gt;&#xA;" OwnerUserId="2868" LastActivityDate="2011-05-19T10:25:50.807" Title="What really are some distant astronomical bodies that seems like dust clouds?" Tags="&lt;astronomy&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10016" PostTypeId="2" ParentId="10012" CreationDate="2011-05-18T16:22:31.960" Score="2" ViewCount="" Body="&lt;p&gt;At this moment, there are no widely known explanations that are as &quot;ready to be used&quot; as the anthropic would-be explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are of course vague ideas how the problem could be solved but no full framework how to incorporate particle physics and predict a tiny C.C. has been proposed. The partial proposals include:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The supersymmetric cancellations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Low-scale supersymmetry reduces the natural prediction of the C.C. from $O(1)$ to $O(10^{-60})$ in Planck units which seems progress but it's still very far from the required $O(10^{-123})$. Despite the fact that SUSY seems to &quot;solve&quot; one half of the problem on the log scale, in some sense, it makes the problem qualitatively worse because one can be &quot;more certain&quot; than the $O(10^{-60})$ piece which is still way too large survives. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No-scale supergravity is a framework to guarantee that the leading term to the cosmological constant cancels. In general, spontaneously broken supersymmetry hasn't been able to explain the smallness of the C.C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mirror matter&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various highly speculative articles suggesting that there exists a &quot;mirror world&quot; whose zero-point energies cancel against those in our world. Such worlds are probably unstable and suffer from many other problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Link to the neutrino masses&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The C.C. expressed as a vacuum energy density is comparable to the fourth power of the neutrino masses and the neutrinos are the lightest massive particles we know. It could be that there is something wrong about the ways how we apply - or fail to apply - the Renormalization Group on the running of the C.C. It may be that there is some mechanism that makes a theory inconsistent if the C.C. at the neutrino scale - which is also the C.C. at zero energies - is much larger than the neutrino scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Seesaw mechanisms&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It just happens that the C.C. is approximately $m_{TeV}^8 / m_{Planck}^4$, which is linked to the formula for the neutrino masses. So on the log scale, the C.C. is on the opposite side from the natural TeV-scale energy density (e.g. predicted by broken SUSY) than the Planck density on the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exist highly speculative proposals how to derive this seesaw formula from a deeper mechanism, for example, from some mixing between different vacua, tunneling, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Others and appraisal&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There also exist various papers on self-adjusting cosmological constants that don't work, as far as I know. If one looks at what is known, it's fair to say that the anthropic explanation of the C.C. is the only available game in town. However, in this particular case, it's totally plausible that a much better explanation will be found in the future. In this case, there's no real evidence that all the alternative explanations have been mapped or ruled out.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-18T16:22:31.960" />
  <row Id="10017" PostTypeId="2" ParentId="10015" CreationDate="2011-05-18T16:24:55.983" Score="4" ViewCount="" Body="&lt;p&gt;It's the Eagle Nebula as pictured by the Hubble Space Telescope, also known as the &quot;Pillars of Creation&quot; picture.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://images.google.cz/images?q=eagle-nebula&amp;amp;biw=1063&amp;amp;bih=802&quot; rel=&quot;nofollow&quot;&gt;http://images.google.cz/images?q=eagle-nebula&amp;amp;biw=1063&amp;amp;bih=802&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Eagle Nebula is a young cluster of stars in the Constellation Serpens that began to clump just 5.5 million years ago:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Eagle_Nebula&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Eagle_Nebula&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The distance is 7,000 light years, so it's in the Milky Way (whose diameter is about 100,000 light years), of course. So yes, there is gas and dust but it's going to collapse into stars at some point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of the nebula is just 70 x 55 light years or so. The linear dimensions are about 10 times larger than the distance between the Sun and another closest star so such a volume of space could normally be expected to contain $O(1,000)$ stars or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted the same apparent object to have distance e.g. 7 million instead of 7 thousand light years, which would be approximately needed for your &quot;galactic&quot; interpretation, the cloud would have to contain a million times more matter or so - to send the right number of photons here. Such an object would be comparably heavy as a galaxy but it would also move away from us and produce a detectable redshift.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-18T16:38:26.580" LastActivityDate="2011-05-18T16:38:26.580" CommentCount="4" />
  <row Id="10018" PostTypeId="1" CreationDate="2011-05-18T16:41:40.330" Score="-2" ViewCount="135" Body="&lt;p&gt;Got a new idea regarding the unification of light wave and sound wave into a single entity ,may be probably a wave.is it possible to proceed in that manner,and has it been done previously,like i have an idea of projecting.The light wave and sound wave and synchronizing their speeds,and let them meet at a point,forming interference, or merging at a point will it create a new wave.&#xA;If it have been implemented in past can i know the way of unification of sound and light and its applications?&lt;/p&gt;&#xA;" OwnerUserId="3590" LastEditorUserId="3590" LastEditorDisplayName="" LastEditDate="2011-05-19T02:43:41.107" LastActivityDate="2011-05-19T02:43:41.107" ClosedDate="2011-05-18T17:44:37.387" Title="A new idea on unifying sound and light" Tags="&lt;light&gt;&lt;sound&gt;&lt;general-physics&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="10019" PostTypeId="1" CreationDate="2011-05-18T17:00:45.790" Score="2" ViewCount="148" Body="&lt;p&gt;Sound waves are normally mechanical wave i want to know whether the sound wave bends if it bends does it undergo diffraction just like light wave?&lt;/p&gt;&#xA;" OwnerUserId="3590" LastEditorUserId="2843" LastEditorDisplayName="" LastEditDate="2011-05-18T17:38:20.463" LastActivityDate="2011-05-18T18:50:03.923" Title="Do sound waves bend and/or diffract?" Tags="&lt;sound&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="10020" PostTypeId="2" ParentId="10018" CreationDate="2011-05-18T17:24:36.820" Score="2" ViewCount="" Body="&lt;p&gt;Sound and light are inherently different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sound is produced in a form similar to that of a 1 dimensional wave: air is compressed and then stretched. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light is of an inherently different nature: electricity and magnetism interact in a fashion that propagates their energy through space in the the form of light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light and sound do not interfere with one another. Why? A molecule of gas (that is an element of a sound wave) may absorb photons of certain wavelengths. The atom has certain electromagnetic properties that allow it to interact with and absorb light. Sound (the mere changes in air density) lacks the ability to interact with the electromagnetic essence of light waves. Apples and Oranges. &lt;/p&gt;&#xA;" OwnerUserId="2843" LastActivityDate="2011-05-18T17:24:36.820" CommentCount="6" />
  <row Id="10021" PostTypeId="1" CreationDate="2011-05-18T17:25:30.797" Score="9" ViewCount="448" Body="&lt;p&gt;I'm aware of Sidney Coleman's 1975/76 sequence of &lt;a href=&quot;http://www.physics.harvard.edu/about/Phys253.html&quot; rel=&quot;nofollow&quot;&gt;54 lectures&lt;/a&gt; on Quantum Field Theory. Are there any other high-quality QFT lecture series available online?&lt;/p&gt;&#xA;" OwnerUserId="1394" LastActivityDate="2011-05-19T09:02:07.540" CommunityOwnedDate="2011-05-19T15:52:07.370" Title="Online QFT video lectures" Tags="&lt;quantum-field-theory&gt;&lt;lecture&gt;" AnswerCount="4" FavoriteCount="5" />
  <row Id="10022" PostTypeId="2" ParentId="10019" CreationDate="2011-05-18T17:27:47.520" Score="5" ViewCount="" Body="&lt;p&gt;Sure sound waves diffract. That's why two people can have a conversation, even if there's an object in between them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I teach this stuff, I often walk out of the classroom, to a place where the students can't see me, just to show that they can still hear me. That illustrates that sound waves diffract, and moreover that they diffract much &lt;em&gt;more&lt;/em&gt; than light waves under these circumstances. That's because the wavelength of sound waves is comparable to the other length scales in the situation, whereas the wavelength of light waves is much shorter.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-05-18T17:27:47.520" CommentCount="1" />
  <row Id="10023" PostTypeId="2" ParentId="10019" CreationDate="2011-05-18T17:29:58.110" Score="2" ViewCount="" Body="&lt;p&gt;Sound waves do, indeed, bend.  In an underwater thermal gradient the sonar waves can bend enough to create blind spots, so that you can not &quot;see&quot; a target that would ordinarily be well within sonar range.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-05-18T17:29:58.110" CommentCount="1" />
  <row Id="10024" PostTypeId="1" CreationDate="2011-05-18T18:13:30.447" Score="2" ViewCount="131" Body="&lt;p&gt;I currently make my living as an electrophysicist, so am a bit embarassed at having not thought this one through, before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnetically, the North pole is.... North.  We get out our compass, and wait for the North-seeking needle to line up.  I've always thought of this end of the needle as the north end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a current carrying loop, if we use the right-hand rule and point our thumb in the direction of conventional current, the fingers show the direction of the mag field lines around the wire.  The direction the mag field lines point &quot;north&quot; when they pass through the interior of the loop.  By stacking many such loops we can create a cylindrical electromagnet, for which it is easy to see which end is the &quot;north&quot; end.  I resume permanent physical magnets follow the same naming convention: that is, mag field lines emanate from the north end and cycle around to the south.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that like poles of a magnet repel and opposites attract.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, now, hanging a magnet from a string (or going back and using our trusty compass), if the north pole of the magnet points towards teh magnetic north pole of Earth, doesn't that mean that the pole near the Arctic circle is really magnetically-speaking the south pole?  Or conversely, when we say &quot;north-seeking&quot; pole of a magnet, is that really the south pole of the magnet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is it?&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-05-18T18:26:24.603" Title="Figuring out North and South Magnetic Poles for Earth" Tags="&lt;electromagnetism&gt;" AnswerCount="4" />
  <row Id="10025" PostTypeId="2" ParentId="10024" CreationDate="2011-05-18T18:19:03.000" Score="1" ViewCount="" Body="&lt;p&gt;In the context of magnets, &quot;north&quot; actually means &quot;north-seeking,&quot; because the names were chosen before people knew much about magnetism. So yes, the magnetic pole of the Earth which is geographically north does in fact have &quot;south&quot; polarity (it's a sink of magnetic field lines). &lt;a href=&quot;http://en.wikipedia.org/wiki/Earth%27s_magnetic_field&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; explains this in some detail.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-05-18T18:19:03.000" CommentCount="4" />
  <row Id="10026" PostTypeId="2" ParentId="10021" CreationDate="2011-05-18T18:21:01.517" Score="5" ViewCount="" Body="&lt;p&gt;The Perimeter Institute has recorded all the courses of their one-year Master's in Theoretical Physics. The courses are given by professors and lecturers from different universities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a full list of the courses they recorded in 2010/w011 go to &lt;a href=&quot;http://pirsa.org/&quot; rel=&quot;nofollow&quot;&gt;Pirsa.org&lt;/a&gt;, and under &quot;4. Collections&quot; select &quot;Course&quot; and &quot;2010&quot;. The courses are labeled by the tag &quot;10/11 PSI&quot;. (I couldnt find a decent link which neatly lists all the courses)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll find multiple courses on QM, QFT, string theory, Beyond the standard Model, General Relativity, condensed matter physics, Quantum Information and many more. The format is also very neat. The Flash presentation is a recording of the class in low-quality, combined with decent photos of the blackboard. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some direct links:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pirsa.org/C10030&quot; rel=&quot;nofollow&quot;&gt;QFT part I -- by K. Zarembo&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pirsa.org/C10033&quot; rel=&quot;nofollow&quot;&gt;QFT part II --  by F. David&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also check out the courses given in 09/11 (search for &quot;2009&quot; instead), where the curriculum was a bit different (also different lecturers). You have for instance David Tong from Cambridge who gives an introductary course on QFT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pirsa.org/C09033&quot; rel=&quot;nofollow&quot;&gt;QFT part I -- by D. Tong&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(part II was also given by F. David)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The site also has a large collection of recorded seminars. &lt;/p&gt;&#xA;" OwnerUserId="1287" LastActivityDate="2011-05-18T18:21:01.517" CommunityOwnedDate="2011-05-19T15:52:07.370" CommentCount="2" />
  <row Id="10027" PostTypeId="1" AcceptedAnswerId="10034" CreationDate="2011-05-18T18:21:33.787" Score="1" ViewCount="73" Body="&lt;p&gt;If I have a strong vessel that I've cooled to say -20°C, and lowered the pressure below  0.006 atmospheres, the water in food placed inside will sublimate and freeze dry the food.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what happens to the water vapor? Does the moisture rise to the top or fall to the bottom of the vessel itself? Or does it just expand to fill the container equally?&lt;/p&gt;&#xA;" OwnerUserId="3594" LastActivityDate="2011-05-18T18:54:09.257" Title="In a low pressure (near vacuum), -20°C vessel, would sublimated moisture rise to the top or fall to the bottom?" Tags="&lt;pressure&gt;&lt;vacuum&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10028" PostTypeId="2" ParentId="10024" CreationDate="2011-05-18T18:22:51.367" Score="2" ViewCount="" Body="&lt;p&gt;The problem is that the terms &quot;north pole&quot; and &quot;south pole&quot; (and even &quot;magnetic north&quot; and &quot;magnetic south&quot;) refer not to any magnetic polarities, per se, but to locations. I'm sure that you are aware that there is a good deal of evidence that the magnetic polarity of the Earth reverses itself every now and then. When it does that, the &quot;north pole&quot; doesn't change its name. North, as a cardinal direction, would still be toward the Arctic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnetically speaking, though, you answered your own question. Indeed, the Earth's north pole is south-seeking. This question has been &lt;a href=&quot;http://wiki.answers.com/Q/Why_is_the_earth%27s_magnetic_north_pole_really_a_magnetic_south_pole&quot; rel=&quot;nofollow&quot;&gt;asked before&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3395" LastActivityDate="2011-05-18T18:22:51.367" />
  <row Id="10029" PostTypeId="2" ParentId="10024" CreationDate="2011-05-18T18:23:50.180" Score="2" ViewCount="" Body="&lt;p&gt;Yes, north was defined as a geographic location long before magnets were studied in detail. Face the sun rise in the north hemisphere  on the 21st of March, your left hand points north, your right south.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is the convention: a north pole is the pole of a magnet pointing to geographic north, and and ditto for south. It was the geography that interested the first compass users, not the physics. So magnetically earth's geographic north pole is a magnetic south pole :), as you pointed out.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-18T18:23:50.180" />
  <row Id="10030" PostTypeId="2" ParentId="10024" CreationDate="2011-05-18T18:26:24.603" Score="2" ViewCount="" Body="&lt;p&gt;You are correct in saying that the pole near the Arctic circle is really the south pole of the Earth's magnetic field because the north pole of a bar-magnet points towards it. (&lt;a href=&quot;http://en.wikipedia.org/wiki/Earth%27s_magnetic_field&quot; rel=&quot;nofollow&quot;&gt;Wiki&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, it is purely convention what you define on a bar magnet to be north and south - we've named the side that points towards the Arctic circle to be the north pole of the bar magnet, and since people don't immediately identify this seeking behavior as opposite poles attracting, we tend to think the earth's field at the Arctic circle is also a north pole when it is strictly speaking, a south magnetic pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I would add that the magnetic poles don't coincide with the geographic poles - the latter define the points of intersection of the Earth's rotational axis with the surface of the Earth. We know for a fact that the magnetic poles have shifted (even reversed) with respect to the geographic poles during the history of the Earth. (&lt;a href=&quot;http://en.wikipedia.org/wiki/Geomagnetic_reversal&quot; rel=&quot;nofollow&quot;&gt;Wiki&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-05-18T18:26:24.603" />
  <row Id="10031" PostTypeId="2" ParentId="9983" CreationDate="2011-05-18T18:28:21.697" Score="0" ViewCount="" Body="&lt;p&gt;The answer to this question is likely no, because the &lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric&quot; rel=&quot;nofollow&quot;&gt;Robertson-Walker spacetime&lt;/a&gt; with a radiation equation of state satisfies Einstein's equation for that stress energy tensor, and, famously, it only has a static solution in the case of the &lt;a href=&quot;http://www.google.com/search?sourceid=chrome&amp;amp;ie=UTF-8&amp;amp;q=einstein+static+universe&quot; rel=&quot;nofollow&quot;&gt;Einstein Static Universe&lt;/a&gt;, where the cosmological constant balances out the gravitational attraction of the expanding universe.  The staticity of the Einstein Static universe is known to be unstable under the influence small perturbations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only reason I say 'likely no' and not no is that it may be possible to save a solution like this by relaxing the requirement that space be homogenous to just spherical symmetry, but I'm inclined to say that it's unlikely that this will end up working out.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-05-18T18:28:21.697" />
  <row Id="10032" PostTypeId="1" AcceptedAnswerId="10037" CreationDate="2011-05-18T18:29:54.520" Score="1" ViewCount="110" Body="&lt;p&gt;Is there a reason that magnetic field lines are drawn from north to south or is this a purely arbitrarily defined protocol (for convenience)? Essentially what I am asking is: without a reference magnet (like the earth) is there a way determining what pole of a magnet is north and which is south? &lt;/p&gt;&#xA;" OwnerUserId="2843" LastActivityDate="2011-05-18T18:51:35.700" Title="What is the difference between north and south magnetically?" Tags="&lt;electromagnetism&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10033" PostTypeId="2" ParentId="10032" CreationDate="2011-05-18T18:32:19.843" Score="0" ViewCount="" Body="&lt;p&gt;The direction of the field lines is pure convention. Magnetic field lines are defined with respect to north magnetic &quot;test charges&quot; placed in the field (and how they move in the field). If they were defined with respect to test south magnetic test charges, the field lines would be reversed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as Lagerbaer points out, it is of course possible to distinguish between the two poles without using magnets - that is partly the point of &quot;electromagnetism&quot; :)&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-05-18T18:51:35.700" LastActivityDate="2011-05-18T18:51:35.700" />
  <row Id="10034" PostTypeId="2" ParentId="10027" CreationDate="2011-05-18T18:36:37.300" Score="2" ViewCount="" Body="&lt;p&gt;I do believe that freeze-dryers have to have a condenser that actively removes the water vapor.  It would depend on the size of the vessel, but the water vapor from the food is likely to be enough to raise the pressure to near atmospheric, and certainly enough to destroy the low pressure required.  So a condenser is actively removing the moisture by exposing a surface at a much lower temperature than the atmosphere.  But come to think of it, I'm not quite sure how this works considering that the condensed water would be ice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Directly to your question, I am fairly positive it fills the entire vessel.  It is, after all, a gas.  There is little in the way of other competing gases and there are negligible effects from things like gravity.  I would imagine that the gas would be a near homogeneous mixture of the water vapor and other sparse gases.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-05-18T18:36:37.300" CommentCount="2" />
  <row Id="10036" PostTypeId="2" ParentId="10002" CreationDate="2011-05-18T18:46:58.210" Score="0" ViewCount="" Body="&lt;p&gt;Well, I will try some simplification as I assume you are a student and have not yet started a serious course on the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The word &quot;quantum&quot; means &quot;a specific quantity&quot;. The verb &quot;quantized&quot; means that some physically measurable  quantities, as energy, angular momentum, and others, have been observed to come in quanta rather than continuously. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Easy examples are colour spectral lines in the spectrum of excited atoms, which instead of showing a continuous frequency spectrum have sharp peaks. These peaks indicate that at that frequency excessively  many atoms change energy levels by emitting a quantum of energy which is observed as &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectral_line&quot; rel=&quot;nofollow&quot;&gt;that line in the spectrum,&lt;/a&gt; way over the background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum mechanics theory was created to explain and predict quantized behaviours in physical observables at the atomic level and below, but also in super fluids and super conductors which have macroscopic extent but quantized behaviours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it is a theory which deals with probabilities and needs a mathematical background to comprehend.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-06-21T03:47:20.707" LastActivityDate="2011-06-21T03:47:20.707" CommentCount="6" />
  <row Id="10037" PostTypeId="2" ParentId="10032" CreationDate="2011-05-18T18:49:24.843" Score="5" ViewCount="" Body="&lt;p&gt;Lagerbaer's comment is correct: you can tell which way the magnetic field vectors point (i.e., which end of your bar magnet is &quot;north&quot;) using a moving charge rather than a reference magnet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's worth adding one more point: the original decision to draw magnetic field lines going away from &quot;north&quot; poles of magnets and towards &quot;south&quot; poles was an arbitrary one. We could have decided to draw all magnetic field lines the other way around, in which case the direction of the Lorentz force would be $-q({\bf v}\times{\bf B})$ instead of $q({\bf v}\times{\bf B})$. We'd either replace the right-hand rule with a left-hand rule, or more likely just reverse the order of ${\bf v}$ and ${\bf B}$ in the cross product, and keep using the same right-hand rule but with ${\bf v},{\bf B}$ interchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, we could also have decided to draw all electric fields the other way around too, by flipping the signs of all electric charges. Arguably some things would be simpler if the electron charge had originally been taken to be positive -- current would flow in metals in the same direction that the charges were moving, for instance -- but it's too late now. &lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-05-18T18:49:24.843" CommentCount="4" />
  <row Id="10038" PostTypeId="2" ParentId="10019" CreationDate="2011-05-18T18:50:03.923" Score="0" ViewCount="" Body="&lt;p&gt;The other example is the difference in how well sound carries up/down wind. Wind velocity increases with height (at least in the near surface boundary layer), this shear means that soundwaves traveling upwind are diffracted upwrds, but waves travelling downwind are diffracted downward, so that in the downwind direction the sound can travel in a channel near the ground.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-05-18T18:50:03.923" />
  <row Id="10039" PostTypeId="1" AcceptedAnswerId="10047" CreationDate="2011-05-18T20:34:02.710" Score="4" ViewCount="216" Body="&lt;p&gt;Does anyone know how to derive the following identity for the integral of the product of three spherical harmonics?:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_0^{2\pi}\int_0^\pi Y_{l_1}^{m_1}(\theta,\phi)Y_{l_2}^{m_2}(\theta,\phi)Y_{l_3}^{m_3}(\theta,\phi)\sin(\theta)d\theta d\phi =$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{\frac{(2l_1+1)(2l_2+1)(2l_3+1)}{4\pi}} \left( {\begin{array}{ccc} l_1 &amp;amp; l_2 &amp;amp; l_3  \\ 0 &amp;amp; 0 &amp;amp; 0  \\ \end{array} } \right) \left( {\begin{array}{ccc} l_1 &amp;amp; l_2 &amp;amp; l_3  \\ m_1 &amp;amp; m_2 &amp;amp; m_3  \\ \end{array} } \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the $Y_{l}^{m}(\theta,\phi)$ are spherical harmonics.  Or does anyone know of a reference where the derivation is given?&lt;/p&gt;&#xA;" OwnerUserId="3597" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-05-18T20:40:13.837" LastActivityDate="2011-05-19T02:03:03.630" Title="Integral of the product of three spherical harmonics" Tags="&lt;mathematical-physics&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="10040" PostTypeId="1" CreationDate="2011-05-18T21:14:35.967" Score="4" ViewCount="165" Body="&lt;p&gt;Wired magazine ran an &lt;a href=&quot;http://www.wired.com/magazine/2011/04/st_equation_sodapop/&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; this month on carbonation in soft drinks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all soft drinks are manufactured effectively identically, why do some types fizz more than others?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, root beer is always extremely fizzy (and &lt;a href=&quot;http://billybrew.com/beer-lacing&quot; rel=&quot;nofollow&quot;&gt;laces&lt;/a&gt; well).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In similar fashion, lemon-lime drinks like Sprite and Sierra Mist are very fizzy - but do not lace like root beer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare those to diet colas, which fizz slowly and/or minimally, and one is left to ponder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is it about some types of drinks that make them hold and release their carbonation differently?&lt;/p&gt;&#xA;" OwnerUserId="2021" LastEditorUserId="2021" LastEditorDisplayName="" LastEditDate="2011-06-30T02:41:07.187" LastActivityDate="2011-06-30T02:41:07.187" Title="Why do some carbonated drinks fizz more than others?" Tags="&lt;fluid-dynamics&gt;&lt;pressure&gt;" CommentCount="5" FavoriteCount="1" />
  <row Id="10041" PostTypeId="1" CreationDate="2011-05-18T21:25:39.837" Score="-2" ViewCount="164" Body="&lt;p&gt;Is it possible to write down step-by-step instructions for inventing a new theory? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been wondering if there exists some 'recipe' or proceedure for inventing a new theory. Presumably some theorists must have different approaches to others but are there some general pointers that would always apply? Realistically, I should point out, I'm not anticipating stumbling upon anything more significant than an interesting little toy model but it would certainly be good practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 1&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of QFT, I'm familiar with Lagrangian-mechanics, extremization-of-action and the Euler-Lagrange equations. I've also got a vague appreciation of what renormalization and gauge invariance imply but I don't really know how one would go about testing either condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I had to guess I'd say, within the construct of QFT at least, that it would proceed like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1/&lt;/strong&gt; add a new term to the Lagrangian,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2/&lt;/strong&gt; check if it breaks the gauge or Lorentz invariance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3/&lt;/strong&gt; fiddle to see what new properties have manifested themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this roughly how one would proceed? Futhermore, if I'm embarrissingly honest, I'm not really sure exactly &lt;em&gt;how&lt;/em&gt; to do any of these things. If anyone could offer some elucidation I'll gladly split the Nobel Prize money with you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, if one wanted their theory to be a little more revolutionary (ie quantum gravity of course), what could we say about possible approaches to finding such a theory? Are there some similarities between the processes-of-invention that occurred for GR, QFT, String-theory, etc, that we can draw on to serve as a guide for inventing new theories? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, GR required differential geometry and tensor calculus. Whilst these areas of mathematics existed pre-Einstein, it took a man of his extra-ordinary genius to use them to invent a completely new perspective of gravity that facilitated a model that does indeed describe nature. Here it seems his initial gut-feeling was that, since inertial motion was preferred in SR, there should exist a theory which prefers no state of motion, from the outset. It seems his 'hunch', when applied to gravity led him straight to what later became the equivalence principle. Oh, what I would do for a 'hunch' like that!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The invention of string theory (as a potential TOE), originally, appears to have been an accident. Veneziano, followed closely by Nambu, Nielsen and Susskind noticed that nuclear interactions of point-particles viewed instead as 1D stings could be modelled by the Euler-Beta function. Then followed a number of couple 'revolutions' and fantastic number of incredible breakthoughs leading to what is now, collectively, refered to as M-theory (thanks to Witten.) In this case it appears the starting point was Veneziano noticing that certain properties of the strong force resembled something he had used in another area of physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please forgive my brutally selective account of these histories but I do so in an effort to explain my query. I've only (very briefly) mentioned the invention of GR and string-theory here as I know less-still about the stories of QM, QFT, EM, LQG,... I imagine almost every physicist has once dreamed that they might be the next Newton/Einstein/Dirac/Feynman etc.. I know there are some very learned people that frequent this site and I'd really love to hear you thoughts on this matter.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="61" LastEditorDisplayName="" LastEditDate="2011-05-19T00:38:06.760" LastActivityDate="2011-05-19T00:38:06.760" ClosedDate="2011-05-19T17:19:58.283" Title="How to invent a theory?" Tags="&lt;mathematical-physics&gt;&lt;soft-question&gt;&lt;theory&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="0" />
  <row Id="10042" PostTypeId="2" ParentId="10021" CreationDate="2011-05-18T21:36:10.763" Score="2" ViewCount="" Body="&lt;p&gt;... and from Lenny Susskind too there is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://itunes.apple.com/us/itunes-u/supersymmetry-grand-unification/id384233338&quot; rel=&quot;nofollow&quot;&gt;http://itunes.apple.com/us/itunes-u/supersymmetry-grand-unification/id384233338&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/view_play_list?p=202191442DB1B300&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/view_play_list?p=202191442DB1B300&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2751" LastActivityDate="2011-05-18T21:36:10.763" CommunityOwnedDate="2011-05-19T15:52:07.370" />
  <row Id="10043" PostTypeId="2" ParentId="10041" CreationDate="2011-05-18T21:51:52.087" Score="-1" ViewCount="" Body="&lt;p&gt;A theory is a systematization and generalization of some experimental data, first of all. If it is not, then it is a mathematical exercise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are still problems in classical subjects like electrodynamics of a single charge. The energy-momentum is not conserved in the current version of CED.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-05-18T22:04:06.090" LastActivityDate="2011-05-18T22:04:06.090" CommentCount="1" />
  <row Id="10044" PostTypeId="1" CreationDate="2011-05-18T22:23:14.950" Score="4" ViewCount="93" Body="&lt;p&gt;More specifically, in quantum field theory books, we usually have this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;Z = \int D(\bar{\psi}, \psi) e^{-S + \int_0^\beta d\tau \sum_l [\bar{\eta}_l (\tau) \psi_l (\tau) + \bar{\psi}_l (\tau) \eta_l (\tau) ]}  \hspace{10mm} (1)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S = \int_0^\beta d\tau \sum_l (\bar{\psi}_l (\tau) \partial_\tau \psi_l (\tau ) +H(\bar{\psi},\psi))$ is the action, and $H$ is the Hamiltonian of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I instead start off with the Hamiltonian in diagonal form and include the source term in, say, a momentum basis? So that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;Z = \int D(\bar{\psi}, \psi) e^{-S + \int_0^\beta d\tau \sum_k [\bar{\eta}_k (\tau) \psi_k (\tau) + \bar{\psi}_k (\tau) \eta_k (\tau) ]}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S$ is now $\int_0^\beta d\tau \sum_k (\bar{\psi}_k (\tau) [\partial_\tau +\epsilon_k]\psi_k (\tau ))$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or do I have to start off with equation (1) and apply various transformations to the Hamiltonian and the rest of the exponent until $H$ it is in a diagonal basis? - If this is the case, the resulting source term in the momentum basis could be extremely complicated after the transformations are performed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! &lt;/p&gt;&#xA;" OwnerUserId="873" LastActivityDate="2011-05-18T22:23:14.950" Title="When can the source term of a partition function be put in?" Tags="&lt;quantum-field-theory&gt;&lt;path-integral&gt;" CommentCount="4" FavoriteCount="4" />
  <row Id="10045" PostTypeId="2" ParentId="10039" CreationDate="2011-05-18T23:04:20.267" Score="0" ViewCount="" Body="&lt;p&gt;Richard Zare's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471858927&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; has a derivation, or at least all the pieces of one. Section 3.9 has the key results, although you have to refer back to previous sections to complete the derivation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if you could also do it by induction, using the recurrence relations for the spherical harmonics on the left and for the 3-$j$'s on the right. I've never seen anyone do anything like this, maybe it doesn't work.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-05-18T23:04:20.267" />
  <row Id="10046" PostTypeId="2" ParentId="10021" CreationDate="2011-05-19T01:25:03.647" Score="1" ViewCount="" Body="&lt;p&gt;For the beginner, there is a great series of lectures by Anthony Zee, author of &lt;em&gt;Field Theory in a Nutshell&lt;/em&gt;, available from ASTI here:&lt;a href=&quot;http://www.asti.ac.za/lectures.php&quot; rel=&quot;nofollow&quot;&gt;http://www.asti.ac.za/lectures.php&lt;/a&gt;. Some of the other lectures are quite good as well. &lt;/p&gt;&#xA;" OwnerUserId="3601" LastActivityDate="2011-05-19T01:25:03.647" CommunityOwnedDate="2011-05-19T15:52:07.370" />
  <row Id="10047" PostTypeId="2" ParentId="10039" CreationDate="2011-05-19T02:03:03.630" Score="2" ViewCount="" Body="&lt;p&gt;Sakurai, Modern Quantum Mechanics, 2nd Ed. p.216&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In his derivation the product of the first two spherical harmonics is expanded using the Clebsch-Gordan Series (which is also proved) to get the following equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y_{l_1}^{m_1}(\theta,\phi)Y_{l_2}^{m_2}(\theta,\phi)(\theta,\phi)\ =$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\sum\limits_{l} \displaystyle\sum\limits_{m} \sqrt{\frac{(2l_1+1)(2l_2+1)(2l+1)}{4\pi}} \left( {\begin{array}{ccc} l_1 &amp;amp; l_2 &amp;amp; l  \\ 0 &amp;amp; 0 &amp;amp; 0  \\ \end{array} } \right) \left( {\begin{array}{ccc} l_1 &amp;amp; l_2 &amp;amp; l  \\ m_1 &amp;amp; m_2 &amp;amp; -m  \\ \end{array} } \right)(-1)^m Y_{l}^{m}(\theta,\phi)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which makes the integral much easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Final Note: Sakurai writes his derivation in Clebsch-Gordan coefficients so the equation was changed to fit with the question asked.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="800" LastActivityDate="2011-05-19T02:03:03.630" />
  <row Id="10049" PostTypeId="1" AcceptedAnswerId="10056" CreationDate="2011-05-19T02:47:55.177" Score="1" ViewCount="93" Body="&lt;p&gt;what is the ram-facing side of  a satellite? What does it mean and why is it called &quot;ram&quot;-facing? &#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3604" LastActivityDate="2011-05-19T03:56:03.547" Title="What is the ram-facing side of spacecraft?" Tags="&lt;astrophysics&gt;" AnswerCount="1" />
  <row Id="10050" PostTypeId="1" CreationDate="2011-05-19T02:58:20.860" Score="0" ViewCount="82" Body="&lt;p&gt;When we pass current to the permanent magnet does its magnetic flux, magnetic property of attraction and repulsion increases? if possible&#xA;Is there any other way for increasing the magnetic flux?&lt;/p&gt;&#xA;" OwnerUserId="3590" LastActivityDate="2011-05-19T02:58:20.860" Title="Can we increase the magnetic flux for a permanent magnet?" Tags="&lt;electromagnetism&gt;" />
  <row Id="10051" PostTypeId="1" CreationDate="2011-05-19T03:17:03.013" Score="2" ViewCount="93" Body="&lt;p&gt;When we are cooling the permanent magnet what property of magnet changes?&#xA;normally the superconductivity of magnets occurs at critical temperature(negative degrees).It is having a long life period and it exhibits huge magnetic field.&#xA;if we keep the magnets at some temperature which is nearer to critical temperature can we get any properties of that superconducting magnet&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is temperature inversely proportional to the magnetic field, magnetic flux density?&lt;/p&gt;&#xA;" OwnerUserId="3590" LastActivityDate="2011-05-19T18:05:36.730" Title="What will happen when we cool the permanent magnet?" Tags="&lt;electromagnetism&gt;&lt;superconductivity&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="10052" PostTypeId="1" AcceptedAnswerId="10054" CreationDate="2011-05-19T03:22:42.243" Score="0" ViewCount="84" Body="&lt;p&gt;I've read on NASA's page on neutron star that one teaspoonful of that star would weigh over 20 billion ton on Earth. If it was somehow possible to bring it to earth would it:&#xA;A) Burn and disappear during earth atmosphere entry?&#xA;B) Assuming we bave 20 billion ton of mass in size of a teaspoon here on earth, would it fall through the ground under its own weight?&lt;/p&gt;&#xA;" OwnerUserId="3605" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-05-19T03:39:48.377" LastActivityDate="2011-05-19T03:43:36.890" Title="Neutron star material" Tags="&lt;neutron-stars&gt;" AnswerCount="1" />
  <row Id="10053" PostTypeId="1" AcceptedAnswerId="10057" CreationDate="2011-05-19T03:37:36.747" Score="2" ViewCount="105" Body="&lt;p&gt;I am reading Introduction to Quantum Mechanics by David Griffiths and I am in Ch2 page 59. He starts out writing the time dependent Schrödinger equation and the solution for $\psi(x,t)$ for the free particle, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\psi(x,t) = A e^{ik(x-(\hbar k/2m)t)} + B e^{-ik(x + (\hbar k/2m)t)}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then he goes and says the following, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now, &lt;em&gt;any&lt;/em&gt; function of $x$ and $t$ that depends on these variables in the special combination $x \pm vt$ (for some constant $v$) represents a wave of fixed profile, traveling in the $\pm x$-direction, at speed $v$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What does this sentence mean?&lt;/p&gt;&#xA;" OwnerUserId="2578" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-19T03:42:21.527" LastActivityDate="2011-05-19T05:18:52.717" Title="Simple Quantum Mechanics question about the Free particle, (part1)" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="10054" PostTypeId="2" ParentId="10052" CreationDate="2011-05-19T03:38:10.857" Score="2" ViewCount="" Body="&lt;p&gt;The reason that the density is so high is because the pressures are so immense. If we somehow teleported a teaspoonful of neutron star material to earth, it would &lt;em&gt;very&lt;/em&gt; rapidly inflate because the pressures aren't high enough to crush it into its dense form. This would effectively be an enormous explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is difficult to describe what it would inflate out into - the neutron star material can be imagined as an incredibly dense soup of neutrons with some protons and leptons in small numbers. The protons and leptons would make neutron-rich elements like &lt;a href=&quot;http://en.wikipedia.org/wiki/Deuterium&quot; rel=&quot;nofollow&quot;&gt;deuterium&lt;/a&gt;, but most of the matter would consist of free neutrons. These free neutrons would undergo &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_decay&quot; rel=&quot;nofollow&quot;&gt;beta decay&lt;/a&gt; to produce neutrinos, protons, and electrons, which would likely recombine to make very large amounts of hydrogen, some helium, and a few heavier atoms. In all of these cases, the atoms would be neutron-rich isotopes, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The behavior would look most like a very rapidly expanding gas. It would explode with such force that it wouldn't even need to &quot;fall through the ground&quot; - it would obliterate the floor entirely.&lt;/p&gt;&#xA;" OwnerUserId="1655" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-05-19T03:43:36.890" LastActivityDate="2011-05-19T03:43:36.890" CommentCount="1" />
  <row Id="10055" PostTypeId="1" AcceptedAnswerId="10058" CreationDate="2011-05-19T03:47:42.993" Score="2" ViewCount="258" Body="&lt;p&gt;Continuing from my first question titled, S&lt;a href=&quot;http://physics.stackexchange.com/questions/10053/simple-quantum-mechanics-question-about-the-free-particle-part1&quot;&gt;imple Quantum Mechanics question about the Free particle, (part1)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Griffiths goes on and says,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;A fixed point on the waveform corresponds to a fixed value of the argument and hence to x and t such that,&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;x +- vt = constant, or x = -+ vt + constant&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What does this mean? I am so confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He goes on and says that psi(x,t) might as well be the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\psi(x,t) = A e^{i(kx-(hk^2/2m)t}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because the original equation,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\psi(x,t) = A e^{ik(x-(\hbar k/2m)t)} + B e^{-ik(x + (\hbar k/2m)t)}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;only differs by the &lt;em&gt;sign&lt;/em&gt; in front of k. He then says let k run negative to cover the case of waves traveling to the left, $k = \pm sqrt(2mE)/hbar$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then after trying to normalize psi(x,t), you find out you can't do it! He then says the following,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;A free particle cannot exist in a stationary state; or, to put it another way, &lt;em&gt;there is no such thing as a free particle with a definite energy.&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How did he come to this logical deduction. I don't follow. Can someone please explain Griffith's statement to me?&lt;/p&gt;&#xA;" OwnerUserId="2578" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-05-19T04:48:03.750" LastActivityDate="2011-08-28T22:55:36.507" Title="Simple Quantum Mechanics question about the Free particle, (part2)" Tags="&lt;quantum-mechanics&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="2" />
  <row Id="10056" PostTypeId="2" ParentId="10049" CreationDate="2011-05-19T03:56:03.547" Score="4" ViewCount="" Body="&lt;p&gt;The following picture is from a simulation for a satellite moving through the ionospheric plasma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/46sb5.png&quot; alt=&quot;ionospheric plasma&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;Ram&lt;/strong&gt; side is the side that points in the direction of the satellite's motion. It is called the ram side because it is the side impacting/ramming into the fluid that the satellite moves through. In this case it is the ionosphere. If the fluid's velocity is comparable to the satellite then it is a little more complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;Wake&lt;/strong&gt; side is exactly the opposite. It faces away from the the satellite's motion and thus does not impact the fluid as much. The satellite's motion creates a wake behind it like a boat through the water.&lt;/p&gt;&#xA;" OwnerUserId="800" LastActivityDate="2011-05-19T03:56:03.547" />
  <row Id="10057" PostTypeId="2" ParentId="10053" CreationDate="2011-05-19T04:58:10.043" Score="5" ViewCount="" Body="&lt;p&gt;It means there are many possible shapes for waves, not just pure sine waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi(x,t) = A\textrm{e}^{-k^2(x-vt)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a possible wavefunction.  It represents a Gaussian wave packet that travels down the x-axis in the positive direction at speed $v$.  The important part is that you can make the substitution $u = x-vt$ into $\psi$ and get a function of a single variable $u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, start with any function $f$ of a single variable $u$.  Now make the substitution $u = x - vt$.  $f$ has now become a wave that travels down the x-axis at speed $v$ with some funky shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mathematically-important thing is that such functions can be represented as a superposition of sinusoidals of continuously-varying frequencies all traveling in tandem down the x-axis (by &quot;traveling&quot; I mean &quot;have phase velocity&quot;).  The sinusoidals that go with a given $f$ are found through &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_analysis&quot; rel=&quot;nofollow&quot;&gt;fourier analysis&lt;/a&gt;. This is important because the sinusoidals are the eigenfunctions of the Hamiltonian for a free particle.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-05-19T05:18:52.717" LastActivityDate="2011-05-19T05:18:52.717" />
  <row Id="10058" PostTypeId="2" ParentId="10055" CreationDate="2011-05-19T05:17:27.460" Score="4" ViewCount="" Body="&lt;p&gt;The first part about velocities says that we're looking at a function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi(x,t) = \psi(u)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $u=x-vt$.  For example, $\psi = \cos(x-vt)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now pick some fixed value for $\psi$, say $0.4$.  Find a place where $\psi = 0.4$ such as $x=1.16$.  If you let some small amount of time $\textrm{d}t$ go by, then look at $x=1.16$ again, $\psi$ has changed a little, but if you look at the point $x = 1.16 + v\textrm{d}t$, you will find $\psi = 0.4$ there, so we could say that the place where $\psi$ is $0.4$ is moving at speed $v$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not the same as saying that the particle is moving at that speed.  It is saying that $v$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Phase_velocity&quot; rel=&quot;nofollow&quot;&gt;phase velocity&lt;/a&gt; of the wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part about normalizability says that a wavefunction must be an element in a vector space called a Hilbert space (by physicists; I think mathematicians call it $L_2$).  The Hilbert space consists of wavefunctions that are square-normalizable; you can square them, integrate from negative infinity to infinity, and get a finite value.  Things that die off exponentially on their tails do this, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sine function doesn't die off exponentially, or at all.  If you square it and integrate from negative infinity to infinity, you get something infinite.  Thus, the sine wave doesn't represent a reasonable probability density function for the location of a particle.  The particle would be &quot;infinitesimally likely to be observed everywhere in an infinite region&quot;, which physically does not make sense.  Instead, for a real particle, we must have a normalizable wavefunction.  Since a free particle with definite energy would have a pure sinusoidal wavefunction, a free particle with definite energy is physically not possible.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-05-19T05:17:27.460" CommentCount="5" />
  <row Id="10059" PostTypeId="1" AcceptedAnswerId="10067" CreationDate="2011-05-19T06:26:56.300" Score="2" ViewCount="112" Body="&lt;p&gt;I am a graduate level student, interested in String theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading a paper on  &quot;String Theory and Einstein's Dream&quot;  published in Current Science, vol. 89, No. 12, p 2045, Dec. 25, 2005. and I came across to know that The Standard model of Quantum physics have failed to explain the interaction of gravitons in the space .One gets a result infinity in his calculation which is wrong as per our experimental values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I really do not understand that Is it the peculiar nature of graviton is creating the problem or the whole model lacks a fundamental drawback in unifying the all four forces of physics? &lt;/p&gt;&#xA;" OwnerUserId="3607" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-05-19T08:21:05.157" LastActivityDate="2011-05-19T08:21:05.157" Title="Drawbacks of Standard model" Tags="&lt;string-theory&gt;&lt;string&gt;" AnswerCount="1" />
  <row Id="10060" PostTypeId="1" CreationDate="2011-05-19T06:45:20.840" Score="0" ViewCount="64" Body="&lt;p&gt;Is there any simple proof of the no-ghost theorem in string theory?&lt;/p&gt;&#xA;" OwnerUserId="3608" LastActivityDate="2011-05-19T06:45:20.840" Title="Is there any simple proof of the no-ghost theorem?" Tags="&lt;string-theory&gt;" CommentCount="2" />
  <row Id="10061" PostTypeId="2" ParentId="9201" CreationDate="2011-05-19T06:51:20.067" Score="1" ViewCount="" Body="&lt;p&gt;This is a very misleading claim which keeps being touted in popular science books and the media.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked about this earlier over here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/2838/total-energy-of-the-universe&quot;&gt;Total energy of the Universe&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I can see, there is a lot of controversy over what total energy even means in the context of general relativity, so claims about &quot;negative energy of gravity&quot; balancing the positive energy of matter are basically vacuous.&lt;/p&gt;&#xA;" OwnerUserId="3335" LastActivityDate="2011-05-19T06:51:20.067" />
  <row Id="10062" PostTypeId="1" CreationDate="2011-05-19T06:56:57.257" Score="9" ViewCount="518" Body="&lt;p&gt;David Deutsch is very fond of pointing out Shor's integer factorization algorithm is a demonstration of the many worlds interpretation. As he often asked, where else did all the exponentially many combinations happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other alternative interpretations of quantum mechanics which can explain Shor's algorithm, and the Deutsch-Jozsa and Simon's algorithm?&lt;/p&gt;&#xA;" OwnerUserId="3609" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-19T19:08:31.260" LastActivityDate="2011-06-06T03:29:16.857" Title="Is Shor's algorithm a demonstration of the many worlds interpretation?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-information&gt;&lt;quantum-interpretations&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="10063" PostTypeId="1" AcceptedAnswerId="10065" CreationDate="2011-05-19T07:04:03.010" Score="0" ViewCount="59" Body="&lt;p&gt;I get the general idea about cosmic inflation, but the numbers associated with it seem to be pulled out at random.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in The Elegant Universe, Brian Greene says that the universe doubled every &#xA;10^(-34) seconds about a hundred times to an overall factor of 10^30.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another astronomy text I read recently gives the doubling time as 10^(-35) secs and the overall expansion factor as 10^50.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where are all these numbers coming from ? For example, why couldn't the universe have doubled every 10^(-40) seconds many billions of times during inflation ?&lt;/p&gt;&#xA;" OwnerUserId="3335" LastActivityDate="2011-05-19T07:42:58.473" Title="Inflation factor and doubling time" Tags="&lt;cosmology&gt;&lt;cosmological-inflation&gt;" AnswerCount="1" />
  <row Id="10064" PostTypeId="2" ParentId="10062" CreationDate="2011-05-19T07:26:03.450" Score="10" ViewCount="" Body="&lt;p&gt;Every working interpretation of quantum mechanics, and the Copenhagen interpretation and Consistent Histories in particular, correctly predicts that Shor's algorithm works much like the other algorithms. The question&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As he often asked, where else did all the exponentially many combinations happen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is loaded because the computational power of a quantum computer is being implicitly compared with the computational power of a classical computer. When it comes to calculations such as those in Shor's algorithm, the former is exponentially larger than the latter, so a way to &quot;imagine&quot; this exponential increase of the speed is to imagine that there are parallel worlds where the &quot;component&quot; (essentially classical) calculations take place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this argument is totally wrong for a simple reason: the real Universe - our Universe - is a quantum system, not a classical system. So it is normal for quantum systems in a single Universe to behave just like the quantum computer running Shor's algorithm. On the contrary, if we only use the classical computers, we exponentially slow down the computer relatively to what it could do. In this sense, Deutsch's &quot;argument&quot; shows that the many-worlds interpretation is just another psychological aid for the people who can't resist to incorrectly think about our world as being a classical world of a sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classical computers may still be very useful, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one more lethal conceptual problem with the &quot;many worlds&quot; explanation of the Shor's algorithm's speed: the whole quantum computer's calculation has to proceed in a completely coherent way and you're &lt;em&gt;not&lt;/em&gt; allowed to imagine that the world splits into &quot;many worlds&quot; as long as things are coherent i.e. before the qubits are measured. Only when the measurement is completed - e.g. at the end of the Shor's algorithm calculation - you're allowed to imagine that the worlds split. But it's too late because by that moment, the whole calculation has already been done in a single (quantum) world, without any help from the parallel worlds. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, when the quantum computer works correctly, the final result after the measurement is pretty much totally determined by the initial data. The probability that Shor's algorithm announces the right unique result is nearly 100% - so there is only one possible outcome so in the quantum computer context, parallel worlds are not really created at all, even in the MWI interpretation. It's a widespread misconception that &quot;everything&quot; about quantum mechanics is more fuzzy, less concentrated, less correlated. But the reality is that quantum predictions may be as sharp - and close to 100% - as those in classical physics and correlations allowed in quantum predictions are often higher than what would be possible in any classical theory (think about Bell's inequalities and other setups). Quantum mechanics is just different (GHZM etc. locate the qualitative difference nicely) but its difference can't be summarized by oversimplified slogans that it's more fuzzy and that the fuzziness is &quot;tamed&quot; by a mechanism. &lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-19T16:15:30.783" LastActivityDate="2011-05-19T16:15:30.783" CommentCount="5" />
  <row Id="10065" PostTypeId="2" ParentId="10063" CreationDate="2011-05-19T07:42:58.473" Score="1" ViewCount="" Body="&lt;p&gt;Dear Cosmonut, the overall factor is usually expressed as a number of e-foldings. You need at least 60 e-foldings or so to explain why the current Universe is apparently so flat and homogeneous. And $e^{60}\approx 10^{26}$ or so. That's where the $10^{30}$ estimate comes from. Less than 60 e-foldings of inflation would mean that inflation doesn't solve a significant part of the problems it is supposed to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is nothing wrong with inflation that lasts for a much longer time. There could have been 50, 100, or 10,000 e-foldings as well and there are papers that find circumstantial evidence why such a very long inflation could be preferred. We don't know how many e-foldings have taken place but if inflation is good for the things we think it is, it should be more than 60 e-foldings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time scale when the inflation occurred is also unknown but it had to be a tiny fraction of a second after the true beginning of our Universe. The higher energy scale - typical &quot;inflaton mass&quot; etc. - you use, the shorter figure for the time after the big bang you get. The typical figure is that inflation began $10^{-36}$ seconds after the big bang and ended around $10^{-33}$ or $10^{-32}$ seconds after the big bang. That's the high-scale, GUT-scale inflation. Note that the shorter time scale is just a little bit longer than the Planck time $10^{-43}$ seconds. It's unlikely that inflation occurred earlier, closer to the Planck scale, because that would predict bigger-than-observed non-uniformities of the cosmic microwave background temperature (they're relatively small because the time when inflation occurred is larger than the Planck time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, inflation could have occurred later than that. Of course, it had to occur before the epoch when the temperatures were close to the electroweak scale, e.g. before $10^{-15}$ seconds or so, because we know that the inflaton isn't too light (lighter than the electroweak scale) and because of other reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What inflation achieves is mostly of a qualitative character - making the space flat and uniform, diluting bad exotic objects, and so on - and this qualitative job may be done at many different time scales and with many different durations of the cosmic inflation. The inflaton couldn't have been directly detected so far so you shouldn't be surprised that all the numerical details about inflation remain largely unknown even though they're constrained by inequalities and they become sharper in particular models.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-19T07:42:58.473" CommentCount="1" />
  <row Id="10066" PostTypeId="1" CreationDate="2011-05-19T08:03:10.940" Score="1" ViewCount="67" Body="&lt;p&gt;90Sr has a half life of 28.5y. It is chemically similar to Ca and enters the body through the food  chain and collects in bones. It is a serious health hazard. How long in (years) will it take for    99.99% of Sr released in a nuclear reactor accident     to disappear?&lt;/p&gt;&#xA;" OwnerUserId="3611" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-05-19T08:36:42.437" LastActivityDate="2011-05-19T09:24:51.897" Title="Nuclear physics Radioactivity" Tags="&lt;homework&gt;&lt;nuclear-physics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="10067" PostTypeId="2" ParentId="10059" CreationDate="2011-05-19T08:13:39.117" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What I really do not understand that Is it the peculiar nature of graviton is creating the problem or the whole model lacks a fundamental drawback in unifying the all four forces of physics?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;both because it's the same thing. The Standard Model - and any quantum field theory - may only contain fundamental fields with spins at most 1. Spin equal to 1 is the right spin of photons, W-bosons, Z-bosons, and gluons - more generally, gauge bosons. But gravitons have and must have the spin equal to 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incorporating the gravitational force is the same thing as incorporating interactions involving the gravitational field - the metric tensor - or, equivalently, incorporating the particles that are quanta of this field, the gravitons. This is not possible in quantum field theory. Of course, it's not possible in the Standard Model either because the Standard Model is a quantum field theory and moreover, it is one defined by a particular spectrum of matter and force particles which don't include (and can't include) gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity may still be added but only as an effective field. While the Standard Model may be made finite to arbitrarily high energies, the addition of gravity reduces the &quot;quality&quot; of the model qualitatively because it may only be reliably used for de facto classical calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, this is not a problem because in particle physics experiments, gravity between the particles is negligible. In principle, it is a huge problem and string theory is the only known - and, arguably, the only mathematically possible way - how to reconcile gravity with quantum field theory.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-19T08:13:39.117" />
  <row Id="10068" PostTypeId="1" CreationDate="2011-05-19T08:20:00.307" Score="4" ViewCount="229" Body="&lt;p&gt;The collapse of the wave function by measurements is one of the most mysterious properties of quantum mechanics. At what scale does the wave function collapse? What are the conditions for a collapse? Can it be made precise? Does the collapse happen instantaneously, or is it spread out in time? What is the precise equation for the collapse? Why is the collapse time-asymmetric, and can this explain the arrow of time?&lt;/p&gt;&#xA;" OwnerUserId="3612" LastActivityDate="2011-08-31T03:33:24.343" Title="On the nature of the collapse of the wave function" Tags="&lt;quantum-mechanics&gt;&lt;quantum-interpretations&gt;" AnswerCount="5" FavoriteCount="3" />
  <row Id="10069" PostTypeId="2" ParentId="10062" CreationDate="2011-05-19T08:27:05.610" Score="16" ViewCount="" Body="&lt;p&gt;No. Shor's algorithm isn't demonstration of MWI. MWI &lt;em&gt;is a way to think&lt;/em&gt; about Shor's algorithm, just like other interpretations of QM. Deutsch advocates MWI as a way to think about quantum algorithms because it is an easy way. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;where else did all the exponentially many combinations happen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why do you need many worlds for exponentially many combinations to occur? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there any other alternative&#xA;  interpretations of quantum mechanics&#xA;  which can explain Shor's algorithm,&#xA;  and the Deutsch-Jozsa and Simon's&#xA;  algorithm?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any correct interpretation of QM can explain these algorithms. Quantum computer scientists use whichever feels comfortable for the task at hand. Read &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Interpretations of quantum mechanics,&#xA;  unlike Gods, are not jealous, and thus&#xA;  it is safe to believe in more than one&#xA;  at the same time. So if the&#xA;  many-worlds interpretation makes it&#xA;  easier to think about the research&#xA;  you’re doing in April, and the&#xA;  Copenhagen interpretation makes it&#xA;  easier to think about the research&#xA;  you’re doing in June, the Copenhagen&#xA;  interpretation is not going to smite&#xA;  you for praying to the many-worlds&#xA;  interpretation. At least I hope it&#xA;  won’t, because otherwise I’m in big&#xA;  trouble.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is comment by &lt;a href=&quot;http://www.scottaaronson.com/blog/?p=218&quot; rel=&quot;nofollow&quot;&gt;Peter Shor on Scott Aaronson's blog&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2011-05-19T08:48:56.200" LastActivityDate="2011-05-19T08:48:56.200" />
  <row Id="10070" PostTypeId="2" ParentId="10068" CreationDate="2011-05-19T08:44:54.380" Score="4" ViewCount="" Body="&lt;p&gt;Dear Jack, there is no physical phenomenon that could be called the collapse. The collapse of the wave function, as first emphasized by Werner Heisenberg and then many others, is just the event when we learn something about a physical property of a physical system. When we learn that Osama bin Laden is located in a building in Pakistan, his wave function - that could have included possible positions at many other places - suddenly &quot;collapses&quot; because we learned about the position. That was pretty much Heisenberg's description of the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wave function is not an actual wave - like an electromagnetic wave. It is a collection of numbers that summarizes our knowledge about the physical system and that can be used to make predictions. Any attempt to &quot;overinterpret&quot; the wave function and &quot;visualize&quot; it as a real wave that objectively exists etc. is fundamentally flawed. The collapse of the wave function is just a process in our brain when we learn that the physical quantity $A$ has taken one of its values, $A=A_i$. When we learn it, we must replace the calculation of probabilities such as $P(B=B_j)$ by the calculation of the conditional probabilities such as $P(B=B_j|A=A_i)$. This replacement may be visualized as a collapse but no actual collapse occurs - and by this statement, I mean that all questions about the detailed &quot;mechanism&quot; of the collapse are based on an incorrect assumption that there is a mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question is at what sense it's possible to say that $A=A_i$ was &quot;perceived&quot; by &quot;something&quot; or &quot;somebody&quot;. Quantum mechanics surely prevents one from declaring such a measurement too early. You never make an error if you continue to use the whole wave function and only &quot;collapse&quot; it in your mind when you want to learn some particular answers. So you may evolve the electrons, hammers, cats, and other people according to Schrödinger's equations so that the wave function spreads into many possibilities, and only when you know what you feel, you may make the &quot;collapse&quot; in your brain according to what you felt, and continue with the simplified wave function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, to simplify your thinking, you may imagine that the wave function for the other objects collapsed earlier than that - you're allowed to imagine that there was a collapse after a sufficiently &quot;macroscopic&quot; interaction. What it exactly means? It means an interaction in which the different values of various properties have &quot;decohered&quot; from each other. Decoherence is a derivation of the classical-quantum boundary - of the point at which it becomes pretty much legitimate to imagine that the objects have well-defined properties such as positions, just like in classical physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one doesn't &lt;em&gt;need&lt;/em&gt; to imagine that this reduction ever takes place. &lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-19T08:44:54.380" CommentCount="5" />
  <row Id="10071" PostTypeId="2" ParentId="10021" CreationDate="2011-05-19T09:02:07.540" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://indico.cern.ch/categoryDisplay.py?categId=72&quot; rel=&quot;nofollow&quot;&gt;http://indico.cern.ch/categoryDisplay.py?categId=72&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The INDICO servers by CERN have &lt;strong&gt;tons&lt;/strong&gt; of video lectures on many different topics, related to particle physics and cosmology. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-05-19T09:02:07.540" CommunityOwnedDate="2011-05-19T15:52:07.370" />
  <row Id="10072" PostTypeId="1" AcceptedAnswerId="10119" CreationDate="2011-05-19T09:07:32.393" Score="1" ViewCount="518" Body="&lt;p&gt;The damping of a spring is calculated with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\zeta] = \frac{[c]}{\sqrt{[m][k]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where c is the 'viscous damping coefficient' of the spring, according to Wikipedia. m is the mass, k is the spring constant, and zeta is the damping ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is the value of c calculated though? Is it a constant for the air through which the spring is moving or does it depend on the spring itself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What data is required to calculate it and how can it be done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just looking at the oscillation of a spring vertically, and I have data for its decreasing amplitude, and the velocity of the spring at all points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the value of the damping ratio, and I'm trying to find the value of 'c' in order to prove the above equation in an investigation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is almost no information about this online.&lt;/p&gt;&#xA;" OwnerUserId="3511" LastEditorUserId="3511" LastEditorDisplayName="" LastEditDate="2011-05-20T11:43:24.933" LastActivityDate="2011-05-22T02:12:56.497" Title="How to calculate viscous damping coefficient?" Tags="&lt;harmonic-oscillator&gt;" AnswerCount="3" CommentCount="15" />
  <row Id="10073" PostTypeId="2" ParentId="10055" CreationDate="2011-05-19T09:17:18.157" Score="2" ViewCount="" Body="&lt;p&gt;I will answer this part of the question since it is important to distinguish what we are doing as physicists.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;A free particle cannot exist in a stationary state; or, to put it another way, there is no such thing as a free particle with a definite energy.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As Mark says in his comment above:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;it's pretty standard to find a solution to an equation and then ask whether that solution is physically meaningful.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which is fine. When the answer is &quot;no&quot;, as in this case, one goes ahead to find physically meaningful solutions that will describe a free particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is ample experimental evidence that particles exist, from atomic physics to nuclear physics to particle physics. Solving Schroedinger's equation for a potential well not only gives well defined state functions, the solutions describe the behaviour of the particle quite well to excellently. One then solves the &quot;no potential&quot; equation and finds the plane waves .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First thought: no potential means free particle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;second thought: are these plane wave functions well behaved so that they can be used to describe quantum mechanically a free particle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third thought should be: No, they cannot be normalized to a probability function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fourth thought: Could one use these solutions as a complete set to find a function that can be normalized and its probability describes a physical free particle ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bingo: wave packets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement &quot;there is no such thing as a free particle with a definite energy&quot; is wrong for a physical particle. It should be : plane waves cannot describe free particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Free particles ( experiment) trump plane waves ( theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You were right to be confused. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-19T09:17:18.157" CommentCount="1" />
  <row Id="10074" PostTypeId="2" ParentId="10066" CreationDate="2011-05-19T09:24:51.897" Score="3" ViewCount="" Body="&lt;p&gt;If initial mass is $M$. Amount remaining after $n$ half lives is $M/2^{n}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask yourself, &lt;strong&gt;How much remains after 99.99% is gone?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equate. &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2011-05-19T09:24:51.897" />
  <row Id="10075" PostTypeId="1" CreationDate="2011-05-19T09:36:30.693" Score="0" ViewCount="162" Body="&lt;p&gt;As the only way String theory can respect the principles of quantum mechanics and Special theory of relativity, by formulating it in hypothetical nine dimensional space.&#xA;You could use string compactification for d number of non compact direction, all it requires to have d non compact direction is to choose an appropriate compact space of dimension (9-d).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question arises here, Then Why our Universe is in Three dimensional space?&lt;/p&gt;&#xA;" OwnerUserId="3607" LastActivityDate="2011-06-24T14:05:06.300" Title="Compactification of Dimensions" Tags="&lt;string-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10076" PostTypeId="2" ParentId="10015" CreationDate="2011-05-19T10:25:50.807" Score="0" ViewCount="" Body="&lt;p&gt;The above image was taken in optical wavelengths, in the inra red large portions are transparent, see : &lt;a href=&quot;http://en.wikipedia.org/wiki/File:M16_-_Eagle_nebula.jpg&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/File:M16_-_Eagle_nebula.jpg&lt;/a&gt;,&#xA;which describes the same object as viewed by NASA's Spitzer space telescope.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-05-19T10:25:50.807" />
  <row Id="10077" PostTypeId="2" ParentId="10055" CreationDate="2011-05-19T10:34:23.630" Score="2" ViewCount="" Body="&lt;p&gt;I would normally just add this as a comment, but I still can not comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say the author is just saying that because the &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_argument&quot; rel=&quot;nofollow&quot;&gt;arguments&lt;/a&gt; are effectively the same, and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_value#Complex_numbers&quot; rel=&quot;nofollow&quot;&gt;modulus&lt;/a&gt; of each number is arbitrary, one can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Distributive_law&quot; rel=&quot;nofollow&quot;&gt;distributive law&lt;/a&gt; to simplify the expression.  If one tries to &lt;a href=&quot;http://en.wikipedia.org/wiki/Normalisable_wave_function&quot; rel=&quot;nofollow&quot;&gt;normalize&lt;/a&gt; using standard procedure, one gets: $$A^2$$  If one does not specify finite boundary conditions, one must find the integral of this constant value from -infinity to infinity which gives an infinite value, and thus is not normalizable.&lt;/p&gt;&#xA;" OwnerUserId="3460" LastActivityDate="2011-05-19T10:34:23.630" CommentCount="2" />
  <row Id="10078" PostTypeId="1" AcceptedAnswerId="10082" CreationDate="2011-05-19T11:50:35.557" Score="3" ViewCount="203" Body="&lt;p&gt;Hubble's constant $a(t)$ appears to be changing over time. The fine stucture constant $\alpha$, like many others in QFT, is a running constant that varies, proportional to energy being used to measure it. Therefore, it could be agued that all running constants have 'evolved' over time as the Universe has expanded and cooled. Both the local and global curvature of the Universe changes over time implying that so too does the numerical value of $\pi$. All these things are however constants (well, let's say parameters since they are not really 'constant'.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://www.youtube.com/watch?v=uNOghidK2TY&quot; rel=&quot;nofollow&quot;&gt;a discussion&lt;/a&gt; with astronomer Sir Fred Hoyle, Feynman said &quot;&lt;em&gt;what today, do we not consider to be part of physics, that may ultimately become part of physics?&lt;/em&gt;&quot; He then goes on to say &quot;&lt;em&gt;..it's interesting that in many other sciences there's a historical question, like in geology - the question how did the Earth evolve into the present condition? In biology - how did the various species evolve to get to be the way they are? But the one field that hasn't admitted any evolutionary question - is physics.&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So have the laws of physics remained form-invariant over the liftetime of the Universe? Does the recent understanding of the aforementioned not-so-constant constants somehow filter into the actual form of the equations being used? Has advances in astronomical observations, enabling us to peer back in time as far back as the CMB, given us any evidence to suggest that the laws of nature have evolved? If Feynman thinks that &quot;&lt;em&gt;It might turn out that they're not the same all the time and that there is a historical, evolutionary question.&lt;/em&gt;&quot; then this is surely a question worth asking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB/ To be clear: this is a question concerning purely physics, whether the equations therein change as the Universe ages, and whether there is any observational evidence for this. It is not intended as an oportunity for a philosophical discussion.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="1696" LastEditorDisplayName="" LastEditDate="2011-05-19T13:32:51.623" LastActivityDate="2011-05-19T14:19:12.453" Title="Do the laws of physics evolve?" Tags="&lt;cosmology&gt;&lt;theory&gt;&lt;cosmological-constant&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="10079" PostTypeId="2" ParentId="10068" CreationDate="2011-05-19T12:48:08.920" Score="2" ViewCount="" Body="&lt;p&gt;An often good on-line source for the interpretation of quantum theory is the Stanford Encyclopedia of Philosophy, which has &lt;a href=&quot;http://plato.stanford.edu/entries/qm-collapse/&quot; rel=&quot;nofollow&quot;&gt;a page on &quot;collapse theories&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a lot of literature on whether one needs collapse if one takes the wave function seriously, as opposed to the mainline Physicist's approach of taking a more empiricist view, as outlined well by Luboš, but the non-collapse alternative is epitomized for most people by the many worlds interpretation.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-05-19T12:48:08.920" />
  <row Id="10080" PostTypeId="2" ParentId="10068" CreationDate="2011-05-19T13:37:32.213" Score="3" ViewCount="" Body="&lt;p&gt;There are currently two different accounts that give a larger picture of what happens when a quantum system is measured. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_decoherence&quot; rel=&quot;nofollow&quot;&gt;One of them&lt;/a&gt; is the fact that many random interactions between the system (which might be a 1-body or N-body quantum system) and the environment (which is considered for most purposes a pseudo-classical system with infinite degrees of freedom) produce interactions that tend to average out interference terms (which are represented by diagonals in the density matrix), leaving only a pure diagonal matrix. the pure diagonal matrix is what one would use to describe a classical system with a number of classical probabilities of being in several states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Many-worlds_interpretation&quot; rel=&quot;nofollow&quot;&gt;The other one&lt;/a&gt; is based on the idea that once you take into account the should-be-obvious fact that any observer of quantum systems must be described by the same quantum laws as any other physical system, becomes clear that the phenomena of collapse is not an &lt;em&gt;ad hoc&lt;/em&gt; postulate you need to plug into the theory by hand, but it is the unescapable consequence of interaction between (quantum) observers and quantum systems; any time a quantum system in superposition of states interacts with a quantum observer, the whole system + observer will evolve into a superposition of entangled states between each eigenstate in the original superposition of the system $\times$ a new observer state where the observer is aware of a different eigenvalue of the system he perceives to have measured by the interaction&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-05-19T13:37:32.213" />
  <row Id="10081" PostTypeId="1" CreationDate="2011-05-19T13:42:42.237" Score="0" ViewCount="97" Body="&lt;p&gt;I've been reading Lee Smolin's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0195126645&quot; rel=&quot;nofollow&quot;&gt;Life of the Cosmos&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Great book and it makes a lot of sense that the conditions in black holes are the same as conditions at the big bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question is, has his theory about &lt;a href=&quot;http://evodevouniverse.com/wiki/index.php/Cosmological_natural_selection_%28fecund_universes%29&quot; rel=&quot;nofollow&quot;&gt;Cosmological Natural Selection&lt;/a&gt; been disproved as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lee_Smolin#Fecund_universes&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt; states?&lt;/p&gt;&#xA;" OwnerUserId="3316" LastActivityDate="2011-05-19T14:32:20.913" Title="Has Cosmological Natural Selection been disproved?" Tags="&lt;astrophysics&gt;&lt;cosmology&gt;&lt;black-holes&gt;&lt;big-bang&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10082" PostTypeId="2" ParentId="10078" CreationDate="2011-05-19T14:09:30.060" Score="5" ViewCount="" Body="&lt;p&gt;For many (most? all?) physicists, it's something like an axiom (or an article of faith, if you prefer) that the true laws don't change over time. If we find out that one of our laws does change, we start looking for a deeper law that subsumes the original and that can be taken to be universal in time and space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good example is Coulomb's Law, or more generally the laws of electromagnetism. In a sense, you could say that Coulomb's Law changed form over time: in the early Universe, when the energy density was high enough that electroweak symmetry was unbroken, Coulomb's Law wasn't true in any meaningful or measurable sense. If you thought that Coulomb's Law today was a fundamental law of nature, then you'd say that that law changed form over time: it didn't use to be true, but now it is. But of course that's not the way we usually think of it. Instead, we say that Coulomb's Law was never a truly correct fundamental law of nature; it was always just a special case of a more general law, valid in certain circumstances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more interesting example, along the same lines: Lots of theories of the early Universe involve the idea that the Universe in the past was in a &quot;false vacuum&quot; state, but then our patch of the Universe decayed to the &quot;true vacuum&quot; (or maybe just another false vacuum!). If you were around then, you'd definitely perceive that as a &lt;em&gt;complete&lt;/em&gt; change in the laws of physics: the particles that existed, and the ways those particles interacted, were completely different before and after the decay. But we tend not to think of that as a change in the laws of physics, just as a change in the circumstances within which we apply the laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is just that when you try to ask a question about whether the fundamental laws change over time, you have to be careful to distinguish between actual physics questions and merely semantic questions. Whether the Universe went through one of these false vacuum decays is (to me, anyway) a very interesting physics question. I care much less whether we describe such a decay as a change in the laws of physics.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-05-19T14:09:30.060" CommentCount="3" />
  <row Id="10084" PostTypeId="2" ParentId="10078" CreationDate="2011-05-19T14:19:12.453" Score="0" ViewCount="" Body="&lt;p&gt;Not exactly in line with your question, but you might wanna have a look at the theory of &lt;a href=&quot;http://evodevouniverse.com/wiki/index.php/Cosmological_natural_selection_%28fecund_universes%29&quot; rel=&quot;nofollow&quot;&gt;Cosmological Natural Selection&lt;/a&gt;, which says that a new universe is spawned within each black hole with parameters inherited from its parent universe, only slightly mutated. In this sesne the laws of physics would be &lt;em&gt;evolving&lt;/em&gt; as parameters change. Also, I just asked a question about it &lt;a href=&quot;http://physics.stackexchange.com/questions/10081/has-cosmological-natural-selection-been-disproved&quot;&gt;here&lt;/a&gt; :)&lt;/p&gt;&#xA;" OwnerUserId="3316" LastActivityDate="2011-05-19T14:19:12.453" CommentCount="2" />
  <row Id="10085" PostTypeId="2" ParentId="10081" CreationDate="2011-05-19T14:27:13.647" Score="3" ViewCount="" Body="&lt;p&gt;Yes, of course, it's trivial to disprove this idea because it's childish nonsense that may sell popular books for readers who don't understand the subject but that contributes nothing to to the actual science. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very easy to adjust the parameters of the cosmological and particle physics model we use to describe the Universe so that the number of black holes that are produced will be vastly higher than in our Universe - in which black holes are very rare. For example, move the inflation scale closer to the Planck scale to create higher non-uniformities in the CMB etc. That will also create a higher number of black holes. So our Universe is demonstrably not optimized for &quot;black hole fertility&quot;. See also&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0407266&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0407266&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://motls.blogspot.com/2006/10/on-cosmic-natural-selection.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2006/10/on-cosmic-natural-selection.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the first paper, Susskind also argues that Smolin needs to assume that the reproduction of universes through black holes beats eternal inflation which it almost certainly doesn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, the counting of black holes is really ill-defined because of the quantum nature of our world - in particular, tiny black holes are indistinguishable from ordinary heavy elementary particles. The very notion that the interior of a black hole may give rise to a new Universe contradicts the holographic principle (and related holographic bounds; you can't have excessively high entropy emerging out of a small region) and there are many other ways to see that the idea is both ill-defined as well as demonstrably wrong in all of its conceivable versions.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-19T14:32:20.913" LastActivityDate="2011-05-19T14:32:20.913" CommentCount="1" />
  <row Id="10086" PostTypeId="1" CreationDate="2011-05-19T14:28:07.787" Score="-1" ViewCount="153" Body="&lt;p&gt;Quantum mechanics is very mysterious. Consciousness is often brought into play to explain quantum phenomena. Is this only a matter of convenience, or is consciousness inherent to interpreting quantum mechanics? What is the role of the mind in affecting the wave function? What counts as a conscious being? Is the wave function all in our minds, and can we change it just with our thoughts?&lt;/p&gt;&#xA;" OwnerUserId="3619" LastActivityDate="2011-05-19T15:20:34.203" ClosedDate="2011-05-19T17:17:43.277" Title="Consciousness and quantum mechanics" Tags="&lt;quantum-interpretations&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="10087" PostTypeId="2" ParentId="10086" CreationDate="2011-05-19T14:57:08.863" Score="0" ViewCount="" Body="&lt;p&gt;Dear Cupcake, the only special role of a conscious being - and indeed, it's an important one - is that it can verify the validity of the laws of physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When this object or subject ;-) wants to predict the future, it will use quantum mechanics or some simplified approximation of it and get the predictions. If it uses the right theory, it will find out that the theory works. Of course, to do such a thing, it inevitably has to have a sufficient number of interacting building blocks so that the approximate classical framework to think about the events is as legitimate as it is for humans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, your sharp separation of objects to conscious and unconscious isn't used by any law of physics. You may imagine that any property of a physical system that may imprint itself into the environment - so that it has decohered - is being &quot;realized&quot; by the corresponding physical system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, we can change the wave function in our thoughts. That's what people are doing when they imagine that the wave function has collapsed, for example. Other people may continue to evolve the wave function without any collapse - so that each person will use a different wave function - and they will only make the collapse once &lt;em&gt;they&lt;/em&gt; are aware of a measured property. It's important to realize that when the wave function collapses in someone's mind, it's just his simplification of a calculation; it doesn't affect any actual observations that will be done or their calculations as long as the calculations are done properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In particular, it's not true that one may single out a subclass of &quot;conscious beings&quot; who know some special password and who have some special supervisor rights to the quantum Universe so that they may interfere with the predictions of quantum mechanics or modify the future facts in any physical way. Every object, including people, obeys the laws of physics without any modification. The conscious people are just those who can realize that.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry that the answer is vague and has no formulae to make the points quantitative but it is because the question is vague and unquantitative. Physics is simply not about consciousness. The aspects of consciousness that are available to the scientific method are studied by neurobiology and similar disciplines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Physics is a systematic method to reconcile, explain, and predict perceptions from the physicist's observations of all other objects in the world and all such objects have to be treated as if they were dead and dull conglomerates of particles and fields.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the observer himself who is using physics may think that he is something more, and he arguably is. But that doesn't affect anything about physics predictions whatsoever. In particular, other physicists may consider the first physicist to be a dull bound state of particles, too. They will get the right probabilistic predictions for any question they may ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's legitimate to employ any idea about who has consciousness or who doesn't. For example, it's totally OK to use solipsism - a paradigm according to which I am the only conscious being. Despite the apparent subjective character of this prescription, it leads to no contradiction. For example, using equations of quantum mechanics, it's possible to show that if another person (treated as a dull bound state of particles) clearly sees something in the skies, I will see it, too.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-19T15:10:28.557" LastActivityDate="2011-05-19T15:10:28.557" />
  <row Id="10088" PostTypeId="1" AcceptedAnswerId="10090" CreationDate="2011-05-19T15:08:28.823" Score="7" ViewCount="248" Body="&lt;p&gt;The electroweak and strong forces seem to be completely different types of forces to gravity. The latter is geometric while the former are not (as far as I'm aware!). So why should they all be described in the same way? - Why are gravitons needed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, I suppose my question is why a theory of everything has to be some form of quantum gravity. Is there a reason why a theory of everything requires gravity to be quantised? Is it possible for the quantum world and gravity to be described by some overarching theory which does not require gravity to be quantised? Or is there some reason why this is not the case?&lt;/p&gt;&#xA;" OwnerUserId="3622" LastEditorUserId="3622" LastEditorDisplayName="" LastEditDate="2011-06-08T18:07:47.770" LastActivityDate="2011-06-08T18:07:47.770" Title="Why does gravity need to be quantised?" Tags="&lt;quantum-gravity&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="4" />
  <row Id="10089" PostTypeId="1" AcceptedAnswerId="10092" CreationDate="2011-05-19T15:10:45.803" Score="1" ViewCount="102" Body="&lt;p&gt;I would like to know what happens with time dilation (relative to surface) at earth's center .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a way to calculate it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is time going faster at center of earth? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've made other questions about this matter and the answers refers to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta\Phi$ (difference in Newtonian gravitational potential between the locations) as directly related, but I think those equation can't be applied to this because were derived for the vecinity of a mass but not inside it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any clues? Thanks&lt;/p&gt;&#xA;" OwnerUserId="1916" LastActivityDate="2011-05-19T15:37:38.050" Title="Gravitational time dilation at the earth's center" Tags="&lt;gravity&gt;&lt;spacetime&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="10090" PostTypeId="2" ParentId="10088" CreationDate="2011-05-19T15:18:51.903" Score="10" ViewCount="" Body="&lt;p&gt;Dear Calvin, if any portion of the world is described by probabilistic wave functions, then the whole world has to be. It's easy to show it. Take a decaying nucleus, connect it to a hammer that kills a cat a that also makes the Sun explode into 2 pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nucleus is evolving into a linear superposition of &quot;decayed&quot; and &quot;not yet decayed&quot; states. Correspondingly, because of the mechanism, the Sun has to evolve into a linear superposition of &quot;exploded&quot; and &quot;not yet exploded&quot;. These two states have different gravitational fields. It proves that in general, the evolution produces linear superpositions of states with different gravitational fields, so the gravitational fields - and any other physical properties of the world - have to be described by linear operators just like any other physical property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If some building blocks can only be predicted probabilistically, it's clear that everything else that may be affected by these building blocks can only be predicted probabilistically, too. The whole world, including gravitational fields, interacts with itself, so clearly the gravitational waves have to be ultimately described by quantum physics, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravitons are physical particles that are quanta of gravitational waves. They have to exist because&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;gravitational waves exist&lt;/li&gt;&#xA;&lt;li&gt;the energy stored in a frequency $f$ classical state is always a multiple of $E=hf$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first point was indirectly, but pretty much conclusively, proved by the observation of the binary pulsar that changes its frequency in the right way, as it emits the gravitational waves and loses energy. Everything agrees with general relativity beautifully. The 1993 physics Nobel prize was given for this confirmation of the gravitational waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second point is a trivial consequence of Schrödinger's equation. Take a classical gravitational wave of frequency $f$ - e.g. similar to what is emitted by the binary pulsar. Divide the corresponding quantum state to energy eigenstate components. They go like&#xA;$$c_n\exp(E_n t/i\hbar)$$ &#xA;If all classical observables evaluated in this state are periodic with periodicity $1/f$, it's trivial to see that all the energy differences $E_m-E_n$ must be multiples of $E_0=hf$, so the energy can only be added to the gravitational waves by quanta, the gravitons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One may also derive the gravitons and their polarizations from the linear approximation of the quantized general relativity. While general relativity has problems at higher-loop level - strongly quantum effects that affect spacetime - this approximation has to work very well, is consistent, and implies the existence of gravitons, spin 2 massless particles, too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum gravity, virtual gravitons are the messengers of the gravitational force in the same way as photons are messengers of the electromagnetic force. Gravitons and photons have a different spin but otherwise the analogy between them is - and must be - much tighter than the wording of your question is willing to admit. They're described as quantum fields and the effective quantum field theory with Feynman diagrams etc. has to work, with the same interpretation, at least in some approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it's not true that the non-gravitational forces can't be geometric. The Kaluza-Klein theory explains the electromagnetic field as &quot;twists&quot; that include an extra dimension of space. Whether or not the extra dimensions of space have this simple form, string theory generalizes the Kaluza-Klein lesson and every field and particle species may be viewed as a component of a generalized geometry field - because the geometry is generalized in such a way that it includes all other fields as well. Some of the string scenarios are very close to the Kaluza-Klein theory, some of them are far from it, but all of them confirm that gravity ultimately comes from the same underlying physics as everything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term &quot;string theory&quot; may sound too narrow-minded in this context. There are very many different ways how to describe its physics, including gauge theory (just an old quantum field theory!) that lives on the holographic boundary of the AdS space. All of them agree that gravitons have to exist and they're just one particle species among many that share a common origin. So the term &quot;string theory&quot; in all these discussions really means &quot;everything we have ever learned about quantum gravity that has worked&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-19T15:26:22.490" LastActivityDate="2011-05-19T15:26:22.490" CommentCount="5" />
  <row Id="10091" PostTypeId="2" ParentId="10086" CreationDate="2011-05-19T15:20:34.203" Score="0" ViewCount="" Body="&lt;p&gt;the introduction of Consciousness into the interpretation of quantum theory is more because of desire for the world to be a particular way than of convenience. It perhaps particularly wants human agency to be unfettered, or at least to be less fettered by hard rules. Consciousness is certainly not inherent to the interpretation of quantum theory, except insofar as someone does have to &lt;em&gt;decide&lt;/em&gt; to do an experiment or to try to construct a new theory and to publish the results in a journal, but this decision process falls outside Physics &lt;em&gt;qua&lt;/em&gt; Physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum mechanics is a theory in Physics that deals with statistics. No-one has found that the statistics of results change in a statistically significant way when someone wills for the result of an experiment to come out one way or another. It can be said that very few Physicists look for such effects, so of course they haven't been found, but AFAIK what searches there have been have not shown significant results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physicists have been able to find extensive systematic order in the statistics of our records of experimental results, but any event that happens only once is largely outside the scope of Science because of the almost absolute requirement that experiments must be reproducible for them to be considered to be Science, essentially as a matter of definition. The idea of reproducibility has come to be one of &lt;em&gt;statistical&lt;/em&gt; reproducibility, which gives up all unique events to non-Science. Of course every apple is unique, but we can still talk about a dozen apples that are the same enough that they are at the same time unique and not unique, so &lt;a href=&quot;http://store.xkcd.com/xkcd/#StandBackScience&quot; rel=&quot;nofollow&quot;&gt;stand back while we try Science&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for what counts as a conscious being, I'd say that's outside Physics &lt;em&gt;qua&lt;/em&gt; Physics. &lt;a href=&quot;http://area51.stackexchange.com/proposals/2578/philosophy&quot;&gt;Philosophy Stack Exchange is almost there!&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Insofar as the wave function is a concept in our heads, we can have as many of them as we want and we can change them as we like, but constructing or finding a piece of the world for whatever wave function we think of may not be easy to do. The Map is not the Territory, however good the map is, and drawing a map is no guarantee that there is a country that has that shape. Unless we're living in the Matrix, in which case the territory is a map. Not Physics, probably.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-05-19T15:20:34.203" />
  <row Id="10092" PostTypeId="2" ParentId="10089" CreationDate="2011-05-19T15:33:29.847" Score="3" ViewCount="" Body="&lt;p&gt;The rule I mentioned in &lt;a href=&quot;http://physics.stackexchange.com/questions/9921/voyager-local-time-dilation-caused-by-gravity&quot;&gt;another question&lt;/a&gt;, that the time dilation factor is $1+\Delta\Phi/c^2$, applies here. The derivation (found in various textbooks) depends only on the assumptions that fields are weak and matter is nonrelativistic, both of which are true for the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modeling the Earth as a uniform-density sphere (not true, of course, but I don't care), we find that $g(r)=GMr/R^3$ where $R$ is the radius of the Earth. So&#xA;$$&#xD;&#xA;\Delta\Phi={GM\over R^3}\int_0^Rr\,dr={GM\over 2R}.&#xD;&#xA;$$&#xA;That means that&#xA;$$&#xD;&#xA;{\Delta\Phi\over c^2}={GM\over 2Rc^2}={1\over 4}{R_s\over R}.&#xD;&#xA;$$&#xA;Here $R_s=2GM/c^2$ is the Schwarzschild radius corresponding to the Earth's mass. Numerically, $R_s$ is about 9 mm, and $R$ is about 6400 km, so $\Delta\Phi/c^2=3\times 10^{-10}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sign of the effect is that clocks tick slower when they're deeper in the potential well. That is, a clock at the Earth's surface ticks 1.0000000003 times faster than one at the center.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-05-19T15:33:29.847" CommentCount="3" />
  <row Id="10093" PostTypeId="2" ParentId="10089" CreationDate="2011-05-19T15:37:38.050" Score="1" ViewCount="" Body="&lt;p&gt;Dear HDE, it's not hard to estimate the gravitational potential at the Earth's center. Of course, it's smooth. Let me assume that the Earth's mass density is uniform which is an OK estimate - up to factors of two or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational acceleration at distance $R$ from the center is $GM/R^2$ if $R$ is greater than the Earth's radius $R_E$. However, for smaller values of $R$, you have to use Gauss' law &#xA;$$\int d\vec S\cdot \vec g \sim GM_{inside}$$&#xA;and determine the total mass inside a smaller sphere. Because $M_{inside}$ goes like $R^3$ for $R&amp;lt;R_E$, and this $R^3$ is still divided by $R^2$ from $\int d\vec S$, it follows that the gravitational acceleration inside the Earth is pretty much proportional to $R$:&#xA;$$ g(R) = g(R_E)\cdot \frac{R}{R_E} $$&#xA;In particular, the gravitational acceleration at the Earth's center is zero and near the center, a particle would oscillate like in a harmonic oscillator, $\vec F\sim -k\vec x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also trivial to calculate the extra decrease of the gravitational potential you get if you go from the surface to the center. On the surface, the gravitational potential is $-GM/R_E$, as you know, because the derivative of $-GM/R$ over $R$ gives the right acceleration. However, the potential is getting even more negative. If you integrate $g(R_E)\cdot R/R_E$ over $R$ from $0$ to $R_E$, you will get $g(R_E) R_E/2$. This has to be taken with the negative sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the potential at the center, assuming uniformity, is&#xA;$$ \Phi = -\frac{GM}{R_E} - g(R_E) \frac{R_E}{2} = -\frac 32 \frac{GM}{R_E} = -\frac 32 g(R_E) R_E $$&#xA;This gravitational potential determines the slowing of time, too. In SI units, $g(R_E)=10$ Newtons per meter and $R_E=6,378,000$. The product, with the $3/2$ factor added, is almost exactly $10^{8}$. Divide it by $c^2=10^{17}$ to get about $10^{-9}$ - the relative red shift from the center of the Earth to infinity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you spend 1 billion years at the center of the Earth, your twin brother outside the gravitational field will get 1 billion and one years older. If you wish, you may interpret it by saying that it's healthy to live at the center of the Earth. Good luck.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-19T15:37:38.050" CommentCount="1" />
  <row Id="10094" PostTypeId="1" AcceptedAnswerId="10123" CreationDate="2011-05-19T15:39:01.443" Score="4" ViewCount="134" Body="&lt;p&gt;In particular, why do we need both of these to find the volume? And should I be thinking of it as an actual volume or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Hilbert space volume is talked about in &lt;a href=&quot;http://tid.uio.no/~ovrum/articles/Volume%20of%20the%20set%20of%20separable%20states%20I.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; paper. It says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There exist the unique, uniform measure $\nu$ on $P$ induced by the Haar measure on the group of unitary matrices $U(N)$. Integration over the set $P$ thus amounts to an integration of the corresponding angles and phases in N-dimensional complex space that determine the families of orthonormal projectors. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since the simplex $\Delta$ is a subset of the ($N-1$)-dimensional hyperplane, there exist a natural measure on $\Delta$ which is deﬁned as a usual normalized Lebesgue measure $\mathcal{L}_{N-1}$ on $R^{N-1}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The volume of the Hilbert space is then defined to be $\mu=\nu \times \mathcal{L}_{N-1}$, but why?&lt;/p&gt;&#xA;" OwnerUserId="3622" LastEditorUserId="3622" LastEditorDisplayName="" LastEditDate="2011-05-20T12:00:39.617" LastActivityDate="2011-05-20T12:00:39.617" Title="Why is the Haar measure times the volume of the eigenvalue simplex considered a good measure of Hilbert space volume?" Tags="&lt;hilbert-space&gt;&lt;volume&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10095" PostTypeId="1" AcceptedAnswerId="10096" CreationDate="2011-05-19T15:41:30.120" Score="1" ViewCount="88" Body="&lt;p&gt;NASA recently announced (http://www.nasa.gov/topics/universe/features/planet20110518.html) that orphan planets, planets no longer orbiting a star, may be more numerous than the stars in our galaxy. The Sun's Hill sphere, its gravitational influence within the Milky Way, is ~2.2 ly. So it's possible, however unlikely, that a Jupiter size orphan planet could attain solar system orbit, say with a semi-major axis of one light year and a ~1.6E7 year period. The barycenter of this body and the Sun would therefore be ~60.3 AU from the Sun. Could we detect the Sun's motion around the hypothetical Jupiter barycenter by its gradual anomalous dispacement versus distant stars and therefore detect the existence of such a far-off massive planet?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-05-19T18:40:25.267" Title="Solar System Capture of Orphan Planets" Tags="&lt;planets&gt;&lt;solar-system&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="10096" PostTypeId="2" ParentId="10095" CreationDate="2011-05-19T16:04:51.637" Score="1" ViewCount="" Body="&lt;p&gt;Dear Michael, yes, you could do it by observing the Sun and stars for those 16 million years or so (the exotic big planet's &quot;year&quot;). Then it would be as visible as the effect of the ordinary Jupiter. Alternatively, you could observe it for 8-16 years only but you would need a million times higher resolution than the resolution for which we may observe the Sun's motion because of Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, this estimate is way too optimistic. The motion of the barycenter would look linear for periods of time much shorter than those 16 million years and to show that the motion is due to an exotic planet, you would surely have to demonstrate some acceleration of the barycenter and maybe even a few higher derivatives to show that it's moving around the curve expected from the extra planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, I think it is impossible in practice.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-19T16:04:51.637" />
  <row Id="10097" PostTypeId="1" AcceptedAnswerId="10099" CreationDate="2011-05-19T17:00:59.203" Score="3" ViewCount="162" Body="&lt;p&gt;I saw this equation on a coffee mug, and I am wondering what it means. I am not far enough along in physics to understand it. Does it actually mean anything or is it gibberish? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ifd4q.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't embed the &lt;a href=&quot;http://images0.cpcache.com/product_zoom/481658830v0_480x480_Front_Color-Black_padToSquare-true.jpg&quot; rel=&quot;nofollow&quot;&gt;image&lt;/a&gt; because I am a new user.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(But I can, L.M.)&lt;/p&gt;&#xA;" OwnerUserId="3626" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-19T17:13:37.350" LastActivityDate="2011-05-19T17:43:35.357" Title="Does this equation mean anything?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10098" PostTypeId="1" CreationDate="2011-05-19T17:01:58.657" Score="3" ViewCount="89" Body="&lt;p&gt;In a perpetual round of reformulations, I've put quantized electromagnetism into a 1-form notation. I'm looking for references that do anything similar, both to avoid reinventing the wheel and perhaps to cite. Taking the electromagnetic field $\hat A$ to be an (operator-valued distributional) 1-form, we can write a smeared operator $\hat A_H$ in terms of the inner product on the exterior algebra as $\bigl&amp;lt;\hat A,H\bigr&amp;gt;$ (taking the test function $H$ also to be a 1-form, but satisfying Schwartz space-like smoothness conditions in real space and in fourier space instead of being a distribution). $\hat A$ projects into annihilation and creation parts, $\hat A^+$ and $\hat A^-$ respectively, $\hat A=\hat A^+ +\hat A^-$, for which the action on the vacuum is defined by $\hat A^+\left|0\right&amp;gt;=0$, and we have the commutation relations $\Bigl[\bigl&amp;lt;\hat A^+,H\bigr&amp;gt;,\bigl&amp;lt;\hat A^-,J\bigr&amp;gt;\Bigr]=\left&amp;lt;H,E(J)\right&amp;gt;$, where $\widetilde{E(J)}(k)=2\pi\delta(k^2)\theta(k_0)\tilde J(k)$ projects to the positive frequency forward light-cone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The commutation relations are not positive semi-definite for arbitrary test functions $H$, $\Bigl[\bigl&amp;lt;\hat A^+,H^*\bigr&amp;gt;,\bigl&amp;lt;\hat A^-,H\bigr&amp;gt;\Bigr]\not\ge 0$, which is fixed by the Gupta-bleuler condition, which can be stated in this formalism as $\delta\hat A^+\left|\psi\right&amp;gt;=0$, for &lt;em&gt;all&lt;/em&gt; states $\left|\psi\right&amp;gt;$, not just for the vacuum state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the Hodge decomposition theorem, we can write the test function $H$ uniquely as $H=d\phi+\delta F+\omega$, where $\phi$ is a 0-form, $F$ is a 2-form, and $\omega$ is a harmonic 1-form, so we can write&#xA;$$\bigl&amp;lt;\hat A^+\!,H\bigr&amp;gt;\left|\psi\right&amp;gt;\!&#xD;&#xA;  =\bigl&amp;lt;\hat A^+\!,d\phi+\delta F+\omega\bigr&amp;gt;\left|\psi\right&amp;gt;\!&#xD;&#xA;  =\Bigl(\!\bigl&amp;lt;\delta\hat A^+,\phi\bigr&amp;gt;+\bigl&amp;lt;\hat A^+\!,\delta F+\omega\bigr&amp;gt;\!\Bigr)&#xD;&#xA;              \left|\psi\right&amp;gt;\!&#xD;&#xA;  =\bigl&amp;lt;\hat A^+\!,\delta F+\omega\bigr&amp;gt;\left|\psi\right&amp;gt;.$$&#xA;The harmonic 1-form $\omega$ has to be zero to satisfy the Schwartz space condition, and by the left action of $\hat A^-$ on arbitrary states we have the same projection of the arbitrary test function $H$ to $\delta F$, so all we are left with is the electromagnetic field observable $\hat\Phi_F=\hat A_{\delta F}=\bigl&amp;lt;\hat A,\delta F\bigr&amp;gt;$. Unsurprisingly, in the free field case, in the absence of charges, the electromagnetic potential observable is exactly equivalent to just the electromagnetic field observable, for which we can verify that $\Bigl[\bigl&amp;lt;\hat A^+,\delta F^*\bigr&amp;gt;,\bigl&amp;lt;\hat A^-,\delta F\bigr&amp;gt;\Bigr]\ge 0$, using which we can use the GNS construction of a free field Hilbert space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is a reference request. Is there any literature that uses this kind of mathematical formalism for the quantized electromagnetic field? Even &lt;em&gt;vaguely&lt;/em&gt; the same! My sense is that AQFT has moved to more abstract methods, while more practical interacting QFT has become historically committed to index notations that are little changed from 50 years ago, even though the methods of interacting quantum fields have changed in many other ways, and that mathematicians who take on the structures of QFT, as Folland does in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821847058&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Quantum-Theory-Mathematical-Surveys-Monographs/dp/0821847058&lt;/a&gt;, make strenuous efforts not to let their notation and methods stray too far from the mainline.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-05-19T17:01:58.657" Title="1-form formulation of quantized electromagnetism" Tags="&lt;quantum-field-theory&gt;&lt;electromagnetism&gt;&lt;mathematical-physics&gt;&lt;reference-request&gt;" CommentCount="2" />
  <row Id="10099" PostTypeId="2" ParentId="10097" CreationDate="2011-05-19T17:14:29.533" Score="7" ViewCount="" Body="&lt;p&gt;Dear Ganesh, it's a totally valid and legitimate interaction Lagrangian of the electron field as it interacts via the electroweak interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this is just a part of the overall Lagrangian and only includes the electron - and its neutrino - as it interacts by exchanging photons, W-boson, and Z-bosons. Note that the &quot;parts&quot; of the electron that are spinning differently, the left-handed $\psi_{eL}$ and right-handed $\psi_{eR}$ electron, are treated differently. In this fundamental way of writing it, the gauge fields responsible for all the forces are the fields $\vec A_\mu$ for the $SU(2)$ group and $B_\mu$ for the $U(1)$ factor. Other particles and kinetic terms for electrons, the gauge bosons, and other particles have to be added.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particle physicists would probably not write the explicit $\hbar$ factors because whoever studies the electroweak theory usually uses units in which $\hbar=1$. Also, the claim that the interaction Lagrangian is legit doesn't imply that the creator of the mug understands particle physics. It's more likely that he copied it from somewhere - or, more likely, asked a physicist to give him the text.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-19T17:43:35.357" LastActivityDate="2011-05-19T17:43:35.357" CommentCount="2" />
  <row Id="10100" PostTypeId="2" ParentId="10097" CreationDate="2011-05-19T17:14:43.043" Score="2" ViewCount="" Body="&lt;p&gt;It's the electroweak Lagrangian. In laymen's terms, it describes how photons, electrons and neutrinos interact with one another. Each term represents a possible interaction. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Electroweak_interaction&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; wiki page for more info. You will see a similar, equivalent Lagrangian expression there with some more detailed explanations, should you want it.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastActivityDate="2011-05-19T17:14:43.043" />
  <row Id="10101" PostTypeId="1" AcceptedAnswerId="10104" CreationDate="2011-05-19T17:25:11.193" Score="0" ViewCount="62" Body="&lt;p&gt;The wave is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bar{E} = E_{0} sin(\frac{2\pi z}{\lambda} + wt) \bar{i} + E_{0} cos(\frac{2 \pi z}{\lambda}+wt) \bar{j}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's simplify with $z = 1$. Now the xy-axis is defined by parametrization $(sin(\frac{2\pi }{\lambda}+wt), cos(\frac{2\pi }{\lambda} + wt)$ where $t$ is time and $\lambda$ is wavelength. This parametrization satisfy the equation $1^2=x^{2}+y^{2}$, a circle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's variate the value of $z$. We know now that it cannot move into x or y coordinates or do we? Not really, the latter simplification is naive -- $x-y$ parametrization depends on the dimension $z$ -- but can we see something from it? If so, how to proceed now?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is that the wave moves along the $z$ -axis to the negative direction as $t$ increases, a thing I cannot see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I am trying to solve this kind of problems is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Parametrize the equation&lt;/li&gt;&#xA;&lt;li&gt;suppose other things constant and change one dimension, observe&lt;/li&gt;&#xA;&lt;li&gt;check other variable&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;...now however I find it hard to parametrize the $z$ so a bit lost. So how can I visualize the wave with pen-and-paper?&lt;/p&gt;&#xA;" OwnerUserId="623" LastEditorUserId="623" LastEditorDisplayName="" LastEditDate="2011-05-19T17:46:17.913" LastActivityDate="2011-05-19T17:49:25.773" Title="Help me to visualize this wave equation in time, to which direction it moves?" Tags="&lt;waves&gt;&lt;wavefunction&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10102" PostTypeId="2" ParentId="10095" CreationDate="2011-05-19T17:42:23.143" Score="0" ViewCount="" Body="&lt;p&gt;Here is a way measurement of the Sun's anomalous movement could theoretically take place. Assuming a circular orbit (eccentricity = 0) for the distant Jupiter, the Sun's motion around its barycenter would be (2 pi (60.3) AU) / 1.6E7 years or ~3550 km/year. If a detector was placed at the barycenter, 60.3 AU from the Sun, it would find the Sun's annual displacement to be ~0.08 arcseconds. Over 10 years, the ~0.8 arsecond movement would be readily measurable. However, as Luboš pointed out, it's probably impossible to do in practice.&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-05-19T17:42:23.143" CommentCount="4" />
  <row Id="10103" PostTypeId="1" AcceptedAnswerId="10106" CreationDate="2011-05-19T17:42:56.560" Score="1" ViewCount="94" Body="&lt;p&gt;In Hamiltonian mechanics, we show $\{H,L_z\}=0$, which can be interpreted as the conservation of angular momentum around $Oz$. Following the same idea, how can we interprete $\{H,L^2\}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the interpretation the same (or &lt;em&gt;only&lt;/em&gt; similar) as in quantum mechanics for $[H,L_z]$ and $[H,L^2]$?&lt;/p&gt;&#xA;" OwnerUserId="2499" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-19T17:56:52.773" LastActivityDate="2011-05-19T18:02:43.173" Title="interpretation of $\{H,L^2\}$" Tags="&lt;quantum-mechanics&gt;&lt;hamiltonian-mechanics&gt;" AnswerCount="1" />
  <row Id="10104" PostTypeId="2" ParentId="10101" CreationDate="2011-05-19T17:43:49.587" Score="1" ViewCount="" Body="&lt;p&gt;Would you agree that $\vec{E}$ depends only on $\frac{2 \pi z}{\lambda} + \omega t$ (taking $E_0$ to be a constant)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, we can imagine picking some spot it space and time, taking note of the value of $\vec{E}$ at that point and looking to see how we have to move to keep the value constant in time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{2 \pi z}{\lambda} + \omega t = C $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where C is determined entirely by our initial choice of space--time location. So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ z  = z(t) =  \frac{ \lambda }{2 \pi} \left( C - \omega t \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;represents a locus of $z$-positions as a function of time where $\vec{E}$ continues to have the same value it had at our starting point. And those positions move in the negative $z$ direction as time increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question for the studuent: how fast do they move?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to answer by inspection.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-19T17:49:25.773" LastActivityDate="2011-05-19T17:49:25.773" CommentCount="1" />
  <row Id="10105" PostTypeId="1" AcceptedAnswerId="10111" CreationDate="2011-05-19T17:51:58.190" Score="1" ViewCount="174" Body="&lt;p&gt;It is known that good DI water have resistance ~20 MΩ·cm&#xA;But how can I measure that? Using good vanilla ohmmeter (with 2000 MΩ range) showed crazy results (too low, not much dependent from distance between probes).&#xA;I need that to compare DI water from 2 sources.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-05-19T19:05:00.220" Title="How to measure resistance of DI water" Tags="&lt;water&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="10106" PostTypeId="2" ParentId="10103" CreationDate="2011-05-19T18:02:43.173" Score="3" ViewCount="" Body="&lt;p&gt;Dear Isaac, yes, $\{H,L^2\}=0$ holds for spherically symmetric Hamiltonians and it means that the magnitude of the angular momentum will be conserved in time. More precisely, the squared magnitude of the angular momentum is conserved but classically, it's the same thing. The time derivative of any observable, whether it's composite or not, is given by its Poisson bracket with the Hamiltonian - all these things are just a classical limit of &quot;commutators&quot; (Poisson bracket) with the corresponding &quot;operators&quot; (observables).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\{H,L_x\}=\{H,L_y\}=\{H,L_z\}=0$, then one may also prove $\{H,L^2\}=0$. This is simplest to prove in the quantum mechanical language. If $H$ commutes with $L_z$, it also commutes with $L_z^2$, and similarly it may commute with $L_x^2$ and $L_y^2$ - and with the sum of these three terms which is $L^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interpretation of classical physics and quantum mechanics is always different but the maths of the Poisson brackets directly reflects the quantum commutators. In quantum mechanics, all these things can only be measured probabilistically and so on; this portion of quantum mechanics is universal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the angular momentum, it's useful to talk about $L^2$ and $L_z$ in quantum mechanics - and we usually no longer add $L_x$ or $L_y$. The pair $L^2$ and $L_z$ commutes with each other, $[L^2,L_z]=0$, but commutators such as $[L_x,L_z]$ are not zero but rather $-i\hbar L_y$, and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, $L^2$ has eigenvalues which are always a bit greater than the maximum eigenvalue of $L_z^2$: you can't ever rotate a nonzero angular momentum exactly in the $z$-direction. Equivalently, the uncertainty principle always guarantees that $L_x$ and $L_y$ can't be simultaneously zero when $L_z$ is nonzero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eigenvalues of $L^2$ are $l(l+1)\hbar^2$ for integer values of $l$ while the eigenvalues of $L_z$ are integers between $-l$ and $+l$, where the latter $l$ is the same $l$ used in the formula for the eigenvalue of $L^2$.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-19T18:02:43.173" />
  <row Id="10107" PostTypeId="2" ParentId="10051" CreationDate="2011-05-19T18:05:36.730" Score="1" ViewCount="" Body="&lt;p&gt;No.  You do not get more or less magnetism by cooling a permanent magnet made out of ferromagnetic material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are thinking of superconduction in some materials at certain (low) temperatures.  This phenomenon affects resistivity of the material, and has nothing to do directly with magnetism.  The zero resistivity can be exploited to set up a never-ending current which can produce an electromagnet. &lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-05-19T18:05:36.730" CommentCount="1" />
  <row Id="10108" PostTypeId="2" ParentId="10095" CreationDate="2011-05-19T18:15:17.213" Score="1" ViewCount="" Body="&lt;p&gt;There is a big error in Your argumentation: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Sun's Hill sphere, its&#xA;  gravitational influence within the&#xA;  Milky Way, is ~2.2 ly. So it's&#xA;  possible, however unlikely, that a&#xA;  Jupiter size orphan planet could&#xA;  attain solar system orbit, say with a&#xA;  semi-major axis of one light year and&#xA;  a ~1.6E7 year period.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A body coming from &quot;outside&quot; (more so if it has some speed relative &#xA;to the sun), cannot access a orbit around the sun without leaving &#xA;the excess energy somewhere. So, in addition to the unlikely meeting &#xA;with such a lonely planet, You need a even more unlikely swing-by &#xA;braking eg by passing one of the outer heavy planets at &#xA;the right distance and direction. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-05-19T18:40:25.267" LastActivityDate="2011-05-19T18:40:25.267" CommentCount="3" />
  <row Id="10109" PostTypeId="5" CreationDate="2011-05-19T18:15:42.257" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-19T18:15:42.257" LastActivityDate="2011-05-19T18:15:42.257" />
  <row Id="10110" PostTypeId="4" CreationDate="2011-05-19T18:15:42.257" Score="0" ViewCount="" Body="Questions that ask about some aspect of physics research or study which doesn't involve the actual physics, such as career advice or historical trivia. In other words, soft questions can be answered without using physical reasoning." OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-29T21:57:48.307" LastActivityDate="2011-07-29T21:57:48.307" />
  <row Id="10111" PostTypeId="2" ParentId="10105" CreationDate="2011-05-19T18:27:29.983" Score="1" ViewCount="" Body="&lt;p&gt;That DI water has some &lt;strong&gt;specific&lt;/strong&gt; resistance, not resistance! &#xA;To measure that, You need a apropriate probe, inserting &#xA;the tips of a vanilla ohmmeter into the water, excuse, &#xA;is ridiculous! Conductivity measurements have to be made using &#xA;alternating currents. The frequency is about some kHz to some dozen kHz. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this purpose You can buy special meters, called conductometers, &#xA;(In electrochemistry one deals with conductance, not resistance) &#xA;at a range of cheap to luxurious. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-05-19T18:33:03.427" LastActivityDate="2011-05-19T18:33:03.427" CommentCount="3" />
  <row Id="10112" PostTypeId="2" ParentId="10105" CreationDate="2011-05-19T18:27:51.390" Score="0" ViewCount="" Body="&lt;p&gt;Resistance (that's what your meter reads) is related to resistivity (that's the 20M Ohm x cm term) via the geometry of the problem.  If you have a body of the measured substance with UNIFORM CROSS SECTION (doesn't matter what shape) between two parallel and highly conductive plate electrodes (one plate on each end of the measured body, and in full contact with it), then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resistance = Resistivity x (L/A), where A is the area of the uniform cross section, and L is the distance between the electrodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to bring the resistance down to something within the range of you multimeter's accuracy, simply construct the measuring set up to have A &gt;&gt; L, then do the math from your resistance measurement to get back to resistivity.  For instance, if you pick a circular cross seciton, with D = 2r = 20 cm (about 8 inches), then A = 315 cm^2.  If you have 1.5 cm of liquid between the two plates, then A/L = 210.  If your meter reads 100K Ohms, then you know the resistivity is 210 x 100K = 21M Ohms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the conductivity I found for &quot;ultra-pure&quot; water is 18.2M (Ohm x cm).  I forget if that is before or after exposure to air.  The CO2 in the air dissolves into the H2O, changing its Ph and conductivity.  Don't be surprised, if you are working with highly purified water, when its conductivity changes after exposure to air.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-05-19T19:05:00.220" LastActivityDate="2011-05-19T19:05:00.220" CommentCount="3" />
  <row Id="10113" PostTypeId="2" ParentId="10105" CreationDate="2011-05-19T18:27:58.877" Score="0" ViewCount="" Body="&lt;p&gt;Two things come to mind. One a matter of experimental technique, and the other one of definitions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Firstly: how clean were/are the probes? If you introduce a water soluble material from the surface of the probes you will have true screwed up your measurement.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Secondly: Are you sure that value is not for the &lt;em&gt;resistivity&lt;/em&gt; of water. Resistivity is &#xA;generally symbolized with $\rho$ and is related to the resistance by&#xA;$$ R = \int \frac{\rho}{A} dr $$&#xA;where $A$ is the cross-sectional area presented along the path element $dr$. This &#xA;implies that the resistance is dependent on the geometry of the measurement. Just &#xA;plunging the probes into a glass of water isn't going to cut it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The units you give are proper for resistivity, and not for resistance.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-05-19T18:27:58.877" CommentCount="1" />
  <row Id="10114" PostTypeId="2" ParentId="10105" CreationDate="2011-05-19T18:28:01.950" Score="0" ViewCount="" Body="&lt;p&gt;Try instead measuring the conductivity. You'll need to use a &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrical_conductivity_meter&quot; rel=&quot;nofollow&quot;&gt;conductivity meter&lt;/a&gt;. Once you know the conductivity, you'll know the resistivity, as it is the reciprocal: $\rho = \frac{1}{\sigma}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to take a look at &lt;a href=&quot;http://cr4.globalspec.com/thread/15373&quot; rel=&quot;nofollow&quot;&gt;this thread&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3395" LastActivityDate="2011-05-19T18:28:01.950" />
  <row Id="10115" PostTypeId="1" AcceptedAnswerId="10117" CreationDate="2011-05-19T20:38:43.023" Score="1" ViewCount="112" Body="&lt;p&gt;Is acoustic pressure a statistical term? Also, what about pressure in a liquid or a gas?&lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="125" LastEditorDisplayName="" LastEditDate="2011-05-20T02:37:15.167" LastActivityDate="2011-05-20T02:37:15.167" Title="Is acoustic pressure a statistical term?" Tags="&lt;pressure&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10116" PostTypeId="2" ParentId="7339" CreationDate="2011-05-19T20:39:37.767" Score="1" ViewCount="" Body="&lt;p&gt;A cable cannot transport radioactivity away from the reactor using electrical current as a transport mechanism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless - if a material is exposed to a source of particles of sufficiently high energy (high energy particle beam, or neutron beam, for instance), the material can become activated (meaning that some of the atoms in the material will become radioactive), and may remain activated for a very long time, so use of materials that had been previously used in particle beam applications must be carefully considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the answer is yes, in the sense that material can become radioactive from exposure to high energy particles as one would find in a reactor/particle beams and that activation can be long lived and could potentially migrate through reuse/re-purposing of components, but no in the sense that the continued activation is dependent on the local presence of such a high energy source. &lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2011-05-19T20:39:37.767" />
  <row Id="10117" PostTypeId="2" ParentId="10115" CreationDate="2011-05-19T20:55:45.723" Score="1" ViewCount="" Body="&lt;p&gt;Yes, any notion of pressure at least normally assumes that there are many, $N\to\infty$, particles that are the sources of the pressure, so in this sense it is a statistical term. But just like with the temperature, one may assign pressure or temperature to a single particle, too, although for a single particle, we would usually say that it's an accident that it has one momentum or energy or another, so there's still some reason to say that it's a statistical term.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-19T20:55:45.723" />
  <row Id="10118" PostTypeId="1" AcceptedAnswerId="10121" CreationDate="2011-05-19T21:02:07.713" Score="1" ViewCount="64" Body="&lt;p&gt;Wikipedia mentions that the word self in the word &quot;self-inductance&quot; is to differentiate it from &quot;mutual inductance&quot;. But it does not state whether the two things are the same thing. So do the both terms mean that when I have a change in current in some thing such as coil or wire, there will be a voltage generated according to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V = L \frac{dI}{dt}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $L$ measures the inductance of the material.&lt;/p&gt;&#xA;" OwnerUserId="623" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-19T21:16:50.340" LastActivityDate="2011-05-19T21:38:20.843" Title="Are inductance and self-inductance synonyms?" Tags="&lt;electromagnetism&gt;&lt;definition&gt;&lt;inductance&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="10119" PostTypeId="2" ParentId="10072" CreationDate="2011-05-19T21:18:50.570" Score="1" ViewCount="" Body="&lt;p&gt;OK, I will assume you have the under-damped case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you continue reading &lt;a href=&quot;http://en.wikipedia.org/wiki/Damping&quot; rel=&quot;nofollow&quot;&gt;the wikipedia article in question&lt;/a&gt; you'll find the solution for a underdamped oscillator writen as&#xA;$$ x(t) = e^{- \zeta \omega_0 t} (A \cos(\omega_\mathrm{d}\,t) + B \sin(\omega_\mathrm{d}\,t )) $$&#xA;with $A$ and $B$ constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, take you data, and plot all the maxima (or minima) as a function of time, fit an exponential {*} to that and $\zeta \omega_0$ pops right out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you also need to get $\omega_o$ from the data use&#xA;$$     \omega_\mathrm{d} = \omega_0 \sqrt{1 - \zeta^2 } $$&#xA;where you get $\omega_\mathrm{d}$ by extracting the average period (i.e. time from peak to peak) in the data and noting that the period is $T_d = \frac{2 \pi}{\omega_\mathrm{d}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have two equations for two unknowns, so all you have left is a bit of algebra.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;{*} Or plot amplitude versus time on semi-log paper if you are doing this the old-school way. Or plot log(amplitude) versus time on linear--linear graph paper. Then extract the slope.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-19T21:56:26.170" LastActivityDate="2011-05-19T21:56:26.170" CommentCount="7" />
  <row Id="10120" PostTypeId="1" AcceptedAnswerId="10122" CreationDate="2011-05-19T21:22:43.343" Score="3" ViewCount="110" Body="&lt;p&gt;So I learned that the de Broglie wavelength of a particle, lambda = h/p, where h is Planck's constant and p is the momentum of the particle. I also learned that a quantum mechanics description of a particle is a wave packet. I learned that a wave packet is a summation of different basis functions overlapped over each other at say x = 0, and these basis functions are the wave functions, psi(x,t). Or is it the probability density, |psi(x,t)|^2 ??? Please correct me on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learned that the more localized the wave packet is in position space, the more un-localized or uncertain you are about the spread of momentum functions. Please also edit my statement I just said because I don't think I stated it in the best way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, you have a particle that is represented by a wave packet that is localized, thus, it has a spread of momentum, so how can you then know it's de Broglie wavelength? Do you average together all the different momentum the particle has and then plug that average momentum into lambda = h/p_average ??&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2011-05-19T22:05:35.043" Title="Confusion between the de Broglie wavelength of a particle and wave packets" Tags="&lt;quantum-mechanics&gt;&lt;momentum&gt;&lt;particles&gt;&lt;wavefunction&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="10121" PostTypeId="2" ParentId="10118" CreationDate="2011-05-19T21:38:20.843" Score="2" ViewCount="" Body="&lt;p&gt;&quot;Mutual inductance&quot; and &quot;self-inductance&quot; are not the same thing, no. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have two loops (of wire, for example), loop A and loop B. When you run a current through loop A, it produces a magnetic field that fills the space around both loops. The mutual inductance between the two loops, denoted $M_{AB}$, is defined as the ratio of the magnetic flux through loop B to the current running around loop A:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M_{AB} = \frac{\Phi_B}{I_A}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do this calculation with loop A and loop B being two different loops, or you can do it using the &lt;em&gt;same&lt;/em&gt; loop as both A and B (i.e. calculate the flux through the same loop that the current runs around). If you use two different loops, it's called the mutual inductance and labeled $M$ (as above); if you use just one loop, it's called the self-inductance and labeled $L$. &quot;Inductance&quot; is technically a general term that encompasses both possibilities, although it's common to use the word to mean self-inductance because that is the usual meaning.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-05-19T21:38:20.843" />
  <row Id="10122" PostTypeId="2" ParentId="10120" CreationDate="2011-05-19T22:05:35.043" Score="2" ViewCount="" Body="&lt;p&gt;I think it would help you to study the theory of Fourier transforms! Then this momentum/position duality becomes much more apparent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A wavepacket is, like you write, a sum of many momentum states (not probability densities). If you look in the momentum space, the wider the spread of the momentum states, the smaller the spread of the position states will be. This is quite obvious from the Fourier theory which is why I recommend studying this. The Heisenberg uncertainty relation of position/momentum is related to this duality - when you make the momentum spread thinner, the position spread will increase and vice versa so there is a non-zero minimum of uncertainty (width of the distribution) where you have made both position and momentum as localized as you can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The de Broglie relations simply relate momentum p to wavelength lambda - actually not more interesting than the simple observation that a sinusodial wave has a frequency and a wavelength. Short wavelength means higher momentum. So if you have a state with an uncertainty in momentum you also have an uncertainty in the de Broglie wavelength. If you want to take an average of it, go ahead as long as you know it's an average of a distribution of a certain width. This will be OK for many applications, but for some the detailed spread will be crucial also of course. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum: Also please note that the position and momentum descriptions are a duality. You cannot specify both, the full information of the state is in either one of them and then you can Fourier-transform between them to extract a better understanding of the problem or extract some numerical predictions etc. This point is lost in some introductions so I'll take the opportunity here to mention it :)&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-05-19T22:05:35.043" />
  <row Id="10123" PostTypeId="2" ParentId="10094" CreationDate="2011-05-19T22:26:50.480" Score="2" ViewCount="" Body="&lt;p&gt;1) First of all, $\mu$ is &lt;em&gt;not&lt;/em&gt; a measure on the $N$-dimensional Hilbert space $H\cong\mathbb{C}^N $, but on the space $C$ of &lt;a href=&quot;http://en.wikipedia.org/wiki/Density_matrix&quot; rel=&quot;nofollow&quot;&gt;density operators&lt;/a&gt; $\rho:H\to H$, i.e., positive operators $\rho\in B(H)\equiv {\cal L}(H,H)$ with trace $\mathrm{tr}(\rho)=1$. Let us below sketch an argument for the choice of the measure $\mu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) A positive operator $\rho$ can always be diagonalized &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U^{-1}\rho U=  \mathrm{diag}(\lambda_1, \ldots,  \lambda_N) ,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by an unitary operator $U\in U(H)\cong U(N)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The positivity yields &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda_1, \ldots,  \lambda_N\geq 0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the trace condition yields &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1=\mathrm{tr}(\rho)=\sum_{i=1}^N  \lambda_i.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) It is not hard to show that &lt;em&gt;if&lt;/em&gt; the eigenvalues $\lambda_1, \ldots,  \lambda_N$ are all different (which is the generic situation; more precisely, below we shall say that this is true &lt;a href=&quot;http://en.wikipedia.org/wiki/Almost_everywhere&quot; rel=&quot;nofollow&quot;&gt;almost everywhere&lt;/a&gt; wrt. the Lebesgue measure $d^N\lambda$), then &lt;em&gt;any&lt;/em&gt; other unitary operator $V\in U(H)$, that diagonalizes $\rho$, must be related to $U$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ VU^{-1}=  P_{\pi} \ \mathrm{diag}(e^{i\varphi_1}, \ldots, e^{i\varphi_N}),\qquad \pi\in S_{N},\qquad \varphi_1, \ldots, \varphi_N \in \mathbb{R}, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where  $P_{\pi}:H\to H$ is a permutation operator corresponding to a permutation $ \pi\in S_{N}$. So the unitary operator $U\in U(H)$ is unique modulo phase factors and permutations of the eigenvalues &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\lambda_1, \ldots,  \lambda_N)\qquad\rightarrow \qquad(\lambda_{\pi(1)}, \ldots,  \lambda_{\pi(N)}),\qquad \pi\in S_{N}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can thus generically identify a density operator $\rho$ with a unitary operator $U$ and a set of eigenvalues $(\lambda_1, \ldots,  \lambda_N)$ modulo the above mentioned relations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Note that there is not a preferred orthonormal basis in $H\cong\mathbb{C}^N$. Therefore it is natural to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Haar_measure&quot; rel=&quot;nofollow&quot;&gt;Haar measure&lt;/a&gt; $\nu$ on $U(H)\cong U(N)$. Hence, a natural class of measures $\mu$ on the space $C$ of density operators $\rho$ is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu=\nu \times  d^N\lambda\ f(\lambda_1, \ldots,  \lambda_N) \ \delta(1-\sum_{i=1}^N \lambda_i)\ \prod_{j=1}^n \theta(\lambda_j),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\delta$ is the Dirac delta distribution; $\theta$ is the Heaviside step function;  $d^N\lambda$ is the $N$-dimensional Lebesgue measure on the space $\mathbb{R}^N$ of real eigenvalues, and $f: \mathrm{R}^n\to [0,\infty[ $ is a totally symmetric &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(\lambda_{\pi(1)}, \ldots,  \lambda_{\pi(N)})=f(\lambda_1, \ldots,  \lambda_N),\qquad \pi\in S_{N},$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;measurable function, that is integrable &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int d^N\lambda \ f(\lambda_1, \ldots,  \lambda_N) \ \delta(1-\sum_{i=1}^N \lambda_i)\prod_{j=1}^N \theta(\lambda_j) ~&amp;lt;~ \infty$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;on the simplex &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta_{N-1}=\left\{\vec{\lambda}\in \mathbb{R}^N \mid \lambda_1\geq 0 , \ldots,  \lambda_N\geq 0; \sum_{i=1}^N  \lambda_i=1\right\},$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a compact set in $\mathbb{R}^N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) Note that the Haar measure $\nu$ on the compact group $U(H)$ is finite, $\nu(U(H))&amp;lt;\infty$, and the volume of the simplex &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm{Vol}(\Delta_{N-1})=&#xD;&#xA;\int d^N\lambda  \ \delta(1-\sum_{i=1}^N \lambda_i)\prod_{j=1}^N \theta(\lambda_j)~&amp;lt;~ \infty,$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and hence the total volume $\mu(C)&amp;lt;\infty$ is finite. Therefore, one may normalize the measure by dividing with $\mu(C)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;7) A choice $f(\vec{\lambda})=1$ is just a convenient choice, but not necessary. Another natural choice, from the perspective of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_matrix&quot; rel=&quot;nofollow&quot;&gt;Gaussian unitary ensemble (GUE),&lt;/a&gt; is the square of the &lt;a href=&quot;http://www.proofwiki.org/wiki/Vandermonde_Determinant&quot; rel=&quot;nofollow&quot;&gt;Vandermonde&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Vandermonde_matrix&quot; rel=&quot;nofollow&quot;&gt;determinant&lt;/a&gt; $f(\vec{\lambda})=\Delta^2(\vec{\lambda})$.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-20T11:25:31.163" LastActivityDate="2011-05-20T11:25:31.163" CommentCount="5" />
  <row Id="10125" PostTypeId="1" AcceptedAnswerId="10148" CreationDate="2011-05-20T01:25:55.520" Score="1" ViewCount="83" Body="&lt;p&gt;I believe most people are familiar with &lt;a href=&quot;http://en.wikipedia.org/wiki/Bell%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Bell's Theorem&lt;/a&gt;.  If I understand correctly, the violations of Bell's inequalities are possible due to the existance of entangled quantum states.  Although popular discussion often circles around nonlocality questions and doubts about what the experiments say with regards to local realism, we rarely see lengthy discussions about the implications of creating the apparatus itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if it is possible to reasonably view the apparatus as a sort of quantum state detector?  In asking this, I am not thinking about a detector as a device that detects a particle, but as a device that has detected the existance of a possible state of particles.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one thinks of the spin correlation &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aStraightLines.svg&quot; rel=&quot;nofollow&quot;&gt;diagram&lt;/a&gt; in the article, one might interpret the possible outcomes as two possible system response curves (or possibly zero or one).  It seems that if the apparatus is properly tuned to a particular type of particle system, one should always see the quantum result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some clarity that can be brought to this issue?&lt;/p&gt;&#xA;" OwnerUserId="3460" LastActivityDate="2011-05-20T09:58:12.443" Title="Apparatus used to test violations of Bell's inequalities" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" />
  <row Id="10126" PostTypeId="1" AcceptedAnswerId="10132" CreationDate="2011-05-20T02:18:47.317" Score="4" ViewCount="203" Body="&lt;p&gt;As I understand it, string theory (incorporating bosons and fermions) &quot;works&quot; in $9+1=10$ spacetime dimensions.  In the context of dual resonance theory, I've read descriptions of why that is &quot;physically&quot;, i.e., how in the process of formulating a quantum theory of relativistic interacting (closed and open) strings, the only way one can do things like get rid of ghosts, for example, is to have ten dimensions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a way to appreciate more intuitively this restriction on dimensionality.  Is there something special about ten-dimensional spaces purely from the point of view of mathematics?  Without reference to the particular detailed exigencies of building a physical theory based on strings, does ten-dimensional topology have some unique features that are lost when one dimension is added or subtracted?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to pose the question:  For purposes of this discussion, let's stipulate by fiat that the universe has $8+1=9$ dimensions.  Would we humans simply be back to the drawing board in terms of coming up with a unified theory of the fundamental forces?  Would anything in string theory be salvagable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that M-theory adds a dimension.  So if someone wants to answer the question by telling me what's unique or special about eleven dimensions, that's fine.&lt;/p&gt;&#xA;" OwnerUserId="1445" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-20T12:11:09.753" LastActivityDate="2011-05-20T12:11:09.753" Title="Mathematically rather than physically speaking, is there something &quot;special&quot; about 10 (or 11) dimensions?" Tags="&lt;string-theory&gt;" AnswerCount="1" />
  <row Id="10127" PostTypeId="1" AcceptedAnswerId="10131" CreationDate="2011-05-20T02:19:36.987" Score="6" ViewCount="206" Body="&lt;p&gt;This is really a follow up to the &lt;a href=&quot;http://physics.stackexchange.com/questions/2916/shape-of-the-universe/2918#2918&quot;&gt;Shape of the universe&lt;/a&gt; question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first answer to the question, Ted Bunn says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, the best available data seem&#xA;  to indicate that the Universe is very&#xA;  nearly spatially flat. This means&#xA;  that, if we do live in a 3-sphere&#xA;  Universe, the radius of the sphere is&#xA;  very large, and the distance you'd&#xA;  have to travel is much larger than the&#xA;  size of the observable Universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question is, can the available data be used to set a lower bound to the size of the Universe? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, suppose the Universe was actually shaped like a 3 sphere. Then based on our best data, what would the &quot;radius&quot; of that sphere be? How much larger is that than the observable universe?&lt;/p&gt;&#xA;" OwnerUserId="3335" LastEditorUserId="125" LastEditorDisplayName="" LastEditDate="2011-05-20T02:36:44.060" LastActivityDate="2011-05-20T05:31:57.790" Title="Size of the universe" Tags="&lt;cosmology&gt;" AnswerCount="1" />
  <row Id="10128" PostTypeId="2" ParentId="6997" CreationDate="2011-05-20T02:55:54.600" Score="1" ViewCount="" Body="&lt;p&gt;I have often thought of the possibility of using a liquid inside the capsule (with the astronaut in an appropriate breathing apparatus of course).  In addition to the increased drag induced on movement, one might be able to cleverly design circulation cells that would induce a force pushing the astronaut towards a grated floor.  I think arm motion would be the biggest issue to compensate for. One more radical idea would be to use liquid as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Liquid_breathing#Space_travel&quot; rel=&quot;nofollow&quot;&gt;breathing medium&lt;/a&gt; as well.&lt;/p&gt;&#xA;" OwnerUserId="3460" LastActivityDate="2011-05-20T02:55:54.600" CommentCount="1" />
  <row Id="10129" PostTypeId="2" ParentId="10068" CreationDate="2011-05-20T03:03:12.707" Score="1" ViewCount="" Body="&lt;p&gt;Contrary to the general tone of the answers I have read so far, I feel there are indeed real difficulties in quantum mechanics which fall under the general category of &quot;collapse of the wave function&quot;. I think part of the problem is the tendency of people to talk in generalities instead of dealing with specific issues. I am therefore going to list a few of what I would consider some of the more glaring instances of wave function collapse.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The flecks of silver on a photographic plate when exposed to the light of a distant star. Whether the wave function is &quot;real&quot; or simply a form of information, by any conceivable calculation it is awfully spread out by the time it reaches the photographic plate. How does the energy concentrate itself enough to activate a highly localized chemical transition?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Radioactive decay. As far as I know, according to the equations there is a superposition of uranium and lead, or radium and radon, and based on this superposition one can calculate that there is a slow but steady emission of gamma waves and alpha waves. The problem is the clicks in the geiger counter. You don't need to hook up a detonator to a box with a cat in it to see that there is a problem here. (There is also a description of the emission whereby it is a tunneling process of alpha particles trapped in the potential well of the nucleus; I think this is equivalent to the way I described it except for a change of basis states, but if I am incorrect it nevertheless remains the case that there is a wave function describing the slow but steady emission process.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The tracks in a could chamber. The emission of charged particles from a decaying nucleus, according to almost any kind of wave equation, ought to be if not spherically symmetric then at least some superposition of spherical harmonics. Yet the observed tracks in a cloud chamber are alwyas straight as an arrow.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Jack asks several questions which I cannot answer, except to some degree where he asks if the process is instantaneous or spread out in time. The problem of course is that the collapse seems to be instantaneous. A mechanism whereby the process can be explained through normal time-evolution of the wave function would make the notion of collapse unnecessary. It is my personal opinion that eventually such explanations will be found.&lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-05-20T03:03:12.707" />
  <row Id="10130" PostTypeId="1" CreationDate="2011-05-20T03:04:53.267" Score="0" ViewCount="146" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/4669/what-are-the-most-important-discoveries-breakthroughs-in-physics-recently&quot;&gt;What are the most important discoveries/breakthroughs in physics recently?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;The last two decades have seen significant advances in mathematics, where long standing conjectures have been proved, leading to much further research.&#xA;Two great examples would be Wiles' proof of the Taniyama-Shimura conjecture (and hence, Fermat's Last Theorem) and Perelman's proof of the Poincare and Thurston conjectures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have there been any comparable advances in theoretical physics in the last 20 years ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there have been very important experimental discoveries such as the accelerating universe and neutrino oscillations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am asking specifically about theoretical discoveries in the last 20 years where, say, some long unexplained phenomenon has been understood and predictions have matched quantitatively with experiment to high accuracy.&lt;/p&gt;&#xA;" OwnerUserId="3335" LastActivityDate="2011-05-20T03:04:53.267" ClosedDate="2011-05-20T03:27:09.240" Title="Physics breakthroughs in the last two decades" Tags="&lt;soft-question&gt;" CommentCount="3" />
  <row Id="10131" PostTypeId="2" ParentId="10127" CreationDate="2011-05-20T04:05:09.270" Score="6" ViewCount="" Body="&lt;p&gt;Yes, of course. First, the radius of the visible Universe is 46 billion light years. Note that it's more than 13.7 billion light years because while the light from the most distant places of the Universe was travelling here for 13.7 billion light years but all &quot;meters&quot; in the light's trajectory continued to expand so that the current radius is larger, 46 billion light years, where &quot;current&quot; means &quot;measured on the slice of spacetime where all observers are 13.7 billion light years separated from the Big Bang by proper time&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, cosmological measurements such as &lt;a href=&quot;http://www.astro.cornell.edu/academics/courses/astro2201/cosmoparms.htm&quot; rel=&quot;nofollow&quot;&gt;WMAP&lt;/a&gt; have measured the density to be very close to the critical density so the curvature to be something like $1.02\pm 0.02$. So if we assume that the topology is a sphere, its own curvature radius has to be about 50 times larger than the radius of the visible universe - or, approximately, the characteristic curvature radius of spacetime as measured now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I don't know the exact figure - which sensitively depends on the error margin you assign to the measured total $\Omega$ - but it's of order 1 trillion light years.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This lower bound is of course much higher than what you can get if you allow the universe to be e.g. a flat 3-torus. In that case, you have to rule out double images etc. and the lower bound will be comparable to the radius of the visible Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, note that the lower bound on the 3-sphere radius doesn't mean that the minimum volume has to be the volume of the 3-sphere even if the 3-sphere the correct local shape. There are other topologies that are locally 3-sphere, such as the &quot;lens spaces&quot; - the quotient of the 3-sphere with respect to the discrete $ADE$ groups such as the symmetry group of an icosahedron (translated from a subgroup of $SO(3)$ to a subgroup of $SU(2)$ which is embedded to $SO(4)$ so that it can naturally act on a 3-sphere). In those ADE cases, the total volume would be divided by the order of the group.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-20T04:05:09.270" CommentCount="2" />
  <row Id="10132" PostTypeId="2" ParentId="10126" CreationDate="2011-05-20T04:19:26.603" Score="2" ViewCount="" Body="&lt;p&gt;Dear Andrew, the derivation of the critical dimension - which can be done in many &quot;physical&quot; ways - is ultimately a reflection of mathematics and a rather deep one. All of string theory is studied at the level of &quot;mathematics&quot; (as opposed to &quot;experiments&quot;, for example). Maybe you need some &quot;substantially simpler mathematics&quot; or &quot;one that doesn't depend on quantum mechanics&quot;? Well, if the maths is too simple, it works in any dimension and there's nothing special in $D=10$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other, in principle non-stringy properties of 10 dimensions. Some of them are clearly not independent of string theory - they refer to a subset of properties of string theory. For example, 10 dimensions are the dimensionality in which the minimally supersymmetric gauge theory with a Majorana Weyl (real chiral) gaugino exists. This gauge theory is embedded in the $N=1$ $D=10$ vacua of string theory as the low-energy limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also other properties of 10 dimensions which, at least at this moment, look totally independent of string theory. For example, the allowed topologies of black hole solutions undergo some transformation as the dimension changes and 10 is an interesting turning point. Unfortunately, I have forgotten the details - I've heard only one talk about it. The result doesn't seem to imply that higher-dimensional gravity would be &quot;inconsistent&quot;, so that's why I say that it doesn't seem to be a strictly stringy result, but there's still something special about $D=10$ here, too.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-20T04:19:26.603" CommentCount="2" />
  <row Id="10133" PostTypeId="1" CreationDate="2011-05-20T05:13:33.953" Score="2" ViewCount="166" Body="&lt;p&gt;I was watching this video on YouTube: &lt;a href=&quot;http://www.youtube.com/watch?v=DxYwdgHpbKM&quot; rel=&quot;nofollow&quot;&gt;2 Spiral Galaxies w/Supermassive Black Holes Collide&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Around half way, and again almost at the end, the black holes seem to suddenly give off some sort of force which blasts away a lot of the matter around them. What is this effect known as or caused by?&lt;/p&gt;&#xA;" OwnerUserId="3258" LastActivityDate="2011-05-20T21:00:48.680" Title="What is the strange event in this simulation of a galactic collision?" Tags="&lt;black-holes&gt;&lt;astronomy&gt;&lt;collision&gt;&lt;galaxy&gt;" AnswerCount="3" />
  <row Id="10134" PostTypeId="1" CreationDate="2011-05-20T05:21:03.260" Score="1" ViewCount="64" Body="&lt;p&gt;It can be shown the sum over all Feynman diagrams in quantum field theory leads to a unitary S-matrix. Where can I find a proof that the sum over string worldsheets with varying genuses also leads to a unitary S-matrix?&lt;/p&gt;&#xA;" OwnerUserId="3634" LastActivityDate="2011-05-20T07:56:56.527" Title="Why is the sum over string worldsheets of varying genuses unitary?" Tags="&lt;string-theory&gt;" AnswerCount="1" />
  <row Id="10137" PostTypeId="1" CreationDate="2011-05-20T05:56:12.387" Score="5" ViewCount="112" Body="&lt;p&gt;The current zeitgeist here is on interpretations of quantum mechanics, so let me add my own two cents here. As you may know, consistent histories is an alternative interpretation proposed in a series of papers 1990-1994 by Gell-Mann and Hartle. In a shocking article, &lt;a href=&quot;http://arxiv.org/abs/gr-qc/9412067&quot; rel=&quot;nofollow&quot;&gt;Dowker and Kent&lt;/a&gt; proved there are many distinct consistent histories for most quantum systems. Even if we constrain ourselves to quasiclassical histories, i.e. those histories which are close to classical, being quasiclassical now does not guarantee being quasiclassical in the future, or in the past. This is disturbing. Do histories which are quasiclassical for all times exist? If they do, are they unique? Is the nightmare scenario of two everywhere quasiclassical consistent histories which are mutually incompatible possible? &lt;/p&gt;&#xA;" OwnerUserId="3637" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-20T13:46:14.127" LastActivityDate="2011-05-22T16:45:21.453" Title="Uniqueness of quasiclassical consistent histories" Tags="&lt;quantum-mechanics&gt;&lt;quantum-interpretations&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="10138" PostTypeId="1" CreationDate="2011-05-20T06:05:02.207" Score="1" ViewCount="110" Body="&lt;p&gt;What is the difference between a divisor and a homology cycle? What is the difference between a D-brane wrapped around a divisor and a D-brane wrapped around a cycle?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1531" LastActivityDate="2011-06-19T08:00:57.260" Title="D-branes wrapping divisors and/or cycles" Tags="&lt;string-theory&gt;" AnswerCount="1" />
  <row Id="10139" PostTypeId="1" CreationDate="2011-05-20T06:46:03.613" Score="2" ViewCount="108" Body="&lt;p&gt;If I know the:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Frequency&lt;/li&gt;&#xA;&lt;li&gt;Effective Radiated Power&lt;/li&gt;&#xA;&lt;li&gt;Height Above Average Terrain&lt;/li&gt;&#xA;&lt;li&gt;Radiation Center Above Mean Sea Level&lt;/li&gt;&#xA;&lt;li&gt;Radiation Center Above Ground Level&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;of a radio signal / tower, is there a way to determine the rough range of the radio signal?&lt;/p&gt;&#xA;" OwnerUserId="3638" LastEditorUserId="3638" LastEditorDisplayName="" LastEditDate="2011-05-20T07:22:42.847" LastActivityDate="2011-06-19T14:01:01.823" Title="Determining a radio signal's range" Tags="&lt;radio&gt;" AnswerCount="2" />
  <row Id="10140" PostTypeId="1" CreationDate="2011-05-20T06:47:39.817" Score="4" ViewCount="402" Body="&lt;p&gt;Not too long ago, Susskind and Bousso uploaded the article &quot;&lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1105/1105.3796v1.pdf&quot; rel=&quot;nofollow&quot;&gt;The Multiverse Interpretation of Quantum Mechanics&lt;/a&gt;&quot; with the thesis that the many-worlds interpretation and the multiverse of eternal inflation are one and the same thing. The parallel worlds of one are exactly the same thing as the parallel worlds of the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, they claim decoherence can't happen over a complete description of the future light-cone of the measurement. Then, they apply that principle to eternal inflation. Without decoherence, superpositions of nucleating bubbles and metastable vacua can't decohere. According to the anthropic principle, most bubbles have no conscious observers, but an exponentially small minority do. Apply black hole complementarity to causal horizons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, somehow, in a way I can't follow, they combine causal diamond worlds into a global multiverse.  Then they claim decoherence is reversible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My head is spinning. What are your opinions on this paper?&lt;/p&gt;&#xA;" OwnerUserId="3639" LastActivityDate="2011-05-25T06:10:02.790" Title="Are many-worlds and the multiverse really the same thing?" Tags="&lt;cosmology&gt;&lt;quantum-interpretations&gt;&lt;anthropic-principle&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="1" />
  <row Id="10141" PostTypeId="2" ParentId="10140" CreationDate="2011-05-20T07:48:24.800" Score="2" ViewCount="" Body="&lt;p&gt;No, they're not the same thing and the paper is meaningless pretty much at all levels, see&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2011/05/bousso-susskind-hypermultiverse.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2011/05/bousso-susskind-hypermultiverse.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-20T07:48:24.800" CommentCount="3" />
  <row Id="10142" PostTypeId="2" ParentId="10138" CreationDate="2011-05-20T07:51:32.130" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Divisor_(algebraic_geometry&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Divisor_(algebraic_geometry&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is a particular submanifold of complex codimension one that may be described by a holomorphic equation in the complex coordinates parametrizing the original manifold. It's usually homologically nontrivial as a cycle. But general cycles are much more general; not every cycle (not even a cycle of the right dimension), is a divisor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, Calabi-Yau three-folds are 6-real dimensional. Divisors are holomorphic cycles with 4 real dimensions, so to say. D4-branes wrapped on such cycles will give you supersymmetric states in simplest cases, so they're stable which shows that their stability has to be protected by a nontrivial homology of the cycle. It's likely that all supersymmetric 4-cycles are divisors; you have to ask a better geometer to be sure. However, there are clearly many 4-cycles that are not supersymmetric and that are consequently not divisors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there are also 3-cycles and 2-cycles in Calabi-Yau three-folds which are surely not divisors because they even have wrong dimensions.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-20T07:51:32.130" CommentCount="1" />
  <row Id="10143" PostTypeId="2" ParentId="10134" CreationDate="2011-05-20T07:56:56.527" Score="1" ViewCount="" Body="&lt;p&gt;A way to prove it is to establish the equivalence of this Riemann-surfaces formula with the formulae you get from the light-cone-gauge calculation of the S-matrix that is derived from a Hermitian Hamiltonian and is, therefore, manifestly unitary. See e.g. Chapter 11 (if I remember well) of Green-Schwarz-Witten for a flavor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you may try to prove the unitarity directly in the Euclidean signature Riemann surfaces. You will be led to the formulae for gluing and cutting of the Riemann surfaces, see the end chapters of Volume I of Polchinski's String Theory for a flavor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also make the proof as similar to the analogous proof in QFT as possible - by using string field theory (either light-cone gauge, as in the first paragraph above, or the covariant one). The action's being real is the key to the unitarity in the latter case.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-20T07:56:56.527" />
  <row Id="10144" PostTypeId="2" ParentId="10139" CreationDate="2011-05-20T08:35:14.870" Score="0" ViewCount="" Body="&lt;p&gt;Yes, there is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go to an experienced Radio Ham and ask. &#xA;To get an answer, prepare in addition to the above: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Receiver Antenna design, location, height. &lt;/li&gt;&#xA;&lt;li&gt;Receiver sensitivity&lt;/li&gt;&#xA;&lt;li&gt;Modulation system &lt;/li&gt;&#xA;&lt;li&gt;Kriterium of &lt;strong&gt;acceptable connection&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Alternatively buy book(s) on radio technology and read. &#xA;Preferably older than 60ties, because the topic is no longer &#xA;popular enough to write books about. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be prepared, that for every decade of frequency You have to learn &#xA;new/differrent facts on propagation of EM waves.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-05-20T08:35:14.870" CommentCount="3" />
  <row Id="10145" PostTypeId="1" AcceptedAnswerId="10149" CreationDate="2011-05-20T08:48:53.767" Score="0" ViewCount="116" Body="&lt;p&gt;Consider two people living on two different smooth 1-manifolds $S$ and $T$ as shown in figure 1. The manifold $S$ is a bump function joining the points $A$ and $B$ and the manifold $T$ is formed by joining two identical bump functions at the points $P$ and $Q$. Let the points marked as $H$ denote the homes of the two guys $X$ and $Y$ living on the manifolds $S$ and $T$ respectively with distances (along the curve) from $H$ and $B$ (and $H$ and $Q$) being 40 km. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $X$ and $Y$ are equipped with cars which do not have a reverse gear. Every time $X$ starts on a drive from his home $H$ it is certain that he reaches back home after a 80Km drive. But same is not the case with $Y$. It is not sure how much long a drive does he need to return back to his home or will he even ever return back to his home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is what kind of a manifolds are $S$ and $T$ and how are they related. Is there any paradox here which needs to be resolved. Life of $Y$ can be made more complicated by joining the points $P$ and $Q$ with some more bump functions each lying on different planes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/jnuOv.jpg&quot; alt=&quot;figure 1&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-22T03:01:47.520" LastActivityDate="2011-05-22T03:01:47.520" Title="A question on smooth 1-manifolds" Tags="&lt;mathematical-physics&gt;&lt;mathematics&gt;&lt;topology&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10146" PostTypeId="2" ParentId="10133" CreationDate="2011-05-20T09:07:43.477" Score="1" ViewCount="" Body="&lt;p&gt;I don't know if that video is pointing to a faithful simulation or not, but orbiting and merging black holes give off &lt;em&gt;massive&lt;/em&gt; amounts of gravitational waves (theoretically) which I guess could cause matter to move like that, especially at the merging moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See this &lt;a href=&quot;http://www.nasa.gov/vision/universe/starsgalaxies/gwave.html&quot; rel=&quot;nofollow&quot;&gt;other simulation from NASA which discusses simulating black hole merging&lt;/a&gt; and the corresponding &lt;a href=&quot;http://www.youtube.com/watch?v=GAwO1okR074&quot; rel=&quot;nofollow&quot;&gt;YouTube video&lt;/a&gt; where you can see the ejection of gravitational waves (note that the video you linked visualizes matter, not the gravitational waves, although the latter will of course perturb the former).&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-05-20T09:07:43.477" CommentCount="3" />
  <row Id="10147" PostTypeId="2" ParentId="10072" CreationDate="2011-05-20T09:34:04.523" Score="1" ViewCount="" Body="&lt;p&gt;The viscous damping coefficient is the coefficient $c$ in the formula &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F=-cv$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $F$ is the damping force and $v$ is the velocity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c$ depends on what causes the damping.  If it is a spring in air, then it is likely to be proportional both to the viscosity of the air and to the relevant area of the the spring leading to the damping.  For low speeds in air you can probably use &lt;a href=&quot;http://en.wikipedia.org/wiki/Drag_%28physics%29#Very_low_Reynolds_numbers_.E2.80.94_Stokes.27_drag&quot; rel=&quot;nofollow&quot;&gt;Stokes' approximation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2714" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-20T21:53:28.620" LastActivityDate="2011-05-20T21:53:28.620" CommentCount="4" />
  <row Id="10148" PostTypeId="2" ParentId="10125" CreationDate="2011-05-20T09:58:12.443" Score="1" ViewCount="" Body="&lt;p&gt;When one discusses with experimentalists building an apparatus creating entangled state, one quickly understands that violating Bell's inequalities is a key benchmark for them in order to be sure to have created entanglement. So the Bell's theorem is actually used as &quot;quantum state detector&quot;, or more technically, as an entanglement witness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exists other quantum witnesses, and they are also used, but a Bell's inequality violation is mainly used by experimentalist because:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;it is historically the first entanglement witnesses : it is well known and has been used for almost 3 decades. &lt;/li&gt;&#xA;&lt;li&gt;it is device-independent. If your apparatus is not well tuned (&lt;em&gt;i.e.&lt;/em&gt; if you don't measure what you think you are measuring), many entanglement witness may report entanglement when the is none. For example, if you forget to rotate your polarizer between measurement, it is easy to be in a situation where you think to have perfect quantum correlations while these correlation are classical. However, due to the minimal assumptions needed to derive Bell's theorem, the violation of Bell's inequality is a proof that there is entanglement somewhere in the apparatus, even if is misaligned. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I don't know if I answer your question, but I hope I brought some clarity.&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2011-05-20T09:58:12.443" CommentCount="1" />
  <row Id="10149" PostTypeId="2" ParentId="10145" CreationDate="2011-05-20T10:00:45.937" Score="1" ViewCount="" Body="&lt;p&gt;The examples you are giving of S and T don't leave a lot of room for debate: S is an open interval (say $]0,1[$) and T is a circle, $S^1$. Already, if you're adding the points A and B to S, you're turning your manifold into something which is called a &lt;em&gt;manifold with boundary&lt;/em&gt; (which isn't a manifold in the technical sense of the term).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what you're looking for, but the 'loopiness' of T is captured in the homology/homotopy of the manifold. S has trivial homology, while $H_1(T) = \mathbb{Z}$ and its fundamental group $\pi_1(T) = \mathbb{Z}$ as well. If you add more 'roads' between P and Q, you're indeed making these groups even larger. (An extra road would yield $H_1(T) = \mathbb{Z}^2$, for example.) The fact that S is 'boring' is because it's &lt;em&gt;contractible&lt;/em&gt;. Algebraic topology is the domain which tries to capture these phenomena; I suggest you read the articles &lt;a href=&quot;http://en.wikipedia.org/wiki/Homology_theory&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Homology_theory&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_group&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fundamental_group&lt;/a&gt;. However, in algebraic topology you're usually not very interested in the exact manifold structure, just in the topology of the space you're describing. A branch of mathematics which does care about manifolds is differential geometry, and in that case you have very similar tools to compute the number of loops etc., see &lt;a href=&quot;http://en.wikipedia.org/wiki/De_Rham_cohomology&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/De_Rham_cohomology&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1471" LastActivityDate="2011-05-20T10:00:45.937" CommentCount="3" />
  <row Id="10150" PostTypeId="1" CreationDate="2011-05-20T11:29:17.640" Score="1" ViewCount="55" Body="&lt;p&gt;Can anyone describe the mathematics behind the calculation of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_keyhole&quot; rel=&quot;nofollow&quot;&gt;resonance keyhole&lt;/a&gt; (for a two-body model)?  It seems like the size and position of the keyhole should be a function only of mass and relative velocity, but I'm having difficulty deriving a formulaic solution.&lt;/p&gt;&#xA;" OwnerUserId="3642" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-20T12:55:25.130" LastActivityDate="2011-05-20T12:55:25.130" Title="Calculation of a Gravity Resonance Keyhole" Tags="&lt;gravity&gt;&lt;orbital-motion&gt;&lt;newtonian-gravity&gt;&lt;resonance&gt;&lt;celestial-mechanics&gt;" FavoriteCount="1" />
  <row Id="10151" PostTypeId="1" CreationDate="2011-05-20T13:24:11.747" Score="0" ViewCount="231" Body="&lt;p&gt;How radio waves create the current in antenna in terms of photons? If it is Compton scattering then why is not changed the freuency of photons?&lt;/p&gt;&#xA;" OwnerUserId="3644" LastActivityDate="2011-05-20T17:59:16.753" Title="Radio waves and frequency of photon" Tags="&lt;photons&gt;&lt;antennas&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="10152" PostTypeId="2" ParentId="10139" CreationDate="2011-05-20T13:28:09.907" Score="2" ViewCount="" Body="&lt;p&gt;There is, but it takes a lot of data. Further, depending on the frequency, the signal can bounce around. Commercial radio stations are required by the relevant regulating agencies to have a survey done. This survey produces a report explaining exactly what you are looking for. However, this is usually done by a consultant who specializes in this kind of survey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, you can make guesstimates. These are very rough. ~100mW, a few houses, maybe a block. Five watts I'd say a couple miles, give or take. I can hit the local repeater on 145.37 on five watts and a good antenna from roughly a 30 minute drive away. 100 watts is good enough for DX work, can go quite far. Under the right conditions, you can go across country on five watts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commercial radio stations use a &lt;em&gt;lot&lt;/em&gt; more power, upwards of a kilowatt. This is chiefly because radios for the FM and AM broadcast bands lack an RF pre-amplifier. Ham radio equipment and the like has an RF pre-amp, and as such can pick up much weaker signals. Further, remember that the intensity of the signal drops off with the square of the distance. It's not uncommon to have the antenna having only a few picowatts to work with when receiving.&lt;/p&gt;&#xA;" OwnerUserId="3643" LastActivityDate="2011-05-20T13:28:09.907" CommentCount="1" />
  <row Id="10153" PostTypeId="1" CreationDate="2011-05-20T13:31:23.960" Score="0" ViewCount="114" Body="&lt;p&gt;X-Ray jets emit from galaxies as a result of spinning mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking about the ties between electricity and magnetism here, and how Gauss law ties all this stuff together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, so is there a way to spin X-Rays?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm talking something that could create a weightlessness affect. Turn a device like this on, and it would repel &lt;strong&gt;mass&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/R2jyH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3599" LastActivityDate="2011-05-20T13:31:23.960" ClosedDate="2011-05-20T16:34:16.487" Title="Spinning X-Rays" Tags="&lt;gravity&gt;&lt;theory&gt;&lt;mass&gt;" CommentCount="6" />
  <row Id="10154" PostTypeId="1" CreationDate="2011-05-20T13:59:14.030" Score="0" ViewCount="119" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/7713/predicting-decay-rates-via-the-standard-model&quot;&gt;Predicting Decay Rates via the Standard Model&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;More specifically, does (any) current theory allow for approximate or exact predictions of atomic decay rates and types based purely on nuclear structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if a nucleus has 'p' protons and 'n' neutrons, is there any equation that predicts what the mean lifetime of the radioactive decay will be?&lt;/p&gt;&#xA;" OwnerUserId="3646" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-20T14:15:12.633" LastActivityDate="2011-05-20T14:15:48.337" ClosedDate="2011-05-20T16:09:51.687" Title="Does the Standard model allow for radioactive decay prediction?" Tags="&lt;nuclear-physics&gt;&lt;standard-model&gt;&lt;radioactivity&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10155" PostTypeId="2" ParentId="10133" CreationDate="2011-05-20T14:06:45.170" Score="1" ViewCount="" Body="&lt;p&gt;Material (like gas) in disorganized orbits around any massive body (or even just a common center of gravity) can collapse inward over time (the mechanism is collisions between the bits). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This results in heating the material. Look up &quot;Virial Heating&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the small size and large local field of black holes gases trapped near them can heat up &lt;em&gt;a lot&lt;/em&gt;, and there is the possibility of jets as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Large gas clouds from one galaxy could introduce a lot of new matter to the vicinity of the central black hole in the other resulting in a period of intense radiation.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-05-20T14:06:45.170" />
  <row Id="10156" PostTypeId="2" ParentId="10154" CreationDate="2011-05-20T14:15:48.337" Score="-1" ViewCount="" Body="&lt;p&gt;Not everything is a matter of &quot;Standard Model&quot;! &#xA;Why do You think there is a connection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google for &quot;Valley of Stability&quot;, and You will find &#xA;a lot on radioactivity. Not long ago there was/were &#xA;threads here on this topic.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-05-20T14:15:48.337" CommentCount="2" />
  <row Id="10157" PostTypeId="2" ParentId="2605" CreationDate="2011-05-20T14:24:11.547" Score="0" ViewCount="" Body="&lt;p&gt;Here's a good way to think of it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider your body like a leather wine bag... with a hole in the bottom.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you add less that what is leaking out, the bag will get smaller.&lt;/li&gt;&#xA;&lt;li&gt;If you add more than is leaking out, the bag will get larger.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As your body extracts the nutrients it requires out of stored cells (some carried in the blood stream and some stored as fat), those cells are disposed of (think poop).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With exercise, your body requires more fuel so it extracts nutrients out of more cells. If you are not putting enough in to replenish the supply, your body weight goes down.&lt;/p&gt;&#xA;" OwnerUserId="3599" LastActivityDate="2011-05-20T14:24:11.547" CommentCount="3" />
  <row Id="10158" PostTypeId="2" ParentId="10151" CreationDate="2011-05-20T14:29:10.400" Score="0" ViewCount="" Body="&lt;p&gt;No, it is not a Compton scattering - the electrons in antenna are not really free. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A radio wave is a flux of coherent photons, they act together, not one by one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any EMW makes charges move but bound charges may make a work, so the incident wave may be partially absorbed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides, a low-frequency Compton scattering is quite the same as the classical EMW scattering. So the &quot;scattered&quot; part of the resulting wave is nearly of the same frequency.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-05-20T14:39:18.927" LastActivityDate="2011-05-20T14:39:18.927" CommentCount="12" />
  <row Id="10159" PostTypeId="1" AcceptedAnswerId="10173" CreationDate="2011-05-20T15:14:03.130" Score="1" ViewCount="147" Body="&lt;p&gt;I remember the following fact about a rigid pendulum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point when the pendulum weight is lowest is a stable equilibrium while the point where the pendulum weight is highest is an unstable one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you perturb the suspension point of the pendulum with small, quick oscillations, the upper point becomes a stable equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can someone give me a reference for the stability of the second equilibrium point of a pendulum with small perturbation?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Has this been actually replicated in an experiment (preferably on youtube :) ) or are the necessary parameters impractical?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3415" LastActivityDate="2011-05-20T19:50:07.990" Title="Upper equilibrium point of a pendulum with small perturbations" Tags="&lt;classical-mechanics&gt;&lt;stability&gt;&lt;equilibrium&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="10160" PostTypeId="2" ParentId="5872" CreationDate="2011-05-20T15:30:36.757" Score="0" ViewCount="" Body="&lt;p&gt;There have been conjectures and implicit assumptions in physics that have been disproved with extremely &lt;em&gt;small&lt;/em&gt; counterexamples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for the spirit of the mathematical question, I think an equivalent would be computationally costly simulations that find unsuspected stable configurations, or accelerator experiments at high energies that shatter conjectures in particle physics.&lt;/p&gt;&#xA;" OwnerUserId="3415" LastActivityDate="2011-05-20T15:30:36.757" />
  <row Id="10161" PostTypeId="1" AcceptedAnswerId="10164" CreationDate="2011-05-20T15:39:24.787" Score="1" ViewCount="89" Body="&lt;p&gt;Is there a good chance that gravitational waves will be detected in the next years?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretical estimates on the size of the effect and the sensitivity of the newest detectors should permit a forecast on this.&lt;/p&gt;&#xA;" OwnerUserId="3415" LastActivityDate="2011-05-20T16:01:57.700" Title="Is there a good chance that gravitational waves will be detected in the next years?" Tags="&lt;gravity&gt;&lt;waves&gt;" AnswerCount="2" />
  <row Id="10162" PostTypeId="2" ParentId="10161" CreationDate="2011-05-20T15:42:55.587" Score="1" ViewCount="" Body="&lt;p&gt;Yes, they expect to see a signal when advanced LIGO is up and running.  Unless there is something supressing the Bh-bh, bh-ns and ns-ns merger rate, if gravitational waves exist (and the hulse-taylor binary makes this almost undeniable), then they should have a positive detection by 2020, unless there is something unknown modifying the way that gravitational waves propogate.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-05-20T15:42:55.587" />
  <row Id="10163" PostTypeId="1" CreationDate="2011-05-20T15:45:49.267" Score="2" ViewCount="142" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/9354/question-about-time-dilation&quot;&gt;Question about Time Dilation..&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I have a question about special relativity which was bothering me for a while now. I know that as one approaches the speed of light, time moves slower for him. So, if I start moving as fast as 99% of the speed of light and travel away from Earth for 1 day and come back, I'll see that (suppose) about 1 year has passed on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my question is, if everything is relative, then how can we say that I was the one moving and Earth was the one staying? I mean if I consider myself as the center o the coordinate system, then I can say that Earth went on a travel with the speed of light and came back. So why is that the time slows down for me and not for the Earth? How do we distinguish between the moving and the static object according to relativity?&lt;/p&gt;&#xA;" OwnerUserId="3648" LastActivityDate="2011-05-21T00:18:41.533" ClosedDate="2011-05-21T02:12:51.703" Title="Special Relativity and time" Tags="&lt;special-relativity&gt;&lt;time&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="10164" PostTypeId="2" ParentId="10161" CreationDate="2011-05-20T16:01:57.700" Score="1" ViewCount="" Body="&lt;p&gt;Yes, most likely, unless there is something fundamentally wrong with our understanding of gravity. The most promising candidate for detection is &lt;a href=&quot;http://en.wikipedia.org/wiki/LIGO#Advanced_LIGO&quot; rel=&quot;nofollow&quot;&gt;Advanced LIGO&lt;/a&gt;, which is currently in the process of being designed and built. The &lt;a href=&quot;https://www.advancedligo.mit.edu/&quot; rel=&quot;nofollow&quot;&gt;website&lt;/a&gt; has some really interesting information listed, including the &lt;a href=&quot;https://www.advancedligo.mit.edu/files/schedule.pdf&quot; rel=&quot;nofollow&quot;&gt;construction schedule&lt;/a&gt; (PDF), and the upgrades, such as upgrading from a 10W to a 200W laser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia, they are expecting to start operations sometime in 2014, which will be after they have completed construction and calibrated the instrument. Of particular note is that the higher power laser will make calibrating the mirrors more challenging, so right now they still have one interferometer (the shorter one) in operation and are performing a &lt;a href=&quot;http://www.universetoday.com/14910/can-light-be-squeezed-to-improve-sensitivity-of-gravitational-wave-detectors/&quot; rel=&quot;nofollow&quot;&gt;squeeze test&lt;/a&gt;. Once Advanced LIGO is complete, they are expecting a sensitivity increase by a factor of 10, pushing the detection rate to possibly daily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/NmvF5.jpg&quot; alt=&quot;Advanced LIGO&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may also be good to note that they are still processing the data from the old data runs (by means of &lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein@Home&quot; rel=&quot;nofollow&quot;&gt;Einstein@Home&lt;/a&gt;), so it is still possible that a detection will turn up within the data, although it will be be of a different type.&lt;/p&gt;&#xA;" OwnerUserId="3395" LastActivityDate="2011-05-20T16:01:57.700" CommentCount="6" />
  <row Id="10165" PostTypeId="2" ParentId="10151" CreationDate="2011-05-20T16:18:27.297" Score="1" ViewCount="" Body="&lt;p&gt;An elementary explanation, at high school level:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The beam of radio wave photons are coherent, as Vladimir said. Coherent means that the electric and magnetic field  of each individual photon has a fixed phase with all the others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the wave reaches an antenna,  some of the photons are absorbed, pushing the electrons to a slightly higher energy level (energy h*nu) in the conduction band. Thus it is not scattering but absorption that generates the current with the frequency of the incoming beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is coherence that , as the photon is absorbed, pushes or repulses the electrons in step, so that  a current that has the frequency of the impinging beam is built up.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-20T16:18:27.297" CommentCount="1" />
  <row Id="10166" PostTypeId="2" ParentId="10163" CreationDate="2011-05-20T16:37:47.380" Score="2" ViewCount="" Body="&lt;p&gt;This is known as the Twin Paradox. It isn't really a paradox, and as stated above, this is because only one of the frames is accelerating. The point of view of the guy on the rocket isn't an inertial frame of reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a Wikipedia link with more information:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Twin_paradox&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Twin_paradox&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3626" LastActivityDate="2011-05-20T16:37:47.380" CommentCount="1" />
  <row Id="10167" PostTypeId="2" ParentId="10151" CreationDate="2011-05-20T17:51:32.833" Score="0" ViewCount="" Body="&lt;p&gt;The beam of radio waves with coherent photons may be produced only by MASER. Without that photons do not have correlated phases, e.g. famous 21 cm hydrogen line in radio-astronomy is not consist from coherent photons. For radio waves, produced by usual transmitters situation even “worst” - here there is hardly even to suggest a strict justification to find photons with frequency related with frequency of transmitter. I do not have exact calculations, but to understand the reason of the problem it is enough to recall, that radio-transmitter may emit waves of different shapes and simply described by classical physics, but photon is emitted due to quantum transition between some energy levels and it is very different process. After all, photon by definition is described via time dependent function like $exp(-i\omega t)$ and we may not emit “triangle” photons instead of “sinusoidal” one using transmitter with electric current with tricky time dependance.&lt;/p&gt;&#xA;" OwnerUserId="3351" LastEditorUserId="3351" LastEditorDisplayName="" LastEditDate="2011-05-20T17:59:16.753" LastActivityDate="2011-05-20T17:59:16.753" />
  <row Id="10168" PostTypeId="2" ParentId="9523" CreationDate="2011-05-20T18:02:25.747" Score="1" ViewCount="" Body="&lt;p&gt;Well, the Oregon Medical site does give you (almost) the power on a plane. There's a linear plot of Magnitude vs. angle that you can convert to a plane using x=arctan(angle). On the other hand, calculating Mie scattering is rather simple. Just check Boren &amp;amp; Huffman &quot;Absorption and Scattering of Light by Small Particles&quot;, where they give the explicit formulas and several approximations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I was wondering if what you're trying to do is right. Are you trying to calculate the transmission through clear, calm water? Because in that case, you are using an incorrect approach. Mie scattering is the electromagnetic solution for a single spherical (or elliptical) particle. It works ok if you have many particles, but not too many. If multiple scattering (waves that scatter on more than one particle) becomes important, it's not longer valid. I'm not sure what is the right approach for liquids or solids, but calculating individual particles isn't. Possibly mean-field theory or effective medium, depending on what are you exactly after.&lt;/p&gt;&#xA;" OwnerUserId="3649" LastActivityDate="2011-05-20T18:02:25.747" CommentCount="1" />
  <row Id="10169" PostTypeId="2" ParentId="9523" CreationDate="2011-05-20T19:13:31.013" Score="1" ViewCount="" Body="&lt;p&gt;If multiple scattering happens, the empirical method &#xA;of &lt;a href=&quot;http://www.graphics.cornell.edu/~westin/pubs/kubelka.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Kubelka-Munk&quot;&lt;/a&gt; is the right thing. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-08-10T19:33:25.390" LastActivityDate="2011-08-10T19:33:25.390" CommentCount="1" />
  <row Id="10170" PostTypeId="2" ParentId="5872" CreationDate="2011-05-20T19:20:18.373" Score="1" ViewCount="" Body="&lt;p&gt;Most of the examples named here are/were not &quot;conjectures&quot;. &#xA;Those were(are) solid theories. They had to be adjusted &#xA;in some conditions, but they are not &quot;wrong&quot; in the sense &#xA;of a falsified mathematical conjecture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All movements of bodys in solar system are still calculated &#xA;by Newtons laws, except perihelion shift of mercury. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Nasa &quot;aims&quot; for the moon, epicycles are still used &#xA;to calculate, afaik.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-05-20T19:20:18.373" />
  <row Id="10172" PostTypeId="1" AcceptedAnswerId="10722" CreationDate="2011-05-20T19:49:32.157" Score="5" ViewCount="185" Body="&lt;p&gt;Would the vast and seemingly diffuse clouds of dark matter floating around our galaxy (and most others) absorb gravitational waves? Is this perhaps why we haven't detected any yet?&lt;/p&gt;&#xA;" OwnerUserId="3653" LastEditorUserId="3653" LastEditorDisplayName="" LastEditDate="2011-06-04T16:36:07.540" LastActivityDate="2011-06-04T16:36:07.540" Title="Would dark matter absorb gravitational waves?" Tags="&lt;general-relativity&gt;&lt;gravity&gt;&lt;dark-matter&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="10173" PostTypeId="2" ParentId="10159" CreationDate="2011-05-20T19:50:07.990" Score="2" ViewCount="" Body="&lt;p&gt;I believe you are referring to Kapitza's Pendulum. The Wikipedia site is a decent introduction: &lt;a href=&quot;http://en.wikipedia.org/wiki/Kapitza%27s_pendulum&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Kapitza%27s_pendulum&lt;/a&gt; and has references to further reading (like his original paper).&lt;/p&gt;&#xA;" OwnerUserId="3601" LastActivityDate="2011-05-20T19:50:07.990" CommentCount="1" />
  <row Id="10174" PostTypeId="1" AcceptedAnswerId="12190" CreationDate="2011-05-20T20:13:49.483" Score="6" ViewCount="153" Body="&lt;p&gt;The conceptual model I have been introduced to for cooper pairs in a bulk superconductor is what I would call the &quot;wake&quot; model, where one electron deforms the positively charged lattice, changing the potential energy landscape and causing the second electron to be drawn closer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems like a terrible conceptual picture both because the electrons in a superconductor should have a well defined localized wave function and because the ground state of the BCS theory has zero net momentum, which would suggest they orbit around each other or something... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone can explain how I should be thinking about this I would be very grateful.  &lt;/p&gt;&#xA;" OwnerUserId="3601" LastEditorUserId="2433" LastEditorDisplayName="" LastEditDate="2011-07-12T22:08:59.803" LastActivityDate="2011-07-12T22:08:59.803" Title="A better conceptual model for cooper pairs in a superconductor" Tags="&lt;quantum-mechanics&gt;&lt;condensed-matter&gt;&lt;superconductivity&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="10175" PostTypeId="2" ParentId="10133" CreationDate="2011-05-20T21:00:48.680" Score="3" ViewCount="" Body="&lt;p&gt;Also, in addition to all of this said above, galaxies are full of dust.  When they collide, they will dump a bunch of dust onto all of the stars in the other galaxy.  This then accelerates the rate of fusion in the stars--this means that you will get a bunch of time-correlated supernovae, which will then eject out a bunch of hot gas.  Several of these happening at once can form a shockwave.  Since it is the very last thing that happens, I assume that this is what your'e seeing.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-05-20T21:00:48.680" />
  <row Id="10176" PostTypeId="1" AcceptedAnswerId="10180" CreationDate="2011-05-20T21:52:42.230" Score="1" ViewCount="91" Body="&lt;p&gt;During the winter, my office gets a little cold. I have a unit (essentially a space heater) that heats the area around it to help keep me warm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that it's summer, I realize the unit also has a cool setting. I don't understand how this can work. It takes energy from the wall, and removes energy from the air. That energy has to go somewhere, doesn't it? Where does that energy go if not to the surrounding air?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a picture of the unit. It's model number is 02044. Extensive investigation (i.e. shaking and listening) show no signs of water inside. Apparently it was recalled in 2008 for starting fires. Hopefully this is a newer model? :) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/BDej5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1954" LastEditorUserId="1954" LastEditorDisplayName="" LastEditDate="2011-05-20T22:37:12.943" LastActivityDate="2011-05-20T22:58:40.450" Title="How does my air-cooling unit work?" Tags="&lt;energy&gt;&lt;temperature&gt;&lt;household-physics&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="10177" PostTypeId="1" AcceptedAnswerId="10205" CreationDate="2011-05-20T22:14:52.767" Score="0" ViewCount="119" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Bowen_ratio&quot; rel=&quot;nofollow&quot;&gt;Bowen Ratio&lt;/a&gt; is the ratio of sensible heat flux to latent heat flux, so presumably it gives some information about the relative importance of these processes. But it is not clear how this information can be used to make inference about a system (e.g. the land-air interface of a corn field). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some uses of the Bowen Ratio?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are typical values, and what is the range of meaningful values (e.g. under what conditions does $B_o\to\pm \infty$?)&lt;/p&gt;&#xA;" OwnerUserId="2809" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-21T02:11:30.253" LastActivityDate="2011-05-21T13:50:49.993" Title="Uses and interpretation of the 'Bowen Ratio' ($B_o=SH/LE$)" Tags="&lt;thermodynamics&gt;&lt;energy&gt;&lt;water&gt;&lt;atmosphere&gt;" AnswerCount="1" />
  <row Id="10178" PostTypeId="2" ParentId="10176" CreationDate="2011-05-20T22:33:16.743" Score="2" ViewCount="" Body="&lt;p&gt;I could not find that exact model number, but &lt;a href=&quot;http://www.northerntool.com/shop/tools/product_200307620_200307620&quot; rel=&quot;nofollow&quot;&gt;this looks identical&lt;/a&gt; and is a fan-forced ceramic heater which can operate in fan-only mode.&lt;/p&gt;&#xA;" OwnerUserId="2714" LastEditorUserId="2714" LastEditorDisplayName="" LastEditDate="2011-05-20T22:44:35.097" LastActivityDate="2011-05-20T22:44:35.097" CommentCount="4" />
  <row Id="10179" PostTypeId="2" ParentId="9635" CreationDate="2011-05-20T22:45:41.660" Score="2" ViewCount="" Body="&lt;p&gt;See arXiv:1104.3145, 1103.6035 and 1104.4127&lt;/p&gt;&#xA;" OwnerUserId="3655" LastActivityDate="2011-05-20T22:45:41.660" CommentCount="1" />
  <row Id="10180" PostTypeId="2" ParentId="10176" CreationDate="2011-05-20T22:58:40.450" Score="4" ViewCount="" Body="&lt;p&gt;I did a quick check and came up with a name of Air Tech Fan Heater Model 02044.  Without knowing more, it is likely that in the cool mode it merely acts as a fan and moves air.  Some describe the cooling sensation simply as the &quot;&lt;a href=&quot;http://science.howstuffworks.com/question22.htm&quot; rel=&quot;nofollow&quot;&gt;wind chill effect&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;" OwnerUserId="3460" LastActivityDate="2011-05-20T22:58:40.450" CommentCount="1" />
  <row Id="10181" PostTypeId="1" CreationDate="2011-05-20T23:03:10.350" Score="0" ViewCount="100" Body="&lt;p&gt;I have a question about the gravity probe B experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this site:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2011/04may_epic/&quot; rel=&quot;nofollow&quot;&gt;http://science.nasa.gov/science-news/science-at-nasa/2011/04may_epic/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The measurements they made confirm Einsteins theory. What I'm wondering is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Did they use a control in the experiment? That is did they put gyroscopes in some place where space time was not supposed to be twisted and get a measurement that was not deflected?&lt;/li&gt;&#xA;&lt;li&gt;If they didn't measure a negative result too, how does this the experiment prove the theory correct?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3656" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-21T02:01:14.693" LastActivityDate="2011-05-21T02:01:14.693" Title="Question about gravity probe B" Tags="&lt;general-relativity&gt;&lt;gravity&gt;&lt;experimental-physics&gt;" CommentCount="3" />
  <row Id="10182" PostTypeId="1" CreationDate="2011-05-20T23:47:50.093" Score="3" ViewCount="100" Body="&lt;p&gt;First, a quick remark: I'm a mathematician, now working on some problems coming from physics (in particular Ising models on quasiperiodic chains). A few things I find rather mysterious. I would appreciate your help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the purpose of generality, let's consider the following Ising model on a chain of $N$ nodes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_N = - \sum_{i = 1}^N J_i\sigma_i^{(x)}\sigma_{i+1}^{(x)} - \sum_{i = 1}^N\sigma_i^{(z)},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $J_i$ depending on the node $i$ (we assume no particular order for generality), and $\sigma_i^{(x),(z)}$ the Pauli matrices. By Jordan-Wigner, we can consider the corresponding Fermionic operator given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\widehat{H}_N = \sum_{i,j}\left[c_i^{\dagger}A_{ij}c_j + \frac{1}{2}\left(c_i^{\dagger}B_{ij}c_j^{\dagger}+ H.c.\right)\right],$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c_i$, $1\leq i \leq N$ are anticommuting Fermionic operators and $\left\{A_{ij}\right\},\left\{B_{ij}\right\}$, $1\leq i, j \leq N$ are the elements of appropriately chosen matrices $A, B$, which depend on $\left\{J_i\right\}_{1\leq i \leq N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use periodic boundary conditions. Now we can extend $\widehat{H}_N$ to a lattice of infinite size, by gluing the unit cell of size $N$ infinitely many times. Let us call this new extension $\tilde{H}_N$. Now the questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What is $H.c.$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I am interested in the thermodynamic limit $N\rightarrow\infty$. Is it obvious whether the sequence of operators $\left\{\tilde{H}_N\right\}$ converges, say in strong operator topology, to some well-defined operator $\tilde{H}$ as $N\rightarrow\infty$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me motivate the second question: For a certain sequence $\left\{J_i\right\}$, constructed deterministically with certain properties (so-called &lt;em&gt;quasi-periodic sequence&lt;/em&gt;), I believe I can say something about what Physicists call the &quot;energy-spectrum in the thermodynamic limit&quot;. I'm interested to know whether this energy spectrum is the spectrum (in the usual functional-analytic sense) of some operator $\tilde{H}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help!&lt;/p&gt;&#xA;" OwnerUserId="3657" LastEditorUserId="3657" LastEditorDisplayName="" LastEditDate="2011-05-23T23:49:12.307" LastActivityDate="2011-05-23T23:49:12.307" Title="Convergence of periodic single fermion operators" Tags="&lt;mathematical-physics&gt;&lt;ising-model&gt;&lt;lattice-model&gt;" CommentCount="6" />
  <row Id="10183" PostTypeId="2" ParentId="10163" CreationDate="2011-05-21T00:18:41.533" Score="-1" ViewCount="" Body="&lt;p&gt;It is the Twin Paradox. Browse the net and almost always the return home phase is said to clear the issue.&lt;br&gt;&#xA;But there is no need to return home, nor ...&lt;br&gt;&#xA;They can comunicate using light signaling. They will receive the signals (encoding a message with their own local time ;-) equally red shifted and they can not decide wich one is older, except if they decode the message.&lt;br&gt;&#xA;&lt;strong&gt;In this way I think that the problem is undecidable under the light of Einstein Special Relativity.&lt;/strong&gt;  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-05-21T00:18:41.533" />
  <row Id="10184" PostTypeId="1" AcceptedAnswerId="10343" CreationDate="2011-05-21T01:12:25.873" Score="5" ViewCount="400" Body="&lt;p&gt;&lt;a href=&quot;http://physics.nist.gov/cgi-bin/Compositions/stand_alone.pl?ele=Cu&amp;amp;ascii=html&amp;amp;isotype=some&quot; rel=&quot;nofollow&quot;&gt;This URL&lt;/a&gt; lists the mass of Copper-63 as 62.9295975(6) and &lt;a href=&quot;http://physics.nist.gov/PhysRefData/Handbook/Tables/coppertable1.htm&quot; rel=&quot;nofollow&quot;&gt;this other URL&lt;/a&gt; lists the mass as 62.939598.  These values differ by almost exactly 0.01 which seems hard to explain by experimental error.  Why is it that these values differ in a significant digit but have the same less-significant digits?  Is one of them a typo of the other?  What is the correct value?  What is the origin of both of these values?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This discrepancy was noted by commentators in &lt;a href=&quot;http://www.theoildrum.com/node/7942&quot; rel=&quot;nofollow&quot;&gt;this article about a supposed cold-fusion reactor&lt;/a&gt;.  According to those commentators, this value is relevant to the cold-fusion debate because it makes all the difference as to whether or not the supposed reaction is energetically feasible.  The linked article concerns the same cold-fusion claim discussed in &lt;a href=&quot;http://physics.stackexchange.com/questions/3799/why-is-cold-fusion-considered-bogus&quot;&gt;this previous physics.SE question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: user9325, voix, and user3673 have indicated that the correct answer is 62.929...  I have started a bounty on the &lt;em&gt;origin&lt;/em&gt; of both the correct and incorrect values.&lt;/p&gt;&#xA;" OwnerUserId="2072" LastEditorUserId="2072" LastEditorDisplayName="" LastEditDate="2011-05-24T11:33:42.293" LastActivityDate="2011-05-24T21:28:09.480" Title="Atomic mass of Copper-63?" Tags="&lt;mass&gt;&lt;atoms&gt;&lt;reference-request&gt;&lt;experimental-evidence&gt;&lt;cold-fusion&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="1" />
  <row Id="10186" PostTypeId="1" AcceptedAnswerId="10198" CreationDate="2011-05-21T02:34:09.687" Score="1" ViewCount="191" Body="&lt;p&gt;So as far as I know, nuclear fission uses high weight atomic elements to manufacture power. If the risk of runaway reactions are a major reason for not expanding this technology, why don't we use elements with lower atomic weight (eg. less energy) or lower amounts of fuel (eg. less total mass)?&lt;/p&gt;&#xA;" OwnerUserId="3658" LastEditorUserId="3415" LastEditorDisplayName="" LastEditDate="2011-05-21T08:24:59.480" LastActivityDate="2011-05-22T20:15:40.267" Title="Noob nuclear physics problem, Why do we use high weight atomic elements?" Tags="&lt;nuclear-physics&gt;&lt;nuclear-engineering&gt;&lt;power&gt;" AnswerCount="3" />
  <row Id="10187" PostTypeId="2" ParentId="10186" CreationDate="2011-05-21T03:22:48.860" Score="0" ViewCount="" Body="&lt;p&gt;Nuclear fission uses large atoms splitting apart to generate energy. Only large atomic nuclei are unstable enough for this to happen at any appreciable rate&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-05-21T03:22:48.860" CommentCount="3" />
  <row Id="10188" PostTypeId="2" ParentId="10186" CreationDate="2011-05-21T03:40:26.667" Score="2" ViewCount="" Body="&lt;p&gt;This is a graph of Binding Energy per nucleon for the various nuclei. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/HqE9U.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fission of heavier nuclei would lead to daughters with higher binding energy than the parent (graph slopes downward towards the high mass end). Daughters will be stable, hence energetics favors fission for heavy nuclei. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In Response to comment:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer address fissionable nuclei, i.e elements heavier than iron. It has received justified criticism that this actually does not explain why certain nuclei are &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fissile&quot; rel=&quot;nofollow&quot;&gt;fissile&lt;/a&gt;&quot; (i.e nuclei that undergo fission when struck by slow thermal neutrons). After reading the question, I realize that the OPs concern was really with fissile materials. So I apologize for the initial answer which concentrated on fissionable nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer, btw, is still energetics. &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_fission#Energetics&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; is a good starting point to get an explanation. &lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="2599" LastEditorDisplayName="" LastEditDate="2011-05-22T20:15:40.267" LastActivityDate="2011-05-22T20:15:40.267" CommentCount="4" />
  <row Id="10189" PostTypeId="1" CreationDate="2011-05-21T03:59:17.677" Score="1" ViewCount="125" Body="&lt;p&gt;Is it impossible for a universe that only contains particles/fields with no rest mass to develop life/intelligence? Assume there is no mechanism to generate a rest mass (Higgs, symmetry breaking, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems ridiculous, but I can't think of a good reason why it can't happen since the life/intelligence may not resemble us. Maybe there is an argument using entropy, information theory, etc. that would be persuasive.&lt;/p&gt;&#xA;" OwnerUserId="1040" LastActivityDate="2011-05-21T04:37:29.350" ClosedDate="2011-05-22T02:47:09.017" Title="Could a universe with only massless particles develop life/intelligence?" Tags="&lt;special-relativity&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10190" PostTypeId="2" ParentId="10189" CreationDate="2011-05-21T04:31:25.487" Score="1" ViewCount="" Body="&lt;p&gt;The &quot;no rest mass for particles&quot; happens, according to Big Bang theory at very high energies where everything is flying around and the forces responsible for the binding of quarks into nucleons and nucleons  bound to nuclei are not in effect. No nuclei, no DNA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once these forces jell, the particles have acquired their mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be simpler to say that massless particles move with the speed of light and cannot be bound at those speeds; Binding is needed in order to have  groupings that could organize into life, if not our DNA, some &quot;DNA&quot;. I replied assuming that you were thinking of the time in the evolution of the universe where symmetries were unbroken.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-05-21T04:37:29.350" LastActivityDate="2011-05-21T04:37:29.350" CommentCount="4" />
  <row Id="10191" PostTypeId="1" CreationDate="2011-05-21T04:53:47.027" Score="3" ViewCount="91" Body="&lt;p&gt;What energy is being transformed to heat when you burn a candle?&lt;/p&gt;&#xA;" OwnerUserId="3659" LastEditorUserId="3415" LastEditorDisplayName="" LastEditDate="2011-05-21T08:25:13.693" LastActivityDate="2011-05-21T08:25:13.693" Title="What energy is transformed to heat when a candle is burned?" Tags="&lt;energy&gt;&lt;heat&gt;" AnswerCount="1" />
  <row Id="10192" PostTypeId="2" ParentId="10191" CreationDate="2011-05-21T05:08:07.297" Score="2" ViewCount="" Body="&lt;p&gt;This would be chemical energy that is converted to heat. The heat from the wick melts the wax which gets absorbed in the wick and then gets burnt (which is really oxidation) to produce heat energy.&lt;/p&gt;&#xA;" OwnerUserId="3660" LastActivityDate="2011-05-21T05:08:07.297" CommentCount="4" />
  <row Id="10194" PostTypeId="1" CreationDate="2011-05-21T05:54:13.463" Score="1" ViewCount="76" Body="&lt;p&gt;I'm trying to calculate Stoke's drag constant using the equation found &lt;a href=&quot;http://en.wikipedia.org/wiki/Drag_%28physics%29#Very_low_Reynolds_numbers_.E2.80.94_Stokes.27_drag&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system I wish to calculate it for is a mass oscillating vertically on a spring through air. The temperature is 25 degrees celsius. The area of the base of the mass is 4.9x10^-4 square metres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to find the 'fluid viscosity' in the linked equation, or what 'Stoke's radius' is.&lt;/p&gt;&#xA;" OwnerUserId="3511" LastActivityDate="2011-05-21T05:59:22.480" Title="Stoke's Drag Constant" Tags="&lt;friction&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" />
  <row Id="10196" PostTypeId="2" ParentId="10184" CreationDate="2011-05-21T06:57:58.333" Score="7" ViewCount="" Body="&lt;p&gt;Since both references give the same percentages and the same overall atomic weight, an easy calculation shows that this only works out for the number in the first link, therefore the second decimal should be 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And I think it would be nice to contact the webmasters of the second link.)&lt;/p&gt;&#xA;" OwnerUserId="3415" LastActivityDate="2011-05-21T06:57:58.333" CommentCount="6" />
  <row Id="10197" PostTypeId="2" ParentId="8188" CreationDate="2011-05-21T08:31:02.800" Score="-2" ViewCount="" Body="&lt;p&gt;I was browsing old questions and noticed this one. I think I ought to take issue with the idea you can measure the spin of a single electron. Suppose I prepare an electron in a definite spin state and send it into another room; I don't think there is any way someone else can tell what state I prepared the electron in. Putting it through a Stern Gerlach apparatus certainly won't do. Isn't saying that you can measure the spin of an electron the same as saying you can measure its position and momentum simultaneously?&lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-05-21T08:31:02.800" CommentCount="2" />
  <row Id="10198" PostTypeId="2" ParentId="10186" CreationDate="2011-05-21T10:01:24.690" Score="1" ViewCount="" Body="&lt;p&gt;The nuclei used are the ones which &lt;strong&gt;can&lt;/strong&gt; be fissioned in a chain reaction!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lighter ones You think of will not split. That's simply a result of experiments.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-05-21T10:01:24.690" CommentCount="5" />
  <row Id="10199" PostTypeId="1" AcceptedAnswerId="10901" CreationDate="2011-05-21T10:08:38.363" Score="1" ViewCount="100" Body="&lt;p&gt;Consider a system of many point particles each having a certain mass and electric charge and certain initial velocity. This system is completely governed by the laws of gravitation and electromagnetic field. If this system is left on its own without any external influence, any physical quantity like velocity of a particle or electric field at a point, of this system will vary smoothly with time and space. (keeping aside any singularities that could occur eventually). My question is whether any such system which is governed by such laws of physics (except ofcourse QM) would remain smooth until the occurrence of a singularity, in the absence of any external influence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS : please feel free to edit/suggest changes to the question in case if it could make it more meaningful.&lt;/p&gt;&#xA;" OwnerUserId="540" LastActivityDate="2011-06-08T03:30:16.563" Title="A question on a system of particles governed by laws of gravity and electromagnetic field" Tags="&lt;gravity&gt;&lt;electromagnetism&gt;&lt;mathematical-physics&gt;&lt;many-body&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="10201" PostTypeId="1" CreationDate="2011-05-21T12:06:29.370" Score="3" ViewCount="82" Body="&lt;p&gt;Spin echo experiments have been able to reverse the motions of all the molecules in a gas in statistical mechanics in the manner of Loschmidt. The Fermi-Ulam-Pasta model has solutions with a single mode dispersing, only to recohere after quite some time has elapsed. Can the same thing happen for decoherence? What are the conditions fyor decoherence to be irreversible?&lt;/p&gt;&#xA;" OwnerUserId="3664" LastActivityDate="2011-05-21T19:06:50.517" Title="What are the conditions for decoherence to be irreversible?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="10202" PostTypeId="2" ParentId="9992" CreationDate="2011-05-21T12:39:09.700" Score="1" ViewCount="" Body="&lt;p&gt;Because of the proliferation of notation across multiple physical specialties, I don't know if there is a concise reference.  I don't know what level you are at, so I don't know where you feel comfortable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would start:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Table_of_mathematical_symbols&quot; rel=&quot;nofollow&quot;&gt;table of mathematical symbols&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then move to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0964350408&quot; rel=&quot;nofollow&quot;&gt;Who is Fourier?&lt;/a&gt; as an introduction and desk reference on Fourier Analysis (and there are PHDs I know who keep this on their desk as well, especially once they discover who the lead advisor for the english edition was).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0071455469&quot; rel=&quot;nofollow&quot;&gt;Quantum Mechanics Demystified&lt;/a&gt; by McMahon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0071455450&quot; rel=&quot;nofollow&quot;&gt;Relativity Demysitified&lt;/a&gt; by McMahon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do not be fooled by the cover art, granted these are not textbooks, but they will give you the basic training needed to step into more rigorous books.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would then take time to understand what the Action is and what it means to vary it.  One quick reference is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e) &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0071543821&quot; rel=&quot;nofollow&quot;&gt;Quantum Field Theory Demystifed&lt;/a&gt; by McMahon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Followed by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f) &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691010196&quot; rel=&quot;nofollow&quot;&gt;Quantum Field Theory in a Nutshell&lt;/a&gt; by Zee&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend starting with just the first few chapters of Zee before hitting the best kept dirty little secret paper on ArXiv that I have found to date:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g) &lt;a href=&quot;http://arxiv.org/abs/0810.3328&quot; rel=&quot;nofollow&quot;&gt;A Simple Introduction to Particle Physics&lt;/a&gt; by Robinson, Bland, Cleaver and Dittman&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Garanteed to turn you into a bed wetter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here you will have some interesting choices, but for recreation I would spend 4 weeks and watch and take copious notes (quite seriously) on all the lectures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;h) &lt;a href=&quot;http://see.stanford.edu/see/lecturelist.aspx?coll=84d174c2-d74f-493d-92ae-c3f45c0ee091&quot; rel=&quot;nofollow&quot;&gt;The Fourier Transform and its Applications&lt;/a&gt; by Brad Osgood at Stanford&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could probabaly actually just start with this, but it is important to make sure you take it seriously before doing so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all that, then I would start in on Supersymmetry, which has its own host of deep dark notation, as discussed in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0071636412&quot; rel=&quot;nofollow&quot;&gt;Supersymmetry Demystified&lt;/a&gt; by Labelle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then you need to tie up everything you know and start in on string theory:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;j)  &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0071498702&quot; rel=&quot;nofollow&quot;&gt;String Theory Demystified&lt;/a&gt; by McMahon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here you should be able to read things directly in the arxiv, or dig into more rigorous textbooks.  &lt;/p&gt;&#xA;" OwnerUserId="3460" LastEditorUserId="3460" LastEditorDisplayName="" LastEditDate="2011-05-21T13:05:41.893" LastActivityDate="2011-05-21T13:05:41.893" CommentCount="1" />
  <row Id="10203" PostTypeId="2" ParentId="10201" CreationDate="2011-05-21T12:57:26.393" Score="0" ViewCount="" Body="&lt;p&gt;Decoherence in a quantum mechanical system appears because the phase information is lost. If it were recorded/known then, as in your first example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Spin_echo&quot; rel=&quot;nofollow&quot;&gt;one could reverse it.&lt;/a&gt; The answer for me seems to be &lt;em&gt;complexity&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some sense there exists a single quantum mechanical state function of the universe. To record it one would need oogleplexes of numbers and functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, decoherence serves the same function that the transition from quantum statistical mechanics to thermodynamics does: it describes measurable experimentally quantities for the system under observation. If we knew all the parameters of the statistical ensemble we would get the same answers, but thermodynamics reduces the complexity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly working with a density matrix, the complexity of the quantum mechanical system is reduced to measurable quantities. So it is the number of parameters that have to be reversed that will decide whether decoherence is reversible or irreversible, and this will be time dependent as tools develop. Who would have thought that thousands of Feynman diagrams could be calculated as they are now doing for the needs of LHC experiments.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-21T12:57:26.393" />
  <row Id="10204" PostTypeId="1" CreationDate="2011-05-21T13:45:17.573" Score="1" ViewCount="106" Body="&lt;p&gt;Is there software available that can analyse a 5MHz RF pulse to give a plot of frequency spectrum.&#xA;The signal data is visible on a LCD screen or a print out could be obtained.&lt;/p&gt;&#xA;" OwnerUserId="3666" LastActivityDate="2011-05-24T20:40:54.273" Title="5MHz RF pulse frequency analysed in software" Tags="&lt;frequency&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="10205" PostTypeId="2" ParentId="10177" CreationDate="2011-05-21T13:50:49.993" Score="1" ViewCount="" Body="&lt;p&gt;I haven't seen it for years, but I think it was just a simple assertion to allow some primitive models to be built without solving the microphysics and biology of evaporation/trnaspiration. I think typical values were about a half. One could presumably go one step further and obtain empirical measurements and build those into models.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-05-21T13:50:49.993" />
  <row Id="10206" PostTypeId="1" CreationDate="2011-05-21T15:02:27.780" Score="2" ViewCount="179" Body="&lt;p&gt;On page 1 of this recent &lt;a href=&quot;http://arxiv.org/abs/1105.3796&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; by Bousso and Susskind we read.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This question is not about philosophy.&#xA;  Without a precise form of decoherence,&#xA;  one cannot claim that anything really&#xA;  &quot;happened&quot;, including the specific&#xA;  outcomes of experiments. And without&#xA;  the ability to causally access an&#xA;  infinite number of precisely decohered&#xA;  outcomes, one cannot reliably verify&#xA;  the probabilistic predictions of a&#xA;  quantum-mechanical theory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_large_numbers&quot; rel=&quot;nofollow&quot;&gt;Law of Large Numbers&lt;/a&gt; there is no requirement that I need to perform an infinite number of experiments in order to verify that results will asymptotically approach an expected value.  Why would Quantum Mechanics be any different?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE:  I wanted to find an relatively simple proof of the strong law of large numbers, best I can come up with is the combination of the following two wikipedia entries:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Kolmogorov%27s_three-series_theorem&quot; rel=&quot;nofollow&quot;&gt;Kolmogorov's Three series Theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Kronecker%27s_lemma&quot; rel=&quot;nofollow&quot;&gt;Kronecker's lemma&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional links are available in articles.&lt;/p&gt;&#xA;" OwnerUserId="3460" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-22T15:24:34.473" LastActivityDate="2011-05-22T15:24:34.473" Title="Quantum Physics and the Law of Large Numbers" Tags="&lt;quantum-mechanics&gt;&lt;quantum-interpretations&gt;" CommentCount="7" />
  <row Id="10207" PostTypeId="2" ParentId="8951" CreationDate="2011-05-21T15:37:56.907" Score="0" ViewCount="" Body="&lt;p&gt;To find the centre of mass of an object, you find the centre of gravity which is at the same point within acceptable error for your camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hang any apex of your steadycam and a plumb-line from a point in space. Trace the path the plumb line takes from one apex to another point. Repeat this for another two apexes. Where all three traces intersect is the centre of mass of your steadycam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google &quot;centre of gravity plumbline&quot; for more details.&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-05-21T15:37:56.907" />
  <row Id="10208" PostTypeId="1" AcceptedAnswerId="10209" CreationDate="2011-05-21T15:56:07.273" Score="4" ViewCount="168" Body="&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: are they mathematically possible at all? physically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with finite mass systems, usually the binding energy contributes to the rest-mass of the system. It would seem that even if you could bind two massless fields/particles, the coupled system would still have a finite rest mass because of the binding energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by any chance, is this how higgs boson gives 'mass' to particles?&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-05-22T05:59:20.203" Title="bound states of massless fields?" Tags="&lt;quantum-field-theory&gt;&lt;mass&gt;&lt;higgs-boson&gt;&lt;massless&gt;&lt;rest-energy&gt;" AnswerCount="3" />
  <row Id="10209" PostTypeId="2" ParentId="10208" CreationDate="2011-05-21T16:02:03.967" Score="2" ViewCount="" Body="&lt;p&gt;Dear lurscher, a good question. But a bound state is something that has a negative binding energy - otherwise it wouldn't be bound. If the rest mass of a bound state AB of massless objects A,B is positive, it can clearly decay to A,B with some energy without violating any conservation law. So AB surely can't be quite stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metastable bound states are a different issue. I think that in some moral sense, it is true that a black hole is a bound state of gravitons, and it becomes very long-lived if it is large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is very misleading to interpret massive particles that obtain mass from the Higgs mechanism as bound states with the Higgs boson. The vacuum itself contains a Higgs condensate - but the individual particles such as the electron don't contain any Higgs excitations, so it is not really a bound state. It is the original electron that has different properties because it is moving in a different environment.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-21T16:02:03.967" CommentCount="6" />
  <row Id="10210" PostTypeId="2" ParentId="10184" CreationDate="2011-05-21T16:22:41.390" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://amdc.in2p3.fr/web/masseval.html&quot; rel=&quot;nofollow&quot;&gt;The 2003 Atomic Mass Evaluation&lt;/a&gt;:&#xA;Cu(63) -  62.929597474&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://amdc.in2p3.fr/web/mass95.html&quot; rel=&quot;nofollow&quot;&gt;The 1995 Update to Atomic Mass Evaluation&lt;/a&gt;:&#xA;Cu(63) - 62.929601079&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://amdc.in2p3.fr/masstables/Ame1993/filel.html&quot; rel=&quot;nofollow&quot;&gt;The 1993 Update to Atomic Mass Evaluation&lt;/a&gt;:&#xA;Cu(63) - 62.929600748&lt;/p&gt;&#xA;" OwnerUserId="238" LastActivityDate="2011-05-21T16:22:41.390" CommentCount="3" />
  <row Id="10211" PostTypeId="1" AcceptedAnswerId="10221" CreationDate="2011-05-21T16:51:59.620" Score="-3" ViewCount="145" Body="&lt;p&gt;I had this doubt bugging my mind for a long time about singularities in Physics. I heard that R.Penrose and S.Hawking have proposed that there could be singularities at Blackholes and at the time of the origin of universe (at the time of Bigbang). The concept of singularity is essentially a mathematical one. (unless you call anything that you cannot explain physically as a singularity). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is (which i want to get it resolved once for all so that i stop bugging people with it :-))....We know that the gravitational field has a singularity at points where there are point masses, so is the case with point charges and electric field. So what is it they have proved about existence of singularities. I assume that they most probably have started with a universe which doesn't have point masses. If they did'nt start with point masses, then what did they actually started with. I want to know what they have started with in mathematical terms and physical terms, once for all. I am open to read some references as well to resolve this doubt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am more puzzled as to why I am not getting an answer to the question I've asked which doesn't involve any analysis but a plain request for information. Did they (R.P and S.H) assume existence of point masses in their model of universe (or objects or whatever it is) or they have assumed smooth matter distributions ? This is a fairly straight forward request for information....I do not understand why I don't get the answer for the question I've asked...rather I get something else.&lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="540" LastEditorDisplayName="" LastEditDate="2011-05-22T04:49:51.483" LastActivityDate="2011-05-22T04:49:51.483" Title="a question about singularities in gravity and Physics in general" Tags="&lt;general-relativity&gt;&lt;singularity&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10212" PostTypeId="2" ParentId="10208" CreationDate="2011-05-21T16:51:59.807" Score="1" ViewCount="" Body="&lt;p&gt;If a bound system is complicated, it may have &quot;massless&quot; excitations. They are quasi-particles. I think all that we observe are quasi-particles rather than free particles in an empty space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solitons may be considered as bound states of different &quot;massless&quot; harmonics in a non-linear dispersive medium.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-05-21T17:20:48.207" LastActivityDate="2011-05-21T17:20:48.207" CommentCount="2" />
  <row Id="10213" PostTypeId="5" CreationDate="2011-05-21T17:03:39.563" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-05-21T17:03:39.563" LastActivityDate="2011-05-21T17:03:39.563" />
  <row Id="10214" PostTypeId="4" CreationDate="2011-05-21T17:03:39.563" Score="0" ViewCount="" Body="Questions about publications, preferably classical papers or books done in the past, but can also include recent publications." OwnerUserId="-1" LastEditorUserId="3064" LastEditorDisplayName="" LastEditDate="2011-05-21T17:12:12.567" LastActivityDate="2011-05-21T17:12:12.567" />
  <row Id="10215" PostTypeId="2" ParentId="10211" CreationDate="2011-05-21T18:13:05.477" Score="0" ViewCount="" Body="&lt;p&gt;I don't think there is any doubt on point masses being a singularity --their very definition asserts that. I will be leaving Hawkings out of my post as has retracted from his original position on the big bang being a singularity (general relativity breaks at times &amp;lt; planck times).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both types of singularities that are under investigation are for real objects. As your question specifically asks about the assumptions Penrose made, so I think that perhaps the best answer can be found in &lt;a href=&quot;http://www.ias.ac.in/jarch/jaa/17/213-231.pdf&quot; rel=&quot;nofollow&quot;&gt;a paper he authored&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Penrose has not proven the existence of singularities. What he has shown is that assuming Einstein's equations hold inside a black hole and if the observed matter densities &gt; 0 for future pointing time like vectors, then we will end up with some sort of a singularity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can of course be other, yet undiscovered, interactions that act at closer distances and prevent singularities but that's outside the scope of his theorem (theorems prove something based on some theories and they act to demonstrate what those theories predict).&lt;/p&gt;&#xA;" OwnerUserId="3669" LastActivityDate="2011-05-21T18:13:05.477" />
  <row Id="10216" PostTypeId="2" ParentId="10140" CreationDate="2011-05-21T18:59:55.547" Score="1" ViewCount="" Body="&lt;p&gt;The standard global picture of eternal inflation doesn't predict domain walls, vacua and metric take on definite classical values! It predicts a quantum superposition, and what's wrong with that?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Definition I&lt;/strong&gt; Consider an instance of decoherence (or &quot;collapse of a wave function&quot;)&#xA;  in a Hilbert space $\mathcal{H}_S$, which occurs as a result of entanglement with another Hilbert space $\mathcal{H}_E$. The event will be said to happen if the entanglement between $\mathcal{H}_E$ and $\mathcal(H)_S$ is irreversible; and the system S can then be treated as if it was in one of the pure states that constitute the basis that diagonalizes the density matrix obtained by tracing over the Hilbert space $\mathcal(H)_E$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Postulate I&lt;/strong&gt; Things happen.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In other words, there exist some entanglements in Nature that will not be reversed with&#xA;  any finite probability.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And by finite probability, he means nonzero, and by nonzero, $\exp(-\exp(10^{123}))$ is considered nonzero because it's not &lt;em&gt;exactly&lt;/em&gt; zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WOW! That's all I can say!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is followed by another bizarre claim that to make any sense of probabilities, the number of trials needs to be infinite. Not the limit of infinite size, or very large sizes, but infinite itself. They insist upon infinitely many trials. Even defenders of the frequentist camp of probabilities don't make such an extreme claim.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And by &lt;em&gt;observable&lt;/em&gt;, I mean that the world is big enough that the observable can be measured infinitely many times by irreversible entanglement.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Please take note of the qualifier infinitely many times, and irreversible entanglement, by which they mean exactly zero probability of reversal. Such absolutism!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reminds me of some cosmologists who some time back, came up with the absolute postulate that the universe has to be a computer with infinite resources, with infinite memory, and able to compute infinitely many time steps. From such a dubious postulate, they concluded the universe must end in an Omega point big crunch with some future civilization taking over the entire universe, and constructing a gigantic computer which can run faster and faster with no limit as time approaches the big crunch singularity.&lt;/p&gt;&#xA;" OwnerUserId="3671" LastActivityDate="2011-05-21T18:59:55.547" />
  <row Id="10217" PostTypeId="2" ParentId="10201" CreationDate="2011-05-21T19:06:50.517" Score="1" ViewCount="" Body="&lt;p&gt;An article you might be interested in: &lt;a href=&quot;http://www.physics.arizona.edu/~cronin/Research/Lab/some%20decoherence%20refs/RBH97.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.physics.arizona.edu/~cronin/Research/Lab/some%20decoherence%20refs/RBH97.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3672" LastActivityDate="2011-05-21T19:06:50.517" />
  <row Id="10218" PostTypeId="2" ParentId="10184" CreationDate="2011-05-21T19:36:16.680" Score="1" ViewCount="" Body="&lt;p&gt;I've started &lt;a href=&quot;http://en.wikipedia.org/wiki/Talk%3aIsotopes_of_copper&quot; rel=&quot;nofollow&quot;&gt;a discussion on this at the Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I had to guess at the origin of this discrepancy it is a typo in one of the old CRC's where someone typed '3' instead of the correct digit of '2'.  If that's the case there may have been merely an internal memo correcting the typo -- if that.&lt;/p&gt;&#xA;" OwnerUserId="3673" LastActivityDate="2011-05-21T19:36:16.680" />
  <row Id="10219" PostTypeId="2" ParentId="10208" CreationDate="2011-05-21T19:54:18.300" Score="1" ViewCount="" Body="&lt;p&gt;In the comments here two issues are being confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is different if the question is : can massless fields be bound?&#xA;It is another story if the question is: can massless fields make a bound state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gluons are massless and are bound in a nucleon, with virtual exchanges with quarks. Photons are massless and are bound in an atom with virtual exchanges with the electrons. So the answer to the first question is, yes, massless fields can be bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly the photons cannot form a bound state, virtual or not, because they do not carry charge with which to exchange another photon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gluons because they carry colour can, within an interaction,( say proton proton,)  form a glueball, which would then decay into quarks and gluons with end result the mesons  we know. There can be no stable glueball because of the form of QCD . It seems from lattice QCD calculations that glueballs can exist. Here is a reference that offers &lt;a href=&quot;http://prd.aps.org/abstract/PRD/v70/i11/e114033&quot; rel=&quot;nofollow&quot;&gt;a glue ball candidate&#xA;,&lt;/a&gt; the eta(1440) which decays to mesons with a 20MeV width. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is massless fields cannot form a stable bound state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Gluons carry colour charge and thus can exchange gluons among themselves, which is how the glueball is mathematically created .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason it is not stable is because there are lower energy states to which it can decay, which is the same reason as for all non stable particles' decays. In order for it to be stable it would have to have a mass smaller than the combined mass of two quarks, orderof Mev. Calculations on the lattice QCD give &lt;a href=&quot;http://prd.aps.org/abstract/PRD/v29/i5/p928_1&quot; rel=&quot;nofollow&quot;&gt;masses that are of the order&lt;/a&gt; of the rho(770Mev) meson,  for the lowest glue ball state. So it is a matter of the lagrangian and the constants entering the calculations. We live in a universe where there are no stable glue balls.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-05-22T05:59:20.203" LastActivityDate="2011-05-22T05:59:20.203" CommentCount="4" />
  <row Id="10220" PostTypeId="1" CreationDate="2011-05-21T21:03:37.090" Score="9" ViewCount="259" Body="&lt;p&gt;The speed of light in a material is defined as $c = \frac{1}{\sqrt{\epsilon \mu}}$. There are metamaterials with negative permittivity $\epsilon &amp;lt; 0$ and permeability $\mu &amp;lt; 0$ at the same time. This leads to a negative refractive index of these materials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But do (meta-) materials exist with only negative $\epsilon &amp;lt; 0$ and positive $\mu &amp;gt; 0$ or vice versa? This would lead to a complex speed of light inside such materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the consequences of a complex speed of light? Could particles reach unlimited speed inside these materials? Would there still be Cherenkov radiation? &lt;/p&gt;&#xA;" OwnerUserId="1648" LastActivityDate="2011-06-08T22:09:06.960" Title="Can the speed of light become complex inside a metamaterial?" Tags="&lt;speed-of-light&gt;&lt;metamaterial&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="10221" PostTypeId="2" ParentId="10211" CreationDate="2011-05-21T22:18:43.630" Score="0" ViewCount="" Body="&lt;p&gt;What they proved is that in general relativity time-like world-lines (i.e. real observer paths) terminate in finite proper time i.e. the time experienced by the observer on the object is finite. This occurs forward in time if a trapped surface exists in space time. Such a surface has all light rays converging from it whatever direction they point and is a sign of a black hole. Backwards in time it occurs cosmologically because of the expanding universe. Terminating world-lines usually mean singularities on the space time manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Singularities like this occur in regions of space-time where curvature is enormous (and ultimately infinite) and as a result one might suspect therefore that classical general relativity is inapplicable there. Quantum gravity theories can avoid the singularities although I am unaware of a general proof they always do.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The singularity proof relies on so-called positive energy conditions of various kinds which classical matter is assumed to satisfy.  &lt;/p&gt;&#xA;" OwnerUserId="3670" LastActivityDate="2011-05-21T22:18:43.630" />
  <row Id="10222" PostTypeId="1" CreationDate="2011-05-21T22:21:34.790" Score="1" ViewCount="75" Body="&lt;p&gt;There is a linear electro-optic effect called Pockels Effect&#xA;The brief is that refractive index changes due to electric field.&#xA;If there is an anisotropy (like birefringence) and electric field is in perpendicular to optical axis then refractive index changes for different polarization of ordinary rays. &#xA;It gets eigendirections 45 degrees to electric field. &#xA;Why so? Why it gets min 45 degrees but not exactly towards direction of E?&lt;/p&gt;&#xA;" OwnerUserId="3674" LastActivityDate="2011-08-08T10:20:59.230" Title="direction of Pockels Effect. Refractive index eigendirections" Tags="&lt;optics&gt;&lt;non-linear-optics&gt;" AnswerCount="1" />
  <row Id="10224" PostTypeId="1" CreationDate="2011-05-21T23:59:25.773" Score="0" ViewCount="100" Body="&lt;p&gt;Why does an object sink when filled with water, even if the same object would float without water inside?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, put an empty glass cup into water, and it floats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you put a plastic container filled with water in water, it'll sink. Why is that?&lt;/p&gt;&#xA;" OwnerUserId="3675" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-22T02:39:25.217" LastActivityDate="2011-05-22T02:39:25.217" Title="Why does an object when filled with water sink, but without water inside float (in a body of water)?" Tags="&lt;water&gt;&lt;buoyancy&gt;" AnswerCount="1" />
  <row Id="10225" PostTypeId="1" AcceptedAnswerId="10227" CreationDate="2011-05-22T00:06:36.550" Score="2" ViewCount="177" Body="&lt;p&gt;Please help me with the following. I want to know if there is an equation/set of equations to find out the projected area of a (3-D) cube when it is oriented at different angles of attack to the fluid flow, rendering the velocity vector of the impinging fluid on the cube surface to change with each angle of attack. In particular, what is the projected  area when the cube is oriented at such an angle that the flow velocity vector passes through one of the vertex and the geometric centre of the cube? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3604" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-22T02:40:30.910" LastActivityDate="2011-05-22T02:40:30.910" Title="How to calculate the projected area at different angles/vectors?" Tags="&lt;fluid-dynamics&gt;&lt;geometry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10226" PostTypeId="2" ParentId="10224" CreationDate="2011-05-22T00:12:46.003" Score="5" ViewCount="" Body="&lt;p&gt;For a given volume, light things float and heavy things sink.  The cup sinks when you fill it with water because it becomes heavier, and therefore more dense.  When the cup becomes more dense than water, it sinks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cup would sink just as well if you filled it with rocks, lead, etc.  The condition for the cup to sink is that its weight must be greater than the weight of the water it displaces (i.e. its weight must be greater than the weight of a cup exactly the  same size, but made out of water and filled with water.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read a number of other questions on similar material here: &lt;a href=&quot;http://physics.stackexchange.com/questions/tagged/buoyancy&quot;&gt;http://physics.stackexchange.com/questions/tagged/buoyancy&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-05-22T00:12:46.003" />
  <row Id="10227" PostTypeId="2" ParentId="10225" CreationDate="2011-05-22T00:21:36.763" Score="3" ViewCount="" Body="&lt;p&gt;The projected area is $\sqrt{3} \ l^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine the long diagonal of a  cube.  This diagonal is the same direction as the velocity vector you're interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length of the diagonal is $\sqrt{3}l$, with $l$ the side length of the cube.  This come from the Pythagorean Theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we project this diagonal onto a side, we get simply $l$.  Therefore, the projection operation at this angle divides by $\sqrt 3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cube has three faces exposed to a fluid flowing coming in from the direction of the long diagonal, each with area $l^2$.  Each of these is projected onto the plane perpendicular to the fluid flow, so the projected area is $(3/\sqrt{3})\ l^2 = \sqrt{3} \ l^2$.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-05-22T00:21:36.763" CommentCount="2" />
  <row Id="10228" PostTypeId="2" ParentId="10072" CreationDate="2011-05-22T02:12:56.497" Score="0" ViewCount="" Body="&lt;p&gt;For the case of a real spring, damping (energy loss) comes from three areas:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Structural Damping. As the material flexes in cycles, there is internal losses that occur due to a hysteresis effect on the force-deflection relationship. This is a small but noticeable effect.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Contact friction. The spring is not floating in space by itself, but is in contact with other objects (like spring retainers and tappets). Where there is contact there is energy loss due to friction. If it is dry friction there is an equivalent damping coefficient calculated that depends in the frequency and amplitude of the oscillation (any vibrations book has it). If it is viscous friction then the damping coefficient depends on the laminar shearing of the fluid (any fluid dynamics book at some point relates viscous coefficient to damping based on geometry).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An finally as you mentioned there is aerodynamic drag that contributes to damping. This is the most difficult to calculate as you need to run a CFD simulation as the spring moves.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The combined effect can be measured (and tested) with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Logarithmic_decrement&quot; rel=&quot;nofollow&quot;&gt;log decrement&lt;/a&gt; method. Hit the spring hard and measure the amplitude as a function of time. If you count the relative decrease and the # of cycles you can use that to calculate and overall effective damping coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you want in the end is a relationship of the form $F_{\rm{damping}} = c\,\dot{x}$ and estimating the $c$ value experimentally or with a simulation. (This was first mentioned by other answers already)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. Be careful with real springs because not all of their mass is in motion and thus any $m$ value used in calculation is going to be inaccurate. Sometimes with vibrations the effective mass of a spring in one end of it is a low as 33% of the total mass.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-05-22T02:12:56.497" />
  <row Id="10229" PostTypeId="1" CreationDate="2011-05-22T11:11:47.360" Score="2" ViewCount="218" Body="&lt;p&gt;I wonder what were the main experiments that led people to develop the concept of wave function collapse? (I think I am correct in including the Born Rule within the general umbrella of the collapse paradigm.) Are there any instances where cases once thought to be examples of collapse have since been explained as the normal time-evolution of the wave function? &lt;/p&gt;&#xA;" OwnerUserId="3253" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-22T19:23:42.850" LastActivityDate="2011-05-22T19:23:42.850" Title="Historical background of wave function collapse" Tags="&lt;quantum-mechanics&gt;&lt;wavefunction&gt;" AnswerCount="1" CommentCount="12" />
  <row Id="10230" PostTypeId="1" AcceptedAnswerId="10259" CreationDate="2011-05-22T12:16:14.783" Score="15" ViewCount="301" Body="&lt;p&gt;Operators can be cyclically interchanged inside a trace:&lt;br&gt;&#xA;${\rm Tr} (AB)={\rm Tr} (BA)$&lt;p&gt;&#xA;This means the trace of a commutator of any two operators is zero:&lt;br&gt;&#xA;${\rm Tr} ([A,B])=0$&lt;p&gt;&#xA;But what about the commutator of the position and momentum operators for a quantum particle?&lt;p&gt;&#xA;On the one hand: ${\rm Tr}([x,p])=0$,&lt;br&gt;&#xA;while on the other hand: $[x,p]=i\hbar$&lt;p&gt;&#xA;How does this work out?&lt;/p&gt;&#xA;" OwnerUserId="975" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-05T21:02:41.203" LastActivityDate="2011-07-05T21:02:41.203" Title="Trace of a commutaor is zero - but what about the commutator of $x$ and $p$?" Tags="&lt;quantum-mechanics&gt;&lt;mathematical-physics&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="5" />
  <row Id="10231" PostTypeId="2" ParentId="10230" CreationDate="2011-05-22T12:44:49.243" Score="15" ViewCount="" Body="&lt;p&gt;$x$ and $p$ do not have finite-dimensional representations. In particular, $xp$ and $px$ are not &quot;trace-class&quot;. Loosely, this means that the traces of $xp$ and $px$ are both infinite, although it's best to take them both to be undefined. Again loosely, if you subtract $\infty-\infty$, you can certainly get $i\hbar$. But you shouldn't. Everything works out if you think of $p$ as a complex multiple of the derivative operator, for which $\frac{\partial}{\partial x}$ and $x$ act on the &lt;em&gt;infinite&lt;/em&gt; dimensional space of polynomials in $x$.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-05-22T12:44:49.243" CommentCount="5" />
  <row Id="10232" PostTypeId="2" ParentId="10062" CreationDate="2011-05-22T12:45:33.403" Score="0" ViewCount="" Body="&lt;p&gt;An alternative view, not much discussed in the literature in this connection, is the quantum logical approach, which emphasizes the non-Boolean structure of properties of quantum systems. (The properties of a classical system form a Boolean algebra, essentially the abstract characterization of a set-theoretic structure. This is reflected in the Boolean character of classical logic, and the Boolean gates in a classical computer.) From this perspective, the picture is entirely different. Rather than ‘computing all values of a function at once,’ a quantum algorithm achieves an exponential speed-up over a classical algorithm by avoiding the computation of any values of the function at all. A crucial difference between quantum and classical information is the possibility of selecting an exclusive disjunction, representing a global property of a function, among alternative possible disjunctions — for example, the ‘constant’ disjunction asserting that the value of the function (for both arguments) is either 0 or 1, or the ‘balanced’ disjunction asserting that the value of the function (for both arguments) is either the same as the argument or different from the argument — without determining the truth values of the disjuncts. Classically, an exclusive disjunction is true if and only if one of the disjuncts is true. This is is redundant information in a quantum computation but essential information classically: an exclusive classical disjunction is true if and only if one of the disjuncts is true. In effect, Deutsch's quantum circuit achieves its speed-up by exploiting the non-Boolean structure of quantum properties to efficiently distinguish between two disjunctive properties, without determining the truth values of the relevant disjuncts (representing the association of individual arguments with corresponding function values). The point of the procedure is to avoid the evaluation of the function in the determination of the global property, in the sense of producing a value in the range of the function for a value in its domain, and it is this feature — impossible in the Boolean logic of classical computation — that leads to the speed-up relative to classical algorithms. For some recent work by Giuntini and others on logics associated with quantum gates, see under ‘quantum computational logics’ in the Other Internet Resources. (For quantum logic not specifically in relation to quantum computation, see the entry on quantum logic and quantum probability).&lt;/p&gt;&#xA;" OwnerUserId="3676" LastActivityDate="2011-05-22T12:45:33.403" />
  <row Id="10233" PostTypeId="1" AcceptedAnswerId="10235" CreationDate="2011-05-22T14:40:38.160" Score="3" ViewCount="121" Body="&lt;p&gt;If I have two system of an Ideal gas A and B Each of these system has a  partition function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z_{A,B} = \left ( \frac{V_{A,B}}{\lambda_T} \right )^{N_{A,B}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda_T = \left ( \frac{m}{2\pi\beta \hbar } \right )^{\frac{1}{2}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The free energy is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{A,B} = -kT \ln \left ( Z_{A,B} \right ) = -kT N_{A,B}\ln \left (\frac{V_{A,B}}{\lambda_T}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the free energy to be extensive the following must be true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{A} + F_B = F_{A+B} \Rightarrow Z_A \cdot Z_B = Z_{A+B}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z_A \cdot Z_B = \left (\frac{V_A}{\lambda_T} \right )^{N_A} \left ( \frac{V_B}{\lambda_T} \right)^{N_B}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z_{A+B} = \left ( \frac{V_{A+B}}{\lambda_T} \right )^{N_A + N_B}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for $Z_A + Z_B = Z_{A+B}$ to be true $V_A^{N_A} V_{B}^{N_B} = \left(V_A + V_B \right)^{N_A + N_B}$ must be true as well but this isn't true for any system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we cannot create energy by mixing two containers of Argon in the same pressure and temperature, something in my understanding is wrong. Where is my fault? &lt;/p&gt;&#xA;" OwnerUserId="3677" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2011-05-22T19:06:24.203" LastActivityDate="2011-06-01T16:28:50.207" Title="proving that the free energy is extensive" Tags="&lt;thermodynamics&gt;&lt;statistical-mechanics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" />
  <row Id="10234" PostTypeId="1" AcceptedAnswerId="10899" CreationDate="2011-05-22T14:45:18.820" Score="2" ViewCount="116" Body="&lt;p&gt;Imagine one have an ideal sensor, which can convert the emission to some kinds of signal (typically, voltage, and suppose no noise at all), then what process can describe the measure data? Is it related to &lt;a href=&quot;http://en.wikipedia.org/wiki/Stochastic_electrodynamics&quot; rel=&quot;nofollow&quot;&gt;Stochastic electrodynamics&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="3678" LastEditorUserId="3678" LastEditorDisplayName="" LastEditDate="2011-06-08T06:24:49.777" LastActivityDate="2011-06-08T06:24:49.777" Title="What stochastic process is associated with spontaneous electromagnetic radiation?" Tags="&lt;electromagnetic-radiation&gt;&lt;electrodynamics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10235" PostTypeId="2" ParentId="10233" CreationDate="2011-05-22T15:20:41.793" Score="4" ViewCount="" Body="&lt;p&gt;What you observe is &lt;a href=&quot;http://en.wikipedia.org/wiki/Gibbs_paradox&quot; rel=&quot;nofollow&quot;&gt;Gibbs' paradox&lt;/a&gt;. The resolution comes about by postulating that the particles are indistinguishable (and thereby introducing a factor $1/N!$). Then the entropy and the free energy becomes extensive (in the thermodynamic limit $N,V\to\infty$, $N/V=\text{const}$).&lt;/p&gt;&#xA;" OwnerUserId="2962" LastEditorUserId="2962" LastEditorDisplayName="" LastEditDate="2011-06-01T16:28:50.207" LastActivityDate="2011-06-01T16:28:50.207" CommentCount="2" />
  <row Id="10236" PostTypeId="2" ParentId="10229" CreationDate="2011-05-22T16:30:01.587" Score="1" ViewCount="" Body="&lt;p&gt;Short answer: there is a review paper written only five years ago&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experimental motivation and empirical consistency in minimal no-collapse quantum mechanics, Maximilian Schlosshauer, Ann. Phys. 321, 112-149 (2006)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a cite from outlook: “We have analyzed three important experimental domains —namely, SQUIDs, molecular diffraction, and Bose-Einstein condensation [...] These experiments have provided powerful examples for the validity of unitary Schroedinger dynamics and the superposition principle on increasingly large length scales.”&lt;/p&gt;&#xA;" OwnerUserId="3351" LastActivityDate="2011-05-22T16:30:01.587" CommentCount="9" />
  <row Id="10237" PostTypeId="2" ParentId="10137" CreationDate="2011-05-22T16:34:45.217" Score="1" ViewCount="" Body="&lt;p&gt;Dear Billy, it is not quite clear why you add the emotional words such as &quot;shocking&quot; and &quot;disturbing&quot;. It is a manifestation of the essence of quantum mechanics that the classical limit or the spectrum of allowed questions is not God-given. The classical limit occurs in situations that depend on the state of the system as well as its Hamiltonian. The process of decoherence also depends on these two things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it shouldn't be surprising that properties that behave classically generally cease to behave classically in the future (or ceased/started in the past). Consider an analogy in statistical physics and thermodynamics. The thermodynamics description works if there are many particles etc. But you may produce lots of new particles e.g. by the LHC proton-proton collisions. Thermodynamics is not applicable to the initial state of 2 particles - too few - but it is useful to describe the quark-gluon plasma and other things that may be formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case of decoherence and consistency of histories is analogous. These things are simply not eternal and God-given. Classical limits only work in some situations, the situations are described by some inequalities involving values of parameters, and because parameters evolve with time, so does the applicability of the classical limit - and so does the consistency of various histories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's really a feature of quantum mechanics, not a vice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the non-uniqueness, the situations in which the questions are non-unique are unrealistic, but yes, in principle, quantum mechanics allows one to ask different types of questions with different alternative answers that can't be reduced to the same fine-grained set of alternatives. Why do you think it's bad? Again, the right questions depend on the Hamiltonian and the state. They're not God-given. And classical physics isn't a &quot;skeleton&quot; of quantum mechanics that quantum mechanics is only allowed to &quot;slightly disturb&quot;. The world is quantum mechanical and quantum mechanics is a qualitatively different description than classical physics, and classical physics is a limit. So classical limit is not a skeleton of quantum physics you may rely upon; it is the sweat on the skin, a derived product that only works on the boundaries of the parameter spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend you a paper by Zurek&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/quant-ph/0105127&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/quant-ph/0105127&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;that addresses and summarizes many issues. Search for &quot;Dowker&quot; in the paper above. The paper has 1,000+ citations. You will see that there's really no problem in understanding classicality etc.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-22T16:45:21.453" LastActivityDate="2011-05-22T16:45:21.453" />
  <row Id="10238" PostTypeId="2" ParentId="10204" CreationDate="2011-05-22T17:13:26.183" Score="0" ViewCount="" Body="&lt;p&gt;First step: gather the data.&lt;br&gt;&#xA;I use G3DATA (free).&lt;br&gt;&#xA;googling : &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;g3data is used for extracting data&#xA;  from graphs. In publications graphs&#xA;  often are included, but the actual&#xA;  data is missing. g3data makes the&#xA;  extracting...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You will need two axis graduated with min, max values. The g3data will use your clicks over the graph to collect data in a datafile.&lt;br&gt;&#xA;The knowledge of the time basis is important if you want the precise frequencies. If you are interested in the proportions between the main frequency and the harmonics it is less important.&lt;br&gt;&#xA;Next steps, in short: make several datafiles, mix, remove outliars, filter,... ,..., fft  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit add:&lt;br&gt;&#xA;One can have a thousand osciloscopes and signal analyzers but yet one has to have acess to specific original 'signal source' to fed them. Obviously you dont have it and any OScope is useless in that situation. That's why I avoided to mention such a useless solution. &#xA;Here a &lt;a href=&quot;http://cleveland2.ce.ttu.edu/software/free_ware/g3data/g3data_HowToUse/g3data.pdf&quot; rel=&quot;nofollow&quot;&gt;how-to-use&lt;/a&gt; the g3data software.&lt;br&gt;&#xA;To treat the data search for FFT software. There are a lot of choices even a DIY.&#xA;I use the free online package &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sage_%28mathematics_software%29&quot; rel=&quot;nofollow&quot;&gt;sagenb&lt;/a&gt;&lt;/strong&gt; (&lt;strong&gt;&lt;a href=&quot;http://www.sagemath.org/&quot; rel=&quot;nofollow&quot;&gt;website&lt;/a&gt;&lt;/strong&gt;) to crunch numbers.&lt;br&gt;&#xA;Give a try on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler_%28software%29&quot; rel=&quot;nofollow&quot;&gt;Euler math toolbox&lt;/a&gt; (&lt;a href=&quot;http://eumat.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;website&lt;/a&gt;)&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Comparison_of_numerical_analysis_software&quot; rel=&quot;nofollow&quot;&gt;WP Comparison_of_numerical_analysis_software&lt;/a&gt;&lt;br&gt;&#xA;Note: FFT and DFT work usually(a) on equally spaced data. You will have to construct such array using interpolation over the gathered data.&lt;br&gt;&#xA;In the filtering and smoothing phase, using a suitable moving average window, one must be careful not to destroy the features you are interested.&lt;br&gt;&#xA;(a) there are algorithms that work the approx DFT in non-equispaced data.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ps : if you are confortable using a programming lang give a try to PyOpenCL. There exists a &lt;a href=&quot;http://packages.python.org/pyfft/&quot; rel=&quot;nofollow&quot;&gt;PyFFT for PyOpenCL&lt;/a&gt; and you graphics board (GPU) will outperform any CPU for large datasets.&lt;br&gt;&#xA;Edit add end.  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-05-24T00:26:21.277" LastActivityDate="2011-05-24T00:26:21.277" CommentCount="2" />
  <row Id="10239" PostTypeId="1" CreationDate="2011-05-22T17:40:54.763" Score="3" ViewCount="99" Body="&lt;p&gt;Quantum mechanics has a peculiar feature, entanglement entropy, allowing the total entropy of a system to be less than the sum of the entropies of the individual subsystems comprising it. Can the entropy of a subsystem exceed the maximum entropy of the system in quantum mechanics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have in mind is eternal inflation. The de Sitter radius is only a few orders of magnitude larger than the Planck length. If the maximum entropy is given by the area of the boundary of the causal patch, the maximum entropy can't be all that large. Suppose a bubble nucleation of the metastable vacuum into another phase with an exponentially tiny cosmological constant happens. After reheating inside the bubble, the entropy of the bubble increases significantly until it exceeds the maximum entropy of the causal patch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is described by entanglement entropy within the bubble itself, when restricted to a subsystem of the bubble, we get a mixed state. In other words, the number of many worlds increases exponentially until it exceeds the exponential of the maximum causal patch entropy. Obviously, the causal patch itself can't possibly have that many many-worlds. So, what is the best way of interpreting these many-worlds for this example?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="3681" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-22T19:21:00.603" LastActivityDate="2011-05-23T09:52:49.990" Title="Can the entropy of a subsystem exceed the maximum entropy of the system in quantum mechanics?" Tags="&lt;quantum-mechanics&gt;&lt;cosmology&gt;&lt;entropy&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="10240" PostTypeId="1" CreationDate="2011-05-22T17:46:35.027" Score="0" ViewCount="164" Body="&lt;p&gt;Here is a question I am curious about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the wave function objective or subjective, or are we not allowed to even ask such a heretical question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If subjective, does it depend upon the mind? If so, can the mind &quot;believe&quot; in the &quot;wrong&quot; wave function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not philosophy, so please pretty please don't close this question.&lt;/p&gt;&#xA;" OwnerUserId="3682" LastActivityDate="2011-05-22T21:52:04.480" ClosedDate="2011-05-22T19:30:21.020" Title="Is the wave function objective or subjective?" Tags="&lt;quantum-interpretations&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="10241" PostTypeId="2" ParentId="10240" CreationDate="2011-05-22T17:55:18.207" Score="-1" ViewCount="" Body="&lt;p&gt;A wave function is as objective as a photo picture of somebody. It encodes an objective information about something. It consist, as any information, of many bits of information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same time is is subjective because it depends on our perception. For one some information tells nothing, for somebody else it may be very speaking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning our minds, yes, everybody believes in the wrong wave function due to lack of knowledge.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-05-22T21:52:04.480" LastActivityDate="2011-05-22T21:52:04.480" />
  <row Id="10242" PostTypeId="2" ParentId="10240" CreationDate="2011-05-22T17:57:44.960" Score="1" ViewCount="" Body="&lt;p&gt;The wave function is not an observable, so you can characterize it however you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, any alternative explanation you want to put forth must agree with experimental reality, which in the end will mean it must be mathematically equivalent to the usual approach and will accordingly have some object that is isomorphic to the wave function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That makes it real enough for &lt;em&gt;my&lt;/em&gt; purposes.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-05-22T17:57:44.960" CommentCount="5" />
  <row Id="10244" PostTypeId="1" CreationDate="2011-05-22T18:20:34.277" Score="5" ViewCount="171" Body="&lt;p&gt;In derivations of decoherence, there eventually comes a point when we are asked to take the partial trace over the environment. Why should this be valid for an entangled system? Why should taking the partial trace give the correct description of the system? Taking the partial trace turns a pure density matrix into a mixed density matrix, but a pure state is always a pure state. In what sense can the reduced mixed density matrix be assigned probabilities?&lt;/p&gt;&#xA;" OwnerUserId="3684" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-22T19:17:46.317" LastActivityDate="2011-05-22T20:48:34.010" Title="Why is it valid to take the partial trace to describe a subsystem?" Tags="&lt;quantum-mechanics&gt;&lt;entanglement&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="10245" PostTypeId="2" ParentId="10244" CreationDate="2011-05-22T18:50:48.213" Score="1" ViewCount="" Body="&lt;p&gt;Practically speaking, without the tracing operation you wouldn't have any way of describing an entangled system. You probably already know that if you have a Bell-like entangled state of two systems, you can't write down a state vector to describe either system independently -- only for the combined system does a state vector exist. The natural extension of an object that describes a system when it is entangled is the reduced density matrix (with a trace in it) with its interpretation as a probability distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's even more obvious why you should use a trace when you realize that the analog of the reduced density matrix in mathematics/statistics is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Marginal_distribution&quot; rel=&quot;nofollow&quot;&gt;marginal probability&lt;/a&gt;. The very idea of writing down a description for a system when you don't care about what the other system does is captured in the marginal distribution and the associated summing over all possible states of the other system/environment (the trace). There's just no other consistent way of thinking about writing down the description for an entangled system.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-05-22T18:50:48.213" />
  <row Id="10246" PostTypeId="1" CreationDate="2011-05-22T19:14:51.480" Score="3" ViewCount="77" Body="&lt;p&gt;So I was reading the EF experiment that's used at the MIT to measure the vacuum permittivity and I was thinking about trying it just to see how it works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ocw.mit.edu/courses/physics/8-02x-physics-ii-electricity-magnetism-with-an-experimental-focus-spring-2005/labs/experiment_ef.pdf&quot; rel=&quot;nofollow&quot;&gt;http://ocw.mit.edu/courses/physics/8-02x-physics-ii-electricity-magnetism-with-an-experimental-focus-spring-2005/labs/experiment_ef.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some questions about it and I hope you can help me with them:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It says &quot;To find the electric force on the foil, assume that the charge density, σ , on the foil, is the same as that of the lower washer&quot;. How good that assumption is?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It says &quot;Charges on the foil feel only horizontal forces from other charges on the bottom plate, so the vertical force on the foil is due to the electric field of just the top charge sheet&quot;. What horizontal forces are they talking about? I thought both the bottom plate and the upper plate were all exerting a vertical force on the foil and that's why I don't understand why the force is due just to the electric field of the upper plate $(V/2d)$ and not from the one of the two plates $(V/d)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, why washers? I think it really doesn't matter if they're just two discs, does it?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3680" LastEditorUserId="3680" LastEditorDisplayName="" LastEditDate="2011-05-22T19:19:55.190" LastActivityDate="2011-06-15T20:01:25.727" Title="Measuring the vacuum permittivity" Tags="&lt;electromagnetism&gt;&lt;experimental-physics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="10247" PostTypeId="2" ParentId="10240" CreationDate="2011-05-22T19:18:40.100" Score="4" ViewCount="" Body="&lt;p&gt;The wave function is a solution of an equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is as subjective and as objective as all mathematical solutions to equations describing physical fields, whether classical or quantum mechanical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly as a solution it is objective, a formula written on paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subjectivity enters in the choice of the equation to be solved and thus the choice of the specific solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objectivity enters again because the particular equation was chosen due to its being appropriate to the problem, having fitted previous experimental observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I consider all this as navel gazing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the solutions of Maxwell's equations objective or subjective? What about the gravitational orbit solutions?&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-22T19:18:40.100" CommentCount="1" />
  <row Id="10248" PostTypeId="1" CreationDate="2011-05-22T19:38:48.833" Score="3" ViewCount="176" Body="&lt;p&gt;I am learning sailing on a 5m catamaran (Nacra 5).&#xA;I am familiar with basic aerodynamics and the physics of the sail and keel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We learned that when sailing closed hauled, too tight a mainsail tends to bring the boat up to the wind.  And that the opposite is true for the jib. For example, one may steer up to the wind to come about, using the mainsail alone, by trimming it tight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why the trimmed mainsail in the above setup gives a larger torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also seems opposed to the rule: &quot;The more the mainsail is sheeted out the more the boat tends to come up.&quot;, as explained in &lt;a href=&quot;http://www.sailtheory.com/mandf.html#sailsteering&quot; rel=&quot;nofollow&quot;&gt;http://www.sailtheory.com/mandf.html#sailsteering&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my understanding, there are several possible competing effects involved:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Effects that tend to INCREASE the mainsail torque to head up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1A) The direction of the sail force becomes more perpendicular to the boat. This increases the heeling torque. Since heeling motion happens faster than turning, the boat will heel more. This moves the sail force out, which increases the lever arm and the torque to head up. see diagram in: &lt;a href=&quot;http://www.sailtheory.com/mandf.html#hellingstuur&quot; rel=&quot;nofollow&quot;&gt;http://www.sailtheory.com/mandf.html#hellingstuur&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1B) The sail force moves backwards since the sail is stretched backwards. This increases the lever arm and the torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Effects that tend to DECREASE the mainsail torque to head up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2A) The direction of the sail force becomes more perpendicular to the boat. Assuming that the force is perpendicular to the sail, and that the center of rotation is between the mast and the center of force, one sees that pulling the sail in, reduces the torque to head up, as can be seen in the following diagram:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/9I2TX.jpg&quot; alt=&quot;diagram of the sail force torque&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Theta gets smaller, the torque Tau is reduced. This is opposed to what is stated in some of the answers below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2B) The sail force is reduced since the sail was pulled beyond its optimal angle of attack, thus losing lift and reducing torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2C) Due to (2B) heeling torque is also decreased. With a similar reasoning to (1A) this decreases the lever arm and torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2D) The sail force moves forward since the aft part of the sail is too flat and stalls. This shortens the lever arm and reduces torque. This is described in &lt;a href=&quot;http://www.sailtheory.com/mandf.html#sailsteering&quot; rel=&quot;nofollow&quot;&gt;http://www.sailtheory.com/mandf.html#sailsteering&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know from reality that 1A+1B &gt; 2A+2B+2C+2D. Considering heeling alone, we also know that 1A &gt; 2C. It remains to be explained why in general this is the case.&lt;/p&gt;&#xA;" OwnerUserId="3687" LastEditorUserId="3687" LastEditorDisplayName="" LastEditDate="2011-05-30T14:22:51.173" LastActivityDate="2011-05-30T14:22:51.173" Title="Why trimming the mainsail gives a larger torque to head up" Tags="&lt;mechanics&gt;&lt;forces&gt;&lt;aerodynamics&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="10249" PostTypeId="2" ParentId="10248" CreationDate="2011-05-22T20:18:52.397" Score="0" ViewCount="" Body="&lt;p&gt;From my sailing experience (which does not include catamarans specifically) a tight sail always wants to turn the boat up into the wind. It is the partial cancellation of this force with the stability provided by a keel and/or centerboard that allows you to move forward at a close haul. Because the force on the boat from the sail always acts at 90 degrees from wherever the boom line is, pulling the sail in close creates a stronger torque that needs to be canceled if you want your boat to stay right side up. Usually this means using your body weight as a counter balance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forcing the jib out can increase the air flow around the foil changing the apparent wind direction, though I'm not sure if this is what you are talking about..&lt;/p&gt;&#xA;" OwnerUserId="3601" LastEditorUserId="3601" LastEditorDisplayName="" LastEditDate="2011-05-22T20:30:41.917" LastActivityDate="2011-05-22T20:30:41.917" CommentCount="2" />
  <row Id="10250" PostTypeId="2" ParentId="10244" CreationDate="2011-05-22T20:48:34.010" Score="4" ViewCount="" Body="&lt;p&gt;Dear Gunther, a partial trace contains all the information about possible predictions made for the subsystem that you haven't traced over. It's trivial to see why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, take the whole system A+B. They may be entangled but we're only interested in future measurements of A. If the description for A+B is a pure state $|\psi\rangle$, then define &#xA;$$\rho = |\psi\rangle \langle \psi |$$&#xA;The expectation values of the operator $P$ in $|\psi\rangle$ are trivially equal to  ${\rm Tr}\,(\rho P)$, do you see why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $\rho$, the density matrix, is an operator acting on the total Hilbert space which is the tensor product $H_A\otimes H_B$. And operators $L$ that only act on the system $A$ have the form&#xA;$$ L = L_A \otimes {\bf 1} $$&#xA;They're just tensor product of an operator acting on $H_A$ only and the identity operator acting on the Hilbert space $H_B$. Do you understand why? Then it's obvious that&#xA;$$ {\rm Tr} \,( \rho L) = {\rm Tr}_{ab} \,( \rho (L_A\otimes {\bf 1}))   $$&#xA;where $ab$ indicates that the tracing goes over indices corresponding to the basis of $H_A$ as well as those of $H_B$. You can see that the tracing over both types of the indices may be made one by one. You may first trace over the $b$-type indices. However, only $\rho$ depends on $b$-type indices while $L_A$ doesn't and ${\bf 1}$ doesn't make a difference. So&#xA;$$ {\rm Tr} \,( \rho L) = {\rm Tr}_{a} {\rm Tr}_{b} \,( \rho (L_A\otimes {\bf 1}))   $$&#xA;But $L_A$ doesn't contain any $b$-type indices, so it's a constant that may be pulled in front of the trace over $b$-type indices:&#xA;$$ {\rm Tr} \,( \rho L) = {\rm Tr}_{a} L_A {\rm Tr}_{b} \,( \rho  {\bf 1})  = {\rm Tr}_a L_A \rho_{\rm traced\,over\,b}  $$&#xA;So the trace over $b$ may be made first and the reduced density matrix may then be traced to find properties of the observables linked to the subsystem A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that this reduction must be possible if you only want to predict A should be intuitive. In classical physics, you also don't have to know another subsystem to write down probabilities of different properties of A - despite the fact that A,B may be correlated. You just integrate over all possible positions and momenta of B. The tracing is the exact quantum analogy of the integration over $x_B,p_B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you have an EPR experiment, A and B may be 100% correlated. But if you only measure A, you won't see the correlation because you don't have any results from B. So you will see that 50% of the photons on the A side are left-handed and 50% are right-handed: the density matrix will be proportional to the unit matrix (times one half) even though the original state for both A,B was pure - but entangled.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-22T20:48:34.010" />
  <row Id="10252" PostTypeId="1" CreationDate="2011-05-22T23:44:28.063" Score="1" ViewCount="187" Body="&lt;p&gt;I don't fully understand what would happen if we could travel at the speed of light. But I saw somewhere here that it would mean events happen out of order. But why is this a problem. It is said that cause has to happen before the effect, but why does this have to be linear? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And why is it that the speed of light is the maximum? I take it that light (photons?) are massless, so why can't they travel faster?&lt;/p&gt;&#xA;" OwnerUserId="4" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-22T23:46:07.570" LastActivityDate="2011-05-23T04:31:34.170" Title="Why is travelling around the speed of light a problem?" Tags="&lt;special-relativity&gt;&lt;speed-of-light&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="10253" PostTypeId="2" ParentId="10252" CreationDate="2011-05-23T00:09:07.467" Score="3" ViewCount="" Body="&lt;p&gt;Before we even consider this, consider the kinetic energy of a relativistic particle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E=mc^{2}\left(\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}-1\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This represents the amount of energy required to accelerate the particle from rest (with respect to a given reference frame) to the speed $v$.  It should be immediately obvious that this quantity becomes infinite as $v\rightarrow c. {}^{1}$  Since it is impossible to generate an infinite amount of energy, this then means that no particle with $m &amp;gt; 0$ can be accelerated to the speed of light.  In fact, since high-energy particles have speeds that are generally extremely close to thespeed of light in a laboratory frame, it has become custom to quote the speed of the particles in terms of their energy--particle physicists routinely talk about &quot;1 GeV&quot; electrons to refer to the kinetic + rest mass energy of the electron in question, rather than bothering to translate this into a translational velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{1}$ This is true unless $m=0$.  In the latter case, you can get a finite answer out of the $0\cdot \infty$ in the case of massless particles, but ONLY if they travel at exactly at the speed of light.  Were they to travel faster than this, the formula would no longer be the indeterminate form $0 \cdot \infty$, but rather $0\cdot ({\rm complex \,\,number})$, which would be a definite, but nonsensical answer.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-05-23T00:09:07.467" />
  <row Id="10254" PostTypeId="2" ParentId="3075" CreationDate="2011-05-23T00:46:21.567" Score="-1" ViewCount="" Body="&lt;p&gt;Nigel asks among other things if there is a semi-classical mechanism. This is an interesting point which has not been addressed so I am going to give it a try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classical equilibrium between the electromagnetic field and a harmonic oscillator occurs when the energy &lt;em&gt;per mode&lt;/em&gt; of the field is equal to the energy per mode of the oscillator. This is true classically and it is also true quantum mechanically. The real problem is that in a classical system of balls connected by springs, every mode gets the same energy no matter its frequency. This is true from the equipartition theorem which is based on maximizing entropy, and it is also true from a strictly mechanical argument of tracking collisions. This is the source of the ultraviolet catastrophe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can evade the ultraviolet catstrophe in an ad hoc manner by restricting the mode energies of the higher frequency e-m field components. Planck's formula is usually explained this way. But a reading of Planck's paper shows that he really solved the problem by redefining the way entropy is calculated for an oscillator. Applying this entropy calculation directly to the electromagnetic field gives the desired results. But one could just as well apply the entropy calculation to the purely mechanical system. If one does, it turns out that the higher frequency oscillators are not excited to the extent one would expect from analyzing them as classical billiard balls connected by springs. The very high frequency oscillations are suppressed. If one then allows the oscillators to carry a small amount of charge, and calculates the resulting radiation according to the classical laws of antenna theory, it turns out you get the correct black-body spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system breaks down if you allow two or three high-frequency oscillators with the properties of a classical billiard ball on a spring to be added to the system. They will quickly equilibrate with the rest of the gas in terms of x,y and z velocities; and once the do this, they will share this energy with their oscillatory mode as well. It is the laws of quantum mechanics that forbid the existence of such oscillators, thereby preserving the thermal equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-05-23T00:46:21.567" />
  <row Id="10256" PostTypeId="2" ParentId="10252" CreationDate="2011-05-23T04:31:34.170" Score="1" ViewCount="" Body="&lt;p&gt;Photons can't travel faster than $c$ and they also cannot travel slower than $c$ &lt;em&gt;because&lt;/em&gt; they are massless. The rest mass is defined as&#xA;$$ m_0^2 = E^2 / c^4 - p^2 / c^2 $$&#xA;and if this is zero, like for photons, you may easily show that $pc^2/E = c$ and this is how the speed is defined for an energy-momentum vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More conceptually, &lt;em&gt;nothing&lt;/em&gt; can move faster than light in special relativity because according to special relativity, it would be equivalent to traveling backwards in time which would violate causality. A trajectory &#xA;$$ t=\tau, \quad x=V\tau$$&#xA;where $V&amp;gt;c$ may be Lorentz-transformed to another trajectory where&#xA;$$ t = a\tau, \quad x = b\tau$$&#xA;where $a$ is negative, so the trajectory goes backwards in time. Because the laws of physics must have the same form in this frame, it means that the identity of the cause and the effect may be switched. Because there's no sharp ordering between the cause and the effect, you may kill your grandfather before he met your grandmother, making your own existence inconsistent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such things can only be avoided if the cause preceded its effect from the viewpoint of &lt;em&gt;all&lt;/em&gt; inertial systems which is equivalent to the effect's belonging to the future light cone of the cause, i.e. to the speed limit $c$ on any velocity that any signal or material object may have.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-23T04:31:34.170" />
  <row Id="10258" PostTypeId="2" ParentId="10140" CreationDate="2011-05-23T07:21:23.683" Score="1" ViewCount="" Body="&lt;p&gt;In Figure 6 of their article, they draw a causal diamond, and divide the future null boundary into $B^+$ and $B^-$. They implicitly assumed that the Hilbert space of $B$ is the tensor product of the Hilbert space of $B^+$ with the Hilbert space of $B^-$. If $B$ were a spacelike surface, that would be true, but it's not. It's null, and null separated operators don't commute. The operation of taking the partial trace over $B^-$ to get the density matrix over $B^+$ is not correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors also can't agree upon whether time evolution between the different $\beta$ slices are unitary or not. They claim it isn't, but cite another article &quot;The Census Taker's Hat&quot; written by Susskind claiming the contrary. One gets the impression Bousso and Susskind aren't in agreement on this issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Section 2.5, the authors forgot their own insight that distinct maximal causal diamonds are complementarity to each other. Because of this, it makes no sense to combine the local states of different overlapping causal diamonds together to get a more global description of a &quot;superobserver&quot; state.&lt;/p&gt;&#xA;" OwnerUserId="3691" LastActivityDate="2011-05-23T07:21:23.683" />
  <row Id="10259" PostTypeId="2" ParentId="10230" CreationDate="2011-05-23T08:08:18.727" Score="7" ViewCount="" Body="&lt;p&gt;After reading @Peter Morgan's answer, and giving it some more thought, I think this is actually simpler than it seems at first.&lt;p&gt;&#xA;For finite-dimensional spaces the trace of a commutator is indeed always zero. For infinite-dimensional spaces the trace is not always defined, since it takes the form of an infinite sum (for countable dimension) or an integral (for continuous dimension) which do not always converge.&lt;br&gt;&#xA;When the trace is defined, it obeys the same rules as in finite dimension, specifically the trace of a commutator is zero. For operators such as $x$, $p$ and their products, the trace is simply not defined, so there is no sense in asking questions about it.&lt;br&gt;&#xA;When computing thermal averages, the factor $e^{-\beta H}$ makes sure the trace converges, since the energy is always bound from below (otherwise the system is unphysical).&lt;p&gt;&#xA;I'm sure the concepts mentioned by @Peter Morgan are important in this context (boundedness, KMS-condition), but I don't know anything about them, and I think the answer I just provided suffices for practical purposes.&lt;/p&gt;&#xA;" OwnerUserId="975" LastEditorUserId="975" LastEditorDisplayName="" LastEditDate="2011-05-23T08:14:14.860" LastActivityDate="2011-05-23T08:14:14.860" />
  <row Id="10260" PostTypeId="1" CreationDate="2011-05-23T08:51:09.397" Score="0" ViewCount="127" Body="&lt;p&gt;A completely mixed state is a statistical mixture with no interference terms, and (QMD, McMahon, pg 229):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho = \dfrac{1}{n}I$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Tr(\rho^2) = \dfrac{1}{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are blackhole quantum states completely mixed, meaning that they each have the same probability of being observed if we could in fact observe a black hole?&lt;/p&gt;&#xA;" OwnerUserId="3460" LastEditorUserId="3460" LastEditorDisplayName="" LastEditDate="2011-05-23T09:28:32.933" LastActivityDate="2011-08-21T10:33:07.053" Title="Are black hole states completely mixed?" Tags="&lt;black-holes&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10261" PostTypeId="2" ParentId="10260" CreationDate="2011-05-23T08:59:41.533" Score="0" ViewCount="" Body="&lt;p&gt;All microstates with the same energy. Otherwise, we ought to have a Gibbs distribution $e^{-\beta E}$.&lt;/p&gt;&#xA;" OwnerUserId="3692" LastActivityDate="2011-05-23T08:59:41.533" CommentCount="1" />
  <row Id="10262" PostTypeId="1" CreationDate="2011-05-23T09:08:18.290" Score="6" ViewCount="163" Body="&lt;p&gt;The topic of &lt;a href=&quot;http://en.wikipedia.org/wiki/Lattice_QCD&quot; rel=&quot;nofollow&quot;&gt;Lattice QCD&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Lattice_gauge_theory&quot; rel=&quot;nofollow&quot;&gt;Lattice gauge theory&lt;/a&gt; or even &lt;a href=&quot;http://en.wikipedia.org/wiki/Lattice_field_theory&quot; rel=&quot;nofollow&quot;&gt;Lattice field theory&lt;/a&gt; is quite old now. And the main reason for the interest in the topic is the ability to calculate nonperturbative stuff on a computer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that to do research with lattice you need an access to some supercomputer. But now everyone can afford something that is as powerful as supercomputers 20 years ago. Maybe it possible to redo some of the results of that time? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any project that is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Relatively simple.&lt;/li&gt;&#xA;&lt;li&gt;Allows to calculate some real-world quantity (like mass of the proton).&lt;/li&gt;&#xA;&lt;li&gt;Can be done with an average home computer.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-05-23T17:03:39.493" Title="Home-made lattice calculation?" Tags="&lt;computational-physics&gt;&lt;lattice-model&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" />
  <row Id="10263" PostTypeId="2" ParentId="10239" CreationDate="2011-05-23T09:52:49.990" Score="1" ViewCount="" Body="&lt;p&gt;Take state $|\psi\rangle=(|00\rangle+|11\rangle)/\sqrt{2}$. It is a pure state, so its (von Neumann) entropy is 0. But both of its one-particle states have entropy equal to 1 bit, as they are completely mixed states of the dimension two. &lt;/p&gt;&#xA;" OwnerUserId="184" LastActivityDate="2011-05-23T09:52:49.990" />
  <row Id="10264" PostTypeId="2" ParentId="9830" CreationDate="2011-05-23T10:31:34.137" Score="2" ViewCount="" Body="&lt;p&gt;I will try to answer by using  simple physical arguments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first, the approximate formula for the capacitance of a ring is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C=\frac{\pi R}{ln (\frac{8R}{r})}(CGS-system)$$ where R is a radius of the ring and r is a radius of the wire. ($R&gt;&gt;r$)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $q$ be the total charge(distributed uniformly) of the ring. Then the electrostatic energy of the ring:&#xA;$$W_q=\frac{q^2}{2C}$$ Now, the radial force (due to the electrostatic repulsions), applied to the ring, can be obtained from a simple differentiation:&#xA;$$F_R=-\frac{dW_q}{dR}$$ Thus $$F_R=\frac{q^2}{2\pi}\frac{ln(\frac{8R}{r})}{R^2}$$&#xA;The tension $T$ in the wire:&#xA;$$T=\frac{F_R}{2\pi}= \left(\frac{q}{2\pi}\right)^2\frac{ln(\frac{8R}{r})}{R^2}$$ &#xA; Tensile stress of the wire:&#xA;$$\sigma=\frac{T}{S}=\frac{T}{\pi r^2} =\frac{1}{\pi}\left(\frac{q}{2\pi}\right)^2\frac{ln(\frac{8R}{r})}{(rR)^2}$$&lt;/p&gt;&#xA;" OwnerUserId="552" LastActivityDate="2011-05-23T10:31:34.137" CommentCount="1" />
  <row Id="10265" PostTypeId="2" ParentId="10262" CreationDate="2011-05-23T11:04:51.147" Score="2" ViewCount="" Body="&lt;p&gt;You may try to download QCD software for CUDA - using GPU etc. - at&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://nemea.ist.utl.pt/~ptqcd/&quot; rel=&quot;nofollow&quot;&gt;http://nemea.ist.utl.pt/~ptqcd/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See also&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.google.cz/search?q=cuda+qcd&quot; rel=&quot;nofollow&quot;&gt;http://www.google.cz/search?q=cuda+qcd&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I don't have any experience and I don't expect the work to be a user-friendly smooth sailing. ;-)&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-05-23T15:23:07.750" LastActivityDate="2011-05-23T15:23:07.750" CommentCount="5" />
  <row Id="10266" PostTypeId="1" CreationDate="2011-05-23T11:43:36.793" Score="3" ViewCount="137" Body="&lt;p&gt;&lt;a href=&quot;http://www.bbc.co.uk/news/science-environment-13469924&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; says that they are only able to achieve such extremely high fiberoptic data rates because the multiplex light and then use a Fourier Transform to split it up again.  But they say that they perform the FT &quot;optically&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I understand the math behing a FT, but I still can't wrap my mind around how this would be done in analog with light.  Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="922" LastActivityDate="2011-05-23T19:35:26.843" Title="&quot;Optically performed&quot; Fourier Transform" Tags="&lt;light&gt;&lt;fourier-transform&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="10267" PostTypeId="1" AcceptedAnswerId="10287" CreationDate="2011-05-23T13:42:42.620" Score="2" ViewCount="163" Body="&lt;p&gt;It is known that nowadays most light bulbs are filled with gas in order to minimise the evaporation rate of tungsten. Is e.g. argon put into the bulb with reduced pressure, so that the glass doesn't explode while heating the wire?&lt;/p&gt;&#xA;" OwnerUserId="2752" LastActivityDate="2011-05-23T17:50:42.417" Title="Pressure inside a light bulb" Tags="&lt;thermodynamics&gt;&lt;everyday-life&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="10268" PostTypeId="2" ParentId="10266" CreationDate="2011-05-23T14:08:25.183" Score="1" ViewCount="" Body="&lt;p&gt;Any positive lens will act on a wavefront as a mechanical optical computer, see any fourier optics book for the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shouldn't be too hard to process when you consider that light can travel as a wave composed of several frequencies, and in certain limits (see above books or 'Optics' by Hecht for a more basic approach.) a lens can act to seperate these frequencies in a power spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically light from a point within an emitting region will expand as a circular wave front, as will all neighbouring points.  At a distance from the source the light intensity will have mixed together into the fourier transform of the source distribution, a focussing lens will carry out the opposite operation and reform the image.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-05-23T14:08:25.183" CommentCount="1" />
  <row Id="10269" PostTypeId="2" ParentId="10267" CreationDate="2011-05-23T14:11:22.827" Score="-1" ViewCount="" Body="&lt;p&gt;The basic idea is to replace the air inside the bulb to avoid ignition of the element.  At ~5000 K the element would become blackened from carbon deposition if any air was present and so this is replaced by something inert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would guess this is used under atmospheric pressure to avoid problems with expansion under heating.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-05-23T14:11:22.827" CommentCount="4" />
  <row Id="10270" PostTypeId="1" CreationDate="2011-05-23T14:53:53.980" Score="0" ViewCount="147" Body="&lt;p&gt;Before asking, I swear that, for how it may seem strange, this is all true. Very briefly, when my pc is off and I am near it, it turns on. I do not touch anything. Nobody else have the same effect. I just have to still near it for 3-5 seconds and it will boot. It begun 1-2 weeks ago. The first time was at night, and I was sleeping, so I wasn't really near to it. It is not &quot;scheduled&quot;. It doesn't happen at 3:00 every day, just random.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some days before it started I had a radiography on my head. May I have absorbed some X-rays, which are now creating some sort of magnetic field around me? I'm quite scared from this thing, even if at first it seemed funny...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance, and excuse me for my poor english.&lt;/p&gt;&#xA;" OwnerUserId="3693" LastActivityDate="2011-05-23T16:03:36.237" ClosedDate="2011-05-23T16:54:19.890" Title="My computer turns on when it &quot;sees&quot; me" Tags="&lt;electromagnetism&gt;&lt;soft-question&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="0" />
  <row Id="10271" PostTypeId="1" AcceptedAnswerId="10772" CreationDate="2011-05-23T15:15:34.410" Score="1" ViewCount="197" Body="&lt;p&gt;I wonder if the refractive index of acetone, ethanol, methanol, heptane was measured in the range above 830 nm. The site &lt;a href=&quot;http://refractiveindex.info&quot; rel=&quot;nofollow&quot;&gt;refractiveindex.info&lt;/a&gt; provides it only for visible, and I couldn't find any other data so far.&lt;/p&gt;&#xA;" OwnerUserId="3694" LastActivityDate="2011-06-05T03:36:25.543" Title="Refractive index of organic solvents in near-infrared" Tags="&lt;optics&gt;&lt;refraction&gt;&lt;data&gt;" AnswerCount="1" />
  <row Id="10272" PostTypeId="2" ParentId="10267" CreationDate="2011-05-23T15:21:06.487" Score="1" ViewCount="" Body="&lt;p&gt;Most classical light bulbs that are produced today - in countries where they haven't been banned - are filled with (chemically inert) argon or nitrogen at a reduced pressure so that the bulb doesn't explode when it's heated up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light bulbs filled with vacuum have been produced, too. Of course, in both cases, the glass has to sustain pressures that are comparable to a significant fraction of one atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-23T15:21:06.487" />
  <row Id="10273" PostTypeId="2" ParentId="10262" CreationDate="2011-05-23T15:45:33.277" Score="3" ViewCount="" Body="&lt;p&gt;You &lt;em&gt;don't&lt;/em&gt; need a really big computer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peter LePage used to do talks where he'd ask the audience fro a &quot;random&quot; number as the beginning of the talk (but not 7, 17, 42, or 69 'cause he'd already done those) and start a simulation on one screen with that number as a seed. Then he'd give a talk on how to speed up LQCD calculations on the other screen while his PII 200 MHz laptop crunched a $b\bar{b}$ state &lt;em&gt;during the talk&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But lets talk about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, heavy flavor is easier to do than anything involving light quarks. Especially $u$ or $d$ quarks. You certainly want to learn this business with a faster problem than the proton mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, &quot;relatively simple&quot; in LQCD is still pretty complicated. Peter's 45 minute talk was intended to give people who knew roughly what a naive LQCD calculation looked like an idea of how his very clever trick for optimizing the business worked. I also sat through four hours of the same thing at a graduate summer school where he was teaching the technique and I only &lt;em&gt;sort of&lt;/em&gt; got it (because it was targeted at students who were &lt;em&gt;doing&lt;/em&gt; LQCD).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a big project.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-23T17:03:39.493" LastActivityDate="2011-05-23T17:03:39.493" CommentCount="1" />
  <row Id="10274" PostTypeId="2" ParentId="10266" CreationDate="2011-05-23T15:50:39.160" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Upon reading the question, I leapt to the conclusion that the asker was curious about the Fourier transforming properties of lenses. I am embarrassed to say that I did not immediately read the article. Upon doing so, I realize that the question is indeed about an all-optical FFT for temporal, not spatial frequencies. I've left my old answer below, but I've added a more appropriate answer here.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The article you linked is, like most popular reporting about science, very vague. I've done some checking, and it appears to me that the journal article being reported on is this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.opticsinfobase.org/oe/abstract.cfm?URI=oe-18-9-9324&quot; rel=&quot;nofollow&quot;&gt;D. Hillerkuss, M. Winter, M. Teschke, A. Marculescu, J. Li, G. Sigurdsson, K. Worms, S. Ben Ezra, N. Narkiss, W. Freude, and J. Leuthold, &quot;Simple all-optical FFT scheme enabling Tbit/s real-time signal processing,&quot; Opt. Express 18, 9324-9340 (2010) &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors describe a method for rapidly isolating signals sent over optical fibers on different wavelengths. This basic idea has been used for quite a long time, and the simpler approach is called Wavelength Division Multiplexing (WDM). The idea is simple enough -- encode optical signals on light of different wavelengths, send them over a fiber, and use a prism or diffraction grating to separate the signals at the receiving end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is a limited range of wavelengths that are useful for transmission over long distances in optical fibers, so the more tightly spaced the signal wavelengths are, the more data can ultimately be sent. Modern techniques use the very closely spaced spectral lines produced by mode-locked lasers. This technique is called Orthogonal Frequency Division Multiplexing (at least that is what I gather. I am not an expert on telecom technology). At this level, it becomes impractical to isolate the signals with a diffraction grating, as the spatial separation between them is too small. Current technology apparently deals with this by digitizing the optical signals and performing Fourier analysis off line (i.e. not in real time) which limits the data rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The optical FFT developed by the authors is a solution to this problem. They have developed an optical apparatus which takes the incoming signal and splits it multiple times. Each copy of the signal is delayed by a specific amount, and then fed through a series of fiber-optic junctions which combine or phase shift as needed to implement a shift-and-add FFT algorithm. The key figure from the paper is reproduced below (this is allowed, I hope?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/jhZYp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The all-optical computation is carried out in real-time, allowing very high data rates.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Old Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the conditions that the medium of propagation is isotropic, and the light is monochromatic and coherent, the electric and magnetic fields along any chosen vector will differ by at most a proportionality constant, so the field can be expressed as a complex scalar field which specifies the magnitude and phase of the EM wave. This is called the &quot;scalar approximation&quot; and provides a mathematical basis for an approximate description of wave optics which applies in the vast majority of situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this scalar approximation, an optical field at one plane can be used to compute the field after some propagation distance. In general, this can be done using the Rayleigh-Sommerfeld Diffraction Integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ U(P_0) = \frac{1}{i \lambda} \int\int_\Sigma U(P_1) \frac{\mathrm{exp}(i k r_{01})}{r_{01}} \cos{\theta} \;\; \mathrm{ d}s$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $U(P_0)$ is the optical field at plane $P_0$, $\lambda$ is the wavelength, $k$ is the waveneumber, $r_{01}$ is the vector from the point being integrated over in $P_1$ to the point being calculated in $P_0$. The region of integration, $\Sigma$, is the aperture in the input plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most cases this integral cannot be solved analytically, and direct numerical solutions are very time consuming. However, under certain conditions (which are easily obtained in the real world) this integral can be approximated by &lt;a href=&quot;http://en.wikipedia.org/wiki/Fraunhofer_diffraction&quot; rel=&quot;nofollow&quot;&gt;Fraunhofer Diffraction&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Fresnel_diffraction&quot; rel=&quot;nofollow&quot;&gt;Fresnel Diffraction&lt;/a&gt;. I will limit myself here to the latter for simplicity, but the two situations are computationally very similar, so the answer to your question would be almost the same if I presented Fraunhofer Diffraction instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fresnel Diffraction is obtained by expanding $r_{01}$ by a Taylor series and keeping terms up to second order. It is accurate when the propagation distance is sufficiently larger than the aperture. The exact definition of this condition varies from author to author, and results do not depend strongly on where this limitation is chosen. Under this assumption, the diffraction integral is reduced to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ U(x,y) \propto \int\int^{\infty}_{-\infty} U(\xi, \eta) \mathrm{exp}\left[\frac{i \pi}{\lambda z}((x-\xi)^2 + (y-\eta)^2)\right] \;\; \mathrm{d}\xi \; \mathrm{d}\eta $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is in the form of a convolution, so convolution theorem allows this to be expressed as a Fourier transform as well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ U(x,y) \propto \int\int^{\infty}_{-\infty} U(\xi, \eta) \mathrm{exp}\left[\frac{i k}{\lambda z} (\xi^2 +\eta^2)\right] \mathrm{exp}\left[\frac{-i 2 \pi}{\lambda z} (x\xi + y\eta)\right] \;\; \mathrm{d}\xi \; \mathrm{d}\eta $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be an equality rather than a proportionality equation, but I have left of some complex factors of unity magnitude for simplicity. My notation has changed here to a less compact and more practical form because this integral is usually intended to be used for actual computation, as opposed to mathematical derivations; $x$ and $y$ are output coordinates, and $\xi$ and $\eta$ are input plane coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, this is simply a Fourier transform of $U(\xi, \eta)$, with some appropriately chosen scaling applied (specifically, $f_X \rightarrow \frac{x}{\lambda z}$, and $f_Y \rightarrow \frac{y}{\lambda z}$. $f_X$ and $f_Y$ are called &quot;spatial frequencies&quot; to avoid confusion with the more common temporal frequencies that we talk about in the context of Fourier analysis of time series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As might be obvious by the scaling I described, this integral is easiest to compute in terms of the &lt;em&gt;angular&lt;/em&gt; position in the output plane ($\frac{x}{z}$) when z is very large. In other words, if an optical field is allowed to propagate to infinity, its angular distribution will be the Fourier transform of the initial field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a lens of focal length $f$, let us modify the initial field $U(\xi, \eta)$ by a quadratic phase term $\mathrm{exp}\left[\frac{-i\pi}{\lambda f}(\xi^2 + \eta^2)\right]$. We are interested in the field at the focal plane of the lens, so we will propagate a distance $z = f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Immediately, by looking at the Fourier transform formulation of the diffraction integral, it is obvious that the quadratic phase term inside the integral is exactly opposed by the phase term due to the lens (note that $k=\frac{2\pi}{\lambda}$). The resulting field is simply a Fourier transform of the input field, expressed in spatial coordinates rather than angular coordinates. &lt;/p&gt;&#xA;" OwnerUserId="869" LastEditorUserId="869" LastEditorDisplayName="" LastEditDate="2011-05-23T19:35:26.843" LastActivityDate="2011-05-23T19:35:26.843" CommentCount="0" />
  <row Id="10275" PostTypeId="1" AcceptedAnswerId="10280" CreationDate="2011-05-23T16:01:08.737" Score="6" ViewCount="236" Body="&lt;p&gt;The following question is about chapter 2 of Sakurai's &lt;em&gt;Modern Quantum Mechanics&lt;/em&gt;. I wish I could link to the Google book, but it doesn't seem to have a satisfactory preview to be able to read the section I'm talking about, so I'll do my best to write out the part I'm talking about...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the section about propagators and Feynman path integrals (p. 113 in my edition) he gives the following example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G(t) \equiv \int d^3 x&amp;#39; K(x&amp;#39;, t; x&amp;#39;,0) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\int d^3 x&amp;#39; \sum_{a&amp;#39;} |\langle x&amp;#39;|a&amp;#39;\rangle|^2 \textrm{exp} \left(\frac{-iE_{a&amp;#39;}t}{\hbar}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=\sum_{a&amp;#39;}\textrm{exp} \left( \frac{-iE_{a&amp;#39;}t}{\hbar} \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He goes on to say that this is equivalent to taking the trace of the time evolution operator in the $\{|a&amp;#39;\rangle\}$ basis, or a &quot;sum over states&quot;, reminiscent of the partition function in statistical mechanics. He then writes $\beta$ defined by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\beta=\frac{it}{\hbar}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;real and positive, but with with $t$ purely imaginary, rewriting the last line of the previous example as the partition function itself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z=\sum_{a&amp;#39;} \textrm{exp} \left( -\beta E_{a&amp;#39;} \right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is this: What is the physical significance (if any) of representing time as purely imaginary? What does this say about the connection between thermodynamics and quantum? The fact that you get the partition function exactly, save for the imaginary time, here seems  too perfect to be just a trick. Can someone explain this to me? &lt;/p&gt;&#xA;" OwnerUserId="3601" LastEditorUserId="3601" LastEditorDisplayName="" LastEditDate="2011-05-23T17:03:52.590" LastActivityDate="2011-06-24T20:19:51.447" Title="Imaginary time in quantum and thermodynamics" Tags="&lt;quantum-mechanics&gt;&lt;thermodynamics&gt;&lt;cosmology&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="10276" PostTypeId="2" ParentId="10270" CreationDate="2011-05-23T16:03:36.237" Score="4" ViewCount="" Body="&lt;p&gt;Are you referring to some kind of sleep mode, that it comes back if you move the mouse or press any key? If so, the mouse may be moved by perturbations on the table be walking nearby, for example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean it's actually turning the power on with your presence then I can think of two possibilities: 1- you're lying; 2- someone is playing a prank on you :P&lt;/p&gt;&#xA;" OwnerUserId="301" LastActivityDate="2011-05-23T16:03:36.237" CommentCount="2" />
  <row Id="10277" PostTypeId="1" CreationDate="2011-05-23T16:05:14.187" Score="2" ViewCount="263" Body="&lt;p&gt;Specifically, how can decoherence explain the appearance of flecks of metallic silver on a photographic plate when exposed to the very weak light of a distant star?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Perhaps the advocates of decoherence need some context for this question. There is a certain definite quantity of energy on the order of one or two eV to drive the chemical reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2AgBr -&gt; 2Ag +  Br2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the reaction responsible for the fleck of silver on the photographic plate. The amount of energy is far greater than can be accounted for in any realistic time frame by the classical e-m wave energy of the light of a distant star. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any explanation must explain where this energy comes from. How does &quot;decoherence&quot; claim to do this? I have heard over and over again that there is a matrix which is diagonalized, but no one has so much as volunteered to say just what matrix they are talking about. Is it, for example, the matrix of position states of the photon? Or perhaps it is the oxidation states of the silver atom? And I would really like a better explanation of how the matrix is &quot;diagonalized&quot; than to simply repeat that it is in &quot;thermal contact with the environment.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I have reviewed the comments again and I find that no one has come close to dealing with the question. I cannot find anything wrong in the way I have asked it so far, so I am posting this edit as my only means to prompt people to attempt an answer.&lt;/p&gt;&#xA;" OwnerUserId="3253" LastEditorUserId="3253" LastEditorDisplayName="" LastEditDate="2011-08-01T14:02:35.120" LastActivityDate="2011-08-01T14:02:35.120" Title="Does decoherence explain all instances of wave function collapse?" Tags="&lt;quantum-mechanics&gt;&lt;mechanics&gt;" AnswerCount="1" CommentCount="14" />
  <row Id="10278" PostTypeId="1" CreationDate="2011-05-23T16:18:52.897" Score="0" ViewCount="128" Body="&lt;p&gt;This question is based upon the gedankenexperiment presented in &lt;a href=&quot;http://physics.stackexchange.com/questions/9904/incommensurability-between-different-observers-describing-the-same-universe&quot;&gt;this question&lt;/a&gt;. Being a gedankenexperiment, it is imaginary of course, and its purpose is to elucidate underlying principles. As such, it is sufficient that it is possible to carry out in principle, but not in practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, Alice falls into a black hole, while Bob stays out. Bob has detectors collecting all the Hawking radiation from start till end, and feeds them into a quantum computer which reconstructs a simulated Alice with the simulation running backward in time at some moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me pause to point out I'm far from the first person to propose this gedankenexperiment. The only difference being others neglected the need for a quantum computer to unscramble the information about Alice, and also the inevitability for that quantum computer to simulate Alice and much of the black hole interior just to extract the qubit she holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two incommensurable versions of the story according to black hole complementarity. In the one according to Bob, the interior of the black hole doesn't exist, and there is a stretched horizon at its boundary. Alice only exists as a highly scrambled delocalized holographic encryption on the stretched horizon. The scrambled information contained there is transmitted as Hawking radiation where it is processed and decrypted by a quantum computer into a simulated Alice far in the future. As Alice is a quantum system, the no Xerox, i.e. no cloning theorem, tells us from a purely quantum information theoretic point of view, the holographic and simulated Alice are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the version according to Alice, she's very much real and localized, and existing inside a very real black hole interior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, I have just presented three versions of the same Alice; the version inside a real interior, a holographic version, and a simulated version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make sense of both stories, we can discard those parts which are inherently incompatible and focus upon the invariants common to both. And that invariant is the quantum informational description of Alice, but not her location in space, or the time she exists in. In fact, even if we take into account the well-known fact that there is no absolute space or time in general relativity, the more relational concepts of coincidence (at the same location and the same time), spatial extent, or time-ordering aren't invariants either. Remember that the simulation is running time in reverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this herald a new paradigm in quantum gravity whereby quantum information is more fundamental than space or time or matter? Are traditional approaches to quantum gravity starting from spacetime, or even space and time separately, or even time but no space missing the point? Do space and time emerge as patterns within the quantum information describing the universe? Not only that, given the same quantum information description of the universe, can we construct wildly different space-time patterns into which the invariant quantum information can fit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Landauer once said information is physical. Could it be space and time are not physical in the same manner, and quantum information is far more physical than space or time? By physical, I mean physical in the usual sense of physical, like when physicists state gauge-invariant quantities are physical, or coordinates are unphysical, or the absolute phase of the wave function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantum information itself contains all the information about our universe, but in an encrypted form. To extract information about a subsystem requires decryption on our part. However, due to complementarity, we can't extract information about all the regions of the universe simultaneously. In decrypting information about a subsystem, the space and time it is most naturally embedded in is also decrypted. For our example, it would be Alice in the black hole interior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is decryption of a subsystem always accompanied by a local neighborhood of space and time it is embedded in, and does this have anything to do with the low entropy of our universe, and can that be traced back to eternal inflation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Decrypting information about a subsystem at different times also doesn't commute with each other, and so, we can only decrypt one time at a time.&lt;/p&gt;&#xA;" OwnerUserId="3500" LastEditorUserId="3500" LastEditorDisplayName="" LastEditDate="2011-05-24T15:01:16.973" LastActivityDate="2011-05-24T15:01:16.973" ClosedDate="2011-05-23T16:48:40.610" Title="Is quantum information more fundamental than space, time and matter in quantum gravity?" Tags="&lt;black-holes&gt;&lt;quantum-gravity&gt;&lt;quantum-information&gt;" CommentCount="1" />
  <row Id="10280" PostTypeId="2" ParentId="10275" CreationDate="2011-05-23T16:38:31.577" Score="0" ViewCount="" Body="&lt;p&gt;The Green's functions in QM and in the heat conduction equations describe the relaxation of a $\delta$-like perturbation at some point at t=0. In QM it is a superposition of waves, in the heat conduction (or diffusion) equation it is spreading out the perturbation over the system. In the latter case the &quot;regular regime&quot; of the relaxation is described &quot;only&quot; with the slowest exponential $exp(-t/\tau_0)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statistical sum $Z$, however makes sense only at finite T. Nobody considers large variations of T. Instead, it is interesting to study $Z(T)$ as a function of the system parameters involved in $E_n$.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-05-23T16:38:31.577" CommentCount="2" />
  <row Id="10281" PostTypeId="2" ParentId="10275" CreationDate="2011-05-23T17:26:19.393" Score="1" ViewCount="" Body="&lt;p&gt;I'm going to &lt;em&gt;venture&lt;/em&gt; a way to think about this, hoping that I'm at a stage in my research to give a moderately coherent picture, however what follows is not a satisfactory detailed analysis. I'm not aware of an intuitive way to think about this being in textbooks or in the literature more generally; AFAIK thermal states are dealt with purely in terms of their algebraic structure (I note that I will be very happy to be corrected on this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hamiltonian operator $\hat H$ describes the time-like evolution of the state (or of the operators if we take the Heisenberg PoV).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In relativistic models, a thermal equilibrium state is associated with a &lt;em&gt;particular&lt;/em&gt; time-like direction. A thermal equilibrium state is invariant under space-like and time-like translations and under rotations orthogonal to a given time-like 4-vector. This &lt;em&gt;definition&lt;/em&gt; applies as well to classical systems as to quantum systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relativistic model makes it clear that a thermal state has &lt;em&gt;inertia&lt;/em&gt;, because it requires energy and momentum to boost the equilibrium state to a different time-like direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In relativistic quantum theory, we have to introduce different Hamiltonian operators for different time-like directions, $\hat H(T^\mu)=\hat P_\mu T^\mu$. This operator commutes with generators of translations and of rotations orthogonal to $T^\mu$, but it does not commute with boosts, just as we require. From a symmetry group perspective, the only resources we have available are the six generators of the Lorentz group, and the definition of the thermal state is intimately connected with boosts (and with no other geometrical structure, such as, for example, conformal transformations). This is where I (very much) regret that I can't make the next step. We want to transform the vacuum from zero inertia to non-zero inertia (or perhaps transform infinite undirected vacuum zero-point energy into time-like directed inertia), which the Gibbs operator seems ideal for. It has felt as if it's on the tip of my tongue for well over five years, which is long enough that, of course, I've never really been close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like I say, this is too fragmentary. I'm sorry if it's not Useful to you in its current form. It's certainly frustrating to see for myself that this attempt to articulate the ideas looks like this. This way of thinking is partly grounded in a paper of mine that I have previously mentioned a few times on Physics SE, &quot;A succinct presentation of the quantized Klein–Gordon field, and a similar quantum presentation of the classical Klein–Gordon random field&quot;, &lt;a href=&quot;http://arxiv.org/abs/quant-ph/?0411156&quot; rel=&quot;nofollow&quot;&gt;quant-ph/0411156&lt;/a&gt;, Phys. Lett. A 338, 8-12(2005), which discusses QFT free fields in sufficiently weird terms that it will twist your head a little, if you want that.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-05-23T17:26:19.393" CommentCount="3" />
  <row Id="10283" PostTypeId="1" AcceptedAnswerId="10289" CreationDate="2011-05-23T17:36:53.557" Score="1" ViewCount="134" Body="&lt;p&gt;We all know that subatomic particles exhibit quantum behavior. I was wondering if there's a known order of magnitude cutoff in &quot;size&quot; where we stop exhibiting such behavior. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, I imagine that there's no absolute number as to when quantum effects vanish, otherwise that would probably be in violation of the Uncertainty Principle [not 100% sure of this].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I have read, it seems to me that we still see quantum effects up to the nanometer level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By quantum behavior, I mean when measured, the wave function collapses. &lt;/p&gt;&#xA;" OwnerUserId="3270" LastEditorUserId="3270" LastEditorDisplayName="" LastEditDate="2011-05-23T18:42:46.807" LastActivityDate="2011-05-23T18:42:46.807" Title="Is there a &quot;Size&quot; Cutoff to Quantum Behaviour?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="3" FavoriteCount="0" />
  <row Id="10284" PostTypeId="2" ParentId="10248" CreationDate="2011-05-23T17:38:30.823" Score="1" ViewCount="" Body="&lt;p&gt;There are two main reasons why close hauling brings your bow to the wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reason is that the trimmed mainsail gives a larger torque: The equivalent force in a freebody diagram is larger for a trimmed (close hauled) mainsail.  Note that the acutal force does not have to change in order to accomplish this, since |torque| = |F X L|; where F = force, L = length of lever arm, both are vector quantities, and 'X' is the vector cross product operator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the main mast as the pivot point for the boat (this is almost certainly not exactly true, but consider it so for the moment).  The lever arm which the mainsail uses to pivot the boat about this point is the distance from the mainsail mast to the belaying point or winch for the line which stays the mainsail in whatever position you want it.  This length does not change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, L does not change.  F does not change.  What changes?  The angle between the two vectors.  When F and L are at almost 90 degree angles, &quot;F x L&quot; is at its largest.  This occurs when close hauled. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A way to visualize the equivalent torque is to picture the orientation of the boat and mast from above.  Assume that the mainsail cross member is as long as the distance from the mast to the belaying point.  Assume also that the mainsail is let out to where it forms a 90 degree angle with the keel line.  That means the mainsail belaying line will form a 45 degree angle with the keel line and the mainsail.  Lets say that 'F' is 1000 lbs.  That tension can be broken into a force and a torque. sine(45) = cosine(45) = 1/1.414 = 0.707.  So the boat experiences 1000sin(45) = 707 lbs of forward thrust through the water.  This force is being aplied to the boat at the belaying point of the mainsail line.  The boast also experiences 1000cos(45) = 707 lbs. of force as the 'F', in a pure (F)(L) torque.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets close haul:  'F' and 'L' are the same, and the mainsail makes a 30 degree angle with the keel line.  The forward thrust to the boat is 1000sin(30) = 500 lbs.  The 'F' in the torque component is 1000cos(30) = 866 lbs.  That is almost a 24% increase in 'F', which will result in a 24% increase in torque (since L remains constant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second reason that close hauling brings the bow into the wind is to think of the mainsail as a weathervane, with the wind coming from just a few degrees aft of a ray which is co-axial with the mainsail cross arm and pointing forward through the mast.  This never changes (significantly, in the steady state, and assumes you are not &quot;running before the wind&quot;).  So, on any given tack, the position of your mainsail relative to the wind does not change.  Given this, what will happen if you haul the mainsail closer to your boat?  Relative to the wind, you are not really hauling the mainsail anywhere, since it stays fixed relative to the wind.  What you are really doing is hauling your keel line (boat's axis) closer to the mainsail's axis, which brings the bow closer to the wind.  &lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-05-23T17:48:27.877" LastActivityDate="2011-05-23T17:48:27.877" CommentCount="1" />
  <row Id="10286" PostTypeId="2" ParentId="10283" CreationDate="2011-05-23T17:43:19.030" Score="2" ViewCount="" Body="&lt;p&gt;There are definitely circumstances where we see quantum behaviors at rather large scales like, for example, in superconductors. In BCS theory cooper pairs are described by a macroscopic wave-function which, to my knowledge is valid for the bulk superconductor no matter how large it is. &lt;/p&gt;&#xA;" OwnerUserId="3601" LastActivityDate="2011-05-23T17:43:19.030" CommentCount="2" />
  <row Id="10287" PostTypeId="2" ParentId="10267" CreationDate="2011-05-23T17:50:42.417" Score="4" ViewCount="" Body="&lt;p&gt;A great link that details some experiments that looked for the answer to this is &lt;a href=&quot;http://iopscience.iop.org/0031-9120/41/6/F04/pdf/pe6_6_f04.pdf&quot; rel=&quot;nofollow&quot; title=&quot;Experiment tests pressure in light bulbs&quot;&gt;Experiment Tests Pressure in light bulbs&lt;/a&gt;.  But first we need to cover some basis of the gas law to do comparisons.  Firstly, most the time a bulb is off, it is near room temperature and most of the time it is on it is at a high temperature but the quantity of gas in the bulb remains the same.  Engineering-wise, it is most desirable to have the pressure close to atmosphere when on, because at that time the glass will be hot and more prone to failure.  Denoting these states as &lt;em&gt;on&lt;/em&gt; and &lt;em&gt;off&lt;/em&gt;, I will write the gas law for the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$PV=nRT$$&#xA;$$\frac{ P_{on} }{ T_{on} } = \frac{ P_{off} }{ T_{off} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The link I am referencing does an experiment where the bulb is submerged in water and the pressure boundary is broken.  Since the bulb has a lower pressure than atmosphere water rushes in, this water mass can be measured and used to find the pressure before it was broken.  This state I will denote as &lt;em&gt;test&lt;/em&gt; and it is compared to the &lt;em&gt;off&lt;/em&gt; state as a constant temperature process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{off} V_{off} = P_{test} V_{test}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this experiment they found two 25 W incandescent lamps to have a pressure of about $\frac{6}{1000} atm$, which is consistent with the low pressure inert atmosphere most people expect.  For higher wattage lamps, they found the pressure to be close, but slightly lower, than atmospheric pressure.  Temperature of the filament, when on, is in the neighborhood of 2700-2900 K, but this is not the temperature of the gas.  Any reasonable assumptions will put the $P_{on}$ value for the low pressure bulbs at much smaller than atmosphere.  For the higher power, higher pressure bulbs, I will address it more specifically using the surface temperature numbers from &lt;a href=&quot;http://board.interfire.org/cgi-bin/ultimatebb.cgi?ubb=get_topic;f=5;t=000120&quot; rel=&quot;nofollow&quot;&gt;this source with clearly no credibility&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_{off} \approx 298 K$$&#xA;&lt;strong&gt;40 W&lt;/strong&gt;&#xA;$$P_{off} \approx 0.914 atm$$&#xA;$$T_{on} \approx 395 K$$&#xA;$$P_{on} = \frac{ T_{on} P_{off} }{ T_{off} } = 1.21 atm$$&#xA;&lt;strong&gt;60 W&lt;/strong&gt;&#xA;$$P_{off} \approx 0.768 atm$$&#xA;$$T_{on} \approx 400 K$$&#xA;$$P_{on} = \frac{ T_{on} P_{off} }{ T_{off} } = 1.03 atm$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To conclude, there are two types of constructions of incandescent light bulbs discussed here, one entailing a low vacuum and one that objectively reaches near atmospheric pressure in the &lt;em&gt;on&lt;/em&gt; state.  There is likely significant variation in products you can buy, and sample numbers given here can't be generalized to any given bulb you're looking at.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-05-23T17:50:42.417" CommentCount="1" />
  <row Id="10288" PostTypeId="2" ParentId="10204" CreationDate="2011-05-23T18:08:27.957" Score="0" ViewCount="" Body="&lt;p&gt;The answer to your question is, &quot;Yes, there is.&quot;   For capturing and analyzing real time signals, often that functionality is sold as a frequency analyzer, in conjuction with an O'scope, or other HW/SW combination, for about the same cost as a modest home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a cheaper solution, you can feed a time domain plot of the signal into a Fourier transform.  You will need to pick points separated by no more than 1/(2F), where F is the highest frequency component you hope to discover.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The SW we used to perform digital signal analysis in my graduate class, was MATLAB.  I'm sure that you could get the same result in just about any language, or even just an Excel spreadsheet, as long as you understand Fourier Transforms.  The G3 package, mentioned by Helder seems intriging.  I may go check it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pulse of a purely sinusoidal 5MHz wave will, indeed, show other frequency components in addition to 5MHz.  In order to not show components other than 5MHz, the pulse would have to be infinitely long, or at least longer than the sampling window.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-05-23T18:13:34.103" LastActivityDate="2011-05-23T18:13:34.103" />
  <row Id="10289" PostTypeId="2" ParentId="10283" CreationDate="2011-05-23T18:20:09.503" Score="1" ViewCount="" Body="&lt;p&gt;The classic experiment demonstrating quantum effects, the 2 slit experiment, has been preformed with subsequently larger and larger particles as our technology available to do it advanced.  Originally, it was performed with electrons, which are just as much matter as any other matter, but are extremely small.  The largest particle it has been demonstrated with are Buckminsterfullerene, which contain 60 Carbon atoms.  For size comparison:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_e = 5.485 x 10^{-4}u$$&#xA;$$m_{buckyball} = 720.642 u$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a good reason that the experiment gets more difficult with increasing mass, and to be sure, the buckball experiment was quite an accomplishment.  To the basics of quantum mechanics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta x\, \Delta p \ge \frac{\hbar}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, the de Broglie wavelength is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda = \frac{h}{p} = \frac{h}{m v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that in order to obtain the same wavelength with the same mass you have to decrease the velocity.  The reason this could be problematic for such experiments is that it is hard to successfully create the conditions needed with a large and slower moving particle, such as needing a better vacuum.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-05-23T18:20:09.503" CommentCount="4" />
  <row Id="10290" PostTypeId="2" ParentId="10283" CreationDate="2011-05-23T18:31:18.277" Score="2" ViewCount="" Body="&lt;p&gt;There is no known cutoff in &quot;size&quot; where systems stop exhibiting &quot;quantum behavior&quot;.  I put those words in quotes, because &quot;size&quot; can mean different things to different behavior, and you weren't explicit about what you mean by &quot;quantum behavior&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Folks have seen double-slit interference with molecules made up of ~60 atoms (I think recent experiments may have increased this number).  If you were to try to do the same thing with a baseball, as far as anyone knows there's no fundamental reason why it wouldn't work, but the experiment is far beyond our technical abilities (it would take a length of time greater than the age of the universe to do the experiment, nobody knows how to isolate the baseball from the environment for that length of time, etc. etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Superconducting rings have been built with diameters of a few centimeters (probably bigger) and still show flux quantization, which is a quantum-mechanical effect due to the electron wavefunction.  Those rings are definitely bigger than a few nanometers, and I'd consider it &quot;quantum behavior&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Folks have shown quantum-mechanical entanglement between two light beams that were separated by kilometers (in EPR tests).  A kilometer sounds like a big &quot;size&quot; to me, but maybe that's not what you meant?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take your pick as to what's the biggest &quot;size&quot; here, but as far as anyone knows, there's no size cutoff to quantum mechanics.  However, depending on the kind of experiment (2-slit interferometer, etc.) it definitely becomes more and more technically difficult to perform the experiment as the system is scaled up.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-05-23T18:31:18.277" CommentCount="5" />
  <row Id="10291" PostTypeId="2" ParentId="5615" CreationDate="2011-05-23T18:33:35.750" Score="-3" ViewCount="" Body="&lt;p&gt;Well it is possable that if the 2 collide a massive explotion could happen. &#xA;Or the 2 would cancle each other out.&lt;/p&gt;&#xA;" OwnerUserId="3698" LastActivityDate="2011-05-23T18:33:35.750" />
  <row Id="10292" PostTypeId="2" ParentId="10277" CreationDate="2011-05-23T18:43:40.693" Score="1" ViewCount="" Body="&lt;p&gt;Almost phenomenologically, &quot;the appearance of flecks of metallic silver on a photographic plate&quot; is a thermodynamic transition that happens at different rates depending on the details of how the plate is prepared and on details of the exposure of the photographic plate. Such thermodynamic transitions are often correlated in nontrivial ways. All QM has to do to be Useful is to model or describe the statistics of the thermodynamic transitions. [Note that my introduction of the idea of a &lt;em&gt;thermodynamic transition&lt;/em&gt; makes my claim here &quot;theory-laden&quot;, not quite phenomenological, at least to that extent.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Explanation&lt;/em&gt; is not &lt;em&gt;necessary&lt;/em&gt; for Usefulness. One topic of research in Philosophy of Physics has been to try to determine what makes a model &quot;explanatory&quot;, which IMO has been rather inconclusive. Models may be more or less Useful for many different reasons, including tractability and directness of reference between elements of the theory and elements of experimental signal data. Note that a class of models may &lt;em&gt;seem&lt;/em&gt; explanatory for 50 years even if it is the phlogiston theory, if the model is superficially nice in whatever ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which brings me to my Answer, which I'm pretty sure you won't find Useful, which is that Decoherence doesn't explain particularly well, whatever that means, partly because it's not a very tractable approach. Decoherence seems to have fairly direct referents, which perhaps is what makes it appeal to some people quite strongly. The same is true of &quot;wave function collapse&quot;: it's possible to structure experimental data taking wave function collapse as a fundamental modeling strategy, but so far no-one has produced a mathematization that is enough more Useful than just dealing with the statistics of thermodynamic events. There are people who think it illuminates what we're doing with QM in ways that might lead to a better mathematical formulation of the whole theory, but, I think, nothing yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a similar vein, you may notice that Particle Physics is more often called High Energy Physics than it used to be, which seems to me to reflect the realization, not uniformly acknowledged, that the explanation of tracks of obviously related thermodynamic events in detectors as &quot;caused by particles&quot; is weakened by the many low-energy experiments that show that the concept of a particle cannot be that simple. As of now, Quantum fields are as likely to be the locus of descriptions of experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious whether you can knock down this argument, such as it is. I think you're looking at this all wrong, but of course it may be me. That I've worked on this for a long time doesn't guarantee much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT (a long comment, in response to Marty's comment that first mentions &quot;Quantum Siphoning&quot;): &lt;em&gt;I&lt;/em&gt; take the Wave Function and operators to be a good way to generate probability measures. The empirical success comes from the probabilities being able to be good models for (or descriptions of) statistics of raw experimental data. I take it that probabilities do &lt;em&gt;not&lt;/em&gt; cause individual events, they describe &lt;em&gt;sets&lt;/em&gt; of events (propensity interpretations of probability notwithstanding). [&lt;em&gt;If&lt;/em&gt; we go the Wigner function route --which I &lt;em&gt;don't&lt;/em&gt;, except as a mathematical equivalence, because I think it obscures the relationship to empirical data-- the wave function is just a generalized probability function that sometimes has negative values.] If one wants to change probability distributions as a result of experience, instead of taking other approaches to statistics, then one should use something like Bayes' rule, which in general doesn't just change the probability from 0.615802 to 0 or to 1. &quot;collapse&quot; of the wave function adds an extra level of structure to the concept of a probability distribution that I think just doesn't fit well, as Mathematics. If people want to use &quot;collapse&quot;, I think it has to be done somehow differently. It's possible that a propensity interpretation could work for you, but I think we would then quite quickly get far enough apart that we can't talk to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I prefer my description of individual events (and we may just have to accept that this is a sticking point)-- that we should say that the individual events are &quot;thermodynamic transitions&quot;, whatever that means, leaving a causal account of how that happens for the future. The concept of thermodynamic transitions is the historical concept from Physics that I think fits the case. A thermodynamic event implicitly invokes at least a large number, perhaps an infinite number of degrees of freedom, to explain what happens when there is an apparent discontinuity, it introduces a degree of complexity that is hard to manage mathematically, which definitely has its problems. Decoherence also introduces an infinite number of degrees of freedom, but I think by introducing the environment in the way it does it doesn't adequately embrace the complexity of the photographic plate. I think your description of what happens in a photographic plate accepts that complexity, but then looks to make &quot;collapse&quot; of a quantum state, which has nowhere near as much structure as the photographic plate, be an explanation of what is happening. It's important that it not be brushed under the table, but we can measure where and when thermodynamic events happen without knowing how they happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope &lt;em&gt;that&lt;/em&gt;'s helpful. I expect no-one else is listening!&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-05-24T10:50:41.977" LastActivityDate="2011-05-24T10:50:41.977" CommentCount="6" />
  <row Id="10293" PostTypeId="1" CreationDate="2011-05-23T18:54:57.423" Score="5" ViewCount="171" Body="&lt;p&gt;Kind of an extension to &lt;a href=&quot;http://physics.stackexchange.com/questions/425/whats-the-relationship-between-thermal-radiation-and-johnson-thermal-noise&quot;&gt;this question&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you heat up an object, and put it in contact with a colder object, in an ideal insulated box, the heat from one will transfer to the other through thermal conduction and they will eventually reach an equilibrium temperature at the midpoint, correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you have a hot resistor (electrical component) and a cold resistor, and connect them by their leads, so that they make a circuit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/E1MCl.png&quot; alt=&quot;hot and cold resistors&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there will be the same conduction and radiation heat transfers.  But also, the hotter resistor will have a larger noise current, right?  So will there additionally be a transfer of electrical energy from one resistor to the other?  Would completing the circuit allow them to reach equilibrium temperature faster than if they were just touching through an insulator with the same thermal conductivity?&lt;/p&gt;&#xA;" OwnerUserId="176" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-06-09T16:59:00.210" LastActivityDate="2011-06-09T16:59:00.210" Title="What happens if you connect a hot resistor to a cold resistor?" Tags="&lt;thermodynamics&gt;&lt;electricity&gt;&lt;heat&gt;&lt;fluctuation-dissipation&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="10294" PostTypeId="2" ParentId="5615" CreationDate="2011-05-23T19:14:53.143" Score="-1" ViewCount="" Body="&lt;p&gt;And now an experimentalist chimes in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Baryon number conservation depends on the lifetime of the proton at low energies, admittedly quite large but finite in many theories. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At high energies, high enough for a quark gluon plasma to form, the preponderance of baryons becomes a slight excess of quarks versus antiquarks in the soup. This because quark antiquark pairs continuously form and annihilate in the sea, and if one would want to keep a conserved number that would be the excess number of quarks over antiquarks in a matter plasma and vice verso in an antimatter one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One supposes that in the collapse of a hole the energies are sufficient for a quark gluon plasma to form and thus make moot the point of baryon number and antibaryon number assumed in the question. The excess of quarks in one hole versus the excess of antiquarks in the other would be proportionately very diluted to assume that the hole could be characterized as matter or antimatter anyway, even if all  other valid  physics arguments already presented by others were not present.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-23T19:14:53.143" CommentCount="2" />
  <row Id="10295" PostTypeId="1" AcceptedAnswerId="10299" CreationDate="2011-05-23T19:35:31.687" Score="1" ViewCount="300" Body="&lt;p&gt;The wave function determines the probabilities of an outcome, and the wave function is parameterized &quot;shaped&quot; depending on the measurement set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. two different measurement set would lead to two different wave function description.&#xA;Let's call A measurement set=&gt;wave function A, and B measurement set=&gt;wave function B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having two entangled particles X and Y,  there is something I can't understand, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supose we make successive measurement with A experimental set&#xA;to successive generated entangled particles X :&#xA;X, X',X'', X''' ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So measurements should match with the expected probabilities outcomes predicted by the wave function for A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the entangled counterpart measured with B experimental set&#xA;to successive generated entangled particles Y :&#xA;Y, Y',Y'', Y''' ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So measurements should match with the expected probabilities outcomes predicted by the wave function for B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can the entangled particles satisfy both wave equation and at the same time, satisfy the entanglement condition? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it can't satisfy both, so decoherence of the entangled counterpart could be detected by a deviation of the outcomes and the expected probabilities.&#xA;If it can satisfy both, then there is no effect in being entangled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is the problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any help about understanding this&lt;/p&gt;&#xA;" OwnerUserId="1916" LastEditorUserId="1916" LastEditorDisplayName="" LastEditDate="2011-05-23T19:45:57.523" LastActivityDate="2011-06-01T14:33:28.857" Title="Can two entanglement particles satisfy at same time two different wave functions?" Tags="&lt;quantum-mechanics&gt;&lt;entanglement&gt;&lt;measurement&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="10296" PostTypeId="2" ParentId="10248" CreationDate="2011-05-23T20:20:46.937" Score="2" ViewCount="" Body="&lt;p&gt;A lot of what I begin with here will already be known by the asker, but just to give some background for my answer...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The action of the wind on the sails can be decomposed into two components: One which acts to move the boat forwards, and another which acts to push it sideways. The sideways force is countered by the keel/centerboard/daggerboard, which leaves the forward force to propel the boat. For this discussion I'll ignore the resulting torque which acts to heel the boat over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force of the water on the keel can be treated as if it acts laterally at one position along the length of the boat, and likewise for the lateral forces from each sail. These lateral forces each produce a torque that would tend to turn the boat, and they are not, in general, going to balance each other. Once the sails are trimmed for a given angle of attack to the wind, this residual torque is countered by the rudder. In this way the boat reaches a steady state where all torques are in equilibrium and it is propelled in the desired direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sail trim is then adjusted, the force from that sail will change in both magnitude and direction, both of which will alter the lateral component of force from that sail. The overall force from a sail is presumably in a direction close to (but I'd guess not exactly) perpendicular to the sail surface. Therefore, when the sail is pulled in close, more of its force is directed laterally. With all other trim being held constant, this change in torque would begin to turn the boat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the excess force is acting at the rear of the boat (i.e. the mainsail is pulled in), then the rear of the boat would be pushed upwind, which would obviously mean that the boat is turning in to the wind. Similarly, if the jib is pulled in, the extra lateral force would push the bow away from the wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As other answers have said, this likely violates your rule: &quot;The more the mainsail is sheeted out the more the boat tends to come up.&quot; because that rule is given in the context of a different rig. In a single-sail rig, for example, the center of lateral force from the mainsail may be well forward of the center of lateral force from the centerboard, while on your catamaran, the daggerboard appears to be even with the mast, which would put the center of lateral force from the mainsail well behind the daggerboard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good illustration of this effect is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Yawl&quot; rel=&quot;nofollow&quot;&gt;Yawl rig&lt;/a&gt;. The small mizzen doesn't provide much propulsion, but it's position at the extreme read of the boat allows it to provide fine adjustment of the torque on the boat. This is what allows the yawl to be trimmed to sail without a hand on the tiller.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-05-23T20:20:46.937" CommentCount="3" />
  <row Id="10297" PostTypeId="1" AcceptedAnswerId="10298" CreationDate="2011-05-23T20:58:25.370" Score="6" ViewCount="135" Body="&lt;p&gt;My secondary school physics textbook has mentioned that protons and neutrons are made up of down and up quarks in different amounts. It has also mentioned that other quarks exist. It states that particles from these quarks are unstable. It also lists the charges of these quarks. It doesn't provide much more detail beyond this. My question is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is (for example) a particle made up of ccs (Wikipedia tells me this is called a Double charmed Omega baryon) any less stable than a proton which has the same charge?&lt;/p&gt;&#xA;" OwnerUserId="3699" LastActivityDate="2011-05-23T21:49:04.090" Title="Why are higher generation of matter unstable?" Tags="&lt;particle-physics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="10298" PostTypeId="2" ParentId="10297" CreationDate="2011-05-23T21:33:36.317" Score="7" ViewCount="" Body="&lt;p&gt;Generally speaking, the lowest mass particles are stable against decay because they can't spontaneously decay into more massive particles (energy wouldn't be conserved in the rest frame). So particles not made up of up and down quarks (least massive of the quarks; see figure) will quickly decay into particles made up of only up and down quarks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/bWzur.png&quot; alt=&quot;Look at those quark masses... the up and down quarks are dwarfed by the others, hence they're more stable&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves the neutron (udd) and proton (uud) as the only stable particles of which the neutron is more massive and can hence decay into a proton (by conversion of a more massive down quark to an up quark through the weak interaction) - this is called Beta decay and is an important and well established radioactive process. There is nothing that the proton can spontaneously decay into while conserving energy in the Standard Model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond the Standard Model, most Grand Unified Theories that try to unify the three non-gravitational forces require proton decay since quarks and leptons can now change into each other. This means that the proton will have a finite (albeit very large) lifetime. Some GUTs, such as SU(5), have already been ruled out because the proton lifetimes they predict are smaller than the observed lifetime.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-05-23T21:49:04.090" LastActivityDate="2011-05-23T21:49:04.090" CommentCount="2" />
  <row Id="10299" PostTypeId="2" ParentId="10295" CreationDate="2011-05-23T21:35:58.337" Score="5" ViewCount="" Body="&lt;p&gt;Suppose the entangled particles are generated in a singlet state, $\frac{1}{\sqrt{2}} (\vert 01 \rangle - \vert 10 \rangle)$. If measurement set A is the standard basis, $\{ \vert 0 \rangle, \vert 1 \rangle \}$, then we would expect the result to be &quot;zero&quot; with probability $\tfrac{1}{2}$ and &quot;one&quot; with probability $\tfrac{1}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I've read your question correctly, it seems that your confusion comes from believing that the measurement results of entangled particles must always be identical. Barring experimental error, this is only the case when the measurement bases, A and B, are identical. If A and B were both the standard basis, then there would be perfect correlation between the A measurements and the B measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If on the other hand, B is a basis like $\{ \vert + \rangle, \vert - \rangle \}$ where $\vert + \rangle = \frac{1}{\sqrt{2}}(\vert 0 \rangle + \vert 1 \rangle)$, and $\vert - \rangle = \frac{1}{\sqrt{2}}(\vert 0 \rangle - \vert 1 \rangle)$, then we only expect a 50% correlation between the results of A measurements on X and B measurements on Y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This correlation is still quite good, and is the foundation of Bell-type inequalities, but it is not so good that either party learns anything about the other's choice of basis (or in your terminology: sees a deviation in expected probability based on the decoherence of their counterpart). Both parties always expect to see the &quot;zero&quot; and &quot;one&quot; states with 50% probability - the entanglement only affects the correlation between the remote parties results were they to compare them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; @HDE, With respect to the results of measurements in a fixed basis, non-entangled particles are much like weighted coin flips. So the experiment in which the parties have non-entangled particles is not much different from if we let two parties choose arbitrarily weighted coins and had them flip those coins to generate their results. Now, the probability that both parties get the same result is a function of which coin they each chose, if they each chose a coin that's weighted 90% towards heads, then they'll get the same result around 82% of the time. 81% of the time they'll both get heads, and 1% of the time they'll both get tails.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at entanglement, here the probability that both parties get the same result is given by $\cos^2({\theta})$ where $\theta$ is the &quot;angle&quot; between their bases. In my example with singlets above the angle was $\pi/4$ and so there was a 50% correlation between the parties' results. With an angle of $\pi/8$ we see an 85% correlation. This is the key difference, the correlation is governed by the similarity of the bases, not the frequency of particular outcomes. An outcome that occurs 10% of the time will yield a coincidence with probability $\cos^2({\theta})$, which will generally not be 10%. To emphasize this, let's think about the weighted coin flips. If Alice gets a tails on a 90% heads coin, then even in an optimistic case, where Bob chooses the same coin, she only expects Bob to get a tails 10% of the time. With entangled particles, when Alice sees a tails, even if it only occurs one in a million times, she expects Bob to see a tails with certainty assuming he chose the same basis, or with probability .85 if he chose one $\pi/8$ radians off, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps&lt;/p&gt;&#xA;" OwnerUserId="3632" LastEditorUserId="3632" LastEditorDisplayName="" LastEditDate="2011-05-27T23:36:22.043" LastActivityDate="2011-05-27T23:36:22.043" CommentCount="3" />
  <row Id="10300" PostTypeId="1" AcceptedAnswerId="10302" CreationDate="2011-05-23T21:55:36.917" Score="0" ViewCount="73" Body="&lt;p&gt;I have the following circuit. All the inverters are ideal CMOS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Zu0TZ.jpg&quot; alt=&quot;Circuit&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to draw the waveforms for each point (A, B, D, E, Vo), given that waveform at input.&lt;br&gt;&#xA;What I want to ask is if what I did so far is correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/XKBvr.jpg&quot; alt=&quot;Waveforms&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time. (And sorry for the poor quality of the images)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is this correct? (It is how I understood your response)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/80qC8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3459" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-24T11:16:59.840" LastActivityDate="2011-05-24T11:16:59.840" Title="Waveforms for a given ideal inverters circuit" Tags="&lt;homework&gt;&lt;electric-circuits&gt;&lt;electronics&gt;&lt;electrical-engineering&gt;&lt;resistors&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10301" PostTypeId="1" AcceptedAnswerId="10305" CreationDate="2011-05-23T21:59:39.477" Score="5" ViewCount="195" Body="&lt;p&gt;In Newton's model of light as being composed of particles, it is easy to imagine reflection as being the rebounding of individual corpuscles off of a surface. However, since light can also behave like a wave, it poses a challenge in visualizing reflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does a wave reflect off of a surface, whether it be specular reflection or diffuse reflection? Must the wave be first absorbed, and then re-emitted? Or is there a different mechanism? &lt;/p&gt;&#xA;" OwnerUserId="3395" LastActivityDate="2011-05-24T00:45:18.163" Title="How does reflection work?" Tags="&lt;electromagnetism&gt;&lt;light&gt;&lt;reflection&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" />
  <row Id="10302" PostTypeId="2" ParentId="10300" CreationDate="2011-05-23T23:07:57.527" Score="2" ViewCount="" Body="&lt;p&gt;I'm not sure why the resistor to ground from B is there, but you are incorrect at point D, the capacitor doesn't pass the DC level as you've indicated. It's a high-pass filter with C and R, so basically you need to move the DC-level on the Vd plot to ground - but keep the two transients like you've plotted them. That is, the curve should start at ground and dip at the down-transition of Vb then make a similar transient upswing after Vb goes up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question should be on the electronics SE instead..&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-05-23T23:07:57.527" CommentCount="5" />
  <row Id="10303" PostTypeId="2" ParentId="10301" CreationDate="2011-05-23T23:28:02.770" Score="1" ViewCount="" Body="&lt;p&gt;Actually, I consider reflection of a wave easier than refection of a particle: say we have a medium in which the wave can propagate easily, that is, its amplitude can vary freely, fullfilling some kind of wave equation. You can imagine it as a sequence of oscillators, where each of them always passes its energy to the next one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we now put a brickwall in the wave's way, we essentially just create a region where there are no, or much fewer, oscillators to take over the energy. So what does the wave do? It can't go on in the original direction, there's no way it could get rid of the energy. So the oscillators have no choice but to send the energy back through the medium.&lt;/p&gt;&#xA;" OwnerUserId="3540" LastActivityDate="2011-05-23T23:28:02.770" />
  <row Id="10304" PostTypeId="2" ParentId="10293" CreationDate="2011-05-23T23:40:39.073" Score="2" ViewCount="" Body="&lt;p&gt;Actually, that's exactly what happens: heat is transported by freely moving electrons. - but you wouldn't call it noise current, you would call it heat transfer, regardless of whether it's phonons or electrons which carry the heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole thing really has nothing to do with the resistors, it would work equally well with any passive component (proof: otherwise you could use that to build a 2nd order perpetuum mobile).&lt;/p&gt;&#xA;" OwnerUserId="3540" LastActivityDate="2011-05-23T23:40:39.073" CommentCount="3" />
  <row Id="10305" PostTypeId="2" ParentId="10301" CreationDate="2011-05-24T00:45:18.163" Score="4" ViewCount="" Body="&lt;p&gt;I'll just go ahead and write this down even though it's already been covered in the other thread.. but I didn't post there so :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, if you think of light as a wave (which is really a semi-classical way of thinking but might be enough to answer your question) you can invoke the Huygen-Fresnel principle which in this case boils down to considering every point on the reflecting surface as an origin of a re-emitted spherical wave with a starting phase directly related to the phase the point got from the incident wavefront.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The superposition of these wavefronts, after you let them destructively interfere with each other, will amount to a new combined wavefront which propagates according to Snells law (angle of incidence = angle of reflectance). See this image for the corresponding illustration of refraction (which is very similar, I couldn't quickly find a good image of reflection):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/X8LO9.png&quot; alt=&quot;Image of wavefront refraction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, light really doesn't behave &quot;sometimes as a particle, sometimes as a wave&quot;. It is always detected as quanta (particles) but the probability amplitudes (phases) propagate in a wavelike fashion. One way of expressing the propagation is to say that a photon is sort of split and takes every possible path between A and B (or, in the case of a reflector, from A to any point on the reflector and then from there to point B by any means). Every path gets a phase contribution, and all the indistinguishable paths are summed. Most paths are simply cancelling each other but some constructively interfere, creating a large contribution (in case you don't know QM, the probability amplitude squared is the probability of the described event so a large contribution means this result will most probably occur). There is a VERY good image and description of this process in Feynman - QED The strange theory of light and matter (as I wrote in the comment above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the reflector, the large contribution occurs at the classical angle of reflectance (Snell's law again). Notice the similarity between this formulation (called the path integral approach) and the semi-classical principe outlined above, this is not a coincidence of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also to briefly digress on your implied question on the per-atom non-zero &quot;reflectance&quot; time - saying that an electron-orbit absorbs the photon energy for a while the re-emits it a non-zero time later is of course also a slight simplification. In reality the electron interacts with the a photon, changes its momentum a bit, the re-emits (interacts) with the new photon and changes its momentum again. This scattering process occurs at all allowed momenta and intermediate times, which are then all superpositioned like above and hence I'm not sure it's meaningful to talk about any appreciable time of reflectance. Notice that this scattering is in practice very different from scattering which can excite the electron to another orbit.&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-05-24T00:45:18.163" CommentCount="1" />
  <row Id="10306" PostTypeId="2" ParentId="1566" CreationDate="2011-05-24T01:41:18.147" Score="2" ViewCount="" Body="&lt;p&gt;No experiments prove any theory. Experiments can only refute theories.&lt;/p&gt;&#xA;" OwnerUserId="3702" LastActivityDate="2011-05-24T01:41:18.147" CommunityOwnedDate="2011-05-24T01:41:18.147" CommentCount="5" />
  <row Id="10307" PostTypeId="2" ParentId="7652" CreationDate="2011-05-24T05:43:51.593" Score="0" ViewCount="" Body="&lt;p&gt;I feel quite sure that we will have technological breakthroughs but i think the most important findings will successfully link perception/awareness/consciousness with the so called external universe.&lt;/p&gt;&#xA;" OwnerUserId="3704" LastActivityDate="2011-05-24T05:43:51.593" CommentCount="1" />
  <row Id="10308" PostTypeId="1" AcceptedAnswerId="10310" CreationDate="2011-05-24T06:41:12.347" Score="7" ViewCount="352" Body="&lt;p&gt;This question is motivated by &lt;a href=&quot;http://xkcd.com/356/&quot; rel=&quot;nofollow&quot;&gt;this xkcd comic strip&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://xkcd.com/356/&quot; rel=&quot;nofollow&quot; title=&quot;I first saw this problem on the Google Labs Aptitude Test. A professor and I filled a blackboard without getting anywhere. Have fun.&quot;&gt;&lt;img src=&quot;http://imgs.xkcd.com/comics/nerd_sniping.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is indeed interesting, and my first recollection upon reading this was a similar problem in the book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/5030008004&quot; rel=&quot;nofollow&quot;&gt;Problems in General Physics&lt;/a&gt; by I.E.Irodov(which, in my humble opinion,is a masterpiece).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question I wanted to ask is, are there any &lt;strong&gt;generalizations&lt;/strong&gt; of the problem known? Can be find the resistance between &lt;em&gt;any&lt;/em&gt; two nodes of the grid as a function of the distance between the nodes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="3705" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-05-24T07:51:52.437" LastActivityDate="2011-05-24T07:51:52.437" Title="The &quot;Nerd Sniping&quot; problem. Generalizations?" Tags="&lt;electric-circuits&gt;&lt;resistors&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="4" />
  <row Id="10309" PostTypeId="1" CreationDate="2011-05-24T07:07:01.990" Score="5" ViewCount="435" Body="&lt;p&gt;Did the law of conservation of energy apply to the earliest moments of the Big Bang? If so, what theoretical physics supports this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hear that Einstein's theory of relativity disputes the law of conservation of energy- so does that mean the law is false, or only some aspect of it?&lt;/p&gt;&#xA;" OwnerUserId="3706" LastEditorUserId="2072" LastEditorDisplayName="" LastEditDate="2011-05-25T10:42:40.970" LastActivityDate="2011-05-25T23:26:16.357" Title="Conservation law of energy and Big Bang?" Tags="&lt;general-relativity&gt;&lt;energy-conservation&gt;&lt;big-bang&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="4" />
  <row Id="10310" PostTypeId="2" ParentId="10308" CreationDate="2011-05-24T07:09:57.917" Score="5" ViewCount="" Body="&lt;p&gt;Yes, it is possible.  For example Kevin Brown did &lt;a href=&quot;http://www.mathpages.com/home/kmath668/kmath668.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.mathpages.com/home/kmath669/kmath669.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; including this table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/aFnYj.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so for the xkcd problem the answer is $-\frac{1}{2}+\frac{4}{\pi} \approx 0.773$.&lt;/p&gt;&#xA;" OwnerUserId="2714" LastActivityDate="2011-05-24T07:09:57.917" CommentCount="3" />
  <row Id="10311" PostTypeId="1" AcceptedAnswerId="10314" CreationDate="2011-05-24T07:26:40.373" Score="3" ViewCount="127" Body="&lt;p&gt;I'm looking for a software or software package (for example C/C++) that can simulate a lot (say thousands at least) of molecules in action (ie. in movement or attached to say static walls).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found several but they seems to be either 'gas only', 'water only', or not really scientific (a lot of small balls dodging around moved by an unique formula). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that exist outside of secret laboratories :) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Valmond&lt;/p&gt;&#xA;" OwnerUserId="3707" LastActivityDate="2011-05-24T09:09:37.670" Title="Does it exist a free good molecule / atom simulation software?" Tags="&lt;simulation&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="10312" PostTypeId="2" ParentId="10309" CreationDate="2011-05-24T07:51:28.333" Score="7" ViewCount="" Body="&lt;p&gt;Yes, the energy conservation law fails not only right after the Big Bang but in any cosmological evolution. See e.g.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2010/08/why-and-how-energy-is-not-conserved-in.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/08/why-and-how-energy-is-not-conserved-in.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The time-translational invariance is broken, so via Noether's theorem, one doesn't expect a conserved quantity. Also, if one defines the &quot;total&quot; stress energy tensor as a variation of the action with respect to the metric tensor, it vanishes in GR because the metric tensor is dynamical and the variation has to vanish because it's an equation of motion (Einstein's equations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the space is asymptotically flat or AdS or similarly simple, a conservation law - for the ADM energy - may be revived.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-24T07:51:28.333" CommentCount="13" />
  <row Id="10313" PostTypeId="2" ParentId="10309" CreationDate="2011-05-24T08:22:07.023" Score="-5" ViewCount="" Body="&lt;p&gt;Certainly yes, and &lt;a href=&quot;http://en.wikipedia.org/wiki/General_relativity#Einstein.27s_equations&quot; rel=&quot;nofollow&quot;&gt;Einstein's theories&lt;/a&gt; support this.  Conservation of energy is a fundamental principle and it is true essentially by definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/History_of_energy#Conservation_of_energy&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's article on the history of energy&lt;/a&gt; includes this quote from Feynman:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;There is a fact, or if you wish, a law, governing natural phenomena that are known to date. There is no known exception to this law—it is exact so far we know. The law is called conservation of energy; it states that there is a certain quantity, which we call energy that does not change in manifold changes which nature undergoes. That is a most abstract idea, because it is a mathematical principle; it says that there is a numerical quantity, which does not change when something happens. It is not a description of a mechanism, or anything concrete; it is just a strange fact that we can calculate some number, and when we finish watching nature go through her tricks and calculate the number again, it is the same.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Physical_cosmology#Energy_of_the_cosmos&quot; rel=&quot;nofollow&quot;&gt;The Wikipedia article on physical cosmology&lt;/a&gt; has this to say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is no unambiguous way to define the total energy of the universe in the current best theory of gravity, general relativity. As a result it remains controversial whether one can meaningfully say that total energy is conserved in an expanding universe. For instance, each photon that travels through intergalactic space loses energy due to the redshift effect. This energy is not obviously transferred to any other system, so seems to be permanently lost. Nevertheless some cosmologists insist that energy is conserved in some sense.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I interpret this to mean that while in some cases it is not meaningful to say that energy is conserved, in these same cases it is no more meaningful to say that it isn't.  The issue is the well-definedness of energy, not whether or not it is conserved.&lt;/p&gt;&#xA;" OwnerUserId="2072" LastEditorUserId="2072" LastEditorDisplayName="" LastEditDate="2011-05-25T16:33:28.813" LastActivityDate="2011-05-25T16:33:28.813" CommentCount="12" />
  <row Id="10314" PostTypeId="2" ParentId="10311" CreationDate="2011-05-24T08:58:05.963" Score="4" ViewCount="" Body="&lt;p&gt;It most certainly exist outside secret labs :) Like Gerben wrote, the fields are called molecular dynamics (MD) and quantum chemistry which, as computers grow faster, will be essential tools of nanotechnology and medicine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecular Dynamics is currently implemented by making certain approximations in that electron motion is not explicitely modelled. In practice, empirical forcefields are matched to experimental data and molecules are essentially modelled by summing the forces on each atom and using $F=ma$, then integrating the acceleration over time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's far from perfect - you usually can't match all measurable physical properties of a medium like water at the same time with the same forcefields. You can't normally model shifts of covalent bonds either since that involves changing the molecular composition which breaks the forcefield definitions. On the other hand, the methods are relatively quick and you can easily simulate systems of hundreds of thousands of atoms over a timescale of at least nanoseconds (microseconds if you have access to a supercomputer :). To interpret results you need some understanding of the deficiencies of the algorithms though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some interesting large simulations to date was of the protein factory, the ribosome, probably the evolutionary oldest complex part of life, and of a &lt;a href=&quot;http://www.ks.uiuc.edu/Research/STMV/&quot; rel=&quot;nofollow&quot;&gt;complete virus, the Satellite Tobacco Mosaic Virus&lt;/a&gt;. Below is a snapshot of a simulation I did of ion-channels in a cell membrane, you see a drug molecule bound in white in the center:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/IBfGN.jpg&quot; alt=&quot;forest of ion channels&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think that the most used (free) simulators are &lt;a href=&quot;http://www.gromacs.org/&quot; rel=&quot;nofollow&quot;&gt;GROMACS&lt;/a&gt; and NAMD, with somewhat differing strengths but both are probably OK to start with. Usually paired with decent graphical tools to visualize what happens like PyMOL or VMD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not particularly difficult to learn about this by writing your own simple simulator as well, if you're so inclined. In either case, simulating a 2x2x2 nm box of argon atoms or water-molecules for example illustrates many of the concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum Chemistry involves getting one step lower into the motion and shifts of the electrons. Thus you can simulate the breaking and bonding of covalent bonds and you don't have to rely on empirical forcefields for every molecule/atom you need in your simulation. However it's a lot slower computationally obviously (a factor of 100 at least? I'm not sure). You might use this to model in detail the mechanism of an active site in an enzyme for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason these are important fields for nanotech and medicine is that the simulations are the equivalents of the act of compiling and testing a program in computer software design, when it comes to designing new drugs and nano-scale machinery or materials. Essentially you want methods to test your ideas without manufacturing them first, and computer simulations take advantage of Moore's law which says that computing power doubles every 18 months...&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-05-24T08:58:05.963" CommentCount="7" />
  <row Id="10315" PostTypeId="1" CreationDate="2011-05-24T08:59:29.287" Score="0" ViewCount="95" Body="&lt;p&gt;Consider image below. The weight of the fire-fighter is 840 N. &lt;strong&gt;What is the torque of the fire-fighter's weight about P and what is the value of the force C which cancels out the torque?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/xs1QM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2727" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-24T09:20:57.967" LastActivityDate="2011-05-24T16:38:27.983" ClosedDate="2011-05-24T20:14:43.320" Title="Moment calculation" Tags="&lt;homework&gt;&lt;newtonian-mechanics&gt;&lt;torque&gt;&lt;statics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10316" PostTypeId="2" ParentId="10311" CreationDate="2011-05-24T09:09:37.670" Score="2" ViewCount="" Body="&lt;p&gt;Simulating the time-evolution of thousands of molecules interacting is generally the domain of molecular dynamics. MD codes usually dramatically simplify calculation by modelling atoms classically, generally with predefined connectivity, heavy parameterisation and bonds modeled by harmonic potentials. Whilst such approaches can give decent results for things like protein conformations they completely neglect quantum mechanics and as such fail to describe chemical reactivity. If this is what you mean by 'not really scientific', well, unfortunately QM calculations are quite computationally expensive**.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even MM calculations get computationally expensive, depending on the time step size, the specific level of theory used and the size, number and degrees of freedom of the molecules you're modelling, but it is certainly possible to do useful MM calculations on desktop computers (this is the whole basis of Folding@Home, which aggregates small jobs done by idle desktop computers around the world).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two very well known FOSS MM codes are &lt;a href=&quot;http://www.gromacs.org/&quot; rel=&quot;nofollow&quot;&gt;GROMACS&lt;/a&gt; and &lt;a href=&quot;http://dasher.wustl.edu/tinker/&quot; rel=&quot;nofollow&quot;&gt;TINKER&lt;/a&gt;. GROMACS is now optionally GPU accelerated using CUDA, making it &lt;em&gt;very, very fast&lt;/em&gt; even on desktop computers (if you have a decent GPU).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** There exist hybrid QM/MM methodologies that combine an embedded QM region in a MM calculation - these are usually for looking at things like explicit solvent effects on molecules of interest. There are also &lt;em&gt;ab initio&lt;/em&gt; MD codes such as Car-Parinello molecular dynamics (CPMD) which go some way to bridging the QM/MM scale divide.&lt;/p&gt;&#xA;" OwnerUserId="1876" LastActivityDate="2011-05-24T09:09:37.670" />
  <row Id="10317" PostTypeId="1" AcceptedAnswerId="10344" CreationDate="2011-05-24T11:39:34.433" Score="1" ViewCount="101" Body="&lt;p&gt;This should be a trivial calculation but somehow I have managed to get myself confused about this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The grand partition function is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathcal Z = \sum_{N=1}^\infty \sum_{r(N)} {\text e}^{-\beta E_r +\beta N \mu}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the average number of particles:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left \langle N \right \rangle = \frac{1}{\mathcal Z} \sum_{N} \sum_{r(N)} N{\text e}^{-\beta E_r + \beta N \mu } = \frac{1}{\beta} \frac{\partial}{\partial \mu} \ln \mathcal Z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I take the derivative of that by the chemical potential? I thought about taking the inverse of the derivative of the chemical potential by the number of particles by I suspect that I'll be in the same position.&lt;/p&gt;&#xA;" OwnerUserId="3677" LastEditorUserId="3677" LastEditorDisplayName="" LastEditDate="2011-05-24T12:09:54.740" LastActivityDate="2011-05-24T23:31:48.223" Title="Calculating the derivative of the average number of particles by the chemical potential" Tags="&lt;statistical-mechanics&gt;&lt;chemical-potential&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10318" PostTypeId="2" ParentId="10140" CreationDate="2011-05-24T14:18:10.560" Score="1" ViewCount="" Body="&lt;p&gt;I see Lenny has done it again. I have to say I disagree with this paper. For the sake of argument, let's grant him the hat complementarity conjecture, which he has yet to justify or derive in any convincing way. By his own admission, complementarity scrambles states in the most efficient manner possible. Consequently, the pointer basis states for the many-worlds to split into do not align in any way with the basis states describing macroscopic multiverse bubbles. I rest my case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should also add the Bondi mass in the terminal supersymmetric bubble with zero cosmological constant, i.e. the &quot;hat&quot;, is small enough it probably can't accommodate all the holographic multiverses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As another poster before me had pointed out, insisting upon irreversible decoherence beyond the level of one in $e^{e^{10^{-123}}}$ is ridiculous. If by &quot;unhappen&quot;, Lenny means no recoherence at that probability, let me point out a lot of more drastic things will happen before then, like Boltzmann brains and the decay of our vacuum into a supersymmetric vacuum.&lt;/p&gt;&#xA;" OwnerUserId="3710" LastActivityDate="2011-05-24T14:18:10.560" />
  <row Id="10319" PostTypeId="2" ParentId="7441" CreationDate="2011-05-24T15:01:33.697" Score="1" ViewCount="" Body="&lt;p&gt;Rather than using material, perhaps magnetic fields configured in stages. Imagine a stack of plates separated at a distance on the order of a meter. Magnetic fields, from superconducting magnetics repeal the plates above or below. Sensors and an electronic system dynamically adjust the fields. I wonder if the fields would need to become attractive at some point due to centifigual forces from the earths rotation. &lt;/p&gt;&#xA;" OwnerUserId="3711" LastActivityDate="2011-05-24T15:01:33.697" CommentCount="1" />
  <row Id="10320" PostTypeId="1" CreationDate="2011-05-24T15:14:13.983" Score="4" ViewCount="160" Body="&lt;p&gt;There are several videos of the reaction, where some amount of burning &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermite&quot; rel=&quot;nofollow&quot;&gt;thermite&lt;/a&gt; explodes on a contact with ice.&lt;br&gt;&#xA;An &quot;original&quot; video: &lt;a href=&quot;http://www.youtube.com/watch?v=IuPjlYxUWc8&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=IuPjlYxUWc8&lt;/a&gt;&lt;br&gt;&#xA;A Mythbusters confirmation: &lt;a href=&quot;http://www.youtube.com/watch?v=BnHR4cMXiyM&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=BnHR4cMXiyM&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several speculative explanations of the phenomenon, summarised in the end of the latter video. And none of them sounds to me satisfactory enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any detailed investigation and/or explanation of the effect?  &lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-05-24T16:34:44.597" LastActivityDate="2011-05-24T19:26:58.930" Title="Explanation of &quot;thermite vs ice&quot; explosion" Tags="&lt;thermodynamics&gt;&lt;experimental-physics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10321" PostTypeId="2" ParentId="10140" CreationDate="2011-05-24T15:58:16.617" Score="0" ViewCount="" Body="&lt;p&gt;I will not say yeah or nay on this.  Besides I have only read part of the paper.  A lot of negativity is being thrown at this, which frankly I can understand.  I am rather guarded on this.  The one problem I have is that the MWI is an interpretation of quantum mechanics, and is not what I would call a theory.  There is no way of testing it.  There does not appear to be any way of falsifying quantum interpretations, so hanging a theory of quantum cosmology on that seems problematic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are four levels of so called multiverse.  If all of these are real, using the word real in a difference sense than reality is used in QM issues of Bell or GHZ states etc, these are the Linde pocket universes, brane-world or quantum fluctuation universes, MWI universes and the rather speculative ideas of an ultimate ensemble by Tegmark.  Lumped all in one pile we might suspect these all fit into a single scheme.  So maybe how branes and strings are wrapped and give rise to a vacuum in a universe is decoherence and this fits into MWI.  However, if MWI is the “bird’s eye” perspective on the universe, it seems difficult to say the splitting off of world due to an electron double slit experiment is the same as bubble nucleation of the vacuum or brane-brane interactions and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be honest some of the post-selected state work in QM to my mind brings question to the whole MWI conjecture.  If post selected states determine an entanglement in the past, then this gives me some pause on the whole worlds splitting off idea of MWI with decoherence or measurement.  Further MWI, and frankly with all quantum interpretations, is just a way to trying to make some aspects of quantum mechanics fit into our classical or macroscopic perceptual bias of the world.  Quantum mechanics is just plain weird in some ways and it butts up against our intuitive ideas about the world.  So we have come up with various ways of trying to transduce this strangeness into our biases.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-05-24T15:58:16.617" />
  <row Id="10322" PostTypeId="1" AcceptedAnswerId="10324" CreationDate="2011-05-24T16:13:58.807" Score="1" ViewCount="105" Body="&lt;p&gt;I've never taken a quantum mechanics class, but I find myself now using principles developed in the quantum theory of angular momentum. One particularly confusing aspect that I'm struggling with is the notation.  Could someone please explain the meaning of a ket (or bra) with two arguments, i.e.: $\left|j_1m_1\right&amp;gt;$? I'm mainly interested in the mathematical meaning (i.e. is it a vector with two components?, are $j_1$ and $m_1$ multiplied), but a physical explanation would be helpful too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen other similar notation as well which perplexes me. For instance, in the definition of the Clebsch-Gordan coefficients: $C_{j_1m_1j_2m_2}^{jm}=\left&amp;lt;j_1m_1j_2m_2\mid j_1j_2jm\right&amp;gt;=\left&amp;lt;j_1j_2jm\mid j_1m_1j_2m_2\right&amp;gt;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what does a bra (or ket) with four arguments written adjacent to each other mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly I've seen $\left|S_x;+\right&amp;gt;$ which, as far as I can tell indicates that the spin component in the x-direction is $+\frac{1}{2}$, but could this be equivalently written simply as $\left|+\right&amp;gt;$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please any help would be very illuminating, every text and resource seems to use similar but slightly different notation and they all seem to assume you already know the conventions.&lt;/p&gt;&#xA;" OwnerUserId="3597" LastActivityDate="2011-05-24T16:28:13.323" Title="state vector notation" Tags="&lt;quantum-mechanics&gt;&lt;notation&gt;" AnswerCount="1" />
  <row Id="10323" PostTypeId="2" ParentId="10315" CreationDate="2011-05-24T16:25:07.253" Score="-1" ViewCount="" Body="&lt;p&gt;The answer to both your questions are hinted at or can be worked out from the equation: (Horizontal_Force_C) x 12m = (FF_weight) x 5m.  You get this equation by summing up the moments around P and setting that sum = 0.  You could pick any number of points about which to sum these moments, but point 'P' is the obvious choice for ease of setting up the problem and performing calculations.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-05-24T16:38:27.983" LastActivityDate="2011-05-24T16:38:27.983" />
  <row Id="10324" PostTypeId="2" ParentId="10322" CreationDate="2011-05-24T16:28:13.323" Score="2" ViewCount="" Body="&lt;p&gt;It is just a label. Each element of the set $\left|j_1m_1\right&amp;gt;$ is a unique state vector. For example, $\left|1,1\right&amp;gt;$ and $\left|1,-1\right&amp;gt;$ are two independent angular momentum state vectors, just like any other state vector. Whenever you see the indices $j$, $m$, etc., see them as referring to a set of state vectors each labelled with particular values and put some numbers into the $j$s and $m$s in an equation to get an idea of what one of the equations belonging to that set is like.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-05-24T16:28:13.323" CommentCount="9" />
  <row Id="10325" PostTypeId="1" AcceptedAnswerId="10353" CreationDate="2011-05-24T16:31:53.120" Score="16" ViewCount="567" Body="&lt;p&gt;Need to brush up on my late-undergrad and early-grad physics and was wondering if anyone can recommend books or lecture notes (hard copy, or on-line) that also have solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two that I have come across are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Princeton Problems in Physics with Solutions&lt;/strong&gt; - Nathan Newbury&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;University of Chicago Graduate Problems in Physics with Solutions&lt;/strong&gt; - Jeremiah A. Cronin&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Spacetime Physics&lt;/strong&gt; - Taylor &amp;amp; Wheeler (favorite book on special relativity; has a lot of problems with solutions at the back; a lot of the problems really enforce the material and discuss paradoxes)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible, please also provide a reason why you like the books as opposed to just listing them.&lt;/p&gt;&#xA;" OwnerUserId="3270" LastEditorUserId="3270" LastEditorDisplayName="" LastEditDate="2011-05-25T03:19:07.707" LastActivityDate="2011-08-16T10:35:01.577" CommunityOwnedDate="2011-05-25T04:01:29.620" Title="Graduate Physics Problems Books" Tags="&lt;soft-question&gt;&lt;education&gt;&lt;books&gt;&lt;reference-request&gt;&lt;big-list&gt;" AnswerCount="9" CommentCount="2" FavoriteCount="9" />
  <row Id="10326" PostTypeId="2" ParentId="10325" CreationDate="2011-05-24T16:49:50.450" Score="1" ViewCount="" Body="&lt;p&gt;Some review/problem set books that I like are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solid State Physics: problems and solutions - Laszalo Mihaly and Michael C. Martin&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problems in Quantum Mechanics with Solutions - Gordon Leslie Squires (The reviews are bare bones but I find the questions to be very good at making you think)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps. I am interested to see what other people come up with. &lt;/p&gt;&#xA;" OwnerUserId="3601" LastActivityDate="2011-05-24T16:49:50.450" CommunityOwnedDate="2011-05-25T04:01:29.620" />
  <row Id="10327" PostTypeId="2" ParentId="10320" CreationDate="2011-05-24T17:13:58.740" Score="2" ViewCount="" Body="&lt;p&gt;This is probably a question for the proposed Chemistry Stack Exchange. Here is one explanation given by Jamie Hyneman of MythBusters. The MythBusters video showed: Thermite place on top of ice → Ignited → Exothermic reaction begins → Big Explosion. Once the thermite has been ignited, a highly exothermic reaction follows it. This exothermic reaction releases a huge amount of energy so fast that the ice becomes &lt;a href=&quot;http://en.wikipedia.org/wiki/Superheating&quot; rel=&quot;nofollow&quot;&gt;superheated&lt;/a&gt; and turns into vapor. This vapor condenses around the fine thermite particles forming a steam-thermite &lt;a href=&quot;http://en.wikipedia.org/wiki/Aerosol&quot; rel=&quot;nofollow&quot;&gt;aerosol&lt;/a&gt;.The thermite mixture continues to react further, resulting in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Steam_explosion&quot; rel=&quot;nofollow&quot;&gt;steam explosion&lt;/a&gt; throwing particles around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another explanation is that the high temperature produced during the thermite reaction results in the following reaction:&#xA;$$3Fe(s) + 4H_{2}O(g) → Fe_{3}O_{4}(s) + 4H_{2}(g)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydrogen released will explode exothermically in air:&#xA;$$H_2 + \frac{1}{2}O_2 → H_{2}O$$&lt;/p&gt;&#xA;" OwnerUserId="966" LastEditorUserId="966" LastEditorDisplayName="" LastEditDate="2011-05-24T19:26:58.930" LastActivityDate="2011-05-24T19:26:58.930" CommentCount="1" />
  <row Id="10328" PostTypeId="2" ParentId="10325" CreationDate="2011-05-24T17:27:22.537" Score="1" ViewCount="" Body="&lt;p&gt;&quot;Problem Book on Relativity and Gravitation&quot; - A. Lightman, W. Press, R. Price, S. Teukolsky&lt;/p&gt;&#xA;" OwnerUserId="1382" LastActivityDate="2011-05-24T17:27:22.537" CommunityOwnedDate="2011-05-25T04:01:29.620" />
  <row Id="10329" PostTypeId="1" AcceptedAnswerId="10331" CreationDate="2011-05-24T17:27:27.943" Score="1" ViewCount="159" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;A four-dimensional differentiable (Hausdorff and paracompact) manifold $M$ will be called a space time if it possesses a pseudo-Riemannian metric of hyperbolic normal signature $(+,-,-,-)$ and a time orientation. There will be no real loss of generality in physical applications if we assume that $M$ and its metric are both $\mathcal{C}^{\infty}$&#xA;  .&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The above is an excerpt is taken from this &lt;a href=&quot;http://rspa.royalsocietypublishing.org/content/314/1519/529.full.pdf+html&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt;. I'd like to know how the assumption that &#xA; $M$ and its metric are both $\mathcal{C}^{\infty}$ be made with out any real loss of generality in physical applications. Any intuitive answer is also appreciated.&lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="540" LastEditorDisplayName="" LastEditDate="2011-05-24T18:20:39.327" LastActivityDate="2011-05-24T18:20:39.327" Title="A question on an assumption of space-time" Tags="&lt;general-relativity&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10330" PostTypeId="1" CreationDate="2011-05-24T18:04:18.283" Score="1" ViewCount="41" Body="&lt;p&gt;I have a quick question from thermodynamics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember that we take kT/2 as the kinetic energy per degree of freedom in kinetic theory of gases. But when we do langevin forces (for example in cavity dynamics), do we take the average thermal energy as kT/2 or kT ?&lt;/p&gt;&#xA;" OwnerUserId="3714" LastActivityDate="2011-05-24T18:04:18.283" Title="thermal energy while calculating Langevin Forces" Tags="&lt;quantum-mechanics&gt;&lt;thermodynamics&gt;&lt;quantum-electrodynamics&gt;" />
  <row Id="10331" PostTypeId="2" ParentId="10329" CreationDate="2011-05-24T18:06:44.197" Score="4" ViewCount="" Body="&lt;p&gt;Dear Rajesh, in reality, physics sometimes works with continuous functions that are not infinitely differentiable - for example look at the energy of the beam at atlas.ch (click at the &quot;Status&quot; button in the middle) when they ramp it up - there are all kinds of discontinuities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But an arbitrary function that is smooth almost everywhere - and this is a description of functions that really covers everything that a physicist would use - may be approximated by infinitely differentiable functions with an arbitrary accuracy. So it doesn't really hurt if your theorems assume that all the spacetime fields including the metric tensor are infinitely differentiable; you may solve the situations involving functions that are not infinitely differentiable by taking a limit of the infinitely differentiable ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May we ask a physics question whether the fields in electromagnetism are infinitely differentiable? Well, we may but it is a meaningless question because the real world is &lt;em&gt;not&lt;/em&gt; described by classical physics. So classical physics itself is just an approximation, so both &quot;classical physics with all differentiable functions&quot; and &quot;classical physics with infinitely differentiable functions only&quot; are just approximations of the reality, with none of them being more &quot;physically real&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum physics, we don't use classical fields but we may use wave functions. The time evolution is dictated by Schrödinger's equations - but may we ask whether $\psi(x,y,z)$ should be an infinitely differentiable function of $x,y,z$? Well, in this case, we usually don't make this assumption. Instead, we use all $L^2$ functions which are much more natural at the level of the Hilbert space, Fourier transforms, and so on. The $L^2$ condition surely doesn't require infinite differentiability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, a finite expectation value of the kinetic energy does force $\psi(x,y,z)$ to be a continuous function. Nevertheless, the infinite differentiability condition isn't ever natural in quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to emphasize that all these extra conditions - whether something should be smooth or infinitely smooth etc. - are only a matter of mathematical taste. There can't exist an operational &quot;physics&quot; way to determine whether the world allows functions that are not infinitely differentiable. It's because the infinitely smooth and not-infinitely smooth functions can mimic each other with an arbitrary accuracy but the accuracy of any measurement in physics is always limited. So the restrictions on the &quot;nice behavior&quot; of the mathematical functions is always a matter of mathematical taste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exotic differentiable structures could be counterexamples to what I just wrote - because they're &quot;qualitatively&quot; different and their pathological behavior is the very point of their existence (so in some important sense, the ability of them to mimic the normal functions disappears) - but their role in physics remains very confusing and limited as of today. &lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-24T18:06:44.197" CommentCount="0" />
  <row Id="10332" PostTypeId="1" AcceptedAnswerId="10334" CreationDate="2011-05-24T18:19:35.733" Score="3" ViewCount="225" Body="&lt;p&gt;I was driving down the road at roughly the speed of traffic.  I saw a police officer parked on the side of the road, and also noticed that a Semi was traveling in the lane right next to him.  This got me thinking, is it possible to avoid being seen from the officer by using the Semi Truck and how fast would I have to go?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say the officer is parked on the side of the road and the semi is traveling on the lane next to the officer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/7EWmd.png&quot; alt=&quot;My little Paint Drawing&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's also assume that the truck is traveling at roughly constant 65 mph.  Also assume that I see the officer and line up my car to block the officer's view via the semi at roughly a mile away. (Ya I know a little far).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to know:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;How fast do I have to go to avoid being seen by the officer?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is this speed constant or variable?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: I am in NO way advocating speeding on the highways and breaking the law.  This was just was just sparked as it somewhat happened while I was driving down the road.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2011-05-25T20:58:55.940" Title="How fast do I need to go in order to avoid being seen by the Police?" Tags="&lt;mechanics&gt;&lt;speed&gt;&lt;visualization&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="10333" PostTypeId="2" ParentId="10332" CreationDate="2011-05-24T18:37:25.063" Score="1" ViewCount="" Body="&lt;p&gt;If the three lanes have the same width, the distance cop-semi is the same as &#xA;semi-me. So You have to go at the double speed of the semi, constantly. &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Intercept_theorem&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Here&lt;/strong&gt;&lt;/a&gt; You find Reason for that. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-05-24T18:37:25.063" />
  <row Id="10334" PostTypeId="2" ParentId="10332" CreationDate="2011-05-24T18:56:49.453" Score="10" ViewCount="" Body="&lt;p&gt;What you want to do is keep the angle between your direction of motion and the line of sight to police car the same as the angle between the truck's direction of motion and the truck's line of sight to the police car.&lt;img src=&quot;http://i.stack.imgur.com/5e1Fo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, we want to keep $a1=a2$ in the picture above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a problem in similar triangles. The answer will be that the ratio of your motion to your distance to the cop will have to be the same as the ratio of the truck's motion to the truck's distance to the cop. This can be seen from the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We note that $$\cot{(a1)}=\frac{\textrm{Truck&amp;#39;s Speed}}{d1}$$&#xA;$$\cot{(a2)}=\frac{\textrm{Car&amp;#39;s Speed}}{d2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $a1=a2$ so we are always hidden behind the truck and solve for $\textrm{Car&amp;#39;s Speed}$:&#xA;$$a1=a2$$&#xA;$$\cot{(a1)}=\cot{(a2)}$$&#xA;$$\frac{\textrm{Truck&amp;#39;s Speed}}{d1}=\frac{\textrm{Car&amp;#39;s Speed}}{d2}$$&#xA;$$\textrm{Car&amp;#39;s Speed}=\frac{\textrm{Truck&amp;#39;s Speed}\times d2}{d1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, suppose the lanes are the same size (10 feet wide, say), and the cop is 5 feet off of the highway. Then, $d1=5\textrm{ feet}$, $d2=15\textrm{ feet}$. The speed you need if the truck is going 65 miles per hour is $$\textrm{Car&amp;#39;s Speed}=\frac{65 \textrm{mph}\times 15}{5}=195\textrm{ mph}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Some concerns were raised in the comments that this treats the truck as a point. This turns out not to matter. Here's a second picture like the first, but now we have a zone (colored in green) which the truck covers. &lt;img src=&quot;http://i.stack.imgur.com/1zkG1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The green triangle gives you a little bit of wiggle room, since you can be covered by the front of the truck or the back or anything in between. However, the total size of your wiggle room does not change while you move (in other words, it doesn't depend on a4). As a result, it should be pretty clear that this doesn't change things much at all - we can think of it as two point-size trucks going at the same speed, and we have to stay between them. Of course, this will give exactly the same answer as the first case - it's really just like hiding behind one point-sized truck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is actually one small change, as David notes: If you start out covered by the front of the truck, you can go a &lt;em&gt;little&lt;/em&gt; slower than the 195 mph cited above, because you can slowly slide back until covered by the back of the truck. However, if the length of the truck is $L_{tr}=40\textrm{ feet}$ (say), then this change in the velocity is quite small. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose that we slide back 40 feet from the front of the truck to the back over the course of a mile. We're going 195 miles per hour, so it takes us 18.5 seconds to go one mile. In those 18.5 seconds, we move 40 feet relative to the truck; this is a speed of about 1.5 mph. So, we can go 1.5 mph slower if we start at the front and go to the back over a mile; taking this into consideration, we get that the speed needed is actually &lt;strong&gt;193.5 mph&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1655" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-05-25T20:58:55.940" LastActivityDate="2011-05-25T20:58:55.940" CommentCount="4" />
  <row Id="10335" PostTypeId="2" ParentId="10332" CreationDate="2011-05-24T18:57:57.330" Score="0" ViewCount="" Body="&lt;p&gt;I would rather the cop be drawn on the left.  Cop is the origin.  Then both lanes intersect with the x-axis, at point $a$ for the truck's lane and $b$ for your lane and $a&amp;lt;b$.  Now, some line of $y=m x$ intersects all 3 participants.  The truck is moving with motion that can be written as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_{truck}(t) = m(t) a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.  We know &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d y_{truck}}{dt}=65mph$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;, so we know (differentiate both sides) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d m}{dt} = \frac{65 mph}{a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;, and it's easy to then see that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d y_{you}}{dt} = 65 mph \frac{b}{a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;done.  Yes it is constant and is larger by the most obvious fraction you could think of.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-05-24T18:57:57.330" />
  <row Id="10336" PostTypeId="2" ParentId="10332" CreationDate="2011-05-24T19:03:03.740" Score="0" ViewCount="" Body="&lt;p&gt;It reduces down to triangle similarity.!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i.stack.imgur.com/l6LZY.jpg&quot; rel=&quot;nofollow&quot;&gt;illustration of triangles&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X$ and $x$ are functions of time and $a$ and $b$ aren't. So we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{X}{x}=\frac{a}{b}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X=\frac{a}{b}x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deriving in relation to time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v=\frac{b}{a}V$&lt;/p&gt;&#xA;" OwnerUserId="301" LastActivityDate="2011-05-24T19:03:03.740" />
  <row Id="10338" PostTypeId="2" ParentId="10325" CreationDate="2011-05-24T19:10:54.677" Score="0" ViewCount="" Body="&lt;p&gt;Problems and Solutions on Electromagnetism (Major American Universities Ph.D. by Yung-Kuo Lim&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problems and Solutions on Mechanics: Major American Universities Ph.D. by Yung-Kuo Lim&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problems and Solutions on Thermodynamics (Major American Universities Ph.D. by U.S.T. of China Physics Coaching Class and Yung-Kuo Lim&lt;/p&gt;&#xA;" OwnerUserId="3509" LastActivityDate="2011-05-24T19:10:54.677" CommunityOwnedDate="2011-05-25T04:01:29.620" />
  <row Id="10339" PostTypeId="2" ParentId="10320" CreationDate="2011-05-24T19:12:01.017" Score="1" ViewCount="" Body="&lt;p&gt;1st - that idea of water split in oxygen/hydrogen or reduced to hydrogen &#xA;by iron or aluminium is not an explanation! This would happen at temperatures well &#xA;above 1000°C, this means those gases would burn as they would form. For an explosion You need a volume of gas/air mixed and prepared, then ignited. Here ignition is always &#xA;present, from the very first microsecond, therefore an explosion cannot be result &#xA;from that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd - &quot;steam-thermite aerosol&quot; . This theory lacks all reality of thermite reaction. &#xA;The reaction is between molten aluminium and molten iron oxide, An obstacle to the reaction &#xA;is formation of aluminium oxide (melting point above 2000C). It is important, to keep &#xA;the reaction heat in the crucible, to make the aluminium oxide staying as a melt. This results in &#xA;separation of iron at the bottom. aluminium oxide on top, whereas the unreacted mixture &#xA;of solid aluminium and iron oxide particles in the middle &quot;goes on&quot;.  If You blow the mixture, aluminium and iron oxide particles are separated, the thermit reaction will &#xA;cease. Molten aluminium would burn vigorously when contacted to air, but that is not convincing to me as an explanation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3rd I'd try this experiment with real thermite mixture (the mixture in both examples was  extremely fine, the iron oxide was pigment quality) made from granules about some millimeters. Another experiment I'd try was placing the thermit bucket just on top of a water surface (instead of ice). Then one could exclude some &quot;Reasons&quot; maybe. &#xA;And last, but not least I would place a professional thermit crucible (from railroad welding) above (some 30 cm)  a bucket full of water, let it react, and then open the &#xA;bottom valve. If my idea is correct, this experiment would lead to a similar explosion, &#xA;caused by melted iron at 2000° splashing into water. If You ever saw a drop of water on a red hot oven plate, You know that I think of Leidenfrosts phemomen. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-05-24T19:12:01.017" CommentCount="1" />
  <row Id="10340" PostTypeId="2" ParentId="10325" CreationDate="2011-05-24T19:56:14.717" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0852745133&quot; rel=&quot;nofollow&quot;&gt;Thinking Like a Physicist: Physics Problems for Undergrads&lt;/a&gt;:  I love this book because it fosters a real sense of physical understanding, so it's not just mathematics, but actual physical reasoning.  Plus, I found the problems challenging and interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, there is always the MIT &lt;a href=&quot;http://ocw.mit.edu/courses/physics/&quot; rel=&quot;nofollow&quot;&gt;Open Course Ware in Phsyics&lt;/a&gt;, which has undergrad and graduate courses with assignments, lecture notes, tests, problems, solutions, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If its qualifying exam questions and problems that you're after, a lot of universities will post examples of past versions online, its just a matter of looking at the department website hard enough. &lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2011-05-24T19:56:14.717" CommunityOwnedDate="2011-05-25T04:01:29.620" CommentCount="3" />
  <row Id="10341" PostTypeId="2" ParentId="10325" CreationDate="2011-05-24T20:38:24.510" Score="1" ViewCount="" Body="&lt;p&gt;The two you have were my favorites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For completeness, a couple of others, which also have worked solutions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;L.A. Sena, A Collection of Questions and Problems in Physics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Constantinescu, Problems in Quantum Mechanics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many problems and solutions sets available for Caltech's comprehensive &quot;Applications of Classical Physics&quot; course here:&#xA;&lt;a href=&quot;http://www.pma.caltech.edu/Courses/ph136/yr2008/&quot; rel=&quot;nofollow&quot;&gt;http://www.pma.caltech.edu/Courses/ph136/yr2008/&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-05-24T20:38:24.510" CommunityOwnedDate="2011-05-25T04:01:29.620" />
  <row Id="10342" PostTypeId="2" ParentId="10204" CreationDate="2011-05-24T20:40:54.273" Score="1" ViewCount="" Body="&lt;p&gt;Thanks for your solutions they have confirmed my initial thoughts.&#xA;A 5MHz pulse of 3 or 4 cycles will have a bandwidth of 1 or 2 MHz but it is the profile of the spectrum that I need.&#xA;I also  need a windows utility to digitise the pulse, so I may try Getdata or Dagra either can produce a file for input to excel but the latter can produce a file for MATLAB.&lt;/p&gt;&#xA;" OwnerUserId="3716" LastActivityDate="2011-05-24T20:40:54.273" CommentCount="3" />
  <row Id="10343" PostTypeId="2" ParentId="10184" CreationDate="2011-05-24T21:28:09.480" Score="7" ViewCount="" Body="&lt;p&gt;Okay, so I did some poking around and the 66th-75th editions of the CRC Handbook of Chemistry and Physics all have the incorrect atomic mass of Cu-63 [62.939598], and from 76th edition on they seem to have figured it out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those isotope mass tables are put together from a number of sources, so it's hard (time consuming) to tell exactly where the error came from. I did notice however that starting in the 76th edition of the CRC, where they get it right, they start citing G. Audi and A.H. Wapstra, &quot;The 1993 atomic mass evaluation&quot;, &lt;em&gt;Nuc. Phys.&lt;/em&gt; A &lt;strong&gt;565&lt;/strong&gt; (1993) 1-65. In editions 66-75, they were citing Audi and Wapstra's &quot;The 1983 atomic mass table&quot; which appeared in &lt;em&gt;Nuc. Phys&lt;/em&gt; &lt;strong&gt;432&lt;/strong&gt;, 1 (1985). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I looked at the 1993 version, and it has the correct 62.929.. mass, but I have not been able to find Wapstra and Audi's 1983 version of the same table, so I don't know if it was one error by the CRC which got carried over year after year, or if 62.939.. is in fact the value given in that paper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did find at least one piece of evidence which points to an error by the editors of the CRC in the 2nd edition of the Encyclopedia of Physics [Lerner and Trigg, 1991]. Their table of isotopes lists the correct 62.929... value and also cites Wapstra and Audi's 1983 paper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that satisfies everyone's curiosity, because I don't think I can do any better then that. ;-)&lt;/p&gt;&#xA;" OwnerUserId="3601" LastActivityDate="2011-05-24T21:28:09.480" />
  <row Id="10344" PostTypeId="2" ParentId="10317" CreationDate="2011-05-24T21:41:31.467" Score="0" ViewCount="" Body="&lt;p&gt;We want to find $$\frac{\partial\langle N\rangle}{\partial\mu}=\frac{1}{\beta} \frac{\partial^2}{\partial \mu^2} \ln \mathcal Z=(\frac{\mathcal Z&amp;#39;}{\beta \mathcal Z})&amp;#39;=(\frac{-\mathcal Z&amp;#39;^2}{\beta \mathcal Z^2}+\frac{\mathcal Z&amp;#39;&amp;#39;}{\beta \mathcal Z})$$ where the prime denotes differentiation w.r.t. $\mu$. In order for this to be zero we must have $$\mathcal Z\mathcal Z&amp;#39;&amp;#39;=\mathcal Z&amp;#39;^2$$ which written out in full is $$\sum_{NN&amp;#39;rr&amp;#39;}{\text e}^{-\beta (E_r+E_{r&amp;#39;}) +\beta (N+N&amp;#39;) \mu}\beta^2N^2=\sum_{NN&amp;#39;rr&amp;#39;}{\text e}^{-\beta (E_r+E_{r&amp;#39;}) +\beta (N+N&amp;#39;) \mu}\beta^2NN&amp;#39;$$ where the summation on both sides is now over $N&amp;#39;$ and $r&amp;#39;$ as well. This requires $N=N&amp;#39;$, which clearly doesn't hold always - this is the same as saying that the square of a sum is not equal to the sum of squares. Therefore, $\frac{\partial\langle N\rangle}{\partial\mu}$ is not zero at finite temperature. You would expect this since $\frac{\partial\langle N\rangle}{\partial\mu}$ is proportional to the variance of the number of particles $\langle (\delta N)^2 \rangle$.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-05-24T23:31:48.223" LastActivityDate="2011-05-24T23:31:48.223" CommentCount="1" />
  <row Id="10345" PostTypeId="2" ParentId="9069" CreationDate="2011-05-24T23:15:00.130" Score="2" ViewCount="" Body="&lt;p&gt;This was the first question I posted on StackExchange and nobody answered it so I will have to try. Looking at the ideal case, there are two plausible answers: 1200 or 600. Assume that the source produces only entangled photons and the detectors are 100% efficient. Without polarizers, every detection is a coincidence. Given that there are 300 coincidences per second with two aligned polarizers,  how many are there with no polarizers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hypothesis 1: there are 1200 coincidences per second. Put a polarizer on one side and only half the photons get through. The number of coincidences goes down to 600. Put a polarizer on the other side: half the photons get through and the number of coincidences goes down to 300, as observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hypothesis 2: there are 600 coincidences per second. Put polarizers on both sides. Since the polarizers are aligned, they knock out photons always in pairs. Since they each knock out half the photons, there are 300 coincidences per second at the detectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot easily choose between these two hypotheses. &lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-05-24T23:15:00.130" />
  <row Id="10346" PostTypeId="1" AcceptedAnswerId="10348" CreationDate="2011-05-24T23:26:36.940" Score="0" ViewCount="181" Body="&lt;p&gt;I need to spec a motor to turn a mixing barrel. The barrel contains loose earth and can be filled to a maximum of 50% of its interior volume. It is a horizontal cylinder, and will rotate through its central axis. What effort/torque is required to start the barrel turning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my case, the mass of the barrel when filled to its maximum is 50 kg, its length is 1.2m, diameter is 700mm. Friction in the bearings and the drive is negligible.&lt;/p&gt;&#xA;" OwnerUserId="3717" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-25T01:16:57.840" LastActivityDate="2011-05-25T18:15:43.860" Title="Torque required to turn a drum/barrel" Tags="&lt;newtonian-mechanics&gt;&lt;torque&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="10347" PostTypeId="2" ParentId="10346" CreationDate="2011-05-25T00:05:19.433" Score="1" ViewCount="" Body="&lt;p&gt;Theoretically, zero torque is required to start the barrel turning as long as it starts out in equilibrium.  To keep it turning at a constant speed, the required torque varies periodically with a time-average of zero.  This assumes that there is no friction in the barrel and that the dirt doesn't roll around inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Realistically, it depends on a number of factors, including the friction in the bearings and other sources of friction, the size of the barrel, the speed at which it rotates, and the extent to which the dirt tumbles down on itself inside the barrel.  If the  barrel has a large radius and small thickness, more torque is required.  If the barrel has low friction, less torque is required.  If the dirt all tumbles inside, the effect is complicated, probably increasing the average power input required but decreasing the peak torque.  There is simply a lot of information missing, without which the torque can't be calculated.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible estimate is to find the maximum torque required to counter gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the dirt is 50 kg, so its weight is 500 N.  If the 50%-full cylinder is split vertically, with the left half full and the right half empty, we would maximize the torque from gravity, which is 500 N multiplied by the distance of the center of mass of the barrel from the central axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The center of mass of the cylinder &lt;a href=&quot;http://arilizi.tripod.com/CM_OF_A_SEMI_CIRCLE.htm&quot; rel=&quot;nofollow&quot;&gt;is $4/3\pi$ the way to the edge of the barrel&lt;/a&gt;, so multiplying this length by the 500 N gives the maximum torque that can be exerted by gravity.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-05-25T00:05:19.433" CommentCount="3" />
  <row Id="10348" PostTypeId="2" ParentId="10346" CreationDate="2011-05-25T00:19:58.743" Score="1" ViewCount="" Body="&lt;h2&gt;BoTE&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Assume the pile will, on average have a slope of 45 degrees. Take $\theta = 0$ to be the horizontal on the side where the pile accumulates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then integrate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \bar\tau = \int_0^R dr \int_{-\frac{3\pi}{4}}^{\frac{\pi}{4}} d\theta \rho l (-g) r^2 \cos(\theta) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \bar\tau = g \rho l \frac{R^3}{3}  \left[ \sin(\theta) \right]_{\theta=-\frac{3\pi}{4}}^{\frac{\pi}{4}}   $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \bar\tau = g \rho l \frac{R^3}{3} [\frac{\sqrt{2}}{2} - -\frac{\sqrt{2}}{2} ]  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \bar\tau = g \rho l \frac{\sqrt{2}}{3} R^3 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\rho$ represents the mass density of the mixture, $l$ the depth of the drum, and $R$ the radius of the drum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\rho l \frac{\pi R^2}{2} = 50\text{ kg}$ so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \bar\tau = g \frac{2\sqrt{2}}{3 \pi} (50\text{ kg}) R $$&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-25T01:49:24.910" LastActivityDate="2011-05-25T01:49:24.910" CommentCount="8" />
  <row Id="10349" PostTypeId="1" CreationDate="2011-05-25T01:08:13.587" Score="0" ViewCount="159" Body="&lt;p&gt;If one lets an empty microwave running for about 10 minutes, just for the sake of the argument, is there any danger at stake? May the microwave damage itself, or destroy something around it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: To further expand the question as David proposed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a joke we use to tell a lot, it goes like, &lt;em&gt;&quot;I need to go home now, I just remembered I kept the microwave running&quot;&lt;/em&gt;, which is kind of a stupid reason, because:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a microwave has a timer, which when expires, it turns itself off automatically and there's no need to rush back home to turn it off&lt;/li&gt;&#xA;&lt;li&gt;a running microwave should not do any harm to anything ? &amp;lt;- this is what I'm not sure about&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I didn't want to provide this story as I believe it to be off-topic, but as I was asked to, I did now. I believe this question can be answered generally by physicists and I really see no reason for this to get closed, but whatever.&lt;/p&gt;&#xA;" OwnerUserId="2834" LastEditorUserId="2834" LastEditorDisplayName="" LastEditDate="2011-05-25T01:28:40.490" LastActivityDate="2011-05-25T01:28:40.490" ClosedDate="2011-05-25T01:19:46.233" Title="Would an empty running microwave cause damage to itself or something else?" Tags="&lt;waves&gt;" CommentCount="3" />
  <row Id="10350" PostTypeId="1" AcceptedAnswerId="10354" CreationDate="2011-05-25T01:17:37.253" Score="2" ViewCount="220" Body="&lt;p&gt;This probably includes a little of biology, but I believe it's mainly physics, so I hope it's ok to ask here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine 2 persons: Person A weights 120 pounds and person B weights 180 pounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine they both fall from a big height:&lt;br&gt;&#xA;a) landing on their knees (a good jump)&lt;br&gt;&#xA;b) falling on their back (a bad fall)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For both of those persons, in both situations, for which of them the fall is easier to handle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I reckon:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;when falling down, the speed is not relative to weight, therefore both those persons will hit the ground at the same speed&lt;/li&gt;&#xA;&lt;li&gt;but, I assume, as the heavier person will create bigger force to the ground, the ground will hit him back with the same force, therefore a heavier person should get a bigger blow when hitting the ground&lt;/li&gt;&#xA;&lt;li&gt;now, this one I'm not sure about, but I assume that a heavier person (considering it's not an unhealthy, fat person, but a healthy one, with muscles, etc) is able to withstand a greater force/blow&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Assuming the last assumption is correct, how are assumptions 2 and 3 related? Is there anything I missed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, which person can withstand falls from greater heights: a lighter one, or a heavier one?&lt;/p&gt;&#xA;" OwnerUserId="2834" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-25T02:07:53.357" LastActivityDate="2011-05-26T18:46:51.060" Title="Which person can handle falls from big heights better: lighter or heavier?" Tags="&lt;forces&gt;&lt;free-fall&gt;&lt;scaling&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="1" />
  <row Id="10351" PostTypeId="2" ParentId="10350" CreationDate="2011-05-25T01:43:17.720" Score="-1" ViewCount="" Body="&lt;p&gt;The lighter person.  The degree of injury is related to surface area.  This is why insects don't mind being dropped and tumbling babies just smile.  I wonder what the dinosaurs were like in action?&lt;/p&gt;&#xA;" OwnerUserId="2072" LastActivityDate="2011-05-25T01:43:17.720" CommentCount="3" />
  <row Id="10352" PostTypeId="1" AcceptedAnswerId="10374" CreationDate="2011-05-25T02:03:14.560" Score="0" ViewCount="72" Body="&lt;p&gt;The formula used in Gyrochronology that relates a star's Period of Rotation-Mass-Age is empirical?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This news &lt;a href=&quot;http://kepler.nasa.gov/news/nasakeplernews/index.cfm?fuseaction=ShowNews&amp;amp;NewsID=129&quot; rel=&quot;nofollow&quot;&gt;How to Learn a Star’s True Age&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;“A star’s rotation slows down&#xA;  steadily with time, like a top&#xA;  spinning on a table, and can be used&#xA;  as a clock to determine its age,&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;pointed to this paper &lt;a href=&quot;http://arxiv.org/abs/1104.2912&quot; rel=&quot;nofollow&quot;&gt;THE KEPLER CLUSTER STUDY: STELLAR ROTATION IN NGC6811&lt;/a&gt;&lt;br&gt;&#xA;and later I found &lt;a href=&quot;http://www.lowell.edu/media/content/release_supplements/gyro_background.pdf&quot; rel=&quot;nofollow&quot;&gt;this one (gyro_background)&lt;/a&gt; with original work.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I couldnt find a justification for the formula. Is the Period proportional to $age^{{1/2}}$ only an empirical result?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that the formula is a data fit and not a direct result of a calculation of the stellar intrinsic dynamics. The rate of mass loss by radiation must have a 'word to say' in the formula.&lt;br&gt;&#xA;Any help is welcome.  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-05-25T09:49:52.203" LastActivityDate="2011-05-25T14:07:29.570" Title="Gyrochronology, the formula is empirical ? " Tags="&lt;astronomy&gt;&lt;dating&gt;&lt;star&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10353" PostTypeId="2" ParentId="10325" CreationDate="2011-05-25T02:45:51.713" Score="4" ViewCount="" Body="&lt;p&gt;(I have a suggestion to make this question a CW.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;General Physics&lt;/strong&gt;: (Early undergrad and advanced high school)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/5030008004&quot; rel=&quot;nofollow&quot;&gt;Problems in Physics I.E Irodov&lt;/a&gt; - (Highly recommended)&lt;/li&gt;&#xA;&lt;li&gt;Problems in Physics S S Krotov - (Once again, highly recommended but out of print)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/144950728X&quot; rel=&quot;nofollow&quot;&gt;Physics Olympiad Books&lt;/a&gt; - (Haven't read but saw some olympiad problems back in the day)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521449758&quot; rel=&quot;nofollow&quot;&gt;Physics by example&lt;/a&gt; (like this book a lot, lower undergrad)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0805390634&quot; rel=&quot;nofollow&quot;&gt;Feynman's Tips on Physics&lt;/a&gt; (Exercises to accompany the famous lectures )&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;General Qualifying exam books:&lt;/strong&gt;&#xA;The following books are a part of a series dedicated to the qualifying exams in American Universities and has a large compilation of problems of all levels. Others in the series include &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/9810212984&quot; rel=&quot;nofollow&quot;&gt;Mechanics&lt;/a&gt;, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/9810206267&quot; rel=&quot;nofollow&quot;&gt;Electromagnetism&lt;/a&gt;, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/9810231334&quot; rel=&quot;nofollow&quot;&gt;Quantum Mechanics&lt;/a&gt;, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/9810200560&quot; rel=&quot;nofollow&quot;&gt;Thermodynamics&lt;/a&gt;, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/9810204396&quot; rel=&quot;nofollow&quot;&gt;Optics&lt;/a&gt; and &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/9810218931&quot; rel=&quot;nofollow&quot;&gt;Solid State Physics&lt;/a&gt;. Unlike other compilation of exercises for qualifiers (such as Princeton or Chicago Problems, or the one mentioned below), they make no excuse for economy and include as many problems from all levels for each subtopic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another good book that I read recently for my exam is the two volume series: &lt;a href=&quot;http://books.google.com/books?id=JtDCeZRMiDEC&amp;amp;printsec=frontcover&amp;amp;dq=A%20Guide%20to%20Physics%20Problems&amp;amp;hl=en&amp;amp;ei=3mzcTbGeLZP6sAOXo43ADg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCoQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;A Guide to Physics Problems&lt;/a&gt; (Part 2 has some relatively easy but interesting problems. I haven't gone through the first part, which is much much more challenging.)&lt;/p&gt;&#xA;" OwnerUserId="1671" LastEditorUserId="1671" LastEditorDisplayName="" LastEditDate="2011-05-25T03:26:54.703" LastActivityDate="2011-05-25T03:26:54.703" CommunityOwnedDate="2011-05-25T04:01:29.620" CommentCount="1" />
  <row Id="10354" PostTypeId="2" ParentId="10350" CreationDate="2011-05-25T02:46:29.113" Score="4" ViewCount="" Body="&lt;p&gt;It is difficult to determine which will fare better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small mammals can survive a fall from arbitrary distances.  Here's &lt;a href=&quot;http://www.straightdope.com/columns/read/1143/do-cats-always-land-unharmed-on-their-feet-no-matter-how-far-they-fall&quot; rel=&quot;nofollow&quot;&gt;one article&lt;/a&gt; I found talking about cats.  A chief contributor to small mammals' survival is that they have a lower terminal velocity due to the way wind resistance scales.  Wind resistance scales with the area of the animal, while weight scales with the volume, so large animals fall faster because they have a higher volume-surface area ration.  On a long fall (hundreds of times the animal's body length), body size influences impact velocity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even without this effect, small animals may have an advantage.  In humans, we can study the statistics of plane crash survivors.  According to Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_sole_survivors_of_airline_accidents_or_incidents&quot; rel=&quot;nofollow&quot;&gt;&quot;Since 1970, two-thirds of lone survivors of airline crashes have been children or flight crew.&quot;&lt;/a&gt;  Children make up a  small percentage of passengers, so it follows they have a better chance of surviving.  It's no great leap to attribute this to their body size.  However, &lt;a href=&quot;http://news.bbc.co.uk/2/hi/uk_news/magazine/8679572.stm&quot; rel=&quot;nofollow&quot;&gt;a BBC news story&lt;/a&gt; has an expert saying there is no physiological advantage to being a child when it comes to surviving a plane crash.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have occasionally seen this sort of question addressed with dimensional analysis, but the difficulty is that it's difficult to pin down what you want to try to scale.  The peak force or pressure, the energy dissipated per unit mass, the peak power dissipation per unit mass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example argument:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume the two people are impacting at the same speed, they need to dissipate the same amount of energy per unit mass.  Assume that people can dissipate a certain amount of energy per unit mass in a given time without harm.  Then whoever can make the impact last for a longer time will fare better.  A taller person can bend their legs through a longer distance, and therefore can make the impact take a longer time, and therefore can fare better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the assumptions in this argument would need to be verified before we can take it very seriously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's another argument:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When two people hit the ground at the same speed, the time it takes them to stop is proportional to their linear dimension because this time is roughly their height divided by the  speed that mechanical waves move through their body.  Their acceleration is inversely proportional to height.  Their mass is proportional to the cube of their height, so the force is proportional to the square of their height.  That makes the pressure independent of height, so large and small people will fare equally well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's another:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same as above, but the deceleration time is proportional to the square root of height because they're flexing their knees, and so the stopping distance is proportional to height.  This now favors short people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same as above, but mass scales with the square of height, because people are not scale invariant (the BMI uses an exponent of two).  This favors tall people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same as above, but mass is independent of height because we're considering a skinny twerp and a muscular jock.  This now favors light people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My conclusion is that the problem is indeterminate.  It depends on whether we're talking about a scaled-down version of the same person, or a single guy who starts taking steroids to prepare for a parachute jump.  It also depends on various material properties of the human body, and on what sorts of things cause injury.  Ultimately I think it's an empirical question, or at least one that requires extensive computer modeling.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-05-25T02:46:29.113" CommentCount="2" />
  <row Id="10355" PostTypeId="2" ParentId="10350" CreationDate="2011-05-25T04:04:43.907" Score="0" ViewCount="" Body="&lt;p&gt;Simple impact damage models of humans (crash-test dummies) are usually based upon peak acceleration. Average acceleration during impact will scale inversely with collision time. If the distance for force absorption scales with height, the impact is self-similar, then peak acceleration scales as the inverse square-root of height. Note a thicker body/bone/muscle should be stronger in proportion to area, so force may scale with surface area, but this is irrelevant. So the taller person wins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arguments about insects don't apply, the terminal velocity of insects is low because of their small size.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-05-25T04:04:43.907" />
  <row Id="10356" PostTypeId="1" AcceptedAnswerId="10358" CreationDate="2011-05-25T04:09:26.437" Score="-1" ViewCount="56" Body="&lt;p&gt;I understand that an event, in a four dimensional space-time, produces a light cone. As time increases the cones gets larger on either side of the event (past and future). For example the if the sun where to &quot;go out&quot; it would take 8 minuets for the earth to be affected by it simply because it takes approximately 8 minuets for light from the sun to reach the earth due to its location in the future light cone of the event (the sun being the event). Einstein made a suggestion that space-time is warped (vs flat) by the distribution of mass and energy and that bodies (like earth, jupiter, etc.) are meant to follow straight paths but cannot in a warped space (or appear not to because space is warped). However light supposedly follows these warped paths (called geodesics) as well even tho light is energy. How can space bend light? and why does light have to follow a specific path which is warped by space? Things with mass (like water) have to go around things (like rocks), but light can go through certain things or expand, but it doesn't move to the side like water does to a rock. Does it?&lt;/p&gt;&#xA;" OwnerUserId="3688" LastActivityDate="2011-05-25T04:29:02.293" Title="future light cones and light paths" Tags="&lt;gravity&gt;&lt;light&gt;&lt;spacetime&gt;&lt;mass&gt;" AnswerCount="1" />
  <row Id="10357" PostTypeId="2" ParentId="2765" CreationDate="2011-05-25T04:25:29.753" Score="0" ViewCount="" Body="&lt;p&gt;well i cannot give the mathematics however (and it is an &quot;idea&quot; assuming it not ridiculous) i would like to point out that water is a polar molecule, and therefore has to give off (an incredibly small) magnetic field. If your cells could fluctuate the magnetic fields of all the water in that region they could give off a signal... However thats just a thought and truth be told probably not likely. &lt;/p&gt;&#xA;" OwnerUserId="3688" LastActivityDate="2011-05-25T04:25:29.753" CommentCount="1" />
  <row Id="10358" PostTypeId="2" ParentId="10356" CreationDate="2011-05-25T04:29:02.293" Score="2" ViewCount="" Body="&lt;p&gt;The underlying space is curved.  There is no 'straight' path.  It's like asking why planes have to travel on great arcs to get the shortest path between cities on the Earth's surface, rather than flying on a straight line on a Mercator projection.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-05-25T04:29:02.293" CommentCount="15" />
  <row Id="10359" PostTypeId="1" CreationDate="2011-05-25T05:41:51.920" Score="3" ViewCount="57" Body="&lt;p&gt;For the purpose of this question, let's restrict ourselves to BKL singularities. BKL cosmologies are homogeneous Bianchi type XIII and IV cosmologies which exhibit oscillatory chaotic behavior, although that's not relevant to this question. Most generic singularities can be approximated locally by a BKL solution. The volume of a BKL universe decreases linearly with time as the singularity approaches. If the matter falling toward the singularity has a nonzero entropy, and the second law of thermodynamics is satisfied, the entropy density will increase without limit as the singularity approaches. Can the entropy density exceed the inverse Planck volume? Being inside a black hole, the holographic bound does not apply. If entropy densities beyond the inverse Planck volume are forbidden, is the second law violated?&lt;/p&gt;&#xA;" OwnerUserId="3718" LastActivityDate="2011-05-25T05:41:51.920" Title="Can the entropy density of a spacelike singularity arbitrarily exceed the inverse Planck volume?" Tags="&lt;quantum-gravity&gt;&lt;statistical-mechanics&gt;" CommentCount="1" />
  <row Id="10360" PostTypeId="2" ParentId="10140" CreationDate="2011-05-25T06:10:02.790" Score="0" ViewCount="" Body="&lt;p&gt;The authors are spouting pure drivel about decoherence. The choice of which environment to trace over is not arbitrary. Decoherence can happen for closed systems. Even if the entire solar system right up to Pluto is included in the system, decoherence does not take minutes to happen. It happens in less than a microsecond. The immediate environment of the measurement already contains degrees of freedom which can decohere the system. Placing mirrors at Pluto is nowhere near enough to reverse decoherence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What decoherence is needs to be spelled out clearly. Decoherence is the exponential suppression of off-diagonal terms in the reduced density matrix. Nothing more, and nothing less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors are under the mistaken impression they can draw any boundary between the system and the environment that they like. You can't just take a causal diamond and call anything outside or the boundary itself the environment, and anything inside the system. The environment is determined dynamically from the interactions. A sharp boundary which is at most a Planck length thick is a bad choice for dividing the environment from the system. The vacuum in quantum field theory has quantum correlations, i.e. quantum entanglement, right up to the correlation length and this can be infinite if massless fields like electromagnetism or gravity exist. A dynamically determined system-environment boundary has to be smeared over the vacuum correlation length at the very least. With a Planck thick boundary, tracing over the environment will lead to a density matrix with most of the von Neumann entropy comming from Planck temperature Unruh radiation near the boundary itself. Defining many-worlds in this manner promotes Planck scale short-lived quantum fluctuations into many-worlds. I don't think that is what the authors have in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, I give this article a thumbs down.&lt;/p&gt;&#xA;" OwnerUserId="3719" LastActivityDate="2011-05-25T06:10:02.790" />
  <row Id="10361" PostTypeId="1" CreationDate="2011-05-25T06:53:23.610" Score="1" ViewCount="95" Body="&lt;p&gt;The anthropic principle is based upon Bayesian reasoning applied to the ensemble of universes, or parts thereof, conditioned upon the existence of conscious observers. That still leaves us with the problem of determining the prior probabilities. This is the famous measure problem. How is this measure determined? Does the ensemble include models with different fundamental laws of physics as in Tegmark's ultimate ensemble? Or should it be restricted to the metastable and stable vacua of the string theory landscape over a particular state with eternal inflation? Or should it be over all possible states in string theory? Should we also weigh universes by the number of Everett branches with conscious observers? Is this weight proportional to the square of the amplitude of the branch? For a particular metastable vacuum and Everett branch, should we also weigh by the population of conscious observers in it? Should we also weigh by the degree of consciousness, or the lifespan? Which measure would favor evolved life over Boltzmann brains?&lt;/p&gt;&#xA;" OwnerUserId="3720" LastActivityDate="2011-06-24T16:03:11.543" Title="The measure problem in the anthropic principle" Tags="&lt;anthropic-principle&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="10362" PostTypeId="1" AcceptedAnswerId="10370" CreationDate="2011-05-25T07:14:09.013" Score="10" ViewCount="344" Body="&lt;p&gt;I read that the non-commutativity of the quantum operators leads to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Uncertainty_principle#Mathematical_derivations&quot; rel=&quot;nofollow&quot;&gt;uncertainty principle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is how both things hang together. Is it that when you measure one thing first and than the other you get a predictably different result than the measuring the other way round?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know what non-commutativity means (even the minus operator is non-commutative) and I think I understand the uncertainty principle (when you measure one thing the measurement of the other thing is kind of blurred - and vice versa) - but I don't get the connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you could give a very easy everyday example with non-commuting operators (like subtraction or division) and how this induces uncertainty and/or give an example with commuting operators (addition or multiplication) and show that there would be no uncertainty involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2011-06-02T15:41:21.433" Title="How does non-commutativity lead to uncertainty?" Tags="&lt;uncertainty-principle&gt;&lt;intuition&gt;&lt;observables&gt;" AnswerCount="3" FavoriteCount="5" />
  <row Id="10363" PostTypeId="2" ParentId="10350" CreationDate="2011-05-25T07:20:05.463" Score="0" ViewCount="" Body="&lt;p&gt;I agree with Mark. If I were betting, weight would not be enough information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the better physically trained person has a better chance of surviving. It is analogous to picking up somebody who weighs 30 kilos. A child, and children are mostly muscle, hangs on. A sack of dead weight might break your back. In effect a well muscled child seems lighter than its dead weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the datum muscle/flab should be given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well muscled people will break a fall in various instinctive ways, aiding survival. Now of course  heavier folks tend to be badly trained and the muscle to flab ratio is smaller, so in that case the light one wins. For two equally muscled people with the same terminal velocity, it will be a tie.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-25T07:20:05.463" CommentCount="1" />
  <row Id="10364" PostTypeId="1" CreationDate="2011-05-25T07:55:21.007" Score="7" ViewCount="148" Body="&lt;p&gt;Do quantum states contain exponentially more information than classical states? It might seem so at first sight, but what about in light of &lt;a href=&quot;http://www.scottaaronson.com/talks/advucsb.ppt&quot; rel=&quot;nofollow&quot;&gt;this talk&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="3721" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-25T18:29:01.880" LastActivityDate="2011-05-25T18:29:01.880" Title="Do quantum states contain exponentially more information than classical states?" Tags="&lt;quantum-information&gt;" AnswerCount="1" FavoriteCount="4" />
  <row Id="10365" PostTypeId="2" ParentId="10350" CreationDate="2011-05-25T08:12:30.330" Score="1" ViewCount="" Body="&lt;p&gt;Let's proceed step by step:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Assuming you drop two balls of the same material and density but different radii (hence, different mass) from the same height. Which one is more likely to break?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The terminal velocity of the smaller ball will be lower and hence the acceleration it will face on impact will be lower than the bigger ball. Everything else is the same (ball material, strength etc.) and hence the bigger ball is more likely to break. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now what if I add elasticity to the equation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Imagine you are dropping an ice ball and a tennis ball. Even if both hit the ground with the same velocity, the ice ball is more likely to shatter into pieces. Why? Because, ice is more inelastic than fiber. The tennis ball is more likely to bounce back. More elasticity increases the time it takes to come to a complete halt (thereby reducing the acceleration). For the same reason, insects and other invertebrates are more likely to sustain fewer injuries than vertebrates of &lt;em&gt;comparable size&lt;/em&gt; (bone vs flesh).&lt;/p&gt;&#xA;" OwnerUserId="3669" LastEditorUserId="3669" LastEditorDisplayName="" LastEditDate="2011-05-25T08:22:45.983" LastActivityDate="2011-05-25T08:22:45.983" CommentCount="1" />
  <row Id="10366" PostTypeId="2" ParentId="10364" CreationDate="2011-05-25T08:32:52.287" Score="6" ViewCount="" Body="&lt;p&gt;The answer depends on what you mean by the information contained in a quantum state.&lt;p&gt;&#xA;If you think of the wavefunction itself as a real physical entity then the answer is yes.&lt;br&gt;&#xA;The problem with this view is that the information in a wavefunction is not accessible to us. We can only learn a small fraction of it through a measurement, while destroying the rest of it in the process. In fact the &lt;a href=&quot;http://en.wikipedia.org/wiki/Holevo_bound&quot; rel=&quot;nofollow&quot;&gt;Holevo bound&lt;/a&gt; states that the amount of information accessible to us is the same as a classical system of the same size. If this is your criteria for the information contained in a quantum state then the answer to your question is no.&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with the second view is that under some conditions it is possible to choose in advance which part of the information in the wavefunction we want (see &lt;a href=&quot;http://portal.acm.org/citation.cfm?id=581771.581773&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; for example). For example, imagine storing a whole phone book in a single quantum state, for which the accessible information is only the size of a single phone number. After retrieving a phone number of our choice, the phone book is destroyed.&lt;br&gt;&#xA;On the one hand, we only got one phone number. On the other hand, we could have chosen any number we wanted, so in some sense the quantum state contained the entire phone book.&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is up to you to choose the view you prefer.&lt;/p&gt;&#xA;" OwnerUserId="975" LastActivityDate="2011-05-25T08:32:52.287" CommentCount="1" />
  <row Id="10367" PostTypeId="2" ParentId="10362" CreationDate="2011-05-25T08:50:43.163" Score="3" ViewCount="" Body="&lt;p&gt;Dear vonjd, you just want to look at a proof of the &quot;uncertainty principle&quot;, namely an inequality that has the uncertainties on the left hand side and the commutator on the right hand side, e.g. at&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Uncertainty_principle#Mathematical_derivations&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Uncertainty_principle#Mathematical_derivations&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your comment that &quot;subtraction and division&quot; are examples of non-commuting operators doesn't indicate that you will actually understand anything above. In quantum mechanics, we always talk about linear operators on a Hilbert space. Even the non-commuting operators have to be linear operators on the Hilbert space, like the position, momentum, angular momentum, or another observable. There is no &quot;division&quot; and not even &quot;subtraction&quot; among them.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-25T08:50:43.163" />
  <row Id="10368" PostTypeId="2" ParentId="10362" CreationDate="2011-05-25T09:28:28.373" Score="4" ViewCount="" Body="&lt;p&gt;A simple example of non-commutativity is rotations in 3D, cf. figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.lightandmatter.com/html_books/genrel/ch07/figs/noncommuting-rotations.png&quot; alt=&quot;Image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physically, the rotations around the $x$- and the $y$-axis are generated by angular momentum operators $\hat{L}_x$ and $\hat{L}_y$, respectively, which do not commute. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the mathematical expressions for $\hat{L}_x$ and $\hat{L}_y$, you may proceed with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Uncertainty_principle#Mathematical_derivations&quot; rel=&quot;nofollow&quot;&gt;mathematical derivation&lt;/a&gt;, that you already linked to, and derive the corresponding uncertainty relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, let me mention that subtraction and division are examples of &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_operation&quot; rel=&quot;nofollow&quot;&gt;binary operations&lt;/a&gt;, while, for starters, the operators in quantum mechanics have only one input, they are so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Unary_operation&quot; rel=&quot;nofollow&quot;&gt;unary operations.&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2451" LastActivityDate="2011-05-25T09:28:28.373" />
  <row Id="10369" PostTypeId="1" CreationDate="2011-05-25T10:16:18.637" Score="3" ViewCount="58" Body="&lt;p&gt;Witten's 1981 paper &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0550321381900213&quot; rel=&quot;nofollow&quot;&gt;&quot;Search for a realistic Kaluza-Klein theory&quot;&lt;/a&gt; is frequently cited for its observation that, in a compactification of d=11 supergravity on a manifold with SU(3) x SU(2) x U(1) isometry, left-handed and right-handed fermions would transform in the same way under the gauge group, in opposition to experiment. Years later, in the era of M-theory, &lt;a href=&quot;http://arxiv.org/abs/hep-th/0109152&quot; rel=&quot;nofollow&quot;&gt;Acharya and Witten&lt;/a&gt; were able to get chiral fermions by compactifying on a manifold with singularities. But Witten 1981 also mentioned, at the end, that coupling to a torsion field might produce chiral asymmetry. (Subsequent explorations of the idea: &lt;a href=&quot;http://prd.aps.org/abstract/PRD/v33/i2/p363_1&quot; rel=&quot;nofollow&quot;&gt;for&lt;/a&gt; and &lt;a href=&quot;http://jmp.aip.org/resource/1/jmapaq/v25/i9/p2696_s1&quot; rel=&quot;nofollow&quot;&gt;against&lt;/a&gt;.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm wondering: &lt;strong&gt;Can torsion in M-theory revive the 1970s version of Kaluza-Klein theory?&lt;/strong&gt; A &lt;a href=&quot;http://arxiv.org/abs/1105.2299&quot; rel=&quot;nofollow&quot;&gt;very recent paper&lt;/a&gt; constructed CFT duals for M-theory on AdS&lt;sub&gt;4&lt;/sub&gt; x M&lt;sup&gt;111&lt;/sup&gt; with torsion flux, M&lt;sup&gt;111&lt;/sup&gt; being one of the &quot;M&lt;sup&gt;pqr&lt;/sup&gt; manifolds&quot; with SU(3) x SU(2) x U(1) isometry group that was considered in Witten 1981. (The M&lt;sup&gt;pqr&lt;/sup&gt; notation is Witten's; in a more recent notation this manifold is designated M&lt;sup&gt;3,2&lt;/sup&gt; or sometimes Y&lt;sup&gt;2,3&lt;/sup&gt;.) Can M-theory on such a background give rise to chiral fermions? &lt;/p&gt;&#xA;" OwnerUserId="1486" LastActivityDate="2011-05-25T10:16:18.637" Title="Chiral fermions from torsion flux in M-theory?" Tags="&lt;string-theory&gt;&lt;standard-model&gt;&lt;fermions&gt;&lt;chirality&gt;" CommentCount="2" />
  <row Id="10370" PostTypeId="2" ParentId="10362" CreationDate="2011-05-25T10:19:43.753" Score="12" ViewCount="" Body="&lt;p&gt;There is a fair amount of background mathematics to this question, so it will be a while before the punch line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum mechanics, we aren't working with numbers to represent the state of a system.  Instead we use &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_space&quot; rel=&quot;nofollow&quot;&gt;vectors&lt;/a&gt;.  For the purpose of a simple introduction, you can think of a vector as a list of several numbers.  Therefore, a number itself is a vector if we let the list length be one.  If the list length is two, then $(.6, .8)$ is an example vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_space#Linear_maps_and_matrices&quot; rel=&quot;nofollow&quot;&gt;operators&lt;/a&gt; aren't things like plus, minus, multiply, divide.  Instead, they are functions; they take in one vector and put out another vector.  Multiplication isn't an operator, but multiplication by two is.  An operator acts on a vector.  For example, if the operator &quot;multiply by two&quot; acts on the vector $(.6, .8)$, we get $(1.2, 1.6)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commutativity is a property of two operators considered together.  We cannot say &quot;operator $A$ is non-commutative&quot;, because we're not comparing it to anything.  Instead, we can say &quot;operator $A$ and operator $B$ do not commute&quot;.  This means that the order you apply them matters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let operator $A$ be &quot;switch the two numbers in the list&quot; and operator $B$ be &quot;subtract the first one from the second&quot;.  To see whether these operators commute, we take the general vector $(a,b)$ and apply the operators in different orders.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example of notation, if we apply operator $A$ to $(a,b)$, we get $(b,a)$.  This can be written $A(a,b) = (b,a)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$BA(a,b) = (b,a-b)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$AB(a,b) = (b-a,a)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we apply the operators in the different orders, we get a different result.  Hence, they do not commute.  The commutator of the operators is defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\textrm{commutator}(A,B) = [A,B] = AB - BA$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a new operator.  Its output for a given input vector is defined by taking the input vector, acting on it with $B$, then acting on the result with $A$, then going back to the original vector and doing the same in opposite order, then subtracting the second result from the first.  If we apply this composite operator (to wit: the commutator) to $(a,b)$, we get (by subtraction using the two earlier results)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(AB - BA)(a,b) = (a,-b)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the commutator of $A$ and $B$ is the operator that multiplies the second entry by minus one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An eigenvector of an operator is a vector that is unchanged when acted on by that operator, except that the vector may be multiplied by a constant.  Everything is an eigenvector of the operator &quot;multiply by two&quot;.  The eigenvectors of the switch operator $A$ are $\alpha(1,1)$ and $\beta(1,-1)$, with $\alpha$ and $\beta$ any numbers.  For $(1,1)$, switching the entries does nothing, so the vector is unchanged.  For $(1,-1)$, switching the entries multiplies by negative one.  On the other hand if we switch the entries in $(.6,.8)$ to get $(.8,.6)$, the new vector and the old one are not multiples of each other, so this is not an eigenvector.  The number that the eigenvector is multiplied by when acted on by the operator is called its eigenvalue.  The eigenvalue of $(1,-1)$ is $-1$, at least when we're talking about the switching operator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum mechanics, there is uncertainty for a state that is not an eigenvector, and certainty for a state that is an eigenvector.  The eigenvalue is the result of the physical measurement of the operator.  For example, if the energy operator acts on a state (vector) with no uncertainty in the energy, we must find that that state is an eigenvector, and that its eigenvalue is the energy of the state.  On the other hand, if we make an energy measurement when the system is not in an eigenvector state, we could get different possible results, and it is impossible to predict which one it will be.  We will get an eigenvalue, but it's the eigenvalue of some other state, since our state isn't an eigenvector and doesn't even have an eigenvalue.  Which eigenvalue we get is up to chance, although the probabilities can be calculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The uncertainty principle states roughly that non-commuting operators cannot both have zero uncertainty at the same time because there cannot be a vector that is an eigenvector of both operators.  (Actually, we will see in a moment that is is not precisely correct, but it gets the gist of it.  Really, operators whose commutators have a zero-dimensional &lt;a href=&quot;http://en.wikipedia.org/wiki/Kernel_%28matrix%29&quot; rel=&quot;nofollow&quot;&gt;null space&lt;/a&gt; cannot have a simultaneous eigenvector.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only eigenvector of the subtraction operator $B$ is $\gamma(0,1)$.  Meanwhile, the only eigenvectors of the switch operator $A$ are $\alpha(1,1)$ and $\beta(1,-1)$.  There are no vectors that are eigenvectors of both $A$ and $B$  at the same time (except the trivial $(0,0)$), so if $A$ and $B$ represented physical observables, we could not be certain of them both $A$ and $B$ at the same time.  ($A$ and $B$ are not actually physical observables in QM, I just chose them as simple examples.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would like to see that this works in general - any time two operators do not commute (with certain restrictions), they do not have any simultaneous eigenvectors.  We can prove it by contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $(a,b)$ is an eigenvector of $A$ and $B$.  Then $A(a,b) = \lambda_a(a,b)$, with $\lambda_a$ the eigenvalue.  A similar equation holds for  $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$AB(a,b) = \lambda_a\lambda_b(a,b)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$BA(a,b) = \lambda_b\lambda_a(a,b)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $\lambda_a$ and $\lambda_b$ are just numbers being multiplied, they commute, and the two values are the same.  Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(AB-BA)(a,b) = (0,0)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the commutator of $A$ and $B$ gives zero when it acts on their simultaneous eigenvector.  Many commutators can't give zero when they act on a non-zero vector, though.  (This is what it means to have a zero-dimensional null space, mentioned earlier.)  For example, our switch and subtract operators had a commutator that simply multiplied the second number by minus one.  Such a commutator can't give zero when it acts on anything that isn't zero already, so our example $A$ and $B$ can't have a simultaneous eigenvector, so they can't be certain at the same time, so there is an &quot;uncertainty principle&quot; for them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the commutator had been the zero operator, which turns everything into zero, then there's no problem.  $(a,b)$ can be whatever it wants and still satisfy the above equation.  If the commutator had been something that turns some vectors into the zero vector, those vectors would be candidates for zero-uncertainty states, but I can't think of any examples of this situation in real physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum mechanics, the most famous example of the uncertainty principle is for the position and momentum operators.  Their commutator is the identity - the operator that does nothing to states.  (Actually it's the identity times $i \hbar$.)  This clearly can't turn anything into zero, so position and momentum cannot both be certain at the same time.  However, since their commutator multiplies by $\hbar$, a very small number compared to everyday things, the commutator can be considered to be almost zero for large, energetic objects.  Therefore position and momentum can both be very nearly certain for everyday things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the angular momentum and energy operators commute, so it is possible for both  of these to be certain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most mathematically accessible non-commuting operators are the spin operators, represented by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pauli_matrices&quot; rel=&quot;nofollow&quot;&gt;Pauli spin matrices&lt;/a&gt;.  These deal with vectors with only two entries.  They are slightly more complicated than the $A$ and $B$ operators I described, but they do not require a complete course in the mathematics of quantum mechanics to explore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the uncertainty principle says more than I've written here - I left parts out for simplicity.  The uncertainty of a state can be quantified via the standard deviation of the probability distribution for various eigenvalues.  The full uncertainty principle is usually stated &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta A \Delta B \geq \frac{1}{2}\mid \langle[A,B]\rangle \mid$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Delta A$ is the uncertainty in the result of a measurement in the observable associated with the operator $A$ and the brackets indicate finding an &lt;a href=&quot;http://en.wikipedia.org/wiki/Expectation_value_%28quantum_mechanics%29&quot; rel=&quot;nofollow&quot;&gt;expectation value&lt;/a&gt;.  If you would like some details on this, I wrote some notes a while ago that you can access &lt;a href=&quot;http://www.scribd.com/doc/15501464/CauchySchwarz-Inequality-and-Heisenbergs-Uncertainty-Principle&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-06-02T15:41:21.433" LastActivityDate="2011-06-02T15:41:21.433" CommentCount="5" />
  <row Id="10371" PostTypeId="2" ParentId="10325" CreationDate="2011-05-25T11:39:01.620" Score="1" ViewCount="" Body="&lt;p&gt;It might be worth it to look at some past problems from physics olympiads. In particular:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;International Physics Olympiad website &lt;a href=&quot;http://www.jyu.fi/ipho/&quot; rel=&quot;nofollow&quot;&gt;http://www.jyu.fi/ipho/&lt;/a&gt; for high-school students, all problems with solutions. Used to be tricky problems, not its more step-by-step, but still challenging and interesting&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moscow City Physics Olympiad &lt;a href=&quot;http://olympiads.mccme.ru/mfo/&quot; rel=&quot;nofollow&quot;&gt;http://olympiads.mccme.ru/mfo/&lt;/a&gt; there is a pdf in russian, but I managed to understand almost every single problem with google translator &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course BAUPC &lt;a href=&quot;http://liquids.seas.harvard.edu/oleg/competition/&quot; rel=&quot;nofollow&quot;&gt;http://liquids.seas.harvard.edu/oleg/competition/&lt;/a&gt; which has few, but (in my opinion) wonderful problems. Some of them are repeated in the classical mechanics book by David Morin ( &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521876222&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Introduction-Classical-Mechanics-Problems-Solutions/dp/0521876222&lt;/a&gt; )&lt;/p&gt;&#xA;" OwnerUserId="2752" LastActivityDate="2011-05-25T11:39:01.620" CommunityOwnedDate="2011-05-25T11:39:01.620" />
  <row Id="10372" PostTypeId="1" CreationDate="2011-05-25T12:55:03.137" Score="4" ViewCount="72" Body="&lt;p&gt;I am doing a quick calculation on how to calculate the pressure needed to inflate a perfectly spherical balloon to a certain volume, however I have difficulties with the fact that the balloon (rubber) has resistance to stretching and how this affects the pressure needed. It has to do with the E-modulus of the material I think, but I can not think of a proper way to calculate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;Looking for a few hours in my materials books now, but after posting it here I found an answer online already:&#xA;&lt;a href=&quot;http://www.colorado.edu/engineering/CAS/courses.d/Structures.d/ASEN3112.Lect05.d/ASEN3112.Lect05.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.colorado.edu/engineering/CAS/courses.d/Structures.d/ASEN3112.Lect05.d/ASEN3112.Lect05.pdf&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one has an example in it which actually answers my question.&lt;/p&gt;&#xA;" OwnerUserId="3725" LastEditorUserId="3725" LastEditorDisplayName="" LastEditDate="2011-05-25T13:03:39.817" LastActivityDate="2011-05-25T13:03:39.817" Title="Inflating a balloon (expansion resistance)" Tags="&lt;thermodynamics&gt;&lt;pressure&gt;&lt;material-science&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="10373" PostTypeId="1" AcceptedAnswerId="10375" CreationDate="2011-05-25T13:43:43.480" Score="-3" ViewCount="137" Body="&lt;p&gt;If &lt;strong&gt;flow of electrons&lt;/strong&gt; analogy can be imagined as flow of water, how to imagine electricity, that comes in this whole picture?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When plug from bottom of sink is taken out, gravity pulls water molecules, creating pressure difference, in sink there is larger pressure than in open hole, making water to move to place, where is less pressure, until pressure is equal in all places, water can flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I have understood, electricity is some field around each electron. Power of field is called &lt;strong&gt;electric charge&lt;/strong&gt;, mesured in &lt;em&gt;columbs&lt;/em&gt;, every electron has some charge, eaven those, who are sitting tight, that is called static electricity? That charge is generated when electrons rub/hit against each other, when they are &quot;made to move&quot;, when &lt;strong&gt;electrical potencial&lt;/strong&gt;, or pressure difference is made, by connecting to medium, another medium, that has... What that connected medium has, that makes electrons flow? Less electrons? Less electrons with electric field around them? If that is field, that can have some maximum ammount of charge, can charge vary, how that affect flow? I don't think, it can be explained with water, as water isn't carrying... anything...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is mesured in amperes? What is rate of flow, is it a speed of water/electrons flow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are electrons same as water, in manner of, they are just some sort of matter, that is flowing in another &lt;strong&gt;medium&lt;/strong&gt;, but water has much larger &lt;strong&gt;particle size&lt;/strong&gt;, so its hard to make is analogy in my mind - there is no medium, in which water can flow, water is medium by itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I forget, that water is medium, but pretend, that water can only flow in... air, places where air can go in... Then how does wire connections work? If There is one wire, that divides in two, as some sort of connection, &lt;strong&gt;does in each wire, goes same ammount of electrons&lt;/strong&gt; (if we pretend ideal/same conditions in both wires)? Does that mean, there will be half of electrons in one wire, and other half in another wire? The more wires I connect, the less electrons will be in each wire? If one of wires will have damaged lattice or resistor sitting middle of it, will electrons &quot;run way back&quot; to connection place and go into other wires, where is less resistance or they will stay at same place, just building up pressure, constantly making denser electron cloud in resistance area... Ehh...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know, I'm asking too many questions, that touches too many sides of electricity, but please, I want some analogy explanation, that would allow me better understand this whole &quot;stuff&quot;. :) Maybe, someone can suggest some visual materials (video, pictures), for this?&lt;/p&gt;&#xA;" OwnerUserId="3724" LastActivityDate="2011-05-26T21:21:53.530" Title="Flow of water and flow of electrons, how this analogy works?" Tags="&lt;quantum-mechanics&gt;&lt;electricity&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="10374" PostTypeId="2" ParentId="10352" CreationDate="2011-05-25T14:07:29.570" Score="1" ViewCount="" Body="&lt;p&gt;Stars lose angular momentum via stellar winds, which because of the coupling of magnetic fields to the stellar surface exhibit a drag on the stars rotation. So we would expect them to slow over time, the real question is a what rate? Stellar magnetic activity which largely drives coronal activity is expected to be stronger with high rotation rates, but does anyone have any decent quantitative theory on this? In any case, another issue is the distribution of initial rotation rates.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-05-25T14:07:29.570" />
  <row Id="10375" PostTypeId="2" ParentId="10373" CreationDate="2011-05-25T14:55:01.907" Score="3" ViewCount="" Body="&lt;p&gt;To begin with, a good physical analogy matches physical quantities of one system to physical quantities of another system that &lt;em&gt;act mathematically equivalent&lt;/em&gt; in some (limited) context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example of &lt;strong&gt;electric current&lt;/strong&gt; (Amperes C/s) and &lt;strong&gt;electric potential&lt;/strong&gt; (Volts or J/C) makes a good comparison to water &lt;strong&gt;flow&lt;/strong&gt; (kg/s) and &lt;strong&gt;gravitational potential&lt;/strong&gt; (J/kg).  However, in order to use this comparison, you need to back off from the concept of pressure, as well as a few others.  Consider a lazy river type situation where a pump imparts the gravitational potential to the water and allows it to flow through the resistance of the rest of the closed loop, and maybe a water wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you could consider a loop of water employing the analogy with quantities of &lt;strong&gt;flow&lt;/strong&gt; (kg/s) and &lt;strong&gt;hydrostatic pressure&lt;/strong&gt; (J/kg or psi or many other units).  An example of this could be something like a flow loop in a nuclear plant, for instance.  There exist components that impart pressure to the flow, then there are components that have friction (again, like a resistor) that balance against the work of the pump.  The reason this might not be as good of an example is because it requires a pressurized loop, which is further from our common everyday experience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that the water does not exist in a medium in the same way the electrons in an electric circuit do.  This is just one of the many many reasons why the analogs are not perfect.  The thing to focus on is that the &lt;em&gt;equations&lt;/em&gt; can be the same with different quantities in place.  Right now, I'm focused on the most simple laws possible.  For instance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V = I R$$&#xA;$$P = I V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The examples I discussed before can certainly be put in terms of these laws.  All that these do is illustrate a case where there is a flow with a unit attached to it, and that flow has a per-unit energy value which dictates how much flow will occur over static components (the resistor case) and then concepts of power (among others) come naturally.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-05-25T15:24:05.067" LastActivityDate="2011-05-25T15:24:05.067" CommentCount="1" />
  <row Id="10376" PostTypeId="1" CreationDate="2011-05-25T14:57:04.450" Score="0" ViewCount="150" Body="&lt;p&gt;Light traveles a straight line write down the principle in which this is manitested.&lt;/p&gt;&#xA;" OwnerUserId="3727" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-05-25T18:27:33.533" LastActivityDate="2011-06-24T19:03:13.987" Title="Light travels a straight line write down the principle in which this is manitested" Tags="&lt;electromagnetism&gt;&lt;optics&gt;&lt;light&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="10377" PostTypeId="2" ParentId="10376" CreationDate="2011-05-25T15:00:12.463" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Principle_of_least_action&quot; rel=&quot;nofollow&quot;&gt;Principle of least action&lt;/a&gt;; or more specifically &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat%27s_principle&quot; rel=&quot;nofollow&quot;&gt;Fermat's principle of least time&lt;/a&gt;.  The least time traveled at a constant speed (first postulate of special relativity) between two points in a flat geometry is a straight line as can be simply shown with variational calculus.&lt;/p&gt;&#xA;" OwnerUserId="1467" LastEditorUserId="1467" LastEditorDisplayName="" LastEditDate="2011-05-25T16:05:09.670" LastActivityDate="2011-05-25T16:05:09.670" CommentCount="6" />
  <row Id="10378" PostTypeId="2" ParentId="10361" CreationDate="2011-05-25T15:18:12.817" Score="3" ViewCount="" Body="&lt;p&gt;One approach is there are no a priori probabilities, but every probability is a function of the prior knowledge. See e.g.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Carlton M. Caves, ''Resource Material for Promoting the Bayesian View of Everything'', &lt;a href=&quot;http://info.phys.unm.edu/~caves/thoughts2.2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://info.phys.unm.edu/~caves/thoughts2.2.pdf&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;BTW, what does mean that &quot;There is $p=\frac{3}{4}$ for the creation of (a very different universe from ours) and  $p=\frac{1}{4}$ for the creation of our universe&quot;. For me it makes totally no sense, as you can put any probabilities u like (justifying it or not) and it won't change anything.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And going into calculations of phase-space of consciousness beings sounds like a crackpottery (or at least - an unjustified philosophical claim), unless it gives any experimental prediction (or simplifies any theory).&lt;/p&gt;&#xA;" OwnerUserId="184" LastEditorUserId="184" LastEditorDisplayName="" LastEditDate="2011-05-25T15:23:40.783" LastActivityDate="2011-05-25T15:23:40.783" />
  <row Id="10379" PostTypeId="1" CreationDate="2011-05-25T15:44:08.337" Score="7" ViewCount="260" Body="&lt;p&gt;From the vantage point of a physicist and the kind of problems he would like a computer program to solve, what are the essential programming languages that a physicist should know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know C++ and I have worked in ROOT. I have also found Mathematica quiet essential, however since shifting to a linux machine I've been trying Sage which is also good, though not quiet at the same level right now. I have found C++ to be really fast, efficient, and best suited for stuff like Monte Carlo simulations. However, it appears that scimpy is also quiet popular. Would it be worthwhile to invest time in learning python, i.e would it have significant gains?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also find the concept of functional programming languages much appealing. I think physicists &#xA;unlike many programmers are more inclined to think this way rather than iteratively. Haskell sounds like a good candidate and I've also heard that it is ideal for simulating quantum algorithms on classical machines. Is anyone familiar with it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also what about scripting languages? I do not know any of those, except basic directory navigation in bash. Is there any utility in Perl and whatever is out there (I know little about these)&lt;/p&gt;&#xA;" OwnerUserId="1671" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-25T16:14:31.527" LastActivityDate="2011-05-25T16:41:01.387" CommunityOwnedDate="2011-05-25T16:54:34.670" ClosedDate="2011-05-25T16:52:03.937" Title="What programming languages would be helpful for a physicist to know?" Tags="&lt;soft-question&gt;&lt;software&gt;&lt;computation&gt;&lt;computational-physics&gt;&lt;big-list&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" />
  <row Id="10380" PostTypeId="1" CreationDate="2011-05-25T15:52:55.900" Score="6" ViewCount="123" Body="&lt;p&gt;What will happen when sound wave hits the metals with very high frequency can it make the electrons to exited state. If it happens then what will be the state of that metal&lt;/p&gt;&#xA;" OwnerUserId="3590" LastActivityDate="2011-06-25T21:03:52.223" Title="If sound wave is going to hit any metals can it make that metal electrons into exited state?" Tags="&lt;sound&gt;&lt;material-science&gt;&lt;metals&gt;" AnswerCount="2" />
  <row Id="10382" PostTypeId="2" ParentId="10380" CreationDate="2011-05-25T16:24:31.703" Score="3" ViewCount="" Body="&lt;p&gt;Your question is very general, so I will give a general answer: Yes, sound waves at very high frequency can excite metals.  It's called, &quot;sonic welding,&quot; and I've personally seen machines which do it. &lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-05-25T16:24:31.703" CommentCount="3" />
  <row Id="10383" PostTypeId="2" ParentId="10379" CreationDate="2011-05-25T16:41:01.387" Score="2" ViewCount="" Body="&lt;p&gt;In computational physics (and other computational disciplines), not enough attention is paid (much of the time) to reproducibility issues and too much is paid to execution speed.  The main advantage of &quot;literate&quot; languages like Python is that it is pretty easy to write readable, easily documentable, and testable code, and to do so quickly.  That makes it more likely that the code is &quot;correct&quot;, and more likely that others will be able to understand what you have done (basic requirements in experimental work).  At the very least, python and other high level languages like it are great for prototyping algorithms before spending the effort to fine tune C++ or fortran code -- and for many tasks where execution speed doesn't matter at all, one might as well do all the work &quot;literately&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out the Software Carpentry project (http://software-carpentry.org/) for more on scientific computing, reproducibility, and Python.&lt;/p&gt;&#xA;" OwnerUserId="1013" LastActivityDate="2011-05-25T16:41:01.387" CommunityOwnedDate="2011-05-25T16:54:34.670" CommentCount="1" />
  <row Id="10386" PostTypeId="2" ParentId="8983" CreationDate="2011-05-25T17:38:09.710" Score="5" ViewCount="" Body="&lt;p&gt;I am not sure why you are rejecting the static friction on the basis on long the parts are in contact. A &quot;bond&quot; is not a chemical bond that might take time, but rather an interaction between adjacent molecules, or atoms. It propagates at the speed of light, so there is plenty of time for the adjacent molecules to &quot;bond&quot; when sufficiently close enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In real life though, pairing down the tire/road contact into a friction coefficient is the wrong approach. It is a non-linear contact, where the higher the normal load the wider the contact patch is and the distribution of contact pressures changes. In addition, some parts have micro sliding as only 1 point in the contact patch is truly stationary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is something called the &quot;Pacejka Magic Formula&quot; which is a well established model of a tire contact, and there are newer ones out there which minor and major refinements to it. In the end, it depends on what you want out of it, in order for you to decide what contact/traction model to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[ref: &lt;a href=&quot;http://www.tut.fi/plastics/tyreschool/moduulit/moduuli_10/hypertext/8/8_1.html&quot; rel=&quot;nofollow&quot;&gt;Magic Formula&lt;/a&gt;]&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-05-25T17:38:09.710" CommentCount="1" />
  <row Id="10387" PostTypeId="1" CreationDate="2011-05-25T17:39:31.703" Score="3" ViewCount="47" Body="&lt;p&gt;Perhaps you know the books:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concepts of Mass in Contemporary Physics and Philosophy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concepts of Force: A Study in the Foundations of Dynamics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by Max Jammer, which discusses mass and force from a perspective which include much historical as well as epistemological, conceptual and axiomatic elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any good books which discuss the energy concept in a similar way as Max Jammer discussed for example mass and force?&lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-25T18:16:03.747" LastActivityDate="2011-05-25T18:16:03.747" Title="Historical and philosophical reflexions about the concept of energy?" Tags="&lt;energy&gt;&lt;books&gt;&lt;history&gt;&lt;reference-request&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="10388" PostTypeId="2" ParentId="10346" CreationDate="2011-05-25T17:41:08.910" Score="0" ViewCount="" Body="&lt;p&gt;The answer is: Torque = Force x Distance = (Mg) x ((4Pi/3)R sin(Theta)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M = Mass.  For simplicity, assume the barrel weight is 0 and the CONTENTS are 50kgs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g = acceleration due to gravity = 9.8 m/(s^2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force = Mg = 490N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R = max radius of the INSIDE of the barrel, which I assume to be 350mm = 0.35m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'Theta' is angle of rotation.  Theta = Angular displacement from load level in the barrel.  For the puposes of this problem, we constrain Theta to the interval 0 to 90 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distance = horizontal displacement of the Center of Gravity (CG) of the load from the center of the barrel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The TOTAL distance of the CG of the load from the axis of barrel rotation is (4/3Pi)(R). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The HORIZONTAL distance of the CG of the load from the axis of barrel rotation is the TOTAL distance (which is (4/3Pi)(R)) times COS(90-Theta).  On our 0 to 90 degree interval, COS(90-Theta) is the same as SIN(Theta).  So the horizontal torque arm length is ((4/3Pi)(R))(SIN(Theta)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mark is correct: the theoretical torque to start the barrel moving (disregarding friction) is zero.  The max torque to keep it moving is given by using the formula above (first line of this post) and choosing the appropriate Theta.  I would pick Theta = 90 degrees, which makes SIN(Theta) = 1, and collapses the formula to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max torque = (Mg)(4/3Pi)R = 490N x (4/3Pi)(0.35m) = 72.8 (Nm) of torque.  This is roughly 53.6 foot-lbs.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-05-25T18:15:43.860" LastActivityDate="2011-05-25T18:15:43.860" />
  <row Id="10389" PostTypeId="1" CreationDate="2011-05-25T18:42:14.307" Score="3" ViewCount="231" Body="&lt;p&gt;I have a mechanical design with LEDs that generate heat.  I want to estimate the temperature at the LED junction vs. time, but especially at steady state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing the LED voltage drop and current, I can estimate the power dissipated via heat (some of the power goes to light, but to be conservative, I thought to estimate heat power as simply I*V).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LED junction temperature needs to be held below a certain temperature.  The thermal system will consist of the LED junction to thermal pad, the circuit board, and the mechanical design.  The thermal resistance is given for the junction to thermal pad (about 10 K/W), and the manufacturer provides a couple different circuit board designs, each with their own thermal resistance.  The best design can achieve 3-4 K/W, but a less expensive design may be an option.  So I need the thermal resistance of the mechanical housing, which is pre-existing, and cannot be changed except for how to attach the LEDs to the housing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The geometry is somewhat complicated, so I thought I'd start with a simple aluminum block (the circuit board mounts directly to the block), to make sure my modeling is correct, and then move to more complicated geometries.  The LEDs are arranged in a line, so I'm going to assume the heat transfer is constant along the LED line (call it the z-axis).  I also don't want to consider convection or radiation right now, along the length of the block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's say I attach a heat sink to the end of the block opposite the LEDs, with a low enough thermal resistance to ambient so the heat sink end of the block is at ambient.  The thermal resistance of an aluminum block is L/(kA), where L is the length, k is the thermal conductivity (0.25 W/(mm*K)), and A is the cross-sectional area.  So if the length is 20 mm (along x-axis), the cross-sectional area is about 300 mm^2 (about 6 mm x 50 mm, with 6 mm along y-axis, and 50 mm along z-axis), the thermal resistance is about 0.27 K/W.  The LED power per unit area is 0.0896 W/mm^2 (total power of 26.88 W, spread across the 300 mm^2).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:  I like to work in mm for this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem is 1-D, and from Fourier's conduction law, the temperature drop across the block at steady state should therefore be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta T = L \cdot P/(kA) = 7.17 K.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to be able to model this and come up with the same result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the final geometry is more complex, I believe I need to use the Finite Element Method for the final solution, but for starters, it'd be nice to know how to start with the heat equation and end up with the result obtained from Fourier's law (since the former can be derived from the latter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found freefem++ (&lt;a href=&quot;http://www.freefem.org/ff++/index.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.freefem.org/ff++/index.htm&lt;/a&gt;), which can approximate a solution to the heat equation using the Finite Element Method, but variational formulation is beyond my understanding, and none of the examples deal with a constant heat source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to setup the heat equation and boundary conditions.  I especially want the steady-state result, but I'm also interested in the temperature evolution vs. time.  The (3-D) heat equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${{\partial u} \over {\partial t}} = \alpha \nabla ^2 u + {Q \over {c \rho}},$$ where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u = u(x,y,z,t)$ = temperature $(K)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha = {k \over c \rho}$ = thermal diffusivity $({mm^2 \over s})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c$ = specific heat $({J \over {kg \cdot K}})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho$ = mass density $({kg \over mm^3})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q$ = power per unit volume $({W \over mm^3})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the simple block problem is 1-D, the heat equation becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${{\partial u} \over {\partial t}} = \alpha {{\partial ^2 u} \over {\partial x^2}} + {Q \over {c \rho}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_t = \alpha u_{xx} + {Q \over {c \rho}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, Q is a function of $(x, t)$.  This paper provides an analytical solution for this 1-D problem:&#xA;&lt;a href=&quot;http://people.math.gatech.edu/~xchen/teach/pde/heat/Heat-Duhamel.pdf&quot; rel=&quot;nofollow&quot;&gt;http://people.math.gatech.edu/~xchen/teach/pde/heat/Heat-Duhamel.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to use Duhamel's Principle, I first reformulate the problem so the heat source is at $L/2$ the temperature at $x=0$ and $x=L$ is held to zero (whether zero or ambient temperature, the solution is the same).  Then the heat source is the Dirac-delta function at $x=L/2$, and the problem becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_t = \alpha u_{xx} + {Q_i \over {c \rho}} \delta(x-{L \over 2}), 0 &amp;lt; x &amp;lt; L, t &amp;gt; 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the boundary conditions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u(0,t) = 0, u(L, t) = 0, t &amp;gt; 0$$&#xA;$$u(x,0) = 0, 0 \le x \le L$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $Q_i = 0.896 W/mm^3$ is the heat source.  I was able to approximate the solution given in the paper using Matlab (see below), and as $ds \rightarrow 0$, the solution becomes closer to a triangle function centered at $x=L/2$ and increasing asymptotically over time to about 7 K.  Note that because the heat source is in the center, I doubled the power and block length, and the solution for the above problem is one half of the block ($x \ge L/2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now what happens when I go to 2-D?  As I said, the real geometry is complex (but can be modeled as a 2-D problem, since the geometry along the z-axis is essentially constant), so I don't expect an analytical solution.  Perhaps my simple 1-D example is not sufficient to demonstrate how to solve the problem with the Finite Element Method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if the heat source came in from the left and the bottom of the block was held at constant temperature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the Duhamel Principle extend to 2-D?  If so, and I approximate the auxiliary homogeneous problem with FEM, then how do I transform that approximation into the nonhomogenous solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately, how would I formulate the nonhomogenous problem using variational formulation, in order to be able to use do FEM analysis directly?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Matlab approximation to 1-D analytical solution&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here is my Matlab code that approximates the solution using Duhamel's Principle.  I did the approximation using both Fourier series and Green's function.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function ThermalModel&#xA;&#xA;% Approximate the analytical solution of the heat equation with a heat&#xA;% source in the center of a block.&#xA;&#xA;% System parameters.&#xA;H = 6;               % the block height (mm)&#xA;L = 40;              % the block length (mm)&#xA;W = 50;              % the block width (mm)&#xA;kAl = 0.25;          % Aluminum thermal conductivity (W/(mm*K))&#xA;c = 897;             % Aluminum specific heat capacity (J/(kg * K)).&#xA;rho = 2.7E-6;        % Density (kg/mm^3).&#xA;alpha = kAl / (c * rho);   % Thermal diffusivity (mm^2/s).&#xA;Qi = 2 * 27 / 300;          % Input power per unit volume length (?).&#xA;&#xA;dx = 0.2;&#xA;dt = .2;&#xA;x = 0:dx:L;&#xA;tmax = 10;&#xA;t = 0:dt:tmax;&#xA;&#xA;% Approximate heat equation using Fourier series and Duhamel's Principle.&#xA;ds = 0.1;&#xA;N = 200;&#xA;n = 1:N;&#xA;b = 2*Qi*sin(n*pi/2)/(c*rho*L);&#xA;% As N goes to infinity, the solution&#xA;% approximates a triangle function centered on L/2.  Because we can't go to&#xA;% infinity, there will always be a sharp spike at x = L/2.&#xA;u = zeros(length(x), length(t));&#xA;for xi = 1:length(x)&#xA;   for ti = 1:length(t)&#xA;      tc = t(ti);&#xA;      for ni = 1:length(n)&#xA;         s = 0:ds:tc;&#xA;         sint = 0;&#xA;         for si = 1:length(s)&#xA;            sint = sint + b(ni)*exp(-alpha*(n(ni)*pi/L)^2*(tc-s(si)))*ds;&#xA;         end&#xA;         u(xi, ti) = u(xi, ti) + sin(n(ni)*pi*x(xi)/L) * sint;&#xA;      end&#xA;   end&#xA;end&#xA;&#xA;figure;&#xA;mesh(t, x, u);&#xA;ylabel('x (mm)');&#xA;xlabel('t (s)');&#xA;zlabel('Temperature (deg C)');&#xA;title('Approximation to heat equation solution with constant heat source at L/2, using Fourier series');&#xA;&#xA;% Approximate solution using Green's function.  Note that as ds -&amp;gt; zero,&#xA;% the solution approximates a triangle function centered at L/2, and&#xA;% increasing asymptotically over time.&#xA;u = zeros(length(x), length(t));&#xA;N = 40;&#xA;n = -N:N;&#xA;ds = 0.01;&#xA;for xi = 1:length(x)&#xA;   for ti = 1:length(t)&#xA;      tc = t(ti);&#xA;      if tc == 0&#xA;         continue;&#xA;      end&#xA;      s = 0:ds:(tc-ds);&#xA;      for si = 1:length(s)&#xA;         nint = 0;&#xA;         for ni = 1:length(n)&#xA;            nint = nint + exp(-(x(xi)-2*n(ni)*L-L/2)^2/(4*alpha*(tc-s(si)))) - ...&#xA;               exp(-(x(xi)-2*n(ni)*L+L/2)^2/(4*alpha*(tc-s(si))));&#xA;         end&#xA;         u(xi, ti) = u(xi, ti) + ...&#xA;            (Qi/(c*rho)) * nint * ds / sqrt(4*pi*alpha*(tc-s(si)));&#xA;      end&#xA;   end&#xA;end&#xA;&#xA;figure;&#xA;mesh(t, x, u);&#xA;ylabel('x (mm)');&#xA;xlabel('t (s)');&#xA;zlabel('Temperature (deg C)');&#xA;title('Approximation to heat equation solution with constant heat source at L/2, using Green''s function');&#xA;&#xA;return;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="3697" LastEditorUserId="3697" LastEditorDisplayName="" LastEditDate="2011-05-26T16:01:18.383" LastActivityDate="2011-08-24T16:34:13.240" Title="LED Thermal Modeling (How to solve heat equation with constant heat source)" Tags="&lt;thermodynamics&gt;&lt;heat&gt;&lt;models&gt;&lt;led&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="10390" PostTypeId="2" ParentId="10309" CreationDate="2011-05-25T20:24:21.750" Score="1" ViewCount="" Body="&lt;p&gt;Energy is indeed conserved perfectly well in General Relativity, even in the extreme conditions of the early big bang. It is not merely true by definition, it is true as a result of the dynamics of the field equations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question has been answered before e.g. at &lt;a href=&quot;http://physics.stackexchange.com/questions/2597/energy-conservation-in-general-relativity/&quot;&gt;Energy conservation in General Relativity&lt;/a&gt; and should probably be closed as repetition,  so I'll just add a link to a long discussion on the subject at &lt;a href=&quot;http://blog.vixra.org/category/energy-conservation/&quot; rel=&quot;nofollow&quot;&gt;http://blog.vixra.org/category/energy-conservation/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1341" LastEditorUserId="1341" LastEditorDisplayName="" LastEditDate="2011-05-25T20:31:44.720" LastActivityDate="2011-05-25T20:31:44.720" CommentCount="0" />
  <row Id="10391" PostTypeId="1" AcceptedAnswerId="10393" CreationDate="2011-05-25T21:00:19.390" Score="3" ViewCount="122" Body="&lt;p&gt;In quantum mechanics, what exactly is a coherent state, and how does it differ from other states?&lt;/p&gt;&#xA;" OwnerUserId="202" LastActivityDate="2011-05-26T06:22:21.283" Title="What is a coherent state?" Tags="&lt;quantum-mechanics&gt;&lt;coherent-state&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="10392" PostTypeId="1" CreationDate="2011-05-25T21:17:32.620" Score="7" ViewCount="177" Body="&lt;p&gt;Is it possible to spontaneously break Lorentz invariance, i.e., have a Lagrangian that respects LI but a vacuum which does not? If it is possible, why isn't there even the slightest hint of the Minkowski vacuum not being Lorentz invariant? (Or in other words, would this pose another fine-tuning problem?)&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-05-26T06:17:40.320" Title="Spontaneous breaking of Lorentz invariance" Tags="&lt;quantum-field-theory&gt;&lt;special-relativity&gt;&lt;symmetry-breaking&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="10393" PostTypeId="2" ParentId="10391" CreationDate="2011-05-25T21:25:28.393" Score="3" ViewCount="" Body="&lt;p&gt;A coherent state is a state that is not an energy eigenstate, so the observables vary in time, but the &lt;em&gt;uncertainty&lt;/em&gt; of the canonical conjugates (position and momentum) stays constant - unlike, for instance, in an ordinary wave packet moving in free space: this &quot;floats away&quot;, the position uncertainty gets ever bigger and bigger.&lt;/p&gt;&#xA;" OwnerUserId="3540" LastEditorUserId="3540" LastEditorDisplayName="" LastEditDate="2011-05-25T21:59:53.003" LastActivityDate="2011-05-25T21:59:53.003" CommentCount="2" />
  <row Id="10394" PostTypeId="2" ParentId="679" CreationDate="2011-05-25T22:16:13.300" Score="1" ViewCount="" Body="&lt;p&gt;It always makes me laugh when people hope they can learn physics online. Yeah, sure you can, just as you can learn languages online but you're not going to be articulate compared to someone that is on a full time course. They live physics for years with like minded people, day in, day out. If you want to learn physics the same way someone learns a foreign language by carrying around a translation dictionary because they lack the effort required, then go right ahead. Just don't delude yourself in believing you can know what a graduate physicist knows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me put it another way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A physicist on a full time course might spend at least 50 hours a week learning about physics. How much time do you have to spare?&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-05-25T22:16:13.300" />
  <row Id="10395" PostTypeId="1" AcceptedAnswerId="10404" CreationDate="2011-05-25T22:54:10.853" Score="4" ViewCount="121" Body="&lt;p&gt;What properties of Germanium make it suitable for Dark Matter detectors?&#xA;I tried googling but there was too many results describing the use of Germanium Chrystals at low tempretures for Dark Matter detaction but not the why of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or alternatively what are the properties that make a good dark matter detector? in this case Germanium just gets to fit the requirement and have a look at other possible candidates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming Germanium is rare and hard to get I prompted this question, on the other hand if it was copper that was being used for such a task, it would have gone completely unnoticed.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="3414" LastEditorDisplayName="" LastEditDate="2011-05-26T05:47:39.397" LastActivityDate="2011-05-27T04:14:36.207" Title="What properties of Germanium make it suitable for Dark Matter detectors?" Tags="&lt;particle-physics&gt;&lt;experimental-physics&gt;&lt;dark-matter&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10396" PostTypeId="2" ParentId="10309" CreationDate="2011-05-25T23:26:16.357" Score="2" ViewCount="" Body="&lt;p&gt;Conservation laws are established in general relativity if there is a Killing vector $K_a$, where for some values of the index $a$ there may be zero entries, so that for a momentum vector $p^a$ $=~(p_t,~{\vec p})$ the inner product $p^aK_a~=~constant$.  The Killing vector is then an isometry such that a vector along a parallel translation defines a conserved quantity relative to the Killing vector.  That conserved quantities are variables conjugate to the components of the Killing vector.  How to find Killing vectors is somewhat involved, but as a rule, if a metric coefficient $g_{ab}~=~g_{ab}(t)$ then there is no Killing vector with a component along that coordinate direction.  The general line element for a cosmology involves a scale factor $a~=~a(t)$, &#xA;$$&#xD;&#xA;ds^2~=~-dt^2~+~a^2(t)(dr^2~+~r^2d\Omega^2&#xD;&#xA;$$&#xA;which is a pretty good clue that this spacetime has not fundamental conservation of energy.  There is no timelike directed part of a Killing vector, therefore conservation of the energy conjugate variable can’t be established fundamentally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is energy absolutely not conserved in our universe?  The answer to this depends upon upon some other conditions; for it does turn out that our universe may have a unique condition which recovers energy conservation..  The FLRW equation for the scale parameter $a~=~a(t)$ is&#xA;$$&#xD;&#xA;\big(\frac{\dot a}{a}\Big)^2~=~8\pi G\rho/3~–~\frac{k}{a^2}&#xD;&#xA;$$&#xA;where ${\dot a}/a~=~H$, the Hubble parameter, and flatness means $k~=~0$, spherical geometry is $k~=~1$ and hyperbolic geometry is $k~=~-1$.  There is an equation of state for the mass-energy and pressure in the spacetime &#xA;$$&#xD;&#xA;\frac{d(\rho a^3)}{dt}~+~p\frac{da^3}{dt}~=~0&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will consider an approximate de Sitter spacetime, which has $\rho~=~constant$, and is identified in ways not entirely understood with the quantum field vacuum.  The FLRW equation for $k~=~0$ is then&#xA;$$&#xD;&#xA;\frac{da}{dt}~=~\sqrt{8\pi G\rho/3}~a&#xD;&#xA;$$&#xA;which has the solution $a~=~\sqrt{3/8\pi G\rho}~exp(\sqrt{8\pi G\rho /3}t)$.  Using the Einstein field equation we then have that the stress energy is $T^{00}~=~8\pi G\rho~=~\Lambda$, which is the cosmological constant.  Returning the first equation, the FLRW equation, we then see that $H^2~=\Lambda/3$. The dynamical equation for the dS spacetime with $\rho~=~const$ gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ρd(a^3)/dt + pda^3/dt = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or $p~=~-\rho$.  This is the equation of state for $p~=~w\rho$, and $w~=~-1$.  This corresponds to a case where the total energy is zero and the first law of thermodynamics is $dF~=~dE~–~pdV~=~0$ means the energy that is increased in a unit volume of the universe under expansion is compensated for by a negative pressure which removes work from the system.  Further $pdV~=~d(NkT)$, and for a constant thermal energy for the vacuum and $Nk~=~S$ the entropy of the universe.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this particular special case we do have an equation of state which gives a conservation of energy.  Another way of seeing this is this spacetime has a time dependent conformal factor $a(t)$, and this metric is conformally equivalent to a flat spacetime, where one can define an ADM mass that is conserved.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the question might be raised whether this pertains to our physical universe.  The inflationary period had a huge exponential acceleration, or equivalently a scale factor which grew extremely rapidly.  This period should then have had conservation of energy.  As for the time period before then, who knows?  After the reheating period the universe became radiation dominated and energy conservation is not immediately apparent.  Energy conservation may only then be established in our universe in the very distant future as it approaches an empty de Sitter vacuum state.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-05-25T23:26:16.357" CommentCount="5" />
  <row Id="10397" PostTypeId="2" ParentId="10389" CreationDate="2011-05-26T00:14:49.953" Score="0" ViewCount="" Body="&lt;p&gt;If your goal is practical, and you're looking for a numerical answer, why not use some numerical method to solve the equation? Mathematica, Maple, and Matlab (I think) all have tools for doing this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the steady state solution, you can simplify the problem greatly since as t goes to infinity, the derivative of u with respect to t goes to zero, giving you a simple Poisson Equation.&lt;/p&gt;&#xA;" OwnerUserId="3018" LastActivityDate="2011-05-26T00:14:49.953" CommentCount="3" />
  <row Id="10398" PostTypeId="1" AcceptedAnswerId="10408" CreationDate="2011-05-26T01:50:30.897" Score="2" ViewCount="244" Body="&lt;p&gt;This question is in reference to my downvoted &lt;a href=&quot;http://physics.stackexchange.com/questions/10309/conservation-law-of-energy-and-big-bang/10313#10313&quot;&gt;answer to this active physics.SE question&lt;/a&gt;.  More than one user has indicated that it is simply &lt;em&gt;wrong&lt;/em&gt; and I am having trouble understanding why.  My point of view is that the fact that energy is conserved is essentially part of its definition.  In particular I don't understand why it is necessary to invoke Noether's theorem, GR, or BB in order to &quot;prove&quot; the conservation of energy.  I can't see how a universe without conservation of energy is even possible in principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A thought experiment (of a sociological nature) to illustrate my point: suppose a perpetual motion machine were discovered or invented that seems to create energy continuously from nothing in defiance of all known physical theories.  So we create a new theory to explain it by saying that the energy comes from &quot;somewhere else&quot;, i.e. a &quot;different universe&quot;.  But now we have expanded our concept of &quot;universe&quot; to include this other place; isn't it still fair to believe that energy is globally conserved within this newly-defined larger universe?  I know that numerous discoveries have been made on the basis of missing energy, e.g. the existence of the neutrino, but seems like it is always preferable to violate Occam's razor by postulating new entities than it is to question the law of conservation of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone help clear this up for me?  I am not so much interested in knowing the mathematics as I am in getting my scientific intuition straightened out.  How is a universe without conservation of energy even possible in principle?  If energy is &lt;em&gt;not&lt;/em&gt; globally conserved as some users have suggested in response to the linked question, does this mean that some sort of cosmological-scale perpetual motion device is actually possible?  If so, why not just define it away as I have done in my thought experiment?&lt;/p&gt;&#xA;" OwnerUserId="2072" LastEditorUserId="2072" LastEditorDisplayName="" LastEditDate="2011-05-26T04:06:20.520" LastActivityDate="2011-05-26T11:39:39.950" Title="Why is global conservation of energy not considered a tautology?" Tags="&lt;energy-conservation&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="1" />
  <row Id="10399" PostTypeId="1" CreationDate="2011-05-26T01:53:45.400" Score="1" ViewCount="52" Body="&lt;p&gt;Is there any mechanism by which the vapor pressure of a liquid substance (lets say in a vacuum) is measured to be greater than the expected saturated vapor pressure at that temperature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, in my work I'm measuring a number density of alkali atoms in an evacuated container, the measured number density exceeds what would be the expected number density due to the saturated vapor pressure of the alkali metal (Cs and K, both of which are liquid at the temperatures I'm measuring) &lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2011-05-26T11:37:42.413" Title="Vapor pressure higher than equilibrium vapor pressure" Tags="&lt;thermodynamics&gt;&lt;physical-chemistry&gt;" AnswerCount="0" CommentCount="4" />
  <row Id="10400" PostTypeId="1" AcceptedAnswerId="10409" CreationDate="2011-05-26T01:54:31.420" Score="1" ViewCount="98" Body="&lt;p&gt;Why we call &quot;constant&quot; to the Hubble constant?,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if the universe were really expanding then the Hubble &quot;constant&quot; should change, being variable, smaller and smaller..with &quot;time&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other example/view of this question:&#xA;if Hubble constant were constant then the universe &quot;age&quot; would be constant too, &#xA;so let's say 13.7 billion years, and stop growing. That's not an expanding universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Hubble constant and universe age were not so constant then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is some property that is changing like a &quot;time&quot; that -everyone in universe- could get same number, so there would be an absolute time.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see the Hubble Constant, and even the age of the universe, a really too speculative theory (perhaps it is as speculative as the posibility of measuring), having more accuracy will lead to better theories...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because some theoric impossibility in measuring that kind of variation? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see a lot of inconsistency. Any light at this respect is welcome, thanks&lt;/p&gt;&#xA;" OwnerUserId="1916" LastActivityDate="2011-05-26T05:36:00.680" Title="Universe Expansion as an absolute time reference" Tags="&lt;astrophysics&gt;&lt;spacetime&gt;&lt;universe&gt;&lt;big-bang&gt;&lt;time&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="10401" PostTypeId="1" AcceptedAnswerId="10405" CreationDate="2011-05-26T02:13:21.057" Score="-1" ViewCount="110" Body="&lt;p&gt;If I were to try to find pi using a ruler and a compass, I would first try to find out how many rational line segments of the diameter I could fit around the interior circumference and then continue to refine until I began converging on some number.  Before the foundations of Non-Euclidean geometry were made, people would not have even questioned that the euclidean pi had a unique nature.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why not think of the standard model parameters in the same light?  Just simply as ratios that might change with the geometry of space or some something similar?&lt;/p&gt;&#xA;" OwnerUserId="3460" LastActivityDate="2011-05-26T03:53:48.887" Title="How is fine tuning of standard model conceptual different than the fine tuning of PI?" Tags="&lt;standard-model&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="10402" PostTypeId="2" ParentId="10398" CreationDate="2011-05-26T02:45:59.877" Score="2" ViewCount="" Body="&lt;p&gt;The issue is &lt;b&gt;finding&lt;/b&gt; that conserved energy.  What Noether's theorem does is give us a nice, clean way to discover what the conserved energy &lt;b&gt;is&lt;/b&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that for certain models in general relativity, there is no clear global notion of conserved energy.  In particular, most of the &quot;nice&quot; ways that we would try and define energy are totally coordinate-dependent--both the components of the metric tensor and their first derivatives are right out as local components of a potential &quot;energy of the gravitational field,&quot; since their value at a point can be set arbitrarily by a coordinate transformation, meaning that if I were to construct an energy density out of the metric and its first derivatives, its value would depend on whether I used (for example) spherical or rectangular coordinates in $\mathbb{R}^{3}$  There are some cases where we can be saved by Noether-type logic, though--if we either have a metric that has a global timelike translation symmetry, or at least a 3-surface that has a timelike or null translational symmetry along it, we can construct something that looks like a conserved energy for the dynamics in that spacetime.  These different 'masses' go by different names, depending on the details of the problem that you're dealing with.  For the sake of posterity, I'll refer to &lt;a href=&quot;http://emis.impa.br/EMIS/journals/LRG/Articles/lrr-2009-4/download/lrr-2009-4Color.pdf&quot; rel=&quot;nofollow&quot;&gt;a pretty technical paper on this&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, it turns out that there are a wide variety of spacetimes where none of this works out, and that expansionary cosmologies containing matter other than the cosmological constant are one of these type of spacetimes.${}^{1}$.  In these cases, there is just no conserved energy or energy density that we can define.  If the universe expands forever, and the photons just keep on redshifting, where is the energy of those redshifting photons going to, after all?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We just have to deal with the fact that the only energy conservation we have is in the local sense, where $T^{ab};{}_{b}=0$.  Globally, energy is not conserved.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{1}$In much the same way that there is no conserved energy for the Lagrangian $L=\frac{1}{2}m{\dot x}^{2} + a_{0}xe^{t}$.  It's just that in this case, we know that it's because we have a time-dependent external potential driving the system, so we don't expect the energy to be conserved.&lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2011-05-26T02:59:09.007" LastActivityDate="2011-05-26T02:59:09.007" CommentCount="3" />
  <row Id="10403" PostTypeId="1" CreationDate="2011-05-26T03:21:11.533" Score="3" ViewCount="161" Body="&lt;p&gt;A similar question was asked &lt;a href=&quot;http://physics.stackexchange.com/questions/2228/what-is-the-definition-of-colour-the-quantum-state&quot;&gt;here&lt;/a&gt;, but the answer didn't address the following, at least not in a way that I could understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electric charge is simple - it's just a real scalar quantity. Ignoring units and possible quantization, you could write $q \in \mathbb{R}$. Combination of electric charges is just arithmetic addition: $ q_{net} = q_1 + q_2 $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to color charge. Because there are three &quot;components&quot;, I am tempted to conclude that color charges are members of $\mathbb{R}^3$. However, I've read that &quot;red plus green plus blue equals colorless&quot;, which seems to rule out this idea. I can only think that either:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;red, green and blue are not orthogonal, or&lt;/li&gt;&#xA;&lt;li&gt;&quot;colorless&quot; doesn't mean zero color charge (unlikely), or&lt;/li&gt;&#xA;&lt;li&gt;color charges don't combine in a simple way like vector addition&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In formulating an answer, please consider that I know some mathematics (vectors, matrices, complex numbers, calculus) but almost nothing about symmetry groups or Lie algebras.&lt;/p&gt;&#xA;" OwnerUserId="1793" LastActivityDate="2011-05-26T15:43:51.133" Title="Mathematically, what is color charge?" Tags="&lt;quantum-chromodynamics&gt;&lt;color&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="10404" PostTypeId="2" ParentId="10395" CreationDate="2011-05-26T03:42:20.430" Score="4" ViewCount="" Body="&lt;p&gt;WIMP's are hypothesized to interact only by the weak force, thus will interact only with the nucleus. The WIMP will hit the nucleus, slow down and change direction. The nucleus will recoil, and introduce vibrations (phonons) into the lattice. Those vibrations create heat which is measured. The crystal must me kept at 10 mK in order to reduce thermal noise. Additionally, the recoil of the nucleus lifts some electrons into the conduction band where they too can be measured.&#xA;Interactions by alpha, beta and gamma rays generate more ionization than temperature rise. WIMP interactions will generate a large nuclear recoil and many phonons but will not interact directly with the shell electrons thus very few electrons will enter the conduction band. By looking for a large phonon signal and a relatively small ionization signal, WIMPS can be sorted out.&#xA;But why Germanium?  Because Ge has the smallest band gap of the semiconductors. An incident particle will lift more electrons out of the ground state and put them into the conduction state. The signal reaching the preamp is larger with Ge than any other semiconductor. Germanium is the gold standard for the detection of gamma rays. Silicon has a larger band gap thus is not as sensitive. &lt;/p&gt;&#xA;" OwnerUserId="3735" LastActivityDate="2011-05-26T03:42:20.430" CommentCount="2" />
  <row Id="10405" PostTypeId="2" ParentId="10401" CreationDate="2011-05-26T03:53:48.887" Score="1" ViewCount="" Body="&lt;p&gt;The number $\pi$ is a mathematical constant that can be calculated. And indeed, as expected from naturalness, it is of order one, so it is clearly &lt;em&gt;not&lt;/em&gt; fine-tuned. So we wouldn't be talking about fine-tuning even if it were just a physical parameter and not a mathematical constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fine-tuning only occurs when a theory requires at least 2 parameters with the same units whose numerical value differs by a big ratio. That's not the case of $\pi$. It is the case of the Higgs mass in the Planck units which is $10^{-15}$, and especially cosmological constant in Planck units which is $10^{-123}$. There are many other less extreme examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if your suggestion is that all such constants should finally be as calculable as $\pi$, that's indeed a classic goal of theoretical physics, one that string theory has the capacity to resolve because everything in string theory is calculable. At the current state of progress, however, there exists a very high - but finite or countable - set of possible vacua where the results of the calculations are different. That's why some people defend the anthropic reasoning.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-26T03:53:48.887" CommentCount="1" />
  <row Id="10406" PostTypeId="2" ParentId="10392" CreationDate="2011-05-26T04:02:49.197" Score="3" ViewCount="" Body="&lt;p&gt;Dear D-brane, yes, in principle, one could have theories that spontaneously break the Lorentz symmetry. Just add a vector field (or another non-scalar field) and some potential of the form&#xA;$$(V_\mu V^\mu - v^2)^2 $$&#xA;which will drive $V_\mu$ towards a vector of the right length, i.e. $(v,0,0,0)$ which would pick a preferred reference frame at each point much like the Higgs boson picks the &quot;right&quot; direction in the space of doublets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lorentz symmetry has been checked at an incredible accuracy, so because the breaking above would break the Lorentz symmetry and introduce Lorentz-breaking coefficients for &lt;em&gt;all&lt;/em&gt; terms in physics, it is not just one fine-tuning. It would be a collection of dozens of fine-tuned quantities - for each term in the Lagrangian that we can measure, there would be at least one fine-tuned quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can say, this is clearly excluded, at least for any microscopic value of $v$.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-26T04:02:49.197" CommentCount="1" />
  <row Id="10407" PostTypeId="2" ParentId="10398" CreationDate="2011-05-26T04:17:23.977" Score="3" ViewCount="" Body="&lt;p&gt;Yes, we can imagine a universe without energy conservation.  Just imagine that there's a dewar that's completely isolated from everything else, and there's an ideal gas inside.  The gas suddenly gets hotter without any chemical change.  Or imagine that all the electrons in the universe are gaining mass.  Or imagine that there's an Energizer Bunny that never, ever stops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More simply, imagine a universe with two states - one high energy and one low energy.  If it ever changes states, then energy is not conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These situations really do violate energy conservation.  They would not be explained by bringing in some previously-unknown source of energy unless that source had some other, verifiable physical meaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing a hypothetical perpetual motion machine to the neutrino is specious.  The neutrino, a single particle, explained conservation of momentum, energy, and angular momentum in the reactions where it was involved - not energy alone.  Still, people were indeed skeptical about neutrinos until they were &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutrino#Direct_detection_from_induced_beta_decay&quot; rel=&quot;nofollow&quot;&gt;directly detected&lt;/a&gt;.  So the science behind neutrinos is not similar to a magical energy repository that exists simply to feed a perpetual motion machine.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a perpetual motion machine were discovered, scientists would work very hard to find out where the energy is coming from.  There's a good reason for that, which is that the laws behind everyday things are known, and those laws conserve energy, so most likely the perpetual motion machine really is drawing energy from some already-known source.  If we ever discovered a true perpetual motion machine, though, we would not invent a new, unknown source of energy to add on to the universe to explain it.  We would can the conservation of energy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something similar happened when it was discovered that in certain circumstances, the universe &lt;a href=&quot;http://en.wikipedia.org/wiki/CP_violation&quot; rel=&quot;nofollow&quot;&gt;violates CP symmetry&lt;/a&gt;.  Scientists did not invent otherwise-undetectable particles or in some other way invoke a deus ex machina to try to patch the situation up.  It is simply accepted that the universe violates CP symmetry.  Similarly, if we found energy is not conserved, we would have to acknowledge that time-translational symmetry is not perfectly true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of energy conservation is to make useful predictions and help our actual understanding of the universe.  Beginning with high school physics and continuing on through all the electromagnetism, quantum mechanics, analytical mechanics, relativity, and especially statistical mechanics that I've learned, energy conservation is an incredibly useful tool.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, it and other conservation laws provide constraints.  If we know the state of a system now but don't know all the physics of that system, we might still at least calculate its energy.  Then, either assuming the system is isolated or finding all the energy that goes in and out of it (all of which is transferred by known physical mechanisms), we know what the energy of the system will be in the future, even if we don't know exactly what the state will be.  That's useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If energy could simply materialize in the system, then we could say &quot;no worries; it just got some energy from the energy gods, who by the way are an established part of the scientific universe&quot;, but we'd be fooling ourselves because we'd have sacrificed the predictive power we used to have.  I can't think of any case where energy is used in that way.  Each time we claim energy enters of exits a system, it does it through a known mechanism that has a physical meaning.  In the case of neutrinos, we could detect the neutrino leaving the system (with a certain probability).  In the case of electromagnetic radiation leaving the system, we can detect  that.  In the case of the system losing thermal energy by conduction, we can detect the temperature change in the surrounding environment, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, it is not correct to say that we sacrifice Occam's razor to preserve energy conservation.  In fact the opposite is true - positing the existence of the neutrino is a simple explanation for many different observed effects, not just the loss of energy when you ignore neutrinos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not knowledgeable enough to speak to your concerns with GR, so I'll leave that to the more experienced users.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-05-26T04:17:23.977" CommentCount="5" />
  <row Id="10408" PostTypeId="2" ParentId="10398" CreationDate="2011-05-26T05:34:40.163" Score="1" ViewCount="" Body="&lt;p&gt;I will address the physical intuition, and I will accept as a premise that Neother's theorem is not necessary for a conserved quantity to exist, which is what you are saying. Sufficient but not necessary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will remold your question and answer  as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Why is global conservation of mass not considered a tautology?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass is conserved in our everyday life, and even in classical microsystems. It is only when we enter the realm of atomic that the anti-intuitive surprise of Quantum mechanical systems catches up with physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the butcher to the baker/ to the candlestick maker, we knew mass was conserved. Weight is money and money is a serious question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happened with quantum mechanics? Mass became connected with momentum and energy in an intimate way, and our strong intuition that mass is conserved gave way to the observation, that this was not always true. And we fell back to mathematics to bring order out of chaos, hence the essential for logic Noether's theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no simple way, going back to the intent of  your real question, of simply adding up mass, (as we try to do when we measure the smoke from the candle and still find mass missing),  so as to conserve it by enlarging the universe. We needed to develop an intuition for  quantum mechanics and relativity.  The underlying structure of the universe revealed that conservation of mass is a concept very useful for human life, but applicable  in specific four dimensional coordinate systems only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true for conservation of baryon number (intimately connected with conservation of mass), as was discussed in another question on this site. When energies become large enough to turn everything to a quark gluon plasma, baryon number conservation becomes meaningless. We could define a  quark number that would describe the plasma but it would have little meaning as most of the energy would be in the sea and the effective mass of the plasma would have little to do with the number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus people were telling you, despite Feynman's interesting quote in your previous answer, that in a similar way, energy conservation can be well defined and very useful in various coordinate systems but cannot be a general law  of the General Relativity universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps on the intuition front.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-26T05:34:40.163" CommentCount="6" />
  <row Id="10409" PostTypeId="2" ParentId="10400" CreationDate="2011-05-26T05:36:00.680" Score="4" ViewCount="" Body="&lt;p&gt;Actually, you're right: the Hubble constant is not really constant. At least, it's not constant in &lt;em&gt;time&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason it's called a constant is that, when Edwin Hubble originally compared the recession velocities of galaxies with their distances in 1929, there was no reason to expect any particular pattern. After all, just a few years prior, people had thought there were no other galaxies. But what Hubble found was that, except for a small amount of random variation, the velocities of galaxies were proportional to their distances; in other words, the ratio $v/d$ was roughly the same for all galaxies he observed. The value of this ratio came to be known as Hubble's constant, $H_0 \equiv \frac{v}{d}$, because it was constant from one galaxy to the next, rather than varying randomly as one might have guessed at the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it wasn't long before people realized that if the recessional velocity of each galaxy was proportional to its distance, you could extrapolate back to some point in the past at which $d = 0$: all the galaxies would have started out in the same place. This gives you an effective age for the universe. If you use a simple linear extrapolation, from basic kinematics you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t = \frac{d}{v} = \frac{1}{H_0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the Hubble &quot;constant&quot; is not constant in time, but rather is inversely related to the age of the universe. As the universe gets older, the Hubble constant gets smaller, as you would expect. This happens because the distance $d$ to any given galaxy increases with time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the fact that the universe has an age doesn't create an absolute time. Sure, different observers at different points in spacetime will measure different values for the age of the universe. And sure, you could &lt;em&gt;define&lt;/em&gt; a time coordinate system by specifying that the time coordinate for any observer is the age of the universe as measured by that observer. This is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Comoving_distance&quot; rel=&quot;nofollow&quot;&gt;comoving time&lt;/a&gt;, and it is a useful and sensible way to set up a coordinate system in time. But it is not the only possibility, and there is certainly nothing so special about it that it deserves to be called &quot;absolute.&quot; Any observer who is moving with respect to the universe as a whole (i.e. relative to the Hubble flow) would not measure time at the same rate as this comoving time.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-05-26T05:36:00.680" CommentCount="3" />
  <row Id="10410" PostTypeId="1" CreationDate="2011-05-26T06:06:38.337" Score="6" ViewCount="285" Body="&lt;p&gt;If I stand exactly in front of a colorful wall, I imagine the light waves they emit, and they receive should randomly double or erase out each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as a result, I imagine I should see a weird combination of colors, or a full-black/full-white/very lightly perception of the wall, when all the light waves that the wall receives and emits cancel out each other or double each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why doesn't that actually happen? Any time I look into a wall, I never see the wall &quot;cancel out&quot; of my perception. Same for radio waves. Shouldn't radio waves not work at all? There are so many sources where they could reflect and cancel out or annoy each other...&lt;/p&gt;&#xA;" OwnerUserId="3737" LastActivityDate="2011-07-12T16:10:48.280" Title="Why don't waves erase out each other when looking onto a wall?" Tags="&lt;electromagnetism&gt;&lt;waves&gt;" AnswerCount="5" CommentCount="7" FavoriteCount="1" />
  <row Id="10411" PostTypeId="2" ParentId="10392" CreationDate="2011-05-26T06:17:40.320" Score="1" ViewCount="" Body="&lt;p&gt;It's easy to write down such models, but they don't describe our universe. We can study models which don't describe our universe, but we have to make clear they are not about our universe.&lt;/p&gt;&#xA;" OwnerUserId="3738" LastActivityDate="2011-05-26T06:17:40.320" />
  <row Id="10412" PostTypeId="2" ParentId="10391" CreationDate="2011-05-26T06:22:21.283" Score="1" ViewCount="" Body="&lt;p&gt;A coherent state isn't the same thing as a squeezed state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a|z\rangle = z|z\rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a coherent state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\alpha x + i\beta p)|z\rangle = z| z \rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a squeezed state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For both, the product of uncertainties is minimized, but the former is more related to the Hamiltonian.&lt;/p&gt;&#xA;" OwnerUserId="3739" LastActivityDate="2011-05-26T06:22:21.283" />
  <row Id="10413" PostTypeId="1" CreationDate="2011-05-26T06:54:55.190" Score="1" ViewCount="65" Body="&lt;p&gt;Newton's work was related to Enlightenment philosophy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Einstein was influenced by Mach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The founders of quantum mechanics had strong philosophical opinions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the role of metaphysics in the development of physics?&lt;/p&gt;&#xA;" OwnerUserId="3740" LastActivityDate="2011-05-26T06:54:55.190" ClosedDate="2011-05-26T08:03:52.960" Title="What is the role of metaphysics in physics?" Tags="&lt;soft-question&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="10414" PostTypeId="2" ParentId="10410" CreationDate="2011-05-26T07:55:18.920" Score="4" ViewCount="" Body="&lt;p&gt;1) First let us separate colour perception from frequency. Individual frequencies have a color correspondence but the colour the human eye perceives &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/vision/colper.html&quot; rel=&quot;nofollow&quot;&gt;is another story&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) White light, such as sunlight, is composed of many frequencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the impinging wave hits a wall it can be&#xA;a) reflected&#xA;b) absorbed&#xA;c) scattered incoherently&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for the light waves to &lt;em&gt;cancel out each other or double each other&lt;/em&gt; the photons have to be, within the uncertainty principle, superimposed in time and space. Sometimes it happens, but the probability is small. That is one of the reasons why a reflected beam can never have the same strength as its original beam. If the frequency is the same the probability will be higher than if the frequencies come from a random palette.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This superposition can be achieved with lasers, where there is control of frequency and the beam is coherent, i.e. the phases are preserved upon reflection. A hologram is an example of superposition of same-frequency light to create a three dimensional shape by peaks and dips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: From a disappeared question the following comment is worth adding:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You can perceive all colours even if only two frequencies are shining on an object. Also in this decade, &lt;a href=&quot;http://en.wikipedia.org/wiki/Edwin_H._Land&quot; rel=&quot;nofollow&quot;&gt;Land&lt;/a&gt; first discovered a two-color system for projecting the entire spectrum of hues with only two colors of projecting light (he later found more specifically that one could achieve the same effect using very narrow bands of 500 nm and 557 nm light).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="4283" LastEditorDisplayName="" LastEditDate="2011-07-12T16:10:48.280" LastActivityDate="2011-07-12T16:10:48.280" CommentCount="3" />
  <row Id="10415" PostTypeId="1" CreationDate="2011-05-26T08:08:57.123" Score="3" ViewCount="118" Body="&lt;p&gt;This question is about Godel's theorem, continuity of reality and the Luvenheim-Skolem theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that all leading physical theories assume reality is continuous. These are my questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Is reality still continuous according to string theory? what is the meaning of continuous if there is a finite size for a string? (My understanding of string theory is very limited :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) According to the Luvenheim-Skolem theorem, any enumerable first order logic theory (physics?) has an enumerable model. This means we must be able to have a discrete theory of physics which is as good as the continuous ones. Is this accurate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Why haven't we found one? I've heard some people say that Heisenberg's formulation of quantum mechanics doesn't speak of anything continuous... is this accurate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) If we can find one / have found one, does this mean that our understanding of physical laws can always be reduced into a Turing machine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) If we can find one / have found one, how can we explain the use of real constants in physics? (such as Pi)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) Does Godel's theorem mean that we could always &quot;notice&quot; that some things are true but not be able to prove them from within the universe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-05-31T22:54:50.733" LastActivityDate="2011-05-31T22:54:50.733" Title="Can we have non continuous models of reality? Why don't we have them?" Tags="&lt;quantum-mechanics&gt;&lt;string-theory&gt;&lt;continuum-mechanics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10416" PostTypeId="2" ParentId="10415" CreationDate="2011-05-26T08:17:27.537" Score="-1" ViewCount="" Body="&lt;p&gt;Physics isn't first order logic!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you want to do is write down an axiomatization of physics using first order logic. If you do that, even the real numbers will need to be encoded using first order logic, say ZFC. The nonstandard model you are looking for requires the axiom of choice. Now, we are really far out from the territory of physics. with the axiom of choice, don't expect computational decidability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question has nothing to do with physics.&lt;/p&gt;&#xA;" OwnerUserId="3741" LastActivityDate="2011-05-26T08:17:27.537" CommentCount="2" />
  <row Id="10417" PostTypeId="2" ParentId="10398" CreationDate="2011-05-26T09:23:21.087" Score="2" ViewCount="" Body="&lt;p&gt;If you observe that energy is not conserved as in the perpetual motion example you give, you cannot just proclaim that there is somewhere else that the energy comes from. That would provide an empty concept of no value. You must find a dynamical system such as a new type of particle or wave and define energy as a function of the variables of the new dynamical entities in such a way that it accounts for any missing or excess energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give an example, binary pulsars are observed to be losing energy when you add up the known forms of energy within the system. This is resolved by showing that the energy in radiated gravitational waves predicted by general relativity can precisely account for the missing energy. This is verified experimentally and will be confirmed more directly when gravitational waves are detected and found to transfer energy back to matter in accordance with the predictions. If they don't then energy conservation will be in trouble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noether's Theorem merely gives a way of showing that energy conservation laws of this sort exist when the dynamical laws are derived from a principle of least action that has time invariance. Einstein first formulated a correct law of energy conservation for gravity without using Noether's theorem and his result is still correct, although there was a long history of arguments before the matter was finally agreed. GR has time translation invariance as a subgroup of diffeomorphism invariance when you treat the gravitational field as a dynamical entity itself. Noether's theorem applies in this case and energy conservation can be derived in that way too. &lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-05-26T09:23:21.087" CommentCount="0" />
  <row Id="10418" PostTypeId="2" ParentId="10410" CreationDate="2011-05-26T09:26:03.800" Score="1" ViewCount="" Body="&lt;p&gt;I will like to point out that the walls do not emit any light but instead they reflect what shines on them. You may also be interested in reading more about &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffuse_reflection&quot; rel=&quot;nofollow&quot;&gt;diffuse reflection&lt;/a&gt; which roughly and non mathematically attempts at explaining some of the physics at work here. If you are looking for a more mathematical treatment, &lt;a href=&quot;http://www.physics.buffalo.edu/phy207/lc/lc37.pdf&quot; rel=&quot;nofollow&quot;&gt;this buffalo lecture on interference&lt;/a&gt; may help you --esp. the coherence bit.&lt;/p&gt;&#xA;" OwnerUserId="3669" LastEditorUserId="3669" LastEditorDisplayName="" LastEditDate="2011-05-26T14:09:55.207" LastActivityDate="2011-05-26T14:09:55.207" CommentCount="2" />
  <row Id="10419" PostTypeId="1" AcceptedAnswerId="10424" CreationDate="2011-05-26T10:07:49.157" Score="0" ViewCount="41" Body="&lt;p&gt;When I perform parametric modeling, if there is significant multicollinearity between variables I think should be independent, but in fact are not, I run into the case where one or more of the coefficients becomes exceeding small (or large) relative to the others.  How is that different than what occurs in fine tuning problems of the standard model?&lt;/p&gt;&#xA;" OwnerUserId="3460" LastActivityDate="2011-05-26T12:03:23.963" Title="Fine tuning and parametric modelings" Tags="&lt;standard-model&gt;" AnswerCount="1" />
  <row Id="10420" PostTypeId="1" AcceptedAnswerId="10472" CreationDate="2011-05-26T10:32:39.520" Score="3" ViewCount="215" Body="&lt;p&gt;Can someone provide an expanded explanation on the statement that the Higgs field is already 3/4 detected?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.staff.science.uu.nl/~hooft101/&quot; rel=&quot;nofollow&quot;&gt;Link to ref&lt;/a&gt; (@nic, sorry I left it off, do a quick search on Higgs to find the right location on page)&lt;/p&gt;&#xA;" OwnerUserId="3460" LastEditorUserId="3460" LastEditorDisplayName="" LastEditDate="2011-05-26T22:32:07.857" LastActivityDate="2011-05-27T04:38:55.693" Title="Is the Higgs 3/4 detected already? [Updated]" Tags="&lt;higgs-boson&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="10422" PostTypeId="2" ParentId="10398" CreationDate="2011-05-26T11:39:39.950" Score="1" ViewCount="" Body="&lt;p&gt;I'm sympathetic to this Question. A historical way to approach it is to note the mutations of Poincar&amp;eacute;'s conventionalism, from which I take the general point that can be extracted to be that the experimental evidence fits into some conceptual systems better than into others. If we take that better fit to mean something, we come to the usual way we do Physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might consider the SEP entry on convention, Section 1.2 particularly, &lt;a href=&quot;http://plato.stanford.edu/entries/convention/&quot; rel=&quot;nofollow&quot;&gt;http://plato.stanford.edu/entries/convention/&lt;/a&gt;. You could certainly find other resources on Conventionalism, including &lt;a href=&quot;http://philsci-archive.pitt.edu/&quot; rel=&quot;nofollow&quot;&gt;http://philsci-archive.pitt.edu/&lt;/a&gt; for preprints on Philosophy, if you don't mind risking never again getting an upvote here if you don't keep your head on straight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I take the description of GR given by Jerry Schirmer partly to cede your point. On the cosmological and gravitational scales there is no global concept of energy because the models are not translation invariant. To make the models translation invariant would require us to introduce a flat background coordinate system, which I think we could do, but we would have to introduce endless epicycles to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cosmological and gravitational scales, however, are not the only component of your Question. You also introduce the somewhat epicyclic nature of the introduction of new particle types such as neutrinos to preserve translation and Lorentz invariance and hence energy and angular momentum conservation. Here I think the answer can only lie in detailed modeling. We &lt;em&gt;might&lt;/em&gt; be able to introduce a model that is not translation and Lorentz invariant on meter or nanometer scales that justifies itself by a need for far fewer quantum field types, but that model has to be exhibited explicitly and its correspondence to the experimental results that are taken to support the current Standard Model of Particle Physics cashed out in detail. Remember the Correspondence Principle? For most working Physicists, put up or shut up is a reasonable response to this, because such conceptual musings will more often put you in a graveyard than not. As you've discovered, some ways of talking about these issues can get you bullied, your current Question worked better, but it's always nerve-wracking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's always difficult to know which concepts are getting in the way of progress and which we should regard as inviolate. That, FWIW, is my view of the Quine-Duhem thesis in a nutshell (somewhat curiously mixed up with other arguments in the SEP, &lt;a href=&quot;http://plato.stanford.edu/entries/mathphil-indis/&quot; rel=&quot;nofollow&quot;&gt;http://plato.stanford.edu/entries/mathphil-indis/&lt;/a&gt;). People sometimes superficially behave as if their favorite concept to drop or modify is obviously the only alternative, although conversation with them will usually see a relatively more subtle appreciation of other people's ideas.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-05-26T11:39:39.950" />
  <row Id="10423" PostTypeId="2" ParentId="9995" CreationDate="2011-05-26T11:56:46.827" Score="0" ViewCount="" Body="&lt;p&gt;I meanwhile found the constant by intensive googling (for example in &lt;a href=&quot;http://www.cse.scitech.ac.uk/ceg/papers/Gu_et_al_Nanoscale_Microscale_Thermophysical_Eng.pdf&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;). It is (depending on the source) around 144 K.&lt;/p&gt;&#xA;" OwnerUserId="2480" LastEditorUserId="2480" LastEditorDisplayName="" LastEditDate="2011-05-27T09:54:13.110" LastActivityDate="2011-05-27T09:54:13.110" CommentCount="0" />
  <row Id="10424" PostTypeId="2" ParentId="10419" CreationDate="2011-05-26T12:03:23.963" Score="0" ViewCount="" Body="&lt;p&gt;The difference is, that if a quantity in your model is secretly theoretically zero, in practice, you will typically still get a small non-zero number, whose size is determined by the scale of uncertainties in measurements, rounding errors, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Naturalness_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;naturalness&lt;/a&gt; problem in fundamental physics, we typically have a dimensionless quantity, whose measured value is significantly bigger than the error-bars (and therefore a non-zero quantity!) but much much smaller than $1$. &lt;/p&gt;&#xA;" OwnerUserId="2451" LastActivityDate="2011-05-26T12:03:23.963" CommentCount="1" />
  <row Id="10425" PostTypeId="2" ParentId="10420" CreationDate="2011-05-26T13:24:34.017" Score="-2" ViewCount="" Body="&lt;p&gt;Yes, it means that 3/4 of 1 data has already been collected. This 3/4 is a theoretical background. The remaining 1/4 is the experimental evidences (not collected yet).&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-05-26T13:24:34.017" />
  <row Id="10426" PostTypeId="1" CreationDate="2011-05-26T13:26:53.763" Score="4" ViewCount="137" Body="&lt;p&gt;The operators $J_1^2$, $J_2^2$, $J_{1z}$, and $J_{2z}$ are mutually commuting operators. Likewise, $J_1^2$, $J_2^2$, $J^2$, and $J_z$ are mutually commuting operators.  The two groups are incompatible, and the simultaneous eigenkets along with their eigenvalues are given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${J_1}^2 \left|j_1,j_2;m_1,m_2\right&amp;gt; = j_1 \left(j_1+1\right) \hbar^2 \left|j_1,j_2;m_1,m_2\right&amp;gt;$&#xA;${J_2}^2 \left|j_1,j_2;m_1,m_2\right&amp;gt; = j_2 \left(j_2+1\right) \hbar^2 \left|j_1,j_2;m_1,m_2\right&amp;gt;$&#xA;$J_{1z} \left|j_1,j_2;m_1,m_2\right&amp;gt; = m_1 \hbar \left|j_1,j_2;m_1,m_2\right&amp;gt;$&lt;br&gt;&#xA;$J_{2z} \left|j_1,j_2;m_1,m_2\right&amp;gt; = m_2 \hbar \left|j_1,j_2;m_1,m_2\right&amp;gt;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${J_1}^2 \left|j_1,j_2;j,m\right&amp;gt; = j_1 \left(j_1+1\right) \hbar^2 \left|j_1,j_2;j,m\right&amp;gt;$&#xA;${J_2}^2 \left|j_1,j_2;j,m\right&amp;gt; = j_2 \left(j_2+1\right) \hbar^2 \left|j_1,j_2;j,m\right&amp;gt;$&lt;br&gt;&#xA;$J^2 \left|j_1,j_2;j,m\right&amp;gt; = j \left(j+1\right) \hbar^2 \left|j_1,j_2;j,m\right&amp;gt;$&lt;br&gt;&#xA;$J_z \left|j_1,j_2;j,m\right&amp;gt; = m \hbar \left|j_1,j_2;j,m\right&amp;gt;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read that each set of eigenkets are mutually orthogonal &lt;a href=&quot;http://farside.ph.utexas.edu/teaching/qm/lectures/node47.html&quot; rel=&quot;nofollow&quot;&gt;[1]&lt;/a&gt; (for eigenkets corresponding to different sets of eigenvalues).  This is what I don't understand.  In principle it makes sense, but when I plug in numbers I don't get zero for the inner product.  For example take the first eigenket: $\left|j_1,j_2;m_1,m_2\right&amp;gt;$.  If I choose different eigenvalues for this eigenket (e.g. let $j_1 = 0$ and then let $j_1 = 1$) I get the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $j_1 = 0$ I can have:&lt;br&gt;&#xA;$\left|0,j_2;0,m_2\right&amp;gt;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $j_1 = 1$ I can have any of the following, since $\left|m_1\right| \leq j_1$:&lt;br&gt;&#xA;$\left|1,j_2;-1,m_2\right&amp;gt;$&lt;br&gt;&#xA;$\left|1,j_2;0,m_2\right&amp;gt;$&lt;br&gt;&#xA;$\left|1,j_2;1,m_2\right&amp;gt;$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I take the inner-product of the $j_1 = 0$ eigenket with any of the $j_1 = 1$ eigenkets I do not get zero, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left&amp;lt;0,j_2;0,m_2 \mid 1,j_2;-1,m_2\right&amp;gt; = {j_2}^2+{m_2}^2$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is non-zero unless $j_2 = 0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I misunderstanding here? How do you show that eigenkets with different eigenvalues are orthogonal?&lt;/p&gt;&#xA;" OwnerUserId="3597" LastActivityDate="2011-05-26T17:09:32.967" Title="Why are two eigen-state-kets with different eigenvalues orthogonal?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="10427" PostTypeId="2" ParentId="10426" CreationDate="2011-05-26T14:05:20.883" Score="5" ViewCount="" Body="&lt;p&gt;Eigenvalues of a Hermitian operator or their set - such as components of $\vec J$ and/or $J^2$ - corresponding to different eigenvalues are always orthogonal to each other because&#xA;$$\langle \psi | M | \phi \rangle = m_\psi \langle \psi | \phi \rangle = m_\phi \langle \psi |  \phi \rangle $$&#xA;I could get any eigenvalue $m_\phi$ or $m_\psi$ by acting with $M$ on the two sides. Because those two things are equal, we have&#xA;$$ (m_\psi-m_\phi) \langle \psi  | \phi \rangle = 0 $$&#xA;which implies - because the eigenvalues differ&#xA;$$\langle \psi |  \phi \rangle = 0$$&#xA;Your derivation of a nonzero inner product is incorrect and you couldn't have derived it from the formulae above your final result because none of them contains any information about the inner product - in fact, they contain no bra vector whatsoever, so your reversal of a ket vector and its interpretation of an inner product was clearly some beginner's misunderstanding of what the symbols mean.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-26T14:05:20.883" CommentCount="2" />
  <row Id="10428" PostTypeId="2" ParentId="10426" CreationDate="2011-05-26T15:25:15.237" Score="3" ViewCount="" Body="&lt;p&gt;your line $\left&amp;lt;0,j_2;0,m_2 \mid 1,j_2;-1,m_2\right&amp;gt; = {j_2}^2+{m_2}^2$ is making an assumption about what an inner product must look like. A technical way to express your problem is to note that the sequence of eigenvalues identify an element in a &lt;em&gt;tensor product&lt;/em&gt; of vector spaces, &lt;em&gt;not&lt;/em&gt; a direct sum. If it was a direct sum, your sum of squares would be right, but it isn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luboš's Answer is completely right, enough so that I upvoted it because he needs all the rep he can get, but it looks like it needs you to know what you're doing already for you to understand it. When you use objects like $\left|j_1,j_2;m_1,m_2\right&amp;gt;$ to represent a state, you implicitly claim that the operators that have the eigenvalues $j_1,j_2;m_1,m_2$ are self-adjoint and mutually commutative. In elementary terms, we can take this to &lt;em&gt;define&lt;/em&gt; the inner product on the Hilbert space. We don't know what the inner product is until we've defined it. If the eigenvalues are different, the inner product is defined to be $0$, if the eigenvalues are the same, the inner product is defined to be $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Luboš says, all the lines above your first introduction of a bra are only about operators acting on a vector space, you haven't got a Hilbert space until you've defined an inner product (and more than that, closure in the norm). Once you've defined an inner product, $(\left|U\right&amp;gt;,\left|V\right&amp;gt;)$, you can &lt;em&gt;define&lt;/em&gt; a bra as the object that acts on a vector to get this value, $\left&amp;lt;U \mid V\right&amp;gt;=(\left|U\right&amp;gt;,\left|V\right&amp;gt;)$. There's a theorem that says we can do this if we're properly careful, &lt;a href=&quot;http://en.wikipedia.org/wiki/Riesz_representation_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Riesz_representation_theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-05-26T15:25:15.237" CommentCount="6" />
  <row Id="10429" PostTypeId="1" AcceptedAnswerId="10432" CreationDate="2011-05-26T15:30:29.350" Score="1" ViewCount="117" Body="&lt;p&gt;This is my first question here. Hope it not too dull to you guys :p&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;http://www.youtube.com/watch?v=DfPeprQ7oGc&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt; on youtube. After 3:24 it says if there is an observer detecting which slit the electron goes through, there won't be an interference pattern. Has there ever been any experiment (not thought experiment) demonstrating this phenomenon really using microscopes or other detectors to see where the electron goes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My physics teacher don't know the answer, but he took an alternative route, said if we suddenly close one of the slits, we know the electron can only go through the other, and it turns out no interference pattern shown. It sounds convincing to me and the experiment is much easier to perform. But if one can really shine a light and see the electron, he or she can alter the intensity or energy of the light to find a critical value below which the interference pattern won't be destroyed. Sounds even cooler :)&lt;/p&gt;&#xA;" OwnerUserId="3743" LastEditorUserId="238" LastEditorDisplayName="" LastEditDate="2011-05-26T18:51:38.037" LastActivityDate="2011-05-26T19:05:05.650" Title="Double slit experiment - how to see an electron going through a slit?" Tags="&lt;quantum-mechanics&gt;&lt;double-slit-experiment&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10430" PostTypeId="2" ParentId="8895" CreationDate="2011-05-26T15:42:04.750" Score="0" ViewCount="" Body="&lt;p&gt;See Tipler, The physics of immortality. There you can read an approach.&lt;/p&gt;&#xA;" OwnerUserId="3744" LastActivityDate="2011-05-26T15:42:04.750" />
  <row Id="10431" PostTypeId="2" ParentId="10403" CreationDate="2011-05-26T15:43:51.133" Score="-2" ViewCount="" Body="&lt;p&gt;I'm not sure about color charges, but in real colors, red, green, and blue aren't orthogonal, but they are linearly independent, so they can be used to span $\mathbb{R}^3$ (i.e. every color can be written as a linear combination of $R, G, B$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps a diversion to $L^*a^*b^*$ color space will help.  In this color space, the color difference between two colors is $\Delta E = \sqrt{\Delta L^2 + \Delta a^2 + \Delta b^2}$.  One $\Delta E$ is supposed to be the minimum (human) perceptible color change, anywhere in color space.  Actually they found later that this wasn't quite right, and you can move about $4 \Delta E$ for yellow or about $0.5 \Delta E$ for gray before you see a color change.  So they came up with the 2000 formula for $\Delta E$, which linearizes the color difference in $L^*a^*b^*$ space.  But let's see what red, green, and blue look like in this color space:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Red:  (L^*,a^*,b^*) = (47, 66, 50)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Green:  (L^*,a^*,b^*) = (49, -66, 33)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Blue:  (L^*,a^*,b^*) = (20, 25, -48)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is for a print example.  The actual $(L^*,a^*,b^*)$ value for red, green, and blue depends on the media or device.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you can see that the colors are not orthogonal.  For example, the dot product of red and green is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Red \cdot Green = -403$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a perceptual sense, all colors have some brightness ($L^*$), so when you add them together, the $L^*$ increases, making the resulting color brighter.  This is what happens on your computer monitor.  To get a neutral color, you want $a^* = b^* = 0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $L^*$ is always positive, so while mathematically, you can span $\mathbb{R}^3$, true colors (whether in $L^*a^*b^*$ or RGB space) only span the &quot;upper half&quot; of $\mathbb{R}^3$.  The only physical way to get a &quot;zero&quot; color is with $(L^*,a^*,b^*) = (0, 0, 0)$ or $(R, G, B) = (0, 0, 0)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in terms of real colors, colorless means that $R \approx G \approx B$ (or $a^* = b^* = 0$).  Color charges may be combined using $R, G, B$ as basis vectors, but those basis vectors are not orthogonal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the other question, I get the sense that the connection to real color is somewhat of a stretch.&lt;/p&gt;&#xA;" OwnerUserId="3697" LastActivityDate="2011-05-26T15:43:51.133" CommentCount="3" />
  <row Id="10432" PostTypeId="2" ParentId="10429" CreationDate="2011-05-26T15:53:53.240" Score="1" ViewCount="" Body="&lt;p&gt;This or a similar video was topic here already: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/783/is-dr-quantums-double-slit-experiment-video-scientifically-accurate&quot;&gt;Is Dr Quantum&amp;#39;s Double Slit Experiment video scientifically accurate?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This: &quot;&quot;But if one can really shine a light and see the electron, &quot;&quot; is the problem, &#xA;in practice one had to &quot;shoot&quot; electrons at the electrons to &quot;see&quot; them. Or on could use &#xA;gamma rays (Compton effect), but always this would disturb the experiment. &#xA;Ther is no &quot;eye&quot; in the quantum world to &quot;watch&quot; those balls without interaction. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-05-26T15:53:53.240" CommentCount="1" />
  <row Id="10433" PostTypeId="1" AcceptedAnswerId="10447" CreationDate="2011-05-26T15:58:46.490" Score="9" ViewCount="336" Body="&lt;p&gt;It's &lt;a href=&quot;http://www.sciencedaily.com/releases/2011/05/110525131707.htm&quot; rel=&quot;nofollow&quot;&gt;announced&lt;/a&gt; that researchers at Imperial College London has found that the electron is almost a perfect sphere. The popular articles all have a nice photo of a billiard ball, etc. It is reported that they found this by measuring the &quot;wobble&quot; as the electron spins (and finding none).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does it exactly mean that the electron is a sphere? Is it the wave function that is spherical? Measuring the spin wobble brings to mind a solid object, which I think the electron surely isn't. So, what would have been wobbling if it didn't have perfect spherical symmetry?&lt;/p&gt;&#xA;" OwnerUserId="852" LastActivityDate="2011-05-28T03:59:53.757" Title="What does it mean to say that the electron is a near-perfect sphere?" Tags="&lt;soft-question&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="10434" PostTypeId="2" ParentId="10426" CreationDate="2011-05-26T16:01:13.227" Score="5" ViewCount="" Body="&lt;p&gt;The error is most likely that you are using &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left&amp;lt;j_1,j_2;m_1,m_2 \mid j&amp;#39;_1,j&amp;#39;_2;m&amp;#39;_1,m&amp;#39;_2\right&amp;gt; = j_1j&amp;#39;_1+j_2j&amp;#39;_2+m_1m&amp;#39;_1+m_2m&amp;#39;_2,&#xD;&#xA;\quad \mathrm{(Wrong!)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where you should be using &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left&amp;lt;j_1,j_2;m_1,m_2 \mid j&amp;#39;_1,j&amp;#39;_2;m&amp;#39;_1,m&amp;#39;_2\right&amp;gt; = \delta_{j_1,j&amp;#39;_1}\delta_{j_2,j&amp;#39;_2}\delta_{m_1,m&amp;#39;_1}\delta_{m_2,m&amp;#39;_2},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\delta_{k,\ell}$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kronecker_delta&quot; rel=&quot;nofollow&quot;&gt;Kronecker delta function.&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the $j$'s and $m$'s are not the coefficients $v^i$ of a vector $\vec{v}=\sum_i v^i \vec{e}_i$ in a Hilbert space, where $\vec{e}_i$ is an orthonormal basis, so that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left&amp;lt;\vec{v}\mid\vec{v}&amp;#39;\right&amp;gt; =  \sum_i (v^i)^*v&amp;#39;^i, \qquad \left&amp;lt;\vec{e}_i\mid\vec{e}_{i&amp;#39;}\right&amp;gt; = \delta_{i,i&amp;#39;}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather, the $j$'s and $m$'s correspond to the $i$-labels of the basis $\vec{e}_i$. For brevity, we often write $\left&amp;lt; i \mid i&amp;#39;\right&amp;gt;$ instead of $\left&amp;lt;\vec{e}_i\mid\vec{e}_{i&amp;#39;}\right&amp;gt;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, to give a complete answer, let me include my comment above that it is a general property of eigenvectors for different eigenvalues of a Hermitian operator, that they are orthogonal to each other, see e.g., Lubos Motl's answer or &lt;a href=&quot;http://en.citizendium.org/wiki/Hermitian_operator&quot; rel=&quot;nofollow&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-26T17:09:32.967" LastActivityDate="2011-05-26T17:09:32.967" CommentCount="3" />
  <row Id="10435" PostTypeId="1" CreationDate="2011-05-26T10:00:39.857" Score="4" ViewCount="223" Body="&lt;p&gt;I was reading this article this morning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.bbc.co.uk/news/science-environment-13539914&quot; rel=&quot;nofollow&quot;&gt;http://www.bbc.co.uk/news/science-environment-13539914&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scientists have discovered a gamma-ray burst whose light has taken 13.14 billion years to reach Earth. This much is cool and interesting. However, the assumption is then stated that this is &quot;the most distant single object yet spied by a telescope&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But hold on a minute. It is also known that galaxies are moving away from each other at incredible speeds, most faster than the speed of light, because the space/time between them is expanding (or something like that... I'm no scientist!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So these so-called amazingly distant objects, well for starters, they don't exist any more... They are things that happened billions of years ago. But not only that, they are probably objects that were only a few million light years away from us when they actually took place. So surely then, the objects themselves aren't the most distant ones, but the light from them has been distorted such that the light has taken that long to reach us?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, if the galaxies are spreading out faster than the speed of light, who is to say this explosion actually happened 13.1 billion years ago? Isn't it possible that the light was created say 5 billion years ago, but has taken much longer to reach us because of the expanding space between the galaxies?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure this stuff has already been considered by scientists, but I find it weird the way news articles always assume that just because light travels at a specific speed, that it's always going to take the same amount of time to reach us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or am I getting it wrong? I'd love to know!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;:-Joe :)&lt;/p&gt;&#xA;" OwnerUserId="3896" LastActivityDate="2011-05-27T15:38:46.330" Title="Are scientists missing the point with distant cosmic objects, or is it just me?" Tags="&lt;astronomy&gt;" AnswerCount="3" CommentCount="11" FavoriteCount="1" />
  <row Id="10436" PostTypeId="2" ParentId="10435" CreationDate="2011-05-26T11:33:28.290" Score="2" ViewCount="" Body="&lt;p&gt;One of the reasons why scientists like to took at distant objects is because they know that, by doing so, they're looking back in time: a distant object is an object from the past. A very distant object (like the one you cited) is an object from near the &lt;a href=&quot;http://en.wikipedia.org/wiki/Age_of_the_universe&quot; rel=&quot;nofollow&quot;&gt;beginning of the universe&lt;/a&gt; (which is interesting to people who want to know about the beginning of the universe).&lt;/p&gt;&#xA;" OwnerUserId="3773" LastActivityDate="2011-05-26T11:33:28.290" CommentCount="12" />
  <row Id="10437" PostTypeId="2" ParentId="6655" CreationDate="2011-05-26T16:17:24.667" Score="0" ViewCount="" Body="&lt;p&gt;to be clear, this is NOT the sort of dynamic casimir effect i'm after in this question, but definitely worth a read: &lt;a href=&quot;http://arxiv.org/abs/1105.4714&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1105.4714&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in particular, it is interesting that they are modulating the inductance with a SQUID, which is not the same but it is quite related to the permittivity&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-05-26T17:06:34.290" LastActivityDate="2011-05-26T17:06:34.290" />
  <row Id="10438" PostTypeId="2" ParentId="10433" CreationDate="2011-05-26T16:24:24.570" Score="6" ViewCount="" Body="&lt;p&gt;Well, one strong piece of evidence for this is the fact that the electric dipole moment&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\vec D} \equiv \int d^{3}x&amp;#39; \rho({\vec r}-{\vec r_{0}})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of the electron is nearly zero.  The electric dipole moment would measure a reflection asymmetry in a potential charge distribution of the electron.  The fact that it takes on a zero means that the electron charge distribution is symmetric about all planes.  Now, it is still possible that the electron may have nonzero values for it's higher multipole moments, while having a zero electric dipole moment which would indicate non-sphericity as well, but as far as I know, there is no evidence for this either, and finding these moments would be an even harder experiment.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-05-26T16:24:24.570" CommentCount="7" />
  <row Id="10439" PostTypeId="2" ParentId="10410" CreationDate="2011-05-26T16:32:00.477" Score="2" ViewCount="" Body="&lt;p&gt;I'd like to take this in three parts, and I am going to stick to a classical description, though the quantum mechanical description is in some ways easier and just as valid.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Superposition&lt;/strong&gt; is the principle that the amplitudes due to two waves incident on the same point in space at the same time can be naively added together, but &lt;em&gt;the waves do not affect each other&lt;/em&gt;.{#}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That last bit is important enough to write again: &lt;em&gt;the waves do not affect each other&lt;/em&gt;. They can, in fact pass right through one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the purposes below we can treat every wave as being a plane wave and thus being described by an amplitude $A$, a wave number $\vec{k}$, and a phase $\phi$. The wave number encodes both the direction of propagation (that why it's a vector) &lt;em&gt;and&lt;/em&gt; the frequency (in the absolute value) $|k| = \frac{\omega}{c} = \frac{f}{2 \pi c}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This description is not complete because (a) the planar nature is only approximate and (b) even allowing for that the phase can be a function of position on the wave front; but we can safely neglect these issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at any given point in space and time we have a big jumble of different waves all added together&#xA;$$ \sum_{\text{directions}} \int d\omega \int d\phi A(\vec{k},\phi) e^{i(\vec{k}\cdot\vec{x} - \omega t + \phi)} $$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;You eye is a camera:&lt;/strong&gt; it has a small opening with a focusing element (lens) at the front and a light sensitive projection surface (retina) at the back. I sorts out incoming light waves two ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the lens focuses light to a particular point on the retina according to it's direction. That makes the sum over direction go away, because each patch on the retina sees only one direction (well, a very small range of directions). If the lens isn't the right shape you lose some of this and the image becomes blurry. Then you need glasses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly individual patches of the retina are sensitive to light of different frequency ranges{*}. So for each rod or cone, the integral over angular frequencies $\omega$ is reduced to quite moderate limits (human vision extends over roughly one octave in the electromagnetic spectrum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That still leaves us with the sum over the phases, and I am going to elide this part of the problem by saying that they are roughly constant over the small dimensions of the eye's lens and time-scale of you retinal response; if that doesn't make you happy I will refer you to genneth's comment: $\langle E^2 \rangle \neq \langle E \rangle^2$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Radios live in Fourier space:&lt;/strong&gt;{+} They typically accept signals from a very wide set of direction, so we don't get the spacial filtering that you get with the eye. However, the electronics that back them select only a very narrow band of frequencies. In essence inspecting the incoming signals in Fourier space. And the signal you are looking for has an additional periodicity (either amplitude in AM or frequency in FM) in the audible range as well, so another round of filtering is done and you generally have a dominate remaining signal. Intentional signals have constant phases, while unintentional signals (i.e. background or noise) have random phases and they tend to cancel out.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;{#} &lt;a href=&quot;http://physics.stackexchange.com/questions/10410/why-dont-waves-erase-out-each-other-when-looking-onto-a-wall/10414#10414&quot;&gt;In another answer anna talks a little bit about the conditions under which this classical statement breaks down.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{*} The mechanism by which the rods and cones respond to light in inherently quantum mechanical, the means by which those signals are propagated to the brain is inherently biochemical and the post processing done by the brain is huge topic in and of itself; all of which is beyond the cope of this discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{+} Traditional radios. Like the AM/FM job in your grandfather's car (since we're all 21th century people and have spiffy combined GPS-mp3 player widget or something). Ultra-wide band like your wireless base station is another matter.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-26T18:22:05.543" LastActivityDate="2011-05-26T18:22:05.543" CommentCount="1" />
  <row Id="10440" PostTypeId="1" AcceptedAnswerId="10441" CreationDate="2011-05-26T17:12:07.193" Score="3" ViewCount="124" Body="&lt;p&gt;Can someone explain these 3 concepts into a unified framework.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Randomness : Randomness as seen in a coin toss, where the system follows known and deterministic (at the length and scale and precision of the experiment) physics but, the complexity allows us to treat the system as random. (i.e we can always run an exact physical simulation to predict the coin's fall)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Chaotic randomness: randomness as seen in weather or fluid flow, where the system is highly sensitive to initial conditions and as a result no matter what the precision of measurement/simulation is, we cannot predict the out come. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Quantum randomness: The inherent probability distributions that we study in QM, where the electron's position probablity distribution is set by QM. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;how do the concepts of randomness in coin toss, randomness in chaos and randomness in QM fit into a unified picture. Can someone please elaborate ? &lt;/p&gt;&#xA;" OwnerUserId="3714" LastEditorUserId="3714" LastEditorDisplayName="" LastEditDate="2011-05-26T17:36:12.240" LastActivityDate="2011-05-26T18:24:11.193" Title="Randomness, Chaos, Quantum mechanical probability functions" Tags="&lt;quantum-mechanics&gt;&lt;chaos-theory&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="10441" PostTypeId="2" ParentId="10440" CreationDate="2011-05-26T18:16:43.127" Score="1" ViewCount="" Body="&lt;p&gt;You pretty much know it already. &quot;Random&quot; is a broad word that we use to mean that we can't predict behavior. Each of the three cases of randomness that you cite is unpredictable for a different reason, though - that's the difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dice are random because they are complicated, chaotic pendulums are random because we aren't good enough to measure their initial position perfectly, and quantum systems are random because they aren't deterministic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expanding on that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The randomness of a coin toss or a dice roll is based on an imprecise model. In principle, if we knew everything about the coin (its initial position, the forces applied, the density of air that slows it down, etc) then you could predict whether it winds up heads or tails with certainty. In the real world, nobody bothers, since constructing this model is very difficult. It depends sensitively on the height you are flipping the coin from, its initial spot on your thumb, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chaotic randomness is due to imprecision in initial measurements alone. Different initial conditions do not cause smooth changes in the final outcome. A good example is the chaotic motion of the planets - if we try to predict the position of Saturn in 500,000,000 years, we get a certain position based on where it is now and all of the forces acting on Saturn. But if we choose a slightly different initial condition - say, 10cm further along in its orbit to start - then we get a totally different answer potentially hundreds of thousands of km off. Then we look at an initial condition in between, 5cm further along - and the deviation is even &lt;em&gt;worse&lt;/em&gt; - it's now millions of kilometers off! In other words, it chaotic randomness arises in systems where &lt;em&gt;improving your accuracy of measurement does not help.&lt;/em&gt; The only way to get the &quot;true&quot; answer is to have the &lt;em&gt;exact&lt;/em&gt; initial value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum randomness is due to fundamental laws of nature. Quantum particles behave randomly on their own because that is just what they do, axiomatically. There is no initial measurement which could even &lt;em&gt;be&lt;/em&gt; exact. The outcome is fundamentally nondeterministic, not a limit based on our models or our measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some sense, quantum randomness guarantees that we can never &quot;beat&quot; chaos by getting a perfect initial measurement. But it arises from a fundamentally different origin. &lt;/p&gt;&#xA;" OwnerUserId="1655" LastActivityDate="2011-05-26T18:16:43.127" />
  <row Id="10442" PostTypeId="2" ParentId="10440" CreationDate="2011-05-26T18:18:57.330" Score="0" ViewCount="" Body="&lt;p&gt;You should check out these two papers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://astro.temple.edu/~powersmr/vol9no1.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;The Nature of Randomness: Part 1 -&#xA;Knowable or Unknowable?&quot; 2008,&#xA;Journal of Risk Finance, 9, 1.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://astro.temple.edu/~powersmr/vol9no2.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;The Nature of Randomness: Part 2 -&#xA;Cognitive Constraints,&quot; 2008, Journal&#xA;of Risk Finance, 9, 2.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They provide a framework for at least two of your randomness concepts (your second one, i.e. deterministic chaos, would be subsumed under the first one) and connect it with concepts from algorithmic complexity theory (Kolmogorov/Chaitin) + plus they are considerably easy to understand, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Esp. the drawn consequence is mind boggling - but you see for yourself...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Highly recommended!&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2011-05-26T18:18:57.330" />
  <row Id="10443" PostTypeId="2" ParentId="10410" CreationDate="2011-05-26T18:20:38.000" Score="3" ViewCount="" Body="&lt;p&gt;You asked two questions.  First, the light:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As has been pointed out, typically (there are exceptions) the colorful wall is not emitting anything.  Let's take blue, for an example.  White light shines on the wall.  White light is composed of many different frequencies/wavelengths, which we would perceive, if they were busted out with a prism, as many different colors.  The blue wall absorbs all these colors except the blue.  So there are not a lot of competing wavelengths coming off the wall; the blue paint allowed the wall to absorb all the competing wavelengths of light, and reflect only blue.  Another way of looking at it is that the wall is covered with a &quot;blue-reject&quot; absorbtion filter.  We get to see the one color that the wall ISN'T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, there may be many competing light sources, with random orientations and intensities, illuminating the blue wall.  Again, everything is absorbed except blue.  The blue light we see is randomly oriented and phased.  No problem.  The sum of the blue light vectors from any given point will still add up to a resultant wave that is of blue frequency.  Yes, if you planned it very, very carefully (e.g. used polarization equipment and lasers with controllable phase offsets) you could engineer some strange effects.  Normal (non-laser) light is much more random than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the radio:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radios run the same way: they receive E-M energy from many different sources of random orientation, polarization, intensity, and frequency.  The radio antenna acts as a band select element, which cuts out the vast majority of &quot;junk&quot; E-M radiation. (For instance, your car's radio antenna is tuned to pick up the waves from about 0.54MhZ (bottom end of the AM band) to 108MHz (top end of the FM band).  It isn't that great for picking up microwave oven radiation at 2450.MHz, nor the 60Hz from the AC lines which power said uWave oven.  Still, some of that E-M junk does get onto the antenna.  Modern car radios use superhetrodyning to mix all this junk with an internally generated frequency.  The result is ANOTHER spectrum of junk which is frequency shifted.  The math is done perfectly so that only the radio frequencies of interest get shifted into a very narrow slot (band select filter) that the radio then processes for you to hear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes, ESPECIALLY with AM radios, you will get fade.  Sometimes this is just because of blind spots due to terrain and atmospheric bending.  Sometimes it is one of the phenomena you described: multipath interference.  The radio wave which reaches your antenna has followed two or more separate paths, with at least one of those paths being longer than the other(s) by odd multiples (1, 3, 5, ...) of one-half a wavelength.  The waves hitting the antenna add to zero, and you hear nothing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-05-26T18:40:24.153" LastActivityDate="2011-05-26T18:40:24.153" />
  <row Id="10444" PostTypeId="2" ParentId="10440" CreationDate="2011-05-26T18:24:11.193" Score="0" ViewCount="" Body="&lt;p&gt;Firstly, I love this question and I wish that others will provide more insights so I can learn more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would classify #1 and #2 as the same underlying substance, unless the system scaling is such that the flows can be affected by quantum behavior in which case #2 and #3 would be the same.  The idea behind a coin toss in the first place is that sensitivity to the Initial Conditions (ICs) is high enough (and could be said to be chaotic) so it is useful as a random number.  We don't know very well where the variation in the ICs come from, but we generally don't care because the coin flip experiment subjects those ICs to a transformation that produces a &lt;em&gt;psuedorandom&lt;/em&gt; number in terms of heads and tails, just like the famous &lt;em&gt;RANDU&lt;/em&gt; computer function.  &lt;em&gt;RANDU&lt;/em&gt; notoriously does not produce a perfect random number (or even a good one), but this same argument applies for the coin flip, although it is chaotic randomness, meaning that the mathematics has a transition to chaos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contrast this to a radioactive decay experiment where the outcome is actually &lt;em&gt;random&lt;/em&gt;.  Let me be clear that I do not understand Quantum Mechanics (QM), and I don't need to.  In fact, it seems like I can't make a statement about QM without being corrected by a physicist.  Radioactive decay appears to be the collapse of the wave function of a coherent state outside of the nucleus.  This is wrong and I don't know why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm fairly sure we will all agree that in the pre-decay state there is a quantifiable probability for the emitted particle to exist outside the nucleus in $1/s$ units, which leads to the decay constant, and combined with the detection efficiency gets the average (or &quot;expected&quot;) rate of the counting.  No one has ever provided me a sufficient explanation for how QM gives the described result.  In a literal interpretation of QM, I think that the particle would simultaneously be both inside the nucleus and flying away in all possible directions and times of decay.  Since the $\Delta m$ of the decay, and thus energy of the particle, is a very accurately known number, maybe the uncertainty of the position of the particle is great.  It wouldn't matter anyway because the entire point is that we're not predicting the position of individual particles and are, in fact, using the fact that we don't know as an engineering convenience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A Defined QM Example&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the question lacks a clear definition of a quantum experiment I will provide one.  Take a GM Tube radiation detector with a small dead time compared to the counting time.  Establish a counting time based on an expected value of 100 counts, then look at the count and record if it is odd or even.  This functionally does the same thing as a heads or tails coin toss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, some purist is going to comment on this and tell me that the evens are more likely in said experiment.  Yes, that's right, I calculate one extra even for every $2.7 x 10^{44}$ experiments.  So while there is some imperfection in the experiment design, the experiment gives a random outcome in the true sense of randomness.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-05-26T18:24:11.193" />
  <row Id="10445" PostTypeId="2" ParentId="10435" CreationDate="2011-05-26T18:27:09.347" Score="3" ViewCount="" Body="&lt;p&gt;Of course the expansion of space is being considered by astronomers. In fact, it's pretty much the &lt;em&gt;only&lt;/em&gt; thing they are considering. The redshift due to expansion of space is the &lt;em&gt;way&lt;/em&gt; that astronomers know that it came from 13.14 billion years ago. What you do is look at the lightwaves very carefully. They will be stretched out (which looks like redshifting) due to the expansion of the universe. The longer they have been flying along, the more stretched out/redshifted they will be. The group cited in the article measured a redshift (a measure of this stretching) of 9.4, which is the largest we've ever observed; we conclude that this light has been traveling longer than any other light we've observed from a single source (the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_microwave_background_radiation&quot; rel=&quot;nofollow&quot;&gt;cosmic background radiation&lt;/a&gt; is way more redshifted).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A number of very clever methods allow us to identify just how long the light must have been traveling for it to received a particular amounts of redshift. This is an application of the relation known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble%27s_law&quot; rel=&quot;nofollow&quot;&gt;Hubble's Law&lt;/a&gt;. If you use it, you find that light with a redshift of 9.4 has been traveling for about 13.14 billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method is used so routinely that you'll hear cosmologists talk about time in terms of redshift, like &quot;Ionization occurred at redshift 17&quot; rather than &quot;Ionization occured 13.5 billion years ago&quot; (those numbers are made up).&lt;/p&gt;&#xA;" OwnerUserId="1655" LastEditorUserId="1655" LastEditorDisplayName="" LastEditDate="2011-05-26T18:36:51.247" LastActivityDate="2011-05-26T18:36:51.247" CommentCount="3" />
  <row Id="10446" PostTypeId="2" ParentId="10410" CreationDate="2011-05-26T18:29:48.777" Score="-1" ViewCount="" Body="&lt;p&gt;The waves do interfere with each other as you expect, but the interference pattern alternates between constructive interference and destructive interference in such a rapid fashion that your eye can't perceive it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to be able to see the interference, take a laser pointer and use that to illuminate the wall.  You should see a pattern called &quot;speckle&quot;.  (It's easiest to see if the beam spot is big; use a lens to do this if you have one.  Moving your head back and forth transversely also helps to make the speckle obvious.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speckle pattern comes from the constructive and destructive interference of the light reflecting from different parts of the wall.  The reason why it looks static (if you can hold your laser pointer still enough) is because the laser has a very narrow spread of frequencies (and, to get more technical, also has transverse phase-coherence).  If you were to change the wavelength of the laser, the pattern would change.  If you were to change the wavelength back and forth very rapidly, the pattern would change very rapidly, your eye would average it out, and you wouldn't be able to see the speckle anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what's going on with broadband light - it contains a whole bunch of different frequencies, so the interference &quot;averages out&quot;. That's why (unless you're using a laser) you never see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for radio waves: yeah, they interfere too.  That's why your microwave has a turntable in it.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-05-26T18:29:48.777" />
  <row Id="10447" PostTypeId="2" ParentId="10433" CreationDate="2011-05-26T18:39:02.470" Score="3" ViewCount="" Body="&lt;p&gt;There's a Nature article that describes the experiment and the results, &lt;a href=&quot;http://www.nature.com/nature/journal/v473/n7348/full/nature10104.html&quot; rel=&quot;nofollow&quot;&gt;http://www.nature.com/nature/journal/v473/n7348/full/nature10104.html&lt;/a&gt;, but that's behind a paywall. The experiment is described in some detail in &quot;Prospects for the measurement of the electron electric dipole moment using YbF&quot;, &lt;a href=&quot;http://arxiv.org/abs/1103.1566&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1103.1566&lt;/a&gt; (I've only scanned the latter, but it looks to be quite informative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the Nature article, &quot;In an atom or molecule with an unpaired valence electron, the interaction of the electron EDM [Electric Dipole Moment] with an applied electric field results in an energy difference between two states that differ only in their spin orientation. This energy difference is proportional to $d_e$ and changes sign when the direction of the field is reversed. A sensitive method of measuring this energy difference is to align the spin perpendicular to the field and measure its precession rate, which is proportional to the energy difference. An alternative description of the method is in terms of an interferometer. There is quantum interference between the two spin states, and the EDM appears as an interferometer phase shift that changes sign when the electric field is reversed.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An electron is not either a sphere or not-a-sphere, but we can introduce more or fewer internal degrees of freedom into the quantum fields that are used to describe experiments results that we attribute to the electron field. Introducing different degrees of freedom has consequences for the geometrical configurations of recorded experimental results. The Nature article is explicit in saying that this is intended to distinguish between different speculative quantum field theories, &quot;many extensions to the standard model naturally predict much larger values of $d_e$ that should be detectable&quot;. This is an experimentalists' article, however, so they link to a theory paper on the subject (which I cannot access directly). If these fields give better descriptions than the standard model of particle physics, we expect to see different, less geometrically symmetrical statistics of events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of the problems of reference here can be avoided if we talk about electron fields instead of about electrons. An &quot;electron field&quot; is less likely to be misrepresented as spherical or not spherical, but it can be associated with (representation spaces of) space-time symmetry groups (which describe in a systematic way how something deviates from being symmetrical). Care is needed because a quantum field is a more elaborate mathematical object than a classical field, but we can loosely think of a quantum field as a way to generate &lt;em&gt;probabilities&lt;/em&gt; that the configuration of a classical field is one thing or another &lt;em&gt;at any single time&lt;/em&gt;, while the details of quantum measurement are such that we &lt;strong&gt;&lt;em&gt;can't&lt;/strong&gt; talk about such probabilities at multiple times&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-05-26T18:39:02.470" CommentCount="2" />
  <row Id="10448" PostTypeId="2" ParentId="10350" CreationDate="2011-05-26T18:41:38.577" Score="0" ViewCount="" Body="&lt;p&gt;I'm assuming the two subjects are made out of the same material(s), and have the same proportions.  I will also assume that the barrier they collide with is unyielding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these assumptions the lighter person is better off.  Here's why:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First consider them in zero g, moving at the same velocity and colliding with the rigid barrier.  If the length of a person is L, then their momentum scales as L^3.  The distance over which they can stop scales as L, so the force of the collision scales as L^2.  The strength of each person scales as L^2 (cross-section), so they're equivalently well off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a fun aside: this is the same reasoning that explains why all animals pretty much jump the same height.  Note that a jumping 10-pound cat raises its center of mass by about 1 meter, a half-ton horse can raise its center of mass by about 1 meter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the zero-g, constant velocity case, the big person and little person are equally well off.  However, for the &quot;falling from big heights&quot; problem, the little person has two advantages:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Air resistance will result in a smaller velocity for the smaller person at impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) With gravity, at impact, the force on the person include both the force from deceleration (which scales as L^2 for a given velocity, see above) but also the force from counteracting static gravity at the surface, which scales as L^3.  This is probably not a big deal for the two masses mentioned in the problem, but starts to become important as the animal becomes very large.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastEditorUserId="1918" LastEditorDisplayName="" LastEditDate="2011-05-26T18:46:51.060" LastActivityDate="2011-05-26T18:46:51.060" />
  <row Id="10449" PostTypeId="1" CreationDate="2011-05-26T18:53:26.913" Score="1" ViewCount="78" Body="&lt;p&gt;What is the temperature of objects in Low Earth Orbit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider LEO to be 600km to 800km.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="3748" LastActivityDate="2011-05-26T19:13:47.703" Title="What are the temperatures of objects in LEO" Tags="&lt;space&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10450" PostTypeId="2" ParentId="8983" CreationDate="2011-05-26T19:01:07.730" Score="0" ViewCount="" Body="&lt;p&gt;We can theorize about why static or dynamic coefficient of friction is more appropriate, and why.  However, the bottom line on why we use the static coefficient is that it has been proven, experimentally, to be the correct one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a thought experiment, try this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a car sitting in a flat road.  What is the coefficient of friction, static or dynamic?  The answer is &quot;Static.&quot;  Now, roll the car very, VERY slowly (like one inch per hour).  What is the coefficient of friction?  Still static.  Now roll it just a little faster and ask the same question.  Now faster.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How fast do you have to go before the answer is that the coefficient of friction is no longer, &quot;static?&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to that question, in my mind, is, &quot;fast enough for something(s) in the chassis/strut/spring/shock/axle/wheel/tire/road system to excite the rubber/concrete interface enough that radial vibration modes of the rubber in the tire were strong enough to lift the wheel off the road and keep it off.&quot; (&quot;vibro-planing&quot; if I may be allowed to coin a word).  The speed required to do this, on a concrete surface in good repair, is probably an order of magnitude beyond the capability of any street-legal conveyance.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-05-26T19:01:07.730" />
  <row Id="10451" PostTypeId="2" ParentId="10429" CreationDate="2011-05-26T19:05:05.650" Score="-1" ViewCount="" Body="&lt;p&gt;I believe that if you just do the experiment with an imperfect vacuum in the first place it won't work.  The random particles in the air interact with the electrons (thus 'observing' them), so you don't even have to.  Granted, if you're using electrons they wouldn't hit the screen in the first place if they scattered so I don't know if what I'm saying applies for that experiment anyway.  But then again, there is still action at a distance, and electrons could interact weakly with charged particles and with uncharged particles through screening.  But someone who actually knows the field theories would have to comment on that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frankly it sounds expensive to set up a device that could measure an electron passing through one gate versus another.  I wonder if people who originally did the experiment actually did that, or just muddled with the setup until it stopped giving the inference patterns and then said that it had been 'observed' in that case.  The latter would obviously be the cost-effective route.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-05-26T19:05:05.650" />
  <row Id="10452" PostTypeId="2" ParentId="10449" CreationDate="2011-05-26T19:13:47.703" Score="2" ViewCount="" Body="&lt;p&gt;It depends which object. If the object is a human being in a spaceship, the temperature is approximately 37 °C. If the object is a spherical object that absorbs and re-emits the solar radiation, it's closer to -15 °C in average, like the Earth without the greenhouse effect, and maybe even lower because the solar radiation is shielded by the Earth much of the time. If it's totally shielded from solar radiation, it may acquired the temperature 2.7 K from the cosmic microwave background. Did you expect some universal temperature? Just to be sure, the atmosphere at those high altitudes is de facto non-existent, so this non-existent atmosphere doesn't enforce any unified &quot;air temperature&quot; over there.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-26T19:13:47.703" />
  <row Id="10453" PostTypeId="2" ParentId="10380" CreationDate="2011-05-26T20:05:29.143" Score="1" ViewCount="" Body="&lt;p&gt;The speed of sound in a metal is dictated by the nuclear lattice, and vibrations thereof, which are referred to as phonons.  The energy of the highest energy phonons is still at least an order of magnitude below the typical electron energy, but it is possible through electron phonon coupling that a phonon could excite an electron in a narrow region of energies around the Fermi energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the &lt;em&gt;state&lt;/em&gt; of the metal - in this case I've discussed above, the metal is still a metal, and solid. &lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2011-05-26T20:05:29.143" CommentCount="2" />
  <row Id="10454" PostTypeId="1" CreationDate="2011-05-26T21:10:48.700" Score="1" ViewCount="130" Body="&lt;p&gt;I am using some Dust Off brand cans to clean out a computer after an accident with concrete dust... long story.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside one of these cans is a flurocarbon which quickly turns into a gas at room temperature. It is not ordinary air as that would be too difficult or expensive to store compressed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I noticed that when orientated with the spray head upwards, the stuff which exited the nozzle was mostly vaporised immediately with a small amount of liquid residue. However, when orientated downwards some kind of liquid which quickly froze exited the nozzle. In fact such ice is still frozen on my computer case and is slowly melting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What causes the two distinct states? And why does the frozen stuff that sprays out upside down take ages to melt, but the frozen stuff which exits when orientated upwards melts, and subsequently evaporates very quickly?&lt;/p&gt;&#xA;" OwnerUserId="104" LastActivityDate="2011-05-27T00:19:17.363" Title="What causes compressed air (from &quot;Dust Off&quot; cans) to freeze, but only when turned upside down?" Tags="&lt;states-of-matter&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10455" PostTypeId="2" ParentId="10373" CreationDate="2011-05-26T21:21:53.530" Score="1" ViewCount="" Body="&lt;p&gt;Picture this, A tank of water with a valve at the bottom, and a hose attached to the valve. The tank represents a battery, the valve represents resistance, and the hose is the wire. If the tank is on the ground and the valve is wide open water will flow through the hose, but instead of water picture marbles. If you raise the tank up high, the water will flow faster, representing more voltage (higher tank) and more current(faster flow). Okay, you can put the tank down now. Close the valve half way, that represents higher resistance. Half the rate of flow comes out the hose. Raise the tank, and more water flows, but at half the rate as before. Now picture a &quot;Y&quot; fitting with a valve on each split. Shut one valve off and run water through it. The water up against the closed valve can't go anywhere so it just sits there waiting for the valve to open. When the valve opens, some water flowing out the other open valve flows through the newly opened one. Open both valves halfway, equal amounts flow out of each. Now close one 3/4 of the way more water flows out of the one that's half way closed. One more point, the bigger the hose the more water can flow, the bigger the wire the more current can flow.&#xA;And remember, unplug it before you work on it.&lt;/p&gt;&#xA;" OwnerUserId="3749" LastActivityDate="2011-05-26T21:21:53.530" CommentCount="1" />
  <row Id="10456" PostTypeId="1" AcceptedAnswerId="10482" CreationDate="2011-05-26T21:27:21.620" Score="0" ViewCount="98" Body="&lt;p&gt;I'm trying to understand wave diffraction and I found this wikipedia &lt;a href=&quot;http://cs.wikipedia.org/wiki/Difrakce&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;. It's in Czech so I'll explain a bit. I'm interested in the 4 images I couldn't find on english wikipedia. The first one is diffraction at large slit, second on large obstacle, third on tiny slit and last on slit which size is comparable with wavelength - happens with light on diffraction grating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got few question to those however. I don't know why, but I think I heard that wave can't pass through slit, which size is smaller than wavelength. For example imagine microwave oven. On doors is some kind of texture with those small slits. Microwaves have wavelength from 1 to 0.001 metres, so those slits should be sufficient and block the wave - that's why they are there in first place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how is then possible that third case with tiny slit? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I'm missing cases what happens at obstacle, which size is comparable to wavelength and if I'm wrong in the above situation, also on a tiny obstacle? (And if I'm right with that microwave thing, does the wave pass as there was no obstacle when hitting something tiny? ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a slit, I can use Huygens principle to create envelope, but what to do on obstacles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And can someone tell me what range of sizes compared to wavelength is considered for comparable obstacle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if it matters I imagine waves as always coming from source point, the way Kirchhoff described them as shown &lt;a href=&quot;http://en.wikipedia.org/wiki/Kirchhoff%27s_diffraction_formula&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, but I don't need to understand those equations and so on. I'd also appreciate if someone got a good website, images, documents,... well whatever that tryes to explain diffraction without too many formulas. I'm not going to compute anything in the end, just trying to get the best possible image of it all for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3662" LastActivityDate="2011-05-27T09:31:26.477" Title="Wave diffraction explanation" Tags="&lt;waves&gt;&lt;interference&gt;" AnswerCount="2" />
  <row Id="10458" PostTypeId="1" AcceptedAnswerId="10460" CreationDate="2011-05-26T21:35:33.063" Score="1" ViewCount="60" Body="&lt;p&gt;If there were a light cone centered at some point $P$, and you were to look at that light cone from different reference frames, would it change its shape? I know that points inside and outside the light cone would remain inside/outside of the light cone in every frame, but does the light cone itself shift?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it does, how would it shift?&lt;/p&gt;&#xA;" OwnerUserId="202" LastActivityDate="2011-05-26T21:44:56.837" Title="Does a light cone look the same from all reference frames?" Tags="&lt;relativity&gt;&lt;light-cone&gt;" AnswerCount="1" />
  <row Id="10459" PostTypeId="2" ParentId="7911" CreationDate="2011-05-26T21:39:51.917" Score="-3" ViewCount="" Body="&lt;p&gt;It's all about capacity, not internal resistance. Compare a shotgun shell to a .22.&lt;/p&gt;&#xA;" OwnerUserId="3749" LastActivityDate="2011-05-26T21:39:51.917" />
  <row Id="10460" PostTypeId="2" ParentId="10458" CreationDate="2011-05-26T21:44:56.837" Score="6" ViewCount="" Body="&lt;p&gt;No, the light cone does not depend on the frame in which it is viewed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The light cone is a collection of events that are lightlike-separated from $P$.  This collection of points is the same in all reference frames because in special relativity the interval is invariant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you swept out a light cone from $P$ by having a source at $P$ emit a spherical electromagnetic wave and noting where the wavefront was at future times, you would get the same result in all reference frames - that the edge of the wavefront was a sphere centered on $P$ with radius $ct$, with $t$ the time since the pulse was emitted.  This is true even if source at $P$ is moving at the time the wavefront is emitted.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-05-26T21:44:56.837" />
  <row Id="10462" PostTypeId="1" AcceptedAnswerId="10506" CreationDate="2011-05-26T22:41:14.960" Score="6" ViewCount="218" Body="&lt;p&gt;The most important property of an antenna, beyond its conductivity(?) seems to be it's length (or more generally, its shape).  But the antenna device is connected to the receiver by an arbitrarily-long connector.  What is the fundamental difference between an antenna and the connector?  Why doesn't the connector's length, straying from some 1/2, 1/4 length of the targeted band, interfere with the reception?&lt;/p&gt;&#xA;" OwnerUserId="2910" LastActivityDate="2011-08-25T10:06:08.160" Title="What makes an Antenna special?" Tags="&lt;rf&gt;" AnswerCount="4" FavoriteCount="3" />
  <row Id="10463" PostTypeId="1" AcceptedAnswerId="10466" CreationDate="2011-05-26T22:47:00.990" Score="2" ViewCount="161" Body="&lt;p&gt;I've read it many places that Amplitude Modulation produces sidebands in the frequency domain.  But as best as I can imagine it, modulating the amplitude of a fixed-frequency carrier wave just makes that &quot;louder&quot; or &quot;quieter&quot;, not higher-frequency or lower-frequency.  That is, I believe I could sketch, on graph paper, a path of a wave function that touches a peak or a trough exactly every 1/f increments, regardless of the &quot;volume&quot;.  Why do the sidebands appear?&lt;/p&gt;&#xA;" OwnerUserId="2910" LastActivityDate="2011-05-27T21:10:08.713" Title="Understanding the cause of sidebands in Amplitude Modulation" Tags="&lt;radio&gt;&lt;rf&gt;" AnswerCount="4" />
  <row Id="10464" PostTypeId="1" AcceptedAnswerId="10465" CreationDate="2011-05-26T23:39:06.883" Score="7" ViewCount="437" Body="&lt;p&gt;Often, I'll be driving down the road on a summer day, and as I look ahead toward the horizon, I notice that the road looks like there's a puddle of water on it, or that it was somehow wet. Of course, as I get closer, the effect disappears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that it is some kind of atmospheric effect. What is it called, and how does it work?&lt;/p&gt;&#xA;" OwnerUserId="3395" LastActivityDate="2011-05-27T22:42:00.027" Title="Why does the road look like it's wet on hot days?" Tags="&lt;optics&gt;&lt;atmosphere&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" />
  <row Id="10465" PostTypeId="2" ParentId="10464" CreationDate="2011-05-26T23:59:03.423" Score="12" ViewCount="" Body="&lt;p&gt;The phenomenon is called Mirage (EDIT: I called it Fata Morgana earlier, but a Fata Morgana is a special case of mirage that's a bit more complex). The responsible effect is the dependence of the refractive index of air on the density of air, which, in turn, depends on the temperature of the air (hot air being less dense than cold air).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ZSSUI.png&quot; alt=&quot;Mechanism Leading to a Fata Morgana&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A non-constant density leads to refraction of light. If there's a continuous gradient in the density, you get a bent curve (i) as opposed to light coming straight at you (d). Your eye does not know, of course, that the light (i) coming at it was bent, so your eye/brain continues the incoming light in a straight line (v).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This mirroring of the car (or other objects) then tricks you into thinking the road is wet, because a wet street would also lead to a reflection. In addition, the air wobbles (i.e. density fluctuations), causing the mirror image to wobble as well, which adds to the illusion of water.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditorDisplayName="" LastEditDate="2011-05-27T15:17:50.880" LastActivityDate="2011-05-27T15:17:50.880" CommentCount="4" />
  <row Id="10466" PostTypeId="2" ParentId="10463" CreationDate="2011-05-27T00:07:29.993" Score="5" ViewCount="" Body="&lt;p&gt;Let your carrier signal be $A_0 \cdot \cos(\omega_c t)$ with amplitude $A_0$ and carrier frequency $\omega_c$. &#xA;Let your signal be a simple wave, $\phi(t) = A_s \cdot \cos(\omega_s t)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the modulated signal becomes&#xA;$$A_0 A_s \cdot \cos(\omega t) \cdot \cos(\omega_s t)$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, as pointed out by George in the comments, the carrier also gets transmitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using simple trigonometry gives you as the final signal&#xA;$$\frac{1}{2} A_0 A_s \cdot ( \cos((\omega - \omega_s) t) + \cos((\omega + \omega_s) t)) + A_0 \cdot \cos(\omega t)$$ &#xA;Hence, the frequency becomes changed, you get the carrier frequency in the middle (at $\omega$) and two side-bands at $\omega \pm \omega_s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in reality your signal is not a simple cosine, but you could do a Fourier decomposition of the signal and treat each frequency independently. The two frequencies then get smeared out and you get the two sidebands.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditorDisplayName="" LastEditDate="2011-05-27T15:22:29.243" LastActivityDate="2011-05-27T15:22:29.243" CommentCount="7" />
  <row Id="10467" PostTypeId="2" ParentId="10454" CreationDate="2011-05-27T00:19:17.363" Score="4" ViewCount="" Body="&lt;p&gt;The fluorocarbon in the can is a liquid under pressure. When you spray the can upright, only the fluorocarbon vapor at the top of the can is released. When you turn the can upside-down, the liquid is forced out instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you have noted, this liquid quickly evaporates because is is a gas at room temperature. In other words, it begins to boil. This carries heat away from the liquid very rapidly, cooling it to its boiling point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also cools whatever you are spraying the liquid on, thus condensing water out of the air and freezing it.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-05-27T00:19:17.363" CommentCount="2" />
  <row Id="10468" PostTypeId="2" ParentId="10464" CreationDate="2011-05-27T00:25:47.693" Score="3" ViewCount="" Body="&lt;p&gt;It is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Mirage#Highway_mirage&quot; rel=&quot;nofollow&quot;&gt;mirage&lt;/a&gt;: in particular it is caused by hot air near the road and less hot air above it  creates a gradient in the refractive index of the air and so making a virtual image of the sky appear to be on or below the road. Air currents make this shimmer, similar to a reflection of the sky on water, hence causing the illusion of wetness.&lt;/p&gt;&#xA;" OwnerUserId="2714" LastActivityDate="2011-05-27T00:25:47.693" CommentCount="5" />
  <row Id="10469" PostTypeId="1" CreationDate="2011-05-27T02:39:17.263" Score="0" ViewCount="41" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/10464/why-does-the-road-look-like-its-wet-on-hot-days&quot;&gt;Why does the road look like it&amp;#39;s wet on hot days?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;When we are traveling in highways that too in summer we can observe the liquid at far to our eyes on roads. But normally there will be no liquid, sun light makes it look like liquid or something like that what makes it to look like? which property of light makes that roads look like? When the roads are not flat we can easily observe this i observed this many times in India.! i am not having 10 reputation too post images for this image have a look at this link&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[http://www.tumblr.com/tagged/long+roads]&lt;/p&gt;&#xA;" OwnerUserId="3590" LastActivityDate="2011-05-27T02:39:17.263" ClosedDate="2011-05-27T02:43:06.417" Title="What makes the roads looks like it is having an liquid on it on summer?" Tags="&lt;light&gt;&lt;atmosphere&gt;&lt;atmospheric-science&gt;" CommentCount="0" />
  <row Id="10470" PostTypeId="1" AcceptedAnswerId="10471" CreationDate="2011-05-27T02:53:57.960" Score="6" ViewCount="314" Body="&lt;p&gt;When the sun is out after a rain, I can see what appears to be steam rising off a wooden bridge nearby. I'm pretty sure this is water turning into a gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I thought water had to reach 100 degrees C to be able to turn into a gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an edge case, for small amounts of water perhaps, that allows it to evaporate?&lt;/p&gt;&#xA;" OwnerUserId="3751" LastActivityDate="2011-05-28T17:06:07.860" Title="How does water evaporate if it doesn't boil?" Tags="&lt;water&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="10471" PostTypeId="2" ParentId="10470" CreationDate="2011-05-27T03:32:20.323" Score="7" ViewCount="" Body="&lt;p&gt;Evaporation is a different process to boiling. The first is a surface effect that happens at any time, while the latter is a bulk transformation that only happens when the conditions are correct.&#xA;Technically the water is not turning into a gas, but random movement of the surface molecules allows some of them enough energy to escape from the surface into the air. The rate at which they leave the surface depends on a number of factors - for instance the temperature of both air and water, the humidity of the air, and the size of the surface exposed.&#xA;When the bridge is 'steaming': the wood is marginally warmer than the air (due to the sun shine), the air is very humid (it has just been raining) and the water is spread out to expose a very large surface area. In fact, since the air is cooler and almost saturated with water, the molecules of water are almost immediately condensing into micro-droplets in the air - which is why you can see them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW - steam is completely transparent. If you can see it then it is water vapour. Consider a kettle boiling - the white plume only occurs a short distance above the spout. Below that it is steam, above it has cooled into vapour.&lt;/p&gt;&#xA;" OwnerUserId="3752" LastActivityDate="2011-05-27T03:32:20.323" CommentCount="6" />
  <row Id="10472" PostTypeId="2" ParentId="10420" CreationDate="2011-05-27T03:48:50.590" Score="3" ViewCount="" Body="&lt;p&gt;The link you provided is a page of G t'Gooft on an official site and is quite interesting reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the statement you are wondering about is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;consider the scientific facts concerning the Standard Model. Fact is that the W+, W‾ and the Z boson each carry three spin degrees of freedom, whereas the Yang-Mills field quanta, which describe their interactions correctly in great detail, each carry only two. Those remaining modes come from the Higgs field. What this means is that three quarters of the field of the Higgs have already been found. The fourth is still missing, and if you calculate its properties, it is also clear why it is missing: it is hiding in the form of a particle that is difficult to detect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He is stating that the Higgs' existence is based on the very strong verification of the Standard Model. There is strong experimental verification of the existence of W+ W- and Z  The construction of the theory is such that since the gauge bosons exist one has  3/4 of the data needed to confirm the model, of which the existence of the Higgs is 1/4, counting up dimensions, of the necessary evidence/prediction. He is equating the existence of the Higgs to the validity of the SM in this statement.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-27T03:48:50.590" CommentCount="2" />
  <row Id="10473" PostTypeId="2" ParentId="10395" CreationDate="2011-05-27T04:07:51.927" Score="1" ViewCount="" Body="&lt;p&gt;Given that they only expect one WIMP per year, lattice damage is not going to be a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, if there are local single atom dislocations to the lattice then annealing should put them right again. I have read in the literature (Nuclear Instrumentation Methods) of the machinations they went through in the early sixties to deal with noise in the nascent low purity Ge crystal detectors.  I read many instances of annealing being used to quiet a crystal. &lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="3735" LastEditorDisplayName="" LastEditDate="2011-05-27T04:14:36.207" LastActivityDate="2011-05-27T04:14:36.207" CommentCount="1" />
  <row Id="10474" PostTypeId="2" ParentId="10420" CreationDate="2011-05-27T04:38:55.693" Score="4" ViewCount="" Body="&lt;p&gt;The Higgs field in the Standard Model is a scalar field that is a complex $SU(2)$ doublet, so it has 4 real scalar degrees of freedom. Electroweak symmetry breaking converts 3 of these 4 scalar degrees of freedom into longitudinal modes of 3 gauge bosons, which then become massive. These 3 massive gauge bosons all have been observed in collider experiments at CERN ($W^\pm$, $Z$ bosons). So we know for sure that they exist. The fourth degree of freedom - the Higgs particle - has not been observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this sense it is true that 3/4 of the Higgs field have been observed, but it is not true that 3/4 of the Higgs particle have been observed, since the latter is precisely the missing quarter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This statement reminds me a bit about a joke on the &quot;early evidence for supersymmetry&quot;: nearly half of the particles of the minimal supersymmetric Standard Model have been observed already ;-)&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-05-27T04:38:55.693" />
  <row Id="10475" PostTypeId="1" CreationDate="2011-05-27T05:06:41.833" Score="-1" ViewCount="77" Body="&lt;p&gt;I am a budding novelist, and im researching a few things for a fictional narrative. the characters are from other dimensions and i want to present a coherent scientific structure. any suggestions orhelp would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3753" LastActivityDate="2011-05-27T05:47:02.707" ClosedDate="2011-05-27T08:01:10.243" Title="purely hypothetically, could a quantum being appear to be human? I ask for the purpose of research on a novel" Tags="&lt;quantum-mechanics&gt;&lt;mechanics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10476" PostTypeId="2" ParentId="10475" CreationDate="2011-05-27T05:47:02.707" Score="0" ViewCount="" Body="&lt;p&gt;Humans &lt;em&gt;are&lt;/em&gt; quantum beings. Decohered quantum beings.&lt;/p&gt;&#xA;" OwnerUserId="3755" LastActivityDate="2011-05-27T05:47:02.707" />
  <row Id="10477" PostTypeId="1" CreationDate="2011-05-27T06:03:45.057" Score="4" ViewCount="144" Body="&lt;p&gt;String theory has at least $10^{500}$ 6D compactifications. &lt;a href=&quot;http://arxiv.org/abs/hep-th/0602072&quot; rel=&quot;nofollow&quot;&gt;Denef and Douglas&lt;/a&gt; proved the computational complexity of finding a compactification which fits the parameters of the low energy effective action like the cosmological constant, the electroweak scale, the Higgs self-coupling, Yukawa couplings and gauge coupling strengths is NP-complete. There are limits on the energies future particle accelerators can reach, and cosmic ray frequencies. The amount of data gleanable from the cosmic microwave background anisotropies is limited, and can only probe up to the energy scale of slow roll inflation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is beyond our abilities to determine the exact string compactification describing our universe, is it not wrong to call the compactification &lt;em&gt;unobservable&lt;/em&gt;? Only things which are experimentally observable, possibly with the aid of theoretical interpolations, and can make experimentally verifiable predictions count as science. Within the realm of physics, are we forbidden to ask which string compactification describes our universe? Must we stick to the effective field theory if we wish to remain physicists and not philosophers?&lt;/p&gt;&#xA;" OwnerUserId="3756" LastActivityDate="2011-06-26T21:02:43.460" Title="Is it outside science to figure out the precise string compactification of our universe?" Tags="&lt;string-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" />
  <row Id="10478" PostTypeId="2" ParentId="8895" CreationDate="2011-05-27T06:12:33.487" Score="1" ViewCount="" Body="&lt;p&gt;Bits, or qubits? Classical or quantum computer? Exact simulation accuracy, or good enough accuracy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If perfect accuracy, the computer can't be part of the universe because no finite entity can simulate itself with perfect accuracy. The measurements also have to be performed at the meta level. &lt;/p&gt;&#xA;" OwnerUserId="3757" LastActivityDate="2011-05-27T06:12:33.487" CommentCount="1" />
  <row Id="10479" PostTypeId="1" CreationDate="2011-05-27T06:42:33.590" Score="3" ViewCount="129" Body="&lt;p&gt;Classically, probability distributions are nonnegative real measures over the space of all possible outcomes which add up to 1. What they mean is open to debate between Bayesians, frequentists and ensemble interpretations. A degenerate distribution is the least random distribution with a probability of 1 for a given fixed event, and 0 for everything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the analog of a classical probability distribution in quantum mechanics? Is it a wave function augmented with the Born interpretation for probabilities, or is it the density matrix? Does a pure density matrix correspond to a degenerate distribution?&lt;/p&gt;&#xA;" OwnerUserId="3758" LastActivityDate="2011-05-27T18:00:35.257" Title="Is the quantum analog of a probability distribution the wave function or the density matrix?" Tags="&lt;quantum-interpretations&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10480" PostTypeId="2" ParentId="10462" CreationDate="2011-05-27T08:59:04.487" Score="1" ViewCount="" Body="&lt;p&gt;It does! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two generally different ways to make such a &quot;connector&quot;. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;tuned , the cable is a kind of a Lecher-line. and has to be tuned to &#xA;the frequency in length. Such cables were used to be used by radio hams &#xA;long ago, today they are outdated because of the need to tune anew &#xA;after any change of frequency.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;matched , here cables with a certain wave resistance are used, this &#xA;resistance has to be the same for the antenna and the &quot;other&quot; end. &#xA;That is the thing laymen are used to. 60 ohms or nowadays often &#xA;50 ohms coaxial cables are used. For such connections length only matters &#xA;due to attanuation. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the fundamental difference&#xA;  between an antenna and the connector?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a rather easy question, an antennas purpose is to &lt;strong&gt;radiate&lt;/strong&gt;, &#xA;whereras the conectors purpose is to lead the energy to/from antenna &#xA;&lt;strong&gt;without radiation&lt;/strong&gt; .&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-05-27T08:59:04.487" CommentCount="2" />
  <row Id="10481" PostTypeId="2" ParentId="10456" CreationDate="2011-05-27T09:26:05.297" Score="0" ViewCount="" Body="&lt;p&gt;I will try to use as less mathematics as I can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Microwave ovens use a &lt;a href=&quot;http://en.wikipedia.org/wiki/Faraday_cage&quot; rel=&quot;nofollow&quot;&gt;Faraday cage&lt;/a&gt; to block the waves from coming out of the oven. Waves can of course pass through slits that are smaller than their wavelength. You can demonstrate this in a water tank with a plastic slit for yourself, which in my opinion is a very good way to explore different kinds of diffraction without using such expensive instruments as LASERS and diffraction gratings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An obstacle is a very general term --people use obstacles describe objects that can reflect waves, absorb waves, change the wavelength of waves etc. &lt;em&gt;Any&lt;/em&gt; physical phenomenon that will cause wavefronts to superimpose (mix up) will result in diffraction, the &quot;obstacle&quot; size and the wavelength do not have to be comparable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best regards,&lt;/p&gt;&#xA;" OwnerUserId="3669" LastActivityDate="2011-05-27T09:26:05.297" />
  <row Id="10482" PostTypeId="2" ParentId="10456" CreationDate="2011-05-27T09:31:26.477" Score="0" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I don't know why, but I think I heard&#xA;  that wave can't pass through slit,&#xA;  which size is smaller than wavelength.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is simply wrong. The energy passing through a slit/hole smaller than &#xA;wavelength is less than transmitted through an comparable area in free field,&lt;br&gt;&#xA;but it is not zero! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real world, being wavelike and particle-like at the same time &#xA;is not on/off, black/white, yes/no, true/untrue. Microwaves from &#xA;Your oven are damped at that grating, but not erased totally. &#xA;(BTW, my MW oven operates at about 12 cm wavelength, not &quot;&quot;1 to 0.001 metres,&quot;&quot;&#xA;blocking the waves were almost impossible then) &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On a slit, I can use Huygens principle&#xA;  to create envelope, but what to do on&#xA;  obstacles?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You could use Huygens principle on both edges of the obstacle, going &#xA;outward for some length of say, two to three obstacle diameters, or &#xA;read something about complementarity of slits and obstacles. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-05-27T09:31:26.477" CommentCount="1" />
  <row Id="10483" PostTypeId="2" ParentId="10479" CreationDate="2011-05-27T12:53:18.737" Score="1" ViewCount="" Body="&lt;p&gt;I'll reproduce my comment from above here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The point is perhaps that quantum theory states are not analogs of&#xA;  probability distributions -- at least not exactly. I suggest&#xA;  arxiv.org/abs/quant-ph/0101012v4 as an interesting attempt to squeeze the two&#xA;  as close together as they can be (it's 34 pages, but Lucien Hardy is&#xA;  relatively easy reading). I'm not as happy as I'd like with this response,&#xA;  which is why it's a comment, not an Answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I decided I was being lazy, and looked at Lucien's paper for what I might take to be its relevance to your Question. Lucien takes quantum pure states to be analogous &lt;em&gt;enough&lt;/em&gt; to degenerate probabilities to &lt;em&gt;use&lt;/em&gt; the idea. Where that becomes interesting is the way he can then &lt;em&gt;characterize&lt;/em&gt; the difference between classical probabilistic states and quantum states, given this starting point. The distinguishing feature is his fifth axiom,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Axiom 5 Continuity. There exists a continuous reversible transformation&#xA;  on a system between any two pure states of that system.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;IMO, this is definitely a curious way to construct things. It has a distinct failing, that it's limited to finite-dimensional Hilbert spaces and probability distributions over a finite set of outcomes, and AFAIK no-one has extended Lucien's analysis to infinite dimensional Hilbert spaces and probability spaces, which somewhat diminishes its interest unless you in any case work only with finite dimensional Hilbert spaces (as you might if you work in quantum information).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point I'd make about this is that this is an interesting partial analogy, although I do not know that any more directly related-to-experiment &lt;em&gt;use&lt;/em&gt; has been made of it. It may well be worth thinking in terms of this partial analogy some more, but my personal assessment has been that this is not something worth hanging my hat on exclusively. On the other hand, it's only if one immerses oneself in a way of thinking in a committed way that new results come from anywhere, and it's because people have made different choices than I make that we get different results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, I think Lucien's paper is relevant for you. It has, to me, the same &lt;em&gt;feel&lt;/em&gt; as the way you have asked your Question.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-05-27T12:58:35.313" LastActivityDate="2011-05-27T12:58:35.313" />
  <row Id="10484" PostTypeId="1" CreationDate="2011-05-27T13:32:05.337" Score="4" ViewCount="164" Body="&lt;p&gt;As a casual science reader, I've always found the implications of relativity (inconsistent clocks after near-light-speed travel and various space-time paradoxes) to be confusing and magical-sounding. Yet I know it's accepted as foundational to modern physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some of the experiments that have borne out Einstein's theory?&lt;/p&gt;&#xA;" OwnerUserId="3764" LastActivityDate="2011-05-27T13:44:14.580" Title="What experimental proof has been found of Einstein's theory?" Tags="&lt;general-relativity&gt;&lt;experimental-evidence&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="10485" PostTypeId="2" ParentId="10484" CreationDate="2011-05-27T13:40:09.460" Score="4" ViewCount="" Body="&lt;p&gt;If you're asking for experimental tests of special relativity, arguably the first one is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Michelson%E2%80%93Morley_experiment&quot; rel=&quot;nofollow&quot;&gt;Michelson-Morley experiment&lt;/a&gt;. Subsequent to that, there have been an enormous number of tests that verify it to incredible precision - a comprehensive account with references can be found here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html&quot; rel=&quot;nofollow&quot;&gt;http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you've tagged your question with general relativity - for which there's a Wikipedia page that summarizes the main predictions (and confirmations) like the precession of Mercury's perihelion, deflection of light by the sun, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tests_of_general_relativity&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Tests_of_general_relativity&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-05-27T13:40:09.460" />
  <row Id="10486" PostTypeId="2" ParentId="10484" CreationDate="2011-05-27T13:44:14.580" Score="2" ViewCount="" Body="&lt;p&gt;You can find an extended list of experiments on the Special Theory of Relativity on the following link:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html&quot; rel=&quot;nofollow&quot;&gt;What is the experimental basis of Special Relativity?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2011-05-27T13:44:14.580" />
  <row Id="10487" PostTypeId="2" ParentId="10295" CreationDate="2011-05-27T13:49:54.237" Score="1" ViewCount="" Body="&lt;p&gt;Suppose we put two people in a room and ask them to toss coins. We expect them to get half heads and half tails. We also expect that when we look at both their coins together they will get opposites, head-and-tail, as often as they get the same, head-and-head or tail-and-tail. If the two coins come up randomly, but they're always either both heads or both tails, we never see head-and-tail, that's a correlation, and we might look for an explanation of that, even though we wouldn't have to look for any explanation of the way each coin came up. The &lt;a href=&quot;http://plato.stanford.edu/entries/physics-Rpcc/&quot; rel=&quot;nofollow&quot;&gt;Reichenbach common cause principle&lt;/a&gt; expresses precisely this requirement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point I'll point out (coarsely) the difference between an empiricist and a realist (and noting that there are instrumentalists, naive or structural realists, etc.). An empiricist looks at the correlations and, finding that those correlations are always there, figures out a way to use them, without much caring what explanation or causes there might be for the correlations. Some people are temperamentally quite happy not to care how or why, they're happy to use it. A realist cares about how and why it happens. The proportions of Physicists who think in one way or the other, and who think in both ways depending on the situation they face in their research, goes in cycles. Of course an empiricist is still &lt;em&gt;curious&lt;/em&gt;, it's just that their curiosity runs in different pathways. It's important that there are both ways of thinking and that there's an interplay between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose now that we do a different experiment. A psychologist gives two people two buttons each, and asks them to press either one or the other, and then looks to see both what the patterns of the individual button presses are and what correlations there are between them. The psychologist carries out lots of tests, for a whole day at a time, and pays the subjects more if they maintain an average of 100 button presses per minute. That's 50000 events per day, say, in contrast to that many events per second, or far more still, in a Physics experiment. Nonetheless, lots of data. In each test, different feedback is given about what the other person has pressed. In one test, we give audible feedback immediately after the button is pressed, in another half a second after the button is pressed, in another we give visual feedback, perhaps of different colors and perhaps in different positions. We try lots of different alternatives, and discover that there are different correlations in each case. People are not very good at being random, but after an hour of something as boring as this someone who is taking the project seriously would get into a vibe with the other person. Finally, we could also introduce a third person whose job it is to try to get the two other people to be synchronized. They can control the feedback lights however they like. Imagine any configuration and different kinds of feedback you like and speculate on what the results would be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physics does just this. There are two detectors and a central source. If there's no source, the detectors &quot;trigger&quot; every now and then, which is called the &quot;dark rate&quot;. When we introduce different central sources, we see different statistics for both the rate at which we see events and for the correlations between the events. We also see different rates and correlations if we introduce different detectors. &lt;em&gt;Without knowing why&lt;/em&gt; there are different rates and correlations, we can see that there &lt;em&gt;are&lt;/em&gt; different rates and correlations. We can use those different rates and correlations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where quantum theory differs from classical probability is that we describe the results we get for a single type of source for all different types of detectors, and, separately, the results for a single type of detector for all different types of sources, whereas for classical probability we describe the results we get for each single type of source and detector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are so many ramifications that I'm uncertain where to stop. For example, a &quot;detector&quot; can have a compound structure. It can be two detectors, your A and B, either close together or far apart, perhaps set to trigger as a pair only when they trigger at the same time. The electronics in real experiments can be byzantine, although it's the job of experimentalists to be able to tell other experimentalists what their electronics does so they can do the same thing, ideally in a different way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notoriously, there are certain kinds of explanations for the statistics we observe in Bell-inequality violating experiments that are not possible. They can't be explained by the idea that the central source sends what might be called classical particles to the two detectors. People have proposed other kinds of causal models, but Physicists have for the most part not found those models useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, this is embarrassingly incomplete, but the nature of your Question and of your response to John Schanck's more-or-less passable account in terms of the nitty gritty of quantum mechanics suggested to me that this type of narrative might be more useful for you.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-05-27T13:49:54.237" CommentCount="1" />
  <row Id="10488" PostTypeId="2" ParentId="10463" CreationDate="2011-05-27T14:15:21.710" Score="2" ViewCount="" Body="&lt;p&gt;In the comments I explained how one obtains three different-frequency addenda in the total signal.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-05-27T14:15:21.710" CommentCount="2" />
  <row Id="10489" PostTypeId="2" ParentId="10462" CreationDate="2011-05-27T14:38:22.963" Score="1" ViewCount="" Body="&lt;p&gt;If you apply an alternating electric current on a piece of wire,  it will radiate. For example we live in a 50 Herz background in our houses because of the wires bringing in the electricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A receiving antenna is oriented and has the size necessary to receive signals for  the TV or radio. In order to shield the wire connecting the television to the antenna, so there is no interference of unwanted frequencies, one uses special wires called coaxial. They have a core wire which carries the signal and are covered outside with wire mesh or aluminium foil in order to shield from any ambient signals the small wanted signal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one uses a transmitting antenna the same logic holds. The wires carrying the signal to the aerial are coaxial.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-27T14:38:22.963" CommentCount="4" />
  <row Id="10490" PostTypeId="2" ParentId="8895" CreationDate="2011-05-27T15:15:50.997" Score="-1" ViewCount="" Body="&lt;p&gt;Anyway you look at it, you need an infinite number of bits. This is because if you have only a finite number, then you can't describe the description, as Konard stated. That is unless the description happens from outside of the universe. In that case the question is simply weather the universe has an infinite amount of information in it or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know any proofs that the universe is composed of infinite information, and some views relate energy and information (a proof that it is possible to convert information to energy can be found in Maxwell's demon thought experiment which was allegedly proved here &lt;a href=&quot;http://www.livescience.com/8944-maxwell-demon-converts-information-energy.html&quot; rel=&quot;nofollow&quot;&gt;http://www.livescience.com/8944-maxwell-demon-converts-information-energy.html&lt;/a&gt;), so if you believe there is a finite amount of energy, then maybe this means that there is also a finite amount of information, so only a finite amount of information can be used to describe it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as far as I know, most physical theories today require continuity and make use of it, so this number of required bits is obviously infinite. So if you believe that, you may be able to describe the universe from within (like Borges' short story &quot;The Aleph&quot; &lt;a href=&quot;http://www.phinnweb.org/links/literature/borges/aleph.html&quot; rel=&quot;nofollow&quot;&gt;http://www.phinnweb.org/links/literature/borges/aleph.html&lt;/a&gt;). It seems to me that in this case the interesting question is weather we have a continuous universe or not. Weather space is enumerable or not. I asked this question yesterday here &lt;a href=&quot;http://physics.stackexchange.com/questions/10415/can-we-have-non-continuous-models-of-reality-why-dont-we-have-them&quot;&gt;Can we have non continuous models of reality? Why don&amp;#39;t we have them?&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="72" LastEditorUserId="72" LastEditorDisplayName="" LastEditDate="2011-05-28T17:35:15.970" LastActivityDate="2011-05-28T17:35:15.970" />
  <row Id="10491" PostTypeId="2" ParentId="10435" CreationDate="2011-05-27T15:38:46.330" Score="0" ViewCount="" Body="&lt;p&gt;Looking at a distant object, billions of light years distant, has many connotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By being so incredibly far away, you're seeing past events. But this does not necessarily mean the object no longer exists. It may still be around, but certainly could be very different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend checking out some of the starter Astronomy videos like &lt;a href=&quot;http://www.khanacademy.org/video/intergalactic-scale?playlist=Cosmology%20and%20Astronomy&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt; at Khan Academy, it helps me get a handle on the scale of things and the physics behind it.&lt;/p&gt;&#xA;" OwnerUserId="1638" LastActivityDate="2011-05-27T15:38:46.330" />
  <row Id="10492" PostTypeId="1" CreationDate="2011-05-27T17:41:20.847" Score="0" ViewCount="49" Body="&lt;p&gt;Some time far into the future, humans have made the technology breakthroughs to construct decoherence-proof reversible quantum computers with quantum fault tolerance. They have also solved the hard AI problem. They program a quantum conscious being on such a computer and also a two state quantum system. The quantum conscious being is programmed to entangle with the quantum system, measuring it. The rules of the Copenhagen interpretation tell us the state of the quantum system happened. The programmers reverse the program, undoing the measurement, with all the memory wiped clean. Did the quantum system unhappen?&lt;/p&gt;&#xA;" OwnerUserId="3767" LastActivityDate="2011-05-27T17:41:20.847" ClosedDate="2011-05-28T05:13:49.210" Title="Can events unhappen for a conscious program on a reversible quantum computer?" Tags="&lt;quantum-interpretations&gt;" CommentCount="3" />
  <row Id="10493" PostTypeId="2" ParentId="10479" CreationDate="2011-05-27T17:51:18.843" Score="1" ViewCount="" Body="&lt;p&gt;This question is rather broad, but I will try to address the main issue.  A state $|\psi\rangle$ is  alinear sum of eigenvectors $|n\rangle$ so that&#xA;$$&#xD;&#xA;|\psi\rangle~=~\sum_nc_n|n\rangle.&#xD;&#xA;$$&#xA;The probabilities maybe directly computed this way from $\langle\psi|\psi\rangle$ $=~1$ as&#xA;$$&#xD;&#xA;\langle\psi|\psi\rangle~=~\sum_{mn}c^*_mc_n\langle m|n\rangle&#xD;&#xA;$$&#xA;and since $\langle m|n\rangle~=~\delta_{mn}$ and the amplitudes define the probabilities as their modulus squared $ c^*_nc_n~=~P_n$ this expectation is then just a sum of the probabilities.  The born rule then tells that for an observable ${\cal O}$ diagonal in this basis with ${\cal O}|n\rangle~=~o_n|n\rangle$ the expected value of the observable for $|\psi\rangle$ is then &#xA;$$&#xD;&#xA;\langle\psi|{\cal O}\psi\rangle~=~\sum_{mn}c^*_mc_no_n \langle m| n\rangle~=~\sum_no_nP_n.&#xD;&#xA;$$&#xA;This is just the sum of the eigenvalues of the observable times their probabilities for being observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density matrix is really a form generalization of this.  We consider the outer product of the states as the density operator&#xA;$$&#xD;&#xA;{\hat\rho}~=~|\psi\rangle\langle\psi|~=~\sum_{mn}c^*_mc_n|n\rangle\langle m|&#xD;&#xA;$$&#xA;which is a matrix representation of this operator.  The trace of the density matrix, $Tr{\hat\rho}~=~\sum_n\langle n|{\hat\rho}|n\rangle$ then gives the sum of probabilities.  I leave it as an exercise to evaluate an observable in the trace $Tr{\cal O}{\hat\rho}$ to derive the Born rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The modulus squares of the amplitudes define the probabilities, which are real valued.  So the connection between the states and the probabilities is not one to one.  One reason people like the density matrix is that there is more of a connection between a linear quantum operator and the probabilities in the trace operation.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="1352" LastEditorDisplayName="" LastEditDate="2011-05-27T18:00:35.257" LastActivityDate="2011-05-27T18:00:35.257" />
  <row Id="10494" PostTypeId="2" ParentId="10477" CreationDate="2011-05-27T18:06:16.810" Score="0" ViewCount="" Body="&lt;p&gt;The precise string compactification is not something we can find out. To ask about it isn't forbidden, but it's a meaningless question. The same thing goes for questions like &quot;is there really a definite string compactification&quot;. Until and unless we can find out, this is a meaningless question.&lt;/p&gt;&#xA;" OwnerUserId="3768" LastActivityDate="2011-05-27T18:06:16.810" />
  <row Id="10495" PostTypeId="1" CreationDate="2011-05-27T18:16:07.150" Score="0" ViewCount="117" Body="&lt;p&gt;String theory - for example - requires extra spatial dimension. Say for example in 10 dimensional string theory, what theoretically prevents clustering of the extra 6 dimensions in 2 timeless 3 dimensional (infinite) spaces?  &lt;/p&gt;&#xA;" OwnerUserId="2942" LastEditorUserId="2942" LastEditorDisplayName="" LastEditDate="2011-06-02T12:34:41.147" LastActivityDate="2011-06-02T12:51:46.300" Title="Could extra dimensions be or become clustered?" Tags="&lt;dimensionality&gt;" AnswerCount="1" />
  <row Id="10496" PostTypeId="1" CreationDate="2011-05-27T18:23:11.343" Score="13" ViewCount="641" Body="&lt;p&gt;The media are reporting the commercially sold 128-bit quantum computer from D-Wave&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://news.google.com/news?ned=us&amp;amp;hl=us&amp;amp;q=d-wave+quantum&amp;amp;cf=all&amp;amp;scoring=n&quot; rel=&quot;nofollow&quot;&gt;http://news.google.com/news?ned=us&amp;amp;hl=us&amp;amp;q=d-wave+quantum&amp;amp;cf=all&amp;amp;scoring=n&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which of course sounds amazing. The gadget is described as something capable of doing quantum annealing&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_annealing&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Quantum_annealing&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which looks less convincing. I want to ask you what classes of problems the D-Wave computer can actually solve or perform. It can't run Shor's algorithm on 128 qubits, can it?&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-28T00:11:57.350" Title="What can the D-Wave quantum computer do?" Tags="&lt;quantum-information&gt;&lt;quantum-computer&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="7" />
  <row Id="10497" PostTypeId="2" ParentId="3862" CreationDate="2011-05-27T18:37:11.103" Score="0" ViewCount="" Body="&lt;p&gt;No, in the many worlds interpretation, every parallel universe is real, but in consistent histories, once you choose your projection operators, only one possibility is real, and all the others are imaginary. This makes consistent histories a lot more like Bohmian mechanics with the world the Bohmian particle sits in being more real than the rest. Why should one world be more real than the others? There is no reason. To copies of you living in a parallel world, they are more real than you are.&lt;/p&gt;&#xA;" OwnerUserId="3769" LastActivityDate="2011-05-27T18:37:11.103" />
  <row Id="10498" PostTypeId="2" ParentId="10470" CreationDate="2011-05-27T19:07:59.863" Score="-1" ViewCount="" Body="&lt;p&gt;Steam  rising from a warm bridge is vaprization of water.  Boiling water is vaporization of water.  Getting cooled down by a breeze after a sweaty workout is vaporization of water.  All result in the same phase change with the same latent heat of vaporization of 540 cal./gram, which is a very powerful cooling effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boiling water is a subset of vaporization of water, wherein the heating of the water is fast enough that vaporization is forced to occur very rapidly AND there is enough water such that the vaporization occurs under water.  &lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-05-27T19:07:59.863" CommentCount="2" />
  <row Id="10499" PostTypeId="2" ParentId="10463" CreationDate="2011-05-27T19:38:47.843" Score="0" ViewCount="" Body="&lt;p&gt;You are correct that the modulated signal touches a peak every 1/f and a trough every 1/f.  But if you looked at the waveform of a carrier being amplitude modulated by a much lower frequency pure sinewave, it is intuitive that there is another frequency component.  (That is the &quot;feel&quot; part.)  The math in other posts shows you what those frequency relationships are.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would add that commercial AM stations do not have signals which conform to these examples:  The examples have zero crossings other than that which would be expected from the carrier (e.g., whenever the cosine term in the modulating signal = 0).  Commercial AM stations do not allow their carriers to get modulated down to zero amplitude.  &lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-05-27T19:38:47.843" CommentCount="2" />
  <row Id="10500" PostTypeId="1" AcceptedAnswerId="10509" CreationDate="2011-05-27T20:50:13.263" Score="4" ViewCount="97" Body="&lt;p&gt;A &lt;a href=&quot;http://www.universetoday.com/85927/australian-student-uncovers-the-universes-missing-mass/&quot; rel=&quot;nofollow&quot;&gt;recent article from a popular astronomy website&lt;/a&gt; tells of discovery of missing mass (not dark matter) that has puzzled astronomers for some time. Apparently, the discovery involves enhanced electron density in filaments associated with superclusters of galaxies. How were astronomers able to determine that this baryonic mass was missing in the first place, and what percentage of total baryonic mass did it entail?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-27T23:12:35.987" LastActivityDate="2011-05-27T23:18:05.780" Title="Baryonic Missing Mass" Tags="&lt;baryons&gt;&lt;galaxies&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" />
  <row Id="10501" PostTypeId="1" CreationDate="2011-05-27T20:55:19.437" Score="1" ViewCount="109" Body="&lt;p&gt;I'm going to be brief, I just saw a Discovery Channel show that showed a lot of interesting phenomena around lightning (like &lt;a href=&quot;http://en.wikipedia.org/wiki/Lightning#Elves&quot; rel=&quot;nofollow&quot;&gt;elves&lt;/a&gt;, how cool is that(!)), and got me wondering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Thinking of lightning as a purely mechanical phenomenon, I would think the elves are the &quot;other side of the momentum balance&quot;. What I mean is this: somewhere in some cloud formation, an event happens that triggers a lightning flash. This means a ton of electrons (and other, associated particles) that start a &lt;em&gt;very&lt;/em&gt; fast journey to the surface of the Earth. Their momentum must be balanced by particles going in the opposite direction, hence &lt;a href=&quot;http://en.wikipedia.org/wiki/Lightning#Elves&quot; rel=&quot;nofollow&quot;&gt;elves&lt;/a&gt;. Am I right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Taking this further, is it possible that the &quot;trigger&quot; for the lightning flash could be atomic/molecular fusion forced by extremely high electrical fields, that may only exist for a nanosecond, causing a ton of energy to be transferred to the electrons around the atoms/molecules, and we have lightning. I would then think of the energy necessary to force the fusion to occur as a quantum fluctuation as in $$\Delta E \Delta t \leq \hbar$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Is every arc caused by a particle collision, somewhat like in a particle accelerator? It seems logical to me: there are more than enough particles in the air to collide with, and all the light could well be some form of Brehm or Cherenkov radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question I seem to be asking is where to find good scientific theory/information about lightning. Some say we know a lot about it, but I haven't found any good papers explaining it. I have a pretty good background in physics (1st year Master student) and am not afraid of serious literature. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="317" LastActivityDate="2011-05-28T04:03:47.820" Title="Lightning and nuclear fusion" Tags="&lt;fusion&gt;&lt;lightning&gt;&lt;qed&gt;" AnswerCount="2" CommentCount="11" />
  <row Id="10502" PostTypeId="2" ParentId="10501" CreationDate="2011-05-27T21:08:51.630" Score="1" ViewCount="" Body="&lt;p&gt;no, that is not how momentum conservation works; the cloud gain a counter moment briefly which is transferred later to the earth, triggering vibrational modes but making net earth momentum conserved. no need to reach a balance. It might be plausible however, that positive ions travel in the opposite direction than electrons to balance the same electric potential, so the elves are probably these bursts of plasma ions moving upward &lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-05-27T21:08:51.630" CommentCount="1" />
  <row Id="10503" PostTypeId="2" ParentId="10463" CreationDate="2011-05-27T21:10:08.713" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;a path of a wave function that touches&#xA;  a peak or a trough exactly every 1/f&#xA;  increments, regardless of the &quot;volume&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First of all, that's not correct. A peak or trough in some signal $U(t)$ is defined by $\frac{\partial U}{\partial t}|_{t_{\mathrm{peak}}}=0$. If $U$ is now a product of some carrier $U_{\mathrm{c}}(t)$ and some modulator $M(t)$,&#xA;$$&#xD;&#xA;  U(t) = M(t)\cdot U_{\mathrm{c}}(t)&#xD;&#xA;$$&#xA;then&#xA;$$&#xD;&#xA;  \frac{\partial U}{\partial t} = M\cdot \partial_t U_{\mathrm{c}} + (\partial_t M)\cdot U_{\mathrm{c}}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $U_{\mathrm{c}}$ itself will certainly &lt;em&gt;not&lt;/em&gt; be zero in a peak, so for $U$ to have a peak at the same time as $U_{\mathrm{c}}$ had, the time-derivative of $M$ needs to be zero - in other words, you need a constant amplitude!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, your idea was quite correct: the frequency of the carrier in a purely amplitude-modulated signal can always be measured exactly, because even though the peaks and dips are moved by the modulation, the &lt;em&gt;zeroes&lt;/em&gt; are &lt;em&gt;not&lt;/em&gt;! So provided that your carrier signal is always positive and your carrier is a simple sine (or other simple) wave, you can just count the zeroes in $U(t)$ over a long time $T\gg\tfrac1f$, divide the number by $2\cdot T$ (two zeroes per period), and always get exactly the frequency of $U_{\mathrm{c}}$.&#xA;&lt;sub&gt;(As a sidenote: you can in fact do this for arbitrary modulators, but it's in general more complicated.)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is: in real applications, you're mainly going through all this modulation/demodulation trouble so that you can have multiple signals transmitted simultaneously, without one getting in the way of the others. That means you can't actually measure any zero passings at all any more, because you don't know where $0$ is! You first have to filter out our specific signal, but this very filtering again only works perfectly if $M$ is constant. If it isn't and you try to filter out your carrier frequency too narrowly, you will filter away any information transmitted in $M$ too. In order to avoid this, you need to space your multiple carrier frequencies far enough apart so you can use a more tolerant filter, and the necessary spacing is just the width of the side bands.&lt;/p&gt;&#xA;" OwnerUserId="3540" LastActivityDate="2011-05-27T21:10:08.713" />
  <row Id="10504" PostTypeId="2" ParentId="9523" CreationDate="2011-05-27T22:15:09.977" Score="-1" ViewCount="" Body="&lt;p&gt;For small particles, compared to light wavelength, start here &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rayleigh_scattering&quot; rel=&quot;nofollow&quot;&gt;WP Rayleigh scattering&lt;/a&gt;&lt;/strong&gt;&#xA;Use a montecarlo method and search for: raytrace, photon mapping, rendering underwater, etc, etc,&#xA;In the simulation of light a whole field of searching and reading is waiting for you.&lt;br&gt;&#xA;There are a lot of free packages and one that is unbiased, physically based, very complete and opensource is the &lt;strong&gt;LuxRender&lt;/strong&gt; one. Find there 'fog' and 'dust' topics.&lt;br&gt;&#xA;Render the view from under water, looking above, is even more complicated because the waves will make it much more difficult to simulate. Sometimes, to render the wavy nature of the ocean surface is used a fractal approach. Rendering the atmhosphere in some sunsets also needs Rayleigh scattering. There exist some approaches that simulate the fog, flames, dust, with an approximation easier to implement although not physically correct.&lt;br&gt;&#xA;I think that you will not need to account for polarization (missing in LuxRender). Mie theory is for larger, in relation to wavelength, spheroid particles, like rain, etc.&lt;br&gt;&#xA;The current version of LuxRender is able to use the graphics card computation power to boost the simulation (OpenCL in a decent card).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;some links:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.physicallybasedrendering.com/&quot; rel=&quot;nofollow&quot;&gt;physically based rendering&lt;/a&gt; Development of an unbiased physically-based ray trace&#xA;&lt;a href=&quot;http://www.luxrender.net/en_GB/index&quot; rel=&quot;nofollow&quot;&gt;LuxRender&lt;/a&gt; is an open source software rendering system for physically correct image synthesis. &#xA;&lt;a href=&quot;http://www.refractivesoftware.com/features.html&quot; rel=&quot;nofollow&quot;&gt;Octane GPU render, unbiased&lt;/a&gt; (payed)&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_ray_tracing_software&quot; rel=&quot;nofollow&quot;&gt;List_of_ray_tracing_software&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-05-27T22:15:09.977" />
  <row Id="10505" PostTypeId="2" ParentId="10495" CreationDate="2011-05-27T22:26:12.823" Score="2" ViewCount="" Body="&lt;p&gt;I will here only comment on the traditional superstring theory story, say, from the first superstring revolution in the 1980s, and leave it to others to include more recent developments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traditionally, the $10$-dimensional target space $(M^{10},g^{(10)})$ with a metric $g^{(10)}$ is viewed as a product $M^{10}=M^4 \times K^6$ with metric $g^{(10)}=g^{(4)}\oplus g^{(6)}$, where $(M^4,g^{(4)})$ is the $4$-dimensional spacetime with a $4$-metric $g^{(4)}$, which we see and observe; and $(K^6,g^{(6)})$ is a compact $6$-dimensional Riemannian manifold, whose characteristic length scales are so small that it has avoided experimental detection so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will assume that the word &lt;em&gt;clustering&lt;/em&gt; in the question (v2) essentially refers to if $(K^6,g^{(6)})$ could be a &lt;em&gt;product&lt;/em&gt; $K^6=K^3\times L^3$ with metric $g^{(6)}=g^{(3)}\oplus h^{(3)}$ of two $3$-dimensional manifolds $(K^3,g^{(3)})$ and $(L^3,h^{(3)})$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, to have avoided experimental detection, the two $3$-dimensional manifolds $K^3$ and $L^3$ must both be &lt;em&gt;compact.&lt;/em&gt; Now, another bit of traditional string wisdom is, that to have unbroken $N=1$ supersymmetry i $4$ dimensions, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Holonomy&quot; rel=&quot;nofollow&quot;&gt;holonomy group&lt;/a&gt; of $(K^6,g^{(6)})$ must be the $8$-dimensional &lt;a href=&quot;http://en.wikipedia.org/wiki/Lie_group&quot; rel=&quot;nofollow&quot;&gt;Lie group&lt;/a&gt; $SU(3)$, see e.g., Green, Schwarz and Witten, &quot;Superstring theory&quot;, chap. 15. See also this &lt;a href=&quot;http://physics.stackexchange.com/q/4972/2451&quot;&gt;question&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the biggest holonomy group that a  $3$-dimensional Riemannian manifold can have, is the 3-dimensional Lie group $O(3)$, so $K^6=K^3\times L^3$ can at most have holonomy group $O(3)\times O(3)$, which is $6$-dimensional, and therefore too small to be $SU(3)$. Hence a product manifold $K^6=K^3\times L^3$ is ruled out.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-02T12:51:46.300" LastActivityDate="2011-06-02T12:51:46.300" CommentCount="1" />
  <row Id="10506" PostTypeId="2" ParentId="10462" CreationDate="2011-05-27T22:37:34.503" Score="5" ViewCount="" Body="&lt;p&gt;The pair of wires which leads signals to/from the antenna uses a simple trick: In order not to radiate they are configured so to carry &lt;strong&gt;opposite&lt;/strong&gt; currents. The radiation from one wire cancels the radiation from the other wire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also helps in the other way: The radiation picked up by one wire cancels the radiation picked up by the other wire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest version is the &quot;twisted pair&quot;. Two parallel wires with anti parallel currents still have some remaining radiation mainly in the plane which intersects the wires. This can be improved by twisting the wires around each other to better cancel radiation over 360 degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better, but more expensive, is the coax (co-axial) cable in which the outer wire is a cylinder surrounding the inner wire. The radiation from the anti-parallel currents is compensated equally well in all directions.&lt;/p&gt;&#xA;" OwnerUserId="3197" LastActivityDate="2011-05-27T22:37:34.503" CommentCount="1" />
  <row Id="10507" PostTypeId="2" ParentId="10464" CreationDate="2011-05-27T22:42:00.027" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Bend the light beam at the interface between a cold air mass with a mass of hot air&lt;/strong&gt; as often &#xA;seen in the mirage of the pavement.&#xA;&lt;img src=&quot;http://i.stack.imgur.com/c1neN.png&quot; alt=&quot;Light Bend&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we look into the hot air region we see the light comming from the colder region, &lt;strong&gt;the sky&lt;/strong&gt;, as &lt;em&gt;metalic&lt;/em&gt; surface.&#xA;&lt;img src=&quot;http://i.stack.imgur.com/hBvTf.jpg&quot; alt=&quot;Highway_mirage&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Physical explanation: The &lt;a href=&quot;http://en.wikipedia.org/wiki/Refractive_index&quot; rel=&quot;nofollow&quot;&gt;index of refraction&lt;/a&gt; of air exhibits a temperature dependence and makes the light rays bend more or less in the layer of separation of hot and cold layers&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-05-27T22:42:00.027" />
  <row Id="10508" PostTypeId="1" CreationDate="2011-05-27T22:46:33.607" Score="4" ViewCount="142" Body="&lt;p&gt;A stationary observer very close to the horizon of a black hole is immersed in a thermal bath of temperature that diverges as the horizon is approached. $$T^{-1} = 4\pi \sqrt{2M(r-2M)}$$ The temperature observed by a stationary observer at infinity can then be obtained through the gravitational redshift formula (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Hawking_radiation#Emission_process&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hawking_radiation#Emission_process&lt;/a&gt;) to be $$T^{-1} = 8 \pi M$$ which is what is often quoted as the temperature of a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As QGR points out &lt;a href=&quot;http://physics.stackexchange.com/questions/3698/flat-space-limit-of-the-schwarzschild-metric-and-hawking-temperature&quot;&gt;here&lt;/a&gt; in an answer to my related question &lt;a href=&quot;http://physics.stackexchange.com/q/3698/520&quot;&gt;here&lt;/a&gt;, the resulting non-zero stress-energy tensor at infinity is incompatible with the asymptotic flatness of the Schwarzschild spacetime. What exactly is going wrong here?&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-05-29T11:35:15.957" LastActivityDate="2011-05-29T11:35:15.957" Title="How can one reconcile the temperature of a black hole with asymptotic flatness?" Tags="&lt;general-relativity&gt;&lt;black-holes&gt;&lt;quantum-gravity&gt;&lt;hawking-radiation&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10509" PostTypeId="2" ParentId="10500" CreationDate="2011-05-27T23:18:05.780" Score="2" ViewCount="" Body="&lt;p&gt;The article answers the question, though it does not lean on it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;His evidence came from the orbital velocities of galaxies in clusters, rotational speeds, and gravitational lensing of background objects. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nor is this a surprise, orbital velocities and lensing are really the only reliable tools for weighing things at a very long distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to note is that the dark matter (what ever it may be) interacts gravitationally, too. So just seeing evidence of more mass than you can account for does not distinguish between some unexpected ordinary matter and dark matter, which is why the x-ray band observation is so interesting.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-05-27T23:18:05.780" CommentCount="5" />
  <row Id="10510" PostTypeId="2" ParentId="3698" CreationDate="2011-05-27T23:26:47.693" Score="1" ViewCount="" Body="&lt;p&gt;I'm not particularly satisfied with the answers that cite topology change or breakdown of semi-classicality. Particularly the latter doesn't explain why a metric which appears more and more flat seems to produce hotter and hotter Hawking radiation well before the Planckian regime is approached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I've thought about this for a while and finally have found a way to be at peace with this behaviour (save for the incompatibility of asymptotic flatness and a temperature at infinity that QGR pointed out, which I've posed as a &lt;a href=&quot;http://physics.stackexchange.com/q/10508/1228&quot;&gt;separate question&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how I'm thinking about it - do leave a comment if you disagree:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Close to the horizon, a stationary observer sees the Rindler metric and the Unruh temperature here is equal to the Hawking temperature - the observer has to fire their engines with a proper acceleration equal to the appropriate proper acceleration in the Unruh case. This also means that a free-falling observer sees no radiation at all (this is true at least close to the horizon, which I think is enough).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as I send the limit $ M \rightarrow 0 $, the appropriate observer that I'm mapping on to as I approach flat spacetime is &lt;em&gt;not&lt;/em&gt; the inertial observer in flat spacetime, but the eternally accelerating observer with proper acceleration $ 1/M $. (Unruh proper acceleration is identified with the surface gravity)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;em&gt;of course&lt;/em&gt; I should expect to see a thermal state of infinite temperature in this limit to flat spacetime since I'm not being mapped to an inertial observer in Minkowski, but to an eternally accelerating Rindler observer with infinite acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were always considering geodesic (free-falling/inertial) observers, at least close to the horizon, this limit makes perfect sense because there is no radiation (thermal or otherwise) during and after taking the limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Note that after asking the question, I confused this limit taking process with black hole evaporation - which is not really what I was asking - I just wanted to know what the temperature of the spacetime looks for flatter and flatter spacetimes - and not the complicated stuff that happens when a black hole actually evaporates. Or in other words, I wanted to know what the $M=0$ element in the set of eternal black holes looks like, and what it looks like is the spacetime seen by a Rindler observer with infinite acceleration in flat spacetime.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditorDisplayName="" LastEditDate="2011-05-28T17:45:48.937" LastActivityDate="2011-05-28T17:45:48.937" />
  <row Id="10511" PostTypeId="2" ParentId="10496" CreationDate="2011-05-28T00:11:57.350" Score="2" ViewCount="" Body="&lt;p&gt;It mentions: &lt;em&gt;Finding a global minimum&lt;/em&gt; where the function of jumping from one minimum to another is handled by quantum tunneling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the feeling that in order to just get an idea of what it can practically do one could look at the mentioned example of spin glasses. In other words, the spin coupling physics close to the actual hardware implementation itself &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Spin_glass&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Spin_glass&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relevant may be the work of Giorgio Parisi (Yes, the one of the Altarelli–Parisi parton evolution equations) and his co-workers Mezard and Virasoro.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the text of the Boltzmann 1992 Medal:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Parisi's deepest contribution concerns the solution of the&#xA;  Sherrington-Kirkpatrick mean field&#xA;  model for spin glasses. After the&#xA;  crisis caused by the unacceptable&#xA;  properties of the simple solutions,&#xA;  which used the &quot;replication trick&quot;,&#xA;  Parisi proposed his replica symmetry&#xA;  breaking solution, which seems to be&#xA;  exact, although much more complex than&#xA;  anticipated. Later, Parisi and&#xA;  co-workers Mezard and Virasoro&#xA;  clarified greatly the physical meaning&#xA;  of the mysterious mathematics involved&#xA;  in this scheme, in terms of the&#xA;  probability distribution of overlaps&#xA;  and the ultrametric structure of the&#xA;  configuration space. This achievement&#xA;  forms one of the most important&#xA;  breakthroughs in the history of&#xA;  disordered systems. &lt;strong&gt;This discovery&#xA;  opened the doors to vast areas of&#xA;  application. e.g., in optimization&#xA;  problems&lt;/strong&gt; and in neural network&#xA;  theories.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Giorgio_Parisi#Awards&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Giorgio_Parisi#Awards&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3197" LastActivityDate="2011-05-28T00:11:57.350" />
  <row Id="10512" PostTypeId="2" ParentId="10433" CreationDate="2011-05-28T00:56:02.787" Score="0" ViewCount="" Body="&lt;p&gt;As the comment to your post says: This spherical charge distribution picture hasn't really much to do with modern QM physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact we are talking about different quantities. The charge distribution is given mathematically by $\bar{\psi}\gamma^o\psi$ while the relevant quantities are the components of the polarization/magnetization tensor $\bar{\psi}\sigma^{\mu\nu}\psi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electric polarization components are zero in the current (Dirac) theory (in the local restframe) while the magnetization components are non zero. The latter represent the magnetic moment of the electron. It is the Dirac equation which mathematically describes the precession of the electron searched for in the experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A non-zero electric dipole moment would give the electron the very special property of either going one of two different ways in a non-homogeneous electric field. Just like the famous Stern Gerlach experiment shows in the case of the magnetic moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stern%E2%80%93Gerlach_experiment&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Stern%E2%80%93Gerlach_experiment&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a behavior can of course never be explained by a classical non spherical charge distribution...&lt;/p&gt;&#xA;" OwnerUserId="3197" LastActivityDate="2011-05-28T00:56:02.787" />
  <row Id="10513" PostTypeId="2" ParentId="10433" CreationDate="2011-05-28T03:59:53.757" Score="0" ViewCount="" Body="&lt;p&gt;It is probably better to talk about deviation from spherical symmetry.  If you want to think of it as a sphere then the classical radius computed from equating the mass-energy of the electron to its field&#xA;$$&#xD;&#xA;mc^2 = \frac{1}{4\pi\epsilon_0}\frac{e^2}{r}&#xD;&#xA;$$&#xA;and with variations on this way of computing this.  If there were some deviation from spherical symmetry it turns out to be on part $10^{16}$ of this classical electron radius.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-05-28T03:59:53.757" />
  <row Id="10514" PostTypeId="2" ParentId="10501" CreationDate="2011-05-28T04:03:47.820" Score="1" ViewCount="" Body="&lt;p&gt;Yes and no. There is a large scale movement back and forth of electrons, but it is not their mechanical momentum that is causing the strange phenomena, it is this HUGE displacement of charge over a very long distance causing the problem. The vertical movement of ice crystals rubs electrons off and transports them to the bottom of the cloud. At the top of the cloud is a huge ball of missing electrons thus a huge positive charge exists here. With massive balls of separated charge displaced over large distances you have the attendant very large electric fields. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electric fields extend far above the thunderstorms and into space. In the upper reaches of the atmosphere there are few atoms per unit volume and quite a few ions. Mix those with very large electric fields and strange things happen.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastActivityDate="2011-05-28T04:03:47.820" />
  <row Id="10515" PostTypeId="2" ParentId="3698" CreationDate="2011-05-28T07:04:57.343" Score="0" ViewCount="" Body="&lt;p&gt;Classically the story is as follows.&#xA;Flat space has no horizon and thus no Hawking temperature.&#xA;A tiny black hole is very different from flat space, no matter how small its mass, since it does have a horizon.&#xA;Thus, the limit $M\to 0$ differs essentially from the limiting point $M=0$, and you cannot conclude anything about the latter by studying the former.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to see that the conclusion that flat space has infinite Hawking temperature obviously is incorrect is to consider different limits of black hole space-times to Minkowski space. There are black holes with positive specific heat (take e.g. BTZ in 3 dimensions or the exact string black hole or Schwarzschild-AdS above the Hawking-Page transition), and if you take any of these black holes and take appropriate limits of mass (and possibly other parameters) to zero then also Hawking temperature drops to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the issue that confuses you is the negative specific heat of semi-classical Schwarzschild black holes, which is responsible for the non-smoothness of the limit $M\to 0$. One may argue on general grounds (or with some simple models) that quantum effects should convert the negative specific heat into a positive one for tiny black holes. Then the limit to vanishing mass would again be smooth.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-05-28T07:04:57.343" />
  <row Id="10516" PostTypeId="2" ParentId="10508" CreationDate="2011-05-28T07:24:21.113" Score="3" ViewCount="" Body="&lt;p&gt;Dear D-brane, indeed, a uniform thermal radiation would curve the Universe. Even if one doesn't immerse the black hole in a thermal bath, the outgoing Hawking radiation may violate the asymptotically flat conditions at any finite time, although just mildly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, an evaporating black hole that is not surrounded in the thermal bath ultimately evaporates and the Hawking radiation dilutes arbitrarily, so that the Universe will be asymptotically flat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And a black hole immersed in a thermal bath of the same temperature does curve the Universe, but the curvature is much smaller than the curvature near the black hole as long as the black hole is much greater than the Planck length (or Planck mass). There is a parametric gap here. In the Planck units, if the radius is $R$, then the mass is also $M=R$ (in four dimensions), but the temperature is $1/R$, the density of radiation is $1/R^d$ i.e. $1/R^4$ in four dimensions, and the amount of radiation (energy per unit time) above the horizon is $R^{d-2}/R^d = 1/R^2$, in any dimension. That's $R^3$ times smaller than $R=M$, in $d=4$, so the Hawking radiation will evaporate the black hole mass in time $R^3$  - more generally, $R^{d-1}$, which is still $R^{d-2}=R^2$ times longer than the characteristic time scale of the black hole (orbital time for light, for example). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bigger a black hole is, the more you can neglect those things. The factors $R^2$ or $R^3$ are huge because, for example, the black hole at the center of the Milky Way has 3+ million solar masses which is almost $10^{37}$ kilograms or $M=10^{45}$ Planck masses. The energy carried by the Hawking radiation is smaller by a factor that is a positive power of $10^{45}$. It's small, indeed.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-28T07:24:21.113" />
  <row Id="10517" PostTypeId="2" ParentId="10508" CreationDate="2011-05-28T07:25:06.867" Score="2" ViewCount="" Body="&lt;p&gt;It is simply not true that a non-vanishing stress tensor is incompatible with asymptotic flatness. The Schwarzschild spacetime is asymptotically flat, period. The semiclassical Hawking-calculation does not in any way change this background, unless you consider effects from backreactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you take into account backreactions the Hawking radiation does change the background, but it does so by essentially diminishing the Bondi mass at scri$^+$. (You can think of the Bondi mass as measuring the amount of energy stored in the black hole; as opposed to the ADM mass, which is defined at spatial infinity and which measures the total energy, including radiation to or from scri, it is defined on light-like infinity and can thus change with the advanced or retarded time, depending on whether you are on scr$^\pm$.) The spacetime remains asymptotically flat in this process.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-05-28T07:25:06.867" CommentCount="4" />
  <row Id="10518" PostTypeId="2" ParentId="10470" CreationDate="2011-05-28T10:16:28.157" Score="1" ViewCount="" Body="&lt;p&gt;(This is a comment to your question, but I can only post it as an answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from kinetic considerations, what people often misunderstand is the thermodynamic description of the process. If you have a pure liquid (say water) then it can only boil (transform into vapour in the bulk) and not evaporate. At an external pressure of 1 atm, water boils at 373 K approx. The phase rule tells you that boiling temperature can only depend on external pressure. When you have a mixture, such as the two-phase problem of water and air (regard for simplicity the air is a pure substance as well) then you have to consider the composition of both phases. There will be some water vapour in the gas phase that will be given by the vapour pressure, and some air mixed in the liquid (given by Henry's law). This is the reason why in equilibrium, at any temperature below 373K you have humidity in the air and some air in the water. It is precisely this air what you observe as forming bubbles in a container when you start heating, quite before the liquid itself boils. (As an aside, the amount of CO2 in the oceans is a major component of the balance of atmospheric CO2 at a given temperature.)&lt;/p&gt;&#xA;" OwnerUserId="3779" LastEditorUserId="3779" LastEditorDisplayName="" LastEditDate="2011-05-28T17:06:07.860" LastActivityDate="2011-05-28T17:06:07.860" CommentCount="5" />
  <row Id="10519" PostTypeId="1" CreationDate="2011-05-28T14:37:13.033" Score="1" ViewCount="35" Body="&lt;p&gt;I read that the field that have simultaneously minimum divergence and minimal transversal extension is the fundamental Gaussian mode. What are the good references in this subject, to prove the assertion above?&lt;/p&gt;&#xA;" OwnerUserId="1618" LastEditorUserId="1618" LastEditorDisplayName="" LastEditDate="2011-05-28T21:03:08.237" LastActivityDate="2011-05-28T21:03:08.237" Title="EM Field having simultaneously minimum divergence and minimal transversal extension" Tags="&lt;electromagnetic-radiation&gt;" />
  <row Id="10520" PostTypeId="2" ParentId="10470" CreationDate="2011-05-28T14:40:49.147" Score="3" ViewCount="" Body="&lt;p&gt;For every temperature, there is some amount of water vapor that can exist as gas mixed in with the air. This is called the &lt;em&gt;saturation pressure&lt;/em&gt; of water at that temperature. The &lt;em&gt;relative humidity&lt;/em&gt; is the amount of water vapor, expressed as a percentage of the saturation pressure. As you increase the temperature, the saturation pressure increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steam is water in its gaseous phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't see water vapor, you can't see steam, but you can see mist, which is water droplets suspended in the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you boil water on the stove, you get steam. This then cools when it comes into contact with the air, increasing the relative humidity above 100%, so the water vapor condenses into mist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the relative humidity is bigger than 100%, water vapor will condense from the air, becoming dew and/or mist. If the relative humidity is less than 100%, water will evaporate into the air, becoming water vapor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the wooden bridge is warmer than the surrounding air, and the relative humidity is around 100%, then water will evaporate off of the wooden bridge, turning into water vapor (the relative humidity is lower right next to the bridge, because the bridge is warmer). When the air containing this water vapor rises and cools, water condenses out of it, turning into the mist that you see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://docs.engineeringtoolbox.com/documents/689/air-moisture-temperature-saturation-diagram-SI.png&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a graph of the saturation pressure (from &lt;a href=&quot;http://www.engineeringtoolbox.com/water-vapor-saturation-pressure-air-d_689.html&quot; rel=&quot;nofollow&quot;&gt;this website&lt;/a&gt;). Note that at 100&amp;deg;C, the pressure is around 101000 Pascals, which is roughly atmospheric pressure. This means that at 100&amp;deg;C, you can have pure water vapor at atmospheric pressure. This is why water boils at 100&amp;deg;C at sea level&amp;mdash;a bubble of steam can form below the surface of the water. At higher altitudes, the boiling point can be substantially lower.&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2011-05-28T14:40:49.147" />
  <row Id="10521" PostTypeId="2" ParentId="9161" CreationDate="2011-05-28T15:07:32.550" Score="0" ViewCount="" Body="&lt;p&gt;Most physical systems modeled by PDEs can be transformed to an integral weak form where the smoothness requirements of the unknowns are lower (*). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, although in the original PDE the density field has to be differentiable (and thus very smooth because derivatives of the density appear), you can write this PDE in a different equivalent form in which the density field only needs to be integrable (and as such accepts kinks and discontinuities). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find examples of the process by googling for weak solutions of the Euler eqts, Navier-Stokes, Maxwell, convection-diffusion-reaction, Stokes eqt...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) by multiplying the PDEs with spaces of sufficiently smooth test functions, integrating over the domain, and shifting the derivatives from the unknowns to the test functions by means of the divergence Theorem.&lt;/p&gt;&#xA;" OwnerUserId="3782" LastActivityDate="2011-05-28T15:07:32.550" />
  <row Id="10522" PostTypeId="1" AcceptedAnswerId="10524" CreationDate="2011-05-28T16:36:29.247" Score="3" ViewCount="127" Body="&lt;p&gt;Helmholtz theorem states that given a smooth vector field $\pmb{H}$, there are a scalar field $\phi$ and a vector field $\pmb{G}$ such that&#xA;$$\pmb{H}=\pmb{\nabla} \phi +\pmb{\nabla} \times \pmb{G},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&#xA;$$\pmb{\nabla} \cdot \pmb{G}=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this decomposition unique? That is, given $\pmb{H}$, are the fields $\phi$, $\pmb{G}$ satisfying the above equations unique?&lt;/p&gt;&#xA;" OwnerUserId="3064" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-29T15:57:30.310" LastActivityDate="2011-05-29T15:57:30.310" Title="Uniqueness of Helmholtz decomposition?" Tags="&lt;mathematical-physics&gt;&lt;mathematics&gt;&lt;vectors&gt;&lt;vector-fields&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="10523" PostTypeId="1" AcceptedAnswerId="10526" CreationDate="2011-05-28T17:16:20.177" Score="2" ViewCount="158" Body="&lt;p&gt;In Physics class, we were building parallel circuits, and as more lights were attached in parallel, they got brighter (as more power was being provided to the lights, and the resistance decreases). So, when I charge a phone, why doesn't the battery charge faster when it is in use (eg. more devices powered on within the phone), than when it is not in use?&#xA;&lt;br /&gt;Clarification: All lights at once became brighter&lt;/p&gt;&#xA;" OwnerUserId="3366" LastEditorUserId="3366" LastEditorDisplayName="" LastEditDate="2011-05-28T17:23:42.940" LastActivityDate="2011-05-29T14:37:02.037" Title="Why doesn't a phone charge faster, rather than slower when it is in use" Tags="&lt;electricity&gt;&lt;power&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="10524" PostTypeId="2" ParentId="10522" CreationDate="2011-05-28T17:18:29.150" Score="5" ViewCount="" Body="&lt;p&gt;With suitable boundary conditions, the decomposition is unique. Without them, it's not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that $(\phi,{\bf G})$ and $(\phi&amp;#39;,{\bf G}&amp;#39;)$ are two different decompositions for the same function. Then &#xA;$$&#xD;&#xA;\nabla(\phi-\phi&amp;#39;)+\nabla\times({\bf G}-{\bf G}&amp;#39;)=0.&#xD;&#xA;$$&#xA;Take the divergence of both sides to find that&#xA;$$&#xD;&#xA;\nabla^2(\phi-\phi&amp;#39;)=0.&#xD;&#xA;$$&#xA;So for any two distinct decompositions, the scalar field $\phi$ must differ by a harmonic function $f$ (that is, one with $\nabla^2f=0$). Moreover, any harmonic function will work -- that is, there will be a way to choose a ${\bf G}&amp;#39;$ to go along with this $\phi&amp;#39;$. To see this, note that we have to choose ${\bf G}&amp;#39;$ to satisfy&#xA;$$&#xD;&#xA;\nabla\times({\bf G}&amp;#39;-{\bf G})=\nabla f.&#xD;&#xA;$$&#xA;The right side of this expression is divergence-free (because it's $\nabla^2f$), and any divergence-free vector field can be expressed as the curl of some other divergence-free vector field, so ${\bf G&amp;#39;}-{\bf G}$ exists. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A couple of notes: This latter fact is the one that lets us define the vector potential for a given magnetic field, specifically in Coulomb gauge. To be honest, I don't remember the proof that there exists a function ${\bf G}$ whose curl is ${\bf B}$ for any divergence-free ${\bf B}$. I do remember how you show that, having gotten such a ${\bf G}$, you can make it divergence-free: Just subtract off $\nabla q$ where $\nabla^2q=\nabla\cdot{\bf G}$. The new ${\bf G}$ will have the same curl as the old one and will be divergence-free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other thing: complications arise if the domain we're considering isn't simply connected. Let's say it is.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is that, to make the decomposition unique, you have to impose strong enough boundary conditions to make it so that no harmonic functions exist. For a compact domain without boundary (such as the surface of a sphere), you don't need any boundary conditions: there are no non-constant harmonic functions on such domains. (Slick proof of this: you can prove that harmonic functions never have local maxima or minima, but a nonconstant function on such a domain must have them -- in particular, it must have a global maximum and a global minimum somewhere.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a compact region with boundary, you need to specify either $\phi$ or the normal component of $\nabla\phi$ on the boundary. For good old infinite space, you need to specify that $\phi$ approach zero (or some other given function) as you tend to infinite distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to check that without such boundary conditions, you get into trouble. For instance, take the functions&#xA;$$&#xD;&#xA;\phi=x, {\bf G}=z\hat{\bf j}.&#xD;&#xA;$$&#xA;They give rise to&#xA;$$&#xD;&#xA;{\bf H}=\nabla\phi+\nabla\times{\bf G}=\hat{\bf i}-\hat{\bf i}=0.&#xD;&#xA;$$&#xA;So this pair can be added to any Helmholtz decomposition without changing the original vector field.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-05-29T15:06:58.073" LastActivityDate="2011-05-29T15:06:58.073" CommentCount="2" />
  <row Id="10525" PostTypeId="1" AcceptedAnswerId="10534" CreationDate="2011-05-28T17:52:45.873" Score="2" ViewCount="100" Body="&lt;p&gt;Prove or disprove the following proposition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any smooth plane vector field $\mathbf{H}=\left(H_x,H_y\right)$, there exist scalar potentials $\phi$, $\psi$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_x=\frac{\partial \phi }{\partial x}+\frac{\partial \psi }{\partial y}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_y=\frac{\partial \phi }{\partial y}-\frac{\partial \psi }{\partial x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3064" LastActivityDate="2011-05-29T19:07:10.720" Title="Helmholtz decomposition in the plane" Tags="&lt;mathematical-physics&gt;&lt;mathematics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="10526" PostTypeId="2" ParentId="10523" CreationDate="2011-05-28T18:02:32.420" Score="3" ViewCount="" Body="&lt;p&gt;When you add components in parallel, the voltage across each component is the same.  Thus, in a simple AC circuit, when you add an extra light bulb in parallel, the  brightness of the light bulbs should stay the same, not increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might seem wrong because there is less total resistance in the parallel circuit, and hence more total current, but the current is divided between the different branches.  The results is that the brightness of a single light bulb is constant.  The total brightness of the light bulbs combined increases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anything, there might be some nominal decrease in brightness per light bulb due to the internal resistance of the battery - when you add more light bulbs in parallel, their effective resistance decreases, so the battery's internal resistance becomes more important and the bulbs are slightly dimmer.  This should be a very small effect for a realistic circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogy to the phone is a non-sequitur.  Presumably it charges slower when in use because the phone is powered by the battery at all times, so using it drains the charge.  I don't know for sure this is how it works, but absent more detailed information about the circuitry of the phone, knowing how a simple DC circuit with light bulbs in parallel works simply does not speak to the workings of the phone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit:&lt;/b&gt; Two people have said that the charger supplies a fixed current, not a fixed voltage, and that this is the real explanation for why the phone charges slower when you are using it.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-05-29T14:37:02.037" LastActivityDate="2011-05-29T14:37:02.037" />
  <row Id="10527" PostTypeId="1" CreationDate="2011-05-28T18:53:47.380" Score="2" ViewCount="391" Body="&lt;p&gt;why should one take more than four dimensions seriously? I mean why shouldn't I simply ignore it as BS? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe is complex enough with 4 dimensions and why do we need more dimensions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before taking any theory with more than 4 dimensions seriously, do we ask for proof of these extra dimensions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a simple layman's explanation for the need [or strong hint] for extra dimensions?&lt;/p&gt;&#xA;" OwnerUserId="1643" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-31T10:03:02.377" LastActivityDate="2011-06-01T00:51:28.257" Title="Is there any theoretical proof for more than 4 spacetime dimensions?" Tags="&lt;dimensionality&gt;&lt;group-representations&gt;&lt;unified-theories&gt;" AnswerCount="5" CommentCount="8" FavoriteCount="0" />
  <row Id="10528" PostTypeId="2" ParentId="7121" CreationDate="2011-05-28T19:05:06.633" Score="0" ViewCount="" Body="&lt;p&gt;A typical control rod will be stainless steel plus boron carbide, boron steel, a silver/cadmium containing metal rod or halfmium. If these were heated up then they would be likely to melt along with the fuel. It is likely that the molten pool of fuel would have mixed with it the control rod material if a &quot;melt down&quot; occurs.&lt;/p&gt;&#xA;" OwnerUserId="3783" LastActivityDate="2011-05-28T19:05:06.633" CommentCount="1" />
  <row Id="10529" PostTypeId="2" ParentId="10525" CreationDate="2011-05-28T19:41:18.067" Score="5" ViewCount="" Body="&lt;p&gt;I ended up answering this on my own. We can certainly find a $\psi$ that solves the following partial differential equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial ^2\psi }{\partial x^2}+\frac{\partial ^2\psi }{\partial y^2}=\frac{\partial H_x}{\partial y}-\frac{\partial H_y}{\partial x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It then follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial }{\partial y}\left(H_x-\frac{\partial \psi }{\partial y}\right)-\frac{\partial }{\partial x}\left(H_y+\frac{\partial \psi }{\partial x}\right)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that there's a scalar field $\phi$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_x-\frac{\partial \psi }{\partial y}=\frac{\partial \phi }{\partial x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_y+\frac{\partial \psi }{\partial x}=\frac{\partial \phi }{\partial y}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_x=\frac{\partial \phi }{\partial x}+\frac{\partial \psi }{\partial y}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_y=\frac{\partial \phi }{\partial y}-\frac{\partial \psi }{\partial x}$&lt;/p&gt;&#xA;" OwnerUserId="3064" LastEditorUserId="3064" LastEditorDisplayName="" LastEditDate="2011-05-29T19:07:10.720" LastActivityDate="2011-05-29T19:07:10.720" />
  <row Id="10531" PostTypeId="1" CreationDate="2011-05-28T20:25:02.830" Score="4" ViewCount="115" Body="&lt;p&gt;How would the quantum mechanical treatment of the moon as a gravitationally-bound object differ from the usual treatment of the hydrogen atom using Schrödinger's equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[The earth's gravitational potential V(r) = -GM/r where M is the mass of the earth: 6 x 10&lt;sup&gt;24&lt;/sup&gt; kg. The mass of the moon is 7.4 x 10&lt;sup&gt;22&lt;/sup&gt; kg and its orbital velocity is 1,023 metres/sec. The moon's mean distance from the earth is 385 million metres.]&lt;/p&gt;&#xA;" OwnerUserId="1394" LastEditorUserId="1394" LastEditorDisplayName="" LastEditDate="2011-05-28T20:42:16.990" LastActivityDate="2011-05-28T22:07:45.183" Title="The Orbiting Moon as a Quantum Object" Tags="&lt;quantum-mechanics&gt;&lt;moon&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="10532" PostTypeId="1" CreationDate="2011-05-28T20:31:31.427" Score="4" ViewCount="230" Body="&lt;p&gt;Let me justify my question before I go on. In string theory, gravitons are strings extended over space. Longitudinal gravitons are pure gauge modes of the diffeomorphism group. However, in string theory, longitudinal gravitons are also extended objects. A condensate of longitudinal gravitons is equivalent to a diffeomorphism, but this diffeomorphism has to be smeared out over the string scale. Is the diffeomorphism group in string theory a quantum deformation smeared out over space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the weak field small string coupling limit, the string theory algebra and the classical diffeomorphism algebra ought to coincide, but in this limit, all such algebras over a Poincare invariant background are isomorphic. Away from this limit, what is the form of this quantum deformation?&lt;/p&gt;&#xA;" OwnerUserId="3785" LastActivityDate="2011-08-29T07:36:25.883" Title="Do we need a quantum deformation of the diffeomorphism group in string theory?" Tags="&lt;string-theory&gt;&lt;quantum-gravity&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" />
  <row Id="10533" PostTypeId="2" ParentId="2229" CreationDate="2011-05-28T20:46:28.480" Score="-4" ViewCount="" Body="&lt;p&gt;dear all &#xA;I  support the theory that photon have a mass please refer the essay on &lt;a href=&quot;http://www.fqxi.org/community/forum/topic/794&quot; rel=&quot;nofollow&quot;&gt;http://www.fqxi.org/community/forum/topic/794&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes&#xA;Bashir &lt;/p&gt;&#xA;" OwnerUserId="3786" LastActivityDate="2011-05-28T20:46:28.480" CommentCount="3" />
  <row Id="10534" PostTypeId="2" ParentId="10525" CreationDate="2011-05-28T22:06:29.043" Score="8" ViewCount="" Body="&lt;p&gt;Your proof is right (and I voted it up accordingly). But this is a result that's worth proving a few different ways, because the different ways lead to different insights. So I'll give some alternative proofs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof 2&lt;/em&gt; (yours is proof 1):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking linear combinations of the equations we're trying to solve, we get an equivalent pair of equations.&#xA;$$&#xD;&#xA;H_x+iH_y=\left({\partial\over\partial x}+i{\partial\over\partial y}\right)(\phi-i\psi)&#xD;&#xA;$$&#xA;$$&#xD;&#xA;H_x-iH_y=\left({\partial\over\partial x}-i{\partial\over\partial y}\right)(\phi+i\psi).&#xD;&#xA;$$&#xA;If we define some shorthand, &#xA;$$&#xD;&#xA;\partial_\pm={\partial\over\partial x}\pm i{\partial\over\partial y},&#xD;&#xA;$$&#xA;$$&#xD;&#xA;f_{\pm}=\phi\pm i\psi,&#xD;&#xA;$$&#xA;$$&#xD;&#xA;H_{\pm}=H_x\pm iH_y,&#xD;&#xA;$$&#xA;we can write these as&#xA;$$&#xD;&#xA;H_+=\partial_+f_-,&#xD;&#xA;$$&#xA;$$&#xD;&#xA;H_-=\partial_-f_+.&#xD;&#xA;$$&#xA;Now we can think of these as two uncoupled equations for the two unknowns $f_\pm$. Once we've solved for them, we can get $\phi,\psi$ from them.  So we have to prove that each of these two equations has a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve for $f_-$, we start by finding a solution $g$ to&#xA;$$&#xD;&#xA;\nabla^2g=H_+.&#xD;&#xA;$$&#xA;(This is Poisson's equation, so it has a solution.) Then set $f_-=\partial_-g$. Because of the identity&#xA;$$&#xD;&#xA;\nabla^2=\partial_+\partial_-,&#xD;&#xA;$$&#xA;we have $$\partial_+f_-=\partial_+\partial_-g=\nabla^2g=H_+,&#xD;&#xA;$$&#xA;which is what we wanted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar construction works for $f_-$. In fact, every step of the argument for $f_-$ is just the complex conjugate of the corresponding step for $f_+$ (as long as the $H$'s are real). That's the way it has to be for $\phi,\psi$ to end up real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a personal reason I like this argument. I work a lot with maps of linear polarization, which are spin-2 fields rather than spin-1 (vector) fields. The equivalent of the Helmholtz theorem for spin-2 fields is called the $E$-$B$ decomposition (in the cosmology literature anyway). The above argument generalizes in a nice way to spin-2 (and presumably higher spins).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proof 3&lt;/em&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, it seems like we ought to be able to get the 2-D result from the 3-D Helmholtz theorem, and it turns out that we can. Here's one way to think about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extend the vector field ${\bf H}$ to be a function of $(x,y,z)$: &#xA;$$&#xD;&#xA;{\bf H}_{3D}(x,y,z)={\bf H}(x,y).&#xD;&#xA;$$&#xA;For simplicity, imagine that $z$ extends only over a finite interval, say 0 to $2\pi$, and has periodic boundary conditions (i.e., points with $z=0$ are identified with those with $z=2\pi$). It's not necessary to do this, but it makes things cleaner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Helmholtz's theorem says that there are functions $\phi,{\bf G}$ such that&#xA;$$&#xD;&#xA;{\bf H}_{3D}=\nabla\phi+\nabla\times{\bf G}.&#xD;&#xA;$$&#xA;If we knew that $\phi,{\bf G}$ were independent of $z$ and that ${\bf G}$ pointed in the $z$ direction, we'd be done. But we don't (yet) know that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the trick. Both $\phi$ and ${\bf G}$ are nice, smooth functions, so they can be expanded in Fourier series in $z$:&#xA;$$&#xD;&#xA;\phi(x,y,z)=\sum_{n=-\infty}^\infty \phi_n(x,y)e^{inz},&#xD;&#xA;$$&#xA;and similarly for ${\bf G}$. Substituting these into the previous equation, we get&#xA;$$&#xD;&#xA;{\bf H}_{3D}=\sum_n\left(\nabla(\phi_ne^{inz})+\nabla\times({\bf G}_ne^{inz})\right).&#xD;&#xA;$$&#xA;Each term on the right has $z$-dependence of the form $e^{inz}$ (even after writing out the derivatives). But the left side is independent of $z$. By the uniqueness of Fourier series, it follows that all the terms with $n\ne 0$ on the right vanish! So&#xA;$$&#xD;&#xA;{\bf H}_{3D}=\nabla\phi_0+\nabla\times{\bf G}_0.&#xD;&#xA;$$&#xA;We know that $\phi_0,{\bf G}_0$ depend only on $(x,y)$, not $z$. Setting $\phi(x,y)=\phi_0(x,y)$ and $\psi(x,y)=G_z(x,y)$ gives the desired result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one could almost certainly be phrased more compactly in more formal mathematical language, probably involving symmetry groups. The basic idea is that the problem we're solving is invariant under translations in $z$, and the operations we're performing (in some sense) &quot;commute&quot; with these translations. That means it's possible to find a solution that respects that symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Proof&quot; 4:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mathematicians won't like this one, but I think it's a nice way to think about it anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take 2-D fourier transforms of everything in sight:&#xA;$$&#xD;&#xA;{\bf H}(x,y)=\int \tilde{\bf H}(k_x,k_y)e^{i{\bf k}\cdot{\bf r}}d^2k,&#xD;&#xA;$$&#xA;and similarly for $\phi,\psi$. When you do, the equations you're trying to solve decouple -- that is, each value of ${\bf k}$ can be solved independently:&#xA;$$&#xD;&#xA;\tilde{\bf H}({\bf k})=i{\bf k}\tilde{\phi}({\bf k})+i\hat{\bf z}\times{\bf k}\tilde\psi({\bf k}).&#xD;&#xA;$$&#xA;(There may be a sign error in the above, but it's &quot;morally&quot; true.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation can be solved algebraically for each ${\bf k}$. In fact, the solution has a nice physical meaning: decompose $\tilde{\bf H}$ into components parallel and perpendicular to ${\bf k}$. $\tilde\phi$ is the parallel component, and $\tilde\psi$ is the perpendicular component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like this one because it too generalizes to the spin-2 case nicely, and provides better intuition than anything else I can think of about the &quot;meaning&quot; of the $E$-$B$ decomposition for spin-2 fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I say the mathematicians won't like it is because not everything has a Fourier transform, and the convergence of Fourier transforms isn't normal pointwise convergence. But for physics applications, it's generally a fine way to think about things.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-05-28T22:24:43.733" LastActivityDate="2011-05-28T22:24:43.733" CommentCount="1" />
  <row Id="10535" PostTypeId="2" ParentId="10531" CreationDate="2011-05-28T22:07:45.183" Score="1" ViewCount="" Body="&lt;p&gt;Dear Nigel,&#xA;your question is about the analogy between the mathematical form of potentials Coulomb and Newton, but physical situations are very different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, you could do it, use wave functions on the Hydrogen atom for solve your problem, but where is the sense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do it if you consider Earth and Moon like material points, but their magnitude is very bigger than their quantum wave lenght. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you can try to enjoy yourself applying quantum mechanics for problem like this, but in this situation you must consider magnitude of Earth and Moon...so your problem is very difficult. And the wave lenghts in this problem make you consider the particles (electrons, protons, neutrons,...) that constitute Earth and Moon too...But thi is like use Relativistical Theory for calculate the motion when you walk on the road...An unuseful and terrible complication ;-) &lt;/p&gt;&#xA;" OwnerUserId="558" LastActivityDate="2011-05-28T22:07:45.183" CommentCount="2" />
  <row Id="10536" PostTypeId="1" AcceptedAnswerId="10542" CreationDate="2011-05-28T22:17:44.220" Score="7" ViewCount="172" Body="&lt;p&gt;Why when we quantize EM field, whe quantize the vector potential $A^\mu$ obtaining vectorial particles (photons) like the elastic field (phonons) and we can't quantize directly the EM-field tensor $F^{\mu\nu}$? We in that situation should obtain tensorial particles of spin 2 like graviton..It's wrong? Why?&lt;/p&gt;&#xA;" OwnerUserId="558" LastActivityDate="2011-05-29T02:27:17.637" Title="Quantizing EM field" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;electromagnetism&gt;&lt;quantum-electrodynamics&gt;&lt;field-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="4" />
  <row Id="10537" PostTypeId="2" ParentId="10527" CreationDate="2011-05-28T23:13:14.243" Score="0" ViewCount="" Body="&lt;p&gt;In an abstract sense, a &quot;dimension&quot; is just a component of a state vector.  For example, one might talk about a 10-dimensional phase-space consisting of 3 components for position, 3 for linear momentum, 3 for angular momentum, and 1 for energy.  Or one might have an &quot;event&quot; vector which includes an additional dimension representing time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are good reasons to believe that there is no 4th spatial dimension completely analogous to the 3 spatial dimensions that we are familiar with: if there were any way to move perpendicularly to space, then this would be happening all the time as a result of interacting with any object that was already moving in that direction.  For example, consider that a 4-body system (gravitational or electromagnetic) will never stay within a plane once disrupted because it is an unstable equilibrium. Perhaps such a 4th dimension exists, but it would have to have either a different topology, or there would have to be some sort of restorative force which keeps us confined to our hyperplane.  The latter case is illustrated by a pool table -- there is a third dimension perpendicular to the table but the balls are glued to the table because of gravity and the counteracting force is provided by the table itself.  There is an &lt;a href=&quot;http://www.archive.org/details/flatlandromanceo00abbouoft&quot; rel=&quot;nofollow&quot;&gt;excellent book called Flatland that you can download for free&lt;/a&gt; which addresses these issues in an intuitive and accessible way.&lt;/p&gt;&#xA;" OwnerUserId="2072" LastEditorUserId="2072" LastEditorDisplayName="" LastEditDate="2011-05-29T02:29:14.463" LastActivityDate="2011-05-29T02:29:14.463" CommentCount="7" />
  <row Id="10538" PostTypeId="1" AcceptedAnswerId="10541" CreationDate="2011-05-28T23:21:01.447" Score="2" ViewCount="114" Body="&lt;p&gt;Why is using a wheatstone bridge such an accurate way of calculating an unknown resistance? What are the benefits of using it over ohm's law?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that it has something to do with the wires heating up during ohm's law calculations, and the fact that an ammeter/voltmeter still draw some current and provide some resistance which makes results unreliable.&lt;/p&gt;&#xA;" OwnerUserId="3044" LastActivityDate="2011-05-29T00:18:47.283" Title="Wheatstone Bridge" Tags="&lt;electric-circuits&gt;" AnswerCount="2" />
  <row Id="10539" PostTypeId="2" ParentId="10527" CreationDate="2011-05-28T23:30:35.467" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;3+1 only describes the empty 'stage' where things happen.&lt;br&gt;&#xA;An empty stage is insufficient to describe the world.&lt;/strong&gt;&lt;br&gt;&#xA;As an answer the above two lines appears to be enough, but I will say a little more.&lt;br&gt;&#xA;It seems so difficult to prove, as to disprove, your radical approach. &#xA;If anyone could prove that 3+1 is sufficient this question was not here. I think that you are &lt;em&gt;presuming&lt;/em&gt; that 3+1 is all the story. Presuming is no good.&lt;br&gt;&#xA;The objective of physics is to explore the better representation of the world and sometimes it is needed to 'complicate' a little to see clearer. &#xA;Kaluza-Klein theory has one more dimension: &lt;strong&gt;charge&lt;/strong&gt;. This has to be seen as a property inherent to the 'actors' (ultimately field), and to me this kind of representation makes sense.&lt;br&gt;&#xA;How sure we are that 3+1+1 is not enough? I think that this path has more to explore before I can consider going to higher dimensions, but this is no more than a personal feeling.&lt;br&gt;&#xA;As an example: Electromagnetism happens in the 3+1 world (and the set of 'real' numbers is enough: real amplitudes, real freqs, real phases, etc..) but the mathematical representation is much more convenient when we use 'complex' numbers instead of 'reals'.&lt;br&gt;&#xA;I'm not talking in defense of any 'strangy' theory, but it seems unwise to be so radically simplistic and say 3+1 is sufficient. &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-05-28T23:30:35.467" />
  <row Id="10540" PostTypeId="2" ParentId="10538" CreationDate="2011-05-29T00:16:17.147" Score="2" ViewCount="" Body="&lt;p&gt;The reasons can be found &lt;a href=&quot;http://www.efunda.com/designstandards/sensors/methods/wheatstone_bridge.cfm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;: &#xA;Even in the most simplistic case when all Rs (they are reference resistors with low tolerance error) are equal the sensitivity to any deviation is very large (last eq in the page). See the structure of the eq put some numbers to see that the numerator goes near zero. It is temperature balanced because the effects in the ensemble cancel (the ratio of resistance changes are near 1).&lt;br&gt;&#xA;A lot of photos here, &lt;a href=&quot;http://physics.kenyon.edu/EarlyApparatus/Electrical_Measurements/Wheatstone_Bridge/Wheatstone_Bridge.html&quot; rel=&quot;nofollow&quot;&gt;interesting&lt;/a&gt;.&#xA;Years ago I saw a very sophisticated Wheatstone bridge used to measure the parameters of a transatlantic submarine communications cable. The maintenance team had to know where (at what distance in the ocean) the cable was faulty.    &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-05-29T00:16:17.147" />
  <row Id="10541" PostTypeId="2" ParentId="10538" CreationDate="2011-05-29T00:18:47.283" Score="2" ViewCount="" Body="&lt;p&gt;A Wheatstone bridge is only a very sensitive way of finding a match between two resistances. You have got to have a nice, well calibrated variable resistance on the opposite side to the unknown resistance. The accuracy with which you can judge the unknown is limited by the calibration accuracy of the variable resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Ohm's law to find an unknown resistance is a great way to do it. You just match the voltage you apply to the unknown based on the range of resistance you expect to encounter. Suppose you had a short piece of heavy copper buss bar you wanted to measure the resistance of. I have a nice AC welder that can put out 400 amps for brief periods. I will short the output with the buss bar, put 400 amps to it and while the current is flowing, I will measure the voltage difference between each end of the bar. With such a small resistance in the unknown resistor, it takes a lot of amps to generate a nice measurable voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, the mega high resistance of thick layers of insulation can be measured by putting a couple of tens of kilovolts to it and reading the milliamperage. I use neon sign transformers and large resistor banks for stepping down and for amp measurements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a series of calibrated shunts that are useful. They run a millivolt per amp. You read .5 volts of potential from one end to the other of the shunt - that is 500 amps. I have a big brass shunt on my car dash that gives me a reading of starting amps. I ran old welding cable to bring the starter amps into the passenger compartment. I have a giant switch on the dash, custom made of 1/4&quot; x 2&quot; copper bussbar in several pieces up to a foot long. No auto thief would ever have the balls to close that switch. I have skulls and red warnings. High voltage, etc.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastActivityDate="2011-05-29T00:18:47.283" CommentCount="3" />
  <row Id="10542" PostTypeId="2" ParentId="10536" CreationDate="2011-05-29T01:41:47.053" Score="7" ViewCount="" Body="&lt;p&gt;Just because $F^{\mu\nu}$ has two indices does not mean that it represents a spin-2 particle. Note that the metric $g^{\mu\nu}$ is a symmetric two indexed object while the EM field strength $F^{\mu\nu}$ is antisymmetric.  In fact, the metric $g^{\mu\nu}$ is analogous to potential $A^\mu$ in EM and the field strength of gravity is the four indexed Riemann tensor $R_{\mu\nu\rho\lambda}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What spin the field represents depends on the symmetries of the indices and the field equations that it obeys.  In particular, the physical degrees of freedom of a massless field of spin $(A/2,B/2)$ can be written in terms of dotted and undotted indices $G_{\alpha_1,\dots,\alpha_A,\dot\beta_1,\dots,\dot\beta_B}$ totally symmetric in the dotted and undotted indices. &#xA;In order for it to be a representation of the Poincare group it must satisfy the supplementary conditions&#xA;$$ \begin{align}&#xD;&#xA;\partial^{\gamma\dot\gamma}G_{\alpha_1,\dots,\alpha_{A-1}\gamma,\dot\beta_1,\dots,\dot\beta_B} &amp;amp;= 0 \\&#xD;&#xA;\partial^{\gamma\dot\gamma}G_{\alpha_1,\dots,\alpha_{A},\dot\beta_1,\dots,\dot\beta_{B-1}\dot\gamma} &amp;amp;=0 \ .&#xD;&#xA;\end{align}$$&#xA;These supplementary conditions imply that each component of $G$ satisfies the Klein-Gordon equation. (The only exception is the scalar case where there are no supplementary conditions because there's no indices. In this case there's just the KG equation $\Box G = 0$.)&#xA;It is also shown that these conditions imply that the field has a definite helicity of $(A-B)/2$ and there is only one degree of freedom in such a field. Physical fields are made of the sum of two such fields of opposite helicity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is clearly the case for the EM field strength $F_{\mu\nu}$ when written in the form $F_{\alpha\beta\dot\alpha\dot\beta}=(\sigma^\mu)_{\alpha\dot\alpha}(\sigma^\nu)_{\beta\dot\beta}F_{\mu\nu}=2\varepsilon_{\alpha\beta}\bar{F}_{\dot\alpha\dot\beta} + 2\varepsilon_{\dot\alpha\dot\beta}F_{\alpha\beta}$, so the field strength decomposes into the sum of two massless fields carrying helicity $\pm1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a good description of the field representations of the Poincare group in section 1.8 of &lt;a href=&quot;http://books.google.com.au/books?id=jZ1Y1Z3BoicC&quot; rel=&quot;nofollow&quot;&gt;Ideas and Methods in Superspace and Supergravity&lt;/a&gt;. Section 1.8.3 deals with the massless representations applicable in the two cases raised in your question.&#xA;Section 1.8.4 has the examples of massless scalar, spin-1/2, EM (spin-1), spin-3/2 and linearized gravity (spin-2).&lt;/p&gt;&#xA;" OwnerUserId="429" LastEditorUserId="429" LastEditorDisplayName="" LastEditDate="2011-05-29T01:47:05.283" LastActivityDate="2011-05-29T01:47:05.283" CommentCount="1" />
  <row Id="10543" PostTypeId="2" ParentId="10536" CreationDate="2011-05-29T02:27:17.637" Score="3" ViewCount="" Body="&lt;p&gt;It's possible to construct a free quantum field on the basis of the EM field instead of on the basis of the EM potential, however if we do so then we cannot use minimal coupling to define an interaction between a spinor field and the EM field. Whatever alternative model one constructs would have to be empirically closely similar to QED. That's moderately killing, I believe, although I have been actively pursuing algebraic constructions in this general area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the EM field, the commutation relations for the creation and annihilation operators can be presented as&#xA;$$[a_{\mu\alpha}(x),a_{\nu\beta}^\dagger(y)]=\int k_{[\mu}g_{\alpha][\nu}k_{\beta]} 2\pi\delta(k^2)\theta(k_0)e^{-ik\cdot (x-y)}\frac{\mathrm{d}^4k}{(2\pi)^4},$$&#xA;antisymmetrizing on the $\mu,\alpha$ and $\nu,\beta$ components, in contrast to the massless Klein-Gordon free field, where there are no space-time indices, for which&#xA;$$[a(x),a^\dagger(y)]=\int 2\pi\delta(k^2)\theta(k_0)e^{-ik\cdot (x-y)}\frac{\mathrm{d}^4k}{(2\pi)^4}.$$&#xA;The first equation is what one obtains by taking the derivatives of the commutation relations of the EM potential. One doesn't see it much in the literature, AFAIK, although a closely related statement can be found in R. Menikoff and D. H. Sharp, &lt;em&gt;J. Math. Phys.&lt;/em&gt; &lt;strong&gt;18&lt;/strong&gt; (1977) 471.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, perhaps, the greater simplicity of the EM field commutation relations, which are already positive semi-definite on the space of test functions as stated here without requiring the Gupta-Bleuler or other machinery to ensure that there is no negative norm sector. It may be that this approach might be helpful in quantum optics, or in other Physics that does not implement interactions by minimal coupling, because of this simplification, however, again AFAIK, I do not believe it is much used. I would be somewhat surprised if it's often used, because the simplification is more conceptual than helpful for calculation, and there is always some pressure to use conventional notations and formalisms.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-05-29T02:27:17.637" />
  <row Id="10544" PostTypeId="1" AcceptedAnswerId="10546" CreationDate="2011-05-29T03:31:10.830" Score="0" ViewCount="80" Body="&lt;p&gt;certain particles (i.e: certain bosons like the photon) do not have an anti-particle, or rather, they are they own anti-particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;lets assume that such symmetry is only approximate and these particles are actually different than their antiparticles, just that the broken symmetry is very small and hard to detect&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt; what experimental facts would we expect to change if such symmetries were only approximate? what physical consequences would they show?&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-05-29T05:51:18.310" Title="self-antiparticles and broken symmetries" Tags="&lt;experimental-physics&gt;&lt;photons&gt;&lt;antimatter&gt;&lt;poincare-symmetry&gt;&lt;broken-symmetries&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10546" PostTypeId="2" ParentId="10544" CreationDate="2011-05-29T04:50:49.840" Score="0" ViewCount="" Body="&lt;p&gt;Within the Standard Model of particle physics it is  the gauge bosons that are their own antiparticle, Z,W,gamma,  the elusive Higgs, plus the gluon of QCD. The SM is consistent with all data gathered up to now, bar the Higgs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any consequences from a hypothesis that there exists an antiphoton to a photon, one would need much more detail than the statement, i.e. the difference consisting in what? Quantum numbers? That there exists a higher symmetry that gives a different quantum number to gauge bosons than to gauge antibosons? One would have to posit a higher symmetry, certainly necessitated by new solid data, and then, if it predicts such a strange situation, check for existence of antiphotons according to the predictions of this new symmetry to falsify or prove its consistency .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;anti&quot; label has to be defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot even think of an experiment that would distinguish between putative antiphotons from photons given the world view we have of particle physics now.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-29T04:50:49.840" CommentCount="1" />
  <row Id="10547" PostTypeId="2" ParentId="10544" CreationDate="2011-05-29T05:27:16.797" Score="1" ViewCount="" Body="&lt;p&gt;Dear lurscher, in quantum mechanics - as demonstrated in quantum field theory - particles of the same species are identical so their wave functions are symmetric (for bosons) or antisymmetric (for fermions). If your new hypothetical antiparticle species were physically different, this symmetry or antisymmetry would have to be broken, and this would not be a small change of the physics - it would be a huge and qualitative change of physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, there would have to exist two independent gravitational fields if you said that the gravitons were not the same thing as the antigravitons. In a similar way, there would be too independent electromagnetic fields - one excited by matter and one excited by antimatter (and containing photons that would be different). Those possibilities are excluded both theoretically and experimentally.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-29T05:27:16.797" />
  <row Id="10548" PostTypeId="2" ParentId="10532" CreationDate="2011-05-29T05:34:54.483" Score="0" ViewCount="" Body="&lt;p&gt;Dear Gustav, nope, there is no deformation of the gauge symmetries in string theory; there is just an infinitesimal extension because aside from the usual massless gauge symmetries we know from quantum field theory, are are - at least in some formulations such as string field theory - a whole infinite tower of additional gauge symmetries that combine into a stringy gauge symmetry (with a whole string field being the parameter of the gauge transformation). The string field theory only works properly for the open strings which can't gravitate but the discussion is analogous for closed strings as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the massless gauge symmetries form a subgroup of the stringy gauge symmetry algebra and it is completely undeformed. The internal structure of a string is extended in space but it's still true that the unphysical model of the particle that emerges out of the strings knows about a particular point in spacetime - essentially the center-of-mass of the string - so the spatial extension of the string doesn't make the theory nonlocal in this simple sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a sense in which much of the nonlocality that you would expect from the extended character of the strings is spurious. For example, perturbative string theory's amplitudes still obey (but saturate) the inequalities one may derive for local quantum field theories. Also, you may imagine that the pieces of the string that are &quot;very far from its center of mass&quot; are associated with high-frequency vibrations of the string, so all their physical effects get rapidly averaged out over the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for many calculations, this is not just approximate: some &quot;derived fields&quot; and &quot;derived gauge symmetries&quot; from string theory behave &lt;em&gt;exactly&lt;/em&gt; locally. So string theory, although it may superficially look like a theory of &quot;generic nonlocal objects&quot;, is actually much closer to a strictly local quantum field theory than how it looks like.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-29T05:34:54.483" CommentCount="1" />
  <row Id="10549" PostTypeId="1" AcceptedAnswerId="10561" CreationDate="2011-05-29T09:10:19.347" Score="1" ViewCount="86" Body="&lt;p&gt;I find the derivation of the Karman-Howarth-Monin relation in the book Turbulence from Frisch (1995) a bit to short. Can someone point me to a more detailed derivation of that relation, if possible in a publicly available source (lecture notes, thesis, arXiv paper)? Or can someone explain to me how Frisch obtains the equation that follows the sentence on page 78:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Starting from the Navier-Stokes equation (6.6), we obtain&#xA;  $$\begin{align} &#xD;&#xA;\partial_t \frac{1}{2}\langle v_i v&amp;#39;_i \rangle =&amp;amp; &#xD;&#xA;- \frac{1}{2} \partial_j \langle v_i v_j v&amp;#39;_i\rangle &#xD;&#xA;- \frac{1}{2} \partial&amp;#39;_j \langle v&amp;#39;_i v&amp;#39;_j v_i\rangle \\&#xD;&#xA;&amp;amp; - \frac{1}{2} \langle v&amp;#39;_i \partial_i p \rangle &#xD;&#xA;- \frac{1}{2} \langle v_i \partial&amp;#39;_i p&amp;#39; \rangle\\&#xD;&#xA;&amp;amp;+ \frac{1}{2} \langle v&amp;#39;_i f_i \rangle &#xD;&#xA;+ \frac{1}{2} \langle v_i f&amp;#39;_i \rangle \\&#xD;&#xA;&amp;amp; + \frac{1}{2} \nu \left( \partial_{jj} + \partial&amp;#39;_{jj} \right) \langle v_iv&amp;#39;_i\rangle.&#xD;&#xA;\qquad (6.11) \end{align} $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1648" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-29T23:24:03.810" LastActivityDate="2011-05-29T23:24:03.810" Title="How to derive the Karman-Howarth-Monin relation for anisotropic turbulence?" Tags="&lt;fluid-dynamics&gt;&lt;reference-request&gt;&lt;turbulence&gt;&lt;navier-stokes&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10550" PostTypeId="2" ParentId="10523" CreationDate="2011-05-29T09:42:47.030" Score="1" ViewCount="" Body="&lt;p&gt;Assuming your phone is typical, the charger can only generate a limited current. When the phone is off, all the current from the charger goes to the battery. When the phone is on, the current from the charger goes partly to charging the battery and partly to running the phone. Hence the battery takes longer to charge.&lt;/p&gt;&#xA;" OwnerUserId="1325" LastActivityDate="2011-05-29T09:42:47.030" CommentCount="2" />
  <row Id="10551" PostTypeId="1" CreationDate="2011-05-29T11:57:16.697" Score="3" ViewCount="239" Body="&lt;p&gt;If I think of a photon as a particle, I think a parallel wire filter should transmit proportionally to the uncovered area.  (and reflect proportionally to the covered area).  Obviously polarization adds complexity by introducing a second factor to be considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if I consider the photons as waves, I think the transmission should be less sensitive to the area covered. Both a fairly thick and a fairly thin filter should transmit most of the light in one polarization and reflect most of the light in the opposite polarization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more precise, consider a spectrum of filters with thick and thin wires. (on the same spacing).&#xA;At one extreme end of the spectrum the thinest wires approach an empty space or vacuum, which transmits everything and reflects nothing.&#xA;At the other end the thickest wires merge together and form a continuous surface, in effect a mirror which reflects everything and transmits nothing.&#xA;My question is what happens in between these two extremes?  Of course there are two answers, one for parallel polarization and one for perpendicular polarization.  What are the two formulas for transmission as a function of wire thickness, or percent of space blocked?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is the answer for the transmitted polarization  is close to a step function, only rounded off in the corners. Other possibilities is a linear function decreasing from one hundred percent to zero, or from fifty percent to zero.  What are the two correct equations for the two complementary polarizations?  Or more specifically,  what is the correct formula for the transmitted polarization?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For part credit, just give a reference or a convincing derivation showing whether this function is closer to a step function or closer to a linear approximation.&#xA;For full credit, provide a reasonably accurate formula with source or derivation showing the full transition from zero to one hundred percent wire coverage.&#xA;It should be accurate enough to get the round off in the corners right to a percent or two (at least 5%)  or prove that there is no significant round off.&#xA;A good answer will have either a derivation or a reference to a reputable source.&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-08-08T11:53:34.517" Title="Particles, waves and parallel wire filters. Transmission formula?" Tags="&lt;electromagnetism&gt;&lt;photons&gt;" AnswerCount="5" CommentCount="10" FavoriteCount="1" />
  <row Id="10552" PostTypeId="1" CreationDate="2011-05-29T14:36:19.823" Score="7" ViewCount="115" Body="&lt;p&gt;Various masses for the Higgs field are compatible with experiment, but is it possible that the Higgs field is not observable because it has a continuous mass spectrum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work in the 60s and 70s on free fields that have a continuous mass spectrum, often called generalized free fields, established that there is no simple particle interpretation if the mass spectrum is not delta function valued, which I take it can be described as due to destructive interference between the different mass components. [I'm taking that largely from R. F. Streater, &lt;em&gt;Rep. Prog. Phys.&lt;/em&gt;, &lt;strong&gt;38&lt;/strong&gt;, 771-846 (1975).]&#xA;Is the possibility that the Higgs has a continuous mass spectrum to be found in the literature, or is there some sufficiently obvious anomaly or other reason why it is not considered in any detail?&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-09-04T01:48:00.570" Title="Is there literature on a continuous mass spectrum for the Higgs field?" Tags="&lt;quantum-field-theory&gt;&lt;particle-physics&gt;&lt;standard-model&gt;&lt;higgs-boson&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="10553" PostTypeId="1" AcceptedAnswerId="10554" CreationDate="2011-05-29T15:13:49.840" Score="0" ViewCount="120" Body="&lt;p&gt;If I have a force, say 24 kg/cm what would that equate to at 2cm? I would like to know the formulae for calculating this. &#xA;For example. If a motor can hold an object of 24kg at 1cm from its pivot point, what is it cable of holding at 5cm? And how is it calculated?&lt;/p&gt;&#xA;" OwnerUserId="3788" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-29T15:26:54.453" LastActivityDate="2011-05-29T19:59:46.937" Title="if a force is 1 newton metre, what is it at 2 meters?" Tags="&lt;homework&gt;&lt;forces&gt;&lt;torque&gt;&lt;unit-conversion&gt;&lt;spring&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="10554" PostTypeId="2" ParentId="10553" CreationDate="2011-05-29T18:08:27.573" Score="2" ViewCount="" Body="&lt;p&gt;The units you mean are probably kg*cm (sometimes written kg.cm in robotics). Your original specification of 24 kgcm is a torque and not a force. The difference in practice is that, as the units imply, your resulting force at a point a distance from the &quot;pivot point&quot; decreases by the distance. So 24 kg*cm means that it can hold 24 kg at 1 cm or 12 kg at 2 cm etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that strictly speaking kg is not a standard unit of force in physics since the force of gravity on a kilogram varies. In robotics, this seems to be the standard unit of servo torque though.&lt;/p&gt;&#xA;" OwnerUserId="2936" LastEditorUserId="2936" LastEditorDisplayName="" LastEditDate="2011-05-29T19:59:46.937" LastActivityDate="2011-05-29T19:59:46.937" CommentCount="3" />
  <row Id="10555" PostTypeId="1" AcceptedAnswerId="10557" CreationDate="2011-05-29T18:11:48.917" Score="0" ViewCount="73" Body="&lt;p&gt;Could you check me on this: if I'm interested in total power (summed over all frequency) of CMBR absorbed by a black body in empty space, would that be proportional to the surface of BB?&lt;/p&gt;&#xA;" OwnerUserId="1600" LastActivityDate="2011-05-31T02:45:15.793" Title="Total power of CMBR" Tags="&lt;cmb&gt;" AnswerCount="1" />
  <row Id="10556" PostTypeId="2" ParentId="10551" CreationDate="2011-05-29T18:15:01.913" Score="0" ViewCount="" Body="&lt;p&gt;An infinite conducting sheet with a resistance of 377 ohms (per square) is a matched impedance to an em wave, so it reflects a quarter the power, transmits a quarter, and absorbs half. Your parallel wires will do the same when their equivalent resistance is 377 ohms per square. If you're thinking copper, don't forget to use the skin effect when calculating the resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For much lower resistances, you have a simple reflector; as the resistance becomes much higher, it simply ceases to influence the wave at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is very seldom any benefit to be gained by trying to analyze these kinds of things in terms of photons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I see now that I have done something very embarassing, for which my only excuse is that I have never before heard of these egg-slicer type polarizers. I meant to calculate the transmission in the PASSING orientation, so I allowed my egg-slicer to be parallel to the &quot;arrows&quot; of the electric field. Of course, this is the BLOCKING orientation! The passing case is much harder for me to analyze. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, I find myself second-guessing even my calculation for the blocking case, so I am going to try to do a better job of justifying it. It was easy to believe in 100% blocking when the electric field &quot;arrows&quot; where slamming into the wires of the egg slicer and just falling down. Now that I see they are lined up so they might pass between the wires, I have to ask again: How do I get 100% blocking? Here is my revised justification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analysis is based on estimating the far field based on the current flowing in the strips. It turns out that conservation of energy limits us in surprising ways. Since the strips are perfect conductors, they can have no losses. All power must be accounted in the reflected and transmitted waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the width of the strips be x (as a fraction of 1) and the current density be I. On the reflected side, the far field is proportional to Ix (the total current per strip), so the reflected power is (Ix)^2. For x=1, the reflected power is 100% (no slits) so the proportionality becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(reflected)  =  (Ix)^2  = 1  (for unit incident power)    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I=1 when x=1: this means that a unit current density generates exactly enough transmitted field to cancel the incident field. We can therefore allow our incident field to have the numerical value of 1. Note that a current density of 1 over the full extent of the sheet gives us a far field of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the transmitted side, one might think that a current I over an average area x should also generate a far field of Ix, giving a transmitted power of (Ix)^2. But we must remember to add it to the incident field before we square. Since the incident field is 1, and the phase must be such as to cancel the incoming wave immediately behind the strips, the average far field is therfore 1-Ix. So the transmitted power is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(transmitted) =  (1-Ix)^2     (again, for unit incidet power)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can try and add up the reflected and transmitted powers to see if they equal one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(reflected)  +  P(transmitted)  = (Ix)^2  +  (1-Ix))^2  =  1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no solutions for this equation except for 100% reflection or 100% transmission. If any current flows at all in the strips, it must be sufficient to reflect all incoming power. In other words, the current in each strip is a constant, regardless of how thin you cut the strips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check my methodology, I attempted to apply the same logic of energy conservation to a conducting sheet of resistance 377 ohms (the matched case). It is a useful case because we can check it against the simple case of the full sheet, with half the power absorbed, a quarter reflected and a quarter transmitted. I will spare everyone the calculation (unless I am asked for it) and simply present the results in simplest form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where x is the percentage width of the strips, the powers are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(reflected)    = (x^2)/k&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(absorbed)     = (2x)/k&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(transmitted)  = 1/k&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where k= (x+1)^2. You can see for x=1 (full sheet) you get the proportions I originally suggested.&lt;/p&gt;&#xA;" OwnerUserId="3253" LastEditorUserId="3253" LastEditorDisplayName="" LastEditDate="2011-05-30T21:54:06.867" LastActivityDate="2011-05-30T21:54:06.867" CommentCount="5" />
  <row Id="10557" PostTypeId="2" ParentId="10555" CreationDate="2011-05-29T18:27:56.880" Score="3" ViewCount="" Body="&lt;p&gt;Yes, for a black body with emissivity of one, it is proportional to the area of the collector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any other object it is also proportional to the emissivity of the object.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="3735" LastEditorDisplayName="" LastEditDate="2011-05-31T02:45:15.793" LastActivityDate="2011-05-31T02:45:15.793" CommentCount="6" />
  <row Id="10558" PostTypeId="2" ParentId="10551" CreationDate="2011-05-29T20:26:47.620" Score="0" ViewCount="" Body="&lt;p&gt;Here is a 2004 paper.....(Sorry, link died and I cannot find it again) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try: &lt;a href=&quot;http://arxiv.org/abs/0906.2139&quot; rel=&quot;nofollow&quot;&gt;&quot;Enhanced transmission of transverse electric waves through subwavelength slits in a thin metallic film&quot;&lt;/a&gt; Yu Qian Ye, Phys. Rev. E 80, 036606 (2009) [3 pages]&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="1656" LastEditorDisplayName="" LastEditDate="2011-08-08T11:53:34.517" LastActivityDate="2011-08-08T11:53:34.517" CommentCount="4" />
  <row Id="10559" PostTypeId="2" ParentId="2175" CreationDate="2011-05-29T21:06:17.967" Score="3" ViewCount="" Body="&lt;p&gt;Here is an interesting site about this idea and similar ideas:&#xA;&lt;a href=&quot;http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/FTL.html#3&quot; rel=&quot;nofollow&quot;&gt;http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/FTL.html#3&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially the problem with this idea is there are no such thing as perfectly rigid bodies. So as you push, it sends a little compression wave through the material, which travels at the speed of sound in the material, as sound is just a type of propagating compression.&lt;/p&gt;&#xA;" OwnerUserId="3018" LastActivityDate="2011-05-29T21:06:17.967" CommentCount="1" />
  <row Id="10561" PostTypeId="2" ParentId="10549" CreationDate="2011-05-29T22:32:14.333" Score="2" ViewCount="" Body="&lt;p&gt;Let me here just derive the equation (6.11) that follows the sentence, you mention.&#xA;The Navier-Stokes equation (6.6a) reads&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\partial_t  v_i +  v_j\partial_j v_i = -\partial_i p + f_i  + \nu~\partial_j\partial_j v_i.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The incompressibility condition (6.6b) reads&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\partial_jv_j=0. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we have in the unprimed and the primed point that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\partial_t  v_i = -\partial_j(v_j v_i) -\partial_i p + f_i  + \nu~\partial_j\partial_j v_i,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\partial_t  v&amp;#39;_i =-\partial&amp;#39;_j(v&amp;#39;_j v&amp;#39;_i) -\partial&amp;#39;_i p&amp;#39; + f&amp;#39;_i  + \nu~\partial&amp;#39;_j\partial&amp;#39;_j v&amp;#39;_i,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;respectively. Therefore, averaging yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align} &#xD;&#xA;\partial_t \langle v_i v&amp;#39;_i \rangle &#xD;&#xA;&amp;amp;=  \langle  v&amp;#39;_i \partial_t v_i \rangle + \langle v_i \partial_t v&amp;#39;_i \rangle  \\&#xD;&#xA;&amp;amp;= - \langle v&amp;#39;_i\partial_j(v_j v_i) \rangle - \langle v_i \partial_j&amp;#39;(v&amp;#39;_j v&amp;#39;_i) \rangle \\&#xD;&#xA;&amp;amp; -\langle v&amp;#39;_i \partial_i p \rangle -\langle v_i \partial&amp;#39;_i p&amp;#39; \rangle\\&#xD;&#xA;&amp;amp; +\langle v&amp;#39;_i f_i \rangle +\langle v_i f&amp;#39;_i \rangle \\&#xD;&#xA;&amp;amp; +\nu \langle v&amp;#39;_i\partial_j\partial_jv_i\rangle + \nu \langle v_i\partial&amp;#39;_j\partial&amp;#39;_j v&amp;#39;_i\rangle \\&#xD;&#xA;&amp;amp;= -\partial_j \langle v_i v_j v&amp;#39;_i\rangle -\partial_j&amp;#39; \langle v&amp;#39;_i v&amp;#39;_j v_i\rangle \\&#xD;&#xA;&amp;amp; -\langle v&amp;#39;_i \partial_i p \rangle -\langle v_i \partial&amp;#39;_i p&amp;#39; \rangle\\&#xD;&#xA;&amp;amp; +\langle v&amp;#39;_i f_i \rangle +\langle v_i f&amp;#39;_i \rangle \\&#xD;&#xA;&amp;amp; +\nu\left( \partial_j\partial_j + \partial&amp;#39;_j\partial&amp;#39;_j\right)\langle v_iv&amp;#39;_i\rangle,&#xD;&#xA;\end{align} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where we have used that averaging and differentiation commute, and also that primed velocities are independent of unprimed derivatives, and vice-versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The full Karman-Howarth-Monin relation is derived by Marc Brachet on p.8-9 &lt;a href=&quot;http://www.lps.ens.fr/~brachet/Publications_&amp;amp;_Reprints_files/Cours_Chili3.pdf&quot; rel=&quot;nofollow&quot;&gt;here,&lt;/a&gt; essentially following the &lt;a href=&quot;http://books.google.com/books?id=K-Pf7RuYkf0C&amp;amp;pg=PA79&amp;amp;lpg=PA79&amp;amp;dq=Karman-Howarth-Monin&amp;amp;source=bl&amp;amp;ots=BKNqDGF9w5&amp;amp;sig=AVvAmvc7FOSRVqEKfk5GQV6D2wM&amp;amp;hl=en&amp;amp;ei=fajiTdDoGoqSOqDWhb4G&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CBgQ6AEwAA#v=onepage&amp;amp;q=Karman-Howarth-Monin&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; by Uriel Frisch (1995).&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-29T23:05:56.643" LastActivityDate="2011-05-29T23:05:56.643" />
  <row Id="10562" PostTypeId="1" CreationDate="2011-05-29T23:36:42.260" Score="1" ViewCount="83" Body="&lt;p&gt;Anyone knows where I can find an errata (or any related material, such as solution sheets, etc) for this book? Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: This is not a physics question, but this book is so popular among physicists that I can't think of anyone else better to answer it.&lt;/p&gt;&#xA;" OwnerUserId="3064" LastActivityDate="2011-05-29T23:36:42.260" CommunityOwnedDate="2011-05-29T23:46:10.900" ClosedDate="2011-05-31T12:05:38.777" Title="errata for Morse &amp; Feshbach - Methods of Theoretical Physics" Tags="&lt;mathematical-physics&gt;&lt;soft-question&gt;&lt;books&gt;&lt;reference-request&gt;" CommentCount="1" />
  <row Id="10563" PostTypeId="2" ParentId="26" CreationDate="2011-05-30T01:19:14.467" Score="0" ViewCount="" Body="&lt;p&gt;This source discusses Ampere's Force Law. The force between two parallel conductors is equal to two times a constant, times the current in wire A, times the current in wire B, divided by the spacing between wires. The constant is known as the magnetic force constant and is 1 x 10E-7 newtons in the SI system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_force_law&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For parallel conductors one meter in length, one meter apart and carrying one ampere, the force between them is exactly 2 x 10E-7 newtons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force in a rail gun will be directly proportional to its current and inversely proportional to the spacing between its rails. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what length should you assume when calculating a rail gun? The length of the conductor should be considered to be the width of the projectile. This is equal to the distance between rails. As the distance between rails decreases, the unit forces increase, but the projectile gets smaller. Those two considerations fall out and it makes no difference what the rail spacing is. In other words, make the rail spacing as small as possible in order to accommodate your projectile. &lt;/p&gt;&#xA;" OwnerUserId="3735" LastActivityDate="2011-05-30T01:19:14.467" />
  <row Id="10564" PostTypeId="2" ParentId="26" CreationDate="2011-05-30T02:10:03.313" Score="0" ViewCount="" Body="&lt;p&gt;The Lorentz Force Law could also help you out,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F= q(E+ v X B) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where E is the electric field, B is the magenetic field strength, v is the velocity, and q is the charge of the object. &lt;/p&gt;&#xA;" OwnerUserId="3018" LastActivityDate="2011-05-30T02:10:03.313" />
  <row Id="10565" PostTypeId="1" CreationDate="2011-05-30T02:20:11.883" Score="9" ViewCount="313" Body="&lt;p&gt;When you release a magnetic dipole in a nonuniform magnetic field, it will accelerate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that for current loops (and other such macroscopic objects) the magnetic moment comes from moving charges, and since magnetic fields do no work on charges (F perpendicular to v) it follows that the work done on the dipole (that caused its gain in kinetic energy)  must have come from somewhere other than the magnetic forces (like electric forces in the material).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what about a pure magnetic moment? I'm thinking of a particle with intrinsic spin. Of course such a thing should be treated with quantum mechanics, but shouldn't classical electrodynamics be able to accommodate a pure magnetic dipole? If so, when I release the pure dipole in a nonuniform B-field and it speeds up, what force did the work? Is it correct to say that magnetic fields DO do work, but only on pure dipoles (not on charges)? Or should we stick with &quot;magnetic forces never do work&quot;, and the work in this case is done by some other force (what?)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to anyone who can alleviate my confusion!&lt;/p&gt;&#xA;" OwnerUserId="1429" LastEditorUserId="1429" LastEditorDisplayName="" LastEditDate="2011-05-30T05:26:16.913" LastActivityDate="2011-06-03T16:48:23.000" Title="Does a magnetic field do work on an intrinsic magnetic dipole?" Tags="&lt;electromagnetism&gt;&lt;classical-electrodynamics&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="10566" PostTypeId="1" CreationDate="2011-05-30T03:53:33.900" Score="3" ViewCount="168" Body="&lt;p&gt;Does there exist a laser that is tunable over the whole visible spectrum?&#xA;If so does there exist one that is solid state?&#xA;This is to settle a score with my dumb workmates.&lt;/p&gt;&#xA;" OwnerUserId="3791" LastActivityDate="2011-05-31T14:21:11.353" Title="Does there exist a laser that is tunable over the whole visible spectrum?" Tags="&lt;laser&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="10567" PostTypeId="2" ParentId="10565" CreationDate="2011-05-30T04:14:42.297" Score="5" ViewCount="" Body="&lt;p&gt;Yes, of course that if a field - magnetic field - is able to make a bar magnet rotate or move, it is doing work. The statement that magnetic fields don't do any work only applies to point-like pure electric charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnetic moments may be visualized as objects with a forced motion of charges (solenoids have the same magnetic field as bar magnets), and if something is moving, the magnetic force is becoming a force that does work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of formulae, the magnetic force on a charge is $q\vec v\times \vec B$ which is identically perpendicular to $\vec v$ and that's why it does no work. However, forces on magnetic dipoles and more general objects don't have the form $\vec v\times$ - they're not perpendicular to $\vec v$, so they do work in general.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-30T04:14:42.297" CommentCount="11" />
  <row Id="10568" PostTypeId="2" ParentId="9935" CreationDate="2011-05-30T06:00:44.957" Score="3" ViewCount="" Body="&lt;p&gt;You seem to have the right mathematical background to take QM head on. But the following readings will make it more enjoyable.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The Feynman Lectures on Physics&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/OAWQM.jpg&quot; alt=&quot;The Feynman Lectures on Physics&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Classical Mechanics&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/op411.jpg&quot; alt=&quot;Classical Mechanics&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CM will introduce you to a lot of new ways to look at Newtonian mechanics. This is not solving pulley problems and sliding weights off wedges. No, this introduces you to linear algebra and powerful principles such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrangian_mechanics&quot; rel=&quot;nofollow&quot;&gt;Lagrangian mechanics&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Calculus_of_variations&quot; rel=&quot;nofollow&quot;&gt;calculus of variations&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Hamiltonian_mechanics&quot; rel=&quot;nofollow&quot;&gt;Hamiltonian mechanics&lt;/a&gt;, chaos theory etc. I found the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hamiltonian_mechanics#Generalization_to_quantum_mechanics_through_Poisson_bracket&quot; rel=&quot;nofollow&quot;&gt;generalization of Hamiltonian mechanics to QM&lt;/a&gt; very elegant.&lt;/p&gt;&#xA;" OwnerUserId="3669" LastActivityDate="2011-05-30T06:00:44.957" CommunityOwnedDate="2011-05-30T06:00:44.957" />
  <row Id="10569" PostTypeId="2" ParentId="10566" CreationDate="2011-05-30T06:00:54.940" Score="1" ViewCount="" Body="&lt;p&gt;Georg  is correct. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Free-electron_laser&quot; rel=&quot;nofollow&quot;&gt;free electron laser&lt;/a&gt; is tunable:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Unlike gas, liquid, or solid-state lasers such as diode lasers, in which electrons are excited in bound atomic or molecular states, FELs use a relativistic electron beam as the lasing medium which moves freely through a magnetic structure, hence the term free electron.[ The free-electron laser has the widest frequency range of any laser type, and can be widely tunable, currently ranging in wavelength from microwaves, through terahertz radiation and infrared, to the visible spectrum, to ultraviolet, to X-rays.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it is not a lasing that can be used in a pointer or a portable method. It needs accelerated electrons. So it depends what sort of argument you have with your friend.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Today, a free-electron laser requires the use of an electron accelerator with its associated shielding, as accelerated electrons are a radiation hazard. These accelerators are typically powered by klystrons, which require a high voltage supply. The electron beam must be maintained in a vacuum which requires the use of numerous vacuum pumps along the beam path. While this equipment is bulky and expensive, free-electron lasers can achieve very high peak powers, and the tunability of FELs makes them highly desirable in several disciplines, including medical diagnosis and non-destructive testing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-30T06:00:54.940" CommentCount="2" />
  <row Id="10570" PostTypeId="1" AcceptedAnswerId="10637" CreationDate="2011-05-30T06:18:30.943" Score="1" ViewCount="119" Body="&lt;p&gt;As piece of paper is folded and unfolded, the stiffness of the sheet may seem to be greatly increased. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To those of you who don't recognise it: take a sheet of A4 paper, grasp one of the edges and move your hand vertically. Now fold the paper in half, unfold it and repeat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is anyone aware of any formal or heuristic explanation of what changed?&lt;/p&gt;&#xA;" OwnerUserId="2752" LastActivityDate="2011-05-31T20:08:22.573" Title="Paper stiffness" Tags="&lt;material-science&gt;&lt;everyday-life&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10571" PostTypeId="1" CreationDate="2011-05-30T07:03:55.957" Score="0" ViewCount="83" Body="&lt;p&gt;Assume I have the Hamiltonian for a 1D topological insulators as:&#xA;$H=sin(P_x) \sigma_x+i \Delta \sigma_{y}+(1-m-cos(P_x)) \sigma_z $ where $m$ is the mass term, $P_x$ is the momentum and $\Delta$ is an extra term to make the Hamiltonian non-Hermitian. If I choose the time reversal operator $T=i \sigma_y K$ and the parity operator $P=\sigma_{y}$, will the Hamiltonian be PT symmetric?&lt;/p&gt;&#xA;" OwnerUserId="2397" LastActivityDate="2011-05-30T07:03:55.957" Title="1D Topological insulator with PT symmetry" Tags="&lt;condensed-matter&gt;&lt;topological-insulators&gt;" CommentCount="1" />
  <row Id="10572" PostTypeId="1" AcceptedAnswerId="10575" CreationDate="2011-05-30T07:25:14.297" Score="6" ViewCount="199" Body="&lt;p&gt;Why does the sound of thunder last several seconds even when lightning lasts for only fraction of a second?&lt;/p&gt;&#xA;" OwnerUserId="3636" LastEditorUserId="3669" LastEditorDisplayName="" LastEditDate="2011-05-30T23:50:42.373" LastActivityDate="2011-05-30T23:50:42.373" Title="Extended sound of thunder" Tags="&lt;atmosphere&gt;&lt;lightning&gt;" AnswerCount="3" />
  <row Id="10574" PostTypeId="1" CreationDate="2011-05-30T08:15:34.557" Score="1" ViewCount="168" Body="&lt;p&gt;If the multiverse of eternal inflation and the many-worlds of quantum mechanics don't exist, how should we interpret the meaning of probabilities? If there is only one copy of the universe out there, what does it mean to say a radioactive nucleus has a 1/2 probability of decaying? Does it mean if we have 1,000,000 similar radioactive nuclei, 500,000 $\pm$ 1,000 of them will decay? How do we make sense of the $\pm$ 1,000 error bar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="3792" LastActivityDate="2011-05-30T15:05:12.613" Title="If the multiverse and many-worlds don't exist, how should we interpret probabilities?" Tags="&lt;cosmology&gt;&lt;quantum-interpretations&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="10575" PostTypeId="2" ParentId="10572" CreationDate="2011-05-30T08:41:29.220" Score="3" ViewCount="" Body="&lt;p&gt;This is because the different &lt;strong&gt;parts of a stroke&lt;/strong&gt; are be very &lt;strong&gt;different in distance&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Reverberations and echoes of sound&#xA;  last longer than those of light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is obviously wrong. Which walls or mountains are the reflectors of those echos?&#xA;Which rooms do the reverberations? &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-05-30T08:48:06.153" LastActivityDate="2011-05-30T08:48:06.153" CommentCount="6" />
  <row Id="10576" PostTypeId="2" ParentId="9982" CreationDate="2011-05-30T09:08:50.663" Score="1" ViewCount="" Body="&lt;p&gt;The answer to your first question is yes.&#xA;If you have only trace-free matter (e.g. only electromagnetic fields or plain vanilla vacuum with no cosmological constant) then the first of your field equations is solved by constant $\phi$, which then plays the role of Newton's constant, as evident from the second field equation, which is then nothing but GR coupled to some traceless energy momentum tensor (or vacuum GR).&#xA;Therefore, any solution to the vacuum Einstein equations is also a solution to the Brans-Dicke equations (but not vice versa, of course). This contains in particular black hole and curvature singularity solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your second question depends on which vacuum you choose. If you are in a constant dilaton vacuum (as in my answer to question 1) and have only trace-free matter then you recover GR and its singularities, with no essential change of interpretation. Otherwise you can view the dilaton just as yet-another matter source as far as singularities are concerned. There may be some qualitative changes as compared to GR if you choose generalized Brans-Dicke models with potentials where the dilaton does not obey the energy conditions that you need in the proofs of singularity theorems. But otherwise there is again no essential change in the story of singularities. Another complication arises if you define singularities not through unboundedness of some curvature invariants, but through completeness properties of geodesics (as is done commonly). Then you have to specify if your geodesics couple to the metric $g$ appearing in the Brans-Dicke action, or to some conformally related metric $\Omega(\phi)g$, with a $\phi$-dependent conformal factor. If $\Omega$ has some zeros or singularities this can change the singularity structure, even if the solution looks formally the same as in GR. You can then convert loci that are singular in GR into loci that are regular in Brans-Dicke theory and vice versa. &lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-05-30T09:08:50.663" CommentCount="3" />
  <row Id="10577" PostTypeId="1" AcceptedAnswerId="10581" CreationDate="2011-05-30T09:11:07.663" Score="1" ViewCount="88" Body="&lt;p&gt;So I was eating a plate of food one day and thought of entropy. As I understand the definition of entropy, it is the logarithm of the number of arrangements or states the object in question can be in, where the object is composed of N number of particles/food particles. Please edit my definition if this is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'm eating a plate of food, and I have diced up chicken, rice and vegetables on the plate all mixed together. I thought, the current system I have, the food is arranged in a state that is of high entropy because this arrangement is common compared to the number of other more rare arrangements, ie 1/3 chicken, 1/3 vegetables, and 1/3 rice in an pie chart looking fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like to eat neatly, and naturally, I've conditioned myself to pick up the food with my fork such that while I pick up chicken and rice, rice falls off one side of my fork in a certain direction. As I eat, I sometimes end up with a more neater, higher entropy state than what I began with. I didn't spend extra energy making it neater, I just pick up my food in strategic ways because that's how I normally eat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how is this possible? I end up with a more neat arrangement on my plate than I started with. I know food gets eaten, so the number of particles decreases with time. So because the number of food particles decreases, the number of arrangements also decrease!! Can someone continue my chain of thought and correct me if I'm wrong? What do I need to think of next? Why is it that it appears that the entropy of my plate of food seems to lower with time?&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2011-05-30T10:31:11.213" Title="Entropy and how it applies to everyday activities like eating food" Tags="&lt;entropy&gt;&lt;everyday-life&gt;&lt;food&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="10578" PostTypeId="1" CreationDate="2011-05-30T09:12:19.147" Score="3" ViewCount="199" Body="&lt;p&gt;There are many articles published in physics journals about &lt;a href=&quot;http://en.wikipedia.org/wiki/Flocking_%28behavior%29&quot; rel=&quot;nofollow&quot;&gt;flocking&lt;/a&gt;. Is there a physical reason for these phenomena or is it just because physics methods are being used to study collective motion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems there is no true mechanics in models of flocking, there are no Hamiltonians defined and so on, only rules of motion, like alignment rule and constant velocity of &lt;a href=&quot;http://en.wikipedia.org/wiki/Self_propelled_particles&quot; rel=&quot;nofollow&quot;&gt;self-propelled particles&lt;/a&gt;. Yet physicists debate the validity of &lt;a href=&quot;http://en.wikipedia.org/wiki/Mermin%E2%80%93Wagner_theorem&quot; rel=&quot;nofollow&quot;&gt;Mermin-Wagner theorem&lt;/a&gt; in simple flocking models. Isn't this fundamentally problematic because of absence of real dynamics in these systems?&lt;/p&gt;&#xA;" OwnerUserId="1800" LastActivityDate="2011-06-30T16:18:25.423" Title="Is there any physics behind flocking?" Tags="&lt;statistical-mechanics&gt;&lt;general-physics&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="10579" PostTypeId="2" ParentId="10574" CreationDate="2011-05-30T09:13:16.633" Score="4" ViewCount="" Body="&lt;p&gt;Dear student, yes, probability always has the same interpretation. If you observe 1 million events - they don't have to be similar (each of the events may be totally different!) - and each of them has the probability 1/2 to produce a &quot;Yes&quot; answer, then you will get 500,000 plus minus 1,000 &quot;Yes&quot; outcomes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an error margin because the predictions are probabilistic and they make complete sense. This is nothing specific to quantum mechanics. If you throw dice 1 million times, you get a number greater than three 500,000 times plus minus 1,000 as well. This makes complete sense and is calculable. What is hard to make sense of is your question.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-30T09:13:16.633" />
  <row Id="10580" PostTypeId="1" CreationDate="2011-05-30T09:17:09.470" Score="1" ViewCount="47" Body="&lt;p&gt;Three resistances AB, BC, CD are joined in series and placed across the terminals of a galvanometer. The resistance of the galvanometer is 15 ohm and it gives full scale deflection for a current of 5 milliampere. Find the value of each of the resistances so that the instrument will serve as a multi-range ammeter, using external connections A and B for the 0-1 ampere range, to A and C for the range 0-0.1 ampere, and to A and D for the range 0-10milliampere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume this needs some kind of simultaneous equation solving, but I can't figure it out!&lt;/p&gt;&#xA;" OwnerUserId="3044" LastActivityDate="2011-05-30T09:17:09.470" ClosedDate="2011-05-30T23:52:34.480" Title="Creating an ammeter?" Tags="&lt;homework&gt;&lt;electric-circuits&gt;" CommentCount="3" FavoriteCount="1" />
  <row Id="10581" PostTypeId="2" ParentId="10577" CreationDate="2011-05-30T10:31:11.213" Score="2" ViewCount="" Body="&lt;p&gt;QEntanglement, the answer is simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your plate of food is not a closed system. For the system to close you need the plate, and the whole of you. with your chewing and digestion. Entropy increases in closed systems.&#xA;Otherwise, think about it, could DNA exist, and the whole caboodle of organic life? We are factories that defy entropy, until we die, because we are open systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In inorganic systems, crystals crystallizing out of a solution are highly ordered, but they are not a closed system. When the solution is considered to close the system, entropy increases.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-30T10:31:11.213" CommentCount="1" />
  <row Id="10582" PostTypeId="1" AcceptedAnswerId="10584" CreationDate="2011-05-30T10:43:36.560" Score="2" ViewCount="119" Body="&lt;p&gt;Maybe another stupid question, but what's the energy of a graviton?&#xA;Is it $\hbar \omega$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it emit gravitons when an apple falls onto the ground, like photons be emitted when an electron transits from a higher energy level to a lower one?&lt;/p&gt;&#xA;" OwnerUserId="3743" LastActivityDate="2011-05-30T16:22:57.823" Title="The energy of a Graviton" Tags="&lt;quantum-mechanics&gt;&lt;gravity&gt;" AnswerCount="2" />
  <row Id="10583" PostTypeId="2" ParentId="10572" CreationDate="2011-05-30T13:35:21.017" Score="6" ViewCount="" Body="&lt;p&gt;I have always assumed it could be put down to a small set of causes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The stroke may have a physical extent and geometry to leads to a perceptible duration between the arrival of the sound due to the near park of the strike and that due to the far part. That is, &lt;a href=&quot;http://physics.stackexchange.com/posts/10575/edit&quot;&gt;what Georg said&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If the geometry of the terrain is right you may be hearing echos off of cliff faces, hill-sides, building, etc. These obviously come by a longer path and therefore arrive later.&lt;/li&gt;&#xA;&lt;li&gt;An active storm has masses of air at different temperature and humidities. &lt;a href=&quot;http://en.wikipedia.org/wiki/Speed_of_sound#Details&quot; rel=&quot;nofollow&quot;&gt;The speed of sound will differ slightly between these regions&lt;/a&gt;, leading to refraction at the boundaries. As such there may be lensing going on &lt;em&gt;in the air&lt;/em&gt; and again you hear the same sound arriving by multiple different paths at different times.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Add to that the fact that most strokes are in fact multiple discharges, and you can hear sounds that last much longer than the observed light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not gone to the trouble of taking precise data, but my impression has been that the rumble tends to be longer for strikes from further away suggesting that the last two points are the dominate effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other things to look for include&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Frequency &lt;a href=&quot;http://en.wikipedia.org/wiki/Dispersion_relation&quot; rel=&quot;nofollow&quot;&gt;dispersion&lt;/a&gt;. But if this was important you would expect the pitch of the sound to have a characteristic dependence on time, which I have never noticed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-30T15:21:32.340" LastActivityDate="2011-05-30T15:21:32.340" CommentCount="7" />
  <row Id="10584" PostTypeId="2" ParentId="10582" CreationDate="2011-05-30T14:00:01.430" Score="2" ViewCount="" Body="&lt;p&gt;It is assumed, but not measured that e =hw.  See this reference about the difficulty of measuring a single graviton.  &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0601043&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/gr-qc/0601043&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is an interesting exercise to compute the amount and wavelength of gravitons from a falling apple.  As a first approximation, one can compute the gravitational waves emitted by an apple orbiting the earth and use the formulas developed for binary black holes and neutron stars by the LIGO VIRGO etc experiments.  In low earth orbit, the apple will radiate at a wavelength of approximately 90 light minutes, give or take a factor of two.  This is very roughly 10^12 meters.  Each graviton will then carry about 10^-30 ergs, a very small amount.  According to the Wikipedia gravitational wave article the sun earth system emits 200 watts of gravitational radiation, but this would typically emit 10^-34 erg gravitons.  200 watts is 2 10^9 erg seconds, so the sun earth system is emitting 10^43 gravitons per second.  Using the formula from the Wikipedia article, the earth-apple system with a one tenth kilogram apple   would emit 10^46 times less gravitational wave power, or 10^-42 watts or 10^-35 ergs/second.  This implies an average of one graviton every 10^5 seconds, or about once every twenty 5400 second orbits.  If your falling apple falls for about one second, it should emit one graviton once out of every one hundred thousand tries.  To be more analagous to the orbital picture, your apple should be thrown horizotally like a baseball, rather than falling vertically&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-05-30T14:00:01.430" CommentCount="2" />
  <row Id="10585" PostTypeId="2" ParentId="10566" CreationDate="2011-05-30T14:12:39.253" Score="1" ViewCount="" Body="&lt;p&gt;For comparison, this company &lt;a href=&quot;http://www.radiantis.com/products/radiantis-opium-hf.html&quot; rel=&quot;nofollow&quot;&gt;Radiantis&lt;/a&gt; sells OPOs that are tunable over almost the whole visible spectrum. OPOs are not lasers, but they &lt;em&gt;are&lt;/em&gt; sources of coherent light which use a solid state gain medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Full disclosure: I do not work for and have never done business with Radiantis, but I have toured their lab once.&lt;/p&gt;&#xA;" OwnerUserId="107" LastEditorUserId="107" LastEditorDisplayName="" LastEditDate="2011-05-31T14:21:11.353" LastActivityDate="2011-05-31T14:21:11.353" CommentCount="2" />
  <row Id="10586" PostTypeId="2" ParentId="10574" CreationDate="2011-05-30T14:48:30.063" Score="7" ViewCount="" Body="&lt;p&gt;This is essentially a philosophical rather than a physical question. (I don't mean that as a pejorative statement, by the way, just a descriptive one.) There are &lt;a href=&quot;http://plato.stanford.edu/archives/sum2003/entries/probability-interpret/&quot; rel=&quot;nofollow&quot;&gt;different philosophical approaches&lt;/a&gt; regarding the meaning of probability. Broadly speaking, some people think of probability in a frequentist sense, meaning that probabilities refer to the frequency of outcomes in an ensemble. Other people think of probability as a description of our knowledge, rather than a description of the system itself. In this way of thinking about things, often called &quot;Bayesian,&quot; probabilities don't require one to postulate or even to imagine ensembles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://blog.richmond.edu/physicsbunn/2009/01/27/what-is-probability/&quot; rel=&quot;nofollow&quot;&gt;Personally&lt;/a&gt;, I think that the frequentist attitude is self-evidently absurd. When I look at the weather report and see that there's a 70% chance of rain tomorrow, I am not required to, and certainly do not, imagine an ensemble of infinitely many Earths, 70% of which will contain rain tomorrow. Rather, this statement simply means that I don't know whether it'll rain or not, but I'll be somewhat more surprised if it doesn't than if it does. E.T. Jaynes's &lt;a href=&quot;http://books.google.com/books?id=tTN4HuUNXjgC&amp;amp;printsec=frontcover&amp;amp;dq=jaynes&amp;amp;hl=en&amp;amp;src=bmrr&amp;amp;ei=86zjTYGFGanm0QGL1uDABw&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCoQ6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; is a standard (and in my opinion excellent) manifesto in favor of this attitude towards probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weather forecasts are not quantum probabilities (at least not entirely, or in any obvious way), but I don't think that matters. I see no reason I can't regard all probabilities in exactly the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By the way, the terms frequentist and Bayesian also refer to different sets of statistical techniques. Although that usage is related to the philosophy-of-probability usage, they're not the same. In particular, I am not claiming (nor do I believe) that frequentist statistical techniques are absurd or invalid. I actually use both sets of techniques in my own work, depending on which is more convenient for any given problem.)&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-05-30T15:05:12.613" LastActivityDate="2011-05-30T15:05:12.613" CommentCount="6" />
  <row Id="10587" PostTypeId="2" ParentId="10582" CreationDate="2011-05-30T16:22:57.823" Score="2" ViewCount="" Body="&lt;p&gt;This is the case with a caveat: It only works for a weak linearized theory.  If gravitons are a small field perturbation on a flat background in a linear Einstein field equation this is the case.  In greater generality this is less certain.  Energy is not localizable in general relativity.  So what I am about to outline below is a linear quantum gravity for long wavelength gravitons in the IR limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gravity wave is a perturbation on a background metric $\eta_{ab}$ with the total metric&#xA;$$&#xD;&#xA;g_{ab}~=~\eta_{ab}~+~h_{ab}.&#xD;&#xA;$$&#xA;The flat background metric has zero Ricci curvature so that to first order in the perturbation expansion&#xA;$$&#xD;&#xA;R_{ab}~=~\delta R_{ab},&#xD;&#xA;$$&#xA;which enters into the Einstein field equation $R_{ab}~-~1/2Rg_{ab}~=~\kappa T_{ab}$, where $\kappa~=~8\pi G/c^4$ is the very small coupling constant between the momentum-energy source and the spacetime configuration or field.  The Ricci curvature to first order is then&#xA;$$&#xD;&#xA;R_{ab}~=~{1\over 2}\Big(\partial_c\partial_a{h^c}_b~+~\partial_c\partial_b{h^c}_a~-~\partial_a\partial_bh~-~\partial_c\partial^ch_{ab}\Big).&#xD;&#xA;$$&#xA;To within first order the harmonic gauge $\partial_c{h^c}_a~=~1/2\partial_mu h$ the Einstein field equation gives&#xA;$$&#xD;&#xA;\partial^c\partial_ch_{ab}~-~\frac{1}{2}\eta_{ab}\partial^c\partial_ch~=~{{16\pi G}\over c^4}T_{ab},&#xD;&#xA;$$&#xA;which is well defined for the traceless metric term ${\bar h}_{ab}~=~h_{ab}~-~{1\over 2}\eta_{ab}h$ with the simple wave equation&#xA;$$&#xD;&#xA;\partial^c\partial_c{\bar h}_{ab}~=~{{16\pi G}\over c^4}T_{ab}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the wave in vacuum the momentum energy source is set to zero, and the wave equation is $\partial^c\partial_c{\bar h}_{ab}~=~0$.  This is a bi-vector analogue to the simple wave equation for an electromagnetic wave in free space.  The wave is a transverse traceless wave $h_{ab}~=~A^{TT}_{ab}exp(ik_cx^c)$ with&#xA;$$&#xD;&#xA;A^{TT}_{ab}~=~\left(\matrix{0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0\cr&#xD;&#xA;0 &amp;amp; A_{xx} &amp;amp; A_{yx} &amp;amp; 0\cr&#xD;&#xA;0 &amp;amp; A_{xy} &amp;amp; -A_{xx} &amp;amp; 0\cr&#xD;&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0&#xD;&#xA;}\right).&#xD;&#xA;$$&#xA;The terms $A_{xx}$ and $A_{xy}$ represent a polarization direction.  The linearized gravity wave then has a helicity of two, which has its quantum analogue in the di-photon state in quantum optics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogue with a photon in this linearized approximation is for the metric perturbation $h_{ab}~=~\phi^c_a\phi_{bc}$ expanded according to a field.  To take this forwards let us expand the field $\phi^a_b$ according to harmonic oscillator operators $b,~b^\dagger$.  The fields are then expanded as&#xA;$$&#xD;&#xA;\phi^a_b~=~\frac{1}{\sqrt{2}}\sum_k E^a_b (b(k)e^{i\theta(k)} + b^\dagger e^{-i\theta(k)})&#xD;&#xA;$$&#xA;where $E^a_b$ tetrad, which is discussed more below.  The metric perturbation is then&#xA;$$&#xD;&#xA;\phi^a_c\phi_{cb}~=~\frac{1}{2}\sum_{kk’}\eta^a_b(b(k)b^\dagger(k’)e^{i\theta(k) – i\theta(k’)}~+~b^\dagger(k)b(k’) e^{-i\theta(k’) – i\theta(k)})&#xD;&#xA;$$&#xA;$$&#xD;&#xA;+~\frac{1}{2}\sum_{kk’}\eta^a_b(b(k)b^(k’)e^{i\theta(k)+i\theta(k’)}~+~b^\dagger(k)b^\dagger(k’)e^{-\theta(k)-i\theta(k’)}),&#xD;&#xA;$$&#xA;The first of these terms is a rotating wave, while the second is counter rotating.  For the sake of simplicity we ignore this for now.  Laplacian operator on the first term gives a wave equation in terms of these fields, and the Hamiltonian is $H~=~\partial_d\phi^c_a\partial^d\phi_{bc}$ is a sum over $\sim~\hbar\omega$ terms.  The reader may fill in the details.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2011-05-30T16:22:57.823" />
  <row Id="10588" PostTypeId="2" ParentId="10572" CreationDate="2011-05-30T16:23:30.617" Score="3" ViewCount="" Body="&lt;p&gt;The 3D shape of the lightning channel over many kilometers of extent can be teased out of three simultaneous sound recordings made from three different spots. The coding is similar to that which produces CAT scans, very abstruse, lots of matrix algebra. Similar systems are used to localize gun shots in urban areas using microphone networks.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastActivityDate="2011-05-30T16:23:30.617" CommentCount="1" />
  <row Id="10589" PostTypeId="2" ParentId="10565" CreationDate="2011-05-30T16:39:14.347" Score="1" ViewCount="" Body="&lt;p&gt;I'm going to take a risk and try to answer this, even though my answer is different to Lubos's and he does have a reputation that is overwhelming right compared to mine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Static Magnetic fields don't don't do work, period (us), full stop (uk). So the work comes from the magnetic dipole itself whose internal energy is affected by the external force that positioned it in the static magnetic field in the first place.&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-05-30T16:39:14.347" CommentCount="4" />
  <row Id="10590" PostTypeId="2" ParentId="1302" CreationDate="2011-05-30T16:41:38.707" Score="1" ViewCount="" Body="&lt;p&gt;I cannot see, above, the correct answer to the first part of the question. They all have the second part right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first part: Why does shaking the bottle make it fizz when you open it? &#xA;The gas is in solution in the closed, unshaken bottle. The solubility of that gas in that liquid at that temperature and that pressure dictates the saturation level of the gas in the liquid. Any more gas than that and it bubbles out, increasing the pressure inside the closed bottle and forcing more gas in solution thus reducing the pressure. An equilibrium is reached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why does shaking cause a problem?&#xA;Because when you shake it, you slosh the contents around. The liquid flows quickly from one end of the container to the other. As it does so, it flows fast enough to become turbulent. In the turbulence, the curlicues, the intricacies, the eddies and the complex flow there are many locations in the liquid where the local pressure is forced lower than the saturation pressure. This is because any packet of liquid that is accelerating in any direction will leave a low pressure zone in its wake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the pressure is forced below the equilibrium level, the gas (given sufficient nucleating seeds) will come of of solution instantly given that the operation is occurring in the bulk and gas bubbles can be created anywhere in 3D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when the gas wants to go back into solution it is severely limited by the relatively small surface area of the liquid in the neck of the bottle. There is only a tiny surface available for gas transport through the liquid/air interface. We are severely dissolution rate limited. It can take hours for a shaken can to quiet down. It can be hastened by chilling as more gas can always be dissolved in a cooler liquid. &lt;/p&gt;&#xA;" OwnerUserId="3735" LastActivityDate="2011-05-30T16:41:38.707" CommentCount="1" />
  <row Id="10591" PostTypeId="1" AcceptedAnswerId="10593" CreationDate="2011-05-30T19:41:06.163" Score="0" ViewCount="77" Body="&lt;p&gt;I'm thinking about a project to tackle, and I'd like to make a simulation that allows the user to define a rope or chain of length L, pin it at arbitrary points r1, r2.... etc. and draw the resulting curve in real time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I'd like the user to be able to alter the field that the rope exists in, for example more complex vector fields than just a straight gravitational field. This is a bonus, however, and I'd like to get the basic example working.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you recommend some resources, preferably free/online, for me to learn the physics involved (I'm not exactly sure what to CALL this area of classical mechanics)? I learn through concepts and then math so a resource that is concept-heavy would be nice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! &lt;/p&gt;&#xA;" OwnerUserId="3390" LastActivityDate="2011-05-30T20:11:29.630" Title="Simulation of physics of chains/ropes in force fields resources?" Tags="&lt;classical-mechanics&gt;&lt;education&gt;&lt;vector-fields&gt;" AnswerCount="1" />
  <row Id="10592" PostTypeId="1" CreationDate="2011-05-30T19:52:30.573" Score="0" ViewCount="178" Body="&lt;p&gt;I am not exactly sure how the physics work here. If we take a car tow rope, which is manufactured for towing a 3500 kg car and we actually start towing such a car on this rope, what force do we really apply to the rope?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could the same rope &lt;strong&gt;safely&lt;/strong&gt; withhold an adult person (100 kg) hanging on it?&lt;/p&gt;&#xA;" OwnerUserId="2834" LastActivityDate="2011-06-02T10:54:28.917" Title="Would a car tow rope withhold an adult person?" Tags="&lt;forces&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="10593" PostTypeId="2" ParentId="10591" CreationDate="2011-05-30T20:02:02.963" Score="2" ViewCount="" Body="&lt;p&gt;Looks like what you are trying to simulate is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Catenary&quot; rel=&quot;nofollow&quot;&gt;Catenary&lt;/a&gt;. And this comes under statics - a part of Classical Mechanics. &lt;a href=&quot;http://ztolk.interestingnonetheless.net/skippingrope.pdf&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; (so does the Wikipedia article) introduces to the basics of catenary analysis. Also see &lt;a href=&quot;http://mathoverflow.net/questions/29247/catenary-curve-under-non-uniform-gravitational-field&quot; rel=&quot;nofollow&quot;&gt;a similar question&lt;/a&gt; on MathOverflow.&lt;/p&gt;&#xA;" LastActivityDate="2011-05-30T20:11:29.630" CommentCount="1" />
  <row Id="10594" PostTypeId="2" ParentId="10592" CreationDate="2011-05-30T20:07:33.917" Score="2" ViewCount="" Body="&lt;p&gt;That is not a question to &quot;physics&quot;, You have to ask the ropemaker! &#xA;I recently looked for some ropes in a mall, all had tags saying &quot;XYZ kg&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force needed to tug a car is absolutely irrelevant. The forces &#xA;when accelerating from stillstand or when the tugged car is braked &#xA;suddenly are far more relevant. For such reasons a rope for tugging has &#xA;to withstand many hundreds of kg at least. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &#xA;I had a look on car tow ropes offered by amazon.de : They name the &#xA;mass of the vehicle the rope may be used for. No mention of actual &#xA;max force the rope will withstand. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-06-02T10:54:28.917" LastActivityDate="2011-06-02T10:54:28.917" CommentCount="6" />
  <row Id="10595" PostTypeId="2" ParentId="10578" CreationDate="2011-05-30T20:33:24.290" Score="2" ViewCount="" Body="&lt;p&gt;For your first question: Yes, there is a physical reason for this phenomenon as is (or should be) for every observable phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your second part: Depends on who is debating the validity or applicability of that theorem. Can you please add a reference, then perhaps someone may be able to answer it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your third question:&#xA;Yes there is a lot of &quot;real&quot; dynamics involved here. In fact, because a lot of objects are interacting with each other, it is highly involved as well. Different models to explain/simulate flocking behavior use almost the same physical phenomena in their calculations --gravitational pull from earth, fluid mechanics, wing flapping. What they debate is usually one or more of the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is a bird's comfort level of independence? Meaning, how far will a bird on the edge will go away from the flock before it senses danger. This has nothing* to do with physics and has everything to do with such factors as the predators around, the evolutionary history of the birds, etc.&lt;/li&gt;&#xA;&lt;li&gt;How close do birds like flying to each other? At what distance they sense an impending collision? Again, nothing to do with physics really.&lt;/li&gt;&#xA;&lt;li&gt;How perfectly do birds prefer to align to the group? Is there a single leader or multiple?&lt;/li&gt;&#xA;&lt;li&gt;Does a change in the wind's direction change the leader's course?&lt;/li&gt;&#xA;&lt;li&gt;etc...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One can of course abstract out the biology and social behaviour out of this model and empirically define a force and construct laws (similar to Newton's laws of gravity, Maxwell's equations etc.) for that force. Perhaps then that person may be able to reuse some of the results that physicists have derived for other problems (&lt;strong&gt;note&lt;/strong&gt;: this depends a lot on how your force laws are defined). Basic physical laws that apply universally shall still be applicable.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Footnotes&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;*You can say everything is physics deep deep down but that is not the point here obviously.&lt;/p&gt;&#xA;" OwnerUserId="3669" LastActivityDate="2011-05-30T20:33:24.290" CommentCount="1" />
  <row Id="10598" PostTypeId="2" ParentId="10578" CreationDate="2011-05-31T00:13:49.087" Score="3" ViewCount="" Body="&lt;p&gt;I think that this &lt;a href=&quot;http://arxiv.org/abs/1010.3656&quot; rel=&quot;nofollow&quot;&gt;Order-of-magnitude &lt;strong&gt;enhancement of wind farm power density&lt;/strong&gt; via counter-rotating vertical-axis wind turbine arrays&lt;/a&gt; is a result of a collective hidrodynamic behaviour somehow similar to a &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/V_formation&quot; rel=&quot;nofollow&quot;&gt;flock of birds&lt;/a&gt;&lt;/strong&gt; (it's physics as anna v. noted)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The V formation &lt;strong&gt;greatly boosts the&#xA;  efficiency and range of flying birds&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or &#xA;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Shoaling_and_schooling&quot; rel=&quot;nofollow&quot;&gt;Shoaling_and_schooling&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is also likely that fish benefit&#xA;  from shoal membership through&#xA;  &lt;strong&gt;increased hydrodynamic efficiency&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The whole issue of colective behaviour is very interesting because the result is more than the sum of the components and we learn how to solve complex problems, like increasing the efficiency of motion or lowering the collective risk, using simple 'inteligent' agents that cooperate. The package &lt;a href=&quot;http://www.cs.gmu.edu/~eclab/projects/mason/#Screenshots&quot; rel=&quot;nofollow&quot;&gt;MASON&lt;/a&gt; Multiagent based simulation (IA) is very interesting.&lt;br&gt;&#xA;A system of Rules can be translated to a system of equations, if it makes you feel more comfortable.&lt;br&gt;&#xA;The theorem you mention, related to statistics (and QM), has nothing to do with 'flocking', imo .  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-05-31T00:13:49.087" />
  <row Id="10600" PostTypeId="2" ParentId="3228" CreationDate="2011-05-31T03:12:25.530" Score="0" ViewCount="" Body="&lt;p&gt;Sir Isaac Newton struggled with exactly this question, in the context of optics, and the best he could come up with was a theory of superluminal waves associated with light particles.&#xA;(The question Newton asked was essentially the same question as yours: If a light beam is 20% absorbed into a glass surface and 80% reflected, and if light is particular, then how can one particle acting alone make a correct decision?)&#xA;The whole business of selecting one from a possible set for no apparent reason can be used for quantum computing. It can effectively compute a &quot;diagonalisation of a matrix&quot;. &#xA;Lets look more closely at the meaning of random.&#xA;Random is the limit of compressibility of a pattern, the removal of all predictability.&#xA;The pattern as a whole has this character, so it is a feature of the set.&#xA;A radioactive particle does not have the knowledge of which you speak, in fact it is missing information. It is forbidden to carry predictive knowledge. No information can be projected from any part of the sequence, present or future, to determine any part of the sequence. In this sense, every event in the sequence has no extractable knowledge about its position in the sequence. It is a safe spy, unable to betray its fellows.&lt;br&gt;&#xA;This does not even answer the question, but gives a better way to think about it. What we see about randomness as contrived should more be seen as a natural default. Einsten protests that God does not play dice, but he is possibly complaining if God does plays dice then that means that God will always be with-holding information. &lt;/p&gt;&#xA;" OwnerUserId="3797" LastActivityDate="2011-05-31T03:12:25.530" />
  <row Id="10601" PostTypeId="2" ParentId="10532" CreationDate="2011-05-31T06:31:01.507" Score="0" ViewCount="" Body="&lt;p&gt;If the gauge group isn't deformed, how do we explain T-dualities and mirror symmetry? Is there a much larger gauge group which contains as subgroups two distinct, but overlapping diffeomorphism groups?&lt;/p&gt;&#xA;" OwnerUserId="3798" LastActivityDate="2011-05-31T06:31:01.507" CommentCount="1" />
  <row Id="10602" PostTypeId="1" CreationDate="2011-05-31T07:02:54.397" Score="3" ViewCount="155" Body="&lt;p&gt;{..everything that follows is in the domain of relativistic kinematics..}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say a particle A collides with a particle B at rest and produces particles C and D. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly is the definition of &quot;threshold energy&quot; for a reaction ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it the energy that A should have so that so that the heavier of C and D is produced at rest? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the masses of all the particles, how does one calculate the threshold energy that A must have to cause this reaction ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I intuitively feel that a threshold energy scenario would mean that C and D are moving along the same direction in which A was coming in. I think it is an unnecessary &quot;waste&quot; of energy for C and D to develop momentum in the two transverse directions of A's motion. But I can't prove this in general. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know how this situation is analyzed. &lt;/p&gt;&#xA;" OwnerUserId="1637" LastActivityDate="2011-05-31T12:00:17.813" Title="Calculating threshold energy of particle reactions" Tags="&lt;special-relativity&gt;" AnswerCount="2" />
  <row Id="10603" PostTypeId="2" ParentId="10551" CreationDate="2011-05-31T07:56:50.887" Score="0" ViewCount="" Body="&lt;p&gt;I can't comment on the blocked polarization (it was also covered in another answer) but the transmitted polarization effectively sees a periodical pattern of apertures right (the wires and the spacings)? Thus at least as a first approximation the standard reasoning of diffractive apertures should help. The light field after passing through an aperture is modelled by the Fourier transform of the aperture. The intensity is the square of the field and its spatial integral is the total transmitted intensity. In the case of the variable aspect-ratio apertures (variable width 0-100% at a fixed spacing) the transmitted total intensity will vary linearily from 0 to a maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The form of each aperture is a &quot;top hat&quot; function, i.e. a hat-like shape whose width varies according the aperture ratio. Its Fourier-transform is a sinc function:&#xA;$$F(k)=a sin(ka/2)/(ka/2)$$&#xA;given that a is the aperture width. Since the right part is the normalized sinc-function with a constant integral both at powers of 1 and 2, you see that the total intensity will vary linearily with a (for a single aperture).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An infinitely repeating array of apertures is modelled by a Dirac-comb (an infinite train of Dirac spikes) convolved with the top-hat aperture. When Fourier-transforming this, the convolution turns into a multiplication and the transform of the Dirac-comb is another Dirac-comb. Thus, the resulting field is a repeating array of Dirac spikes whose height (envelope) is modulated by the aperture-shape and its transform. By a handwaving argument, this means the same linear dependency as for the single aperture (you just have the same sinc-shape but &quot;chopped up&quot; and stretched with the Dirac comb).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, if you restrict the array to a certain width, this is the same as multiplying the infinite array with another top-hat function. Multiplication turns to convolution in the Fourier transform, and again the top-hat turns into a sinc shape which is convolved with the Dirac spike structure, replacing every spike with the shape of the sinc. Again due to the constant integrals of these sinc's, this doesn't change the linear outcome either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry I didn't have the time to write the tex-equations for the above two convolution and multiplication fourier-pairs, I'll see if I can fix that later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, making the total polarizer width thinner increases the width of the individual diffraction lines. Changing the aspect ratio of the apertures changes the overall shape of the envelope over the spikes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that there is a hidden dependency on the light wavelength, the Fourier coordinate k above depends inversely on the wavelength, so for longer wavelengths the intensity distribution widens. When the wavelength approaches the grid period or goes below it, I guess you'll get some special cases here though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some nice examples including applets to see the intensity distributions &lt;a href=&quot;http://cnyack.homestead.com/files/afourtr/ftdiffraction.htm&quot; rel=&quot;nofollow&quot;&gt;see here&lt;/a&gt; and examples of Fourier theory are &lt;a href=&quot;http://www.roymech.co.uk/Useful_Tables/Maths/fourier/Maths_Fourier_transforms.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-05-31T07:56:50.887" CommentCount="7" />
  <row Id="10604" PostTypeId="2" ParentId="10602" CreationDate="2011-05-31T08:27:33.983" Score="1" ViewCount="" Body="&lt;p&gt;Dear Anirbit, the threshold energy is the minimum energy for which the production of a particular collection of particles in the final state may occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that this minimum energy occurs when all the particles - if there are many - of the final state are moving with the same velocity, i.e. if they have the same values of $\vec p / E$. Equivalently, there exists an inertial system in which all the final particles are rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as you correctly pointed out, it doesn't mean that this inertial system where the final particles are at rest has to agree with a particular inertial frame of the initial state. If one considers head-on collisions of two equally heavy particles going in the opposite direction, then it's true: the threshold energy means that the final particles are produced at rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if one considers e.g. collisions with a fixed target - a moving proton with a $\vec p=0$ proton - it is not the case. Instead, the final velocity will be going in the same direction as the initial moving proton. That's dictated by the momentum conservation. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Threshold_energy&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; for a particular calculation of the threshold energy.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-05-31T08:27:33.983" CommentCount="2" />
  <row Id="10605" PostTypeId="2" ParentId="10565" CreationDate="2011-05-31T08:30:37.053" Score="4" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Is it correct to say that magnetic fields DO do work?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes! I show this quantitatively:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each charged particle experiences action of magnetic force. This force is transmitted to a conductor in which the charges move. As a result, the magnetic field acts with a certain force on the current-carrying conductor. Let the volume charge density, (electrons in a metal, for example) is equal to $\rho$. Let distinguish a mental element of volume $dV$ of the conductor. There is a charge equal to $\rho dV$. Then the force acting on the element $dV$ of the conductor can be expressed by the Lorenz formula $\overrightarrow{F}=q(\overrightarrow{v}\times\overrightarrow{B})$ in the form:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\overrightarrow{dF}=\rho (\overrightarrow{v}\times\overrightarrow{B})dV$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\overrightarrow{j}=\rho\overrightarrow{v}$ where $ \overrightarrow{j}$ is the current density vector we can write:$$\overrightarrow{dF}=(\overrightarrow{j}\times\overrightarrow{B})dV$$ If the current flows through a thin conductor, then the following holds:$$\overrightarrow{j}dV= \overrightarrow{dl}I$$ where I is a current in a thin conductor(wire) and $\overrightarrow{dl}$ is the vector of an element of the wire in direction of the current. Thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\overrightarrow{dF}=I(\overrightarrow{dl}\times\overrightarrow{B})$$ This is nothing more than Ampere's force. So the resulting Ampere's force acting on the contour of the current (current loop) in the magnetic field is determined as a line integral along the current loop:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\overrightarrow{F}=I\oint(\overrightarrow{dl}\times\overrightarrow{B})$$ If the magnetic field is nonuniform, then the integral is generally different from zero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A conclusion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows directly from the Lorenz law that magnetic fields do work on a current loop.&lt;/p&gt;&#xA;" OwnerUserId="552" LastActivityDate="2011-05-31T08:30:37.053" CommentCount="10" />
  <row Id="10606" PostTypeId="1" AcceptedAnswerId="10628" CreationDate="2011-05-31T09:05:27.070" Score="5" ViewCount="181" Body="&lt;p&gt;Are there any references that present the explicit variation of the Hilbert-Einstein action plus the Hawking-Gibbons-York boundary term, and demonstrate the cancellation of the normal derivatives of metric variations? I have tried to read the original papers by York and Gibbons&amp;amp;Hawking, but they are not that pedagogical to me...&#xA;Thanks to anyone who can help me!&lt;/p&gt;&#xA;" OwnerUserId="2928" LastActivityDate="2011-05-31T17:27:21.267" Title="Explicit Variation of Gibbons-Hawking-York Boundary Term" Tags="&lt;general-relativity&gt;&lt;variational-principle&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="10607" PostTypeId="2" ParentId="10527" CreationDate="2011-05-31T09:38:09.310" Score="0" ViewCount="" Body="&lt;p&gt;The short answer is: there is no proof (that is no experimental evidence) so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason for considering theories with additional dimensions is that (many) theories that are complicated in 4D can be reformulated in simpler terms as a theory with additional dimensions, which are rolled up in tiny circles (or more generally tiny &lt;em&gt;manifolds&lt;/em&gt;) so that we don't experience them as the other &quot;big&quot; dimensions (called &quot;uncompact&quot;). What is meant by &quot;simpler&quot; is that, for example, a theory with only one (vector or tensor) &lt;em&gt;field&lt;/em&gt; (think particle) in higher dimensions manifests itself as several fields of different kinds in lower dimensions, and their complicated interactions are described geometrically by the shape of the &lt;em&gt;compactification manifold&lt;/em&gt;. In physics people like geometrisation as one can argue that it is more intuitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In trying to formulate a theory that describes particle interactions accurately, one is faced with many possibilities and those that can be formulated with extra dimensions are somewhat simpler. So this is often used as a &lt;em&gt;guiding principle&lt;/em&gt; for formulating a &lt;strong&gt;correct&lt;/strong&gt; theory - that is a theory which isn't contradicted by experiments. There are several examples that fullfill these requirements. But it might turn out that none of these theories (with extra dimensions) will survive when more experimental data will be gathered and compared to the predictions of these theories.&lt;/p&gt;&#xA;" OwnerUserId="1844" LastActivityDate="2011-05-31T09:38:09.310" />
  <row Id="10608" PostTypeId="2" ParentId="10566" CreationDate="2011-05-31T10:44:23.530" Score="4" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://www.coherent.com/products/index.cfm?1557/Chameleon-Family&quot; rel=&quot;nofollow&quot;&gt;Coherent Chameleon&lt;/a&gt; titanium-sapphire laser tunes over a pretty enormous range, mostly in the IR (680 to 1080 nm). With a frequency doubling crystal, it &lt;a href=&quot;http://www.coherent.com/downloads/ChameleonOPO_Vis_DS_0511_7.pdf&quot; rel=&quot;nofollow&quot;&gt;covers most of the visible&lt;/a&gt; (340 to 540 nm), and if you add an OPO, it can tune over the entire visible range with frequency doubling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.fianium.com/supercontinuum.htm&quot; rel=&quot;nofollow&quot;&gt;Supercontinuum sources&lt;/a&gt; cover the entire visible spectrum simultaneously, and are often operated with post-filtering if you want narrowband light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.toptica.com/uploads/media/toptica_BR_iChrome_2008_12.pdf&quot; rel=&quot;nofollow&quot;&gt;Toptica's iChrome&lt;/a&gt; is a narrowband source that tunes over the entire visible by following supercontinuum generation with frequency doubling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these lasers are &lt;a href=&quot;http://en.wikipedia.org/wiki/Solid-state_laser&quot; rel=&quot;nofollow&quot;&gt;solid-state&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, regarding the spirit of your question. Does your bet with your dumb co-workers allow nonlinear optics like supercontinuum generation or frequency doubling? Does it allow mixtures of different gain media? Are you specifically curious about semiconductor lasers? Do you care about the coherence of the output, the bandwidth, or the brightness?&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2011-05-31T10:44:23.530" CommentCount="2" />
  <row Id="10609" PostTypeId="2" ParentId="10602" CreationDate="2011-05-31T12:00:17.813" Score="1" ViewCount="" Body="&lt;p&gt;Luboos has already explained that even at threshold the products are produced at rest &lt;em&gt;only&lt;/em&gt; in the Center of Momentum frame of the initial system, but this is a nice problem to demonstrate the power of calculating with relativistic invariants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the threshold is a naive way, we could calculate the velocity of the center of momentum frame as a function of the beam energy, then boost both beam and target into that frame and calculate the total energy. It is not a particularly difficult task.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we can do better than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total four-momentum vector of the products at threshold, in the CoM from is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbf{P} = \mathbf{p}_C + \mathbf{p}_D = (m_C,\vec{0}) + (m_D,\vec{0}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and squaring that we obtain the (relativisticly invariant) mass of the system at threshold:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M_f = (m_C + m_D, \vec{0})^2 =  m_C^2 + 2 m_C m_D + m_D^2  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can calculate the same quantity for the precursors &lt;em&gt;in the lab frame&lt;/em&gt; of reference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathbf{p}_A = (E_A,\vec{p}_A) \approx (p_A, \vec{p}_A) $$&#xA;$$ \mathbf{p}_B = (E_B,\vec{p}_B) = (m_B, \vec{0}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where I have assumed that the beam is fully relativistic (hey, I spent a lot of time at an electron accelerator facility). This makes the mass squared of the initial system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M_i = (p_A + m_b, \vec{p}_A) = (p_A^2 + 2 p_A m_B + m_B^2) - p_A^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which we simplify and set equal to the mass squared of the final products to find the beam momentum needed for threshold production:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ p_{A,\text{threshold}} = \frac{ (m_C + m_D)^2 - m_B^2}{2 m_b} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One half of this calculation was done in the CoM frame and the other half in the lab frame, but because I only compared Lorentz invariant quantities I was able to combine them with impunity.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-05-31T12:00:17.813" />
  <row Id="10610" PostTypeId="1" CreationDate="2011-05-31T13:26:05.463" Score="1" ViewCount="67" Body="&lt;p&gt;Suppose a large (rigid) block is sitting on top of two smaller blocks of equal height $1$, both of which rest on the ground. We wish to find the position of the block (easy) and the forces of constraint on the block coming from the two smaller blocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we write the Lagrangian along with the equations of constraint $z=1$, we can find the motion of the object using Lagrange multipliers, and we can also find the force of constraint. If there were constraints $\{G_n\}$, then the force of constraint in the $z$ direction would be $F_{n} = \lambda_n \frac{\partial G_n}{\partial z}$ on the constraint $G_n$. However, this tells us nothing about how much force each individual block is exerting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know from experience that the upward force from the smaller blocks must equal $mg$, where $m$ is the mass of the larger block, and furthermore, that the force is evenly distributed, so that each of the two smaller blocks exerts an equal amount of force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems natural to indicate the existence of two blocks by writing down the constraint $z=1$ twice and then trying to solve the resulting system of equations. However, this doesn't seem to get us very far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how is it possible to detect the fact that the force on the two smaller blocks is distributed equally?&lt;/p&gt;&#xA;" OwnerUserId="2591" LastActivityDate="2011-05-31T13:26:05.463" Title="What are the forces of constraint if there are multiple equivalent constraints?" Tags="&lt;forces&gt;&lt;lagrangian-mechanics&gt;" CommentCount="2" />
  <row Id="10611" PostTypeId="1" CreationDate="2011-05-31T14:00:04.057" Score="1" ViewCount="160" Body="&lt;p&gt;One simple way of detailing the very basic structure of the nuclear shell model involves placing the nucleons in a 3D isotropic oscillator.  It's easy to show that the energy eigenvalues are $E = \hbar\omega\left(n+\frac{3}{2}\right)$ with a degeneracy of $D(n) = \frac{1}{2}(n+1)(n+2)$, where $n = n_1 + n_2 + n_3 = a^\dagger_1 a_1 + a^\dagger_2 a_2 + a^\dagger_3 a_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the book I am using presently (Ring, The Nuclear Many-Body Problem) and in several other texts, the authors write the quantum numbers as $n=2(n_r-1)+\ell$ (or, some places, $n=2n_r+\ell$).  This seems odd, since we're trading 3 quantum numbers ($n_1, n_2, n_3$) for 2 ($n_r, \ell$).  I suppose this is fine if we're only using $n_r$ and $\ell$ to enumerate our hypothesized nuclear energy levels, but in any case it's not clear to me &lt;em&gt;why&lt;/em&gt; you would make the choice $n = 2n_r + \ell$ to do this (besides, obviously, that it does an okay job of reproducing the lower magic numbers).  These authors trot the expression out without any explanation, and I'd like to know why this expression is used.  I'm still not clear on the significance of $n_r$ and $\ell$ in nuclei, so it's puzzling.&lt;/p&gt;&#xA;" OwnerUserId="3750" LastActivityDate="2011-05-31T14:00:04.057" Title="Energy Levels of 3D Isotropic Harmonic Oscillator (Nuclear Shell Model)" Tags="&lt;quantum-mechanics&gt;&lt;nuclear-physics&gt;&lt;harmonic-oscillator&gt;&lt;oscillators&gt;" CommentCount="2" />
  <row Id="10612" PostTypeId="1" AcceptedAnswerId="10613" CreationDate="2011-05-31T14:07:16.930" Score="2" ViewCount="149" Body="&lt;p&gt;I understand that mass-energy equivalence is often misinterpreted as saying that mass can be converted into energy and vice versa.  The reality is that energy is always manifested as mass in some form, but I struggle with some cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Understood Nuclear Decay Example&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a simple nuclear reaction, for instance, the total system mass remains the same since the mass deficit (in rest masses) is accounted for in the greater relativistic masses of the products per $E=\Delta m c^2$.  When a neutron decays and you are left with a fast proton and a relativistic electron.  If you could weigh those two without slowing them down, you would find it weighed as much as the original neutron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Light in a Box&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This becomes more difficult for me when moving to massless particles like photons.  Photons can transmit energy from one heavy particle to another.  When a photon is absorbed the relativistic mass (not the rest mass) of the (previously stationary) particle that absorbs it increases.  But if my understanding is correct, the energy must still be manifested as mass &lt;em&gt;somehow&lt;/em&gt; while the photon is in-flight, in spite of the fact that the photon does not have mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's consider a box with the interior entirely lined with perfect mirrors.  I have the tare weight of the box with no photons in it.  When photons are present the box has an additional quantifiable amount of energy (quantified below) due to the in-flight photons.  Say there are $N$ photons... obviously assume $N$ is large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta m = \frac{ E }{ c^2 } = \frac{ N h }{ \lambda c}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interactions are limited to reflections with the wall, which manifest as a constant pressure on the walls.  If I hold this box in a constant gravitational field (like the surface of Earth) then there will be a gradient in the pressure that pushes down slightly.  Is this correct?  Wouldn't there still technically be mass as the photons are in-flight, which would cause its own gravitational field just as all matter does?  How is this all consistent with the assertion that photons are massless?  Is it really correct to say that photons don't have mass?  It seems to be a big stretch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please offer a more complete and physically accurate account of this mirror-box.  Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-05-31T17:02:38.243" Title="Explain how (or if) a box full of photons would weigh more due to massless photons" Tags="&lt;energy&gt;&lt;light&gt;&lt;mass-balance&gt;" AnswerCount="2" />
  <row Id="10613" PostTypeId="2" ParentId="10612" CreationDate="2011-05-31T14:29:09.060" Score="4" ViewCount="" Body="&lt;p&gt;The statement that photons are massless means that photons do not have &lt;b&gt;rest mass&lt;/b&gt;.  In particular, this means that, in units where $c=1$, the magnitude of the photon 3-momentum must be equal to the total energy of the photons, rather than the standard relationship where $m^{2} = E^{2}-p^{2}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, you can create multi-photon systems where the &lt;b&gt;net&lt;/b&gt; momentum is zero, since momentum adds as a vector.  When you do this, however, since the energy of a non-bound state is always non-negative, the energies just add.  So, this system looks just like the rest frame of a massive particle, which has energy associated with its mass and nothing else.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement about gravity is a little bit more subtle, but all photon states will interact with the gravitational field, thanks to the positive results of the light-bending observations that have been made over the past century.  So you don't even need a construction like this to get photons &quot;falling&quot; in a gravitational field. &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-05-31T14:29:09.060" CommentCount="8" />
  <row Id="10614" PostTypeId="2" ParentId="10592" CreationDate="2011-05-31T15:19:17.740" Score="1" ViewCount="" Body="&lt;p&gt;It depends on rolling resistance of wheels, friction resistance in bearings, elevation and acceleration of pulled car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ F=F_r+F_f+F_e+F_a $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$F$ is total force,&lt;/li&gt;&#xA;&lt;li&gt;$F_r=\frac{N_fb}{r}$ is rolling resistance, $b$ is rolling resistance coefficient, $r$ is radius of the wheel and $N_f$ is normal force,&lt;/li&gt;&#xA;&lt;li&gt;$F_f=\mu N_f$ is friction force and $\mu$ is friction coefficient,&lt;/li&gt;&#xA;&lt;li&gt;$F_e=mg*\sin\alpha$ is part of gravitation force, $\alpha$ is angle of slope, $m$ is mass,&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$F_a=ma$ is acceleration force, $a$ is acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it depends on bearings condition, tires and surface conditions, and many other variables...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you want to know whether 3.5 ton rope can lift 100 kg ask sports specialist, especially climbers, whether 100 kg safe rope can pull 3.5 ton car.&lt;/p&gt;&#xA;" OwnerUserId="2307" LastActivityDate="2011-05-31T15:19:17.740" />
  <row Id="10615" PostTypeId="1" AcceptedAnswerId="10626" CreationDate="2011-05-31T15:27:20.870" Score="4" ViewCount="197" Body="&lt;p&gt;I'm a tutor. This is a high school level problem. In high school, every one have might have solved a problem of &lt;a href=&quot;http://www.crbond.com/papers/ent2-3.pdf&quot; rel=&quot;nofollow&quot;&gt;effective resistance of a ladder of resistors having infinite steps.&lt;/a&gt; Now the problem is little different. what if it has &lt;code&gt;n&lt;/code&gt; steps instead of infinite steps. How to calculate effective resistance in that case?&lt;img src=&quot;http://i.stack.imgur.com/n1eIX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="447" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-05-31T15:35:15.880" LastActivityDate="2011-06-02T06:07:35.960" Title="What would be the effective resistance of the ladder of resistors having n steps" Tags="&lt;homework&gt;&lt;electric-circuits&gt;&lt;electrical-engineering&gt;&lt;electronics&gt;&lt;resistors&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="10616" PostTypeId="5" CreationDate="2011-05-31T15:48:24.903" Score="0" ViewCount="" Body="&lt;p&gt;See also:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/tagged/particle-physics&quot;&gt;particle physics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/tagged/qcd&quot;&gt;qcd&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-31T15:48:24.903" LastActivityDate="2011-05-31T15:48:24.903" />
  <row Id="10617" PostTypeId="4" CreationDate="2011-05-31T15:48:24.903" Score="0" ViewCount="" Body="High energy physics is particle physics carried out at interaction energies where the strong coupling constant $\alpha_s$ is much less than one so that QCD is a perturbutive theory." OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-31T15:48:24.903" LastActivityDate="2011-05-31T15:48:24.903" />
  <row Id="10618" PostTypeId="5" CreationDate="2011-05-31T16:06:44.503" Score="0" ViewCount="" Body="&lt;p&gt;Here &quot;linear&quot; means that the force, $F$ has a form like $F = -k(x - x_0)$ where $x$ represents the position of the system, $x_0$ the position of the equilibrium, and $k$ is a positive constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such systems are very common in mechanics. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Generalization&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If we allow &quot;force&quot; and &quot;positions&quot; to take on generalized meaning, such systems are very common in most fields of physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term is generally applied to any system which can be modeled with the mathematics used for the mechanical system, in part because almost any stable arrangement can be modeled as a harmonic oscillator for small displacements.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Solutions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The solutions are oscillatory in nature and described by sinusoids.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Damped and/or driven harmonic oscillators&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;An even larger class of problems can be described by adding resistive terms proportional to the &quot;velocity&quot; of the system and external driving forces.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-31T16:06:44.503" LastActivityDate="2011-05-31T16:06:44.503" />
  <row Id="10619" PostTypeId="4" CreationDate="2011-05-31T16:06:44.503" Score="0" ViewCount="" Body="The term &quot;harmonic oscillator&quot; is used to describe any system with a &quot;linear&quot; restoring force that tends to return the system to a equilibrium state." OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-31T16:06:44.503" LastActivityDate="2011-05-31T16:06:44.503" />
  <row Id="10620" PostTypeId="5" CreationDate="2011-05-31T16:15:22.673" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-31T16:15:22.673" LastActivityDate="2011-05-31T16:15:22.673" />
  <row Id="10621" PostTypeId="4" CreationDate="2011-05-31T16:15:22.673" Score="0" ViewCount="" Body="Fermions are particles with an intrinsic angular momentum (i.e. spin) equal to a &quot;half integer&quot; number of fundamental units: $\frac{(2n+1)}{2} \hbar$ for integer $n$. Fermions are required to be in a quantum state that is globally anti-symmetric, which leads to the Fermi Exclusion Principle barring identical fermions from occupying the same quantum state." OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-05-31T16:15:22.673" LastActivityDate="2011-05-31T16:15:22.673" />
  <row Id="10622" PostTypeId="2" ParentId="10615" CreationDate="2011-05-31T16:16:37.963" Score="5" ViewCount="" Body="&lt;p&gt;For any given $n$, you  can work it out via the rules for series and parallel resistors, but to get a general formula, valid for all $n$, doesn't look easy to me. The best way I know of is to get a recursive relationship giving the resistance of an $n$-step ladder in terms of an $(n-1)$-step ladder. If I'm not mistaken, the $n$-step ladder can be thought of as a single resistor $R$ in series with a parallel combination consisting of another $R$ and an $(n-1)$-step ladder. The rules for series and parallel resistance give&#xA;$$&#xD;&#xA;R_n=R+{RR_{n-1}\over R+R_{n-1}}.&#xD;&#xA;$$&#xA;The initial one-step ladder has $R_1=2R$, and you can use the above formula to work your way up the chain from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made Mathematica work out the first steps, and they seem to be related to the Fibonacci numbers, specifically&#xA;$$&#xD;&#xA;R_n=R{F_{2n+1}\over F_{2n}}.&#xD;&#xA;$$&#xA;I'm sure it's possible to prove that by induction.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-05-31T16:16:37.963" CommentCount="1" />
  <row Id="10623" PostTypeId="2" ParentId="10615" CreationDate="2011-05-31T16:22:00.043" Score="1" ViewCount="" Body="&lt;p&gt;Is interesting. I think you want to proceed as follows:&#xA;(1) Without loss of generality assume the input voltage is 1, and the resistors are single ohm.&#xA;(2) Define r = resistance of the infinite ladder.&#xA;(3) Now look at the circuit where we add one more step on the lefthand side. We have to solve&#xA;for the voltage where its connected, as a function of r.&#xA;(4) The new current is 1 minus this voltage. And the total resistance of the configuration is&#xA;r'=1/(1-V(r)).&#xA;(5) Adjust r until r=r'&#xA;(6) Scale by the R given in the original problem. I.E. the result is r*R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit of algebra in steps 4 and 5.....&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-05-31T16:22:00.043" />
  <row Id="10626" PostTypeId="2" ParentId="10615" CreationDate="2011-05-31T16:46:34.793" Score="5" ViewCount="" Body="&lt;p&gt;I'll take it step by step here.  First I'll write the answer for the first few cases with circuit analysis.  Then I'll apply a reduction to show the pattern that the problem arrives at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N=1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z = R+R=2R$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N=2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z = R+\frac{1}{\frac{1}{R}+\frac{1}{R + R}} = R \left( 1+\frac{1}{1+\frac{1}{1 + 1}} \right)=\frac{5}{3} R$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N=3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z = R+\frac{1}{\frac{1}{R}+\frac{1}{R + \frac{1}{\frac{1}{R}+\frac{1}{R + R}}}} = R \left( 1+\frac{1}{1+\frac{1}{1 + \frac{1}{1+\frac{1}{1 + 1}}}} \right) =\frac{13}{8} R$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point the continued fraction is clearly identifiable.  I do find it a little troubling because it doesn't scale with $N$, but the continued fraction grows as $2N$.  This could be remedied by not considering adding 2 resistors on the end each time, but instead add one resistor in parallel, one in series to that one, then a new in parallel between the two new ones, and use a different index for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computational programs should have some easy function to write continued fraction to some $N$ number of fractions.  Unfortunately I can not find this for Maple, but I need such a procedure in order to give an answer for your question.  I will define such a thing right here.  I intentionally use $n$ in the definition and not $N$ to avoid the inevitable confusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(1)=1$$&#xA;$$F(n+1) = 1+\frac{1}{F(n)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this, I can answer your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z = R  F(2 N) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I can give you a sample of the function that I just created.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(1..6) = [1, 2, \frac{3}{2}, \frac{5}{3}, \frac{8}{5}, \frac{13}{8}]$$&#xA;$$F(\infty) = \frac{ \sqrt5+1}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is the best form with which the problem can be answered.  The fraction can not be reduced easily into some concise algebraic form because the entire point of this exercise is to not introduce assumptions, and the large fraction &lt;em&gt;is&lt;/em&gt; the algebraic form of the answer.  The finite values of the fraction, however, are about as easy to implement procedurally as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Closed Form Solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be my final edit, and this expression pretty much wraps up the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(n) = \frac{\varphi^{n+1}-(1-\varphi)^{n+1}}{\varphi^n-(1-\varphi)^n}$$&#xA;$$\varphi = \frac{ \sqrt5+1}{2}$$&#xA;$$Z = R \frac{\varphi^{2 N+1}-(1-\varphi)^{2 N+1}}{\varphi^{2 N}-(1-\varphi)^{2 N}}$$&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-05-31T18:40:25.947" LastActivityDate="2011-05-31T18:40:25.947" CommentCount="2" />
  <row Id="10627" PostTypeId="2" ParentId="10612" CreationDate="2011-05-31T17:02:38.243" Score="1" ViewCount="" Body="&lt;p&gt;I think that there is some confusion in your understanding of relativistic physics in the statement here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In the case of a simple nuclear reaction, for instance, the total system mass remains the same since the mass deficit (in rest masses) is accounted for in the greater relativistic masses of the products per E=Δmc2. When a neutron decays and you are left with a fast proton and a relativistic electron. If you could weigh those two without slowing them down, you would find it weighed as much as the original neutron.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct statement is that the summed four vector of all the decay products would have the effective mass of a neutron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Masses are not conserved in relativistic physics, in an analogous way that lengths are not conserved when adding vectors in three dimensions. What is conserved is energy and momentum, &lt;a href=&quot;http://en.wikipedia.org/wiki/Four-momentum&quot; rel=&quot;nofollow&quot;&gt;a four vector&lt;/a&gt; whose measure is m*c, where m is the effective mass of the system, similar to the length of a three-vector after the addition of three-vectors .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a pi0 goes into two photons, it is true that the available energy for each gamma in the centre of mass system   of the pio is half the mass of the pio, and the measure of the invariant mass of those two gammas will be the mass of the pio. When more particles are involved , lets say two pio's then the invariant mass of the four gammas four momenta is not additive to two pio masses. It is better to forget about convoluted arguments with masses and think of four momenta when in the relativistic regime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a box of photons will have a four momentum sum in measure equal to E*&lt;em&gt;2/c&lt;/em&gt;*2 - |p|*&lt;em&gt;2=m&lt;/em&gt;*2*c**2. (please see the wiki link for clear terminology)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the three vector momentum sums up to zero, the effective mass of the photons in the box will be E/c**2. Small but there to be weighed.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-05-31T17:02:38.243" />
  <row Id="10628" PostTypeId="2" ParentId="10606" CreationDate="2011-05-31T17:19:49.617" Score="4" ViewCount="" Body="&lt;p&gt;I've never seen a paper where the calculation is performed in a manifestly covariant manner. However, I've posted a set of reference notes on my website (http://jacobi.luc.edu/notes.html) that contains the variations needed to carry out the calculation. Let me summarize the calculation here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The action for gravity on a compact region $M$ with boundary $\partial M$ is&#xA;$$I_{EH} + I_{GHY} = \frac{1}{2 \kappa^2} \int_{M}d^{d+1}x \sqrt{-g} R + \frac{1}{\kappa^2} \int_{\partial M} d^{d}x \sqrt{-h} K ~.$$&#xA;The metric on $M$ is $g_{\mu\nu}$, and $R = g^{\mu\nu} R_{\mu\nu}$ is the Ricci Scalar. The induced metric on the boundary $\partial M$ is $h_{\mu\nu} = g_{\mu\nu} - n_{\mu} n_{\nu}$, where $n^{\mu}$ is the (spacelike) unit vector normal to $\partial M \subset M$. Now consider a small variation in the metric: $g_{\mu\nu} \to g_{\mu\nu} + \delta g_{\mu\nu}$. The quantities appearing in the Einstein-Hilbert part of the action change in the following manner:&#xA;$$ \delta \sqrt{-g} = \frac{1}{2} \sqrt{-g} g^{\mu\nu} \delta g_{\mu\nu}$$&#xA;$$ \delta R = -R^{\mu\nu} \delta g_{\mu\nu} + \nabla^{\mu}\left(\nabla^{\nu} \delta g_{\mu\nu} - g^{\nu\lambda} \nabla_{\mu} \delta g_{\nu\lambda} \right)$$ &#xA;Thus, the change in $I_{EH}$ is &#xA;$$\begin{aligned}\delta I_{EH} = &amp;amp; \frac{1}{2\kappa^{2}}\int_{M} d^{d+1}x \sqrt{-g} \left(\frac{1}{2} g^{\mu\nu} R - R^{\mu\nu} \right)\delta g_{\mu\nu}\\&#xD;&#xA;&amp;amp; + \frac{1}{\kappa^2} \int_{\partial M} d^{d}x \sqrt{-h} \frac{1}{2} n^{\mu} \left(\nabla^{\nu} \delta g_{\mu\nu} - g^{\nu\lambda} \nabla_{\mu} \delta g_{\nu\lambda}\right)~,\end{aligned}$$&#xA;with the boundary term coming from the volume integral of the total derivative in $\delta R$. The variations of the quantities in the GHY term are a bit more complicated to work out, but they all basically follow from standard definitions and this result for the variation of the normal vector:&#xA;$$\delta n_{\mu} = \frac{1}{2} n_{\mu} n^{\nu} n^{\lambda} \delta g_{\nu\lambda} = \frac{1}{2} \delta g_{\mu\nu} n^{\nu} + c_{\mu}~.$$&#xA;In the second equality I've introduced a vector $c_{\mu}$ that is orthogonal to $n^{\mu}$; it is given by&#xA;$$c_{\mu} = - \frac{1}{2} h_{\mu}{}^{\lambda} \delta g_{\nu\lambda} n^{\nu} ~.$$&#xA;The reason I've introduced this vector is that the variation in the trace of the extrinsic curvature can be written as&#xA;$$\delta K= - \frac{1}{2} K^{\mu\nu} \delta g_{\mu\nu} - \frac{1}{2} n^{\mu}\left(\nabla^{\nu} \delta g_{\mu\nu} - g^{\nu\lambda} \nabla_{\mu} \delta g_{\nu\lambda} \right) + D_{\mu} c^{\mu}$$&#xA;where $D_{\mu}$ is the covariant derivative along $\partial M$ that is compatible with the induced metric $h_{\mu\nu}$. So, the change in the GHY part of the action is&#xA;$$\delta I_{GHY} = \frac{1}{\kappa^2} \int_{\partial M} d^{d}x \sqrt{-h}\left(\frac{1}{2}h^{\mu\nu} \delta g_{\mu\nu} K + \delta K \right)~.$$ &#xA;Combining this with $\delta I_{EH}$ we see that the several terms cancel, leaving&#xA;$$\begin{aligned} \delta I = &amp;amp; \frac{1}{2\kappa^2}\int_{M} d^{d+1}x \sqrt{-g}\left(\frac{1}{2} g^{\mu\nu} R - R^{\mu\nu} \right)\delta g_{\mu\nu}\\&#xD;&#xA;&amp;amp; + \frac{1}{\kappa^2}\int_{\partial M} d^{d}x \sqrt{-h}\left(\frac{1}{2}(h^{\mu\nu} K - K^{\mu\nu})\delta g_{\mu\nu} + D_{\mu} c^{\mu} \right) ~.\end{aligned}$$&#xA;We discard the term $D_{\mu} c^{\mu}$, which is a total boundary derivative. &lt;/p&gt;&#xA;" OwnerUserId="1727" LastEditorUserId="1727" LastEditorDisplayName="" LastEditDate="2011-05-31T17:27:21.267" LastActivityDate="2011-05-31T17:27:21.267" CommentCount="3" />
  <row Id="10629" PostTypeId="2" ParentId="10615" CreationDate="2011-05-31T17:39:28.550" Score="1" ViewCount="" Body="&lt;p&gt;Work backwards to make the math easier and also make the pattern emerge more quickly. I did this, but will show results in forward order &lt;/p&gt;&#xA;&#xA;&lt;p&gt;R(1) = 2R = (1+1/1)R&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R(2) = (1+(2/3)) R&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R(3) = (1+(5/8)) R&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R(4) = (1+(13/21)) R&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R(5) = (1+(34/55)) R&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Call this general form ---&gt; R(N) = (1+((a(N)/b(N))) R&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define a(1) = b(1) = 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For N &gt; 1 ---&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a(N) = a(N-1) + b(N-1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b(N) = a(N) + b(N-1) = a(N-1) + 2 b(N-1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You still need to crank through the math for each R(N), since one must know a(N-1) and b(N-1) to get the next term; but it is more straightforward.  Maybe someone up on their mathmatics of series can get that into a closed form for us.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-05-31T17:39:28.550" CommentCount="1" />
  <row Id="10630" PostTypeId="2" ParentId="10527" CreationDate="2011-05-31T18:08:22.100" Score="0" ViewCount="" Body="&lt;p&gt;similar meta-physical question came up on &lt;a href=&quot;http://skeptics.stackexchange.com/questions/3814/is-31-spacetime-privileged&quot;&gt;skeptics.stackexchange&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a physicists point of view &lt;strong&gt;theories can only experimentally falsified&lt;/strong&gt;. As stated in the link atom lasers might be able to measure directly properties of space. Currently one can only indirectly prove, that the concept of spacetime as a real physical entitiy/object fits measurement data (e.g. gravitational waves from Doppelpulsars predicted by ART). But u cant measure space like using a atomic force microscope to &quot;touch&quot; atoms&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It remains currently a meta-physical question whether u assume space is just a describing mathematical concept or a real-world object. Protophysics argues, that geometry is just a set of math. tools to describe universe, spacetime is no object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good question might be, can u describe a 4dim spacetime with a more dimensional spacetime consistently, is there some kind of mathematical redundance? Or is the math. concept of dimensionality so unique, that it would force very specific properties of the physics we try to model within it. This is kind of Ockhams Razor principle. An physical explanation should be as simple as possible. Currently these multi-dim theories of everything (TOE) need this additional dimensions to model the physics they try to describe, but they also give out alot of not falsifiable solutions, the more, the more dimensions they use afaik. So this is the borderline between laboratory physics, meta-physics/philosophy. U cant really logically argue here without ad-hoc hypothesis and subjective assumptions. &lt;/p&gt;&#xA;" OwnerUserId="3715" LastActivityDate="2011-05-31T18:08:22.100" />
  <row Id="10631" PostTypeId="1" CreationDate="2011-05-31T18:17:14.567" Score="-2" ViewCount="168" Body="&lt;p&gt;When tackling a physics problem, An Engineer will manipulate the axes/coordinate system where a Mathematicians and/or Physicists will use the original coordinate system and math. Why do Engineers think differently? I know its likely because that is how they are taught, but why are they taught that way?&lt;/p&gt;&#xA;" OwnerUserId="2843" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-05-31T19:30:47.930" LastActivityDate="2011-05-31T21:16:33.237" ClosedDate="2011-05-31T21:25:39.057" Title="Why do Engineers manipulate coordinate systems?" Tags="&lt;soft-question&gt;&lt;coordinate-systems&gt;&lt;engineering&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="10632" PostTypeId="2" ParentId="10631" CreationDate="2011-05-31T18:21:34.297" Score="-1" ViewCount="" Body="&lt;p&gt;Because engineers like making things simple - it it's easier to work in the coordinate system of the aircraft (rather than galactic coordiantes) then they will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand the physicists will redefine all the constants to 1 to simplify the sums.&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-05-31T18:21:34.297" />
  <row Id="10633" PostTypeId="2" ParentId="10631" CreationDate="2011-05-31T18:49:43.703" Score="0" ViewCount="" Body="&lt;p&gt;Engineers and Physicists have different requirements so they use different tools, and sometimes use the same tools with different approaches&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineers usually are after solving differential equations, or doing resonance analysis on some structure, which mostly involves doing Laplace transforms of complicated systems of equations, these equations might become significant easier to solve in specific coordinate systems. Some coordinate systems are better than others for certain problems&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physics also use this for solving equations (think how easier is to solve Schrödinger equation for the hydrogen atom in spherical coordinates rather than, say, cartesian). However in theoretical physics one usually does not want to focus how the equations look in specific coordinates; one actually wants to see what part of a equation does not change (or change in a preescribed manner) when a coordinate system is changed, since the most interesting theoretical quantities are usually the ones that transform in particular simple and elegant ways&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-05-31T18:49:43.703" />
  <row Id="10634" PostTypeId="2" ParentId="10631" CreationDate="2011-05-31T18:50:16.367" Score="-1" ViewCount="" Body="&lt;p&gt;There are a few reasons. The first is that most engineers do project work, so a coordinate system is usually developed to suit the project making everything simple and easy to input into calculations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second reason is that engineers like to look at solutions to problems by comparing the results calculations and designs with other designs at different stages. It is far easier to compare the dimensions of items when the units and origin of the coordinate system are suited to the problems. For example if you had to compare the depth of bridge girders but the bridge girders were measured as offsets from a origin at the support of the bridge rather than simply the depth of the girder it would be far more difficult to do a simple comparison. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final reason is simply that one engineering structure often interfaces with another. If you take the example of the London Underground it has it's own coordinate system for x, y and z coordinates. This means that it is easy for a new project to connect to an existing project&lt;/p&gt;&#xA;" OwnerUserId="3799" LastActivityDate="2011-05-31T18:50:16.367" CommentCount="1" />
  <row Id="10635" PostTypeId="1" AcceptedAnswerId="10672" CreationDate="2011-05-31T19:05:31.840" Score="0" ViewCount="247" Body="&lt;p&gt;I enjoy reading up on new topics in physics and astronomy, and learning the concepts and ideas in the major theories, like relativity and quantum mechanics. Unfortunately, one thing that I lack completely (and I envy others for) is the mathematical understanding. I would like to start studying the mathematics as well, from Newtonian gravity to more advanced problems, but I was wondering what the best progression of topics would be. How should I learn the mathematics of physics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that there are other questions on how to learn physics, and I apologize if this seems like a duplicate. However, I don't think that the mathematics have specifically been addressed in previous questions.&lt;/p&gt;&#xA;" OwnerUserId="3395" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-02T06:49:39.907" LastActivityDate="2011-06-23T17:11:52.497" Title="Recommended progression with which to learn physics for fun" Tags="&lt;soft-question&gt;&lt;education&gt;&lt;books&gt;&lt;reference-request&gt;&lt;big-list&gt;" AnswerCount="5" CommentCount="4" />
  <row Id="10636" PostTypeId="1" CreationDate="2011-05-31T19:20:53.007" Score="0" ViewCount="86" Body="&lt;p&gt;Typically when I think of a spin moment in an external field, I visualize it classically in terms of a magnetic moment precessing around the external field vector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how is this described quantum mechanically? My guess would be that the external field causes transitions between the up-state and down-state at the Larmor frequency. However if the Zeeman energy was much greater than the thermal energy, wouldn't the spin need to stay in the low energy state? Does that mean the spin  ceases precession in the classical picture?&lt;/p&gt;&#xA;" OwnerUserId="1939" LastActivityDate="2011-06-01T00:28:40.413" Title="QM interpretation of Larmor precession" Tags="&lt;angular-momentum&gt;&lt;spin&gt;&lt;magnetic-moment&gt;" AnswerCount="3" />
  <row Id="10637" PostTypeId="2" ParentId="10570" CreationDate="2011-05-31T20:02:01.557" Score="1" ViewCount="" Body="&lt;p&gt;You have stiffened the paper by greatly increasing its bending moment of intertia.  This can occur in at least two ways:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) When you unfolded the paper, it still had some residual bend which made the paper form a very shallow 'V'.  Even though shallow. it is much deeper than the thickness of the original paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) When you unfolded the paper, it still had some residual bend which made the paper form a very shallow 'V' which was very localized to the area next to the bend.  Even if you refold the paper in the opposite way, to take out or minimize the original fold, there is still a shallow 'V'.  Even though it might be much shallower than in case 1, it is much deeper than the thickness of the original paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases, the increased resistance to bending comes from the new geometry of the paper, more specifially, the geometry of a cross-section of the paper which goes through the bend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of the unbent paper as a beam.  Its resistance to bending is proportional to b(d^3), where 'b' is the width of the beam and 'd' is the depth.  If you take a piece of 8.5&quot; x 11&quot; piece of paper and lay it flat over a pencil on the table, the paper will flop so that both ends touch the table.  The paper forms a beam: 'b' is 8.5&quot; or 11&quot; (depending on how you laid the paper) and 'd' is the thickness of the sheet (say, about one one-hundreth of an inch).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to improve the stiffness and strength of this beam?  Fold the paper, accordion-style, with sharp 1/2&quot; folds.  Then lay it across the pencil, so that the folds are perdendicular to the pencil.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that you folded the paper into an accordion that was 1.1&quot; wide (one tenth of the original unfolded 11&quot;), had about 22 folds and was 8.5&quot; long.  The 'b' for this new beam is 0.5&quot;, which is 50 times more than the (1/100) of an inch thickness of the paper.  The new beam is (50)(50)(50)/10 = 12500 times stiffer than the unfolded sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's go back to your sheet with the one bend.  When you folded that sheet, you increased 'd'.  The amount is hard to quantify, but I would ballpark it at a factor of 5, even if you tried to straighten out the bend by refolding it the opposite way. Again, guesstimate that 'b' of the localized area affected by the bend is 1/10 of the original sheet width.  So the new resistance to bending increased by a factor of (5)(5)(5)/10 = 12.5 times the unfolded sheet.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-05-31T20:08:22.573" LastActivityDate="2011-05-31T20:08:22.573" />
  <row Id="10638" PostTypeId="1" CreationDate="2011-05-31T20:08:26.550" Score="1" ViewCount="84" Body="&lt;p&gt;Say there are $N$ atoms of type $A$ in a box of volume $V$ and they are undergoing a reversible association-dissociation reaction $A + A = A_2$. &#xA;Let an $A$ atom have mass $m$, and hence the molecule $A_2$ has mass $2m$. &#xA;Assume that the temperature $T$ is high-enough that Boltzmann statistics applies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say $N_1$ be the number of $A$ atoms and say $N_2$ be the number of $A_2$ molecules at equilibrium. Then $N_1 + 2N_2 = N$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there had to be $N$ $A$-type atoms in the the box then I guess the partition function $Z$ would be given by,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z = \frac{1}{N!} [ \int e^{-\frac{\beta p^2}{2m}} \frac{d^3p d^3q}{h^3} ] ^N = \frac{1}{N!} [ \frac{V}{h^3} (2\pi m kT)^{\frac{3}{2}} ] ^N .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say that the dissociation energy of the reaction is $E_0$.  Say $N_1$ be the number of $A$ atoms and say $N_2$ be the number of $A_2$ molecules at equilibrium. Then $N_1 + 2N_2 = N$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to write the partition function for the above mixed system in equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the expression for $Z$ as stated above, I am guessing that this partition function (say $Z_N$)  would look like, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z_N = \sum _ {N_1 + 2N_2 = N} e^{-\beta N_2 E_0} \frac{1}{N_1 !}  [ \frac{V}{h^3} (2\pi m kT)^{\frac{3}{2}} ] ^{N_1}  \frac{1}{N_2 !}  [ \frac{V}{h^3} (4\pi m kT)^{\frac{3}{2}} ] ^{N_2}.    $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In the above I am using the idea that for $N_2$ molecules to have formed $N_2$ $E_0$ (binding) energy would have been released and hence that much of extra energy has to be accounted for which is not captured in the partition functions over the $N_1$ atoms and $N_2$ molecules...) &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is the above expression for the partition function correct ? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One further introduces the grand canonical partition function (say $Z_{\mu}$ )  for the above system with a chemical potential $\mu$ to &quot;take care of the constraint $N_1 + 2N_2 = N$&quot;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z_{\mu} = \sum _ {N=0} ^\infty e^{\beta \mu N} Z_N.$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Firstly I don't understand the meaning of the $\mu$ in the above. (I can't relate it to what I would usually think as chemical potential) I mean..isn't the above expression for $Z_{\mu}$ independent of $N_1$ and $N_2$? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How does the $Z_{\mu}$ help find the expectation value for $N_1$ and $N_2$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If my expressions for $Z_N$ is correct then does $Z_{\mu}$ have a closed form expression? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All this is to understand the claim that as expectation values the following relationship apparently holds, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{N_1 ^2} {N_2} \sim e ^{-\frac{|E_0|}{kT}} .$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be glad to know of other insights along the above. &lt;/p&gt;&#xA;" OwnerUserId="1637" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-01T18:33:47.957" LastActivityDate="2011-06-02T11:09:05.863" Title="Statistical physics of molecular dissociation of a diatomic gas" Tags="&lt;statistical-mechanics&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="10639" PostTypeId="1" CreationDate="2011-05-31T20:09:58.390" Score="0" ViewCount="128" Body="&lt;p&gt;I'm trying to find something Einstein (I think) said about time...It&#xA;was something like..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If you removed every particle from space and were left with only one&#xA;pocket watch (clock, timepiece?), time would still exist and the watch&#xA;would be ticking?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is not exactly it so I can't find it in google.&#xA; Do you know of anything he said similar to that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not this one...    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;“It was formerly believed that if all material things disappeared,&#xA;time and  space would be left. According to the relativity theory,&#xA;however, time and  space disappear together with the things.”&lt;/p&gt;&#xA;" OwnerUserId="3800" LastActivityDate="2011-06-28T12:52:09.317" Title="If you removed every particle from space...?" Tags="&lt;general-relativity&gt;&lt;cosmology&gt;&lt;time&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10640" PostTypeId="2" ParentId="10635" CreationDate="2011-05-31T20:32:04.560" Score="1" ViewCount="" Body="&lt;p&gt;I would start with any good high school or introductory college text on physics.  The textbook will not only show you conceptually what is going on, but will, out of neccessity of fulfilling the purpose for which it was written, give you the math along with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One caveat to that is that physics will not teach you calculus, and a college-level text will assume some calculus exposure.  If you have not had differential and integral calculus, then stick to a high school level textbook, or... take a calculus course (or get the book and self-study).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any good textbook will be structured to take you through the topics in a logical order.  You will encounter free-body mechanics before you get to hydrostatics, hydrostatics before aerodynamics, electricity and magnetism before optics, and etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;General guide for picking a GOOD text book: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) Look for texts which have more than one author (a collaborative effort) and are NOT the first edition. Something that is in its, say, fifth edition, is tested and proven and still in demand: an excellent sign of underlying quality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) Look up community college physics courses on the web and write down what text they use, then surf for book rating comments on Amazon and elsewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) Make sure the text book has LOTS of pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.) Make sure the text book has answers to odd-numbered problems (or answers to even-numbered... whatever) in the book.  There is NO better way of verifying you grasped the material than working out a problem.  You want to know that you got the right answer; and you have no teacher to go to.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-06-01T05:46:33.690" LastActivityDate="2011-06-01T05:46:33.690" />
  <row Id="10641" PostTypeId="1" AcceptedAnswerId="10645" CreationDate="2011-05-31T20:39:15.700" Score="2" ViewCount="184" Body="&lt;p&gt;My understanding of an electrical current is that it is the flow of electrons through a material. The only magnitude I can fathom for this process is the number of electrons which are flowing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know ohms law, and how some of the basic components of a circuit work (resistors and capacitors, specifically). If there is a circuit attached to a 120V power source, the voltage at the beginning of the circuit will be 120V, and at the end of it will be 0V. When the current experiences resistance, voltage is lost, so more current has to be supplied to the circuit so that exactly all of the voltage is lost by the end of the circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: what is happening when voltage is lost? When current is increased? When learning about capacitors in university, I understood that voltage is built up on each plate by electrons flowing in and spreading out across the surface of the plate. But if voltage is the number of electrons in a circuit, what is current? And if all voltage is dropped at the end of a circuit, where do the electrons go? I figured that since electrons won't build up ad infinitum throughout the circuit (obviously since everything would build up an increasingly negative charge), the magnitude of the current must be the number of the electrons. But then what is voltage? What is lost? How should I interpret the stored charge on a capacitor?&lt;/p&gt;&#xA;" OwnerUserId="417" LastActivityDate="2011-06-03T12:19:05.307" Title="How can I physically interpret voltage and current?" Tags="&lt;electromagnetism&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="10642" PostTypeId="2" ParentId="10631" CreationDate="2011-05-31T21:16:33.237" Score="1" ViewCount="" Body="&lt;p&gt;Choosing an appropriate coordinate system often vastly simplifies a problem.  Anyone who wants to solve a problem expediently will try to find a coordinate system that simplifies the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your professors told you that physicists do not do this, then your professors told you a falsehood.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-05-31T21:16:33.237" />
  <row Id="10643" PostTypeId="2" ParentId="10635" CreationDate="2011-05-31T21:20:24.360" Score="1" ViewCount="" Body="&lt;p&gt;Basic areas of mathematical physics are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;analysis&lt;/li&gt;&#xA;&lt;li&gt;linear algebra&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Depending on specific physics area like QM or ART u have to enhance ur knowledge with functional analysis or differential geometry. But without basic areas knowledge u will not see any light at the end of the tunnel ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sites like &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html#hph&quot; rel=&quot;nofollow&quot;&gt;hyperphysics&lt;/a&gt; try to keep the math as simple as possible. Also watch some online lectures on MIT for example for basic areas courses. Some universities teach thier physics student by physics Profs., so the examples and lecture is not too abstract as hearing a math. Prof lecture.&lt;/p&gt;&#xA;" OwnerUserId="3715" LastActivityDate="2011-05-31T21:20:24.360" />
  <row Id="10644" PostTypeId="2" ParentId="10641" CreationDate="2011-05-31T21:24:21.857" Score="1" ViewCount="" Body="&lt;p&gt;Without going into the depths of things like displacement current and semiconductors (which is when we learn that everything we were taught in Electricity 101 was a lie), you have it basically right: Current is the flow of electrons.  More quantitatively: 1 Amp = a charge flow of 1 Coulomb/second = 6.25 x 10^18 electrons/second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Voltage is the force with which the electrons are forced through a circuit.  Voltage can be thought of as an electromotive force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Voltage is NOT the number of electrons in a circuit nor on a capacitor.  However, the number of electrons on a plate of a capacitor can produce a voltage which (typically) resists current flow.  That is: the electrons in a capacitor create a voltage which opposes the voltage which forced the electrons onto the capacitor plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When an electron gets to the end of a circuit, it really hasn't gotten to the end of the circuit.  It has only gotten to the part of the circuit you see on the schematic in your text book.  After that it has a long journey back to the power plant, where generators pump it back out to somebody else's hair dryer or toaster oven.  Or, in the case of battery powered electronics, it fuels a chemical reaction.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-05-31T21:45:42.970" LastActivityDate="2011-05-31T21:45:42.970" CommentCount="4" />
  <row Id="10645" PostTypeId="2" ParentId="10641" CreationDate="2011-05-31T21:36:03.343" Score="3" ViewCount="" Body="&lt;p&gt;Current is defined as the total amount of charge that passes through a surface per unit time, and voltage is defined as the electrical potential energy per unit charge. So, for the usual case in which electrons are the charge carriers, the amount of current corresponds to the number of electrons, and the voltage corresponds to the potential energy that each electron has.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When voltage is lost (by which I presume you're talking about the voltage decrease as you move along a circuit), it just means that the electrons lose potential energy as they move through the circuit. By the time they get to the end of the circuit (the positive terminal of the battery), they've lost all their potential energy to collisions with atoms in the wire. Then they go through the battery and get their energy replenished, so they can start the cycle over. Keep in mind that the same electrons keep cycling through the circuit over and over again - they don't go anywhere when they reach the end.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-05-31T21:36:03.343" CommentCount="7" />
  <row Id="10646" PostTypeId="2" ParentId="10636" CreationDate="2011-05-31T21:52:17.933" Score="0" ViewCount="" Body="&lt;p&gt;Normally this is best explained by NMR. U have a population of spins in an thermal equity. How the energy states of this ensemble are populated is defined by Maxwell-Boltzmann statistics. In case of deuterium atmoms, the nucleus of a single deuterium atom is either in spin down or spin up state, ratio of spin up to down is set by MB statistics. There is no vanishing of precession, the spin is a fundamental property like mass, it doesnt cease! This would mean zero energy, in QM u have the ground state energy != 0. And the energy states are discrete as the spin is discrete! It can add to total spin/angular momentum = 0, but this is another topic when looking on several electron atoms and molecules. Actually there is no classical match to QM Spin. The correspondence principle doesnt work here like for impulse operator/observable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to NMR: if u put the deuterium ensemble into an external rf field with rf Larmor-frequency of the deuterium atom, u excite the ensemble out of thermal equity and relax back after a while. Look at NMR Spectroscopy, there is especially the need to implement very high magnetic fields (Zeeman-Energy) to trick out the Boltzmann-Statistics and yield a good signal-noise ratio.&lt;/p&gt;&#xA;" OwnerUserId="3715" LastActivityDate="2011-05-31T21:52:17.933" />
  <row Id="10647" PostTypeId="1" AcceptedAnswerId="10650" CreationDate="2011-05-31T22:00:25.643" Score="3" ViewCount="149" Body="&lt;p&gt;What is the dominant cause for ocean waves at a beach? Are they the result of wind/pressure difference? If so, the waves do seem to exist in similar intensity even during relative quiet times of the day. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a simple mathematical model that we can quickly explain the intensity/frequency of waves with? Does the strength of the waves (say the variance and mean of the amplitude of waves) relate to a simple physical quantity (temperature, off shore wind, pressure difference)?&lt;/p&gt;&#xA;" OwnerUserId="3714" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-06-02T10:39:58.943" LastActivityDate="2011-06-02T10:39:58.943" Title="What is the dominant cause for ocean waves at a beach?" Tags="&lt;fluid-dynamics&gt;&lt;waves&gt;&lt;geophysics&gt;" AnswerCount="2" />
  <row Id="10648" PostTypeId="2" ParentId="10647" CreationDate="2011-05-31T22:10:31.043" Score="0" ViewCount="" Body="&lt;p&gt;The normal dominant cause of most waves you see (i.e., not a Tsunami) is frictional force between the wind and water over very large distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though the wind may not be blowing at your particular beach, it is ALWAYS blowing somewhere out in the ocean.  The waves produced by this effect can (and do) travel thousands of miles away from the source.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-05-31T22:20:24.587" LastActivityDate="2011-05-31T22:20:24.587" CommentCount="1" />
  <row Id="10649" PostTypeId="2" ParentId="10636" CreationDate="2011-05-31T23:37:32.723" Score="1" ViewCount="" Body="&lt;p&gt;Larmor precession is quantum mechanically described by the interaction of the spinor with the magnetic field. Without magnetic field the spin-up eigenstates states in the x and y direction are given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_x^\uparrow ~~=~~ \frac{1}{\sqrt{2}}\left(\begin{array}{r}1 \\ 1\end{array}\right) \exp(-iEt)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_y^\uparrow ~~=~~ \frac{1}{\sqrt{2}}\left(\begin{array}{r}1 \\ i\end{array}\right)\exp(-iEt)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The direction of the spin in the xy-plane is given by the angle between the two spinor components in the complex plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two components get an extra energy factor from the magnetic field B but with a different sign:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S^\uparrow ~~=~~ \frac{1}{\sqrt{2}}\left(\begin{array}{r}\exp(-iE_Bt) \\ \exp(+iE_Bt)\end{array}\right)\exp(-iEt)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is that the spin direction rotates (precesses) in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards, Hans&lt;/p&gt;&#xA;" OwnerUserId="3197" LastEditorUserId="3197" LastEditorDisplayName="" LastEditDate="2011-06-01T00:10:53.227" LastActivityDate="2011-06-01T00:10:53.227" />
  <row Id="10650" PostTypeId="2" ParentId="10647" CreationDate="2011-05-31T23:52:50.347" Score="3" ViewCount="" Body="&lt;p&gt;Yes, primarily wind.  It's called the Kelvin-Helmholtz instability.  Strong winds in an area will excite a range of wavelengths, the longer wavelengths will go faster according to the deep water dispersion relation ( speed proportional to square root of wavelength).  So if you see a train of waves with decreasing wavelength over time, you could in principle infer a common point of origin (this is a common textbook or qualifying exam type problem).&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditorDisplayName="" LastEditDate="2011-06-01T00:13:06.607" LastActivityDate="2011-06-01T00:13:06.607" CommentCount="1" />
  <row Id="10651" PostTypeId="1" CreationDate="2011-05-31T23:58:54.197" Score="12" ViewCount="435" Body="&lt;p&gt;I've often heard the argument that having 3 spatial dimensions is very special. Such arguments are invariably based on certain assumptions that do not appear to be justifiable at all, at least to me. There is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Spacetime#Privileged_character_of_3.2B1_spacetime&quot; rel=&quot;nofollow&quot;&gt;summary of arguments&lt;/a&gt; on Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a common argument for why &gt;3 dimensions is too many is that the gravitational law cannot result in stable orbital motion. A common argument for &amp;lt;3 dimensions being too few is that one cannot have a gastrointestinal tract, or more generally, a hole that doesn't split an organism into two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I being overly skeptical in thinking that while the force of &lt;em&gt;gravity&lt;/em&gt; may not be able to hold objects in stable orbits, there most certainly exist sets of physical laws in higher dimensions which result in formation of stable structures at all scales? It may be utterly different to our universe, but who said a 4D universe must be the same as ours with one extra dimension?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, isn't it very easy to conceive of a 2D universe in which organisms can feed despite not having any holes, or not falling apart despite having them? For example, being held together by attractive forces, or allowing certain fundamental objects of a universe to interpenetrate, and thus enter a region of the body in which they become utilized. Or, conceive of a universe so incomprehensibly different to ours that feeding is unnecessary, and self-aware structures form through completely different processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I realise that this is sort of a metaphysical question, is 3+1 dimensions really widely acknowledged to be particularly privileged by respected physicists?&lt;/p&gt;&#xA;" OwnerUserId="786" LastActivityDate="2011-08-29T20:26:53.720" Title="Is 3+1 spacetime as privileged as is claimed?" Tags="&lt;spacetime&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" />
  <row Id="10652" PostTypeId="2" ParentId="10635" CreationDate="2011-06-01T00:01:34.117" Score="1" ViewCount="" Body="&lt;p&gt;One great place to start (assuming a knowledge of calculus) is a college upperclassman level E&amp;amp;M book like &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/013805326X&quot; rel=&quot;nofollow&quot;&gt;Griffiths&lt;/a&gt;.  While Electrodymamics is an interesting field in its own right, and gives a whole lot of motivation for later topics, its place in a physics curriculum is also the first place that students are introduced to the machinery of solving partial differential equations in several dimensions, including vector calculus, separation of variables, spherical harmonics, etc.  A grounding in these things is also essential before moving on to something like differential geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As stated above, linear algebra is something that is also univerally necessary in higher-level physics.  You can find a ton of resources on this by searching around (I can't recommend a good textbook, I grabbed a library textbook and just did all the problems that had answers in the back of the book, and that was years ago).  Linear algebra is pretty much a prerequisite for a lot of the higher level math, too--you're not going to make much sense of tensors or mapping operators in topology if you don't have some sort of grounding in linear algebra.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-06-01T05:45:05.247" LastActivityDate="2011-06-01T05:45:05.247" />
  <row Id="10653" PostTypeId="2" ParentId="10636" CreationDate="2011-06-01T00:28:40.413" Score="0" ViewCount="" Body="&lt;p&gt;It is a bit confusing in the wave-function (spinor) picture.  However one can just as easily work with the density matrix.  This is the outer product of the wave function with itself.  For a spin-1/2 (two component) wave-function, the pure part of the density matrix is a spin-1 object, a little vector pointing in some direction.  This vector precesses exactly as in the classical case.  (The non-pure part is the spin-0 irreducible rep, totally invariant under rotations.  This shouldn't be too surprising -- if you don't know what direction it is, rotating it shouldn't change anything.)&lt;/p&gt;&#xA;" OwnerUserId="1233" LastActivityDate="2011-06-01T00:28:40.413" />
  <row Id="10654" PostTypeId="2" ParentId="10527" CreationDate="2011-06-01T00:51:28.257" Score="1" ViewCount="" Body="&lt;p&gt;Actually, let's give this a shot.  This isn't evidence &lt;b&gt;for&lt;/b&gt; extra dimensions (the non-obesrvation of extra dimensions/supersymmetry is one of the big reasons string theory is not accepted universally as true, after all), but this is an argument as to why small extra dimensions are unobservable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a particle in a box in quantum mechanics in n spatial dimensions..  If you do this, then Schrödinger's equation for a pure state becomes (inside the box):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E\psi = - \frac{\hbar^{2}}{2m}\nabla^{2}\psi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And where you force $\psi$ to be zero everywhere outside the box, and on the boundary of the box.  Using a bunch of PDE machinery involving separation of variables, we find that the unique solution to this equation is a an infinite sum of terms that look like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi=A\Pi_{i=1}^{n}\sin\left(\frac{m_{i}\pi x_{i}}{L_{i}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where all of the $m$ are integers, and the $\Pi$ represents a product with one $\sin$ term for each dimension in our space${}^{1}$.  Plugging this back into Schrödinger's equation tells us that the energy of this state is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E=\frac{\hbar^{2}}{2m}\left(\sum_{i=1}^{n} \frac{m_{i}^{2}}{L_{i}^{2}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's assume that in the first $d$ dimensions, our box has a large width $L$, while in the last $n-d$ dimensions, our box has a small width $\ell$.  Then, we can split this sum into &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E=\frac{\hbar^{2}}{2m}\left(\sum_{i=1}^{d} \frac{m_{i}^{2}}{L^{2}}+\sum_{i=d+1}^{n} \frac{m_{i}^{2}}{\ell^{2}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, now we can see what's happening--if $L \gg \ell$, there is a much greater energy cost associated with moving in the more constrained or smaller $n-d$ directions than there is in moving in the less constrained $d$ dimensions--the smallest transitions cost an energy proportional to the inverse square of the size of the dimension.  By making these dimensions small enough, we can guarantee that no experiment humans have done has even approached the energy threshold required to induce this transition, meaning that the portion of a particle's wavefunction associated with these extra dimensions is constrained to stay the way they are, making them unobservable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{1}$So, if n=2, a typical state would look something like $\psi=A\sin(\frac{2\pi x}{L_{x}})\sin(\frac{5\pi y}{L_{y}})$&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-06-01T00:51:28.257" />
  <row Id="10655" PostTypeId="2" ParentId="10592" CreationDate="2011-06-01T01:28:44.900" Score="0" ViewCount="" Body="&lt;p&gt;Actually there's three kinds of ropes one might hang an adult human from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rock climbing rope is designed with a great deal of stretch. Thus when someone falls the rope decelerates them slowly, thereby lowering the peak force applied to the rope. To find these google &quot;dynamic climbing rope&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ropes used in rescue are not intended to catch falls. Accordingly they are designed with much less stretch. This makes for a more comfortable ride. To find these, google &quot;low stretch rescue rope&quot; or &quot;static rescue rope&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ropes used for execution are the third type. With these, safety is not a great concern.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-01T01:28:44.900" CommentCount="1" />
  <row Id="10656" PostTypeId="1" CreationDate="2011-06-01T03:11:59.423" Score="3" ViewCount="303" Body="&lt;p&gt;Person A in reference frame A watches person B travel from Star 1 to Star 2 (a distance of d).  Of course, from person B's reference frame, he is at rest and is watching Star 2 traveling to him.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know from the principle of relativity, each one will measure the other one’s clock as running slower than his own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let’s say that Person A measures Person B’s speed to be v, and that Person A measures 10 years for person B to make it to Star 2.  Let’s also say that person B is moving at the speed so that the Gamma Factor is 2.  This means person A observes person’s B’s clock to have elapsed a time of 5 years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let’s look at this from Person B’s perspective:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Person B observes Star 2 approaching (and Star 1 receding from) him also at speed v.  Since the two stars are moving, the distance between them is length contracted (after all, if there were a ruler in between the stars, the moving ruler would be contracted) by a factor of 2.  Since person B measures the initial distance to Star 2 to be d/2 and its speed v, he calculates the time to Star 2’s arrival to be 5 years.  Since he observes person A’s clock as running slow (since Person A is moving also at speed v), when Star 2 arrives, he measures Person A’s clock to have elapsed a time of 2.5 years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you see why I’m confused?  Person A measures Person B’s elapsed time to be the same as Person B measures Person B’s elapsed time (both 5 years), but Person B does not measure Person A’s elapsed time to be the same as Person A measures Person A’s elapsed time (Person B get’s a measurement of 2.5 years while Person A measured 10 years).  This is asymmetrical, which probably means it is wrong.  But I’m not sure what the error is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect if I had done this correctly, each person should measure his own elapsed time to be 10 years and measure the other’s elapsed time to be 5 years.  This would be symmetrical and would make the most sense, but again, I can’t seem to justify how person B wouldn’t measure his trip time to be 5 years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's my mistake?&lt;/p&gt;&#xA;" OwnerUserId="3805" LastActivityDate="2011-06-02T12:46:28.027" Title="Misunderstanding of Special Relativity" Tags="&lt;special-relativity&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="10657" PostTypeId="2" ParentId="10551" CreationDate="2011-06-01T03:13:39.907" Score="0" ViewCount="" Body="&lt;p&gt;I've posted an answer for the blocking case and now I'm going to post an answer for the transmitting case, since the argument for the linear transition still seems to be holding sway. I'm going to argue for the step function much the way I did in the blocking case. The only sticky points are the transitions, where the wires become infinitely thin (blocking case: how do such thin wires block everything?) and where the slits become infinitely thin (transmitting case: how do such thin slits transmit everything?). I'll come back to those later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First the ideal blocking case: since the strips are parallel to the electric field, they function as a perfect conductor and block all transmission. Similarly, in the ideal transmitting case, the strips are perpendicular to the electric field, so they block current flow, acting as a perfect insulator. Hence, no reflection and total transmission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've already argued that for the blocking case, the strips carry the same amount of current as they are made thinner and thinner, thereby functioning as an perfect reflector. But it's harder to see how the complimentary procedure works in the transmitting case. How does power get through the thin little spaces in the venetian blinds? Answer: the slats of the venetian blinds are perfect conductors, so the current induced on the front surface loops around the back. It makes a complete loop, which it must in order to completely cancel the magnetic field inside the superconductor. And since it makes a very thin loop (because the slats are presumed to be very thin), it does not radiate at all because it has negligible magnetic dipole moment. Since it has no external electromagnetic effect, it might as well not be there, and the power seems to flow right through it. (Although technically you might argue that the power is annihilated on the front face and regenerated on the back face so it does not actually penetrate the superconductor.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do these arguments of perfect reflection/transmission break down for the limiting case of very thin wires/gaps? I've already argued that the very thin wires eventually show non-zero resistivity. What about the very thin spaces? By analogy, we can argue that they are not perfect insulators, and as their non-zero conductivity becomes significant, things break down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These notions of non-zero resitivity/conductivity are of course classical, but the actual mechanisms might actually be quantum-mechanical. When do very thin superconducting wires become resistive? when the magnetic field exceeds its quantum-mechanical threshold. And when do very thin insulating gaps become conductive? When they are small enough for the current to tunnel through. &lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-06-01T03:13:39.907" />
  <row Id="10658" PostTypeId="2" ParentId="437" CreationDate="2011-06-01T04:40:27.600" Score="1" ViewCount="" Body="&lt;p&gt;Yup. Like this? &#xA;&lt;a href=&quot;http://www.nist.gov/pml/div684/fcdc/upload/preprint.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.nist.gov/pml/div684/fcdc/upload/preprint.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3797" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-06-01T04:55:34.117" LastActivityDate="2011-06-01T04:55:34.117" CommentCount="2" />
  <row Id="10659" PostTypeId="2" ParentId="10656" CreationDate="2011-06-01T05:25:25.413" Score="3" ViewCount="" Body="&lt;p&gt;Everything you have said describing the situation in your question is correct; Person A and Person B disagree about how much time elapses on person A's clock between the two events.   (The first event is Person B leaving Star 1 and the second event is Person B arriving at Star 2) This is not a logical contradiction.  It stems from the relativity of simultaneity and the fact that the time between two events is different in different reference frames.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time between two events is minimized when the spatial separation between them is zero because the interval &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta s^2 = \Delta t^2 - \Delta x^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is invariant (the same for everyone).  Person B therefore perceives the minimum possible time between the two events, which is 5 years.  Person A perceives some spatial separation between  the events, and so perceives a longer time between them (10 years).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can use this information to work out the speed $v$.  For Person A, $\Delta t^2 - \Delta x^2 = 5^2$ because that's the answer for Person B, and it must be the same for A.  We know $\Delta t^2 = 100$, so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$100 - \Delta x^2 = 25$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta x = \sqrt{75} = 5\sqrt{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v$ is then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = \frac{\Delta x}{\Delta t} = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation is not symmetric with respect to A and B because A is not moving relative to the stars, but B is.  The existence of the stars breaks the symmetry.  A symmetric situation would be if A and B start at their own stars, then meet in the middle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another symmetric scenario would be to let B start moving away from A.  When A's clock reads 10 years, ask her what B's clock reads.  When B's clock reads 10  years, ask him what A's clock reads.  In that case, both would say that the other's clock reads 5 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the setup of the problem is symmetric with respect to A and B, their answers should be, also.  Because this problem does not have that symmetry, the answers A and B give do not have the symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you might be concerned that Person A thinks the time between the two events is 10 years, but according to Person B, Person A's clock reads only 2.5 years elapsed.  This is due to the relativity of simultaneity.  According to Person B, he is arriving at Star 2 and checking Person A's clock simultaneously.  Those events have a big spatial separation, though.  According to Person A, they are not simultaneous.  Person A thinks Person B has checked her clock too soon.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-06-01T05:34:06.103" LastActivityDate="2011-06-01T05:34:06.103" CommentCount="4" />
  <row Id="10660" PostTypeId="1" CreationDate="2011-06-01T05:48:51.497" Score="1" ViewCount="115" Body="&lt;p&gt;Suppose there's a radioactive material and a 1/2 quantum probability of detecting it by a Geiger counter. This puts the system in a superposition. Also suppose you are in the same room, and the walls of the room are perfectly decoherence-proof. You observe the Geiger counter and get a definite result. Either it had gone off, or it hadn't. The Copenhagen interpretation tells you the decay or nondecay only became real when it was measured, and the result reached you. Your friend is waiting outside the room. After a very long delay, you open the door, and he observes whether or not you had seen the Geiger counter go off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He then has the nerve to tell you &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is only one observer, and that's me, not you. You are nothing more than a dull bound state of electrons and protons. Before the door opened, you never had the property of knowing whether the Geiger counter went off. Only when the door opened did you acquire that property.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unnerving, but you reason, your friend is just like you. Both of you are humans and made up of a bound state of electrons and protons. What applies to you ought to apply to him and vice versa. Or are you special and inherently different from him?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You object that you remember very distinctly whether or not you heard the Geiger counter go off. He counters&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Your memory of having known whether or not the Geiger counter went off only came into being when the door opened. Just because you have that memory now does not mean it was real before the door opened. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You object that applying the Copenhagen interpretation to yourself tells you you did have a definite knowledge before the door opened. He counters&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It's a meaningless question whether or not you knew the state of the Geiger counter before the door opened and I measured you because there is no way in principle for me to find out. There is no inconsistency here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't all observers have to agree? Both of you agree on your current memory of having had a definite state and that is observable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did you have a definite knowledge of the state of the Geiger counter before the door opened? Change the story a little. You haven't opened the door yet, but you know your friend is waiting outside. Do you have a definite knowledge now? You think you do, but does that make it so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some further reflection, you realize the present you is a different observer from your past selves. How definite was the past?&lt;/p&gt;&#xA;" OwnerUserId="3806" LastActivityDate="2011-06-01T08:01:16.477" Title="Wigner's friend and intersubjectivity in quantum mechanics" Tags="&lt;quantum-interpretations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10661" PostTypeId="1" CreationDate="2011-06-01T06:38:03.463" Score="3" ViewCount="199" Body="&lt;p&gt;Let us consider Bloch wave function solutions for a particle confined to a 2D square lattice with a potential of the form &#xA;$V=V(x) + V(y)$&#xA;(that is, one that can be factorized).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case we can factorize the Hamiltonian to &#xA;$\hat{H} = \hat{H}_x + \hat{H}_y$&#xA;and the band spectrum structure (the energy dispersion for states) can be presented as&#xA;$E( \vec{q}) = E_x(q_x) + E_y(q_y)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One-dimensional dispersion spectrum has gaps between the bands for quasimomentum values of $q_{x} = n \frac{\pi}{a}$,  here $n$ is integer. It means that the energy gaps between the bands are situated at the quasimomentum border values of $q_{x,y} = n \frac{\pi}{a}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct? &#xA;Is it a specific case of some general principle regarding the positions of band gaps I need to know about? What can be said about the band gap positions generally?&lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="1543" LastEditorDisplayName="" LastEditDate="2011-06-01T17:35:11.167" LastActivityDate="2011-09-02T17:37:09.497" Title="bandgaps for 2D square lattice with potential of the form V=V(x) + V(y) - what are the general properties?" Tags="&lt;solid-state-physics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10662" PostTypeId="2" ParentId="10660" CreationDate="2011-06-01T07:39:12.713" Score="1" ViewCount="" Body="&lt;p&gt;The &quot;inner&quot; observer has observed the outcome of the experiment well before the &quot;outer&quot; observer but there is no contradiction whatsoever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, if the &quot;outer&quot; observer properly defines the projection operator that determines whether the &quot;inner&quot; observer had  a well-defined feeling or perception of knowledge (with some allowed error margin for confusion) after he did his measurement, the &quot;outer&quot; observer will find out that indeed, the &quot;inner&quot; observer had a well-defined feeling. So they will surely agree about that. However, the &quot;outer&quot; observer couldn't have predicted which of the outcomes was actually measured by the &quot;inner observer&quot;: he can just prove that the &quot;inner&quot; observer had a well-defined perception.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the outer observer is allowed - and encouraged - to use quantum mechanics uniformly for all objects, including bound states of protons and electrons that are sometimes called &quot;other people&quot;. If he does so, the wave function - encoding the probability amplitudes sufficient to make any future measurements - is uniformly spreading and linear superpositions of macroscopically distinct states are omnipresent as the intermediate states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was discussed nicely e.g. in Sidney Coleman's &quot;Quantum Mechanics In Your Face&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2010/11/sidney-coleman-quantum-mechanics-in.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/11/sidney-coleman-quantum-mechanics-in.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let me mention that quite typically, the &quot;inner&quot; observer could be &quot;less reliable&quot; and &quot;less decoherent&quot; than the outer one. So he could &quot;perceive&quot; that some other decohered histories have already disappeared as possibilities, but the interference could get restore later etc. which could ultimately refute the &quot;inner&quot; observer's classical memory. That's why a more reliable, more decohered &quot;outer&quot; observer shouldn't be imagining that the wave function of the observer system collapsed before the measurements are actually done - and/or perceived - by the &quot;outer&quot; observer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This comment becomes completely obvious if you imagine that the inner observer is just a collection of a small number of atoms that is decohering pretty slowly. Still, some perception of classical outcomes and decoherence may start to emerge. But it's still fuzzy enough for the classical approximation to be inaccurate. It's clear that in such a situation, the outer observer - who is much more reliably &quot;classical&quot; - can't believe the would-be classical concepts used by the &quot;inner&quot; observer because the &quot;outer&quot; observer knows very well that the &quot;inner&quot; observer is just a quantum mechanical system for which the classical approximation is highly imperfect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that the reason why you and others have problems with these simple things is that you still fundamentally refuse to accept that at the fundamental level, the world is quantum mechanical, and that all classical concepts - such as well-defined positions or completely well-defined results of experiments that become &quot;absolute facts&quot; - are just approximate. They are approximate because classical physics is not the way how the world works at the deepest level. If one wants to describe any system totally accurately, he has to use the full quantum mechanical theory where everything is allowed to have a nonzero chance to interfere and &quot;recohere&quot;, whether or not other &quot;conscious beings&quot; are parts of this world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same moment, every observer who uses a quantum mechanical description and who wants to convert some observations into &quot;facts&quot; must make sure that the &quot;facts&quot; are mutually decohered or &quot;consistent&quot; histories. Still, everyone should acknowledge that all such facts are always an approximation assuming that decoherence is sufficiently strong to make it de facto impossible for the historical facts to be &quot;reverted&quot; or &quot;retroactively modified&quot; by the future evolution. Quantum, small enough particles do many non-classical things. Quantum tunneling is an example. A small observer's perception that a measurement has become a &quot;fact&quot;, but this fact is later &quot;undone&quot; by subsequent recoherence, is another example. That's why one must never assume &quot;too early&quot; that some components of the wave functions have disappeared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, you also incorrectly suggest that the lab with the inner observer may be &quot;screened against decoherence&quot;. This is just a misunderstanding of what decoherence means. Decoherence depends on the separation of the degrees of freedom of the studied system and the environmental degrees of freedom. But that doesn't mean that the &quot;environment&quot; has to be physically separated from the &quot;interesting system&quot;. Quite generally, many of the environmental degrees of freedom are geometrically located &quot;inside&quot; the objects we are interested in. The environment is composed of all the useless, &quot;high-entropy&quot; information we're not interested in - surely things like lots of infrared photons traveling through our bodies, reflecting from the walls of the lab, and so on. You may isolate the lab from the external world but you will not stop the fact that it's still valid to describe the phenomena inside the lab by decoherence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the &quot;outer&quot; observer knows how to define the degrees of freedom in the lab to the interesting and environmental ones, he will also find out that the different states of the &quot;inner&quot; observer's brain have quickly decohered from each other. That means that for the purpose of all predictions about the &quot;inner&quot; observer only, the quantum probability waves may be replaced - with a good enough accuracy - by a nearly diagonal density matrix. The diagonal entries are the probabilities but again, the &quot;outer&quot; observer is not forced to imagine that the (later unobserved) a priori possible outcomes have seen their density matrix elements collapse to zero. He naturally does it only when he actually makes the measurement at the very end. When the decoherence happens in the middle, and it's reliable enough, it's consistent for the outer observer to assume that one of the results was &quot;already real&quot; after the &quot;inner&quot; observer made his measurement. But the &quot;outer&quot; observer surely doesn't have to assume that such a premature collapse has taken place. Why would he do it? Because of the respect to the &quot;inner&quot; observer as a fellow human with consciousness? No such &quot;respect&quot; exists in science. Science doesn't care whether someone may claim to be an observer - it still allows accurate predictions for any conglomerates of particles.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-06-01T08:01:16.477" LastActivityDate="2011-06-01T08:01:16.477" />
  <row Id="10663" PostTypeId="1" CreationDate="2011-06-01T10:32:05.677" Score="0" ViewCount="87" Body="&lt;p&gt;How many virtual gravitons do the sun and earth exchange in one year?&lt;br&gt;&#xA;What are their wavelengths?&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-06-01T11:15:16.127" Title="Sun-Earth Virtual Gravitons?" Tags="&lt;quantum-mechanics&gt;&lt;gravity&gt;&lt;quantum-gravity&gt;" AnswerCount="1" />
  <row Id="10665" PostTypeId="2" ParentId="10663" CreationDate="2011-06-01T11:15:16.127" Score="4" ViewCount="" Body="&lt;p&gt;Dear Jim,&#xA;Let us suppose that gravitons exist, even though a quantum field theory of gravity per se does not. Still gravitons are bosons. There is no conservation of &quot;boson&quot; number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no way to count them, even so when virtual. This is also true for the repulsion of two electrons, which happens with virtual photons. They are not countable.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-06-01T11:15:16.127" CommentCount="1" />
  <row Id="10666" PostTypeId="1" AcceptedAnswerId="10668" CreationDate="2011-06-01T13:53:03.243" Score="0" ViewCount="92" Body="&lt;p&gt;Here's an example to describe the issue&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supossed a high power laser (eg a 100 kW laser, ie, electromagnetic weapons)&#xA;is fired to a target, then it will receive energy and move.&#xA;(and likely to burn or explode, but that's not the point)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, statting that photons have no mass&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the laser transmitter get a kickback/recoil like in a &quot;normal&quot; gun?&lt;/p&gt;&#xA;" OwnerUserId="1916" LastEditorUserId="1916" LastEditorDisplayName="" LastEditDate="2011-06-01T14:05:05.037" LastActivityDate="2011-06-01T15:28:08.323" Title="Action - Reaction pair, through photons" Tags="&lt;forces&gt;&lt;newtonian-mechanics&gt;&lt;photons&gt;&lt;laws-of-physics&gt;" AnswerCount="3" />
  <row Id="10667" PostTypeId="2" ParentId="10666" CreationDate="2011-06-01T13:55:52.037" Score="2" ViewCount="" Body="&lt;p&gt;Yes, recoil makes part of such a process. You may consider this as a decay of an excited laser into two parts: a photon beam goes in one direction, a deactivated laser goes in the opposite one. This happens during emitting photons. The beam fate is a quite another story.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-06-01T14:13:24.200" LastActivityDate="2011-06-01T14:13:24.200" CommentCount="2" />
  <row Id="10668" PostTypeId="2" ParentId="10666" CreationDate="2011-06-01T14:29:56.567" Score="3" ViewCount="" Body="&lt;p&gt;Dear HDE, The laser beam obviously has energy and momentum so the laser transmitter gets recoiled due to the conservation of energy and momentum. See also:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Poynting_vector&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Poynting_vector&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The property we call mass is expressed by: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m ~=~ \sqrt{\frac{E^2}{c^4} - \frac{p^2}{c^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is zero for photons even though the energy $E$ and momentum $p$ are not zero. The mass m is zero if energy relates to momentum as $E^2=p^2c^2$. This shows you that photons can't be at rest because in that case both $p$ and $E$ are zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards, Hans &lt;/p&gt;&#xA;" OwnerUserId="3197" LastActivityDate="2011-06-01T14:29:56.567" CommentCount="5" />
  <row Id="10669" PostTypeId="2" ParentId="10666" CreationDate="2011-06-01T15:28:08.323" Score="1" ViewCount="" Body="&lt;p&gt;I asked a related question recently, &lt;a href=&quot;http://physics.stackexchange.com/questions/10612/explain-how-or-if-a-box-full-of-photons-would-weigh-more-due-to-massless-photon/10613#10613&quot;&gt;Explain how (or if) a box full of photons would weigh more due to massless photons&lt;/a&gt;.  Both of our questions are concerned with the accounting for physical properties when a photon is in-flight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transmitter is pushed back as the photons are emitted.  A given force $F$ comes from the laser at all times it is on, which can be calculated from the momentum (addressed in other answers).  If the laser is on, pointing in a single direction for a given time $t$, a quantifiable &lt;em&gt;impulse&lt;/em&gt; is imparted to the laser apparatus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mass-Energy Balance&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The challenging part comes when you consider that the laser apparatus (obviously including the energy source) looses mass through this process.  If you take the laser apparatus to be at rest, no energy is imparted to it since power is force times the velocity $P=F v$, and $v=0$.  That means that all mass that leaves the laser apparatus system ($\Delta m$) is accounted for by energy in the laser beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E=\Delta m c^2 = N p c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N$ is the number of photons emitted in $t$ time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Center of Mass (CM) and Net Velocity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider an even more challenging component - the center of mass of the total system.  Consider the laser and connected devices to be at some point A, isolated in empty space.  Then consider that the laser fires directly toward another heavy isolated system in empty space B for time $t$.  Take the distance between them to be $d$ and assume that $d&amp;gt;&amp;gt;c t$.  That assumption is to make the beam like a &quot;batch&quot; single emission.  The A system is moving away from B after it is fired.  Both A and B are at rest before firing, $m_A=m_B$, and the system including both masses, system AB, has a CM at $d/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you include the photon in your definition of system AB after firing but before absorption by B, the system MUST have a net velocity of zero and the CM must not move.  This is an extremely challenging concept, and to the extent of my understanding, you must simply consider the photon's relativistic mass.  it has no rest mass, but the location of the photons must have some weighting in calculation of the system CM for the typical principles to still hold, this is $m=\frac{N h}{\lambda c}$.  The momentum of the photons balance the recoil momentum of system A, and the position of the photon balances the recoil movement of system A.  In this way, the photon is very very similar to if the process were instead a bullet.  The main difference is that it has no rest mass and moves at the constant speed of light.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-06-01T15:28:08.323" CommentCount="1" />
  <row Id="10670" PostTypeId="1" AcceptedAnswerId="11018" CreationDate="2011-06-01T16:19:00.687" Score="15" ViewCount="225" Body="&lt;p&gt;It is common knowledge among the educated that the Earth is not exactly spherical, and some of this comes from tidal forces and inhomogeneities but some of it comes from the rotation of the planet itself.  The deformation from the rotational effect makes it more oblate spheroid-like, or as I would prefer, &quot;like a pancake&quot;.  Here is &lt;a href=&quot;http://www.josleys.com/show_gallery.php?galid=313&quot; rel=&quot;nofollow&quot;&gt;one site illustrating the behavior&lt;/a&gt;, and image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/gV7Ay.jpg&quot; alt=&quot;Earth's shape from rotational distortion from Mathematical Imagery&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Literature exists detailing the mathematical expectations for a rotating planet using just hydrostatic forces, for example, see &lt;a href=&quot;http://md1.csa.com/partners/viewrecord.php?requester=gs&amp;amp;collection=TRD&amp;amp;recid=A8242274AH&amp;amp;q=&amp;amp;uid=&amp;amp;setcookie=yes&quot; rel=&quot;nofollow&quot;&gt;Hydrostatic theory of the earth and its mechanical implications&lt;/a&gt;.  I like to imagine a ball of water in space held together by its own gravity.  I also don't want to deviate from consideration of only hydrostatic (and gravitational) forces because I think it is sufficient for this discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would seem that the solution of the described problem is in terms of a small change in radius as a function of the azimuth angle, or z-coordinate if you take the axis of rotation to be the z-axis.  This is using rotational symmetry.  In other words, Earth's deformation due to rotation does not depend on longitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to ask about the extreme case.  Imagine a planet rotating so fast that it is a very thin pancake.  What will occur in this case?  I am curious:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&lt;li&gt;Will the center hollow out, creating a donut shape?&#xA;&lt;li&gt;Will it break up into a multi-body system?&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It seems to me that it would be logical for the high-rotation case to break up into 2 or more separate bodies.  The reason is that a 2 body system is stable an can host a very large angular momentum.  But would it be an instability that leads to this case?  When would such an instability occur and could a rotating planetary body deform in a different kind of shape from the beginning, such as a dumbbell-like shape, which would transition into a 2-body system more logically than the pancake shape?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/aHZ9N.jpg&quot; alt=&quot;Would THIS happen?&quot;&gt;&#xA;&lt;a href=&quot;http://www.tektites.co.uk/primary-shapes.html&quot; rel=&quot;nofollow&quot;&gt;image link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum up, how would a pancake shape transition into a dumbbell shape?  Or would it?  What are the possibilities for the described system?&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-06-12T11:02:21.287" Title="What nonlinear deformations will a fast rotating planet exhibit?" Tags="&lt;astrophysics&gt;&lt;hydrodynamics&gt;&lt;planets&gt;" AnswerCount="1" CommentCount="13" FavoriteCount="3" />
  <row Id="10671" PostTypeId="2" ParentId="10638" CreationDate="2011-06-01T17:55:29.857" Score="1" ViewCount="" Body="&lt;p&gt;Yes, the Maxwell-Boltzmann partition functions of the two subsystems are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z_1(N_1) ~=~ \frac{1}{N_1!}\left[\frac{V}{h^3} \left(\frac{2\pi m}{\beta}\right)^{\frac{3}{2}}\right] ^{N_1},   $$&#xA;$$Z_2(N_2) ~=~ \frac{1}{N_2!}\left[\frac{V}{h^3} \left(\frac{4\pi m}{\beta}\right)^{\frac{3}{2}} e^{-\beta E_0} \right] ^{N_2},   $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where we ignore the rotational and vibrational modes of the molecule $A_2$. Here $E_0&amp;lt;0$ is the binding energy of the chemical reaction $2A \leftrightarrows A_2$. The full partition function, for a fixed number $N=N_1+2N_2$, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Z(N) ~=~ \sum_{N_2=0}^{[\frac{N}{2}]}Z_1(N-2N_2)Z_2(N_2). $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding grand partition functions for the two subsystems are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z_1(\mu)~:=~\sum_{N_1=0}^{\infty}Z_1(N_1)e^{\beta\mu N_1} ~=~\exp\left[\frac{V}{h^3} \left(\frac{2\pi m}{\beta}\right)^{\frac{3}{2}}e^{\beta\mu } \right] ,   $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z_2(\mu)~:=~\sum_{N_2=0}^{\infty}Z_2(N_2)e^{2\beta\mu N_2} ~=~\exp\left[\frac{V}{h^3} \left(\frac{4\pi m}{\beta}\right)^{\frac{3}{2}}e^{\beta(2\mu-E_0) } \right] ,   $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where we have inserted a factor of $2$ to prepare for the implementation of the constraint $N_1+2N_2=N$ of the chemical reaction $2A \leftrightarrows A_2$. The full grand partition function, that is relevant for the chemical reaction $2A \leftrightarrows A_2$, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z(\mu)~:=~\sum_{N=0}^{\infty}Z(N)e^{\beta\mu N}~=~Z_1(\mu)Z_2(\mu).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average occupation numbers are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left&amp;lt; N_1\right&amp;gt;_{\mu} ~=~ \frac{1}{\beta}\frac{\partial\ln Z_1(\mu)}{\partial\mu} = \ln Z_1(\mu) , $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left&amp;lt; N_2\right&amp;gt;_{\mu} ~=~ \frac{1}{2\beta}\frac{\partial\ln Z_2(\mu)}{\partial\mu} = \ln Z_2(\mu) . $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following fraction is independent of the chemical potential $\mu$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\left&amp;lt; N_1\right&amp;gt;^2_{\mu=0}}{\left&amp;lt; N_2\right&amp;gt;_{\mu=0}}~=~\frac{\left&amp;lt; N_1\right&amp;gt;^2_{\mu}}{\left&amp;lt; N_2\right&amp;gt;_{\mu}} ~=~\frac{V}{h^3} \left(\frac{\pi m}{2\beta}\right)^{\frac{3}{2}}e^{\beta E_0} ~\sim~ e^{\beta E_0}~=~e^{-\beta |E_0|} ,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the &quot;$\sim$&quot; sign means &quot;is proportional to&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-02T11:09:05.863" LastActivityDate="2011-06-02T11:09:05.863" CommentCount="4" />
  <row Id="10672" PostTypeId="2" ParentId="10635" CreationDate="2011-06-01T18:17:16.587" Score="2" ViewCount="" Body="&lt;p&gt;One particularly fun book that I really enjoy is &quot;Road to Reality&quot; by Roger Penrose. However it is most certainly not as self contained as it describes itself, and should be supplemented by other sources (both math and physics) if you want to have a good understanding of the material. But still, I personally love his writing style and he still keeps the material interesting even if you get lost in the math. Also, I find the book very inspiring and motivating as each topic I get stuck on, I delve into with other books and once I have a decent understanding, I am one chapter closer to the front lines of physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic idea of the book is the first half is dedicated to mathematics, starting from the most basic definitions of real and complex numbers, through calculus, complex analysis, group theory, and differential geometry. From here he goes on to the physics, starting with spacetime, Lagrangian mechanics, quantum mechanics, the standard model, and going to QFT, Cosmology, Supersymmetry, string theory, and, of course, Penrose's favorite, twistor theory.&lt;/p&gt;&#xA;" OwnerUserId="3018" LastEditorUserId="3018" LastEditorDisplayName="" LastEditDate="2011-06-01T18:22:27.663" LastActivityDate="2011-06-01T18:22:27.663" CommentCount="2" />
  <row Id="10673" PostTypeId="1" CreationDate="2011-06-01T20:18:04.717" Score="1" ViewCount="51" Body="&lt;p&gt;Consider metal, and its reciprocal lattice representation with Fermi surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the correct way to represent a plasmon in this system? M.b. rotating points on the surface? Or 3d membrane-like oscillation of surface? Or what?&lt;/p&gt;&#xA;" OwnerUserId="3809" LastActivityDate="2011-06-01T20:18:04.717" Title="What does plasmon look like in 3D band structure graph?" Tags="&lt;solid-state-physics&gt;" FavoriteCount="1" />
  <row Id="10674" PostTypeId="1" AcceptedAnswerId="10678" CreationDate="2011-06-01T21:32:38.343" Score="4" ViewCount="146" Body="&lt;p&gt;What is the best way to focus (sun)light using flat mirrors? My goal is to start a fire. Cutting the mirrors is easy.&lt;/p&gt;&#xA;" OwnerUserId="2843" LastActivityDate="2011-06-01T23:16:46.517" Title="Focusing Light with Flat Mirrors" Tags="&lt;mirror&gt;&lt;reflection&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="10675" PostTypeId="2" ParentId="10674" CreationDate="2011-06-01T21:55:19.550" Score="1" ViewCount="" Body="&lt;p&gt;You need a high degree of concentration to start a fire. With flat mirrors the max concentration factor is equal to the number of mirrors. Is that last statement true? Or can complicated geometries with multiple reflections do better? But in any case, without curvature you don't get true focusing and would need a large number of small flat mirrors.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-06-01T21:55:19.550" CommentCount="3" />
  <row Id="10676" PostTypeId="2" ParentId="10674" CreationDate="2011-06-01T22:04:16.353" Score="1" ViewCount="" Body="&lt;p&gt;You will want to make the cross section of your device approximate a parabola.  This should be relatively easy to do with a large device (say, using 3cm wide strips of mirror to build something the size of an outdoor barbeque); but tougher as you get smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the parabola is of the form y=(a)(x^2); then the focal length is 1/(4a).  By choosing your scaling factors wisely, you can control whether the focal length falls inside or beyond the physical device.  The break point between these two regions is picking a=1 and letting x range from -1 to 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, as has been pointed out in another post, the best effective sunlight magnification you get will be 'z', where 'z' is the number of flat mirrors.  If you were to do something like mold aluminun flashing to a parabolic surface, the theoretically achievable magnification is infinity, and the practical magnification could easily be in the hundreds.  &lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-06-01T22:14:49.757" LastActivityDate="2011-06-01T22:14:49.757" />
  <row Id="10677" PostTypeId="2" ParentId="10674" CreationDate="2011-06-01T22:44:15.273" Score="5" ViewCount="" Body="&lt;p&gt;For an example, take a look at this guy who glued 5000 flat mirrors onto a satellite dish to &quot;start a fire&quot;, and some more :) Ironically, it was destroyed in an unrelated fire accident later...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=TtzRAjW6KO0&quot; rel=&quot;nofollow&quot;&gt;Solar Death ray with 5000 flat 1x1 cm mirrors&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The R5800 is my latest and greatest solar creation. Made from an ordinary fiberglass satellite dish, it is covered in about 5800 3/8&quot; (~1cm) mirror tiles. When properly aligned, it can generate a spot the size of a dime with an intensity of 5000 times normal daylight. This intensity of light is more than enough to melt steel, vaporize aluminum, boil concrete, turn dirt into lava, and obliterate any organic material in an instant. It stands at 5'9&quot; and is 42&quot; across.&#xA;  Unfortunately, the R5800 was completely destroyed in a storage shed fire on December 14, 2010.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-06-01T22:44:15.273" />
  <row Id="10678" PostTypeId="2" ParentId="10674" CreationDate="2011-06-01T22:58:54.537" Score="5" ViewCount="" Body="&lt;p&gt;Hi Joe: There are many ways to concentrate solar light with flat mirrors but the easiest way would be to lay them out on a flat plane and only adjust the angles towards the focal point. One could call this a &quot;Fresnel mirror concentrator&quot; like in here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.findingnew.com/wp-content/uploads/2011/03/solar-thermal-concentrates.jpg&quot; rel=&quot;nofollow&quot;&gt;http://www.findingnew.com/wp-content/uploads/2011/03/solar-thermal-concentrates.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an overview to give you some more inspiration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.builditsolar.com/Projects/Concentrating/concentrating.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.builditsolar.com/Projects/Concentrating/concentrating.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards, Hans&lt;/p&gt;&#xA;" OwnerUserId="3197" LastActivityDate="2011-06-01T22:58:54.537" CommentCount="1" />
  <row Id="10679" PostTypeId="1" AcceptedAnswerId="10693" CreationDate="2011-06-01T23:11:57.700" Score="1" ViewCount="173" Body="&lt;p&gt;If there is a steel plate floating in space, for ever, within the solar system, how is it going to degrade? Of course without oxygen it is not going to oxidize (rust), but how are cosmic rays, UV light, alpha particles and the general space environment going to affect the structure of steel on the long term?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Let's imagine a plane of steel on orbit around the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just curious.&lt;/p&gt;&#xA;" OwnerUserId="392" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2011-06-02T22:02:58.497" LastActivityDate="2011-06-03T04:57:31.533" Title="How would steel degrade in space" Tags="&lt;space&gt;&lt;physical-chemistry&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="10680" PostTypeId="2" ParentId="10674" CreationDate="2011-06-01T23:16:46.517" Score="1" ViewCount="" Body="&lt;p&gt;Googling around, I find that wood can ignite when struck by 4.3 kiloWatts per square meter, and that the solar constant is 1.3 kiloWatts per square meter.  This implies 4 confocal solar images would be good enough to start a fire under ideal conditions.  I bet it takes more than that, but maybe not too many more.  Good luck and be careful. &lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-06-01T23:16:46.517" CommentCount="6" />
  <row Id="10681" PostTypeId="2" ParentId="7066" CreationDate="2011-06-01T23:38:54.450" Score="0" ViewCount="" Body="&lt;p&gt;A rag-doll is mostly connected with ball joints, which constrain the position of one point of the rigid body, and allow all three degrees of freedom. There are no easy answers, although the equations needed can be found in college level dynamics book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://synthetica.eng.uci.edu/~ataneja/ugrad/Robotics/related%20articles/Geometry%20and%20Screw%20Theory%20for%20Robotics.pdf&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt; that explains the science of rigid bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then again searching for manikin simulation yields a lot of &lt;a href=&quot;http://www.softoxi.com/manikin-simulation/download.html&quot; rel=&quot;nofollow&quot;&gt;hits&lt;/a&gt;, with some game related &lt;a href=&quot;http://www.nada.kth.se/utbildning/grukth/exjobb/rapportlistor/2004/rapporter04/gastrin_johan_04008.pdf&quot; rel=&quot;nofollow&quot;&gt;ones&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-06-01T23:38:54.450" />
  <row Id="10682" PostTypeId="2" ParentId="6027" CreationDate="2011-06-02T02:23:06.560" Score="0" ViewCount="" Body="&lt;p&gt;As dmckee, pointed out in his comment, these equations come from (or derived from) some basic assumptions. To understand &lt;em&gt;how&lt;/em&gt; we obtain these equations from the basic assumptions, you will need to read &lt;em&gt;derivations&lt;/em&gt; of these equations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Watch this video for &lt;a href=&quot;http://youtu.be/N6NPDoIoJWY&quot; rel=&quot;nofollow&quot;&gt;basic understanding&lt;/a&gt;. &#xA;Then watch Khan Academy lectures on projectile motion. (only 3 lectures).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May the force be with you! &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2011-06-02T02:23:06.560" />
  <row Id="10683" PostTypeId="1" CreationDate="2011-06-02T02:23:39.103" Score="2" ViewCount="77" Body="&lt;p&gt;Shlomo Sternberg (math professor at Harvard) wrote a book called &quot;Group theory and physics&quot;. On p156 (&lt;a href=&quot;http://books.google.com/books?id=k2Fp3JA93oYC&amp;amp;pg=PA156&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;) there's a strange offhand comment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Experiments done in 1964 by Fitch and Cronin seem to indicate that CP is not conserved. I do not fully understand the issues involved in the correct interpretation of this experiment, which clearly shows that CP and CPT are not &lt;em&gt;both&lt;/em&gt; conserved. It follows from the locality axioms, that quantum field theory implies that CPT is a symmetry of nature, and hence that CP is violated in the Fitch-Cronin experiment. But other, group theoretical, hypotheses might favor CP. I have my own views on the subject, which I will not expand on in this book.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot find any other publication where Prof. Sternberg elaborated his views on this. But he obviously suspects that CP is an exact symmetry while CPT is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is: (1) Does the Fitch-Cronin experiment (or any other) disprove CP? Or does it only &quot;show that CP and CPT are not &lt;em&gt;both&lt;/em&gt; conserved&quot;? (2) If the latter, is it remotely plausible that CPT is false? (How universal is the CPT theorem? Does it hold even in string theory?)&lt;/p&gt;&#xA;" OwnerUserId="3811" LastActivityDate="2011-06-02T04:52:08.230" Title="Does kaon decay etc prove &quot;CP violation&quot; or just &quot;CP or CPT violation&quot;" Tags="&lt;quantum-field-theory&gt;&lt;string-theory&gt;&lt;particle-physics&gt;&lt;cp-violation&gt;&lt;cpt-violation&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10684" PostTypeId="2" ParentId="193" CreationDate="2011-06-02T02:41:25.067" Score="1" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;One, pretty mathy, but classic book about Riemannain manifolds is: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0125267401&quot; rel=&quot;nofollow&quot;&gt;Semi-Riemannian Geometry by O'Neill&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Some approachable Lie Algebra notes are available here, they are designed to require little background: &lt;a href=&quot;http://arxiv.org/abs/1104.1106&quot; rel=&quot;nofollow&quot;&gt;Lecture Notes on Lie Algebra&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;My personal favorite book about Algebraic/Differential Topology is : &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521589568&quot; rel=&quot;nofollow&quot;&gt;Calculus to Cohomology&lt;/a&gt;. This book is extremely approachable, requiring only multivariable calculus and linear algebra to completely understand it. I cannot recommend it enough, particularly for physics.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also I third Road to Reality. It is a very fun/interesting book!&lt;/p&gt;&#xA;" OwnerUserId="3018" LastActivityDate="2011-06-02T02:41:25.067" CommunityOwnedDate="2011-06-02T02:41:25.067" />
  <row Id="10686" PostTypeId="2" ParentId="10683" CreationDate="2011-06-02T04:52:08.230" Score="1" ViewCount="" Body="&lt;p&gt;CPT is a &lt;a href=&quot;http://en.wikipedia.org/wiki/CPT_symmetry&quot; rel=&quot;nofollow&quot;&gt;general theorem&lt;/a&gt; of quantum field theories:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Specifically, the CPT theorem states that any Lorentz invariant local quantum field theory with a Hermitian Hamiltonian must have CPT symmetry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Questioning CPT invariance is questioning the foundations of modern physics theory, which is probably the reason you cannot find anything on this. The label &quot;crackpot&quot; is easily stuck on theories  denying Lorenz invariance, for example, because is is  founded on solid experimental evidence. Hermitian Hamiltonians describe the plethora of physics data and are a basis of the quantum mechanical description of nature. Locality is also a sine qua non in mainstream physics theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are now a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cp_Violation&quot; rel=&quot;nofollow&quot;&gt;number of experiments&lt;/a&gt; showing CP violation, {a particle is interchanged with its antiparticle (C symmetry), and left and right is swapped (P symmetry).}, not needing to invoke the CPT theorem, except to point out that time reversal does not hold for these reactions.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-06-02T04:52:08.230" CommentCount="2" />
  <row Id="10687" PostTypeId="2" ParentId="10679" CreationDate="2011-06-02T05:00:13.843" Score="1" ViewCount="" Body="&lt;p&gt;I would probably think that it would depend a lot on where it is and what sort of time frame you are looking at. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a place where it would be more &quot;susceptible&quot; to the forms of radiation you described (near a star) there are probably more important environmental factors. Not only will be be gravitationally attracted to a star or planet, but it will be slowly eroded by micro-meteor impacts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course at the longest of the long time ranges, you will have to worry about proton decay or more complicated effects like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the chemistry of steel in particular, I don't have any idea.&lt;/p&gt;&#xA;" OwnerUserId="3018" LastActivityDate="2011-06-02T05:00:13.843" CommentCount="1" />
  <row Id="10688" PostTypeId="2" ParentId="10615" CreationDate="2011-06-02T06:07:35.960" Score="0" ViewCount="" Body="&lt;p&gt;Here is a way an electrician solves the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $U$ be a voltage at the input of the ladder of the resistors. Let $I_k$ be current in the horizontal arm of k-th section, $I_0$ be current in the horizontal arm of 0-th section (initial section), $I_n$ be current in the horizontal arm of n-th section (last section) &lt;strong&gt;So i am working actually with $n+1$ sections.&lt;/strong&gt; Successively applying to closed circuits Kirchhoff's voltage law, we find:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$RI_0+R(I_0-I_1)=U$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R(I_0-I_1)-RI_1-R(I_1-I_2)=0$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$..............................$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-R(I_{n-1}- I_n)+2RI_n=0$$ After simplification:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2I_0-I_1=\frac{U}{R}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3I_k-I_{k+1}-I_{k-1}=0$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3I_n-I_{n-1}=0$$ Now, the effective resistance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_e=\frac{U}{I_0}$$ Next step is to solve the recurrence relation. Surprisingly, Wolfram Alpha does a good job: &lt;a href=&quot;http://www.wolframalpha.com/input/?i=3%2af%28k%29%3Df%28k%2B1%29%2Bf%28k-1%29%2C+2%2af%280%29%3DU%2Bf%281%29%2C+3%2af%28n%29%3Df%28n-1%29&quot; rel=&quot;nofollow&quot;&gt;Link here&lt;/a&gt;. There is $f(k)$ instead of $I_k$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $R_e=R\frac{(7-3\sqrt 5)(3-\sqrt 5)^n+(3+\sqrt 5)^{n+1}}{(1+\sqrt 5)(3+\sqrt 5)^n-2(\sqrt 5-2)(3-\sqrt 5)^n}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer is consistent with that of Zassounotsukushi&lt;/p&gt;&#xA;" OwnerUserId="552" LastActivityDate="2011-06-02T06:07:35.960" CommentCount="1" />
  <row Id="10689" PostTypeId="2" ParentId="10656" CreationDate="2011-06-02T06:57:29.183" Score="1" ViewCount="" Body="&lt;p&gt;Layman's answer here. First we need to clean up the thought experiment a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In relativity texts it's common for these kind of thought experiments to use observers stationed at the points in question, so that the travel delay of light can be ignored. As part of the setup of the thought experiment Person A would have someone whose clock is synchronized with A's clock stationed at Star 2; let's call that someone A2. When you say &quot;Person A measures 10 years for person B to make it to Star 2&quot;, a relativity text would take this to mean that B notes that A2's clock shows 10 years elapsed when B passes by A2. And when you say &quot;for person B to make it to Star 2 ... person A observes person’s B’s clock to have elapsed a time of 5 years&quot;, a relativity text would take this to mean that A2 notes that B's clock elapsed 5 years when B passes by A2. (A relativity text could assume that both A's and B's clock were reset to 0 when they passed each other, as was A2's clock in A/A2's frame.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we wonder, if A2's clock runs at half-speed as measured in B's frame, how could A2's clock elapse double the time that B's clock did, by when B makes it to A2? Wouldn't A2's clock elapse half the time (2.5 years) instead? The answer is that A2's clock didn't elapse double the time that B's clock did; it indeed elapsed half, 2.5 years. If B had a helper B2 at rest in B's frame and passing by A2 at the moment in B's frame that B passes by A, B2 would note that A2's clock is not at 0 but rather at 7.5 years. The difference between t=0 and t=7.5 is explained by relativity of simultaneity. As recorded by observers at rest with respect to B's frame and recorded simultaneously in that frame, nearby clocks at rest with respect to A's frame (and synchronized in A's frame) show greater elapsed times the further away they are in B's direction of motion.&lt;/p&gt;&#xA;" OwnerUserId="2314" LastActivityDate="2011-06-02T06:57:29.183" />
  <row Id="10690" PostTypeId="1" AcceptedAnswerId="10691" CreationDate="2011-06-02T07:42:57.713" Score="6" ViewCount="316" Body="&lt;p&gt;Are there &lt;strong&gt;any analytical proofs for the 2nd law of thermodynamics&lt;/strong&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it based &lt;strong&gt;entirely on empirical evidence&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1512" LastActivityDate="2011-06-02T08:46:10.817" Title="Is there any proof for the 2nd law of thermodynamics?" Tags="&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="4" />
  <row Id="10691" PostTypeId="2" ParentId="10690" CreationDate="2011-06-02T08:38:29.340" Score="7" ViewCount="" Body="&lt;p&gt;It's simple to &quot;roughly prove&quot; the second law in the context of statistical physics. The evolution $A\to B$ of macrostate $A$, containing $\exp(S_A)$ microstates, to macrostate $B$, containing $\exp(S_B)$ microstates, &lt;a href=&quot;http://motls.blogspot.com/2010/10/arrow-of-time-part-101010.html&quot; rel=&quot;nofollow&quot;&gt;is easily shown&lt;/a&gt; by the formula for the probability &quot;summing over final outcomes, averaging over initial states&quot;, to be $\exp(S_B-S_A)$ higher than the probability of the inverse process (with velocities reversed). Because $S_B-S_A$ is supposed to be macroscopic, such as $10^{26}$ for a kilogram of matter, the probability in the wrong direction is the exponential of minus this large difference and is zero for all practical purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more rigorous versions of this proof are always variations of the 1872 proof of the so-called H-theorem by Ludwig Boltzmann:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/H-theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/H-theorem&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This proof may be adjusted to particular or general physical systems, both classical ones and quantum ones. Please ignore the invasive comments on the Wikipedia about Loschmidt's paradoxes and similar stuff which is based on a misunderstanding. The H-theorem is a proof that the thermodynamic arrow of time - the direction of time in which the entropy increases - is inevitably aligned with the logical arrow of time - the direction in which one is allowed to make assumptions (the past) in order to evolve or predict other phenomena (in the future).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every Universe of our type has to have a globally well-defined logical arrow of time: it has to know that the future is being directly evolving (although probabilistically, but with objectively calculable probabilities) from the past. So any universe has to distinguish the future and the past logically, it has to have a logical arrow of time, which is also imprinted to our asymmetric reasoning about the past and the future. Given these qualitative assumptions that are totally vital for the usage of logic in any setup that works with a time coordinate, the H-theorem shows that a particular quantity can't be decreasing, at least not by macroscopic amounts, for a closed system.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-06-02T08:46:10.817" LastActivityDate="2011-06-02T08:46:10.817" CommentCount="3" />
  <row Id="10693" PostTypeId="2" ParentId="10679" CreationDate="2011-06-02T10:50:08.370" Score="3" ViewCount="" Body="&lt;p&gt;As previously mentioned, micrometeorite pitting will occur over sufficiently long time scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steel is composed of microcrystalline grains, and as such it is likely that cosmic rays will cause the formation of crystallographic defects, leading to embrittlement of at least the periphery of the steel in the long term. Inelastic collision between the steel and ions or molecules floating in space may result in interesting chemistry on the surface of the steel, however this absolutely depends on the particular species the steel impacts with. Note also that space is very empty and as such collisions will happen with a very low frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultraviolet light will cause &lt;a href=&quot;http://en.wikipedia.org/wiki/Photoelectric_effect#Spacecraft&quot; rel=&quot;nofollow&quot;&gt;photoelectronic emission&lt;/a&gt;. As such, the steel will develop a positive charge until the electron work function exceeds the energy of the light that is hitting it. This shouldn't degrade the metal, however very high energy photons (such as gamma rays) can cause crystallographic defects as discussed in &lt;a href=&quot;http://sti.srs.gov/fulltext/WSRC-RP-89-619.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; report by N.P. Baumann.&lt;/p&gt;&#xA;" OwnerUserId="1876" LastEditorUserId="1876" LastEditorDisplayName="" LastEditDate="2011-06-03T04:57:31.533" LastActivityDate="2011-06-03T04:57:31.533" />
  <row Id="10694" PostTypeId="2" ParentId="10679" CreationDate="2011-06-02T11:08:50.700" Score="2" ViewCount="" Body="&lt;p&gt;&quot;Steel&quot; is a very misused word today. &#xA;Exept cast iron any alloy containing some iron is called steel, &#xA;irrespective of composition. On the other hand the composition &#xA;makes the &quot;steel&quot; very different in properties, from brittle &#xA;like glass to soft like &quot;mild&quot; steel (this expression is really silly!). &#xA;Corrosion is much dependent whether You have stainless &quot;steel&quot; &#xA;or a carbon steel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is &lt;strong&gt;known&lt;/strong&gt; about &quot;steel&quot; in space is that meteorites &#xA;containing some iron/nickel alloy circled our sun since billions &#xA;off years before falling down on earth. This means that what &#xA;ever happens out there to that alloy, is does not ruin it. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-06-02T11:08:50.700" CommentCount="4" />
  <row Id="10695" PostTypeId="1" AcceptedAnswerId="10699" CreationDate="2011-06-02T12:28:08.533" Score="3" ViewCount="175" Body="&lt;p&gt;I find functioning of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Laser_rangefinder&quot; rel=&quot;nofollow&quot;&gt;laser rangefinder&lt;/a&gt; confusing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The explanation usually goes like this: &quot;you shine a laser beam onto the object, the laser beam gets reflected and gets back to the device and time required for that is used to calculate the distance&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay. But the object surface can be uneven and not perpendicular to the laser beam so only a tiny fraction of beam energy is reflected back to the device. And there's plenty of other radiation around, sunlight included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does a rangefinder manage to &quot;see&quot; that very weak reflected signal in a reliable manner?&lt;/p&gt;&#xA;" OwnerUserId="3816" LastActivityDate="2011-06-02T13:13:08.503" Title="How do laser rangefinders work when the object surface is not perpendicular to the laser beam?" Tags="&lt;optics&gt;&lt;energy&gt;&lt;light&gt;&lt;laser&gt;" AnswerCount="3" />
  <row Id="10697" PostTypeId="2" ParentId="10695" CreationDate="2011-06-02T12:55:04.890" Score="3" ViewCount="" Body="&lt;p&gt;The amount of laser energy reflected back will be the limiting factor of its effective range. However, since the laser's radiation is of a specific wavelength, it won't be confused by extraneous radiation from ambient sources.&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-06-02T12:55:04.890" />
  <row Id="10698" PostTypeId="2" ParentId="10695" CreationDate="2011-06-02T12:55:15.130" Score="8" ViewCount="" Body="&lt;p&gt;Some laser rangefinding uses a &lt;a href=&quot;http://en.wikipedia.org/wiki/Retroreflector&quot; rel=&quot;nofollow&quot;&gt;retroreflector&lt;/a&gt;, which will bounce the laser light back in the direction it came regardless of orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, lasers operate at a very specific frequency, so the signal/noise ratio only needs to be strong enough to be detectable at that frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you shine a normal laser pointer on a wall, even if the wall is pretty far away, you can see the spot it makes.  That means your eye can pick out the reflected laser light.  The electronics can be made better than your eye, so it's not too hard to see reflected laser light..&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-06-02T12:55:15.130" CommentCount="5" />
  <row Id="10699" PostTypeId="2" ParentId="10695" CreationDate="2011-06-02T12:56:10.107" Score="6" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;so only a tiny fraction of beam energy&#xA;  is reflected back to the device.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This tiny fraction is enough. With respect to ambient light: &#xA;One can modulate the laser beam, and filter the the voltage &#xA;of the receiving photodiode for this modulation frequency and phase. &#xA;Another precaution is to have a light filter in front of the &#xA;receiving photodiode which only lets the wavelength of the laser pass. &#xA;I think both precautions are used. &#xA;And of course the receiving photodiode is focussed to a spot of some &#xA;centimeters diameter around the laser spot. &#xA;Try to point the range finder to a mirror, in that case the &#xA;range finder should fail, exept the mirror happens to reflect &#xA;precisely back to the range finder (which is rather unlikely). &#xA;Reason is that from a (clean) mirror You don't get a stray reflection. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-06-02T13:13:08.503" LastActivityDate="2011-06-02T13:13:08.503" />
  <row Id="10700" PostTypeId="2" ParentId="9830" CreationDate="2011-06-02T14:45:29.717" Score="3" ViewCount="" Body="&lt;p&gt;J.M. Hure addressed parts of the problem in &lt;a href=&quot;http://www.aanda.org/index.php?option=com_article&amp;amp;access=standard&amp;amp;Itemid=129&amp;amp;url=/articles/aa/full/2005/16/aa0194-03/aa0194-03.right.html&quot; rel=&quot;nofollow&quot;&gt;Solutions of the axi-symmetric Poisson equation from elliptic integrals&lt;/a&gt; (2005), giving a form of the electric field in the vicinity of a loop charge.  I will post his published result here with his notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These equations are for a mass distributed in a loop with linear mass $\lambda$, radius $a$, and located elevated to a height of $z$ (as far as I'm concerned $z=0$).  Then the coordinates used are $R$ for distance to vertical axis and $Z$ for the vertical position.  Note that this notation is very different from what I use.  Two intermediary values are introduced which are $k$ and $k&amp;#39;$.  The answer is for the gravitational field using the case of Newtonian gravity (standard use of $G$) and is denoted $&amp;lt; \kappa_R, \kappa_\Phi, \kappa_Z&amp;gt;$.   The equations follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k^2 = \frac{ 4 a R }{ (a+R)^2 + (z-Z)^2 }$$&#xA;$$k&amp;#39;^2 = 1-k^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\kappa_R = \frac{G}{R} \sqrt{\frac{a}{R} } k \left( E(k)-K(k)+\frac{(a-R) k^2 E(k)}{2 a k&amp;#39;^2 } \right)$$&#xA;$$\kappa_\Phi = 0$$&#xA;$$\kappa_Z = \frac{G (z-Z)}{ 2 R \sqrt{a R} } \frac{k^3 E(k) }{k&amp;#39;^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, this is what I had.  I gave the potential in my question because it was the most simple expression that could convey the mathematics for a loop geometry.  Hure also noted a paper by Durand in 1964 in regarding these equations, &lt;em&gt;Electrostatique. I. Les distributions&lt;/em&gt;, which may have the first published form for loop geometry.  Hure also discusses self-gravitation of the loop geometry and introduces the concept of a &lt;strong&gt;loop singularity&lt;/strong&gt;.  I think he means to say that a loop singularity implies infinite self-gravitation but I think his wording does not explicitly say that.  Nonetheless, he avoids the problem by writing an integral for self gravitation using a mass density, $\rho$, and not linear mass density, $\lambda$, (similar to my $F_{effective}$ below) indicating that the issue of infinite self-gravitation was well understood.  Hure's paper then goes on to offer empirical methods for the self-gravitation as well as some other problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before I move on to my own solution, I should note one difference about the gravitational problem.  This question is about the electric wire force, in which case the charge is a surface charge.  In the case of gravitation, the entire volume contains a mass density, meaning that ultimately I address a surface integral here but the gravitational problem requires a volume integral.  The volume integral is not easy with the $r&amp;lt;&amp;lt;R$ method I apply here.  I've tried.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will write my own equation for the field here which is mathematically equivalent to the above one.  I will make one new definition, which is the closest distance to the loop from a point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r = \sqrt{ (\rho-R)^2 + z^2 }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&amp;lt; F_\rho, F_\theta, F_z&amp;gt; = \nabla E(\rho,z)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_\rho = \frac{2 k R \lambda }{ l \rho } \left( K \left( 2 \sqrt{ \frac{ x R }{l} } \right) + \frac{ \rho^2-R^2-z^2}{r^2 x} E\left( 2 \sqrt{ \frac{ x R }{l} } \right) \right)$$&#xA;$$F_\theta = 0$$&#xA;$$F_z = \frac{ 4 k R \lambda z }{r^2 l} E\left( 2 \sqrt{ \frac{ x R }{l} } \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting from here I will directly address the question I asked.  Recall the equation for the field around an infinite line charge above.  The potential then follows a $ln(r)$ form, but keep in mind that the potential is a relative measure, but also, since the potential around a ring charge given above was found by integrating the $1/r$ differential potentials, it will have the same offset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E(r) = - 2 \lambda k ln(r) = 2 k \lambda ln \left( \frac{C}{r} \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objectively, we will find a simplified approximation of the previously given $E(\rho,z)$ as $r$ goes to zero and a part of it will be identified as being equivalent to $E(r)$ above, along with another, yet unknown, component.  I did just that, and it was no easy task.  Here is what I obtained.  The two parts are identified and written separately to make the result crystal clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E(\rho,z) \approx 2 k \lambda \left( ln( \frac{8 R}{r} ) + \frac{\rho-R}{R} \left( 1-ln( \frac{8 R}{r} ) \right) \right)$$&#xA;$$E_{line}(\rho,z) = 2 k \lambda \left( ln( \frac{8 R}{r} ) \right)$$&#xA;$$E_{loop}(\rho,z) = 2 k \lambda \left( \frac{\rho-R}{R} \left( 1-ln( \frac{8 R}{r} ) \right) \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, this is an approximation of the previous equation for $E(\rho,z)$ as $r$ goes to zero.  I found it to be a good approximation for $r/R&amp;lt;0.1$ or so, and the approximation becomes more correct as $r$ decreases further.  It is almost exact for wire radius of $cm$ or $mm$ with a loop radius of $m$.  Next, the electric field due to the global loop geometry, $E_{loop}$, is used to answer the problem.  The gradient of this quantity gives the field, $\vec{F}_{loop}$, which can then be used to obtain the force.  The problem is that this field acts very non-uniformly at a certain $r$.  Note that $E_{line}$ acts uniformly and both the potential and field for the infinite line approximation will dominate in magnitude compared to the global loop effects, but results in no net force on the wire while $E_{loop}$ does, this means that a uniform surface charge distribution is a valid approximation.  The average value of $\vec{F}_{loop}$ at some $r$ is obtained by integrating, and the vector nature is dropped while it is noted that the field is directed outward, or in the positive $\rho$ direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho(r,\psi) = R+r \cos{\psi}$$&#xA;$$z(r,\psi) = r \sin{\psi}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{effective} = \frac{1}{2 \pi} \int_0^{2 \pi} \left( \frac{d}{d\rho} E_{loop}(\rho,z) |_{(\rho,z)=\rho(r,\psi),z(r,\psi)} \right) d\psi = \frac{\lambda k}{R} ( ln( \frac{8 R}{r} ) - \frac{3}{2} ) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This point is very close to the final answer.  The math for finding the tension in the wire is returned to and an expression for written for it in terms of the desired quantities.  I also answer the problem for the previously stated values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T = \lambda R F_{effective} = \lambda^2 k ( ln( \frac{8 R}{r} ) - \frac{3}{2} ) = 1.7 GN$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the other part of the question was about what happens as the wire thickness goes to zero.  For this case (constant charge) the force goes to infinity.  Well well all knew that (otherwise why would I have asked the question).  But what about the case of constant voltage and constant surface charge density?  The previous is written using $V$ for the wire voltage and $\sigma$ for the surface charge density.  Note that the voltage of the wire of radius $r$ can validly be found by the infinite line charge approximation as per my prior arguments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T = \frac{V^2}{16 k} \frac{ ln( \frac{8 R}{r} ) - \frac{3}{2} }{ln( \frac{8 R}{r})^2} = 4 \sigma^2 \pi^2 r^2 k ( ln( \frac{8 R}{r} ) - \frac{3}{2} )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the limits can be addressed.  Here is what happens to the force in the wire as the radius goes to infinity with the quantities kept constant.  For completeness, I address the same things in terms of tensile stress.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&lt;li&gt;Constant charge --&gt; infinity, infinity&#xA;&lt;li&gt;Constant voltage --&gt; zero, infinity&#xA;&lt;li&gt;Constant surface charge density --&gt; zero, infinity&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is an interesting result.  I should note that due to the tensile stress result this means that no curved wire with an extraordinarily small radius, a curve, and a charge can exist without tearing itself apart.  I welcome disagreement on that point.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-06-02T14:57:26.040" LastActivityDate="2011-06-02T14:57:26.040" />
  <row Id="10701" PostTypeId="1" CreationDate="2011-06-02T14:47:14.773" Score="2" ViewCount="126" Body="&lt;p&gt;I'm a software developer, and I need to calculate the estimated amount of force expended typing stored text. Preferrably in some interesting way. (i.e. the force exerted on keys thus far is enough to push a car 5 miles) (or:  equivalent to 100 kg of TNT)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We're not going to worry about deletes or moving the cursor or anything, I'm just counting characters of stored text.&lt;/li&gt;&#xA;&lt;li&gt;I don't really care if the space requires more force or not, this is more of a &quot;fun fact&quot; than anything.&lt;/li&gt;&#xA;&lt;li&gt;From what I've found online, &lt;strong&gt;the mean force required for a keystroke is about 12.9N&lt;/strong&gt; &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2665300/&quot; rel=&quot;nofollow&quot;&gt;(source)&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Hundreds of millions of characters have been typed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Questions&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is a good way to make this something people can relate to?&lt;/li&gt;&#xA;&lt;li&gt;How can I calculate it?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thank you all in advance for your valuable time and input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I thought my original post would make this pretty clear: I realize I only have the &lt;em&gt;force&lt;/em&gt; required to push each keystroke. I'm looking for a way to demonstrate that force &lt;em&gt;applied&lt;/em&gt; to something to help people quantify it, hypothetical energy in terms people can understand.&lt;/p&gt;&#xA;" OwnerUserId="3819" LastEditorUserId="3819" LastEditorDisplayName="" LastEditDate="2011-06-02T20:18:55.490" LastActivityDate="2011-06-02T20:18:55.490" Title="How much force is in a keystroke? (estimated, of course)?" Tags="&lt;soft-question&gt;&lt;forces&gt;&lt;spring&gt;&lt;fun-stuff&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="10702" PostTypeId="2" ParentId="10701" CreationDate="2011-06-02T15:20:17.087" Score="4" ViewCount="" Body="&lt;p&gt;@HDE's comment's experimental approach done. Answer between about 1 and 2 ounces = 1/4 to 1/2 Newtons (&lt;a href=&quot;http://www.wolframalpha.com/input/?i=convert%202%20ounces%20to%20newtons&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=convert+2+ounces+to+newtons&lt;/a&gt;). Key travel 3mm, say, = 1.5 milliJoules = $3.59&amp;#215;10^{-7}$ dietary Calories (&lt;a href=&quot;http://www.wolframalpha.com/input/?i=convert%200.5%20newtons%20times%203%20millimeters%20to%20Calories&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/input/?i=convert+0.5+newtons+times+3+millimeters+to+Calories&lt;/a&gt;).&#xA;So, for every 3,000,000 keystrokes, say, that's one Calorie. Not a large proportion of the daily Calorie consumption of the person who's pressing the keys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this shows, Wolfram alpha is your good friend for this kind of Question.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-06-02T15:20:17.087" CommentCount="2" />
  <row Id="10703" PostTypeId="2" ParentId="10701" CreationDate="2011-06-02T15:23:54.783" Score="1" ViewCount="" Body="&lt;p&gt;You seem to be more interested in energy than force.  It is not possible to say &quot;the force exerted on the keys thus far is enough to push a car five miles, or is equivalent to 100 kg of TNT&quot; because pushing a car over a certain distance and exploding a certain amount of TNT aren't examples of a set force.  The TNT has a certain amount of energy.  Pushing the car also requires energy, but the amount of energy required depends on the car and on how fast you push it.  There is a certain minimum force to push the car at a given speed, but once you apply that force, you can push the car one foot or one thousand miles, all while applying the same force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force also doesn't add in the way your question implies.  If I exert one Newton of force on something for one hour, that is not 3600 times more force than if I exert one Newton of force for one second.  Instead, the momentum that I impart to the object will be 3600 times as much.  The force is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if I exert five Newtons of force on a key, release, and exert five Newtons again, that is not ten total Newtons.  It's just five, then zero, then five Newtons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to calculate energy by the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = Fd$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E$ is the energy expended.  $F$ is the force.  $d$ is the distance over which the object travels while the force is exerted.  So, if a keystroke usually involves 10 Newtons and the depth of the keystroke is 1cm, then you are exerting $10 N * 1cm = 0.1 J$ per keystroke.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the 10 N (or 12.9N) figure is almost certainly wrong.  That's the weight of a heavy textbook, not a computer keystroke.  &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/15677060&quot; rel=&quot;nofollow&quot;&gt;This ergonomics paper&lt;/a&gt; discusses the force of keystrokes, but I don't have access to it and no force measurements are cited in the abstract.  &lt;a href=&quot;http://www.tactuskeyboard.com/keymech.htm&quot; rel=&quot;nofollow&quot;&gt;This page&lt;/a&gt; cites forces in the 0.25 - 1.5 N range.  The total energy per keystroke might be about $0.5N * 0.5 cm = 2.5*10^{-3}J$.  That means that in typing this response, I expended around 5J of energy on the keyboard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typing out the complete works of Shakespeare would take about as one Calorie (one one-hundredth the energy you get from a banana).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As dmckee pointed out in a comment, this is the work done on the keyboard, so it's a minimum amount of energy expended assuming the springs in the keyboard don't return any energy to you.  The actual human user is less efficient.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-06-02T15:28:57.687" LastActivityDate="2011-06-02T15:28:57.687" CommentCount="2" />
  <row Id="10704" PostTypeId="2" ParentId="10679" CreationDate="2011-06-02T15:26:07.687" Score="1" ViewCount="" Body="&lt;p&gt;I will address this problem from a very &quot;classical physics&quot; and engineering perspective.  Physicists may offer some failure mechanism for the steel as the universe itself evolves significantly, but as far as I'm concerned &lt;em&gt;it will never degrade&lt;/em&gt;, although &quot;degrade&quot; is a very subjective term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you're looking observe in the sample is long-term &lt;em&gt;radiation embrittlement&lt;/em&gt; as well as &lt;em&gt;swell&lt;/em&gt; and some other effects.  The specifics will depend on the environment, which differs greatly from one point in space to another.  Are we talking about interstellar space?  Intergalactic space?  Wherever it is, it will experience some radiation damage, and expect for a small volume of exceptions we are talking about a service life entailing:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&lt;li&gt;Low temperature&#xA;&lt;li&gt;Radiation&#xA;&lt;li&gt;Long life&#xA;&lt;li&gt;Basically no surface chemistry&#xA;&lt;li&gt;Load bearing or not unspecified&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first thing to understand is that critical metrics for radiation damage come from the time frame times the radiation energy deposition rate.  From there, displacements per atom (dpa) is common metric representing the number of time every atom in the material is displaced from its lattice site on average.  Values greater than 1 are common for discussions on this subject, but starting at around that magnitude effects like embrittlement and swell start to become very major.  The steel will literally swell and becomes very brittle.  If it is a load bearing member, the integrity of the structure it is a part of needs to be carefully considered for the long service life.  However, effects like Stress Corrosion Cracking (SCC) will not be a major problem like they are in things like nuclear reactors because that entails a tipple attack of 1) high temperature 2) chemistry and 3) radiation.  For the steel in space we have the radiation, low temperature, and no chemistry attack from the environment.  Over a long time, self-annealing is also major, but that has an interplay with the temperature, and low temperature may slow that effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally though, the steel will maintain its structure.  You need to ask someone familiar with space travel to quantitatively talk about the radiation environment, but since humans survive in LEO, the radiation in that kind of environment is probably not going to change the material structure of the steel for an extremely long time, at which point it's relatively accurate to say that it doesn't degrade.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-06-02T15:26:07.687" CommentCount="1" />
  <row Id="10706" PostTypeId="2" ParentId="10661" CreationDate="2011-06-02T17:17:00.457" Score="-1" ViewCount="" Body="&lt;p&gt;$q_{x} = n \frac{\pi}{a}$ is only correct solution here for a free electron model afaik, but depending of the actual form of your potential and the direction you view the crystal, you can and will get different positions of the gap or not at all one. Also bands can overlap if they get energetically wide enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm no expert in this, as calculation of electronic band structures is a highly complex topic with lots of different electron models and approximation methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my phenomenological knowledge you distinguish into halfmetals, semiconductors, isolators, metals. Metals like Cu or Al dont have at all energy gaps when you view their band structure scheme. So i highly doubt there are general properties of positions one could derive. The term &quot;bandgap&quot; mostly relates to materials showing a continuous gap like in Germanium independent of directional view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isnt really deeply coverd in most exp. orientated standard solid states books IMO, but search for some theoretical physics solid state scripts on the net if you are deeply intersted. Enough available IMO. Maybe there are some general properties derived by lattice symmetries and potential forms.&lt;/p&gt;&#xA;" OwnerUserId="3715" LastEditorUserId="3970" LastEditorDisplayName="" LastEditDate="2011-07-04T06:03:02.543" LastActivityDate="2011-07-04T06:03:02.543" CommentCount="1" />
  <row Id="10707" PostTypeId="1" AcceptedAnswerId="10708" CreationDate="2011-06-02T17:24:41.940" Score="4" ViewCount="100" Body="&lt;p&gt;Well this is pretty noobish question and I am not sure how to ask.&lt;br /&gt;&#xA;When We talk we don't talk in an uniform frequency. Then how can one measure frequency of ones sound/voice ? &lt;br /&gt;&#xA;I am asking this cause several times I've heard people to say this is his/her frequency of voice. Is this a vague term ? or it has any real meaning ?&lt;/p&gt;&#xA;" OwnerUserId="3822" LastActivityDate="2011-06-02T17:54:11.520" Title="How can a Human voice or animal voice have unique frequency" Tags="&lt;sound&gt;&lt;frequency&gt;" AnswerCount="1" />
  <row Id="10708" PostTypeId="2" ParentId="10707" CreationDate="2011-06-02T17:54:11.520" Score="6" ViewCount="" Body="&lt;p&gt;Your voice, like any sound, is a combination of many frequencies.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physically, your voice consists of pressure waves.  If we plot the pressure as a function of time, we see that it goes up and down in a way that looks somewhat random.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can measure these pressure waves with a microphone, then visualize them with an oscilloscope.  &lt;a href=&quot;http://www.youtube.com/watch?v=HOHSr1YMlNQ&quot; rel=&quot;nofollow&quot;&gt;Here's&lt;/a&gt; a Youtube video where they do this, starting 4:50 into the video.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be able to do this at home using the microphone on your computer and some software like &lt;a href=&quot;http://audacity.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;Audacity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data collected by your microphone is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Time_series&quot; rel=&quot;nofollow&quot;&gt;time series&lt;/a&gt;.  The pressure is a function of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you sang a pure note (or a reasonable approximation thereof), like you hear from an electronic tuner, the pressure would just be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Sine_wave&quot; rel=&quot;nofollow&quot;&gt;sine wave&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could imagine a more complicated sound that was two sine waves on top of each other.  This could produce &lt;a href=&quot;http://en.wikipedia.org/wiki/Beat_%28acoustics%29&quot; rel=&quot;nofollow&quot;&gt;beats&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you add more and more frequencies, more and more complicated sounds become possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a remarkable result that in fact any sound can be represented as a sum of infinitely-many sines waves of different periods added up on top of each other.  This is &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_series&quot; rel=&quot;nofollow&quot;&gt;Fourier's Theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A human voice thus consists of many sine waves combined simultaneously.  Presumably, each individual voice has some special patterns to the way these frequencies are combined, assisting us in recognizing voices.  However, speaker recognition is probably based on other information as well.  I don't know too much about it, but you can check out &lt;a href=&quot;http://en.wikipedia.org/wiki/Speaker_recognition&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We frequently try to isolate the different frequencies in a sound.  This is done electronically through &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronic_filter&quot; rel=&quot;nofollow&quot;&gt;electronic filters&lt;/a&gt;.  A crude example is &quot;turning up the bass&quot; - amplifying the low-frequency components of a sound.  Of course, a professional music studio has far more sophisticated control of the various frequencies.  This control can also be mimicked digitally through &lt;a href=&quot;http://en.wikipedia.org/wiki/Music_sequencer&quot; rel=&quot;nofollow&quot;&gt;music sequencers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a cruder level, you could simply talk directly into an open piano.  The string in the piano will be excited by your voice.  The strings each have a specific frequency, so the strings that are excited the most tell you that their particular frequency is present the most in your voice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your ear accomplishes a similar task.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cochlea&quot; rel=&quot;nofollow&quot;&gt;cochlea&lt;/a&gt; has many small hairs, similar to piano strings, which are tuned to different frequencies.  When they vibrate, they mechanically trigger an ion channel to open, beginning an action potential that is eventually interpreted as sound by your brain.  So, in essence, you are distinguishing the various frequencies in people's voices already.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-06-02T17:54:11.520" />
  <row Id="10709" PostTypeId="1" AcceptedAnswerId="10713" CreationDate="2011-06-02T18:23:19.390" Score="3" ViewCount="105" Body="&lt;p&gt;Is there a fundamental way to look at the universal constants ? can their orders of magnitude be explained from a general points of view like stability, causality, information theory, uncertainty? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for example, what sets the relative magnitudes of Planck's constant compared to say charge of electron or is it just a matter of choice of units. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the physics become more or less cumbersome, insightful if we set all fundamental constants to 1 in appropriate units.&lt;/p&gt;&#xA;" OwnerUserId="3714" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-02T19:28:51.527" LastActivityDate="2011-06-02T20:22:09.480" Title="Deriving or justifying fundamental constants" Tags="&lt;physical-constants&gt;&lt;dimensional-analysis&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="10710" PostTypeId="1" AcceptedAnswerId="10711" CreationDate="2011-06-02T18:32:08.020" Score="2" ViewCount="156" Body="&lt;p&gt;I can't seem to figure out the relationship between $E_k$ and $p$ or $F$.&#xA;I understand that the units are pretty different. But for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bullet with a mass of &lt;em&gt;10.0g&lt;/em&gt; is moving at the speed of &lt;em&gt;1000m/s&lt;/em&gt;. A bull with a mass of &lt;em&gt;400kg&lt;/em&gt; is charging at the speed of &lt;em&gt;5.00m/s&lt;/em&gt; Which has the greater kinetic energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;5000J&lt;/em&gt; for the bullet and &lt;em&gt;5000J&lt;/em&gt; for the bull&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That worked fine but now if you calculate momentum you get:&lt;br /&gt;&#xA;&lt;em&gt;$p = 10 kg\cdot m/s$&lt;/em&gt; for the bullet and &lt;em&gt;$p = 2000 kg\cdot m/s$&lt;/em&gt; for the bull.&lt;br /&gt;&#xA;Which does not make sense!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the bull hit harder than the bullet or the same amount as the bullet?&lt;/p&gt;&#xA;" OwnerUserId="3821" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-02T18:39:07.243" LastActivityDate="2011-06-02T19:28:48.680" Title="What is the relationship between kinetic energy and momentum?" Tags="&lt;homework&gt;&lt;energy&gt;&lt;energy-conservation&gt;&lt;kinematics&gt;&lt;momentum&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="10711" PostTypeId="2" ParentId="10710" CreationDate="2011-06-02T18:50:12.940" Score="8" ViewCount="" Body="&lt;p&gt;If you write down the formulas for kinetic energy,&#xA;$$E_k = \frac{1}{2} m v^2$$&#xA;and momentum&#xA;$$p = mv$$&#xA;you see that you can write the energy in terms of momentum via&#xA;$$E_k = \frac{p^2}{2m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if two objects have the same energy $E_k$, they &lt;em&gt;only&lt;/em&gt; have the same momentum if they also have the same mass. Since the bull has a much larger mass than the bullet, it must therefore have a much larger momentum than the bullet to arrive at the same kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force is change of momentum with time, $F = \dot p$. If we assume that the bullet and the bull hit a target and come to rest in the same time, the bull hits much harder, as it has the higher momentum.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-06-02T18:59:50.237" LastActivityDate="2011-06-02T18:59:50.237" CommentCount="3" />
  <row Id="10712" PostTypeId="2" ParentId="10710" CreationDate="2011-06-02T19:28:48.680" Score="1" ViewCount="" Body="&lt;p&gt;If we apply a given force $F$ for a given time $t$, it changes the momentum by $Ft$. If in that time the force has moved whatever it's pushing by a distance $s$, it changes the energy by $Fs$. If we push a light thing and a heavy thing equally hard for the same amount of time --so they both have the same momentum-- the light thing will be pushed &lt;em&gt;farther&lt;/em&gt; in the same time because it accelerates more, so it has more energy. Equally, if we push them both equally hard until they have the same energy --that is, for the same distance, remembering that in this kind of Physics the bull puts up no resistance other than its mass (and it may even be a Bull Sphere)-- the light thing will have less momentum, because it will take less time to get there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to stopping the bull, you have to keep up a force that's trying to stop it for much more time, the bull &lt;em&gt;keeps on coming&lt;/em&gt;, whereas the same force will stop the bullet in the same distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a fourth concept that also has to be considered, which is &lt;em&gt;pressure&lt;/em&gt;. Sorry! Force, Energy, Momentum, and Pressure are all different. To apply a force to the small area that a bullet has is different from applying the same force to the big area that the bull has. It's more difficult, it needs more pressure, to apply the same force to a smaller area.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-06-02T19:28:48.680" />
  <row Id="10713" PostTypeId="2" ParentId="10709" CreationDate="2011-06-02T20:22:09.480" Score="6" ViewCount="" Body="&lt;p&gt;Your last couple of comments, about units, are incredibly important. It only makes sense to compare two things if they have compatible units. So, to use your example, it doesn't really make sense to talk about the size of Planck's constant relative to the electron charge, but it does make sense to talk about the mass of the muon relative to the mass of the electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of fundamental &lt;a href=&quot;http://math.ucr.edu/home/baez/constants.html&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;dimensionless&lt;/em&gt; constants&lt;/a&gt; in physics. Some are ratios of similar things, such as $m_\mu/m_e$. Others are not so obviously expressed in that form, such as the &quot;fine-structure constant&quot; $\alpha=e^2/(4\pi\epsilon_0\hbar c)$. It makes sense to wonder why the dimensionless constants have the values they do -- one could imagine that a future theory would explain these values. (As Georg says, there is no such theory right now -- by definition the &quot;fundamental&quot; constants are those that don't have a deeper explanation.) On the other hand, one would not expect any fundamental theory to explain why a dimensionful constant, such as the speed of light, had the value that it did: the fact that $c=2.9979\times 10^8$ m/s is an artifact of our definitions of the meter and second. (Actually &lt;a href=&quot;http://physics.nist.gov/cuu/Units/meter.html&quot; rel=&quot;nofollow&quot;&gt;just the meter&lt;/a&gt; in that particular case.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your last question, most people who work in fundamental physics do choose their units so as to set as many fundamental constants equal to 1 as possible. This makes the equations look cleaner, and also makes it easier to see what the interesting dimensionless constants are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes people try to &quot;explain&quot; the values of various physical constants via &lt;a href=&quot;http://en.wikipedia.org/wiki/Anthropic_principle&quot; rel=&quot;nofollow&quot;&gt;anthropic reasoning&lt;/a&gt;, pointing out that if the values were quite different from what they are the Universe would be inhospitable to life. Other people find those arguments to be logically suspect or even morally reprehensible.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-06-02T20:22:09.480" />
  <row Id="10714" PostTypeId="1" CreationDate="2011-06-02T21:17:25.827" Score="3" ViewCount="141" Body="&lt;p&gt;Suppose we have a wheel moving on an horizontal surface, with constant velocity $v$, without sliding. This latter condition implies that the wheel rotates around its centre with angular speed $\omega = v / r$, $r$ being the radius. The rotation of the wheel is caused by the torque of the (static) friction force in the contact point between the wheel and the surface. Because static friction only exists when an opposing force is applied on the object, this means that in order to maintain a constant velocity, an external force must exist. My question is, what happens when this force is no longer applied? I imagine that the wheel will slow down until it stops, but from the equations of kinetic energy, I do not see how to deduce this behaviour. The potential energy is constant because we move horizontally, and the kinetic energy is: $E_k = E_{k_r} + E_{k_t} = \frac{1}{2}I\omega^2 + \frac{1}{2}mv^2$ which equals the total mechanical energy. If the external force stops, then both $v$ and $\omega$ decrease, but why? After all, the mechanical work of the friction force is zero, so why does the wheel slow down?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="3824" LastActivityDate="2011-06-02T21:54:57.713" Title="Wheel moving without sliding" Tags="&lt;newtonian-mechanics&gt;&lt;friction&gt;" AnswerCount="2" />
  <row Id="10715" PostTypeId="2" ParentId="10714" CreationDate="2011-06-02T21:44:57.883" Score="3" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The rotation of the wheel is caused by the torque of the (static) friction force in the contact point &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's just not true.  Newtons's laws of motion dictate that a body in motion continues to stay in motion in the absence of forces acting on it.  Take away both the ground and gravity in your mental picture of the wheel rolling.  It continues to both move forward and spin, as angular momentum is conserved by the same law that conserves momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be some confusions as to where the friction of a rolling wheel comes from.  There is certainly friction with the ground but it's hard to pin it down as static of kinetic.  In the case of a car, for instance, the deformation of the elastic tire dissipates energy and creates friction.  I would hesitate to classify that as either static or kinetic friction as typically defined in physics classes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll offer a specific case.  Consider a wheel rolling with some fiction with the ground that is balanced by a force that acts at the center of the wheel in the direction of motion.  Friction decreases both the linear momentum and rotational momentum.  It has to, the observation of a wheel rolling to a stop wouldn't make any sense otherwise.  The exact direction of the force is something I had to struggle with myself a little bit.  The normal force is generally below the axle of the wheel, but when being slowed down by friction the force from the ground acts slightly in front of the axle.  This causes a small amount of torque opposite the direction of the spinning, which is expected since it must be slowing down the spinning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a wheel with many spokes of springs around it.  When pushed in they don't spring back to their original position right away - this is its means of dissipating energy.  That means that the springs are longest right before they make contact with the ground, which places the normal force slightly in front of directly below the wheel axle.  As they leave the ground they are not quite as long and do not exactly balance the front force.  This is how wheel friction contributes to retarding both the forward motion and the forward spin.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-06-02T21:44:57.883" CommentCount="1" />
  <row Id="10716" PostTypeId="2" ParentId="10714" CreationDate="2011-06-02T21:54:57.713" Score="6" ViewCount="" Body="&lt;p&gt;The problem with your argument lies with this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The rotation of the wheel is caused by the torque of the (static) friction force in the contact point between the wheel and the surface.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That isn't quite correct. When the wheel is spinning up, then yes, static friction is what accelerates it, but once it is rotating, no torque is required to keep the rotation going. So once the wheel is moving at some constant velocity, there is no static friction, and in fact no external force at all, thus there are no changes in energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, every wheel has some amount of dissipative force acting on it, usually due to deformation. The collective effect of the dissipative forces acting on a rolling wheel is sometimes called &quot;rolling friction,&quot; and it acts on the wheel in much the same way as normal friction: it removes energy. Given the constraint that $v = \omega r$, the energy of a wheel is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = \frac{1}{2}(I + mr^2)\omega^2 = \frac{1}{2}\biggl(\frac{I}{r^2} + m\biggr)v^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so when energy is removed, both the angular and linear speeds will decrease. This is what causes the wheel to slow down and eventually stop. Note that in this case, there will be some nonzero amount of static friction as required to prevent slipping.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-02T21:54:57.713" CommentCount="3" />
  <row Id="10717" PostTypeId="1" CreationDate="2011-06-02T22:28:54.500" Score="2" ViewCount="89" Body="&lt;p&gt;I'm trying to numerically solve the differential equations of motion in a system with multiple springs of very high frequency. Because the solution is often a combination of rapidly-oscillating sine and cosine waves, it takes a very long time for a numerical solver to find the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the fourier transform of the solution is probably fairly simple - it's nearly a delta function peaked at the frequency (or frequencies) at which the system oscillates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Fourier transforms turn $\frac{df}{dx}$ into $2 \pi i \xi \hat{f}(\xi)$, it would seem that we could turn all of Newton's equations of the form $F_k=m\frac{d^2}{dt^2} x_k$ (representing, say, the $k$th particle or rigid body in a complicated system) into $\hat{F_k} = m (2 \pi i t)^2 \hat{x_k}$ (where the Fourier transform is taken with respect to time). Then, we could solve all of these simpler equations for the values of the $\hat{x_k}$, then take the Fourier transform (numerically).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, do people ever apply Fourier transforms to the case of many-body systems with some harmonic oscillators?&lt;/p&gt;&#xA;" OwnerUserId="2591" LastActivityDate="2011-06-02T22:39:44.903" Title="Using Fourier Transforms to Solve Systems with springs of high frequency" Tags="&lt;harmonic-oscillator&gt;&lt;fourier-transform&gt;&lt;spring&gt;&lt;numerics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="10718" PostTypeId="1" CreationDate="2011-06-02T22:31:44.027" Score="7" ViewCount="188" Body="&lt;p&gt;I'm studying finite temperature many-body perturbation theory, and am trying to understand The Dyson equation. In particular, I'm using Mattuck - A guide to Feynman diagrams in the many body problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There appear to be two types of Dyson equation: one which is valid in the special case of no external potential and with diagrams calculated in $(k,w)$-space, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;G(k,w) = \frac{1}{G_0 (k,w)^{-1} - \Sigma (k,w)}.&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And one valid at other times which is an integral equation for $G$ which can be different depending on what space you're in and whether you have an external potential or not, but is something like &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;G(k,\tau-\tau') = G_0(k,\tau-\tau') + \int_0^\beta d\tau_1 d\tau_2 G_0(k,\tau- \tau_2) \Sigma(k,\tau_2-\tau_1) G(k,\tau_1-\tau').&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What is an external potential? I can't find an actual definition in the book. At first I assumed it meant the part of the system you want to treat perturbatively, I.e. the Hamiltonian is $H=H_0+H_1$ where $H_1$ is small and can there for be considered a perturbation. But obviously this can't be the case otherwise there's no expansion to use in a Dyson equation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why is the second Dyson equation useful? - How does it allow partial summations? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Every book I've looked at seems to give slightly different rules for calculating the Feynman diagrams. Any thoughts on which is correct? (for example, Negele - Quantum many-particle systems is different to Mattuck).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="873" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-04T05:47:25.630" LastActivityDate="2011-06-04T05:47:25.630" Title="Questions about the Dyson equation" Tags="&lt;quantum-field-theory&gt;&lt;path-integral&gt;&lt;many-body&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="10719" PostTypeId="2" ParentId="10717" CreationDate="2011-06-02T22:39:44.903" Score="2" ViewCount="" Body="&lt;p&gt;Maybe they do; but Fourier transforms have an inherent flaw which makes them less than useful for such cases.  That flaw is that they are transforms of steady state conditions: The initial and final conditions of the system are assumed to be the same; and transients are not considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transform for which you seek is the Laplace transform.  Laplace is a very close cousin to Fourier, but takes into account initial conditions and allows you to inspect transients and the final state of the system.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-06-02T22:39:44.903" />
  <row Id="10720" PostTypeId="1" AcceptedAnswerId="10721" CreationDate="2011-06-03T06:39:03.380" Score="1" ViewCount="74" Body="&lt;p&gt;You must start and end with the same pose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove it if you think you can't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/6CVJf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3828" LastEditorUserId="3828" LastEditorDisplayName="" LastEditDate="2011-06-03T06:48:48.173" LastActivityDate="2011-06-03T07:02:37.663" Title="Is it possible to rotate an angle on perfectly smooth ice?" Tags="&lt;kinematics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10721" PostTypeId="2" ParentId="10720" CreationDate="2011-06-03T07:02:37.663" Score="4" ViewCount="" Body="&lt;p&gt;Yes, it is possible.  This is an example of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Falling_cat_problem&quot; rel=&quot;nofollow&quot;&gt;cat-righting problem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You simply must change your moment of inertia during the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Model your body as two cylinders of equal mass which can exert forces between each other to start spinning.  The bottom cylinder has an adjustable radius, initially set equal to the top cylinder's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set the top cylinder spinning CW.  To conserve angular momentum, the  bottom cylinder spins CCW at the same angular frequency.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wait a short time, then suddenly increase the radius of the bottom cylinder and simultaneously make the top cylinder change directions to start spinning CCW at the same angular frequency as before.  The bottom cylinder will start spinning CW, but since its radius is increased, its moment of inertia is higher, and its angular frequency will be smaller than before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the two cylinders are lined up relative to each other, stop them both, and return the bottom cylinder to its original radius.  The entire apparatus must have rotated because the top cylinder's angular frequency had only a one absolute value the entire time, but the bottom cylinder had two.  They therefore had different total angular displacements, so the entire thing must have rotated.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-06-03T07:02:37.663" CommentCount="5" />
  <row Id="10722" PostTypeId="2" ParentId="10172" CreationDate="2011-06-03T08:32:29.083" Score="2" ViewCount="" Body="&lt;p&gt;It is very difficult to detect gravitational waves because gravity is such a weak force. For the same reason though it's very difficult to dampen gravitational waves. We know that dark matter behaves similar to 'normal matter' for what its equation of state and coupling is concerned, so it won't be able to absorb gravitational waves any more efficiently, which is essentially not at all. (Also: The expression 'gravity waves' describes a phenomenon seen in cloud formation and has nothing to do with General Relativity as Google will tell you.)&lt;/p&gt;&#xA;" OwnerUserId="1458" LastActivityDate="2011-06-03T08:32:29.083" CommentCount="4" />
  <row Id="10723" PostTypeId="2" ParentId="10651" CreationDate="2011-06-03T08:46:47.410" Score="1" ViewCount="" Body="&lt;p&gt;No. While there are some arguments for why 3 spatial dimensions are a good place to live in, the answer to the question why our universe has 3 large spatial dimensions is presently not known. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Karch &amp;amp; Randall wrote a paper on the issue some years back: &lt;a href=&quot;http://xxx.lanl.gov/abs/hep-th/0506053&quot; rel=&quot;nofollow&quot;&gt;http://xxx.lanl.gov/abs/hep-th/0506053&lt;/a&gt; They consider some higher dimensional space filled with objects of different dimensions that have some interactions among each other and argue that 3 dimensional ones are among those most likely to dominate. It's an argument though that is not widely accepted due to the assumptions they have to make for this to work.&lt;/p&gt;&#xA;" OwnerUserId="1458" LastActivityDate="2011-06-03T08:46:47.410" />
  <row Id="10724" PostTypeId="1" CreationDate="2011-06-03T12:08:48.343" Score="0" ViewCount="108" Body="&lt;p&gt;When putting fermions on the string, we have to choose boundary conditions for our spinor fields - Ramond or Neveu-Schwarz. NS conditions on the closed string have antiperiodic conditions such as $\psi_L(\sigma) = -\psi_L(\sigma+\pi)$ (where $\pi$ is the periodicity of the string). I'm confused by this - shouldn't $\psi$ be single-valued?&lt;/p&gt;&#xA;" OwnerUserId="3831" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-06-03T15:09:04.533" LastActivityDate="2011-06-05T18:50:50.150" Title="Do Neveu-Schwarz conditions make sense?" Tags="&lt;string-theory&gt;&lt;string&gt;&lt;boundary-condition&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10725" PostTypeId="2" ParentId="10641" CreationDate="2011-06-03T12:19:05.307" Score="0" ViewCount="" Body="&lt;p&gt;electrons dont go through the ac cicuit. they just vibrate around (50Hz), almost steady&lt;/p&gt;&#xA;" OwnerUserId="3832" LastActivityDate="2011-06-03T12:19:05.307" CommentCount="1" />
  <row Id="10726" PostTypeId="1" AcceptedAnswerId="10742" CreationDate="2011-06-03T12:19:39.063" Score="2" ViewCount="82" Body="&lt;p&gt;I wonder what's the way to regenerate O2 in air without using consumable chemicals (where one can use electricity through electrolisis or using UV lamps)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can dissolve water into O2 &amp;amp; H2, but it won't remove CO2 from air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-04T09:57:28.277" LastActivityDate="2011-06-04T09:57:28.277" Title="Air Regeneration in Closed Systems" Tags="&lt;water&gt;&lt;physical-chemistry&gt;&lt;air&gt;&lt;chemical-compounds&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="10727" PostTypeId="2" ParentId="10724" CreationDate="2011-06-03T13:49:04.167" Score="1" ViewCount="" Body="&lt;p&gt;Dear James, there is no reason why $\psi$ should be periodic. First, if you have a problem, imagine that $\psi$ are just auxiliary variables but the true ones are the bilinears $\psi_i \psi_j$ which are still periodic. Only things such as the world sheet stress-energy tensor $T_{++}$ and $T_{--}$ have to be periodic and they are because they're bilinear in fermions and their derivatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take an analogy with the winding strings. On those strings, $X$ at $\sigma+\pi$ is $X$ at $\sigma$ shifted by the winding number times the circumference. It is not single-valued, either. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a similar way, the fields on the string may be twisted by the action of an orbifold. The only condition is that the operation equivalent to the monodromy around the closed string is a symmetry of the theory. It means that it commutes with the Hamiltonian - so that $T_{ab}$ constructed out of the fields remains periodic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the winding numbers as well as general periodic/antiperiodic conditions for fermions may be viewed as special cases of an orbifold. For each possible group action allowed when you go around the closed string, there is a corresponding GSO-like projection on the string.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the antiperiodic NS boundary conditions on the string are the &quot;primary ones&quot;, the more natural ones than the periodic R boundary conditions. If you map the closed string or a cylinder to a complex plane with an operator at the center by  conformal transformations, the NS sector gets mapped to normal operators such as identities that make the fermions single-valued around the origin, while the R boundary conditions periodic on the cylinder become antiperiodic on the plane, and are associated with spin fields that create branch cuts in the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the NS sector is the normal one while the R sector is the twisted one. But the R sector is allowed, in fact, it is required for modular invariance. In type 0, only NS NS and R R sectors exist. In type II theories, there is a separate orbifold and separate GSO projection on left-moving and right-moving side.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-06-03T15:08:44.953" LastActivityDate="2011-06-03T15:08:44.953" CommentCount="1" />
  <row Id="10728" PostTypeId="2" ParentId="10726" CreationDate="2011-06-03T14:06:16.343" Score="0" ViewCount="" Body="&lt;p&gt;Rather than just splitting water, you want to react it with the CO2 to make O2 and carbohydrates. I think this can be done with an inorganic catalyst (I'd have to Google for the details) but of course the easiest solution is to use photosynthesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;JR&lt;/p&gt;&#xA;" OwnerUserId="1325" LastActivityDate="2011-06-03T14:06:16.343" />
  <row Id="10729" PostTypeId="2" ParentId="7483" CreationDate="2011-06-03T14:13:36.780" Score="0" ViewCount="" Body="&lt;p&gt;Roger Penrose discusses this in chapter 27 of his book The Road to Reality. His conclusion is that you can't just assume inflation will smooth out any initial conditions, and that the pre-inflation universe must already have had a very low entropy.&lt;/p&gt;&#xA;" OwnerUserId="1325" LastActivityDate="2011-06-03T14:13:36.780" />
  <row Id="10730" PostTypeId="2" ParentId="10172" CreationDate="2011-06-03T14:19:46.267" Score="0" ViewCount="" Body="&lt;p&gt;To start this discussion I present a few facts about gravity waves.  A weak linear gravity wave is a perturbation on a background metric $\eta_{ab}$ with the total metric&#xA;$$&#xD;&#xA;g_{ab}~=~\eta_{ab}~+~h_{ab}.&#xD;&#xA;$$&#xA;The Ricci curvature to first order is &#xA;$$&#xD;&#xA;R_{ab}~=~{1\over 2}\Big(\partial_c\partial_a{h^c}_b~+~\partial_c\partial_b{h^c}_a~-~\partial_a\partial_bh~-~\partial_c\partial^ch_{ab}\Big).&#xD;&#xA;$$&#xA;The harmonic gauge $\partial_c{h^c}_a~=~1/2\partial_a h$ gives the Einstein field equation&#xA;$$&#xD;&#xA;\partial^c\partial_ch_{ab}~-~\frac{1}{2}\eta_{ab}\partial^c\partial_ch~=~\frac{16\pi G}{c^4}T_{ab},&#xD;&#xA;$$&#xA;for the traceless metric term ${\bar h}_{ab}~=~h_{ab}~-~(1/2)\eta_{ab}h$ with the simple wave equation&#xA;$$&#xD;&#xA;\partial^c\partial_c{\bar h}_{ab}~=~\frac{16\pi G}{c^4}T_{ab}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gravity wave interacts with a set of test masses by inducing a quadrupolar motion.  Let us suppose we have two such masses.  These masses are on independent geodesics which will deviate from each other according to the variation of a vector connecting the masses $x^a$ by the equation&#xA;$$&#xD;&#xA;\frac{d^2x^a}{ds^2}~=~{R^a}_{bcd}X^cU^bU^d.&#xD;&#xA;$$&#xA;For weak gravity we can set $U^b~\simeq~(1,~0,~0,~0)$, a pure time directed 4-vector and the geodesic deviation equation is approximately&#xA;$$&#xD;&#xA;\frac{d^2x^a}{ds^2}~\simeq~{R^a}_bX^b.&#xD;&#xA;$$&#xA;One can then as an exercise input the Ricci curvature into this equation.  Now let us assume there is a connecting spring between the two masses so that&#xA;$$&#xD;&#xA;\frac{D^2x^a}{ds^2}~\simeq~\frac{d^2x^a}{ds^2}~-~{R^a}_bX^b,&#xD;&#xA;$$&#xA;where now the equation describes a deviation between two nongeodesic moving particles.  Since the perturbing force is a spring we then have $D^2x^a/ds^2~=~-kx^a$, which is just the spring equation familiar from Newton’s second law of motion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address this question about dark matter interacting with gravity waves we think of this spring as the mutual interaction between particles.  The spring constant for DM is very small, for DM is extremely weakly interacting.  The result is that any heat which might be generated by gravity waves, $E~=~(3/2)\langle kx^2\rangle$ $=~kT$ will be very small.  So using DM as a way of detecting gravity waves will likely prove to be frustrating.&lt;/p&gt;&#xA;" OwnerUserId="3833" LastActivityDate="2011-06-03T14:19:46.267" CommentCount="1" />
  <row Id="10731" PostTypeId="1" CreationDate="2011-06-03T14:49:16.513" Score="1" ViewCount="157" Body="&lt;p&gt;There's ample direct evidence for the existence of galactic and stellar mass black holes. However, there is no such direct evidence of primordial black holes, those formed after the Big Bang. A recent paper [http://arxiv.org/abs/1106.0011] describes oscillations the Sun might undergo if it encountered a primordial black hole. The theory that primordial black holes exist hasn't had experimental backing. Why should one believe they actually exist? Is this science or &quot;guesstimating?&quot;&lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="1924" LastEditorDisplayName="" LastEditDate="2011-08-09T21:32:47.860" LastActivityDate="2011-09-04T07:38:12.227" Title="Primordial Black Hole Detection" Tags="&lt;black.holes&gt;&lt;big.bang&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="10732" PostTypeId="1" CreationDate="2011-06-03T14:56:17.107" Score="1" ViewCount="67" Body="&lt;p&gt;Engineering textbooks constantly use the concept of 'phase space' (see e.g. &lt;a href=&quot;http://www.cs.cmu.edu/~baraff/sigcourse/notesc.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.cs.cmu.edu/~baraff/sigcourse/notesc.pdf&lt;/a&gt;). That is, they think of the state of a mechanical system as a high-dimensional geometric space. For example, for a system of particles, the phase space has $6n$ dimensions, where $n$ is the number of particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that the notion of phase space initially developed among theoretical physicists (possible in conjunction with quantum mechanics? or possibly before that point?). Therefore, I'm wondering when and how engineers began to adopt this terminology and concept from physicists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might have something to do with differential equations, since the concept of representing the state of a mechanical system as a point in a space becomes obvious when you view ODE's in terms of flows on euclidean spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: The phase space concept was invented by theoretical physicist and mathematician Willard Gibbs.&lt;/p&gt;&#xA;" OwnerUserId="2591" LastEditorUserId="2591" LastEditorDisplayName="" LastEditDate="2011-06-03T15:23:45.137" LastActivityDate="2011-06-03T15:23:45.137" Title="History of the use of the concept of phase space in engineering" Tags="&lt;history&gt;&lt;engineering&gt;&lt;numerics&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="10733" PostTypeId="1" CreationDate="2011-06-03T14:59:06.273" Score="5" ViewCount="118" Body="&lt;p&gt;This is prompted by the strong claims made in &lt;a href=&quot;http://www.sciencemag.org/content/332/6034/1170.full.html&quot; rel=&quot;nofollow&quot;&gt;Science 332, 1170 (2011)&lt;/a&gt; to have observed trajectories of photons, &quot;something all of our textbooks and professors had always told us was impossible&quot;. I'm suspicious of this claim because they work entirely within quantum theory, and it seems to me that in making weak measurements at two points on what they say is a trajectory, they do nothing else than make unsharp measurements of incompatible observables, a concept that was first introduced, to my knowledge, by Busch and Lahti in &lt;a href=&quot;http://prd.aps.org/abstract/PRD/v29/i8/p1634_1&quot; rel=&quot;nofollow&quot;&gt;Phys. Rev. D 29, 1634–1646 (1984)&lt;/a&gt;, and that is now much used in quantum information. Certainly observables that are at time-like separation are in general incompatible, so an unsharp measurement approach would seem applicable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A brief account of the Science paper mentioned above can be found &lt;a href=&quot;http://www.physorg.com/news/2011-06-quantum-physics-photons-two-slit-interferometer.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, amongst other places, and a few quotes can be found &lt;a href=&quot;http://physicsandphysicists.blogspot.com/2011/06/observing-average-trajectories-of.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, but, this being Science, I couldn't find a preprint. I was first alerted to this paper by &lt;a href=&quot;http://www.bbc.co.uk/news/science-environment-13626587&quot; rel=&quot;nofollow&quot;&gt;a page at the BBC&lt;/a&gt;. The claim highlighted by ZapperZ, &quot;Single-particle trajectories measured in this fashion reproduce those predicted by the Bohm–de Broglie interpretation of quantum mechanics (8), although the reconstruction is in no way dependent on a choice of interpretation&quot;, seems to me particularly egregious, insofar as both their choice of what experiment to do and of what to do with their raw data seem partly driven by the form they want to present their results in. It seems entirely possible to present the data in terms of unsharp measurements and POVMs, which would give relatively less support to the de Broglie-Bohm interpretation.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-06-03T14:59:06.273" Title="Is a weak measurement the same as an unsharp measurement or POVM?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-information&gt;&lt;quantum-interpretations&gt;" CommentCount="2" FavoriteCount="2" />
  <row Id="10734" PostTypeId="2" ParentId="10731" CreationDate="2011-06-03T14:59:20.477" Score="5" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this science or &quot;guesstimating?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's theory.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;General relativity allows for black holes at essentially any mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the theory is strongly supported on several fronts, including the apparent observation of large and huge compact object which fit the bill a black holes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Big Bang cosmology allows for the conditions necessary to produce &quot;small&quot; ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again the theory has some experimental evidence to back it up, though in this case some of the fine detail needed to be convinced that primordial black hole did or did not form is lacking.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Why do you imagine that every reasonable theory will have or not have experimental justification right away? Some experiments or observations are &lt;em&gt;hard&lt;/em&gt; or just take a lot of time.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-06-03T14:59:20.477" CommentCount="3" />
  <row Id="10735" PostTypeId="1" CreationDate="2011-06-03T15:09:30.383" Score="3" ViewCount="126" Body="&lt;p&gt;Can retrocausality resolve the paradoxes of quantum mechanics? The Copenhagen interpretation presumes something has no property until it is measured, but retrocausal interpretations explain that away by claiming that thing has a definite property before measurement, but the choice of which property is affected by the configuration of the measuring apparatus in its future. The philosopher Huw Price has written a lot about this. Nonlocal entanglements are also explained away because entangled pairs are connected by going to the past, then back up to the future. Or going to the future, then back to the past without travelling faster than light at any time.&lt;/p&gt;&#xA;" OwnerUserId="3834" LastActivityDate="2011-06-04T00:25:38.503" Title="Retrocausal interpretations of quantum mechanics" Tags="&lt;quantum-interpretations&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="10736" PostTypeId="2" ParentId="10735" CreationDate="2011-06-03T15:32:19.740" Score="3" ViewCount="" Body="&lt;p&gt;Not &quot;resolve&quot;. Give an interesting account, perhaps. This kind of interpretation suffers from the usual problem for interpretations, that it doesn't add enough that is useful enough for many Physicists to have wanted to use it, at least in the relatively elaborated form proposed by Huw Price. On the other hand, understanding this kind of interpretation and its position relative to other interpretations of quantum theory is essential to anyone working on the Philosophy or Foundations of Physics. It's also notable that backward trajectories do appear in some readings of Feynman diagrams, which could be said to make this kind of interpretation rather mainstream, however I would say that Physicists' use of this kind of language is not quite enough to justify calling it an interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explaining &quot;nonlocality&quot; by introducing &quot;backward propagation&quot; could well be said to be no explanation at all. Neither concept is exactly classically intuitive, for example. Nonetheless, we can hope that understanding that a tradeoff can be made between nonlocality and backward propagation is worthwhile.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-06-03T15:32:19.740" />
  <row Id="10737" PostTypeId="2" ParentId="7506" CreationDate="2011-06-03T16:29:08.523" Score="0" ViewCount="" Body="&lt;p&gt;This may help a lot to judge the merit of Joy Christians papers:&#xA;&lt;a href=&quot;http://www.science20.com/alpha_meme/quantum_crackpot_randi_challenge_help_perimeter_physicist_joy_christian_collect_nobel_prize-79614&quot; rel=&quot;nofollow&quot;&gt;http://www.science20.com/alpha_meme/quantum_crackpot_randi_challenge_help_perimeter_physicist_joy_christian_collect_nobel_prize-79614&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3835" LastActivityDate="2011-06-03T16:29:08.523" CommentCount="1" />
  <row Id="10738" PostTypeId="2" ParentId="10735" CreationDate="2011-06-03T16:33:24.340" Score="1" ViewCount="" Body="&lt;p&gt;initial state: a pair of electron and a positron moving toward each other&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pair of electron and a positron have certain probability of annihilating to each other in certain amount of time, the resulting photon amplitudes are described by some Feynman diagram (the sum of all Feynman diagram with the same In and Out states of course)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case we could interpret the electron as bouncing from a backward propagating photon, and becoming a backward propagating electron (a positron), but such interpretation is pointless because of conditional probabilities and entanglement&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me explain that last phrase above: the initial state assumes that i've already make measurements which couple me (the observer) to a definite state at time $t_0$, so, if i accept that any retrocausal interpretation needs to produce consistent histories (i.e: if i measure A at $t_0$, i always will measure A at $t_0$, regardless of what happens before or after) then all probabilities can only define what happens at other times that are not $t_0$, and such probabilities are given by propagators and a well known state at $t_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so, i hope to have made clear from the above paragraph that before we can speak consistently of any &quot;retrocausality&quot;, we need an alternative way to represent dynamics evolutions to the usual propagator which assumes 100% knowledge of state at $t_0$ and produces amplitude probabilities of states at $t_1$. So we need some formalism that might take a partial state at $t_0$, compute amplitudes at another partial state at $t_1$, then from making measurements at both times, and measure correlations (which could be interpreted in principle as entanglement correlations nonetheless) one could infer that a definite result at the measurement in $t_1$ changed part of the state at $t_0$ that was unknown to us, but it obviously had to be measured by a separate apparatus in order for us to measure such a correlation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should not be a surprise that even if we had such a formalism, it would only differ from the Propagator formalism in the interpretation. I could still assume that the state at $t_0$ was perfectly known, only by different apparatuses, and the propagator produced the observed amplitudes at $t_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the whole &quot;retrocausality&quot; concept becomes non-trivial only if there are &lt;em&gt;fate conditions&lt;/em&gt;, that is, boundary conditions in the future that are unavoidable (for example, imagine a universe with a entirely time simmetrical big-crunch, where the universe seems to have an arrow of time in the first half, but when approaching the half life of the universe, matter slowly starts becoming antimatter by seemingly random fluctuations). Our current universe does not manifest any such flagrant violations of the second law, so the concept of retrocausality is not useful to model physics.&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-06-03T16:41:11.707" LastActivityDate="2011-06-03T16:41:11.707" />
  <row Id="10740" PostTypeId="2" ParentId="10726" CreationDate="2011-06-03T17:24:45.480" Score="0" ViewCount="" Body="&lt;p&gt;There is no inorganic method of which I am aware that is not &quot;messy.&quot;  They all consume and/or produced something extraneous to the clean reaction you want ---&gt; CO2 + energy = C + O2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NASA (these are the people who should know) use electrolysis on the ISS to split 2 H2O into 2 H2 + O2.  The H2 is vented overboard, the O2 is vented to the cabin.  CO2 is captured by a separate system and also vented overboard.  With the cost of upmass being on the order of $10K/lb., you can bet that if there was an easy way to rip the O2 out of CO2, that they'd be doing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using photosynthesis sounds like the easiest way to re-use O2 that is bound in CO2.  The NET equation is 6 CO2 + 6 H2O + energy (light) ---&gt; C6H12O6 + 6 O2.  This uses half the water per liter of O2 produced as electrolysis.  If the sugar produced can be consumed, then that upmass is saved also (reduces the weight of food that needs to be shipped to orbit).&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-06-03T17:49:05.813" LastActivityDate="2011-06-03T17:49:05.813" CommentCount="3" />
  <row Id="10741" PostTypeId="2" ParentId="9341" CreationDate="2011-06-03T18:22:41.317" Score="1" ViewCount="" Body="&lt;p&gt;Take the initial rocket mass to be $M$ and the cumulative quantity of propellant ejected to be $m$, which will be used interchangeably as a dynamic variable and for the final answer.  Unit-wise, I will use exclusively $\beta=v/c$ for representation of velocities of all types.  I am intentionally avoiding using time in my equations.  The essence of the problem is that the velocity depends on the quantity of propellant ejected up until that point (and problem parameters), so I'm going to be finding a solution for $\beta(m)$, that is, the speed of the craft as a function of the mass ejected up until that point.  I'm going to consider the problem in the &quot;rest&quot; reference frame, which is more formally defined by the inertial reference for which it is true that $v(0)=0$.  I'll use the shorthand of the $\gamma$ function, $\gamma(\beta) = (1-\beta^2)^{-1/2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first real physics I'll do here is to dissect a specific reaction.  I define the specific point-in-time I'm talking about as the point when the moving rocket has mass of $M-m+dm$.  It then ejects $dm$ at speed $\beta_e$ relative to itself in the direction opposite of motion.  This ejection increases the rocket's momentum by $dp$ as the ejected mass' momentum changes from $p_{dm}$ to $p_{dm}&amp;#39;$ at speeds of $\beta$ and $\beta_{dm}&amp;#39;$.  Relativistic velocity addition must be done to obtain the latter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\beta_{dm}&amp;#39; = \frac{\beta-\beta_{e}}{1-\beta \beta_e}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the reaction momentum balance is written.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dp = p_{dm} - p_{dm}&amp;#39; = dm \gamma(\beta) \beta - dm \gamma(\beta_{dm}&amp;#39;) \beta_{dm}&amp;#39;$$&#xA;$$\frac{dp}{dm} = \gamma(\beta) \beta - \gamma(\beta_{dm}&amp;#39;) \beta_{dm}&amp;#39; = \frac{\beta}{\sqrt{1-\beta^2}} - \frac{\beta-\beta_e}{1-\beta \beta_e} \frac{1}{\sqrt{1-\left(\frac{\beta-\beta_e}{1-\beta \beta_e}\right)^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I hope that by this point it's blatantly obvious what I'm doing.  I'm attempting to formulate a differential equation where $m$ is the independent variable and we solve for $\beta(m)$.  But we still need the left hand side of that equation.  In order to find this we must continue thinking about the balances of the stated interaction and find the change in momentum of the non-ejected part of the rocket, $dp$, after that approximations can be made.  The momentum of the &lt;em&gt;non-ejected part of the rocket&lt;/em&gt; is $p$ before the ejection and $p&amp;#39;$ after the ejection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dp = p&amp;#39; - p = (M-m) (\beta&amp;#39; \gamma(\beta&amp;#39;) - \beta \gamma(\beta))$$&#xA;$$\beta&amp;#39;-\beta = d\beta &amp;lt;&amp;lt; \beta_e$$&#xA;$$dp = (M-m) ((\beta+d\beta) \gamma(\beta+d\beta) - \beta \gamma(\beta))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd order series expansion about $d\beta=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dp = (M-m) d\beta \gamma(\beta)^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, this can be found by differentiating.  The reason simple differentiating is so hard is that you have to identify &lt;em&gt;what&lt;/em&gt; it is you take the derivative of.  In order to be consistent with the physics of the situation I had to introduce a special $m&amp;#39;&amp;#39;$ variable, which is an &lt;em&gt;invariant&lt;/em&gt; form of $m$, although having the same value.  Basically, $m&amp;#39;&amp;#39;$ is not affected by the loss of $dm$ but $m$ is.  Starting from here I have to start writing $\beta$ in terms of $\beta(m)$ as well, which is the objective.  Pardon the sudden change in notation.  Here is the calculus approach to $dp$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\beta = \beta(m)$$&#xA;$$\frac{dp}{dm} = \frac{d}{dm} \left( (M-m&amp;#39;&amp;#39;) \beta(m) \gamma(\beta(m)) \right) = (M-m&amp;#39;&amp;#39;) \frac{d\beta}{dm} \gamma(\beta)^3 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either approach gives the needed expression for the next step, which is to simply write the differential equation that is the solution to the problem.  Pardon the switch back to suppressing the $m$ dependence again (so it fits on the line), just know that it's really $\beta(m)$ and that $\beta_e$ is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(M-m) \frac{d\beta}{dm} \gamma(\beta)^3 = \frac{\beta}{\sqrt{1-\beta^2}} - \frac{\beta-\beta_e}{1-\beta \beta_e} \frac{1}{\sqrt{1-\left(\frac{\beta-\beta_e}{1-\beta \beta_e}\right)^2}}$$&#xA;$$\beta(0)=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we're done.  This is your answer.  With $M$ and $\beta_e$ specified you can find $\beta(m)$ which is the speed of the rocket as a function of the mass ejected, but remember that $m&amp;lt;M$.  I'll give a sample plot.  This is showing the function of $\beta$, which again is the fraction of the speed of light the rocket is going.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\beta_e=0.1$ and $M=1$&#xA;&lt;img src=&quot;http://i.stack.imgur.com/liSwJ.gif&quot; alt=&quot;Rocket Beta as function of propellent&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some approximations you can get, of course.  Doing a 2nd order taylor expand of the RHS of the diff equation above will result in the following solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\beta = tanh \left( \beta_e ln(\frac{M}{M-m}) \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you simplify even further ($tanh(x)=x$) you'll get the classical version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\beta = \beta_e ln(\frac{M}{M-m}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first of these seems to be a fairly good approximation, but only for $\beta_e&amp;lt;&amp;lt;1$. Obviously $\beta_e$ and $m$ are relevant to the answer, unless you made some different assumptions I did.  I find it most likely, however, that whoever argued the question only has 1 answer out of a, b, c does not have a coherent argument for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: had $\sqrt{2}$ factor incorrectly because I typed in an equation wrong.  Fixed it now, consistent with Wikipedia for 1st approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When $\beta=1$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, even the very first equation written for $\beta_{dm}&amp;#39;$ is invalid, and we must return to the drawing board for calculating $dp$.  I'll have to approach this considering the emission of a single photon, so my notation will be that $dp$ and $dm$ refer to the change in momentum and mass &lt;em&gt;of the rocket according to the stationary observer&lt;/em&gt; due to the emission of a single photon.  The frequency of the light according to the spacecraft will be $\lambda_e$ and in stationary frame, $\lambda_o$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dp = \frac{h}{\lambda_o}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p c = \frac{h c}{\lambda_o} = E =  c^2 dm \rightarrow dm = \frac{h}{c \lambda_o}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dp}{dm} = c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turns out we don't need to do anything with redshift, or even need to know the light frequency!  I actually expected that, so it's ok.  My previous $dp$ by the way, was the wrong units.  I should have written $p=c \beta \gamma(\beta)$, but if you're find with the answer being in terms of $\beta$, then why bother?  So I'll just fudge it so that $dp/dm=1$.  Now we can take the previous expression for that quantity and set it equal to 1 to find the DE for $\beta(m)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dp}{dm} = (M-m) \frac{d\beta}{dm} \gamma(\beta)^3 = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\beta(0) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \beta = ln{ \frac{M}{M-m} } \sqrt{ \frac{1}{ 1+ ln{ \frac{M}{M-m} }^2 } }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll plot this with all the others discussed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M = 1.0, beta_e = 0.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/OciSI.gif&quot; alt=&quot;All plotted together&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&lt;li&gt;Zassou - My answer in the form of the complete diff equation.  Numerically I was only able to evaluate up to $m=0.8$&#xA;&lt;li&gt;Wikipedia - the tanh( ln( .. )) solution that is also in the wikipedia article for this, it's valid for $\beta_e&amp;lt;&amp;lt;1$, and shows some deviation from the actual here because of it, and when $\beta_e$ gets close to 1 you can see a lot more bending of the curve&#xA;&lt;li&gt;Jerry - formula he gave in his answer&#xA;&lt;li&gt;Newton - obviously the classical case, flies off to super-luminary speeds of course&#xA;&lt;li&gt;Photon - equation I just gave - note this DOESN'T solve the same problem since $\beta_e$ is different, which is evidenced by the initial slope&lt;/ul&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-06-06T20:15:25.323" LastActivityDate="2011-06-06T20:15:25.323" CommentCount="10" />
  <row Id="10742" PostTypeId="2" ParentId="10726" CreationDate="2011-06-03T18:39:27.913" Score="1" ViewCount="" Body="&lt;p&gt;Here is a Babelfish translation from German Wiki on Life support systems. (Including NASA!)&#xA;(English WIKI on the topic is very bad) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In order to make for humans possible a&#xA;  surviving, the supply of breathable&#xA;  gas (thus air) must be guaranteed. But&#xA;  air must be as free of harmful&#xA;  substances as possible and determined&#xA;  parameters keep. Thus on board EAT a&#xA;  total pressure from 97,9 to 102.7 kPa,&#xA;  an oxygen partial pressure from 19,5&#xA;  to 23.1 kPa, a nitrogen partial&#xA;  pressure of less as 80 kPa and a&#xA;  carbon dioxide partial pressure of&#xA;  less as 1 kPa than acceptable&#xA;  regarded. The air temperature in EAT&#xA;  is appropriate (adjustable) between&#xA;  18,3 and 26,7 °C. By the air condition&#xA;  technology an air humidity between 25&#xA;  and 75% and a constant air movement&#xA;  between 0,05 and 1,0 m/s are secured,&#xA;  in order to avoid microbe growth and&#xA;  fungus growth on the one hand and a&#xA;  too dry air (danger of the sparking)&#xA;  on the other hand. Here usually&#xA;  conventional air conditioning systems&#xA;  with refrigerants (e.g. ammonia or&#xA;  Freon) are used. Around air thereby&#xA;  condensing heat exchangers will&#xA;  dehumidify begun. [2] For carbon&#xA;  dioxide connection re-usable zeolites&#xA;  or solid amines, in space suits are&#xA;  used also lithium hydroxide. The&#xA;  production of oxygen takes place&#xA;  usually via the electrolysis from&#xA;  water and partly via recuperation from&#xA;  the carbon dioxide by Sabatier process&#xA;  and the following methane pyrolysis.&#xA;  As Backup or with short term&#xA;  employment oxygen also compressed or&#xA;  chemical reactions is used to the&#xA;  supply of oxygen. Pollutants are&#xA;  constantly supervised by appropriate&#xA;  measuring methods like for example&#xA;  mass spectrometers and gas&#xA;  chromatography and by molecular sieve,&#xA;  activated charcoal or lithium&#xA;  hydroxide filtered. In submarines&#xA;  similar values are valid, and partly&#xA;  also similar processes are used. [3]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Think: NASA did/does very different depending on situation! On a spacestation with ample electricity You can do electrolysis, for Apollo they made electricity from oxygen/hydrogen in fuel cells! And cost of upmass includes the cost of energy generators or storage batteries. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But topic here is some (realistic) method for barsmonster to survive the Moscow summer. This is best done by cooling the air intake, demoist it (which would hopefully remove some of the organic &quot;smog&quot; particles) and one could try some PP microfibre filters. The exhaust air would be used to cool down the air taken in.  Any separation of carbon dioxide by ad(ab)sorbents would be much more expensive than this air exchange.   &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-06-03T18:39:27.913" />
  <row Id="10743" PostTypeId="2" ParentId="2605" CreationDate="2011-06-03T18:53:28.540" Score="-1" ViewCount="" Body="&lt;p&gt;This is simple.  The answer is it leaves your body in the form of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your body burns sugars while producing things like heat (you get hot when exercising) and motion (you exercise).  So, we physically convert mass into energy which gets transfered mechanically into whatever we are doing.  That could be a distance traveled (walking/biking etc) or an increase in another objects potential energy (lifting weights etc)&lt;/p&gt;&#xA;" OwnerUserId="3843" LastActivityDate="2011-06-03T18:53:28.540" CommentCount="3" />
  <row Id="10744" PostTypeId="2" ParentId="2605" CreationDate="2011-06-03T18:55:43.990" Score="0" ViewCount="" Body="&lt;p&gt;Additionally, some fats (cholesterols, etc) are converted to bile and excreted by the liver into the digestive track.  On top of this many of the byproducts of cell death are excreted in this fashion.  That brown color to your poop is the product of red blood cells being broken down, and expelled by the liver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some minerals and other substances that are water soluble may be excreted by the kidneys.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2011-06-03T18:55:43.990" />
  <row Id="10745" PostTypeId="1" AcceptedAnswerId="10749" CreationDate="2011-06-03T20:10:22.453" Score="3" ViewCount="213" Body="&lt;p&gt;Say you want to store hot coffee in a container surrounded by a vacuum. To remove all sources of conductive energy loss the container is suspended in the vacuum by a magnetic field and does not have a physical connection to the sides of the vacuum chamber,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is would the magnetic field be a path for energy to be conducted out of the suspended container?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to look at this question would be two magnets are suspended in a vacuum with their poles aligned. A heat source is attached to one of the magnets. Would the second magnet show a corresponding increase in temperature, excluding radiated heat transfer?&lt;/p&gt;&#xA;" OwnerUserId="3854" LastActivityDate="2011-06-04T15:52:27.537" Title="Can heat be transfered via magnetic field in a vacuum?" Tags="&lt;electromagnetism&gt;&lt;energy&gt;&lt;heat&gt;&lt;vacuum&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" />
  <row Id="10746" PostTypeId="2" ParentId="10745" CreationDate="2011-06-03T20:29:19.140" Score="0" ViewCount="" Body="&lt;p&gt;Simply, no. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only transfer path left over in Your scenario is radiation of IR waves, &#xA;which is electromagnetic radiation with wavelength from about 1 to 100 µm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This radiation is independent of all sorts of magnets, nevertheless &#xA;the word magnetic occurs in its name. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But: Suspension by magnets is rather difficult, and if You heat a magnet &#xA;above its Curie point temperature, the magnet will loose its magnetism. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-06-03T20:29:19.140" />
  <row Id="10747" PostTypeId="2" ParentId="10745" CreationDate="2011-06-03T20:30:42.350" Score="1" ViewCount="" Body="&lt;p&gt;No, a static magnetic field does not conduct heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way for thermal energy to cross a vacuum is by radiation. Of course, radiative heat transfer actually does consist of electric and magnetic fields: hot objects give off electromagnetic waves, which travel through the vacuum and are absorbed by, say, the walls of the vacuum chamber. But those are waves, not static fields, and besides, they are actually emitted by the hot object - it's not like it takes advantage of preexisting fields to move the energy.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-03T20:30:42.350" CommentCount="2" />
  <row Id="10748" PostTypeId="2" ParentId="10745" CreationDate="2011-06-03T20:38:08.080" Score="1" ViewCount="" Body="&lt;p&gt;I would say the first point to stress here is that &lt;strong&gt;heat&lt;/strong&gt; &lt;em&gt;by definition&lt;/em&gt; is the &lt;strong&gt;transfer of energy&lt;/strong&gt;. 'Heat loss' can occur via convection, conduction or radiation. Clearly, the first two can be essentially reduced to zero by surrounding your coffee (or whatever) with a perfect vacuum. Creating a &lt;em&gt;perfect&lt;/em&gt; vacuum is pretty much impossible though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All objects at temperatures above &lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_zero&quot; rel=&quot;nofollow&quot;&gt;absolute zero&lt;/a&gt; will radiate heat via emission of electromagnetic waves. (For more details, see this wiki article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_body&quot; rel=&quot;nofollow&quot;&gt;black-body radiation&lt;/a&gt;.) This is a &lt;em&gt;matter-of-fact&lt;/em&gt;, with or without a vacuum, hence why flasks don't keep your coffee warm indefinitely. A reflective layer (normally some sort of foil) is used to contain some of this radiated energy but it works better for some wavelengths, and worse for others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question directly: &lt;strong&gt;no&lt;/strong&gt; - The magnetic field used to suspend the vessel will not interfere with the energy being radiated, nor indeed does it constitute a medium through which heat can excape via any other means.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastActivityDate="2011-06-03T20:38:08.080" />
  <row Id="10749" PostTypeId="2" ParentId="10745" CreationDate="2011-06-03T21:06:42.633" Score="2" ViewCount="" Body="&lt;p&gt;Heat consists of random vibrations in a material. If a magnetic field connects two objects, then it creates a mechanical coupling between the two objects. Such a coupling will couple vibrations, therefore heat will be conducted by the magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, the effect will be very small, but given enough time, heat will be conducted by the magnetic field. The same applies to suspension by a combination electric and magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not entirely incompatible with some of the other (wrong) answers. For example, since the suspended object is not at absolute zero, it is not possible for the magnetic field suspending it to be entirely static.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As an example, let's consider a superconducting metal container of gas suspended in a magnetic field:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/9Xfhr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that the gas is at some non-zero temperature. Then there is a non-zero probability that the gas will concentrate onto one side:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/cgPTJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for this to happen, assuming that the center of mass of the container remained constant, the metal part of the container had to have shifted in the opposite direction:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/zltZX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the levitation is stable, a motion of the superconducting container must be opposed by the magnetic field. Thus the above motion is opposed by a force from the magnetic field. However, for every action there is an opposite and equal reaction (Newton's 3rd law), therefore there will be a corresponding opposing force on the suspension:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/SfMP8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore thermal motion in the suspended container will induce thermal motion in the base, hence there will be conduction of heat.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-06-04T00:08:12.920" LastActivityDate="2011-06-04T00:08:12.920" CommentCount="10" />
  <row Id="10750" PostTypeId="2" ParentId="10718" CreationDate="2011-06-03T22:52:46.247" Score="1" ViewCount="" Body="&lt;p&gt;The two equations are equivalent (at least in the physicist's sense, I won't try to make the derivation rigorous).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second equation, written in the condensed form reads&#xA;$$G = G_0 + G_0 \Sigma G$$&#xA;or, when iterated (plug in the RHS into the second $G$, ad infinitum),&#xA;$$G = G_0 + G_0 \Sigma G_0 + G_0 \Sigma G_0 \Sigma G_0 + \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This last form is actually what one starts with when solving the evolution equation $$(i\partial_t - H_0 - \Sigma) G = \delta$$&#xA;If the $\Sigma$ wouldn't depend on time, you'd just write $G = \exp(-it(H_0 + \Sigma))$ but if it does, you have to integrate -- either by iterative process above, or by time ordered exponential, which is equivalent but I won't go into that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, one can formally sum the above infinite series to obtain&#xA;$$G = G_0 {1 \over 1 - \Sigma G_0} = {1 \over G_0^{-1} - \Sigma}$$&#xA;This relation actually doesn't say anything else than what we have already seen in the evolution equation for $G$, namely that it is inverse to the evolution operator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to your questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;External potential means part of the potential that is not self-energy. Consider the Fermi liquid. There electrons interact by Coulomb potential and you obtain lots of terms for self-energy when you try to compute the Feynman diagrams (Hartree, Fock exchange, potential screening, etc., etc.). By external potential we usually understand the rest of the world. The primary representative would be a laser pulse, which in the simplest case you can model as classical monochromatic EM plane wave perturbation. Or you could include photons + full QED. Depends on what you need.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This should be obvious now: iterate. To any given order (in powers of $\Sigma$) you need. But to higher order you get, the more mess you'll receive. The complexity can be tamed a little by Feynman diagrams but even number of those grows fast and in practice you'll see terms like LO, NLO and at most NNLO, meaning leading order, next-to-leading order, and so on.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Not sure what you mean, you'd need to be more explicit. Obviously you are free to move lots of complex $i$s and minuses between various places (i.e. whether to put them into propagators, vertices, integrals, etc.) and also various combinatorial factors (coming from graph counting). You won't really understand why this works unless you'll derive the whole machinery yourself. I mean, try to calculate some process by hand (meaning evaluating integrals and vacuum expectations explicitly) and then compare that with the diagramatic approach. In any case be sure that diagrams are nothing else than shorthand notation for the real calculations.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-06-03T22:52:46.247" CommentCount="9" />
  <row Id="10751" PostTypeId="2" ParentId="10735" CreationDate="2011-06-04T00:25:38.503" Score="0" ViewCount="" Body="&lt;p&gt;The ordinary Retrocausal interpretation of quantum mechanics, was Cramers transactional&#xA;interpretation, where an event is formed as a single point in space time, where an incoming&#xA;wavefunction from the past, and an incoming (advanced wave) wavefunction from the future meet.&#xA;The theory them a fixed past and future points, which can then predict observation in the&#xA;time between. It remains mathematical correct, but of course is hardly useful for predictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interesting to me, is between an transactional interpretation with one fated future, and&#xA;a manly worlds interpretation with all possible futures, are there any conditions which would&#xA;allow us to removed possible futures, or truncate the calculation of probility amplitude at&#xA;low levels of likelihood.&lt;/p&gt;&#xA;" OwnerUserId="3864" LastActivityDate="2011-06-04T00:25:38.503" />
  <row Id="10752" PostTypeId="2" ParentId="10731" CreationDate="2011-06-04T01:11:10.967" Score="3" ViewCount="" Body="&lt;p&gt;The question might be asked as to why the universe did not produce a gas of black holes?  The universe started out at a very low entropy, and black holes have entropy proportional to their horizon areas.  So there could not have been a lot of primordial black holes in the early universe.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lifetime of a black hole is&#xA;$$&#xD;&#xA;t~=~\frac{5120\pi G^2M^3}{\hbar c^4}.&#xD;&#xA;$$&#xA;A black hole generated at the start of the universe and a mass of $1.7\times 10^{11}$ kg would Hawking evaporate about now.  The final explosion would involve the quick conversion of mass into energy, where $10^5$ kg would be converted into energy in around a second.  This is a considerable explosion, but it is less than conversion rate of matter by a star. So this is not as easy as looking at supernovae if primordial black holes are very distant.  Also the energy would largely be in the gamma ray domain.  The energy output would be a chirp function in its frequency increase and energy.  The primordial black hole needs to explode in our local neighborhood.  The actual dynamics of the explosion needs further examination, which I am sure is in the literature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Chandra satellite could pick up primordial black hole explosion, but so far nothing like this has been found.  The lack of such detection gives weight to the estimate that the universe had a very low entropy at the start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LC&lt;/p&gt;&#xA;" OwnerUserId="3833" LastActivityDate="2011-06-04T01:11:10.967" CommentCount="5" />
  <row Id="10753" PostTypeId="2" ParentId="3228" CreationDate="2011-06-04T01:37:25.390" Score="-1" ViewCount="" Body="&lt;p&gt;It seems to be widely assumed that the observed click in the geiger counter corresponds to the instantaneous decay of a particular tritium atom. I don't know if I'm just pointing out the obvious, but I'm quite sure this correspondence has never been explicitly demonstrated. Quantum mechanics tells us there is a certain flux of electrons emanating from the vial of tritium; that there are is a certain frequency of clicks in a geiger counter; and that if analyzed, the sample of tritium may be separated into two streams, one of which turns out to be helium. These are three different phenomena, none of which can be easily correlated with any of the others. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it plainly, all we can say about your sample of tritium is that the atoms are in a superposition of states. When they are observed individually, they are found to be in one or the other atomic state - tritium or He3. There is no experiment I know of where we can identify the moment when a particular tritium atom changed state.&lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-06-04T01:37:25.390" />
  <row Id="10754" PostTypeId="1" CreationDate="2011-06-04T04:04:46.880" Score="1" ViewCount="331" Body="&lt;p&gt;I have read that string theory predicts (or requires ?) the existence of gravitons.&#xA;So, would that make it a quantum theory of gravity ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, I have also read that quantum gravity would allow us to understand what happens at the singularity inside a black hole. What does string theory say about this ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If string theory can't make any predictions about this, what are we missing ? &#xA;Is the theory incomplete ? Or is it that we have multiple string theories - hence, multiple ways to quantize gravity - and don't know which is the right one ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a physicist, so just looking for an explanation at the Brian Greene popular book level. :)&lt;/p&gt;&#xA;" OwnerUserId="3335" LastActivityDate="2011-06-18T01:41:49.467" Title="Is string theory a quantum theory of gravity?" Tags="&lt;string-theory&gt;&lt;quantum-gravity&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="10755" PostTypeId="2" ParentId="10754" CreationDate="2011-06-04T04:59:36.993" Score="2" ViewCount="" Body="&lt;p&gt;String theories are in the forefront of particle theory at the moment because they are the only theories that can quantize gravity and at the same time allow the  incorporation of the Standard Model of particle physics within the theory. The SM can be thought as an efficient mathematical description of all the data gathered up to now for particle physics, there fore any theory of everything should incorporate it in some form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;String theories are a work in progress and in wait for data that will allow determination of constants on what exactly is nature's choice. We might be lucky with the LHC to make progress on this, we might not and will need patience and decades  of experiments and theoretical studies. One can study what happens in black holes using string theory models but the reason they are popular is that they pave the way to a Theory Of Everything. Or so many physicists hope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, string theories are quantum theories of gravity and include the graviton and a host of other particles that can be counted like angels on the head of a pin. :) . It is data that is needed to solidify a specific string theory model.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-06-04T04:59:36.993" CommentCount="15" />
  <row Id="10756" PostTypeId="1" AcceptedAnswerId="10770" CreationDate="2011-06-04T06:58:33.143" Score="2" ViewCount="197" Body="&lt;p&gt;Other than one having a positive invariant scalar product and the other a negative one, what are the actual physical differences between these vectors?&lt;/p&gt;&#xA;" OwnerUserId="202" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-06-09T17:05:50.067" LastActivityDate="2011-06-09T17:05:50.067" Title="Difference between timelike and spacelike vectors" Tags="&lt;special-relativity&gt;&lt;relativity&gt;&lt;four-space&gt;" AnswerCount="4" />
  <row Id="10757" PostTypeId="1" CreationDate="2011-06-04T07:50:55.103" Score="5" ViewCount="108" Body="&lt;p&gt;I do have a question about an assumption made in the very interesting Hayden-Preskill paper of black holes as informational mirrors. Alice throws her top secret quantum diary which is $k$ qubits long into a black hole hoping to get rid of it. Unfortunately, the top forensic scientist Bob was already maximally entangled with the black hole, and in a sense, &quot;knows&quot; the full microstate of the black hole before Alice dumped her diary into it. If $c$ is the scrambling time, and Bob waits to collect $k+c$ qubits worth of Hawking radiation plus a few more to reduce the margin of error up to the desired level, then in principle, Alice's quantum diary can be read by Bob. This argument is purely quantum informational theoretic. What are the computational resources needed for Bob to decrypt Alice's diary? By assumption, the dynamics of black hole evolution are described by a reversible unitary transformation, and not by Krauss operators. So, if Bob is willing to wait until the black hole has completely evaporated away and reverse the computations, he can read Alice's diary in polynomial time. However, Bob is too impatient for that, and he wants to read it after only getting $k+c$ plus a few more qubits worth of information. If Bob is willing to wait exponentially long (and he's not), he can create exponentially many identical black holes one by one and go over all possible $k$ qubit combinations to dump into the identical black holes and wait until he finds one which gives Hawking radiation that matches the Hawking radiation he actually got from the original hole. Bob is also unable to reverse the data either because he doesn't know the microstate of the black hole after collecting $k+c$ qubits worth of Hawking radiation. And apparently, neither can Alice if she doesn't already have a copy of her diary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an analogous situation in classical computation. A cryptographic hash function is a polynomially computable function which takes in as arguments a seed which can be publicly known, and some unknown string to be encrypted.  It's a one-way function. It's not possible to compute to original string from the encrypted string using polynomial resources. However, if we are given another string and asked to test if it's identical to the original string, we can compute its hash function using the same seed and check if it matches. If it doesn't, they're not identical. If they do, there's a very high chance that they are indeed identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are black holes quantum cryptographic hash functions? What is the best lower bound on the time it would take for Bob to read Alice's diary?&lt;/p&gt;&#xA;" OwnerUserId="3868" LastActivityDate="2011-07-29T00:44:11.117" Title="Hayden-Preskill informational mirrors and decryption" Tags="&lt;black-holes&gt;&lt;quantum-gravity&gt;&lt;quantum-information&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="10758" PostTypeId="2" ParentId="10756" CreationDate="2011-06-04T09:22:20.867" Score="2" ViewCount="" Body="&lt;p&gt;In certain reference frames these intervals are just time (duration of a process) and a distance ( length of a still rod). &lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-06-04T10:17:21.577" LastActivityDate="2011-06-04T10:17:21.577" />
  <row Id="10759" PostTypeId="1" AcceptedAnswerId="10761" CreationDate="2011-06-04T10:22:13.510" Score="3" ViewCount="146" Body="&lt;p&gt;Looking at the relevant wikipedia page, one can read that the graviton should be massless. Is it 100 % certain that it is massless or is there room in any &quot;nonstandard&quot; models for a tiny non-zero mass (which could lead to a similar surprise as the detection of the neutrino oscillation) such that the graviton (if it exists ...) could maybe selfinteract and form something like &quot;graviballs&quot;?&#xA;(My present knowledge of GR and QFT is at the &quot;Demystified-Level&quot;)&lt;/p&gt;&#xA;" OwnerUserId="2751" LastEditorUserId="2751" LastEditorDisplayName="" LastEditDate="2011-06-04T23:02:59.773" LastActivityDate="2011-06-04T23:02:59.773" Title="Possibility of &quot;graviballs&quot;?" Tags="&lt;gravity&gt;&lt;quantum-gravity&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="10761" PostTypeId="2" ParentId="10759" CreationDate="2011-06-04T12:15:00.203" Score="3" ViewCount="" Body="&lt;p&gt;Gravitons do self-interact. That's because gravity couples to energy-momentum tensor which is non-zero even for massless particles (e.g. consider that gravity influences light). So it's certainly a possibility that they might form a bound state. But you can't see this in a linearized theory of gravity because it is a free theory (similar to electrodynamics). So you'd also have to include higher order curvature terms and quantize those, which would in principle resemble something like chromodynamics (although much harder). It is also not clear whether the theory would be consistent without including the rest of the standard model (and beyond). In any case you'd definitely need some form of quantum gravity to answer this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the first question: there are experimental upper bounds on masses of all massless particles. Obviously it's impossible to distinguish whether something is strictly zero or infinitesimally small. Suffice it to say that all experiments conform to the fact that the masses are zero and there is no theoretical justification whatsoever to consider other models.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-06-04T12:15:00.203" CommentCount="2" />
  <row Id="10762" PostTypeId="2" ParentId="10745" CreationDate="2011-06-04T13:09:23.120" Score="2" ViewCount="" Body="&lt;p&gt;The various comments on the earlier Answers to this Question point to its moderately fundamental misconception. Because of special relativity there is no such thing as a magnetic field separately from the electromagnetic field. Thermal fluctuations of a body, including vibrations, are transferred to a remote body by interactions between the bodies, the chief of which is likely to be electromagnetism at scales at which magnetism has been engineered to be a specific constraint on movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first part of the Question obscures matters by introducing a magnetic field without introducing its sources --permanent magnets or electromagnets, say. I think this idealization is unhelpful. If there were really a magnetic field without any source for the magnetic field, then thermodynamics would be inapplicable unless there are &lt;em&gt;other&lt;/em&gt; heat sources; it would further have to be specified whether there are any interactions between the various heat sources, and what those interactions might be. EDIT: An &lt;em&gt;electromagnetic&lt;/em&gt; field is dynamical, in contrast to a magnetic field, so the electromagnetic field itself can be considered to be a separate thermodynamic system, in which case the heat of the confined container will dissipate into the electromagnetic field or vice versa, depending on the relative temperatures of the electromagnetic field and of the container. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final sentence, &quot;Would the second magnet show a corresponding increase in temperature, &lt;em&gt;excluding radiated heat transfer&lt;/em&gt;?&quot;[my emphasis] sets things up so that the answer is no &lt;em&gt;by definition&lt;/em&gt;. If radiated heat transfer is excluded --an impossible idealization according to experience-- then there will be no change of temperature. This, however, is again not a helpful idealization.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-06-04T15:52:27.537" LastActivityDate="2011-06-04T15:52:27.537" CommentCount="2" />
  <row Id="10763" PostTypeId="1" CreationDate="2011-06-04T13:56:40.770" Score="2" ViewCount="76" Body="&lt;p&gt;How would the non-interacting electron orbitals from a perfect DFT solution for a given potential shape differ from the 'true' electron wavefunctions? Or can you only really talk about the total wavefunction? Would they be less localised as they do not include interaction effects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Hohenberg-Kohn showed that a functional of the wavefunction could be transformed into a functional of the density, any 'measurements' on the total density would have to yield the same results as measurements on the total wavefunction, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, if we imagine a 3-electron system, the ground state consists of spin-up and -down electrons in the lowest state, and a singlet electron in the next state. Since there is correlation and exchange, the total wavefunction would be a superposition of the various possible combinations of these. &lt;/p&gt;&#xA;" OwnerUserId="670" LastActivityDate="2011-08-17T18:37:15.200" Title="How would Kohn-Sham orbitals differ from 'true' elecron wavefunctions?" Tags="&lt;quantum-mechanics&gt;&lt;electron&gt;&lt;correlation-function&gt;&lt;exchange-interaction&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" />
  <row Id="10764" PostTypeId="1" CreationDate="2011-06-04T14:05:01.377" Score="1" ViewCount="62" Body="&lt;p&gt;I have some questions about Nanotechnologies. I'm not physicist and my knowledges about the topic are limited by the &quot;Engines of Creation&quot; book by Eric Drexler. So I think that my questions are related mostely to popular science.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;As far as I know the molecular mechanisms like nanoassemblers and nanofactories exist in nature(I mean alive cells). Why have they not been created artificially by now?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In case that they have been artificially created yet, do they have any limitations in developing of macro objects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, macromechanisms which exist in nature like animals are grow very slowly(months and years) and the initial program(i.e. DNA encryption) doesn't allow to make accurate copies. I mean that the twins who are growing from the same genetic material have a significant difference in adulthood(i.e. different phenotypes). Is this phenomena the result of some fundamental low of nature like Heisenberg uncertainty principle or any other that can't be resolved by modern technological methods?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What are the top well known laboratories and organisations in the world which are engaged to resolve such kind of problems?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="3872" LastActivityDate="2011-06-04T16:12:23.560" Title="Nanotechnologies: current problems and general limitations" Tags="&lt;nanoscience&gt;" AnswerCount="2" />
  <row Id="10765" PostTypeId="2" ParentId="10759" CreationDate="2011-06-04T14:19:27.887" Score="2" ViewCount="" Body="&lt;p&gt;There exists a self-consistent, self-interacting gravitational solitons called &quot;gravitational geon&quot; within general relativity. As Marek said above, gravitational field is non-linear and self-interacting so energy momentum of gravity wave itself will produce bound states. You can read a bit more about it in a following Wiki page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Geon_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Geon_%28physics%29&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One important contribution that the Wiki page missed is that you can get a half-integer spin geon from pure gravity, a work by Friedman and Sorkin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://prl.aps.org/abstract/PRL/v44/i17/p1100_1&quot; rel=&quot;nofollow&quot;&gt;http://prl.aps.org/abstract/PRL/v44/i17/p1100_1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know there are not much development in quantum mechanical aspect of them.&lt;/p&gt;&#xA;" OwnerUserId="2058" LastActivityDate="2011-06-04T14:19:27.887" CommentCount="1" />
  <row Id="10766" PostTypeId="2" ParentId="10756" CreationDate="2011-06-04T15:23:20.160" Score="1" ViewCount="" Body="&lt;p&gt;In every reference frame, a time-like vector has a time component; a space-like vector has a space component.&lt;/p&gt;&#xA;" OwnerUserId="1345" LastActivityDate="2011-06-04T15:23:20.160" />
  <row Id="10767" PostTypeId="2" ParentId="10764" CreationDate="2011-06-04T15:47:01.730" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;Laws of classical mechanics cannot just be scaled down to the nano-regime, as alot of additional forces come up (chemical interaction, friction must be inspected newly, cannot be described by rough material parameters as in classical mechanics). Also microfabrication is a highly tricky process, especially if u want to do it in a parallel style fabrication process (e.g. integrated circuits only with limited resolution possible currently) or serial process (e.g. e-beam lithography). Industrial mass production for nano-machines is still a major problem to be solved. &lt;/li&gt;&#xA;&lt;li&gt;Evolution is a process of trial &amp;amp; error under external circumstances on a carbon basis. Synthetic biology already is able to synthesize distinct molecule chains. Generally, &quot;Self organisation&quot; is a major topic in physics, but mostly ruled by non-linear processes which will lead to chaotic results often without heavy controlling and deeply understanding the distinct system. So engineering complex machines build from the nano bottom up and not top down is highly tricky thing. Be patient and give us more time plz ;) Having nano-machines that reproduce and organize themselves out of a few with some raw material is still a bit science fiction&lt;/li&gt;&#xA;&lt;li&gt;We have the the century of NANOPHYSICS. Every major physics department will do research in it. Depending what science field you prefer, from spintronics - photonics, NEMS, microfabrication methods, material research. Most department do specialize on specific topics, as laboratory hardware is pretty expensive. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3715" LastActivityDate="2011-06-04T15:47:01.730" />
  <row Id="10768" PostTypeId="2" ParentId="10764" CreationDate="2011-06-04T16:12:23.560" Score="0" ViewCount="" Body="&lt;p&gt;Your questions may be confusing your enquiry by too closely comparing DNA-based life and nanotech. Although they both operate at a nanoscale to create much larger objects they are very different in approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DNA is the end result of many trillions of random events. Various simple molecules that tended to create more of themselves eventually resulted in more complex chemical mixtures that eventually resulted in the first very very basic cells and the resulting explosion of activity we call &quot;life&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way that you would not call a tree blown down across a stream a &quot;designed, engineered and constructed&quot; bridge you cannot compare DNA with mature nanotech.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even basic nanoassembler doesn't exist yet because the engineering techniques have not progressed that far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There will be limitations on what can be created but they are not in the direction you are wondering about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mostly they will be related to the physical limitations of molecules. They can only withstand so much force/heat etc before changing their structure. Heisenberg's uncertainty, like all physical laws, will of course apply but that law doesn't cover large scale uncertainty such as the twins. It effectively covers utterly tiny quantum events (well below the nano scale but still possible effecting engineering of it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limitation I believe you are interested in is the complexity of the object created.&#xA;A life form is highly complex (once built) because of the vast chaotic detail within it.&#xA;This complexity would require vast computational resources to analyse and reconstruct.&#xA;Compare that to using nanotech to build a kilometer cube of diamond. The instructions for that would be simple enough for a child to memorise.&#xA;Note that the DNA molecules themselves are relatively simple compared to the entire adult organisim.&#xA;Note that the vast computational resources required to handle the design of lifeforms (particularly their brains) may be available if nanotech is used to build extremely powerful computers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practically every large science research group is involved in nanotech.&#xA;Current research has a long way to go to get to the stage of Drexler style mature nanotech. Most research these days is around using techniques such as molecular chemistry and integrated circuit technology to create very large volumes of material which has nanoscale features. Microelectromechanical systems (MEMS) is also a promising area.&lt;/p&gt;&#xA;" OwnerUserId="3829" LastActivityDate="2011-06-04T16:12:23.560" />
  <row Id="10769" PostTypeId="2" ParentId="1453" CreationDate="2011-06-04T17:18:20.393" Score="-3" ViewCount="" Body="&lt;p&gt;Given a small ball sitting on top of a large ball, when the small ball falls off the large ball, at what point on the large ball does the small ball depart?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was a long time ago, but neither us students nor the prof in an undergraduate dynamics course could figure it out. The question sought a general equation for the departure point as a function of the two radii and the mass of the smaller ball, and assumed no friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bonus was to be given if one could do the same for the frictional case with a ball that rolled as it fell off, thus having to include the moment of inertia of the ball. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps it is a triviality and we were just blind, but I would love to see the answer.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastActivityDate="2011-06-04T17:18:20.393" CommunityOwnedDate="2011-06-04T17:18:20.393" CommentCount="2" />
  <row Id="10770" PostTypeId="2" ParentId="10756" CreationDate="2011-06-04T18:18:21.023" Score="3" ViewCount="" Body="&lt;p&gt;A timelike vector connects two events that are causally connected, that is the second event is in the light cone of the first event. A spacelike vector connects two events that are causally disconnected, that is the second event is outside the light cone of the first event.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that sense, the timelike vector can be considered to define a four-velocity direction of an observer and thus the time axis of that observer (to be a four-velocity it should be normalised). On the other hand a spacelike vector can be considered as defining a space axis (a spatial direction) of an observer. Seen like that the two vectors define a time interval or a length on the appropriate inertial frame, as Vladimir said.&lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2011-06-04T18:18:21.023" CommentCount="2" />
  <row Id="10771" PostTypeId="2" ParentId="10756" CreationDate="2011-06-04T19:31:47.010" Score="3" ViewCount="" Body="&lt;p&gt;Although there are already three answers, they don't really answer anything. Each and every one of them fails to distinguish between vectors as part of a tangent space (e.g. four-velocity) and vectors which are present due to linear structure Minkowski possesses because it is &lt;em&gt;flat&lt;/em&gt;. The second kind of vectors are obtained as differences of the space-time events. It's very dangerous to conflate these two notions and I daresay it's a source of some of the standard SR paradoxes. The second kind of vector structure is not present in general curved space-times and one instead has to talk about curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it's not clear which type of vector you ask about. If only about the space-time event differences then this is basically what Vladimir talks about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are instead interested in elements of the tangent space (to some event) then we are talking about notions such as four-velocity and four-momentum (for the purpose of this answer I won't distinguish cotangent space from tangent space) whose physical significance is entirely different from the above notions. The squared norm of four-velocity can be one of $-c^2, 0, c^2$ depending on whether it is time-like, light-like, or space-like. The corresponding squared norm of four-momentum is squared rest mass, which respectively splits into massive observers (with subluminal speed of propagation), massless observers (with luminal speed) and particles with imaginary mass known as tachyons that propagate superluminally.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-06-05T05:45:16.267" LastActivityDate="2011-06-05T05:45:16.267" CommentCount="2" />
  <row Id="10772" PostTypeId="2" ParentId="10271" CreationDate="2011-06-04T20:11:02.263" Score="2" ViewCount="" Body="&lt;p&gt;I can't find any tabulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do a Sellmeier equation fit from visible data and extrapolate it to near-IR. BUT, the visible index is determined by the UV absorptions, so the extrapolation will only be reliable as long as you are much closer to the UV absorption bands than to the IR absorption bands. Since these solvents all have pretty strong high-frequency IR bands (CH stretch and OH stretch around 3um), you probably shouldn't extrapolate past 1um or so. (That's just a guess.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't really say any more unless I know how accurate you need (what application) and what wavelength range you're interested in. You can figure everything out with an FTIR, but it can be a bit complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADDENDUM: If you calculate the index of refraction in the range 850nm-2.5um by extrapolating from the visible, you will overestimate it. The longer the wavelength, the more severe the overestimation. That's because the effect of a resonance, e.g. the CH stretch resonance, is to increase the refractive index for wavelengths longer than the resonance wavelength, and decrease the refractive index for wavelengths shorter than the resonance wavelength. Since your extrapolation from the visible would ignore the IR resonances, it would be an overestimate.&lt;/p&gt;&#xA;" OwnerUserId="3811" LastEditorUserId="3811" LastEditorDisplayName="" LastEditDate="2011-06-05T03:36:25.543" LastActivityDate="2011-06-05T03:36:25.543" CommentCount="2" />
  <row Id="10773" PostTypeId="1" AcceptedAnswerId="10775" CreationDate="2011-06-04T20:22:23.817" Score="3" ViewCount="309" Body="&lt;p&gt;I read recently that if you hold a bullet in one hand and a pistol in the other, both hands at the same height, and subsequently fired the pistol at the same time as dropping the bullet, both bullets would hit the ground at the same time (assuming the fired bullet encounters no obstacles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true, and if so can you explain the reasoning behind it with as few equations as possible? I assume it is essentially that gravity is unaffected by momentum, but it just sounds counter-intuitive.&lt;/p&gt;&#xA;" OwnerUserId="3875" LastActivityDate="2011-06-07T21:07:47.167" Title="How does gravity affect bullets?" Tags="&lt;gravity&gt;&lt;momentum&gt;&lt;motion&gt;" AnswerCount="6" CommentCount="2" />
  <row Id="10774" PostTypeId="2" ParentId="4491" CreationDate="2011-06-04T20:24:41.190" Score="1" ViewCount="" Body="&lt;p&gt;About wavelength change: In general, if you have light at frequency f1 and it's modulated (in any way for any reason) at frequency f2, you'll get sidebands of light at frequencies f1+f2 and f1-f2. So yes, if you spin a half-wave plate at a constant speed, the light traveling through it will acquire sidebands at a shifted frequency and wavelength.&lt;/p&gt;&#xA;" OwnerUserId="3811" LastActivityDate="2011-06-04T20:24:41.190" />
  <row Id="10775" PostTypeId="2" ParentId="10773" CreationDate="2011-06-04T20:40:56.797" Score="3" ViewCount="" Body="&lt;p&gt;This is indeed true.  The easiest way to think about it is in planes.  You have horizontal and vertical planes for velocity and acceleration.  For both bullets, there is initially no vertical velocity and the only net force acting upon them is gravity.  You would then expect them to act similarly in the vertical plane, which is why they both hit the ground at the same time.  There is nothing in the forward motion of the shot bullet that counteracts the pull downward from gravity (such as lift).  I hope that explanation isn't too vague and helps you understand better what is going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I should specify that this only holds true for ideal situations (being in a vacuum, no air resistance, etc.)&lt;/p&gt;&#xA;" OwnerUserId="3876" LastEditorUserId="3876" LastEditorDisplayName="" LastEditDate="2011-06-04T21:16:00.973" LastActivityDate="2011-06-04T21:16:00.973" CommentCount="5" />
  <row Id="10776" PostTypeId="2" ParentId="7251" CreationDate="2011-06-04T20:49:18.733" Score="2" ViewCount="" Body="&lt;p&gt;The simplest &quot;light diode&quot; (optical isolator) (&lt;a href=&quot;http://en.wikipedia.org/wiki/Optical_isolator&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;) is two polarizing sheets 45 degrees apart, with a 45 degree Faraday rotator in between. The polarizing sheets absorb light with one polarization and transmit it with the opposite polarization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check the second law of thermodynamics, you need to assume the polarizing sheets are at the same temperature as the other things you're thinking about. Therefore the polarizing sheets will emit their own blackbody radiation, always polarized in the direction that they absorb  light. The device violates the second law of thermodynamics ONLY if you FORGET the blackbody radiation of the polarizing sheets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things get more complicated when you replace &quot;polarizing sheets&quot; (which emit blackbody radiation) with &quot;polarizing beamsplitters&quot; (which don't). A polarizing beamsplitter sends each light-polarization in a different direction, for example it might back-reflect one polarization and transmit the other. But no matter where you send the non-transmitted light polarization, you will always end up causing problems for your perpetual motion machine. You can try drawing it out to see the various ways this happens. &lt;a href=&quot;http://www.usna.edu/Users/physics/mungan/Scholarship/FaradayIsolators.pdf&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; (linked by Andrew above) goes through some of the possibilities in detail. :-)&lt;/p&gt;&#xA;" OwnerUserId="3811" LastActivityDate="2011-06-04T20:49:18.733" />
  <row Id="10777" PostTypeId="1" AcceptedAnswerId="10784" CreationDate="2011-06-04T20:52:43.903" Score="0" ViewCount="101" Body="&lt;p&gt;Is it possible to determine the position of a particle undergoing circular motion, in x-y coordinates, at any given time and velocity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking it has something to do with $\pi$&lt;/p&gt;&#xA;" OwnerUserId="3821" LastEditorUserId="3821" LastEditorDisplayName="" LastEditDate="2011-06-14T17:23:48.013" LastActivityDate="2011-06-14T17:23:48.013" Title="Position in circular motion" Tags="&lt;kinematics&gt;&lt;motion&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="10778" PostTypeId="1" AcceptedAnswerId="11110" CreationDate="2011-06-04T20:55:06.933" Score="0" ViewCount="91" Body="&lt;p&gt;i having trouble with this formula (asked first in math.Se, i Didn't know the existence of physics.se)&#xA;$$&#xD;&#xA;Z1(t) = Ze+(\sqrt{Z1-Ze}-\frac{2S0}{S1}\sqrt{2g(1+\frac{S1}{S2})}.t)&amp;#178;&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z1 and Z2 are the heights of the vessels.&#xA;S1 and S2 are the sections of the vessels.&#xA;S0 is the section of the tube between them (for the exchange).&#xA;Ze is the final height of the two vessels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z1(t) is the height at t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but i only get incorrect values and don't know if the formula is wrong of if its me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for testing is use Z1 = 45&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z2 = 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S0=2$\pi$0.3=1.884&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S1=2$\pi$10=62.8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S2=2$\pi$10=62.8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Ze = \frac {S1.Z1 + S2.Z2} {S1+S2}=\frac {62.8.45 + 62.8.5} {62.8+62.8}=25$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="3877" LastActivityDate="2011-06-15T08:28:11.687" Title="communicating vessels formula" Tags="&lt;fluid-dynamics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10779" PostTypeId="1" CreationDate="2011-06-04T20:59:02.417" Score="4" ViewCount="159" Body="&lt;p&gt;The title says it all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that a cubit is a superposition of $|0&amp;gt;$ and $|1&amp;gt;$, i.e. the answer to a binary question. So I imagine that specifying an event in spacetime would require an infinite sequence of qubits. I'm thinking of a binary search tree. Is there some other way to do it with a finite amount of qubits?&lt;/p&gt;&#xA;" OwnerUserId="2953" LastActivityDate="2011-08-27T06:35:21.243" Title="How many qubits does it take to specify an event in spacetime?" Tags="&lt;quantum-information&gt;" AnswerCount="2" />
  <row Id="10780" PostTypeId="2" ParentId="10773" CreationDate="2011-06-04T21:06:25.197" Score="4" ViewCount="" Body="&lt;p&gt;The bullets don't hit the ground at the same time exactly because it is very difficult to fire horizontally, there is air resistance to account for, the ground may be sloped, there are Coriolis forces, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, when people refer to this phenomenon, they're referencing the principle of &lt;a href=&quot;http://en.wikipedia.org/wiki/Galilean_invariance&quot; rel=&quot;nofollow&quot;&gt;Galilean relativity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read the famous excerpt from Galileo's book &lt;a href=&quot;http://en.wikipedia.org/wiki/Galileo%27s_ship&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  He considered sailing on a smoothly-sailing ship, and noted that if you're in the hull, you don't notice the ship's motion in any physics experiment.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, things seem to fall completely normally when you're in a moving ship.  Someone standing on the shore watching would think that objects in the ship are falling in parabolic trajectories with constant forward speed equal to the ship's speed, but someone in the ship thinks thinks they fall straight down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine two bullets.  The first is dropped by someone on the ship.  The second is fired from a gun on the shore, but fired at exactly the same speed as the ship's speed (it's either a slow bullet or a fast ship, or both).  The bullets start out with exactly the same trajectory.  They have no way of knowing whether they're supposed to keep up with the ship or the shore or what - they only know their location and velocity.  So the two bullets must fall in the same way.  By Galilean relativity, the bullet on the ship falls in the same amount of time as a bullet that's simply dropped, so the bullet fired from the gun falls in the same amount of time as a bullet that's simply dropped, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to view the math is as a consequence of Newton's laws of motion, which state&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = m \frac{\textrm{d}v}{\textrm{d}t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we transfer reference frames from the shore (frame S) to the ship (frame S'), we're transforming the velocities according to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v&amp;#39; = v + v_s$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $v_s$ the constant speed of the ship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F&amp;#39; = m \frac{\textrm{d}v&amp;#39;}{\textrm{d}t} = m \frac{\textrm{d}v}{\textrm{d}t} = F$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so physics appears to work the same way in both frames (this assumes that mass is invariant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the relativity principle is deeper than Newton's laws, and holds in theories of physics that are more fundamental than Newtonian mechanics, so it also makes sense to take relativity as more of a starting point than a consequence.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-06-04T21:06:25.197" CommentCount="5" />
  <row Id="10781" PostTypeId="2" ParentId="10773" CreationDate="2011-06-04T21:08:44.743" Score="1" ViewCount="" Body="&lt;p&gt;Yes that is true. The earth's surface gravity accelerates every object at approximately $9.8 m/s^2$ ($32.2 ft/s^2$). Objects sometimes will fall slower because of air resistance or friction (like a feather).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you can manage to fire a projectile at more than 18,000mi/hr it will go in orbit of the earth and not fall!&lt;/p&gt;&#xA;" OwnerUserId="3821" LastEditorUserId="3821" LastEditorDisplayName="" LastEditDate="2011-06-04T21:42:08.507" LastActivityDate="2011-06-04T21:42:08.507" CommentCount="5" />
  <row Id="10782" PostTypeId="1" AcceptedAnswerId="10897" CreationDate="2011-06-04T21:29:24.657" Score="5" ViewCount="193" Body="&lt;p&gt;If an observer were to rotate around a point at near light speeds, what sort of length contraction would he observe the universe undergo?&lt;/p&gt;&#xA;" OwnerUserId="202" LastEditorUserId="202" LastEditorDisplayName="" LastEditDate="2011-06-04T23:31:54.693" LastActivityDate="2011-06-08T00:06:52.653" Title="Length contraction in rotating frame?" Tags="&lt;general-relativity&gt;&lt;relativity&gt;&lt;length-contraction&gt;" AnswerCount="1" />
  <row Id="10783" PostTypeId="2" ParentId="10777" CreationDate="2011-06-04T21:32:10.877" Score="1" ViewCount="" Body="&lt;p&gt;Of course. How do you represent a position in 2d space? as a couple of numbers, right? Just make them variables, functions in respect to time. The following functions describe circular motion (with radius a):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x(t)=a \sin(t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y(t)=a \cos(t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, go to this site: &lt;a href=&quot;http://fooplot.com&quot; rel=&quot;nofollow&quot;&gt;http://fooplot.com&lt;/a&gt; , change the style to parametric (they use $s$ instead of $t$) and use the above functions. Try different functions :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if I answered your question.. What level are you?&lt;/p&gt;&#xA;" OwnerUserId="301" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-06-04T21:43:47.690" LastActivityDate="2011-06-04T21:43:47.690" CommentCount="2" />
  <row Id="10784" PostTypeId="2" ParentId="10777" CreationDate="2011-06-04T21:37:18.967" Score="1" ViewCount="" Body="&lt;p&gt;Yes, unless you want to be very picky.  The coordinates are described by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Trigonometric_functions&quot; rel=&quot;nofollow&quot;&gt;trig functions sine and cosine&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose something is moving on a circle with radius $R$, so its $x$ and $y$ coordinates obey&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 + y^2 = R^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it moves at a uniform velocity of $v$, then the angle its path subtends is linear in time (by dimensional analysis), so we can define the angle $\theta$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta = vt/R$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One full rotation turns out to be an angle of about 6.28, but it's really a transcendental number, &lt;a href=&quot;http://en.wikipedia.org/wiki/Pi&quot; rel=&quot;nofollow&quot;&gt;$2\pi$&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you draw a circle, then draw the angle from the equation above, you find a unique intersection point that tells you where the object is.  If you project that point onto the $x$ and $y$ axes, you've graphically determined the coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/g23n1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coordinate you obtain, for a given angle, is linear in $R$ (again by dimensional analysis, coupled with a feature of Euclidean geometry - that it has no innate length scale), so if we divide the coordinates by $R$ we get two functions that map $[0,2\pi) \to [-1,1]$ because $-1$ and $1$ are the minimum and maximum values, since $-R$ and $R$ are the minimum and maximum values of the coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These functions are give the names &quot;cosine&quot; and &quot;sine&quot;, and we say&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x/R = \cos\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y/R = \sin\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sine and cosine functions can be represented as infinite series,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos\theta = 1 - \theta^2/2 + \theta^4/24 - \ldots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sin\theta = \theta - \theta^3/6 + \theta^5/120 - \ldots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know many of the properties of these functions, and they are used extensively in mathematics and physics.  You can calculate their values to high precision with a computer.  However, we cannot simply write down exactly what $\sin(1)$ is in any simple way that you can explain to an elementary-school student.  If you want to know what it is, you can simply use a calculator, like &lt;a href=&quot;http://www.google.com/search?q=sine%281%29&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-06-04T21:37:18.967" CommentCount="4" />
  <row Id="10786" PostTypeId="2" ParentId="1074" CreationDate="2011-06-04T22:04:13.130" Score="0" ViewCount="" Body="&lt;p&gt;Quantum effects, as others have pointed out are negligible in most wires. If you want to derive from a particle like description of the wires, you'd be best off thinking in terms of gas/fluid dynamics (which is believed to in the continuous limit turn into the transport diffusion equations). You can simulate this on a computer by having an many electrons move ballistically through the material, and randomly scatter with some probability $p$ each time step. This model will give you Ohm's law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now granted, if you actually wanted to figure out what the probability $p$ is, and the effective resistance of the wire, things get really complicated and do indeed require quantum mechanics. If you want the math, you'd want to look into the theory of quantum mechanical scattering. The main idea is that an electron will be in a pure momentum p eigenstate, and get randomly placed into a new p' momentum eigenstate. Averaged over a bunch semi-classical particles, the I-V characteristics work out. The justification for this is in truth, somewhat sketchy (the above described simulation has not been formally proved to be the same as using the Boltzmann Transport and Diffusion Equations, we just think it does for appropriate limits).&lt;/p&gt;&#xA;" OwnerUserId="668" LastActivityDate="2011-06-04T22:04:13.130" />
  <row Id="10787" PostTypeId="2" ParentId="10779" CreationDate="2011-06-05T00:38:26.853" Score="0" ViewCount="" Body="&lt;p&gt;You'd never need a infinite sequence as of course there's a limit to the accuracy of the&#xA;a position measurement. Given either by plancks length (but no measurement gets near planck&#xA;length, because no imaging beam get near planks energy), in the non-planckian regime, you can measure a position by scattering a beam of light energy E= hv, and get a position accurate to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta x = \frac{\hbar}{2p} = \frac{\hbar c}{2E} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your starting in a volume V a box you already know the particle is in, then you need,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ B = \log_2 \left ( \frac{V}{\Delta x^3}\right) = 3\log_2 \left ( \frac{ VE^3 }{\hbar^3c^3} \right) $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bits of information to describe where you measured the particle to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Qubits are more complicated though, using qubit instead of digital bits, you can describe&#xA;a destribution of the probabilities of the particle being at each location inside the volume. A measurement yielding Qubits might does not need to collapse the wavefunction, so&#xA;may describe the amplitude of each location.&lt;/p&gt;&#xA;" OwnerUserId="3864" LastActivityDate="2011-06-05T00:38:26.853" CommentCount="1" />
  <row Id="10788" PostTypeId="2" ParentId="10773" CreationDate="2011-06-05T01:24:44.877" Score="2" ViewCount="" Body="&lt;p&gt;To simplify this question and equalizing the atmospheric conditions, instead of bullets falling to the ground, consider yourself jumping up and falling back to the ground. Compare yourself (i) jumping on a stationary platform, and (ii) jumping on a fast moving train, do you think the two cases would be different? Do you feel lighter/heavier when you're on a train than when you're not? Why do you think so?&lt;/p&gt;&#xA;" OwnerUserId="3879" LastActivityDate="2011-06-05T01:24:44.877" />
  <row Id="10789" PostTypeId="1" AcceptedAnswerId="10790" CreationDate="2011-06-05T03:44:01.937" Score="3" ViewCount="179" Body="&lt;p&gt;&lt;em&gt;A while ago, I was hiking near the Hollywood Sign in Los Angeles. When I got to the sign, I found out it was fenced off; with several signs and a security camera promising prompt enforcement. As I was pondering the likelihood it was a bluff, I ran into this sign.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ly75I.jpg&quot; alt=&quot;Sign&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radio waves aren't ionizing like X-rays or Gamma-rays. So assuming an arbitrarily large dose, in what manner do the radio waves interact with the human-body? &lt;/p&gt;&#xA;" OwnerUserId="800" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-05T09:08:52.787" LastActivityDate="2011-06-05T10:56:37.170" Title="Adverse Health Effects of Radio Waves" Tags="&lt;electromagnetic-radiation&gt;&lt;biophysics&gt;&lt;radio&gt;&lt;antennas&gt;&lt;rf&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="10790" PostTypeId="2" ParentId="10789" CreationDate="2011-06-05T08:31:45.197" Score="4" ViewCount="" Body="&lt;p&gt;They heat it, by different degrees depending on the polarization of molecules in the tissues and liquids. The molecules try to re-align after the radio-wave field and the movement dissipates as general heat. Think microwaves.. a consumer-grade &lt;a href=&quot;http://en.wikipedia.org/wiki/Microwave_oven&quot; rel=&quot;nofollow&quot;&gt;microwave oven&lt;/a&gt; operates at the same radio wave spectrum as your home WiFi network (2.4 GHz) but much stronger. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The SAR, &lt;a href=&quot;http://en.wikipedia.org/wiki/Specific_absorption_rate&quot; rel=&quot;nofollow&quot;&gt;Specific Absorption Rate&lt;/a&gt; measures the amount of RF-energy deposited in the body in a well specified way (in W/kg) from a radio product. Cellphones have to be rated with this to be sold. I don't think the Hollywood Sign is rated with a SAR though... :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, the SAR for cellphones is measured using a dummy human head with similar dielectric properties as a real human head. This can pose problems when engineers try to bring these dummies as carry-on luggage through airport security screenings, a friend who works at a cellphone company mentioned ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been reports now and then on RF-waves doing more complicated, specific things with tissues like opening the blood-brain barrier etc. As far as I know none of these have been conclusively proved (in a human model) but this area of course features incredible pressure from health organizations, governments and huge industrial companies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There usually is an &quot;old story&quot; from people in the military service where they heat water-bottles by hanging them in front of their mobile radar station.. I don't know if that is true in particular, but it can't be very good to hang out in front of strong omnidirectional radio-emitters even for no other reason than the tissue-heating :)&lt;/p&gt;&#xA;" OwnerUserId="2936" LastEditorUserId="2936" LastEditorDisplayName="" LastEditDate="2011-06-05T10:56:37.170" LastActivityDate="2011-06-05T10:56:37.170" CommentCount="3" />
  <row Id="10791" PostTypeId="2" ParentId="10754" CreationDate="2011-06-05T10:10:00.103" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I have read that string theory predicts (or requires ?) the existence of gravitons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;String theory models contain massless spin-2-particles called gravitons, therefore they &lt;em&gt;postdict&lt;/em&gt; their existence. The fact that the exchange of such particles results in a GR-like gravitational force was discovered some time before string theory came into being, around 1950, and made popular by Richard Feynman in his lectures on gravitation (see &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0813340381&quot; rel=&quot;nofollow&quot;&gt;amazon&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So, would that make it a quantum theory of gravity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Classical wavelike phenomena correspond to quantum mechanical particles (&quot;wave-particle duality&quot;), and since GR describes waves, many people believe that a quantum version should describe gravitons. On the other hand, string theory works with a classical GR background, so that it can describe quantum corrections to a classical situation only. It is an open problem if that is already all of the story, or if a quantum theory of gravity would need to alter the concept of spacetime of classical GR on a more fundamental level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To be fair: string theorists know and think about this problem, too, and are working on versions of string theory that incorporate a more fundamental change to spacetime.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If so, I have also read that quantum gravity would allow us to understand what happens at the singularity inside a black hole. What does string theory say about this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A basic assumption is this: In a consistent theory of gravitation there should not be any singularities with respect to physically relevant phenomena, and the existence of black hole singularities in GR tell us that classical GR breaks down and should be replaced by a quantum version of gravitation. String theory does not say anything about this, because, today, all formulations of string theory assume the existence of a classical spacetime. A full theory of quantum gravity may be able to explain what happens when a black hole forms in classical GR, for example. Maybe string theory will be able to do so in some future version, but not now (see my previous comment). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If string theory can't make any predictions about this, what are we missing?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Right know there are more answers to this questions than there are heads on this planet thinking about it. The problem is that string theory starts with an unmotivated and unexplainable assumption about the nature of particles, so there is basically no guidance at all what to do when you run into problems. It's not like that you have a set of axioms derived from a well established theory, run into inconsistencies, and go back to the drawing board to see what axiom may have to be replaced...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Or is it that we have multiple string theories - hence, multiple ways to quantize gravity - and don't know which is the right one?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All string theory variants try to do the quantization of gravity the same way (exchange of gravitons on a classical background), so the existence of a bunch of different &quot;string theories&quot; should not be mixed up the the existence of multiple approaches to the quantization of gravity. The question if it is possible to extract the &quot;right&quot; string theory from the plethora of possible ones is a topic of ongoing debate. As for now, it's basically up to you what you believe.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2011-06-05T10:10:00.103" CommentCount="1" />
  <row Id="10792" PostTypeId="2" ParentId="8454" CreationDate="2011-06-05T10:11:33.590" Score="1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Theatrical_smoke_and_fog&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Theatrical_smoke_and_fog&lt;/a&gt; describes the methods used to create smoke. Also I think, in a room filled with people (like at evening parties), the humidity gets high enough to make light beams visible.&lt;/p&gt;&#xA;" OwnerUserId="3694" LastActivityDate="2011-06-05T10:11:33.590" />
  <row Id="10793" PostTypeId="1" AcceptedAnswerId="10810" CreationDate="2011-06-05T14:10:02.540" Score="7" ViewCount="254" Body="&lt;p&gt;I am interested in why many small animals such as ants can lift many times their own weight, yet we don't see any large animals capable of such a feat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been suggested to me that this is due to physics, but I am not even sure what to search for. Could someone explain why indeed it is easy for smaller objects/lifeforms to support several times their own weight, but this is harder as objects/animals become larger?&lt;/p&gt;&#xA;" OwnerUserId="3885" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-06-05T14:23:32.693" LastActivityDate="2011-06-07T23:59:33.193" Title="Why are smaller animals stronger than larger ones, when considered relative to their body weight?" Tags="&lt;biology&gt;&lt;dimensional-analysis&gt;&lt;scaling&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="10794" PostTypeId="1" CreationDate="2011-06-05T14:30:32.940" Score="2" ViewCount="93" Body="&lt;p&gt;This comes from one of my physics course at university (many years ago!), I hope I recall it right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We were studying optics, and were given some polarizer filters to experiment. They were disc- shaped (like a photographic filter). I stacked two polarizers, face to face, and I verified that, when rotating one of them, a particular orientation (and its 180 degrees rotation) blocked the light. That seemed to make perfect sense to me, because I pictured the polarizer as a wire grid (conceptually like this image) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/snSEH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that only let pass light in a given polarization direction: then, adding a second polarizer to the other behind had no effect when the direction of the wires coincided, and result in high light blocking if they were orthogonal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We were all satisfied with this explanation... but then I tried to reverse the front filter (I mean, I turned it face down, always keeping it in front of the other, and keeping its &quot;orientation&quot;). To  my surprise, keeping the orientation did not keep the block-pass behaviour. Say, if originally the filters were at the &quot;let it pass&quot; position (paralells wires, in my mental picture), putting the filter face down (which would let the imaginary wires in the same orientation) resulted in light blocking, and I had to turned it 90 degrees to let pass the light. &#xA;This behaviuor puzzled me a little, as also puzzled my teaching assistant (who didn't come with any explanation). I just assumed that my &quot;grid wire&quot; picture was not very apt, but that was all (not enough scientific spirit on my part, I guess).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody make some sense of this? &lt;/p&gt;&#xA;" OwnerUserId="1064" LastEditorUserId="1064" LastEditorDisplayName="" LastEditDate="2011-06-05T16:35:30.647" LastActivityDate="2011-06-05T20:23:43.887" Title="Intrigued about a polarizer effect" Tags="&lt;optics&gt;&lt;polarization&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="10795" PostTypeId="2" ParentId="10794" CreationDate="2011-06-05T14:56:47.977" Score="1" ViewCount="" Body="&lt;p&gt;Your &quot;grid wire&quot; picture is essentially correct.  One possible explanation for the weird observation is that the polarization of the filter was not aligned with the edges of the piece of polarizer.  That is, you imagined that the wires in the polarizer run parallel to the edge of the polarizer, but they were actually tilted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the wires run at a 45-degree angle to the edge, then turning the polarizer around will make the wires go from tilting up at 45 degrees to tilting down at 45 degrees - an effective 90 degree rotation overall.  This would explain why it blocked light after being flipped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could achieve the same effect with the more typical polarizers in your picture by rotating around the axis of the diagonal, rather than around an axis parallel to a side.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-06-05T14:56:47.977" CommentCount="3" />
  <row Id="10796" PostTypeId="2" ParentId="10794" CreationDate="2011-06-05T15:46:00.770" Score="1" ViewCount="" Body="&lt;p&gt;What you are seeing is likely the result of the the light becoming polarized by virtue of the fact that it is now entering the surface of the tilted glass filter at something other than a 90° angle. When light passes into a refractive medium at 90° all polarizations of the incident beam are treated equally. But when the beam of light enters at an angle to the normal to the surface, the reflected portion starts to favor those photons whose electric field vector is parallel to the surface. As a result the transmitted light is deficient in those polarizations and becomes polarized at right angles to the reflected portion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now you have two polarizing grids competing with a polarization component due to the tilted filter. Strange things happen.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-06-05T20:23:43.887" LastActivityDate="2011-06-05T20:23:43.887" CommentCount="4" />
  <row Id="10797" PostTypeId="2" ParentId="9805" CreationDate="2011-06-05T16:00:29.867" Score="1" ViewCount="" Body="&lt;p&gt;Forgive me for answering my own question, but with some inspiration from the other answers here and some child like thought I believe I have found a great way to help my son gain an intuitive understanding of why spinning tops don't fall over. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, I asked my son to imagine a single marble contained within a box. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then asked him to guess what might happen to the marble if the box was pushed in a straight line from left to right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I asked him what would happen to the marble if the moving box suddenly stopped. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was to encourage him to think about the marbles apparent reluctance to start moving and reluctance to stop moving once it had started. I told him that scientists like to call this inertia. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then asked him to imagine a set of boxes, each containing a marble, attached to the circumference of a wheel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time I asked him to guess what might happen to the marbles in each box when the wheel started spinning, when it remained spinning at a constant speed and finally when the wheel stopped spinning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was to stimulate his thinking about the reluctance of the marbles to start moving, reluctance to stop moving once they had started and the marbles apparent desire to push outwards from the centre of the wheel. I told him that scientists like to call these phenomena rotational inertia and the centrifugal force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, at this point I asked him to imagine what would happen to the marbles in the boxes if we tried to twist the rotating wheel about its axis. Twisting the wheel about its axis would effectively pull two sides of the wheel closer to the axis of rotation of the marbles. The key here is to appreciate the independence of the marbles axis of rotation from that of the boxes attached to the wheel. The outward force of the marbles away from their central axis of rotation would resist this change, thereby causing a resistance to the twisting action on the axis of the wheel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, thinking about individual atoms within a solid body as tiny little marbles locked within their own tiny little boxes made up of adjacent atoms gives him an intuitive way to appreciate why the spinning top does not fall over. No formulas required! &lt;/p&gt;&#xA;" OwnerUserId="3525" LastActivityDate="2011-06-05T16:00:29.867" CommentCount="1" />
  <row Id="10798" PostTypeId="1" CreationDate="2011-06-05T17:23:19.373" Score="2" ViewCount="271" Body="&lt;p&gt;I'm developing a program where I've a ball/sphere rolling in a bowl from the side at top, till the center at bottom, and I'm trying to get the formulas for:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The &lt;strong&gt;rotation angle&lt;/strong&gt; and the &lt;strong&gt;position of the sphere&lt;/strong&gt;, depending on the starting angle measured in radians (0.000 - 6.283), the current seconds and the maximum seconds that the ball will keep rotating and considering its deccelerated.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All I knew about the needed physics is almost already forgotten, and I'm getting horrible headaches to get a properly working formula, so hope anyone can help me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="3547" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-08T08:41:40.667" LastActivityDate="2011-08-07T10:20:15.803" Title="Formulas for ball rolling in a bowl?" Tags="&lt;kinematics&gt;&lt;acceleration&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" />
  <row Id="10799" PostTypeId="1" CreationDate="2011-06-05T17:51:45.183" Score="0" ViewCount="60" Body="&lt;p&gt;I need to construct a universal quantum circuit decomposition for a three-qubit operation where one qubit is the control bit, and a two-qubit unitary operator acts on the other two&#xA;depending on the control bit. I can do the decomposition when there's a one-qubit unitary operation with as many control bits as needed; this two-qubit unitary with one control bit I am not clear about. Anyone has any clue what the recipe for this construction is?&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2011-06-05T17:51:45.183" Title="Quantum circuit decomposition" Tags="&lt;quantum-mechanics&gt;&lt;quantum-computation&gt;" CommentCount="1" />
  <row Id="10800" PostTypeId="1" AcceptedAnswerId="10858" CreationDate="2011-06-05T18:28:49.227" Score="2" ViewCount="216" Body="&lt;p&gt;The Hall effect can be used to determine the sign of the charge carriers, as a positive particle drifting along the wire and a negative particle drifting the other direction get deflected the same (as $F = q \vec{v}\times\vec{B} = (-q) (-\vec{v})\times\vec{B}$). But I don't understand how positive charge carriers are ever possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I've understood, a positive hole is nothing more than the absence of an electron. As all the electrons in the valence band are still negatively charged, why would this hole behave in a magnetic field as if it were positive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a hole is created if an electron is excited into the conductance band. If there is always the same numbers of holes as electrons, how can any Hall effect ever occur?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much&lt;/p&gt;&#xA;" OwnerUserId="3234" LastActivityDate="2011-06-07T08:22:02.890" Title="How can the Hall effect ever show positive charge carriers?" Tags="&lt;solid-state-physics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="10802" PostTypeId="2" ParentId="10793" CreationDate="2011-06-05T19:56:46.367" Score="-2" ViewCount="" Body="&lt;p&gt;To lift anything, a life form on this planet needs muscles. If you want to lift heavier things you need&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;more muscles and&lt;/li&gt;&#xA;&lt;li&gt;a stronger body/legs to support that additional weight.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Stronger muscles and bones need to be supplied with more oxygen, nutrion and so on which leads to needing a stronger heart, a better digestive system and so on. Or simpler: if you want to be stronger, you will get heavier and you need to support your own weight too. And the additional weight increases faster then the additional force you can get out of that.&lt;/p&gt;&#xA;" OwnerUserId="3889" LastActivityDate="2011-06-05T19:56:46.367" CommentCount="1" />
  <row Id="10804" PostTypeId="1" CreationDate="2011-06-05T21:31:34.057" Score="7" ViewCount="327" Body="&lt;p&gt;When a system is expressed in terms of creation and annihilation operators for bosonic/fermionic modes, what exactly is the physical meaning of the order in which the operators act?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for a fermionic system with states $i$ and $j$, $c_i c_j^\dagger$ is different from $c_j^\dagger c_i$ by a sign change, due to anticommutativity. I understand the mathematics of this, but what does it mean intuitively?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The former would be described as destroying a particle in state $j$ &quot;before&quot; creating one in state $i$, but what does &quot;before&quot; actually mean in this context, since there's no notion of time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As another (bosonic) example, $a_i^\dagger a_i$ is clearly different from $a_ia_i^\dagger$, since acting the former on a vacuum state $|0\rangle$ gives zero while the for the latter, $|0\rangle$ is an eigenstate, but again, what is the physical interpretation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My normal interpretation of commutativity as a statement regarding the effect of a measurement on a state fails here since creation/annihilation are obviously not observables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope the question makes sense and isn't too abstract!&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="354" LastEditorDisplayName="" LastEditDate="2011-06-05T22:06:47.570" LastActivityDate="2011-07-28T09:03:11.647" Title="What does the ordering of creation/annihilation operators mean?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;second-quantization&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="10805" PostTypeId="2" ParentId="615" CreationDate="2011-06-05T21:40:31.800" Score="0" ViewCount="" Body="&lt;p&gt;The &quot;speed&quot; of a photographic lens, and the brightness of the image produced by any lens is determined by a single number - the focal ratio. That is the ratio of 1) the distance between the lens and the image (of an infinitely distant object) and 2) the diameter of the lens. It is basically the ratio of collecting area to image area. You can increase it by moving the image closer to the lens but only up to a point where you reach the lens itself. This principle is the reason that no concentrator lens can produce a spot of light on a surface and make that surface achieve a temperature greater than that of the visible surface of the sun.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastActivityDate="2011-06-05T21:40:31.800" />
  <row Id="10807" PostTypeId="2" ParentId="10773" CreationDate="2011-06-05T21:45:46.377" Score="3" ViewCount="" Body="&lt;p&gt;Mark came the closest &lt;a href=&quot;http://physics.stackexchange.com/questions/10773/how-does-gravity-affect-bullets/10780#10780&quot;&gt;in his answer&lt;/a&gt;, but I want to address some of the deviations from idealism in greater detail.  People have mentioned the angle the gun is at and air resistance, so I'll try to touch on both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a height $h$, say the velocity of the bullet is $\vec{V_b} = &amp;lt; V_{bx}, V_{by}&amp;gt;$ (add $t$ dependence when necessary).  The time it takes to fall will be $\sqrt{2h/g}$ in a vacuum based on ideal Newtonian kinematics.  The idea in the frictionless world is that the horizontal velocity of the bullet remains $V_p$ and the resultant velocity over time will be a combination of that and the vertical velocity caused exclusively by gravity, $g t$.  I will use $&amp;lt;x, y&amp;gt;$ vector notation, and I'll just use a big vector arrow to indicate what is shorthand for a vector.  For the other ball that just falls, I'll denote it $\vec{V_f} = &amp;lt;V_{fx},V_{fy}&amp;gt;$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{V_b}(t) = &amp;lt; V_p , -g t&amp;gt;$$&#xA;$$\vec{V_f}(t) = &amp;lt; 0, -g t&amp;gt;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, air resistance (drag) is usually taken to be some proportional relationship to velocity, in the opposite direction of the velocity vector.  I'll take that relationship to be the common form of a constant times the magnitude of velocity squared (linearly $a_b = -C V^2$).  Then I'll write the full differential equations for the problem... in some fairly extreme vector shorthand notation.  Take $r$ to be the position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{a_b}(t) = -\frac{\vec{V_b}(t)}{|\vec{V_b}(t)|} C |\vec{V_b}(t)|^n = -\vec{V_b}(t) C |\vec{V_b}(t)|^{2} $$&#xA;$$ \vec{r_b}&amp;#39; = \vec{V_b}$$&#xA;$$ \vec{V_b}&amp;#39; = \vec{a_b}$$&#xA;$$ \vec{r_b}(0) = &amp;lt;0,h&amp;gt;$$&#xA;$$ \vec{V_b}(0) = &amp;lt;0,0&amp;gt;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very good assumption for this experiment if it was performed at a small height (which is any height you could practically set up the experiment at), the velocity of the bullet will be much greater than the final velocity due to gravity.  The convenient simplification that arises is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\vec{V_b}(t)| = (V_{bx}^2 + V_{by}^2)^{\frac{1}{2}} \approx V_{by}(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This allows a very nice closed form solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{V_b}(t) = \left&amp;lt; \frac{V_p }{ C t V_p + 1 } , -g \frac{ \left( \frac{C t V_p }{2} + t \right) }{C t V_p + 1} \right&amp;gt;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, I can say the bullet will hit the ground at the same time as something dropped at the same time as firing when a certain condition is satisfied, which is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ C t V_p \ll 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up until now I've used a non-standard drag coefficient.  So I'll convert the equation to using the standard drag coefficient and then to ballastic coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{ C_d \rho A t V_p }{ 2 m} =\frac{ \rho t V_p }{ 2 BC} \ll 1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is the first of my answers.  For the other, I'll very simply address the angular sensitivity.  The horizontal distance traveled by the bullet is much greater than the vertical distance it falls, but it still forms a triangle.  We require that the displacement from the horizontal at the end of the flight due to the inclination of the gun must be much less than the height formally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  V_p t \sin{\theta} \approx V_p t \theta \ll h$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \theta \ll \frac{h}{ V_p t} = \frac{1}{V_p} \sqrt{\frac{g h}{2}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say that the gun is at a height $h=20 m$.  Bullet speed would fairly be $V_p=500 m/s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \theta \ll 0.028 rad = 0.802^{\circ} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this value is just a value sufficient to &lt;em&gt;completely screw up the experiment&lt;/em&gt; so you would ideally want it 10x or more smaller than this.  For the other part, dealing with the drag, I'll take some very general numbers.  A drag coefficient for a generic cone is about $C_d = 0.5$, so I'll use that.  For a bullet we'll say $m=10 g$.  Air, $\rho=1.2 kg/m^3$.  Bullet area we'll take as $A=0.2 m^2$.  For 20 m height we have $t=2 s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{ C_d \rho A t V_p }{ 2 m} = 0.006$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it seems my answer is that the gun must be held very still, but provided it's an &quot;ordinary&quot; gun and my numbers are correct, yes, the bullet will hit the ground at pretty much the same time.  Hopefully I've formalized a little bit exactly &lt;em&gt;when&lt;/em&gt; this is true.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-06-05T21:45:46.377" />
  <row Id="10808" PostTypeId="2" ParentId="10800" CreationDate="2011-06-05T21:59:43.973" Score="0" ViewCount="" Body="&lt;p&gt;&quot;...why would this hole behave in a magnetic field as if it were positive?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct that a positively charged hole is nothing but a lack of an electron. It is positively charged because there is a proton in that atom that does not have an electron to balance its charge.&#xA;Holes move around by having an electron cross the hole to get to the other side. For example. A hole is located at 200 Elm Street. There is no house there. Just an empty lot. The house from 198 Elm street is moved onto the 200 lot. A house has moved to the right and a hole has moved to the left. This is how a hole behaves in a magnetic field as if it were positive.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-06-05T23:35:06.907" LastActivityDate="2011-06-05T23:35:06.907" CommentCount="6" />
  <row Id="10809" PostTypeId="2" ParentId="10804" CreationDate="2011-06-05T22:18:42.620" Score="1" ViewCount="" Body="&lt;p&gt;Products of these operators describe transitions of particles from one state to another, nothing else. QFT is about evolution of populations of particle states due to interactions. Consider a regular potential theory of scattering of non-relativistic particles (with no creation/annihilation operators) and then rewrite it formally with help of those operators to see what happens. Keep in mind that only the whole expression makes sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; In a usual description the particle momentum changes in course of scattering: $p = p(t)$. In the operator description the particle with $p(t_1)$ disappears and the particle with $p(t_1+dt)$ appears in course of interaction.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-06-06T09:44:12.220" LastActivityDate="2011-06-06T09:44:12.220" CommentCount="2" />
  <row Id="10810" PostTypeId="2" ParentId="10793" CreationDate="2011-06-06T01:13:40.120" Score="5" ViewCount="" Body="&lt;p&gt;Zasso pointed it already out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scaling up a ant to human size means volume (weight) increasing by length proportional $l^{3}$, but the force of muscles is determined by cross section (not muscle weight), so muscle force goes proportioal to $l^{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smaller factors are likely:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;stiffness (or strentgh of the skeleton)&lt;/li&gt;&#xA;&lt;li&gt;balance point (center of mass)&lt;/li&gt;&#xA;&lt;li&gt;leverage (human skeleton is &quot;sub-optimal&quot; for this, we are afaik best optimized by evolution for long runs, more than any other animal)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;i did some quick further search on &quot;robot insects&quot; on this interesting topic. &lt;a href=&quot;http://www.nanowerk.com/spotlight/spotid=2231.php&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; is quite worth reading and relating biological to technological limits as well as current state of the art in nanobionics:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Interestingly, the force generated&#xA;  from a wide variety of actuator&#xA;  materials and devices has been found&#xA;  to be surprisingly invariant when&#xA;  compared with the actuator mass. A few&#xA;  years back, &lt;strong&gt;a comparison of the&#xA;  force-to-weight ratio of various&#xA;  organisms and machines found a&#xA;  striking similarity, with the force&#xA;  scaling linearly with mass over 20&#xA;  orders of magnitude – from individual&#xA;  protein molecules to rocket engines&lt;/strong&gt;&#xA;  (&quot;Molecules, muscles, and machines:&#xA;  Universal performance characteristics&#xA;  of motors&quot;). Remarkably, this finding&#xA;  indicates that most of the motors used&#xA;  by humans and animals for&#xA;  transportation have a common upper&#xA;  limit of mass-specific net force&#xA;  output that is independent of&#xA;  materials and mechanisms. &lt;strong&gt;Therefore&#xA;  any actuating device produces the same&#xA;  force per mass regardless of the&#xA;  material from which it is constructed&#xA;  and the mechanism by which it&#xA;  operates&lt;/strong&gt;. This study also makes clear&#xA;  that biological systems dominate at&#xA;  the small mass, small force, range. In&#xA;  contrast, human-made machines dominate&#xA;  at the large mass range.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;short example as Sonny asked for in comment:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ant with 10 mm length &amp;amp; 10 mg mass&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow$ lets scale up to human size (2m) $\Rightarrow$ means a factor of 200. So the mass scales with 200x200x200=8000000 (Volume $\propto$ $l^{3}$ ) $\Rightarrow$ human sized ant=80 kg. But muscle forces scales only by factor 200x200=40000. The small ant can carry 100x10mg of her own mass=1g, the human sized ant should be able to carry 1g x 40000=40 kg. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conclusion: pretty comparable to a avg. 80 kg human man able to carry 40 kg! &lt;/p&gt;&#xA;" OwnerUserId="3715" LastEditorUserId="3715" LastEditorDisplayName="" LastEditDate="2011-06-07T23:59:33.193" LastActivityDate="2011-06-07T23:59:33.193" CommentCount="11" />
  <row Id="10811" PostTypeId="1" AcceptedAnswerId="10854" CreationDate="2011-06-06T01:22:45.707" Score="7" ViewCount="186" Body="&lt;p&gt;As I understand it, black holes have an absolute event horizon and an apparent horizon specific an observer.  In addition to black holes, an apparent horizon can come from any sustained acceleration.  Firstly, there is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Event_horizon&quot; rel=&quot;nofollow&quot;&gt;&quot;particle horizon&quot; of the universe&lt;/a&gt;, which is the furthest co-moving distance from which light can reach us before the acceleration of the expansion of the universe prohibits it, then there is the case of a constantly accelerating observer in just regular flat space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My zeroth question is if the Hawking radiation from a black hole could be said to come from the apparent or absolute horizon - which is it?  Obviously the former would be specific to each observer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the case of the universal acceleration, dark energy, cosmological constant (please help me with terminology as appropriate):  One could ask &quot;do we observe radiation from the horizon?&quot;, for which we already have an available answer in the form of the cosmic microwave background (CMB).  But I am curious if this would match the temperature and intensity you would expect if you treated the edge of the universe as a black hole horizon.  Are these two be equivocal in some sense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I'm most interested to ask what would occur for the case of a constantly accelerating observer.  An apparent horizon exists behind the observer beyond which light may never reach.  Could it be possible the Hawking radiation is observed as coming from behind?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/IJRj1.png&quot; alt=&quot;Accelerating Observer&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Could P possibly observe Hawking radiation coming from behind?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Granted, if you took the simple version of the temperature of the Hawking radiation ($T = \frac{\hbar c^3}{8 \pi G M k_B}$) and plugged in some super large mass, you will correspondingly get some super small temperature which will not agree with the 2.7 K of the CMB.  Of course, the two cases I'm discussing do not have any definable mass that can be used.  I would imagine that in the case of a simply accelerating particle, a greater acceleration would correspond with greater Hawking radiation, which is why it's not an issue for normal stuff, but if we accelerated something small with great force, could it observe radiation from that (false) event horizon?&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-06-06T03:22:47.657" LastActivityDate="2011-06-07T19:47:02.510" Title="Do apparent event horizons have Hawking radiation?" Tags="&lt;hawking-radiation&gt;&lt;event-horizon&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="10812" PostTypeId="2" ParentId="10062" CreationDate="2011-06-06T03:29:16.857" Score="0" ViewCount="" Body="&lt;p&gt;No, not really, at least not from Shor's algorithm. Essentially, Shor's algorithm is based upon the discrete Fourier transform and period finding. What Shor's algorithm exploits is not some random or decoherent properties of the various components of the wave function, but coherent periodic phase variations. Basically, some generalization of the Fourier transform. All Shor's algorithm can shed on quantum mechanics is that quantum mechanics can perform Fourier transforms upon the components of the wave function!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case for the many worlds interpretation will be much stronger if there is some quantum algorithm which can exploit and manifest decoherent properties of the various components in a variable manner. Such an algorithm has yet to be discovered, and might not exist at all.&lt;/p&gt;&#xA;" OwnerUserId="3892" LastActivityDate="2011-06-06T03:29:16.857" />
  <row Id="10813" PostTypeId="1" AcceptedAnswerId="10815" CreationDate="2011-06-06T04:29:15.627" Score="6" ViewCount="339" Body="&lt;p&gt;Forgetting Hooke's law for a minute why, from a microscopic perspective (preferably quantum) on up to a macroscopic one, does a spring under tension exert a force?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking that there might be an analogy between the high and low pressure states of an air wave and the density of the mass distribution of the compressed and stretched states of an iron spring, but I don't know nearly enough about solid state physics to even guess.&lt;/p&gt;&#xA;" OwnerUserId="2820" LastActivityDate="2011-06-07T06:51:38.820" Title="Forget Hooke's law. Why does a spring exert a force?" Tags="&lt;quantum-mechanics&gt;&lt;mechanics&gt;&lt;spring&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="10814" PostTypeId="2" ParentId="10813" CreationDate="2011-06-06T05:00:05.043" Score="4" ViewCount="" Body="&lt;p&gt;Any atom in a solid is &quot;kept in place&quot; by some chemical or physical interaction. (various sorts of &quot;chemical bonds&quot;, or some sort of &quot;nonbonding&quot; interaction eg. van der Waals) &#xA;Irrespective of the details, this always results in some &quot;force law&quot;, a curve depicting potential energy over distance of neighbouring atoms. Those &quot;potentials&quot; look all the same, qualitative, zero at big distance and some minimum at balance distance, and a steep rise at distances closer than the latter.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This in turn means You have to do work if You want to stretch or compress a solid, because that means atoms had to move &quot;uphill&quot; out of that minimum at equilibrium distance.  &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-06-06T05:40:24.217" LastActivityDate="2011-06-06T05:40:24.217" />
  <row Id="10815" PostTypeId="2" ParentId="10813" CreationDate="2011-06-06T05:19:37.893" Score="5" ViewCount="" Body="&lt;p&gt;You could make an analogy between the pressure distribution of a sound wave and the mass density distribution of a realistic spring undergoing vibrations, but it wouldn't give you the explanation you're looking for. As a matter of fact, that would be more like explaining a sound wave in terms of springs, rather than what you're trying to do, i.e. explaining a spring in terms of waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I'm not intimately familiar with the details, basically what goes on at the microscopic level of a spring is that, when the spring is at equilibrium, the atoms are set in some sort of rigid structure. Any given pair of atoms has a potential energy which is a function of the distance between those two atoms, so the entire spring has a potential energy determined by all the distances between every possible pair of atoms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U = \sum_{i,j} U_{ij}(r_{ij})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In equilibrium, the spring will take a shape which minimizes this total potential energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think about it, a metal spring might typically be formed by heating some metal to make it malleable (or even melting it), and then forming it into the desired shape before it cools. The heat allows the atoms to move around relatively freely so that they can reach the equilibrium configuration that minimizes their potential energy, then once the spring cools, they are frozen in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the atoms are not &lt;em&gt;completely&lt;/em&gt; frozen in place. As I see that Georg has already written in his answer, the potential energy between two atoms ($U_{ij}(r_{ij})$) has a minimum at their equilibrium distance and goes up on either side. If you add some energy into the system, say by exerting a force on it, you can get the atoms to move closer together or further apart. When you stretch or compress a string, you are really just doing this to all the (pairs of) atoms in the spring simultaneously. The atoms will, of course, &quot;try&quot; to return to their equilibrium position, i.e. they will &quot;try&quot; to minimize their potential energy, and this is what you feel as the restoring force of a spring under tension.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-06T05:19:37.893" />
  <row Id="10816" PostTypeId="2" ParentId="10220" CreationDate="2011-06-06T05:39:37.917" Score="1" ViewCount="" Body="&lt;p&gt;complex quantities always denote loss. So if the velocity is imaginary, it is impossible to for a wave to travel from one point to another. If you look at the Drude model, for some certain frequency the signal will pass so it behaves like a dielectric at that time, but for frequencies lower than the Plasma frequency it will behave like a metal where no transmission is possible and at that time permittivity is less than zero, so at that time the velocity of the wave is imaginary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in my opinion, imaginary velocity means no transmission.&lt;/p&gt;&#xA;" OwnerUserId="3893" LastEditorUserId="661" LastEditorDisplayName="" LastEditDate="2011-06-08T16:56:10.087" LastActivityDate="2011-06-08T16:56:10.087" />
  <row Id="10817" PostTypeId="2" ParentId="10813" CreationDate="2011-06-06T08:57:36.987" Score="1" ViewCount="" Body="&lt;p&gt;Expanding on my comment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A spring, or anything that can be modeled by a string is any interaction that can be described by a  potential that is proportional to $x^2$ the displacement. Why is this potential important? Qualitatively, it is the solutions to this potential (i.e. the motion of a particle that sits in this potential) are the solutions of a simple harmonic oscillator i.e. Sine Waves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a very hand wavy approach to this. You can think that to the lowest order the universe is host to 2 ubiqutious potentials - the $1/x$ potential that is responsible for our inverse square laws and the $x^2$ potential responsible for oscillations and harmonic motions. A constant potential will exert no force and an $x^1$ potential will exert a force on an object that is constant at all distances - clearly unphysical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, a simple equation to model the potential &quot;felt&quot; by two neutral atoms clsoe to each other is given in the Lennard-Jones potential A graph of which is shown below. This type of potential is somewhat ubiquitous (in one form or another) when taking about intermolecular interactions. There are refinements, expansions, better models, etc. But they all have one thing in common a &quot;minimum&quot; or a &quot;cup&quot; in the potential which represents the stable configuration. A particle/atom/crystal/etc. that exists in that region of the potential is &quot;stable&quot;. Nudge it in one direction and it will naturally &quot;fall&quot; towards the low point of the potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this important? Well, to &lt;em&gt;first approximation&lt;/em&gt; the region near the low point of the potential is &lt;em&gt;effectively&lt;/em&gt; a $x^2$ potential - a simple harmonic oscillator. Thus any system that is modeled by this potential will respond like a harmonic oscillator in the limit of a &lt;em&gt;really&lt;/em&gt; small nudge. I.e. it behaves like a spring. Due to the magic of ignoring all mathematical rigor, effectively any potential that &lt;em&gt;vaguely&lt;/em&gt; looks like the L-J potential, in fact any potential that has a local minimum like such, will to first order look like a $x^2$ potential in the small region around the minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/2Rsyu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="95" LastEditorDisplayName="" LastEditDate="2011-06-06T09:04:08.137" LastActivityDate="2011-06-06T09:04:08.137" CommentCount="3" />
  <row Id="10818" PostTypeId="1" CreationDate="2011-06-06T10:15:57.400" Score="2" ViewCount="109" Body="&lt;p&gt;It has always been claimed decoherence solves the problem of the preferred-basis for observed probability distributions, but why should this be the case? If there is only one world, and there are probabilities for certain outcomes, why should the basis in which the probabilities are observed coincide with the pointer basis determined dynamically by decoherence?&lt;/p&gt;&#xA;" OwnerUserId="3895" LastActivityDate="2011-06-06T15:29:41.790" Title="Why should the observed probability distributions in quantum mechanics always align with the pointer basis of decoherence?" Tags="&lt;quantum-interpretations&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10819" PostTypeId="1" AcceptedAnswerId="10820" CreationDate="2011-06-06T10:17:35.813" Score="0" ViewCount="78" Body="&lt;p&gt;In aircraft literature, what does cross-track and along-track wind directions mean? Please explain in terms of the aircraft's motion relative to the wind direction. I can hazard a guess: along-track is wind in the direction of the aircraft's velocity, and cross-track would be perpendicular side-to-side (or is it top-bottom) wind direction wrt to aircraft velcoity direction? Not sure...some explanation and clarification would help. Thanks. &lt;/p&gt;&#xA;" OwnerUserId="3604" LastActivityDate="2011-07-06T13:18:45.680" Title="What does cross-track wind mean?" Tags="&lt;aerodynamics&gt;&lt;airplanes&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10820" PostTypeId="2" ParentId="10819" CreationDate="2011-06-06T13:16:41.380" Score="2" ViewCount="" Body="&lt;p&gt;The &quot;track&quot; is the intended travel path over the ground from the origin to the destination.  Cross-track wind is the wind perpendicular to this line (crosswind), and along-track wind is the headwind/tailwind component.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-06-06T13:16:41.380" CommentCount="4" />
  <row Id="10821" PostTypeId="2" ParentId="10798" CreationDate="2011-06-06T14:02:06.003" Score="0" ViewCount="" Body="&lt;p&gt;There are several ways you could do this. Below I will outline a couple of possible approaches, but it may be that a simpler / more elegant solution exists. Here are my 2cents nonetheless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, it would be a lot easier if you first consider the motion of a zero-radius ball and worry about making it bigger and appearing as if it's 'rolling' later on. Here's how I think you could approach the motion of a zero-radius ball, in the bowl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would need to program time-evolution equations of motion for each of the three dimensions individually. You could do this in your choice of coordinate system. For the geometry of the apperatus you are trying to model, I think &lt;a href=&quot;http://mathworld.wolfram.com/CylindricalCoordinates.html&quot; rel=&quot;nofollow&quot;&gt;cylindrical coordinates&lt;/a&gt; $(r,\theta,z)$ would be easier than cartesian coordinates $(x,y,z)$. In which case, you would need just a circular motion equation (relating $r$ and $\theta$ with time,) with the addition of a resistive term to simulate the energy loss and therefore reduction of the 'orbital' radius. The fall in $z$ could then be determined from the (varying) $r$-value with respect to the geometry of your bowl. I.e. A smaller radial position corresponds to a position nearer the centre of the bowl and therefore a lower height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In cartesian coordinates you would probably need a damped harmonic oscillator equation for each of the $x$- and $y$-coordinates. The change in $z$ could then be determined in the some way as above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Alternatively&lt;/em&gt;, as opposed to having an educated guess at the equations of motion as described above, you could derive them 'from scratch' by considering the potential and kinetic energy exchange (again for each coordinate individually.) If you did this in a vector formalism (using the coordinate system of your choice) you would just have to solve the equations for the velocity $\vec{v}(t)$, then integrate to find the position $\vec{x}(t)$. Again though, you would also need an additional term to account for the loss of energy due to friction. (Otherwise the ball's speed will not decay and therefore neither will its 'orbit'.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you've got your dynamics set up for the motion of a zero-radius ball (a point I guess,) you can then program a modification to adjust how the ball is displayed so that it looks like an object of the desired size. One option would be (for a ball of radius &lt;em&gt;r&lt;/em&gt;) to display the point as moving around a constant distance &lt;em&gt;r&lt;/em&gt; higher than the surface of the bowl, then program a few lines to depict the ball as an object extended by the distance &lt;em&gt;r&lt;/em&gt; around the point. How complex this gets depends on how realistic you want it to appear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: In your question you refer to &quot;rotation angle&quot; and &quot;position&quot; - if I understand you correctly, you mean the initial position and velocity of the ball. These values would be put in by the programmer or program user as initial conditions, after which the dynamics equations would evolve the system.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="1696" LastEditorDisplayName="" LastEditDate="2011-06-06T16:27:46.487" LastActivityDate="2011-06-06T16:27:46.487" CommentCount="3" />
  <row Id="10823" PostTypeId="1" AcceptedAnswerId="10824" CreationDate="2011-06-06T15:17:35.307" Score="2" ViewCount="102" Body="&lt;p&gt;100% of the energy from the sun is reflected back into space, it's just shifted from a low-entropy state to a high-entropy state, and from a high frequency (ultraviolet) to a low frequency (infrared). But does the total number of photons change as a result? Are there more low frequency photons, then, in order to make the same amount of energy?&lt;/p&gt;&#xA;" OwnerUserId="993" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-06T23:15:58.760" LastActivityDate="2011-06-06T23:15:58.760" Title="If 100% of the energy from the sun is reflected back into space" Tags="&lt;energy&gt;&lt;energy-conservation&gt;&lt;photons&gt;&lt;entropy&gt;&lt;sun&gt;" AnswerCount="1" />
  <row Id="10824" PostTypeId="2" ParentId="10823" CreationDate="2011-06-06T15:23:28.550" Score="2" ViewCount="" Body="&lt;p&gt;Reflection, by definition, sends the same photons back. So it can't change the frequency of the photons or their number or the entropy (at least locally, it can't).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you probably want to talk about is complete absorption and black-body re-radiation of the energy. That's something else than reflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, if the planet is in balance, then, of course, it has to emit a higher number of photons that have a lower energy and lower frequency each because the energy of a single photon is related by $E=hf$ to the frequency. The entropy is essentially proportional to the number of photons, so it emits more entropy than it absorbs, and the entropy is increasing.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-06-06T15:23:28.550" CommentCount="1" />
  <row Id="10825" PostTypeId="2" ParentId="10818" CreationDate="2011-06-06T15:29:41.790" Score="3" ViewCount="" Body="&lt;p&gt;It's because the probability that a given mixed state $\rho$ is found in a particular normalized state $\psi$ is only &quot;sharp&quot; and well-defined if $\psi$ is actually an eigenstate of $\rho$. So whatever basis in which $\rho$ is diagonal the decoherence produces, is also the basis of the states that have well-defined probabilities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all their general linear superpositions, one may calculate the expectation values of the probabilities but they're not probabilities that can be measured or &quot;perceived&quot;. The prescription above was described e.g. at&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2011/06/density-matrix-and-its-classical.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2011/06/density-matrix-and-its-classical.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It treats the density matrix in a similar way as the observables - even though the density matrix is not an observable. The rule that only the eigenstates of the density matrix may be &quot;perceived&quot; is pretty much equivalent to the &quot;consistent histories&quot; approach to quantum mechanics that tells you which questions may be asked in quantum mechanics and which can't.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-06-06T15:29:41.790" CommentCount="2" />
  <row Id="10826" PostTypeId="1" CreationDate="2011-06-06T15:45:24.167" Score="1" ViewCount="338" Body="&lt;p&gt;The lubricant is needed only on the hair surface? How can the blade cut the slippery surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does soap  make shaving with razor less painful?&lt;/p&gt;&#xA;" OwnerUserId="652" LastEditorUserId="652" LastEditorDisplayName="" LastEditDate="2011-06-06T17:22:00.927" LastActivityDate="2011-06-15T10:07:29.067" Title="Why does soap make shaving with razor less painful?" Tags="&lt;friction&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="10827" PostTypeId="1" AcceptedAnswerId="10832" CreationDate="2011-06-06T16:20:49.210" Score="3" ViewCount="256" Body="&lt;p&gt;From high school, I remember that Aluminium has 13 electrons and thus has an unpaired electron in the 3p shell. This should make Aluminium magnetic. However, the wiki page of Aluminium says its non-magnetic at one place(with a citation needed tag though) and at another place says it's paramagnetic. Doing a google search shows up some contradictory results. So what is the truth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:The context of the question is &lt;a href=&quot;http://scifi.stackexchange.com/questions/3998/is-magnetos-power-magnetism-or-can-he-manipulate-all-metals/4000#4000&quot;&gt;this answer on scifi.SE about magneto&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3417" LastActivityDate="2011-06-07T19:38:22.220" Title="Is aluminium magnetic?" Tags="&lt;electromagnetism&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" />
  <row Id="10829" PostTypeId="1" AcceptedAnswerId="10830" CreationDate="2011-06-06T17:34:43.433" Score="-2" ViewCount="171" Body="&lt;p&gt;&lt;strong&gt;Intent:&lt;/strong&gt; Not looking to make-up something, I seek explanations which are possible to duplicate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Description of Phenomenon:&lt;/strong&gt;  The halved side of the bubble would be horizontal to Earth's surface, and the bowl shape would be closer to the ground. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image of Shape (not the bubble itself):&lt;/strong&gt; Here's a steel version of the &lt;a href=&quot;http://www.stainless-steel-sphere.com/maoping/UploadFile/image/stainless%20steel%20hemisphere07.JPG&quot; rel=&quot;nofollow&quot;&gt;bubbles shape&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Related Physics:&lt;/strong&gt; &#xA;Here's the best explanation I've found related to the physics know to relate to the issue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We consider half of the soap bubble.&#xA;  The forces on the hemisphere will be&#xA;  the surface tensions on the two&#xA;  circles and the net force from the&#xA;  excess pressure between the inside and&#xA;  the outside of the bubble. This net&#xA;  force is the sum of all of the forces&#xA;  perpendicular to the surface of the&#xA;  hemisphere, but must be parallel to&#xA;  the surface tension. Therefore we can&#xA;  find it by finding the force on the&#xA;  circle that is the base of the&#xA;  hemisphere. The total force must be&#xA;  zero, so we have&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2(2¹r)g = (¹r2) ÆP, which gives ÆP =&#xA;  4g/r.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.kean.edu/~gkolodiy/2091/giach/ch10/CHAP10/giach10.htm&quot; rel=&quot;nofollow&quot;&gt;SOURCE&lt;/a&gt; and &lt;a href=&quot;http://www.kean.edu/~gkolodiy/2091/giach/ch10/CHAP10/Image614.gif&quot; rel=&quot;nofollow&quot;&gt;RELATED IMAGE&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;CURRENT-STATUS:&lt;/strong&gt; I'm out of possible reasons why this would happen, except for one, that being, that any half-bubbles observed might in fact be a whole free-floating bubble with three walls, two of which are refracting near zero light, while the third is highly refractive. The three walls form a single sphere divided in half by single wall; meaning that there's a top-hemisphere, and lower-hemisphere, and the dividing wall. Would this be possible? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BACKGROUND:&lt;/strong&gt; This phenomenon is based on a question I saw years ago posted on another physics community forum. I'm unable to locate that page, but I've searched for it. On that page, more than three(3) people had independently observed the phenomenon; which to me gives some validity to it existence in some form; that, or someone was just having fun, and gaming the forum.&lt;/p&gt;&#xA;" OwnerUserId="2283" LastEditorUserId="2283" LastEditorDisplayName="" LastEditDate="2011-06-07T12:30:29.070" LastActivityDate="2011-07-26T18:58:18.790" Title="Can a &quot;bowl&quot; shaped liquid half-bubble be free-floating in the air?" Tags="&lt;surface-tension&gt;" AnswerCount="4" CommentCount="14" />
  <row Id="10830" PostTypeId="2" ParentId="10829" CreationDate="2011-06-06T18:16:06.907" Score="5" ViewCount="" Body="&lt;p&gt;No. Free-floating bubbles form essentially spherical shapes. In general, they form shapes that &lt;a href=&quot;http://chemistry.about.com/od/bubbles/a/bubblescience.htm&quot; rel=&quot;nofollow&quot;&gt;minimize surface area&lt;/a&gt;, subject to constraints such as the bubble having to enclose a fixed volume. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to see why the bowl shape you imagine wouldn't work is to consider a small element of the surface right near the &quot;corner&quot; of the bowl (where the spherical and planar surfaces meet). The surface tension on either side of the corner would both pull that corner in towards the center. The sharp corner would therefore become rounded. &lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-06-06T18:16:06.907" CommentCount="9" />
  <row Id="10831" PostTypeId="2" ParentId="10829" CreationDate="2011-06-06T19:02:03.600" Score="0" ViewCount="" Body="&lt;p&gt;You ask:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Can a “bowl” shaped liquid half-bubble be free-floating in the air?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and detail:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The halved side of the bubble would be horizontal to Earth's surface, and the bowl shape would be closer to the ground.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have never seen such a bubble. The bubbles floating in air I have made are soap bubbles, i.e. an elastic surface enclosing air. The shape you propose is non-physical even for these physically attainable bubbles for the reasons given by Ted in the previous answer: the surface tension at the circle of the horizontal cut would be large and destroy any cohesion in shape of the soap bubble, at best turning it into a sphere at worst bursting it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a &lt;em&gt;liquid&lt;/em&gt; half bubble &lt;em&gt;free floating&lt;/em&gt; in air would be even more non-physical than a soap bubble, as liquids are generally heavier than air and cannot float whether half sphere  or whole, as rain shows. I am not aware of a liquid lighter than air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;:  From you insistence  in the comments I think you must have observed  something that you interpreted in the words you use, i.e you thought you saw  a floating bowl of liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, this would be a mirage, similar to seeing water on hot pavement: the refraction index different in the volume of the bubble than in the surrounding air will give the impression of liquid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now on hot days, heated bubbles of air can be created and start to rise ( in large scale that is the way inversion layers happen) , giving the possibility that a mirage effect of looking like liquid might be seen. This bubble will be spherical as per the other two answers, as the shapes of self sustained systems are constrained by the equations involved. &lt;strong&gt;BUT&lt;/strong&gt; it may be that the bubble as it rises it finds a cooler air surface layer  against which it is compressed and ends up looking like the bowl you describe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will be very rare, but if I saw a &quot;bowl of liquid in the air&quot; that is the interpretation in physics. A mirage detail.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-06-07T05:22:19.927" LastActivityDate="2011-06-07T05:22:19.927" CommentCount="6" />
  <row Id="10832" PostTypeId="2" ParentId="10827" CreationDate="2011-06-06T19:31:24.950" Score="11" ViewCount="" Body="&lt;p&gt;It really depends on what you mean by &quot;magnetic,&quot; because there are different kinds of magnetic properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Materials like iron are &lt;a href=&quot;http://en.wikipedia.org/wiki/Ferromagnetism&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;ferromagnetic&lt;/em&gt;&lt;/a&gt;, which means that once you align the individual magnetic dipoles in the material, they will tend to stay aligned even without an external magnetic field. Ferromagnetic materials are the ones that permanent magnets are made out of, and they are probably what most people think of when they imagine a magnetic material. There are only three elements (as far as I know) that are ferromagnetic: iron, cobalt, and nickel, although other elements can be combined to make ferromagnetic polyatomic crystals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other materials that aren't ferromagnetic can (and typically do) have interesting magnetic properties, though - in other words, just because a material isn't a ferromagnet doesn't mean it doesn't interact magnetically at all. &lt;a href=&quot;http://en.wikipedia.org/wiki/Paramagnetism&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Paramagnetism&lt;/em&gt;&lt;/a&gt; is one such interaction. When you put a paramagnetic material in a magnetic field, its individual dipoles tend to align with the magnetic field, and thus with each other, thereby making the material magnetic. When this happens, the paramagnetic material is attracted to the magnetic field. The difference is that when you take the external magnetic field away, the individual dipoles in a paramagnetic material don't retain their orientation. Instead, thermal motion takes over and reorients them randomly. So a paramagnetic material only has a net magnetic moment while it is in an external magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Magneto is able to control magnetic fields, then that would potentially allow him to control all sorts of magnetic materials - not just ferromagnets (iron etc.) but also all paramagnetic and perhaps &lt;a href=&quot;http://en.wikipedia.org/wiki/Diamagnetism&quot; rel=&quot;nofollow&quot;&gt;diamagnetic&lt;/a&gt; materials, since he can create the external field necessary to magnetize those materials. In fact, all materials, even non-metals, are diamagnetic to some (small) extent. However, paramagnetism and especially diamagnetism are generally much weaker effects than ferromagnetism, so it stands to reason that Magneto would have a harder time controlling non-ferromagnetic materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest thing to a scientific explanation for Magneto's abilities that I can come up with is that he's able to generate magnetic fields that are strong enough to have a significant effect on ferromagnetic and some of the more paramagnetic materials, but with diamagnetic materials, the magnetic fields he can produce are not strong enough to override other natural forces acting on those materials. Of course, I'm sure that wouldn't really hold up if you really look at the comics or the movie closely... but with comic books you probably don't want to ask too many questions ;-)&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-06T19:31:24.950" CommentCount="1" />
  <row Id="10833" PostTypeId="1" AcceptedAnswerId="10840" CreationDate="2011-06-06T20:17:05.120" Score="0" ViewCount="127" Body="&lt;p&gt;Could someone explain the calculation required to answer this question. It is from a text book and the answer is recorded as 585Hz but I cannot replicate the answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 1931 Michelson used a rotating prism with 32 faces to measure the time for light to travel through a 1.6km evacuated tube to a distant mirror and back. From his results he calculated that the speed of light is 299774000 m/s.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Assuming the prism is rotated by only one face during the time of flight of light calculate the rotation frequency of the prism.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is the book incorrect or am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt:&#xA;$(1/(1600/299774000))/32=5855Hz$ which is suspiciously similar to the suggested answer 585Hz. But then I remembered about the round-trip (3200m) which threw all the numbers again.&lt;/p&gt;&#xA;" OwnerUserId="3900" LastEditorUserId="3900" LastEditorDisplayName="" LastEditDate="2011-06-06T21:17:53.313" LastActivityDate="2011-06-07T18:16:00.807" Title="Michelson rotating mirror experiment" Tags="&lt;optics&gt;&lt;homework&gt;&lt;light&gt;&lt;speed-of-light&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="10834" PostTypeId="2" ParentId="10826" CreationDate="2011-06-06T20:29:01.430" Score="-1" ViewCount="" Body="&lt;p&gt;This is pure speculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Removing soap/shaving cream causes sensations that distract from the discomfort of actually shaving. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evaporation is separated from the skin by a layer of soap bubbles. Removing the soap occurs during shaving results in evaporation occurring directly on the skin, which produces a cool fresh feeling not unlike the cool side of a pillow. The bubbles themselves exert a clinging force - the release of which likely contributes to the distraction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using soap/water/shaving cream likely does not affect the actual mechanics significantly - besides forcing one to shave slowly. &lt;/p&gt;&#xA;" OwnerUserId="2843" LastActivityDate="2011-06-06T20:29:01.430" CommentCount="2" />
  <row Id="10835" PostTypeId="1" CreationDate="2011-06-06T21:23:35.070" Score="2" ViewCount="97" Body="&lt;p&gt;In a previous question, &lt;a href=&quot;http://physics.stackexchange.com/questions/10565/does-a-magnetic-field-do-work-on-an-intrinsic-magnetic-dipole&quot;&gt;Does a magnetic field do work on an intrinsic magnetic dipole?&lt;/a&gt;, one highly rated answer suggested that static magnetic fields do work on intrinsic magnetic dipoles in a non-uniform magnetic field. I can visualise the change in kinetic energy of the nucleus of an atom coming from a change in the configuration of the electrons around the nucleus. But for an electron, since it's truly fundamental, I'm scratching my head over where the energy comes from to change its kinetic energy. If it does, then it really must come from the static magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the experimental evidence that shows the kinetic energy of an electron changes in a static non-uniform magnetic field?&lt;/p&gt;&#xA;" OwnerUserId="2146" LastEditorUserId="2146" LastEditorDisplayName="" LastEditDate="2011-06-06T23:24:04.313" LastActivityDate="2011-06-08T18:02:26.140" Title="Experimental evidence showing the kinetic energy of an electron changes in a static non-uniform magnetic field?" Tags="&lt;electrodynamics&gt;&lt;magnetic-moment&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10836" PostTypeId="2" ParentId="10826" CreationDate="2011-06-06T22:09:45.997" Score="0" ViewCount="" Body="&lt;p&gt;My guess about the pain factor is that the soap residue gets in the way of the razor and your skin.&lt;/p&gt;&#xA;" OwnerUserId="3821" LastActivityDate="2011-06-06T22:09:45.997" CommentCount="1" />
  <row Id="10837" PostTypeId="1" AcceptedAnswerId="10846" CreationDate="2011-06-06T22:21:22.563" Score="1" ViewCount="140" Body="&lt;p&gt;In my current quantum mechanics, course, we have derived in full (I believe?) the wave equations for the time-independent stationary states of the hydrogen atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are told that the Pauli Exclusion principle is a consequence of two electrons not being able to share the same wave equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in our derived equation, we did not have anything including the spin.  We defined $\psi (r,\theta,\phi)$ as $\psi_{n,l,m} (r,\theta,\phi) = R_{n,l}(r) Y_{l,m}(\theta,\phi)$, where $Y_{l,m}(\theta,\phi) = f_{l,m}(\theta) e^{i m \phi}$.  We then were given well-defined $R_{n,l}$ and $f_{l,m}$  that satisfied the partial differential equations in the Schroedinger Equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowhere in our final $\psi$ do we find anything that varies depending on a fourth degree of freedom, not to mention one that behaved as $m_s$ should.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I missing the point of the Pauli Exclusion Principle?  Is there a part of the solutions for $\psi$ that I am not understanding?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;I am referring to an $H^-$ ion, where there are two electrons each with their own wave equation.  If we imagine the case where both have the same quantum numbers n,l,m, but different spin $m_s$, would not their wave equations be exactly the same, and therefore not allowed?&lt;/p&gt;&#xA;" OwnerUserId="71" LastEditorUserId="71" LastEditorDisplayName="" LastEditDate="2011-06-06T22:35:47.720" LastActivityDate="2011-06-07T10:09:28.160" Title="Where is spin in the wave equation of an electron in the hydrogen atom?" Tags="&lt;quantum-mechanics&gt;&lt;electron&gt;&lt;schrodinger-equation&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="10838" PostTypeId="1" AcceptedAnswerId="10924" CreationDate="2011-06-06T22:21:32.813" Score="4" ViewCount="86" Body="&lt;p&gt;COBE, WMAP, and now PLANCK, have or are in the process of measuring anisotropies in the cosmic microwave background. WMAP has dectected E-mode polarization from electron Thomson scattering, but not B-mode polarization. One of PLANCK's objectives is to detect B-mode polarization as a means to determine the scale of inflation. The B-mode is expected to originate from large scale gravitational waves created by inflation, but gravtational lensing is also a B-mode source.&#xA;My question is, how is B-mode polarization generated by gravity waves, and how is it used to determine the scale of inflation? Also, can B-mode detection be a verification of the inflation theory?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-06-08T18:41:34.127" Title="Inflation Scale via CMB Polarization" Tags="&lt;polarization&gt;&lt;planck&gt;&lt;cmb&gt;&lt;inflation&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="10839" PostTypeId="2" ParentId="10829" CreationDate="2011-06-06T22:23:44.160" Score="3" ViewCount="" Body="&lt;p&gt;There is no need to ask the question about a half-sphere because a general answer regarding any surface edge of a substance with the desired properties (I'll call &quot;bubble surface&quot;) will apply for that case.  This is a problem about surface tension (ST).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there are cases where a non-enclosed bubble surface exists, for instance in the case of a bubble blower apparatus.  In this case the edges of the surface are attracted (on contact) to the plastic loop that people blow into.  It's attracted for the same reason that water has a meniscus that reaches up on the side of a container, and it is a lower energy state.  I mean that the surface attached to a wall, like in this example, has a lower surface tension energy than if that same surface area existed elsewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the stable bubble surfaces are local minimums for the ST problem.  The shape with the lowest ST energy is just a sphere of solid water, although for things like water and the bubble liquid, there is an energy penalty/advantage to adhesion to an external surface, and there can be other forces as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case of a bubble surface with the edges unbounded and exposed to air is not a local minimum at all.  An infinitesimally small withdrawal of the edge closer to the rest of the fluid results in a lower energy state, as such there exists a force on that part of the fluid and there is no other force balancing it.  This is a bit more general than surface problems, but a slope (synonymous with gradient) in an energy field generally results in a force, $\vec{F}=\nabla E$.  So whenever such an unbounded surface edge exists, that ST force is actively destroying the shape.  If you pop a bubble this will occur.  Only a small break in a part of the bubble causes the edge all around that point to recede faster than what our eyes can observe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape in your question has edges, so everything I've said about such surfaces applies.  It doesn't exist except for a fraction of a second during which it's destroying itself.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-06-06T22:23:44.160" CommentCount="9" />
  <row Id="10840" PostTypeId="2" ParentId="10833" CreationDate="2011-06-06T22:26:31.437" Score="2" ViewCount="" Body="&lt;p&gt;If I enter your numbers I get 585,49 Hz, so I think you've just missed a zero when you entered it in your calculator :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the 1.6 km or 3.2 km etc, maybe the problem-designers missed it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://oisc.net/Speed_of_Light.htm&quot; rel=&quot;nofollow&quot;&gt;description of the original experiment&lt;/a&gt; with the mile-long tube, but there it seems as if the beam bounces several times back and forth in the tube so maybe neither total length is correct.. &lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-06-06T22:26:31.437" CommentCount="3" />
  <row Id="10842" PostTypeId="5" CreationDate="2011-06-06T22:27:32.420" Score="0" ViewCount="" Body="&lt;p&gt;'Mathematical physics' is the application of mathematics to problems in physics and mathematical methods suitable for such applications, e.g., partial differential equations (PDEs), functional analysis, variational calculus, and potential theory. 'Mathematical physics' is also the study of problems inspired by physics within a mathematically rigorous framework, such as, e.g., rigorous derivation of an atomic energy spectrum, rigorous construction of a quantum field theoretic model, and rigorous description of a phase transition. The 'mathematical physics' tag should not be used for any quantitative physics question. See also &lt;a href=&quot;http://en.wikipedia.org/wiki/Mathematical_physics&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mathematical_physics&lt;/a&gt; .&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-06T22:56:37.913" LastActivityDate="2011-06-06T22:56:37.913" />
  <row Id="10843" PostTypeId="4" CreationDate="2011-06-06T22:27:32.420" Score="0" ViewCount="" Body="'Mathematical physics' is the application of mathematics to problems in physics and mathematical methods suitable for such applications, e.g., partial differential equations (PDEs), functional analysis, variational calculus, and potential theory. 'Mathematical physics' is also the study of problems inspired by physics within a mathematically rigorous framework. The 'mathematical physics' tag should not be used for any quantitative physics question. " OwnerUserId="-1" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-06T22:56:41.143" LastActivityDate="2011-06-06T22:56:41.143" />
  <row Id="10844" PostTypeId="2" ParentId="10837" CreationDate="2011-06-06T22:32:03.400" Score="0" ViewCount="" Body="&lt;p&gt;For the Pauli principle to be explicitly seen/implemented, one has to have at least two electrons. It is not the case in H.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-06-06T22:32:03.400" CommentCount="2" />
  <row Id="10845" PostTypeId="2" ParentId="10837" CreationDate="2011-06-06T22:43:43.753" Score="2" ViewCount="" Body="&lt;p&gt;A description of electron spin and the Pauli exclusion principle needs to go beyond the Schrödinger-equation to the spinor-valued Dirac-equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't remember my atomic physics course very well but at the level of your analysis I think you just add the rule of the two parallel spin orbitals explicitly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a discussion on this topic the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_atom&quot; rel=&quot;nofollow&quot;&gt;wikipedia-page&lt;/a&gt; is helpful.&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-06-06T22:43:43.753" />
  <row Id="10846" PostTypeId="2" ParentId="10837" CreationDate="2011-06-06T22:45:50.630" Score="5" ViewCount="" Body="&lt;p&gt;What you've written down is the spatial part of the electron wavefunction.  The spin state is not included.  The full wavefunction of the electron involves both the spatial part and the spin part.  Sometimes in quantum mechanics books the full electron wavefunction is written as the tensor product of the spatial and spinor parts, sometimes you'll just see it as one written after the other.   Sometimes you'll just see it written as $|n,l,m_l, s,m_s \rangle$.  I'm not sure what book you're using.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you derived the solutions of the TISE for the hydrogen atom Hamiltonian, you probably neglected the relativistic corrections, fine-structure interactions, and hyperfine interactions, leaving you with just the Coulomb potential.  Since the Coulomb potential doesn't depend on electron spin, you could ignore that part of the wavefunction.  (In other words, since the spin operators all commute with the $1/r$ potential, the spin eigenstates are all eigenstates of the $1/r$ Hamiltonian, so it's Kosher to just throw in the $|s, m_s \rangle$ stuff afterwards).&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-06-06T22:45:50.630" CommentCount="5" />
  <row Id="10847" PostTypeId="2" ParentId="214" CreationDate="2011-06-06T23:27:31.953" Score="4" ViewCount="" Body="&lt;p&gt;The reason that jumping can make a relatively large difference is that the kinetic energy is proportional to the square of the velocity. Thus relatively small changes to the velocity can result in relatively large changes to the kinetic energy. In addition, the velocity which a human can achieve in jumping is a substantial percentage of the velocity of fatal falls.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let the human weigh $m$, let him jump with upward velocity $v$ and let the elevator fall from a height $H$. Then the human's initial potential energy will be $10mH$. What fraction of this potential energy can he avoid having turned into kinetic energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any given point before jumping, the human's kinetic energy and potential energy add up to $10mH$. If he jumps at a height of $h$, his potential energy will be $mgh$ and his kinetic energy will be $mg(H-h) = 0.5mV^2$ where $V$ is the elevator (and human before jumping) velocity, taken as a positive number so that $V=\sqrt{2g(H-h)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment of jumping, he will not reduce potential energy, but instead will decrease his velocity. So his kinetic energy decreases from $0.5mV^2$ to $0.5m(V-v)^2$. Therefore his total energy will become:&lt;br&gt;&#xA;$$mgh + 0.5m(V-v)^2$$&#xA;$$= mgh + 0.5m(\sqrt{2g(H-h)}-v)^2$$&#xA;$$= mgH +0.5mv^2 - mv\sqrt{2g(H-h)}.$$&#xA;The terms have a simple interpretation. $mgH$ is the energy in the absence of any jumping. $0.5mv^2$ is the energy of the jump (in the frame of reference of the human). And the remaining term is the reduction in energy due to the reference frame conversion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We wish to make the third term as negative as possible. This occurs when h is small so we put $h=0$ (as our intuition suggests, indeed, the best time to jump is just as the elevator impacts). Then the remaining kinetic energy is:&#xA;$$mgH +0.5mv^2 -mv\sqrt{2gH}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of a height $H$ which is generally fatal for a human is $H=10m$. A maximum velocity for a very athletic human jump is on the order of $v=3.64$ m/s. Such a jump would give a maximum height of 0.66 meters. See: &lt;a href=&quot;http://www.exrx.net/Calculators/VerticalJump.html&quot; rel=&quot;nofollow&quot;&gt;Vertical Jump Test calculator&lt;/a&gt; for data on human jumping capabilities by sex, age, and athletic ability. Using $g=10$ and $m=50$, the kinetic energy before and after jumping are:&#xA;$$mgH = 5000J$$&#xA;$$mgH+0.5mv^2-mv\sqrt{2gH} = 2757J$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, in fact, jumping could reduce the kinetic energy suffered by a factor of two. The final collision with the floor would be reduced from a height of 10m = 32.8 feet, to a height of 5.5m = 18 feet.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-06-09T19:46:31.223" LastActivityDate="2011-06-09T19:46:31.223" />
  <row Id="10848" PostTypeId="1" AcceptedAnswerId="10853" CreationDate="2011-06-07T01:28:29.523" Score="2" ViewCount="175" Body="&lt;p&gt;Ok, so on Science channel was a special about Hawking/Susskind debating black holes, which can somehow remove information from the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) In stars, fusion converts 4 hydrogen into 1 helium, with a small fraction of the mass being converted to energy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happens all the time, and nobody is that amazed about it anymore.   So, I have to assume that science is ok with the idea of fusion happening, and so, converting matter into energy doesn't destroy the information of the matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take that as Assuption 1:&lt;br&gt;&#xA; ** 1: Converting matter to energy is OK.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) In empty space, quantum fluctuations of the energy levels are presumed to allow &quot;pair creation&quot; where a particle and its anti-particle can be created out of the really tiny amounts of energy that exist in empty space.  For quantum fluctutations, if I understand it, the uncertaincy principle actually dictates that the amount of energy can fluctuate, allowing a pair to be created even if there &quot;shouldn't be enough energy&quot;, because the amount of energy in a small space is uncertain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tiny amount of energy becomes a particle + anti-particle, for a tiny instant, and then they fall back into each  other, and are converted back to the same tiny amount of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways, ignoring the fact that I'm fuzzy on all the requirements for pair creation -- it is generally accepted as part of quantum physics that it can happen.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, again, I have to assume that science is ok with the idea of pair creation.  And so, converting energy into matter, must also not destroy the information value of the energy, and that information value must remain even after the 2 particles annihilate and go back to being energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take that as Assumptions 2 + 3:&#xA;  ** 2: Converting energy into matter is OK.&#xA;  ** 3: Pair creation, doesn't mess up anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C) Ok, now black holes.   My (weak) understanding of Hawking's idea on how black holes emit energy is that pair creation can happen close to the black hole.  Since the position of energy is uncertain at the quantum level, some of the energy that &quot;should be&quot; inside the black hole's event horizon, can be treated as uncertainly outside the black hole, and then that extra energy can cause pair creation.   If the black hole then pulls in both particles, then nothing special happens -- the mass/energy stays inside the black hole.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if one or both of the particles moves away from the black hole, and somehow escapes, that can cause the energy to leave the black hole.  This is assumed to be a very unlikely event, but still within possibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways, that's my (very rough) understanding of Hawking's theory on how black holes can loose mass over time, when in theory &quot;nothing can escape a black hole&quot;.  Let's assume Hawking's correct...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a black hole continues to loose mass over cosmic time, it will eventually disappear, and the problem is stated that when it vanishes: All of the information which the black hole had formerly consumned, will be lost.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which contradicts the Law of Conservation of Information.   At least, that's how the Science channel presented it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my idea/question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** Every time the black hole looses mass, it's through the processes of energy-&gt;matter, or matter-&gt;energy-&gt;matter, involved in pair creation.  Based on assumptions 1, 2 + 3 above, none of these should violate the Law of Conservation of Information.   It seems that the very slow process of loosing mass, should also be a very slow process of &quot;leaking&quot; information back into the rest of the universe.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;*** So, WHAT's the problem? ***&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;By my understanding: By the time the black hole vanishes, it's OK -- because all of it's information should also have leaked back into the rest of the universe, and no information will be lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, so please tell me why this is wrong.   It seems intuituve to me, but really smart people have spent years looking into this.   So, am I missing something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&#xA;-Chuck&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ps.  Also, if my statement of fusion, pair creation, and/or hawking radition is wrong -- BUT NOT IN A WAY THAT MATERIALLY AFFECTS THE QUESITON -- please overlook that and just focus on the question.&lt;/p&gt;&#xA;" OwnerUserId="3902" LastActivityDate="2011-06-07T05:07:31.990" Title="Why isn't black hole information loss this easy (am I missing something basic)?" Tags="&lt;black-holes&gt;&lt;hawking-radiation&gt;&lt;information&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="10849" PostTypeId="1" AcceptedAnswerId="10986" CreationDate="2011-06-07T02:02:00.397" Score="3" ViewCount="140" Body="&lt;p&gt;In an infinite plane elastic plate of thickness $d$, it is shown that the modes of oscillation corresponding to a fixed time-frequency $\omega$ have wave-numbers given by solutions of the Rayleigh-Lamb equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\tan (pd)}{\tan (qd)}=-\left[\frac{4k^2pq}{\left(k^2-q^2\right)^2}\right]^{\pm 1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the +1 exponent corresponds to symmetric modes and the -1 exponent to antisymmetric modes, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p^2=\frac{\omega ^2}{c_L^2}-k^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q^2=\frac{\omega ^2}{c_T^2}-k^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c_L^2=\frac{2\mu +\lambda }{\rho }$ is the longitudinal wave-speed and $c_T^2=\frac{\mu }{\rho }$ is the transverse wave-speed ($\mu$ and $\lambda$ being the Lame constants). From this equation one obtains certain dispersion curves, relating the wave-number of each mode to the frequency. My question is: For each value of the frequency, the Rayliegh-Lamb equations give a discrete number of wave-numbers. How do you decide which wave-number belongs to which mode? That is, if I have a set of wave-numbers corresponding to one value of the frequency, say $\omega_1$, and then I get a new set of wave-numbers corresponding to a different frequency, say $\omega_2$, then how do I identify wave-numbers corresponding to different frequencies as belonging to the same mode?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope I've explained myself adequately. You can see that my question can also be framed for many other trascendental equations with multiple roots that show up in mathematical physics. I chose this particular problem to make my point. One might think that it is just a matter of ordering the wave-numbers in increasing order. But I've seen dispersion curves in which the curves for different modes intersect, so an ordering criterion is not applicable (not sure if this happens for the Rayleigh-Lamb modes, though)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. And sorry for the brick of text.&lt;/p&gt;&#xA;" OwnerUserId="3064" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-08-12T04:06:38.850" LastActivityDate="2011-08-12T04:06:38.850" Title="Rayleigh-Lamb dispersion curves" Tags="&lt;mathematical-physics&gt;&lt;rayleigh-lamb-equations&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="10850" PostTypeId="2" ParentId="10835" CreationDate="2011-06-07T02:11:43.377" Score="2" ViewCount="" Body="&lt;p&gt;The Stern Gerlach experiment is such an example (although we have an extra complication because the wave function splits)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) There is a change in the EM Field energy $\tfrac12 B$ because the total magnetic field (static field + electron's own magnetic field) becomes less. A &lt;em&gt;static&lt;/em&gt; field in isolation doesn't change per definition and its energy contents doesn't change either. To do work it needs to provide energy. The energy is provided by the total magnetic field and not the static field in isolation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The equivalent view is that the work is provided by the internal energy -$\mu\cdot B^2$. This is also true, both views are equivalent, even if the energies actually come from different parts of the Lagrangian since ultimately -$\mu\cdot B$ comes from the interaction term $j_\mu A^\mu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equivalence also hold classically even though the math for dipoles is somewhat involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards, Hans &lt;/p&gt;&#xA;" OwnerUserId="3197" LastActivityDate="2011-06-07T02:11:43.377" CommentCount="7" />
  <row Id="10851" PostTypeId="2" ParentId="9386" CreationDate="2011-06-07T02:37:39.987" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Real LC circuits have some resistance,&#xA;  which wastes some of the energy as&#xA;  thermal radiation, and the cycling&#xA;  eventually dies. I think they also&#xA;  have some other non-idealities that&#xA;  allow energy to escape as far field&#xA;  electromagnetic radiation, correct?&#xA;  What are these non-idealities? Are&#xA;  they independent of the resistive&#xA;  component?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Energy will also be lost due to electromagnetic radiation. The amount of energy loss depends on the size of the circuitry. In the limit of infinitely small inductors and capacitors the loss to radiation falls to zero. In general, you want the circuit to be smaller than the natural wavelength of the radiation it produces. For example, if the frequency is 1GHz the period is 1 nsec. With the speed of light $3\times 10^8$ m/s, the wavelength is $(3\times 10^8)/10^9 = 0.3$ meters so you want your circuitry to be much smaller than 0.3 meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to far field radiation, all circuit components have many other non-idealities and some of these result in thermal losses. Mostly it's about inductors, capacitors, and wires having resistance as the other circuit elements do not create heat.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If another identical LC circuit is&#xA;  brought near this oscillating circuit,&#xA;  will energy transfer from one circuit&#xA;  to the other due to the changing&#xA;  magnetic fields?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Will this happen even with ideal&#xA;  components that don't have far field&#xA;  radiation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. To get near field radiation you need to have an antenna and these are not included in ideal circuit components. In the limit of infinitely small circuit components the near field is infinitely small. (This is an instinct from many years of circuit design; I'll check it more carefully and remove this comment if I'm assured of the answer.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there no energy transfer if the&#xA;  coils are perpendicular and the axis&#xA;  of one is coplanar with the other,&#xA;  like this?: | —&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Correct.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Will the energy distribute itself&#xA;  evenly among the circuits, so they&#xA;  reach an equilibrium where each is&#xA;  circulating half of the original&#xA;  energy? Or will the energy oscillate&#xA;  back and forth between the two&#xA;  circuits, with one alternately being&#xA;  depleted or full?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a wonderful question! What you have is a pair of coupled oscillators. In physics we analyze these sorts of things by looking at what kind of activity would NOT alternate between the two circuits. These are called the &quot;normal modes&quot; and since there are two harmonic oscillators here, there will be two normal modes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at the problem from a general point of view. Let $(1,a)$ and $(1,b)$ be the two normal modes where $a$ and $b$ are complex numbers. That is, I've normalized the normal modes to the first oscillator. These two normal modes each has an exponential (i.e. cosine plus sine) dependence. That is, the first would actually look like $(1,a)\cos(\omega_1 t)$ and the second with an unequal $\omega_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get all the energy into the second oscillator (at time t=0), requires taking equal and opposite amounts of the two modes. Thus the values would be&#xA;$$(\cos(\omega_1 t) - \cos(\omega_2 t), a\cos(\omega_1 t)-b\cos(\omega_2 t))$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, we look at the 2nd part of the above. That is, I've arranged to match the phases and sum to zero for the first part at time t=0. At some later time will the second part also match phase and sum to zero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that this can only happen if the magnitudes of $a$ and $b$ are the same. So in general, the answer to your question is that a long term oscillation which sees all the energy in one of the circuits will not, in general, eventually get to a point where all the energy transfers to the other circuit.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-07T02:37:39.987" CommentCount="7" />
  <row Id="10852" PostTypeId="1" AcceptedAnswerId="10920" CreationDate="2011-06-07T02:48:28.017" Score="0" ViewCount="116" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/6646/physics-of-simple-collisions&quot;&gt;Physics of simple collisions&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I have 2 photos of the balls, one before the collision and one after the collision. They do a elastic collision. I want to know how is the reflected value calculated. I know it uses dot product and such stuff. But it all is very vague to me. Can anybody show me a working calculation and explain me based on this figures? I have been trying to understand this for past 2 days but I can't get how it all works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the image before collision:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/nNpCv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the image after collision:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/vlq45.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance :)&lt;/p&gt;&#xA;" OwnerUserId="3904" LastActivityDate="2011-06-08T18:37:10.387" ClosedDate="2011-06-08T17:57:14.540" Title="How are these balls reflected after they hit each other?" Tags="&lt;collision&gt;&lt;reflection&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10853" PostTypeId="2" ParentId="10848" CreationDate="2011-06-07T05:07:31.990" Score="0" ViewCount="" Body="&lt;p&gt;Hawking thought - and could &quot;rigorously&quot; deduce from semiclassical gravity - that the information has to be lost because it can't get out of the black hole interior once it gets there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/clVeX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see why, look at this &quot;Penrose causal diagram&quot; that may be derived for a black hole solution. Diagonal lines at 45 degrees are trajectories of light, more vertical lines are time-like (trajectories of massive objects), more horizontal lines are space-like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look e.g. at the yellow surface of the star (its world line), you see that it ultimately penetrates through the green event horizon into the purple black hole interior. Once the object - and the information it carries - is inside, it can no longer escape outside (don't forget: time is going up), into the light green region, because it would have to move along spacelike trajectories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the information carried by the star that collapsed to the black hole inevitably ends at the violet horizontal singularity and it may never be seen in a completely different region outside the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The information loss of course depends on the detailed geometry of the black hole that is not shared by a helium nucleus, so it shouldn't be surprised that helium nuclei and black holes have different properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we know today, the information is allowed to &quot;tunnel&quot; along spacelike trajectories a little bit in quantum gravity (i.e. string/M-theory). This process is weak but this weak &quot;non-local process&quot; allows the information to be preserved.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-06-07T05:07:31.990" CommentCount="3" />
  <row Id="10854" PostTypeId="2" ParentId="10811" CreationDate="2011-06-07T05:34:21.650" Score="3" ViewCount="" Body="&lt;p&gt;Yes, P will observe a thermal radiation that is locally identical to the Hawking radiation and it is called the Unruh radiation. Its temperature is $a/2\pi$ in the $\hbar=c=k_B=1$ units.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Unruh_radiation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Unruh_radiation&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A few more words about the relationship of Unruh and Hawking radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unruh found his derivation after Hawking found his - even though Unruh's radiation is kind of simpler and more elementary. This anomaly shows that Hawking didn't have any problems with doing more complicated calculations than others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the magic of the black hole background is that the killing vector field $\partial/\partial t$ where $t$ is the ordinary Schwarzschild time coordinate has the character of a &quot;non-accelerating&quot; observer at infinity (away from the black hole) but near the horizon, it is severely accelerating (not to fall to the black hole). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter fact means that in this reference frame, the observer sitting above a black hole horizon (by using his jets) will see the what looks like the Unruh radiation, but because his $t$ coordinate is continuously interpolated to the &quot;static time&quot; at infinity, this radiation will become real at infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important to realize that a star without an event horizon doesn't emit any Hawking radiation. The boundary conditions at the star surface are completely different than at the event horizon. In the case of a star, we can't argue that a &quot;freely falling&quot; observer will see a nice local empty vacuum - because a freely falling observer gets killed when he collides with the star's surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, let's mention that OP's original picture - a way of dividing the Minkowski space by an event horizon for an accelerating observer - is known as the Rindler space:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rindler_coordinates&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Rindler_coordinates&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The event horizon for the observer is observer-dependent and is also known as the Rindler horizon.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-06-07T19:47:02.510" LastActivityDate="2011-06-07T19:47:02.510" CommentCount="2" />
  <row Id="10858" PostTypeId="2" ParentId="10800" CreationDate="2011-06-07T08:22:02.890" Score="4" ViewCount="" Body="&lt;p&gt;There are two essential facts that make a hole a hole:&#xA;&lt;strong&gt;Fact (1)&lt;/strong&gt; The valence band is almost full of electrons (unlike the conduction band which is almost empty);&#xA;&lt;strong&gt;Fact (2)&lt;/strong&gt; The dispersion relation near the valence band maximum curves in the opposite direction to a normal electron or a conduction-band electron.&#xA;Fact (2) is often omitted in simplistic explanations, but it's crucial, so I'll elaborate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;STEP 1:&lt;/strong&gt; Dispersion relation determines how electrons respond to forces (via the concept of effective mass)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EXPLANATION:&lt;/strong&gt; A dispersion relation is the relationship between wavevector (k-vector) and energy in a band, part of the band structure. Remember, in quantum mechanics, the electrons are waves, and energy is the wave frequency. A localized electron is a wavepacket, and the motion of an electron is given by the formula for the group velocity of a wave. An electric field affects an electron by gradually shifting all the wavevectors in the wavepacket, and the electron moves because its wave group velocity changes. Again, the way an electron responds to forces is entirely determined by its dispersion relation. A free electron has the dispersion relation $E=\frac{\hbar^2k^2}{2m}$, where m is the (real) electron mass. In the conduction band, the dispersion relation is $E=\frac{\hbar^2k^2}{2m^*}$ ($m^*$ is the &quot;effective mass&quot;), so the electron responds to forces &lt;em&gt;as if&lt;/em&gt; it had the mass $m^*$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;STEP 2:&lt;/strong&gt; Electrons near the top of the valence band behave like they have negative mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EXPLANATION:&lt;/strong&gt; The dispersion relation near the top of the valence band is $E=\frac{\hbar^2k^2}{2m^*}$ with &lt;em&gt;negative&lt;/em&gt; effective mass. So electrons near the top of the valence band behave like they have negative mass. When a force pulls the electrons to the right, these electrons actually move left!! I want to emphasize again that this is &lt;em&gt;solely&lt;/em&gt; due to &lt;strong&gt;Fact (2)&lt;/strong&gt; above, not &lt;strong&gt;Fact (1)&lt;/strong&gt;. If you could somehow empty out the valence band and just put one electron near the valence band maximum (an unstable situation of course), this electron would really move the &quot;wrong way&quot; in response to forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;STEP 3:&lt;/strong&gt; What is a hole, and why does it carry positive charge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EXPLANATION:&lt;/strong&gt; Here we're finally invoking &lt;strong&gt;Fact (1)&lt;/strong&gt;. A hole is a state without an electron in an otherwise-almost-full valence band. Since a full valence band doesn't do anything (can't carry current), we can calculate currents by starting with a full valence band and &lt;em&gt;subtracting&lt;/em&gt; the motion of the electrons that &lt;em&gt;would be&lt;/em&gt; in the hole state if it wasn't a hole. Subtracting the current from a negative charge moving is the same as adding the current from a positive charge moving on the same path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;STEP 4:&lt;/strong&gt; A hole near the top of the valence band move the same way as an electron near the top of the valence band &lt;em&gt;would&lt;/em&gt; move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EXPLANATION:&lt;/strong&gt; This is blindingly obvious from the definition of a hole. But many people deny it anyway, with the &quot;parking lot example&quot;. In a parking lot, it is true, when a car moves right, an empty space moves left. But electrons are not in a parking lot. A better analogy is a bubble underwater in a river: The bubble moves the same direction as the water, not opposite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;STEP 5:&lt;/strong&gt; Put it all together. From Steps 2 and 4, a hole responds to electromagnetic forces in the exact opposite direction that a normal electron would. But wait, that's the same response as it &lt;em&gt;would have&lt;/em&gt; if it were a normal particle with positive charge. Also, from Step 3, a hole in fact carries a positive charge. So to sum up, holes (A) carry a positive charge, and (B) respond to electric and magnetic fields as if they have a positive charge. That explains why we can completely treat them as real mobile positive charges in their response to both electric and magnetic fields. So it's no surprise that the Hall effect can show the signs of mobile positive charges.&lt;/p&gt;&#xA;" OwnerUserId="3811" LastActivityDate="2011-06-07T08:22:02.890" CommentCount="2" />
  <row Id="10859" PostTypeId="1" AcceptedAnswerId="10862" CreationDate="2011-06-07T09:38:31.927" Score="2" ViewCount="101" Body="&lt;p&gt;(Theoretically)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any differences between photons emited /absorbed by antimatter atoms to photon in usual atoms?&#xA;for example, is it impossible to tell the difference between a photon emmited by an atom and it's anti matter counter part? Assuming the same orbital transitions for their electorn and positrons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the absorptions lines the same as well? or are there absoption lines for anti hydrogen are expected to be different than the hydrogen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is convention for anti elements? e.g is anti matter counter part of He is just denoted as -He or $\bar{He}$?&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-06-07T10:19:50.143" LastActivityDate="2011-06-07T13:18:43.647" Title="Are there any differences between photons emited /absorbed by antimatter atoms to photon in usual atoms?" Tags="&lt;antimatter&gt;" AnswerCount="2" />
  <row Id="10861" PostTypeId="2" ParentId="10837" CreationDate="2011-06-07T10:09:28.160" Score="1" ViewCount="" Body="&lt;p&gt;There is spin of proton and electron in a hydrogen atom. Astronomers know the 21 cm wave, but &#xA;when teaching hydrogen atom, Bohr and Schrödinger etc, spins are omitted. &#xA;I think this is to avoid complications in introductory level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_line&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hydrogen_line&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-06-07T10:09:28.160" />
  <row Id="10862" PostTypeId="2" ParentId="10859" CreationDate="2011-06-07T10:13:31.363" Score="5" ViewCount="" Body="&lt;p&gt;Yes, photons are the same if produced by matter or antimatter. A simple way to see this is that electromagnetism is CP, C and P invariant. Then, changing particles by antiparticles in any process one should obtain exactly the same &quot;amplitude&quot; (i.e. energy levels, probability of absortion emmission, etc...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weak interactions are neitehr CP nor C or P invariant,and there you can find processes where the probability is different if you change all particles by antiparticles. Take a look to &lt;a href=&quot;http://en.wikipedia.org/wiki/CP_violation&quot; rel=&quot;nofollow&quot;&gt;CP violation&lt;/a&gt; in wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About antielemtns, as far as I know, it is difficult to produce them experimentally, and I seriously doubt that something like anti-Helium have ever been measured. Only the nuclei of antihelium have been produced, but the full atom with positrons orbiting around is by far other history. Take a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Antimatter&quot; rel=&quot;nofollow&quot;&gt;wikipedia again&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3830" LastActivityDate="2011-06-07T10:13:31.363" CommentCount="3" />
  <row Id="10863" PostTypeId="1" AcceptedAnswerId="10874" CreationDate="2011-06-07T11:27:14.450" Score="3" ViewCount="155" Body="&lt;p&gt;Consider &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/alptun.html#c2&quot; rel=&quot;nofollow&quot;&gt;this explanation of the alpha decay&lt;/a&gt;: It says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Coulomb barrier faced by an alpha&#xA;  particle with this energy is about 26&#xA;  MeV, so by classical physics it cannot&#xA;  escape at all. Quantum mechanical&#xA;  tunneling gives a small probability&#xA;  that the alpha can penetrate the&#xA;  barrier.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/RF04N.gif&quot; alt=&quot;diagram of the barrier&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The explaining sentence and the attached picture suggest that the alpha particle has to overcome the coulomb barrier in order to escape - and as is hasn't  enough energy to do that, it must tunnel through it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the Coulomb force is a &lt;em&gt;repulsive&lt;/em&gt; force in this case as both the nucleus and the alpha particle are positively charged. So why is there any Coulomb barrier at all, shouldn't the Coulomb force &quot;help&quot; the alpha particle to escape? (The other way round, it's obvious - the particle wants to enter the nucleus, but it is hindered by the repulsive force. But when it wants to leave the nucleus, it is only retained by the nuclear force, as far as I can see.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why does the alpha particle have to go trough the coulomb barrier - or am I just misinterpreting the explanation and the tunneling occurs somewhere else?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="3908" LastActivityDate="2011-06-07T17:03:10.137" Title="Tunneling of alpha particles" Tags="&lt;quantum-mechanics&gt;&lt;nuclear-physics&gt;&lt;radiation&gt;&lt;tunneling&gt;" AnswerCount="2" />
  <row Id="10864" PostTypeId="2" ParentId="9341" CreationDate="2011-06-07T12:03:41.450" Score="0" ViewCount="" Body="&lt;p&gt;Because there is no correct answer for a relativistic case i decided to submit an answer which is correct, in my opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We start from the  rocket motion relation which is correct for both, relativistic and non-relativistic, cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dt}M\overrightarrow v=\overrightarrow u\frac{dM}{dt};\overrightarrow u= \overrightarrow {u&amp;#39;}+\overrightarrow v$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $M$ is the total mass of a rocket(including fuel), $\overrightarrow v$ is the velocity vector of the rocket and $\overrightarrow u$ is the velocity vector of gas jet. $\overrightarrow {u&amp;#39;}$ is velocity vector of the ejected mass&#xA;with respect to the missile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\overrightarrow v$ and $\overrightarrow u$ are taken relative to the inertial coordinate system, which deals with the motion (rather than relative&#xA;with respect to the missile).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a relativistic case we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M=\frac{M&amp;#39;}{\sqrt{1-\frac{v^2}{c^2}}}$$  where  $M&amp;#39;$ is a variable rest mass of the rocket in the the moving coordinate system attached to the rocket. After substituting and skipping the math, the reativistic motion equation displays as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{M&amp;#39;}{\sqrt{1-\frac{v^2}{c^2}}}\frac{d\overrightarrow v}{dt}=(\overrightarrow u- \overrightarrow v)\frac{d}{dt}\left (\frac{M&amp;#39;}{\sqrt{1-\frac{v^2}{c^2}}}\right )$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let assume that the acceleration occurs in the positive direction of the $x$-axis. Then the last equation becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{M&amp;#39;}{\sqrt{1-\frac{v^2}{c^2}}}\frac{dv}{dt}=(u_x-v)\frac{d}{dt}\left (\frac{M&amp;#39;}{\sqrt{1-\frac{v^2}{c^2}}}\right )$$  By the reativistic law of velocity-addition we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u&amp;#39;_x=\frac{u_x-v}{1-\frac{vu_x}{c^2}}$$ where $u&amp;#39;_x$ is velocity of the ejected mass&#xA;with respect to the missile.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After substituting and skipping the math we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dM&amp;#39;}{M&amp;#39;}=-\frac{1}{u&amp;#39;}\frac{dv}{1-\frac{v^2}{c^2}}$$ Here we took $u&amp;#39;_x=-u&amp;#39;$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After integrating we have finally:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{M&amp;#39;}{M&amp;#39;_0}=\left(\frac{c-v}{c+v}\right)^{\frac{c}{2u&amp;#39;}}$$ where $M&amp;#39;_0$ is total mass of the rocket at rest ($v=0$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a photon rocket case it is sufficient the substitution $u&amp;#39;=c$ here.&lt;/p&gt;&#xA;" OwnerUserId="552" LastActivityDate="2011-06-07T12:03:41.450" />
  <row Id="10865" PostTypeId="2" ParentId="9731" CreationDate="2011-06-07T12:08:07.000" Score="-1" ViewCount="" Body="&lt;p&gt;All the three above answers basically are correct and saying almost the same thing, i.e. the end product of matter-antimatter collision is photon at low energies and massive particles such as mesons... at high energies. However,by thinking of the reverse action namely production of matter from a form of energy ( radiation/light) and the source of that unimaginable huge energy,simply using E=mc*c,supports the above mentioned answers firmly. &lt;strong&gt;AA&lt;/strong&gt; 7 June 20011 at10pm&lt;/p&gt;&#xA;" OwnerUserId="3909" LastActivityDate="2011-06-07T12:08:07.000" />
  <row Id="10866" PostTypeId="1" CreationDate="2011-06-07T12:20:06.903" Score="0" ViewCount="101" Body="&lt;p&gt;While the first order metric variation of Hilbert-Einstein action plus Gibbons-Hawking-York boundary term is well-known and takes the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta S_{HE}+\delta S_{GHY}=-\frac{1}{16\pi G}\int d^3x \sqrt{-\gamma}~(K^{\mu\nu}-K\gamma^{\mu\nu})\delta \gamma_{\mu\nu}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($K_{\mu\nu}$ and $\gamma_{\mu\nu}$ are the extrinsic curvature and the projection metric), does anybody know the second order variation of this combined action? Thanks to anyone who can help me!&lt;/p&gt;&#xA;" OwnerUserId="2928" LastActivityDate="2011-06-07T15:44:11.910" Title="2nd order variation of Hilbert-Einstein action + Gibbons-Hawking-York boundary term" Tags="&lt;general-relativity&gt;&lt;variational-principle&gt;" AnswerCount="1" />
  <row Id="10867" PostTypeId="2" ParentId="10859" CreationDate="2011-06-07T12:30:06.173" Score="2" ViewCount="" Body="&lt;p&gt;According to the theory, there is no difference between matter and antimatter radiation of photons. Experimentally, of course, there is a big difference - antielements are much harder to obtain and thus make an antimatter radiation source ;-). &lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-06-07T13:18:43.647" LastActivityDate="2011-06-07T13:18:43.647" />
  <row Id="10868" PostTypeId="2" ParentId="10863" CreationDate="2011-06-07T12:36:55.713" Score="3" ViewCount="" Body="&lt;p&gt;As you state the Coulomb force is repulsive, this is why the potential increases with smaller spearation. But at small enough distances the strong force takes over which has an attracting nature. The strong force is stronger (hence the name) than the Coulomb force, which is why in total the alpha particle in the nucleus sits at a lower potential than the Coulomb force at the edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strong nuclear force can be modeled as a rectangular potential well to first order.&lt;/p&gt;&#xA;" OwnerUserId="2151" LastActivityDate="2011-06-07T12:36:55.713" />
  <row Id="10869" PostTypeId="2" ParentId="10811" CreationDate="2011-06-07T12:58:35.623" Score="0" ViewCount="" Body="&lt;p&gt;The apparent and particle horizons are mathematical artifacts that come about when you consider one object traveling at a fixed speed vs another that's traveling at an increasing speed.&#xA;For example, if you're trying to run (fairly fixed speed) to catch up to someone on a bike, you'd better be pretty close to them while they're first starting (accelerating).  You can catch up at first, but if they get a chance to get up to speed a little, then you'll never catch up.&#xA;These two situations (can catch up / can't catch up) are on two sides of the &quot;bike horizon&quot;.&#xA;The usual meaning of &quot;event horizon&quot; is in the context of a black hole.  There the horizon is a very local effect, and it exists in a very specific place (very different).  An accelerating observer won't see the horizon in a different place.&#xA;If the horizon moved from the perspective of an observer, then you'd start getting contradictions between different observers.  Namely, one person could see particles being emitted from a &quot;lower horizon&quot; that's below where someone else sees the horizon.  However, the other person must also see those particles (they're there, after all), but would see them coming through their &quot;higher horizon&quot; (Hawking radiation is emitted from just above the black hole's horizon, nothing passes through).&lt;/p&gt;&#xA;" OwnerUserId="2760" LastActivityDate="2011-06-07T12:58:35.623" CommentCount="2" />
  <row Id="10870" PostTypeId="1" AcceptedAnswerId="10888" CreationDate="2011-06-07T14:18:50.390" Score="3" ViewCount="97" Body="&lt;p&gt;Suppose two pointed masses are given in space. Suppose further that one of the masses has a given velocity at (local) time 0. Is there a way to compute its position in a future time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neglecting general relativity, I will simply compute an integral, but with general relativity, we see that the metric of the space changes with time, so I need to compute an integral with respect to a measure that changes along time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this be done? If so, how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="3913" LastActivityDate="2011-06-07T18:49:59.617" Title="Calculating position in space assuming general relativity" Tags="&lt;general-relativity&gt;" AnswerCount="1" />
  <row Id="10871" PostTypeId="1" AcceptedAnswerId="10890" CreationDate="2011-06-07T14:35:35.787" Score="3" ViewCount="124" Body="&lt;p&gt;How long does it take for a photon to be reflected? Starting with the photon being absorbed by some atom to the point it's reemitted?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what's the same point with pressure waves, like sound?&lt;/p&gt;&#xA;" OwnerUserId="3914" LastActivityDate="2011-06-07T19:40:53.043" Title="How long do reflections take?" Tags="&lt;electromagnetic-radiation&gt;&lt;sound&gt;&lt;reflection&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="10872" PostTypeId="2" ParentId="10866" CreationDate="2011-06-07T15:44:11.910" Score="1" ViewCount="" Body="&lt;p&gt;A long time ago, when I was in graduate school, a friend and I went to Bryce DeWitt with a similar question. He thought about it for a while, and then said to us:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; &quot;Do you know what your problem is? Too much book learning.&quot;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;He explained that the answer could almost certainly be found in a book or paper, if we looked hard enough, but we would learn much more by working out the answer ourselves. This may be the best advice I ever received as a graduate student. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why am I telling you this? Because the answer to &lt;a href=&quot;http://physics.stackexchange.com/questions/10606/explicit-variation-of-gibbons-hawking-york-boundary-term/10628#10628&quot;&gt;your question about the first variation of the action&lt;/a&gt; (and &lt;a href=&quot;http://jacobi.luc.edu/notes.html&quot; rel=&quot;nofollow&quot;&gt;the notes&lt;/a&gt; I linked to) provides you with everything you need to solve this problem. Give it a try! If you get stuck on a specific step in the derivation, come back here and ask about it -- I'll be happy to work through the details with you. But this is a great chance for you to apply what you learned from your last question.&lt;/p&gt;&#xA;" OwnerUserId="1727" LastActivityDate="2011-06-07T15:44:11.910" CommentCount="1" />
  <row Id="10873" PostTypeId="1" CreationDate="2011-06-07T15:46:47.223" Score="0" ViewCount="40" Body="&lt;p&gt;when i am walking in our railway station at night i observed a train coming with  lights i thought that two trains are coming aside to each other.But it was only one train having two lights .Which property of light makes it look like so, what is the reason of that one &lt;/p&gt;&#xA;" OwnerUserId="3590" LastActivityDate="2011-06-07T15:46:47.223" ClosedDate="2011-06-07T16:59:12.450" Title="When we see two lights behind one another it will look like there are two lights aside what is the reason?" Tags="&lt;light&gt;" CommentCount="2" />
  <row Id="10874" PostTypeId="2" ParentId="10863" CreationDate="2011-06-07T15:59:07.103" Score="3" ViewCount="" Body="&lt;p&gt;This model is something that I've seen a lot of students (including myself) struggle with, and a big part of it is that calling it the &quot;Coulomb barrier&quot; is quite over-simplified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going back to fundamental physics principles, the Coulomb force is repulsive as you say, so I'll write this as $F\propto 1/r^2$, where $F$ is the force and $r$ is the distance from the center of the nucleus and the positive sign indicates that the force is in the positive $r$ direction, away from the nucleus.  The potential, which is what's shown on the graph, is the negative of the integral of the field (which is proportional to the force), this results in $E\propto 1/r$.  For an analog, imagine lifting a weight where &quot;up&quot; is the positive z-axis.  The path involves an increase in z-coordinate and the force from the field (gravitational) is negative.  The negative integral of the field over distance is the potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The missing part is why they &quot;chop off&quot; this $1/r$ function at some radius, presumably the radius of the nucleus.  Well, this allows us to infer some things about the &lt;em&gt;other&lt;/em&gt; force present, which is the nuclear force.  It would appear that the force acts weakly beyond the radius of the nucleus, since the $1/r$ form is unaltered beyond that point.  It also &lt;em&gt;absolutely must&lt;/em&gt; have a potential that increases faster than $1/r$ as $r \rightarrow 0$.  Both of these would be loosely satisfied if the nuclear potential was $-1/r^n$ (note the negative) where $n&amp;gt;1$.  The alpha particle, while in the nucleus, by the way, has kinetic energy which is the reason its energy is higher than the hypothetically lowest energy level possible for it.  This has to do with the fact that quantum mechanics only allows certain (quantized) energy levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just For an example of how this is possible, I'll say $E_{nuclear}=-1/r^4$ (I am not saying this is how the force acts, it's just for utility).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E(r) = E_{nuclear}(r) + E_{Coulomb}(r)$$&#xA;$$E(r) = 1/r - 1/r^4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(r) = -\frac{d}{dr} \left( E_{nuclear}(r) + E_{Coulomb}(r) \right)  = F_{nuclear}(r) + F_{Coulomb}(r) $$&#xA;$$F(r) = 1/r^2 - 1/r^5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/MDMjL.gif&quot; alt=&quot;My sample &amp;quot;edge&amp;quot; for the Coulomb barrier&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intent is that this answer allows you to explicitly answer to yourself what the potential and force contributions are, and give some very rudimentary example for how the &quot;edge&quot; can appear.  Beyond that I hope it's clear how someone may take the shape of the above curve and lump it into a &quot;wall&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-06-07T17:03:10.137" LastActivityDate="2011-06-07T17:03:10.137" CommentCount="3" />
  <row Id="10876" PostTypeId="1" AcceptedAnswerId="10929" CreationDate="2011-06-07T17:53:29.017" Score="3" ViewCount="73" Body="&lt;p&gt;I want to create a set of silicon based materials that have been doped with different materials and/or different amounts of dopants.  The purpose of this is to see how the spring constant of silicon changes with the different electrical properties, specifically, with the change in resistivity (because we are all set up to measure resistivity to a highly accurate degree).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I know that the physical deposition of the interstitial dopants will, by themselves, change the spring constant.  What I am looking for is effects beyond that, effects that are due to changes in the electron mobility as quantified via resistivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, the particular material (silicon) and the various dopants are actually irrelevant to this question.  They were included for context only.  My true question is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any formulas or concepts that will explain or predict how does the spring constant of a given material change as its resistivity changes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any papers cited and/or books (Kittel, Mermin, Dieter, etc) would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the only way to understand this is through the densities of the dopants vs substrate, or any other of the usual suspects, that's fine.  But what I would really like is a more &quot;magical&quot; argument I guess:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a material with a certain resistivity.  I wave a magic wand and change its resistivity.  What is the new spring constant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hmm, I hope that's clear.  Let me try one more:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for an Ohm's &quot;law&quot; like equation that relates spring constant to resistivity with a few physical constants thrown into a constant somewhere in the equation.  Does such a thing exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, well, any help is appreciated!&lt;/p&gt;&#xA;" OwnerUserId="1037" LastActivityDate="2011-06-08T22:49:11.370" Title="How does spring constant change with resistivity changes" Tags="&lt;experimental-physics&gt;&lt;condensed-matter&gt;&lt;spring&gt;&lt;resistors&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10877" PostTypeId="2" ParentId="10833" CreationDate="2011-06-07T18:16:00.807" Score="1" ViewCount="" Body="&lt;p&gt;Speed of light ---&gt; about 3E8 m/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Round trip length ---&gt; 3200 m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time for round trip ---&gt; 3200/(3E8) = about 1.066E-5 = 10.66 uS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mirror turns 1/32 of a full turn in 10.66 uS; so it makes a full turn in (32 x 10.66) uS = 341uS = 0.341mS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/(0.341E-3 s) = 2932/s = 2932 Hz (based on c = 3E8 m/s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the authors made the mistake of not factoring in the round trip, they should have come up with an answer in the rough vicinity of 2 x 2932/s = 5864 Hz.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is your third data point:  The authors made TWO errors: They forgot the round trip (as you did) making them off by a factor of two (to the high side) ~AND~ made a simple decimal error, which reduced their final answer by a factor of 10.  The end result is that they are off to the low side by a factor of five.  5 x 585 Hz = the correct answer of 2925 Hz.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-06-07T18:16:00.807" />
  <row Id="10878" PostTypeId="5" CreationDate="2011-06-07T18:16:23.240" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-06-07T18:16:23.240" LastActivityDate="2011-06-07T18:16:23.240" />
  <row Id="10879" PostTypeId="4" CreationDate="2011-06-07T18:16:23.240" Score="0" ViewCount="" Body="Electrostatics is concerned with the field and potential of stationary electrical charges and electric charge distributions.  Problems are this type are almost exclusively concerned with mathematics of geometries using the inverse-square law." OwnerUserId="-1" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-06-07T18:50:22.300" LastActivityDate="2011-06-07T18:50:22.300" />
  <row Id="10880" PostTypeId="5" CreationDate="2011-06-07T18:21:10.260" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-06-07T18:21:10.260" LastActivityDate="2011-06-07T18:21:10.260" />
  <row Id="10881" PostTypeId="4" CreationDate="2011-06-07T18:21:10.260" Score="0" ViewCount="" Body="Radiation that comes from pair production quantum effects in close vicinity to an event horizon, leading to the potential for eventual evaporation of black holes.  Two mirror particles are created with one falling behind the horizon, becoming casually lost to the rest of the universe, including its neighbor, allowing it to escape the gravitational potential." OwnerUserId="-1" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-06-07T18:50:31.817" LastActivityDate="2011-06-07T18:50:31.817" />
  <row Id="10882" PostTypeId="5" CreationDate="2011-06-07T18:26:23.647" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-06-07T18:26:23.647" LastActivityDate="2011-06-07T18:26:23.647" />
  <row Id="10883" PostTypeId="4" CreationDate="2011-06-07T18:26:23.647" Score="0" ViewCount="" Body="A force is an action on a body.  Formalization of the properties of force were accomplished in Newtonian mechanics which established F=ma as a law to describe motion based on acceleration which comes from the vector sum of forces on the object.  Problems about force are often concerned with the balancing of forces and description of physical systems using this concept." OwnerUserId="-1" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-06-07T18:50:28.247" LastActivityDate="2011-06-07T18:50:28.247" />
  <row Id="10884" PostTypeId="5" CreationDate="2011-06-07T18:33:43.910" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-06-07T18:33:43.910" LastActivityDate="2011-06-07T18:33:43.910" />
  <row Id="10885" PostTypeId="4" CreationDate="2011-06-07T18:33:43.910" Score="0" ViewCount="" Body="A black hole is a volume from which photons, or any matter, can not escape.  More formally, the coordinate speed of light at the event horizon - the boundary of a black hole - is zero, as measured by a sufficiently separated observer." OwnerUserId="-1" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-06-07T18:50:20.883" LastActivityDate="2011-06-07T18:50:20.883" />
  <row Id="10887" PostTypeId="2" ParentId="10827" CreationDate="2011-06-07T18:39:59.573" Score="3" ViewCount="" Body="&lt;p&gt;Short answer:  Aluminum is considered &quot;paramagnetic.&quot;  Most laymen would consider that to mean &quot;non-magnetic.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longer answer (and greatly simplifying): Materials fall into three categories, dependent upon their relative permeability (called 'Ur' in this post).  Almost all materials have a relative permeability which is very, very close to the relative permeability of free space --&gt; Ur = 1.  The classification is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diamagnetic: Ur is VERY close to 1, but slightly less than 1 (e.g., Ur = 0.99997).&#xA;Paramagnetic: Ur is VERY close to 1, but slightly more than 1 (e.g., Ur = 1.00003).&#xA;Ferromagnetic: Ur can be orders of magnitude greater than 1 (e.g., Ur = 7283.5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;~ALL~ materials have a diamagnetic effect.  This is a small magnetic field, which is only present if another magnetic field is induced on the material, and serves to OPPOSE the induced field.  If no other effect swamps out this effect, then the material is considered diamagnetic.  Some examples of diamagnetic materials are: bismuth, copper, lead, mercury, silver, gold, and diamond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some materials have a paramagnetic effect which overcomes the diamagnetic effect.  This is a small magnetic field, which is only present if another magnetic field is induced on the material, and serves to STRENGTHEN the induced field.    Paramagnetism (unlike diamagnetism) is temperature dependent, being stronger at lower temperatures.  A material which has this, but not ferromagnetism, is considered paramagnetic.  Examples of paramagnetic materials are: aluminum; magnesium; tungsten; and titanium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ferromagnetic materials can have their own magnetic field, independent of an applied magnetic field.  In addition, they can have very large relative permeabilities.  Examples of ferromaagnetic materials are: iron; nickel; and cobalt.  When ferromagnetic materials are heated above their &quot;curie temperature,&quot; they lose their ferromagnetic properties and become paramagnetic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there are ferrimagnetic (used in ferrite beads) and anti-ferromagnetic materials, which we won't get into.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the big breakpoint is between ferromagnetic materials and anything else.  It does not matter much to most laymen whether Ur is 0.99999 (diamagnetic) or 1.00001 (paramagnetic).  Both types of materials have a Ur of almost exactly 1.  They do not interrupt or divert magnetic lines of force.  Unless you have a very, VERY highly concentrated magnetic force (as in an MRI exam), there is no observable effect on these materials.  The layman considers them &quot;non-magnetic.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magneto:  Something to think about is that most metals do conduct electricity very well.  Inducing a circular current in a piece of metal would make that piece produce its own magnetic field (electromagnetism).  That field could then be acted upon by external forces to manipulate the object.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-06-07T19:38:22.220" LastActivityDate="2011-06-07T19:38:22.220" />
  <row Id="10888" PostTypeId="2" ParentId="10870" CreationDate="2011-06-07T18:49:59.617" Score="2" ViewCount="" Body="&lt;p&gt;If the moving mass is small enough, you can do this using the geodesic equation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}^2x^\lambda}{\mathrm{d}t^2} + \Gamma^{\lambda}_{\mu\nu}\frac{\mathrm{d}x^\mu}{\mathrm{d}t}\frac{\mathrm{d}x^\nu}{\mathrm{d}t} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is essentially the general relativistic equivalent of Newton's second law: it's a differential equation that governs how a test particle's position changes in time. The connection coefficients $\Gamma^{\lambda}_{\mu\nu}$ (also called Christoffel symbols) can be calculated from the metric. So if the metric is known, even if it's time-dependent, you can calculate the connection coefficients in your desired reference frame, plug them in, and find a solution to the geodesic equation that tells you how the particle will move. There may not be an analytic solution, but in all but the most extreme cases you can either solve the equation numerically, or make some approximation that might make it analytically solvable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the moving particle is not small enough to be considered a test particle, then the situation becomes more complicated because the metric, and thus the connection coefficients, will depend on the motion of the test particle itself. So you wind up with a coupled system of three equations: the geodesic equation for the moving particle, the geodesic equation for the other particle, and the Einstein equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G^{\mu\nu} = 8\pi T^{\mu\nu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which tells you how the metric changes in response to the motion of the two particles. In this case it's highly unlikely that you could find an analytic solution, but you could potentially still use a numeric differential equation solver, at least for some range of time. (All the equations are nonlinear so it's likely that your solution would lose accuracy quickly.)&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-07T18:49:59.617" />
  <row Id="10889" PostTypeId="1" CreationDate="2011-06-07T19:28:08.977" Score="0" ViewCount="180" Body="&lt;p&gt;Given the relationships between mass and energy in relativity, and given that particles with mass can be created given energy over the threshold energy, and vice-versa, can we say that mass is simply an extremely dense form of energy? Or is there a deceptive parallel between the two?&lt;/p&gt;&#xA;" OwnerUserId="202" LastActivityDate="2011-06-09T18:25:12.773" Title="Mass in terms of energy" Tags="&lt;energy&gt;&lt;mass&gt;&lt;relativity&gt;&lt;mass-energy&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="10890" PostTypeId="2" ParentId="10871" CreationDate="2011-06-07T19:40:53.043" Score="4" ViewCount="" Body="&lt;p&gt;If we're talking reflection, then it's best not to think of the photon being absorbed and then re-emitted by an atom.  Rather, think of the photon as having very many virtual interactions with the various electrons in the reflecting surface, the complex phases of which all get added up in a coherent way.  This will be described by the evolution of the photon's wave function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we are dealing with wave phenomenon, which are not localized in time, it is best to ask the question what happens to a wave packet being reflected from the surface.  The reflected wave packet will be delayed and stretched out.  The delay of the packets peak, or group delay, is a reasonable thing to identify with the time it takes to be reflected. This group delay will in turn be related to the skin depth of the reflecting surface.  For a metal say at optical frequencies, the skin depth will be on the order of several nanometers, and so the group delay will be on the order of 10E-17 seconds.  This reflection time will be much smaller then the uncertainty in the photon's time of arrival.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-06-07T19:40:53.043" CommentCount="3" />
  <row Id="10891" PostTypeId="2" ParentId="10293" CreationDate="2011-06-07T19:47:18.517" Score="4" ViewCount="" Body="&lt;p&gt;A resistor at a temperature T has a fluctuating voltage. This is a consequence of the fluctuation dissipation theorem which you can use to calculate the spectrum of the voltage. The wikipedia article on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fluctuation-dissipation_theorem&quot; rel=&quot;nofollow&quot;&gt;Fluctuation Dissipation Theorem&lt;/a&gt; has a section on resistor thermal noise. When measured over a bandwidth $\Delta\nu$, the average squared voltage is:&#xA;$$\langle V^2\rangle = 4Rk_BT\Delta\nu$$&#xA;where $k_B= 1.38\times 10^{-23}$J/K is &lt;a href=&quot;http://en.wikipedia.org/wiki/Boltzmann_constant&quot; rel=&quot;nofollow&quot;&gt;Boltzmann's constant&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have a resistor $R$ maintained at a temperature T, and hooked up to a resistor $R_0$ initially at absolute zero. The thermal noise of the warm resistor will be as given above. When you apply a voltage $V$ to a resistor $R_0$, the dissipation (in watts) will be given by $IV = V^2R_0$, so the watts applied to the resistor initially at absolute zero, over a bandwidth $\Delta\nu$ will be&#xA;$$\langle V^2\rangle = 4R_0Rk_BT\Delta\nu.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As that resistor warms up to temperature $T_0$, it will apply a fluctuating voltage on the warm resistor. Following the above, but with the two resistors swapped, the power applied to the warm resistor by the colder resistor at temperature $T_0$ will be:&#xA;$$\langle V^2\rangle = 4RR_0k_BT_0\Delta\nu.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system will be in balance when the above two powers are equal. This happens algebraically when $T=T_0$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A possible source of paradoxical confusion is that the above calculation was done over a limited bandwidth range. But the calculation does not depend on frequency; instead the power transmitted is simply proportional to the range of bandwidths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the usual physical system, we consider frequencies that run from 0 to infinity. Thus the total bandwidth is infinite. This suggests that the power flow in the above should be infinite. This paradox is avoided by noting that physical resistors have a limited bandwidth. There is always a parasitic capacitance so that the bandwidth is limited on the high side. Thus the power transfer rate depends on how ideal your resistors are.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As an example calculation, suppose that a resistor has a maximum frequency of 100 GHz $= 10^{11}$ Hz, a (room) temperature of 300K, and a resistance of 1000 ohms. Then the power transfer rate is:&#xA;$$ 4\times 1000 \times 1.38\times 10^{-23} 10^{11} \times 300 = 1.66\;\;\textrm{uWatts}$$&#xA;Given the heat capacity of the resistor, you can compute the relaxation time with which the colder resistor exponentially approaches an equal temperature.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-07T19:47:18.517" CommentCount="2" />
  <row Id="10892" PostTypeId="2" ParentId="10889" CreationDate="2011-06-07T20:21:02.730" Score="1" ViewCount="" Body="&lt;p&gt;I'll offer my take on this, which is very specific to me.  As an alternative to the vague notion that matter and energy are &lt;em&gt;interchangeable&lt;/em&gt; I claim the following&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&lt;li&gt;All energy has mass&#xA;&lt;li&gt;All mass has energy&lt;/li&gt;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To specify further, I will describe a process commonly thought as changing matter into energy, then a process commonly thought of as changing energy into mass, then show that neither of them &quot;loose&quot; either matter of energy in favor of creating the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start off with a nuclear reactor.  During operation the nuclear energy is changed into heat and electrical energy.  Say that this is done in a system that is thermally isolated.  It doesn't matter specifically what form that energy takes, because should it go into batteries, the weight of the batteries would increase, should be be stored as heat, the weight of the medium holding the heat increases.  This energy exists either in chemical bonds and kinetic motion respectively.  Both increase the mass of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For changing energy into mass, we can look at particle creation as you mention.  In order to create the particle, that energy had to exist previously, and sure enough, whatever reservoir held the energy before it was used experienced a decrease in mass corresponding with the movement of &quot;energy&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A recent question I wrested with was &lt;a href=&quot;http://physics.stackexchange.com/questions/10612/explain-how-or-if-a-box-full-of-photons-would-weigh-more-due-to-massless-photon&quot;&gt;Explain how (or if) a box full of photons would weigh more due to massless photons&lt;/a&gt;.  Even though photons are massless, if they are somehow confined they will increase the mass of the system they are a part of.  This is because no matter what transitions occur, the measured mass within a boundary that doesn't exchange mass or energy will remain constant.  Likewise, the energy of that system will remain constant.  This is &lt;em&gt;in spite of the fact&lt;/em&gt; that matter-energy transitions are apparently occurring.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-06-07T20:21:02.730" CommentCount="7" />
  <row Id="10893" PostTypeId="2" ParentId="10889" CreationDate="2011-06-07T20:42:56.417" Score="-4" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;and given that particles with mass can&#xA;  be created given energy over the&#xA;  threshold energy, and vice-versa,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, You can't really &lt;strong&gt;create&lt;/strong&gt; particles. You can make pairs of particle/antiparticle, &#xA;which will annihilate back soon. This means, You never can add an additional proton &#xA;to the number of existing protons in this universe. You can create mass, eg by some &#xA;nuclear reaction tranforming eg iron atoms in stable atoms with more nucleons. (Reverse process of fission or radioactive decay) &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-06-07T20:42:56.417" CommentCount="2" />
  <row Id="10894" PostTypeId="2" ParentId="10773" CreationDate="2011-06-07T20:48:33.497" Score="1" ViewCount="" Body="&lt;p&gt;You were told something which was not quite true, to illustrate a principle which is very true:  In simple Newtonian motion, the equations of motion in one axis of a coordinate system do not have any effect on motion in any orthogonal axis.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me restate the case to make it just a bit more truthful.  Lets say the bullet dropping to earth is motion in the 'z' direction.  motion in the horizontal direcion could be 'x' or 'y' or a comination of the two.  let's align our shooter so that he shoots exactly in the 'x' direction.  We further define our coordinate system such that the 'x' direction remains locked in the direction of fire after the bullet is fired (does not turn or move with the Earth.  Now lets place our shooter at the north (or south) pole, in a shooting range which is perfectly flat for at least 2000 feet.  Our shooter is in a space suit, and the enclosed shooting range is pumped down to a vacuum.  Finally, a mechanism is set up to drop a bullet at the exact time (t=0), and from the exact same height, as the shooter's bullet leaves the perfectly level gun barrel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both bullets will hit the range floor at exactly the same time (or so close you can't tell the difference).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dropped bullet has a z component of velocity... and nothing else.  It obeys the equation z(t) = z(0) - (a/2)(t^2); where z(0) = 6 feet, and 'a' = about 32 ft/(s^2).  From a height of, say, 6 feet, it will hit the floor in about 0.25 seconds (&quot;about 0.25...&quot; --&gt; I'm trying to keep the math simple).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shot bullet has a z and an x component of velocity.  The x component will be x(t) = (v)(t), where v = muzzle velocity = maybe 3000 ft/s.  The motion in the 'z' direction is exactly yhe same as for the dropped bullet: z(t) = z(0) - (a/2)(t^2), and z(0) = 6 feet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here is the tricky part: when does the shot bullet hit the floor?  That is motion in the z direction.  Of the two equations governing motion of the bullet, only one has anything to do with vertical motion, and its soluiton is exactly the same as for the dropped bullet: It hits the floor in about 0.25 seconds.  In that amount of time it will have traveled 750 feet in the x direction and will still have a forward velocity of 3000 ft/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The truth that a shot bullet and a dropped bullet fall at the same rate (neglecting secondary effects) is an important phenomenon, which must be taken into account for long range firing.  If a rifle is &quot;sighted in&quot; for a 100 yard target, it will shoot noticably lower (roughly 4 inches lower for a medium power rifle) on a 200 yard target.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-06-07T21:07:47.167" LastActivityDate="2011-06-07T21:07:47.167" />
  <row Id="10895" PostTypeId="1" CreationDate="2011-06-07T21:45:48.447" Score="3" ViewCount="58" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/2178/how-efficient-is-a-desktop-computer&quot;&gt;How efficient is a desktop computer?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I believe the energy conversion efficiency of a modern CPU is extremely low, because it dissipates pretty much all of it as heat, which is not a desired product. But what's the order of magnitude of the fraction that gets spent on the actual computation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there absolute maximums imposed on this efficiency by the known laws of physics, or can information processing occur without any waste heat generation as a by-product?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a bit out of my depth here, having no idea quite how information processing relates to heat and energy (other than suspecting that there is a very deep connection), so apologies if this is nonsensical or vague.&lt;/p&gt;&#xA;" OwnerUserId="786" LastActivityDate="2011-06-07T21:45:48.447" ClosedDate="2011-06-08T00:14:22.100" Title="What is the energy conversion efficiency of a computation device like a modern CPU?" Tags="&lt;energy&gt;&lt;energy-conservation&gt;&lt;information&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="10896" PostTypeId="2" ParentId="389" CreationDate="2011-06-07T23:33:57.267" Score="1" ViewCount="" Body="&lt;p&gt;A very simple picture for understanding the propagation of a positive hole would be those puzzles consisting of 15 square tiles, held together tongue-and-groove fashion in a square frame. There's always one empty space big enough to slide a tile into. This allows one to solve the puzzle by rearranging the tiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this picture, the &quot;hole&quot; is the empty space.  If the empty space is in position 16 at bottom right, it's easy to see that the &quot;hole&quot; can be shifted to position 4 at the top-right by three successive downward moves of the tiles occupying positions 12, then 8, then 4. Although the &quot;virtual&quot; space may not have the same physical reality as the &quot;real&quot; tile, anyone attempting to solve the puzzle quickly adapts to thinking in terms of &quot;shifting the space&quot; by a series of retrograde moves. Although the space has no surface that we can touch, it has a location, which can be manipulated indirectly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For many purposes, a real electron and a virtual positive hole can be thought of as similar, oppositely charged particles. This equivalency breaks down when one imagines building a cathode ray tube for positive holes.  An electron gun can easily fire an electron across the vacuum. But how could a gun propagate a positive hole? There's no real mass to accelerate across the wide gap between gun and phosphor screen. For the positive hole to breach the gap, there would have to be (as within a crystal) a whole chain of filled electron states, stretching unbroken from gun to phosphor, and each capable of sequentially yielding to the hole's forward progress by individual retrograde motions. Not gonna happen!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even so, it's fun to think about this: A &quot;virtual&quot; positive hole has no rest mass.  Yet it &lt;em&gt;can&lt;/em&gt; exert a force and have a measurable velocity. And so, it follows that we can calculate a &quot;virtual&quot; mass!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These simple pictures of course do not convey quantum reality. When a positive hole is activated within an oxide crystal, for instance, its absence isn't necessarily localized as in the tile puzzle as &quot;the absence of a negative charge where one would normally expect one to be.&quot; The effect of that absent charge might be smeared across the nearest 100 or 1,000 atoms, affecting conductivity of the whole neighborhood along the valence band. So that instead of one atom missing a single electron, we might think of a hundred atoms missing .01 electrons, or a thousand atoms missing .001 electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which depending on your point of view may be amusingly counterintuitive, spookily surreal, or just &lt;em&gt;the way things are&lt;/em&gt;. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Of course, the positron is like the proverbial 800 lb. gorilla by comparison to the hole. It has a very real mass equal to the electron. And one could imagine, with proper containment, accelerating a beam of positrons across a CRT, where by virtue of being an anti-particle, they would commence to obliterate themselves and the screen.&lt;/p&gt;&#xA;" OwnerUserId="3917" LastEditorUserId="3917" LastEditorDisplayName="" LastEditDate="2011-06-07T23:49:19.763" LastActivityDate="2011-06-07T23:49:19.763" />
  <row Id="10897" PostTypeId="2" ParentId="10782" CreationDate="2011-06-08T00:06:52.653" Score="4" ViewCount="" Body="&lt;p&gt;The formula for &lt;a href=&quot;http://en.wikipedia.org/wiki/Length_contraction&quot; rel=&quot;nofollow&quot;&gt;length contraction&lt;/a&gt; involves velocity only:&#xA;$$L&amp;#39; = L\sqrt{1-v^2/c^2}.$$&#xA;Note that acceleration does not contribute. So the answer to your question is obtained by simply taking the instantaneous velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the object goes around the circle, the length contraction changes according to the change in direction of the velocity.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-08T00:06:52.653" />
  <row Id="10898" PostTypeId="2" ParentId="10551" CreationDate="2011-06-08T00:25:50.770" Score="0" ViewCount="" Body="&lt;p&gt;The answer depends very much on the wavelength of the photons considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the wire spacing (not thickness) is large compared to the wavelength of the photons we are in the &quot;classical particle&quot; (as opposed to quantum wave) limit; then the amount of absorption is simply the percentage of area blocked. This limit applies when buildings block sunlight. The amount of sunlight blocked is proportional to the size of the buildings. In this limit polarization effects are negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the wire spacing is small compared to the wavelength of the photons we are in the wave limit. This is the limit of a perfect polarization filter. If we assume that the wires are perfect conductors, then the polarization effect does not decrease as the wire size decreases. In practice, wires of excessively small diameter become resistive so the limit is not achieved.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So I'm convinced that (when you assume perfect conductors) there will be a discontinuous change between an infinitesimal wire and no wire at all. Similarly there will be a discontinuous change between an infinitesimal slot and no slot at all. In both cases, it is the presence or absence of electrical conduction that distinguishes the change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In between the thin and thick slit limit, I believe there will be no change in transmission and reflection coefficients (in the limit of closely spaced wiring). This is because what is happening is that the electric field is either getting shorted out by wires, or it is not. There is no intermediate condition (unless one takes into account finite wire impedance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may seem bizarre that one could have full transmission when the slots are very thin. One can think of the slots as each being at some voltage. As the wires become wider, that voltage is maintained over the width of the wire. This means that the voltage difference between adjacent wires does not depend (much) on the thickness of the wires. (We are in the long wavelength limit.) Consequently, as the separations get narrower, the electric field (i.e. volts/meter) becomes larger. As the separation begins to close, the driving electric field goes to infinity. So it is not unintuitive that the transmission is still 100% when the wires are very fat.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-06-08T22:05:42.763" LastActivityDate="2011-06-08T22:05:42.763" CommentCount="4" />
  <row Id="10899" PostTypeId="2" ParentId="10234" CreationDate="2011-06-08T00:45:48.640" Score="2" ViewCount="" Body="&lt;p&gt;The usual physics assumption for spontaneous emission of radiation (which I think is what you're talking about here) is that of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson_process&quot; rel=&quot;nofollow&quot;&gt;Poisson process&lt;/a&gt;; that is, if one has a large number of possible emitters, the times between consecutive emissions is not correlated. On the other hand, if we have only a single emitter, then the time at which it emits follows the &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_distribution&quot; rel=&quot;nofollow&quot;&gt;exponential probability distribution&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These things seem to have nothing to do with Stochastic electrodynamics (but I admit I'd never heard of it and am not inclined to study it).&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-08T00:45:48.640" CommentCount="3" />
  <row Id="10900" PostTypeId="2" ParentId="10220" CreationDate="2011-06-08T01:00:24.203" Score="0" ViewCount="" Body="&lt;p&gt;Since there's only one square root, one would end up with an imaginary velocity rather than the more general case of a complex velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In physics, one occasionally deals with complex energies. I don't know what imaginary velocities will mean and there's not a lot of references in the literature. One paper is &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0004086v1&quot; rel=&quot;nofollow&quot;&gt;&quot;Velocities&quot; in Quantum Mechanics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-06-08T22:09:06.960" LastActivityDate="2011-06-08T22:09:06.960" CommentCount="2" />
  <row Id="10901" PostTypeId="2" ParentId="10199" CreationDate="2011-06-08T01:07:54.387" Score="1" ViewCount="" Body="&lt;p&gt;I think that singularities can occur in the electromagnetic fields (other than the ones occurring at the location of particles), if one allows the particles to move at speed $c$. That is, one can arrange for sonic booms or &lt;a href=&quot;http://en.wikipedia.org/wiki/Cherenkov_radiation&quot; rel=&quot;nofollow&quot;&gt;Cherenkov radiation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This raises the question &quot;is Cherenkov radiation smooth?&quot; I suspect that the non smoothness only occurs at distances larger than the distance between the individual atoms; that is, that the lack of smoothness is in the math approximation rather than the physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, if one does not allow the particles to approach the speed of light, I believe that the sum of all their electromagnetic fields will be a sum of smooth functions and so will be mathematically smooth.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-08T01:07:54.387" />
  <row Id="10902" PostTypeId="2" ParentId="10199" CreationDate="2011-06-08T03:30:16.563" Score="1" ViewCount="" Body="&lt;p&gt;You can always create discontinuous waves (which are solutions that exists even in the total absence of sources (be them gravitational or electromagnetic)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can always write a discontinuous function and use it to define, say, the electric field at some time $t_1$, and ask what do the fields look at time $t_0 &amp;lt; t_1$. Of course such solutions are a bit &lt;em&gt;ad hoc&lt;/em&gt; in the sense that they are simply a wave solution with no actual sources&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You certainly can generate fields that approach a discontinuous one by adding enough particles that you can approximate them by discontinuous functions $\rho$ for densities and currents, but of course the discontinuity is only an approximation&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-06-08T03:30:16.563" />
  <row Id="10903" PostTypeId="2" ParentId="10798" CreationDate="2011-06-08T04:31:09.957" Score="0" ViewCount="" Body="&lt;p&gt;If I understand the problem correctly there is a bowl (like a large salad bowl) and a ball rolling inside of it under the influence of gravity. Here is how to derive the equations of motion. I parametrize the position of the ball center using the quasi-coordinates $q_1$ and $q_2$ as:&#xA;  $$ \vec{r}(q_1,q_2) = [0,0,R]-{\rm Rot}_{X}(q_{1})\cdot{\rm Rot}_{Y}(q_{2})\cdot[0,0,R]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R$ is the radius of the bowl minus the radius of the ball, equals the pitch radius of motion. My coordinate system has the bowl laying on the $x$-$y$ plane and &lt;em&gt;up&lt;/em&gt; being the +$z$ direction. ${\rm Rot}_X()$ and ${\rm Rot}_Y()$ are the $3\times3$ rotation matrices respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiating I get the velocity vector of the center of gravity&#xA;  $$ \vec{v}=\begin{bmatrix}0 &amp;amp; \cos q_{2}\\&#xD;&#xA;\cos q_{2}\cos q_{1} &amp;amp; -\sin q_{2}\sin q_{1}\\&#xD;&#xA;\cos q_{2}\sin q_{1} &amp;amp; \sin q_{2}\cos q_{1}\end{bmatrix}\begin{bmatrix}R\,\dot{q}_{1}\\&#xD;&#xA;R\,\dot{q}_{2}\end{bmatrix} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and differentiating again, I get the acceleration of the c.o.g.&#xA;  $$ \vec{a}=\begin{bmatrix}0 &amp;amp; \cos q_{2}\\&#xD;&#xA;\cos q_{2}\cos q_{1} &amp;amp; -\sin q_{2}\sin q_{1}\\&#xD;&#xA;\cos q_{2}\sin q_{1} &amp;amp; \sin q_{2}\cos q_{1}\end{bmatrix}\begin{bmatrix}R\,\ddot{q}_{1}\\&#xD;&#xA;R\,\ddot{q}_{2}\end{bmatrix}+\begin{bmatrix}R\sin q_{2}\,\dot{q}_{2}^{2}\\&#xD;&#xA;\mbox{-}R(\dot{q}_{1}^{2}+\dot{q}_{2}^{2})\cos q_{2}\sin q_{1}-2R\dot{q}_{1}\dot{q}_{2}\sin q_{2}\cos q_{1}\\&#xD;&#xA;R(\dot{q}_{1}^{2}+\dot{q}_{2}^{2})\cos q_{2}\cos q_{1}-2R\dot{q}_{1}\dot{q}_{2}\sin q_{2}\sin q_{1}\end{bmatrix} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the fun part. The forces acting on the ball are gravity $\vec{W}=[0,0,-m\,g]$ and the contact force &#xA;$$ \vec{N}=F\,\begin{bmatrix}\sin q_{2}\\&#xD;&#xA;\cos q_{2}\sin q_{1}\\&#xD;&#xA;\cos q_{2}\cos q_{1}\end{bmatrix} $$ where $F$ is the magnitude of the force (&lt;em&gt;unknown&lt;/em&gt;), $m$ is the mass of the ball and $g$ is acceleration of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To verify that $\vec{N}$ is a reaction force check with $\vec{N}\cdot\vec{v}=0$, thus providing zero &lt;em&gt;power&lt;/em&gt; to the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations of motion (ignoring the rotational components and friction) are&#xA;$$ \vec{W}+\vec{N} = m\;\vec{a} $$ The solution of which yields $F$, $\ddot{q}_2$ and $\ddot{q}_2$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ddot{q}_{1}=2\dot{q}_{1}\dot{q}_{2}\tan q_{2}-\frac{g}{R}\,\frac{\sin q_{1}}{\cos q_{2}}$$&#xA;$$\ddot{q}_{2}=\mbox{-}\dot{q}_{1}^{2}\sin q_{2}\cos q_{1}-\frac{g}{R}\,\sin q_{2}\cos q_{1}$$&#xA;$$F=m\, g\,\cos q_{2}\cos q_{1}+R\, m\,\left(\dot{q}_{1}^{2}\cos^{2}q_{2}+\dot{q}_{2}^{2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whats left to be done, is put it through a numerical integrator scheme like &lt;code&gt;Runge-Kutta&lt;/code&gt; [&lt;a href=&quot;http://www.myphysicslab.com/runge_kutta.html&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;] [&lt;a href=&quot;http://en.wikipedia.org/wiki/Runge%E2%80%93Kutta_methods&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;] and watch the variables $q_1$ and $q_2$ evolve over time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To add friction, add a component of force in the direction opposite of $\vec{v}$ with magnitude $\mu\,F$ where $\mu$ is the coefficient of friction (like &lt;code&gt;0.02&lt;/code&gt;-&lt;code&gt;0.10&lt;/code&gt; for rolling).&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-06-08T04:31:09.957" CommentCount="2" />
  <row Id="10904" PostTypeId="2" ParentId="1019" CreationDate="2011-06-08T04:42:42.250" Score="0" ViewCount="" Body="&lt;p&gt;I know it's an old question, but it's too much fun to pass this one up. There are quite a few examples of wrong ideas that have taken hold, but for the centrality to physics and the degree to which it is wrong and misleading, it is hard to find a better example than the Bohr atom. When it came out, it electrified the physics world on account of its revolutionary viewpoint and stunning success. But eleven years later it became completely obsolete. Nevertheless, in that short time it gripped the imagination of both the physics world and the general public to the point where it remains today the iconic picture of the atom in the minds of just about everyone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The misleading influences of this model are extremely far reaching. One can begin with the idea of the &quot;quantum leap&quot;, which begat wave function collapse, which begat multiple universes et cetera. But perhaps the most persistent holdover of the Bohr atom is the notion that the ordinary laws of electromagnetics must be suspended at the atomic level, otherwise the atom would be unstable. Of course, nothing of the sort is true in modern quantum mechanics, least of all for the hydrogen atom. The greatest triumph of the Schroedinger equation was to show that the motion of electric charge could be tracked through time, and that the stable atomic configurations were precisely those with no accelerating charge distributions. This was an immediate and obvious consequence of Schroedinger's solution for the hydrogen atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would claim that it is not merely true that the stable configurations in QM are those with no accelerating charges. I would go further and suggest that in those cases where the charges do accelerate, the rate at which the kinetic energy of motion is converted into electromagnetic energy is exactly in agreement with that which would be calculated in the ordinary way by applying Maxwell's equations. So, for example, in the case of a black body radiator, if one simply takes the vibrating atoms, accounting for the accelerating motion of the charges, and applies classical antenna theory, you will get the correct black-body spectrum.&lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-06-08T04:42:42.250" CommunityOwnedDate="2011-06-08T04:42:42.250" />
  <row Id="10905" PostTypeId="2" ParentId="3498" CreationDate="2011-06-08T05:02:31.047" Score="0" ViewCount="" Body="&lt;p&gt;Brilliance is as stated in literature: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;number of photons per second per mm$^2$ per rad$^2$ per $0.001BW$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $BW = \Delta\omega/\omega$ is the like the &quot;binning&quot; size over which the equation was integrated over. So the number of photons that will arrive at your sample depends on what frequency range you are measuring over. If you are measuring the number of photons over an infinitely narrow window of frequencies then the number of photons/second approaches ZERO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore (for a rough average if the brilliance curve is ~constant over your frequency range)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;number of photons/second/mm$^2$ = Brilliance * A/L$^2$ * 0.001 * $\Delta\omega/\omega_0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Delta\omega$ = the window of frequencies you are detecting and $\omega_0$ the center frequency. If however the brilliance curve changes over the frequency range you are detecting you will have to integrate the function.&lt;/p&gt;&#xA;" OwnerUserId="2011" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2011-06-10T11:06:31.793" LastActivityDate="2011-06-10T11:06:31.793" CommentCount="1" />
  <row Id="10906" PostTypeId="2" ParentId="10889" CreationDate="2011-06-08T06:34:25.410" Score="0" ViewCount="" Body="&lt;p&gt;Please spend some time &lt;a href=&quot;http://en.wikipedia.org/wiki/Four-vector&quot; rel=&quot;nofollow&quot;&gt;reading about four vectors.&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept of a four vector is an extrapolation to an extra dimension of the three vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say that you have two three vectors, the three momenta of two particles.&#xA; Each vector has a length: p1, p2. When you add them vectorially, the length of the vectors do not add linearly.  It will depend at what angle they are added. For example, if p1=p2 and the addition is head on, the length of the resultant vector will be 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the same with four vectors where the mass is the measure, what you get from the dot product, of the four vector. If you read the link above you will see that a system of four vectors will have masses which will depend on the values of the summed energies and the summed three vector momenta, in no way linear. When $E^2=P^2$ (in a system where $c=1$) the mass is zero, as happens with the photons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have a system of massive particles the lowest mass you can have is the sum of the  rest masses, if all particles are at rest. These are the masses that enter the $E=mc^2$ (in a system where $c=1$) which says that masses have a minimum energy content. It comes naturally in the four vector representation, since at rest, the three momentum vectors are zero, and only energy exists, thus the &quot;length&quot; of this four vector, which is its mass, is also its energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can say that mass is a form of energy, depending on the coordinate system  where the particles are studied. There is no density concept, just &quot;length&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy is a four vector component. Mass is a scalar identifying the length of the four vector. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="202" LastEditorDisplayName="" LastEditDate="2011-06-09T18:25:12.773" LastActivityDate="2011-06-09T18:25:12.773" CommentCount="5" />
  <row Id="10907" PostTypeId="1" CreationDate="2011-06-08T11:18:21.933" Score="1" ViewCount="77" Body="&lt;p&gt;Suppose an aircraft is moving at a certain fixed altitude above the ground. It follows a path defined by latitude and longitude. Now if we want to define the position of an aircraft at any point in the air, three variables are required for example X for latitude, Y for longitude and Z for the altitude. Suppose an aircraft flies and reach a certain fixed altitude Z, it then follows a route defined by X and Y. Now suppose that at any stage during the flight the aircraft decides to take a turn.The aircraft does a maneuvre in such a way that it turns either to the left or right and increases or decreases it's altitude in the same time. Assuming knowing the speed and the current three Dimensional position of the aircraft, how can the future position of the aircraft after a known time t can be predicted? Also assume that other aircraft related constant parameters are also known like radius of the turn, Bank angle etc&lt;/p&gt;&#xA;" OwnerUserId="3921" LastActivityDate="2011-06-10T13:55:00.773" Title="Given a radius and velocity calculate position of an aircraft banking to make a turn in three dimensional space" Tags="&lt;motion&gt;" AnswerCount="0" CommentCount="5" />
  <row Id="10908" PostTypeId="1" CreationDate="2011-06-08T11:24:07.773" Score="3" ViewCount="123" Body="&lt;p&gt;When trying to intensify a beam of light by refracting it through a lens, (as in a lighthouse fresnel system or similar railroad style switch lamp from years past), is the beam intensity increased by moving the lens closer to the beam? That is, refracting it as close as possible to the light source? Or of a similar condition, would an old kerosene barn lantern be brighter if the globe was an inch away from the flame instead of 3 to 4 inches? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;TE&lt;/p&gt;&#xA;" OwnerUserId="3922" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-20T21:41:56.487" LastActivityDate="2011-07-20T21:41:56.487" Title="intensifying a light source" Tags="&lt;optics&gt;&lt;refraction&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="10909" PostTypeId="1" AcceptedAnswerId="10919" CreationDate="2011-06-08T11:58:04.877" Score="3" ViewCount="82" Body="&lt;p&gt;diodes are made of silicon, which have been enriched in phosphorus and boron to create the PN junction. Since the electrons are always going in the same direction, from N to P in the junction, I was wondering whether a diode had a given lifetime after which there is no electrons anymore to move in the PN junction (no deficiency nor excess in electrons in any side of the PN junction) ?&#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3923" LastActivityDate="2011-06-08T16:52:27.073" Title="lifetime of a PN junction in a diode" Tags="&lt;electricity&gt;&lt;electronics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="10910" PostTypeId="1" CreationDate="2011-06-08T12:48:21.207" Score="2" ViewCount="76" Body="&lt;p&gt;I'm working on a research project involving absorption spectra of particulates in solution. I was curious if someone could clarify or direct me to a resource that explains broadening mechanisms &lt;em&gt;specifically&lt;/em&gt; for absorption spectra. For example, the oft-cited Heisenberg and Doppler broadening effects both make sense to me in the context of emission spectra, but I don't see the mechanism by which they would be influential in absorption spectra. But I also can't find a source to verify this, since emission and absorption spectra are often lumped together in the familiar statement &quot;the absorption spectrum is simply the inverse of the emission spectrum&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks very much for any advice.&lt;/p&gt;&#xA;" OwnerUserId="3924" LastActivityDate="2011-06-09T08:15:45.470" Title="Absorption Spectral Broadening" Tags="&lt;spectroscopy&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="10911" PostTypeId="1" AcceptedAnswerId="10925" CreationDate="2011-06-08T13:24:32.430" Score="2" ViewCount="95" Body="&lt;p&gt;When light it transmitted and reflected from a dielectric surface, the incident and reflected. For so-called Brewster's angle, the reflected light lineary polarized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: how does this law work in case of mirror-like surface, when (ideally) all the light is reflected?&lt;/p&gt;&#xA;" OwnerUserId="2752" LastActivityDate="2011-06-08T20:12:10.790" Title="Polarization and mirrors" Tags="&lt;light&gt;&lt;polarization&gt;" AnswerCount="2" />
  <row Id="10912" PostTypeId="2" ParentId="10911" CreationDate="2011-06-08T13:32:31.687" Score="-2" ViewCount="" Body="&lt;p&gt;I assume &quot;mirror-like&quot; is Your private expression for metallic surface? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case that &quot;law&quot; (which is an observation, not a law!) does &#xA;not apply at all. Reflection at metallic surfaces has totally different &#xA;background. I bet that You will find that at Wikipedia. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-06-08T13:32:31.687" />
  <row Id="10913" PostTypeId="2" ParentId="10910" CreationDate="2011-06-08T13:39:23.130" Score="0" ViewCount="" Body="&lt;p&gt;Is it really difficult to apply the reasons behind those boadiening mechanisms &#xA;to absorption?  And: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;the absorption spectrum is simply the&#xA;  inverse of the emission spectrum&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is a simple fact, which helps when reasoning about the &quot;why&quot;. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-06-08T13:39:23.130" CommentCount="1" />
  <row Id="10914" PostTypeId="2" ParentId="10910" CreationDate="2011-06-08T13:43:47.370" Score="1" ViewCount="" Body="&lt;p&gt;The reason the statement is made &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;the absorption spectrum is simply the inverse of the emission spectrum&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is because  broadening mechanisms depend on kinematics, and kinematics is reversible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By this I mean that if an emitting atom  because of its thermal kinetic energy emits a photon of energy  which, instead of E, is E+delta(E), where the extra energy comes because of its motion, the same atom instead of absorbing E will absorb E+delta(E) again because of its motion when the photon hits it. Statistically the broadening curves will be the same.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-06-08T13:43:47.370" CommentCount="1" />
  <row Id="10915" PostTypeId="1" CreationDate="2011-06-08T14:26:07.587" Score="4" ViewCount="256" Body="&lt;p&gt;Linear antenna directed along z, photons (EM waves) propagate along x. Momentum of photons have only x component. Why electrons in antenna have z component of momentum?&lt;/p&gt;&#xA;" OwnerUserId="3926" LastActivityDate="2011-08-09T14:22:52.010" Title="Why photons transfer to electrons perpendicular momentum?" Tags="&lt;photons&gt;&lt;electron&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="10916" PostTypeId="2" ParentId="214" CreationDate="2011-06-08T14:57:07.243" Score="4" ViewCount="" Body="&lt;p&gt;While everyone agrees that jumping in a falling elevator doesn't help much, I think it is very instructive to do the calculation.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;General Remarks&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The general nature of the problem is the following: while jumping, the human &lt;em&gt;injects muscle energy&lt;/em&gt; into the system. Of course, the human doesn't want to gain even more energy himself, instead he hopes to transfer most of it onto the elevator. Thanks to &lt;em&gt;momentum conservation&lt;/em&gt;, his own velocity will be reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should clarify what is meant by momentum conservation. Denoting the momenta of the human and the elevator with $p_1=m_1 v_1$ and $p_2=m_2 v_2$ respectively, the equations of motion are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dot p_1 = -m_1 g + f_{12} $$&#xA;$$ \dot p_2 = -m_2 g + f_{21} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $f_{21}$ is the force that the human exerts on the elevator. By Newton's third law, we have $f_{21} = -f_{12}$, so the &lt;em&gt;total&lt;/em&gt; momentum $p=p_1+p_2$ obeys&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d}{dt} (p_1 + p_2) = -(m_1+m_2) g $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, this is not a conserved quantity, but the point is that it only depends on the external gravity field, not on the interaction between human and elevator.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Change of Momentum&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;As a first approximation, we treat the jump as &lt;em&gt;instantaneous&lt;/em&gt;. In other words, from one moment to the other, the momenta change by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ p_1 \to p_1 + \Delta p_1, \qquad p_2 \to p_2 + \Delta p_2 .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to momentum &quot;conservation&quot;, we can write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta p := -\Delta p_1 = \Delta p_2 .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that trying to find a force $f_{12}$ that models this instantaneous change will probably give you a headache.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much energy did this change of momentum inject into the system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta E = \frac{(p_1-\Delta p)^2}{2m_1} + \frac{(p_2+\Delta p)^2}{2m_2} - \frac{p_1^2}{2m_1} - \frac{p_2^2}{2m_2} .$$&#xA;$$ = \Delta p(\frac{p_2}{m_2} - \frac{p_1}{m_1}) + (\Delta p)^2(\frac1{2m_1}+\frac1{2m_2}) .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we make use of the fact that before jumping, the velocity of the elevator and the human are equal, $p_1/m_1 = p_2/m_2$. Hence, only the quadratice term remains and we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (\Delta p)^2 = \frac2{\frac1{m_1}+\frac1{m_2}} \Delta E .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the mass of the elevator is important, but since elevators are usually very heavy, $m_1 \ll m_2$, we can approximate this with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (\Delta p)^2 = 2m_1 \Delta E .$$&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Energy reduction&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;How much did we manage to reduce the kinetic energy of the human? After the jump, his/her kinetic energy is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E&amp;#39; = \frac{(p_1-\Delta p)^2}{2m_1} = \frac{p_1^2}{2m_1} - 2\frac{\Delta p\cdot p_1}{2m_1} + \frac{(\Delta p)^2}{2m_1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Writing $E$ for the previous kinetic energy, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E&amp;#39; = E - 2\sqrt{E \Delta E} + \Delta E = (\sqrt E - \sqrt{\Delta E})^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{E&amp;#39;}{E} = (1 - \sqrt{\Delta E / E})^2 .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very useful to estimate the energy $\Delta E$ generated by the human in terms of the maximum height that he can jump. For a human, that's roughly $h_1 = 1m$. Denoting the total height of the fall with $h$, we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{E&amp;#39;}{E} = (1 - \sqrt{h_1/h})^2 .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if a human is athletic enough to jump $1m$ in normal circumstances, then he might hope to reduce the impact energy of a fall from $16m$ to a fraction of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{E&amp;#39;}{E} = (1 - \sqrt{1/16})^2 \approx 56 \% .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then again, jumping while being weightless in a falling elevator is likely very difficult...&lt;/p&gt;&#xA;" OwnerUserId="188" LastActivityDate="2011-06-08T14:57:07.243" CommentCount="1" />
  <row Id="10918" PostTypeId="2" ParentId="10915" CreationDate="2011-06-08T16:17:16.050" Score="3" ViewCount="" Body="&lt;p&gt;Photons are quanta of electromagnetic waves which are transverse so if the momentum of the photon goes in the $x$-direction, then the magnetic and electric fields are in the transverse $yz$-plane, for example $B$ may be in the $y$ direction and $E$ may be in the $z$ direction. Because the electric field accelerates charged particles in the same (or opposite) direction, electrons will be accelerated in the $z$ direction as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/zCsKK.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the wave is unpolarized, about 50 percent of its energy will be composed of the wave above and 50 percent will be composed of the other way where the directions of $B,E$ are interchanged. This other wave won't be able to shake the electrons.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-06-08T16:17:16.050" CommentCount="4" />
  <row Id="10919" PostTypeId="2" ParentId="10909" CreationDate="2011-06-08T16:52:27.073" Score="1" ViewCount="" Body="&lt;p&gt;The electrons which flow in a PN junction do not come from the device itself.  Diodes (and other electrical components) are designed to work in a circuit.  The circuit provides the electrons.  Just as Georg said: your homes wiring does not supply elecrtons, it only carries them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of electro-optic devices, light energy can excite a device into a polarized state, where electrons prefer to flow in a certain direction (and thus develop a voltage).  But if you do not provide a closed circuit loop in which the electrons can travel, you get no current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to think about it: IF you have an unbalance of charge, you unavoidably get an electric field built up.  The field strengths per electron are relatively HUGE per electron (as compared to other forces, like gravity).  Can you imagine the charge which would build up after running a theoretical device which could donate its electrons without replenishing them? &lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-06-08T16:52:27.073" CommentCount="4" />
  <row Id="10920" PostTypeId="2" ParentId="10852" CreationDate="2011-06-08T17:30:49.677" Score="2" ViewCount="" Body="&lt;p&gt;Anthony, even after your parlay with Georg and Mark, I am still not confident I know exactly what you want, but I'll give it a stab...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the the velocities of the balls, both before and after collision, are broken into two components: velocities tangential to the point of contact and velocities normal to the point of contact.  What happens in each of these orthogonal axes does not affect what happens to the other.  This is the first really important concept to grasp: For momentum, what happens in tangential stays in tangential; and what happens in normal stays in normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speaking of tangential:  Notice that the tangential velocity of each ball is exactly the same, both before and after collision.  This would be expected if we assume (and we do) no friction.  Another way to look at it:  If the tangential velocity of each ball was the ONLY component of velocity, and they barely gave each other a frictionless kiss in passing, wouldn't we expect the velocity of each ball to remain unchanged?  Yes.  OK, then, we have established that the tangential velocity (and momentum, by the way) of each ball remains unchanged throughout the collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we go on to the normal component of velocities:  We know the velocities prior to collision, and we know the masses throughout the collision.  We also know that kinetic energy and momentum are conserved (since this is an elastic collision).  Furthermore, we also know that the tangential components of velocities have remained unchanged.  We are left with a boat load of equations and only two unknowns: Normal velocities of the masses (m1 and m2) after the collision (v1an and V2an, respectively).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Velocities before collision: v1b = SQRT((v1bn^2)+(v1bt^2)) and v2b = SQRT((v2bn^2)+(v2bt^2)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Velocities after collision: v1a = SQRT((v1an^2)+(v1at^2)) and v2a = SQRT((v2an^2)+(v2at^2)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since momentum is conserved --&gt; (m1)(v1bn) + (m2)(v2bn) = (m1)(v1an) + (m2)(v2an) ... and the normal velocities must remain normal, it is a short step of reasoning to deduce, for m1=m2, that |v1an| = |v1bn|, but are in the opposite direction, and likewise for |v2an| = |v2bn|.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the more general case, where m1 does not equal m2 and |v1| does not equal |v2|, the equations are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(v1an) = (v1bn)(((m1-m2)+((2m2)(v2bn))/(m1+m2))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(v2an) = (v2bn)(((m2-m1)+((2m1)(v1bn))/(m1+m2))&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-06-08T18:37:10.387" LastActivityDate="2011-06-08T18:37:10.387" />
  <row Id="10921" PostTypeId="1" AcceptedAnswerId="10930" CreationDate="2011-06-08T18:00:19.060" Score="4" ViewCount="209" Body="&lt;p&gt;The Lehman representation of the frequency-dependent single particle Green's function is&#xA;$$G(k,\omega) = \sum_n \frac{|c_k|^2}{\omega - E_n + i\eta}$$&#xA;where $n$ enumerates all the eigenstates of the system, $c_k$ is the overlap between &#xA;$|\Psi_n\rangle$, the eigenstate to eigenenergy $E_n$, and the state $|k\rangle = &#xD;&#xA;c_k^\dagger | 0 \rangle$. And $\eta$ is a very small convergence factor, sometimes also written as $0^+$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding spectral function is&#xA;$$A(k,\omega) = \sum_n |c_l|^2 \frac{\eta}{\pi} \cdot \frac{1}{(\omega - E_k)^2 + \eta^2)}$$&#xA;This means we have Lorenzian with width $\eta$ at the eigenergies of the system, i.e., for small $\eta$&#xA;we see that the spectral function has sharp peaks for all values of $\omega$ that correspond to the system's eigenergies. (Note that for $\eta \rightarrow 0$, the Lorenzian becomes a $\delta$-function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, so good. Now we can also write the Green's function using the &lt;strong&gt;self energy&lt;/strong&gt;. If $\varepsilon(k)$ is the dispersion of the non-interacting particle, then the self energy is defined via&#xA;$$G(k,\omega) = \frac{1}{\omega - \varepsilon(k) - \Sigma(k,\omega) + i\eta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is mentioned on many occasions that a self-energy with a non-zero imaginary part corresponds to a quasi-particle with a finite lifetime. In the spectral function, an imaginary self energy leads to a &lt;em&gt;broad&lt;/em&gt; peak, and the width of that peak is related to the imaginary part of $\Sigma(k,\omega)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; At first, I have shown that the spectral function has infinitely sharp peaks at precisely the system's eigenergies. But later, I have shown that an imaginary self energy leads to &lt;em&gt;broad&lt;/em&gt; peaks. How can I reconcile these two facts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Related Question:&lt;/strong&gt; If I find a &quot;continuum&quot; of states in my spectral function, how can I tell if these form a continuum of &quot;true&quot; eigenstates of the system or a bunch of decaying quasiparticles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motivation: In the Holstein model, we have a single electron on a lattice in the tight-binding approximation interacting with Einstein phonons of frequency $\Omega$. In the non-interacting limit, we have the non-interacting electron band $\varepsilon(k) = -2t \cos(ka)$ and then, beginning at an energy $\Omega$ above $-2t$, a continuum of a states corresponding to an electron of momentum $k - q$ and a phonon of momentum $q$. These are &quot;true&quot; eigenstates of the (non-interacting) system. But if I then turn on the interaction of the type&#xA;$$ \frac{g}{\sqrt{N}} \sum_q c_{k-q}^\dagger c_k (b_q^\dagger + b_{-q})$$&#xA;I still have a continuum an energy $\Omega$ above the ground state, but now these are not true eigenstates any more, since the additional phonon can decay via the interaction term. But when they are not true eigenstates any more, then they shouldn't show up in the Lehman representation and hence shouldn't contribute to the spectral function?&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditorDisplayName="" LastEditDate="2011-06-08T22:39:30.247" LastActivityDate="2011-06-09T02:53:41.077" Title="How to get an imaginary self energy?" Tags="&lt;many-body&gt;&lt;quasiparticles&gt;&lt;phonons&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="10922" PostTypeId="2" ParentId="10835" CreationDate="2011-06-08T18:02:26.140" Score="0" ViewCount="" Body="&lt;p&gt;There is no evidence that the kinetic energy of an electron changes in a static magnetic field. You'd end up with a perpetual motion machine otherwise.&lt;/p&gt;&#xA;" OwnerUserId="3929" LastActivityDate="2011-06-08T18:02:26.140" />
  <row Id="10923" PostTypeId="1" AcceptedAnswerId="10937" CreationDate="2011-06-08T18:15:21.903" Score="3" ViewCount="147" Body="&lt;p&gt;Can the electroweak and strong forces be written as geometric theories? - Why and why not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can quantum mechanics in general?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the Kaluza-Klein theory explains the electromagnetic field as &quot;twists&quot; that include an extra dimension of space. (As written by Lubos Motl in a previous question). &lt;/p&gt;&#xA;" OwnerUserId="3622" LastActivityDate="2011-06-09T06:25:15.340" Title="Can the electroweak/strong forces, and/or quantum mechanics be thought of as geometric?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;geometry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="10924" PostTypeId="2" ParentId="10838" CreationDate="2011-06-08T18:41:34.127" Score="3" ViewCount="" Body="&lt;p&gt;I'll try to hit the main points, but for details you need to read something longer than this post. I think I'd start with the various tutorials at &lt;a href=&quot;http://background.uchicago.edu/&quot; rel=&quot;nofollow&quot;&gt;Wayne Hu's web site&lt;/a&gt;. The &lt;a href=&quot;http://www.nsf.gov/mps/ast/tfcr_final_report.pdf&quot; rel=&quot;nofollow&quot;&gt;Weiss report&lt;/a&gt;, which lays out the case for search for $B$ modes, might also be a good place to look.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The general picture:&lt;/strong&gt; (You may already know this, in which case skip ahead.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both temperature anisotropy and polarization are sourced by inhomogeneities in either the matter distribution or the spacetime geometry. These inhomogeneities can be categorized into scalar, vector, and tensor perturbations. This categorization essentially refers to the way the perturbation transforms under rotations: scalars are invariant, vectors transform with spin 1, tensors transform with spin 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dominant kind of perturbations are scalars, especially &quot;ordinary&quot; density perturbations and the gravitational potential perturbations arising from them. These contribute to both temperature anisotropy and $E$-type polarization, but not to $B$-type polarization. The reason is essentially a symmetry argument. $B$-type polarization can be sourced by pseudoscalars but not scalars -- that is, any process that's invariant under reflections can't act as a source of $B$ polarization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be precise, the above is true to linear order in perturbation theory. At higher order, you can get $B$-type polarization from scalar perturbations. That's the source of the gravitational lensing signal you refer to. The $B$ modes produced by gravitational lensing result from $E$ modes (produced by the &quot;usual&quot; process) interacting with other inhomogeneities along the line of sight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tensor perturbations (perhaps more accurately called spin-2 perturbations) can and do cause temperature anisotropy, $E$-type polarization, and $B$-type polarization. The type of tensor perturbations that are most likely to be found in our Universe are those arising from a background of gravitational waves. Such a background must produce a smaller contribution than scalar perturbations -- if it produced a signal that was comparably large, we'd have seen it by now. That means that searching for this signal in either temperature anisotropy or $E$-type polarization will probably be impossible: there'll be no good way to separate the small contribution from the large one. But since scalars don't produce $B$-type polarization, the $B$ modes are a &quot;clean&quot; channel to use in searching for these gravitational waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why do gravitational waves cause B modes?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the best exposition I've seen of this is &lt;a href=&quot;http://background.uchicago.edu/~whu/intermediate/Polarization/polar4.html&quot; rel=&quot;nofollow&quot;&gt;Wayne Hu's&lt;/a&gt;. The big idea is that anything that transforms like a scalar can't couple to a pseudoscalar. Since gravitational waves are spin-2, they can. One way to think about this is that gravitational waves can be right- or left-circularly polarized. On average, you expect both polarizations to be equally strong, but at any given point, you might see one dominating over the other, just by chance. That is, you might see &quot;handedness&quot; at any given place. That handedness is what you need to get $B$ modes. To be more precise, you can think of everything in Fourier space and consider one Fourier mode at a time. For scalar perturbations, each Fourier mode will have azimuthal symmetry about the direction of the wave vector, but for tensor modes they won't necessarily. That extra degree of freedom makes it mathematically possible to excite $B$ modes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's confusing, check out Wayne's pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Gravitational waves from inflation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During inflation, the energy in the Universe is dominated by a scalar field known as the inflaton. At the end of inflation, the inflaton decays into &quot;ordinary&quot; matter. The density perturbations in the Universe are due to quantum fluctuations of the inflaton laid down during inflation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same time, according to the theory, the spacetime metric itself should be undergoing quantum fluctuations. These fluctuations also get &quot;frozen in&quot; as classical perturbations after the end of inflation. They propagate (in part, anyway) as gravitational waves. These gravitational waves are the source of the $B$-type polarization that people are searching for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that, when you solve the equations for the quantum fluctuations of the inflaton, the amplitude depends on things like the rate at which the field is changing, but the amplitude of the &lt;em&gt;metric&lt;/em&gt; perturbations (which give rise to the gravitational waves) depends only on the energy scale at the time. So measuring the amplitude of density perturbations tells you something complicated about the dynamics of the inflaton field, but measuring the amplitude of gravitational waves tells you something relatively simple about the energy scale of inflation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chief prediction of any specific inflation scenario is a set of angular power spectra, which essentially means the amplitudes of temperature, $E$, and $B$ modes as a function of wavenumber in Fourier space (multipole of spherical harmonics, to be a bit more precise). See, for instance, Figure 2.2 of the &lt;a href=&quot;http://www.nsf.gov/mps/ast/tfcr_final_report.pdf&quot; rel=&quot;nofollow&quot;&gt;Weiss report&lt;/a&gt;. The predicted amplitude of the $B$ signal, especially on large angular scales, essentially depends only on the energy scale of inflation. On small scales (high wavenumbers / multipoles), the dominant cause of $B$ modes is the gravitational lensing signal, which is interesting in itself but which doesn't tell you about inflation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can $B$-modes be used to verify inflation?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll say yes on this. The mere presence of $B$ modes isn't necessarily proof of inflation: you can imagine other mechanisms that might produce them. But inflationary scenarios predict the shape of the $B$ mode power spectrum. If we observe $B$ modes whose power spectrum matches this prediction, that's strong evidence in support of the theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, a failure to detect $B$ modes does not falsify inflation. The amplitude of the signal depends on the energy scale of inflation, and it's easy to construct inflationary models in which that amplitude is undetectably low. So a failure to detect $B$ modes would rule out some of the parameter space for inflationary models, but it wouldn't rule out the entire parameter space.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-06-08T18:41:34.127" CommentCount="2" />
  <row Id="10925" PostTypeId="2" ParentId="10911" CreationDate="2011-06-08T20:12:10.790" Score="2" ViewCount="" Body="&lt;p&gt;The easy answer is to say that Brewster's law only applies to reflection from the interface with a transparent medium, and a mirror isn't transparent.  Indeed for an ideal perfect mirror, all light of both polarizations is reflected perfectly, so there is nothing to say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an actual real-world mirror, the metal mirror surface will have a finite skin depth, and can be considered a dielectric medium with a very large, complex index of refraction.  This does lead to a small polarization dependence of the reflection coefficient for near grazing incidence angles.  The analogue of Brewster's angle occurs at an angle given by 2*pi*(skin depth) / wavelength above grazing, where the parallel polarized reflection coefficient will reach a minimum ( but not zero, still close to 1).  For more details, see for example Landau and Lifshitz, volume 8, section 87.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-06-08T20:12:10.790" />
  <row Id="10926" PostTypeId="1" CreationDate="2011-06-08T21:35:35.117" Score="3" ViewCount="80" Body="&lt;p&gt;I have a question concerning transverse (solenoidal) current in the Coulomb gauge. This current is the one that enables the radiation, since we have a wave equation for the vector potential:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nabla^2 \vec A - \mu_0\epsilon_0\frac{\partial^2\vec A}{\partial t^2} = -\mu_0\vec{J}_{\perp}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the transverse current can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$J_\perp = \nabla \times \nabla \times\int\frac{\vec J(\vec x&amp;#39;)\textrm{d}^3x&amp;#39;}{4\pi|\vec x - \vec x&amp;#39;|}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This transverse current extends to all space. We can solve the wave equation using Green's function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec A(\vec x, t) = \frac{\mu_0}{4\pi}\int\frac{\vec J_\perp(\vec x&amp;#39;, t&amp;#39;)}{|\vec x - \vec x&amp;#39;|} \delta(|\vec x- \vec x&amp;#39;|/c-(t-t&amp;#39;))\textrm{d}^3x&amp;#39;\textrm{d}t&amp;#39;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that, when we want to keep only terms that will result in fields falling like $1/r$, neglecting $1/r^2$ and so on, it seems that we don't have to consider the complicated expression for the transverse current. It seems to be enough in some cases to keep the projection of the current perpendicular to the observation vector $\vec x$ and derivate the vector potential with respect to $t$, though this is not obvious, as $\vec J_\perp$ covers all the space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I checked two different cases -an instantaneous dipole at the origin and an instantaneously accelerating point charge, and the field that goes as $1/r$ is created,&#xA;as expected, by currents transverse to the line of sight. However, I do not have a rigorous explanation to ignore the non-perpendicular currents in the integral for $\vec A$. I've asked my boss, checked some books and searched in the Internet, but I couldn't find any hint or proof. Do you know of any place?&#xA;Thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="3931" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-06-09T07:08:26.910" LastActivityDate="2011-07-11T12:10:11.247" Title="Transverse current equivalence in Coulomb gauge" Tags="&lt;radiation&gt;&lt;classical-electrodynamics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="10927" PostTypeId="2" ParentId="10923" CreationDate="2011-06-08T22:27:28.503" Score="3" ViewCount="" Body="&lt;p&gt;By &quot;geometric&quot; I'm going to assume that you mean &quot;having to do with the geometry of the usual 3+1 dimensions&quot;, that is, geometric in the sense that electricity and magnetism is geometric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a question that was researched deeply in the 1950s especially by Coleman and Mandula after which the &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Coleman%E2%80%93Mandula_theorem&quot; rel=&quot;nofollow&quot;&gt;Coleman-Mandula&lt;/a&gt;&quot; theorem is named. As wikipedia puts it, &quot;It states that &quot;space-time and internal symmetries cannot be combined in any but a trivial way&quot;. Thus the internal symmetries are not related to &quot;geometry&quot; in the sense of our usual world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Coleman-Mandula theorem depends on a lot of complicated mathematics. One can imagine a bunch of ways that one could get around it. If you &lt;a href=&quot;http://www.google.com/search?q=%22coleman-mandula%22%20site%3aarxiv.org&amp;amp;ie=utf-8&amp;amp;oe=utf-8&amp;amp;aq=t&amp;amp;rls=org.mozilla%3aen-US%3aofficial&amp;amp;client=firefox-a&quot; rel=&quot;nofollow&quot;&gt;google Coleman-Mandula&lt;/a&gt; on arXiv.org, you can find papers on this subject, i.e. extensions and consequences of the Coleman-Mandula theorem, as well as papers proposing how one might get around it:&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-08T22:27:28.503" CommentCount="1" />
  <row Id="10928" PostTypeId="2" ParentId="10915" CreationDate="2011-06-08T22:36:57.813" Score="1" ViewCount="" Body="&lt;p&gt;The electrons in the antenna are moving perpendicular to the direction of motion of the emitted radiation. This is natural for a transverse wave. Since light is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Transverse_wave&quot; rel=&quot;nofollow&quot;&gt;transverse wave&lt;/a&gt;, as shown by Luboš Motl's beautiful graphic, there's nothing surprising here.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-08T22:36:57.813" />
  <row Id="10929" PostTypeId="2" ParentId="10876" CreationDate="2011-06-08T22:49:11.370" Score="1" ViewCount="" Body="&lt;p&gt;Re:  I am looking for an Ohm's &quot;law&quot; like equation that relates spring constant to resistivity with a few physical constants thrown into a constant somewhere in the equation. Does such a thing exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a specific material with a specific dopant, you can get a relationship between the two, but there is no general relationship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While a &lt;em&gt;component&lt;/em&gt; of the spring constant is related to the electron &quot;pressure&quot; -- and hence related to the Fermi energy -- the exact nature of the relationship (and how big a role it would play in the overall spring constant) will depend on the specific material.  The resistivity is also related to the Fermi energy, but how much and in what manner will depend on the specific material.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-06-08T22:49:11.370" CommentCount="2" />
  <row Id="10930" PostTypeId="2" ParentId="10921" CreationDate="2011-06-09T02:53:41.077" Score="2" ViewCount="" Body="&lt;p&gt;This is a great question. The point is we've made a huge structural change in going from Lehmann representation to the, um, other one. In Lehmann representation, we've chosen to write $G$ as a sum over an &lt;em&gt;infinite number&lt;/em&gt; of &lt;em&gt;real&lt;/em&gt; poles, which is fine of course, but when you start adding infinite numbers of delta functions together, things can get tricky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Aside: consider for a moment if we integrated Im$G$ over all $k$ -- we're left with the density of states $\rho(\omega)=\sum_n \delta(\omega-E_n)$... and yet when you sit down and work it out for, say, free particles in continuous 3-space, you get a nice smooth function of $\omega$. Why? Well because the eigenvalues $E_n$ become arbitrarily close to one another, the sum becomes an integral, under which delta functions immediately switch from foe to friend.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with that in mind, the spectral function you got from the Lehmann representation will have structure at all of the eigenvalues of the full many-body system (which will merge into smooth functions in an infinite system), but this is silly since those eigenvalues don't generally correspond to excitations that we could ever think about in any intuitive way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, instead of ever thinking about the real eigenstates of many-body systems, we give up and introduce &lt;em&gt;quasiparticles&lt;/em&gt; and work with their Green functions, which, as you showed, have &lt;em&gt;one complex&lt;/em&gt; pole related to $\Sigma$, with the imaginary part giving the spread on the real-frequency axis that the real eigenvalues have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I think this should help with your related question too, but it's late -- let me know if I've been opaque.)&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2011-06-09T02:53:41.077" CommentCount="5" />
  <row Id="10931" PostTypeId="2" ParentId="10915" CreationDate="2011-06-09T04:47:41.590" Score="1" ViewCount="" Body="&lt;p&gt;Maybe interesting for some to take this a bit further:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This classical behavior goes all the way down to single cycle laser pulses which ionize atoms by accelerating them in the direction transverse to the motion of the laser pulse. Make sure to look at the very interesting (Realplayer) video...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cfa.harvard.edu/itamp/attosecondpdfs/paulus.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.cfa.harvard.edu/itamp/attosecondpdfs/paulus.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cfa-www.harvard.edu/dvlwrap/itamp/0311/paulus.ram&quot; rel=&quot;nofollow&quot;&gt;http://cfa-www.harvard.edu/dvlwrap/itamp/0311/paulus.ram&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Why is this behavior still classical? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(1)&lt;/strong&gt; Because in this case even the single cycle of the ultrashort laser pulse has a wavelength much longer as the size of the field of the (bounded) electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(2)&lt;/strong&gt; The single cycle laser pulse kicks the electron in one direction.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;When, how and why do we get quantum mechanical Compton scattering effects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When&lt;/strong&gt;: Mainly if the wavelength of radiation becomes smaller (e.g. x-ray) as the volume in which the electron is contained. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How&lt;/strong&gt;: The electron gets kicked in the direction parallel to the momentum of the photon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why&lt;/strong&gt;: The electron can self-interfere when going from an &lt;em&gt;initial state&lt;/em&gt; $\psi_i$ to a &lt;em&gt;final state&lt;/em&gt; $\psi_f$. The interference current is a sinusoidal pattern of alternating &lt;em&gt;charge&lt;/em&gt; and &lt;em&gt;spin&lt;/em&gt; density with the same wavelength and direction as the incoming photon. The direction of the electron spin is essential in the interaction because it is the effective current from the alternating spin density which is the source of an alternating &lt;em&gt;transverse&lt;/em&gt; electromagnetic field which compensates the electromagnetic field of the incoming photon: The photon is absorbed. Finally, the state of the electron after absorbing the photon isn't that of a free electron and another transition takes place under the emission of an outgoing photon which leaves the electron in a free (on-the-mass-shell) state. The math which describes these processes corresponds to the tree-level Feynman diagrams of the interaction &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards, Hans&lt;/p&gt;&#xA;" OwnerUserId="3197" LastEditorUserId="3197" LastEditorDisplayName="" LastEditDate="2011-06-09T05:03:54.517" LastActivityDate="2011-06-09T05:03:54.517" CommentCount="4" />
  <row Id="10932" PostTypeId="1" CreationDate="2011-06-09T04:52:09.823" Score="14" ViewCount="236" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;br&gt;&#xA;How can you show the inconsistency of regularization by dimensional reduction in the $\mathcal{N}=1$ superfield approach (without reducing to components)?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background and some references:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regularization by dimensional reduction (DRed) was introduced by &lt;a href=&quot;http://inspirebeta.net/record/140227&quot; rel=&quot;nofollow&quot;&gt;Siegel in 1979&lt;/a&gt; and was shortly after seen to be inconsistent &lt;a href=&quot;http://inspirebeta.net/record/152280&quot; rel=&quot;nofollow&quot;&gt;Siegel in 1980&lt;/a&gt;.  Despite this, it is commonly used in supersymmetric calculations since it has most of the advantages of (normal) dimensional regularization (DReg) and (naively) preserves supersymmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The demonstration of the inconsistency of DRed is based on the combination of 4-dimensional identities, such as the product of epsilon-tensors&#xA;$$ \varepsilon^{\mu_1\mu_2\mu_3\mu_4} \varepsilon^{\nu_1\nu_2\nu_3\nu_4}&#xD;&#xA;   \propto \det\big((g^{\mu_i\nu_j})\big) &#xD;&#xA;$$&#xA;and the d-dimensional projections of 4-dimensional objects.&#xA;Details can be found in the references above and below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various proposals have been made on how to consistently use DRed and most involve restrictions on the use of 4-dimensional identities using epsilon-tensors and $\gamma_5$ matrices. (Note that the treatment of $\gamma_5$ in DReg is also a little tricky...). This means we also have to forgo the use of Fierz identities in the gamma-matrix algebra (which is also a strictly 4-dimensional thing - or whatever integer dimension you're working in). This means we lose most of the advantages that made DRed attractive in the first place - maintaining only the fact that it's better than DReg in SUSY theories.&#xA;The latest such attempt is &lt;a href=&quot;http://inspirebeta.net/record/678311&quot; rel=&quot;nofollow&quot;&gt;Stockinger 2005&lt;/a&gt;, but it's also worth looking at the earlier discussions of &lt;a href=&quot;http://inspirebeta.net/record/154499&quot; rel=&quot;nofollow&quot;&gt;Delbourgo and Jarvis 1980&lt;/a&gt;, and &lt;a href=&quot;http://inspirebeta.net/record/155960&quot; rel=&quot;nofollow&quot;&gt;Bonneau 1980&lt;/a&gt;. The pragmatic discussion in &lt;a href=&quot;http://arxiv.org/abs/hep-ph/9707278&quot; rel=&quot;nofollow&quot;&gt;Jack and Jones 1997&lt;/a&gt; is also worth reading - it also contains a fairly complete set of references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, all of the &quot;fixes&quot; are hard to do when using superfields, since the $D$-algebra has all of the &quot;bad&quot; 4-dimensional algebra built in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question is:&lt;/strong&gt; What is the easiest way of showing the inconsistency of DRed in the superfield approach? (I want an answer that does not rely on reducing to components!).&#xA;I'm guessing that it should somehow follow from the $D$-algebra acting on dimensionally reduced superfields.&lt;/p&gt;&#xA;" OwnerUserId="429" LastEditorUserId="429" LastEditorDisplayName="" LastEditDate="2011-06-18T04:10:05.017" LastActivityDate="2011-06-18T04:10:05.017" Title="Superfields and the Inconsistency of regularization by dimensional reduction" Tags="&lt;quantum-field-theory&gt;&lt;supersymmetry&gt;&lt;renormalization&gt;&lt;tensor-calculus&gt;&lt;superalgebra&gt;" CommentCount="6" FavoriteCount="3" />
  <row Id="10933" PostTypeId="1" CreationDate="2011-06-09T05:04:32.867" Score="4" ViewCount="594" Body="&lt;p&gt;In history we are taught that the Catholic Church was wrong, because the Sun does not move around the Earth, instead the Earth moves around the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then in physics we learn that movement is relative, and it depends on the reference point that we choose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Woulnd't the Sun (and the whole universe) move around the Earth if I place my reference point on Earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was movement considered absolute in physics back then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I'm not religious&lt;/p&gt;&#xA;" OwnerUserId="3935" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-06-09T07:07:39.080" LastActivityDate="2011-06-13T04:12:26.190" Title="Why do we say that the earth moves around the sun?" Tags="&lt;classical-mechanics&gt;&lt;astronomy&gt;" AnswerCount="7" CommentCount="2" />
  <row Id="10934" PostTypeId="2" ParentId="10933" CreationDate="2011-06-09T05:29:30.140" Score="10" ViewCount="" Body="&lt;p&gt;yes, you may describe the motion from any reference frame, including the geocentric one, assuming that you add the appropriate &quot;fictitious&quot; forces (centrifugal, Coriolis, and so on).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the special property of the reference frame associated with the Sun - more precisely, with the barycenter (center of mass) of the Solar System, which is just a solar radius away from the Sun's center - is that this system is inertial. It means that there are no centrifugal or other inertial forces. The equations of physics have a particularly simple form in the frame associated with the Sun.&#xA;$$ M_1 d^2 / dt^2 \vec x = G M_1 M_2 (\vec r_1-\vec r_2) / r^3 + \dots $$&#xA;There are just simple inverse-squared-distance gravitational forces entering the equations for the acceleration. For other frames, e.g. the geocentric one, there are many other inertial/centrifugal &quot;artificial&quot; terms on the right hand side that can be eliminated by going to the more natural solar frame. In this sense, the heliocentric frame is more true.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-06-09T05:29:30.140" CommentCount="5" />
  <row Id="10935" PostTypeId="2" ParentId="10933" CreationDate="2011-06-09T05:34:47.800" Score="2" ViewCount="" Body="&lt;p&gt;Yes, the proposition: &quot;the sun moves around the earth&quot; had the earth immobile. This suited the theology of the times which was completely anthropocentric and that is why it prevailed over other theories coming from antiquity, like Aristarchos', who had a heliocentric proposal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relativity of motion was explored, as Lubos describes, when equations could be written down, and one chooses the heliocentric for its beauty and simplicity. The epicycles exist if one plots the solutions in a geocentric system, but they are so cumbersome and &quot;ugly&quot; as a shorthand of physics.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-06-09T05:34:47.800" CommentCount="4" />
  <row Id="10936" PostTypeId="2" ParentId="10933" CreationDate="2011-06-09T05:39:52.417" Score="8" ViewCount="" Body="&lt;p&gt;Imagine two donut-shaped spaceships meeting in deep space.  Further, suppose that when a passenger in ship A looks out the window, they see ship B rotating clockwise.  That means that when a passenger in B looks out the window, they see ship A rotating clockwise as well (hold up your two hands and try it!).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From pure kinematics, we can't say &quot;ship A is really rotating, and ship B is really stationary&quot;, nor the opposite.  The two descriptions, one with A rotating and the other with B, are equivalent.  (We could also say they are both rotating a partial amount.)  All we know, from a pure kinematics point of view, is that the ships have some relative rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, physics does not agree that the rotation of the ships is purely relative.  Passengers on the ships will feel &lt;a href=&quot;http://physics.stackexchange.com/questions/7680/artificial-gravity-on-a-rotating-space-stations&quot;&gt;artificial gravity&lt;/a&gt;.  Perhaps ship A feels lots of artificial gravity and ship B feels none.  Then we can say with definity that ship A is the one that's really rotating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So motion in physics is not all relative. There is a set of reference frames, called inertial frames, that the universe somehow picks out as being special.  Ships that have no angular velocity in these &lt;a href=&quot;http://physics.stackexchange.com/questions/3193/what-determines-which-frames-are-inertial-frames&quot;&gt;inertial frames&lt;/a&gt; feel no artificial gravity.  These frames are all related to each other via the &lt;a href=&quot;http://en.wikipedia.org/wiki/Poincar%C3%A9_group&quot; rel=&quot;nofollow&quot;&gt;Poincare group&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general relativity, the picture is a bit more complicated (and I will let other answerers discuss GR, since I don't know much), but the basic idea is that we have a symmetry in physical laws that lets us boost to reference frames moving at constant speed, but not to reference frames that are accelerating.  This principle underlies the existence of &lt;a href=&quot;http://en.wikipedia.org/wiki/Inertia&quot; rel=&quot;nofollow&quot;&gt;inertia&lt;/a&gt;, because if accelerated frames had the same physics as normal frames, no force would be needed to accelerate things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Earth going around the sun and vice versa, yes, it is possibly to describe the kinematics of the situation by saying that the Earth is stationary.  However, when you do this, you're no longer working in an inertial frame.  Newton's laws do not hold in a frame with the Earth stationary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was dramatically demonstrated for Earth's rotation about its own axis by &lt;a href=&quot;http://en.wikipedia.org/wiki/Foucault_pendulum&quot; rel=&quot;nofollow&quot;&gt;Foucalt's pendulum&lt;/a&gt;, which showed inexplicable acceleration of the pendulum unless we take into account the fictitious forces induced by Earth's rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if we believed the Earth was stationary and the sun orbited it, we'd be at a loss to explain the Sun's motion, because it is extremely massive, but has no force on it large enough to make it orbit the Earth.  At the same time, the Sun ought to be exerting a huge force on Earth, but Earth, being stationary, doesn't move - another violation of Newton's laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the reason we say that the Earth goes around the sun is that when we do that, we can calculate its orbit using only Newton's laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, in an inertial frame, the sun moves slightly due to Earth's pull on it (and much more due to Jupiter's), so we really don't say the sun is stationary.  We say that it moves much less than Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This answer largely rehashes Lubos' above, but I was most of the way done when he posted, and our answers are different enough to complement each other, I think.) &lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-06-09T07:46:22.087" LastActivityDate="2011-06-09T07:46:22.087" CommentCount="2" />
  <row Id="10937" PostTypeId="2" ParentId="10923" CreationDate="2011-06-09T06:25:15.340" Score="3" ViewCount="" Body="&lt;p&gt;From Qmechanic's comment, maybe something like the noncommutative geometry approach would appeal to you. See e.g., &lt;a href=&quot;http://en.wikipedia.org/wiki/Noncommutative_standard_model&quot; rel=&quot;nofollow&quot;&gt;Noncommutative standard model on Wikipedia&lt;/a&gt; and &lt;a href=&quot;http://www.alainconnes.org/&quot; rel=&quot;nofollow&quot;&gt;Alain Conne's homepage&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="429" LastActivityDate="2011-06-09T06:25:15.340" CommentCount="1" />
  <row Id="10938" PostTypeId="1" AcceptedAnswerId="10939" CreationDate="2011-06-09T07:33:00.433" Score="1" ViewCount="139" Body="&lt;p&gt;A light switches on and off at a fast rate, with equal time off and on, why do we see a light that appears only on rather than only off?&lt;/p&gt;&#xA;" OwnerUserId="3937" LastActivityDate="2011-06-09T18:13:03.997" Title="A light flickers at a fast rate why do humans only see light on?" Tags="&lt;light&gt;" AnswerCount="2" />
  <row Id="10939" PostTypeId="2" ParentId="10938" CreationDate="2011-06-09T07:54:31.517" Score="5" ViewCount="" Body="&lt;p&gt;This is a question for  a biology board. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is because of the way the cells in our retina work. Have you heard of the &quot;after image&quot;? If one sees a bright enough objects and shuts ones eyes, an after image appears with eyes closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same biological mechanism keeps the sense of &lt;em&gt;light&lt;/em&gt;, rather than &lt;em&gt;dark&lt;/em&gt; and bridges over the gap. I am guessing at the physics here, but it must be due to the chemistry of the organic molecules and how it is used to transmit to the head the information &lt;em&gt;On&lt;/em&gt; and &lt;em&gt;Off&lt;/em&gt;. Obviously once excited to an &lt;em&gt;on&lt;/em&gt; state, there is a lifetime to decay to the &lt;em&gt;off&lt;/em&gt;, and depending on the frequency from a certain point onward it bridges the gap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something like that happens also with the change of frames in the film of movies and other  optical motions.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-06-09T07:54:31.517" CommentCount="6" />
  <row Id="10940" PostTypeId="2" ParentId="10910" CreationDate="2011-06-09T08:15:45.470" Score="1" ViewCount="" Body="&lt;p&gt;It should be emphasized that you are measuring absorption spectra &lt;strong&gt;in solutions&lt;/strong&gt;. This means that the spectral broadening is dominated by interactions with solvent molecules and lines will be much broader compared to gas-phase spectra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is nearly impossible to calculate time-dependent interactions of absorbing species with hundreds of solvent molecules but there are some empirical rules. For example, polar solvents (methanol) will perturb the electronic state of a molecule stronger compared to unpolar solvents (hexane), so, generally, linewidth in hexane is smaller. Also, this interaction is generally stronger for the excited state (because the electron cloud &quot;expands&quot;) and thus absorption bands will be shifted in wavelength compared to gas-phase spectra - normally, to longer wavelength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find these rules in any &lt;em&gt;old&lt;/em&gt; textbook on UV-Vis spectroscopy. I say old to exclude books on high-resolution gas-phase spectroscopy which is the major topic of the last 30 years. &lt;/p&gt;&#xA;" OwnerUserId="701" LastActivityDate="2011-06-09T08:15:45.470" CommentCount="3" />
  <row Id="10941" PostTypeId="2" ParentId="7218" CreationDate="2011-06-09T08:30:41.020" Score="1" ViewCount="" Body="&lt;p&gt;Nonzero Momentum in static field configuration is actually a good thing. Consider a coaxial cable carrying DC current and voltage. Internally it has a constant E and H field. The Poynting flux is nonzero and shows that there is energy transport. The energy flow is indeed in the direction ExH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the field configuration shows no movement is irrelevant. Momentum is moving energy, not something else moving!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- &#xA;Jos&lt;/p&gt;&#xA;" OwnerUserId="3939" LastActivityDate="2011-06-09T08:30:41.020" />
  <row Id="10942" PostTypeId="2" ParentId="10222" CreationDate="2011-06-09T08:31:37.540" Score="1" ViewCount="" Body="&lt;p&gt;Pavel, I think an excellent and concise description of the Pockels effect is provided here: &lt;a href=&quot;http://www.rp-photonics.com/pockels_effect.html&quot; rel=&quot;nofollow&quot;&gt;http://www.rp-photonics.com/pockels_effect.html&lt;/a&gt;. If you need some more information, please provide an illustration to your question, it will help to understand it.&lt;/p&gt;&#xA;" OwnerUserId="3694" LastActivityDate="2011-06-09T08:31:37.540" />
  <row Id="10943" PostTypeId="2" ParentId="10933" CreationDate="2011-06-09T11:15:36.073" Score="-5" ViewCount="" Body="&lt;p&gt;I'm putting down a much shorter answer:&#xA;The sun doesn't move, the earth (together with the other revolving planets in our galaxy) does. The earth basically rotates around the sun in a ring (and its axis, but that's besides the point).&#xA;Besides, the churches have always made many wrong claims (especially in the middle ages), such as that the earth was flat. They though so because we 'stand on top of the earth' whilst there really is no up or down.&lt;/p&gt;&#xA;" OwnerUserId="3943" LastActivityDate="2011-06-09T11:15:36.073" CommentCount="3" />
  <row Id="10944" PostTypeId="1" CreationDate="2011-06-09T13:22:34.783" Score="2" ViewCount="59" Body="&lt;p&gt;I'd like to know the how the number density of LO and LA photons varies as a function of temperature of the material. Is there a simple expression for these two cases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing that this would work,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N_{LO} = \int g_{LO}(E) f(E, T) dE$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N_{LA} = \int g_{LA}(E) f(E, T) dE$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $g(E)$ is the density of states for LO and LA phonons and $f$ is the Bose-Einstein distribution. What would be appropirate limits for the integrals. Does anybody know a reference where the density of states for these two modes is given?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best regards,&lt;/p&gt;&#xA;" OwnerUserId="1194" LastEditorUserId="1194" LastEditorDisplayName="" LastEditDate="2011-06-10T08:26:02.473" LastActivityDate="2011-06-10T08:26:02.473" Title="Number density of LO and LA phonons as a function of temperature?" Tags="&lt;statistical-mechanics&gt;&lt;solid-state-physics&gt;&lt;phonons&gt;" CommentCount="6" />
  <row Id="10948" PostTypeId="2" ParentId="10933" CreationDate="2011-06-09T16:43:14.953" Score="1" ViewCount="" Body="&lt;p&gt;The sun, moon, earth (and so on) all move around each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason we say the earth moves around the sun is because the effects are more visible on a macro scale, and easier to predict with reasonable precision. Yes, it's most correct to say that all motion is relative, but it gets a lot more complicated to explain it if you're speaking to a layman.&lt;/p&gt;&#xA;" OwnerUserId="2637" LastActivityDate="2011-06-09T16:43:14.953" />
  <row Id="10949" PostTypeId="2" ParentId="10938" CreationDate="2011-06-09T17:43:58.817" Score="2" ViewCount="" Body="&lt;p&gt;The presupposition of your question is incorrect.  Normal household lighting is not off half the time.  household voltage is an AC voltage, provided at 60Hz (in the U.S.; 50Hz in many countires outside the U.S.).  Incandescent and flouresccent lighting use both the positive and negative going portions of the wave, and therefore would appear (if viewed in slow motion) to flicker at 120Hz.  But the percentage of time that the voltage is at zero (off) is much less than 1% of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For incandescent lighting, the light comes from a heated filament.  When current is removed, this filament cools (and gives less light) at an exponential decay rate, with a time constant on the order of 10mS.  But the period of the 120 Hz power lobes is less than 10ms.  That is roughly 1 time constant of the decay rate.  So, you see, the filament never gets a chance to cool off and stop giving light:  The light is on 100% of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not an expert on flourescents, so I will skip the opportunity to embarrass myself, and go on to something which will make that discussion moot: LEDs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LEDs ~DO~ turn on and off instantly.  So say you had a string of 50 to 70 LEDs in series, and another identical string run in parallel but with polarity reversed.  Put a current limiting resistor on the string, and (with careful engineering) you have something which DOES turn ON and stay OFF about 50% of the time.  In that case you STILL would perceive it to be on all the time, since your eye cannot track the lighting changes and acts as an averaging device.  You perceive the light to be ON all the time, but at half the brightness that you would perceive if they were, indeed, ON all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we vary the duty cycle, say 10% ON and 90% OFF, you would still perceive the LEDs to be ON all the time, just much dimmer.  By varying the amount of time ON (the width of the &quot;ON&quot; pulse), we can make the string appear any brightness between 0 and 100%.  This is called Pulse Width Modulation (PWM) and is a common technique for digitally varying the power to a device to mimic analog scaling.  It works extremely well when the target device is the human eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that the ON/OFF cycles are about 120 Hz, but the human eye-brain system can not track much over 20 Hz.  When you see a motion picture (a movie) you are not watching motion at all, but a series of stills flashed up in rapid succession, somewhere in the neighborhood of 30 pictures a second.  Do you remember all 30 pics in that second?  No.  You perceive that as 1 second of fluid motion.  How much more so, then, will you perceive light flickering at 120 Hz to be a smooth and constant light level?&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-06-09T18:13:03.997" LastActivityDate="2011-06-09T18:13:03.997" CommentCount="2" />
  <row Id="10951" PostTypeId="2" ParentId="10933" CreationDate="2011-06-09T19:10:59.977" Score="1" ViewCount="" Body="&lt;p&gt;There may be a confusion : &lt;strong&gt;it is wrong to say that the Earth is the centre of the Universe&lt;/strong&gt;, that is, the (unique) point from which the Universe is to be (fundamentally) described (the fact that the Sun turns around the Earth is only a consequence of this) ; &lt;strong&gt;what actually matters is that there is no centre of the Universe&lt;/strong&gt; : there is no such point ; the description of the Universe from any point is equivalent to the description of the Universe from any other (then you are allowed to describe motions either from the Earth or from the Sun).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, in classical mechanics, the Universe is said to be an &lt;strong&gt;affine&lt;/strong&gt; space.&lt;/p&gt;&#xA;" OwnerUserId="2499" LastEditorUserId="2499" LastEditorDisplayName="" LastEditDate="2011-06-12T17:48:28.017" LastActivityDate="2011-06-12T17:48:28.017" />
  <row Id="10952" PostTypeId="1" AcceptedAnswerId="10957" CreationDate="2011-06-09T20:11:50.697" Score="2" ViewCount="83" Body="&lt;p&gt;One of my work colleagues told me that a cable he is sending a signal through is losing power at high frequencies.  So he recommends the signal should be amplified before being sent.  The explanation was given for the power loss is that higher frequency signals are lossier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a newcomer to signal processing, I'd like to understand more about why or how this effect occurs.  How is the frequency of the signal causing the power loss to occur?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Wikipedia's article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Coaxial_cable&quot; rel=&quot;nofollow&quot;&gt;Coaxial cable&lt;/a&gt; I found this, which seems promising:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If an ordinary wire is used to carry&#xA;  high frequency currents, the wire acts&#xA;  as an antenna, and the high frequency&#xA;  currents radiate off the wire as radio&#xA;  waves, causing power losses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is understanding how antennae work key to understanding why the high frequency results in power loss?&lt;/p&gt;&#xA;" OwnerUserId="3950" LastActivityDate="2011-06-10T00:29:32.523" Title="Loss of Power at high frequencies" Tags="&lt;power&gt;&lt;signal-processing&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="10953" PostTypeId="1" AcceptedAnswerId="10969" CreationDate="2011-06-09T20:23:33.530" Score="7" ViewCount="315" Body="&lt;p&gt;OK, before I ask my question, let me frame it with a few (uncontroversial?) statements:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The low-field-limit plane-wave solution to Einstein's equations is helicity-2.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the early days of string theory, when people were factorizing the Veneziano amplitude and its generalizations in order to find a spectrum of physical states, they could never get rid of the massless spin-2 state (i.e., closed loop) that kept cropping up, although they tried mightily to do so.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In string theory, one can only formulate a physical theory of interacting massless spin-2 states (i.e., closed loops) if the gauge symmetry of those states can be deformed into the diffeomorphism symmetry of GR.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Conformal invariance of the 2-D worldsheet action of a closed loop in string theory leads via state-operator correspondence to gravitational-wave solutions of GR.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Points (3) and (4) were shamelessly lifted from Lubos Motl's blog.  All this goes to say is that string theory &quot;postdicts&quot; gravity, as Ed Witten put it in a talk I saw back in the 90s.  In other words, it seems that as soon as one starts writing down a quantum theory of interacting strings, gravity drops out of it like magic and there's nothing you can do about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This brings me to my question.  If a closed loop is intrinsically spin-2 and by its very nature must satisfy the symmetry requirements of general relativity, why can't one work backwards from these seemingly stringent symmetry requirements and demonstrate that a massless spin-2 point particle can NEVER satisfy them?  Is it possible that only an extended object can serve as the carrier of a force which manifests itself as the curvature of spacetime?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think that such a demonstration would show that not only does string theory postdict gravity, but gravity requires strings.&lt;/p&gt;&#xA;" OwnerUserId="1445" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-09T21:04:22.013" LastActivityDate="2011-08-20T21:37:40.537" Title="Does gravity require strings?" Tags="&lt;general-relativity&gt;&lt;string-theory&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="10955" PostTypeId="1" AcceptedAnswerId="10976" CreationDate="2011-06-09T21:30:18.880" Score="3" ViewCount="210" Body="&lt;p&gt;What is quantum discord? I stumbled upon this term on &lt;a href=&quot;http://www.nature.com/news/2011/110601/full/474024a.html?WT.mc_id=TWT_NatureNews&amp;amp;utm_source=feedburner&amp;amp;utm_medium=twitter&amp;amp;utm_campaign=Feed%3a%20yahoo/qUaz%20%28Nature%20news%29&quot; rel=&quot;nofollow&quot;&gt;Quantum Computing: The power of discord&lt;/a&gt;, but have never heard of it before. Can you give a bit more mathematical explanation of the term here? &lt;/p&gt;&#xA;" OwnerUserId="1648" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2011-06-10T15:04:02.653" LastActivityDate="2011-06-10T16:41:28.023" Title="What is quantum discord?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-information&gt;&lt;quantum-computation&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="10956" PostTypeId="1" AcceptedAnswerId="10958" CreationDate="2011-06-09T21:52:55.787" Score="3" ViewCount="251" Body="&lt;p&gt;I'm a little familiar with the physics and thermodynamics of the wind chill effect, but this question seems to come up from time to time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why, given two temperature sensors or thermometers in the same environment, do both report the same temperature if one is exposed to wind when the other is shielded from it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People often ask because the temperature reported by, for example, their vehicle, doesn't seem to change as they drive at different speeds, etc. (Other than of course the changes from one actual temperature to another as they change geography.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that inert devices aren't endothermic like we are, so the effects of wind chill don't apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you explain this in layman's terms?&lt;/p&gt;&#xA;" OwnerUserId="1638" LastActivityDate="2011-06-17T06:17:59.993" Title="Why does a thermometer in wind not show a lower temperature than one shielded from it?" Tags="&lt;thermodynamics&gt;&lt;temperature&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="10957" PostTypeId="2" ParentId="10952" CreationDate="2011-06-09T22:02:59.380" Score="0" ViewCount="" Body="&lt;p&gt;Basically a high frequency means a high change in the electric field sourrounding the conductor. That results in a higher emittance of electromagnetic waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want an easy to understand analogon: Suppose you are wiggling a rope. That creates a standing wave in the rope. The faster you wiggle, the more wavelength you will see in the rope and the more you will sweat, while you do it :D&lt;/p&gt;&#xA;" OwnerUserId="3952" LastActivityDate="2011-06-09T22:02:59.380" CommentCount="1" />
  <row Id="10958" PostTypeId="2" ParentId="10956" CreationDate="2011-06-09T22:13:41.820" Score="12" ViewCount="" Body="&lt;p&gt;Its really pretty simple, the thermometer measures temperature, wind chill measures heat loss for a body warmer than the air. Wind makes more unheated air available to conduct heat away from a hot body, but with a body at air temperature no heat is being condicted away from the thermometer. If you asked a secondary question, when I put a thermometer outside, how long must I wait until it has reached air temperature so that its reading if meaningful? In that case, the windy conditions will decrease the time it takes to equilibrate.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-06-09T22:13:41.820" />
  <row Id="10959" PostTypeId="2" ParentId="10956" CreationDate="2011-06-09T22:54:19.837" Score="3" ViewCount="" Body="&lt;p&gt;I maintain that this isn't an absolute fact but is probably true under the outside conditions you are talking about.  We should all agree that wind blowing does, in fact, increase the thermal coupling between the thermometer and the air.  However, the device &lt;em&gt;has no heat production&lt;/em&gt;.  If there is no heat production, it doesn't matter how strong or weak the thermal connection between two isolated items is because they will eventually equilibrate to the same temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A major caveat to that is that fact that the device and the air are not in an isolated system.  So &lt;em&gt;what else is there&lt;/em&gt;?  Well, there are heat potential sources/sinks that exist at different temperatures.  These include the night sky for instance, the sun, and the ground.  There exists radiative heat transfer between all of these and they all constitute a different temperature than the air.  To summarize in an equation, there is no heat production, $\dot{Q}$, but there is a heat flow, denoted by $\dot{q}_c(..)$ for conductive and $\dot{q}_r(...)$ for radiative, from everything that can thermally interact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{Q} = 0 = \dot{q}_c(T_\text{thermometer},T_\text{air})+\dot{q}_r(T_\text{thermometer},T_\text{sky})+\dot{q}_r(T_{\text{thermometer}},T_\text{ground})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And basically $q_c$ dominates, which could be written with Newton's law of cooling, proportional to $\Delta T=T_\text{thermometer},T_\text{air}$, which means that $T_{\text{thermometer}}=T_\text{air}$ ultimately.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="3887" LastEditorDisplayName="" LastEditDate="2011-06-17T06:17:59.993" LastActivityDate="2011-06-17T06:17:59.993" />
  <row Id="10960" PostTypeId="2" ParentId="10908" CreationDate="2011-06-09T23:32:35.167" Score="1" ViewCount="" Body="&lt;p&gt;A point source of light radiates that light spherically outward, and the total power radiated by the light remains the same at all distances.  As one gets further away, there is a greater total surface area, and the intensity of the light (power/area) is reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lenses don't &quot;intensify&quot; light, they only change its direction (and possibly attenuate it).  Rays of light that would have been diverging from each other can be made parallel (to form a beam) or converging (to focus the light in one spot).  There is still the same total amount of light, it has just been concentrated into a smaller area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making a beam requires that the rays of light exiting the lens have to be parallel or nearly parallel.  This requires that the light source be at the focus of the lens.  If the lens were moved relative to the light source, it would no longer form a beam and instead either converge or diverge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting a far-away lens for a nearer one with the focus in the same position would not necessarily intensify the beam.  The light source radiates the same amount of power per unit of solid angle, and a larger, far-away lens could do the same work of a smaller, nearby lens.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2011-06-09T23:32:35.167" CommentCount="1" />
  <row Id="10961" PostTypeId="1" CreationDate="2011-06-09T23:43:01.470" Score="1" ViewCount="58" Body="&lt;p&gt;Ok, digging into Nielson-Ninomiya Theorem (1) and doubler fermions, as well as solutions to these problems using Domain Wall Fermions (2) and overlap lattice fermions (3), both of which make effective use of 5th dimension, isn't that a pretty solid case that our universe is either of continuous 3+1 dimension space, or a space that is 4+1 dimension space that can be defined on a lattice? Should I buy a shirt that says: 5DLQCD or STFU?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) :http://latticeqcd.blogspot.com/2005/12/nielsen-ninomiya-theorem.html&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) :http://arxiv.org/PS_cache/hep-lat/pdf/9410/9410018v1.pdf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) :http://arxiv.org/PS_cache/hep-lat/pdf/0405/0405024v1.pdf&lt;/p&gt;&#xA;" OwnerUserId="3954" LastEditorUserId="3954" LastEditorDisplayName="" LastEditDate="2011-06-09T23:58:43.303" LastActivityDate="2011-06-10T01:22:11.173" Title="GDIWTFU with lattice QCD and the 5th dimension" Tags="&lt;quantum-chromodynamics&gt;" AnswerCount="0" CommentCount="2" />
  <row Id="10962" PostTypeId="2" ParentId="10952" CreationDate="2011-06-10T00:09:08.933" Score="1" ViewCount="" Body="&lt;p&gt;Your wiggle analogy contains too many poorly defined terms to be absolutely sure, but I think it is correct and may well help you understand what is going on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coaxial cables have, as does every other circuit element in the universe, a high frequency cutoff. It comes gradually as the frequency increases but eventually you can't get any power through the coax. The main reason is that each type of coax has a characteristic impedence that is a function of frequency. Another reason is that most losses in the system are frequency dependent. There are dielectric losses inside the plastic that separates the central conductor from the outer conductive shield. With some plastics, high frequencies can make it through the plastic and reach the center conductor.  There are stray currents, parasitics, leaks, losses, radiating elements and pretty much all of them get worse at higher frequencies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solutions include bigger coax, swapping aluminum for copper, upgrading to teflon wherever possible.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastActivityDate="2011-06-10T00:09:08.933" />
  <row Id="10963" PostTypeId="2" ParentId="10908" CreationDate="2011-06-10T00:26:09.543" Score="4" ViewCount="" Body="&lt;p&gt;Surprisingly, the maximum intensity one can obtain from a light source does not depend on the distance to the light source nor on the distance between the light source and the imaging equipment. Instead, it depends on the distance between the imaging equipment and the object being heated (assuming that the imaging system is designed to be the best possible for these distances). What you want to do is &quot;refract as close as possible&quot; to the object being heated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By thermodynamics, it's not possible for the light emitted by an object to heat up another that is already hotter than the first. For example, this means that we can't use an imaging system (or a non imaging power concentrating system such as a collection of mirrors) to use sunlight to heat a spot on the earth to a temperature higher than that of the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when you use an imaging system to magnify the effect of the sun, you do not get an image that is hotter than the sun. The best you can do is to make the image of the sun appear larger. And indeed, this is what a magnifying glass does. If you look through one (from the distance appropriate to frying an ant), you will see that objects in the direction of the image appear larger than they actually are. So when you apply a magnifying glass to an ant, what the ant sees is that the sun has increased in size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best you can do is to make the sun appear over all the sky, plus, with a mirror, to appear in the direction of the ground as well. Alternatively, you can arrange for the bottom side of the ant to be an insulator. (Uh, even concrete is a pretty good insulator if you try this at home.) This will heat the ant up until it approaches the temperature of the sun (with optimum optics and insulation). At that time the ant will radiate as much heat back to the sun as the sun radiates to the ant. And so the heat flows will balance and the ant will cease heating up. (All this assumes black body type radiation which is generally an okay approximation for this sort of thing.)&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-10T00:26:09.543" CommentCount="2" />
  <row Id="10964" PostTypeId="2" ParentId="10952" CreationDate="2011-06-10T00:29:32.523" Score="2" ViewCount="" Body="&lt;p&gt;Losses in coaxial cable are resistive. For low frequencies, one uses the full thickness of the coaxial cable and resistance is low. As frequencies increase, the signal is unable to penetrate as deeply into the conductor. This is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Skin_effect&quot; rel=&quot;nofollow&quot;&gt;skin effect&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as frequencies increase, the amount of metal that is used to carry the signal decreases. The result is an increase in the resistance and hence higher losses.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-10T00:29:32.523" />
  <row Id="10965" PostTypeId="2" ParentId="6928" CreationDate="2011-06-10T00:58:42.690" Score="0" ViewCount="" Body="&lt;p&gt;The GE Mark V containment system used at Fukushima has a design basis that calls for the concrete containment vessel to withstand a complete meltdown. This is typical of reactors in the US. There are many penetrations of the vessel and some of them may be leaking, but that does not necessarily mean the vessel was breached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pebble bed reactors can be shut down completely and not melt down. The next reactors in the US will also be able to lose all coolant flow and not melt down.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastActivityDate="2011-06-10T00:58:42.690" />
  <row Id="10967" PostTypeId="1" CreationDate="2011-06-10T02:22:10.640" Score="4" ViewCount="185" Body="&lt;p&gt;In a question titled &lt;a href=&quot;http://physics.stackexchange.com/questions/2041/how-are-classical-optics-phenomena-explained-in-qed-snells-law&quot;&gt;How are classical optics phenomena explained in QED (Snell's law)?&lt;/a&gt; Marek talked about the probability amplitude for photons of a given path. He said that it was $\exp(iKL)$, and that &quot;...this is very simplified picture but I don't want to get too technical so...&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know how it arises, even if it is technical. I find it very strange. If we compare it to the case of a particle obeying the Schrodinger equation, we have $\exp(iS/h)$ where $S$ is the action of a given path. $S$ is what we want to minimize(in the classical limit). In this case the path is a space-time path. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the other case, of the photon, where $L$ is that we have to minimize(in the classical limit or if you want in the geometrical optics limit) the paths are only in space, and I can't find any temporal dependence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;when I check any book about QED, I can read about the photon propagator (about space-time paths) but I never have found out about the expression $\exp(iKL)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general terms I find hard to relate what Feynman teaches in his book with what I have read in the &quot;formal QED books&quot; like Sokolov, Landau, Feynman or Greiner.&lt;/p&gt;&#xA;" OwnerUserId="3956" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-10T04:01:55.827" LastActivityDate="2011-06-10T22:09:05.777" Title="How is the path integral for light explained, or how does it arise?" Tags="&lt;quantum-electrodynamics&gt;" AnswerCount="4" FavoriteCount="2" />
  <row Id="10968" PostTypeId="1" CreationDate="2011-06-10T02:31:21.613" Score="3" ViewCount="77" Body="&lt;p&gt;In organic light emitting diodes, only singlet excitons contribute to the light emission (this is flouresence). Everything I read, say that triplet excitons cannot recombine on the same timescale as the singlets because spin selection rules forbid it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone expand on how the spin selection rules allow for singlet and not triplet emission? &lt;/p&gt;&#xA;" OwnerUserId="1939" LastActivityDate="2011-07-19T06:49:33.683" Title="selection rule used in singlet/triplet recombination in LEDs" Tags="&lt;quantum-mechanics&gt;&lt;spin&gt;&lt;semiconductor-physics&gt;" AnswerCount="1" />
  <row Id="10969" PostTypeId="2" ParentId="10953" CreationDate="2011-06-10T03:36:44.030" Score="2" ViewCount="" Body="&lt;p&gt;First, a correction to (3). It is not really incorrect but the diffeomorphisms are required for any spin-2 particles in any relativistic theory, not just string theory. It's because one needs to get rid of the negative-normed time-like modes of the spin-0 particles, excitations of $g_{0i}$, roughly speaking (where $g$ is a name of the spin-2 field, not necessarily a priori the metric tensor). A gauge symmetry is needed to do so. Because it is a symmetry, it must have a conserved current. Because it has Lorentz indices and other choices would be too constraining, this conserved quantity has to coincide with the stress-energy tensor of the remaining matter. The stress-energy tensor is the density and flux of the energy and momentum which is by Noether's theorem associated with spacetime translations. By densitizing it, we clearly get diffeomorphisms. The only semi-consistent modification of the procedure above would be to make the symmetry spontaneously broken in some way, obtaining massive gravitons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, a closed string is not &quot;just&quot; a spin-2 particle. A closed string may carry any integral value of the spin, depending on the state (and for a closed superstring, also all half-integral values are allowed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, your attempt to eliminate ordinary spin-2 particles in this simple way doesn't work because the closed string in the graviton spin-2 vibration mode &lt;em&gt;is&lt;/em&gt; the graviton. They have indistinguishable properties at low energies. Indeed, the consistent completion of the graviton's interactions at high energies requires the graviton to be completed to the whole &quot;string multiplet&quot; with the corresponding symmetry structure but this statement has not been rigorously proved and your observations are not enough to prove it because the spin-2 massless strings and gravitons are indistinguishable from the low-energy physics viewpoint.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-06-10T03:36:44.030" CommentCount="1" />
  <row Id="10970" PostTypeId="2" ParentId="10967" CreationDate="2011-06-10T04:31:21.320" Score="3" ViewCount="" Body="&lt;p&gt;Ordinary massive particles have the action equal to&#xA;$$ S = -m_0\int d\tau_{\rm proper} $$&#xA;which is negative and equal to the proper time along the world line multiplied by the rest mass. However, photons classically move along time-like geodesics and all of them have a vanishing proper duration. So one couldn't say which of these &quot;zigzag&quot; timelike trajectories is the right one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Snell's law needs another step to be derived. We need to assume a constant frequency of the photon. Because the frequency is specified, the velocity of the wave packet is determined by the local wave number. This reduces the selection to trajectories in space - Snell's law only addresses light's journey through static environments - because the direction of the trajectory in time is determined at each point by the known frequency. Also, the phase contributed to the path integral is simply the phase of the light&#xA;$$\exp(iKL), \quad K = 2\pi / \lambda$$&#xA;where $\lambda$ is the wavelength of the light in a given environment. If there are many environments along the path, $KL$ should be replaced by $\sum_i K_i L_i$. However, the thing we're minimizing isn't really the action, at least I don't see how to derive Snell's law directly from the principle of least action and the concepts of particles. However, if you study the action for the electromagnetic field,&#xA;$$ S = -\frac{1}{4} \int d^4x F^{\mu\nu}F_{\mu\nu} $$&#xA;then I believe that if you make the Ansatz that $F$ describes a constant-frequency electromagnetic wave of a unit intensity, the action $F^2$ could be perhaps reduced to $\sum_i K_i L_i$ simply because the same Snell's law follows from the principle of least action in electromagnetism. However, this derivation wouldn't be &quot;straightforward&quot;. For example, the description via Snell's law knows nothing about the two transverse polarizations included in Maxwell's theory.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-06-10T04:31:21.320" CommentCount="4" />
  <row Id="10971" PostTypeId="1" CreationDate="2011-06-10T04:51:22.203" Score="1" ViewCount="114" Body="&lt;p&gt;It is known that for a classical system the amount of information needed to store its state is the same as the amount of information that can be stored in that system. This amount is equal to Shannon's entropy or its classical generalizations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand the amount of information needed to describe the state of a quantum system is greater than the amount of information that can be stored in such quantum system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My impression is that to describe the state of a quantum system of $N$ qubits one needs $2^N$ bits of information, on the other hand one can only store $N$ bits of information inside such system, which is in my impression equal to von Neumann's entropy of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus I have two possible explanations of such discrepancy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) The true entropy of a quantum system is exactly von Neumann's entropy. The difference is only due the fact that a classical storage can only independently operate with digits not less than one bit, thus to store, say 4 variables of 1/4 bit each a classical system needs 4 bits (at least one bit for each variable) while a quantum system can use parts of a digit independently and thus store all 4 variables in one binary digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) The true entropy of a quantum system of $N$ qubits is exactly $2^N$ bits but the amount of &lt;strong&gt;extractable&lt;/strong&gt; information is only $N$ bits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which explanation do you consider more correct or are there possible other explanations?&lt;/p&gt;&#xA;" OwnerUserId="1186" LastActivityDate="2011-06-10T05:37:52.477" Title="Relation between classical and quantum information" Tags="&lt;quantum-mechanics&gt;&lt;quantum-information&gt;&lt;entropy&gt;" AnswerCount="1" />
  <row Id="10972" PostTypeId="2" ParentId="10971" CreationDate="2011-06-10T05:37:52.477" Score="4" ViewCount="" Body="&lt;p&gt;The entropy of $N$ qubits is always $N\log(2)$. That's pretty much by definition. One must carefully refine the definition in quantum mechanics. When $N$ qubits are measured, what we get is the same amount of information as in the case when we measure $N$ classical bits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, when we're preparing a state of $N$ qubits, we may choose $2^N$ complex coefficients of the individual $2^N$ states. Well, only $2^N - 1$ of these complex numbers are physical because one (real) overall normalization has to be fixed to be one and one (real) overall angle (phase) has no physical impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is no sense in which even a conceptually misleading calculation could produce $2^N$ bits from $N$ qubits. You may have incorrectly assumed that all the amplitudes in front of the $2^N$ basis vectors of the Hilbert space for $N$ qubits have to be either $0$ or $1$. But it's not the case: they're arbitrary complex numbers - only constrained by the overall normalization of the state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to precisely specify the state, you need to choose a point in ${\mathbb{CP}}^{2^N - 1}$. That's a continuous space so &quot;classically&quot; you need an infinite amount of information to specify a point in it. However, there's no physical sense in which the state actually carries an infinite amount of information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything that can be measured on one copy of such a system only reveals the information of $N$ bits. Their being quantum means that the initial state may be prepared in continuously many ways but this &quot;information&quot; we put into the preparation of the initial state can't be gotten back. Being quantum also means that the measurement may in principle measure the qubits with respect to a generally rotated basis. And the operations of the quantum computer may be arbitrary unitary transformations, too.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-06-10T05:37:52.477" CommentCount="3" />
  <row Id="10973" PostTypeId="2" ParentId="10967" CreationDate="2011-06-10T09:10:04.423" Score="1" ViewCount="" Body="&lt;p&gt;Like other areas in physics the resolution of EM/light phenomena can be described at many different levels and when you try to combine the terminology or equations from different levels, you usually end up confused :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like Lubos implies, the simple phase-equation that relates a pathlength to a phase depends on an assumption of a wave propagating with a fixed speed and a fixed wavelength. Except for polarization, this is enough to describe the propagation of light in a classical way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in more general descriptions, a Lagrangian formulation is used where the concept of a wave itself doesn't really arise until the equations are extremalized. That is, if you write out the &quot;path integral&quot; as a sum of the $exp(iS/\hbar)$ contributions where each $$S=-\int L dx^4$$ as you've probably seen in the textbooks (Lubos example above with the EM field tensor F is the kinetic part of the EM Lagrangian), you see that you do the integral over spacetime (the L here is the Lagrangian, not the pathlength). The Lagrangian formulation is designed so that the extremal value of this corresponds to the usual wave-propagations, but you actually formally sum over any configuration of the spacetime field in question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in a flash of confusing circularity, the photon propagator you've read about which is used to perturbatively find solutions to the above, is actually just a Green's function of the &lt;em&gt;already extremalized&lt;/em&gt; solution of the free photon field per the above general formulation.. I wish I could dig deeper into this here but I'm just learning how these different formulations fit together myself. Suffice to say, if you want to build a completely particle-based approach from scratch, you need to postulate the shape of the propagator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Feynman notes in his simple QED book he makes a couple of shortcuts. Apart from leaving out polarization, after talking about the simple path-length-dependent phase in 10 chapters he then casually drops the note that this is simply an approximation and tries to explain the framework of the above in 2 pages, much to the confusion of the reader :)&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-06-10T09:10:04.423" />
  <row Id="10974" PostTypeId="2" ParentId="10967" CreationDate="2011-06-10T09:25:25.760" Score="0" ViewCount="" Body="&lt;p&gt;To use a popular buzzword, I used a renormalization group to move from a scale of microscopic physics to a scale of macroscopic classical physics :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To describe what I've actually done: I've summed over the microscopic degrees of freedom. In a full theory of a single photon particle (i.e. &lt;em&gt;not&lt;/em&gt; QED) the particle travels through all trajectories and receives a phase proportional to their length. But as you know, under certain assumptions most of these trajectories are unimportant because they destructively interfere. So we may as well sum them up. What we will be left with is just a classical trajectory. Therefore the major contribution to the probability amplitude is the phase along classical trajectory which is precisely just $\exp(iKL)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approximation certainly breaks down when $L \to 0$ because lots of quantum mechanical effects will start to mess up the picture. Not only that, we will need to take full QED and even standard model into account, as there is always a possibility that the photon will create an electron-positron pair, etc and all of these modify the phase. Nevertheless, you can again sum over all these higher-order effects to obtain an effective propagator, this is all part of renormalization. These were the technicalities I wanted to avoid.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-06-10T09:25:25.760" CommentCount="9" />
  <row Id="10975" PostTypeId="2" ParentId="9715" CreationDate="2011-06-10T13:02:19.513" Score="1" ViewCount="" Body="&lt;p&gt;OK Georg, I am replying to this because it is coming around once more, and good answers exist in &lt;a href=&quot;http://en.wikipedia.org/wiki/Directed-energy_weapon&quot; rel=&quot;nofollow&quot;&gt;wikipedia.&lt;/a&gt; as I am too rusty to sit and calculate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Laser beams begin to cause plasma breakdown in the air at energy densities of around a megajoule per cubic centimeter. This effect, called &quot;blooming,&quot; causes the laser to defocus and disperse energy into the atmosphere. Blooming can be more severe if there is fog, smoke, or dust in the air.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the effect of strong energy in a laser beam ( the one known coherent electromagnetic weapon grade) turns air into plasma after a certain energy density. It will melt whatever it hits at those energies, which after all is the purpose of the weapon. Mirrors are out because the reflecting surface absorb part of the radiation and will be destroyed at such energy densities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle a coherent high energy density beam of soft photons could also pair produce by higher order diagrams, but one would have to calculate at what energy density this would destroy the beam even in vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the question is really &quot;how to defend a target from a laser weapon&quot;, the answer is &quot;smoke&quot;, aluminum confetti and such.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-06-10T13:02:19.513" />
  <row Id="10976" PostTypeId="2" ParentId="10955" CreationDate="2011-06-10T13:32:19.837" Score="5" ViewCount="" Body="&lt;p&gt;It is basically a measure of the quantumness of some correlations, which is not vanishing for some separable state. It was introduced by Ollivier and Zurek (&lt;a href=&quot;http://prl.aps.org/abstract/PRL/v88/i1/e017901&quot; rel=&quot;nofollow&quot;&gt;PRL&lt;/a&gt;/&lt;a href=&quot;http://arxiv.org/abs/quant-ph/0105072&quot; rel=&quot;nofollow&quot;&gt;arXiv&lt;/a&gt;). It is the difference between to different generalizations of the classical (Shannon) conditional entropy to the quantum world,  ant is 0 for pure bipartite separable state. It has then be proven to be the amount of entanglement needed in the task of state-merging (&lt;a href=&quot;http://pra.aps.org/abstract/PRA/v83/i3/e032324&quot; rel=&quot;nofollow&quot;&gt;PRA&lt;/a&gt;/&lt;a href=&quot;http://arxiv.org/abs/1008.3205&quot; rel=&quot;nofollow&quot;&gt;arXiv&lt;/a&gt; and &lt;a href=&quot;http://pra.aps.org/abstract/PRA/v83/i3/e032323&quot; rel=&quot;nofollow&quot;&gt;PRA&lt;/a&gt;/&lt;a href=&quot;http://arxiv.org/abs/1008.4135&quot; rel=&quot;nofollow&quot;&gt;arXiv&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definition&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://prl.aps.org/abstract/PRL/v88/i1/e017901&quot; rel=&quot;nofollow&quot;&gt;PRL&lt;/a&gt;/&lt;a href=&quot;http://arxiv.org/abs/quant-ph/0105072&quot; rel=&quot;nofollow&quot;&gt;arXiv&lt;/a&gt;)&#xA;Classically the conditional entropy $H(A|B)$ is a measure of the uncertainty one has on the  variable $A$ once we know the variable $B$. Of course, the definition of &quot;knowing&quot; $B$ becomes problematic when $B$ is quantum. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Classically, one can define $H(A|B)$ as the average $H(A|B)=\sum_b {\mathcal P}(B=b)H(A|B=b)$, each $H(A|B=b)$ being the entropy of $A$ given that the random variable has the value $b$. If one generalizes this to the quantum world, the $B=b$ part implies a quantum measurement (a POVM) which should be specified. A natural choice is the &quot;best&quot; measurement, the one which minimizes the entropy. The Shannon $H$ entropy is replaced by the Von Neumann entropy, and we define &#xA;$S(A|B_c)=\min_{\text{POVM}} \sum_{b}\mathcal{P}(\text{POVM applied to B gives } b) S(A|\text{POVM applied to B gives }b)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The previous definition leads classically to a redefinition of the conditional entropy&#xA;as an entropy difference : $H(A|B)=H(A,B)-H(B)$, which is always positive. Its quantum version, $S(A|B)=S(AB)-S(B)$ can be negative (in contrast with $S(A|B_c)$). Its negativity is a sufficient condition for entanglement. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The discord is defined as $S(A|B_v)-S(A|B)$ and is always positive. You can maybe see it&#xA;as the amount of correlation between $A$ and $B$ which is destroyed by a classical measurement of $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Link with state merging&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://pra.aps.org/abstract/PRA/v83/i3/e032324&quot; rel=&quot;nofollow&quot;&gt;PRA&lt;/a&gt;/&lt;a href=&quot;http://arxiv.org/abs/1008.3205&quot; rel=&quot;nofollow&quot;&gt;arXiv&lt;/a&gt; and &lt;a href=&quot;http://pra.aps.org/abstract/PRA/v83/i3/e032323&quot; rel=&quot;nofollow&quot;&gt;PRA&lt;/a&gt;/&lt;a href=&quot;http://arxiv.org/abs/1008.4135&quot; rel=&quot;nofollow&quot;&gt;arXiv&lt;/a&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The state merging primitive is the following. Suppose Alice, Bob and Charly share a 3-party pure entangled state. Alice want to send her part to Bob without destroying the quantum correlations  between $AB$ and $C$. Basically, she has to teleport $A$ to Bob, and the minimal amount of entanglement Alice and Bob need to perform this task is given by the quantum discord.&lt;/p&gt;&#xA;" OwnerUserId="373" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2011-06-10T16:41:28.023" LastActivityDate="2011-06-10T16:41:28.023" CommentCount="2" />
  <row Id="10977" PostTypeId="1" CreationDate="2011-06-10T13:48:07.680" Score="1" ViewCount="156" Body="&lt;p&gt;To calculate the work done by a gravitational field, the equation is written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W=GMm(r_\text{initial}^{-2} - r_\text{final}^{-2})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose a small mass has distance $R$ from a big mass. So here a gravitational force exists between the 2 masses. If we displace the small mass a very short distance $\mathrm{d}r$, it will need an opposite equal force against the gravitational force. Considering this fact, my question is, if these two forces are equal how we will displace the thing?  &lt;/p&gt;&#xA;" OwnerUserId="3962" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-10T16:25:55.000" LastActivityDate="2011-07-10T18:39:46.110" Title="Work done by gravitational field" Tags="&lt;forces&gt;&lt;newtonian-mechanics&gt;&lt;newtonian-gravity&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="10979" PostTypeId="1" CreationDate="2011-06-10T15:32:39.280" Score="1" ViewCount="63" Body="&lt;p&gt;Suppose particle A interacts with particle B by means of an exchange particle E. Presumably, since this is quantum physics, the exchange particle emission and receipt are just one of many possible paths by which the two particles can interact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that mean that A is in a continual state of superposition of having emitted an exchange particle in every direction? And that B then is in a continual state of superposition of having absorbed it (and thus moved if there was a force involved) and not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somehow that seems like the most sensible option, because the alternative is that A knows where B is and sends it E at some frequency determined by the strength of the fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process above would also entangle A and B as they cannot disagree over whether they exchanged E.&lt;/p&gt;&#xA;" OwnerUserId="670" LastActivityDate="2011-06-12T15:24:21.360" Title="Superposition and particle mediated interactions" Tags="&lt;quantum-mechanics&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="10980" PostTypeId="2" ParentId="9715" CreationDate="2011-06-10T15:36:18.417" Score="2" ViewCount="" Body="&lt;p&gt;You have some information about photon-photon scattering in vacuum on the &lt;a href=&quot;http://www.extreme-light-infrastructure.eu/High-field_5_2.php&quot; rel=&quot;nofollow&quot;&gt;Extreme-Light Infrastructure's website&lt;/a&gt;. In particular, they say &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For electric fields $E\sim mc^2/λ_C ~ 10^{29} {\mathrm W}/{\mathrm cm}^2$, where $λ_C$ is the Compton wavelength (Schwinger (1951)), virtual electron-positron pairs will be able to separate and become real.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which means that various interesting effects happens here, even in vacuum, limiting the power of the light beam. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While ELI's website shows that it these effects will probably seen in not too futuristic experiment, this density is expected to be reached in very short (attosecond) pulses at a low repetition rate. For a weapon, the relevant quantity is the average power, and making&#xA;a weapon which concentrates the power of $10^{20}$ nuclear reactor in one  cm² doesn't seam very realistic, except if you build a &lt;a href=&quot;http://en.wikipedia.org/wiki/Death_Star&quot; rel=&quot;nofollow&quot;&gt;Death Star&lt;/a&gt; ;-)&lt;/p&gt;&#xA;" OwnerUserId="373" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2011-07-27T09:54:35.283" LastActivityDate="2011-07-27T09:54:35.283" CommentCount="6" />
  <row Id="10981" PostTypeId="1" CreationDate="2011-06-10T15:44:03.010" Score="2" ViewCount="95" Body="&lt;p&gt;Please can someone help me with a equation for how much energy a person generates when swinging on a swing?&lt;/p&gt;&#xA;" OwnerUserId="3966" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-10T16:17:18.943" LastActivityDate="2011-06-10T17:41:34.140" Title="Energy produced by a swing?" Tags="&lt;energy&gt;&lt;newtonian-mechanics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="10982" PostTypeId="2" ParentId="10979" CreationDate="2011-06-10T16:10:31.200" Score="2" ViewCount="" Body="&lt;p&gt;If we are in QM, we have to admit that Particle A is &quot;everywhere&quot; and particle B is &quot;everywhere&quot; so their &quot;exchanges&quot; also happen everywhere. Their &quot;exchanges&quot; lead to correlation of relative position of these particles. Both particles are in mixed states but the system quasi-particles (relative motion and center of inertia motion) may be in pure states, as if they belonged to non interacting and &quot;independent&quot; systems.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-06-10T16:10:31.200" />
  <row Id="10983" PostTypeId="2" ParentId="10826" CreationDate="2011-06-10T17:07:23.853" Score="7" ViewCount="" Body="&lt;p&gt;It helps to be a bit clear about what we mean when we say something is slippery or lubricated. Slipping occurs when only weak intermolecular attachments occur at an interface. Soap lubricates many solid surfaces because a soap/water mixture forms micelles, little balls of the soap &quot;molecules&quot; with a hydrophilic outer shell and a hydrophobic inner area (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Soap&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Soap&lt;/a&gt;). These micelles can break and reform to create coatings on a surface in such a way as to lower intermolecular interactions between two objects (say, your foot and the floor). (Or, the little balls and water molecules just slide past each other, which produces the same effect.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This effect is only slightly relevant to shaving in that it allows the razor to slide more smoothly across your skin. (Remember that the razor is almost parallel to your skin, so the physical situation is very close to two flat surfaces coming into contact. That's also why it can help to pull your skin taut.) Lubrication does very little to the hair because the razor, as a sharp edge, presents many more molecules per unit surface area when it encounters a hair than two flat surfaces coming into contact. The sharp edge of the razor is much more likely to find microscopic roughness in the hair and, from that starting point, cleave through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why does soapy water, or shaving cream, actually make a shave less painful, if it's not the lubrication? It's actually because the hair absorbs moisture and becomes softer. From a molecular perspective, water penetrates the outer layer of the hair (the cuticle), assisted by the soap (because the cuticle is hydrophobic), and gets into the interior of the hair (specifically, the cortex). (Hair structure: &lt;a href=&quot;http://en.wikipedia.org/wiki/Hair#Description&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hair#Description&lt;/a&gt;) This swells the hair and helps separate the keratin strands, lowering the density and making it easier for the razor to cut. So it's not so much the lubricating effect of the soap, but the hydrating effect, that provides the shaving assistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, the swelling of the hair when wet helps to lift it up slightly from the skin, which lets you cut the hair closer to the root. This effect can be sufficient to cut the hair so that, when dry, it sits slightly below the surface of the skin. That's why you also get a smoother shave when using some form of lubricant.&lt;/p&gt;&#xA;" OwnerUserId="1196" LastEditorUserId="1196" LastEditorDisplayName="" LastEditDate="2011-06-11T17:12:26.023" LastActivityDate="2011-06-11T17:12:26.023" CommentCount="2" />
  <row Id="10984" PostTypeId="2" ParentId="10981" CreationDate="2011-06-10T17:16:36.667" Score="3" ViewCount="" Body="&lt;p&gt;To calculate the maximum energy of the &lt;em&gt;swinger&lt;/em&gt; one could use either the gravitational potential energy at the &lt;em&gt;swinger's&lt;/em&gt; highest position, $GPE=mg(h_{max}-h_{min})$, or the kinetic energy, $KE=mv^{2}/2$, at the lowest of the &lt;em&gt;swinger's&lt;/em&gt; positions; where the speed (and therefore KE) would be greatest. Experimentally, this would involve measuring either the change in &lt;em&gt;verticle&lt;/em&gt; height or the speed attained at the low-point, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy in the system is converted from one form to the other for every quarter-wavelength, e.g. high behind swing position to bottom of swing position. Equating the kinetic energy at the minimum height with the change in gravitational potential energy from the maximum height one obtains: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$mv^{2}/2=mg(h_{max}-h_{min})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that the the maximum speed one can obtain is in fact indepenent of the mass of the person (it appears on both sides of the equation so it cancels out.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $v$ we get: $v=\sqrt{2g(h_{max}-h_{min})}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a normal garden-variety swing, the change in &lt;em&gt;vertical&lt;/em&gt; height would be about $1m$. This would give a maximum speed of $v=\sqrt{2g(1)}=4.4ms^{-1}$. Where I used $g=9.81ms^{-2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bare in mind that this approach does not account for wind resistance which, IMO, would be pretty much negligable at these speeds anyway.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="1696" LastEditorDisplayName="" LastEditDate="2011-06-10T17:41:34.140" LastActivityDate="2011-06-10T17:41:34.140" CommentCount="3" />
  <row Id="10985" PostTypeId="2" ParentId="10977" CreationDate="2011-06-10T18:14:09.127" Score="1" ViewCount="" Body="&lt;p&gt;okay the topic says about work done by the gravitational force and i assume that by &lt;em&gt;&quot;displace the small mass a very short distance dr, it will need an opposite equal force against the gravitational force&quot;&lt;/em&gt; , the writer meant an equal external force acting on small mass which made dr displacement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes if a equal external force as gravitational force act on a body it wont move as the total force on that body would be zero . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compute work done by gravitational force the key idea is that we have to consider &lt;strong&gt;only the gravitational force&lt;/strong&gt; . There may be other force acting on the body but those wont be counted for calculating the work by gravitational force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example , lets consider a boy is lifting a book of mass m. now , he is lifting it with F' force and F is the gravitational force . Surely, F'&gt;F or he wont be able to lift the book. If he made a displacement dr then &lt;strong&gt;work done by him&lt;/strong&gt; is  W'=F'.dr&#xA;But the &lt;strong&gt;work done by the gravitational force&lt;/strong&gt; is W&quot;=F.dr=-mgdr&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;total work done on the book&lt;/strong&gt;  will be W=W&quot;+W'=F'.dr-mgdr&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it can be seen that if we want to calculate the work done by gravitational field we can do this even if an external force is acting on the body.the external force must be grater or less than gravitational force or else no displacement will occur and hence there wont be any work done by any force.&lt;/p&gt;&#xA;" OwnerUserId="3967" LastActivityDate="2011-06-10T18:14:09.127" CommentCount="1" />
  <row Id="10986" PostTypeId="2" ParentId="10849" CreationDate="2011-06-10T20:08:56.877" Score="3" ViewCount="" Body="&lt;p&gt;Let me see if I understand the question correctly: in general our solutions for a system will involve for every wavevector k a set of frequencies $\omega(k)$. When these curves giving $\omega(k)$ do not cross, there is in obvious sense in which we can separate our solutions into different modes. But when they do cross how do we assign modes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the answer is in general we don't. (Or we do but we shouldn't.) Unless the solutions have different symmetry properties, which is common enough, our label don't have meaning except convenience.  I don't think this is not entirely about definitions. Consider adiabatic motion: we think that if we perturb our solution slowly enough it will just change its frequency and wavenumber, but stay on the same mode. But in the vicinity of a crossing of two modes our solution will evolve to have components in both modes, barring symmetry. So is there is no sense to think of them as separate modes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The only case I can think where the labeling of modes actually matters is in topological insulators, a case which requires no [bulk] band touching. Other than that it is just a convenience or it labels symmetry properties)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies if I misunderstood your question&lt;/p&gt;&#xA;" OwnerUserId="3970" LastActivityDate="2011-06-10T20:08:56.877" CommentCount="3" />
  <row Id="10987" PostTypeId="2" ParentId="10967" CreationDate="2011-06-10T22:09:05.777" Score="1" ViewCount="" Body="&lt;p&gt;This is not meant as a comprehensive answer, but perhaps is a helpful addition to the above answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a less well-known Hamiltonian formulation of the path integral in which all classical paths through phase space are considered, not just over real space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have &#xA;$ S = \int L(q,\dot{q}) dt = \int (P dq - H(P,q) dt) $.  We integrate $e^{iS/\hbar}$ over all paths $(q(t),P(t))$.  If we restrict ourselves to paths satisfying $H(P,q) = const $, then the $H dt$ term is a constant and the $P dq$ term obviously produces the $e(iKL)$ factor.  &lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-06-10T22:09:05.777" CommentCount="1" />
  <row Id="10988" PostTypeId="2" ParentId="10979" CreationDate="2011-06-10T22:51:26.880" Score="1" ViewCount="" Body="&lt;p&gt;You are correct in your reasoning but I'd want to warn about thinking of A as being in a superposition of having sent a messenger particle and not, and B in a similar superposition. Like you write, the process &quot;entangles&quot; A and B, so the entire process is in a superposition - A+B and A+B+E, for every moment of E which is compatible with your measurements momenta of the incoming and outgoing A and B's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that usually a spacetime (4D) analysis is done so you don't have to think conceptually about A having sent out an E into &quot;thin air&quot;.. either there is an exchange (completely described in the 4D volume) or not.&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-06-10T22:51:26.880" />
  <row Id="10989" PostTypeId="1" CreationDate="2011-06-11T00:28:06.617" Score="3" ViewCount="129" Body="&lt;p&gt;This might sound like the silliest question ever, but can someone shed some light on what is actually happening in the following situation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;There is an asteroid in outer space what is rotating with high speed. There is an astronaut standing on its surface.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Does the astronaut feel lighter compared to if the asteroid wasn't rotating?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can he jump higher compared to if the asteroid wasn't rotating?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What happens if the astronaut makes a big jump vertically? Is he going to land on the exact same place, or the asteroid could &quot;rotate under&quot; the jumping astronaut?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3975" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-11T01:40:23.547" LastActivityDate="2011-06-11T03:45:28.893" Title="Astronaut on a rotating asteroid" Tags="&lt;gravity&gt;&lt;rotational-dynamics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="10990" PostTypeId="1" CreationDate="2011-06-11T01:15:00.927" Score="1" ViewCount="67" Body="&lt;p&gt;I almost all movies where you could see an animation about an asteroid, they move in a very distinct way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to explain better, but I think what we can see in the movies is that the asteroid is rotating around the x axis with constant speed, around the y axis with constant speed and around the z axis with constant speed, where x, y and z axes are of a reference frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this kind of motion? Is it correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the Euler's rotation theorem any 3D rotation that has a fixed point also has a fixed axis. So if they are rotating around more then one axis, does it mean that they don't have any fixed point? But then what is the motion what their center of mass is doing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suppose, that the correct movement would be when asteroids are rotating like planets, around one fixed axis. But to me the rotation commonly used is not like that. What is the correct way?&lt;/p&gt;&#xA;" OwnerUserId="3975" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-11T01:40:38.927" LastActivityDate="2011-06-11T01:59:58.800" Title="The rotating movement of an asteroid" Tags="&lt;rotational-dynamics&gt;" AnswerCount="1" />
  <row Id="10991" PostTypeId="1" AcceptedAnswerId="10995" CreationDate="2011-06-11T01:44:58.543" Score="1" ViewCount="161" Body="&lt;p&gt;If I have a car (with a particular engine) optimized (shape &amp;amp; weight distribution-wise) for attaining the top speeds possible, and I put that engine into a car which is heavier (but otherwise the same shape &amp;amp; design), will the heavier car have the same top speed in the real world?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing that the heavier car will accelerate at a slower rate, but I am not sure whether it would eventually hit the same top speed as the lighter car. Factors such as air resistance &amp;amp; the way racing cars are designed to hug the ground (as I understand it) might cause them to not have the same max speed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they don't have the same top speed, would it be possible to re-design the heavier car (i.e. change its shape and weight distribution) so that it has the same max speed as (or a higher max speed than) the lighter car? My thinking is that if the heavier car doesn't need to use the air resistance to &quot;hug the ground&quot; then it might be able to be designed more aerodynamically?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Update 1&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Okay, $F_{ground}$ increases with $m$, which decreases $|v|_{max}$. That makes sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But could the heavier car go as fast or faster with a different design? Here's my reasoning: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Speed increases while the car's $F_{engine}$ is greater than friction's $F_{ground} + F_{air}$. &lt;/li&gt;&#xA;&lt;li&gt;$F_{air}$ increases as $|v|$ increases.&lt;/li&gt;&#xA;&lt;li&gt;&quot;Upside down wings&quot; are used to provide extra $F_{downwards}$ (lets call it $F_{wings}$).&lt;/li&gt;&#xA;&lt;li&gt;Having too little $F_{downwards}$ decreases $F_{engine}$.&lt;/li&gt;&#xA;&lt;li&gt;Bigger 'wings' in #3 increases $F_{wings}$ but also increases $F_{air}$&lt;/li&gt;&#xA;&lt;li&gt;$F_{downwards} = F_{gravity} + F_{wings}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Based on this logic, a lighter car will need bigger 'wings' (#6) to maintain traction (#3) in order to maintain speed (#4), but increasing $F_{wings}$ increases $F_{air}$ by #5, which decreases $|v|_{max}$ (#1 + #2). However, as $m$ increases, $F_{gravity}$ increases, therefore less $F_{wings}$ is needed (#6), and therefore less $F_{air}$ is experienced. So we have&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;heavier car would have greater $F_{ground}$ which decreases $|v|_{max}$ by a constant amount&lt;/li&gt;&#xA;&lt;li&gt;lighter car would have greater $F_{air}$ which increases as $|v|$ increases&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So following this reasoning, wouldn't it be possible to build a heavier car which has greater $|v|_{max}$ than a lighter car?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Update 2&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Clarification: #4 is supposed to mean &quot;when there's too little force pushing the car down, the wheels will slip, which reduces the amount of force the engine can provide&quot;. Is that correct?&lt;/p&gt;&#xA;" OwnerUserId="3976" LastEditorUserId="3976" LastEditorDisplayName="" LastEditDate="2011-06-12T00:51:51.850" LastActivityDate="2011-06-13T14:54:46.810" Title="Will a lighter car have a higher top speed than a heavier car with an equal power engine?" Tags="&lt;fluid-dynamics&gt;&lt;forces&gt;&lt;aerodynamics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="10992" PostTypeId="2" ParentId="10990" CreationDate="2011-06-11T01:59:58.800" Score="2" ViewCount="" Body="&lt;p&gt;Any motion of a rigid object can be decomposed into translation (straight-line motion) of the center of mass, and rotation around the center of mass. In a movie scene, an asteroid would probably be both rotating and translating with respect to the camera, because it makes the motion look more complicated that way. But if you imagine a different camera moving along with the asteroid, it would see the motion as only rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, any rotation can always be expressed as a rotation around a fixed axis with a fixed speed. Even if it appears that the asteroid is rotating around several different axes simultaneously, there really is just one axis, according to Euler's theorem. In movie scenes, they tend to orient that axis at an odd angle to the camera, again to make it look more complicated, but given enough information about the characteristics of the motion, you can always find the one axis of rotation and identify the speed of the rotation around it.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-11T01:59:58.800" CommentCount="3" />
  <row Id="10993" PostTypeId="2" ParentId="5567" CreationDate="2011-06-11T02:11:40.810" Score="1" ViewCount="" Body="&lt;p&gt;thanks for your comments.I recently have come across with two key-concepts:shot noise and Back-action noise in qunatum optics.This is very important for me to know first that are shot noise and back-action the same? and then I am wondering whether back-action is the nature of light beam in vaccum or it's related basically to measurement detector like homodyne?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really like to know exclusively the characteristics of Back-action noise to be able to model it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any comments can be a help and very welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best,&#xA;MS&lt;/p&gt;&#xA;" OwnerUserId="3977" LastActivityDate="2011-06-11T02:11:40.810" />
  <row Id="10994" PostTypeId="1" CreationDate="2011-06-11T02:16:30.443" Score="3" ViewCount="44" Body="&lt;p&gt;I recently have come across with two key-concepts: shot noise and back-action noise in quantum optics. This is very important for me to know: first, are shot noise and back-action the same? Please let me know if there is any other equivalent term for back-action noise in quantum optics community.&#xA;I am also wondering whether back-action is the nature of light beam in vacuum, or is it related basically to measurement detector like homodyne?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really like to know exclusively the characteristics of Back-action noise to be able to model it.&lt;/p&gt;&#xA;" OwnerUserId="3977" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-11T02:34:01.387" LastActivityDate="2011-06-11T02:34:01.387" Title="back-action noise" Tags="&lt;quantum-mechanics&gt;&lt;quantum-optics&gt;" />
  <row Id="10995" PostTypeId="2" ParentId="10991" CreationDate="2011-06-11T02:29:26.743" Score="3" ViewCount="" Body="&lt;p&gt;The problem you've formulated is that these two cars are identical aside from the mass difference, so let's just limit this to two identical cars where one has an added weight in it.  The heavier car will accelerate slower, based on simple $F=ma$, where $F$ is the same, so $a$ must be smaller for a larger $m$.  Friction, which determines the max speed along with the force, is a little bit messier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The air resistance is basically only affected by the shape of the car, so it will be completely unchanged.  The heavier car, however, will have greater ground friction from the contact between the wheels and road and because of that will have a slower max speed.  One way you can convince yourself that the tire friction depends on the weight of the car is to just consider that the the tire deforms and creates heat, and greater mass will cause it to deform more every turn.  For super fast cars tire friction is actually extremely significant.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Update&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Due to increased formalism by the question, I can offer a little more detail.  Here is a really basic force diagram I made for this.  It's not perfect, but I think it's sufficient.  And by the logic of the question, $F_{drag}$ is broken up into 2 parts where one is from the body and one is from the wing.  Furthermore, the wing aerodynamic forces are really the wind specific drag plus $F_{wing}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Pb1nX.png&quot; alt=&quot;Car forces&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we were talking about increasing the weight of the car.  That increases $F_g$ and $F_{ground}$ because in real life the wheel friction force has a significant dependence on the weight.  The question mentioned the need for the wing in order to maintain traction.  So, traction is a tricky point, because it is related to the weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_g=M g$$&#xA;$$traction = \frac{M g+F_{wing}}{M g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is my belief that if a given turning radius without slipping was needed for a race car, this quantity is what would need to be kept constant.  I think this gets to a little of what the question wanted, which is that if the mass, $M$, changed, some redesign would be necessary to maintain the same traction (see the above equation).&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-06-13T14:54:46.810" LastActivityDate="2011-06-13T14:54:46.810" CommentCount="5" />
  <row Id="10996" PostTypeId="2" ParentId="10989" CreationDate="2011-06-11T03:29:42.380" Score="3" ViewCount="" Body="&lt;p&gt;Because of the centrifugal force countering gravity of the rotating asteroid, the astronaut would weigh less than on a non-rotating body. Also, the astronaut would be able to jump higher since his muscular strength would be unaffected. Finally, if the astronaut jumped vertically, he would not land on the same spot due to Coriolis force effects.&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-06-11T03:29:42.380" />
  <row Id="10997" PostTypeId="2" ParentId="10989" CreationDate="2011-06-11T03:45:28.893" Score="3" ViewCount="" Body="&lt;ol&gt;&lt;li&gt;Yes he feels lighter unless he's on the North or South pole, and on most of the asteroid it would feel like partly sideways gravity.  Since we can assume the asteroid is rigid, it is distinguished in such a way from my recent [hydrostatic self-gravitation problems][1] question.&#xA;&lt;li&gt;Yes he can jump higher... unless he's standing on a pole.&#xA;&lt;li&gt;This part is more complicated, I'll address more detail below&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;On Equator&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the astronaut jumps from the equator he leaves the ground from a point that has an apparent gravity (gravity combined with rotational acceleration) normal to the surface.  For small jumps on a large asteroid, yes, he will make it back to the same spot.  But what's the cutoff?  I believe it would be the point at where orbital dynamics started to matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g = \frac{a}{r^2} + b \frac{V_x^2}{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this would be the relevant equation, since $V_x$, the horizontal velocity in the CM frame, would be the invariant quantity.  Say he jumps a distance $d$ upwards, then to the extent that $d (2 a / r^3 + b V_x^2/r^2) \ll a/r^2+b V_x^2/r $, he would land about in the same place.  If this is not true he would land in a different place.  I'm not entirely sure about this, but it's my best shot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Poles&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, if he's on the poles the rotation doesn't matter, and he can jump to infinity and make it back to the exact same spot as an academic exercise.  In reality, the jump speed would have to be very carefully produced to go far and make it back without reaching escape velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Elsewhere&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, the apparent gravity is non-normal.  So the astronaut already feels like he's on a slope, and obviously he would have to jump in the direction that felt like &quot;up&quot; for this to make sense at all.  But if he did that, similar rules to the Equator would apply.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-06-11T03:45:28.893" CommentCount="2" />
  <row Id="10998" PostTypeId="1" AcceptedAnswerId="11001" CreationDate="2011-06-11T04:29:20.897" Score="3" ViewCount="185" Body="&lt;p&gt;I'm doing old exam questions, and here is one that on first glance seemed rather simple to me, but I just can't get it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given are two operators $A$ and $B$, and all we know about them is that &#xA;$$[A,B] = B$$&#xA;and &#xA;$$B^\dagger B = 1 - A^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, I must find the &quot;hermiticity properties&quot; of $A$ and $B$. So far, the only progress I have made is to note that $A^2$ must be hermitian, because for any operator, $B^\dagger B$ is hermitian. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this does not suffice to determin the hermiticity of $A$ itself, as $A$ could be hermitian, anit-hermitian or even have one hermitian part $A_h^\dagger = A_h$ and one antihermitian part $A_a^\dagger = -A_a$ as long as $A_h A_a = -A_a A_h$, i.e. the hermitian and anti-hermitian part of $A$ anti-commute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sure that there must be a really simple trick, but I just don't get it...&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-06-12T15:22:22.887" Title="Commutators and Hermiticity - Exam question" Tags="&lt;operators&gt;" AnswerCount="3" FavoriteCount="0" />
  <row Id="10999" PostTypeId="2" ParentId="10998" CreationDate="2011-06-11T07:40:13.410" Score="0" ViewCount="" Body="&lt;p&gt;You can say a bit more if for instance $A$ and $B$ are &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_operators&quot; rel=&quot;nofollow&quot;&gt;normal&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Compact_operator&quot; rel=&quot;nofollow&quot;&gt;compact&lt;/a&gt; operators on a complex Hilbert space $H$. One may prove that such operators are always diagonalizable. Under the assumption that $B$ is normal $BB^{\dagger}=B^{\dagger}B$, we calculate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0=-[B^{\dagger}B,B]=[A^2-1,B]=A[A,B]+[A,B]A=AB+BA, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0=[AB+BA,B]=[A,B]B+B[A,B]=2B^2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $B$ under above assumptions is diagonalizable, this implies $B=0$, so $B$ is then Hermitian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover $A^2=1$ is an involution. Assuming $A$ is diagonalizable, it must have eigenvalues $\pm 1$, so $A$ is then also Hermitian.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-11T12:47:01.277" LastActivityDate="2011-06-11T12:47:01.277" CommentCount="4" />
  <row Id="11000" PostTypeId="1" CreationDate="2011-06-11T10:39:11.773" Score="5" ViewCount="425" Body="&lt;p&gt;Browsing an archive of problems of a local physics olympiad, i stumbled upon a problem which seems not a very trivial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given two identical metal spheres in vacuum, with mass $m$ and radius $R$. One sphere with a charge $Q$ and other with no charge(neutral). At initial moment, they are very far from each other. After releasing, due to the electrostatic attraction, the spheres come together and collide. Find the velocity of the spheres after the collision if the collision is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) perfectly elastic&lt;br&gt;&#xA;b) inelastic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignore gravity forces and a possible spark discharge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part b) seems especially tricky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, there was probably a typo. The part b) must be simply &quot;inelastic collision&quot; that means kinetic energy is not conserved and the spheres do not stick. Sorry for confusion!&lt;/p&gt;&#xA;" OwnerUserId="552" LastEditorUserId="552" LastEditorDisplayName="" LastEditDate="2011-06-13T06:09:00.357" LastActivityDate="2011-06-14T20:25:25.153" Title="Two spheres (A physics olympiad problem)" Tags="&lt;classical-mechanics&gt;&lt;electrostatics&gt;" AnswerCount="2" CommentCount="20" FavoriteCount="2" />
  <row Id="11001" PostTypeId="2" ParentId="10998" CreationDate="2011-06-11T11:29:26.543" Score="5" ViewCount="" Body="&lt;p&gt;The left hand side of the second equation is Hermitian, and therefore so is the right hand side. Therefore $A^2$ is Hermitian. Its eigenvalues are real. It follows that the eigenvalues of $A$ are either real or pure imaginary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More importantly, the first commutator is known from raising and lowering operators of the $SU(2)$ algebra,&#xA;$$ A = J_z, \quad B = k J_+ $$&#xA;is a representation of the commutator. By the way, the second equation is also very similar to what $J_\pm, J_z$ satisfy except for the wrong coefficient in front of $A^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let me prove that $A$ actually has no nonzero pure imaginary eigenvalues. Take an eigenstate $\psi$ of $A$ with eigenvalue $a$. The first commutator implies that $B\psi$ is also an eigenstate of $A$ with eigenvalue $a+1$. The commutator means that $B$ carries the $A$-charge equal to one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if $a$ is nonzero and pure imaginary, $a+1$ is neither pure imaginary nor real. It follows that the only way how to satisfy it is to have $B\psi = 0$ which is a trivial eigenstate $A$ with any eigenvalue. However, it is not possible to have $B\psi=0$ because of the second condition: the expectation value of $B^\dagger B$ in the state $\psi$ is the same as the expectation value of $1-A^2$. The former is zero because $B\psi=0$. However, the latter is not zero because the eigenvalue of $A^2$ in $\psi$ is negative, so $1-A^2$ can't vanish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we may assume that $A$ only has real eigenvalues. If the eigenvalues of $A$ are nonzero, it's clear that $A$ can't have any non-trivial Jordan blocks with these nonzero eigenvalues because $A^2$ would have them as well so it couldn't be Hermitian. So all the nonzero-eigenvalue portion of $A$ has to be diagonalizable. We've almost proved that $A$ is Hermitian. However, there's a subtle problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the vanishing eigenvalues of $A$? Well, I clearly can't prove that $A$ isn't allowed to have the nilpotent $2\times 2$ blocks. For example, we may set $B=0$ and $A$ equal to a nilpotent nonzero $2\times 2$ matrix, $((0,1),(0,0))$, and all equations are going to be satisfied. This $A$ is not Hermitian. So the allowed $2\times 2$ nontrivial nilpotent Jordan blocks with a vanishing eigenvalue are the only obstacle in proving that $A$ is Hermitian.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-06-11T11:29:26.543" CommentCount="3" />
  <row Id="11002" PostTypeId="2" ParentId="10926" CreationDate="2011-06-11T11:53:55.363" Score="1" ViewCount="" Body="&lt;p&gt;Ok, I'll answer my own question, since I've just found a solution thanks to Mr. Jackson. We can find in this article: &lt;a href=&quot;http://arxiv.org/pdf/physics/0204034v2&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/physics/0204034v2&lt;/a&gt;, that the equation (3.16) implies that every field with dependence $1/r$ is created by a current that is transverse to the direction of observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec{A}_{Coulomb-radiation} = \frac{\mu}{4\pi}\int \frac{\textrm{d}^2 x&amp;#39;}{r}&#xD;&#xA;[\vec{J}(\vec x&amp;#39;,t&amp;#39;)-\hat r(\hat r \cdot \vec{J}(\vec x&amp;#39;,t&amp;#39;)]_{retarded}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And precisely the current present in the integral is the perpendicular component to the&#xA;line of sight.&lt;/p&gt;&#xA;" OwnerUserId="3931" LastActivityDate="2011-06-11T11:53:55.363" />
  <row Id="11003" PostTypeId="1" AcceptedAnswerId="11013" CreationDate="2011-06-11T12:10:04.920" Score="4" ViewCount="211" Body="&lt;p&gt;It seems to be related to &lt;a href=&quot;http://en.wikipedia.org/wiki/Exchange_interaction&quot; rel=&quot;nofollow&quot;&gt;exchange interaction&lt;/a&gt;, but I can't penetrate the Wikipedia article. What has the Pauli exclusion principle to do with indistinguishability?&lt;/p&gt;&#xA;" OwnerUserId="2824" LastActivityDate="2011-06-11T19:19:20.977" Title="What causes the Pauli exclusion principle (and why does spin 1/2 = fermion)?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11004" PostTypeId="2" ParentId="11003" CreationDate="2011-06-11T12:25:26.313" Score="4" ViewCount="" Body="&lt;p&gt;The exchange interaction is, on the contrary, one of the approximate low-energy consequences of the Pauli exclusion principle and the identical character of the particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fundamental justification of these facts is offered by quantum field theory. Relativity requires observables to be linked to regions of spacetime - so that they don't communicate over spacelike separations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that the observables must be - or at least, in the simplest cases such as the Standard Model and its subsets, they are - functionals of quantum fields. Spacelike separated fields either commute or anticommute with each other - these are the only two ways how to make sure that they don't interact faster than light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Pauli exclusion principle only applies to half-integral-spin particles because of Pauli's spin-statistics theorem,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Spin-statistics_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Spin-statistics_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The page above also has a convincing bogus proof and a proof.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-06-11T12:25:26.313" />
  <row Id="11005" PostTypeId="1" CreationDate="2011-06-11T13:07:17.747" Score="0" ViewCount="92" Body="&lt;p&gt;I would like to use something like a kinematic equation to know where the person will be in the jump at a given time.  However since I think acceleration is not constant because of the jump force it doesn't seem I can use that equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like this is a summation of forces with a kinematic equation but it has been a few years since I've had to do this kind of thing.  Any help is greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="2596" LastActivityDate="2011-06-11T17:24:35.923" Title="Can someone explain the physics of a standing jump?" Tags="&lt;forces&gt;&lt;newtonian-mechanics&gt;&lt;kinematics&gt;" AnswerCount="2" />
  <row Id="11006" PostTypeId="2" ParentId="11005" CreationDate="2011-06-11T13:58:32.167" Score="0" ViewCount="" Body="&lt;p&gt;when we toss a ball vertically into the air from our palm, we do work on it. in this way we provide it the necessary energy for it to reach a certain height. similar is your jump. here your feet provide the required force on the floor and the floor pushes you back with an equal and opposite force(Newton's 3rd law). this enables you to attain a certain height above ground. this problem is very easy when we apply conservation of energy principle. the KE that you gain when you jump up = your PE when you have reached the maximum height. equating these two energies we get v=(2gh)^(1/2).&#xA;why do you think acceleration is not constant? always remember that when a resultant unbalanced external force acts on a body it accelerates. here at height h your body weight mg is the resultant force. thus g here is acceleration which is constant.&lt;/p&gt;&#xA;" OwnerUserId="3938" LastActivityDate="2011-06-11T13:58:32.167" CommentCount="2" />
  <row Id="11007" PostTypeId="1" CreationDate="2011-06-11T15:05:38.033" Score="1" ViewCount="173" Body="&lt;p&gt;I was having a debate the other day with a work colleague where I explained that gravity is a weak force because it is easily broken. Then I remembered a lecture by someone, I forget who, that explained gravity is very weak because you can break its influence just by jumping or lifting a pencil, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He countered that with something along the lines of 'that even though the pencil or your body is being moved away from the source of gravity it is still affected by gravity and thus it has weight'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is jumping a good example of gravity being a weak force?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. You can probably tell, my colleague and I are not physicists but we enjoy our little debates, we just need to get our facts straight.&lt;/p&gt;&#xA;" OwnerUserId="3983" LastEditorUserId="2499" LastEditorDisplayName="" LastEditDate="2011-06-12T18:09:23.107" LastActivityDate="2011-06-12T18:26:59.370" Title="Can the effects of gravity be broken by jumping?" Tags="&lt;gravity&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="11008" PostTypeId="2" ParentId="11007" CreationDate="2011-06-11T15:26:33.790" Score="0" ViewCount="" Body="&lt;p&gt;Well, keep in mind gravity isn't being &quot;broken&quot; when you jump, it is still exerting a force of approximately F_g=G * m_1 * m_2 / r^2, where G is the gravitational constant (6.677 * 10^(-11)), m_1 and m_2 are the two masses, and r is the distance between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of gravity being a weak force, relative the electromagnetic force is shown by jumping as the electromagnetic forces moving your muscles are able to overcome (a better word than &quot;broken&quot;) the gravity of the entire earth for a period of time, so F_g &amp;lt; F_jump. Since you cannot jump again midair very effectively, it is temporary, as F_g is present still in the air. But assuming you had lots of energy and a space suit you could climb a very long ladder out to where gravities effects would be minimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another good example that I like of electromagnetic force &gt; gravity is the fact that when you jump out of a window (first story please!), land on the pavement. There the electromagnetic interactions between your electrons and the pavements electrons are stopping you from going all the way to the center of the earth.&lt;/p&gt;&#xA;" OwnerUserId="3018" LastActivityDate="2011-06-11T15:26:33.790" />
  <row Id="11009" PostTypeId="2" ParentId="11007" CreationDate="2011-06-11T15:56:18.747" Score="2" ViewCount="" Body="&lt;p&gt;I think what you heard in that lecture is this argument:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Gravitation is by far the weakest of the four interactions. Hence it is always ignored when doing particle physics. &lt;strong&gt;The weakness of gravity can easily be demonstrated by suspending a pin using a simple magnet (such as a refrigerator magnet). The magnet is able to hold the pin against the gravitational pull of the entire Earth.&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Yet gravitation is very important for macroscopic objects and over macroscopic distances for the following reasons. Gravitation:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;is the only interaction that acts on all particles having mass;&lt;/li&gt;&#xA;  &lt;li&gt;has an infinite range, like electromagnetism but unlike strong and&#xA;  weak interaction&lt;/li&gt;&#xA;  &lt;li&gt;cannot be absorbed, transformed, or shielded against;&lt;/li&gt;&#xA;  &lt;li&gt;always attracts and never repels.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Jumping, or lifting a pencil, is in your example &quot;breaking&quot; the influence of gravity because the electromagnetic interactions between your feet and the ground are able to counteract the gravitational force of the entire planet, thus demonstrating that gravity is a weak force, so I'd say yes, it's a good example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_interaction#Gravitation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fundamental_interaction#Gravitation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3951" LastActivityDate="2011-06-11T15:56:18.747" />
  <row Id="11010" PostTypeId="2" ParentId="11007" CreationDate="2011-06-11T16:18:20.307" Score="0" ViewCount="" Body="&lt;p&gt;Gravity is a very &quot;weak&quot; force, but there's something special about it in how it is truly cumulative.  All matter (as far as I'm concerned) has the same sign of gravity pull.  By that I mean the force is &quot;inward&quot;, or written as $-\vec{r}/|\vec{r^3}|$, so you could call it &quot;negative&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;user599884 gave a good reason the force can be called weak, which is true by all means.  In fact, two balls containing exactly $1 C$ (Coulomb) of charge placed $1 m$ away will repel each other with $9 GN$ of force, which is something like a million tons of force.  BUT, a ball with a $1 C$ charge on it has an imbalance of electrons versus protons of about $10^{18}/10^{23} Z\approx 0.00001$ times the total, or about 0.001%.  That is some mighty force created by unthinkably small amount of matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact of life, however, is that we don't experience significant bulk forces from the electromagnetic or nuclear forces.  I should clarify, however, that those forces give the form to everything around you on small scales.  In fact, before humans started making magnets and electric machines, there was very little bulk E&amp;amp;M force in nature that could exert significant forces on macroscopic objects, even though those forces are so powerful!  Why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a very profound distinction between between gravity and other forces.  Regarding electrostatic forces, there are 2 kinds of charges and matter will tend to try to balance those (and does a superb job at it actually).  For magnetism, charges will ultimately move as a result of magnetic fields in a way that decreases the strength of the magnetic field.  Not so for gravity.  Gravity clumps matter together with doesn't degrade it's gravitational strength on large scales, making it truly cumulative.  This is why over large time scales gravity &quot;wins&quot;, the galaxies, planets, and stars are a result of gravitational clumping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravity of $1 g$ you experience around you is the resultant force from every single atom in all of the planet.  Just a tiny tiny tiny fraction of all those atoms exerting a different force like E&amp;amp;M or nuclear would be able to easily counter the $1 g$, but, the difference is that those other forces are balanced perfectly and don't give a resultant force that can operate on you.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-06-11T16:18:20.307" CommentCount="4" />
  <row Id="11011" PostTypeId="2" ParentId="11000" CreationDate="2011-06-11T16:23:07.340" Score="3" ViewCount="" Body="&lt;p&gt;There is an electrostatic attraction between the two spheres even though one has a net zero charge. On the side of the uncharged sphere that faces toward the (say) positively charged sphere, there will be felt a strong electromagnetic field arising from the charged sphere. This field will attract conduction band electrons and cause them to rush to the side facing the charged sphere. As a result, the far side of the uncharged sphere will have a shortage of electrons thus will be positively charged.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastActivityDate="2011-06-11T16:23:07.340" CommentCount="1" />
  <row Id="11012" PostTypeId="2" ParentId="11005" CreationDate="2011-06-11T17:24:35.923" Score="1" ViewCount="" Body="&lt;p&gt;You are correct to say that the acceleration during the jump will not be constant. However, once the jumper is in the air, only gravity (ignoring air resistance and other effects) acts to accelerate the body. The details of the jump are not important to applying kinematic equations to the motion, so long as you know the initial velocity (angle and magnitude). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, note that this assumes you're working with a point mass, i.e. no rotation or internal energy, which isn't strictly accurate for the real world. Insofar is you can approximate things to this ideal case, the kinematic equations will track the center of mass of the jumper. Of course, real long jumpers rotate their bodies so their feet land farther from the starting point than their center of mass, but if you're only concerned with where the center of mass is, not the jump distance that would be measured in a competition, kinematics will serve you acceptably well.&lt;/p&gt;&#xA;" OwnerUserId="1196" LastActivityDate="2011-06-11T17:24:35.923" />
  <row Id="11013" PostTypeId="2" ParentId="11003" CreationDate="2011-06-11T19:19:20.977" Score="6" ViewCount="" Body="&lt;p&gt;Indistinguishableness of particles is formulated in QM in terms of the total wave function symmetry. Then the wave functions can be symmetric or antisymmetric on their arguments. In case of antisymmetry ($\psi(x_1,x_2) = -\psi(x_2,x_1))$ the particles are called fermions and $\psi(x_1,x_2=x_1) = 0$. They say &quot;the particles cannot occupy the same state&quot;. It's a feature of nature, an experimental fact.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-06-11T19:19:20.977" CommentCount="2" />
  <row Id="11014" PostTypeId="1" AcceptedAnswerId="11034" CreationDate="2011-06-11T22:09:36.303" Score="4" ViewCount="276" Body="&lt;p&gt;I have read that the night sky should have been bright because every spot should end up pointing to a star in the infinite universe but this is not the case because the universe is expanding. I am thinking if the universe cannot be expanding with speed faster than light then the dark night sky is also due to the universe is relatively young and when it gets older the night sky will eventually end up bright. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct or am I missing something? &#xA;Will the speed of universe expansion will make the sky bright but the red shift make it invisible to our eyes ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also if it's correct, does this mean that cosmologists do detect new stars appearing from nowhere from time to time? Does this really happen?&lt;/p&gt;&#xA;" OwnerUserId="3241" LastActivityDate="2011-06-13T23:54:28.230" Title="Will the night sky eventually be bright?" Tags="&lt;cosmology&gt;&lt;space-expansion&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11015" PostTypeId="2" ParentId="11014" CreationDate="2011-06-11T22:51:07.623" Score="1" ViewCount="" Body="&lt;p&gt;The way that you ask your question confuses the answer because you say &quot;Will the speed of universe expansion will make the sky bright but the red shift make it invisible to our eyes&quot;, because the sky is already &quot;bright&quot; in certain wavelengths, particularly the centimeter cosmic background radiation (CMB).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than this revision, yes, the observation that the night sky is dark has been a clear argument against an infinitely old universe since long ago.  The evidence for the big bang in the form of a consistently increasing red shift pretty much seals the deal for myself, regarding the fact that the universe has an age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, over time, you are entirely correct in the assertion that the &lt;em&gt;number of observable objects&lt;/em&gt; will increase drastically, and quite possibly infinitely.  Consider that we only see $x$ distance away which terminates at the CMB, thus limiting the number of galaxies we can see, with the furthest galaxies being the earliest evolutionary stage of galaxies.  The number of &quot;young&quot; galaxies we can see will progressively increase as more of the veil from the CMB is pulled back through the arrival of the new light.  The &quot;young&quot; galaxies we can see now will mature and the total number will increase.  Whether or not this will increase forever is disputable since dark energy pulls space apart could prevent it but we can't claim to know exactly what the behavior of dark energy far in to the future will be.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Additions&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I started thinking about the problem more and I wanted to formalize things a bit better.  Take the most basic case, we'll deal with a flat Newtonian space for now.  As before, take $x$ to be the distance to a certain galaxy we are current seeing.  Take the present time (after the big bang) to be $t$ and that we're observing that galaxy at $t&amp;#39;$.  It follows...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=c (t-t&amp;#39;)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine the universe has a galaxy density of $\rho$ galaxies per unit volume.  Then knowing that, we can actually write the rate $r$ at which galaxies older than $t&amp;#39;$ are appearing into our view.  This is done knowing the surface of a sphere is $4\pi r^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r=4 \rho c \pi x^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's fascinating to consider that in a line connecting every object in the night sky and us, there exists the entire history of the object encoded in the light waves making their way to us.  One way to talk about the acceleration of the universe is to say that there is a slowdown in the rate at which we are receiving this information.  We are watching the far off objects &lt;em&gt;in slow motion&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we make the obviously incorrect but useful assumption that all objects emit light at the same rate at all times, then the intensity we see will be proportional to $1/x^2$, and given some $S$ which is, say, the number of photons emitted total per unit time, then the intensity of light we receive from a given body would be $S/(4 \pi x^2)$.  Multiplying this by the rate, we can get a very nice equation for $s(x)$ which is the contribution to the number of photons we receive from the differential &quot;shell&quot; of stars at $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s(x) = S \rho c $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation is important because it is cumulative from time at $t&amp;#39;$ to $t$, meaning that the objects that entered our field of vision from the &quot;genesis&quot; of that type of object are still contributing to the population of photons reaching us today.  So the number of photons we are receiving could be said to be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{t&amp;#39;}^t S \rho c dt = S \rho c (t-t&amp;#39;)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more advanced view of the situation simply notes that the &quot;movie&quot; for each of these stars is being played in slow motion.  We'll just define a factor for that and put it in the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$l(x) = \frac{\Delta t_{object}}{\Delta t_{Earth}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should preface this by saying that this isn't actually saying time is going slower for that object, and this &lt;em&gt;isn't even&lt;/em&gt; the time dilation as defined by general relativity, this is the time dilation you would measure by watching a clock in a galaxy far away with a space telescope and comparing it to the local time.  Yes, these two are different, and yes, I am avoiding advanced relativistic concepts by making it an accounting problem.  Now the total # of photons we receive per unit time is the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{t&amp;#39;}^t S \rho c l(x) dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I won't use any calculus chain rules because there's no guarantee that $l(t)$ is any more helpful to you than $l(x)$!  But I should also note that the final $x$ you get in this equation at $t$ will be meaningless.  It is not the general relativity distance, it's some bastardization of that by using $c t$, which is clearly not how it actually works.  Nonetheless, there is some usefulness in the above equation.  We can even identify the radiative energy being received by considering the energy of the photon being proportional to it's frequency, with $E_e$ being the energy of the emitted photon and $E_o$ the observed photon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{E_e}{E_o} = l(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the total energy would then be the following with $h$ the familiar plank constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = \int_{t&amp;#39;}^t S h \rho c l(x)^2 dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, my intent is for these to be instructive &quot;kindergarten&quot; equations for the subject.  The bottom line is still clear from them - that the # of photons reaching us would increase linearly over time but it's less since $l(x)\le 1$.  Similarly, the radiative energy reaching us would be less by even a smaller factor due to the redshift.  I hope this is a clear picture.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-06-13T23:54:28.230" LastActivityDate="2011-06-13T23:54:28.230" CommentCount="7" />
  <row Id="11016" PostTypeId="2" ParentId="7433" CreationDate="2011-06-12T07:05:41.140" Score="-1" ViewCount="" Body="&lt;p&gt;in liquefaction of gases we reduce temperature of gases in order to convert them into liquid.&lt;/p&gt;&#xA;" OwnerUserId="3989" LastActivityDate="2011-06-12T07:05:41.140" />
  <row Id="11017" PostTypeId="1" AcceptedAnswerId="11025" CreationDate="2011-06-12T07:39:49.093" Score="5" ViewCount="216" Body="&lt;p&gt;I'm very new to quantum mechanics. I'm thinking of writing a quantum computer simulator, would the following work?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each qubit is stored as a single bit,&lt;/li&gt;&#xA;&lt;li&gt;For each operation, the qubits involved are transformed into a complex vector of amplitudes of size 2^N. This will result in a vector containing 2^N-1 0s and one 1.&lt;/li&gt;&#xA;&lt;li&gt;This vector is multiplied by the unitary matrix representing the operation.&lt;/li&gt;&#xA;&lt;li&gt;The resulting vector is squared element-wise.&lt;/li&gt;&#xA;&lt;li&gt;An outcome is picked using the elements of the result as probabilities and the qubits are set according to this outcome.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm not concerned with the running time or the memory cost of the algorithm. What I am concerned about is whether this would result in a physically accurate simulation.&lt;/p&gt;&#xA;" OwnerUserId="832" LastActivityDate="2011-06-12T23:08:50.833" Title="Is this algorithm for simulating a quantum computer accurate?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="11018" PostTypeId="2" ParentId="10670" CreationDate="2011-06-12T10:24:41.367" Score="1" ViewCount="" Body="&lt;p&gt;For an experimental test see:&#xA;Liquid marbles &#xA;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2001Natur.411..924A&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/2001Natur.411..924A&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(paywall &lt;a href=&quot;http://www.nature.com/nature/journal/v411/n6840/full/411924a0.html&quot; rel=&quot;nofollow&quot;&gt;http://www.nature.com/nature/journal/v411/n6840/full/411924a0.html&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a paper in General Relativity see:&#xA;Accurate simulations of the dynamical bar-mode instability in full general relativity&#xA;&lt;a href=&quot;http://adsabs.harvard.edu/abs/2007PhRvD..75d4023B&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/2007PhRvD..75d4023B&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically once you start increasing rotation the pancake will become unstable and &#xA;will make a transition to a rotating bar (dumbbell).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway I think that nobody ever saw the bar actually breaking into 2 pieces.&#xA;Typically you loose matter from the external regions, redistribute angular momentum&#xA;and go back to axisymmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers &lt;/p&gt;&#xA;" OwnerUserId="3229" LastEditorUserId="3229" LastEditorDisplayName="" LastEditDate="2011-06-12T11:02:21.287" LastActivityDate="2011-06-12T11:02:21.287" CommentCount="4" />
  <row Id="11019" PostTypeId="2" ParentId="11017" CreationDate="2011-06-12T10:55:30.580" Score="2" ViewCount="" Body="&lt;p&gt;The information about an $N$-qubit computer is given by $2^N$ complex numbers. A single operation means to multiply this vector by a unitary matrix which means doing something between $2^N$ and $4^N$ operations (usually the first one because the operations are &quot;localized&quot; on the qubits, so they're not quite generic matrices of the same size).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end, you measure the qubits as $N$ ordinary bits. That's when the vector gets reduced to $2^{N}-1$ zeros and $1$ entry equal to one. It's not quite clear from your wording whether you realize that this special form only occurs once, at the very end of the calculation - after you do the &quot;measurement&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem in turning this to a practical solution is that $2^N$ operations is a very high number for useful quantum computers that would have at least $N=128$ or much more. A quantum computer would do the operation in one step instead of $10^{38}$ steps. Also, it wouldn't need the classical memory $10^{38+}$ - just 128 spins etc. - to remember all those quantum amplitudes haha.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-06-12T10:55:30.580" CommentCount="1" />
  <row Id="11021" PostTypeId="1" CreationDate="2011-06-12T13:07:49.203" Score="3" ViewCount="276" Body="&lt;p&gt;Sorry if this is a trivial question, but I do not know enough physics to answer this on my own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can there exist a satellite that revolves around a planet so large such that the planet covers the entire sky as seen from at least one point on the satellite?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 1&lt;/strong&gt;&#xA;I am sorry for being unclear earlier. But assume human eyes can't resolve less than $0.02^\circ$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added 2&lt;/strong&gt;&#xA;Assuming both the planet and satellite (ignoring atmosphere to make things simpler) are spheres and since spheres are locally euclidean, for large enough radii a planet could cover the whole sky as it appears to the eye. Consider this picture:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/h5Emi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3991" LastEditorUserId="3991" LastEditorDisplayName="" LastEditDate="2011-06-15T14:40:52.893" LastActivityDate="2011-06-18T13:09:14.787" Title="Can a planet cover the whole sky as seen from its satellite" Tags="&lt;gravity&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="11023" PostTypeId="2" ParentId="11021" CreationDate="2011-06-12T13:24:26.847" Score="4" ViewCount="" Body="&lt;p&gt;No, because, if you assume spherical shapes for both satellite and main body, this would only be (geometrically) possible if they touch at the point where you stand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be possible if you stand in a hole on the satellite, the walls of which obscure part of the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the case of main body and satellite orbiting each other and nearly touching, I think this is theoretically possible (if neither has an atmosphere), but not a stable configuration. I'm no astronomer, though.&lt;/p&gt;&#xA;" OwnerUserId="3951" LastEditorUserId="3951" LastEditorDisplayName="" LastEditDate="2011-06-12T13:30:50.200" LastActivityDate="2011-06-12T13:30:50.200" />
  <row Id="11024" PostTypeId="2" ParentId="11021" CreationDate="2011-06-12T13:32:41.787" Score="7" ViewCount="" Body="&lt;p&gt;If the planet and satellite are perfect spheres, or any convex shape then the answer is NO. If the planet is close to the satellite and you stand on the point nearest to it then you will only be able to see a small band around the horizon, but the whole horizon cannot be blocked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the satellite is not convex and you stand in a hole or crater on the satellite so that your view of the sky is reduced then the planet can fill the rest of the visible sky, so the answer is YES.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the satellite itself is a sphere then the planet cannot obscure the whole horizon unless the satellite somehow passes through a tunnel or at least a gorge in the planet, not very likely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: Since the question has been updated there are a couple of extra points to be made&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the observer on the satellite is limited by the resolution of his vision then a sufficiently big planet could obscure the whole sky with the only band of sky left near the horizon being too small for him to see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, regarding Carl's comment on the effect of GR. If the planet is sufficiently massive, but not the satellite, then light will bend towards the centre of the planet. In this case it is possible for the planet to cover the whole sky in the most extreme circumstances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned in the comments on this answer, an atmosphere on the satellite would bend the light towards the satellite making it harder for a plenet to cover the whole sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the atmosphere is on the planet it could only help if the satellite was inside the atmosphere, in which case the drag would slow it down so that it would not remain in orbit for very long.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastEditorUserId="1341" LastEditorDisplayName="" LastEditDate="2011-06-18T13:09:14.787" LastActivityDate="2011-06-18T13:09:14.787" CommentCount="10" />
  <row Id="11025" PostTypeId="2" ParentId="11017" CreationDate="2011-06-12T15:05:29.493" Score="7" ViewCount="" Body="&lt;p&gt;Yes, this will essentially work, although there are some pitfalls in the way you've phrased things. For one, you cannot store each qubit as a single bit, but you can work things out so that any computation you do starts from a state which can be represented by $N$ classical bits, i.e. the zero state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following this your second and third bullet points are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the fourth, element-wise squaring implies that your measurement is in the computational basis. If you need to measure in, say, the Fourier basis, you will need to convert to a computational basis measurement by including a Fourier transformation in step 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bullet 5, like bullet 1, conflates the classical information encoded onto/extracted from qubits with the qubits themselves. Everything here is fine assuming you mean for step 5 to be the absolute end of the computation, but you cannot, in general, loop from step 5 back to step 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately the reason your simulation works is that any computation on $N$ qubits can be represented as a single ($2^N\times 2^N$) unitary acting on the zero state followed by measurement in the computational basis. You cannot, however, use a simulation such as yours to do gate-by-gate simulation of a quantum algorithm.&lt;/p&gt;&#xA;" OwnerUserId="3632" LastActivityDate="2011-06-12T15:05:29.493" CommentCount="3" />
  <row Id="11026" PostTypeId="2" ParentId="9772" CreationDate="2011-06-12T15:18:26.587" Score="1" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;install virtualbox in your mac&lt;/li&gt;&#xA;&lt;li&gt;create a windows image&lt;/li&gt;&#xA;&lt;li&gt;install IRIS on your windows image&lt;/li&gt;&#xA;&lt;li&gt;work as usual&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-06-12T15:18:26.587" CommentCount="1" />
  <row Id="11027" PostTypeId="2" ParentId="10998" CreationDate="2011-06-12T15:22:22.887" Score="0" ViewCount="" Body="&lt;p&gt;You can say a bit more if you for instance assume that the Hilbert space $H$ is a &lt;em&gt;finite dimensional complex&lt;/em&gt; vector space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Spectrum. As Lubos Motl observes, $A^2=1-B^{\dagger}B$ must be Hermitian. In fact, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectrum_of_a_matrix&quot; rel=&quot;nofollow&quot;&gt;spectrum&lt;/a&gt; $\mathrm{Spec}(A^2)$ of $A^2$ must satisfy &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm{Spec}(A^2) ~\subseteq~ ]-\infty,1], $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and therefore the spectrum $\mathrm{Spec}(A)$ of $A$ must satisfy &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm{Spec}(A) ~\subseteq~ [-1,1] \cup i\mathbb{R}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Lubos Motl explains, the equation $[A,B]=B$ implies one of two alternatives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lambda\in\mathrm{Spec}(A) \qquad \Leftrightarrow \qquad\exists v\in H\backslash\{0\}: (A-\lambda)v=0  $$&#xA;$$\Rightarrow\qquad(A-\lambda-1)Bv=0 \qquad\Rightarrow\qquad \lambda+1\in\mathrm{Spec}(A) \qquad\mathrm{or} \qquad Bv=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second alternative $Bv=0$ implies that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda^2v=A^2v=(1-B^{\dagger}B)v=v \qquad \Rightarrow \qquad \lambda =\pm 1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is very restrictive. We conclude that the spectrum $\mathrm{Spec}(A)$ of $A$ must satisfy &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm{Spec}(A) ~\subseteq~ \{-1,0,1\}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) We can write $A$ on Jordan canonical block form by a choice of basis. Note that the basis is not necessarily orthonormal. The diagonal elements satisfy $A_{ii}\in \{-1,0,1\}$. Now square the Jordan canonical block form to get $A^2$ on triangular form. The diagonal elements satisfy $(A^2)_{ii}\in \{0,1\}$. From the characteristic polynomial for $A^2$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_{A^2}(\lambda)=\det(A^2-\lambda)= \prod_i ((A^2)_{ii}-\lambda),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we deduce that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm{Spec}(A^2) ~\subseteq~ \{0,1\}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Define eigenspaces &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_0 :=\mathrm{ker}(A^2)=\mathrm{ker}(B^{\dagger}B-1),$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_1 :=\mathrm{ker}(A^2-1)=\mathrm{ker}(B^{\dagger}B)=\mathrm{ker}(B),$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;each of which may be zero. (The equality $\mathrm{ker}(B^{\dagger}B)=\mathrm{ker}(B)$ can be deduced from, e.g., &lt;a href=&quot;http://en.wikipedia.org/wiki/Polar_decomposition#Matrix_polar_decomposition&quot; rel=&quot;nofollow&quot;&gt;polar decomposition&lt;/a&gt; of $B$.) Since $A^2$ is Hermitian, we have that $H=H_0\oplus H_1$ is an orthogonal decomposition of $H$. Moreover, the operator $A^2$ is the orthogonal projection on the subspace $H_1$, and $B^{\dagger}B$ is the orthogonal projection on the subspace $H_0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Next, the equation $[A,B]=B$ implies that $A\mathrm{ker}(B) \subseteq\mathrm{ker}(B)$, i.e., that $AH_1 \subseteq H_1$. Furthermore, $A\mathrm{ker}(A^2)\subseteq\mathrm{ker}(A)\subseteq\mathrm{ker}(A^2)$, i.e., that $AH_0 \subseteq H_0$. So $A$ is stable under the decomposition $H=H_0\oplus H_1$. Let us introduce operators &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_1 := A |_{H_{1}} \qquad \mathrm{and} \qquad A_0 := A |_{H_{0}}, $$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;with spectra&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm{Spec}(A_1) ~\subseteq~ \{\pm 1\} \qquad \mathrm{and} \qquad \mathrm{Spec}(A_0) ~\subseteq~ \{0\},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;respectively. It follows from the fact that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (A_1\pm 1)^2 = \underbrace{(A_1^2-1)}_{=0} \pm 2(A_1\pm 1), $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that the series of &lt;a href=&quot;http://planetmath.org/encyclopedia/GeneralizedEigenspace.html&quot; rel=&quot;nofollow&quot;&gt;generalized eigenspaces&lt;/a&gt; are just ordinary eigenspaces&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{ker}(A\pm 1)=\mathrm{ker}(A_1\pm 1)\subseteq\mathrm{ker}(A_1\pm 1)^2\subseteq\mathrm{ker}(A_1\pm 1)^3\subseteq\ldots =\mathrm{ker}(A_1\pm 1). $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the operator $A_1:H_1\to H_1$ is diagonalizable (but not necessarily in an orthonormal basis), &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H_1 = \mathrm{ker}(A+1)+\mathrm{ker}(A-1). $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, it follows from the fact that $A_0^2 = 0$, that the series &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{ker}(A)=\mathrm{ker}(A_0)\subseteq\mathrm{ker}(A_0^2)\subseteq\mathrm{ker}(A_0^3)\subseteq\ldots =\mathrm{ker}(A_0^2)=H_0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;stabilizes after at most two steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Next, the equation $[A,B]=B$ implies after a straightforward calculation that $B(H_0) \subseteq H_1$, and hence $B^2=0$ nilpotent. In fact, a refined argument yields that $B(H_0) \subseteq \mathrm{ker}(A-1)$. Therefore $BA_0=0$, and since $B|_{H_{0}}$ is injective on $H_0$, we finally conclude that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_0 = 0\qquad \Leftrightarrow \qquad H_0=\mathrm{ker}(A).$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the operator $A:H\to H$ is diagonalizable (but not necessarily in an orthonormal basis), &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H = (\mathrm{ker}(A-1)+\mathrm{ker}(A+1))\oplus\mathrm{ker}(A). $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) If we choose ortonormal bases for $\mathrm{ker}(A-1)$ and $\mathrm{ker}(A)$, and then extend with an ortonormal basis for&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H&amp;#39;_{-1}:= \left( \mathrm{ker}(A-1)\oplus\mathrm{ker}(A) \right)^{\perp}, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is not hard to see that the operators $A$ and $B$ are of the block form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A= \left[\begin{array}{ccc}1&amp;amp;T&amp;amp;0 \\ 0&amp;amp;-1&amp;amp;0 \\ 0&amp;amp;0&amp;amp;0  \end{array}\right], \qquad B= \left[\begin{array}{ccc}0&amp;amp;0&amp;amp;U \\ 0&amp;amp;0&amp;amp;0 \\ 0&amp;amp;0&amp;amp;0  \end{array}\right], \qquad U^{\dagger}U=1,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the row and columns are ordered as $\mathrm{ker}(A-1)\oplus H&amp;#39;_{-1}\oplus\mathrm{ker}(A)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7) Two special cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A= \left[\begin{array}{cc}1&amp;amp;T \\ 0&amp;amp;-1 \end{array}\right], \qquad B= \left[\begin{array}{cc}0&amp;amp;0 \\ 0&amp;amp;0 \end{array}\right].$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A= \left[\begin{array}{cc}1&amp;amp;0 \\ 0&amp;amp;0 \end{array}\right], \qquad B= \left[\begin{array}{cc}0&amp;amp;U \\ 0&amp;amp;0 \end{array}\right], \qquad U^{\dagger}U=1.$$&lt;/p&gt;&#xA;" OwnerUserId="2451" LastActivityDate="2011-06-12T15:22:22.887" />
  <row Id="11028" PostTypeId="2" ParentId="10979" CreationDate="2011-06-12T15:24:21.360" Score="0" ViewCount="" Body="&lt;p&gt;the only interesting sublety here that i think its worth mentioning is that, indeed, the state is in superposition of (many) states of the two charges and a carrier, but &lt;em&gt;the carrier momentum 4-vector does not need to be on-shell&lt;/em&gt;. You can say that in general, this is what makes different real field bosons from their virtual exchange carrier counterparts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a quick reminder, being off-shell basically means that $p_{\mu}p^{\mu}$ can be anything, rather than the usual value of zero for photons&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-06-12T15:24:21.360" />
  <row Id="11029" PostTypeId="1" CreationDate="2011-06-12T16:19:20.987" Score="-2" ViewCount="162" Body="&lt;p&gt;I tried to post a good question with images, but I was stopped by the 10 pt rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is now posted at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://d0vscdf.blogspot.com/2011/06/maybe-i-just-dont-know-enough.html&quot; rel=&quot;nofollow&quot;&gt;http://d0vscdf.blogspot.com/2011/06/maybe-i-just-dont-know-enough.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just want to know if anyone can explain what I see.&lt;/p&gt;&#xA;" OwnerUserId="3993" LastActivityDate="2011-06-12T18:59:11.917" ClosedDate="2011-06-12T19:57:08.073" Title="Why I like the DO data" Tags="&lt;experimental-evidence&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="11030" PostTypeId="1" CreationDate="2011-06-12T16:59:19.330" Score="0" ViewCount="36" Body="&lt;p&gt;I have been wondering about how a hard drive actually stores data.I have read articles and they all point to tiny magnetized regions being used to store the actual data.If indeed this is how data gets stored,what is this material in specific?.&lt;/p&gt;&#xA;" OwnerUserId="3994" LastActivityDate="2011-06-12T16:59:19.330" ClosedDate="2011-06-12T17:05:49.200" Title="Data storage question" Tags="&lt;material-science&gt;" CommentCount="2" />
  <row Id="11031" PostTypeId="1" AcceptedAnswerId="11033" CreationDate="2011-06-12T18:11:40.143" Score="5" ViewCount="131" Body="&lt;p&gt;If I'm correct a quark can change color by emitting a gluon. For example a blue up quark $u_b$ can change into a red up quark by emitting a gluon:&#xA;$$u_b \longrightarrow u_r + g_{b\overline{r}}$$&#xA;(Here, the subscript indicates color and $\overline{r}$ means anti-red). This is needed to keep the color balance (left hand side: $b$, right hand side $r+b+\overline{r}=b$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is then, do color-neutral gluons exist? Eg a gluon that is blue-anti-blue?&#xA;If it would, it could be created by any quark then:&#xA;$$u_b \longrightarrow u_b + g_{b\overline{b}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm learning about the Standard Model in school, but the text isn't always that clear. &lt;/p&gt;&#xA;" OwnerUserId="3515" LastActivityDate="2011-06-12T18:39:26.323" Title="Do color-neutral gluons exist?" Tags="&lt;standard-model&gt;&lt;weak-force&gt;&lt;quark-model&gt;" AnswerCount="1" />
  <row Id="11032" PostTypeId="2" ParentId="11007" CreationDate="2011-06-12T18:26:59.370" Score="0" ViewCount="" Body="&lt;p&gt;More simply, I think you just need to understand what you were meant by &quot;the effects of gravity broken by jumping&quot;. By &lt;strong&gt;jumping&lt;/strong&gt;, that is, using your muscles (which work thanks to electric forces), you can compete with the gravitational effects of the entire Earth. But you are not going to escape from them ; &lt;strong&gt;gravity is still acting on you&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gravity is actually the only force that non-scientific people think of; they are rarely conscious that electric forces are either existent and much more strong that gravity, and for this reason, they need to be explained why it can often be neglected.&lt;/p&gt;&#xA;" OwnerUserId="2499" LastActivityDate="2011-06-12T18:26:59.370" />
  <row Id="11033" PostTypeId="2" ParentId="11031" CreationDate="2011-06-12T18:39:26.323" Score="5" ViewCount="" Body="&lt;p&gt;Color-neutral gluons that have the component blue-antiblue do exist, much like red-antired and green-antigreen. However, the sum of these three possible kinds of gluons is unphysical, so there are only two &quot;diagonal&quot; types of gluons. None of these two types of gluons are &quot;genuinely color-blind&quot; or &quot;completely color-neutral&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is more manifest if you realize that the color dependence of the gluon field may be written as a traceless $3\times 3$ Hermitian matrix. It is traceless because the gauge group is $SU(3)$ rather than $U(3)$ whose dimension is 8 rather than 9. (There are 3 complex entries strictly above the diagonal, which are copied in the complex conjugate way beneath the diagonal, plus 2 or 3 real entries on the diagonal, depending on whether we require the trace to vanish.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Completely color-neutral gluons, if they were added, would be proportional to the identity matrix and they would couple to all three colors of the quarks equally. In other words, the interactions mediated by such gluons would only depend on the baryon number of the quarks. Experimentally, this interaction doesn't exist. In beyond-the-Standard-Model physics, one may try to extend $SU(3)$ to $U(3)$ in this way (this is very common in braneworld models) but because no new baryon-charge long-distance interaction is seen, the $U(1)$ in the $U(3)$ has to be spontaneously broken at a pretty high energy scale.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-06-12T18:39:26.323" CommentCount="2" />
  <row Id="11034" PostTypeId="2" ParentId="11014" CreationDate="2011-06-12T18:55:28.273" Score="5" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;: As Zassounotsukushi correctly points out in the comments, my original answer was wrong. I said before that objects move out across our horizon, but they don't. Sorry about that. I hope I've fixed things now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's best to avoid phrases like &quot;the Universe is expanding faster than the speed of light.&quot; In general relativity, notions like distances and speeds of faraway objects become hard to define precisely, with the result that sentences like that have no clear meaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we leave that terminological point aside, your questions are perfectly well-posed and physically meaningful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's true that we can only see out to a finite distance, due to the Universe's finite age, and that this is an explanation of &lt;a href=&quot;http://en.wikipedia.org/wiki/Olbers%27_paradox&quot; rel=&quot;nofollow&quot;&gt;Olbers's paradox&lt;/a&gt;, which is the name for the old puzzle of why the night sky is dark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the Universe expands, more objects pop into view, since that &quot;horizon distance&quot; is continually getting bigger, at least in principle. In fact, though, that's a very small effect and would not lead to the night sky becoming brighter in practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we should note that we wouldn't expect to see stars popping into view as our horizon expands. The reason is that objects right near the edge of the horizon are so far away that we would see them as they were long ago, around the time of the Big Bang. In practice, we can't see all the way back to $t=0$, because the early Universe was opaque, but still, we can see back in time to long before there were discrete objects like stars. When we look back near the horizon, we see a nearly-uniform plasma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there's a much more important point. The further away we try to look, the more redshifted the light from a given object is. Even if we could see an object near our horizon, the radiation from it would be shifted to extremely long wavelengths, which also means that it would carry extremely little energy. In practice, this just means that things near our horizon become unobservably faint. In a practical sense, our ability to see faraway objects actually decreases with time: although in principle our horizon grows, the redshift causes any given object to become unobservably faint much faster than the rate at which new stuff is brought in across the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lawrence Krauss has written a bunch about this stuff. The details are in &lt;a href=&quot;http://arxiv.org/pdf/astro-ph/9902189&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;, and he has a &lt;em&gt;Scientific American&lt;/em&gt; article (&lt;a href=&quot;http://www.scientificamerican.com/article.cfm?id=the-end-of-cosmology&quot; rel=&quot;nofollow&quot;&gt;paywalled&lt;/a&gt;). Dennis Overbye wrote about this stuff in the &lt;a href=&quot;http://www.nytimes.com/2007/06/06/health/06iht-05essa.6017776.html?scp=45&amp;amp;sq=lawrence%20krauss&amp;amp;st=cse&quot; rel=&quot;nofollow&quot;&gt;NY Times&lt;/a&gt; a while back too. (If you read the pop stuff, tread carefully. Some of it seems to be saying the incorrect thing I said before, namely that things that are currently inside our horizon move outside of it. The technical article is correct, but the nontechnical ones can be misleading. That's my excuse for messing things up in my original answer, but it's not a very good excuse, because I should've known better.)&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-06-13T21:11:13.237" LastActivityDate="2011-06-13T21:11:13.237" CommentCount="5" />
  <row Id="11036" PostTypeId="2" ParentId="11029" CreationDate="2011-06-12T18:59:11.917" Score="1" ViewCount="" Body="&lt;p&gt;OK, I have figured out what you are after with the Delta. You attribute significance to the dip in the data just before the peak of the di-boson curve (red), at the bin of 65GeV .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are missing is the errors. If you look at the errors, the dip is less than two sigma from the di-boson curve, so it is not significant and any further analysis is a study of randomness.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-06-12T18:59:11.917" />
  <row Id="11037" PostTypeId="1" AcceptedAnswerId="11038" CreationDate="2011-06-12T19:34:48.880" Score="11" ViewCount="337" Body="&lt;p&gt;Under Newtonian model of gravity, a perfect sphere creates the same gravitation field as a  point mass in its center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General Relativity describes gravitation differently. How much this difference affect the above equivalence? If it does not hold, what kind of difference (qualitatively) there is?&lt;/p&gt;&#xA;" OwnerUserId="1206" LastEditorUserId="1206" LastEditorDisplayName="" LastEditDate="2011-06-12T22:39:01.550" LastActivityDate="2011-06-12T22:39:01.550" Title="Does gravitation of a sphere equal gravitation of a point?" Tags="&lt;general-relativity&gt;&lt;gravity&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11038" PostTypeId="2" ParentId="11037" CreationDate="2011-06-12T20:23:57.960" Score="10" ViewCount="" Body="&lt;p&gt;Actually, the same result is true in general relativity: any spherically symmetric mass distribution gravitates in exactly the same way as a point mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the more precise statement: any spherically symmetric spacetime, in which all of the matter is concentrated within some radius (i.e., in which the stress-energy tensor vanishes outside of a certain radius), has the same geometry, outside of that radius,as the Schwarzschild metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This result is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Birkhoff%27s_theorem_%28relativity%29&quot; rel=&quot;nofollow&quot;&gt;Birkhoff's theorem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-06-12T20:23:57.960" CommentCount="2" />
  <row Id="11039" PostTypeId="1" AcceptedAnswerId="11045" CreationDate="2011-06-12T20:54:53.810" Score="6" ViewCount="161" Body="&lt;p&gt;The quark composition of the neutral pion ($\pi^0$) is $\frac{u\bar{u} - d\bar{d}}{\sqrt{2}}$. What does this actually mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's bizarre that a particle doesn't have a definite composition. There's a difference of 2 MeV between the quark masses and I don't understand how this can be ignored. If I were to somehow manage to make a bound state of an up and an anti-up quark, what would it be? Would it be a variation on the neutral pion or would it somehow transform into the mixture?&lt;/p&gt;&#xA;" OwnerUserId="3234" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-12T22:47:39.523" LastActivityDate="2011-08-14T19:33:36.820" Title="What does it mean that the neutral pion is a mixture of quarks?" Tags="&lt;quantum-mechanics&gt;&lt;particle-physics&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="11040" PostTypeId="2" ParentId="11017" CreationDate="2011-06-12T23:08:50.833" Score="3" ViewCount="" Body="&lt;p&gt;No, this won't work. The last step of picking an outcome by squaring the elements of the result and resetting the qubits are set according to this outcome is equivalent to measuring the state of your quantum compute. If you measure the state of your quantum computer after every step, this decoheres any coherent superposition. You will get a computation which can easily be simulated on a classical computer, and which you won't be able to use for any interesting quantum computations. Pretty much any quantum algorithm that takes more than one step will give an example of a case where this won't give the right answer. &lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2011-06-12T23:08:50.833" CommentCount="3" />
  <row Id="11041" PostTypeId="1" CreationDate="2011-06-13T00:13:42.950" Score="2" ViewCount="89" Body="&lt;p&gt;This is more of a theoretical thought-experiment question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, how much geothermal energy can we extract before the loss of the magnetic field makes it a terribly bad idea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the geothermal energy inside the earth now, which is lost through &quot;natural processes&quot;, last until the end of the Sun (say, another 5billion years?)  And if so, how much excess is leftover?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) do we know how much thermal energy there is?  Call this: A&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) do we know at what point the core would stop producing a magnetic field? (ie: is it when the iron in the core is no longer molten?)  Call this: B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) do we know the rate that heat is radiated out naturally?  Call the annual rate: C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I guess:   A-B - (5.0 * 10^9 * C) is the amount of heat for &quot;safe&quot; geothermal extraction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also: Considering such extreme timescales -- are there any other factors adding heat into the Earth?  Maybe tidal interactions with Sun/Moon/etc add anything over billions of years?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3902" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-13T10:09:27.850" LastActivityDate="2011-07-13T10:09:27.850" Title="Limit on geothermal energy that could be extracted before the earth's magnetic field collapsed?" Tags="&lt;energy&gt;&lt;heat&gt;&lt;renewable-energy&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11042" PostTypeId="1" AcceptedAnswerId="11101" CreationDate="2011-06-13T00:27:46.723" Score="6" ViewCount="198" Body="&lt;p&gt;I have been taught in school that atoms cannot have more than 8 electrons in the outer shell. Palladium atom's electron configuration is 2,8,18,18. Why isn't it 2,8,18,17,1 like the case of Platinum 2,8,18,32,17,1 or Nickel 2,8,17,1 and they are all in the same group?&lt;/p&gt;&#xA;" OwnerUserId="3241" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-06-14T17:24:08.607" LastActivityDate="2011-06-14T23:08:26.417" Title="Is Palladium an exception?" Tags="&lt;atoms&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="11043" PostTypeId="2" ParentId="11041" CreationDate="2011-06-13T01:12:39.457" Score="4" ViewCount="" Body="&lt;p&gt;The amount of energy we can capture as geothermal is only a very tiny fraction of the amount that leaks naturally through the earth's crust. In addition, the distance from the earth's surface to the generator is so very long that the time constant for heat movement at one end to cause heat change at the other is very long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get an idea of the calculation for this last point, note that we can measure the &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0921818101000911&quot; rel=&quot;nofollow&quot;&gt;climate for going back 80,000 years&lt;/a&gt; by looking at the temperatures in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kola_Superdeep_Borehole&quot; rel=&quot;nofollow&quot;&gt;Ural superdeep borehole which goes down 12km&lt;/a&gt;. When digging such a hole, the temperatures at the top of the hole is the current temperature. Deeper temperatures correspond to averages of temperatures over longer periods of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature T at depth z at time t &lt;a href=&quot;http://www.kilty.com/pdfs/t-d.pdf&quot; rel=&quot;nofollow&quot;&gt;is given by&lt;/a&gt;:&#xA;$$T(z,t) = \int_{-\infty}^{t}\frac{z\;\exp(-z^2/4\alpha(t-\tau))\;f(\tau)\;d\tau}{\sqrt{4\pi\alpha(t-\tau)^3}}$$&#xA;where $f(\tau)$ is the surface temperature and $\alpha$ is the thermal diffusivity. From this, we see that the temperature change at the surface is exponentially damped at a depth $z$ by $\sqrt{t}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the borehole paper above, we have that it takes about 80,000 years to change the temperature of 12km of rock through conduction. The distance to the earth's center is around 6400 km, maybe 6000 km to the magnetic core. So the earliest time we can expect to see a (just barely measurable) change to the earth's magnetic core is around&#xA;$$80,000\;\textrm{years}\;\times \frac{6000^2}{12^2} = 20\; \textrm{billion years}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In actual fact, the earth's core will cool down a lot sooner than this. The reason for the discrepancy is that I've ignored convection. But in any case, there's nothing to worry about.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-13T01:12:39.457" CommentCount="2" />
  <row Id="11044" PostTypeId="2" ParentId="11041" CreationDate="2011-06-13T02:19:26.883" Score="2" ViewCount="" Body="&lt;p&gt;Pretty much an echo of Carl. We can really only mine heat from the top few kilometers. The impedence from the deep earth will be dominated by the vast distances to the deep. There really is a huge amount of heat down there, the cores temperature is similar to the surface tempoerature of the sun. Also we still have radioactive decay going on. I have read, that a limit that will hit earlier (with or without geothermal heat mining) is that mantle convection will decrease and no longer support plate tectonics. With plate tectonics, the recycling of important chemical elements stops, and import ones (for life) would get trapped in sediment. But that is a few billion years off. Solar evolution (its growing more luminous at about 10% per billion years at the moment) will cause a runaway greenhouse long before then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't say geothermal extract can't exceed the current rate of geothermal flux. It is heat mining, i.e. using up heat that took millions of years to accumulate in say a hundred years (at a given locality). It really should be considered more as an energy store, then as a renewable source, as any aggressive rate of exploitation would greatly exceed the rate of recharge.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-06-13T02:19:26.883" />
  <row Id="11045" PostTypeId="2" ParentId="11039" CreationDate="2011-06-13T03:00:35.717" Score="6" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I think it's bizarre that a particle doesn't have a definite composition.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yeah, it is. As qftme said, that's quantum mechanics for you. It really doesn't make sense until you immerse yourself in the subject for long enough (and even then, only somewhat). But it does appear to be the way the universe works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, just so everyone is on the same page, let me start from the basics. If you're familiar with linear algebra, you know that a vector in a 2-dimensional vector space, for example, can be written as a linear combination $\alpha|0\rangle + \beta|1\rangle$ of two basis elements $|0\rangle$ and $|1\rangle$. For example, a direction vector of length 1 that points northeast can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{|\text{north}\rangle + |\text{east}\rangle}{\sqrt{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or it could be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\text{northeast}\rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha|\text{north-northeast}\rangle + \beta|\text{east-southeast}\rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;etc. You could figure out what the coefficients $\alpha$ and $\beta$ are in that last case, but it doesn't matter. The point is, there are an infinite number of ways to decompose any vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pion state is an example of such a vector. It's often considered to be a member of a three-dimensional vector space. One possible basis for that vector space is $u\bar{u}$, $d\bar{d}$, and $s\bar{s}$. But another possible basis is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi^0 = \frac{u\bar{u} - d\bar{d}}{\sqrt{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta = \frac{u\bar{u} + d\bar{d} - 2s\bar{s}}{\sqrt{6}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta&amp;#39; = \frac{u\bar{u} + d\bar{d} + s\bar{s}}{\sqrt{3}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This basis is useful because these particular combinations happen to be relatively stable; in other words, when a particle consisting of any combination of $u\bar{u}$, $d\bar{d}$, and $s\bar{s}$ is detected in a cloud chamber (if you're old-school) or a calorimeter or something like that, it will behave like one of these three particles. It's possible that what was actually emitted was the quantum state $u\bar{u}$, but in terms of the &quot;stable&quot; states, that is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u\bar{u} = \frac{1}{\sqrt{2}}\pi^0 + \frac{1}{\sqrt{6}}\eta + \frac{1}{\sqrt{3}}\eta&amp;#39;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(hopefully I did the math right). So you would have a probability of $\frac{1}{2}$ that it acts like (or technically, collapses to) a pion, $\frac{1}{6}$ that it collapses to an eta meson, and $\frac{1}{3}$ that it collapses to an eta prime meson. One of those three possibilities is what you'd actually observe in your detector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do this the other way around, too: suppose that instead of $u\bar{u}$, you started with a pion, and instead of measuring the &quot;stable&quot; meson type, you were able to directly measure the quark content. Since the pion state contains equal components of $u\bar{u}$ and $d\bar{d}$, your hypothetical quark flavor measurement would give you one of those outcomes with 50% probability each: half the time you'd find that you had an up quark and an anti-up quark, and the other half of the time you'd find a down and anti-down quark. That's what the state $\frac{u\bar{u} - d\bar{d}}{\sqrt{2}}$ actually means: it governs the probabilities that the pion will interact with a quark flavor measurement as each particular quark type.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-13T03:00:35.717" CommentCount="12" />
  <row Id="11047" PostTypeId="2" ParentId="10933" CreationDate="2011-06-13T04:12:26.190" Score="2" ViewCount="" Body="&lt;p&gt;This was going to be a comment on Luboš Motl's answer, but it would be more appropriate as a full answer now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His answer says:  Laws of physics can be written more simply for the solar system's center of mass (barycenter) than for a point on Earth (geocentric).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just one thing!  One mustn't neglect the non-idealities of the barycenter itself, which has a location in the Milky Way that biases it gravitationally at least.  On the surface this is splitting hairs, but the greater point is that &lt;strong&gt;the idealness of any reference frame is also relative&lt;/strong&gt;, and no &quot;ultimate&quot; frame exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, choosing a point on the skin of an elephant over a geocentric point is sacrificing universality just as much as choosing a geocentric point over the barycenter is.  To a flee however, consideration of physics formulated at a point beyond the surface of the elephant may be just &quot;academic&quot;.  Sound familiar?&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-06-13T04:12:26.190" />
  <row Id="11048" PostTypeId="2" ParentId="11039" CreationDate="2011-06-13T05:10:25.790" Score="3" ViewCount="" Body="&lt;p&gt;David gives a complete answer on the mechanism. I will tackle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There's a difference of 2 MeV between the quark masses and I don't understand how this can be ignored. If I were to somehow manage to make a bound state of an up and an anti-up quark, what would it be? Would it be a variation on the neutral pion or would it somehow transform into the mixture?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You seem to imply that assuming the probabilistic interpretation, if the two photons coming from the pi0 decay are measured, half of the time their effective mass should be smaller because the u quark has a smaller mass than the d quark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Within a bound state particles are virtual. Virtual means that their mass is not constrained to be the invariant mass they would have as a free particle. Think of the nucleons in the nucleus, the proton and the neutron in deuterium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of a proton is 938.272013(23) MeV/c&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;          neutron 939.565378(21) MeV/c &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The mass of deuteron       1875.612 859&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the simple addition gives    1877.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the difference is called binding energy, but the point is that neither the proton nor the neutron can have their invariant mass within the bound nucleus, they have a virtual mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) even though mass and energy are connected through E=mc**2, mass is not a conserved quantity in special relativity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going back to vector addition in three dimensions: when operating  a vector addition, the lengths are not conserved. Two vectors may add up to a zero length vector if they are in opposite direction  and of the same magnitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass is the equivalent measure in special relativity four vectors, it is the &quot;length&quot; of the four vector and follows vector algebra. It is not conserved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument: since d quark has a larger invariant mass than a u quark the combination of down  anti down must  have a larger invariant mass than the up antiup one is wrong. The four vector algebra comes out that both have the mass of the pi0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) As the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quark&quot; rel=&quot;nofollow&quot;&gt;quarks&lt;/a&gt; are always bound in hadrons, there are two definitions of a quark's mass, the &lt;em&gt;current mass&lt;/em&gt;, the one entering the QCD equations which are the ones you quote,  and the &lt;em&gt;constituent mass&lt;/em&gt;. That last is the mass with the accompanying gluons within the hadron and is the same for up and down. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-06-13T07:31:55.520" LastActivityDate="2011-06-13T07:31:55.520" />
  <row Id="11049" PostTypeId="1" CreationDate="2011-06-13T05:26:14.623" Score="3" ViewCount="103" Body="&lt;p&gt;Gell-Mann and Hartle came up with the &lt;a href=&quot;http://arxiv.org/abs/1106.0767&quot; rel=&quot;nofollow&quot;&gt;extended probability ensemble interpretation&lt;/a&gt; over &lt;a href=&quot;http://arxiv.org/abs/0801.0688&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, they extended probability theory to include real numbers which may be negative or greater than 1. This necessitates an unconventional interpretation of probabilities, and for that, they used the &quot;no Dutch book&quot; interpretation of probability theory. For the &quot;no Dutch book&quot; interpretation to even make any sense, they have to introduce cybernetic agents, which they call Information Gathering and Utilizing Systems (IGUS). These agents can make bets, and a Dutch book against them is a collection of bets made with the agent such that individually, the agent would prefer to make the bet, but collectively, the agent will lose a fixed amount &lt;em&gt;no matter what the outcome of the bet is&lt;/em&gt;. The condition that no Dutch book can be made against an agent is equivalent to the agent's belief system satisfying the axioms of classical probability theory. An agent with negative probabilities or a probability greater than one is susceptible to Dutch books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only bets on outcomes which can be recorded can be settled, but such outcomes aren't susceptible to Dutch books, or so they claim. Otherwise, don't bet. This already brings up a whole host of issues. Do the bets really have to happen, or is it enough to merely imagine them? Or is it enough that the agent is prepared to bet should the need arise? Or does the bet happen at the meta level? Is the agent free to choose whether or not to bet?&#xA;Besides, an agent which always loses isn't necessarily an inconsistent agent, merely one which fails to satisfies its goals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An agent can have a belief system not susceptible to Dutch books but is still lousy anyway. Another more knowledgeable agent can exploit their better knowledge to offer a series of bets, which while not guaranteeing a profit all the time, is likely to bring a profit most of the time, and only a loss rarely (whatever &quot;most of the time&quot; and &quot;rarely&quot; really mean).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors state that the preferred basis is the basis of the Feynman path integral, but what about S-dualities? Even without S-dualities, a quantum field theory can be described equivalently either as a path integral over field configurations, or a path integral over Feynman diagrams with edges and vertices. Which basis is preferred?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fine-grained probability distribution can have negative probabilities, but there are coarse-grainings with only non-negative probabilities, and at least one such coarse-graining always exists; the coarse-graining with no resolution. A maximal coarse-graining is one associated with non-negative marginal probabilities for which any extension of it will contain negative probabilities. Such coarse-grainings are referred to as realms. Unfortunately, mutually incompatible realms are the norm, and not the exception.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No problem, they write. There &quot;really&quot; is a &quot;real&quot; fine-grained history, and whichever realm we choose, the &quot;real&quot; history is the one containing the &quot;real&quot; fine-grained history, which is &lt;em&gt;unobservable&lt;/em&gt;. We may choose incompatible realms if we so wish, and whichever one we choose, there is a corresponding &quot;real&quot; history. But what is the probability distribution for the &quot;real&quot; fine-grained history? It can be &lt;strong&gt;negative&lt;/strong&gt; or greater than 1! How can something with &lt;em&gt;negative&lt;/em&gt; probability happen? So just how real is it anyway?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As they so clearly pointed out, the past history is unobservable. Only present records are observable. But there's always the possibility that the present record is misleadingly deceptive about the past history, assuming an actual past history even exists. When should we choose to measure the records? Now? When is &quot;now&quot;? Why isn't now one second ago or a year from now?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The records are only approximately recorded and approximately decoherent anyway. So, we can in principle make a Dutch book with exponentially small payoff against such agents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Realms which are &quot;typical&quot; are preferred over realms which are not. Typicality means the log of the coarse-grained probability for the &quot;real&quot; coarse-grained history is much less than the Shannon entropy. But why is typicality preferred?&lt;/p&gt;&#xA;" OwnerUserId="4003" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-06-13T12:36:03.500" LastActivityDate="2011-06-13T12:36:03.500" Title="Does the extended probability ensemble interpretation of quantum mechanics make any sense?" Tags="&lt;quantum-interpretations&gt;" CommentCount="1" FavoriteCount="2" />
  <row Id="11050" PostTypeId="1" AcceptedAnswerId="11051" CreationDate="2011-06-13T08:45:58.230" Score="3" ViewCount="170" Body="&lt;p&gt;I'm studying vector analysis and it is hard for me to understand what divergence of a vector field really is. I know that $divF=\nabla\cdot F$ but I don't understand what kind of quantity it gives and what it is used for.&#xA;Could anyone give me a good example of what it is used for (in real physics)? &lt;/p&gt;&#xA;" OwnerUserId="4004" LastActivityDate="2011-06-13T10:04:49.783" Title="Concrete example of divergence of a vector field" Tags="&lt;vector-fields&gt;" AnswerCount="3" />
  <row Id="11051" PostTypeId="2" ParentId="11050" CreationDate="2011-06-13T08:56:11.720" Score="6" ViewCount="" Body="&lt;p&gt;It is just &#xA;$$\partial F_x/\partial x + \partial F_y/\partial y + \partial F_z/\partial z $$&#xA;and measures whether the field is a source or sink at a given place. A basic introduction is here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Divergence&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Divergence&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the most important relationship that gives the divergence an &quot;intuitively comprehensible&quot; meaning is Gauss' theorem&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Gauss_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The flux of the field $F$ over the surface $S$ of a small volume - the integral $\oint F\cdot dS$ where $dS$ is a vector normal to the infinitesimal area $dS$, with the same magnitude - may be written as ${\rm div} F$ times $dV$. The latter is the volume of the interior bounded by the area $S$. If the volume is non-infinitesimal, one has to replace ${\rm div}(F)dV$ by the integral of this.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-06-13T08:56:11.720" CommentCount="4" />
  <row Id="11052" PostTypeId="2" ParentId="11050" CreationDate="2011-06-13T09:30:53.573" Score="4" ViewCount="" Body="&lt;p&gt;The most obvious example from physics is the Maxwell equation&#xA;$$&#xD;&#xA;  \nabla\cdot \mathbf{E} = 4\pi\,\rho&#xD;&#xA;$$&#xA;which simply states that the electric field $\mathbf{E}$ &quot;comes out of&quot; any charged particle (where there is a finite charge density $\rho$), and does not have any source at places where $\rho$ is zero.&lt;/p&gt;&#xA;" OwnerUserId="3540" LastActivityDate="2011-06-13T09:30:53.573" />
  <row Id="11053" PostTypeId="2" ParentId="11050" CreationDate="2011-06-13T10:04:49.783" Score="-4" ViewCount="" Body="&lt;p&gt;I learnt to understand it as a measure of ' the rate of change of magnitude of a vector field.'  So that if a field 'has' divergence, $Div F$ is non zero, then the magnitude of vectors within some region will be changing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A physical example I encountered is a can of compressed air that has been recently opened (alright this is a little artificail but it serves a purpose!).  Just after opening the air near the back of the can can't escape as there is air in the way and so the velocity is low in that region.  Near the opening the air can readily escpe and so the velocity in that region is fast.  Thus measuring the divergence from the back to the front gives a non zero value and hence we have 'divergence'.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-06-13T10:04:49.783" CommentCount="6" />
  <row Id="11054" PostTypeId="1" AcceptedAnswerId="11113" CreationDate="2011-06-13T11:12:59.123" Score="8" ViewCount="241" Body="&lt;p&gt;Assuming only Newtonian gravity, suppose that the universe consists of an infinite number of uniform planets, uniformly distributed in a two-dimensional grid infinite in both directions and not moving relative to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any reason to believe that this is not an equilibrium state? A friend of mine who knows much more applied maths than me assures me that this system is not in equilibrium, but I'm unable to find any evidence for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to be able to say that the net force on each planet in the grid is zero. We approach this by summing the forces exerted on a planet &lt;em&gt;P&lt;/em&gt; by each other planet over the whole space, but note that we can pair off the other planets that exert equal and opposite forces on &lt;em&gt;P&lt;/em&gt; to show that the total sums to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I going mad?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For bonus points, presumably the same result holds for any number of dimensions to the grid.&lt;/p&gt;&#xA;" OwnerUserId="4006" LastActivityDate="2011-06-15T13:39:14.447" Title="infinite grid of planets with newtonian gravity" Tags="&lt;gravity&gt;&lt;newtonian-mechanics&gt;&lt;mathematics&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="11055" PostTypeId="2" ParentId="11000" CreationDate="2011-06-13T11:15:25.237" Score="4" ViewCount="" Body="&lt;p&gt;This is an edited answer following the remark by Qmechanic pointing out that the capacitance used in the original answer assumes electrical contact between the spheres.&#xA;Here I assume that the actual contact time is very small, so no discharge occurs. On, the other hand it is assumed that the motion is slow enough that the laws of electrostatics apply. The capacitance formulas are taken from the &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0304388610001312&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;: &quot;Capacitance coefficients of two spheres&quot; by John Lekner. The same notation is used, where the charged sphere is denoted with the subscript $a$ and the other sphere with the subscript $b$. Also CGS units are used (The capacitance has units of length).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting $ Q_a = Q$, $Q_b = 0$, we obtain the following formula for the effective capacitance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ C_s \equiv \frac{V_a}{Q} = \frac{C_{aa}C_{bb}-C_{ab}^2}{C_{bb}},$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $s$ is the separation distance. The expressions for the capacitance matrix elements at infinite separations $s=\infty$ are given in the text eq. (8) resulting the following result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ C_{\infty} = R.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is just the capacitance of the charged sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The capacitance matrix elements become singular when the spheres approach contact $s\to 0$, but using the asymptotic formulas (16)-(18), the effective capacitance has a well defined limit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_0 = \lim_{u \to 0}\frac{R}{2} \frac{(\ln(\frac{2}{u})-\psi(\frac{1}{2}))^2-(\ln(\frac{2}{u})+\gamma)^2}{(\ln(\frac{2}{u})-\psi(\frac{1}{2}))}= -R (\psi(\frac{1}{2})+\gamma) = 2 R \ln 2,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\gamma$ is the Euler constant and $\psi$ is the digamma function, and the following identity was used&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\gamma +\psi(\frac{1}{2}) = -2 \ln 2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By energy conservation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2 \frac{m v^2}{2} + \frac{Q^2}{C_0}=  \frac{Q^2}{C_{\infty}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = \sqrt{\frac{Q^2 (2\ln 2-1)}{2m R \ln 2}},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the speed just after collision in the case of an elastic collision. Of course, the speed is zero in the case (b) of inelastic collision.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-14T20:25:25.153" LastActivityDate="2011-06-14T20:25:25.153" CommentCount="14" />
  <row Id="11056" PostTypeId="2" ParentId="11054" CreationDate="2011-06-13T11:28:32.520" Score="6" ViewCount="" Body="&lt;p&gt;This is a very subtle limit of general relativity in which the evaluation of indefinite forms becomes important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, note that because of the uniform matter density, the gravitational potential won't be well-behaved. It would have to satisfy &#xA;$$\Delta \Phi \sim \rho_{\rm planets} $$&#xA;but because the right hand side is uniform (at distances longer than the grid spacing), you would need something like&#xA;$$\Delta = C\vec x\cdot \vec x$$&#xA;That's bad because the gravitational potential clearly goes to infinity at infinity and there's only one place at which the gravitational potential is stationary. All other planets would feel force. However, you may also ignore the gravitational potential and consider its gradient only. Equivalently, you may add a Newtonian &quot;cosmological constant term&quot; by&#xA;$$\Delta \Phi \sim \rho_{\rm planets} -\rho_0 $$&#xA;Here, the cosmological constant term $\rho_0$ may be chosen to cancel the average value of $rho_{\rm planets}$. This extra addition of $\rho_0$ wouldn't affect the accelerations and forces because it's completely uniform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you insist on the standard formulae for the gravitational potential and you ban any kind of cosmological constant, this setup can't be in equilibrium. However, if you allow a constant vacuum energy to be added and to subtract the average density coming from the planetary grid, you may get a stationary state similar to a non-relativistic version of the Einstein static Universe that Einstein wanted to believe in general relativity (which was proved wrong when the expansion of the Universe was observed by Hubble).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like the Einstein static universe, this uniform configuration of matter is unstable. A small deviation will start to destroy the grid and make the planets clump, converting a part of their potential energy to kinetic energy and chaotic motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, if you didn't want to use the gravitational potential at all, you could just compute the accelerations. Your reasoning - ending up with an &quot;equilibrium&quot; answer - is based on symmetry. The acceleration from the other planets is proportional to the integral&#xA;$$ \int   \frac{\vec r}{r^3} r^2 dr $$&#xA;written in the spherical coordinates. By symmetry, you could argue that it vanishes as the integrand is an odd function. However, that's only true around an origin that has to be chosen differently for each planet-probe. If you shift the center of the spherical coordinates, you will inevitably get a different result. This problem is mathematically isomorphic to the &quot;anomalies&quot; that arise from linear divergences in quantum field theory. In some sense, the theory is ill-defined with your planetary grid.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-06-13T11:28:32.520" CommentCount="4" />
  <row Id="11057" PostTypeId="1" AcceptedAnswerId="11060" CreationDate="2011-06-13T13:48:42.150" Score="0" ViewCount="99" Body="&lt;p&gt;Imagine the folllowing situation: A coherent light source is attached to a car such that the emitted light beam path is &quot;being crossed over&quot; by the car i.e. the long parallel light beams are struck by the moving car such that their path wrt exach other is a 'X'. Say the car is going at a speed of 3000m//s and the coherent light source is attached to the car. do the light particles experience doppler shift? do they photon get absored by th car body?  does the path of the photons change in any velcoity due to a gain in momentum? what about the angle of being hit - does it influence any of the above properties of light particles?&lt;/p&gt;&#xA;" OwnerUserId="3604" LastEditorUserId="3604" LastEditorDisplayName="" LastEditDate="2011-06-13T14:19:08.030" LastActivityDate="2011-06-13T15:32:28.147" Title="What happens when light moves perpendicular to a moving object?" Tags="&lt;particle-physics&gt;&lt;speed-of-light&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11058" PostTypeId="1" CreationDate="2011-06-13T15:16:08.323" Score="1" ViewCount="100" Body="&lt;p&gt;If nuclear decay rate were affected by sun, &#xA;then emission probabilities would be subject to sun state and its influence,&#xA;so quantum randomness would depend on it,&#xA;Would it still be truly random?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could argue that randomness would keep intact when consider the whole system..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about a non-isolable effect? what if there is no way to &quot;define a system&quot; without a sun/star influence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a variation source were found (for example neutrino emission) then perhaps it could be generated intentionally to control decay rates. Could we be sure that emission probabilities would never could be controled to reach 0 or 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for reading&lt;/p&gt;&#xA;" OwnerUserId="1916" LastActivityDate="2011-06-13T16:15:24.987" Title="Nuclear decay rate affected by sun and quantum randomness" Tags="&lt;quantum-mechanics&gt;&lt;radiation&gt;&lt;radioactivity&gt;&lt;sun&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" />
  <row Id="11059" PostTypeId="2" ParentId="11058" CreationDate="2011-06-13T15:30:04.293" Score="2" ViewCount="" Body="&lt;p&gt;If decay rates are found could be reduced to 0, then we are thoroughly wrong in what we thought about nuclear decay, and that is unlikely to happen.  The term &quot;nuclear decay&quot; would probably be obsolete in that case.  Quantum mechanics, itself, may have answering to do given the strong form of what you are speculating about.  But there is little case to believe that will happen.  Quantum tunneling is extremely unlikely to be disproved or even modified in any meaningful way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a diagram from a &lt;a href=&quot;http://physics.stackexchange.com/questions/10863/tunneling-of-alpha-particles/&quot;&gt;previous question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/2J0TE.gif&quot; alt=&quot;Alpha decay&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gist of nuclear decay is quantum tunneling.  Now, how could neutrinos from the sun affect this?  Well it could be an interaction that is allowable or not allowable under the standard model, although I'm not an expert on that.  Whatever it is, however, it's almost certainly not changing the fundamental nature of the nuclear decay by quantum tunneling, which is a truly quantum random process from our view of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever field or neutrino flux that causes the effect (assuming it's verified) won't be fundamentally changing nuclear decay.  It will just be another interaction in the book of nuclear interactions.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-06-13T15:30:04.293" />
  <row Id="11060" PostTypeId="2" ParentId="11057" CreationDate="2011-06-13T15:32:28.147" Score="3" ViewCount="" Body="&lt;p&gt;I am guessing that the setting you are talking about is the following:&#xA;You have a reference frame where the x axis is horizontal and the y axis is vertical and there is a stream of photons moving along the y axis. Then there is a car moving on the x axis with some velocity u, heading to the stream of photons. And the question is, what will the driver see? What will be the angle of attack of the photons and what will be their energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a standard special relativity exercise. First of all, the Doppler formula doesn't apply in this case. The correct way to handle this problem is to calculate the four-momentum of the photon in the first reference frame and then transform it to the frame of the car. The time-component of the four-momentum will give you the change in energy of the photon and the ratio of the spatial components (x and y) will give you the angle of attack where you will notice that there is an aberration effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, the energy will be increased by a gamma factor ($E'=\gamma E$) and the angle of attack will be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan\theta&amp;#39;=\frac{1}{\gamma \beta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\gamma$ and $\beta$ are the parameters for the frame of the moving car. Of course the speed of the photons is always the speed of light $c$.&lt;/p&gt;&#xA;" OwnerUserId="383" LastActivityDate="2011-06-13T15:32:28.147" />
  <row Id="11061" PostTypeId="1" CreationDate="2011-06-13T16:06:29.367" Score="4" ViewCount="126" Body="&lt;p&gt;In Feynman's simple QED book he talks about the probability amplitude P(A to B) ,where A and B are events in spacetime, and he says that it depends of the spacetime interval but he didn't put the expression. I would like to know what expression was he referring to. &#xA;He said then that the amplitudes for a photon to go faster or slower than c are very smaller than the contribution from speed c. How is that  this contributions cancel out for long distances?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like also that you explain how it is related to the path integral of the form exp(iKL) of the question &lt;a href=&quot;http://physics.stackexchange.com/questions/10967/how-is-the-path-integral-for-light-explained-or-how-does-it-arise&quot;&gt;How is the path integral for light explained, or how does it arise?&lt;/a&gt;&#xA;I think this path integral is very strange because it doesn't have spacetime fixed points but space fixed points. Its paths doesn't have the same time if we assume speed constant so they wont reach the same event.&lt;/p&gt;&#xA;" OwnerUserId="3956" LastActivityDate="2011-09-05T19:55:22.510" Title="how is shown that photon speed is constant using QED?" Tags="&lt;quantum-electrodynamics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11062" PostTypeId="2" ParentId="11058" CreationDate="2011-06-13T16:15:24.987" Score="4" ViewCount="" Body="&lt;p&gt;If you are referring to a &lt;a href=&quot;http://science.slashdot.org/story/10/08/24/0155229/The-Strange-Case-of-Solar-Flares-and-Radioactive-Decay-Rates&quot; rel=&quot;nofollow&quot;&gt;disputed experiment&lt;/a&gt; that showed a correlation with the sun's activity , &lt;a href=&quot;http://arxiv.org/abs/1006.5071&quot; rel=&quot;nofollow&quot;&gt;another lab experiment disproved&lt;/a&gt; any neutrino flux connection with higher neutrino fluxes than the sun provides . Neutrinos are weakly interacting and the sun too far away for them to set up an effective field to change potential barriers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been experiments that find a tiny difference under different temperatures for various decays, and attributed to the differences in the &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v98/i25/e252501&quot; rel=&quot;nofollow&quot;&gt;electron density&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The electron-capture (EC) decay rate of 7Be in C60 at the temperature of liquid helium (T=5  K) was measured and compared with the rate in Be metal at T=293  K. We found that the half-life of 7Be in endohedral C60 (7Be@C60) at a temperature close to T=5  K is 52.47±0.04  d, a value that is 0.34% faster than that at T=293  K. In this environment, the half-life of 7Be is nearly 1.5% faster than that inside Be metal at room temperature (T=293  K). We then interpreted our observations in terms of calculations of the electron density at the 7Be nucleus position inside the C60; further, we estimate theoretically the temperature dependence (at T=0  K and 293 K) of the electron density at the Be nucleus position in the stable center inside C60. The theoretical estimates were almost in agreement with the experimental observations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are also experiments that say that the casimir effect changes decay rates. All this is research area and liable to errors not well controlled,but @Zassounotsukushi is correct in stating that any such observations will just need different theoretical modelling and will not dispute the randomness of the specific new decay mode.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-06-13T16:15:24.987" CommentCount="2" />
  <row Id="11063" PostTypeId="1" CreationDate="2011-06-13T16:39:53.147" Score="3" ViewCount="77" Body="&lt;p&gt;It is known that there is a famous quantum factorization algorithm by Peter Shor. The algorithm is thought to be suitable only for quantum gate computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But can a an adiabatic quantum computer especially that which is capable of quantum annealing be used for factorization?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asking this because it seems that Geordie Rose &lt;a href=&quot;http://dwave.wordpress.com/2011/05/11/learning-to-program-the-d-wave-one/#comment-22008&quot; rel=&quot;nofollow&quot;&gt;claims&lt;/a&gt; in his blog that they have a quantum factorization algorithm that is somehow &quot;better than Shor&quot;. But the details are unavailable as of now.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastActivityDate="2011-06-13T16:39:53.147" Title="Can quantum annealing be used for factorization?" Tags="&lt;quantum-information&gt;&lt;quantum-computation&gt;&lt;quantum-computer&gt;" CommentCount="1" />
  <row Id="11064" PostTypeId="1" AcceptedAnswerId="11125" CreationDate="2011-06-13T18:35:37.140" Score="7" ViewCount="104" Body="&lt;p&gt;I've heard mentioned in various classes that neutron stars, like superconductors, are described by BCS theory. I know that in superconductors a key element in forming cooper pairs is a net attractive force between the electrons which would normally repel one another. That attractive force is accounted for via lattice vibrations (phonons) created and &quot;absorbed&quot; by electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: what provides the attractive force between neutrons? just gravity? If it is true that neutron stars follow BCS theory, by what means was someone able to verify that?&lt;/p&gt;&#xA;" OwnerUserId="3601" LastActivityDate="2011-06-15T13:47:40.857" Title="BCS wave function in Neutron stars" Tags="&lt;cosmology&gt;&lt;condensed-matter&gt;&lt;superconductivity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="11065" PostTypeId="1" AcceptedAnswerId="11069" CreationDate="2011-06-13T19:10:32.913" Score="5" ViewCount="101" Body="&lt;p&gt;There is a &lt;a href=&quot;http://io9.com/5811236/this-is-the-coolest-science-experiment-youll-see-all-week&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt; of an experiment from University of Mexico using corn syrup (highly viscous) and water. They are &quot;mixed together&quot; in a container by turning a crank but when the crank is turned in the opposite direction they neatly un-mix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the two liquids actually do not blend, but how can the syrup droplets be returned to very close to their initial state. It looks as if the entropy is greatly reduced. Is the increase in temperate of the liquids (through the turning of the crank) the answer?&lt;/p&gt;&#xA;" OwnerUserId="852" LastActivityDate="2011-06-14T00:10:42.457" Title="Entropy decrease using Stoke's flow?" Tags="&lt;fluid-dynamics&gt;&lt;couette-flow&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11066" PostTypeId="1" CreationDate="2011-06-13T19:52:08.453" Score="1" ViewCount="76" Body="&lt;p&gt;I was blown away by the abundance of good info here, and I thought maybe you could help me out with this query. I am in the process of developing a product, and I need to find a clear plastic resin (visually clear) that is also transparent to x-rays (medical application). The only only other characteristic that I need is durability (scratch-resitant). Any advice?&lt;/p&gt;&#xA;" OwnerUserId="4008" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-13T20:01:52.937" LastActivityDate="2011-06-13T20:01:52.937" ClosedDate="2011-06-13T19:57:48.410" Title="What properties of certain plastics make them x-ray transparent?" Tags="&lt;electromagnetism&gt;&lt;optics&gt;&lt;material-science&gt;" CommentCount="3" />
  <row Id="11067" PostTypeId="1" AcceptedAnswerId="11071" CreationDate="2011-06-13T21:42:33.097" Score="0" ViewCount="75" Body="&lt;p&gt;In the &lt;a href=&quot;http://www.bipm.org/en/CGPM/db/3/2/&quot; rel=&quot;nofollow&quot;&gt;Resolution 2 of the 3rd meeting of the CGPM&lt;/a&gt;, defining the &lt;a href=&quot;http://www.bipm.org/en/si/si_brochure/chapter2/2-1/kilogram.html&quot; rel=&quot;nofollow&quot;&gt;kilogram&lt;/a&gt;, “the International Service of Weights and Measures” is mentioned (the &lt;a href=&quot;http://www.bipm.org/fr/CGPM/db/3/2/&quot; rel=&quot;nofollow&quot;&gt;French original text&lt;/a&gt; reads “le Service international des Poids et Mesures”). I wonder which organization this should be, and I am unable to find any other mention on Google except quotations of the resolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry if this question is considered off-topic here, I thought Physics would be the best match.)&lt;/p&gt;&#xA;" OwnerUserId="2793" LastActivityDate="2011-06-14T01:03:37.817" Title="What is “International Service of Weights and Measures”?" Tags="&lt;si-units&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11069" PostTypeId="2" ParentId="11065" CreationDate="2011-06-13T23:33:45.173" Score="3" ViewCount="" Body="&lt;p&gt;Entropy never increases during this &quot;mixing process&quot; as it normally would, because it is &lt;em&gt;not&lt;/em&gt; really a mixing process. The only thing that happens is a huge distortion of the distribution of the colors, but no information about it is lost. Compare it to performing a Fourier transform on some function: this also results in something that looks completely different, sometimes chaotic and high-entropy-like, but it's really just an invertible transformation.&lt;/p&gt;&#xA;" OwnerUserId="3540" LastActivityDate="2011-06-13T23:33:45.173" CommentCount="2" />
  <row Id="11070" PostTypeId="2" ParentId="11065" CreationDate="2011-06-14T00:10:42.457" Score="1" ViewCount="" Body="&lt;p&gt;My simple explanation is that the entropy never increased in the first place, and like a good magic trick, they showed you a state where it looked like mixing had occurred, but it in fact had not.  In the false &quot;mixed&quot; state the colors were extremely systematically layered, looping around the apparatus several times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia actually does a far better job than I could with this, but I'll reference their equations anyway.  Here is the typical Navier-Stokes equation beautifully written from &lt;a href=&quot;http://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article&lt;/a&gt;.  You find in the article about &lt;a href=&quot;http://en.wikipedia.org/wiki/Stokes_flow&quot; rel=&quot;nofollow&quot;&gt;Stokes flow&lt;/a&gt; that in the case of a very low Reynolds number, the effective derivative part, the left hand side, is all equal to zero.  Also, I think that $\mathbf{f}=-\rho g$ downward, but I'd like someone to confirm that fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\overbrace{\rho \Big(&#xD;&#xA;\underbrace{\frac{\partial \mathbf{v}}{\partial t}}_{&#xD;&#xA;\begin{smallmatrix}&#xD;&#xA;  \text{Unsteady}\\&#xD;&#xA;  \text{acceleration}&#xD;&#xA;\end{smallmatrix}} +&#xD;&#xA;\underbrace{\mathbf{v} \cdot \nabla \mathbf{v}}_{&#xD;&#xA;\begin{smallmatrix}&#xD;&#xA;  \text{Convective} \\&#xD;&#xA;  \text{acceleration}&#xD;&#xA;\end{smallmatrix}}\Big)}^{\text{Inertia (per volume)}} =&#xD;&#xA;\overbrace{\underbrace{-\nabla p}_{&#xD;&#xA;\begin{smallmatrix}&#xD;&#xA;  \text{Pressure} \\&#xD;&#xA;  \text{gradient}&#xD;&#xA;\end{smallmatrix}} +&#xD;&#xA;\underbrace{\mu \nabla^2 \mathbf{v}}_{\text{Viscosity}}}^{\text{Divergence of stress}} +&#xD;&#xA;\underbrace{\mathbf{f}}_{&#xD;&#xA;\begin{smallmatrix}&#xD;&#xA;  \text{Other} \\&#xD;&#xA;  \text{body} \\&#xD;&#xA;  \text{forces}&#xD;&#xA;\end{smallmatrix}} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to this, the mass balance in an incompressible flow is the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\boldsymbol{\nabla}\cdot\mathbf{v}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution to these equations includes $\mathbf{v}$ (vector velocity) and $p$ (scalar pressure).  The important part to understand is that the time derivative is gone.  In this case, the second you stop putting torque into the machine, the fluid is stationary.  So it's actually entirely the boundary conditions that define the movement of the fluid at any given time.  I think it's fair to them make the following statement about the position and velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^{t_1} \mathbf{v} dt = \mathbf{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the &quot;mapping&quot; of one particle in the position to the position at the end of the experiment.  The entire idea is that they can produce the conditions such that the following is also met.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{t_1}^{t_2} \mathbf{v} dt = -\mathbf{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ordinarily this would be the exact thing prevented by entropy.  Basically, the equation leads to the creation of too much information to be exactly reversed by simple set of boundary conditions.  I don't have any good way to write this, but the boundary conditions are sufficiently reversed in time and the rate at which it's turned also doesn't matter as a result of the form of the equations.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-06-14T00:10:42.457" />
  <row Id="11071" PostTypeId="2" ParentId="11067" CreationDate="2011-06-14T00:53:32.410" Score="3" ViewCount="" Body="&lt;p&gt;It was an early name of what became the trinity of the &lt;a href=&quot;http://www.bipm.org/&quot; rel=&quot;nofollow&quot;&gt;International Bureau of Weights and Measures (BIPM)&lt;/a&gt;, the General Conference on Weights and Measures (CGPM) and the International Committee for Weights and Measures (CIPM).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phrase &quot;international service of weights and measures&quot; or its French equivalent seems to have been used earlier at the CIPM meeting of 1887 related to adoption of the centigrade scale of the hydrogen thermometer.&lt;/p&gt;&#xA;" OwnerUserId="2714" LastActivityDate="2011-06-14T00:53:32.410" CommentCount="2" />
  <row Id="11072" PostTypeId="2" ParentId="11067" CreationDate="2011-06-14T01:03:37.817" Score="4" ViewCount="" Body="&lt;p&gt;CGPM, in French is &quot;de Conference Generale des Poids et Measures&quot;, known as the &quot;General Conference of Weights and Measures&quot; in English.  It was formed in 1875 when a group of industrialized nations signed the &quot;Treaty of the Meter&quot;, after much debate and dicussion in order that standards of measurements, both physical and treaty based, would be on a common basis between the signatory countries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The BIPM, Bureau International des Poids et Measures (International Bureau of Weights and Measures, located in Serves, France, is the scientific branch of a three part organisation that includes the CGPM and the CIPM. The CGPM holds regular meetings to regulate the measurement side of world commerce through updates to the Treaty as well as activities such as international intercomparisons of physical standards and more mundane things like bugeting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CIPM is the Comite International des Poids et Measures, or International Committee of Weights and Measures made up of 18 members out of body of the CIPM.  The CIPM receives, organises and presents proposed items to the CGPM for a vote by all signatory members.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be of interest that of the seven basic units of measurements, from which all other units may be derived, the kilogram is the last to be based on an artifact rather than a repeatable natural phenomena.  The artifact is called &quot;Le Grande K&quot; and is kept under tightly controlled conditions in a bomb proof safe in Serves.  Experiments are underway to rectify this situation using Avagadro's Number or the watt balance (a variation of the balance used to determine the ampere).  An uncertainty of less than 1 part in 10 billion must be realized to replace the comparison methods that have an uncertainy at least equal to that or less if done in vacuo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The United States has been a member since the inception of the Treaty of the Meter and responsibility for measurements reside within the National Institute of Standards and Technology (NIST).  In particular, the maintenence and dissemination of the mass unit is a function of the Mass and Force Group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much more information can be gleaned from the BIPM.org and the NIST.gov websites.&lt;/p&gt;&#xA;" OwnerUserId="4014" LastActivityDate="2011-06-14T01:03:37.817" CommentCount="1" />
  <row Id="11073" PostTypeId="2" ParentId="353" CreationDate="2011-06-14T01:10:23.823" Score="1" ViewCount="" Body="&lt;p&gt;The short answer is that all other physical units can be derived from these seven.&lt;/p&gt;&#xA;" OwnerUserId="4014" LastActivityDate="2011-06-14T01:10:23.823" CommentCount="2" />
  <row Id="11075" PostTypeId="1" CreationDate="2011-06-14T04:14:28.583" Score="0" ViewCount="124" Body="&lt;p&gt;An electron is shot towards a target that is negatively charged. While the electron is traveling, the target makes an abrupt move towards the electron. While the information that the target moved is traveling from the target to the electron, the electron behaves like an electron that is moving towards a target that is in the original position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can energy be conserved when an electron that is moving towards a nearby charge behaves like it was moving towards a far away charge? Seems we end up with electron being at 2 meters distance from the target, while the electron had enough energy to travel to at most 4 meters distance from the target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also seems to me that &quot;moving the target requires energy&quot; is not a solution to this problem.&lt;/p&gt;&#xA;" OwnerUserId="4017" LastActivityDate="2011-06-14T18:55:21.347" Title="How is energy conserved when a moving charge has false ideas about positions of other charges" Tags="&lt;special-relativity&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="11076" PostTypeId="2" ParentId="11075" CreationDate="2011-06-14T05:06:05.283" Score="4" ViewCount="" Body="&lt;p&gt;Moving the target requires energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the target is an infinite plane with constant charge density $\sigma$.  It will not radiate when you move it because the electric field is constant everywhere.  Suppose the test charge $q$ is small enough that its radiation is negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electric field of the plane is $2\pi\sigma$ in the direction perpendicular to the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The charge begins a distance $d$ from the plane.  The potential energy in the system is $-2\pi\sigma q d$ (define to be zero when $d=0$).  The force on the charge is $2\pi\sigma q$ and by Newton's third law there is an equal and opposite force on the plane.  (We are assuming there is no radiation, so momentum of the charge carriers is conserved, and Newton's third law holds.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The charge also has some kinetic energy $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We move the plane towards the charge a distance $\Delta d$.  This takes energy because the force of the charge on the plane does negative work.  The energy required is the force multiplied by the distance, or $2\pi\sigma q \Delta d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The new distance of the charge from the plane is $d - \Delta d$, so the new potential energy is $-2\pi\sigma q (d-\Delta d)$.  The potential energy has increased by $2\pi\sigma q \Delta d$, exactly the amount of work that had to be put into the system to move the plane.  The charge still has kinetic energy $T$, so energy is conserved in that the change in energy of the system is equal to the energy that was used to move the plane.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-06-14T05:06:05.283" CommentCount="2" />
  <row Id="11077" PostTypeId="1" AcceptedAnswerId="11079" CreationDate="2011-06-14T07:16:05.553" Score="0" ViewCount="145" Body="&lt;p&gt;Every morning, when I ride my bicycle, there is a very loud squeak coming from the brakes. I have an aluminum rim (I think that they are of the sprint type) and rubber pad brakes. A couple of brake events afterward end that and after that the breaks are silent. I suspect it has something to do with the morning plumpness but yet:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why would a damp rim give a noise under friction?&lt;/li&gt;&#xA;&lt;li&gt;Why is it so loud? There is no amplification mechanism I can think of in the braking system.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Some photos:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aBicycle_rim_diagrams_04_.png&quot; rel=&quot;nofollow&quot;&gt;sketch&lt;/a&gt; of what I believe is my rim is.&lt;br&gt;&#xA;A &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/7/78/Roller_Cam_Bicycle_Brake_Front_crop.JPG&quot; rel=&quot;nofollow&quot;&gt;braking mechanism&lt;/a&gt;, Not mine but quite similar. The rim also looks similar.&lt;br&gt;&#xA;A closeup on a &lt;a href=&quot;https://store.nexternal.com/reparto/images/c-br-pad.jpg&quot; rel=&quot;nofollow&quot;&gt;brake pad&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Photos 1 and 2 are taken from wikipedia, photo 3 is taken from a reparto corse&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3677" LastActivityDate="2011-06-14T11:01:36.983" Title="Why does my bicycle brakes squeak in the morning?" Tags="&lt;sound&gt;&lt;bicycles&gt;" AnswerCount="2" />
  <row Id="11078" PostTypeId="1" CreationDate="2011-06-14T09:50:53.870" Score="3" ViewCount="750" Body="&lt;p&gt;According to the Newton mechanics, the force is responsible for changing the body velocity, and the body mass is the body inertia - a property to resist to the applied force. These two things make a clear sense in the Newton equation (Second Law) $ma=F$. Both &quot;parameters&quot; (the body mass and the force strength) are observable and measurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I am reading an &lt;strong&gt;&lt;a href=&quot;http://www.staff.science.uu.nl/~hooft101/gthpub/GtH_Yukawa_06.pdf&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;&lt;/strong&gt; of G. 't Hooft where he states that &quot;The interactions among particles have the effect of modifying masses and coupling strengths&quot;. So not only velocities are modified with forces but the particles masses and the forces themselves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see here a contradiction with the Newton's definition of force and mass. As soon as the QFT is more fundamental than Classical Mechanics, does it mean that the 't Hooft's law is stronger than the Newton's ones? What is the impact (effect) of this stronger law on Classical Mechanics? How to define now the masses and forces if they are self-modifiable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; OK, is there a classical physics example to demonstrate that a potential interaction can modify mass and charge?&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-06-21T18:49:42.110" LastActivityDate="2011-08-27T01:59:53.100" Title="Stronger than Newton's laws?" Tags="&lt;quantum-field-theory&gt;&lt;classical-mechanics&gt;" AnswerCount="6" CommentCount="35" FavoriteCount="4" />
  <row Id="11079" PostTypeId="2" ParentId="11077" CreationDate="2011-06-14T10:52:52.567" Score="0" ViewCount="" Body="&lt;p&gt;It's because your brake pads are parallel to the rim. When you brake your pads starts oscillating (forward and backward) and those vibrations are amplified by the rim. Therefore ideal brake setup is to have small angle between pads and the rim.&lt;/p&gt;&#xA;" OwnerUserId="2307" LastActivityDate="2011-06-14T10:52:52.567" CommentCount="5" />
  <row Id="11080" PostTypeId="2" ParentId="11077" CreationDate="2011-06-14T11:01:36.983" Score="2" ViewCount="" Body="&lt;p&gt;1.: I figure that the dampness on the rim changes the friction coefficients in such a way that they are right for &lt;a href=&quot;http://en.wikipedia.org/wiki/Stick-slip_phenomenon&quot; rel=&quot;nofollow&quot;&gt;stick-slip&lt;/a&gt; to occur. When the rims are dry after a few braking actions, the parameters change, stick-slip disappears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.: The whole rim is your body of resonance.&lt;/p&gt;&#xA;" OwnerUserId="3951" LastActivityDate="2011-06-14T11:01:36.983" CommentCount="1" />
  <row Id="11081" PostTypeId="1" CreationDate="2011-06-14T11:52:01.757" Score="3" ViewCount="59" Body="&lt;p&gt;This is a question for the physical chemists out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given chemical reaction there is a barrier to traverse in order to proceed from reactants to products. Reactants e.g. an organic base abstracting a proton can approach each other at different orientations - attack angles. At times, this orientation will lead to better orbital overlap between the lone pair on the base and the s orbital of the hydrogen.  I have heard people talk about better orbital overlap, lowering/narrowing the barrier to reaction. Is this a valid way of conceptualizing this? Does better orbital overlap change the barrier? I personally have always associated a better overlap as starting further along the reaction coordinate/barrier, not the actual barrier itself being altered?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts from the physchems?&lt;/p&gt;&#xA;" OwnerUserId="3424" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-06-14T17:24:52.333" LastActivityDate="2011-07-14T18:19:07.493" Title="Physical Chemistry: What's the relationship between orbital overlap and barrier shape?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" />
  <row Id="11082" PostTypeId="2" ParentId="353" CreationDate="2011-06-14T12:37:33.077" Score="3" ViewCount="" Body="&lt;p&gt;SI units are pragmatic so their definition must be easily realized. That is why ampere is the basic unit, since it is easier to measure. Theoretically, the elementary charge is much more fundamental than an arbitrarily chosen amount of current, but that cannot be reproduced with equal or higher precision, at least not now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the candela is chosen more arbitrarily, since its definition doesn't give a specific instruction to measure it. One can also choose lux as the fundamental unit and define it as &quot;The lux is the illuminance of monochromatic radiation of frequency $540\times 10^{12}\,\text{Hz}$  and that has a radiant intensity in that direction of 1683 watt per steradian.&quot; I guess this has more to do with historic reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And by the way, if you don't need to take into account the human vision, you don't need candela; watt is better.&lt;/p&gt;&#xA;" OwnerUserId="3887" LastActivityDate="2011-06-14T12:37:33.077" />
  <row Id="11083" PostTypeId="2" ParentId="9411" CreationDate="2011-06-14T13:22:39.423" Score="2" ViewCount="" Body="&lt;p&gt;If you are interested in minimal mechanical work required to move the system from one state to another it can be roughly arbitrary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, during the process the system should change its state (and so its entropy and energy), however this change can be due to mechanical work or heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can just cool the mixture so the constituents will separate (like cooling air). Maybe it is also valid for liquids --- one of the constituent will turn solid earlier. No mechanical work is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you can say for sure is the change of the entropy of the outside world --- it is no less then the difference of entropies of mixed and unmixed states (2nd law). And of course you are sure about the energy change of the outside world --- it is exactly the energy difference of mixed and unmixed states (1st law).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't say anything specific about work or heat required, they are not functions of state. Functions of state are entropy and energy, they are specified.&lt;/p&gt;&#xA;" OwnerUserId="4020" LastActivityDate="2011-06-14T13:22:39.423" />
  <row Id="11084" PostTypeId="2" ParentId="11081" CreationDate="2011-06-14T13:32:53.457" Score="2" ViewCount="" Body="&lt;p&gt;According to the typical (covalent) bond theory, when two orbitals approach, the forming &quot;bonding&quot; orbital has lower energy and the &quot;anti-bonding&quot; orbital has higher energy. The stabilization depends on the overlap between the orbitals (among other things). Therefore if you have up to two electrons available, the energy of the system will decrease by forming a bond. Now if you have two pathways to proceed in a reaction, the pathway implying maximal overlap between the orbitals will be the minimal energy reaction pathway, approximately.&lt;/p&gt;&#xA;" OwnerUserId="3572" LastActivityDate="2011-06-14T13:32:53.457" />
  <row Id="11085" PostTypeId="2" ParentId="9262" CreationDate="2011-06-14T13:52:24.460" Score="0" ViewCount="" Body="&lt;p&gt;The derivative-like line shape is a result of the use of field modulation.  In order to get sufficient signal to noise, the B_0 field (large, static field) is modulated (usually at 100kHz) and a lock-in amplifier (or equivalent) is used to reject any frequencies beside 100kHz.  The result of this field modulation is that the signal that is obtained is not the value of the absorbance, rather the difference in the absorbances between the ends of the modulation amplitude (more or less) i.e. ~d(absorption)/dB. A slightly more detailed explanation (with a &lt;a href=&quot;http://www.bruker-biospin.com/uploads/pics/fig2-7.jpg&quot; rel=&quot;nofollow&quot;&gt;picture&lt;/a&gt;, figure 2.7) is &lt;a href=&quot;http://www.bruker-biospin.com/cwpractice.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of caveats and such due to the fact that it's not a true derivative, and there is, therefore, some signal distortion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is currently (again) development of detection techniques which are not dependent on the modulation of the field for S/N.&lt;/p&gt;&#xA;" OwnerUserId="4021" LastActivityDate="2011-06-14T13:52:24.460" />
  <row Id="11086" PostTypeId="2" ParentId="4995" CreationDate="2011-06-14T13:54:57.563" Score="0" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Solving this for p gives me a second degree polynomial with two solutions for p. I am not sure yet how to discriminate between the two.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Guess that might be ruled out by the second law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle{\frac{\partial}{\partial x} \left [\rho v s \right] + \frac{q}{T} \geqslant 0}$ (if you don't take into account heat conductivity)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is indeed depends on the temperature profile, which you assume to be known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle{\rho s = n k \left[ \frac{5}{2} + \ln \left[\frac{T^{3/2}}{n} \right] + \text{const} \right]}$&lt;/p&gt;&#xA;" OwnerUserId="4020" LastEditorUserId="4020" LastEditorDisplayName="" LastEditDate="2011-06-14T17:29:43.517" LastActivityDate="2011-06-14T17:29:43.517" CommentCount="4" />
  <row Id="11087" PostTypeId="1" AcceptedAnswerId="11090" CreationDate="2011-06-14T14:58:58.820" Score="1" ViewCount="89" Body="&lt;p&gt;It is said to avoid staring at sun as it can damage eyes, but more then that it is also said that do not come out in sun during eclipse as it emits dangerous rays?&#xA;Is that true? If yes, Why?&lt;/p&gt;&#xA;" OwnerUserId="847" LastActivityDate="2011-06-14T17:22:23.693" Title="Does dangerous rays emit during Solar Eclipse?" Tags="&lt;sun&gt;" AnswerCount="3" />
  <row Id="11088" PostTypeId="2" ParentId="11075" CreationDate="2011-06-14T15:08:15.200" Score="4" ViewCount="" Body="&lt;p&gt;The heart of your concern is the non-locality, right? The way to relieve your mind is to think of everything locally instead of globally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We first learn about energy conservation as a global law: The total energy in some (large) region remains constant as a function of time. But there's another, local formulation of energy conservation: The rate of change of the energy &lt;em&gt;density&lt;/em&gt; at any given point equals the rate of energy flow into that point. Formally,&#xA;$$&#xD;&#xA;{\partial\rho\over\partial t}=-\nabla\cdot{\bf J},&#xD;&#xA;$$&#xA;where $\rho$ is energy density and ${\bf J}$ is energy current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Technical note: Actually, $\rho$ is part of a 4-vector and ${\bf J}$ is part of a rank-2 tensor, and they're usually called other things rather than $\rho,{\bf J}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly speaking, the second version says that the energy conservation accounting works out correctly point by point in space, which is a stronger statement than global energy conservation. In fact, global energy conservation follows from local energy conservation, but the reverse isn't true. A world in which energy spontaneously disappeared in this room and appeared in Antarctica, with nothing flowing from here to there, would satisfy global but not local energy conservation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory of electricity and magnetism satisfies local energy conservation, not just global energy conservation. That is, when you take into account the energy stored in the electromagnetic fields, you can keep track of the energy density and energy current at each point, and the books balance everywhere. Moreover, the energy density and energy current are well-defined functions of the electric and magnetic fields, so they change smoothly and causally with time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you adopt that view of energy conservation, it becomes obvious that energy is conserved, and there are no non-locality or causality issues, even in situations such as the ones you describe. The energy flows around smoothly and causally, in such a way that the total is conserved.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-06-14T15:08:15.200" CommentCount="3" />
  <row Id="11089" PostTypeId="2" ParentId="4995" CreationDate="2011-06-14T15:12:09.100" Score="1" ViewCount="" Body="&lt;p&gt;I think you were on the right path in your original post.  For one, let's at least talk about the boundary conditions.  I'll rewrite your equations with a more useful construct for the &quot;constant&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho v = \rho_{in} v_{in}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p + \rho v^2 = p_{in} + \rho_{in} v_{in}^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And state of course&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p=\rho r T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $v$ will only have one direction and will never change sign, so no need to be dealing with vector math.  The unknowns are fully stated as $v$, $T$, and $\rho$, which can be a function of the position down the pipe, $x$, but I suggest that the easy way is to find an answer in terms of temperature instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving the equations you can arrive at the following in order to define $\rho$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0=\rho^2 r T - (p_{in} +\rho_{in} v_{in}^2) \rho + (\rho_{in} v_{in})^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now stop and think about this equation for a little bit, and consider the part of $-b \pm \sqrt{b^2-4ac}$, and that the term $ac$ is positive, and $-b$ is positive.  That means that the only way for this to be positive is for the sign to be positive, or for us to take the largest root.  Forget entropy arguments, in the world I live in, density is clearly positive.  We have now defined $\rho(T)$ as the positive root for the preceding equation.  The rest follows easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v=\frac{\rho_{in} v_{in}}{\rho(T)}$$&#xA;$$p = \rho(T) r T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the above, and plug in $T_{out}$ and you're finished.  For those of you who will yell at me for not being 100% explicit, here you go:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho(T) = \frac{p_{in} +\rho_{in} v_{in}^2 + \sqrt{(p_{in}+\rho_{in}v_{in}^2)^2-4 r T (\rho_{in} v_{in})^2}}{2 r T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to recap, you should have known quantities for $p_{in}$, $T_{in}$, and $v_{in}$.  If you don't, then you need to figure out what the question is.  It follows that $\rho_{in} = p_{in}/(r T_{in})$, and these all go into the above equations.  Then if you know how much it was cooled by the time it reaches the end of the pipe, plug in that for $T$ and the above equations are the answer to your problem.  You may have the information in some different form, like knowing the length of $x$ and having basically $dT/dx$, although that quantity might have to come after applying different physical laws.  Whatever, you have what I think is an agreeable solution giving a $T$, the rest is up to whatever it is you need.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-06-14T15:17:34.013" LastActivityDate="2011-06-14T15:17:34.013" />
  <row Id="11090" PostTypeId="2" ParentId="11087" CreationDate="2011-06-14T15:18:43.530" Score="4" ViewCount="" Body="&lt;p&gt;There's certainly no problem being out in the Sun during an eclipse: There's nothing being emitted then that's not being emitted at other times. The danger is just that the relative darkness near totality may make it seem safe to look at the Sun, even when it's not. But as long as you don't look directly at the Sun, you're fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the time when an eclipse is total, it is safe to look at it with the naked eye. But &lt;strong&gt;&lt;em&gt;only&lt;/em&gt;&lt;/strong&gt; during totality. Even when the eclipse is almost total -- a thin crescent or even just &lt;a href=&quot;http://en.wikipedia.org/wiki/Baily%27s_beads&quot; rel=&quot;nofollow&quot;&gt;Bailey's beads&lt;/a&gt; or the &lt;a href=&quot;http://en.wikipedia.org/wiki/Diamond_ring_effect&quot; rel=&quot;nofollow&quot;&gt;diamond ring&lt;/a&gt; visible -- you can damage your eyes by looking at it. So if you're planning to view an eclipse, be sure you can identify totality before looking with the naked eye. Viewing with an experienced eclipse-watcher is a good idea if you can arrange it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;a href=&quot;http://www.mreclipse.com/Totality/TotalityCh11.html&quot; rel=&quot;nofollow&quot;&gt;Mr. Eclipse&lt;/a&gt; puts it&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Permanent eye damage can result from&#xA;  looking at the disk of the Sun&#xA;  directly, or through a camera&#xA;  viewfinder, or with binoculars or a&#xA;  telescope even when only a thin&#xA;  crescent of the Sun or Baily's Beads&#xA;  remain. The 1 percent of the Sun's&#xA;  surface still visible is about 10,000&#xA;  times brighter than the full moon.&#xA;  Staring at the Sun under such&#xA;  circumstances is like using a&#xA;  magnifying glass to focus sunlight&#xA;  onto tinder. The retina is delicate&#xA;  and irreplaceable. There is little or&#xA;  nothing a retinal surgeon will be able&#xA;  to do to help you. Never look at the&#xA;  Sun outside of the total phase of an&#xA;  eclipse unless you have adequate eye&#xA;  protection.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-06-14T15:18:43.530" />
  <row Id="11091" PostTypeId="2" ParentId="11087" CreationDate="2011-06-14T16:08:45.743" Score="1" ViewCount="" Body="&lt;p&gt;It's amazing how long the solar eclipse myth persists in spite of attempts to educate the public. There are still school principals in many countries that deny children the opportunity to view totality, one of the most wonderful sights in nature. During partial eclipse phases, solar filters that reduce luminosity by a factor of 10^5 will permit safe viewing. However, during totality, no filtering is required since the luminosity approximates that of a full moon. In my eclipse trips we often passed out filters to the interested public.&lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="1924" LastEditorDisplayName="" LastEditDate="2011-06-14T16:28:41.803" LastActivityDate="2011-06-14T16:28:41.803" />
  <row Id="11092" PostTypeId="1" CreationDate="2011-06-14T16:30:38.853" Score="1" ViewCount="106" Body="&lt;p&gt;First, I give an absurd example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A conscious observer lives on Earth at time t. A light-year away, at a space-like separation, a nuclear bomb chain reaction goes off. A nuclear bomb chain reaction is quantum all the way and leads to a splitting into expo-exponentially many parallel worlds. In one reference frame, the explosion happens months after the observer at time t. Let's say at time t, we have N copies of the observer out there distributed across N parallel worlds. In another boosted reference frame, the explosion happens before the observer at time t. What this means is, at time t, the observer is split into N times expo-exponentially many copies, and for each version of the observer, there are expo-exponentially many exactly identical copies lying about. By anthropic reasoning, and anthropic reasoning is necessary in the many worlds interpretation, the latter reference frame would be preferred by an expo-exponential factor in complete violation of Lorentz invariance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to save the many worlds interpretation from this absurdity is to find a manifestly Lorentz invariant version of splitting. How can this be done?&lt;/p&gt;&#xA;" OwnerUserId="4022" LastActivityDate="2011-07-14T20:19:08.543" Title="Is there a manifestly Lorentz invariant formulation of the many worlds interpretation?" Tags="&lt;quantum-interpretations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11094" PostTypeId="2" ParentId="11087" CreationDate="2011-06-14T17:22:23.693" Score="-1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Looking at a solar eclipse is just as dangerous as looking at the sun under normal conditions (a very dangerous thing to do), &lt;strong&gt;because the sun light still bends around the moon&lt;/strong&gt;. The sun is constantly emitting dangerous UV light that can cause great harm to your eyes.&quot; -P. Erik Gundersen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3821" LastActivityDate="2011-06-14T17:22:23.693" CommentCount="2" />
  <row Id="11095" PostTypeId="2" ParentId="11092" CreationDate="2011-06-14T17:29:07.853" Score="2" ViewCount="" Body="&lt;p&gt;In the so-called &quot;many worlds interpretation,&quot; the &quot;splitting of worlds&quot; is not an actual physical process. In this interpretation, all there is is the wavefunction of the Universe, and all it ever does is evolve smoothly according to the appropriate equation of motion. The whole theory is manifestly Lorentz-invariant from end to end. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;standard&quot; (Copenhagen) interpretation is the one that's not manifestly Lorentz-invariant, since a measurement at one place causes &quot;collapse&quot; everywhere all at once.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-06-14T17:29:07.853" />
  <row Id="11096" PostTypeId="1" AcceptedAnswerId="11884" CreationDate="2011-06-14T18:30:34.870" Score="7" ViewCount="173" Body="&lt;p&gt;Changing Electric Field causes Magnetic filed and changing Magnetic Field causes Electric Field. Is there anything similar in relation to Gravitational Field? What sort of field is created by varying Gravitational Field?&lt;/p&gt;&#xA;" OwnerUserId="4023" LastEditorUserId="4023" LastEditorDisplayName="" LastEditDate="2011-07-06T05:19:46.020" LastActivityDate="2011-07-06T05:19:46.020" Title="Field created by varying Gravitational field" Tags="&lt;electromagnetism&gt;&lt;gravity&gt;&lt;vector-fields&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="11097" PostTypeId="2" ParentId="11075" CreationDate="2011-06-14T18:55:21.347" Score="-2" ViewCount="" Body="&lt;p&gt;As I mention in the comments, gravitational effects take care of conservation of energy when gravitating masses are accelerated non-locally, as described in the comment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surprisingly, when charged objects are accelerated non-locally, gravitational effects take care of conservation of energy. Energy gravitates after all, and energy is needed to accelerate charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electro-magnetics effects have some role in here too: they ensure that a lot of energy is required to very rapidly accelerate a charge to speed 0.99999999999 c.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I bet this is how it works. &lt;/p&gt;&#xA;" OwnerUserId="4017" LastActivityDate="2011-06-14T18:55:21.347" />
  <row Id="11098" PostTypeId="1" CreationDate="2011-06-14T19:23:58.833" Score="2" ViewCount="120" Body="&lt;p&gt;If the charge is large, considering charge density is usually meaningful despite the discrete nature of electrical charge. The following sentence is part of a problem in a textbook on electromagnetics:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Imagine a sphere of radius &lt;em&gt;a&lt;/em&gt; filled with negative charge of uniform density, the total charge being equivalent to that of two electrons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The charge is only 2e, and so it seems that the model wouldn't work at all. Is there any real-world application of such a calculation?&lt;/p&gt;&#xA;" OwnerUserId="1778" LastActivityDate="2011-06-15T09:00:10.947" Title="Is it meaningful to imagine a sphere uniformly charged with 2e?" Tags="&lt;electrostatics&gt;&lt;models&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="11099" PostTypeId="2" ParentId="11098" CreationDate="2011-06-14T19:42:27.703" Score="2" ViewCount="" Body="&lt;p&gt;No, I would suspect there is no particle with such a uniform charge density. As you state, at this level, charges are discrete with quarks having multiples of $$\pm\frac{1}{3}e$$ and leptons having either +/-e or zero charge. The charge density of a proton falls off nearly exponentially, see&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://prl.aps.org/abstract/PRL/v99/i11/e112001&quot; rel=&quot;nofollow&quot;&gt;http://prl.aps.org/abstract/PRL/v99/i11/e112001&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I found was the charge density of Helium ions which seems to be very roughly constant. But the total charge would be +1. see&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ae.utexas.edu/~lraja/APL_80(4)_Kothnur_et_al.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ae.utexas.edu/~lraja/APL_80(4)_Kothnur_et_al.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is probably not intended to be a real world problem but just a homework problem which keeps the numbers simple. At this scale you won't be able to use classical electrodynamics anyways.&lt;/p&gt;&#xA;" OwnerUserId="2151" LastActivityDate="2011-06-14T19:42:27.703" />
  <row Id="11100" PostTypeId="2" ParentId="353" CreationDate="2011-06-14T20:01:37.473" Score="0" ViewCount="" Body="&lt;p&gt;I wanted to add this as a comment to the question itself, but guess I don't have that &quot;privilege&quot; yet. So for whatever it's worth, here goes...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.independent.co.uk/news/science/scientists-fear-kilograms-dont-weigh-as-much-as-they-used-to-2192481.html&quot; rel=&quot;nofollow&quot;&gt;http://www.independent.co.uk/news/science/scientists-fear-kilograms-dont-weigh-as-much-as-they-used-to-2192481.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3998" LastActivityDate="2011-06-14T20:01:37.473" />
  <row Id="11101" PostTypeId="2" ParentId="11042" CreationDate="2011-06-14T23:08:26.417" Score="6" ViewCount="" Body="&lt;p&gt;As suggested by M.Sameer I convert my comment into an answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dear M.Sameer: It seems that you are missing that the $n+\ell$  Madelung rule is not an exact result derived from first principles, but rather a rule of thumb, that holds for, say, approximately 95 percent of all the elements, with important exceptions, cf. this &lt;a href=&quot;http://en.wikipedia.org/wiki/Madelung_rule#The_Madelung_energy_ordering_rule&quot; rel=&quot;nofollow&quot;&gt;wikipedia page.&lt;/a&gt; Nevertheless, there are semi-rigorous theoretical &lt;a href=&quot;http://en.wikipedia.org/wiki/Madelung_rule#The_n_.2B_l_energy_ordering_rule&quot; rel=&quot;nofollow&quot;&gt;arguments&lt;/a&gt; why such a rule of thumb should hold to a good approximation.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastActivityDate="2011-06-14T23:08:26.417" CommentCount="1" />
  <row Id="11102" PostTypeId="2" ParentId="11021" CreationDate="2011-06-15T02:53:34.297" Score="0" ViewCount="" Body="&lt;p&gt;I think if the satellite gets that close you have to consider tidal drag (as well as atmospheric drag it there is one present), which would mean any super close orbit would decay on a timescale that is small compared to astronomical timescales. The planet must be rotating slower than the orbital speed at its equatorial radius, otherwise it would be unstable, and mass/angular momentum would be shed -presumably forming a ring which might condense into a moon. Also if there are any other bodies nearby, so that it isn't governed by two body dynamics, the orbit will vary with time, and unless the near grazing condition was the innermost approach of an oscillation, it probably wouldn't avoid an impact for long. Now if you want to select a special time, just before the satellite impacts the planet, then yes it can happen. In fact a satellite in a wrong way orbit will lose momentum due to tidal drag and spiral in until it impacts. I think that is your best scenario for reaching your scenario.&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-06-15T02:53:34.297" />
  <row Id="11103" PostTypeId="1" CreationDate="2011-06-15T03:07:43.430" Score="0" ViewCount="34" Body="&lt;p&gt;According to Steven Hawking in his book &quot;A Brief History of Time&quot; time moves slower the further it is into the earths atmosphere. This is because a light wave loses energy the further it goes into earth atmosphere making the wavelengths longer, therefor slowing down time. However if the sun &quot;turned off&quot; time would not stop and we (pretending we lived) would age probably at the same rate (although i dont know for sure). Could someone please explain this?&lt;/p&gt;&#xA;" OwnerUserId="3688" LastActivityDate="2011-06-15T03:07:43.430" ClosedDate="2011-06-15T03:43:52.557" Title="time slowing further into the earths atmosphere" Tags="&lt;light&gt;&lt;time&gt;" CommentCount="2" />
  <row Id="11104" PostTypeId="1" AcceptedAnswerId="11123" CreationDate="2011-06-15T03:32:40.003" Score="3" ViewCount="66" Body="&lt;p&gt;I've read that maximally entangled qubit states are a good &quot;unit&quot; of bipartite entanglement since it is possible to create any other entangled state from them using local operations and classical communication (LOCC) provided sufficiently many copies are available. What would the protocol be to construct a maximally entangled qutrit ($\vert \psi \rangle_{AB} = \frac{1}{\sqrt{3}}(\vert 00 \rangle + \vert 11 \rangle + \vert 22 \rangle)$) between two space separated parties from a set of $n$ Bell states ($\vert \Phi^+ \rangle^{\otimes n}_{AB} = 2^{-n/2} (\vert 00 \rangle + \vert 11 \rangle)^{\otimes n}$) initially shared by those parties using only LOCC?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can, please include in your answer why local operations alone would not be sufficient.&lt;/p&gt;&#xA;" OwnerUserId="3632" LastActivityDate="2011-06-15T19:10:39.243" Title="Constructing a maximally entangled qutrit state from $n$ Bell states" Tags="&lt;quantum-mechanics&gt;&lt;quantum-information&gt;&lt;entanglement&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="11105" PostTypeId="1" CreationDate="2011-06-15T03:54:48.863" Score="0" ViewCount="87" Body="&lt;p&gt;is it possible to have length, width and time. A 2D or 1D object with the 4th dimension (although then there would not be 4 dimensions so would it b 3D?)                   .&lt;/p&gt;&#xA;" OwnerUserId="3688" LastActivityDate="2011-06-15T10:28:03.847" Title="length, width, and time" Tags="&lt;time&gt;&lt;dimensionality&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11106" PostTypeId="2" ParentId="11105" CreationDate="2011-06-15T04:34:55.473" Score="1" ViewCount="" Body="&lt;p&gt;Sure, a 2D object that exists at more than one point in time would have those three dimensions. It'd be called 2+1-dimensional and there are various physical theories that could be adapted to describe it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our own universe, of course, all actual objects are 3+1-dimensional, because as far as we know, the universe itself is 3+1-dimensional (3 dimensions of space and one of time).&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-15T04:34:55.473" CommentCount="7" />
  <row Id="11107" PostTypeId="1" AcceptedAnswerId="11150" CreationDate="2011-06-15T06:10:26.213" Score="-1" ViewCount="95" Body="&lt;p&gt;what is the key difference between the windings of the electric motor and electric generator?&lt;/p&gt;&#xA;" OwnerUserId="4026" LastActivityDate="2011-06-15T20:46:01.960" Title="How does the electric motor and generator vary?" Tags="&lt;electromagnetism&gt;&lt;electricity&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="11108" PostTypeId="1" CreationDate="2011-06-15T06:20:01.423" Score="1" ViewCount="65" Body="&lt;p&gt;Can we scatter the lightning before it hits the ground?Can we scatter the lightning by connecting many wires to a lightning rod?&lt;/p&gt;&#xA;" OwnerUserId="4026" LastEditorUserId="4026" LastEditorDisplayName="" LastEditDate="2011-06-27T05:33:38.607" LastActivityDate="2011-06-27T05:33:38.607" ClosedDate="2011-06-15T07:11:18.030" Title="Can the lightning be scattered with out hitting the ground?" Tags="&lt;electricity&gt;" CommentCount="2" />
  <row Id="11109" PostTypeId="2" ParentId="11105" CreationDate="2011-06-15T08:14:30.373" Score="0" ViewCount="" Body="&lt;p&gt;There is a nice animation about the 10th dimensions which goes through the various dimensions. Check it out: &lt;a href=&quot;http://www.youtube.com/watch?v=JkxieS-6WuA&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=JkxieS-6WuA&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3677" LastEditorUserId="3677" LastEditorDisplayName="" LastEditDate="2011-06-15T10:28:03.847" LastActivityDate="2011-06-15T10:28:03.847" />
  <row Id="11110" PostTypeId="2" ParentId="10778" CreationDate="2011-06-15T08:28:11.687" Score="0" ViewCount="" Body="&lt;p&gt;like the comment aheadd said , i was doing it wrong , using perimeter instead of area (2piR instead of piR²).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you wan to watch the result look &lt;a href=&quot;http://www.wolframalpha.com/input/?i=plot%2025%2b%28Sqrt%5B45-25%5D-%282Pi0.3%C2%B2%29/%28Pi10%C2%B2%29Sqrt%5B2%2a980%281%2b%28Pi10%C2%B2%29/%28Pi10%C2%B2%29%29%5Dt%29%C2%B2%20from%200%20to%2038&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3877" LastActivityDate="2011-06-15T08:28:11.687" />
  <row Id="11111" PostTypeId="2" ParentId="11098" CreationDate="2011-06-15T09:00:10.947" Score="2" ViewCount="" Body="&lt;p&gt;Solid conductive metals have a pool of electrons that are naturally delocalized, i.e., it is energetically favorable for these electrons to spread out somewhat, so they don't have a well-defined location.  This may be difficult to grasp if you think of electrons as objects having well-defined locations at all times, but such an intuition is imported from experiences in the macroscopic classical world, and does not approximate quantum mechanical behavior well.  That is, while electrical charge takes on a discrete set of values, the position of electrons does not.  There is a brief description in &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrical_conduction#Metals&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article on conduction&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The delocalized electrons are free to move about when acted on by potentials, and this is what allows such metals to conduct electricity.  If you add two electrons to a neutral conductive metallic sphere with no potentials applied, it will be energetically favorable for the charge density to be uniform, and there are no obstructions to realizing such a configuration.  In summary, the answer to your title question is &quot;yes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding real-world applications, if the radius of the sphere is small enough (read: nanoscale), the charge density will be high enough to affect the outcome of chemical reactions.  Even if the radius is larger, small static charges may contribute noise to the readings of sensitive instruments.&lt;/p&gt;&#xA;" OwnerUserId="761" LastActivityDate="2011-06-15T09:00:10.947" CommentCount="1" />
  <row Id="11112" PostTypeId="1" AcceptedAnswerId="11135" CreationDate="2011-06-15T09:19:34.820" Score="1" ViewCount="80" Body="&lt;p&gt;Being a fan of stackoverflow, I thought I could try this question here, hope its not too off topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wanting to build a free standing climbing wall in my garden and dont know where to start working out its strength and so construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will be made from interlocking (using bolts) 50cmx50cmx50cm mild steel cubes stacked 3m tall with a base of 1mx1m. The 4 base cubes will then be bolted to a concrete foundation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I then bolt other cubes to the side of the main stack to create an overhang how can I work out the stress put onto the joints between the overhang and the main stack and between the stack and the foundations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bolts between the cubes will be 3cm long and 2cm diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I havent decided what fixings to use between the stack and concrete as I believe this is where the problem will lie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help its much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="4027" LastEditorUserId="4027" LastEditorDisplayName="" LastEditDate="2011-06-15T14:46:52.230" LastActivityDate="2011-06-17T16:34:52.457" Title="Stresses put through joints on a steel climbing wall" Tags="&lt;forces&gt;&lt;stress-strain&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11113" PostTypeId="2" ParentId="11054" CreationDate="2011-06-15T09:57:07.973" Score="5" ViewCount="" Body="&lt;p&gt;In order to say that a configuration is in an equilibrium state, you need to say that the net force acting on each planet is zero.  However, in order to define net force (if you're a mathematician), you need to choose and justify a rule for regularizing a sum that is not absolutely convergent.  The rule that you seem to have chosen takes a limit over concentric balls centered at the planet in question.  This method gives a convenient answer of zero net force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might ask why this is preferable to some other regularization method.  For example, if you chose a limit over increasingly large ellipsoids of equal eccentricity, for which one focus is the planet in question, and the other focus moved away in a certain direction in the plane, you'd get an attractive force in that direction.  This is a valid summation method, in the sense that the contributions from all of the planets are eventually considered when computing the net force on a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One advantage of the spherical regularization over the ellipsoidal method is that it is invariant under rigid motions, and in fact any regularization method that is invariant under reflections or rotations will give you an answer of zero.  This makes the equilibrium an aesthetically natural answer, but that doesn't necessarily mean it gives the correct answer.  In particular, I think the setup is sufficiently separated from realistic physical situations that a preference for symmetry doesn't justify a regularization, so I'd say that this question does not have a well-defined answer (but physicists might disagree).&lt;/p&gt;&#xA;" OwnerUserId="761" LastActivityDate="2011-06-15T09:57:07.973" CommentCount="1" />
  <row Id="11114" PostTypeId="2" ParentId="10826" CreationDate="2011-06-15T10:07:29.067" Score="0" ViewCount="" Body="&lt;p&gt;I guess answer from #Micheal is correct.&#xA;I would just like to add that the &quot;pain&quot; factor in shaving comes from friction of blade with the skin, removing some skin tissue causing pain. The fact that shaving&#xA;action of shaving is performed periodically(for most people) causes the thin hair(previously) when shaved to grow from thick base further resulting in thickening of the hair. This results in tightening of the hair in the follicle (http://en.wikipedia.org/wiki/Hair_follicle) which gives the feel of hardness of the regrown hair. Now the soap is a surfactant and plays the role of cleansing agent for skin and lubricating agent for hair. Since the oils in the follicle are cleansed by soap the water effectively hydrates the skin, resulting in softening of the hair and follicular clamp. This could result in swelling due to hydration is also the reason why hot water works better than cold water. End result is an easy shave with minimal skin damage and hence less pain.&lt;/p&gt;&#xA;" OwnerUserId="4028" LastActivityDate="2011-06-15T10:07:29.067" />
  <row Id="11115" PostTypeId="1" AcceptedAnswerId="11118" CreationDate="2011-06-15T10:28:11.017" Score="2" ViewCount="63" Body="&lt;p&gt;To help settle an office discussion could someone please explain why this is/is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that materials could be found that are light and strong enough for its construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to feed the back-end of a wind tunnel to the front forming a torus and circulating the air back around could you then place a wing in the this tunnel and produce enough lift to fly the torus?&lt;/p&gt;&#xA;" OwnerUserId="4029" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-06-15T13:05:42.137" LastActivityDate="2011-06-15T17:29:41.623" Title="Effects on the lift of a Wing in a hermetically sealed torus" Tags="&lt;aerodynamics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11116" PostTypeId="1" CreationDate="2011-06-15T10:36:18.657" Score="0" ViewCount="46" Body="&lt;p&gt;Not exactly physics question but physicist must have encountered this problem before. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my code I generate a table of six tab separated columns (x,y,z,type,charge,id). I want to create a 3d visualization of the particles and to color them base them by some property (charge, type, size). What (Linux) program can do this easily?&lt;/p&gt;&#xA;" OwnerUserId="3677" LastActivityDate="2011-06-15T10:36:18.657" ClosedDate="2011-06-15T15:30:33.670" Title="How to display particles" Tags="&lt;simulation&gt;&lt;particles&gt;&lt;visualization&gt;" CommentCount="3" />
  <row Id="11117" PostTypeId="1" AcceptedAnswerId="11122" CreationDate="2011-06-15T11:18:01.693" Score="4" ViewCount="140" Body="&lt;p&gt;The value of the fine structure constant is given as&#xA;$$&#xD;&#xA;\alpha = \frac{e^2}{4\pi\varepsilon_0\hbar c} = \frac{1}{137.035\,999..}&#xD;&#xA;$$&#xA;It's value is only dependent on physical constants (the elementary charge $e$, speed of light $c$, Plancks constant $\hbar$), the vacuum permitivvity $\varepsilon_0$) and the mathematical constant $\pi$, which are considered to be constant under all cirumstances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the Wikipedia article &lt;a href=&quot;http://en.wikipedia.org/wiki/Coupling_constant#QED_and_the_Landau_pole&quot; rel=&quot;nofollow&quot;&gt;Coupling constant&lt;/a&gt; states&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In particular, at low energies, α ≈ 1/137, whereas at the scale of the Z boson, about 90 GeV, one measures α ≈ 1/127.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't understand how this can be possible, except that one of the physical constants above or even $\pi$ are actually not constant, but dependent on the energy scale. But that seems nonsense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what do physicists mean when they say that the fine structure constant $\alpha$ increases with energy? Can you perhaps reformulate the quoted sentence above so that it makes more sense?&lt;/p&gt;&#xA;" OwnerUserId="1648" LastActivityDate="2011-06-15T13:24:50.920" Title="Is the fine structure constant actually a constant or does its value depend on the energy scale?" Tags="&lt;quantum-electrodynamics&gt;&lt;physical-constants&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11118" PostTypeId="2" ParentId="11115" CreationDate="2011-06-15T11:24:28.653" Score="4" ViewCount="" Body="&lt;p&gt;No. Concerning aerodynamic forces, the wind tunnel is a closed system (I.e. no exchange of momentum with the outside is possible), so it won't move.&#xA;Crowley's comment is an alternative and equally valid formulation/argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Btw. most wind tunnels are closed systems to begin with.&lt;/p&gt;&#xA;" OwnerUserId="3951" LastActivityDate="2011-06-15T11:24:28.653" />
  <row Id="11119" PostTypeId="2" ParentId="11117" CreationDate="2011-06-15T11:28:42.727" Score="1" ViewCount="" Body="&lt;p&gt;Do not worry, it is a constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a sloppiness and misunderstanding in physics when the energy dependence of the cross sections is wrongly attributed to the &quot;fundamental constant&quot; whereas it is a cross section feature.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-06-15T11:28:42.727" />
  <row Id="11120" PostTypeId="1" AcceptedAnswerId="11140" CreationDate="2011-06-15T13:08:33.277" Score="1" ViewCount="52" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Scientists say rare drop in sunspot activity could cause global cooling&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.globalpost.com/dispatch/news/business-tech/science/110615/science-news-solar-flares-sunspots-global-warming&quot; rel=&quot;nofollow&quot;&gt;http://www.globalpost.com/dispatch/news/business-tech/science/110615/science-news-solar-flares-sunspots-global-warming&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Last week's event is almost certain to be just the start of a cycle that is expected to peak in 2013. It is during this rising activity that we can expect the Earth to be buffeted by some devastating solar storms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://timesofindia.indiatimes.com/home/science/Solar-storm-rising-Earth-may-be-hit-by-cosmic-hurricane/articleshow/8858351.cms&quot; rel=&quot;nofollow&quot;&gt;http://timesofindia.indiatimes.com/home/science/Solar-storm-rising-Earth-may-be-hit-by-cosmic-hurricane/articleshow/8858351.cms&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Solar Flares Could Cripple Earth's Tech Infrastructure in 2013&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.pcworld.com/article/229876/solar_flares_could_cripple_earths_tech_infrastructure_in_2013.html&quot; rel=&quot;nofollow&quot;&gt;http://www.pcworld.com/article/229876/solar_flares_could_cripple_earths_tech_infrastructure_in_2013.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't the first article contradictory with the other two? Which one is closer to the truth?&lt;/p&gt;&#xA;" OwnerUserId="2868" LastActivityDate="2011-06-15T18:28:27.717" Title="Will the sun cool and produce a disrupting EMP?" Tags="&lt;astrophysics&gt;&lt;electromagnetic-radiation&gt;&lt;sun&gt;" AnswerCount="1" />
  <row Id="11122" PostTypeId="2" ParentId="11117" CreationDate="2011-06-15T13:24:50.920" Score="5" ViewCount="" Body="&lt;p&gt;Expanding on what Vladimir said: the thing that is changing with energy is $e$ (the others are not constants so much as conversion factors between length and time, time and energy, etc.). The reason the charge can vary is that the vacuum is not entirely empty. Sloppily speaking, near a charge, the electric field interacts with virtual (electron/positron) pairs and the effect is that the virtual pairs screens the &quot;raw&quot; electric field. Thus, if you're far away, you see one value, but as you get closer the electric field raises faster than $1/r^2$. With scattering experiments, how close you get to a charge is directly related to the in-going energy of the particles. Now, in modern physics, we account for this by saying that the charge $e$ changes with energy scale; this sounds bizarre in the form I just explained (since you might expect that we just declare the force to be not $1/r^2$), but it turns out that this is the neatest way intellectually to understand it, due to a convergence of issues to do with wanting to preserve observed symmetries in the theory at all scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, for things like colour charges in QCD, the vacuum &lt;em&gt;anti-screens&lt;/em&gt;, which is to say that the observed field strength increases as you get further away. Heuristically, this is what leads to confinement of quarks in the &quot;normal&quot; phase. &lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-06-15T13:24:50.920" CommentCount="11" />
  <row Id="11123" PostTypeId="2" ParentId="11104" CreationDate="2011-06-15T13:34:40.527" Score="2" ViewCount="" Body="&lt;p&gt;Basically, when you have $n$ Bell-states, you already have a maximally entangled $2^n$ dimensional state. For example, when $n=2$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{align}&#xD;&#xA;\lvert\Phi^+\rangle_{AB}^{\otimes 3}&#xD;&#xA;&amp;amp;= \frac12 (\lvert0\rangle_A\lvert0\rangle_B + \lvert1\rangle_A\lvert1\rangle_B)(\lvert0\rangle_A\lvert0\rangle_B + \lvert1\rangle_A\lvert1\rangle_B)\\&#xD;&#xA;&amp;amp;=\frac12 (\lvert00\rangle_A\lvert00\rangle_B + \lvert01\rangle_A\lvert01\rangle_B +&#xD;&#xA;\lvert10\rangle_A\lvert10\rangle_B + \lvert11\rangle_A\lvert11\rangle_B\\&#xD;&#xA;&amp;amp;=\frac12 (\lvert0\rangle_A\lvert0\rangle_B + \lvert1\rangle_A\lvert1\rangle_B +&#xD;&#xA;\lvert2\rangle_A\lvert2\rangle_B + \lvert3\rangle_A\lvert3\rangle_B&#xD;&#xA;)&#xD;&#xA;\end{align}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really want to confert thes ququarts to qutrits, then you have to do this probabilistically. If you are not to picky about the success probability, you just have to measure whether the sate written above is in the $\lvert3\rangle$ state or not. If not (which happens with probability 3/4, you have converted 2 qubit maximally entangled pairs into 1 maximally entangled qutrit pairs. You need therefore on average $\frac2{3/4}=8/3=2.667$ qubit pairs per produced qutrit pairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, processing more pairs simultaneously allows you to approach the asymptotic rate of $\log_2 3=1.585$ qubit pair per qutrit pairs.&lt;/p&gt;&#xA;" OwnerUserId="373" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2011-06-15T19:10:39.243" LastActivityDate="2011-06-15T19:10:39.243" />
  <row Id="11124" PostTypeId="2" ParentId="11054" CreationDate="2011-06-15T13:39:14.447" Score="2" ViewCount="" Body="&lt;p&gt;Well, the main question about a 2D grid in a 3+1 dimensional space-time has already been answered by Lubos Motl and Scott Carnahan with the main conclusion that both the potential sum and the force sum are not &lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_convergence&quot; rel=&quot;nofollow&quot;&gt;absolutely convergent,&lt;/a&gt; and therefore not mathematically well-posed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the bonus question about a grid of dimension $d$, clearly a 3D grid would only make things worse, which leaves us with a 1D grid, say, a single row of planets along the $x$-axis at position $x\in a\mathbb{Z}$, where $a$ is the grid size. The corresponding potential sum is logarithmically divergent, but the force sum is actually absolutely convergent, so here the calculation makes rigorous sense. If we call the unit-vector in the $x$-direction for $\hat{x}$, then the gravitational force on the planet $P$ at $\vec{r}=\vec{0}$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{F}~=~\frac{Gm^2}{a^2}\hat{x}\sum_{n\in\mathbb{Z}\backslash\{0\}}\frac{n}{|n|^3}~=~\vec{0},$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is zero by $n\leftrightarrow -n$ symmetry, as expected.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastActivityDate="2011-06-15T13:39:14.447" />
  <row Id="11125" PostTypeId="2" ParentId="11064" CreationDate="2011-06-15T13:47:40.857" Score="7" ViewCount="" Body="&lt;p&gt;I'm afraid the actual situation is much more complicated than you've been told. For one thing, the superconductivity does not occur between neutrons, but between quarks themselves. The topic of high density QCD is a very cool interplay of condensed matter and high energy physics, and a &lt;a href=&quot;http://arxiv.org/abs/hep-ph/0011333&quot; rel=&quot;nofollow&quot;&gt;very nice review is available by Frank Wilczek&lt;/a&gt;. However, that article does need some background in QCD and superconductivity simultaneously to appreciate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A shortened version might go something like this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Inspiration: free fermions are incredibly unstable to superconductivity, in that &lt;em&gt;any&lt;/em&gt; attractive interaction will cause it (in fact, there's an old theorem by Pierls (?) that almost all interactions (even repulsive ones) will cause superconductivity if you cool far enough). In QCD, quarks &lt;em&gt;naturally&lt;/em&gt; attract already! So at sufficiently high density and low temperature, we can imagine that QCD can cause a strong attractive instability to a Fermi gas of quarks.&lt;/li&gt;&#xA;&lt;li&gt;Complications: 6 flavours, chiralities, masses of quarks are different, etc.&lt;/li&gt;&#xA;&lt;li&gt;Simplification by complication: realise that the normal state of QCD (i.e. 3- and 2-quark combinations) is just that: one possible state. Other phases of QCD exist, and we can study the phase boundaries and so forth even if we can't compute things exactly (universality saves the day!). We find that at &lt;em&gt;really&lt;/em&gt; high densities, quarks pair up to give a background diquark condensate, through which single quarks move, and through the Anderson-Higgs mechanism gains a large mass by eating some Goldstone modes. All gluons become gapped (again, Anderson-Higgs), apart from one which mixes with the photon. The symmetries of this solution is actually the same as that of normal matter --- replace baryons with quarks (+ their diquark condensed background) and mesons with diquarks; this suggests that they are really the same phase in theory. In practise, getting from one to another requires some other phases in the middle, which are more complicated and arise due to the quark masses and the number of flavours, etc.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-06-15T13:47:40.857" CommentCount="2" />
  <row Id="11126" PostTypeId="1" AcceptedAnswerId="11131" CreationDate="2011-06-15T14:12:27.357" Score="2" ViewCount="49" Body="&lt;p&gt;I have a very practical question where I've calculated the mechanical work needed by a simple mechanical system by solving the line integral $W = \int_C \ F \ dx$. However, since I have a black spot in my brain for electrical calculations I figured I could just (considering a 1.5V 2700mAh Alkaline battery) calculate the current required by $I = P / V$ and using that calculate how many batteries I need or how long my mechanical system could work. Is this correct or am I missing something crucial here?&lt;/p&gt;&#xA;" OwnerUserId="3725" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2011-06-15T14:27:28.357" LastActivityDate="2011-06-15T17:19:03.683" Title="Mechanical work to required battery power" Tags="&lt;mechanics&gt;&lt;electronics&gt;" AnswerCount="2" />
  <row Id="11127" PostTypeId="2" ParentId="11126" CreationDate="2011-06-15T14:33:38.183" Score="2" ViewCount="" Body="&lt;p&gt;From the specifications of your battery, that is 1.5V and 2700mAh, you can compute that there is $14580$ Joules of energy stored in your batteries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula $P=U\cdot I$ relates power to voltage and current. You battery specs give voltage and capacity (that is total charge stored). The former is in Volt, the latter in milli-Ampère-hour. The product is therefore not a power but an energy. And that corresponds to the $14580$ Joules I mentioned. Of course, you should also factor in possible losses, so in practice, you'll never draw that much work from the batteries.  &lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2011-06-15T14:33:38.183" CommentCount="3" />
  <row Id="11128" PostTypeId="1" CreationDate="2011-06-15T16:17:31.810" Score="2" ViewCount="58" Body="&lt;p&gt;This experimental Question is a result of reading a particular article on Bell violations. I addressed the e-mail below to the corresponding authors —because who knows, they might reply— but it is not specific to them, it's just that their article crystallized the Question, and perhaps someone on Physics SE can tell me what the results of the experiment would be without doing it (although I think these time-dependent properties are not part of the usual data sheets for off the shelf components).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a subsidiary question: have such experiments been done, and I haven't seen them in the literature?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dear Johannes and Anton,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&amp;nbsp;&amp;nbsp;I've read your PNAS&#xA;  article &quot;Violation of local realism&#xA;  with freedom of choice&quot; as a result of&#xA;  a blog posting by Sabine Hossenfelder&#xA;  at&#xA;  &lt;a href=&quot;http://backreaction.blogspot.com/2011/06/nonlocal-correlations-between-canary.html&quot; rel=&quot;nofollow&quot;&gt;http://backreaction.blogspot.com/2011/06/nonlocal-correlations-between-canary.html&lt;/a&gt;&#xA;  &lt;em&gt;[Physics SE readers can find PNAS and arXiv links there]&lt;/em&gt;.&#xA;  I was impressed by your three key&#xA;  remarks on the second page, which seem&#xA;  to me nicely balanced.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&amp;nbsp;&amp;nbsp;Reading your article I wondered as I&#xA;  have many times whether Bell&#xA;  inequality violations appear &lt;em&gt;instantly&lt;/em&gt;&#xA;  when the source is turned on, or&#xA;  to what extent they increase over time. I am partly led to&#xA;  the questions below because it seems that&#xA;  Bell violating experiments run&#xA;  continuously, whereas it seems&#xA;  that technological applications may&#xA;  well be intermittent.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&amp;nbsp;&amp;nbsp;I imagine, specifically,&#xA;  an experiment &lt;em&gt;based on&lt;/em&gt; an experiment&#xA;  in which we observe Bell violations routinely&#xA;  (presumably in CHSH form). The crucial&#xA;  physical modification is to block both&#xA;  paths from the source to the two&#xA;  detectors physically, at a point near&#xA;  the source, on a time scale of, say,&#xA;  about a second, so that light travels&#xA;  from the source only for half a second&#xA;  at a time. A steady state condition of&#xA;  the source, the quantized electromagnetic&#xA;  field, and the detectors will&#xA;  presumably not be instantly&#xA;  established, but, I suppose, would be&#xA;  established in half a second, so that&#xA;  if we considered data only from the&#xA;  last quarter second of the on-phase we&#xA;  would see the usual violation of Bell&#xA;  inequalities.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&amp;nbsp;&amp;nbsp;The crucial modification&#xA;  of the analysis is to consider the way&#xA;  Bell violation changes over time,&#xA;  microsecond by microsecond (or more or&#xA;  less finely time-sliced as experience&#xA;  indicates), after the moment that the&#xA;  physical block is removed. Clearly&#xA;  this is a statistical analysis, since&#xA;  we would expect to see approximately&#xA;  one photon pair every four&#xA;  microseconds (at a local production&#xA;  rate comparable to the 250,000 photon&#xA;  pairs per second that you report in&#xA;  your PNAS article).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&amp;nbsp;&amp;nbsp;I suppose that a number of&#xA;  questions emerge, some of which may be&#xA;  characteristics of the source and&#xA;  detectors; others might be&#xA;  characteristic of the experimental&#xA;  apparatus more generally considered.&#xA;  Firstly, most elementarily, how does&#xA;  the rate of single photon detections&#xA;  change over time in the microseconds&#xA;  after the physical block is removed?&#xA;  We would expect that there would be a&#xA;  rapid approach to the steady state,&#xA;  but not instantaneous.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&amp;nbsp;&amp;nbsp;Secondly, how does the&#xA;  rate of photon pair detections change&#xA;  over time in the microseconds after&#xA;  the physical block is removed? Does&#xA;  the approach to the steady state have&#xA;  the same relaxation rate as the&#xA;  relaxation rate for the single photon&#xA;  approach to the steady state?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&amp;nbsp;&amp;nbsp;Thirdly, how does the&#xA;  violation of Bell inequalities vary&#xA;  over time in the microseconds after&#xA;  the physical block is removed? Again,&#xA;  does this have the same relaxation&#xA;  rate for the approach to the steady&#xA;  state?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&amp;nbsp;&amp;nbsp;It seems a matter of&#xA;  technological interest as well as of&#xA;  foundational interest whether&#xA;  relaxation of Bell inequality&#xA;  violations towards the steady state&#xA;  immediately after a light source is&#xA;  exposed is the same as the relaxation&#xA;  rates of single photon detections&#xA;  and/or double photon detections, or&#xA;  whether we have to wait longer for&#xA;  usable Bell violations to emerge.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&amp;nbsp;&amp;nbsp;Because I suppose these&#xA;  questions to be of general interest, I&#xA;  have also posted them at&#xA;  &lt;a href=&quot;http://physics.stackexchange.com&quot;&gt;http://physics.stackexchange.com&lt;/a&gt;. I don't imagine the results of such an experiment will go&#xA;  against quantum theoretical expectations, but I'm curious whether characterizing these&#xA;  time-dependent properties of the sources and detectors might give some surprises.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-06-15T16:17:31.810" Title="Do Bell inequality violations appear instantly when the source is turned on, or do they increase over time?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-optics&gt;&lt;bells-inequality&gt;" FavoriteCount="2" />
  <row Id="11129" PostTypeId="1" AcceptedAnswerId="11130" CreationDate="2011-06-15T16:28:19.773" Score="2" ViewCount="173" Body="&lt;p&gt;How would i go about magnetizing something?  specifically, a whole buch of bb's.  i'm wanting to try this so that i could make a (hopefully) cheaper version of &lt;a href=&quot;http://www.thinkgeek.com/interests/dads/bbe8/&quot; rel=&quot;nofollow&quot;&gt;bucky balls&lt;/a&gt;.  but if not it'd at least be a cool experiment to do at home.&lt;/p&gt;&#xA;" OwnerUserId="4030" LastActivityDate="2011-06-20T19:24:59.890" Title="How do you magnetize an object?" Tags="&lt;electromagnetism&gt;&lt;home-experiment&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="11130" PostTypeId="2" ParentId="11129" CreationDate="2011-06-15T16:47:21.640" Score="4" ViewCount="" Body="&lt;p&gt;To permanently magnetize them, you need to start with bb's that are made of a ferromagnetic material, e.g., iron (or some kinds of steel), nickel or cobalt.  This is the challenging part.  When I was a kid they made bb's out of copper.  Nowadays they probably make them out of some other less expensive but equally non-ferromagnetic material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you can find a ferromagnetic bb, the next step is to subject it to a very strong magnetic field, which you could do with a homemade electromagnet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can slightly magnetize the right object simply by subjecting it to high shock, for example, by striking it with a hammer; but a blow strong enough to be of magnetic benefit might squash your bb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:  Was doing a little research to answer a comment, and discovered that those &quot;copper&quot; BBs I shot as a kid were actually copper-coated steel BBs.  So, your magnetic BBs may be much easier to come by than I thought.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-06-20T19:24:59.890" LastActivityDate="2011-06-20T19:24:59.890" CommentCount="5" />
  <row Id="11131" PostTypeId="2" ParentId="11126" CreationDate="2011-06-15T17:19:03.683" Score="1" ViewCount="" Body="&lt;p&gt;You have (1.5V)(2.7A)(3600s) = 14580 Watt-seconds = 14580 Joules of energy in the battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BUT you can only get that much energy out if you extract it the same way the manufacturer did it during their optimized tests, which is typically at a much lower amperage than what you want to drive your motor.  Also, remember that there are friction and other losses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you parallel enough batteries to keep the draw from any one battery to somewhere near a value that would result in total discharge at 100 hours (27mA in your case), and factor in a total system efficiency rate of 50%, my guess is that you will be close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, with batteries properly arranged and sized (quantity of batteries) for your system, you should expect to get (14580J)(50%) = 7290J of actual work out of EACH battery.  That is about 5368 ft.lb. of work (lift 100 pounds a height of 53.7 feet). &lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-06-15T17:19:03.683" />
  <row Id="11132" PostTypeId="1" AcceptedAnswerId="11134" CreationDate="2011-06-15T17:22:14.127" Score="0" ViewCount="85" Body="&lt;p&gt;I am studying about forces as vectors. And they give me this equation:&lt;br /&gt;&#xA;$c^2 = a^2 + b^2 - 2ab \cos C$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody explain me the second part of the equation? I perfectly understand $c^2 = a^2 + b^2$ but what's with the $-2ab \,\mathrm{cos}\, C$?&lt;/p&gt;&#xA;" OwnerUserId="3821" LastEditorUserId="3821" LastEditorDisplayName="" LastEditDate="2011-06-15T17:54:02.850" LastActivityDate="2011-06-15T17:54:02.850" Title="The resultant of two forces acting at any angle?" Tags="&lt;vectors&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="11133" PostTypeId="2" ParentId="11115" CreationDate="2011-06-15T17:29:41.623" Score="1" ViewCount="" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand why, mentally draw a freebody diagram of the system in space.  You have a torus.  You do not know (or care) what is inside the torus.  The torus is motionless.  Now someone starts a process inside the torus (again, you do not know, nor care, what).  That process is allowed to continue to its end (or some steady state).  Nothing leaves nor enters the system during this time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The end result, per the law of conservation of momentum, is that the center of mass of this system has not moved.  Also, per Newtons law of F=ma, since the outside forces acting upon this system sum to zero, we must assume 'a' is zero; therfore the center of the mass of the system did not move.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-06-15T17:29:41.623" />
  <row Id="11134" PostTypeId="2" ParentId="11132" CreationDate="2011-06-15T17:35:50.053" Score="4" ViewCount="" Body="&lt;p&gt;This is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_cosines&quot; rel=&quot;nofollow&quot;&gt;law of cosines&lt;/a&gt;! The Pythagorean theorem is just a special case of this more general result - in the Pythagorean theorem we are dealing with a right triangle &lt;em&gt;only&lt;/em&gt;, and $C=90$ degrees. This equation is true in the more general case of &lt;em&gt;any&lt;/em&gt; triangle.&lt;/p&gt;&#xA;" OwnerUserId="1655" LastActivityDate="2011-06-15T17:35:50.053" CommentCount="2" />
  <row Id="11135" PostTypeId="2" ParentId="11112" CreationDate="2011-06-15T17:40:20.700" Score="4" ViewCount="" Body="&lt;p&gt;Dave, to answer your question, we need to know the wall thickness and weight of each cube.  BUT I can tell you that if you give any reasonable answer (like wall thickness = 1/8 of an inch = about 3.2mm), and choose steel bolts with right-sized steel washers, that it will be pretty hard to fail.  Also, assuming your bolts are 60ksi tensile strength steel (which is not a very high quality bolt), they will hold about 29K pounds each (tensile strength).  That is over 13 metric tons PER BOLT : Certainly not the weak link in your design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the major sress-strain &quot;weak point&quot; for the interface between the overhanging cubes and the rest of the structure will be shear stress on the bolts converting into tearing forces in the cube wall (think of sticking a nail into aluminum foil, and then pulling it towards you, tearing the foil).  If you used 2cm diameter bolts and 3.2mm thick steel wall in the cubes, each cube would be about 38.4 kg. (84.5 lbs.).  If we assume 60ksi steel for the cubes as well as the bolts, each bolt would support about (0.099 inch^2)(60ksi) = 5950 lbs. = 26500N of shear before irreversibly deforming the hole.  One bolt in each corner gives 4 x 5950 lbs. = about 24000 pounds.  Enough to support well over 100 large men hanging off the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the other cube-to-cube interfaces, assuming the bolt holes are near the cube edges, the weak point will be tension in the bolt converting into a force which tends to pull the nut or bolt out of the cube, thereby ripping the cube wall (think of a nail inserted into foil, and then pulled through so that the head of the nail rips through the foil).  Good choices for washers will greatly strengthen these weak points.  Assuming you get the right washers (say, 25-30mm thick steel, ID = 2.1cm and OD = 4cm) and attach the cubes very near the corners, then the cube wall would be expected to support (2/3)(3.1416)(0.04m)(0.0032m)((39.36in/1m)^2)(60ksi) = 24920 lbs = 111kN PER BOLT.  (The 2/3 is a fudge factor, since the steel opposite the bolt from the cube corner will not aid as much in resisting pull-through.)  Four bolts --&gt; about 100K lbs or 444kN.  That is roughly the same force as needed to support 500 large men.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The metal to concrete connection does not need to be the weak point of your structure, but it does require a little thought for structural integrity over time (rust prevention).  Chooing the right size, shape and material for the bolts and embedding them deeply enough into rebar-reinforced concrete during the pour will ensure that the mode of failure is NOT having the bolt rip out of the concrete.  Certainly four 2-cm diameter bolts total, one at each of the four corners of the 1m x 1m structure, would do the trick.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resistance of the structure to leaving the ground would be the weight of the structure, including the concrete.  The resistance to tipping would be about (2)(111kN)(0.9m) = about 200kNm, or ((width of concrete foundation)/2)(weight of concrete + structure), whichever is less.  It is almost guaranteed that the least of these will be governed by the weight of your concrete.  You can greatly increase resistance to tipping (from, say, a hurricane or tornado) by digging post holes at the four corners of your foundation, widen them at the very bottom, and fill the holes with rebar-reinforced concrete as part of the monolithic pour of your foundation.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-06-17T16:34:52.457" LastActivityDate="2011-06-17T16:34:52.457" />
  <row Id="11136" PostTypeId="1" AcceptedAnswerId="11143" CreationDate="2011-06-15T17:55:14.457" Score="6" ViewCount="430" Body="&lt;p&gt;Ok so the universe is in constant expansion, that has been proven, right? And that means that it was smaller in the past.. But what's the smallest size we can be sure the universe has ever had?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just want to know what's the oldest thing we are sure about.&lt;/p&gt;&#xA;" OwnerUserId="3935" LastActivityDate="2011-07-25T10:42:17.307" Title="What has been proved about the big bang, and what has not?" Tags="&lt;universe&gt;&lt;big-bang&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" />
  <row Id="11137" PostTypeId="1" CreationDate="2011-06-15T17:56:57.733" Score="6" ViewCount="180" Body="&lt;p&gt;Einstein predicted that gravitational force can act on light. This was verified in one of solar eclipse that light from a star near to the sun's disc bent due to Sun's gravity as predicted. Since Sun's corona is very hot, there should be hot gases filled beyond the corona through thousands of miles forming a convex-lens. Can it be refraction through convex lens that causes bending of light from the star rather than Sun's gravity?&lt;/p&gt;&#xA;" OwnerUserId="3636" LastEditorUserId="829" LastEditorDisplayName="" LastEditDate="2011-06-16T18:12:51.267" LastActivityDate="2011-06-16T18:12:51.267" Title="Effect of gravitation on light" Tags="&lt;general-relativity&gt;&lt;light&gt;&lt;sun&gt;&lt;refraction&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="11138" PostTypeId="1" CreationDate="2011-06-15T18:05:43.850" Score="4" ViewCount="115" Body="&lt;p&gt;Is transparency of material has something to do with inter- or intra-molecular bonding? E.g. both graphite and diamond are carbon, but graphite is opaque and diamond transparent.&lt;/p&gt;&#xA;" OwnerUserId="3636" LastEditorUserId="3636" LastEditorDisplayName="" LastEditDate="2011-06-19T04:59:23.800" LastActivityDate="2011-06-19T04:59:23.800" Title="Transparency of materials" Tags="&lt;light&gt;&lt;material-science&gt;&lt;crystals&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="11139" PostTypeId="2" ParentId="9983" CreationDate="2011-06-15T18:10:20.410" Score="1" ViewCount="" Body="&lt;p&gt;You may be interested in the &lt;a href=&quot;http://scienceworld.wolfram.com/physics/Bertotti-RobinsonSolution.html&quot; rel=&quot;nofollow&quot;&gt;Bertotti-Robinson&lt;/a&gt; solution to the field equations. It is the only conformally flat solution to the electrovacuum equations with a non-null (in the sense of the electromagnetic invariants) Faraday tensor. (See also Chapter 7 of Griffiths and Podolsky, &lt;em&gt;Exact space-times in Einstein's general relativity&lt;/em&gt;.) (Note that in the form presented in the ScienceWorld link it is not obviously static. See below.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This solution, however, has a uniform electromagnetic field, and is not asymptotically flat. In the asymptotically flat case, using the Jebsen-Birkhoff-Eiseland rigidity theorem, a spherically symmetric solution (as you postulated in the first line of your question) must admit an additional symmetry, and the field equations reduce to ODEs. Asymptotic flatness allows you to prescribe asymptotic data &quot;at infinity&quot; and integrate back. In this case you have a stronger result that for &lt;em&gt;any&lt;/em&gt; (possibly nonlinear) model of electromagnetism that reduces to Maxwell theory in the weak field limit, a spherically symmetric solution must be static and must also be singular along the symmetry axis. For a review of this fact and other related results, see &lt;a href=&quot;http://arxiv.org/abs/1012.1400&quot; rel=&quot;nofollow&quot;&gt;the recent paper by Shadi Tavildar-Zadeh&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar to the Bertotti-Robinson solution is also the Melvin Universe, which is a static non-singular, cylindrical symmetric space-time with an electric field directed along the axis of the cylinder. (You can find more about this solution in Griffiths and Podolsky's book mentioned above.) This solution isn't really physical: it was shown by Thorne, and by Melvin and Wallingford that the geometry is very strange in the Melvin Universe: time-like geodesics with only radial components to the velocity cannot escape to radial infinity ever; the geodesics oscillate and repeatedly cross the central symmetry axis. In otherwords, the gravitational effects induced by the electric field in this solution is &lt;em&gt;very&lt;/em&gt; strong. (For a non-linear electrodynamics version of the Melvin Universe, there's &lt;a href=&quot;http://iopscience.iop.org/0264-9381/18/9/305&quot; rel=&quot;nofollow&quot;&gt;a paper of Gibbons and Herdeiro&lt;/a&gt; on the subject.) &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Since you also asked how one coms up with these solutions: a good method is guess and check. &lt;em&gt;Guess&lt;/em&gt; an &lt;em&gt;ansatz&lt;/em&gt; (for example, a symmetry class), reduce the equation, and try to solve. For example, this would be how I derive the Bertotti-Robinson solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will make the horribly unphysical ansatz that the solution is spherically symmetric &lt;strong&gt;and&lt;/strong&gt; that the space-time is a direct product space-time. So in particular the geometry decomposes into $\mathbb{S}^2 \times (\mathbb{R}^2,g)$. Jebsen-Birkhoff-Eiseland still applies, so we can take one coordinate of $\mathbb{R}^2$ to be the Killing coordinate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The metric then takes the canonical form&#xA;$$ -A(r) dt^2 + B(r) dr^2 + ds^2_{\mathbb{S}^2}$$&#xA;As a direct product metric, the Ricci curvatures are just the sums of the component Ricci curvatures. So using the standard formulae for curvatures of two dimensional surfaces, we quickly compute that in the spherical directions the Ricci curvature is that of the standard sphere, which is equal to the metric (2 dimensional manifold with constant curvature 1). In the $\mathbb{R}^2$ direction, the Gaussian curvature is &#xA;$$ - \frac{1}{2\sqrt{A(r)B(r)}} \left( \frac{A&amp;#39;(r)}{\sqrt{A(r)B(r)}}\right)&amp;#39; $$&#xA;The trace-free condition from Einstein-Maxwell requires that the Gaussian curvature equal to -1, which you can either solve explicitly for by integrating in $r$, or by &quot;knowledge&quot; that the only 2 dimensional Lorentzian manifold with constant negative curvature is AdS2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you take $r$ to be the &quot;inverse radial&quot; coordinate (where $r = 0$ corresponds to spatial infinity), the usual presentation of AdS2 gives that $A(r) = B(r) = r^{-2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Maxwell field, you observe that by spherical symmetry $F$ only has $dr\wedge dt$ component (if you toss out magnetic monopoles; you can gain back the other term using a duality rotation anyway). So $F$ is given by a scalar function $f$. The curvature condition implies that $R_{rr} = T_{rr} = F_{rt}F_{rt}g^{tt} = -f^2 A^{-1} = -1g_{rr} = -1/r^2$ (observe that you can say the same thing using $R_{tt}$ or the spherical components). This means that $f^2 = r^{-4}$, or $f = r^{-2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In remains to check that such a solution verifies Maxwell's equation. $dF = f&amp;#39; dr\wedge dr\wedge dt = 0$ clearly. while the Hodge dual $*F = \sqrt{|g|}g^{tt}g^{rr} f dvol_{\mathbb{S}^2} = - dvol_{\mathbb{S}^2}$, so $d*F = 0$ also. So the Bertotti-Robinson solution is indeed a static solution to the field equations. That the solution is non-singular follows from the fact that it is a direct product of two constant curvature spaces, so the curvature is bounded, and that the electromagnetic invariant $g(F,F)$ is constant valued, so the Maxwell field does not blowup either. &lt;/p&gt;&#xA;" OwnerUserId="829" LastActivityDate="2011-06-15T18:10:20.410" />
  <row Id="11140" PostTypeId="2" ParentId="11120" CreationDate="2011-06-15T18:22:55.173" Score="3" ViewCount="" Body="&lt;p&gt;The first is front line research &lt;a href=&quot;http://wattsupwiththat.com/2011/06/14/all-three-of-these-lines-of-research-to-point-to-the-familiar-sunspot-cycle-shutting-down-for-a-while/&quot; rel=&quot;nofollow&quot;&gt;announced &lt;/a&gt;at a conference, and is news, and closer to the current truth as far as the way cycle 24 is manifesting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of a large solar eruption is another story, one large eruption was manifested even though the cycle is not yet at its maximum; there is no direct contradiction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now as far as cooling goes, there is no solid model nor long range data to be able to predict whether low intensity sun cycles lead to Maunder minimum type cooling. The energy depletion is of the order of 1.4 watts/meter^2, not enough to make a large direct temperature effect.  In 30 years we may have the data to know whether it is a coincidence that the world cooled during the Maunder minimum or somehow the various hypotheses of the sun's indirect influence lead to a causal connection. ( magnetic fields, cosmic rays, UV and planckton, etc)&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-06-15T18:28:27.717" LastActivityDate="2011-06-15T18:28:27.717" />
  <row Id="11141" PostTypeId="2" ParentId="11137" CreationDate="2011-06-15T18:34:56.780" Score="7" ViewCount="" Body="&lt;p&gt;Although the corona's temperature is very high, its density is extremely small, in the order of ~10^-11 kg/m^3. Therefore coronal refraction would be negligeable.&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-06-15T18:34:56.780" CommentCount="2" />
  <row Id="11142" PostTypeId="1" CreationDate="2011-06-15T18:36:06.157" Score="3" ViewCount="163" Body="&lt;p&gt;What is/are the properties of a liquid (e.g. viscosity, surface tension) which is conducive to formation of stable bubbles floating in air (not the bubble inside the liquid)? E.g., if soap dissolved water is bubbled through a straw, big-small air bubbles are formed which comes out of the liquid and keep floating in the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;br&gt;&#xA;This Wikipedia page (http://en.wikipedia.org/wiki/Surface_tension#Surfactants) attributes this to low surface tension of the liquid, but I can enumerate several liquids like spirit, petrol which have quite low surface tension, still if you blow through a straw into it, you won't get bubbles. &lt;/p&gt;&#xA;" OwnerUserId="3636" LastEditorUserId="3636" LastEditorDisplayName="" LastEditDate="2011-06-19T05:02:42.353" LastActivityDate="2011-08-18T17:26:53.957" Title="Properties of liquid and air bubbles" Tags="&lt;surface-tension&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="11143" PostTypeId="2" ParentId="11136" CreationDate="2011-06-15T18:38:04.433" Score="26" ViewCount="" Body="&lt;p&gt;Spencer's comment is right: we never &quot;prove&quot; anything in science. This may sound like a minor point, but it's worth being careful about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I might rephrase the question like this: &lt;em&gt;What's the smallest size of the Universe for which we have substantial observational evidence in support of the standard big-bang picture?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People can disagree about what constitutes substantial evidence, but I'll nominate the epoch of nucleosynthesis as an answer to this question. This is the time when deuterium, helium, and lithium nuclei were formed via fusion in the early Universe. The observed abundances of those elements match the predictions of the theory, which is evidence that the theory works all the way back to that time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The epoch of nucleosynthesis corresponds to a redshift of about $z=10^9$. The redshift (actually $1+z$) is just the factor by which the Universe has expanded in linear scale since the time in question, so nucleosynthesis occurred when the Universe was about a billion times smaller than it is today. The age of the Universe (according to the standard model) at that time was about one minute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other people may nominate different epochs for the title of &quot;earliest epoch we are reasonably sure about.&quot; Even a hardened skeptic shouldn't go any later than the time of formation of the microwave background ($z=1100$, $t=400,000$ years). In the other direction, even the most credulous person shouldn't go any earlier than the time of electroweak symmetry breaking ($z=10^{15}$, $t=10^{-12}$ s.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I vote for the nucleosynthesis epoch because I think it's the earliest period for which we have reliable astrophysical evidence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nucleosynthesis evidence was controversial as recently as about 10 or 15 years ago, but I don't think it is anymore. One way to think about it is that the theory of big-bang nucleosynthesis depends on essentially one parameter, namely the baryon density. If you use the nucleosynthesis observations to &quot;measure&quot; that parameter, you get the same answer as is obtained by quite a variety of other techniques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument for an earlier epoch such as electroweak symmetry breaking is that we think we have a good understanding of the fundamental physical laws up to that energy scale. That's true, but we don't have direct observational tests of the cosmological application of those laws. I'd be very surprised if our standard theory turns out to be wrong on those scales, but we haven't tested it back to those times as directly as we've tested things back to nucleosynthesis.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-06-15T20:52:34.343" LastActivityDate="2011-06-15T20:52:34.343" CommentCount="1" />
  <row Id="11144" PostTypeId="2" ParentId="11137" CreationDate="2011-06-15T18:43:13.833" Score="8" ViewCount="" Body="&lt;p&gt;Michael Luciuk's answer is right, but there's an even stronger reason for rejecting this hypothesis: refraction in the corona would be wavelength-dependent, but the gravitational bending due to the Sun has been measured over a wide range of wavelengths (at least from visible to radio) and has been found to be independent of wavelength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clifford Will's &lt;a href=&quot;http://relativity.livingreviews.org/Articles/lrr-2006-3/&quot; rel=&quot;nofollow&quot;&gt;review of tests of general relativity&lt;/a&gt; is a good source for details on this. This particular test is in Section 3.4.1, but the whole thing is worth reading if you want to think seriously about this subject.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-06-15T18:43:13.833" />
  <row Id="11145" PostTypeId="2" ParentId="11129" CreationDate="2011-06-15T19:36:16.737" Score="2" ViewCount="" Body="&lt;p&gt;Here is a great paper on how to magnetize. &lt;a href=&quot;http://www.oersted.com/magnetizing.PDF&quot; rel=&quot;nofollow&quot;&gt;http://www.oersted.com/magnetizing.PDF&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes you are correct it is a hard paper to “weed threw” but all the necessary information is there. If you go down to the explanation on the apparatus for magnetizing you can essentially deduce that. &#xA;1)  You need impact the material being magnetized.&#xA;2)  You need to put the material in a strong magnetic field&lt;br&gt;&#xA;3)  So the magnetic field and the material must be aligned just right&#xA;   “A ferromagnet, like a paramagnetic substance, has unpaired electrons. However, in addition to the electrons' intrinsic magnetic moment's tendency to be parallel to an applied field, there is also in these materials a tendency for these magnetic moments to orient parallel to each other to maintain a lowered energy state. Thus, even when the applied field is removed, the electrons in the material maintain a parallel orientation.&#xA;   Every ferromagnetic substance has its own individual temperature, called the Curie temperature, above which it loses its ferromagnetic properties. This is because the thermal tendency to disorder overwhelms the energy-lowering due to ferromagnetic order.&#xA;   Some well-known ferromagnetic materials that exhibit easily detectable magnetic properties (to form magnets) are nickel, iron, cobalt, gadolinium and their alloys.”&lt;/p&gt;&#xA;" OwnerUserId="4034" LastEditorUserId="4034" LastEditorDisplayName="" LastEditDate="2011-06-16T00:50:40.563" LastActivityDate="2011-06-16T00:50:40.563" CommentCount="2" />
  <row Id="11146" PostTypeId="2" ParentId="10246" CreationDate="2011-06-15T20:01:25.727" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The experiment calls for noting when the foil just lifts off the bottom plate. Since the foil and the washer are both presumably good conductors, the charge density should be the same until they stop touching. So the &quot;goodness&quot; of this assumption depends on the variation in the voltage across the capacitor after the foil lifts off.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The foil should feel forces from both plates, so I think your source is wrong in this. Horizontal forces come from the fact the plates are not infinite, and therefore some of the field goes beyond the plates.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Washers are both readily available and fairly good conductors, so they're used in a lot of &quot;do it at home&quot; type experiments. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4033" LastActivityDate="2011-06-15T20:01:25.727" />
  <row Id="11147" PostTypeId="1" AcceptedAnswerId="11205" CreationDate="2011-06-15T20:07:01.980" Score="1" ViewCount="80" Body="&lt;p&gt;the answer given by classical quantum mechanic to the for atomica levels does not provide that an electron in an excited levels can radiate a photon and move to a lower level. How QED justifies this process?&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2011-06-17T14:51:20.457" Title="Photon emission from excited atoms" Tags="&lt;photons&gt;&lt;quantum-electrodynamics&gt;&lt;atoms&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="11148" PostTypeId="2" ParentId="4118" CreationDate="2011-06-15T20:37:48.157" Score="-1" ViewCount="" Body="&lt;p&gt;Takins string theory seriously-- the visible universe is a 9-d object, 3 d's of (10^27)m, and 6 d's of (10^-35)m. The 9th root of the product of these is the diameter of the VU in the 9-d 'bulk'--about (10^14.67)m, about the diameter of a proton (!!!). I call this the bogglesomely factoidal deduction, or the BFD-- mysterious and rich, so rich in implications.It concerns the question only tangentially; but I would certainly like input/feedback. As astonishing as it is, I have encountered the BFD nowhere, despite hours of googling. Is it too obvious to be noticed,or too trivial to mention? Back to the question: the planckon (not my coinage) is the smallest thinkable or possible dimension of space, and to speak of points in space,as STists do, makes no sense: a category error, like 'an atom of mayonnaise'. Perhaps the 'purpose' of planckons is to contain information (extractable in principle) so that 'quantum determinism' can work. The universe expands sothat seemingly empty planckons can store information?&lt;/p&gt;&#xA;" OwnerUserId="4037" LastActivityDate="2011-06-15T20:37:48.157" />
  <row Id="11149" PostTypeId="1" CreationDate="2011-06-15T20:37:55.187" Score="2" ViewCount="138" Body="&lt;p&gt;The idea is to give examples of processes that deal with properties of a particle that have clear wave analogues.&lt;/p&gt;&#xA;" OwnerUserId="4036" LastActivityDate="2011-09-04T03:06:22.240" Title="What are examples of Solutions of Newton's Laws that have analogs to Solutions of the Wave Equation?" Tags="&lt;classical-mechanics&gt;&lt;waves&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" />
  <row Id="11150" PostTypeId="2" ParentId="11107" CreationDate="2011-06-15T20:46:01.960" Score="1" ViewCount="" Body="&lt;p&gt;Electric motors vary... a lot: size; shape; purpose; AC or DC; linear, rotational, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electric generators, likewise, vary... a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, a DC motor can be thought of as a DC generator with the energy conversion process running in reverse:  A DC motor converts electrical energy into mechanical energy, and the generator converts mechanical energy into electrical energy.  AND the neat thing is that this process can run either direction in the very same device.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electric cars are doing this now: the electric motor which powers the wheels also acts as an elecrodynamic brake when you want to slow down.  The motors use the car's kinetic energy to power each wheel motor as a generator, reclaiming electrical energy to recharge the car's batteries which got the wheels rolling in the first place.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-06-15T20:46:01.960" CommentCount="1" />
  <row Id="11151" PostTypeId="2" ParentId="11147" CreationDate="2011-06-15T20:49:21.443" Score="2" ViewCount="" Body="&lt;p&gt;An electron is never alone. It is always coupled to the quantum field oscillators. Preparing an atom in an excited state means the entire excitation energy is concentrated in the electron motion. Due to permanent interaction with quantum (photon) oscillators, the occupation numbers of some oscillators start to grow from zero to one. At the same time the probability amplitude for the electron to stay on the same level decays and that to appear on a lower level grows too. So, in the end you have an electron on a lower energy level and one or more photons created (oscillators excited). This process of the energy redistribution in the system with many coupled degrees of freedom takes some time (life time of the level).&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-06-15T20:49:21.443" />
  <row Id="11152" PostTypeId="2" ParentId="7412" CreationDate="2011-06-15T20:55:58.143" Score="3" ViewCount="" Body="&lt;p&gt;I would say Knill and Laflamme's work on Efficient Quantum Computation with just LINEAR optics was a pretty big breakthrough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It allowed for renewed interest and thinking about the role of photons in quantum computation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See--&gt; &lt;a href=&quot;http://www.nature.com/nature/journal/v409/n6816/abs/409046a0.html&quot; rel=&quot;nofollow&quot;&gt;http://www.nature.com/nature/journal/v409/n6816/abs/409046a0.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4036" LastActivityDate="2011-06-15T20:55:58.143" />
  <row Id="11154" PostTypeId="1" CreationDate="2011-06-16T00:14:47.307" Score="28" ViewCount="2369" Body="&lt;p&gt;One of the major impediments to the widespread adoption of electric cars is a shortage of lithium for the batteries.  I read an article a while back that says that there is simply not enough lithium available on the entire planet to make enough batteries to replace every gasoline-powered car with one electric car.  And that confuses the heck out of me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Big Bang theory says that in the beginning, there was a whole bunch of hydrogen, and then lots of hydrogen started to clump together and form stars, and those stars produced lots of helium through fusion, and then after helium, all the rest of the elements.  That's why hydrogen is the most common element in the universe by far, and helium is the second most common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, lithium is #3 on the periodic table.  By extrapolation, there ought to be several times more lithium around than, say, iron or aluminum, which there is definitely enough of for us to build plenty of cars with.  So why do we have a scarcity of lithium?&lt;/p&gt;&#xA;" OwnerUserId="4039" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-16T15:21:53.430" LastActivityDate="2011-07-29T14:47:48.437" Title="Why is there a scarcity of lithium?" Tags="&lt;nuclear-physics&gt;&lt;big-bang&gt;&lt;geophysics&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="4" />
  <row Id="11155" PostTypeId="2" ParentId="11154" CreationDate="2011-06-16T01:08:38.047" Score="41" ViewCount="" Body="&lt;p&gt;Actually, what you've read about the production of nuclei is not quite correct. There are several &lt;em&gt;different&lt;/em&gt; processes by which atomic nuclei are produced: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Big_Bang_nucleosynthesis&quot; rel=&quot;nofollow&quot;&gt;Big Bang nucleosynthesis&lt;/a&gt; is the fusion of hydrogen nuclei to form heavier elements in the early stages of the universe, as it cooled from the big bang. There are rather specific thermal requirements for this process to occur, so there was only a short time window in which heavier elements could form, meaning that the only fusion to actually happen in significant amounts was the conversion of hydrogen (and deuterium) to helium, and an &lt;em&gt;extremely&lt;/em&gt; tiny amount of lithium.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_nucleosynthesis&quot; rel=&quot;nofollow&quot;&gt;Stellar nucleosynthesis&lt;/a&gt; is the fusion of hydrogen and other nuclei in the cores of stars. This is something separate from big bang cosmology, since stars didn't form until millions of years into the universe's lifetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, contrary to what you might have read, not &lt;em&gt;all&lt;/em&gt; elements are formed in stellar nucleosynthesis. There are specific &quot;chains&quot; of nuclear reactions that occur, and only the elements that are produced by those reactions will exist in a star in appreciable quantities. Most stars produce their energy using either the &lt;a href=&quot;http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction&quot; rel=&quot;nofollow&quot;&gt;proton-proton chain&lt;/a&gt; (in lighter stars) or the &lt;a href=&quot;http://en.wikipedia.org/wiki/CNO_cycle&quot; rel=&quot;nofollow&quot;&gt;CNO cycle&lt;/a&gt; (in heavier stars), both of which consume hydrogen and form helium. Once most of the hydrogen has been consumed, the star's temperature will increase and it will start to fuse helium into carbon. When the helium runs out, it will fuse carbon into oxygen, then oxygen into silicon, then silicon into iron. (Of course the actual process is more complicated - see the Wikipedia articles for details.) Several other elements are produced or involved along the way, including neon, magnesium, phosphorous, and others, but lithium is not among them. In fact, stars have a tendency to &lt;a href=&quot;http://en.wikipedia.org/wiki/Lithium_burning&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;consume&lt;/em&gt; lithium&lt;/a&gt;, rather than producing it, so stars actually tend to have only small amounts of lithium.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Supernova_nucleosynthesis&quot; rel=&quot;nofollow&quot;&gt;Supernova nucleosynthesis&lt;/a&gt; is the fusion of atomic nuclei due to the high-pressure, high-energy conditions that arise when a large star explodes in a type II supernova. There are certain similarities between this and big bang nucleosynthesis, namely the high temperatures and pressures, but the main difference is that an exploding star will have &quot;reserves&quot; of heavy elements built up from a lifetime of nuclear fusion. So instead of just forming a lot of helium as occurred just after the big bang, a supernova will form a whole spectrum of heavy elements. In fact supernovae are the only natural source of elements heavier than iron, since it actually requires an input of energy to produce those elements as fusion products. I believe some amount of lithium would be formed in a supernova along with all the other elements, but since a large star would have used up its hydrogen and helium in the central region where most of the action takes place, lithium is probably not a particularly common reaction product.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-16T15:24:49.123" LastActivityDate="2011-06-16T15:24:49.123" CommentCount="4" />
  <row Id="11156" PostTypeId="2" ParentId="11138" CreationDate="2011-06-16T01:25:28.307" Score="2" ViewCount="" Body="&lt;p&gt;Neither nor. Inter-, intra-molecular bonding types (van der waals, ionic, covalent, metallic bonding) are all playing minor or major role in a solid depending on its material (element mixture, inter-atomic distance, crystal lattice type...). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end the &lt;strong&gt;dielectric function&lt;/strong&gt; will theoretically define the reflectivity, transparency, absorption of a solid for a specific wavelength of light for these types of bonding (ionic/electronic polarisation). Look at the picture and formula &lt;a href=&quot;http://en.wikipedia.org/wiki/Dielectric_function#Complex_permittivity&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Transparency is mainly defined by free electrons, position of Fermi Energy &amp;amp; type of bandstructure in a solid&lt;/strong&gt;. For instance, &lt;strong&gt;metals&lt;/strong&gt; (metallic &quot;intermolecular&quot; bonding) are often reflective and non transparent due to free electrons, their Fermi Energy is outside of eventual band gaps. Many &lt;strong&gt;isolators&lt;/strong&gt; (ionic &quot;intermolecular&quot; bonding) have Fermi Energy within an energy band gap of &gt;4 eV , but &lt;strong&gt;optical photon energy is around 1 to 3 eV&lt;/strong&gt;. So electrons in valence band beneath the band gap cannot absorb these optical photons, the material is transparent. Many &lt;strong&gt;semiconductors&lt;/strong&gt; have a band gap around only 1 eV, so their valence electrons can absorb photons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the main specific bonding types in an semiconductor will co-define Fermi Energy, band structure. But they are necessary, not sufficient factors for optical properties of solids. You can roughly classify solid types by their dominant bonding types (isolators-ionic/covalent, metals-metallic,...). &lt;/p&gt;&#xA;" OwnerUserId="3715" LastActivityDate="2011-06-16T01:25:28.307" />
  <row Id="11157" PostTypeId="1" CreationDate="2011-06-16T01:45:45.460" Score="2" ViewCount="94" Body="&lt;p&gt;Inspired by the wiki-article on atom-lasers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Another application, which might also&#xA;  benefit from atom lasers, is atom&#xA;  interferometry. In an atom&#xA;  interferometer an atomic wave packet&#xA;  is coherently split into two wave&#xA;  packets that follow different paths&#xA;  before recombining. Atom&#xA;  interferometers, which can be more&#xA;  sensitive than optical&#xA;  interferometers, could be used to test&#xA;  quantum theory, and have such high&#xA;  precision that they may even be able&#xA;  to detect changes in space-time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;, i would ask you to list more examples of &lt;strong&gt;CURRENT measurement techniques getting developed &amp;amp; tested, that will likely drastically enlarge our knowledge and falsification options of physical theories. But not Sci-Fi and Star Trek Physics ;)&lt;/strong&gt;. Mainly i can think of improvements that will allow&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;more accurate measurement of physical constants/parameters (higher particle energies CERN, better atomic clocks/time definition by Pulsar periods)&lt;/li&gt;&#xA;&lt;li&gt;direct measurement of so far only indirect definable physical quantities (e.g. atom laser)&lt;/li&gt;&#xA;&lt;li&gt;at all new concepts of (real-time) measurement techniques due to improvements e.g. in computational physics (e.g. quantum computing) or better visualisation of phenomena (adaptive optics in telescopes)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Please name the technique and give short example/link to the physics field profiting from it. Examples i mentioned can also be explained further, as im no expert in these fields&lt;/p&gt;&#xA;" OwnerUserId="3715" LastEditorUserId="3715" LastEditorDisplayName="" LastEditDate="2011-07-04T20:10:00.580" LastActivityDate="2011-07-04T20:10:00.580" CommunityOwnedDate="2011-06-16T04:07:48.147" Title="What near-future measurement technique improvements will extend our physical knowledge?" Tags="&lt;measurement&gt;" CommentCount="6" FavoriteCount="2" />
  <row Id="11158" PostTypeId="1" CreationDate="2011-06-16T02:00:43.760" Score="2" ViewCount="132" Body="&lt;p&gt;Generalizing Born's Rule for 4-dimensions  $x_4$, write  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle a\rangle = \int\Psi A\Psi^* \mathrm{d}x_4$$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is this consistent with quantum mechanics? &lt;/li&gt;&#xA;&lt;li&gt;Is this a generalized form of the Born's Rule for the phase space of the wave function?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;@DavidZas &#xA;I was thinking of a four dimensional space where Time is not a parameter. There been recent work on Time Reversed light pulses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://prl.aps.org/abstract/PRL/v106/i19/e193902&quot; rel=&quot;nofollow&quot;&gt;http://prl.aps.org/abstract/PRL/v106/i19/e193902&lt;/a&gt; &#xA;and the PHYSORG description &lt;a href=&quot;http://www.physorg.com/news/2011-05-physicists-time-reversed-pulses.html&quot; rel=&quot;nofollow&quot;&gt;http://www.physorg.com/news/2011-05-physicists-time-reversed-pulses.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this leads me to wonder about abstract phase space, and I mean phase both in the epoch angle of a wave function and the phase space of a Hamiltonian (which yes, is very confusing).&#xA;The way I see it, the wave function implictly has the Action of phase space built into it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\psi  = e^{i(\omega  t - k x)} = e^{\frac{i}{\hbar }(\mathbb{E} t - \mathbf{p} x)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the phase and the action are connected by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\mathbb{E} t - \mathbf{p} x}{\hbar }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine dividing the phases by the action of the particle&#xA;It's a bit like there's a blob (note the very precise language) of phase space wobbling about like the liquid drop model of the nuclear physics. Only this is a more abstract phase space that's oscillating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, again, would Born's Rule apply to such a system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually after rereading what I just written, I'll be surprised if anyone replies this time! :)&lt;/p&gt;&#xA;" OwnerUserId="3588" LastEditorUserId="3588" LastEditorDisplayName="" LastEditDate="2011-06-16T11:57:33.243" LastActivityDate="2011-06-16T13:07:11.133" Title="Is there a four dimensional form of Born's Rule -redub" Tags="&lt;quantum-mechanics&gt;&lt;operators&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="11159" PostTypeId="2" ParentId="11158" CreationDate="2011-06-16T03:13:34.993" Score="6" ViewCount="" Body="&lt;p&gt;Of course this is consistent -- the calculation of an expectation value is defined in fact for operators in a totally generic Hilbert space. If you want that Hilbert space to describe 4 spatial dimensions, more power to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you are imagining $x_4$ as a time dimension, it's important to remember that Q.M. treats space and time asymmetrically. (Or more concretely, time is a &lt;em&gt;parameter&lt;/em&gt; and not an &lt;em&gt;operator&lt;/em&gt;, which is really just to say that there's nothing special at all about time averaging in Q.M.!)&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2011-06-16T03:13:34.993" CommentCount="4" />
  <row Id="11160" PostTypeId="1" AcceptedAnswerId="11186" CreationDate="2011-06-16T04:42:35.227" Score="7" ViewCount="197" Body="&lt;p&gt;How does a spinning windmill produce electricity?What is the principle behind the windmill?&lt;/p&gt;&#xA;" OwnerUserId="4026" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-12T15:28:32.290" LastActivityDate="2011-07-12T15:28:32.290" Title="How do you produce electricity from a wind mill?" Tags="&lt;mechanics&gt;&lt;electricity&gt;&lt;renewable-energy&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="2" />
  <row Id="11161" PostTypeId="2" ParentId="11154" CreationDate="2011-06-16T04:43:27.143" Score="20" ViewCount="" Body="&lt;p&gt;The key word in what you've heard is &quot;available&quot; because there is quite a lot of lithium in the earth that is not so easy to obtain.  The notion of &quot;available Lithium&quot; probably means known land reserves, which &lt;a href=&quot;http://www.ioes.saga-u.ac.jp/ioes-study/li/lithium/occurence.html&quot; rel=&quot;nofollow&quot;&gt;according to this page&lt;/a&gt; amount to 14 million tons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount dissolved in seawater is estimated at 230 billion tons (which is enough for lots of batteries).  Seawater extraction does not seem to be economically viable yet, but people are studying it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The estimated concentration of lithium in the Earth's crust ranges from 1 to 31 ppm, so if we excavate the whole crust, we'll get between 20 and 600 trillion tons.  In other words, if our civilization ever came to a point where we really needed lots of lithium, we wouldn't have to go too far to find it.&lt;/p&gt;&#xA;" OwnerUserId="761" LastActivityDate="2011-06-16T04:43:27.143" CommentCount="4" />
  <row Id="11162" PostTypeId="1" CreationDate="2011-06-16T05:52:43.827" Score="2" ViewCount="121" Body="&lt;p&gt;Supposed a lens arrangement is prepared where light from an object is collimated, focused and recollimated etc. before entering a CCD array. Given that we can calculate the diffraction-limited resolution for each lens in the system, how do we measure the diffraction limited resolution for the whole setup?&lt;/p&gt;&#xA;" OwnerUserId="4032" LastActivityDate="2011-07-09T09:49:06.470" Title="Calculating diffraction-limited resolution for a lens setup" Tags="&lt;optics&gt;" AnswerCount="2" />
  <row Id="11163" PostTypeId="1" CreationDate="2011-06-16T06:36:11.427" Score="0" ViewCount="52" Body="&lt;p&gt;I'm looking for a low-cost way to play with images projected from my pico projector.  I want to orient the projector horizontally, but project the image downward onto the table.  I was going to purchase two lenses and a mirror and test the results against my rough calculations, but then I saw that vendors on ebay sell self-contained units with a 90 degree prism and optics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/aheFA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume it makes the most sense to purchase a single unit with the optics and prism inside, but are there any downsides?  What specs should I look out for with these products?  I've only had limited experience with machine vision lenses, where my imager is looking straight down through the optics, and working distance and field of view are the main specifications that I have to check.&lt;/p&gt;&#xA;" OwnerUserId="1722" LastEditorUserId="1722" LastEditorDisplayName="" LastEditDate="2011-06-16T06:50:05.540" LastActivityDate="2011-06-16T06:50:05.540" ClosedDate="2011-06-16T07:06:52.640" Title="Projecting an image 90 degrees from a projector" Tags="&lt;optics&gt;" CommentCount="6" />
  <row Id="11164" PostTypeId="2" ParentId="11162" CreationDate="2011-06-16T06:37:49.350" Score="4" ViewCount="" Body="&lt;p&gt;(Edited based on your comments)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to briefly clarify what exactly is meant when we talk about being &quot;diffraction limited.&quot; As light is focused, it will reach some minimum spot size before it begins to expand again. The size of this spot depends on how much the light beam is distorted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A perfectly collimated beam (with perfectly planar wavefront) passing through a perfect lens would come out of the lens with perfectly spherical wavefront, and all of the rays in the beam would be converging to a single point. In this case, the spot size is determined solely by the angle occupied by the converging cone of light*. This is what we call &quot;diffraction limited.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the beam is abberated, for example by a poorly manufactured lens, then the beam will not have perfectly spherical converging wavefronts, and the resulting focal spot will be spread out over a larger area. The magnitude of these abberations is what determines the resolution of an optical system when it is &lt;strong&gt;not&lt;/strong&gt; diffraction limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of the diffraction limited spot is a function of the f-number at the image plane. So, if you know the beam diameter after the last lens element, and the back focal distance, you can compute the diffraction limited spot size just like you would for any other lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*this is the case only assuming that the beam is always the same shape. In practice most beams are circular, so all we need to worry about is its diameter. If the beam is a different shape, then its diffraction limited spot size (and shape!) will change.&lt;/p&gt;&#xA;" OwnerUserId="869" LastEditorUserId="869" LastEditorDisplayName="" LastEditDate="2011-06-16T15:31:35.270" LastActivityDate="2011-06-16T15:31:35.270" CommentCount="4" />
  <row Id="11165" PostTypeId="2" ParentId="11160" CreationDate="2011-06-16T06:48:51.967" Score="6" ViewCount="" Body="&lt;p&gt;First kind of principles is 'How the linear motion is transferred into revolutionary':&#xA;There are horizontal- and vertical-axis windmills, horizontal are more common, vertical ones doesn't care about the wind direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second way how to sort windmills is by the principle of transfer energy conserved in axial flow into momentum. Easier way is based on resistance - the blade is usually steel plate - and is used in small and/or old windmills, Sophisticated windmills are based on wing lift - shape is like airplane's wing - and those are made of composites like fiberglass or carbon fiber. Vertical-axis turbines are based different resistance of the profile when the profile is moving against the wind and with it - easiest is S shaped plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And revolutionary motion is transferred through braking system (security), gearbox (to obtain exact frequency) into a generator (Alternator, dynamo, 3-phase generator). This part is almost the same as in any other power-plant. Except solar ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays most wind turbines are horizontal and based on wing-lift. They suppose that the velocity of the wind is 1/3 of the velocity in front of the turbine. Theoretical efficiency is about 46%, real one barely reach 29% (Output wattage / Energy flow into the windmill)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want some equations to support my numbers and/or describe the relation between profile parameters, wind characteristics and windmill design add comment, I'll find it out in my papers...&lt;/p&gt;&#xA;" OwnerUserId="2307" LastEditorUserId="2307" LastEditorDisplayName="" LastEditDate="2011-06-16T06:54:17.503" LastActivityDate="2011-06-16T06:54:17.503" CommentCount="1" />
  <row Id="11166" PostTypeId="2" ParentId="2160" CreationDate="2011-06-16T07:05:04.403" Score="1" ViewCount="" Body="&lt;p&gt;Yes, you can deposit SiO2 with PVD, ion plating.&#xA;The lower the pressure the lower the evaporation temp. needed.&#xA;The question is what do you want to deposit the SiO2 onto and how thick?&#xA;You can use an RF plasma (13.56 MHz at about 5-15 W per cubic meter, Ar gas) with DC bias on the part.&#xA;Use powdered SiO2 in a tunsten or Ta boat with negative melting current.&#xA;Shoot at about 5 x 10-4 millliTorr.&#xA;Bias the part, on an isolated steel or copper tray, with 30-40V DC + at up to 7 amps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other side of the DC circuit is the grounded vacuum chamber wall.&#xA;You must ground everything to the chamber and then ground the chamber.&#xA;Look at www.lesker.com or svc.org for details on setup.&#xA;Or you could send it to me or any number of coaters and they can coat it for you.&#xA;Look up vacuum coating PVD.&lt;/p&gt;&#xA;" OwnerUserId="4045" LastActivityDate="2011-06-16T07:05:04.403" CommentCount="1" />
  <row Id="11167" PostTypeId="2" ParentId="11160" CreationDate="2011-06-16T07:22:58.363" Score="0" ViewCount="" Body="&lt;p&gt;Is it electromagnetic induction what you are asking for?&lt;/p&gt;&#xA;" OwnerUserId="3887" LastActivityDate="2011-06-16T07:22:58.363" CommentCount="1" />
  <row Id="11168" PostTypeId="2" ParentId="11160" CreationDate="2011-06-16T07:52:30.393" Score="3" ViewCount="" Body="&lt;p&gt;Generating Electricity from wind uses the same technology that any generator does. BUT unlike steam, water pressure or gas wind is not continuous so its input is not constant.  1st of all the speed governor is critical if not the most important part of the device. 2nd the high wind failure mechanism is critical you ether design for say 60 mile H. or have the blades collapse. 3rd and not to be forgotten you need knowledge of wind patterns to calculate wind shadows. after all this you can calculate blade design and aerodynamics.  &lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-06-16T07:52:30.393" />
  <row Id="11169" PostTypeId="1" AcceptedAnswerId="11250" CreationDate="2011-06-16T08:53:29.530" Score="2" ViewCount="162" Body="&lt;p&gt;How can I approach this task:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to design diffraction-limited lens for monochromatic light (~1nm line width, so I guess no need to correct chromatic aberrations), geometry aberrations are also irrelevant BUT maximal lpmm required over as wide as possible area (i.e. coma need to be corrected for example) and NA should be at least 0.3-0.5. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I approach this task?&#xA;Is Zemax is state of art software for that kind of tasks? What kinds of lens designs should I start to look at?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Manufacturing is not a problem (provided that there are no aspherical elements).&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-06-25T18:44:08.663" Title="Lens design - max lpmm, monochromatic light" Tags="&lt;optics&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="11170" PostTypeId="1" AcceptedAnswerId="11174" CreationDate="2011-06-16T10:04:47.163" Score="2" ViewCount="185" Body="&lt;p&gt;I decided to dig a tunnel inside the Earth. In equatorial plane. It should be designed in such a way that it follows the Coriolis effect. That means if, say a stone is dropped from rest into the tunnel, then it travels in the tunnel without touching the walls of the tunnel until it reaches the tunnel's lip on the other side somewhere near the surface of the Earth. So the question itself is an equation of the curve of the tunnel. I suspect that there is no nice closed form for the curve. But who knows? Nevertheless, it would be good to know how deep the tunnel goes and where it rises again to the surface and the travel time.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could say that it would be impossible to do this on Earth, but you actually could do this on the Moon.&lt;/p&gt;&#xA;" OwnerUserId="552" LastActivityDate="2011-06-16T14:19:26.080" Title="A tunnel inside the Earth (but not an ordinary tunnel)" Tags="&lt;classical-mechanics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="11171" PostTypeId="2" ParentId="11154" CreationDate="2011-06-16T10:34:38.410" Score="5" ViewCount="" Body="&lt;p&gt;The common mistake behind this statement is that lithium is seen as a fuel that's consumed and discarded. After all, that's how oil works. There's not enough cheap lithium for that. But like steel, lithium will be recycled.. &lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-07-29T10:08:33.007" LastActivityDate="2011-07-29T10:08:33.007" CommentCount="2" />
  <row Id="11172" PostTypeId="2" ParentId="11170" CreationDate="2011-06-16T11:06:00.707" Score="0" ViewCount="" Body="&lt;p&gt;Well...&#xA;In that measure you have to notice central gravitation field.&#xA;So define all the forces applied to the stone during the fall - Gravitation force $F_g$ as function of distance to Earth/Moon core, Centrifugal force $F_r$ as function of distance to axis and Coriolis Force $F_c$. Sum of them is mass times acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ ma=m\frac{d^2r}{dt^2}=F_g+F_c+F_r $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrate twice and here's your result.&#xA;I'd rather use simulink or another program to solve it numerically...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Special conditions:&lt;br&gt;&#xA; North/south pole: No Coriolis effect, no rotation: The stone shall touch the opposite ground and return as linear harmonic oscillator.&lt;/p&gt;&#xA;" OwnerUserId="2307" LastActivityDate="2011-06-16T11:06:00.707" CommentCount="1" />
  <row Id="11173" PostTypeId="2" ParentId="11158" CreationDate="2011-06-16T13:07:11.133" Score="3" ViewCount="" Body="&lt;p&gt;Your starting point is &lt;em&gt;not&lt;/em&gt; a &quot;generalization of the Born rule&quot;. Assuming that you properly defined a Hilbert space that contains your $\Psi$ as a vector, your starting point is an &lt;em&gt;instance&lt;/em&gt; of the Born rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Born rule takes a vector $V$, in a Hilbert space, $H$, on which there is an inner product $(\cdot,\cdot)$, by definition, and tells us that the expected value of an observable represented by the operator $\hat A$, which acts on the vector $V$ in that Hilbert space, is $\bigl&amp;lt;\hat A\bigr&amp;gt;=(V,\hat A V)$. There's no mention, as you see, of 3-dimensional space. I suspect you're thinking in terms of the Schrodinger equation, where the Hilbert space is the infinite-dimensional Hilbert space of complex-valued functions on $\mathbb{R}^3$, $L^2(\mathbb{R}^3)$, and seeking to generalize to space-time by introducing the Hilbert space $L^2(\mathbb{R}^4)$. Since this is just another Hilbert space, the Born rule applies to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can do this as Mathematics, but to my knowledge this construction is not used in Physics. Instead, the usual route is to introduce operator-valued distributions as part of quantum field theory. The approach that I have suggested you are proposing &lt;em&gt;could&lt;/em&gt; be seen as the one-particle subspace of the many-particle Hilbert space of a quantum field theory (I usually try to avoid all talk of particles, but I think putting this statement in field-theoretic terms wouldn't be helpful here). If you want to move to Minkowski space, I believe you will not be able to avoid moving to quantum field theory. Or, at least, that you will have to be able to present very explicitly what the relationship is between quantum field theory and whatever you move to.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-06-16T13:07:11.133" />
  <row Id="11174" PostTypeId="2" ParentId="11170" CreationDate="2011-06-16T13:15:09.470" Score="7" ViewCount="" Body="&lt;p&gt;I am not sure whether you meant initially at rest relative to the universe, or to the surface of the Earth. Here are the answers to both versions:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Universe&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the latitude be $\theta_0$. In the non-rotating reference frame (of the universe), the motion of the stone is in simple harmonic motion. So $r(t) = r_0 \sin (\omega t)$, where $2\pi / \omega$ is the time it takes to get to the center of the earth. The latitude is constant. And the longitude $\phi_0$ (in the non-rotating reference frame) is also constant. But the earth spins with constant angular velocity $\Omega_0 = 2\pi$ radians per 24 hours. So in the rotating coordinate $\phi&amp;#39;(t) = \phi_0 - \Omega t$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So measured relative to the surface of the earth in spherical coordinates you have the parametric description&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ r(t) = r_0\sin(\omega t) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $r_0$ is the radius of the Earth. (Note: this assumes that the Earth is perfectly spherical and of uniform density inside, which is obviously not quite physical. Of course, digging a tunnel like that is also not quite physical...) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \theta(t) = \theta_0, \phi(t) = \phi_0 - \Omega t $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding $\omega$ from the assumption that Earth is perfectly spherical and of uniform density is left as an exercise to the reader. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second interesting exercise is to find the conditions on the density of Earth and on the rate of revolution that allows for the stone to travel in a closed orbit (again, this is much much simpler when considered in the non-rotating reference frame...)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Surface of the earth&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again we work in the non-rotating reference frame. We have again, via the constant density assumption, (and assuming that the mass of stone is 1) that the potential energy is $P = \alpha r^2$. The kinetic energy is $2K = \dot{r}^2 + r^2\dot{\theta}^2 + (r\cos\theta)^2\dot{\phi}^2$. The conservation of angular momentum means that angular component of the velocity is of size $L / r$, where $L$ can be computed from the rate of rotation of the Earth. So the conserved energy is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E = \dot{r}^2 + \frac{L^2}{r^2} + \alpha r^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives an ODE for $r$. Similarly using a tilted coordinate system you can solve for the angles using the conservation of angular momentum by integrating an ODE and plugging in the solution for $r$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the maximum depth, however, you don't need to explicitly solve the ODE: The energy is known initially: $E_0 = 0+ \frac{L^2}{r_0^2} + \alpha r_0^2$, where $L$ depends only on the rate of revolution for Earth, and $\alpha$ on the mass of Earth (assuming uniform density). $r_0$ is the radius of Earth. At the maximum depth, $\dot{r}$ is again 0. So you are down to finding the &quot;other&quot; positive root of the quartic polynomial &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E_0 r^2 = L^2 + \alpha r^4 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which you can solve explicitly using the quadratic formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ r^2 = \frac{E_0 \pm \sqrt{ E_0^2 - 4 \alpha L^2}}{2\alpha } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the + solution is the radius of the earth, and the - solution is the depth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging in physical numbers: at the initial drop, $\dot{\theta} = 0$ and $\dot{\phi} = \Omega = \frac{2\pi}{86400} s^{-1}$. The radius of Earth we take to be $6.38 \times 10^6 m$, the mass $6\times 10^{24} kg$. So $\alpha = G M / r_0^3 = 1.5 \times 10^{-6} s^{-2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conserved angular momentum is initially&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ L_0^2 = r_0^4 \cos(\theta_0)^4 \Omega^2 = 8.8\times 10^{18}\cos\theta_0^4 m^4 s^{-2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the conserved Energy is initially&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E_0 = L_0^2 r_0^{-2} + \alpha r_0^2 = (\cos\theta_0)^4 2.15 \times 10^5 + 6.1\times 10^7 m^2 s^{-2} \sim 6.1\times 10^7 m^2 s^{-2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the initial angular momentum contribution is very small). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $E_0^2 = 3.7 \times 10^{15}$ and $4\alpha L_0^2 = 5.28\times 10^{13}\cos(\theta_0)^4$, so the answer is that the maximum depth is very close to the center of the earth! Using the binomial expansion we get that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ r^2 \sim \frac{L_0^2}{E_0}  \implies r \sim 3.9\times 10^5 \times \cos(\theta_0)^2 m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you start at the equator where $\theta_0 = 0$, the hole will get about 94% to the center of the Earth. &lt;/p&gt;&#xA;" OwnerUserId="829" LastEditorUserId="829" LastEditorDisplayName="" LastEditDate="2011-06-16T14:19:26.080" LastActivityDate="2011-06-16T14:19:26.080" CommentCount="2" />
  <row Id="11175" PostTypeId="1" CreationDate="2011-06-16T13:20:55.043" Score="2" ViewCount="75" Body="&lt;p&gt;In an electron gun, the heating filament heats the cathode, releasing electrons by thermionic emission. I've read that &lt;em&gt;&quot;electrons are negatively charged particles and the positively charged cylindrical anode develops a strong electric field that exerts a force on the electrons, accelerating them along the tube&quot;&lt;/em&gt;. However, I don't think that this explanation is very clear, and I was wondering specifically how the &quot;strong electric field&quot; inside the cylindrical anode is able to accelerate the electrons?&lt;/p&gt;&#xA;" OwnerUserId="1846" LastActivityDate="2011-08-16T14:25:23.790" Title="Electron gun - electron in cylindrical anode" Tags="&lt;electromagnetism&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11176" PostTypeId="1" CreationDate="2011-06-16T14:44:57.020" Score="1" ViewCount="88" Body="&lt;p&gt;I get confused when I see expressions like &quot;the universe is $x$ years old&quot; or &quot;$10^{-2}$ seconds after the big bang&quot; since it seems to me that relativity shows such statements don't have meaning. Is it assumed or experimentally verified or proved that space-time is equipped with a projection to the real numbers whose differential is non-zero on tangent vectors with a non-zero time component? If assumed, why is this a reasonable assumption? and if proved, what are the initial axioms (causality)?  Looking at a similar question, perhaps the point is that all geodesic given by an initial point and negative time-pointing tangent vector must converge to some given point in finite (backwards) time?  And to get completely cranky, are there good scientific reasons to assume no closed geodesics?&lt;/p&gt;&#xA;" OwnerUserId="4058" LastEditorUserId="829" LastEditorDisplayName="" LastEditDate="2011-06-16T15:10:15.643" LastActivityDate="2011-06-16T15:10:51.570" Title="Does space-time have a projection to time?" Tags="&lt;general-relativity&gt;&lt;arrow-of-time&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="11177" PostTypeId="1" AcceptedAnswerId="11179" CreationDate="2011-06-16T15:04:20.187" Score="0" ViewCount="122" Body="&lt;p&gt;When I was student I was told that time is defined by the requirement that the physical laws are simple. For example, in classical mechanics time can be defined by the requirment that the velocity of an isolated body is constant. One could generalize this approach by assuming that &lt;em&gt;space-time&lt;/em&gt; is defined by the requirement that the physical laws are simple. This can be easily expressed in mathematical language as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general relativity the universe is represented by the triple $(M, g, T)$ where $M$ is a four-dimensional differentiable manifold, and $g$ and $T$ are a Lorentzian metric and a tensor field on $M$ satisfying Einstein's equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the above approach, we could say that the &quot;true&quot; physical reality of the universe is  represented by the pair $(M, T)$, while the metric $g$ emerges as an appeareance from the requirement that evolution appears simple, i.e., that $g$ and $T$ satisfy Einstein's equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the mathematical point of view, this approach poses for example the following problems: does any tensor field  $T$ admit a metric $g$ such that Einstein's equation is satisfied? To what extent a tensor field $T$ univocally determines the metric $g$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this approach make sense, at least from the mathematical point of view? &lt;/p&gt;&#xA;" OwnerUserId="4059" LastEditorUserId="4059" LastEditorDisplayName="" LastEditDate="2011-06-16T15:59:41.240" LastActivityDate="2011-06-16T15:59:41.240" Title="Can space-time be defined by the requirement that the physical laws are simple?" Tags="&lt;general-relativity&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="11178" PostTypeId="2" ParentId="11176" CreationDate="2011-06-16T15:10:51.570" Score="3" ViewCount="" Body="&lt;p&gt;The statements of the age of the universe timescale are related to the cosmic time, a timescale derived from the expansion of the universe in general relativity of a roughly homogenous universe (the &lt;a href=&quot;http://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric&quot; rel=&quot;nofollow&quot;&gt;Friedmann-Lemaitre&lt;/a&gt; universe/metric). Different homogenous densities of the universe define different cosmic times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The assumption is a homogenous expanding/contracting universe that adheres to GR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, this type of solution to the GR field-equations contains a global &quot;preferred&quot; time, but it doesn't say anything about the small-scale structure of spacetime or your closed timelike curves (if that was what you were thinking about).&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-06-16T15:10:51.570" CommentCount="6" />
  <row Id="11179" PostTypeId="2" ParentId="11177" CreationDate="2011-06-16T15:43:47.833" Score="1" ViewCount="" Body="&lt;p&gt;Addressing just the last two paragraphs of your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $T$ is &quot;non-degenerate&quot; (at every point as a linear transformation from $T_pM \to T_p^*M$), then modulo some issues with the constraint equations $g$ can be solved from $T$, at least locally in time, after prescribing it in a compatible way on a space-like hypersurface. See &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=700744&quot; rel=&quot;nofollow&quot;&gt;this article by De Turck&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumeably the issue with degeneracy has more to do with the method involved in solving the problem, then with it being a critical condition, since the evolution for Ricci-flat metrics (Einstein vacuum equations) is also well-posed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main difficulty is about the &quot;univocal&quot; part: due to diffeomorphism invariance, the &quot;local&quot; problem for prescribing Ricci curvature is under-determined. (The global problem may have topological constraints; that I am not sure about.) So one expects that for the local problem the issue is more likely the overabundance of solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the prescription of initial/boundary data may be crucial. Roughly speaking the prescribing Ricci curvature equation is analogous to the inhomogeneous wave equation with a source term. In which case the hyperbolic nature of the equations shows that for any initial configuration there exists a different evolution. So it may be that this will lead to multiple different metrics compatible with your $T$ (indeed consider the case where you restrict your manifold to be $M = \mathbb{R}^4$; the &lt;a href=&quot;http://academic.research.microsoft.com/Publication/3903740/the-global-nonlinear-stability-of-the-minkowski-space&quot; rel=&quot;nofollow&quot;&gt;global nonlinear stability of Minkowski space&lt;/a&gt;, combined with classical results on &lt;a href=&quot;http://www.springerlink.com/content/qr4950g0j2x7567n/&quot; rel=&quot;nofollow&quot;&gt;classical results on the existence of solutions to the vacuum constraint equations&lt;/a&gt; shows that on $M = \mathbb{R}^4$ there are many incongruent (since for these other solutions the Weyl tensor is non-vanishing) solutions to Einstein's equation when $T = 0$; I would expect something similar for the case of prescribed $T$). &lt;/p&gt;&#xA;" OwnerUserId="829" LastActivityDate="2011-06-16T15:43:47.833" CommentCount="1" />
  <row Id="11180" PostTypeId="2" ParentId="11175" CreationDate="2011-06-16T16:01:55.107" Score="1" ViewCount="" Body="&lt;p&gt;any electric field will accelerate electrons, according to the Lorentz force law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec{F} = q \vec{E} + \vec{v} \times \vec{B}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where q is the charge of the electron, and $\vec{E}$ is the electric field.  I assume there is no magnetic field in this example, so the second term is zero.&lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2011-06-16T16:01:55.107" CommentCount="1" />
  <row Id="11181" PostTypeId="2" ParentId="10004" CreationDate="2011-06-16T16:20:48.820" Score="4" ViewCount="" Body="&lt;p&gt;I'm getting the impression that a good part of this question (and perhaps also &lt;a href=&quot;http://physics.stackexchange.com/q/11158/2451&quot;&gt;this&lt;/a&gt; physics.SE question?) arises from a wrong presumption that time and position should be on equal footing in quantum mechanics. They are not. The position $\hat{\bf r}$ is an &lt;em&gt;operator&lt;/em&gt;, while time $t$ is a &lt;em&gt;parameter.&lt;/em&gt;  (Notation: In the following boldface denotes a vector quantity, and a hat denotes an operator quantity.) In fact, Pauli's Theorem shows, under mild assumptions, that time cannot be an operator in quantum mechanics, cf. Ref. (1). See also &lt;a href=&quot;http://physics.stackexchange.com/q/6584/2451&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://physics.stackexchange.com/q/5268/2451&quot;&gt;this&lt;/a&gt; physics.SE questions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the position space Schroedinger representation of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Schr%C3%B6dinger_picture&quot; rel=&quot;nofollow&quot;&gt;Schroedinger picture,&lt;/a&gt; we have a wave function $\Psi({\bf r},t)=\left&amp;lt;{\bf r}|\Psi(t)\right&amp;gt;$, the position operator $\hat{\bf r}$ becomes multiplication with ${\bf r}$, and the momentum operator $\hat{\bf p}= \frac{\hbar}{i}{\bf\nabla}$ is differentiation wrt. position. Both $\hat{\bf r}$ and $\hat{\bf p}$ are Hermitian operators. The Hamiltonian $\hat{H}=H(\hat{\bf r},\hat{\bf p})$ is composed out of the operators $\hat{\bf r}$ and $\hat{\bf p}$ in such a way that it is Hermitian. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is wrong to claim that the Hamiltonian $\hat{H}$ is $i\hbar \partial_{t}$. (If it was, for starters, the Schroedinger equation $i\hbar \partial_{t}\Psi({\bf r},t)=\hat{ H}\Psi({\bf r},t)$ would become an empty statement without any content.) This should be compared with the fact that (in the position space Schroedinger representation) the momentum operator $\hat{\bf p}= \frac{\hbar}{i}{\bf\nabla}$ is a differentiation wrt. position. In particular, time and position differentiations are, in this sense, not on equal footing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the frequency operator may be defined as $\frac{1}{h}\hat{H}=\frac{1}{h}{H}(\hat{\bf r},\hat{\bf p})$, as David Zaslavsky observes in a comment to lurscher's answer. The corresponding expectation value is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left&amp;lt;\Psi(t) \left|\frac{1}{h}\hat{H} \right| \Psi(t)\right&amp;gt; ~=~ \int\! d^{3}r \ \Psi^{*}({\bf r},t)\ \frac{1}{h}\hat{H}\Psi({\bf r},t). $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Wolfgang Pauli 1933, in &lt;em&gt;Handbuch der Physik (Encyclopedia of Physics)&lt;/em&gt; (ed. S. Fluegge), vol. 5, pp.1-168, Springer Verlag, 1958.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-17T10:51:22.930" LastActivityDate="2011-06-17T10:51:22.930" />
  <row Id="11182" PostTypeId="2" ParentId="11154" CreationDate="2011-06-16T16:50:09.140" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.npr.org/2011/05/09/136029474/lithium-battery-industry-keeps-going-and-going&quot; rel=&quot;nofollow&quot;&gt;According to this NPR story&lt;/a&gt;, there is no shortage of lithium for batteries:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;I don't know of any serious person in&#xA;  the automotive industry or in the&#xA;  lithium industry who believed that&#xA;  there is a serious, long-term supply&#xA;  problem,&quot; he says. &quot;In fact, for the&#xA;  next 10 years there will probably be&#xA;  an oversupply of lithium because so&#xA;  many companies have now moved into the&#xA;  market.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;And unlike the impact of mining other&#xA;  natural resources, concentrating&#xA;  lithium is an &quot;environmentally benign&quot;&#xA;  process, Fletcher says. &quot;It's about as&#xA;  low-impact as mining can get. They're&#xA;  really just pumping water up ... and&#xA;  there are really no toxic chemicals in&#xA;  a lithium-ion battery.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4063" LastActivityDate="2011-06-16T16:50:09.140" CommentCount="1" />
  <row Id="11183" PostTypeId="2" ParentId="11154" CreationDate="2011-06-16T16:51:15.957" Score="0" ViewCount="" Body="&lt;p&gt;We may have a scarcity of lithium due to the fact that lithium reacts with almost everything. It's so willing to shed it's electrons, that is why it is such a great candidate for batteries. Actually, it's so volatile it doesn't exist in nature in its pure form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(paraphrased from a radio interview on NPR)&lt;/p&gt;&#xA;" OwnerUserId="4064" LastActivityDate="2011-06-16T16:51:15.957" CommentCount="1" />
  <row Id="11184" PostTypeId="2" ParentId="11154" CreationDate="2011-06-16T16:55:14.000" Score="3" ViewCount="" Body="&lt;p&gt;&lt;em&gt;This is a small complement to David's and Scott's answers&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As usual &lt;a href=&quot;http://en.wikipedia.org/wiki/Lithium&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's page on Lithium&lt;/a&gt; contains useful information :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Both natural isotopes have anomalously low nuclear binding energy per nucleon compared to the next lighter and heavier elements, helium and beryllium, which means that alone among stable light elements, lithium can produce net energy through nuclear fission. The two lithium nuclei have lower binding energies per nucleon than any other stable compound nuclides other than deuterium, and helium-3. As a result of this, though very light in atomic weight, lithium is less common in the solar system than 25 of the first 32 chemical elements. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;[...]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;⁷Li is one of the primordial elements (or, more properly, primordial nuclides) produced in Big Bang nucleosynthesis. A small amount of both ⁶Li and ⁷Li are produced in stars, but are thought to be burned as fast as produced. Additional small amounts of lithium of both ⁶Li and ⁷Li may be generated from solar wind, cosmic rays hitting heavier atoms, and from early solar system ⁷Be and ¹⁰Be radioactive decay. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, basically, Lithium is (barely) produced as David Zaslavsky told you in his answer, and the reason the production is low because Lithium is barely stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as @Scott Carnahan tells in his answer, the notion of lithium scarcity is linked with its repartition on earth. And the reason it is difficult to obtain is ultimately its  high chemical reactivity, which means that it is basically diluted everywhere, and is rarely concentrated in easy to mine deposits.  On the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lithium&quot; rel=&quot;nofollow&quot;&gt;same wikipedia page as above&lt;/a&gt;, they say :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Although lithium is widely distributed on Earth, it does not naturally occur in elemental form due to its high reactivity.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;[...] &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;According to the Handbook of Lithium and Natural Calcium, &quot;Lithium is a comparatively rare element, although it is found in many rocks and some brines, but always in very low concentrations. There are a fairly large number of both lithium mineral and brine deposits but only comparatively a few of them are of actual or potential commercial value. Many are very small, others are too low in grade.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="373" LastActivityDate="2011-06-16T16:55:14.000" />
  <row Id="11185" PostTypeId="1" CreationDate="2011-06-16T19:26:28.400" Score="1" ViewCount="88" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/10933/why-do-we-say-that-the-earth-moves-around-the-sun&quot;&gt;Why do we say that the earth moves around the sun?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I should preface by stating that I'm not a physicist and my knowledge of the theory of relativity is limited to what Encarta told me in the 90's (does everyone know what Encarta was?) and I don't even understand the Wikipedia entry!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Encarta had a video on the theory of relativity that basically explained in part that (i suppose in vacuum) there's no way to tell which body is in motion relative to another (without observing from outside the space in which they are moving i suppose).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so my question is whether this warrants a reassessment of our reliance on a heliocentristic model.  is it possible the planets do not revolve the way we've modeled them?  i'm not necessarily talking geocentric, just different from our current model.  should we be exploring this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ps -- while researching this question i came across geocentricity.com, the first sentence of which starts &quot;Of all the sciences, the Holy Bible...&quot; I'm really not interested in starting a religious debate so please keep answers strictly physics related (though i hope i don't have to even say that on &lt;strong&gt;&lt;em&gt;physics&lt;/em&gt;&lt;/strong&gt;.stackexchange.com!)&lt;/p&gt;&#xA;" OwnerUserId="4065" LastActivityDate="2011-06-16T19:26:28.400" ClosedDate="2011-06-16T19:55:11.357" Title="heliocentricity and the theory of relativity" Tags="&lt;orbital-motion&gt;&lt;relativity&gt;" CommentCount="6" />
  <row Id="11186" PostTypeId="2" ParentId="11160" CreationDate="2011-06-16T20:57:48.037" Score="2" ViewCount="" Body="&lt;p&gt;The short answer, as Karsus Ren says, is that it's electromagnetic induction that generates the electricity: the relative movement of a conductor and a magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Vladimir Kalitvianski points out, strictly speaking, windmills don't generate electricity - wind turbines generate electricity, and windmills mill grain. Still, it's become reasonably common to refer to wind turbines as windmills.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Crowley says, a wind turbine takes the wind's horizontal motion, turns it into rotary motion, which in some (but not all) turbines then goes through a gearbox. The motion then (depending on the nacelle design) either:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) rotates either a conducting coil (typically copper) through a magnetic field; or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) rotates magnets around a coil;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;either way, this generates electricity through electromagnetic induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnets may be permanent magnets (typically a neodymium alloy) or electromagnets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theoretical limit on how much of the wind's energy can be captured by the blades is called the Betz limit, and is an efficiency of ~59.3% (16/27).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Danish wind-turbine manufacturer Vestas has some very accessible stuff on their e-learning site at &lt;a href=&quot;http://www.vestaselearning.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.vestaselearning.com/&lt;/a&gt; It might seem banal at times, but don't be fooled - there is a lot of decent material in there.&lt;/p&gt;&#xA;" OwnerUserId="4066" LastActivityDate="2011-06-16T20:57:48.037" />
  <row Id="11187" PostTypeId="2" ParentId="10956" CreationDate="2011-06-16T21:15:30.237" Score="2" ViewCount="" Body="&lt;p&gt;The thermometer in wind will (at equilibrium) show a lower temperature than one in still air, if and only if it is a wet-bulb thermometer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wind's cooling effect comes from the wind evaporating moisture away. Your skin is moist, so a wind evaporates some of that moisture, taking away (on average) the higher-energy water, lowering the average energy of the remaining moisture - hence the cooling effect of sweating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A dry-bulb thermometer has no moisture to evaporate away, and so records no wind-chill effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A wet-bulb thermometer (e.g. a normal mercury thermometer with its bulb wrapped in waterlogged muslin) will record a wind-chill effect.&lt;/p&gt;&#xA;" OwnerUserId="4066" LastActivityDate="2011-06-16T21:15:30.237" CommentCount="1" />
  <row Id="11188" PostTypeId="1" CreationDate="2011-06-17T03:02:56.770" Score="7" ViewCount="268" Body="&lt;p&gt;I understand that a particle approaching a finite potential barrier with $E &amp;lt; V_0$, there still is a probability of finding the particle on the other side of the barrier due to quantum tunneling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, since the wavefunction is nonzero inside the barrier region, is it possible to actually make a position measurement and locate a particle inside the barrier?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean, if we can say that &quot;there is a nonzero probability that the particle is inside the barrier&quot;, surely this would suggest we can do so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, why not? Am I understanding the whole wavefunction / probability distribution thing right?&lt;/p&gt;&#xA;" OwnerUserId="2557" LastActivityDate="2011-06-18T12:59:37.063" Title="Can a particle be *physically* observed inside a quantum barrier?" Tags="&lt;quantum-mechanics&gt;&lt;tunneling&gt;" AnswerCount="7" CommentCount="8" FavoriteCount="2" />
  <row Id="11189" PostTypeId="2" ParentId="11188" CreationDate="2011-06-17T03:26:58.730" Score="-1" ViewCount="" Body="&lt;p&gt;Actuality as i understand it the problem is that as you approach the particle in any way to measure it IT wont be there, further more just looking at it will change the nature of the experiment. &lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-06-17T03:26:58.730" />
  <row Id="11190" PostTypeId="2" ParentId="11188" CreationDate="2011-06-17T05:15:47.557" Score="5" ViewCount="" Body="&lt;p&gt;If you have been following particle physics, you will know that a proton is composed of three &lt;a href=&quot;http://en.wikipedia.org/wiki/Quarks&quot; rel=&quot;nofollow&quot;&gt;quarks&lt;/a&gt; that are never able to come out of the nucleon formation, i.e. the potential well in a quantum mechanical picture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we know they exist if not by measurement inside that potential well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our tool is scattering. When one scatters an electron against a proton, in the potential barrier formulation, the trajectory of the electron after scatter reflects the wave function of the particles inside the nucleus because the electron wave function penetrates, i.e. has a solution, inside the potential barrier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first experiments decided that there are &lt;em&gt;partons&lt;/em&gt; inside the nucleus, because the scattering showed a hard core. Eventually these were studied in various experiments and the Standard Model of physics was established.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you are correct in your surmises.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: from thinking on the answer of Peter Morgan I will add that Xrays are an even simpler example. Even simple xrays of one's hand. The skin is a potential barrier and the xray photons wave function is scattered within, conveying information when they come out by their scattering angle and density,  on the bones etc. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-06-17T15:08:28.207" LastActivityDate="2011-06-17T15:08:28.207" CommentCount="4" />
  <row Id="11191" PostTypeId="1" CreationDate="2011-06-17T05:42:23.173" Score="0" ViewCount="34" Body="&lt;p&gt;How can we construct a windmill for household purpose? What are the components required to construct?&lt;/p&gt;&#xA;" OwnerUserId="4026" LastActivityDate="2011-06-17T05:42:23.173" ClosedDate="2011-06-17T06:15:12.300" Title="Can we build a windmill for a home purpose?" Tags="&lt;mechanics&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="11192" PostTypeId="2" ParentId="11188" CreationDate="2011-06-17T06:08:39.370" Score="3" ViewCount="" Body="&lt;p&gt;Yep, it is indeed possible for a measurement of the particle's position to reveal that it is inside the barrier, because the wavefunction there is nonzero - or more precisely, because the quantity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_\text{barrier} \langle\psi|x\rangle\langle x|\psi\rangle\mathrm{d}^n x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which represents the probability of the particle being measured within the barrier, is nonzero. Keep in mind that a &quot;measurement&quot; can be any interaction with another particle, it doesn't have to be performed by an actual measuring device. So if the particle interacts with the barrier itself, that counts. In that case you could have something like the particle tunneling partway through the barrier and possibly getting stuck in the middle (but you wouldn't see it because your measuring device can't get in there).&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-17T06:08:39.370" CommentCount="3" />
  <row Id="11193" PostTypeId="1" AcceptedAnswerId="11219" CreationDate="2011-06-17T08:20:14.753" Score="3" ViewCount="94" Body="&lt;p&gt;Feynman talking about electricity: &lt;a href=&quot;http://www.youtube.com/watch?v=qhh32JYkQPk&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=qhh32JYkQPk&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When brushing your hair, Feynman mentioned that a &quot;few&quot; electrons were transferred to the brush (or vice versa, can't remember off the top of my head how the convention goes). Clearly this will vary depending on how much you comb your hair, so my question is what is the approximate charge of the brush which would be required so that it would be noticeable (i.e. I could make pepper dance with the brush, or jiggle a piece of paper at close range).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if the question is silly :P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rofler&lt;/p&gt;&#xA;" OwnerUserId="4069" LastActivityDate="2011-06-17T20:58:58.497" Title="How many electrons are displaced when combing hair?" Tags="&lt;electrostatics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="11194" PostTypeId="1" AcceptedAnswerId="11199" CreationDate="2011-06-17T09:33:22.153" Score="18" ViewCount="444" Body="&lt;p&gt;Why commercial wind generators usually have just 2-3 blades?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having more blades would allow to increase power OR decrease diameter.&#xA;Decreased diameter would also reduce stress due to different wind speed on different height...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But despite that commercial generators have few blades...&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-12T15:39:00.457" LastActivityDate="2011-07-12T15:39:00.457" Title="Wind generators - why so few blades?" Tags="&lt;electricity&gt;&lt;fluid-dynamics&gt;&lt;renewable-energy&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="5" />
  <row Id="11196" PostTypeId="2" ParentId="11175" CreationDate="2011-06-17T12:11:08.090" Score="1" ViewCount="" Body="&lt;p&gt;Where ever You read that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;electrons are negatively charged&#xA;  particles and the positively charged&#xA;  cylindrical anode develops a strong&#xA;  electric field that exerts a force on&#xA;  the electrons, accelerating them along&#xA;  the tube&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;its plain nonsense. If a heated cathode (called &quot;thermionic&quot; for strange reasons&#xA;in the anglosaxon literature) is placed within a cylinder with positive &#xA;voltage, the electrons will fly to that cylinder along the field lines (at first) between &#xA;the emitting surface and the cylinder. Of course this is not the way to &#xA;make an &quot;electron gun&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those, the emission spot is enclosed in a cylinder with a hole opposite &#xA;to the emitting spot, and this Wehnelt cylinder is &lt;strong&gt;negatively&lt;/strong&gt; charged. &#xA;It acts like grid in a vacuum tube and as a electron lens, focusing the beam. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a sketch look here: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://de.wikipedia.org/wiki/Elektronenkanone&quot; rel=&quot;nofollow&quot;&gt;http://de.wikipedia.org/wiki/Elektronenkanone&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and for the text look at the english (less informative) version. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-06-17T14:14:35.377" LastActivityDate="2011-06-17T14:14:35.377" />
  <row Id="11197" PostTypeId="1" AcceptedAnswerId="11198" CreationDate="2011-06-17T12:12:41.270" Score="2" ViewCount="67" Body="&lt;p&gt;If I've got an instance of a fundamental particle, how can I separate out the measurements of these three concepts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I think) I understand the theory behind them, and why the particles in the standard model are predicted to have the values they do. However, the process of validation of these numbers eludes me. How have they been historically measured?&lt;/p&gt;&#xA;" OwnerUserId="2235" LastActivityDate="2011-06-17T12:52:51.470" Title="How does one experimentally determine chirality, helicity and spin?" Tags="&lt;spin&gt;&lt;chirality&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="11198" PostTypeId="2" ParentId="11197" CreationDate="2011-06-17T12:46:00.913" Score="1" ViewCount="" Body="&lt;p&gt;For spin measurements the original experiment was the Stern-Gerlach experiment in which you will see that a prior unpolarized beam will split up in two (Spin up and down) orientations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;see: &lt;a href=&quot;http://en.wikipedia.org/wiki/Stern%E2%80%93Gerlach_experiment&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Stern%E2%80%93Gerlach_experiment&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For helicity, a very ingenious and fascinating experiment is the famours Goldhaber experiment that uses a very peculiar set of elements and also the Mößbauer effect to measure the helicity of neutrinos. The helicity is the projection of the Spin onto the momentum direction, and thus if you measure spin and momentum, you can compute helicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Link to original paper here: &lt;a href=&quot;http://www.bnl.gov/nh50/&quot; rel=&quot;nofollow&quot;&gt;http://www.bnl.gov/nh50/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People often confuse helicity and chirality they are &lt;em&gt;only&lt;/em&gt; the same in the case of massless particles. This is also the reason why we know that (at least interacting) neutrinos are always left-handed (the famous $$SU(2)_L$$ ), at least if we assume that neutrinos are massless (which is almost the case, but not strictly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast to helicity for massive particles (where you can always boost into a frame where you change the sign of the momentum direction and thus change helicity), chirality is a Lorentz-invariant property of the particle.&lt;/p&gt;&#xA;" OwnerUserId="2151" LastEditorUserId="2151" LastEditorDisplayName="" LastEditDate="2011-06-17T12:52:51.470" LastActivityDate="2011-06-17T12:52:51.470" />
  <row Id="11199" PostTypeId="2" ParentId="11194" CreationDate="2011-06-17T13:09:29.247" Score="14" ViewCount="" Body="&lt;p&gt;More blades give you more cost, but very little increase in efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three blades turns out to be the optimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With four or more blades, costs are higher, with insufficient extra efficiency to compensate. Edit: prompted by a comment, here's some elucidation - this is more expensive per unit electricity generated, if you go for more, but shorter, blades: if you have 4 shorter blades (rather than three longer ones), the blades are sweeping through a smaller volume of air (i.e. an amount of air with a lot less energy), swept area being proportional to the square of the radius. And the efficiency is only a few percent higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You get higher mechanical reliability with three blades than with two: with two blades, the shadowing effect of tower &amp;amp; blade puts a lot of strain on the bearings. So although it costs more to make a three-bladed turbine, they tend to have a longer life, lower maintenance needs, and thus on balance reduce the unit cost of electricity generated, as the increased availability and reduced maintenance costs outweigh the extra cost of the third blade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit2: For the nitty-gritty of wind-turbine aerodynamics, wikipedia isn't a bad place to start: &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Wind_turbine_aerodynamics&amp;amp;oldid=426555179&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/w/index.php?title=Wind_turbine_aerodynamics&amp;amp;oldid=426555179&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4066" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-06-17T15:42:20.277" LastActivityDate="2011-06-17T15:42:20.277" CommentCount="4" />
  <row Id="11200" PostTypeId="2" ParentId="11188" CreationDate="2011-06-17T13:13:55.883" Score="2" ViewCount="" Body="&lt;p&gt;If one can observe individual events on the other side of the potential barrier from the source, but only when the source is turned on, then one might say that &lt;em&gt;something&lt;/em&gt; must have caused the events. On the macroscopic scale, it must have been the source because it's only when the source is turned on that there are events. One might say, further, on the microscopic scale, that whatever it was that caused the events must have come &lt;em&gt;through&lt;/em&gt; the space between the source and the apparatus in which the events happened, it can't have just jumped the gap. In this sense, we might say that detecting a particle on the other side of the barrier shows that a particle must have passed through the barrier, so this &lt;em&gt;is&lt;/em&gt; a detection of a particle in the barrier. We know, however, that thinking in terms of something like classical particles is problematic, so that it's best not to think of there being a particle that travels from one place to another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I liked Anna v's Answer enough to upvote it, but I point out that it is only by a specific kind of semi-classical inference that we can say that an event on the other side of the lab (which we say is caused by a scattered particle) means that there was another particle that caused the scattering. In this case, where there is a potential barrier, whatever material apparatus implements the potential barrier will get in the way of being able to do such a scattering experiment in any clean way. It would only be by subtracting the background caused by the material apparatus that we could tell that there were events that were caused by the &quot;particles that classically couldn't be there because of the potential barrier&quot;. The significance of the potential barrier for the interpretation of the recorded events in terms of particles is that there is &lt;em&gt;more&lt;/em&gt; background. [There's never &lt;em&gt;no&lt;/em&gt; background to subtract, but often the background can be made small enough that we don't have to pay much attention to it. A common failing of interpretations, IMO, is to ignore the background in principle.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we come to the statistical character of QM. Because we have to subtract the background, we cannot say that any &lt;em&gt;individual&lt;/em&gt; event is caused by an &lt;em&gt;individual&lt;/em&gt; particle scattering another particle. In that sense I think we cannot say that we can &quot;locate [an individual] particle inside the barrier&quot; by scattering (or by other interactions inside the barrier), because any individual event may have been caused by the background. Where there is more background, as there must be when we are trying to detect inside a potential well, we are less able to attribute individual events to identifiable individual particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, we &lt;em&gt;can&lt;/em&gt; establish that a given wave function is a very effective tool for predicting just how many detection events we would record if we put a given type of detection apparatus at any given point, and if we ensure that there is very little background. Then, we can extrapolate to say what we &lt;em&gt;would&lt;/em&gt; observe if we put that type of apparatus in a place where we in fact cannot put it. That's not &quot;physical&quot; in the sense that I think you mean it, but it's mathematically reasonable just up to the point that we think about actually implementing the measurement, and it lets us imagine how we would engineer new experiments in a useful way. I take this to be the basis of David Zaslavsky's Answer (which is more-or-less OK, but I did not upvote it because I think it does not enough address your question).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IMO, however, without explication here, it's good to &lt;em&gt;start&lt;/em&gt; to think in terms of the quantum field being the intermediary when modelling an experiment, instead of particles, even when you're doing 2nd year undergraduate Physics.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-06-17T13:13:55.883" CommentCount="2" />
  <row Id="11201" PostTypeId="2" ParentId="11188" CreationDate="2011-06-17T13:45:39.293" Score="5" ViewCount="" Body="&lt;p&gt;Yes you appear to &quot;understand the whole wavefunction / probability distribution thing right&quot;, and it is possible to measure a probability of presence in the forbidden region, but this probability is usually small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know whether any experiment has shown it with &quot;real particle&quot; like atoms, electrons or neutron, but it has been definitely shown with photons. The &quot;classically forbidden region&quot; in the particle pictures corresponds to &lt;a href=&quot;http://en.wikipedia.org/wiki/Evanescent_waves&quot; rel=&quot;nofollow&quot;&gt;evanescent waves&lt;/a&gt;, and there have been experiment where evanescent waves have been used to excite the fluorescence of atoms. Each photon emitted by an atom in the evanescent wave can then be seen as a position measurement of a photon in the evanescent wave region.&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2011-06-17T13:45:39.293" />
  <row Id="11202" PostTypeId="2" ParentId="11194" CreationDate="2011-06-17T14:00:25.637" Score="0" ViewCount="" Body="&lt;p&gt;&quot;If a turbine has an efficiency of 50 percent, the theoretical power output is found by multiplying the square of the turbine radius by the cube of the wind speed&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read more: Wind Turbine Capacity Factors | eHow.com &lt;a href=&quot;http://www.ehow.com/info_8484534_wind-turbine-capacity-factors.html#ixzz1PXduF0g2&quot; rel=&quot;nofollow&quot;&gt;http://www.ehow.com/info_8484534_wind-turbine-capacity-factors.html#ixzz1PXduF0g2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you will want to create the largest radius but keep the weight to a minimum (so that it will actually start to turn in less than 100mph wind) so lighter design is the key hence fewer blades&lt;/p&gt;&#xA;" OwnerUserId="4073" LastActivityDate="2011-06-17T14:00:25.637" CommentCount="2" />
  <row Id="11203" PostTypeId="2" ParentId="11194" CreationDate="2011-06-17T14:38:16.530" Score="1" ViewCount="" Body="&lt;p&gt;I am not an aerodynamical engineer, but off the top of my head there are several additional factors worth considering&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Economics&lt;/em&gt; The optimisation maybe based on most kWh capacity per dollar invested, which would put additional constraints.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Mechanical stress of the tower&lt;/em&gt; More area presented to the wind means more stress on the tower holding the turbine up. Furthermore, presumeably there are issues related to how frequently the blade passes in front of the tower. &lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Turbulence/wake&lt;/em&gt; the closer the various blades are together, the more likely small scale effects like turbulence and wake of the adjacent blades will effect the aerodynamics. Turbulence, for example, can lead to inefficiencies and also differential stress along the body of the blade (compared to laminar flow). Having a few slow-moving, long blades can therefore be a plus to having a lot of fast-moving, short blades. (for the same mass/area/force, many short blades have smaller angular momentum and hence will spin faster in the end, compared to few long blades)&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Mechanical stress on the blades&lt;/em&gt; to fit more blades at the hub, each blade would have to be thinner in cross section (or you can make the hub really big, but that would just present a huge area for the wind to strike that does no work), so mechanically you probably cannot fit too many blades on the turbine. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="829" LastActivityDate="2011-06-17T14:38:16.530" />
  <row Id="11204" PostTypeId="2" ParentId="11188" CreationDate="2011-06-17T14:50:07.850" Score="0" ViewCount="" Body="&lt;p&gt;Frédéric's answer is the best one. I myself was going to write something like scattering light in the forbidden region by admixtures or photo-reactions in the forbidden region. Note, to cause a photo-reaction, you have to have sufficient energy. In a quasi-stationary case the energy is $E=\hbar \omega$. So it is the same as outside the barrier. No energy conservation violation occurs. You just observe a particle in a classically forbidden region, but quantum mechanically it is not forbidden. That is why the probability is not zero.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-06-17T14:50:07.850" />
  <row Id="11205" PostTypeId="2" ParentId="11147" CreationDate="2011-06-17T14:51:20.457" Score="1" ViewCount="" Body="&lt;p&gt;If your atom were alone, it would be described by its Hamiltonian $H_{a}$. The eigenstate of its Hamiltonian correspond to the energy level of the atoms and would be, by definition, steady states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as stated by @Vladimir Kalitvianski, the atom is not alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you are performing a CQED (cavity QED) experiment with your atom. You put your atome in a single-mode optical cavity. The field of the cavity is described by its Hamiltonian $H_c=\sum_{n\ge0} n\hbar\omega|n\rangle\langle n|$, where $|n\rangle$ is tha n-photons Fock state. If the atom and the cavity interact, you need to describe this interaction by another term $H_i$ such that the global Hamiltonian is now&#xA;$$ H=H_a+H_c+H_i.$$&#xA;Because of the $H_i$ term, neither the eigenstates of $H_a$ (a.k.a the atomic energy levels)&#xA;nor the eigenstates of $H_c$ (a.k.a the Fock states) are eigenstate of $H$. They are not steady state anymore. For example, for a two-level atom described by its ground state $|g\rangle$ and its excited state $|e\rangle$, two of these steady states are quantum superpositions of $|e\rangle_a|0\rangle_c$ and $|g\rangle_a|1\rangle_c$. One of the consequences is that we can have &lt;a href=&quot;http://en.wikipedia.org/wiki/Vacuum_Rabi_Oscillation&quot; rel=&quot;nofollow&quot;&gt;vacuum Rabi oscillation&lt;/a&gt;: If you put an excited atom in a cavity with no light inside ($|e\rangle_a|0\rangle_c$), the state of the atom+cavity ensemble oscillates between $|e\rangle_a|0\rangle_c$ and $|g\rangle_a|1\rangle_c$. In other words, the atom releases the photon and recaptures it periodically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want more details, look for Jaynes-Cummings model on the internet, on &lt;a href=&quot;http://en.wikipedia.org/wiki/Jaynes%E2%80%93Cummings_model&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; or in your favourite quantum mechanics textbook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have several cavities, and/if your cavity is big enough, the atom is coupled to several modes, and the recapture of the photon takes longer. You can then make your cavity grow such that your atom &quot;sees&quot; many modes, which you can describe by a continuum, with a density of mode. From the atom's point of view, there is no difference between sitting of the middle of many infinitely large cavities (one per direction of the space) and sitting in free space.&#xA;In that limit, the oscillation becomes an exponential decay, the recapture of the photon being postponed to an infinitely distant time. &#xA;To compute that decay, you should use &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermi_golden_rule&quot; rel=&quot;nofollow&quot;&gt;Fermi Golden Rule&lt;/a&gt;, invented by Dirac [!] to get the decay rate.&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2011-06-17T14:51:20.457" />
  <row Id="11206" PostTypeId="1" AcceptedAnswerId="11207" CreationDate="2011-06-17T15:44:17.420" Score="2" ViewCount="135" Body="&lt;p&gt;An entry level confusion about spacetime. I understand that a 4-vector describes a point or event in spacetime. But I've also read (Bertschinger, 1999) that re spacetime &quot;we are discussing tangent vectors that lie in the tangent space of the manifold at each point&quot;. If a point/event is described by a single 4-vector, what are all these tangent vectors that lie on the same point? Do they have different co-ordinates to the 'point/event' 4-vector? Could I also ask how a 4-vector 'contains' any sense of direction (I'm thinking here of a vector having direction and magnitude)? &lt;/p&gt;&#xA;" OwnerUserId="4075" LastActivityDate="2011-06-18T17:08:05.987" Title="Nature of spacetime 4-vector and tangent space?" Tags="&lt;spacetime&gt;&lt;relativity&gt;&lt;vectors&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="11207" PostTypeId="2" ParentId="11206" CreationDate="2011-06-17T16:23:55.293" Score="6" ViewCount="" Body="&lt;p&gt;Your confusion comes from the difference between &lt;em&gt;special&lt;/em&gt; and &lt;em&gt;general&lt;/em&gt; relativity. In &lt;em&gt;special&lt;/em&gt; relativity, the space-time manifold is assumed to carry the structure of 4-dimensional Minkowski space, which has the nice property that it is canonically identified with its own tangent space at the origin (since it is a vector space). So in &lt;em&gt;special&lt;/em&gt; relativity you can speak of a space-time event as a 4-vector, and you can also speak of &lt;em&gt;global&lt;/em&gt; Lorentz transformations (by doing the &lt;em&gt;local&lt;/em&gt; Lorentz transformation in the tangent space and propagating the transformation to the whole space-time using the canonical identification). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;em&gt;general&lt;/em&gt; relativity, however, the space-time is allowed to be an arbitrary Lorentzian manifold (what we do is break the &lt;em&gt;global&lt;/em&gt; Lorentz symmetry of Minkowski space and require it to only hold infinitesimally, i.e. on the tangent space), and you don't have a canonical way of identifying the entire space-time with the tangent space of a fixed point. Therefore you cannot speak of a space-time event (now just a point in your space-time manifold) as a 4-vector!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; Let me try to make the difference between an affine and non-affine space more apparent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us start by considering a two dimensional manifold. In fact, we'll just compare the usual flat plane and the sphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the plane, we can fix an arbitrary point and call it the origin. Starting from this origin, pick a direction, and call it the $x$ direction. Consider the following two paths. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;First you go 1 unit in the $x$ direction. Turn left for 90 degrees. travel another unit.&lt;/li&gt;&#xA;&lt;li&gt;First you face the $x$ direction. Now turn left for 90 degrees, travel one unit. Turn right now for 90 degrees and travel another unit. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These two operations take you to the same place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the sphere, however, you can again fix an arbitrary point and call it the &quot;origin&quot;, and a direction which we call &quot;$x$&quot;. Now consider&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;First you go a quarter the way around the sphere in the $x$ direction. Turn left for 90 degrees, travel another quarter of the circumference. &lt;/li&gt;&#xA;&lt;li&gt;First you face $x$ direction, turn left for 90 degrees, travel a quarter of the circumference. Now turn right and travel a quarter of the circumference. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(Try to trace this out on a tennis ball or on an orange.) You don't end up at the same place! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, going back to the plane, if instead you start by going &quot;face $x$. Turn left for 45 degrees. Travel for $\sqrt{2}$ units.&quot; You will again end up at the same place as the previous two tries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But on the sphere, if you start by &quot;face $x$. Turn left for 45 degrees. Travel for $\sqrt{2}$ times a quarter circumference.&quot; You will end up somewhere else entirely again compared to the previous two tries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this mean? On a flat plane, from the above demonstration, after fixing an origin and a direction $x$, we can uniquely describe every point on the plane as &quot;certain units in the $x$ direction, and then certain units to the left of that&quot;. And it doesn't matter at all in which orders you zigzag to the destination. This identifies the plane with a vector space. That is, you can say that &quot;the displacement from point $p$ to point $q$ is the vector $\vec{v}$, and the displacement from point $q$ to point $r$ is the vector $\vec{w}$, so the displacement from the point $p$ to point $r$ is the sum of vectors $\vec{v} + \vec{w} = \vec{w} + \vec{v}$&quot;. So after fixing a point as the origin, we can describe all other points as a vector displacement relative to this origin, and add and subtract the vectors as appropriate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the surface of the sphere (a curved manifold), however, the above example shows that the intuition we learned from studying classical mechanics about using vector addition for the &quot;displacement vector&quot; cannot hold for the positions of points. So you shouldn't think about space-time events in general relativity (which is a point on some possibly curved manifold), as vectors relative to a fixed origin. (Because with vectors you would be tempted to add them and so forth.)&lt;/p&gt;&#xA;" OwnerUserId="829" LastEditorUserId="829" LastEditorDisplayName="" LastEditDate="2011-06-18T17:08:05.987" LastActivityDate="2011-06-18T17:08:05.987" CommentCount="6" />
  <row Id="11208" PostTypeId="1" CreationDate="2011-06-17T18:22:15.820" Score="2" ViewCount="158" Body="&lt;p&gt;What is the best form for a kettle, that is, to have the water boil the fastest? I am particularly interested in the following case: for one given kettle (whose volume is constant) containing different volumes of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not like to wait for my coffee in the morning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT] Suppose that the heat source is uniform and flat only under the kettle.&lt;/p&gt;&#xA;" OwnerUserId="2499" LastEditorUserId="2499" LastEditorDisplayName="" LastEditDate="2011-06-20T12:36:39.087" LastActivityDate="2011-08-19T19:32:44.890" Title="the form of a kettle" Tags="&lt;thermodynamics&gt;&lt;fluid-dynamics&gt;&lt;water&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="11209" PostTypeId="2" ParentId="11206" CreationDate="2011-06-17T18:59:08.873" Score="1" ViewCount="" Body="&lt;p&gt;Think of a liquid that is flowing in two dimensions on a flat surface. At every point we can draw a vector that points in the direction that the fluid is flowing in at that point. That is a vector in the vector space at that point, the tangent space. There is a tangent space at every point, because at every point the fluid &lt;em&gt;could&lt;/em&gt; be flowing in any direction, although an equation of motion might require, for example, continuity from point to point. An example of a vector flow (found through Google images for &quot;fluid flow&quot;) is &lt;a href=&quot;http://www.dstu.univ-montp2.fr/geofracnet/Images/connollyflownet2.jpg&quot; rel=&quot;nofollow&quot;&gt;http://www.dstu.univ-montp2.fr/geofracnet/Images/connollyflownet2.jpg&lt;/a&gt;. Graphically, of course, the tangent vectors to the flow can only be displayed at a finite number of points, but from the point of view of a differential equation there's a flow vector at every point. All the tangent spaces are exactly the same as all the others. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;vectors&quot; used to describe the points are vectors in the vector space at one particular point, the center of the coordinate system, with the vector describing the location of the other point relative to the center of the coordinate system as distances in two directions that aren't parallel. The four-dimensional case is no different provided we concern ourselves only with vector spaces, but the idea of a Lorentzian distance is of course different from the idea of a Euclidean distance. In coordinate terms, we could talk of two functions, $V_1(x,y)$ and $V_2(x,y)$, where $x$ and $y$ are components of the position of a point, and $V_1(x,y)$ and $V_2(x,y)$ are components of a vector at that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A vector in 2 or in 4 dimensions can of course represent something other than a velocity field. In due course you will get straight what can be represented by vectors and what requires tensors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine a fluid flowing on the surface of a sphere. Again there is a tangent space at every point on the surface of the sphere, and all the tangent spaces are exactly the same as the others, but now it's harder to describe explicitly the way in which the many tangent spaces are the same as each other. Saying which point is which by using vectors in a tangent space at a specific point is also more problematic.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-06-17T18:59:08.873" CommentCount="3" />
  <row Id="11210" PostTypeId="1" CreationDate="2011-06-17T19:02:29.497" Score="0" ViewCount="51" Body="&lt;p&gt;I'm starting Physics, and I don't understand why the acceleration along the x-axis is zero for an object thrown near the surface of the Earth. This may be problem specific, but I wouldn't know since I am just beginning.&lt;/p&gt;&#xA;" OwnerUserId="4078" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-17T20:42:19.003" LastActivityDate="2011-06-17T20:42:19.003" Title="In a 2D problem with a thrown object, why is the acceleration along the x-axis equal to 0?" Tags="&lt;mechanics&gt;&lt;kinematics&gt;&lt;acceleration&gt;&lt;projectile&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11211" PostTypeId="1" AcceptedAnswerId="11228" CreationDate="2011-06-17T19:09:48.007" Score="3" ViewCount="87" Body="&lt;p&gt;This comes from a discussion forum, where a friend of mine asked the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We can see objects in space billion of light years away, right? I started wondering about that.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If you take 2 objects in space, the other should be able to see the other no matter what angle in degrees you position it at. That would almost seem to imply that light is being sent out in an infinite number of degrees/angles from the source. But that cannot be true because energy cannot be infinite.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If the observer goes out far enough from the source, would there be gaps in the light? Could you pick a viewing angular degree (of extremely high angular precision) where there's no light?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm actually quite curious about this question myself and really have no answer, and the discussion hasn't really yielded a satisfying answer.  So I figured I would bring it here on behalf of my friend and to sate my own curiosity.&lt;/p&gt;&#xA;" OwnerUserId="4079" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-17T20:36:57.443" LastActivityDate="2011-06-18T10:53:21.803" Title="Distant bodies emitting photons" Tags="&lt;electromagnetism&gt;&lt;light&gt;&lt;photons&gt;" AnswerCount="3" />
  <row Id="11212" PostTypeId="2" ParentId="11210" CreationDate="2011-06-17T19:30:33.563" Score="3" ViewCount="" Body="&lt;p&gt;Because gravity points purely along the y-axis. If you ignore air drag or thrusters or engines (as you would in a projectile problem), there are no forces along the x-direction, and hence no acceleration along the x-direction.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-06-17T19:30:33.563" />
  <row Id="11213" PostTypeId="2" ParentId="11211" CreationDate="2011-06-17T19:38:16.737" Score="1" ViewCount="" Body="&lt;p&gt;You're looking at this the wrong way. The total energy being sent out is finite. And the total energy contained within a solid angle $d\Omega$ is proportional to $d\Omega$. So the energy contained in a very tiny solid angle gets tinier as the angle is decreased and approaches zero as the angle approaches zero. In other words, the smaller a cone you count photons in, the lesser photons you'll find. This ensures that the total energy (obtained by integrating over all solid angles) is finite as we started out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how almost any integral you'll find in physics works.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-06-17T19:38:16.737" />
  <row Id="11214" PostTypeId="2" ParentId="11210" CreationDate="2011-06-17T20:05:30.323" Score="2" ViewCount="" Body="&lt;p&gt;The acceleration in the x-axis can be non-zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most 2-D physics problems using Cartesian coordinates, the convention is that the y-axis is vertical and the x-axis is horizontal (doesn't need to be that way; it is just the convention used).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the y-axis is TYPICALLY the name given to the vertical axis.  And gravity on a free falling or stationary object TYPICALLY works only in this vertical plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to modify your physics problems, say, to make a ball roll freely down a ramp in the earth's gravitational field, you would find that you get acceleration in both the x and y axes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you get to 3-D physics problems, the convention changes: Typically x and y are the assigned names for the two horizontal axes of the Cartesian coordinate system, and the z-axis is the vertical one.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-06-17T20:12:24.790" LastActivityDate="2011-06-17T20:12:24.790" />
  <row Id="11215" PostTypeId="1" CreationDate="2011-06-17T20:23:44.217" Score="1" ViewCount="35" Body="&lt;p&gt;The full MSSM contains 120 parameters. In SUSY searches, one usually picks a model like MSUGRA which makes a few assumptions and only has 5 free parameters like $m_0$, $m_{1/2}$, ....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm wondering if there is a way to look at supersymmetry in a more phenomenological way, and parametrize it by (low-energy) coupling constants or branching ratios, cross sections of particles, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way it's done now is that you, for example, make a couple of assumptions at the GUT scale and then work your way down using the renormalization group equations. Also you make specific assumptions about the breaking mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'd like to do is to say: Give me a model / a subset of parameter space where we have light chargino like particles decaying such and such. Or where those particles have couplings in such and such region. Or where the lightest squark like particle is heavier than X GeV. One would probably have to speak of a &quot;lightest strongly interacting superpartner&quot; etc. since the parametrisation would not be sufficent to determine the model completely. As an experimentalist, I could live with it - it would even be interesting - if the theory (after implementation in a MC generator) would give me several different sets of MC for a given query, say one with a lowest cross section bound, one with the highest cross section, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this is to simply assess what kinds of SUSY are findable by our accelerators, and also to have a parametrisation that gives certain experimentally similar final states similar model parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is someone working on this, or is it a pipe dream?&lt;/p&gt;&#xA;" OwnerUserId="825" LastActivityDate="2011-06-17T20:23:44.217" Title="Parametrisation of general MSSM/SUSY based on collider experiment observables" Tags="&lt;particle-physics&gt;&lt;supersymmetry&gt;" />
  <row Id="11216" PostTypeId="1" CreationDate="2011-06-17T20:34:54.933" Score="0" ViewCount="57" Body="&lt;p&gt;Pop-sci ST books say, &quot;At every point of space, there is a Calabi-Yau space...&quot; or similar. But uncertainty p. and noncommgeom say, no points in space, only planckons (not my coinage). Is 'point of space' (meaning real, not abstract/mathematical space) inexact terminology, or a category error, like &quot;an atom of mayonnaise'- only worse? So I speculate that C-Y space can not be a continuum either, but must consist of discrete calabicles (my coinage). &lt;/p&gt;&#xA;" OwnerUserId="4080" LastActivityDate="2011-06-17T20:34:54.933" Title="the search for calabicles" Tags="&lt;spacetime&gt;" />
  <row Id="11218" PostTypeId="2" ParentId="11211" CreationDate="2011-06-17T20:36:24.700" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If the observer goes out far enough from the source, would there be gaps in the light? Could you pick a viewing angular degree (of extremely high angular precision) where there's no light?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, you can't. Light is a wave, and accordingly it spreads out in all directions and fills all of space (if you assume there are no other objects that might cast shadows). So no matter how far out you go, you will never fully escape the light wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the intensity of the light (energy per unit detector area per unit time) does become smaller as you move further away. So for any given intensity $I_0 &amp;gt; 0$, you can put your detector far enough away that it will detect an intensity less than $I_0$. But the intensity never actually drops to zero.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;That would almost seem to imply that light is being sent out in an infinite number of degrees/angles from the source. But that cannot be true because energy cannot be infinite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In short, your friend's misconception is that it would require an infinite amount of energy for light to be sent out in an infinite number of directions. That is not the case.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-17T20:36:24.700" />
  <row Id="11219" PostTypeId="2" ParentId="11193" CreationDate="2011-06-17T20:58:58.497" Score="3" ViewCount="" Body="&lt;p&gt;I'm assuming you just want an order of magnitude answer, because there's no specific numbers that would allow us to be more precise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Static_electricity#Energies_involved&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Static_electricity#Energies_involved&lt;/a&gt; a common voltage for everyday static electricity is 10,000 V, and the amount of energy released in a discharge is less than a joule, usually something like 0.1 J. A simple calculation tells us that the charge involved is $Q = 2 E / V$ or 20 microcoulombs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many electrons in 20 microcoulombs? A single electron is $2 \times 10^{-19}$ C, so to make $2 \times 10^{-5}$ C you need about $10^{14}$ electrons, or a hundred trillion. So give or take a few factors of ten, this is the number of electrons transferred from the hair to the brush. It's much more than a billion, but much much less than Avogadro's number.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-06-17T20:58:58.497" CommentCount="1" />
  <row Id="11221" PostTypeId="2" ParentId="11188" CreationDate="2011-06-17T22:05:25.563" Score="0" ViewCount="" Body="&lt;p&gt;To be close to second year undergrad physics, consider a non-relativistic electron with energy $E$, bounded to a double well potential $V({\bf r})$ with a classically forbidden tunneling region with potential energy $V_0$ in between, i.e., $E&amp;lt;V_0$. (Let us assume for simplicity that the full potential profile $V({\bf r})\leq V_0$, i.e., $V_0$ is a global maximum for the profile.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.physics.orst.edu/~minote/COURSES/ph427/lib/exe/fetch.php?media=double_square_well2.png&quot; alt=&quot;An example of a double well&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure 1: An example of a double well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I read the question, Josh Chen is not disputing that an electron prepared in one well can reappear in the other well. Instead, the question is that since the integral of the square of the wave function over the classically forbidden tunneling region&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{\{{\bf r}\in\mathbb{R}^3\mid V({\bf r}) &amp;gt; E\}} d^{3}r \ |\Psi({\bf r},t)|^2~&amp;gt;~0,$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is strictly non-zero, does that actually mean experimentally that there is a non-negative probability to find the electron inside the classically forbidden tunneling region as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Born_rule#The_rule&quot; rel=&quot;nofollow&quot;&gt;Born rule&lt;/a&gt; tells us, and how would one measure that probability, at least in principle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, the Born rule holds also in this situation. To measure the position of the electron, we will here use a photon with wave length $\lambda$ and of energy &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_{\lambda}~=~hf~=~\frac{hc}{\lambda}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will assume that the energies involved&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|E|, |V|, E_{\lambda} ~\ll~ E_0=m_0 c^2$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;are much smaller than the rest energy $E_0$ of the electron, so we can treat the electron using non-relativistic quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electron's wave function $\Psi({\bf r},t)$ is exponentially decaying in the classically forbidden tunneling region with a characteristic tunneling penetration depth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta ~\sim~ \frac{h}{\sqrt{2m_0(V-E)}}~=~\frac{hc}{\sqrt{2E_0(V-E)}}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Since we are not really interested in the possibility that the electron could reach the other well, let us for simplicity assume that the electron penetration depth $\delta&amp;lt;\Delta$ is smaller than the separation $\Delta$ of the two wells, i.e. we are effectively studying a single well.) To use the photon as a 'microscope' in order to claim that we have detected the electron inside the classically forbidden tunneling region, the 'microscope' should have a resolution better than the electron penetration depth. In other words, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda \ll \delta \qquad \Leftrightarrow \qquad E_{\lambda} \gg \sqrt{E_0(V_{0}-E)} $$&#xA;$$ \Rightarrow \qquad \frac{E_{\lambda}}{E_0} \gg \sqrt{\frac{V_{0}-E}{E_0}} &amp;gt; \frac{V_{0}-E}{E_0}&#xD;&#xA;\qquad \Rightarrow \qquad E+ E_{\lambda}\gg V_{0},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e., the photon could knock the electron completely out of the well profile, so that the electron continues to spatial infinity. In principle, the incoming photon could be aimed at the classically forbidden tunneling region, and we could have prepared detectors in a $4\pi$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Solid_angle&quot; rel=&quot;nofollow&quot;&gt;solid angle&lt;/a&gt; to capture and measure energy and momentum of all outgoing particles (the electron plus photons), and then calculate backwards to determine that a scattering event must have taken place inside the classically forbidden tunneling region. The missing energy between the incoming and the outgoing particles will be equal to the classically forbidden energy $E-V_{0}&amp;lt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if we had used soft photons with energy $E_{\lambda}&amp;lt;V_{0}-E$, the above inequalities get reversed, and the resolution will be too poor to determine whether the electron is inside or outside the classically forbidden tunneling region, cf. the uncertainty principle.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-18T12:59:37.063" LastActivityDate="2011-06-18T12:59:37.063" />
  <row Id="11222" PostTypeId="1" AcceptedAnswerId="11230" CreationDate="2011-06-17T22:41:34.830" Score="1" ViewCount="134" Body="&lt;p&gt;Let foo be some unit of distance and bar be some unit of time which have been chosen so that the speed of light c = 1 foo/bar.  Position several observers along a line each separated by one foo, and place light sources some distance apart amongst the observers, initially off.  All observers and light sources are stationary with respect to one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/3GK3n.gif&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in my diagram the tick marks are the observers, and the scale is one foo.  For the example, the light sources are ten foo apart at points L and R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now at time 0 bars, ignite both light sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the diagram, at time 4 bars, A has observed source L but not R, C has observed R but not L, and B has not yet observed either light source.  At time 5 bars, B observes both light sources, and at time 6 bars all three labeled observers have observed both light sources.  At time 7 bar, a total of five observes (A, B, C, and two unlabeled) have observed both sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let r be the length of the line segment containing all observers that have observed both light sources.  In other words, r is the length of the intersection of the event horizons on the two ignition events (I think I am using that term right, if not, please let me know).  It seems that dr/dt = 2 foo/bar; for instance, in the 1 bar interval from 6 to 7, the change in r was 2 foo.  As c = 1 foo/ bar, r is growing at twice the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since r is not a particle and cannot be used to transmit information amongst the observers, is there any contradiction here?  Have I made any logical falicies?&lt;/p&gt;&#xA;" OwnerUserId="4083" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-18T05:15:56.807" LastActivityDate="2011-06-18T05:49:18.050" Title="Thought experiment that seems to involve something growing at twice the speed of light. Is anything wrong?" Tags="&lt;speed-of-light&gt;&lt;velocity&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11223" PostTypeId="2" ParentId="11222" CreationDate="2011-06-17T23:16:08.993" Score="5" ViewCount="" Body="&lt;p&gt;The theory of relativity is often expressed as &quot;&lt;strong&gt;Nothing&lt;/strong&gt; can be faster than the speed of light&quot;. This is wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can always define certain things that seem to &quot;move&quot; faster than the speed of light. For example, if you shine a laser pointer at the moon and wiggle it around, the little red dot on the moon's surface can &quot;move&quot; faster than the speed of light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better way to express that the speed of light poses a certain limit is to say that two space-time events $(x_1, y_1, z_1, t_1)$ and $(x_2, y_2, z_2, t_2)$ can only be in a &lt;em&gt;causal relationship&lt;/em&gt; if the distance of the space-points is so low that light can move from one to the other in time at most $t_2 - t_1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between two events in space-time is defined as &#xA;$$c^2 (t_1 - t_2)^2 - (x_1 - x_2)^2 - (y_1 - y_2)^2 - (z_1 - z_2)^2 - $$.&#xA;Distances can be &lt;strong&gt;space-like&lt;/strong&gt; and &lt;strong&gt;time-like&lt;/strong&gt;, with space-like distances being those where the distance above is negative and time-like distances being those where the distance above is positive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is positive, this means that the time-difference $t_1 - t_2$ is enough for the light to move from $(x_1, y_1, z_1)$ to $(x_2, y_2, z_2)$, and if it's negative, then, well, light cannot move from one point to the other point in time $t_1 - t_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right way to express that &quot;nothing&quot; can be faster than the speed of light is to say that two space-time events $e_1$ and $e_2$ can only be in a &lt;em&gt;causal relationship&lt;/em&gt; if their distance is time-like. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, your paradox vanishes if you realize that the growth of $r$ is related to two events that are not in a causal relationship because their distance is space-like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this becomes easier to imagine if you think of a more extreme example: Think of $n$ light bulbs all a distance $\Delta$ apart, and each of them is connected to a timer. The timers are synchronized and programmed so that timer number $i$ turns the light bulb on at time $i * dt$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you'll see is that lightbulb $i$ is turned on at time $i * dt$. The location of the last switched on light-bulb therefore &quot;moves&quot; with speed $\Delta / dt$. If you make $\Delta$ large enough and $dt$ small enough, this speed will be faster than the speed of light. But the way we have constructed this chain of bulbs already tells you that the switching on of the light-bulbs is not related causally, since they all have their own chip switching them on. Puh, I hope this makes sense :)&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-06-17T23:16:08.993" CommentCount="1" />
  <row Id="11224" PostTypeId="1" AcceptedAnswerId="11242" CreationDate="2011-06-18T01:07:45.003" Score="2" ViewCount="75" Body="&lt;p&gt;Consider a system of particles where the kinetic energy of the system is varying with time. I'd like to know the significance (or meaning) of the time derivative of the kinetic energy being zero at a point. What is the significance of time instances where the kinetic energy has maxima and minima ?&lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="540" LastEditorDisplayName="" LastEditDate="2011-06-18T17:10:26.400" LastActivityDate="2011-06-18T19:15:20.757" Title="significance of maxima and minima of time varying kinetic energy of a system" Tags="&lt;classical-mechanics&gt;&lt;mathematical-physics&gt;&lt;mechanics&gt;&lt;many-body&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="11225" PostTypeId="2" ParentId="10002" CreationDate="2011-06-18T01:11:22.133" Score="1" ViewCount="" Body="&lt;p&gt;I don't know how you can understand it as photon by Englert–Greenberger duality relation, which is in itself derived from the wave nature. When light behaves like a wave, consider it as a wave; don't try to comprehend it as photons.&lt;/p&gt;&#xA;" OwnerUserId="3887" LastActivityDate="2011-06-18T01:11:22.133" />
  <row Id="11226" PostTypeId="2" ParentId="10754" CreationDate="2011-06-18T01:41:49.467" Score="0" ViewCount="" Body="&lt;p&gt;A paper with precisely the same title as this question has appeared on the archive:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1105/1105.6359v1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/arxiv/pdf/1105/1105.6359v1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4085" LastActivityDate="2011-06-18T01:41:49.467" CommentCount="1" />
  <row Id="11227" PostTypeId="1" AcceptedAnswerId="11229" CreationDate="2011-06-18T01:54:50.297" Score="2" ViewCount="111" Body="&lt;p&gt;Consider a solid ball of radius $r$ and mass $m$ rolling without slipping in a hemispherical bowl of radius $R$ (simple back and forth motion). Now, I assume the oscillations are small and so the small angle approximation holds. I wish to find the period of oscillation and I analyze the motion in two ways, first using conservation of energy and secondly using dynamics. However, I receive two inconsistent answers. One or both of the solutions must be wrong, but I cannot figure out which one and more importantly, I cannot figure out why. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Method 1&lt;/strong&gt;: We write the energy conservation equation for the ball&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$mgh + \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = Constant$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from the center of mass, we take the height as $h = R-(R-r)cos\theta$ where $\theta$ is the angle from the vertical. Applying the no slip condition $v = r\omega$ and taking the moment of inertia for a solid sphere $I = \frac{2}{5}mr^2$ we can write the energy equation as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$mg(R-(R-r)cos\theta) + \frac{7}{10}mr^2\omega^2 = Constant$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiating with respect to time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$mg(R-r)sin\theta\cdot\omega + \frac{7}{5}mr^2\omega\cdot\alpha = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;taking the small angle approximation $sin\theta = \theta$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g(R-r)\theta + \frac{7}{5}r^2\alpha=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-\frac{5g(R-r)}{7r^2}\theta = \alpha$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which we can get $T = 2\pi\sqrt{\frac{7r^2}{5g(R-r)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Method 2&lt;/strong&gt;: The only torque acting on the ball at any point in its motion is the friction force $f$. So we can write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tau = I\alpha = fr$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;again using the rolling condition $a = r\alpha$ and the moment of inertia for a solid sphere,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{2}{5}ma = f$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net force acting on the system is the tangential component of gravity and the force of friction, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = ma = mgsin\theta - f$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{7}{5}a = gsin\theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;taking the small angle approximation and converting $a$ to $\alpha$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha = \frac{5g}{7r}\theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and a corresponding period of $T = 2\pi\sqrt{\frac{7r}{5g}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the solutions are very different and I would appreciate it if someone would point out where I went wrong.&lt;/p&gt;&#xA;" OwnerUserId="3762" LastActivityDate="2011-06-18T05:48:08.223" Title="Analyzing the motion of a ball rolling without slipping inside a hemispherical bowl" Tags="&lt;classical-mechanics&gt;&lt;newtonian-mechanics&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="11228" PostTypeId="2" ParentId="11211" CreationDate="2011-06-18T05:41:58.977" Score="5" ViewCount="" Body="&lt;p&gt;Well, and in the two wave answers nobody considers the quantum mechanical picture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photons compose these classical waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photons carry momentum equal to energy if we set c=1, p=hnu&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the distance from the source becomes large enough that individual photons can be counted in a counter, there will be a point where gaps will exist and no photons will be counted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking this  solution from &lt;a href=&quot;http://answers.yahoo.com/question/index?qid=20110501044609AA6qcO0&quot; rel=&quot;nofollow&quot;&gt;Yahoo answers&lt;/a&gt; one can see that for a given wavelength and intensity, a delta(x) between two detected photons  can be found where individual photons will be very rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer depends on the original intensity,which falls with the distance as 1/r**2 ,the distance, the wavelength observed and the time available for the detection. If one waited an infinite time, the answer  &quot;there is no gap would hold probabilistically. For any reasonable delta(t) there will be gaps that cannot be predicted in r,theta,phi because they will depend on the probability function of the photons. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-06-18T10:53:21.803" LastActivityDate="2011-06-18T10:53:21.803" CommentCount="1" />
  <row Id="11229" PostTypeId="2" ParentId="11227" CreationDate="2011-06-18T05:48:08.223" Score="3" ViewCount="" Body="&lt;p&gt;Your first derivation, using energy, uses two different meanings for the same symbol $\omega$.  In one place, you interpret it as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega = \dot{\theta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the time derivative of the angle of the line from the center of the ball to the center of the bowl with the vertical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In another place, you interpret $\omega$ as the time derivative of the unnamed angle through which the ball itself has rotated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two angles are related to each other by the $r/(R-r)$ factor by which you are off.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-06-18T05:48:08.223" CommentCount="0" />
  <row Id="11230" PostTypeId="2" ParentId="11222" CreationDate="2011-06-18T05:49:18.050" Score="2" ViewCount="" Body="&lt;p&gt;Lagerbaer gives an excellent detailed explanation, but to put it more simply: there is no contradiction, no logical fallacy, the distance $r(t)$ as you have defined it does indeed grow at a rate of $2c$ as measured by any of the observers A, B, C, etc. But $r$ does not correspond to a real thing. It does not measure the position of a particle or an information signal relative to that of an observer, so there's no reason it can't grow faster than light.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-18T05:49:18.050" />
  <row Id="11231" PostTypeId="1" AcceptedAnswerId="11267" CreationDate="2011-06-18T06:27:14.247" Score="3" ViewCount="90" Body="&lt;p&gt;In flat space-time the electric potential energy between two charges is $\frac{k Q_1 Q_2}{r_{12}}$, where $Q$'s are charges and $r_{12}$ is the distance between them. What would happen if the two charges are placed in a strongly curved space-time, which makes &quot;distance&quot; and &quot;duration&quot; measures different from place to place? Will the two feel the same strength of electric force/potential energy?&lt;/p&gt;&#xA;" OwnerUserId="3743" LastActivityDate="2011-06-19T15:08:43.927" Title="Electric potential energy in curved space-time" Tags="&lt;gravity&gt;&lt;electromagnetism&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="11232" PostTypeId="1" AcceptedAnswerId="11237" CreationDate="2011-06-18T06:29:08.063" Score="1" ViewCount="56" Body="&lt;p&gt;If I had say a dozen or so small bar magnets and I wanted to used them to trip the traffic light sensors (which are a coil embedded in the road where a vehicle would stop for the lights), what would be the best layout? Would it be best to put them all end-to-end to create one long 'super' magnet? Or should I separate them into say a 3 by 4 grid with them all orientated the same way? Or is there some other optimal layout?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By 'optimal layout' I mean, 'most likely to trip the sensor', which I'm guessing is the same as, 'would induct the most current in the sensor coil'.&lt;/p&gt;&#xA;" OwnerUserId="1147" LastActivityDate="2011-06-20T09:12:14.423" Title="Optimum magnet layout for tripping a coil sensor" Tags="&lt;electromagnetism&gt;" AnswerCount="2" />
  <row Id="11233" PostTypeId="1" AcceptedAnswerId="11751" CreationDate="2011-06-18T07:23:20.100" Score="3" ViewCount="211" Body="&lt;p&gt;What are the possible outcomes if earth slows down spinning on its own &#xA;axis? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be specific:&#xA;Can the decrease in the internal centrifugal (or centripetal) force due to slowing down earth's spin: &lt;br&gt;&#xA;a. change earth's structure given the fact that major part of earth is not solid and the earth's oblong shape may be in dynamic equilibrium of gravity vs this internal centrifugal force.&lt;br&gt;&#xA;b. alter the sense of the &quot;vertical&quot; as indicated by the plumb line especially when we are off equator, given the fact that the direction of centrifugal force acting on a physical body has a non-zero component on the horizontal direction except on equator and poles? &#xA;If this is true, will the altered horizontal component be sufficient enough to affect high rise buildings?&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="3636" LastEditorUserId="3636" LastEditorDisplayName="" LastEditDate="2011-06-28T16:53:11.483" LastActivityDate="2011-06-30T13:49:28.453" CommunityOwnedDate="2011-06-28T16:49:22.870" Title="Outcomes of earth slowing down spinning on its own axis" Tags="&lt;geophysics&gt;&lt;rotational-dynamics&gt;" AnswerCount="5" FavoriteCount="0" />
  <row Id="11234" PostTypeId="1" AcceptedAnswerId="11236" CreationDate="2011-06-18T07:57:52.467" Score="5" ViewCount="63" Body="&lt;p&gt;Can the sum of angular momentum of all rotating electrons in all the aligned atoms in a permanent magnet have a significant contribution to the macro angular momentum of the magnet? If yes, why does it not behave like a spinning top , like, resisting a change in its orientation?&lt;/p&gt;&#xA;" OwnerUserId="3636" LastActivityDate="2011-06-18T09:49:44.273" Title="Sum of angular momentum of all electrons in a magnet" Tags="&lt;electromagnetism&gt;&lt;classical-mechanics&gt;&lt;mechanics&gt;&lt;angular-momentum&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11235" PostTypeId="2" ParentId="11232" CreationDate="2011-06-18T08:47:09.950" Score="-1" ViewCount="" Body="&lt;p&gt;The road coils at stoplights very in diameter and number if windings. They work bay sensing the change of inductance in the coils caused bay the car within the field. Now how to calculate the max effect of putting magnets on the road depends on many local factors. Can’t see a way to do this unless you have some more data or do some trial and error.&lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-06-18T08:47:09.950" CommentCount="1" />
  <row Id="11236" PostTypeId="2" ParentId="11234" CreationDate="2011-06-18T09:49:44.273" Score="0" ViewCount="" Body="&lt;p&gt;They do. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein%E2%80%93de_Haas_effect&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Einstein%E2%80%93de_Haas_effect&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-06-18T09:49:44.273" />
  <row Id="11237" PostTypeId="2" ParentId="11232" CreationDate="2011-06-18T09:53:34.397" Score="1" ViewCount="" Body="&lt;p&gt;&quot;small Magnets&quot; will not trip such sensors at all,&#xA;because those measure the self inductance of the coil. &#xA;Tripping is done by either increasing inductance from a big &#xA;ferroelectric car body, or from lowered inductance caused &#xA;by some eg big/extended enough aluminium mass. &#xA;Magnets, even when &quot;big&quot; are borderline. The magnetic field &#xA;is useless, and the response to the test field (AC) is &#xA;very small, because the coercitive field strenth of a magnet is high. &#xA;A big metallic magnet might act like a non-ferromagnetic metal. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-06-20T09:12:14.423" LastActivityDate="2011-06-20T09:12:14.423" />
  <row Id="11238" PostTypeId="2" ParentId="11233" CreationDate="2011-06-18T09:58:49.683" Score="2" ViewCount="" Body="&lt;p&gt;I am not shure, what You are asking for, because I can't imagine You &#xA;do not know the answer: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Days will become longer and longer.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is all, and it is nothing but the continuation of &#xA;a trend which lasts since billions of years. Days were &#xA;about 11 hours when earth was yong, afaik.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-06-18T16:03:36.873" LastActivityDate="2011-06-18T16:03:36.873" CommentCount="3" />
  <row Id="11239" PostTypeId="1" CreationDate="2011-06-18T10:17:22.663" Score="2" ViewCount="112" Body="&lt;p&gt;I mean total inertial on my body siting here at my computer &quot;California US.&quot; Ok so the earth is rotating on its axis and in turn around the sun and the sun around the galaxy. The object the question how much kinetic energy is there in a litter of water &quot;at rest&quot; on my desk. &lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-06-18T13:11:13.377" Title="Do we know exactly how fast we are going relative to the center of the galaxy" Tags="&lt;classical-mechanics&gt;&lt;astrophysics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11240" PostTypeId="2" ParentId="11239" CreationDate="2011-06-18T13:11:13.377" Score="6" ViewCount="" Body="&lt;p&gt;The standard value for the speed of the solar system relative to the Galactic center is 220 km/s. If you want to start digging into the literature on this, here's &lt;a href=&quot;http://adsabs.harvard.edu/abs/1989ApJ...342..272F&quot; rel=&quot;nofollow&quot;&gt;one article&lt;/a&gt;. This value is found by measuring radial velocities (via Doppler shifts) of many objects around the Galaxy and fitting to a model of Galactic rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get your velocity at any given location and time of year / day, you have to add in the velocity of the Earth relative to the solar system center of mass (essentially, the velocity of Earth relative to the Sun), and the velocity of you relative to the center of the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnitudes of those are easy to calculate, at least approximately. The Earth-Sun velocity is roughly $2\pi R/T$, where $R=1.5\times 10^{11}$ m is the radius of Earth's orbit and $T$ is a year. Similarly, the velocity of a point on Earth's surface relative to Earth's center is $2\pi r\cos l/t$, where $l$ is your latitude, $r=6400$ km is Earth's radius, and $t$ is a sidereal day (4 minutes less than a regular day). The two speeds work out to about 30 km/s and (460 m/s)$\cos l$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To add them up, you need to know the directions of the various vectors. That's a mildly annoying calculation. The Sun-Galaxy velocity is simple in Galactic coordinates (where the direction to the Galactic center is one of the coordinate directions). The Earth-Sun velocity is simple in ecliptic coordinates. The you-Earth velocity is simple celestial coordinates. The rotation matrices required to convert all these into each other are known and can be looked up somewhere, no doubt.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-06-18T13:11:13.377" CommentCount="5" />
  <row Id="11241" PostTypeId="1" AcceptedAnswerId="11278" CreationDate="2011-06-18T15:56:31.523" Score="6" ViewCount="213" Body="&lt;p&gt;This is a thought I asked myself often, but never did real efforts &#xA;to get an answer. Barsmonsters question about number of fans of a &#xA;wind turbine made me think of it again&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do blades of aircraft propellers or wind turbines cover only a small fraction of the area  they circle? Propellers of ships or Kaplan turbine propellers cover almost the entire circle. (Independent of number of blades). Same for steam turbines and jet turbines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am shure that the forms of those propellers are very close to optimum, due to decades of &#xA;experience. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-12T15:39:17.070" LastActivityDate="2011-07-12T15:39:17.070" Title="Why do blades of windturbine propellers as comp. to propellers of ships cover  very different areas?" Tags="&lt;aerodynamics&gt;&lt;hydrodynamics&gt;&lt;renewable-energy&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="1" />
  <row Id="11242" PostTypeId="2" ParentId="11224" CreationDate="2011-06-18T19:15:20.757" Score="2" ViewCount="" Body="&lt;p&gt;As Anna pointed out in her comment, it really depends on the system. In general, there's not much you can say except that the minimum of kinetic energy corresponds to the minimum speed and the maximum of kinetic energy corresponds to the maximum speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many systems in which the minimum speed (and thus the minimum kinetic energy) is zero, and in those systems, finding the times at which the kinetic energy is a minimum tells you when the object stops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, if you actually take the derivative of kinetic energy with respect to time, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}K}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}\biggl[\frac{1}{2}mv^2\biggr] = m\vec{v}\cdot\vec{a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(using the nonrelativistic expression for $K$). There are three ways this can be equal to zero:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$v = 0$: the object is not moving. This corresponds to a minimum of kinetic energy.&lt;/li&gt;&#xA;&lt;li&gt;$a = 0$: the object is not accelerating (and by $F = ma$ is also experiencing no net force). This corresponds to either constant velocity motion, or a maximum of kinetic energy, in a simple harmonic oscillator for example.&lt;/li&gt;&#xA;&lt;li&gt;$\vec{v}\perp\vec{a}$: the object is experiencing pure centripetal acceleration, which means it's moving with a momentarily constant radius of curvature. This can be either a minimum or a maximum of kinetic energy, in an orbit for example.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Using the relativistic expression, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}K}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}\bigl[(\gamma - 1)mc^2\bigr] = \frac{m\vec{v}\cdot\vec{a}}{\bigl(1-\frac{v^2}{c^2}\bigr)^{\frac{3}{2}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which the same conclusions follow.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-18T19:15:20.757" />
  <row Id="11243" PostTypeId="1" CreationDate="2011-06-18T20:14:16.180" Score="3" ViewCount="101" Body="&lt;p&gt;I would like to understand what is meant when one introduces a generator $G(z)$ as the superpartner of the energy-momentum tensor $T(z)$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;How does one decide that this $G(z)$ should have a &quot;conformal weight&quot; of $\frac{3}{2}$ ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How are the following OPEs for it derived ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$T(z)G(w) = \frac{\frac{3}{2}}{(z-w)^2}G(w) + \frac{ \partial_w G(w) } {z-w} + ...   $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G(z)G(w) = \frac{ \frac{2c}{3} }{(z-w)^3} + \frac{2T(w)}{z-w} + ... $&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If one wants to here create a $\cal{N}=2$ superconformal algebra then one apparently needs to introduce two such conformal weight $\frac{3}{2}$ supercurrents say $G^1(z)$ and $G^2(z)$  (with OPEs as above) and another &quot;$U(1)$ current&quot; $J(z)$ such that,&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$ G^1(z) G^2(z) = \frac{  \frac{2c}{3}}{(z-w)^3} + \frac{2T(w)}{(z-w)} + i( \frac{2J(w)}{(z-w)^2} + \frac{\partial _w J(w)} {(z-w)} ) +... $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T(z)J(w) = \frac{J(w)}{(z-w)^2} + \frac{ \partial _w J(w)} {(z-w)} + ...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$  J(z)G^{1/2}(w) = \pm \frac{ iG^{2/1}(w)}{(z-w)} + ... $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$J(z) J(w) = \frac{ \frac{c}{3} }{ (z-w)^3 } + ... $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be grateful if someone can give some explanations about how the above construction is made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Especially regarding the need and motivation to introduce the field $J$ (..and its OPEs..)&lt;/p&gt;&#xA;" OwnerUserId="1637" LastActivityDate="2011-06-20T13:52:13.370" Title="Superpartner for the stress-energy tensor" Tags="&lt;quantum-field-theory&gt;&lt;supersymmetry&gt;&lt;conformal-field-theory&gt;" AnswerCount="1" />
  <row Id="11244" PostTypeId="2" ParentId="11241" CreationDate="2011-06-18T20:32:23.233" Score="1" ViewCount="" Body="&lt;p&gt;I simple answer for a simple question. The main difference between Hydrodynamic and Aerodynamic is that a gas “Air” is spongy in other words it compresses. This is the main reason a propeller on a plain gets propulsion threw the air by having a high speed, narrow and long blade. Where the prop in the water gets it from a much lower speed but a wider and shorter blade&lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-06-18T20:32:23.233" CommentCount="3" />
  <row Id="11245" PostTypeId="2" ParentId="11138" CreationDate="2011-06-18T21:55:30.497" Score="-1" ViewCount="" Body="&lt;p&gt;Can anyone be satsified with an answer which does not provide a mechanism for absorption of energy, as opposed to scattering? Any explanation which relies on the excitation of certain energy levels must also explain why the light is not simply scattered when that same energy level decays. This point is seldom addressed in discussions I have seen on this topic.&lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-06-18T21:55:30.497" CommentCount="4" />
  <row Id="11246" PostTypeId="1" CreationDate="2011-06-18T22:48:33.857" Score="2" ViewCount="80" Body="&lt;p&gt;Please please help me out with this. I am trying to find a software/program that I could use the simulate the scattering effect of light when it strikes gas particles moving at supersonic speeds? Please help me as I have no idea what to use. I have googled several searches but I could find nothing. Please help. thanks a lot. &lt;/p&gt;&#xA;" OwnerUserId="3604" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-19T00:36:02.427" LastActivityDate="2011-07-20T09:23:30.597" Title="What software can I use to simulate supersonic gas particle flow?" Tags="&lt;optics&gt;&lt;fluid-dynamics&gt;&lt;speed&gt;&lt;computational-physics&gt;&lt;cfd&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11247" PostTypeId="1" CreationDate="2011-06-18T23:18:52.530" Score="6" ViewCount="639" Body="&lt;p&gt;This is a question on gauge invariance in quantum mechanics.  I do some simple math on a 1D wave-function with periodic boundary conditions, and get that gauge invariance is violated.  What am I doing wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider one coordinate dimension configured as a ring. The gauge dependent momentum operator can be written: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_{op}=-i \frac{\partial}{\partial x} - k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Units have been chosen so that $\hbar = 1$, $k$ is an arbitrary real constant different for each gauge and $x$ represents the coordinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gauge dependent eigenfunction can be written&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\psi(x)= Ae^{i(n+k)x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where A is a constant determined by normalization.  As is well known in quantum mechanics, an operator applied to one of its eigenfunctions should yield a real constant eigenvalue multiplying the same eigenfunction:  Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[-i \frac{\partial}{\partial x} - k]  Ae^{i(n+k)x}= nAe^{i(n+k)x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that the real number n is the eigenvalue, which must be determined by the boundary conditions.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boundary condition for this periodic system must be that the wave function should join onto itself smoothly everywhere.  Thus, if the coordinate is chosen such that x extends from –$\pi$ around the ring to $\pi$ then the eigenfunction in equation 3 must have (n + k) = m, where m is an integer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under these conditions, the eigenvalue n in equation 3 will be n = m – k.  This eigenvalue depends explicitly on k, and so is not gauge invariant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming this simple situation should be gauge invariant, but I don't see where I goofed.  I'd appreciate any help. &lt;/p&gt;&#xA;" OwnerUserId="4092" LastEditorUserId="3234" LastEditorDisplayName="" LastEditDate="2011-06-19T00:35:36.407" LastActivityDate="2011-08-01T19:31:37.127" Title="proof of gauge invariance for quantum 1D ring" Tags="&lt;quantum-mechanics&gt;&lt;gauge-theory&gt;" AnswerCount="5" CommentCount="20" FavoriteCount="1" />
  <row Id="11248" PostTypeId="2" ParentId="9708" CreationDate="2011-06-19T01:27:22.627" Score="0" ViewCount="" Body="&lt;p&gt;It's a bit like asking if Concrete is Matter or Energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fire is composed of several states of energy and matter, like sublimated matter turning into gases; or photons released on oxidization of matter; or sound is released via turbulence from convection; plasma occurs whenever the temperature ionizes atomic bonds and Sooty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several papers on the subject e.g. &quot;Yule Log Science&quot; &lt;a href=&quot;http://www.sciencentral.com/articles/view.php3?type=article&amp;amp;article_id=218392881&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencentral.com/articles/view.php3?type=article&amp;amp;article_id=218392881&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3588" LastEditorUserId="3588" LastEditorDisplayName="" LastEditDate="2011-06-19T01:34:26.823" LastActivityDate="2011-06-19T01:34:26.823" />
  <row Id="11249" PostTypeId="2" ParentId="11243" CreationDate="2011-06-19T02:29:17.150" Score="1" ViewCount="" Body="&lt;p&gt;I don't know much about supersymmetry but in absent of any other answers, maybe you will benefit infinitesimally from my guesses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets think in terms of a non-interacting SUSY theory with a bosonic and a fermionic field $S \sim \int\text d^2z\left(\partial X\bar{\partial}X - \left(\psi\bar{\partial}\psi + \bar{\psi}\partial\bar{\psi}\right)\right)$. Then from space-time symmetries we have a conserved energy-momentum tensor $T(z) = T_{boson}(z) + T_{fermion}(z)$ (no coupling, since they don't interact), and similarity for the anti-holormorphic part. It is then very natural to guess that your $G(z)$ is nothing but the conserved charge coming from supersymmetry (which could be to derive from Noether's theorem). Now, we know that we have a primary field $\psi(z)$ (the fermion) with $h_{\psi}=\frac 12$ and $j(z)=i\partial X(z)$ (the boson) with $h_j=1$. It is natural to guess that $G(z)$ is a combination of $\psi(z)$ and $j(z)$, which potentially could make it a primary field with conformal dimension $h_G = h_{\psi} + h_j = \frac 32$. The $T(z)G(w)$ OPE is nothing but the statement that $G(z)$ is an primary field with $h_G=\frac 32$. The $G(z)G(w)$ OPE should then be derived by using the OPE's of $\psi(z)$ and $j(z)$ together with Wicks theorem, as usual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the $\mathcal N = 2$ case, I have no idea. If you are less lazy than me and actually carry out this calculation, please write a comment on whether it works or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WARNING: the content of this answer is not thought through and could be completely wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Regarding why $G(z)$ is called a superpartner to $T(z)$. I vaguely remember that one usually do supersymmetry in terms of superspaces and superfields. My guess would be that one can construct a super (field) energy-momentum tensor, probably of the form $\sim G(z) + \theta T(z)$, and then they are &quot;superpartners&quot; in this sort of way.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT2: @Arnirbit, in my answer I assumed knowledge of some basic aspects of non-supersymmetric 2d CFT. In particular the free boson and fermion (I must declare myself as newcomer to CFT and therefore not an expert).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to your questions. A primary field is a field which transforms in a particularly &quot;nice&quot; way under conformal transformations $(z,\bar z)\rightarrow (f(z), \bar f(\bar z))$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \phi(z,\bar z) \rightarrow \left(\frac{\partial f}{\partial z}\right)^h\left(\frac{\partial \bar f}{\partial \bar z}\right)^{\bar h} \phi\left(f(z),\bar f(z)\right),$$&#xA;where $\phi(z,\bar z)$ is a primary field with conformal weight $(h,\bar h)$. There are a connection between transformation properties of fields and OPE's, one can for example show that a primary field of weight $(h,\bar h)$ have the following OPE with the energy-momentum tensor&#xA;$$ T(z)\phi(w,\bar w) = \frac h{(z-w)^2}\phi(w,\bar w) + \frac 1{z-w}\partial_w\phi(w,\bar w) + \text{regular terms},$$&#xA;and similar for the anti-holormorphic part (one way to derive this is by using Ward-identities). This equality must of course be interpreted as valid under a time-ordered correlation function. Another property of primary fields is that their two-point function is by symmetry fixed to be&#xA;$$ \left&amp;lt;\phi(z,\bar z)\phi(w,\bar w)\right&amp;gt; = \frac C{(z-w)^{2h}(\bar z-\bar w)^{2\bar h}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, lets concentrate on the free Bosonic theory $$S[X] = T\int\text d^2z\;\partial X(z,\bar z)\bar{\partial}X(z,\bar z).$$ The equations of motion is $\partial\bar{\partial}X(z,\bar z) = 0$, meaning that $j(z)\equiv i\partial X(z,\bar z)$ is holormorphic (depends only on $z$, not $\bar z$) and similarly for $\bar j(\bar z)\equiv i\bar{\partial}X(z,\bar z)$. One can calculate the two-point function $\left&amp;lt;X(z,\bar z)X(w, \bar w)\right&amp;gt; \propto  \log\left|z-w\right|^2$. This does not have the above form, and therefore $X(z,\bar z)$ is not a primary field! However $$\left&amp;lt;j(z)j(w)\right&amp;gt; = -\left&amp;lt;\partial_zX(z,\bar z)\partial_wX(w,\bar w)\right&amp;gt;\propto -\partial_z\partial_w\log\left|z-w\right|^2 = \frac 1{(z-w)^2}.$$ This says that $j(z)$ is a (chiral) primary field with conformal dimension $(1,0)$, with the OPE $j(z)j(w) = \frac 1{(z-w)^2} + \dots$ (if you like fancy words, this OPE gives you a $\mathfrak{u}(1)$ affine Kac-Moody algebra). You can calculate the energy-momentum tensor which will be of the form $T(z) = -\gamma :\partial X\partial X: = \gamma :jj:(z)$, for some appropriate constant $\gamma$ and normal ordering $:\; :$ is necessary on the quantum level. Using Wicks theorem one finds&#xA;$$ T(z)j(w) = \frac 1{(z-w)^2}j(w) + \frac 1{z-w}\partial_wj(w) + \text{regular terms}, $$&#xA;which again says that $j(z)$ has conformal dimensions $(h,\bar h) = (1,0)$. A similar analysis for the free fermion will tell you that $\psi$ is a primary field with $(h,\bar h) = (\frac 12, 0)$ and give you the OPE $\psi(z)\psi(w) = \frac 1{z-w} + \dots$ (and similarly for $\bar{\psi}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are not already familiar with all this, then my quick sketch will not be sufficient and I recommend you to consult a textbook. Finally, I think that when you combine the free bosonic and free fermionic theory in a supersymmetric way, then you can find out why $G(z)$ with conformal weight $h = \frac 32$ appears.&lt;/p&gt;&#xA;" OwnerUserId="1469" LastEditorUserId="1469" LastEditorDisplayName="" LastEditDate="2011-06-20T13:52:13.370" LastActivityDate="2011-06-20T13:52:13.370" CommentCount="1" />
  <row Id="11250" PostTypeId="2" ParentId="11169" CreationDate="2011-06-19T03:04:34.497" Score="4" ViewCount="" Body="&lt;p&gt;Lens design is an art as much as a science, and is heavily based on experience. To do this, you would need at least a graduate level background in optical design, and preferably also years of experience. If you have that, Zemax would be quite acceptable, although I personally use Code V. Either one should be more than adequate. Oslo would also work. Generally lens desingers prefer to use what they already know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of a starting direction, it is best to begin with some standard design that has specs as close as possible to your goal. Given that you are imaging between finite conjugates (I'm assuming this because you are doing lithography) you would likely use something symmetrical about the stop, like a double gauss lens. The double gauss includes some elements for chromatic correction, but you can remove those for simplicity. You may need to add more to improve your performance later on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, a good desk reference for finding a starting design is &quot;Practical Optical System Layout: And Use of Stock Lenses&quot; or &quot;Modern Lens Design: A Resource Manual&quot; by Warren J. Smith.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course because you are designing lenses, I assume you already own &quot;Modern Optical Engineering&quot; by Warren J. Smith, because anybody who wants to do any optical design should already be comfortable with that book. If you don't, own it, you're not really an optical engineer :)&lt;/p&gt;&#xA;" OwnerUserId="869" LastEditorUserId="869" LastEditorDisplayName="" LastEditDate="2011-06-19T03:12:55.867" LastActivityDate="2011-06-19T03:12:55.867" />
  <row Id="11251" PostTypeId="1" CreationDate="2011-06-19T03:07:01.393" Score="1" ViewCount="44" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/4071/popular-depictions-of-electromagnetic-wave-is-there-an-error&quot;&gt;Popular depictions of electromagnetic wave: is there an error?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;The electric and magnetic components of light increase and decrease at the same time, pass through zero magnitude and then increase in the opposite direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When |electric component|= 0 =|magnetic component|, where does the energy go? I asked my physics professor this and he said that it went into momentum, which allowed the fields to disappear into and emerge from seemingly nothing. I remain skeptical. Is there an equation somewhere that describes this momentum? If not, is there another explanation? &lt;/p&gt;&#xA;" OwnerUserId="2843" LastEditorUserId="2843" LastEditorDisplayName="" LastEditDate="2011-06-19T03:20:27.283" LastActivityDate="2011-06-19T03:20:27.283" ClosedDate="2011-06-19T03:43:42.433" Title="Where does light energy go when the electric and magnetic components equal zero?" Tags="&lt;light&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="11252" PostTypeId="2" ParentId="11142" CreationDate="2011-06-19T05:17:35.407" Score="0" ViewCount="" Body="&lt;p&gt;OK, here is a complicated and clear explanation of what you are  asking, which is also new to me: &lt;a href=&quot;http://www.exploratorium.edu/ronh/bubbles/soap.html&quot; rel=&quot;nofollow&quot;&gt;Soap is a complicated chemically molecule&lt;/a&gt; it breaks the high surface tension to allow bubbles, and also once in the air part of it  protects the bubble from evaporation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So  it is an interplay between two components: surface tension and, as Georg points out,&lt;a href=&quot;http://en.wikipedia.org/wiki/Amphiphilic&quot; rel=&quot;nofollow&quot;&gt;amphiphilic&lt;/a&gt; effects . It is not just the lower surface tension that creates bubbles, so for the other liquids you list an additive should be found that would work the same way soap works for water, not allowing evaporation.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-06-19T17:15:34.760" LastActivityDate="2011-06-19T17:15:34.760" CommentCount="3" />
  <row Id="11253" PostTypeId="2" ParentId="11241" CreationDate="2011-06-19T05:41:55.507" Score="5" ViewCount="" Body="&lt;p&gt;An engineering answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that it's not just about water vs air. It depends on a lot of things: the density &amp;amp; compressibility of the fluid, and tradeoffs between torque, efficiency, cost, materials, maintenance needs, fouling hazards, and so on. Below is a water turbine blade, the &lt;a href=&quot;http://www.marineturbines.com/&quot; rel=&quot;nofollow&quot;&gt;SeaGen&lt;/a&gt;, that's not much different from a wind turbine blade, because it's doing a similar job, with similar constraints, but in water rather than air. But bear in mind that this is (AFAIK) the &lt;strong&gt;first&lt;/strong&gt; tidal turbine to achieve commercial grid operation, so later designs could diverge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some blades are indeed close to optimised &lt;em&gt;given the materials available at the time&lt;/em&gt;, thanks to decades of experience. However, new materials can enable further optimisations to blade design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/SH88g.jpg&quot; alt=&quot;SeaGen Turbine&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4066" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-06-19T06:12:37.013" LastActivityDate="2011-06-19T06:12:37.013" />
  <row Id="11254" PostTypeId="1" AcceptedAnswerId="11263" CreationDate="2011-06-19T05:53:53.383" Score="2" ViewCount="118" Body="&lt;p&gt;Consider a system whose generalized co-ordinates are $q_i$ and is under the constraints $\dot{q_i} = K_i \forall i = 1,2,3,...$ where $K_i$ are constants. I have a problem in writing the Lagrange's equation for this system as $\dot{q_i}$ is a constant. What would be $\frac{\partial L}{\partial \dot{q_i}}$ ?&lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="540" LastEditorDisplayName="" LastEditDate="2011-06-19T06:23:30.170" LastActivityDate="2011-06-19T11:37:47.443" Title="a question on Lagrange's equation when the time derivative of the generalized co-ordinates is constant " Tags="&lt;classical-mechanics&gt;&lt;mathematical-physics&gt;&lt;mechanics&gt;&lt;newtonian-mechanics&gt;&lt;mathematics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="11255" PostTypeId="2" ParentId="11233" CreationDate="2011-06-19T05:58:19.750" Score="0" ViewCount="" Body="&lt;p&gt;Yarkovsky Effect&lt;/p&gt;&#xA;&#xA;&lt;p&gt;quoting wikipedia - &quot;The Yarkovsky effect is a force acting on a rotating body in space caused by the anisotropic emission of thermal photons, which carry momentum.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Yarkovsky_effect&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Yarkovsky_effect&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, the Earth absorbs momentum from Sun photons and remits this momentum as the Earth rotates, for retrograde planetary motion this causes the planet to move Sunwards, and oppositely for prograde motion (Earth has a prograde motion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this effect is minimal over a short period of time, it will sum to measurable effects over a great period (to wit millions of years).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be easily imagined that for sufficient time frames, any rotating-orbiting body will move closer or further from the Sun until it reaches an orbit where the momentum released via the Yarkovsky effect will balance that gained from Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in answer to your question (from the Yarkovsky effect) any change in rotation will cause an eventual change in orbit, and depending on the planets retrograde or prograde rotation will cause the planet to either heat up or cool down depending on the new orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noting once more - this may take 'billions of years'&lt;/p&gt;&#xA;" OwnerUserId="3588" LastEditorUserId="3588" LastEditorDisplayName="" LastEditDate="2011-06-19T06:31:09.587" LastActivityDate="2011-06-19T06:31:09.587" CommentCount="2" />
  <row Id="11256" PostTypeId="1" CreationDate="2011-06-19T06:37:50.890" Score="3" ViewCount="122" Body="&lt;p&gt;I know this is a bad question to most serious Physics but I have a question about what is considered “Perpetual motion.” The Foucault pendulum in the UN consists of sphere that passes directly over a raised metal ring at the centre that contains an electromagnet, which induces a current in the copper inside the ball. This supplies the necessary energy to overcome friction and air resistance and keeps it swinging uniformly. Now the swing of the pendulum is induced but the 36h 45m clockwise shift generated bay the earths rotation is perpetual or as long as the earth rotates.  Is this assumption correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does a generator that works on tidal movements not fall in to the same assumption?&lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-06-19T07:37:49.487" Title="What do we consider &quot;Perpetual Motion&quot;" Tags="&lt;classical-mechanics&gt;&lt;forces&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="11257" PostTypeId="2" ParentId="11256" CreationDate="2011-06-19T07:06:38.033" Score="2" ViewCount="" Body="&lt;p&gt;The term &quot;perpetual motion&quot; signifies a machine that can keep moving indefinitely without any additional energy inputs, without losing kinetic energy. The Foucault pendulum you describe has additional energy inputs (from the electromagnet), to overcome friction and air resistance, as you say, so it's not a perpetual-motion machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perpetual-motion machines cannot exist, according to the laws of thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By extension, the phrase &quot;perpetual motion machine&quot; now refers to any machine that claims to break / bypass the law of entropy, whether or not it actually includes motion.&lt;/p&gt;&#xA;" OwnerUserId="4066" LastActivityDate="2011-06-19T07:06:38.033" />
  <row Id="11258" PostTypeId="2" ParentId="11256" CreationDate="2011-06-19T07:13:10.867" Score="4" ViewCount="" Body="&lt;p&gt;No, neither of the systems you've brought up is an example of what physicists mean when they say &quot;perpetual motion.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The catch is that in physics, perpetual motion doesn't mean what you might literally interpret it to mean - that is, it doesn't just refer to something that is able to keep moving forever. The laws of physics have no problem with the idea that something can be in motion perpetually; for example, a rock moving at constant velocity through an otherwise empty universe. What a physicist means by perpetual motion is a system that is able to maintain its original state of motion, without being subject to a driving force, &lt;em&gt;in the presence of dissipative forces&lt;/em&gt; like friction or &lt;a href=&quot;http://en.wikipedia.org/wiki/Bremsstrahlung&quot; rel=&quot;nofollow&quot;&gt;bremsstrahlung radiation&lt;/a&gt;. A driving force adds energy to the system, whereas dissipative forces take energy away, so if there are dissipative forces but no driving forces, the system's energy has to decrease over time, and thus it can't keep moving in its original pattern. Therefore this particular meaning of perpetual motion is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, all physical systems are subject to dissipative forces, so it is impossible to construct a machine that maintains its original motion without an external source of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Perpetual_motion&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article on perpetual motion&lt;/a&gt; does a very nice job of explaining this in more detail. I suggest you take a look at it.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-19T07:13:10.867" CommentCount="1" />
  <row Id="11259" PostTypeId="2" ParentId="11256" CreationDate="2011-06-19T07:24:21.173" Score="0" ViewCount="" Body="&lt;p&gt;From the webster.com dictionary , &lt;em&gt;perpetual&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a : continuing forever : everlasting  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In our universe as far as we have learned, &lt;strong&gt;nothing continues forever&lt;/strong&gt;. So the definition of &lt;em&gt;perpetual motion&lt;/em&gt; is a relative definition in time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a human lifetime the earth will be continuously turning on its axis,&#xA;as well as with the planets around the sun.&#xA;An atom will have its electrons in continuous &quot;motion&quot; around the nucleus,( until the expansion of the universe destroys everything,) etc &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that sense, a generator absorbing the energy of the tides is taking advantage of this, limited in definition, perpetual motion. In detail the energy turned into electricity is taken infinitesimally out of the angular momentum of the earth, so the perpetual motion of the earth is delayed by a tiny bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;perpetual motion machines&lt;/em&gt; on the other hand that crank inventors try to demonstrate, would  get more energy out than the one coming in, thus breaking the law of conservation of energy.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-06-19T07:24:21.173" />
  <row Id="11260" PostTypeId="2" ParentId="11256" CreationDate="2011-06-19T07:37:49.487" Score="0" ViewCount="" Body="&lt;p&gt;From Wikipedia: &quot;Perpetual motion describes hypothetical machines that operate or produce useful work indefinitely and, more generally, hypothetical machines that produce more work or energy than they consume, whether they might operate indefinitely or not.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is not moving indefinitely, but DOING WORK. People didn't build perpetual motion machines to observe inertial movement, but try to power factories, transportation, etc without any energy source.&lt;/p&gt;&#xA;" OwnerUserId="3887" LastActivityDate="2011-06-19T07:37:49.487" />
  <row Id="11261" PostTypeId="2" ParentId="11254" CreationDate="2011-06-19T07:52:41.163" Score="-2" ViewCount="" Body="&lt;p&gt;You already have the relation of $q$ in terms of $t$, what are you solving?&lt;/p&gt;&#xA;" OwnerUserId="3887" LastActivityDate="2011-06-19T07:52:41.163" CommentCount="1" />
  <row Id="11262" PostTypeId="2" ParentId="11247" CreationDate="2011-06-19T09:15:30.100" Score="1" ViewCount="" Body="&lt;p&gt;I believe you're seeing a problem where there isn't one. Indeed we have $n + k \in \mathbb{Z},$ so if you shift $k$, say $$k \mapsto k + \delta,$$ you need to adjust $$n \mapsto n - \delta.$$ I wouldn't say that $n$ explicitly depends on $k$, but shifting $k$ by an arbitrary &lt;em&gt;real&lt;/em&gt; (= non-integer) constant does force an adjustment for $n$ - not very surprising. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I think you can set up the problem in a more natural way. Forget about the gauge part, just consider the wave function $\phi(x) = e^{i n x}$; to satisfy the p.b.c., you find that $n \in \mathbb{Z}.$ Now add a gauge part, $U_k(x) = e^{ikx};$ you then find that the derivative changes to $\partial_x \mapsto D_x = \partial_x - ik$ (modulo a sign error) etc. The total wave function changes to $\phi(x) \mapsto U_k(x) \phi(x) = \Psi(x)$; again, you find that $k \in \mathbb{Z}$ etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this new set-up, you find that you can change $k$ by an integer without punishment, and I think this is also meant in your problem; once you restrict yourself to the integers, you have integer freedom in your choice of gauge.&lt;/p&gt;&#xA;" OwnerUserId="1471" LastActivityDate="2011-06-19T09:15:30.100" CommentCount="4" />
  <row Id="11263" PostTypeId="2" ParentId="11254" CreationDate="2011-06-19T11:26:06.203" Score="3" ViewCount="" Body="&lt;p&gt;I will assume that we are talking about a classical (as opposed to a quantum) system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, Karsus Ren is of course right that $\dot{q}^i=K^i$ implies that the solution $q^i(t)$ is an affine function of time $t$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the heart of the question (v2) is not so much the solution itself, but more that if the Lagrangian $L=L(q,\dot{q},t)$ is given, and we have constraints $\dot{q}^i=K^i$, should we substitute $\dot{q}^i\to K^i$ in (1) &lt;em&gt;none,&lt;/em&gt; (2) &lt;em&gt;all,&lt;/em&gt; or (3) &lt;em&gt;some&lt;/em&gt; of the $\dot{q}$ appearances in $L=L(q,\dot{q},t)$ before differentiating wrt. $\dot{q}^i$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that it doesn't matter. Normally we treat constraints by introducing Lagrange multipliers $\lambda_i$ and a new Lagrangian&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \tilde{L}(q,\dot{q},\lambda,t) = L(q,\dot{q},t) + \lambda_i (\dot{q}^i-K^i). $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lagrange eqs. wrt. $\lambda_i$ yield the constraints $\dot{q}^i=K^i$.&#xA;This is the evolution equation for $q^i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the Lagrange eqs. wrt. $q^i$ yield&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial L}{\partial q^i} = \frac{d}{dt} \left[ \frac{\partial L}{\partial \dot{q}^i} + \lambda_i \right]. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the evolution equation for $\lambda_i$. We see from the last equation that a change in the definition of the derivative $\frac{\partial L}{\partial \dot{q}^i}$ leads to a corresponding change in $\lambda_i$, but this has no consequences for $q^i$, as expected.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-19T11:37:47.443" LastActivityDate="2011-06-19T11:37:47.443" CommentCount="2" />
  <row Id="11264" PostTypeId="1" AcceptedAnswerId="11266" CreationDate="2011-06-19T13:40:38.803" Score="3" ViewCount="176" Body="&lt;p&gt;For an observable $A$ and a Hamiltonian $H$, &lt;a href=&quot;http://en.wikipedia.org/wiki/Heisenberg_picture&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; gives the time evolution equation for $A(t) = e^{iHt/\hbar} A e^{-iHt/\hbar}$ in the Heisenberg picture as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dt} A(t) = \frac{i}{\hbar} [H, A] + \frac{\partial A}{\partial t}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From their derivation it sure looks like $\frac{\partial A}{\partial t}$ is supposed to be the derivative of the original operator $A$ with respect to $t$ and $\frac{d}{dt} A(t)$ is the derivative of the transformed operator. However, the Wikipedia derivation then goes on to say that $\frac{\partial A}{\partial t}$ is the derivative with respect to time of the &lt;em&gt;transformed&lt;/em&gt; operator. But if that's true, then what does $\frac{d}{dt} A(t)$ mean? Or is that just a mistake?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I need to know which term to get rid of if $A$ is time-independent in the Schrodinger picture. I think it's $\frac{\partial A}{\partial t}$ but you can never be too sure of these things.) &lt;/p&gt;&#xA;" OwnerUserId="1046" LastEditorUserId="1046" LastEditorDisplayName="" LastEditDate="2011-06-19T13:58:56.300" LastActivityDate="2011-06-19T16:03:56.963" Title="Which derivative with respect to time is which in the Heisenberg picture of quantum mechanics?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="11265" PostTypeId="2" ParentId="11264" CreationDate="2011-06-19T14:36:19.853" Score="3" ViewCount="" Body="&lt;p&gt;It's easiest to derive this from the Schrödinger picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $B(t)$ be a time-dependent operator in the Schrödinger picture. The corresponding operator in the Heisenberg picture is $A(t) = e^{iHt/\hbar} B(t) e^{-iHt/\hbar}$. Differentiation with respect to $t$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d}{dt} A(t) = e^{iHt/\hbar} \left(\frac{i}{\hbar} H B(t) + \frac{\partial}{\partial t}B(t) - \frac{i}{\hbar} B(t) H) \right) e^{-iHt/\hbar} $$&#xA;$$ = e^{iHt/\hbar} \left(\frac{i}{\hbar} [H,B(t)] + \frac{\partial}{\partial t}B(t)\right) e^{-iHt/\hbar}&#xD;&#xA;= \frac{i}{\hbar} [H,A(t)] + \frac{\partial A}{\partial t}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the last partial derivative is to be understood in the sense that you take the operator $\frac{\partial B}{\partial t}$ and &quot;evolve it in time&quot; via the Schrödinger equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;Useful non-example: the velocity operator $\vec v$. The velocity operator is the derivative of the position operator, but it's the total derivative as the system evolves. Hence,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec v = \frac{i}{\hbar} [H,\vec r] .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Schrödinger picture, the position operator is, of course, time independent. Since $H$ is time independent as well, this is also the right velocity operator in the Schrödinger picture.&lt;/p&gt;&#xA;" OwnerUserId="188" LastActivityDate="2011-06-19T14:36:19.853" CommentCount="0" />
  <row Id="11266" PostTypeId="2" ParentId="11264" CreationDate="2011-06-19T14:59:27.600" Score="4" ViewCount="" Body="&lt;p&gt;There is no mistake on the Wikipedia page and all the equations and statements are consistent with each other. In&#xA;$$A_{\rm Heis.}(t) = e^{iHt/\hbar} A e^{-iHt/\hbar}$$&#xA;the letter $A$ in the middle of the product represents the Schrödinger picture operator $A = A_{\rm Schr.}$ that is not evolving with time because in the Schrödinger picture, the dynamical evolution is guaranteed by the evolution of the state vector $|\psi\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this doesn't mean that the time derivative $dA_{\rm Schr.}/dt=0$. Instead, we have&#xA;$$ \frac{dA_{\rm Schr.}}{dt} =  \frac{\partial A_{\rm Schr.}}{\partial t} $$&#xA;Here, $A_{\rm Schr.}$ is meant to be a function of $x_i, p_j$, and $t$. In most cases, there is no dependence of the Schrödinger picture operators on $t$ - which we call an &quot;explicit dependence&quot; - but it is possible to consider a more general case in which this explicit dependence does exist (some terms in the energy, e.g. the electrostatic energy in an external field, may be naturally time-dependent). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Schrödinger's picture, $dx_{i,\rm Schr.}/dt=0$ and $dp_{j,\rm Schr.}/dt=0$ which is why the total derivative of $A_{\rm Schr.}$ with respect to time is given just by the partial derivative with respect to time. Imagine, for example, &#xA;$$ A_{\rm Schr.}(t) = c_1 x^2 + c_2 p^2 + c_3 (t) (xp+px) $$&#xA;We would have&#xA;$$ \frac{dA_{\rm Schr.}(t)}{dt} = \frac{\partial c_3(t)}{\partial t} (xp+px).$$&#xA;These Schrödinger's picture operators are called &quot;untransformed&quot; on that Wikipedia page. The transformed ones are the Heisenberg picture operators given by &#xA;$$A_{\rm Heis.}(t) = e^{iHt/\hbar} A_{\rm Schr.}(t) e^{-iHt/\hbar}$$&#xA;Their time derivative, $dA_{\rm Heis.}(t)/dt$, is more complicated. An easy differentiation gives exactly the formula involving $[H,A_{\rm Heis.}]$ that you quoted as well. &#xA;$$\frac{d}{dt} A_{\rm Heis.}(t) = \frac{i}{\hbar} [H, A_{\rm Heis.}(t)] + \frac{\partial A_{\rm Heis.}(t)}{\partial t}.$$&#xA;The two terms in the commutator arise from the $t$-derivatives of the two exponentials in the formula for the Heisenberg $A_{\rm Heis.}(t)$ while the partial derivative arises from $dA_{\rm Schr.}/dt$ we have always had. (These simple equations remain this simple even for a time-dependent $A_{\rm Schr.}$; however, we have to assume that the total $H$ is time-independent, otherwise all the equations would get more complicated.) The two exponentials on both sides never disappear by any kind of derivative, so obviously, all the appearances of $A$ in the differential equation above are $A_{\rm Heis.}$. The displayed equation above is the (only) dynamical equation for the Heisenberg picture so it is self-contained and doesn't include any objects from other pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Heisenberg picture, it is no longer the case that $dx_{\rm Heis.}(t)/dt=0$ (not!) and the similar identity fails for $p_{\rm Heis.}(t)$ as well. $A_{\rm Heis.}(t)$ is a general function of all the basic operators $x_{i,\rm Heis.}(t)$ and $p_{j,\rm Heis.}(t)$, as well as time $t$.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="1236" LastEditorDisplayName="" LastEditDate="2011-06-19T16:03:56.963" LastActivityDate="2011-06-19T16:03:56.963" CommentCount="10" />
  <row Id="11267" PostTypeId="2" ParentId="11231" CreationDate="2011-06-19T15:08:43.927" Score="1" ViewCount="" Body="&lt;p&gt;In a general curved spacetime, the problem is complicated and obviously you can't end up with the &quot;same&quot; formula, especially not if you interpret $r_{12}$ in a random way, e.g. as the proper distance between the two points. Such a formula is surely incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve the problem, you have to solve the field equations such as&#xA;$$\nabla\cdot \vec D = \rho $$&#xA;in a given curved spacetime. Well, you should write these equations using $F_{\mu\nu}$ with the general relativistic treatment of the indices - you should use the full Einstein-Maxwell equations. Note that in general relativity, you should actually also accept that the electric field itself will contribute to the curvature of spacetime, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result will depend on the precise shape of the space. You may determine the total mass/energy (note that $E=mc^2$ in relativity) either from $Q_1 \phi_2$, calculated in one way or another, and taking the proper redshifts into account, or as the ADM energy i.e. from the asymptotic curvature induced by the configuration of the charges.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-06-19T15:08:43.927" />
  <row Id="11268" PostTypeId="1" CreationDate="2011-06-19T16:23:09.440" Score="2" ViewCount="86" Body="&lt;p&gt;The oil drop experiment performed by Robert Millikan in 1909 enables us to calculate the elementary charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following forces have effect on the droplet:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$F_\mathrm{G} = m g = \frac{4}{3} \, \pi r^3 \rho_\mathrm{Oil} g$&lt;/li&gt;&#xA;&lt;li&gt;$F_\mathrm{A} = \frac{4}{3} \, \pi r^3 \rho_\mathrm{Air} g$&lt;/li&gt;&#xA;&lt;li&gt;$F_\mathrm{E} = q E = \frac{qU}{d}$&lt;/li&gt;&#xA;&lt;li&gt;$F_\mathrm{R} = 6\,\pi\eta r v$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, which steps of calculation are necessary to finally calculate the elementary electric charge (charge of the electron) or at least a multiple of it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q = \frac{9 \pi d}{2 U} \sqrt{\frac{\eta^3 (v_1 - v_2)}{\rho^* g}} (v_1 + v_2)$$&lt;/p&gt;&#xA;" OwnerUserId="3262" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-06-19T16:46:57.343" LastActivityDate="2011-06-20T07:38:54.633" ClosedDate="2011-09-05T21:14:35.737" Title="Derivation of the oil drop experiment" Tags="&lt;homework&gt;&lt;experimental-physics&gt;" AnswerCount="0" CommentCount="1" />
  <row Id="11269" PostTypeId="1" AcceptedAnswerId="11272" CreationDate="2011-06-19T19:59:19.427" Score="2" ViewCount="46" Body="&lt;p&gt;I'm preparing for a grade 11 physics exam but having trouble with a question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a 19.9kg mass slides in a straight 28.8m line&#xA;experiencing friction with a coefficient of 0.105, how much thermal energy is produced?&lt;/p&gt;&#xA;" OwnerUserId="4090" LastEditorUserId="4090" LastEditorDisplayName="" LastEditDate="2011-06-19T20:16:31.413" LastActivityDate="2011-06-19T21:25:47.707" Title="calculating thermal energy [exam prep]" Tags="&lt;energy&gt;&lt;forces&gt;&lt;friction&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="11270" PostTypeId="1" CreationDate="2011-06-19T20:37:02.760" Score="5" ViewCount="57" Body="&lt;p&gt;[sorry, this way below the level of this forum -- flames are most welcome]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a photon is absorbed by a piece of matter that does not reflect it -- where does the photon &quot;go&quot;? Eg, one shines light at a black object -- clearly the photon arrives at the object and then, well, vanishes.  Or does it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks very much.&lt;/p&gt;&#xA;" OwnerUserId="4091" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-20T02:12:14.940" LastActivityDate="2011-06-20T02:12:14.940" Title="Photon absorption" Tags="&lt;electromagnetism&gt;&lt;photons&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11271" PostTypeId="2" ParentId="11247" CreationDate="2011-06-19T21:22:53.460" Score="2" ViewCount="" Body="&lt;p&gt;1) Let us for simplicity put various constants to one: Speed of light $c=1$; Planck constant $\hbar=1$; Mass  $m=1$ of non-relativistic scalar (Bosonic) particle in $1+1$ dimensions; charge of particle $q=1$; Circumference of spatial circle $\ell=2\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The mechanical momentum (sometimes called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_momentum&quot; rel=&quot;nofollow&quot;&gt;kinetic momentum&lt;/a&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{v}~=~\hat{p}-A_x~=~\frac{1}{i}\partial_x-A_x~=~\frac{1}{i}D_x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(or equivalently, the covariant derivative) commutes with the Hamiltonian $\hat{H}=\frac{1}{2}\hat{v}^2+\Phi$ in the temporal gauge $\Phi=A^t=0$, which we will assume from now on. (Recall that $\Phi=A^t$ is the temporal component of the gauge potential.)  So we can find common eigenstates. Suppressing the time dependence in what follows, we want to solve the mechanical momentum eigenvalue equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{v}\psi_v(x)=v \psi_v(x),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the mechanical momentum eigenvalue $v\in\mathbb{R}$ is related to the energy&#xA;$E=\frac{1}{2}v^2$. The solution is $e^{ivx}$ times a Wilson line:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi_v(x)~=~ \psi_v(0)\exp\left[ivx+i\int_0^x A_x(x&amp;#39;)dx&amp;#39;\right] ,\qquad v\in\mathbb{R}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Under a local gauge transformation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi(x)\longrightarrow \tilde{\psi}(x):=e^{i\alpha(x)}\psi(x), \qquad A_x(x)\longrightarrow \tilde{A}_x(x):=A_x(x)+\partial_x\alpha(x), $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is well-known that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauge_covariant_derivative&quot; rel=&quot;nofollow&quot;&gt;covariant derivative&lt;/a&gt; (or mechanical momentum) transforms covariantly,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D_x\psi(x)\longrightarrow e^{i\alpha(x)}D_x\psi(x), \qquad &#xD;&#xA;D_x \longrightarrow \tilde{D}_x ~=~ e^{i\alpha(x)}D_x e^{-i\alpha(x)}, &#xD;&#xA;\qquad \hat{v}\longrightarrow e^{i\alpha(x)}\hat{v} e^{-i\alpha(x)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that they are not gauge invariant. They are only gauge covariant. However, we may easily construct manifestly gauge invariant quantities, for instance,  $|\psi(x)|^2$; $\psi^*(x) \hat{v}\psi(x)$. In particular, the mechanical momentum eigenvalue &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v~=~\frac{\hat{v}\psi_v(x)}{\psi_v(x)}~=~\pm\sqrt{2E}~\in~\mathbb{R}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a gauge invariant (still assuming temporal gauge $A^t=0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Finally, we assume for simplicity that $A_x(x)=A_x\in\mathbb{R}$ and $\tilde{A}_x(x)=\tilde{A}_x\in\mathbb{R}$ are independent of $x$. This corresponds to a partial gauge fixing. We still have residual gauge transformations left where $\alpha(x)$ is an affine function of $x$. The eigenfunction becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi_v(x)~=~ \psi_v(0)e^{i(v+A_x)x},\qquad&#xD;&#xA;\tilde{\psi}_v(x)~=~ \tilde{\psi}_v(0)e^{i(v+\tilde{A}_x)x},\qquad v\in\mathbb{R}. $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now recall that the $x$-coordinate is periodic $x\sim x +2\pi$. The wave function should be single-valued &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi_v(x+2\pi)=\psi_v(x),\qquad\tilde{\psi}_v(x+2\pi)=\tilde{\psi}_v(x),$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v+A_x,v+\tilde{A}_x~\in~\mathbb{Z},$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;as OP observes. In other words, $A_x$ and $\tilde{A}_x$ belong to the same shifted lattice $\mathbb{Z}-v$. The residual affine function $\alpha(x)$ must have $x$-independent integer-valued derivative &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\partial_x\alpha~=~\tilde{A}_x-A_x~=~(v+\tilde{A}_x)-(v+A_x)~\in~\mathbb{Z}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) So what have we learned? &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;On one hand, we may write $v\in\mathbb{Z}-A_x$, so we see that the energy $E=\frac{1}{2}v^2$ and the mechanical momentum $v$ &lt;em&gt;depend on the gauge&lt;/em&gt; potential $A_x\in\mathbb{R}$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On the other hand, we saw in section 3 that $E$ and $v$ &lt;em&gt;are gauge invariants,&lt;/em&gt; i.e., they are invariant under gauge transformations (still assuming temporal gauge $A^t=0$). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The two statements (1.) and (2.) do not clash in terms of physics, only in terms of semantics. In particular, if we perform a gauge transformation on the formula $v\in\mathbb{Z}-A_x$, we cannot claim that $v$ changes since a gauge transformation changes $A_x$ by an integer. ($A_x$ itself is not necessarily an integer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) Another example is the canonical momentum operator $\hat{p}=\frac{1}{i}\partial_x$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;On one hand, it is &lt;em&gt;independent of the gauge&lt;/em&gt; potential $A_x$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On the other hand, it is &lt;em&gt;not a gauge covariant quantity,&lt;/em&gt; i.e., it does not transform covariantly under a gauge transformation &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\hat{p}\longrightarrow  e^{i\alpha(x)}\hat{p} e^{-i\alpha(x)}~=~\hat{p}-\partial_x\alpha(x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, the two statements (1.) and (2.) do not clash in terms of physics, only in terms of semantics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7) Finally, let us connect to Marek's comment. We have a Hilbert space $L^2(\mathbb{R}/\mathbb{Z})$ of wave functions on a circle $\mathbb{R}/\mathbb{Z}$, and a Heisenberg algebra $[\hat{x},\hat{v}]=i$. (More precisely, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stone%E2%80%93von_Neumann_theorem&quot; rel=&quot;nofollow&quot;&gt;Stone-von Neumann theorem&lt;/a&gt; is a statement about the corresponding &lt;a href=&quot;http://en.wikipedia.org/wiki/Heisenberg_group&quot; rel=&quot;nofollow&quot;&gt;Heisenberg group&lt;/a&gt; to avoid issues with &lt;a href=&quot;http://en.wikipedia.org/wiki/Unbounded_operator&quot; rel=&quot;nofollow&quot;&gt;unbounded operators.&lt;/a&gt;) I interpret Marek's comment as roughly saying that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{x}=x, \qquad \hat{v}=\frac{1}{i}\partial_x-A_x,$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;yields inequivalent irreducible representations of the Heisenberg algebra labeled by a continuous label $A_x\in[0,1[$. Changing $A_x \to A_x+1$ yields an equivalent representation due to gauge symmetry.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-05T13:42:17.977" LastActivityDate="2011-07-05T13:42:17.977" CommentCount="9" />
  <row Id="11272" PostTypeId="2" ParentId="11269" CreationDate="2011-06-19T21:25:47.707" Score="1" ViewCount="" Body="&lt;p&gt;For the force of gravity use $F = m a$ where $a$ is the acceleration of gravity. For the frictional force, multiply this $F$ by the coefficient to get $F_f = \mu F = 0.105 F$. For the thermal energy use $W = F_f\;\; d$, where $W$ is the work and $d$ is the distance.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-19T21:25:47.707" />
  <row Id="11273" PostTypeId="2" ParentId="11270" CreationDate="2011-06-19T21:28:25.223" Score="2" ViewCount="" Body="&lt;p&gt;One way or another, the photon is absorbed by the material. If the photon has a high enough energy it could raise the energy level of an electron or even liberate one completely from an atom. This is the basis of the photo-electric effect that Einstein explained (and got the Noble prize for if I recall).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A low energy photon can excite a phonon, that is, a vibration in the solid. These increase the heat of the solid and with enough of them, you could get a detectable increase in temperature.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-19T21:28:25.223" />
  <row Id="11274" PostTypeId="2" ParentId="11241" CreationDate="2011-06-19T21:30:38.003" Score="3" ViewCount="" Body="&lt;p&gt;Apart from fluid dynamics there are also practical considerations.&lt;br&gt;&#xA;Ships propellers have to fit under ships - it would be inconvenient if they were 100m in diameter, however efficent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Model aeroplane propellers, were size is pretty much a free choice, ideally have single blade propellers (just one side with a counter weight) which are much to large to let the aeroplane take off/land on wheels&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-06-19T21:30:38.003" />
  <row Id="11275" PostTypeId="2" ParentId="11270" CreationDate="2011-06-19T21:43:10.580" Score="5" ViewCount="" Body="&lt;p&gt;You could think of a photon in this case as simply a little bundle of energy. That energy is absorbed by the material, and  the photon ceases to exist. An analogy would be waves crashing on a shore -- the wave carries energy, which is absorbed by the shore. Where did the wave go?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this is a vastly simplified answer. A photon is the quanta of energy in the electromagnetic field, while ocean waves are not quantized at all. The important thing to understand is that a photon is not a material object like something you can hold in your hand, and there should be no reason to expect that it should continue to exist after an interaction.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-06-19T21:43:10.580" />
  <row Id="11276" PostTypeId="1" CreationDate="2011-06-19T21:48:00.557" Score="2" ViewCount="73" Body="&lt;p&gt;I have a differential equation of the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{d^2 y}{dt^2} + f(t) \frac{dy}{dt} + g(t) y = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f$ and $g$ are known functions of time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a systematic (or otherwise) way of finding the conserved quantities, if there are any?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to google this topic, but haven't had much success yet. Perhaps someone with more mathematical knowledge than myself (i.e. almost everyone) can point me in the right direction. Even phrases to search for would help.&lt;/p&gt;&#xA;" OwnerUserId="1429" LastActivityDate="2011-06-20T00:21:32.210" Title="How do you find conserved quantities for linear second order ODEs?" Tags="&lt;classical-mechanics&gt;&lt;mathematical-physics&gt;&lt;mathematics&gt;&lt;conservation-laws&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="11277" PostTypeId="2" ParentId="11276" CreationDate="2011-06-19T23:13:28.883" Score="3" ViewCount="" Body="&lt;p&gt;There is a very general &lt;a href=&quot;http://en.wikipedia.org/wiki/Lie_theory&quot; rel=&quot;nofollow&quot;&gt;Lie theory&lt;/a&gt; that is most useful for partial differential equations but it can also be applied here. The full theory is not easy to explain and I am only familiar with it from one course I attended, so I cannot give a reference from the top of my ahead. I will try to provide it later. Try to look for keywords &lt;em&gt;symmetries and conserved quantities&lt;/em&gt; and &lt;em&gt;contact transformation&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic point is finding infinitesimal transformations $(t,y) \to (t,y) + \epsilon(t,y)$ that preserve the equation. Under special circumstances (like the possibility of variational formulation of the problem) some of these give rise to conserved quantities via Noether's theorem. But there need not be any associated conserved quantities (indeed, the space of infinitesimal transformations that preserve the equation is infinite-dimensional).&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-06-19T23:13:28.883" />
  <row Id="11278" PostTypeId="2" ParentId="11241" CreationDate="2011-06-19T23:56:16.200" Score="4" ViewCount="" Body="&lt;p&gt;This &lt;a href=&quot;http://www.aerospaceweb.org/question/propulsion/q0039.shtml&quot; rel=&quot;nofollow&quot;&gt;question of the week&lt;/a&gt; article describes the tradeoffs of aircraft propeller blades - number, angle, shape, length etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final paragraph gives a good summary and relate directly to your question about why the blades do not cover the entire circle. It turns out that they do cover as much as is needed for a given engine to transfer the available energy to the air, which is I guess why you see everything from simple 2-blade props to jet-turbines.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We are now left with the final two options, increasing the blade chord or the number of blades. Both have the effect of increasing the solidity of the propeller disk. Solidity simply refers to the area of the propeller disk occupied by solid componenets (the blades) versus that area open to the air flow. As solidity increases, a propeller can transfer more power to the air.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;While increasing the blade chord is the easier option, it is less efficient because the aspect ratio of the blades is decreased resulting in some loss of aerodynamic efficiency. Thus, increasing the number of blades is the most attractive approach. As the power of engines increased over the years, aircraft designers adopted increasingly more propeller blades. Once they ran out of room on the propeller hub, designers adopted twin contra-rotating propellers on the same engine. Two good examples are the Tu-95 bomber and Tu-114 airliner. These Russian aircraft were equipped with the most powerful turboprop engines ever built, and both designs feature a total of eight propeller blades per engine. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-06-19T23:56:16.200" />
  <row Id="11279" PostTypeId="1" CreationDate="2011-06-20T00:16:05.717" Score="2" ViewCount="83" Body="&lt;p&gt;The world &lt;em&gt;A Song of Fire and Ice&lt;/em&gt; series is set on is characterized by seasons that last several years (how years are measured on this planet is not specified to my knowledge).  However, they are of unpredictable, variable lengths, though seemingly in balance (long summers are followed by long winters).  Rather than some sort of massive El Ni&amp;ntilde;o, it seems to be a result of something astrophysical, as winter can be forecast by the shortening of days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some possible mechanic for this or is it outside the realm of possibility?&lt;/p&gt;&#xA;" OwnerUserId="431" LastActivityDate="2011-06-20T20:50:05.357" Title="What could cause a planet to have apparently random season lengths?" Tags="&lt;astrophysics&gt;&lt;climate-science&gt;" AnswerCount="4" />
  <row Id="11280" PostTypeId="2" ParentId="11276" CreationDate="2011-06-20T00:21:32.210" Score="3" ViewCount="" Body="&lt;p&gt;+1 to Marek's answer. I will provide you with some references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The canonical reference is without much doubt&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://books.google.com/books?id=sI2bAxgLMXYC&amp;amp;hl=da&amp;amp;source=gbs_navlinks_s&quot; rel=&quot;nofollow&quot;&gt;Peter J. Olver - Applications of Lie groups to differential equations&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I bought this book a few years ago, but haven't found much time to study it thoroughly. But if you know modern differential geometry well (manifolds, vector fields, pull-back, differential forms, ...), then this book is a pleasure to read. For easier/less formal introductions, take a look at&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://books.google.com/books?id=XZ-oyMF8uqcC&amp;amp;hl=da&amp;amp;source=gbs_navlinks_s&quot; rel=&quot;nofollow&quot;&gt;Brian Cantwell - Introduction to symmetry analysis&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://books.google.com/books?id=wVu0hEWLKuAC&amp;amp;hl=da&amp;amp;source=gbs_navlinks_s&quot; rel=&quot;nofollow&quot;&gt;Peter E. Hydon and Peter Ellsworth Hydon - Symmetry methods for differential equations: a beginner's guide&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I don't know these two books very well, but I took a quick look at them a few years ago and they seemed quite useful.&lt;/p&gt;&#xA;" OwnerUserId="1469" LastActivityDate="2011-06-20T00:21:32.210" CommentCount="1" />
  <row Id="11281" PostTypeId="2" ParentId="11279" CreationDate="2011-06-20T00:37:42.383" Score="1" ViewCount="" Body="&lt;p&gt;N-body problem should pretty well cover this. Solving the almost intractable problem of three gravitating mass goes back to Newton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/N-body_problem#Three-body_problem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/N-body_problem#Three-body_problem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoting from the above article: &quot;Specific solutions to the three-body problem result in chaotic motion with no obvious sign of a repetitious path.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So assuming three orbiting bodies in chaotic motion, one could predict aperiodic seasons. Hence, yes, there should exist somewhere in this universe a planetary system with random season lengths.&lt;/p&gt;&#xA;" OwnerUserId="3588" LastActivityDate="2011-06-20T00:37:42.383" CommentCount="2" />
  <row Id="11282" PostTypeId="2" ParentId="11279" CreationDate="2011-06-20T00:45:13.083" Score="2" ViewCount="" Body="&lt;p&gt;The seasons on earth are caused by the rotation of the earth around the sun in addition with the inclination of the axis of the earth. This causes the angle of incident sunlight to be lower in winter and higher in summer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the inclination of the axis of the earth is pretty constant. But if that axis would shift in some funny way, you could have any season any time. So, for a summer that lasts several years, the axis would have to shift so that the light incident on a given hemisphere remains at a high angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, even defining a year might be hard on such a planet. On earth, we think of a year as the time it takes for the earth to circle the sun once. But that is not how ancient civilizations measured it. They looked at the height of the sun in zenith and noted that there was a day when the sun stood highest, and a day where the sun stood lowest, and these are summer and winter solstice, respectively. With the earth's axis's angle fixed, the time span from one summer solstice to the next is the same as the time it takes for the earth to completely circle the sun, but in our imaginary planetary system, these two movements would not be related anymore. But with random axis movement, it might be impossible to tell when a circle around the sun is completed...&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-06-20T00:45:13.083" CommentCount="1" />
  <row Id="11284" PostTypeId="1" CreationDate="2011-06-20T01:31:47.240" Score="3" ViewCount="186" Body="&lt;p&gt;Is there any restriction in what we know of physics to the existence of other type of forces that obey the inverse square law in 3 dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean other than electromagnetic and gravitational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also if there were some other kind of force ..would it follow that it would have a dual like electric and magnetic forces are duals of each other.&lt;/p&gt;&#xA;" OwnerUserId="4094" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-20T02:14:07.343" LastActivityDate="2011-08-19T15:36:59.843" Title="existence of other forces obeying inv square law" Tags="&lt;gravity&gt;&lt;electromagnetism&gt;&lt;forces&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="11286" PostTypeId="2" ParentId="11284" CreationDate="2011-06-20T02:11:22.227" Score="0" ViewCount="" Body="&lt;p&gt;No, there's no reason that the electromagnetic and gravitational forces necessarily &lt;em&gt;have&lt;/em&gt; to be the only ones with $\frac{1}{r^2}$ dependence. Or rather, there's no reason that we know of, and I don't think there are any theories being seriously considered that do predict those to be the only two inverse-square-law forces. They are simply the only ones that have been observed. (Note that gravity is actually not quite an inverse-square force once you take general relativity into account.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If another inverse-square force did exist,it would not necessarily have a dual force, the way the electric and magnetic forces do.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-20T02:11:22.227" CommentCount="5" />
  <row Id="11287" PostTypeId="1" CreationDate="2011-06-20T04:04:27.963" Score="2" ViewCount="127" Body="&lt;p&gt;After some discussions with my friend about some &quot;popular&quot; aspects of quantum mechanics, my friend asked me whether there exist any books that could convey the basic ideas in a non-technical way (my friend is not in any technical field). I am in mathematics, so I'm not aware of any such texts in physics. Your suggestions would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3657" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-20T06:31:16.870" LastActivityDate="2011-06-20T13:41:23.723" CommunityOwnedDate="2011-06-20T16:04:44.980" Title="Popular books on QM" Tags="&lt;quantum-mechanics&gt;&lt;soft-question&gt;&lt;books&gt;&lt;reference-request&gt;&lt;big-list&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="11288" PostTypeId="1" CreationDate="2011-06-20T04:12:24.767" Score="2" ViewCount="174" Body="&lt;p&gt;There's no guarantee that supersymmetry (or more exotic new physics) will be seen at the LHC. Meanwhile, it's standard lore that a Higgsless standard model &lt;a href=&quot;http://research.kek.jp/group/www-theory/theory_center/TC_LHC091028/tanabashi.pdf&quot; rel=&quot;nofollow&quot;&gt;becomes nonunitary&lt;/a&gt; somewhere in the vicinity of 1 TeV. Are we at least guaranteed to get &lt;em&gt;some&lt;/em&gt; positive information about new physics, in the form of deviations from the predictions of this &quot;higgsless standard model&quot;? &lt;/p&gt;&#xA;" OwnerUserId="1486" LastActivityDate="2011-06-20T06:38:31.867" Title="Might the LHC see nothing new at all?" Tags="&lt;standard-model&gt;&lt;large-hadron-collider&gt;&lt;higgs-boson&gt;&lt;electroweak&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="11289" PostTypeId="1" CreationDate="2011-06-20T05:01:55.410" Score="0" ViewCount="28" Body="&lt;p&gt;Recently i got to know that there is a machine to produce artificial thunders?Does the machine to produce artificial thunders exist?If it exist what is the principle behind the machine?How does the machine produce artificial thunders?&lt;/p&gt;&#xA;" OwnerUserId="4026" LastActivityDate="2011-06-20T05:01:55.410" ClosedDate="2011-06-20T06:13:18.963" Title="Is there any macine that generates the artificial thunders?" Tags="&lt;electricity&gt;" CommentCount="1" />
  <row Id="11290" PostTypeId="2" ParentId="11287" CreationDate="2011-06-20T05:08:25.300" Score="2" ViewCount="" Body="&lt;p&gt;There's the book &quot;Quantum enigma&quot;... It discusses the various interpretations of quantum mechanics for people with little background.&lt;/p&gt;&#xA;" OwnerUserId="4095" LastActivityDate="2011-06-20T05:08:25.300" CommunityOwnedDate="2011-06-20T16:04:44.980" />
  <row Id="11292" PostTypeId="2" ParentId="414" CreationDate="2011-06-20T05:11:35.797" Score="3" ViewCount="" Body="&lt;p&gt;There's a fantastic article on the relationship between the Riemann Hypothesis and &quot;quantum chaos&quot; at &lt;a href=&quot;http://www.msri.org/ext/Emissary/EmissarySpring02.pdf&quot; rel=&quot;nofollow&quot;&gt;www.msri.org/ext/Emissary/EmissarySpring02.pdf&lt;/a&gt; (starts on page 1, continues on page 12). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an excerpt (recall that &lt;a href=&quot;http://en.wikipedia.org/wiki/Montgomery%27s_pair_correlation_conjecture&quot; rel=&quot;nofollow&quot;&gt;Montgomery's Conjecture&lt;/a&gt; is a conjecture about the expected number of zeros of the Riemann zeta function that follow a zero in an interval of a certain length):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Montgomery was taken aback to discover&#xA;  that Dyson knew very well the rather&#xA;  complicated function appearing in&#xA;  Montgomery’s conjecture, and even knew&#xA;  it in the context of comparing gaps&#xA;  between points with the average gap.&#xA;  However — here’s the amazing thing: It&#xA;  wasn’t from number theory that Dyson&#xA;  knew this function but from quantum&#xA;  mechanics. It is precisely the&#xA;  function that Dyson himself had found&#xA;  a decade earlier when modelling energy&#xA;  levels in complex dynamical systems&#xA;  when taking a quantum physics view-&#xA;  point. It is now believed that the&#xA;  same statistics describe the energy&#xA;  levels of chaotic systems; in other&#xA;  words, quantum chaos!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The article describes some other surprising connections as well, between different zeta functions and the energy levels of other kinds of chaotic systems. Instead of copying those out here (I can't summarize, since I don't understand it well myself), I'll just end with a quote from the article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In summary, the more intuitive&#xA;  development of quantum chaos allows&#xA;  more fruitful predictions about the&#xA;  distribution of primes (and beyond).&#xA;  On the other hand the more cautious&#xA;  development of prime number theory&#xA;  leads to more accurate predictions in&#xA;  quantum chaos.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3199" LastActivityDate="2011-06-20T05:11:35.797" />
  <row Id="11293" PostTypeId="1" CreationDate="2011-06-20T05:18:55.370" Score="1" ViewCount="73" Body="&lt;p&gt;I am trying to follow a book (Introduction to Ligand Field Theory by Ballhausen in 1962 on pg 15), but it isn't clear how they make a particular leap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&#xA;I want to find the wave function for the term $^1 D$. We know that $\psi(L,M_L,S,M_S) = \psi (2,2,0,0)$ is made up of three micro states: $(2^+,0^-)$, $(2^-,0^+)$, $(1^+,1^-)$. Thus, a linear combination must be taken of these micro states. Now in the book they say they must be orthogonal to $\psi(3,2,1,0)$ and $\psi(4,2,0,0)$. Presumably these are picked because we just determined those via lowering operators in the previous section. I don't take it that there is another reason those two wave functions are mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2^+,0^−)$ means that electron 1 has $m_s=+1/2$ and $m_l=2$ and that electron 2 has $m_s=−1/2$ and $m_l=0$. We write that $\Psi=|(\psi^+_1)(\psi^−_2)|$ where we have written the short form of the determinantal antisymmetrized normalized wave function, which comes from the diagonal element in the Slater determinant AND separated the orbital- and spin-dependent parts of the wave function (spin is denoted by super plus or minus sign).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;&#xA;Anyway, I understand that $\psi(2,2,0,0) = a (2^+,0^-) + b (2^-,0^+) + c(1^+,1^-)$.&#xA;Then they write &quot;and we get $a \sqrt{3} - b \sqrt{3} + c \sqrt{8} = 0$ and $a + b = 0$.&quot; I'm totally lost how they made this leap and why these must be equal to zero. Does this have to do with the orthogonal wave functions just mentioned? Where did the numbers in the sqrt come from? And the minus sign! Greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4096" LastEditorUserId="4096" LastEditorDisplayName="" LastEditDate="2011-06-20T18:11:42.967" LastActivityDate="2011-06-21T03:17:00.007" Title="Determining wave function for term symbol 1D" Tags="&lt;quantum-mechanics&gt;&lt;wavefunction&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="11294" PostTypeId="2" ParentId="11288" CreationDate="2011-06-20T05:28:36.673" Score="1" ViewCount="" Body="&lt;p&gt;The LHC opens a new frontier in energy exploration. The results will be &quot;new&quot; by definition, as with all explorations. Now whether there will be dragons, tigers or just moose is something we have to wait for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we enter a desert as far as specific new resonances/particles go, which are predicted by various models, that will also be new physics information. Supersymmetry will disappear on the horizon, the Higgs will be rescued by composite Higg models until the next machine and higher energy levels are explored. This rescue will come, because the Standard Model has been very successful in encapsulating all known high energy data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion we will find new physics, maybe unexpected, but new. Already the asymmetry in the jets of &lt;a href=&quot;http://motls.blogspot.com/2011/04/is-top-antitop-asymmetry-due-to-stop.html&quot; rel=&quot;nofollow&quot;&gt;top production&lt;/a&gt; at the Tevatron is an indication that something new is going on at high energies. The researchers have to keep an open mind, not run after two and three sigma deviations, be very careful in systematic error estimates, and &lt;em&gt;&quot;something is coming , don't know what, ought to be grand&quot;&lt;/em&gt; as the song from the &lt;em&gt;west side story&lt;/em&gt; says.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-06-20T05:28:36.673" CommentCount="2" />
  <row Id="11295" PostTypeId="1" CreationDate="2011-06-20T05:28:58.707" Score="7" ViewCount="196" Body="&lt;p&gt;The idea of “hadronic supersymmetry” originated in the mid-1960s and derives from the observation that baryons and mesons have similar Regge slopes, as if antiquarks and diquarks are superpartners. This is most efficiently explained by supposing a QCD string model of baryons in which a diquark substitutes for an antiquark at one end of the string, but there are &lt;a href=&quot;http://arxiv.org/abs/0901.4508&quot; rel=&quot;nofollow&quot;&gt;superstring models in which the baryon-meson relationship is a genuine supersymmetry&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2005, Alejandro Rivero &lt;a href=&quot;http://arxiv.org/abs/hep-ph/0512065&quot; rel=&quot;nofollow&quot;&gt;proposed&lt;/a&gt; that mesons are to &lt;em&gt;leptons&lt;/em&gt; as diquarks are to quarks. If the top quark is excluded from consideration, as too heavy to hadronize, then one has enough quark-antiquark pairings to match the electromagnetic charges of all the leptons, with a few leftover u,c pairings with charge ±4/3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The construction is a little complicated, but the possibility that supersymmetry is already right in front of us is so amazing that it just cries out to be investigated. I have an &lt;a href=&quot;http://www.physicsforums.com/showthread.php?t=485247&quot; rel=&quot;nofollow&quot;&gt;ongoing discussion&lt;/a&gt; with Rivero, about whether his correspondence might be realized in a preon model, or a &quot;partially composite&quot; model, or a sophisticated string-theory construction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, I'd like to know, &lt;strong&gt;is there some reason why this can't be realized?&lt;/strong&gt; The closest thing to such a reason that I've seen is found in comment #11 in that thread: &quot;For a SUSY theory, not just the spectrum must be supersymmetric, but also the interactions between the particles.&quot; So far, all Rivero has is the observation that some quantum numbers of diquarks/mesons can be correlated with some quantum numbers of fundamental quarks/leptons, just as if they were elements of a single superfield. I suspect (but I'm such a susy newbie that I can't even confirm this) that you could realize the correspondence in a QCD-like supersymmetric theory, if it was first expressed in terms of meson and diquark variables, and then supersymmetry was completely broken (see comments #40 through #44), but that this would somehow be trivial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, I think the possibility that supersymmetry is already right in front of us would be important enough, that someone should investigate even such a &quot;trivial&quot; realization of Rivero's correspondence, with a view to understanding (i) whether such a &quot;hard&quot; form of susy breaking might realistically occur (ii) how it would affect the various roles and problems associated with supersymmetry in contemporary physical thought (protect the Higgs mass, supply dark matter candidates; technical model-building issues like &lt;a href=&quot;http://arxiv.org/abs/hep-th/9810155&quot; rel=&quot;nofollow&quot;&gt;&quot;the supersymmetric flavor problem, the gaugino mass problem, the supersymmetric CP problem, and the mu-problem&quot;&lt;/a&gt;...). &lt;/p&gt;&#xA;" OwnerUserId="1486" LastActivityDate="2011-06-24T23:33:39.410" Title="Alejandro Rivero's correspondence: diquarks and mesons as superpartners of quarks and leptons" Tags="&lt;supersymmetry&gt;&lt;high-energy-physics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="4" />
  <row Id="11296" PostTypeId="2" ParentId="11287" CreationDate="2011-06-20T06:08:36.300" Score="-2" ViewCount="" Body="&lt;p&gt;Without question you need to read Macschrödinger's Cat by the somewhat cryptically name Reed de Buch&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://farm6.static.flickr.com/5022/5851536199_dbd36f1b19_m.jpg&quot; alt=&quot;Cover Macschrödingers-Cat&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.amazon.com&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com&lt;/a&gt;/Macschrödingers-Cat-strange-tale-universe/dp/1460965019/ref=sr_1_2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It wasn't until I found this bizarre book that I finally understood how it was possible to understand modern physics without equations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't treat the reader like an idiot, yet at the same time it's filled with idiocy. You could also look at the author's blog &lt;a href=&quot;http://reeddebuch.blogspot.com/&quot; rel=&quot;nofollow&quot;&gt;http://reeddebuch.blogspot.com/&lt;/a&gt; but he hasn't written much this year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although, how the author turned Erwin Schrödinger into a deranged hebephrenic Scotsman beats me.&lt;/p&gt;&#xA;" OwnerUserId="3588" LastEditorUserId="3588" LastEditorDisplayName="" LastEditDate="2011-06-20T13:41:23.723" LastActivityDate="2011-06-20T13:41:23.723" CommunityOwnedDate="2011-06-20T16:04:44.980" />
  <row Id="11298" PostTypeId="2" ParentId="11288" CreationDate="2011-06-20T06:38:31.867" Score="5" ViewCount="" Body="&lt;p&gt;It is possible that the LHC could find just a standard model scalar Higgs. It would have to fit in the windows between 135 GeV and 157 GeV or 173 GeV to 185 GeV. In this case no new physics is needed and other effects such as large CP violations and dark matter go unexplained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is below 135 GeV the standard model would predict an instability of the vacuum so something else such as SUSY would have to be there to correct. This would already be indirect evidence for beyond standard model physics, but it is just about possible that the new physics would not be seen yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Higgs mass above 185 GeV is inconsistent with other measurements, but only when assuming the standard model applies, so again a heavy Higgs would be indirect evidence of new physics. There is no certainty that the new physics would be seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Higgs was excluded over all its mass range that would also be an indication of beyond standard model physics. It would mean one of two things, either the Higgs really does not exists and something else would have to be breaking electro-weak symmetry, or the Higgs exists but some new physics hides it from us, e.g. it may have a dominant decay mode into new particles that form jets that are hard to see against the background. Such scenarios are possible but it would be as if Nature was being devious to make life awkward for us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something needs to explain dark matter but that could easily be a particle with mass 10 TeV, or a lighter particle that does not interact strongly enough with standrd model particles to be seen at the LHC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in short, yes it is quite possible that nothing, or just the standard Higgs may be seen. &lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-06-20T06:38:31.867" />
  <row Id="11299" PostTypeId="1" CreationDate="2011-06-20T06:59:29.787" Score="0" ViewCount="40" Body="&lt;p&gt;Do spacelike singularities really exist in quantum gravity? If the memory of anything which falls into a black hole can't get out, is there any sense in which the interior of the black hole is real? Similarly, if we don't have any direct records of the spacelike singularity in the past, is there any sense in which it is real?&lt;/p&gt;&#xA;" OwnerUserId="4098" LastActivityDate="2011-06-20T06:59:29.787" Title="Do spacelike singularities really exist in quantum gravity?" Tags="&lt;quantum-gravity&gt;" CommentCount="2" />
  <row Id="11300" PostTypeId="1" CreationDate="2011-06-20T07:00:44.793" Score="0" ViewCount="57" Body="&lt;p&gt;If there are no localized observables in quantum gravity, does spacetime really exist, or might spacetime really be an illusion?&lt;/p&gt;&#xA;" OwnerUserId="4098" LastActivityDate="2011-06-20T07:00:44.793" Title="Does spacetime really exist in quantum gravity?" Tags="&lt;quantum-gravity&gt;" CommentCount="6" />
  <row Id="11301" PostTypeId="1" CreationDate="2011-06-20T07:19:05.877" Score="4" ViewCount="72" Body="&lt;p&gt;In a previous question on &lt;a href=&quot;http://physics.stackexchange.com/questions/3773/shape-of-the-higgs-branching-ratio-to-zz&quot;&gt;Higgs branching ratios&lt;/a&gt;, I find this image&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/eljaD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(originally from page 15 &lt;a href=&quot;http://arxiv.org/abs/hep-ph/9704448&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am VERY intrigued by the fact that decays to WW, gg, and ττ are almost equally probable, for a standard model Higgs with a mass in the vicinity of 115 GeV. I've &lt;a href=&quot;http://physics.stackexchange.com/questions/8999/115-gev-170-gev-and-the-noncommutative-standard-model&quot;&gt;noted previously&lt;/a&gt; that this is a special value: it lies in a narrow range of values for which the SM vacuum is metastable. Is there a connection? &lt;/p&gt;&#xA;" OwnerUserId="1486" LastActivityDate="2011-06-20T07:19:05.877" Title="Why are WW gg ττ branching ratios so similar for a 115 GeV SM Higgs?" Tags="&lt;standard-model&gt;&lt;higgs-boson&gt;" CommentCount="5" />
  <row Id="11303" PostTypeId="1" AcceptedAnswerId="11368" CreationDate="2011-06-20T07:28:50.423" Score="4" ViewCount="141" Body="&lt;p&gt;Imagine you have a pile of snow and a pile of ice shards. You put a soda bottle which has a room temperature into both piles. Which bottle is going to cool down faster?&lt;/p&gt;&#xA;" OwnerUserId="4100" LastActivityDate="2011-06-20T22:19:42.627" Title="what cools bottle of water faster: ice or snow" Tags="&lt;temperature&gt;&lt;ice&gt;" AnswerCount="4" />
  <row Id="11304" PostTypeId="1" AcceptedAnswerId="11308" CreationDate="2011-06-20T07:53:09.537" Score="0" ViewCount="76" Body="&lt;p&gt;let's say we pour some water to the Sun. Or boil water on 2000 degrees Celsius on Earth.&lt;/p&gt;&#xA;" OwnerUserId="4101" LastActivityDate="2011-06-20T08:42:20.730" Title="What happen when a gas is being heated far beyond its boiling point?" Tags="&lt;material-science&gt;" AnswerCount="2" />
  <row Id="11305" PostTypeId="2" ParentId="11304" CreationDate="2011-06-20T08:13:27.110" Score="3" ViewCount="" Body="&lt;p&gt;Well, take a look at liquid nitrogen - it just boils rapidly, and then it's just 'hot' gas, no matter how high is the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you pour water on sun - it will evaporate instantly, and you will have just steam at 6000 temperature (but in that range there will be molecular dissociation, but it's out of scope I guess). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, at high temperatures gas is likely to go into plasma state (i.e. became ionized).&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-06-20T08:13:27.110" />
  <row Id="11306" PostTypeId="1" AcceptedAnswerId="11960" CreationDate="2011-06-20T08:15:57.263" Score="1" ViewCount="121" Body="&lt;p&gt;&lt;strong&gt;Problem/Purpose of me asking this question to you people who know more than me:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm doing a science project where I'm collimating a beam of light to a focus point in a light medium (water vapor or fog) and I want to calculate the intensity near it. I can't seem to find an equation that describes this problem. I want to know two things. If you know anything that can be related in solving this problem, it is much appreciated! :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Issues/Things I need help to figure out!:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) If I focus a collimating beam of light with a lens, (say a hand held magnifying glass), into a relatively uniformly dispersed light medium, (water vapor or theatrical fog) Can this focus point be seen in &lt;strong&gt;ANY direction&lt;/strong&gt;, (say like 5 feet away from the focus point)? Doesn't light scatter isotropically in this case? IF NOT, What is the preferred direction of scattering of the light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) If an equation exists (And light scatters isotropically in the light medium used), can this equation say given the parameters, (light frequency used, index of refraction of medium, density of medium, size of collimated beam, lens dimensions used, etc) give the intensity of light in terms of the distance a person is away from the focal point? I am aware of the &lt;strong&gt;inverse square law&lt;/strong&gt;, but in my case, its a bit different, isn't it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wouldn't my situation involve some type of directionality? How do you find the best viewpoints from the focal point in which the intensity of the focal point is most profound?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MORE relevant or related questions that need to be addressed:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does the particle size matter?, (the particles that make up the light medium)&lt;/li&gt;&#xA;&lt;li&gt;How do I determine the correct density of the given medium to produce the most profound effect, (Having the focal point illuminate as brightly as possible)&lt;/li&gt;&#xA;&lt;li&gt;How do I determine the right intensity of the initial column of light that is focused?&lt;/li&gt;&#xA;&lt;li&gt;Combining the previous two bullets, How do I determine the right combination of the density of light medium and intensity of light used to illuminate the focus point as bright as possible?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Gist:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm trying to do is to create a &quot;point of light&quot; inside a suspended light medium (Ideally viewable in all directions) and I'm trying to figure a way to figure this out with equations before buying a whole ensemble of things (Fog machine, light source/laser, magnifying glass or multiple lenses, etc) to test it. (If not viewable in all directions, and it has directionality, then I'm just gonna combine multiple systems pointing multiple columns of focused light in different directions to the same point in space to obtain an acceptable looking &quot;point of light&quot; in a medium).&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2011-07-06T22:47:55.657" Title="How do you calculate the intensity of light around the focal point from a focused collimated beam of light?" Tags="&lt;optics&gt;&lt;light&gt;&lt;experimental-physics&gt;&lt;scattering&gt;&lt;reflection&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="11307" PostTypeId="2" ParentId="11303" CreationDate="2011-06-20T08:16:56.997" Score="5" ViewCount="" Body="&lt;p&gt;This depends on contact area between bottle &amp;amp; ice/snow, and consistency of snow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is not much air in the snow, it should have bigger contact area with bottle, and thus heat will be transferred faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ice will contact with the bottle mainly at shards edges, so contact area is small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. Adding water will change everything, as contact area would be maximized in both cases.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-06-20T08:16:56.997" CommentCount="4" />
  <row Id="11308" PostTypeId="2" ParentId="11304" CreationDate="2011-06-20T08:42:20.730" Score="1" ViewCount="" Body="&lt;p&gt;The water will probably turn into plasma really fast. I'm too lazy here to look up exactly how to figure this out but I can provide an outline to answering your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Figure out how much energy is needed to split water into its atomic constituents. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Figure out how much energy is needed to pull two hydrogens off of oxygen. Then see if 2000 degrees C is more or less energy using the Temperature to Energy relationship, E = 0.5*kT, (E is average kinetic energy of particles in question, k is Boltzmann's constant, and T is temperature in Kelvins.) NOTE. This will give an approximate value of the energy in terms of temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IF you have more than enough energy to do this, then use the remaining energy you have left and go onto part b.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Figure out how much energy is needed to ionize an electron from Hydrogen. Multiply that by 2, because you have two hydrogen atoms. Then figure out how much energy is needed to ionize all of Oxygen (Thats a lot of freaking energy). Then again, see how much energy you have left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IF you still have more than enough energy, then move onto part c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) So by now you have 2 protons, 2 electrons from the 2 Hydrogens, and an Oxygen nucleus and its how ever many electron. You can then figure out how much energy is needed to break apart the Oxygen nucleus into free protons and neutrons, (THIS IS A LOT OF FREAKING ENERGY, and I don't know how to calculate this... good luck finding that.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IF you still have more than enough energy, then move onto part d.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) Figure out how much energy is needed to split protons into quarks, LOL. And see if you still have enough energy left. This is kinda overkill. Again, don't know how to do this. Sorry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you figured out how much energy is needed to break one water molecule into whatever part you get to, b, c, or d. Multiply that by the number of water molecules you have to figure out the total amount of energy needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in conclusion, its just simply comparing how much energy is needed to break apart water molecules and seeing what it turns into.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2011-06-20T08:42:20.730" CommentCount="1" />
  <row Id="11309" PostTypeId="2" ParentId="11246" CreationDate="2011-06-20T09:18:13.877" Score="1" ViewCount="" Body="&lt;p&gt;I would think you can decouple those two problems, since the respective timescales are very different - particles at supersonic speed are still as good as static for photons moving at lightspeed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'd say find separate software for both - Fluent or OpenFoam for the CFD as per BarsMonster's linked question, and some raytracing package for the scattering simulation. I'm not knowledgeable in that regard, but &lt;a href=&quot;http://www.blender.org&quot; rel=&quot;nofollow&quot;&gt;Blender&lt;/a&gt; may be worth a shot.&lt;/p&gt;&#xA;" OwnerUserId="3951" LastActivityDate="2011-06-20T09:18:13.877" />
  <row Id="11310" PostTypeId="2" ParentId="11303" CreationDate="2011-06-20T09:22:37.567" Score="1" ViewCount="" Body="&lt;p&gt;For one you must define room temperature. because if the bottle is worm enough to to slightly melt the show first and then refreeze then this one will cool 1st of not the air in the snow can act as an insulator and this one will take longer. Now the shape and sise of the ice will make a large contribution/small to the surface contact on the bottle this will also make a difference on the air flow around the pieces of ice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to make the experiment MUCH more specific to get a correct answer.&lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-06-20T09:22:37.567" CommentCount="4" />
  <row Id="11311" PostTypeId="1" AcceptedAnswerId="11372" CreationDate="2011-06-20T09:54:30.567" Score="2" ViewCount="69" Body="&lt;p&gt;&lt;strong&gt;transition probabilites&lt;/strong&gt; of atomic systems prone to some time-varying electromagnetic field are very often calculated using &lt;strong&gt;perturbation theory&lt;/strong&gt; leading to expressions including the vector potential $\mathbf{A}$. This approach seems to be very complicated and I am wondering:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h3&gt;Is using the electrostatic potential $U$ to calculate &quot;electric transitions&quot; (approximately) sufficient?&lt;/h3&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This was the short version of my question; below you will find some further notes that (hopefully) contain all information needed to clearify it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;strong&gt;relativistic quantum mechanics&lt;/strong&gt;, the action of an an &lt;strong&gt;electrodynamic field&lt;/strong&gt; on an electron can be realized by &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauge_theory#An_example%3a_Electrodynamics&quot; rel=&quot;nofollow&quot;&gt;minimal coupling&lt;/a&gt; using the replacement $\partial_\mu \rightarrow \partial_\mu - \mathrm{i}eA_\mu$. Thus, in a certain sense, (quantum) electrodynamics can be derived from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_equation&quot; rel=&quot;nofollow&quot;&gt;Dirac equation&lt;/a&gt; using the invariance of the free-electron action with respect to a phase change.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Electromagnetic Fields and the Schrödinger Equation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, often one must ignore the beauty of &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauge_theory&quot; rel=&quot;nofollow&quot;&gt;gauge theory&lt;/a&gt; to be able to &lt;strong&gt;practically calculate&lt;/strong&gt; something under certain simplifications using the non-relativistic &lt;a href=&quot;http://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation&quot; rel=&quot;nofollow&quot;&gt;Schrödinger-equation&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There, one can incorporate an external electromagnetic field applying the replacements&#xA;$$H\left(\mathbf{p},\mathbf{x}\right) \rightarrow H\left(\mathbf{p}-e\mathbf{A}\left(t,\mathbf{x}\right),\mathbf{x}\right)+eU\left(t,\mathbf{x}\right)$$ to the Hamiltonian of the system where $\mathbf{A}$ is the known &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_potential&quot; rel=&quot;nofollow&quot;&gt;vector potential&lt;/a&gt; and $U$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electric_potential&quot; rel=&quot;nofollow&quot;&gt;potential corresponding to the electric field&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &lt;a href=&quot;http://en.wikipedia.org/wiki/Coulomb_gauge#Coulomb_gauge&quot; rel=&quot;nofollow&quot;&gt;Coulomb gauge&lt;/a&gt; $\mathrm{div}\mathbf{A}=0$ one can always achieve $U\equiv 0$ if no sources are present and the overall Hamiltonian will look like $$H = H_0 -\frac{e}{m}\mathbf{A}\left(t,\mathbf{x}\right)\mathbf{p}$$&#xA;where $H_0 = \frac{\mathbf{p}^2}{2m} + V(\mathbf{x})$ and the term in $\mathbf{A}^2$, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ponderomotive_force&quot; rel=&quot;nofollow&quot;&gt;ponderomotive potential&lt;/a&gt; has been neglected.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Time-Dependent Perturbation Theory&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The last expression for $H$ has the form $$H = H_0(\mathbf{p},\mathbf{x}) + H_t(\mathbf{p},\mathbf{x},t)\ .$$&#xA;Using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dyson_series&quot; rel=&quot;nofollow&quot;&gt;Dyson series&lt;/a&gt;, one can formally write down the time-dependent solution in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Interaction_picture&quot; rel=&quot;nofollow&quot;&gt;interaction picture&lt;/a&gt; as &#xA;$$|\psi_I(t)&amp;gt; = T\exp\left[\frac{1}{\mathrm{i}\hbar}\int_{t_{0}}^{t}H_{t,I}\left(t^{\prime}\right)dt^{\prime}\right]|\psi_I(t_0)&amp;gt; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, one takes only the first two terms into account to find that &lt;strong&gt;transition amplitudes&lt;/strong&gt; are approximately given by $$A_{n\rightarrow m}\approx\frac{1}{\mathrm{i}\hbar}\int_{t_{0}}^{t}\left\langle m\right|H_{t,I}\left(t^{\prime}\right)\left|n\right\rangle dt^{\prime}\ \mathrm{with}$$&#xA;$$H_{t,I}\left(t\right) = e^{\mathrm{i}\left(t-t_{0}\right)H_{0}/\hbar}H_t\left(t\right)e^{-\mathrm{i}\left(t-t_{0}\right)H_{0}/\hbar}\ .$$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Relation to the Question&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;In the given case we have $$H_t = -\frac{e}{m}\mathbf{A}\left(t,\mathbf{x}\right)\mathbf{p}$$&#xA;and assuming that $$\mathbf{A}\left(\mathbf{x},t\right) = \mathbf{A}\left(\mathbf{x}\right) T(t)\ ,$$&#xA;and further using&#xA;$$\mathbf{p}=\frac{\mathrm{i}m}{\hbar}\left[H_{0},\mathbf{x}\right]\ ,$$&#xA;one will have to solve $$A_{n\rightarrow m}\propto &amp;lt;m|\mathbf{A}\left(\mathbf{x}\right)\cdot \mathbf{x}|n&amp;gt;\int_{t_0}^t e^{-\mathrm{i}\omega_{mn}\tau}T(\tau) d\tau\ +\ \mathrm{h.c.}$$&#xA;with $\hbar \omega_{mn} = E_m - E_n$, where the $E_i$ are energy levels of the unperturbed $H_0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using here the vector potential $\mathbf{A}$ is the correct way to incorporate the field, of course. For me, it seems &lt;strong&gt;extremely complicated&lt;/strong&gt; to do so since to calculate the transition probabilities for an &lt;strong&gt;arbitrary field distribution&lt;/strong&gt; one will have to expand $\mathbf{A}$ into &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_spherical_harmonics&quot; rel=&quot;nofollow&quot;&gt;vector spherical harmonics&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the use of a scalar potential like $U$ would greatly simplify computations and my question can now be re-asked as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h3&gt;Why can't we just simply use $U(\mathbf{x})\cdot T(t)$ to calculate the transition amplitudes $A_{n\rightarrow m}$ for &quot;electric excitations&quot;?&lt;/h3&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thank you very much&lt;/strong&gt; for insights, comments and corrections.&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2011-06-21T01:20:00.310" Title="Using the Scalar Electrostatic Potential to Calculate Transition Probabilities" Tags="&lt;quantum-mechanics&gt;&lt;electrodynamics&gt;&lt;approximations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11312" PostTypeId="1" CreationDate="2011-06-20T10:05:05.563" Score="1" ViewCount="90" Body="&lt;p&gt;Hi I've done an experiment in physics using a basic circuit. Can anybody explain how come the data logging equipment I was using continued to register a current when there isn't a positvie voltage in the circuit. &lt;/p&gt;&#xA;" OwnerUserId="4103" LastActivityDate="2011-06-20T11:55:00.807" Title="No Voltage in a Circuit but still a Current, How?" Tags="&lt;electric-circuits&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="11313" PostTypeId="1" CreationDate="2011-06-20T11:27:32.427" Score="3" ViewCount="107" Body="&lt;p&gt;Noether's theorem needs the lagrangian to be invariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, given a lagrangian $L$, we know that the lagrangians $\alpha L$ (where $\alpha$ is any constant) and $L + \frac{df}{dt}$ (where $f$ is any function) lead to the same equations of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we then consider that the lagrangian is invariant under a transformation if we find $\alpha L$ or $L + \frac{df}{dt}$ instead of $L$?&lt;/p&gt;&#xA;" OwnerUserId="2499" LastActivityDate="2011-06-20T13:04:00.480" Title="invariance of lagrangian in Noether's theorem" Tags="&lt;classical-mechanics&gt;&lt;lagrangian-mechanics&gt;&lt;noethers-theorem&gt;" AnswerCount="1" />
  <row Id="11314" PostTypeId="1" AcceptedAnswerId="11315" CreationDate="2011-06-20T11:28:10.163" Score="1" ViewCount="63" Body="&lt;p&gt;Which symbols are usually used to denote an arbitrary operator in quantum mechanics, such as &lt;em&gt;O&lt;/em&gt; in the following example?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$O \mbox{ is Hermitian} \Leftrightarrow \Im{\left&amp;lt; O \right&amp;gt;} = 0$&lt;/p&gt;&#xA;" OwnerUserId="1778" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-20T16:58:04.173" LastActivityDate="2011-06-21T14:48:01.877" Title="How is an arbitrary operator usually denoted in quantum mechanics?" Tags="&lt;quantum-mechanics&gt;&lt;soft-question&gt;&lt;operators&gt;&lt;notation&gt;" AnswerCount="2" />
  <row Id="11315" PostTypeId="2" ParentId="11314" CreationDate="2011-06-20T11:43:28.410" Score="6" ViewCount="" Body="&lt;p&gt;It's common to put a hat over anything that's an operator instead of a c-number, so that $\hat A$ is an operator, $A$ is a c-number. Then we can use any letter as either an operator or as a c-number. Your $\hat O$ or $O$ to some extent suggests something that is specifically an observable quantity, just as $\hat H$ suggests a Hamiltonian operator, although one would usually expect such uses to be explicitly stated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of the day, however, it's a matter of art to get a paper to look recognizably like the other papers in a field. The use of a particular symbol for a particular purpose can come almost to define a particular area of Physics, and people in that field may stop reading a paper for flagrantly breaking such a rule unless a good enough reason is given for doing something else. To have those nuances down pat requires that you read enough papers in the field you want to write in carefully enough to notice what is used consistently for what. &lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-06-21T14:48:01.877" LastActivityDate="2011-06-21T14:48:01.877" CommentCount="4" />
  <row Id="11316" PostTypeId="2" ParentId="11312" CreationDate="2011-06-20T11:55:00.807" Score="2" ViewCount="" Body="&lt;p&gt;If you have a short circuit you may have just few millivolts, and still a significant current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option - when measuring voltage you probably not connected ground properly of your voltmeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, small current might be just calibration error.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-06-20T11:55:00.807" CommentCount="1" />
  <row Id="11317" PostTypeId="2" ParentId="11313" CreationDate="2011-06-20T12:07:44.590" Score="4" ViewCount="" Body="&lt;p&gt;Here I would like to mention the notion of &lt;em&gt;quasi-symmetry.&lt;/em&gt; In general, if the Lagrangian (resp. Lagrangian density) is only invariant up to a total time derivative (resp. space-time divergence) when performing a certain off-shell$^1$ variation, one speaks of a &lt;em&gt;quasi-symmetry,&lt;/em&gt; see, e.g., J.V. Jose and E.J. Saletan, &quot;Classical Dynamics: A Contemporary Approach&quot;, p. 565.   &lt;a href=&quot;http://en.wikipedia.org/wiki/Noether%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Noether's first Theorem&lt;/a&gt; does also hold for quasi-symmetries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^1$ Here the word &lt;em&gt;off-shell&lt;/em&gt; means that the Lagrangian eqs. of motion are not assumed to hold under the specific variation. If we assume the Lagrangian eqs. of motion to hold, any variation of the Lagrangian is trivially a total derivative.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-20T13:04:00.480" LastActivityDate="2011-06-20T13:04:00.480" CommentCount="2" />
  <row Id="11318" PostTypeId="2" ParentId="11208" CreationDate="2011-06-20T12:49:55.443" Score="0" ViewCount="" Body="&lt;p&gt;Well, I'd say have a thin flat base of a well-heat-conducting material as big as your heat source (to maximize heat transfer to the water) and a well-insulated sphere cap (for minimum surface area to minimize heat loss) of your specified volume on top. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If everything but your base is adiabatic you don't even have to care about the form of the kettle :-P&lt;/p&gt;&#xA;" OwnerUserId="3951" LastEditorUserId="3951" LastEditorDisplayName="" LastEditDate="2011-06-20T12:56:33.290" LastActivityDate="2011-06-20T12:56:33.290" CommentCount="2" />
  <row Id="11319" PostTypeId="2" ParentId="11287" CreationDate="2011-06-20T12:52:34.803" Score="2" ViewCount="" Body="&lt;p&gt;Try &quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B0048ELDQY&quot; rel=&quot;nofollow&quot;&gt;How to teach Physics to your dog&lt;/a&gt;&quot;, &lt;a href=&quot;http://physics.stackexchange.com/users/166/chad-orzel&quot;&gt;Chad Orzel&lt;/a&gt;, which I think is still in bookshops as well as at Amazon.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-06-20T12:52:34.803" CommunityOwnedDate="2011-06-20T16:04:44.980" />
  <row Id="11320" PostTypeId="1" CreationDate="2011-06-20T13:01:41.083" Score="2" ViewCount="95" Body="&lt;p&gt;Is there a particle called &quot;tachyons&quot; that can travel faster than light?  If so, would Einstein's relativity be wrong?  According to Einstein no particle can travel faster than light.&lt;/p&gt;&#xA;" OwnerUserId="4105" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-06-20T17:11:43.903" LastActivityDate="2011-06-20T17:11:43.903" Title="Tachyon and Photons" Tags="&lt;special-relativity&gt;&lt;speed-of-light&gt;&lt;photons&gt;&lt;tachyon&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="11321" PostTypeId="1" AcceptedAnswerId="11323" CreationDate="2011-06-20T13:04:02.697" Score="2" ViewCount="161" Body="&lt;p&gt;I'm far from being a physics expert and figured this would be a good place to ask a beginner question that has been confusing me for some time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Galileo, two bodies of different masses, dropped from the same height, will touch the floor at the same time in the absence of air resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BUT Newton's second law states that $a = F/m$, with $a$ the acceleration of a particle, $m$ its mass and $F$ the sum of forces applied to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that acceleration represents a variation of velocity and velocity represents a variation of position. I don't comprehend why the mass, which is clearly affecting the acceleration, does not affect the &quot;time of impact&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain this to me? I feel pretty dumb right now :)&lt;/p&gt;&#xA;" OwnerUserId="4104" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-20T16:10:02.763" LastActivityDate="2011-06-20T22:19:04.343" Title="Confused about the role of mass" Tags="&lt;gravity&gt;&lt;newtonian-mechanics&gt;&lt;acceleration&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="11322" PostTypeId="2" ParentId="11321" CreationDate="2011-06-20T13:15:25.150" Score="3" ViewCount="" Body="&lt;p&gt;Because force 'pushing' object closer to earth is proportionally bigger for 'heavier' object. &#xA;But heavier object is also have higher gravitation force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So these two factors perfectly compensate each other: Yes you need more force for a set acceleration, but more force is here due to heavier mass.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-06-20T13:15:25.150" />
  <row Id="11323" PostTypeId="2" ParentId="11321" CreationDate="2011-06-20T13:16:16.050" Score="4" ViewCount="" Body="&lt;p&gt;it is because the Force at work here (gravity) is also dependent on the mass&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gravity acts on a body with mass m with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = mg$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you will plug this in to $$F=ma$$ and you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ma = mg$$&#xA;$$a = g$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this is true for all bodies no matter what the mass is. Since they are accelerated the same and start with the same initial conditions (at rest and dropped from a height h) they will hit the floor at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a peculiar aspect of gravity and underlying this is the equality of inertial mass and gravitational mass (here only the ratio must be the same for this to be true but Einstein later showed that they're really the same, i.e. the ratio is 1)&lt;/p&gt;&#xA;" OwnerUserId="2151" LastActivityDate="2011-06-20T13:16:16.050" CommentCount="1" />
  <row Id="11324" PostTypeId="2" ParentId="11321" CreationDate="2011-06-20T13:18:02.533" Score="5" ViewCount="" Body="&lt;p&gt;Newton's gravitational force is proportional to the mass of a body, $F=\frac{GM}{R^2}\times m$, where in the case you're thinking about $M$ is the mass of the earth, $R$ is the radius of the earth, and $G$ is Newton's gravitational constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, the acceleration is $a=\frac{F}{m}=\frac{GM}{R^2}$, which is independent of the mass of the object. Hence any two objects that are subject only to the force of gravity will fall with the same acceleration and hence they will hit the ground at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I think you were missing is that the force $F$ on the two bodies is not the same, but the accelerations &lt;em&gt;are&lt;/em&gt; the same.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-06-20T13:18:02.533" />
  <row Id="11325" PostTypeId="2" ParentId="11321" CreationDate="2011-06-20T13:25:39.067" Score="-1" ViewCount="" Body="&lt;p&gt;Lets say two separate mass $M_1$ and $m_2$ where $M_1$ &gt;&gt; $m_2$, both fall, from the same instant in a gravitational field &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force on $M_1$ is $F1$ = G $M_{\text{earth}}$    $M_1$/ $R^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force on $m_2$ is $F2$ = G $M_{\text{earth}}$  $m_2$/ $R^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the forces are  $F1$ &gt;&gt; $F2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, most people think $M_1$ should accelerate much faster than $m_2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as you wrote above a = F/m and substituting $F1$, $F2$, $M_1$ and $m_2$ into that formula we find:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F1$/$M_1$ = $F2$/$M_2$ = G $M_{\text{earth}}$/ $R^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the acceleration is independent of the masses we drop, and is a constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&#xA;Bother, by the time I had written this down, it was answered, obviously the other authors had a different acceleration in their typing.&lt;/p&gt;&#xA;" OwnerUserId="3588" LastActivityDate="2011-06-20T13:25:39.067" />
  <row Id="11327" PostTypeId="5" CreationDate="2011-06-20T13:56:04.807" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-06-20T13:56:04.807" LastActivityDate="2011-06-20T13:56:04.807" />
  <row Id="11328" PostTypeId="4" CreationDate="2011-06-20T13:56:04.807" Score="0" ViewCount="" Body="Photons are electromagnetic waves.  They exhibit particle-like qualities in many situations and have zero rest mass." OwnerUserId="-1" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-06-20T16:05:25.743" LastActivityDate="2011-06-20T16:05:25.743" />
  <row Id="11329" PostTypeId="5" CreationDate="2011-06-20T13:58:05.663" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-06-20T13:58:05.663" LastActivityDate="2011-06-20T13:58:05.663" />
  <row Id="11330" PostTypeId="4" CreationDate="2011-06-20T13:58:05.663" Score="0" ViewCount="" Body="The speed of light is fundamental universal constant that marks the maximum speed at which information can propagate. Its value is $299792458\frac{\mathrm{m}}{\mathrm{s}}$." OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-20T16:06:43.810" LastActivityDate="2011-06-20T16:06:43.810" />
  <row Id="11331" PostTypeId="2" ParentId="11306" CreationDate="2011-06-20T14:00:58.767" Score="1" ViewCount="" Body="&lt;p&gt;Anything you do involving a focused beam of light is just not going to scatter the way you would like it to. If you are using a fog machine to scatter the light, then it will make the focusing beam just as visible as the focal spot itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are attempting to produce a point of light that seems to float in space, maybe you could consider some sort of fiber optic solution. Perhaps you could illuminate a glass bead with light from a fiber. Get the glass bead sandblasted so it has a rough finish, and it will scatter light (relatively) uniformly into all directions. &lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-06-20T14:00:58.767" CommentCount="1" />
  <row Id="11332" PostTypeId="5" CreationDate="2011-06-20T14:06:57.580" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-06-20T14:06:57.580" LastActivityDate="2011-06-20T14:06:57.580" />
  <row Id="11333" PostTypeId="4" CreationDate="2011-06-20T14:06:57.580" Score="0" ViewCount="" Body="Quantum field theory is a theoretical framework for quantum mechanics where fields or particles requiring infinite degrees of freedom can be represented.  In QTF fields have a fundamental field quanta representing the smallest interaction allowable by the uncertainty principle." OwnerUserId="-1" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-06-20T16:05:31.563" LastActivityDate="2011-06-20T16:05:31.563" />
  <row Id="11334" PostTypeId="1" CreationDate="2011-06-20T14:13:41.130" Score="4" ViewCount="124" Body="&lt;p&gt;Inspired by &lt;a href=&quot;http://physics.stackexchange.com/q/6419/3816&quot;&gt;this question&lt;/a&gt; and specifically &lt;a href=&quot;http://physics.stackexchange.com/questions/6419/electric-engine-transmission/6425#6425&quot;&gt;this answer to it&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my experience there's always some very specific limit to how much torque an electric motor can output. For example, an electric drill will often have a manually switched mechanical transmission - if one needs to drill some relatively weak materials (like wood) he will use the setting that outputs lets torque at higher RPM and if one want to drill steel or mix cement mortar he will use the settings for more torque at lower RPM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relation between RPM and torque is more or less clear if one imagines a set of two gears of different diameters and thinks that their radiuses are lever arms - three times more RPM automatically induces three times lower torque and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But where does the limit to any given electric motor torque come from? Say I have some specific motor right now in front of me and it can output 40 Newton-meters at 500 RPM. Why exactly 40 and not more?&lt;/p&gt;&#xA;" OwnerUserId="3816" LastActivityDate="2011-08-05T03:48:32.407" Title="Why and how exactly is electric motor torque limited?" Tags="&lt;electromagnetism&gt;&lt;electricity&gt;&lt;torque&gt;&lt;electrical-engineering&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="11338" PostTypeId="1" CreationDate="2011-06-20T15:01:45.197" Score="4" ViewCount="99" Body="&lt;p&gt;Can one show that in quantum field theory &lt;em&gt;at least some example&lt;/em&gt; massive particles propagate with speed less than speed of light, while massless travel at speed of light? Well, motion is a different thing in QM than in classical mechanics, and question might be formulated differently.&lt;/p&gt;&#xA;" OwnerUserId="4107" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-20T16:08:23.917" LastActivityDate="2011-06-20T22:08:39.597" Title="Massive particles and speed of their propagation" Tags="&lt;quantum-field-theory&gt;&lt;speed-of-light&gt;&lt;mass&gt;" AnswerCount="4" />
  <row Id="11339" PostTypeId="2" ParentId="11334" CreationDate="2011-06-20T15:12:16.753" Score="2" ViewCount="" Body="&lt;p&gt;Let's imagine that we live in beautiful world of brush-less motors without friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Limitation must come from coil resistivity - the more power you pump into coils, the more losses you have due to their non-0 resistivity. At high RPM resistance also increases due to skin effect (reducing effective cross-section of the wire). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if one want to have more force at higher RPM - he would need thicker wire in coils (copper, or even silver), and in case of high RPM - made of litz wire (this would be incredibly expensive).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you can make coils out of superconductor - problems solved, and your power is limited only by mechanics &amp;amp; coil commutation.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-06-20T15:12:16.753" CommentCount="1" />
  <row Id="11340" PostTypeId="1" AcceptedAnswerId="11342" CreationDate="2011-06-20T15:12:26.297" Score="5" ViewCount="96" Body="&lt;p&gt;I'm reading (I'm trying to read) Schutz's &quot;A first course in general relativity&quot; (1985). On page 126 he mentions that a small change in angle theta in polar coordinates is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/07Kjg.gif&quot; alt=&quot;small change in theta - polar coordinates&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't see why this is. Can anyone please explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="4075" LastActivityDate="2011-06-20T22:56:58.617" Title="Small change in theta - polar coordinates" Tags="&lt;general-relativity&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="11342" PostTypeId="2" ParentId="11340" CreationDate="2011-06-20T15:31:40.987" Score="9" ViewCount="" Body="&lt;p&gt;The polar angle is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta = \arctan(y/x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the total derivative (&quot;a small change in theta&quot;) of theta is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d\theta = \frac{\partial \theta}{\partial x}dx + \frac{\partial \theta}{\partial x}dy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this works out to &#xA;$$\frac{\partial \theta}{\partial x} = \frac{-y}{x^2+y^2} = \frac{-y}{r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial \theta}{\partial y} = \frac{x}{x^2+y^2} = \frac{x}{r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therfore the result is as given in the book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: in case it is not clear: $$\frac{\partial}{\partial x}\arctan x = \frac{1}{x^2+1}$$&lt;/p&gt;&#xA;" OwnerUserId="2151" LastActivityDate="2011-06-20T15:31:40.987" CommentCount="2" />
  <row Id="11343" PostTypeId="2" ParentId="11338" CreationDate="2011-06-20T15:46:54.770" Score="4" ViewCount="" Body="&lt;p&gt;I am not sure what kind of argument do you want. There are many, each with different amount of hand-waving. The most simple is: in the long distance limit QFT has to reduce to classical physics, so the particles have to move according to special relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With little less hand-waving, the propagator of a free scalar particle with mass $m$ is ${1 \over p^2 - m^2}$. This has a pole at $p^2 = m^2$ and therefore this is the major contribution to the propagation. So the particle needs to be close to being on-shell (note that nobody is saying that it needs to be &lt;em&gt;precisely&lt;/em&gt; on-shell and indeed, it need not; this is especially clear in calculation of loop diagrams).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thing holds for any other (i.e. non-scalar) particle, the propagator will again contain a pole at mass $m$; the only difference is that it will carry some additional Lorentz/Dirac structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the most non-trivial part is: what if the theory is not free? Well, then obviously all the usual problems of the QFT set in, one has to renormalize, etc. For one thing, mass of the particle gets screwed up by loop contributions and needs to be fixed by counterterms (indeed, if one computes mass of the electron in a naive QED, it comes out infinite, so it wouldn't be able to even move, contrary to the observations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure any of this satisfies you. You should be more precise about what kind of argument you want to hear.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-06-20T15:46:54.770" CommentCount="2" />
  <row Id="11344" PostTypeId="2" ParentId="11338" CreationDate="2011-06-20T16:46:39.940" Score="2" ViewCount="" Body="&lt;p&gt;And here is an experimentalist's point of view:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When particles are virtual, anything goes as long as the equations allow it. QFT is a mathematical system that successfully predicts the crossections and angular distributions of particle interactions. It is full of virtual particles, i.e. off mass shell and whatever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only zero mass on shell particles we can experiment with are photons , and they do travel with the speed of light . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nonzero stable (or semi) ones like  electrons and protons  and muons etc we have experimental evidence that they go with less than the speed of light. All the unstable ones with lifetimes of 10^-14 sec or less, are not accessible to measurement of the speed so the question is moot.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-06-20T16:46:39.940" CommentCount="1" />
  <row Id="11345" PostTypeId="2" ParentId="11314" CreationDate="2011-06-20T16:47:51.313" Score="3" ViewCount="" Body="&lt;p&gt;In my experience, which covers basic QM and quantum field theory to some extent, people tend to use $A$ or $O$ for a generic operator. The convention that $O$ indicates an observable is not universal - at least, not universally used, although it's probably at least familiar to most physicists. (Same with the hat convention that Peter mentioned; everyone understands it but not everyone uses it all the time.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a paper refers to multiple generic operators, then it's typical (again, in my experience) to denote them using sequential letters starting from $A$. So for example, one might talk about a commutator $[A,B]$ or $[[A,B],C]$ or something like that. Other than that, though, I would consider it unusual to see letters other than $A$ or $O$ used to denote generic operators.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-20T16:47:51.313" />
  <row Id="11346" PostTypeId="2" ParentId="11320" CreationDate="2011-06-20T17:09:55.373" Score="8" ViewCount="" Body="&lt;p&gt;As a matter of fact, Einstein's theory of relativity is what predicts the &lt;em&gt;possibility&lt;/em&gt; of tachyons (faster-than-light particles) - or rather, relativity is what makes tachyons special. Without relativity, particles moving faster than light would just be really fast particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although it's often said that Einstein's relativity says that nothing can move faster than light, that's not actually true. What special relativity actually says (among many other things) is that anything which moves slower than light will always move slower than light, anything which moves at the speed of light will always move at the speed of light, and anything which moves faster than light will always move faster than light. Effectively, nothing can cross the &quot;barrier&quot; of light speed. So there is nothing in relativity that forbids the existence of tachyons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, based on relativity and other theories, we can predict certain properties that tachyons would have, if they exist, and they turn out to be rather strange. For example, their mass would have to be represented by an imaginary number. Also, a tachyon would get faster and faster as it loses energy, as opposed to normal particles, which need to gain energy to increase their speed. If you follow this sort of reasoning far enough, eventually you wind up with nonsensical results. Combined with the fact that no sign of a real tachyon has never been observed, this leads most physicists to believe that tachyons either do not exist, or are prohibited (by some as-yet-unknown law of physics) from interacting with normal matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're interested in learning more, I would second the suggestion that you check out the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tachyon&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; that anna linked to in a comment.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-20T17:09:55.373" CommentCount="1" />
  <row Id="11347" PostTypeId="1" AcceptedAnswerId="11360" CreationDate="2011-06-20T17:19:35.307" Score="1" ViewCount="52" Body="&lt;p&gt;This question is from the 1975 Canadian Association of Physicists Exam. No solutions are posted and I am quite lost on how to proceed with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A particle is constrained to move along the x-axis of a Cartesian co-ordinate system and an identical particle is constrained along the y-axis. Show that if the particles are originally at rest and attract each other according to any law which depends only on the distance between them, then they will reach the origin simultaneously.&lt;/p&gt;&#xA;" OwnerUserId="3762" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-20T17:52:23.343" LastActivityDate="2011-06-20T20:23:28.893" Title="Constrained particles under distance dependent force" Tags="&lt;classical-mechanics&gt;&lt;homework&gt;&lt;forces&gt;&lt;newtonian-mechanics&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="11348" PostTypeId="2" ParentId="11247" CreationDate="2011-06-20T17:36:10.257" Score="4" ViewCount="" Body="&lt;h2&gt;&lt;strong&gt;2nd Update&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If you just want to think about a 1D universe with periodic boundary conditions, there is simply the point that there is no gauge transformation that takes from one you value of $A$ constant to another. Recall a gauge transformation takes $A_\mu\rightarrow A_\mu +\partial_\mu f$, where $f$ is an arbitrary &lt;em&gt;smooth&lt;/em&gt;  function. To go from $A = A_1$ to $A=A_2$ you would want to gauge transform with $f(x) = (A_1-A_2)x$. But since your system is periodic in $x$ this is not a smooth function. That would be okay if $\exp(i f(x)$ was a single valued function but that only happens when $A_1 -A_2$ is an integer. It is an interesting fact about some topologically nontrivial spacetimes that there are gauge fields which give the same electromagnetic fields but are not equivalent by gauge transformation. And Quantum Mechanics actually cares about the gauge fields. So it does not break gauge invariance for different $A$ to give different physical answers, since they are not connected by a gauge transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The update below is mainly concerned with thinking about your 1D quantum ring as embedded in actual space. Still worth reading&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Arthur, I applaud the fact you're still attempting to understand this. So I'm trying to re-explain what I wrote here. QMechanic and Marek have both provided good explanations, but they don't hit exactly the source of what seems to be your confusion. Let me try again to explain.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Your question is: I have this simple system with a gauge field, and my answer seems to be depend on my choice of gauge. How can this be? I thought nothing physical can depend on your choice of gauge.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Let's step back and remember where we first met gauge invariance. We had these $E$ and $B$, the electromagnetic fields, and these are observable physical quantities. But these are a pain to work with so we introduce the gauge fields $A$. But these are massively redundant - I can gauge transform my fields $A$ to something completely different and still have the same $E$ and $B$.  Since $E$ and $B$ are what we care about in the first place we demand our theories be &quot;gauge invariant&quot; - nothing physical can change when I do a gauge transformation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Gauge invariance is created by the fact that different gauge fields lead to &lt;strong&gt;physically equivalent&lt;/strong&gt; situations. There are lots of things that look like gauge fields - but only in those situations where gauge transformation connect &lt;strong&gt;physically equivalent&lt;/strong&gt; situations  must gauge invariance be respected. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the situations you mention yourself, the rigid rotator, there is something that looks a lot like a gauge field. We have a hamiltonian $H = \frac{1}{2m}(p - m\omega R)^2$, where $\omega$ is the angular velocity. This looks just like a gauge field $A= m\omega R$. But different values of $A$ and $\omega$ do &lt;strong&gt;not&lt;/strong&gt; correspond to the same physical situation. The rotator is spinning at a different speed. This is visibly different. Because the different $A$ are physically inequivalent, there is no reason to expect the answer to be independent of the value of $A$. And thats what you find. Correctly.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the case where $A$ actually is the electromagnetic vector potential the situation is slightly more subtle, since thats really supposed to gauge invariant. But I show below in the original post that, here as well, different choices of $A$ correspond to different physical situations. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bottom Line:&lt;/strong&gt; Just because something looks like a gauge field doesn't mean we must have gauge invariance. Gauge invariance is only a must when gauge transformations connect &lt;strong&gt;physically equivalent&lt;/strong&gt; situations. In the systems that realize a 1D quantum ring different choices of $A$ are not physically equivalent.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Original Post&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Your original calculation is correct. The energy levels do depend on the magnitude of the gauge vector $A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this not break gauge invariance? Calculate $\oint A(r)\cdot dr$, that is the integral of your gauge field around the loop. This is $\int dS\cdot \nabla \times A$ by Stokes theorem. This is $\int dA \cdot B$ by the  definition of the gauge field. So its equal to the magnetic flux $\Phi$ threading through your ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you take $A$ to be constant like you did, then we have $A = \frac{\Phi}{2\pi R}$, where $R$ is the radius of your ring. But $\Phi$ is a physical quantity -  it can't change under gauge transformations. There is no gauge transformation that keeps $A$ constant and changes its magnitude.  So its okay that different $A$ give different energy levels - they correspond to different physical situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the basis for many physical phenomena: the Aharanov-Bohm effect, flux quantization, Little-Parks effect, weak localization etc.. It's all on wikipedia. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the weird part here is not so much that your answer depends on the guage, since the different gauges are not equivalent. The weird part is that if I take a ring a light year in radius and put a magnetic field through the a meter-squared patch in the center of the ring, all the particles know about this magnetic field even though they are a light year away. But if I cut a meter long chunk out of my ring they suddenly forget all about that magnetic field.  (Only in the limit where everything is impossibly clean, but still, its wacky).&lt;/p&gt;&#xA;" OwnerUserId="3970" LastEditorUserId="3970" LastEditorDisplayName="" LastEditDate="2011-07-06T13:03:21.107" LastActivityDate="2011-07-06T13:03:21.107" CommentCount="11" />
  <row Id="11349" PostTypeId="2" ParentId="11338" CreationDate="2011-06-20T17:53:45.157" Score="2" ViewCount="" Body="&lt;p&gt;Although people often talk about particles in Particle Physics, nonetheless it's as tendentious to say that there are particles in QFT as it is to say so in non-relativistic QM. There are only events. In general, we can't tie multiple events tightly together as caused by one particle unless the events are strung out close together on an obvious single track. In general, however, we &lt;em&gt;can&lt;/em&gt; compute correlations between events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant mathematics here is the &lt;em&gt;propagator&lt;/em&gt; in its real space representation instead of in its Fourier space representation. Formulae and graphs of the way in which correlations between events that are caused by a free quantum field change at various space-like and time-like separations &lt;a href=&quot;http://en.wikipedia.org/wiki/Propagator&quot; rel=&quot;nofollow&quot;&gt;can be found on Wikipedia&lt;/a&gt; as formulae and graphs for the Feynman propagator. There are correlations at space-like and time-like separation even for zero-mass propagators. If we say that such correlations are caused by particles even when they are less than 100%, then particles do travel at less than the speed of light, but then you may feel that you have to say that particles also travel faster than light because there are correlations at space-like as well as at time-like separation (but I'm suggesting that you resist talking about particles, talk about fields instead; talk about High Energy Physics instead of about Particle Physics). For interacting quantum fields, there are higher-order correlations that do not appear for free fields, which are typically small perturbations of the free field case for quantum electrodynamics, but which may be qualitatively different for hadrons. This, however, does not change there being correlations at space-like and at time-like separation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QFT computes the evolution of probability distributions from an initial time to a final time, the S-matrix, as part of which there are various conserved integer charges. This discreteness justifies some kinds of particle thinking, but is not enough to justify fully classical ideas. What we observe at the final time are tracks in enormous detectors, but the presence of these tracks does not justify believing that when there is no enormous detector there still are tracks when we're not making them happen by putting somewhat exotic metastable states of matter in strategic places. Even for High Energy Physics, there are interference effects when there is no enormous detector present (although interference effects at &lt;em&gt;high&lt;/em&gt; energy happen at &lt;em&gt;small&lt;/em&gt; scales).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What &lt;em&gt;is&lt;/em&gt; required to be causal in QFT are the effects of measurements. If we measure some quantity in a given region of space-time, it should not change the results of measurement in space-like separated regions of space-time. This, however, does not impose a hard and fast causality for states. In particular, vacuum states of quantum field theories require correlations between measurements at all space-like separations to be the same at all points of space-time.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-06-20T17:53:45.157" CommentCount="6" />
  <row Id="11350" PostTypeId="1" AcceptedAnswerId="11408" CreationDate="2011-06-20T18:02:53.533" Score="2" ViewCount="79" Body="&lt;p&gt;In a book I am reading on special relativity, the infinitesimal line element is defined as $dl^2=\delta_{ij}dx^idx^j$ (Einstein summation convention) where $\delta_{ij}$ is the euclidean metric. Next, if we have some curve C between two points $P_1$ and $P_2$ in this space then the length of the curve is given as $\Delta L = \int_{P_1}^{P_2}dl$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having trouble deriving the next statement, which I quote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A curve in D-dimensional Euclidean&#xA;  space can be described as a subspace&#xA;  of the D-dimensional spce where the D&#xA;  co-ordinates $x^i$ are given by single&#xA;  valued functions of some parameter&#xA;  $t$, in which case the length of the&#xA;  curve from $P_1=x(t_1)$ to&#xA;  $P_2=x(t_2)$ can be written&#xA;  $$\Delta L = \int_{t_1}^{t_2}\sqrt{\delta_{ij}&#xD;&#xA;\dot{x}^i \dot{x}^j} dt \qquad&#xD;&#xA; \mbox{where}\; \dot{x}^i\equiv&#xD;&#xA; \frac{dx^i}{dt}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="2599" LastEditorDisplayName="" LastEditDate="2011-06-20T18:41:09.693" LastActivityDate="2011-06-21T20:41:42.067" Title="Length of a curve in D dimensional euclidean space" Tags="&lt;mathematical-physics&gt;&lt;differential-geometry&gt;" AnswerCount="3" CommentCount="9" />
  <row Id="11351" PostTypeId="2" ParentId="11347" CreationDate="2011-06-20T18:08:13.663" Score="2" ViewCount="" Body="&lt;p&gt;The operative word is &quot;attract,&quot; which means that the force exerted on each particle points toward the other particle. Of course, not all of that force actually affects the motion of the particle, since each particle is constrained to move only along one axis. So a good first step would be to find the component of force on each particle that acts in the direction that particle is free to move in. Once you do that, try playing around with the equations to relate the motions of the two particles to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that you don't know how the force between the two particles depends on the distance between them, so you'll need to write it as an unknown function, $f(r)$ for example.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-20T18:08:13.663" />
  <row Id="11352" PostTypeId="2" ParentId="11279" CreationDate="2011-06-20T18:10:08.023" Score="0" ViewCount="" Body="&lt;p&gt;IANAA, but a way to do this could be to have the planet have a very long orbit (centruries) but have the axis of rotation precess chaoticly on the time scale of a decade. To have chaotic precession on the decade long scale we need, there would have to be a very large object nearby. For example, Hyperion (a moon of Saturn) has an axis of rotation that precesses chaoticly on the scale of 40 days. The planet of A Song of Ice and Fire has a moon, but this would only be a weak perturbation, which would lead to chaos on a time scale much longer than a decade if its chaotic at all. The earth's axis does flop around but on a very long time scale (and some random Google hits say its not chaotic, for what thats worth)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other issue is that if the axis is precessing randomly then you would expect to have it end up everywhere, say with north pole stuck pointing at the sun. This would lead to half the planet being in a decades long day and the other half in a decades long night. Which is not in the book. So you need to find some mechanism that keeps the axis of rotation approximately perpendicular to the plane of orbit. Maybe two similarly sized planets with approximately aligned axes of rotation perpendicular to the plane of orbit? Conservation of angular momentum would make large deviations hard I would guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So its not too physical plausible... but its more physically plausible than a wight?&lt;/p&gt;&#xA;" OwnerUserId="3970" LastActivityDate="2011-06-20T18:10:08.023" />
  <row Id="11353" PostTypeId="2" ParentId="11350" CreationDate="2011-06-20T18:14:55.430" Score="0" ViewCount="" Body="&lt;p&gt;I think you should take that as the definition of the word &quot;length&quot;.  I wouldn't try to derive it at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is basically saying that if, for example, you want to know the length of the unit circle in the first quadrant, set&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^1 = \cos t$$&#xA;$$x^2 = \sin t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{x}^1 = -\sin t$$&#xA;$$\dot{x}^2 = \cos t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and do&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^{\pi/2} \sqrt{(-\sin t)^2 + (\cos t)^2}dt = \pi/2$$&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-06-20T18:14:55.430" CommentCount="3" />
  <row Id="11354" PostTypeId="2" ParentId="11347" CreationDate="2011-06-20T18:15:39.283" Score="-1" ViewCount="" Body="&lt;p&gt;Let the distance (d) related force between the two particles be (K)(d^Q), where K is a constant and Q is any power (e.g., for gravity K = MxMy times the gravitational constant, and Q = (-2)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x, Vx, Ax, and Mx is the positon, velocity, acceleration, and mass, respectively, of the particle constrained to the x-axis.  X0 = position at t=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y, Vy, Ay, and My is the positon, velocity, acceleration, and mass, respectively, of the particle constrained to the y-axis. Y0 = position at t=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d = SQRT(x^2 + y^2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the x and y particles start out somewhere on the positive x and y axes, the force on the particle constrained to the x-axis, which results in motion is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(x) = -(K)(d^Q)cos(tan^-1(y/x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the x and y particles start out somewhere on the positive x and y axes, the  force on the particle constrained to the y-axis, which results in motion is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(y) = -(K)(d^Q)sin(tan^-1(y/x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mx = My = M.  f = MA&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(y)/f(x) = MyAy/MxAx = Ay/Ax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(y)/f(x) = [-(K)(d^Q)sin(tan^-1(y/x)] / [-(K)(d^Q)cos(tan^-1(y/x)]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(y)/f(x) = tan(tan^-1(y/x)) = y/x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:  Ay/Ax = y/x.  Ay = Ax(y/x).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT:  The below is incorrect (see comments section against this post)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x = (Ax)(t^2)/2 + X0.  When x=0 ---&gt; t = SQRT(-2(X0)(Ax))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y = (Ax)(y/x)(t^2)/2 + Y0.  When y=0 ---&gt; t = SQRT(-2(Y0)(Ax)(y/x))&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You wanted to know how to get started.  Armed with the fact that Ay/Ax = y/x, I think you can take from there. &lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-06-20T20:23:28.893" LastActivityDate="2011-06-20T20:23:28.893" CommentCount="4" />
  <row Id="11355" PostTypeId="2" ParentId="11350" CreationDate="2011-06-20T18:29:32.950" Score="0" ViewCount="" Body="&lt;p&gt;Well, $\mathrm{d}l$ represents an infinitesimal length along the curve. $\mathrm{d}t$ also represents an infinitesimal length along the curve, although if the parametrizations $t$ and $l$ are different, the two infinitesimal lengths are not going to be the same. You can write the identity $\mathrm{d}l = \frac{\mathrm{d}l}{\mathrm{d}t}\mathrm{d}t$ and substitute in the definition of $\mathrm{d}l$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}l}{\mathrm{d}t}\mathrm{d}t = \frac{\sqrt{\delta_{ij}\mathrm{d}x^i\mathrm{d}x^j}}{\mathrm{d}t}\mathrm{d}t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since $\mathrm{d}x^{i(j)}$ and $\mathrm{d}t$ are positive lengths, you can do a little algebraic manipulation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}l}{\mathrm{d}t}\mathrm{d}t = \sqrt{\frac{\delta_{ij}\mathrm{d}x^i\mathrm{d}x^j}{\mathrm{d}t^2}}\mathrm{d}t = \sqrt{\delta_{ij}\dot x^i\dot x^j}\mathrm{d}t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not 100% mathematically rigorous, but in physics we think of derivatives and differentials as the limit of finite lengths and ratios, so it makes physical sense at least. And as long as your parametrization isn't singular, the math should hold up. (If you do have a singular parametrization, then I believe the result you're asking about still holds, although you need to resort to more precise math to prove it.)&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-20T18:29:32.950" CommentCount="3" />
  <row Id="11356" PostTypeId="1" CreationDate="2011-06-20T18:32:18.067" Score="1" ViewCount="79" Body="&lt;p&gt;I am trying to understand a paper in which they mention;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot; We non-dimentionalize the problem hereafter by selecting mass, length and time units such that the fluid density &#xA;$\rho\equiv1$&#xA;the gravitational acceleration g&#xA;$\equiv1$&#xA;and initial tank depth&#xA;$h_0\equiv1$&#xA;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reson for doing this is to calculate a pressure from a plot of y vs &#xA;$P/\rho g h_0$&#xA;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the value in the plot is 20, what is the value in Pascal for a &#xA;$\rho = 1025 kg/m^3$&#xA;,&#xA;$g = 9.81 m/s^2$&#xA;and&#xA;$h = 50 m$&#xA;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="4108" LastEditorUserId="4108" LastEditorDisplayName="" LastEditDate="2011-06-21T10:14:47.333" LastActivityDate="2011-06-21T14:23:11.543" Title="Non-Dimentionalization of equations" Tags="&lt;pressure&gt;&lt;units&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11358" PostTypeId="2" ParentId="11303" CreationDate="2011-06-20T18:53:10.390" Score="1" ViewCount="" Body="&lt;p&gt;The first thing to realize is that &quot;ice&quot; is a pile of small ice shards and snow is a pile of itty-bitty ice shards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the snow and ice are at the same temperature, the answer to your question with come down to which one has more contact area and (to a much lesser extent) how that contact area is distributed.  Also note that the contact area and its distribution could change over time, as the can melts the ice/snow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'd have to run the experiment.  My guess is that ice would be the winner.  This is because ice would maintain a much larger contact area over time.  The can in the snow would melt the snow beneath it and drop away from the &quot;snow roof&quot; over it; whereas I would expect the &quot;ice roof&quot; to collapse with the can, continuing to cool it.  The can contents in contact with the roof would be expected to be the warmest, thereby giving maximum cooling effect from any contact on the top side of the can.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-06-20T18:53:10.390" CommentCount="5" />
  <row Id="11359" PostTypeId="2" ParentId="11208" CreationDate="2011-06-20T18:54:56.130" Score="-1" ViewCount="" Body="&lt;p&gt;Industrial steam generators in power applications do, in fact, look rather different from your ordinary steam kettle.  Here is an example image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/RvMOU.gif&quot; alt=&quot;Steam generator for thermal plant&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a lot going on here, but the important part to focus on are the thick vertical lines that represent a &quot;baffle&quot; that separates the inflowing feedwater + recirculated saturated liquid from the steam generating region.  At this point I'm almost sure you have no idea what I'm saying so I should start from the top.  The &quot;hot side&quot; is the &quot;primary coolant&quot; marked on the image.  The function of the steam generator is to boil the &quot;secondary&quot; water with the heat from the hot side.  The incoming secondary water is called the feedwater.  This flows over tubes containing the hot liquid and partially boils.  After it partially boils, the steam bubbles are separated from the liquid part and the steam is taken to where it will be used and the liquid is sent back to flow over the hot tubes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible, although not assured until engineering analysis is done, that a &quot;baffle&quot; type of structure could improve the performance of a kettle as well.  It would be a cylindrical structure close to the walls, resting above the bottom of the kettle, with a top below the water surface.  The reason has to do with the circulation.  In a kettle, gravity performs the function of steam separation and circulation.  As I'm pointing out here, industrial steam generators use gravity (sometimes with the aid of pumps) for the task of circulation and steam separation but the functions are distinguished in the geometric design of the apparatus, which is carefully optimized to obtain better performance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a ordinary kettle, no internal structures are present so the liquid, powered by density difference between hot and cold, mixes however nature intends.  It is still possible in this configuration that fairly local isolated hot and cold eddies will form which is less efficient for mixing, and the baffle structure could largely prevent that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A drawback, however, is that using a baffle causes the steam to separate more efficiently which is actually undesirable in a kettle.  The function is usually to produce hot liquid, not steam, and designs with a baffle would produce steam sooner while leaving the rest of the liquid further from boiling point.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-06-20T18:54:56.130" CommentCount="4" />
  <row Id="11360" PostTypeId="2" ParentId="11347" CreationDate="2011-06-20T18:58:45.083" Score="0" ViewCount="" Body="&lt;p&gt;If we call the particles $A$ and $B$, we can write the force that $B$ exerts upon $A$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{F}_{AB} = F\left(\|\mathbf{r}_{AB}\|\right)\frac{\mathbf{r}_{AB}}{\|\mathbf{r}_{AB}\|}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathbf{r}_{AB} = \mathbf{r}_B - \mathbf{r}_A$ is the displacement vector between $B$ and $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As only of the &lt;em&gt;x&lt;/em&gt; coordinate of $A$ and the &lt;em&gt;y&lt;/em&gt; coordinate of $B$ take nonzero values, we can simplify the expression of the force to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{F}_{AB} = F\left(\sqrt{x_A^2 + y_B^2}\right) \frac{(-x_A, y_B)}{\sqrt{x_A^2 + y_B^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this expression is not the total force acting over $A$, because $A$ is constrained to move over the &lt;em&gt;x&lt;/em&gt; axis. This means that we need to keep only the projection of $\mathbf{F}_{AB}$ over the &lt;em&gt;x&lt;/em&gt; axis to get the total force over $A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{F}_A = F\left(\sqrt{x_A^2 + y_B^2}\right) \frac{(-x_A, 0)}{\sqrt{x_A^2 + y_B^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By an identical process, we can arrive to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{F}_B = F\left(\sqrt{x_A^2 + y_B^2}\right) \frac{(0, -y_B)}{\sqrt{x_A^2 + y_B^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this expressions for the forces, together with Newton's 2nd Law, we get the following system of differential equations for $x_A$ and $y_B$ (the only nonzero coordinates of the particles)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d^2x_A}{dt^2} = \frac{1}{m} F\left(\sqrt{x_A^2 + y_B^2}\right)\frac{-x_A}{\sqrt{x_A^2 + y_B^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d^2y_B}{dt^2} = \frac{1}{m} F\left(\sqrt{x_A^2 + y_B^2}\right)\frac{-y_B}{\sqrt{x_A^2 + y_B^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two differential equations are almost identical, with only one difference. Lets rewrite the equations in a way that emphasizes this difference&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G(x_A, y_B) = \frac{F\left(\sqrt{x_A^2 + y_B^2}\right)}{m\sqrt{x_A^2 + y_B^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d^2x_A}{dt^2} = -x_A G(x_A, y_B)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d^2y_B}{dt^2} = -y_B G(x_A, y_B)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that this is enough (maybe more than enough?) of a starting point. It's essentially the first part of David's answer translated to equations. :-)&lt;/p&gt;&#xA;" OwnerUserId="1656" LastActivityDate="2011-06-20T18:58:45.083" CommentCount="2" />
  <row Id="11361" PostTypeId="7" CreationDate="2011-06-20T19:00:03.230" Score="0" ViewCount="" Body="&lt;p&gt;Physics relies heavily on math. Many posts include mathematical notation, written using LaTeX commands and delimited by dollar signs. If you see dollar signs with what appears to be gibberish between them, it just means that &lt;a href=&quot;http://www.mathjax.org&quot; rel=&quot;nofollow&quot;&gt;MathJax&lt;/a&gt;, the math rendering we use, isn't working properly for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some basics for people new to LaTeX. Inline formulas are surrounded by single dollar signs (&lt;code&gt;f(x) = ax^2 + bx + c&lt;/code&gt; renders as $ f(x) = ax^2 + bx + c $ ), and block formulas by double dollar signs so that &lt;code&gt;x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}&lt;/code&gt; renders as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} .$$&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-06-23T18:19:24.093" LastActivityDate="2011-06-23T18:19:24.093" />
  <row Id="11362" PostTypeId="2" ParentId="11356" CreationDate="2011-06-20T19:42:11.853" Score="3" ViewCount="" Body="&lt;p&gt;As you note, the quantity on the vertical axis of the plot is $P/\rho g h_0$. So if this quantity equals 20, then &#xA;$$&#xD;&#xA;P/\rho g h_0=20,&#xD;&#xA;$$&#xA;or&#xA;$$&#xD;&#xA;P=20\rho g h_0.&#xD;&#xA;$$&#xA;The units on both sides of this expression match, so this expression is correct in any system of units. If you want to know the pressure in SI units (pascals), just use SI units for the stuff on the right. If I'm not mistaken, $\rho gh_0=5.03 \times 10^5$ Pa, so $20\rho g h_0=1.01\times 10^7$ Pa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to think about it: If you want, you can actually work out the sizes of the various units required to make the various given values equal 1. That is, you can assume that lengths, masses, and times are measured in units called L, M, T. Then you want to know how many meters there are in an L, how many kg in an M, and how many seconds in a T. You know that&#xA;$$&#xD;&#xA;1025 {\rm kg/m^3}=1 {\rm M/L^3},&#xD;&#xA;$$&#xA;and similarly for the other two quantities to be set equal to 1. Solve these three equations for M,L,T, and you'll know the size of the various &quot;new&quot; units in SI units (kg,m,s). The pressure is 20 of the new units, and now that you know what the new units are you can convert back to SI.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: I'd originally, incorrectly written M/T$^3$ instead of M/L$^3$ in the above equation. Someone edited it to Tonnes/m$^3$, which is not what I meant at all. In that expression, the right-hand side is supposed to be in the &quot;made-up&quot; units M,L,T, which don't correspond to any standard units. M/L$^3$ is correct.)&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-06-21T14:23:11.543" LastActivityDate="2011-06-21T14:23:11.543" />
  <row Id="11363" PostTypeId="1" AcceptedAnswerId="11448" CreationDate="2011-06-20T19:47:21.540" Score="5" ViewCount="91" Body="&lt;p&gt;I need to implement a random force in my code according to the fluctuation dissipation theorem. I have a Gaussian distribution function ready width average 0 and distribution 1 and I know I need to multiply it by something but I'm not sure what. The fluctuation dissipation theorem gives (in one dimension):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left \langle A (t_1) A (t_2) \right \rangle = 2 m \gamma k_B T \delta (t_1 - t_2 )$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't convince my self if I should multiply my Gaussian with my $\sqrt{dt}$ or divide by it. I can find justification for both direction:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The force should act like delta function (no correlation between times) hence, I should divide by it. &lt;/li&gt;&#xA;&lt;li&gt;The units are ok when multiplying by $\sqrt{dt}$.&lt;/li&gt;&#xA;&lt;li&gt;The larger the time step, the more collision happened hence the force should be larger.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I remember that in some place I read or heared I should divide by the $\sqrt{dt}$ but I couldn't find where. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should note that I tried both approach, when I divide I suspect the forces to be too large (I only see noise) and when I multiply, I almost can't see any random effect.&lt;/p&gt;&#xA;" OwnerUserId="3677" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-20T20:15:42.493" LastActivityDate="2011-06-22T20:23:05.993" Title="How to include random force in the simulation (Classical Molecular Dynamics)" Tags="&lt;simulation&gt;&lt;computational-physics&gt;&lt;fluctuation-dissipation&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="11364" PostTypeId="2" ParentId="11363" CreationDate="2011-06-20T20:34:40.567" Score="1" ViewCount="" Body="&lt;p&gt;Think about how your discrete time variables are being generated from continuous time variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $A_i =\int_{t_i}^{t_i+dt}A(t)$ then $&amp;lt;A_i A_j&amp;gt; \propto dt * \delta_{ij}$.&#xA;On the other hand if $A_i = {1 \over{dt}}\int_{t_i}^{t_i+dt}A(t)$ then $&amp;lt;A_i A_j&amp;gt; \propto {1\over{dt}} \delta_{ij}$&lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-06-20T20:34:40.567" />
  <row Id="11365" PostTypeId="2" ParentId="11279" CreationDate="2011-06-20T20:50:05.357" Score="0" ViewCount="" Body="&lt;p&gt;I don't think orbital dynamics allows for it. You could postulate that it's star might be an irregular variable star, whose intensity chaotically changes. Then the seasons would be more or less global. I don't know if any chaotic variable stars would have had an evolutionary history such that an inhabitted planet is a possibilty (i.e. a stable enough climate long enough for life to evolve), most likely not. But, it is a work of fiction not fact.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-06-20T20:50:05.357" />
  <row Id="11366" PostTypeId="2" ParentId="11338" CreationDate="2011-06-20T22:08:39.597" Score="0" ViewCount="" Body="&lt;p&gt;One can do the same thing you can do in ordinary quantum mechanics: show that  $\frac{d}{dx}\langle\hat{X}\rangle \approx \frac{1}{m}\langle\hat{P}\rangle$ in the non-relativistic limit where $\hat{P}$ and $\hat{X}$ are the average momentum and average position operators respectively. You can do this for the free Klein-Gordon field with basic tools. I can elaborate if you want.&lt;/p&gt;&#xA;" OwnerUserId="3970" LastActivityDate="2011-06-20T22:08:39.597" />
  <row Id="11367" PostTypeId="2" ParentId="11321" CreationDate="2011-06-20T22:19:04.343" Score="2" ViewCount="" Body="&lt;p&gt;There are two ways that mass could effect the time of impact:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) An object which is very massive has a stronger attraction to the earth. Logically, this might make the object fall faster and so reach the ground sooner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) An object which is very massive is difficult to get moving. (I.e. it has very high inertia.) Thus one might logically expect the very massive object to be more difficult to get moving and so to lose the race.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The miracle is that in the world we live in, these two effects exactly balance and so the heavier mass reaches the ground at the same time.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now let me give a simple explanation for why it's natural that this comes about. Suppose we have two very heavy masses. If we drop them separately they take some time T to fall. On the other hand, if we attach them together, will they take the same length of time? Think about a sphere split into two halves:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/FibKE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two two halves of the sphere would fall at the same speed as each other. So if you dropped them next to each other, they'd fall together. And dropping them next to each other isn't going to be any different from screwing them together and dropping them together. That is, there won't be any force on the screws. So the combined (or screwed together) sphere has to fall at the same rate as the split sphere.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-20T22:19:04.343" />
  <row Id="11368" PostTypeId="2" ParentId="11303" CreationDate="2011-06-20T22:19:42.627" Score="-1" ViewCount="" Body="&lt;p&gt;Thermal conductivity of ice is higher than thermal conductivity of snow, so ice is the winner. Ice has higher thermal conductivity because it is more dense than snow. When snow is as dense as ice then it is no longer snow, it is ice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your answers, without them I wouldn't guessed it right :) .&lt;/p&gt;&#xA;" OwnerUserId="4100" LastActivityDate="2011-06-20T22:19:42.627" CommentCount="0" />
  <row Id="11369" PostTypeId="2" ParentId="11340" CreationDate="2011-06-20T22:35:21.833" Score="6" ViewCount="" Body="&lt;p&gt;I'll give what I think is a more beautiful explanation. First of all, calculate $dr$ in terms of $dx$ and $dy$:&lt;br&gt;&#xA;$$r^2 = x^2 + y^2$$&#xA;$$2r\;dr = 2x\;dx + 2y\;dy$$&#xA;$$dr = \frac{x\;dx}{r} + \frac{y\;dy}{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regard to the above, note that $(x/r,y/r)$ is a unit vector in the $dr$ direction. Since $d\theta$ is perpendicular to $dr$; we first look for a unit vector that is perpendicular to $(x/r, y/r)$. The answer is clearly $(y/r,-x/r)$ or its negative. To get the right sign, note that when $y=0$, we are on the x-axis and we want $d\theta$ to point in the $+y$ direction so we use $(-y/r,x/r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we need an overall scale. The circumference of the unit circle is $2\pi$ and this is what you get from integrating $d\theta$ from $0$ to $2\pi$. For larger radii, we need to multiply by the radius. So we get:&lt;br&gt;&#xA;$$r\;d\theta \;\;=\;\;   - \frac{y\;dx}{r} + \frac{x\;dy}{r},$$&#xA;$$d\theta   \;\;=\;\; - \frac{y\;dx}{r^2} + \frac{x\;dy}{r^2}.$$&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-06-20T22:56:58.617" LastActivityDate="2011-06-20T22:56:58.617" CommentCount="2" />
  <row Id="11370" PostTypeId="2" ParentId="11241" CreationDate="2011-06-20T22:56:19.680" Score="3" ViewCount="" Body="&lt;p&gt;From the point of view of aerodynamics, a smaller number of longer blades is preferred. From the point of view of engineering strength, a larger number of shorter blades is better. What is actually chosen is a trade-off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be obvious why shorter blades are stronger. Longer and thinner blades are more efficient aerodynamically for the same reason that longer thinner wings are more efficient -- the longer thinner wings have less drag compared to the amount of lift produced. Hence high efficiency &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aK21_glider.jpg&quot; rel=&quot;nofollow&quot;&gt;gliders&lt;/a&gt; have long thin wings. The wikipedia photo:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/nnyIc.jpg&quot; alt=&quot;Glider picture&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-20T22:56:19.680" />
  <row Id="11371" PostTypeId="1" CreationDate="2011-06-20T23:00:36.947" Score="2" ViewCount="55" Body="&lt;p&gt;The problem is simple, I suppose...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a beverage at room temperature (let's say 23°C) in an aluminum can (it's a Barq's root beer.) I want to put it in a freezer (let's say it's at -17°C). How long do I have to wait for it to reach a chilled temperature (let's say 8°C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to know for a couple reasons. One, I don't want the can to explode. Two, I don't want it to get too cold for my teeth (perhaps I shouldn't be drinking pop? :-) ) Three, I want don't my coworkers to steal my nice, cold beverage on the first hot day of the year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I could find the right equations, I would just plug in the numbers, but I can't seem to find the right ones.&lt;/p&gt;&#xA;" OwnerUserId="1954" LastActivityDate="2011-06-20T23:38:31.420" Title="How can I calculate how long to keep my beverage chilled?" Tags="&lt;temperature&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="11372" PostTypeId="2" ParentId="11311" CreationDate="2011-06-20T23:02:25.947" Score="1" ViewCount="" Body="&lt;p&gt;If the case is purely electrostatic, one does not need time dependent perturbation theory, since its static. So you can just do the usual time independent perturbation theory. If the field is not static then you can't have &lt;strong&gt;A&lt;/strong&gt; = 0, and its not time independent, so you need all this technology. I don't think there is anything deeper going on.&lt;/p&gt;&#xA;" OwnerUserId="3970" LastEditorUserId="3970" LastEditorDisplayName="" LastEditDate="2011-06-21T01:20:00.310" LastActivityDate="2011-06-21T01:20:00.310" CommentCount="3" />
  <row Id="11373" PostTypeId="2" ParentId="11293" CreationDate="2011-06-20T23:20:42.167" Score="1" ViewCount="" Body="&lt;p&gt;Apparently I figured it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ANSWER&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Previously on pg. 14 we determined using ladder operators that $\psi(4,2,0,0) = \sqrt{3/14} (2^+, 0^-) + \sqrt{8/14} (1^+,1^-) - \sqrt{3/14}(2^-,0^+)$ and $\psi(3,2,1,0) = sqrt{1/2} [(2^+,0^-) + (2^-,0^+)]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the singlet D term, we have $\psi(2,2,0,0) = a (2^+,0^-) + b(2^-,0^+) + c(1^+,1^-)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if we take $\psi(2,2,0,0) \cdot \psi(4,2,0,0) = \sqrt{3/14}a + \sqrt{8/14}c - \sqrt{3/14}b = 0$ because he has said that these micro states must be orthogonal. Multiply by $\sqrt{14}$ to obtain his result. It also follows that $\psi(2,2,0,0) \cdot \psi(3,2,1,0) = \sqrt{1/2} a + \sqrt{1/2} b = 0$. Multiplication by $\sqrt{2}$ will give his result of $a + b = 0$. Also realize that $a^2 + b^2 + c^2 = 1$ must be true as a normalizing condition. Now we have three equations and three unknowns, so solving it is simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\psi(3,2,1,0)$ and $\psi(4,2,0,0)$ were picked in the first place because we had three unknowns and needed three equations to solve it. These wave functions must be orthogonal in the first place, therefore it was a matter of convenience AND that they were made up of the corresponding micro states in the first place.&lt;/p&gt;&#xA;" OwnerUserId="4112" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-06-21T03:17:00.007" LastActivityDate="2011-06-21T03:17:00.007" />
  <row Id="11374" PostTypeId="2" ParentId="11371" CreationDate="2011-06-20T23:32:07.320" Score="1" ViewCount="" Body="&lt;p&gt;I would probably use Newton's Law of Cooling;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dT}{dt}=-k(T - T_0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $T_0$ is the ambient temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as what k is, you could just do a little experiment. Find the temperature of the beer, put it in for 5 minutes, take it out, take the temperature again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you aren't the experimental type, $k = A \cdot H$ where $A$ is the surface area (a nice cylinder) and $H$ is the heat transfer coefficient.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For water to air, through metal, you get about $11 \frac{W}{m^2K}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I bet non-linear effects will cause enough issues that you will probably have a pretty wide error, (Specific type of metal of a can (aluminum), the shape of it, thickness, fact that beer is not just water, etc.). Probably an experimental approach of finding k will give you the most accurate way of going about it.&lt;/p&gt;&#xA;" OwnerUserId="3018" LastEditorUserId="3018" LastEditorDisplayName="" LastEditDate="2011-06-20T23:38:31.420" LastActivityDate="2011-06-20T23:38:31.420" CommentCount="3" />
  <row Id="11375" PostTypeId="1" AcceptedAnswerId="11377" CreationDate="2011-06-21T00:46:47.903" Score="0" ViewCount="66" Body="&lt;p&gt;I have exam tommorow and cant get this figured out :( dont blame me but please answer this question. I Want to simplify this term:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 times ((e topowerof 5jt) + (e topowerof -5jt))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4113" LastActivityDate="2011-06-21T01:04:27.200" ClosedDate="2011-06-21T01:02:45.733" Title="How to simplify e to power of j.t" Tags="&lt;mathematics&gt;" AnswerCount="1" />
  <row Id="11376" PostTypeId="1" CreationDate="2011-06-21T00:52:22.857" Score="2" ViewCount="46" Body="&lt;p&gt;Is this following proposal plausible, worth considering, or dismissable as lunatic fringe science?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if the universe isn't really what we think it is but some universal quantum computer where we read off one output at a time? This universal quantum computer is initialized to a superposition of program instructions -- which are identified with the laws of physics -- and after reading off one output after another, one finds there are in general two different kinds of questions we can ask; general relativity type questions, where we get GR type answers, and quantum mechanical type questions, where we get QM type answers. But we have to choose between one or the other, and whichever we choose, we get the answer appropriate for that domain. The place where both are united is the Planck scale, but unfortunately, the Planck scale as way too inaccessible to us, and is hence unobservable. Can this explain the confusion about UV/IR mixings?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;String compactifications are computer programs determining the parameters of the effective field theory describing the domain of answerable experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A complementarity between the laws of physics themselves. GR/QM complementarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any flaw with my proposal?&lt;/p&gt;&#xA;" OwnerUserId="3769" LastActivityDate="2011-06-21T00:52:22.857" ClosedDate="2011-06-21T01:08:01.597" Title="Complementarity between the laws of physics?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-gravity&gt;&lt;quantum-interpretations&gt;" CommentCount="2" />
  <row Id="11377" PostTypeId="2" ParentId="11375" CreationDate="2011-06-21T00:53:29.970" Score="0" ViewCount="" Body="&lt;p&gt;6cosine(5t) ; use eulers formula for sine and cosine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;exp(i*theta)=cos(theta)+i*sin(theta)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from this you can show, &#xA;sin(theta)=(1/2i)*(exp(i*theta)-exp(-i*theta))&#xA;cos(theta)=(1/2)*(exp(i*theta)+exp(-i*theta))&lt;/p&gt;&#xA;" OwnerUserId="3714" LastEditorUserId="3714" LastEditorDisplayName="" LastEditDate="2011-06-21T01:04:27.200" LastActivityDate="2011-06-21T01:04:27.200" CommentCount="2" />
  <row Id="11378" PostTypeId="1" CreationDate="2011-06-21T01:11:59.240" Score="1" ViewCount="102" Body="&lt;p&gt;(I recently asked this on &lt;a href=&quot;http://math.stackexchange.com/questions/46447/how-to-determine-how-much-more-energy-is-needed-per-extra-kilogram-of-weight-i-ca&quot;&gt;maths&lt;/a&gt; but was directed here)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have recently become a runner and having a keen interest in kinematics I'm very interested in the maths/physics of my running.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please help me determine a function that will enable me to determine how much harder I have to work to carry my 10kg overweight body over a given distance? I was pretty sure I could just use &lt;code&gt;work=weight x distance&lt;/code&gt; for the flat roads, but someone has pointed out that my time would be meaningless in such a scenario, but I'm having difficultly determining this for the hills too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition I need to include time as a variable so I can work out how my time affects my energy expenditure.&lt;/p&gt;&#xA;" OwnerUserId="4114" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-06-21T03:50:23.237" LastActivityDate="2011-07-07T07:11:14.937" Title="Running: Determine how much more energy is needed per extra kilogram of weight" Tags="&lt;kinematics&gt;&lt;biophysics&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="11380" PostTypeId="1" AcceptedAnswerId="11382" CreationDate="2011-06-21T03:11:49.287" Score="3" ViewCount="108" Body="&lt;p&gt;Can someone explain in both layman's terms and also technically why when I pulled my glass filled with liquid soda from the freezer, the liquid soda quickly froze?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doesn't this violate the 2nd law of thermodynamics since heat moved away from the glass with soda and to the ambient?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="4115" LastActivityDate="2011-06-21T03:30:22.477" Title="Why did my liquid soda freeze once I pulled it out of the fridge?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" />
  <row Id="11381" PostTypeId="2" ParentId="11378" CreationDate="2011-06-21T03:28:07.927" Score="1" ViewCount="" Body="&lt;p&gt;This really is a question about metabolism, you could refer to &lt;a href=&quot;http://www.brianmac.co.uk/energyexp.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.brianmac.co.uk/energyexp.htm&lt;/a&gt; and use the tables and calculator to find what your energy expenditure is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets assume you are ten kilograms overweight and wish to find the energy expenditure (EE) for this excess mass. Find your EE for your excess weight say 110 kilos, next find your EE for your normal weight say 100 kilos, and substract the difference and that will give your EE for the excess 10 kilos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the EE for going up a hill, find the difference in height:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work = m g $\Delta$h&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the only real way of knowing your energy expenditure for such a messy physical system is to measure your metabolism, and I know this sounds silly but what if you counted the number of breaths you took as your ran? Your oxygen intake is directly proportional to your metabolism, that might give the best estimate of EE. Good luck.&lt;/p&gt;&#xA;" OwnerUserId="3588" LastActivityDate="2011-06-21T03:28:07.927" CommentCount="1" />
  <row Id="11382" PostTypeId="2" ParentId="11380" CreationDate="2011-06-21T03:30:22.477" Score="10" ViewCount="" Body="&lt;p&gt;Your soda was in a supercooled state. Being in the freezer, it was at a temperature &lt;em&gt;below&lt;/em&gt; its freezing point however it remained as a liquid as the glass was too smooth to allow ice crystals to start to form (in technical terms, the phase transition requires a nucleation site). When you removed it from the freezer, you gave it the disturbance necessary to catalyse the transition from supercooled liquid into a block of ice. No violation of the 2nd law occurred.&lt;/p&gt;&#xA;" OwnerUserId="1876" LastActivityDate="2011-06-21T03:30:22.477" CommentCount="2" />
  <row Id="11385" PostTypeId="2" ParentId="11378" CreationDate="2011-06-21T09:52:28.370" Score="1" ViewCount="" Body="&lt;p&gt;Most good 100 m runners are &quot;compact&quot; athletes in general, whareas &#xA;long distance runnners look &quot;hungry&quot; :=)&#xA;From this You see that there is no &quot;function&quot; for that, there are a lot of &#xA;factors to consider. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, one had to know what that &quot;additional kilogram&quot; is: muscles, fat, &#xA;lead in the pockets, bones, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, for every step a runner does, he lifts his center of gravity &#xA;for about 10 or 20 cmters and falls down again. How much of the energy invested in &#xA;lift can be recovered by elastic processes is a debate in sports since &#xA;decades. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-06-21T09:52:28.370" CommentCount="1" />
  <row Id="11386" PostTypeId="1" CreationDate="2011-06-21T12:30:20.073" Score="7" ViewCount="172" Body="&lt;p&gt;In a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cathode_ray_tube&quot; rel=&quot;nofollow&quot;&gt;CRT&lt;/a&gt;, where do the ejected electrons go after they cause fluorescence on the screen, have they lost most of their energy, or do they actually go through the glass?&lt;/p&gt;&#xA;" OwnerUserId="4119" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-07T18:47:27.563" LastActivityDate="2011-08-14T16:50:19.483" Title="Electrons in CRT" Tags="&lt;electron&gt;" AnswerCount="3" />
  <row Id="11388" PostTypeId="1" AcceptedAnswerId="11528" CreationDate="2011-06-21T13:07:40.477" Score="4" ViewCount="127" Body="&lt;p&gt;The metric for the BTZ black hole is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ds^2=-N^2dt^2+N^{-2}dr^2+r^2(N^\phi dt +d\phi)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $N^2=-M+\frac{r^2}{l^2}+\frac{J^2}{4r^2}$ and $N^\phi=-\frac{J}{2r^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is often said that BTZ black hole is asymptotically AdS$_3$, but if I take $r\rightarrow \infty$ limit, then the BTZ metric, to the leading order of each component, becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ds^2\rightarrow-\frac{r^2}{l^2}dt^2+\frac{l^2}{r^2}dr^2+r^2d\phi^2-Jdtd\phi,$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while the AdS$_3$ metric reads, asymptotically,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ds^2_{AdS_3}\rightarrow-\frac{r^2}{l^2}dt^2+\frac{l^2}{r^2}dr^2+r^2d\phi^2.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is that since the off-diagonal term $-Jdtd\phi$ doesn't approach to zero at infinity, how can one claim that BTZ is asymptotically AdS$_3$? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2928" LastActivityDate="2011-06-24T06:50:27.150" Title="Why is BTZ black hole asymptotically AdS_3?" Tags="&lt;general-relativity&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="11389" PostTypeId="1" CreationDate="2011-06-21T14:23:51.070" Score="1" ViewCount="28" Body="&lt;p&gt;I'm trying to simulate the degenerative Anderson model. So depending on an energy difference first orbital and afterwards spin magnetism occurs. First i try to solve an easier ansatz with a limitation of only two orbitals to $l_z = \pm 1$.&#xA;In this case i end up with a set of 4 equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta \tan^{-1} (1/\pi * n_{m \sigma}) = E_0-E_f+(U-J)n_{\overline{m}\sigma}+U(n_{1\overline{\sigma}}+n_{2\overline{\sigma}})$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where here $\sigma$ is the direction of the spin of certain electron in the orbital $m$, here only $1,2$ and the overline always is the opposite. U is a Hubbard potential and $J$ an intraatomic exchange. &#xA;So we gathered a self consistent set of non-linear equation and i want to find solution in terms of $n_{m \sigma}$ satisfying all 4 equations. I already did some simulation and were able to plot it on dependancy of the energy difference with respect to the fermi energy, which is my parameter. So i change it in a certain range and one finds magnetic solution for $E_0 &amp;lt; E_f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a good method to solve such a problem with not too much calculation power? And in the end i want to play around with both potentials and see how the number of electron is changing for each situation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any advice,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;/no&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: i already asked a question concerning to my code i wrote in SO, but i was rethinking the method of mine for solving this problem. I did an iterative attempt and reformulated the equation for a root searching algorithm and varied the energy difference $E_0 - E_f$.  The programming stuff is on &lt;a href=&quot;http://stackoverflow.com/questions/6269212/ipython-solve-set-of-self-consistent-equation&quot;&gt;ipython&lt;/a&gt; plus a result of one simualtion run.&lt;/p&gt;&#xA;" OwnerUserId="4120" LastActivityDate="2011-06-21T14:23:51.070" Title="Degenerated Anderson Model Simulation" Tags="&lt;electromagnetism&gt;&lt;simulation&gt;&lt;numerics&gt;" />
  <row Id="11390" PostTypeId="1" AcceptedAnswerId="11393" CreationDate="2011-06-21T14:48:42.860" Score="3" ViewCount="156" Body="&lt;p&gt;Is all of the visible Universe exactly the same age everywhere? What about the Universe beyond what we can see?&lt;/p&gt;&#xA;" OwnerUserId="2540" LastActivityDate="2011-06-21T15:48:41.143" Title="Is the entire Universe the same age?" Tags="&lt;cosmology&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="11392" PostTypeId="2" ParentId="11390" CreationDate="2011-06-21T14:54:05.590" Score="0" ViewCount="" Body="&lt;p&gt;No, the further away we look, the younger the Universe becomes. That's why we're able to see &lt;a href=&quot;http://www.nytimes.com/2011/01/27/science/space/27galaxy.html&quot; rel=&quot;nofollow&quot;&gt;galaxies that were forming when the Universe was approximately a half billion years old&lt;/a&gt;. Although it's &lt;em&gt;possible&lt;/em&gt; there is an unknown portion of the Universe beyond what we can see, we currently have no reason to believe such a region exists. Rather, the boundaries of what we can see are dictated by &lt;a href=&quot;http://en.wikipedia.org/wiki/Timeline_of_the_Big_Bang#Dark_ages&quot; rel=&quot;nofollow&quot;&gt;when light decoupled from matter&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3624" LastEditorUserId="3624" LastEditorDisplayName="" LastEditDate="2011-06-21T15:01:08.533" LastActivityDate="2011-06-21T15:01:08.533" />
  <row Id="11393" PostTypeId="2" ParentId="11390" CreationDate="2011-06-21T15:10:17.387" Score="10" ViewCount="" Body="&lt;p&gt;Ben Hocking is right that the Universe we observe is not the same age everywhere: the further away we look, the younger things are, simply because the light from faraway objects takes time to reach us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you may be wondering about more than that: what if we could remove those light-travel-time delays and consider the Universe as it is &lt;em&gt;now&lt;/em&gt;, as opposed to the way we &lt;em&gt;see&lt;/em&gt; it (with greater delays for more distant objects)? Would it all be the same age in that case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That turns out to be a more subtle question than you might think, because of relativity. Relativity says, among other things, that different choices of reference frame lead to different notions of time -- things that are simultaneous in one reference frame are not simultaneous in another. When you consider what a distant object is like &lt;em&gt;now&lt;/em&gt;, you have to be careful to specify what reference frame you're talking about, because different frames lead to different &quot;nows.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At each point in spacetime, there is a reference frame that seems like the &quot;most natural&quot; one to use, namely the one in which the expansion of the Universe looks roughly the same in all directions. Cosmologists tend to use that reference frame to define and synchronize their clocks. That is, the time coordinate at any point in spacetime is, by definition, the amount of time that would have elapsed according to an observer who'd been at rest in that reference frame since the Big Bang. That time coordinate is often called &quot;cosmic time.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use cosmic time as your time coordinate, then it is true that, at any fixed moment of time, all points in the Universe have the same age. But it's true pretty much because of the way we defined our time coordinate, so this is kind of a vacuous statement!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Based on the comments, I realize I should have explained some things explicitly. I'm considering here &quot;cosmological&quot; effects, meaning I'm ignoring things that are due to small-scale inhomogeneities and thinking about large scales, on which the Universe is approximately homogeneous. If, for instance, you hang around near the horizon of a black hole, you'll age at a different rate from someone else. Even if you avoid such extreme cases, small-scale variations in the gravitational potential will lead to small-amplitude age variations. In fact, the definition of cosmic time only really makes sense in the approximation where we're willing to average over small-scale inhomogeneities. If you're not willing to do that, then different prescriptions for synchronizing your clocks, no one of which is obviously more natural than another, will lead to small differences in ages of things.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-06-21T15:48:41.143" LastActivityDate="2011-06-21T15:48:41.143" CommentCount="16" />
  <row Id="11394" PostTypeId="1" CreationDate="2011-06-21T15:42:45.597" Score="1" ViewCount="184" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/14255/experimental-evidence-for-parallel-universes&quot;&gt;Experimental evidence for parallel universes&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Is there an universe similar to ours somewhere else, I mean I had heard from Einsteins theories that,actually when I am typing this question here, there is an another world somewhere else where I would be typing this question simultaneously. Is this true, if so any evidences.&lt;/p&gt;&#xA;" OwnerUserId="4105" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-09-02T00:19:14.460" LastActivityDate="2011-09-02T00:19:14.460" ClosedDate="2011-09-02T00:24:55.230" Title="Presence Of an Another Universe" Tags="&lt;cosmology&gt;&lt;universe&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" />
  <row Id="11396" PostTypeId="1" CreationDate="2011-06-21T15:59:53.670" Score="4" ViewCount="238" Body="&lt;p&gt;The use of imaginary and complex values comes up in many physics/engineering derivations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it making the process of derivation easier or is it essential without which it would be impossible to derive some results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.It doesnt look like it is mandatory for a Newtonian to gen relativistic results and electrodynamics..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.Can we say the same thing about quantum mechanics either way for sure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could this be a difference in quantum mechanics over the classical picture? &lt;/p&gt;&#xA;" OwnerUserId="4094" LastActivityDate="2011-06-21T19:46:26.140" Title="Physics math without sqrt(-1)" Tags="&lt;mathematical-physics&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="11397" PostTypeId="1" AcceptedAnswerId="11413" CreationDate="2011-06-21T17:00:06.757" Score="3" ViewCount="161" Body="&lt;p&gt;I was playing a table-top rpg with a friend. It was set... well it was pretty weird, so I won't get into it. Suffice it to say the bad guy had a stick of dynamite and we were on a volcano. I hit the dynamite with my slingshot, knocking it into the lava directly behind the villain. My DM said the dynamite melts away. I said it should have blown him up. Needless to say, as a couple nerds (we were, after all, playing a fantasy game and then arguing about the physics of it) we were pretty upset with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would have happened in this... totally realistic scenario? Would the stick of dynamite exploded? Or simply melt away?&lt;/p&gt;&#xA;" OwnerUserId="1954" LastActivityDate="2011-06-21T23:59:18.050" Title="If you throw dynamite into a pool of lava, will it explode or melt?" Tags="&lt;explosions&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="11398" PostTypeId="1" AcceptedAnswerId="11402" CreationDate="2011-06-21T17:03:12.983" Score="3" ViewCount="219" Body="&lt;p&gt;Let's say we have 2 participles facing each other and traveling at speed of light&#xA;&lt;img src=&quot;http://i.stack.imgur.com/ONNCa.png&quot; alt=&quot;example&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I'm sitting on #1 participle so in my point of view #2 participle's speed is c+c=2c, double light speed? Please say why I am incorrect :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: About sitting me is just example, so in point of view of #1 participle, the second one moves at c+c=2c speed?&lt;/p&gt;&#xA;" OwnerUserId="3992" LastEditorUserId="3992" LastEditorDisplayName="" LastEditDate="2011-06-21T17:12:43.137" LastActivityDate="2011-06-21T17:23:30.747" Title="Double light speed" Tags="&lt;speed-of-light&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="11399" PostTypeId="2" ParentId="11396" CreationDate="2011-06-21T17:04:56.440" Score="2" ViewCount="" Body="&lt;p&gt;about 2. check &lt;a href=&quot;http://physics.stackexchange.com/questions/1323/madelung-transformation&quot;&gt;this question&lt;/a&gt; about an alternative formalism for Quantum Mechanics with equations where only &lt;em&gt;real&lt;/em&gt; probability densities and currents appear. The relevant wikipedia article is &lt;a href=&quot;http://en.wikipedia.org/wiki/Madelung_equations&quot; rel=&quot;nofollow&quot;&gt;this one about Madelung equations&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know any attempts to extend the same to QFT. Since complex residues are the butter and bread of most Feynmann loop diagrams, i would doubt it would be easy, or rewarding &lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-06-21T17:17:47.927" LastActivityDate="2011-06-21T17:17:47.927" CommentCount="1" />
  <row Id="11400" PostTypeId="2" ParentId="11398" CreationDate="2011-06-21T17:09:49.603" Score="1" ViewCount="" Body="&lt;p&gt;Easy. you cannot sit on a particle moving at the speed of light. If you could then you would be massless, and unable to sum properly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, there is no reference-frame moving with either photon, so no operational way to measure relative velocities between them. relative velocities have meaning only from a inertial frame. &lt;strong&gt;There are no inertial frames moving with the photon&lt;/strong&gt;, otherwise this frame would measure that photon to be at rest&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-06-21T17:16:59.083" LastActivityDate="2011-06-21T17:16:59.083" CommentCount="3" />
  <row Id="11401" PostTypeId="2" ParentId="11398" CreationDate="2011-06-21T17:19:55.603" Score="6" ViewCount="" Body="&lt;p&gt;This is what special relativity is all about..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In special relativity you cannot simply state that particle 2 is moving at c+c=2c in a reference frame where particle 1 is at rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speeds add like this (easily found in wikipedia):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_2^&amp;#39; = \frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. the speed of particle 2 $v_2^&amp;#39;$ in a reference frame where particle 1 is at rest is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_2^&amp;#39; = \frac{c+c}{1+1} = c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you cannot move faster than at the speed of light in the vacuum.&lt;/p&gt;&#xA;" OwnerUserId="2151" LastActivityDate="2011-06-21T17:19:55.603" />
  <row Id="11402" PostTypeId="2" ParentId="11398" CreationDate="2011-06-21T17:23:30.747" Score="13" ViewCount="" Body="&lt;p&gt;One of the results of special relativity is that a particle moving at the speed of light does not experience time, and thus is unable to make any measurements. In particular, it cannot measure the velocity of another particle passing it. So, strictly speaking, your question is undefined. Particle #1 does not have a &quot;point of view,&quot; so to speak. (More precisely: it does not have a rest frame because there is no Lorentz transformation that puts particle #1 at rest, so it makes no sense to talk about the speed it would measure in its rest frame.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But suppose you had a different situation, where each particle was moving at $0.9999c$ instead, so that that issue I mentioned isn't a problem. Another result of special relativity is that the relative velocity between two particles is &lt;em&gt;not&lt;/em&gt; just given by the difference between their two velocities. Instead, the formula (in one dimension) is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_\text{rel} = \frac{v_1 - v_2}{1 - \frac{v_1v_2}{c^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you plug in $v_1 = 0.9999c$ and $v_2 = -0.9999c$, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_\text{rel} = \frac{1.9998c}{1 + 0.9998} = 0.99999999c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is still less than the speed of light.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-21T17:23:30.747" CommentCount="2" />
  <row Id="11403" PostTypeId="2" ParentId="14255" CreationDate="2011-06-21T17:31:47.447" Score="2" ViewCount="" Body="&lt;p&gt;There is no evidence that another universe exists. There are various physical theories that allow for the existence of parallel universes, and as far as I know there are no widely accepted theories that prohibit their existence outright, so strictly speaking, I suppose it's possible. But it's unlikely that we would ever be able to detect them if they do exist.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-21T17:31:47.447" CommentCount="4" />
  <row Id="11404" PostTypeId="1" AcceptedAnswerId="11406" CreationDate="2011-06-21T18:03:14.643" Score="1" ViewCount="70" Body="&lt;p&gt;I am part of a school project, &lt;a href=&quot;http://projectstratos.com/&quot; rel=&quot;nofollow&quot;&gt;Project Stratos&lt;/a&gt; to send a balloon to the edge of space (the closer side :P) and was wondering how you would work out the accent rate of a large balloon (roughly 1m^3 of helium with 100g of mass) and the size of it as it increases its Altitude. I am creating a live map (that will be based on predictions rather than its actual location) and want to know the speed it will float up into the atmosphere. Currently we are assuming the ascent rate will be about 5m/s but I doubt that is very accurate and would this speed increase as it gets higher?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I would also quite like to know the burst height of the balloon.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditorDisplayName="" LastEditDate="2011-06-21T21:01:33.133" LastActivityDate="2011-06-21T21:01:33.133" Title="Ascent rate and size of balloon" Tags="&lt;kinematics&gt;&lt;atmosphere&gt;&lt;speed&gt;" AnswerCount="1" />
  <row Id="11405" PostTypeId="2" ParentId="11396" CreationDate="2011-06-21T19:46:26.140" Score="9" ViewCount="" Body="&lt;p&gt;The use of complex numbers is never really &lt;em&gt;essential&lt;/em&gt;, but if applicable it is almost always more &lt;em&gt;convenient&lt;/em&gt; than the equivalent representation in a 2d real vector space (in fact, one typically learns the formal properties of complex number manipulations by their effect on $(a,b) = a+ib.$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mention that complex numbers don't seem necessary for classical electrodynamics, and I agree -- however I can't imagine any clear-minded person forgoing their use. In fact it is in classical E&amp;amp;M that I think complex numbers really exhibit their gracefulness in the description of physical phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, as lurscher has mentioned, there are formulations of QM that avoid explicit reference to complex numbers -- they are equivalent mathematical &lt;em&gt;representations&lt;/em&gt;, but the manipulations have an added degree of bookkeeping that we had already built into complex numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's the rub. Complex numbers are a tool for describing a theory, not a property of the theory itself. Which is to say that they can not be the fundamental difference between classical and quantum mechanics. The real origin of the difference is the non-commutative nature of measurement in QM. Now this is a &lt;em&gt;property&lt;/em&gt; that can be captured by all kinds of beasts -- even real-valued matrices.&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2011-06-21T19:46:26.140" CommentCount="1" />
  <row Id="11406" PostTypeId="2" ParentId="11404" CreationDate="2011-06-21T19:55:54.267" Score="1" ViewCount="" Body="&lt;p&gt;Ascent rate of a balloon (assuming spherical symmetry) depends on the following forces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) The upward buoyant force $F_B=\frac{4}{3}\pi r^3\rho_{air}g$&lt;br/&gt;&#xA;(2) The gravitational pull downwards: $F_G=\frac{4}{3}\pi r^3\rho g$&lt;br/&gt;&#xA;(3) The drag force acting: $F_D\, =\, \tfrac12\, \rho_{air}\, v^2\, C_D\, A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a first glance it might appear that the balloon soon reaches a &lt;a href=&quot;https://secure.wikimedia.org/wikipedia/en/wiki/Terminal_velocity&quot; rel=&quot;nofollow&quot;&gt;terminal velocity&lt;/a&gt;. But the quantities involved in these equations aren't all independent of each other or remain constant. For example the density of air &lt;a href=&quot;https://secure.wikimedia.org/wikipedia/en/wiki/Density_of_air#Altitude&quot; rel=&quot;nofollow&quot;&gt;changes with altitude&lt;/a&gt;. And the atmospheric pressure drops as you ascent, causing the balloon to increase in volume, thereby increasing the drag on it. Thus to analyse the motion of the balloon carefully, one has to resort to numerical methods and computers. But if you are looking for an approximation the ascent rate could be taken as the terminal velocity and could be obtained by setting,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_B=F_G+F_D$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;leading to,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = \sqrt{\frac{8 g r}{3 C_D} \left( \frac{\rho_{air} - \rho}{\rho_{air}} \right)}$$&lt;/p&gt;&#xA;" LastActivityDate="2011-06-21T20:27:59.837" CommentCount="6" />
  <row Id="11407" PostTypeId="2" ParentId="11078" CreationDate="2011-06-21T20:11:00.510" Score="0" ViewCount="" Body="&lt;p&gt;Here comes an experimentalist's view:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I accept as premise that four vectors and special relativity exist and are well documented to hold for elementary particles.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I accept that the same holds true for the Heisenberg uncertainty principle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In special relativity mass is the &quot;scalar length&quot; of the four vector. When one has a specific particle, a photon, one knows the mass is 0. If one has two photons, their combined mass is not additive, it follows four vector algebra and will depend on the angles . The same will be true for two electrons, their combined mass is not additive but forms their invariant mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the meaning of mass is different  between Newtonian physics and relativistic physics, since in Newtonian masses are additive in relativistic they are not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have the Heisenberg uncertainty principle and within a specific region delta(p)*delta(x)&gt;h,  momentum is not well defined nor position, thus an experiment to get at the force, F=dp/dt will be indeterminate to that extent, and even for a single particle it will measure a different mass depending on the x within the HP. More so when there is an interaction of two particles, and I cannot think of an experiment where a second particle will not be involved in the measurement, where the combination of the indeterminacy of momentum and the vector addition will combine to even higher uncertainty of the combined mass measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my simple minded answer  without going into QED and virtual photons etc., where still four vector algebra will hold. There is a limit to the range in which Newtonian physics is valid. It is not possible to mix classical and relativistic and quantum mechanical  concepts of mass, so there can be no example.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-06-21T20:18:25.603" LastActivityDate="2011-06-21T20:18:25.603" CommentCount="7" />
  <row Id="11408" PostTypeId="2" ParentId="11350" CreationDate="2011-06-21T20:41:42.067" Score="1" ViewCount="" Body="&lt;p&gt;You can derive the correct results when you use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_%28infinitesimal%29&quot; rel=&quot;nofollow&quot;&gt;key property of differentials&lt;/a&gt; $$dx_i=\dot{x}_i dt.$$ Note that&#xA;$\Delta L$ is invariant under reparameterization $t&amp;#39;=f(t)$ as you can check easily (this is in fact the reason why you can write it as $\int d\ell$ without any reference to a parametrization). However, to calculate the length $\Delta L$ it is advisable to introduce some (arbitrary) parameterization. If you are interested in unique parameterizations: there exists also a unique parameteriziation with respect to arclength which has some nice feature.&lt;/p&gt;&#xA;" OwnerUserId="2962" LastActivityDate="2011-06-21T20:41:42.067" />
  <row Id="11409" PostTypeId="1" CreationDate="2011-06-21T20:45:23.503" Score="3" ViewCount="101" Body="&lt;p&gt;I've done a little research but I can't seem to find the answers to these questions in regards to a spinning space station simulating gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If we imagine the space station as a hollow doughnut and I'm floating in the center of the tube while the station is stopped what happens when the station starts? How are forces translated to my body that push me towards the floor or would I just remain floating there while the station flies by?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Does the spinning station rely on friction to simulate gravity? If the inner surface of the station were frictionless would I still move along with the station or would the station simply fly by me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4124" LastActivityDate="2011-06-21T23:49:22.613" Title="Artificial Gravity - Spinning Station Questions" Tags="&lt;gravity&gt;" AnswerCount="1" />
  <row Id="11411" PostTypeId="2" ParentId="11409" CreationDate="2011-06-21T21:10:26.107" Score="4" ViewCount="" Body="&lt;p&gt;1) If you were perfectly positioned and there was no air in that doughnut (you'd have to be in a space suit); then you would float free.  That is, you would stay still; and, from your perspective, the station walls would appear to race by you at the speed the station was spinning.  This would continue until microgravity forced you into contact with the inner wall (the side of the tube closest to the center of the station's center of mass).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)  No.  friction plays no part, except that it might be used to impart tangential acceleration to you.  If the station inside walls were smooth and frictionless (and that includes no air friction), the station would continue to slide by you, and you would remain weightless. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to note that this is only a thought experiment.  In reality, air friction (if nothing else) would QUICKLY accelerate you tangentially.  That tangential acceleration would throw you against the outer space station walls, where the non-zero friction there would help the air friction in accelerating you tangentially even more.  Once you gain any tangential velocity, it is ineveitable that you will contact the outer wall and gain radial acceleration, which you would percieve as weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To throw some numbers at this:  A theoretical space station of radius 900 meters, doing a complete rotation every 60 seconds (in order to generate about 1g at the outer rim) would have an outer tube tangential velocity of about 90 m/s.  The air in the tube would have the same tangential velocity.  If you were instantly instantiated in this tube, weightless, with zero tangential velocity, you would experience an air blast somewhat similar to standing on the hood of a Ferrari, screaming down the Autobahn at 324 km/hr (roughly 190 mph). &lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-06-21T21:27:10.133" LastActivityDate="2011-06-21T21:27:10.133" CommentCount="4" />
  <row Id="11412" PostTypeId="1" AcceptedAnswerId="11416" CreationDate="2011-06-21T21:24:37.980" Score="5" ViewCount="114" Body="&lt;p&gt;I have a question concerning a possible derivation of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Landau_quantization&quot; rel=&quot;nofollow&quot;&gt;Landau quantization&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our lecture notes (and some other places as well), the following ansatz is used:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\Psi(x,y,z) = \exp{\left(-j(\beta y + k_z z)\right)}u(x)&#xD;&#xA;$$,&#xA;which then leads to the known results (u(x) is identified as an harmonical oszillator,...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm unhappy with the ansatz itself: If I calculate the mean momentum in y direction, I obtain $\hbar \beta$. From classical mechanics, I know the solution can be nothing but spiral-like movement (should be compatible with quantum mechanics, as I could just use a particle with a huge momentum, wich could be directly observed...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what am I missing?? I would also expect some symmetry around some path center?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for my maybe rough english, but I'm not too familiar with using it in physical context...&lt;/p&gt;&#xA;" OwnerUserId="4127" LastEditorUserId="4127" LastEditorDisplayName="" LastEditDate="2011-06-21T21:42:09.777" LastActivityDate="2011-06-21T23:57:52.217" Title="Free particle in magnetic field / Landau quantization" Tags="&lt;quantum-mechanics&gt;&lt;condensed-matter&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="2" />
  <row Id="11413" PostTypeId="2" ParentId="11397" CreationDate="2011-06-21T21:56:39.687" Score="5" ViewCount="" Body="&lt;p&gt;The most likely answer is that the nitroglycerin portion of the dynamite would deflagrate (burn) and the diatomaceous earth would melt. Neither constituent would detonate.  Dynamite needs a shockwave to detonate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this is not to say that the deflagration of the nitro would be a tame thing.  Bullets are propelled out of guns by deflagration of materials behind said bullets.  It appears as an explosion because it is confined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dynamite would burn, most likely very rapidly.  Rapidly enough to make it appear as an explosion?  Probably not, since it is not confined; but I would not try viewing the experiment unprotected from anywhere closer than 30 yards. &lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-06-21T21:56:39.687" CommentCount="3" />
  <row Id="11415" PostTypeId="2" ParentId="11397" CreationDate="2011-06-21T23:35:57.757" Score="1" ViewCount="" Body="&lt;p&gt;Dynamite may be detonated by heat using a fuse and doesn't require another high explosive to detonate unlike TNT for example. It also doesn't require confinement in order to explode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dynamite will not melt away and the lava will definitely cause it to explode.&lt;/p&gt;&#xA;" OwnerUserId="4128" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-06-21T23:59:18.050" LastActivityDate="2011-06-21T23:59:18.050" CommentCount="1" />
  <row Id="11416" PostTypeId="2" ParentId="11412" CreationDate="2011-06-21T23:57:52.217" Score="5" ViewCount="" Body="&lt;p&gt;The key insight is that whenever (electro-)magnetism and Hamiltonians are involved, there is always a &lt;em&gt;gauge fixing&lt;/em&gt; happening somewhere. The basic issue is that the Hamiltonian is something like $$H = \left|p-eA\right|^2,$$ and for any given magnetic field $B$ there is a variety of potentials $A$ which can fit. For a uniform magnetic field, a frequently used one is the &lt;em&gt;Landau gauge&lt;/em&gt; where $$A = \left(0, Bx, 0\right).$$ In this gauge, the Hamiltonian is $$H = p_x^2 + (p_y - eBx)^2.$$ You then stick your ansatz in and get what you already know. The point to remember is that $p_x = i \frac{\partial}{\partial x}$ is &lt;em&gt;not&lt;/em&gt; the mechanical momentum. It is the &lt;em&gt;canonical&lt;/em&gt; momentum, which is defined to be the one that obey the canonical commutation relation $\left[ x, p_x \right] = i$. Ditto for $y$ and $p_y$. The mechanical momentum is exactly what's in the Hamiltonian: $p - eA$. Fairly sure if you follow that through, you get exactly what you expect, i.e. no average motion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a lot of applications, it's nicer to use the &lt;em&gt;symmetric gauge&lt;/em&gt;, where $A = \frac{1}{2}\left(-By, Bx, 0\right)$. In that case, you retain the explicit $x$-$y$ symmetry and things like angular momentum keeps making sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that things like $p$ change depending on the gauge you choose. Physical variables such as actual mechanical momentum does not. Indeed, in this case of a magnetic field in the $z$ direction, the mechanical momentum in the $x$ and $y$ directions &lt;em&gt;do not commute&lt;/em&gt;! You cannot simultaneously know both.&lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-06-21T23:57:52.217" CommentCount="2" />
  <row Id="11418" PostTypeId="2" ParentId="3189" CreationDate="2011-06-22T01:17:30.730" Score="0" ViewCount="" Body="&lt;p&gt;I've found another recent paper(18 Mar 2011) on an experiment to test the difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs//1103.3559&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs//1103.3559&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trouble is, I can't see how the experiment proves in favor of Minkowski. As far as I can make out, Zhong-Yue. Wang1 , Pin-Yu Wang, Yan-Rong Xu write out some theory, pronounce Minkowski is correct &lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g.&quot;Abraham-&#xA;Minkowski controversy is now resolved.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then do a Cherenkov effect experiment, then write some gobblygook about the soul of physics. All this is written up in Wikipedia as a crucial experiment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Abraham-Minkowski_controversy#Crucial_Experiment&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Abraham-Minkowski_controversy#Crucial_Experiment&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My trouble is, I can't see how it's so?&lt;/p&gt;&#xA;" OwnerUserId="3588" LastActivityDate="2011-06-22T01:17:30.730" />
  <row Id="11419" PostTypeId="1" AcceptedAnswerId="11422" CreationDate="2011-06-22T01:46:45.010" Score="1" ViewCount="107" Body="&lt;p&gt;I know that in general, interacting fermions cannot necessarily be described by a single Slater determinant.  Can anyone provide a simple example of a state that has no such representation?&lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-21T00:16:56.297" LastActivityDate="2011-07-21T09:41:29.560" Title="Example of a wavefunction that cannot be represented by a single Slater determinant" Tags="&lt;quantum-mechanics&gt;&lt;exchange-interaction&gt;&lt;spin-statistics&gt;" AnswerCount="2" />
  <row Id="11420" PostTypeId="2" ParentId="11386" CreationDate="2011-06-22T01:49:44.383" Score="5" ViewCount="" Body="&lt;p&gt;Typical commercial CRTs for televisions or computer monitors have electron guns that work in the range $5-30\text{ keV}$. Such electrons have a very short penetration depth in solids and will essentially all be stopped in the collimator or the glass of the tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the Particle Physics Data Book chapter on the &lt;a href=&quot;http://pdg.lbl.gov/2011/reviews/rpp2011-rev-passage-particles-matter.pdf&quot; rel=&quot;nofollow&quot;&gt;Passage of Particle Through Matter (PDF link!)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is that the screen will develop a net negative charge. This is the cause of the build-up of dust on the screen and of the low level discharge that you can feel if you run your hand over a CRT surface after it has been running for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Left to itself the charge bleeds off. Mostly into the surrounding atmosphere, but also into the bulk of the CRT's case.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-07-07T12:51:29.683" LastActivityDate="2011-07-07T12:51:29.683" CommentCount="2" />
  <row Id="11421" PostTypeId="1" AcceptedAnswerId="11425" CreationDate="2011-06-22T02:04:42.153" Score="1" ViewCount="69" Body="&lt;p&gt;In an answer to &lt;a href=&quot;http://physics.stackexchange.com/questions/11409/artificial-gravity-spinning-station-questionscorioles&quot;&gt;Artificial Gravity - Spinning Station Questions&lt;/a&gt; Vintage wrote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A theoretical space station of radius 900 meters, doing a complete rotation every 60 seconds (in order to generate about 1g at the outer rim) would have an outer tube tangential velocity of about 90 m/s. The air in the tube would have the same tangential velocity. If you were instantly instantiated in this tube, weightless, with zero tangential velocity, you would experience an air blast somewhat similar to standing on the hood of a Ferrari, screaming down the Autobahn at 324 km/hr.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The original question was from slight_disregard, I'm just expanding on it.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If I’m standing inside the station as Vintage states I’m going 324 km/hr if I jump straight up at 1 m/sec where do I land? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Does the air 10m up from where I’m standing going at the same speed or does the Coriolis effect make it go slower? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If I get in my Ferrari and go 324 km/hr opposite to the spin direction does my car become weightless?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4034" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-22T02:10:08.917" LastActivityDate="2011-06-22T05:53:24.293" Title="Artificial Gravity - Spinning Station Questions II" Tags="&lt;classical-mechanics&gt;&lt;rotational-dynamics&gt;&lt;centrifugal-force&gt;&lt;coriolis-force&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="11422" PostTypeId="2" ParentId="11419" CreationDate="2011-06-22T03:27:00.753" Score="4" ViewCount="" Body="&lt;p&gt;How about $\Psi(x_1,x_2) \equiv (x_1 -x_2)e^{-(x_1-x_2)^2}$?&lt;/p&gt;&#xA;" OwnerUserId="3970" LastActivityDate="2011-06-22T03:27:00.753" />
  <row Id="11423" PostTypeId="1" CreationDate="2011-06-22T03:27:02.860" Score="3" ViewCount="149" Body="&lt;p&gt;Is there an upper limit to the number of orbits an electron can have around say a proton?&#xA;Arent there states that are unstable(for n!=1) with corresponding mean/half lives and therefore uncertainty in energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do we differentiate between 2 n values (say v1,v2) esp if the uncertainty associated with their energy levels gets larger to the order of the energy difference associated with the transition? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also is it known that we can predict the following data with only computation resources being a bottleneck?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The half lives of each state(decay rate).&#xA;The transition rate(from v2 to v1 given v2&gt;v1)&lt;/p&gt;&#xA;" OwnerUserId="4094" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-22T04:16:09.353" LastActivityDate="2011-07-23T18:45:22.843" Title="electron orbits" Tags="&lt;quantum-mechanics&gt;&lt;uncertainty-principle&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="11424" PostTypeId="2" ParentId="4108" CreationDate="2011-06-22T04:16:03.903" Score="1" ViewCount="" Body="&lt;p&gt;Generalized momentum and momentum are two separate things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an electromagnetic field, the mechanical (also known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_momentum&quot; rel=&quot;nofollow&quot;&gt;kinetic&lt;/a&gt;) momentum of a non-relativistic charged particle is still $\vec{p}=m\vec{v}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec{P}=m\vec{v}+q\vec{A}$ is called the canonical momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in &lt;a href=&quot;http://en.wikipedia.org/wiki/Analytical_mechanics&quot; rel=&quot;nofollow&quot;&gt;analytical mechanics,&lt;/a&gt; the &lt;a href=&quot;http://en.wikipedia.org/wiki/Canonical_coordinates#Generalized_coordinates&quot; rel=&quot;nofollow&quot;&gt;generalized&lt;/a&gt; momentum can be almost everything, including position coordinates like $x,y,z$ or $r,\theta,\phi$.&lt;/p&gt;&#xA;" OwnerUserId="3887" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-06-22T14:17:48.233" LastActivityDate="2011-06-22T14:17:48.233" CommunityOwnedDate="2011-06-22T04:16:03.903" CommentCount="2" />
  <row Id="11425" PostTypeId="2" ParentId="11421" CreationDate="2011-06-22T05:53:24.293" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You land slightly ahead of where you jumped.  As mentioned in the comments, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Momentum#Momentum_in_electromagnetism&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Coriolis effect only applies to things that are moving in the rotating reference frame.  If the air is stationary in the rotating frame, it feels only the centrifugal force.  There will be a pressure gradient, creating buoyancy, just like on Earth, but all the air will have the same angular velocity.  (This means the air closer to the center of rotation has lower linear velocity, and you will feel a slight wind when you jump.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes, you become weightless.  As seen from an inertial frame comoving with the center of the ship, you are now stationary.  Of course there would be aerodynamic effects to contend with.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-06-22T05:53:24.293" />
  <row Id="11426" PostTypeId="2" ParentId="11386" CreationDate="2011-06-22T06:04:39.363" Score="2" ViewCount="" Body="&lt;p&gt;CRTs have an &quot;aquadag coating&quot; of fine carbon particles that allow the built-up negative charge to bleed off to ground (or chassis, at least).  Wikipedia: &lt;a href=&quot;http://en.wikipedia.org/wiki/Aquadag&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Aquadag&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="353" LastActivityDate="2011-06-22T06:04:39.363" CommentCount="1" />
  <row Id="11428" PostTypeId="1" AcceptedAnswerId="11433" CreationDate="2011-06-22T07:42:30.513" Score="-1" ViewCount="156" Body="&lt;p&gt;If a fan that is using 50 watts is moving 1 m³/min of air. Lets say the walls are the same temperature as the air so there’s no heat dissipation there. How do I know if the fan is putting in more energy in to the room than its dissipating?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the walls are 10 deg cooler than the air will this be sufficient for the molecules to dissipate the heat in to them. Where’s the point of balance.&lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-06-22T22:33:38.173" Title="A fan in a hot room at what point does it put in more energy that it dissipates" Tags="&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11429" PostTypeId="1" CreationDate="2011-06-22T08:08:54.637" Score="1" ViewCount="54" Body="&lt;p&gt;Consider a capillary tube (say from a liquid / capillary thermometer), that means a tube of small internal diameter which holds liquid by capillary action . The tube is fulfilled with water and closed at one end. How long it would take for water to completely evaporate out through the open end of the tube?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be specific, let the diameter of the tube be $d=1$mm, the length of the tube $L=1$m, the environmental temperature $T=300$ K and pressure $P=10^5$ Pa, the relative humidity $f=60$ %, &lt;a href=&quot;http://en.wikipedia.org/wiki/Contact_angle&quot; rel=&quot;nofollow&quot;&gt;the contact angle&lt;/a&gt; a) $\theta=0^{\circ}$ and b) $\theta=180^{\circ}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it have any effect on the time when the tube is positioned horizontally or vertically(open end above)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/1akdv.gif&quot; alt=&quot;A capillary tube with a liquid&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="552" LastActivityDate="2011-06-22T08:08:54.637" Title="Evaporation from a capillary tube" Tags="&lt;thermodynamics&gt;" CommentCount="3" />
  <row Id="11430" PostTypeId="2" ParentId="11428" CreationDate="2011-06-22T09:39:33.273" Score="1" ViewCount="" Body="&lt;p&gt;From my empirical experience, to dissipate 2kW of power in a room, one need 18C temperature difference above walls with one fan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to dissipate 0.05kW, walls need to be cooler by ~0.45C only.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-06-22T09:39:33.273" CommentCount="1" />
  <row Id="11431" PostTypeId="1" AcceptedAnswerId="11432" CreationDate="2011-06-22T11:01:36.833" Score="6" ViewCount="140" Body="&lt;p&gt;Is it possible to find directions just by looking at moon(not full bright)? I was curious to use the clue that if moon is half lit then the direction in which it is lit will it be east?&lt;/p&gt;&#xA;" OwnerUserId="4131" LastEditorUserId="4131" LastEditorDisplayName="" LastEditDate="2011-06-28T07:21:34.303" LastActivityDate="2011-06-28T07:21:34.303" Title="Finding Directions using moon" Tags="&lt;astronomy&gt;" AnswerCount="1" />
  <row Id="11432" PostTypeId="2" ParentId="11431" CreationDate="2011-06-22T12:36:24.393" Score="9" ViewCount="" Body="&lt;p&gt;If you know what time of day it is, then this works fine. A half-full moon is either waxing (becoming more full), in which case it's known as &lt;a href=&quot;http://earthsky.org/moon-phases/first-quarter&quot; rel=&quot;nofollow&quot;&gt;first quarter&lt;/a&gt;, or waning (becoming less full), in which case it's known as &lt;a href=&quot;http://earthsky.org/moon-phases/last-quarter&quot; rel=&quot;nofollow&quot;&gt;last quarter&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A first quarter moon rises in the East at approximately noon and sets in the West at approximately midnight, and its bright side will be towards the Sun, which will be towards the &lt;em&gt;West&lt;/em&gt;, because it is trailing the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A last quarter moon rises in the East at approximately midnight and sets in the West at approximately noon, and its bright side will be towards the Sun, which will be towards the &lt;em&gt;East&lt;/em&gt;, because it is leading the Sun.&lt;/p&gt;&#xA;" OwnerUserId="3624" LastEditorUserId="3624" LastEditorDisplayName="" LastEditDate="2011-06-24T01:47:12.933" LastActivityDate="2011-06-24T01:47:12.933" CommentCount="2" />
  <row Id="11433" PostTypeId="2" ParentId="11428" CreationDate="2011-06-22T12:43:00.890" Score="5" ViewCount="" Body="&lt;p&gt;Assuming no heat transfer from a cooler outside to a warmer inside, a fan always adds more energy than it dissipates (because it dissipates no energy). However, by introducing a breeze, it accelerates evaporation off your skin, which in turn cools you off. (This is related to the wind chill factor meteorologists talk about.) How much cooler you will feel will be a function of how much you are sweating. It's important to note that pets, who don't sweat from the majority of their skin like we do, thus aren't cooled as much by a fan as we are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method by which a fan can cool (as mentioned by Mitchell) is to move hot air away from a hot object, thus reducing the air's insulatory nature. This is why fans help cool off CPUs even though they don't sweat.&lt;/p&gt;&#xA;" OwnerUserId="3624" LastEditorUserId="3624" LastEditorDisplayName="" LastEditDate="2011-06-22T22:33:38.173" LastActivityDate="2011-06-22T22:33:38.173" CommentCount="5" />
  <row Id="11434" PostTypeId="1" CreationDate="2011-06-22T13:31:06.123" Score="1" ViewCount="61" Body="&lt;p&gt;I'm looking for a good introductory text on solid state devices for particle detection.&lt;br&gt;&#xA;Thanks a lot.&lt;/p&gt;&#xA;" OwnerUserId="4132" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-22T17:09:13.673" LastActivityDate="2011-08-04T00:13:25.233" Title="Reference for solid state particle detector" Tags="&lt;experimental-physics&gt;&lt;books&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11435" PostTypeId="1" CreationDate="2011-06-22T13:31:48.757" Score="0" ViewCount="88" Body="&lt;p&gt;What is the function of electric water pump, i.e does it push or pull water or both?&lt;/p&gt;&#xA;" OwnerUserId="4133" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-22T17:04:53.847" LastActivityDate="2011-06-22T22:32:43.650" Title="Function of electric water pump" Tags="&lt;electricity&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="11436" PostTypeId="1" CreationDate="2011-06-22T13:51:53.193" Score="0" ViewCount="56" Body="&lt;p&gt;A particle of charge $q$ and mass $m_0$&#xA;is subject to the action of a uniform electro-magnetic field $\vec{E}=(0,E,0), \vec{B}=(0,0,B)$ and at &#xA;$t = 0$ moves with velocity $\vec{v}=(v_1,v_2,v_3)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the acceleration at $t = 0$ if $v_1 = v_3 = 0$?&lt;/p&gt;&#xA;" OwnerUserId="4134" LastEditorUserId="4134" LastEditorDisplayName="" LastEditDate="2011-06-22T14:32:47.087" LastActivityDate="2011-06-22T16:19:20.877" Title="The acceleration of a particle in uniform electro-magnetic field" Tags="&lt;electromagnetism&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="11437" PostTypeId="2" ParentId="14255" CreationDate="2011-06-22T13:53:13.123" Score="1" ViewCount="" Body="&lt;p&gt;There are speculative theories that suggest areas of cold/hot patches in the cosmic microwave background are 'bruises' caused by collisions of expanding bubble universe... These are not 'parallel' in your sense though, more they are just different regions of space-time.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-06-22T13:53:13.123" />
  <row Id="11438" PostTypeId="1" AcceptedAnswerId="11458" CreationDate="2011-06-22T14:42:39.513" Score="4" ViewCount="96" Body="&lt;p&gt;I am using the standard symbols of $V_\mu$ for the gauge field, $\lambda$ for its fermionic superpartner and $F$ and $D$ be scalar fields which make the whole thing a $\cal{N}=2$ vector/gauge superfield in $2+1$ dimensions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the non-Abelian super-Chern-Simon's lagrangian density would be,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Tr[\epsilon^{\mu \nu \rho}(V_\mu \partial_\nu V_\rho - \frac{2}{3}V_\mu V_\nu V_\rho) +i\bar{\lambda_a}\lambda_a - 2FD]$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly this is classically scale invariant.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I would like to know of the argument as to why this is also quantum theoretically conformal   (..may be there is some obvious symmetry argument which I am missing..)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Also is it true or obvious that if the above is perturbed by a $\lambda Tr[\Phi ^4]$ potential then this might flow to a fixed point which is $\cal{N}=3$ ? And then will it still be superconformal ? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would like to know of a way of understanding this phenomenon of supersymmetry enhancement by renormalization flow. If someone could point me to some beginner friendly expository reference regarding this. &lt;/p&gt;&#xA;" OwnerUserId="1637" LastActivityDate="2011-06-22T21:24:22.473" Title="Argument for quantum theoretic conformality of $\cal{N}=2$ super-Chern-Simon's theory in $2+1$ dimensions" Tags="&lt;quantum-field-theory&gt;&lt;supersymmetry&gt;&lt;conformal-field-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="11439" PostTypeId="2" ParentId="11435" CreationDate="2011-06-22T16:12:38.197" Score="0" ViewCount="" Body="&lt;p&gt;I have a simple water pump that is an attachement to the Black end Decker electric tool. It is a small, 15cm or so disk, evidently rotating chambers inside to create a suction vacuum. So this one pulls the water.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-06-22T16:12:38.197" />
  <row Id="11440" PostTypeId="2" ParentId="11436" CreationDate="2011-06-22T16:19:20.877" Score="1" ViewCount="" Body="&lt;p&gt;As the comments said, this is a basic application of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lorentz_force&quot; rel=&quot;nofollow&quot;&gt;Lorentz force&lt;/a&gt;.  Written in your notation, we have the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{F} =  q[\vec{E} + (\vec{v} \times \vec{B})]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{v} = \left( 0, v_2, 0 \right)$$&#xA;$$\vec{E} = \left( 0, E, 0 \right)$$&#xA;$$\vec{B} = \left( 0, 0, B \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most likely thing you had trouble with would be the cross product I think.  You can use a formula for this or your right hand.  For the latter, in my own head, the magnetic field is &quot;up&quot; and the y-axis is toward my chest when I hold out my right hand.  My middle finger points toward me, my index finger points up, and my thumb points to the right.  So the answer is in the positive x-axis!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/l2zeq.png&quot; alt=&quot;RHR&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{v} \times \vec{B} = \left( v_2 B , 0, 0 \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will stop here because any more would certifiably be me doing it &quot;for you&quot;.  The idea of something being trivial is relative to the person, but everything that remains to answer your question should be trivial unless you are missing some fundamental part of physics instruction.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-06-22T16:19:20.877" />
  <row Id="11441" PostTypeId="1" CreationDate="2011-06-22T16:19:52.617" Score="1" ViewCount="55" Body="&lt;p&gt;Decoherence apparently only works if a lot of degrees in the environment are measured with infinitely many copies reaching to all eternity. If too few memories and records survive with infinitely many copies for all eternity, would not their sparsity not contain enough coarse-grained information to determine the past history of humans? Can we have a macroscopic superposition of humans? If I don't remember something in my past, was I in a superposition in the past? The brain is hidden inside the skull. Decoherence is significant only if indirect records of internal brain states survive for all eternity? Does that mean the human brain is a quantum computer?&lt;/p&gt;&#xA;" OwnerUserId="4135" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-22T17:03:59.370" LastActivityDate="2011-06-22T17:03:59.370" ClosedDate="2011-06-22T17:04:20.337" Title="Macroscopic superposition of humans and past thoughts?" Tags="&lt;quantum-mechanics&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="11442" PostTypeId="1" CreationDate="2011-06-22T16:53:04.117" Score="1" ViewCount="226" Body="&lt;p&gt;I've had argument on a forum about this. People think, that:&#xA;if the same object is magnetised, it has less energy. so by $E = mc^2$ it weighs less than an object that is not magnetised (hence spin of its particles is chaotic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've thought about experiment :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we have square of 4 protons and another square of 4 antiprotons. If we magnetise each square and annihilate them, do we get the same energy that we would get if only the antiprotons were magnetised?&lt;/p&gt;&#xA;" OwnerUserId="4136" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-23T05:17:21.287" LastActivityDate="2011-06-23T05:50:35.520" Title="Does magnetised material lose weight?" Tags="&lt;electromagnetism&gt;&lt;mass&gt;" AnswerCount="1" />
  <row Id="11444" PostTypeId="1" CreationDate="2011-06-22T18:36:49.680" Score="3" ViewCount="51" Body="&lt;p&gt;How does atmospheric CO2 and other Greenhousgases (GHG) affect the incoming (from sun) and outgoing (from earth) radiation. I understand that at certain wavenumbers (or areas of wavenumbers) in the Infrared (IR) these molecules (amongst others) absorb an IR-photon, get a bit &quot;warmer&quot; (means perhaps rotate faster or similar) and after a while reemit another photon with less energy (because the enhanced movement costs a bit of energy) ... resulting in a further warming. The emitted photon either hits another photon or will be going to space or will go back to earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have lots of molecules (N) and statistically these three mechanisms lead to a warming of the surface of the earth (let aside scattering etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &#xA;1. is this picture correct&#xA;2. when yes how are the different mechanisms in terms of photonic energies&lt;/p&gt;&#xA;" OwnerUserId="4138" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-22T19:03:12.900" LastActivityDate="2011-07-22T19:26:09.233" Title="How is the mechanism of greenhouse gases interacting with IR radiation?" Tags="&lt;photons&gt;&lt;atmospheric-science&gt;&lt;climate-science&gt;" AnswerCount="1" />
  <row Id="11445" PostTypeId="2" ParentId="11444" CreationDate="2011-06-22T19:12:10.010" Score="1" ViewCount="" Body="&lt;p&gt;To make things simple you want to compare two atmospheres that are identical, except that one has infinitessimally more of some IR absorber. In the second case, some IR photons that were traveling upwards are absorbed, and thermalized (i.e. the energy is transferred to the surrounding gas molecules, before being re-emitted). The IR photons are basically transferring energy, in this case from the surface (or perhaps a lower level of the atmosphere), to wherever they are absorbed. If the surface is warmer than the atmosphere where our absorber molecule resides, then that layer is on net absorbing more energy than it was in the unperturbed case, so that layer of the atmosphere will warm (with respect to the control case). Eventually once we let the atmosphere respond (but then the two atmospheres have different temperature profiles), we see that there is also more downgoing IR radiation below our absorbed level, so the layers of atmosphere below, and the ground will see a net warming effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To think of it in another way, the radiative impedence of the system (surface cum atmosphere) to space is higher with the addition of the greenhouse gases, so given a fixed energy input (absorbed sunlight) the equilibrium temperature will rise. Beyond that first order effect, the details of atmospheric structure and flow begin to change, and then the hard part of the problem begins.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-06-22T19:12:10.010" />
  <row Id="11446" PostTypeId="2" ParentId="11363" CreationDate="2011-06-22T19:32:09.487" Score="0" ViewCount="" Body="&lt;p&gt;I just happened to simulate the exact problem. Firstly, this falls under the topic of stochastic differential equations, as it will help us choose the right mathematics to work with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every physical process with dissipation, we can in-general associate a noise force to explain the physics of diffusion and noise dynamics. This is captured under the fluctuation dissipation theorem. There is a more deeper reason for this when treated at the quantum mechanical level (to maintain commutation relationship of quantized fields under going dissipation for example in a cavity). &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fluctuation-dissipation_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fluctuation-dissipation_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to answer your question, the square root term appears when discretizing the differential equation.We need to write down the derivative as (F(n+1)-F(n))/DT and the variance equation in discrete form by replacing the continuous time delta function (which has units of inverse time) with Kronecker delta function divided by the time step. The variance of the noise for fixed time step then has a sqrt(DT) in the denominator and will produce the form as you show in the question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: that the delta function has units of inverse time. so there is no discrepancy in units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few key points to keep in mind. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;use a first order integration method such as forward euler or midpoint method. &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Backward_Euler_method#The_backward_Euler_method&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Backward_Euler_method#The_backward_Euler_method&lt;/a&gt; . We can easily show that, higher order methods converge to wrong noise variances. In general, stochastic differential equations do not admit the higher order integration methods like higher order runge kutta. see Phys. Rev. E 74, 068701 (2006) Comment on “Numerical methods for stochastic differential equations” &lt;a href=&quot;http://pre.aps.org/abstract/PRE/v74/i6/e068701&quot; rel=&quot;nofollow&quot;&gt;http://pre.aps.org/abstract/PRE/v74/i6/e068701&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the noise is additive any first order integration will give a correct answer. But, if the noise is multiplicative, appropriate calculus must be employed (Ito or stratonovich). &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stratonovich_integral&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Stratonovich_integral&lt;/a&gt;&#xA;http://en.wikipedia.org/wiki/Ito_Calculus&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A quick check for the SDE will be to implement Brownian motion of a particle at temperature T and verify the average energy to be 0.5*kT.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3714" LastActivityDate="2011-06-22T19:32:09.487" />
  <row Id="11448" PostTypeId="2" ParentId="11363" CreationDate="2011-06-22T20:23:05.993" Score="0" ViewCount="" Body="&lt;p&gt;So... the delta function in time is defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta (t - t_0) = \int _0^\infty \delta (t) dt = \int_t^{t_0} \delta (t) dt \equiv 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which means that the delta function has the unit of $sec^{-1}$ hence I should divide by the time step.&lt;/p&gt;&#xA;" OwnerUserId="3677" LastActivityDate="2011-06-22T20:23:05.993" CommentCount="2" />
  <row Id="11449" PostTypeId="1" AcceptedAnswerId="11460" CreationDate="2011-06-22T20:25:55.170" Score="2" ViewCount="177" Body="&lt;p&gt;Is mass converted into energy in exothermic chemical / nuclear reactions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My (A Level) knowledge of chemistry suggests that this isn't the case. In a simple burning reaction, e.g. $C+O_2\to \text{CO}_2$, energy is released by the CO bonds forming; the atoms lose potential energy when they pull themselves towards each other, in the same way that a falling object converts GPE to KE. There are the same number of protons, electrons etc. in both the reactants and products. I would have assumed that this reasoning extends to nuclear fission/fusion as well, but one physics textbook repeatedly references very small amounts of mass being converted into energy in nuclear reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I just wanted to know if I was wrong about either of these types of reactions, and if so, what mass is lost exactly?&lt;/p&gt;&#xA;" OwnerUserId="4140" LastEditorUserId="4140" LastEditorDisplayName="" LastEditDate="2011-06-23T07:42:24.857" LastActivityDate="2011-06-24T03:41:13.520" Title="Conversion of mass to energy in chemical/nuclear reactions" Tags="&lt;nuclear-physics&gt;&lt;physical-chemistry&gt;&lt;mass-energy&gt;" AnswerCount="5" />
  <row Id="11450" PostTypeId="2" ParentId="11449" CreationDate="2011-06-22T20:31:12.823" Score="-3" ViewCount="" Body="&lt;p&gt;Chemical reactions of any kind ---&gt; No conversion.  All mass (and energy) is conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nuclear reactions ---&gt; Yes, mass is converted to energy, in both fission and fusion reactions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What mass?  Well, going by my college instruction, I would have said that a neutron is lost (converted to energy).  But a quick surf of the internet shows me that what I learned in college is now considered an old wives' tale.  The new answer, though obviously not what I learned, defies my efforts to understand, let alone condense into easy reader format for you.  Maybe someone else will give it a go.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-06-22T20:57:57.760" LastActivityDate="2011-06-22T20:57:57.760" CommentCount="9" />
  <row Id="11451" PostTypeId="1" CreationDate="2011-06-22T20:32:07.550" Score="4" ViewCount="114" Body="&lt;p&gt;I've been following the news around the work they are doing at the LHC particle accelerator in CERN. I am wondering what the raw data that is used to visualize the collisions looks like. Maybe someone can provide a sample csv or txt?&lt;/p&gt;&#xA;" OwnerUserId="4141" LastActivityDate="2011-06-24T03:03:07.127" Title="What does the raw data from a particle collision look like?" Tags="&lt;particle-physics&gt;&lt;large-hadron-collider&gt;&lt;cern&gt;&lt;particle-detectors&gt;&lt;data&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="11452" PostTypeId="2" ParentId="11449" CreationDate="2011-06-22T20:46:47.037" Score="1" ViewCount="" Body="&lt;p&gt;The potential energy of the chemical bonds &lt;em&gt;do&lt;/em&gt; correspond to an increase of mass proportionate to that energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for example the energy in a CO-O bond is 110 kcal / mol or about 7.6 x 10^-19 Joules per bond. Dividing by c^2 gives us a mass of about 8.5 x 10^-33 grams. In a mole of CO2, that amounts to 5.1 x 10^-9 grams. Since a mole of CO2 has a mass of about 44g, you're going to have a hard time measuring the mass difference.&lt;/p&gt;&#xA;" OwnerUserId="3624" LastActivityDate="2011-06-22T20:46:47.037" />
  <row Id="11453" PostTypeId="1" AcceptedAnswerId="11469" CreationDate="2011-06-22T20:47:34.150" Score="3" ViewCount="82" Body="&lt;p&gt;I was watching a few sci-fi movies and was wondering the real science explaining what would happen if you were to be subject to the conditions of outerspace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read the wikipedia article on space exposure, but was still confused. If a person was about the same distance from the sun as earth is, would they still freeze to death? (as shown in the movie Sunshine)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm reading from all sorts of sites with conflicting information about what would actually happen when a person is exposed to the vacuum of space...&lt;/p&gt;&#xA;" OwnerUserId="2383" LastActivityDate="2011-06-23T02:00:23.703" Title="Direct exposure to the vacuum of space" Tags="&lt;space&gt;&lt;vacuum&gt;" AnswerCount="4" />
  <row Id="11454" PostTypeId="2" ParentId="11451" CreationDate="2011-06-22T20:48:03.057" Score="4" ViewCount="" Body="&lt;p&gt;Different pieces of equipment will produce somewhat different looking data, but typically it consists of voltages defined as a function of time. In some cases (spark chambers, for example) the &quot;voltage&quot; is digital, and in others it is analog.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traditionally, the time series for the data is slower than the times required for the (almost light speed) particles to traverse the detector. Thus one had an effective photograph for a single experiment. More modern equipment is faster but they still display the data that way. Here's an LHC example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/pk7iV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above, the data has been organized for display according to the shape and geometry of the detector. The raw data itself would be digitized and just a collection of zeroes and ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are typically two types of measurements, &quot;position&quot; and &quot;energy&quot;. The position measurements are typically binary, that is, they indicate that a particle either came through that (very small) element or did not. In the above, the yellow lines are position measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that some of the yellow lines are curved. Actually all of them are curved at least some. This is because there is a strong magnetic field. The curvature of the particle tracks helps determine what particles they are. For example, given the same speed and charge, a heavier particle will run straighter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclotron#Relativistic&quot; rel=&quot;nofollow&quot;&gt;radius of curvature&lt;/a&gt; is given by:&lt;br&gt;&#xA;$$r = \frac{m\gamma E}{pB}$$&#xA;where $\gamma = 1/\sqrt{1-(v/c)^2}$ is the Lorentz factor, $E$ is the energy, and $p$ is the momentum. This helps determine the particle type and energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy measurements are generally analog. In them, one gets an indication of how much energy was deposited by the particle as it went by. In the above, the light blue and red data are energy measurements. For these measurements, one doesn't get such a precise position, but the amplitude is very accurate.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-06-24T03:03:07.127" LastActivityDate="2011-06-24T03:03:07.127" CommentCount="8" />
  <row Id="11455" PostTypeId="2" ParentId="11449" CreationDate="2011-06-22T20:49:14.227" Score="1" ViewCount="" Body="&lt;p&gt;Theoretically, even chemical reactions do convert mass to energy or vice versa. But the amount of mass converted is so small that we are generally unable to detect it.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-22T20:49:14.227" />
  <row Id="11456" PostTypeId="2" ParentId="11442" CreationDate="2011-06-22T20:53:08.357" Score="2" ViewCount="" Body="&lt;p&gt;(1) Magnetized objects typically have higher energy than before they were magnetized. There is energy in the magnetic field, and the magnetic field for a magnetized object is larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to see this is to imagine a large collection of tiny magnets. If you piled them into a mound they wouldn't organize themselves into all pointing in the same direction. Instead, they'd energetically prefer to be pointing in random directions so their magnetic fields cancel better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) So magnetizing an object increases its energy and therefore increases its mass. But with the usual magnets available to mankind, the change in mass is so small as to be undetectable.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-22T20:53:08.357" />
  <row Id="11457" PostTypeId="2" ParentId="11453" CreationDate="2011-06-22T21:03:37.023" Score="1" ViewCount="" Body="&lt;p&gt;They would freeze.&lt;br&gt;&#xA;They wouldn't freeze to death - since they would die of something else (lack of oxygen) first&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although a small part of you is facing a hot sun at 6000K most of your surface is facing cold dark space at 3K.  You can work out what temperature you will reach - it depends only on how reflective you are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you have the same reflectivity as the earth (35%) and you aren't close enough to the earth to receive any significant heat from it then you would end up at about the same temperature the Earth would have without the greenhouse effect of it's atmosphere = which is about -20C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were made of much darker material like the moon - you would get much colder. Parts of the moon not facing the sun or earth get down to around -150C&lt;/p&gt;&#xA;" OwnerUserId="2525" LastEditorUserId="2525" LastEditorDisplayName="" LastEditDate="2011-06-22T21:14:14.070" LastActivityDate="2011-06-22T21:14:14.070" />
  <row Id="11458" PostTypeId="2" ParentId="11438" CreationDate="2011-06-22T21:24:22.473" Score="3" ViewCount="" Body="&lt;p&gt;The usual argument for why the Chern-Simons action is exactly conformal is that the action is gauge invariant only if the coupling constant, or Chern-Simons level $k$ (which you have not included in your Lagrangian) is integer valued. If the theory was not conformal there would be a non-zero beta-function which would make the coupling depend continuously on the renormalization scale $\Lambda$. But the integer $k$ can not be a continuos function of $\Lambda$, and hence the beta-function has to vanish. This argument doesn't depend on supersymmetry and hence holds equally well for the $\mathcal{N}=2$ case in your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the extension to $\mathcal{N}=3$: If I remember correctly you need to add two chiral multiplets $Q$ and $\tilde{Q}$ in conjugate representations of the gauge group and with a kinetic term plus a superpotential of the form you give (something like $W = \frac{1}{k} (\tilde{Q}T^aQ) (\tilde{Q}T^aQ)$). By $\mathcal{N}=3$ supersymmetry the coupling is the same as the Chern-Simons level, and hence the theory is again conformal. Unfortunately I don't know of any good reference which gives a general introduction to this theory. For some recent applications of it see eg papers by Gaiotto and Yin (arXiv:0704.3740) and Aharony, Bergman, Jafferis and Maldacena (arXiv:0806.1218). These references also contain discussions about how the $\mathcal{N}=3$ Chern-Simons theory appears as a conformal fixed point starting from $\mathcal{N}=2$ or $3$ Chern-Simons-matter theory or Yang-Mills-Chern-Simons.&lt;/p&gt;&#xA;" OwnerUserId="3131" LastActivityDate="2011-06-22T21:24:22.473" CommentCount="9" />
  <row Id="11459" PostTypeId="2" ParentId="11435" CreationDate="2011-06-22T21:27:38.717" Score="1" ViewCount="" Body="&lt;p&gt;Depending on the pump design it can push, pull or do both.  In reality, all pumps are pushers, since even a pulling pump (e.g., sucking water up a drinking straw) is simply removing pressure on the pump side so that greater pressure on the pumped fluid at the other end of the line can push it into the pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Positive displacement pumps (e.g., gear pumps, peristaltic pumps and piston pumps) are typically considered &quot;pushers,&quot; even though they are pulling the pumped material into the pump before pushing it out.  You could use this intake to pump materials, in which case, I guess, you would have a &quot;puller.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A centrifugal pump could be considered a pusher or puller, depending on which end the longest line is connected to.  Then there is the electrodynamic pump, with no moving parts, which forces electrical current through the pumped media at right angles to an applied magnetic field, to induce a forward force in the conductive fluid itself: hard to characterize that one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If any pump could be classified as a pulling pump it would be a pump working on the Venturi/Bernoulli effect.  Pulling pumps which are open to the atmosphere have a theoretical pulling limit of 14.7 psi.  This is because the real action is pushing by the atmosphere on the end away from the pump, and atmospheric pressure at sea level is 14.7 psi.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-06-22T22:32:43.650" LastActivityDate="2011-06-22T22:32:43.650" />
  <row Id="11460" PostTypeId="2" ParentId="11449" CreationDate="2011-06-22T21:36:45.963" Score="5" ViewCount="" Body="&lt;p&gt;This is actually a more complex question than you might think, because the distinction between mass and energy kind of disappears once you start talking about small particles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is mass exactly? There are two common definitions: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The quantity that determines an object's resistance to a change in motion, the $m$ in $\sum F = ma$&lt;/li&gt;&#xA;&lt;li&gt;The quantity that determines an object's response to a gravitational field, the $m$ in $F_g = mg$ (or equivalently, in $F_g = GMm/r^2$)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The thing is, energy actually satisfies both of these definitions. An object that has more energy - of any form - will be harder to accelerate, and will also respond more strongly to a given gravitational field. So &lt;em&gt;technically&lt;/em&gt;, when computing the value of $m$ to plug into $\sum F = ma$ or $F_g = mg$ or any other formula that involves mass, you &lt;em&gt;do&lt;/em&gt; need to take into account the chemical potential energy, thermal energy, gravitational binding energy, and many other forms of energy. In this sense it turns out that &quot;mass&quot; is effectively just a word for the total energy of an object (well, divided by a constant factor: $m_\text{eff} = E/c^2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If mass is just another word for energy, why do we even talk about it? Well, for one thing, people got used to using the word &quot;mass&quot; before anyone knew about all its subtleties ;-) But seriously: if you really look into all the different forms of energy that exist, you'll find that figuring out how much energy an object actually has can be very difficult. For instance, consider a chemical compound - $\mathrm{CO}_2$ for example. You can't just figure out the energy of a $\mathrm{CO}_2$ molecule by adding up the energies of one carbon atom and two oxygen atoms; you also have to take into account the energy required to make the chemical bond, any thermal energy stored in vibrational modes of the molecule or nuclear excitations of the atoms, and even slight adjustments to the molecular structure due to the surrounding environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For most applications, though, you can safely ignore all those extra energy contributions because they're extremely small compared to the energies of the atoms. For example, the energy of the chemical bonds in carbon dioxide is one &lt;em&gt;ten-billionth&lt;/em&gt; of the total energy of the molecule. Even if adding the energies of the atoms doesn't quite get you the exact energy of the molecule, it's often close enough. When we use the term &quot;mass&quot;, it often signifies that we're working in a domain where those small energy corrections don't matter, so we can just add the masses of the parts to get the mass of the whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, whether the &quot;extra&quot; energies matter or not depends on what sort of process you're dealing with, and specifically what energies are actually affected by the process. In chemical reactions, the only changes in energy that really take place are those due to breaking and forming of chemical bonds, which as I said are a miniscule contribution to the total energy of the particles involved. But on the other hand, consider a particle accelerator like the LHC, which collides protons with each other. In the process, the chromodynamic &quot;bonds&quot; between the quarks inside the protons are broken, and the quarks then recombine to form different particles. In a sense, this is like a chemical reaction in which the quarks play the role of the atoms, and the protons (and other particles) are the compounds, but in this case the energy involved in the bonds (by this I mean the kinetic energy of the gluons, not what is normally called the &quot;binding energy&quot;) is fully &lt;em&gt;half&lt;/em&gt; of the energy of the complete system (the protons) - in other words, about half of what we normally consider the &quot;mass&quot; of the proton actually comes from the interactions between the quarks, rather than the quarks themselves. So when the protons &quot;react&quot; with each other, you could definitely say that the mass (of the proton) was converted to energy, even though if you look closely, that &quot;mass&quot; wasn't really mass in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nuclear reactions are kind of in the middle between the two extremes of chemical reactions and elementary particle reactions. In an atomic nucleus, the binding energy contributes anywhere from 0.1% up to about 1% of the total energy of the nucleus. This is a lot less than with the color force in the proton, but it's still enough that it needs to be counted as a contribution to the mass of the nucleus. So that's why we say that mass is converted to energy in nuclear reactions: the &quot;mass&quot; that is being converted is really just binding energy, but there's enough of this energy that when you look at the nucleus as a particle, you need to factor in the binding energy to get the right mass. That's not the case with chemical reactions; we can just ignore the binding energy when calculating masses, so we say that chemical reactions do not convert mass to energy.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-24T03:41:13.520" LastActivityDate="2011-06-24T03:41:13.520" CommentCount="1" />
  <row Id="11461" PostTypeId="1" CreationDate="2011-06-22T21:45:22.333" Score="2" ViewCount="72" Body="&lt;p&gt;Is there a way to take a moving source already into account when one derives the wave equation for sound waves and derive from that using only math the Doppler effect for moving sources?&lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-06-23T13:09:02.877" LastActivityDate="2011-08-22T13:33:25.613" Title="Wave equation for sound waves and moving source" Tags="&lt;waves&gt;&lt;doppler-effect&gt;" AnswerCount="1" />
  <row Id="11462" PostTypeId="2" ParentId="11388" CreationDate="2011-06-22T21:48:31.993" Score="2" ViewCount="" Body="&lt;p&gt;The BTZ black hole metric has a constant curvature, and hence the space is &lt;em&gt;locally&lt;/em&gt; isometric to AdS$_3$. Moreover, the global space can be written as a quotient space of AdS$_3$ by a discrete group. However, I don't see a simple way of seeing this from the coordinated in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That the space is &lt;em&gt;asymptotically&lt;/em&gt; AdS$_3$ means that it satisfies the Brown-Henneaux boundary conditions. Change coordinates to $\rho = e^r$. Then the asymptotic metric in the question takes the form&#xA;$$&#xD;&#xA;ds^2 = l^2 d\rho^2 + e^{2\rho}\left(-\frac{1}{l^2} dt^2 + d\phi^2\right) - J\,dt\,d\phi&#xD;&#xA;$$&#xA;The general Fefferman- Graham expansion takes the form&#xA;$$&#xD;&#xA;ds^2 = l^2 d\rho^2 + \left(e^{2\rho} \gamma^{(0)}_{ij} + \rho \gamma^{(1)}_{ij} + \gamma^{(2)}_{ij} + \dotsb \right) dx^i dx^j&#xD;&#xA;$$&#xA;but for a metric satisfying the Brown-Henneaux boundary conditions the $\gamma^{(1)}$ term vanishes. This is enough to ensure that the asymptotic isometry algebra is the same as in AdS$_3$, namely that of two Virasoro algebras.&lt;/p&gt;&#xA;" OwnerUserId="3131" LastEditorUserId="3131" LastEditorDisplayName="" LastEditDate="2011-06-23T09:14:33.113" LastActivityDate="2011-06-23T09:14:33.113" CommentCount="3" />
  <row Id="11463" PostTypeId="2" ParentId="11453" CreationDate="2011-06-22T21:53:03.277" Score="1" ViewCount="" Body="&lt;p&gt;Remember the pressure on your body is 14.7 lb at sea level so 1st whatever air you have is now at that pressure in you longs. An inexperience diver experiences this when he dives 10m takes a lung full of air and surfaces without exhaling he will be lucky if he don't die from it. So you exhale... now if something does not burst in your body because of internal pressure or nitrogen gassing in your blood well now you can die from lack of oxygen. Dissipation of heat is one of the biggest problems to a body in space. If the body is not rotating one side will cook and the other will freeze. If its rotating just right it will become a new moon&amp;lt;-- he he. &lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-06-22T21:53:03.277" />
  <row Id="11464" PostTypeId="2" ParentId="8768" CreationDate="2011-06-22T22:57:09.953" Score="2" ViewCount="" Body="&lt;p&gt;The BPS conditions in $\mathcal{N}=4$ SYM is just another name for the shortening conditions in the $psu(2,2|4)$ symmetry algebra of the theory. This algebra has 32 odd generators: 16 supercharges $Q_{\alpha a}$ and $\tilde{Q}_{\dot{\alpha}}^a$ and 16 superconformal charges $S^a_\alpha$ and $\tilde{S}_{\dot\alpha a}$. Here the indices $\alpha,\dot\alpha=1,2$ are Weyl spinor indices and $a=1,\dotsc,4$ is an index in the fundamental representation of the $SU(4)$ R-symmetry group. The generators $S$ and $\tilde{S}$ have dimension $-1/2$ and hence annihilates a primary state. The highest weight state of a BPS multiplet is furthermore annihilated by one or more of the supercharges. In particular a $\frac{1}{2}$-BPS state, or chiral primary, is annihilated by half of the charges $Q$ and $\tilde{Q}$. Similarly $\frac{1}{4}$-BPS ($\frac{1}{8}$-BPS) states are annihilated by a quarter (an eighth) of these charges. The exact details of what fields make up the operators in the various subsectors of course depend on the choice of simple roots (or lowering/raising operators) in the algebra, but the notation in the question seems fairly standard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The details of the symmetry algebra should be covered in any review of $\mathcal{N}=4$ SYM or the AdS/CFT-correspondence. See eg arXiv:1012.3938, which however only discusses the planar theory where only single trace operators are relevant. From an algebra point of view the more general case should however be essentially the same.&lt;/p&gt;&#xA;" OwnerUserId="3131" LastActivityDate="2011-06-22T22:57:09.953" CommentCount="12" />
  <row Id="11465" PostTypeId="1" AcceptedAnswerId="11526" CreationDate="2011-06-22T23:15:41.527" Score="3" ViewCount="55" Body="&lt;p&gt;I would like a suggestion on a price-efficient way of experimenting with spontaneous down-conversion and single-photon counting. The simple dual-slit experiment could be one part of an application although there are many, but I'm mainly asking about the key components which are similar - the laser that pumps a SPDC-crystal, the crystal itself, and photon-counters (the other standard optical bench items I'm more familiar with).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, is a 405 nm (blue) diode-laser module the way to go, with a BBO-crystal, and what to use for detectors - are avalance photo-diodes good enough? What kind of quality does the laser need to have (wavelength stability etc)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there is a Journal of american physics teachers or something which have articles covering these fundamental experimental setups now and then, but I thought I'd ask here in case anyone who works at a lab (with limited funding :) with a good setup could share some ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this gets many answers, it should be turned into a community wiki entry I guess.&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-06-24T04:27:58.427" Title="The most price-efficient experimental setup involving SPDC, single-photon counting etc" Tags="&lt;quantum-mechanics&gt;&lt;experimental-physics&gt;&lt;quantum-optics&gt;&lt;laser&gt;&lt;spdc&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="11467" PostTypeId="1" CreationDate="2011-06-22T23:41:33.213" Score="1" ViewCount="20" Body="&lt;p&gt;In relation to atmospheric spectroscopy and some other gaseous spectroscopy applications where organic molecules are present in the vapor, does anyone know of any papers or resources on pressure broadening due to hydrocarbons or other organic molecules in the vapor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Broadening due to noble gases seems to have been well studied in various AMO sources,but it seems like slim pickings when you want to talk about larger molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically I'm looking at the line-broadening of alkali vapor absorption spectra due to the presence of short length alkenes in the vapor (Short means &amp;lt; 30 carbons)&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-06-23T13:07:26.763" LastActivityDate="2011-06-23T13:07:26.763" Title="Linewidth pressure broadening due to organic molecules" Tags="&lt;spectroscopy&gt;" />
  <row Id="11468" PostTypeId="1" AcceptedAnswerId="11476" CreationDate="2011-06-23T00:53:41.213" Score="2" ViewCount="87" Body="&lt;p&gt;I keep wondering how a Rubbia thorium reactor would handle a natural disaster of Fukushima level intensity. As I understand it the nuclear chain reaction would stop instantly if the power is cut, but nasty waste products such as Uranium-233 would still require substantial cooling and hence it would seem that this reactor is just as unsafe as any other type of reactor with definite risk of meltdown, radiation and other problems?&lt;/p&gt;&#xA;" OwnerUserId="4145" LastActivityDate="2011-06-23T06:16:03.860" Title="Is a Rubbia thorium reactor safer than other modern reactor types?" Tags="&lt;nuclear-engineering&gt;" AnswerCount="2" />
  <row Id="11469" PostTypeId="2" ParentId="11453" CreationDate="2011-06-23T01:13:23.513" Score="3" ViewCount="" Body="&lt;p&gt;You'd freeze to death faster in the Atlantic ocean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space has essentially no thermal conductivity.  All the heat you lose will be radiated away.  According to the Stefan-Boltzman law, $W = \sigma T^4$, you would lose at most 500 watts per square meter of body surface area.  By contrast, the convective heat transfer coefficient in water is about 12,500 watts/square meter / degree Kelvin temperature difference.  So, I think freezing would be the least of your concerns.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditorDisplayName="" LastEditDate="2011-06-23T01:20:03.397" LastActivityDate="2011-06-23T01:20:03.397" />
  <row Id="11470" PostTypeId="1" AcceptedAnswerId="11513" CreationDate="2011-06-23T01:34:51.210" Score="4" ViewCount="131" Body="&lt;p&gt;The human brain is said to produce a magnetic field resulting from the action potentials released inside the brain. What's the nature of such a field in terms of size and strength, and what is the potential for manipulation of brain functions by interfering with it by means of electromagnetic radiation?&lt;/p&gt;&#xA;" OwnerUserId="4145" LastActivityDate="2011-06-23T18:36:54.297" Title="What are the characteristics of the magnetic field surrounding a human brain?" Tags="&lt;electromagnetism&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11471" PostTypeId="2" ParentId="11468" CreationDate="2011-06-23T01:49:25.663" Score="2" ViewCount="" Body="&lt;p&gt;Fukushima incident has reached current scale due to human fails (both at the engineering, construction, operation &amp;amp; disaster mitigation stages), not natural disaster. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No matter how safe is the reactor, if humans will continue to ignore safety for the sake of profit, such fails will continue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thorium reactor is indeed 'safer' to a some degree, due to the fact that combined activity of wastes is lower, but If we had Thorium reactor at Fukushima, we might have had similar result (i.e. destruction of all cores &amp;amp; release of comparable(maybe some 2 times less) amount of radioactivity into environment). Yes, cooling is also needed for thorium reactor is shutdown state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although, it is possible(and it's being done for modern reactors) to have both thorium &amp;amp; usual uranium reactor built which are able to stay cooled passively for extended period - but still probably not long enough for fukushima case. &lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-06-23T01:49:25.663" />
  <row Id="11472" PostTypeId="2" ParentId="11453" CreationDate="2011-06-23T02:00:23.703" Score="1" ViewCount="" Body="&lt;p&gt;Just want to add that freezing is 'not an issue'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that water boiling temperature gets lower at lower pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In vacuum blood boils even at 36.6, so all your blood circulation is stopped immediately due to bubbles of blood (water) vapor. So you loose consciousness in few seconds, and die in minutes due to lack of oxygen in brain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you would rotate around yourself at ~60RPM and not wearing white, you should nor freeze nor fry in short term (minutes). If you would freeze - then earth would also freeze. As earth is at equilibrium at 20C, you should be near it too if you are not wearing white.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-06-23T02:00:23.703" CommentCount="1" />
  <row Id="11474" PostTypeId="2" ParentId="11449" CreationDate="2011-06-23T04:48:48.630" Score="1" ViewCount="" Body="&lt;p&gt;First you have to understand, there are two definitions for mass: rest mass and motional mass. Rest mass is the mass measured in the frame where the object is stationary. Motional mass is $m_1=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$. The old school calls the motional mass mass, while the new school call rest mass mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rest mass is not additive. The rest mass of a molecule is not the sum of its atoms, but the sum plus the bond energy divided by $c^2$. Therefore in a chemical reaction, though the number of atoms (or electrons, protons, neutrons) are conserved, the rest mass is not, since the bond energy is changed. And this can be called mass being converted into energy. In&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motional mass, however, is just another name for energy, and therefore always conserved. In combustion, the molecular bond energy is converted to the additional kinetic energy of $\mathrm{CO_2}$ molecules, so the mass(or energy) is conserved. In nuclear reaction, the electromagnetic and strong potential energy between nucleons are converted into the kinetic energy of new nuclei and (sometimes) flying neutrons. Sometimes the new nuclei are in excited state, and will soon emit gamma radiation, which are basically high energy photons with zero rest mass but nonzero motional mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of old school books I read use double standards. When they refer to mass generically, they mean motion mass, which increases with speed; but when they refer to mass in nuclear reaction, they mean rest mass. I guess this is how you got confused.&lt;/p&gt;&#xA;" OwnerUserId="3887" LastActivityDate="2011-06-23T04:48:48.630" />
  <row Id="11475" PostTypeId="1" AcceptedAnswerId="11525" CreationDate="2011-06-23T05:36:55.343" Score="1" ViewCount="102" Body="&lt;p&gt;What is the centripetal acceleration and angular velocity of a child located 8.2 m&#xA;the center of a carousel? The speed (size of the tangential velocity) of the child is&#xA;2.1 m / s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A train moves in a straight path north until it turns to west.&#xA;If the road segment used to change direction is shaped like a quarter circle of radius 30 m and the train takes 30 seconds to traverse that part of the road,&#xA;What is it the speed (size of the velocity vector) and the centripetal acceleration acts on the train as it traverses the curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am reviewing some concepts like centripet force, &lt;code&gt;ar = ( v^2 ) / r&lt;/code&gt;&#xA;also this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The direction of the centripital&#xA;  acceleration is always inwards along&#xA;  the radius vector of the circular&#xA;  motion. The magnitude of the&#xA;  centripetal acceleration is related to&#xA;  the tangential speed and angular&#xA;  velocity as follows:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/XGojw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please guide me to solve the 2 problems above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the fisrt one is it only :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;(2.1 m / s)^2 /  8.2 m&lt;/code&gt;  ?&lt;/p&gt;&#xA;" OwnerUserId="4147" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-06-23T13:07:07.387" LastActivityDate="2011-06-24T03:52:03.603" Title="Doubt concerning centripetal acceleration" Tags="&lt;homework&gt;&lt;kinematics&gt;&lt;rotation&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="11476" PostTypeId="2" ParentId="11468" CreationDate="2011-06-23T06:16:03.860" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;As I understand it the nuclear chain reaction would stop instantly if the power is cut, but nasty waste products such as Uranium-233 would still require substantial cooling&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are focusing on entirely the wrong thing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reactor burning Thorium produces fission products (like all reactors), including the majority of the heat source as well as radioactive Iodine and Cesium isotopes.  The problem with Fukushima was mostly the release of such isotopes into the atmosphere (ultimately, isn't all of nuclear safety concerned with this?).  However much heat U-233 produces is completely beside the point when you ask to compare it to Fukushima, it matters, just not for this question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I keep wondering how a Rubbia thorium reactor would handle a natural disaster of Fukushima level intensity. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The key design feature of the Rubbia proposed reactor is that it is a &lt;strong&gt;subcritical accelerator driven reactor&lt;/strong&gt;.  To the extent that it uses a solid fuel form that does not have active fission product removal it is subject to the aforementioned problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problems of sustained removal of decay heat in even the most dire of circumstances, however, have been addressed very well just with the next generation of reactors we are building today (although the concern can never be completely eliminated).  The &quot;passive&quot; designs share the same fuel type with Fukushima (and almost all other reactors in the world), and just use natural forces to cool the reactor for weeks to months after an accident so loss of power isn't a concern.  These same passive design principles could (and would) be applied to the Rubbia reactor design, because the Rubbia idea addresses concerns very different from the problems at Fukushima, which include sustainability and criticality safety.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reduced radioactivity of accelerator driven subcritical reactors and Thorium reactors is not the short-lived stuff that creates problems during accidents.  Mostly they reduce the long term waste and the unused fuel.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-06-23T06:16:03.860" CommentCount="4" />
  <row Id="11477" PostTypeId="1" AcceptedAnswerId="11496" CreationDate="2011-06-23T07:08:37.133" Score="3" ViewCount="198" Body="&lt;p&gt;This brand new published result (nature):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Experimental non-classicality of an indivisible quantum system&lt;/strong&gt;&lt;br&gt; by&#xA;  Radek Lapkiewicz,    Peizhe Li,  Christoph Schaeff,  Nathan K. Langford,     Sven Ramelow,   Marcin Wieśniak     &amp;amp; Anton Zeilinger&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(see &lt;a href=&quot;http://www.nature.com/nature/journal/v474/n7352/full/nature10119.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, for a more popular article about it see &lt;a href=&quot;http://www.newscientist.com/article/dn20600-quantum-magic-trick-shows-reality-is-what-you-make-it.html?DCMP=OTC-rss&amp;amp;nsref=physics-math&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;; also for a pre-print see the ArXiv &lt;a href=&quot;http://arxiv.org/abs/1106.4481&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;seems to support the Copenhagen interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question&lt;/strong&gt; &lt;br&gt;Does this definitely rule out the many-worlds interpretation - or are there still loopholes? How could a many-worlds interpretation of this experiment possibly look like (if possible)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;br&gt;&#xA;Because I obviously created some confusion, how I came to that question: at the end of the NewScientist-article it says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Niels Bohr, a giant of quantum&#xA;  physics, was a great proponent of the&#xA;  idea that the nature of quantum&#xA;  reality depends on what we choose to&#xA;  measure, a notion that came to be&#xA;  called the Copenhagen interpretation.&#xA;  &quot;This experiment lends more support to&#xA;  the Copenhagen interpretation,&quot; says&#xA;  Zeilinger.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="171" LastEditorUserId="171" LastEditorDisplayName="" LastEditDate="2011-06-23T15:42:30.483" LastActivityDate="2011-06-23T15:42:30.483" Title="Does this new quantum experiment rule out the possibility of a many-worlds interpretation?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-interpretations&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="11478" PostTypeId="1" CreationDate="2011-06-23T07:42:04.327" Score="1" ViewCount="73" Body="&lt;p&gt;We typically say that forces cause acceleration inversely proportionate to mass. Would it be any less correct to say that acceleration causes forces proportionate to mass? Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that the underlying question in my mind - essentially, what distinguishes cause from effect - is far more general. But this seems like a good place to start.)&lt;/p&gt;&#xA;" OwnerUserId="4149" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-23T09:56:01.843" LastActivityDate="2011-06-23T09:56:01.843" ClosedDate="2011-06-23T09:53:08.960" Title="What, if anything, makes forces the &quot;cause&quot; and acceleration the &quot;effect&quot;?" Tags="&lt;mechanics&gt;&lt;forces&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11479" PostTypeId="1" CreationDate="2011-06-23T07:44:03.750" Score="2" ViewCount="37" Body="&lt;p&gt;From a paper on tunnel design I've been reading: (&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0886779887900113&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencedirect.com/science/article/pii/0886779887900113&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the present application, the solu-&#xA;  tion corresponding to a sinusoidal load&#xA;  in an infinite elastic medium is sought.&#xA;  Since no closed-form solution to this&#xA;  problem exists, a numerical procedure&#xA;  should be used. The procedure involves,&#xA;  first, deriving the solution to the case of&#xA;  a constant pressure applied to a finite&#xA;  strip in an infinite body. The solution&#xA;  for a sinusoidal distribution of loading&#xA;  then can be found by dividing the&#xA;  wavelength into several segments and&#xA;  assuming the pressure on each segment&#xA;  to be constant. In the present case, this&#xA;  procedure is applied to calculate the&#xA;  displacements under a sinusoidal line&#xA;  load. Each wavelength was divided in 10&#xA;  and 20 segments and a line load of 4, 6, 8&#xA;  and 10 wavelengths were considered. It&#xA;  was found that the calculated displace-&#xA;  ments became insensitive to the number&#xA;  of wavelengths when the latter exceeded&#xA;  6, and that 10 segments were enough to&#xA;  represent each wavelength.&#xA;  As a result of this analysis, the vertical&#xA;  displacement under a sinusoidal load&#xA;  may be approximated by&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$u_y = \frac{(3-4\nu)}{16\pi(1-4\nu)G}\sigma L\sin\frac{2\pi x}{L}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The authors refer to the plane strain solution to Kelvin's problem before this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you suggest how I would go about deriving this as mentioned in the paper—from going for a finite strip in an infinite body to applying a &lt;em&gt;numerical&lt;/em&gt; solution and getting the result for displacement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: $\nu$ poisson's ratio, $L$ Length of strip, $G$ is the shear modulus.&lt;/p&gt;&#xA;" OwnerUserId="4148" LastActivityDate="2011-06-23T07:44:03.750" Title="Displacement due to sinusoidal load on a finite strip in an infinite plane" Tags="&lt;mechanics&gt;&lt;elasticity&gt;" />
  <row Id="11480" PostTypeId="1" CreationDate="2011-06-23T07:52:34.303" Score="2" ViewCount="79" Body="&lt;p&gt;I have often heard neighbours talking things like in a multi storied apartment, the upper flats are more hot in summer then lower flats (or vice versa?) and similarly for some comparison in winter? and similarly something for basement..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have never understood these things well, even though people talk as if this is obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what all factors cause such differences?&lt;/p&gt;&#xA;" OwnerUserId="1093" LastActivityDate="2011-06-23T12:30:32.083" Title="In summer are the upper storey flats more hot or the lower storey flats?" Tags="&lt;thermodynamics&gt;&lt;energy&gt;&lt;heat&gt;&lt;space&gt;&lt;temperature&gt;" AnswerCount="2" />
  <row Id="11481" PostTypeId="1" CreationDate="2011-06-23T07:56:27.263" Score="1" ViewCount="114" Body="&lt;p&gt;Trying to understand Schutz's AFCIGR, where does the minus sign appear from in the metric tensor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/jXilu.gif&quot; alt=&quot;metric&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that this expresses the invariance of the spacetime interval. Schutz says (I think) that the metric is a (0,2) tensor. I assume that means it is the product of two one-forms, so presumably one of these one-forms has a -ve time component. What does that mean? Why don't both one-forms have a -ve time component? Looking at a Minkowski diagram what is a straightforward way to understand/visualise those one-forms? At my level, two things have been multiplied together to give a 4x4 matrix which has -1 in the top left corner. What are those two things and why do they give a -1 time component?&lt;/p&gt;&#xA;" OwnerUserId="4075" LastActivityDate="2011-06-23T11:35:33.007" Title="Where does the minus sign appear from in the metric tensor?" Tags="&lt;special-relativity&gt;&lt;relativity&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="11482" PostTypeId="2" ParentId="11478" CreationDate="2011-06-23T08:07:53.963" Score="1" ViewCount="" Body="&lt;p&gt;Newton's first law states that an object will keep doing what it is doing if left alone, in other words - The natural state of an object is static - unchanging - motion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton's second law &lt;em&gt;clarifies&lt;/em&gt; the first. Acceleration, or any change in motion, is an unnatural state for an arbitrary object left to its laurels, however it is a state that clearly exists all around us. Newton defines the &quot;thing&quot; that forces an object to change its state of being - a force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this most rigorous sense, a force is &lt;em&gt;defined&lt;/em&gt; to be that which causes a change in motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The observation of a change in momentum necessitates that there is some force driving that change, so in this sense the two are equivalent (there is an equals sign there after all) - wherever you see a (net) force you will see an acceleration, wherever you see an acceleration you will find a force responsible for it. However, going back to the first law, acceleration is a change in the (kinetic) state of an object, an objects natural tendency is to statically maintain its state. The observation of an unnatural state of being would logically imply that there is a cause.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively it seems unnatural that accelerations would happen spontaneously and that the universe will invent a force  just to balance the books if you will. &lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2011-06-23T08:07:53.963" />
  <row Id="11483" PostTypeId="2" ParentId="11480" CreationDate="2011-06-23T08:54:12.583" Score="2" ViewCount="" Body="&lt;p&gt;It depends very much on the insulation. The top floor can be very hot in a cement apartments flat  without an attic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second consideration is orientation. East facing and west facing flats are hotter, without extra shading and insulation because in the summer the sunlight is strong from morning  until it sets. North facing flats are the best for summer ( and for painters). South facing in temperate regions, where the sun at noon is very high, are OK too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basement flats are cooler because the earth two meters down has a steady temperature less than 20C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a north looking basement flat will be the coolest. Depressing too:no view . A south facing basement flat will  be the warmest in winter and as cool in the summer,since the sun seldom reaches down to basement level.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-06-23T10:55:52.760" LastActivityDate="2011-06-23T10:55:52.760" />
  <row Id="11484" PostTypeId="2" ParentId="11480" CreationDate="2011-06-23T09:03:17.270" Score="2" ViewCount="" Body="&lt;p&gt;It will depend on a lot of factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, otherwise-identical flats near the top, will be hotter than those near the bottom, for two reasons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) heat rises  - so heat will rise from lower flats to upper flats. More accurately, the density of air decreases as temperature rises, so hotter air will tend to rise up through buildings, where convection is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) overshading is likely to be less, higher up: in summer, most of the heat in a typical flat will come from solar gain (rather than, say, from internal gains from cooking, people, appliances). The more that windows are overshaded, the lower this solar gain is. Flats low down will have their windows overshaded by neighbouring buildings, trees, and so on. Flats higher up will see more sky from their windows; so will have higher solar gain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Winter&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much of the stuff above, particularly about heat rising, but also about solar gain, still applies in winter: although heating systems may now be the single largest source of heat, solar gains can still be relevant, if there are large south-facing windows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Basements&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, there's little or no solar gain - there may be some small windows at the footway ground level, but not much. However, as @anna-v says, there is the moderating effect of the ground itself, which acts as a large thermal store. This large thermal mass will act as a seasonal buffer, heating very slowly through spring and summer, and cooling slowly through autumn and winter, thus typically moderating both the hottest summer temperatures and the coldest winter temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are software packages, such as &lt;a href=&quot;http://apps1.eere.energy.gov/buildings/energyplus/energyplus_about.cfm&quot; rel=&quot;nofollow&quot;&gt;Energyplus&lt;/a&gt;, and &lt;a href=&quot;http://www.passivhaustagung.de/Passive_House_E/PHPP.html&quot; rel=&quot;nofollow&quot;&gt;PHPP&lt;/a&gt; that can model solar gain and the effects of thermal mass, at any time of year, for any location; but note that will need a &lt;strong&gt;lot&lt;/strong&gt; of input parameters to do a decent job of it.&lt;/p&gt;&#xA;" OwnerUserId="4066" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-06-23T12:30:32.083" LastActivityDate="2011-06-23T12:30:32.083" CommentCount="2" />
  <row Id="11485" PostTypeId="2" ParentId="11478" CreationDate="2011-06-23T09:05:59.340" Score="0" ViewCount="" Body="&lt;p&gt;I hope @Toph is not jesting whit this question but I will bite. There are 2 actions and effects when a force is applied. Say Object A pushes on object B the 1st cause/effect is that B resist moving “Static Resistance.” It’s apparent here that A courses the effect that B will resist. If B does not move then the effect is all on A and the energy is dissipated in heat and A stops. If B does move then is because A transfers the force to B or B takes the force from A. The only reason I contemplate this is that the more I learn of quantum-mechanics the more I wonder if physics of the big are what they appear to be.&lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-06-23T09:05:59.340" />
  <row Id="11486" PostTypeId="2" ParentId="11481" CreationDate="2011-06-23T09:24:46.120" Score="4" ViewCount="" Body="&lt;p&gt;The metric is a symmetric bilinear form&#xA;$$&#xD;&#xA;ds^2 = \eta_{ij} dx^i dx^j = -dt^2 + dx^2 + dy^2 + dz^2&#xD;&#xA;$$&#xA;Hence the minus sign is not a property of the one-forms $dx^i$ but of the coefficients $\eta_{ij}$&lt;/p&gt;&#xA;" OwnerUserId="3131" LastActivityDate="2011-06-23T09:24:46.120" CommentCount="3" />
  <row Id="11487" PostTypeId="1" AcceptedAnswerId="11498" CreationDate="2011-06-23T09:31:39.813" Score="4" ViewCount="102" Body="&lt;p&gt;According to a book I'm reading, the Fourier transform is widely used in QM. That came as a huge surprise to me. (Unfortunately, the book doesn't go on to give any simple examples of how it's used!) So can someone provide one please?&lt;/p&gt;&#xA;" OwnerUserId="993" LastActivityDate="2011-06-23T15:31:53.597" Title="Very simple example of the way the Fourier transform is used in quantum mechanics?" Tags="&lt;fourier-transform&gt;" AnswerCount="3" />
  <row Id="11489" PostTypeId="2" ParentId="11487" CreationDate="2011-06-23T09:47:48.243" Score="2" ViewCount="" Body="&lt;p&gt;In quantum mechanics, the momentum and position wave functions are Fourier transform pairs, to within a factor of Planck's constant. With this constant properly taken into account, the inequality above becomes the statement of the Heisenberg uncertainty principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find more here:&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_uncertainty_principle#Uncertainty_principle&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fourier_uncertainty_principle#Uncertainty_principle&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="171" LastActivityDate="2011-06-23T09:47:48.243" />
  <row Id="11490" PostTypeId="2" ParentId="11487" CreationDate="2011-06-23T10:39:53.583" Score="2" ViewCount="" Body="&lt;p&gt;Apart from its physical meaning (canonically conjugated observables (energy and time, position and momentum...) are fourier transforms of one another), the fourier transform is an important tool in &lt;em&gt;any&lt;/em&gt; field that deals with linear differential equations. In quantum mechanics, everything is linear, so it is quite often useful to perform a fourier transform to make solving the equations easier.&lt;/p&gt;&#xA;" OwnerUserId="3540" LastActivityDate="2011-06-23T10:39:53.583" />
  <row Id="11491" PostTypeId="2" ParentId="11481" CreationDate="2011-06-23T11:35:33.007" Score="1" ViewCount="" Body="&lt;p&gt;Although one can think of the metric as about infinitesimal distances, it can also be thought of in a dual way, as about differential equations. Thus, with your sign convention for the metric, the wave equation is&#xA;$$g^{ij}\frac{\partial^2\phi(x)}{\partial x^i\partial x^j}=-\frac{\partial^2\phi(x)}{\partial t\partial t}+\frac{\partial^2\phi(x)}{\partial x\partial x}+\frac{\partial^2\phi(x)}{\partial y\partial y}+\frac{\partial^2\phi(x)}{\partial z\partial z}=0.$$&#xA;In this POV, the metric determines the differential equations that are satisfied by the fields that are introduced in the space-time. The Laplace equation has very different properties than does the wave equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are formalisms in which d$t$ is replaced by id$t$, so that the metric can be uniformly positive. However most people keep this as a mathematical trick up one's sleeve, not as a fundamental part of the construction of a theory. There was a period in the development of GR when imaginary time was much more commonly used than it is now.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-06-23T11:35:33.007" />
  <row Id="11492" PostTypeId="1" CreationDate="2011-06-23T12:08:35.393" Score="0" ViewCount="57" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/759/does-quantum-physics-really-suggests-this-universe-as-a-computer-simulation&quot;&gt;Does Quantum Physics really suggests this universe as a computer simulation?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Might our universe described by &quot;quantum gravity&quot; really be a finite but superastronomical quantum computer simulation which is barely inconsistent? The only way to describe the universe completely is to step outside the universe. Being barely inconsistent, we sentient beings can't find out the inconsistency for a very very long time, maybe ages, but as the looming inconsistency approaches, the computer simulation is stopped, and end information about the universe is saved and read to be used by another computer simulation at a higher level, and this process goes on ad infinitum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Might our world really be an infinite hierarchy of computer simulations?&lt;/p&gt;&#xA;" OwnerUserId="4152" LastActivityDate="2011-06-23T12:08:35.393" ClosedDate="2011-06-23T13:06:05.983" Title="Might quantum gravity be a barely inconsistent theory running as a quantum computer simulation?" Tags="&lt;quantum-gravity&gt;" CommentCount="2" />
  <row Id="11493" PostTypeId="2" ParentId="11461" CreationDate="2011-06-23T12:15:01.860" Score="0" ViewCount="" Body="&lt;p&gt;If I understand you correctly, you want to desribe sound waves from the viewpoint of a moving observer. To do this, you just have to take the standard wave equation and perform a variable substitution $\mathbf{x}\mapsto \mathbf{y}:=\mathbf{x}-\mathbf{v}\cdot t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You start with&#xA;$$&#xD;&#xA;  \partial_t^2 p(\mathbf{x},t) = c^2\nabla^2 p(\mathbf{x},t)&#xD;&#xA;  = c^2\sum_i\partial_{\!x_i}^2 p(x_1,\ldots,t)&#xD;&#xA;$$&#xA;The right side transforms as&#xA;$$&#xD;&#xA;  c^2\sum_i\partial_{\!x_i}^2 p(x_1,\ldots,t)&#xD;&#xA;  = c^2\sum_i\partial_{\!x_i}^2 p(y_1+v_1t,\ldots,t)&#xD;&#xA;$$&#xA;where we yet have to replace the derivative WRT $x_i$ against the derivative WRT $y_i$, but they are the same:&#xA;$$&#xD;&#xA;  \frac{\partial}{\partial x_i} = \frac{\partial y_i}{\partial x_i}\frac{\partial}{\partial y_i} = (1+0)\cdot\frac{\partial}{\partial y_i};&#xD;&#xA;$$&#xA;so&#xA;$$&#xD;&#xA;  c^2\nabla_{\!\!\mathbf{x}}^2p(\mathbf{x},t) = c^2\nabla_{\!\!\mathbf{y}}^2p(\mathbf{y}+\mathbf{v}t,\ t).&#xD;&#xA;$$&#xA;The left side transforms as&#xA;$$&#xD;&#xA;  \partial_t^2 p(\mathbf{x},t) = \partial_t^2 p(\mathbf{y}+\mathbf{v}\cdot t,\ t).&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, we would like to drop the $+\mathbf{v}t$, but before we can do this we need to take into account the derivative of this term:&#xA;$$&#xD;&#xA;  \partial_t p(\mathbf{y}+\mathbf{v}\cdot t,\ t)&#xD;&#xA;  = \left.\bigl(\partial_t p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t}&#xD;&#xA;              + \mathbf{v}\cdot\nabla_{\!\!\mathbf{y}}p(\mathbf{y}+\mathbf{v}\cdot t,\ t)&#xD;&#xA;$$&#xA;now the second derivative:&#xA;$$&#xD;&#xA;  \partial_t\left.\bigl(\partial_t p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t}&#xD;&#xA;  = \left.\bigl(\partial_t\partial_\tau p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t}&#xD;&#xA;  + \left.\bigl(\partial_t^2 p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t}&#xD;&#xA;$$&#xA;where&#xA;$$&#xD;&#xA;  \left.\bigl(\partial_t\partial_\tau p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t}&#xD;&#xA;   = \left.\bigl(\partial_t\,\mathbf{v}\!\cdot\!\nabla_{\!\!\mathbf{y}}p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t}&#xD;&#xA;$$&#xA;and&#xA;$$&#xD;&#xA;  \partial_t\,\mathbf{v}\!\cdot\!\nabla_{\!\!\mathbf{y}}p(\mathbf{y}+\mathbf{v}\cdot t,\ t)&#xD;&#xA;  = \mathbf{v}\!\cdot\!\nabla_{\!\!\mathbf{y}}&#xD;&#xA;    \left(\left.\bigl(\partial_t p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t}&#xD;&#xA;              + \mathbf{v}\cdot\nabla_{\!\!\mathbf{y}}p(\mathbf{y}+\mathbf{v}\cdot t,\ t)\right)&#xD;&#xA;$$&#xA;$$&#xD;&#xA;  = \left.\bigl(\partial_t\,\mathbf{v}\!\cdot\!\nabla_{\!\!\mathbf{y}}p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t}&#xD;&#xA;  + \mathbf{v}\cdot\nabla_{\!\!\mathbf{y}}^2\,\mathbf{v}\,p(\mathbf{y}+\mathbf{v}\cdot t,\ t)&#xD;&#xA;$$&#xA;All together, we get&#xA;$$&#xD;&#xA;  \partial_t^2p(\mathbf{x},t)&#xD;&#xA;  = \left.\left(&#xD;&#xA;    \partial_t^2 p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)&#xD;&#xA;   + 2 \partial_t\,\mathbf{v}\!\cdot\!\nabla_{\!\!\mathbf{y}}p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)&#xD;&#xA;   + \mathbf{v}\cdot\nabla_{\!\!\mathbf{y}}^2\,\mathbf{v}\,p(\mathbf{y}+\mathbf{v}\cdot t,\ t)&#xD;&#xA;  \right)\right|_{\tau=t}.&#xD;&#xA;$$&#xA;&lt;em&gt;Now&lt;/em&gt; we may drop the $+\mathbf{v}t$ on both sides of the equation, and get&#xA;$$&#xD;&#xA;  \partial_t^2 p(\mathbf{y},t)&#xD;&#xA;   + 2 \partial_t\,\mathbf{v}\!\cdot\!\nabla p(\mathbf{y},t)&#xD;&#xA;   + \mathbf{v}\cdot\nabla^2\,\mathbf{v}\,p(\mathbf{y},t)&#xD;&#xA;  = c^2\nabla^2p(\mathbf{y},t)&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="3540" LastActivityDate="2011-06-23T12:15:01.860" CommentCount="3" />
  <row Id="11495" PostTypeId="2" ParentId="11295" CreationDate="2011-06-23T13:53:11.527" Score="3" ViewCount="" Body="&lt;p&gt;Ha, by opening a huge bounty (which I promise I will concede if there is at least an answer beyond mine) I have lost all my privileges, including to do comments (EDIT: I recovered most, thanks everyone :-D Now please give a nice pretty answer so I will be able to pay the bounty happyly) So take this as a note of extra information:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First, &lt;a href=&quot;http://dftuz.unizar.es/~rivero/research&quot; rel=&quot;nofollow&quot;&gt;I am&lt;/a&gt; not Mitchell, of course :-D  I thank him the interest on the issue.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-ph/0512065&quot; rel=&quot;nofollow&quot;&gt;arxiv:hep-ph/0512065&lt;/a&gt;, named in the question. is only a piece of a saga, the &quot;&lt;a href=&quot;http://dorigo.wordpress.com/2007/10/16/guest-post-alejandro-rivero-sbootstrap/&quot; rel=&quot;nofollow&quot;&gt;sBootstrap&lt;/a&gt;&quot;, with different ideas in each preprint, not everything at the same &quot;confidence degree&quot;&lt;/li&gt;&#xA;&lt;li&gt;Besides the named preprint, see &lt;a href=&quot;http://arxiv.org/abs/0910.4793&quot; rel=&quot;nofollow&quot;&gt;arXiv:0910.4793&lt;/a&gt; &lt;a href=&quot;http://arxiv.org/abs/0710.1526&quot; rel=&quot;nofollow&quot;&gt;arXiv:0710.1526&lt;/a&gt;, as well as the PF threads mentioned by Mitchell, and even a brief note at &lt;a href=&quot;http://vixra.org/abs/1102.0034&quot; rel=&quot;nofollow&quot;&gt;Vixra&lt;/a&gt; :-)&lt;/li&gt;&#xA;&lt;li&gt;It is not only that, as Mitchell says, &quot; one has enough quark-antiquark pairings &quot;. With three generations, one has exactly the needed number of pairings, not more not less. In some sense the idea predicts three generations, as it only works nicely with them.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;While the main argument is the algebraic match, I have collected too a lot of collateral evidence that seems to indicate that some connections do exist between electroweak and QCD mass scale. Some of them are well known (eg, why the yukawa of the top quark is equal to one? Why do the muon and the tau lurk in the QCD natural and chiral scales?) and others are surprising: the &lt;a href=&quot;http://dorigo.wordpress.com/2006/09/14/a-mistery-behind-the-z-width/&quot; rel=&quot;nofollow&quot;&gt;equality between the decay width of pions and Z0 particles&lt;/a&gt;, when scaled via the cube of the mass (&lt;a href=&quot;http://arxiv.org/abs/hep-ph/0603145&quot; rel=&quot;nofollow&quot;&gt;arXiv:hep-ph/0603145&lt;/a&gt;,&lt;a href=&quot;http://arxiv.org/abs/hep-ph/0507144&quot; rel=&quot;nofollow&quot;&gt;arXiv:hep-ph/0507144&lt;/a&gt;). Or the coincidende that the basic mass scale of Koide formula is 313 MeV, equal to the constituient mass of QCD.&lt;/p&gt;&#xA;" OwnerUserId="1335" LastEditorUserId="1335" LastEditorDisplayName="" LastEditDate="2011-06-24T23:33:39.410" LastActivityDate="2011-06-24T23:33:39.410" CommentCount="4" />
  <row Id="11496" PostTypeId="2" ParentId="11477" CreationDate="2011-06-23T14:04:46.683" Score="7" ViewCount="" Body="&lt;p&gt;They don't make any claim in the paper about interpretations of quantum theory, either for the Copenhagen interpretation or against many-worlds interpretations. Nor does the Phys. Rev. Lett. 101, 20403 (2008) that they cite as their principal theoretical source. The Vienna group's stated intention here, as, I think, in a number of papers over the last few years, has been to try to rule out &lt;em&gt;contextual&lt;/em&gt; classical particle models for experiments. This is to me something of a straw man, but they have been hacking away at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was going to ask that you expand your Question to say why you think this experiment supports the Copenhagen interpretation over other interpretations, because I do not see that to be the case, but I finally saw the off the cuff remark to that effect from Zeilinger at the very end of the New Scientist article [&lt;em&gt;you should have cited this&lt;/em&gt;]. That's definitely not enough to rule out many-worlds interpretations without a much more substantial argument. There isn't, so far, enough of an argument to have loopholes in it.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-06-23T14:04:46.683" CommentCount="5" />
  <row Id="11497" PostTypeId="1" AcceptedAnswerId="11502" CreationDate="2011-06-23T15:20:56.887" Score="2" ViewCount="187" Body="&lt;p&gt;I'm confused how&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{\mathbf{r}}_{j}=\sum_{k}\frac{\partial\mathbf{r}_{j}}{\partial q_{k}}\dot{q}_k+\frac{\partial\mathbf{r}_{j}}{\partial t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;leads to the relation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial\dot{\mathbf{r}}_{i}}{\partial\dot{q}_{j}}=\frac{\partial\mathbf{r}_{i}}{\partial q_{j}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://farside.ph.utexas.edu/teaching/336k/lectures/node77.html&quot; rel=&quot;nofollow&quot;&gt;Sources suggest&lt;/a&gt; that while differentiating the first equation the generalized velocity and generalized coordinates are considered independent of each other i.e, $(\dot{q_1},\dot{q_2},...,\dot{q_n})$ is independent of $(q_1,q_2,...,q_n)$. I don't understand how they are independent. &lt;/p&gt;&#xA;" OwnerUserId="966" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-06-23T17:00:01.870" LastActivityDate="2011-06-25T21:54:52.947" Title="Generalized Coordinates" Tags="&lt;classical-mechanics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="11498" PostTypeId="2" ParentId="11487" CreationDate="2011-06-23T15:31:53.597" Score="1" ViewCount="" Body="&lt;p&gt;Given that &lt;a href=&quot;http://physics.stackexchange.com/questions/11487/very-simple-example-of-the-way-the-fourier-transform-is-used-in-quantum-mechanics/11490#11490&quot;&gt;leftaroundabout&lt;/a&gt; and &lt;a href=&quot;http://physics.stackexchange.com/questions/11487/very-simple-example-of-the-way-the-fourier-transform-is-used-in-quantum-mechanics/11489#11489&quot;&gt;vonjd&lt;/a&gt; have addressed the fundamental place of the Fourier transform in the formalism, let me talk a little about an experimental application.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;What is the shape and size of a atomic nucleus?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;From Rutherford we learned that the nucleus is rather a lot smaller than the atom as a whole. Now, electron microscopy can just about provide vague picture of a medium or large atom as a out-of-focus ball, but there is no hope of employing that technique to something orders of magnitude smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we do is scatter things off of the component parts of the nucleus. A nice reaction here is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ e + A \to e + p + B $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ represents that target nucleus and $B$ the remnant after we bounce a proton out. (This is what nuclear physicists call &quot;quasi-elastic scattering&quot;.) Now, if (1) we are shooting a beam of electrons as a stationary target, (2) we have a precision measurement of the momenta of the incidence and scattered electrons and the ejected proton, (3) we are willing to neglect excitation energy of the remnant nucleus, and (4) we assume that $p$ mostly did not interact with the remnant &lt;em&gt;after&lt;/em&gt; being scattered, we know the momentum of the proton &lt;em&gt;inside&lt;/em&gt; the nucleus at the time it was struck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Collect enough statistics on this and we have sampled the proton momentum distribution of the nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here's the fun part: you can show that the &lt;em&gt;spacial&lt;/em&gt; distribution of protons in the nucleus is the Fourier transform of the &lt;em&gt;momentum&lt;/em&gt; distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And bingo, &lt;em&gt;a measurement of the size of the nucleus&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do it with a polarized target and you can get info on the shape as well.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-06-23T15:31:53.597" />
  <row Id="11499" PostTypeId="2" ParentId="11497" CreationDate="2011-06-23T16:11:18.017" Score="1" ViewCount="" Body="&lt;p&gt;If $q$ is a known function of time, then $\dot q$ is not independent. But before writing and solving the equations (before finding $q(t)$, you may have unknown and independent variables (like $x$ and $y$), which is suggested by your source.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-06-23T16:11:18.017" CommentCount="0" />
  <row Id="11500" PostTypeId="1" AcceptedAnswerId="11504" CreationDate="2011-06-23T17:06:19.927" Score="6" ViewCount="105" Body="&lt;p&gt;I'm trying to understand the units in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$mx&amp;#39;&amp;#39;+kx=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the general sol: $x(t)=A \cos(w_0 t)+B \sin(w_0 t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We let $w_0 =\sqrt{\frac{k}{m}}$ the units for the spring constant k are  kg*meters/sec^2&#xA;or N/m, m is kg, so the units of w0 seem to be &quot;per seconds&quot; that is 1/sec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, later we put w0 in to the cos and sin functions which will return dimensionless ratios. The constants A,B must be in meters, since x is in meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is why my book says w0 has units radians/sec, I get that the input for cosine is radians or some other angle measure, but, where did the radians come from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My analysis of units only found 1/sec as the units!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have just been informed that radians are unitless. So that answers part of this question, yet I still don't know why we can't say w0 is in degrees or rotations... How do I know what kind of cosine and sine table to use with this unitless number?&lt;/p&gt;&#xA;" OwnerUserId="4158" LastEditorUserId="4158" LastEditorDisplayName="" LastEditDate="2011-06-23T17:26:40.957" LastActivityDate="2011-06-23T18:33:27.747" Title="Simple harmonic motion what are the units for $\omega_0$?" Tags="&lt;classical-mechanics&gt;&lt;units&gt;" AnswerCount="4" />
  <row Id="11501" PostTypeId="2" ParentId="10635" CreationDate="2011-06-23T17:11:52.497" Score="1" ViewCount="" Body="&lt;p&gt;You might also try working through the math-physics relationship from the other end: start with a math book. Larson and Stewart's Calculus textbooks are some pretty general books that cover basic calculus--starting from algebraic/geometric principles--to elementary multi-variable calculus and differential equations, with some other cool stuff thrown in between. Tons of problems/answers/diagrams/examples, and generally some very good explanations, as long as you're willing to think about the ideas. You may even find these books too easy--I don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although these books will not necessarily give you much about the physics concepts that you are interested in, I mention this entryway into studying physics because:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The understanding of many mathematical concepts, especially those involving higher-level calculus and linear algebra, is so very helpful to a thorough understanding of physics. However, I am not talking about pure number-crunching or calculation power--but rather, an understanding of the relationships in orders of magnitude, or the nature of functions' behavior over time (differential equations), or the effects of dimensions and units. The concept of simple harmonic motion--in an oscillating spring, for example--grows so much clearer when you have a good mathematical conceptual understanding of sinusoidal functions, even without any numbers or physical parameters to plug in. And that is one of many examples.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The books that I mentioned, along with most general college math textbooks, have some great example sections with applications in physics. With these problems as a springboard, perhaps you will be able to use your conceptual knowledge of a physical phenomenon to understand a similar idea in math--and vice versa.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;One more suggestion--One, Two, Three...Infinity by George Gamow has some great explanations of all sorts of physical phenomena, with some good math to back them up (though the math in this book is often more conceptual than technical). It is one of my favorite books of all-time.&lt;/p&gt;&#xA;" OwnerUserId="4159" LastActivityDate="2011-06-23T17:11:52.497" />
  <row Id="11502" PostTypeId="2" ParentId="11497" CreationDate="2011-06-23T17:12:12.387" Score="8" ViewCount="" Body="&lt;p&gt;Starting with the following definition of $\mathbf{r}_j$ in terms of generalized coordinates&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{r}_{j}=\mathbf{r}_{j}(q_1, q_2, \ldots, q_n,t) \; ,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is clear that taking the total derivative with respect to time, we arrive at the following relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{\mathbf{r}}_{j}=\sum_{k}\frac{\partial\mathbf{r}_{j}}{\partial q_{k}}\dot{q}_k+\frac{\partial\mathbf{r}_{j}}{\partial t} \; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But because of the appearance of $\dot{q}_k$, $\dot{\mathbf{r}}_{j}$ also depends on them. Therefore, it makes sense to take partial derivatives of $\dot{\mathbf{r}}_{j}$ with respect to the $\dot{q}_k$ and from the relation above, we indeed get the relation you find.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While $\dot{q}_k$ is the time derivative of $q_k$, it can not be expressed as a function of the $q_j$. The derivative is an operator, not a function from tuples of real numbers to real numbers so that does not count. Physically, you could say the difference is in the fact that &lt;strong&gt;a real function on n-tuples only involves information of what happens in that point at that specific instant. A derivative however involves information about what happens at a point at an instant but also at a different point an infinitesimal instant before that&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or another way of seeing this is that &lt;strong&gt;the state of a classical system is completely specified when one gives the positions and velocities at one instant&lt;/strong&gt;. If only positions where sufficient, the equations of motion would be first order equations and the time derivative would really be dependent on the positions at one instant, because it would be fully determined by them. This is not the case with the classical second order equations we have for most classical systems of point particles.&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2011-06-23T17:12:12.387" CommentCount="2" />
  <row Id="11503" PostTypeId="2" ParentId="11500" CreationDate="2011-06-23T17:21:54.937" Score="5" ViewCount="" Body="&lt;p&gt;Radians are kind of a funny unit from the dimensional analysis perspective: radians are dimensionless. That means that rad/s and 1/s are equivalent from the point of view of dimensional analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to think about this is that angular measures in radians are really just ratios of like quantities: $\theta$ in radians is, by definition, the ratio of the length of a circular arc subtending $\theta$ to the radius of the circle. So a radian is really a meter per meter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, when doing dimensional analysis in physics, this means that you can slip radians into and out of your units with wild abandon. For instance, if a circle of radius $r$ is rotating at angular speed $\omega$, then the speed of a point on the rim is &#xA;$$&#xD;&#xA;v=r\omega.&#xD;&#xA;$$&#xA;The right side of this expression has units of m rad/s, and the left side has units of m/s. But the units balance, because a radian (or a m/m if you prefer) is dimensionless.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-06-23T17:21:54.937" CommentCount="2" />
  <row Id="11504" PostTypeId="2" ParentId="11500" CreationDate="2011-06-23T17:22:18.963" Score="4" ViewCount="" Body="&lt;p&gt;Ah, good question. The radian is actually a &quot;fake unit.&quot; What I mean by that is that the radian is &lt;a href=&quot;http://en.wikipedia.org/wiki/Radian&quot; rel=&quot;nofollow&quot;&gt;defined&lt;/a&gt; as the ratio of distance around a circle (arclength) to the radius of a circle - in other words, it's a ratio of one distance to another distance. For an angle of one radian specifically, the arclength $s$ is equal to the radius $r$, so you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1\text{ rad} = \frac{s}{r} = \frac{r}{r} = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The units of distance (meters or whatever) cancel out, and it turns out that &quot;radian&quot; is just a fancy name for 1!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, this also implies that &quot;degree&quot; is just a fancy name for the number $\frac{\pi}{180}$, and &quot;rotation&quot; is just a fancy name for the number $2\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This actually addresses the edit to your question. Suppose that you had some object oscillating at $\omega = \pi/4\frac{\mathrm{rad}}{\mathrm{s}} = 0.785\frac{\mathrm{rad}}{\mathrm{s}}$, and you wanted to evaluate its position after 10 seconds. To get the cosine term, you would plug the numbers in, getting&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos\bigl(0.785\tfrac{\mathrm{rad}}{\mathrm{s}}\times 10\mathrm{s}\bigr) = \cos(7.85\text{ rad}) = \cos(7.85)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then you would go to a trig table in radians (or your calculator in radian mode) and look up 7.85.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, suppose that you were measuring $\omega_0$ in degrees per second instead of radians per second. You would instead have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos(45^\circ/\mathrm{s}\times 10\mathrm{s}) = \cos(450^\circ)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you go look this up in a trig table given in degrees, you will get the same answer as $\cos(7.85)$. Why? Well, remember that the unit &quot;degree&quot; is just code for $\pi/180$, so this is actually equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\cos\bigl(450\times\tfrac{\pi}{180}\bigr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $450\times\frac{\pi}{180} = 7.85$, which is just $450^\circ$ converted to radians. So now you have the same value in the cosine, $\cos(7.85)$. Trig tables listed in degrees already have this extra factor of $\frac{\pi}{180}$ built into them as a convenience for you; basically, if you look up any number $\theta$ in a table that uses degrees, what you get is actually the cosine (or sine, or whatever) of $\theta\times\frac{\pi}{180}$.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-23T18:33:27.747" LastActivityDate="2011-06-23T18:33:27.747" />
  <row Id="11505" PostTypeId="2" ParentId="11500" CreationDate="2011-06-23T17:35:59.683" Score="0" ViewCount="" Body="&lt;p&gt;$\cos{(w_0t)}$ will be a solution of the given equation with $w_0=\sqrt{\frac{k}{m}}$ only if $\frac{d}{dx}\cos{x}=-\sin(x)$, which it will only be if $x$ is in units radians. If $x$ is in units degrees or grads there will be a dimensionless conversion constant, $\frac{d}{dx}\cos{x}=-C\sin(x)$. The convention that we use radians is so geometrically natural that it's close to hard-wired for mathematicians and physicists. I think the two Answers that appeared just as I started to write this are wrong in principle to take 1 radian=1, but in practice taking the radians unit to be dimensionless is unlikely to get you into trouble. Nonetheless, there's a reason why this is an international standard. Nice Question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I'm not entirely happy with the above. It depends whether we &lt;em&gt;define&lt;/em&gt; $\cos{}$ to map an angle to a dimensionless ratio, or to map dimensionless numbers to a dimensionless ratio. The second definition would make the earlier Answers right and me wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT(2): Rather than writing $\sin{(60^o)}$ or $\sin{(\frac{60\pi}{180}^r)}$, we might write $\sin^{[o]}(60)$ or $\sin^{[r]}(\frac{60\pi}{180})$. When we use a table of sines, we check to see whether it's the sine-of-degrees function or the sine-of-radians function, then we give it the number $60$ or the number $\frac{60\pi}{180}$ as appropriate. There are different sine functions, which are related by linear transformations of their arguments. If we include translations, then $\cos$ becomes also a different sine function.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-06-23T18:21:16.573" LastActivityDate="2011-06-23T18:21:16.573" CommentCount="3" />
  <row Id="11506" PostTypeId="1" AcceptedAnswerId="11509" CreationDate="2011-06-23T17:55:05.177" Score="4" ViewCount="96" Body="&lt;p&gt;As the question in the title states I am wondering what material can be effectively used to shield gamma rays apart from lead? I believe concrete is often used, but it is nowhere near as effective as lead (6 cm to match 1 cm of lead as I understand it). I also hear significant bodies of water helps, as does tightly packed dirt, but surely there must be other materials that shield nearly as effectively as lead?&lt;/p&gt;&#xA;" OwnerUserId="4145" LastActivityDate="2011-06-23T19:35:01.580" Title="What other shielding material than lead is effective against gamma rays?" Tags="&lt;electromagnetic-radiation&gt;&lt;radiation&gt;&lt;radioactivity&gt;" AnswerCount="1" />
  <row Id="11507" PostTypeId="2" ParentId="11500" CreationDate="2011-06-23T18:00:39.323" Score="0" ViewCount="" Body="&lt;p&gt;The importance of radians is &quot;not&quot; in its dimensions. It is necessary to mention whether $\omega_0$ is in radian/sec (and not just 1/sec) because angles can be expressed in various units - degrees, radians, etc (all which are dimensionless) and unless we know in which unit $\omega_0$ is expressed, we cannot do mathematical operations on it, like finding its sine or cosine, etc.&lt;br&gt;&#xA;For example: $sin(60^o)$, where 60 is in degrees, is totally different from $sin(60^r)$ where 60 is in radians. So, it is &quot;very&quot; important to always mention $\omega_0$ in radians/sec rather than just 1/sec.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is the reason why $\omega_0$ is called as angular frequency and not just frequency, although both have the same dimensions.)&lt;/p&gt;&#xA;" OwnerUserId="1494" LastActivityDate="2011-06-23T18:00:39.323" />
  <row Id="11509" PostTypeId="2" ParentId="11506" CreationDate="2011-06-23T18:12:15.067" Score="5" ViewCount="" Body="&lt;p&gt;There is nothing magical about lead for this purpose. The driving factor is the number of electrons per unit volume, which reduces (to a first approximation) to the mass density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You get very good (better than lead) shielding performance from gold, tungsten, mercury, etc; and quite reasonable performance from iron or copper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Question for the student:&lt;/em&gt; why is lead such a common choice for this application?&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-06-23T19:35:01.580" LastActivityDate="2011-06-23T19:35:01.580" CommentCount="4" />
  <row Id="11510" PostTypeId="2" ParentId="11423" CreationDate="2011-06-23T18:12:34.473" Score="0" ViewCount="" Body="&lt;p&gt;The decay rate from a certain state depends on the environment.  If the atom is in an optical cavity which is strongly detuned from all of the decay modes of the atom, it will stay in the excited state much longer.   Because &lt;em&gt;how&lt;/em&gt; much longer it remains excited&#xA; depends on the properties of the cavity, I don't think there's a hard &quot;states with $n&amp;gt;something$ cannot be distinguished from other states&quot; limit, at least in non-relativistic QM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Density of states factors pop up in statistical physics, so you can still tell there's a bunch of states there even if you can't distinguish them.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2011-06-23T18:12:34.473" />
  <row Id="11511" PostTypeId="2" ParentId="11475" CreationDate="2011-06-23T18:16:08.160" Score="3" ViewCount="" Body="&lt;p&gt;cMinor.  You are good to go on the first problem.  Just crank through the math.  You should get an answer just over 0.5 (m/(s^2)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the train problem, find the path length of the quarter circle, then divide it by 30 seconds to get train speed.  You will then have all you need to do the (v^2)/r thing for that problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Angular velocity does not need to be found to solve either of these problems.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-06-23T18:16:08.160" CommentCount="1" />
  <row Id="11512" PostTypeId="1" CreationDate="2011-06-23T18:26:47.687" Score="4" ViewCount="61" Body="&lt;p&gt;The Nobel Prize in Physics in 1980 was awarded to James Cronin and Val Fitch for their discovery of a violation of Charge and Parity Invariance, in which the neutral kaon's quarks change into their corresponding antiquarks, and vice-versa. However, the vice-versa does not occur with the same probability and shows a reason for the matter-antimatter imbalance in the universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every article nowadays in magazines and websites seems to say that there has been no observed asymmetry between matter and antimatter, except there was already a Nobel Prize awarded for one such observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this violation of CP invariance not sufficient enough for certain researchers and articles to describe the dominance of matter over antimatter? Wouldn't this imbalance in the nature of quarks and antiquarks be extremely important and consequential in the moments after the Big Bang, especially in the moments where the universe was a quark-gluon plasma?&lt;/p&gt;&#xA;" OwnerUserId="4162" LastActivityDate="2011-06-23T21:33:11.227" Title="Why is the decay of neutral kaons (violates CP invariance) seemingly not sufficient enough for certain people to describe matter-antimatter imbalance?" Tags="&lt;antimatter&gt;&lt;cp-violation&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="11513" PostTypeId="2" ParentId="11470" CreationDate="2011-06-23T18:36:54.297" Score="2" ViewCount="" Body="&lt;p&gt;The human body produces a wide range of bioelectromagnetic signals from various electrical impulses in the brain. The origin of the magnetic field is the charge exchange in the muscular and neural tissues (i.e no magnetic material is usually present in the body with very rare exceptions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The brain's magnetic field varies from 10s of fT to 100s of fT [1]. The frequency varies from 0.1 Hz to &amp;lt;100 Hz. Measurement of brain magnetism is limited by the complexity of signal due to the overlap of the signals from various parts of the brain. In magentoencephelograpahy this is done using SQUIDS [2].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part of your question is about magnetic excitement of brain functions. This is in fact possible and some testing/use is reported with limited clinical success. The strength of the applied magnetic fields are 1-5T for trans-cranial excitation [3].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://www.bem.fi/book/12/12.htm#02&quot; rel=&quot;nofollow&quot;&gt;http://www.bem.fi/book/12/12.htm#02&lt;/a&gt;&#xA;[2] &lt;a href=&quot;http://www.scholarpedia.org/article/Magnetoencephalogram&quot; rel=&quot;nofollow&quot;&gt;http://www.scholarpedia.org/article/Magnetoencephalogram&lt;/a&gt;&#xA;[3] &lt;a href=&quot;http://en.wikipedia.org/wiki/Transcranial_magnetic_stimulation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Transcranial_magnetic_stimulation&lt;/a&gt;&#xA;[4] &lt;a href=&quot;http://www.bem.fi/book/22/22.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.bem.fi/book/22/22.htm&lt;/a&gt;&#xA;An excellent source for this topic is [1]&lt;/p&gt;&#xA;" OwnerUserId="3714" LastActivityDate="2011-06-23T18:36:54.297" CommentCount="2" />
  <row Id="11514" PostTypeId="1" AcceptedAnswerId="11522" CreationDate="2011-06-23T19:13:35.997" Score="2" ViewCount="72" Body="&lt;p&gt;I am a math grad student attempting to read Bernard Whiting's paper &quot;Mode Stability of the Kerr Black Hole.&quot; If you are in a university network, the paper should be easily found by a google search. At the beginning of the paper Whiting refers to the &quot;master equation of Teukolsky&quot; from the paper &quot;Rotating Black Holes: Separable Wave Equations for Gravitational and Electromagnetic Perturbations.&quot; However, the equation Whiting writes down appears (to me) to be different than Teukolsky's. I assume I am missing something silly and was hoping that someone else has read this paper and can easily point out my mistake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One needs to look at the paper for all the relevant definitions, but for ease of reference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whiting writes down&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{\partial_r\Delta\partial_r - (1/\Delta)\{(r^2+a^2)\partial_t + a\partial_{\phi} - (r-M)s\}^2 - 4s(r+ia\cos\theta)\partial_t + $&#xA;$\partial_{\cos\theta}\sin^2\theta\partial_{\cos\theta} + (1/\sin^2\theta)\{a\sin^2\theta\partial_t + \partial_{\phi} + i\cos\theta\cdot s\}^2\}\psi = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Delta = r^2 - 2Mr + a^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while Teukolsky writes down&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{(\frac{(r^2+a^2)^2}{\Delta} - a^2\sin^2\theta)\partial^2_t + \frac{4Mar}{\Delta}\partial^2_{t,\phi} + (\frac{a^2}{\Delta} - \frac{1}{\sin^2\theta})\partial^2_{\phi} - \Delta^{-s}\partial_r(\Delta^{s+1}\partial_r) - $&#xA;$\frac{1}{\sin\theta} \partial_{\theta}(\sin\theta\partial_{\theta}) - 2s(\frac{a(r-M)}{\Delta} + \frac{i\cos\theta}{\sin^2\theta})\partial_{\phi} - 2s(\frac{M(r^2-a^2)}{\Delta}-r-ia\cos\theta)\partial_t$&#xA;$ + (s^2\cot^2\theta - s)\}\psi = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I expanded out Whiting's equation out I did not get Teukolsky's equation. For example, it is clear that the $\partial_r$ terms are different. I also got that the constant terms (no derivative terms) did not agree.&lt;/p&gt;&#xA;" OwnerUserId="4163" LastActivityDate="2011-06-24T02:25:11.700" Title="Notational(?) Question in Whiting's Paper &quot;Mode Stability of the Kerr Black Hole&quot;" Tags="&lt;general-relativity&gt;&lt;black-holes&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="11515" PostTypeId="1" CreationDate="2011-06-23T19:57:00.233" Score="1" ViewCount="91" Body="&lt;p&gt;This is a follow-up my &lt;a href=&quot;http://physics.stackexchange.com/questions/11449/conversion-of-mass-to-energy-in-chemical-nuclear-reactions&quot;&gt;previous question&lt;/a&gt;. Several of the answers indicated that the mass of a particle (atom, proton, etc.) &lt;em&gt;increase&lt;/em&gt; with the &quot;binding energy&quot; of it's component particles - the energy needed to split them apart - which doesn't make sense to me. I'm convinced that, at least in an exothermic chemical reaction (where the product bond energies are larger) the product's particles will lose mass (proportionally to the heat dissipated) or at least have no change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use a larger-scale analogy, if an object, a &quot;particle&quot;, is 100m above the Earth's surface, it has potential energy from gravity. As it falls, this energy is lost, converted into KE. Overall, the two &quot;particles&quot;, the object and the Earth, end up with the total energy, and therefore the same total mass. There is no room for a &quot;binding energy&quot; that adds mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reasoning is that this extends to particles, with electrostatic or nuclear forces taking the place of gravity. Potential energy of component particles becomes KE of the bonded particle, they end up with the same mass. Actually, if this KE is dissipated (as in a burning / nuclear fusion reaction) the particles should actually have more mass in their uncombined/unreacted state, thanks to their PE. Surely it isn't possible for the mass to increase without an external input of energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, answerers of my &lt;a href=&quot;http://physics.stackexchange.com/questions/11449/conversion-of-mass-to-energy-in-chemical-nuclear-reactions&quot;&gt;energy in chemical reactions question&lt;/a&gt; said that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the energy involved in the bonds is ... half of what we normally consider the &quot;mass&quot; of the proton - David Zaslavsky&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;potential energy of the chemical bonds do correspond to an increase of mass - Ben Hocking&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, how can this be, and where is my reasoning incorrect? What exactly is the binding energy (if not just the energy needed to break the bond), and where does it come from?&lt;/p&gt;&#xA;" OwnerUserId="4140" LastActivityDate="2011-06-24T03:41:02.617" Title="The &quot;binding energy&quot; of bonded particles adds mass?" Tags="&lt;nuclear-physics&gt;&lt;energy-conservation&gt;&lt;physical-chemistry&gt;&lt;binding-energy&gt;" AnswerCount="3" CommentCount="6" />
  <row Id="11516" PostTypeId="2" ParentId="11515" CreationDate="2011-06-23T21:06:25.057" Score="5" ViewCount="" Body="&lt;p&gt;A bound system will indeed have a lower mass than it's constituent free parts.  The binding energy is usually understood to be the energy it would take to separate the bound system to free constituent parts. Then we would say $M_{tot} = M_1 + M_2 - E_{bind}/c^2$  Or we could by convention make the binding energy a negative number and say it &quot;increases&quot; mass by a negative amount.  I think people speaking a little loosely is what's causing the confusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we are talking about a nucleus, then it is impossible because of color confinement to separate the constituent parts (quarks) to independent free particles, so what we call the &quot;binding energy&quot; in this case is not as clear.  &lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-06-23T21:06:25.057" CommentCount="2" />
  <row Id="11517" PostTypeId="2" ParentId="11515" CreationDate="2011-06-23T21:21:49.240" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm convinced that, at least in an exothermic chemical reaction (where the product bond energies are larger) the product's particles will lose mass (proportionally to the heat dissipated) or at least have no change.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It would be helpful to have a more explicit reaction, like $A\Rightarrow B+C+energy$, because otherwise it's hard to define what kind of energy it is and then identify if it stays in the system or not.  In the example I present, it is true that $m_A&amp;gt;m_B+m_C$.  This is simple enough and I think everyone will agree on that.  Now, if you are considering a chemical reactor (where the mass change is small anyway), then energy manifested as kinetic energy of $B$ and $C$ will still be measured as a part of the weight of the total system until the thermal energy is dissipated out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A system that undergoes an exothermic reaction and then expels the released energy (through emission or conduction) will have a lower mass.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To use a larger-scale analogy, if an object, a &quot;particle&quot;, is 100m above the Earth's surface, it has potential energy from gravity. As it falls, this energy is lost, converted into KE. As it falls, this energy is lost, converted into KE. Overall, the two &quot;particles&quot;, the object and the Earth, end up with the total energy, and therefore the same total mass. There is no room for a &quot;binding energy&quot; that adds mass.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ah ha!  You have discovered an argument for photon emission from quantum mechanical state transitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your example seems nonsensical precisely because there is no obvious path for released energy to leave the system.  It is dissipated as thermal energy and the total mass of the system remains constant (although thermal energy contributes a negligible fraction anyway).  In the case of elementary particles energy must be conserved and there is no obvious analog for a force like friction.  In these cases where QM physics rules, transitions are allowed (nuclear excited states, electron orbital transitions) by the release of energy through photon emission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Binding energy is generally negative&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope it's obvious that the system of you + the Earth does, in fact, have binding energy.  This would account for a very small mass difference between the two states you discussed assuming everything else is equal (and everything else is clearly not equal).  But in general, new particles are most often formed through exothermic reactions (thermo 2nd law arguments apply here), and in all these reactions energy leaves the system, provided that you start with everything at rest and end with everything at rest and don't complicate it with things like thermal energy.  That means that the mass change will be negative.  Over time the universe likes to see these systems loose rest mass and dissipate that energy through high-entropy photon emission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means the Earth right now is lighter than all its constituents would be if they remained scattered throughout space.  This is the case for most things in the universe.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-06-23T21:21:49.240" />
  <row Id="11518" PostTypeId="2" ParentId="11512" CreationDate="2011-06-23T21:33:11.227" Score="4" ViewCount="" Body="&lt;p&gt;1) Does antimatter-matter symmetry exist? Yes there is a CP violation and the whole Nobel prize thing. On the other hand there is CPT symmetry which is very protected. So call it what you want. As for the popsci articles... I would express my thoughts, but this is a family site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)Does CP violation explain matter-antimatter imbalance? It's certainly necessary for this to happen. You also need to violate Baryon number (the number of quarks minus the number of antiquarks) and you need to be out of thermal equilibrium.  You can engineer all these things within the Standard Model plus inflationary cosmology, but I think you get an imbalance which is way too tiny. Like only one part in a trillion. So the whole thing is still an ongoing research problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll try to find you a decent reference article.&lt;/p&gt;&#xA;" OwnerUserId="3970" LastActivityDate="2011-06-23T21:33:11.227" CommentCount="1" />
  <row Id="11519" PostTypeId="1" CreationDate="2011-06-23T22:58:26.840" Score="0" ViewCount="116" Body="&lt;p&gt;I made a fusor once, like the easy science project deuterium deuterium ones, but they're really inefficient. I was wondering if it would be possible to make a small tokamak. Not one that will make a ton of power, I just want to make one for the sake of making one. Like a tokamak the size of a pizza box. What materials would I need and how would I go about assembling it? Assume it is deuterium-deuterium or p-B11 fusion. Maybe He3-He3.&lt;/p&gt;&#xA;" OwnerUserId="4144" LastActivityDate="2011-07-31T03:42:45.893" Title="how would you make a small tokamak?" Tags="&lt;energy&gt;&lt;nuclear-physics&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="11520" PostTypeId="1" CreationDate="2011-06-23T23:07:51.623" Score="2" ViewCount="57" Body="&lt;p&gt;I keep seeing documentaries describing the pile-up of nuclear waste from nuclear power stations in the US and in other countries. Countries such as Finland are working to deposit waste in old mines in geologically stable regions, and I was under the impression that the US was heading down that track as well, but this has now apparently been put on hold. A case of NIMBY I presume (not in my back yard).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The media story is that the waste is piling up, and is being stored in pools at nuclear power stations all over the US - which sounds dangerous considering the prevalence of natural disasters. I wonder what the real truth is? I am not quite sure what the situation is in France and England, but I got the impression they are experimenting with methods of re-use of the waste products as well as storing some material under ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the current state-of-the-art in terms of nuclear waste reuse? Are new reactors capable of burning discarded fuel from older reactors? Are there realistic expectations to be able to eliminate deposition of nuclear waste altogether in the foreseeable future?&lt;/p&gt;&#xA;" OwnerUserId="4145" LastEditorUserId="4145" LastEditorDisplayName="" LastEditDate="2011-06-24T15:41:51.340" LastActivityDate="2011-06-24T15:41:51.340" ClosedDate="2011-06-24T06:49:48.973" Title="How viable is reuse of nuclear waste at this point, and what are future prospects?" Tags="&lt;nuclear-engineering&gt;" CommentCount="7" />
  <row Id="11521" PostTypeId="1" CreationDate="2011-06-24T00:17:59.453" Score="2" ViewCount="101" Body="&lt;p&gt;I know $K(a,b,t)$ is the probability amplitude of find a particle that starts at point a in b in a time t later.  There is also an expression that sometimes is called green function:&lt;br&gt;&#xA;$$G(a,b,E)=(i/\hbar)\int_{-\infty}^\infty\;\exp(iEt/\hbar)\;K(a,b,t)\;dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or Fourier transform of Feynman propagator&lt;br&gt;&#xA;See: Grosche. Handbook of Feynman path integrals page 149.&#xA;     Keller. the feynman integral page 461.&lt;br&gt;&#xA;     &lt;a href=&quot;http://arxiv.org/abs/cond-mat/0304290v1&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/cond-mat/0304290v1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know if $G(a,b,E)$ could be the amplitude that a particle of energy $E$ at the initial point $a$ will appear at some (arbitrary) time at $b$. It seems that Martin Schaden and Larry Spruch use this interpretation in &lt;a href=&quot;http://arxiv.org/abs/cond-mat/0304290v1&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/cond-mat/0304290v1&lt;/a&gt; but I have not found this in any book of quantum mechanics.&lt;/p&gt;&#xA;" OwnerUserId="3956" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-06-24T16:47:50.733" LastActivityDate="2011-06-24T16:47:50.733" Title="What is the meaning of the Fourier transform of Feynman propagator?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-interpretations&gt;&lt;path-integral&gt;&lt;fourier-transform&gt;" AnswerCount="1" />
  <row Id="11522" PostTypeId="2" ParentId="11514" CreationDate="2011-06-24T02:25:11.700" Score="2" ViewCount="" Body="&lt;p&gt;Probably the definition of $\psi$ are slightly different? Judging by the fact that you get a different $\partial_r$ term and a different constant term, I am guess that Whiting's equation is Teukolsky's but commuted against $\Delta^l$ for some power $l$. (In other words, in Whiting's equation, replace $\psi \to \Delta^l\psi$ and divide the entire equation by $\Delta^l$, you should be able to find a $l$ (which depends on $s$, I think) that makes the two equation appear the same.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Observe that $\Delta^l$ commutes through the entire equation except for terms which have derivatives in $r$.)&lt;/p&gt;&#xA;" OwnerUserId="829" LastActivityDate="2011-06-24T02:25:11.700" CommentCount="2" />
  <row Id="11523" PostTypeId="2" ParentId="11521" CreationDate="2011-06-24T03:24:19.493" Score="3" ViewCount="" Body="&lt;p&gt;Yes, the Schaden and Spruch interpretation is correct. The interpretation is not used much because it's not as well connected into how experiments are run.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the usual text books, the Fourier transform is taken over position and time $(\vec{x},t)$ to get energy and momentum $(E,\vec{p})$. This is done by four integrations, one each getting rid of one of the four variables $x_1, x_2, x_3, t$, and replacing with the corresponding one of the four Fourier transformed variables $p_1, p_2, p_3, E$. Mathematically, nothing is wrong with only doing one of the four transforms, and the interpretation is clear.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-06-24T03:35:01.327" LastActivityDate="2011-06-24T03:35:01.327" />
  <row Id="11524" PostTypeId="2" ParentId="11515" CreationDate="2011-06-24T03:41:02.617" Score="1" ViewCount="" Body="&lt;p&gt;Maybe an easier way to see what is going on is to consider the bound state. What will you have to do to take it apart?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it's bound, separating the components means you're going to have to do work against a potential. That work is energy that you're adding to the system. And so, as you separate the particles, their energy increases and so does their mass.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-24T03:41:02.617" />
  <row Id="11525" PostTypeId="2" ParentId="11475" CreationDate="2011-06-24T03:52:03.603" Score="1" ViewCount="" Body="&lt;p&gt;On this class of problems, there's a secret that allows you to get the answer without having to memorize any formulas. (By the way, I got a 990 on the physics GRE so you can trust that I use this sort of thing.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have three things that contribute to the problem: $m$, $r$ and $v$. These have three different units, $[kg], [m], [m/s]$. Therefore, anything you want can be obtained from these in only one way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Centripetal acceleration is in units of $[m/s^2]$, correct? from this we see that mass does not enter into it. We have velocity $[m/s]$ and radius $[m]$ and the only way we can get $[m/s^2]$ from these is by $v^2/r$ which is what you wrote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one caveat on using this trick is that you want to avoid things associated with the period because you'll need a factor of $2\pi$. For example, instead of computing the &lt;a href=&quot;http://en.wikipedia.org/wiki/Frequency&quot; rel=&quot;nofollow&quot;&gt;frequency&lt;/a&gt; in Hz or $[1/s]$, you will get the &lt;a href=&quot;http://en.wikipedia.org/wiki/Angular_frequency&quot; rel=&quot;nofollow&quot;&gt;angular frequency&lt;/a&gt; (i.e. the rate at which the object eats up angles). To get this sort of thing, just recall that the distance around the circle is $2\pi r$.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-24T03:52:03.603" CommentCount="2" />
  <row Id="11526" PostTypeId="2" ParentId="11465" CreationDate="2011-06-24T04:27:58.427" Score="4" ViewCount="" Body="&lt;p&gt;The traditional method is with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Photomultiplier&quot; rel=&quot;nofollow&quot;&gt;photomultiplier tube&lt;/a&gt; or PMT. These are very expensive but are easily recognized and you might try hanging around the auctions at the local research university to see what comes up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More modern is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Avalanche_photodiode&quot; rel=&quot;nofollow&quot;&gt;Avalanche photodiode&lt;/a&gt; or APD. Amazingly, my favorite electronics supplier, &lt;a href=&quot;http://search.digikey.com/scripts/DkSearch/dksus.dll?vendor=0&amp;amp;keywords=avalanche+photodiode&quot; rel=&quot;nofollow&quot;&gt;Digikey, has them&lt;/a&gt;, at this time as cheap as $942. These might be harder to spot at the auction (as they are quite small).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Never fear, this is another alternative. The maker of the Digikey APDs is &lt;a href=&quot;http://www.advancedphotonix.com//&quot; rel=&quot;nofollow&quot;&gt;Advanced Photonix Inc&lt;/a&gt; which offers them &lt;a href=&quot;http://www.advancedphotonix.com/ap_products/standard_SiAvalanche.asp?from=leftnav&quot; rel=&quot;nofollow&quot;&gt;only through Digikey&lt;/a&gt;, at least according to their website.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were an aspiring young scientist, I might consider contacting API and seeing if they were willing to loan me a &quot;solder sample&quot; (out of spec) APD. Before contacting them I would do the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Get &lt;a href=&quot;http://www.advancedphotonix.com/ap_products/pdfs/SD197-70-72-591.pdf&quot; rel=&quot;nofollow&quot;&gt;the data sheet for their most common and cheapest part&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Study the data sheet enough to completely understand the part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Design a circuit that will use the part in a manner so as not to damage it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) Send the proposed design to an &lt;a href=&quot;http://www.advancedphotonix.com/ap_contact/contact_form.asp&quot; rel=&quot;nofollow&quot;&gt;API sales / support engineer using their contact form&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) Apply prayer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(6) If it didn't work, repeat with another APD maker.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The sad fact of engineering is that the sales reps give away free samples to engineers who almost never use them. If you happen to known someone in an industry which uses APDs, they may have some sitting around in their desks that they couldn't care less about. I'd be willing to bet that there are LHC engineeers / physicists who would have a few lying around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get one, you might consider using your contacts to find someone connected to you who works at one of the colliders. If you're at a big school, it should be easy to use google to find a prof who's in a collaboration.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-24T04:27:58.427" CommentCount="8" />
  <row Id="11527" PostTypeId="1" AcceptedAnswerId="11531" CreationDate="2011-06-24T06:36:55.320" Score="1" ViewCount="165" Body="&lt;p&gt;As of this writing it has been made clear to me that classical physics' Luminiferous aether was a terriblly poor discriptor of space. With the advent of Special Relativity and General Relativity, that aether ceased to be applicable. Today I began reading on the Higgs boson, and the Higgs field. How are those theories different from or similar to the seemingly irrational Luminiferous aether?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.&#xA;I am a first year student at Bakersfield Community College. If this question seems naive, you now have an explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(EDIT)&#xA;To me the similarity rests upon their fundamental influences regarding static positions in space. I currently only accept models in which static spacial positions are impossible. Contrastingly relative positions seem quite reasonable. In my humble opinion my naivatisms inhibit me from absorbing and understanding the Higgs boson. If they actually are similar, public protest would probably be more apparant. However, I have put forth my best effort to pose an direct question here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. The questions Where does space go when it falls? and What happens to time lost when matter accelerates? come much earlier in my studies, and they seem relevant.&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="4097" LastEditorDisplayName="" LastEditDate="2011-06-24T07:23:08.023" LastActivityDate="2011-06-24T11:14:45.643" Title="The Higgs field a new Luminiferous aether?" Tags="&lt;higgs-boson&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11528" PostTypeId="2" ParentId="11388" CreationDate="2011-06-24T06:50:27.150" Score="3" ViewCount="" Body="&lt;p&gt;While there are more mathematical definitions available, an appropriate working definition of locally asymptotically $\textrm{AdS}_3$ is that the line-element in Gaussian normal coordinates for $\rho\to\infty$ must take the form&#xA;\begin{equation}&#xA;ds^2 = d\rho^2 + (\gamma_{ij}^{(0)} e^{2\rho} + {\rm subleading}_{ij}) dx^i dx^j&#xA;\end{equation}&#xA;where $\gamma_{ij}^{(0)}$ is the 2-dimensional Minkowski metric and &quot;subleading&quot; means &quot;not diverging as fast as $e^{2\rho}$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can easily convince yourself that the BTZ black hole (and any other asymptotic AdS$_3$ space-time you might have encountered) can be written in this form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 3-dimensional vacuum Einstein gravity the asymptotic solutions to the equations of motion imply that the subleading terms must fall off as in the Brown-Henneaux boundary conditions (BH bc's) provided in Olof's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is not true that BH bc's are necessary for asymptotically $\textrm{AdS}_3$ behavior. You can obtain as asymptotic symmetry algebra (ASA) two copies of the Virasoro algebra even for weaker boundary conditions. The question of whether or not such boundary conditions are useful depends on the theory under consideration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Permit me to given an example that I know quite well: In topologically massive gravity at a certain critical point one should allow logarithmic violations of the BH bc's in order to accommodate the full normalizable spectrum of linearized fluctuations around $\textrm{AdS}_3$ and to avoid the elimination of otherwise valid classical solutions. See &lt;a href=&quot;http://arxiv.org/abs/arXiv:0808.2575&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/arXiv:0808.2575&lt;/a&gt;, in particular Eq. (8), which displays violations of the BH bc's in two entries of the metric. Nevertheless, one can show that one has two copies of the Virasoro algebra as ASA, and that the asymptotic charges are finite and conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite generally, a reasonable strategy to find the &quot;right&quot; boundary conditions is to weaken them as much as possible, but without creating inconsistencies, like infinite or non-conserved charges. This can be sometimes a bit of an art and may require physical input.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-06-24T06:50:27.150" CommentCount="3" />
  <row Id="11530" PostTypeId="2" ParentId="11527" CreationDate="2011-06-24T09:05:58.050" Score="1" ViewCount="" Body="&lt;p&gt;Actually, the luminiferous ether was a very good idea which turned out to be wrong.   The Higgs field (or at least the Higgs boson) is another very good idea that may or may not turn out to be wrong.  In this sense they could be similar.&lt;br&gt;&#xA;Lots of very smart people believed in the luminiferous ether for many years.  The difference between an ether and a field is fairly subtle.  The difference in Maxwell's theory before and after Einstein is even more subtle.  Maxwell's equations did not change, but the prediction (retrodiction) for the Michelson-Morley experiment did change, because other equations did change (Shift from Galileo transform to Lorentz  transform.)&lt;br&gt;&#xA;The luminiferous ether was a perfectly good mathematical theory.  The Higgs field is also a perfectly good mathematical theory.  In this way they are similar.  I am not sure about your idea that they both represent some kind of fixed point or maybe an absolute rather than relative spacetime.&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-06-24T09:05:58.050" />
  <row Id="11531" PostTypeId="2" ParentId="11527" CreationDate="2011-06-24T11:14:45.643" Score="5" ViewCount="" Body="&lt;p&gt;Here comes a simple minded answer by an experimentalist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The luminiferous ether was discarded because it violated special relativity. It presupposed a fixed reference frame of the ether against which everything moved.  In special relativity there exists no absolute frame of reference, and special relativity has been vindicated many times experimentally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Higgs field, as also the vacuum sea in general, comes from quantum field theory formulations of the interactions of elementary particles. All quantum field theories are consistent with  special relativity , and thus the Higgs field is also consistent with special relativity. It therefore cannot play the role of the luminiferous ether, and also the same holds for the vacuum sea, which is seething with virtual pairs of particle/antiparticle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the level of your studies this should suffice. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-06-24T11:14:45.643" CommentCount="5" />
  <row Id="11532" PostTypeId="1" AcceptedAnswerId="11551" CreationDate="2011-06-24T11:53:34.333" Score="2" ViewCount="132" Body="&lt;p&gt;This has always puzzled me. Yesterday (in London) it started hailing despite it being about $20^oC$. A couple of years ago I experienced hail in Sicily when it was about $35^oC$ in the shade!. How is this possible?&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-26T00:05:37.923" LastActivityDate="2011-06-26T00:05:37.923" Title="Hail stones in summer?" Tags="&lt;temperature&gt;&lt;atmosphere&gt;&lt;atmospheric-science&gt;&lt;meteorology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11533" PostTypeId="1" AcceptedAnswerId="11552" CreationDate="2011-06-24T13:00:53.070" Score="5" ViewCount="170" Body="&lt;p&gt;For one of my experimental setup I need to place a mirror perfectly parallel to a wall. It can be placed at any distance from the wall. I would like to use any method other than direct measurement. I am free to use the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a webcam&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a secondry mirror&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: It's not necessary to use all or either of them.&lt;/p&gt;&#xA;" OwnerUserId="4131" LastEditorUserId="4131" LastEditorDisplayName="" LastEditDate="2011-06-26T18:23:06.190" LastActivityDate="2011-06-27T08:49:34.770" Title="How to place a mirror parallel to a wall?" Tags="&lt;optics&gt;&lt;geometric-optics&gt;" AnswerCount="2" />
  <row Id="11534" PostTypeId="2" ParentId="10075" CreationDate="2011-06-24T13:03:46.533" Score="2" ViewCount="" Body="&lt;p&gt;M - Theory suggests that the universe we live in is a D-brane. D branes comes in different dimensionalities from the D25 to D0. There is no known compelling reason why a particular configuration of D branes, compact spaces and the laws of physics are prefered over the others. The current view is that all possibilities can happen in the multiverse, so maybe is kind of lucky that this particular configuration, that we call the universe, is stable and have all the constants tuned for us to observe it.&lt;/p&gt;&#xA;" OwnerUserId="4173" LastEditorUserId="4173" LastEditorDisplayName="" LastEditDate="2011-06-24T14:05:06.300" LastActivityDate="2011-06-24T14:05:06.300" />
  <row Id="11535" PostTypeId="2" ParentId="11295" CreationDate="2011-06-24T13:27:56.973" Score="2" ViewCount="" Body="&lt;p&gt;I will expand on my comment here, which is an answer to part of the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Nonetheless, I think the possibility that supersymmetry is already right in front of us would be important enough, that someone should investigate even such a &quot;trivial&quot; realization of Rivero's correspondence, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I said above : &lt;em&gt;When I first met super-symmetry, decades ago, I was surprised to learn that it is used in &lt;a href=&quot;http://iopscience.iop.org/1742-6596/237/1/012005&quot; rel=&quot;nofollow&quot;&gt;nuclear physics models&lt;/a&gt;. Whenever bosons and fermions gather together super-symmetry can be invoked.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am old enough to have started particle physics with the tau/theta puzzle and to have been entranced by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Eightfold_Way_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;eightfold way&lt;/a&gt;, when it was clearly presented to us. That was the way we met SU(3). Meeting SU(3) as the symmetry governing the strong interaction, at a later date, opened my perception that in a similar way that when one has a complicated crystal structure, various symmetries can be found, but there is one basic symmetry of the crystal, in the symmetries involved in elementary particles various groups can be expressed as a consequence, as projections/consequences  without being the fundamental symmetry underlying the physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The poetic &quot;eightfold way&quot; is just a &quot; cute&quot; projection of deeper underlying symmetries, which depend on similar groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now super-symmetry is a symmetry between bosons and fermions. As my link shows this exists in nuclear physics states too, and actually I have the impression that that is where super-symmetry usage originated in physics, when they were trying to organize nuclear levels and their similarities and differences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have found  super-symmetric groupingss between diquarks and mesons, it is interesting, but unless it addresses fundamental questions of model building, one would tend to see it as an interesting analog super-symmetric eightfold way  representation, showing nothing more fundamental than that super-symmetry is a viable representation for elementary particles. IMO ofcourse. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-06-24T13:27:56.973" CommentCount="4" />
  <row Id="11536" PostTypeId="2" ParentId="5801" CreationDate="2011-06-24T13:49:19.460" Score="1" ViewCount="" Body="&lt;p&gt;My answer to your question is in the context of brane inflation. The mechanism is as follows: A pair of D3 and Anti D3 separated by a warped dimension (Bulk), move closer until the separation of the branes is about the string length scale, then a stretched tachyon forms creating an instability that destroys both branes, releasing all the energy to closed string states. Since radiation is included in the spectrum of closed strings, both particles and radiation are produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0610221&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0610221&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4173" LastActivityDate="2011-06-24T13:49:19.460" />
  <row Id="11537" PostTypeId="2" ParentId="11533" CreationDate="2011-06-24T14:00:03.073" Score="2" ViewCount="" Body="&lt;p&gt;Fasten the secondary mirror to the wall, facing out, and support the webcam centered in front of it, looking into it. Turn on the camera. Rotate the webcam up and down, left and right, until it's image of itself in the secondary mirror is centered. Put the other mirror behind the webcam: you will see what looks like a great many images of the camera if the mirrors are not parallel to each other. Try it: you should see right away what you have to do to make the mirrors parallel to each other (and to the wall). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You said the mirror can be any distance to the wall, so an alternative method is to set that distance to zero.  &lt;/p&gt;&#xA;" OwnerUserId="1462" LastActivityDate="2011-06-24T14:00:03.073" CommentCount="5" />
  <row Id="11538" PostTypeId="2" ParentId="8976" CreationDate="2011-06-24T14:02:03.993" Score="-1" ViewCount="" Body="&lt;p&gt;The quantum conformal anomaly is not present in string theory if the dimension of space time is fixed to 26 for bosonic string theory and 10 for superstring theory. In string theory conformal symmetry in the world sheet is necessary for the theory to be consistent, so the trace of the energy momentum tensor vanishes. If not we don't have string theory at all.&lt;/p&gt;&#xA;" OwnerUserId="4173" LastActivityDate="2011-06-24T14:02:03.993" CommentCount="4" />
  <row Id="11539" PostTypeId="1" CreationDate="2011-06-24T14:54:56.610" Score="0" ViewCount="157" Body="&lt;p&gt;What velocity would a small body moving towards a large body reach under gravity alone?&#xA;More importantly, is there a 'terminal velocity' for the small body due to the increasing energy required to accelerate the body as its velocity increases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumptions: No other forces, gravity acts over infinite distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parameters: Large mass=$10^{24}$kg Small mass=$100$kg, Distance between them=$10^{12}$m.&lt;/p&gt;&#xA;" OwnerUserId="4175" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-06-24T16:42:07.897" LastActivityDate="2011-06-25T07:43:51.490" Title="Terminal velocity of a body in a vacuum" Tags="&lt;gravity&gt;" AnswerCount="2" />
  <row Id="11540" PostTypeId="2" ParentId="11539" CreationDate="2011-06-24T15:16:50.490" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;is there a 'terminal velocity' for the small body due to the increasing energy required to accelerate the body as its velocity increases?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No there is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the conditions you specify, the only limiting force is the one when they literally collide.  There is a maximum speed that will be reached, of course, right before they hit.  In the absence of any other retarding force, however, the gravitational potential energy will be converted to kinetic energy with 100% efficiency.  If it misses, for instance, then you enter the realm of orbital motion where the energy is constantly swapping back and forth between kinetic and gravitational potential energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.  I think you have a solid intuition about the situation.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Limitations&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The question can be formalized in the following way, there is a small mass $m$ and a large mass $M$ for which $m\ll M$.  Say $M$ is at the origin and $m$ starts on the positive x-axis at a distance sufficiently large such that we can say the gravitational potential is zero.  The acceleration of $m$ is written for the non-relativistic case as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a = - \frac{G M}{x^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will follow that the kinetic energy as $m$ reaches any given point will be the following, then we can write the velocity.  The conceptual question that follows is whether it can exceed the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ KE = m \frac{G M}{x} = \frac{1}{2} m v^2$$&#xA;$$ v = \sqrt{ \frac{ 2 G M }{ x} }$$&#xA;$$ v \stackrel{?}{&amp;gt;} c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An obvious limitation is the case where $x$ reaches the value such that we are on the surface of $M$ if we take it to be a planet for instance.  We can wave away this limitation by saying that M is a point.  However, the formula for kinetic energy is invalid for when $v$ approaches $c$ (as well as the formula for potential energy!) so that's not very helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of common knowledge about general relativity, yes, it actually is the case that the speed exceeds the speed of light going inward by some formalizations in some reference frames.  The equations above, however, do nothing more than give you and idea when &lt;em&gt;when&lt;/em&gt; such concepts are going to become important.  Any mass, no matter how small, can form the conditions of a black hole provided that the density is high enough (just decrease $x$ in that equation).&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-06-24T15:57:51.250" LastActivityDate="2011-06-24T15:57:51.250" CommentCount="5" />
  <row Id="11542" PostTypeId="1" AcceptedAnswerId="11548" CreationDate="2011-06-24T16:43:27.263" Score="4" ViewCount="356" Body="&lt;p&gt;Why is the gravitation force always attractive? Is there a way to explain this other than the curvature of space time? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: If the simple answer to this question is that mass makes space-time curve in a concave fashion, I can rephrase the queston as why does mass make space-time always curve with concavity?&lt;/p&gt;&#xA;" OwnerUserId="3714" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-06-24T16:48:58.343" LastActivityDate="2011-07-11T16:00:18.540" Title="Why is gravitation force always attractive?" Tags="&lt;gravity&gt;&lt;spacetime&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="4" />
  <row Id="11543" PostTypeId="1" AcceptedAnswerId="11583" CreationDate="2011-06-24T16:51:05.167" Score="5" ViewCount="140" Body="&lt;p&gt;I am wondering what will be the physics to explain how two neutral, chemically nonreactive objects stick. I know that using van der Waals formalism, we can treat neutral body electrodynamic forces and arrive with attractive forces that pull the objects together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, once the objects touch (say a mechanical cantiliver in a MEMS sensor like the one used in an iPhone), what happens to the forces? A quantitative answer or some estimate on how strong the attractive force is for simple cases will be very appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in response to anna's comment :&#xA;Let us consider what happens in vacuum for ultra smooth surfaces, with no residual electrical charge and fully chemically stablized surfaces (example, silicon crystals with stabilised surface bonds).  &lt;/p&gt;&#xA;" OwnerUserId="3714" LastEditorUserId="3714" LastEditorDisplayName="" LastEditDate="2011-06-26T05:26:36.730" LastActivityDate="2011-06-26T07:32:54.543" Title="What happens when two smooth surfaces touch?" Tags="&lt;electrodynamics&gt;&lt;friction&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11544" PostTypeId="1" AcceptedAnswerId="11554" CreationDate="2011-06-24T16:56:42.963" Score="2" ViewCount="94" Body="&lt;p&gt;Does one need to invoke quantum mechanics to explain casimir force or vander waals force. I see that textbooks show derivation of vander waal force with no QM but casimir force is typically described with QM. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a difference between vanderwaal and casimir forces ? Are there distinct examples of these two forces in real life. Is there a way to prove a given force is vanderwaal and not casimir or vice versa. &lt;/p&gt;&#xA;" OwnerUserId="3714" LastActivityDate="2011-06-24T22:05:52.213" Title="vander waals and casimir forces" Tags="&lt;electromagnetic-radiation&gt;&lt;casimir-effect&gt;" AnswerCount="2" />
  <row Id="11545" PostTypeId="2" ParentId="11542" CreationDate="2011-06-24T17:25:04.507" Score="2" ViewCount="" Body="&lt;p&gt;It is not true at all that gravitation is always attractive. In fact, if it were always attractive, universe would not be expanding at an accelerated rate right now and an inflationary period would not have ocurred.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;currently the only known source of expansion components of gravity is the cosmological constant, which, incidentally, is precisely the physical quantity that our theories fail to predict by the largest amount: 120 or 60 orders of magnitude, depending on whom you ask&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-06-24T17:25:04.507" CommentCount="2" />
  <row Id="11546" PostTypeId="1" CreationDate="2011-06-24T17:58:44.987" Score="3" ViewCount="67" Body="&lt;p&gt;I have a question about the relation:&#xA;$\exp(-i \vec{\sigma} \cdot \hat{n}\phi/2) = \cos(\phi/2) - i \vec{\sigma} \cdot \hat{n} \sin(\phi/2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my texts, I see $\phi\hat{n}$ always as c-numbers. My question is whether or not this relation can be generalized for $\hat{n}$ being an operator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so how exactly would the expression be different?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1939" LastActivityDate="2011-06-24T22:11:13.387" Title="generalizing spin rotations" Tags="&lt;quantum-mechanics&gt;&lt;angular-momentum&gt;&lt;spin&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11547" PostTypeId="2" ParentId="11542" CreationDate="2011-06-24T18:26:35.890" Score="1" ViewCount="" Body="&lt;p&gt;If gravity is entropic, &lt;a href=&quot;http://en.wikipedia.org/wiki/Entropic_gravity&quot; rel=&quot;nofollow&quot;&gt;as suggested recently by Verlinde and earlier by others&lt;/a&gt;, gravity might be expected to be mostly attractive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My speculative imaginary view of this has been that if the evolution of quantum physical objects is not &lt;em&gt;perfectly&lt;/em&gt; conservative, then there might be processes which convert energy between scales. One possibility is that fields at the scales that determine gravitational interactions would be converted to lower scale matter that is currently not detectable. The result would be both an inflow and an outflow from regions that contain large masses, but on different scales and with different effects, attractive for matter that interacts more with the inflow, repulsive for matter that interacts more with the outflow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the human scale I suppose we have to imagine that we are pressed down by a flow of something that isn't the same as matter, it's the gravitational field if you will, and of course not an aether, metaphorically being sucked into the earth as food for a fundamental entropic process. Whatever exhaust there is from this process interacts with us enough less to be essentially undetectable on anywhere close to human scales, but would be detectable on either very large or very small scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like I say, speculative, and also only a very small part of a whole entropic process. I can take a few or even a lot of downvotes, but pretend for a moment we're shooting the breeze at coffee. I haven't followed the Verlinde and other entropic gravity literature at all closely, so I don't know whether something like this of model has been suggested in a mathematical form, which would be required for it to be publishable (although being published definitely wouldn't be enough to make this not speculative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you ask for &lt;em&gt;explanations&lt;/em&gt; for the established mathematics of General Relativity, I think the only currently possible response is speculation. GR is more grounded in empirical principles than in models that could be taken to be explanatory (which I say without prejudice insofar as I take empirically supported principled theories to be preferable to more-or-less ad-hoc models, except for the elusive question of how to imagine effective new empirically supported principles). I note that lurscher doesn't address your request for &lt;em&gt;explanation&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-06-24T18:26:35.890" CommentCount="5" />
  <row Id="11548" PostTypeId="2" ParentId="11542" CreationDate="2011-06-24T19:11:05.920" Score="3" ViewCount="" Body="&lt;p&gt;Gravity is mediated by a spin two particle. Electromagnetism by spin 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is &lt;a href=&quot;http://www.quora.com/Quantum-Field-Theory/Why-do-particles-of-odd-integer-spin-generate-forces-which-can-be-both-attractive-and-repulsive-whereas-particles-of-even-integer-spin-only-attract&quot; rel=&quot;nofollow&quot;&gt;a link&lt;/a&gt; that answers your question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;even and odd spin do differ in that they require a product of charges with different signs to get attraction or repulsion:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;spin even:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;$q_1 q_2 &amp;gt; 0$: attractive&lt;/li&gt;&#xA;  &lt;li&gt;$q_1 q_2 &amp;lt; 0$: repulsive&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;spin odd:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;$q_1 q_2 &amp;lt; 0$: attractive&lt;/li&gt;&#xA;  &lt;li&gt;$q_1 q_2 &amp;gt; 0$: repulsive&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;In the case of gravity, mediated by spin 2 particles, charge is mass, which is always positive.  Thus, $q_1 q_2$ is always greater than zero, and gravity is always attractive.  For spin 0 force mediators, however, there is no restriction on the charges and you can very well have repulsive forces.  A better rephrasing of the question is: &quot;Why do particles of odd spin generate repulsive forces between like charges, while particles of even spin generate attractive forces between like charges?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Goes on to derive this&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="4283" LastEditorDisplayName="" LastEditDate="2011-07-11T16:00:18.540" LastActivityDate="2011-07-11T16:00:18.540" CommentCount="4" />
  <row Id="11549" PostTypeId="2" ParentId="11544" CreationDate="2011-06-24T19:53:07.763" Score="0" ViewCount="" Body="&lt;p&gt;I think people tend to equate both and present them as &quot;alternative explanations of the same phenomenon&quot; simply because both forces tend to have the same order of magnitude and dimensional dependency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't heard a proper argument why this should be so. Casimir is related to missing modes lowering the inter-layer vacuum pressure, while Van Der Waals are dipole-dipole interactions. So it seems to me they are different effects, which very similar behaviours in the same orders of magnitude, so probably the experimental measurements are actually subtle contributions from both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know of any practical way of discerning between both macroscopically, i would be glad to hear what others have to say on this&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-06-24T19:53:07.763" CommentCount="2" />
  <row Id="11550" PostTypeId="2" ParentId="10275" CreationDate="2011-06-24T20:19:51.447" Score="0" ViewCount="" Body="&lt;p&gt;The appearance of imaginary numbers in quantum mechanics, relativity, quantum cosmology, electromagnetism, etc. etc. etc., can be consistently explained by the presence of an interpenetrating 4-brane, the fourth spatial dimension of which is imaginary and rotated to real time in our 3-brane. See &lt;a href=&quot;http://physics.esotec.org/Unification_of_Physics.pdf&quot; rel=&quot;nofollow&quot;&gt;http://physics.esotec.org/Unification_of_Physics.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4177" LastActivityDate="2011-06-24T20:19:51.447" />
  <row Id="11551" PostTypeId="2" ParentId="11532" CreationDate="2011-06-24T21:25:59.160" Score="7" ViewCount="" Body="&lt;p&gt;Temperature generally gets cooler as you go higher in altitude (which is one reason why you have snow on mountain peaks long after it has melted away in the foothills).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hail develops in thunderstorms.  A thunderstorm ~BY DEFINITION~ is a storm which has developed through the freezing layer.  So think about this:  If you see lightning or hear thunder from a storm, you can be absolutely sure that at least part of that storm is above the freezing altitude, and at least part of the storm is below it.  Thunderstorms often develop on warm days; and they also often develop on the edge of a cold front, where cold air is overtaking and riding over the less dense warmer air.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A typical thunderstorm case:  It is a very warm day, and the air might be quite humid.  The warm moist air tries to rise to get higher than the cooler drier (and therefore heavier) air above it.  The moist air rises.  As it rises it cools past its dew point and moisture droplets precipitate.  These droplets can be blown up by the rising air, higher than the freezing altitude, where they freeze.  They start falling through the moist air, freezing a layer of moisture onto the small iceball, then may be blown up again by updrafts.  The process can repeat many times as the little iceball goes up and down past the freezing level, each time growing by adding another layer of frozen moisture.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually the little iceball is too heavy for the updrafts to keep it aloft in the freezing zone, and it falls to earth as hail. &lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-06-24T21:57:57.713" LastActivityDate="2011-06-24T21:57:57.713" CommentCount="1" />
  <row Id="11552" PostTypeId="2" ParentId="11533" CreationDate="2011-06-24T21:44:25.993" Score="1" ViewCount="" Body="&lt;p&gt;Forget the webcam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;attach the secondary mirror to the wall, at a height that is near the height of the center of the primary mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then adjust the horizontal and vertical tilt of the primary mirror to center the multiple images of the secondary mirror within each other on the primary mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a cheap laser pointer and a carpenter's square, you can set up the laser pointer so that it is exactly square to the wall (vertically and horizontally) and then adjust the primary mirror such that the laser beam goes back to the laser.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-06-24T21:44:25.993" CommentCount="5" />
  <row Id="11553" PostTypeId="1" AcceptedAnswerId="11557" CreationDate="2011-06-24T20:20:56.960" Score="7" ViewCount="309" Body="&lt;p&gt;Recently, I've been watching &quot;The Big Bang Theory&quot; again and as some of you might know, it's a series with a lot of scientific jokes in it - mostly about Physics or Mathematics. I understand most of the things mentioned in the series and whenever I don't understand a joke, I just look up the knowledge I'm missing on wikipedia - e.g. I learnt about Schrödinger's cat in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, at one point, I did not know how to proceed, which is why I'm asking this question. At a Physics quiz, the participants are asked to &quot;solve&quot; the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/GXfQe.jpg&quot; alt=&quot;&amp;quot;Solve the equation!&amp;quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution turns out to be $-8 \pi \alpha$. My questions are: What is the meaning of this equation? How does one obtain it? And of course, how does one solve it and is the solution given by the university janitor (in the series) correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I am sorry if this is an equation found in Physics, I didn't know.&lt;/p&gt;&#xA;" OwnerUserId="977" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-25T00:09:10.517" LastActivityDate="2011-06-25T17:31:16.857" Title="Where does this equation originate from? (found in the Big Bang Theory)" Tags="&lt;quantum-field-theory&gt;&lt;mathematics&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="11554" PostTypeId="2" ParentId="11544" CreationDate="2011-06-24T22:05:52.213" Score="3" ViewCount="" Body="&lt;p&gt;The Casimir effect and the Van der Waals force between two conducting plates are one and the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, consider the boundary conditions postulated for the Casimir effect. The electric field has to be exactly zero at the plates. Because of this, it is said, the zero point energy of the vacuum is lower in between the plates than outside, which causes the interaction. But these references to the vacuum and virtual particles are mere heuristics. What does it mean for the electric field to be zero at the plates? The charges in the plate will have to redistribute and polarize the plate to generate a corresponding field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the interactions between fluctuating polarizations are precisely the &lt;a href=&quot;http://en.wikipedia.org/wiki/London_dispersion_force&quot; rel=&quot;nofollow&quot;&gt;dispersion forces&lt;/a&gt; that are responsible for the Van der Waals interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus they are two explanation of the same phenomenon.&lt;/p&gt;&#xA;" OwnerUserId="3234" LastActivityDate="2011-06-24T22:05:52.213" CommentCount="5" />
  <row Id="11555" PostTypeId="2" ParentId="11546" CreationDate="2011-06-24T22:11:13.387" Score="2" ViewCount="" Body="&lt;p&gt;In your expression, $\hat n$ is a unit vector. For a general vector $\vec{n}$ of length $|\vec{n}|$, the expression would be&#xA;$$\exp{(-i\vec{\sigma}\cdot\vec{n}\phi/2)}=\cos(|\vec{n}|\phi/2)-i\frac{\vec{\sigma}\cdot\vec{n}}{|\vec{n}|}\sin(|\vec{n}|\phi/2).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a matrix case, a reasonably natural assumption would be that $\vec{\sigma}\cdot\vec{N}$ is a matrix constructed as a sum of tensor products, $\vec{\sigma}\cdot\vec{N}=\sigma_1\otimes N_1+\sigma_2\otimes N_2+\sigma_3\otimes N_3$. Other constructions are possible, but the end result will presumably be a matrix except in special cases. The matrix obtained would not necessarily be Hermitian, in which case the expansion of $\exp{(-i\vec{\sigma}\cdot\vec{N}\phi/2)}$ would include $\cosh$ and $\sinh$ components as well as $\cos$ and $\sin$ components. For a matrix $M$ that has eigenvalues $m_i$, $\exp{(M)}$ has eigenvalues $\exp{(m_i)}$, in the same basis, which can be used to expand the matrix $\exp{(M)}$ with $M=-i\vec{\sigma}\cdot\vec{N}\phi/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expression for the vector $\hat{n}$ comes out so nicely because the matrix $\vec{\sigma}.\hat{n}$ only has eigenvalues $\pm 1$. If one ensures by one's choice of matrices $\vec{N}$ and by one's choice of the construction $\vec{\sigma}\cdot\vec{N}$ that the resulting matrix only has eigenvalues $\pm 1$, one would have the same happy simplicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are probably enough distinct constructions that one would like to have some external motivation, either from Physics or from some other branch of Mathematics, for introducing a specific choice. &lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-06-24T22:11:13.387" CommentCount="1" />
  <row Id="11556" PostTypeId="1" CreationDate="2011-06-24T22:26:03.437" Score="1" ViewCount="92" Body="&lt;p&gt;I'm only in high school (junior) so goes easy on me. But, how do physicist refute these paradoxes? Considering there are a number of theories regarding the origin of the universe and some postulate that the universe is infinite, I assume they've come across this. Just looking for a direct answer to what should be a relatively simple question. Here are the two paradoxes (for those unaware):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zeno's paradox &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A similar paradox arises if the past is infinite. If there exists an infinite past, we would never have the present day. If there was an infinite set of past events and each event requires the previous event to occur, would we ever have the present? Of course not. This is because if today is dependent upon the fact that yesterday happened, and there is an infinite set of these dependencies (i.e. forever) - today will have not occurred. This is similar to the library book example mentioned earlier.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hilbert's hotel&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If the universe did not have a beginning, then the past would be infinite, i.e. there would be an infinite number of past times. There cannot, however, be an infinite number of anything, and so the past cannot be infinite, and so the universe must have had a beginning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why think that there cannot be an infinite number of anything? There are two types of infinites, potential infinites and actual infinites. Potential infinites are purely conceptual, and clearly both can and do exist. Mathematicians employ the concept of infinity to solve equations. We can imagine things being infinite. Actual infinites, though, arguably, cannot exist. For an actual infinite to exist it is not sufficient that we can imagine an infinite number of things; for an actual infinite to exist there must be an infinite number of things. This, however, leads to certain logical problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most famous problem that arises from the existence of an actual infinite is the Hilbert’s Hotel paradox. Hilbert’s Hotel is a (hypothetical) hotel with an infinite number of rooms, each of which is occupied by a guest. As there are an infinite number of rooms and an infinite number of guests, every room is occupied; the hotel cannot accommodate another guest. However, if a new guest arrives, then it is possible to free up a room for them by moving the guest in room number 1 to room number 2, and the guest in room number 2 to room number 3, and so on. As for every room n there is a room n + 1, every guest can be moved into a different room, thus leaving room number 1 vacant. The new guest, then, can be accommodated after all. This is clearly paradoxical; it is not possible that a hotel both can and cannot accommodate a new guest. Hilbert’s Hotel, therefore, is not possible.&quot;&lt;/p&gt;&#xA;" OwnerUserId="4178" LastActivityDate="2011-06-24T22:26:03.437" ClosedDate="2011-06-25T00:10:49.063" Title="Is it possible for the universe to be infinite considering Zeno's and Hilbert's paradox?" Tags="&lt;cosmology&gt;" CommentCount="2" />
  <row Id="11557" PostTypeId="2" ParentId="11553" CreationDate="2011-06-24T22:51:09.943" Score="16" ViewCount="" Body="&lt;p&gt;This is a so called Feynman diagram you see on the board. It is a suggestive way to write the formula written below the diagram. Each aspect of the diagram directly translates to part of the formula via the so called &quot;Feynman Rules&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With Feynman diagrams you can calculate the &quot;amplitude&quot; (that is related to the quantum mechanical probability for a process to happen). In this case the process it electron muon scattering. As you can see initially there is a electron and a muon (the lower part of the diagram where the time arrow starts). The electron is on the left and carries momentum $p_1$ and spin $s_1$. This translated to a so called spinor $u^{(s_1)}(p_1)$ in the formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial muon carries momentum $p_2$ and spin $s_2$ therefore its spinor is $u^{(s_2)}(p_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since electrons and muons are electrically charged (both are have negative charge), they will repel each other via the electromagnetic force. The &quot;force carrier&quot;, also called &lt;em&gt;gauge boson&lt;/em&gt; of the EM foce is the photon $\gamma$, the quantum of the EM &lt;em&gt;vector field&lt;/em&gt; $A_\mu$. The electron and muon exchange a photon carrying momentum $q$ and represened by the term $\frac{-ig_{\mu\nu}}{q^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outgoing electron with momentum $p_3$ and spin $s_3$ corresponds to the term $\bar{u}^{(s_3)}(p_3)$ and likewise the outgoing muon to $\bar{u}^{(s_4)}(p_4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strength with which the electrons and muons couple to the EM field is given by $-ie\gamma^\mu$, or as written here using the fine structure constant $\alpha$:  $-i\sqrt{4\pi\alpha}\gamma^\mu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From these parts you construct an electron &quot;current&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bar{u}^{(s_3)}(p_3) (-i\sqrt{4\pi\alpha}\gamma^\mu) u^{(s_1)}(p_1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and muon current&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bar{u}^{(s_4)}(p_4) (-i\sqrt{4\pi\alpha}\gamma^\nu) u^{(s_2)}(p_2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which get couple by the photon. I.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bar{u}^{(s_3)}(p_3) i\sqrt{4\pi\alpha}\gamma^\mu u^{(s_1)}(p_1) \frac{-ig_{\mu\nu}}{q^2} \bar{u}^{(s_4)}(p_4) i\sqrt{4\pi\alpha}\gamma^\nu u^{(s_2)}(p_2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since the exchanged momentum can be anything you have to integrate over all possible momenta&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int d^4q$ but make sure to conserve moentum via Dirac delta functions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta^{(4)}(p_1-p_3-q)\delta^{(4)}(p_2+q-p_4)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is why you get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int {d^4q \bar{u}^{(s_3)}(p_3) i\sqrt{4\pi\alpha}\gamma^\mu u^{(s_1)}(p_1) \frac{-ig_{\mu\nu}}{q^2} \bar{u}^{(s_4)}(p_4) i\sqrt{4\pi\alpha}\gamma^\nu u^{(s_2)}(p_2) \delta^{(4)}(p_1-p_3-q)\delta^{(4)}(p_2+q-p_4)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lorentz indices and sign of the photon propagator in the screenshot are wrong btw.&lt;/p&gt;&#xA;" OwnerUserId="2151" LastEditorUserId="2151" LastEditorDisplayName="" LastEditDate="2011-06-25T17:31:16.857" LastActivityDate="2011-06-25T17:31:16.857" CommentCount="2" />
  <row Id="11558" PostTypeId="2" ParentId="11078" CreationDate="2011-06-24T22:59:16.867" Score="0" ViewCount="" Body="&lt;p&gt;Hydrodynamics is a whole classical world when you can find most of the effects you ask about; from Archimedes change of weight to renormalisation of turbulent fluid. Were you born in the XIXth century, you would be militant against Hydrodynamics I guess :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are asking for examples without fields, with only a finite number of degrees of freedom, you are reducing the scope a lot.  But generically, the point is that you get some parameter to express departure from Newton's law. Typically, c and h. You need a finite c, besides a non zero h. Then you use many body mechanics to approach to a situation that happens to be near a finite model with small h and very big c, and this we call the classical approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of setting h=c=1 is computationally very practical, but a lot of people gets into problems because of it. For instance, fermionic lagrangians have really a $h$ somewhere multiplying them, so in the classical limit they dissapear; this is a reason to have classical bosonic fields only.&lt;/p&gt;&#xA;" OwnerUserId="1335" LastActivityDate="2011-06-24T22:59:16.867" CommentCount="1" />
  <row Id="11559" PostTypeId="2" ParentId="11542" CreationDate="2011-06-24T23:53:24.577" Score="0" ViewCount="" Body="&lt;p&gt;When I was a schoolboy, our teacher of physics asked once one of our brilliant students (a girl), something like: &quot;What is the nature of gravity?&quot;. She thought for a moment and answered: &quot;I do not know. And what is it? Our teacher replied: &quot;If I had known...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as gravity is not derived from other nature features, it should be considered as a fundamental law. We may study its properties but we may not ask such questions - just by definition of a fundamental law. It is as an axiom - it is given as such.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-06-24T23:53:24.577" />
  <row Id="11560" PostTypeId="2" ParentId="11542" CreationDate="2011-06-24T23:59:53.067" Score="-1" ViewCount="" Body="&lt;p&gt;artwork:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dipole &lt;strong&gt;+-&lt;/strong&gt; &amp;lt;--- some distance ---&gt;  &lt;strong&gt;+-&lt;/strong&gt; dipole&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;two dipoles are always attractive&lt;/strong&gt; (or a dipole and another charge). &#xA;if they are like this +- ... -+ or -+ ...+- the dipoles will rotate and the configuration become attractive +- ... +- or -+ ... -+ .&lt;br&gt;&#xA;they obey a 1/r³ relation.&lt;br&gt;&#xA;if you can consider that inside the baryons (neutron, proton) can exist a configuration of dipoles you have an answer.&lt;br&gt;&#xA;(read the book of  book of Douglas Pinnow 'Our Resonant Universe'. It is a monography of a model of particles where this happens).&lt;br&gt;&#xA;how to go from 1/r³ to 1/r² ? integrate along the path.&lt;br&gt;&#xA;why ? explore the concept of polarizable vaccum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the consensus is that gravitation is not electromagnetism but, in that way it is always attractive. and I like it.&lt;br&gt;&#xA;but the order of magnitude of gravitation is $10^{-35}$ (more or less, by memory) of EM so how can it be EM ?&lt;br&gt;&#xA;can you figure out two EM radiators in each dipole in opposition of phase (one the +, other the -) extremely near one of the other? yes the radiated EM field has to be extremely faint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The connection with EM is much more compelling, imo, that a connection with thermodynamics or other ...)  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-06-24T23:59:53.067" />
  <row Id="11561" PostTypeId="1" CreationDate="2011-06-25T00:11:19.843" Score="3" ViewCount="116" Body="&lt;p&gt;I was trying to understand Witten's proof of the Positive Energy Theorem in General Relativity by reading the &lt;a href=&quot;http://projecteuclid.org/DPubS?service=UI&amp;amp;version=1.0&amp;amp;verb=Display&amp;amp;handle=euclid.cmp/1103919981&quot; rel=&quot;nofollow&quot;&gt;original argument given by Witten&lt;/a&gt;. I am comfortable with the overall argument, but I would like to understand the following statement, made on the second paragraph of page 394 in the previous link:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The only invariants that can be formed from the $1/r$ term in the metric tensor are the total energy and the total momentum.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These invariants (which I will call &quot;$1/r$-built&quot; for short) refer to an asymptotically flat initial value (spacelike smooth) 3-surface in spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an obvious reason why this should be true? Looking at the definition of the ADM-energy and momentum, it seems plausible because the only data we have are the first and second fundamental form, and it should be possible to write any invariant as a combination of these, and consequently, any $1/r$-built invariant as a function of ADM energy and momentum. However, this reasoning is too hand-wavy, so I wonder if there is a clear cut explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My interest in this fact is that although it is not really a logical step in the proof, if true it is probably one of the best ways to motivate a spinorial proof (I was thinking of something along the lines of &quot;if we can construct a manifestly non-negative $1/r$-built invariant by means of the asymptotic behavior of spinors, then it should be possible to prove that energy is non-negative by writing this invariant as a function of ADM energy and momentum&quot;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a purely physical perspective, if it were true that ADM energy and momentum suffice to specify the system (to order $1/r$), then they should be the only independent invariants. However, this suggests that an asymptotic observer who knew the total energy and momentum could reconstruct the metric up to order $1/r$. I was thinking if it is not possible to construct a counterexample by defining an axisymmetric spacetime in which the Killing field of azimuthal symmetry is build up only with $1/r$ terms and showing that it is asymptotically distinguishable from its non-rotating counterpart. In this spirit, I think it is worth asking a broader version of my question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What kinds of &quot;physically interesting&quot; boundary terms can appear in a Hamiltonian formulation for an asymptotically flat spacetime manifold in General Relativity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="565" LastActivityDate="2011-06-25T00:11:19.843" Title="Asymptotic Invariants in General Relativity" Tags="&lt;general-relativity&gt;" CommentCount="5" FavoriteCount="1" />
  <row Id="11562" PostTypeId="1" AcceptedAnswerId="11573" CreationDate="2011-06-25T04:02:05.497" Score="4" ViewCount="195" Body="&lt;p&gt;Is nature symmetric with respect to presence of particles? Do we have an antiparticle for every particle thought of? Are there any proven examples where we don't have an antiparticle? And what about antiparticle of a photon (we know it can also behave as a particle)? &lt;/p&gt;&#xA;" OwnerUserId="4131" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-25T07:03:42.647" LastActivityDate="2011-06-25T19:53:17.730" Title="Is nature symmetric between particles and antiparticles?" Tags="&lt;particle-physics&gt;&lt;antimatter&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="11563" PostTypeId="2" ParentId="11562" CreationDate="2011-06-25T07:23:29.847" Score="4" ViewCount="" Body="&lt;p&gt;Theoretically, for every known particle there is an anti-particle. But &lt;a href=&quot;http://www2.slac.stanford.edu/vvc/theory/antiquarks.html&quot; rel=&quot;nofollow&quot;&gt;bosons are their own antiparticle&lt;/a&gt;, for example, the $\pi^+$ which is (mostly) composed of an up-quark and a down-antiquark $u\bar{d}$, has as an anti particle the $\pi^- = d\bar{u}$. The antiparticle of the photon is just the photon, while the antiparticle of the $W^+$ is the $W^-$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Composite particles, such as the hydrogen atom, also have corresponding antiparticles, but not all of them have been observed. So far the heaviest antimatter atom observed is &lt;a href=&quot;http://en.wikipedia.org/wiki/Antimatter#Antihelium&quot; rel=&quot;nofollow&quot;&gt;anti-helium&lt;/a&gt;, but theoretically, it seems clear that anti-uranium could be produced.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-25T07:23:29.847" CommentCount="13" />
  <row Id="11564" PostTypeId="2" ParentId="11539" CreationDate="2011-06-25T07:43:51.490" Score="3" ViewCount="" Body="&lt;p&gt;The answer is one of general relativity and there are a bunch of possible answers depending on how you interpret &quot;the speed of light&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual interpretation is the same as the definition of speeds in our everyday world, i.e. the distance divided by the time. In GR we usually have (locally) a time coordinate and three space coordinates. So the speed of light means the maximum amount a space coordinate can change divided by a time interval in which the change can happen. For example, if the position can change at most by 1 light-year in a time period of one year, then we say that the speed of light is 1 light-year per year. In this sense, the maximum speed a particle can reach is the (local) speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for calling it the &quot;local&quot; speed of light is that it doesn't make sense to compare speeds at two points that are far apart. In such a case, the relative speeds depend on our choice of coordinates.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-25T07:43:51.490" CommentCount="1" />
  <row Id="11565" PostTypeId="1" AcceptedAnswerId="11566" CreationDate="2011-06-25T15:43:40.873" Score="1" ViewCount="120" Body="&lt;p&gt;Everyone knows how the surface of a spinning bucket of water would look like on earth - parabolic. But what if we turned off gravity (for instance by doing the experiment in a freely falling lift)? Would the surface be still parabolic? I'll explain my confusion in more detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The velocity of the spinning bucket is transferred to the water by means of frictional forces arising in the boundary between the bucket and water. But these frictional forces exist no matter whether there is gravity or not. So if I consider the whole bulk of water inside the bucket as a single system, this frictional force would give it a positive torque. Thus the water has to rotate. For the sustained rotation of water, a centripetal force has to exist. In normal gravity, the water surface changes its shape into a paraboloid so that there is a net force on any particle directed inward. But in free fall, there is no pressure on a particle inside the liquid. Thus the only force that can supply the centripetal acceleration is inter-molecular force between the particles which is weak to sustain huge velocities. So what exactly happens?&lt;/p&gt;&#xA;" OwnerUserId="966" LastActivityDate="2011-06-25T16:48:15.653" Title="Spinning bucket of water in zero gravity" Tags="&lt;classical-mechanics&gt;&lt;mechanics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11566" PostTypeId="2" ParentId="11565" CreationDate="2011-06-25T16:48:15.653" Score="5" ViewCount="" Body="&lt;p&gt;Assuming the bucket has a lid you will end up with most of the water lining the outside of the bucket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how your basic artificial-gravity, spinning habitat works, after all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some water could, in principle, remain floating in the center but it is not stable.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If the bucket has no lid, the water oozes up against the sides and runs out the open end.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-06-25T16:48:15.653" />
  <row Id="11567" PostTypeId="1" CreationDate="2011-06-25T18:02:01.020" Score="5" ViewCount="259" Body="&lt;p&gt;If I remember correctly, I heard some people saying that the transistors on CPUs today are so small, that they have to use quantum physics to make CPUs. Is that correct?&lt;/p&gt;&#xA;" OwnerUserId="2679" LastActivityDate="2011-06-25T18:57:29.860" Title="Is Quantum Physics used in manufacturing CPUs?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="11568" PostTypeId="2" ParentId="11567" CreationDate="2011-06-25T18:08:40.467" Score="8" ViewCount="" Body="&lt;p&gt;Quantum mechanics is part and parcel of understanding the function of &lt;em&gt;any&lt;/em&gt; transistor, and therefore any integrated circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As component sizes have gotten smaller, tunneling has become an increasingly important limiting factor in the design and layout of chips.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-06-25T18:08:40.467" CommentCount="0" />
  <row Id="11569" PostTypeId="2" ParentId="11567" CreationDate="2011-06-25T18:34:27.017" Score="9" ViewCount="" Body="&lt;p&gt;Well, they are not using it, but taking into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tunneling is one of the most important:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) As gate oxide thickness is just 1-2nm, electrons can tunnel through it -&gt; power consumption increases (or transistor might turn itself on if gate is not connected at the moment). So if you replace usual SiO2 with high-K dielectric (like HfO2) you would be able to increase gate oxide thickness (=dramatically reduce tunneling) but electromagnetic field will remain the same (i.e. transistor would work exactly as with thin oxide)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Flash memory directly rely on quantum tunneling effect - electrons into strong electromagnetic field tunnel right into middle of dielectric, and form 'trapped' charge, which may stay there for years.&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="930" LastEditorDisplayName="" LastEditDate="2011-06-25T18:57:29.860" LastActivityDate="2011-06-25T18:57:29.860" CommentCount="4" />
  <row Id="11570" PostTypeId="2" ParentId="222" CreationDate="2011-06-25T18:49:01.353" Score="0" ViewCount="" Body="&lt;p&gt;I used this software in my energy transfer courses, I never used it for symbolic, so I don't know if you can do symbolic computation, however it is very good at solving equations. As well as for conversions. It is not free, however you can download a student version, which I used the whole semester without problems. It is called ees. The company I think is called f chart. I know it is not exactly what you asked for, however it's a useful software to have around, especially when working with a lot of equations, since the software actually warns you about any inconsistency in the units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also useful if you want to calculate say for example entropy, the software can do it for you if you have the pressure or temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mhhe.com/engcs/mech/ees/download.html&quot; rel=&quot;nofollow&quot;&gt;http://www.mhhe.com/engcs/mech/ees/download.html&lt;/a&gt;&#xA;http://www.fchart.com/ees/&lt;/p&gt;&#xA;" OwnerUserId="4181" LastEditorUserId="4181" LastEditorDisplayName="" LastEditDate="2011-06-25T19:00:34.353" LastActivityDate="2011-06-25T19:00:34.353" CommunityOwnedDate="2011-06-25T18:49:01.353" />
  <row Id="11571" PostTypeId="1" CreationDate="2011-06-25T18:53:13.370" Score="0" ViewCount="103" Body="&lt;p&gt;Ignoring the whole movie situation along with the power demands, what do you think Iron Man's repulsors use to &quot;repulse&quot; his targets? I don't think they are lasers or electric beams, or &quot;reverse magnetism&quot;, but I'm guessing its some sort of focused sonic waave or a shock wave generator.&lt;/p&gt;&#xA;" OwnerUserId="4144" LastActivityDate="2011-06-26T17:17:50.700" ClosedDate="2011-06-26T18:51:59.847" Title="Iron Man repulsor question" Tags="&lt;particle-physics&gt;" AnswerCount="2" />
  <row Id="11572" PostTypeId="2" ParentId="11571" CreationDate="2011-06-25T18:55:13.853" Score="2" ViewCount="" Body="&lt;p&gt;I bet it's imagination. In imagination-land everything is possible :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality them most compact &amp;amp; efficient are conventional weapons.&#xA;Then probably only 10kW CO2 laser, which is real, but not as cool as in movie.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-06-25T18:55:13.853" CommentCount="3" />
  <row Id="11573" PostTypeId="2" ParentId="11562" CreationDate="2011-06-25T19:53:17.730" Score="6" ViewCount="" Body="&lt;p&gt;The anti-particle for any particle is obtained by charge C and parity P conjugation. C is the operation that interchanges positive and negative charges and P is the operation that reflects in a mirror. The combined operation of CP must produce a particle of the same mass. This is a theorem of relativistic quantum field theory due to CPT symmetry. This other particle is either the same particle or an antiparticle with opposite charge and/or chirality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some particles such a photons, gluons, Z bosons, pions, Higgs, graviton etc, do not have anti-particles because they are invariant under the CP transformation. You can say that they are their own anti-particle. This can only happen for particles without electric charge and with no chirality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle the QCD color charge is also reversed for an anti-particle. This suggests that a gluon should not be regarded as its own anti-particle, but since colourless states are never seen the distinction cannot really be made in any operational sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All known particles which are their own anti-particle are bosons, but it is also possible for a fermion to be its own anti-particle if it is a Majorana spinor rather than a Dirac spinor. No known fermions are of this type (unless neutrinos are Majorana) but they exist in SUSY models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observed elementary particles that do have anti-particles include all the quarks and leptons (except possibly the neutrinos) and the charged W bosons. Any composite particle also has an antiparticle made of the anti-particles of its constituents. This can only be its own anti-particle if all its constituents are (e.g. a glueball), or if it is made of particle/anti-particle combinations as is the case for pions.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-06-25T19:53:17.730" CommentCount="3" />
  <row Id="11574" PostTypeId="1" CreationDate="2011-06-25T20:57:41.617" Score="0" ViewCount="31" Body="&lt;p&gt;I understand that physical bodies can be measured in terms of the volume that their convex hull occupies.  If that volume was plotted on a histogram for all the physical bodies in the universe, that would that graph look like?&lt;/p&gt;&#xA;" OwnerUserId="4184" LastEditorUserId="4184" LastEditorDisplayName="" LastEditDate="2011-06-25T21:35:40.460" LastActivityDate="2011-06-25T21:35:40.460" ClosedDate="2011-06-26T01:53:48.420" Title="frequency distribution of physical body volume" Tags="&lt;frequency&gt;&lt;volume&gt;" CommentCount="5" />
  <row Id="11576" PostTypeId="1" CreationDate="2011-06-25T21:54:58.860" Score="8" ViewCount="180" Body="&lt;p&gt;This has been bugging me for some time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, Hawking radiation is the result of the mismatch between the vacuum state of a quantum field as seen by a free falling observer (falling directly toward the black hole) and one that is sitting at a constant radius far away from the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is perhaps the result of my naive understanding of the subject, but it doesn't make sense to me to say that a black hole radiates, because it's observer-dependent. How do we know that the thermal bath of particles that we see when we sit at a constant radius actually causes the black hole to lose mass? Is it just a simple energy conservation argument, or is there some subtle process here that I'm missing?&lt;/p&gt;&#xA;" OwnerUserId="4185" LastActivityDate="2011-06-26T06:45:59.077" Title="How do we know that black holes evaporate?" Tags="&lt;black-holes&gt;&lt;hawking-radiation&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="11577" PostTypeId="1" AcceptedAnswerId="11578" CreationDate="2011-06-26T00:07:35.427" Score="3" ViewCount="112" Body="&lt;p&gt;I have a circuit whit a AC source a capacitor and a resistance all in series. I find that the difference of potential between the capacitor leads begin to change after some instants as it should. But my question is, if the resistance is very high, say infinite, the capacitor have no tension per se, but it experiences high voltage as a whole in respect to the ground. What i don’t understand is why there is not electric field between the plates.&#xA;I’m sorry if the question isn’t clear but I really cannot understand the implication of the definition of “potential”.&lt;/p&gt;&#xA;" OwnerUserId="4186" LastEditorUserId="4186" LastEditorDisplayName="" LastEditDate="2011-06-26T01:31:14.357" LastActivityDate="2011-06-26T01:31:14.357" Title="How does capacitance work?" Tags="&lt;electromagnetism&gt;&lt;homework&gt;" AnswerCount="3" />
  <row Id="11578" PostTypeId="2" ParentId="11577" CreationDate="2011-06-26T00:21:21.117" Score="2" ViewCount="" Body="&lt;p&gt;If you choose $R\gg\omega/C$ , then most the voltage will drop off in the resistor. You may now argue that, for any voltage to drop off in a resistor, there needs to be some current in it. And there is, in fact, such a current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine the condensator plates as two antennas. Yes, pretty strange antennas, but antennas they are. What you essentially do is: you send an AC signal from one of these antennas over to the other one. If the receiver antenna has no significant load to feed (that is, if $R$ is big) then its potential will pretty much exactly follow the potential of the sender antenna. In other words,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;All the voltage you send into the circuitry will end up at the resistor.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The voltage along the capacitor will be very small.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3540" LastActivityDate="2011-06-26T00:21:21.117" CommentCount="3" />
  <row Id="11579" PostTypeId="2" ParentId="11577" CreationDate="2011-06-26T01:02:04.730" Score="5" ViewCount="" Body="&lt;p&gt;I'd like to expand on leftaroundabout's answer as the question actually was phrased &quot;what is capacitance&quot; and this seems to be the core of the OP's problem (your feeling of the meaning of potential is probably correct though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The antenna-analogy is a bit vague. You could see the capacitor as two reservoirs of electrons, linked so that when one fills up, the other is emptied and vice versa. In reality for example you have two plates near each other, each one can store electrons and the other feels the field from the other plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you start with one reservoir empty and the other full.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then send in a current into the first end of the capacitor for a short time, filling up the empty reservoir, and during this time the full reservoir in the other end will empty out into the other wire. Thus, during this short time, it &lt;em&gt;looks&lt;/em&gt; as if a current is passing straight through although no electrons are passing the di-electric!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the first reservoir is full though, the current stops. You then need to reverse the current, emptying it while the other reservoir pulls back its released electrons to refill, and now it looks as if our reversed current is passing straight through the capacitor again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why AC (an alternating current) seemingly pass through a capacitor - an AC source simply &quot;wiggles&quot; the current back and forth so you continously fill and empty the reservoirs at both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your AC will pass through the cap and you get the same measured potential on both ends of the cap, in the case of infinite load (infinite R).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Important things I've left out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cap doesn't fill up the &quot;reservoir&quot; at a constant rate of course. As it is filled, the field from the stored electrons oppose the current (their potential is increased) so they fill up slower and slower, the reverse when it's emptied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the frequency of the AC has to be compared to the size of the reservoirs (the capacitance!) when you consider how easy the current will cross (the impedance!).&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-06-26T01:02:04.730" />
  <row Id="11580" PostTypeId="2" ParentId="11577" CreationDate="2011-06-26T01:28:07.743" Score="0" ViewCount="" Body="&lt;p&gt;I think I got this. The capacitor it’s not even involved in the process! It’s like a dangling  wire. The real thing is that the AC source drain electrons from the ground, lowering the ground potential, but the ground is conventionally 0, so the potential of whatever is on the other side must go up!&lt;/p&gt;&#xA;" OwnerUserId="4186" LastActivityDate="2011-06-26T01:28:07.743" CommentCount="2" />
  <row Id="11581" PostTypeId="1" CreationDate="2011-06-26T02:04:45.080" Score="0" ViewCount="75" Body="&lt;p&gt;From what I have read, electromagnetic waves travel at a speed relative from their source. Given &lt;code&gt;x&lt;/code&gt; speed at a source, light travels at a speed of &lt;code&gt;c+x&lt;/code&gt;. Given the additional conundrum regarding the discrepancies of whatever position being inside or outside whatever physical system, could those two possibly be a link between an arbitrary speed of light and any measurements make of electromagnetism? To me the difference, between is and is not, results from an agreement on a what particular threshold of measure dictates something being at some arbitrary location. In astrophysics, for example, red shift pays a significant role in resolving a bodies speed relative to ours, but fails to determine the bodies' velocity. (extra points) When considering how relative light is, how is it bound to our universe? Where does space go when gravity occurres? Is a loss in time during acceleration altering the future while absorbing it?&lt;/p&gt;&#xA;" OwnerUserId="4097" LastActivityDate="2011-06-26T02:04:45.080" ClosedDate="2011-06-26T02:59:02.367" Title="nonrelative speed of light" Tags="&lt;special-relativity&gt;" CommentCount="1" />
  <row Id="11582" PostTypeId="2" ParentId="11576" CreationDate="2011-06-26T05:29:55.163" Score="3" ViewCount="" Body="&lt;p&gt;Well, here is a simple scenario :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantum field theoretic vacuum is seething with pair production of virtual particles everywhere. Take one such virtual pair at the horizon of the black hole. One part of the pair if it is going down is grabbed by the hole and disappears while the other is with its high momentum runs away from the horizon, on shell. Where did it find the energy ? &lt;a href=&quot;http://en.wikipedia.org/wiki/Hawking_radiation&quot; rel=&quot;nofollow&quot;&gt;read on&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A slightly more precise, but still much simplified, view of the process is that vacuum fluctuations cause a particle-antiparticle pair to appear close to the event horizon of a black hole. One of the pair falls into the black hole whilst the other escapes. In order to preserve total energy, the particle that fell into the black hole must have had a negative energy (with respect to an observer far away from the black hole). By this process, the black hole loses mass, and, to an outside observer, it would appear that the black hole has just emitted a particle. In another model, the process is a quantum tunneling effect, whereby particle-antiparticle pairs will form from the vacuum, and one will tunnel outside the event horizon.&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The far away observer is utilized to define the &quot;negative energy&quot; of the eaten up  partner.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-06-26T06:45:59.077" LastActivityDate="2011-06-26T06:45:59.077" CommentCount="15" />
  <row Id="11583" PostTypeId="2" ParentId="11543" CreationDate="2011-06-26T07:32:54.543" Score="2" ViewCount="" Body="&lt;p&gt;I will take a stab, though it needs a solid state physics background, in which I am weak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) in the case of crystals, as the two surfaces touch the crystal symmetry will take over with maybe some dislocation plane. i.e. the van der Waals forces will make the two one and you will not be able to easily separate them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) if amorphous,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) inorganic  example glass on glass, the attraction is not strong enough to create one body. The two will be separable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) organic surfaces with strong static  fields  example two cling film layers ,would adhere as one body, depending on the material .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it will depend on the specific materials under study.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-06-26T07:32:54.543" />
  <row Id="11584" PostTypeId="1" AcceptedAnswerId="11587" CreationDate="2011-06-26T09:18:48.663" Score="3" ViewCount="124" Body="&lt;p&gt;Has there been any cases where the only explanation has been that at least one of the physical constants must have changed to explain an experiment/event/obervation? I am not intrested in large scale obsorvations such Gaolaxies holding together with insufficient obsorvable mass or stars seem to be moving at faster than speed of the light phenomenons. Just simple instances of experiments/events that the only possible logical explenation can be obtained if one had to assume a known fundamental physical constant must have varied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to have this question to be only for non &lt;a href=&quot;http://physics.stackexchange.com/questions/704/are-the-physical-laws-scale-dependent&quot;&gt;scale dependent situations&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to give you an idea, consider that in mathematics the value of constant $\pi$ changes with the geometry.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-06-26T15:24:40.410" Title="Cases of any known fundamental physical constants changing within our locality?" Tags="&lt;physical-constants&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" />
  <row Id="11585" PostTypeId="2" ParentId="11584" CreationDate="2011-06-26T11:50:24.807" Score="0" ViewCount="" Body="&lt;p&gt;Pi doesnt change in math, because it's defined for a circle in euclidean geometry. Constants dont change because then they wouldnt be constant. &lt;/p&gt;&#xA;" OwnerUserId="1708" LastActivityDate="2011-06-26T11:50:24.807" CommentCount="1" />
  <row Id="11586" PostTypeId="1" CreationDate="2011-06-26T11:54:25.093" Score="0" ViewCount="22" Body="&lt;p&gt;I know that there are certain adsorbents which increase in volume when they adsorb gas (due to a structural change).  I was wondering.  I know that the geometry of adsorbents in a packed bed is really important for design.  However, I am unsure how an expanding adsorbent will work here.  Does any one know of an example where this type of adsorbent is used in a packed bed?&lt;/p&gt;&#xA;" OwnerUserId="3311" LastActivityDate="2011-06-26T11:54:25.093" Title="Use adsorbent with expanding volume in a column" Tags="&lt;thermodynamics&gt;" />
  <row Id="11587" PostTypeId="2" ParentId="11584" CreationDate="2011-06-26T12:50:54.723" Score="4" ViewCount="" Body="&lt;p&gt;It's somewhat glib to put quantum mechanics this way, but one could say that Planck's constant was zero before Planck introduced it, insofar as the classical limit of QM is $\hbar\rightarrow 0$ (which is not so very far).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something I've played with, to absolutely no good effect so far, is the idea that quantum fluctuations might be greater or less from place to place and from time to time, which fits a little better with your Question as you have put it. However, one would then describe quantum fluctuations at different space-time points &lt;em&gt;relative&lt;/em&gt; to the &lt;em&gt;unchanged&lt;/em&gt; constant $\hbar$, which would now be defined as &quot;the amplitude of quantum fluctuations under such and such conditions&quot;. That's essentially Solomoan's Answer, however his apparent assumption that the idea is obvious is a little too fast for someone with almost any philosophical leanings. [The boiling point of water is a constant, 100°C, if one defines the conditions carefully enough. A very interesting account of the historical and philosophical process associated with the definition of temperature scales is given by Hasok Chang, in a book that won a major Philosophy of Science prize, &lt;a href=&quot;http://www.oup.com/us/catalog/general/subject/Philosophy/Science/~~/dmlldz11c2EmY2k9OTc4MDE5NTMzNzM4OQ==&quot; rel=&quot;nofollow&quot;&gt;&quot;Inventing Temperature&quot;&lt;/a&gt;. If you read that —it's very accessible as these things go— you will ask a different Question.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to approach this Question may be to suppose that new theories introduce &lt;em&gt;new&lt;/em&gt; constants, in terms of which old constants can be expressed, together with the conditions in which they accurately describe the Physical world. Thus, in my speculative example as I put it above, one might define the amplitude of quantum fluctuations as $\hbar$ at the surface of the earth, and describe it's variation as a function of altitude —this would at least change our theory of gravity, and likely lots else besides, so the whole of Physics becomes part of the discussion. &lt;em&gt;That&lt;/em&gt; is a thread in Philosophy of Science that is known as the Quine-Duhem thesis, or, as Wikipedia has it, &lt;a href=&quot;http://en.wikipedia.org/wiki/Duhem%E2%80%93Quine_thesis&quot; rel=&quot;nofollow&quot;&gt;the Duhem-Quine thesis&lt;/a&gt;, which can be loosely summarized as the idea that a theory stands or falls as a whole, a point that I take to be vitiated by the always present possibility of changing any theory in very small ad-hoc ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your Question to some extent opens a philosophical Pandora's box, which you should be cautious about opening. My view is that anyone who wants to change Physics significantly must consider these and similar ideas at length, but for Physicists trying to do the everyday job of Physics it is perhaps as counterproductive to spend time on this as it is to spend time learning to play the violin.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-06-26T12:50:54.723" />
  <row Id="11588" PostTypeId="1" AcceptedAnswerId="11596" CreationDate="2011-06-26T13:42:53.513" Score="1" ViewCount="174" Body="&lt;p&gt;I've been reading about the work of Marie Curie recently after a friend filled me in on what she did (never having had much of an idea previously) and it's all very interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I can't understand however is what real impact her discovery of radium had for the development of physics in general? I mean in the textbooks and websites I've been browsing, Curie is made out to be some sort of physics pioneer - but I can't see where her discoveries specifically moved the study of physics forward (don't get me wrong, I'm sure she was extremely influential, I'm just trying to understand &lt;strong&gt;why&lt;/strong&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did the discovery of radium/radioactivity impact the study of physics? What impact did Marie Curie have in actually moving towards modern physics? I can only seem to find a variety of claims that she did impact physics, rather than how specifically - it all seems a bit odd.&lt;/p&gt;&#xA;" OwnerUserId="3044" LastActivityDate="2011-06-26T21:39:03.463" Title="Work of Marie Curie?" Tags="&lt;radioactivity&gt;&lt;history&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="11589" PostTypeId="2" ParentId="11584" CreationDate="2011-06-26T15:24:40.410" Score="4" ViewCount="" Body="&lt;p&gt;I guess from your use of the phrase &quot;within our locality&quot; that this is not what you mean, but various people have tried to measure astrophysically whether fundamental constants have evolved over time. A while back, there were some claims (e.g., &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v87/i9/e091301&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; and others by the same people) that observations of spectral lines in very distant objects could best be explained by assuming that fundamental constants (specifically the fine-structure constant) had changed with time. There is, to say the least, &lt;a href=&quot;http://arxiv.org/pdf/astro-ph/0301507v3&quot; rel=&quot;nofollow&quot;&gt;no consensus&lt;/a&gt; that these observations and their interpretation are correct, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People certainly try to measure changes in physical constants locally, but as far as I know there's no evidence for such variation.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-06-26T15:24:40.410" />
  <row Id="11590" PostTypeId="1" CreationDate="2011-06-26T15:34:23.510" Score="2" ViewCount="63" Body="&lt;p&gt;Does anybody know a way to create (in laboratory, but with no super-sophisticated equipement) a beam of light with partial linear polarization whose degree of polarization can be adjusted (at least over the range 0-0.5)?&lt;/p&gt;&#xA;" OwnerUserId="2752" LastActivityDate="2011-06-27T06:08:46.840" Title="Partially polarized light in laboratory" Tags="&lt;home-experiment&gt;&lt;polarization&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="11591" PostTypeId="2" ParentId="11590" CreationDate="2011-06-26T15:47:39.510" Score="2" ViewCount="" Body="&lt;p&gt;Start with unpolarized let. Let it &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/polar.html#c1&quot; rel=&quot;nofollow&quot;&gt;scatter off a surface&lt;/a&gt; at a variable angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also try scattered sunlight (i.e., light from the blue sky). The degree of polarization depends on the angle between the direction of the Sun and the direction the light is coming from. Apparently under the right circumstances you can get u&lt;a href=&quot;http://www.polarization.com/sky/sky.html&quot; rel=&quot;nofollow&quot;&gt;p to 75% polarization&lt;/a&gt; this way..&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-06-26T15:47:39.510" />
  <row Id="11592" PostTypeId="2" ParentId="11588" CreationDate="2011-06-26T16:13:24.177" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Her achievements include a theory of&#xA;  radioactivity (a term that she&#xA;  coined[2]), techniques for isolating&#xA;  radioactive isotopes, and the&#xA;  discovery of two elements, polonium&#xA;  and radium. Under her direction, the&#xA;  world's first studies were conducted&#xA;  into the treatment of neoplasms, using&#xA;  radioactive isotopes. She founded the&#xA;  Curie Institutes: the Curie Institute&#xA;  (Paris) and the Curie Institute&#xA;  (Warsaw).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;her husband and his brother had&#xA;  invented the electrometer, a sensitive&#xA;  device for measuring electrical&#xA;  charge. Using the Curie electrometer,&#xA;  she discovered that uranium rays&#xA;  caused the air around a sample to&#xA;  conduct electricity.[16] Using this&#xA;  technique, her first result was the&#xA;  finding that the activity of the&#xA;  uranium compounds depended only on the&#xA;  quantity of uranium present. She had&#xA;  shown that the radiation was not the&#xA;  outcome of some interaction of&#xA;  molecules, but must come from the atom&#xA;  itself. In scientific terms, this was&#xA;  the most important single piece of&#xA;  work that she conducted.[17]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;--copied from wikipedia&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess fills some of the void&lt;/p&gt;&#xA;" OwnerUserId="4131" LastActivityDate="2011-06-26T16:13:24.177" />
  <row Id="11593" PostTypeId="2" ParentId="11590" CreationDate="2011-06-26T16:54:55.277" Score="1" ViewCount="" Body="&lt;p&gt;Why not just take polarizing filter, and mix beam of 100% polarized light with variable ammount of non-polarized light using semi-transparent mirror? One should be careful not to polarize light accidentally :-)&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-06-26T16:54:55.277" CommentCount="2" />
  <row Id="11594" PostTypeId="1" CreationDate="2011-06-26T17:07:36.570" Score="3" ViewCount="182" Body="&lt;p&gt;I need to find the principal point in an image. Its a point where the principal axis intersects the image sensor.&#xA;Due to misalignment this point is not at the center of image always(or image sensor). I need to precisely determine its location using any of the optical methods available(if any). Any suggestions are greatly appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More information: The camera is giving me a live feed over a screen and I am able to store, analyze any part of it in real time or later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/JeXNa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4131" LastEditorUserId="4131" LastEditorDisplayName="" LastEditDate="2011-06-27T12:19:14.107" LastActivityDate="2011-07-06T23:18:46.457" Title="How to find the principal point in an image?" Tags="&lt;optics&gt;&lt;geometric-optics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11595" PostTypeId="2" ParentId="11571" CreationDate="2011-06-26T17:17:50.700" Score="0" ViewCount="" Body="&lt;p&gt;As pointed out by BarsMonster ,newton's Third law says the iron man will be acted upon by the equal and opposite force which he is exerting on any body. Hence its just mere imagination.&lt;/p&gt;&#xA;" OwnerUserId="4131" LastActivityDate="2011-06-26T17:17:50.700" CommentCount="5" />
  <row Id="11596" PostTypeId="2" ParentId="11588" CreationDate="2011-06-26T17:19:07.080" Score="0" ViewCount="" Body="&lt;p&gt;Not an expert here but I think it went something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Her research established the techniques and methods that other physicists like Rutherford needed to probe the internal structure of the atom.  The concept of an atom (I believe) was postulated at the time but not really taken seriously.  Certainly there was no real concept of an atom's internal structure.  The energy contained in the radioactivity had to come from somewhere.  Finding out where it came from through experimental investigation of the radioactivity was really the start of atomic physics and paved the way for the development of modern phyiscs.  The subsequent research changed the way we think about matter.&lt;/p&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2011-06-26T17:19:07.080" />
  <row Id="11597" PostTypeId="2" ParentId="9786" CreationDate="2011-06-26T17:45:38.500" Score="2" ViewCount="" Body="&lt;p&gt;As genneth said in a comment, any metal around (or maybe somewhat above) room temperature should have a higher heat capacity than $3k_B$ per atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each vibrational degree of freedom (a.k.a. phonon mode) has a heat capacity of $k_B$ as long as the temperature $T$ and vibrational frequency $\nu$ satisfy $k_BT\gg h\nu$ (if this is not satisfied, the heat capacity is less than $k_B$). There are three phonon modes per atom, so phonons give you $3k_B$, as long as the temperature is high enough. For example, in gold, all the phonon frequencies are &lt;a href=&quot;http://dx.doi.org/10.1103/PhysRevB.8.3493&quot; rel=&quot;nofollow&quot;&gt;less than 5 THz&lt;/a&gt;; 5 THZ corresponds to 240K; so at room temperature the phonon heat capacity is almost $3k_B$ (but a bit less).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I chose gold as an example because its atoms are heavy so they vibrate slowly. Metals with lighter atoms have higher vibration frequencies so a higher required temperature to get the full $3k_B$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On top of the phonons, a metal also has heat capacity from kinetic energy of the free electrons. So altogether it can be more than $3k_B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I looked up gold's heat capacity (.128 or .129 J/gK) and atomic mass (196.97) and got $3.03k_B$ to $3.06k_B$ per atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm a bit surprised it's not higher, since each atom should contribute at least one free electron, and a free electron would be expected to have $1.5k_BT$ of translational kinetic energy. I guess it's too simplistic to treat the electrons like non-interacting free particles. For example, maybe there is a ceiling on electron kinetic energy because of the band structure, or because of velocity-dependent phonon scattering? I'm not sure.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other possible degrees of freedom that provide extra heat capacity in some solids include plasmons, magnons, excitons, polaronic excitations, and many others. :-)&lt;/p&gt;&#xA;" OwnerUserId="3811" LastActivityDate="2011-06-26T17:45:38.500" CommentCount="1" />
  <row Id="11598" PostTypeId="2" ParentId="8538" CreationDate="2011-06-26T18:30:57.107" Score="0" ViewCount="" Body="&lt;p&gt;I strongly believe that both the images are right. Even the eyes have got lens and you can replace the whole eye with a lens and retina. It is only the way they are representing the information. The first picture is a bit deceptive, in the sense that it hides the basic principle of a point getting focused by a lens on a sensor plane. &lt;/p&gt;&#xA;" OwnerUserId="4131" LastActivityDate="2011-06-26T18:30:57.107" CommentCount="1" />
  <row Id="11599" PostTypeId="2" ParentId="10477" CreationDate="2011-06-26T20:17:00.190" Score="2" ViewCount="" Body="&lt;p&gt;There is an real life analogy that provides a nice answer to this question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1999 Lord Monckton marketed a toy called the &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Eternity_puzzle&quot; rel=&quot;nofollow&quot;&gt;Eternity Puzzle&lt;/a&gt;&quot;. It was a tiling problem with 209 pieces. He thought it was sufficiently complex that nobody would solve it even with a very powerful computer, so he offered £1 million to anyone who could. One year later it was solved by some Cambridge mathematicians. How many combinations did he estimate people would have to try to solve it? $10^{500}$ of course!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem of finding the correct string vacuum for the universe won't be solved so quickly. It will probably require some more input from experiments and cosmology, but there is no reason to think it is not soluble simply because of the large number (assuming the universe is really constructed that way). It may (or may not) also be necessary to understand the foundations of string theory better first.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastEditorUserId="1341" LastEditorDisplayName="" LastEditDate="2011-06-26T21:02:43.460" LastActivityDate="2011-06-26T21:02:43.460" />
  <row Id="11600" PostTypeId="1" CreationDate="2011-06-26T20:59:01.713" Score="7" ViewCount="110" Body="&lt;p&gt;For a system of N particles with k holonomic constraints, their Cartesian coordinates are expressed in terms of generalized coordinates as $$\mathbf{r}_1 = \mathbf{r}_1(q_1, q_2,..., q_{3N-k}, t)$$ $$...$$ $$\mathbf{r}_N = \mathbf{r}_N(q_1, q_2,..., q_{3N-k}, t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each particle in space can be uniquely identified by 3 independent variables, so why aren't the above of the form $$\mathbf{r}_i = \mathbf{r}(q_{i1}, q_{i2}, q_{i3}) ?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note there is only one transformation $\mathbf{r}$ for all $\mathbf{r}_i$, a function of only three generalised coordinates and independent of t.&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-06-27T16:46:43.600" Title="Why so many arguments for the transformation equations of generalized coordinates?" Tags="&lt;classical-mechanics&gt;&lt;lagrangian-mechanics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11601" PostTypeId="2" ParentId="11600" CreationDate="2011-06-26T21:07:06.247" Score="14" ViewCount="" Body="&lt;p&gt;The generalized coordinates of a system of $N$ particles apply to the system as a whole, not the individual particles, and accordingly they can (and often do) combine the coordinates of multiple particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One common example is that of two-body orbital motion: one generalized coordinate is the position of the center of mass of the system,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{q}_1 = \frac{m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2}{m_1 + m_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the other is the displacement vector between the two bodies,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{q}_2 = \mathbf{r}_2 - \mathbf{r}_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the generalized coordinates depends on physical coordinates from &lt;em&gt;both&lt;/em&gt; objects, and if you invert the transformation you'll find that the physical coordinates of each object depend on &lt;em&gt;both&lt;/em&gt; generalized coordinates. So you can't just express $\mathbf{r}_i$ in terms of $q_{i(1,2,3)}$ alone.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-26T21:07:06.247" />
  <row Id="11602" PostTypeId="2" ParentId="11588" CreationDate="2011-06-26T21:39:03.463" Score="0" ViewCount="" Body="&lt;p&gt;To quote Cornell University professor L. Pearce Williams (was on wikipedia):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The result of the Curies' work was epoch-making. Radium's radioactivity was so great that it could not be ignored. It seemed to contradict the principle of the conservation of energy and therefore forced a reconsideration of the foundations of physics. On the experimental level the discovery of radium provided men like Ernest Rutherford with sources of radioactivity with which they could probe the structure of the atom. As a result of Rutherford's experiments with alpha radiation, the nuclear atom was first postulated. In medicine, the radioactivity of radium appeared to offer a means by which cancer could be successfully attacked.&quot; -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Radium Emanations&quot; or radon was also used as a source of x-rays for medical imaging. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of the reason for her continued scientific stature results from her fame in popular culture as pre-dating the feminist movement and became a martyr. Marie Curie died of aplastic anemia contracted from exposure to radiation. I know this last part has little to do with physics, but name recognition contributes to her influence.&lt;/p&gt;&#xA;" OwnerUserId="2843" LastActivityDate="2011-06-26T21:39:03.463" />
  <row Id="11603" PostTypeId="2" ParentId="11594" CreationDate="2011-06-26T21:48:08.677" Score="1" ViewCount="" Body="&lt;p&gt;To do this you would need to take an image of some sort of reference object, like a regular grid pattern. You would also have to know that your imaging system has some measurable field aberrations which are radially symmetric, such as distortion. Then you could do some simple image processing to locate the center of the distortion pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a regular grid pattern, you would plot the positions of the grid lines in the image. With third order distortion, the grid lines should be spaced quadratically (ie. with the space between each grid line increasing linearly). The grid spacing will reach a max (or min) at the &quot;principal point.&quot; (I use scare quotes around &quot;principal point&quot; because the principal point has a specific meaning in geometrical optics, and it is not the meaning used in the question here.)&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-06-26T21:48:08.677" CommentCount="12" />
  <row Id="11604" PostTypeId="1" CreationDate="2011-06-26T22:47:37.787" Score="1" ViewCount="84" Body="&lt;p&gt;There are 11 gaseous elements and two liquid elements at standard temperature and pressure. The rest are solid. Can phase be predicted from quantum mechanical principles?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-07-06T02:37:28.840" Title="Phase of Elements" Tags="&lt;elements&gt;&lt;quantum.mechanics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="11605" PostTypeId="1" CreationDate="2011-06-26T23:42:29.047" Score="9" ViewCount="153" Body="&lt;p&gt;The &quot;minus sign problem&quot; in quantum simulation refers to the fact that the probability amplitudes are not positive-definite, and it is my understanding that this leads to numerical instability when for example summing over paths, histories or configurations since large amplitudes of one phase can completely negate other large contributions but with the reverse phase. Without controlling this somehow, the sampling of alternative configurations has to be extremely dense, at least for situations where interference is expected to be important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming I haven't misunderstood it so far, my main question is when this is a show-stopper and when it can be ignored, and what the best workarounds are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, let's say that I want to simulate Compton scattering or something similar, numerically, to second order. I could evaluate the Feynman diagrams numerically to a certain resolution and sum them. I assume this won't work well. In Lattice QCD, complete field configurations are randomly generated and the action calculated and amplitude summed I guess (I have only superficial knowledge of Lattice QCD unfortunately). &lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-06-27T13:25:44.607" Title="When is the &quot;minus sign problem&quot; in quantum simulations an obstacle?" Tags="&lt;quantum-mechanics&gt;&lt;simulation&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="11606" PostTypeId="2" ParentId="11605" CreationDate="2011-06-27T00:38:28.600" Score="0" ViewCount="" Body="&lt;p&gt;As for the lattice part: here, you're interested in doing Monte Carlo Markov chains, which consists of summing over configurations $s$, each having a certain weight $p(s)$ and for which your observable $O$ takes the value $O(s)$. Normally, you propose a new configuration $s&amp;#39;$ at each step, calculate it's probability relative to your current configuration $p(s&amp;#39;)/p(s)$, pull a random number $q \in [0,1]$ out of a hat and you switch configurations $s \mapsto s&amp;#39;$ iff $p(s&amp;#39;)/p(s) &amp;gt; q.$ Thus if you start with a 'positive' configuration, you'll never pass by a 'negative' configuration, so that a significant part (typically ~half) of your phase space is never seen by your Monte Carlo run; your final estimate for $\langle O \rangle$ will thus be nonsense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Note: you could propose another scheme (the above one is called Metropolis), but fundamentally you run into problems because a Markov chain with negative weights doesn't make sense.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a lot of literature on the sign problem in lattice QCD, as it is still open in general. A typical way out is &lt;em&gt;reweighting&lt;/em&gt;; by noting that $$p(s) = \frac{p(s)}{|p(s)|} |p(s)|,$$ you can safely choose a new weight $\pi(s) = |p(s)| &amp;gt; 0$ and multiply every observable by the sign of the weight, $$O(s) \mapsto  \frac{p(s)}{|p(s)|} O(s).$$ However, this might be expensive and even though formally the method is correct, it's not very elegant or insightful. However, since it's an open problem, not everyone will agree on its status or a way out...&lt;/p&gt;&#xA;" OwnerUserId="1471" LastActivityDate="2011-06-27T00:38:28.600" />
  <row Id="11607" PostTypeId="1" AcceptedAnswerId="11612" CreationDate="2011-06-27T03:23:13.790" Score="3" ViewCount="152" Body="&lt;p&gt;Can the &lt;a href=&quot;http://en.wikipedia.org/wiki/Church%E2%80%93Turing_thesis&quot; rel=&quot;nofollow&quot;&gt;Church–Turing Thesis&lt;/a&gt; be proved assuming classical mechanics, how is the proof or disproof?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited: I was looking for a proof of &quot;everything computable by a device obeying CM is computable by a Turing machine&quot;, but it seems obvious now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My followup question is why does it hold for QM?&lt;/p&gt;&#xA;" OwnerUserId="1708" LastEditorUserId="1708" LastEditorDisplayName="" LastEditDate="2011-06-27T21:03:16.033" LastActivityDate="2011-06-27T21:03:16.033" Title="Church–Turing Thesis" Tags="&lt;quantum-mechanics&gt;&lt;classical-mechanics&gt;&lt;mathematical-physics&gt;&lt;computation&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="11608" PostTypeId="1" AcceptedAnswerId="11609" CreationDate="2011-06-27T04:12:55.523" Score="0" ViewCount="203" Body="&lt;p&gt;A Lightning rod is fixed to the building and it is not given earthing then what happens? How does the lightning reach the ground through the building?&lt;/p&gt;&#xA;" OwnerUserId="4026" LastActivityDate="2011-08-31T17:28:47.517" Title="What happens when a lightning rod is not earthed?" Tags="&lt;electricity&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="11609" PostTypeId="2" ParentId="11608" CreationDate="2011-06-27T04:27:53.293" Score="7" ViewCount="" Body="&lt;p&gt;A Lightning rod without earthing is as good as any other conductor laying stray in open. If its not earthed then it wont work at all. There is nothing very fancy about the whole process. You need to connect both ends of a circuit to make current flow. An improper grounding connection can cause disaster sometimes as the lightening may find another route like plumbing to reach to the ground, which will produce the same effect as that of a lightning striking any building and can result in fire. Even simple grounding is not advisable for such a system. The area at the place of grounding is treated with saline water to increase conductivity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many times improper grounding renders the whole system inappropriate. Even sharp curves are avoided in the conductor connecting to the ground to reduce inductance. If ever curves are made then the radius is kept large to decrease inductance.&lt;/p&gt;&#xA;" OwnerUserId="4131" LastEditorUserId="4131" LastEditorDisplayName="" LastEditDate="2011-06-29T17:45:41.477" LastActivityDate="2011-06-29T17:45:41.477" CommentCount="2" />
  <row Id="11610" PostTypeId="2" ParentId="11590" CreationDate="2011-06-27T06:08:46.840" Score="1" ViewCount="" Body="&lt;p&gt;Start with a narrow beam of unpolarized light (e.g. an incandescent bulb shining through a small hole.) Put a sheet of ordinary window glass (or a glass slide or whatever) into the beam, and I'm going to talk about the light transmitted through the glass. If the glass is normal to the light beam, the transmitted light beam will be unpolarized. Then rotate the glass so the light goes through at a shallower and shallower grazing angle. The transmitted light gets more and more linearly polarized. At the shallowest (most grazing) angle possible, 31% of the transmitted intensity is one polarization, and 69% is the other (if I calculated correctly), so the light is 38% linearly polarized. [Don't forget, in your experiments, that the overall intensity of the transmitted light also changes as you rotate the glass.] The nice thing is that the transmitted beam stays in more or less the same place as you rotate the glass, which is convenient for an experiment. The beam only shifts a little bit due to the refraction. If you put two identical pieces of glass in the beam (one after the other) and counter-rotate them, then the beam stays in exactly the same place (the refraction shifts cancel out) and you can get more polarization as well (up to 67% polarized if I calculated correctly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need more polarization than that, of course, you can use the beam of light reflected off the glass, which goes from unpolarized at near-normal incidence, to 100% polarized at the Brewster angle.&lt;/p&gt;&#xA;" OwnerUserId="3811" LastActivityDate="2011-06-27T06:08:46.840" CommentCount="0" />
  <row Id="11612" PostTypeId="2" ParentId="11607" CreationDate="2011-06-27T10:03:32.617" Score="2" ViewCount="" Body="&lt;p&gt;This question was asked and &lt;a href=&quot;http://mathoverflow.net/questions/54820/physics-and-churchturing-thesis&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;answered on mathoverflow&lt;/strong&gt;&lt;/a&gt;. You should read it there. We could have migrated it but that site is independent of SE network. &lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2011-06-27T10:03:32.617" CommentCount="1" />
  <row Id="11613" PostTypeId="2" ParentId="11605" CreationDate="2011-06-27T10:50:01.623" Score="1" ViewCount="" Body="&lt;p&gt;I think that there is a slight misrepresentation of a problem in your question. The fact that quantum-mechanical amplitudes are generally complex numbers is not the root of the problem. It is usually solved by going into &quot;imaginary time&quot;. By changing $t\to -it$ you effectively obtain the statistical mechanics problem instead of the quantum mechanical one:&lt;br&gt;&#xA;$$ t \to -it,\quad p_0\to -ip_0$$&#xA;$$\int\!\!\!\cal{D}\phi\; exp\left[i\int\!\!\!dt\; \cal{L}[\phi]\right]&#xD;&#xA;\to \int\!\!\!\cal{D}\phi\; exp\left[-\int\!\!\!dt\; H[\phi]\right]$$&#xA;Note that Lagrangian is effectively replaced by Hamiltonian. And, instead of summing some complex numbers we've got the real and positive numbers to sum. Which is the work for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Metropolis%E2%80%93Hastings_algorithm&quot; rel=&quot;nofollow&quot;&gt;Metropolis algorithm&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The good news are that this actually works well. The bad news is that this only works for bosonic fields. For fermions your path integral is over Grassmanian-number-valued field. And there is no way of (well, effectively) represent the Grassmaian number in numerical calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of tricks to deal with fermions and this is a rapidly developing field and I'm not an expert there at all. One of the ways to go is to calculate the integral over the fermions analytically, leaving only (sometimes, effective) bosonic degrees of freedom. Which gives you something like:&#xA;$$\int\!\!\!\cal{D}\phi\; exp\left[-\int\!\!\!dt\; H[\phi]\right]\rho[\phi]$$&#xA;with $\rho[\phi]$ being not necessarily positive. That is where Metropolis ceases to work effectively and/or even properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further you can read &lt;a href=&quot;http://en.wikipedia.org/wiki/Numerical_sign_problem&quot; rel=&quot;nofollow&quot;&gt;the article on Wikipedia&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-06-27T10:50:01.623" CommentCount="6" />
  <row Id="11614" PostTypeId="1" AcceptedAnswerId="11620" CreationDate="2011-06-27T12:36:41.143" Score="3" ViewCount="217" Body="&lt;p&gt;Do the laws of physics change anywhere in the universe?Or will they change from place to place in the universe?&lt;/p&gt;&#xA;" OwnerUserId="4026" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-27T19:39:01.300" LastActivityDate="2011-06-27T19:39:01.300" Title="Do the laws of physics work everywhere in the universe?" Tags="&lt;laws-of-physics&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="11615" PostTypeId="2" ParentId="11614" CreationDate="2011-06-27T13:12:25.917" Score="3" ViewCount="" Body="&lt;p&gt;The aim of physics theoretical research  is to formulate mathematically the &quot;laws of physics&quot; in such a form that the formulae will work anywhere in the observed/known  universe. We can test well our immediate neighbourhood, analyse signals coming from the beginning of the Big Bang and expect that the final Theory Of Everything (TOE) will be consistent through the whole. Physics is not there yet.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-06-27T13:12:25.917" />
  <row Id="11617" PostTypeId="2" ParentId="11605" CreationDate="2011-06-27T13:25:44.607" Score="4" ViewCount="" Body="&lt;p&gt;The sign problem is an obstacle in two cases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Simulating spin or boson hamiltonians with geometric frustration (Although here we can rely on the Marshall Sign Rule to tell us what hamiltonians will always have positive-definite ground-state wavefunctions)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Simulating fermi hamiltonians away from special symmetry points (in reference to Kostya's answer, we &lt;em&gt;always&lt;/em&gt; integrate the Grassmann fields analytically -- the tool for this is usually called a &quot;Hubbard-Stratonovich transformation,&quot; which leaves you with an auxillary bosonic field that you sample through Monte Carlo -- the sign problem shows up in that your probability measure is, again, not positive definite &lt;em&gt;necessarily&lt;/em&gt;. Sometimes, by chance, you get lucky though -- the Hubbard model with an attractive interaction, or in general at half filling, has no sign problem - but this seems to be fortuitous at best.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good introduction to the sign problem can be found here: &lt;a href=&quot;http://arxiv.org/abs/cond-mat/0408370&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/cond-mat/0408370&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main problem with &quot;reweighting&quot; is tackled in equation 7 of that paper: &lt;em&gt;exponential&lt;/em&gt; growth of error in both particle number and inverse temperature (both of which we're trying to scale to infinity...)&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2011-06-27T13:25:44.607" />
  <row Id="11618" PostTypeId="1" CreationDate="2011-06-27T13:28:29.503" Score="2" ViewCount="180" Body="&lt;p&gt;Are there any examples of fermionic particles or quasiparticles for which the interaction potential is a globally smooth function? i.e. no singularities or branch points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, in Flügge's Practical Quantum Mechanics, problem 148 has two repulsive particles on a circle. This is supposed to model the two helium electrons in the ground state. The equation he gives is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ -\frac{\hbar^2}{2mr^2}\left(\frac{\partial^2 \psi}{\partial x_1^2}+\frac{\partial^2 \psi}{\partial x_2^2}\right)+V_0\cos(x_1-x_2)\psi=E\psi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't quite follow why this potential does not have a singularity when $x_2\rightarrow x_1$. Are there other such examples?&lt;/p&gt;&#xA;" OwnerUserId="4197" LastEditorUserId="4197" LastEditorDisplayName="" LastEditDate="2011-06-27T15:17:55.980" LastActivityDate="2011-08-29T14:36:29.313" Title="Fermionic interaction potentials" Tags="&lt;quantum-mechanics&gt;&lt;interactions&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11619" PostTypeId="1" CreationDate="2011-06-27T14:02:08.483" Score="1" ViewCount="60" Body="&lt;p&gt;Is there a way of calculating the mass of a galaxy, or even a nebula from the luminosity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm deleting this, and moving the question to Astronomy Stack Exchange - thanks david&lt;/p&gt;&#xA;" OwnerUserId="3588" LastEditorUserId="3588" LastEditorDisplayName="" LastEditDate="2011-06-28T00:39:53.490" LastActivityDate="2011-06-28T00:39:53.490" ClosedDate="2011-06-28T00:51:06.503" Title="Mass of a galaxy via Luminosity" Tags="&lt;astrophysics&gt;&lt;mass&gt;&lt;luminosity&gt;&lt;galaxies&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11620" PostTypeId="2" ParentId="11614" CreationDate="2011-06-27T14:26:27.327" Score="7" ViewCount="" Body="&lt;p&gt;There's &lt;a href=&quot;http://physics.stackexchange.com/questions/10078/do-the-laws-of-physics-evolve/10082#10082&quot;&gt;another question&lt;/a&gt; on this site about whether the laws of physics change over time. I think that the answers to that one (including mine) apply pretty much perfectly to this question about whether the laws change in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We &lt;em&gt;expect&lt;/em&gt; the fundamental laws of physics to be the same throughout space. In fact, if we found that they were not, we would strongly expect that that meant that the laws we had discovered were not the fundamental ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very sensible to ask whether the laws as we currently understand them vary with respect to position. People do try to test these things experimentally from time to time. For instance, some experiments to test whether fundamental constants change with time are also sensitive variations in the fundamental constants with position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some cosmological theories, especially some of those that come under the heading of &quot;multiverse&quot; theories do allow for the possibility that the laws are different in different regions of space, although generally only on scales much larger than what we can observe. In general, in such theories, the &lt;em&gt;truly&lt;/em&gt; fundamental laws are the same everywhere, but the way the evolution of the Universe played out in different regions is so different that the laws &lt;em&gt;appear&lt;/em&gt; quite different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way this can happen is by the mechanism of spontaneous symmetry breaking. When the Universe cooled down from very high temperatures, it probably underwent various transitions, more or less like phase transitions, in which an initially symmetric state turns into a less-symmetric state. In those transitions, there may be different ways that the final state can come out, and they may be quite dramatically different -- completely different sorts of particles may exist, for instance. There could be different regions of the Universe in which the symmetry breaking went different ways, in which case the &quot;apparent&quot; laws would be utterly different in different regions, but probably only on scales many orders of magnitude larger than what we can see.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-06-27T14:26:27.327" />
  <row Id="11621" PostTypeId="2" ParentId="11619" CreationDate="2011-06-27T14:34:12.373" Score="2" ViewCount="" Body="&lt;p&gt;There's no way to calculate such a thing from first principles: the luminosity of a galaxy depends on how efficiently it managed to form stars out of the matter available to it. Star formation is far too complicated to admit of a first-principles calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are empirical relationships between mass and luminosity, which have been used and refined for many years. When trying to understand the large-scale structure of the Universe, we want to know where the mass is, but what we can measure is where the light is. Hence the great interest in mass-luminosity relations. The link you provide in your edit is an example of this work. To find others, a good buzzword to search for is &quot;mass-to-light ratio.&quot; For instance, &lt;a href=&quot;http://iopscience.iop.org/0004-637X/569/2/720&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; might be a good place to start (although I haven't read it).&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-06-27T14:34:12.373" />
  <row Id="11622" PostTypeId="1" CreationDate="2011-06-27T14:59:42.027" Score="7" ViewCount="135" Body="&lt;p&gt;Is there a physical example of a field that would have the following Lagrangian density&#xA;$$&#xD;&#xA;L= \sqrt{1+\phi_x^2 +\phi_y^2+\phi_z^2}&#xD;&#xA;$$ &#xA;where the subscripts denote partial derivatives and $\phi$ is a scalar field?&lt;/p&gt;&#xA;" OwnerUserId="4198" LastActivityDate="2011-06-27T21:39:26.727" Title="What corresponds to this Lagrangian density?" Tags="&lt;lagrangian-mechanics&gt;&lt;field-theory&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11623" PostTypeId="2" ParentId="11594" CreationDate="2011-06-27T15:45:30.663" Score="1" ViewCount="" Body="&lt;p&gt;I suggest shining a narrow laser beam along the optical axis. Just don't fry your sensor! Hopefully, the antireflective coating of the lens is not perfect, and you will be able to see on the object side &lt;strong&gt;two&lt;/strong&gt; reflections of the beam: one from each side of the lens. Align the laser in such a way that both reflections come back exactly at the output port of the laser: the beam is then perpendicular to both faces of the lens, and therefore it is exactly along it's optical axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now look at the image: you have one bright spot exactly at the focus.&lt;/p&gt;&#xA;" OwnerUserId="4202" LastActivityDate="2011-06-27T15:45:30.663" CommentCount="4" />
  <row Id="11624" PostTypeId="2" ParentId="11622" CreationDate="2011-06-27T16:13:27.747" Score="7" ViewCount="" Body="&lt;p&gt;This looks a lot like soap film statics, but with an extra dimension. Consider a soap film glued to a ring. The film is described by a function $z = \phi(x,y)$, with $z$ the height of the film above the xy-plane. We want to minimise the potential energy of the film, which means to a good approximation minimising the surface area. The total area of the film is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ A = \int \sqrt{1 + \phi_x^2 + \phi_y^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the integration is over the entire xy-plane. This functional looks very similar to your Lagrangian density. The solution can be derived by applying the Euler-Lagrange equations and the solution is confusingly also called &lt;a href=&quot;http://mathworld.wolfram.com/LagrangesEquation.html&quot; rel=&quot;nofollow&quot;&gt;Lagrange's equation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ (1 + \phi_y)^2 \phi_{xx} + 2 \phi_x \phi_y \phi_{xy} + (1 + \phi_x)^2 \phi_{yy} = 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give you an idea of what the solutions look like, if the field doesn't change too rapidly we can make some approximations ($\phi_x \approx \phi_y \approx \phi_{xy} \approx 0$) and get &lt;a href=&quot;http://en.wikipedia.org/wiki/Laplace%27s_equation&quot; rel=&quot;nofollow&quot;&gt;Laplace's equation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \phi_{xx} + \phi_{yy} = \Delta \phi = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also provides an interesting interpretation for solutions to the Laplace equation, as they are approximate minimal-area surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of this carries over three dimensions but the minimal-surface interpretation isn't so clear anymore.&lt;/p&gt;&#xA;" OwnerUserId="3234" LastEditorUserId="3234" LastEditorDisplayName="" LastEditDate="2011-06-27T17:59:00.823" LastActivityDate="2011-06-27T17:59:00.823" CommentCount="6" />
  <row Id="11625" PostTypeId="2" ParentId="8538" CreationDate="2011-06-27T16:13:44.207" Score="1" ViewCount="" Body="&lt;p&gt;Both drawings are correct for both the eye and the camera, as optically&#xA;the eye behaves like a camera. The difference is that the first drawing&#xA;shows the principal rays while the second shows the marginal rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The marginal rays are interesting mostly when discussing focus and depth&#xA;of field. Compared to most cameras, the eye has a very large depth of&#xA;field, and a very good autofocus system. Then, the issues of focus and&#xA;depth of field are less important for the eye than for the camera.&#xA;That's why the right picture is just slightly less relevant for the eye.&lt;/p&gt;&#xA;" OwnerUserId="4202" LastActivityDate="2011-06-27T16:13:44.207" />
  <row Id="11626" PostTypeId="2" ParentId="11600" CreationDate="2011-06-27T16:46:43.600" Score="2" ViewCount="" Body="&lt;p&gt;The $k$ holonomic constraints are used to eliminate $k$ $q$s, so reducing their number from $3N$ to $3N-k$. This then introduces the dependence of some of the transformation equations on t and other $q$s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have k holonomic constraints of the form $$\mathbf{f}_1(q_1, q_2,..., q_{3N},t) = 0$$ $$...$$ $$\mathbf{f}_k(q_1, q_2,...,q_{3N}, t) = 0$$ $3N$ q coordinates for the $N$ particles $$(q_1,q_2,q_3), (q_4,q_5,q_6),..., (q_{3N-2},q_{3N-1},q_{3N})$$ $3N$ transformation equations relating cartesian to generalised coordinates $$\mathbf{r}_1 = \mathbf{r}_1(q_1,q_2,q_3)$$ $$...$$ $$\mathbf{r}_N = \mathbf{r}_N(q_{3N-2},q_{3N-1},q_{3N})$$ Using the first constraint to eliminate $q_1$ gives $\mathbf{r}_1= \mathbf{g}_1(q_2,q_3,..,q_{3N},t)$ which is of the same form as one of the transformation equations you quoted. Which $q$ can be eliminated depends upon the constraints of the problem and so in general, the transformation equations are of the form $$\mathbf{r}_i= \mathbf{r}_i(q_1,q_2,..,q_{3N-k},t)$$&lt;/p&gt;&#xA;" OwnerUserId="1345" LastActivityDate="2011-06-27T16:46:43.600" CommentCount="0" />
  <row Id="11627" PostTypeId="2" ParentId="8782" CreationDate="2011-06-27T17:26:37.157" Score="-1" ViewCount="" Body="&lt;p&gt;fs=-kx this is Hooke's law if you look up Hooke's law you should find what the spring coefficient is as it is part of the Hooke theory&lt;/p&gt;&#xA;" OwnerUserId="4204" LastActivityDate="2011-06-27T17:26:37.157" CommentCount="0" />
  <row Id="11628" PostTypeId="2" ParentId="9150" CreationDate="2011-06-27T17:26:57.200" Score="1" ViewCount="" Body="&lt;p&gt;Yes the fish will swim normally.  Most people who have fish in an aquarium would agree.  Would those fish lose the ability to swim if I lowered plate glass onto the water surface to make a totally encapulated swimming space?  No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to look at it:  If you had a totally encased mixing bowl, which was totally filled with water, and manufactured so as to have the mixing blades (inside the bowl) driven by shafts which passed through hermetically sealed openings in the lid of the bowl to the electric blender motor.  When you turn the blender on, do the blades move?  you bethcha, they do.  So do fish fins.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-06-27T17:26:57.200" CommentCount="3" />
  <row Id="11630" PostTypeId="2" ParentId="9111" CreationDate="2011-06-27T18:07:33.483" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1008.0129&quot; rel=&quot;nofollow&quot;&gt;This paper by Borcherds&lt;/a&gt; offers a dense treatment of perturbative renormalization in an altered version of Wightman's axioms.  There is no failure of linearity in the test functions anywhere - see especially the bottom of page 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the problem in your reasoning is your main assumption that test functions define absolute energy scales.  At least in the context of Borcherds's treatment, Feynman measures are essentially made from the data of energy cutoffs (which form an infinite dimensional space of possibilities), and the Wightman distributions are constructed from a choice of a Lagrangian and a Feynman measure, so the energy scale is an input to the distribution.  One may then probe the distributions with test functions, but these test functions are not connected to the energy scale that you chose when constructing the distribution.  In fact, there is an infinite dimensional ultraviolet group (also called the group of renormalizations) that acts both on the space of Lagrangians and on the space of Feynman measures, such that the resulting Wightman distributions are fixed.  This group does not act on the space of test functions.&lt;/p&gt;&#xA;" OwnerUserId="761" LastActivityDate="2011-06-27T18:07:33.483" CommentCount="1" />
  <row Id="11632" PostTypeId="2" ParentId="11608" CreationDate="2011-06-27T18:18:35.410" Score="3" ViewCount="" Body="&lt;p&gt;Any lightning protection system which includes a lightning rod will have the lightning rod galvanically connected to ground.  Usually this is done with very heavy gauge copper or aluminum wire which goes directly to a buried metal rod (typically copper).  But some designs use the steel in steel-girder construction buildings as the path to ground.  Some U.S. government/military buildings have many grounding points (reference MIL-HDBK-419, Volumes I and II).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lightning rod, whether grounded or ungrounded, which is higher than the rest of the building, will attract more lightning to that building.  A lightning rod which is otherwise correctly installed, but which is ungrounded, serves the purpose of attracting more lightning to an unprotected building.  The lightning ~WILL~ find a path from the rod to ground.  And, since a proper path has not been provided, the path found will almost certainly be through places in the building which you don't want it to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You ask how lightning passes through the building to ground.  That is a huge questions with many answers.  You might have better asked how it crosses the vast expanse of sky to get to the rod in the first place.  That, believe it or not, is the easier question to answer.  The practical answer to your question is that lighning does a lot of very wierd stuff.  It does NOT necessarily follow the path of least resistance.  I'd be guessing, but I'd like to say it follows the path of least impedance.  In any case, if you do not give it a low impedance path to ground, it will find the next lowest... even if the next lowest is not low at all.  That's where you find stories of lightning passing through doors, windows, bathtubs, plumbing, people, pets, mattresses, wooden studs, concrete, and the like.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-06-27T18:32:07.300" LastActivityDate="2011-06-27T18:32:07.300" />
  <row Id="11633" PostTypeId="1" AcceptedAnswerId="11655" CreationDate="2011-06-27T18:26:42.703" Score="3" ViewCount="153" Body="&lt;p&gt;Physics me this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;equivalence principle&lt;/strong&gt; has rigorous physical definitions that say, for one, that the laws of physics are the same in all inertial reference frames.  This is to say that &lt;em&gt;the behavior of the universe&lt;/em&gt; is the same for any system regardless of velocity $(\vec{v})$ and position $(\vec{r})$ according to special relativity.  General relativity even establishes some sorts of parallels for acceleration $(\vec{a})$ and spacetime curvature, although it still leaves some reference frames with an obviously more complicated geometry of spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;Copernican principle&lt;/strong&gt; is the &quot;notion&quot; or &quot;philosophy&quot; that neither humans or any given group examining the universe has a privileged view.  The idea has been generalized to extend from the original intention applying to the Earth to extend inward as far as a human individual, or outward as far as our local group of galaxies.  At the largest scale we have found the distribution and characteristic of matter throughout the universe to be roughly homogenous, as sort of the ultimate example of this principle, leaving no location or collection of matter privileged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hubble's constant correlates $\vec{r}~$ and $\vec{v}$ between any two points in space, which defines the comoving coordinates, which identifies a point within the constantly ballooning and accelerating mesh of matter that fills the universe.  Without loss of generality I'll say that $\vec{r}$ is truly independent and non-privileged, but $\vec{v}$ has a clearly preferred value, which is that which matches the average flow of matter around $\vec{r}$.  Reflect for a second to note that $\vec{v}$ is a privileged view of the universe &lt;em&gt;even though&lt;/em&gt; it is non-privileged regarding the physical laws of the universe.  Why why why?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect quick dismissals of the &quot;problem&quot; due to the fact that the Copernican principle is not a physical principle and can thus be comfortably wrong.  But it still seems like the implications would be non-trivial and would keep physicists up at night.  Ultimately if we truly find the &quot;Grant Unified Theory&quot; it should fully allow the Lorentz Invariance AND provide some way for the big bang to create matter that is roughly stationary to other (close-by) matter.  If not, why didn't the big bang create matter with $\vec{v}=-c\dots c$ and momentum spanning $\vec{p}=\infty \dots \infty $ for a given $\vec{r}$?  That would obviously be non-nonsensical as collisions would be releasing infinite energy.  But wouldn't it be more consistent with the Copernican principle while at the same time being allowable by consistent physical laws to have galaxies flying by at $0.99999c$, allowed by some kind of reduced probability of interaction?  Then matter would could uniformly occupy both $\vec{v}$ and $\vec{r}$.  I have never heard this even mentioned as an issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why did the big bang create matter along the tapestry of the comoving coordinates?  Can any physical theories be said to predict or deny this?&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="2307" LastEditorDisplayName="" LastEditDate="2011-06-27T19:29:58.763" LastActivityDate="2011-06-28T03:44:13.633" Title="How are the comoving coordinates NOT a prefered reference frame?" Tags="&lt;equivalence-principle&gt;&lt;general-relativity&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="11634" PostTypeId="2" ParentId="11633" CreationDate="2011-06-27T19:10:16.523" Score="1" ViewCount="" Body="&lt;p&gt;Cosmological expansion is entirely &quot;Copernican&quot; in this sense: any observers in any far away galaxy will see our galaxy moving at an opposite $\vec{v}$ approximately symmetric (not exact because the model and real universe have inhomogeinity discrepancies) but at a cosmological scale the law is fully &lt;em&gt;covariant&lt;/em&gt; in different points of space&lt;/p&gt;&#xA;&#xA;&lt;p&gt;any rough variations in a presumably entirely covariant momentum density of matter near the planck era should have been able to thermalize after the inflation stage ended, Although we certainly don't know how was matter and energy distributed near this planck era&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i hope i'm clearing the confusion, or maybe i misunderstood your question?&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-06-27T19:10:16.523" CommentCount="1" />
  <row Id="11635" PostTypeId="2" ParentId="11633" CreationDate="2011-06-27T19:29:06.080" Score="1" ViewCount="" Body="&lt;p&gt;The universe of today is not at all like the universe just after the big bang. What was actually &quot;created by&quot; the big bang (or rather, what our current physical theories predict existed very shortly after the big bang) was a plasma of elementary particles at an exceedingly high temperature (billions, trillions, etc. of Kelvins depending on how far back you go). This high-temperature sea of particles would have filled phase space, just as you're suggesting: the particles would be in random positions undergoing random motions, and many of them would indeed be passing each other at high speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as space expanded, the particles spread out and the energy density of the universe decreased. Once the energy density dropped enough, the particles started interacting, first via the strong force, then the electromagnetic force, then gravity. These interactions, which are necessary for the formation of large structures like galaxies, tend to slow down the high-velocity massive particles and transfer their energy to photons (the CMB) and gravitons. To put it another way, it's basically impossible to put together the number of particles required to form something like a galaxy (or a cluster, or a universe) if they're all passing each other at high speed.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-27T19:29:06.080" CommentCount="1" />
  <row Id="11636" PostTypeId="1" CreationDate="2011-06-27T19:55:35.537" Score="8" ViewCount="116" Body="&lt;p&gt;The source of Sun's incessant energy is hydrogen; which is continuously converting to helium through nuclear fusion reaction releasing energy. Why does not all hydrogen convert into helium in one big explosion just like a man-made bomb or cracker explodes?&lt;/p&gt;&#xA;" OwnerUserId="3636" LastEditorUserId="3636" LastEditorDisplayName="" LastEditDate="2011-06-29T00:21:36.823" LastActivityDate="2011-06-29T00:21:36.823" Title="Hydrogen as a fuel in Sun" Tags="&lt;nuclear-physics&gt;&lt;sun&gt;&lt;fusion&gt;" AnswerCount="1" />
  <row Id="11637" PostTypeId="1" CreationDate="2011-06-27T19:58:40.387" Score="2" ViewCount="106" Body="&lt;p&gt;I am conducting an experiment that requires comparison to models for a water droplet's change in size due to evaporation.  I've found plenty of papers, but they all seem to require measurement of the drop's radius and/or height.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the purposes of the experiment, I need a rate of change of the drops volume - $ \frac{dV}{dt} $ - related to temperature, humidity, local pressure, and surface tension.  The contact angle starts at $ \approx 90 $ and a volume less than 60$\mu L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of such a paper?  If so, please post the citation and/or a link.  I suspect that the lack of papers would be due to the great age of the discovery, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks   &lt;/p&gt;&#xA;" OwnerUserId="1037" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-27T20:26:46.363" LastActivityDate="2011-07-01T20:51:37.600" Title="Water droplet evaporation due to humidity and temperature" Tags="&lt;fluid-dynamics&gt;&lt;water&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11638" PostTypeId="1" CreationDate="2011-06-27T20:04:56.597" Score="2" ViewCount="121" Body="&lt;p&gt;Shouldn't it be possible for an incoming photon to excite one of the 1s electrons to a 2p state (or one of even higher energy) and then for the excited electron to drop back to 1s and kick out the 2s electron?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some fundamental reason that prohibits this process, or is it just very unlikely, or is it not called &quot;Auger process&quot; because the first electron is excited instead of kicked out?&lt;/p&gt;&#xA;" OwnerUserId="4206" LastEditorUserId="4206" LastEditorDisplayName="" LastEditDate="2011-06-30T21:57:16.993" LastActivityDate="2011-08-31T11:37:36.003" Title="Why can free lithium atoms not take part in an Auger process?" Tags="&lt;spectroscopy&gt;&lt;atoms&gt;&lt;electron&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="11640" PostTypeId="2" ParentId="11633" CreationDate="2011-06-27T20:52:18.213" Score="0" ViewCount="" Body="&lt;p&gt;In order to gain more specificity in this discussion, I want to entertain a certain case, and hoping that it may provide &quot;part&quot; of the answer, I will put it here.  Let's say we are a particle in the electroweak epoch, where temperatures were $10^{28} K$.  You might have guessed it, but this would put particle speeds close to the speed of light.  But how close?  I will try that here.  Firstly, by the simple definition of temperature we find the following for kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$KE=\frac{3}{2} k T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a high $T$ we will have relativistic speeds and it will be sufficient to set the above value, found for the electroweak epoch, to just $E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E\approx KE = 207,000 J$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, these are units of Joules... for a single particle.  We are pretty thoroughly relativistic.  Now let's find velocity, assuming it's an electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E^2=(pc)^2+(m c^2)^2$$&#xA;$$p=\gamma \beta c m$$&#xA;$$\beta = \frac{ \sqrt{ E^2-(m c^2)^2}}{E}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For highly relativistic particles you will find:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\beta \approx 1- \frac{(m c^2)^2}{2 E^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;numerically, I can only write this slightly indirectly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\beta = 1 - 10^{-37}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this is only to demonstrate that the particles in the high-energy soup of the early universe were going so fast that they were effectively indistinguishable from the speed of light, although small differences in speed make a large difference in the energetics of a collision at this speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the funny thing about this is that the early universe actually fits rather well my concept of matter traveling at almost all possible speeds at all locations (almost).  In such a early universe plasma, it would be difficult to distinguish which particle was closest to the net-zero momentum vector.  The greater philosophical question, however, seems to be what dynamics could possibly create such a balancing of momentum.  Could the universe somehow figure out a way to create a universe where no $\vec{p}$ was preferred, and then later things balance out to make it so?  If we consider the universe coming from a true singularity, then it maybe it could be that originally all possible $\vec{p}$ existed and this then evolved into all possible $\vec{r}$ existing.  I have just as much trouble imaging that as I do imagining an infinite universe in the first place.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-06-27T20:52:18.213" CommentCount="1" />
  <row Id="11641" PostTypeId="2" ParentId="11636" CreationDate="2011-06-27T20:53:07.323" Score="4" ViewCount="" Body="&lt;p&gt;The reaction rate doesn't increase that quickly with temperature, but pressure does. If you perturb a solar model, making a zone near the core marginally hotter, the increased pressure will rapidly (at roughly the soundspeed divided by a characteristic length) cause it to expand. That lowers the pressure and temperature enough to substantially quench the reaction. So for normal stars during the long early phases of their evolution, the stability of thermonuclear burning is actually very high. There are some cases in stellar eveolution where pressure becomes insensitive to temperature, generally when the density is so high that electrons average kinetic energy is (almost) independent of temperature. In which case, fusion reactions can accelerate dramatically, but in most cases once the temperature is high enough to increase the pressure, the affected parts of the star rapidly expand and quench the reaction. When a sunlike star intitiates Helium fusion such a runaway reaction referred to as a Helium flash ocurrs, but it is easily contained. On the other hand, when a white dwarf accumulates a critical mass of material to intiate carbon burning, the reaction isn't contained and the star is destroyed in a class 1-a supernova. As dmkee points out, this differs from the core collapse supernova of massive stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A useful thing to remember is the virial theorem. If the star (or a portion of it) expands because of an increase in temperature, then once the system reaches hydrodynamic equilibrium the temperature must decrease, as the virial theorem determines that kinetic energy is proportional to gravitational binding energy. Thus unless the reaction=&gt;heating=&gt;reaction feedback is faster than the time to reach hydrostatic equilibrium, the star should be stable.&lt;/p&gt;&#xA;" OwnerUserId="377" LastEditorUserId="377" LastEditorDisplayName="" LastEditDate="2011-06-28T21:46:53.117" LastActivityDate="2011-06-28T21:46:53.117" CommentCount="3" />
  <row Id="11642" PostTypeId="2" ParentId="11618" CreationDate="2011-06-27T20:56:12.310" Score="0" ViewCount="" Body="&lt;p&gt;Well there's no particular reason for a textbook problem to actually model a physical system... But one can certainly write something like this as a completely valid approximation. Take Flugge's example, with He3 so its fermionic[fn.2]. Say the size of the atoms is very small, much smaller than the scales on which the ground state wavefunction varies, which is reasonable enough.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there should really be a term $V_{repulse}(x_1-x_2)\psi$ where $V_{repulse}$ gets really big when $|x_1-x_2|\rightarrow 0$ to capture the fact that you can't put the two atoms on top of each other [fn.2]. But this is going to be really short ranged, almost zero if $|x_1-x_2|$ is significantly bigger then the size of the atom. On the other hand we know that $\psi(x_1,x_2)\rightarrow 0$ when $x_1 \rightarrow x_2$. So in precisely the region where $V_{repulse}$ would matter, $\psi$ is basically zero. So we can basically ignore $V_{repulse}\psi$. More exactly, the term is proportional to (size of atom)/(size of circle) squared, which could be very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you don't always have to include a repulsive term. It can actually be quite negligible, even though it seems like a fact you can't ignore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[fn. 1] There are also times when you can ignore the repulsive interactions of bosons, although its not suppressed like the fermions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[fn. 2] Its not really true that it should diverge as $x_1\rightarrow x_2$. If you really got the two atoms on top of each they would stop behaving like pointlike atoms, so your model would stop being applicable, rather than anything going to infinity. &lt;/p&gt;&#xA;" OwnerUserId="3970" LastActivityDate="2011-06-27T20:56:12.310" CommentCount="5" />
  <row Id="11645" PostTypeId="1" CreationDate="2011-06-27T22:36:06.007" Score="1" ViewCount="41" Body="&lt;p&gt;These questions are in reference to this beautiful review article by Minahan - &#xA;&lt;a href=&quot;http://arxiv.org/pdf/1012.3983v2&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/1012.3983v2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I gained a lot by reading some of its sections but not everything is clear to me.&#xA;I would like to ask a few questions to clarify some of the things in it. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;On page 5 between equation 3.7 and 3.8 it says that the R-charge representation of the $S$ are &quot;reversed&quot; compared to that of the $Q$. What does it exactly mean ? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There it does not explicitly specify the commutation relationship between $M_{\mu \nu}$ and $S^a_\alpha$ and $\bar{S}_{\dot{\alpha}a}$. Should I assume that its similar to that with $Q_{\alpha a}$ and $\bar{Q}^a_\dot{\alpha}$ ? Like if I may think - &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$[M^{\mu \nu},S^a_\alpha] = i \gamma ^{\mu \nu}_{\alpha \beta} \epsilon ^{\beta \gamma} S^a_{\gamma }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[M^{\mu \nu},\bar{S}_{\dot{\alpha} a} ] = i \gamma ^{\mu \nu}_{\dot{\alpha} \dot{\beta}} \epsilon ^{\dot{\beta} \dot{\gamma}} \bar{S}_{\dot{\gamma} a}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Comparing equation 3.12 to 3.9 I see some possible discrepancies. Is there a factor of $\frac{1}{2}$ missing with the term containing $D$  on the RHS of equation 3.12 ? In the same RHS of equation 3.12 in the $M_{\mu \nu}$ term why has the $\gamma ^{\mu \nu}$ of equation 3.9 become $\sigma ^{\mu \nu}$ ? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the statement just below equation 3.14 it says &quot;Hence a primary operator with R-charges $(J_1,0,0)$ is annihilated by $Q_{\alpha 1}$ and $Q_{\alpha 2}$ if $\Delta = J_1$&quot;...Is this a consistency statement? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From this how does it follow that thesame operator is also annihilated by $\bar{Q}^3_{\dot{\alpha}}$ and $\bar{Q}^4_{\dot{\alpha}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same strain can one also say that $\Delta = -J_1$ is consistent with the primary operator being annihilated by $Q^3_\alpha$ and $Q^4 _ \alpha$ ? &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I guess the above conclusions follow from taking different values of $a$ and $b$ in the equation 3.13. But one would get an extra factor of $\frac{1}{2}$ on the RHS of 3.13 if one puts in the factor of $\frac{1}{2}$ with the term containing $D$ on the RHS of equation 3.12 (..as I think it should be ..)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On page 8 one creates bispinors $F_{+\alpha \beta}$ and $F_{-\dot{\alpha} \dot{\beta}}$ out of $F_{\mu \nu}$. I would like to know what is the intuition/motivation/reason for doing this ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is the bispinor version of $F$ still have the meaning of a field strength ? If so then how does it relate to the bispinor version (?) $D_{\alpha \dot{\beta}}$ of the covariant derivative ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I have already put in too many questions for one question. May be I will ask some more about this review in a separate question. &lt;/p&gt;&#xA;" OwnerUserId="1637" LastActivityDate="2011-06-27T22:36:06.007" Title="Some questions about chapter I.1 (by Minahan) of the &quot;Review of AdS/CFT Integrability&quot;" Tags="&lt;superconformality&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="11646" PostTypeId="1" AcceptedAnswerId="11649" CreationDate="2011-06-27T23:28:25.437" Score="4" ViewCount="181" Body="&lt;p&gt;I read that thermodynamic entropy is a measure of the number of microenergy states. What is the derivation for $S=k\log N$, where $k$ is Boltzmann constant, $N$ number of microenergy states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is the logarithmic measure justified?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does thermodynamic entropy have anything to do with information entropy (defined by Shannon) used in information theory?&lt;/p&gt;&#xA;" OwnerUserId="3714" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-06-29T09:49:41.920" LastActivityDate="2011-06-29T09:49:41.920" Title="How is thermodynamic entropy defined? What is its relationship to information entropy?" Tags="&lt;thermodynamics&gt;&lt;entropy&gt;&lt;information&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="11647" PostTypeId="1" CreationDate="2011-06-27T23:30:51.377" Score="0" ViewCount="188" Body="&lt;p&gt;Can someone compare the energy efficiency of human brain as a computer ? What is the energy in joules / flop ? may be some reasonable assumptions on the computational load of common tasks such as pattern recognition or speech synthesis can be used.&lt;/p&gt;&#xA;" OwnerUserId="3714" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-28T01:39:26.527" LastActivityDate="2011-07-28T15:20:27.273" Title="how does human brain compare to a modern CPU in energy per bit?" Tags="&lt;thermodynamics&gt;&lt;energy&gt;&lt;computation&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="11648" PostTypeId="1" AcceptedAnswerId="11650" CreationDate="2011-06-27T23:44:29.717" Score="1" ViewCount="89" Body="&lt;p&gt;Is it right that all units in physics can be defined in terms of only mass, length and time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it so? Is there some principle that explains it or is it just observational fact?&lt;/p&gt;&#xA;" OwnerUserId="1708" LastActivityDate="2011-06-28T00:41:53.527" Title="Fundamental units" Tags="&lt;si-units&gt;&lt;unit-conversion&gt;&lt;units&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="11649" PostTypeId="2" ParentId="11646" CreationDate="2011-06-28T00:24:00.617" Score="7" ViewCount="" Body="&lt;p&gt;I think that the best way to justify the logarithm is that you want entropy to be an extensive quantity -- that is, if you have two non-interacting systems A and B, you want the entropy of the combined system to be &#xA;$$&#xD;&#xA;S_{AB}=S_A+S_B.&#xD;&#xA;$$&#xA;If the two systems have $N_A,N_B$ states each, then the combined system has $N_AN_B$ states. So to get additivity in the entropy, you need to take the log.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might wonder why it's so important that the entropy be extensive (i.e., additive). That's partly just history. Before people had worked out the microscopic basis for entropy, they'd worked out a lot of the theory on macroscopic thermodynamic grounds alone, and the quantity that they'd defined as entropy was additive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the number of states available to a macroscopic system tends to be absurdly, exponentially large, so if you don't take logarithms it's very inconvenient: who wants to be constantly dealing with numbers like $10^{10^{20}}$?&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-06-28T00:24:00.617" CommentCount="3" />
  <row Id="11650" PostTypeId="2" ParentId="11648" CreationDate="2011-06-28T00:40:00.870" Score="6" ViewCount="" Body="&lt;p&gt;Which units are fundamental and which are derived is pretty much a matter of arbitrary convention, not an objective fact about the world. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might think that the &lt;em&gt;number&lt;/em&gt; of fundamental units would be well-defined, but even that's not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take electric charge for example. In the SI system of units (i.e., the &quot;standard&quot; metric system), charge cannot be expressed in terms of mass, length, and time: you need another independent unit. (In the SI, that unit happens to be the Ampere; the unit of charge is defined to be an Ampere-second.) But sometimes people use different systems of units in which charge can be expressed in terms of mass, length, and time. By decreeing that the proportionality constant in Coulomb's Law be equal to 1,&#xA;$$&#xD;&#xA;F={q_1q_2\over r^2},&#xD;&#xA;$$&#xA;you can define a unit of charge to be (if I've done the algebra right) $(ML^3/T^2)^{1/2}$,&#xA;where $M,L,T$ are your units of mass, length, time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether charge is defined in terms of mass, length, time, or whether it's an independent unit, is a matter of convenience, not a fact about the world. People can and do make different choices about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, some people choose to get by with &lt;em&gt;fewer&lt;/em&gt; independent units than the three you mention. The most common choice is to decree that length and time have the same units, using the speed of light as a conversion factor. You can even go all the way down to &lt;em&gt;zero&lt;/em&gt; independent units, by working in what are often called &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck_units&quot; rel=&quot;nofollow&quot;&gt;Planck units&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, you can dial up or down the number of &quot;independent&quot; units in your system at will. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more example, which seems silly at first but is actually of some historical interest. You can imagine using different, independent units of measure for horizontal and vertical distances. That'd be terribly inconvenient for doing physics, but for many applications it's actually quite convenient. (In aviation, altitudes are often measured in feet, while horizontal distances are measured in miles. In seafaring, leagues are horizontal and fathoms are vertical. &lt;a href=&quot;http://blog.richmond.edu/physicsbunn/2008/03/26/a-yard-of-snow/&quot; rel=&quot;nofollow&quot;&gt;Yards are pretty much always used for horizontal distance&lt;/a&gt;.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds absurd to think of using different units for different directions, but in the context of special relativity, using different units for space and time (different directions in spacetime) is sort of similar. If we had evolved in a world in which we were constantly zipping around near light speed, so that special relativity was intuitive to us, we'd probably think that it was obvious that distance and time &quot;really&quot; came in the same units.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-06-28T00:40:00.870" CommentCount="2" />
  <row Id="11651" PostTypeId="2" ParentId="11648" CreationDate="2011-06-28T00:41:53.527" Score="0" ViewCount="" Body="&lt;p&gt;It is mostly by convention, since any unit could be derived from any three (independent) units. Typically we like our base units to be measurable in some sense. Distance, Time and Mass are perfectly reasonable, but not required. You can make up a unit system based on Energy, Acceleration and Density to derive all the other units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. Look up dimensional analysis to figure out how to related together different units.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-06-28T00:41:53.527" />
  <row Id="11652" PostTypeId="1" CreationDate="2011-06-28T02:19:54.827" Score="-1" ViewCount="241" Body="&lt;p&gt;Andromeda is made of antimatter. Am I wrong? Why?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course I do not know that Andromeda is made of antimatter.&lt;br&gt;&#xA;&lt;strong&gt;_&lt;em&gt;but&lt;/em&gt;____&lt;/strong&gt; I do not know that Andromeda is made of matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know what is the correct sentence?  Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT&lt;br&gt;&#xA;I used Andromeda as a substitute for any non-local galaxy.&lt;br&gt;&#xA;Lets put some geometry (due to the first two comments):&lt;br&gt;&#xA;Antimatter galaxies (AMGs) have in general AMGs in the neighborough. MGs only MGs.&lt;br&gt;&#xA;For instance suppose the hurricanes in the northern hemisphere rotate X way and in the southern rotate anti-X way, and they dont meet ;).&lt;br&gt;&#xA;That is to say, it is not mandatory that AMGs colide at all with MGs.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that we have no way to decide if the light we receive is originated in an anti-atom or from an atom.  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-06-28T10:21:22.187" LastActivityDate="2011-06-28T10:22:59.070" Title="Andromeda is made of antimatter. Am I wrong? Why?" Tags="&lt;antimatter&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="11653" PostTypeId="1" CreationDate="2011-06-28T02:35:03.593" Score="3" ViewCount="55" Body="&lt;p&gt;I'm wondering if there's any simple way to define and solve for a null cone for a general spacetime geometry in $n+1$ dimensions, given its vertex $p^\mu$. I can't seem to find a simple way to do it that doesn't involve solving for null geodesics. It seems like it's something that should be in a lot of GR textbooks, but I can't seem to find a simple alternative (perhaps I'm not looking hard enough, but that's why I'm asking).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, the best I can do is to define it as the $n$-surface formed by the family of null geodesics $y_{\bf\Omega}^\mu$ passing through some spacetime point $p^\mu$, where $\bf\Omega$ is a combination of $n-1$ angles that distinguishes the curves (at $p^\mu$, it's the collection of angles specifying the direction of the null curve).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, I'm interested in doing a foliation of spacetime using future or past null cones with respect to some timelike curve $\gamma$, and would like to define the null cones in the form of a constraint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{\gamma}(x^\mu)=\tau$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\tau$ is the proper time along the timelike curve $\gamma$. This constraint satisfies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U(C_{\gamma})=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $U$ is the vector field defined using the tangent vectors to $y_{\bf\Omega}^\mu$ (this can be defined for all spacetime, if you take the union of all the null cones along $\gamma$). In any case, I still have to solve for the null geodesics, given what I have so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a fan of solving for all the null geodesics at every point along the curve, so I'm looking for a simpler way to do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the feeling that there should be a simpler way of defining and solving for null cones--it seems like they've been extensively studied in GR. Is there any way to do this without resorting to solving for the null geodesics? In particular, has anyone written down a general constraint form for the null cone?&lt;/p&gt;&#xA;" OwnerUserId="4185" LastEditorUserId="4185" LastEditorDisplayName="" LastEditDate="2011-06-28T02:40:11.097" LastActivityDate="2011-06-28T11:59:24.677" Title="Is there a simple way to define/solve for a null cone for a general spacetime geometry?" Tags="&lt;general-relativity&gt;&lt;differential-geometry&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="11654" PostTypeId="2" ParentId="11647" CreationDate="2011-06-28T02:49:54.297" Score="2" ViewCount="" Body="&lt;p&gt;Human power consumption &lt;a href=&quot;http://hypertextbook.com/facts/2001/JacquelineLing.shtml&quot; rel=&quot;nofollow&quot;&gt;can be guesstimated&lt;/a&gt; as 100W, similar to the power consumption of an ordinary computer, plus or minus a few orders of magnitude depending on one's idea of &quot;ordinary&quot;.  A computer can do billions of flops per second, and it would take me many seconds or minutes to perform one with pen and paper, and furthermore I will make many more errors.  If we assume that there is some other task which is stacked the opposite way, i.e. a human can perform it a billion times faster than a computer, and that both of these are in some sense extreme cases, then given some more &quot;fair&quot; test we can say that ratio of the efficiency is probably somewhere between $10^{-9}$ and $10^9$.&lt;/p&gt;&#xA;" OwnerUserId="2072" LastEditorUserId="2072" LastEditorDisplayName="" LastEditDate="2011-06-28T04:34:42.820" LastActivityDate="2011-06-28T04:34:42.820" CommentCount="5" />
  <row Id="11655" PostTypeId="2" ParentId="11633" CreationDate="2011-06-28T03:44:13.633" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Comoving coords are relative to a universe were matter shrinks.&lt;/strong&gt;&lt;br&gt;&#xA;Any measure we do is a ratio between a quantity and a chosen standard.&lt;br&gt;&#xA;Any measure is a dimensionless quantity. &#xA;If we measure an expansion it may signal that the unit of measure is shortening.&lt;br&gt;&#xA;All system of units we use are &lt;strong&gt;atomic units&lt;/strong&gt;.&lt;br&gt;&#xA;But we can not only squeeze the atom length, also its mass has to decrease as time goes by. Also the time unit vary with atom size, provided that c, G, $\varepsilon$, properties of space (and fields) are really constants.&lt;br&gt;&#xA;Now we have to decide: is space expanding or matter shrinking?&lt;br&gt;&#xA;It happens that we are anthropocentric and we think that the atom is invariant.&lt;br&gt;&#xA;Do we prefer to admit that 'space' expands or, like I do, that 'matter shrinks' ?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have no basis, a priori, to assert one or the other.&lt;br&gt;&#xA;A study must be done to study how can the laws of physics hold in a system of units, the comoving one, that are not derived from 'our' atomic standards were laws hold.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A preliminary study was done that shows that laws of physics hold good in that scenario:&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/astro-ph/0208365&quot; rel=&quot;nofollow&quot;&gt;A relativistic time variation of matter/space fits both local and cosmic data&lt;/a&gt;&lt;br&gt;&#xA;No one dared to say a single argument (until now) against that novel theory. In my viewpoint there are no arguments to deny the correctness of the study.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the very near future a definitive study will be published, revealing a &lt;em&gt;&lt;strong&gt;scaling universe&lt;/strong&gt;&lt;/em&gt;, self-similar.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But 'prefered' reference frame is strong, like absolute, and physics dont like this notion. In a surprising way no one realized that the notion of an invariant atom is also an absolute reference.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sun and the stars are the beacons (references) of the navigators. Now lets suppose that all the stars from the sky are erased. Are we lost? NO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately there are still a reference available, like no other, common to all observers, which is : &lt;strong&gt;The Referential of the Light, the CMB&lt;/strong&gt;. It is not an Einsteinien    referential because it it not attached to any observer. &#xA;With this referential alone we can say: the Earth is moving at 1 rev/day (sideral, local time, i.e. defined with our atoms) and at 540Km/s (+- see the WP) in direction to Leo constellation. &#xA;Why is this CMB referential so special, so absolute? It is the only one where a released photon propagates isotropically. The CMB can provide a length and time unit common to all observers. Even if we dont like the word absolute, it is a prefered reference frame. I've no doubt.  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-06-28T03:44:13.633" CommentCount="4" />
  <row Id="11656" PostTypeId="1" CreationDate="2011-06-28T04:33:23.007" Score="3" ViewCount="162" Body="&lt;p&gt;According to string landscape theory, our vacua with a cosmological constant of $10^{-123}$ is a metastable vacua which can decay to a supersymmetric vacua with either a zero or negative cosmological constant. However, such a decay is destined to happen sooner or later, but if we wait for it, it will happen at an unpredictable random location and time. Suppose we have some computer memories and computations that we wish to preserve through a bubble nucleation. Is it possible to engineer a controlled bubble nucleation to a supersymmetric vacua with an exactly zero cosmological constant, and construct computers with data storage such that the memories contained therein can survive the bubble transition to an entirely new law of physics unscathed, and the computers can also continue their computations?&lt;/p&gt;&#xA;" OwnerUserId="4211" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-06-28T04:34:45.487" LastActivityDate="2011-08-21T12:31:34.637" Title="Can computers survive bubble nucleations?" Tags="&lt;string-theory&gt;&lt;cosmology&gt;&lt;vacuum&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="11657" PostTypeId="2" ParentId="11652" CreationDate="2011-06-28T05:03:05.903" Score="5" ViewCount="" Body="&lt;p&gt;I believe the current modern theory is that there cannot exist anti-matter galaxies unless anti-matter is discovered to have a repulsive interaction with matter via gravity (as opposed to the normal attractive force), which would have interesting applications to the structure of the universe. However, to my understanding, this is not generally theoretically supported as most models have anti-matter interacting normally, and also if Andromeda was an anti-matter galaxy then it would not be colliding with the Milky Way anytime soon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there were anti-matter galaxies with anti-matter behaving like matter gravitationally then we would occasionally see HUGE explosions across the sky as anti-matter galaxies collide with matter galaxies and anti-matter stars collide with matter stars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just some quick napkin calculations: lets take two solar mass stars, one matter and one anti-matter and see how much energy is released when they annihilate. $2M = 4 \times 10^{30}$ so since $E=mc^2$ we have $E= 3.6 \times 10^{47}$ joules which is about four times the energy released by the most intense gamma ray burst. &#xA;Although the dynamics would be interesting as I wonder how much a repulsion effect annihilations with inter-stellar medium would be, as well as wether or not when the two stars start to annihilate they would &quot;explode&quot; away from each other...&lt;/p&gt;&#xA;" OwnerUserId="3018" LastActivityDate="2011-06-28T05:03:05.903" CommentCount="6" />
  <row Id="11658" PostTypeId="2" ParentId="11656" CreationDate="2011-06-28T05:45:38.367" Score="1" ViewCount="" Body="&lt;p&gt;No-one knows, but see &lt;a href=&quot;http://arxiv.org/abs/1007.5054&quot; rel=&quot;nofollow&quot;&gt;Louis Clavelli on &quot;Susyria&quot;&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1486" LastActivityDate="2011-06-28T05:45:38.367" />
  <row Id="11659" PostTypeId="2" ParentId="5961" CreationDate="2011-06-28T07:51:51.980" Score="0" ViewCount="" Body="&lt;p&gt;The magnetic moment of the sphere equals it's volume times it's average&#xA;magnetization. Now you have to estimate the magnetization, and&#xA;here is where the problem gets hard: it depends on the magnetic&#xA;properties of your material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how a sphere of 100% pure iron would behave, but I can tell&#xA;that the behavior of a sphere of almost-100% pure iron can be hard to&#xA;predict. That's because magnetic properties can be strongly dependent on&#xA;microscopic details of the material, including impurities and&#xA;crystallographic defects. The magnetization can also depend on the&#xA;magnetic history of the sample, e.g. it's orientation relative to the&#xA;Earth's magnetic field last time it was cooled below it's Curie&#xA;temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the microscopic scale, you should have the saturation magnetization&#xA;of iron, which is 1.72 MA/m (2.16 T$/\mu_0$) at room temperature. But&#xA;it's unlikely that the macroscopically averaged magnetization would be&#xA;anywhere near that, unless your iron is instead a very strong iron-based&#xA;magnet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your iron is sufficiently pure, I would expect quite a soft magnetic&#xA;behavior from it. If we approximate it by a perfectly soft material&#xA;(linear with infinite susceptibility), then it's magnetization should be&#xA;3 $H_0$, where the factor 3 comes from the spherical shape and $H_0$ is&#xA;the field externally applied to your sphere, for example the Earth's&#xA;magnetic field ($\mu_0H_0 \approx 50$ µT).&lt;/p&gt;&#xA;" OwnerUserId="4202" LastActivityDate="2011-06-28T07:51:51.980" />
  <row Id="11660" PostTypeId="2" ParentId="11656" CreationDate="2011-06-28T09:10:09.580" Score="1" ViewCount="" Body="&lt;p&gt;OK, once again from &lt;a href=&quot;http://en.wikipedia.org/wiki/False_vacuum&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The possibility that we are living in a false vacuum has never been a cheering one to contemplate. Vacuum decay is the ultimate ecological catastrophe; in the new vacuum there are new constants of nature; after vacuum decay, not only is life as we know it impossible, so is chemistry as we know it. However, one could always draw stoic comfort from the possibility that perhaps in the course of time the new vacuum would sustain, if not life as we know it, at least some structures capable of knowing joy. This possibility has now been eliminated.&#xA;  Sidney Coleman &amp;amp; F. de Luccia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;further:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The possibility that we are living in a false vacuum has been considered. If a bubble of lower energy vacuum were nucleated, it would approach at nearly the speed of light and destroy the Earth instantaneously, without any forewarning. Thus, this vacuum metastability event is a theoretical doomsday event. This was used in a science-fiction story in 1988 by Geoffrey A. Landis,&lt;a href=&quot;http://en.wikipedia.org/wiki/Noah%27s_Ark&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt; in 2000 by Stephen Baxter,[3] and in 2002 by Greg Egan.[4]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In a nutshell, no you cannot build an &lt;a href=&quot;http://en.wikipedia.org/wiki/Noah%27s_Ark&quot; rel=&quot;nofollow&quot;&gt;Ark&lt;/a&gt; to carry a computer because the whole structure and order of physical constructs as we know them will be destroyed, and computers and the structure carrying them are made out of matter as we know it.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-06-28T13:40:45.417" LastActivityDate="2011-06-28T13:40:45.417" />
  <row Id="11661" PostTypeId="1" CreationDate="2011-06-28T09:26:51.943" Score="1" ViewCount="69" Body="&lt;p&gt;The orbit of the earth seems to be very predictable. But as it is a many-body problem having sun, earth, moon, jupiter and so on, is it really that stable or will it start making strange movements sooner or later?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there asteroids or moons in our solar system with obviously chaotic orbits?&lt;/p&gt;&#xA;" OwnerUserId="4212" LastActivityDate="2011-06-28T09:47:34.337" Title="Is the orbit of earth around the sun chaotic?" Tags="&lt;astronomy&gt;&lt;chaos-theory&gt;" AnswerCount="1" />
  <row Id="11662" PostTypeId="2" ParentId="11661" CreationDate="2011-06-28T09:47:34.337" Score="3" ViewCount="" Body="&lt;p&gt;I think that this &lt;a href=&quot;http://en.wikipedia.org/wiki/Stability_of_the_Solar_System&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; pretty much summarizes all of this.&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-06-28T09:47:34.337" CommentCount="1" />
  <row Id="11663" PostTypeId="2" ParentId="11646" CreationDate="2011-06-28T09:48:38.673" Score="4" ViewCount="" Body="&lt;p&gt;The thermodynamic entropy is &lt;em&gt;not&lt;/em&gt; what you wrote. That is the information entropy. Microstates and the counting of are not thermodynamic concepts, but rather statistical mechanics ones. Thermodynamics proceeds by identifying reversible processes, and goes on to study them. A key point is noticing that there are quantities called &lt;em&gt;state variables&lt;/em&gt; which are intrinsic to a substance, which do not depend on the path used to obtain it, such as temperature, or pressure. Heat transfer is explicitly (and experimentally) not one such quantity. However, by considering a perfect reversible heat engine, it is possible to show that the quantity $dQ/T$ is a state variable. Integrating it gives $$\Delta S = \int \frac{dQ}{T}$$ and is the change of entropy. Thermodynamically one can only talk of changes, and the zero is not particularly well defined --- the 3rd law of thermodynamics tries to define it, but it is not always applicable. &lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-06-28T09:48:38.673" />
  <row Id="11664" PostTypeId="1" CreationDate="2011-06-28T10:04:36.090" Score="6" ViewCount="63" Body="&lt;p&gt;I'm currently preparing for an examination of course in introductory (experimental) particle physics. One topic that we covered and that I'm currently revising is the universality in weak interactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I don't really understand the point my professor wants to make here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me show you his exposure to the topic: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We stark by looking at three different weak decays: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\beta$-decay in a nuclei: $^{14}O \rightarrow ^{14}N^{*} + e^{+} + \nu_{e}$ (Lifetime $\tau$=103sec)  &lt;/li&gt;&#xA;&lt;li&gt;$\mu$-decay: $\mu^{+}\rightarrow e^{+} + \nu_{e} + \overline{\nu_{\mu}}$ ($\tau$=2.2$\mu$sec)&lt;/li&gt;&#xA;&lt;li&gt;$\tau-decay$: $\tau^{+}\rightarrow$ $\mu^{+} + \nu_\mu + \overline{\nu_{\tau}}$ or $\tau^{+}\rightarrow$ $e^{+} + \nu_e + \overline{\nu_{\tau}}$ ($\tau = 2 \cdot 10^{13}$ sec).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now he points out that the lifetimes are indeed quiet different. &#xA;Never the less the way the reactions behave is the same? (Why? Okay we always have a lepton decaying into another leptons and neutrinos? But what's the deal?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then he writes down Fermi's golden rule given by: $W=\frac{2\pi}{\hbar} |M_{fi}|^{2} \rho(E&amp;#39;)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now he says that universality means that the matrix element $|M_{fi}|$ is the same in all interactions. The phase space however is the same (Why? First of all I have often read on  the internet that universality means that that certain groups of particles carry the same &quot;weak charge&quot;? And secondly: What do particle physicists mean when they talk about greater phase-space? Three dimensional momentum space? But how do you see or measure that this space is bigger? And bigger in what respect? More momentum states?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now he says that the different phase spaces come from the different lifetimes. He calculates &#xA;$\rho(E&amp;#39;)=\frac{dn}{dE}=\int \frac{d^{3}p_{\nu} d^{3}p_{e}}{dE} = p_{max}^{5} \cdot \int \frac{d^{3}(p_{\nu}/p_{max}) d^{3}(p_{e}/p_{max})}{d(E/p_{max})}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the last integral is supposed to be identical for all decays. &#xA;And $p_{max}$is suppose to be different in all decays . But why? &#xA;And what is the definition of $p_{max}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now he has $\tau = \frac{1}{M}$. So he gets $\tau = const \cdot \frac{1}{|M_{fi}|^{2} p_{max}^5}$. Hence he gets in the ln($\tau$)-ln($p_{max}$) diagram a line with slope -5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now he claims that this &quot;proofs&quot; that  $|M_{fi}|$ is constant in all processes. &#xA;Again why? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please give me some overview and explain to me why he is doing all that stuff? I din't really have much background when it comes to particle physics. So can someone explain it to me in a clear and easy way?&lt;/p&gt;&#xA;" OwnerUserId="4213" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-06-28T13:35:46.043" LastActivityDate="2011-06-28T15:02:54.800" Title="Universality in Weak Interactions" Tags="&lt;particle-physics&gt;&lt;phase-space&gt;&lt;fermis-golden-rule&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="11665" PostTypeId="2" ParentId="11652" CreationDate="2011-06-28T10:14:38.510" Score="-1" ViewCount="" Body="&lt;p&gt;One way of looking at this is thru time. Andromeda is either &quot;made of matter&quot; or &quot;made of antimatter&quot; if we experience it using a direction of time we exhibit on our galaxy. If we experience it in reverse time, it changes from matter to antimatter and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is whether the direction of time is absolute. If it's not, Andromeda is either made of matter or antimatter, depending on the direction of time of the observer.&lt;/p&gt;&#xA;" OwnerUserId="4214" LastEditorUserId="4214" LastEditorDisplayName="" LastEditDate="2011-06-28T10:22:59.070" LastActivityDate="2011-06-28T10:22:59.070" CommentCount="4" />
  <row Id="11666" PostTypeId="2" ParentId="11653" CreationDate="2011-06-28T11:59:24.677" Score="1" ViewCount="" Body="&lt;p&gt;Let me answer the &lt;em&gt;source&lt;/em&gt; of your question: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm interested in doing a foliation of spacetime using future or past null cones with respect to some timelike curve $\gamma$, and would like to define the null cones in the form of a constraint ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;That&lt;/strong&gt; is a priori impossible. In an arbitrary space-time geometry, there can be &lt;a href=&quot;http://en.wikipedia.org/wiki/Conjugate_points&quot; rel=&quot;nofollow&quot;&gt;focal and conjugate points&lt;/a&gt;. Unless you are willing to assume from the start some bounds on the null injectivity radius (to be infinite), you can not in general foliate an entire space-time with null cones (mostly because the cones will likely cease to be cones after some affine parameter). &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;With that said, if you are defining null hypersurfaces, you &lt;em&gt;can&lt;/em&gt; solve instead for the eikonal equation. That is, let $u$ be a parameter describing your null cones (in the sense that $u:M\to \mathbb{R}$ is a real valued function on your manifold, and that its level sets are &quot;null cones&quot;). Then a necessary equation satisfied by the function $u$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ g^{\mu\nu}\partial_\mu u\partial_\nu u = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the difficulty of this approach is to &quot;match&quot; the tip of the cone to be precisely along the curve $\gamma$, but regularity issues like this has to manifest itself somewhere or another. &lt;/p&gt;&#xA;" OwnerUserId="829" LastActivityDate="2011-06-28T11:59:24.677" CommentCount="2" />
  <row Id="11667" PostTypeId="2" ParentId="11646" CreationDate="2011-06-28T12:05:47.857" Score="1" ViewCount="" Body="&lt;p&gt;The reason for the logarithm in the statistical definition of entropy is that it makes the entropy additive and the number of microstates multiplicative. This means that if you have subsystems A &amp;amp; B, total number of microstates is the product of the subsystem’s number of microstates, and the entropy is additive over subsystems due to the properties of the logarithm. A way to see the relation of thermodynamic entropy and statistical entropy is to picture the volume occupied by a system of classical particles as partitioned in cells of very small volume. Then you start counting the number of ways to distribute N particles in the cells. The number of microstates goes like the numbers of cells to the power of the number of particles. The number of cells is the volume divided by the number of partitions. Taking the logarithm of this expression you get number of particles multiplied by the logarithm of the volume over the number of partitions. Multiplying by k you get an expression for the entropy. Then the change of entropy, for a process where volume changes, is obtained by subtracting the final entropy from the initial entropy. The number of partitions is cancelled from the equation by the properties of logarithms. So this is another reason for the logarithm in the expression. What you get finally is the Boltzmann constant multiplied by the number of particles and the logarithm of final volume over initial volume. The way this result is obtained makes clear that the change in entropy is path independent. The same expression for the change in entropy can be obtained, using thermodynamics, for the reversible isothermal expansion of an ideal gas, by dividing the heat absorbed over the temperature. If a process is not isothermal then you can break it in small nearly isothermal infinitesimal paths, then integrate the change in entropy. This is the easiest way to answer your question. Similar arguments can be used for quantum systems but they are more complex.     &lt;/p&gt;&#xA;" OwnerUserId="4173" LastActivityDate="2011-06-28T12:05:47.857" />
  <row Id="11668" PostTypeId="2" ParentId="2625" CreationDate="2011-06-28T12:14:41.393" Score="1" ViewCount="" Body="&lt;p&gt;Thermodynamics is the study of thermal and mechanical properties of equilibrium states of macroscopic systems, using the empirical laws of thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statistical mechanics uses the laws of classical mechanics or the postulates of quantum mechanics, and the principles of statistics to reproduce the thermodynamical properties of macroscopic systems from the microscopical constituents. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statistical mechanics aims to obtaining expresions for state functions of the systems from microscopic physics.&lt;/p&gt;&#xA;" OwnerUserId="4173" LastActivityDate="2011-06-28T12:14:41.393" />
  <row Id="11669" PostTypeId="2" ParentId="10639" CreationDate="2011-06-28T12:52:09.317" Score="2" ViewCount="" Body="&lt;p&gt;Did you ever wondered why momentum (a dynamical quantity that involves mass and velocity) is conjugated to position and energy is conjugated to time? When you were taught in mathematics that space is tridimensional, did you ever wondered how empty space have to be tridimensional if it contains nothing! If empty space do not contains anything why not describe it by, let's say 10 dimensions. To answer your question you first must understand the difference between physical space and mathematical space. Mathematical space is abstract; it exists but just in our minds. To detect space and measure is properties you have to use signals. The propagation of those signals is what defines space. The paths taken by this signals can curve so the signals again are defining space. Without the signal there is no space. Matter also occupies space. But if you take matter away from space there is nothing left. So matter is not occupying space, matter is space! That is why you were taught in mathematics that space is tridimensional, because you professor was really thinking latter to put surfaces and volumes in this artificial construction that he called space. Space can curve but only in the presence of energy and mass. This means that you can't separate them you cannot separate time and space from matter and energy. You need time because you have to parameterize the trajectories of the signals that define space. There is no space nor there time if you take matter and energy away. They are linked as one and the same. If string theory, or it's latter incarnation M-Theory, is correct the world volume of p-branes is space and time as we know it, there is nothing outside matter and energy. So to answer your question if you remove every particle, you are removing space and time because they are the same!&lt;/p&gt;&#xA;" OwnerUserId="4173" LastActivityDate="2011-06-28T12:52:09.317" />
  <row Id="11670" PostTypeId="1" AcceptedAnswerId="12010" CreationDate="2011-06-28T12:57:49.573" Score="1" ViewCount="74" Body="&lt;p&gt;This is an extension of my previous question &lt;a href=&quot;http://physics.stackexchange.com/questions/11588/work-of-marie-curie&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Marie Curie isolated radium in 1903, which paved the way for the development of the theory of radioactivity. In regards to the techniques she used however, what was the impact of the establishment of those techniques? Were the techniques used by Curie new? Are they used today? What impact did they have on modern physics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can find a lot of information online about the actual isolation of radium, but not regarding the significance of the techniques used.&lt;/p&gt;&#xA;" OwnerUserId="3044" LastActivityDate="2011-08-15T03:33:03.757" Title="Significance of isolation of radium?" Tags="&lt;radioactivity&gt;&lt;history&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11671" PostTypeId="2" ParentId="11664" CreationDate="2011-06-28T13:34:59.010" Score="5" ViewCount="" Body="&lt;p&gt;The phase-space represents the &quot;number&quot; of allowed final states (think of a discrete quantum system with degenerate final states, except that here we have a continuum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More final states makes the transition more likely to happen and thus gives it a shorter lifetime.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Each of the reactions that you show has a three body final state and a single body initial state. Without loss of generality, we consider the system in the rest frame of the initial particle and write it's energy and momentum as $\mathbf{\mathcal{P}} = (\mathcal{M},\vec{0})$. For each of the products write $\mathbf{p}_i = (E_i,\vec{p}_i)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves us 12 unknown parameters at this point, but each product particle has a definite mass, which puts three constraints on the system, and energy and momentum are conserved putting another four constrains on the system. Only some of these may be redundant (in particular one of the masses is fixed by the other concerns).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we visualize the parameter space?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Assume (again, without loss of generality) that the particle indexed 1 will have it's momentum parallel with $+\hat{z}$. So the first parameter is $p_1 = \vec{p}_1 \cdot \hat{z}$. This is the choice of a kinetic energy (1 parameter).&lt;/li&gt;&#xA;&lt;li&gt;Each of the other two particles must have $\vec{p}_2 = (p_\perp, p_\|)_2$. This is equivalent to choosing a angle with respect to $\hat{z}$ and a kinetic energy (2 parameters).&lt;/li&gt;&#xA;&lt;li&gt;The final particle's energy and momentum are fixed by the choices we have made so far.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We have a three parameter phase space $\langle p_{1,z}, p_{2,\perp}, p_{2,\|} \rangle$ (and a three fold ambiguity of orientation for the whole system) and these parameters are linked by the necessity of conserving energy and momentum (which you represent in the integrals as a bunch of Dirac $\delta$ functions), so they map out an allowed volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the important part: &lt;em&gt;the size of that volume depends in large measure on the energy excess represented by $\mathcal{M} - \sum_i m_i$,&lt;/em&gt; and on the relative masses of the products.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I one wrote a Mote Carlo generator for a reaction with un-measured branching ratios using this kind of phase-space calculation to compute the relative likelihood of each final state. It got branching ratios in as good agreement with the data we eventually took as those based on a many-body calculation I found.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-06-28T15:02:54.800" LastActivityDate="2011-06-28T15:02:54.800" CommentCount="1" />
  <row Id="11672" PostTypeId="2" ParentId="11078" CreationDate="2011-06-28T13:41:43.117" Score="0" ViewCount="" Body="&lt;p&gt;When a mass is accelerated unruh radiation is produced. An accelerated particle moves in rindler space so boundary conditions due to the presence of an horizon causes the instability of the vacuum that produces a thermal atmosphere of particles that radiates. Also the accelerating particle produce gravitons. The very nature of inertia, and therefore mass, is veiled in mystery. The Higgs fiels is atributed with the property that it gives mass to particles. The Higgs mechanism is nothing that a well-defined mathematical procedure to give mass to Yang Mills fields while preserving renormalizability. But what is the origin of the Higgs field? Is it gravitational? Is it an artificial construction? If the Unruh radiation scenario is correct when you accelerate a mass you have to subs tract the back reaction due to the emission of gravitons and unruh rad. In response to your answer, I would say that Newtons Law is actually and approximate law of nature that cannot be used rigorously to define mass anymore. The problem is that QFT, String Theory, or M-Theory cannot be used either because there are not universally accepted. The nature of the Higgs boson is unknown, and unruh radiation or gravitons have not been detected. In this sense, the meaning of mass is still an open question.&lt;/p&gt;&#xA;" OwnerUserId="4173" LastActivityDate="2011-06-28T13:41:43.117" CommentCount="3" />
  <row Id="11673" PostTypeId="1" CreationDate="2011-06-28T14:03:23.323" Score="4" ViewCount="129" Body="&lt;p&gt;Given an array of charge for a given area (2D or 3D), what algorithm would describe the path that lightning takes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example algorithm would be from the highest charge of the cloud, find the lowest charge adjacently below and move there. Repeat until ground reached.&lt;/p&gt;&#xA;" OwnerUserId="4217" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-05T21:02:57.150" LastActivityDate="2011-07-05T21:02:57.150" Title="Algorithm of Lightning Strikes?" Tags="&lt;electricity&gt;&lt;electrostatics&gt;&lt;lightning&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11674" PostTypeId="1" CreationDate="2011-06-28T14:07:16.957" Score="2" ViewCount="101" Body="&lt;p&gt;How flux compactification solves the moduli space problem in string theory? Please provide some details and, if posible, an example. Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="4173" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-06-28T17:35:13.253" LastActivityDate="2011-06-28T18:27:19.047" Title="Flux compactification" Tags="&lt;string-theory&gt;&lt;string&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="11675" PostTypeId="1" AcceptedAnswerId="11703" CreationDate="2011-06-28T15:46:34.327" Score="6" ViewCount="179" Body="&lt;p&gt;it has been said that the electron is the fundamental representation of the Poincare group, with only two conmuting observables, $( \sigma , p_{\mu})$. This question regards what is usually called the &lt;em&gt;superselection rules&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what i don't understand why states that are superpositions of different $C$ like $e + \overline{e}$ (and electron and a positron superposition) are supressed and how. What about superpositions like $e + \mu$ (an electron and a muon)? and finally, why neutrino flavors are different and in this regard, can be mixed in superposition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what about superpositions of particles with different mass? what are the currently accepted mechanisms for this supression? are there other alternatives worth pursuing? are there high-decoherence impedances in some state combinations that make them decohere quicker than others or are they &lt;strong&gt;fundamentally&lt;/strong&gt; supressed?&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-06-28T16:23:40.330" LastActivityDate="2011-06-29T18:47:23.430" Title="why certain superpositions of quantum states are supressed?" Tags="&lt;quantum-field-theory&gt;&lt;particle-physics&gt;&lt;poincare-symmetry&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" />
  <row Id="11676" PostTypeId="2" ParentId="11673" CreationDate="2011-06-28T16:25:02.133" Score="1" ViewCount="" Body="&lt;p&gt;Plasma physics is usually computed with particle-in-cell methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Particle-in-cell&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Particle-in-cell&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1670" LastActivityDate="2011-06-28T16:25:02.133" />
  <row Id="11677" PostTypeId="2" ParentId="11078" CreationDate="2011-06-28T16:31:18.343" Score="0" ViewCount="" Body="&lt;p&gt;The proper force on a particle is clearly defined and independent of its motion in any frame. Its rest mass can be calculated from $F = ma$ in its proper frame. There is no important contradiction to Newton's law of inertia when used this way.&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-06-28T16:31:18.343" />
  <row Id="11678" PostTypeId="1" CreationDate="2011-06-28T17:13:52.310" Score="2" ViewCount="110" Body="&lt;p&gt;I'm interrested in making computer simulation but I've run into rather physics oriented problem. I have to choose how to propagate my wave. Though I've found technique called FDTD (finite-difference time-domain), I could find only explanations how to work with this method while solving electromagnetic waves but I'm more interested in mechanical waves - water, earth,... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there's first problem. EM waves are nicely discribed with Maxwells equations, but I couldn't find anything like that for mechanical ones. Only equation I could find is $y = A \sin(k x - \omega t + O)$. This is fine, one can choose a distance and a time and he gets height of that point (considering that we are working on a plane). But doesn't wave gets also damped - in time as well as distance? And what formula describes this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the second part of a question - in school I was teached a propagation theory called Huygens principle, and it goes like: at every point where wave envelope propagate it creates elementary waves, and frontface of those again create envelope. I don't fully understand how it should be imagined, because creating waves all around a circle like on this image: &lt;a href=&quot;http://kr.cs.ait.ac.th/~radok/physics/fig300.jpg&quot; rel=&quot;nofollow&quot;&gt;http://kr.cs.ait.ac.th/~radok/physics/fig300.jpg&lt;/a&gt; (sorry, but I got a little problem with tags..), there is nice envelope but inside of that is mess of curves that would gave us interference. So it's just a model and we don't care what's inside as long as we have envelope, or am I missing something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as I said, I'm creating computer simulation and propagating wave like that would be probably slow, so is there any other theory on how wave propagates or this is the best and most used one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="3662" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-06-28T18:05:04.533" LastActivityDate="2011-06-30T16:59:06.590" Title="Wave equations &amp; propagation theories" Tags="&lt;mechanics&gt;&lt;oscillators&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="11679" PostTypeId="2" ParentId="11674" CreationDate="2011-06-28T18:27:19.047" Score="3" ViewCount="" Body="&lt;p&gt;Nonzero fluxes are required because of some equations of motion linking them to a nonzero Euler character. Once they're there, they induce a superpotential that stabilizes some moduli, usually the complex structure moduli (the very &quot;stabilizes&quot; means that the allowed values of these moduli at which the total potential has a local minimum is discrete, assuming fixed values of other moduli).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dilaton-axion field is stabilized by the Gukov-Vafa-Witten superpotential while nonperturbative effects are typically needed to stabilize the Kähler moduli.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2006/04/flux-compactifications-of-m-theory-and.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2006/04/flux-compactifications-of-m-theory-and.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is of course a big technical topic and the literature on stabilized flux vacua is large. One can't describe everything in a single comment. The Becker-Becker-Schwarz textbook probably has a good treatment of these matters.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-06-28T18:27:19.047" />
  <row Id="11680" PostTypeId="1" AcceptedAnswerId="11863" CreationDate="2011-06-28T18:29:31.903" Score="5" ViewCount="129" Body="&lt;p&gt;In terms of dollars per watt, using theoretical efficiency limits, what technology holds the most promise to become the primary solar energy capture technology? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My hunch is carbon-based modules, since materials are abundant and relatively easy to manufacture; however their efficiency currently is nowhere near silicon-based semiconductors. In the future, assuming that scientists will push very close to the theoretical limits, what will provide the most power for the lowest cost?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is relevant at all, I am trying to decide where to focus my graduate studies in physics, and I want to contribute in getting the most promising technology to its limit. If you can at least point me in the right direction to figure out the answer myself, I would greatly appreciate it.&lt;/p&gt;&#xA;" OwnerUserId="4218" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-13T10:05:18.083" LastActivityDate="2011-07-13T17:30:23.740" Title="What type of solar energy technology has the most future potential?" Tags="&lt;energy&gt;&lt;power&gt;&lt;renewable-energy&gt;&lt;graphene&gt;&lt;technology&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" />
  <row Id="11681" PostTypeId="2" ParentId="11656" CreationDate="2011-06-28T18:29:59.013" Score="0" ViewCount="" Body="&lt;p&gt;I bring you good news! There is hope! The Coleman-de luccia analysis assumes a vacuum state for the original and final vacua. This is a very good approximation for the inflationary epoch, but not for our current phase with its extremely small cosmological constant. Remember it's the difference in pressure which pushes the domain wall outward at close the speed of light in the Coleman-de Luccia analysis, but with such a tiny cosmological constant, even the cosmic microwave background or intergalactic gas is sufficient to counteract the difference in pressure. Ordinary matter and photons in our phase will bounce off the domain wall instead of passing through it, unless they have enormous kinetic energies. A future civilization will only need to create a very high grade vacuum with pressure less than the cosmological constant sealed in a box and create the new vacuum in it. This vacuum will expand until the domain wall hugs the walls of the box, and then, it will stop expanding. With a suitable choice of bombardment of ultra high energy particles to penetrate the domain wall, a computer can be reconstructed on the other side. Computations require a steady supply of free energy, and this requires the bubble to expand slowly. But the outside universe is safe. The bubble can't expand to reach galactic clusters because even dilute intergalactic gas has enough pressure to resist it. The bubble can only expand in the void between galactic clusters in the far future when the cosmic microwave background pressure has gone really down.&lt;/p&gt;&#xA;" OwnerUserId="4219" LastActivityDate="2011-06-28T18:29:59.013" CommentCount="1" />
  <row Id="11682" PostTypeId="2" ParentId="11233" CreationDate="2011-06-28T19:14:47.907" Score="1" ViewCount="" Body="&lt;p&gt;For point b: Yes, but very slightly. (I can't properly answer a, since I'm not a geophysicist). Let's do some back of the envelope calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The centripetal acceleration at the equator is given by $\omega^2 r$, where $\omega$ is the angular velocity of the earth, and $r$ is the radius. $\omega$ is pretty small, somewhere around $7 \times 10^{-5} \frac{rad}{s}$, $r$ is about $6.4 \times 10^6 m$, which makes the centripetal acceleration somewhere around $4 \frac{cm}{s^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the gravitional acceleration is $980 \frac{cm}{s^2}$&lt;/li&gt;&#xA;&lt;li&gt;the slowing down is &lt;em&gt;very&lt;/em&gt; gradual (the change in Earth's rotational period (a day in other words) is 2 milliseconds per century &lt;a href=&quot;http://en.wikipedia.org/wiki/Leap_second&quot; rel=&quot;nofollow&quot;&gt;according to Wikipedia&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;you're not going to notice anything over your lifetime, and I doubt whether it could be measured experimentally over the course of a hundred years. At the very least, high rise buildings won't notice it as much as the forces they are continually subjected to; think thermal expansion/contraction, wind (the Empire State Building sways up to 3 cm in bad storms), earthquakes and even passing traffic!.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If&lt;/strong&gt; the earth's rotation would suddenly scream to a halt, things are different. For starters, the rotational energy is $\frac{m \omega^2 r^2}{5}$, about $2.9 \times 10^{29} J$ (if I didn't mess up too much :) which is 'equivalent' to $2 \times 10^{11}$ Japan 2011 earthquakes, or 'enough' to raise the sealevel temperature by several thousand degrees. All that energy has to go &lt;em&gt;somewhere&lt;/em&gt;, but fortunately it also has to come from somewhere.&lt;/p&gt;&#xA;" OwnerUserId="2445" LastEditorUserId="2445" LastEditorDisplayName="" LastEditDate="2011-06-29T13:25:06.360" LastActivityDate="2011-06-29T13:25:06.360" CommunityOwnedDate="2011-06-28T19:14:47.907" CommentCount="1" />
  <row Id="11683" PostTypeId="2" ParentId="11678" CreationDate="2011-06-28T19:49:16.590" Score="1" ViewCount="" Body="&lt;p&gt;Your mechanical vibration expression can be rewritten (k subscript added by me) as the real part of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Ae^{j(k_1x-\omega t+\phi)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This describes a standing wave, or one component of a plane wave, created by a source which is constantly on, in a lossless medium.  Notice that the entire exponent is multiplied by the imaginary factor 'j'.  (EDIT: Scratch the 'standing wave' part: That would take two wave components, one with a -jkx factor in the exponent and another with a +jkx factor (traveling in the opposite direction).)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a wave set up in a lossy medium by a constantly ON source, and you want to describe how it attenuates with distance, you need to add in a non-imaginary loss factor, $e^{-k_2x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So then, assuming your source is at the origin, and the plane wave travels only in the positive 'x' direction, your expression for finding the instantaneous wave height at any point in the path of the plane wave would be the real part of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Ae^{-k_2x}e^{j(k_1x-\omega t+\phi)} = Ae^{(-k_2x+j(k_1x-\omega t+\phi))}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be easy enough to code up.  The tough part will be defining $k_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, on the Huygens thing: Don't worry about it too much if you are just doing plane waves or trying to analyze sound traveling down a lossy channel.  Where Huygens really shines is explaining diffraction patterns or why light bends around a sharp corner.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-06-30T16:59:06.590" LastActivityDate="2011-06-30T16:59:06.590" CommentCount="2" />
  <row Id="11684" PostTypeId="2" ParentId="11678" CreationDate="2011-06-28T19:52:21.097" Score="1" ViewCount="" Body="&lt;p&gt;The simplest wave-like phenomena are governed by a differential equation aptly named the &quot;wave equation&quot;. In one dimension this is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial^2 \psi(x,t)}{\partial x^2} = \frac{1}{c^2}\frac{\partial^2 \psi(x,t)}{\partial t^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\psi(x,t)$ is the height of wave at position $x$ and time $t$. Here $c$ is the speed of your waves. If you want your wave to be damped you can alter the equation to be: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial^2 \psi(x,t)}{\partial x^2} = \frac{1}{c^2}\frac{\partial^2 \psi(x,t)}{\partial t^2}-r\frac{\partial \psi(x,t)}{\partial t}$ where $r$ is a parameter that controls how fast your wave is damped &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Wikipedia on the Wave Equation for more info. There's also a section on how to implement this as a computer program if my symbols mean nothing to you.&lt;/p&gt;&#xA;" OwnerUserId="3970" LastActivityDate="2011-06-28T19:52:21.097" CommentCount="1" />
  <row Id="11685" PostTypeId="2" ParentId="3527" CreationDate="2011-06-28T20:43:58.287" Score="1" ViewCount="" Body="&lt;p&gt;Given that the initial balls do not bounce back, the original question can be answered directly as the top response shows, but he needed to specify that he was assuming the other balls do not bounce back or otherwise move. I can explain it without math: more or fewer balls can't come off at a slower or faster speed because energy and momentum must be conserved, and they are each a different function of velocity.  If you change the velocity, and if no other balls move, then you would be violating one of the principles of conservation.  For example, if only 1 ball came off with twice the velocity, momentum would be the same (conserved) as the incoming 2 balls, but there would be twice as much energy in the outgoing ball as there was in the 2 incoming balls. Conversely, if 4 balls came off with 1/2 the velocity, momentum would be conserved, but the energy would be half of what was put it in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is much more difficult to answer the question if other balls are allowed to bounce back in the other direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers based on separation are not correct and one poster hinted at how the device can really be explained. It is only a happen accident that the series of independent strikes calculation is correct for multiple balls of equal weight. Imagine 3 balls with the one in the middle that is much heavier. The first ball will bounce off, which is not what happens when they are touching. So the calculation is wrong for a fundamental reason.  When they are touching, the middle heavy ball is in a compressed state, not yet pushing back on the first ball to make it bounce back before it is releasing energy to the last ball. This can't be modeled with one striking pair followed by another.  To see everything more clearly, model the balls as big, weak springs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From wiki: &quot;At the point of collision, two shock waves propagate, one forward and one backward. They are carrying all the kinetic energy of the formerly-moving balls as potential energy in the elastic compression of the metal. The wave going backward is reflected forward when it reaches the beginning of the chain of balls, coming behind the wave that was moving forward first. The first wave reaches the end and is reflected back and it collides with the delayed wave at a point symmetrical to the initial point of collision, forcing the balls apart, releasing the potential energy as kinetic energy.[8] This explanation is more complicated if balls of equal weight are given different lengths or if the balls have different weights, but the solution for the final velocities can be found by examining the compression and expansion of the metals as a result of the shock waves. The conservation of energy and momentum are adequate to explain the system if the potential energy and transition time is included in the conservation of energy calculations.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to answer the original question: As the other poster said, the shock wave meets at a point symmetrical to the original collision.  However, the answer is more complicated if the balls at the end are of equal weight but different in length.  The shock waves will not meet to begin with, but the expanding system of metal will eventually cause the equal-weight balls to separate so that the initial balls do not bounce back.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Important Update/Correction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In looking at this in much more detail, I've found that no one understands how the cradle works.  The reflected shock wave theory comes from: &#xA;&lt;a href=&quot;http://www.physikdidaktik.uni-karlsruhe.de/publication/ajp/Ball-chain_part1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.physikdidaktik.uni-karlsruhe.de/publication/ajp/Ball-chain_part1.pdf&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but these researchers followed this paper up with this: &#xA;&lt;a href=&quot;http://www.physikdidaktik.uni-karlsruhe.de/publication/ajp/Ball-chain_part2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.physikdidaktik.uni-karlsruhe.de/publication/ajp/Ball-chain_part2.pdf&lt;/a&gt;&lt;br&gt;&#xA;and then others:&#xA;&lt;a href=&quot;http://www-astro.physics.ox.ac.uk/~ghassan/newton_cradle_2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www-astro.physics.ox.ac.uk/~ghassan/newton_cradle_2.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in which they show the collision time is 10 times slower than sound (shock) wave propagation which indicates &quot;hertzian compression&quot; (not simply shock waves) describes the cradle by including the springiness of the interfaces between the balls. There are 4 interfaces in 5 balls, so energy and momentum give the solution for the final velocity of 2 balls and then the ratio of the 3 subsequent interfaces to the first interface gives the 3 variables needed to solve for the other 3 balls.  They apply F=ma to each ball where it is in contact with its 1 or 2 neighbors, subject to a psuedo-spring force (hertzian) in the surface contact of F=kx^1.5 rather than F=kx.  This gives a set interdependent differential equations which I'll state for the most general case of different masses and different surface spring constants (youngs modulus) at the end.  This may agree with experiment for the case of 1 ball striking 4 that are in contact, but it gives a solution for the case of 2 striking 3 that is not correct.  I solved this system of equations in excel and got the same result as this paper:&#xA;&lt;a href=&quot;http://www.damtp.cam.ac.uk/user/hinch/publications/PRSLA455_3201.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.damtp.cam.ac.uk/user/hinch/publications/PRSLA455_3201.pdf&lt;/a&gt;&#xA;in which ball 4 and 5 velocities would theoretically be 0.80 and 1.14 times the velocity of the 2 incoming balls.  The paper asserts that you can observe the difference in velocity, but they fail to point out that this is a 2 times difference in kinetic energy and therefore the height reached by the 5th ball is 2 times more than the 4th, which is not what happens at all. (If 5th ball comes out 30 degrees max, the 4th ball would theoretically reach only 21 degrees, which is not what happens: they both reach 30 degrees at the same time).&#xA;Here are the F=ma equations that have to be solved with runge-kutta, if the Hertzian theory is correct. I did it in excel.&lt;br&gt;&#xA;x is displacement from at-rest point &lt;/p&gt;&#xA;&#xA;&lt;p&gt;m1a1= - A*(x1-x2)^1.5 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;m2a2 = A*(x1-x2)^1.5 - B*(x2-x3)^1.5 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;m3a3 = B*(x2-x3)^1.5 - C*(x3-x4)^1.5 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;m4a4 = C*(x3-x4)^1.5 - D*(x4-x5)^1.5 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;m5a5 = D*(x4-x5)^1.5 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A=(2/((1/k1)^(2/3)+(1/k2)^(2/3)))^1.5 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;B=(2/((1/k2)^(2/3)+(1/k3)^(2/3)))^1.5 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;C=(2/((1/k3)^(2/3)+(1/k4)^(2/3)))^1.5 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;D=(2/((1/k4)^(2/3)+(1/k5)^(2/3)))^1.5 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For plugging into runge-kutta: &#xA;a=dV(t)/dt and x=t*V(t) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A, B, C, D are the net effective spring constant between each pair of balls for the surface compression. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for 2 balls or more balls striking the others, no one knows how it works.  Even as late as 2004, these papers were incorrectly described as the complete solution:&#xA;&lt;a href=&quot;http://www.maths.tcd.ie/~garyd/Publications/Delaney_2004_AmJPhys_Rocking_Newtons_Cradle.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.maths.tcd.ie/~garyd/Publications/Delaney_2004_AmJPhys_Rocking_Newtons_Cradle.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4220" LastEditorUserId="4220" LastEditorDisplayName="" LastEditDate="2011-07-07T16:29:34.717" LastActivityDate="2011-07-07T16:29:34.717" CommunityOwnedDate="2011-07-07T16:29:34.717" CommentCount="2" />
  <row Id="11686" PostTypeId="1" AcceptedAnswerId="12289" CreationDate="2011-06-28T21:18:05.250" Score="4" ViewCount="180" Body="&lt;p&gt;Take the game of pong as a simple example. When you hit the ball with a paddle that has a frictional surface, the ball will spin as well as change direction according to the coefficient of kinetic friction on the paddle and the velocity of the paddle. The ball will then spin, and when it hits another surface this spin will cause a change in direction as well. Assuming that the velocity of the paddle is never low enough to use the static coefficient of friction of the paddle, how would i find the generated angular velocity on the ball and the new velocity of the ball? &lt;/p&gt;&#xA;" OwnerUserId="2404" LastActivityDate="2011-07-18T00:47:23.253" Title="finding angular velocity and regular velocity when bouncing off a surface with friction" Tags="&lt;forces&gt;&lt;friction&gt;&lt;torque&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="11687" PostTypeId="1" CreationDate="2011-06-28T22:23:11.243" Score="6" ViewCount="161" Body="&lt;p&gt;For example, must the shock wave in each body be of a particular form which influences the shape and material properties of the bodies?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect part of the the answer is that the objects must be spherical, and the round-trip of the shock wave in each body must be the same, but this is just a guess.&lt;/p&gt;&#xA;" OwnerUserId="2146" LastEditorUserId="2146" LastEditorDisplayName="" LastEditDate="2011-06-28T23:05:34.313" LastActivityDate="2011-08-01T01:10:38.490" Title="What are the properties of two bodies for their collision to be elastic?" Tags="&lt;classical-mechanics&gt;&lt;momentum&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="11689" PostTypeId="1" CreationDate="2011-06-28T23:20:32.970" Score="3" ViewCount="139" Body="&lt;p&gt;I'm trying to take advantage of a particular identity for the sum of the product of three Clebsch-Gordan coefficients, however, the present form of my equation is slightly different.  Is there a symmetry relation that will allow me to change:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{\alpha\beta\delta}C_{a\alpha b\beta}^{c\gamma}C_{d\delta b\beta}^{e\epsilon}C_{d\delta f\phi}^{a\alpha}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Into:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{\alpha\beta\delta}C_{a\alpha b\beta}^{c\gamma}C_{d\delta b\beta}^{e\epsilon}C_{a\alpha f\phi}^{d\delta}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice I need to swap j2m2 with jm in the last Clebsh-Gordan coefficient.  Does anyone know a way to do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: My notation follows that of Varshalovich, $C_{j_1 m_1 j_2 m_2}^{jm}$&lt;/p&gt;&#xA;" OwnerUserId="3597" LastActivityDate="2011-08-07T14:20:35.320" Title="Clebsch-Gordan Identity" Tags="&lt;quantum-mechanics&gt;&lt;angular-momentum&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="11690" PostTypeId="2" ParentId="11689" CreationDate="2011-06-28T23:57:28.037" Score="1" ViewCount="" Body="&lt;p&gt;Notice that $C^{22}_{1111}=1$ but $C^{11}_{2211}=0$. I don't think that this is true unless $a=d$ and the sums over $\alpha$ and $\delta$ have the same range.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-08T13:26:46.003" LastActivityDate="2011-07-08T13:26:46.003" />
  <row Id="11692" PostTypeId="2" ParentId="11233" CreationDate="2011-06-29T01:13:09.730" Score="1" ViewCount="" Body="&lt;p&gt;EDIT: Answer below is talking about a radical sudden slow-down of angular velocity, not a gradual one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question a is actually pretty interesting. We can see a very extreme example of this in the case of a neutron star, some of which rotate with surface speeds of 1/3 the speed of light due to conservation of angular momentum, making it an oblate shape. The basic structure of a neutron star is analogous to earth, it has a hard &quot;crust&quot; (which is significantly stronger than earth's crust) and a fluid like core. As it slows down, the crust holds firm against the inner fluid, until suddenly it ruptures and the neutron star undergoes a &quot;stellar quake,&quot; which create small gamma ray bursts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine if for some reason or other the earth is significantly slowed down, a similar event would occur. At first there would be rather little seismic activity, but sometime the pressure would be so great that somewhat catastrophic fissures would occur in the earth's crust as it ruptures and re-shapes itself to be more spherical.&lt;/p&gt;&#xA;" OwnerUserId="3018" LastEditorUserId="3018" LastEditorDisplayName="" LastEditDate="2011-06-29T20:32:54.020" LastActivityDate="2011-06-29T20:32:54.020" CommunityOwnedDate="2011-06-29T01:13:09.730" CommentCount="2" />
  <row Id="11693" PostTypeId="1" CreationDate="2011-06-29T01:28:59.970" Score="1" ViewCount="69" Body="&lt;p&gt;I am building a preheater for a maple syrup evaporator and am going to use the steam generated by the heating process to pre-heat the incoming sap from, say 5 degrees C to (hopefully) something on the order of 80-90 C.  (the goal is to heat the sap/fluid before it gets into the evaporating pan so that we are more efficient - the steam used is &quot;free&quot; - it is a waste product of the process and capturing the heat and subsequent condensation for hot water is desirable)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My flow rate through the copper tubes will be on the order of 35 to 56 liters per hour (10 to 15 gallons per hour) (this is for the end points of the assembly)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only know the outside diameter of the copper pipes - they will be 1/2&quot; OD.&lt;br&gt;&#xA;There will be 5 parallel lines inside a hood and at each end a manifold - the incoming cold sap comes into the low side manifold and makes its way up to the high side manifold in one of the 5 pipes.  The manifolds are 20 inches long and are perpendicular to the other pipes which are 40 inches long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat exchanger will sit in a hood to provide a little back pressure for the steam and to collect the steam and force it to contact the pipes.  The boiling surface area of the sap underneath the heat exchanger is larger than the heat exchanger and the hood but most of the steam rises through the hood and will transfer some heat to the copper which in turn heats the fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how can I calculate the some likely heat exchange/output temperature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that the issues are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;efficiency of heat transfer from steam to copper&lt;/li&gt;&#xA;&lt;li&gt;efficiency of heat transfer from copper to fluid&lt;/li&gt;&#xA;&lt;li&gt;how much time the fluid is in the lines&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I guess I am looking for an upper bounds and lower bounds (I know I won't get the fluid over 100 C - that's about all I know for sure - well - I guess I can also be sure that the temperature will rise some - but not sure how much.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many people report output temps on the order of 95C but I am skeptical.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can I do to improve the heat transfer to maximize the scavenged heat?  I can try to find fins from a baseboard radiator, however this is for food product - so I am unsure if I will do that.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some pictures of something I would like to do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://s662.photobucket.com/albums/uu349/bkm_photos/Oil%20Tank%20Evaporator/?action=view&amp;amp;current=evap-096.jpg&amp;amp;newest=1#!oZZ80QQcurrentZZhttp%3A%2F%2Fs662.photobucket.com%2Falbums%2Fuu349%2Fbkm_photos%2FOil%2520Tank%2520Evaporator%2F%3Faction%3Dview%26current%3Devap-087.jpg%26newest%3D1&quot; rel=&quot;nofollow&quot;&gt;http://s662.photobucket.com/albums/uu349/bkm_photos/Oil%20Tank%20Evaporator/?action=view&amp;amp;current=evap-096.jpg&amp;amp;newest=1#!oZZ80QQcurrentZZhttp%3A%2F%2Fs662.photobucket.com%2Falbums%2Fuu349%2Fbkm_photos%2FOil%2520Tank%2520Evaporator%2F%3Faction%3Dview%26current%3Devap-087.jpg%26newest%3D1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.leaderevaporator.com/p-286-parallel-flow-sap-pre-heater.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.leaderevaporator.com/p-286-parallel-flow-sap-pre-heater.aspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wood is used as the heat source and one can assume a constant fire/heat source that will be providing 10 to 15 gallons of evaporation per hour from a 24&quot; by 60&quot; pan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT (answering a comment)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wall thickness of the copper pipe is 0.569 inches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is only one fluid being heated - that is the sap from a maple tree.  It is water with sugar in it - and by boiling it we can remove most of the water until it turns to maple syrup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The level of the fluid (sap) in the evaporator pans is kept constant by a float valve.  The sap in the pan boils - that is where the steam comes from.  So the rate of evaporation will be the same (approximately) as the flow through the heating tubes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat is generated by a wood fueled fire in a firebox just below the evaporator pans that hold the sap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maple sap is held in a storage tank above the whole contraption.  Gravity feeds it into the evaporator (a 5'x2' pan) through the preheater/heat exchanger.  The fluid flow/level is regulated by a float valve.  Steam is a by-product of the process - and using the wasted heat from the steam (allowing it to condense on the copper pips and transferring heat then drawing that condensation away so as not to have to re-heat it) makes the process a little more efficient.  One study that was done shows that there is approximately 15% improvement in boiling rate (best case) for preheating the sap before it goes directly into the pan that gets heated.&lt;/p&gt;&#xA;" OwnerUserId="4221" LastEditorUserId="4221" LastEditorDisplayName="" LastEditDate="2011-06-29T04:17:52.113" LastActivityDate="2011-06-29T14:12:04.187" Title="How to calculate heat exchange/deltaT in a moving volume of fluid?" Tags="&lt;fluid-dynamics&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11694" PostTypeId="1" CreationDate="2011-06-29T02:13:47.860" Score="4" ViewCount="59" Body="&lt;p&gt;As I understand it fusion inside a sun can produce heavier and heavier elements until some sort of &quot;nucleus size limit&quot; is reached. As far as I understand, the limit is thought to be reached with the element iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What accounts for this &quot;fusion element limit&quot;? Is it the battle between the nuclear strong force and the electromagnetic force which limits the size of the nucleus that can exist in a fusion environment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the limit of what elements can be created the same for all types of suns?&lt;/p&gt;&#xA;" OwnerUserId="4145" LastActivityDate="2011-07-04T21:44:50.383" Title="What elements can be created in the fusion process of different types of suns?" Tags="&lt;electromagnetism&gt;&lt;nuclear-physics&gt;&lt;sun&gt;&lt;fusion&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" />
  <row Id="11697" PostTypeId="1" CreationDate="2011-06-29T03:45:32.203" Score="2" ViewCount="85" Body="&lt;p&gt;This is a question about terminology. To me, it's clear that the nuclei of an atom is still an atom. But a comment by Willie Wong at &lt;a href=&quot;http://physics.stackexchange.com/questions/11562/is-nature-symmetric-between-particles-and-antiparticles/11563#comment-28303&quot;&gt;Is nature symmetric between particles and antiparticles?&lt;/a&gt; raises the question. Similarly, arguments go either way at the physics forums question: &quot;is it possible for an atom to have no electrons?&quot; &lt;a href=&quot;http://www.physicsforums.com/showthread.php?t=50333&quot; rel=&quot;nofollow&quot;&gt;http://www.physicsforums.com/showthread.php?t=50333&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-06-29T04:10:38.623" CommunityOwnedDate="2011-06-29T03:54:48.547" Title="If an atom is fully ionized by removing all electrons, is it still an atom?" Tags="&lt;atoms&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="11698" PostTypeId="2" ParentId="11697" CreationDate="2011-06-29T03:58:21.490" Score="2" ViewCount="" Body="&lt;p&gt;In most cases I don't see it behaving as &quot;an atom&quot; in any meaningful way (i.e. in most circumstances it is not going to engage in chemistry but in magnetohydrodynamics). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Low energy may make an interesting borderline case.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-06-29T03:58:21.490" CommunityOwnedDate="2011-06-29T03:58:21.490" CommentCount="1" />
  <row Id="11699" PostTypeId="2" ParentId="11697" CreationDate="2011-06-29T03:58:51.770" Score="2" ViewCount="" Body="&lt;p&gt;No, it's just a nucleus at that point, although we should admit that the distinction is semantics.  Because it is semantics, I look to other presidents and clearly find that it would not be consistent to call them atoms.  Consider the case of twice ionized Helium, $He^{+2}$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&lt;li&gt;When the particle has high energy, moving among ordinary matter we call it an alpha particle&#xA;&lt;li&gt;When the particle has high energy among a sea of other high energy (disassociated) particles we call it part of a plasma&#xA;&lt;li&gt;When the particle has thermal energy among other disassociated particles with thermal energies... that doesn't make any sense, why would they be disassociated?&#xA;&lt;li&gt;When the particle is sitting by its lonesome in the middle of space, I would call it an alpha particle&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It's pretty much impossible to ask the question without considering the context in which the nucleus/atom exists, but I can't think of a context in which it should clearly be called an atom.  So for the general sense, I would say it shouldn't be called an atom.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-06-29T03:58:51.770" CommunityOwnedDate="2011-06-29T03:58:51.770" />
  <row Id="11700" PostTypeId="2" ParentId="11697" CreationDate="2011-06-29T04:10:38.623" Score="1" ViewCount="" Body="&lt;p&gt;If we are thinking of low atomic number atoms, as &#xA;Zassounotsukushi  mentions, scientists use different names for the bare nuclei of atoms. Mainly, though,  because they were encountered as unknown entities in experiments and then identified with the nucleus of an atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we are thinking of the periodic table, it would be hard and confusing to come up with a different name for each electronless representative. I have not heard of experiments with stripped atoms with high atomic numbers, but presumably there exists such a possibility. Common sense dictates that an adjective should be found accompanying the periodic table name: &quot;stripped iron&quot; or some such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all terminology should reduce confusion not increase it.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-06-29T04:10:38.623" CommunityOwnedDate="2011-06-29T04:10:38.623" />
  <row Id="11701" PostTypeId="1" CreationDate="2011-06-29T04:14:48.763" Score="0" ViewCount="90" Body="&lt;p&gt;I have read in many places &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Entropy of an isolated systems never&#xA;  decreases&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And as a corollary:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;As &lt;strong&gt;Universe is an isolated system&lt;/strong&gt;(I) then its entropy is constantly &lt;strong&gt;increasing&lt;/strong&gt;(II)&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I) If there are boundaries, how can we know about what happens there with entropy? and if there aren't any boundaries, what does it mean to be isolated? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;II) Why it couldn't remain constant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are experimental evidences of &lt;strong&gt;I&lt;/strong&gt; and/or &lt;strong&gt;II&lt;/strong&gt; ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1916" LastActivityDate="2011-06-29T05:02:26.197" Title="Why it is said that Universe is an isolated system?" Tags="&lt;astrophysics&gt;&lt;universe&gt;&lt;entropy&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="11702" PostTypeId="1" CreationDate="2011-06-29T04:20:59.210" Score="4" ViewCount="222" Body="&lt;p&gt;The question aims to this issue : if there is some technological arrangement (or action) to take over the particle/system in order to keep it in a coherent state, then the field, (force or whatever) keeping it away from  interacting with &quot;an external system&quot; isn't it itself an interaction? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean, supposing you reach  enough isolation to avoid decoherence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you know the particle is still there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1916" LastEditorUserId="1916" LastEditorDisplayName="" LastEditDate="2011-07-05T14:43:08.133" LastActivityDate="2011-07-05T14:43:08.133" Title="How could a particle be isolated to avoid decoherence?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-information&gt;&lt;measurement&gt;&lt;measurement-problem&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" />
  <row Id="11703" PostTypeId="2" ParentId="11675" CreationDate="2011-06-29T04:50:20.230" Score="4" ViewCount="" Body="&lt;p&gt;The fact that combinations such as $(e^-+e^+)/\sqrt{2}$, (or other combinations such as a proton with an electron) are suppressed is a little mysterious. It falls outside of quantum mechanics. Any &quot;theory of everything&quot; needs to answer this sort of question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can't form combinations of electrons with positrons because they have different electric charges (and lepton number). Since electric charge is conserved, the formation of such a combination is impossible. If we could produce such a state, a measurement of it would result in the total charge of the universe changing from an undetermined state to one of two incompatible possibilities. Thus the superselection rule arises from an exact symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that it is possible to create a state where either an electron or a positron is present at a particular point. For example, the electron could have gone left and the positron right, or alternatively, the electron went right and the positron left. Those kind of combinations are possible and we write them as: $(e^-_Le^+_R + e^-_Re^+_L)/\sqrt{2}$. If we make a measurement on this state according to the natural basis, we either end up with the electron or the positron on the right, but the total charge is still zero.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The case of $e^-+\mu^-$, an electron and a muon (both negatively charged but with vastly different masses) allows linear superposition. There is no violation of charge symmetry, and unlike say a proton and a positron, there is not a violation of lepton or baryon number. In this case there is no superselection rule and these combinations can be obtained. They are observed in weak interactions. One sees these combinations in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pontecorvo%E2%80%93Maki%E2%80%93Nakagawa%E2%80%93Sakata_matrix&quot; rel=&quot;nofollow&quot;&gt;MNS matrix&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of a weak interaction is a $W^-$ particle which decays into a combination of a charged lepton (an electron, muon or tau: $e^-, \mu^-,\tau^-$) and an anti-neutrino (i.e. $\bar{\nu}_1,\bar{\nu}_2,\bar{\nu}_3$). Which of these is produced is determined only by measurement so until then, one has a linear superposition of an electron, muon, and tau state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an introduction to the subject see &quot;Neutrino Oscillations for Dummies&quot; by Chris Waltham, Am.J.Phys. 72 (2004) 742-752 &lt;a href=&quot;http://arxiv.org/abs/physics/0303116&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/physics/0303116&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A good question is &quot;how long can superpositions last?&quot; For the case of the superpositions of the anti-neutrinos, observations of anti-neutrinos emitted from the sun indicate that the superpositions last until they are observed on the earth. This is observed as the interference between anti-neutrinos (or neutrinos) with different masses and is usually described as &quot;neutrino oscillation&quot;, but can also be described as &quot;&lt;a href=&quot;http://carlbrannen.wordpress.com/2008/06/26/neutrino-oscillation-the-calculation/&quot; rel=&quot;nofollow&quot;&gt;neutrino interference&lt;/a&gt;&quot; between the three mass eignstates of neutrinos. So in this case at least, the superposition lasts about 8 minutes, but there is no theoretical maximum time for a superposition to last.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the mass differences between neutrinos are extremely small, around 0.05 eV or less than a millionth of the mass of an electron. So this isn't a counterexample to Penrose's proposal that quantum measurement results from differences in mass between different superpositions. See &quot;Testing Gravity-Driven Collapse of the Wavefunction via Cosmogenic Neutrinos&quot;, Joy Christian, Phys.Rev.Lett. 95 (2005) 160403 / &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0503001&quot; rel=&quot;nofollow&quot;&gt;quant-ph/0503001&lt;/a&gt; for a discussion. The abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is pointed out that the&#xA;  Diosi-Penrose ansatz for&#xA;  gravity-induced quantum state&#xA;  reduction can be tested by observing&#xA;  oscillations in the flavor ratios of&#xA;  neutrinos originated at cosmological&#xA;  distances. Since such a test would be&#xA;  almost free of environmental&#xA;  decoherence, testing the ansatz by&#xA;  means of a next generation neutrino&#xA;  detector such as IceCube would be much&#xA;  cleaner than by experiments proposed&#xA;  so far involving superpositions of&#xA;  macroscopic systems. The proposed&#xA;  microscopic test would also examine&#xA;  the universality of superposition&#xA;  principle at unprecedented&#xA;  cosmological scales.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-06-29T18:47:23.430" LastActivityDate="2011-06-29T18:47:23.430" CommentCount="10" />
  <row Id="11704" PostTypeId="2" ParentId="11675" CreationDate="2011-06-29T04:59:26.773" Score="2" ViewCount="" Body="&lt;p&gt;The total electric flux at infinity will always decohere states with different charges exactly. This is the cause of the superselection rules.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-06-29T04:59:26.773" CommentCount="2" />
  <row Id="11705" PostTypeId="2" ParentId="11701" CreationDate="2011-06-29T05:02:26.197" Score="1" ViewCount="" Body="&lt;p&gt;For I) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A system is isolated when there is no energy exchange with the surroundings. For example the contents of a vacuum flask are isolated. In practice there is an error in the isolation parameter, i.e. delta(energy) that is why after some time the contents will be found to be in room temperature. Experiments are designed so as to work within the errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe is by definition of the word isolated, because it contains everything by definition, and thus there can be no exchange of energy with anything. If there could be, it would be counted within the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now models may be found where &quot;universe&quot; is defined with some error in this energy exchange. Even speculated as many isolated ones, but there would be no way of knowing since no energy would be exchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For II)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has to do with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Entropy&quot; rel=&quot;nofollow&quot;&gt;definition of entropy&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/7xHUG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P_i is the probability that the system is in the ith microstate, and k_b is the Boltzmann constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of microstates increase in time when there are energy exchanges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the isolated universe the microstates increase in time because we have &lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble_constant&quot; rel=&quot;nofollow&quot;&gt;experimental evidence&lt;/a&gt; that the universe is expanding. As the volume gets larger the number of possible microstates increases&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-06-29T05:02:26.197" CommentCount="2" />
  <row Id="11706" PostTypeId="1" CreationDate="2011-06-29T05:19:17.863" Score="0" ViewCount="36" Body="&lt;p&gt;The traditional formalism for andreev reflection deals with what happens at normal metal, super conductor interface.http://en.wikipedia.org/wiki/Andreev_reflection&#xA;(i.e when an electron from normal metal is incident on the super conductor interface, a hole is generated in the normal metal which lets the formation of a cooper pair in the SC which propagates in the appropriate direction)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we apply a similar formalism when an electron of say up-spin is incident from a normal metal to half-metal (magnet) ? what happens when this electron comes spin polarized at an angle to the interface ? note that unlike in superconductivity, there is a direction associated with the half metal (which is basically a magnet with an orientation)&lt;/p&gt;&#xA;" OwnerUserId="3714" LastActivityDate="2011-06-29T08:04:36.763" Title="How to apply Andreev reflection formalism to ferromagnet ,normal metal interface?" Tags="&lt;electromagnetism&gt;&lt;spin&gt;&lt;superconductivity&gt;&lt;reflection&gt;&lt;spinors&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="11707" PostTypeId="1" AcceptedAnswerId="11716" CreationDate="2011-06-29T07:28:12.993" Score="3" ViewCount="135" Body="&lt;p&gt;I was visiting the Australian Synchrotron earlier today as part of a tour group; as the guide was going over the booster and storage rings I was reminded of something I learnt of quantum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I know my quantum well enough, every so often, there are spontaneous pairs of electrons and positrons created and destroyed everywhere. Now I know that they don't last very long, but I got to thinking, what would happen if a pair was created at just the right instant for the positron to collide with one of the super-high-energy electrons in the booster or storage ring of a synchrotron?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I asked our guide he said that the probability was pretty low since there were only 1x10^6 electrons in there and naturally the pair would have to generate at just the right spot at just the right time. Ultimately he wasn't quite sure what would happen if the circumstances were correct for it though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been doing some further thinking, if I remember correctly, these pairs form as an electron spiralling in towards a central positron. I think that therefore at ordinary energies any nearby electrons would be repelled by the electron around the positron (which is a little reminiscent of the atomic stuff of first year chemistry).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question is, could an electron, at the sorts of energy levels of the order of giga-electron-volts, get past the spiralling electron and collide with the positron instead of its paired electron doing so? As this energy could not simply vanish, would you then get a lonely (now unpaired) electron and an explosion? or would the energy somehow transfer itself to the other electron?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My interest here is towards what would actually happen if all circumstances were just right; also appreciated would be anyone who might be able to tell me the theoretical frequency of electron-positron pairs being spontaneously generated, as well as helping me to think up a way of determining the probability of finding an electron within one of the rings (for instance, perhaps if electrons can be considered to have an effective 'volume' within which they are most likely to appear and comparing that to the volume of the ring?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your time, I really appreciate it.&lt;/p&gt;&#xA;" OwnerUserId="4223" LastActivityDate="2011-06-29T19:49:42.783" Title="What happens where an electron is annihilated by a spontaneously generated positron-electron pair?" Tags="&lt;quantum-mechanics&gt;&lt;electron&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="11708" PostTypeId="2" ParentId="11706" CreationDate="2011-06-29T08:04:36.763" Score="1" ViewCount="" Body="&lt;p&gt;There is no Andreev reflection in this case. You have to quantify the&#xA;electron spin on the direction of the half-metal. The quantization&#xA;direction is up to you, the electron does not decide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the electron is up, it will be transmitted with some probability&#xA;depending on the precise band-structure matching. If it is down, it will&#xA;be 100% reflected (assuming an ideal half-metal). If it happens to be at&#xA;an angle, then you will just describe it as a quantum combination of up&#xA;and down states, and you will get an angle-dependent transmission&#xA;probability.&lt;/p&gt;&#xA;" OwnerUserId="4202" LastActivityDate="2011-06-29T08:04:36.763" />
  <row Id="11709" PostTypeId="1" CreationDate="2011-06-29T08:28:44.180" Score="4" ViewCount="127" Body="&lt;p&gt;There lies a homogeneous rigid rod of mass $M$ and of length $H$ on a frictionless table at rest. A small bullet of mass $m$ moves toward the rod with velocity $v_0$, perpendicular to the rod and collides absolutely elastically with the rod at a point, located at a distance $x$ from the center of the rod. Before they finally move away from each other, they may still collide each other a few times.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the maximum number of possible collisions of the bullet with the rod before they leave each other forever?&lt;/p&gt;&#xA;" OwnerUserId="552" LastActivityDate="2011-06-29T13:39:27.910" Title="Collision between a rod and a bullet" Tags="&lt;classical-mechanics&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="11710" PostTypeId="2" ParentId="11646" CreationDate="2011-06-29T09:35:40.857" Score="3" ViewCount="" Body="&lt;p&gt;The connection between thermodynamic and information theoretic entropies is a deep connection, and has been an active research area for more than a century. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get more information, you can look at&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://plato.stanford.edu/entries/information-entropy/&quot; rel=&quot;nofollow&quot;&gt;This well written comprehensive review&lt;/a&gt; at Stanford Encyclopedia of Philosophy, including the link with macroscopic thermodynamical entropy.&lt;/li&gt;&#xA;&lt;li&gt;This &lt;a href=&quot;http://lcni.uoregon.edu/~mark/Stat_mech/thermodynamic_entropy_and_information.html&quot; rel=&quot;nofollow&quot;&gt;webpage&lt;/a&gt; and this &lt;a href=&quot;http://en.wikipedia.org/wiki/Entropy_in_thermodynamics_and_information_theory&quot; rel=&quot;nofollow&quot;&gt;wikipedia page&lt;/a&gt; on the subject give shorter overviews of some of the links between these entropies. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Various Researches on Maxwell's demon,Szilard's engine and Landauer's principle can also be helpful. &lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2011-06-29T09:35:40.857" />
  <row Id="11711" PostTypeId="1" AcceptedAnswerId="11749" CreationDate="2011-06-29T10:53:53.597" Score="1" ViewCount="65" Body="&lt;p&gt;As I see it, as light comes through approximately same angular area of the lens, it should have less aberrations, and at the same time, it should still have high θ -&gt; diffraction limit should stay as without this diaphragm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing?&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-07-01T07:54:52.240" Title="Does 'ring' diaphragm reduce lens spherical abberations?" Tags="&lt;optics&gt;" AnswerCount="1" />
  <row Id="11712" PostTypeId="2" ParentId="11675" CreationDate="2011-06-29T11:05:10.977" Score="3" ViewCount="" Body="&lt;p&gt;Superselection is about relative phases. In the particular case of $e+\bar{e}$, it concerns the U(1) symmetry. Both observables and states have to be invariant under symmetry transformations (for the symmetry to &lt;em&gt;be&lt;/em&gt; a symmetry), but $e$ and $\bar{e}$ transform differently under U(1) transformations, so $e+\bar{e}$ cannot be invariant because it transforms to $\mathrm{e}^{i\alpha}e+\mathrm{e}^{-i\alpha}\bar{e}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;definition&lt;/em&gt; of different particle wave functions is that they transform differently under the symmetries of the system, so we can't take superpositions of different particles by definition. Of course we &lt;em&gt;can&lt;/em&gt; construct tensor product states and mixtures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A separate question, of course, is why one symmetry and not another, but once you introduce a symmetry into quantum mechanics, a lot follows.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-06-29T11:10:22.583" LastActivityDate="2011-06-29T11:10:22.583" />
  <row Id="11713" PostTypeId="2" ParentId="11702" CreationDate="2011-06-29T11:25:54.803" Score="3" ViewCount="" Body="&lt;p&gt;There's a lot of creativity in constructing an experiment, exploiting many different interactions in many different configurations, so there isn't a one size fits all Answer to your Question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One source of decoherence, however, are thermal fluctuations of the EM field (the immediate environment), which are driven by the thermal fluctuations of whatever surrounds the system of experimental interest. To reduce the effects of decoherence, we can surround the system by something cold, so that the EM field is more driven by the cold surrounding, and much less driven by the hotter surroundings that are further away. How we keep the cold surroundings from heating up, and how we create the cold surroundings in the first place, are relatively modern miracles of invention, which in Physics experiments are likely to be multi-stage exploits. To some considerable extent, as refrigeration technology improves, so improves Physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can't totally eliminate thermal fluctuations of the EM field, insofar as the 3rd law of thermodynamics is empirically supported, so yes, the surroundings still affect the system of experimental interest, but &lt;em&gt;less&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-06-29T11:25:54.803" CommentCount="6" />
  <row Id="11714" PostTypeId="2" ParentId="11709" CreationDate="2011-06-29T12:58:42.643" Score="2" ViewCount="" Body="&lt;p&gt;I don't know if there can be a closed form solution. Here's an estimate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in the centre of mass frame, the radial distance to the particle is hyperbolic in the sense that once the radial velocity of the particle is positive, it can not decrease any more (between collisions this is obvious from the conversion between Cartesian and polar coordinates, and at collisions this follows from the conservation of linear momentum). So after the initial impact (which is perpendicular to the rod, and hence starts with zero radial velocity), the radial velocity will be no less than the equivalent linear motion with no collisions. This allows us to get an estimate on the maximum time the particle remains in &quot;batting distance&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, by conservation of energy, we can get an absolute upper bound on the relative angular velocity of the rod versus the particle, which combined with the maximum time we have, we can get a bound of the maximum number of collisions (between collisions, the angular position of the rod versus the particle in the frame of the centre of rod's mass must change by at least $\pi$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we implement this (very crude) estimate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we compute the situation after the first collision. For convenience we re-scale $M+m = 1$, and we re-define the &lt;em&gt;radius&lt;/em&gt; of the rod to be $H$. Let $v&amp;#39;_0$ be the initial velocity of the particle in the centre-of-mass frame (so $v_0' = Mv_0$). In the COM frame, letting $v_m$ be the velocity of the particle, the total translational kinetic energy is $T = \frac12 \frac{m}{M} v_m^2$. Letting $\omega$ be the angular velocity of the rod about its mass centre, the rotational kinetic energy is $R = \frac16 MH^2\omega^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the COM frame, at the moment of the perpendicular collision (when the COM is on the rod), the total angular momentum is given by&#xA;$$ L  = m v_m x + \frac{1}{3}M\omega H^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So conservation of energy and conservation of momentum says that after the initial impact, $v_m$ satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{m}{M}(1 + \frac{m}{M}) \left( v&amp;#39;_0^2 - v_m^2\right) = \frac13 \left(\frac{3mx}{MH}\right)^2 \left(v&amp;#39;_0 - v_m\right)^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which one can in principle solve using the quadratic formula. The total time within batting range would be bounded above by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T_B = \frac{\sqrt{H^2 - x^2}}{v_m} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using just the conservation of energy, a very crude upper bound for the relative angular speed between the spinning rod and the particle (using the fact that after the first collision, the centres of masses of the particle and the rod only increase in distance, so subsequent impact spots will get further and further away from the centre of the rod) can be written as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |\Omega| \leq \sqrt{2} \frac{|v&amp;#39;_0|}{Mx} \max (1, 2 \frac{x}{H}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the bound on the maximum number of collisions will be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T_B |\Omega| \pi^{-1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this estimate is very wasteful when $x\to 0$, since in that limit, we expect the elastic collision to send the rod off not spinning, and there would be only one collision. This estimate is also completely inadequate to treat the edge case where $v_m = 0$ is the solution to the first collision, in which case the total number of collisions should be two. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle someone more clever can take this and get a much more refined estimate. (Or perhaps &quot;intuit&quot; an answer otherwise.)&lt;/p&gt;&#xA;" OwnerUserId="829" LastEditorUserId="829" LastEditorDisplayName="" LastEditDate="2011-06-29T13:39:27.910" LastActivityDate="2011-06-29T13:39:27.910" CommentCount="3" />
  <row Id="11715" PostTypeId="1" AcceptedAnswerId="11723" CreationDate="2011-06-29T13:22:43.930" Score="1" ViewCount="120" Body="&lt;p&gt;I'm looking for a list of immiscible (and miscible) liquids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I currently developing a game that will use the property of differents liquid, some of them will mix, other will not (positioning in the recipient is important so density is).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm searching but not found anything very relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course I know water, oil (and mercury). But I'm looking for more than two (at least 8 differents elements in three or four groups).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can always invent them but I'd like to keep things real as far as I can. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elements must be at the liquid state in normal conditions.&lt;/p&gt;&#xA;" OwnerUserId="3877" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-06-29T13:34:53.550" LastActivityDate="2011-06-30T10:44:41.273" Title="List of immiscible liquids" Tags="&lt;physical-chemistry&gt;" AnswerCount="1" />
  <row Id="11716" PostTypeId="2" ParentId="11707" CreationDate="2011-06-29T13:42:40.230" Score="3" ViewCount="" Body="&lt;p&gt;Welcome to the community WizzPhiz. You do not state your age in your profile, but I would think &amp;lt;20?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no spontaneously physical electron positron pairs created from the vacuum. The reason is called &quot;conservation of energy&quot; It would take energy to create such a pair. The vacuum sea consists of &lt;a href=&quot;http://en.wikipedia.org/wiki/Virtual_particle&quot; rel=&quot;nofollow&quot;&gt;&quot;virtual particles&quot;&lt;/a&gt; and the electrons going around the accelerator cannot &quot;see&quot; them  as in addition to being virtual they are created and annihilated in  a small delta(time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://viper.princeton.edu/~mcdonald/e144/adrianfestdoc.pdf&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt;, which is a festschrift for a scientist, in paragraphs 2 and later there are a lot of explorations of electrons scattering off radiation, even very low black body radiation that exists in a vacuum, but these are not the Dirac sea pairs you are asking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vacuum pairs do have experimental signatures in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Casimir_effect&quot; rel=&quot;nofollow&quot;&gt;Casimir&lt;/a&gt; effect, and in the widening of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lamb_shift&quot; rel=&quot;nofollow&quot;&gt;Lamb shift&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one goes to General Relativity the vacuum particles theoretically may become physical in accelerated systems but there is no solid experimental evidence of this. The beam energies in the accelerators we have are not in that ball park of acceleration.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-06-29T19:49:42.783" LastActivityDate="2011-06-29T19:49:42.783" CommentCount="8" />
  <row Id="11717" PostTypeId="2" ParentId="11693" CreationDate="2011-06-29T14:12:04.187" Score="2" ViewCount="" Body="&lt;p&gt;What you need to apply is the engineering equations for a heat exchanger.  In the below equation, $\dot{Q}$ is the heat transfer rate, $UA$ is the heat transfer coefficient times the area and $\Delta T_m$ is the log mean temperature difference (LMTD).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{Q}=U A \Delta T_m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this problem there are 3 barriers to the heat transfer in series.  You have the convective heat transfer coefficient of the syrup $h_c$, the convective heat transfer of the steam $h_s$, and that of the pipe metal $h_R$.  Take $n$ to be the number of pipes, $r_i$ to be the inner radius of the pipe, and $r_0$ to be the outer radius of the pipe, and $R$ the heat transfer coefficient of the Copper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$UA = \frac{2\pi n}{\frac{1}{h_s r_i}+\frac{1}{R L} ln \frac{r_o}{r_i} +\frac{1}{h_c r_0} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LMTD is the average temperature difference, but for the specific case where one part of the heat exchanger is saturated, and thus constant temperature, just know that it is the following, where $T_{s}$ is the saturation temperature of the steam, or 100 degrees C.  Then $T_h$ and $T_c$ are the temperatures after and before preheating respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta T_m = \frac{T_h-T_c}{ln\frac{T_h-T_s}{T_h-T_s} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hard part of the above equations is the $h_c$ and the $h_s$.  You will probably use things like the Dittus-Boelter correlation.  But it's more important that for the moment we address $\dot{Q}$ itself.  For the described hood, I see two possibilities.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&lt;li&gt;The steam is being provided at a faster rate than it is being condensed&#xA;&lt;li&gt;The steam is being provided at a slower rate that it is being condensed&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In the first case, you will see steam leaking out of the hood.  In this case, the equilibrium operation see the air mostly evacuated because it is pushed out.  In the other case, steam is present &lt;em&gt;with&lt;/em&gt; air and the air reduces the heat transfer.  In case #2 the given is the rate of steam condensation, which directly determines $\dot{Q}$ and $h_c$ adjusts to compensate.  In case #1 $h_c$ is a given based on the assumption of your geometry and the atmospheric steam heat transfer properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $h_s$ use &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_transfer_coefficient&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h_s={{k_w}\over{D_H}}Nu$$&#xA;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k_w$ - thermal conductivity of the liquid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D_H$ - $D_i$ - Hydraulic diameter (inner), Nu - Nusselt number&#xA; $Nu = {0.023} \cdot Re^{0.8} \cdot Pr^{n}$  (Dittus-Boelter correlation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pr - Prandtl number, Re - Reynolds number, n = 0.4 for heating (wall hotter than the bulk fluid) and 0.33 for cooling (wall cooler than the bulk fluid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't have right now are the numbers for applying the above for Maple Syrup and the approach for the steam side of the tubes.  This is all I have time for right now, but I think this amount will still be helpful.  I can try to look up these things later, maybe you can specify what you understand the least first.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-06-29T14:12:04.187" CommentCount="1" />
  <row Id="11718" PostTypeId="2" ParentId="11707" CreationDate="2011-06-29T14:24:15.300" Score="3" ViewCount="" Body="&lt;p&gt;The electron is annihilated by the positron of the pair, and the other electron stays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;normal&quot;  virtual creation/annihilation process is &#xA;$$\text{nothing}&amp;#160;\rightarrow e^- + e^+ \rightarrow \text{nothing}$$ &#xA;If you add an electron nearby which does not interact with the pair, this becomes&#xA;$$e^- \rightarrow 2e^- + e^+ \rightarrow e^-$$&#xA;and if the electron is annihilated by the virtual pair's, it becomes&#xA;$$e^- \rightarrow 2e^- + e^+ \rightarrow e^-.$$&#xA;Since the electrons are undistinguishable, the two last equations are the same:  there is no difference to which electron the positron &quot;choses&quot; to annihilate. And actually, they are the same as the boring equation where no virtual pair appears : &#xA;$$e^-\rightarrow e^-$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a more technical side, the 3 processes correspond to 3 different Feymann diagrams, but the indistinguishability of their outputs make them interfere and and doesn't allow us to infer what has &quot;really&quot; happened. And the question of &quot;what happened in reality&quot; does not have a sense (that's why the pair is called &quot;virtual&quot;). We can only compute the contribution of the the various processes to movement of the electron. &lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2011-06-29T14:24:15.300" CommentCount="2" />
  <row Id="11719" PostTypeId="1" CreationDate="2011-06-29T15:19:32.237" Score="1" ViewCount="33" Body="&lt;p&gt;The dynamic viscosity of air is $1.78\cdot 10^{-5}Pas = 1.78\cdot 10^{-5}\frac{kg}{ms}$. How do I convert this to the minute scale? I see a reason for multiplying the value with 60 (because of the $\frac{kg}{ms}$) or dividing it by 60 (because of the $Pa s$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Background: I need the value in a numerical CFD simulation, where I have all my timesteps in the minute scale.&lt;/p&gt;&#xA;" OwnerUserId="2480" LastActivityDate="2011-06-29T17:36:06.020" Title="dynamic viscosity in the minute scale" Tags="&lt;unit-conversion&gt;" AnswerCount="1" />
  <row Id="11720" PostTypeId="1" CreationDate="2011-06-29T16:43:58.633" Score="0" ViewCount="138" Body="&lt;p&gt;When we are in beach we can see waves coming when the sun light goes of the waves amplitude will increase what is the reason &#xA;Some people said me that for moon light the frequency and velocity of waves will increase &#xA;What is the reason behind that? but only at night these waves frequency will increase what makes it so&lt;/p&gt;&#xA;" OwnerUserId="3590" LastEditorUserId="3590" LastEditorDisplayName="" LastEditDate="2011-06-29T16:52:21.953" LastActivityDate="2011-08-29T01:36:18.040" Title="What makes waves at beaches to come with high velocity and frequency at nights?" Tags="&lt;waves&gt;&lt;water&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="11721" PostTypeId="1" CreationDate="2011-06-29T16:58:42.900" Score="-2" ViewCount="69" Body="&lt;p&gt;normally at ocean beaches the waves are normal but they make lot of that a person standing at far can even listen that sound&lt;br&gt;&#xA;What makes that sound what going to happen a&lt;/p&gt;&#xA;" OwnerUserId="3590" LastActivityDate="2011-06-30T17:43:54.150" ClosedDate="2011-06-30T19:02:29.737" Title="What makes sound when a huge wave is coming in an ocean?" Tags="&lt;waves&gt;&lt;water&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11722" PostTypeId="1" CreationDate="2011-06-29T17:04:37.517" Score="2" ViewCount="76" Body="&lt;p&gt;This is in continuation to what I was asking here earlier -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/11438/argument-for-quantum-theoretic-conformality-of-caln-2-super-chern-simons-th&quot;&gt;Argument for quantum theoretic conformality of $\cal{N}=2$ super-Chern-Simon&amp;#39;s theory in $2+1$ dimensions&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or one can look at this paper - &lt;a href=&quot;http://arxiv.org/pdf/0704.3740&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/0704.3740&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am mainly having in mind the $\cal{N}=2$ lagrangian that would be obtained by combining the equations 2.4, 2.5 and 2.7 of that paper.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In that light I would like to understand the meaning of the comment made in that paper just below equation 2.9 &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I have heard that one way of arguing the existence of the $\cal{N}=3$ fixed point in the flow of the matter couplings of the $\cal{N}=2$ theory is to see that for large values of the coupling the theory is effectively behaving like the Wess-Zumino model and for small values it is like pure-super-Chern-Simons theory. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is known about the signs and the derivatives of the beat functions of these two limiting theories?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like if it known that the sign of the beta function is different for these two cases then it might suffice to argue by continuity that the beta function does go to zero somewhere in between and then that is a strong candidate to be the $\cal{N}=3$ theory. &lt;/p&gt;&#xA;" OwnerUserId="1637" LastActivityDate="2011-07-01T17:50:25.793" Title="Argument for quantum theoretic conformality of $\cal{N}=2$ super-Chern-Simon's theory in $2+1$ dimensions -Part 2" Tags="&lt;quantum-field-theory&gt;&lt;supersymmetry&gt;&lt;conformal-field-theory&gt;&lt;superconformality&gt;" FavoriteCount="1" />
  <row Id="11723" PostTypeId="2" ParentId="11715" CreationDate="2011-06-29T17:05:44.580" Score="0" ViewCount="" Body="&lt;p&gt;There is a &quot;common&quot; chart in most chemistry labs that covers exactly this topic. One version of it can be found here: &lt;a href=&quot;http://chemistry.kenyon.edu/getzler/Research%20files/miscibility_elutropicity.pdf&quot; rel=&quot;nofollow&quot;&gt;http://chemistry.kenyon.edu/getzler/Research%20files/miscibility_elutropicity.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference's sake, this was found by a Google Image search using the term &quot;miscibility.&quot; A clear case of having to know something exists in order to find it.&lt;/p&gt;&#xA;" OwnerUserId="1196" LastActivityDate="2011-06-29T17:05:44.580" CommentCount="1" />
  <row Id="11724" PostTypeId="1" CreationDate="2011-06-29T17:09:25.657" Score="2" ViewCount="60" Body="&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/mELg2.jpg&quot; alt=&quot;two muppets&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After Yield point on stress strain diagram the under curve occurs what does it mean what will happen for the mild steel at that particular time and again why the curve goes to up and reaches ultimate point  i am not getting that WHY that under curve comes&lt;/p&gt;&#xA;" OwnerUserId="3590" LastActivityDate="2011-06-30T03:01:35.427" Title="Stress vs Strain for mild steel" Tags="&lt;classical-mechanics&gt;&lt;mechanics&gt;" AnswerCount="2" />
  <row Id="11725" PostTypeId="1" CreationDate="2011-06-29T17:11:03.240" Score="0" ViewCount="62" Body="&lt;p&gt;From my understanding, the Big Bang is theorized to have been a result of matter ejecting from a decayed white hole space/time singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.infidels.org/library/modern/quentin_smith/uncaused.html&quot; rel=&quot;nofollow&quot;&gt;http://www.infidels.org/library/modern/quentin_smith/uncaused.html&lt;/a&gt; provides an explanation for the beginning of the universe. Assuming I understand it correctly, there are distortions of space, time, and matter called singularities. A black hole is one, and in it is theorized to be of infinite mass, infinite density, infinite space, and most importantly, infinite &lt;em&gt;time&lt;/em&gt;. So, there essentially is no measure of time inside of them. It is theorized that before the universe began there existed a singularity, and out of the singularity came the universe. So, considering that there is no measure of time in a singularity, it could have always existed and have had no beginning. Furthermore, there would have existed nothing outside of this singularity, so there could have been no measure of time outside of it either. At some point, matter and energy were ejected from the white hole, and this event was the &quot;big bang,&quot; the beginning of our universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this accurate, and if not, what of my interpretation needs to change?&lt;/p&gt;&#xA;" OwnerUserId="4225" LastActivityDate="2011-06-29T17:11:03.240" ClosedDate="2011-06-29T18:22:24.610" Title="Was the Big Bang a result of a decayed white hole singularity?" Tags="&lt;big-bang&gt;&lt;singularity&gt;" CommentCount="5" />
  <row Id="11726" PostTypeId="1" CreationDate="2011-06-29T17:11:31.657" Score="0" ViewCount="83" Body="&lt;p&gt;In reading another question about &lt;a href=&quot;http://physics.stackexchange.com/questions/5462/does-the-potential-energy-for-a-given-photon-increase-or-decrease-in-quanta&quot;&gt;gravity's effects on a photon&lt;/a&gt;, I wondered if it were possible for a photon to ever be redshifted to zero wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that black holes have a gravity field strong enough to keep gravity from escaping but what about the affect on the photon that is escaping.  Say we have a single photon escaping on a path directly away from a singularity.  Also conjecture that we can release a single photon at various distances from the singularity (still on a trajectory directly away) so as to select the strength of the gravity field it must traverse.  I know the distance has already been calculated (the event horizon I believe) but what of the photon itself?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a point where the amount of redshifting (e.g. gravitational) causes the wavelegth of the photon to become zero? &lt;a href=&quot;http://physics.stackexchange.com/questions/4821/redshifted-photon-energy/4822#4822&quot;&gt;This answer mentions the change in potential and kinetic energy.&lt;/a&gt;  Is it possible for all of a photon's kinetic energy be transformed to potential energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the implications, both it it were possible and/or impossible? More importantly if impossible, what makes it impossible?  This would beg the question, what is the smallest amount of energy possible (or quanta) for a photon? If that minima-photon attempted to escape from within a gravity field, it can't slow down and can't lose wavelength, so what does it do? &lt;/p&gt;&#xA;" OwnerUserId="4224" LastActivityDate="2011-06-29T18:20:39.930" Title="Can a photon exiting from a gravity well ever reach a wavelength of zero?" Tags="&lt;photons&gt;&lt;gravitational-redshift&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11727" PostTypeId="1" CreationDate="2011-06-29T17:22:48.287" Score="0" ViewCount="50" Body="&lt;p&gt;What is the structure of the AL and c.i. These strength do depend fully on cores grain and fine grain structure or is there any other difference between the two metals What is the structure of these two metals what makes AL such good metal?  &lt;/p&gt;&#xA;" OwnerUserId="3590" LastActivityDate="2011-06-30T17:40:20.847" Title="What makes the aluminium metal having good strength and lighter in weight when compare to cast iron?" Tags="&lt;metals&gt;&lt;meteorology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11728" PostTypeId="2" ParentId="11719" CreationDate="2011-06-29T17:36:06.020" Score="1" ViewCount="" Body="&lt;p&gt;If you just want to do a straight up unit conversion, do the proper steps for dimensional analysis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;1.78\times 10^{-5}\frac{kg}{m\dot\ sec} * \frac{60\ sec}{1\ min} = (1.78 \times 10^{-5}) * 60 \frac{kg}{m\dot\ min} &#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the seconds cancel and you're left with the term on the &quot;minute scale&quot;, if this is what you mean.&lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2011-06-29T17:36:06.020" CommentCount="3" />
  <row Id="11729" PostTypeId="2" ParentId="11726" CreationDate="2011-06-29T18:20:39.930" Score="3" ViewCount="" Body="&lt;p&gt;Ted's comment is correct, redshift makes the wavelength increase. But if you actually meant zero &lt;em&gt;frequency&lt;/em&gt; (which corresponds to infinite wavelength), then it's almost possible, if you emit a photon from &lt;em&gt;just&lt;/em&gt; outside the event horizon of a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, suppose the photon is emitted from a small coordinate distance $\epsilon R_s$ ($\epsilon \ll 1$) outside the event horizon. According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_redshift&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, the redshift of that photon once it reaches an infinite distance away from the hole would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{r\to +\infty}z(r) = \frac{1}{\sqrt{1-\left(\frac{R_s}{R^*}\right)}}-1 = \frac{1}{\sqrt{1-\left(\frac{R_s}{R_s(1 + \epsilon)}\right)}}-1 \approx \sqrt{\frac{1}{\epsilon}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Redshift relates to wavelength as $z = \frac{\lambda_o - \lambda_e}{\lambda_e}$ ($\lambda_o$ is the observed wavelength, $\lambda_e$ is the wavelength at emission), so given that $\epsilon$ is very small,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda_o \approx \frac{\lambda_e}{\sqrt{\epsilon}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, by emitting a photon close enough to the event horizon, you can arrange for it to be redshifted to as large a positive wavelength (and thus as small a frequency) as you want. But there's no value of $\epsilon$ that will actually give you an infinite wavelength (zero frequency).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to go all the way down to $\epsilon = 0$, i.e. emit the photon from right on the event horizon, then it would just be stuck there, since there are no outgoing null geodesics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;*Everything here applies only to Schwarzschild black holes, although I don't know of anything that would allow a photon to be redshifted to infinite wavelength from any other kind of black hole either.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-06-29T18:20:39.930" CommentCount="4" />
  <row Id="11730" PostTypeId="2" ParentId="11724" CreationDate="2011-06-29T18:51:42.243" Score="2" ViewCount="" Body="&lt;p&gt;I think you're getting confused between stress and engineering stress. As a material has more and more strain, it starts to thin (called necking), and the engineering stress declines. However, measured relative to cross sectional area, the stress continues to increase.&lt;/p&gt;&#xA;" OwnerUserId="4033" LastActivityDate="2011-06-29T18:51:42.243" />
  <row Id="11731" PostTypeId="2" ParentId="11720" CreationDate="2011-06-29T20:20:32.853" Score="0" ViewCount="" Body="&lt;p&gt;It doesn't, all acepted models of wave generation base solely on force transmission by wind, current, sea states and gravity. Any &quot;light inducted effect&quot; if ever exist is neglected by comparasion with the other forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards&lt;/p&gt;&#xA;" OwnerUserId="3836" LastActivityDate="2011-06-29T20:20:32.853" />
  <row Id="11732" PostTypeId="1" AcceptedAnswerId="11733" CreationDate="2011-06-29T20:53:10.143" Score="1" ViewCount="95" Body="&lt;p&gt;I'm studying a lightbulb and its variable resistance, given by the expression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R(T) = Ro[1 + α(T-T_0)]$, where $R_0$ is the resistance of the lamp at $T_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, $R$ is not given by Ohm's law ($V=Ri$). So, which expression can I use to calculate the power consumed by a lightbulb?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$P = R\cdot i^2$&lt;/li&gt;&#xA;&lt;li&gt;$P = \frac{V^2}{R}$&lt;/li&gt;&#xA;&lt;li&gt;$P = V\cdot i$ (I don't think I can use this one)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I have measured the current and the e.p.d. at the bulb using instruments.&lt;/p&gt;&#xA;" OwnerUserId="2613" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2011-07-01T14:30:16.623" LastActivityDate="2011-07-01T14:30:16.623" Title="How do I calculate the power consumed by a lightbulb?" Tags="&lt;electromagnetism&gt;&lt;light&gt;&lt;electricity&gt;&lt;power&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="11733" PostTypeId="2" ParentId="11732" CreationDate="2011-06-29T20:56:52.217" Score="4" ViewCount="" Body="&lt;p&gt;All those expressions are equivalent and can be used, if you simultaneously measure, the voltage, current (and know the resistance from the temperature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do you think ohm's law doesn't apply?  Resistance is only dependent on temperature in your case (not current or voltage).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general I would use P = IV, if Ohm's law actually doesn't apply.&lt;/p&gt;&#xA;" OwnerUserId="1467" LastActivityDate="2011-06-29T20:56:52.217" CommentCount="1" />
  <row Id="11734" PostTypeId="2" ParentId="8246" CreationDate="2011-06-29T21:02:54.053" Score="1" ViewCount="" Body="&lt;p&gt;I can't vouch for the answer &lt;a href=&quot;http://www.goshen.edu/merrylea/sugar/physics.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, but it purports to be the reason.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The evaporator is perfectly level. If&#xA;  that's the case, visitors often wonder&#xA;  how it is that sap moves through the&#xA;  pan. Forgotten is the fact that rising&#xA;  steam is the equivalent of dipping or&#xA;  draining liquid from the pan. A level&#xA;  bottom means that sap moves under the&#xA;  pull of gravity as water leaves the&#xA;  pan as steam. (That, of course, is the&#xA;  whole objective of the process!) The&#xA;  sugar stays behind forming a syrup of&#xA;  increasing sweetness to make the&#xA;  finished product. At that point, the&#xA;  sugar concentration is at 60% and the&#xA;  boiling point is elevated about 7&#xA;  degrees. The difference in viscosity&#xA;  coupled with the baffles in the pan&#xA;  allows finished syrup to accumulate in&#xA;  the last section without our ever&#xA;  having to mechanically separate the&#xA;  final product. At the end of the&#xA;  season or if, for some reason, we wish&#xA;  to shut down the process and empty the&#xA;  pans, it is possible to draw off&#xA;  virtually all the syrup. One could&#xA;  not, of course, simply drain the pan.&#xA;  The fire would cause melt-down in a&#xA;  matter of seconds. Instead, soft water&#xA;  from the farm pond takes the place of&#xA;  the entering raw sap and supplies the&#xA;  necessary cooling liquid. Boiling&#xA;  continues, and because of their&#xA;  different viscosities, the syrup and&#xA;  pond water do not mix significantly.&#xA;  This same situation occurs when oils&#xA;  of different densities flow&#xA;  sequentially in a transcontinental&#xA;  pipeline. Very little mixing occurs at&#xA;  the interface between the two&#xA;  products. Virtually all the syrup can&#xA;  be drawn off before the pond water&#xA;  arrives at the discharge valve.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suppose the big question is why doesn't the liquid become homogeneous?  That's probably due to the baffles limiting the interface significantly - reducing the the ability of the most concentrated fluid to mix with the most diluted.  &lt;/p&gt;&#xA;" OwnerUserId="4221" LastActivityDate="2011-06-29T21:02:54.053" CommentCount="1" />
  <row Id="11735" PostTypeId="2" ParentId="11720" CreationDate="2011-06-29T22:33:34.720" Score="0" ViewCount="" Body="&lt;p&gt;Light and heat from the sun causes the water to worm up and evaporate causing air currents which in turn changes the frequency of the waves. The divergence of heat absorption of the land and the ocean courses a temperature difference which in turn causes large masses of air to flow and change the wave size and frequency. At night when the temperature somewhat equalizes the wind dies down the waves mellow out and the frequency changes. Now by earth standards these are local effects. You need to study the major air currents and the colories effect of the earth spinning has on global wave currents and air flow&lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-06-29T22:33:34.720" />
  <row Id="11736" PostTypeId="1" CreationDate="2011-06-29T23:15:34.047" Score="3" ViewCount="174" Body="&lt;p&gt;This question &lt;em&gt;is&lt;/em&gt; cross-posted from Bicycles.SE, but it is really one for those that know a bit about physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the back tyre of a bicycle wear out quicker than the front tyre?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have my uneducated suspicions but I would appreciate an educated answer.&lt;/p&gt;&#xA;" OwnerUserId="4227" LastActivityDate="2011-06-30T16:19:38.627" Title="On a bicycle, why does my back tyre wear so much more quickly than the front?" Tags="&lt;friction&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="11737" PostTypeId="2" ParentId="11720" CreationDate="2011-06-30T00:52:31.950" Score="0" ViewCount="" Body="&lt;p&gt;I too have heard this phenomenon. At night, the waves will suddenly go lap-lap-lap-lap-lap-lap very quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At night, things are very calm, and the only waves are very tiny. At my local beach, the day time waves might be two or three feet tall. Nice body surfing. At night and early the next morning, the water is calm and the waves are small maybe 6 inches. Being small, they have a short distance between them so they come quickly. Since it is calm, they are very regular and precise. There are various interference phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have two sources of such waves, their action will depend upon the sum of the two waves. Since the source frequencies cannot be identical, there is a &quot;beat&quot; that develops. You can lay in bed hearing nothing at all. No breaking waves anywhere. And then a constructive interference begins to form and when they agree you will hear lap-lap-lap quickly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle of the wave front to the beach is normally zero, but when the direction is off then little curls will form as the wave breaks at different stations along the beachfront.&#xA;You can then hear a lap-lap-lap that travels from right to left in stereo.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastActivityDate="2011-06-30T00:52:31.950" />
  <row Id="11738" PostTypeId="1" CreationDate="2011-06-30T01:16:31.367" Score="2" ViewCount="57" Body="&lt;p&gt;Suppose I have a gas of particles that is initially uniformly distributed so that the number density is $n_0$ (number of particles per unit volume), and then I displace the particles by the vector field $\vec{d}(\vec{x})$ (i.e. the particle initially at position $\vec{x}$ is displaced by the vector $\vec{d}$). How is the resulting number density $n(\vec{x})$ related to the displacement vector $\vec{d}(\vec{x})$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure this must be done somewhere in a standard textbook but I can't find where.&lt;/p&gt;&#xA;" OwnerUserId="1429" LastActivityDate="2011-06-30T12:06:45.320" Title="In a gas of particles, how is the displacement vector related to the number density?" Tags="&lt;general-relativity&gt;&lt;classical-mechanics&gt;&lt;thermodynamics&gt;&lt;mathematical-physics&gt;&lt;vector-fields&gt;" AnswerCount="2" />
  <row Id="11739" PostTypeId="2" ParentId="11736" CreationDate="2011-06-30T01:58:59.363" Score="1" ViewCount="" Body="&lt;p&gt;Unless the cyclist is leaning forward most of the weight is on the back tire. Therefore it has more contact with the road and when you turn there is more lateral force on the back. You do more braking whith the back so you don’t go head first. And finally you are applying your entire forward force on it so that contributes to more wear.&lt;/p&gt;&#xA;" OwnerUserId="4034" LastEditorUserId="2525" LastEditorDisplayName="" LastEditDate="2011-06-30T16:19:38.627" LastActivityDate="2011-06-30T16:19:38.627" CommentCount="9" />
  <row Id="11740" PostTypeId="1" AcceptedAnswerId="11765" CreationDate="2011-06-30T02:57:26.813" Score="4" ViewCount="197" Body="&lt;p&gt;A book by C. J. Ballhausen led me to believe that a quick way to check that I performed step operators properly was by observing that the &quot;wave function should appear normalized,&quot; but I have found some issue applying this in practice and believe it is due to my misunderstanding of the underlying physics; I'm trying to understand what C.J.B. meant by that and if it applies in my case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In his case he was observing two equivalent electrons. Let's say they are two equivalent  $p$ electrons for the sake of example. (He was actually considering $d$ electrons and I can provide the specifics of that if it will help.) There are several unperturbed functions which can be described by symbols such as $(1^+ 0^+)$, in which case the first number means $m_l$ of the 1st 2p electron, the next symbol indicates its spin and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the term $ ^1 D : M_L = 2, M_S=0: (1^+ 1^-) $ is an eigenfunction of $p^2$ configuration that is known. Using a step down operator on the angular momentum gives: $M_L=1 : (2)^{(-1/2)} [ (1^+ 0^-) - (1^- 0^+) ]$. Here I get the impression that what we observe appears normalized because squaring the coefficients gives unity. I realize one could in principle perform $\int \psi^* \psi d \tau$, but I do not suspect that is what he means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we apply the step operator again we get $M_L=0: (6)^{(-1/2)} [ (1^+ -1^-) - (1^- -1^+) + 2(0^+ 0^-) ]$. Here the square of the coefficients is decidedly unity. His examples given in the book also happen to go to unity; is this just coincidence or is it going to always be true? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My specific example is as follows, starting with a $d^3$ configuration: $$\psi(L,M_L,S,M_S)=\psi(5,5,\frac{1}{2},\frac{1}{2})=(2^+,2^-,1^+)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the lowering operators gives $\sqrt{10} \psi(5,4,\frac{1}{2},\frac{1}{2}) = -\sqrt{4} (2^+,1^+,1^+) - \sqrt{4} (2^+,1^+,1^-) + \sqrt{6}(2^+,2^-,0^+)$ and the coefficients go to unity as expected. Above you will notice that the ordering in the first and third terms has changed and an odd permutation brings about a change in sign. Also notice the first term must be equal to zero by Pauli Principle. Dividing out we get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi(5,4,\frac{1}{2},\frac{1}{2}) = \sqrt{3/5} (2^+,2^-,0^+) - \sqrt{2/5} (2^+,1^+,1^-)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll notice that the coefficients squared sum up to 1, so all appears normalized and well. Now we apply the lowering operator again to give $\sqrt{(L-M_L+1)(L+M_L)} = \sqrt{(5-4+1)(5+4)} = \sqrt{(2)(9)} = \sqrt{18}$ times the function for $M_L=3$ $^1 H$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying to the RHS using $\sqrt{(l-m_l+1)(l+m_l)}$. We are working with $d$ orbitals, therefore $l=2$. So for the case of $m_l=2$ we get $\sqrt{(2-2+1)(2+2)}=\sqrt{4}$ and for $m_l=1$ we get $\sqrt{(2-1+1)(2+1)} = \sqrt{6}$ and finally for $m_l=0$ we get $\sqrt{(2-0+1)(2+0)}=\sqrt{(3)(2)} = \sqrt{6}$. Applying this gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sqrt{18} \psi(5,3) = \sqrt{3/5} [ \sqrt{4} (1^+, 2^-, 0^+) + \sqrt{4} (2^+,1^-,0^+) + \sqrt{6} (2^+, 2^-, -1^+)] $$ &lt;br/&gt;&#xA;$ - \sqrt{2/5} [ \sqrt{4} (1^+,1^+,1^-) + \sqrt{6} (2^+,0^+, 1^-) + \sqrt{6} (2^+,1^+,0^-) ]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplification results in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sqrt{18} \psi(5,3) = \sqrt{12/5} (1^+, 2^-, 0^+) + \sqrt{12/5} (2^+,1^-,0^+) + \sqrt{18/5} (2^+, 2^-, -1^+)$$ &lt;br/&gt; $ - \sqrt{8/5}(1^+,1^+,1^-) - \sqrt{12/5} (2^+,0^+, 1^-) - \sqrt{12/5}(2^+,1^+,0^-) $ &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fourth term cannot exist by the Pauli Principle, so we have instead,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sqrt{18} \psi(5,3) = \sqrt{12/5} (1^+, 2^-, 0^+) + \sqrt{12/5} (2^+,1^-,0^+)$$ &lt;br/&gt; $ + \sqrt{18/5} (2^+, 2^-, -1^+) - \sqrt{12/5} (2^+,0^+, 1^-) - \sqrt{12/5}(2^+,1^+,0^-) $ &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need to fix the ordering of the first term and the fourth term to give,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sqrt{18} \psi(5,3) = -\sqrt{12/5} (2^-, 1^&#xD;&#xA;+, 0^+) + \sqrt{12/5} (2^+,1^-,0^+) + \sqrt{18/5} (2^+, 2^-, -1^+)$$ &lt;br/&gt; $ + \sqrt{12/5} (2^+,1^-, 0^+) - \sqrt{12/5}(2^+,1^+,0-) $ &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we divide thru by $\sqrt{18} $ the like terms yielding,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \psi(5,3) = -\sqrt{12/90} (2^-, 1^&#xD;&#xA;+, 0^+) + \sqrt{12/90} (2^+,1^-,0^+) + \sqrt{18/90} (2^+, 2^-, -1^+)$$ &lt;br/&gt;  $+ \sqrt{12/90} (2^+,1^-, 0^+) - \sqrt{12/90}(2^+,1^+,0^-) $ &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our problem is that 12/90 + 12/90 +18/90 + 12/90 +12/90 =11/15, instead of 15/15. I'm sure my mistake is stupid somewhere, can somebody point out where I've gone wrong?&lt;/p&gt;&#xA;" OwnerUserId="4112" LastEditorUserId="4112" LastEditorDisplayName="" LastEditDate="2011-06-30T23:00:12.303" LastActivityDate="2011-07-02T07:44:53.627" Title="Wave function normalization" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="0" />
  <row Id="11741" PostTypeId="2" ParentId="11724" CreationDate="2011-06-30T03:01:35.427" Score="0" ViewCount="" Body="&lt;p&gt;The physics here get quite tricky. At point E the material has not changed in its molecular structure then there’s a small transition gap to point Y. You can compare this gap to the ice to liquid water transition. I make this analogy because from point Y on the material will act like elastic seamy liquid. The other thing that happens is the molecular alignment changes and the material actually gets harder in layman’s terms we call it “Forging”. This is why the material is getting thinner “necking” yet it takes more force until it gets to the U point. At point U the material is at its hardest and past this the molecular cohesion starts to brake down. After U we apply more force to break it “doted line” in actuality the material takes less force to failure R.&lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-06-30T03:01:35.427" />
  <row Id="11742" PostTypeId="2" ParentId="2071" CreationDate="2011-06-30T03:30:48.943" Score="1" ViewCount="" Body="&lt;p&gt;I have driven enough cars on ice ponds to have a good feel for it. Most of my driving was done with a 1974 Datsun pickup with a clutch. You simply start in low and give it just enough gas to not break free, sensing that point with engine tone. The breakaway spot is sensed by mashing and letting up on the accelerator pedal. You must deliberately and continuously cause a breakaway condition every two or three seconds so that you know how hard you can push it. You work on up through the gears, shifting higher as soon as possible. Higher gears are better because the wheels will not break away as easily. The maximum acceleration will be at a torque just below breakaway. I drove a K-Whopper cab over with a  21 speed Spicer for awhile. It is all in that sweet spot.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastActivityDate="2011-06-30T03:30:48.943" />
  <row Id="11743" PostTypeId="2" ParentId="11738" CreationDate="2011-06-30T07:28:06.403" Score="1" ViewCount="" Body="&lt;p&gt;Take a point A.  Find where it's been translated to, say A'.  Add to the density at A' the original density at A divided by the Jacobian of the transformation at A.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrate this over all A.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-06-30T07:28:06.403" CommentCount="1" />
  <row Id="11744" PostTypeId="1" AcceptedAnswerId="11754" CreationDate="2011-06-30T08:04:41.223" Score="7" ViewCount="237" Body="&lt;p&gt;In a &lt;a href=&quot;http://en.wikipedia.org/wiki/Tug_of_war&quot; rel=&quot;nofollow&quot;&gt;tug-of-war&lt;/a&gt; match today, my summer camp students were very concerned about putting the biggest people at the back of the rope.  Is there any advantage to this strategy?&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-07-02T18:50:04.393" Title="Does it matter how you order your tug-of-war participants?" Tags="&lt;newtonian-mechanics&gt;&lt;everyday-life&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="11745" PostTypeId="2" ParentId="11744" CreationDate="2011-06-30T09:28:34.960" Score="2" ViewCount="" Body="&lt;p&gt;I can see two advantages:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) the more heavy set the stronger the muscles, so it means a lot of the force is coming on a long arm. The tug is supposed to be linear but rotations will happen, and there will be amplification  of the effect as a lever arm. If the rope were solid this is obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Think of  two people of equal weight,on  a see-saw. The one with the shorter distance from the fulcrum is at a disadvantage and has to put in more force to swing.&#xA;A tug of war has a &quot;fulcrum&quot; at the center where the two teams meet.&#xA; Putting the heavy players  at the end will shift the centre of mass towards the end and will give an advantage in the impulse transfer by a longer distance from the center.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-07-02T18:50:04.393" LastActivityDate="2011-07-02T18:50:04.393" CommentCount="4" />
  <row Id="11747" PostTypeId="1" CreationDate="2011-06-30T11:02:46.987" Score="-4" ViewCount="86" Body="&lt;p&gt;I learned yesterday that the inverse of square roots is used to calculate the vectors of surface normals in 3d graphics. It seems like such a mind-bogglingly simple idea, and it leads me to wonder if it hasn't found use elsewhere. Is there anywhere in physics where one might need to calculate the reciprocal of a number's square root?&lt;/p&gt;&#xA;" OwnerUserId="993" LastActivityDate="2011-06-30T15:50:36.320" Title="What is a simple physics application of the reciprocal of a square root?" Tags="&lt;vectors&gt;" AnswerCount="1" />
  <row Id="11748" PostTypeId="2" ParentId="11738" CreationDate="2011-06-30T12:06:45.320" Score="2" ViewCount="" Body="&lt;p&gt;A simple 1D calculation gives, at first order, something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\frac{1}{n} = \frac{1}{n_0}\left(1+\vec{\nabla}\cdot\vec{d}\right)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but only if $\vec{d}$ is small enough. Otherwise, for calculating&#xA;$n(\vec{x})$, you need to evaluate the divergence at a point $\vec{x&amp;#39;}$&#xA;such that $\vec{x&amp;#39;}+\vec{d}(\vec{x&amp;#39;}) = \vec{x}$.&lt;/p&gt;&#xA;" OwnerUserId="4202" LastActivityDate="2011-06-30T12:06:45.320" CommentCount="3" />
  <row Id="11749" PostTypeId="2" ParentId="11711" CreationDate="2011-06-30T12:13:02.197" Score="2" ViewCount="" Body="&lt;p&gt;Less spherical aberration, yes, but not less coma, nor less astigmatism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diffraction would look like on a telescope with a very important&#xA;central obstruction. It will not have the usual diffraction pattern, but&#xA;the diffraction-limited MTF will be bad for low spatial frequencies&#xA;(rapid decay of contrast) and then flatten out in such a way that the&#xA;ultimate resolution (extinction of contrast) will be the same as for the&#xA;full aperture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I found this page with nice simulations of the effect of a&#xA;central obstruction on both the MTF and the images:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.damianpeach.com/simulation.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.damianpeach.com/simulation.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scroll down to &quot;Part 2. Performance effects of different size central obstructions&quot;.&#xA;As you can see, although the resolution (at MTF = 0) is the same, the&#xA;image with 50% obstruction looks bad because of the loss of contrast at&#xA;mid and low frequencies.&lt;/p&gt;&#xA;" OwnerUserId="4202" LastEditorUserId="4202" LastEditorDisplayName="" LastEditDate="2011-07-01T07:54:52.240" LastActivityDate="2011-07-01T07:54:52.240" CommentCount="3" />
  <row Id="11750" PostTypeId="2" ParentId="11736" CreationDate="2011-06-30T12:51:23.037" Score="3" ViewCount="" Body="&lt;p&gt;Certainly my experience when cycling has been broadly the same. Supposing we're cycling at a constant speed on a flat surface, there are, at the lowest level of thinking about this problem, four forces acting: (1) wind resistance, mostly horizontal and to the rear, but with smaller components acting on the wheel spokes that I will discuss in a moment; (2) gravity, vertical and downwards; (3) A force acting at the front tyre (about which more in a moment); (3) a force acting at the rear tyre (again, more in a moment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vertical force of gravity is balanced by the upward vertical forces acting at the tyres. The vertical forces acting through the front and rear tyres will be different depending on the mass distribution of the rider and of the bike components, but the vertical component of the force will not contribute as much to wear as will the horizontal component of the force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizontal forces at the tyres are the most interesting for this Question. Both tyres are at rest relative to the ground in the region of contact with the ground. There is frictional resistance to any attempt to make the tyres not be at rest relative to the ground wherever they are in contact. Tyre resistance starts and stops when the vertical motion of the tyre material takes it in and out of contact with the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a frictional force that acts &lt;em&gt;backwards&lt;/em&gt; on the front wheel that is sufficient to balance the air and axle resistance of the front wheel. I think this can best be thought of in terms of torque. The wheel is rotating at a constant speed, which is resisted by friction in the axle and by wind resistance, both of which act against the direction of rotation of the wheel. The force (properly, the torque) acting in the region of contact of the wheel with the ground has to act to keep the wheel rotating, which means that the force has to act backwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a forward force acting at the rear wheel, that balances the backward forces of wind resistance and at the front wheel. The upper part of the chain is under more tension than the lower part of the chain (because there is more force acting on one pedal than on the other), which acts as a torque on the rear wheel that is &lt;em&gt;much more&lt;/em&gt; than enough to balance the air and axle resistance acting on the rear wheel. In the absence of frictional resistance provided by the tyre at the ground (and &lt;em&gt;transmitted&lt;/em&gt; by the spokes, but I'm not going into how that could be said in more detail), the rear wheel would accelerate to much higher speed (turning the bike upside down shows how fast the rear wheel turns with only a little arm strength if there is only wind and axle resistance). The frictional resistance at the rear wheel therefore acts to keep the rear wheel from rotating too fast, so it acts in the forward direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the frictional resistance at the rear wheel balances both the frictional resistance at the front wheel and the wind resistance, the frictional resistance at the rear wheel is strictly larger in amplitude than the frictional resistance at the front wheel unless the bicycle is not moving, so there must always be greater wear of the rear tyre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, when we accelerate, the details of the above description don't change. When we brake, the resistance of the brake materials on the wheel rim have to be included. Not going over the handlebars dictates that we should use the front brakes more judicially than we use the rear brakes, but there is another Question of whether the front or the rear brakes wear more quickly that I suspect is more personal. One might as a matter of habit never use the front brakes. The recent fad for fixed wheel bikes must result in more wear on the rear tyre, although this will only be significant if traffic forces many stops and starts.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-06-30T12:51:23.037" />
  <row Id="11751" PostTypeId="2" ParentId="11233" CreationDate="2011-06-30T13:18:17.657" Score="2" ViewCount="" Body="&lt;p&gt;I would say: a - yes, it will change the shape of the Earth and b - no,&#xA;it will not change the vertical, because of a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this needs a clarification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a - Yes, because the Earth is globally in hydrostatic equilibrium, and&#xA;the change in rotational speed is so slow that it cannot drive the Earth&#xA;out of this equilibrium. Faster excitations, like plate tectonics or&#xA;glaciations, do drive some out of equilibrium behavior, but the rotation&#xA;slow down is just too slow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to get an idea of how slow this is: The earth flattening is now&#xA;$f = 1/298$. Assuming it is proportional to the rotational speed of the&#xA;Earth, and assuming a slowdown of 2 ms/d/century as suggested before, we&#xA;get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\frac{df}{dt} = - f \times 2.3\times 10^{-8}/\text{cy}&#xD;&#xA;              = - 7.8\times10^{-11}/\text{cy}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that the equatorial radius is getting shorter by 0.16 mm per&#xA;century, while the polar radius is increasing at twice this rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b - It actually depends on what you are comparing the vertical to. If&#xA;you compare the new vertical with what the old vertical was at the same,&#xA;fixed geocentric latitude, then yes, it will change. If you compare at&#xA;the same position on the crust, it will also change, but for another&#xA;reason: plate tectonics will move your reference point far from where&#xA;it was originally. If you compare at a fixed geographic latitude, it&#xA;will not change, simply because the geographic latitude is defined by&#xA;the direction of it's vertical. In any case, the vertical will stay&#xA;perpendicular to the geoid because of the definition of the vertical&#xA;(gradient of potential) and of the geoid (equipotential). It will also&#xA;stay roughly perpendicular to the Earth's crust because of point a.&#xA;Comparing with tall buildings will not be possible because the buildings&#xA;we build today will not be here in a billion years.&lt;/p&gt;&#xA;" OwnerUserId="4202" LastEditorUserId="4202" LastEditorDisplayName="" LastEditDate="2011-06-30T13:49:28.453" LastActivityDate="2011-06-30T13:49:28.453" CommunityOwnedDate="2011-06-30T13:18:17.657" CommentCount="2" />
  <row Id="11752" PostTypeId="2" ParentId="11618" CreationDate="2011-06-30T14:22:07.860" Score="0" ViewCount="" Body="&lt;p&gt;Apparently the &lt;a href=&quot;http://prd.aps.org/abstract/PRD/v30/i8/p1712_1&quot; rel=&quot;nofollow&quot;&gt;Gaussian Effective Potential&lt;/a&gt; is used a fair amount. This would be something like $$V(x_1,x_2)=V_0 \exp(-\alpha(x_1-x_2)))$$. Thanks for the responses. &lt;/p&gt;&#xA;" OwnerUserId="4197" LastActivityDate="2011-06-30T14:22:07.860" />
  <row Id="11753" PostTypeId="2" ParentId="11747" CreationDate="2011-06-30T15:38:30.627" Score="1" ViewCount="" Body="&lt;p&gt;Multiplying by the reciprocal of the squareroot of the length squared, is a common method of normalizing vectors (in the L2 norm). Of course dividing by the square root is algebraically equivalent, but on a computer eliminates a rounding operation. Vectors that are normalized, are (somewhat) easier to work with.&lt;/p&gt;&#xA;" OwnerUserId="377" LastEditorUserId="377" LastEditorDisplayName="" LastEditDate="2011-06-30T15:50:36.320" LastActivityDate="2011-06-30T15:50:36.320" />
  <row Id="11754" PostTypeId="2" ParentId="11744" CreationDate="2011-06-30T15:43:41.423" Score="1" ViewCount="" Body="&lt;p&gt;Think of how much horizontal force you can apply. The puller, is leaning horizontally, and pushing (at an angle on the ground), and vertcal forces and rotational torque must separately balance. The more the puller can lean backwards, and the lower the rope, the more horizontal force he can apply. Presumably he can lean more, and get the rope lower from the back of the line.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-06-30T15:43:41.423" CommentCount="1" />
  <row Id="11755" PostTypeId="2" ParentId="10578" CreationDate="2011-06-30T16:18:25.423" Score="1" ViewCount="" Body="&lt;p&gt;Depends what do you understand by 'physics'. If in narrow sense only first-principle theory and experiment, then - no. Obviously birds are not particles. But hey, modeling folding of proteins (or even - orbitals of atoms) you consider an effective model, nowhere near to the particle physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when you by 'physics' understands mathematical description of real phenomena, then surely analysis of 'collective behavior' (or 'flocking' to be more specific) is physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As many mathematical tools came from statistical mechanics (or is related), it is very natural they publish in physical journals.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course, the motion need not to obey any fundamental laws (e.g. conservation principles, Hamiltonian dynamics, ...) - it is only an effective description.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the reference see e.g.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Tamás Vicsek, Anna Zafiris, &lt;a href=&quot;http://arxiv.org/abs/1010.5017&quot; rel=&quot;nofollow&quot;&gt;Collective motion&lt;/a&gt;, arXiv:1010.5017v1 [cond-mat.stat-mech]&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="184" LastActivityDate="2011-06-30T16:18:25.423" />
  <row Id="11756" PostTypeId="2" ParentId="11744" CreationDate="2011-06-30T16:40:02.540" Score="3" ViewCount="" Body="&lt;p&gt;Yes, if you put the strongest people in the back the rope will be straighter, making  it more likely everyone is pulling in the same direction.&lt;/p&gt;&#xA;" OwnerUserId="394" LastActivityDate="2011-06-30T16:40:02.540" />
  <row Id="11757" PostTypeId="2" ParentId="11732" CreationDate="2011-06-30T17:21:45.050" Score="0" ViewCount="" Body="&lt;p&gt;Ohm's law applies.  Looks like the easiest form to use would be ${V^2}/{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part which you are missing, and which makes it impossible to solve this even in the steady state, is the equation for heat dissipation versus temperature in the lamp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve the steady state problem, you need thermal resistance (or conductivity) to ambient $T_0$ (as well as knowing ambient $T_0$).  You need all this to come up with a steady state T, which allows you to solve for R.  The real world solution involves radiant, convective and conductive losses, so will appear to be a non-linear equation.  Once you have that equation in hand, you will almost certainly have to iterate to close in on a solution: guess at R; get power; solve for T; use that T to calculate a new R; repeat; repeat; repeat...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve the transient problem, you additionally need the specific heat (heat storage capacity per $T_{delta}$) of all elements in the thermal path.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-06-30T17:30:05.703" LastActivityDate="2011-06-30T17:30:05.703" />
  <row Id="11758" PostTypeId="2" ParentId="11727" CreationDate="2011-06-30T17:40:20.847" Score="-1" ViewCount="" Body="&lt;p&gt;ONE thing which makes Aluminum better than steel in SOME applications is its higher Strength/Weight ratio.  A big contributor to this better ratio is the much lower density of Aluminum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some classes of aluminum alloys which are stronger than steel alloys, but, in general, aluminum is actually a bit weaker than steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, aluminum is MUCH lighter (less dense) than steel: $2.8g/cm^3$ for Al as opposed to $7.8g/cm^3$ for steel, making Al a clear winner in the strength/weight arena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lower density of Al is due to its lower atomic weight.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-06-30T17:40:20.847" />
  <row Id="11759" PostTypeId="2" ParentId="11721" CreationDate="2011-06-30T17:43:54.150" Score="-1" ViewCount="" Body="&lt;p&gt;What makes the sound is the wave crashing into the beach: many tons of falling water slapping the ground.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-06-30T17:43:54.150" />
  <row Id="11760" PostTypeId="1" AcceptedAnswerId="11766" CreationDate="2011-06-30T17:56:10.730" Score="3" ViewCount="269" Body="&lt;p&gt;Landau writes &quot;It is found, however, that a frame of reference can always be chosen in &#xA;which space is homogeneous and isotropic and time is homogeneous.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does he mean that we can prove the existence of an inertial frame or does he want to say that it is assumed by doing enough number of experiments?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we start with some axioms and definitions of properties of space and time and then deduce the existence of such a frame in which space is homogeneous and isotropic and time is homogeneous?&lt;/p&gt;&#xA;" OwnerUserId="4235" LastEditorUserId="4235" LastEditorDisplayName="" LastEditDate="2011-07-01T09:47:08.287" LastActivityDate="2011-07-02T14:23:25.957" Title="Landau's ambiguous statement about the existence of inertial frames" Tags="&lt;inertial-frames&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="11762" PostTypeId="1" CreationDate="2011-06-30T18:21:43.510" Score="5" ViewCount="188" Body="&lt;p&gt;I am wondering if there is a limit to rotational speed of an object just like there is one for translation speed ? what are the implications of general relativity for rotating objects ?&lt;/p&gt;&#xA;" OwnerUserId="3714" LastActivityDate="2011-07-05T00:26:34.163" Title="maximum rotational speed" Tags="&lt;general-relativity&gt;&lt;angular-momentum&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="11763" PostTypeId="1" CreationDate="2011-06-30T18:31:29.207" Score="9" ViewCount="136" Body="&lt;p&gt;There is a recent paper on arxiv receiving lot of acclaim &lt;a href=&quot;http://arxiv.org/abs/1105.4714&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1105.4714&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors experimentally show that moving a mirror of a cavity at high speeds produces light from high vacuum. The usual doubts about the experimental techniques seem to be very clearly addressed and reviewed (as per Fred Capasso's comments) &lt;a href=&quot;http://www.nature.com/news/2011/110603/full/news.2011.346.html&quot; rel=&quot;nofollow&quot;&gt;http://www.nature.com/news/2011/110603/full/news.2011.346.html&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is :&#xA;Can someone explain how correlated/squeezed photons are generated in this process ? I can get a feel for how a moving mirror can generate real photons by imparting energy to the vacuum (correct me if this is not consistent with the detailed theory). But, I don't see how photons are generated in pairs. Could someone describe the parametric process happening here ? &lt;/p&gt;&#xA;" OwnerUserId="3714" LastEditorUserId="3714" LastEditorDisplayName="" LastEditDate="2011-06-30T21:20:21.410" LastActivityDate="2011-07-07T00:04:49.463" Title="how does dynamic casimir effect generate correlated photons" Tags="&lt;quantum-electrodynamics&gt;&lt;non-linear-optics&gt;&lt;casimir-effect&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" />
  <row Id="11764" PostTypeId="2" ParentId="11744" CreationDate="2011-06-30T19:14:34.763" Score="1" ViewCount="" Body="&lt;p&gt;In the pure physics sense, if 3 people that pull 40Lb 30Lb and 20lb the total pull will be 90lb no matter where you put them. Now in the strategy is the last person in line is called the anchor he/she is usually the biggest and strongest of the bunch the next 3 are the pullers they do a straight yank as hard as they can and they are the pre-anchor the rest are heavers. The heavers pull in a rhythm or cycle called by the 1st person in line or the captain and they are arranged by height.  The basic strategy depends on the 3 types of human force. The marathon runner “Constant force applied over time.” The sprinter “Grate force over a short time.” And the weight lifter “Exert and hold great power.”&lt;/p&gt;&#xA;" OwnerUserId="4034" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-06-30T21:24:49.347" LastActivityDate="2011-06-30T21:24:49.347" />
  <row Id="11765" PostTypeId="2" ParentId="11740" CreationDate="2011-06-30T20:56:47.017" Score="5" ViewCount="" Body="&lt;p&gt;It was just an arithmetic error:&lt;br&gt;&#xA;$$\psi(5,3) = -\sqrt{12/90} (2^-, 1^+, 0^+) + \sqrt{12/90} (2^+,1^-,0^+) + \sqrt{18/90} (2^+, 2^-, -1^+)$$&lt;br&gt;&#xA;$$+ \sqrt{12/90} (2^+,1^-, 0^+) - \sqrt{12/90}(2^+,1^+,0^-)$$&lt;br&gt;&#xA;needs to be simplified as the second and fourth terms are the same. One has:&lt;br&gt;&#xA;$$\psi(5,3) = -\sqrt{12/90} (2^-, 1^+, 0^+) + 2\sqrt{12/90} (2^+,1^-,0^+) + \sqrt{18/90} (2^+, 2^-, -1^+)$$&lt;br&gt;&#xA;$$ - \sqrt{12/90}(2^+,1^+,0^-)$$&lt;br&gt;&#xA;which is normalized:&lt;br&gt;&#xA;$$12/90 + 4(12/90) + 18/90 + 12/90 = 1.$$&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="1272" LastEditorDisplayName="" LastEditDate="2011-07-02T07:44:53.627" LastActivityDate="2011-07-02T07:44:53.627" CommentCount="4" />
  <row Id="11766" PostTypeId="2" ParentId="11760" CreationDate="2011-06-30T21:14:34.730" Score="3" ViewCount="" Body="&lt;p&gt;I believe this is just a restatement of the first Newton's law.&lt;/p&gt;&#xA;" OwnerUserId="4237" LastActivityDate="2011-06-30T21:14:34.730" CommentCount="10" />
  <row Id="11767" PostTypeId="2" ParentId="11763" CreationDate="2011-06-30T21:29:46.570" Score="4" ViewCount="" Body="&lt;p&gt;Ref [19] in the arXiv paper, C. M. Caves and B. L. Schumaker, Phys Rev A 31, 3068 (1985), gives a clean description of a parametric amplifier as the prototype of a two-photon device, at the bottom of its second page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a [parametric amplifier], an intense laser beam at frequency&#xA;  $2\Omega$ —the pump beam— illuminates a suitable nonlinear medium. The&#xA;  nonlinearity couples the pump beam to other modes of the electromagnetic&#xA;  field in such a way that a pump photon at frequency $2\Omega$ can be&#xA;  annihilated to create &quot;signal&quot; and &quot;idler&quot; photons at frequencies&#xA;  $\Omega\pm\epsilon$ and, conversely, signal and idler photons can be&#xA;  annihilated to create a pump photon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One way to think of the present situation would be as a dual of this description. That is, the medium, the Josephson junction, oscillates at a pump frequency $2\Omega$, and interacts nonlinearly with the vacuum state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the arXiv paper itself, there is a clear parallel,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Quantum theory allows us to make more detailed predictions than just that&#xA;  photons will simply be produced. If the boundary is driven sinusoidally at&#xA;  an angular frequency $\omega_d = 2\pi f_d$, then it is predicted that&#xA;  photons will be produced in pairs such that their frequencies, $\omega_1$ and&#xA;  $\omega_2$, sum to the drive frequency, i.e., we expect $\omega_d = \omega_1 + \omega_2$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of the comments on the Nature News page, Edward Schaefer at 2011-06-06 12:39:04 PM, makes a point that I think is worth highlighting, that &quot;The mirror transfers some of its own energy to the virtual photons to make them real.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-06-30T21:29:46.570" />
  <row Id="11768" PostTypeId="1" CreationDate="2011-06-30T21:50:46.327" Score="7" ViewCount="228" Body="&lt;p&gt;For a static electric field $E$ the conservation of energy gives rise to $$\oint E\cdot ds =0$$ Is there an analogous mathematical expression the conservation of momentum gives rise to?&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="1345" LastEditorDisplayName="" LastEditDate="2011-08-24T22:16:57.573" LastActivityDate="2011-08-25T16:32:08.560" Title="Does a static electric field and the conservation of momentum give rise to a relationship between $E$, $t$, and some path $s$?" Tags="&lt;classical-electrodynamics&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" />
  <row Id="11770" PostTypeId="1" CreationDate="2011-07-01T01:36:29.523" Score="3" ViewCount="214" Body="&lt;p&gt;i have a very basic question from school days. what does it mean to say an object is moving with uniform speed? it seems to me now that it should be an unit dependent concept.&#xA;for example if speed is the derivative of distance traveled, i.e. $X&amp;#39;(t)$ , and I decide to measure distance on a new scale $F(X)$, a monotonic function of $X$, but not a linear multiple. Then, speed in that scale at a point t would be $F&amp;#39;(X(t))$. $X&amp;#39;(t)$ , which will not be constant&#xA;as $F$ can be an arbitrary increasing function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if indeed &quot;uniformness of speed&quot; is a unit dependent concept, what does it mean to&#xA;say, light travels at a constant speed ? also this would mean &quot;uniform acceleration&quot; is also a&#xA;unit dependent concept. how is then, &quot;acceleration due to gravity&quot; a universal constant ?&#xA;would it cease to be a constant if I measured acceleration in $\log(m/s^2)$ ?&lt;/p&gt;&#xA;" OwnerUserId="4238" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-26T01:28:39.400" LastActivityDate="2011-07-26T01:28:39.400" Title="what is uniform velocity?" Tags="&lt;classical-mechanics&gt;&lt;velocity&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11771" PostTypeId="2" ParentId="9560" CreationDate="2011-07-01T02:01:23.903" Score="-1" ViewCount="" Body="&lt;p&gt;The strong force is gravity belive it or not yet it's range does not exceed the diameter of the nuclious , how ever the power of the strong force falls off by magtitudes with distance but  in fact it extends almost infinity as gravity gravity is the strong force manifest threw another demension one day they will fidn this out  when that is found out the will be able to control gravity like in an electrical manor further more what is happening is you havea board filled with holes  and another filled with pegs once you line them up you willhave the unified field threoy complete right now we are just not lining things up right that is why many and all equations fail at unification&#xA; trust me it isnt balony &#xA; how do i know?&#xA; I have done it&lt;/p&gt;&#xA;" OwnerUserId="4239" LastActivityDate="2011-07-01T02:01:23.903" CommentCount="1" />
  <row Id="11772" PostTypeId="2" ParentId="11770" CreationDate="2011-07-01T02:01:55.447" Score="4" ViewCount="" Body="&lt;p&gt;Neat trick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately, you're not actually changing anything physical. The graph drawn in those units would not be a straight line but the particle would still move the same distance in any fixed amount of time, i.e. it would still be uniform velocity. So it might move 8 units of distance in a 1 second interval and 16 units in the next 1 second interval but the actual distance moved would be the same in both intervals of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you're doing is more like a change of coordinates than a change of units.&lt;/p&gt;&#xA;" OwnerUserId="2953" LastActivityDate="2011-07-01T02:01:55.447" CommentCount="13" />
  <row Id="11773" PostTypeId="2" ParentId="9560" CreationDate="2011-07-01T02:05:32.150" Score="-2" ViewCount="" Body="&lt;p&gt;A quetsion came to mind  what is the  frequency that is as fast as it propagation speed&#xA; in other words what is a frequency = to c answer  144,000,000,000,000 htz approx  at 1 v&#xA; I call it wav speed a 60htz wave speed is 16,000,000 mph  yet 3100 miles long .&lt;/p&gt;&#xA;" OwnerUserId="4239" LastActivityDate="2011-07-01T02:05:32.150" CommentCount="1" />
  <row Id="11774" PostTypeId="1" AcceptedAnswerId="11776" CreationDate="2011-07-01T02:43:50.030" Score="6" ViewCount="104" Body="&lt;p&gt;Is there a general algorithm to find the spectrum of $S S^\dagger$, where $S$ is a homogenous polynomial (of degree $n$) of the annihilation operators (of $d$ variables)?&lt;/p&gt;&#xA;" OwnerUserId="184" LastActivityDate="2011-07-28T08:50:52.443" Title="Finding the spectrum of a polynomial of the creation and annihilation operators" Tags="&lt;quantum-mechanics&gt;&lt;quantum-optics&gt;&lt;second-quantization&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="11775" PostTypeId="1" AcceptedAnswerId="11778" CreationDate="2011-07-01T06:13:05.730" Score="2" ViewCount="113" Body="&lt;p&gt;The standard equation for the doppler effect from a frame of reference with constant velocity is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_0$ initial frequency;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v$ is the velocity of waves in the medium;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_r$ is the velocity of the receiver relative to the medium; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_s$ is the velocity of the source relative to the medium; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f = \left(\frac{v + v_r}{v + v_s}\right)f_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'd like to know what's the equivalent for an accelerating frame, I have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f = \left(\frac{v&amp;#39;\text{dt} + v_r}{v&amp;#39;\text{dt} + v_s}\right)f_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it just doesn't seem correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mea Culpa - I was thinking of the signal being emitted from an accelerating train B as being the frame of reference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, second attempt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)You are standing at a platform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)The wind is stationary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)A train is accelerating towards you - blowing its whistle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the formula for the frequency received at the station? Where the standard formula is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f = \left(\frac{v }{v + v_s}\right)f_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;rewriting for an accelerating frame of reference for the source&#xA;$$f = \left(\frac{v }{v + (dv_s/dt)dt}\right)f_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I don't think this goes anywhere, and yes a non-linear solution is probably the answer but it's like a three legged black dog with rabies - it's better than no dog at all.&lt;/p&gt;&#xA;" OwnerUserId="3588" LastEditorUserId="3588" LastEditorDisplayName="" LastEditDate="2011-07-02T00:15:12.223" LastActivityDate="2011-07-03T01:02:40.903" Title="Doppler effect from an accelerating frame of reference" Tags="&lt;doppler-effect&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="11776" PostTypeId="2" ParentId="11774" CreationDate="2011-07-01T07:49:09.250" Score="5" ViewCount="" Body="&lt;p&gt;The supspaces $V_n = Span \{ (a_1^{\dagger})^{n_1}, . . . (a_d^{\dagger})^{n_d} |0&amp;gt;\}$, $n_i \ge 0$,  $ n_1 + . . . n_d = n$, constitute of invariant subspaces of the operator $ S S^{\dagger}$ action. The dimension of $V_n$ is $ \frac{(d+n-1)!}{(d-1)! n!}$. Thus the operator can be represented on each of these subspaces as a square matrix of size $ \frac{(d+n-1)!}{(d-1)! n!}$ for which the spectrum can be found by elementary linear algebra. The spectrum on the whole of the Fock space is the union of the spectra over $V_n$, $ n = 0, 1, . . .$&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorUserId="2190" LastEditorDisplayName="" LastEditDate="2011-07-28T06:32:13.790" LastActivityDate="2011-07-28T06:32:13.790" CommentCount="5" />
  <row Id="11777" PostTypeId="2" ParentId="11775" CreationDate="2011-07-01T10:05:26.727" Score="0" ViewCount="" Body="&lt;p&gt;The observed &quot;frequency&quot; in an accelerated reference frame should be equal at least to the &quot;frequency&quot; in an inertial co-moving reference frame. It means replacing the RF velocity with a time-dependent velocity $v_r (t)$.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-07-01T10:05:26.727" />
  <row Id="11778" PostTypeId="2" ParentId="11775" CreationDate="2011-07-01T13:09:12.027" Score="2" ViewCount="" Body="&lt;p&gt;In the classical Doppler effect, you should think of $v_s$ and $v_r$ to be the velocities of the source and receiver &lt;strong&gt;relative to the medium&lt;/strong&gt;, &lt;strong&gt;at the time when the signal was sent and received&lt;/strong&gt;. In an accelerated reference frame, it is essentially the medium that is moving. Remember that the Doppler shift concerns the frequency sent by the source compared to the frequency seen by the receiver. So it does not depend on which reference frame you, as a third party, is in, but only depends on the intrinsic wave speed of the medium $v$ (which we assume to be constant, when measured in a frame in which the medium is at rest), and the velocities relative to the medium of the source and receivers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that said, that means you can compute relative to the unaccelerated reference frame by doing a coordinate change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you are in a reference frame in which the medium itself is moving with variable velocity $\tilde{v}(t)$, and you measure the source and receivers to be moving with velocity $\tilde{v}_s(t)$ and $\tilde{v}_r(t)$. Then the Doppler effect can be computed by looking at the reference frame in which the medium is not moving (so the corresponding velocities are $v_s(t) = \tilde{v}_s(t) - \tilde{v}(t)$ and similarly $\tilde{v}_r(t)$). If in this reference frame $v_s$ and $v_r$ are not constant in time, then given a time $t_r$ when the signal was received, you will have to compute (using that the wave travels with constant speed $v$ and using the trajectory of the receiver versus the source) the time $t_s$ at which the signal was sent, and plug those values into your Doppler shift expression (with $v_s \to v_s(t_s)$ and $v_r \to v_r(t_r)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the wave speed itself $v$ depends on the macroscopic of the entire medium, then you will have to solve the very complicated system for elastodynamics/acoustic in the medium, for which the final answer would depend on how you model the medium. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit for the new clarification&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the particular case, assume you, the receiver at the platform is measuring the signal at time $t_0 = 0$. And choose coordinate system so that you are at the origin. Let $x(t)$ denote the position of the train that is approaching the station. (or leaving the station, you just need to adjust the sign appropriately.) The signal you observe will have been emitted by the train at time $t_s &amp;lt; 0$, when $|x(t_s)| = v \cdot (0-t_s)$ where $v$ is the speed of sound in air. (For the time being we'll assume you have a non-supersonic train, so there is a unique time $t_s$ that satisfies the requirements.) Then the expression for the doppler shift would be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f = \left( \frac{v}{v + \dot{x}(t_s)}\right) f_0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, solving for $t_s$ is generally hard. But we can get a good estimate on $f$ if we assume that our train is reasonably close to the station when you are making the observation, and if the train is not accelerating too much, by using Newton's method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the zero'th order you'd get $t_s = 0$. To the first order in Newton's method, you can solve and get $t_s ~ x(0) / (v - \dot{x}(0))$. Then we can approximate the velocity $\dot{x}(t_s) \sim \dot{x}(0) + \ddot{x}(0) t_s$ which leads to a first order correction to the doppler shift formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f \sim \left( \frac{v}{v+ \dot{x}(0) + \frac{\ddot{x}(0) x(0)}{v - \dot{x}(0)}}\right) f_0 $$&lt;/p&gt;&#xA;" OwnerUserId="829" LastEditorUserId="829" LastEditorDisplayName="" LastEditDate="2011-07-02T17:01:16.147" LastActivityDate="2011-07-02T17:01:16.147" CommentCount="3" />
  <row Id="11779" PostTypeId="1" CreationDate="2011-07-01T14:55:08.107" Score="6" ViewCount="194" Body="&lt;p&gt;I am working on a software library for Units of Measure. To represent dimensions, I need to know the required range of exponents for each of the seven base units (precisely, I need to know the required maximum number of bits for each exponent). After reading many related questions, I still have some questions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understand, all physical quantities can be expressed as the product of the seven base units, with each base unit raised to the power of 0, +-1, +-2, +-3, and so on. So the dimension of a quantity is represented by a vector / array of 7 integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. for volume (length ^ 3) that array would be (3, 0, 0, 0, 0, 0, 0), for acceleration (length / (time ^ 2)) it would be (1, -2, 0, 0, 0, 0, 0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I could calculate the square of a volume (with dimension 6,0,0,..), it doesn't make much sense, because I have never encountered any physical quantity that needs (length ^ 6). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) What is the maximum required exponent for Length ? Do you know any example where +-3 is not enough ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Maximum exponent for Time ? Any example beyond +-4 ?&lt;br&gt;&#xA;&lt;em&gt;(update: raised to 4 for capacitance)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) Maximum exponent for Mass ? Any example beyond +-1 ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) Maximum exponent for Electric Current ? Any example beyond +-1 ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e) Maximum exponent for Temperature ? Any example beyond +-4 ?&lt;br&gt;&#xA;&lt;em&gt;(update: raised to 4 for Stefan-Boltzmann constant)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f) Maximum exponent for Amount of Substance ? Any example beyond +-1 ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g) Maximum exponent for Luminous Intensity ? Any example beyond +-1 ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.. plus three bonus questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;h) Is there some required order of the seven base units ? I have seen more than one ..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) Are those seven base units enough for everything, or is there some bleeding edge science that needs more ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;j) Are the exponents always integral ? Never ever something like ^ 1.5 or ^ e ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(update)&lt;br&gt;&#xA;I guess I should clarify:&lt;br&gt;&#xA;I am aware that &lt;strong&gt;values&lt;/strong&gt; might need rational or real exponents.&lt;br&gt;&#xA;I am aware that &lt;strong&gt;intermediate&lt;/strong&gt; results might need a wider range of integer exponents.&lt;br&gt;&#xA;I have read the relevant SI standard document, and lots of websites about the topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are just about the base units exponents that define the dimension of (base or derived) units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most precise statement i found so far was: &lt;em&gt;the exponents are very small integers&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to exploit all filthy speed tricks of CPU or GPU assembly programming, I need to know the exact number of bits required.&lt;/p&gt;&#xA;" OwnerUserId="4243" LastEditorUserId="4243" LastEditorDisplayName="" LastEditDate="2011-07-01T19:27:58.823" LastActivityDate="2011-07-02T11:50:09.683" Title="Whats the range of dimension exponents in the SI Units system?" Tags="&lt;si-units&gt;&lt;units&gt;" AnswerCount="4" CommentCount="8" />
  <row Id="11780" PostTypeId="2" ParentId="11775" CreationDate="2011-07-01T15:14:43.367" Score="1" ViewCount="" Body="&lt;p&gt;Your first formula can be rewritten as $\frac{f_r}{c_r+v_r}=\frac{f_s}{c_s+v_s}$, where the equivalences with your notation are $f_s=f_0$, $f_r=f$, and $c_s=v=c_r$ are the velocities of the medium at the sender and at the receiver. This equation tells us that the number of wave crests per unit length is the same for the sender and for the receiver. We should preserve this equality in an expression for the number of wave crests per unit length in the accelerating frame of reference. I take it we're working in 1+1 dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the accelerating frame of reference, we can take $v_s$ and $v_r$ to be constant, but the velocities $c_r$ and $c_s$ to be functions of time, $c_r(t)=c_r+at$, $c_s(t)=c_s-at$, say, with the difference of sign because the direction in which the sender is sending waves is opposite to the direction in which the receiver receives (this classically, not SR, of course). Presuming that the wavelength is short enough relative to the characteristic lengths introduced by the ratios of the acceleration relative to the velocities in the problem, the equation for the frequency relationship would be, approximately, $\frac{f_r}{c_r(t)+v_r}=\frac{f_s}{c_s(t)+v_s}$. You have to get the relative signs of the acceleration and of the various velocities right, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is so far from a generally covariant description that there doesn't seem much point to consider the question of accelerated coordinates, but whatever. The elementary idea of counting wave crests, for example, does not extend to a general situation in which there is a (continuous or discrete) distribution of energy at different wave numbers.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-07-01T15:14:43.367" />
  <row Id="11781" PostTypeId="1" CreationDate="2011-07-01T16:12:52.120" Score="1" ViewCount="83" Body="&lt;p&gt;A body point charge $Q$ moves in relation to the reference system $\Sigma$ according to the law of motion $x(t)=v_0 t$, $y(t)=0$, $z(t)=0$.&lt;/p&gt;&#xA;" OwnerUserId="4244" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-01T17:35:53.833" LastActivityDate="2011-07-01T17:35:53.833" ClosedDate="2011-07-01T17:31:17.220" Title="Show that the electric field E in the system at rest is $E=\frac{Q}{4 \pi \epsilon_0 \sqrt{(x^2+y^2+z^2)^3}} (x,y,z)$" Tags="&lt;electromagnetism&gt;&lt;special-relativity&gt;&lt;homework&gt;" CommentCount="5" />
  <row Id="11782" PostTypeId="1" AcceptedAnswerId="11788" CreationDate="2011-07-01T16:30:14.587" Score="2" ViewCount="187" Body="&lt;p&gt;Just passed a helicopter on my way to work. We have read in some detail how an airplane gets forward thrust and lift by deflecting air. How does a helicopter with horizontal fans achieve that ? &lt;/p&gt;&#xA;" OwnerUserId="3714" LastEditorUserId="3714" LastEditorDisplayName="" LastEditDate="2011-07-01T20:37:29.747" LastActivityDate="2011-07-01T22:08:02.567" Title="how does a helicopter get forward thrust?" Tags="&lt;aerodynamics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="11783" PostTypeId="2" ParentId="11782" CreationDate="2011-07-01T17:00:18.663" Score="7" ViewCount="" Body="&lt;p&gt;the fans are not fixed to be always horizontal; the pilot is able to control the deflection angle of all the fans, this deflection angle makes air to be pushed in one preferential horizontal direction which gives the thrust&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-07-01T17:00:18.663" CommentCount="2" />
  <row Id="11784" PostTypeId="2" ParentId="11779" CreationDate="2011-07-01T18:51:14.290" Score="1" ViewCount="" Body="&lt;p&gt;I don't think you can just use integers in all cases.  For example the statcoulomb, the gaussian unit of charge, is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 \text{ statcoulomb} = \frac{g^{1/2} cm^{3/2}} {s}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also,the Stefan-Boltzmann constant's units are proportional to $T^{-4}$. Specifically, &lt;a href=&quot;http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law&quot; rel=&quot;nofollow&quot;&gt;from Wikipedia&lt;/a&gt;:&#xA;$$\sigma = 5.670400 \times 10^{-8}J s^{-1}m^{-2}K^{-4}$$&lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-01T19:17:31.577" LastActivityDate="2011-07-01T19:17:31.577" CommentCount="6" />
  <row Id="11785" PostTypeId="2" ParentId="11779" CreationDate="2011-07-01T19:46:24.357" Score="4" ViewCount="" Body="&lt;p&gt;At least one common quantity I can think of has dimension with a non-integer exponent. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Specific_detectivity&quot; rel=&quot;nofollow&quot;&gt;specific detectivity, $\text{D}^*$&lt;/a&gt; is a common descriptor of photodiodes, and I'm sure one could make an analogous figure of merit for other types of sensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unit of $\text{D}^*$ is the &quot;Jones,&quot; which is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{cm \cdot \sqrt{Hz}}{W}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Watts in SI decompose into&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Kg \cdot m^2 \cdot s^{-3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which makes one Jones equal to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{s^{2.5}}{Kg \cdot m} \times 10^{-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with dimension:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{time}^{2.5} \text{mass}^{-1} \text{length}^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other quantities related to $\text{D}^*$, such as Noise Equivalent Power (NEP), which come up a lot in radiometry. Basically, any measurement which is normalized by frequency bandwidth will end up with that $\sqrt{Hz}$ in the units.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-07-01T19:46:24.357" CommentCount="11" />
  <row Id="11786" PostTypeId="2" ParentId="11637" CreationDate="2011-07-01T20:51:37.600" Score="2" ViewCount="" Body="&lt;p&gt;Well, I found a paper that works for my particular situation (a drop evaporating of PTFE) that does the trick.  I doubt that very many people need this particular question answered, but here it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Drop Evaporation on Solid Surfaces: Constant Contact Angle Mode&quot; &#xA;H. Ebril, G. McHale, and M Newton&#xA;Langmuir 2002, 18, 2636-2641&lt;/p&gt;&#xA;" OwnerUserId="1037" LastActivityDate="2011-07-01T20:51:37.600" />
  <row Id="11787" PostTypeId="1" CreationDate="2011-07-01T21:10:30.603" Score="1" ViewCount="77" Body="&lt;p&gt;I'm reading a paper that examines the evaporation rates of water.  In the final formula, it has the following constant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c_s - c_\infty $ where $c_s$ is the concentration of the vapor at the sphere surface and $c_\infty$ is the concentration of the vapor at infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm fairly confident in how to derive $c_\infty$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Calculate water vapor pressure $p_s = 610.78 e^{\frac{17.2694 T}{T+238.3}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Actual vapor pressure is then: $ p=(Relative Humidity)p_s$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Using ideal gas law gives concentration at infinity: $c_\infty = \frac{(Molar Mass of Water)p}{RT}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This generally looks like it gives me answers consistent with the paper's values.  But, if I screwed up, please let me know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is with $c_s$.  It's weird, because I recognize how easy it should be to get this, but just can't.  I've asked some of the other grad students here and they basically all are befuddled.  I think this is the physicist's version of a &quot;tip of my tongue&quot; experience.  So, can anyone give me a link or method by which I can get this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1037" LastActivityDate="2011-09-03T17:41:02.513" Title="How to calculate concentration of vapor at the surface of a water drop" Tags="&lt;fluid-dynamics&gt;&lt;water&gt;&lt;formula&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11788" PostTypeId="2" ParentId="11782" CreationDate="2011-07-01T22:08:02.567" Score="6" ViewCount="" Body="&lt;p&gt;In general, the forward thruswt is achieved by tilting the entire helicopter forward.  This converts some of the lift produced by the main rotor into a forward component of force.  The same is true for turns.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, the swashplate may be used to create a thrust imbalance; but this imbalance does not provide any horizontal forces.  It merely creates a force imbalance which tips the helo slightly, so that a component of the upward force of the main rotor gets converted into a lateral force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at any helo moving with any kind of rapid forward velocity and you will see that the entire craft is tipped noticably forward.  Same with sharp turns: the helo tilts noticably in the direction of the turn.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-07-01T22:08:02.567" CommentCount="2" />
  <row Id="11789" PostTypeId="2" ParentId="11762" CreationDate="2011-07-01T22:15:20.563" Score="1" ViewCount="" Body="&lt;p&gt;The maximum rotational velocity of a solid body is governed by the allowable material stresses.  These forces reach a point which tears the body apart long before the tangential velocities at the rim approach any significant fraction of the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a ratio that describes allowable max rotational speed of a disk in terms of strength of material it is built out of.  Been a while and I forget the name of the guy this was named after, but a Google search should turn something up for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could construct a disk of sufficient material strength, such that you only had to worry about relativity, then the limit would be when the tangential velocity at the rim = c.  Thus, the rotational angular velocity, in radians, of a disk of radius r, would be $c/(2rPi)$&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-07-01T22:27:36.653" LastActivityDate="2011-07-01T22:27:36.653" CommentCount="0" />
  <row Id="11791" PostTypeId="2" ParentId="11779" CreationDate="2011-07-01T23:47:42.667" Score="2" ViewCount="" Body="&lt;p&gt;Regardless of whether you really need it, here's what I have found to be usable:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said, I represent all units internally as cgs:&#xA;$$&#xD;&#xA;  \mathrm{U} = \mathrm{cm}^c\cdot \mathrm{g}^g\cdot\mathrm{s}^s&#xD;&#xA;$$&#xA;Where, in general, $c,g,s\in\mathbb{Q}$.&#xA;For this, I have 3 &lt;code&gt;int&lt;/code&gt;s, defined by&#xA;$$&#xD;&#xA;  i_c = 39916800 \cdot c&#xD;&#xA;$$&#xA;etc., so that, $39916800$ being $11!$, I can be almost completely sure that nobody will ever come up with a unit such that $i_c\not\in\mathbb{Z}$. It also supports exponents up to $\frac{2^{31}}{11!}\approx 53$, which I'm quite sure should be enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may argue that using full integer variables is a great waste of resources, but I actually think (haven't tried it, though) that it's faster than a naïvely implemented array of &lt;code&gt;char&lt;/code&gt;s, because the processor does not even directly calculate those, if you ask it to: it first needs to cast them to &lt;code&gt;int&lt;/code&gt;, do the calculation there, and cast them back to &lt;code&gt;char&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as I said, you probably don't need to worry &lt;em&gt;at all&lt;/em&gt; about performance if you do the unit stuff at compile time. The quantities do not need any knowledge about their SI exponentials whatsoever then, and if you need some way to uniquely identify units, you can just put all the units that actually occur during the preprocessing in a hash map or something and let the calculable quantities store only pointers to the unit objects.&lt;/p&gt;&#xA;" OwnerUserId="3540" LastEditorUserId="3540" LastEditorDisplayName="" LastEditDate="2011-07-01T23:53:07.407" LastActivityDate="2011-07-01T23:53:07.407" CommentCount="2" />
  <row Id="11792" PostTypeId="1" CreationDate="2011-07-02T01:50:55.137" Score="0" ViewCount="64" Body="&lt;p&gt;What the theta in schwinger function and what is theta formula?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is theta formula general solution of klein gordon equation? if so, what is its coefficient of exp(-i*p*x)?&lt;/p&gt;&#xA;" OwnerUserId="4247" LastActivityDate="2011-09-06T01:42:17.960" Title="What the theta in schwinger function and what is theta formula?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="11793" PostTypeId="2" ParentId="11702" CreationDate="2011-07-02T07:35:16.833" Score="1" ViewCount="" Body="&lt;p&gt;As an experimentalist, I will first address this last summary of your question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do you know the particle is still there?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let us define the terms of the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) In particle physics we know a particle was there by the tracks it leaves in a bubble chamber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b)By the signals it sets off as it passes and ionizes &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The measurements have shown us that we deal with very small dimensions in all quantities, mass, size etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also have found that particles follow quantum dynamics and the solutions of the appropriate equations of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally: can I trap one particle and &quot;know&quot; it is there? I have not done it, but it is being done billions of times a second at the accelerators. If I went to the trouble to design an experiment that has trapped a single proton in a magnetic configuration, I would know it was there from the radiation it would emmit as it oscillated in the magnetic trap. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually though, because of the very small values accompanying the existence of a particle one deals with a flux of them at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now coherence. Coherence is the term describing the quantum mechanical solution of the equations of more than one particle, and refer to the phase differences between those particles : i.e. coherence means that those  phase differences remain constant . Described as quantum mechanical waves, the particles are &quot;in step&quot;. If you only have one particle, as in my gedanken experiment above, the quantum mechanical solution is known and phases can only be defined with respect to the field. As long as energy is supplied to my proton this description will hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;know the particle&quot; phrase should become &quot;know the particles&quot; phrase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coherence is observed macroscopically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in laser light&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in superconducting magnets, over kilometers of wire length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in superfluidity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these require zillions of particles and no question should arise if they are there or not. The answer by Peter Morgan addresses the question of stability of such systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I suspect you are asking the question from statements of coherence and the density matrix formulation. This has to do with the quantum mechanical statistical behaviour&#xA; of many particles, so again, your one particle question does not compute.&#xA;You should maybe clarify in your head what you really want to learn about coherence. Maybe the &lt;a href=&quot;http://en.wikipedia.org/wiki/Density_matrix&quot; rel=&quot;nofollow&quot;&gt;density matrix&lt;/a&gt; formalism confuses you?&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-07-02T07:35:16.833" CommentCount="1" />
  <row Id="11794" PostTypeId="1" CreationDate="2011-07-02T10:34:25.027" Score="1" ViewCount="39" Body="&lt;p&gt;It is known that when a beam of lineary polarized light falls perpendicularly on a linear polarizer, the intensity of polarization changes according to Malus' law and the direction of polarization changes as cosine of angle between polarization vector and polarizer vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: is anyone familiar with mathematical treatment of how the direction and intensity of polarization changes when the angle of incidence changes?&lt;/p&gt;&#xA;" OwnerUserId="2752" LastActivityDate="2011-07-02T11:12:00.183" Title="Linear polarizer and the angle of incidence" Tags="&lt;light&gt;&lt;polarization&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="11795" PostTypeId="2" ParentId="11794" CreationDate="2011-07-02T11:06:31.217" Score="1" ViewCount="" Body="&lt;p&gt;It depends on the polarizing material.&#xA;I assume below that it is about layer of absorptive material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When wave travels through an anisotropic material, its phase shift and absorption is polarization-dependent. So you should be able to treat it in similar way as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Birefringence&quot; rel=&quot;nofollow&quot;&gt;Birefringence&lt;/a&gt; (giving a complex refraction index to take into account the absorption). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to consider a simple case of very thin polarizer, with the 100% absorption of one polarization, and 0% of the another - it simplifies. Just you need to say if polarization in the direction $z$ (i.e. the perpendicular to the polarizer) is absorbed or not. Of course for other efficiencies it is still relatively simple, as long as you study the case of the very thin polarizer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in analyzing different polarizers (e.g. polarizing cubes, basing on the birefringence and total internal reflection - then you can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fresnel_equations&quot; rel=&quot;nofollow&quot;&gt;Fresnel equations&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="184" LastEditorUserId="184" LastEditorDisplayName="" LastEditDate="2011-07-02T11:12:00.183" LastActivityDate="2011-07-02T11:12:00.183" />
  <row Id="11796" PostTypeId="2" ParentId="11775" CreationDate="2011-07-02T11:07:39.530" Score="0" ViewCount="" Body="&lt;p&gt;Constraints: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;non relativistic, i.e. all speeds are small compared to the speed&#xA;of the waves in the medium, the receiver stay still in the medium,&#xA;and the problem of taking measures at distance is not important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be $\frac{f}{f_{0}}=\frac{v+v_{r}}{v+v_{s}}$, lets use dimensionless&#xA;units, be $v=1,\: v_{s}\leftarrow v_{s}/v,\: v_{r}\leftarrow0$ (the&#xA;platform is still).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{f}{f_{0}}=\frac{1}{1+v_{s}(t_{E})}$, where $v_{s}(t_{E})$&#xA;is the speed of the emitter at the time of the emission, measured&#xA;by the receiver. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed of the emitter can be represented by f(t), measured by the receiver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say for instance $f(t)=t^{2}$ (dimensionless), as if the emitter&#xA;is moving subjected to a central force like gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{f}{f_{0}}=\frac{1}{1\pm t_{E}^{2}}$, and the $\pm$ sign account&#xA;for the relative direction of movement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A complete study of Doppler effect, like no other, is needed to understand light properties under the CMB referential, is presented here: &lt;a href=&quot;http://arxiv.org/abs/physics/0205033&quot; rel=&quot;nofollow&quot;&gt;Cosmological Principle and Relativity - Part I&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT add:&lt;br&gt;&#xA;   the motion of the emitter is along the direction emitter-receiver (angle = 0)  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-07-03T01:02:40.903" LastActivityDate="2011-07-03T01:02:40.903" />
  <row Id="11797" PostTypeId="2" ParentId="11638" CreationDate="2011-07-02T11:19:42.723" Score="0" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Conversely, AES is sensitive to the&#xA;  lighter elements, and unlike X-ray&#xA;  fluorescence, Auger peaks can be&#xA;  detected for elements as light as&#xA;  lithium (Z = 3). Lithium represents&#xA;  the lower limit for AES sensitivity&#xA;  since the Auger effect is a &quot;three&#xA;  state&quot; event necessitating at least&#xA;  three electrons. Neither H nor He can&#xA;  be detected with this technique.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is found here: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Auger_electron_spectroscopy&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Auger_electron_spectroscopy&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-07-02T11:19:42.723" CommentCount="2" />
  <row Id="11798" PostTypeId="2" ParentId="11779" CreationDate="2011-07-02T11:50:09.683" Score="2" ViewCount="" Body="&lt;p&gt;In a paper, dated yesterday, where units are studied I can see wavelength raised to -5 , temperature raised to 4.&lt;br&gt;&#xA;I do not know the actual maximum exponents in all circumstances.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Constant or Quantity --- Dimension&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planck Constant &lt;em&gt;h&lt;/em&gt;  --- $ML^{2}T^{-1}$&lt;br&gt;&#xA;Stefan Constant $\sigma$ ---  $MT^{-3}\theta^{-4}$&lt;br&gt;&#xA;Boltzmann Constant &lt;em&gt;k&lt;/em&gt; --- $ML^{2}T^{-2}\theta^{-1}$&lt;br&gt;&#xA;Temperature $\theta$  --- $\theta$&lt;br&gt;&#xA;Energy  --- $ML^{2}T^{-2}$&lt;br&gt;&#xA;Force  ---  $MLT^{-2}$&lt;br&gt;&#xA;Pressure  --- $ML^{-1}T^{-2}$&lt;br&gt;&#xA;Luminosity --- $MT^{-3}$&lt;br&gt;&#xA;Power  --- $ML^{2}T^{-3}$&lt;br&gt;&#xA;Velocity  --- $LT^{-1}$&lt;br&gt;&#xA;Electron charge  --- $Q$&lt;br&gt;&#xA;Proton mass ---  $M$&lt;br&gt;&#xA;Bohr radius  --- $L$   &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-07-02T11:50:09.683" />
  <row Id="11799" PostTypeId="1" AcceptedAnswerId="11805" CreationDate="2011-07-02T13:55:37.180" Score="3" ViewCount="194" Body="&lt;p&gt;Currently I am stuck, trying to derive the self-inductance of a long wire. According to literature it should be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L=\frac{\mu_r\mu_0l}{8\pi}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and in literature its derived by looking at the energy of the magnetic field. I tried to derive this formula via the magnetic flux and I am getting $4\pi$ instead of $8\pi$. This are my considerations:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   _ _&#xA; /  |R \&#xA;|   '   | a wire with radius R and length l&#xA; \ _ _ /    &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The magnetic flux density $B(r)$ is given according to Ampere's law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(r) 2\pi r=\mu_0\mu_r\frac{r^2}{R^2}I$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B(r)=\mu_0\mu_r\frac{r}{2\pi R^2}I$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $r$ is the distance from the center of the wire, $R$ is its radius and $I$ is the total current through the wire. Now I know that the magnetic flux $\phi$ through the upper part of a longitudinal section is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi = \int_A B dA = \int_0^R B(r)l dr =  \frac{\mu_0\mu_rIl}{4\pi}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $l$ is the wire's length. No I use $\phi = LI$ and arrive at $4\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong? Where is the mistake in my considerations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover I have the following problem. If I look at an entire longitudinal section of the wire and not only at its upper half the magnetic flux is zero:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   _ _&#xA; /  |  \&#xA;|   |2r | =&amp;gt; Magnetic flux is zero (the magnetic field &#xA; \ _|_ /     penetrating the upper half of the longitudinal cross section is &#xA;             exactly opposite to the magnetic field penetrating the lower&#xA;             half)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hopefully I formulated my problem clear enough. If not please ask me for further details.&lt;/p&gt;&#xA;" OwnerUserId="393" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-03T04:34:09.100" LastActivityDate="2011-07-22T17:58:05.663" Title="Derivation of self-inductance of a long wire" Tags="&lt;electromagnetism&gt;&lt;homework&gt;&lt;inductance&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="3" />
  <row Id="11800" PostTypeId="1" CreationDate="2011-07-02T14:02:56.060" Score="5" ViewCount="299" Body="&lt;p&gt;There have been recent questions about the vacuum. In my simplified knowledge the vacuum is like a ground state energy level, and also that there might even exist other lower energy levels than the vacuum we find ourselves in. The sea is a soup of created and annihilated pairs of virtual particles with virtual energies and momenta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a normal sea on earth, which also represents a ground state of water, energy can be taken out of random waves by the clever construction of valves that allow only one way motion of water. Is it conceivable that a  gadget of similar function could be found for the vacuum sea, or is it forbidden by conservation laws?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition tells me that it might be possible if GR is taken into account, but my physics knowledge does not stretch to support this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; : An explanation of why I am asking this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me expand on the example of the sea. The energy from the waves comes from either tides, i.e. gravitational forces, or wind (temperature differentials). If these were missing the oceans would be like glass representing a unique ground state of the gravitational well of the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an analogy, a gravitational wave going through the vacuum  would be supplying energy to the vacuum sea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been thinking of this analogy ever since cold fusion surfaced and refuses to die out, the most &lt;a href=&quot;http://physics.stackexchange.com/questions/3799/why-is-cold-fusion-considered-bogus&quot;&gt;recent one being discussed here&lt;/a&gt; too. Approached from nuclear physics orders of magnitude the claims seem preposterous. There are people though who believe they have  results of extra energy over input energy, much more than chemical reactions could supply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This set me thinking on vacuum energy and the analogy with getting energy from the sea. A crystal is a prime candidate for any exploration of such concepts and in all cold fusion &quot;successful&quot; results crystals have been used. Now if the effect depended on the vacuum and how much distorted it was by a passage of a gravitational wave at the time of the experiment, or the exact orientation of the crystal, or the type of impurities in the crystal ( F centers etc) one would expect to get haphazard results, and non repeatable by other experimenters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this would be the first experimental evidence of gravitational waves :).&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-07-03T13:09:42.717" LastActivityDate="2011-08-02T19:37:04.993" Title="Can energy be taken out of the QFT vacuum?" Tags="&lt;quantum-field-theory&gt;" AnswerCount="2" CommentCount="12" FavoriteCount="1" />
  <row Id="11801" PostTypeId="2" ParentId="11760" CreationDate="2011-07-02T14:23:25.957" Score="1" ViewCount="" Body="&lt;p&gt;The edition Mechanics was 1st published at 1960 and was written earlier than that. &#xA;Landau died in 1968 aged 60.&lt;br&gt;&#xA;Just 4 years before, in 1964, the CMB was discovered and the properties of it was unknown for years after.&lt;br&gt;&#xA;It appears that Landau's  saying was vindicated with a referential with the properties of the CMB.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Earth (we, and the labs) is moving in relation to the CMB and the universe appear to us to be non isotropic. From the perspective of CMB, the referential of light, i.e. where light propagates equally in all directions, the universe is isotropic.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition every observer in the universe can use (share) a common length and time base using the CMB properties for the calibration purposes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This special referential is not attached to any observer as with the Einstein ones.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-07-02T14:23:25.957" CommentCount="8" />
  <row Id="11802" PostTypeId="2" ParentId="2071" CreationDate="2011-07-02T14:59:44.487" Score="0" ViewCount="" Body="&lt;p&gt;On snowy roads you have reduced friction (duh!) which translates to reduced available traction on the driving wheels. Driving in 1st gear is a bad idea, because with just a little extra throttle you can exceed traction (due to the torque multiplication on the wheels). Driving in 5th gear is a bad idea also, because you have to maintain higher speeds (&gt;35 mph typically) for the engine not to stall. So the answer is in-between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to maintain control of the car and dampen the gas pedal response as much as possible, I would recommend driving in whatever gear keeps the engine happy at about 2000-2500 rpm (gas engines, lower for diesels).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick is keep the momentum going on slopes (no acceleration / deceleration ) and brake early &lt;strong&gt;before&lt;/strong&gt; you start cornering. Do *&lt;em&gt;not&lt;/em&gt; trail brake (apply brakes and steering at the same time). If you feel the driving wheels spinning let go of the gas slowly and wait for the car to settle.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-07-02T14:59:44.487" />
  <row Id="11803" PostTypeId="1" AcceptedAnswerId="11810" CreationDate="2011-07-02T15:57:58.213" Score="6" ViewCount="267" Body="&lt;p&gt;In a NPR News story from a few years back:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;A gamma-ray burst from about 13&#xA;  billion light years away has become&#xA;  the most distant object in the known&#xA;  universe.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a layman when it comes to physics, so cut me some slack if this is an ignorant question, but assuming the universe is around 14B years old, and has been expanding since the Big Bang, how is it that we can see events so far back in time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand how it would work if you had a static universe and the GRB happened 13B years ago at 13B Light years away and the light just arrived. However, at the time of the burst wouldn't we (or at least the matter that we are made from) have been much closer to the source of that burst, and wouldn't the light have blown by us eons ago? How is it we are seeing it now? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were expanding away from it at close to light speeds it would seem to make sense for why it took so long for it to get here, except for that whole notion that light moves at the same speed relative the to the observer, which I think would blow that idea out of the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps gamma rays travel at sub-light speeds? But, I'd still think the math would require that they travel MUCH slower than light for this scenario to play out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another, possibility is that the light has wrapped around a finite universe a few times before reaching us. Of course if that were the leading theory, there wouldn't be any remaining controversy about the finite vs. infinite universe models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing here? &lt;/p&gt;&#xA;" OwnerUserId="625" LastActivityDate="2011-07-07T15:34:57.997" Title="How is it possible for astronomers to see something 13B light years away?" Tags="&lt;astrophysics&gt;&lt;astronomy&gt;&lt;universe&gt;" AnswerCount="4" />
  <row Id="11804" PostTypeId="1" AcceptedAnswerId="11994" CreationDate="2011-07-02T16:31:28.227" Score="3" ViewCount="104" Body="&lt;p&gt;Let us consider a Wightman field theory for the free scalar neutral field $\phi$, and let $O\mapsto\mathfrak F(O)$ be the net of local von Neumann field algebras. If we take a non-empty bounded open subset $O$ of $\mathbb R^4$, then according to the Reeh-Schlieder theorem the vacuum vector $\Omega_0$ of the Fock construction is cyclic. This implies that the set of vectors $\{W\Omega_0\ |\ W\in\mathfrak F(O)\}$ is total in the Fock Hilbert space $\Gamma(\mathscr H)$, i.e. it is dense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is $\mathfrak F(O)$ not irreducible in this case? If we take $C$ in the commutant of $\mathfrak F(O)$, i.e. $C\in\mathfrak F(O)&amp;#39;$, then we can show that $\Omega_0$ is an eigenvector of $C$, i.e. $\exists\lambda_C\in\mathbb R$ s.t. $C\Omega_0=\lambda_C\Omega_0$. Moreover, using commutativity, we deduce that $(\psi,CW\Omega_0)=\lambda_C(\psi,W\Omega_0)$ for any $\psi\in\Gamma(\mathscr H)$ and $W\in\mathfrak F(O)$. But since the $W\Omega_0$s are elements of a dense subset of the Fock Hilbert space, then we must conclude that $C=\lambda_C\text{Id}$ for any $C\in\mathfrak F(O)&amp;#39;$, i.e. $\mathfrak F(O)$ is irreducible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's wrong with the above proof?&lt;/p&gt;&#xA;" OwnerUserId="4248" LastActivityDate="2011-07-07T17:46:22.327" Title="Why is the local field algebra $\mathfrak F(O)$ associated with a bounded non-empty open region $O$ of space-time not irreducible?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="11805" PostTypeId="2" ParentId="11799" CreationDate="2011-07-02T17:00:20.873" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Edit 1:&lt;/strong&gt; I think I just understood you question: you are actually&#xA;trying to calculate some sort if “internal” inductance, i.e. the&#xA;contribution to the inductance of only the field &lt;em&gt;inside&lt;/em&gt; the conductor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When calculating the flux, you have to choose a closed path over which&#xA;you would want the electromotive force, and then integrate the magnetic&#xA;flux over the surface limited by this path. Normally the path would be&#xA;the whole electrical circuit, but since you are only interested in the&#xA;contribution of the internal field, you chose the return path along the&#xA;edge of the wire, which is fine. Now you have to choose the forward&#xA;path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The forward path should be along the lines of current. The problem is&#xA;that, different lines of current give different fluxes. Then you can&#xA;calculate the flux as a function of where, in the conductor's&#xA;cross-section, you take the forward path. But since you are using the&#xA;low-frequency approximation (no skin effect, then uniform current&#xA;density), you can just average the forward-path dependence over the&#xA;whole cross-section. Then you get the missing factor two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A somewhat different argument is given in this old &lt;a href=&quot;http://www.g3ynh.info/zdocs/refs/NBS/Rosa1908.pdf&quot; rel=&quot;nofollow&quot;&gt;bulletin of the&#xA;Bureau of Standards&lt;/a&gt;:&#xA;the author instead weights individual flux lines as per the fraction of&#xA;the conductor they enclose. This gives the same factor two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2:&lt;/strong&gt; As requested, a few clarifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By “integrate the magnetic flux” I really mean “calculate the magnetic&#xA;flux”. I used “integrate” because the calculation involves an integral:&#xA;$$&#xD;&#xA;\phi = \int_A \mathbf{B}\cdot\mathbf{n}\; \mathrm{d} A&#xD;&#xA;$$&#xA;where $\mathbf{n}$ is the unit normal to the surface. It's not exactly&#xA;the same as “integrate the magnetic field” because of the dot product&#xA;with $\mathbf{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I talked about “forward path” and “return path” because, if it's not an&#xA;antenna (as the low-frequency approximation suggests), a wire is usually&#xA;part of a transmission line which consists of at least &lt;em&gt;two&lt;/em&gt; conductors. &#xA;Assume for example that you use a pair of wires to connect a source to a&#xA;load, like in the figure below (I hope everyone can see Box Drawing&#xA;characters):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;╔════════╗                 ╔════════╗&#xA;║        ╟→→→→→→→→→→→→→→→→→╢        ║&#xA;║ source ║   (flux here)   ║  load  ║&#xA;║        ╟←←←←←←←←←←←←←←←←←╢        ║&#xA;╚════════╝                 ╚════════╝&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where the arrows represent the electric current. I assume the wire you&#xA;are interested in is the top one, which I called “forward path”. The&#xA;bottom wire, which I called “return path”, brings the current back to&#xA;the source. Taken together, these two wires form a loop and the current&#xA;will make some magnetic flux through the loop. Then, if you try to&#xA;change the current, some electromotive force will appear because of this&#xA;flux, and you will be able to model this as the effect of an inductor&#xA;along the transmission line, as below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;╔════════╗                 ╔════════╗&#xA;║        ╟────(inductor)───╢        ║&#xA;║ source ║                 ║  load  ║&#xA;║        ╟─────────────────╢        ║&#xA;╚════════╝                 ╚════════╝&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is the self inductance of the transmission line, and is what I&#xA;first thought you where trying to calculate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The self inductance of a bare wire is somewhat ill-defined. Well, it&#xA;&lt;em&gt;is&lt;/em&gt; defined, but with some assumptions about the surface over which to&#xA;integrate the flux, and it scales as $l\log\frac{l}{r}$, which makes&#xA;it's value per unit length diverge logarithmically when considering an&#xA;arbitrarily long wire, as pointed out by Zassounotsukushi and mmc. Once&#xA;you add the second wire, the surface over which you have to integrate&#xA;the flux is clearly defined, and the inductance of the line scales like&#xA;$l\log\frac{d}{r}$, where $d$ is the distance between the wires. No more&#xA;logarithmic divergence with respect to $l$. On the other hand, it&#xA;depends logarithmically on the distance between the wires, therefore you&#xA;cannot just assume that the return path is just far enough to be&#xA;ignored. BTW, the return path is not necessarily a wire, it could be,&#xA;e.g., a ground plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the particular calculation you are doing (only the contribution of&#xA;the field &lt;em&gt;inside&lt;/em&gt; the conductor), you use a very narrow loop where the&#xA;return path is replaced by a line along the edge of the conductor, in&#xA;order to enclose only the internal field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Original answer below&lt;/strong&gt;, which is somewhat bogus, as I thought you&#xA;where after the total self-inductance (including external field) per&#xA;unit length of an infinite wire. The comments of Georg refer to this&#xA;original version.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You cannot assign an inductance to a long wire alone: you have to consider the whole circuit. The current carried by the wire has to come back in some way, and you need to know how far from your wire is the way back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume for a moment that the wire is actually the inner conductor of a coaxial cable. You can easily calculate the linear inductance of the cable as a function of the inner an outer conductor radii. Now make the outer radius go to infinity and you have a diverging self-inductance! This means that in practice you can never assume that the way back is “far enough” to ignore it.&lt;/p&gt;&#xA;" OwnerUserId="4202" LastEditorUserId="4202" LastEditorDisplayName="" LastEditDate="2011-07-04T08:57:28.207" LastActivityDate="2011-07-04T08:57:28.207" CommentCount="6" />
  <row Id="11806" PostTypeId="1" AcceptedAnswerId="11813" CreationDate="2011-07-02T17:06:49.777" Score="6" ViewCount="244" Body="&lt;p&gt;I'm a mathematician with hardly any knowledge of physics. Before I start reading volumes of physics books, I have a few questions that have been bugging me and that will help me start reading physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's forget Quantum Mechanics for the purpose of the discussion, and focus on Relativity. I always fail to understand what the objects are and how to relate them intuitively to perceivable space and time. What are the objects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I mean: we begin with spacetime being a pseudo-Riemannian manifold with some metric (is this metric assumed to be locally Lorentzian?). For each point in this manifold, there should be some parameters, right? Like whether there's a particle there, what particle is it (information that includes, for example, mass), is there a magnetic field there and so forth. Together, this pseudo-Riemannian manifold, with a pre-specified metric, and with a set of parameters for each point, is what described the universe in relativity, right? What &lt;em&gt;precisely&lt;/em&gt; are those parameters in General Relativity?&lt;/p&gt;&#xA;" OwnerUserId="4249" LastActivityDate="2011-07-02T18:56:48.947" Title="Objects in Physics as a mathematician would see them" Tags="&lt;general-relativity&gt;&lt;mathematics&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="11807" PostTypeId="2" ParentId="11803" CreationDate="2011-07-02T17:14:14.740" Score="1" ViewCount="" Body="&lt;p&gt;I believe the quasar you're refering to was observed to have a z, or redshift of 10. This implies it is receding from us at a velocity of 0.984c, very close to light speed. With a Hubble constant of 74 km/s/Mpc, its distance is indeed about 13 billion light years. There's no reason to believe we were close to the event when it occurred. For example, the CMB's redshift is about 1100, implying a much higher recession velocity and greater distance, and that was initiated about 400,000 years after the Big Bang.&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-07-02T17:14:14.740" CommentCount="1" />
  <row Id="11808" PostTypeId="2" ParentId="11803" CreationDate="2011-07-02T17:16:43.050" Score="2" ViewCount="" Body="&lt;p&gt;The answer to this question goes back to the idea of cosmological inflation. Often, particular in popular media, expansion is viewed as two galaxies going farther and farther away from one another. This is only sorta true. What is really happening is the space between them is expanding (the balloon analogy is often used, imagine placing two dots on a deflated balloon, and then blowing that balloon up, the dots are now farther from one another), and because of this, it is very hard to judge what distance &quot;means&quot; at such great scales. Even terms like relative velocity get extremely hard to define when looking billions of light years away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, when talking about &quot;distances&quot; in astronomy, one uses the redshift of the light, which is usually a more functional description. For example this object had a redshift of z=10, which has more of a redshift than any other astronomical object, and we are looking at an image of how it would have looked like if we were right next to it 485 million years after the big bang. However our proto-galaxy was probably some distance from it to begin with and as the light was traveling, space itself expanded considerably, making the light only reach us now, a dozen or so billion years later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as &quot;looping around a finite universe,&quot; this is very unlikely and there is no evidence for it. Gamma rays will travel at the speed of light through a vacuum.&lt;/p&gt;&#xA;" OwnerUserId="3018" LastEditorUserId="3018" LastEditorDisplayName="" LastEditDate="2011-07-02T17:37:54.380" LastActivityDate="2011-07-02T17:37:54.380" />
  <row Id="11809" PostTypeId="2" ParentId="11800" CreationDate="2011-07-02T17:41:16.423" Score="1" ViewCount="" Body="&lt;p&gt;The ground state (=vacuum) is an eigenstate of the system Hamiltonian. The wave function is unique, no energy uncertainty exist. In this sense there is no fluctuation of energy to harvest. &quot;Fluctuations&quot; exist for non commuting with the Hamiltonian variables but it cannot be used to get some energy for the reason given above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In GR the energy is not conserved and this is not connected to the vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: in heat machines one tries to increase the temperature difference $T_{hot}-T_{cold}$ to increase the machine efficiency. Vacuum is the coldest body ever known. So the vacuum &quot;cleaner&quot; efficiency cannot exceed zero.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-07-03T19:10:40.283" LastActivityDate="2011-07-03T19:10:40.283" CommentCount="3" />
  <row Id="11810" PostTypeId="2" ParentId="11803" CreationDate="2011-07-02T17:47:40.443" Score="10" ViewCount="" Body="&lt;p&gt;There are at least two ideas involved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First is that the expansion of the universe is not linear. While the Big Bang happened around 14B years ago, that does not mean that 13B years ago, the Universe is 1/14th of its present size. Current theory suggests that a large portion of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Inflationary_epoch&quot; rel=&quot;nofollow&quot;&gt;cosmological inflation&lt;/a&gt; (where the Universe increased by 26 or more orders of magnitude in linear dimensions) happened within much, much less than a second after the Big Bang. And as another example, the current theory estimates that at the time that the cosmic microwave background was emitted (which was about 0.5 million years after the birth of the Universe, placing it about 1/30000 the current age of the Universe), the universe is already about 1/1000 its current size (in length). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second is that the apparent recession of far away objects from us is not so much objects flying apart from each other. Rather, it is &lt;em&gt;space&lt;/em&gt; being added in between objects. Imagine you being the photon, and two turtles (moving slower than you) being the galaxies. Put turtle one in the first carriage of a train, and put turtle two on the 10th carriage of a train. And you start walking. Say it takes you 1 minute to traverse a carriage, and it take the turtles 10 minutes. Then in the case where the turtles walk away from each other, it will take you a bit under 12 mintues to get from the first turtle to the second (you walk 10 minutes to the tenth train, and the turtle has gotten to the 11th. You walk another minute to the 11th train. The turtle is just a few steps in front of you.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that's not how the universe expands. The expansion of the universe is more like the following: suppose every 6 minutes, all the carriages decouple, and between each pair of the original carriages plops one more car! So you walk for 6 minutes (having traversed 6 cars), and you look up, and see that the second turtle is 8 cars in front of you (and the first turtle is 12 cars behind). And you walke another 6 minutes. Plop comes the extra cars, and now you are 4 cars from the second turtle and 36 cars from the turtle behind. And finally after another 4 mintues you catch up to the second turtle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the point of view of the second turtle though, you would have travelled from a turtle that is &lt;strong&gt;now&lt;/strong&gt; 40 cars away from him, while taking only 16 minutes! This ties back into the funny idea that &lt;em&gt;light emitted from an object 13B lightyear away&lt;/em&gt; can take &lt;em&gt;quite a bit less than 13B years&lt;/em&gt; to get here, due to the inflationary Universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why cosmologists and astronomers use red-shift to measure distance, because there is &lt;a href=&quot;http://en.wikipedia.org/wiki/Comoving_distance&quot; rel=&quot;nofollow&quot;&gt;no reasonable intrinsic notion of distance&lt;/a&gt; that is free from ambiguity: should distance be described by how far away the turtles are when you started walking? or when you finished walking? or the number of carriages you (the photon) have traversed? Instead of that, they measure it using &lt;strong&gt;red-shifts&lt;/strong&gt;, which can roughly fit into this turtle-you framework as &lt;strong&gt;how flushed your cheek is from all that walking&lt;/strong&gt; when you reached turtle number two. Based on the redness of your cheeks, the turtles can calculate how much you exerted yourself, and thus for how long you've been traveling, and using known rules of the addition of new cars (the value of Hubble constant), the turtles can estimate the distances to other turtles. &lt;code&gt;:-)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm going to skip discussion of &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_candles&quot; rel=&quot;nofollow&quot;&gt;standard turtles&lt;/a&gt;, which are turtles from which you will always depart well rested and not flushed, nor how the turtle simiano-ferroequinologists found out about their rates of locomotive expansion.)&lt;/p&gt;&#xA;" OwnerUserId="829" LastEditorUserId="829" LastEditorDisplayName="" LastEditDate="2011-07-02T19:20:44.950" LastActivityDate="2011-07-02T19:20:44.950" CommentCount="4" />
  <row Id="11811" PostTypeId="2" ParentId="11806" CreationDate="2011-07-02T17:52:23.483" Score="3" ViewCount="" Body="&lt;p&gt;I would say, as far as general relativity is concerned, &quot;object&quot; is a fairly vacuous term. What one is really looking at is some perturbation in the stress-energy tensor. Einstein's Equations (without cosmological constant) are $G_{\mu\nu}= \frac{8 \pi G}{c^4} T_{\mu\nu}$, which defines a relation between Einstein curvature, and therefore the metric, and the stress-energy tensor at a given point. I think this tensor is what these &quot;parameters&quot; you are looking for are, as it defines the energy density, the energy flux, the shear stress, and the pressure. &lt;/p&gt;&#xA;" OwnerUserId="3018" LastEditorUserId="3018" LastEditorDisplayName="" LastEditDate="2011-07-02T17:58:17.460" LastActivityDate="2011-07-02T17:58:17.460" />
  <row Id="11812" PostTypeId="2" ParentId="11806" CreationDate="2011-07-02T17:53:36.167" Score="3" ViewCount="" Body="&lt;p&gt;In GR, the &quot;objects&quot; (whatever they might be for your problem) need to be mapped to the stress-energy tensor, which the GR field equations relate to the curvature of the manifold. That mapping is not really part of GR though but it is surely part of the everyday application of GR, although it will differ depending on the problem and what level of description of the physical objects is used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One introductory physical object which is often used as an example in introductory GR texts as far as I know is &quot;dust&quot;, to quote Schutz: &quot;'dust' is defined to be a collection of particles, all of which are at rest in some one Lorentz frame&quot;. You define a density of the particles in a volume, and the momentum flux component $\alpha$ of the particles across each perpendicular surface of constant $\beta$ in spacetime is one of the components of the stress-energy tensor $T^{\alpha\beta}$. So $T^{00}$ is the energy-components of the dust particles across a surface of constant time, that is, the energy density. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there you can keep adding more complex objects like fluids or electromagnetic fields etc. as long as you map them to the stress-energy tensor.&lt;/p&gt;&#xA;" OwnerUserId="2936" LastEditorUserId="2936" LastEditorDisplayName="" LastEditDate="2011-07-02T18:05:53.633" LastActivityDate="2011-07-02T18:05:53.633" />
  <row Id="11813" PostTypeId="2" ParentId="11806" CreationDate="2011-07-02T18:26:15.147" Score="5" ViewCount="" Body="&lt;p&gt;General relativity is a mathematical theory that generalizes the special theory of relativity and mechanics. It turns out that one can treat the gravitational effects as non-inertial forces, thanks to one of the principles that governs this theory, that is also considered the most important one, namely the principle of equivalence postulated by Einstein. That the metric of space-time must be locally flat, i.e. Minkowskian, is a consequence of this principle. In fact, the principle of equivalence states that in a free falling frame of reference, i.e. a reference where there are no inertial forces, the laws of physics are those of special relativity. As you are a mathematician there is no need for me to recall that if a manifold has a null curvature tensor, then there are global coordinates on the manifold such that the components of the metric tensor, say $g$, are exactly those of the Minkowski metric $\eta = \operatorname{diag}(1,-1,-1,-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The metric $g$ on the manifold $M$ describing the space-time is not a-priori given, but it is determined by the distribution of matter. Now comes your question on the objects of general relativity. These objects are all sorts of geometrical object that you can construct on a (smooth) manifold. Thus you'll have tensors of any ranks, and even spinors of any rank. What makes them physical objects is just their interpretation. Einstein's field equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Ric}-\frac12\operatorname{Tr}(\text{Ric})g = \chi T,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\text{Ric}\in T^*M\otimes T^*M$ is the Ricci tensor, $g\in T^*M\otimes T^*M$ the metric on $M$, and $T\in T^*M\otimes T^*M$ the energy-stress tensor, gives you the metic $g$ in terms of the energy-matter distribution $T$ on $M$, that is just a tensor density on $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mathematical problem of studying the geodesics on a manifold $M$ described by a metric $g$ is the equivalent of the physical problem of determining the motion of a particle that is freely falling in the gravitational field generated by a matter distribution $T$ such that the resulting metric is $g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the most spectacular prediction of the general theory of relativity is the existence of so-called black holes. Again you find these object by studying the mathematical properties of the solutions of the Einstein's equations and then give them a physical interpretation. There are some tough situation in physics when you can't base your theories on observations and experiments. Therefore you must start with a mathematical theory and develop it as further as you can. Each result you obtain is then physically interpreted. This is quite the situation of general relativity (see, e.g. the problem of the detection of gravitaional waves, or the observation of naked singularities).&lt;/p&gt;&#xA;" OwnerUserId="4248" LastActivityDate="2011-07-02T18:26:15.147" CommentCount="1" />
  <row Id="11814" PostTypeId="2" ParentId="11806" CreationDate="2011-07-02T18:30:29.297" Score="3" ViewCount="" Body="&lt;p&gt;Those set of parameters are usually called &lt;em&gt;matter fields&lt;/em&gt;. They contribute to GR on the &quot;right hand side&quot; of the equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ G_{\mu\nu} = T_{\mu\nu} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you mentioned that you are a mathematician, let me be slightly more technical and verbose. The &lt;em&gt;space-time&lt;/em&gt; is given by a Lorentzian manifold $(M,g)$. The matter-fields should be considered to be a collection $\{\Phi_A\}_{A\in \mathcal{A}}$ where $\mathcal{A}$ is some indexing set, with each individual $\Phi_A$ being a section of some fibre bundle $(E_A,M,\pi_A)$ over $M$. (There are more general possibilities for matter fields, but let us stick to these now.) For example, the scalar field is given by a trivial complex line bundle over $M$, while the usual description of Maxwell's theory of electro-magnetism admits the formulation of the field (the vector potential) as a section of the cotangent bundle $T^*M$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;dynamics&lt;/em&gt; of the fields are generally prescribed by some equations of motions, and their contribution to gravity is taken to be their contribution to the energy momentum tensor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T_{\mu\nu} = T_{\mu\nu}(\{ \Phi_A\}_{A\in \mathcal{A}}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in such a way that the condition &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \nabla^\mu T_{\mu\nu} = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\nabla$ is the covariant derivative of the metric $g$ is satisfied. (This is due to that the identity must hold for the left hand side of Einstein's equation according to the contracted Bianchi identity.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that general relativity by itself is a theory of gravity. It doesn't really specify what the matter fields are. It only requires that the matter fields, when they do exist, have dynamics that obey the conservation law given by the divergence condition on the energy-momentum tensor. To get an actual physical model of the world, you will have to come-up with some rules that govern the behaviour of the matter fields. In modern physics, this is generally through some sort of &lt;em&gt;action principle&lt;/em&gt;, since the Euler-Lagrange equations will automatically be compatible with the divergence condition above, if you take $T_{\mu\nu}$ to be the Einstein-Hilbert stress-energy generated from the action principle. &lt;/p&gt;&#xA;" OwnerUserId="829" LastActivityDate="2011-07-02T18:30:29.297" CommentCount="1" />
  <row Id="11815" PostTypeId="2" ParentId="11806" CreationDate="2011-07-02T18:56:48.947" Score="2" ViewCount="" Body="&lt;p&gt;The 'objects' that are studies in relativistic physics are the same as what people are studying in non-relativistic physics.  Consider a few examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If you are concerned with relativistic mechanics one can consider extended objects like rods of length L and calculate moments of intertia. (Here the object is the rod)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you are studying quantum mechanics then you will study the Dirac equation rather than the Schrodenger equation (Here the objects being studies are electrons)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you are studying statistical mechanics you may study the equation of state using a gas of photons rather than some non-relativistic gas. (not if object here applies unless we're talking about the collection of photons themselves)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The basic things which are studied are still the same things you'd study in non-relativistic physics.  Nothing really changes there.  What does change is the geometry of the underlying manifolds where the objects of study live.  This usually adds two things regardless of wether you're talking special or general relativity:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Extra terms in the potentials used to describe interactions between the 'objects'&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Extra terms in the equations of motions which arise from the geometry of the space-time the objects live in.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;General relativity is still physics so look to physics first to understand what you are trying to study.  The questions being asked are still the same.  I know sometimes it's hard to see that when you pull open a book that says general relativity on it and the discussion revolves around connections, curvature, exterior algebra, and so forth but the objects of study are still there as you correctly assert.  :)  My suggestion is to crack open a basic physics book and then try to find the relativistic analogs of what you see there in more advanced books like Misner, Thorne, and Wheeler's Gravitation or Robert Wald's General Relativity.  They both have very physical approaches to GR and you may have an easier time getting at the 'objects' that way.&lt;/p&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2011-07-02T18:56:48.947" />
  <row Id="11816" PostTypeId="2" ParentId="11800" CreationDate="2011-07-02T19:07:34.073" Score="0" ViewCount="" Body="&lt;p&gt;Peter Milonni has devoted a lot of work to this area, i will recommend this book: &lt;a href=&quot;http://avaxhome.ws/ebooks/science_books/physics/The_Quantum_Vacuum_An_Introduction_to_Quantum_Electrodynamics_-_Peter_W_Milonni.html&quot; rel=&quot;nofollow&quot;&gt;The Quantum Vacuum&lt;/a&gt;, it basically gives a overview on lot of physical systems where vacuum field effects are dominant, and how certain boundary conditions can effect them to produce unexpected effects&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the expansion of fields around the vacuum in harmonic modes with quantized particle population of the modes is only valid when the physics can be fairly approximated in terms of the vacuum eigenmodes (which are the regular field wave functions). Of course, when there are special boundary conditions, this doesn't change too much since we still have eigenmodes (which correspond to the boundary geometry) but still can talk about particle populations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when the system (meaning, the boundary conditions) are dynamic (&lt;a href=&quot;http://physics.stackexchange.com/questions/11763/how-does-dynamic-casimir-effect-generate-correlated-photons&quot;&gt;moving mirrors&lt;/a&gt;, or &lt;a href=&quot;http://physics.stackexchange.com/questions/6705/what-do-we-know-of-superconductivity-in-thin-layers&quot;&gt;superconductor walls moving over the superconductivity phase&lt;/a&gt;) i don't think there is an authoritative answer about the validity of these approximations. So in short, &lt;em&gt;there could be interesting things to say about the vacuum dynamics when a more friendly framework exists to make computations in highly dynamical regimes&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made a somewhat relevant question a while ago &lt;a href=&quot;http://physics.stackexchange.com/questions/6655/dynamic-casimir-effect&quot;&gt;about casimir walls that melted and froze back with an oscillatory transversal magnetic field&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i'm sorry, i'm not addressing your larger question, as to what relevant things change when taking GR in consideration. Hopefully someone more knowledgeable on the matter will. &lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-07-02T19:07:34.073" CommentCount="5" />
  <row Id="11817" PostTypeId="1" CreationDate="2011-07-02T19:09:55.973" Score="4" ViewCount="188" Body="&lt;p&gt;Heat or thermal energy as understood is nothing but motion of molecules of the matter. If the molecules are tightly bound (in case of solids), it is to-and-fro molecular vibrations, otherwise it is continuous random motion of molecules (in case of liquids/gases/plasma).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sound, being waves, is also vibrations in matter. Why, then, if we heat one end of a solid rod, assuming rod is at least few feet in length, does it take ages for the heat to reach the other end, whereas sound reaches in no time ? (sound travels at 1400 m/s in solid)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doesn't it show that heat is more an intra-atomic feature rather than an atomic or molecular motion? Observation that electrical good conductors are also good conductors for heat, can we assume that heat is chaotic motion of electrons (the &quot;electron gas&quot;)? (Of course, all the known phenomena related to heat will be needed to be explained with this hypothesis.) &lt;/p&gt;&#xA;" OwnerUserId="3636" LastEditorUserId="3636" LastEditorDisplayName="" LastEditDate="2011-07-02T19:44:07.330" LastActivityDate="2011-07-05T01:42:06.323" Title="Understanding Heat" Tags="&lt;heat&gt;&lt;temperature&gt;&lt;kinetic-theory&gt;&lt;vibration&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="11818" PostTypeId="2" ParentId="11817" CreationDate="2011-07-02T19:25:34.597" Score="10" ViewCount="" Body="&lt;p&gt;The analogy is a very good one, because heat transfer is in fact modelled by phonons, which you could also use to describe sound waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crucial difference is that sound waves have a much longer wavelength (at least in the range of some millimetres) than thermal phonons (not more than a few orders of magnitude bigger than the atomic lattice scale). These small-wavelength phonons can easily scatter at any lattice impurities, while the sound waves need macroscopic pertubations (like air gaps in an insulated glazing) to do so.&lt;/p&gt;&#xA;" OwnerUserId="3540" LastActivityDate="2011-07-02T19:25:34.597" CommentCount="3" />
  <row Id="11819" PostTypeId="2" ParentId="11817" CreationDate="2011-07-02T19:40:50.090" Score="-1" ViewCount="" Body="&lt;p&gt;I feel that when a physicist speak about Heat he/she has a flow of energy in mind. Suppose that you have a rod and that the two extrema are held at different temperatures. Then the Fourier law states that there must be a flow of Heat from the hotter end to the cooler. When a physicist speak, instead, of the molecular motion he/she is thinking to the internal energy of the body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now when molecules and atoms are involved, it more likely that we must enter into the quantum world. By the way, we can make some few heuristic semiclassical consideration, namely we can apply Boltzmann statistics to the quantum structure of atomic and molecular spectra. A body that is immersed in a certain environment will be in thermal equilibrium state. Atoms and molecules receives energy from the thermal bath, but they also radiate energy in such a way the the total balance is &quot;no energy exchange&quot;, therefore no energy flow, i.e. no heat flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We must note though that when we deal with atomic or molecular excitation levels, we are considering relatively tiny amounts of energy. Take as a reference the binding energy of the electron in the hydrogen atom, this being roughly 13.6 eV. Sounds excitations involves way more energy than this and in this case you can forget that the body has a quantum nature. You can treat it as a continuum and apply the laws of classical mechanics, i.e. the theory of elasticity and forth.&lt;/p&gt;&#xA;" OwnerUserId="4248" LastActivityDate="2011-07-02T19:40:50.090" CommentCount="9" />
  <row Id="11820" PostTypeId="1" CreationDate="2011-07-02T21:31:18.183" Score="7" ViewCount="405" Body="&lt;p&gt;I know that if we solve the maxwell equation, we will end up with the phase velocity of light is related to the permeablity and the permitivity of the material. But this is not what I'm interested in, I want to go deeper than that. We know that the real speed of light is actually not changing, the decrease in speed is just apparent. Material is mostly empty, the light will still travel with c in the spacing. The rare atoms will disturb the light in some way. So I am interested in how the atoms affect the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some textbooks that I read explain it in a way kind of like this:&#xA;In a material the photons is absorbed by atom and then re-emitted a short time later, it then travels at a short distance to the next atom and get absorbed&amp;amp;emitted again and so on. How quickly the atoms in a material can absorb and re-emit the photon and how dense the atoms decides the speed of light in that material. So the light appears slower because it has a smaller “drift speed”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But recently I realize an alternative explanation:&#xA;Atoms respond to the light by radiating electromagnetic wave. This “new light” interferes with the “old light” in some way that result in delayed light(advanced in phase),this can easily shown by using simple phasor diagram. Consequently effectively the light covers a smaller phase each second, . Which gives the impression of a lower phase velocity . However the group velocity is changing in a complicated way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that the first explanation does not explain the change in phase velocity of light. if we consider light travelling into a slab of negative refractive index non-dispersive material, let’s say the light is directed perpendicular to the slab. The phase velocity’s direction will be flipped, but group velocity’s direction in the material will not change. Only the second explanation can explain the flipped phase velocity direction. I guess that the velocity that we get in the first explanation is actually belongs to the group velocity. It makes sense to me that the front most of the photon stream determines the first information that the light delivers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is What really cause the phase velocity of light to be decreased?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&quot;drift velocity&quot; of photons ( they aren't the same photons, they are re-emitted all the time)&#xA; 2. phase difference between absorbed and emitted light&#xA; 3. something else&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also, I still don't really understand about the detail of the absorption-emission process for small light's wavelength ( for large lambda compare to the atoms spacing, the photons will be absorbed by the phonons). The dispersion relation that we know is continous and also ome material is non-dispersive, therefore the absorption process must accurs in all frequency for a certain range. So definitely it doesn't involve the atomic transition, otherwise it will be quantized. My guess is that the relevant absorption process gets smooth out by the dipole moment. What makes the spectrum continous?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: link for dispersion relation:&#xA;&lt;a href=&quot;http://refractiveindex.info/?group=CRYSTALS&amp;amp;material=Si&quot; rel=&quot;nofollow&quot;&gt;http://refractiveindex.info/?group=CRYSTALS&amp;amp;material=Si&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4252" LastEditorUserId="4252" LastEditorDisplayName="" LastEditDate="2011-07-04T21:33:10.680" LastActivityDate="2011-07-04T21:33:10.680" Title="What really cause light to appear slower in media?" Tags="&lt;quantum-mechanics&gt;&lt;electromagnetism&gt;&lt;optics&gt;&lt;speed-of-light&gt;&lt;atoms&gt;" AnswerCount="3" CommentCount="10" FavoriteCount="2" />
  <row Id="11821" PostTypeId="2" ParentId="2071" CreationDate="2011-07-03T02:10:07.453" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;After the edit&lt;/strong&gt;&#xA;Lets get some nomenclature out of the way. Tire radius is $R$, total gear ratio is $\gamma$ such that the engine rotational speed $\omega_E$ is a function of tire speed $\omega_E = \gamma \omega_T$ and the vehicle speed is $v=\omega_T R$. The maximum traction (force) applied on the car allowed by friction is $F_{max}=\mu\, m_T\, g$ where $m_T$ is the total mass supported by the driving wheels ($m_T = m_{car}$ for AWD, and $m_T = 0.6 m_{car}$ for typical FWD).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at any given car speed $v$ the engine speed is $\omega_E = \gamma\, \frac{v}{R}$ while the engine is making $T$ torque. The torque at the wheels is $T_{wheel} = \gamma\, T$ with forward thrust of $F = \gamma \frac{T}{R}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check that power is conserved through the gears, $F\,v = T\,\omega_E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If thrust is equal traction, then $F=F_{max}$ or $ \gamma \frac{T}{R} = \mu\, m_T\, g $ which is solved for torque as $T =\mu \dfrac{m_T\,g\,R}{\gamma}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the more friction, mass, or tire radius you have the more torque you can apply. The more gear ratio you have the less torque. Note that with this notation 1st gear has the highest $\gamma$. This is found by multiplying the final drive ratio (like 3.2:1) to the first gear ratio (like 3.75:1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It all makes sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;NOTE:&lt;/em&gt; the reason I went to trouble to establish the engine speed first is because typically the max. available torque is a function of engine speed. Also the max. engine speeds limits what is the lowest gear available at any speed.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-07-03T02:10:07.453" />
  <row Id="11822" PostTypeId="2" ParentId="5670" CreationDate="2011-07-03T02:42:32.030" Score="0" ViewCount="" Body="&lt;p&gt;This statement is true and a direct consequence of the 5th Law of Thermodynamics, the Onsager Relations for which Lars Onsager of Yale received the Nobel Prize in 1968.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an electrical circuit, for DC, current takes the path of least reisitance;&#xA;For AC it takes the path of least inductance (impedance).&#xA;So a pulse of voltage will cause the current distribution to be determined by path inductance and then finish with the distribution determined by resistance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a circuit board, this is critical as the ground plane insures the differences between these two distribution paths are minimal.&lt;/p&gt;&#xA;" OwnerUserId="4258" LastActivityDate="2011-07-03T02:42:32.030" />
  <row Id="11823" PostTypeId="2" ParentId="11820" CreationDate="2011-07-03T03:40:25.913" Score="4" ViewCount="" Body="&lt;p&gt;Looks like you are already familiar with the classical explanation but are still curious about the quantum version of it. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2.phase difference between absorbed and emitted light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yeah, this is essentially the lowest order contribution to the phase shift in the photon-electron scattering. Here is the sloppy way to visualize it continuously (this is basically the 'classical EM wave scattering' point of view): you can imagine that the &quot;kinetic energy&quot; (-&gt; frequency) of the &quot;photon&quot; increases as it approaches the atom's potential well and then it goes back to its normal frequency upon leaving the atom. This translates to a net increase in the phase ($(n-1)\omega/c$).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&quot;drift velocity&quot; of photons ( they aren't the same photons, they are re-emitted all the  time)&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By &quot;drift velocity&quot; do you mean a pinball-like, zigzag motion of the photon? This won't contribute that much because it requires more scattering (basically it is a higher order process).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And also, I still don't really understand about the detail of the absorption-emission process.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes the absorption will still occur in all range of the frequency. The hamiltonian of the atom will be modified by the field (by $- p \cdot E$ where p is the dipole moment of the atom and E is the electric field component of the light). This will give us the required energy level to absorb the photon momentarily, which will be re-emitted again by stimulated+spontaneous emission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: clarification, the term 'energy level' is misleading, since the temporarily 'excited' atom is not in an actual energy eigenstate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the diagram here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Raman_scattering&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Raman_scattering&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1814" LastEditorUserId="1814" LastEditorDisplayName="" LastEditDate="2011-07-04T16:53:23.957" LastActivityDate="2011-07-04T16:53:23.957" CommentCount="1" />
  <row Id="11824" PostTypeId="1" CreationDate="2011-07-03T04:07:26.077" Score="1" ViewCount="143" Body="&lt;p&gt;suppose i have a space-time where we have one point-like object*  which we will call movement space probe or $\mathbf{M}_{A}$ for short, and it will be moving with constant velocity $V^A_{\mu}$ in some referential frame. We don't know (so far) how flat or twisted is the overall space-time, but we know that as far as our probe goes, it seems to be moving with total uniformity (its velocity is constant in our coordinates)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Goal:&lt;/strong&gt; understand the true physical degrees of freedom in which we have to express the space-time and matter kinematics, assuming space-time is not necessarily flat, but momentarily forgetting to demand strictly $G_{\mu \nu} = T_{\mu \nu}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in the case where we demand that the geometry is completely flat and Minkowski everywhere, it seems we have to require momenta for the probe to be of the form $M^A V^A_{\mu}$. We call this description the &lt;em&gt;flat description&lt;/em&gt; or representation $W_0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but for our premise, space will be allowed to expand and/or contract arbitrarily in any geometrically meaningful way that comply with the observed movement of our probe, even if such geometries are not required to satisfy the Einstein equations dynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example of the above, lets consider another interesting representation (lets call it $W_1$), which is a geometry description where all the displacement i observe of our probe is due to space-time geometry to be expanding behind the probe and contracting in front of it.&#xA;In this representation, if i consider geodesic curves joining my origin of coordinates and the probe along its worldline, and then i proceed to parallel-transport the velocity vector $V_{\mu}$ of the probe along these geodesics, &lt;em&gt;the coordinates of the velocity of the probe transported in such way are zero.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a part of my stated goal, i want to consider both representations $W_0$ and $W_1$ to belong to the same equivalence class under some (unknown to me) equivalence relation, which i'll refer as &lt;em&gt;the geomorphism&lt;/em&gt;, in lack of a cuter name &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that this is &lt;strong&gt;NOT&lt;/strong&gt; a simple diffeomorphism, since i'm stating that space-time might actually be curved differently in each representation of the geomorphism, instead a diffeomorphism is a lot more strict than that, demanding that covariant geometric quantities transform &quot;nicely&quot; everywhere. The geomorphism only cares about &lt;em&gt;keeping invariant across representations my actual observables (the coordinates of my probe), leaving the rest of space-time (i.e: its curvature and metric) to vary otherwise arbitrarily&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, a trivial corollary of the above is that two diffeomorphically equivalent space-times &lt;em&gt;are&lt;/em&gt; geomorphically equivalent. However, two geomorphically equivalent space-times &lt;em&gt;are not necessarily&lt;/em&gt; diffeomorphically equivalent&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) is there a way to represent the above equivalence relation? what group structure does it have? how does it relate to &lt;strong&gt;diff(M)&lt;/strong&gt;? &lt;strong&gt;I'm hoping to get an idea about what sort of techniques i can learn and exercise to compute the group structure given the above definition&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) can i split or recast the Einstein equations in such a way that each component transforms differently under the above equivalence relation? obviously, if the whole Einstein equations do not change when making such equivalence transformations, then in this case the answer is that we do not need any such split. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;*. &lt;em&gt;for simplicity i am considering a single object with uniform movement, although everything said should apply to many point-like moving masses as well, not necessarily with uniform movement&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-07-04T17:21:27.350" LastActivityDate="2011-07-05T20:37:36.253" Title="composition of space expansion and movement as a gauge invariance" Tags="&lt;general-relativity&gt;&lt;space-expansion&gt;&lt;group-representations&gt;&lt;covariance&gt;&lt;diffeomorphism-invariance&gt;" />
  <row Id="11825" PostTypeId="2" ParentId="11820" CreationDate="2011-07-03T04:24:48.703" Score="1" ViewCount="" Body="&lt;p&gt;I don't believe it is generally helpful to try and analyze these things in terms of photons, so I'm going to try and point out a few things about the classical picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big difficulty from the mathematical perspective is that you're working in a continuous medium where the phase of the wave is changing continuously. It makes the visualisation much easier to start off with if we restrict ourselves to a thin slab, where &quot;thin&quot; means small with respect to the wavelength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that there is a dielectric constant which represents the tendency for charges to displace themselves in response to an external field. But how fast to the charges respond? Is it a quasi-static case, where the maximum field strength coincides with the maximum charge displacement? I think we will find that this is the case, for example, when light is travelling through glass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in this case the displacement current is leading the incident field by 90 degrees. This makes sense: as the frequency of light approaches the resonant frequency of the material, the phase lags more and more; when the phase difference goes to zero, you have resonant absorption. (EDIT: To be more clear, I choose to define the phase difference in terms of its far-field relation to the incident field!) In the case of the thin slab, you can see that the transmitted wave is the sum of the incident wave and a wave generated by the displacement current. Because you are absorbing, the phase in the far field must be opposite so that energy is removed from the incident wave. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is instructive to do the energy balance. Let's say the displacement current generates a wave equal to 2% of the incident wave. Then the amplitude of the reflected wave is 2%, and the transmitted wave is 98%. It is easy to calculate (by squaring amplitudes) that almost 4% of the energy is missing. Where does it go? It continuously builds up the amplitude of the displacement current until the resistive losses in the material are equal to the power extracted from the incident wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's now go back to the case of the transparent medium. Take the same value for the displacement current, namely 2%. The reflected wave is the same, but the transmitted wave is different because now you are adding phasors that are at 90 degrees to each other, so the amplitude of the transmitted wave is, to the first order, unchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the phase that's confusing. Because we are in the quasi-static regime, the phase is leading. In any case it must be leading in comparison to the absorptive case. Don't we want a lagging phase in order to slow down the wave? This is where you have to be very careful. Because we are adding a leading phase, the wave peaks occur sooner than otherwise...in other words, they are close together. This is indeed the condition for a wave to travel slower. It's all very confusing, which is why I took the case of a thin slab so the math would be simpler. Let the incident wave be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sin(kx-wt)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the wave generated by the slab will be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.02*cos(kx-wt)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the cosine wave leads the sine wave by 90 degrees. If you draw these two waves on a graph and add them together, you can see that the peaks of the sine wave are pushed slightly to the left. This makes the wave appear slightly delayed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The continous case is harder to do mathematically but you can see that it ought to follow by treating it as a series of slabs.&lt;/p&gt;&#xA;" OwnerUserId="3253" LastEditorUserId="3253" LastEditorDisplayName="" LastEditDate="2011-07-03T05:53:29.247" LastActivityDate="2011-07-03T05:53:29.247" CommentCount="1" />
  <row Id="11826" PostTypeId="1" AcceptedAnswerId="11865" CreationDate="2011-07-03T06:02:52.423" Score="9" ViewCount="358" Body="&lt;p&gt;Why should two sub-atomic (or elementary particle) - say electrons need to have identical static properties - identical mass, identical charge? Why can't they differ between each other by a very slight degree? Is there a theory which proves that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine an alien of size of order of Milky-way galaxy examining our solar system with a probe of size of 10's of solar system dimension and concludes that all planets are identical.&lt;/p&gt;&#xA;" OwnerUserId="3636" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-20T19:00:19.437" LastActivityDate="2011-07-20T19:00:19.437" Title="Are all electrons identical?" Tags="&lt;particle-physics&gt;&lt;electron&gt;&lt;subatomic&gt;&lt;exchange-interaction&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="11827" PostTypeId="1" CreationDate="2011-07-03T06:24:21.067" Score="0" ViewCount="51" Body="&lt;p&gt;I've been fascinated by the LHC for a number of years now and I had always wanted to see it in person. Last year I decided to do something about it and planned a European vacation around a visit to CERN. I got a really wonderful tour of the facility, including the testing/assembly building for the beam lines, the Alpha Experiment (they recently made anti-hydrogen) the LHC and ATLAS control rooms and had coffee in the cafeteria. I felt like a total science groupie, but it was an experience I'll never forget.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm trying to plan a science experiment/lab themed road trip of the US this summer and I'm wondering if anyone here has any recommendations for fun places to visit? I live in Portland, OR and am planning on starting here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The places I'm currently looking into visiting are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;LIGO Hanford&lt;/li&gt;&#xA;&lt;li&gt;SLAC&lt;/li&gt;&#xA;&lt;li&gt;NIF (Lawrence Livermore in general)&lt;/li&gt;&#xA;&lt;li&gt;Soudan Mine (If Minnesota manages to restart the state government)&lt;/li&gt;&#xA;&lt;li&gt;Tevatron&lt;/li&gt;&#xA;&lt;li&gt;JPL&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'd also be love to check out some big telescopes (optical or radio).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So where else should I go on my nerdcation? Where would you go?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S: I know this isn't quite a &quot;Physics&quot; question, but this felt like the best place on Stack Exchange to ask. If there is a better place to post it, please let me know and I'll move it. &lt;/p&gt;&#xA;" OwnerUserId="1479" LastActivityDate="2011-07-03T06:24:21.067" ClosedDate="2011-07-03T14:17:23.260" Title="What Experiments Would you Most Like to Visit?" Tags="&lt;experimental-physics&gt;" CommentCount="3" />
  <row Id="11828" PostTypeId="2" ParentId="11820" CreationDate="2011-07-03T08:35:13.707" Score="2" ViewCount="" Body="&lt;p&gt;In addition to everything said, I'd like to comment on the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We know that the real speed of light is actually not changing, the decrease in speed is just apparent. Material is mostly empty, the light will still travel with c in the spacing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1) The speed of light &lt;em&gt;does change&lt;/em&gt;. It's the speed of light &lt;em&gt;in vacuum&lt;/em&gt; that doesn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) A very short answer to your question: Light is, so to say, ‘larger’ than the inter-atomic space. Therefore I wouldn't speak of it travelling in the spacing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This situation is similar to a human running through bushes as opposed to running in a forest. Because you are larger than the individual branches of a bush, you interact with the bush in a different manner than with trees in the forest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;May be an illustration with a subwavelength-diameter optical fibre could help to see the problem from another side. This fibre is thinner than the wavelength of light. For example, half a micron diameter for a 1 um light. As the light propagates through such a fibre (which can be done with basically 100 % transmission), the light field doesn't ‘fit’ into the fibre, and about half of energy propagates actually outside the fibre. However, it is not the case that the outer part of light goes faster than the one inside the fibre. The wave remains single, travelling with the speed of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v = \frac{c}{n_\text{eff}}, $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $ n_\text{eff}$ is the effective refractive index, which is somewhere between 1 (the apparent refractive index for the outer part of the wave) and $n_\text{glass}$ (for the inner one). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, light is ‘larger’ than the inter-atom distance, and therefore it will continuously ‘see’ the atoms.&lt;/p&gt;&#xA;" OwnerUserId="3694" LastActivityDate="2011-07-03T08:35:13.707" CommentCount="1" />
  <row Id="11829" PostTypeId="1" AcceptedAnswerId="11830" CreationDate="2011-07-03T09:59:12.327" Score="1" ViewCount="115" Body="&lt;p&gt;me and my friend has a discussion last night, and he argued that both an electromagnetic field and gravititonal field are infinite in their area of effect, but with diminishing effects as you get farther away from the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I argued that only gravity field is infinite in range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So which one of us is correct, and please explain in laymen's terms, as we're not physicists :)&lt;/p&gt;&#xA;" OwnerUserId="4259" LastActivityDate="2011-07-04T11:44:33.793" Title="Gravititonal fields compared to electromagnetic fields - are they infinite in range?" Tags="&lt;electromagnetism&gt;&lt;gravity&gt;" AnswerCount="2" />
  <row Id="11830" PostTypeId="2" ParentId="11829" CreationDate="2011-07-03T10:31:55.677" Score="2" ViewCount="" Body="&lt;p&gt;Actually, your friend is right! The electromagnetic field has an infinite range, for example for a charged point particle the electric field is proportional to &lt;a href=&quot;http://en.wikipedia.org/wiki/Coulomb%27s_law#Basic_equation&quot; rel=&quot;nofollow&quot;&gt;$\frac 1{r^2}$&lt;/a&gt;, where $r$ is the distance to the particle. However in very special circumstances, like inside a superconductor, the electromagnetic field will be short-range because the photons will effectively be massive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See this &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_force#Overview&quot; rel=&quot;nofollow&quot;&gt;table&lt;/a&gt;, where you can see both the electromagnetic and gravitational fields have infinite range of interaction, but the Weak doesn't. It's the famous Higgs boson which renders the weak field short range (by making the corresponding gauge bosons massive), very similar to what happens inside a superconductor.&lt;/p&gt;&#xA;" OwnerUserId="1469" LastEditorUserId="1469" LastEditorDisplayName="" LastEditDate="2011-07-04T11:44:33.793" LastActivityDate="2011-07-04T11:44:33.793" CommentCount="3" />
  <row Id="11831" PostTypeId="2" ParentId="11829" CreationDate="2011-07-03T10:44:03.147" Score="1" ViewCount="" Body="&lt;p&gt;The key difference is that there are positive and negative electric charges and the universe is basically neutral.  The opposite effects of the charges cancel out so on large scales, ie astronomical, the net electrical forces are much less than the gravitational forces, which are all one sign and hence never &quot;cancel out.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-07-03T10:44:03.147" />
  <row Id="11832" PostTypeId="1" AcceptedAnswerId="11833" CreationDate="2011-07-03T13:58:01.370" Score="-1" ViewCount="73" Body="&lt;p&gt;If a planet is directly between two stars &lt;em&gt;y&lt;/em&gt; distance apart (so it's &lt;em&gt;y/2&lt;/em&gt; from each planet), with velocity &lt;em&gt;x&lt;/em&gt; such that &lt;em&gt;x&lt;/em&gt; is sufficient to continue orbiting at that distance (&lt;em&gt;y/2&lt;/em&gt;), which does it orbit, and how does it continue?&lt;/p&gt;&#xA;" OwnerUserId="4260" LastActivityDate="2011-07-04T00:35:11.630" Title="What is the activity of a planet in a double star system?" Tags="&lt;gravity&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="11833" PostTypeId="2" ParentId="11832" CreationDate="2011-07-03T14:26:02.393" Score="4" ViewCount="" Body="&lt;p&gt;Given that the mass is at the CoM of the system, there is a solution in which it has no velocity (i.e. it just sits there while the stars orbit around their common center which happens to include a planet). I believe the planet is positionally unstable in that case (that is, it won't stay there in the real world).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from that there is no circular or elliptical orbit at all. I don't know if there are any stable orbits that pass through that point (in fact I tend to doubt it, there will be a sensitive dependence), but if there are they will certainly be strange.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-07-03T23:36:04.537" LastActivityDate="2011-07-03T23:36:04.537" />
  <row Id="11834" PostTypeId="2" ParentId="11762" CreationDate="2011-07-03T16:16:35.693" Score="2" ViewCount="" Body="&lt;p&gt;Short answer, yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer, there is always a limit, but what sets the limit differs by what you are spinning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black hole, which can't 'fly apart' as far as we know, can't rotate any faster than the angular velocity that would make its event horizon rotate at the speed of light. That is a limit set by relativity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objects other than black holes have limits set by the binding forces holding them together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a neutron star is held together by gravitation. If you spin it fast enough (fast enough in this case being a few thousand times per second) it will simply 'come apart' when the gravitation is roughly balanced by the rotation (the surface is moving at a substantial fraction of the speed of light at that point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For objects held together by electromagnetism (ordinary matter), I believe the record is currently held by &lt;a href=&quot;http://news.oneindia.in/2010/10/01/graphenebecomes-fastest-spinning-object-with-record-60mrpm.html&quot; rel=&quot;nofollow&quot;&gt;micrometer sized graphite flakes spinning at 60,000,000 rpm&lt;/a&gt; (as of Oct 2010). The group at the University of Maryland that did it said that the theoretical limit for graphene was about 1,000 times that fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even an atomic nucleus (held together by the Strong Nuclear Force) will come apart &lt;a href=&quot;http://ns.ph.liv.ac.uk/ns-highspin.html&quot; rel=&quot;nofollow&quot;&gt;if spun hard enough&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4262" LastActivityDate="2011-07-03T16:16:35.693" CommentCount="2" />
  <row Id="11835" PostTypeId="1" AcceptedAnswerId="11836" CreationDate="2011-07-03T17:52:42.723" Score="4" ViewCount="178" Body="&lt;p&gt;I have been doing Landau and due to its concise style been facing a few problems. I hope you can help me out here somehow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)Does the &quot;homogeneity of space and time&quot; essentially talk about the invariance of the Lagrangian with co-ordinate system or time invariance? Is this the basis of Landau's proof of the Law of Inertia?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The Lagrangian for a closed system of particles is described by adding a  function of the co-ordinates. This essentially means that the same trajectory satisfies the extremal principle.(I am a bit confused about this though). So why the function added is not of time too?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) On the subject of &quot; the Lagrangian being defined only to within an additive total time derivative of any function of co-ordinates and time&quot;, is it true that the function can be of velocity too ( Though this would make the entire exercise useless, since we would know everything)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The disadvantage of trying to teach oneself physics without anyone to refer to is that I am not completely sure that what I am saying makes sense.&lt;/p&gt;&#xA;" OwnerUserId="4263" LastActivityDate="2011-07-05T03:35:25.417" Title="Invariance and forms of the Lagrangian" Tags="&lt;classical-mechanics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11836" PostTypeId="2" ParentId="11835" CreationDate="2011-07-03T20:03:24.200" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt; Yes it is about invariance with respect to time and space translations: $t\to t+t_0$ and $\vec{r}\to\vec{r}+\vec{r}_0$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.&lt;/strong&gt; I see no &lt;em&gt;fundamental&lt;/em&gt; problen in introducing the time dependence of the potential  energy. But it would mean that the way our particles interact with each other changes with time somehow. Seems to me that it is pressuposed that the interaction cannot change in closed systems -- rather natural assumption for me...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3.&lt;/strong&gt; The problem with a function of speed is that the new &quot;lagrangian&quot; will depend on second derivative of coordinates:&#xA;$$L&amp;#39;(q,\dot{q},\ddot{q},t)=L(q,\dot{q},t)+\frac{d}{dt}f(q,\dot{q},t)$$&#xA;Which contradicts to the experimentally established statement that a system fully described by coordinates and velocities (begining of chap.1). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... trying to teach oneself...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd stronlgy discourage to study mechanics by means of Landau &lt;em&gt;only&lt;/em&gt;. He sometimes is too fast -- there are some subtleties that he skips without paying attention. Also in some places he uses non-standard terminology, which can be misleading when dealing with other sources.&lt;br&gt;&#xA;I'd recommend &lt;strong&gt;Mathematical Methods of Classical Mechanics&lt;/strong&gt; by V.I. Arnold. &#xA;Personally I find it to be very nice complement to Landau.&lt;/p&gt;&#xA;" OwnerUserId="386" LastEditorUserId="4235" LastEditorDisplayName="" LastEditDate="2011-07-04T18:29:37.643" LastActivityDate="2011-07-04T18:29:37.643" />
  <row Id="11838" PostTypeId="2" ParentId="11702" CreationDate="2011-07-04T00:52:42.747" Score="5" ViewCount="" Body="&lt;p&gt;If you're talking about building a quantum computer, then there are some modes of the system which you need to keep isolated so that you can make sure that any coherence of these modes is preserved, but there are other modes of the system that you use to control the system, and these aren't isolated. This idea is also used in &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_error_correction&quot; rel=&quot;nofollow&quot;&gt;quantum error correction&lt;/a&gt;. This process uses active control on certain modes of a system to suppress the decoherence of other modes of the system. You can be sure that the system is still there by observing the modes that don't need to be isolated. A similar idea is used in building a  &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_clock#Quantum_clocks&quot; rel=&quot;nofollow&quot;&gt;quantum logic clock&lt;/a&gt;, which is the most accurate clock &lt;a href=&quot;http://www.sciencemag.org/content/329/5999/1630.abstract&quot; rel=&quot;nofollow&quot;&gt;ever built&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditorDisplayName="" LastEditDate="2011-07-04T00:58:47.717" LastActivityDate="2011-07-04T00:58:47.717" CommentCount="5" />
  <row Id="11839" PostTypeId="1" AcceptedAnswerId="11851" CreationDate="2011-07-04T00:54:49.133" Score="4" ViewCount="187" Body="&lt;p&gt;I've arrived at this question because I've been reading Weinberg's Quantum Theory of Fields Volume I, and I'm in the second chapter about relativistic quantum mechanics. Weinberg discusses the representations of the Poincare group, and after deriving the structure constants he gets the structure constants for the Galilean group as well. I have a few questions about this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He introduces these operators $M$ and $W$--where do they come from? (This question may make sense only to those reading Weinberg, I'm not sure, so it's less important).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we conclude that in the Galilean Algebra&#xA;\begin{equation}&#xA;[K_i, P_j] = -iM\delta_{ij}&#xA;\end{equation}&#xA;Shouldn't boosts and translations commute? Why is this equal to $M$ instead of total energy? Is it because $M$ is a &quot;central charge&quot;? &lt;strong&gt;What exactly is a central charge?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it have to do with the fact that $M$ is in the Lie Algebra, it is a generator of a symmetry transformation, and because it is in the center and commutes, its charge is conserved. But how do we decide what its conserved charge is? What is to prevent us from adding more of these central charges?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, look at the product of a translation and a boost (K is the generator of boosts):&#xA;\begin{equation}&#xA;e^{-i\vec{K}\cdot \vec{v}}e^{-i\vec{P}\cdot \vec{a}}=e^{iM\vec{a}\cdot \vec{v}/2}e^{-i(\vec{K}\cdot \vec{v} + \vec{P}\cdot \vec{a})}&#xA;\end{equation}&#xA;Weinberg says that this phase factor (with the M) shows we have a projective representation, &quot;with a superselection rule forbidding the superposition of states of different mass&quot;. Where did this phase factor come from? &lt;strong&gt;What are these superselection rules about?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1635" LastActivityDate="2011-07-07T13:53:51.653" Title="Representation of the Galileo Group and Central Charges" Tags="&lt;quantum-field-theory&gt;&lt;group-representations&gt;" AnswerCount="1" FavoriteCount="3" />
  <row Id="11840" PostTypeId="1" CreationDate="2011-07-04T01:55:09.117" Score="5" ViewCount="142" Body="&lt;p&gt;I understand that if a pair quark/antiquark, out of the vacuum, tries to separate then the energy increases, another pair is produced, and we finish with two mesons or generically two hadron jets. But in numerical simulations we can just put a single quark and calculate the field, can we? What does it happen, does the total energy diverge? Or does it increases up to, say, some &quot;constituient&quot; mass? &lt;/p&gt;&#xA;" OwnerUserId="1335" LastActivityDate="2011-09-02T18:04:48.103" Title="What happens to a single quark in lattice QCD simulations?" Tags="&lt;quantum-field-theory&gt;&lt;quantum-chromodynamics&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11841" PostTypeId="1" CreationDate="2011-07-04T05:20:58.070" Score="3" ViewCount="40" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;How does unitarity require that every scalar operator in a $2+1$ SCFT will have to have a scaling dimension $\geq \frac{1}{2}$ ? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why is an operator with scaling dimension exactly equal to $\frac{1}{2}$ said to be &quot;free (i.e decoupled from the rest of the theory)&quot; ?  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let us say that we know in some such theory the R-charge is monotonically decreasing with increasing coupling constant (say $\lambda$). Then define $\lambda _n ^f$ to be that value of the coupling constant at which the R-charge of the operator $Tr[\phi ^n]$ becomes $= \frac{1}{2n}$. Let $\lambda _n ^m$ be that that value of the coupling at which the R-charge of the same operator becomes $= \frac{2}{n}$ (..marginal ?..) Then clearly $\lambda _n ^m &amp;lt; \lambda_n ^f$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How does this imply that there has to exist a $\lambda _c \leq \lambda_2^f$ where the theory might undergo a phase transition? &lt;/p&gt;&#xA;" OwnerUserId="1637" LastActivityDate="2011-07-04T05:20:58.070" Title="About unitarity and R-charge in 2+1 superconformal field theory" Tags="&lt;quantum-field-theory&gt;&lt;supersymmetry&gt;&lt;conformal-field-theory&gt;&lt;superconformality&gt;" FavoriteCount="1" />
  <row Id="11842" PostTypeId="1" AcceptedAnswerId="11843" CreationDate="2011-07-04T07:03:47.153" Score="0" ViewCount="128" Body="&lt;p&gt;I know the rifling in a gun or rifle puts a spin on the bullet along the axis of trajectory. Now I don’t understand exactly what does it make the trajectory more stable and the travel grater?&lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-07-04T16:22:29.690" Title="A spinning bullet" Tags="&lt;mechanics&gt;&lt;momentum&gt;&lt;aerodynamics&gt;" AnswerCount="3" />
  <row Id="11843" PostTypeId="2" ParentId="11842" CreationDate="2011-07-04T07:41:13.713" Score="3" ViewCount="" Body="&lt;p&gt;Bullet is not of perfect shape, and these imperfections will force bullet to flip. When bullet spins, these imperfections work in all directions equally, so bullet does not flip. This cause near-perfect trajectory. Flipped bullet have higher air drag -&gt; reduced range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, please note, that even bullet spin does not guarantee that bullet would not flip when going from supersonic to subsonic speed.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-07-04T07:41:13.713" CommentCount="2" />
  <row Id="11844" PostTypeId="2" ParentId="11842" CreationDate="2011-07-04T08:28:06.203" Score="1" ViewCount="" Body="&lt;p&gt;It's called precession.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The air friction and other hydrodynamical forces a bullet experiences often exert a net torque that flip the bullet, if it is not spinning. When it spins, however, it will precess and thus remain stable, like a spinning top resisting gravity.&lt;/p&gt;&#xA;" OwnerUserId="3887" LastActivityDate="2011-07-04T08:28:06.203" />
  <row Id="11845" PostTypeId="2" ParentId="11817" CreationDate="2011-07-04T09:38:13.257" Score="1" ViewCount="" Body="&lt;p&gt;leftaroundabout gave an excellent explanation for the thermal conduction&#xA;of insulators. However, in the case of metals, a significant amount of&#xA;energy is carried by the excitations of electrons (the width of their&#xA;Fermi-Dirac distribution). The thermal conductivity is then related to&#xA;how far an excited electron can travel before being scattered, and is&#xA;therefore related to the electrical conductivity. In most metals, the&#xA;electrons will have a greater contribution to the thermal conductivity&#xA;than the phonons.&lt;/p&gt;&#xA;" OwnerUserId="4202" LastActivityDate="2011-07-04T09:38:13.257" CommentCount="1" />
  <row Id="11846" PostTypeId="1" AcceptedAnswerId="11848" CreationDate="2011-07-04T09:45:09.730" Score="2" ViewCount="147" Body="&lt;p&gt;Is the assumption of space time to be a continuum is just a matter of mathematical taste ? Isn't there any physical significance associated with it.&lt;/p&gt;&#xA;" OwnerUserId="540" LastActivityDate="2011-07-04T20:18:25.657" Title="Is the assumption of space time to be a continuum is just a matter of mathematical taste?" Tags="&lt;mathematical-physics&gt;&lt;spacetime&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11847" PostTypeId="2" ParentId="11842" CreationDate="2011-07-04T09:59:30.850" Score="2" ViewCount="" Body="&lt;p&gt;There is a basic misunderstanding here: &#xA;The rifling does not change anything for a spherical bullet. &#xA;So, in times before rifling bullets were spherical (&quot;rounds&quot;) &#xA;always. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a rifled gun, the bullet can be made elongated, flying like a &#xA;top, the bullet will keep its orientation with a small precesession. &#xA;(Which is unavoidable, but not sought for with rasant weapons, &#xA;contrary to artillery grenades) &#xA;Now this prolate form gives  the bullet more mass per cross sectional &#xA;area, this in turn causes the longer travel/stronger impact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sectional_density&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sectional_density&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-07-04T16:22:29.690" LastActivityDate="2011-07-04T16:22:29.690" />
  <row Id="11848" PostTypeId="2" ParentId="11846" CreationDate="2011-07-04T12:00:49.090" Score="9" ViewCount="" Body="&lt;p&gt;Underlying this question is an erroneous assumption about what physics is and what it isn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that physics is a &lt;em&gt;description&lt;/em&gt; of reality. However, it does not say anything about whether the description &lt;em&gt;equals&lt;/em&gt; reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, nobody knows whether space-time &lt;em&gt;truly is&lt;/em&gt; continuous or not. But continuity sure &lt;em&gt;describes&lt;/em&gt; it extremely well. Also note that this description might change in the future when we can do even more precise measurements.&lt;/p&gt;&#xA;" OwnerUserId="188" LastActivityDate="2011-07-04T12:00:49.090" CommentCount="2" />
  <row Id="11849" PostTypeId="1" CreationDate="2011-07-04T12:24:52.183" Score="1" ViewCount="68" Body="&lt;p&gt;We know that CMB is isotropic when viewed outside of the spinning and revolving earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it homogeneous?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we relate CMB to an inertial frame in the Newtonian sense (in which space and time are homogeneous and isotropic)? Or can it just provide an idea to build upon a new theory in which global inertial frames exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please feel free to see beyond Einstein's General Theory of Relativity and to question it.&lt;/p&gt;&#xA;" OwnerUserId="4235" LastActivityDate="2011-07-04T12:24:52.183" Title="Cosmic Microwave Background (CMB) and its relation to Inertial Frames" Tags="&lt;general-relativity&gt;&lt;inertial-frames&gt;" CommentCount="3" />
  <row Id="11851" PostTypeId="2" ParentId="11839" CreationDate="2011-07-04T12:45:15.950" Score="3" ViewCount="" Body="&lt;p&gt;The boosts and translations do not commute in neither relativistic or nor relativistic systems, please see for example the case of the &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Poincar%C3%A9_group&quot; rel=&quot;nofollow&quot;&gt;Poincare group&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $K_i = M_{0i}$ , $(i=1,2,3)$, we get for the Poincare group:&#xA;$[K_i, P_j] =i (\eta_{0j}P_i-\eta_{ij}P_0) = -i \delta_{ij}P_0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, The Galilean group can be obtained from the Poincare group by means of Wigner-Inonu contraction, see for example &lt;a href=&quot;http://lecospa.ntu.edu.tw/upload/sgtalk/20100209000459_Wigner-Inonu%20Contraction.pdf&quot; rel=&quot;nofollow&quot;&gt;this link.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the non relativistic limit the rest momentum is dominated by the mass term thus under the contraction the above relation becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[K_i, P_j] =-i \delta_{ij}M$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above reasoning shows that the parameter $M$ is the mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considered as an Lie algebra element M commutes with all other generators and cannot be removed by a smooth redefinition of the generators,&#xA;this is the reason that it is called a central extension, see for example chapter 3 of &#xA;&lt;a href=&quot;http://books.google.com/books?id=sHJRFHz1rYsC&amp;amp;printsec=frontcover&amp;amp;dq=Ballentine&amp;amp;hl=en&amp;amp;ei=DHURTvSsOYiXhQfoptj-DA&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=book-thumbnail&amp;amp;resnum=1&amp;amp;ved=0CCoQ6wEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Ballentine's book&lt;/a&gt;, where several representations of the Galilean group are constructed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The consequence is that in each irreducible representation of The Galilean group $M$ must be represented by a scalar (the particle's mass), and two representations with different masses are&#xA;unitarily inequivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula of the product of a translation and a boost can be directly obtained the cation of the Baker–Campbell–Hausdorff formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula means that the wavefunctions acquires a phase factor in a transformed inertial frame (in other words, the wavefunction representation is projective). In quantum mechanics a global phase is not measurable, thus no physical consequence is caused&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, Representations corresponding to different masses fall into different superselection sectors, this means that linear superpositions&#xA;of wavefunctions of particles with different masses are unphysical, because in this case each component would acquire a different phase which contradicts experiment because differential phases are observable in quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worthwhile to mention that this is exactly the example which was originally analyzed by V. Bargmann in his paper &quot;On unitary ray representations of continuous groups&quot;, Ann. of Math.,59 (1954), 1–46.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This update contains the answers to the questions appearing in Aruns comments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The noncommutativity of the momentum and boost operators is quantum mechanical. The commutators are written in $\hbar= 1$ units, working in general units, the boost-momentum commutator has the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[K_i, P_j] =i \hbar \delta_{ij}M$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The noncommutativity is observable only in the action of the operators on the wave functions. Let me elaborate this point for the case a free particle considered in Ballentine's book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the phase space (i.e., the manifold of initial data) is $\mathbb{R}^6$ covered with the coordinates $\{ \mathbf{q}, \mathbf{v} \}$, which upon quantization satisfy the commutation relations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[ q_i, v_j] = \frac{i}{M} \delta_{ij}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the operators:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The finite translation operator $\mathcal{D}_{q_0} =\exp(i \frac{ M\mathbf{v}. \mathbf{q_0}}{\hbar})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The finite boost operator  $\mathcal{B}_{v_0} = \exp(i \frac{\mathbf{K}. \mathbf{v_0}}{\hbar}) = \exp(i \frac{M \mathbf{q}. \mathbf{v_0}}{\hbar})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a wave function on the phase space $\psi(\mathbf{q})$. It is not difficult to see that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathcal{B}_{v_0} \mathcal{D}_{q_0} \psi(\mathbf{q})=  \exp(\frac{i M \mathbf{v_0}.\mathbf{q_0}}{\hbar}) \mathcal{D}_{q_0} \mathcal{B}_{v_0} \psi(\mathbf{q})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fourth relativistic momentum components is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ P_0 = \sqrt{M^2 c^4 + p^2 c^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the non-relativistic limit $c \rightarrow \infty$, the first term dominates and we get $ P_0 \approx M c^2$. Now, you absorb the factor $c^2$ into the definition of the generators, (or work in units c=1). This is essentially&#xA;the Wigner-Inonu contraction of the Poincare group to the Galilean group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll give you here the standard argument of the uniatry inequivallence between the quantization of two free particles with different masses $ M_1 \ne M_2$.&#xA;Let us use the subscript 1 and 2 for the individual particle operators (Each acting on a different Hilbert space ). If the two representations are unitarily equivalent means that there is an isometry $U$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{K_2} =  U \mathbf{K_1}U^{*} $  and $\mathbf{P_2} =  U \mathbf{P_1}U^{*} $, thus we obtain $[\mathbf{K_2}, \mathbf{P_2}] = U [\mathbf{K_1}, \mathbf{P_1}]U^{*}$ , which means $M_1 = M_2$, a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unitary inequivalence is similar to representations of $SU(2)$ having different $J$, consquently having different dimensions. In the finite dimensional case, it is obvious that we cannot find an isometry between different dimensional Hilbert spaces. &#xA;One might think that all infinite dimensional representations of the Galilean are unitarily equivalent which is not the case.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorUserId="2190" LastEditorDisplayName="" LastEditDate="2011-07-07T13:53:51.653" LastActivityDate="2011-07-07T13:53:51.653" CommentCount="7" />
  <row Id="11852" PostTypeId="1" AcceptedAnswerId="11858" CreationDate="2011-07-04T15:16:47.450" Score="0" ViewCount="70" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Automatic_quartz&amp;amp;oldid=432281459&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article on automatic quartz watch&lt;/a&gt; describes the watch mechanism as follows: a rotating pendulum is attached to a pinion and when the wearer moves his hand the pinion is rotated at &lt;strong&gt;&lt;em&gt;up to 100 thousand revolutions per minute&lt;/em&gt;&lt;/strong&gt; and that pinion rotates an electric generator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AFAIK the fastest hard disk drives currently reach only 15K RPM and they have serious problems with heat dissipation - a fan is required for cooling the drive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can a miniature pinion inside a watch be rotated at such speed without the whole assembly being worn out by friction in no time?&lt;/p&gt;&#xA;" OwnerUserId="3816" LastActivityDate="2011-07-04T19:30:32.630" Title="How could pinion in automatic quartz watch be rotated at 100K RPM?" Tags="&lt;mechanics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="11853" PostTypeId="1" CreationDate="2011-07-04T15:31:14.547" Score="1" ViewCount="115" Body="&lt;p&gt;If we have an &lt;strong&gt;isolated&lt;/strong&gt; systems $Sb$ with thermodynamic entropy $Eb=X$ &#xA;(and growing by 2nd law of thermodynamic)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could define an abstract system $Sa$ (cointaining the $Sb$ system) but defining the posibles microstates as a single bit, being $1(one)$ when $Eb&amp;gt;=X$, and $0(zero)$ when $Eb &amp;lt; X$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The microstate will tend to be $1$, so Entropy of the $Sa$ system tends to zero.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;po--&gt;0 &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;p1--&gt;1&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Eb = - k [p0 * log (p0) + p1 * log (p1)] --&gt; 0 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How could it be possible that an isolated system have a decrescent entropy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the growing (or not growing) of the entropy depends of the microstates choosen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if there is a restriction on how to choose the microstates definition, then what is that restriction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1916" LastEditorUserId="1916" LastEditorDisplayName="" LastEditDate="2011-07-26T13:04:46.150" LastActivityDate="2011-07-26T17:19:04.630" Title="Restrictions on defining microstates (Entropy)" Tags="&lt;thermodynamics&gt;&lt;entropy&gt;&lt;information&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="11855" PostTypeId="2" ParentId="11853" CreationDate="2011-07-04T17:34:16.387" Score="2" ViewCount="" Body="&lt;p&gt;The entropy $E_b$ is a &lt;em&gt;macroscopic&lt;/em&gt; thermodynamic quantity. Therefore, any quantity derived from it will also be a macroscopic quantity. Therefore, your states are not microstates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general restriction for a microstates is that once someone tells you what microstate a system is in, you know &quot;everything&quot; there is to know about the system. In classical mechanics, this means that a microstate specifies all positions and momenta of all particles in the system. In quantum mechanics, it means that you specify the complete many-body wavefunction or many-body state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your curious state does &lt;em&gt;not&lt;/em&gt; tell you what state your system is in: There are many ways the system can be  in state $1$, and many ways it can be in state $0$. Because of that, it is not a microstate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the general restriction is that a microstate has a unique realization in the system.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-07-04T17:34:16.387" CommentCount="8" />
  <row Id="11856" PostTypeId="2" ParentId="9570" CreationDate="2011-07-04T17:45:25.377" Score="2" ViewCount="" Body="&lt;p&gt;Landau's Vol 3 on non-relativistic QM (first edition -- at least I think that's what I got here in front of me..) has it in Ch. XVI (Motion in a magnetic field), sec. 128 (the current density in a magnetic field), pg. 486 (the last problem in the chapter).  As usual the L&amp;amp;L stuff isn't super explicit but it's there.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look in 'Introduction to Elementary Particles' by Griffiths in sec 5.5 on Hyperfine structure you'll see the hyperfine splitting carried out in excruciating detail for hydrogen and positronium.  That has the usage of F = I +/- 1/2 in there explicitly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the same argument carries through for all the alkalis.  It has been a while tho so don't bet the farm on it just yet but take a look.  &lt;/p&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2011-07-04T17:45:25.377" />
  <row Id="11858" PostTypeId="2" ParentId="11852" CreationDate="2011-07-04T18:17:25.250" Score="4" ViewCount="" Body="&lt;p&gt;Why is 100K RPM an issue? Dental drills works at around 10 kHz, with is about 600K RPM. &lt;code&gt;:-)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the &lt;a href=&quot;http://www.google.com/patents?id=F5EoAAAAEBAJ&amp;amp;zoom=4&amp;amp;pg=PA26#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Seiko Patent&lt;/a&gt; for the design of the Kinetic, the drive-train steps up the rotation of the spinning weight by 100 fold for the generator. When I charge my watch I shake it at around 5 - 10 Hz, which means 0.5-1 kHz for the small pinion, which gives 30 - 60k RPM indeed. So I don't think the Wikipedia article is that far off the mark. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When things are really small and light and precision made, I think you don't have to worry as much about heat dissipation and wear. (A typical hard-drive disk is about a couple inches in diameter. The pinion can't be more than a 10th of an inch, if you allow two steps to up the rotation speed.) &lt;/p&gt;&#xA;" OwnerUserId="829" LastEditorUserId="829" LastEditorDisplayName="" LastEditDate="2011-07-04T18:22:39.013" LastActivityDate="2011-07-04T18:22:39.013" CommentCount="2" />
  <row Id="11859" PostTypeId="2" ParentId="11852" CreationDate="2011-07-04T19:30:32.630" Score="1" ViewCount="" Body="&lt;p&gt;The rotation of the counterweight in the watch moves at that speed for a fraction of a second. Say if you hit a tennis ball the wait will only go 2 or tree turns and the initial speed can very well be at that speed. When your arm changes direction the weight will most likely stop. Now I don't know if calling the counterweight a &quot;pendulum&quot; is correct because all the math on a pendulum are based on gravity and the counterweight depends on changes of direction and momentum. &lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-07-04T19:30:32.630" />
  <row Id="11860" PostTypeId="1" CreationDate="2011-07-04T19:42:52.847" Score="4" ViewCount="152" Body="&lt;p&gt;Newton spent much of his life in the fruitless pursuit of alchemy, but along the way discovered and worked-out much else that was real, practical science. And I was thinking about why it is that we can safely dismiss anyone who claims to have invented a perpetual motion machine, but then I thought: even when we fail to achieve a goal, sometimes we succeed in something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any record of new science or engineering being discovered by someone who was otherwise trying to build the impossible, even if they failed at said goal? Maybe they invented a new low-friction ball bearing, or discovered a new phenomenon with energy transfer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely out of all the cranks and all of the gadgets they built, one of them must have had something good that was salvagable. I'm interested in this for historical and rhetorical purposes; I'd love to have a good example of how a crackpot can be redeemed, even if by accident.&lt;/p&gt;&#xA;" OwnerUserId="961" LastActivityDate="2011-09-03T20:50:46.543" CommunityOwnedDate="2011-07-04T20:36:16.600" Title="Accidental benefits of seeking perpetual motion? (Science history)" Tags="&lt;history&gt;&lt;perpetual-motion&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="11861" PostTypeId="2" ParentId="11846" CreationDate="2011-07-04T20:18:25.657" Score="-1" ViewCount="" Body="&lt;p&gt;Space and time are inclusive notions. Many do not know it but it is so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you tend the number of information bits (measurement points, pixels, if you like) to infinity, then certainty emerges. Space and time are those certainties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As inclusive entities, they are not fundamental, primary but secondary. In other words, they are illusions. Elementary, fundamental are those bits of information which constitute the whole picture.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-07-04T20:18:25.657" />
  <row Id="11862" PostTypeId="2" ParentId="6140" CreationDate="2011-07-04T20:31:59.663" Score="1" ViewCount="" Body="&lt;p&gt;The clock speed is reduced by the amount of the Lorentz factor.&#xA;The waste heat is more energetic by the amount of the Lorentz factor.&#xA;So the power that the computer draws is unchanged. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When speed is 0.8 c, waste heat photons that have two times the normal energy are produced at rate that is half of the normal rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an important point missing in this answer:&#xA;Let us consider a laptop. When we accelerate the laptop to relativistic speed, it takes some energy to accelerate the energy in the laptop battery. When this relativistic laptop computes, the kinetic energy of the chemical energy in the battery turns into kinetic energy of waste heat. &lt;/p&gt;&#xA;" OwnerUserId="4017" LastEditorUserId="4017" LastEditorDisplayName="" LastEditDate="2011-07-05T12:03:30.723" LastActivityDate="2011-07-05T12:03:30.723" />
  <row Id="11863" PostTypeId="2" ParentId="11680" CreationDate="2011-07-04T21:26:33.070" Score="5" ViewCount="" Body="&lt;p&gt;As an ex-physicist who now works as a quant in power markets I think it's safe to say the physics of the matter will be swamped by the economics in commodities and how power markets work.  Two things to note:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;power prices are set by markets and not by the viability of the technology (prime mover)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;solar is hard to make money with w/o a long term Power Purchase Agreement (PPA) and those are SUPER hard to come by.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are many viable technologies available right now that from a physics perspective look very promising but from a market perspective are dead in the water.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider flywheels from Beacon power:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.beaconpower.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.beaconpower.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They approached my company a year or so ago looking to put together a deal with our origination group.  The technology is sound, proven, and is clearly efficient.  It's a really sweet idea that has tons of promise.  Did we do the deal?  No.  Unfotunately pricing out regulation is hard and the rules that are followed by ISOs for payouts on regulation services are unclear.  The ISOs are interested in the technology but they are making up the rules as they go and companies that try to make money off the flywheels are currently going back and forth with them to create a market structure that can allow someone who owns this technology to make a reasonable return.  By reasonable I mean one that beats out gas, coal, oil, etc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My point is that the market structure (or lack of one) is killing the deals for new tech like this.. Not the efficiency. Same goes for the solar (and the wind). The prime mover that will succeed is the one that's pushed by the companies that understand how best to work the power markets. Siting and development (meaning leaning how to best construct the installation) are huge factors as well. It's a bad scene when your dev guys don't know how to properly handle the solar modules or manage a solar installation project.  It is indeed true that the physics of the technology must create a low-cost solution but the final word on success for a particular tech will be more based on the market acumen of the policy groups involved and the marketing/origination function than anything else. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Storage tech like flywheels from Beacon are a huge part of this puzzle too 'cause the sun never seems to be out exactly when you need it and the wind blows best during off peak hours. :)  You need a good way to store the power so you can release it during favorable times of the day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all, figuring out what tech is 'most promising' is more than just making a choice based on physics.  It's a good question tho.  Ur best bet is to talk to someone in power-markets from a big company like Constellation, PG&amp;amp;E, or perhaps TVA and ask them about the hurdles those technologies face.  Then marry that up with a physics assessment on efficiency and you'll get a good hint on which tech to help push forward with your studies.&lt;/p&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2011-07-04T21:26:33.070" CommentCount="3" />
  <row Id="11864" PostTypeId="2" ParentId="11694" CreationDate="2011-07-04T21:44:50.383" Score="3" ViewCount="" Body="&lt;p&gt;From a stat mech point of view one can view a star's goal as throwing off entropy over it's life span.  Iron is as stable as things get from an entropy perspective.  This results in the fusion limit.  The entropy that comes along with enough temp and pressure to continue this process is not favorable for the continuation of fusion beyond iron.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this is true for all stars until things get to the point gravitationally where Fermi pressure is important for balancing.&lt;/p&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2011-07-04T21:44:50.383" />
  <row Id="11865" PostTypeId="2" ParentId="11826" CreationDate="2011-07-04T22:14:34.450" Score="17" ViewCount="" Body="&lt;p&gt;One good piece of evidence that all particles of a given type are identical is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Exchange_interaction&quot; rel=&quot;nofollow&quot;&gt;exchange interaction&lt;/a&gt;. The exchange symmetry (that one can exchange any two electrons and leave the Hamiltonian unchanged) results in the Pauli exclusion principle for fermions. It also is responsible for all sorts of particle statistics effects (particles following the Fermi-Dirac or Bose-Einstein distributions) depending on whether the particles are fermions or bosons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the particles were even slightly non-identical, it would have large, observable effects on things like the allowed energies of the Helium atom.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-06T19:21:32.387" LastActivityDate="2011-07-06T19:21:32.387" CommentCount="5" />
  <row Id="11868" PostTypeId="2" ParentId="11136" CreationDate="2011-07-05T01:40:13.120" Score="-5" ViewCount="" Body="&lt;p&gt;BBT is not proved and it can not be proved, and I point in this answer to a new model that will knock it down.&lt;br&gt;&#xA;While a physical law was never broken, and is expected to be unbreakable; a theory come and go as time goes by.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the begining of BBT (Big Bang Theory) it was expected that gravity will, in due time, make the universe to collapse. A few years ago, against the expectations, it was measured an acceleration of the expansion.&lt;br&gt;&#xA;Because the matter content of Universe is not working to bring an end to the expansion we can deduce that Bariogenesis can not be an argument in favor of BBT (matter plays a minor role).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cosmological constant and the Dark Energy appeared magically to explain the feature, without explaining. Its only a parameter inserted in the equations, and proceed as usual.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain the observed, i.e. measured, expansion of the space we have to consider what stands behind 'making a measure', what referential can we chose, etc,... and one have to search for more than the naive explanation that you know of.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ΛCDM model (BBT) has 6 parameters: H0 , w, ΩΛ , ΩM , ΩR and ΩK   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://sites.google.com/site/scalinguniverse2011/self-similar-model&quot; rel=&quot;nofollow&quot;&gt;You can find here a killing theory with only one parameter (H0)&lt;/a&gt;&lt;/strong&gt;; this paper was dated July 1, 2011 and submitted to PRX; it is easy reading and any undergrad student can understand it. It is my hope that the teachers will follow. I'm very proud to see my name in this foundational paper, and I will stand up to argue in favor of it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://sites.google.com/site/scalinguniverse2011/self-similar-model&quot; rel=&quot;nofollow&quot;&gt;A self-similar model of the Universe unveils the nature of dark energy&lt;/a&gt;&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The abstract  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This work presents a critical yet&#xA;  previously unnoticed property of the&#xA;  units of some constants, able of&#xA;  supporting a new, self-similar, model&#xA;  of the universe. This model displays a&#xA;  variation of scale with invariance of&#xA;  dimensionless parameters, a&#xA;  characteristic of self-similar&#xA;  phenomena displayed by cosmic data.&#xA;  The model is deducted from two&#xA;  observational results (expansion of&#xA;  space and invariance of constants) and&#xA;  has just one parameter, the Hubble&#xA;  parameter. Somewhat surprisingly,&#xA;  classic physical laws hold both in&#xA;  standard and comoving units, except&#xA;  for a small new term in the angular&#xA;  momentum law that is beyond present&#xA;  possibilities of direct measurement.&#xA;  In spite of having just one parameter,&#xA;  the model is as successful as the ΛCDM&#xA;  model in the classic cosmic tests, and&#xA;  a value of H0 = 64 km s−1 Mpc−1 is&#xA;  obtained from the fitting with&#xA;  supernovae Ia data from Union&#xA;  compilation. It is shown that in&#xA;  standard units the model corresponds&#xA;  to Big Bang cosmologies, namely to the&#xA;  ΛCDM model, unveiling what dark energy&#xA;  stands for. This scaling (dilation)&#xA;  model is a one-parameter model that&#xA;  seems able of fitting cosmic data,&#xA;  that does not conflict with&#xA;  fundamental physical laws and that is&#xA;  not dependent on hypotheses, being&#xA;  straightforwardly deducted from the&#xA;  two observational results above&#xA;  mentioned.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will try to explain this novel theory in a few words:&lt;br&gt;&#xA;All system of units we use are 'atomic units' because are based in the properties of the atoms around us. And we only knew how to apply the laws of nature if we use those units.&lt;br&gt;&#xA;We are 'anthropocentric' and we can only imagine that the atom is absolute, invariant in time. Yes, we are using an 'absolute reference' and also denying that an 'absolute reference' can exist.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A measure is an adimensional quantity, a ratio of two quantities: e.g. if we are measuring a distance in the numerator is the amount of length and in the divisor is the amount corresponding to the unit of measure.&lt;br&gt;&#xA;As we are measuring an expansion we have to consider : the space expands, the unit of measure is decreasing (our atom) or a mix of both.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the paper it is shown that the laws of nature can support a scaling property of the universe, where the atom is no more invariant. It is shown that by choosing a  proper referential, a comoving one, the dark energy and the cosmological constant are artifacts of BBT. The BB is no more needed; the Inflation era is no more needed, etc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unexpected: (quote)   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Atomic measures are number counts&lt;/strong&gt;&lt;br&gt;&#xA;  We will now see that the measures of&#xA;  bodies’ properties using atomic units&#xA;  are independent of the base quantities&#xA;  and dependent on the number of&#xA;  particles or atoms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-07-05T01:40:13.120" CommentCount="8" />
  <row Id="11869" PostTypeId="2" ParentId="11817" CreationDate="2011-07-05T01:42:06.323" Score="0" ViewCount="" Body="&lt;p&gt;Here is a numerical simulation of heat transfer in a thread which is only one atom thick, each number is an atom, 8 is an atom with 8 units of kinetic energy, and so on: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First left side has more heat energy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8        0        0        0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After one collision between neighbor atoms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4        4        0        0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After one more collision between neighbor atoms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4        2        2        0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After one more collision between neighbor atoms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3        3        1        1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you see, there is some heat traveling at speed of sound. At the beginning of the simulation, when there was a heat difference of 8 between neighbor atoms, large fraction of heat energy traveled at speed of sound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Smell does not travel at speed of sound. If some smell was transferred when molecules touch, then smell would travel like heat energy)&lt;/p&gt;&#xA;" OwnerUserId="4017" LastActivityDate="2011-07-05T01:42:06.323" CommentCount="7" />
  <row Id="11870" PostTypeId="2" ParentId="10230" CreationDate="2011-07-05T01:47:59.370" Score="-4" ViewCount="" Body="&lt;p&gt;Whenever something goes tricky, do not always rely on &quot;infinite, undefined, etc&quot; to get over contradictions easily. Whatever you have been told may not be correct forever. So go brave, you may discover something new or find a better explanation. I don't have a good answer here. But what I think is this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the $\infty_1-\infty_2$ is giving you an $i\hbar$, then the Tr{$\infty_1$}-Tr{$\infty_2$} should also give you something consistent. But unfortunately it doesn't in this case. Then I start to wonder whether all linear operators have well-defined matrix form. Which may not be neccesarily true. Suppose they have at least Hilbert space matrix form, infinite dimensional matrices may have some differences in algebraic operations from finite dimensional matrices. But the trace operator is only well-defined on finite dimensional matrices. Does it or should it behave the same way over infinite dimensional matrices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the original definition of some concept is missing in the bigger space, you may consider extend its definition to try to fit everything nicely!-:)&lt;/p&gt;&#xA;" OwnerUserId="4275" LastActivityDate="2011-07-05T01:47:59.370" CommentCount="1" />
  <row Id="11871" PostTypeId="1" AcceptedAnswerId="11874" CreationDate="2011-07-05T03:29:26.533" Score="8" ViewCount="382" Body="&lt;p&gt;I have noticed that obstructing a spotlight typically results in a blurred shadow unlike the crisp batman symbol in the comics of batman. Is there a way to create a spotlight with a crisp batman symbol?&lt;/p&gt;&#xA;" OwnerUserId="2843" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-07-05T05:54:16.240" LastActivityDate="2011-07-05T09:56:15.880" Title="Batman spotlight in the sky" Tags="&lt;optics&gt;&lt;light&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" />
  <row Id="11872" PostTypeId="2" ParentId="11835" CreationDate="2011-07-05T03:35:25.417" Score="0" ViewCount="" Body="&lt;p&gt;(2) This is essentially the assumption that the laws of universe don't change with time. It is still the assumption in all of the theories till present (correct me if I am wrong) including String Theory.&lt;/p&gt;&#xA;" OwnerUserId="4235" LastActivityDate="2011-07-05T03:35:25.417" />
  <row Id="11873" PostTypeId="2" ParentId="11871" CreationDate="2011-07-05T03:36:22.813" Score="0" ViewCount="" Body="&lt;p&gt;If you use a huge laser (or maybe several smaller ones) you would keep the light from spreading too much. Also, it only work with clouds to reflect it! and the smoother the better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if there is another solution.&lt;/p&gt;&#xA;" OwnerUserId="301" LastActivityDate="2011-07-05T03:36:22.813" CommentCount="2" />
  <row Id="11874" PostTypeId="2" ParentId="11871" CreationDate="2011-07-05T04:11:54.353" Score="6" ViewCount="" Body="&lt;p&gt;You need to have a focusing lens built into your projector. After that, it's just like any other movie projector sort of thing, you can focus it provided you have an object to project it onto (like a flat cloud). Physically, the clarity of the image is limited by two or three things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) The size of your optics (the large diameter lens) gives a limitation in that it's impossible for a lens of, say, 2 meters, to focus light in a direction more accurately than the wavelength of light divided by that diameter. For the 550 nm that the human eye adapts to at night, this is about 550nm/2m = 0.000000275 radians = 0.00002 degrees. This is probably not going to be a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) There's also a limit based on the size of your light source (the bright thing from which the light emits, for example, the filament of a clear glass light bulb). This is a thermodynamic limit; you can't focus your light source so as to produce a temperature hotter than the light source. In practice, this means that if your light source has a size of 1cm, then you can't focus the beam from it to a size smaller than 1cm. By the way, this restriction is the limiting presence in the number of pixels displayed by a projection system. High intensity (small size high light output) sources are expensive and so projector bulbs are sometimes &lt;a href=&quot;http://www.projectorcentral.com/replacement_lamps.htm&quot; rel=&quot;nofollow&quot;&gt;almost as expensive as the rest of the projector&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) If the image is viewed off-axis, a non-flat projection screen (or a translucent screen such as a cloud) will make the image blurred. But this won't be such a big deal from near where the image is projected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither of these restrictions will significantly degrade your image. Good hunting!&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-07-05T04:11:54.353" CommentCount="6" />
  <row Id="11875" PostTypeId="1" CreationDate="2011-07-05T07:58:23.500" Score="0" ViewCount="47" Body="&lt;p&gt;I'm working with a megapixel camera and lens that needs to be focused for an OCR application. In order to measure the focus quality during the set-up, I've built a tool that gives the contrast value between two pixels. In very simply words, more the contrast is high more the focus is good...&#xA;Due to optical distortions, the focus value in the sides of my field of view, is not the same as in the center.&#xA;My question is how could I calculate the distortion between center and sides, in %.&#xA;For examples fcenter = 62 ; fleftside = 42.&#xA;             Is that correct to say Leftdistortion = 1-(42/62)*100 ??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Raphael&lt;/p&gt;&#xA;" OwnerUserId="2861" LastActivityDate="2011-07-07T09:09:51.770" Title="Optical distortions and focus losses calculation" Tags="&lt;optics&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="11876" PostTypeId="2" ParentId="11875" CreationDate="2011-07-05T08:04:20.060" Score="1" ViewCount="" Body="&lt;p&gt;You cannot figure that via pure math. It 100% depend of specific lens design you have - you could have 0 difference, you could have sphere-like focal plane, you could have anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your best bet is to measure focal plane differences across the frame, and interpolate.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-07-05T08:04:20.060" CommentCount="2" />
  <row Id="11877" PostTypeId="2" ParentId="11871" CreationDate="2011-07-05T08:05:56.313" Score="2" ViewCount="" Body="&lt;p&gt;You can use a Laser. Just track it whit moving mirrors like the beam of a TV tube. There are computers that do an outline but its totally possible to track horizontal lines that appear and disrepair at the proper moments and the smoke from fire woks can make a perfect screen. O ya and a BIG laser.  &lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-07-05T08:05:56.313" CommentCount="1" />
  <row Id="11878" PostTypeId="1" CreationDate="2011-07-05T09:31:29.567" Score="16" ViewCount="417" Body="&lt;p&gt;I found Sean Carroll's &quot;A No Nonsense Introduction to General Relativity&quot; (about page &lt;a href=&quot;http://preposterousuniverse.com/grnotes/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  pdf &lt;a href=&quot;http://preposterousuniverse.com/grnotes/grtinypdf.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;), a 24-page overview of the topic, very helpful for beginning study.  It all got me over the hump of learning the meaning of various terms associated with GR, most of which I had heard before without understanding.  It also outlined the most important examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a similar document for quantum field theory, which presents the main equations, briefly describes the main ideas, and summarizes the most important applications and results so that the reader can feel the lay of the land before studying in depth?&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-07-05T09:36:50.000" LastActivityDate="2011-08-09T01:35:30.197" CommunityOwnedDate="2011-07-05T11:57:09.840" Title="A No-Nonsense Introduction to Quantum Field Theory" Tags="&lt;quantum-field-theory&gt;&lt;reference-request&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="8" />
  <row Id="11879" PostTypeId="2" ParentId="11871" CreationDate="2011-07-05T09:56:15.880" Score="1" ViewCount="" Body="&lt;p&gt;Some 10 or 20 years ago many &quot;discos&quot; in Germany used &#xA;to project a strong laser beam to the sky, (As a kind of ad) &#xA;Afaik they used some holographic projection. &#xA;(Similar to the &quot;picture lenses&quot; You get along with &#xA;laser pointers, but all much stronger. &#xA;I have seen that from far (some 15 km), so I cant say &#xA;about the quality of that pictures. &#xA;Today this lasers no longer exist, maybe this is  forbidden meanwhile. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-07-05T09:56:15.880" CommentCount="2" />
  <row Id="11880" PostTypeId="2" ParentId="11136" CreationDate="2011-07-05T09:58:29.980" Score="0" ViewCount="" Body="&lt;p&gt;For an alternate answer, consider the inflationary universe. (Look up Inflation (Cosmolgy) in Wikipedia.)&lt;br&gt;&#xA;Cosmologists say we have substantial evidence for inflation.  The time of inflation is somewhat uncertain, but with high probability it is very early, if it occurred at all.  Wikipedia says 10^-36 to 10^-32 seconds. &lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-07-05T09:58:29.980" CommentCount="2" />
  <row Id="11881" PostTypeId="1" AcceptedAnswerId="11942" CreationDate="2011-07-05T12:00:52.193" Score="5" ViewCount="186" Body="&lt;p&gt;I'm wondering why the weak interaction only affects left-handed particles (and right-handed antiparticles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before someone says &quot;because thats just the way nature is&quot; :-), let me explain what I find needs an explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the limit of massless fermions, chirality (handedness) becomes helicity $(\vec S \cdot \hat p)$. Now, helicity is a property of the state of motion of an object in space. It is pretty unobvious to me how the internal symmetry SU(2) × U(1) would &quot;know&quot; about it, and be able to distinguish the two different helicity states of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a more technical level, IIRC, left and right handed spinors are distinguished by their transformation properties under certain space-time transformations, and are defined independent of any internal symmetry. If we want to get the observed V-A / parity violating behavior, we have to plug in a factor of $(1 - \gamma^5)$ explicitly into the Lagrangian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any reason this has to be like this? Why is there no force coupling only to right handed particles? Why is there no $(1 + \gamma^5)$ term? Maybe it exists at a more fundamental level, but this symmetry is broken?&lt;/p&gt;&#xA;" OwnerUserId="825" LastActivityDate="2011-07-07T11:03:07.937" Title="Why does the weak force distinguish left and right handedness?" Tags="&lt;quantum-field-theory&gt;&lt;particle-physics&gt;&lt;weak-force&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="11882" PostTypeId="1" AcceptedAnswerId="11886" CreationDate="2011-07-05T13:23:34.447" Score="7" ViewCount="550" Body="&lt;p&gt;So there appears to be quite a bit of misinformation on the web as to why people should stay in their cars during a thunderstorm. So I'd like to clear some things up. One such non-nonsensical answer is that cars have rubber tires which insulate you from the ground. I believe this contributes little to nothing to the actual reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;correct&quot; answer appears to be because the car acts like a Faraday cage. The metal in the car will shield you from any external electric fields and thus prevent the lightning from traveling within the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what happens if you have an imperfect Faraday cage around you? Say for example, you had a window open. I think the car would still protect you a) because it still acts as a Faraday cage, albeit a bit not perfect and b) because electrons will travel the path of least resistance which would be through the body of the car and not through you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now going with my b) reasoning, wouldn't you be just as safe standing next to a giant conductive pole (i.e. a lightning rod)? Wouldn't the lightning just go through the lightning rod and you'd be 100% safe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also a side question: lightning is essentially just an huge electric arc from the clouds to the ground, correct?&lt;/p&gt;&#xA;" OwnerUserId="4278" LastActivityDate="2011-07-08T19:00:14.093" Title="Why stay in the car during thunderstorms?" Tags="&lt;lightning&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="11883" PostTypeId="1" CreationDate="2011-07-05T13:57:45.027" Score="2" ViewCount="47" Body="&lt;p&gt;I'm reading a review about membranes properties and I have reach a section about fluid membranes. The section discuss the principal curvatures ($c_1, c_2$) and the spontaneous curvatures ($c_0$). After stating some properties of $c_0$, the following two sentences appear:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The membrane also is assumed to be incompressible. Hence, all the contributions to the surface tension vanish.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure about the meaning of these two sentences, I think it means that by bending the membrane I can not generate surface tension, is that correct?&lt;/p&gt;&#xA;" OwnerUserId="3677" LastActivityDate="2011-07-05T13:57:45.027" Title="A question about surface tension of membranes and their curvature" Tags="&lt;elasticity&gt;&lt;curvature&gt;" CommentCount="1" />
  <row Id="11884" PostTypeId="2" ParentId="11096" CreationDate="2011-07-05T14:22:04.247" Score="3" ViewCount="" Body="&lt;p&gt;The only way this would work is to have a set of equations describing gravitation that are structurally the same as Maxwell's equations.  Then you can have something similar to the concept of induction and induced fields.  Ted Bunn above has pointed out already that gravitomagnetism is the name for this setup.  Check out the wikipedia page on this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitomagnetism&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Gravitomagnetism&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has a nice summary of the GEM equations and puts them right up next to Maxwell's equations so you can see the equivalences. &lt;/p&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2011-07-05T14:22:04.247" />
  <row Id="11885" PostTypeId="1" AcceptedAnswerId="12172" CreationDate="2011-07-05T14:25:46.893" Score="4" ViewCount="98" Body="&lt;p&gt;My question is in reference to Landau's Vol. 1 Classical Mechanics. On Page 6, the starting paragraph of Article no. 4, these lines are given:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If an inertial frame $К$ is moving with an infinitesimal velocity $\mathbf\epsilon$ relative to another inertial frame $K&amp;#39;$, then $\mathbf v&amp;#39; = \mathbf v+\mathbf \epsilon$. Since the equations of motion must have the same form in every frame, the Lagrangian $L(v^2)$ must be converted by this transformation into a function $L&amp;#39;$ which differs from $L(v^2)$, if at all, only by the total time derivative of a function of co-ordinates and time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1) Doesn't this hold for same frame? Why is Landau changing the Lagrangian of frame $K$, $L$, to $L&amp;#39;$ with the change satisfying this condition? So, how can he assume that the action would be minimum for the same path in $K&amp;#39;$ as there was in $K$? In two frames the points $q_1$ and $q_2$ aren't same which are at $t_1$ and $t_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) How did he assume that this is the one and only way to change the Lagrangian without changing the path of least action? Can we prove this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to first question, I feel that there is something fundamentally amiss from my argument as the Lagrangian is dependent only on magnitude of velocity, so $q_1$ and $q_2$ won't matter. I have made an explanation myself that since the velocity is changed infinitesimally, it should essentially be the same path governed by the previous Lagrangian, the path it took with constant velocity $v$. But, still I am not convinced. The argument isn't concrete in my head. Please build upon this argument or please provide some alternative argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the question (1) and argument above are very poorly framed but I am reading Landau alone without any instructor and so have problems forming concrete ideas.&lt;/p&gt;&#xA;" OwnerUserId="4235" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-05T19:14:21.593" LastActivityDate="2011-07-12T02:20:37.260" Title="Invariance of Lagrange on addition of total time derivative of a function of coordiantes and time" Tags="&lt;classical-mechanics&gt;&lt;lagrangian-mechanics&gt;" AnswerCount="1" CommentCount="20" FavoriteCount="0" />
  <row Id="11886" PostTypeId="2" ParentId="11882" CreationDate="2011-07-05T14:53:24.007" Score="5" ViewCount="" Body="&lt;p&gt;Yes, if it is not a plastic covered car it is an effective Faraday cage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the tires are such that the car is insulated electrically, if it is hit it will take some time to discharge to the ground, but still the passengers would be safer than standing next to it outside. I have learned that modern tires are particularly constructed so that the static charge generated by the friction on the road is discharged so that would also help. ( in olden times they used to have chains trailing from the trucks in order to discharge the static. Recently I saw a car with a discharger too, trailing on the road!).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;lightning is essentially just an huge electric arc from the clouds to the ground, correct?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wrong, the current actually &lt;a href=&quot;http://en.wikipedia.org/wiki/Lightning&quot; rel=&quot;nofollow&quot;&gt;may start from the ground.&lt;/a&gt; That is the rational of the lightning rods, to  create a path for a current to be generated by the potential difference to the cloud and  to meet the current from the clouds in a prefered location instead of a random one. It is not wise to stand next to a rod, read in the link the amount of power  dissipated by a bolt.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The average peak power output of a single lightning stroke is about one trillion watts — one &quot;terawatt&quot; (1012 W), and the stroke lasts for about 30 millionths of a second — 30 &quot;microseconds&quot;.[18]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And it is not wise to stand, because you may also give rise to leaders that will meet the lightning path. If in the open it is best  you fall on the ground as much sheltered as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A colleague once was about 20 meters from a lightning bolt, and he was so shocked by the sound and fury, it took him a week to come down to normal. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-07-05T14:53:24.007" />
  <row Id="11887" PostTypeId="1" CreationDate="2011-07-05T15:15:07.560" Score="1" ViewCount="141" Body="&lt;p&gt;The canonical momentum is a fundamental conserved quantity from Noether's theorem for translational invariance of the Lagrangian. Yet I'm finding it very difficult to see its derivation, or even a statement of what it is for something as fundamental as a relativistically moving charge in a static Coulomb electric field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone state what it is, or even give a derivation if it's not too much trouble?&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="1345" LastEditorDisplayName="" LastEditDate="2011-07-05T15:24:11.467" LastActivityDate="2011-08-26T15:00:32.293" Title="What is the conserved canonical momentum for a relativistically moving charge in a static Coulomb electric field?" Tags="&lt;classical-electrodynamics&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="11888" PostTypeId="2" ParentId="11882" CreationDate="2011-07-05T15:39:17.097" Score="6" ViewCount="" Body="&lt;p&gt;In answer to your second question:&#xA;&quot;Now going with my b) reasoning, wouldn't you be just as safe standing next to a giant conductive pole (i.e. a lightning rod)? Wouldn't the lightning just go through the lightning rod and you'd be 100% safe?&quot;&#xA;   No. A lightning strike can carry a high current, up to 200 kiloamps has been recorded. Not only can the current be high, but the rate of change in this current is high as well. That means the induced magnetic field around the pole and the path of the lightning can be quite high. This can induce high voltages in the vicinity of the strike. In fact many people who are reported to have been &quot;hit&quot; by lightning, have not taken a direct hit, but suffer from induced currents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct about the widely misunderstood thing about rubber tires. A spark that has enough voltage to travel thousands of feet, will not be stopped by the few inch gap between the conductive part of the car and the ground.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-07-05T15:39:17.097" CommentCount="3" />
  <row Id="11889" PostTypeId="2" ParentId="11882" CreationDate="2011-07-05T16:23:58.373" Score="3" ViewCount="" Body="&lt;p&gt;Some remarks: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;open window&quot;:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;this is irrelevant!  Because the window is not metallic,&#xA;it is not a piece of the Faraday cage, irrespective whether open or not. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A spark that has enough voltage to&#xA;  travel thousands of feet, will not be&#xA;  stopped by the few inch gap between&#xA;  the conductive part of the car and the&#xA;  ground.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Right, and another fact: The spark will &quot;avoid&quot; to go via the car, because &#xA;that will produce two plasma/metal interfaces with a certain voltage drop each. &#xA;For that reason short pices of metal are avoided. This is strongest during &#xA;development of the strike, when currents are still small. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have learned that modern tires are&#xA;  particularly constructed so that the&#xA;  static charge generated by the&#xA;  friction on the road is discharged so&#xA;  that would also help.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is something new to me. Bluntly: I doubt it. &#xA;There is no necessity for that, and second making tire rubber conductive &#xA;will damage the rubber material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thunderstorms are associated with rain (most often) , in that case isolation by tyres is &#xA;no longer existent. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-07-06T10:23:41.003" LastActivityDate="2011-07-06T10:23:41.003" CommentCount="5" />
  <row Id="11890" PostTypeId="2" ParentId="10804" CreationDate="2011-07-05T16:36:31.203" Score="3" ViewCount="" Body="&lt;p&gt;OP is basically asking for an &lt;em&gt;intuitive&lt;/em&gt; understanding of operator ordering. Well, the quantum world is something that us Earthlings notoriously do not understand well. Often we start with a classical model with commuting quantities. When we next want to quantize the model, we at first do not know which way we should order the corresponding non-commuting quantum operators. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say for simplicity that the classical Hamiltonian $H=AB$ is a product of two classical quantities $A$ and $B$. And say that the corresponding two quantum operators $\hat{A}$ and $\hat{B}$ have a c-number commutator $[\hat{A},\hat{B}]=\hbar{\bf 1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are initially many ways to choose an operator ordering and to choose a representation (ket-space), which the quantum operators act on. Say that we have chosen a specific notion of ordering that we call normal ordering $:\hat{A}\hat{B}:$, and say that we have chosen a notion of Fock space vacuum. To parametrize our ignorance we now introduce a c-number parameter $c$, and define the quantum Hamiltonian as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{H}~=~:\hat{A}\hat{B}:~+~\hbar c{\bf 1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way, if we have made a wrong choice by normal ordering the operators, we can always absorb the error into the definition of the c-number parameter $c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can often limit the possible choices of $c$ further by demanding Hermiticity of $\hat{H}$ and imposing other physical requirements. For instance, in (Bosonic) string theory, a similar so-called intercept parameter $a$ is &lt;em&gt;completely&lt;/em&gt; fixed by consistency requirements (Lorentz symmetry in the light-cone formulation; nilpotency of the BRST charge in the covariant formulation), see chapter 2 and 3 in Green, Schwarz and Witten, &quot;Superstring theory&quot;, vol. 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar story holds for Fermionic operators. &lt;/p&gt;&#xA;" OwnerUserId="2451" LastActivityDate="2011-07-05T16:36:31.203" />
  <row Id="11891" PostTypeId="2" ParentId="11787" CreationDate="2011-07-05T17:22:17.620" Score="0" ViewCount="" Body="&lt;p&gt;So, it turns out that the answer is incredibly simple.  Which is what I suspected given the mechanism:  Question seems super easy, but can't seem to get answer of the top of my head.  So I go searching for an answer, but because the solution is so obvious, no book has it explicitly.  Ask people, again, no one seems to have it in mind, but can use complicated formulas based on it using already derived numbers.  Ladeeda.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the paper is looking for is the difference between actual vapor pressure at the drop as opposed to vapor pressure in the room.  So all one needs to do is do steps 1 and 3 above and skip 2.  Then multiply the result by (1-H) where H is the relative humidity.  Good Times! &lt;/p&gt;&#xA;" OwnerUserId="1037" LastActivityDate="2011-07-05T17:22:17.620" CommentCount="1" />
  <row Id="11892" PostTypeId="5" CreationDate="2011-07-05T18:01:22.457" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-07-05T18:01:22.457" LastActivityDate="2011-07-05T18:01:22.457" />
  <row Id="11893" PostTypeId="4" CreationDate="2011-07-05T18:01:22.457" Score="0" ViewCount="" Body="Light Amplification by Stimulated Emission of Radiation. A means of creating coherent light by either driving an atomic or molecular transition in an optical cavity or firing a beam of electrons through an undulator." OwnerUserId="-1" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-05T18:57:08.847" LastActivityDate="2011-07-05T18:57:08.847" />
  <row Id="11894" PostTypeId="2" ParentId="11878" CreationDate="2011-07-05T18:05:02.093" Score="3" ViewCount="" Body="&lt;p&gt;Veltman's book Diagrammatica is awesome when it comes to the basics of field theory and what it all means:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521456924&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Diagrammatica-Feynman-Diagrams-Cambridge-Lecture/dp/0521456924/ref=ntt_at_ep_dpt_2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading through that one is in my opinion getting it right from the horse's mouth.  Not heavy on the calculations but more on how the QFT and it's tools all work together to form a coherent picture of the subject.&lt;/p&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2011-07-05T18:05:02.093" CommunityOwnedDate="2011-07-05T18:05:02.093" CommentCount="3" />
  <row Id="11895" PostTypeId="5" CreationDate="2011-07-05T18:07:25.633" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-07-05T18:07:25.633" LastActivityDate="2011-07-05T18:07:25.633" />
  <row Id="11896" PostTypeId="4" CreationDate="2011-07-05T18:07:25.633" Score="0" ViewCount="" Body="Electromagnetic waves, visible to the human eye if between about 400 and 700 nm in wavelength." OwnerUserId="-1" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-05T18:57:27.960" LastActivityDate="2011-07-05T18:57:27.960" />
  <row Id="11897" PostTypeId="1" CreationDate="2011-07-05T18:29:56.627" Score="6" ViewCount="214" Body="&lt;p&gt;Well-known physics establishes that, compared to some reference frame (I'll denote as &quot;stationary&quot; here), one can do several things to start at rest, end at rest, and experience &lt;em&gt;less&lt;/em&gt; time than objects in the stationary reference frame do, although one can not experience &lt;em&gt;more&lt;/em&gt; time given the start and end at rest condition.  The former can be accomplished by the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&lt;li&gt;Take a trip at relativistic speeds&#xA;&lt;li&gt;Approach to an object in a deep gravity well, approaching an event horizon potential&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Can either of these be done for an object with mass $m$ while expending a small amount of energy to do it, formally using an amount of energy $E\ll m c^2$?  To be sufficiently clear, the requirement for traveling to the &quot;distant future&quot; is satisfied if $\Delta t_m \ll \Delta t$, meaning that the time experienced for the object in question is much less than that experienced by the stationary reference frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conventional thinking is that #1 is simply not possible given my requirement.  Any physical ship that let you planet-hop while living a single human lifetime would require spaceships that probably expend much more energy in the propellent than what the rest mass of the spaceship is.  This is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Relativistic_rocket&quot; rel=&quot;nofollow&quot;&gt;relativistic rocket&lt;/a&gt; problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking about #2 is what caused me to ask this question.  Imagine a large black hole and a spaceship that starts off at a great distance from it.  It then uses a small amount of energy to enter into a very highly elliptical orbit that comes close to the event horizon.  I think (although I am not sure) it would seem to the stationary reference frame that the flyby takes a small amount of time.  The spaceship would be going very fast (large time dilatation), but only for a short time and thus could not travel to the distant future and make it back out without using huge amounts of energy.  You can introduce other possibilities, like a stationary planet-like surface where a traveler could sit to pass the time, but the escape potential renders this useless too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a fundamental reality that we can not use time dilation to go to the distance future without using extraordinary energy to do so, or do I simply lack the creativity to think of how?&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-07-11T14:00:57.307" Title="Is travel to the distant future possible without massive energy expenditure?" Tags="&lt;time-travel&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" />
  <row Id="11898" PostTypeId="2" ParentId="11887" CreationDate="2011-07-05T19:00:54.930" Score="2" ViewCount="" Body="&lt;p&gt;Put speed of light $c=1$ and use sign convention $(-,+,+,+)$. To have that the canonical momentum &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{p}~=~\frac{\partial L}{\partial d_t \vec{x}}~=~ \gamma m_{0}d_t \vec{x} +q\vec{A}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is conserved $d_t \vec{p}=0$, and a $\vec{x}$-translational symmetry of the Lagrangian &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L~=~T-U, \qquad T~=~-\frac{m_0}{\gamma},  \qquad U~=~ q (\phi -\vec{A}\cdot d_t \vec{x}) ,\qquad \gamma = \frac{1}{\sqrt{1- d_t \vec{x}\cdot d_t \vec{x}}},$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;we must have that the Lagrangian $L$ is independent of the position $\vec{x}$, and we must therefore have that the 4-gauge potential &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A^{\mu}=(\phi,\vec{A})$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is independent of position $\vec{x}$. Hence $A_{\mu}=A_{\mu}(t)$ is a function of time only. And hence &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{E}~=~ -\vec{\nabla} \phi -\partial_t \vec{A} ~=~-\partial_t \vec{A}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\vec{E}$ is assumed to be static, i.e., independent of $t$. Hence $\vec{A}$ is just an affine function of time, and $\vec{E}$ is independent of both space and time. It follows that the conservation law for the canonical momentum $d_t \vec{p}=0$ is just the relativistic Newton's 2nd law &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ d_t(\gamma m_{0}d_t \vec{x})~=~q \vec{E} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for a charged particle in a constant $\vec{E}$-field.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-05T19:22:32.137" LastActivityDate="2011-07-05T19:22:32.137" CommentCount="1" />
  <row Id="11900" PostTypeId="2" ParentId="11882" CreationDate="2011-07-05T19:59:47.703" Score="2" ViewCount="" Body="&lt;p&gt;Be careful about your definition of &lt;code&gt;giant conductive pole&lt;/code&gt; (ie be exact in you description).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people may consider trees as a conductive pole. The danger here is not only from electrical discharge but from exploding tree (due to sap boiling in an inclosed space).&lt;/p&gt;&#xA;" OwnerUserId="4281" LastActivityDate="2011-07-05T19:59:47.703" />
  <row Id="11901" PostTypeId="2" ParentId="11096" CreationDate="2011-07-05T20:31:11.217" Score="0" ViewCount="" Body="&lt;p&gt;Isn't frame dragging an example of this?&lt;/p&gt;&#xA;" OwnerUserId="4210" LastActivityDate="2011-07-05T20:31:11.217" />
  <row Id="11902" PostTypeId="1" AcceptedAnswerId="11906" CreationDate="2011-07-05T20:53:30.217" Score="1" ViewCount="79" Body="&lt;p&gt;I am trying to simulate a game in Box2D(Physics engine). The game that I am trying to simulate is very simple and can be found here: &lt;a href=&quot;http://www.makaimedia.com/#/speartoss&quot; rel=&quot;nofollow&quot;&gt;http://www.makaimedia.com/#/speartoss&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to know is that, is the spear in that game following real physics principles? In other words, if I went out and threw a spear, will it behave like the spear in this game? Specifically, I am taking about the rotation of the spear. In the game, the spearhead is point up first half of the trajectory and as soon as it hits the vertex, the spear head starts point down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am simulating a spear in Box2d physics engine. The density of the spear is 15x more than the density of the handle. I set an angle (between 0, 90 degrees) and give it a linear velocity like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    b2Vec2 force;&#xA;    force.Set(cos(angle) * 22.0f , sin(angle) * 22.0f);&#xA;    _spearBody-&amp;gt;SetLinearVelocity(force);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What happens is, the spearhead never points downwards. It stays at the exact same angle throughout the trajectory. Is this more realistic or the physics in speartoss game?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4284" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-05T21:01:48.443" LastActivityDate="2011-07-05T21:23:38.707" Title="Is this simulation following real physics?" Tags="&lt;newtonian-mechanics&gt;&lt;rotational-dynamics&gt;&lt;aerodynamics&gt;" AnswerCount="1" />
  <row Id="11903" PostTypeId="2" ParentId="8903" CreationDate="2011-07-05T20:55:01.150" Score="0" ViewCount="" Body="&lt;p&gt;main advantage of lagrangian and hamiltonian mechanics over Newtonian mechanics we can deal with scalar quantities , energy, whereas in the later we have to deal with vector quantities.&#xA;Besides this we can approach easily to any system ( e.g. mechanical, electrical, optical etc.) with the lagrangian and Hamiltonian mechanics. But this easy acess can not be achieved with  NEWTONIAN MECHANICS. &lt;/p&gt;&#xA;" OwnerUserId="4285" LastActivityDate="2011-07-05T20:55:01.150" />
  <row Id="11904" PostTypeId="1" AcceptedAnswerId="11909" CreationDate="2011-07-05T21:03:13.543" Score="5" ViewCount="155" Body="&lt;p&gt;I just started learning about optics, and in the book I'm reading they explain how the electrical field caused by a single charged particle could be described by a series of field lines, and compare them to ropes, to provide an intuition of the concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then they say that and that if we wiggle the particle up and down, that would produce transversal waves in the horizontal field lines, but no waves in the vertical lines. I know that the physical analogy is not to be taken literally, but I don't understand why wouldn't that cause compression waves in the vertical lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean, even though the direction of the field in the points directly above and below the particle doesn't change, the intensity does. And I assume it wouldn't instantly. So what am I missing?&lt;/p&gt;&#xA;" OwnerUserId="4283" LastActivityDate="2011-07-06T04:08:06.667" Title="Why aren't there compression waves in electromagnetic fields?" Tags="&lt;electromagnetism&gt;&lt;optics&gt;&lt;waves&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" />
  <row Id="11905" PostTypeId="1" CreationDate="2011-07-05T21:07:27.897" Score="3" ViewCount="199" Body="&lt;p&gt;I thought the Hamiltonian was always equal to the total energy of a system but have read that this isn't always true. Is there an example of this and does the Hamiltonian have a physical interpretation in such a case?&lt;/p&gt;&#xA;" OwnerUserId="4208" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-20T21:17:44.777" LastActivityDate="2011-07-20T21:17:44.777" Title="When is the Hamiltonian of a system not equal to its total energy?" Tags="&lt;classical-mechanics&gt;&lt;energy&gt;&lt;hamiltonian-mechanics&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="11906" PostTypeId="2" ParentId="11902" CreationDate="2011-07-05T21:10:26.533" Score="2" ViewCount="" Body="&lt;p&gt;It is following a very incidental &quot;special case&quot; of physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spear can be represented as a rigid body with a center of mass. When you impart purely linear velocity, the spear's center of mass will travel according to projectile motion. There is no reason for the spear to change it's orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the spear to change it's orientation, a rotational force, or torque, is required. For the simulation you linked to be accomplished in a natural manner, the thrower would have to impart a specific &quot;twist&quot; with his wrist at the time of release, just enough so that the rotational velocity corresponds exactly to the orientation of the spear's linear velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, the simulation probably just changes the orientation at every step to align with the velocity vector. This is quite common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should note that this &quot;special case&quot; is actually very natural during a real human throw. The rotational velocity at release comes from the way you rotate your forearm during the throw. Humans have a natural intuition about physics like that, same as playing sports.&lt;/p&gt;&#xA;" OwnerUserId="3390" LastEditorUserId="3390" LastEditorDisplayName="" LastEditDate="2011-07-05T21:23:38.707" LastActivityDate="2011-07-05T21:23:38.707" CommentCount="3" />
  <row Id="11907" PostTypeId="2" ParentId="8903" CreationDate="2011-07-05T21:57:11.613" Score="3" ViewCount="" Body="&lt;p&gt;To answer the second part of your question, I will give a classic example of harmonic motion. The potential energy of a spring is $U=\frac{1}{2}kx^2$, where $k$ is the spring constant and $x$ is the displacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Newtonian Mechanics:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F=m\frac{d^2x}{dt^2}=\frac{-dU}{dx}=-kx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $m\frac{d^2x}{dt^2}=-kx$, which is an easy differential equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lagrangian Mechanics:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we know the Euler-Lagrange Equations $\frac{\partial L}{\partial q} = \frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$, we identify coordinates $q=x$, and we define our Lagrangian $L=T-V$ ($T$ is kinetic energy and $V$ is potential energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T=\frac{1}{2}m\dot{x}^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V=\frac{1}{2}kx^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we plug this all into our little Euler Lagrange Equation, and, solving through, you get (drum roll), $m\frac{d^2x}{dt^2}=-kx$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So after all of this we get the same equation as with Newtonian mechanics and with a lot more work right? In this example probably, and most other simple systems. However, Lagrangian Mechanics has some very powerful applications. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following system: You have multiple pendula connected by springs, and each pendulum begins with some initial position and velocity. How do you go about solving this system? In Newtonian Mechanics it will get extremely complex to work out all the forces involved. However, taking a Lagrangian perspective, much of the hard work is taken care of as you can easily define $q_i=\theta_i$, $\theta$ being the angular displacement of each the pendulum. And instead of having to deal with the various forces you just deal with the potential and kinetic energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An even more, exponentially more, important application is in classical field theory (I know it has some important connections to with QFT, but I am not in any position to knowledgeably comment on that). Electromagnetism and General Relativity are two excellent examples. You can derive Maxwell's equations entirely from the electromagnetic Lagrangian ( $L=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+A_{\mu}J^\mu$) and you can derive extremely important results in general relativity from the Hilbert Action ($S_H=\int \sqrt{-g}Rd^nx$) and similar variational principles.&lt;/p&gt;&#xA;" OwnerUserId="3018" LastActivityDate="2011-07-05T21:57:11.613" />
  <row Id="11908" PostTypeId="2" ParentId="11096" CreationDate="2011-07-05T22:14:30.880" Score="0" ViewCount="" Body="&lt;p&gt;The answer is very much yes! In fact, Gravity Probe B detected frame dragging effects by comparing rotation speed of four extremely precise gyroscopes onboard. As one might think, the force is very small for an object like earth rotating so slowly. However for fast rotating massive bodies, is has noticeable effects. Supermassive black holes in active nuclei have enormous jets that are very closely aligned even at megaparsec (millions of light years) distances. As far as the nature of the field created, it is very much congruent to the relationship between the electric field and the magnetic field.&lt;/p&gt;&#xA;" OwnerUserId="3018" LastActivityDate="2011-07-05T22:14:30.880" />
  <row Id="11909" PostTypeId="2" ParentId="11904" CreationDate="2011-07-05T22:45:54.720" Score="3" ViewCount="" Body="&lt;p&gt;If you are going to pursue this physical analogy, which can be useful at times, then you must consider the electric field lines to be of constant tension.  That is, the tension of these lines is a constant no matter how much you stretch them.  This is different from ordinary ropes or strings or whatever, where the more you stretch them, the higher the tension.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More technically, if you examine the Maxwell stress tensor for a pure electric field, you will find a tension term along the direction of the field and a pressure term transverse to the field.  So you in the static case, you can think of the electric field lines as being in balance between tension along field lines and a pressure pushing different lines apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an ordinary stretched string or something like that, if you move the end of the string longitudinally then the stretching or compression changes the tension and the difference in tension will propagate along the string, producing a longitudinal wave.  In the case of electric field lines, there is no change in tension to propagate along the line, since the tension is a fixed constant.  I hope this helps with your intuition.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-07-05T22:45:54.720" CommentCount="1" />
  <row Id="11910" PostTypeId="2" ParentId="11905" CreationDate="2011-07-05T22:56:49.797" Score="4" ViewCount="" Body="&lt;p&gt;Goldstein's Classical Mechanics (2nd Ed.) pg. 349, section 8.2 on cyclic coordinates and conservation theorems' has a good discussion on this.  In his words:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The identification of H as a constant of the motion and as the total energy &#xA;are two separate matters.  The conditions sufficient for one are not &#xA;enough for the other.  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;He then goes on to provide an example of a 1-d system in which he chooses two different generalized coordinate systems.  For the first choice, H is the total energy while for the second choice H ends up being just a conserved quantity and NOT the total energy of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check it out.  It's a very nice example.&lt;/p&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2011-07-05T22:56:49.797" />
  <row Id="11911" PostTypeId="2" ParentId="11905" CreationDate="2011-07-05T23:01:03.990" Score="6" ViewCount="" Body="&lt;p&gt;The Hamiltonian is in general not equal to the energy when the coordinates explicitly depend on time.  For example, we can take the system of a bead of mass $m$ confined to a circular ring of radius $R$. If we define the $0$ for the angle $\theta$ to be the bottom of the ring, the Lagrangian $$L=\frac{mR^2\dot{\theta}^2}{2}-mgR(1-\cos{(\theta)}).$$ The conjugate momentum $$p_{\theta}=\frac{\partial L}{\partial \dot{q}}=mR^2\dot{\theta}.$$ And the Hamiltonian $$H=\frac{p_{\theta}}{2mR^2}+mgR(1-\cos{\theta}), $$ which is equal to the energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt;, if we define the $0$ for theta to be moving around the ring with an angular speed $\omega$, then the Lagrangian $$L=\frac{mR^2(\dot{\theta}-\omega)^2}{2}-mgR(1-\cos{(\theta-\omega t)}). $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conjugate momentum $$p_{\theta}=\frac{\partial L}{\partial \dot{q}}=mR^2\dot{\theta}-mR^2 \omega.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the Hamiltonian $$H=\frac{p_{\theta}}{2mR^2}+p_{\theta}\omega+mgR(1-\cos(\theta-\omega t)), $$ which is &lt;em&gt;not&lt;/em&gt; equal to the energy (in terms of $\dot{\theta}$ it has an explicit dependence on $\omega$).&lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2011-07-05T23:01:03.990" />
  <row Id="11912" PostTypeId="1" AcceptedAnswerId="11917" CreationDate="2011-07-05T23:06:23.520" Score="4" ViewCount="134" Body="&lt;p&gt;My question is very simple, and most likely a stupid one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One observer is at a point in space were the gravitational force form massive bodies (or a single massive body) cancel each-other out. The second observer is in another hypothetical situation where there exists no massive bodies and therefore no gravity. My question is this, is there a relativistic time difference between the two?&lt;/p&gt;&#xA;" OwnerUserId="4287" LastActivityDate="2011-07-06T19:29:01.350" Title="Would there be time dilation at the point where two gravitational fields cancel each other out?" Tags="&lt;general-relativity&gt;&lt;gravity&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" />
  <row Id="11913" PostTypeId="2" ParentId="11912" CreationDate="2011-07-05T23:16:32.823" Score="1" ViewCount="" Body="&lt;p&gt;Yes.  Good example:  First observer right in the middle between two identical massive bodies.  Second observer far away.  In the weak field limit, the redshift is twice as much as if there were only one massive body and otherwise the same set up. &lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-07-05T23:16:32.823" CommentCount="2" />
  <row Id="11914" PostTypeId="5" CreationDate="2011-07-06T00:02:09.663" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-07-06T00:02:09.663" LastActivityDate="2011-07-06T00:02:09.663" />
  <row Id="11915" PostTypeId="4" CreationDate="2011-07-06T00:02:09.663" Score="0" ViewCount="" Body="Negatively charged particle with spin 1/2.  A component of mundane terrestrial matter, and part of all neutral atoms and molecules. It has a mass about 1/1800 that of a proton. Its antiparticle is the positron." OwnerUserId="-1" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-06T04:12:06.703" LastActivityDate="2011-07-06T04:12:06.703" />
  <row Id="11916" PostTypeId="2" ParentId="11904" CreationDate="2011-07-06T00:22:25.573" Score="0" ViewCount="" Body="&lt;p&gt;Let us consider a horizontally pointing laser pointer that is being wiggled up and down. At some distance away the laser beam makes a bright spot on a wall. The spot moves up and down, say 10 meters, when the pointer device is moving up and down two millimeters. This is caused by relativistic aberration:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Relativistic_aberration&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Relativistic_aberration&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, now we wiggle the laser pointer horizontally and in the beam's direction. Now the bright spot on the wall does not move at all, just the size of the spot changes a tiny amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Quantum Electro Dynamics tells us that electric field consists of stream of virtual photons, but I'm not saying that aberration of virtual photons is an explanation of anything at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just saying that transverse wiggle and longitudinal wiggle are very different, because of relativistic effects, that are different in transverse and longitudinal directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh yes I must add that a sensitive instrument can detect a movement of a charge from any direction, and the information travels at the speed of light. &lt;/p&gt;&#xA;" OwnerUserId="4017" LastActivityDate="2011-07-06T00:22:25.573" />
  <row Id="11917" PostTypeId="2" ParentId="11912" CreationDate="2011-07-06T01:45:40.197" Score="9" ViewCount="" Body="&lt;p&gt;In the &quot;weak field limit&quot; where the graviational forces are small (such as anything in the solar system, and basically anything not right next to a black hole), the time dilation relative to a  distant observer is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta T/\Delta T_0 = 1 - \Phi/c^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here $\Delta T_0$ is the time elapsed for an observer at infinity,  $\Delta T$ is that time elapsed at some point in the system,  and $\Phi$ is the gravitational potential at that point. As @leftaroundabout stated correctly the important factor is the gravitational potential not gravitational forces. Since potentials add, instead of cancelling like the forces, we get twice the time dilation with two planets than we get with one, as @Jim Graber said. &lt;/p&gt;&#xA;" OwnerUserId="3970" LastEditorUserId="3970" LastEditorDisplayName="" LastEditDate="2011-07-06T16:50:15.603" LastActivityDate="2011-07-06T16:50:15.603" CommentCount="2" />
  <row Id="11918" PostTypeId="2" ParentId="11905" CreationDate="2011-07-06T02:05:52.180" Score="8" ViewCount="" Body="&lt;p&gt;In an ideal, holonomic and monogenic system (the usual one in classical mechanics), Hamiltonian equals total energy when and only when both the constraint and Lagrangian are time-independent and generalized potential is absent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the condition for Hamiltonian equaling energy is quite stringent. Dan's example is one in which Lagrangian depends on time. A more frequent example would be the Hamiltonian for charged particles in electromagnetic field&#xA;$$H=\frac{\left(\vec{P}-q\vec{A}\right)^2}{2m}+q\varphi$$&#xA;The first part equals kinetic energy($\vec{P}$ is canonical, not mechanical momentum), but the second part IS NOT necessarily potential energy, as in general $\varphi$ can be changed arbitrarily with a gauge.&lt;/p&gt;&#xA;" OwnerUserId="3887" LastActivityDate="2011-07-06T02:05:52.180" CommentCount="5" />
  <row Id="11919" PostTypeId="2" ParentId="11434" CreationDate="2011-07-06T02:14:27.583" Score="1" ViewCount="" Body="&lt;p&gt;In Gilmore's book:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0850667550&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Single-Particle-Detection-Measurement-Gilmore/dp/0850667550&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chapter 5 is on solid state ionization and it's use in detectors.  That was hanging around the lab when I was working in particle physics.  I'd give that one a try.  It's a bit old these days but the principles are there.&lt;/p&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2011-07-06T02:14:27.583" />
  <row Id="11920" PostTypeId="2" ParentId="11604" CreationDate="2011-07-06T02:37:28.840" Score="-1" ViewCount="" Body="&lt;p&gt;Seems to me this can be done in principle using the density operator $\rho$ and driving towards the equation of state by expressing the partition function as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho = \frac{1}{Z_G} exp^{-\beta K}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K = H - \mu N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And write out the Hamiltonian for a system of N atoms.  Once you have the partition function written out you can get at the equation of state.  Once you have the equation of state you can determine phase of the system at STP.  This would be a grand canonical formulation and I'm thinking it's not very easy to do with anything but the simplest of atoms, namely H and He.&lt;/p&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2011-07-06T02:37:28.840" CommentCount="1" />
  <row Id="11921" PostTypeId="2" ParentId="8903" CreationDate="2011-07-06T02:47:21.190" Score="4" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;Lagrangian mechanics can be derived from Principle of Least Action, or from Newtonian mechanics. This is not a fundamental difference.&lt;/li&gt;&#xA;&lt;li&gt;Pretty much. In Newtonian mechanics you start with drawing a bunch of vectors and then list your equations. In Lagrangian mechanics you first identify all constraints, choose generalized coordinates, then write down Lagrangian, and plug it into the Lagrangian equation $\frac{d}{dt}\frac{\partial L}{\partial \dot{q_i}}-\frac{dL}{dq_i}=0$&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;Lagrangian mechanics are better when there are lots of constraints. The more the constraints, the simpler the Lagrangian equations, but the more complex the Newtonian become. Lagrangian mechanics is not very suited for non-ideal or non-holonomic systems, such as systems with friction.&lt;br/&gt;&lt;/li&gt;&#xA;&lt;li&gt;Lagrangian mechanics is also much more extensible. It can remain almost the same form in hydrodynamics, electrodynamics, electric circuits, special and general relativity, etc.&lt;/li&gt;&#xA;&lt;li&gt;Principle of Least Time is closely related to Principle of Least Action, but they are in fact very different. I don't see how you can derive the former from the latter.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3887" LastActivityDate="2011-07-06T02:47:21.190" />
  <row Id="11922" PostTypeId="2" ParentId="11881" CreationDate="2011-07-06T02:48:38.613" Score="1" ViewCount="" Body="&lt;p&gt;There's nothing a priori that says it has to be that way and (I believe) neutrino oscillations have now shown that they have a mass which implies both right and left handed neutrinos exist.  That said, the lagrangian is still seriously skewed towards the left handed interaction.  So it's still very left-handed in a sense but there's nothing that says a coupling to right handed particles is out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We just don't see a strong coupling experimentally.  &lt;/p&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2011-07-06T02:48:38.613" CommentCount="2" />
  <row Id="11923" PostTypeId="2" ParentId="11904" CreationDate="2011-07-06T02:54:02.057" Score="1" ViewCount="" Body="&lt;p&gt;It depends on what you mean by &quot;compression wave&quot;.  When we typically think of compression wave, we think of sound waves, where the air (the medium) has a pressure differential between the peak and trough of the wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Electromagnetism, the wave is not a change in the medium^, it is a change in the electromagnetic field.*  Because of this, we have to ask, what &quot;compresses&quot; in the compression wave?  One possible answer is that the &quot;EM Field&quot; gets more dense, or more strongly positive, at which point we are back where we started: the analogy gets us nowhere, it is neither wrong, nor more insightful.  We also find that it starts to break down (what about &quot;strongly negative&quot; E-Field, this doesn't really work in a pressure analogy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the E-Field doesn't have compression waves because it doesn't modify the medium in which it is traveling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;^In this case, medium is understood to be the vacuum, or space-time, not the macroscopic medium (or dielectric).  In a dielectric, it is kind of possible for EM waves to PRODUCE compression waves (waves of varying density of the medium), but they cannot fundamentally BE compression waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*For a long time, this wasn't well understood, which is why (pre-Einstein), the dominant belief in physics was in a &quot;luminiferous aether&quot; as the medium in which EM waves traveled.  Michelson and Morley actually &quot;disproved&quot; this in 1887 with their &lt;a href=&quot;http://en.wikipedia.org/wiki/Michelson%E2%80%93Morley_experiment&quot; rel=&quot;nofollow&quot;&gt;seminal experiment&lt;/a&gt; (though I believe Michelson spent the rest of his career trying to improve upon his initial measurement and find the aether).  Combined with their null result, and Einstein's Theory of Special Relativity which came out 34 years later, the idea of a &quot;medium&quot; in which EM waves propagate is largely considered false.&lt;/p&gt;&#xA;" OwnerUserId="1256" LastEditorUserId="1256" LastEditorDisplayName="" LastEditDate="2011-07-06T04:08:06.667" LastActivityDate="2011-07-06T04:08:06.667" CommentCount="2" />
  <row Id="11925" PostTypeId="1" CreationDate="2011-07-06T05:59:52.960" Score="0" ViewCount="75" Body="&lt;p&gt;Paper, one of the most common material for scientists that exhibits many strange behaviours.&#xA;Here are some intriguing questions about paper and my own answer for them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. Why wet spots are dark/more transparent?&lt;/strong&gt;&lt;br&gt;&#xA;When you get the paper/cloth wet, the air between the fibers is replaced by water. Water has index of refraction(n=1.33) much closer to that of fiber (cellulose, n=1.48) than air has. Thus, each time a light beam encounters a fiber, it is bent much less than it would be with dry paper. In average, the light beam can pass through many more fiber before it turns all around. So more of the light ends up travelling forward into the fabric. I think this explanation makes lots of sense, we know that it is even possible to make water looks white e.g. in a very brutal wave. I also already did an experiment using photo diode, it turns out that the glycerin(n=1.47) spot is more transparent than the water spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Why is it easier to tear wet paper than dry paper?&lt;/strong&gt;&lt;br&gt;&#xA;Paper is made of small cellulose fibers tangled together like spaghetti, they can stick together by relying on friction force only. If you put water on it, it will lower the friction between those fibers, thus they can slide over easily and therefore much easier to be tore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3. How does paper make sound when it is teared?&lt;/strong&gt;&lt;br&gt;&#xA;-The sound is due to snapping fibers.(sound due to oscillating fibre)&lt;br&gt;&#xA;-The sound is produced by the &quot;stick-slip&quot; movement between fibers(screeching sound).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4. Why does dry paper makes more sound than wet paper?&lt;/strong&gt;&lt;br&gt;&#xA;-Wet-&gt;more massive, lower oscillatory frequency, falls out of human's hearing range&lt;br&gt;&#xA;-Lower friction, small amount of energy converted into sound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5. Why does dry paper produces greater sound if we tear it faster?&lt;/strong&gt;&lt;br&gt;&#xA;-More fiber's snapped/screeching in a short time&lt;br&gt;&#xA;-More frequencies inside our range&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;6. Why does paper wrinkle or become wavy when it is wet?&lt;/strong&gt;&lt;br&gt;&#xA;When the paper is dry, the fibers are pulled one side or the other which then creates tension. The entangled fibers are bent to some radius of curvature, because they are elastic they want to go back to their original shape(larger radius of curvature).They tend to get apart but friction force holds them together. When the paper gets wet, it lowers  its static friction, therefore some strongly pulled fibers that previously unable to defeat the friction force are now able to slip off. Thus the paper will appears stretched and more wavy, and retain its shape even thought it is dried.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect awesome qualitative answers, or even better with some experimental prove that everybody can try himself/herself since it is very easy experimenting with paper, like &quot;if we tear it this way, then&quot; or &quot;if the paper is folded several times, then..&quot;. I already tore so much paper just to help me thinking about it.&lt;/p&gt;&#xA;" OwnerUserId="4252" LastEditorUserId="4252" LastEditorDisplayName="" LastEditDate="2011-07-06T06:31:00.703" LastActivityDate="2011-07-06T06:31:00.703" ClosedDate="2011-07-06T07:46:40.137" Title="The Physics of Paper" Tags="&lt;optics&gt;&lt;sound&gt;&lt;material-science&gt;&lt;friction&gt;&lt;home-experiment&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="11926" PostTypeId="1" AcceptedAnswerId="11937" CreationDate="2011-07-06T08:14:25.347" Score="2" ViewCount="104" Body="&lt;p&gt;One definition of the chemical potential is the change of internal energy of the system with respect to a particle of the added substance with the system entropy, molar volume, and all other species particle numbers remaining constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A thought experiment: Let's suppose I have a completely rigid container with 1 mole of argon in it with an infinite amount of insulation surrounding the container.  Therefore, I have zero heat transfer and the molar volume is constant.  I admit one atom of argon into the container in a reversible manner (same P and T).  The pressure of the system must increase differentially according to the ideal gas law.  Hence the entropy will decrease differentially except that it must stay constant and so temperature will have to increase for this process to be isentropic.  But for an ideal gas, internal energy depends only on temperature.  So dU/dn is positive.  Yet, when I look up values for argon as a gas at STP, its value is 0 (like all the elements in their standard states).  Where is my reasoning wrong?    &lt;/p&gt;&#xA;" OwnerUserId="4289" LastActivityDate="2011-07-06T14:42:07.277" Title="What is wrong with this reasoning on chemical potential?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="11927" PostTypeId="1" CreationDate="2011-07-06T08:43:35.253" Score="6" ViewCount="356" Body="&lt;p&gt;We know that from our experience when we tear up a piece of paper, we can hear a characteristic sound. What is the underlying mechanism behind it? What do the dominant frequencies (edit: I don't mean there is one main frequency of the sound, it is more likely the noise produced contains mostly higher or lower frequencies) and intensity of the sound depend on? The explanation should agree with the real life cases -- here are a few that I observed (correct me if I'm wrong):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Dry paper makes louder sound and higher frequencies than wet paper&lt;/li&gt;&#xA;&lt;li&gt;Paper produces louder sound and higher frequencies if we tear it faster&lt;/li&gt;&#xA;&lt;li&gt;It is hard to tell which one produces louder sound, folded paper or single layer paper&lt;/li&gt;&#xA;&lt;li&gt;Folded paper produces lower frequencies than single layer paper&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Also, is it the same mechanism that explains the sound produced when we are cutting paper using a scissor?&lt;/p&gt;&#xA;" OwnerUserId="4252" LastEditorUserId="4283" LastEditorDisplayName="" LastEditDate="2011-07-12T16:11:20.917" LastActivityDate="2011-07-22T17:42:57.327" Title="How does paper make sound when it is teared?" Tags="&lt;sound&gt;&lt;material-science&gt;&lt;home-experiment&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="11928" PostTypeId="1" CreationDate="2011-07-06T09:34:09.110" Score="1" ViewCount="37" Body="&lt;p&gt;Say I have a sample of some alkali metal - Rubidium for example. A sample of Rubidium consists of approximately 27% $^{87}$Rb and 73% $^{85}$Rb. Does the vapor pressure of each isotope individual &lt;em&gt;necessarily&lt;/em&gt; have to be in the same proportion? More specifically, assuming I have enough Rubidium around that I have enough atoms of each isotope to reach the saturated vapor pressure, is there any physical reason why the ratio of the vapor pressures would match their relative abundances?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess this is coming from this perspective - the linear absorption spectrum for a sample of natural rubidium at a specific fine transition (D2 line - 5S1/2 -&gt;5P3/2 if you are interested) looks like image below, the $^{87}$Rb peaks (outer) are clearly smaller than the $^{85}$Rb peaks (inner), and this amplitude ratio is directly related to their natural abundances. If the atoms are in a gas, what forces their vapor pressure to have the same ratio as their relative abundance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/0WgIG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Along a similar line of questioning. A sample of Rb will naturally have some trace amount of Cs in it. The saturated number density of Cs at room temperature is ~$10^9$ atoms per cm$^3$, and I'm looking at a volume of say 1cm$^3$. If I have a droplet of 1 gram of Rb in the volume, which is .001 moles, even at .001% impurity I have $&amp;gt;10^{17}$ atoms of cesium in the sample, clearly enough to reach saturated density. -Does this imply that you can do spectroscopy on a volume with Rb and obtain the same spectrum as a volume with a droplet of Cs in it? If not, what prevents the Cs impurity in Rb to reach full saturation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Note: Perhaps I should make it clear : The linear absorption spectrum is proportional to the theoretical absorption (for a single atom) times the number density of atoms in the sample. The relative amplitudes of the peaks are therefore a proportional to the relative abundances of the two isotopes.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2011-08-05T23:39:25.883" Title="Relationship of the abundance of an isotope and the vapor pressure" Tags="&lt;thermodynamics&gt;&lt;spectroscopy&gt;" AnswerCount="2" />
  <row Id="11929" PostTypeId="2" ParentId="5928" CreationDate="2011-07-06T10:11:05.163" Score="1" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/g2UJZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;Striking Similarity between a spiral Galaxy and a Hurricane&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-07-06T10:11:05.163" CommunityOwnedDate="2011-07-06T10:11:05.163" CommentCount="1" />
  <row Id="11930" PostTypeId="1" CreationDate="2011-07-06T10:41:37.427" Score="0" ViewCount="57" Body="&lt;p&gt;muon neutrino momentum distribution &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read the public data of T2K ,KEK to find this subject, I'm curiously that it's coincides with my prediction perfectly: The neutrino get its momemtum by its effect-partner, which is obvious in the reactional formula, especialy that by me. You can find the figures&lt;a href=&quot;http://blog.sciencenet.cn/home.php?mod=space&amp;amp;uid=595492&amp;amp;do=blog&amp;amp;id=462424&quot; rel=&quot;nofollow&quot;&gt; Here &lt;/a&gt;( htis link is a blog including the detialed idea) the first is the radium function of e, the second cut from paper from the lab, the third is the momemtum spectrum. Do any one find the simlar distribution of other particle? I just work on Hep just for a few months.&lt;/p&gt;&#xA;" OwnerUserId="4291" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-07-06T14:37:02.633" LastActivityDate="2011-07-06T14:37:02.633" Title="muon neutrino momentum distribution" Tags="&lt;particle-physics&gt;&lt;neutrino&gt;&lt;beam&gt;" AnswerCount="1" />
  <row Id="11931" PostTypeId="2" ParentId="11927" CreationDate="2011-07-06T10:48:15.527" Score="2" ViewCount="" Body="&lt;p&gt;Paper is made from pulp fibres, typically some µmeters in diameter and&#xA;some mmters long. This fibres lie more or less statistically in the &#xA;plane of the paper, entangled like a can of worms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When tearing, You rupture those fibres, each giving a small sound. &#xA;That is what we hear, a rather &quot;blue&quot; noise (noise is the appropriate &#xA;word). Of course there is no predominant frequency, otherwise we would &#xA;hear someting more like a musical sound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a cut by scissors (or a knife) the mechanism is similar, &#xA;but in case of scissors the tool makes some hiss of its own,&#xA;even without paper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hoped that with the basic mechanism of noise-generation Emitabsob &#xA;woud be able to think about his sub-questions. But that did not happen. &#xA;&lt;strong&gt;Look at the edges of the torn parts&lt;/strong&gt; (dry/wet) and rethink again. &#xA;Maybe a magnifying glass might help to see the obvious. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-07-07T09:54:36.963" LastActivityDate="2011-07-07T09:54:36.963" CommentCount="6" />
  <row Id="11932" PostTypeId="1" AcceptedAnswerId="11934" CreationDate="2011-07-06T11:57:40.990" Score="1" ViewCount="103" Body="&lt;p&gt;What's so ingenious in diffractive lenses?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my naive eye they seem to be just Fresnel lenses with smaller features.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes it so magic and why all the fuss about it?&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="4283" LastEditorDisplayName="" LastEditDate="2011-07-11T16:00:35.433" LastActivityDate="2011-07-11T16:00:35.433" Title="What is so good about diffractive optics?" Tags="&lt;optics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="11933" PostTypeId="2" ParentId="11928" CreationDate="2011-07-06T12:31:54.703" Score="1" ViewCount="" Body="&lt;p&gt;From the logic side - if there were much difference in wapor pressure between isotopes, isotope separation would be much easier process than we see now. Surely there is a difference, but not something you can see on spectrum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can't comment on Cs saturation though.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-07-06T12:31:54.703" />
  <row Id="11934" PostTypeId="2" ParentId="11932" CreationDate="2011-07-06T12:44:59.953" Score="3" ViewCount="" Body="&lt;p&gt;Diffractive optics aren't magic, they are simply another tool that can be used in designing an optical system. They can do things that refractive optics cannot, and they are often lighter and smaller than an equivalent refractive optic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to keep in mind, however, that the benefits of a Diffractive Optical Element (DOE) are not free. DOEs have limitations of their own. They are harder to produce, and typically produce the desired results only under very specific conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, lets say you want to produce a circular laser beam with a very uniform intensity profile. What are your options for achieving this?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Most laser sources produce a roughly Gaussian beam, so you could expand this beam heavily with a refractive expander, and then mask out everything but the center of the beam. This will give you a relatively uniform beam, but you will waste a lot of light.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You could use a more complicated refractive design, like a micro-lens array. This is difficult to engineer, and won't give perfect results, but it can do a very good job under a variety of conditions. The beam intensity can be made uniform over a large distance, and the input beam to the micro-lens array will not need to be perfectly collimated. It will also work across a relatively broad wavelength range.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, you could design a DOE beam shaper. These can be designed to give any intensity profile you like, but it will be expensive to produce. It may (depending on what it is doing) have certain flaws characteristic of DOEs, like a strong zero-order beam (where a large fraction of input to the DOE passes through without being shaped). It may be very sensitive to errors in the input beam wavelength or collimation, and it may only produce the desired intensity profile over a short range of distances.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Like any tool available to a lens designer, DOE's have their uses. They can have very strong negative dispersion, which is often useful to correct chromatic aberration, and as I said they can be designed to produce arbitrary illumination patterns which would be outrageously difficult to make with purely refractive optics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, while you can say they are &quot;just Fresnel lens with smaller features,&quot; it is important to understand that a Fresnel lens &lt;strong&gt;is&lt;/strong&gt; a diffractive optic, just a very simple one. In fact, when your understanding of diffraction is deep enough, you will realize that, in some sense, all lenses are diffractive optics. While you can engineer the phase profile of a DOE to produce a highly complex optical field, you could also design one to produce a simple focal spot; the resulting design would be a simple lens!&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-07-06T12:44:59.953" />
  <row Id="11935" PostTypeId="2" ParentId="2670" CreationDate="2011-07-06T13:21:14.570" Score="0" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/vcugE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strict short answer to the Question – the number of particles is invariant (at large).&lt;br&gt;&#xA;It is know that SM does not preserve energy, i.e. Noether is only valid as long as the ratio matter/space is constant.&lt;br&gt;&#xA;From the above Answers we see that you don’t know of any scale-invariant theory that supports the physical laws.&lt;br&gt;&#xA;The main question is: How to show that physical laws hold in a scale-invariant model? Many physicists tried and failed (Dirac, Canuto Hoyle and Narlikar , Maeder and Bouvier, Wesson).&lt;br&gt;&#xA;I will present a resume of  ‘&lt;strong&gt;&lt;a href=&quot;https://sites.google.com/site/scalinguniverse2011/self-similar-model&quot; rel=&quot;nofollow&quot;&gt;A self-similar model of the Universe unveils the nature of dark energy&lt;/a&gt;&lt;/strong&gt;’  by Alfredo G. Oliveira, submitted to PRX in July 1,2011. (Oh my gosh, my name is in the paper!)&lt;br&gt;&#xA;If we attach a referential to a particle, say an atom, above represented in grey, we cannot find any evolution. It is our actual situation; we look around in the labs and we are naturally blinded to any evolution.&lt;br&gt;&#xA;The Question mentions only a modification of the length, Lubos answer also mentioned a varying time, but that procedure is short of what is needed to have a correct self-similar model. It has to be done in a ‘physical’ way:&lt;br&gt;&#xA;Lets &lt;strong&gt;shrink an atom&lt;/strong&gt; (the atom is our reference for Mass, Length, Time) of the Past into the one of Present. As viewed from an external invariant reference ‘S’(Space) the Length Unit changed, and also the Mass Unit changed and also the Time unit changed because the speed of light is the constant c — it’s a property of the field/space.&lt;br&gt;&#xA;It is evident that an atomic observer (bounded to his atomic reference) sees a space expansion.  The cosmologic redshift of the light of galaxies (far away in time and distance) trace the fact that the atomic processes were slower in the past as compared to the ones of the present.&lt;br&gt;&#xA;Be $M(t_S)=Q(t_S)=L(t_S)=T(t_S)=\alpha(t_S)$ the relation that describe the evolution of units thru time, as viewed by S in relation to the units of the atomic observer ( $\alpha(t_S)$ is the scaling law).&lt;br&gt;&#xA;It is derived in the paper, using only the laws of physics, making no hypotheses, and departing only from measured data that &lt;strong&gt;the scaling law is  $\alpha(t_S) = e^{-H_0\cdot t_S}$.&lt;/strong&gt;&lt;br&gt;&#xA;Quoting the &lt;strong&gt;Summary and Conclusions&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Henry Poincaré analyzed how we acquire information, stressing the relative nature of our data and that our choice of units serves the convenience of obtaining the simplest form for physical laws;&lt;br&gt;&#xA;  Einstein analyzed how we calibrate reference frames, how we attribute coordinates to occurrences, what is the kind of time and length units we use;&lt;br&gt;&#xA;   here, the reflection on this subject is extended to the properties of the units, which enabled us to understand that the invariance of particles in standard units is a property of these units and not of the particles; &#xA;  it become also clear how the space expansion may trace a self-similar phenomenon and an important yet previously unnoticed property of the units of field constants was found, which is able of supporting the observed space dilation. &#xA;  From two accepted observational results, the invariance of constants and the scalar space expansion, and considering that the observed space expansion is consequence of a self-similar phenomenon, it is deducted a model that verifies the classic cosmic tests as well as&#xA;  the $\Lambda$CDM model in spite of having just one parameter, the Hubble parameter. This model has surprising features, namely:&lt;br&gt;&#xA;  (1) There is no theoretical conflict with fundamental physical laws but for a new term in one conser- vation law, which is beyond present possibilities of direct measurement.&lt;br&gt;&#xA;  (2) The standard systems of units lose their privileged role, physical laws being valid also in a space, comoving, system of units.&lt;br&gt;&#xA;  (3) In standard units, this model supports the same description of the universe of the $\Lambda$CDM model.&#xA;  Despite this scaling model not being a cosmological model, it gives some contributions to cosmology, namely:&lt;br&gt;&#xA;  (1) Space is older than matter.&lt;br&gt;&#xA;  (2) Matter, field and radiation evanesce in space units.&lt;br&gt;&#xA;  (3) A simple explanation arises for the lack of tendency for gravitational collapse.&lt;br&gt;&#xA;  (4) The roles of dark energy and of cosmological inflation are made clear.&lt;br&gt;&#xA;  This paper is just the first of a set of three; the second paper analyses the consequences of this model at the solar system scale and the third analyses the large-scale structure of the universe.&lt;br&gt;&#xA;  Until now, the knowledge of the universe was established in units where atomic properties are&#xA;  invariant; these units are very convenient for describing systems of bodies but, when used to describe space properties, the result is puzzling. To have surpassed this limitation is a major achievement of this work.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course one can argue 'I do not believe that it can happen to atoms!' and I will counter argue 'How can the space expand?'.&lt;br&gt;&#xA;&lt;a href=&quot;https://sites.google.com/site/scalinguniverse2011/self-similar-model&quot; rel=&quot;nofollow&quot;&gt;The paper is available here&lt;/a&gt; (the arxiv is closed to my friend Alfredo, probably not even Perelman can use arxiv anymore). I know this model since 1991, and a preliminary version can be found in the arxiv dated 2002; back then the public was unprepared to read this model and it is my expectation that we have evolved to a more mature position.  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-07-06T13:21:14.570" />
  <row Id="11936" PostTypeId="2" ParentId="11930" CreationDate="2011-07-06T14:35:10.683" Score="2" ViewCount="" Body="&lt;p&gt;Do you understand how neutrino beams are made for such experiments?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The beam from some more mundane particle accelerator is used to direct high energy particles (often protons) into a target which generates a spray of assorted products containing many pions and kaons. This secondary beam is filtered and selected for momentum by passing it through collimators and dipole magnets; focused using a &quot;horn&quot;; and allowed to propogate down another beam-line where the hadrons decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The decay products include many neutrinos, and relativity means that most of them are going mostly in the direction of the final beam-line. Their momentum distribution is a function of the hadrons beam's momentum (which can be selected), the hadron beams compositions, the decay kinematics of pions and kaons, and the angle relative the deca beam-line at which you place your detector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the momentum distribution of the neutrinos &lt;strong&gt;&lt;em&gt;that react&lt;/em&gt;&lt;/strong&gt; in your detector also has a factor due to the energy dependence of the neutrino interaction cross-section.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-07-06T14:35:10.683" CommentCount="1" />
  <row Id="11937" PostTypeId="2" ParentId="11926" CreationDate="2011-07-06T14:42:07.277" Score="1" ViewCount="" Body="&lt;p&gt;Your argument purports to show the chemical potential is positive. You look it up and find that it's zero. In fact, the chemical potential for an ideal gas is negative!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I'm not following the step where you conclude that the entropy must go down because the pressure goes up. The entropy of an ideal gas is given by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sackur%E2%80%93Tetrode_equation&quot; rel=&quot;nofollow&quot;&gt;Sackur-Tetrode equation&lt;/a&gt;:&#xA;$$&#xD;&#xA;S=Nk\left[\ln\left({V\over N}\left(4\pi m U\over 3Nh^2\right)^{3/2}\right)+{5\over 2}\right]&#xD;&#xA;$$&#xA;Incidentally, I got this particular form of the equation from the textbook by &lt;a href=&quot;http://physics.weber.edu/thermal/&quot; rel=&quot;nofollow&quot;&gt;Schroeder,&lt;/a&gt; which is the one I like best for this sort of thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the circumstances you describe (increasing $N$ by 1 while holding $U/N$ constant), this reduces to&#xA;$$&#xD;&#xA;S=Nk\left[\ln(V/Nv_Q)+ {5\over 2}\right],&#xD;&#xA;$$&#xA;where $v_Q$, the quantum volume per particle, is constant and $Nv_Q\ll V$. (The latter is the condition for the gas to be nondegenerate.) The derivative of this with respect to $N$ is positive: upon adding a particle at constant $T,V$, the entropy goes up by&#xA;$$&#xD;&#xA;\Delta S=k\left[\ln(V/nv_Q)+{3/2}\right]&amp;gt;0.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, where are you looking up the value for $\mu$ for argon at STP? Is it possible that the table you're looking in lists values relative to STP? It's certainly not true that the chemical potential of an ideal gas at STP is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to get the chemical potential of an ideal gas is by imagining adding the new particle in such a way that the total energy (not the temperature) remains fixed. The relevant identity then is&#xA;$$&#xD;&#xA;\mu=-T\left(\partial S\over\partial N\right)_{U,V},&#xD;&#xA;$$&#xA;which comes from $dU=T\,dS-P\,dV+\mu\,dN$. Taking the derivative of the above expression for $S$, we get&#xA;$$&#xD;&#xA;\mu=-kT\ln\left[{V\over N}\left(2\pi mkT\over h^2\right)^{3/2}\right].&#xD;&#xA;$$&#xA;Again, for gases that are far from degenerate, the quantity inside the logarithm is large, so the chemical potential is negative.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-07-06T14:42:07.277" CommentCount="3" />
  <row Id="11938" PostTypeId="1" CreationDate="2011-07-06T15:02:10.003" Score="0" ViewCount="66" Body="&lt;p&gt;Why EM waves having only x momentum transfers to electron z momentum? Electron begins oscillating along z, so will not radiate EM waves along z direction, to compensate its z momentum. It seems that it is violated conservation of momentum?&lt;/p&gt;&#xA;" OwnerUserId="3926" LastActivityDate="2011-07-06T16:15:56.103" Title="Why electromagnetic waves propagating along x transfers to electron momentum along z?" Tags="&lt;electromagnetic-radiation&gt;&lt;electrodynamics&gt;&lt;momentum&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="11939" PostTypeId="1" AcceptedAnswerId="13231" CreationDate="2011-07-06T15:22:33.667" Score="4" ViewCount="212" Body="&lt;p&gt;The usual Lagrangian for a relativistically moving charge, as found in most text books, doesn't take into account the self force from it radiating em energy. So what is the lagrangian for a relativistic charge that includes the self force?&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-08-06T08:38:48.213" Title="What is the Lagrangian for a relativistic charge that includes the self force?" Tags="&lt;classical-electrodynamics&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" />
  <row Id="11940" PostTypeId="1" AcceptedAnswerId="11947" CreationDate="2011-07-06T15:39:21.670" Score="44" ViewCount="19196" Body="&lt;p&gt;From my understanding of light, you are always looking into the past based on how much time it takes the light to reach you from what you are observing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example when you see a star burn out, if the star was 5 light years away then the star actually burnt out 5 years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am 27 years old, if I was 27 light years away from Earth and had a telescope strong enough to view Earth, could I theoretically view myself being born?&lt;/p&gt;&#xA;" OwnerUserId="947" LastActivityDate="2011-07-07T06:37:55.437" Title="If you view the Earth from far enough away can you observe its past?" Tags="&lt;light&gt;&lt;spacetime&gt;" AnswerCount="8" CommentCount="3" FavoriteCount="17" />
  <row Id="11941" PostTypeId="2" ParentId="11940" CreationDate="2011-07-06T15:54:28.040" Score="13" ViewCount="" Body="&lt;p&gt;No. Because you cannot reach the speed of light, even if you had started travelling away from Earth the day you were born, you could never catch the light carrying the image of your being born.&lt;/p&gt;&#xA;" OwnerUserId="3887" LastActivityDate="2011-07-06T15:54:28.040" CommentCount="6" />
  <row Id="11942" PostTypeId="2" ParentId="11881" CreationDate="2011-07-06T16:02:35.897" Score="2" ViewCount="" Body="&lt;p&gt;I think you are sort of reversing the logic of chirality and helicity in the massless limit. Chirality defines which representation of the lorentz group your Weyl spinors transform in.  It doesn't 'become' helicity, helicity 'becomes' chirality in the massless limit. That is, chirality is what it is, and it defines a representation of a group and that can't change. This other thing we have defined called helicity just happens to be the same thing in a particular limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now once you take the massless limit the Weyl fermions are traveling at the speed of light you can no longer boost to a frame that switches the helicity. I think its best to think of a fermion mass term as an interaction in this case and remember that the massive term of a Dirac fermion is a bunch of left and right- handed Weyl guys bumping up into one another along the way. Conversely if you want to talk about a full massive Dirac fermion that travels less than c and you can boost to change the helicity, but that full Dirac fermion isn't the thing carrying weak charge, only a 'piece' of it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.quantumdiaries.org/2011/06/19/helicity-chirality-mass-and-the-higgs/&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt; on helicity and chirality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Left%E2%80%93right_symmetry&quot; rel=&quot;nofollow&quot;&gt;left-right symmetry&lt;/a&gt; being broken people have certainly built models along these lines but I don't think they have worked out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this answer your question?&lt;/p&gt;&#xA;" OwnerUserId="4292" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-07T11:03:07.937" LastActivityDate="2011-07-07T11:03:07.937" CommentCount="2" />
  <row Id="11943" PostTypeId="2" ParentId="11938" CreationDate="2011-07-06T16:15:56.103" Score="1" ViewCount="" Body="&lt;p&gt;In the particle formulation of electromagnetism, light is carried by photons. Momentum is absolutely conserved. If a photon hits an electron, there will be momentum balance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the wave formulation of EM the electric field,( supposed it is a polarized beam),  oscillates in the z direction. An electron that acquires a momentum will take part of the energy of the beam and the beam will acquire a small z component to conserve momentum. On the average,since the electron will be hit in both z directions by the incoming beam, the effect will be that the beam will spread/degrade, to conserve momentum and energy.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-07-06T16:15:56.103" CommentCount="12" />
  <row Id="11944" PostTypeId="2" ParentId="11940" CreationDate="2011-07-06T16:19:06.210" Score="17" ViewCount="" Body="&lt;p&gt;I'd have to say yes - at least theoretically. If there was an observer (not you) 27 light years away, he could see your birth. Awesome as it sounds, it isn't practically feasible. For one thing, the observer would need a large enough telescope to observe you. If the telescope isn't large enough, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Optical_telescope#Angular_resolution&quot; rel=&quot;nofollow&quot;&gt;resolution&lt;/a&gt; would be low, and the observer wouldn't be able to make out things far away. To get an idea, the size (diameter of mirror for instance) of the telescope that'll enable you to see an object of about 10m at that distance is about $10^{8}$km while the diameter of earth is about 12800 km. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But again, you yourself wouldn't be able to make this observation, as the information regarding your birth would have gone at the speed of light, and you wouldn't be able to intercept it.&lt;/p&gt;&#xA;" OwnerUserId="966" LastActivityDate="2011-07-06T16:19:06.210" />
  <row Id="11945" PostTypeId="2" ParentId="11940" CreationDate="2011-07-06T16:47:08.153" Score="7" ViewCount="" Body="&lt;p&gt;I think the other posters are forgetting that nature allows for much more possibilities. True, you can't catch up to light, but that might not be necessary. As General Relativity tells us, space-time is a 4 dimensional manifold. Einstein's equations tie its local structure to the energy-momentum density. But that still leaves open a lot of possibilities for the global structure of the universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that the universe would be something like a torus or a sphere, that is finite but unbounded. This concretely means that if you go long in enough in one direction, you will ultimately end up where you started again. Now, that is also true for light. If then the expansion of the universe did not happen so fast that light going around got behind the cosmological horizon, it is theoretically possible that we could see our own galaxy in an earlier stage of its evolution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The global shape of the universe is still an open problem. The most commonly accepted model though assumes an infinite flat space-time. Other models like &lt;a href=&quot;http://en.wikipedia.org/wiki/Shape_of_the_Universe&quot; rel=&quot;nofollow&quot;&gt;Poincaré dodecahedral space and the Picard horn&lt;/a&gt; do seem to fit current cosmological data. Well, they did last time I looked, maybe they have been refuted by now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if the universe is well described by a Poincaré dodecahedral space FLRW model, then there is a chance that we can see our own past for a limited amount of time, i.e. until it disappears beyond the cosmological horizon. Don't hope to see yourself though.&lt;/p&gt;&#xA;" OwnerUserId="369" LastActivityDate="2011-07-06T16:47:08.153" />
  <row Id="11946" PostTypeId="2" ParentId="11939" CreationDate="2011-07-06T16:57:54.807" Score="-1" ViewCount="" Body="&lt;p&gt;If the charge is point-like then its classical self-energy is infinite. As the self-force comes from the gradient of potentials, the result is undetermined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a theory with a local interaction. Then we must take into account some field, e.g. the electromagnetic field. Now for a massive relativistic particle moving at constant velocity there is always an inertial frame of reference where the particle is not moving. Therefore the magnetic field vanishes and the electric field is that given by Coulomb's law, i.e.&#xA;$$\mathbf E(\mathbf r)=\frac q{4\pi\epsilon}\frac{\mathbf r}{\Vert \mathbf r\Vert^3}.$$&#xA;Since the particle stays at $\mathbf r=0$, the force is undefined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far the only massless particle known is the photon, which is neutral.&lt;/p&gt;&#xA;" OwnerUserId="4248" LastActivityDate="2011-07-06T16:57:54.807" CommentCount="4" />
  <row Id="11947" PostTypeId="2" ParentId="11940" CreationDate="2011-07-06T17:08:39.987" Score="60" ViewCount="" Body="&lt;p&gt;Yes, you can.  And you do not even need to leave the Earth to do it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are always viewing things in the past, just as you are always hearing things in the past.  If you see someone do something, who is 30 meters away, you are seeing what happened $(30\;\mathrm{m})/(3\times10^8\;\mathrm{m}/\mathrm{s}) = 0.1\;\mu\mathrm{s}$&#xA;in the past.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had a mirror on the moon (about 238K miles away), you could see about 2.5 seconds into earth's past.  If that mirror was on Pluto, you could see about 13.4 hours into Earth's past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are relying on hearing, you hear an event at 30 m away about 0.1 s after it occurs.  That is why runners often watch the starting pistol at an event, because they can see a more recent picture of the past than they can hear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To more directly answer the intent of your question:  Yes, if you could magically be transported 27 lightyears away, or had a mirror strategically placed 13.5 lightyears away, you could see yourself being born.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="4202" LastEditorDisplayName="" LastEditDate="2011-07-06T21:09:00.300" LastActivityDate="2011-07-06T21:09:00.300" CommentCount="14" />
  <row Id="11948" PostTypeId="1" AcceptedAnswerId="11950" CreationDate="2011-07-06T17:48:39.973" Score="1" ViewCount="71" Body="&lt;p&gt;This question is sparked by my answer to this question: &lt;a href=&quot;http://physics.stackexchange.com/questions/11902/is-this-simulation-following-real-physics/11906#11906&quot;&gt;Is this simulation following real physics?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After examining the math, I don't see how it is theoretically possible for the situation simulated to happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation in question is whether or not a spear in projectile motion can stay oriented over the entire path such that the angle is equal to the angle of the velocity vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I start with proposing that the rate of change of the angle of the velocity vector must be a constant - there can be no torque applied during the projectile motion, only the initial angular velocity imparted at the release.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So taking a projectile, with say an initial velocity of &amp;lt;20, 20&gt; - this would be at a 45 degree angle, the case where the condition intuitively seems most likely:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dy}{dt} = -9.8t + 20$, $\frac{dx}{dt} = 20$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We find $\frac{dy}{dx}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dy}{dx} = \frac{-9.8t + 20}{20}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And an expression for the angle, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta = \arctan{\frac{dy}{dx}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graphing the derivative of this function over the time span of the projectile shows that it is NOT a constant. This implies that there would have to be some sort of torque being applied mid-air if the angular displacement is directly proportional to $\theta$. However, it IS a very shallow bowl. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, is this analysis correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, if this implies that the condition is impossible, what happens when one throws a javelin or spear perfectly? Why is it so natural to do so?&lt;/p&gt;&#xA;" OwnerUserId="3390" LastEditorUserId="3390" LastEditorDisplayName="" LastEditDate="2011-07-06T17:53:40.560" LastActivityDate="2011-07-06T18:22:21.550" Title="Is it theoretically possible for the orientation angle of a projectile to remain exactly equal to the orientation of velocity?" Tags="&lt;classical-mechanics&gt;&lt;rotational-dynamics&gt;&lt;velocity&gt;&lt;projectile&gt;" AnswerCount="1" />
  <row Id="11949" PostTypeId="1" AcceptedAnswerId="11998" CreationDate="2011-07-06T18:09:16.723" Score="1" ViewCount="65" Body="&lt;p&gt;We have a structure that is sloped at 20 Degrees to the horizontal and weights 160N. We then have a force (wind) which is acting along the horizontal plane which has been calculated to be of about 8000N. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem we are having is that we are unable to find what amount of ballast that needs to be added at the bottom of the structure to stop the structure from being blown away. The structure will not be bolted to the ground. At the moment we do not have an exact amount for the coefficient of friction, but the structure will be placed on a flat concrete surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have tried to find a solution by breaking the forces along the vertical and horizontal components, but we where unsure if we where heading in the right direction. Below is an image which describes the setting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/KQA35.png&quot; alt=&quot;Problem Description&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please advise us as to which mathematical model we should use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="4293" LastActivityDate="2011-07-07T19:06:01.927" Title="Problem with forces acting upon a sloped plane" Tags="&lt;homework&gt;&lt;forces&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11950" PostTypeId="2" ParentId="11948" CreationDate="2011-07-06T18:22:21.550" Score="2" ViewCount="" Body="&lt;p&gt;If you assume that no torque is imparted to the object during flight then you are entirely correct about the fact that the spear direction can not match the velocity direction.  It may cross, perhaps cross twice, and perhaps tangent, but it will be a very different behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A front-loaded weight in significant atmosphere, however, has a strong tendency to orient itself along the direction of motion.  Given a shape with some symmetry, if the center of mass is offset from the center point of the object, then air resistance will tend to point the CM to the front of the object.  In the case of a stick, you can imagine a constant force per unit length from the air, meaning that the effective point of action is in the geometric middle.  If the CM is not in that location, an effective torque will be present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My favorite example of this is lawn darts.  The CM is intentionally place very far in the front and the object has a very significant air resistance to mass ratio.  It orients itself with the tail fin trailing behind no matter how you throw it.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-07-06T18:22:21.550" CommentCount="4" />
  <row Id="11951" PostTypeId="2" ParentId="11949" CreationDate="2011-07-06T18:39:28.570" Score="1" ViewCount="" Body="&lt;p&gt;I would treat the point of contact with the ground as having rotational freedom but no freedom of horizontal or vertical motion.  From there, the moments of all the forces must balance.  I would assign the positive x-axis to be the direction of the slanted object and the origin to be the point of contact with the ground.  You then have 3 forces, all acting at different locations along the object.  Take $l$ to be the length of the object.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&lt;li&gt;$F_g$ force of gravity which acts at distance $\frac{1}{2}l$ and angle $\theta=-\frac{8}{9} \pi$ (downward)&#xA;&lt;li&gt;$F_w$ force of the wind that acts at distance $d_w$ and angle $\theta=\frac{8}{9} \pi$ (horizontal)&#xA;&lt;li&gt;$F_b$ force of the ballast that acts at distance $d_b$ and angle $\theta=-\frac{8}{9} \pi$&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Alternatively, denote these as $F_i$, $d_i$, and $\theta_i$ for i from 1 to 3.  Write the moment balance at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_i F_i d_i sin \theta_i=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If everything else is known then you can solve this for $F_b$.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-07-06T18:39:28.570" />
  <row Id="11952" PostTypeId="2" ParentId="11940" CreationDate="2011-07-06T19:08:22.710" Score="-2" ViewCount="" Body="&lt;p&gt;No, I don't think the concept of time works quite like that.  Yes, if a star is 5 light years away and it burned out - yes it burned out 5 years ago for that point in time.  Space time is relevant for that time period.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example: &#xA;You are 27 Years old (Present Day Earth Time)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are teleported 27 Light Years Away to Planet X (Your Present Age is 27 on Planet X).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You look back at Earth - You are non-existent on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because Earth's time does not bend back.  It took 27 years for Earth's image to reach you at the time you left. So unless you were at both planets at the same time you would be 27 on Planet X while on Earth you would be 54 Years old.  Earth's time does not bend back, it only moves forward.&lt;/p&gt;&#xA;" OwnerUserId="4295" LastActivityDate="2011-07-06T19:08:22.710" CommentCount="1" />
  <row Id="11953" PostTypeId="2" ParentId="11912" CreationDate="2011-07-06T19:29:01.350" Score="2" ViewCount="" Body="&lt;p&gt;The gravitational force may be zero at the mid-point, but the spacetime itself is still stretched out pretty strongly - twice as strongly in fact - and that's all that matters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, yes, time is still skewed.&lt;/p&gt;&#xA;" OwnerUserId="4210" LastActivityDate="2011-07-06T19:29:01.350" />
  <row Id="11954" PostTypeId="2" ParentId="11897" CreationDate="2011-07-06T21:04:32.597" Score="1" ViewCount="" Body="&lt;p&gt;Well, as is mentioned &lt;a href=&quot;http://physics.stackexchange.com/questions/11912/would-there-be-time-dilation-at-the-point-where-two-gravitational-fields-cancel-e&quot;&gt;here&lt;/a&gt;, time dilation is caused by gravitational potentials, but not necessarily gravitational forces.  So, it might be possible to travel between two equally sized black holes to experience extreme time dilation effects.  Traveling through would not cost much energy, however it would potentially have an extreme effect on the time dilation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note however that I have never taken a GR class.  I only noticed this from the other question and thought it would be relevant here.  The equation given by one of the answers there is incorrect because you are no longer in the weak field limit, but I don't know what the strong field limit is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm making this community wiki so that someone else more knowledgeable than me can clarify.&lt;/p&gt;&#xA;" OwnerUserId="1256" LastActivityDate="2011-07-06T21:04:32.597" CommunityOwnedDate="2011-07-06T21:04:32.597" CommentCount="1" />
  <row Id="11955" PostTypeId="2" ParentId="11939" CreationDate="2011-07-06T21:20:16.047" Score="0" ViewCount="" Body="&lt;p&gt;It is the usual CED Lagrangian for a charge and electromagnetic field; it includes the free field Lagrangian, a free particle Lagrangian and the interaction Lagrangian $jA$. Self-action is encoded in it when you consider the system as a whole, as coupled unknown variables. Of course, it yields rubbish: an infinite &quot;electromagnetic mass&quot; that you need to discard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I say &quot;rubbish&quot; because, on one hand, we define the charge as a stable, fixed entity; on the other hand we introduce a self-force to get some dynamics due to it. Our definition contradicts our intentions.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-07-06T21:28:22.447" LastActivityDate="2011-07-06T21:28:22.447" />
  <row Id="11956" PostTypeId="1" CreationDate="2011-07-06T21:36:41.367" Score="0" ViewCount="46" Body="&lt;p&gt;For an system that amplifies strength by about two (press capacity of 400 lbs.), which would be the most practical, or the most compact?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pneumatic system that uses air cylinders or air muscles,&#xA;A hydraulic system that uses water in cylinders, or&#xA;A group of piezoelectric wires as fibers, so that when you run current through them, they all contract?&lt;/p&gt;&#xA;" OwnerUserId="4144" LastActivityDate="2011-07-06T21:36:41.367" ClosedDate="2011-07-06T23:57:51.963" Title="Which actuator is more effective: pneumatic, hydraulic, or electric?" Tags="&lt;mechanics&gt;&lt;fluid-dynamics&gt;" CommentCount="2" />
  <row Id="11957" PostTypeId="1" CreationDate="2011-07-06T21:45:32.540" Score="3" ViewCount="71" Body="&lt;p&gt;I was going to buy this add on, as I almost exclusively use my sunglasses for driving. I'm just stacking them over my normal ones now... but it's quite annoying that I almost paid the extra $$$ for this feature. Found out at the last second, luckily.&lt;/p&gt;&#xA;" OwnerUserId="4304" LastActivityDate="2011-07-06T22:09:16.143" Title="Why do prescription glasses that darken in sunlight fail to darken when the light travels through a car windshield?" Tags="&lt;light&gt;" AnswerCount="1" />
  <row Id="11958" PostTypeId="2" ParentId="11957" CreationDate="2011-07-06T22:09:16.143" Score="9" ViewCount="" Body="&lt;p&gt;These type of &lt;a href=&quot;http://en.wikipedia.org/wiki/Photochromic_lens&quot; rel=&quot;nofollow&quot;&gt;glass/plastic&lt;/a&gt; responds to UV light. Car windshield has UV filters so the glasses does not darken as much inside cars.&lt;/p&gt;&#xA;" OwnerUserId="3360" LastActivityDate="2011-07-06T22:09:16.143" />
  <row Id="11959" PostTypeId="1" AcceptedAnswerId="11977" CreationDate="2011-07-06T22:37:57.100" Score="2" ViewCount="108" Body="&lt;p&gt;As far as I see from wiki, 'consumer'-grade(non-cryogenic) thermal imaging cameras use microbolometer sensors to get integrated IR intensity over some 5-12um range. But I had an impression, that having just integrated IR intensity you can't get surface temperature precisely, especially with 0.1C precision we see on consumer cameras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IR thermometers (pyrometers) for example use 2-band IR measurement, but I don't see this principle being used in thermal imaging. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or there is some sort of 'know-how' matrix of IR band filters in front of the sensor?&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="930" LastEditorDisplayName="" LastEditDate="2011-07-06T22:44:36.953" LastActivityDate="2011-07-07T09:22:07.203" Title="How thermal imaging cameras work" Tags="&lt;thermal-radiation&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="11960" PostTypeId="2" ParentId="11306" CreationDate="2011-07-06T22:47:55.657" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The scatter direction depends on the size of the particle and the wavelength. Very small particles (e.g. nitrogen atoms of the atmosphere) scatter isotropically. There is still an effect on the polarisation of the scattered light (Bees use that to locate the sun if they can't see it directly).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Often Gaussian beams are used to describe how the intensity propagates in an optical train (a system of lenses and mirrors) &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_beam&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Gaussian_beam&lt;/a&gt;. Note that this describes the intensity of a Laser with a Gaussian intensity profile (this is a good approximation for many lasers, especially if you focus them through a pinhole). If you have an extended light source you will have to add the intensities of several such beams.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Once you have a numeric intensity profile (I guess 2D is enough for your usage), you can try an exponential decay law to estimate the effects of scatter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your focus spot will look more intense the higher the numerical aperture of your lens is.&#xA;If you use a lens that has 2.5 cm diameter and f=10cm the spot won't look as intense as with a lens that has f=3cm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you thought of using fluorescence? You could use dissolve some colour in water and use laser protection goggles to see the fluorescent light. Then you don't have to cope with scattering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get polygonal mirrors out of old laser printers. That way you can scan the beam in one direction. If you use a laser diode, you can modulate the intensity very fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently purchased a 405nm Laser with 120mW for $120 from lasever.com. 120mW is very dangerous. If you don't have protection goggles or you share your space with other people don't use lasers&gt;0.5mW!&lt;/p&gt;&#xA;" OwnerUserId="4306" LastActivityDate="2011-07-06T22:47:55.657" />
  <row Id="11961" PostTypeId="2" ParentId="11928" CreationDate="2011-07-06T23:07:19.573" Score="-1" ViewCount="" Body="&lt;p&gt;I posed this question to some coworkers outside of physics.se here is the first order answer that satisfied me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evaporation is effectively a statistical process that is determined by the probability of a certain atom from escaping its liquid bonds and becoming free. The process does not care about what atom evaporates (in the approximation that the two isotopes have the same mass and binding energy - which is reasonable), it effectively becomes a probability problem. You have a bag of atoms, some colored blue and some colored red. The probability of picking a red atom out of the bag (having 85Rb escape the liquid and go free) is given by its relative abundance to blue atoms, assuming that there are enough atoms around that the atoms lost to the vapor do not significantly affect the ratio of isotopes in the the liquid, we would expect, statistically,  the abundance of isotopes in the vapor would match that of the liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Cs impurities, the same logic applies, although you have to modify a little bit due to the heavier weight of the Cs atom and potentially a difference in enthalpy of vaporization.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2011-07-06T23:07:19.573" />
  <row Id="11963" PostTypeId="1" AcceptedAnswerId="11968" CreationDate="2011-07-06T23:38:46.410" Score="3" ViewCount="168" Body="&lt;p&gt;One of the unsolved problems I still have is measuring temperature in furnace at 700-1100C range (I need some +-0.1% repeatability and +-1% precision).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Obviously, contact thermometers are quite rare &amp;amp; expensive for this range, everything I was able to find is rated at 450-500C max (especially as this should work in aggressive oxygen/ozone/chlorine atmosphere).&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consumer grade pyrometers does not work above ~400C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've checked blackbody emission for 700-1100C objects, and it seems that even at 500C objects start to emit at 1um range (not much though), which is easily measurable by cheap silicon photo-diodes (ones I have on the desk have maximum at 880nm, with 70% sensitivity at 1um and 10% at 1.1um, visible light filtered).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As my target is always the same with fixed emissivity, there should be no vital need for 2-band measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any issues with this single-band temperature measurement approach? Alternatively, I might add PbS photoresistor (2.1um maximum), but they are not that precise, quite expensive and scarce. What are your thoughts or suggestions?&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="930" LastEditorDisplayName="" LastEditDate="2011-07-09T18:29:33.110" LastActivityDate="2011-07-17T23:24:52.700" Title="Will this poor-man's single band pyrometer work?" Tags="&lt;thermal-radiation&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="11964" PostTypeId="2" ParentId="11959" CreationDate="2011-07-06T23:42:41.710" Score="1" ViewCount="" Body="&lt;p&gt;Here  &lt;a href=&quot;http://www.jenoptik.com/en_40150_thermographic_cameras&quot; rel=&quot;nofollow&quot;&gt;http://www.jenoptik.com/en_40150_thermographic_cameras&lt;/a&gt;&#xA;they claim +/- 1.5K accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this spec another company claims 50mK Minimum Discernable Temperature Difference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.argusdirect.com/assets/media/files/datasheets/cameras/argussc-specialist-search-camera/argus%20sc%20english.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.argusdirect.com/assets/media/files/datasheets/cameras/argussc-specialist-search-camera/argus%20sc%20english.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The precision of these bolometers is nice but the accuracy is more than 10 times worse:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Accuracy_and_precision&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Accuracy_and_precision&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I couldn't find any datasheets on the sensors on the e2v site.&lt;/p&gt;&#xA;" OwnerUserId="4306" LastActivityDate="2011-07-06T23:42:41.710" CommentCount="2" />
  <row Id="11965" PostTypeId="2" ParentId="11875" CreationDate="2011-07-06T23:59:28.577" Score="0" ViewCount="" Body="&lt;p&gt;If you don't know the specifications of your objective you can't calculate this.&#xA;I describe a method you could try to measure the optical transfer function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Print out gratings 10 (or so) gratings with different periods.&#xA;Capture images of these gratings. Make sure that the central part of the camera is as sharp as possible for the finest grating and make sure that the gratings are oriented in the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your lens has an aperture try opening it as far as possible. This decreases your depth of field and allows you to find the focus more precisely. Conversely for your application of capturing OCR images you should close the aperture as far as possible (use a bright lamp) this improves the point spread function of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your lens doesn't have an aperture you can try to build one with black paper and a hole punch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a raytracer you have to evaluate this equation on every surface:&#xA;$${\bf t}=\eta{\bf i}-\left(\eta{\bf i n}+\sqrt{1-\eta^2(1-({\bf i n})^2)}\right){\bf n}$$&#xA;with $\eta=n_1/n_2$, $n_1$ the index on the left (downstream) side of the surface, ${\bf n}$ the normal of the surface at the intersection with the ray.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its possible to write a write a ray surface intersector for conic surfaces as a 10 line function that you can use among the other things for plane and spherical surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing you might need to look into is the Sellmeier coefficients of the glasses in your system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find the image position by tracing rays from the object into the image plane and look where beams starting from one point intersect. In a good system the focus spot will be smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also draw a Gaussian sphere around each image point and sum up the optical distances along the ray. That way you'll obtain the wave aberrations.&lt;/p&gt;&#xA;" OwnerUserId="4306" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-07-07T09:09:51.770" LastActivityDate="2011-07-07T09:09:51.770" CommentCount="2" />
  <row Id="11966" PostTypeId="1" AcceptedAnswerId="11983" CreationDate="2011-07-07T00:04:19.337" Score="2" ViewCount="197" Body="&lt;p&gt;I ran into this while looking at measures of humongous amounts of data. How does the information (data) collected in an inverse femtobarn exposure compare to a gigabyte of data ?&lt;/p&gt;&#xA;" OwnerUserId="3714" LastEditorUserId="3714" LastEditorDisplayName="" LastEditDate="2011-07-08T20:34:55.953" LastActivityDate="2011-07-08T20:34:55.953" Title="How large is the information collected from an inverse femtobarn of  collisions?" Tags="&lt;high-energy-physics&gt;&lt;si-units&gt;&lt;information&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="11968" PostTypeId="2" ParentId="11963" CreationDate="2011-07-07T01:16:47.713" Score="4" ViewCount="" Body="&lt;p&gt;I am using a CN7600 in my furnace. &lt;a href=&quot;http://www.omega.com/manuals/manualpdf/M1303.pdf&quot; rel=&quot;nofollow&quot;&gt;Omega 76000&lt;/a&gt; This controller is very accurate.  As for all them measurements I have made it keeps the oven 750 C. It fluctuates from 750 to 751 which just over 1% if I take 750.5 as my required temperature. Now whit a little improvement  in the furnace design I can improve this. The problem is eliminating &quot;Hot pockets&quot; in the furnace by having a very mild air/gas flow around the subject. The prove wires can be in a Pirex tube for temps under 1000 C and the controller can compensate for this.&lt;/p&gt;&#xA;" OwnerUserId="4034" LastEditorUserId="4034" LastEditorDisplayName="" LastEditDate="2011-07-07T06:52:44.730" LastActivityDate="2011-07-07T06:52:44.730" CommentCount="5" />
  <row Id="11969" PostTypeId="2" ParentId="11940" CreationDate="2011-07-07T02:09:45.407" Score="1" ViewCount="" Body="&lt;p&gt;Technically, yes, though the idea of an absolute past kinda screws with the whole question. Better would be a light cone-based analysis. But yes, if you were in the past light cone of a past Earth, then you would see that past Earth.&lt;/p&gt;&#xA;" OwnerUserId="4308" LastActivityDate="2011-07-07T02:09:45.407" />
  <row Id="11970" PostTypeId="2" ParentId="11940" CreationDate="2011-07-07T02:57:53.040" Score="-2" ViewCount="" Body="&lt;p&gt;I didn't think light could travel that far. Doesn't it slowly loose energy over time? I'm not an expert, but I'm just saying...&lt;/p&gt;&#xA;" OwnerUserId="4310" LastActivityDate="2011-07-07T02:57:53.040" CommentCount="2" />
  <row Id="11971" PostTypeId="2" ParentId="11927" CreationDate="2011-07-07T05:58:07.933" Score="0" ViewCount="" Body="&lt;p&gt;If you snapped a piece of wood, it would make a noise, as you tear apart a lot of chemical bonds in the wood, releasing waste energy as sound and a little heat. The structure of paper is fairly similar, except ground up, so it makes a noise as well as you rip apart the same cellulose strands.&lt;/p&gt;&#xA;" OwnerUserId="4308" LastActivityDate="2011-07-07T05:58:07.933" CommentCount="2" />
  <row Id="11972" PostTypeId="2" ParentId="11940" CreationDate="2011-07-07T06:37:55.437" Score="1" ViewCount="" Body="&lt;p&gt;Could you see yourself born? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not if you were born indoors&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2011-07-07T06:37:55.437" CommentCount="1" />
  <row Id="11973" PostTypeId="2" ParentId="111" CreationDate="2011-07-07T06:59:21.250" Score="0" ViewCount="" Body="&lt;p&gt;I am just starting with the book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540655581&quot; rel=&quot;nofollow&quot;&gt;Mechanics:  From Newton's Laws to Deterministic-Chaos&lt;/a&gt;. It seems to introduce the mathematics later.&lt;/p&gt;&#xA;" OwnerUserId="4263" LastActivityDate="2011-07-07T06:59:21.250" CommunityOwnedDate="2011-07-07T13:41:37.800" />
  <row Id="11974" PostTypeId="2" ParentId="11378" CreationDate="2011-07-07T07:11:14.937" Score="1" ViewCount="" Body="&lt;p&gt;I have thought about this answer today while cycling. I have absolutely no idea how good is my ``solution'' but I am sure it isn't total b*&lt;em&gt;*&lt;/em&gt;*it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While running, you pay your energy mainly for three different reasons&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;you accelerate to somewhat constant speed (actually you accelerate to an average speed but this should be a good enough approximation)&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You are advancing against air friction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You are not running in a straight line but you are actually bouncing up an down. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In all three cases you are implementing force along a rout to generate energy (work) and you are transforming chemical energy (proton gradient, sugar, ATP etc.) into kinetic energy via muscle contraction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are applying force to generate work, and the force scales like your mass then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = ma \Rightarrow w = ma\Delta r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which means you more energy which is equal to your overweight times some constant. With good enough scales, you can estimate the excess energy which is consumed but running twice the same distance in as similar situation as possible (same time of day, same air conditioning etc.) once with a 10kg weight on your back and once with your clothing alone. weighting yourself before and after the run, mass difference can be translated to energy via some engineering tables (wikipedia states that one g of fat stores ~40kJ of energy)&lt;/p&gt;&#xA;" OwnerUserId="3677" LastActivityDate="2011-07-07T07:11:14.937" CommentCount="1" />
  <row Id="11975" PostTypeId="1" AcceptedAnswerId="11980" CreationDate="2011-07-07T08:27:24.550" Score="3" ViewCount="167" Body="&lt;p&gt;In the time-independent Schrodinger equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(-\frac{\hbar^2}{2m}\Delta+V(\mathbf{r})\right)\psi(\mathbf{r})=E\psi(\mathbf{r})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What functions $V(\mathbf{r})$ are mostly used in research (e.g. Constant potential, Inverse Power-Law Potentials, Finite square well and Infinite square well)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And who can give a brief history of different potential functions studied in Schrodinger equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And can anyone explain to me what is 'Hartree Potential'?&lt;/p&gt;&#xA;" OwnerUserId="4314" LastEditorUserId="661" LastEditorDisplayName="" LastEditDate="2011-08-14T22:33:12.400" LastActivityDate="2011-08-15T15:33:39.050" Title="What potential energy functions are mostly used in Schrodinger equation?" Tags="&lt;quantum-mechanics&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="11977" PostTypeId="2" ParentId="11959" CreationDate="2011-07-07T09:22:07.203" Score="2" ViewCount="" Body="&lt;p&gt;I'm not an expert at thermography per se, but I've designed and built a commercial microbolometer IR camera and looked at several other cameras (for thermography) and as far as I know, they don't have any &quot;funny business&quot; going on. Commercial handheld &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermography&quot; rel=&quot;nofollow&quot;&gt;thermography&lt;/a&gt; cameras require you to input the surface emissivity, air temperature and transmittance parameters etc to get an accurate reading, so obviously in practice for random measurements you won't get anywhere near +/- 1K accuracy, but the noise (precision) can be as low as dozens of mK like quoted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In industrial (repeatable) applications you can probably calibrate this very accurately though.&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-07-07T09:22:07.203" CommentCount="4" />
  <row Id="11978" PostTypeId="1" AcceptedAnswerId="12238" CreationDate="2011-07-07T09:59:13.107" Score="4" ViewCount="69" Body="&lt;p&gt;I've seen microscope lenses optimized for 0.17mm covering glass. I don't see what needs to be optimized here? As glass does not touch the lens (as in case of oil/water immersion) - it should just affect focal distance without introducing any aberrations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that correct, or covering glass will cause aberrations imbalance, and will require recalculating the lens? (I can probably only think of very slight chromatic aberration, but it's not important in my case)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same for water immersion: If we cover sample with 1mm of water, but water does not touch the lens, will it cause any additional aberrations and require recalculating the lens?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same for diffraction-limited laser focusing optics: I've seen some of them are optimized for laser output windows - what is the nature of this optimization?&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-07-13T14:43:36.337" Title="Impact of covering glass on lens performance" Tags="&lt;optics&gt;" AnswerCount="2" />
  <row Id="11979" PostTypeId="2" ParentId="11978" CreationDate="2011-07-07T10:09:50.010" Score="3" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Is that correct,&lt;/strong&gt; or covering glass&#xA;  will cause aberrations imbalance&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A flat will cause a small spherical aberration. &#xA;In old physics books used to be sketches for that. &#xA;You know that rays not perpendicular to a flat will be shifted &#xA;somewhat after passage. Still collinear to the ray before, but shifted.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW: what is aberration imbalance, or what is balanced aberration?&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-07-07T10:09:50.010" CommentCount="4" />
  <row Id="11980" PostTypeId="2" ParentId="11975" CreationDate="2011-07-07T11:05:07.600" Score="1" ViewCount="" Body="&lt;p&gt;The mostly used potentials are the parabolic potentials. You may have seen it in quantum harmonic oscillator with potential energy $\frac{1}{2}m\omega^2 x^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parabolic potential is of the form $\frac{1}{2}k x^2$ and corresponds to the force $F=-\frac{dV}{dx}=-kx$. It can be used as an approximation to potentials at local minima (stable equilibrium points), at which the second derivative is positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose that you have a potential with local minimum at $x=0$ and $V&amp;#39;&amp;#39;(0)&amp;gt;0$. Taking Taylor expansion of $V(x)$ near the $x=0$ you get: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V(x)=V(0)+V&amp;#39;(0)x+\frac{1}{2}V&amp;#39;&amp;#39;(0)x^2+\cdot\cdot\cdot$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can choose $V(0)=0$ and of course $V&amp;#39;(0)=0$ (local minimum at $x=0$). So finally you get $V(x)=\frac{1}{2}V&amp;#39;&amp;#39;(0)x^2=\frac{1}{2}kx^2$, where $k=V&amp;#39;&amp;#39;(0)&amp;gt;0$, as a good approximation of the $V(x)$ near the $x=0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This parabolic approximation can be used for vibrations of diatomic or polyatomic molecules.&lt;/p&gt;&#xA;" OwnerUserId="3066" LastEditorUserId="3066" LastEditorDisplayName="" LastEditDate="2011-08-15T15:33:39.050" LastActivityDate="2011-08-15T15:33:39.050" CommentCount="9" />
  <row Id="11981" PostTypeId="2" ParentId="111" CreationDate="2011-07-07T11:31:01.640" Score="0" ViewCount="" Body="&lt;p&gt;Although not specifically answering the needs of Cedric H. (symplectic geometry) I find Introduction to Dynamics by Percival &amp;amp; Richards one of the best (and simplest) introductions to Lagrangian and Hamiltonian dynamics, particularly canonical transformations and so on. I point out this book because it is probably not so well known.&lt;/p&gt;&#xA;" OwnerUserId="3572" LastActivityDate="2011-07-07T11:31:01.640" CommunityOwnedDate="2011-07-07T13:41:37.800" />
  <row Id="11982" PostTypeId="1" CreationDate="2011-07-07T11:42:02.160" Score="7" ViewCount="183" Body="&lt;p&gt;Are there any &quot;natural&quot; physical observables which have non-empty point spectrum which consists of numbers which are not algebraic numbers?&lt;/p&gt;&#xA;" OwnerUserId="4321" LastEditorUserId="4438" LastEditorDisplayName="" LastEditDate="2011-07-10T06:05:23.657" LastActivityDate="2011-07-10T06:05:23.657" Title="Observables with transcendental eigenvalues" Tags="&lt;quantum-mechanics&gt;&lt;observables&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="11983" PostTypeId="2" ParentId="11966" CreationDate="2011-07-07T12:20:55.517" Score="6" ViewCount="" Body="&lt;p&gt;I think what you're getting at is not some kind of mathematically rigorous equivalence, but more what it means for a particle physics experiment like ATLAS to collect 1 inverse femtobarn of data. And actually this is computable quire easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The design frequency of the LHC is 40Mhz (which corresponds to 25ns bunch spacing, but now i is at 50ns). But since most events are uninteresting background all modern experiments have a system called a &quot;Trigger&quot; which only records events which pass some rough requirements which would render them interesting (maybe a high-momentum electron or jet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ATLAS is routinely recording at 300Hz (a $10^5$ reduction of rate from the initial collision rate). That is 300 events per second. The size of an event in terms of storage space varies from experiment to experiment and depends on the software it uses, but for ATLAS it is something of the order of 1.5MB/event.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently the LHC runs at peak luminosities of 12600 $\mu b^{-1}/s$ (microbarn per second), this decreases over time since the beam intensities decrease, let's just run with 1000$\mu b^{-1}/s$. An inverse femtobarn is $10^9\mu b^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{300 \text{ Events}}{s}\frac{1.5 \text{ MB}}{\text{ Event}}\frac{s}{1000\mu\text{b}^{-1}} \approx 0.5 \frac{\text{MB}}{\mu\text{b}^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so for $10^9 \mu b^{-1}$ we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0.5\cdot10^9\text{MB}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so  500 TB of data&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: this is just an back-of-the-envelope calculation  of course. The rates are constantly changing and the luminosities as well. So collecting 1/fb of data in a low lumi setting requires much more data (since one would still max out the bandwidth of 300Hz recording) than in high lumi settings (where one is still bound by the 300Hz boundary, so the trigger would have to do a tighter selection)&lt;/p&gt;&#xA;" OwnerUserId="2151" LastEditorUserId="2151" LastEditorDisplayName="" LastEditDate="2011-07-08T12:34:22.067" LastActivityDate="2011-07-08T12:34:22.067" CommentCount="3" />
  <row Id="11984" PostTypeId="2" ParentId="11982" CreationDate="2011-07-07T12:39:19.373" Score="8" ViewCount="" Body="&lt;p&gt;I don't have a complete answer, but maybe the following is useful for your purposes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the Laplacian $\Delta$ on a &lt;a href=&quot;http://en.wikipedia.org/wiki/Vibrations_of_a_circular_drum&quot; rel=&quot;nofollow&quot;&gt;circular drum&lt;/a&gt; of unit radius. As explained on the wikipedia page, the axially symmetric eigenvectors $\Delta u(r) = -\lambda^2 u(r) $ are Bessel functions $u(r)=J_0(\lambda r)$. Obviously, the boundary condition requires $J_0(\lambda)=0$. In other words, the eigenvalues of this Laplacian correspond to zeroes of Bessel functions, and I would be very surprised if these numbers are not transcendental. In fact, &lt;a href=&quot;http://mathworld.wolfram.com/TranscendentalNumber.html&quot; rel=&quot;nofollow&quot;&gt;Mathworld&lt;/a&gt; mentions that the first zero has been proven to be transcendental by Le Lionnais.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding quantum mechanical situation would be the Hamilton operator $H=-\hbar^2/2m\cdot \Delta$ of a free 2D electron confined to the unit disk. The boundary conditions are the same, $\psi|_{&#xD;&#xA;\partial\Omega}=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;Of course, the electron has the problem that it's not clear whether $\hbar= h/2\pi$ should be counted as transcendental or algebraic, physicists frequently set $\hbar = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, consider a 1D electron inside a box of length $L$, i.e. $\psi(0)=\psi(L)=0$. The eigenstates are simply standing waves, and the eigenvalues are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E_n = \frac{\hbar^2}{2m_e} \left(\frac{\pi n}{L}\right)^2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you count $h$ as algebraic, then this is algebraic. But if you count $\hbar$ as algebraic, then this is transcendental. Your way out is probably to turn the question into a &lt;em&gt;relative&lt;/em&gt; one: is there a physical observable whose eigenvalues are algebraically independent of the $E_n$? Clearly, you only need to consider the eigenvalues of the Laplace operator now.&lt;/p&gt;&#xA;" OwnerUserId="188" LastEditorUserId="188" LastEditorDisplayName="" LastEditDate="2011-07-09T09:21:24.727" LastActivityDate="2011-07-09T09:21:24.727" CommentCount="7" />
  <row Id="11985" PostTypeId="2" ParentId="11982" CreationDate="2011-07-07T12:52:10.123" Score="4" ViewCount="" Body="&lt;p&gt;Transcendental numbers often pop out in higher-loop Feynman diagram calculations. One example is the spectrum of local operators in $\mathcal{N}=4$ supersymmetric Yang-Mills theory, where the anomalous dimensions at higher loops in general contain $\zeta$-functions. As an example, the dimension of the operator $\mathop{Tr} \phi^a\phi^a$ is given perturbatively by&#xA;$$\gamma = 4 + 12g^2 - 48g^4 + 336g^6 + (-2496+576\zeta(3)-1440\zeta(5))g^8 + \mathcal{O}(g^{10})$$&#xA;(I don't think there is a rigorous proof showing that $\zeta(n)$ is non-algebraic for positive and odd values of $n$, but it seems likely to me...)&lt;/p&gt;&#xA;" OwnerUserId="3131" LastActivityDate="2011-07-07T12:52:10.123" CommentCount="5" />
  <row Id="11986" PostTypeId="1" AcceptedAnswerId="11988" CreationDate="2011-07-07T13:30:55.193" Score="2" ViewCount="126" Body="&lt;p&gt;I just read the wikipedia page&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_potential&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Gravitational_potential&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't understand how to get this formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho(\mathbf{x}) = \frac{1}{4\pi G}\Delta V(\mathbf{x})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone tell me why?&lt;/p&gt;&#xA;" OwnerUserId="4314" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-07T18:54:44.017" LastActivityDate="2011-07-07T18:54:44.017" Title="How can we get this formula for Gravitational potential?" Tags="&lt;newtonian-gravity&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="11987" PostTypeId="2" ParentId="111" CreationDate="2011-07-07T14:24:21.893" Score="1" ViewCount="" Body="&lt;p&gt;This isn't explicitly about just mechanics as it tries to hit a lot of different areas in physics but it covers the material you are asking for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486661695&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Differential-Forms-Applications-Physical-Sciences/dp/0486661695&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will absolutely provide you with a firm basis in the geometrical considerations you mentioned in your question (tangent bundle, cotangent bundle, 1-form, 2-form, etc.).  &lt;/p&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2011-07-07T14:24:21.893" CommunityOwnedDate="2011-07-07T14:24:21.893" />
  <row Id="11988" PostTypeId="2" ParentId="11986" CreationDate="2011-07-07T14:54:47.063" Score="4" ViewCount="" Body="&lt;p&gt;This equation comes most simply from the following two:&#xA;$$&#xD;&#xA;{\bf g}=-\nabla V,&#xD;&#xA;$$&#xA;$$&#xD;&#xA;\nabla\cdot{\bf g}=-4\pi G\rho.&#xD;&#xA;$$&#xA;Plut the first one into the second, and you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the natural next question is where those two equations come from.&#xA;The first one is the relationship between the gravitational field ${\bf g}$ and the gravitational potential $V$. It's pretty much the definition of gravitational potential. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second one is Gauss's Law for gravitational fields. It follows from the inverse-square law. Wikipedia gives a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss%27s_law&quot; rel=&quot;nofollow&quot;&gt;derivation&lt;/a&gt;. (That derivation is for electric fields and potentials, rather than gravitational ones, but if you just replace ${\bf E}$ with $-{\bf g}$ and &quot;charge&quot; with &quot;mass,&quot; it's exactly the same.)&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-07-07T14:54:47.063" />
  <row Id="11989" PostTypeId="2" ParentId="11986" CreationDate="2011-07-07T15:21:44.307" Score="2" ViewCount="" Body="&lt;p&gt;$\Delta=\nabla^2$ is the Laplace operator. So from $V(\textbf{x})=-G\int_{\mathbb{R}^3}\frac{1}{|\textbf{x}-\textbf{r}|}\rho(\textbf{r})dv(\textbf{r})$ we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta V(\textbf{x})=-G\int_{\mathbb{R}^3}\left(\nabla\cdot\nabla\frac{1}{|\textbf{x}-\textbf{r}|}\right)\rho(\textbf{r})dv(\textbf{r})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's easy to show that $\nabla\frac{1}{|\textbf{x}-\textbf{r}|}=-\frac{\textbf{x}-\textbf{r}}{|\textbf{x}-\textbf{r}|^3}$  and $\nabla\cdot\left(\frac{\textbf{x}-\textbf{r}}{|\textbf{x}-\textbf{r}|^3}\right)=4\pi\delta^3(\textbf{x}-\textbf{r})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(tell me if you need help with this)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta V(\textbf{x})=G\int_{\mathbb{R}^3}4\pi\delta^3(\textbf{x}-\textbf{r})\rho(\textbf{r})dv(\textbf{r})=4\pi G \rho(\textbf{x})$&lt;/p&gt;&#xA;" OwnerUserId="3066" LastActivityDate="2011-07-07T15:21:44.307" CommentCount="5" />
  <row Id="11990" PostTypeId="2" ParentId="11803" CreationDate="2011-07-07T15:34:57.997" Score="1" ViewCount="" Body="&lt;p&gt;That quasar's light started its travel to us about 13B years ago, but its distance is NOT 13B light-years. The distance is more something like 50 billion light-years (for afficionados, I am talking about the comoving distance). The reason is that the universe itself expands while the light travels to us, so that the distance is larger than the speed of light multiplied by 13B years (i.e. larger than if the universe were static). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the reader's question, it is quite true that 'we' were close to the burst when it happened, but the space between us and the burst expanded since then, and it takes light time to cover that separation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the answers imply say that the reason is in cosmological inflation, but this is not correct. Inflation happens very early in the history of the universe (approx 10^{-35} seconds after the Big Bang), and cannot be verified just by looking at the quasars and such since it does not affect distances in the 'recent' history of the universe. Instead, verifying inflation involves sophisticated measurements of fluctuations of matter and light in the universe (cosmic microwave background radiation, etc) which were laid out well before the existence of any quasars.  &lt;/p&gt;&#xA;" OwnerUserId="1334" LastActivityDate="2011-07-07T15:34:57.997" CommentCount="2" />
  <row Id="11991" PostTypeId="2" ParentId="2269" CreationDate="2011-07-07T15:46:06.480" Score="1" ViewCount="" Body="&lt;p&gt;This is an excellent question. It indeed &lt;em&gt;may&lt;/em&gt; be true that the universe oscillates, and that we are not experiencing just one of the phases of this oscillation. (This theory, which is plagued by some rather technical problems but is still viable, sometimes goes under the name of 'cyclic universe'). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we do know, mainly from the three pillars of cosmology that Raskolnikov explained, is that the universe was small and hot early on. But we do not factually know whether the Big Bang (ie. singularity) actually happened, and we do not know what happened before inflation (i.e. before about 10^{-35} seconds after the Big Bang). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while the smart money is still betting on the existence of the singularity - being arguably the simplest theory that also passes all observational tests with flying colors - future observations will likely able to distinguish between it and the cyclic scenario. &lt;/p&gt;&#xA;" OwnerUserId="1334" LastActivityDate="2011-07-07T15:46:06.480" />
  <row Id="11992" PostTypeId="1" AcceptedAnswerId="11993" CreationDate="2011-07-07T16:58:09.993" Score="2" ViewCount="100" Body="&lt;p&gt;I'm reading an article which describes a generation of artificial AKG signal. The article states that they use the following equation of motion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \left \lbrace \begin{array} {ccc} \dot x &amp;amp;=&amp;amp; \alpha x - \omega y \\ \dot y &amp;amp;=&amp;amp; \alpha y + \omega x\end{array}\right .$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\alpha = 1- \sqrt{x^2 + y^2}$ and $\omega$ is the angular velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have figured out the easy part where I use the $\theta$ and $r(=1)$ coordinates thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \left \lbrace \begin{array} {ccc} &#xD;&#xA;\dot x &amp;amp;=&amp;amp; \frac{d}{dt} \cos (\theta) =  -\sin (\theta) \frac{d}{dt} = -\omega y \\ \dot y &amp;amp;=&amp;amp; \frac{d}{dt} sin (\theta) = \cos (\theta) \frac{d}{dt} = \omega x\end {array}\right .$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which has made me assume that $r = r(t)$ and $\langle r\rangle = 1$. I have tried to solve this simple deferential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dr}{dt} = 1-r$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but the solution I have found was:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r(t) = 1-c\text{e}^{t} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which has many  problems, units to say the least. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear to me that the units problem is originated from the definition of $\alpha$ but there is no reference to that in the article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was hoping somebody of you has encountered such a problem or is handling such approach. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;I have forgotten to mentioned the very important fact that they are using a unit circle. Also, my main issue is with the $\alpha$ part which, on the unit circle should be 0...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;A &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.65.2929&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; to the article&lt;/p&gt;&#xA;" OwnerUserId="3677" LastEditorUserId="3677" LastEditorDisplayName="" LastEditDate="2011-07-08T06:19:31.653" LastActivityDate="2011-07-08T06:19:31.653" Title="Circular equation of motion" Tags="&lt;kinematics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="11993" PostTypeId="2" ParentId="11992" CreationDate="2011-07-07T17:30:04.850" Score="3" ViewCount="" Body="&lt;p&gt;As far as units are concerned, the initial equations only make sense if everything's dimensionless. $x,y$ must be dimensionless for $\alpha=1-\sqrt{x^2+y^2}$ to make sense. Then $\alpha x$ is dimensionless. The first equation therefore implies that $t$ and $\omega$ are dimensionless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, we can go ahead and solve this system of equations. Start with your initial equations, and make the substitutions $x=r\cos\theta$ and $y=r\sin\theta$. All of $x,y,r,\theta$ are functions of $t$, so that, e.g., $\dot x=\dot r\cos\theta-r\dot\theta\sin\theta$. If you express everything in terms of $r,\theta$ and simplify a bit, the two equations reduce to&#xA;$$&#xD;&#xA;\dot r=r-r^2,&#xD;&#xA;$$&#xA;$$&#xD;&#xA;\dot\theta=\omega.&#xD;&#xA;$$&#xA;The solution for $\theta$ is just $\theta=\theta_0+\omega t$, of course. The solution for $r$ is&#xA;$$&#xD;&#xA;r(t)={1\over 1-Ce^{-t}},&#xD;&#xA;$$&#xA;where $C$ is chosen to match your initial conditions.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-07-07T17:30:04.850" CommentCount="2" />
  <row Id="11994" PostTypeId="2" ParentId="11804" CreationDate="2011-07-07T17:46:22.327" Score="2" ViewCount="" Body="&lt;p&gt;The question (v1) deals with the axiomatic QFT formulation of the real Klein-Gordon field $\phi$ in Minkowski spacetime $M$ with a vacuum state $\Omega_0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $O\subseteq M$ be a fixed non-empty bounded open subset. I would say that the error in OP's proof is the sentence &lt;em&gt;If we take $C$ in the commutant of ${\mathfrak F}(O)$, i.e. $C\in{\mathfrak F}(O)&amp;#39;$, then we can show that $\Omega_0$ is an eigenvector of $C$.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My counterexample goes roughly like this. Since $O\subseteq M$ is bounded, we can choose another non-empty open subset $U$ in $M$ that is causally disconnected from $O$, i.e. such that every pair of points $x\in O$ and $y\in U$ are space-like separated. Consider next a test-function $f\in S(M)$ with support ${\mathrm supp}(f)\subseteq U$ in such a way that the one-particle state $\phi[f]\Omega_0$ does not vanish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From one of the &lt;a href=&quot;http://ncatlab.org/nlab/show/Wightman+axioms&quot; rel=&quot;nofollow&quot;&gt;Wightman axioms&lt;/a&gt;, we know that $C:=\phi[f]$ commutes with the operators in ${\mathfrak F}(O)$. However, a one-particle state $\phi[f]\Omega_0$ cannot be proportional to the vacuum state $\Omega_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, ${\mathfrak F}(O)$ is not irreducible.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastActivityDate="2011-07-07T17:46:22.327" CommentCount="1" />
  <row Id="11995" PostTypeId="1" CreationDate="2011-07-07T17:54:38.753" Score="2" ViewCount="174" Body="&lt;p&gt;I should begin by saying that &lt;em&gt;I am a total newbie when it comes to Quantum Mechanics.&lt;/em&gt; Therefore my question might sound metaphysical to people who know their stuff. So please forgive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am trying to understand is the basic system of an electron in a box. Suppose I place a detector at a fixed point inside and try to detect the presence of an electron. There are two possibilities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) The detector detects an electron. The Copenhagen interpretation explains this by saying that the act of measurement forced the wave function to collapse at that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) The detector did not detect anything. Would this &quot;not detecting an electron&quot; qualify as a true measurement in the Copenhagen sense? I mean, if this qualified as a measurement would it change the wave function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What led me to this confusion is the explanation in &lt;em&gt;Griffiths' Quantum Mechanics.&lt;/em&gt; He states that the same experiment when done soon after possibility (1) gives us the same result i.e, observe the electron (which does not confuse me.) But what happens when you do the experiment soon after possibility (2)? Would I still not observe the electron?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if I extend this to the two-slit experiment and place the detector at only one slit. If I didn't observe the electron, it means that it passed through the other hole, which doesn't lead to the pattern. So again, does this make &quot;not detecting the electron&quot; a true observation which collapses the wave function?&lt;/p&gt;&#xA;" OwnerUserId="966" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-07T17:58:41.303" LastActivityDate="2011-07-09T18:09:31.367" Title="Defining Measurement in Quantum Mechanics" Tags="&lt;quantum-mechanics&gt;&lt;quantum-interpretations&gt;&lt;double-slit-experiment&gt;&lt;measurement-problem&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="11996" PostTypeId="2" ParentId="11897" CreationDate="2011-07-07T17:57:24.383" Score="0" ViewCount="" Body="&lt;p&gt;One problem with the black hole flyby solution is that stable orbits are bounded away from the event horizon - any object free-falling through the photon sphere will wind up passing through the event horizon.  If you don't feel like using a lot of propulsion energy, this bounds the amount of time you can spend in each pass where your time is highly dilated relative to a distant observer.  On the other hand, if the black hole is rotating quickly, you can extract energy for your next flyby using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Penrose_process&quot; rel=&quot;nofollow&quot;&gt;Penrose process&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="761" LastActivityDate="2011-07-07T17:57:24.383" CommentCount="1" />
  <row Id="11997" PostTypeId="2" ParentId="2269" CreationDate="2011-07-07T18:26:00.390" Score="-1" ViewCount="" Body="&lt;p&gt;Until further clarification the answer is NO:&lt;br&gt;&#xA;The claim that 'space expands' must be backed by the claim that 'the atom is invariant'.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bigger atom in the past is observed as a redshift (a longer wavelength) because the standard rule we use to measure it is based in the actual atomic dimension.&lt;br&gt;&#xA;In a figurative way: to measure wavelengths of the past we must use 'inches' and now we use 'cm'. The &lt;em&gt;same&lt;/em&gt; amount of distance gives two different measures. Did the distance increased ? or the unit of measure decreased?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A claim that the atom can vary in size, using the same laws of physics, can be found here: &lt;strong&gt;&lt;a href=&quot;https://sites.google.com/site/scalinguniverse2011/&quot; rel=&quot;nofollow&quot;&gt;A self-similar model of the Universe unveils the nature of dark energy&lt;/a&gt;&lt;/strong&gt;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asserting that the atom is invariant is equal to say: '&lt;em&gt;Here is an absolute reference frame&lt;/em&gt;', against Einstein.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From now on a clarification is needed. imo.  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-07-08T09:54:31.770" LastActivityDate="2011-07-08T09:54:31.770" CommentCount="3" />
  <row Id="11998" PostTypeId="2" ParentId="11949" CreationDate="2011-07-07T19:06:01.927" Score="1" ViewCount="" Body="&lt;p&gt;You have three problems to solve: slipping, tipping around a horizontal axis) and rotationally turning (around a vertical axis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would assume that if you solve slipping and tipping, that you will have generated enough static friction to make turning (spinning like a top) not a problem.  (YOU need to check this, but I would expect it to be so.)  So now you have only tipping and slipping to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For tipping, you pick a convenient point to sum moments.  If there is no motion, then the moments must sum to zero.  You can hypothesize the case where the structure is just about to tip and see what CW moment is needed to counteract the CCW moment generated by the horizontal force.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the diagram you drew, if you pick the point of rotation as the vertex of the $20^o$ angle, and take the length from that angle to the point of application of the force to be $L_1$ meters, then the horizontal force acts through a moment arm of ${L_1}{sin(20^o)m}$, and the total Counter-Clockwise (CCW) moment generated is ${8000L_1}{sin(20^o)N^.m}$.  The CW moment needed to balance that out is ${((160L_2)+(WL_3}{cos(20^o)N^.m}$, where $W$ is the weight you add to the structure.  $L_2$ is the length along the 20 degree slope that would just meet up with a vertical line drawn through the center of the 160N force; and $L_3$ is the distance along the slope that would just meet up with a vertical line drawn through the center of gravity of $W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For slipping, you really do need to know the static coefficient of friction, $u_s$ (or make a conservative guess).  Then you need to solve for $(160N+W)u_s &amp;gt; 8000N$.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-07-07T19:06:01.927" />
  <row Id="11999" PostTypeId="1" CreationDate="2011-07-07T19:16:17.193" Score="2" ViewCount="109" Body="&lt;p&gt;Why is it, that - if displayed in a strain/stress-diagram - the curve will at first be linear, and then change into a &quot;curved shape&quot;, where it will first decrease again, and only later increase again? Is there any tendency, for which materials this holds true and for which it doesn't? Can I identify these by certain characteristics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example graph, please consider this graph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/3aBj6.gif&quot; alt=&quot;sample stress-strain curve&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4328" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-07T19:36:25.663" LastActivityDate="2011-07-22T12:49:20.197" Title="What is the cause of strain hardening and strain softening?" Tags="&lt;material-science&gt;&lt;stress-strain&gt;" CommentCount="2" />
  <row Id="12000" PostTypeId="2" ParentId="11995" CreationDate="2011-07-07T20:08:41.017" Score="2" ViewCount="" Body="&lt;p&gt;Yes, not detecting an electron is also a fully-qualified measurement, &lt;em&gt;provided&lt;/em&gt; you make sure the measurement &lt;em&gt;would&lt;/em&gt; have detected the electron, had it been there. That condition is usually not given in real quantum mechanics experiments. Keep in mind that &quot;measuring an electron&quot; actually means you let something interact with it; it's not possible to register an electron going through a slit in the way a hidden camera can register a car passing a bridge. Pretty much the only way to make sure an electron will be detected if it's there is to use a bulky solid-state detector, which will completely absorb the electron. If you have a box with an electron and such a detector and have not measured the electron yet, you can be sure that it is &quot;still outside of the detector&quot; and you will not measure it if you do another measurement very quickly after the first one, either (you need to be &lt;em&gt;really&lt;/em&gt; quick though: the state &quot;outside of the detector&quot; is not an energy eigenstate and therefore unstable in time). If you &lt;em&gt;have&lt;/em&gt; measured the electron, you can be sure it's not out in the box anymore, simply because it's stuck in the detector!&lt;/p&gt;&#xA;" OwnerUserId="3540" LastActivityDate="2011-07-07T20:08:41.017" CommentCount="2" />
  <row Id="12001" PostTypeId="1" AcceptedAnswerId="12003" CreationDate="2011-07-07T20:33:26.050" Score="2" ViewCount="64" Body="&lt;p&gt;The book Introduction to Solid State Physics by Kittel says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;We have seen that a crystal is invariant under any translation of the form T [...]. Any local physical property of the crystal, such as the charge concentration, electron number density, or magnetic moment density is invariant under T. [...] n(r+T) = n(r)&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this the case? Even though the underlying atoms are invariant under translations in T, why should the electron density also be invariant under the same transformation? Couldn't for example if you have this 1D crystal:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;o   o   o   o   o&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is invariant under transformations $T = {...,-1,0,1,2,3,...}$. Why couldn't the electron densities be like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;o h o l o h o l o h o&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where h is a high density and l is a low density. So they would only be invariant under $T = {...,-2,0,2,4,...}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just an example: why do the local physical properties have to be invariant under T?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose this question could be phrased even more generally: if the causes of some phenomenon have a symmetry, does the phenomenon also always have the same symmetry? If yes, why, and if no, what are the conditions under which the phenomenon does have the same symmetry?&lt;/p&gt;&#xA;" OwnerUserId="4329" LastActivityDate="2011-07-07T21:08:12.690" Title="About symmetry, and about electron density in crystals in particular" Tags="&lt;solid-state-physics&gt;&lt;symmetry&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="12002" PostTypeId="2" ParentId="12001" CreationDate="2011-07-07T21:04:49.207" Score="1" ViewCount="" Body="&lt;p&gt;my guess is that this is just an approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrons in the symmetric grid of the crystal are acting more like a wave than  as cloud of point-like charges. The lowest energy modes for the electron standing-waves are those with long(er) wavelengths and are approximately uniform in the lattice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, wavefunctions that vary significantly between nearby lattice cells can be interpreted as representing high-energy modes, which might occur only far from the equilibrium. However i'm not familiar with the condensed matter specifics of crystals enough to tell you under what circumstances that approximation will hold&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-07-07T21:04:49.207" />
  <row Id="12003" PostTypeId="2" ParentId="12001" CreationDate="2011-07-07T21:05:04.760" Score="2" ViewCount="" Body="&lt;p&gt;Of course you can have configurations which do not respect the underlying symmetry of they dynamics, the issue is that those will not be eigenstates of the Hamiltonian.  Typically when talking solid state people are interested in energy bands which are eigenstates of the Hamiltonian, and these must also be eigenstates of the lattice translations. This follows from elementary operator algebra.  The non-symmetric configurations would correspond to  superpositions of different energy bands and would not be stationary in time.  &lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-07-07T21:05:04.760" CommentCount="4" />
  <row Id="12004" PostTypeId="2" ParentId="12001" CreationDate="2011-07-07T21:08:12.690" Score="1" ViewCount="" Body="&lt;p&gt;This is actually a very deep question. Until very recently scientists believed that yes, if a problem has a certain symmetry, then the solution will share it as well. But then the concept of a spontaneously broken symmetry arose:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Spontaneous_symmetry_breaking&quot; rel=&quot;nofollow&quot;&gt;Wikipedia link for more advanced discussion and examples&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can provide a very simple explanation, since I haven't learned the mathematical aspect of it yet (usually people apply it to particle physics and the Standard Model). Look at the graph on the tab of your web browser, next to &quot;solid state physics - About ...&quot; It's called a Mexican Hat potential, funnily enough, and it's important in dealing with the Higgs Mechanism. Now if you place a ball at the top of the hat, there is rotational symmetry about the axis of the hat. But this doesn't mean that this symmetry exists for all future times; as soon as the ball rolls down, it picks a direction, and we say that the symmetry is &lt;strong&gt;broken&lt;/strong&gt;. Basically if a certain &quot;direction&quot; &lt;em&gt;has&lt;/em&gt; to be picked, but it doesn't matter &lt;em&gt;which one&lt;/em&gt; is picked, spontaneous symmetry breaking has occured.&lt;/p&gt;&#xA;" OwnerUserId="1635" LastActivityDate="2011-07-07T21:08:12.690" CommentCount="2" />
  <row Id="12005" PostTypeId="1" AcceptedAnswerId="12008" CreationDate="2011-07-07T21:42:51.560" Score="1" ViewCount="105" Body="&lt;p&gt;The particle P moves along a space curve. At one instant it has velocity $v = (4i-2j-k)$ $m/s$. The magnitude of the acceleration is 8 $m/s^2$. The angle between the acceleration and the velocity vector is 20 degrees, so one can calculate that the acceleration in the direction of the velocity is 7.52.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the radius of curvature from this information? I am clueless... no formulas have been introduced, as I have seen there are throughout the Internet. The chapter that I'm working on is about three dimensional coordinate systems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of my attempts has been to try to imagine an infinitesimal change in velocity, v = r$\theta$. This implies $\frac{dv}{dt} = r\frac{d\theta}{dt}$. Could I perhaps know somehow what $\frac{d\theta}{dt}$ is?&lt;/p&gt;&#xA;" OwnerUserId="714" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-08T11:02:24.720" LastActivityDate="2011-07-09T01:36:23.137" Title="Find radius of curvature, given a velocity vector and acceleration magnitude?" Tags="&lt;homework&gt;&lt;kinematics&gt;&lt;acceleration&gt;&lt;velocity&gt;&lt;coordinate-systems&gt;" AnswerCount="3" />
  <row Id="12006" PostTypeId="2" ParentId="11882" CreationDate="2011-07-07T21:44:11.870" Score="0" ViewCount="" Body="&lt;p&gt;I would stay in my car for reasons of comfort.  The car will attract a little more lightning than you would if you were standing up; but it might (or might not) protect you from said lightning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lightning strike locations are psuedo-randomly dispersed.  If a thunder storm develops over, say, a 100 square kilometer PERFECTLY FLAT area, and throws out 100 strikes, then the expected strike coverage is 1 strike per sq. kilometer.  Given that a strike will occur in that square kilometer, the chances of it striking any given 1m x 1m spot is (1x1)/(1000x1000) = 1E-6 = 0.0001%.  Literally and exactly one chance in a million.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the earth is not flat.  It has been shown that taller objects attract more lightning.  Taller conductive objects attract lighnting in direct proportion to the area they &quot;protect.&quot;  For a long slim object, that area is the area covered by a cone, which has its apex at the top of the object and which slants out at a $45^o$ angle in all directions.  So, in this example, a 10m tall metal light pole would protect an area of ${(Pi)}{10^2} = 314 m^2$.  It chances of getting struck by lightning in this storm would be 314 in a million, or roughly 0.3 in a thousand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything which is fully within the volume of the cone of this light pole is considered protected (though I would not stand RIGHT NEXT to the light pole).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 10m tall and 100m x 100m warehouse would &quot;protect&quot; an area of about 120mx120m = $14400 m^2$.  Its chances of taking that sole lighning bolt are 14400 in a million, or about 14.4 chances in a thousand: 1.44%.  Now we are talking about real chances of taking a hit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's go back to you in your car, out in flatland.  Assume you are 1.6m tall and your car is about 1.4 m tall x 2m wide x 5m long.  For purposes of a lightning strike, you are considered a conductive object.  If you are out standing in the rain, like a dummy, you &quot;protect&quot; an area of about $8m^2$.  Your car &quot;protects&quot; an area of about 4.8m x 7.8m = about $22m^2$.  So, you have about 3 times greater chance of getting hit by lightning in your car.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If hit by lightning out in the open, you are obviously toast.  If you are in the car... well... lightning does weird things.  It is better than a direct hit to your cranium; but I would definitely not say that being in a car getting hit by lightning is safe.  Let's say the path from the roof of your wet car to ground is 1 Ohm, and the path tfrom the roof of your car to ground THROUGH YOU is about 1000000 Ohms (1M Ohm, includes air gap, since your dry-skin body only provides about 20K-30K resistance).  Say the lightning strike is 30000 Amps.  About 29999.97 amps go through the car structure.  About 0.03 amps = 30mA go through you... and you die.  Oh yeah, you might not be charred and smoking, but dead nonetheless.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I had to take a choice between a 35% chance of getting hit in the head by lighning while standing out in a field, or a 100% chance of my car getting hit while I was in it, I think I would take the standing out in the field option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best protection would be to lie in a ditch in the field.  But that gets you very wet and uncomfortable.  (Can you imagine flying down the freeway with your family, and you run into a thunderstorm; and you say, &quot;Hey, honey, lets pull over and go lay out in the ditch in the pouring rain until this is over.&quot;?)  I like to stay dry, so would take my 22 out of a million shot in the car.  Of course, if I was REALLY concerned, I would try to greatly lower those odds by finding a parking spot about 4m out from the base of that 10m tall light pole (or tree or whatever).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:  I miscalculated the area protected by a standing person... fixed.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-07-08T19:00:14.093" LastActivityDate="2011-07-08T19:00:14.093" CommentCount="2" />
  <row Id="12008" PostTypeId="2" ParentId="12005" CreationDate="2011-07-07T22:12:51.190" Score="2" ViewCount="" Body="&lt;p&gt;Radius of curvature is governed by $a = v^2/r$.  The radius of curvature thus calculated is good at that instant only, since 'v' will continue to increase; and, if 'a' remains constant, change 'r'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 'a' in the equation is the component of total acceleration which is normal to the velocity vector, or $sin(20^o)(8m/s^2)$&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-07-07T22:12:51.190" />
  <row Id="12009" PostTypeId="2" ParentId="12005" CreationDate="2011-07-07T22:48:12.453" Score="0" ViewCount="" Body="&lt;p&gt;The speed vector you gave is constant so there can't be any acceleration. However, for $|\textbf{v}|=u=const$, as Vintage said,  $\textbf{a}=\frac{u^2}{R}\textbf{N}$, where $\textbf{N}$ is the unit vector normal  to the curve. So $|\textbf{a}|=\frac{u^2}{R}|\textbf{N}|\Leftrightarrow R=\frac{u^2}{|\textbf{a}|}$. &lt;/p&gt;&#xA;" OwnerUserId="3066" LastEditorUserId="3066" LastEditorDisplayName="" LastEditDate="2011-07-07T22:55:43.317" LastActivityDate="2011-07-07T22:55:43.317" />
  <row Id="12010" PostTypeId="2" ParentId="11670" CreationDate="2011-07-07T23:30:57.223" Score="1" ViewCount="" Body="&lt;p&gt;It was Henri Becquerel who discovered radioactivity. Ernest Rutherford proved that what was happening was the disintegration and transmutation of atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radioactivity was discovered in Uranium, but the Curies discovered a mineral that was more radiocative.  They correctly deduced that it must contain a new element.  They then used purely chemical methods to separate the various elements, keeping track of which residuum had kept the radioactivity.  It's not my area, but I think that all of the methods were standard apart from using the radioactivity to trace.  And there was an enormous amount of work to be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They actually found two new elements, radium and polonium.  In a sense they were lucky, because they could have been chasing a radioactive isotope of a known element.  Isotopes were only recognised later.&lt;/p&gt;&#xA;" OwnerUserId="4330" LastActivityDate="2011-07-07T23:30:57.223" />
  <row Id="12011" PostTypeId="1" CreationDate="2011-07-08T01:11:52.430" Score="2" ViewCount="64" Body="&lt;p&gt;I'm trying to understand Newton's Shell Theorem (Third)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Shell_theorem&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Shell_theorem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this applies to a sphere of &lt;em&gt;constant density&lt;/em&gt;. How is this formulated for sphere of &lt;em&gt;varying density&lt;/em&gt;, e.g., a ball of gas bound together by gravity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, this requires another question: how does the density of ball of gas bound by gravity diminish with radius? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as DMCKEE pointed out &quot;You can't answer the problem ignoring thermodynamics&quot;, so I've removed that proviso.&lt;/p&gt;&#xA;" OwnerUserId="3588" LastEditorUserId="3588" LastEditorDisplayName="" LastEditDate="2011-07-08T02:23:16.660" LastActivityDate="2011-07-08T18:28:45.037" Title="Non-Linear Density Shell Problem" Tags="&lt;classical-mechanics&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="12012" PostTypeId="1" AcceptedAnswerId="12015" CreationDate="2011-07-08T01:26:11.123" Score="8" ViewCount="211" Body="&lt;p&gt;As I've stated in a prior question of mine, I am a mathematician with very little knowledge of Physics, and I ask here things I'm curious about/things that will help me learn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This falls into the category of things I'm curious about. Have people considered whether spacetime is simply connected? Similarly, one can ask if it contractible, what its Betti numbers are, its Euler characteristic and so forth. What would be the physical significance of it being non-simply-connected?&lt;/p&gt;&#xA;" OwnerUserId="4249" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-08T03:17:59.873" LastActivityDate="2011-07-08T20:41:41.453" Title="Is spacetime simply connected?" Tags="&lt;mathematical-physics&gt;&lt;spacetime&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="12013" PostTypeId="1" CreationDate="2011-07-08T03:17:28.687" Score="3" ViewCount="149" Body="&lt;p&gt;As always the caveat is that I am a mathematician with very little knowledge of physics. I've started my quest for knowledge in this field, but am very very far from having a good grasp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General relativity assumes that spacetime is a $4$-dimensional manifold. Ultimately, I know that physicists deal with Calabi-Yau manifolds (in string theory, I imagine). These are holomorphic complex manifolds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How, and when, is this change of hypotheses justified? Is one way to view the general relativistic space time that spacetime is a variety of dimension $2$ over $\mathbb{C}$? Or is this change done later? Why is it ever done?&lt;/p&gt;&#xA;" OwnerUserId="4249" LastActivityDate="2011-07-08T08:41:02.753" Title="How is the complexification of spacetime justified?" Tags="&lt;general-relativity&gt;&lt;string-theory&gt;&lt;spacetime&gt;&lt;calabi-yau&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="12014" PostTypeId="2" ParentId="12011" CreationDate="2011-07-08T04:07:47.927" Score="3" ViewCount="" Body="&lt;p&gt;The shell theorem applies to spherically-symmetric mass distributions.  If you have a compressible fluid so that the density varies with depth, you can still apply the shell theorem (unless there is something else going on, for example net angular momentum that breaks the spherical symmetry).  Specifically, the gravitational field is the same as that of a point mass at the center of the distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a demonstration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton's law for the gravitational field of a point mass is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_g/m = GM/r^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $F_g/m$ the acceleration of a test particle and M the mass of the gravitating body.  This is linear in M, meaning that we can take two mass distributions and superimpose them, then find the gravitational field of this new mass distribution by superimposing the two associated individual gravitational fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the shell theorem holds for constant density spheres.  Take two spheres, one slightly larger than the other, so their radii are $r$ and $r+dr$.  Give the larger sphere density $\rho$ and the smaller one density $-\rho$.  Superimpose them.  You get a thin shell of radius $r$ and thickness $dr$.  The shell is empty in the middle because the positive and negative density distributions are canceling there.  (It doesn't matter that there is no such thing as negative mass - the shell theorem is purely mathematical and so holds for negative numbers, and the final distribution we're dealing with has no negative mass; it was just a calculation trick.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By superimposing the two gravitational fields of the positive and negative density distributions, we see that the gravitational field of this shell is the same as that of a point mass at the center of the shell with the mass $4\pi r^2 dr \rho$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shell theorem thus holds for these thin shells.  A distribution with varying density at different radii is simply a lot of such shells, so the shell theorem holds for arbitrary $\rho(r)$ as long as $\rho$ is a function of $r$ alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the change in density as a function of radius, we need to balance the hydrostatic pressure with the gravitational attraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a little piece of the mass distribution with volume $dV$.  It is pulled down by a gravitational force&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_g = \frac{GM(r) dm}{r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dm$ is $\rho dV$ and $M(r)$ is the mass located closer to the center of the distribution than the point in question, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M(r) = \int_0^r 4\pi x^2 \rho(x) dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is balanced by the force from pressure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_p = \nabla P dV$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We assume there is a relationship between pressure and density so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla P = \frac{\partial P}{\partial r} = \frac{d P}{d\rho}\frac{d \rho}{dr}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equating the gravitational and pressure forces gives, after some algebra&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4\pi G r^2\rho = \frac{d}{dr} \left( \frac{r^2}{\rho} \frac{dP}{d\rho} \frac{d\rho}{dr} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution depends on $\rho(P)$, and is subject to the boundary condition $P(R) = 0$ with $R$ the radius of the distribution, as well as the condition $M(R) = M$, the total mass.  We will probably find $R = \infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, once we specify a relationship between $\rho$ and $P$ based on the thermodynamics of the system involved, we can integrate the differential equation to find $\rho(r)$.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-07-08T18:28:45.037" LastActivityDate="2011-07-08T18:28:45.037" CommentCount="0" />
  <row Id="12015" PostTypeId="2" ParentId="12012" CreationDate="2011-07-08T05:32:13.097" Score="4" ViewCount="" Body="&lt;p&gt;I suppose there are many aspects to look at this from, anna v mentioned how Calabi-Yao manifolds in string theory (might?) have lots of holes, I'll approach the question from a purely General Relativity perspective as far as global topology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solutions in the Einstein Equations themselves do not reveal anything about global topology except in very specific cases (most notably in 2 (spacial dimensions) + 1 (time dimension) where the theory becomes completely topological). A metric by itself doesn't necessarily place limits on the topology of a manifold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond this, there is one theorem of general relativity, called the &lt;a href=&quot;http://arxiv.org/pdf/gr-qc/9305017v2&quot; rel=&quot;nofollow&quot;&gt;Topological Censorship Hypothesis&lt;/a&gt; that essentially states that any topological deviation from simply connected will quickly collapse, resulting in a simply connected surface. This work assumes an asymptotically flat space-time, which is generally the accepted model (as shown by supernova redshift research and things of that nature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another aspect of this question is the universe is usually considered homogenous and isotropic in all directions, topological defects would mean this wouldn't be true. Although that really isn't a convincing answer per say...&lt;/p&gt;&#xA;" OwnerUserId="3018" LastActivityDate="2011-07-08T05:32:13.097" CommentCount="4" />
  <row Id="12016" PostTypeId="1" CreationDate="2011-07-08T06:48:22.377" Score="2" ViewCount="49" Body="&lt;p&gt;If we have an array of telescopes attached one after another, would the resultant magnification be multiplied?Also would such a contraption be feasible to make telescopes with amazing magnification?&lt;/p&gt;&#xA;" OwnerUserId="4335" LastActivityDate="2011-07-08T07:30:10.437" Title="Magnification multiplication using telescope arrays?" Tags="&lt;optics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12017" PostTypeId="2" ParentId="12016" CreationDate="2011-07-08T07:18:09.730" Score="3" ViewCount="" Body="&lt;p&gt;One of the first things I learnt studying astonomy is that &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnification&quot; rel=&quot;nofollow&quot;&gt;magnification&lt;/a&gt; of your optics is not as important as &lt;a href=&quot;http://en.wikipedia.org/wiki/Angular_resolution&quot; rel=&quot;nofollow&quot;&gt;resolution&lt;/a&gt; of it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The telescope allows you to see tiny details of celectial objects for three reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It magnifies stuff.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It gives picture with higher resolution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It redirects more light into your eye -- so you can see faint objects.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Magnification only will not do the job. Think of a dark computer raster picture. Magnification will give you almost no extra details (&lt;a href=&quot;http://en.wikipedia.org/wiki/Raster_graphics&quot; rel=&quot;nofollow&quot;&gt;for example&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And stacking telescopes doesn't help with #2 and #3. I think that it would actually have negative effect. &lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-07-08T07:18:09.730" CommentCount="1" />
  <row Id="12018" PostTypeId="2" ParentId="12016" CreationDate="2011-07-08T07:30:10.437" Score="3" ViewCount="" Body="&lt;p&gt;If you head down to your local telescope shop and check out their inventory, you might see two telecopes.  One is a refractive telescope with a 2 inch aperture, and it has its magnification listed as 1 million. Another is a reflective telescope with a 6 inch aperture and a magnification of 200.  Which will allow you to see planets better?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Telescopes do two things: collect light and spread it out (magnify it).  How much light they collect is limited by the size of the aperture. The magnification is determined by the optics. Note that since magnification spreads light out, it makes the light dimmer. The reason things usually don't look dim in a telescope is because the light has been concentrated from an aperture that's several times larger than the pupil of a human eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that because of diffraction effects, the amount of detail you can resolve is also limited by the size of the aperture (sometimes; other things can obscure detail more than this, but this is a big one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, you would indeed magnify the images more and more with each telescope, but the reason telescopes are useful is not the magnification so much as the light collection, and that was accomplished by the first telescope.  You would be better off spreading the telescopes out to get the full light collection power of all of their apertures, using cameras instead of eyes, and combining the pictures in a computer.  This is what some actual observatories do.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2011-07-08T07:30:10.437" />
  <row Id="12019" PostTypeId="2" ParentId="12011" CreationDate="2011-07-08T07:43:46.210" Score="0" ViewCount="" Body="&lt;p&gt;A really simple way of thinking about the shell theorem is that it applies to an infinitesimally thin spherical shell of constant surface density. Then, any spherically symmetric ball of stuff (even if it varies in density with distance from the center) can be built of these shells (with different densities for each shell).&lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2011-07-08T07:43:46.210" />
  <row Id="12020" PostTypeId="2" ParentId="12013" CreationDate="2011-07-08T07:45:37.317" Score="4" ViewCount="" Body="&lt;p&gt;The Calabi-Yau manifolds used as compact dimensions in string theory are ordinary real 6-dimensional (or real 8-dimensional, in the case of an F-theory description) manifolds and you don't have to distinguish them from other manifolds over real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, they're special 6- or 8-dimensional manifolds that may also be described as 3-dimensional or 4-dimensional complex manifolds. That's because there are natural complex coordinates and many things are holomorphic etc. In particular, the monodromy group isn't the generic $SO(6)$ or $SO(8)$ but the subgroups of $U(3)$ or $U(4)$ which preserve the merging of the pairs of real dimensions into the complex ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complex manifolds in this sense are just special examples of real manifolds of a doubled dimension. They're special examples with special properties - in particular, such complex manifolds preserve supersymmetry under some additional assumptions, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Independently of that, one may also complexify spacetime dimensions by analytically continuing in them. This is much more natural in the case of momentum spaces - and the continuation to complex values of energies and momenta is possible and useful because the scattering amplitudes are analytic functions of the momenta and energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The continuation to imaginary values of time is useful in thermodynamics - a thermal ensemble is obtained from a periodic Euclidean (imaginary) time - and has new special features in quantum gravity - the black hole solutions may be Euclideanized and these solutions nicely shrink and become regular near the event horizons (like a cigar).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question of complexification has many facets and it seems that you are a bit confused what you're exactly asking about. Different things have different justifications. But the very general idea of yours that one has to have &quot;justification&quot; is wrong. In physics, the justification is that something works. On the contrary, you have to have a justification if you claim that there is a problem with any of these methods and procedures. There is no problem with complexification of space, time, momentum space, or other things. Because you haven't really proposed any conceivable problem, it's not possible to answer your question in more detail. One can't be solving a problem if there's no problem to start with.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-07-08T07:45:37.317" />
  <row Id="12021" PostTypeId="2" ParentId="12013" CreationDate="2011-07-08T08:41:02.753" Score="0" ViewCount="" Body="&lt;p&gt;Only for BPS configurations in supersymmetric theories.&lt;/p&gt;&#xA;" OwnerUserId="1039" LastActivityDate="2011-07-08T08:41:02.753" CommentCount="1" />
  <row Id="12022" PostTypeId="1" AcceptedAnswerId="12023" CreationDate="2011-07-08T08:51:39.407" Score="5" ViewCount="152" Body="&lt;p&gt;I've been trying to gain some intuition about Virasoro Algebras, but have failed so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Mathematical Definition seems to be clear (as found in &lt;a href=&quot;http://en.wikipedia.org/wiki/Virasoro_algebra&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Virasoro_algebra&lt;/a&gt;). I just can't seem to gain some intuition about it. As a central extension to Witt Algebras, I was hoping that there has to be some geometric interpretation, as I can imagine Witt Algebras rather well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone has some nice Geometric or Visual Interpretation of Virasoro Algebra, I'd greatly appreciate it! &lt;/p&gt;&#xA;" OwnerUserId="4337" LastActivityDate="2011-07-08T09:01:24.623" Title="Geometric/Visual Interpretation of Virasoro Algebra" Tags="&lt;string-theory&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="12023" PostTypeId="2" ParentId="12022" CreationDate="2011-07-08T09:01:24.623" Score="1" ViewCount="" Body="&lt;p&gt;The simplest visual representation of the Lie group associated with the Virasoro (Lie) algebra is the group of reparametrizations of a circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that $\sigma$ is a periodic variable with the periodicity $2\pi$. An infinitesimal diffeomorphism is specified by a periodic function $\Delta \sigma(\sigma)$ with the periodicity $2\pi$. So the generators of the reparameterizations may be written as $f(\sigma)\partial / \partial \sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possible functions $f(\sigma)$ may be expanded to the Fourier series, so a natural basis of the generators of the reparametrizations of the circle are&#xA;$$ L_m = i \exp(im\sigma) \frac{\partial}{\partial \sigma} $$&#xA;As an exercise, calculate that the commutator $[L_m,L_n]$ is what it should be according to the Virasoro algebra, namely $(m-n)L_{m+n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Virasoro algebra for a closed string has two copies of the algebra above - and for the open string, it's only one copy but it's different than the &quot;holomorphic&quot; derivatives I used above. There are various related ways to represent the algebra but the reparameterizations of the circle are the simplest example.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-07-08T09:01:24.623" CommentCount="5" />
  <row Id="12024" PostTypeId="1" CreationDate="2011-07-08T09:10:30.633" Score="1" ViewCount="28" Body="&lt;p&gt;I'm trying to trigger a human touch in a capacitive touchscreen without the 'human' part. I've noticed that the negative pole of a battery triggers it if you are currently holding it. Or if you use the edge - why only the edge? - of a wire touching the battery and your hand. I also attached a wire in a metal bottle and connected the battery to it's end (with the battery touching the screen) and it also worked - but sometimes it also didn't. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: How can I trigger a touch on the screen without the human touch, using objects from the hardware store?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have almost no physics knowledge for this, so you guys have to tell me exactly what to do (which material, where to buy and etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="4338" LastActivityDate="2011-07-08T15:51:55.500" ClosedDate="2011-07-09T01:24:21.590" Title="Changing the capacitance of an object to trigger a capacitive touchscreen" Tags="&lt;experimental-physics&gt;&lt;technology&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12025" PostTypeId="1" AcceptedAnswerId="12164" CreationDate="2011-07-08T09:29:55.020" Score="4" ViewCount="82" Body="&lt;p&gt;I am starting to build multistage Peltier cooler at the moment, and it should be able to reach -100C at least (but if I fail I can always get boring &lt;a href=&quot;http://en.wikipedia.org/wiki/Liquid_nitrogen&quot; rel=&quot;nofollow&quot;&gt;LN2&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing some experiments with superconductors would be great, but where can I get superconductor samples or which one is easier to make on my own (have access to all chemicals &amp;amp; lab). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Especially awesome would be to be able to make thin superconductor film.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-11T22:50:07.523" LastActivityDate="2011-07-11T22:50:07.523" Title="What are the easiest to get/make LN2 superconductors?" Tags="&lt;thermodynamics&gt;&lt;experimental-physics&gt;&lt;superconductivity&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="12026" PostTypeId="5" CreationDate="2011-07-08T13:20:36.227" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-07-08T13:20:36.227" LastActivityDate="2011-07-08T13:20:36.227" />
  <row Id="12027" PostTypeId="4" CreationDate="2011-07-08T13:20:36.227" Score="0" ViewCount="" Body="Two Hydrogen atoms covalently bonded to one Oxygen atom. One of the more common compounds on the surface of the earth." OwnerUserId="-1" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-08T17:23:16.840" LastActivityDate="2011-07-08T17:23:16.840" />
  <row Id="12028" PostTypeId="5" CreationDate="2011-07-08T13:24:41.447" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-07-08T13:24:41.447" LastActivityDate="2011-07-08T13:24:41.447" />
  <row Id="12029" PostTypeId="4" CreationDate="2011-07-08T13:24:41.447" Score="0" ViewCount="" Body="Two related pheonomena: the tendency of like charges to repel and unlike charges to attract, and the tendency of parallel currents to attract and antiparallel currents to repel. Classically governed by Maxwell's equations and the Lorentz force law." OwnerUserId="-1" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-08T17:23:14.140" LastActivityDate="2011-07-08T17:23:14.140" />
  <row Id="12030" PostTypeId="1" CreationDate="2011-07-08T13:33:14.653" Score="1" ViewCount="85" Body="&lt;p&gt;What is the equation of motion for a single scalar field Lagrangian density in which the potential explicitly depends on time? For example: $$U(\phi,t)=\frac{1}{2}\phi^2 - \frac{1}{3} e^{t/T}\phi^3 + \frac{1}{8}\phi^4$$ where $T$ is a constant.&lt;/p&gt;&#xA;" OwnerUserId="4361" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-08T14:03:00.073" LastActivityDate="2011-07-09T00:39:39.533" Title="Equation of motion for explicit time dependent potential" Tags="&lt;quantum-field-theory&gt;" AnswerCount="1" />
  <row Id="12031" PostTypeId="1" CreationDate="2011-07-08T13:34:05.950" Score="1" ViewCount="69" Body="&lt;p&gt;There's a caveat, which is often ignored, to the &quot;easy&quot; equation for parallel plate capacitors C = epsilon * A / d, namely that d must be much smaller than the dimensions of the parallel plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an equation that works for large d? I tried finding one and could not. (These two papers talk about fringing fields for disc-shape plates but don't seem to have a valid equation for d -&gt; infinity: &lt;a href=&quot;http://www.santarosa.edu/~yataiiya/UNDER_GRAD_RESEARCH/Fringe%20Field%20of%20Parallel%20Plate%20Capacitor.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.santarosa.edu/~yataiiya/UNDER_GRAD_RESEARCH/Fringe%20Field%20of%20Parallel%20Plate%20Capacitor.pdf&lt;/a&gt; and &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.167.3361&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.167.3361&amp;amp;rep=rep1&amp;amp;type=pdf&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My hand-waving intuition is that as d -&gt; infinity, C should decrease to a constant value (which is the case for two spheres separated by a very large distance, where C = 4*pi*e0/(1/R1 + 1/R2) ), because at large distances from each plate, the electric field goes as 1/R, so the voltage line integral from one plate to the other will be a fixed constant proportional to charge Q.&lt;/p&gt;&#xA;" OwnerUserId="4358" LastActivityDate="2011-07-08T17:59:27.307" Title="capacitance between widely-separated parallel plates" Tags="&lt;electrostatics&gt;" AnswerCount="1" />
  <row Id="12032" PostTypeId="2" ParentId="12025" CreationDate="2011-07-08T14:03:58.667" Score="1" ViewCount="" Body="&lt;p&gt;Making large, good, high-Tc superconductor crystals is usually difficult. Also thin films are normally grown in specialised setups (i.e. Pulsed laser deposition, molecular beam epitaxy labs. etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However blocks of YtBaCuO are available commercially. This transition at around 90K which is above the 77K boiling point of liquid nitrogen.&lt;/p&gt;&#xA;" OwnerUserId="1072" LastActivityDate="2011-07-08T14:03:58.667" />
  <row Id="12033" PostTypeId="2" ParentId="12030" CreationDate="2011-07-08T14:13:49.893" Score="1" ViewCount="" Body="&lt;p&gt;It's just a Klein-Gordon equation with a RHS.&#xA;Explicitly, the E-L equation for a scalar field is&#xA;$$\partial_{\mu} {\partial {\mathcal L} \over \partial \phi_{,\mu}} - {\partial {\mathcal L} \over \partial \phi} = 0$$&#xA;so for your potential we have&#xA;$$\square \phi + \phi - e^{t/T} \phi^2 + {1 \over 2} \phi^3 = 0$$&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-08T14:13:49.893" CommentCount="3" />
  <row Id="12034" PostTypeId="1" AcceptedAnswerId="12493" CreationDate="2011-07-08T14:22:52.583" Score="9" ViewCount="141" Body="&lt;p&gt;In order to calculate the cross-section of an interaction process the following formula is often used for first approximations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\sigma = \frac {2\pi} {\hbar\,v_i} \left| M_{fi}\right|^2\varrho\left(E_f\right)\,V&#xD;&#xA;$$&#xA;$$&#xD;&#xA;M_{fi} = \langle\psi_f|H_{int}|\psi_i\rangle&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very often plane waves are assumed for the final state and therefore the density of states is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\varrho\left(E_f\right) = \frac{\mathrm d n\left(E_f\right)}{\mathrm d E_f} = \frac{4\pi {p_f}^2}{\left(2\pi\hbar\right)^3}\frac V {v_f}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the derivation of this equation in the context of the non relativistic Schrödinger equation. But why can I continue to use this formula in the relativistic limit: $v_i, v_f \to c\,,\quad p_f\approx E_f/c$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very often books simply use this equation with matrix element derived from some relativistic theory, e.g. coupling factors and propagators from the Dirac equation or Electroweak interaction. How is this justified?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Specific concerns:&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Is Fermi's golden rule still valid in the relativistic limit?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Doesn't the density of final states has to be adapted in the relativistic limit?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1096" LastEditorUserId="1096" LastEditorDisplayName="" LastEditDate="2011-07-08T20:15:25.127" LastActivityDate="2011-07-20T05:40:37.870" Title="Cross-section in relativistic limit: Fermi's golden rule still valid?" Tags="&lt;particle-physics&gt;&lt;relativity&gt;&lt;cross-section&gt;&lt;fermis-golden-rule&gt;" AnswerCount="1" FavoriteCount="2" />
  <row Id="12035" PostTypeId="1" CreationDate="2011-07-08T15:07:55.853" Score="3" ViewCount="150" Body="&lt;p&gt;I mean, what is happening at a microscopic level to cause this behavior? Here's what I got from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;On &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Reflection_%28physics%29&amp;amp;oldid=438178514#Reflection_of_light&quot; rel=&quot;nofollow&quot;&gt;Reflection (physics)#Reflection of light&lt;/a&gt; it says that &quot;&lt;em&gt;solving Maxwell's equations for a light ray striking a boundary allows the derivation of the Fresnel equations, which can be used to predict how much of the light reflected, and how much is refracted in a given situation.&lt;/em&gt;&quot;&lt;/li&gt;&#xA;&lt;li&gt;On &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Specular_reflection&amp;amp;oldid=434277452#Explanation&quot; rel=&quot;nofollow&quot;&gt;Specular reflection#Explanation&lt;/a&gt; it says that &quot;&lt;em&gt;for most interfaces between materials, the fraction of the light that is reflected increases with increasing angle of incidence&lt;/em&gt; $\theta_i$&quot; (but doesn't explain why)&lt;/li&gt;&#xA;&lt;li&gt;Finally, on &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Reflection_coefficient&amp;amp;oldid=419455415#Optics&quot; rel=&quot;nofollow&quot;&gt;Reflection coefficient#Optics&lt;/a&gt;, it says basically nothing, redirecting the reader to the Fresnel equations article.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What I'm trying to find, instead, is a basic level explanation that could provide an intuition on why this happens, rather than analytic formulations or equations to calculate these values. Is there a good analogy that explains this behavior?&lt;/p&gt;&#xA;" OwnerUserId="4283" LastActivityDate="2011-07-09T15:35:19.850" Title="Why does light reflect more intensely when it hits a surface at a large angle?" Tags="&lt;optics&gt;&lt;reflection&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" />
  <row Id="12036" PostTypeId="1" CreationDate="2011-07-08T15:20:37.807" Score="2" ViewCount="118" Body="&lt;p&gt;Can windmills allow us to consume (and, eventually, over-consume) the wind as a natural resource somewhat in the same manner that we are over-consuming many other natural resources?&lt;/p&gt;&#xA;" OwnerUserId="4384" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-12T15:38:45.823" LastActivityDate="2011-07-12T15:38:45.823" Title="Is it possible that windmills can consume the wind?" Tags="&lt;energy&gt;&lt;renewable-energy&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="12037" PostTypeId="2" ParentId="12036" CreationDate="2011-07-08T15:28:15.787" Score="2" ViewCount="" Body="&lt;p&gt;Wind is caused by differences in air pressure, where air flows from high pressure areas to low pressure ones. To &quot;consume&quot; all the wind would require air pressure to reach equilibrium, something that won't happen since the effects of solar heating and cooling from bodies of water constantly introduces differences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only effect you might observe if you saturate the environment with windmills is that the upwind ones will spoil it for the down-wind ones, just as a sailing ship can spoil the wind for any directly behind it.&lt;/p&gt;&#xA;" OwnerUserId="4383" LastActivityDate="2011-07-08T15:28:15.787" />
  <row Id="12038" PostTypeId="1" CreationDate="2011-07-08T15:33:06.703" Score="1" ViewCount="39" Body="&lt;p&gt;&lt;strong&gt;Fact:&lt;/strong&gt; We all know that during the day &lt;strong&gt;concrete&lt;/strong&gt; absorbs heat and &lt;strong&gt;releases it during the night&lt;/strong&gt;, making urban areas hotter than rural areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I observed that after sunset the ambient temperature is going down until late at night when   it actually starts to feel hot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would assume that soon after sunset the temperature should decrease slowly until sunrise, but that's not the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I would like to understand which phenomena take place from sunset to sunrise in urban areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take into account that I live in Athens (Greece) where there is a huge urban area and the sea is nearby. &lt;/p&gt;&#xA;" OwnerUserId="4373" LastActivityDate="2011-07-08T17:30:41.217" Title="Is there a timeline in urban area night temperature?" Tags="&lt;thermodynamics&gt;&lt;temperature&gt;" AnswerCount="1" />
  <row Id="12039" PostTypeId="2" ParentId="12035" CreationDate="2011-07-08T15:43:46.553" Score="-1" ViewCount="" Body="&lt;p&gt;Perhaps it is because in the case of grazing incidence a lot more scatterer are in the path of the photon. At least this is why I think we observe X-ray reflection at all and were able to build Chandra and Rosat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I would like to hear a better explanation as well.&lt;/p&gt;&#xA;" OwnerUserId="4306" LastActivityDate="2011-07-08T15:43:46.553" />
  <row Id="12040" PostTypeId="2" ParentId="12038" CreationDate="2011-07-08T15:46:11.700" Score="2" ViewCount="" Body="&lt;p&gt;The phenomena that take place is one aspect of the &lt;a href=&quot;http://dx.doi.org/10.1002/qj.49710845502&quot; rel=&quot;nofollow&quot;&gt;Urban Heat Island effect&lt;/a&gt;. The large thermal mass (aka thermal inertia or heat capacity) of the urban fabric heats up from solar radiation during the day. Once the night-time  air temperatures drop below the temperature of the buildings, they give up some of that heat again, heating the urban air. The rate at which this happens, and the rate at which the heat is kept in the city, &lt;a href=&quot;http://dx.doi.org/10.1016/0004-6981%2873%2990140-6&quot; rel=&quot;nofollow&quot;&gt;will depend on wind speeds, the size of the city&lt;/a&gt; and the urban layout, amongst other things. So the place where you measure temperature, will affect the pattern you see, depending on how close to a building the measurement is, where in the city it is, and what the wind patterns are.&lt;/p&gt;&#xA;" OwnerUserId="4066" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-08T17:30:41.217" LastActivityDate="2011-07-08T17:30:41.217" CommentCount="1" />
  <row Id="12041" PostTypeId="2" ParentId="12036" CreationDate="2011-07-08T15:50:52.617" Score="1" ViewCount="" Body="&lt;p&gt;No, because as long as the sun shines, there'll be wind. That's what's meant by calling it a renewable resource.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any wind moving now, will dissipate to low-grade heat within a few hours or days anyway - the winds of last year have all gone into heat death, and have been replaced by new winds, caused by new temperature differences caused by different solar radiation across the earth, and differential rates of heating between land and sea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a wind turbine doesn't really change the fate of the energy in any given wind - either the wind will turn to low-grade heat through friction; or it will get converted to electricity, consumed as electricity, and that end electricity use will, one way or another, almost always end up as low-grade heat, typically sooner rather than later.&lt;/p&gt;&#xA;" OwnerUserId="4066" LastActivityDate="2011-07-08T15:50:52.617" CommentCount="7" />
  <row Id="12042" PostTypeId="2" ParentId="12024" CreationDate="2011-07-08T15:51:55.500" Score="0" ViewCount="" Body="&lt;p&gt;A tightly rolled anti static bag (like the one memory chips or electronics come in) will work just fine. Easy to keep near the unit.&lt;/p&gt;&#xA;" OwnerUserId="4390" LastActivityDate="2011-07-08T15:51:55.500" />
  <row Id="12043" PostTypeId="2" ParentId="12035" CreationDate="2011-07-08T17:40:01.430" Score="1" ViewCount="" Body="&lt;p&gt;Well, imagine shooting a piece of glass with a gun.  If you shoot at a glancing angle, it is more likely to ricochet off the glass without damaging it.  This is because the impulse required to reflect the bullet is smaller for shallow grazing angles, as most of the bullets momentum is parallel to the interface.  Now obviously the physics of reflecting a photon is very different than reflecting a bullet, but the analogy is that the impulse required to reflect the photon becomes smaller, and it becomes relatively &quot;easier&quot; for the medium to supply that small impulse than to let the photon go through.  &lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-07-08T17:40:01.430" CommentCount="1" />
  <row Id="12044" PostTypeId="2" ParentId="12035" CreationDate="2011-07-08T17:58:04.720" Score="0" ViewCount="" Body="&lt;p&gt;user1631's analogy holds.  You can think of the reflection as the surface generating an electromagnetic wave which cancels the incoming wave on the non-reflecting side and reflects the wave on the reflecting side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a high angle of incidence this wave needs to have a high intensity.  At a low angle of incidence, the wave needs a lower intensity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At higher intensities, more of this &quot;reflecting wave&quot; is going to be transmitted through the surface or turned into wasted energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, at low angles, you only need to change the direction of the EM vector a little bit.  At high angles, it needs to change direction completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/10301/how-does-reflection-work&quot;&gt;How does reflection work?&lt;/a&gt; might help.&lt;/p&gt;&#xA;" OwnerUserId="4403" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-07-08T23:02:33.693" LastActivityDate="2011-07-08T23:02:33.693" CommentCount="1" />
  <row Id="12045" PostTypeId="2" ParentId="12031" CreationDate="2011-07-08T17:59:27.307" Score="2" ViewCount="" Body="&lt;p&gt;There is, of course, a solution for the most simple case of the circular disk parallel plate capacitor. See, for example, the article by G.T. Carlson and B.L. Illman, &lt;a href=&quot;http://ajp.aapt.org/resource/1/ajpias/v62/i12/p1099_s1&quot; rel=&quot;nofollow&quot;&gt;Am. J. Phys. 62, 1099 (1994)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4237" LastActivityDate="2011-07-08T17:59:27.307" CommentCount="2" />
  <row Id="12046" PostTypeId="1" AcceptedAnswerId="12047" CreationDate="2011-07-08T18:27:12.603" Score="3" ViewCount="123" Body="&lt;p&gt;&lt;strong&gt;Observation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I know that all computer screens are able to project many different colors by varying how they display the RGB (Red, Green, Blue) pixels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the difference between say, yellow light (575 nm) from sunlight vs. the same yellow light that's displayed on the computer screen? Aren't they different? Is our brains mixing together the RGB lights together and its being interpreted as yellow light or is yellow light the same (575 nm) light that comes from sunlight actually hits our eyes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Relationship between Fourier Series and light?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) So I know that in Fourier Series, you have basis functions, sines and cosines that make up any other function depending on the coefficients that are in front of the sines and cosines terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I know that RGB colors on a screen are the &quot;basis&quot; for all the other colors that a computer screen makes up and the intensity of each of the RGB are like the &quot;coefficients&quot;. So, am I making the correct relationship here?&lt;/p&gt;&#xA;" OwnerUserId="2578" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-07-08T18:29:50.420" LastActivityDate="2011-07-11T20:04:48.563" Title="Colors from a computer vs. colors from visible spectrum of sunlight" Tags="&lt;light&gt;&lt;color&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="12047" PostTypeId="2" ParentId="12046" CreationDate="2011-07-08T18:46:48.123" Score="9" ViewCount="" Body="&lt;p&gt;Yes, the mixing is happening in the eyes &amp;amp; brain; no, an RGB mix of yellow isn't the same as a pure yellow frequency; but our eyes will see it as the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eyes have 3 (or 2, if you're colour-blind) types of colour sensors, each of which responds with a different signal profile - each peaks at a particular frequency, and trails off for frequencies that differ from that. The brain merges the signals from those 3 (or 2) different sensors, to make sense of the colour signals to create a single colour signal, and it can't tell whether that was a balanced combination of red and green, or a pure yellow frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/1982/is-colour-as-represented-using-primary-colours-accurate-only-to-humans/1983#1983&quot;&gt;See also this answer to a previous, related question.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That explains most colours we see. Except for when we see a combination of red and blue, with no signals in between. There isn't a colour in the spectrum for that - the colours in between red and blue all feature higher signals in the middle, around green. To have signals from red and blue but not green, doesn't map to the spectrum. And our brain won't show a combination of two or more colours for a single point, it always maps a single point to a single colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So our brain creates a new colour, not on the spectrum, for a combination of red and blue.  Hence, purple pigments aren't real, in that sense - purple is the brain's interpolation of red + blue + no green. Purple is just a pigment of our imagination.&lt;/p&gt;&#xA;" OwnerUserId="4066" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-08T18:52:21.850" LastActivityDate="2011-07-08T18:52:21.850" CommentCount="4" />
  <row Id="12048" PostTypeId="1" AcceptedAnswerId="12052" CreationDate="2011-07-08T18:47:18.330" Score="1" ViewCount="105" Body="&lt;p&gt;Based on John Isaacks' question, &quot;&lt;a href=&quot;http://physics.stackexchange.com/questions/11940/if-you-view-the-earth-from-far-enough-away-can-you-observe-its-past&quot;&gt;If you view the Earth from far enough away can you observe its past?&lt;/a&gt;&quot; and the responses, it appears that we could use mirrors to see into the past.  Using Vintage's example, a single mirror on the moon would only allow an observer to see 0.1μs into the past.  Would it therefore be theoretically possible to align a massive series of mirrors that sent a single beam of light back and forth over a larger span of time (e.g., months, years, decades)?  If so, would it be possible to send messages to ourselves in the past via such an arrangement? Or would we always be looking into the past and only able to send messages to the future?&lt;/p&gt;&#xA;" OwnerUserId="4406" LastActivityDate="2011-07-09T01:46:51.820" Title="Could we enable ourselves to send messages to and receive messages from the future?" Tags="&lt;light&gt;&lt;spacetime&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="12049" PostTypeId="1" CreationDate="2011-07-08T19:13:32.113" Score="3" ViewCount="148" Body="&lt;p&gt;I was reading thru the wiki on Quasars and came across the fact that the most distant Quasar is 29 Billion Light years. This is what the article exactly says &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The highest redshift quasar known (as of June 2011[update]) is ULAS_J1120+0641, with a redshift of 7.085, which corresponds to a &lt;strong&gt;proper distance&lt;/strong&gt; of approximately 29 billion light-years from Earth.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I come to understand that the Big Bang singularity is believed to be around 13.75 Billion years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how is this possible? Does the presence of such a quasar negate the Big Bang Theory? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a student of Physics and was reading this out of (whimsical) curiosity. Is there something I'm missing here or the &quot;proper distance&quot; mentioned in the fact is a concept that explains this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: My Bad! Here's how..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple google search led me to this article (sciencedaily.com/releases/2011/06/110629132527.htm) which says the farthest quasar found is 12.9 billion LYs and not 29 billion.&#xA;So in the end we have just proven that wikipedia needs more moderation. Lol, what a waste of time!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I did find out a lot of new stuff from the answers, so thanks for all your responses!&lt;/p&gt;&#xA;" OwnerUserId="4413" LastEditorUserId="4413" LastEditorDisplayName="" LastEditDate="2011-07-08T21:08:03.147" LastActivityDate="2011-07-08T21:59:10.997" Title="A doubt about the age of the universe" Tags="&lt;universe&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="12050" PostTypeId="2" ParentId="12048" CreationDate="2011-07-08T19:33:58.130" Score="3" ViewCount="" Body="&lt;h2&gt;No.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You could not look into the future, all you could do is look at photons that have traveled a long distance, and reach their destination some time after they have been fired, not before they have been fired.&lt;/p&gt;&#xA;" OwnerUserId="503" LastActivityDate="2011-07-08T19:33:58.130" />
  <row Id="12051" PostTypeId="1" CreationDate="2011-07-08T19:35:44.913" Score="-1" ViewCount="155" Body="&lt;p&gt;If there is no gravity on the moon, how could this flag be flapping in the wind?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.stumbleupon.com/su/2wD6eg/hea-www.harvard.edu/~fine/images/desktops/Armstrong.jpg&quot; rel=&quot;nofollow&quot;&gt;http://www.stumbleupon.com/su/2wD6eg/hea-www.harvard.edu/~fine/images/desktops/Armstrong.jpg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4415" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-08T19:50:24.307" LastActivityDate="2011-07-08T20:05:11.953" Title="If there is no gravity on the moon why is the american flag waving?" Tags="&lt;space&gt;&lt;moon&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="12052" PostTypeId="2" ParentId="12048" CreationDate="2011-07-08T19:44:07.233" Score="2" ViewCount="" Body="&lt;p&gt;Not with currently known physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sending a signal to yourself would require sending a faster than light signal to something that could retransmit it to you and that was itself moving almost at the speed of light. This works because of the way that we cut spacetime into $E^3$ slices. The transmitter moving at close to the speed of light would have its slices tilted relative to ours in such a manner that a superluminal particle could move into its future while moving into our past. The book &quot;It's About Time: Understanding Einstein's Relativity&quot; by N. David Mermin has the clearest exposition of this that I've seen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light, by definition, moves at the speed of light. So light could not be used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would require not only particles with superluminal velocities but also the ability to &lt;em&gt;control&lt;/em&gt;  those particals to send a signal with them.&lt;/p&gt;&#xA;" OwnerUserId="2953" LastEditorUserId="2953" LastEditorDisplayName="" LastEditDate="2011-07-09T01:46:51.820" LastActivityDate="2011-07-09T01:46:51.820" />
  <row Id="12053" PostTypeId="2" ParentId="12051" CreationDate="2011-07-08T19:45:31.457" Score="7" ViewCount="" Body="&lt;p&gt;Ignoring the &quot;no gravity&quot; part of your question - there is, it's 1/6th that of the Earth, the flag looks like it's &quot;waving&quot; because the horizontal pole stuck part way out. This meant that the flag didn't &quot;unfurl&quot; fully and is hanging like a curtain rather than being stretched flat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was also rotated several times before Armstrong and Aldrin took the photo so any movement is due to simple inertia - (&lt;a href=&quot;http://science.nasa.gov/science-news/science-at-nasa/2001/ast23feb_2/&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happened on Apollo 11 and they liked the effect so much they replicated it on subsequent missions.&lt;/p&gt;&#xA;" OwnerUserId="3948" LastEditorUserId="3948" LastEditorDisplayName="" LastEditDate="2011-07-08T19:53:21.497" LastActivityDate="2011-07-08T19:53:21.497" CommentCount="2" />
  <row Id="12054" PostTypeId="2" ParentId="12051" CreationDate="2011-07-08T19:48:15.930" Score="0" ViewCount="" Body="&lt;p&gt;Frankly I dont really have an answer for you but your question is a little off. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moon does have a gravity its just pretty weak (abt 1/6th of earth's gravity). Due to this reason moon was unable to hold on to its atmosphere. Moon's gaseous atmosphere escaped its surface during moon's creation which means moon doesn't even have wind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Neil Armstong's footprints (if he really went there) would still be intact since there is no wind on moon to erode away the moon dust of the foot prints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Absence of wind therefore makes the fluttering flag even more questionable.&lt;/p&gt;&#xA;" OwnerUserId="4413" LastActivityDate="2011-07-08T19:48:15.930" />
  <row Id="12055" PostTypeId="2" ParentId="12049" CreationDate="2011-07-08T20:02:45.513" Score="-4" ViewCount="" Body="&lt;p&gt;&quot;Is there something I'm missing here...?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, you are missing a lot, like the fact that astronomers really don't know how far away something is, if it is more than, say, about 3k light-years out.  (Compare that to the size of the our own Milky Way galaxy = 1k ly thick by 100k ly in diameter.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me explain...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our main tool for measuring how far away something that is far away is, is to use parallax:  We see what direction something is from our planet, then wait half a year to take the same measurement from 2AU away.  If the difference in angle is 2 arcseconds (an arcsecond is 1/3600 of a degree), then the object would have been only 1 arcsecond of difference if we could have made our second measurement from the Sun.  That would mean (if you do the trigonometry) that the object is about 206265 AUs away.  That's about 3.26 light-years.  It is also, by definition, a parsec (short for &quot;parallax of an arc-second&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though 1 arcsecond is an eensy-weensy-itty-bitty-very-super-tiny-small angle, we have the equipment to accurately measure it.  Nailing down Alpha Centauri, at about 4 light-years (just over a parsec) away, for instance, is a piece of cake.  But as you go farther out, it gets harder.   Judging the distance of an object by parallax that is 400 light-years away, for instance, would take equipment and methods which are one hundred times more accurate than those which could barely peg Alpha Centauri.  Judging the actual distance of objects at 4000 light-years takes a thousand-fold increase in accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is that at a certain point we can't tell jack using parallax.  We lose 3-D vision of the stars, and everything looks like it was painted on a great curved canvas.  The distance at which this happens is close enough that the vast majority of our own galaxy appears in 2-D.  Certainly any other galaxy is way, WAY beyond the point where 3-D viewing becomes impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do we tell how far away stuff is?  We guess.  Astrononers say, &quot;Hmmm, that star has a color of 'this', and we think it's color should be 'that', and 'this' minus 'that' is the red shift.  And a red shift of 'that' means it must be moving away at this speed.  And our theory says the universe is this old; so (guesstimated speed) x (theoretical time) means it must be this far away.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when someone tells you that something is 29 BILLION light-years away, and you know we start guessing in the 100s or 1000s of light-years, then you KNOW they are flat out pulling a WAG out of their nether-regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that doesn't knock your socks off, this might:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not too long ago, one of these bright, telescope-toting homosapiens realized that the red shift from all these galaxies indicated an initial starting point for the universe which was, galactically speaking, very, very, Very, VERY close to our neck of the Milky Way.  That was absolutely unacceptable, since it might fuel a creationist view of things.  So a theory was invented that EVERYTHING was expanding: the space between galaxies, the space within galaxies, even the space between the atoms in your body.  Then someone realized that doesn't quite hold water.  Don't know where this theory has morphed to in recent years, as I do not try to track fairytales in the birthing process, but I await, with bemusement, the tortured death of it and the move to something even more insane.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-07-08T20:39:34.653" LastActivityDate="2011-07-08T20:39:34.653" CommentCount="5" />
  <row Id="12056" PostTypeId="2" ParentId="12051" CreationDate="2011-07-08T20:05:11.953" Score="0" ViewCount="" Body="&lt;p&gt;First of all, there &lt;em&gt;is&lt;/em&gt; gravity on the moon; it's just weaker than the gravity on Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, gravity has essentially nothing to do with whether a flag flaps. If you managed to create windy conditions in a zero-gravity environment, any flags placed in that wind would flap just as well as they do on Earth (or perhaps even better, since gravity wouldn't be pulling them down when the wind dies down). The reason a flag streams out behind its flagpole is just that the flag reacts to the wind, whereas the flagpole doesn't - it's closely related to the reason that an arrow or a rocket maintains its orientation during flight (the fins react to the wind more than the body or nose). And although I'm not an expert on fluid dynamics, as far as I know the actual flapping has to do with turbulence induced in the wind by the flagpole and the flag itself. None of these behaviors require gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, it's not even clear that the flag is flapping in that photo, because there's a rod along the top holding the flag up. But as I said in my comment, that sort of discussion is better handled on &lt;a href=&quot;http://skeptics.stackexchange.com&quot;&gt;Skeptics&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-07-08T20:05:11.953" />
  <row Id="12058" PostTypeId="1" CreationDate="2011-07-08T20:35:31.620" Score="5" ViewCount="100" Body="&lt;p&gt;We usually calibrate the cameras on our microscopes by capturing 20 images of a blurry (not sharp) fluorescent particle. For each pixel in this stack of 20 images we calculate the intensity variance. By plotting the variance against the intensity we obtain the scale factor that allows us to convert the ADU (arbitrary data unit) into detected photoelectrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this approach relies on a light source with Poisson statistics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is applicable for fluorescence light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But can we also use light of a halogen lamp instead of fluorescent light?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I heard that photons leaving a heated metal surface have a particular statistic. I think the argument was that they are coming out in bunches and it has something to do with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Work_function&quot; rel=&quot;nofollow&quot;&gt;workfunction&lt;/a&gt;. It would be nice if someone could give a good reference on this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/9BX1D.jpg&quot; alt=&quot;Calibration results of an EMCCD camera&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture above shows the calibration curves for an EMCCD camera.&#xA;The left graph corresponds to the readout mode you would use for bright images and has a readnoise of 15 electrons per pixel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph on the right shows the result for the low noise mode. In this mode the charge that has been accumulated under the pixels is transferred through an electron multiplication register. Impact ionization increases the number of electrons. The gain is usually up to 300x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inset on the lower right is how a calibration image looks like. Note that this one is fake.&lt;/p&gt;&#xA;" OwnerUserId="4306" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-07-10T11:09:32.747" LastActivityDate="2011-07-10T23:05:57.880" Title="Photon statistics of an incandescent light source" Tags="&lt;optics&gt;&lt;photons&gt;&lt;statistics&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="12059" PostTypeId="1" AcceptedAnswerId="12071" CreationDate="2011-07-08T20:37:47.603" Score="2" ViewCount="142" Body="&lt;p&gt;I came across the use of the unit barn and inverse barn while reading about the operation of LHC. What is an inverse femtobarn ? What does it tell about the experiment being described ?  &lt;/p&gt;&#xA;" OwnerUserId="3714" LastActivityDate="2011-07-14T19:26:04.773" Title="what is an inverse femto barn?" Tags="&lt;large-hadron-collider&gt;&lt;si-units&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="12060" PostTypeId="2" ParentId="12012" CreationDate="2011-07-08T20:41:41.453" Score="1" ViewCount="" Body="&lt;p&gt;Benjamin Horowitz's answer covered a lot of the key points, but it's worth adding that the question of the topology of the universe has been investigated by astrophysical observations. If the universe is multiply connected, and if the length scale is shorter than the horizon scale, then we should be able to see evidence of it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To take a simple example, imagine that the universe is geometrically flat but has the geometry of a 3-torus. Specifically, take a cubical volume, and identify opposite faces, so that if you &quot;leave&quot; the cube through one face you reenter through the opposite face. If the length of an edge of the cube is sufficiently small, then you could see multiple copies of any given object. Of course, if the length is much larger than the horizon, then there's no way to tell the difference between this model and one in which space is infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way to test these models is the &quot;&lt;a href=&quot;http://www.citebase.org/abstract?id=oai%3aarXiv.org%3agr-qc/9602039&quot; rel=&quot;nofollow&quot;&gt;circles in the sky&lt;/a&gt;&quot; technique, in which you look for correlated circles in different directions in maps of the microwave background radiation. T&lt;a href=&quot;http://prl.aps.org/abstract/PRL/v92/i20/e201302&quot; rel=&quot;nofollow&quot;&gt;he result is negative&lt;/a&gt;: we don't live in a multiply-connected universe with a sufficiently short length scale to be observable.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-07-08T20:41:41.453" CommentCount="2" />
  <row Id="12061" PostTypeId="1" CreationDate="2011-07-08T20:43:12.810" Score="8" ViewCount="131" Body="&lt;p&gt;How is information defined from a thermodynamics point of view ? I came across some definitions using the concept of free energy of a system. If I have information stored in a finite volume of space what does it mean for the free energy and entropy of the space ?&lt;/p&gt;&#xA;" OwnerUserId="3714" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-09T00:08:23.260" LastActivityDate="2011-09-03T06:18:26.033" Title="How is information defined from a thermodynamics point of view?" Tags="&lt;thermodynamics&gt;&lt;information&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="4" />
  <row Id="12062" PostTypeId="2" ParentId="12049" CreationDate="2011-07-08T20:46:37.610" Score="1" ViewCount="" Body="&lt;p&gt;Yes, there is something you're missing. If you're familiar with special relativity you know that velocities don't add the same simple way they do in Newtonian mechanics. If one spaceship is moving at $c/2$ to the right and another is moving $c/2$ to the left, the relative velocity between them is not $c$, as one might expect, but $4c/5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general relativity the same kind of thing applies to distances as well as velocities. In flat space we define the distance between two objects as the length of the (unique) straight line from one to the other, but in general relativity space can be curved and there is no such thing as a &quot;straight line&quot;. The closest analog is a geodesic, which is a smoothly curved path between two points whose length is a (local) minimum. The distance to the quasar you quoted is defined as the length of a certain geodesic connecting it to us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there's no reason to expect geodesic distances to behave just like straight-line distances in flat space. In the FLRW metric (which defines the basic shape of the whole Universe in big bang cosmology), there exist geodesics that are longer than the geodesic from the big bang to us used to define the &quot;age of the universe&quot;. So having a quasar whose &quot;proper distance&quot; or &quot;comoving distance&quot; from us is 29 billion light years is not a contradiction at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, another seemingly contradictory thing you might run into is relative velocities faster than the speed of light. This can happen because relative velocity is also a tricky thing in GR, which is only defined once you decide on a specific path to &quot;parallel transport&quot; the velocity vector along. The relative velocity between two nearby objects is well-defined and can never be greater than $c$, but the relative velocity between two very distant objects (defined a certain way) can be greater than $c$.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-07-08T20:46:37.610" CommentCount="1" />
  <row Id="12063" PostTypeId="2" ParentId="12049" CreationDate="2011-07-08T20:56:01.410" Score="5" ViewCount="" Body="&lt;p&gt;In the expanding universe, you have to be a bit careful to define exactly what you mean by distance. The &quot;proper distance&quot; referred to here in that article means the distance measured at the present time. We have to be careful even to define what we mean by that last phrase -- time is relative, you know. But if the universe is approximately homogeneous, then there is a &quot;natural&quot; choice of time coordinate called &quot;cosmic time.&quot; If you imagine many, many rulers stretched out between you and the quasar, the proper distance the total length of all of them, added up at the present value of cosmic time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's not the same as the distance that the light has traveled, though. There are various fancy general-relativistic reasons why not, but the main idea is a very simple one. That quasar is moving away from us, so it used to be closer to us. The light we're seeing now was emitted when the distance was much shorter, so it didn't have to travel anywhere near 29 billion light-years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The truth is that that 29-billion light-year figure is calculated based on a certain model of the universe; it's not measured directly. The model it's calculated on is based on the theory of general relativity, and includes the best currently-measured age of the universe. So pretty much by definition, there can't be any contradiction between that distance and the age of the universe.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-07-08T21:59:10.997" LastActivityDate="2011-07-08T21:59:10.997" />
  <row Id="12064" PostTypeId="1" CreationDate="2011-07-08T21:19:50.857" Score="5" ViewCount="147" Body="&lt;p&gt;This question concerns the article &lt;a href=&quot;http://amasci.com/elect/mcoils.html&quot; rel=&quot;nofollow&quot;&gt;Right Angle Circuitry&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having passed a grad E&amp;amp;M course, I feel like this should make perfect sense to me and I should be able to answer all the questions it raises. Sadly that is not yet the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the answers to all the unanswered questions here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why does the circuit in Figure 8 not act as a transformer?&lt;/li&gt;&#xA;&lt;li&gt;Is it possible to extract energy from a toroid with capacitor plates as shown in Figure 9? If not, why not?&lt;/li&gt;&#xA;&lt;li&gt;Have structures like that in Figure 10 ever been used in practice as transmission lines?&lt;/li&gt;&#xA;&lt;li&gt;Last question in the article: Is it possible to create &quot;insulating electric circuitry which conducts only displacement current&quot;?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-09-05T05:33:33.000" Title="Right Angle Circuitry" Tags="&lt;electromagnetism&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="12066" PostTypeId="2" ParentId="12035" CreationDate="2011-07-08T22:06:53.423" Score="-1" ViewCount="" Body="&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/FRzMm.png&quot; alt=&quot;REFLECTION under two different angles&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of the intersection of the ray with the surface at A is greater than the area of intersection at B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To a greater area correspond more atoms to reflect the light.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT add&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A justification:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations related to graphs on Steve answer (the reflected coefficients $R_s and R_p$) are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fresnel_equations&quot; rel=&quot;nofollow&quot;&gt;Fresnel equations&lt;/a&gt;&#xA;when I read there $(n_1/n_2\cdot\sin\theta_i)^2$ I see a proportionality to the area.&lt;br&gt;&#xA;The angles are in relation to the normal and sin² traces an area.&#xA;A sensivity of the equations in relation to this factor is apparent in the graphs ($\theta_i$), because $\sin\theta_i, \sin\theta_i^2$ will 'follow' the  shape of ($\theta_i$)&lt;br&gt;&#xA;I think that my viewpoint is justified.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT add end&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the reflected rays are absent in the pic. Its not relevant.  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-07-09T15:35:19.850" LastActivityDate="2011-07-09T15:35:19.850" CommentCount="5" />
  <row Id="12068" PostTypeId="2" ParentId="12048" CreationDate="2011-07-08T22:21:34.180" Score="1" ViewCount="" Body="&lt;p&gt;Bouncing light off a distant mirror allows you to &quot;see into the past&quot; because it takes the light a while to get to the mirror and back.  You can achieve the same goal much more efficiently by just recording the event or information, and then just letting the recording go into the future &lt;a href=&quot;http://xkcd.com/209/&quot; rel=&quot;nofollow&quot;&gt;the usual way&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2011-07-08T22:21:34.180" />
  <row Id="12069" PostTypeId="1" AcceptedAnswerId="12254" CreationDate="2011-07-08T22:24:33.413" Score="6" ViewCount="248" Body="&lt;p&gt;It is very crucial that I ask whether it ``could'' and not whether it does. I do not mean to be the least controversial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my surprise, having read ``Physics for Future Presidents'' by Richard Muller last year, I've come across a sentence of the sort (I'm paraphrasing): cell phone radiation is way too weak to effect molecular structure, and therefore any claim about cell phone radiation causing cancer can be attributed to people blaming cell phones for their cancer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems like a very strong statement! Is this a consensus among physicists that there is absolutely nothing to the claim that cell phone radiation could possibly (in any significant way) cause cancer? Is this really just being thrown around because people don't understand the physics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be very precise, the question is: is there a physics model that would suggest a mechanism by which cell phone radiation can cause any sort of damage that could lead to cancer?&lt;/p&gt;&#xA;" OwnerUserId="4249" LastActivityDate="2011-07-29T20:29:22.763" Title="Could cell-phone radiation cause cancer?" Tags="&lt;electromagnetic-radiation&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="12070" PostTypeId="1" CreationDate="2011-07-08T22:28:30.830" Score="0" ViewCount="69" Body="&lt;p&gt;Muon has mass 100 m_e tha't the up limit of its generated netrino. Even for moving muon with speed close to light the energy increament of the system almost all lands on generated electron, when moving coordinate system to calculate this value.&#xA;However most people evaluates this value to 1000Mev so high (&lt;a href=&quot;http://www.dailymail.co.uk/sciencetech/article-2003842/Researchers-closer-Holy-Grail-science-finding-matter-dominates-universe.html&quot; rel=&quot;nofollow&quot;&gt; here&lt;/a&gt;), How they&#xA;do to them? I'm confused. &lt;/p&gt;&#xA;" OwnerUserId="4291" LastActivityDate="2011-07-09T07:06:36.403" ClosedDate="2011-07-09T08:50:41.870" Title="Is the energy of muon neutrino too great?" Tags="&lt;particle-physics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12071" PostTypeId="2" ParentId="12059" CreationDate="2011-07-08T22:59:48.343" Score="5" ViewCount="" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Barn_%28unit%29&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;barn&lt;/em&gt;&lt;/a&gt; is a unit of area, equal to $10^{-28}\text{ m}^2$. The prefix &lt;em&gt;femto-&lt;/em&gt; signifies $10^{-15}$, so a femtobarn is equal to $10^{-43}\text{ m}^2$. When high-energy physicists talk about inverse femtobarns, they mean &lt;em&gt;collisions per femtobarn&lt;/em&gt; of beam cross-sectional area. The inverse femtobarn is not a measure of information, but rather of the effectiveness of the particle accelerator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the reasoning behind that: imagine two beams of particles coming at each other, and for simplicity let's designate one beam as the &quot;target&quot; and the other as the &quot;probe&quot; (it doesn't matter which is which). The scattering cross section $\sigma$ is the effective size of a single target particle, as &quot;seen&quot; by a probe. So the probability of a collision will be just the fraction of the target beam area that is actually occupied by the target particles; in other words, the scattering cross section divided by the average cross-sectional area per particle,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_\text{collision} \approx \frac{\sigma}{A_t/n_t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($t$ for &quot;target&quot;). Here I'm assuming that the beam is highly diffuse, and that we're only talking about one &quot;bunch&quot; (a finite-length section of a particle beam).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total &lt;em&gt;number&lt;/em&gt; of collisions is just this probability times the number of chances for a collision to happen, namely the number of probe particles in a bunch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_\text{collision} = \sigma \frac{n_t n_p}{A_t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($p$ for &quot;probe&quot;). Finally, since this happens every time two bunches cross paths, to get the overall rate at which collisions occur you need to multiply by the rate of bunch crossings, which is denoted $f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_\text{collision} = \sigma \frac{f n_t n_p}{A_t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantity $\frac{f n_t n_p}{A_t}$ is defined as the two-beam &lt;em&gt;luminosity&lt;/em&gt; $L$. Roughly speaking, it represents the number of &lt;em&gt;potential&lt;/em&gt; collisions per unit area per unit time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the actual number of collisions for any particular process, you multiply by luminosity by the cross-section for that process and integrate over time,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_{X\to Y} = \bigl(\sigma_{X\to Y}\bigr)\times\biggl(\int L \,\mathrm{d}t\biggr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This split is convenient because the first factor depends only on the physical process being considered, and the second factor depends only on the design of the accelerator. So that second factor is a good way to characterize the production capacity of a given accelerator. The units of this value are 1/area, and the inverse femtobarn turns out to be roughly the right magnitude for measuring $\int L \,\mathrm{d}t$ at current particle accelerators. This is what particle physicists mean when they talk about inverse femtobarns.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-14T19:26:04.773" LastActivityDate="2011-07-14T19:26:04.773" CommentCount="1" />
  <row Id="12072" PostTypeId="1" CreationDate="2011-07-08T23:03:08.763" Score="1" ViewCount="46" Body="&lt;p&gt;Lets say I want to design a Nomarski prism that would split the ordinary and extraordinary beam by an angle of 0.32 mrad. I used a raytracer to find the internal angle between the quartz wedges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I couldn't decide how thick the quartz wedges should be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original patent mentions the thickness but I don't understand the sentence:&#xA;&lt;a href=&quot;http://www.google.co.uk/patents?id=QI5kAAAAEBAJ&amp;amp;zoom=4&amp;amp;dq=nomarski%20patent%202924142&amp;amp;pg=PA6#v=onepage&amp;amp;q=Fig.%204&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://www.google.co.uk/patents?id=QI5kAAAAEBAJ&amp;amp;zoom=4&amp;amp;dq=nomarski%20patent%202924142&amp;amp;pg=PA6#v=onepage&amp;amp;q=Fig.%204&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/0Trmk.png&quot; alt=&quot;Picture of Nomarski Patent&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture shows the Nomarski prism from the side. The optical axis of the system is horizontal.&#xA;The internal angle between the wedges (or blades) is $\alpha$. The &lt;em&gt;optic&lt;/em&gt; axis (note the difference between optical axis of the system and optic axis of the crystal) of the first prism Q2 is perpendicular to the screen. The wedge Q1 has its axis inclined in-plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2 is 0.5mm thick at the center.&#xA;Q1 is 1mm thick at the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Both compensate their birefringence.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This last expression isn't clear to me.&#xA;I have the feeling that I understand this only vaguely.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is the birefringence in Q2 stronger&#xA;       because of the orientation of the&#xA;       optic axis? (Probably yes)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why does&#xA;        the birefringence need to be&#xA;            compensated? (must have something to&#xA;            do with the adjusting the bias by&#xA;            shifting the Nomarski prism&#xA;            transversal to the optical axis in&#xA;            the microscope)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How is the total thickness of the prism (here 1.5mm) chosen? Can a thinner Nomarski prism  be used with higher angles?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4306" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-07-09T19:34:10.777" LastActivityDate="2011-07-09T19:34:10.777" Title="What defines the thickness of a Nomarski prism?" Tags="&lt;optics&gt;&lt;polarization&gt;&lt;refraction&gt;&lt;crystals&gt;&lt;polarisation&gt;" />
  <row Id="12074" PostTypeId="2" ParentId="12005" CreationDate="2011-07-09T01:36:23.137" Score="0" ViewCount="" Body="&lt;p&gt;FYI - The formula for decomposing the acceleration vector is&#xA;$ \vec{a} = \dot{v} \vec{e} + \frac{v^2}{r} \vec{n} $ where $v$ is the speed, $\dot{v}$ the rate of change of speed (acceleration along the trajectory),  the vectors $\vec{e}$ and $\vec{n}$ are along the trajectory and normal to it and $r$ is the curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at &lt;a href=&quot;http://www-users.math.umd.edu/~jec/math241/accel.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; article for all the math the accompany this (Fenet Frames), or just look up &lt;a href=&quot;http://en.wikipedia.org/wiki/Acceleration&quot; rel=&quot;nofollow&quot;&gt;Acceleration&lt;/a&gt; in wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-07-09T01:36:23.137" />
  <row Id="12076" PostTypeId="1" CreationDate="2011-07-09T03:11:56.453" Score="2" ViewCount="95" Body="&lt;p&gt;What happen if someone is just close enough to a black hole, and he emits light tangentially out, and the light got bent and circles around, what will he see, can he see the light source from his back?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;             +----------&amp;gt;&#xA;            /            \&#xA;   light   /              \&#xA;          ^                \&#xA;observer  +   O black hole /&#xA;sees light\               /&#xA;           \             /&#xA;             &amp;lt;----------- light path&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4428" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-07-11T16:05:02.223" LastActivityDate="2011-07-11T16:05:02.223" Title="Returning Light" Tags="&lt;black-holes&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12077" PostTypeId="2" ParentId="12035" CreationDate="2011-07-09T04:42:40.617" Score="6" ViewCount="" Body="&lt;p&gt;First, I just want to remind readers that it is NOT true that &quot;more glancing angle always means more reflection&quot;. For p-polarized light, as the angle goes away from the normal, it gets less and less reflective, then at the Brewster angle it's not reflective at all, and then beyond the Brewster angle it becomes more reflective again:&lt;img src=&quot;http://i.stack.imgur.com/yMKp4.png&quot; alt=&quot;image from wikipedia&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, it's certainly true that as the angle approaches perfectly glancing, the reflection approaches 100%. Even though the question asks for non-mathematical answers, the math is pretty simple and understandable in my opinion...here it is for reference. (I don't have any non-mathematical answer that's better than other peoples'.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Maxwell's equations boundary conditions say that certain components of the electric and magnetic fields have to be continuous across the boundary. The situation at almost-glancing angle is that the incoming and reflected light waves almost perfectly cancel each other out (opposite phase, almost-equal magnitude), leaving almost no fields on one side of the boundary; and since there's almost no transmitted light, there's almost no fields on the other side of the boundary too. So everything is continuous, &quot;zero equals zero&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this cannot work at other angles is that two waves cannot destructively interfere unless they point the same direction. (If two waves have equal and opposite electric fields and equal and opposite magnetic fields, then they have to point the same direction, there's a &quot;right-hand rule&quot; about this.) At glancing angle, the incident and reflected waves are pointing almost the same direction, so they can destructively interfere. At other angles, the incident and reflected waves are pointing different directions, so they cannot destructively interfere, so there has to be a transmitted wave to make the boundary conditions work.  :-)&lt;/p&gt;&#xA;" OwnerUserId="3811" LastActivityDate="2011-07-09T04:42:40.617" CommentCount="1" />
  <row Id="12078" PostTypeId="1" CreationDate="2011-07-07T06:53:38.460" Score="1" ViewCount="516" Body="&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; [for aptness of this question to this site, read 'real world applications' as 'applications in Physics']&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept of function (of the form $f : \mathbb{R} \to \mathbb{R}$ ) has been used in  various real world applications of Mathematics. In almost all of the applications the general consideration such as approximating a given function with another function is viewed through a lens of mean square error or one of several other such very similar metrics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is concerning the use of functions which have singularities. I've heard from some Physicists that the use of such functions is just a matter of mathematical taste as such functions could be easily approximated with smooth functions and hence the choice of using such functions is just a matter of mathematical taste and it wouldn't make any conceptual difference. I tend not to agree with such an argument because it all depends on the lens of approximation that is being used. If we do not use the metrics like the ones mentioned above then it could be different. A function which contains singularities is fundamentally conceptually different from a smooth function which approximates it no matter how good the approximation is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is, are there any applications of Mathematics in which functions with singularities are used not just as a matter of mathematical taste but for conceptual reasons ?&lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-29T22:01:31.423" LastActivityDate="2011-07-29T22:01:31.423" Title="real world applications of Mathematics which use functions with singularities, not just as a matter of mathematical taste but for conceptual reasons" Tags="&lt;mathematical-physics&gt;&lt;mathematics&gt;" AnswerCount="6" CommentCount="4" />
  <row Id="12079" PostTypeId="2" ParentId="12078" CreationDate="2011-07-07T07:27:21.523" Score="1" ViewCount="" Body="&lt;p&gt;[I understand you are looking for more of a pure mathematical take on this, but there are many more examples of these sorts of approximations in applied mathematics (a couple I will document below). Which of course must have a pure mathematical basis as many are to do with the differentiability of the underlying functions.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes there are functions in which singularities are approximated by piece-wise continuous functions. I will assume for the sake of argument that you are not limiting this discussion to relativistic-type &quot;singularities&quot; which are understood as a point at which a given mathematical object is not defined, but also to those of a point of an exceptional set which fail to be well-behaved in some particular way, such as differentiability. In the latter case there are many examples in computational mathematics and mathematical physics where the discontinuity or singularity can cause mathematical instability. An example of this to name one would be the modelling of a shock wave or &quot;jump discontinuity&quot; in computational fluid dynamics. A sudden step-change in any given variable can cause the numerical scheme to become unstable and &quot;blow-up&quot; in this instance the evolution equations are smoothed using artificial dissipation to avoid this behaviour. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact there are many examples of such modelling in computational relativistic astrophysics. Not only in the modelling of shock waves etc. but also in the modelling of flow moving across entities such as event horizons when modelling flows from compact objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sure there are many more instances but these are a few I have thought of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this is of some use.&lt;/p&gt;&#xA;" OwnerUserId="4312" LastActivityDate="2011-07-07T07:27:21.523" />
  <row Id="12080" PostTypeId="2" ParentId="12070" CreationDate="2011-07-09T07:06:36.403" Score="-1" ViewCount="" Body="&lt;p&gt;If I believe the data and analysis of KEK or BooNE the decay to generate nu should be a whole instead of phase-linking one Then all will be harmonic.&lt;/p&gt;&#xA;" OwnerUserId="4426" LastActivityDate="2011-07-09T07:06:36.403" CommentCount="1" />
  <row Id="12081" PostTypeId="2" ParentId="12076" CreationDate="2011-07-09T07:20:59.007" Score="3" ViewCount="" Body="&lt;p&gt;It seems that you want a position where light can get emitted and then return to where it started and then continue on forever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such places do exist; what you're looking for is a circular null orbit. Uh, &quot;null&quot; means having no change in proper time, that is, the sort of path (or orbit or geodesic) that is suitable for light. The radius for this orbit is $3r_s/2$ where $r_s$ is the Schwarzschild radius. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_metric#Orbital_motion&quot; rel=&quot;nofollow&quot;&gt;wikipedia, Schwarzschild metric / orbital motion.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spacecraft emitting the light would have to use thrusters (or something else, maybe a rope or what have you) to keep itself stationary, but the value is outside the &lt;a href=&quot;http://en.wikipedia.org/wiki/Event_horizon&quot; rel=&quot;nofollow&quot;&gt;event horizon&lt;/a&gt; at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow&quot;&gt;Schwarzschild radius&lt;/a&gt; (where $r_s=1$), so this is possible.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-07-09T07:20:59.007" />
  <row Id="12082" PostTypeId="1" CreationDate="2011-07-09T08:09:36.583" Score="0" ViewCount="46" Body="&lt;p&gt;What should be the limits of integration for euclidean action S[phi] in 3d and 4d? This action is negatively exponentiated to calculate the decay rate. I suspect that it is variable limit problem. &lt;/p&gt;&#xA;" OwnerUserId="4361" LastActivityDate="2011-07-09T08:09:36.583" Title="Quantum tunneling in Field theory with Time dependent potential" Tags="&lt;quantum-field-theory&gt;&lt;lagrangian-mechanics&gt;" CommentCount="2" />
  <row Id="12083" PostTypeId="2" ParentId="12078" CreationDate="2011-07-09T09:09:18.403" Score="0" ViewCount="" Body="&lt;p&gt;I agree the asker's opinion, an equation has its own category for its solution, a unconcerned breaking point or singularity is deadly to ruin the meaning and sometime &#xA;far away off the real solution.&lt;/p&gt;&#xA;" OwnerUserId="4426" LastActivityDate="2011-07-09T09:09:18.403" />
  <row Id="12084" PostTypeId="2" ParentId="11162" CreationDate="2011-07-09T09:49:06.470" Score="0" ViewCount="" Body="&lt;p&gt;In the spot diagrams of optic designs the ray aberrations are always compared to the size of diffraction limited spot.&#xA;In order to do this you have to find the smallest aperture in your system. This can be a deliberately placed mechanical aperture but it can also be the circumference of a lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I copied the following spot diagram from &#xA;&lt;a href=&quot;http://www.astronomy.net/articles/17/&quot; rel=&quot;nofollow&quot;&gt;http://www.astronomy.net/articles/17/&lt;/a&gt;.&#xA;The black ring is the diffraction limited spot size (3.3 $\mu$m if you convert from inch) the red dots are the spot with ray aberrations. You can see a strong Coma with $1\mu$m spread in x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/nH5Up.gif&quot; alt=&quot;Spot diagram showing comparing diffraction limited spot with strong coma.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read a nice discussion about this topic a while ago but can't remember which book it was in. I tried to implement it in a raytracer. It involved an iteration over all apertures in the optical train. One had to trace all apertures into the image and chose which one produced the smallest opening angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think any decent book on lens design should contain a chapter on the topic. See for example chapter 6 in Warren J. Smith: &quot;Modern optical engineering: the design of optical systems&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4306" LastActivityDate="2011-07-09T09:49:06.470" />
  <row Id="12085" PostTypeId="2" ParentId="12078" CreationDate="2011-07-09T10:04:43.680" Score="1" ViewCount="" Body="&lt;p&gt;One example from physics that springs to mind is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Operator_product_expansion&quot; rel=&quot;nofollow&quot;&gt;Operator Product Expansion&lt;/a&gt; (OPE) particularly used in conformal field theory and string theory. The singular part of the OPE is the most important, while the regular part rarely plays a significant role. Conceptually, the singular part (often called the &lt;em&gt;contraction&lt;/em&gt;) measures the non-commutativity of the operators. &lt;/p&gt;&#xA;" OwnerUserId="2451" LastActivityDate="2011-07-09T10:04:43.680" CommentCount="6" />
  <row Id="12086" PostTypeId="2" ParentId="3327" CreationDate="2011-07-09T10:15:25.790" Score="0" ViewCount="" Body="&lt;p&gt;There is a really nice paper on how they fixed the Hubble Space Telescope. They shifted the CCD by small amounts and captured several different images of the point spread function (which was quite bad at the time). Then they propagated their measurement back into the plane of the mirror and were able to determine the phase error and locate the original manufacturing problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This enabled to fix the telescope by installing a correction mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complicated paper:&#xA;&lt;a href=&quot;http://www.optics.rochester.edu/workgroups/fienup/PUBLICATIONS/AO93_PRComplicated.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.optics.rochester.edu/workgroups/fienup/PUBLICATIONS/AO93_PRComplicated.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some nice animation on the work they do for the new telescope: &#xA;&lt;a href=&quot;http://www.optics.rochester.edu/workgroups/fienup/Tom/tom_research.html&quot; rel=&quot;nofollow&quot;&gt;http://www.optics.rochester.edu/workgroups/fienup/Tom/tom_research.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4306" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-07-09T10:30:30.923" LastActivityDate="2011-07-09T10:30:30.923" />
  <row Id="12087" PostTypeId="2" ParentId="12078" CreationDate="2011-07-09T11:43:57.273" Score="0" ViewCount="" Body="&lt;p&gt;To simplify the computations in N-body simulations it is used to do $\frac{1}{(d+\epsilon)^2}\:(\epsilon&amp;gt;0)$ instead of $\frac{1}{d^2}$ to avoid collisions. In my opinion the singularity is mainly due to the discretisation method. Other way to deal with the problem is the use of a much shorter time unit (adaptative) in the region where stars get closer. In the real world, AFAIK, there are no reports of star collisions, nor when galaxies are merging  neither when they are heavily packed as it happens with clusters.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-07-09T11:43:57.273" />
  <row Id="12088" PostTypeId="2" ParentId="11995" CreationDate="2011-07-09T12:52:10.740" Score="1" ViewCount="" Body="&lt;p&gt;The quantum formalism actually has a very nice feature for exactly the kind of question you're asking. They're called projection operators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, of course, you talk about measuring some Hermitian operator. But you can also take one of the eigenstates of the operator (or any state really) and turn it into a projection operator, which corresponds to asking the system &quot;Are you in state $S$&quot;? Given a state $|S \rangle$ the corresponding projection operator is simply $|S\rangle \langle S|$. You can check that its an observable with eigenvalues 1 (yes) and 0 (no). If you get &quot;yes&quot;, then you know the state is $S$, but if you get &quot;no&quot; then you've projected the system onto the entire subspace of possibilities that are orthogonal to $S$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also ask the system very specific kinds of questions. Say you have an electron in a 1-D well, and you know it is somewhere between x=0 and x=2. Suppose you want to ask &quot;Are you somewhere between 0 and 1?&quot; You can formalize this question as the sum (integral) of the projection operators $|x\rangle \langle x|$ between x=0 and x=1. If you get yes (1), then you know the particle is between 0 and 1, and if you get no (0) then you know the particle is somewhere between 1 and 2. So there is a great deal of freedom to ask the system very detailed questions, neglecting of course the practical difficulties which may be associated with these particular measurements. These are in fact the most general type of quantum mechanical measurement and they're called positive operator-valued measures (POVM).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Sorry, I had to use parentheses for my bras and kets because the site didn't like the proper ones with arrows. They seemed to break everything.&lt;/p&gt;&#xA;" OwnerUserId="4430" LastEditorUserId="966" LastEditorDisplayName="" LastEditDate="2011-07-09T18:09:31.367" LastActivityDate="2011-07-09T18:09:31.367" />
  <row Id="12089" PostTypeId="1" AcceptedAnswerId="12141" CreationDate="2011-07-09T16:38:30.873" Score="1" ViewCount="216" Body="&lt;p&gt;Few days a ago i was soldering small thing which contained diode, battery and some other useless things.&#xA;Unfortunately i soldered diode reversed and it didn't work. When i reversed it again it started to light. &#xA;I was trying to find answer in google, but only thing i could find is fact that electricity in diode can flow only in one direction, but my question is why? How does it work ? Is this fact connected with construction of diode?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers Chris&lt;/p&gt;&#xA;" OwnerUserId="4432" LastActivityDate="2011-07-11T18:57:43.793" Title="Why electricity can flow only in one direction through diode?" Tags="&lt;electricity&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="12090" PostTypeId="1" AcceptedAnswerId="12095" CreationDate="2011-07-09T16:39:49.357" Score="2" ViewCount="59" Body="&lt;p&gt;I was playing a little bit with the basic physics behind water power production but I can't get the numbers right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that I put a windmill that pumps water into a watertank on the top of my house, &#xA;then I connect some kind of pipe with generator and starts to drain the watertank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much electric power (kWh) can I get out from a watertank with size $X\text{ m}^3$ placed $Y\text{ m}$ above the ground?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the formulas look like?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let's put some numbers on this problem, and see where we end up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say the tank is $1\text{ m}^3$, and it is $10\text{ m}$ off the ground so the water will fall $10\text{ m}$ to the generator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's connect the generator with a standard garden hose that has a 1 inch diameter, &#xA;with an area of $2.54\text{ cm}/(2\pi) \sim 5.1\text{ cm}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then I guess we would get a $10\text{ m}$ column of water pressure, that could be transformed with the area into the force the hight is putting on the system. Something like the earths gravity (9.82)*density*height = 9.82*1*10 ~ 98 Newton (???).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then maybe use that we can find that pressure=Force/Area, but how to move from pressure to energy?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Thanks David Zaslavsky for the example, and in theory that would mean that to store 1kWh I need like 40m3 at 10m height. That more or less mean that if one would try to build something like this in real life things need to be quite big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also thanks Fortunato for illustrate the practical problem in extracting the energy, and that even thou it is hard to get hight numbers it can be worth the effort anyway.&lt;/p&gt;&#xA;" OwnerUserId="4431" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-13T10:09:45.383" LastActivityDate="2011-07-13T10:09:45.383" Title="Small scale water power, how does water volume and hight convert into electric energy?" Tags="&lt;homework&gt;&lt;electricity&gt;&lt;energy&gt;&lt;forces&gt;&lt;renewable-energy&gt;" AnswerCount="2" />
  <row Id="12091" PostTypeId="2" ParentId="12089" CreationDate="2011-07-09T19:19:22.373" Score="0" ViewCount="" Body="&lt;p&gt;This really depends on the type of diode you are talking about but for the most common types there is something called a PN-Junction that allows current to flow in only one direction.  This page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/PN_junction&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/PN_junction&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;on wikipedia explains how a PN-junction works.  The short version is the voltage has to be above a certain threshold defined by the materials in the junction before the juice can flow.  This shows up in the I-V characteristic curve as a point in the plot where the derivative is discontinuous.  The details on other features of the I-V curve are up for grabs depending on the materials used but it's the junction and the application of a voltage beyond that threshold that means you have a diode on your hands.&lt;/p&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2011-07-09T19:19:22.373" />
  <row Id="12092" PostTypeId="2" ParentId="12089" CreationDate="2011-07-09T19:37:14.363" Score="1" ViewCount="" Body="&lt;p&gt;The engineer guy explains how a diode works in this video: &lt;a href=&quot;http://www.engineerguy.com/videos/video-transistor-point-contact.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.engineerguy.com/videos/video-transistor-point-contact.htm&lt;/a&gt;&#xA;I'm not going to repeat him here, its better to watch the animation.&#xA;Note that he also explains how a transistor works and you see a nice working (?) model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Afterwards you can watch this video, which explains applications of a diode:&#xA;&lt;a href=&quot;http://www.youtube.com/user/Afrotechmods#p/u/15/cyhzpFqXwdA&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/user/Afrotechmods#p/u/15/cyhzpFqXwdA&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4306" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-07-09T20:06:36.787" LastActivityDate="2011-07-09T20:06:36.787" />
  <row Id="12093" PostTypeId="1" CreationDate="2011-07-09T19:41:03.787" Score="6" ViewCount="189" Body="&lt;p&gt;I would like to know if there are some open mathematical problems in General Relativity, that are important from the point of view of Physics. I mean is there something that still needs to be justified mathematically in order to have solid foundations? Please tell me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; The same question was asked (and answered from a mathematical perspective) on &lt;a href=&quot;http://math.stackexchange.com/questions/50521/open-problems-in-general-relativity&quot;&gt;math.SE&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4434" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-07-09T19:59:26.903" LastActivityDate="2011-07-11T20:10:55.217" CommunityOwnedDate="2011-07-09T19:58:02.787" Title="Open problems in General Relativity" Tags="&lt;general-relativity&gt;&lt;big-list&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="12094" PostTypeId="1" AcceptedAnswerId="12101" CreationDate="2011-07-09T19:42:06.330" Score="4" ViewCount="154" Body="&lt;p&gt;Noether's theorem gives rise to quantities that are conserved over time. But does it also give rise to quantities that are conserved over space?&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-07-10T07:33:36.743" Title="Does Noether's theorem also give rise to quantities conserved over space?" Tags="&lt;noethers-theorem&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="12095" PostTypeId="2" ParentId="12090" CreationDate="2011-07-09T19:50:47.753" Score="3" ViewCount="" Body="&lt;p&gt;What you're looking for is actually &lt;em&gt;energy&lt;/em&gt;, not power, and you can put an upper limit on the amount you can get by computing the gravitational potential energy lost by the water as it drops. If the volume of the tank is $V = X\text{ m}^3$ and its height above the ground (or more precisely, above the point where you extract the energy) is $h = Y\text{ m}$, the amount of energy you get is no greater than&#xA;$$E = \rho V g h$$&#xA;where $\rho$ is the density of water and $g$ is the gravitational acceleration. If you put in all the numbers and unit conversions to get it in kilowatt-hours, that works out to&#xA;$$E = 1000\frac{\mathrm{kg}}{\mathrm{m}^3}\times X\text{ m}^3\times 9.81\frac{\mathrm{m}}{\mathrm{s}^2}\times Y\text{ m}\times\biggl(3.60\times 10^6\frac{\mathrm{kWh}}{\mathrm{J}}\biggr) = 0.00273XY\text{ kWh}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, of course, you wouldn't get that much energy out. In order to do so, you would have to extract &lt;em&gt;all&lt;/em&gt; kinetic energy from the water as it falls - that is, your generator would have to stop the flow of water completely, and no real generator can actually do that. The closest you can get is probably a &lt;a href=&quot;http://en.wikipedia.org/wiki/Water_wheel#Backshot_wheel&quot; rel=&quot;nofollow&quot;&gt;backshot water wheel&lt;/a&gt;, although water wheels suffer from other inefficiencies that reduce the amount of energy you actually get down to about 60% of the theoretical maximum. Although it doesn't actually stop the water, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Water_turbine&quot; rel=&quot;nofollow&quot;&gt;turbine&lt;/a&gt; is mechanically better and actually can get you a higher efficiency overall.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-09T19:56:28.620" LastActivityDate="2011-07-09T19:56:28.620" />
  <row Id="12096" PostTypeId="1" AcceptedAnswerId="12098" CreationDate="2011-07-09T19:56:09.453" Score="3" ViewCount="63" Body="&lt;p&gt;Was wondering if trapping light in a reflective chamber could heat rods sufficiently enough to heat water? Trying to think of new ways in which we can utilise solar energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/WkixY.jpg&quot; alt=&quot;Ref pic A.&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4271" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-12T15:26:12.010" LastActivityDate="2011-07-12T15:26:12.010" Title="Solar chamber trapped light" Tags="&lt;thermodynamics&gt;&lt;light&gt;&lt;heat&gt;&lt;renewable-energy&gt;" AnswerCount="2" />
  <row Id="12097" PostTypeId="2" ParentId="12069" CreationDate="2011-07-09T20:04:16.523" Score="3" ViewCount="" Body="&lt;p&gt;It could due to multi-photon effects. It's already known that cell phones cause changes to brain activity. This was widely reported in the news. For example see &quot;&lt;a href=&quot;http://abcnews.go.com/Health/Wellness/cell-phone-study-cell-ups-brain-activity/story?id=12971636&quot; rel=&quot;nofollow&quot;&gt;Cell Phone Study: Cell Ups Brain Activity&lt;/a&gt;&quot;. As of now, this is an unknown effect. So it might be associated with something that causes cancer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm not worrying.)&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-07-09T20:04:16.523" CommentCount="5" />
  <row Id="12098" PostTypeId="2" ParentId="12096" CreationDate="2011-07-09T20:05:30.393" Score="2" ViewCount="" Body="&lt;p&gt;Yes, harnessing light can be a good way to heat water to usable hot temperatures, i.e. 50 to 80 degrees Celsius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two factors that are important are the surface area of your collector (which looks quite small in your design), and the efficiency of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, we use either flat plates (high area) or (partially-)evacuated tubes (high efficiency).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sunlight is around 1kW per square metre at the surface in full sun, and thermal collector efficiencies might be in the range 50-90% or so. Depending on where on earth you are, you might get 1-7 full sun hours per day. External air temperatures, collector tilt, alignment, and your latitude, will all affect the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a home in the UK, solar thermal energy should be sufficient for a small installation to meet half the home's annual hot water demand. If you can tap into larger-scale seasonal storage, or you're closer to the equator, you can increase that proportion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an Evacuated-tube diagram from &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_thermal_collector&quot; rel=&quot;nofollow&quot;&gt;wikipedia's article on solar thermal collectors&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/pXZLk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4066" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-09T20:11:08.670" LastActivityDate="2011-07-09T20:11:08.670" CommentCount="2" />
  <row Id="12099" PostTypeId="2" ParentId="12094" CreationDate="2011-07-09T20:10:23.337" Score="3" ViewCount="" Body="&lt;p&gt;Yes, in a field-theoretic setting. Consider a $3+1$ dimensional flat space-time for simplicity. &lt;a href=&quot;http://en.wikipedia.org/wiki/Noether%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Noether's theorem&lt;/a&gt; gives rise to a conservation law of the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_{\mu} J^{\mu}(x)=0,$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. the Noether current $J^{\mu}(x)$ is a divergence-less four-current. [We use the symbol $d_{\mu}$ (rather than $\partial_{\mu}$) to stress the fact that the derivative $d_{\mu}$ is a &lt;em&gt;total&lt;/em&gt; derivative, which involves both implicit differentiation through the field variables $\phi^{\alpha}(x)$, and explicit differentiation wrt. $x^{\mu}$.] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say we want to consider a quantity $Q$ that is independence of the $x^1$-coordinate. Define the so-called 'charge' $Q=Q(x^1)$ by integrating $J^1(x)$ over the $2+1$ dimensional plane with fixed $x^1$-coordinate $x^1$. Then it is easy to show using a $2+1$ dimensional &lt;a href=&quot;http://en.wikipedia.org/wiki/Divergence_theorem&quot; rel=&quot;nofollow&quot;&gt;divergence theorem&lt;/a&gt; that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_{1} Q(x^1)=0$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;by imposing pertinent boundary conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above construction may be vastly generalized to a geometrically covariant framework, where the preferred direction is given by a non-vanishing vector field, which can be time-like, space-like, light-like, or some combination.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-10T07:33:36.743" LastActivityDate="2011-07-10T07:33:36.743" />
  <row Id="12100" PostTypeId="1" CreationDate="2011-07-09T20:28:04.623" Score="1" ViewCount="65" Body="&lt;p&gt;Why is it not possible to store windfarm energy in battery banks? A lot of energy isn't being used when wind farms are at peak and we have no way of storing it. This is wasteful. So why don't we use  battery banks?&lt;/p&gt;&#xA;" OwnerUserId="4271" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-12T15:39:28.647" LastActivityDate="2011-07-12T15:39:28.647" Title="Energy storage and battery banks" Tags="&lt;energy&gt;&lt;energy-conservation&gt;&lt;renewable-energy&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="12101" PostTypeId="2" ParentId="12094" CreationDate="2011-07-09T20:56:53.837" Score="7" ViewCount="" Body="&lt;p&gt;In basic Lagrangian mechanics (of the sort that is covered in a sophomore-level classical mechanics class), no it doesn't. The reason is that time plays a special role in the basic Lagrangian theory: it's the only independent parameter, which everything else is expressed as a function of. This is related to the fact that the action is the integral of the Lagrangian over &lt;em&gt;time&lt;/em&gt;, not space, and that in turn means that the &quot;classical&quot; version of Noether's theorem only works for conservation in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when you generalize to field theory, the situation changes: in field theory, both the time and space coordinates are considered to be independent parameters, so that everything else is expressed as a function of both time and space. In particular, instead of the classical Lagrangian, you have a Lagrangian density $\mathcal{L}$, which lets you express the action as a spacetime integral,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S = \int \mathcal{L}\mathrm{d}^4x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the field theory version of Noether's theorem doesn't assign any special status to time. Instead of temporal conservation laws ($\frac{\mathrm{d}Q}{\mathrm{d}t} = 0$), it gives you spacetime conservation laws of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial j^\mu}{\partial x^\mu} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can turn this into a temporal conservation law by integrating the current $j^\mu$ over a spacelike volume (i.e. all of space at a single moment in time):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q = \int_{t=\text{const}} g_{\mu\nu} j^\mu\mathrm{d}x^\nu\quad\to\quad\frac{\mathrm{d}Q}{\mathrm{d}t} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but you can just as well turn it into a spatial conservation law by integrating over a spatiotemporal volume:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q = \int_{x=\text{const}} g_{\mu\nu} j^\mu\mathrm{d}x^\nu\quad\to\quad\frac{\mathrm{d}Q}{\mathrm{d}x} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this way, yes it is possible to create a spatial conservation law using Noether's theorem.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-10T02:50:27.507" LastActivityDate="2011-07-10T02:50:27.507" CommentCount="4" />
  <row Id="12102" PostTypeId="2" ParentId="12090" CreationDate="2011-07-09T21:14:09.363" Score="0" ViewCount="" Body="&lt;p&gt;Water turbines are a matter of pressure and flow if the pipe coming from the tank has 90s in it this will restrict the flow and the efficiency goes down. But the point to consider here is that your method of storing energy “water in a tank” is cheep and efficient in the long run. I helped some friends make a system for storing excess energy this way. When the solar panels have charged the batteries and the electricity is not needed the pump to reservoir is turned on. Late night when the wind turbine kicks in it does that same. This method made it possible to need 0 power from the grid. The point I’m trying to make is that the choice of turbine “efficiency” the height of the reservoir “pressure” and the design of the pipe system “flow” are THE determining factors way before you do the math. In your example a standard garden hose is ½” in diameter you must include the light and the layout of the hose. At lower pressures you will get more flow from a hose that is straight rather than in a coil. You need to monitor all this factors in your experiments.&lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-07-09T21:14:09.363" />
  <row Id="12104" PostTypeId="2" ParentId="12100" CreationDate="2011-07-09T21:50:57.350" Score="0" ViewCount="" Body="&lt;p&gt;If you know how to make a gigawatt battery that does no occupy a football field and does not have sufficient pollutants to kill the state AND where do you dispose of it when it dies. Then YA sure put it in batteries. Sorry for the sarcasm but there is many ways to store energy but if you don't use oil we don't consider them &amp;lt;--&quot;disregard remark it's not physics&quot;.  &lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-07-09T21:50:57.350" CommentCount="1" />
  <row Id="12105" PostTypeId="1" CreationDate="2011-07-09T21:51:27.667" Score="1" ViewCount="84" Body="&lt;p&gt;I read that powerful pulsed lasers can change isotopes:&#xA;&lt;a href=&quot;http://www.springerlink.com/content/89c45n12bxkn075l/&quot; rel=&quot;nofollow&quot;&gt;J. Magill, et. al.: &quot;Laser transmutation of iodine-129&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did anyone estimate what would be the energy costs to transmutate 1 kg of fission product from a conventional reactor using such a laser or an accelerator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it even possible to transmutate a mixture of isotopes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would that be a viable alternative to long term storage if we find a way to get enough energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now I seriously doubt that we could ever process 100s of kilograms using such techniques.&lt;/p&gt;&#xA;" OwnerUserId="4306" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-07-10T11:16:23.733" LastActivityDate="2011-07-10T12:45:58.273" Title="Is it viable to convert nuclear waste into safe isotopes?" Tags="&lt;nuclear-physics&gt;&lt;nuclear-engineering&gt;&lt;isotope&gt;" AnswerCount="2" />
  <row Id="12106" PostTypeId="2" ParentId="12096" CreationDate="2011-07-09T21:56:16.013" Score="0" ViewCount="" Body="&lt;p&gt;Any mirror can reflect light onto an absorbing rod, and increase the&#xA;heat transfer to that rod.  A 'reflecting chamber', however, sounds like&#xA;a black body made with reflecting surfaces and a few absorbers, and THAT is&#xA;less effective in proportion to the number of reflections before light hits&#xA;the target.  Reflecting surfaces may be 95% reflective, but all will absorb&#xA;at least a LITTLE light, which will therefore not produce heat in the&#xA;absorbing rods.   There's also the problem of the light that inevitably&#xA;gets scattered out of the aperture, which might just hit your mirrors&#xA;and exit without ever encountering any target.&lt;br&gt;&#xA;That's why careful geometry and aiming  of reflectors is a feature of a good&#xA;solar furnace; generally incoming sunlight makes one reflection then it's on&#xA;the target.&#xA;The absorptivity of the rod (if it does 95% absorption, some light is lost)&#xA;can be enhanced by using some optical tricks, like making the rod black&#xA;at visible-light wavelengths but white at infrared (this mimics the greenhouse effect).&lt;/p&gt;&#xA;" OwnerUserId="4272" LastActivityDate="2011-07-09T21:56:16.013" />
  <row Id="12107" PostTypeId="2" ParentId="12100" CreationDate="2011-07-09T22:05:14.673" Score="3" ViewCount="" Body="&lt;p&gt;We can. But they don't scale well: to get grid-level storage, you need to be able to scale to gigawatts of power, and gigawatt-hours of energy. And to be able to cycle hundreds, or thousands, of times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To date, we have one technology which will do that, which is pumped storage hydro, which typically has a round-trip efficiency of 75%. So even though that's worse than a decent battery's round-trip efficiency, its scalability means that it dominates grid storage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that's just one small part of the picture. Behind the question of storage, is the physics question of how do you balance an electricity grid. The grid typically has very low capacitance, so electricity in and electricity out must balance at every second. To manage that balance, you can either adjust the amount going in, or the amount going out, or both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of ways to integrate high wind penetrations into the grid: this is a solved problem technically. See, for example, the &lt;a href=&quot;http://dx.doi.org/10.1016/j.enpol.2010.11.040&quot; rel=&quot;nofollow&quot;&gt;work in Energy Policy&lt;/a&gt; by &lt;a href=&quot;http://dx.doi.org/10.1016/j.enpol.2010.11.045&quot; rel=&quot;nofollow&quot;&gt;Delucchi &amp;amp; Jacobson&lt;/a&gt;; or &lt;a href=&quot;http://www.theiet.org/publishing/books/renewable/scenarios.cfm&quot; rel=&quot;nofollow&quot;&gt;the book by Gregor Czisch on renewable scenarios&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are more ways to do virtual storage than direct storage. For example, delaying consumption of 1GWh of electrical energy at 1GW power, for one hour, is equivalent to storing it at 100% efficiency for one hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the UK, a lot of energy is used for domestic hot water use. So thermal storage and delayed heating of that thermal storage, can act as a virtual storage for electrical heating of water. To put some numbers on that, UK domestic hot water storage is currently about equivalent to 60GWh @ 30GW for 24 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, with 20 million cars, if they were all electrified, you might have 20 million 50kWh batteries, which is 1TWh of electrical storage. V2G (vehicle to grid) studies often refer to a round-trip efficiency of about 75%, which comes from 10% loss on charge, 10% loss on discharge, and a few percent on transmission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find out more, there's a wealth of literature on integrating renewables into the grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper &quot;Energy-storage technologies and electricity generation&quot;, &lt;a href=&quot;http://dx.doi.org/10.1016/j.enpol.2008.09.037&quot; rel=&quot;nofollow&quot;&gt;Hall &amp;amp; Bain, Energy Policy 2008&lt;/a&gt; available as a &lt;a href=&quot;http://www.bis.gov.uk/assets/bispartners/foresight/docs/energy/energy-storage-technologies-and-electricity-generation.pdf&quot; rel=&quot;nofollow&quot;&gt;pdf here&lt;/a&gt;, sets out some of the issues of batteries and the grid.&lt;/p&gt;&#xA;" OwnerUserId="4066" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-09T22:17:30.260" LastActivityDate="2011-07-09T22:17:30.260" />
  <row Id="12108" PostTypeId="1" CreationDate="2011-07-09T22:33:51.117" Score="1" ViewCount="31" Body="&lt;p&gt;This question is related to &lt;a href=&quot;http://physics.stackexchange.com/questions/11978/impact-of-covering-glass-on-lens-performance&quot;&gt;Impact of covering glass on lens performance&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use a 63x TIRF objective with a numerical aperture of 1.46 and oil immersion.&#xA;The immersion oil has an index of $n_e=1.518$.&#xA;I'm observing green eGFP fluorescence at around 510 nm.&#xA;Does it matter if I use a coverslip (BK7 glass) with 140 $\mu$m thickness instead of 170 $\mu$m? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The objective is a Zeiss Objective alpha &quot;Plan-Apochromat&quot; 63x/1.46 Oil Corr M27&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/zgRpH.png&quot; alt=&quot;Zeiss Objective&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4306" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-07-09T23:06:21.770" LastActivityDate="2011-08-08T23:21:10.567" Title="Does coverslip thickness matter for high NA oil immersion objectives?" Tags="&lt;optics&gt;&lt;refraction&gt;&lt;microscopy&gt;" AnswerCount="1" />
  <row Id="12109" PostTypeId="1" CreationDate="2011-07-09T22:35:16.723" Score="1" ViewCount="34" Body="&lt;p&gt;Another crazy question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) I will need to generate some ozone for chemical reactions, and what seems to be the most straightforward way is using germicidal lamps, but the question is how to estimate how much ozone is produced (for let's say 100W of lamps) and what should be inlet oxygen flow for maximum ozone concentration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Also, I see that there are ozone-generating &amp;amp; non-ozone generating germicidal lamps - based on whether their glass absorb 193nm line or not. If anyone worked with non-ozone producing lamps - do you ever feel ozone odor when working with them, or ozone generation is 0? For example lamps I have nearby give ozone odor in few seconds after turning on at the distance of 1 meter (with no air movement) - but that might not mean anything as humans are very sensitive to ozone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) And finally, how many centimeters of oxygen under normal conditions is needed to absorb half of 193nm light? If it's more than few centimeters, I guess I will have to build box with reflectors - as far as I see, aluminum foil should reflect 193nm light back, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. Oxygen is to be generated by electrolysis of ultra pure water, so I guess it will be 100% wet. I cannot mess with compressed oxygen, that's a little over my danger tolerance :-) , and chemical oxygen generation might give me too much contaminants in line with other difficulties (hard to maintain low constant flow for hours, especially when not using water).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PPS. I know that both germicidal lamps &amp;amp; ozone is very dangerous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PPPS. I guess this is still not chemistry question, as it mainly focuses on UV light. Anyway, we don't have such stack-exchange site...&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-07-09T22:35:16.723" ClosedDate="2011-07-09T23:22:43.093" Title="Ozone generation" Tags="&lt;electricity&gt;" CommentCount="3" />
  <row Id="12110" PostTypeId="2" ParentId="12108" CreationDate="2011-07-09T22:39:52.777" Score="2" ViewCount="" Body="&lt;p&gt;Yes, you are getting a bit higher aberrations in the corners of the image (i.e. they might be a little more blurry).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as image quality is ok for you, you may continue doing so.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-07-09T22:39:52.777" CommentCount="8" />
  <row Id="12111" PostTypeId="2" ParentId="12105" CreationDate="2011-07-09T22:55:27.780" Score="7" ViewCount="" Body="&lt;p&gt;The funny thing is that in the case described in the paper, pretty safe isotope (I-129) was converted into deadly dangerous I-128, I would prefer to leave it as is :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transmutating random mix of isotopes would give you more random mix of isotopes (garbage in-&gt;garbage out), and it's a big question if this would lead to lower integral activity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to transmutate isotopes sensibly, we would need to separate waste into individual isotopes first. Another funny thing is that it would automagically solve the problem - as separated radioactive isotopes are all precious - part of them may be turned back to fuel (mainly U, Pt) , part of them may be used for radioisotope sources of energy, and the rest sold for medical &amp;amp; industrial equipment, or stored much more compact or safe than it is now. The problem is that it's not very commercially viable to process all fuel at the moment - only relatively small part of waste is processed (on 'Mayak' in Russia for example).&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-07-09T22:55:27.780" CommentCount="1" />
  <row Id="12112" PostTypeId="2" ParentId="6208" CreationDate="2011-07-10T00:38:51.697" Score="1" ViewCount="" Body="&lt;p&gt;For the average person (no science background) I recommend going to &lt;a href=&quot;http://ocw.mit.edu/courses/chemistry/5-111-principles-of-chemical-science-fall-2008/&quot; rel=&quot;nofollow&quot;&gt;MIT ocw&lt;/a&gt;, buying an &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/071675701X&quot; rel=&quot;nofollow&quot;&gt;older edition of their book&lt;/a&gt; (for litterally 5$ - shipping included) off amazon, and doing the course on your own. They have everything there except for homework solutions and a professor/pupils. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: With your background in physics, general chemistry is going to be extremely boring in the areas that overlap with physics. The area of chemistry furthest from physics is probably organic chemistry (it boarders with biology). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vast majority of chemistry that is neither biological nor physics-related is merely computing how chemicals react together. A little of that is touched on in general chemistry, but much more detail is examined in later courses such as Quantitative Analysis (that is what my school called it - a 3rd year course usually). &lt;/p&gt;&#xA;" OwnerUserId="2843" LastEditorUserId="2843" LastEditorDisplayName="" LastEditDate="2011-07-10T02:04:22.577" LastActivityDate="2011-07-10T02:04:22.577" CommunityOwnedDate="2011-07-10T00:38:51.697" />
  <row Id="12114" PostTypeId="2" ParentId="2131" CreationDate="2011-07-10T01:03:38.103" Score="0" ViewCount="" Body="&lt;p&gt;It is true that nearly all of the observational evidence for the accelerating universe - that is, the presence of dark energy - comes from the measurements of the expansion rate (the Hubble parameter H(z), for aficionados). These measurements are usually not direct measurements of the expansion rate, but rather measurements of distances (to type Ia supernovae, for example), preferred length scales in the distribution of galaxies or in the distribution of cold and hot spots of the cosmic microwave background, and other geometrical measurements. All of these things directly depend on the expansion rate, and they have been the fundamental means of how we found out about dark energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A qualitatively different signature of dark energy is its effect on the growth of density fluctuations in the universe - that is, the rate at which galaxies and other objects form in time. Recent measurements (of galaxy distribution at different cosmic epochs etc) indicate that measurements of this growth also indicates dark energy. [In fact, ALL of the measurements in cosmology are in perfect accord with the presence of dark energy.] However, in General Relativity, the growth rate can be shown to ALSO depend mainly on the expansion rate H(z). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the answer to Bruce Connor's excellent question is unfortunately not clear-cut: while many distinct experiments indicate dark energy, essentially all of them are based off of measurements of the rate of expansion - or the Hubble parameter H(z). &lt;/p&gt;&#xA;" OwnerUserId="1334" LastActivityDate="2011-07-10T01:03:38.103" />
  <row Id="12115" PostTypeId="2" ParentId="2131" CreationDate="2011-07-10T02:53:07.133" Score="-1" ViewCount="" Body="&lt;p&gt;Space expansion? yes. It is measured in relation to the atoms around us.&lt;br&gt;&#xA;What if the atom can vary its dimensions thru time?&lt;br&gt;&#xA;I'm pretty sure that no one presented evidence that the atom is invariant.&lt;br&gt;&#xA;And yet, everybody is claiming that the universe expands.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space expansion or matter shrinks ?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The search of a scaling model of the universe, a self-similar one or dilation, has been pursued by the scientific community since Dirac, Hoyle &amp;amp; Narlikar, and others without results.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A scaling model is born, derived from data, using standard physics and making no hypotheses, this model has only one parameter ($H_0$) :  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://vixra.org/pdf/1107.0016v1.pdf&quot; rel=&quot;nofollow&quot;&gt;A Self-Similar Model of the Universe Unveils the Nature of Dark Energy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, from now on I'will ask for proper evidence (theoretical or experimental) that the atom is invariant every-time that I hear someone to say: the universe is expanding.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dark Energy? NO, imo it is an artifact of the reference we use (invariant atomic units).    &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-07-10T02:53:07.133" />
  <row Id="12116" PostTypeId="1" AcceptedAnswerId="12124" CreationDate="2011-07-10T04:35:17.887" Score="6" ViewCount="106" Body="&lt;p&gt;Many things have glow in the dark properties (glow sticks, paint, toys ..), and I am wondering what is the physics behind them. How do these materials store light energy and emit it later when dark?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What dictates the wavelength(s) of the glow? Is there a limit on how much energy can be stored and recovered in these materials? Do they give out heat as well as light?&lt;/p&gt;&#xA;" OwnerUserId="392" LastEditorUserId="4444" LastEditorDisplayName="" LastEditDate="2011-07-10T16:01:07.730" LastActivityDate="2011-07-10T16:01:07.730" Title="How does stuff glow in the dark?" Tags="&lt;light&gt;&lt;chemical-compounds&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12117" PostTypeId="2" ParentId="12116" CreationDate="2011-07-10T06:37:36.737" Score="5" ViewCount="" Body="&lt;p&gt;This phenomenon is called phosphorescence. Electrons in the material are pushed into excited states by light, and then drop back down, emitting light with a characteristic frequency corresponding to the energy of the transition. It is analogous to fluorescence, however whereas fluorescence is a very rapid process, phosphorescence is delayed due to the electronic transition involved being 'forbidden' (in practice, this means low-probability at non-zero temperatures). As such, the process of all of the atoms or molecules in the material returning to their ground state is a drawn out over minutes or hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Glow sticks are a different matter - they employ an organic fluorescent dye and an oxidising agent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wavelength: determined by the energy of the electronic transition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy limit: You will eventually reach a point of saturation where the material is totally excited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat emission: probably.&lt;/p&gt;&#xA;" OwnerUserId="1876" LastActivityDate="2011-07-10T06:37:36.737" CommentCount="0" />
  <row Id="12118" PostTypeId="2" ParentId="12046" CreationDate="2011-07-10T07:32:29.127" Score="3" ViewCount="" Body="&lt;p&gt;If you compare at the oscillating light waveform for the yellow of the sun, and the oscillating light waveform for the similar-looking yellow of the computer screen (made of red and green pixels), the waveforms look entirely different, their Fourier transforms look entirely different, and in fact they have absolutely no mathematical relationship of any kind. The only relationship that the two waveforms have is: &quot;both wavesforms happen to excite the three human &lt;a href=&quot;http://en.wikipedia.org/wiki/Photoreceptor_cell&quot; rel=&quot;nofollow&quot;&gt;photoreceptor cells&lt;/a&gt; in the same X:Y:Z ratio&quot;.&lt;/p&gt;&#xA;" OwnerUserId="3811" LastActivityDate="2011-07-10T07:32:29.127" />
  <row Id="12119" PostTypeId="2" ParentId="3096" CreationDate="2011-07-10T08:21:43.233" Score="1" ViewCount="" Body="&lt;p&gt;Here, is the method to deal with DIVERGENT INTEGRALS (single and multiple) in quantum physics by using the method of DIVERGENT SERIES and Zeta regularization&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://vixra.org/pdf/1009.0047v4.pdf&quot; rel=&quot;nofollow&quot;&gt;http://vixra.org/pdf/1009.0047v4.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://vixra.org/pdf/1009.0047v4.pdf&quot; rel=&quot;nofollow&quot;&gt;http://vixra.org/pdf/1009.0047v4.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for multiple integrals we can apply the method on each variable, so we can get FINITE RESULTS using zeta regularization for multi-loop integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logarithmic divergence is handled with the FUNCTIONAL DETERMINANT or infinite product over all the integers (1+x)(2+x)........ using Hurwitz Zeta function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so we have managed to get only finite results in QFT :) using Zeta regularization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;not only for DIVERGENT SUMS but also for DIVERGENT INTEGRALS we can obtain finite corrections using the Zeta regularization algorithm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my paper in the subject&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://vixra.org/abs/1009.0047&quot; rel=&quot;nofollow&quot;&gt;http://vixra.org/abs/1009.0047&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this can be used to regularize DIVERGENT INTEGRALS in quantum mechanics, and MULTIPLE integrals by doing a term by term integration on each variable, you are free to check it out and give your opinion.&lt;/p&gt;&#xA;" OwnerUserId="4435" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-07-10T22:43:55.607" LastActivityDate="2011-07-10T22:43:55.607" />
  <row Id="12120" PostTypeId="1" AcceptedAnswerId="12133" CreationDate="2011-07-10T08:36:18.917" Score="0" ViewCount="88" Body="&lt;p&gt;I have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Uninterruptible_power_supply&quot; rel=&quot;nofollow&quot;&gt;UPS&lt;/a&gt; of 1000 Volts connected with 2 batteries each of 150 Amp. How much time it will take to consume the whole UPS (after fully charged) when a device of 1Amp is getting electricity form that UPS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please also explain me the calculation.&lt;/p&gt;&#xA;" OwnerUserId="4440" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-10T12:26:59.843" LastActivityDate="2011-07-10T19:02:48.870" Title="Calculating time for a fully charged UPS" Tags="&lt;electricity&gt;&lt;electric-circuits&gt;&lt;electrical-engineering&gt;" AnswerCount="3" />
  <row Id="12121" PostTypeId="2" ParentId="12120" CreationDate="2011-07-10T09:11:21.400" Score="0" ViewCount="" Body="&lt;p&gt;Don't think you can calculate this. I depends on the type, age and history of the batteries. All his factors make a big difference. If you rely need to know the TCT &quot;total charge time&quot; put them threw the cycle once and measure it. And remember this will change with time. &lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-07-10T09:11:21.400" />
  <row Id="12122" PostTypeId="1" CreationDate="2011-07-10T09:20:29.947" Score="6" ViewCount="312" Body="&lt;p&gt;I am following the first volume of the course of theoretical physics by Landau. So, whatever I say below mainly talks regarding the first 2 chapters of Landau and the approach of deriving Newton's laws from Lagrangian principle supposing Hamilton's principle of extremum action. Please keep this view in mind while reading and answering my queries and kindly neglect the systems to which Action Principle is not applicable:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we use homogeneity of space in Euler-Lagrange equations, we obtain a remarkable result i.e. the conservation of momentum for a closed system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this result, using the form of Lagrange for a closed system of particles, transforms into $ \Sigma F = 0 $ . Now, how from this can we conclude that the internal forces that particles exert come in equal and opposite pairs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because for 2 particles this comes out as $ F_{1} + F_{2} = 0 $ and we take the forces exerted by particles on one other to be independent of other particles (i.e. Superposition Principle) as an experimental fact?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I doubt it as whole of Newtonian Mechanics is derivable from Lagrangian Mechanics and supposed Symmetries. So, according to me, a fact like Newton's Third Law should be derivable from it without using an additional experimental fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an idea to prove it rigorously. Consider two particles $i$ and $j$. Let the force on $i$ by $j$ be $F_{ij}$ and on $j$ by $i$ be $k_{ij}F_{ij}$. Now the condition becomes $\Sigma (1+k_{ij})F_{ij}=0$ where the terms to be included and rejected in summation understood. As this must be true for any value of $F_{ij}$, we get $k_{ij}=-1$. I don't know if this argument or refinement of such an argument holds or not. I can see many questions arising in this argument and it's not very convincing to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to hear from you people as to if it is an experimental result used or not? If not, then is the method given above right or wrong? If wrong, how can we prove it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My method of proof uses the fact of superposition of forces itself, so it is flawed. I have assumed that the coefficients $k_{ij}$ are constants and don't change in the influence of all other particles which is exactly what superposition principle says.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the superposition of Forces can be derived by superposition of potential energies at a point in space and potential energy being more fundamental in Lagrangian Mechanics, I restate my question as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Is the principle of superposition of potential energies by different sources at a point in space derivable from the inside of Lagrangian Mechanics or is it an experimental fact used in Lagrangian Mechanics?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I, now, doubt this to be derivable as the fundamental assumption about potential energy is only that it is a function of coordinates of particles and this function may or may not respect superposition.&lt;/p&gt;&#xA;" OwnerUserId="4235" LastEditorUserId="4235" LastEditorDisplayName="" LastEditDate="2011-07-12T11:37:39.523" LastActivityDate="2011-08-17T07:18:10.420" Title="Deriving Newton's Third Law from homogeneity of Space" Tags="&lt;lagrangian-mechanics&gt;&lt;conservation-laws&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="12123" PostTypeId="2" ParentId="12120" CreationDate="2011-07-10T09:26:40.247" Score="1" ViewCount="" Body="&lt;p&gt;Battery capacity is normally rated in ampere-hours (Ah), not ampere. This means what it sounds like - if you have 2 * 150 Ah of batteries, you can (ideally) pull 1A for 300 hours like in your question (*), or 300 amps for 1 hour. This is an intuitive value of battery capacity but is not strictly speaking a measurement of the energy stored since this depends on the voltage which you pull the current at (and both of these vary during the discharge of a real battery).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You wrote that your batteries are rated at 150A, but you might have misread the 150 Ah figure, in which case your answer is above. You could also have quoted the &lt;em&gt;maximum current capacity&lt;/em&gt; of the battery of 150A (typical of a car battery for example), which isn't related to the capacity at all and I agree with the other answer here that you cannot calculate it without further information. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use the old water analogy - the voltage is, roughly speaking, the water pressure of the battery and the ampere is the rate of water in the stream coming from it. You would have no indication of the size of the reservoir so you cannot calculate for how long your device can run.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*)&#xA;This is of course assuming that you can connect your device at all, i.e. an UPS with an output of 1000 volts will not work very well if you connect it to a domestic 110V or 220V piece of equipment :)&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-07-10T09:26:40.247" />
  <row Id="12124" PostTypeId="2" ParentId="12116" CreationDate="2011-07-10T11:51:14.277" Score="2" ViewCount="" Body="&lt;p&gt;When one excites fluorescence with light (e.g. blue light), the incoming photon pushes an electron into a higher state. Generally one always excites vibrational states (movement of the molecules' atoms) as well. Therefore the emitted photon has lower energy (e.g. green light). The lost energy (internal conversion) heats the solvent or lattice. The shift in frequency (from blue to green) is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Stokes_shift&quot; rel=&quot;nofollow&quot;&gt;Stokes shift&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People plot these energy levels in &lt;a href=&quot;http://en.wikipedia.org/wiki/Jablonski_diagram&quot; rel=&quot;nofollow&quot;&gt;Jablonski diagrams&lt;/a&gt;. I stole this from &#xA;&lt;a href=&quot;http://biophys.med.unideb.hu/old/pharmacy/fluorescence_print.pdf&quot; rel=&quot;nofollow&quot;&gt;http://biophys.med.unideb.hu/old/pharmacy/fluorescence_print.pdf&lt;/a&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/ISFHj.png&quot; alt=&quot;Jablonski diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can only excite as many electrons as there are in the material. This means that absorbing laser goggle are not safe for high power pulsed laser. The early photons of the pulse will saturate all the electrons in the goggles and the rest of the pulse can advance into the eyes and do damage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a &lt;a href=&quot;http://en.wikipedia.org/wiki/Glow_stick&quot; rel=&quot;nofollow&quot;&gt;glow stick&lt;/a&gt; the excitation is done by chemical energy. Hydrogen peroxide splits some organic molecule and one of the produced excited carbon dioxide molecules can transfer its energy to a fluorophore. I once saw an experiment where the chemicals from the glow stick were added to tea and the tea's chlorophyll started glowing in red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only reason I can think of right now, why anyone would want to store energy like this, is to be able to access it with the speed of light. In high power pulsed lasers special crystals are pumped into the higher state and the energy can suddenly be converted into photons. That way it is possible to create pulses with Petawatts of power.&lt;/p&gt;&#xA;" OwnerUserId="4306" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-07-10T12:04:30.083" LastActivityDate="2011-07-10T12:04:30.083" />
  <row Id="12125" PostTypeId="2" ParentId="12105" CreationDate="2011-07-10T11:58:44.413" Score="2" ViewCount="" Body="&lt;p&gt;I may illustrate the already mentioned problem with mixture of many isotopes. Why only I-129 is chosen? In the paper it is compared with only two isotopes with life times actual for period more than 100 000 years after incident. If we talk about periods comparable with our time scale, the list of isotopes is rather big, e.g. for melting of a nuclear reactor core isotopes actual for long period of time are likely:&#xA;H-3, Sr-89, Sr-90, Y-91, Nb-93m, Nb-95, Ru-103, Ru-106, Ag-110m, Cd-113m, Cd-115m, Sn-121m, Sn-123, Sb-124, Sb-125, I-129, Cs-134, Cs-137, Ce-141, Ce-144, Pm-147, Tb-160, Pu-238, Pu-239, Pu-240, Am-241, Pu-241, Cm-242, Pu-242, Am-243, Cm-244 (the list was borrowed from a document published even before Fukushima and illustrates level of the problem).&lt;/p&gt;&#xA;" OwnerUserId="3351" LastEditorUserId="3351" LastEditorDisplayName="" LastEditDate="2011-07-10T12:45:58.273" LastActivityDate="2011-07-10T12:45:58.273" />
  <row Id="12126" PostTypeId="2" ParentId="12122" CreationDate="2011-07-10T14:16:49.523" Score="4" ViewCount="" Body="&lt;p&gt;Within the framework of classical mechanics, the Newton's third law is an independent postulate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton's third law in its strong form says that not only are mutual forces of action and reaction &lt;em&gt;equal&lt;/em&gt; and &lt;em&gt;opposite&lt;/em&gt; between two bodies at position $\vec{r}_1$ and $\vec{r}_2$, they are also &lt;em&gt;collinear,&lt;/em&gt; i.e. parallel to $\vec{r}_2-\vec{r}_1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we derive &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation&quot; rel=&quot;nofollow&quot;&gt;Lagrange equations&lt;/a&gt; from &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion&quot; rel=&quot;nofollow&quot;&gt;Newton's laws&lt;/a&gt; (see e.g. Herbert Goldstein, &quot;Classical Mechanics&quot;, chapter 1), it may appear a bit hidden when we actually use Newton's third law. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the derivation, we assume that forces of constraints do no virtual work$^{1}$. Consider now a rigid body. It's a fact that we rely heavily on Newton's third law in its strong form to argue that the internal forces of constraints (which hold the rigid body together) do no virtual work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://en.wikipedia.org/wiki/D%27Alembert%27s_principle&quot; rel=&quot;nofollow&quot;&gt;D'Alembert's principle&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Virtual_work#Principle_of_virtual_work_for_applied_forces_in_static_equilibrium&quot; rel=&quot;nofollow&quot;&gt;principle of virtual work&lt;/a&gt; for more information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^{1}$ This does not hold for, e.g., sliding friction forces, which we therefore have to exclude.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastActivityDate="2011-07-10T14:16:49.523" CommentCount="8" />
  <row Id="12127" PostTypeId="2" ParentId="414" CreationDate="2011-07-10T14:27:07.800" Score="0" ViewCount="" Body="&lt;p&gt;there is a conjecture so called RIEMANN HYPOTHESIS , which have a deep relationship between the roots of the Riemann Zeta function and the eigenvalues of a Hamiltonian&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1101.3116&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1101.3116&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://findarticles.com/p/articles/mi_m1200/is_7_174/ai_n30887057/&quot; rel=&quot;nofollow&quot;&gt;http://findarticles.com/p/articles/mi_m1200/is_7_174/ai_n30887057/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and my humble paper about the subject &lt;a href=&quot;http://vixra.org/pdf/1007.0005v8.pdf&quot; rel=&quot;nofollow&quot;&gt;http://vixra.org/pdf/1007.0005v8.pdf&lt;/a&gt; in fact RIEMANN HYPOTHESIS is juts to use WKB to find a Hamiltonian whose eigenvalues are the square of the RIemann zeros (imaginary part) and its FUNCTIONAL DETERMINANT is just the RIemann Xi function&lt;/p&gt;&#xA;" OwnerUserId="4435" LastActivityDate="2011-07-10T14:27:07.800" />
  <row Id="12128" PostTypeId="1" AcceptedAnswerId="12131" CreationDate="2011-07-10T14:55:35.663" Score="2" ViewCount="99" Body="&lt;p&gt;I've had a discussion with my father today, about the fuel usage of a vehicle at the same rpm, but a different gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He claims that the following situations have the same fuel usage:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Gear: 2&#xA;rpm: 2000&#xA;fuel usage: 5.1l/100km&#xA;&#xA;Gear: 5&#xA;rpm: 2000&#xA;fuel usage: 5.1l/100km&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I say it should be something along the lines of:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Gear: 2&#xA;rpm: 2000&#xA;fuel usage: 4.1l/100km&#xA;&#xA;Gear: 5&#xA;rpm: 2000&#xA;fuel usage: 5.4l/100km&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Who is right and why?&#xA;He is also claims that the force to maintain the speed will be the same across different gears with the same rpm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, I'm not sure if it should be on physics, but it fitted best here.&lt;/p&gt;&#xA;" OwnerUserId="4444" LastActivityDate="2011-07-10T16:10:02.717" Title="Fuel usage at the same constant rpm at different gears" Tags="&lt;energy&gt;&lt;speed&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="12129" PostTypeId="1" AcceptedAnswerId="12130" CreationDate="2011-07-10T15:05:59.297" Score="0" ViewCount="160" Body="&lt;p&gt;I was watching Discovery channel the other night, they were telling that time slows down when you travel at a higher speed. This means there is a difference between the actual speed you travel at, and the perceived speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know what the perceived speed is, the speed it seems you're travelling at?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this imply? Does it have any meaning, in some theories perhaps, that the perceived speed is lower? Does this mean that you can never actually travel at the speed of light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(All this is if we don't account for the fact that's impossible to reach the speed of light.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also not into physics, so try to keep it simple...&lt;/p&gt;&#xA;" OwnerUserId="4444" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-07-10T16:01:26.827" LastActivityDate="2011-07-11T07:50:04.197" Title="Speed of light, observed speed while travelling at the speed of light" Tags="&lt;special-relativity&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="12130" PostTypeId="2" ParentId="12129" CreationDate="2011-07-10T15:57:05.953" Score="1" ViewCount="" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;I apologise, this answer as it stood originally was sloppy and flat out wrong on key points, and probably still needs attention from a physicist.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you travel close to the speed of light, you experience less subjective time, according to the Lorentz transform. Moreover, external distances are contracted by the same transform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The factor by which time slows down for a rapidly moving object is $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$. As velocity $v$ increases it approaches equality with the speed of light, $c$, and the factor approaches infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that if you had an arbitrarily fast spacecraft (I assume you don't) you could tour the universe in (from your reference frame) a few years, however you may return to find humanity extinct and continents in unexpected locations. You would actually perceive yourself as travelling at arbitrarily high speeds, in terms of the subjective time it takes to travel from point to point. However, in making measurements against objects you move past, you would see that your subjective velocity does not exceed the speed of light as the rest of the universe appears length-contracted. There are other limitations in your ability to enjoy your tour of the universe - if you set your sights on a star that is millions of light-years away it may well rapidly age (again, from your reference frame) and die before you get there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lorentz_factor&quot; rel=&quot;nofollow&quot;&gt;Lorentz factor&lt;/a&gt; at wikipedia&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Relativistic_Doppler_effect&quot; rel=&quot;nofollow&quot;&gt;Relativistic Doppler Effect&lt;/a&gt; at wikipedia&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3rd book of Tipler Physics also has a great primer on relativity, albeit with neglect of acceleration, which is critical in generating discrepant clocks between two observers a la the twin paradox.&lt;/p&gt;&#xA;" OwnerUserId="1876" LastEditorUserId="1876" LastEditorDisplayName="" LastEditDate="2011-07-11T07:50:04.197" LastActivityDate="2011-07-11T07:50:04.197" CommentCount="5" />
  <row Id="12131" PostTypeId="2" ParentId="12128" CreationDate="2011-07-10T16:10:02.717" Score="4" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I've had a discussion with my father&#xA;  today, about the fuel usage of a&#xA;  vehicle at the same rpm, but a&#xA;  different gear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fuel injector system does not always inject a constant amount of fuel into the engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are driving down a hill and remove your foot from the gas pedal (a little) the engine will stay the same rpm but with less fuel passing thou the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you are driving up a hill you need to press down the gas pedal to maintain the same rpm and speed, more or less you need to push more fuel into the engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why you will not have a constant fuel consumption at a given rpm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then when you remove some load, let's say you are on the top of the hill, &#xA;you will have a small surplus of &quot;energy released&quot; inside the engine that will push the rpm higher until you remove your foot from the gas pedal and reducing the amount of fuel sent into the engine. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He is also claims that the force to&#xA;  maintain the speed will be the same&#xA;  across different gears with the same&#xA;  rpm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The faster you travel the more wind resistance you will have, therefore you need more power to overcome this resistance. &#xA;And to get more power out from a engine you need to send in more fuel (or use the amount have in there in a more efficient way).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;But why do we need to switch gears up and down then, &#xA;why can't we just inject more fuel into the engine? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where we go into chemistry since you can only make gas burn within a specified rate between fuel and air, and the volume inside the engine has a fix size and therefore has a maximum volume of gas that you can burn per rotation. Send in more fuel than this and the engine will drop in efficiency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And when you then switch down a gear you will increase the rpm at this speed and that way increase the amount of fuel you can burn in the engine within this timeframe and you will in most of the cases more power out from the engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But at the end of the day there is a lot of other constraints in a combustion engine that will limit how it works beside this simplified version I just describe here.&lt;/p&gt;&#xA;" OwnerUserId="4431" LastActivityDate="2011-07-10T16:10:02.717" CommentCount="1" />
  <row Id="12132" PostTypeId="2" ParentId="12122" CreationDate="2011-07-10T16:43:48.583" Score="0" ViewCount="" Body="&lt;p&gt;It might be a little crude, but doesn't time reversal symmetry help to prove this?&#xA;I can't quite think of a rigorous proof, but it seems to me that the matrix Fij should be transposed somehow if you change t-&gt;-t.&lt;/p&gt;&#xA;" OwnerUserId="4445" LastActivityDate="2011-07-10T16:43:48.583" CommentCount="2" />
  <row Id="12133" PostTypeId="2" ParentId="12120" CreationDate="2011-07-10T16:57:25.310" Score="2" ViewCount="" Body="&lt;p&gt;There are some missing data in your question.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What voltage does the batteries have, I'm going to assume 12V since it is common.&lt;/li&gt;&#xA;&lt;li&gt;Battery capacity, you typed it as 150A but I guess it was 150Ah. Please note that a normal car battery on a car like a VW Golf has approx 60Ah so 150Ah is a quite big battery.&lt;/li&gt;&#xA;&lt;li&gt;Output power, you state that you have 1000V (Volts) out, but I guess you are talking about 1000W (Watts). And that should be maximum power.&lt;/li&gt;&#xA;&lt;li&gt;Output voltage should be either 230V (or 110V). I will assume 230V.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first thing is that a device connected to the UPS and is using 1A, that is 1A at 230V or $230V * 1A = 230W.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we go inside the UPS and have a look. We still has the 230W but since it is now a 12V system, we need to reverse the conversion into amps again and get something like $\frac{230W}{12V}=19A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then we divide the battery capacity with this current and get something like, &#xA;$\frac{(150*2)Ah}{19A}=15h$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that those 15 hours is the best value you can expect to see, but since there is losses in the conversion from 12V to 230V we probably loose some 10-20% of the energy, and that translate directly into a shorter time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say 80% of 15h would be approx 12h.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Then I must add that Fortunato does have a point. Batteries degrade over time and can't hold the same charge, so make sure you have some margin and check/service your UPS on yearly basis. &lt;/p&gt;&#xA;" OwnerUserId="4431" LastEditorUserId="4431" LastEditorDisplayName="" LastEditDate="2011-07-10T19:02:48.870" LastActivityDate="2011-07-10T19:02:48.870" CommentCount="1" />
  <row Id="12134" PostTypeId="2" ParentId="12069" CreationDate="2011-07-10T17:03:17.413" Score="2" ViewCount="" Body="&lt;p&gt;This &lt;a href=&quot;http://en.wikipedia.org/wiki/Mobile_phone_radiation_and_health&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt; article  covers most bases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, the connection to cancer is unclear from controlled studies. There is some surface heat that can be generated when next to the head, of the order of 2 watts but not clear how bad that is. Sunbathing hatless heats by many more watts (1300/m^2). Cancer of the skin has been connected with the sun due to ultraviolet radiation ( cell phones are microwave range), but not the head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were a man I would avoid hanging a cell phone next to the family jewels, because heat is deleterious to fertility:).&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-07-10T17:03:17.413" CommentCount="4" />
  <row Id="12137" PostTypeId="1" CreationDate="2011-07-10T21:31:23.563" Score="0" ViewCount="100" Body="&lt;p&gt;Is the quantum measurement problem and the issue of quantum interpretations nothing but a word game? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Steven Weinberg,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So irrelevant is the philosophy of quantum mechanics to its use, that one begins to&#xA;  suspect that all the deep questions about the meaning of measurement are really empty, forced&#xA;  upon us by our language, a language that evolved in a world governed very nearly by classical&#xA;  physics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But what makes the &quot;words&quot; and language &quot;real&quot;? Or is the previous question merely more words? Do &quot;words&quot; &quot;really&quot; &quot;exist&quot;?&lt;/p&gt;&#xA;" OwnerUserId="4449" LastActivityDate="2011-07-10T22:12:08.767" ClosedDate="2011-07-10T22:47:23.827" Title="Is the quantum measurement problem merely a word game?" Tags="&lt;quantum-interpretations&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12138" PostTypeId="2" ParentId="12137" CreationDate="2011-07-10T22:12:08.767" Score="0" ViewCount="" Body="&lt;p&gt;The way I understand it is, that if we had &quot;discovered&quot; Quantum Mechanics before classical mechanics (for some strange reason), that we wouldn't have had to question the meaning of quantum measurement in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a matter of fact, we would probably find measurement in the classical sense rather strange. And we would be questioning that instead.&lt;/p&gt;&#xA;" OwnerUserId="4337" LastActivityDate="2011-07-10T22:12:08.767" />
  <row Id="12139" PostTypeId="2" ParentId="12058" CreationDate="2011-07-10T23:05:57.880" Score="2" ViewCount="" Body="&lt;p&gt;Your methodology assumes that the light source not only has Poissonian statistics but also has the same average intensity for each image. For example if you used a fluorescent tube plugged into a wall socket, its intensity goes up and down at 120Hz (in the US), because that's twice the AC frequency. So if your exposure time (a.k.a. integration time) was less than 10ms, you could big fluctuations not because of photon shot noise, but just because some exposures are taken with a brighter light source and others with a dimmer light source. So you could get the wrong photon conversion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Poisson statistics is not a concern, every macroscopic light source has Poisson statistics. I am quite confident that a halogen bulb sends out photons with Poisson statistics. I don't know what you heard about photons &quot;coming out in bunches&quot;, I haven't heard of that. Anyway even if they do, it would still be Poisson statistics for any given particle. Well, I guess it's important for this measurement that the fluorescent lifetime be much less than the time-separation between photons. So if the light source was really bunchy enough, it could mess up the data--not because the photons are non-Poissonian, but because of a &quot;saturable absorber&quot; effect. I would be surprised if this was an issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frame-to-frame light-intensity fluctuations, on the other hand, may or may not be a problem. I'm not sure what light sources are more or less stable. Of course it certainly depends on the exposure time of each frame and of the whole measurement. You can of course measure this stability directly. But actually in your graph, frame-to-frame-light-intensity-variations would give a variance proportional to the square of intensity, whereas shot noise is linear, and the graph you show is linear. So I guess it's not a problem!&lt;/p&gt;&#xA;" OwnerUserId="3811" LastActivityDate="2011-07-10T23:05:57.880" CommentCount="2" />
  <row Id="12140" PostTypeId="1" CreationDate="2011-07-10T23:51:11.073" Score="8" ViewCount="217" Body="&lt;p&gt;The origin of spin is some what a puzzle to me, every thing spin from galaxies to planets to weather to electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where has all the angular momentum come from? Why is it so natural?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was also thinking do photons spin? we always think of the wave as a standard 2d sin wave but could this rotate in 3d? What implications would this have? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what about spacetime how does all the spinning effect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has always been avoided in all lectures and classes i ever went to.&lt;/p&gt;&#xA;" OwnerUserId="4451" LastActivityDate="2011-07-11T09:40:44.973" Title="Why does every thing spin?" Tags="&lt;cosmology&gt;&lt;angular-momentum&gt;&lt;spin&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" />
  <row Id="12141" PostTypeId="2" ParentId="12089" CreationDate="2011-07-11T00:24:05.180" Score="2" ViewCount="" Body="&lt;p&gt;A diode consists of two materials known as p-type and n-type semiconductors, connected in series which allows current to flow through them differently. In the n-type semiconductor, electrons travel with enough energy such that they're not attached to an atom and are said to be in the conduction energy band. For the p-type semiconductor, electrons &quot;hop&quot; from atom to atom, but lacking the energy to free them, are said to be in the valency energy band.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the interface between the n-type and p-type materials, a travelling electron has to move either from the n-type to the p-type in one direction, the p-type to the n-type in the other, to continue moving. Is there a difference between the two directions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, an electron moving from the n-type to the p-type material can occur spontaneously because the free electron's energy is released as radiation and it can move to a lower energy state, attached to an atom in the p-type semiconductor. But to move from the p-type to the n-type it has to gain energy from somewhere, and this isn't spontaneous because there is no guarantee of some other process providing this energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of a ball at the top of a hill: It can move from the top to the bottom while spontaneously releasing energy, but some other process must provide the energy to take it from the bottom to the top. And this analogy provides a basic explanation for why a diode conducts in one direction, but not in the other.&lt;/p&gt;&#xA;" OwnerUserId="1345" LastActivityDate="2011-07-11T00:24:05.180" />
  <row Id="12142" PostTypeId="2" ParentId="12140" CreationDate="2011-07-11T04:59:07.233" Score="1" ViewCount="" Body="&lt;p&gt;In elementary particles all particles that have spin different than 0, spin, i.e. have angular momentum, so photons are spinning too, they have spin 1. There exist particles and systems with spin 0 (pions as an example), those do not spin :) .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since physics started from macroscopic studies one has to look at the equations that describe motion classically, very successfully. These equations, obey  &lt;a href=&quot;http://en.wikipedia.org/wiki/Noether%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;&quot;Noether's theorem&quot;&lt;/a&gt; that shows there are conserved quantities in the dynamics of motion coming from the symmetries of the system. Energy, momentum and &lt;a href=&quot;http://en.wikipedia.org/wiki/Angular_momentum&quot; rel=&quot;nofollow&quot;&gt;angular momentum&lt;/a&gt; are conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that once a path or a system rotation is established by some interaction, for example by a grazing impact of two asteroids,  &lt;em&gt;if there are no further interactions&lt;/em&gt; the asteroids will keep on spinning because the angular momentum they gave each other will be conserved individually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your question &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where has all the angular momentum come from? Why is it so natural?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is : from conservation laws. It is natural because equations of motion and conservation laws are a description of the mechanics of nature, and that is the way nature works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now space time and angular momentum are another story in General Relativity, where, because a rotating object has acceleration in the radial direction it distorts space time around it. For a simplified view see this &lt;a href=&quot;http://space.about.com/b/2011/05/12/nasa-confirms-two-predictions-of-general-relativity.htm&quot; rel=&quot;nofollow&quot;&gt;experiment.&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-07-11T04:59:07.233" />
  <row Id="12143" PostTypeId="2" ParentId="12129" CreationDate="2011-07-11T06:06:46.173" Score="1" ViewCount="" Body="&lt;p&gt;Your perceived speed is the same as the speed you travel at. That is, if you travel at 0.8c past the earth, people on the earth will measure your speed as 0.8c, and you will measure the speed of the earth going past you also as 0.8c. Thus you will conclude that you are traveling at 0.8c. So there is no difference between the speed you travel at and the speed you perceive that you travel at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the people on the earth will see you as slowed down (relative to themselves). Similarly, you will see the people on the earth as slowed down. The two of you can't come to agreement on this because you don't have a consistent way of comparing your watches. Whichever one of you does the comparison will conclude that the other is slowed down. This is related to the fact that in relativity there is no way to determine time in an absolute fashion. It's always relative. There's no way to figure out which of you is moving and which is standing still. It is only when you get back to the earth that you might decide that it was you who was moving and it was you who was experiencing a slowing in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that when you get back to the earth you determine that you were the one that was slowed down is only due to the fact that you were the one who changed course and returned to the earth. If instead the earth were somehow sped up and caught up with you, everything would be reversed. It would you who grew older than the people on earth.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-07-11T06:06:46.173" />
  <row Id="12144" PostTypeId="2" ParentId="12140" CreationDate="2011-07-11T08:37:52.367" Score="3" ViewCount="" Body="&lt;p&gt;I think that the simplest way to answer this question is to state the apposite. when you put, launch or deliver something in space it is nearly impossible for it not to spin. To have no angular momentum in 0g is nearly impossible. Besides vacuum random spin is the most difficult problem in space. &lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-07-11T08:37:52.367" />
  <row Id="12145" PostTypeId="2" ParentId="12140" CreationDate="2011-07-11T09:40:44.973" Score="2" ViewCount="" Body="&lt;p&gt;The origin of spin can be traced in two fundamental physical postulates:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Einstein's (general) &lt;a href=&quot;http://en.wikipedia.org/wiki/Principle_of_relativity&quot; rel=&quot;nofollow&quot;&gt;relativity postulate &lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Wigner's &lt;a href=&quot;http://en.wikipedia.org/wiki/Particle_physics_and_representation_theory&quot; rel=&quot;nofollow&quot;&gt;principle&lt;/a&gt; stating that elementary particles carry irreducible unitary representations of the symmetries of nature&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;According to the first principle, every local reference frame of space-time is Minkowskian and the laws of physics are the same&#xA;in all local reference frames. Now, the automorphism group of a Minkowski space is the Poincare group, therefore the laws of physics are convariant&#xA;under the Poincare group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second principle allows us to actually identify between elementary particles and irreducible representations of the symmetry groups of nature.&#xA;Applying this principle to the Poincare group, we obtain that elementary particles carry irreducible unitary representations of the Poincare group, and by consequence irreducible unitary representations&#xA;of its subroups, in particular, the rotation group. Now, since elementary representations of the rotation group are classified by the spin, then elementary particles carry spin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a subtlety in this description, in that the representations of the rotation group correspond only to integer spin and as we know half integer spin exists in nature also.&#xA;This issue was also addressed by Wigner, who generalized the correspondence between elementary particles and representations to projective representations as well (see for example Wigner's &lt;a href=&quot;http://books.google.com/books?id=Sc3PRTALFpkC&amp;amp;printsec=frontcover&amp;amp;source=gbs_ge_summary_r&amp;amp;cad=0#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;collected works&lt;/a&gt;). The projective representations of the rotation group correspond to half integer as well as integer spin.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastActivityDate="2011-07-11T09:40:44.973" />
  <row Id="12146" PostTypeId="1" AcceptedAnswerId="12150" CreationDate="2011-07-11T12:48:00.010" Score="0" ViewCount="174" Body="&lt;p&gt;Its makers say that it's a highly efficient motor because it resonates with (what they call) &quot;Essential Energy&quot;, which is free energy that is everywhere in space. (I think that’s just a different name for Dark Energy which somehow they are managing to disturb with the coils of the motor, and when it disturbs back, a bit of power is generated in the coil).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’ve listened to a radio interview of the makers, and they sad that a small motor powered by a 9v battery ran for about 2 hours with enough torque that they couldn’t stop it with their own hands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that what is really happening? Is it revolutionary? Or just bull… just a motor that consumes less energy since its coils are turned off half of the time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, strange things happen on the proximities of a running motor (a similar motor will start running even though it has no power, but it will run slower than the first one). Caused by the same thing that wireless cell phone chargers use to work? &lt;/p&gt;&#xA;" OwnerUserId="4456" LastActivityDate="2011-07-11T14:52:51.170" Title="Is the Keppe Motor &quot;real&quot;?" Tags="&lt;electromagnetism&gt;&lt;dark-energy&gt;" AnswerCount="3" />
  <row Id="12147" PostTypeId="1" CreationDate="2011-07-11T12:51:29.340" Score="5" ViewCount="147" Body="&lt;p&gt;I fly model aircraft. Recently I changed from a &quot;1300kv&quot; motor to a &quot;1500kv&quot; motor. The difference is in speed; the 1300kv can go about 16,400rpm full-throttle and with a fully charged battery and the 1500kv can go nearly 19,000rpm. (This is no load speed, but the prop adds very little load when the aircraft is not flying.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I noticed that when I increased the throttle beyond about 75% the motor and propeller became incredibly loud. Below that they were loud, but no where near as much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I did some ballpark calculations. I'm using a 7x3.5&quot; propeller - which means it has a 7&quot; diameter and 3.5&quot; pitch. In this case we are only interested in diameter. 7&quot; diameter means 3.5&quot; radius, or 8.86 cm. Doing 19,000 rpm means than I'm doing about 317 rps. According to 2 * pi * r the circumfrence of the prop's tip is 55.64 cm. Plug this into 317 rps and we get 17,638 cm/s or 176.4 m/s. Which is about half the speed of sound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although this is not actually the speed of sound it's very close to it, and much higher than I expected. So what does that mean? I figure that although the propeller itself isn't traveling faster than the speed of sound, the air it pushes out of the way is. My physics teacher showed that if you take a metre stick and swing it through the air (avoiding hitting anyone) it makes a wooshing sound - the air is moving faster than the ruler was. So the prop is pushing the air away faster than this and causing this very loud noise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my reasoning correct? Is my prop breaking the sound barrier?!&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="104" LastEditorDisplayName="" LastEditDate="2011-07-11T13:31:01.630" LastActivityDate="2011-07-12T00:28:01.400" Title="Does my model aircraft propeller beat the sound barrier?" Tags="&lt;aircraft&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12148" PostTypeId="2" ParentId="12146" CreationDate="2011-07-11T12:56:36.633" Score="0" ViewCount="" Body="&lt;p&gt;Calling this bull.... is an insult to bulls. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-07-11T12:56:36.633" />
  <row Id="12149" PostTypeId="2" ParentId="12147" CreationDate="2011-07-11T13:24:23.957" Score="4" ViewCount="" Body="&lt;p&gt;In general yes, you may have supersonic flow in some areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But prop is subsonic, and should work fine till 0.9mach (~300 m/s) if it have optimal shape for that speed (some planes even had variable prop's angle of attack to get this for all speeds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in your case you must have excessive noise due to turbulent air flow due to non-optimal prop shape. When flow goes from laminar to turbulent - you should hear significant increase in noise.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-07-11T13:24:23.957" />
  <row Id="12150" PostTypeId="2" ParentId="12146" CreationDate="2011-07-11T13:34:35.673" Score="2" ViewCount="" Body="&lt;p&gt;A total lack of anything in the peer review literature is a warning sign. A website that explains the operating principle of the motor as arising from disinverted Aristotlean metaphysics is something else entirely.&lt;/p&gt;&#xA;" OwnerUserId="1876" LastActivityDate="2011-07-11T13:34:35.673" CommentCount="3" />
  <row Id="12151" PostTypeId="2" ParentId="12076" CreationDate="2011-07-11T13:54:10.717" Score="1" ViewCount="" Body="&lt;p&gt;I found “The mathematical theory of black holes” by S. Chandrasekhar (1992) quite useful for such a purpose. The already mentioned $3 r_s/2 $ orbit for light really exists, but it is not stable, i.e. any small fluctuation or error in initial direction make impossible for light to return in the initial point and walk around black hole forever. It is because the last stable circular orbit is $3 r_s$ (see already cited wikipedia article). Yet in the mentioned book is described null geodesic (i.e. path of a light beam) asymptotically spiralling around the circle $3 r_s/2 $.&lt;/p&gt;&#xA;" OwnerUserId="3351" LastEditorUserId="3351" LastEditorDisplayName="" LastEditDate="2011-07-11T14:25:52.913" LastActivityDate="2011-07-11T14:25:52.913" />
  <row Id="12152" PostTypeId="2" ParentId="12146" CreationDate="2011-07-11T14:52:51.170" Score="2" ViewCount="" Body="&lt;p&gt;I always look up when perpetual motion type machines come up, i.e. more energy out than in. I also asked a question here on the possibility of &lt;a href=&quot;http://physics.stackexchange.com/questions/11800/can-energy-be-taken-out-of-the-qft-vacuum&quot;&gt;milking energy out of the vacuum.&lt;/a&gt; It is always possible that by serendipity somebody may hit the pot of gold of energy, never mind how he/she interprets it. If a machine gives more energy output than input, the theory will be found soon enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a &lt;a href=&quot;http://www.youtube.com/watch?v=SkCg_NBbVy0&amp;amp;feature=related&quot; rel=&quot;nofollow&quot;&gt;demonstration&lt;/a&gt; of a motor, and a &lt;a href=&quot;http://www.keppemotor.com/en/technical-information.php&quot; rel=&quot;nofollow&quot;&gt;web site&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the web site they say that a commercial product will be ready by2009. It is now 2011 and it is the first time I hear of this. Hmmm is all I say. There is nothing like the market to clear the wheat from the chaff, so I would say this is chaff. &lt;a href=&quot;http://keppemotorclub.bestforumpro.com/t2-welcome-to-the-keppe-motor-forum&quot; rel=&quot;nofollow&quot;&gt;This is&lt;/a&gt; supposed to be the users forum, and not many beating a path to the door.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-07-11T14:52:51.170" CommentCount="8" />
  <row Id="12153" PostTypeId="2" ParentId="12147" CreationDate="2011-07-11T16:48:03.303" Score="1" ViewCount="" Body="&lt;p&gt;It is important to note that if you bench test the prop. At a determined speed the prop will not be able to get clean air the air expelled from the back will start to flow to the front due to the very thin air in the front of the prop. This will cause as @BarsMonster states for the air to become turbulent. If the plane is flaying the prop gets clean air all the time and it should not make the noise. If it does the shock wave at the tips of the blade will cause it to actually slow down and speed up again. the plane in flight will make a strange off and on loud humming. Props are designed to not go supersonic at the tips. The rpm and diameter must be just right. This is why on turbo props the blades have flat ends and a shorter diameter. &lt;/p&gt;&#xA;" OwnerUserId="4034" LastEditorUserId="4034" LastEditorDisplayName="" LastEditDate="2011-07-12T00:28:01.400" LastActivityDate="2011-07-12T00:28:01.400" CommentCount="1" />
  <row Id="12154" PostTypeId="1" CreationDate="2011-07-11T17:15:24.840" Score="3" ViewCount="192" Body="&lt;p&gt;I have some trouble understanding exactly what a mole represents. As I understand, one unit mole is 1/12 of the mass of an atom of carbon-12 (thus it is the mass of one nucleon?). What is a mole, then?&lt;/p&gt;&#xA;" OwnerUserId="4457" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-11T18:16:40.770" LastActivityDate="2011-07-14T01:16:19.837" Title="What exactly is a mole?" Tags="&lt;units&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="12155" PostTypeId="2" ParentId="12154" CreationDate="2011-07-11T17:33:53.973" Score="11" ViewCount="" Body="&lt;p&gt;A mole is nothing more than a countable number of things.  Specifically, it is approximated by 6.02 x 10^23 number of things.  Hence, you can have a mole of argon atoms, a mole of electrons, or a mole of house keys all of which contain approximately 6.02 x 10^23 of their respective items.  Note that each of these groups of things will have a &lt;em&gt;different&lt;/em&gt; mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are referring to as 1/12 of a C-12 atoms is not a mole, but an atomic mass unit.&lt;/p&gt;&#xA;" OwnerUserId="4289" LastActivityDate="2011-07-11T17:33:53.973" CommentCount="7" />
  <row Id="12156" PostTypeId="2" ParentId="12064" CreationDate="2011-07-11T17:58:50.397" Score="0" ViewCount="" Body="&lt;p&gt;I'll take a stab at your first two questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.)  The circuit does not act as a transformer because there is no linking current.  The &quot;A-field flux&quot; is there, but no traveling charge.  Putting in a ring of copper (as he does in other pics) allows the A-field to generate current which would link to the next core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.)  Yes, but not in any practical sense as shown.  The capacitor plates need to be very close together, as they would be in a commercial cap.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-07-11T17:58:50.397" CommentCount="2" />
  <row Id="12157" PostTypeId="1" CreationDate="2011-07-11T18:13:45.593" Score="0" ViewCount="72" Body="&lt;p&gt;I mean for undergraduates and graduates on the follweing topics: &#xA;classical mechanics, electromagnetism, thermodynamics, special relativity, general relativty and quantum mechanics.&lt;/p&gt;&#xA;" OwnerUserId="4460" LastActivityDate="2011-07-11T18:13:45.593" ClosedDate="2011-07-11T18:17:16.027" Title="What are the best physics textbooks ever?" Tags="&lt;homework&gt;" CommentCount="2" />
  <row Id="12158" PostTypeId="2" ParentId="12089" CreationDate="2011-07-11T18:57:43.793" Score="0" ViewCount="" Body="&lt;p&gt;Let me try yet one more type of explanation, which I will confine to the PN junction diode (covers virtually all diodes used in modern circuits).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diode consists of a p-doped region (p-type) slapped up against an n-doped region (n-type).  In the p-type, the electron (e-) flow is largely accomplished by electrons moving from hole to hole. This is, electrically, exactly analogous (and is often visualized) as holes moving in a direction opposite to e- flow (although there is no physical movement of postivie charge)  In the n-type, there are loosely bound e- which can be donated (moved).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the PN junction of the diode, loosely bound e- in the n-type fall into the holes of the adjacent p-type.  What you then have is an abundance of e- in a thin layer of the p-type layer at the junction, and a depletion of them (creating a net positive charge) in a thin layer of the n-type.  This sets up a voltage field of positive in the n-type relative to negative in the p-type.  This pushes any free e- in the n-type further away from the junction.  The result is a thin PN layer which has no free holes and no free e-.  The layer becomes an insulator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you apply a positive voltage to the p-type and a negative at the n-type, e- in the p-type are removed, making free holes.  Simultaneously, the positive voltage is conteracting the reverse voltage which had been set up in the PN junction, and e- in the n-type are force closer to the p-type, where they can cross over and fill up the new holes.  Current flows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, you apply positive voltage to the n-type, and negative to the p-type (&quot;reverse-biasing&quot; the diode) you simply reinforce the voltage gradient which was already naturally set up in the PN junction.  The e- are forced even farther away from the PN junction, and the insulative boundary (depletion region) thickens.  No current flows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get more in-depth than that might take a good portion of a graduate course in materials science.  I hope what I have written suffices.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-07-11T18:57:43.793" />
  <row Id="12159" PostTypeId="2" ParentId="12154" CreationDate="2011-07-11T19:08:27.550" Score="2" ViewCount="" Body="&lt;p&gt;The mole is of such great use because of the predictable and unique relationship between the number of atoms (counted in moles) in a sample and the sample's easily measured weight or mass (depending on your unit system). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The mass per mole of a substance is called its molar mass. Since the&#xA;  standard unit for expressing the mass of molecules or atoms (the&#xA;  dalton or atomic mass unit) is defined as 1/12 of the mass of a 12C&#xA;  atom, it follows that the molar mass of a substance, measured in grams&#xA;  per mole, is exactly equal to its mean molecular or atomic mass,&#xA;  measured in daltons; which is to say, to the substance's mean&#xA;  molecular or atomic weight.&#xA;  -Wikipedia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example: one mole (6.02X10^23 atoms) of Carbon atoms has a mass of 12.011 grams, which is also the atomic weight of Carbon in Daltons (&lt;a href=&quot;http://3.bp.blogspot.com/-Q_mcKbGutWw/TfEuTdojy-I/AAAAAAAAGH4/LulenJnl7zQ/s1600/table.gif&quot; rel=&quot;nofollow&quot;&gt;shown in this periodic table&lt;/a&gt;). The molar mass of water (2 hydrogen atoms and 1 oxygen atom) =  2*(1.0079) + 1*(15.999) = 18.0148 grams (I am using a lot decimal places for effect despite that being impractical in real life). &lt;/p&gt;&#xA;" OwnerUserId="2843" LastActivityDate="2011-07-11T19:08:27.550" />
  <row Id="12160" PostTypeId="2" ParentId="12046" CreationDate="2011-07-11T20:04:48.563" Score="0" ViewCount="" Body="&lt;p&gt;When you say “yellow light (575 nm) from sunlight” I assume you mean the&#xA;monochromatic light you get by sending sunlight through a narrow-band&#xA;filter centered at 575 nm. Unfiltered sunlight is not 575 nm, it's very&#xA;wide band.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference is then that the monitor yellow has a complex spectrum,&#xA;which is the combination of the spectra of red and green phosphors.&#xA;Visually both will look very close, although monochromatic yellow is&#xA;slightly more saturated. Its saturation can not be matched by the&#xA;monitor because, as all spectral colors, it lies outside the&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gamut&quot; rel=&quot;nofollow&quot;&gt;monitor gamut&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light spectra are related to the&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;Fourier transform&lt;/a&gt;&#xA;(rather than the Fourier series). The spectrum is the squared modulus of&#xA;the Fourier transform of the electromagnetic oscillation. The basis is&#xA;of infinite dimensions, because there are infinitely many possible&#xA;frequencies. Colors, on the other hand, are not really related to&#xA;Fourier transforms or series. They exist in a 3D space because our eyes&#xA;have three kinds of sensors (setting aside the rods that we use in dark&#xA;situations) called L, M and S cones (for long, medium and short&#xA;wavelengths). The only thing they can see as “color” is the combination&#xA;of how mush each of these sensors are excited. The space of colors is&#xA;much smaller than the space of spectra, therefore, most colors can be&#xA;reproduced by many different spectra.&lt;/p&gt;&#xA;" OwnerUserId="4202" LastActivityDate="2011-07-11T20:04:48.563" />
  <row Id="12161" PostTypeId="2" ParentId="12093" CreationDate="2011-07-11T20:10:55.217" Score="3" ViewCount="" Body="&lt;p&gt;General relativity has extremely solid foundations.  Every experiment testing it has confirmed its predictions to great precision.  For example, recent measurments of binary systems have confirmed the precise energy loss from the system that GR predicts from gravity wave radiation (in the Jupiter-Sun system the predicted loss of energy at a rate of a 40 watt bulb is impossible to measure accurately since the effect is so small).  Predictions regarding gravitational impacts on the passage of time, the bending of light around massive objects, and frame dragging, for example, have also been confirmed with high preceision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like quantum chromodynamics (the mathematical physics of the strong force that binds atomic nuclei), the math of GR is not particularly easy to apply numerically in systems that are not highly stylized with assumptions include to simplify the calculations.  It is fair to say that all of the mathematically easy cases of the implications of GR have been well explored, but that there are plenty of possible physical systems for which we do not have good numerical theoretical predictions - only approximations informed by analytical insights drawn from an understanding of the equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, there are many theoretical predictions of GR that have not been experimentally tested because no one has done the math to make the prediction yet.  In practice, a fair amount of the computer programs used to simulate gravitational mechanics at levels smaller than the entire universe and outside the immediate vicinity of very dense objects like black holes and neutron stars use a Newtonian gravity approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations of quantum mechanics are formulated on a Minkowski background that includes special relativity but not general relativity.  Usually this produces very little conflict - GR is generally applied to dynamics on a scale where quantum mechanical effects have little impact; QM is usually applied in circumstances where gravity is negligible relative to the effects of other forces.  In the few circumstances where there is good reason to think that both GR and QM could have measurable impacts (e.g. in the early Big Bang and around black holes) there have been some ad hoc efforts by Bekenstein, Hawking and others to consider the effects of both, but there is no consensus means of formulating QM in a way that fully respects GR (or visa versa) in general.&lt;/p&gt;&#xA;" OwnerUserId="4461" LastActivityDate="2011-07-11T20:10:55.217" CommunityOwnedDate="2011-07-11T20:10:55.217" CommentCount="1" />
  <row Id="12162" PostTypeId="2" ParentId="11878" CreationDate="2011-07-11T20:17:37.957" Score="2" ViewCount="" Body="&lt;p&gt;Feynman's Book &quot;QED&quot; is probably your best bet.  He spells out what is going on in the equations of quantum electrodynamics in about 120 pages and briefly touches on how these equations generalize in other parts of QFT.&lt;/p&gt;&#xA;" OwnerUserId="4461" LastActivityDate="2011-07-11T20:17:37.957" CommunityOwnedDate="2011-07-11T20:17:37.957" />
  <row Id="12163" PostTypeId="2" ParentId="12154" CreationDate="2011-07-11T20:19:38.627" Score="5" ViewCount="" Body="&lt;p&gt;A mole is like saying &quot;a dozen,&quot; as Luksen points out in a comment.  But why pick $(6.02)(10^{23})$ instead of, say, $(6.07)(10^{24})$?  The reason is that the number was strategically picked such that the weight, in grams, of a mole would be numerically exactly the same as its molecular weight.  It is a number that was formulated for the convenience of the chemist (just like the &quot;dozen&quot; was created for the convenience of... uh... someone... the baker?  No, that was 13...)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for instance, if you know you have a mole of a substance, you can weigh that mole, in grams, to find the molecular weight without doing any math (like 18.0 grams is about a mole of water, which has a molecular weight of about 18.0).  If, on the other hand, you want a mole of a substance and you know its molecular weight (as you almost always do) you can weigh it out in grams to get exactly a mole.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-07-11T20:19:38.627" />
  <row Id="12164" PostTypeId="2" ParentId="12025" CreationDate="2011-07-11T20:24:02.553" Score="2" ViewCount="" Body="&lt;p&gt;The owner of superconductors.org has &lt;a href=&quot;http://www.superconductors.org/Play.htm&quot; rel=&quot;nofollow&quot;&gt;a list of places to find HTSC superconductors, kits, and instructions.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also review the literature for recipes.  HTSC producers usually publish successful and repeatable recipes in refereed journals, and some components are available commercially.&lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2011-07-11T20:24:02.553" />
  <row Id="12165" PostTypeId="1" AcceptedAnswerId="12167" CreationDate="2011-07-11T21:08:33.437" Score="7" ViewCount="143" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://stats.stackexchange.com/questions/12897/how-to-predict-how-much-data-to-collect&quot;&gt;The same question on CrossValidated&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Apologies if I'm being a bit vague in what follows, I've been asked to keep certain aspects of the experiment confidential for the time being. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An analogous experiment would be like trying to 'see' the ebb-and-flow of the tide (0.5 day period) by locating a photon detector at the bottom of the ocean (though of course this wouldn't work and is silly, but the principle is at least quite similar.) Hope that clarifies it a bit, let me know if not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently in the planning stages of this experiment that I am hoping will detect a 0.155% signal variation (relative magnitude) within a resonable time frame (less than 6 months ideally.) I've calculated the rate of (usable) data will be around 68 events per day, though it should be stressed this is a random variable. Now I'm trying to work out - &lt;strong&gt;how many days will I need to run the detector for to see the variation with a confidence level of 3σ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some other details that may (or may not) be relevant include: the variation in the signal is expected to be sinusoidal with a period of 0.5 days. For this reason I reduced my useful event rate to 34 (Ie half) as clearly there is no variation to see when the sinusoidal signal is at or close to the mean value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been googling for a method to predict the size of a data set necessary to see such a small signal variation but have come up with nothing. I would be extremely grateful for any hints / tips anyone could offer.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-15T16:23:31.080" LastActivityDate="2011-08-15T16:23:31.080" Title="How to predict how much data to collect" Tags="&lt;experimental-physics&gt;&lt;statistics&gt;&lt;data-analysis&gt;&lt;confidence-level&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="2" />
  <row Id="12167" PostTypeId="2" ParentId="12165" CreationDate="2011-07-11T22:17:23.050" Score="4" ViewCount="" Body="&lt;p&gt;I don't think there's any way you're going to do this in six months.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll give a calculation below, but first an order of magnitude estimate. If you've detected a total of $N_{\rm events}$ events, your measurement of a modulation will have an error of order $N_{\rm events}^{-1/2}$ -- &#xA; -- these things always do! -- so the number of events required is going to go like $1/f^2$ where $f$ is the modulation level you're looking for. In your case, $f=0.00155$, corresponding to about 400,000 events, which will take decades at the given event rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $N_{\rm events}$ be the total number of events in your data set. Suppose that you bin your data into $N$ bins by time of day. You're assuming that the signal is of the form &#xA;$$&#xD;&#xA;s_j=A+B\cos(t_j),&#xD;&#xA;$$&#xA;where $t_j$ is the time of day corresponding to the $j$th bin, and times of day are measured from the time when the signal is at its maximum. (If you don't know when that is and are planning to fit for it, then that'll change things.) Here $A$ is the average number of vents, so &#xA;$$&#xD;&#xA;A=N_{\rm events}/N,&#xD;&#xA;$$&#xA;and &#xA;$$&#xD;&#xA;B=fA={fN_{\rm events}\over N},&#xD;&#xA;$$&#xA;where $f=0.00155$ is the modulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming further that your data are equally distributed across all times of day, the errors in $s_j$ will all be approximately equal (because $f$ is small). In this case, the best estimator of $B$ is&#xA;$$&#xD;&#xA;\hat B={2\over N}\sum_j s_j\cos(t_j).&#xD;&#xA;$$&#xA;We want to find the variance $\sigma_B^2$ of this estimator. The individual $s_j$ are all independent and have nearly equal variances $\sigma^2$,&#xA;so&#xA;$$&#xD;&#xA;\sigma_B^2={4\sigma^2\over N^2}\sum_j\cos^2(t_j).&#xD;&#xA;$$&#xA;Assuming that $N$ is large enough that that sum can be approximated by an integral, the sum comes out to $N/2$, so&#xA;$$&#xD;&#xA;\sigma_B^2={2\over N}\sigma^2.&#xD;&#xA;$$&#xA;For Poisson distributed events, the variance is equal to the expected value: $\sigma^2=A=N_{\rm events}/N$. Therefore,&#xA;$$&#xD;&#xA;\sigma_B^2={2N_{\rm events}\over N^2}.&#xD;&#xA;$$&#xA;The fractional uncertainty is&#xA;$$&#xD;&#xA;{\sigma_B\over B}={\sqrt{2N_{\rm events}}\over N}{N\over fN_{\rm events}}=\sqrt{2\over f^2N_{\rm events}}.&#xD;&#xA;$$&#xA;For a 3-sigma detection, you want this to be equal to 1/3, so&#xA;$$&#xD;&#xA;N_{\rm events}={18\over f^2}=2.5\times 10^6.&#xD;&#xA;$$&#xA;(My initial guess was off by a factor of 18 -- $3^2$ because of the 3 sigma, and 2 because of the point you noted about data near the zeroes of the modulation not helping.) At 68 events per day, this works out to about 300 years. Sorry.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-07-11T22:44:04.253" LastActivityDate="2011-07-11T22:44:04.253" CommentCount="5" />
  <row Id="12168" PostTypeId="2" ParentId="12165" CreationDate="2011-07-11T22:25:56.580" Score="2" ViewCount="" Body="&lt;p&gt;Back of the envelope calculation. (I'm rushed, hope I got this right.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probability questions like this are best done using probabilities, so first let's convert your estimate to a probability $p$:&lt;br&gt;&#xA;Your signal variation is 0.00155 so:&lt;br&gt;&#xA;$$1-2p = 0.00155$$&#xA;So $p = 0.499225$ and $1-p = 0.500775$. The standard deviation is&lt;br&gt;&#xA;$$\sigma = \sqrt{p(1-p)/N} \approx \sqrt{1/(2N)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want the standard deviation to be 1/3rd of the difference between 0.5 and $p$ so we solve for N:&lt;br&gt;&#xA;$$(0.500775-0.5)/3 = \sqrt{1/(2N)}$$&lt;br&gt;&#xA;to get $N= 7.5\times 10^6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 68 events per day (actually it will be less because of the sine wave), this amounts to 21 thousand days.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-07-11T22:25:56.580" />
  <row Id="12169" PostTypeId="1" AcceptedAnswerId="12171" CreationDate="2011-07-12T00:17:51.480" Score="10" ViewCount="108" Body="&lt;p&gt;By the &quot;No Hair Theorem&quot;, three quantities &quot;define&quot; a black hole; Mass, Angular Momentum, and Charge. The first is easy enough to determine, look at the radius of the event horizon and you can use the Schwarzschild formula to compute the mass. Angular Momentum can be found using the cool little ergosphere Penrose &quot;discovered&quot;. However, I don't know how to determine the charge of the black hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can an electromagnetic field escape the event horizon of a Reissner-Nordström black hole? Is there any experiment we could theoretically do to a black hole to determine its charge?&lt;/p&gt;&#xA;" OwnerUserId="3018" LastActivityDate="2011-07-12T07:33:27.913" Title="Detection of the Electric Charge of a Black Hole" Tags="&lt;general-relativity&gt;&lt;electromagnetism&gt;&lt;black-holes&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" />
  <row Id="12170" PostTypeId="1" CreationDate="2011-07-12T00:35:02.310" Score="4" ViewCount="116" Body="&lt;p&gt;From the &lt;a href=&quot;http://en.wikipedia.org/wiki/Closed_timelike_curve&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;, it seems that physicists tend to view closed timelike curves as an undesirable attribute of a solution to the Einstein Field Equations. Hawking formulated the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chronology_protection_conjecture&quot; rel=&quot;nofollow&quot;&gt;Chronology protection conjecture&lt;/a&gt;, which I understand essentially to mean that we expect a theory of quantum gravity to rule out closed timelike curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am well-aware that the existence of closed timelike curves implies that time travel is technically possible, but this argument for why they should not exist isn't convincing to me. For one, if the minimal length of any closed timelike curve is rather large, time travel would be at least infeasable. Furthermore, this is essentially a philosophical argument, which is based, at least in part, on our desire to retain causality in studying the large scale structure of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, the best argument I've heard against CTCs is that the 2nd law of thermodynamics wouldn't seem to have a meaningful interpretation in such a universe, but this isn't totally convincing. A good answer to this question would be some form of mathematical heuristic showing that in certain naive ways of combining quantum mechanics and gravity, CTCs are at least implausible in some way. Essentially, I'm trying to find any kind of an argument in favor of Hawking's conjecture which is not mostly philosophical.  I realize that such an argument may not exist (especially since no real theory of quantum gravity exists), so other consequences of the (non)existence of CTCs would be helpful.&lt;/p&gt;&#xA;" OwnerUserId="4463" LastActivityDate="2011-07-12T14:24:09.670" Title="Why should one expect closed timelike curves to be impossible in quantum gravity?" Tags="&lt;general-relativity&gt;&lt;quantum-gravity&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="12171" PostTypeId="2" ParentId="12169" CreationDate="2011-07-12T00:50:52.517" Score="8" ViewCount="" Body="&lt;p&gt;A charged black hole does produce an electric field. In fact, at great distances (much larger than the horizon), the field strength is $Q/(4\pi\epsilon_0 r^2)$, just like any other point charge. So measuring the charge is easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for how the electric field gets out of the horizon, the best answer is that it doesn't: it was never in the horizon to begin with! A charged black hole formed out of charged matter. Before the black hole formed, the matter that would eventually form it had its own electric field lines. Even after the material collapses to form a black hole, the field lines are still there, a relic of the material that formed the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A long time ago, back when the &lt;em&gt;American Journal of Physics&lt;/em&gt; had a question-and-answer section, someone posed the question of how the electric field gets out of a charged black hole. Matt McIrvin and I wrote &lt;a href=&quot;https://facultystaff.richmond.edu/~ebunn/ajpans/ajpans.html&quot; rel=&quot;nofollow&quot;&gt;an answer&lt;/a&gt;, which appeared in the journal. It pretty much says the same thing as the above, but a bit more formally and carefully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, I just noticed a mistake in what Matt and I wrote. We say that the Green function has support only on the past light cone. &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0306052&quot; rel=&quot;nofollow&quot;&gt;That's actually not true&lt;/a&gt; in curved spacetime: the Green function has support in the interior of the light cone as well. But fortunately that doesn't affect the main point, which is that there's no support &lt;em&gt;outside&lt;/em&gt; the light cone.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-07-12T00:59:19.230" LastActivityDate="2011-07-12T00:59:19.230" CommentCount="2" />
  <row Id="12172" PostTypeId="2" ParentId="11885" CreationDate="2011-07-12T02:20:37.260" Score="3" ViewCount="" Body="&lt;p&gt;Even if you change frames, the physics is still the same and the particle will follow the same path, no? And there is certainly more than one way to change the Lagrangian without affecting the path of least action--add any combination of total time derivatives to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I said it would follow the same path, I meant the same path after you take into account the fact that you shifted frames. If $q_1$ and $q_2$ label the same point even after you shift frames (so that in new coordinates $q_1=q^{new}_1−ϵt$ and etc.) then the particle will be at $q_2$ at $t_2$ if it was at $q_1$ at $t_1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean that if the particle starts at time $t_1$ at $q_1$ it DOES end up at $q_2$ at time $t_2$ provided you take into account how the points look different because of the new frame. The path taken by the particle must be the same; physics doesn't depend on the inertial frame you are in and this is the point Landau is making. If you think that in frame K' the particle doesn't end up at $q_2$ at time $t_2$ then it is purely because the points are labeled differently in this frame. It also has nothing to do with v being infinitesimal; that doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the other question, you can also multiply the whole Lagrangian by a constant. That's kind of obvious though. Basically you need the kinetic energy term, and the only way you can further modify it is by adding terms right? If you multiplied the Lagrangian by a non-constant term, for instance, the form of the kinetic energy term would change. Then he rules out what terms you can add. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe if you go to this page and look under the section &quot;Is the Lagrangian unique?&quot; it will help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikibooks.org/wiki/Classical_Mechanics/Lagrange_Theory&quot; rel=&quot;nofollow&quot;&gt;en.wikibooks.org/wiki/Classical_Mechanics/Lagrange_Theory&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically you change the Lagrangian by adding terms, and it can be proved that ONLY if this term is a total time derivative of a function of coordinates and time that the action is still extremized. In other words, no, it is not possible to keep the path of least action by adding any other function.&lt;/p&gt;&#xA;" OwnerUserId="1635" LastActivityDate="2011-07-12T02:20:37.260" />
  <row Id="12173" PostTypeId="1" CreationDate="2011-07-12T03:44:50.827" Score="3" ViewCount="95" Body="&lt;p&gt;Many of us may remember the problems from thermodynamics where we had to determine if a pretend air conditioning unit supplier's claims violated the 2nd law.  I want to pose a similar question regarding a new &quot;green&quot; Air Conditioning (AC) technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An ice energy storage unit for AC works by making ice during the night and using the latent heat (mostly) to cool the building during the day.  The major motive for doing this is to move the energy demand from the peak energy use times to the low-demand times at night when it is cheaper.  The cost of electricity is not relevant to this question.  The claims of a vendor of this technology, &lt;strong&gt;Ice Bear Energy Storage System&lt;/strong&gt;, include the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.ice-energy.com/ice-bear-energy-storage-system&quot; rel=&quot;nofollow&quot;&gt;Bottom line: The Ice Bear system reduces total net energy consumption for most buildings under virtually all operating conditions and installations.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is this claim physically plausible, given reasonable assumptions about climate?  There are some example cities below, I imagine that considering an average location or just the extremes would be sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a few things to note, but these are certainly not comprehensive.  Creativity may be important for identifying all the factors at play.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The temperature of ice is $0^{\circ} C$ ($32^{\circ} F$)&lt;/li&gt;&#xA;&lt;li&gt;Room temperature is $22.2^{\circ} C$ ($72^{\circ} F$)&lt;/li&gt;&#xA;&lt;li&gt;The outside ambient temperature is lower at night than during the day.  Exactly how much depends on the location and &lt;a href=&quot;http://en.wikipedia.org/wiki/Diurnal_temperature_variation&quot; rel=&quot;nofollow&quot;&gt;humidity particularly.&lt;/a&gt;  Locations mentioned on their website include &lt;a href=&quot;http://www.ice-energy.com/case-studies&quot; rel=&quot;nofollow&quot;&gt;Glendale, CA; Howell, NJ; Las Vegas, NV; Fort Collins, CO; Orange County, CA; and Redding, CA.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The ice storage duration is 6 hours and since these are multiple MW products, I'll say it's fair to assume almost perfect insulation unless someone wants to challenge this.&lt;/li&gt;&#xA;&lt;li&gt;You &lt;em&gt;can&lt;/em&gt; obtain useful work while exchanging the heat from the office building to the ice.  I believe this would be important considering the wording of &quot;total net energy consumption&quot;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h1&gt;Temperatures&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I have &lt;a href=&quot;http://www.bestplaces.net/climate/&quot; rel=&quot;nofollow&quot;&gt;looked up&lt;/a&gt; the temperatures for most of the mentioned cities.  For some I couldn't get data for I used a nearby city instead.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;(City/State) (night-day difference) (seasonal difference) (hottest month high)&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Redding CA, 27.4, 44, 99&lt;/li&gt;&#xA;&lt;li&gt;Las Vegas NV, 26.5, 48, 104&lt;/li&gt;&#xA;&lt;li&gt;Denver CO, 27.8, 45, 88&lt;/li&gt;&#xA;&lt;li&gt;Lakehurts NJ, 27.8, 45, 88&lt;/li&gt;&#xA;&lt;li&gt;Santa Anna CA, 20.5, 18, 83&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These seem to be fairly consistent for the night-day difference in particular.  I hope that simplifies the problem.  It seems clear that running at night &quot;buys&quot; you about 27.5 degrees F.  Now, the absolute temperatures as well as the seasonal variation will affect the efficiency, but to a much smaller degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My conflict: Can a $27.5$ degree difference really make up for a $72-32=40$ degree penalty on the thermal cycle?  &lt;em&gt;Really&lt;/em&gt;?  It's not a trivial problem, but I am skeptical.  I will certainly send them an email letting them know they violated the 2nd law depending on the outcome here.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-07-12T15:11:01.743" LastActivityDate="2011-07-12T15:11:01.743" Title="Physicality of claims regarding AC ice energy storage units" Tags="&lt;thermodynamics&gt;&lt;weather&gt;" AnswerCount="1" />
  <row Id="12174" PostTypeId="2" ParentId="12173" CreationDate="2011-07-12T05:46:07.283" Score="2" ViewCount="" Body="&lt;p&gt;The technology you describe can actually save &lt;em&gt;some energy&lt;/em&gt; using the fact that the night temperature can be much lower than the day temperature. When you cool the ice using a refrigerator-like unit it's very important how cold the air you use to cool the refrigerator radiator is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you can place the radiator in such way that it is intensely cooled with outside air and use much colder air available during the night you can cool the ice more efficiently than you would cool the inside air during the day with a usual AC system. So not only you shift to cheaper electricity but you &lt;em&gt;could&lt;/em&gt; also use less electricity because with colder outside air cooling can be more efficient during the night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One point to illustrate this: my own electricity usage pattern is mostly constant during the year. Now the electricity bill is about 30% larger in the hottest summer months because of the refrigerator that constitutes the most of my electricity consumption. Of course the refrigerator also heats up faster with warmer room air, but it also cools less efficiently and these two factors noticeably boost energy consumption.&lt;/p&gt;&#xA;" OwnerUserId="3816" LastActivityDate="2011-07-12T05:46:07.283" CommentCount="6" />
  <row Id="12175" PostTypeId="1" CreationDate="2011-07-12T06:04:21.747" Score="9" ViewCount="235" Body="&lt;p&gt;Every once in a while, we get a question asking for a book or other educational reference on a particular topic at a particular level. This is a meta-question that collects all those links together. If you're looking for book recommendations, this is probably the place to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the questions linked below, as well as others which deal with more specialized books, can be found under the tag &lt;a href=&quot;/questions/tagged/books&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged 'books'&quot; rel=&quot;tag&quot;&gt;books&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find a question that should be added to the list, edit it into the existing list answer rather than creating your own answer for it.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-07-31T19:38:01.427" CommunityOwnedDate="2011-07-12T06:05:18.203" Title="Book recommendations" Tags="&lt;books&gt;&lt;faq&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" />
  <row Id="12176" PostTypeId="2" ParentId="12175" CreationDate="2011-07-12T06:05:09.857" Score="7" ViewCount="" Body="&lt;p&gt;This list is pretty rough, but I figured it would be better to post &lt;em&gt;something&lt;/em&gt; rather than keep putting it off. If anyone wants to organize the links a little better, that would help. I'm not sure all the questions currently linked actually need to be here, anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a question to add, please edit it in. However, make sure of a few things first:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The question should be tagged &lt;a href=&quot;/questions/tagged/books&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged 'books'&quot; rel=&quot;tag&quot;&gt;books&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;It should be of the form &quot;What are good books to learn/study [subject] at [level]?&quot;&lt;/li&gt;&#xA;&lt;li&gt;It shouldn't duplicate a topic and level that's already in the list&lt;/li&gt;&#xA;&lt;li&gt;It should be the kind of question that is appropriate as CW, i.e. there is no one correct answer&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Newtonian Mechanics&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;General&lt;/em&gt;: &lt;a href=&quot;http://physics.stackexchange.com/questions/6068/list-of-good-classical-physics-books&quot;&gt;List of good classical physics books&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Foundations&lt;/em&gt;: &lt;a href=&quot;http://physics.stackexchange.com/questions/12964/book-suggestions-for-foundation-of-newtonian-mechanics&quot;&gt;Book suggestions for foundation of Newtonian Mechanics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Classical Mechanics&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Introductory (math)&lt;/em&gt;: &lt;a href=&quot;http://physics.stackexchange.com/questions/9165/which-mechanics-book-out-of-these-is-the-best-for-beginner-in-math-major&quot;&gt;Which Mechanics book out of these is the best for beginner in math major?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Advanced&lt;/em&gt;: &lt;a href=&quot;http://physics.stackexchange.com/questions/111/book-about-classical-mechanics&quot;&gt;Book about classical mechanics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Coordinate-free&lt;/em&gt;: &lt;a href=&quot;http://physics.stackexchange.com/questions/1601/classical-mechanics-without-coordinates-book&quot;&gt;Classical mechanics without coordinates book&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Elasticity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/9544/good-books-on-elasticity&quot;&gt;good books on elasticity&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Quantum Mechanics&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Popular&lt;/em&gt;: &lt;a href=&quot;http://physics.stackexchange.com/questions/11287/popular-books-on-qm&quot;&gt;Popular books on QM&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Quantum Field Theory&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/8441/what-is-a-complete-book-for-quantum-field-theory&quot;&gt;What is a complete book for quantum field theory?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/12385/are-there-books-on-regularization-at-an-introductory-level&quot;&gt;Are there books on Regularization at an Introductory level?&lt;/a&gt; (maybe this should go under &quot;Math&quot;?)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Particle Physics&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Popular&lt;/em&gt;: &lt;a href=&quot;http://physics.stackexchange.com/questions/9389/good-book-about-elementary-particles-for-high-school-students&quot;&gt;Good book about elementary particles for high school students?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Introductory&lt;/em&gt;: &lt;a href=&quot;http://physics.stackexchange.com/questions/312/particle-physics-getting-started&quot;&gt;Particle physics getting started&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;General Relativity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Introductory&lt;/em&gt;: &lt;a href=&quot;http://physics.stackexchange.com/questions/363/getting-started-general-relativity&quot;&gt;Getting started general relativity&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fluid Dynamics&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/9815/boundary-layer-theory-in-fluids-learning-resources&quot;&gt;Boundary layer theory in fluids learning resources&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Group Theory&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/6108/comprehensive-book-on-group-theory-for-physicists&quot;&gt;Comprehensive book on group theory for physicists?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Spectral Theory&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/598/books-for-linear-operator-and-spectral-theory&quot;&gt;Books for linear operator and spectral theory&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Experimental Particle Physics&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/11434/reference-for-solid-state-particle-detector&quot;&gt;Reference for solid state particle detector&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem books&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/10325/graduate-physics-problems-books&quot;&gt;Graduate Physics Problems Books&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Functional Calculus&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/7737/introductory-texts-for-calculus-of-functionals&quot;&gt;Introductory texts for calculus of Functionals&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Condensed Matter&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/5347/soft-condensed-matter&quot;&gt;Soft Condensed Matter&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Miscellaneous&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/884/books-that-every-physicist-should-read&quot;&gt;Books that every physicist should read&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/6157/list-of-freely-available-physics-books&quot;&gt;List of freely available physics books&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/13020/books-that-every-non-physicist-should-read&quot;&gt;Books that every non-physicist should read&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Math&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/193/best-books-for-mathematical-background&quot;&gt;Best books for mathematical background?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Fractals&lt;/em&gt;: &lt;a href=&quot;http://physics.stackexchange.com/questions/12784/fractals-in-physics-suggest-book-titles&quot;&gt;Fractals in physics. Suggest book titles&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-31T19:38:01.427" LastActivityDate="2011-07-31T19:38:01.427" CommunityOwnedDate="2011-07-12T06:05:18.203" CommentCount="1" />
  <row Id="12177" PostTypeId="2" ParentId="12169" CreationDate="2011-07-12T07:33:27.913" Score="1" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Aharonov%E2%80%93Bohm_effect&quot; rel=&quot;nofollow&quot;&gt;Aharanov-Bohm effect&lt;/a&gt; (where an electrically charged particle is effected by the electric and magnetic fields even though it travels only in a region where these fields are zero, i.e. outside of a solenoid) demonstrates that from the point of view of quantum mechanics, the underlying fields are not the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_field&quot; rel=&quot;nofollow&quot;&gt;electromagnetic field&lt;/a&gt;, but instead the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_four-potential&quot; rel=&quot;nofollow&quot;&gt;electromagnetic 4-potential&lt;/a&gt;. This says that forces are not enough to define physics, one must also use potentials (energies). So maybe the question should be not how the electric field gets out of the black hole but instead how the electric / electromagnetic potential gets out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electromagnetic potential $A^\mu$ is not uniquely defined. There is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauge_freedom#Gauge_freedom&quot; rel=&quot;nofollow&quot;&gt;gauge freedom&lt;/a&gt;; one can always add the gradient of a function of space and time to get a different electromagnetic potential $A&amp;#39;^{\mu}$ that has the same electric and magnetic field: $A&amp;#39;^{\mu} = A^\mu + \partial^\mu \Gamma$. For a charged black hole, this means that there can be a non-zero magnetic potential $A^j$, (which still gives a zero magnetic field).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the point is that the question of &quot;how does the electric field get out of a black hole&quot; has an analog in the quantum mechanics of flat space-time; &quot;how does the Aharanov-Bohm effect work?&quot; In both cases, there seems to be a global requirement that things be consistent even though it appears that there might logically be no relationship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To detect an electric field or an electromagnetic potential we use a small test charge.&#xA;In the two cases, we consider interactions between &quot;restricted&quot; electrons and &quot;test&quot; electrons. The restricted electrons are stuck inside the black hole, or are on paths inside a solenoid that generate essentially no electric or magnetic fields external to the solenoid. The test electrons detect the electric field outside the black hole, and the electromagnetic potential outside the solenoid.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-07-12T07:33:27.913" />
  <row Id="12178" PostTypeId="2" ParentId="12170" CreationDate="2011-07-12T08:41:25.943" Score="0" ViewCount="" Body="&lt;p&gt;Closed timelike curves can be used to create paradoxes! That's the reason why they are a problem - because you might go back and kill your grandfather. This is not just a &quot;philosophical&quot; argument, it's a matter of logic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principle that reality is not self-contradictory - i.e., that there are no real paradoxes, only apparent paradoxes - allows you to deduce that, in reality, CTCs can't be used to create paradoxes. So either there are no CTCs, or they are &quot;harmless&quot; in some way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Logic can get you that far: you should expect CTCs to be impossible or to be harmless. But your question goes further and asks why we should specifically expect them to be impossible. Well, maybe we shouldn't. CTCs can show up in anti de Sitter space; they might show up in a gravitational path integral; there might be a CTC at the big bang (as suggested by Gott and Li); there might be a giant CTC connecting cosmological future and past (as suggested by Gödel). I'm not aware of a knockdown &lt;em&gt;technical&lt;/em&gt; argument (i.e. a totally robust physics argument) against any of these. &lt;/p&gt;&#xA;" OwnerUserId="1486" LastActivityDate="2011-07-12T08:41:25.943" CommentCount="1" />
  <row Id="12179" PostTypeId="1" AcceptedAnswerId="12187" CreationDate="2011-07-12T09:01:59.907" Score="1" ViewCount="72" Body="&lt;p&gt;Suppose you do a Stern-Gerlach experiment with atoms in a $j=1$ state. There would be three separate beams ($m_j = -1, 0, 1$) coming out of the apperatus. But what would be the relative distribution of atoms in these beams?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On first thought I would say that the distribution should be $\frac{1}{3}$, $\frac{1}{3}$, $\frac{1}{3}$. But considering the &quot;classical&quot; distribution of the projections of the angular momentum onto one axis I could imagine that there could be more atoms in the $m_j = 0$ state. This would still be compatible with symmetry considerations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To state the question differently: how does the density matrix of an unpolarized beam of $j=1$ atoms look like?&lt;/p&gt;&#xA;" OwnerUserId="4468" LastEditorUserId="4468" LastEditorDisplayName="" LastEditDate="2011-07-12T11:42:18.007" LastActivityDate="2011-07-12T11:42:18.007" Title="Stern-Gerlach-Experiment with j=1 atoms?" Tags="&lt;quantum-mechanics&gt;&lt;spin&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="12180" PostTypeId="2" ParentId="3541" CreationDate="2011-07-12T09:17:58.677" Score="0" ViewCount="" Body="&lt;p&gt;In response to discussion &quot;yes, from a layperson's view ...&quot;&#xA;It is important to notice the supernatural was given a definition, i.e. non-physical&#xA;then it was said that it must not be natural - natural possibly meaning physical.&#xA;One of the axioms of the view was that 'only what is physical can be found' and 'what exists is physical'. although the author make it sounds a conclusion it is an axiom therefore there is circular reasoning.&#xA;Instead of 'by definition' it should be said 'by definition and by the assumpion that something's existance means it is is physical I conclude that the supernatural does not exist'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also important to notice that any definitions given can create axioms and that any axioms given can create undefined terms (as in an axiomatic approach to geometry).&#xA;Also what is the author's definition of 'found'? is it that it is found by experimental physics or is it that is found by theoretical physics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the energy all I have to say is that:&#xA;It is generally accepted that particles would need energy to be created, namelly a photon. The author argues that the photon is matter itself. &#xA;the axioms include: 'everything that we can think of as a particle is matter'. well actually, this is either a axiom that leads (possibly together with other axioms) to the conclusion that everything is physical or a conclusion that is a consequence of the axiom that everything is physical. The arthor does not prove these ideas idependently. Again circular reasoning.&lt;/p&gt;&#xA;" OwnerUserId="4469" LastActivityDate="2011-07-12T09:17:58.677" />
  <row Id="12181" PostTypeId="1" CreationDate="2011-07-12T09:20:57.957" Score="0" ViewCount="43" Body="&lt;p&gt;If the only things which exist are those which are remembered forever, and memories are gappy and sparse, it is possible to have a memory of an event with some of its contextual information forgotten, like the relative time ordering of that event with respect to other recalled events. Without any further contextual information to disambiguate, is it meaningless to presuppose that event has to either lie in the future, past, or spacelike separated with respect to another recalled event? In other words, is time-ordering an illusion with no basis in reality? &lt;/p&gt;&#xA;" OwnerUserId="4470" LastActivityDate="2011-07-12T09:20:57.957" ClosedDate="2011-07-12T13:42:51.040" Title="Is time-ordering an illusion in quantum gravity?" Tags="&lt;quantum-gravity&gt;" CommentCount="1" />
  <row Id="12182" PostTypeId="2" ParentId="12170" CreationDate="2011-07-12T09:31:35.797" Score="0" ViewCount="" Body="&lt;p&gt;It's certainly possible that just as the holographic principle can save unitarity for black holes, that some generalization of the holographic principle coupled with cosmic censorship for closed timelike curves can save unitarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, the interior of a time machine is only real to the extent that memories and records of the interior can get out.&lt;/p&gt;&#xA;" OwnerUserId="4471" LastActivityDate="2011-07-12T09:31:35.797" />
  <row Id="12183" PostTypeId="1" AcceptedAnswerId="12193" CreationDate="2011-07-12T09:36:10.863" Score="3" ViewCount="147" Body="&lt;p&gt;Here in this forum (and elsewhre) I read about &quot;charged&quot; black holes. &#xA;What is the background for this questions? Is this more a question of &#xA;theroretical interest?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &quot;practice&quot; :=) black holes should be neutral, because matter &#xA;is neutral globally. A star is charged a little bit, depending &#xA;on surface temperature and plasma physics in corona. &#xA;But this charge will be blast away (with surface material) &#xA;when the interior of the star collapses to form a black hole. &#xA;Again, when a black hole &quot;eats&quot; matter from around, charge separation &#xA;might occur, but only to a limited extent, until the voltage &#xA;developed would cancel the separation mechanism. (As in case of a star) &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-07-13T17:45:56.203" Title="Charged Black holes - What is the Interest in those?" Tags="&lt;black-holes&gt;" AnswerCount="5" />
  <row Id="12184" PostTypeId="2" ParentId="12183" CreationDate="2011-07-12T10:48:10.910" Score="3" ViewCount="" Body="&lt;p&gt;I found this interesting bit of &lt;a href=&quot;http://www.sciencedaily.com/releases/2003/09/030905072028.htm&quot; rel=&quot;nofollow&quot;&gt;observation&lt;/a&gt; on the black hole in the center of our galaxy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Our infrared detection shows for the first time that the black hole's meal is more like the Grand Rapids, in which energetic glitches from shocked gas are occurring almost continually.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://adsabs.harvard.edu/abs/2006AAS...208.2104W&quot; rel=&quot;nofollow&quot;&gt;theoretical speculation&lt;/a&gt; on interstellar lightning offers  a mechanism how there could be charge separation in interstellar gas. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;. Possible chemical and electromagnetic signatures of interstellar lightning are discussed. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If such a mechanism is detected and established it is conceivable that a black hole could  eat a dust cloud with a, for example negative charge, while the positive charged cloud is attracted to another black hole :). Lo, charged black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think theorists just solve and see that the solutions are physical, and then wait to see whether observations verify the solutions, by the work of hard pressed astronomers and astrophysicists :).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is like the question of &quot;why climb the mountain&quot; answer: &quot;because it is there.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-07-12T10:48:10.910" />
  <row Id="12186" PostTypeId="2" ParentId="12183" CreationDate="2011-07-12T11:13:05.490" Score="3" ViewCount="" Body="&lt;p&gt;To throw also a slightly less mainstream suggestion out there (I don't know too much about it, but I know the field exists and is active), there is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_hole_electron&quot; rel=&quot;nofollow&quot;&gt;proposal&lt;/a&gt; to model an electron as a tiny charged black hole. The main interest in that is, I think, born from some considerations by Einstein of the motion of a particle on a gravitating background (the geodesic hypothesis). The main papers were &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=1503389&quot; rel=&quot;nofollow&quot;&gt;this one by Einstein, Infeld, and Hoffman&lt;/a&gt; and &lt;a href=&quot;http://www.ams.org/mathscinet-getitem?mr=1689&quot; rel=&quot;nofollow&quot;&gt;this follow-up by Einstein and Infeld&lt;/a&gt;, which modelled particles as little singularities in space-time. &lt;/p&gt;&#xA;" OwnerUserId="829" LastActivityDate="2011-07-12T11:13:05.490" />
  <row Id="12187" PostTypeId="2" ParentId="12179" CreationDate="2011-07-12T11:30:32.927" Score="2" ViewCount="" Body="&lt;p&gt;The z-projection of the angular momentum of a completely unpolarized system is distributed uniformly both classically and quantum mechanically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum mechanics, the density matrix of a three state system&#xA;is given by:&#xA;$\rho = \frac{1}{3}(I_3+\overrightarrow{n}.\overrightarrow{\lambda})$, where $I_3$ is the three dimensional unit matrix.&#xA;$\overrightarrow{n}$ is the polarization vector and $\overrightarrow{\lambda}$ is a vector of Gell-mann matrices. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a fully polarized system $\overrightarrow{n}.\overrightarrow{n}= 1$, while in the case of a completely unpolarized system:&#xA;$\overrightarrow{n} = 0$. In this case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho = \frac{1}{3}(I_3)$. Therefore, the three states are equally distributed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In classical mechanics, the phase space of a spin system is the unit sphere. The z-component of a unit angular momentum system is given by: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ J_z = cos(\theta)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\theta$ is the inclination angle. A completely unpolarized classical angular momentum should be distributed uniformly with respect to the surface area of the sphere which is the classical phase space.&#xA;The area of a spherical zone of thickness $h$ is $S = 2 \pi R h$, where $R=1$ is the sphere radius (This result was already known to &lt;a href=&quot;http://descartes.cnice.mec.es/descartes2/previas_web/materiales_didacticos/ing_redondos/arquime1.htm&quot; rel=&quot;nofollow&quot;&gt;Archimedes&lt;/a&gt;). Therefore the z-coordinate is uniformly distributed on the sphere and the z-component of the angular momentum expressed as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ J_z = \frac{z}{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is linear in $z$, therefore also uniformly distributed.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorUserId="2190" LastEditorDisplayName="" LastEditDate="2011-07-12T11:37:24.863" LastActivityDate="2011-07-12T11:37:24.863" CommentCount="7" />
  <row Id="12188" PostTypeId="2" ParentId="12170" CreationDate="2011-07-12T11:43:28.063" Score="1" ViewCount="" Body="&lt;p&gt;It is widely believed (but it has not been rigurously proved so far) that to sustain stable closed timelike curves, you need copious (i.e of the order of the total mass of the universe) quantities of exotic matter. Exotic matter is just a generic term to describe matter for which the stress-energy tensor satisfies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ g^{\mu \nu} T_{\mu \nu} &amp;lt; 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, exotic matter violates the Strong Energy Condition, which is known to hold in all known quantum physical theories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This applies not only to CTC, but as well as stable wormholes, warp drives, or anything actually fun. This is in itself, outstanding evidence of the catholic nature of God, since interestellar travel would raise the need for some akward explaining from the Pope (i know, BS, but let me have my punchline please)&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-07-12T14:24:09.670" LastActivityDate="2011-07-12T14:24:09.670" />
  <row Id="12189" PostTypeId="1" CreationDate="2011-07-12T13:17:44.267" Score="1" ViewCount="68" Body="&lt;p&gt;What is the Jost function in scattering theory? Is it an operator or some kind of determinant? How is it obtained?&lt;/p&gt;&#xA;" OwnerUserId="4435" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-12T20:49:14.897" LastActivityDate="2011-07-12T20:49:14.897" Title="What is the Jost function in scattering theory?" Tags="&lt;quantum-mechanics&gt;&lt;scattering&gt;" AnswerCount="1" />
  <row Id="12190" PostTypeId="2" ParentId="10174" CreationDate="2011-07-12T13:37:43.977" Score="6" ViewCount="" Body="&lt;p&gt;It is an incorrect picture to envision the Cooper pairs as existing as an isolated occurrence in a lattice, since the very existence of Cooper pairs depends on a supporting cast of other electrons.  In his work, Cooper showed that the ground state of a metal is unstable against an arbitrarily small net attraction between two electrons of opposite momentum, with this momentum near the Fermi momentum of the system.  Therefore, the energetically favorable state is the paired state, and the superconducting state exists when all of the conduction electrons are bound into such pairs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, something which is under-emphasized in the &quot;wake model&quot; is the size of the Cooper pairs in &quot;classical&quot; superconductors (ie, superconductors that have the pairing mechanism described by BCS theory). The spatial extent of Cooper pairs is quite large, on the order of thousands of angstroms (many, many lattice spacings), meaning that the pair condensation results in millions (or more) over lapping pairs.&lt;br&gt;&#xA;In effect, in the superconducting state, these millions of paired electrons travel like a superfluid through the material.  &lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2011-07-12T13:37:43.977" />
  <row Id="12191" PostTypeId="1" AcceptedAnswerId="12192" CreationDate="2011-07-12T14:21:31.600" Score="3" ViewCount="79" Body="&lt;p&gt;I am currently working with pressure differences in the milibar range. However I do not have a good feeling for pressure differences. In every day life: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Where does a pressure of say 100 mbar over atmospheric pressure occur?&lt;/li&gt;&#xA;&lt;li&gt;When I blow up my mouth, what is the pressure I get?&lt;/li&gt;&#xA;&lt;li&gt;When I blow up an air balloon, what is the pressure inside?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2480" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-07-13T22:36:07.657" LastActivityDate="2011-07-13T22:36:07.657" Title="Get an intuition for pressure values" Tags="&lt;pressure&gt;&lt;everyday-life&gt;" AnswerCount="2" />
  <row Id="12192" PostTypeId="2" ParentId="12191" CreationDate="2011-07-12T14:36:31.450" Score="3" ViewCount="" Body="&lt;p&gt;Your mouth or balloon blowing are very bad examples, the reason is because there is no obvious reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A McDonalds straw is a better example.  Say that it's 7 inches long.  You know the density, you know gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P=\rho g h$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the full 7 inches will get you about 17 mbar.  From my everyday experience and intuition, I believe that I can maintain a pressure in my mouth or blowing up a balloon several times what is required to suck up a (watery) drink the full length of a straw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;100 mbar is likely to overwhelm your mouth, but it might just be &quot;athletic&quot;.  In order to test it, you'll only need a few pipes from an aquarium store, a tub, and maybe some clamps (unless you have more specific equipment).  Looking at some anecdotal information online, I see some people reporting in the neighborhood of 2 psi (137 mbar), corresponding with reports that &quot;it felt like my eyes were about to pop out of my head&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, safety first!&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-07-12T14:42:52.623" LastActivityDate="2011-07-12T14:42:52.623" />
  <row Id="12193" PostTypeId="2" ParentId="12183" CreationDate="2011-07-12T14:54:31.477" Score="5" ViewCount="" Body="&lt;p&gt;&quot;Realistic&quot; black holes, by which I mean black holes of the sort for whose existence we have significant evidence, are almost certainly very close to electrically neutral, pretty much for the reasons you suggest. The two main types of black holes that seem to exist in nature are those of stellar-mass and supermassive ones at galactic centers. Both of these almost certainly formed out of large amounts of matter that was, to an excellent approximation, electrically neutral. You could get small amounts of charge separation, but I don't think anyone expects that to occur at any measurable level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course it's always possible that this is wrong: either the black holes we know about could be charged for some reason, or there could be whole other populations of charged black holes. So we'll have to do observations and experiments to check. People are always trying to improve our observations of black holes, not specifically to check for charge but to measure other interesting things such as angular momentum. I'd guess that, if these black holes had a large charge, such observations would reveal it, but I've never looked into details.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-07-12T14:54:31.477" />
  <row Id="12194" PostTypeId="1" CreationDate="2011-07-12T15:01:12.407" Score="0" ViewCount="62" Body="&lt;p&gt;Let's say we have 100 grams of chocolate which has 571 kilocalories (so it's about 2.4 megajoules) and we have barbell which weights 100 kilograms and need to lift it to 0.5 meters height. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how many times you can lift it eating only 100 grams of chocolate? Let's use potential formula&#xA;$$E=mgh$$&#xA;n – how many times you lift&#xA;$$E=nmgh$$&#xA;$$n=\frac{E}{mgh}$$&#xA;$$n=\frac{2.4\cdot 10^6J}{100kg\cdot10\frac{m}{s^2}\cdot0.5m}=4800$$&#xA;So I get 4800 times, but I doubt it's true. So where and how many (approximately) our gained energy disappears, I mean becomes other kind of energy?&lt;/p&gt;&#xA;" OwnerUserId="3992" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-12T16:09:58.343" LastActivityDate="2011-07-12T16:09:58.343" Title="People eating and gaining energy" Tags="&lt;energy&gt;&lt;biophysics&gt;&lt;food&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12195" PostTypeId="2" ParentId="12183" CreationDate="2011-07-12T15:29:16.787" Score="3" ViewCount="" Body="&lt;p&gt;Let me add to the already many good answers, that from a string-theoretic perspective, (near)-extremal charged (Reissner-Nordstr\&quot;om) black holes in various dimensions are important, because they are simpler to calculate with (mainly because of supersymmetry). They therefore serve as an important theoretical testbed for ideas concerning microscopic counting of entropy, holography, unitarity, etc.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastActivityDate="2011-07-12T15:29:16.787" />
  <row Id="12196" PostTypeId="5" CreationDate="2011-07-12T15:38:14.387" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-07-12T15:38:14.387" LastActivityDate="2011-07-12T15:38:14.387" />
  <row Id="12197" PostTypeId="4" CreationDate="2011-07-12T15:38:14.387" Score="0" ViewCount="" Body="Energy sources that do not rely on the consumption of an exhaustible fuel. Tidal, geothermal, and the direct and indirect solar energies (direct: photovoltaics, CSP, solar thermal; indirect: wind, wave, biomass) are renewable energy." OwnerUserId="-1" LastEditorUserId="4030" LastEditorDisplayName="" LastEditDate="2011-07-18T20:05:51.603" LastActivityDate="2011-07-18T20:05:51.603" />
  <row Id="12198" PostTypeId="2" ParentId="12194" CreationDate="2011-07-12T15:43:22.200" Score="2" ViewCount="" Body="&lt;p&gt;Yes, if you combust the chocolate to extract all chemical energy from it, and have a machine with perfect energy efficiency, then you can lift 100 kilograms 2.4 kilometers high. But your body is rather far from a machine with perfect energy efficiency. (Your heart needs to pump, your diaphragm goes up and down, in addition to other mechanical and chemical &quot;inefficiencies&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, doing a google search brings up &lt;a href=&quot;http://www.jssm.org/vol2/n3/8/v2n3-8pdf.pdf&quot; rel=&quot;nofollow&quot;&gt;this study&lt;/a&gt; which studies aerobic weight lifting exercises (so presumably the weight is much less than 100 kilograms, though distance is about the same). The measured total aerobic energy consumption (from carbon dioxide content of the exhaled breath) is around 10 kilojoules mer minute. For that exercise, your 100 grams of chocolate can sustain you for around 4 hours. (If you were to do 4800 reps in 4 hours, that would require sustaining 20 reps a minute, a truly fearsome rate...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Basal_metabolic_rate&quot; rel=&quot;nofollow&quot;&gt;basal metabolic rate&lt;/a&gt; for humans, say we have a typical male standing at 170 centimeters, 75 kilograms, and 30 years of age, he would burn about 1600 kilocalories per day doing absolutely nothing. So a 100g chocolate bar gets him through about a third of the day. &lt;/p&gt;&#xA;" OwnerUserId="829" LastActivityDate="2011-07-12T15:43:22.200" CommentCount="1" />
  <row Id="12199" PostTypeId="1" AcceptedAnswerId="12253" CreationDate="2011-07-12T15:57:03.903" Score="7" ViewCount="234" Body="&lt;p&gt;I am trying to numerically solve Schrödinger's equation with Cayley's expansion ($\hbar=1$)&#xA;$$\psi(x,t+\Delta t)=e^{-i H\Delta t}\psi(x,t)\approx\frac{1-\frac{1}{2}i H\Delta t}{1+\frac{1}{2}i H\Delta t}\psi(x,t)$$&#xA;and second order finite difference approximation for space derivative&#xA;$\psi&amp;#39;&amp;#39;(x)\approx\frac{\psi_{j+1}^n-2\psi_j^n+\psi_{j-1}^n}{\Delta x^2}$,&#xA;as described in Numerical Recipes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hamiltonian is that of one-dimensional harmonic oscillator ($m=1$): $H=\frac{\partial^2}{\partial x^2}+\frac{1}{2}kx^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i end up computing is this system of linear equations with a 3-band matrix at every time step:&#xA;$$(1+\frac{1}{2}iH\Delta t)\psi_j^{n+1}=(1+\frac{1}{2}iH\Delta t)^*\psi_j^n$$&#xA;or explicitly&#xA;$${\boldsymbol A}{\boldsymbol \Psi}^{n+1}={\boldsymbol A}^*{\boldsymbol \Psi}^n\;,$$&#xA;where the elements on the main diagonal of ${\boldsymbol A}$ are $d_j=1+\frac{i\Delta t}{2m\Delta x^2}+\frac{i\Delta t}{4}x_j^2$ and the elements on the upper and lower diagonals are $a=-\frac{i\Delta t}{4m\Delta x^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the initial condition i choose the coherent state, for which there is a known analytical solution. While trying to propagate the wavefunction over few periods of oscillation, the wavefunction gets distorted. When timestep is smaller, the distortions appear later.&#xA;I am wondering what is the reason for this process, if it has to do with the Courant condition and if there is a known relation between the size of timestep and start of these distortions.&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://db.tt/91BvmYA&quot; rel=&quot;nofollow&quot; title=&quot;propagation with distortions&quot;&gt;Here&lt;/a&gt; is a video of propagation.&lt;/p&gt;&#xA;" OwnerUserId="1800" LastEditorUserId="1800" LastEditorDisplayName="" LastEditDate="2011-07-13T11:32:02.537" LastActivityDate="2011-07-13T23:43:27.037" Title="Solving Schrödinger's equation with Crank-Nicolson method" Tags="&lt;quantum-mechanics&gt;&lt;computational-physics&gt;&lt;numerics&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="12200" PostTypeId="1" AcceptedAnswerId="12205" CreationDate="2011-07-12T16:26:43.830" Score="8" ViewCount="388" Body="&lt;p&gt;Why is electricity not transmitted wirelessly such that we don't need to span cables on the earth's surface? As in: electricity is transmitted wirelessly from the power plant to the household.&lt;/p&gt;&#xA;" OwnerUserId="652" LastEditorUserId="1638" LastEditorDisplayName="" LastEditDate="2011-07-12T19:42:26.397" LastActivityDate="2011-08-12T17:25:54.917" Title="Why is electricity not transmitted wirelessly?" Tags="&lt;electricity&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="6" CommentCount="4" FavoriteCount="2" />
  <row Id="12201" PostTypeId="1" CreationDate="2011-07-12T16:41:11.087" Score="5" ViewCount="102" Body="&lt;p&gt;The book I'm reading about optics says at some point that &quot;each color (wavelength) contained in the white light interferes only with itself&quot;. But why is this so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I moved the rest of the question &lt;a href=&quot;http://physics.stackexchange.com/q/12208/4283&quot;&gt;elsewhere&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4283" LastEditorUserId="4283" LastEditorDisplayName="" LastEditDate="2011-07-12T17:23:19.197" LastActivityDate="2011-07-12T18:59:21.563" Title="Why don't waves with different wavelengths interfere with each other in white-light?" Tags="&lt;optics&gt;&lt;interference&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12202" PostTypeId="2" ParentId="12200" CreationDate="2011-07-12T16:44:43.190" Score="3" ViewCount="" Body="&lt;p&gt;In French, radio transmitting is named &quot;radio*&lt;em&gt;diffusion&lt;/em&gt;*&quot;, an expression &#xA;much better. The best electromagnetical waves to focus and concentrate to &#xA;some goal are IR and visible lasers, but aside from generating loss those&#xA;will &lt;strong&gt;diffuse&lt;/strong&gt; the radiation, instead of &lt;strong&gt;transmitting&lt;/strong&gt; .&#xA;BTW, &quot;cables on the earth&quot; are used for long distance only, from 20 kV downward &#xA;almost everything is in cables under soil here in Germany. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could answer in this way: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is more and more information transmitted no longer by wireless,&#xA;but by cables? &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-07-13T09:45:28.513" LastActivityDate="2011-07-13T09:45:28.513" CommentCount="3" />
  <row Id="12203" PostTypeId="1" CreationDate="2011-07-12T16:47:29.273" Score="2" ViewCount="16" Body="&lt;p&gt;Could anyone suggest a recent review article or book on gas chromatography instrumentation? My interest is in the devices themselves, rather than using them for a particular application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rsi.aip.org/resource/1/rsinak/v61/i11/p3317_s1&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt;, &quot;Gas Chromatography&quot; in Review of Scientific Instruments is the sort of thing I'm going for; I'm just checking to see if there's a more recent or superior introduction.&lt;/p&gt;&#xA;" OwnerUserId="4458" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-12T17:49:06.167" LastActivityDate="2011-07-12T17:49:06.167" Title="Review of Gas Chromatographs?" Tags="&lt;experimental-physics&gt;&lt;reference-request&gt;&lt;measurement&gt;" />
  <row Id="12204" PostTypeId="2" ParentId="12201" CreationDate="2011-07-12T16:49:11.033" Score="6" ViewCount="" Body="&lt;p&gt;Add any instant in time, light of different wavelengths can be said to interfere.  However, because of the extreme frequencies of optical light, any cross interference will get time-averaged away very quickly unless the two waves are very close in frequency.  &lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-07-12T16:49:11.033" CommentCount="2" />
  <row Id="12205" PostTypeId="2" ParentId="12200" CreationDate="2011-07-12T16:51:12.717" Score="8" ViewCount="" Body="&lt;p&gt;Electricity is the flow of electrical charge - generally electrically charged particles called electrons in a wire. It can't flow through air, except in the form of electrically charged particles of air - as in a spark or lightning stroke.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnetic fields can travel in air, so you can send electricity by using it to make a magnetic field and then using the magnetic field at the other end to make electricity. This is how a transformer works - &lt;strong&gt;but&lt;/strong&gt; it only works efficiently if the two sets of wire making the magnetic field are very close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use it for sending small amounts of electricity a short distance where a wire (or connector) would be difficult, such as charging an electric toothbrush - but it's not efficent for large amounts or a long distance.&lt;/p&gt;&#xA;" OwnerUserId="2525" LastEditorUserId="103" LastEditorDisplayName="" LastEditDate="2011-08-12T17:25:54.917" LastActivityDate="2011-08-12T17:25:54.917" CommentCount="2" />
  <row Id="12206" PostTypeId="1" CreationDate="2011-07-12T16:53:21.327" Score="0" ViewCount="94" Body="&lt;p&gt;For some electric devices, like a fan or air conditioner, I read about their power consumption in watts on their specification guide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it tell about the power at normal or full speed? or Does the speed even affect the power consumption? Can I find the consumption at different speeds?&lt;/p&gt;&#xA;" OwnerUserId="4472" LastActivityDate="2011-08-12T09:09:57.247" Title="Total power consumption of electric device" Tags="&lt;electricity&gt;&lt;power&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12207" PostTypeId="2" ParentId="12200" CreationDate="2011-07-12T17:06:21.260" Score="3" ViewCount="" Body="&lt;p&gt;It can be done via electromagnic waves. The higher the tansmission frequency the better. Tesla wanted to do it. But is is very inefficient, much of the power will be dissapated where it won't do you any good (such as heating the ground), or be poached by non-paying people who have learned how to harvest it. There is some work on antennas that can be used to obtain energy from the environment (mainly radio waves), but these are for very low power remote devices, they are not suitable for bulk power applications.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-07-12T17:06:21.260" />
  <row Id="12208" PostTypeId="1" CreationDate="2011-07-12T17:22:10.257" Score="6" ViewCount="256" Body="&lt;p&gt;The book I'm reading about optics says that an anti-reflective film applied on glass* makes the glass &lt;em&gt;more&lt;/em&gt; transparent, because the air-film and glass-film reflected waves (originated from a paraxial incoming wave) interfere destructively with each other, resulting on virtually no reflected light (therefore the &quot;extra&quot; light gets transmitted). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, &lt;a href=&quot;http://physics.stackexchange.com/questions/10410/why-dont-waves-erase-out-each-other-when-looking-onto-a-wall/10439#10439&quot;&gt;this answer&lt;/a&gt; states that &lt;em&gt;&quot;Superposition is the principle that the amplitudes due to two waves incident on the same point in space at the same time can be naively added together, but the waves do not affect each other.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how does this fit into this picture? If the reflected waves actually continue happily travelling back, where does the extra transmitted light come from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;* the film is described as (1) having an intermediate index of refraction between those of air and glass, so that both the air-film and film-glass reflections are &quot;hard&quot;, i.e., produce a 180º inversion in the phase of the incoming wave, and (2) having a depth of 1/4 of the wavelength of the wave in the film, so that the film-glass reflection travels half its wavelength back and meets the air-film reflection in the opposite phase, thus cancelling it.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="4283" LastActivityDate="2011-07-14T17:25:04.863" Title="How does anti-reflective coating work?" Tags="&lt;optics&gt;&lt;interference&gt;&lt;reflection&gt;" AnswerCount="7" FavoriteCount="2" />
  <row Id="12209" PostTypeId="2" ParentId="12201" CreationDate="2011-07-12T17:41:47.183" Score="2" ViewCount="" Body="&lt;p&gt;Interference is not an instant phenomenon but a time-lasting one. The time should be sufficient to speak of certain frequencies. Two different frequencies can be easily distnguished with the corresponding resonators but this process takes time. It is in a long term condition that two frequencies are separated (resonators pumped). Instantly we cannot even speak of a frequency.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-07-12T18:59:21.563" LastActivityDate="2011-07-12T18:59:21.563" />
  <row Id="12210" PostTypeId="2" ParentId="12208" CreationDate="2011-07-12T17:56:54.547" Score="2" ViewCount="" Body="&lt;p&gt;Perhaps it will help to recall that energy is a nonlinear function of the electromagnetic field.  The superposition principle applies to the electromagnetic field, not the energy or power.    So if two waves are superimposed out of phase, 1 - 1 = 0, we can say they are both happily traveling &quot;independent&quot; of each other (from the point of view of the EM field), but from the point of view of the energy they contain they are not independent.  &lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-07-12T17:56:54.547" CommentCount="1" />
  <row Id="12211" PostTypeId="1" CreationDate="2011-07-12T18:17:30.120" Score="0" ViewCount="105" Body="&lt;p&gt;Before I take P-Chem I would like to understand how physicists view the same material. What Physics course(s) should I take and/or books should I read to learn the same material from a physicist's perspective? Would thermodynamics be equivalent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Background: I have done the math: Calculus 1-3 and differential equations and have already taken general physics 1+2 and 2 semesters of modern physics (relativity, QM...). &lt;/p&gt;&#xA;" OwnerUserId="2843" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-12T20:48:30.510" LastActivityDate="2011-07-13T09:04:00.950" ClosedDate="2011-07-13T12:22:08.703" Title="Physics Equivalent of Physical Chemistry (P-Chem)" Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="12212" PostTypeId="2" ParentId="12211" CreationDate="2011-07-12T18:26:11.187" Score="0" ViewCount="" Body="&lt;p&gt;As I understand the subject, P-Chem is applied quantum mechanics. In other words, P-Chem investigates ways in which chemical behaviors can be derived from the underlying quantum mechanical behavior of atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in that sense, the physical chemistry of physics is... physical chemistry. If you would like to have a better understanding of basic quantum mechanics, I could recommend the text book I used for intro QM as an undergrad in physics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles&quot; by R. Eisberg and R. Resnick&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(EDIT) Actually I think this is a better intro QM text. It is what I used in the more advanced second semester quantum theory class as an undergrad:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Introductory Quantum Mechanics&quot; by R. Liboff&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-07-12T18:26:11.187" CommentCount="5" />
  <row Id="12213" PostTypeId="2" ParentId="12208" CreationDate="2011-07-12T18:28:09.940" Score="0" ViewCount="" Body="&lt;p&gt;The wave reflected from the air-film interface continues happily traveling back, as you say, but so does the wave reflected from the film-glass interface. Since they are the same frequency but in antiphase, they interfere destructively as long as they keep going. Since the superposition principle states that they do not affect each other in any way, they keep on going as long as they like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is no energy transported backwards in the reflected waves, because the energy is proportional to the square of the &lt;em&gt;total&lt;/em&gt; electric field. It is the energy that is conserved, not the electric field, so all the energy (if not absorbed) is transmitted.&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2011-07-12T18:28:09.940" CommentCount="4" />
  <row Id="12214" PostTypeId="2" ParentId="12189" CreationDate="2011-07-12T19:48:01.553" Score="2" ViewCount="" Body="&lt;p&gt;The standard reference to scattering theory is Roger G. Newton, &quot;Scattering Theory of Waves and Particles&quot;, where full information is given. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me here just give the definition of the Jost function in the case of a radial time independent Schrodinger Equation. There are three distinguished solutions&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The regular solution $\varphi(r,k)$ with boundary conditions $\lim_{r\to 0}\varphi =0$ and $\lim_{r\to 0}\frac{\partial \varphi}{\partial r}=1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Two irregular solutions $f_{\pm}(r,k)$ with boundary conditions $\lim_{r\to \infty}e^{\mp ikr}f_{\pm}(r,k)=1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(Since the Schrodinger equation is of second order, only two of the three above solutions are independent.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now define ${\cal F}_{\pm} ~:=~ \mathrm{Wronskian}(f_{\pm},\varphi)&#xD;&#xA;~:=~f_{\pm}\frac{\partial \varphi}{\partial r}-\varphi\frac{\partial f_{\pm}}{\partial r} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Jost function is ${\cal F}_{+}$.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastActivityDate="2011-07-12T19:48:01.553" CommentCount="1" />
  <row Id="12215" PostTypeId="1" CreationDate="2011-07-12T20:49:31.233" Score="1" ViewCount="188" Body="&lt;p&gt;Why isn't pressure used as flight?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard that 2L bottles can hold a pressure of up to 90 PSI safely. Since $F = PA$, if the nozzle of a pressure rocket is about 4 inches squared in area, that would be a thrust of 360 pounds!? Is there something wrong with my math, or why don't we just pressurize air and use it for flight?&lt;/p&gt;&#xA;" OwnerUserId="4144" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-12T23:17:56.083" LastActivityDate="2011-08-25T17:11:17.520" Title="Why isn't pressure used for flight?" Tags="&lt;classical-mechanics&gt;&lt;fluid-dynamics&gt;" AnswerCount="4" CommentCount="4" />
  <row Id="12216" PostTypeId="2" ParentId="12215" CreationDate="2011-07-12T21:11:07.757" Score="1" ViewCount="" Body="&lt;p&gt;The principal disadvantage is the indirect use of energy. Energy is used to compress air, which in turn provides the energy to run the rocket. Any conversion of energy between forms results in loss. Hence it's more economically efficient to burn fuel rather than compress gas and then release it. This idea has been used in &lt;a href=&quot;http://en.wikipedia.org/wiki/Compressed_air_car&quot; rel=&quot;nofollow&quot;&gt;Compressed Air Cars&lt;/a&gt; much before. Also look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Compressed_air_energy_storage&quot; rel=&quot;nofollow&quot;&gt;Compressed Air Energy&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="966" LastActivityDate="2011-07-12T21:11:07.757" CommentCount="6" />
  <row Id="12217" PostTypeId="2" ParentId="12206" CreationDate="2011-07-12T21:15:52.423" Score="0" ViewCount="" Body="&lt;p&gt;The &quot;Wattage&quot; written on an electric device is the maximal &#xA;power the device will take up. &#xA;Exemption is the power of motors when starting from rest. &#xA;In this case the power can be much higher for some seconds/&#xA;a minute than rated. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-07-12T21:15:52.423" CommentCount="1" />
  <row Id="12218" PostTypeId="2" ParentId="12215" CreationDate="2011-07-12T21:29:35.173" Score="1" ViewCount="" Body="&lt;ul&gt;&#xA;&lt;li&gt;1st - &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Compressed air &lt;strong&gt;is&lt;/strong&gt; used for rocket flight, by amateurs! &#xA;Look here: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Water_rocket&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Water_rocket&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2nd - &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Do You have an idea how long it will take to blow out the 2 ltr 90 psi &#xA;through that 4 x 4 inch &quot;nozzle&quot;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was (is still?) a use of compressed carbon dioxide &#xA;as a &quot;fuel&quot; for model aicraft motors. (Cox, afair) &#xA;This cylinder/piston engines were powered from the same &#xA;steel ampoules as used in soda fountains. This is liquid &#xA;carbon dioxide, so energy content is miles higher than &#xA;in case of compressed gas. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-07-13T09:18:24.263" LastActivityDate="2011-07-13T09:18:24.263" />
  <row Id="12219" PostTypeId="2" ParentId="12215" CreationDate="2011-07-12T21:41:39.263" Score="7" ViewCount="" Body="&lt;p&gt;In a way it is. A jet engine will create a high pressure region in the combustion chamber and discharge it through the nozzle. It turns out this is inefficient, and it is better to use the pressure to drive turbine blades that push the cold air around the engine. Remember cold = higher density = higher momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To a lesser degree a propeller will create a high pressure region behind the blades, that results in thrust.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-07-12T21:41:39.263" CommentCount="1" />
  <row Id="12220" PostTypeId="2" ParentId="12208" CreationDate="2011-07-12T22:36:43.870" Score="3" ViewCount="" Body="&lt;p&gt;The thickness of the AR coating is chosen such that the reflections from the two interfaces cancel out (at the wavelength for which the AR coating was designed):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/mxwEa.png&quot; alt=&quot;http://upload.wikimedia.org/wikipedia/commons/8/8c/Optical-coating-2.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Anti-reflective_coating&quot; rel=&quot;nofollow&quot;&gt;Anti-reflective coating&lt;/a&gt; in Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As endolith points out in the comments, to explain how the transmission is enhanced, you have to draw a few more rays in the diagram.  Here's another illustration, from the Wikipedia article for &lt;a href=&quot;http://en.wikipedia.org/wiki/Fabry%E2%80%93P%C3%A9rot_interferometer&quot; rel=&quot;nofollow&quot;&gt;Fabry–Pérot interferometer&lt;/a&gt;, which shows a few higher-order reflections:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/d/db/Etalon-1.svg/297px-Etalon-1.svg.png&quot; alt=&quot;Fabry-Perot Etalon&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the anti-reflective coating, you choose the thickness such that R1 and R2 cancel while T1 and T2 constructively interfere.  Note that this is dependent on the wavelength, the angle of incidence, and the index of refraction of whatever is being coated.  With other thicknesses, you can make a high-reflectivity coating, or a coating of whatever reflectivity you want.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-07-14T17:25:04.863" LastActivityDate="2011-07-14T17:25:04.863" CommentCount="4" />
  <row Id="12221" PostTypeId="1" CreationDate="2011-07-12T22:45:28.683" Score="7" ViewCount="367" Body="&lt;p&gt;Because of my mathematical background, I've been finding it hard to relate the physics-talk I've been reading, with mathematical objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In (say special) relativity, we have a Lorentzian manifold, $M$. This manifold has an atlas with local coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In differential geometry, when people talk about a `change of coordinates' they mean switching from one local coordinate system in this atlas, to another. For example, one coordinate system in this atlas is a map $\phi_1: U_1 \rightarrow V$ where $V$ is an open set of $M$, and $U$ is an open set of $\mathbb{R}^4$; and if another is $\phi_2: U_2 \rightarrow W$ is another ($U_2$ and open in $\mathbb{R}^4$, and $W$ an open in $M$), then $\phi_1^{-1} |_{V\cap W}\circ \phi_2|_{\phi_2^{-1}(V\cap W)}$ is a coordinate change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in physics it seems that the meaning is different. Indeed if $p \in M$ then you can have a reference frame at $p$, but you can also have a reference frame that is &lt;em&gt;accelerated&lt;/em&gt; at $p$. I'm not sure how to interpret this mathematically! What is the mathematical analogue of having an accelerated frame of reference at a point, as opposed to having an inertial frame of reference at a point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help would be much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="4249" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-14T10:05:01.570" LastActivityDate="2011-07-19T21:10:08.683" Title="What does a frame of reference mean in terms of manifolds?" Tags="&lt;general-relativity&gt;&lt;mathematical-physics&gt;&lt;special-relativity&gt;&lt;differential-geometry&gt;&lt;inertial-frames&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" />
  <row Id="12222" PostTypeId="2" ParentId="12208" CreationDate="2011-07-12T23:15:45.490" Score="4" ViewCount="" Body="&lt;p&gt;I think that to really understand this, you have to abandon the idea that there are individual, separate electromagnetic waves. In reality, there's just one global electromagnetic field, $\bigl(\mathbf{E}(\mathbf{x}),\mathbf{B}(\mathbf{x})\bigr)$. It evolves in space and time in a manner determined by &lt;a href=&quot;http://en.wikipedia.org/wiki/Maxwell%27s_equations&quot; rel=&quot;nofollow&quot;&gt;Maxwell's equations&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For certain configurations of the EM field - specifically, those with 2D translational symmetry - the evolution described by Maxwell's equations results in the shape of the field propagating in one direction. It's much like the way waves on the ocean (normally) propagate across its surface in one direction, without changing their shape. For this reason, we call these configurations of the EM field &quot;plane waves.&quot; This is the sort of wave most people usually think of when they imagine a light wave. The key point, though, is that the idea of propagating plane waves really only arises in one particular case: when you have an isolated, 2D-symmetric EM field configuration. In &lt;em&gt;general&lt;/em&gt;, the way the field evolves in time and space is more complicated than simple directional propagation, so in general, you can't always think of the field evolution as a wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of reflection specifically, even just reflection from a &lt;em&gt;single&lt;/em&gt; surface, what this means is that the model of the incident wave reflecting off the boundary to produce a separate reflected wave is too simplistic. A more realistic description would be that the EM field has to satisfy specific conditions at the boundary between the surfaces, and that the only way to do this is for the field on the side of the incident wave to take a different value than it would have based on the incident wave alone. The difference between the actual field and the field that would be produced by the incident wave alone is called the reflected wave, because &lt;em&gt;if&lt;/em&gt; you shut off the incident wave and wait a long time, you'll wind up with a simple plane wave propagating backwards away from the boundary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same holds true (i.e. the wave description is too simplistic) for double reflection with thin film interference; in fact, even more so, because it's a more complicated system. In this case, if you have a particular relationship between the distances and frequencies involved, you can arrange it so that the boundary conditions are satisfied by the incident wave alone, so there's no &quot;extra&quot; contribution to the field to be considered a reflected wave. Or in other words, if you shut off the incident wave and wait a long time, you will wind up with nothing propagating backwards, and thus we say that there is no reflected wave.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-07-12T23:15:45.490" />
  <row Id="12223" PostTypeId="2" ParentId="12208" CreationDate="2011-07-12T23:24:59.607" Score="-1" ViewCount="" Body="&lt;p&gt;In the &lt;a href=&quot;http://en.wikipedia.org/wiki/Anti-reflective_coating&quot; rel=&quot;nofollow&quot;&gt;WP-link&lt;/a&gt;  we can obtain a substantial explanation.&lt;br&gt;&#xA;Read about the Fresnel coefficients &lt;a href=&quot;http://en.wikipedia.org/wiki/Fresnel_equations&quot; rel=&quot;nofollow&quot;&gt;WP-here&lt;/a&gt;&lt;br&gt;&#xA;Play with multilayers at «&lt;a href=&quot;http://thinfilm.hansteen.net/&quot; rel=&quot;nofollow&quot;&gt;thinfilm&lt;/a&gt;» (it uses the &lt;a href=&quot;http://en.wikipedia.org/wiki/Transfer-matrix_method_%28optics%29&quot; rel=&quot;nofollow&quot;&gt;WP-transfer-matrix method&lt;/a&gt;).&lt;br&gt;&#xA;The transmission is increased due to the forward and backward reflections in the coated film (medium 2) that are forward propagated to the medium 3 as seen in this image from &lt;a href=&quot;http://www.ifm.liu.se/courses/TFYY67/Lect13.pdf&quot; rel=&quot;nofollow&quot;&gt;pag 13.7 of B.O Sernelius lectures&lt;/a&gt;&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/JElJo.png&quot; alt=&quot;multilayer&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens to the reflected rays in the layer 1 ?&#xA;They simply cancel (vanish,do not exist) because they are in phase opposition in relation to the incident wave and almost all of the energy in the layer 2 will be forward propagated to 3.&lt;br&gt;&#xA;Lets analyse the following experiment :&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/Chdfd.png&quot; alt=&quot;experiment&quot;&gt;&lt;br&gt;&#xA;The image is a rework of a picture from the first WP-link, that can be misleading because it figures two reflected rays R1 and R2 that probably do not exist.&lt;br&gt;&#xA;If those rays are really there, and we can not measure them because they are in phase opposition, then the observer in the bottom of the image will see the light restored to a nice level because the second coated window do the opposite action of the first window (beam splitter). The particle nature of light may favour this outcome.&lt;br&gt;&#xA;If the observer can not see light then we must conclude that there is no field, energy, photons in the regions ?!?!?! of the experiment. The wave nature of light favour this outcome.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experiment is very easy to perform.&lt;br&gt;&#xA;Can someone post the outcome?  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-07-13T14:10:23.550" LastActivityDate="2011-07-13T14:10:23.550" CommentCount="3" />
  <row Id="12224" PostTypeId="2" ParentId="12211" CreationDate="2011-07-13T03:34:49.597" Score="0" ViewCount="" Body="&lt;p&gt;This really depends on the faculty of your institution - some physical chemistry curricula are heavy on molecular orbital theory and spectroscopy rather than macroscopic phenomena like thermodynamics.  In that case, the toolset and language build off what you see in a quantum mechanics class, but the increased complexity of the systems makes the reasoning necessarily a bit less rigorous.  It's hard to say what the physics equivalent is, because to a reductionist, it is a system of approximation tools that allow one to apply quantum mechanical principles to non-homogeneous many-body systems like molecules.&lt;/p&gt;&#xA;" OwnerUserId="761" LastActivityDate="2011-07-13T03:34:49.597" />
  <row Id="12225" PostTypeId="2" ParentId="12208" CreationDate="2011-07-13T03:47:19.203" Score="-1" ViewCount="" Body="&lt;p&gt;Let's talk about Where the extra transmitted energy comes from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy quantum collapses into that position where it is found when measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the energy is found is a fundamentally random thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, sometimes all energy is found in the area where waves cancel. Where cancellation approaches perfect cancellation, there probability of finding energy approaches zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm saying that you will find photons with the normal photon energy in the cancellation area, and sometimes you find more photons in the cancellation area, than in the area of constructive interference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider a very short pulse of light going through a window pane with anti-reflective coating. When a photon is detected behind the window, energy comes from where the reflected pulse is, and from where the transmitted pulse is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When absence of photon is detected behind the window, energy goes from where the transferred pulse is, to where the reflected pulse is.    &lt;/p&gt;&#xA;" OwnerUserId="4017" LastEditorUserId="4017" LastEditorDisplayName="" LastEditDate="2011-07-14T01:56:23.037" LastActivityDate="2011-07-14T01:56:23.037" CommentCount="1" />
  <row Id="12226" PostTypeId="2" ParentId="12221" CreationDate="2011-07-13T04:32:37.660" Score="3" ViewCount="" Body="&lt;p&gt;I believe the equivalent mathematical terminology to an inertial frame is &quot;normal coordinates&quot; at the point $p$. Meaning, in your coordinates the metric at $p$ is just the flat Minkowski metric and all the first derivatives of the metric vanish at $p$. Conversely an accelerated frame would be any coordinates that are not normal.&lt;/p&gt;&#xA;" OwnerUserId="3970" LastEditorUserId="3970" LastEditorDisplayName="" LastEditDate="2011-07-13T21:01:16.723" LastActivityDate="2011-07-13T21:01:16.723" CommentCount="1" />
  <row Id="12227" PostTypeId="1" AcceptedAnswerId="12228" CreationDate="2011-07-13T04:48:39.347" Score="5" ViewCount="155" Body="&lt;p&gt;How are field theory Langrangians treated when some terms have 2 derivatives but others have only 1? Because the number of derivatives in a Lagrangian term is more easily even than odd, the discussions for newcomers to physics of breaking up Lagrangians into free theories and perturbations often does not give clear instruction on how to conceptualize and handle terms with a single derivative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should a mathematician speaking to physicists refer to the role played by terms with only one derivative (e.g. Chern Simons like terms) in the presence of a Yang Mills kinetic term? Would it be a kinetic term of lower order? A velocity dependent potential term? Would the interpretation change if the second order Yang - Mills like term were discarded? Is an m D m -like term with 1-derivative of a field m treated as part of the perturbation in perturbation theory or as generating a lower order summand of the 2nd order 'free' operator to be inverted?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please feel free to rephrase the question if you understand what is being asked and the confusion is complicating the inquiry. Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="4477" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-07-13T22:36:23.943" LastActivityDate="2011-07-13T22:36:23.943" Title="Lagrangians combining terms with 1 and 2 derivatives" Tags="&lt;quantum-field-theory&gt;&lt;string-theory&gt;&lt;mathematical-physics&gt;&lt;gauge-theory&gt;&lt;path-integral&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12228" PostTypeId="2" ParentId="12227" CreationDate="2011-07-13T05:55:20.803" Score="1" ViewCount="" Body="&lt;p&gt;The OP is asking several questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The kinetic terms are, roughly speaking, the terms with temporal derivatives, e.g.,&#xA;$\frac{m}{2} \dot{q}^2$; or  $p_i\dot{q}^i$; or a symplectic potential term $\vartheta_I(z)\dot{z}^I$, where $z^{I}$ is a phase space variable. (I'm using notation from point mechanics for the recognition value, however it works in field theory as well.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Physicists often prefer a Hamiltonian formulation (also known as a first order formulation) withtime-derivatives (=velocities) appearing at most linearly, cf. the symplectic potential term $\vartheta_I(z)\dot{z}^I$. It is in principle possible to achieve a Hamiltonian form of the Lagrangian density ${\cal L}$ by Legendre transformation in the pertinent sectors at the expense of sometimes introducing first and second constraints via a Dirac-Bergmann analysis. Similarly to the Darboux theorem in finite dimensions, it is often possible to locally achieve a symplectic potential term of the standard form $p_i\dot{q}^i$ by a change of variables. See also the &lt;a href=&quot;http://arxiv.org/abs/hep-th/9306075&quot; rel=&quot;nofollow&quot;&gt;Faddeev-Jackiw method&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Concerning theories with both quadratic and linear dependencies of the same velocity, it is a good idea to first look at the simple example of a charged particle in an electro-magnetic field. Yang-Mills action plus a Chern-Simons action is, a bit oversimplified, just a field-theoretic version of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) A fourth question is related to how to divide a Lagrangian density $\cal L={\cal L}_0+{\cal V}$ into a free part $\cal L_0$ and an interaction part ${\cal V}$. As a rule of thumb, we put as much as we can solve exactly in the free part, which usually restrict us to quadratic (and linear) terms only. In particular quadratic terms that are linear in velocities belong to the free part. Finally, the Hessian of the free part has to be invertible so that we can find the propagator. Zero-modes may cause extra subtleties. We set up the perturbative expansion around a stationary value of the free part. &lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-13T09:42:14.223" LastActivityDate="2011-07-13T09:42:14.223" />
  <row Id="12229" PostTypeId="2" ParentId="12211" CreationDate="2011-07-13T09:04:00.950" Score="0" ViewCount="" Body="&lt;p&gt;One does not need &quot;expertise&quot;, just to know is enough :=)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physical Chemistry (PC) has (as many subjects) two meanings: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The research which stakeholders of PC do. &#xA;This cannot be fixed, the variations are strong.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The contents of textbooks and lectures on PC, &#xA;this is standardized to a certain extent. &#xA;This PC covers the physics which one needs in &#xA;a full chemistry study, but is not teached in &#xA;standard physics lectures. (or is thaught there &#xA;not detailled enough) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The canon is: kinetic gas theory, Thermodynamics (incl. statistical)&#xA;Electrochemistry, Quantum physics and spectroscopy, chemical Kinetics,&#xA;boundary layers, theory of bonds, liquids, solids.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparison to physics: eg. classical and &quot;Mischphasenthermodynamik&quot; &#xA;(couldn't find a English word) are dealt much more thoroughly. &#xA;Quantum physics is only rudmentary, just to allow a more qualitative &#xA;understanding on spectroscopy and theory of bonds. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-07-13T09:04:00.950" />
  <row Id="12230" PostTypeId="2" ParentId="12208" CreationDate="2011-07-13T09:42:15.990" Score="-1" ViewCount="" Body="&lt;p&gt;So many words, but not realizing what Waldirs &#xA;question/problem really is. &#xA;In the linked answer dmckee speaks about superposition and &#xA;says that the wave do not &quot;affect&quot; each other. &#xA;Waldir seems to see some &quot;affection&quot; in the interference working &#xA;in a antireflection coating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question of Waldir is : What is &quot;affect&quot; in this context?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When dmckee speaks of superposition etc, this means: there is &#xA;only linear interaction between waves. Under this conditions two &quot;waves&quot; &#xA;cross each other and do not affect each other: &#xA;Interference, as working in antireflection coatings is not such an &quot;affect&quot;.&#xA;Nonlinear effects are eg photon absorption in a photocathode, &#xA;or heterodyne mixing in a KHDP crystal. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-07-13T09:42:15.990" />
  <row Id="12231" PostTypeId="2" ParentId="12227" CreationDate="2011-07-13T10:02:35.473" Score="0" ViewCount="" Body="&lt;p&gt;It is practically unknown that the perturbation term is also of kinetic nature; that is why it gives increments (shifts, &quot;corrections&quot;) to the fundamental constants. Renormalizations are discarding those increments, and the remaining interaction is the true interaction. It is done perturbatively so it is not &quot;visible&quot;. I have this explained in my detailed &lt;strong&gt;&lt;a href=&quot;http://arxiv.org/abs/0811.4416&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-07-13T13:01:12.707" LastActivityDate="2011-07-13T13:01:12.707" />
  <row Id="12232" PostTypeId="2" ParentId="3151" CreationDate="2011-07-13T11:01:26.447" Score="0" ViewCount="" Body="&lt;p&gt;Since this came up to first page by an edit, I will offer my two cents of the euro.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Solar energy has the problem of most renewable energy resources: lack of storage capacity for the energy. Though one could devise a water storage and recovery for off time use, this is not different than having a grid based on stable sources, oil, coal, gas, nuclear, hydroelectric,  and conserving those sources when sun or wind or tides are available. Lack of good long term electric batteries have &quot;solar cars&quot; possible only as demonstrations whereas in countries on the sun belt they could be  a good alternative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)It is very expensive to install, and is only for a decade or so. I calculated that to power a house in Greece, where sunshine is plenty, at the level of kilowatt/hrs  that a normal household uses would cost about 40.000euro for the panels at the prices available here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both 1 and 2 can be overcome if good cheap batteries are invented, which is a matter of materials science and not theory. If a viable storage exists there will be a demand and the prices will drop.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-07-13T11:01:26.447" />
  <row Id="12233" PostTypeId="2" ParentId="12221" CreationDate="2011-07-13T12:27:31.990" Score="5" ViewCount="" Body="&lt;p&gt;Remarks: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In the following explanation  4-dimensional space-times $M$ equipped with a metric of signature (3,1) are considered.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are several Wikipedia pages treating frames (sometimes called tetrads or Vielbeins) in GR. See for example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cartan_formalism_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrad_formalism&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is a very good introductory chapter on the subject in chapter 5 of these &lt;a href=&quot;http://kolhoz.tiera.ru/P_Physics/PGr_Gravitation/Aldrovandi%20R.,%20Pereira%20J.%20Introduction%20to%20General%20Relativity%20%28web%20draft,%202004%29%28185s%29_PGr_.pdf&quot; rel=&quot;nofollow&quot;&gt;notes&lt;/a&gt; by: R. Aldrovandi and J. G. Pereira.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A frame in GR means a set of four vector fields $\mathbf{e}_a: M \rightarrow TM$, $a=0, 1,2,3$ satisfying the constraint equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{g} = \eta^{ab} \mathbf{e}_a \mathbf{e}_b$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathbf{g}$ is the inverse metric tensor and $\eta^{ab}$ is the flat Lorentzian metric. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These vector fields can be thought of as the mapping of the coordinate vectors of some given Mikowski space through the local coordinate system to the tangent space. In physical terms, we associate each such a frame with a local observer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, basically, we can work with the components of the frame vector fields instead of the metric, but one observes that the frame fields have 16 components, while the metric has (due to its symmetry) only 10 components.&#xA;This redundancy is due to the fact that the frame fields are not unique and a new set of frame fields  $\mathbf{e}^{\prime}_a$ satisfying&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{e}^{\prime}_a = M_a^b(x) \mathbf{e}_b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;satisfies the same constraint, where $M_a^b(x)$ is a Lorentz transformation matrix (i.e. satisfying $M_a^b(x) M_c^d(x) \eta_{bd} = \eta_{ab}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please observe that we can choose a non-constant Lorentz transformation depending on the location on the manifold, for this reason, these transformations are called local Lorentz transformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the dimension count checks: 16 frame components = 10 Metric components + 6 Lorentz transformations at every point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formalism may seem merely a change of variables, but this is not the whole story.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the local Lorentz transformations can be viewed as sections of a principal $SO(3,1)$ bundle over $M$ (This bundle is called $SO(M, \mathbf{g})$. &#xA;Thus this formulation is a formulation of GR as a gauge theory.Now, since we can allow the local Lorentz transformations to depend on the coordinates, this formalism allows to define accelerating frames, simply by taking thelocal Lorentz transformations to depend on time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, in the standard formulation of GR allows to can define classical fields as sections of bundles whose local transformations are functions of the coordinate transformation (diffeomorphisms) of the base manifold.&#xA;These bundles are called natural bundles, for example the coordinate transformation of the tangent bundle is the Jacobian matrix of the coordinate transformations of the base manifold. (Similarly, the inverse Jacobian matrix for the cotangent bundle).&#xA;Thus the standard formulation of GR allows the definition of vector fields, tensor fields etc. but not spinor fields, which are very important in physics.&#xA;Spinor bundles are not natural, but there is no natural way to define a general coordinate transformation of a spinor field given a diffeomorphism of the base manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the base manifold $M$ has a spin structure, then the frame formalism allows to define spinor fields as follows: Since $M$ is spin, $SO(M, \mathbf{g})$ can be lifted to a spin bundle $Spin(M, \mathbf{g})$ , then a spinor bundle is the associated bundle corresponding to a fundamental spinor representation, and spinor fields are sections of the spinor bundle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This  construction can be performed in local coordinates follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we can form the dual frame  $\mathbf{e}^a: M \rightarrow T^{*}M$ by requiring:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\langle \mathbf{e}^a, \mathbf{e}_b \rangle = \delta^a_b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dual frame can be used to define the frame components of any vector field $\mathbf{V}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V^a = \langle \mathbf{e}^a, \mathbf{V} \rangle$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversly, one can form the &quot;curved&quot; components of a vectors using the original frame. For example, consider the Dirac matrices $\{\gamma^a\}$ generating the Clifford algebra $Cl(3,1)$.&#xA;Then their curved components are given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\gamma^{\mu} = \gamma^a e_a^{\mu}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, one uses the metric $\mathbf{g}$ to lower &quot;curved indices&quot;, the inverse metric to raise &quot;curved indices&quot;.&#xA;and similarly, the Lorentz metric $\mathbf{\eta}$ for the flat indices. One uses the frame vectors and their dual to replace curved indices with flat indices and vice-versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next ,the &lt;a href=&quot;http://en.wikipedia.org/wiki/Spin_connection&quot; rel=&quot;nofollow&quot;&gt;spin connection&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega_{\mu}^{ab} = e^a_{\nu}(\partial_{\mu} e^{\nu b}+ e^{\sigma b}\Gamma^{\nu}_{\sigma \nu})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, $\Gamma^{\nu}_{\sigma \nu}$ is the Levi-Civita connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not difficult to verify (by looking at the local Lorentz transformations) that $\omega_{\mu}^{ab}\sigma_{ab}$ is a connection on $Spin(M, \mathbf{g})$, where $\sigma_{ab}$  are the generators of the fundamental spinor representation.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Using the above data, the fully covariant Dirac equation on $M$ takes the form:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$-i \gamma^{\mu} D_{\mu} \psi + m \psi = 0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $D_{\mu}$ is the covariant derivative associated with the spin connection&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ D_{\mu} = \partial_{\mu}-i\omega_{\mu}^{ab}\sigma_{ab}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the fully covariant Dirac equation looks just like the Dirac equation coupled to a gauge field given by the spin connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classical fields where this construction is possible are sections of bundles called &quot;gauge natural bundles&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to mention that the solution of the fully covariant Dirac equation depends on the frame fields, but observable quantities such as the number of bound states for example, depend only on the metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since local observers are identified with points on the fibers of the frame bundle, then all frames are inertial because they can be obtained from the action of a Lorentz transformation on a single frame (i.e. point on the fiber).&#xA;The parameters of the lorentz transformation are the velocity vector and the orientation of the frame. It is explicit in the equations that we can allow variable Lorentz transformations, i.e., Lorentz transformations dependent&#xA;on the local coordinates of the base manifold, in particular on the time coordinate. &#xA;Now I'll divide my answer into two parts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particles: Suppose the four velocity vector of a particle moving on a geodesic is given by $V^{\mu} = \frac{dx^{\mu}}{d\tau}$, ($\tau$ is any parameter along the path) then the frame coordinates of this vector are: $V^a = e^a_{\mu} V^{\mu}$  and the components of the velociy measured by an observer moving with a velocity defined by the Lorenntz matrix $M(x)$ are&#xA;$V^{\prime b} = M^b_a(x) V^a$. Again, a variable $M$ indicates an accelerating frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fields: The equations of motion will be covariant with respect to these transformations, because for sections of natural bundles, the frame vectors&#xA;do not appear in the equations of motion, while in the case of gauge natural sections such as spinors these (variable Lorentz) transformations will appear as gauge transformations and the equations of motion are constructed to be gauge invariant. &#xA;Thus, the equations of motion are not affected by local Lorentz transformations, or in other words, Physics looks the asme to all observers even if they are accelerating.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorUserId="2190" LastEditorDisplayName="" LastEditDate="2011-07-14T14:35:19.510" LastActivityDate="2011-07-14T14:35:19.510" CommentCount="7" />
  <row Id="12234" PostTypeId="2" ParentId="12215" CreationDate="2011-07-13T12:53:44.697" Score="0" ViewCount="" Body="&lt;p&gt;Yes, you can use compressed gas for acceleration. But you will need a light gas ($H_2$ preferably) as the maximum speed a gas can reach leaving such a high pressure device is limited by it's weight and temperature ($u_{max}=\sqrt{2c_p T_o}$). Which is very dangerous and will probably burn up when released anyway. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why not burn it in the first place? Apart from the boost you get from the pressure, you get another boost from the expanding gas. As it turns out, you gain more power from burning it, thus the compression is mostly used to limit the size of the canisters instead of actual propulsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sort of thing is done in older Rockets (V2). Modern day rockets still use similar techniques though.&lt;/p&gt;&#xA;" OwnerUserId="4337" LastEditorUserId="4337" LastEditorDisplayName="" LastEditDate="2011-08-25T17:11:17.520" LastActivityDate="2011-08-25T17:11:17.520" />
  <row Id="12235" PostTypeId="1" CreationDate="2011-07-13T13:39:33.830" Score="6" ViewCount="246" Body="&lt;p&gt;Given a 2-loop divergent integral $\int F(q,p)\,\mathrm{d}p\mathrm{d}q$, can it be solved iteratively? I mean&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I integrate over $p$ keeping $q$ constant&lt;/li&gt;&#xA;&lt;li&gt;Then I integrate over $q$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In both iterated integrals I use dimensional regularization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can it be solved iteratively? I presented a paper to a teacher of mine about regularization of integrals using the Zeta regularization. He told me that for one dimensional integrals (or one loop integrals) it was fine, but that my method could not handle multi loop integrals. I argue back that you could apply the regularization method by introducing a regulator of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(a+qi)^{-s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could make the integral on each variable by iterated regularization, that is applying the algorithm iteratively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: i think they have cheated me :) making up excuses not to put me atention&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by the way , how can i insert math codes on my posts ?? is just setting &lt;code&gt;$&lt;/code&gt; at the beginning and the end of the equation ??&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Edit: 19th of July&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;thanks :) your answer was quite useful :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;however can I not  insert a term $(qi+a)^{-s} $ on each variable and then apply the regularization iteratively ??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I say so because I made a paper &lt;a href=&quot;http://vixra.org/abs/1009.0047&quot; rel=&quot;nofollow&quot;&gt;http://vixra.org/abs/1009.0047&lt;/a&gt; to regularize integrals using the regulator $(q+1)^{-1}$ and tried to extend it to several variables, however that was my doubt, if i could apply my regularization scheme to every variable :) thanks again&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean for a one dimensional integral $\int dx (x+a)^{m-s}$, I know how to regularize it by using the Euler-Maclaurin summation formula plus the Riemann Zeta function $ \zeta (s-m)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then for multiloop integrals, I had thought that i could introduce the $s$-regulators (x+a)^{-s}(y+a)^{-s}.. and so on on each variable, and then apply iterated integration... :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so this is a resume of my method..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) i know how to use Zeta regularization to get finite values for the integral $ \int dx(x+a)^{m-s} $ in terms of the Riemann Zeta function $ \zeta (s-k) $ with k=-1,0,1,....,m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) for a more general 1-dimensional integral $ \int dx f(x)(x+a)^{-s}$ i add substract a Polynomial $K(x+a)(x+a)^{-s} $ to get a finite part and then regularize the divergent integrals  $dx(x+a)^{m-s} $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) for a more complicate 2-loop integral $ \iint dpdq F(q,p)(p+a)^{-s}(q+a)^{-s} $ to obtain a regularization of it i do the sema method , first on 'p' considering 'q' a constant and i treat it as a one dimensional integral over 'p' and then over 'q' by substracting a Polynoamials $ K(q,p+a)(p+a)^{-s} $ and so on&lt;/p&gt;&#xA;" OwnerUserId="4435" LastEditorUserId="429" LastEditorDisplayName="" LastEditDate="2011-07-19T13:49:45.273" LastActivityDate="2011-07-20T17:17:13.613" Title="Iterated dimensional regularization" Tags="&lt;quantum-field-theory&gt;&lt;mathematical-physics&gt;&lt;regularization&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="2" />
  <row Id="12236" PostTypeId="2" ParentId="12221" CreationDate="2011-07-13T13:41:34.933" Score="0" ViewCount="" Body="&lt;p&gt;It is likely mean a consideration of a map to reference frame of accelerated observer discussed, e.g., in ch. 13.6 of Misner, Thorne, Wheeler, “Gravitation”. Maybe it is simple to start with consideration of coordinate system corresponding to motion with constant acceleration in ch. 26 of Pauli “Theory of relativity” about hyperbolic motion. &lt;/p&gt;&#xA;" OwnerUserId="3351" LastActivityDate="2011-07-13T13:41:34.933" />
  <row Id="12237" PostTypeId="1" AcceptedAnswerId="12244" CreationDate="2011-07-13T14:37:46.460" Score="8" ViewCount="173" Body="&lt;p&gt;Recall that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermion_doubling&quot; rel=&quot;nofollow&quot;&gt;fermion doubling&lt;/a&gt; is the problem in taking the $a \to 0$ limit of a naively discretized fermionic theory (defined on a lattice with lattice spacing $a$). After such a limit one finds themselves with an additional amount (precisely $2^d$) of fermionic fields. One can fix this by considering different discretizations of the action that make unwanted fields decouple in the continuum limit. The downside is that the additional terms have to spoil some nice features of the theory (chiral symmetry, locality, lattice symmetry, ...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I wonder what is the true reason for the appearance of new fields. Is it the fermionic nature of the theory? (In other words, is a similar problem ruled out for bosonic fields?) And do all (naive?) fermionic theories (that is, independent of the continuum form of the action) suffer from this problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, how can one tell &lt;em&gt;a priori&lt;/em&gt; what will the field content of a lattice theory in the continuum limit be? Or is the field content fundamentally a continuum limit property that has to be calculated?&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-15T04:49:22.643" Title="What is the fundamental reason of the fermion doubling?" Tags="&lt;quantum-field-theory&gt;&lt;fermions&gt;&lt;lattice-model&gt;" AnswerCount="2" />
  <row Id="12238" PostTypeId="2" ParentId="11978" CreationDate="2011-07-13T14:43:36.337" Score="2" ViewCount="" Body="&lt;p&gt;Any time you know there will be something in your optical system, it is generally good practice to include it in your model during the design process. Of course, in many applications these things will not make a difference. The image quality of your point-and-shoot camera will not become unacceptable if somebody chooses to take a picture through their bedroom window, for example. In the cases you cited, however, the extra aberrations produced by the cover glass or laser window can indeed be significant. I'll explain why, starting with some background.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Aberration Theory&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Wavefront aberrations are usually expressed as a polynomial function &lt;sup&gt;1&lt;/sup&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ W = \sum_{i,j,k} W_{i j k} h^i \rho^j \phi^k $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $h$ is the object height (or angle, in the case of an object at infinity), and $(\rho, \phi)$ are polar coordinates in the pupil, normalized such that $h=1$ is the maximum object height and $\rho = 1$ is the edge of the pupil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, $W_{040}$ is the coefficient of spherical aberration. It is often called &quot;third-order&quot; spherical, because lens designers are concerned with the ray angle errors, which are given by the first derivative of the wavefront error; this also distinguishes it from higher order aberrations, like 5th order spherical, $W_{060}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of ways that aberration coefficients can be computed. Continuing with spherical aberration as our example, the contribution from a single surface can be expressed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ W_{040}= -\frac{1}{8} A^2 Y_a \Delta\left(\frac{u_a}{n}\right) \rho^4 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A \equiv n i_a$, $n$ is the refractive index, $i_a$ is the angle of incidence of the marginal ray (the &quot;a-ray&quot;) on the surface, $Y_a$ is the marginal ray height on the surface, $u_a$ is the (paraxial) marginal ray angle, and $\Delta(x) \equiv x&amp;#39; - x$, where primed variables represent quantities after refraction at the surface, and un-primed variables are quantities immediately before refraction.&lt;sup&gt;2&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of techniques for computing aberration coefficients for complex optical systems, such as &quot;Seidel Sums&quot; or the more modern and efficient technique of &quot;G-sums,&quot;&lt;sup&gt;3&lt;/sup&gt; but the important point here with respect to your question is this: &lt;strong&gt;the aberrations due to a surface depend just as much on the characteristics of the rays at the surface as they do on the shape of the surface.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, you can see that the aberrations introduced by a flat plate can be very significant indeed. In the formalism I have presented here, a high NA microscope system would have very large values for $u_a$, and the design must absolutely include these in balancing aberrations if very high performance is to be achieved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should also help make clear why liquid immersion is helpful in getting to very high NA without aberrations becoming a problem. &lt;strong&gt;However&lt;/strong&gt;, in your question you also ask about the case of a water droplet over the object, without being in contact with the microscope objective as in a true liquid immersion objective. In this case you must keep in mind that the curved surface of the droplet would be a refractive surface and &lt;strong&gt;must&lt;/strong&gt; be included in the optical calculations if the resulting lens is to function at all!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One last note, on the brief discussion in comments about the reason for cover glass. Cover glass serves a variety of purposes, but isn't really intended as a beneficial component of the microscope imaging system. It is primarily used to hold the sample still; protect it from air; and most importantly, to hold it &lt;em&gt;flat&lt;/em&gt;, because a high NA microscope system will have a &lt;em&gt;very&lt;/em&gt; shallow depth of field. It would be quite inconvenient for the microscope user to have micron-scale surface variations in his sample produce focus variations between adjacent features in the image. Cover glass can be accommodated by the optical design, but if it weren't for these practical considerations the lens designer would likely not call for it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Footnotes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;: These are Seidel aberrations; more modern approaches use an orthogonal set of polynomials as their basis functions, such as Zernike polynomials, which are more appropriate for numerical techniques. Seidel aberrations are left over from the time when aberrations had to be calculated by hand, and are still useful and well understood by modern lens designers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2&lt;/sup&gt;: The &lt;em&gt;marginal ray&lt;/em&gt; is the ray from an object point on the optical axis, which goes through the edge of the pupil/aperture stop. Also important is the &lt;em&gt;chief ray&lt;/em&gt;, which is a ray from the edge of the field of view through the center of the pupil/aperture stop. These rays are used heavily in the geometrical analysis of optical systems. For more on this, see &quot;Modern Optical Engineering&quot; by W.J. Smith.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;3&lt;/sup&gt;: Again, these are covered by Smith.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-07-13T14:43:36.337" />
  <row Id="12239" PostTypeId="2" ParentId="6530" CreationDate="2011-07-13T15:02:11.030" Score="-1" ViewCount="" Body="&lt;p&gt;but WHY rigour in physics ???, .. if things go well using formal manipulation and we can make the experiments agree why to spoil  a BEAUTIFUL theory with a RIGOROUS maths, that maybe will the theory 'harder' to understand ??&lt;/p&gt;&#xA;" OwnerUserId="4435" LastActivityDate="2011-07-13T15:02:11.030" CommentCount="1" />
  <row Id="12243" PostTypeId="1" CreationDate="2011-07-13T16:07:14.807" Score="6" ViewCount="111" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Vaneless_ion_wind_generator&quot; rel=&quot;nofollow&quot;&gt;Described here&lt;/a&gt; and in the references, the idea is to use a large fence as a grid electrode.  Supply a high voltage to charge water droplets that pass through it (or ionize air molecules?), and then the wind blows them away, converting wind pressure into electric potential?, then they hit another electrode or the Earth to complete the circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/sCvNH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would such a device actually work?  If so, why isn't it built or used?  Can the energy output be estimated and compared to a traditional wind turbine?&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2011-09-05T03:18:07.510" Title="Would the ionic wind power fence actually work?" Tags="&lt;power&gt;&lt;renewable-energy&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="12244" PostTypeId="2" ParentId="12237" CreationDate="2011-07-13T16:09:50.683" Score="7" ViewCount="" Body="&lt;p&gt;The fermion doubling is manifested through the existence of extra poles in the Dirac propagator on the lattice. These poles cannot be made to disappear at the continuum limit. (The number of doublers can be reduced by different discretizations but not eliminated at all, this is essentially the Nielsen-Ninomiya theorem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for the fermion doubling lies in the existence of chiral anomaly. This anomaly exists in the continuum limit due to the chiral non-invariance of the path integral measure and not because of the non-invariance of Lagrangian. In the lattice formulation of a chiral theory based on a discretization of the Lagrangian, the anomaly is absent and the lattice formulation generates the extra species just to cancel this anomaly in the continuum limit. Since the axial anomaly exists in nature $\pi^0 \rightarrow \gamma \gamma$, this situation is unacceptable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fermion doubling problem is an artifact of the realization of the theory by means of quarks where the axial anomaly is not present in the lagrangian but in the path integral measure. There are approaches of other types of discretizations such as by means of fuzzy spaces in noncommutative geometry, where the quarks are not the basic fields of the theory. In these approaches, the fermion doubling problems do not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an update referring to Marek's first comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For fermions, the axial anomaly can be recovored only by a non-trivial regularization of the path integral measure of the fermions. &#xA;Any finite dimensional approximation of this measure as a product of Berezin-Grassmann integrals does not produce the axial anomaly.&#xA;This is the reason why the Lattice regularization does not produce the axial anomaly and as a consequence the phenomenon of doubling occurs where the&#xA;different species of doublers are of opposite chirality to cancel the anomaly. This is a property of fermion fields represented by Grassmann variables.&#xA;In effective field theories (where the basic fields are pions) such as sigma models, the axial anomaly is manifested through a Wess-Zumino-Witten term in the Lagrangian, but these theories are not even &#xA;perturbatively renormalizable and I think this is the reason why they were not put on a lattice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One approach that I know of which solves the fermion doubling problem is the regularization by means of a fuzzy space&#xA;approximation of the space-time manifold. The philosophy of this approach is explained in the introduction of Mark's Rieffel's &lt;a href=&quot;http://arxiv.org/PS_cache/math/pdf/0108/0108005v2.pdf&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;. The resolution of the fermion doubling using this approach is given nicely in Badis Ydri's &lt;a href=&quot;http://arxiv.org/PS_cache/hep-th/pdf/0110/0110006v1.pdf&quot; rel=&quot;nofollow&quot;&gt;thesis&lt;/a&gt;. (There is also more recent work on the subject by A. Balachandran and B. Ydri in the arxiv).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main idea is that the Poisson algebra of functions over certain spaces (such as the two sphere, and more generally coadjoint orbits of compacl Lie groups) can be approximated by finite dimensional matrices. These approximations are called fuzzy spaces. Gauge fields and Fernmions can be constructed on these fuzzy spaces, which have the correct continuum limit when the matrix dimensions become infinite. This formulation contains the axial anomaly inherently, thus it is free of the doubling problem.&#xA;The only drawback that I can see of this approach is that it is applicable only to some special 4-dimensional manifolds such as $CP^2$ or $S^2 \times S^2$, because the fuzzy manifold is required to be Berezin-quantizable.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorUserId="2190" LastEditorDisplayName="" LastEditDate="2011-07-15T04:49:22.643" LastActivityDate="2011-07-15T04:49:22.643" CommentCount="7" />
  <row Id="12245" PostTypeId="2" ParentId="11680" CreationDate="2011-07-13T17:30:23.740" Score="-1" ViewCount="" Body="&lt;p&gt;Plants such as flowers because of photosynthesis which is efficient.&lt;/p&gt;&#xA;" OwnerUserId="4485" LastActivityDate="2011-07-13T17:30:23.740" CommentCount="1" />
  <row Id="12246" PostTypeId="2" ParentId="12183" CreationDate="2011-07-13T17:45:56.203" Score="2" ViewCount="" Body="&lt;p&gt;Yet another point (if someone do not know yet:)&#xA;S. Chandrasekhar in &quot;The Mathematical Theory of Black Holes&quot;&#xA;wrote, comparing the charged BH (i.e. Reissner-Nordstroem solution) and usual one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And among the differences, the principal one -- from which all others&#xA;  derive -- is the fact that the Reissner-Nordstroem space-time has an&#xA;  inner Cauchy horizon which makes its maximal analytic extension reveal&#xA;  the possibility, in principle, of exploring other worlds and&#xA;  emancipating oneself from one's past.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3351" LastActivityDate="2011-07-13T17:45:56.203" CommentCount="2" />
  <row Id="12247" PostTypeId="2" ParentId="12200" CreationDate="2011-07-13T17:48:47.373" Score="8" ViewCount="" Body="&lt;p&gt;Power is transmitted wirelessly in many applications, just not where transmission efficiency is high importance. And that is the reason it is not done more: efficiency.  At low frequencies, galvanic (that means metal) conduction of electricity is many, many, many orders of magnitude more efficient than, say, air.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At higher frequencies, one can use electricity to create E-M waves which travel well in air (or even space); but the problem is that the medium in which they travel is typically 3D (rather than the functionaly 1D path in an electric circuit), so the stuff spreads out a lot and the receiving station only gets a very small fraction of it (your car or home radio, receiving a signal from a braodcast station, is an example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Techniques have been and are continually being developed to force power to travel in a 1-D manner, instead of 3D, in air without wave guides.  Some successes include phased array RADARs, lasers, microwave antennas for telecommunications, Yagi antennas, and parabolic dish antennas.  However, no commercially viable technique has surfaced to move large amounts of electrical power over long distances without long stretches of metal.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-07-13T17:48:47.373" CommentCount="3" />
  <row Id="12249" PostTypeId="1" AcceptedAnswerId="12251" CreationDate="2011-07-13T19:41:57.373" Score="0" ViewCount="60" Body="&lt;p&gt;I know this is probably some basic math once you have the figures, but my Google-fu is failing me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I have &lt;strong&gt;5 pounds of air&lt;/strong&gt;. Air meaning, the typical composition of the atmosphere of Earth. And by 5 pounds I mean the weight of the mass of the air, not it's pressure or anything like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know approximately how much &lt;strong&gt;volume&lt;/strong&gt; that amount of air would take up at a given pressure -- specifically, sea level pressure. I know that can vary -- lets just say its a typical warm summer day, or just an average. I'm not looking for exact numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, I'd rather a subjective interpretation of the exact figure if you can offer one. Something like 'about the volume of the air in a 12x12 room', or what have you. I have no idea how big 5 pounds of air is.&lt;/p&gt;&#xA;" OwnerUserId="3006" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-13T20:24:08.403" LastActivityDate="2011-07-14T18:55:32.497" Title="What is the volume of a parcel of air of a giving weight at average sea level pressure" Tags="&lt;homework&gt;&lt;pressure&gt;&lt;air&gt;&lt;volume&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="12250" PostTypeId="2" ParentId="12249" CreationDate="2011-07-13T20:07:59.567" Score="0" ViewCount="" Body="&lt;p&gt;One mole of air weights roughly 29 g and it's volume is about 25 L (22.4 L at 0°C IIRC).&lt;/p&gt;&#xA;" OwnerUserId="4202" LastActivityDate="2011-07-13T20:07:59.567" />
  <row Id="12251" PostTypeId="2" ParentId="12249" CreationDate="2011-07-13T20:51:23.433" Score="2" ViewCount="" Body="&lt;p&gt;The easy way to answer this is to just look up the density of air $\rho$ under whatever conditions you need it and then calculate $V=m/\rho$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't have a ready source for the density you need, though, you can use the ideal gas law,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$PV=nRT$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $n$ is the number of moles of the gas, which you can get by dividing the mass by the average molar mass of the gas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n=\frac{m}{\mu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average molar mass of air would be a weighted average of the molar masses of its constituent gases. This is something you look up for the composition of air under your specified conditions, but it'll be around 29 g/mol. This at least does not depend &lt;em&gt;directly&lt;/em&gt; on temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need more accuracy, you could use something like the Van der Waals equation.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-07-13T20:51:23.433" CommentCount="1" />
  <row Id="12252" PostTypeId="2" ParentId="12191" CreationDate="2011-07-13T20:56:16.837" Score="3" ViewCount="" Body="&lt;p&gt;One millibar ($\text{mbar}$) is aproximately the pressure you feel on your hand when you hold a penny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A penny has mass $ m = 0.0025\;\text{kg} $ and diameter $ D = 0.019\;\text{m} $ making the pressure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P = \frac{m\;g}{\pi D^2/4} = 86\;\text{Pa} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $ 100\;\text{Pa} = 1\;\text{mbar} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you stack 116 pennies you will get $100\;\text{mbar}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could try with other coins, or household items, as long as you know their weight, or mass, and the contact area. A marble, for example, is not a good idea because the contact area depends on the load applied.&lt;/p&gt;&#xA;" OwnerUserId="392" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2011-07-13T21:02:17.367" LastActivityDate="2011-07-13T21:02:17.367" CommentCount="1" />
  <row Id="12253" PostTypeId="2" ParentId="12199" CreationDate="2011-07-13T22:30:01.873" Score="4" ViewCount="" Body="&lt;p&gt;Crank-Nicholson method is effectively the average of forward (explicit) Euler&lt;/p&gt;&#xA;&#xA;&lt;p&gt;psi(x,t+dt)=psi(x,t) - i*H psi(x,t)*dt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and backward (implicit) Euler method&lt;/p&gt;&#xA;&#xA;&lt;p&gt;psi(x,t+dt)=psi(x,t) - i*H psi(x,t+dt)*dt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The backward component makes Crank-Nicholson method stable. The forward component makes it more accurate, but prone to oscillations. If you want to get rid of oscillations, use a smaller time step, or use backward (implicit) Euler method. That is all there is to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a rough idea of how to estimate the time it takes these oscillations to show themselves. These oscillations occur because for high-frequency harmonics in your solution, Courant condition is violated. How long it takes for these oscillations to show depends on how much power (in the sense of Fourier series) they contain in the initial condition. I.e., if your time step is dt, then harmonics with frequency above 1/dt (roughly) are prone to instability. Find the length scale L corresponding to frequency 1/dt, make a Fourier expansion of the initial condition psi(x,0) and calculate the fraction of power in the Fourier series in wavelengths shorter than that L. When these harmonics (initially weak) grow large enough to be comparable to the total signal, they will show. Assuming that they grow exponentially, this time scales as the logarithm of that fraction you found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Estimating that time seems too much trouble for this problem. Just use a smaller time step and/or backward Euler method.&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditorDisplayName="" LastEditDate="2011-07-13T23:43:27.037" LastActivityDate="2011-07-13T23:43:27.037" CommentCount="3" />
  <row Id="12254" PostTypeId="2" ParentId="12069" CreationDate="2011-07-13T23:40:13.920" Score="1" ViewCount="" Body="&lt;p&gt;Having co-authored a review of the epidemiologic research*, I believe long term (approx. 10 or more years) cell phone use increases tumor risk in humans, especially gliomas, acoustic neuromas, and tumors of the parotid gland.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree with mbq that the effect is likely due to electromagnetic radiation (EMR) that interferes with bioelectrical processes through non-thermal mechanisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cursory review (unpublished) of 116 published toxicology studies, including animal studies and human cellular studies, suggests that some cell phone carrier systems are more likely to be biologically reactive than others. Namely, systems that pulse the signal at extremely low frequencies (i.e., TDMA, GSM, and UMTS [e.g., AT&amp;amp;T and T-Mobile]) are about three times more likely to yield statistically significant biologically reactive outcomes as compared to carriers that do not pulse the signal (i.e., CDMA and W-CDMA [e.g., Verizon and Sprint]) (36% vs. 12%).  However, when EMR is tested at higher power outputs that exceed the current legal limits for cell phones (based upon a measure called the Specific Absorption Rate), then toxic effects are more likely observed regardless of carrier. However, at these higher power outputs we cannot rule out a thermal mechanism.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Seung-Kwon Myung, Woong Ju, Diana D. McDonnell, Yeon Ji Lee, Gene Kazinets, Chih-Tao Cheng, and Joel M. Moskowitz. Mobile Phone Use and Risk of Tumors: A Meta-Analysis. Journal of Clinical Oncology. 20(27):5565-5572. Published online first Oct 13, 2009. Nov 20, 2009.URL: &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/19826127&quot; rel=&quot;nofollow&quot;&gt;http://www.ncbi.nlm.nih.gov/pubmed/19826127&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Joel M. Moskowitz, Ph.D., UC Berkeley&lt;/p&gt;&#xA;" OwnerUserId="4487" LastActivityDate="2011-07-13T23:40:13.920" CommentCount="4" />
  <row Id="12255" PostTypeId="2" ParentId="12154" CreationDate="2011-07-14T01:16:19.837" Score="1" ViewCount="" Body="&lt;p&gt;I mole is just a way to represent a number of atoms. Instead of saying 12 eggs we can also say a dozen. Similarly one mole of carbon is the same as saying 6.02*10^23 atoms of carbon.&lt;/p&gt;&#xA;" OwnerUserId="4489" LastActivityDate="2011-07-14T01:16:19.837" />
  <row Id="12256" PostTypeId="2" ParentId="2164" CreationDate="2011-07-14T01:31:31.203" Score="6" ViewCount="" Body="&lt;p&gt;This &quot;air glider&quot; works &lt;a href=&quot;http://css.engineering.uiowa.edu/~me_160/Examples%20for%20class/3-160.pdf&quot; rel=&quot;nofollow&quot;&gt;as a hovercraft&lt;/a&gt;, using air pressure to lift itself and its load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.tools4flooring.com/roberts-10-480-dual-air-glider-p-298.html&quot; rel=&quot;nofollow&quot;&gt;a commercial model&lt;/a&gt; we get the following specifications:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;8-1/2 in. x 36 in. (22 cm x 91 cm) dual pads (0.4 m&lt;sup&gt;2&lt;/sup&gt; total area).&lt;/li&gt;&#xA;&lt;li&gt;750 pounds (~3500 N) of lifting capacity.&lt;/li&gt;&#xA;&lt;li&gt;1.75 HP blower.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The required (relative) pressure to lift 3500 N of weight using a 0.4 m&lt;sup&gt;2&lt;/sup&gt; platform is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_{\rm cushion} = \frac{3500\,\mathrm{N}}{0.4\,\mathrm{m^2}} \approx 9000\,\mathrm{Pa}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can take the air inside the &quot;cushion&quot; as stagnated to calculate the velocity of the exit flow. Then the dynamic pressure at the exit of the gap must be about 9000 Pa, giving an air velocity of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{2}\rho_{\rm air}\,v_{\rm gap}^2 = 9000\,\mathrm{Pa}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{2}\cdot 1.205\,\mathrm{kg\cdot m^{-3}}\,\cdot v_{\rm gap}^2 = 9000\,\mathrm{Pa}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_{\rm gap} \approx 135\,\mathrm{m\cdot s^{-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume the gap is about 0.2 mm (just a guess), the flow rate will be about&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q = 4\cdot(0.22\,\mathrm{m} + 0.91\,\mathrm{m})\cdot 0.0002\,\mathrm{m}\cdot 135\,\mathrm{m\cdot s^{-1}} \approx 0.12\,\mathrm{m^3\cdot s^{-1}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the required power&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P = Q\cdot p_{\rm cushion} \approx 0.12\,\mathrm{m^3\cdot s^{-1}}\cdot 9000\,\mathrm{Pa} \approx 1100\,\mathrm{W}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;below the specified 1.75 HP (1305 W).&lt;/p&gt;&#xA;" OwnerUserId="1656" LastActivityDate="2011-07-14T01:31:31.203" CommentCount="3" />
  <row Id="12257" PostTypeId="1" CreationDate="2011-07-14T02:11:09.937" Score="0" ViewCount="29" Body="&lt;p&gt;I am looking for an inertial navigation system for use in &lt;a href=&quot;http://www.usfirst.org/roboticsprograms/frc/default.aspx?id=966&quot; rel=&quot;nofollow&quot;&gt;mobile robotics&lt;/a&gt;. Most of what I know about inertial systems is gleaned from hours on google, so feel free to correct any misunderstandings I have. I am looking for an inertial navigation system, not an inertial measurement unit because it seems that transforming an IMU into a functional INS requires a large investment of time.&#xA;I would like the INS to have the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cost &amp;lt; $800&lt;/li&gt;&#xA;&lt;li&gt;Size &amp;lt; 10&quot;x10&quot;x10&quot;&lt;/li&gt;&#xA;&lt;li&gt;Weight &amp;lt; 8lbs&lt;/li&gt;&#xA;&lt;li&gt;Maximum Rotational Speed &gt; 200deg/sec&lt;/li&gt;&#xA;&lt;li&gt;Maximum Acceleration &gt; 1.8g&lt;/li&gt;&#xA;&lt;li&gt;Drift* &amp;lt; 3&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of a product or kit which meets these specifications?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*The unit will be operating for 15 seconds. If features such as 3DoF lock (the motion of a mobile robot is planar, so you can safely ignore roll, pitch and Y acceleration) or velocity reset at rest are available feel free to count their effects into the drift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in Advance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Alex &lt;/p&gt;&#xA;" OwnerUserId="4486" LastActivityDate="2011-07-14T02:11:09.937" ClosedDate="2011-07-14T12:01:58.807" Title="Choosing an Inertial Navigation System" Tags="&lt;motion&gt;&lt;inertial-frames&gt;&lt;sensor&gt;&lt;relative-motion&gt;" />
  <row Id="12258" PostTypeId="1" AcceptedAnswerId="12260" CreationDate="2011-07-14T02:44:21.140" Score="4" ViewCount="153" Body="&lt;p&gt;Why the log? Is it there to make the growth of the function slower?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this is a common experimental observable, it doesn't seem reasonable to take the range from $[0,\infty)$ to $(-\infty,\infty)$ (For a particle emitted along the beam axis after collision $\theta = 0$ wouldn't be better to have a number that says how close it is to zero rather than one that says how large a number it is. I hope that makes the question clear.)&lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="2599" LastEditorDisplayName="" LastEditDate="2011-07-14T02:52:32.850" LastActivityDate="2011-07-15T04:39:39.857" Title="Why is pseudorapidity defined as $-\log \tan \theta/2$" Tags="&lt;special-relativity&gt;&lt;high-energy-physics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" />
  <row Id="12259" PostTypeId="2" ParentId="12258" CreationDate="2011-07-14T03:42:54.460" Score="2" ViewCount="" Body="&lt;p&gt;Oh god.. I remember stressing over this one when I started in particle physics.  :)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the use of the log:  This is just a hyperbolic trig thing.  First write down the pseudorapidity using momenta perpendicular and parallel to the beam line like they do in the second expression here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pseudorapidity&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Pseudorapidity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then use the inverse hyperbolic identities you find here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperbolic_cosine&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hyperbolic_cosine&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you can get back to the angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding it's use as an experimental observable:  There are two reasons high energy guys prefer to use it (from wikipedia):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Pseudorapidity depends only on the polar angle of its trajectory, and not on the energy of the particle.  (nice sometimes when you're comparing data from one detector to the next)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In hadron collider physics, the rapidity (or pseudorapidity) is preferred over the polar angle θ because, loosely speaking, particle production is constant as a function of rapidity. One speaks of the &quot;forward&quot; direction in a hadron collider experiment, which refers to regions of the detector that are close to the beam axis, at high | η | .&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2011-07-14T03:42:54.460" />
  <row Id="12260" PostTypeId="2" ParentId="12258" CreationDate="2011-07-14T05:50:14.257" Score="8" ViewCount="" Body="&lt;p&gt;The advantage of this particular definition is that differences in pseudorapidity are invariant under boosts along the $z$ axis. Specifically, consider a Lorentz transformation corresponding to a boost by velocity $\beta\hat{z}$. Since the tangent is basically a transverse distance over a longitudinal distance, it transforms under a Lorentz boost along the longitudinal axis with a factor $\gamma$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan\phi \sim\frac{\Delta x_T}{\Delta x_L} \qquad\to\qquad \tan\phi&amp;#39; = \frac{\Delta x_T}{\Delta x_L/\gamma} = \gamma\tan\phi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if you put that in the pseudorapidity formula, you find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta = -\log\biggl(\tan\frac{\theta}{2}\biggr) \qquad\to\qquad \eta&amp;#39; = -\log\biggl(\tan\frac{\theta}{2}\biggr) - \log\gamma$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term is a function of the particle's trajectory, but the second term is a function of the boost parameter only - it doesn't depend on the particle at all. So if you have two particles coming out of a collision with pseudorapidities $\eta_1$ and $\eta_2$, the $\log\gamma$ term is the same for both, and thus when you take the difference it cancels out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta_1 - \eta_2 = \eta_1&amp;#39; - \eta_2&amp;#39;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why it's so important to keep the pseudorapidity difference invariant is that in particle physics, people like to make plots called &quot;&lt;a href=&quot;http://www.google.com/search?q=lego+plots&amp;amp;um=1&amp;amp;tbm=isch&amp;amp;source=og&amp;amp;sa=N&amp;amp;tab=wi&amp;amp;biw=1680&amp;amp;bih=850&quot; rel=&quot;nofollow&quot;&gt;lego plots&lt;/a&gt;&quot; which show the directional distribution of the particles detected after a collision. When you do this, you &lt;em&gt;could&lt;/em&gt; plot the particle detections vs. the polar and azimuthal angles $\theta$ and $\phi$. But if you use pseudorapidity instead of the angle $\theta$, the fact that $\Delta\eta$ is invariant means that you can perform a Lorentz boost on your data by just translating the whole graph along the $\eta$ axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is useful because in &lt;a href=&quot;http://en.wikipedia.org/wiki/Hadron&quot; rel=&quot;nofollow&quot;&gt;hadron&lt;/a&gt; collisions, it's often the case that one of the individual quarks or gluons involved in the collision may have a lot more momentum than the other, so all the particles produced come out near one end of the detector. But by translating the graph appropriately, you can effectively shift to the center-of-mass frame of the colliding quarks or gluons, where the particles come out symmetrically distributed, and it's a lot easier to analyze.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;By the way, the reason it's $\tan\frac{\theta}{2}$ rather than just $\tan\theta$ is that we'd like a jet which comes out of the collision at $\theta \approx \frac{\pi}{2}$, where the resolution is best, to have the same shape in the lego plot as it does in physical space. In particular, a circular jet should appear circular on the graph. This requires that the two coordinates be scaled the same way. If we were using $(\theta,\phi)$ as the directional coordinates, this wouldn't be a problem, since both are measured in radians, so we just need to choose a scaling for $\eta$ such that a small increment in $\theta$ near $\frac{\pi}{2}$ corresponds to the same numerical increment in $\eta$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\biggl|\frac{\mathrm{d}\eta}{\mathrm{d}\theta}\biggr|\biggl(\theta=\frac{\pi}{2}\biggr) = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument in the first part of the post (along with some common-sense conditions) basically requires that psuedorapidity be defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta = -\log(\tan a\theta)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some constant $a$. Plugging into the derivative gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\biggl|\frac{\mathrm{d}\eta}{\mathrm{d}\theta}\biggr|_{\frac{\pi}{2}} = \biggl|\frac{a\,\sec^2(a\theta)}{\tan(a\theta)}\biggr|_{\frac{\pi}{2}} = \biggl|\frac{a}{\sin(a\theta)\cos(a\theta)}\biggr|_{\frac{\pi}{2}} = \frac{2a}{\sin(2a\pi/2)} = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm abusing the notation a bit by using vertical bars for both absolute value and substitution, but hopefully the meaning is clear.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a transcendental equation, so you can't solve it analytically, but with a little mathematical reasoning it's not hard to convince yourself that $a = \pm\frac{1}{2}$ are the only nonzero solutions. Having a negative argument to a logarithm brings in an extra imaginary term, though it would cancel out anyway, so we might as well choose the positive one.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-15T04:39:39.857" LastActivityDate="2011-07-15T04:39:39.857" CommentCount="1" />
  <row Id="12261" PostTypeId="2" ParentId="12237" CreationDate="2011-07-14T06:27:25.367" Score="2" ViewCount="" Body="&lt;p&gt;A density matrix view of the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $p_\mu$ is near zero, one is considering a momentum near zero and the discrete lattice works fine. It's when $p_\mu$ is near $\pm \pi/a$, or more generally, near $n\pi/a$ for $n\neq 0$ an integer, that one finds problems. Instead of having very large momenta, these values of $p_\mu$ essentially give momenta as small as those near $p_\mu=0$, but with signs in the $\gamma$ matrices negated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a type of &lt;a href=&quot;http://en.wikipedia.org/wiki/Aliasing&quot; rel=&quot;nofollow&quot;&gt;aliasing&lt;/a&gt; problem. As with the usual aliasing, the problem at $\pm \pi/a$ goes away when one makes the sample rate faster, that is, replaces $a$ with a smaller value. And just as with aliasing, making $a$ smaller does not eliminate aliasing entirely, but instead pushes the problem to a higher frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This drawing shows the usual aliasing effect. Note that the high frequency black signal (which corresponds to a high momentum) appears as a low frequency red signal (with low momentum):&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/PYUil.jpg&quot; alt=&quot;enter image description here&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The difference with the usual aliasing is what happens when $p_\mu = (2n+1)\pi/a$. These are the values that give a continuous Dirac equation with negated gamma matrices. To understand better what is going on here, let's consider the density matrix form. Density matrices avoid unphysical complex phases. I'll work in 3+1 dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One obtains a density matrix $\rho$ by multiplying a ket by a bra:&lt;br&gt;&#xA;$$\rho = |a\rangle\langle a|.$$&#xA;The reason there are four spin-1/2 particles in 3+1 dimensions is that there are four primitive (i.e. trace=1) solutions to the idempotency equation:&lt;br&gt;&#xA;$$\rho^2 = \rho$$&lt;br&gt;&#xA;One has latitude in how one chooses the four states. In general, one chooses two elements of the Dirac algebra that (a) square to unity, (b) commute, and (c) are independent. These are called a &quot;complete set of commuting roots of unity.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The commuting roots of unity are operators; one chooses them according to what operators one wishes to diagonalize. &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_spinor#Construction&quot; rel=&quot;nofollow&quot;&gt;Following the wikipedia article on the construction of Dirac spinors&lt;/a&gt;, if we choose z-spin and charge $Q$, our complete set of commuting roots of unity is:&lt;br&gt;&#xA;$$\sigma_z = i\gamma^1\gamma^2,\;\;\; Q = -\gamma^0$$&lt;br&gt;&#xA;The four independent states are then:&lt;br&gt;&#xA;$$\rho = (1\pm \sigma_z)(1\pm Q)/4.$$&#xA;To get the spinors from a density matrix, one chooses a nonzero column and normalizes. Thus spinors and density matrices are alternative mathematical representations of wave functions; neither is more fundamental.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If we discretize the density matrix, we will end up with the usual aliasing problem. From the point of view of lattice type calculations this is acceptable; there will be no duplicated particles. But spinors carry an extra degree of freedom; the arbitrary complex phase. This makes their aliasing behavior more complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So consider what happens for a frequency a little larger than $\pi/2$. In the following illustration we color every other sample red or blue:&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/l3OKn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;In the above, the density matrix will see this frequency appropriately as a high frequency. But with a spinor, we have arbitrary complex phase freedom. So we can negate the blue dots; the result is a low frequency. Thus the arbitrary complex phases of spinors naturally give aliasing problems at half frequencies.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-07-14T06:27:25.367" CommentCount="3" />
  <row Id="12262" PostTypeId="1" AcceptedAnswerId="12273" CreationDate="2011-07-14T08:20:20.033" Score="4" ViewCount="129" Body="&lt;p&gt;If we were to build a high speed rail up the side of a mountain like in some ScFi movies, what is the velocity needed at the point of living the mountain excluding angular momentum from earth’s rotation to achieve escape velocity?&lt;/p&gt;&#xA;" OwnerUserId="4034" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-14T22:34:28.490" LastActivityDate="2011-07-14T22:34:28.490" Title="Rocket launch from a mountain" Tags="&lt;gravity&gt;&lt;newtonian-mechanics&gt;&lt;rocket&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="12263" PostTypeId="2" ParentId="12262" CreationDate="2011-07-14T09:24:31.727" Score="3" ViewCount="" Body="&lt;p&gt;Escape velocity has got nothing to do with mass of the rocket. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_e = \sqrt{\frac{2GM}{(r_0+h)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r_0 -$ Radius of the earth&lt;br&gt;&#xA;$h -$ Height of the mountain&lt;/p&gt;&#xA;" OwnerUserId="126" LastActivityDate="2011-07-14T09:24:31.727" CommentCount="3" />
  <row Id="12264" PostTypeId="1" CreationDate="2011-07-14T11:07:32.587" Score="1" ViewCount="96" Body="&lt;p&gt;What effect does the quantum world have on radio waves? For example, if I could shrink myself down and stand on the nucleus (or even smaller sub atomic particles making up the nucleus) with a device which could measure radio waves of the surrounding world (ie with all the signals modern day humanity produces), what readings would I pick up? Would they be the same as normal or would the quantum aspects somehow effect the signals and if so, in what way would they be effected?&lt;/p&gt;&#xA;" OwnerUserId="2247" LastActivityDate="2011-07-14T16:48:15.503" Title="Radio waves within an atom" Tags="&lt;quantum-mechanics&gt;&lt;waves&gt;&lt;radio&gt;&lt;cmb&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12265" PostTypeId="1" AcceptedAnswerId="12269" CreationDate="2011-07-14T12:46:24.237" Score="2" ViewCount="88" Body="&lt;p&gt;I am reading Landau's Volume 2 of the course of theoretical physics. I have a doubt after reading the first few pages of it which I explain below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Landau first defines intervals and on pages 5 and 6 shows that two events having time like interval between them can never occur simultaneously in any reference system. Then he goes on to construct a 2D space-time graph (for visualization) with an event O occurring at (0,0,0,0). Then he considers any event which occurs in future in that frame and is time-like w.r.t. origin and says on page no. 7, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But two events which are separated by a time-like interval cannot&#xA;  occur simultaneously in any reference system. Consequently, it is&#xA;  impossible to find a reference frame in which any of the events in&#xA;  region aOc occurred &quot;before&quot; the event O, i.e. at time t&amp;lt;0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The argument above just proves that because interval square should be positive, i.e. the events can't be simultaneous. But, if I replace the difference in time in the original frame with its negative in my proposed frame and let the space distance between them to be same in both frames, then I get an in my proposed frame an interval which is time like but in it the order of events is changed. Am I making some gross error or Landau has missed some argument?&lt;/p&gt;&#xA;" OwnerUserId="4235" LastActivityDate="2011-07-14T15:37:05.370" Title="Non-interchangeability of time-like intervals" Tags="&lt;special-relativity&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="12266" PostTypeId="2" ParentId="12264" CreationDate="2011-07-14T13:04:03.253" Score="0" ViewCount="" Body="&lt;p&gt;A radio wave is nothing more than a photon. A photon can be envisioned as a fuzzy particle with dimensions about the same as the wavelength. It is thus many tens of meters in size. The probability of interaction between an antenna of nuclear dimensions and such a large photon is extremely small. &lt;/p&gt;&#xA;" OwnerUserId="3735" LastActivityDate="2011-07-14T13:04:03.253" CommentCount="3" />
  <row Id="12267" PostTypeId="1" AcceptedAnswerId="12270" CreationDate="2011-07-14T14:16:05.437" Score="1" ViewCount="86" Body="&lt;p&gt;If a particle in a delta potential well has negative energy, why the particle will be bound in the well? And if it has positive well, why it is free to move in either half-space: x &amp;lt; 0 or x &gt; 0?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just read these in a wikipedia page:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Delta_potential&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Delta_potential&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how the energy of the particle influence its motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And by the way, why $e^{ikx}$ represents a wave traveling to the right, and $e^{-ikx}$ one traveling to the left?&lt;/p&gt;&#xA;" OwnerUserId="4314" LastActivityDate="2011-07-14T15:49:10.740" Title="About the delta potential well" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="12268" PostTypeId="2" ParentId="12265" CreationDate="2011-07-14T14:39:57.580" Score="0" ViewCount="" Body="&lt;p&gt;I don't know what Landau says, but space-time interval doesn't distinguish order. To derive the non-interchangeability of time-like intervals, Lorentz transformation must be explicitly used.&lt;/p&gt;&#xA;" OwnerUserId="3887" LastActivityDate="2011-07-14T14:39:57.580" CommentCount="2" />
  <row Id="12269" PostTypeId="2" ParentId="12265" CreationDate="2011-07-14T15:37:05.370" Score="2" ViewCount="" Body="&lt;p&gt;If you believe that (a) timelike separated events cannot be simultaneous in any reference frame, &lt;em&gt;and&lt;/em&gt; (b) the set of inertial frames is (in some appropriate sense) a continuous set, then L&amp;amp;L's conclusion follows. After all, if there were two frames in which the order of two timelike separated events differed, then by continuously transforming one frame into another, you could find one in which they were simultaneous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But without some such additional assumption, you're right that the conclusion doesn't logically follow. There are coordinate systems that preserve the spacetime interval but flip the direction of time, such as the substitution $t\to -t$ that you mention. As BebopbutUnsteady observes in a comment to Karsus Ren's answer, we often use the term &quot;orthochronous Lorentz transformation&quot; to refer to a transformation that preserves the direction of time. The full group of Lorentz transformations (i.e., of all transformations that preserves the interval) includes both orthochronous and non-orthochronous components, which are not connected to each other. Physically, we usually only consider the orthochronous ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You do have to be careful with the terminology: sometimes people use &quot;Lorentz transformation&quot; to mean just the orthochronous ones; sometimes it's the full group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, pretty much the same thing applies to spatial reflections: is $x\to -x$ (leaving $y,z,t$ unchanged) a Lorentz transformation? After all, it preserves the spacetime interval. Often we refer to non-reflecting Lorentz transformations as &quot;proper.&quot; So when people are being careful with their terminology they often refer to &quot;proper orthochronous Lorentz transformations.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-07-14T15:37:05.370" CommentCount="2" />
  <row Id="12270" PostTypeId="2" ParentId="12267" CreationDate="2011-07-14T15:49:10.740" Score="6" ViewCount="" Body="&lt;p&gt;I'm not totally sure I know what the first questions mean, so let me start with the last one. The wavefunction of a particle is a function of time as well as position. A lot of quantum mechanics is concerned with wavefunctions corresponding to &quot;stationary states.&quot; You might think that such wavefunctions would have no time-dependence, but that's not the case. A stationary state corresponds to a particle with a certain amount of energy $E$, and the wavefunction has time dependence of the form $e^{-i\omega t}$, where $\omega=E/\hbar$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People often leave out that time dependence, just to confuse you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when people refer to a wavefunction of the form $e^{ikx}$, they really mean $e^{ikx}e^{-i\omega t}$, or $e^{i(kx-\omega t)}$. This corresponds to a wave moving to the right. In particular, a point of any given &quot;phase&quot; moves to the right at speed $\omega/k$. For instance, the point $kx-\omega t=0$ is a point where the wavefunction is purely real. This point moves along according to the rule $x=(\omega/k)t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to think about it: a particle has definite momentum if it is an eigenfunction of the momentum operator: &#xA;$$&#xD;&#xA;\hat p\psi = p\psi.&#xD;&#xA;$$&#xA;Here $\hat p=-i\hbar(\partial/\partial x)$ is the momentum operator (specifically its $x$ component), and the number $p$ is the eigenvalue. For a particle with such a wavefunction, a measurement of momentum is guaranteed to give the value $p$. The wavefunction $\psi(x)=e^{ikx}$ obeys this rule:&#xA;$$&#xD;&#xA;\hat pe^{ikx}=-i\hbar{\partial(e^{ikx})\over\partial x}=\hbar ke^{ikx}=\hbar k \psi,&#xD;&#xA;$$&#xA;so $p=\hbar k$. So a particle with positive $k$ will be observed to have positive momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now back to the first couple of questions. When you solve for the wavefunction of a particle in a delta function potential with negative energy, you find that the solutions are of the form $e^{-k|x|}$ for some $k$. These wavefunctions approach zero for large $|x|$. So physically, the particle has essentially zero probability of being far away from the origin. That's what &quot;bound state&quot; means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, with positive energy, there are solutions that look like $e^{\pm ikx}$. These wavefunctions do not die away to zero at large distances: $|\psi|^2$ remains constant as $|x|\to\infty$. So these particles have significant probability to be anywhere.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-07-14T15:49:10.740" />
  <row Id="12271" PostTypeId="2" ParentId="12262" CreationDate="2011-07-14T16:37:08.993" Score="0" ViewCount="" Body="&lt;p&gt;An intuitive way to think about it. Escaping an inverse square law to infinity requires energy (kinetic energy is this example) equivalent to the force of gravity times the radius. So the number od gees you would need is the ratio of track length divided by the radius of the earth. You have the additional complication that you must curve from horizontal velocity to velocity at and angle, and this means you are also dealing with centripetal forces during the transition from plain to mountain.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-07-14T16:37:08.993" CommentCount="1" />
  <row Id="12272" PostTypeId="2" ParentId="12264" CreationDate="2011-07-14T16:48:15.503" Score="1" ViewCount="" Body="&lt;p&gt;If you are an electron in an atom and you have some energetic spectrum, a radio-wave may cause your transitions from one level to another. Your quantum character is manifested in your discrete energy levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are an electron in a solid state, in a metal, for example, then your spectrum may be continuous and any energies are allowed. In this case you behave as a classical charge.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-07-14T16:48:15.503" />
  <row Id="12273" PostTypeId="2" ParentId="12262" CreationDate="2011-07-14T17:05:23.823" Score="1" ViewCount="" Body="&lt;p&gt;The total escape velocity is about 11,200 m/s (approximately 7 miles/sec.) in a direction roughly tangential to the earth's surface.  At 30 degrees north latitude (e.g., somewhere in southern Texas) the eath's spin would contribute about 400 m/s to the tangential velocity, so actual speed, relative to the earth's surface at said launch point would be about 10,800 m/s, if the load was launched in a roughly tangential direction, eastward; closer to the 11,200 m/s if you went straight up with it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The height of the mountain will serve a good purpose in getting you up to where the air is thinner when you leave the track; but, otherwise has negligible on how fast you need to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main problem with this is that 11,200 m/s, at 14000 feet, is about Mach 24; roughly 13 times the muzzle velocity of a 30-06 high-powered rifle.  Heating, shredding, melting, ablation, and all sorts of bad stuff occur at these speeds, even in thin air.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Studies have been done on this, to engineer a ground-power-assisted launch.  The idea is to get the load up to a very high speed --but not high enough to shred or melt it-- and let the rocket engines take over from there.  I believe I read about a track hypothesized for somewhere in the vast wastelands of Texas.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-07-14T18:26:31.897" LastActivityDate="2011-07-14T18:26:31.897" CommentCount="1" />
  <row Id="12274" PostTypeId="1" CreationDate="2011-07-14T17:40:22.477" Score="0" ViewCount="40" Body="&lt;p&gt;I'm sure this is pretty basic, but I'm not very good at electricity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At both ends of a conductor we apply 12 volts. For one minute an electric charge of 72 C passes through the conductor. What is the resistance of the conductor?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4457" LastActivityDate="2011-07-14T17:53:47.247" ClosedDate="2011-07-14T18:00:26.360" Title="Resistance from voltage and charge" Tags="&lt;homework&gt;&lt;electricity&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12275" PostTypeId="2" ParentId="12274" CreationDate="2011-07-14T17:53:47.247" Score="1" ViewCount="" Body="&lt;p&gt;Current is the amount of charge passing though a surface area per second. If in 60 seconds 72C of charge flows through this surface, then in 1 second, 72C/60 flows, and that is your current I = 72/60 Amps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use Ohms law, V = IR to find the resistance R with V = 12v and I = 72/60A&lt;/p&gt;&#xA;" OwnerUserId="4485" LastActivityDate="2011-07-14T17:53:47.247" />
  <row Id="12276" PostTypeId="2" ParentId="12249" CreationDate="2011-07-14T18:55:32.497" Score="2" ViewCount="" Body="&lt;p&gt;Air has a density of 1.29 grams/liter at $0^oC$ $(273^oK)$ at sea level.  Using the formula which David so kindly provided, we readily see that the same mass of air grows in volume in direct proprtion to absolute ($^oK$) temperature.  A warm sunny day is about $77^oF$ which is about $25^C$, or $298^oK$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the volume of 1.29 grams of air at sea level on a 25C day would be (1 liter)(298K/273K) = 1.0916 liters.  This makes the density of air at that warmer condition = 1.29 grams/ 1.0916 liters = 1.182 grams/liter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1 liter air)/(1.182grams) x (454.5 grams/lb.) x (5 lbs.) = (1923 liters air) / (5 lbs. air).  That's 1.923 cubic meters of air: about 106% of the air you could hold in a 4ft x 4ft x 4ft box.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-07-14T18:55:32.497" />
  <row Id="12277" PostTypeId="2" ParentId="2164" CreationDate="2011-07-14T19:18:26.230" Score="-1" ViewCount="" Body="&lt;p&gt;Holding the glider in the air is Bernoulli's Principle. This basic principle of aerodynamics states that fast-flowing air has lower pressure than slow-flowing air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a complement of answer made by mmc&lt;/p&gt;&#xA;" OwnerUserId="4493" LastActivityDate="2011-07-14T19:18:26.230" CommentCount="1" />
  <row Id="12278" PostTypeId="1" AcceptedAnswerId="12281" CreationDate="2011-07-14T20:02:44.840" Score="0" ViewCount="89" Body="&lt;p&gt;Assume it's a standard .45 caliber handgun. Shooter is 10 feet away, and we hold a desk, mini-table, blackboard, etc on our chests(to protect vitals). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you figure out the bullet's decrease in velocity, and likelihood of flying away from you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the idea thinking  whether could we protect ourselves against a madman like the VT or Columbine guys?&lt;/p&gt;&#xA;" OwnerUserId="4429" LastEditorUserId="4495" LastEditorDisplayName="" LastEditDate="2011-07-14T22:43:36.620" LastActivityDate="2011-07-14T22:43:36.620" Title="If someone shoots a gun (45 caliber) at someone holding up a table or desk 10 feet away, does it slow or turn the bullet away?" Tags="&lt;velocity&gt;&lt;projectile&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="12279" PostTypeId="1" CreationDate="2011-07-14T20:31:08.857" Score="1" ViewCount="129" Body="&lt;p&gt;I was wondering how much energy would be required to accelerate 1000kg to 0.99c at 1G.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is what the rate of increase of energy is required as velocity increases. I was looking at the Lorentz factor curve but I'm not sure if this affects the energy required or just time dilation. eg. 0.99c =  Lorentz factor of 7. So travel only takes 1/7th the time from the point of view of the traveler but does that mean you need 7 times as much energy to get to 0.99c than from say 0.0c to 0.1c. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does any body have a link to a graph that shows this rate of increase of energy required or is it the same as the Lorentz factor curve?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a programmer not a physics or math person so go easy on me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thankyou.&lt;/p&gt;&#xA;" OwnerUserId="4494" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-15T18:12:17.320" LastActivityDate="2011-07-15T18:12:17.320" Title="whats the rate of energy increase required for constant acceleration between 0.0c and 0.99c?" Tags="&lt;special-relativity&gt;&lt;energy&gt;&lt;velocity&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="12280" PostTypeId="2" ParentId="12279" CreationDate="2011-07-14T21:15:03.763" Score="6" ViewCount="" Body="&lt;p&gt;The energy of the mass at rest is&#xA;$$&#xD;&#xA;E_0 = m c^2&#xD;&#xA;$$&#xA;At speed $v = 0.99 c$ it becomes&#xA;$$&#xD;&#xA;E =  \frac{m c^2}{\sqrt{1-v^2/c^2}}&#xD;&#xA;$$&#xA;So, the increase in energy is&#xA;$$&#xD;&#xA;E - E_0 =  m c^2\left(\frac{1}{\sqrt{1-v^2/c^2}} - 1\right) \approx 6.09 m c^2.&#xD;&#xA;$$&#xA;irrespective of the acceleration.&lt;/p&gt;&#xA;" OwnerUserId="4202" LastActivityDate="2011-07-14T21:15:03.763" CommentCount="6" />
  <row Id="12281" PostTypeId="2" ParentId="12278" CreationDate="2011-07-14T21:30:38.837" Score="4" ViewCount="" Body="&lt;p&gt;In order to answer this question we must make some more assumptions.  Given your scenario, one of the tragic school shootings which occurred in the States, we can assume the shooter will be within 50 feet.   Since we are at a school, then let us assume typical materials found there such as desks and tables composed of laminate or other glued wooden materials augmented by plastic.  With these assumptions, any bullet from a .22 caliber (~ 2 oz) and larger at close range will most likely penetrate the barrier.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a bullet impacts upon a surface of a material as given above, it penetrates, begins to deform and may fragment in the material. To determine how it slows and deforms, we must consider the size, shape, velocity and composition of the bullet, the interaction of the bullet with atmosphere, such as its dynamic stability, as well as the material response of the material being hit.  Laminates should exhibit an inelastic response to the bullet adding to the difficulty of a priori calculating what will happen besides simply saying that it will go through or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most bullets are made of lead surrounded (jacketed fully or partially) by a harder metal such as copper.  This is done as a trade off between ballistic stability and deformability in order to wound. There are several types of bullets, the 2 most common are full metal jacket bullets, which are conical at the tip and various hollow tip styles which are blunt tipped and designed to expand as they penetrate the material since the segmented jacket peels back at it travels through the material creating a larger wound channel due to the high pressure zone commonly associated with hollow-tip bullets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A  bullets decrease in velocity due to traveling through a material is a nontrivial calculation which is usually not even attempted.  Typically, the velocity given is an average muzzle velocity and the energy in foot pounds is given as the total  amount of energy dumped into a ballistic pendulum, at a given distance, exhibiting an elastic impulsive response to the collision.  The bullet hits it, the pendulum moves, and knowing facts about the bullet and pendulum, we calculate the energy deposited in the collision as an average. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ballistics is an experimental science, in which bullets are created according to various known criteria and then tested against various barriers while measuring performance before and after penetration.  For handguns, after a certain point, bigger is not always better and jacketed blunt bullets (hollow tips) can cause a more lethal wound than a full metal jacket despite the fact that they may not penetrate as far in ballistic gelatin. It is possible that bullets of different sizes but traveling at varying speeds will give similar wound profiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A .45 caliber bullet is capable of penetrating a typical hollow cinderblock if it is jacketed.  Given what you present, desks or tables provide little cover, but better than none at all. Real examples of ballistics tests can be found here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ballisticsbytheinch.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.ballisticsbytheinch.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.theboxotruth.com/docs/bot1.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.theboxotruth.com/docs/bot1.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4495" LastEditorUserId="4495" LastEditorDisplayName="" LastEditDate="2011-07-14T21:35:59.050" LastActivityDate="2011-07-14T21:35:59.050" CommentCount="7" />
  <row Id="12282" PostTypeId="1" AcceptedAnswerId="12391" CreationDate="2011-07-14T22:31:18.160" Score="0" ViewCount="128" Body="&lt;p&gt;Am I correct in assuming it  is the weight of all that liquid + gravity?&lt;/p&gt;&#xA;" OwnerUserId="4429" LastActivityDate="2011-07-18T05:40:55.280" Title="I noticed today that it is easier to open a soda bottle(standard 20 oz) upside-down, why is this?" Tags="&lt;gravity&gt;&lt;fluid-dynamics&gt;" AnswerCount="2" CommentCount="11" />
  <row Id="12283" PostTypeId="1" CreationDate="2011-07-14T22:42:09.307" Score="3" ViewCount="205" Body="&lt;p&gt;By chance(playing around really) I saw that a spring(mainly from a pen) placed on a neodymium hard-disk magnet(and then flicked by your finger at the top) makes a nice-effect (see youtube video ). It appears to oscillate in slow-motion(looks like tornado).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, &quot;slow-motion&quot; is purposely simplistic and unscientific - I am very far from a physicist. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the video - &lt;a href=&quot;http://www.youtube.com/watch?v=n0OJQ1iXZg0&amp;amp;feature=channel_video_title&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=n0OJQ1iXZg0&amp;amp;feature=channel_video_title&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was too impatient in the video though, I should have zoomed in on the spring and waited. Sorry about that.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a page about the magnets used:&#xA;&lt;a href=&quot;http://www.reuk.co.uk/Hard-Disk-Drive-Magnets-For-Wind-Turbines.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.reuk.co.uk/Hard-Disk-Drive-Magnets-For-Wind-Turbines.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the polarities, plus a horizontal profile below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/7krCW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/fPLEl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More details: You really want to use a retractable pen  spring, the thin kind. And Hard-drive magnets are key - I think it doesn't work with others. I think it's partly because of the 4-poles of a neodymium magnet. i.e, it's actually two-magnets-in-one.&#xA; Cigarette lighters also have a long delicate magnet, which is good but too tipsy.&lt;br&gt;&#xA;LBNL, supposedly you can stack these magnets, but they seem impossible to separate from the backing-piece.   I appreciate any tips or advice.&lt;/p&gt;&#xA;" OwnerUserId="4429" LastEditorUserId="4429" LastEditorDisplayName="" LastEditDate="2011-07-16T01:16:05.423" LastActivityDate="2011-08-15T20:57:59.167" Title="If you place a spring on a neodymium hard-drive magnet, it appears to vibrate in slow-motion. Why is that so?" Tags="&lt;electromagnetism&gt;&lt;spring&gt;" AnswerCount="2" CommentCount="14" />
  <row Id="12284" PostTypeId="1" AcceptedAnswerId="12285" CreationDate="2011-07-15T00:26:18.477" Score="2" ViewCount="43" Body="&lt;p&gt;For two particles, $\langle {\mathcal T} a(t_1) a^\dagger (t_2) \rangle = \langle &#xD;&#xA;a(t_1) a^\dagger (t_2)\rangle \theta (t_1-t_2) + \xi \langle a^\dagger (t_2)a(t_1) \rangle \theta (t_2-t_1)$ with $\xi$ is  a plus sign for bosons and a minus sign for fermions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I write, for example,  $\langle {\mathcal T} a(t_1) a^\dagger (t_2) a(t_3) a^\dagger (t_4) \rangle$ ?&lt;/p&gt;&#xA;" OwnerUserId="873" LastActivityDate="2011-07-15T00:48:32.317" Title="What's the definition of the time ordering operator for more than two particles?" Tags="&lt;quantum-field-theory&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12285" PostTypeId="2" ParentId="12284" CreationDate="2011-07-15T00:48:32.317" Score="3" ViewCount="" Body="&lt;p&gt;Just sum over each permutation of [1,2,3,4], for each permutation $[I_1,I_2,I_3,I_4]$you would have a factor of $\theta(t_{I_1}-t_{I_2})\theta(t_{I_2}-t_{I_3})\theta(t_{I_3}-t_{I_4})$ times the corresponding operator product, etc.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-07-15T00:48:32.317" />
  <row Id="12286" PostTypeId="1" AcceptedAnswerId="12293" CreationDate="2011-07-15T01:13:36.740" Score="7" ViewCount="249" Body="&lt;p&gt;Consider that we have two balls, one white and one black, and two distant observers A and B with closed eyes. We give the first ball to the observer A and the second ball to the observer B. The observers don't know the exact color (state) of their balls, they know only the probability of having one or another color, until they look at them (measure). If the observer A looks at his ball he will see its color, which is white, so he immediately knows the color of the second ball. Lets call this “classical entanglement”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: What is the difference between this “classical entanglement” and the quantum entanglement, for example, of two entangled electrons with opposite spins states? Can this analogy be used to explain the quantum entanglement?&lt;/p&gt;&#xA;" OwnerUserId="3066" LastActivityDate="2011-07-16T17:15:48.937" Title="Quantum entanglement vs classical analogy" Tags="&lt;quantum-mechanics&gt;&lt;entanglement&gt;" AnswerCount="5" CommentCount="7" FavoriteCount="4" />
  <row Id="12287" PostTypeId="1" AcceptedAnswerId="12478" CreationDate="2011-07-15T01:22:47.147" Score="1" ViewCount="169" Body="&lt;p&gt;The time is treated differently in special relativity and quantum mechanics. What is the exact difference and why relativistic quantum mechanics (Dirac equation etc.) works?&lt;/p&gt;&#xA;" OwnerUserId="3066" LastActivityDate="2011-07-19T20:05:09.837" Title="Time in special relativity and quantum mechanics" Tags="&lt;quantum-mechanics&gt;&lt;special-relativity&gt;&lt;time&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="12288" PostTypeId="2" ParentId="12286" CreationDate="2011-07-15T03:43:26.757" Score="0" ViewCount="" Body="&lt;p&gt;No, quantum entanglement is not in any way analogous to your “classical entanglement”, or in more technical term, local hidden-variable theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain this, certain level of quantum mechanics and mathematics must be involved, but in a nutshell, such theory has been definitively invalidated by experiments.&lt;/p&gt;&#xA;" OwnerUserId="3887" LastActivityDate="2011-07-15T03:43:26.757" CommentCount="1" />
  <row Id="12289" PostTypeId="2" ParentId="11686" CreationDate="2011-07-15T05:07:27.837" Score="2" ViewCount="" Body="&lt;p&gt;First of all, choose the reference frame co-moving with the paddle and assume that this reference frame is inertial. This is the key to all ball-and-wall problems. Of course, ignore gravity and air drag. Now we have a spinning ball incident on a stationary surface with friction. Let &lt;strong&gt;V&lt;/strong&gt; be the vector of the ball's velocity with respect to the paddle, and &lt;strong&gt;V&lt;/strong&gt;_n and &lt;strong&gt;V&lt;/strong&gt;_t be the normal and the tangential components of the ball's velocity before impact. Let &lt;strong&gt;&amp;omega;&lt;/strong&gt; be the ball's angular velocity before impact. All quantities in &lt;strong&gt;bold&lt;/strong&gt; are vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it gets as complicated as you wish. To make it as simple as possible, let us assume that the ball spends time &lt;em&gt;&amp;tau;&lt;/em&gt; in contact with the surface, and that all forces acting on it during contact are constant in time. Further, assume that the friction coefficient, &lt;em&gt;&amp;mu;&lt;/em&gt;, does not depend on the relative velocity of the ball and the paddle. &lt;em&gt;Important:&lt;/em&gt; This assumption will break down if rotation of the ball ever reaches such speed that the ball will start rolling instead of sliding; at this moment the tangential force vanishes, and such situation must be treated independently. A LOT of assumptions to make it possible to tackle the problem analytically, huh? But the good news is that it's all Mechanics 101 from now on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ball is perfectly elastic, then there are no dissipative forces acting in the direction normal to the surface. This means that when the ball rebounds, its normal component of velocity will just be inverted:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;V&lt;/strong&gt;_n'=-&lt;strong&gt;V&lt;/strong&gt;_n             &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This allows us to calculate the average normal force acting on the ball during contact (just Newton's 2nd law):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;|F_n|=2m|V_n|/&lt;em&gt;&amp;tau;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tangential force acting on the ball due to friction during contact is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;F&lt;/strong&gt;_t=&lt;em&gt;&amp;mu;&lt;/em&gt; |F_n| &lt;strong&gt;e&lt;/strong&gt; = 2 m &lt;em&gt;&amp;mu;&lt;/em&gt; |V_n|/&amp;tau; &lt;strong&gt;e&lt;/strong&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;strong&gt;e&lt;/strong&gt; a &lt;em&gt;unit&lt;/em&gt; vector in the direction of the friction force. It can be found as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;e&lt;/strong&gt;=( [ &lt;strong&gt;R&lt;/strong&gt; x &lt;strong&gt;&amp;omega;&lt;/strong&gt; ] &amp;minus; &lt;strong&gt;V&lt;/strong&gt;_t ) / |[ &lt;strong&gt;R&lt;/strong&gt; x &lt;strong&gt;&amp;omega;&lt;/strong&gt; ] &amp;minus; &lt;strong&gt;V&lt;/strong&gt;_t|, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;strong&gt;R&lt;/strong&gt; is the vector from the center of the ball to the point of contact, &lt;strong&gt;&amp;omega;&lt;/strong&gt; is the ball's angular velocity, and &lt;strong&gt;V&lt;/strong&gt;_t is the tangential speed of the ball. Here | | denote taking a vector's modulus (length).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This force, acting &lt;em&gt;&amp;tau;&lt;/em&gt; seconds, will the change the ball's tangential momentum by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;Delta;&lt;strong&gt;p&lt;/strong&gt;_t = &lt;em&gt;&amp;tau;&lt;/em&gt; &lt;strong&gt;F&lt;/strong&gt;_t = 2 m &lt;em&gt;&amp;mu;&lt;/em&gt; |V_n| &lt;strong&gt;e&lt;/strong&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(note that &lt;em&gt;&amp;tau;&lt;/em&gt; cancels out!), so the ball's tangential speed after impact will be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;V&lt;/strong&gt;_t'=&lt;strong&gt;V&lt;/strong&gt;_t + &lt;strong&gt;e&lt;/strong&gt;|&amp;Delta;p_t|/m = &lt;strong&gt;V&lt;/strong&gt;_t + 2 &amp;mu; |V_n| &lt;strong&gt;e&lt;/strong&gt;. &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the angular momentum picked up by the ball equals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;Delta;&lt;strong&gt;L&lt;/strong&gt; = R &amp;tau; |F_t| &lt;strong&gt;f&lt;/strong&gt; = 2 R m &lt;em&gt;&amp;mu;&lt;/em&gt; |V_n| &lt;strong&gt;f&lt;/strong&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where R is the radius of the ball (also, R=|&lt;strong&gt;R&lt;/strong&gt;|), and &lt;strong&gt;f&lt;/strong&gt; is a &lt;em&gt;unit&lt;/em&gt; vector in the direction in which the angular momentum was picked up. To find &lt;strong&gt;f&lt;/strong&gt;, use&#xA;&lt;strong&gt;f&lt;/strong&gt;=[ &lt;strong&gt;R&lt;/strong&gt; x &lt;strong&gt;e&lt;/strong&gt; ] / R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This change in &lt;strong&gt;L&lt;/strong&gt; will make the ball's angular velocity after impact&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&amp;omega;&lt;/strong&gt;'=&lt;strong&gt;&amp;omega;&lt;/strong&gt; + &lt;strong&gt;f&lt;/strong&gt; |&amp;Delta;L|/I =&lt;strong&gt;&amp;omega;&lt;/strong&gt; + |2 R m &lt;em&gt;&amp;mu;&lt;/em&gt; V_n|/I &lt;strong&gt;f&lt;/strong&gt;. &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; (3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here &lt;em&gt;I=(2/5)mR&amp;sup2;&lt;/em&gt; is the moment of inertia of the ball (assuming that it is a hollow uniform sphere). Note that the mass &lt;em&gt;m&lt;/em&gt; cancels out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equations (1), (2) and (3) give you the relationship between pre- and post-impact linear velocity &lt;strong&gt;V&lt;/strong&gt; and angular velocity &lt;strong&gt;&amp;omega;&lt;/strong&gt;. They are determined in the paddle's reference frame, so to get lab (room) frame quantities, you need to transfer back to that frame. It is tedious to do by hand, but if you are developing a ping pong emulator in which you use a game console's accelerometer to emulate a paddle, the computer will happily do this for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me re-iterate at this point that this calculation does not apply to cases when the ball is rolling (or nearly rolling), as opposed to sliding. This situation will occur if [&lt;strong&gt;R&lt;/strong&gt; x &lt;strong&gt;&amp;omega;&lt;/strong&gt; ] = &lt;strong&gt;V&lt;/strong&gt;_t, or, equivalently, &lt;strong&gt;e&lt;/strong&gt;=0. If this situation occurs (it will if initial conditions are close to it), the ball will start rolling, its tangential momentum and angular velocity will freeze, and it will just rebound. To treat this situation properly, you may want to choose a finite &lt;em&gt;&amp;tau;&lt;/em&gt;, a large integer &lt;em&gt;N&lt;/em&gt;, and break &lt;em&gt;&amp;tau;&lt;/em&gt; into smaller time steps &lt;em&gt;&amp;Delta;&amp;tau;=&amp;tau;/N&lt;/em&gt;. Then calculate all quantities during impact in &lt;em&gt;&amp;Delta;&amp;tau;&lt;/em&gt; increments. Once you detect the rolling condition &lt;strong&gt;e&lt;/strong&gt;=0, you make the ball rebound with the current &lt;strong&gt;V&lt;/strong&gt;_t' and &lt;strong&gt;&amp;omega;&lt;/strong&gt;'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may have messed up the signs here and there, but, basically, it is all very simple. To top it off, here is a diagram (you are correct, I am not an artist!) In my example, &lt;strong&gt;&amp;omega;&lt;/strong&gt; is pointing toward us, but in the solution, its direction can be arbitrary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/RAGiB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;EDIT&lt;/em&gt;&lt;/strong&gt;: I just realized that equations (2) and (3) are only correct if the ball is spinning in the direction it is flying (i.e. if vector &lt;strong&gt;e&lt;/strong&gt; is parallel or anti-parallel to vector &lt;strong&gt;V&lt;/strong&gt;_t). So 2D Pong is OK with equations (2) and (3). However, in 3D problem, if the ball is spinning sideways, then during contact, &lt;strong&gt;&amp;omega;&lt;/strong&gt; and &lt;strong&gt;V&lt;/strong&gt;_t will change in length, and therefore unit vectors &lt;strong&gt;e&lt;/strong&gt; and &lt;strong&gt;f&lt;/strong&gt; will change direction. This means that to get the ball's speed and spin, we need to integrate in time. It can be done analytically (a nice problem to torture graduate students), and it is not difficult to do numerically. One should choose a time of contact duration &lt;em&gt;&amp;tau;&lt;/em&gt; a large integer &lt;em&gt;N&lt;/em&gt;, and small time step &lt;em&gt;&amp;Delta; t=&amp;tau;/N&lt;/em&gt;. And then calculate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;V&lt;/strong&gt;_t(t+&amp;Delta;t)=&lt;strong&gt;V&lt;/strong&gt;_t(t) + &lt;strong&gt;e&lt;/strong&gt;|F_t|/m &amp;Delta; t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&amp;omega;&lt;/strong&gt;(t+&amp;Delta;t)=&lt;strong&gt;&amp;omega;&lt;/strong&gt; + &lt;strong&gt;f&lt;/strong&gt;|F_t|R/I &amp;Delta; t&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and do it &lt;em&gt;N&lt;/em&gt; times, re-calculating the values of &lt;strong&gt;e&lt;/strong&gt; and &lt;strong&gt;f&lt;/strong&gt; every time step. A fringe benefit of this approach, it is easy to control the onset of rolling. If any component of vector &lt;strong&gt;e&lt;/strong&gt; changes sign (i.e., goes through 0) at any time step, sliding stops and rolling begins. The answer should not depend or &lt;em&gt;&amp;tau;&lt;/em&gt; or &lt;em&gt;N&lt;/em&gt;.&#xA;&lt;strong&gt;&lt;em&gt;EDIT END&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditorDisplayName="" LastEditDate="2011-07-18T00:47:23.253" LastActivityDate="2011-07-18T00:47:23.253" CommentCount="7" />
  <row Id="12290" PostTypeId="1" CreationDate="2011-07-15T05:49:46.553" Score="2" ViewCount="46" Body="&lt;p&gt;For astronomy as originally invented it is good for finding  the angular diameter of stars and is not a rapidly fluctuating observable like the amplitude in ordinary interferometry. The same concept in particle physics is not so straightforward. When I measure the 2 particle correlation, what do I get? What does it mean? &lt;/p&gt;&#xA;" OwnerUserId="2599" LastActivityDate="2011-07-15T05:49:46.553" Title="A moderate introduction to Hanbury Brown Twiss interferometry in particle physics" Tags="&lt;particle-physics&gt;" />
  <row Id="12291" PostTypeId="2" ParentId="12286" CreationDate="2011-07-15T06:19:29.497" Score="1" ViewCount="" Body="&lt;p&gt;In classical physics when we have a composite system that consists of two (or more) subsystems, then the state of that system is given by a combination of the states of the individual subsystems. The black-and-white-ball system you mentioned is an example of that; the state of the composite system is that one ball is black and the other white, defined solely by the fact that the state of one subsystem is that the ball is white, and the state of the other that the ball is black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum mechanics there are cases where things behave like in classical physics, and cases where they don't. When we have a composite system in quantum mechanics, there are two options for its state (wavefunction):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;it can be given just by knowing the states of the individual subsystems, or&lt;/li&gt;&#xA;&lt;li&gt;it may be its own thing, i.e. a state that is defined for the composite system and which  cannot be decomposed into states of the individual subsystems.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The first option has a classical counterpart, like your example, and has nothing to do with quantum entanglement. A quantum state with the property of option 1 is called a product state in the quantum-physics literature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second option is what is called an entangled state, and has no counterpart in classical physics. It displays the idea of entanglement clearly: it is the phenomenon where the state of a system is &lt;em&gt;not&lt;/em&gt; given by some combination of the states of the individual subsystems that make it up. In fact, all you can define for entangled systems is the state of the composite system, you cannot assign states to the individual subsystems. Clearly, that can never happen in classical physics.&lt;/p&gt;&#xA;" OwnerUserId="4483" LastActivityDate="2011-07-15T06:19:29.497" />
  <row Id="12293" PostTypeId="2" ParentId="12286" CreationDate="2011-07-15T07:28:19.983" Score="4" ViewCount="" Body="&lt;p&gt;Quantum entanglement is different from the &quot;classical entanglement&quot; in the following way:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In your example, each ball has only one property of interest, namely &quot;color $\in$ {white, black}&quot;.&lt;/li&gt;&#xA;&lt;li&gt;In the traditional examples of quantum entanglement, each ball (particle) has &lt;strong&gt;two&lt;/strong&gt; properties of interest, namely &quot;spin in x-direction $S_x$&quot; and &quot;spin in y-direction $S_y$&quot;. Moreover, the properties are &lt;strong&gt;complementary&lt;/strong&gt;, i.e. you can't actually measure them &lt;em&gt;simultaneously&lt;/em&gt; to arbitrary precision.&lt;/li&gt;&#xA;&lt;li&gt;Still, in the entangled state, each of these properties &lt;em&gt;alone&lt;/em&gt; is perfectly &lt;strong&gt;(anti-)correlated&lt;/strong&gt;: If A makes a measurement of $S_x$ and obtains the value, say +1, then B will always obtain -1 if he also measures $S_x$. Similar for -1 and +1 and for the other spin direction $S_y$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The paradox, now, is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that Alice measures $S_x$ and obtains, say +1. But observer Bob measures $S_y$, obtaining, say +1. Exhulted, Alice proclaims that she has managed to measure two complementary properties simultaneously! After all, her measurement gave her the spin in x-direction, +1, while Bob's measurement allows her to conclude that the particle also has spin -1 in y-direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine her surprise, then, when she tries to confirm her conclusion by measuring $S_y$ herself and obtaining +1 in 50% of the cases. &lt;/p&gt;&#xA;" OwnerUserId="188" LastEditorUserId="188" LastEditorDisplayName="" LastEditDate="2011-07-15T11:46:10.100" LastActivityDate="2011-07-15T11:46:10.100" CommentCount="10" />
  <row Id="12294" PostTypeId="2" ParentId="12287" CreationDate="2011-07-15T08:14:59.893" Score="0" ViewCount="" Body="&lt;p&gt;Time is always time. It is special. Another thing is its involvement in transformations of measured data from one reference system to another. This involvement does not change its meaning. In a given reference frame the time is unique and the space coordinates are multiple - according to the number of particles to be observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning Dirac equation, it took some efforts to make it work after its invention. It works because it was made work, if you like. Besides, it depends what exactly do you mean by &quot;relativistic QM&quot;. QED, for example, is rather difficult to make work. Its sensible results only appear at page 500 or so, when the infrared catastrophe is resolved.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-07-15T08:30:15.487" LastActivityDate="2011-07-15T08:30:15.487" />
  <row Id="12295" PostTypeId="2" ParentId="9776" CreationDate="2011-07-15T10:38:27.400" Score="0" ViewCount="" Body="&lt;p&gt;Ship's use doppler logs to tell them how fast they are moving through the water. They are used to calculate forwards/backwards speed as well as sideways speed. Very large ships, such as super tankers, would not be able to dock safely without this equipment.&lt;/p&gt;&#xA;" OwnerUserId="4498" LastActivityDate="2011-07-15T10:38:27.400" CommunityOwnedDate="2011-07-15T10:38:27.400" />
  <row Id="12296" PostTypeId="2" ParentId="9776" CreationDate="2011-07-15T11:05:18.797" Score="1" ViewCount="" Body="&lt;p&gt;Two suggestions from experience:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Let your students stay near a highway for 10 minutes (car engine works just like siren, but many sound changes challenge their mind)&lt;/li&gt;&#xA;&lt;li&gt;I also notice that talking, while yawning causes me to here my own voice if lower frequencies.   &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4499" LastActivityDate="2011-07-15T11:05:18.797" CommunityOwnedDate="2011-07-15T11:05:18.797" />
  <row Id="12297" PostTypeId="1" CreationDate="2011-07-15T12:39:22.797" Score="2" ViewCount="71" Body="&lt;p&gt;Here's a simple 'derivation' of the Brownian motion law that after N steps of unit distance 1, the total distance from the origin will be sqrt(N) on average. It's certainly not rigorous, but I'm wondering if people think it's reasonable, or possibly even a commonly known.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;An object takes one step from the origin, so is at a distance 1: d = 1 for N=1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On average, the next step will be neither exactly toward or exactly away from the origin, so you compromise and say it steps along a direction that's perpendicular to the vector connecting the origin to its present location - sort of half way between walking backwards and walking forwards. By Pythagoras's theorem, the average distance will then be d = sqrt(1^2+1^2) = sqrt(2) for N=2.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Likewise for N=3, stepping in a normal direction gives d = sqrt( sqrt(2)^2 + 1^2 ) = sqrt(3), so in general d = sqrt(N).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This seems to work in dimensions 2 or higher.&lt;/p&gt;&#xA;" OwnerUserId="4500" LastActivityDate="2011-07-15T14:27:23.657" Title="Is this geometrical 'derivation' of Brownian motion legitimate?" Tags="&lt;statistical-mechanics&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="12298" PostTypeId="2" ParentId="12297" CreationDate="2011-07-15T14:27:23.657" Score="1" ViewCount="" Body="&lt;p&gt;As a heuristic description, this is exactly right and correctly captures the essence of the subject. To turn it from a heuristic &quot;derivation&quot; into an actual derivation, you just need to make precise the notion that the two vectors are perpendicular on average. The precise meaning of &quot;perpendicular on average&quot; that's useful in this context is that the dot product is zero on average. That is, if $\vec r_n$ is the position vector after $n$ steps, and $\vec s_n$ is the vector representing the $n$th step, then&#xA;$$&#xD;&#xA;\langle \vec r_n\cdot\vec s_{n+1}\rangle=0.&#xD;&#xA;$$&#xA;The angle brackets mean an ensemble average -- that is, an average over many trials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This statement is true -- the easiest way to prove it is that the probability distribution for $\vec s_{n+1}$ is symmetric about 0, so positive and negative contributions to the dot product occur equally. And it's sufficient to prove the standard formula. Since $\vec r_{n+1}=\vec r_n+\vec s_{n+1}$, &#xA;$$&#xD;&#xA;r_{n+1}^2=r_n^2+s_{n+1}^2+2\vec r_n\cdot\vec s_{n+1}.&#xD;&#xA;$$&#xA;In the ensemble average, the last term vanishes, so $r^2$ increases, on average, by $s^2$ (i.e., by 1 for unit steps) on each step.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-07-15T14:27:23.657" CommentCount="1" />
  <row Id="12299" PostTypeId="2" ParentId="12078" CreationDate="2011-07-15T14:35:26.077" Score="0" ViewCount="" Body="&lt;p&gt;There are functions with singularities that arise all the time in physics. For example, an undamped (no friction) harmonic oscillator of mass $m$ driven by $F(t)=F_0\sin \omega t$ will have infinite amplitude at the resonance frequency, for the amplitude is given by&#xA;$$&#xD;&#xA;A(\omega)=\frac{F_0/m}{\omega^2-\omega_0^2},&#xD;&#xA;$$&#xA;where $\omega_0=\sqrt{k/m}$ is the resonance frequency. Therefore, $\omega=\omega_0$ is a singularity of $A(\omega)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's true that there are no undamped oscillators in nature, and the above equation is modified:&#xA;$$&#xD;&#xA;A(\omega)=\frac{F_0/m}{\sqrt{(\omega^2-\omega_0^2)^2+\left(\frac{b\omega}{m}\right)^2}},&#xD;&#xA;$$&#xA;where $b$ is a constant that comes in with the friction term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when, for example, mechanics build suspension systems for cars, they make sure that the typical frequency of oscillation of the system is far away from the resonance frequency, so that the car does not start oscillating violently when the driver goes over a bump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example (this involves no idealization) comes from quantum field theory, and it is that physical masses of particles appear as poles of the Fourier-transformed two-point function of the field corresponding to the particle. So there is a function there that has a pole (singularity) when its variable becomes equal to the mass of the physical particle.&lt;/p&gt;&#xA;" OwnerUserId="4483" LastActivityDate="2011-07-15T14:35:26.077" />
  <row Id="12301" PostTypeId="1" AcceptedAnswerId="12304" CreationDate="2011-07-15T15:14:51.220" Score="7" ViewCount="118" Body="&lt;p&gt;I was reading an article regarding the Shuttle's GPCs and how they stack up against commercially-made hardware on &lt;a href=&quot;http://www.nasa.gov/mission_pages/shuttle/flyout/flyfeature_shuttlecomputers.html&quot; rel=&quot;nofollow&quot;&gt;http://www.nasa.gov/mission_pages/shuttle/flyout/flyfeature_shuttlecomputers.html&lt;/a&gt; and was finding it quite interesting.  One thing that caught my attention was that they have a lot of issues caused by radiation with their off-the-shelf laptops (IBM/Lenovo Thinkpads), with the memory in each machine typically getting fried 2-3 times per mission (and a lot more on Hubble missions), with a high probability that it will happen while crossing the South Atlantic Anomaly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That got me thinking.  The Thinkpads, like most off the shelf computers, are plastic.  Sure, they have RF shielding, but that exists mostly to appease regulators regarding RF emissions and would offer little to no radiation protection.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are laptops on the market with metal cases (most notably the Apple laptops, made from a milled block of aluminium).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would an off-the-shelf machine in a metal casing hold up better in space compared to an otherwise identical machine in a plastic case?  I understand some plastics can be quite effective radiation blockers, and metals can actually make the situation worse (though I'm by no means an expert on nuclear physics, I'm just interested).  Are there any cases on record of a matal-clad laptop going into space, and if so how did it hold up compared to the plastic-clad thinkpads? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies if this is the wrong SE site to be asking this on, but while it is to do with computers I think the answer to this question lies in physics so it seems to me that this is the correct place to be asking a question like this.  &lt;/p&gt;&#xA;" OwnerUserId="4502" LastActivityDate="2011-07-15T17:59:36.280" Title="Laptops in space" Tags="&lt;space&gt;&lt;radiation&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="12302" PostTypeId="2" ParentId="3125" CreationDate="2011-07-15T15:56:54.307" Score="0" ViewCount="" Body="&lt;p&gt;I'm here because of a similar question. Can you kiss your reflection in a mirror? I am thinking you cannot, because once your lips touch the mirror, the reflection of your lips disappears. Am I right?&lt;/p&gt;&#xA;" OwnerUserId="4504" LastActivityDate="2011-07-15T15:56:54.307" CommentCount="1" />
  <row Id="12303" PostTypeId="2" ParentId="12078" CreationDate="2011-07-15T16:04:32.703" Score="0" ViewCount="" Body="&lt;p&gt;I can recall solving -1D differential equations in the 70's using the Galerkin method. If solutions that have singularities are present it is possible to incorporate the form of the singularity into one or more basis functions. Of course the weight function has to be choosen carefully, so that for instance all neccessary integrals are finite (usually the intgral of two basis functions times the weight function). As long as the positions and form of the singularities are known beforehand this seemed to work well.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-07-15T16:04:32.703" />
  <row Id="12304" PostTypeId="2" ParentId="12301" CreationDate="2011-07-15T16:18:23.457" Score="4" ViewCount="" Body="&lt;p&gt;The bad news: Space radiation is much harsher compared to boring gamma rays from our primitive nuclear reactors. Space radiation has much higher energy levels, and you cannot completely shield it, even with 10 meters of lead (which is in fact not very effective for neutrons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The good news is that an individual gamma photon, for example, usually would not introduce memory error by itself, even when going right through a memory cell. The real problem is boron. If the stable isotope &lt;sup&gt;10&lt;/sup&gt;B catches a neutron from space radiation, it will decay into lithium and an alpha particle (and a “harmless” gamma photon). This alpha particle is the one causing soft errors, because it’s heavy and causes huge damage to semiconductor when passing though it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is to only use &lt;sup&gt;11&lt;/sup&gt;B in ICs. (Boron is used in packaging and insulation layers on the ASICs.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem, which in some cases needs special ASIC designs (for special prices, of course) is photocurrent - gamma rays are like light, and will generate electricity right inside the silicon chip (like in a solar cell), which might randomly turn transistors on. In some cases, two randomly turned on transistors might short the power rail to ground, and the chip will burn to ashes (= latchup; usually happens with ‘parasitic’ BJT transistors in MOSFETs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, a common solution is using ECC memory everywhere, which can autofix errors on-the-go.&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="4438" LastEditorDisplayName="" LastEditDate="2011-07-15T17:59:36.280" LastActivityDate="2011-07-15T17:59:36.280" CommentCount="4" />
  <row Id="12305" PostTypeId="1" CreationDate="2011-07-15T16:45:52.307" Score="0" ViewCount="77" Body="&lt;p&gt;I was shown a toy, yesterday, which I would like help understanding qualitatively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fellow engineer showed me a kit which included three main parts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) A base (black box), approximately 4 inches square and 2 inches high, weighing about 2 pounds (quite heavy for its size).  No batteries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) A rigid plexiglass window, also 4 inches square, which rested nicely on top of the base. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) A  top.  This object was about 1.5 inches tall.  a flat metal washer (ID = about 0.25&quot;, OD = about 1.0&quot;) was mounted on the top-side of the top, and a plastic washer slid down over the top post of the top, to keep the metal washer securely in place.  The owner said that the system was designed like that so that multiple metal washers could be used. (We only used one.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OPERATION:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) Place the top in the very center of the plexiglass and give it a good spin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.)  With the top spinning in the very center of the plexiglass, gently and slowly lift the plexiglass about 1 inch off of the base.  The top will levitate up off the plexiglass about 4 inches, and hang suspended for about 1/2 a second before it falls off to a side (as if it were sliding off of an invisible 4-inch-tall and 4-inch radius hemisphere.  EDIT:  We must have been doing something wrong.  I found a video of this top staying levitated for almost 2 minutes:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;www.fatbraintoys.com/toy_companies/fascinations/levitron_platinum_pro_combo.cfm?display=video&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.)  I noticed that on one of my trial runs, the washers flew off the top AFTER it had levitated and BEFORE it had fallen to the desk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QUESTIONS:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) (My main question) How does this thing work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.)  Anyone know where this kit (and/or other like-minded novelties) are sold?  (I tried &quot;Think Geek,&quot; but was disappointed in their selection of mag-lev products.)  EDIT:  Found a very similar toy at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;www.fatbraintoys.com/toy_companies/fascinations/levitron_platinum_pro_combo.cfm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-07-15T17:57:14.077" LastActivityDate="2011-07-15T19:18:03.220" Title="Help understanding a Magnetic Levitation &quot;Physics Toy&quot;" Tags="&lt;electromagnetism&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12306" PostTypeId="2" ParentId="9776" CreationDate="2011-07-15T17:24:34.830" Score="-1" ViewCount="" Body="&lt;p&gt;Why do your students need so many examples? Are they really dumb? By reading a standard physics text (I read Physics book written by Resnick/Halliday/Walker, illustrated version) even an average student like me, could fairly easily grasped Doppler's effect! Good luck!&lt;/p&gt;&#xA;" OwnerUserId="3636" LastActivityDate="2011-07-15T17:24:34.830" CommunityOwnedDate="2011-07-15T17:24:34.830" CommentCount="2" />
  <row Id="12307" PostTypeId="1" AcceptedAnswerId="12747" CreationDate="2011-07-15T17:53:49.577" Score="4" ViewCount="136" Body="&lt;p&gt;I wonder, are there any fundamental issues leading to reduced performance of Peltier elements at cryogenic temperatures (-100C and lower)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is theoretical/practical minimum temperature achievable by a cascade of Peltier elements, provided that each next element have about 3 times less power than previous in cascade, so that they are not overwhelmed by selfheating? Let's say first element is water-cooled down to 20C when dissipating 150W.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-07-25T17:26:30.557" Title="Minimal temperature achievable by vanilla Peltier element?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" />
  <row Id="12308" PostTypeId="1" CreationDate="2011-07-15T18:10:26.120" Score="1" ViewCount="22" Body="&lt;p&gt;What is the relation between electric power and the source's internal resistivity? (I'm not sure that's the correct English term.) In my specific problem, I am given the following values and am asked to find the source's internal resistivity, $r.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{P}{P_t} = \frac{1}{2}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R = 2 Ω$&lt;/p&gt;&#xA;" OwnerUserId="4457" LastActivityDate="2011-07-15T18:10:26.120" ClosedDate="2011-07-15T18:13:04.120" Title="Relations between power and a source's resistivity" Tags="&lt;homework&gt;&lt;electricity&gt;" CommentCount="1" />
  <row Id="12309" PostTypeId="1" AcceptedAnswerId="12316" CreationDate="2011-07-15T19:09:02.540" Score="2" ViewCount="159" Body="&lt;p&gt;I know that the electric field outside of a capacitor is 0 and I know it is easy to calculate using Gauss's law. We create cylindrical envelope that holds the same amount of charges (of opposite signs) in each plate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why can't I pick an envelope which include only part of one of the plate? Guess's law states, specifically, that I can pick any envelope I want.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have encountered this question a couple of years a go and I got an answer which I was not completely happy with. &lt;/p&gt;&#xA;" OwnerUserId="3677" LastActivityDate="2011-07-16T06:55:39.627" Title="Electric field outside a capacitor" Tags="&lt;electrostatics&gt;" AnswerCount="2" />
  <row Id="12310" PostTypeId="2" ParentId="12305" CreationDate="2011-07-15T19:18:03.220" Score="1" ViewCount="" Body="&lt;p&gt;This gadget (or maybe one of those) is sold by the name &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Levitron &lt;/p&gt;&#xA;&#xA;&lt;p&gt;here: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.levitron.com/physics.html&quot; rel=&quot;nofollow&quot;&gt;http://www.levitron.com/physics.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You find an explanation. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-07-15T19:18:03.220" CommentCount="2" />
  <row Id="12311" PostTypeId="2" ParentId="12309" CreationDate="2011-07-15T19:23:48.490" Score="4" ViewCount="" Body="&lt;p&gt;You can pick such a box.  However, this box alone does not allow you to determine the electric field.  Gauss' law states that the total flux is equal to the charge enclosed times $4\pi$.  It doesn't tell you where that flux leaves the box.  There could be a lot here, none there, a negative amount at a third place, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some situations, symmetry allows us to use Gauss' law to find the electric field.  A Gaussian box enclosing both plates of a parallel-plate capacitor is symmetric with respect to a reflection through a plane through the middle of the capacitor.  It is also symmetric with respect to rotations about an axis perpendicular to the plates (ignoring edge effects).  The rotational symmetry lets us say that the electric field first points along the axis perpendicular to the plates.  The reflection symmetry tells us that the electric field must be the same through both sides of the box parallel to the plates.  That electric field must be zero because the box has no net charge in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A box around a single plate does not have the same symmetries.  It still has the rotational symmetry, so the electric field must be perpendicular to the plates.  However, we don't have the reflection symmetry any more, so the electric field can be different strengths on different sides of the box.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know there must be net flux through the box because the box encloses net charge.   Using the fact  that the electric field is zero outside the capacitor, we can deduce the he flux through a box that encloses only one plate is all through the side of the box that's inside the capacitor.  Hence, the electric field must be $4\pi\rho$ inside the capacitor.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-07-15T19:37:40.817" LastActivityDate="2011-07-15T19:37:40.817" CommentCount="5" />
  <row Id="12313" PostTypeId="1" CreationDate="2011-07-15T20:35:09.997" Score="1" ViewCount="80" Body="&lt;p&gt;In optics one of &quot;diffraction-limited&quot; criteria is wavefront tolerance: a textbook example is is optical system with 1/4 wavelength limits producing image of a point source with 68% of the energy contained in the airy disk. Some optics books, however introduce a &quot;roughness&quot; idea, implying that certain mirror surface pattern would imply light scattering, and supporting this idea with naive ray-trace diagram. Does it make any sense, I would think one can made quite strong assertion, namely that &lt;em&gt;any mirror surface shape within 1/8 wavelength tolerance of paraboloid would produce an of a point source with 68% of the energy contained in the airy disk area&lt;/em&gt;. Any proofs/refutations?&lt;/p&gt;&#xA;" OwnerUserId="4507" LastActivityDate="2011-07-15T23:59:34.477" Title="Mirror &quot;roughness&quot;" Tags="&lt;optics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="12314" PostTypeId="1" AcceptedAnswerId="12970" CreationDate="2011-07-15T21:17:53.700" Score="8" ViewCount="217" Body="&lt;p&gt;The &quot;Ozma problem&quot; was coined by Martin Gardner in his book &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Ambidextrous_Universe&quot; rel=&quot;nofollow&quot;&gt;&quot;The Ambidextrous Universe&quot;&lt;/a&gt;, based on &lt;a href=&quot;http://en.wikipedia.org/wiki/Project_Ozma&quot; rel=&quot;nofollow&quot;&gt;Project Ozma&lt;/a&gt;. Gardner claims that the problem of explaining the humans left-right convention would arise if we enter into communication (by radio waves, no images allowed) with life on another planet. We can ask the aliens to perform any experiment they want. It is claimed that the classical experiments with magnets, electrical currents, light polarization, gyroscopes etc. can't solve the problem. The only simple experiment that solves the problem is the beta decay in which the parity is not conserved and this can be used to distinguish humans left-right spatial relations. But all this are quite old, the parity violation in weak interactions was experimentally proved in 50s by Chien-Shiung Wu. I was wondering if something has changed since then and if there are some other experiments (maybe some thought experiments) and theories that can solve the Ozma Problem.&lt;/p&gt;&#xA;" OwnerUserId="3066" LastEditorUserId="3066" LastEditorDisplayName="" LastEditDate="2011-07-15T21:35:11.147" LastActivityDate="2011-08-23T06:48:00.940" Title="The Ozma Problem" Tags="&lt;symmetry&gt;&lt;parity&gt;&lt;cp-violation&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" />
  <row Id="12315" PostTypeId="1" AcceptedAnswerId="12317" CreationDate="2011-07-15T21:22:27.260" Score="2" ViewCount="81" Body="&lt;p&gt;I noticed this one day, a lightning/thunder occurred and my Fabulosa Spanish music died for a second. &#xA;But not FM? Thank You&lt;/p&gt;&#xA;" OwnerUserId="4429" LastActivityDate="2011-07-15T21:54:36.003" Title="Why is it that when driving in a car, and a lightning bolt strikes, my AM radio gets cut off for a while, but FM stays on" Tags="&lt;radio&gt;&lt;lightning&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12316" PostTypeId="2" ParentId="12309" CreationDate="2011-07-15T21:48:16.503" Score="2" ViewCount="" Body="&lt;p&gt;Outside two infinite parallel plates with opposite charge the electric field is zero, and that can be proved with Gauss's law using any possible Gaussian surface imaginable. However, it might be extremely hard to show if you don't choose the Gaussian surface in a smart way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual way you'd show that the electric field outside an infinite parallel-plate capacitor is zero, is by using the fact (derived using Gauss's law) that the electric field above an infinite plate, lying in the $xy$-plane for example, is given by&#xA;$$&#xD;&#xA;\vec{E}_1=\frac{\sigma}{2\epsilon_0}\hat{k}&#xD;&#xA;$$&#xA;where $\sigma$ is the surface charge density of the plate. If you now put another plate with opposite charge, i.e. opposite $\sigma$, some distance below or above the first one, then that contributes its own electric field,&#xA;$$&#xD;&#xA;\vec{E}_2=-\frac{\sigma}{2\epsilon_0}\hat{k}&#xD;&#xA;$$&#xA;in the region above it. Since the electric field obeys the principle of superposition, the net electric field above both plates is zero. The same happens below both plates, while between the plates the electric field is constant and nonzero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your way of doing it is a little more tricky, but again gives the same answer. For example, if you choose the Gaussian surface to have an hourglass shape with different radii for the two sides, then indeed the net charge enclosed is not zero. However, when you calculate the total electric flux through that surface, you have to be careful to realize that there is nonzero electric field between the two plates, and therefore there is a nonzero flux through the part of the Gaussian surface that lies between the plates. That flux, of course, has to be accounted for. Assuming that you know the electric field inside the capacitor, $\vec{E}_\text{inside}$, you can do the integral $\oint\vec{E}_\text{inside}\cdot d\vec{A}$ for such a Gaussian surface (it's not that hard actually), and you find that the flux through the part of the surface that lies between the plates is exactly equal to $q_{\text{enclosed}}/\epsilon_0$. Thus, the net flux through the part of the Gaussian surface that lies outside the plates has to be zero, proving, after a little thought, that the electric field outside the capacitor is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final answer for $\vec{E}$ never depends on the Gaussian surface used, but the way to get to it always does. That's why the Gaussian surface has to be chosen in a smart way, i.e. in a way that makes the calculation of $\oint\vec{E}\cdot d\vec{A}$ easy.&lt;/p&gt;&#xA;" OwnerUserId="4483" LastEditorUserId="4483" LastEditorDisplayName="" LastEditDate="2011-07-16T06:55:39.627" LastActivityDate="2011-07-16T06:55:39.627" />
  <row Id="12317" PostTypeId="2" ParentId="12315" CreationDate="2011-07-15T21:54:36.003" Score="7" ViewCount="" Body="&lt;p&gt;AM radio typically transmits at around 1 MHz, FM radio at about 90 MHz.  Measurements of the RF spectrum of lightning strikes show a falloff with frequency of about 20 dB per decade in that frequency range, so with FM about 2 decades above AM, you'd expect AM to have about 40dB higher interference from a lightning strike.  In addition to that, FM signals attenuate faster with range, so depending on your distance from the lightning strike the effective AM/FM interference ratio could be even larger.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-07-15T21:54:36.003" CommentCount="1" />
  <row Id="12318" PostTypeId="2" ParentId="12313" CreationDate="2011-07-15T23:48:02.427" Score="1" ViewCount="" Body="&lt;p&gt;No rigorous proof, but this is the experience of people making telescope optics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Low-amplitude roughness does matter. This is called &quot;dog biscuit&quot; or &quot;orange peel&quot; or &quot;ripple&quot;, depending on the particular variety. You can see it in the Foucault tester when checking the mirror during polishing, the shadows look rough, instead of smooth. In theory, a rippled mirror may still be a λ/4-compliant device. In practice, when you turn it towards the sky, the &quot;dog biscuit&quot; produces less sharp images, you never get the crisp tiny Airy disk at the eyepiece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure why that happens, but I've a few ideas. Imagine the surface of the mirror, instead of flat or paraboloidal, it's a sawtooth shape. Each &quot;tooth&quot; is less than or equal to λ/4, but each one of them, within that limit, is very sharp, maybe a 45 deg at the top. What happens to the wavefront hitting that surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The naive answer is that it takes the same sawtooth shape. However, the size of the sawtooth cell pattern is less than λ/4, so it should not affect the wavefront too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe some low-grade dispersion happens, maybe the wavefront gets muddled a little bit. Honestly, I don't know what's the causal explanation, only the practical reality at the eyepiece; an &quot;orange peel&quot; paraboloid never performs as well as a silky smooth one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Opticians are not relying so much on the Foucault test anymore. You could have a mirror that tests down to λ/12 on Foucault, yet it's ripply and rough. A different test has been created, called the &lt;a href=&quot;http://www.telescope-optics.net/Strehl.htm&quot; rel=&quot;nofollow&quot;&gt;Strehl test&lt;/a&gt; which is simply a measure of how much the diffraction figure deviates from ideal. The reason why some light misses the Airy is irrelevant to this test, all that matters is how big the difference is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An orange peel mirror that tests at λ/12 on Foucault may indeed produce a weaker number on the Strehl tester than a λ/8 mirror that is free of microripple. A weaker Strehl number tends to correlate better with poorer image at the eyepiece than a weaker Foucault number. Yet even Strehl is not perfect. The ultimate tester is the human eye in real observing conditions.&lt;/p&gt;&#xA;" OwnerUserId="4210" LastEditorUserId="4210" LastEditorDisplayName="" LastEditDate="2011-07-15T23:59:34.477" LastActivityDate="2011-07-15T23:59:34.477" CommentCount="1" />
  <row Id="12319" PostTypeId="2" ParentId="12282" CreationDate="2011-07-16T00:04:15.377" Score="0" ViewCount="" Body="&lt;p&gt;The weight &lt;strong&gt;is&lt;/strong&gt; caused by gravity. So, yes, gravity works for you if you turn the bottle upside-down.&lt;/p&gt;&#xA;" OwnerUserId="4210" LastActivityDate="2011-07-16T00:04:15.377" CommentCount="2" />
  <row Id="12320" PostTypeId="1" CreationDate="2011-07-16T02:48:34.623" Score="1" ViewCount="32" Body="&lt;p&gt;I am wondering in Homodyne detector structure with  a beamsplitter, why phase of input beam to beamsplitter can be conjugated?  I mean for example for an input beam operator &lt;strong&gt;b&lt;/strong&gt; onto the beamsplitter at homodyne detector output we will have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ y=j \left[ b \exp(i \phi) - b \exp(-i \phi) \right] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am concerned to know why either incident beam phase or phi can be conjugated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any comment regarding this concept would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3977" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-16T05:04:55.353" LastActivityDate="2011-07-16T05:04:55.353" Title="Quadrature Separation in Homodyne Detector" Tags="&lt;optics&gt;" CommentCount="4" />
  <row Id="12321" PostTypeId="1" AcceptedAnswerId="12323" CreationDate="2011-07-16T05:42:42.210" Score="3" ViewCount="87" Body="&lt;p&gt;Rubber-bands are simple, yet very useful. Old rubber bands(5 years?) get brittle? Why is that?&lt;/p&gt;&#xA;" OwnerUserId="4429" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-16T16:00:34.020" LastActivityDate="2011-07-16T16:41:50.907" Title="Is there any way to increase a rubber-bands lifetime?" Tags="&lt;material-science&gt;&lt;elasticity&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12322" PostTypeId="1" CreationDate="2011-07-16T08:40:03.290" Score="3" ViewCount="112" Body="&lt;p&gt;What is the 'Gelfand-Yaglom' Theorem?  I have heard that it is used to calculate Functional determinants by solivng an initial value problem of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Hy(x)-zy(x)=0$ with $y(0)=0$ and $y&amp;#39;(0)=1$.  Here $H$ is the Hamiltonian and $z$ is a real parameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it that simple?  If $H$ is a Hamiltonian, could I use the WKB approximation to solve the initial value problem and to be valid for $z$ big?&lt;/p&gt;&#xA;" OwnerUserId="4435" LastEditorUserId="3970" LastEditorDisplayName="" LastEditDate="2011-07-17T04:19:22.797" LastActivityDate="2011-08-01T12:55:43.797" Title="Gelfand-Yaglom theorem for functional determinants" Tags="&lt;mathematical-physics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12323" PostTypeId="2" ParentId="12321" CreationDate="2011-07-16T10:29:10.327" Score="2" ViewCount="" Body="&lt;p&gt;If a rubber band is usable for 5 years, it's a very good one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is not much You can do, just store them in a dark and cool place. &#xA;Light plus oxygen is the most dangerous enemy of rubber. Nothing can be done in use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest influence has the producer, because he can choose &#xA;a persistent (expensive) kind of rubber (eg Kalrez®, ask for prizes!)&#xA;Of course there are less extreme rubbers in prize. &#xA;Another way is the color, adding carbon black prolonges life of &#xA;rubber substantially. (Protection from light)  and to add some &#xA;antioxidants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The background of embrittlement of polybutadiene (polyisoprene) based &#xA;rubbers is crosslinking by radical chain reactions with oxygen. &#xA;Cheap urethane rubbers die from similar reactions, but those usually &#xA;become sticky, even liquid by this processes. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-07-16T16:41:50.907" LastActivityDate="2011-07-16T16:41:50.907" CommentCount="3" />
  <row Id="12324" PostTypeId="1" CreationDate="2011-07-16T11:16:53.273" Score="4" ViewCount="182" Body="&lt;p&gt;I have always written the commutation rules of quantum theory as ,&#xA;$[q,p] = i\hbar\delta _{ij}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But seems that some people write this as,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[q^i,v_j]= \frac{i\hbar}{M}\delta^i _{j}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(..this is often done in the context of taking the Galilian group limit of the Poincare group…though I am not sure which aspect of it does it emphasize-- the non-relativstic aspect or the non-quantum aspect?..) &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;But somehow dimensionally the second form doesn't look okay.&#xA;Am I missing something?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the same strain, it seems that the operators for &quot;finite boost by $v$ &quot; and is done by the operator $exp(\frac{iK.v}{\hbar})$ and the &quot;finite translation by $q$&quot; is effected by the operator $exp(\frac{iMv.q}{\hbar})$. (..where $v$, $q$ and $K$ are all $3-vectors$..)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I would like to know how the  above is rationalized. &#xA;To ask again - is the above taking just the non-relativistic limit or is it also a non-quantum limit ? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1637" LastEditorUserId="1637" LastEditorDisplayName="" LastEditDate="2011-07-16T20:48:16.340" LastActivityDate="2011-07-17T06:25:26.563" Title="Commutation of operators in quantum theory" Tags="&lt;quantum-mechanics&gt;&lt;special-relativity&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="12325" PostTypeId="2" ParentId="12286" CreationDate="2011-07-16T12:48:28.493" Score="1" ViewCount="" Body="&lt;p&gt;I am dismayed that Greg's answer continues to get upvotes because I think it is badly off target. I would like to note first that quantum entanglement experiments are not done with electrons but rather with photons. So I have taken the liberty of translating Bob's article word-for-word into the language of photons. It is often said that the case of entangled photons corresponds perfectly to the case of electrons except that all the angles are divided by 2: instead of perfect anti-correlation at 180 degrees, it becomes perfect anti-correlation at 90 degrees, etc. I hope I have transformed Greg's argument correctly; and I believe that it becomes transparently nonsensical when put in these terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is how it goes: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Quantum entanglement is different from the &quot;classical entanglement&quot; in the following way:&#xA;In your example, each ball has only one property of interest, namely &quot;color ∈￼ {white, black}&quot;. &#xA;In the traditional examples of quantum entanglement, each ball (particle) has two properties of interest, namely &quot;ability to penetrate a vertical polarizer&quot; and &quot;ability to penetrate a 45 degree offset polarizer&quot;. Moreover, the properties are complementary, i.e. you can't actually measure them simultaneously to arbitrary precision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Still, in the entangled state, each of these properties alone is perfectly (anti-)correlated: If Alice sees a particle penetrate her vertical polarizer, then Bob will always see a particle blocked.  And similarly if they both offset their polarizers 45 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The paradox, now, is the following:&#xA;Suppose that Alice measures sees a particle go through her vertical polarizer. But observer Bob offsets his polarizer 45 degrees and also sees a particle go through. Exhulted, Alice proclaims that she has managed to measure two complementary properties simultaneously! After all, her measurement showed her the particle had the ability to go through a vertical polarizer, while according to Bob's measurement her particle should with certainty be blocked by a 45 degree offset polarizer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Imagine her surprise, then, when she tries to confirm her conclusion by putting an offset polarizer in series with her vertical polarizer and finding that the particle which got through the first polarizer still gets through the second polarizer fifty percent of the time, even when Bob simultaneously detects a particle penetrating his offset polarizer.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope I have not done Greg's answer any injustice, but I believe this is what it leads to. (I should also mention that I totally disagree with the premise of starting off with colored balls as the classical &quot;straw man&quot;. But that's another question.)&lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-07-16T12:48:28.493" />
  <row Id="12326" PostTypeId="2" ParentId="7153" CreationDate="2011-07-16T13:23:44.327" Score="1" ViewCount="" Body="&lt;p&gt;I think I found a solution in a book &quot;Physics of Optoelectronic devices&quot; by S.L Chuang. It's written there that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\epsilon=\frac{a_{substrate}-a_{layer}}{a_{substrate}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It makes sense - the higher the difference between lattice constants, the higher the VB splitting. It's of order of several meV, which is also a reasonable value. I attach a plot in case someone has a similar problem in the future. Red lines represent an unstrained GaAsSb, LH level is plotted in green, HH and CB splitting in blue. It's interesting to see that the band gap is lower when the layer is stretched and higher when it's &quot;squeezed&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/EpmGw.png&quot; alt=&quot;Strained GaAsSb/InP&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1505" LastActivityDate="2011-07-16T13:23:44.327" />
  <row Id="12327" PostTypeId="2" ParentId="6263" CreationDate="2011-07-16T13:26:58.467" Score="0" ViewCount="" Body="&lt;p&gt;I have been reading a bit about III-V compounds since I posted this question and it seems that the binomial approximation is sufficient in most, if not all, cases.&lt;/p&gt;&#xA;" OwnerUserId="1505" LastActivityDate="2011-07-16T13:26:58.467" />
  <row Id="12328" PostTypeId="2" ParentId="12283" CreationDate="2011-07-16T15:28:29.803" Score="0" ViewCount="" Body="&lt;p&gt;The comments section has already pointed out that eddy currents induced in the spring and acted upon by the magnetic field gives rise to a Waltenhof pendulum type of effect. There is also a more dominant and basic effect which is that the base of the spring isn't fixed and so effectively you have two springs in series. For a spring with spring constant $k$ oscillating longitudinally with a mass m on one end $$\omega_0 = \sqrt{\frac k m}$$ For two springs connected in series $$\omega_0 = \sqrt{\frac{k_1k_2}{(k_1+k_2)m}}$$ Vibrating the spring transversely means the expressons for the frequency of oscillation are far more complicated, but the qualitative explanation you're looking for is the same - there is an effective spring connected in series.&lt;/p&gt;&#xA;" OwnerUserId="1345" LastActivityDate="2011-07-16T15:28:29.803" CommentCount="6" />
  <row Id="12329" PostTypeId="2" ParentId="12286" CreationDate="2011-07-16T17:15:48.937" Score="0" ViewCount="" Body="&lt;p&gt;One another difference that I think is crucial is that the classical analogy is not really probabilistic. When we repeat the experiment in the exactly the same way, the results in the classical analogy will be always the same. The measurement was made at the moment when Alice and Bob chose their balls, and that choice will be always the same if we repeat it in exactly the same way. Conversely, in quantum case the Alice's $S_x$ is not necessary the same, it is always +1 or -1 with equal probability.&lt;/p&gt;&#xA;" OwnerUserId="3066" LastActivityDate="2011-07-16T17:15:48.937" />
  <row Id="12330" PostTypeId="1" AcceptedAnswerId="12332" CreationDate="2011-07-16T18:12:23.003" Score="1" ViewCount="38" Body="&lt;p&gt;Let's say I have a molecular like CH3F (i.e. fluoromethane), and I'm able to measure the angle (theta) between the C-H bonds.  Provided (theta) what is the angle between the C-F bond and the C-H bonds? &lt;/p&gt;&#xA;" OwnerUserId="4512" LastActivityDate="2011-07-16T18:42:08.253" Title="Calculating angles for tetrahedral molecular geometry" Tags="&lt;geometry&gt;" AnswerCount="1" />
  <row Id="12331" PostTypeId="1" AcceptedAnswerId="12387" CreationDate="2011-07-16T18:40:23.667" Score="3" ViewCount="255" Body="&lt;p&gt;In Landau's Classical Theory of Fields, one finds the statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Every rotation in the four-dimensional space can be resolved into six&#xA;  rotations, in the planes $xy,zy,xz,tx,ty,tz$ (just as every rotation&#xA;  in ordinary space can be resolved into three rotations in the planes&#xA;  $xy,zy,xz$).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can I prove this statement? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3064" LastActivityDate="2011-07-20T22:20:23.973" Title="Rotations in space-time" Tags="&lt;special-relativity&gt;&lt;spacetime&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="12332" PostTypeId="2" ParentId="12330" CreationDate="2011-07-16T18:42:08.253" Score="1" ViewCount="" Body="&lt;p&gt;Use a coordinate system with the C at the origin and the F on the $z$ axis, and one of the H's in the $xz$ plane. Let $\alpha$ be the angle between the C-F and C-H bonds and $\theta$ be the angle between two C-H bonds. The Cartesian components of unit vectors in the directions of the 3 H's are&#xA;$$&#xD;&#xA;(\sin\alpha,0,\cos\alpha),(\sin\alpha\cos 2\pi/3,\sin\alpha\sin 2\pi/3,\cos\alpha),(\sin\alpha\cos 4\pi/3,\sin\alpha\sin 4\pi/3,\cos\alpha).&#xD;&#xA;$$&#xA;The dot product of any two of these should equal $\cos\theta$:&#xA;$$&#xD;&#xA;\cos\theta=\sin^2\alpha\cos 2\pi/3+\cos^2\alpha=-{1\over 2}\sin^2\alpha+\cos^2\alpha={3\over 2}\cos^2\alpha-{1\over 2}.&#xD;&#xA;$$&#xA;The solution is&#xA;$$&#xD;&#xA;\cos\alpha=\pm\sqrt{1+2\cos\theta\over 3}.&#xD;&#xA;$$&#xA;The negative root is the relevant one here, since we know $\alpha$ is obtuse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you happen to remember that the angle for methane is $\theta=\cos^{-1}(-1/3)$, you can check that this makes sense: the above formula gives $\alpha=\theta$ in that case, as it should.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-07-16T18:42:08.253" />
  <row Id="12333" PostTypeId="1" AcceptedAnswerId="12337" CreationDate="2011-07-16T18:43:19.617" Score="4" ViewCount="272" Body="&lt;p&gt;We all know the &quot;apples and oranges&quot; rule which says that it's meaningless to add or subtract two different quantities like apples and oranges. But the same rule doesn't hold for the multiplication and division. So my question is, why two quantities with different units cannot be added to one another but the multiplication of them is allowed and how this is expressed mathematically?&lt;/p&gt;&#xA;" OwnerUserId="3066" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-16T18:50:54.907" LastActivityDate="2011-07-20T03:08:12.367" Title="Addition of different physical quantities" Tags="&lt;mathematical-physics&gt;&lt;units&gt;" AnswerCount="10" />
  <row Id="12334" PostTypeId="1" AcceptedAnswerId="12340" CreationDate="2011-07-16T18:45:34.957" Score="3" ViewCount="85" Body="&lt;p&gt;When is a coincidence a coincidence? We know that to identify entangled photons, the electronics is set to look for simultaneous clicks at opposite detectors. The size of the window is to some degree arbitrary. What I am wondering is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you run a Bell-type entanglement experiment where you get, for example, 100 coincidences per second with a 100 picosecond interval, what happens if you double the window to 200 picoseconds? Do you expect, in general, to get 200 coincidences per second? I can think of three possible answers and I wonder which one is most true:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If your window is small enough, doubling the window doesn't matter because you are accurately identifying all available entangled particles. You still get 100 coincidences per second (or possible a few more for truly sporadic coincidence counts.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Double the window, double the coincidences. You get 200 per second.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Double the window and you get four times the coincidences: 100 entirely in the first half of the window, 100 in the seconde half, and 200 &quot;cross-coincidences&quot;. 400 per second in total.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I hope this question is meaningful and I wonder what the answer is.&lt;/p&gt;&#xA;" OwnerUserId="3253" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-16T18:46:45.147" LastActivityDate="2011-07-17T13:03:10.917" Title="Coincidence detectors in Bell tests: How close is close enough?" Tags="&lt;quantum-mechanics&gt;&lt;mechanics&gt;&lt;experimental-physics&gt;&lt;entanglement&gt;" AnswerCount="1" />
  <row Id="12335" PostTypeId="2" ParentId="12333" CreationDate="2011-07-16T19:03:58.533" Score="0" ViewCount="" Body="&lt;p&gt;Multiplication is possible, because you also multiply &lt;strong&gt;units of measurement&lt;/strong&gt;. In other words, if you multiply 2 apples by 3 oranges, what you get is not mere 6, but, &lt;strong&gt;6 orange-apples&lt;/strong&gt;. Likewise, if you divide 6 apples by 2 oranges, you get 3 apple/orange. This does really make sense as &lt;strong&gt;for each orange, there are 3 apples&lt;/strong&gt;. But when adding or subtracting numbers, you actually can't add or subtract &lt;strong&gt;units of measurement&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4499" LastActivityDate="2011-07-16T19:03:58.533" CommentCount="2" />
  <row Id="12336" PostTypeId="2" ParentId="12331" CreationDate="2011-07-16T19:28:54.810" Score="3" ViewCount="" Body="&lt;p&gt;Just to expand on my comment more, essentially what we are looking at is the &lt;a href=&quot;http://en.wikipedia.org/wiki/SO%284%29&quot; rel=&quot;nofollow&quot;&gt;special orthogonal group of order 4&lt;/a&gt;, often abbreviated SO(4). &quot;Special&quot; (also sometimes called normal) as it's determinate is 1, and &quot;orthogonal&quot; since lets have $M \in SO(k)$ then $MM^T=1$. (In special relativity it is actually SO(3,1), cause your looking at Lorentzian space not Euclidean space so you have a sign difference). With SO(K), you can rotate axis 1 into axes 2, 3,...,K. With axis 2, you can rotate it into 3,4,...K. Therefore, SO(K) has $(K-1)+(K-2)+...+2+1 = \frac{K(K-1)}{2}$ generators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively it sorta makes sense that any rotation can be decomposed this way, as, thinking of the 3 dimensional case, any rotation can be seen as rotating one way, and then rotating the other way (very specific, right? :P). 4 dimensions is just another layer on top of that, as you can rotate any of the other 3 dimensions into the 4th (called a boost if the 4th is time).  &lt;/p&gt;&#xA;" OwnerUserId="3018" LastEditorUserId="3018" LastEditorDisplayName="" LastEditDate="2011-07-16T19:42:56.013" LastActivityDate="2011-07-16T19:42:56.013" />
  <row Id="12337" PostTypeId="2" ParentId="12333" CreationDate="2011-07-16T19:47:20.883" Score="3" ViewCount="" Body="&lt;p&gt;I think I figured out how this can be expressed mathematically. The units can be thought as mathematical constants and physical quantities as numbers multiplied by that constants (units). So when we add two same quantities with same units we can add the numbers, for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2\hspace{0.2cm} apples + 5\hspace{0.2cm} apples = 2\cdot a +5\cdot a = (2+5)\cdot a=7\cdot a = 7\hspace{0.2cm} apples $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for different quantities we can't add the numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2\hspace{0.2cm} apples + 5\hspace{0.2cm} oranges = 2\cdot a +5\cdot o $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore we can't write $2\hspace{0.2cm} apples + 5\hspace{0.2cm} oranges = 7\hspace{0.2cm} apples + oranges$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However we can multiply (and divide) them, multiplying the numbers and the units:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2\hspace{0.2cm} apples \cdot 5\hspace{0.2cm} oranges = (2\cdot a) \cdot (5\cdot o) = (2\ \cdot 5) \cdot (a\cdot o) =10 \cdot (a\cdot o)=10\hspace{0.2cm} apples \cdot oranges$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to work.&lt;/p&gt;&#xA;" OwnerUserId="3066" LastEditorUserId="3066" LastEditorDisplayName="" LastEditDate="2011-07-16T19:53:36.470" LastActivityDate="2011-07-16T19:53:36.470" CommentCount="15" />
  <row Id="12338" PostTypeId="2" ParentId="3618" CreationDate="2011-07-16T20:10:24.587" Score="1" ViewCount="" Body="&lt;p&gt;Not a direct answer to your question but still a surprising derivation of Maxwells equations: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://scholar.google.de/scholar?q=Feynman%E2%80%99s%20proof%20of%20the%20Maxwell%20equations&amp;amp;um=1&amp;amp;ie=UTF-8&amp;amp;sa=N&amp;amp;hl=de&amp;amp;tab=ws&quot; rel=&quot;nofollow&quot;&gt;Feynman's proof of the Maxwell equations&lt;/a&gt; (FJ Dyson - Phys. Rev. A, 1989) shows, that it is possible to derive Maxwells equations from  Newtons second law of motion and the uncertainty principle.&lt;/p&gt;&#xA;" OwnerUserId="1648" LastActivityDate="2011-07-16T20:10:24.587" />
  <row Id="12340" PostTypeId="2" ParentId="12334" CreationDate="2011-07-16T20:51:32.337" Score="2" ViewCount="" Body="&lt;p&gt;About $\pm 1$ nanosecond. The data can be visualized by plotting the difference between times of events at the two ends of the Bell experiment. The data from Gregor Weihs's experiment, Phys.Rev.Lett. 81, 5031 (1998), results in the attached graph (from an embedded postscript file, a much clearer PDF version is &lt;a href=&quot;http://pantheon.yale.edu/~pwm22/longdist34-Alice+-3ns.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;), which shows a very distinct drop off in coincidence outside a small range of time differences.&lt;img src=&quot;http://i.stack.imgur.com/bDSth.png&quot; alt=&quot;Time difference from each of 'Alice's events to the nearest 'Bob's event&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph shows the time difference from each of 'Alice's events to the nearest 'Bob's event, with only events within $\pm 3$ nanoseconds shown. In the ten second time-slot shown, that's 14841 of a total of 388660 Alice events. In a very few cases, there are more than one Bob events within 3 nanoseconds of the marked Alice event. As you see, increasing the width of the coincidence window above $\pm 1$ nanosecond will have relatively little impact on the numbers of coincidences. The different graphic symbols indicate Alice's experimental result for the particular Alice event without showing Bob's experimental result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I regret that I don't have comparable data for a different experiment, so I can't tell you whether different experimental apparatuses show different dependencies on the coincidence window width.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: As a response to the comments below, a different graph, done long enough ago that I'm not sure that it's from the same data, but I'm pretty certain that the features are generic.&lt;img src=&quot;http://i.stack.imgur.com/5KU5z.png&quot; alt=&quot;8192 time differences&quot;&gt; These graphs (the 2nd and 3rd graphs are just the first graph enlarged at the origin) plot the $i$'th &lt;em&gt;sorted&lt;/em&gt; time difference between Alice's TOEs (Times Of Events) and the closest Bob's TOE for 8192 Alice events. The sorting of CTDs  (Closest TOE differences) makes the graph monotonely increasing. In this case, there are just 8192 Alice events instead of the 388660 Alice events in the full &quot;&lt;em&gt;longdist34&lt;/em&gt; &quot; dataset. Looking at the second graph, there are just less than 400 Alice events for which the corresponding CTD is less than about 0.1 microseconds; in the 3rd graph, there are 135 events with CTDs less than 1 nanosecond (the jumps are because at that time I was working with CTDs with 1 nanosecond resolution, whereas I later discovered that there is more resolution in the data; the resolution and systematic variations of the two clocks' time stamps at the times of events (which are what are recorded in the Alice and Bob datasets) are definitely issues to understand, however I ignore them here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first graph, one can see an almost linear rise from 350 events until, say, about 5000 events, then there are relatively fewer events that have longer time differences. The cutoff is inevitable, because Bob's events are rarely separated by long time intervals, so that the largest CTD is relatively unlikely to be longer than whatever the average time separation is between Bob's events. In this case, the longest CTD for 8192 Alice events is about 0.14 milliseconds. I think there's an answer in these graphs to your supplementary questions in your Comments below (albeit a graphical explanation, not what I'd call a mathematical explanation).&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-07-17T13:03:10.917" LastActivityDate="2011-07-17T13:03:10.917" CommentCount="12" />
  <row Id="12341" PostTypeId="1" AcceptedAnswerId="12364" CreationDate="2011-07-16T21:38:47.883" Score="8" ViewCount="256" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;One can define the Poincare group as the group of isometries of the Minkowski space. Is its Lie algebra given either by the equations 2.4.12 to 2.4.14 (..as also given in this page - &lt;a href=&quot;http://en.wikipedia.org/wiki&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki&lt;/a&gt;/Poincaré_group..) or equations 2.4.18 to 2.4.24 of Weinberg's volume 1 of his QFT books?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What confuses me is that in deriving the commutation relations between $J^{\mu \nu}$ and $P^\mu$ he did use quantum theoretic arguments about the Hilbert space operator $U$ but I guess there is nothing quantum about the Lie algebra he derives in the aforementioned equations. Is that right? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This quantum confusion steepens when one looks at the $K_i$ (..relatvistic boost along the $i^{th}$ spatial direction..), $P_j$ (..linear momentum along the $j^{th}$ spatial direction..) commutator being non-zero. This is justified by saying that the exponential action of the boosts and the translations on the Hilbert space states do not commute and that is being reflected here. (..they pick up an extra phase proportional to the mass and the dot product of the boost velocity and the displacement vector..) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But if the afore mentioned equations are really the Lie algebra of the isometry group of the Minkowski spacetime then in the Galilean limit shouldn't they be instead reflecting the fact that Galilean boosts and translations when acting on the spacetime coordinates do infact commute. But the $K_i$ and $P_j$ commutation continues to be non-zero even when the Galilean limit is taken on page 62.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes me strongly suspicious that the equations 2.4.12 to 2.4.14 are not the Lie algebra of the isometry group of the Minkowski spacetime but are the Lie algebra of the group whose elements are $U(\Lambda, a)$ (..using Weinberg's notation..)…right? &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;So is the &quot;low velocity&quot; limit taken on page 62 recovering non-relativistic quantum theory ? (and not Newtonian physics)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On page 89 of the same book he derives the topology of the inhomogeneous Lorentz group as being $R^4 \times R^3 \times S^3/Z_2$. Since this is a connected manifold, I guess that by the term ``inhomogeneous Lorentz group&quot; he is meaning only the proper orthochronous component of the full relativistic symmetry group. right? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I can't see how the above topology matches with the semidirect product structure for posibly the same thing as given on this Wikipedia page -http://en.wikipedia.org/wiki/Poincaré_group ? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When people talk of the Poincare group is the full symmetry group of relativity is what is being referred to or is it just its proper orthochronous component (and not the other 3 components) ? &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I am familiar with the notion of &quot;central charge&quot; as in the &quot;first&quot; term on the RHS of the TT OPE of CFT…what also has the interpretation as the zero-point energy when doing the plane-minus-point&amp;lt;-&gt;cylinder conformal transformation.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In this light it is not clear to me as to what is meant when one says that one can add in the &quot;mass&quot; as a central charge to the Galilean group..an extra generator which commutes with all the rest so that with this &quot;central extension&quot; the free particles will lie in the unitary representations of the Galilean group rather than the projective representations before the extension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be grateful if someone can shed light on this issue and help reconcile the two &quot;different&quot; notions of central charge. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When one takes a &quot;low velocity&quot; limit of the Poincare algebra to get the Galilean algebra then is one taking just a non-reltivistic limit or is one also taking a non-quantum limit ? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(..I guess this will depend on my first query about whether what Weinberg calls as the Poincare algebra in the quoted equations has any quantum effect encoded in it {as it seems to be!} or is it just the Lie algebra of the isometry group of the Minkowski spacetime..) &lt;/p&gt;&#xA;" OwnerUserId="1637" LastEditorUserId="1637" LastEditorDisplayName="" LastEditDate="2011-07-17T13:43:04.403" LastActivityDate="2011-07-18T15:35:58.503" Title="Poincare group vs Galilean group" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;special-relativity&gt;&lt;symmetry&gt;&lt;poincare-symmetry&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="3" />
  <row Id="12342" PostTypeId="2" ParentId="12333" CreationDate="2011-07-16T21:45:07.187" Score="1" ViewCount="" Body="&lt;p&gt;In fact, we may add apples and oranges but then we deal with vectors rather than scalars ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean $2a + 5o $  is a vector in two-dimensional space with the independent unit vectors $a$ and $o$. It is just (2,5), no problem.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-07-17T11:22:33.110" LastActivityDate="2011-07-17T11:22:33.110" CommentCount="11" />
  <row Id="12343" PostTypeId="2" ParentId="12333" CreationDate="2011-07-16T22:42:59.480" Score="1" ViewCount="" Body="&lt;p&gt;It's just because the process of multiplying units is well-defined; that's the way with which you define new units. For example, take a square of side $a=1\text{ m}$. To find the area $A$ of the square you square the side, &#xA;$$&#xD;&#xA;A=a^2=(1\text{ m})^2=1 \text{ m}^2.&#xD;&#xA;$$&#xA;That defines the square-meter, a unit of area. If you had a rectangle with a side $b=1\text{ m}$ and another side $c=1\text{ ft}$, then you can still find the area $A^\prime$ of the rectangle simply by multiplying $b$ with $c$:&#xA;$$&#xD;&#xA;A^\prime=bc=(1\text{ m})\cdot(1\text{ ft})=1\text{ m}\cdot\text{ft}.&#xD;&#xA;$$&#xA;This defines a new unit of area, the $\text{m}\cdot\text{ft}$, which is perfectly ok, but not very intuitive to work with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding or subtracting quantities with units is allowed only if the quantities have the same units, simply because in the process of adding or substracting the units are spectators, i.e. they factor out. For example, take a line-segment of length $\ell_1=1\text{ m}$ and a line segment of length $\ell_2=2\text{ m}$. The total length of the two line segments is&#xA;$$&#xD;&#xA;\ell=\ell_1+\ell_2=(1\text{ m})+(2\text{ m})=(1+2)\text{ m}=3\text{ m}.&#xD;&#xA;$$&#xA;If you couldn't factor out the units, you wouldn't be able to go from step 2 to step 3.&lt;/p&gt;&#xA;" OwnerUserId="4483" LastEditorUserId="4483" LastEditorDisplayName="" LastEditDate="2011-07-17T00:50:27.257" LastActivityDate="2011-07-17T00:50:27.257" CommentCount="4" />
  <row Id="12344" PostTypeId="1" CreationDate="2011-07-16T23:09:55.033" Score="5" ViewCount="135" Body="&lt;p&gt;The Maxwell-Boltzmann velocity distribution in 3D space is&#xA;$$&#xD;&#xA;f(v)dv = 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} v^2 \exp\left(-\frac{m v^2}{2k_B T}\right)dv$$&#xA;It gives the probability for a single particle to have a speed in the intervall $[v,v+dv]$. But this probability is not zero for speeds $v &amp;gt; c $ in conflict with special relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a generalization of the Maxwell-Boltzmann velocity distribution which is valid also in the relativistic regime so that $f(v) = 0$ for $v&amp;gt;c$ ? And how can it be derived? Or can a single particle distribution simply not exist for relativistic speeds, because for high energies, we always have pair-production meaning the particle number is not conserved and a single particle distribution can not be defined in a consistent way? &lt;/p&gt;&#xA;" OwnerUserId="1648" LastActivityDate="2011-07-17T21:15:40.433" Title="Is there a relativistic generalization of the Maxwell-Boltzmann velocity distribution?" Tags="&lt;thermodynamics&gt;&lt;special-relativity&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="12345" PostTypeId="2" ParentId="12344" CreationDate="2011-07-16T23:17:32.033" Score="6" ViewCount="" Body="&lt;p&gt;The point of a Boltzmann distribution is that it maximizes entropy given a fixed energy.  The concept applies to systems with other degrees of freedom besides translational kinetic energy.  The general distribution, from &lt;a href=&quot;http://en.wikipedia.org/wiki/Boltzmann_distribution&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ppu6H.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the simple adjustment to the Maxwell-Boltzmann distribution you cited is to replace the Newtonian kinetic energy $\frac{mv^2}{2}$ with the relativistic kinetic energy $(\gamma - 1)mc^2$ everywhere it appears in the distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pair creation is a separate issue that I'll leave to someone else.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-07-16T23:17:32.033" CommentCount="5" />
  <row Id="12346" PostTypeId="1" AcceptedAnswerId="12351" CreationDate="2011-07-17T02:13:34.250" Score="1" ViewCount="114" Body="&lt;p&gt;I would like to know if there is some formula / graph which would provide / show the efficiency of a certain type of propeller in space. Specifically, I'm interested in the acceleration attainable at certain speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm writing a science fiction book and I'm trying to make it as correct as possible, fact wise. The propeller I'm talking about is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ion_drive&quot; rel=&quot;nofollow&quot;&gt;ion drive&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my book takes place in a world where fusion power is finally ours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, please, let us assume that we have unlimited energy so you could power a dozen huge ion drives non stop. OK, there's the question of argon/xenon fuel, let's assume we have 1 year of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So... the question is... what speed could you reach?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a continuous acceleration of $10\frac{\mathrm{m}}{\mathrm{s}}$ is applied (I put that number because it would also constitute an advantage for my crew - living in Earth's gravity), that would mean that a ship would reach the speed of light in just 347 &lt;strong&gt;DAYS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I know that's impossible because the EFFICIENCY of the ion drive would DECREASE as the ship's speed would approach the exhaust speed of the drive's &quot;nozzle&quot; (well, it doesn't have a nozzle per-se, as you can see in Wiki, but anyway...)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Please do not fear to elaborate on top of my question. Let's suppose for example that maybe the ion drives of the future have a much higher thrust/efficiency/nozzle exhaust speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't only about currently POSSIBLE facts but also about THEORETICAL limitations which might be overcome in the future (such as fusion energy).&lt;/p&gt;&#xA;" OwnerUserId="2414" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-17T03:54:21.380" LastActivityDate="2011-07-17T04:57:07.773" Title="Ion Drive Propulsion Top Speed" Tags="&lt;special-relativity&gt;&lt;kinematics&gt;&lt;acceleration&gt;&lt;propulsion&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="12347" PostTypeId="2" ParentId="12333" CreationDate="2011-07-17T03:24:46.070" Score="0" ViewCount="" Body="&lt;p&gt;It's not used in Physics AFAIK, at least not in a formal way, but one could write $$5\,\mathrm{apples} + 2\,\mathrm{oranges}=7 (\mathrm{apple\,or\,orange}\!)\!\mathrm{s},$$ or $$5\,\mathrm{men} + 2\,\mathrm{women}=7\,\mathrm{people}.$$ That is, one invents objects that contain other objects. However, I can imagine that there might be inequivalent extensions of the ordinary axioms of arithmetic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any Mathematics might in the future be useful in Physics. There is a lot of Mathematics that isn't currently used in Physics, but ruling something out outright seems foolhardy.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-07-17T03:24:46.070" />
  <row Id="12348" PostTypeId="1" CreationDate="2011-07-17T03:37:59.893" Score="1" ViewCount="101" Body="&lt;p&gt;Can someone provide a trigonometry/geometry insight to deduce the angle of the plane is the same as the angle of the component of the weight?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/W5OfC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="401" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-17T05:35:14.030" LastActivityDate="2011-07-17T23:41:38.363" Title="How to deduce this free body diagram?" Tags="&lt;geometry&gt;&lt;free-body-diagram&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12349" PostTypeId="2" ParentId="12348" CreationDate="2011-07-17T04:04:28.937" Score="1" ViewCount="" Body="&lt;p&gt;Angles with their sides perpendicular are always equal. In the present example the arrow of $\vec{F}_1$ is perpendicular to the baseline, and the longest dotted line is perpendicular to the incline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can just imagine rotating one of the two triangles to put it on top of the other. Since the sides start off perpendicular, after a 90-degree rotation they will align, and hence show you that the angles are equal.&lt;/p&gt;&#xA;" OwnerUserId="4483" LastActivityDate="2011-07-17T04:04:28.937" />
  <row Id="12350" PostTypeId="2" ParentId="12346" CreationDate="2011-07-17T04:18:27.573" Score="1" ViewCount="" Body="&lt;p&gt;Well, the structure of spacetime (i.e. special relativity) imposes an overall speed limit of the speed of light, $c = 299792458\frac{\mathrm{m}}{\mathrm{s}}$. So whatever maximum speed you do get up to, it's going to be less than that. You've done the calculation to show that, non-relativistically, the ship would pass the speed of light in less than a year; even without doing any calculations, that tells that if the ship accelerates for that long, it's going to wind up at a significant fraction of light speed. So very roughly, you could just say the maximum speed is $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But suppose you want a more precise figure than that. You can calculate it, but first you have to clearly define what you're calculating. (This is what always trips people up when working with relativity) So here's the situation: suppose the ship starts out on a planet. Initially, the planet and ship are at rest with respect to each other, and let's assume that the planet does not accelerate at all (essentially, it remains at rest forever). It will define the &quot;lab frame.&quot; That is, we're going to calculate the speed of the ship as &lt;strong&gt;observed by someone at rest with respect to the planet&lt;/strong&gt;, but after 1 year &lt;strong&gt;as measured by people on the ship&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is actually a well-studied problem, called the &quot;relativistic rocket.&quot; Since I'm a little short on time, I'm just going to refer to a &lt;a href=&quot;http://www.desy.de/user/projects/Physics/Relativity/SR/rocket.html&quot; rel=&quot;nofollow&quot;&gt;solution posted elsewhere&lt;/a&gt;, although if you like I would be happy to come back to this later and edit in an explanation of where this formula comes from. The formula we need is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = c \tanh(a \tau/c)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a$ is the acceleration as measured in the ship's frame and $\tau$ is the proper time (i.e. time also measured in the ship's frame). Plugging in your numbers, $a = 10\frac{\mathrm{m}}{\mathrm{s}^2}$ and $\tau = 1\text{ year}$, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = 0.783c = 2.35\times 10^5\frac{\mathrm{km}}{\mathrm{s}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your scenario, once your ship reaches this speed, it would run out of fuel and would be stuck at that speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, given the way you've defined the problem, the engine parameters are completely irrelevant. In reality, you would measure the amount of fuel by its mass, not by how long it will last, and in that case you would need the engine efficiency, thrust, etc. as well as the mass of the spaceship to figure out how long a given amount of fuel would last.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-07-17T04:18:27.573" CommentCount="2" />
  <row Id="12351" PostTypeId="2" ParentId="12346" CreationDate="2011-07-17T04:57:07.773" Score="3" ViewCount="" Body="&lt;p&gt;The principle of relativity says that we can analyze a physical situation from any reference frame, as long as it moves with some constant speed relative to a known inertial frame.  Thus, the ion drive does not find it more difficult to accelerate the ship when the ship is &quot;going fast&quot; because the ion drive cannot physically distinguish going fast from going slow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the ion drive is going fast in the reference frame of Earth, then when the ion drive burns, say 1 kg of fuel, it picks up less speed in the Earth frame than it does in the rocket frame due to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Velocity-addition_formula&quot; rel=&quot;nofollow&quot;&gt;relativistic velocity addition law&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That velocity addition law is just the angle-addition law for the hyperbolic tangent.  So, suppose the ship accelerates by shooting individual ions out the back.  Each time it does this, it accelerates the same amount from its own comoving frame.  Then from an Earth frame, the $\textrm{arctanh}$ of the rocket's speed increases by the same amount each time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, as a function of the proper time $\tau$ experienced on the rocket, the acceleration of the rocket is $a(\tau)$ in a comoving frame, there is a quantity called the rapidity of the rocket which increases the way velocity does in Newtonian mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rapidity $\theta$ will be $\theta(\tau) = \int_0^\tau a(\tau) d\tau$, and the velocity is then $v(\tau) = \tanh\theta$.  Specifically, if $a = g$, the velocity is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v(\tau) = \tanh(g\tau)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one year of time has passed on the rocket, its velocity relative to Earth will be $\tanh(1.05) = 0.78$, or 78% the speed of light.  The limit of the $\tanh$ function is one as $\tau \to \infty$, so the rocket never gets to light speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more important limiting factor is the fuel.  If the rocket carries all its fuel, then once it burns through it all, it can't go any more.  Fusion isn't a way around this because by $E=mc^2$ there is a limited energy you can get from a given mass of fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a fraction $f$ of the rocket is fuel, when the fuel is all burned, the momentum of the rocket will be $\gamma m (1-f) \beta$, with $m$ the original mass.  The energy of the rocket is $\gamma m (1-f)$.  Similar relations hold for the fuel.  The conservation of momentum and energy give&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = \gamma m (1-f) + E_{fuel}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 = \gamma m \beta (1-f) + p_{fuel}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_{fuel}$ and $p_{fuel}$ are the energy and momentum of the fuel after burning.  Solving for $\beta$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\beta = \frac{-p_{fuel}}{m - E_{fuel}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minus sign shows that the fuel and rocket go opposite directions.  To maximize $\beta$, we want to make $p_{fuel}$ as large as possible subject to a fixed $E_{fuel}$.  This means that we want the speed of the fuel as high as possible, so assume the fuel is massless with $\beta_{fuel} = 1$ and $p_{fuel} = -E_{fuel}$.   Plugging this into the previous equations and doing some algebra, I got&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\beta = \frac{1 - (1-f)^2}{1 + (1-f)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if half the rocket's original mass were fuel, it would only get to 3/5 the speed of light.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-07-17T04:57:07.773" CommentCount="6" />
  <row Id="12352" PostTypeId="2" ParentId="12324" CreationDate="2011-07-17T06:25:26.563" Score="2" ViewCount="" Body="&lt;p&gt;There is nothing wrong about the dimensional analysis. $\hbar$   has the dimension of $x\cdot p$, so $[x,p]=i\hbar$   is the most standard commutator of quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the velocity is $v=p/m$, which is just another way of writing the usual simple definition of the momentum, $p=mv$, so the right commutator $[x,v]$ will obviously include $i\hbar/m$, too. It's the same thing divided by $m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exponentials are just two examples of the most standard way to get a finite transformation from the (infinitesimal, Hermitian) generator $G$. The finite transformation is always&#xA;$$ \lim_{N\to\infty} \left( 1 + \frac GN \right)^{\phi N}  = \exp(i\phi G) $$&#xA;where $\phi$ is the finite amount of the transformation. For Galilean boosts, the generator is $G=(\sum q_i M_i)/M_{\rm total}$ - the center of mass (your statement that the generator is $v$ is just incorrect). For translations, the generator is $G=p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, getting finite elements of a Lie group from the generator of a Lie algebra - by an exponential - is the first thing that an undergrad learns when he hears both about Lie groups and Lie algebras, and if you haven't understood this point before, it just proves that every single question that you have previously posted on this server was totally inappropriate because you are missing at least three critical layers of knowledge that are pre-requisites for the topics discussed in all your previous questions (about advanced topics such as supermultiplets in exotic supersymmetric theories).&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-07-17T06:25:26.563" CommentCount="3" />
  <row Id="12353" PostTypeId="1" AcceptedAnswerId="12355" CreationDate="2011-07-17T06:59:51.947" Score="0" ViewCount="51" Body="&lt;p&gt;For our Classical Mechanics class, I'm reading Chapter 1 of Goldstein, et al. Now I come across Eq. (1.50). To put it in context:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\begin{align*}&#xD;&#xA;\sum_i{\dot{\mathbf{p}_i} \cdot \delta\mathbf{r}_i}&amp;amp;=\sum_i{m_i\ddot{\mathbf{r}}_i \cdot \delta{\mathbf{r}_i}}\\&#xD;&#xA;&amp;amp;=\sum_{i,j}{m_i\ddot{\mathbf{r}}_i} \cdot \frac{\partial\mathbf{r}_i}{\partial q_j} \delta q_j&#xD;&#xA;\end{align*}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Consider now the relation Eq. (1.50):&#xA;  $$\begin{align*}&#xD;&#xA;\sum_{i,j}{m_i\ddot{\mathbf{r}}_i} \cdot \frac{\partial\mathbf{r}_i}{\partial q_j}&amp;amp;=&#xD;&#xA;\sum_i{\left[ \frac{d}{dt} \left( m_i\dot{\mathbf{r}}_i \cdot \frac{\partial\mathbf{r}_i}{\partial q_j} \right) - &#xD;&#xA;m_i\dot{\mathbf{r}}_i \frac{d}{dt} \left( \frac{\partial \mathbf{r}_i}{\partial q_j} \right) \right]}&#xD;&#xA;\end{align*}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm at a loss for how he resolved it that way. He goes on to explain that we can interchange the differentiation with respect to $t$ and $q_j$. My question is: &lt;strong&gt;Why is there a subtraction in Eq. (1.50)?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="293" LastEditorUserId="293" LastEditorDisplayName="" LastEditDate="2011-07-17T07:12:12.967" LastActivityDate="2011-07-17T07:47:31.020" Title="D'Alembert's Principle and the term containing the reversed effective force" Tags="&lt;classical-mechanics&gt;&lt;homework&gt;" AnswerCount="1" />
  <row Id="12355" PostTypeId="2" ParentId="12353" CreationDate="2011-07-17T07:47:31.020" Score="5" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is there a subtraction in Eq. (1.50)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Goldstein is using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Product_rule&quot; rel=&quot;nofollow&quot;&gt;Leibniz rule for differentiation of a product&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d (fg)}{dt}~=~\frac{d f}{dt}g + f\frac{d g}{dt} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f=m_i\dot{\mathbf{r}}_i $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g=\frac{\partial \mathbf{r}_i}{\partial q_j}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minus is caused by moving a term to the other side of the equation.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastActivityDate="2011-07-17T07:47:31.020" CommentCount="2" />
  <row Id="12356" PostTypeId="2" ParentId="12348" CreationDate="2011-07-17T10:59:08.833" Score="1" ViewCount="" Body="&lt;p&gt;We have this theorem in Geometry:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Angles with perpendicular lines are equal.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But why? We can proof it. Consider these angles:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/GjtxK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;EIH + IHE + HEI = 180&#xA;GIB + IBG + BGI = 180&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;thus&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;EIH + IHE + HEI = GIB + IBG + BGI&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But &lt;code&gt;EHI = 90 = IGB&lt;/code&gt;, because lines are perpendicular. Also &lt;code&gt;HIE = GIB&lt;/code&gt; because:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HIE + HIG = 180 = GIB + HIG =&amp;gt; GIB = HIE&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Therefore, we can remove equal values from both sides to get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;**HEI = IBG**&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4499" LastEditorUserId="401" LastEditorDisplayName="" LastEditDate="2011-07-17T23:41:38.363" LastActivityDate="2011-07-17T23:41:38.363" CommentCount="3" />
  <row Id="12357" PostTypeId="2" ParentId="12344" CreationDate="2011-07-17T13:29:44.310" Score="1" ViewCount="" Body="&lt;p&gt;The usual, or text book generalisation, which was first derived in 1911 is not covariant. Presumably a fully covariant distribution would cover all these details assuming that it exists in the first place. The most recent attempt I am aware of is by Ewald Lehmann, &quot;Covariant equilibrium statistical mechanics&quot;, Journa of Mathematical Physics 47, 023303,2006.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;David Sher&lt;/p&gt;&#xA;" OwnerUserId="4516" LastActivityDate="2011-07-17T13:29:44.310" />
  <row Id="12358" PostTypeId="1" AcceptedAnswerId="12398" CreationDate="2011-07-17T13:40:17.443" Score="8" ViewCount="160" Body="&lt;p&gt;As far as I know quarks are never found in isolation, so how can we determine their rest mass?&lt;/p&gt;&#xA;" OwnerUserId="3743" LastEditorUserId="1501" LastEditorDisplayName="" LastEditDate="2011-07-17T20:32:46.273" LastActivityDate="2011-07-18T11:28:38.767" Title="How to determine the mass of a quark?" Tags="&lt;particle-physics&gt;&lt;experimental-physics&gt;&lt;quark-model&gt;" AnswerCount="1" CommentCount="14" FavoriteCount="5" />
  <row Id="12359" PostTypeId="1" AcceptedAnswerId="12482" CreationDate="2011-07-17T13:50:53.573" Score="0" ViewCount="59" Body="&lt;p&gt;What is the characteristic bump height of periodic grating below which diffraction effect cease to exist (let assume a threshold of peaks to valleys intensity of 20% as the minimum detectable by human eye). Is it significantly less than lambda?&lt;/p&gt;&#xA;" OwnerUserId="4507" LastActivityDate="2011-07-19T22:04:41.153" Title="Diffraction pattern threshhold" Tags="&lt;optics&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="12361" PostTypeId="1" AcceptedAnswerId="12363" CreationDate="2011-07-17T13:59:11.980" Score="3" ViewCount="115" Body="&lt;p&gt;This is a follow-up question to &lt;a href=&quot;http://physics.stackexchange.com/q/12353/293&quot;&gt;D'Alembert's Principle and the term containing the reversed effective force&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the second term of Eq. (1.45)&#xA;$$\begin{align*}&#xD;&#xA;\sum_i{\dot{\mathbf{p}}_i \cdot \delta{\mathbf{r}_i}} &amp;amp;=&#xD;&#xA;\sum_i{m_i\ddot{\mathbf{r}}_i \cdot \frac{\partial\mathbf{r}_i}{\partial q_j} \delta q_j}&#xD;&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting from Eq. (1.50), I was able to follow that&#xA;$$\begin{align*}&#xD;&#xA;\sum_i{m_i\ddot{\mathbf{r}}_i \cdot \frac{\partial\mathbf{r}_i}{\partial q_j}} &amp;amp;=&#xD;&#xA;\sum_i{\left[ \frac{d}{dt} \left( m_i\mathbf{v}_i \cdot \frac{\partial\mathbf{v}_i}{\partial\dot{q}_j} \right) - m_i\mathbf{v}_i \cdot \frac{\partial\mathbf{v}_i}{\partial q_j} \right]}&#xD;&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Goldstein substituted the above equation to (1.45) by saying:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... and the second term on the left-hand side of Eq. (1.45) can be expanded to&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By &quot;second term&quot;, I understood it to be the very first equation I mentioned above. Therefore this is how I understood it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}&#xD;&#xA;\sum_i{\dot{\mathbf{p}}_i \cdot \delta{\mathbf{r}_i}} &amp;amp;=&#xD;&#xA;\sum_i{m_i\ddot{\mathbf{r}}_i \cdot \frac{\partial\mathbf{r}_i}{\partial q_j} \delta q_j}\\&#xD;&#xA;&#xD;&#xA;&amp;amp;= \sum_i\left[ \frac{d}{dt}\left[ \frac{\partial}{\partial\dot{q}_j} \left( \frac{1}{2}\sum_i{m_iv_i^2} \right) \right] - \frac{\partial}{\partial q_j} \left( \frac{1}{2}\sum_i{m_iv_i^2} \right) - Q_j \right]\delta q_j\\&#xD;&#xA;&#xD;&#xA;&amp;amp;= \sum_i{\left[ \frac{d}{dt} \left( \frac{\partial T}{\partial\dot{q}_j} \right) - \frac{\partial T}{\partial q_j} -Q_j \right] \delta q_j}&#xD;&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was able to follow&#xA;$$\begin{align*}&#xD;&#xA;T &amp;amp;= \sum_i{m_i\mathbf{v}_i} \cdot \partial\mathbf{v}_i\\ &#xD;&#xA;T &amp;amp;=\frac{1}{2} \sum_i{m_iv_i^2}&#xD;&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am at a loss: &lt;strong&gt;Where does $-Q_j$ come from?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="293" LastEditorUserId="293" LastEditorDisplayName="" LastEditDate="2011-07-17T14:19:04.123" LastActivityDate="2011-07-17T23:07:28.273" Title="D'Alembert's Principle: Where does $-Q_j$ come from?" Tags="&lt;classical-mechanics&gt;&lt;homework&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="12362" PostTypeId="2" ParentId="12361" CreationDate="2011-07-17T14:23:06.653" Score="0" ViewCount="" Body="&lt;p&gt;It should be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}&#xD;&#xA;\sum_i{ (\dot{\mathbf{p}}_i-\mathbf{F}_i) \cdot \delta{\mathbf{r}_i}} &amp;amp;=&#xD;&#xA;\sum_{i,j}{(m_i\ddot{\mathbf{r}}_i-\mathbf{F}_i) \cdot \frac{\partial\mathbf{r}_i}{\partial q_j} \delta q_j}\\&#xD;&#xA;&#xD;&#xA;&amp;amp;= \sum_{i,j}\left[ \frac{d}{dt}\left[ \frac{\partial}{\partial\dot{q}_j} \left( \frac{1}{2}\sum_i{m_iv_i^2} \right) \right] - \frac{\partial}{\partial q_j} \left( \frac{1}{2}\sum_i{m_iv_i^2} \right) - Q_j \right]\delta q_j\\&#xD;&#xA;&#xD;&#xA;&amp;amp;= \sum_{i,j}{\left[ \frac{d}{dt} \left( \frac{\partial T}{\partial\dot{q}_j} \right) - \frac{\partial T}{\partial q_j} -Q_j \right] \delta q_j}&#xD;&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So read it again.&lt;/p&gt;&#xA;" OwnerUserId="3887" LastActivityDate="2011-07-17T14:23:06.653" CommentCount="1" />
  <row Id="12363" PostTypeId="2" ParentId="12361" CreationDate="2011-07-17T14:37:54.707" Score="2" ViewCount="" Body="&lt;p&gt;Similar to Newton's 2nd law, the D'Alembert's principle has both a dynamical and a kinetic term,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ &#xD;&#xA;\sum_i (\mathbf{F}^{(a)}_i - \dot{\mathbf{p}}_i) \cdot \delta \mathbf{r}_i~=~0. \qquad (1.45)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On one hand, the dynamical term &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_i \mathbf{F}^{(a)}_i \cdot \delta \mathbf{r}_i = \sum_j Q_j \delta q_j \qquad (1.48)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;contains the generalized force&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_j=\sum_i\mathbf{F}^{(a)}_i\cdot \frac{\partial \mathbf{r}_i}{\partial q_j}.\qquad (1.49)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the kinetic term &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{\mathbf{p}}_i \cdot \delta \mathbf{r}_i~=~ \sum_j{\left[ \frac{d}{dt} \left( \frac{\partial T}{\partial\dot{q}_j} \right) - \frac{\partial T}{\partial q_j} \right] \delta q_j} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;contains the kinetic energy $T =\frac{1}{2} \sum_i{m_iv_i^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: It is true that the third edition of Goldstein wrongly says &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and the second term on the left-hand side of Eq. (1.45) can be expanded into&#xA;  $$ \sum_i\left[ \frac{d}{dt}\left[ \frac{\partial}{\partial\dot{q}_j} \left( \sum_i{\frac{1}{2}m_iv_i^2} \right) \right] - \frac{\partial}{\partial q_j} \left( \sum_i{\frac{1}{2}m_iv_i^2} \right) - Q_j \right]\delta q_j. \qquad $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second edition does not have the $-Q_j$ term, so an unfortunate mistake was introduced during the update to the third edition. This is not the first time that I have noticed that the second edition is often more carefully written than the third edition in what concerns old material. (The third edition contains a new chapter 11 about classical chaos.)&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-17T15:17:06.183" LastActivityDate="2011-07-17T15:17:06.183" CommentCount="1" />
  <row Id="12364" PostTypeId="2" ParentId="12341" CreationDate="2011-07-17T15:59:37.643" Score="4" ViewCount="" Body="&lt;p&gt;It is important to distinguish between three group actions that are named &quot;Galilean&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-The Galilean transformation group of the Eucledian space (as an automorphism group).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-The Galilean transformation group of the classical phase space (whose Lie algebra constitute a Lie subalgebra of the Poisson algebra of the phase space). This is the classical action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-The Galilean transformations of the wavefunctions (which are infinite dimensional irreducible representations). This is the quantum action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only the first group action is free from the central extension. Both classical and quantum actions include the central extension (which is sometimes called the Bargmann group). &#xA;Thus, the central extension is not purely quantum mechanical, however, it is true that most textbooks describe the central extension for the quantum case.&#xA;I'll explain first the quantum case, then I'll return to the classical case and compare oth cases to the Poincare group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum mechanics, a wavefunction in general is not a function on the configuration manifold, but rather a section of a complex line bundle over the phase space. In general the lift of a symmetry (an automorphism of the phase space) is an automorphism of the line bundle which is therefore a $\mathbb{C}$ extension of the automorphism of the base space. In the case of a unitary symmetry, this will be a $U(1)$ extension. Sometimes, this extension is trivial as in the case of the Poincare group.&#xA;Now, the central extensions of a Lie group $G$ are classified by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_cohomology&quot; rel=&quot;nofollow&quot;&gt;group cohomology&lt;/a&gt; group $H^2(G, U(1))$. In general, it is not trivial to compute these cohomology groups, but the case of the Galilean and Poincare groups can be heuristically understood as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The application of the Galilean group action $\dot{q} \rightarrow \dot{q}+v$ to the non-relativistic action of a free particle: $S = \int_{t_1}^{t_2}\frac{m }{2}\dot{q}^2dt$, produces a total derivative leading to $S \rightarrow S + \frac{m}{2}v^2(t_2-t_1) + mvq(t_2) - mvq(t_1)$:&#xA;Now Since the propagator $G(t_1, t_2)$ transforms as $ exp(\frac{iS}{\hbar})$ and the inner product $\psi(t_1)^{\dagger} G(t_1, t_2) \psi(t_2)$ must be invariant, we get that the wavefunction must transform as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\psi(t,q) \rightarrow exp(\frac{i m}{2\hbar}(v^2 t+2vq) \psi(t,q) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, no application of a smooth canonical transformation can romove the total derivative from the transformation law of the action, this is the indication that the central extension is non-trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case of the Poincare group is trivial. The relativistic &lt;a href=&quot;http://en.wikipedia.org/wiki/Action_%28physics%29#Action_principle_for_single_relativistic_particle&quot; rel=&quot;nofollow&quot;&gt;free particle&lt;/a&gt; action is invariant under the action of the Poincare group, thus the transformation of the wavefunction doen not acquire additional phases and the group extension is trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classically, the phase space is $T^{*}R^3$ and the action of the boosts on the momenta is given by: $p \rightarrow p + mv$, thus the generators of the boosts must have the form&#xA;$K = mvq$, then the action is easily obtained using the Poisson brackets{q, p} = 1, and the Poisson bracket of a Boost and a translation is non-trivial {K, p} = m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that the Lie algebra action acquires the central extension in the classical case is that the action is Hamiltonian, thus realized by Hamiltonian vector fields and vector fields do not commute in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Iwasawa decomposition of the Lorentz group provides the answer to your second question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$SO^{+}(3,1) = SO(3) A N$&#xA;where $A$ is generated by the Boost $M_{01}$ and $N$ is the Abelian group generated by $M_{0j}+M_{1j}$, $j&amp;gt;1$. Now both subgroups $A$ and $N$ are homeomorphic as manifolds to $R$ and $R^2$ respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To your third question: The limiting process which produces the Galilean group from the Poincare group is called the Wingne-Inonu contraction. This contraction produces the non-relativistic limit.&#xA;Its relation to quantum mechanics is that there is a notion of contarction of Lie groups unitary representations, however not a trivial one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In classical mechanics, observables are expressed as functions on the phase space. see for example chapter 3 of Ballentine's book for the explicit classical realization of the generators of the Galilean group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a case where the full geometric quantization recepie can be carried out. See the following &lt;a href=&quot;http://ugcs.caltech.edu/~mainiero/articles/geom_quant_mainiero_hassan.pdf&quot; rel=&quot;nofollow&quot;&gt;two&lt;/a&gt; &lt;a href=&quot;http://math.berkeley.edu/~alanw/GofQ.pdf&quot; rel=&quot;nofollow&quot;&gt;articles&lt;/a&gt; for a review. (The full proof appears in page 95 of the second article. The technical computations are more readable in pages 8-9 of the first article).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The central extensions appear in the process of prequantization.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First please notice that the Hamiltonian vector fields $X_f$ corresponding to the Galilean Lie algebra generators close to the non-centrally extended algebra, &#xA;(because, the hamiltonian vector field of constant functions vanish).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;However, the prequantized operators&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$\hat{f} = f - i\hbar(X_f - \frac{i}{\hbar}i_{X_f} \theta)$, ($\theta$ is a symplectic potential whose exterior derivative equals the symplectic form) close to the centrally extended algebra because their action is isomorphic to the action of the Poisson algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prequantized operators are used as operators over the Hilbert space of the square integrble polarized sections, thus they provide a quantum realization of the centrallly extended Lie algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your second question, the Wingne-Inonu contraction acts on the level of the abstract Lie algebra and not for its specific realizations.&#xA;A given realization is termed &quot;Quantum&quot;, if it refers to a realization on a Hilbert space (in contrast to realization by means of Poisson brackets, which is the classical one).&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorUserId="2190" LastEditorDisplayName="" LastEditDate="2011-07-18T15:35:58.503" LastActivityDate="2011-07-18T15:35:58.503" CommentCount="11" />
  <row Id="12365" PostTypeId="1" CreationDate="2011-07-17T16:13:05.437" Score="2" ViewCount="73" Body="&lt;p&gt;Particle dark matter coherent scattering direct searches rely on the scattering of the DM particle off a nucleus (Xe, Si, Ge, NaI etc). The wisdom is this is coherent scattering: so the rate ~atomic mass squared, since $q^2 \approx zero$. Is this completely sensible? Despite $q^2 \approx 0$ for galactic wind DM particles, thus with a size of interaction supposedly $\approx \hbar/q^2 &amp;gt; R$ nucleus, is it really sensible for a tiny heavy object exchanging a similarly tiny object to extend over the entire nucleus???&#xA;Argument for no:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\approx 100$ GeV particle: Compton wavelength &amp;lt; 1F the size of a single nucleon.&lt;/li&gt;&#xA;&lt;li&gt;object exchanged, Z or H, also on shell size &amp;lt; 1F.&lt;/li&gt;&#xA;&lt;li&gt;neutralino - esp. if it could be a 4th gen. Dirac or Majorana neutrino**, will have an oscillation length &amp;lt; 1 F - i.e. its wavefunctions will be changing on a scale small with the nucleus&lt;/li&gt;&#xA;&lt;li&gt;compared with e-m scattering, like e, mu, proton off nuclei, the e-m field clearly could encompass the entire nucleus at low $q^2$.&lt;/li&gt;&#xA;&lt;li&gt;compared with light neutrino scattering, even though W,Z heavy, the neutrino itself at low q^2 is larger than the nucleus.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;** if the neutrino mixing angles $\sin^2\theta \approx (1/\Delta m^2)^2$ or $(1/\Delta m^2)1.5$, which they do for gen 1-3, then a 4th gen 100 GeV neutrino has a lifetime  $\approx T_{top} \times 10^{44} &amp;gt; T_{universe}$, assuming the charged partner mass $ &amp;gt; M_{ne_u}$.&lt;/p&gt;&#xA;" OwnerUserId="4518" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-07-17T19:27:21.257" LastActivityDate="2011-07-18T17:25:57.183" Title="Particle dark matter coherent scattering?" Tags="&lt;dark-matter&gt;" AnswerCount="0" CommentCount="2" />
  <row Id="12366" PostTypeId="2" ParentId="12331" CreationDate="2011-07-17T16:33:46.793" Score="5" ViewCount="" Body="&lt;p&gt;Take a system of four mutually orthogonal vectors lined up with the four axis x,y,z,t in 4D space. Apply an arbitrary rotation to move the four axis to some other positions. They remain mutually orthogonal of course. We now need to show that we can rotate the vectors back to their starting position using rotations in the six planes. The inverse of these would then be the six rotations we require to match the original rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with the vector that was originally aligned with the x-axis. We need to rotate it back there. First use a rotation in the xy plane to rotate it so that it moves into the xzt hyperplace. Then use a rotation in the xz plane so that it moves into the xt plane. Then use a rotation in the xt plane so that it lines up with the x axis as required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other three orthogonal vectors will have been rotated as well, but since they remain orthogonal to the first vector they are now in the yzt hyperplane orthogonal to the x axis. You can now apply use the three rotations in the remaining planes yz, zt and yt to line them up with their original axis using the solution for 3d space. So it's done.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-07-17T16:33:46.793" />
  <row Id="12369" PostTypeId="1" CreationDate="2011-07-17T18:13:35.887" Score="4" ViewCount="130" Body="&lt;p&gt;Atomic bombs make mushroom. It's well-known and we've seen many images of atomic experiments around the world. My question is, why mushroom? Does TNT and Grenade also make mini-mushrooms? Is it a physical attribute of explosion, or something specific to atomic energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="4499" LastActivityDate="2011-07-17T21:45:50.303" Title="Why atomic bomb makes a mushroom?" Tags="&lt;atoms&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="12370" PostTypeId="2" ParentId="12333" CreationDate="2011-07-17T19:17:52.163" Score="0" ViewCount="" Body="&lt;p&gt;We count, adimensional. We measure, adimensional. Any measure is a ratio of two quantities: The amount we want to measure is divided by the amount of a chosen standard. To the standard we ascribe a name like meter and regulate a precise way to define it. &#xA;When we measure mass we make a count of barions as found here:&lt;br&gt;&#xA;&lt;a href=&quot;http://vixra.org/pdf/1107.0016v1.pdf&quot; rel=&quot;nofollow&quot;&gt;section II. ON UNITS, PHYSICAL LAWS AND SCALING, subsection A. On quantities and units and Atomic measures are number counts&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we add measures we add pure numbers. And the role of the units? Its a remind of what standard was chosen to represent them and the degree of the dimension (Length-L,Time-T,Charge-Q,Mass-M,Temperature). Temperature and Angle measures are pure numbers by definition, not a ratio. As an example $m^3$ has the dimention $L^3$, a volume. It has no meaning the addition of a volume with an area or a temperature with an angle. We can add quantities of the same kind (dimension and degree); a class of fruit or 'object' or 'bag content' allow the counting of both apples and oranges in the same bag. We can add X meters to Y decimeters, the result is still an amount of length. &#xA;Adding quantities of different class, lets say A and B, give a result that belong to a third class (neither A nor B).&lt;br&gt;&#xA;Quoting from the above paper: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Let us begin by the quantities length and time; length is a&#xA;  geometrical, static, concept; time is a concept linked to the ﬂow of&#xA;  occurrences, the contrary of static; they are, clearly, distinct&#xA;  concepts.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can we add length quantity to a time quantity? No.&lt;br&gt;&#xA;&quot;Base units are not independent&quot;, i.e. the length, time, mass and charge units are interdependent; the speed of light and the atom properties interconnect them:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Time and length units are linked through the speed of light.&#xA;  Therefore, while the concepts of length and time are independent,&#xA;  their units are not. This has consequences in the description of the&#xA;  universe; for instance, relativistic space-time is a property of the&#xA;  description of the universe using such units and a reference frame&#xA;  calibrated by the method described by Einstein&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Multiplication is an addition (just repeated N times) e.g. 5m x 3 = 15m, and 5m x 3m = 15 $m^2$ (dimension $L^2$). Division of quantities of the same nature is a measurement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the author of the above recent paper is a friend of mine and comments are welcome and, if pertinent, I will forward them. Sorry for the bad english.)     &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-07-18T06:52:47.120" LastActivityDate="2011-07-18T06:52:47.120" />
  <row Id="12371" PostTypeId="2" ParentId="12333" CreationDate="2011-07-17T19:23:04.643" Score="4" ViewCount="" Body="&lt;p&gt;I guess I might as well collect my comments into an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are actually implicitly asking two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;whether it makes mathematical sense to add different quantities; and&lt;/li&gt;&#xA;&lt;li&gt;if it &lt;em&gt;does&lt;/em&gt; make sense why one doesn't encounter it more often in physics.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The answer to 1. is positive in certain cases, especially when talking about fruits. The structure is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Free_abelian_group&quot; rel=&quot;nofollow&quot;&gt;free abelian group&lt;/a&gt;. It's essentialy one copy of integers for every fruit with addition defined component-wise: $(5a + 2o) + (2a + 3o) = (7a + 5o)$ and so on. One can similarly formalize other concepts of addition of different quantities. One can also introduce multiplication and talk about polymonial rings $K[x,y,\dots,z]$ (where the variables are understood to represent units) or take the field of fractions of that, or even introduce non-commutativity. There are many mathematical structures that can accomodate all of the operations ever needed in physics (and more).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we come to the point 2. We've seen that it's &lt;em&gt;possible&lt;/em&gt; to add different quantities. But that tells you &lt;em&gt;nothing&lt;/em&gt; about whether such an operation is ever useful. In particular, when talking about elementary operations used in physical problems to arrive at a result which is &lt;em&gt;always&lt;/em&gt; a well-defined quantity with units. I claim that this is why we don't use in physics anything else than addition of quantities with same units because we want to have reasonable units at every step of the calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this is also consistent with taking products of different quantities because this operation doesn't spoil the fact that at every step of the derivation we have a well-defined units of the expression.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-07-17T19:32:04.897" LastActivityDate="2011-07-17T19:32:04.897" CommentCount="2" />
  <row Id="12372" PostTypeId="2" ParentId="12341" CreationDate="2011-07-17T20:12:03.487" Score="3" ViewCount="" Body="&lt;p&gt;You are still asking way too many questions at once. So again, consider splitting them next time. I will address only the topology part here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As topological spaces we have&#xA;$$SO(3) = {\mathbb R \mathbb P}^3 = S^3 /\sim ,$$&#xA;$$SO(4) = S^3 \times S^3 / \sim, $$&#xA;$$SO^+(1,3) = {\mathbb R}^3 \times S^3 / \sim $$&#xA; (in all of these cases $\sim$ is an identification of the antipodal points on the relevant spheres). One can get these results by noting that the double cover of $SO(3)$ is $SU(2)$, double cover of $SO(4)$ is $SU(2) \times SU(2)$ and the double cover of $SO^+(1,3)$ is $SL(2, {\mathbb C})$.&#xA; (Note that the above factorizations can be also written as $X/({\mathbb Z} / 2{\mathbb Z})$ which is a factorization of a space by the action of the group on the space; in this case the action is sending a point to its antipodal point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding translations only means using semidirect product on the level of groups, or direct product on the level of topological spaces. In total, we have that the connected component of the Poincaré group is ${\mathbb R}^4 \times {\mathbb R}^3 \times S^3 / \sim$. As for the wikipedia page, I am not sure about your confusion. It defines Poincaré group as ${\mathbb R}^4 \rtimes O(1,3)$ and topologically this is disjoint union of four copies of ${\mathbb R}^4 \times {\mathbb R}^3 \times S^3 / \sim$.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-07-17T21:29:25.307" LastActivityDate="2011-07-17T21:29:25.307" CommentCount="2" />
  <row Id="12373" PostTypeId="1" CreationDate="2011-07-17T21:01:32.620" Score="2" ViewCount="97" Body="&lt;p&gt;my question concerns the interaction of light and matter in a semi-classical approach. &#xA;(Quantized Atoms, Classical Fields)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Coulomb gauge (div A = 0 , $\phi$=0) we have $E = -\frac{1}{c}\partial_{t}A$ and &#xA;$B = rot(A)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's now consider a system of N charged particles (all denoted by an index k). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $H_{0}$ be the unperturbed time-independent Hamiltonian and suppose that the eigenvalue problem corresponding to this Hamiltoninan is solved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $H_{1}=-\sum\limits_{k=0}^{N} \frac{e_{k}}{m_{k} c} A(x_{k},t) \cdot p_{k} + \sum\limits_{k=0}^{N} \frac{e_{k}^2}{2 m_{k} c^2} A(x_{k},t)^{2}$ be the time-dependent pertubation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atom is now placed at the origin. &#xA;The wavelength of the considered light is large compared to the diameter of the atom. &#xA;Hence we have approximately $A_(x_{k},t)\psi(x_{1},...,x_{n}) \approx A_(0,t)\psi(x_{1},...,x_{n})$. (Note that the atom is placed at the origin). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the second term in the perturbed Hamiltonian $H_{1}(t)$ (proportional to A^{2})&#xA;is of the form $\lambda \cdot f(t)$ for some real number $\lambda$ and some scalar time-dependent function f(t) using this approximation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowit is said that this term only causes a phase factor in the wave function $\phi_{I}(t)$&#xA;viewed in the interaaction picture. I don't really understand why that is the case. &#xA;A time -depnedent Hamiltonian does not trigger in genereal a propagator of the form exp( i*...) . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone help me with this question??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for the responses. &lt;/p&gt;&#xA;" OwnerUserId="4520" LastActivityDate="2011-08-16T22:25:37.563" Title="Atomic Transitions - Dipole Approximation" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" />
  <row Id="12374" PostTypeId="2" ParentId="12344" CreationDate="2011-07-17T21:15:40.433" Score="1" ViewCount="" Body="&lt;p&gt;Per Mark's request, I'll provide an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I think it's not possible to obtain such a formula. One can of course naively try to extrapolate various classical formulae but all of these attempts are bound to fail. Here's why: even when constructing relativistic quantum mechanics one finds that the theory is not consistent. For example, in non-relativistic quantum mechanics one has a position operator that can be used to get information on precise position of the particle (up to the uncertainty given by Heisenberg's uncertainty principle). But when one includes relativity into the picture the theory stops being consistent. This reflects the fact that to localize the particle with great precision one has to make experiments with increasingly high energy and at certain point the energy is enough for the appearance of new particles. In fact, creation and annihilation of particles is inevitable in relativistic regime (and in some systems it's not even clear what particles should be and one has to talk about fields instead). The situation is even more pronounced in statistical physics where there are huge number of particles present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More importantly, there's no need to get such a formula. Consider systems for which it would be useful. Such systems would have to be extremely non-classical (like neutron stars, black holes interior, quark-gluon plasma, etc.) and the concept of velocity would have no meaning as there's no way to observe individual particles of these systems; which is in contrast to the classical case where you can test Maxwell distribution by letting the particles out of the box one by one and seeing how fast they are (the actual experiment is lot more sophisticated of course, but that's unimportant here).&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-17T21:15:40.433" CommentCount="1" />
  <row Id="12375" PostTypeId="2" ParentId="12361" CreationDate="2011-07-17T21:32:13.900" Score="0" ViewCount="" Body="&lt;p&gt;You've misinterpreted what Goldstein states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and the second term on the left-hand side of Eq. (1.45) can be expanded into&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To save confusion for some, it could be better expressed as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and the second term on the left-hand side of Eq. (1.45) can be expanded, so that Eq. (1.45) becomes&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It looks repetitive, so the authors probably stuck with the current form, relying on the understanding of the student to see what is obviously meant. The relevant equations are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_i (\mathbf{F}^{(a)}_i - \dot{\mathbf{p}}_i) \cdot \delta \mathbf{r}_i~=~0. \qquad (1.45)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_i{\mathbf{F}_i \cdot \delta{\mathbf{r}_i}} = \sum_{i,j}{\mathbf{F}_i \cdot \frac{\partial\mathbf{r}_i}{\partial q_j} \delta q_j = \sum_{j}{Q}_j \delta q_j}\qquad (1.48)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second term in equation (1.45) is therefore expanded as in (1.48) and then combined with the expansion for the first term $\mathbf{\dot p}_i\cdot\delta\mathbf{r}_i$ elsewhere&#xA;,  to give&#xA;$$\sum_j\left[ \frac{d}{dt}\left[ \frac{\partial}{\partial\dot{q}_j} \left( \sum_i{\frac{1}{2}m_iv_i^2} \right) \right] - \frac{\partial}{\partial q_j} \left( \sum_i{\frac{1}{2}m_iv_i^2} \right) - Q_j \right]\delta q_j$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note (1.45) is multiplied by -1 so the signs in the expression derived elsewhere for $\mathbf{\dot p}_i\cdot\delta\mathbf{r}_i$ remain the same, hence the $-Q_j$ term rather than $Q_j$. The outer summation is over $j$ and not $i$ as you've put down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a link for correctons to this book if you're convinced you're right, &lt;a href=&quot;http://astro.physics.sc.edu/goldstein/&quot; rel=&quot;nofollow&quot;&gt;http://astro.physics.sc.edu/goldstein/&lt;/a&gt; although in this case it looks as if it's a misinterpretation on your part.&lt;/p&gt;&#xA;" OwnerUserId="2146" LastEditorUserId="2146" LastEditorDisplayName="" LastEditDate="2011-07-17T23:07:28.273" LastActivityDate="2011-07-17T23:07:28.273" CommentCount="1" />
  <row Id="12376" PostTypeId="2" ParentId="12369" CreationDate="2011-07-17T21:45:50.303" Score="5" ViewCount="" Body="&lt;p&gt;Its due to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rayleigh%E2%80%93Taylor_instability&quot; rel=&quot;nofollow&quot;&gt;WP-Rayleigh–Taylor instability&lt;/a&gt;: it is an instability of an interface between two fluids of different densities, which occurs when the lighter fluid is pushing the heavier fluid.&lt;br&gt;&#xA;Hot hair raises and colder goes down.&lt;br&gt;&#xA;A Mushroom cloud formed by hot wet air :&lt;br&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/T5Ze5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;google for images or videos of 'Rayleigh–Taylor'  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-07-17T21:45:50.303" CommentCount="3" />
  <row Id="12377" PostTypeId="1" AcceptedAnswerId="12381" CreationDate="2011-07-17T21:50:36.760" Score="1" ViewCount="111" Body="&lt;p&gt;Quantum dots are quantum systems that are confined by definition on the nano scale. Why didn't people study similar systems on a much smaller scale, something as small as the dimension of the nucleus? Is it practically difficult? or useless?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R.&lt;/p&gt;&#xA;" OwnerUserId="4521" LastActivityDate="2011-08-18T13:08:37.723" Title="Quantum dots on the nuclear scale!" Tags="&lt;quantum-mechanics&gt;&lt;nanoscience&gt;" AnswerCount="4" />
  <row Id="12378" PostTypeId="2" ParentId="12377" CreationDate="2011-07-17T22:05:04.567" Score="0" ViewCount="" Body="&lt;p&gt;The term quantum dot is, as you said, used to describe the &lt;strong&gt;need to treat&lt;/strong&gt; a system on the nanoscale &lt;strong&gt;quantum-mechanically&lt;/strong&gt;. This is important for example in nanophotonics where e.g. a coupling of a large system to such a quantum mechanical one leads to &lt;strong&gt;interesting physics of nanoscale devices&lt;/strong&gt;. See e.g. &lt;a href=&quot;http://www.nature.com/nature/journal/v432/n7014/abs/nature03119.html?lang=en&quot; rel=&quot;nofollow&quot;&gt;Vacuum Rabi splitting with a single quantum dot in a photonic crystal nanocavity&lt;/a&gt; (Yoshie et al, Nature, 2004) or &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v95/i6/e067401&quot; rel=&quot;nofollow&quot;&gt;Exciton-Photon Strong-Coupling Regime for a Single Quantum Dot Embedded in a Microcavity&lt;/a&gt; (Peter et al, PRL, 2005).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But bringing this notion back to where it came from - quantum systems which can only be described by quantum mechanics - does not make sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Greets&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2011-07-17T22:05:04.567" CommentCount="3" />
  <row Id="12379" PostTypeId="2" ParentId="12373" CreationDate="2011-07-17T22:07:19.967" Score="1" ViewCount="" Body="&lt;p&gt;If you use the approximation $A(x_k,t) \sim A(0,t)$ then the $A^2$ part of the Hamiltonian is just $f(t) \widehat{\mathbf 1}$ and so is quite uninteresting. This is because the identity operator $\widehat{\mathbf 1}$ commutes with everything and so does the associated evolution operator $$U(t&amp;#39;, t&amp;#39;&amp;#39;) = \exp \left(-{i \over \hbar} \int_{t&amp;#39;}^{t&amp;#39;&amp;#39;} f(t) {\rm d} t \right) \widehat{\mathbf 1}.$$&#xA;This is the complex phase you were looking for.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-17T22:07:19.967" />
  <row Id="12380" PostTypeId="2" ParentId="2328" CreationDate="2011-07-17T22:10:55.880" Score="0" ViewCount="" Body="&lt;p&gt;Here's another example from Griffith's book &quot;Introduction to Electrodynamics&quot; which illustrates phenomena where what we &lt;em&gt;see&lt;/em&gt; is not what we &lt;em&gt;observe&lt;/em&gt;. The &lt;strong&gt;apparent speed&lt;/strong&gt; can be much greater than the speed of light. This speed is just what we see, an illusion, and it's the result of our inability sometimes to see the actual direction of movement of an distant object w.r.t. us  and the fact that the light needs some finite time to get to our eyes. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Problem 12.6&lt;/strong&gt; Every 2 years, more or less, The New York Times publishes&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;an article in which  some astronomer claims to have found an object&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;traveling faster than the speed of light. Many  of these reports&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;result from a failure to distinguish what is seen from what is&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;observed--that  is, from a failure to account for light travel time.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Here's an example: A star is traveling with  speed $v$ at an angle $\theta$ to&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;the line of sight (Fig. 12.6). What is its apparent speed across the&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;sky'? &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/rpn9v.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(Suppose the light signal from $b$ reaches the earth at a time At after&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;the signal from a, and the  star has meanwhile advanced a distance $\Delta s$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;across the celestial sphere; by &quot;apparent speed&quot;  I mean $\Delta s/\Delta t$.) What&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;angle $\theta$ gives the maximum apparent speed? Show that the apparent &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;speed can be much greater than $c$, even if $v$ itself is less than $c$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be easily shown that the apparent speed in this example is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_{app}=\frac{v\sin\theta}{1-\frac{v}{c}\cos\theta}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the angle $\theta$ that gives the maximum apparent speed we just differentiate and solve, for $\theta$, the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d u_{app}}{d\theta}=0 \Leftrightarrow \theta_{max}=\cos^{-1}(\frac{v}{c})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this angle, $u_{app}=\frac{v}{\sqrt{1-v^2/c^2}}=\gamma v$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This result shows that when $v\to c$, $u_{app}\to \infty$, even though $v&amp;lt;c$.&lt;/p&gt;&#xA;" OwnerUserId="3066" LastEditorUserId="3066" LastEditorDisplayName="" LastEditDate="2011-07-17T23:29:30.843" LastActivityDate="2011-07-17T23:29:30.843" CommentCount="3" />
  <row Id="12381" PostTypeId="2" ParentId="12377" CreationDate="2011-07-17T22:12:44.070" Score="1" ViewCount="" Body="&lt;p&gt;normal matter structure is entirely constructed from the electronic bindings, so it is in the realm of the possible to engineer how the atoms are binded together exactly and this is the aim of lower-level nanotechnology. However, it is with current technogy and physics, impossible to create complex structures at lower scales (i.e: nuclear scale). And i don't think that is something that is in principle possible unless some dramatic breakthrough occurs in how we obtain larger-scale nuclear matter, which from the experimental limitations that there are to obtain high Z nucleus that might probe the regions of &lt;a href=&quot;http://en.wikipedia.org/wiki/Island_of_stability&quot; rel=&quot;nofollow&quot;&gt;higher islands of stability&lt;/a&gt;, one can safely infer that this is &#xA;a very challenging problem in of itself&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; this is not entirely related, but it shows an example of how assumptions as the one i've made above about manipulation of small scales being out of engineering reach can be twisted: &lt;a href=&quot;http://www.cesr.fr/~g-wave05/presentations/optics/Halloin.pdf&quot; rel=&quot;nofollow&quot;&gt;this slide about non-homogeneous diffraction crystals&lt;/a&gt; shows how to do something that most physicists have thought for long to be essentially impossible; &lt;em&gt;X-ray and gamma-ray optics&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-08-18T13:08:37.723" LastActivityDate="2011-08-18T13:08:37.723" />
  <row Id="12382" PostTypeId="2" ParentId="12377" CreationDate="2011-07-17T22:19:20.860" Score="2" ViewCount="" Body="&lt;p&gt;Depends on your definition of a quantum dot. Usually they are understood to be a collection of excitons (bound states of electrons and holes) confined to a small volume. But the smaller the volume, the higher the energy of the system and eventually you would destroy the exciton bonds (besides, there would be a technical problem of how to maintain such a localized potential well).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I suppose by quantum dot you mean arbitrary quantum system that is very localized. In this case, nucleus itself should satisfy you well.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-17T22:19:20.860" CommentCount="1" />
  <row Id="12383" PostTypeId="1" AcceptedAnswerId="12392" CreationDate="2011-07-17T23:01:12.583" Score="3" ViewCount="85" Body="&lt;p&gt;Quoting from Wolfram Mathworld: &quot; It is the shape used in the construction of soccer balls, and it was also the configuration of the lenses used for focusing the explosive shock waves of the detonators in the Fat Man atomic bomb (Rhodes 1996, p. 195).&quot; but further insights are not provided. (http://mathworld.wolfram.com/TruncatedIcosahedron.html)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why would the Manhattan project scientists choose this geometry for designing the implosive lenses of the Fat Man atomic bomb? Assuming the explosive charges were placed on the faces of the icosahedron, is this the optimal geometry for focusing the shock waves on a single point (presumably a spherical plutonium core), or was this mainly due to practical/engineering considerations?&lt;/p&gt;&#xA;" OwnerUserId="4257" LastActivityDate="2011-07-18T06:05:19.593" Title="Why was the truncated icosahedron (i.e. soccer ball) geometry chosen for the implosive lenses in the “Fat Man” atomic bomb?" Tags="&lt;mathematical-physics&gt;&lt;geometry&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12384" PostTypeId="1" AcceptedAnswerId="12395" CreationDate="2011-07-17T23:38:54.950" Score="1" ViewCount="40" Body="&lt;p&gt;The GLAP equation describes the evolution of parton distribution under successive branchings. When branching takes place at a finite angle, the outgoing parton carries some finite transverse momentum. However, this is ignored in deep inelastic scattering calculations, where the parton momentum is always assumed to be longitudinal.&#xA;I have trouble understanding why the angle should be ignored. The GLAP equation effectively comes from resummation of collinear branching to all orders in perturbation theory. So you can imagine 100 branchings, each at angle of $\pi/200$, resulting in an outgoing parton at an angle of $\pi/2$, i.e. perpendicular, to the velocity of the incoming nucleon. Why is this not a problem?&lt;/p&gt;&#xA;" OwnerUserId="2643" LastActivityDate="2011-07-18T07:22:12.947" Title="QCD Parton Branching - Is the angle always negligible?" Tags="&lt;quantum-chromodynamics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="12385" PostTypeId="1" AcceptedAnswerId="13260" CreationDate="2011-07-18T00:45:27.313" Score="1" ViewCount="199" Body="&lt;p&gt;&lt;em&gt;Edit in response to a comment&lt;/em&gt;. Are there books on Regularization, in the context of  quantum field theory at an Introductory level? Could you suggest one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Added&lt;/em&gt;: I posted at math.SE the question &lt;em&gt;&lt;a href=&quot;http://math.stackexchange.com/questions/52139/reference-request-introduction-to-mathematical-theory-of-regularization&quot;&gt;Reference request: Introduction to mathematical theory of Regularization&lt;/a&gt;&lt;/em&gt; and accepted &lt;a href=&quot;http://math.stackexchange.com/questions/52139/reference-request-introduction-to-mathematical-theory-of-regularization/52146#52146&quot;&gt;this&lt;/a&gt; answer by Willie Wong.&lt;/p&gt;&#xA;" OwnerUserId="251" LastEditorUserId="251" LastEditorDisplayName="" LastEditDate="2011-07-18T21:30:37.320" LastActivityDate="2011-08-09T12:27:41.377" CommunityOwnedDate="2011-07-18T01:54:19.220" Title="Are there books on Regularization at an Introductory level?" Tags="&lt;mathematical-physics&gt;&lt;books&gt;&lt;regularization&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="4" />
  <row Id="12386" PostTypeId="1" CreationDate="2011-07-18T04:42:10.030" Score="-3" ViewCount="297" Body="&lt;p&gt;I have no formal training at all in relativity. But I was just wondering if our perception of time is altered by different events. Here are some examples I had in mind:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If I touch a stove for 10 seconds versus  enjoying chocolate for 10  seconds would I experience a time distortion? An observer who records the 10 seconds would have an objective sense of time. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If I ran &quot;very quicky&quot; for 10 minutes would I experience a &quot;slowing down&quot; of time since I am exerting more effort?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Do these examples have nothing to do with relativity? In &quot;real relativity&quot; do things actually slow down when we approach the speed of light? For example, if I was going near the speed of light would I notice the hand on my watch go by slower? &lt;/p&gt;&#xA;" OwnerUserId="4523" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-18T05:45:45.543" LastActivityDate="2011-07-19T00:00:04.240" Title="Relativity in Real Life" Tags="&lt;special-relativity&gt;&lt;time&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="12387" PostTypeId="2" ParentId="12331" CreationDate="2011-07-18T04:59:21.763" Score="3" ViewCount="" Body="&lt;p&gt;Revised answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think Landau is referring to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler_angles&quot; rel=&quot;nofollow&quot;&gt;extrinsic Euler angle&lt;/a&gt; parametrization of rotations (also known as the method of &lt;a href=&quot;http://en.wikipedia.org/wiki/Givens_rotation&quot; rel=&quot;nofollow&quot;&gt;Givens rotations&lt;/a&gt; or Jacobi rotations).  Basically, there is an explicit algorithm by which one can achieve any orientation-preserving orthogonal transformation as a (highly non-unique) sequence of rotations in pairs of coordinates.  You can prove the existence of this decomposition in general by induction on the number of coordinates, and this is essentially what Philip Gibbs did in &lt;a href=&quot;http://physics.stackexchange.com/questions/12331/rotations-in-space-time/12366#12366&quot;&gt;his answer&lt;/a&gt;, for the case of dimension 4.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Original answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way I know how to make Landau's statement both precise and correct is to say that the vector space of first-order infinitesimal rotations in 4 dimensions is spanned by infinitesimal rotations in the 6 pairs of axes.  (In particular, the word &quot;resolved&quot; here is a bit of a puzzle to me.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any linear transformation in $n$ dimensions (including any rotation) can be written as an $n \times n$ matrix, where for each $k$ between $1$ and $n$, the $k$th column of the matrix gives the coordinates of where the $k$th basis vector goes.  In order for a transformation to be a rotation, we need the lengths of the vectors to be preserved, and we need the angles between them to stay the same.  We can encode these conditions in a succinct equation asserting that our matrix times its transpose is the identity.  The set of such transformations is given by the solutions to the matrix equation, so it forms an algebraic subset of the $n^2$-dimensional space of matrices.  In fact, it has the structure of a Lie group, called the orthogonal group $O(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can describe infinitesimal rotations by adding an infinitesimal element $\epsilon$ to our number system, which satisfies the properties that $\epsilon \neq 0$ and $\epsilon^2 = 0$.  An infinitesimal transformation is a matrix of the form $I + \epsilon M$, where $I$ is the $n \times n$ identity matrix, and $M$ is any matrix with real (non-infinitesimal) entries.  In order for this to be a rotation, it is necessary and sufficient that the matrix equation $(I + \epsilon M)(I + \epsilon M)^T = I$ is satisfied.  The left side can be expanded as $I + \epsilon M + \epsilon M^T$, so the equation becomes $\epsilon (M + M^T) = 0$.  Since the entries of $M$ and $M^T$ are non-infinitesimal, this is equivalent to $M$ being skew-symmetric, i.e., $M = -M^T$.  That is, the space of first-order infinitesimal rotations is the space of matrices of the form $I + \epsilon M$, where $M$ is skew-symmetric - this is also called the Lie algebra of the group $O(n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It remains to find a set that spans the space of skew-symmetric matrices.  A natural method is given taking all pairs of distinct coordinates, and for each pair, choosing an antisymmetric combination, i.e., $e_{ij} - e_{ji}$, where $e_{ij}$ is the matrix that has a 1 in the $i$th row and $j$th column and zeroes elsewhere.  This forms a linearly independent set of size $\binom{n}{2}$, which is the dimension of the space of skew-symmetric matrices.  If a matrix $M$ has the form $e_{ij} - e_{ji}$, we can think of it as an infinitesimal rotation in the $x_i x_j$ direction, since exponentiating yields the rotation: $$e^{tM} = (e_{ii} + e_{jj}) \cos t + (e_{ij} - e_{ji}) \sin t + \sum_{k \not \in \{i,j \} } e_{kk}.$$&#xA;In dimensions 3 and 4, the rotations you listed are precisely those given by pairs of distinct coordinates.&lt;/p&gt;&#xA;" OwnerUserId="761" LastEditorUserId="761" LastEditorDisplayName="" LastEditDate="2011-07-19T18:58:41.560" LastActivityDate="2011-07-19T18:58:41.560" CommentCount="2" />
  <row Id="12389" PostTypeId="2" ParentId="12386" CreationDate="2011-07-18T05:01:14.320" Score="7" ViewCount="" Body="&lt;p&gt;Relativity has nothing to do with how things &quot;feel&quot;. Time dilation, for example, is not a manifestation of the fact that it's more enjoyable to eat chocolate than to touch a stove. The results of the theory of relativity refer to measurements made by observers with absolutely no... feelings. They describe the fundamental reality of physical processes, a reality that is revealed by measurements of time intervals and distances, among other observables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part of your question, you wouldn't notice anything out of the ordinary for things that you don't move with respect to. It's only when you observe phenomena that happen in reference frames in motion with respect to you that things start looking weird. For example, something that you know has a length of 1 m when it's sitting next to you, might suddenly appear to be shorter if it's flying by you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for moving close to the speed of light, it will certainly make your clocks tick slower, but only as observed by an observer, call him Scott, who you're moving with respect to. You, looking at your clock (which is not moving with respect to you), would see no difference in the way it ticks before and after you started moving. However, if you look at Scott's clock, that would seem to be running slower than yours, simply because, from your perspective, Scott is moving with respect to you. According to you, Scott's clock is running slower, and, according to Scott, your clock is running slower. Furthermore, in your respective reference frames, you are both right. That's the essence of the theory of relativity. &lt;/p&gt;&#xA;" OwnerUserId="4483" LastEditorUserId="4483" LastEditorDisplayName="" LastEditDate="2011-07-18T05:11:45.300" LastActivityDate="2011-07-18T05:11:45.300" CommentCount="2" />
  <row Id="12390" PostTypeId="1" AcceptedAnswerId="12393" CreationDate="2011-07-18T05:34:53.780" Score="-8" ViewCount="145" Body="&lt;p&gt;In my opinion, In pure 2th-dimensional space, There is 2 movable ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in pure 3th-dimensional space, There is 3 movable ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I think in right way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any answers will be appreciated, thank you.&lt;/p&gt;&#xA;" OwnerUserId="4558" LastActivityDate="2011-07-18T13:46:49.457" ClosedDate="2011-07-18T16:55:11.600" Title="How Many Movable Ways(Direction) In Pure Nth-Dimensional Space?" Tags="&lt;space&gt;" AnswerCount="4" CommentCount="8" />
  <row Id="12391" PostTypeId="2" ParentId="12282" CreationDate="2011-07-18T05:40:55.280" Score="0" ViewCount="" Body="&lt;p&gt;The weight of liquid impact will be minimal (since with a screw cap the force will be redirected along the screw threading) - it's the lubricating effect of the liquid that actually helps. You start unscrewing, the liquid gets into the threading and lubricates it.&lt;/p&gt;&#xA;" OwnerUserId="3816" LastActivityDate="2011-07-18T05:40:55.280" />
  <row Id="12392" PostTypeId="2" ParentId="12383" CreationDate="2011-07-18T05:51:16.323" Score="1" ViewCount="" Body="&lt;p&gt;Without some additional context, there is no optimal geometry for focusing shock waves.  There are only those geometries that produce a spherical shock and those geometries that don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wasn't there, but some considerations that may have influenced the design include:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;They wanted a spherical shock, so it is natural to start with an approximately spherical initial geometry, and tweak it into shape.  This consideration favors more lenses roughly evenly distributed on the sphere.  This is also a good strategy for computational reasons, since the resources necessary to model a propagating shock with low accumulated error at the time were quite expensive.  Modeling a bigger lens may lead to greater error.  According to Wikipedia, more modern weapons have far fewer detonators and more exotic lens geometries.  Presumably, this is because the computational resources became less expensive, and reliability of mass-produced detonators became more of a concern than reliability of mass-produced lenses.&lt;/li&gt;&#xA;&lt;li&gt;If you have a distribution of lenses, you need to work out the propagation of the shock not only inside a single lens, but also the interface between different lenses.  This consideration favors having fewer lenses, or at least fewer lens geometries.  In particular, highly symmetric configurations mean fewer computations.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The truncated icosahedron has lots of roughly evenly spaced faces, and is highly symmetric: the symmetry group is transitive on the hexagons, the pentagons, the p-h edges, the h-h edges and the corners.  This means that you only need to calculate the shock wave geometry for a pentagon, a hexagon, two kinds of edges between them, and a three-way intersection.  One could instead consider dodecahedra or icosahedra (which have more transitive symmetry), but I guess that would make the lenses either substantially larger or have sharp angles that make them more difficult to handle.&lt;/p&gt;&#xA;" OwnerUserId="761" LastActivityDate="2011-07-18T05:51:16.323" />
  <row Id="12393" PostTypeId="2" ParentId="12390" CreationDate="2011-07-18T06:03:59.203" Score="2" ViewCount="" Body="&lt;p&gt;I guess that by pure $N$-dimensional space you just mean $N$-dimensional space. Indeed, in an $N$-dimensional space, there are exactly $N$ linearly-independent, i.e. mutually perpendicular, directions in which one can move.&lt;/p&gt;&#xA;" OwnerUserId="4483" LastActivityDate="2011-07-18T06:03:59.203" CommentCount="13" />
  <row Id="12394" PostTypeId="2" ParentId="12383" CreationDate="2011-07-18T06:05:19.593" Score="0" ViewCount="" Body="&lt;p&gt;Yes, it was chosen from practical point of view. Given the fact that surface separating 2 types of explosives is curved, it gives you perfect spherical implosion wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Main advantage of icosahedron is that you need only one set of forms to make parts of tiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Square/tetrahedron for example also needs 1 set of forms, but would require higher curvature of surface separating 2 explosives types which is harder to do with same precision (micron-level CNC would really help back then :-) ), and also would require higher bomb diameter to carry same amount of explosives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As technology advanced, it became possible to reduce number of tiles while maintaining implosion precision, the last I've heard of was just 2 ignitors/tiles in soviet warheads (going below 2 is not safe, as we still want to survive single accidental ignition).&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-07-18T06:05:19.593" />
  <row Id="12395" PostTypeId="2" ParentId="12384" CreationDate="2011-07-18T07:22:12.947" Score="1" ViewCount="" Body="&lt;p&gt;A couple of reasons: first, the whole framework of parton distributions is developed to apply to high-energy hadrons, as they appear in scattering experiments. These hadrons move with a very large longitudinal momentum, typically hundreds or thousands of GeV, and except at very small $x$, the partons will also have large longitudinal momenta. If a parton is going to diverge from the hadron path by any significant angle, it'll have to have a similarly large transverse momentum, and outside of hard scattering events, there just isn't enough energy around to do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At small $x$, however, it's conceivable that a parton could have an amount of transverse momentum comparable to its longitudinal momentum. The reason we don't care about this is that the cross section for parton splitting is roughly inversely related to the transverse momentum transfer ($\hat\sigma \sim 1/p_T^2$ I think), so it exhibits a sharp peak at $p_T^2 = 0$. This means we can consider the only non-negligible contribution to come from $\theta = 0$, i.e. true collinear branching, and any significant deviations of partons from the original longitudinal path are exponentially suppressed.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-07-18T07:22:12.947" />
  <row Id="12397" PostTypeId="2" ParentId="12235" CreationDate="2011-07-18T11:10:22.603" Score="5" ViewCount="" Body="&lt;p&gt;Assuming that you can &lt;em&gt;actually do&lt;/em&gt; the dimensionally regularized integrals, then yes, you can integrate iteratively. Normally, two-loop and higher-loop integrals are quite difficult and you need good tricks like turning them into differential equations or using Mellin-Barnes parametrization of the propagators (or even just Feynman or Schwinger parametrizations).&#xA;The new standard book on evaluating dimensionally regularized multiloop integrals is &#xA;&lt;a href=&quot;http://books.google.com/books?id=uYW6raeFrCUC&quot; rel=&quot;nofollow&quot;&gt;Feynman Integral Calculus&lt;/a&gt; by Vladimir Smirnov (2006). Also see the notes for the Father &amp;amp; Son &lt;a href=&quot;http://www.ippp.dur.ac.uk/Workshops/08/FeynmanIntegrals&quot; rel=&quot;nofollow&quot;&gt;lectures at Durham&lt;/a&gt; a couple of years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only are multiloop integrals hard, but the regularization scheme becomes important once you leave one-loop. This is because you have to make a consistent renormalization / subtraction of counter-terms. Regularization schemes that modify the diagrams/integrals and not the original Lagrangian can lead to messy renormalization analysis. The simplest renormalization analyses are associated with regularizations that change the Lagrangian (or dimension) combined with a mass-independent minimal subtraction scheme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inserting terms like $(q^2)^s$ is a type of analytic regularization, which (like most regularizations) is easy at one-loop, but tricky at higher loops. Analytic regularization was extensively examined by &lt;a href=&quot;http://www.math.rutgers.edu/~speer/homepage/publications.html&quot; rel=&quot;nofollow&quot;&gt;Eugene Speer&lt;/a&gt; in the 60s and 70s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zeta regularization does only work at one-loop, but it has a generalization/extension called &lt;em&gt;Operator Regularization&lt;/em&gt; that was &lt;a href=&quot;http://inspirebeta.net/search?p=f%20a%20McKeon%20and%20t%20Operator%20Regularization&quot; rel=&quot;nofollow&quot;&gt;studied by McKeon and friends&lt;/a&gt;. From memory, the calculations get a bit messy, although there is &lt;a href=&quot;http://arxiv.org/abs/1006.1806&quot; rel=&quot;nofollow&quot;&gt;a new(ish) paper&lt;/a&gt; on using operator regularization and Feynman diagrams that is on my reading list  (previous work used a more Schwinger-like, functional approach to loop calculations). &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I hope that something I said in the above is useful and not too much is wrong. ;)&lt;/p&gt;&#xA;" OwnerUserId="429" LastActivityDate="2011-07-18T11:10:22.603" CommentCount="16" />
  <row Id="12398" PostTypeId="2" ParentId="12358" CreationDate="2011-07-18T11:28:38.767" Score="5" ViewCount="" Body="&lt;p&gt;For the light quarks, one can use chiral perturbation theory to relate the mass of the light hadrons to the mass of the light quarks. These two links give details and caveats of the procedures, as well as the most precise determinations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pdg.lbl.gov/2011/reviews/rpp2011-rev-quark-masses.pdf&quot; rel=&quot;nofollow&quot;&gt;http://pdg.lbl.gov/2011/reviews/rpp2011-rev-quark-masses.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pdg.lbl.gov/2011/listings/rpp2011-list-light-quarks.pdf&quot; rel=&quot;nofollow&quot;&gt;http://pdg.lbl.gov/2011/listings/rpp2011-list-light-quarks.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1337" LastActivityDate="2011-07-18T11:28:38.767" CommentCount="3" />
  <row Id="12399" PostTypeId="2" ParentId="12390" CreationDate="2011-07-18T11:52:36.453" Score="0" ViewCount="" Body="&lt;p&gt;If by movable ways one is to understand degrees of freedom then there are Permutations(1,n) + Permutations(2,n) + ...+Permutations(n-1,n) degrees of freedom in n dimensional space, but the question will be more intresting in the case various fractal or topological dimension. For example how many degrees of freedom are there in a space of complex or matrix valued dimensions? but that would be more for the math forum.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-07-18T11:52:36.453" CommentCount="0" />
  <row Id="12401" PostTypeId="2" ParentId="12390" CreationDate="2011-07-18T12:44:42.730" Score="0" ViewCount="" Body="&lt;p&gt;You are considering only the &lt;a href=&quot;http://en.wikipedia.org/wiki/Taxicab_geometry&quot; rel=&quot;nofollow&quot;&gt;Taxicab geometry&lt;/a&gt;&lt;br&gt;&#xA;Move the original question to the one: How distances are calculated?&lt;br&gt;&#xA;The surface of a sphere (think of the Earth, use latitude and longitude and disregard altitude) is a 2 dimensional one and you can move in any direction between 0º and 360º. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nice document about the &lt;strong&gt;&lt;a href=&quot;http://www.cedar.buffalo.edu/~srihari/CSE555/Chap4.Metrics-TangentDistance.pdf&quot; rel=&quot;nofollow&quot;&gt;Notion of Distance&lt;/a&gt;&lt;/strong&gt; that I will leave here just for reference:&lt;br&gt;&#xA;a few topics: Metric Distance ,Binary Vector Distances,Tangent Distance, Distance Measures, Definition of a Metric, Weighted Euclidean Distance, nearest-neighbor classification, cluster analysis, multi-dimensional scaling, Proximity ,similarity, dissimilarity, Euclidean Distance, Similarities between cups, Sample Covariance between variables X and Y, Correlation Coefficient, Correlation Matrix, Generalizing Euclidean Distance, Minkowski Metric, Vector Space Representation of Document, Cosine Distance between Document Vectors, Distance Measures for Binary Data, Hamming Distance,Tanimoto Metric, k-nn digit recognition accuracy, Tri-Edge Inequality (TEI), Tangent Distance, Learning from a limited set of samples, Euclidean Distance versus Tangent Distance versus Manifold Distance in a picture, (Levenshtein distance is not there).&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-07-18T12:44:42.730" />
  <row Id="12402" PostTypeId="2" ParentId="12390" CreationDate="2011-07-18T13:46:49.457" Score="1" ViewCount="" Body="&lt;p&gt;You seem to be highly uncertain about what an $n$-dimensional space is supposed to be. When we talk about such a thing, we usually mean an $n$-dimensional &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_space&quot; rel=&quot;nofollow&quot;&gt;vector space&lt;/a&gt;&lt;sup&gt;*&lt;/sup&gt;. A vector space is a precisely defined mathmatical structure. What you call &quot;ways&quot; are vectors in this space. The special thing about vectors is that you can&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Add two of them together and get again a vector – this means that you can go one way and then another way, and then describe the total way you've walked again as one single way.&lt;/li&gt;&#xA;&lt;li&gt;&quot;Scale&quot; them. If you can go, say, $6$ feet in some direction, you can as well go $2\cdot 6\:\mathrm{ft}$ or any other multiple in the same direction.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, this means that &lt;em&gt;any&lt;/em&gt; vector space&lt;sup&gt;†&lt;/sup&gt; contains infinitely many &quot;ways&quot;: you can take some vector $\mathbf{v}$ and some vector $\mathbf{w}$, and some number $a$. Then $\mathbf{v}+a\cdot\mathbf{w}$ will always be a vector, and for different numbers $a$ (of which there are of course infinitely many) the result will always be different&lt;sup&gt;‡&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what does &quot;$3$-dimensional&quot; mean now? It obviously does &lt;em&gt;not&lt;/em&gt; mean that there are only 3 ways to go. But, and that's it: it is &lt;em&gt;sufficient&lt;/em&gt; to &quot;know&quot; only 3 vectors to describe any way at all. That is, any vector $\mathbf{v}\in\mathbb{R}^3$ can be written as $x\cdot \mathbf{e}_x+y\cdot\mathbf{e}_y+z\cdot\mathbf{e}_z$, where $x,y,z$ are simply numbers and $\mathbf{e}_x,\mathbf{e}_y,\mathbf{e}_z$ are your known unit-directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;* it could also be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Manifold&quot; rel=&quot;nofollow&quot;&gt;manifold&lt;/a&gt;, which is essentially a (more complicated) generalization of a vector space.&lt;br&gt;† in fact, you can mathematically consider degenerate spaces with finite elements, but these are of no physical interest.&lt;br&gt;‡ provided that $\mathbf{w}$ is not the zero-vector, that is, the way where you just stay at the same place.&lt;/p&gt;&#xA;" OwnerUserId="3540" LastActivityDate="2011-07-18T13:46:49.457" />
  <row Id="12403" PostTypeId="2" ParentId="12386" CreationDate="2011-07-18T14:41:11.300" Score="1" ViewCount="" Body="&lt;p&gt;To my understanding your question seems to be getting at explaining some neurological/psychological phenomenon and seek some possible explanations from Physics. To my knowledge the current research has not come to such a stage ( neither it was intended to, i guess, i don't know much about such things ) where you can attempt to explain neurological/psychological phenomenon from basic laws of Physics such as SR. I hope my answer serves as a compliment to AndyS's.&lt;/p&gt;&#xA;" OwnerUserId="540" LastActivityDate="2011-07-18T14:41:11.300" CommentCount="1" />
  <row Id="12404" PostTypeId="1" AcceptedAnswerId="12412" CreationDate="2011-07-18T15:32:33.350" Score="4" ViewCount="218" Body="&lt;p&gt;Theoretically, What is the difference between a black hole and a point particle of certain nonzero mass. Of-course the former exists while its not clear whether the later exists or not,&#xA;but both have infinite density.&lt;/p&gt;&#xA;" OwnerUserId="540" LastActivityDate="2011-07-19T17:36:06.163" Title="what is the difference between a blackhole and a point particle" Tags="&lt;gravity&gt;&lt;theory&gt;&lt;general-relativity&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="12405" PostTypeId="2" ParentId="12404" CreationDate="2011-07-18T15:41:15.930" Score="-3" ViewCount="" Body="&lt;p&gt;The notion of a point-like particle (electron, for example) is a very bad notion dwelling in Physics despite experiments say contrary. But if an electron tends to &lt;strong&gt;&lt;a href=&quot;https://docs.google.com/leaf?id=0B4Db4rFq72mLNmRjM2FjMzEtNmU1NS00Njk0LWIzOTAtZTk3YTNkZGYyZGUx&amp;amp;hl=en_US&quot; rel=&quot;nofollow&quot;&gt;explode&lt;/a&gt;&lt;/strong&gt;, the gravitating mass tends to &lt;strong&gt;&lt;a href=&quot;https://docs.google.com/leaf?id=0B4Db4rFq72mLZGUxZDJhNGItMTFkMS00YzllLWE0YjYtMjdjM2I2MzJlNzdj&amp;amp;hl=en_US&quot; rel=&quot;nofollow&quot;&gt;collapse&lt;/a&gt;&lt;/strong&gt;. Both features are manifestations of our blunders in describing physical phenomena.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-07-18T15:50:19.987" LastActivityDate="2011-07-18T15:50:19.987" CommentCount="2" />
  <row Id="12406" PostTypeId="2" ParentId="12404" CreationDate="2011-07-18T15:43:48.020" Score="0" ViewCount="" Body="&lt;p&gt;A point particle is a mathematical simplification, not a real thing. In reality, every particle has size, but for most considerations it's ok to neglect that size and treat it as a point particle.&lt;/p&gt;&#xA;" OwnerUserId="4483" LastActivityDate="2011-07-18T15:43:48.020" CommentCount="5" />
  <row Id="12408" PostTypeId="1" AcceptedAnswerId="12432" CreationDate="2011-07-18T16:30:16.290" Score="1" ViewCount="156" Body="&lt;p&gt;Moore's law has succesfully predicted up to now that integrated circuit transister density doubles every two years. However, computer performance is dependent on additional factors like architecture, chip design and software.&#xA;What physics-related factors will enhance or limit ultimate performance of future computers?&lt;/p&gt;&#xA;" OwnerUserId="1924" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-18T17:13:51.757" LastActivityDate="2011-07-20T01:59:12.917" Title="Future Computer Performance" Tags="&lt;computational-physics&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="12412" PostTypeId="2" ParentId="12404" CreationDate="2011-07-18T17:24:48.013" Score="3" ViewCount="" Body="&lt;p&gt;We should probably distinguish between a particle being &quot;point-like&quot; and a particle being &quot;structure-less&quot;. In classical mechanics we talk of &quot;point-like&quot; particles, objects with no extension.  It is the case that in general relativity any &quot;point-like&quot; mass would be inside of its event horizon and so would be a black hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quantum-mechanics even a &quot;structure-less&quot; particle - a particle with no consitituent parts - is wave-like and has extension, though not a fixed size, and it can never be come exactly point-like since that would take an infinite amount of energy. I do not believe it to be the case, therefore, that quantum-mechanically all particles are black holes in any sense.&lt;/p&gt;&#xA;" OwnerUserId="3970" LastEditorUserId="3970" LastEditorDisplayName="" LastEditDate="2011-07-18T18:19:18.003" LastActivityDate="2011-07-18T18:19:18.003" CommentCount="2" />
  <row Id="12414" PostTypeId="2" ParentId="12408" CreationDate="2011-07-18T17:28:59.713" Score="5" ViewCount="" Body="&lt;p&gt;The first thing that comes to mind is that the speed of light limits the rate at which different components of a computer, or even a single chip, can communicate with each other. For example, if you have, say, 10 cm of wire running between your motherboard and hard drive, it will necessarily take a minimum of about a third of a nanosecond to fetch data from the drive. Right now HDD access technology is nowhere near that fast, at least not for personal computers, but it's conceivable that at some point in the future, we could have some sort of super-fast drive access for which this limit becomes an issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could make a similar argument for the CPU itself: if CPU features are sized on the order of 10s of nanometers, it takes a signal a minimum of about $10^{-17}\text{ s}$ to travel across them. So we will not be able to make a computer that runs faster than $10^{17}\text{ Hz}$ without exploiting either subatomic technology or advanced parallel processing ;-) Again, clearly we are nowhere close to this being a problem in PCs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Large supercomputers do have to deal with these issues, though, since different parts of the system can be separated by several meters, which corresponds to tens of nanoseconds of signal travel time. This means that the supercomputer &lt;em&gt;as a whole&lt;/em&gt; would be limited to operating frequencies on the order of hundreds of MHz. In practice, modern supercomputers operate as a large cluster of individual nodes each acting quasi-independently, so the light speed limits aren't an issue.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-07-18T17:28:59.713" CommentCount="3" />
  <row Id="12417" PostTypeId="1" CreationDate="2011-07-18T18:32:29.337" Score="2" ViewCount="141" Body="&lt;p&gt;Let's say you have the ability to shine some light into a perfectly round sphere and the sphere's interior surface was perfectly smooth and reflective and there was no way for the light to escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could observe the inside of the sphere, what would you observe? A glow? And would temperature affect the outcome?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seems silly, it's just something I've always thought about but never spent enough time (until now) to actually find an answer.&lt;/p&gt;&#xA;" OwnerUserId="4531" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-18T20:38:29.103" LastActivityDate="2011-07-22T06:54:48.303" Title="What happens to light in a perfect reflective sphere?" Tags="&lt;optics&gt;&lt;light&gt;&lt;reflection&gt;&lt;mirror&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="12418" PostTypeId="1" AcceptedAnswerId="12477" CreationDate="2011-07-18T18:36:11.653" Score="2" ViewCount="125" Body="&lt;p&gt;As we know, the mass of well known particles are below.&#xA;$$m_{n}\approx0.0000000000000000000000000016749286[kg]$$&#xA;$$m_{p}\approx0.0000000000000000000000000016726231[kg]$$&#xA;$$m_{e}\approx0.00000000000000000000000000000091093897[kg]$$&#xA;I roughly know the way how to measure the electron's mass, but I don't know rest two of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any answers will be apreciated, Thank you.&lt;/p&gt;&#xA;" OwnerUserId="4558" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-18T20:55:46.997" LastActivityDate="2011-07-19T19:57:06.533" Title="How Can We Measure The Mass Of Particle?" Tags="&lt;experimental-physics&gt;&lt;mass&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12419" PostTypeId="2" ParentId="12417" CreationDate="2011-07-18T19:18:54.897" Score="2" ViewCount="" Body="&lt;p&gt;If you are observing the inside of the sphere, you are absorbing light to make the observation.  The light would get dimmer and dimmer very quickly until you could see nothing at all.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2011-07-18T19:18:54.897" CommentCount="2" />
  <row Id="12420" PostTypeId="1" CreationDate="2011-07-18T19:25:20.107" Score="6" ViewCount="127" Body="&lt;p&gt;I am curious what the electric and magnetic field's of light look like when time is stopped. A &quot;photograph&quot; or illustration/description of these fields at a moment in time is what I desire. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, does the picture change at various points in the wave?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that &lt;a href=&quot;http://physics.stackexchange.com/questions/4071/popular-depictions-of-electromagnetic-wave-is-there-an-error&quot;&gt;this&lt;/a&gt; is like a &quot;time exposure&quot; of a light wave.&lt;/p&gt;&#xA;" OwnerUserId="2843" LastEditorUserId="2843" LastEditorDisplayName="" LastEditDate="2011-07-19T03:18:04.673" LastActivityDate="2011-07-19T09:10:50.703" Title="What does an electromagnetic wave look like at a fixed moment in time?" Tags="&lt;electromagnetic-radiation&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" />
  <row Id="12421" PostTypeId="1" AcceptedAnswerId="12424" CreationDate="2011-07-18T19:37:53.243" Score="3" ViewCount="82" Body="&lt;p&gt;Why is straight gasoline (or whetever the mixture was before the introduction of ethanol) more efficient (ie, more miles/gallon) than E85?  I've known since it's introduction that E85 was less efficient, but why is it?&lt;/p&gt;&#xA;" OwnerUserId="4030" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-18T20:52:38.510" LastActivityDate="2011-07-18T20:52:38.510" Title="Why is E85 less efficient than straight gasoline?" Tags="&lt;energy&gt;&lt;physical-chemistry&gt;&lt;renewable-energy&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12422" PostTypeId="1" CreationDate="2011-07-18T20:00:44.647" Score="8" ViewCount="225" Body="&lt;p&gt;Hi physics stack exchange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am new and a mathematicians - so go easy on me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been trying to read up on QFT in the book Diagrammatica: The Path to Feynman Diagrams, and I have a question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The construction there seems very dependent on the choice of basis. Indeed, there is often an incomming basis |0&gt;,|p&gt;,|p'&gt;,|pp'&gt;,... and a similar outgoing basis. Also when considering spin (and anti) structures it fixes some basis elements related to the representation of $\mathbb{R}^4$ to &quot;fatten up&quot; the basis for the Hilbert space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;strong&gt;Where does this choice of basis come from?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://physics.stackexchange.com/questions/10818/why-should-the-observed-probability-distributions-in-quantum-mechanics-always-ali&quot;&gt;this possibly related question&lt;/a&gt;&#xA;it seems that the choice of basis maybe related to Eigen-vectors for the Hamiltonian, but maybe I misunderstood that question - indeed I don't understand the question but the answer seems to suggest this to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now.. None of the two bases (plural basis?) seems to be Eigen-spaces for the Hamiltonian, but is it then true that the incoming basis is given by a trival Hamiltonian (by which I probably means something like purely kinetic energy), which corresponds to no interactions (and similar for the outgoing basis)? Does this even make sense? I can't make this compatible in my head with the fact that e.g. the Hamiltonian in the simplest $\sigma-\pi$-model is $2\sigma \pi^2$ does not seem trivial at time $\pm$infinity - indeed it seems time independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have probably misunderstood something simple that makes some of this gibberish, and I would appreciate any clarifications.&lt;/p&gt;&#xA;" OwnerUserId="3988" LastActivityDate="2011-08-17T05:57:44.533" Title="Where do the choice of basis in QFT come from?" Tags="&lt;quantum-field-theory&gt;" AnswerCount="7" CommentCount="1" FavoriteCount="3" />
  <row Id="12423" PostTypeId="2" ParentId="12377" CreationDate="2011-07-18T20:02:04.893" Score="0" ViewCount="" Body="&lt;p&gt;People &lt;strong&gt;did&lt;/strong&gt; study similar systems on a much smaller scale, and you probably know about those studies. Hint: another name for “quantum dot” is “artificial atom”.&lt;/p&gt;&#xA;" OwnerUserId="4202" LastActivityDate="2011-07-18T20:02:04.893" />
  <row Id="12424" PostTypeId="2" ParentId="12421" CreationDate="2011-07-18T20:02:06.953" Score="2" ViewCount="" Body="&lt;p&gt;There is less chemical energy per unit mass in ethanol than there is in the major chemical components of gasoline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard enthalpies of formation for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ethanol:&lt;/strong&gt; -277.0 kJ/mol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;n-Hexane:&lt;/strong&gt; -40.0 kJ/mol (other hexanes have similar (40-60 kJ/mol) standard enthalpies)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Carbon Dioxide:&lt;/strong&gt;  -393.5 kJ/mol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Water:&lt;/strong&gt; -285.83 kJ/mol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combustion of ethanol: 1365 kJ/mol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combustion of n-Hexane: 4381 kJ/mol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even taking into account that n-Hexane has a molecular mass ~2 times greater than ethanol, you can see that burning hexane releases a lot more energy.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2011-07-18T20:02:06.953" CommentCount="2" />
  <row Id="12425" PostTypeId="2" ParentId="12333" CreationDate="2011-07-18T20:31:49.167" Score="0" ViewCount="" Body="&lt;p&gt;Quantities with different units aren't added together because there aren't any physically interesting equations such as $a + b = c$ that allow us to predict a numerical value for one, given the others. On the other hand, there are plenty where you have to multiply values together such as $F = m\frac{dv}{dt},~ C = q/v$. There's also division and forming exponents of quantities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding apples and oranges together is meaningful since it gives the the total number of objects. On the other hand, multiplying them together is generally meaningless because there isn't a meaningful quantitiy this operation calculates a value for.&lt;/p&gt;&#xA;" OwnerUserId="1345" LastActivityDate="2011-07-18T20:31:49.167" CommentCount="1" />
  <row Id="12426" PostTypeId="2" ParentId="12333" CreationDate="2011-07-18T21:49:32.360" Score="0" ViewCount="" Body="&lt;p&gt;Here is a rhetorical counter-question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q&lt;/strong&gt;: When are 2 feet x 3 feet not 6 square feet (i.e., 6 feet$^2$)? &lt;br&gt;&#xA;&lt;strong&gt;A&lt;/strong&gt;: When the two measurements were not taken at right angles to one another, and from the same origin point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, no you cannot multiple apples and oranges.  The units are not related to each other in a defined way.  You can &lt;em&gt;define&lt;/em&gt; your own dimensional result, but then you've got something meaningless that you've defined mathematically.  I.e., you can define a word to mean &quot;apples x oranges&quot;, but that doesn't mean it has any meaning.&lt;/p&gt;&#xA;" OwnerUserId="4536" LastActivityDate="2011-07-18T21:49:32.360" />
  <row Id="12427" PostTypeId="2" ParentId="12422" CreationDate="2011-07-18T22:31:24.537" Score="4" ViewCount="" Body="&lt;p&gt;The basis at $t=-\infty$ and the basis at $t=+\infty$ are both the free field basis. The free field Hamiltonian has a kinetic energy term proportional to the square of the temporal gradient of the field, but it also has a potential energy term proportional to the sum of the squares of the spatial gradients of the field, and, in the case of $\mathrm{mass} &amp;gt; \mathrm{zero}$, a potential energy term proportional to the square of the amplitude of the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perturbation theory calculates how a given state of the free field is modified by an additional interaction term (which after regularization and renormalization can be loosely thought of as infinitesimal) that acts for an infinite length of time, but that does not act at $t=-\infty$ or at $t=+\infty$. The interaction is modeled, at least in elementary textbooks, as being turned on and off by a smooth function $g(t)$ that we take to satisfy $g(t)=1$ for $|t|&amp;lt;T$ for some large time $T$, which is taken &lt;em&gt;not&lt;/em&gt; to break translation invariance when we take the limit $T\rightarrow \infty$, although of course for $T&amp;lt;\infty$ it does. There are numerous other limits that have to be taken, all of which has to be done &quot;right&quot;, which is the point of regularization and renormalization. Calculating what the eigenstates of the perturbed Hamiltonian are is an intractable problem, so we work with the eigenstates of the free field Hamiltonian, which &lt;em&gt;are&lt;/em&gt; tractable — indeed they are elementary by comparison to the eigenstates of the perturbed Hamiltonian.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-07-18T22:31:24.537" />
  <row Id="12428" PostTypeId="1" CreationDate="2011-07-18T22:40:32.260" Score="0" ViewCount="41" Body="&lt;p&gt;When conducting MIL-STD-810, Method 509 salt Spray test many metals, but especially aluminum come out covered with slime.  The slime usually dries to an easily removed powder.  What is this slime?  I was once told it was some sort of aluminum hydrate, I'm not sure I believe that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one time I did this test with aluminum that did not include galvanically dissimilar metals in contact with the aluminum, I don't recall any slime, but this test was a flat sheet treated with Hexavalent Chromium -- there was no corrosion at all.&lt;/p&gt;&#xA;" OwnerUserId="929" LastActivityDate="2011-07-18T22:40:32.260" ClosedDate="2011-07-18T22:59:01.430" Title="In Aerospace Salt Spray tests aluminum often gets covered in slime, what is this slime chemically?" Tags="&lt;physical-chemistry&gt;" CommentCount="2" />
  <row Id="12429" PostTypeId="2" ParentId="12422" CreationDate="2011-07-18T23:01:01.647" Score="3" ViewCount="" Body="&lt;p&gt;It works roughly as follows. Inspired by quantum mechanics, we'd like to find a space similar to $L^2({\mathbb R})$ that contains enough structure so that we can represent operators $\hat X$ and $\hat P$ (understood as elements of the Lie algebra of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heisenberg_group&quot; rel=&quot;nofollow&quot;&gt;Heisenberg group&lt;/a&gt;) on it as multiplication and derivation operators. One can generalize this to $n$ degrees of freedom by working on $L^2({\mathbb R}^n)$ and this also gives a clue about what to do for fields (by approximating those fields as piecewise constant configurations). Note that there is no discussion of basis whatsoever here due to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stone%3dvon_Neumann_theorem&quot; rel=&quot;nofollow&quot;&gt;Stone-von Neumann theorem&lt;/a&gt; which claims that every two (strongly continuous) representations of the Heisenberg group are equivalent, i.e. it doesn't matter what basis one chooses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is unfortunately not the case for fields. I'll not provide you with complete derivation here (which would be pedagogical but unfortunately too long) but instead I'll just state the result: the right space to work in is the space of $M \equiv L^2(X, \mu_{C, \eta})$ where $X$ is a suitable topological space (e.g. think of it as continuous functionals on ${\mathbb R}^d$ with some standard topology) and $\mu_{C, \eta}$ is a Gaussian measure (e.g. on a Borel $\sigma$-algebra associated to the space $X$) with covariance $C$ and expectation $\eta$. The physical significance of $C$ is that it is (naively) equal to $\Omega^{-1} / 2$ where $\Omega$ is a one-particle Hamiltonian and the physical significance of $\eta$ is that it gives a vacuum expectation value of the field $\hat \Phi(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the fun part starts. One can pick a representation of the field operators $\hat \Phi(x)$ and $\hat \Pi(x)$ on $M$ in the following way $\hat \Phi(x) \leftrightarrow \Phi(x) + \eta(x)$, $\hat \Pi(x) \leftrightarrow -i{ \delta \over \delta \Phi(x)} + \Omega^{-1}(\Phi(x) - \eta(x))$, find a vacuum state $V[\Phi]$ (of the $\Omega$ operator) and construct creation and annihilation operators in the usual fashion and eventually build a Fock space $F(V)$. Now consider a shift operation on the space of functionals $T_{\Psi} : \Phi(x) \mapsto \Phi(x) - \Psi_x$. It turns out that if $\Psi(x)$ is non-regular enough then $T_{\Psi} F(V)$ will be &lt;strong&gt;orthogonal&lt;/strong&gt; to the original Fock space $F(V)$ (contrast this with the quantum mechanical case where the shift by $a$ is realized by the unitary operator $\exp(-ia \hat P)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, it is possible to find a Fock space associated to some other covariance operator $C&amp;#39;$ (or equivalently one-particles Hamiltonian $\Omega'$) that is orthogonal to $F(V)$. So we see that in contrast with quantum mechanics where it was enough to work on one space $L^2({\mathbb R}^n)$ and disregard things like choices of basis or even Hamiltonian until necessary, this is no longer the case when working with fields. The space of continuous functionals is simply too big and carries a huge amount of unitarily inequivalent representations (described as Fock spaces).&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-18T23:01:01.647" CommentCount="9" />
  <row Id="12430" PostTypeId="2" ParentId="12386" CreationDate="2011-07-19T00:00:04.240" Score="1" ViewCount="" Body="&lt;p&gt;You will never &quot;feel&quot; relativistic effects in real life, because the differences are waaay too small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here's a few examples of relativity in real life:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your GPS device, and the satellites it's connecting to, need to apply relativistic corrections to keep their signals in perfect sync and ensure sufficient precision when determining your position. Without those corrections, your GPS device would be less precise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a non-relativistic universe, the metal gold would reflect light in a slightly different way, and it would not exhibit its characteristic color. It would look silvery. Its trademark hue is tributary to relativistic effects in the electron cloud in the gold atom.&lt;/p&gt;&#xA;" OwnerUserId="4210" LastActivityDate="2011-07-19T00:00:04.240" />
  <row Id="12431" PostTypeId="1" AcceptedAnswerId="12443" CreationDate="2011-07-19T01:41:00.837" Score="6" ViewCount="95" Body="&lt;p&gt;I have read in multiple places that the virial coefficients in the virial equation of state, $Z=1+{B \over v}+{C \over v^2}...$, are functions of temperature only and are independent of pressure (or equivalently, molar volume).  Is this purely an empirical observation or is there some deeper statistical mechanics reason why the virial coefficients should not depend on how close the particles are together (i.e. the molar volume)?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that $B$ represents pair-wise interactions, $C$ represent three-particle interactions, etc. but again shouldn't the coefficient, say $B$, depend on how close the two particles are to each other?&lt;/p&gt;&#xA;" OwnerUserId="4289" LastActivityDate="2011-07-19T21:10:01.790" Title="Why are the virial coefficients functions of temperature only?" Tags="&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="12432" PostTypeId="2" ParentId="12408" CreationDate="2011-07-19T01:51:43.333" Score="8" ViewCount="" Body="&lt;p&gt;Starting from 90nm tech processes we've started to see sad signs of stagnation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Most of delay in logic circuits is in interconnect, not transistors&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Most of energy dissipated is due to quantum tunneling, not transistor switching. By far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) As consequence of #2 - transistor gate width scaling has significantly slowed down, as well as dielectric width (it's already at 1.5nm, not much to reduce). We are near quantum limits already at transistor sizing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if we can make 11nm transistors today, it would not make things 3 times faster and 9 smaller than we have now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are few way to significantly improve CPU technology:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) superconductor interconnect - will improve CPU performance by a large margin, and will allow much larger CPU's. The problem here is that Physicists haven't yet discovered suitable superconductors, which could be patterned in 50-500nm lines and does not require some -100C. The one who will find a way to do such interconnect at room temperature would be extremely rich. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) 3D transistor packing: i.e. instead 2D array of 1000x1000 transistors we would have 100x100x100 3D array - and interconnect length is 10 times shorter. This is actively being researched on, the problem is that some stages of manufacturing layer of modern high-performance transistors require temperatures of 1000C and higher, and this would destroy transistors on lower levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Optical interconnect: doing tiny LEDs &amp;amp; photo-diodes and transparent channels out of SiO2 would also allow for faster interconnect. This is also actively being worked on.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;All these fancy things like graphene transistors, quantum dots and fancy HEMT structures indeed are abit faster (and ALOT more expensive or complex to manufacture), but does not solve problems with interconnect, which is major problem. Individual transistors at chips can switch much faster than 4Ghz, but we aren't really limited by transistor switching speed at the moment.&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="930" LastEditorDisplayName="" LastEditDate="2011-07-19T04:03:17.407" LastActivityDate="2011-07-19T04:03:17.407" CommentCount="3" />
  <row Id="12434" PostTypeId="2" ParentId="12420" CreationDate="2011-07-19T04:50:25.580" Score="1" ViewCount="" Body="&lt;p&gt;Have a look at the graphic in &lt;a href=&quot;http://physics.stackexchange.com/questions/10915/why-photons-transfer-to-electrons-perpendicular-momentum/10918#10918&quot;&gt;Lubos Motl's answer&lt;/a&gt; to another electromagnetic question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It shows a  polarised plane wave propagating in time and you can take a time projection that answers your question. Unpolarised plane waves have all possible orientations in space.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-07-19T04:50:25.580" />
  <row Id="12435" PostTypeId="1" AcceptedAnswerId="12438" CreationDate="2011-07-19T05:02:43.067" Score="11" ViewCount="344" Body="&lt;p&gt;What's the cleanest/quickest way to go between Einstein's postulates [&lt;a href=&quot;http://en.wikisource.org/wiki/On_the_Electrodynamics_of_Moving_Bodies&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;] of&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Relativity: Physical laws are the same in all inertial reference frames.&lt;/li&gt;&#xA;&lt;li&gt;Constant speed of light:  &quot;... light is always propagated in empty space with a definite speed $c$ which is independent of the state of motion of the emitting body.&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;to Minkowski's idea [&lt;a href=&quot;http://en.wikisource.org/wiki/Space_and_Time&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;] that space and time are united into a 4D spacetime with the indefinite metric &#xA;$ds^2 = \vec{dx}^2 - c^2 dt^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related to the question of what is the best derivation of the correspondence are:&lt;br&gt;&#xA;Is the correspondence 1:1? (Does the correspondence go both ways?)&lt;br&gt;&#xA;and are there any hidden/extra assumptions?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Edit&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Marek's answer is really good (I suggest you read it and its references now!), but not quite what I was thinking of. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for an answer (or a reference) that shows the correspondence using only/mainly simple algebra and geometry. An argument that a smart high school graduate would be able to understand.&lt;/p&gt;&#xA;" OwnerUserId="429" LastEditorUserId="429" LastEditorDisplayName="" LastEditDate="2011-07-20T00:46:41.980" LastActivityDate="2011-08-16T06:34:08.957" Title="Einstein's postulates &lt;==&gt; Minkowski space. (In layman's terms)" Tags="&lt;special-relativity&gt;&lt;spacetime&gt;&lt;minkowski-space&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="6" />
  <row Id="12436" PostTypeId="1" AcceptedAnswerId="12439" CreationDate="2011-07-19T06:01:24.373" Score="2" ViewCount="82" Body="&lt;p&gt;I'm looking at the solutions to a problem about a uniform thin disk. For the sake of this question, I start with&#xA;$$L=\frac{1}{2}m\left( r\omega \right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we plug it into Lagrange's equations:&#xA;$$\begin{align*}&#xD;&#xA;\frac{d}{dt} \frac{\partial L}{\partial \dot{q}_j} - \frac{\partial L}{\partial q_j} &amp;amp;= Q\\&#xD;&#xA;\frac{d}{dt} \frac{\partial \frac{1}{2}m\left( r\omega \right)^2}{\partial \left( r\omega \right)} - \frac{\partial \frac{1}{2}m\left( r\omega \right)^2}{\partial q_j} &amp;amp;= Q&#xD;&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How is it that $\dot{q}_j = r\omega$?&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;What is $q_j$ then, as well?&lt;/strong&gt; I'm thinking along the lines of&#xA;$$\begin{align*}&#xD;&#xA;\dot{q}_j &amp;amp;= r\omega\\&#xD;&#xA;\dot{q}_j &amp;amp;= r\frac{d\theta}{dt}\\&#xD;&#xA;q_j &amp;amp;= r\theta&#xD;&#xA;\end{align*}$$&lt;/p&gt;&#xA;" OwnerUserId="293" LastEditorUserId="293" LastEditorDisplayName="" LastEditDate="2011-07-19T06:21:38.740" LastActivityDate="2011-07-19T07:07:32.433" Title="Lagrange's equations: What is $\dot{q}_j$?" Tags="&lt;classical-mechanics&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" />
  <row Id="12437" PostTypeId="2" ParentId="10968" CreationDate="2011-07-19T06:49:33.683" Score="2" ViewCount="" Body="&lt;p&gt;The main effect of an electromagnetic wave is basically that the electric field in the electromagnetic wave shoves charged particles around (ions and/or electrons). That's called &quot;Electric dipole coupling&quot;. Electric dipole coupling is almost always much stronger than other effects of the electromagnetic wave. For example, the electromagnetic wave has a magnetic field too, which can exert forces on molecules. This &quot;magnetic dipole coupling&quot; is a much smaller effect than the electric dipole coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shoving an electron around (the electric dipole coupling) does not directly change or rotate the electron's spin. It only changes the spin a little bit, due to &quot;spin-orbit coupling&quot;, an effect related to special relativity (in the &quot;rest frame&quot; of the moving electron, the electric fields get converted to magnetic fields, which torques the spin). Therefore, since spin-orbit coupling is a weak effect (unless the electron is traveling near the speed of light), the coupling between electromagnetic waves and molecules cannot usually involve a change of spin. That means that a spin-singlet molecule absorbing light almost definitely goes into a spin-singlet excited state. Conversely, it is exceedingly unlikely for a spin-triplet excited state to emit light while changing into a spin-singlet ground state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, another possibility is for the magnetic field of the light wave to directly torque the spin of the electron. This effect is normally even weaker than the effect of the wave via the spin-orbit effect.&lt;/p&gt;&#xA;" OwnerUserId="3811" LastActivityDate="2011-07-19T06:49:33.683" />
  <row Id="12438" PostTypeId="2" ParentId="12435" CreationDate="2011-07-19T06:55:34.557" Score="12" ViewCount="" Body="&lt;p&gt;I will first describe the naive correspondence that is assumed in usual literature and then I will say why it's wrong (addressing your last question about hidden assumptions) :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The postulate of relativity would be completely empty if the inertial frames weren't somehow specified. So here there is already hidden an implicit assumption that we are talking only about rotations and translations (which imply that the universe is isotropic and homogenous), boosts and combinations of these. From classical physics we know there are two possible groups that could accomodate these symmetries: the Gallilean group and the Poincaré group (there is a catch here I mentioned; I'll describe it at the end of the post). Constancy of speed of light then implies that the group of automorphisms must be the Poincaré group and consequently, the geometry must be Minkowskian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;em&gt;Sidenote:&lt;/em&gt; how to obtain geometry from a group? You look at its biggest normal subgroup and factor by it; what you're left with is a homogeneous space that is acted upon by the original group. Examples: $E(2)$ (symmetries of the Euclidean plane) has the group of (improper) rotations $O(2)$ as the normal subgroup and $E(2) / O(2)$ gives ${\mathbb R}^2$. Similarly $O(1,3) \ltimes {\mathbb R}^4 / O(1,3)$ gives us Minkowski space.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The converse direction is trivial because it's easy to check that the Minkowski space satisfies both of Einstein postulates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to address the catch: there are actually not two but &lt;strong&gt;eight&lt;/strong&gt; kinematical groups that describe isotropic and uniform universes and are also consistent with quantum mechanics. They have classified in the &lt;a href=&quot;http://usparc.ihep.su/spires/find/hep/www?irn=6756301&quot; rel=&quot;nofollow&quot;&gt;Bacry, Lévy-Leblond&lt;/a&gt;. The relations among them is described in the &lt;a href=&quot;http://www.math.uh.edu/~tomforde/Articles/Missed-Opportunities-Dyson.pdf&quot; rel=&quot;nofollow&quot;&gt;Dyson's Missed opportunities&lt;/a&gt; (p. 9). E.g., there is a group that has absolute space (instead of absolute time that we have in classical physics) but this is ruled out by the postulate of constant speed of light. In fact, only two groups remain after Einstein's postulate have been taken into account: besides the Poincaré group, we have the group of symmetries of the &lt;em&gt;de Sitter space&lt;/em&gt; (and in terms of the above geometric program it is $O(1,4) / O(1,3)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, one could also drop the above mentioned restriction to groups that make sense in quantum mechanics and then we could also have an &lt;em&gt;anti de Sitter space&lt;/em&gt; ($O(2,3) / O(1,3)$). In fact, this shouldn't be surprising as general relativity is a natural generalization of the special relativity so that the Einstein's postulates are actually weak enough that they describe &lt;em&gt;maximally symmetric Lorentzian manifolds&lt;/em&gt; (which probably wasn't what Einstein intented originally).&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-19T06:55:34.557" CommentCount="5" />
  <row Id="12439" PostTypeId="2" ParentId="12436" CreationDate="2011-07-19T07:07:32.433" Score="2" ViewCount="" Body="&lt;p&gt;You have it backwards. When trying to describe some system, you first need to find kinematical parameters that describe possible configurations the system could be in. Here, in the most simple picture, the disk is characterized by a single parameter $q \equiv \phi$, the angle it has rotated (since the beginning of the experiment, say) and the corresponding velocity parameter $\dot q = \dot \phi = \omega$. Once you have your parameters, you can try to express kinetic and potential energy in terms of them and finally you can write a Lagrangian $L = T(\dot \phi) - V(\phi)$. Here you have only kinetic energy and it corresponds (contrary to what you write) to a particle constrained to a circle of radius $r$ (the actual disk would carry a different factor in front of $mv^2 / 2 = m (r \omega)^2 / 2$ due to the fact that its center of mass is at the position $r_0 &amp;lt; r$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, with that out of the way, you have Lagrange's equations&#xA;$${{\rm d} \over {\rm d} t} {\partial L \over \partial {\omega}} - {\partial L \over \partial {\phi}} = 0.$$&#xA;Can you take it from here?&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-19T07:07:32.433" CommentCount="4" />
  <row Id="12440" PostTypeId="1" AcceptedAnswerId="12441" CreationDate="2011-07-19T07:09:03.847" Score="0" ViewCount="46" Body="&lt;p&gt;I'm looking to transfer very simple information using audio waves. One of the approaches I'm looking into is using different frequencies. For example, &quot;command 1&quot; will be transmitted using 500Hz wave, where &quot;command 2&quot; using 700Hz wave. On the other side, the decoder will look if we have a wave with that frequency and then interpret the command.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The challenge appears when I'd like to transmit 2 (or more) commands simultaneously - I get a new wave form. If I remember right from my physics studies - it's a &quot;Wave package&quot;, so I get new frequencies, additional to the original ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is - Is there a way to receive this &quot;wave package&quot; and interpret the original commands from it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks! &lt;/p&gt;&#xA;" OwnerUserId="4540" LastActivityDate="2011-07-19T07:28:01.940" Title="Waves and information" Tags="&lt;waves&gt;&lt;information&gt;" AnswerCount="1" />
  <row Id="12441" PostTypeId="2" ParentId="12440" CreationDate="2011-07-19T07:28:01.940" Score="2" ViewCount="" Body="&lt;p&gt;Actually, when you combine (superpose) waves with different frequencies, they don't produce new frequencies. You can perform a &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;Fourier transform&lt;/a&gt; on the combined wave to get the two original frequencies that went into it. This is a very common operation in signal processing and it can easily be implemented by a computer, or by certain kinds of purely physical systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to give an example, this is how multiple radio stations can be broadcasting at once. Each station has a different frequency, and their signals add together in the air, but your radio is easily able to pick out the individual signal from just one station.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-07-19T07:28:01.940" CommentCount="7" />
  <row Id="12443" PostTypeId="2" ParentId="12431" CreationDate="2011-07-19T07:51:36.657" Score="3" ViewCount="" Body="&lt;p&gt;I will only give a general idea of why this should be so. The concrete derivation for non-ideal gas is a little tricky and not very illuminating in my opinion. So, let us write the grand canonical partition function&#xA;$$\Omega \equiv \sum_n Z_n {\lambda}^n$$&#xA;where $\lambda = e^{\mu \over k_B T}$ and $Z_n$ is $n$-particle canonical partition function.&#xA;Now, the virial expansion is nothing else than &lt;a href=&quot;http://en.wikipedia.org/wiki/Cluster_expansion&quot; rel=&quot;nofollow&quot;&gt;cluster expansion&lt;/a&gt; of $\Omega$. This is just a combinatorial theorem, so that you can write $$\log(\Omega) = \sum_n B_n {\lambda}^n$$ with $B_n = V^{n-1} f_n(Z_1, \cdots, Z_n)$ and $f_n$ determined by the cluster expansion theorem.&#xA;Because none of the $Z_n$ depends on the pressure, $B_n$ can't either. Actually, the pressure enters these formulas only through the thermodynamic identity $\log(\Omega) = {pV \over k_B T}$.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-07-19T09:04:38.693" LastActivityDate="2011-07-19T09:04:38.693" />
  <row Id="12444" PostTypeId="1" AcceptedAnswerId="12449" CreationDate="2011-07-19T08:03:08.563" Score="8" ViewCount="192" Body="&lt;p&gt;I've been seeing many feynman diagrams lately that I can't quite interpret yet. I've heard a basic Quantum Field Theory lecture and so to me, a Feynman diagram is simply a mnemonic picture to quickly write down (and remember) all of the possible terms in the perturbation series of the matrix element in scattering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But they seem to be much more than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider for example &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/b/b8/Pn_scatter_pi0.png&quot; rel=&quot;nofollow&quot;&gt;this diagram of proton-neutron scattering via pion exchange&lt;/a&gt;. The picture seems to have an intuitive meaning, but how can it be a valid Feynman diagram? Wouldn't that imply that there is an unterlying theory with a proper Lagrangian from which Feynman rules can be derived that would assign a number to this diagram?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question: When describing nucleon-nucleon interactions in this way, do people actually write down a Lagrangian and derive those rules?&lt;/strong&gt; (I am confused because I've seen diagrams like that in several books and they usually write down the cross sections simply from analogy arguments to other theories.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any clarification will be greatly appreciated,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;David&lt;/p&gt;&#xA;" OwnerUserId="4541" LastActivityDate="2011-07-19T10:33:49.973" Title="Feynman diagrams in effective theories" Tags="&lt;quantum-field-theory&gt;" AnswerCount="3" />
  <row Id="12446" PostTypeId="2" ParentId="12444" CreationDate="2011-07-19T08:18:51.303" Score="5" ViewCount="" Body="&lt;p&gt;Well, it need not be a Lagrangian, you can work in Hamiltonian formalism too (which is often possible outside of particle physics, e.g. in condensed matter theory). But except for this caveat &lt;em&gt;the answer is a&lt;/em&gt; &lt;strong&gt;resounding yes&lt;/strong&gt;: the theory comes first and diagrams are just mnemonics. Diagram by itself doesn't make any sense if you can't reconstruct the corresponding term in the expansion from it. And you can only do this if you know precisely what theory you are dealing with (so you need to know what types of particles there are and in particular their Lorentz structure and Dirac structure, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you are familiar enough with the diagrammatic approach and comfortable with most of QFT and particle physics (so that you can write down Lagrangian of the most common types of interactions and also &quot;see&quot; what interaction vertices it corresponds to) then there is no problem working directly with Feynman diagrams and never mention any Lagrangian at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me make (by now perhaps superfluous) analogy to the differential calculus. You can learn your ${{\rm d} x^n \over {\rm d} x} = n x^{n-1}$ formulas and product and chain rules and comfortably differentiate all kinds of functions. Yet, at some point you should learn that these are just mnemonics and differentiation is defined in terms of limits (which is what you will have to return to anytime you encounter a function for which you have no rule). Still, most of the time the standard rules are all you need.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-19T08:18:51.303" CommentCount="3" />
  <row Id="12447" PostTypeId="2" ParentId="12420" CreationDate="2011-07-19T08:51:39.807" Score="1" ViewCount="" Body="&lt;p&gt;Here are some animations / pictures which give a &#xA;more complete impression of a electromagnetic wave: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/File:Felder_um_Dipol.jpg&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/File:Felder_um_Dipol.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.elsenbruch.info/ph12_dipol.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.elsenbruch.info/ph12_dipol.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter link has an animated Near-field at the &#xA;bottom of the page.&#xA;The links from Joe and Lubos are limited to a &quot;ray&quot; &#xA;perpendicular to the dipols far-field, they do not &#xA;give the whole wave. This kind of pictures are &#xA;somewhat &quot;dangerous&quot;, because laypeople often take &#xA;a piece of such a ray with the e- and B-field sines &#xA;as a model for photons. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-07-19T09:10:50.703" LastActivityDate="2011-07-19T09:10:50.703" />
  <row Id="12448" PostTypeId="2" ParentId="12444" CreationDate="2011-07-19T10:27:51.680" Score="3" ViewCount="" Body="&lt;p&gt;The effective Lagrangian upon which this Feynman diagram is based is the Yukawa effective theory. See, for example, the following &lt;a href=&quot;http://www.thphys.uni-heidelberg.de/~duo/skripten/wetteric_QFT2.pdf&quot; rel=&quot;nofollow&quot;&gt;exposition &lt;/a&gt;by: Christof Wetterich. The kinetic terms are given in equations 1.3 and 1.4 and the intercation term (Yukawa interaction) is given in equation 1.5 for a charged pion doublet and by 1.19 for a pion triplet including the neutral Pion. Please notice that the difference between the proton and neutron masses explicitely breaks the isospin invariance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an old theory that was used before the standard model was discovered. Although, this is a perturbatively renormalizable theory, there is no profit in performing computation&#xA;beyond the tree level, because the proton, neutron and pions are not elementary. Also both the Yukawa coupling and the particle masses are external parameters.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastActivityDate="2011-07-19T10:27:51.680" />
  <row Id="12449" PostTypeId="2" ParentId="12444" CreationDate="2011-07-19T10:33:49.973" Score="6" ViewCount="" Body="&lt;p&gt;For nucleon-nucleon interaction please keep in mind that in this low-energy regime pertubative QCD breaks down and reactions are not really calculable. For the specific pion exchange you mention have a look at &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/BR7jg.gif&quot; alt=&quot;Nucleon-Nucleon-Interaction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/forces/funfor.html&quot; rel=&quot;nofollow&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/forces/funfor.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as to why this QCD-process can be seen as an exchange of a pion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general you can get Lagrangians for effective (i.e. low-energy) theories by integrating out the high momentum degrees of freedom. For example the W-boson propagator&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{-i(g_{\mu\nu}-\frac{q_\mu q_\nu}{m_W})}{q^2-M^2_W}$ in the limit of small $q^2$ becomes $\frac{ig_{\mu\nu}}{M^2_W}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so instead of the full electro-weak symmetry there would be a new lagrangian with a four point interaction term $\frac{G_F}{\sqrt{2}}J_\mu J^\mu$ where $J_\mu$ is a left-handed Dirac current. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more general terms this integrating out of high momentum degrees of freedom is the point of view on renormalization taken by Wilson. In this view our current theories are effective theories that result from an unknown fundamental Lagrangian from which all the high momentum d.o.f. have been integrated out. (c.f. renormalization group)&lt;/p&gt;&#xA;" OwnerUserId="2151" LastActivityDate="2011-07-19T10:33:49.973" CommentCount="3" />
  <row Id="12450" PostTypeId="1" CreationDate="2011-07-19T11:53:07.113" Score="1" ViewCount="135" Body="&lt;p&gt;Dirac gives the relation: $\exp(iaq)f(q,p) = f(q, p - a\hbar)\exp(iaq)$ where $\hbar$ is Planck's constant. Can anybody give me the corresponding relation when the $\exp$ function is a $\ln$?&lt;/p&gt;&#xA;" OwnerUserId="4543" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-21T10:53:54.213" LastActivityDate="2011-07-21T10:53:54.213" Title="Operator relation involving the logarithm of an operator?" Tags="&lt;quantum-mechanics&gt;&lt;mathematical-physics&gt;&lt;operators&gt;&lt;commutator&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="12451" PostTypeId="2" ParentId="12431" CreationDate="2011-07-19T11:53:30.050" Score="1" ViewCount="" Body="&lt;p&gt;You asked two different questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Why it doesn't depend on the the pressure? Because pressure is not an independent variable. You can calculate it as the derivative of the thermodynamical potential, although, to be honest, in this case at least, the thermodynamical potential is already $-Pv$ because you do this expansion using a grand-canonical ensemble.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why it doesn't depend on how close the particles are? Well, it does not depend because you average over it. When you do thermodynamics, you don't care about specific micro-states, just the ensemble average. If you see the definition of the virial coefficients, you will realize that those are integrals over the positions with a probability density that comes from a perturbative expansion of the Gibbs distribution. This is the average I'm referring to.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1337" LastActivityDate="2011-07-19T11:53:30.050" />
  <row Id="12453" PostTypeId="2" ParentId="12450" CreationDate="2011-07-19T13:19:09.433" Score="6" ViewCount="" Body="&lt;p&gt;You say anticommutator but this relation is a simple consequence of the &lt;strong&gt;commutation&lt;/strong&gt; relations $[q, p] = i \hbar$.&#xA;Let us compute&#xA;$$[q^n, p] = q [q^{n-1}, p] + [q, p] q^{n-1} =$$&#xA;$$ =q(q[q^{n-2}, p] + [q,p] q^{n-2}) + i \hbar q^{n-1} = \cdots = i \hbar n q^{n-1}$$&#xA;Next, suppose a function $g(q) = \sum_{n=0}^{\infty} {a_n \over n!} q^n$ is analytic in $q$ around 0 (it also works for function analytic around other points but it would make the derivation a little messier). Then we have&#xA;$$[g(q), p] = \sum_{n=0}^{\infty} {a_n \over n!} [q^n, p] = i \hbar \sum_{n=1}^{\infty} {a_n \over (n-1)!} q^{n-1} = i \hbar g&amp;#39;(q) $$&#xA;For $g(q) = \exp(iaq)$ we get $[\exp(iaq), p] = - \hbar a \exp(iaq)$ (this is already your relation for $f(p,q) \equiv p$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next we compute (by successive commutations of the exponential to the right using the relation above)&#xA;$$\exp(iaq) p^n = (p - \hbar a) \exp(iaq))p^{n-1} = \cdots = (p - \hbar a)^n \exp(iaq)$$&#xA;Finally, assuming $f(q,p) = \sum_{n=0}^{\infty} {b_n(q) \over n!} p^n$ is analytic around 0 in $p$ we have&#xA;$$\exp(iaq) f(q,p) =  \sum_{n=0}^{\infty} {b_n(q) \over n!} \exp(iaq) p^n= $$&#xA;$$ =\sum_{n=0}^{\infty} {b_n(q) \over n!} (p - \hbar a)^n \exp(iaq) = f(q, p-\hbar a) \exp(iaq) $$&#xA;which is your claimed relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there is no reason to expect that $\log$ will give any reasonable relation. Exponential was special because it maps $q$ (which is a generator of the Lie algebra of the Heisenberg group in this representation) to an element that performs shift in momentum (which is precisely what your relation says).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To demonstrate this, if we instead chose above $g(q) = \log(iaq)$ we would only get $[\log(iaq), p] = - {a \over q} $ but this obviously can't lead to any similar relation like the one we had above.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-19T13:19:09.433" CommentCount="1" />
  <row Id="12454" PostTypeId="2" ParentId="12422" CreationDate="2011-07-19T13:32:50.407" Score="1" ViewCount="" Body="&lt;p&gt;I don't quite understand what you are asking, but I interpret your question to be the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the construction of the Fock space $F(V)$ via multi-particle states $|0\rangle, |n_1\rangle, |n_1n_2\rangle, \dots$ independent of the basis? In other words, does the Fock space depend on the choice of basis for the single particle Hilbert space, $V \simeq \bigoplus_n \mathbb C|n\rangle$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is &quot;no&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is often very convenient to choose a basis of the single particle Hilbert space and use it for labeling states from the Fock space. For instance, the notation $|n_1,n_2\rangle$ is often denotes a state with $n_1$ particles in the (single-particle) quantum state $|1\rangle$ and $n_2$ particles in quantum state $|2\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, choosing a basis is not mandatory for constructing the Fock space. It's simply the direct sum of the tensor powers of $V$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F(V) = \bigoplus_{n=1}^\infty S(V^{\otimes n}) ,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S$ denotes symmetrization (for bosons) or antisymmetrization (for fermions). No choice of basis to be seen anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: for $n_1=1$, $n_2=1$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ |n_1,n_2\rangle = \frac{1}{\sqrt{2}}(|1\rangle\otimes|2\rangle - |2\rangle\otimes|1\rangle)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the notation on the left-hand side depends on a choice of basis, but the expression on the right-hand side gives it a basis-independent meaning.&lt;/p&gt;&#xA;" OwnerUserId="188" LastActivityDate="2011-07-19T13:32:50.407" CommentCount="3" />
  <row Id="12455" PostTypeId="2" ParentId="12450" CreationDate="2011-07-19T14:03:19.803" Score="3" ViewCount="" Body="&lt;p&gt;You can get an integral representation of the result using the following integral representation of the natural logarithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln(x) = \int_0^{\infty}\frac{e^{-s}-e^{-sx}}{s} ds$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This representation converges for $0 &amp;lt; |x|&amp;lt; \infty$, however, extra care must be taken when used as an operator identity.&#xA;At least for plane waves in the momentum space, the result is trivially convergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the identity to the function we obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln(iaq)f(q, p) = \int_0^{\infty}\frac{e^{-s}f(q, p)-f(q, p-sa\hbar)e^{-iasq}}{s}ds$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a technical method to perform the computation of $ \ln(q) \ln(p)$. However, it must be emphasized that&#xA;the logarithm is a multiple valued function and the validity of the expression must be tested on the class of functions you want to apply this operator identity on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The natural logarithm of a number can be defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln(x) = \frac{d}{dt} x^t|_{t=0}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this formula we may write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln(q)\ln(p)  = \frac{d}{ds}\frac{d}{dt} q^s p^t|_{s=t=0}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If s and t were integer we would have the following relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q^s p^t = (i\hbar)^s\frac{t!}{s!} p^{t-s} q^s$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We continue this relation to non-integer exponents as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q^s p^t = (i\hbar)^s\frac{\Gamma(1+t)}{\Gamma(1+s)} p^{t-s} q^s$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Gamma$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_function&quot; rel=&quot;nofollow&quot;&gt;gamma function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By taking the derivatives with respect to s and t for both sides and the limits to zero, we obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln(q)\ln(p)  = (\ln(i\hbar p) - \gamma) (\gamma + \ln(q) - \ln(p))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, $\gamma$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler-Mascheroni_constant&quot; rel=&quot;nofollow&quot;&gt;Euler constant&lt;/a&gt;, and the following identities were used:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d}{dx} \Gamma(x) = \Gamma(x) \psi(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\psi$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Digamma_function&quot; rel=&quot;nofollow&quot;&gt;digamma function&lt;/a&gt;, and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\psi(1) = - \gamma$.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorUserId="2190" LastEditorDisplayName="" LastEditDate="2011-07-21T09:01:08.727" LastActivityDate="2011-07-21T09:01:08.727" CommentCount="6" />
  <row Id="12456" PostTypeId="1" AcceptedAnswerId="12479" CreationDate="2011-07-19T14:05:17.940" Score="10" ViewCount="214" Body="&lt;p&gt;You may have seen the story of the &lt;a href=&quot;http://www.cnn.com/2011/TECH/mobile/07/18/iphone.skydive/index.html?hpt=us_t2&quot; rel=&quot;nofollow&quot;&gt;iPhone which was dropped from perhaps 13,500&lt;/a&gt; feet by a skydiver - it survived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This made me wonder how to work out the terminal velocity for something like that. Obviously calculating terminal velocity for a sphere can be relatively straightforward, but with a flattened oblong, what factors come into play?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;End on will be fast, flat will be slow, but is there a stable configuration?&lt;/p&gt;&#xA;" OwnerUserId="2498" LastEditorUserId="4550" LastEditorDisplayName="" LastEditDate="2011-07-19T17:26:31.763" LastActivityDate="2011-07-21T12:34:57.730" Title="What is the terminal velocity for a mobile phone" Tags="&lt;gravity&gt;&lt;velocity&gt;&lt;air&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="5" />
  <row Id="12457" PostTypeId="1" CreationDate="2011-07-19T14:06:10.203" Score="1" ViewCount="80" Body="&lt;p&gt;What is sector decomposition and how can it be used to 'disentangle' UV and IR divergences?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read about it in the paper &lt;a href=&quot;http://arxiv.org/abs/1011.5493&quot; rel=&quot;nofollow&quot;&gt;SecDec: A general program for sector decomposition&lt;/a&gt;,&lt;br&gt;&#xA;but I have no idea if, with a suitable change of variable $x=1/q$ on each coordinate, we can change the IR divergences to UV divergences or change the UV divergences to IR ones and then apply sector method to disentangle the divergences.&lt;/p&gt;&#xA;" OwnerUserId="4435" LastEditorUserId="429" LastEditorDisplayName="" LastEditDate="2011-08-01T04:16:45.537" LastActivityDate="2011-08-01T05:55:24.603" Title="What is sector decomposition" Tags="&lt;mathematical-physics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="12458" PostTypeId="2" ParentId="12422" CreationDate="2011-07-19T14:37:46.940" Score="2" ViewCount="" Body="&lt;p&gt;The basis for QFT is normally the eigenvectors of some exactly solvable problems, free particles, as example. These eigenvectors (in scattering problems) are exact ones in asymptotic regions where the interaction is negligible. The latter is not always the case so the one encounters mathematical and conceptual problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These basis eigenvectors can describe bound states too, see atom-atomic scattering in atomic physics.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-07-19T14:37:46.940" />
  <row Id="12459" PostTypeId="2" ParentId="12457" CreationDate="2011-07-19T14:49:47.217" Score="0" ViewCount="" Body="&lt;p&gt;UV divergence is due to self-action. It exists even in statics. No self-action is introduced, no UV divergence appears in calculation. Renormalizations discard the self-action contributions perturbatively rather than exactly, at the beginning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IR divergence is due to too distant initial approximation, distant from the exact solution, of course. It is like using a Taylor expansion $f(x) = f(0) + f&amp;#39;(0)\cdot x + ...$ for calculations where $x$ is infinite (distant from zero).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So these divergences are not interchangeable.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-07-19T14:49:47.217" CommentCount="5" />
  <row Id="12460" PostTypeId="1" CreationDate="2011-07-19T14:57:46.310" Score="6" ViewCount="98" Body="&lt;p&gt;I've read &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_computer&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, I've read &lt;a href=&quot;http://www.howstuffworks.com/quantum-computer.htm&quot; rel=&quot;nofollow&quot;&gt;How Stuff Works&lt;/a&gt;, I've read &lt;a href=&quot;http://www.singularity.com/&quot; rel=&quot;nofollow&quot;&gt;The Singularity is Near&lt;/a&gt;, but I still just don't get it. How does a Quantum Computer work? It sounds very intriguing, but I just can't wrap my mind around it. I would greatly appreciate it if someone could explain it to me as they would explain the basics of it to their 7 year old nephew (who apparently is probably about equal to me in their ability to understand this concept!) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically analogies work best when explaining concepts to the unenlightened, but this is quantum theory, so I'm not sure any analogy could exist that can explain such a bizarre concept ;)&lt;/p&gt;&#xA;" OwnerUserId="4545" LastEditorUserId="4545" LastEditorDisplayName="" LastEditDate="2011-07-19T15:09:30.610" LastActivityDate="2011-07-19T17:52:05.797" Title="How Does a Quantum Computer Work?" Tags="&lt;quantum-computer&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="12461" PostTypeId="1" CreationDate="2011-07-19T15:00:41.233" Score="5" ViewCount="110" Body="&lt;p&gt;I want to outline a solid argument (or bulletpoints) to show how weak is the idea of diff(M) being the gauge group of general relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;basically i have these points that in my view are very solid but i want to understand if there are misconceptions on my part that i'm simply not getting and if its so, i ask for help to make the case more solid, or understanding why it doesn't apply (to gravity):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.mth.kcl.ac.uk/~streater/lostcauses.html#XXII&quot; rel=&quot;nofollow&quot;&gt;gauge groups are not the same as a symmetry group&lt;/a&gt; (thanks to Raymond Streater for making that point completely clear)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;gauge invariance in electrodynamics is an observation that physical observables are unchanged after a gauge transformation &lt;strong&gt;without changing coordinate frame&lt;/strong&gt; ( we are ask to believe that in gravity someone did the same? that is, someone made the observation that physical observables are unchanged after a diffeomorphism-gauge-transformation, only to later argue that because of this, that there are no physical observables to begin with, that doesn't make a lot of sense, to not say that its just plain stupid circular argument )&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Maxwell%27s_equations_in_curved_spacetime&quot; rel=&quot;nofollow&quot;&gt;classic electrodynamics is also invariant (as in symmetry invariant, not as gauge-invariant) under Diff(M)&lt;/a&gt;. The invariance is of course broken when the theory is quantized and $\hbar$ makes an appearance, because it assumes a preferred scale for certain energies. The key point here is: &lt;strong&gt;classical gravity is not special regarding having diff(M) as a symmetry group&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;from bulletpoints 2 and 3, if i cannot infer that Diff(M) is a gauge-invariance of electrodynamics, the same should apply to gravity&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For this question, i would say that a valid answer would either disprove any of the arguments as fallacies themselves (hence showing a solid argument why gravity is special and diff(M) is without a doubt its gauge group), or improve the argument for making it bullet-proof (sorry for the pun)&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-07-19T15:06:44.477" LastActivityDate="2011-07-19T20:06:29.447" Title="argument about fallacy of diff(M) being a gauge group for general relativity" Tags="&lt;general-relativity&gt;&lt;gauge-theory&gt;&lt;diffeomorphism-invariance&gt;&lt;symmetry-group&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="12462" PostTypeId="1" CreationDate="2011-07-19T15:08:10.687" Score="8" ViewCount="106" Body="&lt;p&gt;In astrophysics, I often come across the speed of sound. I understand that, broadly, it represents the speed at which perturbations travel through a medium. But there's more than one speed of sound. The most common seem to be &lt;em&gt;isothermal&lt;/em&gt; and &lt;em&gt;adiabatic&lt;/em&gt;, which are defined as $c_s^2=(dp/d\rho)_T$ and $c_s^2=(dp/d\rho)_S$, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, when do these different speeds apply? When do perturbations travel at the adiabatic sound speed, and when the isothermal? Are there any other useful sound speeds?&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2011-07-19T18:52:42.883" Title="What's the difference between different speeds of sound?" Tags="&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" />
  <row Id="12463" PostTypeId="1" CreationDate="2011-07-19T15:35:40.440" Score="1" ViewCount="69" Body="&lt;p&gt;In the spirit of Einstein's arguments using flashes of light, moving trains and mirrors; &#xA;what is the best way to go between Einstein's postulates [&lt;a href=&quot;http://en.wikisource.org/wiki/On_the_Electrodynamics_of_Moving_Bodies&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;] of&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Relativity: Physical laws are the same in all inertial reference frames.&lt;/li&gt;&#xA;&lt;li&gt;Constant speed of light:  &quot;... light is always propagated in empty space with a definite speed $c$ which is independent of the state of motion of the emitting body.&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;to Minkowski's idea [&lt;a href=&quot;http://en.wikisource.org/wiki/Space_and_Time&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;] that space and time are united into a 4D spacetime with the Lorentzian signature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument can only use simple mathematics (algebra, geometry, a little calculus). I want an argument that a smart high school graduate would be able to understand.&#xA;None of the group theory and coset/homogeneous spaces of &lt;a href=&quot;http://physics.stackexchange.com/questions/12435/einsteins-postulates-minkowski-space/12438#12438&quot;&gt;Marek's answer&lt;/a&gt; to &lt;a href=&quot;http://physics.stackexchange.com/q/12435/429&quot;&gt;my previous question&lt;/a&gt;. (Although, if you do have a little knowledge of Lie algebras, you should go read Marek's answer and references now!).&lt;/p&gt;&#xA;" OwnerUserId="429" LastActivityDate="2011-07-19T16:44:00.023" Title="Einstein's postulates &lt;==&gt; Minkowski space. In layman's terms" Tags="&lt;special-relativity&gt;&lt;spacetime&gt;&lt;minkowski-space&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" />
  <row Id="12464" PostTypeId="2" ParentId="12462" CreationDate="2011-07-19T15:45:31.370" Score="3" ViewCount="" Body="&lt;p&gt;Fluids are complicated systems described by non-linear differential equations that can't be reasonably treated in a full generality (certainly not analytically). Just consider the kinds of waves that propagate in the sea -- deep or shallow water, solitons, tsunami and many others (this is not to say that these are sound waves; but as an illustration of complicated wave behavior it should suffice). So, to proceed one often employs some approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By far the most popular one (with many applications) rests on the linearization of the problem around an equilibrium solution where one replaces complicated non-linear equations with second-order wave equation that describes propagation of the perturbation in the system. Now, this places some consistency conditions on how big those perturbation can be so that higher-order effects can be ignored. Depending on the precise form of the equations, this might require that the temperature or some other parameter stays constant (otherwise the induced heat transport effects might destroy the linearization, for example)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly, isothermal processes are slow (so that there is enough time for the transfer of the heat with the environment which keeps the temperature constant) whereas in the adiabatic case the wave propagates so fast that the environment can't catch up with it and so no heat is exchanged (but the temperature can change). I think for the most familiar types of materials where the speed of sound is quite big one uses the adiabatic approximation (certainly for the propagation of sound in the air). I guess for less standard materials (as encountered in astrophysics) you might need isothermal approximation too but it's hard to say more than this without knowing what system you have in mind precisely.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-19T15:45:31.370" CommentCount="0" />
  <row Id="12465" PostTypeId="1" AcceptedAnswerId="12471" CreationDate="2011-07-19T16:21:26.733" Score="4" ViewCount="105" Body="&lt;p&gt;I first heard about the Tokamak in highschool 10 years ago and was wondering how far the technology has come since then.  Can it sustain a reaction for more than a few seconds?  Are these devices still huge or have they been made on a smaller scale?&lt;/p&gt;&#xA;" OwnerUserId="4539" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-19T17:26:55.490" LastActivityDate="2011-07-20T04:49:03.277" Title="Recent news about Tokamak" Tags="&lt;experimental-physics&gt;&lt;fusion&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12466" PostTypeId="1" CreationDate="2011-07-19T16:28:21.773" Score="1" ViewCount="50" Body="&lt;p&gt;Does anyone know the refractive index suppliers? I've found Cargille Labs (which customer service is terrible so far but the liquids may actually be OK), but nothing else comparable. I'd like to have a set of liquids with the refractive index in the range of 1.3...1.4. Are they so hard to produce that almost nobody does it or am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="3694" LastActivityDate="2011-07-19T16:28:21.773" ClosedDate="2011-07-19T17:28:37.413" Title="Refractive index liquids: Why hard to buy?" Tags="&lt;optics&gt;" CommentCount="5" />
  <row Id="12467" PostTypeId="2" ParentId="12435" CreationDate="2011-07-19T16:44:00.023" Score="5" ViewCount="" Body="&lt;p&gt;Let's build up to this. Suppose you know about x and y coordinates in the Euclidean plane, but to you they're just arbitrary labels for points, like zip codes or phone numbers. Then suppose someone tells you that observers can view the plane from different directions, but the laws of geometry stay the same. You now know that x and y aren't really separate. Under a 90-degree rotation, x could become y, and y could become -x. There is a quantity that is conserved under these rotations, which is the distance $\sqrt{\Delta x^2+\Delta y^2}$ defined by the Pythagorean theorem. All observers agree on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's consider relativity according to Galileo and Newton. All observers agree on time intervals. However, they disagree on other things, such as distances. If I tap the &quot;b&quot; key on my keyboard twice, like this, bb, I say that the distance between those two events is zero. But a Martian looking at me through a telescope says the earth is spinning and revolving around the sun, so the distance between b and b was a hundred meters. Only observers at rest relative to one another agree on Pythagorean distances, but they do agree on these regardless of rotation. Just as in the example of the Euclidean plane we saw that rotation could mix x and y, in Galilean relativity we see that an observer's motion along the x axis mixes x and t according to $x&amp;#39;=x-vt$, $t&amp;#39;=t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Learning about Einstein's second postulate is like learning about the rotational symmetry of the Euclidean plane. It tells you that there was a higher degree of symmetry than you believed. It says that there is something that all observers agree on. Just as all observers in the Euclidean plane agree whether points A and B are 1 meter apart, all observers in a relativistic universe agree on whether two events A and B could represent the emission of a ray of light from A and its reception at B. In this case we say that the separation of A and B is lightlike. Let the difference between A and B's x coordinates be $\Delta x$, and so on, and for convenience let's use units of seconds and light-seconds, so that c=1, and we don't have to write factors of c. If a certain observer says A and B are lightlike in relation to one another, then that observer has $\sqrt{\Delta x^2+\Delta y^2+\Delta z^2}=\Delta t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An observer in a different state of motion will measure different values for these deltas, and as in the Euclidean and Galilean examples, the equations relating these deltas have to be linear. (A nonlinear relationship like $x&amp;#39;=x^2$ would violate the homogeneity of space.) The details of the actual equations and how they're derived isn't really the topic here. But we would like to find something that both observes agree on, just as observers in the Euclidean plane agree on distances, and observers in a Galilean universe agree on times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We might hope that they would agree on the difference $\Delta t-\sqrt{\Delta x^2+\Delta y^2+\Delta z^2}$. If they did agree on this, then they would certainly agree on whether events were lightlike. But this conjecture doesn't work. One easy way to see this is with a variation on the well-known thought experiment of the train and the two lightning flashes. If the flashes are simultaneous in the dirt's frame, then observer K in a train going in one direction sees $\Delta t&amp;lt;0$, while observer K' in a train going the opposite way sees $\Delta t&amp;gt;0$. Since K and K' agree on $\Delta x$, ... but disagree on the sign of $\Delta t$, they disagree on the value of the expression conjectured above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does turn out to work is the difference $\Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2$. We can see that it doesn't fall prey to the same counterexample of the trains, because the sign of $\Delta t$ is irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By analogy with the Euclidean unification of the x and y axes, we think of this as a kind of unification of the t axis with the x, y, and z axes in relativity. The occurrence of the + sign in the time term and the - signs on the spatial ones is called the signature. The distinction between a signature like ++++ and one like +--- is real, and says that time is not exactly the same as a spatial dimension. The distinction between +--- and -+++ is not physically significant, and different people use different conventions.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-07-19T16:44:00.023" CommentCount="3" />
  <row Id="12468" PostTypeId="2" ParentId="12404" CreationDate="2011-07-19T17:02:12.580" Score="3" ViewCount="" Body="&lt;p&gt;One big difference is that all electrons, for example, are identical, but all black holes are not. In particular, a black hole can have any mass at all, whereas a particle like an electron has a fixed value for its mass. This property of fundamental particles like electrons is ultimately what allows us to define fixed scales of length and time in the laws of physics. In a universe that didn't have massive fundamental particles, the laws of physics would have a certain kind of symmetry called conformal invariance, which would make it impossible to construct clocks or rulers according to universally standardized rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another difference is that there are fundamental particles such as electrons and neutrinos that are stable (don't spontaneously undergo radioactive decay), whereas it is believed that black holes will ultimately evaporate into fundamental particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say that both have infinite density, but this is probably not actually true. The mass of a particle like an electron is probably attributable to the soup of virtual particles that surrounds it, whereas in general relativity a black hole's mass really is localized at a mathematical point. (Of course this is kind of an unfair comparison, since we know that GR is wrong below the Planck scale. It's possible that GR's singularities aren't really singularities. I'm just trying to give an answer in terms of established physical theories.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's tempting to imagine that fundamental particles &lt;em&gt;are&lt;/em&gt; black holes, but this is not possible. Classically, a spinning, charged black hole has constraints on its angular momentum and its charge in relation to its mass. Otherwise, there is no event horizon, and we have a naked singularity rather than a black hole. An electron violates both of these limits, but we don't observe that electrons have the properties predicted for these naked singularities. For example, naked singularities have closed timelike curves in the spacetime surrounding them, which would violate causality, but there is no evidence that electrons cause causality violation.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastEditorUserId="4552" LastEditorDisplayName="" LastEditDate="2011-07-19T17:36:06.163" LastActivityDate="2011-07-19T17:36:06.163" />
  <row Id="12469" PostTypeId="2" ParentId="9727" CreationDate="2011-07-19T17:15:41.220" Score="7" ViewCount="" Body="&lt;p&gt;I'll stick my neck out and say that the answer to your question is simply &quot;yes.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, these detailed thermal models are complex and hard to do, so we want confirmation from independent groups. We have that: Rievers and Lämmerzahl, &quot;High precision thermal modeling of complex systems with application to the flyby and Pioneer anomaly,&quot; gr-qc/1104.3985&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, we could ask whether these results contradict previous work. The answer is basically no. Previous work was simply sloppy. There is a nice talk on this topic here by Toth: &lt;a href=&quot;http://streamer.perimeterinstitute.ca/Flash/a2cc528b-1d36-4a2e-af73-5f81b8b17477/viewer.html&quot; rel=&quot;nofollow&quot;&gt;http://streamer.perimeterinstitute.ca/Flash/a2cc528b-1d36-4a2e-af73-5f81b8b17477/viewer.html&lt;/a&gt; There is a long history where people did back-of-the-envelope estimates of the thermal effects and said, &quot;Look, the order of magnitude is too small to matter!&quot; It just turns out that the back-of-the-envelope were wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, all of this stuff is very tough to be sure of, because there are so many uncertainties about things like the degradation of the white paint on the RTGs. Therefore it would be good to have independent ways of testing the hypothesis of a gravitational anomaly, without having to use the Pioneer data at all. We do have these independent tests. If the effect obeyed the equivalence principle, it would have had effects on the outer solar system that are not in fact observed: Iorio, &quot;Does the Neptunian system of satellites challenge a gravitational origin for the Pioneer anomaly?,&quot; &lt;a href=&quot;http://arxiv.org/abs/0912.2947&quot; rel=&quot;nofollow&quot;&gt;gr-qc/0912.2947&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's dead, Jim.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2011-07-27T08:03:11.887" LastActivityDate="2011-07-27T08:03:11.887" CommentCount="1" />
  <row Id="12470" PostTypeId="2" ParentId="12460" CreationDate="2011-07-19T17:43:19.873" Score="3" ViewCount="" Body="&lt;p&gt;A quantum computer is just a computer which uses quantum states instead of classical states.  Quantum states are just &lt;em&gt;combinations&lt;/em&gt; of classical states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say a computer takes a certain input $x_i$ to output $f(x_i)$. According to quantum mechanics, you can set up your computer such that its input is a &lt;em&gt;combination&lt;/em&gt; of all of the classical inputs $x_i$, namely $\frac{1}{n} \sum_i^n x_i$. After operating, your computer would then take this combination of all inputs to the combination of all outputs $\frac{1}{n} \sum_i^n f(x_i)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is all there was to it, we could compute &lt;em&gt;all&lt;/em&gt; of the outputs of a function as fast as we could compute &lt;em&gt;one&lt;/em&gt;. The quantum caveat is that when you go to look at your answer, you don't see $\frac{1}{n} \sum_i^n f(x_i)$, you see $f(x_i)$, for some $i$, with probability $\frac{1}{n}$. So it looks like we've only computed the function once after all! However, if we're clever, we can still benefit from our quantum computer. Let's say, instead of wanting to know all the values, $f(x_i)$, we want to know some function $g(f)$, which depends on &lt;em&gt;all&lt;/em&gt; of the answers $f(x_i)$. If we're clever enough, we can sometimes come up with an algorithm which takes $\frac{1}{n}\sum_i^n f(x_i)$ to $g(f)$. The classical computer might have needed to compute every $f(x_i)$ just to find $g$, but the quantum computer can just do it &quot;once.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please Note: this isn't exactly how it works in most cases, but most cases require quite a bit of math to describe accurately. But this is the &quot;general idea&quot; of why quantum computers can outperform classical ones.&lt;/p&gt;&#xA;" OwnerUserId="4280" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-07-19T17:52:05.797" LastActivityDate="2011-07-19T17:52:05.797" />
  <row Id="12471" PostTypeId="2" ParentId="12465" CreationDate="2011-07-19T18:14:43.447" Score="4" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can it sustain a reaction for more than a few seconds? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, at &lt;a href=&quot;http://en.wikipedia.org/wiki/Joint_European_Torus&quot; rel=&quot;nofollow&quot;&gt;JET&lt;/a&gt;.&#xA;&lt;em&gt;Lifetime of the plasma: 20–60 s&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At &lt;a href=&quot;http://www.iter.org/mach/cryo&quot; rel=&quot;nofollow&quot;&gt;ITER&lt;/a&gt;:&#xA;&lt;em&gt;It will operate over a wide range of ITER plasma scenarios, from short plasma pulses (a few hundred seconds) with enlarged fusion power (700 MW) to long plasma pulses of 3,000 s&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are these devices still huge or have they been made on a smaller scale?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They have to be large in order to give out more energy than they consume. This is the &lt;a href=&quot;http://www.iter.org/&quot; rel=&quot;nofollow&quot;&gt;ITER&lt;/a&gt; link, the one planned and already being built, to get the size idea.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-07-20T04:49:03.277" LastActivityDate="2011-07-20T04:49:03.277" />
  <row Id="12472" PostTypeId="2" ParentId="12462" CreationDate="2011-07-19T18:52:42.883" Score="4" ViewCount="" Body="&lt;p&gt;The isothermal sound speed applies when the cooling timescale is very fast compared to the propagation speed of the wave.   Often in astrophysics the temperature of the gas is set by thermal balance between heating sources and radiative cooling and the timescale to get into balance is short compared to the sound wave travel timescale.  One example is spiral density waves traveling in a galactic disk where an effective sound speed would be that of the $10^4$K gas and about that of the velocity dispersion of the molecular clouds.   &lt;/p&gt;&#xA;" OwnerUserId="4554" LastActivityDate="2011-07-19T18:52:42.883" />
  <row Id="12473" PostTypeId="1" CreationDate="2011-07-19T19:03:38.407" Score="5" ViewCount="132" Body="&lt;p&gt;In the late 80's and early 90's, Smalley and others made claims that the C60 fullerene bearing icosahedral symmetry was the most spherical molecule known, and perhaps the most spherical that could exist.  While it makes sense to me that the smallest member of the 60n^2 isocahedral family of buckyballs would be the most spherical (I believe larger members should tend towards and isocahedron?), are there really no other more spherical geometries that are possible or actually experimentally realized?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently took a look at: &quot;Production and isolation of an ellipsoidal C-80 fullerene&quot; by Chun-Ru Wang et. al. (http://pubs.rsc.org/en/content/articlelanding/2000/cc/b000387p/unauth), and it looks like C80 forms an ellipsoid.  However, I'm not sure about C80's endohedral counterparts.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update - To be clearer about what I mean by &quot;more spherical&quot;, I'm looking for a family of buckyballs that converges quickly on the volume of a circumscribing sphere as a function of the number of carbon atoms (or alternatively, as a function of the 'k'th smallest member of its symmetry family).  For example, from this Mathematica document: (&quot;http://reference.wolfram.com/legacy/v5_2/Demos/Notebooks/BuckyballConstruction.html&quot;) we can see that the icosahedral C60 fills ~87% the volume of its circumscribing sphere.  If C-80 filled a larger volume of its circumscribing sphere than C60, as the k=1 member of its family of buckyballs, I would consider that &quot;more spherical&quot; than C60.  The problem with that assumption, however, is that C60 has 20 fewer carbon atoms than C80.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update 2 - Another way to measure how &quot;spherical&quot; a buckyball is could involve looking at the distribution of angles between any two adjacent carbon bonds.  One would particularly want to look at the global minimum angle(s), being careful to avoid problems with large planar faces that approximate graphene and washout outliers.  If the global minimum angle between carbon bonds tended to ~(120 degrees - epsilon), as a function of the kth minimum member of a particular symmetry group, that family of buckyballs would presumably also converge on their circumscribing spheres as a function of the number of carbon atoms. &lt;/p&gt;&#xA;" OwnerUserId="4257" LastEditorUserId="4257" LastEditorDisplayName="" LastEditDate="2011-07-19T20:41:17.280" LastActivityDate="2011-09-05T01:21:21.413" Title="Is C60 really the &quot;most spherical&quot; fullerene?" Tags="&lt;mathematics&gt;&lt;geometry&gt;" AnswerCount="1" CommentCount="9" />
  <row Id="12474" PostTypeId="2" ParentId="12461" CreationDate="2011-07-19T19:32:35.843" Score="3" ViewCount="" Body="&lt;p&gt;I will mostly talk about the classical physics as this is complicated enough already (I might mention something about quantum stuff at the end). So, let us first get all the relevant terms straight, so that we avoid any further confusion. In particular, we need to be precise about what we mean by &lt;em&gt;invariance&lt;/em&gt; because already two different notions have been thrown into one bag.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;A symmetry group&lt;/em&gt; of a physical system is a group of transformations that leave the system invariant. E.g. the electric field of a point charge is invariant under rotations w.r.t. to that point. In other words, we want the group to act trivially. But that means that this immediatelly rules out any equations that carries tensorial indices (i.e. transforms in a non-trivial representation of the rotation group). For these equations, if you perform a rotation, the equation will &lt;strong&gt;change&lt;/strong&gt;. Of course, it will change in an easily describable manner and a different observer will agree. But the difference is crucial. E.g. in classsical quantum mechanics we require the equations to be scalar &lt;strong&gt;always&lt;/strong&gt; (which is reflected in the fact that Hamiltonian transforms trivially under the symmetry group).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Carrying a group action&lt;/em&gt; is a broader term that includes tensorial equations we have left out in the previous bullet point. We only require that equations or states are being acted upon by a group. Note that the group action need not have any relation to the symmetry. E.g. take the point charge and translate it. This will certainly produce a different system (at least if there is some background so that we can actually distinguish points).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;A gauge group&lt;/em&gt; of a system is a set of transformations that leave the states invariant. What this means is that the actual states of the system are equivalence classes of orbits of the gauge group. Explicitly, consider the equation ${{\rm d} \over {\rm d} x} f(x) = 0$. This has a solution $g(x) \equiv C$ for any $C$. But if we posit that the gauge group of the equation consists of the transformations $f(x) \mapsto f(x) + a$ then we identify all the constant solutions and are left with a single equivalence class of them -- this will be the physical state. This is what gauge groups do in general: they allow us to treat equivalence classes in terms of their constituents. Obviously, gauge groups are completely redundant. The reason people work with gauge groups in the first place is that the description of the system may simplify after introduction of these additional parameters that &quot;see&quot; into the equivalence class. Of course historical process went backwards: since gauge-theoretical formulation is simpler, this is whan people discovered first and they only noticed the presence of the gauge groups afterwards.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, having said this, let's look at the electromagnetism (first in the flat space). What symmetries are the Maxwell equations invariant under? One would like to say &lt;em&gt;under Lorentz group&lt;/em&gt; but this actually not the case. Let's look at this more closely. As alluded to previously the equation&#xA;$$\partial_{\mu} F^{\mu \nu} = J^{\nu}$$&#xA;can't really be invariant since it carries a vector index. It transforms in the four-vector representation of the Lorentz group, yes -- but it is certainly not invariant. Contrast this with the Minkowski space-time itself which is left invariant by the Lorentz group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also have ${\rm d} F = 0$ and therefore (in a contractible space-time) also $F = {\rm d} A$ which is obviously invariant w.r.t. to $A \mapsto A + {\rm d} \chi$. In terms of the above discussion the equivalence class $A + {\rm d} \Omega^0({\mathbb R}^{1,3})$ is the physical state and the gauge transformation lets us distinguish between its constituents&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's move to a curved spacetime now. Then we have&#xA;$$\nabla_{\mu} F^{\mu \nu} = J^{\nu}$$&#xA;Again, this is not invariant under ${\rm Diff}(M)$. But it transforms under an action of ${\rm Diff}(M)$. The only thing in sight that is invariant under ${\rm Diff}(M)$ is $M$ itself (by definition).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the very same way, GR is not invariant under ${\rm Diff}(M)$ but only transforms under a certain action of it (different than EM though, since GR equations carry two indices). Also, ${\rm Diff}(M)$ can't possibly be a gauge group of any of these systems since it would imply that almost all possible field configurations get cramped to a single equivalence class possibly indexed by some topological invariant which can't be changed by a diffeomorphism. In other words, theory with ${\rm Diff}(M)$ as a gauge group would need to be purely topological with no local degrees of freedom.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-07-19T20:06:29.447" LastActivityDate="2011-07-19T20:06:29.447" CommentCount="10" />
  <row Id="12475" PostTypeId="1" CreationDate="2011-07-19T19:48:17.380" Score="1" ViewCount="58" Body="&lt;p&gt;In the path integral approach to Quantum Mechanics, can two distinctly different paths of the possible infinite paths have the same phase, i.e can there be a bimodal distribution of the phases associated with each path? And Why?&lt;/p&gt;&#xA;" OwnerUserId="4555" LastActivityDate="2011-08-18T20:26:55.287" Title="Two paths having the same phase in the path integral approach" Tags="&lt;quantum-mechanics&gt;&lt;path-integral&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12476" PostTypeId="2" ParentId="12475" CreationDate="2011-07-19T19:55:30.173" Score="0" ViewCount="" Body="&lt;p&gt;sure they can;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can simply put a half-reflecting mirror in front of a source of light (like a laser), so 50% of the wave amplitude will go straight while the rest is being reflected. Both paths are bimodal, and they can still interfere with each other, if the path difference at the moment of mixing both again is the right one. &lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-07-19T19:55:30.173" CommentCount="2" />
  <row Id="12477" PostTypeId="2" ParentId="12418" CreationDate="2011-07-19T19:57:06.533" Score="4" ViewCount="" Body="&lt;p&gt;The hardest part is to get started by measuring one particle's mass. Once you have that, you can get the others more easily because ratios of masses are easier to measure than absolute ones. You're quoting values that are good to 8 sig figs, and I'm sure the experiments needed in order to get that kind of precision are extremely complex. I'm going to describe how to measure them with much less heroic precision, using the kind of equipment you could find in a high school with a well equipped physics and chem lab. To this kind of precision, the masses of the neutron and proton are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say you know how to measure the electron's mass, but just to be concrete, let's say that we'll do that by measuring the wavelengths in the emission spectrum of hydrogen and solving the equations in the Bohr model to get the electron's mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we can measure the charge to mass ratio of the electron $-e/m_e$ by accelerating electrons through a known potential and then measuring their deflection in a magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now measure the charge-to-mass ratio of some element such as sodium by doing electrolysis and finding the ratio of how much charge flowed around the circuit to how much mass was deposited on the electrode. Since we know how many neutrons and protons there are in sodium, we can infer the charge-to-mass ratios $e/m_p$ and $e/m_n$ (which, to our precision, are equal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, since we know $m_e$, $-e/m_e$, $e/m_p$, and $e/m_n$, it's straightforward to find $m_n$ and $m_p$.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-07-19T19:57:06.533" />
  <row Id="12478" PostTypeId="2" ParentId="12287" CreationDate="2011-07-19T20:05:09.837" Score="1" ViewCount="" Body="&lt;p&gt;Quantum mechanics can be reconciled with special relativity to make quantum field theory, but there are some awkward things going on in that marriage. SR treats time symmetrically with position, but in quantum mechanics, position is an operator and time isn't. Baez at UCR has a nice discussion of that here: &lt;a href=&quot;http://math.ucr.edu/home/baez/uncertainty.html&quot; rel=&quot;nofollow&quot;&gt;http://math.ucr.edu/home/baez/uncertainty.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-07-19T20:05:09.837" CommentCount="2" />
  <row Id="12479" PostTypeId="2" ParentId="12456" CreationDate="2011-07-19T20:55:48.070" Score="2" ViewCount="" Body="&lt;p&gt;From the equilibrium between drag and weight:&#xA;$$&#xD;&#xA;\frac{1}{2} C_x \rho v^2 S = m g&#xD;&#xA;$$&#xA;we can write the terminal velocity as&#xA;$$&#xD;&#xA;v = \sqrt{\frac{2 m g}{C_x \rho S}}&#xD;&#xA;$$&#xA;where $m$ is the mass of the phone, $C_x$ its drag coefficient,&#xA;$S$ its section, $g$ the acceleration of gravity,&#xA;and $\rho$ the density of air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this is not really a good answer, as the big question is how to&#xA;estimate $S$ (depends on the phone orientation) and $C_x$. But at least&#xA;you can compute an order of magnitude, as in most cases $C_x$ of of&#xA;order 1.&lt;/p&gt;&#xA;" OwnerUserId="4202" LastActivityDate="2011-07-19T20:55:48.070" CommentCount="5" />
  <row Id="12480" PostTypeId="2" ParentId="12221" CreationDate="2011-07-19T20:55:59.610" Score="2" ViewCount="" Body="&lt;p&gt;David Bar Moshe has given a very complete answer at a high level of sophistication in both math and physics. If that exactly meets the needs of the OP and others who read this page, that's great. I would just like to take a shot at addressing the OP's question in simpler language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GR does not have global frames of reference the way SR does. (When David Bar Moshe says, &quot;A frame in GR means ...,&quot; what he's defining isn't a global frame of reference, it's a collection of local frames of reference, each defined at a different point.) As a concrete example, in SR we expect to be able to say that in object A's frame of reference, distant object B has some well-defined velocity. In GR, we can't do this. For example if A is our galaxy and B is some cosmologically distant galaxy, then there is no uniquely well-defined way of defining B's velocity in A's frame. We can say that B is moving relative to A, or we can say that both A and B are at rest and the space between them is expanding. GR doesn't say that one of these verbal descriptions is to be preferred over the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The OP is asking about charts and atlases and how they relate to changes of frames of reference. They don't. All of the interesting issues can be discussed without ever having to deal with a manifold that requires more than one chart. For example, in FRW cosmological models, you can cover the entire spacetime with a single chart using coordinates $(t,x,y,z)$, and you would typically do this in such a way that a galaxy at rest relative to the Hubble flow had constant x, y, and z. An example of a change of coordinates would be $t\rightarrow 2t$. In this spacetime, there is no notion of a global frame of reference, and there is nothing that plays the role of a Lorentz boost as there would be in SR. If you try to apply the usual Lorentz transformation equations to $(t,x,y,z)$, you get a set of coordinates $(t&amp;#39;,x&amp;#39;,y&amp;#39;,z&amp;#39;)$ that are perfectly valid but utterly uninteresting physically, and you &lt;em&gt;cannot&lt;/em&gt; interpret the new ones as a frame that is moving at certain speed relative to some frame defined by the old ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Re the OP's question about accelerating versus nonaccelerating, GR is relatively agnostic about this. We do need to distinguish frames from coordinates, however. Any smooth change of coordinates (diffeomorphism) is allowed, and has no effect on the form of the laws of physics (Einstein field equations). For example, you can describe flat spacetime in accelerating coordinates, where it appears to have a gravitational field: &lt;a href=&quot;http://en.wikipedia.org/wiki/Rindler_coordinates&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Rindler_coordinates&lt;/a&gt; In terms of frames, we do have preferred frames in GR, which are the frames of observers who are free-falling. In GR, the frame of a rock sitting on the ground is considered noninertial, and the frame of a falling rock is considered inertial; it's exactly the opposite of what we'd say in Newtonian physics.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastEditorUserId="4552" LastEditorDisplayName="" LastEditDate="2011-07-19T21:10:08.683" LastActivityDate="2011-07-19T21:10:08.683" />
  <row Id="12481" PostTypeId="2" ParentId="12431" CreationDate="2011-07-19T21:10:01.790" Score="0" ViewCount="" Body="&lt;p&gt;I think, this is only a math trick. Take the equation of state $f(P,T,\rho)=0$ and rewrite it as $P=P(T,\rho)$. Multiply it by $V$: $$PV=g(T,\rho)$$ Now expand it into Taylor series:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$PV=g(T,0)+\frac{\partial g(T,0)}{\partial\rho}\rho+ \frac{1}{2}\frac{\partial^2 g(T,0)}{\partial\rho^2}\rho^2+...$$ Here $\rho$ is density. Because $\rho\sim\frac{1}{v}$ we get an expansion like in the question.&lt;/p&gt;&#xA;" OwnerUserId="552" LastActivityDate="2011-07-19T21:10:01.790" CommentCount="1" />
  <row Id="12482" PostTypeId="2" ParentId="12359" CreationDate="2011-07-19T21:58:43.793" Score="0" ViewCount="" Body="&lt;p&gt;Assume sinusoidal grate -- &lt;code&gt;sin(x)&lt;/code&gt; is the simplest possible periodic function. Direct axis &lt;code&gt;x&lt;/code&gt; horizontally, and &lt;code&gt;y&lt;/code&gt; vertically, and lets calculate diffraction at the point &lt;code&gt;(x,y)&lt;/code&gt;. According to Fraunhofer diffraction we must first calculate distance from the observation point &lt;code&gt;(x,y)&lt;/code&gt; to the source of secondary wave &lt;code&gt;(x_1, h*sin(d*x_1)))&lt;/code&gt; which is &lt;code&gt;r=sqrt((y-h*sin(d*x_1))^2+(x-x_1)^2)&lt;/code&gt;. Here the &lt;code&gt;h&lt;/code&gt; is the height of the ripple, and &lt;code&gt;d&lt;/code&gt; its frequency. This r is plugged in into the amplitude integral &lt;code&gt;Integrate(i*exp(-2*pi*i*r/lambda)/(r*lambda),x1=-R..R)&lt;/code&gt;. Mathematica expression:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Integrate[i*exp(-2*pi*i*sqrt((y-h*sin(x1))^2+(x-x1)^2)/lambda)/(sqrt((y-h*sin(x1))^2+(x-x1)^2)*lambda),{x1,-d,d}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Maple:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Int(i*exp(-2*pi*i*sqrt((y-h*sin(d*x1))^2+(x-x1)^2)/lambda)/(sqrt((y-h*sin(d*x1))^2+(x-x1)^2)*lambda),x1=-R..R);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Neither succeeded solving it analytically, so one have to regress to numerics. Therefore, I assigned the following numeric values:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radius of the flat mirror:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R=20 mm&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Wavelength:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;lambda=0.0004 mm (=400 nm)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Defect density/frequency:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;d=10 (10 ripples/mm -- with higher values Wolfram alpha times out)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Defect height&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;h=0.0001 mm (1/1 wave)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Location&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;y=100 mm (10 cm above the mirror).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With all the assignments the only free variable remaining is x, so one should be able to plot intensity graph. Unfortunately, Wolfram Alpha refuse to understand what &lt;code&gt;Plot[Integral[]]&lt;/code&gt; is and suggests some stock market graphs instead. I had to calculate pointwise. Here is an expression which calculates intensity at the axis (&lt;code&gt;x=0&lt;/code&gt;):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Integral[i*exp(-2*pi*i*sqrt((100-0.0001*sin(10*x1))^2+(0.0-x1)^2)/0.0004)/(sqrt((100-0.0001*sin(10*x1))^2+(0.0-x1)^2)*0.0004),x1=-20..20]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which walpfa evaluates to &lt;code&gt;0.62+3.81i&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The characteristic angle of periodic grating is &lt;code&gt;lambda*d&lt;/code&gt; so the diffraction pattern linear dimension is &lt;code&gt;lambda*d*y&lt;/code&gt; which in our case is conveniently &lt;code&gt;1&lt;/code&gt;. Therefore, we could see diffraction pattern by probing only three points: &lt;code&gt;x=0.0&lt;/code&gt;, &lt;code&gt;x=0.5&lt;/code&gt;, and &lt;code&gt;x=1.0&lt;/code&gt;. Here are the intensities:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x=0.0 -&amp;gt; 0.62+3.81*i&#xA;x=0.5 -&amp;gt; 7.76+3.0*i&#xA;x=1.0 -&amp;gt; 3.51+1.8*i&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In other words, the diffraction pattern is quite noticeable when defect size is comparable to lambda. To doublecheck, what if we shrink grating height 10 fold? There:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x=0.0 -&amp;gt; 3.4+3.5*i&#xA;x=0.5 -&amp;gt; 4.1+3.1*i&#xA;x=1.0 -&amp;gt; 3.88+3.33*i &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4507" LastEditorUserId="4507" LastEditorDisplayName="" LastEditDate="2011-07-19T22:04:41.153" LastActivityDate="2011-07-19T22:04:41.153" />
  <row Id="12483" PostTypeId="2" ParentId="12456" CreationDate="2011-07-19T23:47:36.460" Score="1" ViewCount="" Body="&lt;p&gt;Suppose that there was no turbulence in the air.&lt;br&gt;&#xA;It will fall down as a leave, horizontal, slightly oscillating around he vertical axis in relation to the Earth surface, about 6 degrees, the same as the long ice plates do within contrails and cirrus clouds.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the formula of Bonet's answer the section S is maximum (good to minimize the speed). The drag coeff ?  I dont know. Someone can edit this answer to make it more complete.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the physics do not end here, I think. The phone will slide in large horizontal moves, as a sail or wing, much like a skydiver do. It is much more than a simple vertical ballistic  motion.&lt;br&gt;&#xA;Imo, thats the why.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT add&lt;br&gt;&#xA;assuming only an horizontal descent (the mass distribution is centered) :&lt;br&gt;&#xA;exploring the numbers: the best result, for Cx = 1.3, is v= 15.6 m/s (in my oppinion it is too much)&lt;br&gt;&#xA;Cx  0.5;   1;  1.3 ; g  9.8 ms^(-2); Ipod4 137g, dept 9.3mm surface 0.1152*0.0586 m2; Density of air 1.25 Kg/m^(-3)&lt;br&gt;&#xA;Cx=0.5  v=25m/s;  Cx=1 v=17m/s; Cx=1.3 v=15.6 m/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to explore other aerodynamic effects.&lt;br&gt;&#xA;altitude 4100 m , retrieved 800m away from where he landed. The angle of the descent of the Iphone4 probably is greater than 11º (arctan(800/4000)=arctan(1/5)). From this we can conclude that it was not a balistic descent and  aerodynamic effects indeed are present.&lt;br&gt;&#xA;I do not know the mass distribution in the Iphone4. If it can be made similar to a marble seed, &lt;img src=&quot;http://i.stack.imgur.com/8YTny.jpg&quot; alt=&quot;marble seed&quot;&gt;&lt;br&gt;&#xA;with much of the mass near the one top then the cellphone will rotate  and significantly slow down.&lt;br&gt;&#xA;reference EXPLORING THE BIOFLUIDDYNAMICS OF SWIMMING AND FLIGHT by David Lentink&#xA;pages 111-119 chapter 5.1 &lt;a href=&quot;http://library.wur.nl/WebQuery/edepot/16365&quot; rel=&quot;nofollow&quot;&gt;A LEADING EDGE VORTEX PROLONGS DESCENT OF MAPLE SEEDS&lt;/a&gt;&lt;br&gt;&#xA;The package ELMER, among others, can be used to make the numerical aerodynamic simulation with all this variables including, if needed, the building and the wind (any upward air flow in the trajectory of the mobile ?).&lt;br&gt;&#xA;EDIT add  end&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-07-21T12:34:57.730" LastActivityDate="2011-07-21T12:34:57.730" CommentCount="5" />
  <row Id="12484" PostTypeId="1" AcceptedAnswerId="12519" CreationDate="2011-07-20T00:23:43.417" Score="4" ViewCount="82" Body="&lt;p&gt;Cosmological Inflation was proposed by Alan Guth to explain the flatness problem, the horizon problem and the magnetic monopole problem. I think I pretty much understand the first two, however I don't quite understand how a period of exponential expansion fully explains monopole problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Weinburg's Cosmology, the issue is essentially that various grand unified theories predict that the standard models $SU(3)\times SU(2)\times U(1)$ arose from the breaking of an original simple symmetry group. For many of these theories, a crazy particle known as a &quot;magnetic monopole&quot; is created at a certain energy (sometimes quoted at around $M = 10^{16} GeV$). So my question is why does a period of rapid expansion somehow or other result in a low density of magnetic monopoles (assuming they exist/existed at all)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think, like in nucleosynthesis, that the primary factor in monopole creation is energy density, and since inflation is still a &quot;smooth&quot; process, at some point the universe would hit the proper energy density to create magnetic monopoles. How does the rate of expansion at the time they were created effect overall present density?&lt;/p&gt;&#xA;" OwnerUserId="3018" LastEditorUserId="3018" LastEditorDisplayName="" LastEditDate="2011-07-20T03:33:50.767" LastActivityDate="2011-07-20T14:39:39.197" Title="How does Inflation solve the Magnetic Monopole problem?" Tags="&lt;cosmological-inflation&gt;&lt;magnetic-monopoles&gt;" AnswerCount="1" />
  <row Id="12485" PostTypeId="1" AcceptedAnswerId="12507" CreationDate="2011-07-20T01:13:13.950" Score="4" ViewCount="82" Body="&lt;p&gt;So, I've recently been reading up on &lt;a href=&quot;http://en.wikipedia.org/wiki/Wormhole#Schwarzschild_wormholes&quot; rel=&quot;nofollow&quot;&gt;Schwarzschild wormholes&lt;/a&gt; and I've learned that they cannot exist becuase they violate the 2nd Law of Thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm asking is: Why do they violate the Law?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I probably sound like an idiot, but I just can't understand why they violate Thermodynamics.&lt;/p&gt;&#xA;" OwnerUserId="3564" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-20T02:18:23.280" LastActivityDate="2011-07-20T10:14:27.683" Title="Why can't Schwarzschild wormholes exist?" Tags="&lt;general-relativity&gt;&lt;thermodynamics&gt;&lt;wormholes&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12486" PostTypeId="2" ParentId="12408" CreationDate="2011-07-20T01:59:12.917" Score="0" ViewCount="" Body="&lt;p&gt;The other answers detailed addressed the physical part.&lt;br&gt;&#xA;When the physical limits are an obstacle we will see a change of paradigm. &#xA;I will continue with other factors that may be important in the future.&#xA;I suspect that we are yet limited to a 2D (planar) design of chips and even 3D is a repackage of a 2D design.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the work can be done in parallel: physical simulations, database operations, pattern recognition, datafile processing, etc...  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;AMD-ATI and Nvidia have done a great job with their high-end graphics cards. By now they have towsands  of parallel processors on a chip. The software has also progressed with OpenCL, and PyOpenCL, Nvidia CUDA and ATI-Brook+, NoSQL and CouchDb databases, map-reduce algorithms. Clusters of graphics cards behave as a supercomputer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that in the future we will see 3D neuronal networks, implemented in hardware with on-the-fly programmable coefficients that will try to mimic the brain (the most efficient parallel computer - per watt,  and the only one that is really intelligent). &#xA;Heat removal is a major issue in any circumstance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;New technology will make the entrance, may be this one:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.technologyreview.com/blog/arxiv/26560/?ref=rss&quot; rel=&quot;nofollow&quot;&gt;Superconducting Niobium Chip Smashes Silicon Power Consumption Standards&lt;/a&gt; A superconducting logic chip with a clock speed of 6 GHz beats silicon energy efficiency by two orders of magnitude.  &lt;a href=&quot;http://arxiv.org/abs/1103.4269&quot; rel=&quot;nofollow&quot;&gt;arxiv-Ultra-Low-Power Superconductor Logic&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(note: I'm playing around since chip 8080, and I saw a huge evolution and I'll see more, if I manage to keep alive ;)&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-07-20T01:59:12.917" />
  <row Id="12487" PostTypeId="1" CreationDate="2011-07-20T02:26:27.100" Score="8" ViewCount="380" Body="&lt;p&gt;The Earth turns with a very high velocity, around its own axis and around the Sun. &#xA;So why can't we feel that it's turning, but we can still feel earthquake.&lt;/p&gt;&#xA;" OwnerUserId="4556" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-20T21:12:47.010" LastActivityDate="2011-07-20T21:12:47.010" Title="Why can't we feel the Earth turning?" Tags="&lt;gravity&gt;&lt;classical-mechanics&gt;&lt;centrifugal-force&gt;&lt;coriolis-force&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="12488" PostTypeId="1" CreationDate="2011-07-20T02:35:39.633" Score="8" ViewCount="223" Body="&lt;p&gt;We don't normally consider the possibility that massless particles could undergo radioactive decay. There are elementary arguments that make it sound implausible. (A bunch of the following is summarized from Fiore 1996. Most of the rest, except as noted, is my ideas, many of which are probably wrong.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- Normally we state the lifetime of a particle in its rest frame, but a massless particle doesn't have a rest frame. However, it is possible for the lifetime $\tau$ to be proportional to energy $E$ while preserving Lorentz invariance (basically because time and mass-energy are both timelike components of four-vectors).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- The constant of proportionality between $\tau$ and $E$ has units of mass&lt;sup&gt;-2&lt;/sup&gt;. It's strange to have such a dimensionful constant popping up out of nowhere, but it's not impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- We would typically like the observables of a theory to be continuous functions of its input parameters. If X is a particle of mass m, then a decay like X$\rightarrow$3X is forbidden by conservation of mass-energy for m&gt;0, but not for m=0. This discontinuity is ugly, but QFT has other cases where such a discontinuity occurs. E.g., historically, massive bosons were not trivial to incorporate into QFT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4- In a decay like X$\rightarrow$3X, the products all have to be collinear. This is a little odd, since it doesn't allow the clear distinction one normally assumes in a Feynman diagram between interior and exterior lines. It also means that subsequent &quot;undecay&quot; can occur. Strange but not impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what about less elementary arguments? My background in QFT is pretty weak (the standard graduate course, over 20 years ago, barely remembered).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5- The collinearity of the decay products makes the phase-space volume vanish, but amplitudes can diverge to make up for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6- If X is coupled to some fermion Y, then one would expect that decay would correspond to a Feynman diagram with a box made out of Y's and four legs made out of X's. If Y is a massive particle like an electron, P. Allen on physicsforums argues that when the energy of the initial X approaches zero, the X shouldn't be able to &quot;see&quot; the high-energy field Y, so the probability of decay should go to zero, and the lifetime $\tau$ should go to infinity, which contradicts the requirement of $\tau\propto E$ from Lorentz invariance. This seems to rule out the case where Y is massive, but not the case where it's massless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7- If X is a photon, then decay is forbidden by arguments that to me seem technical. But this doesn't forbid decays when X is any massless particle whatsoever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8- There are some strange thermodynamic things going on. Consider a one-dimensional particle in a box of length $L$. If one X is initially introduced into the box with energy $E=nE_o$, where $E_o$ is the ground-state energy, then it undergoes decays and &quot;undecays,&quot; and if I've got my back-of-the-envelope estimate with Stirling's formula right, I think it ends up maximizing its entropy by decaying into about $\sqrt{n}$ daughters at a temperature $\sim \sqrt{hE/L}$. If you then let it out of the box so that it undergoes free expansion, it acts differently from a normal gas. Its temperature approaches zero rather than staying constant, and its entropy approaches infinity. I may be missing something technical about thermo, but this seems to violate the third law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is this: &lt;strong&gt;Is there any fundamental (and preferably simple) argument that makes decay of massless particles implausible?&lt;/strong&gt; I don't think it can be proved completely impossible, because Fiore offers field theories that are counterexamples, such as quantum gravity with a positive cosmological constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&#xA;Fiore and Modanese, &quot;General properties of the decay amplitudes for massless particles,&quot; 1996, &lt;a href=&quot;http://arxiv.org/abs/hep-th/9508018&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9508018&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4552" LastEditorUserId="4552" LastEditorDisplayName="" LastEditDate="2011-08-11T14:34:57.090" LastActivityDate="2011-08-11T22:57:33.160" Title="decay of massless particles" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;thermodynamics&gt;&lt;special-relativity&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" />
  <row Id="12489" PostTypeId="2" ParentId="12487" CreationDate="2011-07-20T02:40:41.500" Score="15" ViewCount="" Body="&lt;p&gt;Because the rotation of the earth is very smooth and doesn't change, the centripetal acceleration we feel is very nearly constant.  This means that the (small) centrifugal force from the rotation gets added to gravity to make up the &quot;background force&quot; we don't notice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earthquakes are not at all smooth and the accelerations involved are large and change direction a lot.  This makes it easy to feel them. &lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2011-07-20T02:40:41.500" CommentCount="1" />
  <row Id="12490" PostTypeId="2" ParentId="5243" CreationDate="2011-07-20T02:56:52.963" Score="1" ViewCount="" Body="&lt;p&gt;The second link you provide clearly indicates that one should try to roll with the punch and should never lean into it. Paul Hewitt, author of the popular textbook &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0321568095&quot; rel=&quot;nofollow&quot;&gt;Conceptual Physics&lt;/a&gt; and former Golden Gloves champ, also teaches that one should roll with a punch. He justifies it in terms of the impulse-momentum theorem: He assumes that the force between the fist and head is what brings the fist to rest whether one rolls with the punch or not, so prolonging the contact time one reduces the average force by the fist on the head. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strangely enough, he also mentions catching a baseball barehanded as another example of the impulse-momentum theorem, claiming it will hurt much less if you give with the ball as you catch it, the opposite of what you state. My experience squares with Hewitt's on this point. Try holding your hand inches in front of a brick wall and catching a baseball. Without being able to give during the catch, it hurts.&lt;/p&gt;&#xA;" OwnerUserId="4208" LastActivityDate="2011-07-20T02:56:52.963" CommentCount="2" />
  <row Id="12491" PostTypeId="2" ParentId="12333" CreationDate="2011-07-20T03:08:12.367" Score="0" ViewCount="" Body="&lt;p&gt;A different way of stating the ideas about units is that when an equation is stated in one set of units, we would like to be able to convert that statement into different units. For example, 3 m x 7 m = 21 m&lt;sup&gt;2&lt;/sup&gt; can be converted into 300 cm x 700 cm = 210000 cm&lt;sup&gt;2&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise 3 km + 700 m = 3700 m can be converted into 300000 cm + 70000 cm = 370000 cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is no sensible way to take 3 gallons + 2 hours = 5 somethings and convert it into, say, a statement in terms of minutes rather than hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One highfalutin' way of expressing all this is that we're dealing with affine geometry. When we multiply 3 newtons by 7 meters to get 21 newton-meters, we're finding the area of a rectangle in the affine plane. In affine geometry, distances measured along non-parallel lines cannot be compared, so we can't say that 7 meters is greater than 3 newtons.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-07-20T03:08:12.367" />
  <row Id="12492" PostTypeId="2" ParentId="11284" CreationDate="2011-07-20T05:11:09.197" Score="1" ViewCount="" Body="&lt;p&gt;The $1/r^2$ behavior is characteristic of a massless particle mediating the interaction. For electromagnetism, that particle is the photon, while for gravity it's the graviton. When an interaction is mediated by a massless particle, then it is of infinite-range. For example, two electrons will repel each other regardless of how far they are (as long as they are in causal contact of course). Another force that behaves like $1/r^2$ would have to also be of infinite-range, and so very easy to observe, unless it was extremely weak. So, most likely, there are no other $1/r^2$ forces.&lt;/p&gt;&#xA;" OwnerUserId="4483" LastActivityDate="2011-07-20T05:11:09.197" />
  <row Id="12493" PostTypeId="2" ParentId="12034" CreationDate="2011-07-20T05:40:37.870" Score="1" ViewCount="" Body="&lt;p&gt;Fermi's golden rule still applies in the relativistic limit, and can be rewritten in a Lorentz invariant fashion.  Starting with the transition probability&#xA;$$ W_{i\rightarrow f} = \frac{2\pi}{\hbar} |m_{if}|^2 \rho(E) \,,$$ to have $W$ Lorentz invariant we'd like both the matrix element $|m_{if}|^2$ and the density of final states $\rho(E)$ to be invariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be done by shifting a few terms around. A little bit of handwaving to motivate it: The wave function $\psi$ (which is in the matrix element) has to be normalized by $\int |\psi|^2 dV = 1$, which gives us a density (of probability to encounter a particle) of $1/V$.  Now, a boosted observer experiences length contraction of $1/\gamma$, which changes the density to $\gamma/V$.  To obtain the correct probability again, we should re-normalize the wave function to $\psi&amp;#39; = \sqrt{\gamma}\,\psi $ by pulling the Lorentz factor out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we intoduce a new matrix element $$|{\cal M}_{if}|^2 = |m_{if}|^2 \prod_{i=1}^n (2 \gamma_i m_i c^2) =|m_{if}|^2  \prod_{i=1}^n (2E_i)^2 $$ (this is for an $n$-body process).  Now the transition probability (here in differential form) becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;  dW = \frac{2\pi}{\hbar} \frac{|{\cal M}_{if}|^2}{ (2E_1)^2  (2E_2)^2  \cdots}&#xD;&#xA;  \cdot  \frac{1}{(2\pi\hbar)^{3n}} &#xD;&#xA;  \, d^3p_1 \, d^3p_2 \, \cdots &#xD;&#xA;  \delta({p_1}^\mu + {p_2}^\mu + \ldots - {p}^\mu )&#xD;&#xA;$$&#xA;The delta function is there to ensure conservation of momentum and energy.  Now we can regroup the terms:&#xA;$$&#xD;&#xA;  \Rightarrow \quad&#xD;&#xA;  dW = \frac{2\pi}{\hbar} \frac{|{\cal M}_{if}|^2}{ 2E_1 2E_2 \cdot \ldots}&#xD;&#xA;  \cdot d_\mathrm{LIPS}&#xD;&#xA;$$&#xA;The density of states/&quot;phase space&quot; $d\rho$ is replaced by a relativistic version, sometimes called the Lorentz invariant phase space $d_\mathrm{LIPS}$, which is given by&#xA;$$&#xD;&#xA;  d_\mathrm{LIPS} =  \frac{1}{(2\pi\hbar)^{3n}} &#xD;&#xA;  \prod_{i=1}^n \frac{d^3p_i}{ 2E_i }&#xD;&#xA;  \delta\left(\prod_{i=1}^n {p_i}^\mu - {p}^\mu \right) \,.&#xD;&#xA;$$&#xA;The nice thing about the relativistic formula for $dW$ is that, in the case you are scattering particles off one another, it immediately shows us &lt;em&gt;three&lt;/em&gt; important contributions: not only the matrix element and phase space, but also the flux factor $1/s$ (where $s = ({p_1}^\mu + {p_2}^\mu)^2$ is the Mandelstam variable, and in case the masses are negligible, $ s \approx 2 E $).  This flux factor is responsible for the general $1/Q^2$ falling slope when you plot cross section over momentum transfer $Q = \sqrt{s}$, which comes entirely from relativistic kinematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this answers your questions. Here is a &lt;a href=&quot;http://www-pnp.physics.ox.ac.uk/~libby/Teaching/Lecture4.pdf&quot; rel=&quot;nofollow&quot;&gt;presentation (PDF)&lt;/a&gt; that sums it up, with an explicit proof that it is Lorentz invariant.&lt;/p&gt;&#xA;" OwnerUserId="825" LastActivityDate="2011-07-20T05:40:37.870" />
  <row Id="12495" PostTypeId="1" AcceptedAnswerId="12496" CreationDate="2011-07-20T08:20:08.420" Score="0" ViewCount="58" Body="&lt;p&gt;I've noticed that during wet weather my washing does not want to dry.&#xA;Is it because of lower temperature, or because of air humidity?&lt;/p&gt;&#xA;" OwnerUserId="4557" LastActivityDate="2011-07-28T12:00:28.733" Title="How air humidity affects dry process of a washing?" Tags="&lt;air&gt;&lt;weather&gt;" AnswerCount="3" />
  <row Id="12496" PostTypeId="2" ParentId="12495" CreationDate="2011-07-20T08:28:53.677" Score="1" ViewCount="" Body="&lt;p&gt;It's both, but humidity have higher impact I believe.&#xA;At 100% humidity washing will never dry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperature does affect drying speed, but even at -20C it will get dry eventually.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-07-20T08:28:53.677" />
  <row Id="12497" PostTypeId="2" ParentId="401" CreationDate="2011-07-20T08:29:27.463" Score="3" ViewCount="" Body="&lt;p&gt;First of all &lt;strong&gt;do not&lt;/strong&gt; use a &lt;a href=&quot;http://en.wikipedia.org/wiki/Raster_graphics&quot; rel=&quot;nofollow&quot;&gt;raster graphics&lt;/a&gt; software like Gimp to draw pictures. This has serious disadvantages when you want to make screen readable documents (the picture pixelizes). For this purpose always use &lt;a href=&quot;http://en.wikipedia.org/wiki/Vector_graphics&quot; rel=&quot;nofollow&quot;&gt;vector graphics&lt;/a&gt;. Wikipedia has a nice &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_vector_graphics_editors&quot; rel=&quot;nofollow&quot;&gt;list of vector graphics software&lt;/a&gt;. Among them, I'd recommend the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Inkscape&quot; rel=&quot;nofollow&quot;&gt;Inkscape&lt;/a&gt; (Cross-platform): Although the learning curve is a bit steep, its worth it. Note that Inkscape was not meant for making scientific diagram. Still you'll find a LaTeX plugin, export to LaTeX/PSTricks etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dia_%28software%29&quot; rel=&quot;nofollow&quot;&gt;Dia&lt;/a&gt; (Cross-platform): Very simple and easy, but not very powerful. &lt;strong&gt;Can't process LaTeX&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ipe_%28program%29&quot; rel=&quot;nofollow&quot;&gt;Ipe&lt;/a&gt; (Cross-platform): Ipe is something in between Inkscape and Dia. Also Ipe was designed for scientific drawing. Can process LaTeX source code and import PDF figures.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/WinFIG&quot; rel=&quot;nofollow&quot;&gt;WinFIG&lt;/a&gt; (Cross-platform): Although commercial, its quite popular in the scientific community.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="966" LastEditorUserId="966" LastEditorDisplayName="" LastEditDate="2011-07-20T19:57:59.497" LastActivityDate="2011-07-20T19:57:59.497" CommunityOwnedDate="2011-07-20T08:29:27.463" />
  <row Id="12498" PostTypeId="2" ParentId="12495" CreationDate="2011-07-20T08:35:36.233" Score="1" ViewCount="" Body="&lt;p&gt;Its because of the higher humidity.  Clothes being left out to dry are undergoing two processes: water from the air condensing on the clothes, and water in the clothes evaporating. The lower the humidity, the less water is condensing on clothes and the greater the net transfer of water to the air is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperature plays a role in how fast this process is, but you can have warm weather even when the humidity is high.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2011-07-20T08:35:36.233" />
  <row Id="12501" PostTypeId="1" AcceptedAnswerId="12508" CreationDate="2011-07-20T09:32:14.813" Score="0" ViewCount="68" Body="&lt;p&gt;What is &lt;strong&gt;&lt;em&gt;&quot;Strong Force&quot;&lt;/em&gt;&lt;/strong&gt; &amp;amp; &lt;strong&gt;&lt;em&gt;&quot;Weak Force&quot;&lt;/em&gt;&lt;/strong&gt; in Astronomy terms?&lt;/p&gt;&#xA;" OwnerUserId="4560" LastActivityDate="2011-07-20T10:28:41.227" Title="What is &quot;Strong Force&quot; &amp; &quot;Weak Force&quot; in Astronomy terms?" Tags="&lt;astrophysics&gt;" AnswerCount="1" />
  <row Id="12503" PostTypeId="2" ParentId="9727" CreationDate="2011-07-20T09:38:31.103" Score="8" ViewCount="" Body="&lt;p&gt;A NASA team recently (july 2011) doubled the amount of available data related to the pioneer anomaly, and they saw an exponentially decreasing acceleration with a 27 year half life, consistent with the thermal scenario. More precisely they say :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The rationale for an exponential model is based on the possibility that the acceleration may be due to thermal recoil forces generated on-board. Due to degradation of the RTG thermocouples and changes in the thermal louver system, the resulting thermal recoil force could have a half-life significantly shorter than the 87.74 year&#xA;  half-life of the ²³⁸Pu fuel, with 27 years being in the acceptable range. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For them &quot;The most likely cause of the Pioneer anomaly&#xA;is the anisotropic emission of on-board heat&quot; and promise a more detailed analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper is &lt;a href=&quot;http://arxiv.org/abs/1107.2886&quot; rel=&quot;nofollow&quot;&gt;arXiv:1107.2886&lt;/a&gt; (now published in &lt;a href=&quot;http://link.aps.org/doi/10.1103/PhysRevLett.107.081103&quot; rel=&quot;nofollow&quot;&gt;PRL&lt;/a&gt;) and there is &lt;a href=&quot;http://www.technologyreview.com/blog/arxiv/27012/&quot; rel=&quot;nofollow&quot;&gt;a short discussion on it&lt;/a&gt; on The Physics arXiv Blog.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my (scientist but non-expert) point of view, this temporal decay looks like an independent confirmation of the thermal origin of the Pioneer anomaly and basically settles the question.&lt;/p&gt;&#xA;" OwnerUserId="373" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2011-08-20T12:24:38.130" LastActivityDate="2011-08-20T12:24:38.130" CommentCount="2" />
  <row Id="12504" PostTypeId="1" CreationDate="2011-07-20T09:40:56.570" Score="5" ViewCount="158" Body="&lt;p&gt;I am a beginner in QED and QFT. &#xA;What is known (or expected to be) about bound states in QED? As far as I understand,&#xA;in non-relativistic QM electron and positron can form a bound state. Should it be true in QED? Most of the basic text books on QFT I known treat QED with perturbative methods which do not help to study bound states. Is there a literature to read about it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Positronium&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;, electron and positron form a positronium. It is an unstable particle which can annihilate to two photons. In particular electron and positron cannot form a bound state. But still there is a logical possibility that several electrons and positrons can form a bound state (though probably once one has at least one electron and one positron, they will necessarily annihilate. Is it true?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if one considers QED with both electrons and muons, then electron and antimuon form muonium. Clearly they cannot annihilate to photons. However&#xA;in &lt;a href=&quot;http://en.wikipedia.org/wiki/Muonium&quot; rel=&quot;nofollow&quot;&gt;another article&lt;/a&gt; in wikipedia it is claimed that muonium is unstable. Is it due to some effects of QED or due to actual presence of other, say weak, interactions?&lt;/p&gt;&#xA;" OwnerUserId="4562" LastEditorUserId="4562" LastEditorDisplayName="" LastEditDate="2011-07-21T08:16:29.443" LastActivityDate="2011-07-21T15:06:49.870" Title="Bound states in QED" Tags="&lt;quantum-field-theory&gt;&lt;quantum-electrodynamics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="12506" PostTypeId="2" ParentId="12504" CreationDate="2011-07-20T10:01:31.850" Score="3" ViewCount="" Body="&lt;p&gt;Yes, in QED there are bound states. Nobody can forbid us to take the Coulomb potential into account exactly and the rest - by the perturbation theory. In this way they obtain the Lamb shift, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are different approaches to bound stated in QED: poles of the scattering matrix, Bethe-Salpeter equation, Schwinger approach, Logunov-Tavkhelidze quasi-potential approach, etc.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-07-20T10:01:31.850" CommentCount="8" />
  <row Id="12507" PostTypeId="2" ParentId="12485" CreationDate="2011-07-20T10:14:27.683" Score="0" ViewCount="" Body="&lt;p&gt;I don't know why they should violate thermodynamics either, but they don't exist because they're static. They cannot be created at any finite time - they must have existed since the beginning of time and will exist forever. The physically realistic Schwarzschild solution is created from collaps and does not have the second asymptotic region. &lt;/p&gt;&#xA;" OwnerUserId="1458" LastActivityDate="2011-07-20T10:14:27.683" CommentCount="1" />
  <row Id="12508" PostTypeId="2" ParentId="12501" CreationDate="2011-07-20T10:28:41.227" Score="3" ViewCount="" Body="&lt;p&gt;I assume you are referring to the strong and weak nuclear forces. These are two fundamental forces (the others being electromagnetism and gravity). Due to the very short range on which these forces are effective, they are less directly related to the intermediate and large scale structure of the universe than electromagnetism and gravity, however they are of key importance to the stability and nuclear structure of matter, and thus changing them much would result in a markedly different universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strong nuclear force is mediated by particles called gluons, and is responsible for the very strong, very short range attraction between nucleons. This is the reason why atomic nuclei stick together despite protons having a positive charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weak nuclear force is mediated by W and Z bosons, and is responsible for certain types of radioactive decay. For instance, iirc, neutron decay involves interconversion of a neutron into a proton, electron and electron neutrino via a W boson.&lt;/p&gt;&#xA;" OwnerUserId="1876" LastActivityDate="2011-07-20T10:28:41.227" CommentCount="1" />
  <row Id="12509" PostTypeId="1" CreationDate="2011-07-20T10:36:25.887" Score="2" ViewCount="81" Body="&lt;p&gt;Environmentally induced decoherence makes wave function collapse unnecessary. But the environment, usually taken to be some heat bath, introduces a preferred frame. (That in which the total (spatial) momentum vanishes.) So, doesn't then the decoherence time depend on the motion of the prepared state relative to the environment? And, doesn't the ultimate environment, all particles in the universe, introduce a preferred frame into quantum mechanics in the sense that the decoherence time is relative to this frame? And would this be measureable, at least in principle? I.e. I could go into a frame with high boost relative to the CMB restframe and notice that the decoherence time changes? &lt;/p&gt;&#xA;" OwnerUserId="1458" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-21T00:06:46.273" LastActivityDate="2011-07-21T04:01:08.440" Title="Does decoherence single out a preferred frame?" Tags="&lt;quantum-mechanics&gt;&lt;special-relativity&gt;&lt;coherence&gt;" AnswerCount="1" />
  <row Id="12510" PostTypeId="1" AcceptedAnswerId="12516" CreationDate="2011-07-20T10:47:12.600" Score="1" ViewCount="73" Body="&lt;p&gt;Imagine I have an sp3 hybridized carbon attached to four separate polyethylene chains.  By pulling on the polyethylene chains in some manner, is it possible for me to mechanically isomerize the chiral center prior to breaking any carbon-carbon bonds?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps a more realistic scenario would involve shining a laser on a compound similar to 2,2-dimethylpropane (i.e. a fully saturated carbon compound where four carbons are attached to an sp3 hybridized carbon), but with asymmetric functionalizations on the bonded carbons to allow for detection of isomerization.  Just as in the previous scenario, can mechanical isomerization occur prior to breaking any carbon-carbon bonds?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Quoting from Georg's answer: &quot;In &quot;practice&quot; there is a big problem, You would need two Laplacian demons (with two hands each) to do the experiment...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not clear to me that this is an impossible experiment.  For example, you could attach one polyethylene chain to a surface and then try to make a covalent bond between the other relevant chain and, say, an AFM tip.  If this is successful, one would then proceed in a manner similar to that reported in &quot;How Strong is a Covalent Bond?&quot; by Michel Grandbois et. al. (http://www.sciencemag.org/content/283/5408/1727.short), which estimates a ~4.0 nanonewton rupture force for a C-C bond (See Fig. 4) and directly measures the rupture of a single Si-C bond to be ~2.0 +/- 0.3 nanonewtons.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case where one wants to look for mechanical isomerization of the sp3-hybridized carbon, one would look for an earlier sub-4.0 nN peak in the force vs. extension curve (or whatever critical force is experimentally measured for breaking a C-C bond), perhaps allow relaxation at some critical force to complete the isomerization process, and then compare the observed displacement with a geometrical model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sounds ridiculously hard, but I see no fundamental reason it couldn't be done.&lt;/p&gt;&#xA;" OwnerUserId="4257" LastEditorUserId="4257" LastEditorDisplayName="" LastEditDate="2011-07-20T21:57:50.327" LastActivityDate="2011-07-20T21:57:50.327" Title="Is it possible to mechanically isomerize an sp3 hybridized carbon center?" Tags="&lt;physical-chemistry&gt;&lt;geometry&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12511" PostTypeId="1" AcceptedAnswerId="12517" CreationDate="2011-07-20T11:10:12.393" Score="1" ViewCount="65" Body="&lt;p&gt;I want to build simplest interferometer which should be able to measure movements down to fraction of wavelength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the simplest scheme for that, and what are the requirements for a laser?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a bunch of laser diode-based ones, and I guess they might be not coherent enough...&#xA;Are green DPSS ones any better?&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-07-20T16:51:53.440" Title="Simplest interferometer" Tags="&lt;interferometry&gt;" AnswerCount="2" />
  <row Id="12512" PostTypeId="1" AcceptedAnswerId="12550" CreationDate="2011-07-20T12:08:10.080" Score="2" ViewCount="187" Body="&lt;p&gt;This is in context of classical Newtonian physics.&#xA;Consider a system of n different point mass particles. Initially all are spread around on one plane. No particle possess any velocity to begin with. There is no external force acting on this system. Each particle has some mass. The only force to consider here is gravitation between any two particles. &#xA;When two particles collide, lets say they form a lump with mass equal to sum of the two. This now behaves like one single particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first assumption is, the centre of mass of this system, throughout time won't move. Right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a sufficient amount of time, will everything collapse into one single particle at the centre of mass? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or it's possible to see a system where few particles are rotating in one direction and other in other direction, so that angular momentum is conserved and net angular momentum is still zero? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, can you give one simple example with points, mass and initial position which will achieve this kind of state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: On similar note, can we have say 3 particles which are attracted to each other but never collide, something like a simple harmonic motion?&lt;/p&gt;&#xA;" OwnerUserId="4564" LastEditorUserId="4564" LastEditorDisplayName="" LastEditDate="2011-07-20T17:25:46.543" LastActivityDate="2011-07-21T04:09:01.557" Title="N particles, will there be any rotation after a period of time or everything will collapse" Tags="&lt;gravity&gt;&lt;newtonian-mechanics&gt;&lt;angular-momentum&gt;&lt;rotational-dynamics&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="12513" PostTypeId="2" ParentId="12487" CreationDate="2011-07-20T12:57:59.150" Score="8" ViewCount="" Body="&lt;p&gt;Dan's answer is essentially good, but miss one effect : the &lt;a href=&quot;http://en.wikipedia.org/wiki/Coriolis_effect&quot; rel=&quot;nofollow&quot;&gt;Coriolis effect&lt;/a&gt;. You can imagine a planet spinning much more rapidly than the earth, but at a constant angular speed. On that quickly rotating planet, the explanation of Dan would still stand, but as soon as on moves, we would feel a lateral Coriolis force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Coriolis acceleration is $2\vec{\Omega}\times\vec v$, where $\vec{\Omega}$ is the (vectorial) angular frequency of the planet's rotation and $\vec v$ the speed of the object moving. For an object moving at the speed of sound (340 m/s) near the Earth's pole, where the effect is maximum, the Coriolis acceleration is&#xA;$$2\frac{2\pi}{24\times60\times60}\times 340 \simeq \frac{12\times 340}{24\times 3600}\sim \frac1{20} = 5\times10^{-2} \mathrm{m}\cdot\mathrm{s}^{-2}.$$&#xA;This corresponds to an acceleration which is half a percent of the gravity acceleration, for a situation which is already quite far from everyday life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This small effect can accumulate over long distance and can have visible effects, notably at meteorological scales. In some sense, we feel the Earth rotation when we feel the dominant wind direction in our region. The parameter characterizing the intensity of the Coriolis effect for a phenomenon is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rossby_number&quot; rel=&quot;nofollow&quot;&gt;Rossby number&lt;/a&gt;, which is big if the Coriolis effect is negligible. If the phenomenon you analyse has a typical speed $v$, occur over a distance $L$, the Rosby number is essentialy proportional to the ration of the rotation period (24 h in our case) over the time $v/L$ it takes to go over the typical distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For meteorological depressions, the wind take several days to go over the thousands of kilometres they span, and the Coriolis effect has an important effect. To really feel the effect in everyday's life, one would need to be on a planet with a day of a few seconds, like the Little Prince's &lt;a href=&quot;http://home.pacific.net.hk/~rebylee/text/prince/14.html&quot; rel=&quot;nofollow&quot;&gt;lamplighter's planet&lt;/a&gt; !  Of course, if you don't live on a rapidly rotating asteroid, you can see the effect on a &lt;a href=&quot;http://youtu.be/_36MiCUS1ro&quot; rel=&quot;nofollow&quot;&gt;carousel&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2011-07-20T12:57:59.150" CommentCount="1" />
  <row Id="12514" PostTypeId="2" ParentId="12512" CreationDate="2011-07-20T13:29:42.720" Score="0" ViewCount="" Body="&lt;p&gt;It sounds like you've got a bunch of particles with $1/r^2$ attractive forces which (perfectly) inelastically collide. If so, the momentum and angular momentum of the system will be conserved, so the centre of mass will not move, and there will be no net rotation. A single &quot;clump&quot; in the middle is certainly a stable situation, and I think there cannot be others --- they will necessarily have finite angular momentum, though a proof escapes me.&lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-07-20T13:29:42.720" CommentCount="4" />
  <row Id="12515" PostTypeId="2" ParentId="12512" CreationDate="2011-07-20T14:03:45.703" Score="2" ViewCount="" Body="&lt;p&gt;Well, the planetary system is an example, though it is not isolated as in your problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are requesting is a simple solution to a many body problem. It does not exist.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Deterministic_chaos&quot; rel=&quot;nofollow&quot;&gt;Deterministic chaos&lt;/a&gt; is one way to study such problems, also numerical solutions in computers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example &lt;a href=&quot;http://arxiv.org/abs/1106.2590&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; is such a study:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Onset of Secular Chaos in Planetary Systems: Period Doubling &amp;amp; Strange Attractors&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also &lt;a href=&quot;http://www.pnas.org/content/98/22/12342.full&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; , from their abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Over the last two decades, there has come about a recognition that chaotic dynamics is pervasive in the solar system. We now understand that the orbits of small members of the solar system—asteroids, comets, and interplanetary dust—are chaotic and undergo large changes on geological time scales. Are the major planets' orbits also chaotic? The answer is not straightforward, and the subtleties have prompted new questions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Numerical solutions are used in the virtual &lt;a href=&quot;http://www.stellarium.org/&quot; rel=&quot;nofollow&quot;&gt;planetaria&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as genneth has said angular momentum and momentum will be conserved, but that is the only sure statement for an N-body gravitational system as the one you postulate. (same is true in N body systems for other forces too). Since you posit initial conditions at rest the total system will have 0 momentum and angular momentum, but some individual orbits will appear as not all solutions will be head on collisions. Rather the opposite, unless there is one very large mass particle to start with at the centre of mass.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-07-20T17:54:47.980" LastActivityDate="2011-07-20T17:54:47.980" CommentCount="12" />
  <row Id="12516" PostTypeId="2" ParentId="12510" CreationDate="2011-07-20T14:19:15.907" Score="2" ViewCount="" Body="&lt;p&gt;The exact answer is a problem of computational chemistry, &#xA;maybe somebody has caculated the energy of such a center &#xA;in a planar configuration, but I don't know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &quot;practice&quot; there is a big problem, You would need two Laplacian demons &#xA;(with two hands each) to do the experiment :=)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, on the level of such a molecule, &quot;mechanic&quot; is a unrealistic simplification. &#xA;Nature on this level is not electric as opposed to mechanic or what ever.&#xA;So forcing the four ligands in a plane (the activated complex of thiś isomerisation) &#xA;will lead to rupture of one of the bonds, either forming a radical pair or &#xA;an ion pair, this depends on the milieu (gas phase, polar liquid, unpolar etc) &#xA;The possible (imaginable) configurations of the four ligands make a energy hypersurface&#xA;and system will go over the lowest pass  to a more stable configuration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why can one be so shure that it impossible? Think of reason for that &quot;hybridisation&quot;!&#xA;Why is the cabon not bound to three neighbours at 90 degrees like PH3 (nearly) by three bonds&#xA;via a pure p-orbital and the forth to a pure s-orbital? Reason is simply space. &#xA;There is not enough room around a carbon , contrary to atoms in higher periods like phosphor or sulfur.  In this sense the hybridisation is forced by &quot;mechanics&quot;. &#xA;Think of Gillespie-Nyholm rules and retropolate &quot;down&quot; for the crowded space &#xA;around a C-Atom.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-07-20T14:19:15.907" />
  <row Id="12517" PostTypeId="2" ParentId="12511" CreationDate="2011-07-20T14:24:12.680" Score="2" ViewCount="" Body="&lt;p&gt;Well, one problem you are likely to encounter is that your setup will likely be vibrating with amplitude on this order.. do you have a floating optical table?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coherence is probably not that important. At a minimum you will need a beamsplitter, two mirrors and a diode or some other way to measure the interference pattern. A lens or two to magnify the pattern will also be helpful.&lt;/p&gt;&#xA;" OwnerUserId="2963" LastActivityDate="2011-07-20T14:24:12.680" CommentCount="4" />
  <row Id="12519" PostTypeId="2" ParentId="12484" CreationDate="2011-07-20T14:39:39.197" Score="3" ViewCount="" Body="&lt;p&gt;Monopoles are still created in inflationary models. They're just created before (or during) inflation, so that the rapid expansion thereafter dilutes their density to unobservably low levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the time when the monopoles are created, they're created at a density of order 1 per Hubble volume -- that is, there's one in each &quot;observable Universe&quot; at that time. In general, when a symmetry breaks, topological defects form that are separated on a length scale of order (speed of propagation of the field)(time scale over which the symmetry breaks). The first is of order $c$, and the second is of order the Hubble time, so monopoles are separated by a distance of order the Hubble length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should take &quot;of order&quot; here very liberally -- I don't actually care if I'm off by factors of $10^5$ or $10^{10}$ or anything measly like that! After all, inflation blows up lengths by something like $10^{20}$ or more. So one monopole per horizon volume becomes one per $10^{60}$ horizon volumes. (Also, the horizon volume continues to change after inflation is over, but not by anything like this sort of factor.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With densities like that, we certainly wouldn't expect to see any monopoles. Problem solved.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-07-20T14:39:39.197" />
  <row Id="12520" PostTypeId="1" AcceptedAnswerId="12522" CreationDate="2011-07-20T14:46:11.173" Score="8" ViewCount="234" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The existence of the antiproton with -1 electric charge, opposite to the +1 electric charge of the proton, was predicted by Paul Dirac in his 1933 Nobel Prize lecture.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Quotation by &lt;a href=&quot;http://en.wikipedia.org/wiki/Antiproton&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question.&lt;/strong&gt; How did Paul Dirac predict existence of antiproton?&lt;/p&gt;&#xA;" OwnerUserId="4558" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2011-07-20T15:56:13.840" LastActivityDate="2011-07-21T16:05:22.007" Title="How Did Paul Dirac Predict The Existence of Antiproton?" Tags="&lt;history&gt;&lt;antimatter&gt;&lt;particles&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="12521" PostTypeId="2" ParentId="401" CreationDate="2011-07-20T15:48:32.110" Score="2" ViewCount="" Body="&lt;p&gt;Sometimes raster graphics is also necessary. I often used &lt;a href=&quot;http://www.povray.org/&quot; rel=&quot;nofollow&quot;&gt;POV-Ray&lt;/a&gt; to make some illustrations. &#xA;&lt;img src=&quot;http://i.stack.imgur.com/Fh0AA.jpg&quot; alt=&quot;black hole&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be prepared with any size using the same script with description of the picture, that avoid scaling problem. But it is not a graphical editor and fast only for preparation of simple pictures.&lt;/p&gt;&#xA;" OwnerUserId="3351" LastActivityDate="2011-07-20T15:48:32.110" CommunityOwnedDate="2011-07-20T15:48:32.110" />
  <row Id="12522" PostTypeId="2" ParentId="12520" CreationDate="2011-07-20T15:50:14.200" Score="7" ViewCount="" Body="&lt;p&gt;You'll find Dirac's 1933 Nobel lecture on the &lt;a href=&quot;http://nobelprize.org/nobel_prizes/physics/laureates/1933/dirac-lecture.html&quot; rel=&quot;nofollow&quot;&gt;Nobelprize.org website&lt;/a&gt;. The &lt;a href=&quot;http://nobelprize.org/nobel_prizes/physics/laureates/1933/dirac-lecture.pdf&quot; rel=&quot;nofollow&quot;&gt;pdf&lt;/a&gt; is quite brief (5 pages long) and speaks on the antiproton at the end (p4). The argument is the following : &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In any case I think it is probable that negative protons can exist, since as far as the theory is yet definite, there is a complete and perfect symmetry between positive and negative electric charge, and if this symmetry is really fundamental in nature, it must be possible to reverse the charge on any kind of particle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In short, the particle-antiparticle symmetry seems to be a law of nature, so the proton should also have an an anti-proton partner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited to expand below&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@ANKU's answer above is the answer to the question &quot;How did Dirac predict the existence of antimatter&quot;, and this work was done for the electron (or the positron). Once the positron has been predicted and and observed, came the intuition that this symmetry was much more general. However, in 1933, Dirac didn't think that this theory could be directly applied to protons. To quote his Nobel lecture :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The theory of electrons and positrons which I have just outlined is a self- consistent theory which fits the experimental facts so far as is yet known. One would like to have an equally satisfactory theory for protons. One might perhaps think that the same theory could be applied to protons. This would require the possibility of existence of negatively charged protons forming a mirror-image of the usual positively charged ones. There is, however, some recent experimental evidence obtained by Stern about the spin magnetic moment of the proton, which conflicts with this theory for the proton. As the proton is so much heavier than the electron, it is quite likely that it requires some more complicated theory, though one cannot at the present time say what this theory is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="373" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2011-07-21T16:05:22.007" LastActivityDate="2011-07-21T16:05:22.007" CommentCount="3" />
  <row Id="12523" PostTypeId="2" ParentId="12488" CreationDate="2011-07-20T16:09:22.787" Score="1" ViewCount="" Body="&lt;p&gt;&quot;&lt;strong&gt;Is there any fundamental (and preferably simple) argument that makes decay of massless particles implausible?&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For decay into massive particles&lt;/strong&gt;: I don't think you need any more than Special Relativity to answer this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Massless particles necessarily travel at the speed of light. Therefore, even if they were unstable, they cannot decay in any reference frame. This is ultimately due to nature of &lt;a href=&quot;http://en.wikipedia.org/wiki/Time_dilation&quot; rel=&quot;nofollow&quot;&gt;time dilation&lt;/a&gt; in Special Relativity. In a sense, one could say massless particles (which must travel at $c$) do not &lt;em&gt;experience&lt;/em&gt; time and therefore cannot decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more QFT-type language:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, consider a tree-level diagram of a hypothetical photon decay: $\gamma\rightarrow e^+e^-$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the translational-invariance of the vacuum means that momentum conservation disallows this transition; since the three-momentum of the virtual pair must be zero whilst that of the photon cannot be zero (since it travels at $c$ in all reference frames.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note however that in Feynman diagrams such as for: $e^+e^-\rightarrow e^+e^-$ via a virtual photon, $\gamma^*$, exchange the situation is somewhat different. A virtual photon has a finite mass, and thus a rest frame, so the three-momentum is conserved at each vertex. This however, should &lt;strong&gt;not&lt;/strong&gt; be refered to as the decay of a photon and does &lt;strong&gt;not&lt;/strong&gt; mean that a &lt;em&gt;real&lt;/em&gt; photon can decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For decay into massless particles&lt;/strong&gt;: ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realise this only addresses part of the question but if no one else contributes a more all-encompassing answer I will endeavour to expand on it once I've done a bit more research.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="1696" LastEditorDisplayName="" LastEditDate="2011-07-20T17:42:20.777" LastActivityDate="2011-07-20T17:42:20.777" CommentCount="5" />
  <row Id="12525" PostTypeId="2" ParentId="12511" CreationDate="2011-07-20T16:51:53.440" Score="2" ViewCount="" Body="&lt;p&gt;You should also consider the laser's frequency stability. My understanding is that the frequency of light can go up and down by some nanometers as temperature (and perhaps some of the electrical inputs) of a laser diode varies. Searching seems to turn up quite a few articles on stabilizing their output, exactly for this purpose.&lt;/p&gt;&#xA;" OwnerUserId="4458" LastActivityDate="2011-07-20T16:51:53.440" />
  <row Id="12527" PostTypeId="2" ParentId="12509" CreationDate="2011-07-20T17:19:53.940" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Environmentally induced decoherence makes wave function collapse unnecessary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As far as I know, that's still a conjecture. I might be wrong about that, though.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But the environment, usually taken to be some heat bath, introduces a preferred frame. (That in which the total (spatial) momentum vanishes.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's the rest frame of the system + reservoir, but it is not &quot;preferred&quot; in the sense that it violates Lorentz invariance.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So, doesn't then the decoherence time depend on the motion of the prepared state relative to the environment?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only the part of the environment that is directly interacting with your system. If your reservoir is a collection of matter interacting with the system by the exchange of blackbody radiation photons, then the decoherence time will depend on how often the system interacts with one of those photons.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And, doesn't the ultimate environment, all particles in the universe,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What make you think that the universe is finite?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;introduce a preferred frame into quantum mechanics in the sense that the decoherence time is relative to this frame?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All decoherence times can be calculated from the decoherence times of the same system in any other inertial frame.  The rest frame of the observable universe is in no way &quot;preferred&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And would this be measureable, at least in principle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That decoherence is affected by time dilation like everything else? Yes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I.e. I could go into a frame with high boost relative to the CMB restframe and notice that the decoherence time changes? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Long story short: you probably aren't going to be able to use decoherence to find some universal preferred frame and disprove special relativity.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-21T04:01:08.440" LastActivityDate="2011-07-21T04:01:08.440" CommentCount="5" />
  <row Id="12528" PostTypeId="1" AcceptedAnswerId="12530" CreationDate="2011-07-20T17:35:05.710" Score="5" ViewCount="129" Body="&lt;p&gt;The preface to Mark Srednicki's &quot;Quantum Field Theory&quot; says that to be prepared for the book, one must recognize and understand the following equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d\sigma}{d\Omega} = |f(\theta,\phi)|^2,  \qquad (1)$$&#xA;$$a^{\dagger}|n\rangle = \sqrt{n+1} \space |n+1\rangle, \qquad  (2)$$&#xA;$$J_{\pm} |j,m \rangle = \sqrt{j(j+1)-m(m\pm 1)} \mid j,m \pm 1 \rangle, \qquad  (3)$$&#xA;$$A(t) = e^{+iHt/\hbar}Ae^{-iHt/\hbar}, \qquad  (4)$$&#xA;$$H = p\dot{q}-L, \qquad (5)$$&#xA;$$ct&amp;#39;=\gamma (ct-\beta x), \qquad  (6)$$&#xA;$$E=(\mathbf{p}^2c^2+m^2c^4)^{1/2}, \qquad  (7)$$&#xA;$$\mathbf{E} =-\mathbf{\dot{A}}/c-\mathbf{\nabla} \varphi. \qquad  (8)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am certainly not ready to dive into this book, so I would like some help identifying these equations and learning more about their fundamental usefulness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't recognize (1), but (2) looks like a quantum mechanical creation operator?  I thought those were only really useful in the context of the harmonic oscillator problem, but maybe everything is just a complicated HO problem?  (3) has to do with angular momentum?  (4) is a plane wave solution to the Schrodinger Eqn?  (5) is the classical mechanics Hamiltonian, with cannonical coordinates?  (6) is the relativistic Lorentz transformation.  (7) is the general form of mass energy equivalence from SR.  (8) is the electric field expressed as vector and scalar potentials?  Is that really the only E&amp;amp;M machinery required?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any insight as to why these particular expressions are relevant / important / useful to QFT is also appreciated.  Also, where are the statistical mechanics ideas hiding?  In the QM?&lt;/p&gt;&#xA;" OwnerUserId="4530" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-20T17:48:09.310" LastActivityDate="2011-07-21T09:31:57.560" Title="Essential background for QFT study" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;electromagnetism&gt;&lt;classical-mechanics&gt;&lt;special-relativity&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12529" PostTypeId="2" ParentId="12520" CreationDate="2011-07-20T17:36:14.290" Score="11" ViewCount="" Body="&lt;p&gt;The Dirac equation implies negative energies as well as positive. This is due to energy-momentum relation $E=\pm \sqrt{m^2+p^2 }$. If we replace $E$ and $p$ by operators $E\to i\frac{\partial }{\partial t}$ and $p\to -i\nabla$ we get the Klein-Gordon equation $(\Box+m^2)\phi=0$ for scalar (spinless) fields $\phi$. The problem with this equation is that it gives solutions with negative probability density and negative energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to overpass the problem with negative probability density, Dirac made the K-G equation linear in time derivative $\frac{\partial }{\partial t}$ (and in space derivative to make it covariant). So he get the following equation (the Dirac equation): $(i\gamma^{\mu}\partial_{\mu}-m)\psi=0$ for particles with 1/2-spin (fermions: electrons, protons etc.). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Dirac equation gives positive probability densities which is good, but the problem with negative energy quantum states remained. To overpass this problem, Dirac postulated that the universe is filled with infinitely dense &quot;sea&quot; of negative energy particles (electrons), the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_sea&quot; rel=&quot;nofollow&quot;&gt;Dirac sea&lt;/a&gt;. Due to Pauli exclusion principle no other electron can fall into the Dirac sea, but sometimes one electron can leave the Dirac sea creating a hole which would act like positive energy electron with opposite charge - the positron, experimentally discovered by Carl Anderson. This holes are called &lt;a href=&quot;http://en.wikipedia.org/wiki/Antiparticle&quot; rel=&quot;nofollow&quot;&gt;antiparticles&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the Dirac sea theory has some problems, like the problem of infinite charge of the universe and the fact that the bosons, which have antiparticles too, do not obey the Pauli exclusion principle and the hole theory doesn't work for them. This problems are solved in quantum field theory.&lt;/p&gt;&#xA;" OwnerUserId="3066" LastEditorUserId="3066" LastEditorDisplayName="" LastEditDate="2011-07-20T22:27:13.577" LastActivityDate="2011-07-20T22:27:13.577" CommentCount="4" />
  <row Id="12530" PostTypeId="2" ParentId="12528" CreationDate="2011-07-20T17:55:53.843" Score="11" ViewCount="" Body="&lt;p&gt;They are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The definition of the scattering cross-section in terms of the scattering amplitude.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;HO/photon creation operator.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Angular momentum raising and lowering operator.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Heisenberg equations of motion for an operator.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Definition of the Hamiltonian.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lorentz transformation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Relativistic equation for energy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Electric field in terms of scalar and vector potentials.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(1)&lt;/strong&gt; is because you calculate a lot of cross-sections, &lt;strong&gt;(2) &amp;amp; (3)&lt;/strong&gt; because you use a lot of raising and lowering operators, &lt;strong&gt;(4)&lt;/strong&gt; because you need to know the difference between the Schrodinger, Hisenberg, and Interaction pictures of QM, &lt;strong&gt;(5)&lt;/strong&gt; becuase you use both Lagrangians and Hamiltonians, &lt;strong&gt;(6) &amp;amp; (7)&lt;/strong&gt; for basic special relativity, and &lt;strong&gt;(8)&lt;/strong&gt; for basic E&amp;amp;M.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="66" LastEditorDisplayName="" LastEditDate="2011-07-21T09:31:57.560" LastActivityDate="2011-07-21T09:31:57.560" CommentCount="1" />
  <row Id="12531" PostTypeId="2" ParentId="391" CreationDate="2011-07-20T18:02:23.150" Score="-1" ViewCount="" Body="&lt;p&gt;For every action there is an equal and opposite reaction, and in my understanding, everything has an equal opposite including time and light.&lt;/p&gt;&#xA;" OwnerUserId="4566" LastActivityDate="2011-07-20T18:02:23.150" />
  <row Id="12532" PostTypeId="1" CreationDate="2011-07-20T18:20:02.033" Score="0" ViewCount="141" Body="&lt;p&gt;Which cosmologies consistent with physics as we currently know it is consistent with Eternal Life lasting forever starting from here on Earth while preserving the memories of human life? With computers and robots colonizing the universe and growing without bound?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tipler had suggested an Omega Point associated with a big crunch in a closed universe, but unfortunately, scales below the Planck length do not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dyson had performed an analysis in the article &quot;TIME WITHOUT END: PHYSICS AND BIOLOGY IN AN OPEN UNIVERSE&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4567" LastActivityDate="2011-08-27T05:35:20.533" Title="Which cosmologies support Eternal Life?" Tags="&lt;cosmology&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12533" PostTypeId="1" CreationDate="2011-07-20T18:45:26.753" Score="4" ViewCount="114" Body="&lt;p&gt;Why zeta regularization only valid at one-loop? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean there are zeta regularizations for multiple zeta sums.&#xA;Also we could use the zeta regularization iteratively on each variable &#xA;to obtain finite corrections to multiple loop diagrams.&lt;/p&gt;&#xA;" OwnerUserId="4435" LastEditorUserId="429" LastEditorDisplayName="" LastEditDate="2011-08-01T07:50:50.470" LastActivityDate="2011-08-01T08:14:24.227" Title="Why Zeta regularization is not valid for multiple-loops?" Tags="&lt;quantum-field-theory&gt;&lt;mathematical-physics&gt;&lt;renormalization&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12534" PostTypeId="5" CreationDate="2011-07-20T19:12:53.427" Score="0" ViewCount="" Body="&lt;p&gt;Entropy is defined to be extensive (that is additive when two systems are considered together).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In statistical mechanics the entropy is identified with the logarithm of the number of microscopic states of a system which has a given set of macroscopic properties.  &lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-07-20T22:24:42.697" LastActivityDate="2011-07-20T22:24:42.697" />
  <row Id="12535" PostTypeId="4" CreationDate="2011-07-20T19:12:53.427" Score="0" ViewCount="" Body="A important property of all systems in thermodynamics and statistical mechanics. Entropy characterizes the degree to which the energy of the system is *not* available to do useful work&#xD;&#xA;&#xD;&#xA;" OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-07-20T22:24:07.580" LastActivityDate="2011-07-20T22:24:07.580" />
  <row Id="12536" PostTypeId="5" CreationDate="2011-07-20T19:16:42.877" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-07-20T19:16:42.877" LastActivityDate="2011-07-20T19:16:42.877" />
  <row Id="12537" PostTypeId="4" CreationDate="2011-07-20T19:16:42.877" Score="0" ViewCount="" Body="A nucleus made of protons and neutrons surrounded by a cloud of electrons equal in number to the protons." OwnerUserId="-1" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-20T23:08:46.410" LastActivityDate="2011-07-20T23:08:46.410" />
  <row Id="12538" PostTypeId="5" CreationDate="2011-07-20T19:18:13.500" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-07-20T19:18:13.500" LastActivityDate="2011-07-20T19:18:13.500" />
  <row Id="12539" PostTypeId="4" CreationDate="2011-07-20T19:18:13.500" Score="0" ViewCount="" Body="A theorem that relates continuous symmetries (continuous transformations that don't affect the value of the lagrangian) to quantities conserved in time." OwnerUserId="-1" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-20T23:08:48.297" LastActivityDate="2011-07-20T23:08:48.297" />
  <row Id="12541" PostTypeId="1" AcceptedAnswerId="12549" CreationDate="2011-07-20T21:10:17.387" Score="6" ViewCount="94" Body="&lt;p&gt;Much is made of finding exoplanets in habitable zones, locations with orbital semi-major axes permitting water in the liquid state. Habitability may be compromised if such bodies become tidally locked, orbiting within the star's tidal lock radius. The illustration below shows the tidal lock radius for various stellar spectral classes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/KcynV.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diagram shows Mercury within the tidally locked zone although its really in a 3:2 spin-orbit resonance. &lt;strong&gt;How is the tidal lock radius calculated?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1924" LastActivityDate="2011-07-21T03:58:41.810" Title="Tidal Lock Radius in Habitable Zones" Tags="&lt;exoplanets&gt;&lt;tidal-effect&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="12542" PostTypeId="2" ParentId="12512" CreationDate="2011-07-20T21:11:14.093" Score="1" ViewCount="" Body="&lt;p&gt;If you are thinking about largeish N (hundreds to millions or more), then it resembles a star cluster. There is the issue raised by more than one other poster, that if they are point particles, they can't combine/stick. Close graviational interactions resemble a sort of collision. Some particles will leave collisions with enough kinetic energy to escape from the system (i.e. fly away, and maintain a nonzero velocity in the limit where time goes to infinity), so you get a clump of gravitationally bound particles, and occasionally a particle acquires enough kinetic energy to escape from the collective graviational embrace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a truly 2D universe, the configuation will stay 2D, but in any real 3D situation any infinitessimal motion in the third dimension will cause the system to spreadout over three dimensions. I believe it reaches a longterm quasi-steady state that resembles a globular cluster, but remember that occasionally a particle will escape, removing energy (and energy per mass) from the cluster. Also the net momementum and angular momentum of the bound particles, equals zero, but nothing constrains these two subsets individually to have zero momentum or angular momentum, so the cluster might end up with nontrivial momentum and angular momentum.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-07-20T21:11:14.093" />
  <row Id="12543" PostTypeId="2" ParentId="12488" CreationDate="2011-07-20T23:10:40.593" Score="5" ViewCount="" Body="&lt;p&gt;I agree with qftme's answer for the case of massive decay products.  By energy conservation alone, $\gamma \rightarrow e^+e^-$ should be allowed, but momentum conservation forbids it (as well as the opposite case, $e^+e^-$ annihilation).  It is only allowed if you have some other particle involved to take care of the photon momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of a massless particle decaying into other massless particles, this only works if your particles have a self-interaction.  As far as I know, you can only have this for Bosons.  QED (photons) has no self interaction, and the weak bosons (W, Z) are massive, but this is well known to occur in QCD in form of $g \rightarrow gg$.  In fact, gluons are more likely to interact at lower energies, than at highest energies where QCD is &lt;em&gt;asymptotically free&lt;/em&gt; - that means the interaction is well-behaved and weak, and we can use perturbation theory.  At low energies, gluons just keep splitting and producing even lower energy quarks or gluons, until there are in a sense &quot;infinitely many&quot; of &quot;infinitely low&quot; energy.  This is what is called &lt;em&gt;infrared divergence&lt;/em&gt;.  It may sound strange, but in practice all observable quantities remain finite, so there is nothing to worry about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your point number 5 is also applies here.  The decay products are not only very low energetic, but tend to be very collinear (&lt;em&gt;collinear divergence&lt;/em&gt;).  This is the source of the famous &quot;jets&quot; that appear in high energy collider experiments.&lt;/p&gt;&#xA;" OwnerUserId="825" LastActivityDate="2011-07-20T23:10:40.593" CommentCount="7" />
  <row Id="12544" PostTypeId="2" ParentId="4094" CreationDate="2011-07-20T23:29:29.410" Score="1" ViewCount="" Body="&lt;p&gt;I don't know if anyone is still watching this thread, but anyway, the 2001 paper referred to by the OP describes an idea called doubly special relativity (DSR). There is a WP article on it, which provides a more current view. Basically my impression is that DSR didn't work out well, and nobody, including Magueijo and Smolin, is really working on it anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an answer to this question in the context of loop quantum gravity, see Rovelli and Speziale, &quot;Reconcile Planck-scale discreteness and the Lorentz-Fitzgerald contraction,&quot; &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0205108&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/gr-qc/0205108&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The OP asked the question in a somewhat naive way, but that doesn't mean that the whole issue is trivial. In SR, we have a constant called c. It's constant by definition. But that doesn't mean that it's a trivial statement that when an observer sees a particle as having a velocity equal to c, that fact can be Lorentz invariant.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-07-20T23:29:29.410" />
  <row Id="12545" PostTypeId="1" AcceptedAnswerId="12546" CreationDate="2011-07-21T01:59:59.793" Score="2" ViewCount="60" Body="&lt;p&gt;The de broglie hypothesis suggests a particle can be associated with a wave of&#xA;momentum  $p = \hbar k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my question is the following: how does one arrive at this concept of the momentum of a wave?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the classical definition of momentum, however I do not see an obvious extension to an object such as a vibrating guitar string.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;De Broglie is saying that if I have a vibrating guitar string and the wavelenth of vibration decreases then the momentum of the guitar string increases?    Is it the case that a guitar string at a higher frequency(time) has higher energy?&#xA;   is the case that a guitar string vibrating at a higher spatial frequency will also have higher energy? etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry if this guitar string example is bad... i'm just trying to understand this properly&lt;/p&gt;&#xA;" OwnerUserId="4570" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-21T03:17:37.647" LastActivityDate="2011-07-21T03:29:15.617" Title="Quantum momentum (De Broglie)" Tags="&lt;quantum-mechanics&gt;&lt;momentum&gt;" AnswerCount="2" />
  <row Id="12546" PostTypeId="2" ParentId="12545" CreationDate="2011-07-21T03:05:32.813" Score="2" ViewCount="" Body="&lt;p&gt;In the de Broglie relation $p=\hbar k$, the left-hand side represents the momentum of a &lt;em&gt;particle&lt;/em&gt;, while the $k$ on the right is the wavenumber of a &lt;em&gt;wave&lt;/em&gt;. The key here is that if the relation is going to be valid, the wave has to have the right amplitude to represent exactly one particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose we have a light wave that represents exactly one photon's worth of light. It has a certain energy and frequency, related by $E=\hbar \omega$, and a certain momentum and wavenumber, related by $p=\hbar k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose we take that light wave and scale up its amplitude (i.e., the strength of its electric and magnetic fields) by a factor of 2. Since the energy of a wave is proportional to the square of its amplitude, this quadruples the energy. It also quadruples the momentum. (You can see this either by using the relativistic relation $E=pc$ for m=0, or by using the Poynting vector.) But we haven't changed $\omega$ or $k$, so this wave violates the de Broglie relations. The key here is that this wave is no longer the right strength to represent one photon. It represents four photons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's play with some numbers in the case of the guitar string. Say $\omega$ is 1000 Hz, and $E$ is 1 J. When you calculate $\hbar\omega$, you get something many orders of magnitude smaller than this $E$. That's because there isn't just one particle in the guitar string, there are many. The same idea applies to momentum.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-07-21T03:05:32.813" />
  <row Id="12547" PostTypeId="2" ParentId="12545" CreationDate="2011-07-21T03:29:15.617" Score="1" ViewCount="" Body="&lt;p&gt;The assumption that Plank made from his &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck%27s_law&quot; rel=&quot;nofollow&quot;&gt;Planck's law&lt;/a&gt; of blackbody radiation was that the electromagnetic radiation is quantized, it takes only integer multiples of the quantity $hf$, where $h$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck%E2%80%93Einstein_equation&quot; rel=&quot;nofollow&quot;&gt;Plank constant&lt;/a&gt; (with units of energy $\times$ time) and $f$ is the frequency of radiation (with units 1/time). This is the energy of the quanta of the electromagnetic radiation - a photon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, firstly the Plank constant $h$ was defined by Plank as a proportionality constant between the energy $E$ of a photon and the frequency of its associated electromagnetic wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E=hf$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be rewritten in terms of the wavelength, instead of the frequency, using the relation $f=c/\lambda$. We get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E=\frac{hc}{\lambda}$ (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, from special relativity energy-momentum relation $E^2=m^2 c^4 + p^2 c^2$ we know that for a photon with zero rest mass, $m=0$, its energy is related to its momentum through the relation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E=pc$ (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining the equations (1) and (2) we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p=\frac{h}{\lambda}=\hbar k$, where $\hbar=\frac{h}{2\pi}$ and $k=\frac{2\pi}{\lambda}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;De Broglie made the assumption that the particles have wave properties too, i.e an frequency and wave length and this has been experimentally proved. They are particles and waves at the same time. The wave properties of particles are given by the de Broglie equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f=\frac{E}{h}$ and $\lambda=\frac{h}{p}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for a given particle's energy $E$ and momentum $p$ we can find its frequency and wave length.&lt;/p&gt;&#xA;" OwnerUserId="3066" LastActivityDate="2011-07-21T03:29:15.617" CommentCount="1" />
  <row Id="12548" PostTypeId="1" CreationDate="2011-07-21T03:56:29.780" Score="-2" ViewCount="124" Body="&lt;p&gt;I know that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p+\bar{p}\to 4\pi^++4\pi^-+(\gamma)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before the collision, the sum of absolute electric charge value is $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left | +1 \right |+\left | -1 \right |=2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the collision, the $p$ collapses to $4\pi^+$, and $\bar{p}$ collapses to $4\pi^-$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, $4\pi^+$ collapse to $4\mu^++(\nu)$, and $4\pi^-$ collapse to $4\mu^-+(\bar{\nu})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi^+\to\mu^++\nu$$&#xA;$$\pi^-\to\mu^-+\bar{\nu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then, $4\mu^+$ collapse to $4e^++(\nu+\bar\nu)$, and $4\mu^-$ collapse to $4e^-+(\nu+\bar\nu)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu^+\to e^++(\nu+\bar\nu)$$&#xA;$$\mu^-\to e^-+(\nu+\bar\nu)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p+\bar{p}\to 4\pi^++4\pi^-+(\gamma)\to 4\mu^++4\mu^-+(\gamma) \to 4e^++4e^-+(\gamma)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this situation, the sum of absolute electric charge value is $8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left | +4 \right |+\left | -4 \right |=8$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question.&lt;/strong&gt; How could it possible that the sum of absolute electric charge value has increased?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added.&lt;/strong&gt; 'So, what's the significance?'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electric force bwtween $p$ and $e^-$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F=k\frac{e^2}{r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electric force bwtween $4e^+$ and $e^-$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{n=1}^{4} k\frac{e^2}{(r_{n})^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p$'s electric charge is +1, but after this collision, $p$ collapses to $4e^+$, it means the total electric charge has increased $+1$ to $+4$.&lt;/p&gt;&#xA;" OwnerUserId="4558" LastEditorUserId="4558" LastEditorDisplayName="" LastEditDate="2011-07-21T19:02:35.087" LastActivityDate="2011-07-21T19:02:35.087" Title="Transition of Electric Charge In Collision Between Proton And Antiproton" Tags="&lt;particle-physics&gt;&lt;particles&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="12549" PostTypeId="2" ParentId="12541" CreationDate="2011-07-21T03:58:41.810" Score="5" ViewCount="" Body="&lt;p&gt;From Wikipedia (which cites the paywalled &lt;a href=&quot;http://dx.doi.org/10.1006/icar.1996.0117&quot; rel=&quot;nofollow&quot;&gt;http://dx.doi.org/10.1006/icar.1996.0117&lt;/a&gt;), we get &lt;a href=&quot;http://en.wikipedia.org/wiki/Tidal_locking#Timescale&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Tidal_locking#Timescale&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a is the semi-major axis (or orbital radius) of the object (I'm not totally sure if the logic changes for elliptical orbits). Anyways, we can easily re-arrange the equation to express a (or the radius) in terms of $t_{lock}$ (or the time) and all the other variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then (using the Wikipedia variable names) we get $$a = (\frac{3t_{lock}Gm_p^2 k_2 r^5}{wIQ})^{1/6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So then if we know t (which is basically the age of the stellar system in question - which we can get a rough idea of since we do have a rough idea of how each star is), then we can get a value of a for that particular t.&lt;/p&gt;&#xA;" OwnerUserId="1576" LastActivityDate="2011-07-21T03:58:41.810" CommentCount="4" />
  <row Id="12550" PostTypeId="2" ParentId="12512" CreationDate="2011-07-21T04:09:01.557" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My first assumption is, the centre of mass of this system, throughout&#xA;  time won't move. Right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, by conservation of linear momentum.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Or it's possible to see a system where few particles are rotating in&#xA;  one direction and other in other direction, so that angular momentum&#xA;  is conserved and net angular momentum is still zero?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even with only three bodies, solutions can be found where they escape to infinity (with respect to the center of mass). This requires a binary system to be formed by energy conservation. See figure 10.b at page 10 of &lt;a href=&quot;http://arxiv.org/abs/1103.3662&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; for an example.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;EDIT: On similar note, can we have say 3 particles which are attracted&#xA;  to each other but never collide, something like a simple harmonic&#xA;  motion?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is also possible. See &quot;Periodic Solution Standish&quot; at page 6 of the &lt;a href=&quot;http://arxiv.org/abs/1103.3662&quot; rel=&quot;nofollow&quot;&gt;same paper&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1656" LastActivityDate="2011-07-21T04:09:01.557" CommentCount="1" />
  <row Id="12551" PostTypeId="2" ParentId="12548" CreationDate="2011-07-21T04:09:13.010" Score="5" ViewCount="" Body="&lt;p&gt;It's certainly possible to have an absolute electric charge value that increases, as long as total charge still sums up to 0. After all - this is why virtual particles (which consist of a pair of positive/negative charges) can be created&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Virtual_particle&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Virtual_particle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any physical significance to the absolute electric charge value? Also - I'm just curious - where did you get the reaction equation from? &lt;/p&gt;&#xA;" OwnerUserId="1576" LastActivityDate="2011-07-21T04:09:13.010" CommentCount="7" />
  <row Id="12552" PostTypeId="1" AcceptedAnswerId="12567" CreationDate="2011-07-21T04:26:17.150" Score="4" ViewCount="65" Body="&lt;p&gt;So, I have a couple questions about transfer orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, an object's, say a satellite, orbit is basically tied to it velocity relative to the object, the planet, that it is orbiting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic premise is that you have the satellite in a circular orbit of X radius, going speed v. Now, if the satellite decelerates, it reduces its velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the new velocity is not enough to maintain its original orbit, the satellite will &quot;fall&quot; towards the planet. Now, even though the satellite was (say) originally at the proper velocity for the new lower orbit due to the deceleration, by the time it actually arrives at the new orbit, its velocity likely changed, due to the different acceleration vector of the planet, the vector causing the actual fall. Even if it was at the proper velocity, it's actual vector would like be wrong, it would no longer be tangential to the planet, it would likely be steeper. This would effectively cause the satellite to fall right through the lower orbit and keep accelerating toward the planet. Minimally it would not be in the stable orbit desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, this is why there is a second acceleration/deceleration necessary to correct the satellites velocity and perhaps it vector once it reaches the new orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the opposite all applies if trying to get to a higher orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is during the initial deceleration, is that simply applied opposite the current orbital velocity vector? And when the satellite arrives at the new orbit, what direction is the correctional acceleration/deceleration made? Is it the difference of the desired tangential direction that it wants to go and the current direction it is going? or simply opposite its current vector?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the point of the Hohmann transfer orbits is to use as little velocity change (and thus fuel) as possible, but can an orbital transfer be faster or slower if you're willing to burn more fuel to affect the velocity change? Will a slower satellite &quot;fall faster&quot; than a faster one? Intuitively that makes sense to me, but at the same time there's the concept that no matter how fast the bullet, it hits the ground at the same time, it just travels farther. So I'm not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you apply more thrust and you're trying to lift to a higher orbit, that seems directly related to the amount of thrust you're willing to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this how the transfer orbits work? There's the math and formulas to calculate the delta v necessary on both ends of the transfer, but I'm just trying to visualize the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if you had unlimited fuel, and time was more a consideration, would you even bother with this process versus something more &quot;point and shoot, turn and burn&quot;?&lt;/p&gt;&#xA;" OwnerUserId="2285" LastActivityDate="2011-07-21T13:40:44.573" Title="Hohmann transfer orbit questions" Tags="&lt;orbital-motion&gt;" AnswerCount="1" />
  <row Id="12553" PostTypeId="1" CreationDate="2011-07-21T05:57:36.077" Score="2" ViewCount="87" Body="&lt;p&gt;Does the axis of Time point into a black hole or away from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you give a reference paper?&lt;/p&gt;&#xA;" OwnerUserId="3588" LastActivityDate="2011-08-14T22:03:57.347" Title="Direction of Time on Event Horizon" Tags="&lt;event-horizon&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="12555" PostTypeId="1" CreationDate="2011-07-21T06:24:21.197" Score="2" ViewCount="40" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The wheel of a car has a radius of 20 cm. It initially rotates at 120&#xA;  rpm. In the next minute it makes 90 revolutions. What is the angular&#xA;  acceleration?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the answer is solved by using one of the angular kinematic equations. More specifically delta theta. The problem I am having trouble understanding is the answer which probably stems from my poor fundamentals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is: (90x2pi) = 4pi(60) + 1/2 (alpha) (60)^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So where does 90x2pi come from?&lt;/p&gt;&#xA;" OwnerUserId="4571" LastActivityDate="2011-07-21T08:01:57.047" Title="So what is happening in this angular kinematics equation?" Tags="&lt;kinematics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12556" PostTypeId="2" ParentId="12555" CreationDate="2011-07-21T08:01:57.047" Score="4" ViewCount="" Body="&lt;p&gt;The equations you're using are equations for $\theta$, the angle through which the wheel rotates.  In physics, we measure angles in radians.  There are $2\pi$ radians  in a circle, so $90$ revolutions is $90*2pi$ radians.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-07-21T08:01:57.047" />
  <row Id="12557" PostTypeId="1" AcceptedAnswerId="12568" CreationDate="2011-07-21T07:10:47.760" Score="5" ViewCount="103" Body="&lt;p&gt;&lt;em&gt;I have encountered this claim while searching for sources answering &quot; &lt;a href=&quot;http://skeptics.stackexchange.com/questions/5310/can-we-see-the-curvature-of-earth-from-the-top-of-worlds-tallest-building&quot;&gt;Can we see the curvature of earth from the top of world&amp;#39;s tallest building?&lt;/a&gt; &quot;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Horizon#Curvature_of_the_horizon&quot; rel=&quot;nofollow&quot;&gt;article on horizon&lt;/a&gt; claims (with no references), that compared to a geometrical model, the the apparent curvature of Earth is less than expected due to refraction of light in the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking about the problem, I fail to see why this could be so. It is easy to see the refraction affects a distance of the apparent horizont, but I fail to see how it could affect its curvature, as the refraction seems to have the same effect in all directions, because your distance to horizon is the same in all directions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that claim just bogus and there is no such effect, or is the effect real?&lt;/p&gt;&#xA;" OwnerUserId="4573" LastEditorUserId="4573" LastEditorDisplayName="" LastEditDate="2011-07-21T10:46:06.270" LastActivityDate="2011-07-21T15:36:07.267" Title="Is apparent horizon curvature lesser due to refraction of light in the atmosphere?" Tags="&lt;optics&gt;&lt;atmosphere&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="12558" PostTypeId="2" ParentId="11419" CreationDate="2011-07-21T09:41:29.560" Score="4" ViewCount="" Body="&lt;p&gt;BebopButUnsteady answer is obviously correct. Actually most wave functions cannot be written as single Slater determinants and certainly no energy eigenstates of N-fermion systems (atom, molecules or solids) interacting by two-body potentials (like the Coulomb potential). The Slater determinant is just a nice simple approximation to ground state (close-shells) wave functions of these systems, but it can be easily shown that the actual wave function is not really a single Slater determinant. The Slater determinants minimize the energy of a quantum system where the fermions interact via an effective average Coulomb potential. They really cannot minimize the energy of the true Coulomb potential. This problem is usually called the lack of dynamic correlation in Quantum Chemistry.&lt;/p&gt;&#xA;" OwnerUserId="3572" LastActivityDate="2011-07-21T09:41:29.560" />
  <row Id="12559" PostTypeId="1" AcceptedAnswerId="12561" CreationDate="2011-07-21T09:59:17.023" Score="14" ViewCount="169" Body="&lt;p&gt;Noether's Theorem is used to related the invariance under certain continuous transformations to conserved currents. A common example is that translations in spacetime correspond to the conservation of four-momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of angular momentum, the tensor (in special relativity) has 3 independent components for the classical angular momentum, but 3 more independent components that, as far as I know, represent Lorentz boosts. So, what conservation law corresponds to invariance under Lorentz boosts?&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2011-07-21T13:26:17.673" Title="What conservation law corresponds to Lorentz boosts?" Tags="&lt;special-relativity&gt;&lt;conservation-laws&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="3" />
  <row Id="12560" PostTypeId="1" AcceptedAnswerId="12565" CreationDate="2011-07-21T10:57:45.303" Score="2" ViewCount="74" Body="&lt;p&gt;I'm looking for something short of a molecular dynamics package, where I can build up simple geometric shapes with flexible linkages/etc and simulate the consequences of electrostatic repulsion between surfaces.  Something, say, that would let me simulate billiards in a magnetic field.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have good recommendations for a 3D Newtonian dynamics software package (free or not) with these sort of capabilities?  &lt;/p&gt;&#xA;" OwnerUserId="4257" LastActivityDate="2011-07-21T13:26:29.750" Title="Software for simulating 3D Newtonian dynamics of simple geometric objects (with force fields)" Tags="&lt;electromagnetism&gt;&lt;software&gt;&lt;models&gt;" AnswerCount="2" />
  <row Id="12561" PostTypeId="2" ParentId="12559" CreationDate="2011-07-21T11:57:52.990" Score="13" ViewCount="" Body="&lt;p&gt;&lt;em&gt;Warning:&lt;/em&gt; this is a long and boring derivation. If you are interested only in the result skip to the very last sentence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noether's theorem can be formulated in many ways. For the purposes of your question we can comfortably use the special relativistic Lagrangian formulation of a scalar field. So, suppose we are given an action&#xA;$$S[\phi] = \int {\mathcal L}(\phi(x), \partial \phi(x), \dots) {\rm d}^4x.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose the action is invariant under some infinitesimal transformation $m: x^{\mu} \mapsto x^{\mu} + \delta x^{\mu} = x^{\mu} + \epsilon a^{\mu}$ (we won't consider any explicit transformation of the fields themselves). Then we get a conserved current&#xA;$$J^{\mu} = {\partial {\mathcal L} \over \partial \phi_{,\mu}} \phi^{,\nu} a_{\nu} - {\mathcal L} a^{\mu} = \left ({\partial {\mathcal L} \over \partial \phi_{,\mu}} \phi^{,\nu} - {\mathcal L} g^{\mu \nu} \right) a_{\nu} .$$&#xA;We obtain a conserved charge from it by letting $Q \equiv \int J^0 {\rm d}^4x$ since from $\partial_{\mu}J^{\mu} =0$&#xA;we have that&#xA;$$ {\partial Q \over \partial t} = \int {\rm Div}{\mathbf J}\, {\rm d}^3 x = 0$$ &#xA;which holds any time the currents decay sufficiently quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the transformation is given by translation $m_{\nu} \leftrightarrow \delta x^{\mu} = \epsilon \delta^{\mu}_{\nu}$ we get four conserved currents&#xA;$$J^{\mu \nu} = {\partial {\mathcal L} \over \partial \phi_{,\mu}} \phi^{,\nu} - {\mathcal L} g^{\mu \nu} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This object is more commonly known as stress energy tensor $T^{\mu \nu}$ and the associated conserved currents are known as momenta $p^{\nu}$.&#xA;Also, in general the conserved current is simply given by $J^{\mu} = T^{\mu \nu} a_{\nu}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a Lorentz transformation we have&#xA;$$m_{\sigma \tau} \leftrightarrow \delta x^{\mu} =&#xD;&#xA;\epsilon \left(g^{\mu \sigma} x^{\tau} - g^{\mu \tau} x^{\sigma} \right)$$ (notice that this is antisymmetric and so there are just 6 independent parameters of the transformation) and so the conserved currents are the angular momentum currents&#xA;$$M^{\sigma \tau \mu} = x^{\tau}T^{\mu \sigma} - x^{\sigma}T^{\mu \tau}.$$&#xA;Finally, we obtain the conserved  angular momentum as&#xA;$$M^{\sigma \tau} = \int \left(x^{\tau}T^{0 \sigma} - x^{\sigma}T^{0 \tau} \right) {\rm d}^4 x . $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for particles we can proceed a little further since their associated momenta and angular momenta are not given by an integral. Therefore we have simply that $p^{\mu} = T^{\mu 0}$ and $M^{\mu \nu} = x^{\mu} p^{\nu} - x^{\nu} p^{\mu}$. The rotation part of this (written in the form of the usual pseudovector) is&#xA;$${\mathbf L}_i = {1 \over 2}\epsilon_{ijk} M^{jk} = ({\mathbf x} \times {\mathbf p})_i$$&#xA;while for the boost part we get&#xA;$$M^{0 i} = \left(t {\mathbf p} - {\mathbf x} E \right)^i $$&#xA;which is nothing else than the center of mass at $t=0$ (we are free to choose $t$ since the quantity is conserved) multiplied by $\gamma$ since we have the relations $E = \gamma m$, ${\mathbf p} = \gamma m {\mathbf v}$. Note the similarity to the ${\mathbf E}$, $\mathbf B$ decomposition of the electromagnetic field tensor $F^{\mu \nu}$.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-07-21T13:26:17.673" LastActivityDate="2011-07-21T13:26:17.673" CommentCount="3" />
  <row Id="12563" PostTypeId="1" AcceptedAnswerId="12564" CreationDate="2011-07-21T12:44:24.813" Score="1" ViewCount="94" Body="&lt;p&gt;I was wondering if it was possible to project a magnified image on a wall without the need of focusing, so just by dimensioning the lenses right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I have to use the principal of Maxwellian View for the illumination of the slide. However, there are a lot of parameters left and I can design a system that requires a very minimal (say 1mm) displacement of the projection lens when projecting on 0.5m and, say, 2.5m. However, I was wondering if I oversee something here. I've seen that it is possible however I was also wondering if some special kind of lens is needed here.&lt;/p&gt;&#xA;" OwnerUserId="3725" LastActivityDate="2011-07-21T18:17:21.400" Title="Projecting image without manual focussing" Tags="&lt;optics&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="12564" PostTypeId="2" ParentId="12563" CreationDate="2011-07-21T13:00:19.323" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I was wondering if it was possible to project a magnified image on a&#xA;  wall without the need of focusing, so just by dimensioning the lenses&#xA;  right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course You can!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real problem is that You do not know the exact position of the slide &#xA;in its holder. &#xA;Second problem is, that it is cheaper to use a standard commercial &#xA;projector including manual focus, as opposed to a custom made &#xA;projector without manual focus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know abot &quot;principal of maxwellian wiev&quot; , but I know why You &#xA;shoud not rely on the method of fixed focus often used in cheap cameras:&#xA;This would be wasting a lot of light shone on the slide. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-07-21T18:17:21.400" LastActivityDate="2011-07-21T18:17:21.400" />
  <row Id="12565" PostTypeId="2" ParentId="12560" CreationDate="2011-07-21T13:08:10.047" Score="0" ViewCount="" Body="&lt;p&gt;I think that the 'agent based' approach is good for your purpose. I've already used MASON to play a little with 'a sort of' gravity.&lt;br&gt;&#xA;(you will have to know java - easier if you work on top of the examples- and the documentation is very good)&lt;br&gt;&#xA;&lt;a href=&quot;http://www.cs.gmu.edu/~eclab/projects/mason/#Screenshots&quot; rel=&quot;nofollow&quot;&gt;MASON&lt;/a&gt;&lt;br&gt;&#xA;  is a fast discrete-event multiagent simulation library core in Java, designed to be the foundation for large custom-purpose Java simulations, and also to provide more than enough functionality for many lightweight simulation needs. MASON contains both a model library and an optional suite of visualization tools in 2D and 3D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;see the examples:&lt;br&gt;&#xA;Bouncing Particles - is a tutorial demo of simple particles bouncing and interacting with one another.&lt;br&gt;&#xA;Balls and Bands (Tutorial 5)  (mass springs between objects)&#xA;simulates Hooke's Law with balls connected by rubber bands of different strengths.&lt;br&gt;&#xA;HeatBugs -- shown in wireframe 3D.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can also explore the &lt;a href=&quot;http://www.spiderland.org/&quot; rel=&quot;nofollow&quot;&gt;BREVE&lt;/a&gt; package.&lt;br&gt;&#xA;Breve A 3d Simulation Environment for Multi-Agent Simulations and Artificial Life (IA)&lt;br&gt;&#xA;objects, joints, gravity.  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-07-21T13:08:10.047" />
  <row Id="12566" PostTypeId="2" ParentId="12560" CreationDate="2011-07-21T13:26:29.750" Score="1" ViewCount="" Body="&lt;p&gt;I've not tried it myself, but you may want to check out the &lt;a href=&quot;http://www.csc.fi/english/pages/elmer&quot; rel=&quot;nofollow&quot;&gt;Elmer multiphysics package&lt;/a&gt;, which looks like it can do time dependendent 3D finite element mechanics + electrostatics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a nice pic they have of a beam being deflected by electrostatic attraction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/53Yzm.gif&quot; alt=&quot;beam being deflected by electrostatic potential&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's F.O.S.S. They have windows binaries and it's in the ubuntu repo.&lt;/p&gt;&#xA;" OwnerUserId="1876" LastActivityDate="2011-07-21T13:26:29.750" CommentCount="1" />
  <row Id="12567" PostTypeId="2" ParentId="12552" CreationDate="2011-07-21T13:40:44.573" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My question is during the initial deceleration, is that simply applied&#xA;  opposite the current orbital velocity vector? And when the satellite&#xA;  arrives at the new orbit, what direction is the correctional&#xA;  acceleration/deceleration made? Is it the difference of the desired&#xA;  tangential direction that it wants to go and the current direction it&#xA;  is going? or simply opposite its current vector?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The two directions are the same. The impulses are done at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Apsis&quot; rel=&quot;nofollow&quot;&gt;apoapsis and the periapsis&lt;/a&gt; of the transfer orbit, where the velocity is also tangential to the circular orbits (the larger one at apoapsis and the smaller at periapsis).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I understand that the point of the Hohmann transfer orbits is to use&#xA;  as little velocity change (and thus fuel) as possible, but can an&#xA;  orbital transfer be faster or slower if you're willing to burn more&#xA;  fuel to affect the velocity change?&#xA;  [snip]&#xA;  Finally, if you had unlimited fuel, and time was more a consideration,&#xA;  would you even bother with this process versus something more &quot;point&#xA;  and shoot, turn and burn&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you have enough &lt;a href=&quot;http://en.wikipedia.org/wiki/Delta-v&quot; rel=&quot;nofollow&quot;&gt;delta-v&lt;/a&gt; and thrust, you can just &quot;point and burn&quot;. You will be doing a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperbolic_trajectory&quot; rel=&quot;nofollow&quot;&gt;hyperbolic transfer orbit&lt;/a&gt; that, in the limit of very large velocities, will tend to a straight line ($e \to \infty$).&lt;/p&gt;&#xA;" OwnerUserId="1656" LastActivityDate="2011-07-21T13:40:44.573" />
  <row Id="12568" PostTypeId="2" ParentId="12557" CreationDate="2011-07-21T14:12:54.320" Score="3" ViewCount="" Body="&lt;p&gt;Yes -- because refraction influences the apparent distance to the horizon, it also has an effect on the curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To visualize this, it might help to think in extreme cases, for example in the case where due to refraction the horizon is at an apparent distance of only 1 meter. In this case, the curvature of the horizon would be extreme (it would be a circle of radius 1 meter around you). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality the curvature effect is much smaller of course, and I doubt that it is visible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As an aside: one result of atmospheric refraction that &lt;em&gt;is&lt;/em&gt; observable is a phenomena called &lt;a href=&quot;http://en.wikipedia.org/wiki/Green_flash&quot; rel=&quot;nofollow&quot;&gt;'the green flash'&lt;/a&gt;. Because refraction is colour-dependent, the red and yellow part of the sun could already have set, while a small part of the 'green sun' is still above the horizon. This can be observed by the naked eye, preferably when the horizon is sharp.)&lt;/p&gt;&#xA;" OwnerUserId="4574" LastEditorUserId="4574" LastEditorDisplayName="" LastEditDate="2011-07-21T14:26:19.243" LastActivityDate="2011-07-21T14:26:19.243" CommentCount="3" />
  <row Id="12569" PostTypeId="2" ParentId="12563" CreationDate="2011-07-21T14:17:44.120" Score="1" ViewCount="" Body="&lt;p&gt;The only possibility to do so is using point light source, illuminate slide directly and you'll get projected image in 'focus' at any distance without any focusing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problems are obvious - point-sources cannot be very strong, especially as you might want 0.5mm size and lower (achievable on LEDs and especially lasers with aspheric optics)&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-07-21T14:17:44.120" />
  <row Id="12570" PostTypeId="1" CreationDate="2011-07-21T14:35:33.507" Score="7" ViewCount="206" Body="&lt;p&gt;This question is closely related to my previous question &lt;a href=&quot;http://physics.stackexchange.com/questions/12504/bound-states-in-qed&quot;&gt;Bound states in QED&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Muonium is a system of electron and anti-muon. &lt;a href=&quot;http://en.wikipedia.org/wiki/Muonium&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; in wikipedia claims that muonium is unstable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;QUESTION: Why it is unstable? Is it due to presence of weak interactions, or it can be explained only in the framework of QED (involving both electrons and muons and their anti-particles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compare with, the positronium, consisting from electron and positron, in an unstable particle (see e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Positronium&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; in wikipedia). It can annihilate to two photons. This effect can be explained by the usual QED. This explanation does not seem to work for the muonium.&lt;/p&gt;&#xA;" OwnerUserId="4562" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-21T20:52:36.897" LastActivityDate="2011-07-21T20:52:36.897" Title="Why muonium is unstable?" Tags="&lt;quantum-field-theory&gt;&lt;particle-physics&gt;&lt;quantum-electrodynamics&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" />
  <row Id="12571" PostTypeId="2" ParentId="12504" CreationDate="2011-07-21T14:56:25.450" Score="2" ViewCount="" Body="&lt;p&gt;You have to be careful about use of the term &lt;em&gt;bound state&lt;/em&gt; in QFT. There is a related notion of &lt;em&gt;resonance&lt;/em&gt;, which is a state that looks approximately like a bound state for a while (possibly a long while!) but eventually decays. For example, if you consider the nonrelativistic hydrogen atom, it has bound states corresponding to the electron orbitals (the n,l,m quantum numbers from any first course in QM). However, if you quantize the electric field, the electron is now free to emit a photon and drop down an energy level. As a result, &lt;em&gt;all bound states except the ground state become resonances once you quantize the EM field&lt;/em&gt;. This is a general feature when you go from N-particle QM to QFT: because QFT allows particles to spontaneously emit photons (or gluons, Z/W, etc.) with some small but non-zero probability, many states that were bound states in QM become resonances in QFT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the articles you are reading that mention positronium or muonium or whatever would be more correct in calling them &lt;em&gt;resonances&lt;/em&gt;. They are states that behave approximately like bound states on a short timescale, but because there is a nonzero probability of emitting photons (or other particles) they eventually do, and hence become &quot;unbound&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book &lt;em&gt;Mathematical Concepts of Quantum Mechanics&lt;/em&gt; by Gustafson and Sigal has a very good treatment of resonances and how they come about in QFT. Indeed, most of the second half of the book is devoted to giving a mostly self-contained proof of the aforementioned theorem about bound states of the hydrogen atom. The book is fairly readable and only assumes prior knowledge of basic classical mechanics and EM, as well as some basic properties of the Fourier transform (though these are reviewed).&lt;/p&gt;&#xA;" OwnerUserId="4160" LastEditorUserId="4160" LastEditorDisplayName="" LastEditDate="2011-07-21T15:06:49.870" LastActivityDate="2011-07-21T15:06:49.870" CommentCount="1" />
  <row Id="12572" PostTypeId="2" ParentId="12570" CreationDate="2011-07-21T15:22:12.353" Score="1" ViewCount="" Body="&lt;p&gt;As Marek said, it is as unstable as any other radioactive atom. It is a weak decay of muon or antimuon that destroys the muonium.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-07-21T15:22:12.353" />
  <row Id="12573" PostTypeId="2" ParentId="12563" CreationDate="2011-07-21T15:26:05.677" Score="2" ViewCount="" Body="&lt;p&gt;You can do it by correctly choosing the &lt;a href=&quot;http://en.wikipedia.org/wiki/Focal_length&quot; rel=&quot;nofollow&quot;&gt;focal length&lt;/a&gt; of the lens(es) you're using. Is that what you mean? The drawback is that you can only use it in one specific setup. (I'm not sure if I understand correctly as this seems a bit trivial).&lt;/p&gt;&#xA;" OwnerUserId="4574" LastActivityDate="2011-07-21T15:26:05.677" />
  <row Id="12574" PostTypeId="2" ParentId="12557" CreationDate="2011-07-21T15:36:07.267" Score="0" ViewCount="" Body="&lt;p&gt;Sometimes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When you are standing on the ground&lt;/strong&gt;, you are in air at roughly one atmosphere of pressure; so are the points on the earth that you see on the horizon; and so is the whole path of the light from the horizon to you. In this case there is no significant boundary where the refractive index (RI) changes, nor is there an appreciable deviation in RI over the path of the light you are seeing. Therefore there will be no bending of the light rays, and no effect on the perceived curvature of the horizon.&lt;sup&gt;1&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When you are at significant altitude&lt;/strong&gt;, you are in air at significantly lower pressure than the air at the surface of the earth. Therefore the light from the horizon moves through a region with a pronounced gradient in RI. This causes the light rays to curve, making the radius of the horizon appear larger (and thus the curvature smaller) than it otherwise would. It is this scenario that the linked Wikipedia page is talking about when it states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;At an altitude of 10 km ... the apparent curvature is less than that due to refraction of light in the atmosphere&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;: There will of course be small deviations in air pressure (and therefore the RI) along the path of the ray, but these would be random and time-varying, so they will manifest as a slight blurring or &quot;wobble&quot; of the image of a point seen on the horizon, not a shift in it's observed location.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2011-07-21T15:36:07.267" />
  <row Id="12575" PostTypeId="2" ParentId="12570" CreationDate="2011-07-21T15:43:05.077" Score="6" ViewCount="" Body="&lt;p&gt;Let's start with nuclear beta decay as a similar example. 12C, with 6 protons and 6 neutrons, is stable. As Marek has pointed out, it's not enough just to say that neutrons are unstable. If all neutrons were unstable with respect to beta decay, then 12C could decay into 12N. The reason this can't happen is that the mass of a 12N nucleus is higher than the mass of a 12C. That means that if the decay were to proceed, the total mass-energy of the products (12N, electron, and antineutrino) would be greater than the initial mass-energy of the 12C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the muonium example, the muon wants to decay into a positron plus some neutrinos. The difference between the mass of a muon and the mass of a positron is 105.147 MeV. (The masses of the neutrinos are negligible.) This decay would be forbidden if muonium's binding energy was -105.147 MeV or stronger. But it's not. The binding energy of muonium is basically the same as the binding energy  of hydrogen, about 14 eV.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-07-21T15:43:05.077" />
  <row Id="12576" PostTypeId="2" ParentId="12570" CreationDate="2011-07-21T15:45:14.013" Score="6" ViewCount="" Body="&lt;p&gt;Here is a rough estimate why the muonium atom is unstable against the muon beta decay (in contrast to a neutron in a deutron). The binding energy of the electron in a muonium atom neglecting the reduced mass effect is approximately equal to its binding energy in a Hydrogen atom according to the Bohr's model, i.e., 13.6 ev. The muon performs a beta decay to a positron and a neutrino. The rest mass difference between the muon and the positron is about 105.6 - 0.5 = 105.1 Mev. Thus the binding energy is so small compared to the mass difference, thus the muon decay mode is much favorable. In contrast the mass of the deutron is 1875.6 MeV while the rest mass of the outputs of a neutron beta decay + a proton is 2(938.27 MeV) + 0.511 MeV = 1877.05. Thus the deutron favors the stable state.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastActivityDate="2011-07-21T15:45:14.013" />
  <row Id="12577" PostTypeId="2" ParentId="12563" CreationDate="2011-07-21T16:00:10.940" Score="1" ViewCount="" Body="&lt;p&gt;You can build a projector that doesn't need any lens.&#xA;The idea is to deflect a laser (or 3 lasers) on a mirror with maybe 1mm diameter:&#xA;&lt;a href=&quot;http://www.youtube.com/watch?v=Z448L94DVt4&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=Z448L94DVt4&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can buy those MEMS devices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ipms.fraunhofer.de/content/dam/ipms/common/products/SAS/Systeme/laserprojmeasure-e.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ipms.fraunhofer.de/content/dam/ipms/common/products/SAS/Systeme/laserprojmeasure-e.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best ones I've seen had the a two-axis mirror oscillate with its eigen frequencies. The beam traces a complex Lissajous pattern and by switching three lasers (red, green, blue) very fast, you can draw a picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can easily run into problems with speckles, though.&lt;/p&gt;&#xA;" OwnerUserId="4306" LastActivityDate="2011-07-21T16:00:10.940" />
  <row Id="12579" PostTypeId="1" CreationDate="2011-07-21T17:26:55.337" Score="0" ViewCount="87" Body="&lt;p&gt;I have 2 pin outlet for A.C current . Now when I connect a TV plug in that and when I touch one of the prongs with my tester, the tester is showing response. But on the other prong the tester shows no response. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using an A.C current(50 amps)(230 volts).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the two prongs will act as a phase for every 1/50th second alternatively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e For the first 1/50 second one hole in the outlet must act as a phase and in the next 1/50 second other outlet hole must act as a phase. And this cycle must continue...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So , If i touch any prongs my tester should show response. But only one is showing response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my concept is wrong please ignore my mistake and explain where Ive gone wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone please explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="4575" LastActivityDate="2011-08-20T22:26:51.673" Title="Why only one prong in the plugs are not showing response to tester pen?" Tags="&lt;electricity&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="12580" PostTypeId="1" AcceptedAnswerId="12587" CreationDate="2011-07-21T18:48:56.300" Score="6" ViewCount="181" Body="&lt;p&gt;I took a bucket, drilled 2 different sized holes on the side near the bottom and filled it with water. The stream of water the proceeded from the larger hole traveled further than the stream from the smaller one. How does the size of the hole affect the distance that the water travels?&lt;/p&gt;&#xA;" OwnerUserId="2843" LastActivityDate="2011-07-22T18:27:05.417" Title="How does a hole's size affect the distance that water will squirt" Tags="&lt;fluid-dynamics&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" />
  <row Id="12582" PostTypeId="2" ParentId="11799" CreationDate="2011-07-21T19:13:41.963" Score="4" ViewCount="" Body="&lt;p&gt;Have a look to this old paper:&#xA;&lt;a href=&quot;http://www.g3ynh.info/zdocs/refs/NBS/Rosa1908.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.g3ynh.info/zdocs/refs/NBS/Rosa1908.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4580" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-07-22T17:58:05.663" LastActivityDate="2011-07-22T17:58:05.663" />
  <row Id="12583" PostTypeId="2" ParentId="12580" CreationDate="2011-07-21T19:21:32.683" Score="3" ViewCount="" Body="&lt;p&gt;If the only forms of energy were the potential energy of the water in the bucket and the kinetic energy of the escaping stream, then by conservation of energy the velocity of the stream would have to equal $\sqrt{2gh}$, where $h$ is the difference in height between the hole and the surface. Since you observe that $v$ (or at least the component of $v$ parallel to the axis of the flow) depends on other factors, it's possible that in your experiment, there were other forms of energy involved. Two possible candidates are: (a) frictional heat, and (b) the kinetic energy of the water flowing toward the hole. We expect effect b to be minimal or zero, both because the velocity of this flow is small and because this KE represents a reservoir of energy that is either constant or only very slowly changing as the water level drops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another issue is that kinetic energy is independent of the direction of motion, and the flow of the stream as it exits may not be entirely parallel, so the velocity of flow in the direction parallel to the axis of flow will in general be less than $\sqrt{2gh}$. The stream contracts significantly on its way out, so quite a bit of its KE is in the form of motion that's transverse to the axis. This leads to something called an efflux coefficient. There is a discussion of both the frictional effect and the non-parallel effect in the Feynman Lectures, section 40-3. The non-parallel effect can be quite large. Feynman gives some examples for different types of holes and discharge tubes, and the efflux coefficients are typically factor-of-two effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to approach this is through conservation of mass. When you put your thumb over the end of a garden hose, the cross-sectional area $A_1$ of the aperture under your thumb is less than the cross-sectional area $A_2$ of the hose. Since water is being neither created nor destroyed, we must have $v_1/v_2=A_2/A_1$. This is the opposite of what you observed, and the two situations are not really analogous. They would be analogous if there was a piston pressing down on the surface of the water at the top of the bucket so that the level was constrained to go down at a fixed rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edited the above answer to improve it based on comments by Omega Centauri.]&lt;/p&gt;&#xA;" OwnerUserId="4552" LastEditorUserId="4552" LastEditorDisplayName="" LastEditDate="2011-07-21T20:02:18.560" LastActivityDate="2011-07-21T20:02:18.560" CommentCount="8" />
  <row Id="12584" PostTypeId="2" ParentId="12580" CreationDate="2011-07-21T19:38:04.057" Score="0" ViewCount="" Body="&lt;p&gt;I like the notation from Ben in his answer.  So we will compare two situations with equal water column height but different areas of the hole, $A_1$ and $A_2$.  We find that the distance the water squirts is different due to friction.  I propose that the friction is from three different sources.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Fluid internal friction as it approaches the hole&lt;/li&gt;&#xA;&lt;li&gt;Boundary layer friction around the perimeter of the hole&lt;/li&gt;&#xA;&lt;li&gt;Air friction after it leaves the hole&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I would actually imagine that the 3rd one would be the most significant.  Some meaningful relations for this could be deduced exclusively from the hole size.  If we assume the first two sources of friction are small, then the initial velocity leaving the hole is the same for both cases, thus we have a certain flow rate and velocity for each.  If we assume the holes to be circular and extend this assumption to the stream itself, we can find the perimeter interacting with air and the linear density of the stream of water (close to the hole).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the problem is similar, but not exactly the same, as that of a projectile with air resistance.  The metric that matters is the ratio of the drag coefficient to the mass, but the linear mass density in the case here.  I'll try to come back and put some equations up here later so we can maybe even get a general figure of merit for where deviation from a parabolic path of the water jet will be observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Local forms loss of the hole&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realized that the parts of 1&amp;amp;2 is really a pretty standard engineering problem.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;K = Local forms loss coefficient&lt;/li&gt;&#xA;&lt;li&gt;H = fluid head, used in place of pressure&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H = \frac{v^2}{2 g} + K \frac{v^2}{2 g}$$&#xA;$$ v = \sqrt{ \frac{2 g h}{1+K} }$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we won't be able to avoid the inevitability of finding Reynolds number, so here we are.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\rho = 1000 kg / m^3$&lt;/li&gt;&#xA;&lt;li&gt;$\mu = 0.001 Pa s $&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$Re = \frac{G D}{\mu} = 2 \frac{v \rho}{\mu} \sqrt{ \frac{A }{\pi} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we're talking about in the neighborhood of a hole $1 cm^2$ and a velocity $1 m/s$ then we find Reynolds number to be around 1,000.  Now, I found no choice to use an actual academic paper to get the loss coefficient.  See &lt;a href=&quot;http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&amp;amp;id=JFEGA4000124000003000685000001&amp;amp;idtype=cvips&amp;amp;gifs=yes&quot; rel=&quot;nofollow&quot;&gt;Accurate Evaluation of the Loss Coefficient and the Entrance Length of the Inlet Region of a Channel&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K = 4.183 \times 10^{-5} Re + 0.152$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So K will be in the neighborhood of 0.2.  This is the major difference for the two cases regarding 1&amp;amp;2 types of losses.  I tried some numbers, like 1 m for the head, and found without the forms loss you get about 4.4 m/s exit velocity, with 1 cm^2 hole, you get 2.8 m/s.  Increasing the size of the hole 10x I get 2.1 m/s and decreasing it 10x I get 3.4 m/s.  This isn't what I expected.  I thought that a small hole would make the fluid velocity smaller, but maybe I've messed something up with the signs although I can't find it right now.  I know how I would approach #3 as well, but I don't have the time.  I certainly don't think anyone has sufficient evidence to say one type of loss is greater until some more work is done.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-07-22T18:27:05.417" LastActivityDate="2011-07-22T18:27:05.417" CommentCount="6" />
  <row Id="12585" PostTypeId="2" ParentId="12579" CreationDate="2011-07-21T19:42:26.083" Score="0" ViewCount="" Body="&lt;p&gt;The answer is that one prong has near zero (ground) volatge, and the other varies. You get three wires coming off the distribution antenna, a floating ground, and two hots, of different sign. The voltage between the two hots is 240, and between a hot and the float ground 120. Most plug pairs utilize opposing hots, so combining both plugs you have (non code compliant) 240V available.&#xA;   The floating ground may drift a few volts from actual ground, thats what the third post is for, so you shouldn't count on it as grounding for safety purposes.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-07-21T19:42:26.083" CommentCount="2" />
  <row Id="12586" PostTypeId="2" ParentId="8227" CreationDate="2011-07-21T19:53:47.680" Score="7" ViewCount="" Body="&lt;p&gt;This common confusion stems from our familiarity with photographs. We forget that we rotate them to face ourselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a picture of yourself and hold it up in front of you. Probably you are holding it so that you can see your image. If so, you &quot;flipped&quot; the image of yourself when you rotated it 180 degrees around the vertical axis. When you look to the left side of the photo, you are looking over the right shoulder of your image. These directions are flipped!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now look in a mirror. When you look to the left, you are looking over the left shoulder of your image. These directions are not flipped!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now pick up the picture again and turn it so it's facing the same direction you are facing. You have removed the 180 degree rotation so that you and your image are &quot;looking&quot; in the same direction. The left side of your image is again to your left. If the picture is transparent enough that you can see your image, you'll see not the back of your head, but your eyes, giving you the impression that you're looking back at yourself. A mirror image! But again, left and right are not flipped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you say the mirror &quot;flips&quot; left and right, you are speaking from the frame of reference of one who is used to the 180 degree rotation that you apply to view an opaque photograph. But that's what we all do because we consider photographs, rotated 180 degrees to face ourselves, as being the &quot;correct&quot; left-right orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What a mirror really flips is the depth dimension. That which is behind you appears to be in front of you.&lt;/p&gt;&#xA;" OwnerUserId="4582" LastActivityDate="2011-07-21T19:53:47.680" CommentCount="2" />
  <row Id="12587" PostTypeId="2" ParentId="12580" CreationDate="2011-07-21T20:05:08.323" Score="4" ViewCount="" Body="&lt;p&gt;I believe the physically relevant parameters here are: 1. internal pressure P at the bottom of the bucket but not too close to the hole, units [$kg*m^{-1} s^{-2}$], the fluid density $\rho$ [$kg*m^{-3}$], hole area $A$ units $m^2$, and viscosity $\eta$ $kg m^{-1} s^{-1}$.  First, let's ignore the viscosity, and find which combinations can give us a velocity, units [$m/s$].  The unique combination is $\sqrt(P/\rho)$, and so in the limit of vanishing viscosity the velocity will be independent of the hole size. Now if we add in viscosity, there is a dimensionless combination available, $\rho \eta^2 P^{-1} A^{-1}$.  We could have any function of this dimensionless ratio, but common sense tells us that at least in the limit of small $\eta$ it should be a decreasing function, since viscosity should take kinetic energy away from the fluid.  From this ratio we see that decreasing the size of the hole is equivalent to increasing the viscosity, and thus the smaller hole  should have a smaller exit velocity than the bigger one.&#xA;I guess this is a long-winded way of saying the big hole will have higher Reynolds number and be less effected by the viscosity.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditorDisplayName="" LastEditDate="2011-07-21T20:10:25.917" LastActivityDate="2011-07-21T20:10:25.917" CommentCount="3" />
  <row Id="12588" PostTypeId="2" ParentId="8227" CreationDate="2011-07-21T20:21:41.557" Score="0" ViewCount="" Body="&lt;p&gt;All you have to do is imagine the line y=x+1 on the cartesian plane. If you reflect the line across a mirror in the vertical direction (across the y-axis), then the line becomes y=-x+1. Therefore, a vertical mirror flips the line's horizontal coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you reflect the lines across a mirror is in the horizontal direction (across the x-axis), then the line becomes y=-x-1. A horizontal mirror flips the line's vertical coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mirror just flips the axis perpendicular to its orientation! &lt;/p&gt;&#xA;" OwnerUserId="4587" LastActivityDate="2011-07-21T20:21:41.557" />
  <row Id="12589" PostTypeId="1" CreationDate="2011-07-21T20:37:45.340" Score="2" ViewCount="56" Body="&lt;p&gt;Is there a manageable formula or set of formulas or simple algorithms that approximate stellar luminosity and effective temperature (or radius) as a function of stellar age?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware that accurate modeling of these attributes is complex and is determined by many factors; what I'm looking for is something that serves as a decent approximation of the sort used in numerous illustrations or applets that show example &quot;paths&quot; taken by representative stars as they trace their evolution across the H-R diagram.&lt;/p&gt;&#xA;" OwnerUserId="2540" LastEditorUserId="2540" LastEditorDisplayName="" LastEditDate="2011-07-22T03:37:12.770" LastActivityDate="2011-07-23T16:54:03.733" Title="Are there formulae for calculating stellar luminosity and effective temperature as a function of age?" Tags="&lt;astrophysics&gt;&lt;stellar-evolution&gt;&lt;stellar-physics&gt;&lt;luminosity&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" />
  <row Id="12590" PostTypeId="1" AcceptedAnswerId="12592" CreationDate="2011-07-21T22:13:14.633" Score="4" ViewCount="98" Body="&lt;p&gt;My guess is that they find a certain tracks coming from a certain source by &quot;combintaorially&quot; selecting all track pairs and finding their invariant mass. If this is true, of which I am not sure, how is the background defined, i.e how far away from the invariant mass peaks? Any references will help too.&lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-12T19:13:02.367" LastActivityDate="2011-08-12T19:13:02.367" Title="What is meant by &quot;combinatorial background&quot; in experimental high energy physics" Tags="&lt;experimental-physics&gt;&lt;high-energy-physics&gt;&lt;data-analysis&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="12591" PostTypeId="2" ParentId="8227" CreationDate="2011-07-21T22:21:45.527" Score="2" ViewCount="" Body="&lt;p&gt;Imagine a mirror placed on the floor. Your face, looking at it, will be pointed downwards; but the face &quot;in the mirror&quot; is &quot;looking&quot; up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other answers here are great — I just wanted to make sure we discard the assumption that the mirror is placed on a wall!&lt;/p&gt;&#xA;" OwnerUserId="4600" LastActivityDate="2011-07-21T22:21:45.527" CommentCount="1" />
  <row Id="12592" PostTypeId="2" ParentId="12590" CreationDate="2011-07-21T23:29:09.567" Score="6" ViewCount="" Body="&lt;p&gt;Especially in hadron colliders, often-times you have a lot going on in a single event. There might be multiple jets (which originate from quarks or gluons, which cannot survive alone, cf. confinement etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of the particles one would be looking for in such events have a very short life-time, such as a top quark, a W or Z-boson. Thus, as opposed to more obvious things in an event, one has to reconstruct the existence of those particles by correctly identifying their decay product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a W-boson can decay into two quarks, which will form jets. Now, if you have many jets in an event, which of those two will you pick? Probably the ones whose combined invariant mass is near the W-mass. But still, since unfortunately the objects in the event don't come with labels, we cannot be sure these jets are the correct &lt;em&gt;combination&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example would be a top-antitop-event. The top decays, almost to a 100%, into a W boson and a b quark (which will become a jet). Each W can either decay leptonically or hadronically. So in a all-hadronic top-antitop event you have 6(!) jets you have to combine correctly to reconstruct the event. Even if you're looking at a pure sample of top-antitop events (like in a Monte-Carlo simulation), you are bound to get those combinations wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These things are what is called combinatorial background.&lt;/p&gt;&#xA;" OwnerUserId="2151" LastActivityDate="2011-07-21T23:29:09.567" />
  <row Id="12593" PostTypeId="1" CreationDate="2011-07-22T00:14:06.363" Score="0" ViewCount="31" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/8081/what-happens-to-chemical-compunds-that-include-radioactive-nuclei-when-those-dec&quot;&gt;What happens to chemical compunds that include radioactive nuclei, when those decay?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Suppose a radioactive atom, which is incorporated in a molecule, decays? What becomes of the bonds? Are there instances in which those bonds affect the decay, either with regard to the half-life, or the action of the decay? Are the resultant products, for example, slowed by the bonds?&lt;/p&gt;&#xA;" OwnerUserId="860" LastActivityDate="2011-07-22T00:14:06.363" ClosedDate="2011-07-22T03:13:19.217" Title="radioactive decay in molecular constituents" Tags="&lt;radioactivity&gt;&lt;radiation&gt;" CommentCount="2" />
  <row Id="12594" PostTypeId="1" AcceptedAnswerId="12890" CreationDate="2011-07-22T01:39:51.117" Score="7" ViewCount="146" Body="&lt;p&gt;Let's say I have a polymer, of contour length $L_p$ and persistence length $P$, positioned between two parallel plates separated by a distance $z$.  I slowly squeeze the plates together until only two-dimensional diffusion is observed.  Under the worm-like chain (WLC) model, how will the force curve, $F(z)$, scale as a function of the distance between the plates? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a sort of addendum... I'm curious how compression of the polymer between two parallel plates contrasts with the force curve associated with packing the polymer in a spherical cage?  In the latter case of the spherical cage, I'm not exactly sure that the WLC model is relevant when the cross-sectional dimensions of the cage approach the polymer's persistence length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update - To make a prediction, the problem of squeezing a WLC or FJC polymer between two plates is probably more straightforward than that of 'crushing' a polymer into a spherical ball (which could perhaps be experimentally accomplished by placing a hydrophilic polymer into increasingly 'bad'/non-polar solvent) since it avoids having to estimate any bending energies.  So, focusing on the 3D --&gt; 2D scenario, my guess is that the energy to accomplish the task of squeezing the polymer until only two dimensional diffusion is observed would be similar to that of reducing one degree of diffusional freedom of a monoatomic gas - i.e. (Number of persistence length units in polymer) * $\frac{1}{2}(kB)(T)$.  I would also predict that the force curve is non-linear, since the probability density for the location of a particular persistence length segment of a polymer is probably not uniform across the gap between the plates.  However, that's simply a guess.&lt;/p&gt;&#xA;" OwnerUserId="4257" LastEditorUserId="4257" LastEditorDisplayName="" LastEditDate="2011-07-25T10:32:45.027" LastActivityDate="2011-07-28T15:55:46.110" Title="Force curve associated with squeezing a worm-like chain (WLC) between two parallel plates" Tags="&lt;statistical-mechanics&gt;&lt;material-science&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="12595" PostTypeId="1" CreationDate="2011-07-22T01:40:14.793" Score="1" ViewCount="43" Body="&lt;p&gt;Is it possible to provide a brief discussion of picture changing operators in string theory in here? If possible then it will be nice to have a discussion of it with some references! Thank you!&lt;/p&gt;&#xA;" OwnerUserId="1349" LastActivityDate="2011-07-22T01:40:14.793" ClosedDate="2011-07-22T03:11:34.740" Title="Picture changing operators" Tags="&lt;string-theory&gt;" CommentCount="1" />
  <row Id="12596" PostTypeId="1" CreationDate="2011-07-22T01:47:35.960" Score="2" ViewCount="88" Body="&lt;p&gt;Suppose I am studying a field theory at finite temperature or some black hole formation scenario from boundary theory perspective in the sense of AdS/CFT. How is it possible to gain information about them from say looking at the two point functions (propagators) of  operators in field theory? I mean will there be some special pole structures etc. in that Green's function? Is there a generic such behavior? Can you suggest me some references?&lt;/p&gt;&#xA;" OwnerUserId="1349" LastActivityDate="2011-07-23T09:47:34.800" Title="Correlation functions in thermal field theory etc" Tags="&lt;quantum-field-theory&gt;&lt;string-theory&gt;" AnswerCount="2" />
  <row Id="12597" PostTypeId="2" ParentId="8227" CreationDate="2011-07-22T02:40:45.040" Score="0" ViewCount="" Body="&lt;p&gt;The mirror is not inverting the image in either direction but rather is acting like a rubber stamp doing a 1:1 transfer from surface to surface. If you wrote a word on a rubber stamp with fresh ink, then pressed the stamp to paper, the word on the paper would be backwards but not upside down. It's just a 1:1 transfer.&lt;/p&gt;&#xA;" OwnerUserId="4612" LastActivityDate="2011-07-22T02:40:45.040" />
  <row Id="12598" PostTypeId="2" ParentId="8227" CreationDate="2011-07-22T03:12:28.257" Score="0" ViewCount="" Body="&lt;p&gt;Think about where a point above, below, left, and right of your point of view are in the reflection.  Your head is still on top, your feet still on the bottom in the mirror.  Likewise, your left hand is still to the left and your right hand to the right.  It seems flipped because, to look behind you, you are used to turning around (which swaps left/right), rather than flipping (which swaps top/bottom).  If we flipped upside down instead of turning around to see behind us, you would experience the opposite effect.&lt;/p&gt;&#xA;" OwnerUserId="4615" LastActivityDate="2011-07-22T03:12:28.257" />
  <row Id="12599" PostTypeId="1" CreationDate="2011-07-22T05:55:35.193" Score="8" ViewCount="227" Body="&lt;p&gt;Suppose you take a metal disc and cut a small, circular hole in the center.&#xA;When you heat the whole thing, will the hole's diameter increase or decrease? and why?&lt;/p&gt;&#xA;" OwnerUserId="4621" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-22T06:09:53.977" LastActivityDate="2011-07-27T22:38:51.087" Title="Will a hole cut into a metal disk expand or shrink when the disc is heated?" Tags="&lt;temperature&gt;&lt;material-science&gt;&lt;elasticity&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="12600" PostTypeId="2" ParentId="12599" CreationDate="2011-07-22T06:22:29.260" Score="6" ViewCount="" Body="&lt;p&gt;Good question! Assuming the disc is uniform and isotropic (the same in different directions), the hole will expand in the same ratio as the metal. You can see this because the thermal expansion equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{d} L = L\alpha\mathrm{d}T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;applies to all lengths associated with the metal, including the circumference of the hole, since the edge of the hole is made out of metal. And if the circumference of the hole expands, so does the diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a disc with different regions that are made of different types of metal, or if the metal that makes up your disc has an anisotropic crystal structure (so that it expands by different factors in different directions), then the analysis is more complicated. But in both cases, I think the hole would still get larger since the overall change in size is still an expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to get the hole to shrink, you would need to use a material with a negative thermal expansion coefficient $\alpha &amp;lt; 0$, which means it gets smaller as the temperature gets higher. In that case the entire disc would shrink as it heats up. I don't know of any specific examples of materials that behave that way, but I think they might exist.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-07-22T06:22:29.260" CommentCount="1" />
  <row Id="12601" PostTypeId="2" ParentId="12417" CreationDate="2011-07-22T06:54:48.303" Score="0" ViewCount="" Body="&lt;p&gt;OK, the inside of the sphere is perfectly-reflecting, and there's an ideal &lt;a href=&quot;http://en.wikipedia.org/wiki/Optical_isolator&quot; rel=&quot;nofollow&quot;&gt;optical diode&lt;/a&gt; to let light in but keep it inside. As you keep the light turned on, the photon density in the sphere goes up and up, of course. It &quot;looks&quot; brighter and brighter, but you don't see that because the light can't escape. After turning  the light off, it stays bright, the photons just keep bouncing around. If you &quot;stick your head in&quot; to look, you see a bright uniform glow that quickly dies away because your head and eyes are absorbing all the photons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But do the photons bounce around forever? No!! Even a perfectly-reflective sphere will still interact with the light, because of &lt;a href=&quot;http://en.wikipedia.org/wiki/Radiation_pressure&quot; rel=&quot;nofollow&quot;&gt;radiation pressure&lt;/a&gt;. Each time a photon bounces off a wall, the wall gets kicked backwards, gaining energy at the expense of the photon (on average). Light can't produce a smooth force, only a series of kicks with &lt;a href=&quot;http://en.wikipedia.org/wiki/Shot_noise&quot; rel=&quot;nofollow&quot;&gt;shot noise&lt;/a&gt; statistics, because one photon hits the wall at a time. These kicks eventually heat up the walls, and cool down the photons. (From the photon's point of view, the photon frequency is going down because of Doppler-shifts during reflection off the moving walls.) Eventually everything equilibrates to a uniform temperature, hotter than the sphere started out. I don't know how long that would take. [In any realistic circumstance this radiation pressure effect can be ignored, because it is much less important than the &quot;reflection is not 100% perfect&quot; effect.]&lt;/p&gt;&#xA;" OwnerUserId="3811" LastActivityDate="2011-07-22T06:54:48.303" CommentCount="3" />
  <row Id="12602" PostTypeId="1" AcceptedAnswerId="12603" CreationDate="2011-07-22T07:09:04.463" Score="5" ViewCount="108" Body="&lt;p&gt;Fairly straightforward question. If not, why not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that if they do, it is not perceived due to the regions of highest dispersion being in one's region of lowest visual acuity.&lt;/p&gt;&#xA;" OwnerUserId="1876" LastActivityDate="2011-07-22T14:22:57.173" Title="Do eyeballs exhibit chromatic aberration?" Tags="&lt;optics&gt;&lt;biophysics&gt;" AnswerCount="3" />
  <row Id="12603" PostTypeId="2" ParentId="12602" CreationDate="2011-07-22T08:20:16.500" Score="5" ViewCount="" Body="&lt;p&gt;Yes, it does. We don't see it because our brain automagically 'correct it' because it always see the same aberration from the childhood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our eye focuses on 'green' wavelength as it's its peak sensitivity, so red and especially violet lines are usually slightly out-of-focus.&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="930" LastEditorDisplayName="" LastEditDate="2011-07-22T14:21:21.930" LastActivityDate="2011-07-22T14:21:21.930" CommentCount="2" />
  <row Id="12604" PostTypeId="1" CreationDate="2011-07-22T08:28:00.893" Score="-7" ViewCount="161" Body="&lt;p&gt;Let the amount of energy in one pulse of (laser) light be $E$, and the wavelength be $\lambda$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This pulse goes straight to the mirror, and it is reflected by the mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the reflectivity of this mirror be $r_{\lambda}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/APbUi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question.&lt;/strong&gt;  How can I get the formula of $r_{\lambda}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any answers (or hints) will be appreciated, thank you.&lt;/p&gt;&#xA;" OwnerUserId="4558" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-07-22T17:58:12.560" LastActivityDate="2011-07-22T18:42:45.880" Title="Light Energy Absorption In Mirror" Tags="&lt;optics&gt;&lt;light&gt;&lt;reflection&gt;&lt;mirror&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="12605" PostTypeId="2" ParentId="12602" CreationDate="2011-07-22T08:37:41.223" Score="2" ViewCount="" Body="&lt;p&gt;Because refractive index is a function of wavelenght, every lens experiences chromatic aberration, and so does a human eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could have severe effect on human vision. However, eye has tuned spectral bandwidth - it is well known eye is most sensible to visible light which has wavelenght of 550 nm. Relative luminosity is also a function of wavelenght (in which more than 70% of the luminous energy is confined to a defocus range of less than 0.25 D defocus on either side of focus (if eye is optimally focused at 550 nm)). The exact spectrum to which we are sensitive, depends on light levels. Rods saturate at moderate and high light levels, so in this region spectral sensitivity is governed by cones. At lower light levels cones are no longer as sensitive and rods dominate the response. All this results in different spectral sensitivity for various luminosity and in this way eye is able to filter out some effects of chromatic aberration.&lt;/p&gt;&#xA;" OwnerUserId="4264" LastActivityDate="2011-07-22T08:37:41.223" CommentCount="1" />
  <row Id="12606" PostTypeId="2" ParentId="12599" CreationDate="2011-07-22T09:58:45.947" Score="6" ViewCount="" Body="&lt;p&gt;Ahh, this is an interesting piece of &quot;intuitive&quot; but wrong thinking. &#xA;(I had similar thougths as a beginner, maybe it's &quot;natural&quot;?) &#xA;The argument from David re circumference is right of course, but &#xA;not so &quot;convincing&quot; intuitively to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try following: Think of a plug of the same metal, exactly fitting into that hole, &#xA;taken in and out at changing temperatures. &#xA;If the hole would shrink when heated, this would mean that the plug &#xA;had to excert a pressure to &quot;keep&quot; its place. Same would apply &#xA;for any piece of the original metal disk, where no hole is, but could be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The keyword is homegeneity, more precise, the persistence of homogeneity &#xA;at different temperatures, if only there is no difference in temperature &#xA;within the disk (no temp. gradient)&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-07-22T10:46:54.347" LastActivityDate="2011-07-22T10:46:54.347" CommentCount="1" />
  <row Id="12607" PostTypeId="2" ParentId="8227" CreationDate="2011-07-22T10:46:34.697" Score="0" ViewCount="" Body="&lt;p&gt;Left and right is relative to you only but not to the rest of your environment (including the mirror), but up and down is static for the whole environment including you and the mirror.&lt;/p&gt;&#xA;" OwnerUserId="4040" LastActivityDate="2011-07-22T10:46:34.697" />
  <row Id="12608" PostTypeId="2" ParentId="12596" CreationDate="2011-07-22T10:54:41.173" Score="1" ViewCount="" Body="&lt;p&gt;I don't have a complete answer but here's something to give you a rough idea.&#xA;We consider a Gaussian scalar field with one-point Hamiltonian $\Omega$, i.e. Hamiltonian of this field is given by&#xA;$$H = : {1 \over 2} \int {\rm d}^d {\mathbf x} \left( \pi(\mathbf x)^2 + \phi(\mathbf x) \Omega^2\phi(\mathbf x)\right): = \int {\tilde {\rm d} \mathbf p} E(\mathbf p)  a^{\dagger}(\mathbf p)a(\mathbf p)$$&#xA;where we introduced $\tilde {\rm d} \mathbf p = {{\rm d} \mathbf p \over (2\pi)^d 2E(\mathbf p)} .$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Euclidean two-point correlation function of this field is given by&#xA;$$\left&amp;lt;{\mathbf x} \right| G_E(t, t&amp;#39;) \left|{\mathbf x&amp;#39;} \right&amp;gt; = &#xD;&#xA;\int {{\rm d}^{d+1} p \over {2 \pi}^{d+1}}&#xD;&#xA;{e^{i p_0(t-t&amp;#39;) + {\mathbf p} \cdot ({\mathbf x -\mathbf x&amp;#39;})} \over p_0^2 + E(\mathbf p)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while for the thermal Green function we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\left&amp;lt;{\mathbf x} \right| G_{\beta}(t, t&amp;#39;) \left|{\mathbf x&amp;#39;} \right&amp;gt; = &#xD;&#xA;{1 \over \beta} \sum_{\omega_n = {2 \pi n / \beta}}&#xD;&#xA;\int {{\rm d}^{d} p \over {2 \pi}^{d}}&#xD;&#xA;{e^{i \omega_n(t-t&amp;#39;) + {\mathbf p} \cdot ({\mathbf x -\mathbf x&amp;#39;})} \over \omega_n^2 + E(\mathbf p)^2}.&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be summed and separated into the contributions from the ground state and from the excited states&#xA;$$\left&amp;lt;{\mathbf x} \right| G_{\beta}(t, t&amp;#39;) \left|{\mathbf x&amp;#39;} \right&amp;gt; = &#xD;&#xA;\left&amp;lt;{\mathbf x} \right| G_E(t, t&amp;#39;) \left|{\mathbf x&amp;#39;} \right&amp;gt; +&#xD;&#xA;\int {{\rm d}^{d} p \over {2 \pi}^{d}}&#xD;&#xA;{1 \over E(\mathbf p)}&#xD;&#xA;{\cosh E(\mathbf p) (t - t&amp;#39;) \over e^{\beta E(\mathbf p)} - 1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two general observations that can be seen from this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the thermal Green function has a pole at $\beta E = 2 \pi n$. This comes from the fact that we have compactified time (and so temperature) to a circle.&lt;/li&gt;&#xA;&lt;li&gt;in the limit $\beta \to \infty$ the contribution from excited states dies off and we are left with the vacuum Green function.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-07-22T17:37:45.847" LastActivityDate="2011-07-22T17:37:45.847" CommentCount="4" />
  <row Id="12609" PostTypeId="2" ParentId="12604" CreationDate="2011-07-22T11:27:47.350" Score="5" ViewCount="" Body="&lt;p&gt;You can obtain the reflection coefficient using the Fresnel equations:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fresnel_equations&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fresnel_equations&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a metal mirror and &lt;strong&gt;perpendicular incidence&lt;/strong&gt; the (intensity) reflectivity is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r_\lambda=|R_\perp|^2=\frac{(1-n(\lambda))^2+\kappa(\lambda)^2}{(1+n(\lambda))^2+\kappa(\lambda)^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $n$ is the real part of the refractive index (e.g. n(530 nm)=0.95 for Aluminium at green (530 nm) light) and $\kappa$ is the imaginary part (e.g. $\kappa(530 nm)=6.4$ for Aluminium). So for Aluminium you obtain $r_\textrm{green}=91\%$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find &lt;strong&gt;values for other metals and wavelengths&lt;/strong&gt; here: &lt;a href=&quot;http://refractiveindex.info/?group=METALS&amp;amp;material=Aluminium&quot; rel=&quot;nofollow&quot;&gt;http://refractiveindex.info/?group=METALS&amp;amp;material=Aluminium&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general $r$ increases with the frequency of light. The electrons can't follow the field for high frequencies. One can measure $n$ and $\kappa$ of a material by Ellipsometry &lt;a href=&quot;http://en.wikipedia.org/wiki/Ellipsometry&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Ellipsometry&lt;/a&gt; . $\kappa$ is related to the electrical conductivity $\sigma(\omega)$ of the metal.&lt;/p&gt;&#xA;" OwnerUserId="4306" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-07-22T18:42:45.880" LastActivityDate="2011-07-22T18:42:45.880" CommentCount="7" />
  <row Id="12610" PostTypeId="2" ParentId="12602" CreationDate="2011-07-22T11:36:16.160" Score="1" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Here is an interesting webpage on the aberrations of the eye:&lt;/strong&gt;&#xA;&lt;a href=&quot;http://www.telescope-optics.net/eye_chromatism.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.telescope-optics.net/eye_chromatism.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://kaschke-medtec.de/SS10/Exercise5.pdf&quot; rel=&quot;nofollow&quot;&gt;http://kaschke-medtec.de/SS10/Exercise5.pdf&lt;/a&gt; the eye has quite a large chromatic aberration of 2 dptr across the visible range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I heard (the rumor?) that for this reason the color bands in the french flag don't have the same width.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also read of the interesting (and probably true) theory that the colour bands around edges actually help you to focus while you reading text. I think that explains why I find reading quite hard with LED illumination of small (non-overlapping) spectral bandwidth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could think it would be useful to wear color corrected glasses but I don't think that would help in any way (the aberrations due to the glasses will be small compared to the aberrations in the eye).&lt;/p&gt;&#xA;" OwnerUserId="4306" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-07-22T14:22:57.173" LastActivityDate="2011-07-22T14:22:57.173" CommentCount="2" />
  <row Id="12611" PostTypeId="1" AcceptedAnswerId="14097" CreationDate="2011-07-22T11:45:13.453" Score="9" ViewCount="269" Body="&lt;p&gt;From what I remember in my undergraduate quantum mechanics class, we treated scattering of non-relativistic particles from a static potential like this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Solve the time-independent Schrodinger equation to find the energy eigenstates.  There will be a continuous spectrum of energy eigenvalues.&lt;/li&gt;&#xA;&lt;li&gt;In the region to the left of the potential, identify a piece of  the wavefunction that looks like $Ae^{i(kx - \omega t)}$ as the incoming wave.&lt;/li&gt;&#xA;&lt;li&gt;Ensure that to the right of the potential, there is not piece of the wavefunction that looks like $Be^{-i(kx + \omega t)}$, because we only want to have a wave coming in from the left.&#xA;&lt;li&gt;Identify a piece of the wavefunction to the left of the potential that looks like $R e^{-i(kx + \omega t)}$ as a reflected wave.&lt;/li&gt;&#xA;&lt;li&gt;Identify a piece of the wavefunction to the right of the potential that looks like $T e^{i(kx - \omega t)}$ as a transmitted wave.&lt;/li&gt;&#xA;&lt;li&gt;Show that $|R|^2 + |T|^2 = |A|^2$.  Interpret $\frac{|R|^2}{|A|^2}$ as the probability for reflection and $\frac{|T|^2}{|A|^2}$ as the probability for transmission.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This entire process doesn't seem to have anything to do with a real scattering event - where a real particle is scattered by a scattering potential - we do all our analysis on a stationary waves.  Why should such a naive procedure produce reasonable results for something like Rutherford's foil experiment, in which alpha particles are in motion as they collide with nuclei, and in which the wavefunction of the alpha particle is typically localized in a (moving) volume much smaller than the scattering region?&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-24T21:07:59.227" LastActivityDate="2011-08-29T07:16:23.183" Title="Why can we treat quantum scattering problems as time-independent?" Tags="&lt;quantum-mechanics&gt;&lt;scattering&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="5" />
  <row Id="12612" PostTypeId="2" ParentId="12596" CreationDate="2011-07-22T12:23:53.577" Score="0" ViewCount="" Body="&lt;p&gt;It's a little old now, but try Hiroomi Umezawa, &quot;Advanced Field Theory; Micro, Macro, and Thermal Physics&quot;, AIP, 1993.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a slightly different notation from that used by Marek, the two point function changes from&#xA;$$\left&amp;lt;0\right|\phi(x)\phi(x+y)\left|0\right&amp;gt;=\hbar\int 2\pi\delta(k^2-m^2)\theta(k_0)\mathrm{e}^{-ik\cdot y}\frac{\mathrm{d}^4k}{(2\pi)^4},$$&#xA;the free Klein-Gordon vacuum state case, to the corresponding thermal state case,&#xA;$$\omega_\mathsf{T}\Bigl[\phi(x)\phi(x+y)\Bigr]=\hbar\int 2\pi\delta(k^2-m^2)\theta(k_0)\coth\left[ \!\frac{\hbar k_0}{2\mathsf{k_BT}}\!\right]\mathrm{e}^{-ik\cdot y}\frac{\mathrm{d}^4k}{(2\pi)^4}.$$&#xA;On the basis of &lt;em&gt;this&lt;/em&gt; presentation, the thermal free field case is no more than a different measure on the forward-light cone, which is, inevitably, not Lorentz invariant (the $k_0$ in the $\coth$ factor picks out a preferred frame). The deformation is no less smooth than the vacuum 2-point function.&#xA;Additionally, although the above equations don't show it, the thermal field state is still Gaussian, like the vacuum state, so the whole structure of the thermal state over the free Klein-Gordon field is completely specified by the 2-point function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could also construct higher-order deformations of the mass-shell measure, by adding extra factors of $\coth\left[ \!\frac{\hbar k&amp;#39;_0}{2\mathsf{k_BT&amp;#39;}}\!\right]$, or otherwise, quite possibly corresponding to different time-like directions and temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A presentation of the Klein-Gordon field that is enough to reconstruct the above can be found in my &quot;A succinct presentation of the quantized Klein–Gordon field, and a similar quantum presentation of the classical Klein–Gordon random field&quot;, quant-ph/0411156, Phys. Lett. A 338, 8-12(2005), although mostly I'm grinding quite a different axe in that paper.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-07-23T09:47:34.800" LastActivityDate="2011-07-23T09:47:34.800" CommentCount="2" />
  <row Id="12613" PostTypeId="1" AcceptedAnswerId="12615" CreationDate="2011-07-22T13:51:15.613" Score="2" ViewCount="90" Body="&lt;p&gt;In his popular book on relativity, in chapter IX, &quot;The relativity of simultaneity&quot;, Einstein describes an experiment in which a flash happens simultaneously on A and B, as defined by the fact that an observer at the middle point M can see the light coming from A and B at the same moment:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              --train--&amp;gt;&#xA;---embankment---A-------M-------B----&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then, he proceeds to say that an observer on the train &quot;is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I find the opposite result, from the train point of view:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the embankment moves to the left (in the diagram) at a speed of v.&lt;/li&gt;&#xA;&lt;li&gt;so, after dt, A moved by v*dt to the left, the light coming from A moved by c*dt to the right, and the distance between A and the light increased by (c+v)*dt.&lt;/li&gt;&#xA;&lt;li&gt;similarly, the distance between B and the light coming from B increased by (c-v)*dt.&lt;/li&gt;&#xA;&lt;li&gt;so, the light coming from A is going to reach the middle point M before the light coming from B, so the observer is going to see the light from A first (if he happens to be at M at the correct moment), which is the opposite of what is said in the book.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Another way to state about the same thing is: the light from both points go towards M, but M is moving to the left, so the light from A is going to reach M before the light from B does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is the error in my reasoning?&lt;/p&gt;&#xA;" OwnerUserId="4631" LastEditorUserId="4631" LastEditorDisplayName="" LastEditDate="2011-07-23T08:13:20.187" LastActivityDate="2011-07-23T08:13:20.187" Title="The relativity of simultaneity" Tags="&lt;special-relativity&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="12614" PostTypeId="1" AcceptedAnswerId="12631" CreationDate="2011-07-22T14:11:37.383" Score="4" ViewCount="161" Body="&lt;p&gt;I just read a paper 'A pocket calculator determination of energy eigenvalues' by J Killingbeck&#xA;(1979).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Link: &lt;a href=&quot;http://iopscience.iop.org/0305-4470/10/6/001&quot; rel=&quot;nofollow&quot;&gt;http://iopscience.iop.org/0305-4470/10/6/001&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some questions about section 2.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Why $R&amp;lt;0$ implies the $E$ estimate was too high and $R&amp;gt;1$ implies too low? And is it possible that $0&amp;lt;R&amp;lt;1$ remains?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why &quot;if the starting point is chosen at $r=h(l+1)$, then $R(hl)$ can be assigned any&#xA;finite value without disturbing the calculation at larger $r$ values&quot; ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How to guess a trial energy $E$ (especially the ground state energy)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(Actually I'm interested in a later paper by Killingbeck (1982) 'Finite-difference methods for eigenvalues'(Link: &lt;a href=&quot;http://iopscience.iop.org/0022-3700/15/6/009&quot; rel=&quot;nofollow&quot;&gt;http://iopscience.iop.org/0022-3700/15/6/009&lt;/a&gt;), in which the author applies the method in 'A pocket calculator determination of energy eigenvalues' to some potentials with highly singularity at origin)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;mod note: copying the text of whoplisp's post here since it isn't actually an answer to the question&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They convert the Schroedinger equation into a finite difference equation&#xA;(instead of a variational approach):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-\frac{1}{2}\Delta\psi +V\psi=E\psi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V$ is a central potential&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ansatz with solid harmonic $\xi_l$ of degree $l$ (e.g. $x$ or $xy$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\psi=\xi_l\phi(r)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;leads to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2r(V-E)\phi=rD^2\phi+(2l+2)D\phi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notation: $D=\frac{d}{dr}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;finite difference approximation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h^2D^2\phi \rightarrow \phi(r+h)+\phi(r-h)-2\phi(r)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2hD\phi \rightarrow \phi(r+h)-\phi(r-h)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;crucial step: &lt;strong&gt;introduce ratio variable&lt;/strong&gt; $R(r)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi(r+h)=R(r)\phi(r)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;convert Schroedinger equation in recursive equation to calculate $R(r)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2r[1+h^2(V-E)]=R(r)[r+h(l+1)]+[r-h(l+1)](R(r-h))^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the corresponding 1D Schroedinger equation is:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$2[1+h^2(V-E)]=R(x)+(R(x-h))^{-1}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;if $r=h(l+1)$ is chosen then $R(hl)$ can be assigned any finite value without&#xA;disturbing the calculation at large $r$ values&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to find the ground states with angular momentum $l$ start with an estimate for $E$,&#xA;some non-zero $R(hl)$ and some small $h$. $R(r)$ as $r$ increases will either become negative or pass through unity from below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;first thing happens when guessed $E$ was too high&lt;/p&gt;&#xA;&#xA;&lt;p&gt;second thing indicates $E$ was too low (because $\psi$ increases with $r$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you just try various values and 'sandwhich' the solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;example for $V=\lambda r-1/r$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda=1$, $h=.1$, state=2p: 1.9759&lt;/p&gt;&#xA;" OwnerUserId="4314" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-07-22T18:35:19.013" LastActivityDate="2011-07-23T08:51:46.300" Title="On numerically solving the Schrödinger equation" Tags="&lt;quantum-mechanics&gt;&lt;computational-physics&gt;&lt;schrodinger-equation&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" />
  <row Id="12615" PostTypeId="2" ParentId="12613" CreationDate="2011-07-22T14:30:12.797" Score="1" ViewCount="" Body="&lt;p&gt;Your result is correct and, of course, there is no problem with relativity. When you're on the train and you see how events unfold for an observer on the platform at point M, then you conclude that he will see the light from A first. When you're on the platform and you see how events unfold for an observer on the train, then you conclude that he sees the light from B first. There is no problem with that&amp;mdash;that's actually the essence of the theory of relativity: things look different from different reference frames, and every observer is right in their respective reference frame. When the two observers compare their conclusions, they will be able to attribute the difference to the fact that the speed of light is constant.&lt;/p&gt;&#xA;" OwnerUserId="4483" LastActivityDate="2011-07-22T14:30:12.797" />
  <row Id="12616" PostTypeId="1" CreationDate="2011-07-22T15:25:29.233" Score="5" ViewCount="120" Body="&lt;p&gt;Everytime I check the weather news and storms are announced, the UV index is very often (if not always) &quot;very high&quot; or &quot;extreme&quot;, why is that?&lt;/p&gt;&#xA;" OwnerUserId="4634" LastActivityDate="2011-09-04T03:41:59.743" Title="Why is UV index so high when storms are announced?" Tags="&lt;weather&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12617" PostTypeId="2" ParentId="12616" CreationDate="2011-07-22T15:58:35.423" Score="2" ViewCount="" Body="&lt;p&gt;If you check the process of storm creation, you can understand that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atmospheric pressure of storm's outside is very high, and the pressure of eye of storm is very low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the atmospheric pressure of storm's outside is very high, the sky is very clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/DGMGg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the UV index goes high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Link Added.&lt;/strong&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Transmittance&quot; rel=&quot;nofollow&quot;&gt;Transmittance&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Opacity_%28optics%29&quot; rel=&quot;nofollow&quot;&gt;Opacity&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For someone who can't understand the reason why if the sky is clear, UV index goes high.)&lt;/p&gt;&#xA;" OwnerUserId="4558" LastEditorUserId="4558" LastEditorDisplayName="" LastEditDate="2011-07-22T16:21:42.903" LastActivityDate="2011-07-22T16:21:42.903" CommentCount="14" />
  <row Id="12619" PostTypeId="2" ParentId="12611" CreationDate="2011-07-22T17:11:56.667" Score="-1" ViewCount="" Body="&lt;p&gt;The answer to this is the same as the answer to why you solve the Time-Independent-Schrodinger-Equation to find the time evolution of a bound particle.  First you solve the TISE to find the stationary states $\psi_n$, then you write the particle's wavefunction $\Psi(t=0)$ in terms of a superposition of the $\psi_n$.  Since you know how the stationary states evolve in time, you now know (at least in principle) how ANY wavefunction evolves in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the same thing for scattering.  You figure out what happens for the energy eigenstates, and now you know what will happen for any wavepacket (which you would write as a superposition of energy eigenstates, of course).  And here it's even easier than the bound states: if all you care about is R and T, and your wavepacket has a narrow range of energies (for which T is nearly constant), then the value of T for your wavepacket is the same as what you just calculated for the energy eigenstate.  Huzzah!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your wavepacket involves a superpostion of a wide range of energies, with a wide range of T's, then your life will be more complicated, of course.  But in scattering experiments, folks usually try to employ nearly monoenergetic beams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because quantum mechanics classes spend so much time mired in the details of solving the TISE (either for scattering or bound states), they often lose sight of one of the motivations for solving the TISE:  it's a tool for finding the time behavior of any initial condition.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-07-22T17:11:56.667" CommentCount="13" />
  <row Id="12620" PostTypeId="1" CreationDate="2011-07-22T17:40:26.953" Score="3" ViewCount="48" Body="&lt;p&gt;I'm looking at Section 2.4.1 of Nielsen and Chuang's &lt;em&gt;Quantum Computation and Quantum Information&lt;/em&gt; were they derive the density operator versions of the evolution and measurement postulates of quantum mechanics and something is bugging me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{(p(i), | x_i \rangle) \colon i=1,2,...,n\}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;be an ensemble and suppose you perform a measurement on this ensemble that results in outcome $m$. Then the post-measurement ensemble is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{(p(i|m), | x&amp;#39;_i \rangle) \colon i=1,2,...,n\}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p(i|m)$ is the probability that the state of the system was originally $| x_i \rangle$ given that the measurement outcome was $m$. It seems natural enough to use $p(i|m)$ for the probabilities of the post-measurement ensemble. However, mathematically, I do not see why that should be the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question then is this: Given only the definition of what an ensemble is and the state vector version of the postulates of QM, is there a way to derive the rules to compute the post-measurement probabilities of an ensemble?&lt;/p&gt;&#xA;" OwnerUserId="4637" LastActivityDate="2011-07-22T17:40:26.953" Title="Mixed state after measurement" Tags="&lt;quantum-mechanics&gt;" CommentCount="1" />
  <row Id="12621" PostTypeId="2" ParentId="11927" CreationDate="2011-07-22T17:42:57.327" Score="2" ViewCount="" Body="&lt;p&gt;I found a better explanation that seems more convincing to me. I tried sliding two edges of paper perpendicularly, surprisingly the sound produced by doing that is pretty similar with the one produced by tearing paper. It seems that the sound produced is due to the transversal vibration of the whole paper, the friction or the fibers snapping merely acts as driving force. and notice if you hold the paper near the sliding point, the frequency of sound produced becomes higher because the shorter the vibrating paper's length the higher the frequency is(only short wavelengths of standing wave are allowed).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now which one gives more dominant driving force, friction or snapping fibers?&lt;br&gt;&#xA;I think friction is more dominant, because sliding the edges gives a similar sound even without involving any snapping process. Also if we slide it faster, the frequency will be higher.&lt;/p&gt;&#xA;" OwnerUserId="4252" LastActivityDate="2011-07-22T17:42:57.327" />
  <row Id="12622" PostTypeId="2" ParentId="14255" CreationDate="2011-07-22T17:48:13.513" Score="4" ViewCount="" Body="&lt;p&gt;There are many different versions of the multiple-universe idea in physics. Scientific American has a nice (skeptical) article on the topic this month, &quot;Does the multiverse really exist?&quot; by the well known relativist G.F.R. Ellis. Some of the possibilities discussed include the many-worlds interpretation of quantum mechanics, chaotic inflation, the string-theory landscape, and cyclic models such as ekpyrotic models and Penrose's conformal cyclic cosmology. You can find information about specific ideas from this list on Wikipedia, etc. There is not likely to be a generic answer to your question that will encompass all of these cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing you have to watch out for is that it is not necessarily easy to come up with a suitable definition of &quot;universe.&quot; For example, you could say that it means the set of all events that a given observer O, who is assumed to be immortal, could in principle observe, subject to general relativity's restriction to the propagation of information at no more than the speed of light. The accelerating expansion of the universe implies that there is a cosmological horizon beyond which O can never see, so by this definition of &quot;universe,&quot; O's universe has a boundary. With this definition, if you have two observers, O and P, there can be events that are in O's universe and also in P's universe, but other events that are in one but not the other. You could say, &quot;Oh no, I don't like that definition, it's goofy. I want the &lt;em&gt;whole&lt;/em&gt; universe to be included, not just the part that a particular observer can ever observe.&quot; But then you have the same philosophical problem as in other incarnations of the multiple-universe idea, which is that you have to talk about things that can never be observed, even in principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will give one example where it seems to me to be valuable to discuss multiple universes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;History offers several examples of cosmological models that were wrong, and that with hindsight we can say should have been considered implausible at the time, because they required fine tuning. Newton envisioned an infinite, homogeneous cosmos so that despite the force of gravitational attraction, the force on any particle would cancel out by symmetry, thereby allowing the universe to exist forever. Einstein had a similar model in which the cosmoogical constant was exactly sufficient to balance gravitational attraction and keep the universe static. Even before Hubble discovered cosmological expansion, there was a problem with both of these models, because they were unstable, and required fine-tuning. In the Newton model, any deviation from homogeneity, no matter how small, causes a snowball effect. Similarly, Einstein's cosmological constant had to be perfectly tuned, or else there would be a vicious cycle of accelerating expansion or collapse. The lesson from history seems to be that we should not believe in cosmological models that require fine-tuning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the current consensus model of cosmology has multiple fine-tuning problems. There is a flatness problem, and also an entropy problem. Inflation is supposed to fix the former, but inflation has a lot of problems, one of which is that inflation is in some sense improbable. To define the notion of its probability, you pretty much have to start talking about the set of all possible universes, and asking how many of them have inflation that could lead to a universe such as ours. This is similar to the Einstein fine-tuning problem, where you basically want to object because the perfectly tuned value of the cosmological constant seems low in probability. You can't talk about probability unless you have a sample space, and here the sample space is some kind of multiverse. If you refuse to talk about the multiverse, then you've closed off an avenue of argument that has been historically fruitful.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-07-22T17:48:13.513" CommentCount="1" />
  <row Id="12624" PostTypeId="2" ParentId="12613" CreationDate="2011-07-22T18:23:12.500" Score="0" ViewCount="" Body="&lt;p&gt;It looks to me that you've left out other important parts of the passage that are needed to understand what's going on here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motion of A and B from the point of view of an observer on the train is irrelevant to determining which flash reaches the midpoint of the train first. What matters is that both flashes travel at speed c, and the times at which they were emitted from A and B. The main problem in your reasonsing is your assumption that they are both emitted from A and B simultaneously for an observer on the train.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you read the passage more carefully, I think you'll find that the event of the two flashes occuring at the midpoint of the train doesn't exist for an observer on the embankment, and so doesn't exist either for a person on the train. Since the light flashes travel at the same speed c on the train, the only conclusion left is that the flashes coudn't have occured at the same time. And this highlights the relativity of simultaneity in a simple way as Einstein was trying to do here: Events which are simultaneous in one frame are not so in any other frame.&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="1345" LastEditorDisplayName="" LastEditDate="2011-07-22T18:34:37.233" LastActivityDate="2011-07-22T18:34:37.233" />
  <row Id="12625" PostTypeId="1" CreationDate="2011-07-22T20:38:44.230" Score="3" ViewCount="93" Body="&lt;p&gt;If Hubble's constant is $2.33 \times 10^{-18} \text{ s}^{-1}$ and the earth orbits the sun with average distance of 150 million kilometers; Does that mean the earth's orbital radius increases approximately $11\text{ m}/\text{year}$?  Does the earth's angular momentum change?  If so, where does the torque come from?  If the angular momentum doesn't change, does the earth's orbital velocity (length of a year) change?  If so, where does the lost kinetic energy go?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Aside: the 11 meters per year figure comes from Hubble expansion of space the distance of the earth's orbital radius integrated over an entire year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(2.33 \times 10^{-18}\text{ s}^{-1}) (1.5 \times 10^{11} \text{ m}) (3.15 \times 10^7 \text{ s}/\text{year}) = 11 \text{ m}/\text{year}$$&lt;/p&gt;&#xA;" OwnerUserId="4639" LastEditorUserId="4639" LastEditorDisplayName="" LastEditDate="2011-07-23T00:03:49.697" LastActivityDate="2011-07-23T18:07:49.997" Title="How does Hubble's constant affect the Earth's orbit" Tags="&lt;angular-momentum&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="12626" PostTypeId="2" ParentId="12625" CreationDate="2011-07-22T20:58:04.670" Score="3" ViewCount="" Body="&lt;p&gt;No. Hubble's constant roughly says how the distance between two objects at rest with the universe grows. It does not say that the distant between everything is growing - the size of the hydrogen atom is not increasing. (My size is increasing, but from dietary rather than cosmological sources.) The size of objects and orbits are maintained by a balance of forces (classically). To whatever extent one can think of the expansion of the universe as pushing the Earth and Sun apart, it is already taken into account in setting the Earth's orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The change in the Hubble constant can effect the orbit, see the paper linked by Ben Crowell. But just taking the Hubble constant and multiplying it by the Earth's radius, as I believe you have done, does not give you anything sensible.&lt;/p&gt;&#xA;" OwnerUserId="3970" LastEditorUserId="3970" LastEditorDisplayName="" LastEditDate="2011-07-22T22:33:35.030" LastActivityDate="2011-07-22T22:33:35.030" CommentCount="12" />
  <row Id="12627" PostTypeId="1" CreationDate="2011-07-22T21:01:59.553" Score="3" ViewCount="177" Body="&lt;p&gt;I am currently stuck with the following problem: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; _________________________         _________________________&#xA;|                 |       |    \  |                         | &#xA;|        V1       |   V2  | ====&amp;gt; |        V1+V2            |&#xA;|                 |       |    /  |                         |      &#xA;|_________________|_______|       |_________________________| &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In a box there are two different gases in two compartments seperated by a solid wall. The Temperature $T$ of both gases is the same and both gases consist of $1mol$ particles. Therefore on the left side of the box, you have a volume of $V_1$ with a pressure of $p_1$ and on the right hand side we have volume $V_2$ and pressure $p_2$. No we remove the separating wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets first assume that we have to different gases on the left and on the right side. My goal is to calculate the change of entropy which occurs due to the removal of the wall. My first Ansatz was:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta S_1 = nR\ln(\frac{V_1+V_2}{V_1})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta S_2 = nR\ln(\frac{V_1+V_2}{V_2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\Delta S$ is the change of entropy of the system, $n$ the number of mols and $R$ the universal gas constant. The total change of entropy would now be the sum of both:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta S_{tot} = S_1 + S_2 = nR\ln(\frac{(V_1+V_2)^2}{V_1V_2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a while I thought about the problem in a different way: First I move the wall isothermally so that $V_1=V_2$. The entropy change due to the movement would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta S_{mov1} = nR\ln(\frac{V_1+V_2}{2V_1})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta S_{mov2} = nR\ln(\frac{V_1+V_2}{2V_2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta S_{mov} = S_{mov1} + S_{mov2} = nR\ln(\frac{(V_1+V_2)^2}{4V_1V_2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now i remove the wall and get a entropy change due to the mixing of the gasses:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta S_{rem} = R(n_1 ln(\frac{n_1+n_2}{n_1}) + n_2 \ln(\frac{n_1+n_2}{n_2})) = 2nR\ln(2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta S_{tot} =  \Delta S_{rem} + \Delta S_{mov} =  nR\ln(\frac{(V_1+V_2)^2}{4V_1V_2}) + 2nR\ln(2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is clearly different from the result I got from my approach in the first Ansatz. However since entropy is a state variable this should not happen. No I am stuck. I tend to the second result, but I am very far from sure. It would be great if you could explain which ansatz is correct (if any^^).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover if I had identical gases in $V_1$ and $V_2$, how would the entropy behave? Since the pressure changes after the removal of the wall, I would say that the entropy also changes. Just like in my second Ansatz, however without the mixing term. Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;||~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Oh, dear. I just realized that $2nRln(2)$ can be written as $nRln(4)$. Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$nR\ln(\frac{(V_1+V_2)^2}{4V_1V_2}) + 2nR\ln(2) = nR\ln(\frac{(V_1+V_2)^2}{V_1V_2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so both ansatzes give the same result. Since this is cleared out, my second question remains. In case of two identical gases does the entropy change according to my second ansatz (without the mixing entropy)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ftiaronsem&lt;/p&gt;&#xA;" OwnerUserId="393" LastEditorUserId="393" LastEditorDisplayName="" LastEditDate="2011-07-23T06:41:29.630" LastActivityDate="2011-08-26T08:40:03.370" Title="Entropy of two expanding and mixing ideal gases" Tags="&lt;thermodynamics&gt;&lt;homework&gt;&lt;entropy&gt;" AnswerCount="2" />
  <row Id="12628" PostTypeId="2" ParentId="12611" CreationDate="2011-07-22T21:15:51.713" Score="0" ViewCount="" Body="&lt;p&gt;First suppose that the Hamiltonian $H(t) = H_0 + H_I(t)$ can be decomposed into free and interaction parts. It can be shown (I won't derive this equation here) that the retarded Green function for $H(t)$ obeys the equation&#xA;$$G^{(+)}(t, t_0) = G_0^{(+)}(t, t_0) - {i \over \hbar} \int_{-\infty}^{\infty} {\rm d} t&amp;#39; G_0^{(+)}(t,t&amp;#39;) H_I(t&amp;#39;) G^{(+)}(t&amp;#39;, t_0)$$&#xA;where $G_0^{(+)}$ is the retarded Green function for $H_0$.&#xA;Letting this equation act on a state $\left| \psi(t_0) \right&amp;gt;$ this becomes&#xA;$$\left| \psi(t) \right&amp;gt; = \left| \varphi(t) \right&amp;gt;  - {i \over \hbar} \int_{-\infty}^{\infty} {\rm d} t&amp;#39; G_0^{(+)}(t,t&amp;#39;) H_I(t&amp;#39;)\left| \psi(t&amp;#39;) \right&amp;gt; $$&#xA;where $\varphi(t) = G_0^{(+)}(t,t&amp;#39;) \left| \psi(t_0) \right&amp;gt;$.&#xA;Now, we suppose that until $t_0$ there is no interaction and so we can write $\left |\psi(t_0) \right&amp;gt;$ as superposition of momentum eigenstates&#xA;$$\left| \psi(t_0) \right&amp;gt; = \int {\rm d}^3 \mathbf p a(\mathbf p) e^{-{i \over \hbar} E t_0} \left| \mathbf p \right&amp;gt;.$$&#xA;A similar decomposition will also hold for $\left| \phi(t) \right&amp;gt;$. This should inspire us in writing $\left| \psi(t) \right &amp;gt;$ as&#xA;$$\left| \psi(t) \right&amp;gt; = \int {\rm d}^3 \mathbf p a(\mathbf p) e^{-{i \over \hbar} E t} \left| \psi^{(+)}_{\mathbf p} \right&amp;gt;$$&#xA;where the states $\left| \psi^{(+)}_{\mathbf p} \right&amp;gt;$ are to be determined from the equation for $\left|\psi(t) \right&amp;gt;$. Now, the amazing thing (which I again won't derive due to the lack of space) is that these states are actually eigenstates of $H$:&#xA;$$H \left| \psi^{(+)}_{\mathbf p} \right&amp;gt; = E \left| \psi^{(+)}_{\mathbf p} \right&amp;gt;$$&#xA;for $E = {\mathbf p^2 \over 2m}$ (here we assumed that the free part is simply $H_0 = {{\mathbf p}^2 \over 2m}$ and that $H_I(t)$ is independent of time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, one can derive advanced eigenstates from advanced Green function&#xA;$$H \left| \psi^{(-)}_{\mathbf p} \right&amp;gt; = E \left| \psi^{(-)}_{\mathbf p} \right&amp;gt;.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in one dimension and for an interaction Hamiltonian of the form $\left&amp;lt; \mathbf x \right| H_I \left| \mathbf x&amp;#39; \right&amp;gt; = \delta(\mathbf x - \mathbf x&amp;#39;) U(\mathbf x)$ it can be further shown that&#xA;$$\psi^{(+)}_p \sim  &#xD;&#xA;\begin{cases}&#xD;&#xA;e^{{i \over \hbar}px} + A(p) e^{-{i \over \hbar}px} \quad x&amp;lt; -a \cr&#xD;&#xA;B(p)e^{{i \over \hbar}px} \quad x&amp;gt; a &#xD;&#xA;\end{cases}$$&#xA;where $a$ is such that the potential vanishes for $|x| &amp;gt; a$ and $A(p)$ and $B(p)$ are coefficients fully determined by the potential $U(x)$.&#xA;Similar discussion again applies for wavefunctions $\psi^{(-)}_p$. Thus we have succeeded in reducing the dynamical problem into a stationary problem by writing the non-stationary states $\psi(t, x)$ in the form of stationary $\psi^{(+)}_p(x)$.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-07-22T21:22:44.297" LastActivityDate="2011-07-22T21:22:44.297" CommentCount="2" />
  <row Id="12629" PostTypeId="1" CreationDate="2011-07-22T18:10:59.100" Score="10" ViewCount="396" Body="&lt;p&gt;When near high tension power lines, particularly after a good rain, the lines themselves emit a buzzing noise. A similar noise can be heard coming out of the electric meters attached to my apartment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard before that this is supposedly from the 60Hz AC current that's running through the lines -- namely, that the buzz is the same 60Hz which is in the lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm skeptical of this though for a couple of reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I don't see any reason the change in electricity would somehow be audible.&lt;/li&gt;&#xA;&lt;li&gt;The noise subjectively sounds relatively high pitch. 60Hz would sound extremely low pitched -- it's near the base of human hearing of 20Hz (typical).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What is the actual cause of that buzzing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I just spent some time playing with a tone generator and the noise I hear from these things sounds closest to 120Hz using a square or triangle wave. (Oddly, not a sine wave, as I would have expected) Perhaps that helps?&lt;/p&gt;&#xA;" OwnerUserId="696" LastEditorUserId="696" LastEditorDisplayName="" LastEditDate="2011-07-23T00:21:29.363" LastActivityDate="2011-07-23T11:11:11.513" Title="Why do power lines buzz?" Tags="&lt;electricity&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="12630" PostTypeId="2" ParentId="12629" CreationDate="2011-07-22T20:50:43.800" Score="0" ViewCount="" Body="&lt;p&gt;Well, it's only a &lt;a href=&quot;http://en.wikipedia.org/wiki/Mains_hum&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; reference, but it says for high voltage power lines, it can be corona discharge. For transformers or other objects containing coils, it can be magnetic. Seems to me, for things carrying AC, it's pretty hard to make them &lt;em&gt;not&lt;/em&gt; hum.&#xA;If you're wondering how it sounds, that page has some recordings, but you might have to double the frequency (go up one octave), because for a 60hz signal, the voltage and/or current reaches a maximum 120 times per second.&lt;/p&gt;&#xA;" LastActivityDate="2011-07-22T21:04:39.563" CommentCount="4" />
  <row Id="12631" PostTypeId="2" ParentId="12614" CreationDate="2011-07-22T22:06:42.123" Score="1" ViewCount="" Body="&lt;p&gt;I can't entirely make sense of some the claims in that letter, but this looks like a version of solving for the bound states of potential which is fairly standard. It's in Griffiths Quantum Mechanics I think. I'll write it down here for you and maybe you can piece together the rest. Anyway it goes like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're trying to find the values of $E$ for which the radial Schrodinger equation &#xA;$-\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial\psi}{\partial r} )+ \frac{\ell(\ell+1)}{r^2}\psi + V(r)\psi = E\psi$&#xA;has normalizable solutions. On the other hand that's a second order ODE for any $E$ so it has to have two solution - they're just not going to be normalizable. Specifically they are going to blow up quite badly as $r\rightarrow\infty$. So we are just going to guess a value of $E$, numerically solve that equation by finite differences and see if the solutions blow up at infinity or not. When we found a value of $E$ where they don't blow up we know we found a bound state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the particular issues that come up when we are looking for the ground state. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$R &amp;lt; 0$ means your wavefunction has a node (it changes sign) and $R &amp;gt; 1$ means your wavefunction is increasing. It is the case that the ground state wavefunction has no nodes so you must be looking in the vicinity of one of the excited states. As for $R &amp;gt; 1$, while asymptotically your wavefunction has to decrease, I don't see why its not possible that it could increase at some point. So this does not seem to be valid termination rule in general (though given a particular potential you might be able to say more).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;See that $R(h(l+1))$ does not depend on $R(hl)$ since the prefactor is zero. After that $R(x)$ depends only on $R(x-h)$ so $R(hl)$ never enters. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This method is very computationally cheap so why not just try a lot of energies? You can always bound the energy from above by the standard variational methods. Also just dimensional analysis could give you the order of magnitude.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3970" LastActivityDate="2011-07-22T22:06:42.123" CommentCount="2" />
  <row Id="12633" PostTypeId="2" ParentId="12629" CreationDate="2011-07-22T23:01:39.033" Score="0" ViewCount="" Body="&lt;p&gt;This is the only explanation I can think off : Lorentz force due to earth's magnetic field&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's do a rough estimation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we have some power line at 1000A (should be ok for 100mm^2 wire), 200 meters between towers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F=200 meters * 1000A * 40 microteslas (some average Earth's magnetic field) = 8 newtons (Not looking at angle at the moment, depend on position &amp;amp; location)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While 8 newtons itself is not much to significantly move heavy wires at 50/60 hertz (this sample wire would weight ~180kg), if line have any mechanical resonance at multiple of 50/60 hertz, vibration will amplify and will be significant to make audible sound.&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-07-22T23:01:39.033" CommentCount="4" />
  <row Id="12634" PostTypeId="1" AcceptedAnswerId="12635" CreationDate="2011-07-22T23:26:25.867" Score="2" ViewCount="146" Body="&lt;p&gt;When a shuttle is in orbit, it is essentially rotating around the &quot;centre&quot; of the Earth at a great speed. So why does there seem to be no centrifugal force sticking them to the 'ceiling' of the shuttle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because the shuttle is not actually being rotated around a point but rather continuously falling or something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="4641" LastActivityDate="2011-08-08T20:19:41.453" Title="Why don't astronauts in orbit get stuck to the &quot;ceiling&quot;?" Tags="&lt;gravity&gt;&lt;orbital-motion&gt;&lt;centrifugal-force&gt;" AnswerCount="2" />
  <row Id="12635" PostTypeId="2" ParentId="12634" CreationDate="2011-07-23T00:09:56.930" Score="10" ViewCount="" Body="&lt;p&gt;By definition an orbit occurs when gravity balances with the &quot;centrifugal&quot; force. It is essentially a &lt;strong&gt;free fall&lt;/strong&gt; situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is the same reason why you don't get stuck to the ceiling of a free falling elevator. Both the spacecraft and the occupants are moving &lt;strong&gt;in-sync&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-07-23T00:09:56.930" CommentCount="3" />
  <row Id="12636" PostTypeId="1" CreationDate="2011-07-23T00:27:10.580" Score="1" ViewCount="119" Body="&lt;p&gt;Can Heisenberg's Uncertainty principle be rewritten in terms of energy density&lt;/p&gt;&#xA;&#xA;&lt;p&gt;writing&#xA;$$\Delta E \Delta T \geqslant  \hbar/2$$&#xA;in factors of energy density &#xA;$\Delta \sigma \text{  }= \frac{3\Delta E}{4\pi  r^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get &#xA;$$\Delta \sigma  \frac{3\text{$\Delta $T}}{4\pi  r^3} \geqslant  \hbar/2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;does $$\frac{3\text{$\Delta $T}}{4\pi  r^3}$$&#xA;represent something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or does it make more sense to write&#xA;$$\text{$\Delta $x} \frac{\text{$\Delta $E}}{c^2} \geqslant  \frac{\hbar}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT after noting Willie's comment I've corrected the form to $\frac{\hbar}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, now I see c is constant and can be moved to the right side  - is it mathematical correct to write&#xA;$$\text{$\Delta $x} \text{$\Delta $E} \geqslant  \frac{\hbar  c^2}{2}$$&#xA;or does this violate the Cauchy–Schwarz inequality? This is the question I was trying to understand at the beginning.&lt;/p&gt;&#xA;" OwnerUserId="3588" LastEditorUserId="3588" LastEditorDisplayName="" LastEditDate="2011-07-24T00:13:51.280" LastActivityDate="2011-08-19T07:17:00.293" Title="Heisenberg's Uncertainty Forms" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="12637" PostTypeId="2" ParentId="12629" CreationDate="2011-07-23T02:47:05.240" Score="5" ViewCount="" Body="&lt;p&gt;The reason for this is something known as &quot;magnetostriction&quot;, which is strain induced in a magnetic lattice due to the magnetic nature of a material.  There is a pretty good explanation of the buzzing noises in transformers &lt;a href=&quot;http://grieg.gotdns.com/blog/?p=474&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, but the summary is that the ferromagnetic domains in the transformer core are subjected to the 60Hz (in the US) oscillations of the magnetic field due to the AC current.  Because of the electrical cycle, there are two impacts on the core per AC oscillation, so the strain is changing at 60 Hz, which produces the 120 Hz noise that we can identify a &quot;great B&quot; note.&lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2011-07-23T02:47:05.240" />
  <row Id="12638" PostTypeId="2" ParentId="12636" CreationDate="2011-07-23T03:23:12.633" Score="0" ViewCount="" Body="&lt;p&gt;it probably depends on what sort of problem you are trying to solve, in general if something is equal to something else, then it can be substituted into an inequality.&lt;/p&gt;&#xA;" OwnerUserId="4570" LastActivityDate="2011-07-23T03:23:12.633" />
  <row Id="12639" PostTypeId="2" ParentId="12614" CreationDate="2011-07-23T08:51:46.300" Score="1" ViewCount="" Body="&lt;p&gt;Bebop answered all the original questions already.&#xA;However it might be useful to see an implementation.&#xA;The following code calculates the energies of the 1s state with lambda=0 and lambda=1 with varying values of $h$ (see table 1 in the paper).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I start with a value of 0.5 for $r=h(l+1)$ and then use the recursive equation to calculate R(r) for increasing r in the function calc-domain. The function returns, when R is either smaller then 0 or bigger than 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function bracket-E calls calc-domain with different guesses for E and tries to home in onto the value where R stays within $[0,1]$ for the longest distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In table 1 the paper gives -0.4815 as the value for E for lambda=0 and h=0.4.&#xA;This code returns -0.48145604506134987.&#xA;Note that the calculation was done using rational bignums and should be correct to 42 digits.&#xA;I rounded the result into a double for convenience.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;;see 0305-4470_10_6_001.pdf, runs in Steel Bank Common Lisp http://www.sbcl.org/&#xA;&#xA;(defun Rr (r h l E Rr-h lam)&#xA;  &quot;Caculate R at position r using a value of R at r-h and a guess E.&quot;&#xA;  (let* ((V (- (* lam r) (/ r)))&#xA;     (lhs (* 2 r (+ 1 (* h h (- V E))))))&#xA;    (/ (- lhs (/ (- r (* h (1+ l)))&#xA;         Rr-h))&#xA;       (+ r (* h (1+ l))))))&#xA;&#xA;(defun calc-domain (E h l lam)&#xA;  &quot;Calculate ratio function over the domain from r0 to big r,&#xA;if Rhl becomes negative, the guess E was too high&quot;&#xA;  (let* ((r0 (* h (1+ l)))&#xA;     (Rhl .5s0))&#xA;    (loop while (&amp;lt; (- (* Rhl Rhl) Rhl) 0) &#xA;       do&#xA;     (setf Rhl (Rr r0 h l E Rhl lam))&#xA;     (incf r0 h))&#xA;    Rhl))&#xA;&#xA;(defun bracket-E (left right h l lam)&#xA;  (let* ((rleft (calc-domain left h l lam))&#xA;     (rright (calc-domain right h l lam))&#xA;     (center (/ (+ left right) 2))&#xA;     (rcenter (calc-domain center h l lam)))&#xA;    (when (&amp;lt; (abs (- left right)) 1e-42)&#xA;      (return-from bracket-E (* 1d0 center)))&#xA;    (cond &#xA;      ((&amp;lt; rcenter 0 rleft)&#xA;       (bracket-E left center h l lam))&#xA;      ((&amp;lt; rright 0 rcenter)&#xA;       (bracket-E center right h l lam))&#xA;      (t (break &quot;error: There is no zero between ~a, try increasing the interval&quot; &#xA;        (list left right))))))&#xA;&#xA;;; reproduce values for the 1s state of table 1:&#xA;(loop for h in '(4/10 2/10 1/10 5/100 1/1000) collect&#xA;     (bracket-E -2 2 h 0 0))&#xA;&#xA;;; lam=0&#xA;;; (-0.48145604506134987d0 -0.49509748816490173d0 -0.49875694513320923d0&#xA;;;     -0.4996879608376049d0 -0.49949414802320075d0)&#xA;&#xA;;; lam=1&#xA;;; (0.6055558981315698d0 0.5841694474220276d0 0.5794421434402466d0&#xA;;;            0.5782957077026367d0 0.5777209093467737d0)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4306" LastActivityDate="2011-07-23T08:51:46.300" CommentCount="7" />
  <row Id="12640" PostTypeId="2" ParentId="12629" CreationDate="2011-07-23T09:41:13.973" Score="3" ViewCount="" Body="&lt;p&gt;&quot;O sancta simplicitas&quot;, what an amount of nonsense a bad Wikipedia page can produce.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Thomas O and Mike Dunlavey &quot;had&quot; it, but then...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &quot;hum&quot; (wrong word, better call it hiss) coming from &#xA;HV power lines is corona discharge, nothing else. &#xA;A DC corona is a straightforward hiss (blueish) when it comes from&#xA;a power line, it is modulated by the 50 or 60 Hz of the voltage. &#xA;This is the spectrum in the wiki page on top. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The wiki page is very bad, inasmuch not saying what kind of source &#xA;is behind that spectrum, and in general mixing up all sorts of &#xA;hum associated with power mains. The sound examples are about &#xA;transformer hum or from a bad audio amplifier. &#xA;(Sounding is predominantly 100 or 120 Hz from frequency doubling) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;HV Power line wires are made from aluminium, but to resist mechanical &#xA;stress they contain a steel cable core. (6 aluminium cables areound a steel cable)&#xA;But this is not the cause for the hiss asked for here. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If some electrostatic, magnetic or magnetostrictive forces were relevant, &#xA;the hiss would not depend on weather, as it does. It's strongest at &#xA;foggy or drizzle conditions. In case You posess a neon sign or a ignition &#xA;transformer of about 10 kV secondary, You can easily produce the &quot;hum&quot; &#xA;by connecting the terminals to some fine wires. A sandwich of a glas plate &#xA;and two metal grids (like in a ozone generator) works very loud. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the sound You produced with that tone generator is similar, because it has the &#xA;mains frequency basis and (from recangular or triangular) a lot of high &#xA;harmonics making the &quot;hiss&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-07-23T11:11:11.513" LastActivityDate="2011-07-23T11:11:11.513" CommentCount="2" />
  <row Id="12641" PostTypeId="1" AcceptedAnswerId="12646" CreationDate="2011-07-23T10:56:37.323" Score="2" ViewCount="93" Body="&lt;p&gt;Most audio signal are described by a curve having two parameters: The amplitude and the frequency (amount of time from positive to negative amplitude value).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to understand is how the sound level and absolute pitch are encoded to these two values. How would amplitude and frequency change if I'd change volume or pitch?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance!&lt;/p&gt;&#xA;" OwnerUserId="4643" LastActivityDate="2011-07-23T16:49:46.467" Title="Understand audio signals" Tags="&lt;sound&gt;" AnswerCount="4" />
  <row Id="12642" PostTypeId="2" ParentId="12641" CreationDate="2011-07-23T11:07:09.860" Score="0" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Most audio signal are described by a curve having two parameters: The&#xA;  amplitude and the frequency&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is wrong. The parameters of &quot;audio curves&quot; are time &#xA;and amplitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether there is any &quot;frequency&quot; hidden in the curves, &#xA;is a matter of Fourier Transformation Analysis &#xA;(normally done by Your ear/brain, I hope).&#xA;&quot;Sound level&quot; is a logarithmic measure of amplitude. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-07-23T11:13:32.243" LastActivityDate="2011-07-23T11:13:32.243" CommentCount="3" />
  <row Id="12643" PostTypeId="2" ParentId="12641" CreationDate="2011-07-23T11:28:06.353" Score="2" ViewCount="" Body="&lt;p&gt;The waveform in the image below is the signal, that the microphone picks up.&#xA;When I whistle, the pressure of the air fluctuates in this manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the each line in the picture below the 2048 samples of the waveform were Fourier transformed and the magnitude is plotted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I whistled, I changed pitch between two values and you can nicely see this in the waterfall plot of the time-slice Fourier transform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The screenshot was taken while I emitted a low pitch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Volume and pitch are just axes in those plots.&#xA;The software is from &lt;a href=&quot;http://www.baudline.com&quot; rel=&quot;nofollow&quot;&gt;http://www.baudline.com&lt;/a&gt;.&#xA;It's not free software but you can download and run it on your Linux.&#xA;I don't know anything easier/better and suggest you try it, its fun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can tune a guitar, figure out the quality of MP3 files or your recording equipment, see where the instruments are in stereo recordings. The coolest thing I did with this was to display an image that I encoded into sound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/NvcOH.png&quot; alt=&quot;screenshot of baudline with me whistling&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4306" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-07-23T11:39:04.410" LastActivityDate="2011-07-23T11:39:04.410" CommentCount="8" />
  <row Id="12644" PostTypeId="1" CreationDate="2011-07-23T14:19:50.400" Score="3" ViewCount="78" Body="&lt;p&gt;I have a question converning multiplets when describing atoms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let &#xA;$H=\sum\limits_{k=1}^{N} (p_{k}^2 - \frac{Z}{|x_{k}|} + \sum\limits_{i&amp;lt;k}^{1..N} \frac{1}{|x_{i} - x_{k}|}$ be a (self-adjoint) operator on the Hilbertspace of totally antisymmetrized wave functions of N particles H^(N)_{a}. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my notes it is written that SO(3) x SU(2) is a symmetry group of this hamiltonian. &#xA;The group SO(3) x SU(2) consists of pairs (R,A)$\in SO(3) x SU(2)$ and has a unitary representation U on H^(N)_{a} given by: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(U(R,A)\phi)(x_{1} s_{1} ,...., x_{N} s_{n}) = As_{1}s&amp;#39;_{N} ... A s_{N} s&amp;#39;_{N}\phi(R^{-1}x_{1} s_{1} s&amp;#39;_{1} ... R^{-1} x_{N} s&amp;#39;_{N}) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here comes my problem. I don't really understand this definition. &#xA;It's not given in my notes what the s' are and I don't understand what $As_{1}s&amp;#39;_{N}$ means...A vague guess might be that you multiply the vector $s_{1}$ in $C^{2}$ is multiplied by the matrix A in SU(2) and then scalar mulitplied by $s&amp;#39;_{N}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Butit's not clear to me..Can anyone help me to understand this definition better?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to that I thought that if a group G is a symmetry group of a Hamiltonian H then it means that [H,U]=0 for ANY representation U of the group G. &#xA;But in the above example one only gives one particular representation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be really happy if someone could help me. &#xA;Thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="4520" LastActivityDate="2011-07-23T15:34:22.413" Title="SO(3) x SU(2) Symmetry of the Hamiltonian" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="12645" PostTypeId="2" ParentId="12644" CreationDate="2011-07-23T15:23:58.133" Score="1" ViewCount="" Body="&lt;p&gt;Hopefully the $SO(3)$ part of the symmetry is clear: it's just a rotation that acts on the position part of the wavefunction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $SU(2)$ part comes from the fact that these are (supposedly) spin 1/2 particles. Since each of these is described by two spin states we act on them by using a special unitary matrix acting on a two-dimensional complex space ${\mathbb C}^2$. But this is precisely a matrix from $SU(2)$ (in other words, this is a tautological representation where the matrix is represented by itself). Now, when acting on $N$ particles we take a tensor product of these representations. Therefore we get a representation ${\mathbb C}^2 \otimes \cdots \otimes {\mathbb C}^2$. I think what you've written must contain some typo but generally it has a correct structure of containing the matrix $A$ $N$ times and each time it acts on a pair of two spin-vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the second question: it's certainly not true that the Hamiltonian commutes with any representation. This is because not every Hilbert space carries every representation of the group. It only contains some of them. The general form of the theorem (due to Wigner) is the following: it the system is invariant under a symmetry then there exists an unitary of anti-unitary operator that represents it on the Hilbert space. In other words, it only asserts the existence of &lt;em&gt;some&lt;/em&gt; representation. Obviously then, since most of the representations aren't even there they surely can't commute with Hamiltonian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you were confused by some standard example like Coulomb potential problem. In this case every representation of $SO(3)$ (indexed by a non-negative integer) is realized on the Hilbert space. But it's easy to construct a Hilbert space where this fails, e.g. the ${\mathbb C}^2$ above only carries spin 0 and spin 1/2 representations of $SU(2)$ and no other (since reps with higher spin are more than two-dimensional).&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-07-23T15:34:22.413" LastActivityDate="2011-07-23T15:34:22.413" />
  <row Id="12646" PostTypeId="2" ParentId="12641" CreationDate="2011-07-23T15:33:12.367" Score="0" ViewCount="" Body="&lt;p&gt;We can think of a sound signal either as a function of time or as a function of frequency. The two ways are related by what is called a Fourier transform, so they are often written as $f(t)$ and as $\tilde f(\omega)$, using the same letter, but with the &quot;tilde&quot; above one of them indicating that one is the Fourier transform of the other. From these functions it is also possible to construct other functions of both time and frequency, such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wigner_quasi-probability_distribution#Uses_of_the_Wigner_function_outside_quantum_mechanics&quot; rel=&quot;nofollow&quot;&gt;Wigner function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would take &quot;pitch&quot; as more-or-less another name for frequency. Audacity help says &quot;Generally synonymous with the fundamental frequency of a note, but in music, often also taken to imply a perceived measurement that can be affected by overtones above the fundamental.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;squared amplitude&lt;/em&gt; of a function is the square of the function, properly speaking the norm, written as $|f(t)|^2$ and $|\tilde f(\omega)|^2$. This function of time or of frequency is always positive or zero, whereas the original function can be positive or negative (or it can be a 2-valued function, which in the time coordinates represent both the signal and its rate of change at a given time). It's common to call the &quot;squared amplitude&quot; the amplitude, just to confuse matters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sound level sometimes refers to the logarithm of the squared amplitude, as Georg says, but sometimes not. Wikipedia, for example, suggests that a &lt;a href=&quot;http://en.wikipedia.org/wiki/Sound_level_meter&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;loudness meter&lt;/em&gt;&lt;/a&gt; reports on a logarithmic scale, whereas &lt;em&gt;sound level&lt;/em&gt; might mean either linear or logarithmic scale. One test for what is meant is whether the signal level is followed by &quot;dB&quot;, which will always indicate a logarithmic scale. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately I'd say that it's the mathematical relationships between these objects that one tries to understand, but a lot can be understood just by experimenting with something like Audacity (which I think is simpler to use than the software mentioned by @whoplisp, although also less powerful), using the &quot;Analyze-&gt;Plot Spectrum&quot; option for various sounds. Here's a plot for me whistling, apparently at about 1300Hz (the fundamental frequency) in the 0.026 seconds that I analyzed here, but with various higher frequencies also represented.&#xA;&lt;img src=&quot;http://i.stack.imgur.com/kcvNw.png&quot; alt=&quot;Audacity screen shot&quot;&gt;&#xA; When reading something about sound or music, there are various technical usages that are not very standardized, and one often has to figure out what's meant by the overall context. There are many different expert groups that have a strong interest in signal analysis of various kinds, and they have often developed specialized languages appropriate to their own interests.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastEditorUserId="1588" LastEditorDisplayName="" LastEditDate="2011-07-23T15:50:08.267" LastActivityDate="2011-07-23T15:50:08.267" CommentCount="2" />
  <row Id="12647" PostTypeId="1" CreationDate="2011-07-23T15:33:25.140" Score="6" ViewCount="263" Body="&lt;p&gt;Can the physics complications of introducing spin 3/2 Rarita-Schwinger matter be put in geometric (or other) terms readily accessible to a mathematician?&lt;/p&gt;&#xA;" OwnerUserId="4477" LastActivityDate="2011-08-23T19:33:44.313" Title="What are the mathematical problems in introducing Spin 3/2 fermions?" Tags="&lt;quantum-field-theory&gt;&lt;particle-physics&gt;&lt;mathematical-physics&gt;&lt;supersymmetry&gt;&lt;fermions&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="12648" PostTypeId="2" ParentId="12641" CreationDate="2011-07-23T16:49:46.467" Score="1" ViewCount="" Body="&lt;p&gt;Some folks have put a lot of work into these answers, but they are at the wrong level for the OP. The OP is asking basic questions about sound waves, and therefore will certainly have no idea about Fourier analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you raise the volume, the amplitude goes up. Physically, the amount of energy in the wave is proportional to the square of the amplitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you raise the pitch, the frequency goes up. In terms of music, if you sing a scale like CDEFGABC, the ratio of the frequencies of the two C's is 2. (This is called an octave in music.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the frequency is not the &quot;amount of time from positive to negative amplitude value.&quot; The time from a positive peak to a negative peak of a sine wave is T/2, half the period. The frequency is 1/T.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-07-23T16:49:46.467" CommentCount="2" />
  <row Id="12649" PostTypeId="2" ParentId="12589" CreationDate="2011-07-23T16:54:03.733" Score="2" ViewCount="" Body="&lt;p&gt;It's a really complicated relationship that depends on the metallicity of the star. There is a paper that &lt;em&gt;does&lt;/em&gt; show this though: See &lt;a href=&quot;http://adsabs.harvard.edu/abs/1992A&amp;amp;AS...96..269S&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;New grids of stellar models from 0.8 to 120 solar masses at Z = 0.020 and Z = 0.001&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the Geneva Grids:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://obswww.unige.ch/~mowlavi/evol/stev_database.html&quot; rel=&quot;nofollow&quot;&gt;http://obswww.unige.ch/~mowlavi/evol/stev_database.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An extensive and homogenous database of stellar evolution models for&#xA;  masses between 0.8 and 120 solar masses and metallicities from Z=0.001&#xA;  to 0.1 is available. In general the models include evolutionary phases&#xA;  from the main sequence up to either the end of carbon burning for&#xA;  massive stars, the early asymptotic giant branch phase for&#xA;  intermediate-mass stars, or core helium flash for low-mass stars.&#xA;  Pre-main sequence tracks, both canonical (i.e. evolved at constant&#xA;  mass) and accretion scenarios are also provided, as well as horizontal&#xA;  banches for low-mass stars. Predictions regarding the spectral&#xA;  evolution of massive stars can further be obtained from the so-called&#xA;  &quot;combined stellar structure and atmosphere models&quot; (CoStar).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In addition to the evolutionary grids we also provide Fortran codes&#xA;  for the calculation of isochrones and stellar population burst models.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Finally, references are also given to the serie of papers on stellar&#xA;  models with rotation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are also older ad hoc models applied to the Sun: see &lt;a href=&quot;http://adsabs.harvard.edu/full/1981SoPh...74...21G&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/full/1981SoPh...74...21G&lt;/a&gt; and &lt;a href=&quot;http://articles.adsabs.harvard.edu/full/1984SSRv...38..243S&quot; rel=&quot;nofollow&quot;&gt;http://articles.adsabs.harvard.edu/full/1984SSRv...38..243S&lt;/a&gt;. They were referenced in 2011 publications though (like Pierrehumbert's 2011 Neoproterozoic Climate paper)&lt;/p&gt;&#xA;" OwnerUserId="1576" LastActivityDate="2011-07-23T16:54:03.733" />
  <row Id="12650" PostTypeId="1" CreationDate="2011-07-23T17:09:28.823" Score="0" ViewCount="107" Body="&lt;p&gt;This question is rather historical one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;kilometres can be defined in metres, metres in centimetres, centimetres in millimetres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There must be some elementary unit (like millimetre or smth.) which cannot be defined in smaller units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is : How does this elementary unit came into being??&#xA;e.g. How did scientists decide about exact distance between point A and point B which is considered millimetre??&lt;/p&gt;&#xA;" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-23T17:42:27.560" LastActivityDate="2011-07-23T18:54:32.970" Title="units of measure" Tags="&lt;si-units&gt;&lt;units&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="12651" PostTypeId="2" ParentId="12650" CreationDate="2011-07-23T17:14:44.340" Score="1" ViewCount="" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Meter&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; (and I have seen it elsewhere) one part in 10^7 of the meridian through Paris from the equator to the north pole won out over a pendulum with a half period of a second.  Nowadays it is based on the speed of light and the definition of the second.&lt;/p&gt;&#xA;" OwnerUserId="3257" LastActivityDate="2011-07-23T17:14:44.340" />
  <row Id="12652" PostTypeId="2" ParentId="12650" CreationDate="2011-07-23T17:17:36.777" Score="2" ViewCount="" Body="&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/SI_base_unit&quot; rel=&quot;nofollow&quot;&gt;SI base unit&lt;/a&gt; (and perhaps also &lt;a href=&quot;http://en.wikipedia.org/wiki/History_of_the_metric_system&quot; rel=&quot;nofollow&quot;&gt;history of the metric system&lt;/a&gt;) on Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, kilometres, centimetres, millimetres and other prefixed SI length units are defined as multiples or fractions of the base unit, which is the metre.  The definition of the metre itself has varied over time: Wikipedia has &lt;a href=&quot;http://en.wikipedia.org/wiki/History_of_the_metre&quot; rel=&quot;nofollow&quot;&gt;an article on that&lt;/a&gt; too.&lt;/p&gt;&#xA;" OwnerUserId="4646" LastActivityDate="2011-07-23T17:17:36.777" CommentCount="5" />
  <row Id="12653" PostTypeId="1" CreationDate="2011-07-23T17:30:13.730" Score="2" ViewCount="76" Body="&lt;p&gt;My car teacher instructs:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;if you drive over 40 kph, use first brake and then use the clutch&lt;/li&gt;&#xA;&lt;li&gt;if you drive under 40kph, use first the clutch and then use the brake&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I have no idea what kind of clutch the Toyota has, Wikipedia lists a number of clutches with different physical methods such as centrifugal clutch. So I am unsure whether there is one answer for all kind of clutches but it is then a challenge! Explain the logic between clutch braking and normal braking with the brake pedals. I am a bit lost with the instructor's guide, I like to slow down by changing the gear lower so not to consume so much fuel (but I am uncertain whether it is a bad idea). Anyway falsify the tip by the instruction, when is it wrong?&lt;/p&gt;&#xA;" OwnerUserId="623" LastActivityDate="2011-07-23T19:36:36.620" ClosedDate="2011-07-28T12:37:20.873" Title="Explain normal braking and clutch braking with motor vehicles such as Cars" Tags="&lt;centrifugal-force&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="12654" PostTypeId="2" ParentId="12625" CreationDate="2011-07-23T18:07:49.997" Score="0" ViewCount="" Body="&lt;p&gt;For objects smaller than cosmic scale, such as atoms, planets and solar systems, the electromagnetic and gravitational forces that hold them together are not changing (as far as we know) and so those objects do not change size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between galaxies, so widely separated, there's just gravity, and that tends to average out due to every galaxy being surrounded by other galaxies in all directions.  On a cosmic scale, galaxies are like a gas, with galaxies being the &quot;molecules&quot; and described by the idea gas equation.  To account for gravity and finite size of the galaxies, we might use the Van der Waals equation or some other variation, but that's beside the point, useful only for increasing accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hubble's constant describes the rate at which the &quot;container&quot; of the galactic gas is expanding, the way the density of galaxies decreases over time.  In an ordinary gas such as air, when in an expanding chamber, certainly the molecules are not expanding.  Likewise, neither are the galaxies changing their sizes, at least not for Hubble-related reasons.&lt;/p&gt;&#xA;" OwnerUserId="353" LastActivityDate="2011-07-23T18:07:49.997" CommentCount="1" />
  <row Id="12655" PostTypeId="2" ParentId="12653" CreationDate="2011-07-23T18:14:28.673" Score="0" ViewCount="" Body="&lt;p&gt;When not fed with gas, the motor compresses air and therefore decelerates the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Decoupling the motor from the wheels at fast driving speed by using the clutch increases fuel consumption (a small amount of gas is injected to keep the motor running) and causes more wear on the breaks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motor has to run at a certain rotation rate. If you break without clutch at slow speeds the electronics might inject extra fuel to prevent the motor from stopping.&lt;/p&gt;&#xA;" OwnerUserId="4306" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-07-23T19:36:36.620" LastActivityDate="2011-07-23T19:36:36.620" CommentCount="8" />
  <row Id="12656" PostTypeId="2" ParentId="8227" CreationDate="2011-07-23T18:24:59.570" Score="0" ViewCount="" Body="&lt;p&gt;Because your eyes are set left and right and how you relate to the mirror.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you turn a picture to reflect in a mirror, you usually turn it left or right.  If you turned it vertically, the up and down reflection will be opposite, but left and right will still be the same.&lt;/p&gt;&#xA;" OwnerUserId="4651" LastActivityDate="2011-07-23T18:24:59.570" />
  <row Id="12657" PostTypeId="1" AcceptedAnswerId="12663" CreationDate="2011-07-23T18:35:27.223" Score="5" ViewCount="139" Body="&lt;p&gt;I have noticed that when I microwave an ice cube it appears to melt more slowly than I would expect. For example, an equal volume of water starting at 0 deg C would probably be at boiling point before an ice cube that was at -15 deg C had melted. I realize there is enthalpy of fusion to take into account in the melting process but I believe there is more to it than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it a microwave oven works by exciting the water molecules in whatever is being cooked and if memory serves the frequency used is one that causes rotation of the molecule. Since the ice cube is solid I'm assuming the molecules aren't free to rotate and therefore the microwaves have a much reduced effect. In fact I'm wondering if a perfect single crystal of water would respond at all to being microwaved. Does this sound right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to rack my brain for a way of testing this theory but I can't think of a way of getting an perfectly dry ice cube into a microwave to see if anything happens. Even a tiny amount of surface water, caused from interaction with a warm atmosphere, would encourage melting.&lt;/p&gt;&#xA;" OwnerUserId="4650" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-07-24T12:51:26.867" LastActivityDate="2011-07-24T16:49:19.013" Title="ice in a microwave" Tags="&lt;water&gt;&lt;ice&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="12658" PostTypeId="1" CreationDate="2011-07-23T18:36:19.063" Score="0" ViewCount="47" Body="&lt;p&gt;If there had been no tsunami would the Japanese Fukushima nuclear plants be operating today? &lt;/p&gt;&#xA;" OwnerUserId="4651" LastActivityDate="2011-07-23T18:36:19.063" ClosedDate="2011-07-23T19:05:17.210" Title="Fukushima Nuclear Plant Running?" Tags="&lt;nuclear-physics&gt;&lt;nuclear-engineering&gt;" CommentCount="5" />
  <row Id="12659" PostTypeId="2" ParentId="11423" CreationDate="2011-07-23T18:45:22.843" Score="2" ViewCount="" Body="&lt;p&gt;Interesting question! This article &lt;a href=&quot;http://accessscience.com/content/Rydberg%20atom/596700&quot; rel=&quot;nofollow&quot;&gt;http://accessscience.com/content/Rydberg%20atom/596700&lt;/a&gt; says that &quot;The natural lifetime of an undisturbed Rydberg atom increases as $n^3$ for a given electronic angular momentum.&quot; That means that each state has a natural width $\Delta E_{w}\propto n^{-3}$. The energy levels go like $n^{-2}$, and differentiation tells us that the spacing between successive levels goes as $\Delta E_s\propto n^{-3}$. This means that $E_w/E_s$ is approximately independent of $n$ for large $n$. Since we know that $\Delta E_w/\Delta E_s &amp;lt;&amp;lt; 1$ for small $n$, it sounds to me like this continues to be true for large $n$. In other words, the states are still well defined and non-overlapping for arbitrarily large $n$. (Of course there will still be degeneracy within each $n$ value.) This is all for a hydrogen atom in free space with zero external electric or magnetic field and no background of blackbody photons.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-07-23T18:45:22.843" />
  <row Id="12660" PostTypeId="2" ParentId="12650" CreationDate="2011-07-23T18:54:32.970" Score="0" ViewCount="" Body="&lt;p&gt;While the metre (or meter if people prefer) is the SI unit of length I believe the poster is essentially is there a length after which nothing can be smaller. The shortest possible length is probably the Plank length which is a tiny 1.616252(81)×10−35 metres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IIRC, the meter is currently determined by a certain number of wavelengths of the fundamental emission of a Caesium atom - but don't quote me on that.&lt;/p&gt;&#xA;" OwnerUserId="4650" LastActivityDate="2011-07-23T18:54:32.970" CommentCount="2" />
  <row Id="12661" PostTypeId="2" ParentId="12647" CreationDate="2011-07-23T18:56:54.593" Score="-5" ViewCount="" Body="&lt;p&gt;Physics complications of &quot;introducing&quot; spin 3/2 matter are the same as for spin 1/2 and spin 0 - the initial approximation in the corresponding &lt;strong&gt;&lt;em&gt;interaction theory&lt;/em&gt;&lt;/strong&gt; is physically wrong and calculations give too big (= just wrong) perturbative corrections. It is a complete failure of physics description and it cannot be casted in &quot;geometric terms&quot;. Most people, however, does not see it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite accessible explanation, both for physicists and mathematicians, is given &lt;strong&gt;&lt;a href=&quot;https://docs.google.com/leaf?id=0B4Db4rFq72mLZDBmMjVhN2QtZWM0Yy00YTBjLWIwMmEtNDU4N2E2ZjY4NjNj&amp;amp;hl=en_US&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit for downvoters:&lt;/strong&gt; While in case of Rarita-Schwinger equation the solution violates even causality and it cannot be repaired with the constant renormalizations, this feature is still not considered as a failure of coupling. Indeed, we cannot be wrong. It is nature who is wrong, especially at short distances.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-07-24T12:24:03.637" LastActivityDate="2011-07-24T12:24:03.637" CommentCount="10" />
  <row Id="12662" PostTypeId="2" ParentId="12657" CreationDate="2011-07-23T19:23:18.723" Score="0" ViewCount="" Body="&lt;p&gt;This is a nice application of $W=F\Delta x$ for mechanical work, or its rotational analogue $W=\tau\Delta\theta$. Molecules in a liquid are free to rotate, so $\Delta\theta$ can be large. If you think of the lattice in an ice crystal as being built out of individual water molecules, then $\Delta\theta$ is limited by the ability of the lattice to deform, and you get a much smaller amount of work done by the electric field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try melting a stick of butter in the microwave, you'll see that it doesn't melt evenly. The melting starts in a certain spot, and then the spot spreads. The idea here is that once you get even a tiny bit of liquid, that liquid becomes efficient at absorbing energy, so the process snowballs. For this reason, I don't think you can test the theory by trying to use ice that has been thoroughly dried. As soon as the tiniest amount of liquid forms, it starts to grow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some possible experimental tests:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The explanation is independent of the detailed nature of the substance, so it should be true for all substances that the solid heats more slowly than the liquid. This seems to be verified by the fact that butter acts the same way as water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An insulating liquid without any electric dipole moment should heat more slowly than one like water that has a dipole moment. I don't know of a safe, easy example, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT] Reading a little more on the web, it turns out that there are four qualitatively different effects by which microwaves can heat matter:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;dielectric heating -- This is the effect I described above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;ionic conductivity&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;electronic conductivity&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;hysteresis&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In most foods, 1 and 2 are of approximately equal strength. 3 can occur in soot particles formed when food burns. In pure water and ice, only 1 is significant.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastEditorUserId="4552" LastEditorDisplayName="" LastEditDate="2011-07-24T16:49:19.013" LastActivityDate="2011-07-24T16:49:19.013" CommentCount="5" />
  <row Id="12663" PostTypeId="2" ParentId="12657" CreationDate="2011-07-23T20:01:17.417" Score="3" ViewCount="" Body="&lt;p&gt;The unusual thing is the really high absorption of microwaves by &#xA;bulk water, whereas the ice behaves more normal like most solids and liquids. &#xA;In liquid water we have an effect of relaxation of orientational polarisation. &#xA;The polarisation is achieved not by rotation (not possible in liquid water) &#xA;but by shift of hydrogen atoms along the hydrogen bonds. &#xA;This is a kind of Kohlrausch conduction mechanism. &#xA;This process is extremely fast, so polarisation of water is &#xA;one of the fastest processes in liquids. &#xA;There is debate, whether tunneling plays a role to enhance the shift of the protons.&#xA;The same mechanism is responsible for the extraordinary (about tenfold) &#xA;mobilities of H+ Ions in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a plot of Water spectrum in microwave domain. Note the incredible &#xA;absorption maximum with k about 3 ! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=bj1EnQPB0CMC&amp;amp;pg=PA184#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=bj1EnQPB0CMC&amp;amp;pg=PA184#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-07-23T20:01:17.417" CommentCount="2" />
  <row Id="12664" PostTypeId="1" AcceptedAnswerId="12858" CreationDate="2011-07-23T20:17:33.977" Score="9" ViewCount="296" Body="&lt;p&gt;In almost all proofs I've seen of the Lorentz transformations one starts on the assumption that the required transformations are linear. I'm wondering if there is a way to prove the linearity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prove that any spacetime transformation $\left(y^1,y^2,y^3,y^4\right)\leftrightarrow \left(x^1,x^2,x^3,x^4\right)$ that preserves intervals, that is, such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(dy^0\right)^2-\left(dy^1\right)^2-\left(dy^2\right)^2-\left(dy^3\right)^2=\left(dx^0\right)^2-\left(dx^1\right)^2-\left(dx^2\right)^2-\left(dx^3\right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is linear (assuming that the origins of both coordinates coincide). That is, show that $\frac{\partial y^i}{\partial x^j}=L_j^i$ is constant throughout spacetime (that is, show that $\frac{\partial L_j^i}{\partial x^k}=0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus far all I've been able to prove is that $g_{ij}L_p^iL_q^j=g_{pq}$ (where $g_{ij}$ is the metric tensor of special relativity) and that $\frac{\partial L_j^i}{\partial x^k}=\frac{\partial L_k^i}{\partial x^j}$. Any further ideas? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3064" LastEditorUserId="3064" LastEditorDisplayName="" LastEditDate="2011-07-25T18:42:24.763" LastActivityDate="2011-07-28T13:08:09.500" Title="Proving that interval preserving transformations are linear" Tags="&lt;special-relativity&gt;" AnswerCount="6" CommentCount="9" FavoriteCount="2" />
  <row Id="12665" PostTypeId="2" ParentId="12664" CreationDate="2011-07-23T21:59:34.313" Score="5" ViewCount="" Body="&lt;p&gt;Let's first assume that the scalar product that is preserved has positive signature to show the main idea. Also, you say you don't want to assume homogeneity but this is already implicit in your equation since to form intervals differences of space-time points are used so we might as well take one of those points to be zero of a vector space (equivalently, you might be talking about the preservation of a scalar product on a tangent space to a point but this is also linear, not affine).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $$f : \mathbb R^2 \to \mathbb R^2, \quad (x,y) \mapsto (A(x,y), B(x,y))$$&#xA;be length-preserving and suppose $f$ is analytic with&#xA;$$A(x,y) = \sum_{n,m=0}^{\infty} {a_{n,m} \over n! m!} x^n y^m, \quad B(x,y) = \sum_{n,m=0}^{\infty} {b_{n,m} \over n! m!} x^n y^m.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have $x^2 + y^2 = A(x,y)^2 + B(x,y)^2$  for all $x,y \in \mathbb R^2$ or explicitly first&#xA;$$ x^2 + y^2 = \left(\sum_{n,m=0}^{\infty} {a_{n,m} \over n! m!} x^n y^m \right)^2 +&#xD;&#xA;\left( \sum_{n,m=0}^{\infty} {b_{n,m} \over n! m!} x^n y^m \right)^2.$$&#xA;This immediately shows that the only non-vanishing coefficients occur when $n+m \leq 1$. We just need to investigate $n=m=0$ case but this is trivial since $f(0,0) = (a_{0,0}, b_{0,0})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the $n$D case the discussion is completely analogous. For arbitrary signature some care needs to be taken since we can't use $x^2 + y^2 = 0 \rightarrow x = y =0$ anymore (perhaps one can work in $\mathbb C$ instead of $\mathbb R$ and use analytic continuation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last remaining ingredient of this argument is the analyticity of $f$. But this is trivial since $||f(x,y)||^2 = x^2 + y^2$ and $||\cdot||^2$ are analytic around any $(x,y) \in \mathbb R^2$.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-07-24T16:12:55.670" LastActivityDate="2011-07-24T16:12:55.670" CommentCount="8" />
  <row Id="12666" PostTypeId="2" ParentId="10651" CreationDate="2011-07-23T23:43:02.230" Score="1" ViewCount="" Body="&lt;p&gt;If you look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Static_forces_and_virtual-particle_exchange#Inverse_square_law&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Static_forces_and_virtual-particle_exchange#Inverse_square_law&lt;/a&gt; you'll see a line of reasoning that doesn't seem to depend on the number of spacelike dimensions, yet still arrives at an inverse &lt;em&gt;square&lt;/em&gt; law.  I realize this isn't exactly a rigorous QED calculation (for which I feel far too stupid) but it makes me reconsider my former belief in non-privilege.  If d = 3 is the only case that allows both radiation and conservation of energy, then that's just... wow.&lt;/p&gt;&#xA;" OwnerUserId="4652" LastActivityDate="2011-07-23T23:43:02.230" CommentCount="1" />
  <row Id="12667" PostTypeId="2" ParentId="12664" CreationDate="2011-07-23T23:51:57.217" Score="6" ViewCount="" Body="&lt;p&gt;Here I just want to mention that there exists a direct proof in $1+1$ dimensions using elementary arguments. Let the two coordinate patches $U_x$ and $U_y$ (which are, say,  both convex sets in $\mathbb{R}^2$, containing the origin) have light-cone coordinates $x^{\pm}$ and $y^{\pm}$, respectively. The metric reads &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ dy^{+}dy^{-} ~=~   dx^{+}dx^{-}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to three PDE's&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\partial y^{+}}{\partial x^{+}} \frac{\partial y^{-}}{\partial x^{+}} ~=~0 \qquad \qquad\Leftrightarrow\qquad \qquad\frac{\partial y^{+}}{\partial x^{+}}~=~0 \qquad\mathrm{or}\qquad \frac{\partial y^{-}}{\partial x^{+}} ~=~0 ,$$&#xA;$$ \frac{\partial y^{+}}{\partial x^{-}} \frac{\partial y^{-}}{\partial x^{-}} ~=~0\qquad \qquad\Leftrightarrow \qquad\qquad\frac{\partial y^{+}}{\partial x^{-}}~=~0 \qquad\mathrm{or}\qquad \frac{\partial y^{-}}{\partial x^{-}} ~=~0,$$&#xA;$$ \frac{\partial y^{+}}{\partial x^{+}} \frac{\partial y^{-}}{\partial x^{-}} +\frac{\partial y^{+}}{\partial x^{-}} \frac{\partial y^{-}}{\partial x^{+}} ~=~1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\det \frac{\partial y}{\partial x}\neq 0$, there are really only two possibilities. Either  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial y^{-}}{\partial x^{+}}~=~0 ~=~\frac{\partial y^{+}}{\partial x^{-}},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial y^{+}}{\partial x^{+}}~=~0 ~=~\frac{\partial y^{-}}{\partial x^{-}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By possibly relabeling $x^{+} \leftrightarrow x^{-}$, we may assume the former. So &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y^{+}~=~f^{+}(x^{+})\qquad \mathrm{and} \qquad y^{-}~=~f^{-}(x^{-}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the third PDE, we conclude that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\partial f^{+}}{\partial x^{+}}\frac{\partial f^{-}}{\partial x^{-}} ~=~1. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By separation of variables, this is only possible if $\frac{\partial f^{\pm}}{\partial x^{\pm}}$ is independent of $x^{\pm}$. It follows that $y^{\pm}~=~f^{\pm}(x^{\pm})$ are affine functions. Q.E.D.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastActivityDate="2011-07-23T23:51:57.217" CommentCount="1" />
  <row Id="12668" PostTypeId="1" AcceptedAnswerId="12672" CreationDate="2011-07-24T00:18:24.240" Score="1" ViewCount="59" Body="&lt;p&gt;I am aware of three apparently different forms of angular momentum: spin, orbital and macroscopic, like a spinning crystal or baseball.&#xA;Can angular momentum be transferred or converted from one of these forms to another?&#xA;If so, how? &lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-07-24T04:30:13.743" Title="Converting Angular Momentum?" Tags="&lt;angular-momentum&gt;&lt;spin&gt;&lt;orbital-motion&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="12669" PostTypeId="1" AcceptedAnswerId="12676" CreationDate="2011-07-24T00:18:28.233" Score="7" ViewCount="265" Body="&lt;p&gt;In particular I am curious if the values of the rest masses of the electron, up/down quark, neutrino and the corresponding particles over the next two generations can be defined as constant or if there is an intrinsic uncertainty in the numbers. I suppose that since there is no (as of yet) mathematical theory that produces these masses we must instead rely on experimental results that will always be plagued by margins of error. But I wonder does it go deeper than this? In nature do such constant values exist or can they be &quot;smeared&quot; over a distribution like so many other observables prior to measurement?  (Are we sampling a distribution albeit a very narrow one?) Does current theory say anything about this? (i/e they must be constant with no wiggle room vs. no comment) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am somewhat familiar with on-shell and off-shell particles- but I must confess I'm not sure if this figures into my question. I'd like to say that as an example I'm talking about the rest mass of the electron as could be determined by charge and charge/mass ratios. But perhaps this very number is itself influenced by off-shell electron masses? Perhaps this makes no sense. Needless to say I'd appreciate any clarification.    &lt;/p&gt;&#xA;" OwnerUserId="1567" LastActivityDate="2011-07-25T13:10:10.583" Title="Are the rest masses of fundamental particles certainly constants?" Tags="&lt;particle-physics&gt;" AnswerCount="7" FavoriteCount="3" />
  <row Id="12670" PostTypeId="2" ParentId="12669" CreationDate="2011-07-24T02:09:00.290" Score="4" ViewCount="" Body="&lt;p&gt;I guess I can think of three possible ways in which masses could be non-constant. (1) They could change due to quantum-mechanical fluctuations, (2) they could be slightly different for different particles at the same time, (3) or they could change over cosmological time intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number 1 seems to be what you had in mind, but I don't think it works. The standard picture is that for a particle of mass m, its momentum p and mass-energy E can fluctuate, but the fluctuations are always such that $m=\sqrt{E^2-p^2}$ (with c=1) stays the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Re #2, here's some good info: &lt;a href=&quot;http://physics.stackexchange.com/questions/11826/are-all-electrons-identical&quot;&gt;Are all electrons identical?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Re #3, one thing to watch out for is that it is impossible, even in principle, to tell whether a unitful fundamental constant is changing. The notion only makes sense when you talk about unitless constants: Duff, &lt;a href=&quot;http://arxiv.org/abs/hep-th/0208093&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/0208093&lt;/a&gt; However, it certainly does make sense to talk about changes in the unitless ratios of fundamental constants, such as the ratio of two masses or the fine structure constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are claims by Webb et al. J.K. Webb et al., &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0012539v3&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/astro-ph/0012539v3&lt;/a&gt; that the fine structure constant has changed over cosmological timescales. Chand et al., Astron. Astrophys. 417: 853, failed to reproduce the result, and IMO it's bogus. I'm not aware of any similar tests for the ratios of masses of fundamental particles. If you change the ratio of masses of the electron and proton, it will change the spectrum of hydrogen, but at least to first order, the change would just be a rescaling of energies, which would be indistinguishable from a tiny change in the Doppler shift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brans-Dicke gravity (Physical Review 124 (1961) 925, &lt;a href=&quot;http://loyno.edu/~brans/ST-history/&quot; rel=&quot;nofollow&quot;&gt;http://loyno.edu/~brans/ST-history/&lt;/a&gt; ) has a scalar field that can be interpreted as either a local variation in inertia or a local variation in the gravitational constant G. This could in some sense be interpreted as meaning that, e.g., electrons at different locations in spacetime had different masses, but all particles would be affected in the same way, so there would be no effect on ratios of masses -- hence the ambiguity between interpreting it as a variation in inertia or a variation in G. B-D gravity has a unitless constant $\omega$, and the limit $\omega\rightarrow\infty$ corresponds to general relativity. Solar system tests constrain $\omega$ to be at least 40,000, so B-D gravity is basically dead these days.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastEditorUserId="4552" LastEditorDisplayName="" LastEditDate="2011-07-24T02:22:20.967" LastActivityDate="2011-07-24T02:22:20.967" CommentCount="2" />
  <row Id="12671" PostTypeId="2" ParentId="5584" CreationDate="2011-07-24T03:22:16.293" Score="0" ViewCount="" Body="&lt;p&gt;In 1974 String theories are shown to require extra dimensions. An object similar to the graviton is found in superstring theories.&#xA;However these alternative dimensions or realities must be exactly the same as each other, even to the last grain of sand. For example There can not be different version of you in those universes, they must exactly be the same. To explain why, one must consider the butterfly effect, any small change can lead to large changes, leading to completely different worlds containing completely different contents. So you and I are unique to this  universe alone or there are exact copy of us in other universes, much like effect of two parallel mirror&lt;/p&gt;&#xA;" OwnerUserId="4653" LastActivityDate="2011-07-24T03:22:16.293" />
  <row Id="12672" PostTypeId="2" ParentId="12668" CreationDate="2011-07-24T04:13:04.477" Score="2" ViewCount="" Body="&lt;p&gt;Are you looking for something like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein%E2%80%93de_Haas_effect&quot; rel=&quot;nofollow&quot;&gt;Einstein-de Haas effect&lt;/a&gt;? The angular momentum of the spins in a ferromagnet is converted to mechanical angular of the bulk magnet.&lt;/p&gt;&#xA;" OwnerUserId="3970" LastActivityDate="2011-07-24T04:13:04.477" />
  <row Id="12673" PostTypeId="2" ParentId="12668" CreationDate="2011-07-24T04:30:13.743" Score="1" ViewCount="" Body="&lt;p&gt;the different &quot;forms&quot; of angular momentum are still the same fundamental thing, they are just generated in different manners. there is a big difference in how spin, orbital and classical angular momenta are created... &#xA;but it's still same units of action for all of them (I believe).&#xA;to convert from one &quot;form&quot; to another therefore would be like asking how to convert from one type of kinetic energy to another... i'm sure there are example of angular momentum transfer--- just examine hydrogen state transitions for example&lt;/p&gt;&#xA;" OwnerUserId="4570" LastActivityDate="2011-07-24T04:30:13.743" CommentCount="1" />
  <row Id="12674" PostTypeId="1" AcceptedAnswerId="12678" CreationDate="2011-07-24T04:43:44.893" Score="2" ViewCount="107" Body="&lt;p&gt;consider a transition for an electron in the Hydrogen atom from the ground state to the 1st excited state.&#xA;   Let's say this transition occurs through absorption of a photon of exactly the energy  required to excite the electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what happens to the photon right after it is absorbed?&#xA;    is it annhilated when it is absorbed?&#xA;    does its wavefunction cease to exist?&#xA;    I'm curious about this process thanks.&lt;/p&gt;&#xA;" OwnerUserId="4570" LastActivityDate="2011-07-26T00:25:14.157" Title="Hydrogen transition and photon behavior" Tags="&lt;quantum-mechanics&gt;&lt;photons&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="12675" PostTypeId="2" ParentId="12669" CreationDate="2011-07-24T05:00:18.143" Score="0" ViewCount="" Body="&lt;p&gt;if you think about it, any time you perform a measuremeant, you must use a device which is limited by quantum mechanical laws.&#xA;   Any device measuring the weight of a particle must record the weight via some state transition internal to the device... (e.g. the device must change in some way when the particle is placed near it so even if the particle is at rest when the observation is made, the device must have some transition in internal position and momenta to record an observation). So I would say that quantum mechanics does impose a limit on how close we can get to finding the actual  rest mass of a particle. That limit is always loosely proportional to plancks constant or plancks constant times 1/2.&lt;/p&gt;&#xA;" OwnerUserId="4570" LastActivityDate="2011-07-24T05:00:18.143" CommentCount="7" />
  <row Id="12676" PostTypeId="2" ParentId="12669" CreationDate="2011-07-24T06:25:04.840" Score="3" ViewCount="" Body="&lt;p&gt;They most certainly &lt;strong&gt;are not&lt;/strong&gt;. You are right that there is no theory that explains masses (these are input as parameters) but note that our current theories used to explain e.g. LHC data (that is, quantum field theories) inevitably come with a scale attached: you need to describe upto what energies you do physics otherwise the theory just doesn't make sense [&lt;em&gt;insert usual story here about renormalization and infinities often told to scare little children before their going to bed&lt;/em&gt;].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this shouldn't come as such a surprise since there are new particles awaiting discovery just behind the corner, so claiming that we have a complete theory would be preposterous. Instead, what we claim is that we have a good theory that works upto some scale. Consequently, all of the parameters that are inserted by hand must depend on the scale. Again, this is because theories at different scales are potentially completely different (e.g. at the &quot;present scale&quot; there is no supersymmetry assumed while it is conceivable that at a little higher scale our theories will have to include it) and so the parameters of the theories that are used to connect the theory with experiment potentially have no relation to each other.  This phenomenon is known as running of coupling constants or, briefly, &lt;a href=&quot;http://en.wikipedia.org/wiki/Coupling_constant#Running_coupling&quot; rel=&quot;nofollow&quot;&gt; the running coupling&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moral is that all the rest masses and interaction &quot;constants&quot; depend on some scale. They shouldn't be thought as something inherently deep about the nature but just as fitting parameters that describe only effective masses and effective coupling. To illustrate why they are just effective: consider an electron in classical physics. We can measure its charge by usual methods. This value is the long-distance low-energy $e(E \to 0)$ limit of the scale dependent coupling $e(E)$. As you increase the energy and try to probe electron at shorter distances you will find that lots of others electron-positron pairs appear, screening the electron, and the charge that you will measure will be &lt;em&gt;different&lt;/em&gt; due to these changed conditions (we talk about the polarization of vacuum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for the sake of completeness: one &lt;em&gt;could&lt;/em&gt; say that $E \to 0$ limit is the most important thing about couplings and that we should take that as definition. If so, then these long-distance couplings are indeed constants as one was used in classical physics. But this point of view is worthless in particle physics where people instead try to make $E$ as high as possible to obtain a theory valid at high scales (since this is what they need at LHC).&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-24T06:25:04.840" CommentCount="3" />
  <row Id="12677" PostTypeId="2" ParentId="9969" CreationDate="2011-07-24T06:35:31.063" Score="0" ViewCount="" Body="&lt;p&gt;A magnetic field arises due charges in motion because when the charges are in motion the electric  field keeps on changing continously and other reason is that when the charges are in motion they behave as tiny magnets and produce magnetic field perpendicular to its direction &lt;/p&gt;&#xA;" OwnerUserId="4655" LastActivityDate="2011-07-24T06:35:31.063" />
  <row Id="12678" PostTypeId="2" ParentId="12674" CreationDate="2011-07-24T06:41:00.903" Score="3" ViewCount="" Body="&lt;p&gt;Well, in the personificated picture that we often have of particles, it indeed ceases to exist. In a less dramatic language this can also be described on the level of fields: electron in the atom is a localized Dirac field while the incoming photon is an EM wave. This process is then morally no different from classical theory of interaction of waves with matter: some of the wave is reflected, some is scattered, some is absorbed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, one could ask a question similar to yours about the creation of particle-antiparticle pairs. Are these brought into existence by some act on magic? It turns out that on the level of fields these are nothing else than quantum mechanical fluctuations that you can imagine as ripples and peaks on the sea surface. The sharp peaks then essentially have properties of particles. Indeed, this is how we think of particles in general: they are well-formed local portions of the field.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-24T06:41:00.903" />
  <row Id="12679" PostTypeId="1" AcceptedAnswerId="12778" CreationDate="2011-07-24T07:20:20.387" Score="2" ViewCount="155" Body="&lt;p&gt;How would one perform an $N$-body simulation in General Relativity (GR) for something like galaxy formation or galactic dynamics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose one wants to simulate the rotation curve $v(r)$ for galaxies with an $N$-body simulation, and wants to do it in the framework of GR / the Einstein equations.&lt;/p&gt;&#xA;" OwnerUserId="47" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-24T17:03:38.017" LastActivityDate="2011-07-28T12:58:27.077" Title="$N$-body simulation in General Relativity" Tags="&lt;general-relativity&gt;&lt;simulation&gt;&lt;galaxies&gt;&lt;n-body&gt;" AnswerCount="1" CommentCount="10" />
  <row Id="12680" PostTypeId="1" CreationDate="2011-07-24T09:37:40.227" Score="1" ViewCount="71" Body="&lt;p&gt;I know that if the integral is convergent we can always make a change of variable to make it better, however what happens with DIVERGENT integrals? can we make a change of variable into a divergent integral after having it regularized?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean we insert the regulator $(q+a)^{-s}$ or similar on each variable and then make the change of variable to polar, cilindrical or other&lt;/p&gt;&#xA;" OwnerUserId="4435" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-07-24T13:43:42.287" LastActivityDate="2011-07-24T13:43:42.287" Title="Can we make a change of variables (for example to polar coordinates) into a divergent integral?" Tags="&lt;mathematics&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="12681" PostTypeId="2" ParentId="12680" CreationDate="2011-07-24T10:34:18.890" Score="3" ViewCount="" Body="&lt;p&gt;Here's some elementary stuff (so that we don't get bogged down in unnecessary technicalities) to think about. Suppose we'd like to integrate something like $$\int_0^{\infty} {1 \over x^2} {\rm d}x.$$&#xA;This of course diverges but we can insert some regularizator $M$ that captures the divergence in the $M \to 0$ limit&#xA;$$I(M) := \int_0^{\infty} {1 \over (x + M)^2} {\rm d}x = \left[- {1 \over x + M} \right]_0^{\infty} = {1 \over M}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We obviously can't change the value of $I(M)$ by using any (smooth) substitution. so, suppose we changed our original integral using $x = y^2$. Then we need to evaluate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;Edit:&lt;/strong&gt; there was a mistake in this calculation that changes the result of the discussion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^{\infty} {2 \over y^3} {\rm d}y$$&#xA;and using the same regularization procedure $x \to x + M$ or $y \to \sqrt{y^2 + M}$ we get&#xA;$$I(M) = \int_0^{\infty} {2 \over (y^2 + M)^{3/2}} {\rm d}y = \int_M^{\infty} {1 \over z^2} {\rm d}y = {1 \over M} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand if we naively performed the regularization $y \to y + M$ we would obtain a different result&#xA;$$I(M) = \int_0^{\infty} {2 \over (y + M)^3} {\rm d}y = {1 \over M^2} .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I guess the point is that one needs to make a regularization that respects the substitution to obtain consistent results.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-07-24T10:46:20.157" LastActivityDate="2011-07-24T10:46:20.157" CommentCount="2" />
  <row Id="12682" PostTypeId="1" CreationDate="2011-07-24T10:47:24.520" Score="0" ViewCount="50" Body="&lt;p&gt;I notice in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_spectrum&quot; rel=&quot;nofollow&quot;&gt;EMS&lt;/a&gt; chart that only half of the IR range will travel through the earth's atmosphere.  So in the range of frequencies that penetrate the atmosphere what frequency travels farthest?&lt;/p&gt;&#xA;" OwnerUserId="4656" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-24T11:11:39.997" LastActivityDate="2011-08-23T13:33:41.667" Title="range of infrared light" Tags="&lt;light&gt;&lt;atmosphere&gt;&lt;atmospheric-science&gt;&lt;frequency&gt;" AnswerCount="1" />
  <row Id="12683" PostTypeId="1" CreationDate="2011-07-24T11:20:00.343" Score="0" ViewCount="62" Body="&lt;p&gt;I have very little understanding of electrostatics stuff, so please bear with me. I have been trying to find an equation (or some solution) of how to calculate the electric field strength (in N/C) of a conducting rectangular (nearly flat) plate which has non-zero potential to it, say the plate has a potential of 5V. I have searched over the internet (and read Griffiths book on the method of images section), and it seems one would need to apply the method of images to solve the problem. However, since almost all examples regarding method of images is about &quot;grounded&quot; plates at 0V, it is very difficult for me to understand how to use method of images to solve for the electric field of a plate at 5V? Also, I would like to mention that within the electric field of the 5V conducting plate, there is a small freely-moving positively charged particle in the vicinity (I assume because the charge on this small particle is the same as the charge on an electron/proton-- it can be regraded as a test charge and therefore, modification to the combined electric field between the conducting plate and the charge is not necessary?-please correct me here if I am wrong.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help. &lt;/p&gt;&#xA;" OwnerUserId="3604" LastActivityDate="2011-08-23T14:33:42.473" Title="Electric field of not-grounded conducting plate with a given potential?" Tags="&lt;electrodynamics&gt;&lt;electrostatics&gt;&lt;electric-circuits&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12684" PostTypeId="2" ParentId="12680" CreationDate="2011-07-24T12:02:41.520" Score="-3" ViewCount="" Body="&lt;p&gt;If an integral is convergent, nothing can make it better ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an integral is divergent, every means is good nowadays to make it convergent and equal to what you want, unfortunately.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-07-24T13:41:02.420" LastActivityDate="2011-07-24T13:41:02.420" />
  <row Id="12685" PostTypeId="2" ParentId="12680" CreationDate="2011-07-24T12:34:03.080" Score="1" ViewCount="" Body="&lt;p&gt;Important point: a change of variables cannot improve convergence! When you regulate an divergent integral $I$ ($=\infty$), you're actually calculating another integral $J$ (that might depend on some parameters $s,t,\ldots$) and you try to 'recover' $I$ by exchanging the order of integration and taking some limits. However, since you &lt;em&gt;know&lt;/em&gt; that $I$ doesn't exist and $J$ does, it's obvious that the two aren't equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same goes for you current question: by construction, a change of variables leaves the value of the integral unchanged and cannot change the convergence of an integral. Of course, you can change the integrand to turn an existing integral $I$ into something divergent, but that doesn't make a lot of sense, mathematically. You're probably interested in common regularization methods (like the one you cite in your problem). They can actually be tested on convergent integrals to see if they work: suppose you regularize the already convergent integral $I$ by $J(\varepsilon)$ and $$\lim_{\varepsilon \rightarrow 0} \, J(\varepsilon) \neq I,$$ then you know something is wrong with your regularization.&lt;/p&gt;&#xA;" OwnerUserId="1471" LastActivityDate="2011-07-24T12:34:03.080" CommentCount="1" />
  <row Id="12686" PostTypeId="2" ParentId="12682" CreationDate="2011-07-24T13:26:35.890" Score="1" ViewCount="" Body="&lt;p&gt;I suppose what you want is the graph at &lt;a href=&quot;http://en.wikipedia.org/wiki/Radio_window&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Radio_window&lt;/a&gt;, provided you are talking about vertical transmission from space to the earth's surface. Results would presumably be different for horizontal transmission because the pressure and chemical composition of the atmosphere would be much closer to constant, but I can't immediately source those transmission coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you see, radio waves transmit best, 10 micron comes through pretty well, etc. &lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-07-24T13:26:35.890" CommentCount="3" />
  <row Id="12687" PostTypeId="2" ParentId="12669" CreationDate="2011-07-24T13:31:46.617" Score="-2" ViewCount="" Body="&lt;p&gt;The rest masses are certain as they correspond to the ground states of QM systems where the energy is certain. It is valid for any system - &quot;elementary&quot; or compound. High temperatures may create the energy uncertainty and it may influence the &quot;measured&quot; mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass &quot;measurement&quot; can be indirect via precise transitions expressed with theoretical formulas involving mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many current theories fail to simply use the experimental data on masses and charges and it is obviously not a nature feature but a human being imperfection. If you read Marek's answer, you will learn that, as soon as Marek does not know all the particles the nature can produce, he and his theory cannot be wrong. So masses and charges are running for him, they run his errands. And the main errand is to describe the experimental data with a wrong theory. Difficult task but fortunately feasible with help of running constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: For those who does not know that most of &quot;our theories&quot; are non renormalizable, the usual blah-blah about &quot;scale dependence&quot; is not applicable at all (it does not help).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 2: I see downvoters do not want to leave the rest masses at rest.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-07-25T13:10:10.583" LastActivityDate="2011-07-25T13:10:10.583" CommentCount="2" />
  <row Id="12688" PostTypeId="2" ParentId="12683" CreationDate="2011-07-24T13:34:15.147" Score="0" ViewCount="" Body="&lt;p&gt;It sounds a bit like you're missing something from the problem description? 5V potential relative to what? It would make sense if you have the field between two conducting plates of different potential for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;0V&quot; they are talking about in the method of mirrors is really a way of saying that any inherent excess or deficit charge in the conductor vanishes (since you assume the plate has a connection to a 0V potential reservoir of vanishing resistance) hence you have a neutral overall charge distribution on the plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This allows you to consider the field-distribution when for example a point-charge with charge q is placed in the vicinity of the grounded plate; by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Method_of_image_charges&quot; rel=&quot;nofollow&quot;&gt;method of images&lt;/a&gt; the solution is equivalent to no plate but a charge of charge -q an equivalent distance behind the plate. By differentiating the potential field you can find the equivalent distribution of charge on the plate. Without reading the actual problem, the test charge if it has a specified charge really sounds like it has to be considered in the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply saying that a plate has a 5V potential doesn't give you the information needed to calculate a charge distribution on it or a potential field strength. It doesn't even mean that the plate is non-neutral, because Volt is a relative unit. It would have made sense if there was an equivalent plate of, say, 0V in parallel. It might also be a red herring to compound the exercise, which might be a &quot;standard&quot; pointcharge vs. plate problem.&lt;/p&gt;&#xA;" OwnerUserId="2936" LastActivityDate="2011-07-24T13:34:15.147" CommentCount="6" />
  <row Id="12689" PostTypeId="1" CreationDate="2011-07-24T13:43:35.097" Score="3" ViewCount="50" Body="&lt;p&gt;I am not sure whether it is some well-known named model in statistical physics. &#xA;I could not find it in any standard text-book that I know of. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let there be $N$ identical classical particles occupying a square lattice of $2N$ sites with at most one particle per site. Let the alternate sites be names $A$ and $B$. Let $p$ be the fraction of particles on $A$-type sites and assume that for a fixed value of $p$, all configurations are equally likely. (mean field approximation?)  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Then how does one calculate the entropy as a function of $p$? I guess there is some simplification that will happen if one works in the limit of $N$ being very large. (thermodynamic limit?)  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let there be a repulsive interaction energy of $E_0$ whenever neighbouring $A$ and $B$ type sites are occupied. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In that case the claim seems to be that the average total energy $E(p)$ is given by $4NE_0p(1-p)$. (..I am not sure whether here any large $N$ limit has been taken..but I guess this too is some asymptotic answer..)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would be glad if someone can help calculate/derive the above two bulleted questions. &#xA;I would like to know if there is some general framework or class of models from which this comes. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Apparently the interesting thing about this model is that it has a second order phase transition.  How does one prove that? Or is it somehow supposed to be obvious? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Will some single unique value of $p$ get picked out in the high and low temperature limit? How does one determine that? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the last/above two of the questions one would obviously need to assume that the system is at a thermal equilibrium at a temperature $T$. Then I guess the value of $p$ gets determined by minimizing the ``free energy&quot; $=F(p) = E(p) - TS(p)$. But to be able to do this calculation one would probably first need to find the answer to the first two questions in the large $N$/ thermodynamic limit. &lt;/p&gt;&#xA;" OwnerUserId="1637" LastActivityDate="2011-07-24T18:01:14.890" Title="Repulsive classical identical particles on a square lattice" Tags="&lt;statistical-mechanics&gt;&lt;solid-state-physics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12690" PostTypeId="1" CreationDate="2011-07-24T14:17:05.690" Score="4" ViewCount="164" Body="&lt;p&gt;I'm going to be choosing a university course soon, and I want to go into a branch of physics. A dream job for me would be to work in research, however, I do realise that this isn't for everyone and is difficult to reach. So what is the best way to go about achieving this aim? What things can I do which will help me?&lt;/p&gt;&#xA;" OwnerUserId="4658" LastEditorUserId="184" LastEditorDisplayName="" LastEditDate="2011-07-26T12:17:36.200" LastActivityDate="2011-07-26T12:17:36.200" CommunityOwnedDate="2011-07-24T15:59:01.740" Title="'Getting in' to research physics?" Tags="&lt;soft-question&gt;&lt;education&gt;&lt;physics-careers&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="12691" PostTypeId="2" ParentId="12664" CreationDate="2011-07-24T14:34:27.943" Score="2" ViewCount="" Body="&lt;p&gt;The first condition implies that the Jacobian matrix $L^i_j=\frac{\partial y^i}{\partial x^j}$ is a Lorentz transformation.&#xA;By substitution of the definition of the Jacobian in this condition, we obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g^{ij}\frac{\partial y^k}{\partial x^i}\frac{\partial y^l}{\partial x^j} = g^{kl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, taking the diagonal equations equating $l=k$, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g^{ij}\frac{\partial y^k}{\partial x^i}\frac{\partial y^k}{\partial x^j} = g^{kk}= \pm 1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The plus sign for the time coordinate and the minus sign for the space coordinates).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is just the Hamilton-Jacobi equation for a free relativistic particle, whose unique solution can be obtained by separation of variables:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y^k = \sum_i f^{(k)}_i(x^i)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Substitution, we obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac {df^{(k)}_i(x^i)}{dx^i} = const$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the new coordinates are linear functions of the old coordinates. The constant coefficients are not independent, since the Jacobian matrix must be a Lorentz transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon lurscher's suggestion, here are two references containing the Hamilton-jacobi equation of a relativistic particle.&#xA;(Both references refer to a particle in an external electromagnetic field. In order to obtain the Hamilton-Jacobi equation for the free particle&#xA;one needs the particualr case with a vanishing vector potential): &lt;a href=&quot;http://arxiv.org/abs/physics/0309059&quot; rel=&quot;nofollow&quot;&gt;reference-1 (by A. granik)&lt;/a&gt;, &lt;a href=&quot;http://www.math.osu.edu/~gerlach/laser/action-angle_dynamical_systemJun20005la/node6.html&quot; rel=&quot;nofollow&quot;&gt;reference-2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The needed version appears in equation (33) of the first reference, the second reference contains the (proper) time dependent version).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, I'll give here an other derivation based on the WKB approximation of the Klein- Gordon equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac {1}{c^2}\frac {\partial^2\psi}{\partial t^2}-\nabla^2 \psi + \frac{m^2 C^2}{\hbar^2}\psi = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plane wave solutions are given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\psi = C \exp(i\frac{\mathbf{p}.\mathbf{x}-\sqrt{m^2c^4+p^2c^2}t}{\hbar})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To perform a WKB approximation, we seek a solution of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\psi = A(x,t)\exp(\frac{iS(x,t)}{\hbar})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and take the leading terms in the limit $\hbar \rightarrow 0$. ($S$ is sometimes called the Hamilton-Jacobi phase function)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By subtitution, we obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$((\frac {1}{c^2}\frac {\partial^2A}{\partial t^2}-\nabla^2A)+\frac{2i}{\hbar}(\frac {1}{c^2}\frac {\partial A}{\partial t} \frac {\partial S}{\partial t} -\mathbf{\nabla}A.\mathbf{\nabla}S) -\frac{A}{\hbar^2}(\frac {1}{c^2}\frac {\partial^2S}{\partial t^2}-\nabla^2S - m^2 c^2 )) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The leading term is the hamilton-Jacobi equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac {1}{c^2}\frac {\partial^2S}{\partial t^2}-\nabla^2S - m^2 c^2 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which can be seen to be equivalent to each equation on the main diagonal of the matrix equation written in the original answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it is also easy to see the uniqueness of the solution. For the free particle, one can see that the non-leading terms actually vanish. i.e., the WKB approximation is exact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hamilton-Jacobi phase function $S$ is just the phase of the plane wave solutions of the Klein-Gordon equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ S = \mathbf{p}.\mathbf{x}-\sqrt{m^2c^4+p^2c^2}t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On $\mathbb{R}^4$, all solutions of the free Klein-Gordon equation in Cartesian coordinates are of the form of the plane waves, which implies that the Hamilton-Jacobi phase function is linear in the Cartesian coordinates.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorUserId="2190" LastEditorDisplayName="" LastEditDate="2011-07-25T09:22:00.030" LastActivityDate="2011-07-25T09:22:00.030" CommentCount="5" />
  <row Id="12692" PostTypeId="2" ParentId="12674" CreationDate="2011-07-24T14:50:54.640" Score="-1" ViewCount="" Body="&lt;p&gt;No, the wave function of a photon does not cease to exist. Technically a photon is an excited state of an oscillator. After absorption the oscillator wave function becomes a wave function of the ground state. The electron wave function, on the contrary, becomes an excited one from the ground state wave function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total wave function is a product of independent state wave functions. None of them ceases to exist while interactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each state can also be described with &quot;occupation numbers&quot; $n_i (t)$. Physical processes are redistributions of the occupation numbers. Transition from $n_{ph} = 1$ to $n_{ph} = 0$ is absorption of a photon but not a photon wave function destruction.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-07-24T15:21:35.123" LastActivityDate="2011-07-24T15:21:35.123" />
  <row Id="12693" PostTypeId="1" CreationDate="2011-07-24T15:16:24.317" Score="3" ViewCount="132" Body="&lt;p&gt;in order to confirm whether I understand the relation ship between amps, columbs and electrons I decided to try out the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my laptop adapter's label reads 3.34 amps.&#xA;I assume that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;amps are the rate of flow of charge (i.e speed of flow measured in coulombs per second) at a point in a circuit.&lt;/li&gt;&#xA;&lt;li&gt;that elementary charge (e) is the charge of one electron&lt;/li&gt;&#xA;&lt;li&gt;that one coulomb is the sum of the charge given by 1/e electrons&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;so the number of electrons that will pass through my laptop (or at least the adapter) per second are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.34 x (1/e) = 2.084664255E19 electrons&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; having seen the answers I have seen that there's a lot of information missing here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of the circuit I would like to focus on is any cross section of the cable that leads from the adapter to the laptop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also assume that the laptop is at full power and drawing the following as labelled on the adapter:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;19.5 V (since this is out of the adapter I'm assuming DC current)&lt;/li&gt;&#xA;&lt;li&gt;3.34 A&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4659" LastEditorUserId="966" LastEditorDisplayName="" LastEditDate="2011-07-26T14:00:49.433" LastActivityDate="2011-07-26T14:00:49.433" Title="number of electrons my laptop uses in a second" Tags="&lt;electricity&gt;&lt;electronics&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="12694" PostTypeId="2" ParentId="12690" CreationDate="2011-07-24T15:31:17.650" Score="0" ViewCount="" Body="&lt;p&gt;I'd say study hard, but also learn about the fundamentals of experiment design, learn how to write proper scientific papers, and learn how publish your work.  These things aren't (in my experience) typically taught in the undergraduate or even graduate level coursework, but are absolutely crucial to &quot;success&quot; in a research environment. &lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2011-07-24T15:31:17.650" CommunityOwnedDate="2011-07-24T15:59:01.740" />
  <row Id="12695" PostTypeId="1" AcceptedAnswerId="12701" CreationDate="2011-07-24T16:08:02.887" Score="3" ViewCount="88" Body="&lt;p&gt;Imagine a perfectly insulated box, placed inside the box is an electric heater. The heater is switched on and the box is left to reach equilibrium with its surroundings. What is the final temperature inside the box?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly a real box would eventually reach an equilibrium temperature with its surroundings at the point where the energy being dissipated by the box matched the energy being added by the heater. If no energy can escape though it seems to imply that the temperature will just keep rising forever which feels wrong to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect the answer has something to do with the maximum temperature the heating element can achieve before failing. If we had a heating element that never burnt out though what would the final temperature then be?&lt;/p&gt;&#xA;" OwnerUserId="4650" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-07-25T06:21:39.010" LastActivityDate="2011-07-25T19:08:47.563" Title="heater in a perfectly insulated box" Tags="&lt;thermodynamics&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12696" PostTypeId="5" CreationDate="2011-07-24T16:10:48.100" Score="0" ViewCount="" Body="&lt;p&gt;The charge of QCD is called &quot;color&quot; and comes in three fundamental values called &quot;red&quot;, &quot;green&quot;, and &quot;blue&quot;, to which are coupled eight distinct states of a single vector boson called a &quot;gluon&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike the photon in &lt;a href=&quot;http://physics.stackexchange.com/questions/tagged/quantum-electrodynamics&quot;&gt;QED&lt;/a&gt;, gluons are themselves color charged objects, which means the theory is complicated by the existence of three- and four-gluon vertexes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A important feature of the strong interaction is confinement, which demands that all all objects able to exist at large distance separation must be color-neutral: either adding equal amounts of red, green, and blue (as in baryons); or by adding color and anti-color (as in mesons). &lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-07-24T16:36:48.710" LastActivityDate="2011-07-24T16:36:48.710" />
  <row Id="12697" PostTypeId="4" CreationDate="2011-07-24T16:10:48.100" Score="0" ViewCount="" Body="Quantum-ChromoDynamics (QCD) is the quantum field theory believed to describe the strong nuclear force. " OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-07-25T01:18:22.997" LastActivityDate="2011-07-25T01:18:22.997" />
  <row Id="12698" PostTypeId="5" CreationDate="2011-07-24T16:33:05.920" Score="0" ViewCount="" Body="&lt;p&gt;In QED, electrically charged are coupled by a uncharged, massless vector boson called the photon (and identified as a quantum of light).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By adding three additional, massive vector bosons (the $Z^0$ and $W^{\pm}$) which couple to the weak hyper-charge ($T_3 - q \ \sin^2 \theta_W$ in which $q$ is electric charge and $T_3$ is the third component of the weak isospin), the theory can be extended to cover the weak nuclear force as well.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-07-24T16:33:05.920" LastActivityDate="2011-07-24T16:33:05.920" />
  <row Id="12699" PostTypeId="4" CreationDate="2011-07-24T16:33:05.920" Score="0" ViewCount="" Body="Quantum-ElectroDynamics (QED) is the quantum field theory believed to describe the electromagnetic interaction (and with some extension the weak nuclear force)." OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-07-24T16:33:05.920" LastActivityDate="2011-07-24T16:33:05.920" />
  <row Id="12700" PostTypeId="1" AcceptedAnswerId="12707" CreationDate="2011-07-24T16:33:16.110" Score="4" ViewCount="214" Body="&lt;p&gt;Where does the top's angular momentum get transferred to? Does it &lt;em&gt;very slightly&lt;/em&gt; change the angular momentum of the table, and then the angular momentum of the Earth? &lt;/p&gt;&#xA;" OwnerUserId="1576" LastEditorUserId="966" LastEditorDisplayName="" LastEditDate="2011-07-26T14:02:05.483" LastActivityDate="2011-07-28T15:26:19.880" Title="How is angular momentum conserved when a spinning top finally stops spinning?" Tags="&lt;classical-mechanics&gt;&lt;angular-momentum&gt;" AnswerCount="3" />
  <row Id="12701" PostTypeId="2" ParentId="12695" CreationDate="2011-07-24T17:15:27.867" Score="8" ViewCount="" Body="&lt;p&gt;You have imagined a system with constant heat input and no output. As long as those conditions pertain there is &lt;em&gt;no&lt;/em&gt; limit to the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice you reach an ultimate temperature when&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Some part of your electric circuit breaks or melts and the heater stops.&lt;/li&gt;&#xA;&lt;li&gt;The insulator melts or the ever increasing interior pressure bursts the box open&lt;/li&gt;&#xA;&lt;li&gt;Your approximation of perfect insulation fails and the box is leaking as much heat as is being generated by the heater.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-07-24T17:15:27.867" CommentCount="2" />
  <row Id="12702" PostTypeId="2" ParentId="12674" CreationDate="2011-07-24T17:19:17.567" Score="-2" ViewCount="" Body="&lt;p&gt;I'm going to go way out on a limb and say that this is a hypothetical experiment that has never been carried out. I believe it is impossible to prepare a hydrogen atom in the ground state, expose it to light energy, and then observe it in the excited state. Never been done, as far as I know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This posting has been accumulating anonymous downvotes since it went up yesterday. I don't believe I've said anything controversial. Certainly, everyone &quot;knows&quot; that this it what happens when you expose hydrogen to light energy. I'm just questioning whether it's actually observable in an experimental sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I happen to believe that nothing of the sort happens. I think that when hydrogen is in the presence of an oscillating electromagnetic field of the right frequency, it is driven into a superposition of the s and p states. In this state, the charge distribution of the atom oscillates like a tiny antenna; and it interacts with the external field exactly as would a classical antenna with the same parameters. There is nothing in the mathematics of the interaction which requires it to end up in a pure p state, and nothing that requires the energy to be exchanged with the external field in specific quantities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that my opinions would be considered to be verging on &quot;crackpot&quot;, but that is not the point. I am simply asking people who think they know better than me: just what is the experimental evidence that their model of quantized energy exchange is superior to my model. I have already said that I do not believe that observation of the s-p transition of hydrogen provides any such evidence.&lt;/p&gt;&#xA;" OwnerUserId="3253" LastEditorUserId="3253" LastEditorDisplayName="" LastEditDate="2011-07-26T00:25:14.157" LastActivityDate="2011-07-26T00:25:14.157" />
  <row Id="12703" PostTypeId="1" AcceptedAnswerId="12710" CreationDate="2011-07-24T17:24:01.533" Score="4" ViewCount="127" Body="&lt;p&gt;I found this interesting problem in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521876222&quot; rel=&quot;nofollow&quot;&gt;Introduction to Classical Mechanics with Problems and Solutions by David Morin&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given a point $P$ in space, and given a piece of malleable material of&#xA;  constant density, how should you shape and place the material in order&#xA;  to create the largest possible gravitational field at $P$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="966" LastActivityDate="2011-07-25T08:04:57.803" Title="Maximal Gravity" Tags="&lt;gravity&gt;&lt;classical-mechanics&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="12704" PostTypeId="2" ParentId="12690" CreationDate="2011-07-24T17:32:01.697" Score="2" ViewCount="" Body="&lt;p&gt;Take the physics courses you plan with your advisor not with an eye towards getting through easily but with a passion for the subject. If you are passionate about the work you do, if you enjoy studying hard problems and finding even harder ones, the road towards research will be easier, because professors are looking for the dedicated spark in future graduate students and post docs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is similar with going into professional athletics: you have to enjoy the challenge and be willing to work to the limit of your abilities. If you think of a research job as a career maybe you should change  your objective.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-07-24T17:32:01.697" CommunityOwnedDate="2011-07-24T17:32:01.697" />
  <row Id="12705" PostTypeId="2" ParentId="12669" CreationDate="2011-07-24T17:39:05.900" Score="-1" ViewCount="" Body="&lt;p&gt;Marek, you know well that the total wave function of any system is a product of wave functions of independent degrees of freedom (or &quot;particles&quot;) of this system. Each particle wave function may be in the ground or excited state and is determined with some specific to it constants, including mass. This fact does not depend on the state of other wave functions in this product so if a heavy particle wave function remains in its ground state (not yet born particle), it does not influence the other constants (masses of already born particles). When we discover a new heavy particle and use its excited wave function, its presence does not influence the previous particle masses. Masses in such a construction are scale and energy independent. For example the photon mass is the same in presence of electron and in absence of it, and vice versa. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy in an experiment is not a scale of the theory but a given energy of the system, here somebody fooled you. At a given total energy the particle populations may be quite different. It's an external parameter to the theory; the theory should work at any energy used in calculations. So the &quot;running constants&quot; is not the physical property of particles or nature but a feature of the human failure do describe the excitation processes properly. Remember the useless self-action we introduce in the theory (for the sake of what?). We are obliged to get rid of its effects in the end. As well, we couple coupled already things, we couple them with a wrong understanding of physics and obtain immediately lots of problems. And you promote all that as &quot;physics&quot;. -1 for &quot;... all of the parameters that are inserted by hand must depend on the scale.&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-07-24T21:45:10.917" LastActivityDate="2011-07-24T21:45:10.917" CommentCount="3" />
  <row Id="12706" PostTypeId="2" ParentId="12690" CreationDate="2011-07-24T17:43:04.847" Score="1" ViewCount="" Body="&lt;p&gt;Make sure that you have some sort of life skill when you're done that isn't research, too.  Teach yourself to program, or work on an experiment where you build things.  It will enhance your research, and make you more employable should reasearch not work out.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-07-24T17:43:04.847" CommunityOwnedDate="2011-07-24T17:43:04.847" CommentCount="1" />
  <row Id="12707" PostTypeId="2" ParentId="12700" CreationDate="2011-07-24T17:46:43.187" Score="10" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Angular_momentum#Conservation_of_angular_momentum&quot; rel=&quot;nofollow&quot;&gt;Conservation of Angular momentum&lt;/a&gt; applies only when you don't have any net torque on the system. For a spinning top, the frictional force between the top and the floor would give it enough torque in the opposite direction so as to make the top stop (this is much like how a block loses its linear momentum due to friction when it slides down a plank). As far as the energy goes, it gets converted into heat, sound etc. Of course if you took an atomic point of view, you'd see that the angular momentum of the top slightly changed the angular momentum of the molecules of the table. It would also have changed the angular momentum of the earth.  But such effects are hardly noticeable. This xkcd comic is relevant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://xkcd.com/162/&quot; rel=&quot;nofollow&quot; title=&quot;With reasonable assumptions about latitude and body shape, how much time might she gain them? Note: whatever the answer, sunrise always comes too soon. (Also, is it worth it if she throws up?)&quot;&gt;&lt;img src=&quot;http://imgs.xkcd.com/comics/angular_momentum.jpg&quot; alt=&quot;xkcd162&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="966" LastEditorUserId="97" LastEditorDisplayName="" LastEditDate="2011-07-28T15:26:19.880" LastActivityDate="2011-07-28T15:26:19.880" CommentCount="3" />
  <row Id="12708" PostTypeId="2" ParentId="12703" CreationDate="2011-07-24T17:49:03.323" Score="0" ViewCount="" Body="&lt;p&gt;My first guess is to shape the material into a unit-radius sphere and place it so that P is on its surface- this way, you don't 'lose' any gravitational field due either to the 1/r^2 term, or to any asymmetric considerations in the direction of the vector pointing from the source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words- here we have a maximally symmetric source, and we are sitting right on top of it. I'd be interested to see if anything more exotic comes up, though!&lt;/p&gt;&#xA;" OwnerUserId="4663" LastActivityDate="2011-07-24T17:49:03.323" />
  <row Id="12709" PostTypeId="2" ParentId="12689" CreationDate="2011-07-24T18:01:14.890" Score="2" ViewCount="" Body="&lt;p&gt;Okay, I think I have some of it figured out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first part of the problem, we need to know number of microstates if the $A$-density is $p$. We need to position $pN$ particles into $N$ $A$-sites and $(1-p)N$ particles into $N$ $B$-sites. Since these numbers are equal the total number of states is &#xA;$$\#_{\rm states}(p) = {N \choose pN}^2$$&#xA;And so the entropy (with $k_B \equiv 1$) can be written as&#xA;$$S(p) = \log(\#_{\rm states}(p)) =　2 \log{N! \over (pN)! ((1-p)N)!} \sim $$&#xA;$$ \sim 2\left( N \log N - (pN)\log(pN) - ((1-p)N \log ((1-p)N)\right) = $$&#xA;$$ = 2N \left( p \log p - (1-p) \log (1-p) \right)$$&#xA;where we have used the Stirling approximation of the factorial $N! \sim \sqrt{2 \pi N} \left({N \over e} \right)^N$ and we discarded the square root since this isn't proportional to $N$ under logarithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average energy is easily computed. First note that since $Np$ is the number of occupied sites the total energy will be $Np$ times site energy. Now, the site energy will be a sum of contributions from each edge and since the problem has lattice symmetry it will get multiplied by 4 (here I suppose you assume the problem is 2D) and since the neighboring site is occupied with probability $(1-p)$ we get $E = 4 E_0 Np (1-p)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One looks at the thermodynamic limit of macroscopic observables and looks for non-analyticies (these signify phase transition). The most standard quantity we look at is the free energy and its derivatives. As for &lt;em&gt;proving&lt;/em&gt; phase-transition, now that is very hard problem and has been rigorously done only for few special models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct in your last paragraph, one can determine the equilibrium by minimizing the free energy. I'll leave this to you since it is a routine computation.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-24T18:01:14.890" CommentCount="2" />
  <row Id="12710" PostTypeId="2" ParentId="12703" CreationDate="2011-07-24T18:07:27.577" Score="5" ViewCount="" Body="&lt;p&gt;This website seems to provide the solution &lt;a href=&quot;http://star.tau.ac.il/QUIZ/02/A02.02.html&quot; rel=&quot;nofollow&quot;&gt;http://star.tau.ac.il/QUIZ/02/A02.02.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2451" LastActivityDate="2011-07-24T18:07:27.577" CommentCount="1" />
  <row Id="12711" PostTypeId="2" ParentId="12703" CreationDate="2011-07-24T18:09:26.773" Score="0" ViewCount="" Body="&lt;p&gt;Hmmm...that's fun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to note that we seek the largest possible &lt;em&gt;field&lt;/em&gt;, and not the lowest potential versus infinity which might yield a lower result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's consider some obvious cases (all of which I will reduce to a dependence only on $\rho$ and $M$.).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A nearby sphere&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We form the material into a sphere of volume $V=M/\rho$ and radius $R = 3/4 V^{1/3}/\pi$, place it next to $p$ and compute the field as $F_G = G M/R^2 = (3/4) G M V^{-2/3} = (3/4) G M^{1/3} \rho^{-2/3}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A nearby plane segment&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We form the material into a disk of thickness $T$ radius $R = 10 t$ (claiming that this is sufficient for a &quot;infinite place&quot; approximation near the center), so that $V = 100 t^3$so that $t = (V/100)^{1/3} = [M/(100 \rho)]^{1/3}$. We place the it such that $p$ borders center on one side. The surface density is $\sigma = \rho * t = 100^{-2/3} \rho^{2/3} M^{1/3}$ and the field is $F_G = 2 \pi G \sigma = 2 \pi 100^{-2/3} G M^{1/3} \rho^{-2/3}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that so far, both distributions result in a form $F_G = C \times G M^{1/3} \rho^{-2/3}$, so we need only compare the constants. Well, $2 \pi 100^{-2/3} = 2 \pi (0.046) = 0.29$, so the spherical distribution is better than the flat one.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-07-24T18:09:26.773" CommentCount="1" />
  <row Id="12712" PostTypeId="2" ParentId="12700" CreationDate="2011-07-24T18:15:08.690" Score="2" ViewCount="" Body="&lt;p&gt;yeah that's right it does exactly that&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2011-07-24T18:15:08.690" CommentCount="1" />
  <row Id="12713" PostTypeId="2" ParentId="12647" CreationDate="2011-07-24T19:06:45.483" Score="0" ViewCount="" Body="&lt;p&gt;You may be interested in following article: &#xA;Thomas-Paul Hack, Mathias Makedonski &quot;A No-Go Theorem for the Consistent Quantization of the Massive Gravitino on Robertson-Walker Spacetimes and Arbitrary Spin 3/2 Fields on General Curved Spacetimes&quot;&#xA;&lt;a href=&quot;http://arxiv.org/abs/1106.6327&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1106.6327&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1611" LastActivityDate="2011-07-24T19:06:45.483" />
  <row Id="12714" PostTypeId="2" ParentId="12695" CreationDate="2011-07-24T19:22:40.817" Score="3" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Imagine a perfectly insulated box, placed inside the box is an&#xA;  electric heater. The heater is switched on and the box is left to&#xA;  reach equilibrium with it's surroundings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is totally contradictionary. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Either the box is perfectly isolated, which means it will never &#xA;reach equilibrium with the surround, &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Or there is some thermal conduction between the box and the sourroundings, &#xA;then You will reach not an equilibrium, but a stationary state. &#xA;The temperature will reach a point, where heat flow to outside is equal to&#xA;to the heat produced by the heater inside. This is called &#xA;a steady state. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-07-25T19:08:47.563" LastActivityDate="2011-07-25T19:08:47.563" />
  <row Id="12715" PostTypeId="2" ParentId="12700" CreationDate="2011-07-24T19:42:20.840" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does it very slightly change the angular momentum of the table, and&#xA;  then the angular momentum of the Earth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, it does. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exactly the same amount but opposite direction of &#xA;the momentum earth received when the top was accelerated first. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-07-24T19:42:20.840" />
  <row Id="12716" PostTypeId="2" ParentId="12627" CreationDate="2011-07-24T19:54:23.163" Score="-1" ViewCount="" Body="&lt;p&gt;If gases 1 and 2 are identical then the final state is macroscopically no different from the initial one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus for identical gases $\Delta S_{tot} = 0$, and your formulae are inapplicable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can only use this approach  whenever the atoms or molecules of mixing gases are different, though this difference was arbitrarily small. Theoretically, we can assume the limit in which the properties of atoms of a gas go infinitely close to the properties of atoms of another gas.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the number of different types of atoms is finite, and therefore there can not be done such a limit in which the properties of one atom are continuously moved into the property of another. (Quantum mechanics!)&lt;/p&gt;&#xA;" OwnerUserId="552" LastActivityDate="2011-07-24T19:54:23.163" CommentCount="15" />
  <row Id="12717" PostTypeId="1" AcceptedAnswerId="12719" CreationDate="2011-07-24T19:58:50.560" Score="0" ViewCount="139" Body="&lt;p&gt;I want to charge a 12V battery of 150Ah with a solar panel. The solar panel specs is&#xA;12V, 25 Watt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone please provide me how to calculate that how much time it will take to charge the battery? Please provide the calculations and formulas.&lt;/p&gt;&#xA;" OwnerUserId="4440" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-25T06:21:44.750" LastActivityDate="2011-07-25T08:51:54.653" Title="Charging 12V 150Ah battery" Tags="&lt;electricity&gt;&lt;power&gt;&lt;renewable-energy&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="12718" PostTypeId="2" ParentId="12717" CreationDate="2011-07-24T21:57:47.513" Score="0" ViewCount="" Body="&lt;p&gt;150 Amper*hour is the battery charge capacity. With power of 25 Watt = 25 A*V at V = 12 V you will supply 25/12 Coulomb each second (Amper*s). So the charge time is equal to 150/(25/12) = 72 hours.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-07-24T21:57:47.513" />
  <row Id="12719" PostTypeId="2" ParentId="12717" CreationDate="2011-07-24T22:47:56.957" Score="2" ViewCount="" Body="&lt;p&gt;Watts (electrical power) = volts * amps,  so 25W = 12V * 2.1A&lt;/p&gt;&#xA;&#xA;&lt;p&gt;150Amp Hour is the total capacity so 150amp * 1hour, 1amp * 150hours, or 2.1amp for 72hours&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's in an ideal world of course, there are heating losses as you charge the battery, the voltage of the solar panel varies with the load and if you entirely empty a 12V lead acid battery you are likely to damage it. But basically you are looking at 10days of full sunshine &lt;/p&gt;&#xA;" OwnerUserId="2525" LastEditorUserId="2525" LastEditorDisplayName="" LastEditDate="2011-07-25T03:22:53.203" LastActivityDate="2011-07-25T03:22:53.203" CommentCount="3" />
  <row Id="12720" PostTypeId="1" AcceptedAnswerId="12722" CreationDate="2011-07-25T02:34:03.720" Score="1" ViewCount="48" Body="&lt;p&gt;A ball is launched as a projectile with initial speed $v$ at an angle $\theta$ above the horizontal. Using conservation of energy, find the maximum height $h_\text{max}$ of the ball's flight.&#xA;Express your answer in terms of $v$, $g$, and $\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am doing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}&#xD;&#xA;\frac{1}{2}mv^2 &amp;amp;= mgh_\text{max} + \frac{1}{2}m(v\cos\theta)^2\\&#xD;&#xA;v^2 &amp;amp;= 2gh_\text{max} + (v\cos\theta)^2 \\&#xD;&#xA;h_\text{max} &amp;amp;= \bigl(v^2 - (v\cos\theta)^2\bigr)/2g \\&#xD;&#xA;h_\text{max} &amp;amp;= v^2\bigl(1 - (\cos\theta)^2\bigr)/2g \\&#xD;&#xA;h_\text{max} &amp;amp;= \frac{v^2\sin^2\theta}{2g}&#xD;&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that correct, or is there a much easier way...?&lt;/p&gt;&#xA;" OwnerUserId="4147" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-25T03:07:11.670" LastActivityDate="2011-07-25T11:42:16.917" Title="Finding the max height of a ball launched as a projectile using work-energy" Tags="&lt;homework&gt;&lt;energy&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12721" PostTypeId="2" ParentId="12669" CreationDate="2011-07-25T02:57:27.850" Score="-1" ViewCount="" Body="&lt;p&gt;The rest masses of fundamental particles certainly are &lt;strong&gt;NOT&lt;/strong&gt; constants !   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will do copy/past from pages 6-9 (of 20) of a recent document by Alfredo (Independent researcher)&lt;br&gt;&#xA;&lt;em&gt;the sufficient information to show how atomic properties can change&lt;/em&gt;, including mass, at a cosmological level.&lt;br&gt;&#xA;Starting only from data and making no hypothesis he formally presents a dilation(scaling) model of the universe where the atoms are not invariant and physical laws hold, without contradiction with GR, and compares the model with FRW and $\Lambda$CDM models. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the whole paper deserves your attention, it is very clear and it only uses basic physics, imo accessible to undergraduated students. He makes a full study on how we measure, units, local and field constants and laws, how can exist a variation and why we are not aware of such, and a lot more)   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;quoting &lt;a href=&quot;http://vixra.org/pdf/1107.0016v1.pdf&quot; rel=&quot;nofollow&quot;&gt;Alfredo&lt;/a&gt;:&lt;br&gt;&#xA;&lt;strong&gt;How the universe can be scaling&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have seen that if space expansion traces a scaling phenomenon,&#xA;we should expect to detect varying field constants; we have now to&#xA;find out why that is not observed. The first thing to do is to look&#xA;up to the dimension functions of field and some other constants:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{array}{ccl}&#xD;&#xA;\left[G\right] &amp;amp; = &amp;amp; M^{-1}L^{3}T^{-2}\\&#xD;&#xA;\left[\varepsilon\right] &amp;amp; = &amp;amp; M^{-1}Q^{2}L^{-3}T^{2}\\&#xD;&#xA;\left[c\right] &amp;amp; = &amp;amp; LT^{-1}\\&#xD;&#xA;\left[h\right] &amp;amp; = &amp;amp; ML^{2}T^{-1}\\&#xD;&#xA;\left[\sigma\right] &amp;amp; = &amp;amp; M^{-3}L^{-8}T^{5}.&#xD;&#xA;\end{array}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations of field constants ($G,\varepsilon$ and &lt;em&gt;c&lt;/em&gt;) display&#xA;a peculiar characteristic: &lt;em&gt;the summation of exponents of the&#xA;dimension function of each field constant is zero!&lt;/em&gt; This is unexpected&#xA;and does not happen with the other constants. It means that if all&#xA;the four base units concerned change by the same factor,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;M=Q=L=T,&#xD;&#xA;$$&#xA;then the measuring units of field constants hold invariant, $[G]=[\varepsilon]=[c]=1$.&#xA;To see the relevance of this, let us consider that the atomic units&#xA;of mass, charge, length and time change all at the same rate in relation&#xA;to the space units. In that case, because of the property shown above,&#xA;the atomic units of the field constants hold invariant in relation&#xA;to the space ones and, therefore, the field constants are invariant&#xA;in both systems (they are invariant in space units by definition of&#xA;these ones). The geometry of space would be scaling in atomic units&#xA;while the value of field constants would hold invariant--- &lt;em&gt;which&#xA;is exactly what cosmic data seems to display&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the dimensions of field constants display null summation&#xA;of exponents can just be a coincidence, but it is also the kind of&#xA;indication we were looking for, a property embedded in physical laws.&#xA;This is the only way we can consider a previously unknown fundamental&#xA;property without conflicting with established physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have now the fundamental understanding that can support a scaling&#xA;(dilation) model of the universe and we will now proceed to the formal&#xA;development of that model.&lt;br&gt;&#xA;...&lt;br&gt;&#xA;Hence, one of the systems of units is defined from matter properties,&#xA;designated here by atomic system and identified by A (&quot;A&quot; from &quot;atomic&quot;)&#xA;and the other is the space system of units, identified by S (&quot;S&quot;&#xA;from &quot;space&quot;); the later is such that space properties (geometry&#xA;and field constants) remain invariant in it, which is required to&#xA;qualify the S system as internally defined in relation to space. Thus,&#xA;the conditions that define the S system are the following:&lt;br&gt;&#xA; - The units of S are such that the S measures of field constants hold&#xA;   invariant;&lt;br&gt;&#xA; - The length unit of S is such that the wavelength of a propagating&#xA;   radiation in vacuum is time invariant.&lt;br&gt;&#xA;The base quantities are Mass (&lt;em&gt;M&lt;/em&gt;), Charge (&lt;em&gt;Q&lt;/em&gt;), Time (&lt;em&gt;T&lt;/em&gt;),&#xA;Length (&lt;em&gt;L&lt;/em&gt;) and Temperature ($\theta$), and the ratio between&#xA;A and S base units is denoted by $M_{AS},Q_{AS},T_{AS},L_{AS},\theta_{AS}$.&#xA;Note that the ratio between the A and S units of any quantity or constant&#xA;is therefore expressed by the respective dimension function;&lt;br&gt;&#xA;...&lt;br&gt;&#xA;&lt;strong&gt;Postulates&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model will be deducted not from hypotheses&#xA;but from relevant observational results, which are stated as postulates:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In atomic units (A), all local and field constants are&#xA;time-independent. &lt;/li&gt;&#xA;&lt;li&gt;$L_{AS}\,$decreases with time.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The first postulate is not fully supported in experience, as we cannot&#xA;state it with the required error margin; however, we have also no&#xA;sound indication from observations that it might be otherwise. The&#xA;second postulate represents the observed phenomenon of space expansion&#xA;in atomic units, stated in this unusual way because it is presented&#xA;as a function of $L_{AS}\,$, i.e., of the ratio between atomic&#xA;and space length units and not the inverse, as usual.&lt;br&gt;&#xA;...&lt;br&gt;&#xA;S units, by definition, are such that  (eq.1)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac{dG_{S}}{dt_{S}}=\frac{d\varepsilon_{S}}{dt_{S}}=\frac{dc_{S}}{dt_{S}}=0.&#xA;\end{equation}&#xA;Since the field constants are time-invariant also in atomic units,&#xA;as stated by postulate 1, and since the two systems of units are identical&#xA;at $t=0$, then the values of these constants are the same in the&#xA;two systems at whatever time moment:   (eq.2)&#xA;$$&#xD;&#xA;\begin{array}{ccccc}&#xD;&#xA;G_{A} &amp;amp; = &amp;amp; G_{S} &amp;amp; = &amp;amp; G\\    &#xD;&#xA;\varepsilon_{A}^{\vphantom{l}^{\vphantom{L}}} &amp;amp; = &amp;amp; \varepsilon_{S} &amp;amp; = &amp;amp; \varepsilon\\  &#xD;&#xA;c_{A}^{\vphantom{l}^{\vphantom{L}}} &amp;amp; = &amp;amp; c_{S} &amp;amp; = &amp;amp; c.  &#xD;&#xA;\end{array}&#xD;&#xA;$$&lt;br&gt;&#xA;The relation between the S and A values of each constant is the one&#xA;between the respective A units and S units, which is given by the&#xA;dimension function,&lt;br&gt;&#xA;therefore  (eq.3)&lt;br&gt;&#xA;$$  &#xD;&#xA;\begin{array}{ccccl}&#xD;&#xA;\dfrac{G_{S}}{G_{A}} &amp;amp; = &amp;amp; \left[G\right]_{AS} &amp;amp; = &amp;amp; {M_{AS}^{-1}}{L_{AS}^{3}}{T_{AS}^{-2}}=1\\  &#xD;&#xA;\dfrac{\varepsilon_{S}^{\vphantom{l}^{\vphantom{L}}}}{\varepsilon_{A}} &amp;amp; = &amp;amp; \left[\varepsilon\right]_{AS} &amp;amp; = &amp;amp; {M_{AS}^{-1}}{Q_{AS}^{2}}{L_{AS}^{-3}}{T_{AS}^{2}}=1\\  &#xD;&#xA;\dfrac{c_{S}^{\vphantom{l}^{\vphantom{L}}}}{c_{A}} &amp;amp; = &amp;amp; \left[c\right]_{AS} &amp;amp; = &amp;amp; L_{AS}{T_{AS}^{-1}}=1.  &#xD;&#xA;\end{array}&#xD;&#xA;$$&#xA;This set of equations implies $M_{AS}=Q_{AS}=T_{AS}=L_{AS}$. By postulate&#xA;2, $L_{AS}$ is a time function, therefore the solution can be presented&#xA;as:  (eq.4)&lt;br&gt;&#xA;$$ &#xD;&#xA;\begin{equation}&#xD;&#xA;M_{AS}(t)\,=Q_{AS}(t)\,=T_{AS}(t)\,=L_{AS}(t)\,.\,&#xD;&#xA;\end{equation}&#xD;&#xA;$$&#xA;Note that temperature is independent of this result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is to define this time function, which is the space&#xA;scale factor law. As all the above four base quantities follow this&#xA;function, it is convenient to identify it by a specific designation;&#xA;in this work this scaling law is identified by the symbol $\mathcal{\alpha}$:  (eq.5)&lt;br&gt;&#xA;$$&#xD;&#xA;\begin{equation}&#xD;&#xA;\alpha(t)\,=\, L_{AS}(t).&#xD;&#xA;\end{equation}&#xD;&#xA;$$&#xA;...&lt;br&gt;&#xA;&lt;strong&gt;The scaling law&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make no hypothesis on the cause of the expansion is to consider&#xA;that expansion is due to a fundamental property; to consider otherwise&#xA;would imply a specific hypothesis on a particular phenomenon driving&#xA;the expansion. Therefore, for this model, the space expansion is due&#xA;to a fundamental property, tracing a self-similar phenomenon. Likewise,&#xA;as no hypothesis is made on how fundamental properties may vary with&#xA;position on space and time, it is assumed that they do not depend&#xA;on it. This implies that the scaling has a constant time rate in some&#xA;physically relevant system of units, i.e., that the scaling&#xA;law is exponential in such system of units. There are only two possibilities&#xA;in the framework established for this model: either space expansion&#xA;is exponential in A units ($L_{SA}(t_{A})=\alpha^{-1}(t_{A})$ is&#xA;exponential) or matter evanesces exponentially in S units ($L_{AS}(t_{S})=\alpha(t_{S})$&#xA;is exponential). The former case does not fit observations; only the&#xA;later case is possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general expression for a scaling law exponential in S units is  (eq.6)&#xA;$$&#xD;&#xA;\begin{equation}&#xD;&#xA;\alpha(t_{S})=k_{1}e^{k_{2}\cdot t_{s}}\,;&#xD;&#xA;\end{equation}&#xD;&#xA;$$&#xA;at the moment $t_{A}=t_{S}=0$ it is $\alpha(0)=L_{AS}(0)=1,$ so &#xA;$k_{1}=1$; note now that  (eq.7)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{equation}&#xD;&#xA;\frac{dt_{S}}{dt_{A}}=T_{AS}=\alpha\,&#xD;&#xA;\end{equation}&#xD;&#xA;$$&#xA;which shows that the variation of the measure of time is inversely&#xA;proportional to the time unit; and that  (eq.8)&#xA;$$&#xD;&#xA;\begin{equation}&#xD;&#xA;r_{A}=r_{S}{L_{AS}^{-1}}=r_{S}\cdot\alpha^{-1},&#xD;&#xA;\end{equation}&#xD;&#xA;$$&#xA;where &lt;em&gt;r&lt;/em&gt; is the distance to some point, or its length coordinate;&#xA;as the rate of space expansion at &lt;em&gt;t&lt;/em&gt;=0 is, by definition, the&#xA;value of Hubble constant, represented by $H_{0}$, then  (eq.9)&lt;br&gt;&#xA;$$&#xD;&#xA;\begin{equation}&#xD;&#xA;H_{0}=\left(\frac{1}{r_{A}}\frac{dr_{A}}{dt_{A}}\right)_{0}=-k_{2},&#xD;&#xA;\end{equation}&#xD;&#xA;$$&#xA;therefore  (eq.10)&lt;br&gt;&#xA;$$&#xD;&#xA;\begin{equation}&#xD;&#xA;\alpha(t_{S})=e^{-H_{0}\cdot t_{S}}.&#xD;&#xA;\end{equation}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hubble constant is the present space expansion rate for an atomic&#xA;observer and is the matter evanescence rate (negative) for a space&#xA;observer.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-07-25T02:57:27.850" CommentCount="4" />
  <row Id="12722" PostTypeId="2" ParentId="12720" CreationDate="2011-07-25T03:10:59.873" Score="1" ViewCount="" Body="&lt;p&gt;Under the constraints of the problem, then yes, what you're doing is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you weren't required to use conservation of energy, then it would probably be easier to calculate the vertical component of the initial velocity and use 1D kinematics.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-07-25T03:10:59.873" CommentCount="1" />
  <row Id="12723" PostTypeId="1" CreationDate="2011-07-25T05:50:10.903" Score="1" ViewCount="63" Body="&lt;p&gt;As photons bounce around and finally find their way into our eyes, are they continually relinquishing the information of the previous thing they &quot;bounced off of&quot;?  Is this the reason why we receive a coherent, continuous image and not a chronologically chaotic image of the past meshed with the present? &lt;/p&gt;&#xA;" OwnerUserId="4667" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-25T06:33:35.727" LastActivityDate="2011-07-25T06:33:35.727" Title="Is there a limit to the visual information a photon can carry?" Tags="&lt;light&gt;&lt;photons&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="1" />
  <row Id="12724" PostTypeId="2" ParentId="12717" CreationDate="2011-07-25T06:20:40.133" Score="0" ViewCount="" Body="&lt;p&gt;The answers you've got so far from Vladimir and Martin give you a good first-order approximation: power = current x voltage. Energy = power x time. So your 12V battery of 150Ah needs 1800Wh of energy (12 x 150). So a 25W PV panel would need 72 hours at full output (1800Wh/25W).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation is:&#xA;Peak-hours required = $\frac{V_{batt} \times capacity_{batt}}{Power_{PV}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's an unusual combination of quite a small panel and a very big battery - is the battery designed to be a multi-week store for a low-power system, by any chance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more accurate answer, you need a &lt;strong&gt;lot&lt;/strong&gt; more information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The output of the panel at any moment will depend on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) the voltage, (which will be determined by the battery, and will change as the battery charges)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) the ambient temperature, and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) the amount light hitting the panel. The amount of light hitting the panel will depend on panel tilt, orientation, overshading, weather, altitude, location, time of year, time of day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (1), you need the panel IV curve, typically presented on the manufacturer's datasheet. That datasheet will also tell you how output varies with panel temperature. You can then estimate the panel's temperature from the ambient temperature - it will (to the first order) be a fairly steady amount above ambient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some example power curves for a 12V PV (and it looks like it's approx 125W peak) panel, from &lt;a href=&quot;http://www.altestore.com/howto/Solar-Power-Residential-Mobile-PV/Off-Grid-Solar-Systems/Electrical-Characteristics-of-Solar-Panels-PV-Modules/a87/&quot; rel=&quot;nofollow&quot;&gt;altestore.com&lt;/a&gt; :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/m8KyD.jpg&quot; alt=&quot;PV panel I-V curve&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need to know about the battery's characteristics as it charges: the heat losses, and how its voltage changes with temperature and % charged. In a more complex system that involves a controller too, you'd need to know how the controller behaves, how it tracks the maximum power point, and what its internal losses are. A controller is very desirable, not only because it will track the PV panel's maximum power point, but also because it will prevent the battery over-charging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various online PV calculators that will combine some of this information to give you estimates of output, such as the &lt;a href=&quot;http://www.nrel.gov/rredc/pvwatts/&quot; rel=&quot;nofollow&quot;&gt;US NREL PVWATTS calculator&lt;/a&gt;. And there are free online GIS insolation resources to give you the raw source information to do detailed calculations yourself, such as the &lt;a href=&quot;http://re.jrc.ec.europa.eu/pvgis/&quot; rel=&quot;nofollow&quot;&gt;EU PVGIS database.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4066" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-25T08:51:54.653" LastActivityDate="2011-07-25T08:51:54.653" CommentCount="1" />
  <row Id="12725" PostTypeId="2" ParentId="12723" CreationDate="2011-07-25T06:33:14.310" Score="4" ViewCount="" Body="&lt;p&gt;It depends on what the photon is bouncing off of. For most materials, it's true that a reflected photon carries no information about what it interacted with prior to the reflection, but that's not the case for all materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be precise, when a photon - actually an electromagnetic wave - is reflected, what really happens is that it gets absorbed by the reflecting material, and then a new EM wave is emitted to &quot;replace&quot; it. (Actually, it's more complicated than that; see &lt;a href=&quot;http://physics.stackexchange.com/questions/12208/how-does-anti-reflective-coating-work&quot;&gt;this other question&lt;/a&gt; for example, but the re-emission explanation will suffice for now.) There are two properties of interest that could apply to the re-emitted wave:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For most materials, the re-emitted wave's frequency is determined by the electronic properties of the material itself; these materials can only emit certain specific frequencies of light, which determines their color. Other materials, though, can emit any frequency of light, which means the frequencies (and thus colors) of the re-emitted EM waves will match those of the absorbed waves. These materials are either reflective or white.&lt;/li&gt;&#xA;&lt;li&gt;For most materials, the absorption and emission are separate events, which means that the emitted EM wave can be in a random direction, regardless of which direction the absorbed wave came from. This is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffuse_reflection&quot; rel=&quot;nofollow&quot;&gt;diffuse reflection&lt;/a&gt; and it means that the appearance of the reflecting material will be determined by the material itself. But for some materials, due to their electronic structure the absorption and emission are &quot;linked&quot; in a sense, such that the emitted wave comes out at an angle which depends on the direction of the absorbed wave and the orientation of the surface. This is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Specular_reflection&quot; rel=&quot;nofollow&quot;&gt;specular reflection&lt;/a&gt; and it is the kind of thing that happens in mirrors. In this case, the &quot;appearance&quot; of the material is determined by the absorbed light. If you think about it, when you look at a mirror, you don't see the mirror itself, but rather some other thing whose reflection appears in the mirror.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In fact, for both these effects, the possibilities listed are only extremes of a continuum. A material can be partially reflective, for example, meaning that part of the absorbed EM wave is re-emitted in a specific direction and part of it is re-emitted randomly in any direction, as in window glass. In a case like this, you could say that the reflected light waves carry partial information about what they had previously been emitted from.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-07-25T06:33:14.310" />
  <row Id="12726" PostTypeId="2" ParentId="12703" CreationDate="2011-07-25T08:04:57.803" Score="0" ViewCount="" Body="&lt;p&gt;The trivial solution is compact the mass to a point object, I guess. In that case the gravity at point P is infinite, because the distance is zero. I'm not sure if this violates the 'constant density' principle though.&lt;/p&gt;&#xA;" OwnerUserId="4574" LastActivityDate="2011-07-25T08:04:57.803" />
  <row Id="12727" PostTypeId="2" ParentId="12669" CreationDate="2011-07-25T08:19:12.533" Score="-2" ViewCount="" Body="&lt;p&gt;@Marek: Wilson never dealt with the true QFT; he used QFT methods in his phase transition problems where QFT methods are not applicable by definition, strictly speaking. That is why his QFT is called an &quot;effective&quot; (smoothed) description. Then QFT people, who could not make up their minds for decades what the value for the cut-off to choose in their RG, decided to call their sorry QFTs &quot;effective&quot; too and choose the cut-off according to the energy. This choice was as good as their previous choices where cut-off tended to infinity but such a choice &quot;explained&quot; renormalizations &quot;a la Wilson&quot;. It is a sloppiness and weakness of QFT theorists that you promote here. None received Nobel prise for cut-off dependent constants in QFT, do not exaggerate. I downvote when I see you do not understand physics, purpose of physics, and repeat as a parrot what sinful people use as a pretext to hide their failures.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-07-25T08:57:03.860" LastActivityDate="2011-07-25T08:57:03.860" CommentCount="5" />
  <row Id="12728" PostTypeId="2" ParentId="12693" CreationDate="2011-07-25T10:02:07.903" Score="-2" ViewCount="" Body="&lt;p&gt;most wall outlets in the united states run at 60HZ. each cycle gives two peaks so if you multiply your result by 120 you should get the number of electrons. though to be fair I think if you actually study conductors you would know that the mean drift velocity is super slow--- so that it's not really correct to say that electrons are passing through the laptop... though I don't really know enough about this to go any further.&lt;/p&gt;&#xA;" OwnerUserId="4570" LastActivityDate="2011-07-25T10:02:07.903" CommentCount="2" />
  <row Id="12730" PostTypeId="2" ParentId="12720" CreationDate="2011-07-25T11:42:16.917" Score="0" ViewCount="" Body="&lt;p&gt;I am not sure what David is pointing at in his second paragraph, but you could also consider only the vertical component of the velocity and ignore the horizontal one. The initial vertical velocity is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;v_{vert} = v \sin(\theta)&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;0.5 m v^2 = m g h&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this leads to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\begin{align*}&#xD;&#xA;0.5 m (v \sin(\theta))^2 = m g h_{max}\\&#xD;&#xA;\frac{0.5 (v \sin(\theta))^2}{g} = h_{max}\\&#xD;&#xA;h_{max} = \frac{v^2 \sin^2(\theta)}{2 g  }&#xD;&#xA;\end{align*}&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="4574" LastActivityDate="2011-07-25T11:42:16.917" CommentCount="1" />
  <row Id="12731" PostTypeId="2" ParentId="12693" CreationDate="2011-07-25T12:10:39.303" Score="1" ViewCount="" Body="&lt;p&gt;I would start from power consumption.  It's a little unclear to me if you're talking about electron use in all of the components or the CPU, but since 3 Volts was used, I'll take the discussion to be limited to CPU.  Let's say 30 W of power consumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I=\frac{P}{V}=\frac{30 W}{3 V} = 10 A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for the electron flux (denoted $\phi$), we can just divide by the electron charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi = \frac{I}{e} = \frac{10 A}{1.9 \times 10^{-19} C} = 5.263 \times 10^{19} \frac{1}{s} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the number of &lt;em&gt;net&lt;/em&gt; electrons passing the power supplying terminals.  I've heard that the drift velocity of electrons is 1000 to 10,000 times less than the actual velocity of electrons.  So... you could just multiply the above number to get $5.3 \times 10^{22}$ electrons participating in current passing through a cross section of the terminal wire, but this is unsatisfactory for a number of reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, at any given point in your circuit there are lots of electrons passing every which way.  Why bother counting these unless we're talking about the role of conduction band electrons in temperature itself?  Additionally, even if we restrict ourselves to the net electron flux in a wire - there are LOTS of wires in the CPU, and the passing of electrons through the terminal wire is not exclusive with electrons passing through other points, and a CPU is a very complicated machines with many many routes.  In order to obtain more objective measures, you might want to look at the rate at which electrons bleed off, or loose voltage in some way (this would be the most analogous to a computer &quot;using&quot; those electrons).  Otherwise, there is a virtually unlimited number of net electrons passing through all points in the circuits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also don't mean to present this as a very thorough answer, but at least I've taken a different approach to get the same order of magnitude of electrons that the question had.  These numbers are just generally representative of a few Amps, but one needs to be specific in labeling what that number represents.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-07-25T13:25:03.283" LastActivityDate="2011-07-25T13:25:03.283" />
  <row Id="12732" PostTypeId="1" AcceptedAnswerId="12750" CreationDate="2011-07-25T13:14:34.603" Score="5" ViewCount="170" Body="&lt;p&gt;Consider the one-dimensional Schrodinger equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{1}{2}D^2 \psi(x)+V(x)\psi(x)=E\psi(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $D^2=\dfrac{d^2}{dx^2},V(x)=-\dfrac{1}{|x|}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to calculate the ground state energy(which is known to be $-0.5$) with finite difference method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose $\psi(-a)=\psi(a)=0$ for sufficiently large $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Split $[-a,a]$ into $N$ sub-intervals with equal length $h$:&#xA;$$-a=x_1&amp;lt;x_2&amp;lt;\cdots&amp;lt;x_{N+1}=a$$&#xA;$x_{i+1}-x_i=h.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D^2\psi(x_n)\to\frac{\psi(x_{n-1})-2\psi(x_{n})+\psi(x_{n+1})}{2h^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{1}{2}D^2 \psi(x_n)+V(x_n)\psi(x_n)=E\psi(x_n)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-y_{n-1}+2(1+V_n h^2)y_n-y_{n+1}=2h^2 Ey_n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $y_n=\psi(x_n)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that $y_1=y_{N+1}=0$, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;y_2 \\ &#xD;&#xA;\vdots \\ &#xD;&#xA;y_N&#xD;&#xA;\end{pmatrix}&#xD;&#xA;=&#xD;&#xA;2h^2E&#xD;&#xA;\begin{pmatrix}&#xD;&#xA;y_2 \\ &#xD;&#xA;\vdots \\ &#xD;&#xA;y_N&#xD;&#xA;\end{pmatrix}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H=\begin{pmatrix}&#xD;&#xA;2(1+V_2h^2) &amp;amp; -1 &amp;amp;  &amp;amp;  \\ &#xD;&#xA;-1 &amp;amp; 2(1+V_3h^2) &amp;amp; \ddots &amp;amp;  \\ &#xD;&#xA; &amp;amp; \ddots &amp;amp; \ddots &amp;amp; -1 \\ &#xD;&#xA; &amp;amp;  &amp;amp; -1 &amp;amp; 2(1+V_Nh^2)&#xD;&#xA;\end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a $N-1$ tridiagonal matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the least eigenvalue of $H$ is $\lambda$, then I expect $\dfrac{\lambda}{2h^2}$ to approximate the ground state energy $E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I write a MATLAB program:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function m = onedimen1(a,M)&#xA;&#xA;N = 2*M+1;&#xA;&#xA;h = 2*a/N;&#xA;t = -a;&#xA;&#xA;for i = 1 : N-2&#xA;    t = t+h;&#xA;    H(i,i) = 2*(1+vp(t)*h*h);&#xA;    H(i+1,i) = -1;&#xA;    H(i,i+1) = -1;&#xA;end&#xA;&#xA;H(N-1,N-1) = 2*(1+vp(t+h)*h*h);&#xA;&#xA;m1=eig(H)/(2*h*h);&#xA;m = m1(1:5);&#xA;&#xA;  function y = vp(x)&#xA;    y = - 1/(abs(x)) ;&#xA;  end&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;but when I try&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;onedimen1(10,100)&#xA;onedimen1(10,1000)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;the least eigenvalue of $H$ divided by $2h^2$ appears to be too small(and tends to negative infinity when $N$ increases), but the second largest divided by $2h^2$ is always near $-0.5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P.S. If I change $V(x)$ to $V(x)=\dfrac{1}{|x|}$ then the least eigenvalue of $H$ is stable)&lt;/p&gt;&#xA;" OwnerUserId="4314" LastActivityDate="2011-07-28T09:22:29.997" Title="Solving one dimensional Schrodinger equation with finite difference method" Tags="&lt;quantum-mechanics&gt;&lt;computational-physics&gt;&lt;schrodinger-equation&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="12733" PostTypeId="1" CreationDate="2011-07-25T13:31:26.393" Score="1" ViewCount="68" Body="&lt;p&gt;I really need some very simple discussions of random walk (probability). Couldn't get anything from class, more so from Reif. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4668" LastActivityDate="2011-07-25T14:26:48.650" ClosedDate="2011-07-25T21:07:41.780" Title="Can somebody provide some sort of crash course on random walk and its problems at the level of a beginning undergraduate student in physics?" Tags="&lt;statistical-mechanics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="12734" PostTypeId="1" CreationDate="2011-07-25T14:15:19.773" Score="1" ViewCount="118" Body="&lt;p&gt;Suppose if a windmill is made torotate in clockwise direction and due to air flow if it rotates in anti clockwise direction then does it produce electricity?&lt;/p&gt;&#xA;" OwnerUserId="4026" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-25T15:13:08.187" LastActivityDate="2011-07-25T15:20:18.583" Title="What happens if a wind mill rotates in opposite direction?" Tags="&lt;electromagnetism&gt;&lt;electricity&gt;&lt;renewable-energy&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="12735" PostTypeId="2" ParentId="12733" CreationDate="2011-07-25T14:26:48.650" Score="1" ViewCount="" Body="&lt;p&gt;Random walk is intimately connected with diffusion, heat equation, Laplacian, harmonic functions, quadratic forms and Gaussian distribution. Here's a sketch of the relationship in the discrete case (continuous case is conceptually similar but requires much more familiarity with probability theory). I'll try to be as consice as possible since there is too much going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider some connected subset $S$ of a square lattice and suppose there is a random process such that in each time step the particle can jump to a neighboring site (with probability of jumps being uniform for simplicity). This can be considered as an operator (more precisely a probability kernel) $P$ acting on probability measures on $S$ (so called transition operator in the theory of Markov processes). As this is essentialy linear algebra we obviously want to find eigenstates of this operator. So, if $f$ is a stationary distribution w.r.t. this walk, it must be the case that $$\sum_{j \in n(i)} {1 \over 4} f(j) = f(i)$$ where $n(i)$ is the set of neighbors of the site $i$. In other words, such a function must be (discretely) harmonic $\Delta f = 0$ where $\Delta$ is a discretized Laplacian. Since we now have eigenstates we can also consider evolution of the distribution under time in a macroscopic description and this will bring us to the heat equation $\partial_t f = k \Delta f$, $k$ being some coefficient of diffusion or conduction or whatever it is we are trying to model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there is another nice connection between this random process and certain difference equation &lt;a href=&quot;http://en.wikipedia.org/wiki/Feynman%2dKac_formula&quot; rel=&quot;nofollow&quot;&gt;Feynman-Kac formula&lt;/a&gt; (here shown in the continuous version -- stochastic process and PDE). This formula tells us that a solution to this equation can be written as a sum over all paths of the random walk. Incidentally, the equation we are interested in is given by a quadratic potential between nearest neighbors (this in turn leads to Gaussian measures since the Hamiltonian $H$ is quadratic and the equilibrium measure is given by usual Boltzmann formula  $p \sim \exp(-\beta H)$). The condition on being a solution is then that the function is harmonic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the problems and topics, one is interested in things such as recurrence (whether the random walk ever returns to origin with positive probability), exit times, etc. If the RW is recurrent one can study the average time of recurrence. Note that these questions are not that hard on square lattice with uniform probability but become harder on arbitrary graphs with arbitrary probability assignments. We can again study stationary measures, evolution of macroscopic observables, Gaussian distributions, etc. (and essentialy these become tools that can tell us something about the properties of graphs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a square lattice in 1D and 2D the random walk is recurrent (&quot;drunkard finds his way home&quot;) but in 3D it isn't. This is connected to the fact that Coulomb potential (recall the relationship with harmonics!) in 2D is logorithmic while in 3D it has a power law. Another very close relation is that there are no equilibrium Gibbs states for a massive free field in dimensions less than 3.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-25T14:26:48.650" CommentCount="1" />
  <row Id="12736" PostTypeId="1" CreationDate="2011-07-25T14:53:29.363" Score="0" ViewCount="89" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/5456/the-speed-of-gravity&quot;&gt;The speed of gravity&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I am real noob in physics, so sorry if this question is really stupid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today in a casual conversation I claimed that if the sun were to instantly disappear, we would have felt it in any way nos ooner than after about 8 minutes(which is the distance in light minutes from Earth to Sun). My friend said that light-wise I am right, but we would instantly have felt the gravitation change. I argued that gravitation doesn't spread faster than light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right or is he right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance and sorry again for the noob Q :)&lt;/p&gt;&#xA;" OwnerUserId="4669" LastActivityDate="2011-07-25T15:08:09.490" ClosedDate="2011-07-25T15:20:05.983" Title="Does gravity spread instantly?" Tags="&lt;gravity&gt;&lt;speed-of-light&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="12737" PostTypeId="2" ParentId="12736" CreationDate="2011-07-25T14:59:52.407" Score="3" ViewCount="" Body="&lt;p&gt;First of all, Sun can not just disappear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answering the question, gravity does change with the speed of light: gravity waves have a speed of light (formally, the answer is more complex as the speed of light is affected by gravity itself, but to make the long answer short it is a speed of light). Thus, if masses and their configuration change the resulting gravity field changes with the speed of light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google gravitation waves.&lt;/p&gt;&#xA;" OwnerUserId="4657" LastActivityDate="2011-07-25T14:59:52.407" />
  <row Id="12738" PostTypeId="2" ParentId="12736" CreationDate="2011-07-25T15:00:35.287" Score="1" ViewCount="" Body="&lt;p&gt;This question has been asked before: &lt;a href=&quot;http://physics.stackexchange.com/questions/5456/the-speed-of-gravity&quot;&gt;The speed of gravity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your friend is definitely wrong, but there's also a problem with your example. General relativity doesn't allow matter to be created or destroyed. (Technically the way to say this is that GR has local conservation of energy-momentum.) If you put in an assumption that matter is destroyed, then the equations of general relativity are not self-consistent, so they don't predict anything. To fix up your example, it would be better to imagine that the sun was suddenly yanked out of the solar system at high speed. In that case, we would feel the change in gravity 8 minutes later, at the same time as the change in sunlight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General relativity predicts that disturbances in the gravitational field propagate as gravitational waves, and that low-amplitude gravitational waves travel at the speed of light. Gravitational waves have never been detected directly, but the loss of energy from the Hulse-Taylor binary pulsar has been checked to high precision against GR's predictions of the power emitted in the form of gravitational waves. Therefore it is extremely unlikely that there is anything seriously wrong with general relativity's description of gravitational waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does it make sense that low-amplitude waves propagate at c? In Newtonian gravity, gravitational effects are assumed to propagate at infinite speed, so that for example the lunar tides correspond at any time to the position of the moon at the same instant. This clearly can't be true in relativity, since simultaneity isn't something that different observers even agree on. Not only should the &quot;speed of gravity&quot; be finite, but it seems implausible that that it would be greater than c; based on symmetry properties of spacetime, one can prove that there must be a maximum speed of cause and effect.[Ignatowsky, Pal] Although the argument is only applicable to special relativity, i.e., to a flat spacetime, it seems likely to apply to general relativity as well, at least for low-amplitude waves on a flat background. As early as 1913, before Einstein had even developed the full theory of general relativity, he had carried out calculations in the weak-field limit that showed that gravitational effects should propagate at c. This seems eminently reasonable, since (a) it is likely to be consistent with causality, and (b) G and c are the only constants with units that appear in the field equations, and the only velocity-scale that can be constructed from these two constants is c itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;High-amplitude gravitational waves need &lt;em&gt;not&lt;/em&gt; propagate at c. For example, GR predicts that a gravitational-wave pulse propagating on a background of curved spacetime develops a trailing edge that propagates at less than c.[MTW, p. 957] This effect is weak when the amplitude is small or the wavelength is short compared to the scale of the background curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;W.v.Ignatowsky, Phys. Zeits. 11 (1911) 972&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Palash B. Pal, &quot;Nothing but Relativity,&quot; Eur.J.Phys.24:315-319,2003, &lt;a href=&quot;http://arxiv.org/abs/physics/0302045v1&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/physics/0302045v1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MTW - Misner, Thorne, and Wheeler, Gravitation&lt;/p&gt;&#xA;" OwnerUserId="4552" LastEditorUserId="4552" LastEditorDisplayName="" LastEditDate="2011-07-25T15:08:09.490" LastActivityDate="2011-07-25T15:08:09.490" />
  <row Id="12739" PostTypeId="2" ParentId="12734" CreationDate="2011-07-25T15:11:17.580" Score="3" ViewCount="" Body="&lt;p&gt;Wind turbines use induction generators, and most commonly, the jargon of doubly-fed induction generators applies.  Now, the terminology I've seen most often in textbooks identifies motors and generators to be the same thing, just operating in different &quot;directions&quot;, and both are called machines.  The next major item to cover is the fact that a wind turbine produces power for a 3-phase alternating current electrical grid.  This is important and I will return to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mechanically, the blades harness the power of the wind through their airfoil shape and orientation.  The common designs, if my understanding is correct, actually rotate in order to face the wind at all times.  Now, the question can still remain of what would happen if this system was not functioning, someone let it face the opposite direction (face opposite the wind direction) and the gearbox and electrical systems were still connected.  In this configuration, the wind would be pushing it to rotate in the opposite direction of what it normally does because the orientation of the blades would be opposite (they actually have control of the orientation of the blades, but turning 180 degrees may or may not be within the engineered range of motion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note that we are going from positive rotation, positive torque (torque from the wind) to negative rotation, negative torque.  It is a generator as long as the direction of rotation and torque are the same, and it is a motor if the directions are opposite.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, we are still talking about the wind imparting energy to the turbine but in the opposite direction (rotation) than usual.  This could work except for one problem, which has to do with the 3-phase power.  Academically, as well as practically, the phases may be labeled A, B, and C.  One phase &quot;leads&quot; another phase by 120 degree electrical angle.  Think of a sinusoid and you won't miss any accuracy in the model.  This means that any given phase will have a phase in front of it and a phase behind it.  The normal ordering of the phases is ABC, logically.  Electrical devices that deliver current in all phases at equal magnitude in that order are said to be delivering positive sequence current.  An alternative exists, in fact two do.  You may want to read up more here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Symmetrical_components&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Symmetrical_components&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A negative sequence current is where all phases get current in equal amounts and the order is reverse, that is, CBA.  Zero sequence is yet another concept where current is delivered through all sequences with no electrical angle between them.  If the generator was working in this backwards direction and it was a synchronous machine (it's not btw) then it would be delivering negative sequence current.  This is not desirable for the operation of the electric grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A problem with this description is that the machine is really an induction generator, and not only would the delivered currents be wrong, but the electronics for energizing the rotor itself would be messed up.  I don't know what would happen in this case, but I hope I've at least impressed the importance of the &quot;handedness&quot; in the electrical systems of generators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, it would still produce electricity, but it would be the wrong kind of electricity.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-07-25T15:11:17.580" CommentCount="3" />
  <row Id="12740" PostTypeId="2" ParentId="12734" CreationDate="2011-07-25T15:20:18.583" Score="0" ViewCount="" Body="&lt;p&gt;In theory, and taking a very simplistic view, yes: it moves a conductor through a magnetic field, and this would induce an electric current, regardless of whether the rotation is clockwise or anticlockwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in practice, the electronics, mechanics and blade design of a wind turbine stop all this from happening, and neither the mechanics nor the electronics are designed to cope with it.&lt;/p&gt;&#xA;" OwnerUserId="4066" LastActivityDate="2011-07-25T15:20:18.583" />
  <row Id="12741" PostTypeId="1" AcceptedAnswerId="12743" CreationDate="2011-07-25T15:20:57.803" Score="0" ViewCount="45" Body="&lt;p&gt;I was wondering whether or not a slide in a slide projector acts as light diffuser? So when I have a light source that does not have a parallel beam on the slide can I expect that the other side of the slide is illuminated with a Lambertian of similar reflectance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering this because I want to know if I have to raytrace the light path of a projector from the slide (assuming Lambertian reflectance of the slide) or from the light source.&lt;/p&gt;&#xA;" OwnerUserId="3725" LastActivityDate="2011-07-25T16:24:50.377" Title="Does a slide in a projector act as a diffuser?" Tags="&lt;optics&gt;&lt;light&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="12742" PostTypeId="1" AcceptedAnswerId="12755" CreationDate="2011-07-25T15:25:56.743" Score="3" ViewCount="100" Body="&lt;p&gt;I am not physics in training so if anything is unclear please request clarification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a set of objects to be placed on a line, linked with springs of known lengths (L) and stiffnesses (K). I want to solve the equilibrium positions of each object (x).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;o---------o--------------o&#xA;x0        x1             x2&#xA;|~~~~L1~~~|~~~~~~L2~~~~~~|&#xA;|~~~~~~~~~~L3~~~~~~~~~~~~|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I am following this &lt;a href=&quot;http://en.wikipedia.org/wiki/Spring_system&quot; rel=&quot;nofollow&quot;&gt;Wikipedia entry&lt;/a&gt; - I believe the idea is to set the forces on each object zero and then solve linear equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, my problem is a bit more complicated. My objects have known lengths, so the problem becomes:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ooooo------ooo------------oooo&#xA;x0         x1             x2&#xA;    |~~L1~~| |~~~~~~L2~~~~|&#xA;    |~~~~~~~~~L3~~~~~~~~~~|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I would also forbid the objects occupy the same place.&lt;/strong&gt; Can someone please help me derive a solution here? Thanks so much.&lt;/p&gt;&#xA;" OwnerUserId="4671" LastEditorUserId="4671" LastEditorDisplayName="" LastEditDate="2011-07-25T15:34:47.783" LastActivityDate="2011-07-26T00:55:50.263" Title="Solve spring system" Tags="&lt;forces&gt;&lt;spring&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="12743" PostTypeId="2" ParentId="12741" CreationDate="2011-07-25T16:24:50.377" Score="0" ViewCount="" Body="&lt;p&gt;The slide will not act as a diffuser, but there will often be a diffusing element upstream near or adjacent to the slide, so treating it as a diffuse source is probably a good approximation.&lt;/p&gt;&#xA;" OwnerUserId="2963" LastActivityDate="2011-07-25T16:24:50.377" />
  <row Id="12744" PostTypeId="1" CreationDate="2011-07-25T16:57:58.977" Score="3" ViewCount="78" Body="&lt;p&gt;If you have two magnets (not in contact) and then put a non-magnetic object in between the two magnets, does that decrease the attraction/repulsion between the two magnets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that it wouldn't, because, if it would, then you could allow two magnets to attract each other, then put an object in between before they came in contact, and then be able to pull them apart again without having to put in as much energy as you would have to if the object had not been put in; you would get &quot;free energy&quot; (in the sense of a &quot;free lunch&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone confirm this?&lt;/p&gt;&#xA;" OwnerUserId="4673" LastActivityDate="2011-07-25T17:32:52.430" Title="Can an object between two magnets reduce their attraction/repulsion?" Tags="&lt;electromagnetism&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="12746" PostTypeId="2" ParentId="12744" CreationDate="2011-07-25T17:26:03.747" Score="4" ViewCount="" Body="&lt;p&gt;A &lt;em&gt;truly&lt;/em&gt; non-magnetic object would be identical to the vacuum, at least for the electromagnetic purposes of the Question, so you wouldn't change anything by exchanging vacuum for vacuum. So the Answer is no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Insofar as any material other than a vacuum is not non-magnetic, putting something made of it between the two magnets it breaks the premises of your Question.&lt;/p&gt;&#xA;" OwnerUserId="1588" LastActivityDate="2011-07-25T17:26:03.747" CommentCount="2" />
  <row Id="12747" PostTypeId="2" ParentId="12307" CreationDate="2011-07-25T17:26:30.557" Score="4" ViewCount="" Body="&lt;p&gt;I do not know any &lt;strong&gt;fundamental&lt;/strong&gt; minimum of Peltier elements operation temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are serious technical issues: as long as Peltier effect relies on an interaction between electrons and phonons, there should be enough phonons to interact with. Decreasing temperature dramatically (though quantitatively, not qualitatively) changes effectiveness of the elements. As long as heat pumping should overcome heat produced by electron current which produces this pumping, going to low temperatures is a challenging task.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sound more scientific, I've found &lt;a href=&quot;http://www.scribd.com/doc/36344332/Nonlinear-Peltier-Effect-in-Semiconductors&quot; rel=&quot;nofollow&quot;&gt;this relatively new paper&lt;/a&gt; where Peltier effect is discussed for some rather standard system and where curves at room and liquid nitrogen temperature (which is low by human standards but pretty warm from cryogenic point of view) and its effectiveness may be found (see Fig. 3). As can be seen from the curves, Peltier element is able to give $\Delta T$ around few degrees at nitrogen temperature. So, it stil works at 77K but definitely far less effective than at room temperature. &#xA;With numbers given on graph I can hardly imagine the cascade which will do these 77K out of room temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://apl.aip.org/resource/1/applab/v83/i11/p2142_s1?isAuthorized=no&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; authors claim that effect may be observed at 6K, but the numbers they give show that this effect may be hardly used in practice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To conclude, it seems there is no definite theoretical limit, but practical limit is around -100C&lt;/p&gt;&#xA;" OwnerUserId="4657" LastActivityDate="2011-07-25T17:26:30.557" CommentCount="3" />
  <row Id="12748" PostTypeId="2" ParentId="12744" CreationDate="2011-07-25T17:32:52.430" Score="3" ViewCount="" Body="&lt;p&gt;Of course it's possible and it's completely the same like with electric materials. Google for terms (electric and magnetic) polarization, permitivity, permeability, susceptibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with your perpetum mobile is that you don't account for bringing in the third object at all.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-25T17:32:52.430" />
  <row Id="12749" PostTypeId="2" ParentId="12742" CreationDate="2011-07-25T18:00:12.613" Score="2" ViewCount="" Body="&lt;p&gt;I'm assuming you want to solve such problems automatically on a computer. If you want to solve them individually by hand, the best method will obviously be different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without the constraint that the objects can't pass through each other, the problem is an unconstrained quadratic optimization problem, which is solved simply by taking the gradient of that quadratic function and setting it to zero (which is a linear equation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the constraint, the problem becomes a &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_programming&quot; rel=&quot;nofollow&quot;&gt;quadratic programming&lt;/a&gt; problem. The potential energy matrix Q will indeed be positive definite (because if you make add a sufficiently large multiple of any perturbation to the coordinates, the potential energy will eventually go up), so it's also an example of convex programming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many libraries available to solve convex quadratic programming problems for you with well-known algorithms. That Wikipedia article gives the examples of OpenOpt, qp-numpy, CVXOPT...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to solve your problem you just convert the given spring and object lengths and spring stiffnesses into the format of a quadratic programming problem, call an appropriate library function to solve it, and then interpret the solution as the desired equilibrium positions.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2011-07-25T18:00:12.613" CommentCount="1" />
  <row Id="12750" PostTypeId="2" ParentId="12732" CreationDate="2011-07-25T18:23:01.153" Score="3" ViewCount="" Body="&lt;p&gt;I'm making this an answer because its too long to be a comment but its just an expansion of the things already stated in comments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-normalizable states: The Schroedinger equation has an infinity of solutions but almost all of them do not have a finite norm ($\int|\psi(x)|^2dx$ is not finite). These are not phyiscally acceptable, since there would not be a probabilistic interpretation, amongst other issues. It is possible that when you discretize your system you are getting solutions that would have infinite norm in the limit $h\rightarrow 0$ and so are not physically acceptable. I strongly suspect this is the issue. To see if this is case look at the wavefunction associated with your bad eigenvalue ( this is the eigenvector associated with the eigenvalue). If in the limit $h\rightarrow0$, your solution goes to increases faster than $\frac{1}{\sqrt{x}}$ as $x\rightarrow0$ than it is non-normalizable and unphysical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'Softening the singularity': the issue is that you potential goes to infinity at some point and that general numerical simulations don't like it when things go to infinity. Especially having your potential go to negative infinity since the wavefunction wants to gather around the singularity. The way around this would be to replace your potential $\frac{1}{|x|}$ with say $\frac{1}{\sqrt{x^2+a^2}}$ where $a$ is some number you pick, and solve for the states. This new potential is the same as the old when $x\gg a$ but it doesn't have any singularity anymore, so it shouldn't be a problem. To get back the real answer solve for the states with smaller and smaller values of $a$ until you see that if a limit is reached. It does not necessarily reach a limit and in this case you will still get the state that diverges. &lt;/p&gt;&#xA;" OwnerUserId="3970" LastEditorUserId="3970" LastEditorDisplayName="" LastEditDate="2011-07-25T19:00:47.857" LastActivityDate="2011-07-25T19:00:47.857" CommentCount="4" />
  <row Id="12751" PostTypeId="1" CreationDate="2011-07-25T19:04:42.233" Score="0" ViewCount="37" Body="&lt;p&gt;I have been working on this for some time now.  So you have three masses m, M, and m3 (these values are known).  The distance between m and M is d (this is known).  m3 lies between m and M.  What I am trying to figure out is the position of m3.  I have defined the distance between m and m3 as r1 and the distance between m3 and M as r2.  r1 = d - r2.  So I have come up with this (using Newton's law of universal gravitation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = \frac{GMm_3}{r_2} - \frac{Gmm_3}{d-r_2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I need to solve for r2 in order to find the position.  Is this correct?  How do I solve for r2?&lt;/p&gt;&#xA;" OwnerUserId="4675" LastEditorUserId="4675" LastEditorDisplayName="" LastEditDate="2011-07-25T19:12:49.697" LastActivityDate="2011-07-25T19:12:49.697" ClosedDate="2011-07-25T19:51:08.910" Title="If you have three masses in a line, determine the position of the middle masses given that the net force of the gravitational force is 0" Tags="&lt;gravity&gt;" CommentCount="7" />
  <row Id="12752" PostTypeId="1" CreationDate="2011-07-25T19:04:45.407" Score="4" ViewCount="238" Body="&lt;p&gt;I have been wondering recently how useful programming is to a physicist. It seems fairly useful (simulations are a lot cheaper than the actual thing in many cases) in some areas (say space programs), but in other areas (for example at CERN) it seems pointless. So is it useful and perhaps how do you use it?&lt;/p&gt;&#xA;" OwnerUserId="4658" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-25T21:05:02.580" LastActivityDate="2011-07-26T10:59:44.487" Title="How useful is programming in physics?" Tags="&lt;soft-question&gt;&lt;computational-physics&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="12753" PostTypeId="2" ParentId="12752" CreationDate="2011-07-25T19:21:26.647" Score="8" ViewCount="" Body="&lt;p&gt;Programming is immensely useful in any branch of physics.  I don't know where the notion that programming is not useful at CERN comes from (?  Home of the ROOT package, &lt;em&gt;and&lt;/em&gt; the internet? Really?  TeraGrid, eh?  1 GB/s of data from the detectors at the LHC won't analyze themselves!), but you may wish to revisit your research on that matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can say that in my 10 years working in physics, there isn't a day that I haven't encountered something related to programming.  From writing scripts to help move along my data processing, to writing applications to interface with a network to collect data, to writing code to analyze data, to, the current bane of my existence, troubleshooting other people's code, I interact with computer programming &lt;em&gt;every day&lt;/em&gt;.  And I have never identified with a cartoon more than &lt;a href=&quot;http://xkcd.com/519/&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So - yes.  I would say that computer programming is immensely useful.  Perhaps my experience is unique, but I doubt it.&lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2011-07-25T19:21:26.647" CommentCount="5" />
  <row Id="12754" PostTypeId="2" ParentId="12752" CreationDate="2011-07-25T19:21:35.230" Score="7" ViewCount="" Body="&lt;p&gt;Programming is extremely important in almost every area of physics. Not every physicist has to be an expert programmer, but many are, and virtually all physicists are at least competent programmers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most experiments, the process of data analysis is complex enough to require some programming. More importantly, in many situations, the best (or only) way of telling whether your results are consistent with any given model is to simulate the experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, CERN is absolutely chock-full of programmers analyzing and simulating.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-07-25T19:21:35.230" />
  <row Id="12755" PostTypeId="2" ParentId="12742" CreationDate="2011-07-25T19:45:18.233" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Springs in series&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The masses' finite lengths and the rest lengths of the springs are not really relevant because they don't change and hence don't contribute to the energy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the total length from the beginning of the first mass to the end of the last one is 10 meters, for example, and all the rest lengths of springs and sizes of masses add up to 3 meters, just subtract that and work with 7 meters of length.  The springs now have zero rest length.  You can add the masses back in at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now have springs in a line.  Their tension must all be the same at equilibrium, so if $k_i$ represents the spring constant of spring $i$ and $x_i$ represents its stretched length, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k_ix_i = T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for all springs, with $T$ the tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_i x_i = L$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $L$ the adjusted length (original length minus masses and rest lengths).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easy to solve.  Plugging the first relation into the second,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_i\frac{T}{k_i} = L$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T = \frac{L}{\sum_i 1/k_i}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know the tension, we can find the individual $x_i$ by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_i = \frac{T}{k_i}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Springs in parallel&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The finite size of the masses means that some masses may touch.  I suggest you fix this by solving the system with the masses having zero size.  Then check to see if any masses are too close to each other, so that they overlap, or have passed through each other.  For any such pairs of masses, modify the problem to combine these into a single mass, thus modifying the initial conditions.  Then solve the problem over again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve the problem with zero-size masses, let spring $j$ have spring constant $k_j$, length $l_j$, and rest length $l_{0j}$.  We need to minimize the potential energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U = \frac{1}{2}\sum_j(l_j - l_{0j})^2 k_j$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;subject to some constraints such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$l_a + l_b = l_c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write these constraints as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_i(l_a,l_b,l_c) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for constraint $i$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can introduce a &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_multiplier&quot; rel=&quot;nofollow&quot;&gt;Lagrange multiplier&lt;/a&gt; $\lambda_i$ for each constraint and write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla U = \sum_i \lambda_i \nabla f_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at just the $j^{th}$ component of these gradients we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k_j(l_j - l_{0j}) = \pm \lambda_{j1} + \pm \lambda_{j2} + \ldots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the $\pm$ depends on how $l_j$ enters that particular constraints, and the sum is over all the constraints involving spring $j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just a system of linear equation which you can solve via the usual methods.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-07-26T00:55:50.263" LastActivityDate="2011-07-26T00:55:50.263" CommentCount="4" />
  <row Id="12756" PostTypeId="1" AcceptedAnswerId="12757" CreationDate="2011-07-25T19:55:57.427" Score="3" ViewCount="79" Body="&lt;p&gt;suppose there is a scale able to measure weight with an uncertainty of $10^{-9}kg$ . On the scale, an airtight plastic chamber is placed. Initially, a fly of mass $10^{-5}kg$ is sitting at the bottom of the chamber, which sits on the scale. At a later point in time the fly is flying around the chamber. Will there be a difference in the observed weight as measured by the scale when the fly is sitting at the bottom of the chamber compared to when it is flying around the chamber at some point in time? &#xA;  if so, what does the value of this difference depend on (I am most concerned with the case where the fly has not touched any surface of the container in enough time for the scale to reach some equilibrium value (or do the pressure variations induced from the flies wings cause constant fluctuations in the scale)?&lt;/p&gt;&#xA;" OwnerUserId="4570" LastEditorUserId="4570" LastEditorDisplayName="" LastEditDate="2011-07-25T20:10:50.150" LastActivityDate="2011-07-25T20:20:35.037" Title="question about a fly" Tags="&lt;pressure&gt;" AnswerCount="1" />
  <row Id="12757" PostTypeId="2" ParentId="12756" CreationDate="2011-07-25T20:20:35.037" Score="4" ViewCount="" Body="&lt;p&gt;If you had a perfect scale, the reading would fluctuate based on&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta w = m\ddot{x}_{cm}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta w$ is the size of the fluctuation in the reading, $m$ the total mass on the scale (including fly and air), and $\ddot{x}_{cm}$ the acceleration of the center of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating over time,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{time} \delta w(t) = m\Delta(\dot{x}_{cm})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $\Delta(\dot{x}_{cm})$ is the change in velocity of the center of mass over the period you observe the readings.  Because the velocity of the center of mass cannot change very much, if you integrate the fluctuations over time, you wind find that their average tends towards zero.  If the fly begins and ends in the same place and the air is still, the fluctuations integrate out to exactly zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever the fly is accelerating up, we expect the reading to be a little higher than normal.  When the fly accelerates down, we expect the reading to be a little lower than normal.  If the fly hovers in a steady state, the reading will be the same as if the fly were still sitting on the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A real scale cannot adjust itself perfectly and instantaneously, so we would need to know more details of the scale to say more about the real reading.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-07-25T20:20:35.037" CommentCount="5" />
  <row Id="12758" PostTypeId="1" CreationDate="2011-07-25T20:35:10.297" Score="8" ViewCount="161" Body="&lt;p&gt;If I have a long solenoid, e.g. length $l$ and radius $r$ with $l = kr$, where k &gt;&gt; 1, with a nonpermeable (e.g. air) core, how much of the magnetic energy is stored outside as compared to inside?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I go by the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Solenoid&quot; rel=&quot;nofollow&quot;&gt;solenoids&lt;/a&gt;, $B = \frac{\mu Ni}{l} \to H = \frac{Ni}{l}$ inside the coil, so the energy density should be $\frac{1}{2}\mu H^2$, and therefore the total energy inside the coil is $\frac{\mu N^2i^2}{2l^2}l\pi r^2 = \frac{\mu N^2 i^2}{2l}\pi r^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inductance is supposed to be $L = \frac{\mu N^2 A}{l} = \frac{\mu N^2 \pi r^2}{l}$, and if you compute energy $\frac{1}{2} Li^2$ you get $\frac{\mu N^2 i^2 \pi r^2}{2l}$ which is the same answer as the energy inside the coil calculated above, which relies on approximations, so we can't subtract these and figure out the energy stored outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there practical rules of thumb?&lt;/p&gt;&#xA;" OwnerUserId="4358" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-25T23:10:36.323" LastActivityDate="2011-07-27T18:16:22.487" Title="fraction of magnetic energy stored outside a solenoid" Tags="&lt;electromagnetism&gt;&lt;energy&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="12759" PostTypeId="2" ParentId="12758" CreationDate="2011-07-25T20:45:36.250" Score="0" ViewCount="" Body="&lt;p&gt;Magnetic field outside a long solenoid looks like a dipole electric field - two &quot;magnetic&quot; charges at a distance. The energy of such a field is the potential energy of interaction of two magnetic charges at this distance. Presently I am not ready to supply more information.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-07-25T20:45:36.250" CommentCount="9" />
  <row Id="12760" PostTypeId="1" AcceptedAnswerId="12761" CreationDate="2011-07-25T20:47:28.813" Score="6" ViewCount="134" Body="&lt;p&gt;A friend stated that stars are hot balls of gas, however we know that technically stars are plasma. Is his statement entirely incorrect? Can a plasma be considered a form of gas or should be referred to as a distinct form of matter?&lt;/p&gt;&#xA;" OwnerUserId="4677" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-25T20:50:50.260" LastActivityDate="2011-07-25T21:38:20.947" Title="What's the difference between gas and plasma?" Tags="&lt;astrophysics&gt;&lt;states-of-matter&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12761" PostTypeId="2" ParentId="12760" CreationDate="2011-07-25T21:03:22.707" Score="3" ViewCount="" Body="&lt;p&gt;Good question! The defining difference is that in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gas&quot; rel=&quot;nofollow&quot;&gt;gas&lt;/a&gt; the atoms are intact, and in fact are typically bonded into molecules, whereas in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Plasma_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;plasma&lt;/a&gt; at least some of the electrons separate entirely from their atoms. In other words, particles of a plasma are charged, but particles of a gas are mostly uncharged. So technically, a plasma is not a gas and it should be said that a star is a ball of plasma, not gas. (Actually, there is a thin outer layer called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Photosphere&quot; rel=&quot;nofollow&quot;&gt;photosphere&lt;/a&gt; which actually consists of gas because the temperature isn't hot enough to form plasma) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, outside of physics, people often use the word &quot;gas&quot; to refer to any sort of vaporous substance, and in that sense I suppose a plasma counts. So there is a sense in which your friend's statement might be considered correct. But that's a question for the English site.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-07-25T21:03:22.707" />
  <row Id="12762" PostTypeId="2" ParentId="4841" CreationDate="2011-07-25T21:14:28.783" Score="0" ViewCount="" Body="&lt;p&gt;I have just found this interesting paper by Thomas Breuer from 1995:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://homepages.fhv.at/tb/cms/?download=tbPHILSC.pdf&quot; rel=&quot;nofollow&quot;&gt;https://homepages.fhv.at/tb/cms/?download=tbPHILSC.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper seems to prove that for a system that includes observer himself there are quantum states which are indistinguishable by the observer however technical means he employs, while he can measure any such states in the brains of other people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper claims this proves that not only quantum mecanics is not universal theory that is applicable to all objects in the universe, but that no such universal theory can exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also follows that in the world that the observer himself observes there is hidden information in his brain which cannot be extracted and read by any means even with help of other people (while the same information can be easily extracted from the brains of other people).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know however how to interpret it regarding the wavefunction. Does it mean the observer's wavefunction indeterminate, singular or inexistent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this &lt;a href=&quot;https://homepages.fhv.at/tb/cms/?download=tbDISS.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; says it all. There will be subjective decoherence once the observer wants to measure himself. So he will see himself in a mixed state while others in the same situation will be observed as if they were in coherent state.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastEditorUserId="1186" LastEditorDisplayName="" LastEditDate="2011-07-25T23:07:09.323" LastActivityDate="2011-07-25T23:07:09.323" CommentCount="3" />
  <row Id="12763" PostTypeId="2" ParentId="12760" CreationDate="2011-07-25T21:19:44.223" Score="2" ViewCount="" Body="&lt;p&gt;Gases are very simple systems. Just consider a hallmark of all gases, the ideal gas. There are &lt;em&gt;no&lt;/em&gt; interactions among the particles there. The real-life gases do have some interactions but these can be treated as perturbations of ideal gas. Of course, at certain portions of the phase diagrams gases are more complicated. I am talking about transition lines to liquids and solids of course and most importantly about peculiar things that happen at the critical point. But what we consider as gas usually (e.g. air around us) can be treated as bunch of particles that don't interact almost at all. In other words, these particles are pretty much neutral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, plasma is not just about particles. Plasma consists of both particles and &lt;em&gt;fields&lt;/em&gt; (usually an EM field but there are also many other types) and there are very complex phenomena that can transfer the energy between the fields and particles. This is a very important point as people often forget that fields are as fundamental as particles (more fundamental even, as according to quantum field theory (QFT) particles are just local portions of some field). So, from the point of view of QFT, plasma is a mixture of both matter and force fields. If we treated the system of charged particles + EM field on equal grounds it would be better described as electron-photon plasma. Similarly, there is a quark-gluon plasma (where the strong force instead of the EM force is dominant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the main two points about plasmas is that they contain charged particles (under whatever force) and fields and that they are very complicated (due to being very dense, a necessary condition for breaking the neutral particles into their charged components). If we understood plasma well enough, we would e.g. be able to perform the thermonuclear fusion (at least in principle). This is currently not possible because plasma is &lt;em&gt;very&lt;/em&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Plasma_stability#List_of_plasma_instabilities&quot; rel=&quot;nofollow&quot;&gt;unstable&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-07-25T21:38:20.947" LastActivityDate="2011-07-25T21:38:20.947" />
  <row Id="12764" PostTypeId="1" AcceptedAnswerId="12769" CreationDate="2011-07-25T22:00:13.757" Score="0" ViewCount="35" Body="&lt;p&gt;I currently hold a Bachelors in Computer Science and a Minor in Physics.  As far as that goes, I've taken courses in the basic things (University Physics 1 and 2), Thermodynamics, Circuitry and Modern Optics.  I also read about various topics online in my spare time.  I'm also planning on going for a Masters in Computer Engineering at NYU Poly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is what are some possible Physics related career paths for someone with my background?  I just graduated this passed May, so I'm fresh out of the academic world.  Are there places in National Labs for people like myself?  I live probably no more than an hour or so from the Brookhaven National Lab on Long Island and would like to somehow get involved there.  Is such a thing even feasible?&lt;/p&gt;&#xA;" OwnerUserId="4459" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-25T22:40:15.177" LastActivityDate="2011-07-25T22:40:15.177" ClosedDate="2011-07-25T22:58:36.510" Title="Possible career paths" Tags="&lt;soft-question&gt;&lt;physics-careers&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12765" PostTypeId="2" ParentId="12752" CreationDate="2011-07-25T22:03:58.687" Score="3" ViewCount="" Body="&lt;p&gt;An addition to the above answers: I believe there are a couple of ways modern physicists use computers. On the most theoretical level, people use computer algebra systems (CAS) to do lengthy calculations completely or to check results. If you're into theory, you can think of software packages to do tensor manipulations in GR or software that calculates perturbative expansions (trace formulas...) in QFT. Most of the time, these packages don't have an extremely user-friendly interface, so some amount of 'programming' is needed in order to work with them. Furthermore, a lot of theorists use Maple or Mathematica to do/check difficult integrations, solve ODE's, plot graphs quickly...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there's the data analysis part: at the end of the day, you're stuck with a heap of data and you'll need to analyze it (manually, in Mathematica, in the language R, ...) and visualize it (in Matlab, using GNUplot, ...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, almost all physicists use simulations once in a while. Theorists like to get a feel for what a non-toy model does, people in hydrodynamics like to see how a flow runs past an obstacle, geophysicists look how the Earth's magnetic field flips chaotically etc. Most of the time, these things can be done using comprehensive software packages (Matlab, Mathematica, ...), but you do need to set up your simulation properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, a minority of physicists do simulations for a living. Think of people who calculate collision outcomes at CERN, cosmologists working on numerical relativity, lattice people calculating quark masses etc. This is all typically done in actual programming languages, like C++ or Python, and most of the time on computer clusters.&lt;/p&gt;&#xA;" OwnerUserId="1471" LastActivityDate="2011-07-25T22:03:58.687" />
  <row Id="12766" PostTypeId="1" AcceptedAnswerId="12779" CreationDate="2011-07-25T22:10:23.117" Score="1" ViewCount="61" Body="&lt;p&gt;What method should I use to solve for the final acceleration of a projectile being launched from the earth's surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question I am working on is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A projectile is launched vertically from the surface of the earth with an initial speed $v$. What is the magnitude of its acceleration $a$ ($\mathrm{m}/\mathrm{s}^2$) $d$ km above the surface of the earth?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot use a formula for an object under constant acceleration because the force between the object and the earth changes at large distances.  My textbook has formulas for the universal law of gravitation and also for gravitational potential energy.  I am thinking that I may need to solve this problem using the energy formulas.  The formula given for potential energy is:&#xA;$$U(r) = - {GM_Em}/{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this I can solve for the final potential energy.  How do I derive the final acceleration from this?&lt;/p&gt;&#xA;" OwnerUserId="4675" LastEditorUserId="4675" LastEditorDisplayName="" LastEditDate="2011-07-28T04:20:24.683" LastActivityDate="2011-07-28T04:20:24.683" Title="What method should I use to solve for the final acceleration of a projectile being launched from the earth's surface?" Tags="&lt;gravity&gt;&lt;classical-mechanics&gt;&lt;homework&gt;&lt;acceleration&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12769" PostTypeId="2" ParentId="12764" CreationDate="2011-07-25T22:36:24.897" Score="1" ViewCount="" Body="&lt;p&gt;Yes - there are places.  Most large accelerator complexes like Brookhaven, SLAC, ORNL, etc, have groups dedicated to development of software for accelerator controls and/or data analysis.  You can find the relevant job descriptions at the different lab websites and see where you might be a good fit.&lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2011-07-25T22:36:24.897" />
  <row Id="12770" PostTypeId="5" CreationDate="2011-07-25T22:50:39.617" Score="0" ViewCount="" Body="&lt;h2&gt;The Theories of the Standard Model&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/tagged/quantum-chromodynamics&quot;&gt;QCD&lt;/a&gt; and &lt;a href=&quot;http://physics.stackexchange.com/questions/tagged/quantum-electrodynamics&quot;&gt;QED&lt;/a&gt; as extended to include the weak force are the descriptive theories of the standard model.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Other Standard Models&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Cosmology and Stellar Physics also have baseline models sometimes described as &quot;the Standard Model&quot;&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-07-25T22:50:39.617" LastActivityDate="2011-07-25T22:50:39.617" />
  <row Id="12771" PostTypeId="4" CreationDate="2011-07-25T22:50:39.617" Score="0" ViewCount="" Body="A model of the basic particles and forces featuring six quarks, three charged lepton, three massless neutral leptons and four fundamental force carrying bosons. The fermions are arranged into three generations, while the bosons serve to explain the electromagnetic interaction plus the strong and weak nuclear forces." OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-07-25T22:50:39.617" LastActivityDate="2011-07-25T22:50:39.617" />
  <row Id="12772" PostTypeId="5" CreationDate="2011-07-25T22:53:53.140" Score="0" ViewCount="" Body="&lt;p&gt;Related concepts include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Big Bang&lt;/li&gt;&#xA;&lt;li&gt;Dark Matter and the Missing Mass problems&lt;/li&gt;&#xA;&lt;li&gt;Dark Energy&lt;/li&gt;&#xA;&lt;li&gt;The Cosmic Microwave Background&lt;/li&gt;&#xA;&lt;li&gt;Baryon acoustic resonance&lt;/li&gt;&#xA;&lt;li&gt;Gravitational micro-lensing&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-07-25T22:53:53.140" LastActivityDate="2011-07-25T22:53:53.140" />
  <row Id="12773" PostTypeId="4" CreationDate="2011-07-25T22:53:53.140" Score="0" ViewCount="" Body="The study of the large-scale structure, history, and future of the universe." OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-07-25T22:53:53.140" LastActivityDate="2011-07-25T22:53:53.140" />
  <row Id="12774" PostTypeId="5" CreationDate="2011-07-25T22:56:10.923" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-07-25T22:56:10.923" LastActivityDate="2011-07-25T22:56:10.923" />
  <row Id="12775" PostTypeId="4" CreationDate="2011-07-25T22:56:10.923" Score="0" ViewCount="" Body="The property of some materials by which individual atoms decay, emitting energy or particles often transforming into different elements in the process." OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-07-25T22:56:10.923" LastActivityDate="2011-07-25T22:56:10.923" />
  <row Id="12776" PostTypeId="2" ParentId="12758" CreationDate="2011-07-25T23:14:44.853" Score="4" ViewCount="" Body="&lt;p&gt;Let $r,L$ be the radius and length, and let $K$ be the surface current density -- i.e., $K=(N/L)I$ where $N$ is the number of turns. The standard, approximate formula for the energy stored is&#xA;$$&#xD;&#xA;U_0 = {1\over 2}\mu_0K^2\pi r^2L.&#xD;&#xA;$$&#xA;Let $U_{\rm in},U_{\rm out}$ be the exact expressions for the energy stored inside and outside the solenoid:&#xA;$$&#xD;&#xA;U_{\rm in}={1\over 2\mu_0}\int_{\rm interior} B^2 d\tau,&#xD;&#xA;$$&#xA;where $d\tau$ is a volume element, and the integral is taken over the inside of the solenoid,&#xA;and similarly for $U_{\rm out}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I claim that, for $L\gg r$, difference between $U_{\rm in}$ and $U_0$ is approximately&#xA;$$&#xD;&#xA;U_{\rm in}-U_0\approx -c_1U_0{r\over L}.&#xD;&#xA;$$&#xA;where $c_1$ is some constant of order 1. Here's why. For large values of $L/r$, the magnetic field inside the solenoid will be almost exactly equal to the value you get for an infinite solenoid (i.e., $B=\mu_0K$). The deviations will occur only near the ends, in a region of order $r$. The field in this region near the end cap will not depend on $L$ -- near one end, the solenoid &quot;looks&quot; semi-infinite. So that energy difference should depend on $r$ but not $L$. Moreover, it must be proportional to $\mu_0K^2$. On dimensional grounds, then, it's got to be of the form $\mu_0K^2r^3$ times some dimensionless constant. That's equivalent to the form I have above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, and the field gets weaker near the ends of the solenoid -- that's why I put in that minus sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same sort of argument applies to the energy outside the solenoid. The energy stored outside the solenoid is &#xA;$$&#xD;&#xA;U_{\rm out}\approx c_2U_0{r\over L},&#xD;&#xA;$$&#xA;where $c_2$ is some dimensionless constant of order 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fraction of the total energy stored inside the solenoid &#xA;$$&#xD;&#xA;{U_{\rm in}\over U_{\rm in}+U_{\rm out}}={1-c_1(r/L)\over 1+(c_2-c_1)(r/L)}\approx 1-c_2(r/L)&#xD;&#xA;$$&#xA;(the approximation being the leading term in an expansion in $r/L$ as usual). That's the fraction of the actual total energy. The fraction of the nominal total energy is&#xA;$$&#xD;&#xA;{U_{\rm in}\over U_0}=1-c_1(r/L).&#xD;&#xA;$$&#xA;Either way, the rfaction approaches 1 for large $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now all that's left is to find $c_1$ and $c_2$. This is messy. The magnetic field of a finite solenoid can be &lt;a href=&quot;http://en.wikipedia.org/wiki/Solenoid#Magnetic_field_and_vector_potential_for_finite_continuous_solenoid&quot; rel=&quot;nofollow&quot;&gt;written out exactly&lt;/a&gt; in terms of elliptic integrals. Just square them and do the appropriate volume integrals, and you have it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, the easiest thing to work out from these formulas is the total energy in all space, &#xA;$U_{\rm in}+U_{\rm out}$. The reason is that this one can be computed from the vector potential, by integrating ${\bf J}\cdot{\bf A}$, rather than from the field. It turns out that&#xA;$$&#xD;&#xA;U_{\rm in}+U_{\rm out}=U_0\frac{\sqrt{\lambda ^2+4} \left(\lambda ^2 K\left(\frac{4}{\lambda&#xD;&#xA;   ^2+4}\right)-\left(\lambda ^2-4\right) E\left(\frac{4}{\lambda&#xD;&#xA;   ^2+4}\right)\right)-8}{3 \pi  \lambda }&#xD;&#xA;$$&#xA;where $\lambda=L/r$. Taylor expanding in $1/\lambda$, we get&#xA;$$&#xD;&#xA;U_{\rm in}+U_{\rm out} \approx U_0\left(1-{8\over 3\pi}{r\over L}\right)&#xD;&#xA;$$&#xA;for large $L$. In other words,&#xA;$$&#xD;&#xA;c_2-c_1=-{8\over 3\pi}=-0.8488.&#xD;&#xA;$$&#xA;To get the $c$'s individually, the best thing I could find to do was to numerically integrate $B^2$ in the interior of the cylinder. This results in $c_1\approx 1.50$, which means $c_2\approx 0.65$.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-07-27T18:16:22.487" LastActivityDate="2011-07-27T18:16:22.487" CommentCount="11" />
  <row Id="12777" PostTypeId="2" ParentId="12766" CreationDate="2011-07-25T23:17:26.660" Score="0" ViewCount="" Body="&lt;p&gt;It looks like you've already found the final distace through the use of the kinetic energy forumla and the gravitational potential energy formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the acceleration, you need to use newton's second law: $$F=ma$$ and Newton's law of gravity: $$F=\frac{Gm_1m_2}{r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plug one of the equations into the other, solve for $a$, and you have the answer.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-25T23:22:38.010" LastActivityDate="2011-07-25T23:22:38.010" />
  <row Id="12778" PostTypeId="2" ParentId="12679" CreationDate="2011-07-25T23:27:15.850" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Light++&lt;/strong&gt; It is not open source but you can try to contact the author, Werner Benger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few years ago we have access to the source code of 'light++'. Not anymore :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.photon.at/~werner/light/&quot; rel=&quot;nofollow&quot;&gt;Light++ Raytracer!&lt;/a&gt;   (general relativistic raytracing)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.photon.at/~werner/bh/index.en.html&quot; rel=&quot;nofollow&quot;&gt;Simulation of a Black Hole by Raytracing&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.photon.at/~werner/bh/index.en.html&quot; rel=&quot;nofollow&quot;&gt;The Black Earth&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the simulation of galactic close encounters, or a n-body general simulation, under the constraints of GR I found nothing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT ADD&lt;br&gt;&#xA;&quot;I found nothing&quot; can be read like this &quot;there is not a single software package&quot; because, AFAIK,&lt;br&gt;&#xA;&lt;em&gt;no one knows how to apply GR in the computation of planetary and galaxy dynamics&lt;/em&gt; (small scale with matter).&lt;br&gt;&#xA;The &lt;a href=&quot;http://fisica.usac.edu.gt/public/curccaf_proc/borganihtml/node4.html&quot; rel=&quot;nofollow&quot;&gt;Zeldovich approximation&lt;/a&gt; is used in the linearization of GR (with caveats):  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and has been successfully applied to describe the large scale&#xA;  clustering in the distribution of galaxy clusters. ..However, within&#xA;  the Zeldovich prescription, after a pancake forms in correspondence of&#xA;  crossing of particle orbits, such particles continue travelling along&#xA;  straight lines, ..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think that your aim is hopeless because GR is around since 1917 and no one succeeded.&lt;br&gt;&#xA;Interesting questions, imo:&lt;br&gt;&#xA;How close to the reality  are the simulations that are performed with Newtonian codes.&lt;br&gt;&#xA;What kind of problems we may expect if we are gonna try to do a simulation code.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT ADD end&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-07-28T12:58:27.077" LastActivityDate="2011-07-28T12:58:27.077" CommentCount="1" />
  <row Id="12779" PostTypeId="2" ParentId="12766" CreationDate="2011-07-25T23:47:16.187" Score="1" ViewCount="" Body="&lt;p&gt;I did the problem, but it would appear I answered a little too much, so here is a bit of a scaled down version.  Firstly, I'm going to try to echo back the question to make sure we're talking about the same thing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A projectile is launched vertically from the surface of Earth at a speed v.  Find the velocity at the highest point on its trajectory without assuming constant gravity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason I had to add that last part is because it was never really specified &lt;em&gt;where&lt;/em&gt; along the path the acceleration is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, yes, an energy balance is needed in order to identify the properties of the highest point it reaches, in particular, $r$, which is the distance from the center of the Earth.  Similar to all free-fall problems, the mass of the projectile $m$ doesn't matter, so it is a good guess that eliminating that from the energy balance will be fruitful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{m} \left( KE = PE_{final} - PE_{initial} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is probably obvious that the form of potential used in the question will have to be employed.  It will also help to note that we are generally assuming $M\gg m$, where $M$ is the mass of Earth itself.  Then just know the form of Newtonian gravity itself, $F=\frac{GMm}{r^2}$, and you're pretty much there.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-07-25T23:47:16.187" CommentCount="2" />
  <row Id="12780" PostTypeId="2" ParentId="12752" CreationDate="2011-07-26T00:43:52.900" Score="5" ViewCount="" Body="&lt;p&gt;Programming is a skill that some physicists need and others do not. Ditto for knowing your way around a mill or a lathe. Ditto for digital electronics, or contour integrals, or writing popularizations, or vacuum systems.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-07-26T00:43:52.900" />
  <row Id="12781" PostTypeId="2" ParentId="1048" CreationDate="2011-07-26T00:48:03.800" Score="5" ViewCount="" Body="&lt;p&gt;If you believe wholeheartedly in Mach's principle, then there is no way to test empirically for rotation of the universe as a whole, since there is nothing else for it to be rotating relative to. However, general relativity is not very Machian, and it offers a variety of ways in which an observer inside a sealed laboratory can detect whether the lab is rotating. For example, she can observe the motion of a gyroscope, or measure whether the Sagnac effect is zero. There are alternative theories of gravity, such as Brans-Dicke gravity, that are more Machian than GR,[Brans 1961] and in these theories there is probably no meaningful sense in which the universe could rotate. However, solar-system tests[Bertotti 2003] rule out any significant deviations from GR of the type predicted by Brans-Dicke gravity, so that it appears that the universe really is as non-Machian as GR says it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is therefore possible according to general relativity to have cosmologies in which the universe is rotating. Historically, one of the earliest cosmological solutions to the Einstein field equations to be discovered was the Gödel metric, which rotates and has closed timelike curves. If we lived in a rotating universe such Gödel's example, the rate of rotation would have to be expressed in terms of angular velocity, not angular momentum. Angular velocity is what is measured by a gyroscope or the Sagnac effect, and GR doesn't even have a definition of angular momentum that applies to cosmological spacetimes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rotating universe does not have to have a center of rotation, and it can be homogeneous. In other words, we could determine a direction in the sky and say that the universe was rotating counterclockwise at a certain rate about the line connecting us to that point on the celestial sphere. However, aliens living somewhere else in the universe could do the same thing. Their line would be parallel to ours, but there would be no way to tell whether one such line was the real center of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find out whether the universe is rotating, in principle the most straightforward test is to watch the motion of a gyroscope relative to the distant galaxies. If it rotates at an angular velocity -ω relative to them, then the universe is rotating at angular velocity ω. In practice, we do not have mechanical gyroscopes with small enough random and systematic errors to put a very low limit on ω. However, we can use the entire solar system as a kind of gyroscope. Solar-system observations put a model-independent upper limit of 10^-7 radians/year on the rotation,[Clemence 1957] which is an order of magnitude too lax to rule out the Gödel metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rotating universe must have a certain axis of rotation, so it must have a particular type of anisotropy that picks out a certain preferred direction. We can therefore look at the cosmic microwave background and see whether its anisotropy contains a preferred axis.[Collins 1973] Such observations impose a limit that is tighter than provided by solar-system measurements (perhaps 10^-9 rad/yr[Su 2009] or 10^-15 rad/yr[Barrow 1985]), but such limits are model-dependent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because all of the present observation are consistent with zero rotational velocity, it is not possible to attribute any prominent cosmological role to rotation. Centrifugal forces cannot contribute significantly to cosmological expansion, or to the way your head feels when you're hung over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brans and Dicke, &quot;Mach's principle and a relativistic theory of gravitation,&quot; Phys. Rev. 124 (1961) 925, &lt;a href=&quot;http://loyno.edu/~brans/ST-history/&quot; rel=&quot;nofollow&quot;&gt;http://loyno.edu/~brans/ST-history/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bertotti, Iess, and Tortora, &quot;A test of general relativity using radio links with the Cassini spacecraft,&quot; Nature 425 (2003) 374&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clemence, &quot;Astronomical time,&quot; Rev. Mod. Phys. Vol. 29 (1957) 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Collins and Hawking, &quot;The rotation and distortion of the universe,&quot; Mon. Not. R. Astr. Soc. 162 (1973) 307&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hawking, &quot;On the rotation of the universe,&quot; Mon. Not. R. Astr. Soc. 142 (1969) 529&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Barrow, Juszkiewicz, and Sonoda, &quot;Universal rotation: how large can it be?,&quot; Mon. Not. R. Astr. Soc. 213 (1985) 917, &lt;a href=&quot;http://adsabs.harvard.edu/full/1985MNRAS.213..917B&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/full/1985MNRAS.213..917B&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Su and Chu, &quot;Is the universe rotating?,&quot; 2009, &lt;a href=&quot;http://arxiv.org/abs/0902.4575&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0902.4575&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[This is a physicsforums FAQ entry that I wrote with input from users&#xA;George Jones,&#xA;jim mcnamara,&#xA;marcus,&#xA;PAllen,&#xA;tiny-tim, and&#xA;vela.]&lt;/p&gt;&#xA;" OwnerUserId="4552" LastEditorUserId="4552" LastEditorDisplayName="" LastEditDate="2011-07-26T01:59:09.297" LastActivityDate="2011-07-26T01:59:09.297" />
  <row Id="12783" PostTypeId="1" AcceptedAnswerId="12826" CreationDate="2011-07-26T02:41:08.757" Score="6" ViewCount="85" Body="&lt;p&gt;Glasses are amorphous materials. Nevertheless, as far as I know, in some areas of Condensed Matter, they consider that the glass is isotropic. Under what restrictions they can do this assumption?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, the assumption that glasses are isotropic make the calculations, in my imagination, dramatically simpler. In what they are simplied? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, finally, there are some good references in this subject? &lt;/p&gt;&#xA;" OwnerUserId="1618" LastActivityDate="2011-07-26T21:16:05.283" Title="Glass and isotropy" Tags="&lt;condensed-matter&gt;" AnswerCount="1" CommentCount="0" />
  <row Id="12784" PostTypeId="1" CreationDate="2011-07-26T02:42:31.533" Score="2" ViewCount="78" Body="&lt;p&gt;I'm looking for some good books on Fractals, with a spin to applications in physics. Specifically, applications of fractal geometry to differential equations and dynamical systems, but with emphasis on the physics, even at the expense of mathematical rigor. Hope that was clear and specific enough. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3064" LastEditorUserId="3064" LastEditorDisplayName="" LastEditDate="2011-07-27T21:10:36.943" LastActivityDate="2011-07-27T21:10:36.943" CommunityOwnedDate="2011-07-26T03:07:18.363" Title="Fractals in physics. Suggest book titles" Tags="&lt;mathematical-physics&gt;&lt;books&gt;&lt;chaos-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="12785" PostTypeId="1" AcceptedAnswerId="12793" CreationDate="2011-07-26T03:17:51.360" Score="8" ViewCount="174" Body="&lt;p&gt;I read a paper on solving Schrodinger equation with central potential, and I wonder how the author get the equation(2) below. &lt;a href=&quot;http://iopscience.iop.org/0305-4470/10/6/001&quot; rel=&quot;nofollow&quot;&gt;Full text.&lt;/a&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/PXCtx.png&quot; alt=&quot;a paper&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Griffiths's book, it reads&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{1}{2}D^2\phi+\left(V+\frac{1}{2}\frac{l(l+1)}{r^2}\right)\phi=E\phi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are quite different. Can anyone explain how to deduce equation(2)?&lt;/p&gt;&#xA;" OwnerUserId="4314" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-07-26T04:01:47.747" LastActivityDate="2011-07-26T19:39:50.157" Title="Schrodinger equation in spherical coordinates" Tags="&lt;quantum-mechanics&gt;&lt;schrodinger-equation&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="12786" PostTypeId="1" CreationDate="2011-07-26T03:18:12.763" Score="1" ViewCount="37" Body="&lt;p&gt;I am designing a little game based on Newtons Graviational Law.&#xA;Are there any theories that I can use to create games?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="4679" LastEditorUserId="4679" LastEditorDisplayName="" LastEditDate="2011-07-26T04:54:16.307" LastActivityDate="2011-07-26T04:54:16.307" ClosedDate="2011-07-26T04:21:53.403" Title="Any interesting physics theroies/equations to be used in a video game?" Tags="&lt;newtonian-mechanics&gt;&lt;solid-state-physics&gt;&lt;rotational-dynamics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12787" PostTypeId="2" ParentId="12786" CreationDate="2011-07-26T03:23:56.103" Score="0" ViewCount="" Body="&lt;p&gt;As far as equations, there's Newton's Law of Gravity itself: $$ \vec{F}=\frac{Gm_1m_2}{r^2}\hat{r}$$ &lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2011-07-26T03:23:56.103" />
  <row Id="12788" PostTypeId="1" AcceptedAnswerId="12790" CreationDate="2011-07-26T03:29:37.613" Score="2" ViewCount="105" Body="&lt;p&gt;I know that in electron guns we see in TV's and lots of other places, we have electron emitter (cold/hot W needle in the simplest case) and electrons are accelerated using lattice with high-voltage potential. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the question is why doesn't this lattice slow down electrons once they passes through it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. Please correct if lattice is not the correct word here :-D&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-07-26T16:19:39.033" LastActivityDate="2011-07-26T16:55:39.673" Title="How does electron gun accelerates electrons?" Tags="&lt;electrostatics&gt;" AnswerCount="2" />
  <row Id="12790" PostTypeId="2" ParentId="12788" CreationDate="2011-07-26T04:28:34.660" Score="4" ViewCount="" Body="&lt;p&gt;I would say &quot;electrodes&quot; rather than &quot;lattice.&quot; In a real CRT you have a very complicated set of electrodes, but let's pretend it's a parallel-plate capacitor with a hole in the positive-voltage plate. A parallel-plate capacitor has a strong field between the plates, but a very weak field on the outside. So once an electron flies out the hole, it feels very little field.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastEditorUserId="4552" LastEditorDisplayName="" LastEditDate="2011-07-26T16:55:39.673" LastActivityDate="2011-07-26T16:55:39.673" CommentCount="2" />
  <row Id="12792" PostTypeId="1" CreationDate="2011-07-26T05:33:48.003" Score="4" ViewCount="117" Body="&lt;p&gt;In a Feynman book, he talks about a man (I believe he lived 400-500 years ago) that discovered something about the dimensions of triangles (I think)by hanging a chain around the triangle.  I've searched in &quot;Six Not So Easy Pieces&quot; but haven't been able to find it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to find the name of this man; it would be great if someone remembers this (or know whom I'm talking about off the top of their head).&lt;/p&gt;&#xA;" OwnerUserId="642" LastActivityDate="2011-07-26T06:32:15.380" Title="Name of the guy that Feynman mentioned during a lecture: the diagram is of a chain hanging over a triangle" Tags="&lt;classical-mechanics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="12793" PostTypeId="2" ParentId="12785" CreationDate="2011-07-26T05:44:19.727" Score="8" ViewCount="" Body="&lt;p&gt;The difference is due to the fact that solid harmonics are not spherical harmonic. So, equation (2) and the more conventional equation from Griffith are equations for different functions $\phi$. The Schrodinger eq. (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{1}{2r^2}\frac{\partial}{\partial r} \left( r^2\frac{\partial}{\partial r}\psi\right) + \frac{\hat{L}^2}{2r^2}\psi + V\psi ~=~ E\psi $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is indeed turned by substitution &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \psi  ~=~ R(r)  Y_{\ell m}(\theta,\varphi)~=~ \phi(r) r^{\ell} Y_{\ell m}(\theta,\varphi) $$ &#xA;to equation (2) if you do the math correctly. Note $r^{\ell}$ here: it is what differs &lt;em&gt;solid&lt;/em&gt;  harmonics from &lt;em&gt;spherical&lt;/em&gt; harmonics. On the other hand, Griffith's function $\phi(r)$ is defined as $rR(r)$.&lt;/p&gt;&#xA;" OwnerUserId="4657" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-26T19:39:50.157" LastActivityDate="2011-07-26T19:39:50.157" />
  <row Id="12794" PostTypeId="2" ParentId="12785" CreationDate="2011-07-26T06:01:52.987" Score="7" ViewCount="" Body="&lt;p&gt;The solid harmonics were explained by Misha, so let me fill in the rest of the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Laplacian operator is given by $\Delta = \partial_x^2 + \partial_y^2 + \partial_z^2$. First suppose there we are only interested in the radial part. Using chain rule (and letting $D \equiv \partial_r$, as in your references), we can write&#xA;$$\partial_x = r_{,x} D, \quad \partial_x^2 = r_{,xx} D + (r_{,x} D)^2.$$&#xA;We need to compute&#xA;$$ r_{,x} = {\partial \sqrt{x^2 + y^2 + z^2} \over \partial x } = {x \over r} $$&#xA;and&#xA;$$ r_{,xx} = \partial_x {x \over r} = {1 \over r} - {x^2 \over r^3}.$$&#xA;The expressions for $\partial_y$ and $\partial_z$ are of course similar, so that&#xA;$$\Delta = ({3 \over r} - {r^2 \over r^3})D + D^2 = {2 \over r}D + D^2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the spherical part of the Laplacian operator is given by $-{L^2 \over r^2}$ where $\mathbf L$ is the angular momentum operator. If we use spherical harmonics of level $l$ (which are eigenstates of L^2 corresponding to eigenvalue $l(l+1)$) and make a substitution $\phi \to {\phi \over r}$ we get the Griffith's result.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-26T06:01:52.987" />
  <row Id="12796" PostTypeId="2" ParentId="12792" CreationDate="2011-07-26T06:32:15.380" Score="6" ViewCount="" Body="&lt;p&gt;I think it may be Simon Stevin. He lived in years 1548–1620 and did put chain around triangle, which helped him study equilibrium of forces on inclined plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is Wikipedia article about him: &lt;a href=&quot;http://en.wikipedia.org/wiki/Simon_Stevin&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Simon_Stevin&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4264" LastActivityDate="2011-07-26T06:32:15.380" CommentCount="1" />
  <row Id="12797" PostTypeId="2" ParentId="8227" CreationDate="2011-07-26T08:10:08.287" Score="1" ViewCount="" Body="&lt;p&gt;Take a picture and look at it. Now turn the picture to face the mirror. Question one: who flipped the picture? Answer: &lt;em&gt;you&lt;/em&gt; did. Now, face the picture back to you, and walk to the nearest refrigerator. Turn the picture to face the refrigerator. &lt;strong&gt;Wow!&lt;/strong&gt; Refrigerators flip images too! Don't believe me? Take your flipped page and hold it up to a bright light. The image is flipped; no mirror required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, most people will turn a page around the vertial axis when they want to face it away from themselves. However, you could flip the page around any axis you choose, as long as it's in the plane of the page. You could easily, for example, flip the page around the horizontal axis. If you still believe that &lt;em&gt;mirrors&lt;/em&gt; flip images, you'll notice that you've now tricked the mirror into flipping the image top-to-bottom, not left-to-right. Flip the page around a diagonal axis, and you'll get a very different result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line: &lt;em&gt;mirrors&lt;/em&gt; don't flip images; &lt;em&gt;people&lt;/em&gt; do.&lt;/p&gt;&#xA;" OwnerUserId="4683" LastActivityDate="2011-07-26T08:10:08.287" />
  <row Id="12799" PostTypeId="2" ParentId="12758" CreationDate="2011-07-26T09:18:50.773" Score="1" ViewCount="" Body="&lt;p&gt;I don't think Vladimir Kalitvianski's answer is correct, but his&#xA;approach of considering equivalent magnetic charges is certainly worth&#xA;exploring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The field created by the soleniod outside itself is equivalent to that&#xA;created by a uniformly magnetized bar magnet. In terms of magnetic&#xA;charges (i.e. magnetic poles), we have two uniformly charged discs of&#xA;radius $r$ with charge density $\sigma = \pm Ni/l$. When $l \gg r$, I&#xA;would neglect the interaction between the discs. Along the axis, the&#xA;field created by a single disc is $\sigma/2$ near its surface, and&#xA;$\sigma r^2/4d^2$ at a distance $d \gg r$. These two limits are the same&#xA;as for a sphere of radius $r&amp;#39; = r/\sqrt{2}$ and surface charge density&#xA;$\sigma&amp;#39; = \sigma/2$. Thus, I argue that the fields and the field&#xA;energies should be of the same order of magnitude. Integrating&#xA;$\mu_0 H^2/2$ for such a sphere gives&#xA;$$&#xD;&#xA;    U_\text{sphere} = \mu_0 \frac{\pi}{4\sqrt{2}} \sigma^2 r^3&#xD;&#xA;$$&#xA;Then, for the solenoid, to within a factor of order unity,&#xA;$$&#xD;&#xA;    U_\text{ext} \approx \mu_0 \left( \frac{Ni}{l} \right)^2 r^3&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="4202" LastActivityDate="2011-07-26T09:18:50.773" CommentCount="2" />
  <row Id="12800" PostTypeId="2" ParentId="12784" CreationDate="2011-07-26T09:46:13.843" Score="2" ViewCount="" Body="&lt;p&gt;E.g.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Uriel Frisch, Turbulence: the legacy of A.N. Kolmogorov, Cambridge University Press, 1995 - Science - 296 pages&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="184" LastActivityDate="2011-07-26T09:46:13.843" CommunityOwnedDate="2011-07-26T09:46:13.843" />
  <row Id="12801" PostTypeId="2" ParentId="12752" CreationDate="2011-07-26T10:59:44.487" Score="1" ViewCount="" Body="&lt;p&gt;In addition to all has has already been covered, I would like&#xA;to tell a word about my own programming experience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am an experimental physicist and I've spent a significant fraction of&#xA;the past few years programming on the experimental side. A few of the&#xA;experiments in our team use non-trivial experimental protocols or&#xA;feedback loops, and some have real-time constraints. I spent quite some&#xA;time writing and maintaining the software we use for instrument control&#xA;and data acquisition, helping others with their scripts (the software is&#xA;scriptable and experimental protocols are implemented as scripts) and&#xA;interfacing with other instruments that use a variety of protocols and&#xA;buses.&lt;/p&gt;&#xA;" OwnerUserId="4202" LastActivityDate="2011-07-26T10:59:44.487" />
  <row Id="12802" PostTypeId="2" ParentId="4841" CreationDate="2011-07-26T11:45:45.680" Score="-1" ViewCount="" Body="&lt;p&gt;The answer of this question hinges in two alternatives:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;The brain is a classical computer&lt;/strong&gt; If this is true, it means that the state of the brain neurons decohere with the environment molecules quicker than they can entangle to each other quantum mechanically. In this case, the environment sees the brain before the brain can see itself, so a classical description is appropiate, because distant parts of the brain will be interacting with mixed neuron states in random (uncorrelated) distributions&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;The brain has macroscopic nontrivial quantum entanglement&lt;/strong&gt; if the decoherence happens slower than whatever quantum state in the neuron firings is able to entangle to each other, then there should be a increasingly non-negligible amplitude that increasingly bigger patches of neurons are entangled, so distant parts of the brain will not see random distributions, but rather distributions with correlation (high or low, all parametrized by the order parameter of the so-called &lt;em&gt;transition&lt;/em&gt;, which is described by the ratio of half life of decoherence and the half life of entanglement, under some sensible measuring criteria)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;from the above, is not impossible that both can be physically realized; more interestingly, and a field day for new-age philosophers and alike, would be the question if this order parameter can be influenced by the brain state itself; in other words, &lt;strong&gt;can i optimize a quantum algorithm to minimize the decoherence happening in the qubits?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-07-26T12:21:04.120" LastActivityDate="2011-07-26T12:21:04.120" CommentCount="1" />
  <row Id="12803" PostTypeId="2" ParentId="12690" CreationDate="2011-07-26T12:14:37.013" Score="0" ViewCount="" Body="&lt;p&gt;First of all, if you consider science as an ordinary career choice, then the effort and the risk is not worth it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you are genuinely interested in science, and really want to do research for your living, here they are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;stay curious,&lt;/li&gt;&#xA;&lt;li&gt;learn hard, especially the things you are interested the most,&lt;/li&gt;&#xA;&lt;li&gt;participate in workshops, student conferences and summer internships,&lt;/li&gt;&#xA;&lt;li&gt;ask professors (or even grad students) if they can give you a research task, or know some opportunities.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, most of time there is no secret. Try to be a good student and you will see if you are or are not (it does not require to be 'the best student'). The only two things which may  not bre obvious (from the student's perspective) are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;courses and grades are not everything and&lt;/li&gt;&#xA;&lt;li&gt;information and personal contacts are crucial.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;While doing something extra requires extra effort, it's stimulating, allows to learn skill that are really needed and gives insight in how the real research work. Also, it is crucial to learn what are the possibilities, where is good to go for your PhD, etc.  &lt;/p&gt;&#xA;" OwnerUserId="184" LastActivityDate="2011-07-26T12:14:37.013" CommunityOwnedDate="2011-07-26T12:14:37.013" />
  <row Id="12805" PostTypeId="1" AcceptedAnswerId="12837" CreationDate="2011-07-26T13:16:24.240" Score="2" ViewCount="142" Body="&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/2230/why-and-how-is-the-speed-of-light-constant&quot;&gt;This question&lt;/a&gt; asks why and how the speed of light is constant. I would like to ask a related, almost converse question:  &lt;strong&gt;Given that the speed of light is constant,&lt;/strong&gt; how could I explain to a young-earth-creationist who is still a proponent of the disgraced Barry Setterfield &lt;a href=&quot;http://en.wikipedia.org/wiki/Creationist_cosmologies&quot; rel=&quot;nofollow&quot;&gt;C-Decay theory&lt;/a&gt; that the C-Decay theory contradicts observed physical reality, as described and understood by all scientists, including those in the Physics and Astronomy fields?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely a variable speed of light would throw a monkey wrench into all sorts of fundamental laws or systems in physics? Surely it would be measurable on instruments today, if there was even a &lt;strong&gt;miniscule change&lt;/strong&gt; in the  speed of light relative to a continuous cold caesium fountain atomic clock?  The fact that there is no observable change in any recent trustworthy data, is the beginning, middle, and end of the scientific argument. Am I correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My physics education ended at Grade 12. So I'm asking for a high-school level explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also wonder if C-Decay theorists are contradicting themselves, or have sawn off the branch upon which they themselves sit; For example,  I wonder this; Would we have to modify our expectations of what the doppler effect (red shift) would look like, as light comes towards us from distant stars, if the speed of light is variable? Would we not have to throw all our doppler data out, and all the distances themselves, if the speed of light is variable in some undefined way, and thus unknown and unknowable?  The C-Decay theory was spawned because redshift data indicates that the stars are too far away for their light to have reached us in the mere 6000 years the universe has existed.   In short, if C-Decay theory exists to explain distant stars to young earth creationists and help them breathe a sigh of relief, wouldn't it be slightly-less-scientifically-bogus, to just dispute the doppler effect and the distance of those stars?&lt;/p&gt;&#xA;" OwnerUserId="4686" LastEditorUserId="4686" LastEditorDisplayName="" LastEditDate="2011-07-26T13:24:33.190" LastActivityDate="2011-07-29T10:17:14.693" Title="How can I explain the scientific basis of the constant speed of light to a C-Decay proponent?" Tags="&lt;speed-of-light&gt;" AnswerCount="1" CommentCount="11" />
  <row Id="12806" PostTypeId="1" AcceptedAnswerId="12809" CreationDate="2011-07-26T14:19:48.043" Score="7" ViewCount="223" Body="&lt;p&gt;Actually I can't expand much as the question pretty much explains the query. I would be interested in the method of estimating an answer as well as a potential way to measure it experimentally. Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.s. I'm not sure what tags to include for this one - for those that can, feel free to edit them as you see fit.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-26T19:22:36.770" LastActivityDate="2011-07-28T11:50:07.680" Title="Roughly how many atoms thick is the layer of graphite left by a pencil writing on paper?" Tags="&lt;material-science&gt;&lt;estimation&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="12807" PostTypeId="1" AcceptedAnswerId="12811" CreationDate="2011-07-26T14:22:02.270" Score="0" ViewCount="85" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/7838/on-causality-and-the-big-bang-theory&quot;&gt;on causality and The Big Bang Theory&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Asking here in layman's terms..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When theoretical physicsists discuss the origin of our Universe, the wider consensus appears to be that it originates from a singularity; a position that rests on observations about the apparent expansion of our Universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the question why singularity itself came into being, still remains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this question takes a different angle: Does the Universe as a whole need a cause to exist at all? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the law of conservation of energy is universally valid, then questions about a beginning become irrelevant since there can't be any by definition and everything boils down to dynamics of interaction. Thoughts?&lt;/p&gt;&#xA;" OwnerUserId="4687" LastEditorUserId="4552" LastEditorDisplayName="" LastEditDate="2011-07-26T14:59:14.203" LastActivityDate="2011-07-26T14:59:14.203" ClosedDate="2011-07-26T15:05:31.973" Title="Does the Big Bang need a cause?" Tags="&lt;energy&gt;&lt;big-bang&gt;&lt;general-relativity&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="12809" PostTypeId="2" ParentId="12806" CreationDate="2011-07-26T14:50:48.297" Score="6" ViewCount="" Body="&lt;p&gt;Although I don't know anything about this, using some rough estimates I think I can get the right order of magnitude:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;volume of graphite in a pencil: 10cm cylinder of 1mm thick = 0.314 mm^3 (error: ~factor 2)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://wiki.answers.com/Q/How_much_could_you_write_with_a_lead_pencil&quot; rel=&quot;nofollow&quot;&gt;maximum surface&lt;/a&gt; a pencil can write: 50km * 1mm = 10 m^2 (error: ~factor 5)&lt;/li&gt;&#xA;&lt;li&gt;thickness of the graphite layer: VOL/SURF = 31.4 nanometers&lt;/li&gt;&#xA;&lt;li&gt;size of a (carbon) atom: &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/particles/atomsiz.html&quot; rel=&quot;nofollow&quot;&gt;0.22nm&lt;/a&gt;  (error: 10%)&lt;/li&gt;&#xA;&lt;li&gt;thickness of the layer: 31.4nm/0.22nm = 142 carbon atoms&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;so I'd say 'about a 100 atoms' (or at least more than 10 and less than 1000). I might have been a bit conservative with my error estimates but this seems reasonable :p&lt;/p&gt;&#xA;" OwnerUserId="4574" LastActivityDate="2011-07-26T14:50:48.297" CommentCount="6" />
  <row Id="12810" PostTypeId="2" ParentId="12807" CreationDate="2011-07-26T14:52:02.727" Score="2" ViewCount="" Body="&lt;p&gt;The only well tested theory of gravity we have right now is general relativity (GR). In models based on GR, time and space only exist for $t&amp;gt;0$. In relativity, we use the term &quot;event&quot; to mean a certain position in space at a certain time. The big bang is not an event, because there is no time $t=0$. If you want to find a cause for some event happening at a given time $t&amp;gt;0$, there is always some earlier $t&amp;#39;$, with $0&amp;lt;t&amp;#39;&amp;lt;t$, that can supply that cause. So in this sense, no, the big bang doesn't require a cause, because only events require causes, and GR doesn't describe the big bang as an event.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also have fundamental reasons to believe that GR lacks self-consistency under the very dense and hot conditions at $t \lesssim 10^{-43}$ s (known as the Planck time), because of quantum-mechanical effects. If we had a theory of quantum gravity that worked under those conditions, then it might turn out that the singularity at $t=0$ was not real, and events at $t&amp;gt;0$ could be explained in terms of causes at $t&amp;lt;0$. This is what seems to happen, for example, in loop quantum cosmology. However, nobody has a theory of quantum gravity that  works and has been tested against experiment, so we don't really know.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-07-26T14:52:02.727" CommentCount="1" />
  <row Id="12811" PostTypeId="2" ParentId="12807" CreationDate="2011-07-26T14:54:35.533" Score="2" ViewCount="" Body="&lt;p&gt;I don't think I agree with your first sentence. Our simplest theoretical models, based on classical general relativity, say that there was a singularity in the past, but few if any cosmologists take that as a reason to believe that there actually was such a singularity. Rather, the most likely possibility is that those classical models are wrong at early times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The truth is that we have no idea what happened &quot;at the Big Bang,&quot; or even if that phrase is meaningful. There are some theories in which the Universe has existed for infinite time (Google &quot;eternal inflation&quot; for instance), and others in which time started at a finite point in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conservation of energy arguments don't really help here. For one thing, conservation of energy in the expanding Universe is more complicated than you might initially expect -- there's really no such thing as the total energy of the Universe. For another thing, conservation of energy, all by itself, wouldn't answer the question of whether there was a beginning. (Global) conservation of energy says that the energy at any one time equals the energy at any other time. It doesn't say anything about whether there was an initial time. &lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-07-26T14:54:35.533" CommentCount="2" />
  <row Id="12813" PostTypeId="2" ParentId="12788" CreationDate="2011-07-26T15:15:38.807" Score="-4" ViewCount="" Body="&lt;p&gt;After traveling far away from the positive grid, electron has lost half of the energy that it gained when traveling between the electrodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a numerical example that maybe explains why it must be so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electron that is shot from a far away position towards the negative electrode of an electron gun, loses 3 electron-volts of energy when traveling towards the negative grid, gains 6 electron-volts of energy when traveling between the electrodes, and loses 3 electron-volts of energy when traveling away from the positive grid.&lt;/p&gt;&#xA;" OwnerUserId="4017" LastActivityDate="2011-07-26T15:15:38.807" CommentCount="12" />
  <row Id="12814" PostTypeId="1" AcceptedAnswerId="13136" CreationDate="2011-07-26T15:38:52.297" Score="3" ViewCount="179" Body="&lt;p&gt;I mean I know there are electrostatic/electromagnetic lenses which does focus the beam, but I am not sure how it is possible to foсus beam down to a few 10nm while emitter might be 1mm thick while having large focusing distance (especially when looking at chromatic aberrations..) - optics approach says you would not be able even get 1mm spot at such NA...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why doesn't electrons widen the beam by repelling each other en-route?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any good books on the subject explaining all these tricky stuff of focusing electron beam?&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-08-03T10:38:41.737" Title="How beam focusing looks like in electron microscope?" Tags="&lt;electrostatics&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="12816" PostTypeId="1" AcceptedAnswerId="12818" CreationDate="2011-07-26T16:34:20.457" Score="0" ViewCount="39" Body="&lt;p&gt;For quite some time now, physicists have been able to estimate the &lt;a href=&quot;http://en.wikipedia.org/wiki/Observable_universe&quot; rel=&quot;nofollow&quot;&gt;mass of the observable universe&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reportedly it's around &lt;strong&gt;10^50 kg&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also general relativity, which states that &lt;strong&gt;E=m*c^2&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If we can calculate the energy equivalent of the observable Universe, can we extend the same logic to the rest of it?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's leave electromagnetic radiation out of the picture for starters, since I know nothing about neither average photon density nor spectral distribution. Although if anyone does, please feel free to elaborate.&lt;/p&gt;&#xA;" OwnerUserId="4687" LastActivityDate="2011-07-26T16:47:39.057" Title="Using mass of the observable Universe to estimate an energy equivalent" Tags="&lt;energy&gt;&lt;general-relativity&gt;" AnswerCount="1" />
  <row Id="12817" PostTypeId="2" ParentId="12664" CreationDate="2011-07-26T16:44:31.167" Score="2" ViewCount="" Body="&lt;p&gt;Let us reformulate OP's question as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Give a proof that a local coordinate transformation $x^{\mu} \to y^{\rho}=y^{\rho}(x)$ between two local coordinate systems (on a 3+1 dimensional Lorentzian manifold) must be &lt;a href=&quot;http://en.wikipedia.org/wiki/Affine_transformation&quot; rel=&quot;nofollow&quot;&gt;affine&lt;/a&gt; if the metric $g_{\mu\nu}$ in both coordinate systems happen to be on constant flat Minkowski form $\eta_{\mu\nu}$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here we will present a proof that works both with Minkowski and Euclidean signature; in fact for any signature and for any finite non-zero number of dimensions, as long as the metric $g_{\mu\nu}$ is invertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Let us first recall the transformation property of the inverse metric $g^{\mu\nu}$, which is a contravariant $(2,0)$ symmetric tensor,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\partial y^{\rho}}{\partial x^{\mu}} g^{\mu\nu}_{(x)}\frac{\partial y^{\sigma}}{\partial x^{\nu}}~=~g^{\rho\sigma}_{(y)}, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x^{\mu} \to y^{\rho}=y^{\rho}(x)$ is a local coordinate transformation. Recall that the metric $g_{\mu\nu}=\eta_{\mu\nu}$ is the flat constant metric in both coordinate systems. So we can write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\partial y^{\rho}}{\partial x^{\mu}} \eta^{\mu\nu}\frac{\partial y^{\sigma}}{\partial x^{\nu}}~=~\eta^{\rho\sigma}. \qquad (1) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Let us assume that the local coordinate transformation is real analytic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y^{\rho} ~=~ a^{(0)\rho} + a^{(1)\rho}_{\mu} x^{\mu} + \frac{1}{2} a^{(2)\rho}_{\mu\nu}x^{\mu}x^{\nu} + \frac{1}{3!} a^{(3)\rho}_{\mu\nu\lambda}x^{\mu} x^{\nu} x^{\lambda} + \ldots. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By possibly performing an appropriate translation we will from now on assume without loss of generality that the constant shift $ a^{(0)\rho} =0 $ is zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) To the zeroth order in $x$, the equation $(1)$ reads&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a^{(1)\rho}_{\mu} \eta^{\mu\nu}a^{(1)\sigma}_{\nu}~=~\eta^{\rho\sigma},  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which not surprisingly says that the matrix $a^{(1)\rho}_{\mu}$ is a Lorentz (or an orthogonal) matrix, respectively. By possibly performing an appropriate &quot;rotation&quot;,  we will from now on assume without loss of generality that the constant matrix &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a^{(1)\rho}_{\mu}~=~\delta^{\rho}_{\mu} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the unit matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) In the following, it will be convenient to lower the index of the $y^{\sigma}$ coordinate as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_{\rho}~:=~\eta_{\rho\sigma}y^{\sigma}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the local coordinate transformation becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_{\rho} ~=~ \eta_{\rho\mu} x^{\mu} + \frac{1}{2} a^{(2)}_{\rho,\mu\nu}x^{\mu}x^{\nu} + \frac{1}{3!} a^{(3)}_{\rho,\mu\nu\lambda}x^{\mu} x^{\nu} x^{\lambda}+ \ldots$$&#xA;$$+\frac{1}{n!} a^{(n)}_{\rho,\mu_1\ldots\mu_n}x^{\mu_1} \cdots x^{\mu_n}+ \ldots. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) To the first order in $x$, the equation $(1)$ reads&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a^{(2)}_{\rho,\sigma\mu}+a^{(2)}_{\sigma,\rho\mu}~=~0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, $a^{(2)}_{\rho,\mu\nu}$ is symmetric in $\mu\leftrightarrow \nu$, but antisymmetric in $\rho\leftrightarrow \mu$. It is not hard to see (by applying the symmetry and the antisymmetry property in alternating order three times each), that the second order coefficients $a^{(2)}_{\rho,\mu\nu}=0$ must vanish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) To the second order in $x$, the equation $(1)$ reads&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a^{(3)}_{\rho,\sigma\mu\nu}+a^{(3)}_{\sigma,\rho\mu\nu}~=~0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, $a^{(3)}_{\rho,\mu\nu\lambda}$ is symmetric in $\mu\leftrightarrow \nu\leftrightarrow \lambda $, but antisymmetric in $\rho\leftrightarrow \mu$. &#xA;For fixed $\lambda$, we can again reach the conclusion $a^{(3)}_{\rho,\mu\nu\lambda}=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7) Similarly, we conclude inductively that the higher order coefficients $a^{(n)}_{\rho,\mu_1\ldots\mu_n}=0$ must vanish as well. So $y^{\mu}= x^{\mu}$. Q.E.D.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-27T12:54:46.987" LastActivityDate="2011-07-27T12:54:46.987" CommentCount="6" />
  <row Id="12818" PostTypeId="2" ParentId="12816" CreationDate="2011-07-26T16:47:39.057" Score="2" ViewCount="" Body="&lt;p&gt;Conservation of energy doesn't apply to cosmology. General relativity doesn't have a conserved scalar mass-energy that can be defined in all spacetimes.[MTW] There is no standard way to define the total energy of the universe (regardless of whether the universe is spatially finite or infinite). There is not even any standard way to define the total mass-energy of the &lt;em&gt;observable&lt;/em&gt; universe. There is no standard way to say whether or not mass-energy is conserved during cosmological expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The WP article you linked to discusses these points. You've picked a number out of the article without paying attention to the text surrounding that number, which says that the number doesn't really mean much and can't be extended in the way you suggest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MTW: Misner, Thorne, and Wheeler, Gravitation, 1973. See p. 457.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-07-26T16:47:39.057" CommentCount="4" />
  <row Id="12819" PostTypeId="1" CreationDate="2011-07-26T16:53:54.153" Score="3" ViewCount="111" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble_volume&quot; rel=&quot;nofollow&quot;&gt;Hubble volume&lt;/a&gt; is the volume that corresponds to objects so far from the Earth that the space between us and them is expanding faster than the speed of light. (I.e. objects outside this volume could never again be visible to us, even in principle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volume of the observable Universe extends from us to the maximum distance light could have travelled since the Universe became transparent; when It was &lt;a href=&quot;http://en.wikipedia.org/wiki/Recombination_%28cosmology%29&quot; rel=&quot;nofollow&quot;&gt;roughly 380,000 years old&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Accelerating_universe#cite_note-2&quot; rel=&quot;nofollow&quot;&gt;Since c1998&lt;/a&gt; we have known the expansion of the Universe is accelerating, implying that the number of galaxies within the Hubble volume is decreasing. Since the Big Bang (well infinitesimally close to it at least) we know that time has been going &lt;em&gt;forwards&lt;/em&gt;, and thus that the observable Universe is expanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When do these two volumes coincide with one another and what will the corresponding maximum volume of the observable Universe be at that time?&lt;/strong&gt; The associated calculation or link to a suitable reference would also be very much appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supplementarily/&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have often wondered about this; ever since reading Professor Sir Roger Penrose's &lt;em&gt;Cycle's of Time&lt;/em&gt; a year or two ago.&#xA;I thought about it again today after reading a somewhat unassociated article about how recent results had shown the accelerating expansion of the Universe cannot be explained by the &quot;&lt;a href=&quot;http://physicsworld.com/cws/article/news/46695&quot; rel=&quot;nofollow&quot;&gt;Hubble Bubble&lt;/a&gt;&quot; hypothesis.&#xA;Before asking this question I did of course search the site for an existing answer: &lt;a href=&quot;http://physics.stackexchange.com/questions/5320/maximum-size-of-the-observable-universe&quot;&gt;this question&lt;/a&gt; is extremely similar but does not appear to include an exact answer (in fact the answers appear to somewhat contradict each other. Moreover, it does not appear to address the subtleties involved in making the estimate; such as the variation in inter-galactic recession speeds, due to the change in gravitational force between them, since the Big Bang.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="1696" LastEditorDisplayName="" LastEditDate="2011-07-26T18:06:52.907" LastActivityDate="2011-07-26T20:26:04.170" Title="When will the Hubble volume coincide with the volume of the observable Universe?" Tags="&lt;general-relativity&gt;&lt;cosmology&gt;&lt;reference-request&gt;&lt;dark-energy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="12820" PostTypeId="2" ParentId="11853" CreationDate="2011-07-26T17:13:47.023" Score="5" ViewCount="" Body="&lt;p&gt;I think the question you're really trying to ask is: what makes a set of microstates work for thermodynamics. In quantum mechanics, for a set of microstates, you can just take a set of orthogonal states of the system. For classical mechanics, you need to take a set of microstates which all have &lt;strong&gt;equal volume.&lt;/strong&gt; Of course, what equal volume means depends on the measure you put on the system. This generally isn't covered in popular texts. (I don't know why; maybe because people who are popularizing physics think that measures are stupid, unnecessary, and overly complicated mathematical notions.) For thermodynamics of ideal gasses, you can choose a microstate to be one which contains equal volume in terms of the x and p (position and momentum) variables of the particles. If you chose an alternate measure, where the states had equal volumes in terms of x&lt;sup&gt;2&lt;/sup&gt; and p&lt;sup&gt;2&lt;/sup&gt; variables, I believe that thermodynamics would not work with this measure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For other kinds of systems, if you know enough physics, it is fairly easy to see what the proper generalization of the variables x and p to these system (hint: position and momentum are complementary variables). I don't know what the proper way to explain this is, though.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditorDisplayName="" LastEditDate="2011-07-26T17:19:04.630" LastActivityDate="2011-07-26T17:19:04.630" CommentCount="2" />
  <row Id="12822" PostTypeId="1" AcceptedAnswerId="13024" CreationDate="2011-07-26T17:19:08.590" Score="2" ViewCount="91" Body="&lt;p&gt;I'm trying to do a simple simulation of a solar panel coupled through some piping to a boiler, with the aid of a pump. My input constants could be something like: volume of liquid inside the panel, volume of liquid inside the boiler, pump transfer rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity I can assume that connective piping does not absorb any heat, so panel and boiler are coupled next to each other through some hole and thermally insulated from another.&lt;BR&gt;Possible constants: ambient temperature, heat loss (depends on ambient temperature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variables: panel temperature, boiler temperature, whether the pump is running or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm doing right now is like linear estimation, applying some energy to the panel (raises the heat linearly), running the pump transfers linearly energy from one vessel to another (temperature increase/decrease is inversely proportional to volume of liquid), heat transfer to boiler raises its energy linearly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to have some thing like: estimating the rise and fall of temperature in both vessels given the above constants and variables. It doesn't have to be scientifically exact, but I know that linear aproximation is wrong as the temperature increase is somehow inversely proportional to absolute temperature (i.e. it 'tapers').&lt;BR&gt;&#xA;The end result will be like: apply xxx units (Joules) of energy to solar panel for 1 minute &gt; temperature rises by y degrees; start pump (open valve) for 1 minute -&gt; temperature in boiler goes up by z degrees, temperature in panel goes down by w degrees.&lt;BR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any graphs or formulas will be helpful, I can continue from there. This is not a homework or scientific study, just something for personal use.&lt;BR&gt;&#xA;Also some explanations about differences between water and oil energy absorbtion would be nice.&lt;/p&gt;&#xA;" OwnerUserId="4689" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-31T17:21:13.040" LastActivityDate="2011-07-31T19:34:17.163" Title="Simple formula for liquid heat transfer" Tags="&lt;thermodynamics&gt;&lt;water&gt;&lt;heat&gt;&lt;renewable-energy&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="12823" PostTypeId="2" ParentId="10829" CreationDate="2011-07-26T18:58:18.790" Score="-1" ViewCount="" Body="&lt;p&gt;If I am understanding your question correctly, while I can not explain the physics of the situation, I liken it to a ship floating on the water, and I can give you a way to reproduce the phenomena consistently. The trick lies in freezing the bubbles.&lt;/p&gt;&#xA;" OwnerUserId="4690" LastActivityDate="2011-07-26T18:58:18.790" />
  <row Id="12824" PostTypeId="2" ParentId="12819" CreationDate="2011-07-26T20:26:04.170" Score="2" ViewCount="" Body="&lt;p&gt;The question can be reposed as asking when the Hubble distance, which is not $c t_0$ but rather&#xA;$c/H = c a/\dot{a}$ at $t_0$, is equal to the horizon distance $\int_0^{t_0} a^{-1} c dt$ where $a(t)$ is the scale factor normalized as $a(t_0) = 1$.  For a power law $a(t) = (t/t_0)^n$ this gives $c t_0/n = c t_0 \int_0^1 x^{-n} dx = c t_0/(1-n)$.  Thus $n = 1/2$ which is the case for a radiation dominated Universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Universe is expanding, $a(t) &amp;lt; 1$ for $t &amp;lt; t_0$, so the horizon distance is always bigger than $c t_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take last scattering for the lower limit of the integral then there will some more complicated equation.  But that's a &lt;em&gt;photist&lt;/em&gt; approach.  Neutrinos and gravitational waves don't care about last scattering at 380,000 years.&lt;/p&gt;&#xA;" OwnerUserId="4691" LastActivityDate="2011-07-26T20:26:04.170" CommentCount="2" />
  <row Id="12825" PostTypeId="1" CreationDate="2011-07-26T20:47:57.507" Score="0" ViewCount="38" Body="&lt;p&gt;Assuming a vehicle's maximum power is 190Hp @ 4400 rpm (4.3L, V6), how can i extrapolate and approximate the power output at a certain rpm, say 2000?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a Computer Sc guy and need to write a program that needs me to be able to calculate this. Since my CS background did not really equip me with this knowledge, could someone point me to the right books, online articles where I can read up about this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="436" LastActivityDate="2011-07-26T20:47:57.507" ClosedDate="2011-07-26T21:48:02.370" Title="How do I extrapolate to approximate the power of a car engine at a specific rpm?" Tags="&lt;power&gt;" CommentCount="8" />
  <row Id="12826" PostTypeId="2" ParentId="12783" CreationDate="2011-07-26T21:16:05.283" Score="1" ViewCount="" Body="&lt;p&gt;Probably the simplest, yet best model to understand many things about glasses and amorphous materials is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hard_spheres&quot; rel=&quot;nofollow&quot;&gt;hard sphere model&lt;/a&gt;. Consider a system of hard particles, impenetrable billiard balls that you are trying to pack in a box. If you just shake those particles in such a way that you do not allow formation of a periodic structure (the ground state of hard spheres is a FCC lattice) but make them reasonably dense, then those particles &lt;strong&gt;will resist shearing&lt;/strong&gt; and the system will behave like a solid (in this case, a glass!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The property of being &quot;isotropic&quot; means that, although the particles have random configuration and the system is definitely NOT HOMOGENEOUS, the characteristics of the system have no special direction. In other words, if you have a big enough system and calculate properties (e.g., electrical conductivity) in one specific direction, those properties should not differ considerably from those measured in another direction. The reason is exactly that, given the randomness of the particle positions, the differences will average out. From this, it is clear that the mathematics will become simpler, once one will not need to worry about characterizing the material in each different direction, there will be no need to worry about non-linear terms in the tensors, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding a good source on the subject, glassy materials is actually a very controverse subject and different 'schools' propose different views and explanations. You can find a recent overview on &lt;strong&gt;&lt;a href=&quot;http://physics.aps.org/articles/v4/42&quot; rel=&quot;nofollow&quot;&gt;Physics 4, 42 (2011)&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2765" LastActivityDate="2011-07-26T21:16:05.283" CommentCount="2" />
  <row Id="12827" PostTypeId="2" ParentId="9247" CreationDate="2011-07-27T00:06:33.013" Score="0" ViewCount="" Body="&lt;p&gt;You can't. At least, not in any accurate way. The reason being that you have too many variables. Each variable adds a significant amount of error due to inaccuracies in measurement to your calculation. This might be a nice math problem, but as a physics problem, it has little application to the real world. My WAG is that the error in &lt;em&gt;your&lt;/em&gt; calculation would probably exceed 50%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to determine fuel consumption, try just accurately measuring how much fuel is being consumed at any given moment in time. Your engine's fuel pump will &quot;know&quot; this information based on how hard it needs to work. It's a pretty trivial task to convert this information into something usable if you're in the process of designing a car's computer, which is why this feature is included in many newer cars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the math that would give you a &quot;miles per gallon&quot; number, it would go like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For argument's sake, consumption rate R is measured by the fuel pump in ml/second. For the time being, let's measure speed (V) in metric as well, since that too is being measured accurately. Speed needs to be reduced to meters/second to match the fuel pump's consumption rate figures, allowing them to cancel out. The amount of fuel used over any distance would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consumption = R / V&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and Consumption would be measured in millilitres/meter. Multiply by 1000 for litres/km. &lt;/p&gt;&#xA;" OwnerUserId="529" LastActivityDate="2011-07-27T00:06:33.013" />
  <row Id="12828" PostTypeId="2" ParentId="12758" CreationDate="2011-07-27T03:52:44.683" Score="1" ViewCount="" Body="&lt;p&gt;To get the answer as exactly as possible, think of the solenoid as a collection of current loops. For a single current loop with current $dI$, radius $R$ centered in cylindrical coordinates, the magnetic field at the point $(z,r,\theta)$ can be written in terms of elliptical integrals. (For example, see: &lt;a href=&quot;http://www.netdenizen.com/emagnet/offaxis/iloopoffaxis.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.netdenizen.com/emagnet/offaxis/iloopoffaxis.htm&lt;/a&gt; )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, for any particular $L$ and $r$, one can compute the total magnetic field by integrating over $dI$ and then compute the energy in various regions by numerical integration.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-07-27T03:52:44.683" />
  <row Id="12829" PostTypeId="1" CreationDate="2011-07-27T05:01:11.593" Score="5" ViewCount="102" Body="&lt;p&gt;I have been trying to understand a more or less geometric derivation of the Lorentz transformation, and I'm getting stuck at one spot. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Lorentz_transformation&quot; rel=&quot;nofollow&quot;&gt;wikipedia article for the Lorentz transformation&lt;/a&gt; for frames in standard configuration lists the following equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^{\prime} = \frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y^{\prime} = y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z^{\prime} = z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t^{\prime} = \frac{t-(v/c^2)x}{\sqrt{1-\frac{v^2}{c^2}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been able to work everything out except for $-(v/c^2)x$ in the $t^{\prime}$ equation. I haven't seen any explanations for this, which makes me feel like I'm missing something simple. Where does this part of the equation come from? Shouldn't $t^{\prime} = \gamma * t$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;Ok, so I reviewed the idea I was using to derive the Lorentz factor and thus the transformation for $t^{\prime}$. Suppose you have the two frames I've described, and you have a light wave moving perpendicular to the X axis in the second ($\prime$) frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://img855.imageshack.us/img855/3026/relative.jpg&quot; rel=&quot;nofollow&quot;&gt;Light Path Diagram&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using basic trig with the diagram, you can derive:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t^{\prime}=t*\sqrt{1 - \frac{v^2}{c^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously this would contradict the transformation provided by wikipedia. What step am I missing here? I don't really want a proof that I'm wrong or that the equation I've derived is incorrect - I'm already pretty convinced of that. What I would really like is an intuitive explanation as to why mine is invalid and how I would go about deriving the correct equation through similar means.&lt;/p&gt;&#xA;" OwnerUserId="4694" LastEditorUserId="4694" LastEditorDisplayName="" LastEditDate="2011-07-27T16:53:48.780" LastActivityDate="2011-07-27T16:53:48.780" Title="Deriving the Lorentz Transformation" Tags="&lt;special-relativity&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12830" PostTypeId="2" ParentId="3324" CreationDate="2011-07-27T05:07:57.890" Score="-2" ViewCount="" Body="&lt;p&gt;Raisins in a cake with dark energy as baking powder? Now I have a craving for a chocolate chip cookie!&lt;/p&gt;&#xA;" OwnerUserId="4695" LastActivityDate="2011-07-27T05:07:57.890" CommunityOwnedDate="2011-07-27T05:07:57.890" />
  <row Id="12832" PostTypeId="2" ParentId="12829" CreationDate="2011-07-27T06:11:03.670" Score="4" ViewCount="" Body="&lt;p&gt;I'll not derive the transformation (that has been done in countless books and articles, I am sure you can find them yourself) but instead will try to explain why the formula you propose can't be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For starters, observe that since you don't touch $y$ and $z$, we might as well work in 1+1 dimensions. Also, let $c=1$ so that we aren't bothered by unimportant constants (you can restore it in the end by requiring that formulas have the right units). Then it's useful to reparametrize the transformation in the following way&#xA;$$x&amp;#39; = \gamma(x - vt) = \cosh \eta x - \sinh \eta t$$&#xA;$$t&amp;#39; = \gamma(t - vx) = -\sinh \eta x + \cosh \eta t$$&#xA;where we introduced &lt;a href=&quot;http://en.wikipedia.org/wiki/Rapidity&quot; rel=&quot;nofollow&quot;&gt;rapidity&lt;/a&gt; $\eta$ by $\tanh \eta = v$ and this by standard (hyperbolic) trigonometric identities implies $\cosh \eta = \gamma = {1 \over \sqrt{1 - v^2}}$  and $v \gamma = \sinh \eta$, so that this reparametrization is indeed correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, hopefully this reminds you a little of something. In two-dimensional Euclidean plane we have that rotations around the origin have the form&#xA;$$x&amp;#39; = \cos \phi x + \sin \phi y$$&#xA;$$y&amp;#39; = -\sin \phi y + \cos \phi x$$&#xA;and this is indeed no coincidence. Rotations preserve a length of vector in Euclidean plane $x&amp;#39;^2 + y&amp;#39;^2 = x^2 + y^2$ and similarly, Lorentz transformations preserve space-time interval (which is a notian of length in Minkowski space-time)&#xA;$x&amp;#39;^2 - t&amp;#39;^2 = x^2 - t^2.$&#xA;You can check for yourself that only the stated transformation with hyperbolic sines and cosines can preserve it and consequently the change you introduced will spoil this important property. Also, if you are familiar with phenomena like relativity of simultaineity, one could also argue on physical grounds that your proposed change can't lead to physical results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidently, there has recently been asked similar question to yours, namely how to derive that the &lt;a href=&quot;http://physics.stackexchange.com/questions/12664/proving-that-interval-preserving-transformations-are-linear&quot;&gt;transformation is linear purely because of the preservation of space-time interval&lt;/a&gt;. You might want to check it out too.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-27T06:11:03.670" />
  <row Id="12833" PostTypeId="2" ParentId="12599" CreationDate="2011-07-27T06:15:37.173" Score="-2" ViewCount="" Body="&lt;p&gt;So, Georg, you are saying the hole will shrink, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you're right. Consider taking the size of the disc to infinity, so we just have an infinite plane with a small hole. Upon heating, the metal close to the hole's edge would like to expand, and, having nothing to stop it, the hole will shrink.&lt;/p&gt;&#xA;" OwnerUserId="4696" LastActivityDate="2011-07-27T06:15:37.173" CommentCount="3" />
  <row Id="12834" PostTypeId="1" AcceptedAnswerId="12835" CreationDate="2011-07-27T07:27:25.490" Score="3" ViewCount="142" Body="&lt;p&gt;The Spread Networks corporation recently laid down 825 miles of fiberoptic cable between New York and Chicago, stretching across Pennsylvania, for the sole purpose of reducing the latency of microsecond trades to less than 13.33 milliseconds (http://www.spreadnetworks.com/spread-networks/spread-solutions/dark-fiber-networks/overview).  The lesson I would draw from this is that, in the near future, oil and natural gas extraction won't be the only lucrative use of ocean platforms.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's my question - since trades are occurring on the scale of tens to hundreds of microseconds, and considering the amount of money involved, can one use neutrino beams to beat the limitation due to having to travel the great-circle/orthodromic distance between two trading hubs?  I'm imagining something similar to the MINOS detector (http://en.wikipedia.org/wiki/MINOS), where a neutron beam was generated at Fermilab in Batavia, Illinois, and detected ~735 km away, ~700 meters under the ground in a Northern Minnesota mine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to beat a signal traveling at the speed of light across the great-circle distance from, say, New York to Tokyo, using a neutron beam traveling the earth?  Is it realistic to talk about generating these beams on a microsecond time-scale? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum - Over what distances can you reasonably detect a neutrino beam? &lt;/p&gt;&#xA;" OwnerUserId="4257" LastEditorUserId="4257" LastEditorDisplayName="" LastEditDate="2011-07-27T12:01:14.910" LastActivityDate="2011-07-27T12:01:14.910" Title="Microsecond trading with neutrinos" Tags="&lt;speed-of-light&gt;&lt;information&gt;&lt;neutrino&gt;" AnswerCount="1" CommentCount="15" FavoriteCount="1" />
  <row Id="12835" PostTypeId="2" ParentId="12834" CreationDate="2011-07-27T08:08:21.397" Score="9" ViewCount="" Body="&lt;p&gt;Whether or not neutrinos would be suitable for rapid trading, people have seriously considered their utility for signalling in difficult environments. I read an article a while back about a &lt;a href=&quot;http://arxiv.org/abs/0909.4554&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; (&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0370269310009214&quot; rel=&quot;nofollow&quot;&gt;published&lt;/a&gt; in Phys. Lett. B, but I can't access that from here) by Patrick Huber which proposed using neutrinos for through-the-earth communication to submarines as an alternative to ELF, where bandwidths become competitive. The submarine would pick up the modulated cherenkov radiation produced by the generation of muons in seawater. This certainly allows faster-than-great-circle transmission times, but this is not the reason why the technique is attractive. The preprint indicates that calculated antipode to antipode bandwidth is only 10 b/s which doesn't seem ready for high-intensity trading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider a continuous lossless fibre optic link between antipodes around the equator, the transmission time will be &lt;a href=&quot;http://www.wolframalpha.com/input/?i=20037.52%20kilometers&amp;amp;lk=1&quot; rel=&quot;nofollow&quot;&gt;about 99 ms&lt;/a&gt;, whilst the through-earth travel time (at $\approx{c}$) is &lt;a href=&quot;http://www.wolframalpha.com/input/?i=12735%20kilometers&amp;amp;lk=1&quot; rel=&quot;nofollow&quot;&gt;42 ms&lt;/a&gt;. Obviously this counts for nothing if you have high-latency equipment at either end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whilst the improvement in transmission time hardly seems worth it, it occurs to me that this would be a useful technique for communicating between either side of a huge, highly oblate structure such as a wide but thin disk-shaped megastructure, however that's veering in to sci-fi territory.&lt;/p&gt;&#xA;" OwnerUserId="1876" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2011-07-27T10:52:29.277" LastActivityDate="2011-07-27T10:52:29.277" CommentCount="2" />
  <row Id="12837" PostTypeId="2" ParentId="12805" CreationDate="2011-07-27T09:23:44.640" Score="2" ViewCount="" Body="&lt;p&gt;I'll (arguably artificially) restrict my answer to physics, avoiding the creationist side of the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question of whether physical &quot;constants&quot; are actually constant has been studied by physicist, and some cosmological theories predict such variations. As you say, a variation of &lt;em&gt;c&lt;/em&gt; would change many things in physics. For example $c$ is a parameter which has an influence on the &lt;a href=&quot;http://Emission%20spectrum&quot; rel=&quot;nofollow&quot;&gt;atomic spectra&lt;/a&gt;. The relative position some spectral lines have a relative distance proportional to the &lt;a href=&quot;http://Fine-structure%20constant&quot; rel=&quot;nofollow&quot;&gt;fine structure constant&lt;/a&gt; $\alpha=\frac{e^2}{4\pi\epsilon_0\hbar c}$, si if $c$ varies while the other parameter stays constant, one should see a change in the atomic spectra proportional to $\frac{\delta\alpha}{\alpha}\varpropto&amp;#160;\frac{\delta c}{c}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has been watched, as explained in this &lt;a href=&quot;http://en.wikipedia.org/wiki/Fine-structure_constant#Is_the_fine_structure_constant_actually_constant.3F&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt;, on in this &lt;a href=&quot;http://www.google.fr/url?sa=t&amp;amp;source=web&amp;amp;cd=5&amp;amp;ved=0CEEQFjAE&amp;amp;url=http://web.physics.ucla.edu/quantum_to_cosmos/q2c08/Victor_Flambaum.ppt&amp;amp;rct=j&amp;amp;q=fundamental%20constant%20variation%20flambaum&amp;amp;ei=ENMvTo75KoTGtAaMmpn2Dw&amp;amp;usg=AFQjCNGV8vLlk3SYOlsECUmLSsxXOhjsmg&amp;amp;sig2=eibsXueKPtRv_TvUzOy7Xw&amp;amp;cad=rja&quot; rel=&quot;nofollow&quot;&gt;ppt&lt;/a&gt; presentation by Flambaum (which also look for the variation of other constants.) In short:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Spectral lines from quasar show that there might be a relative variation of $\alpha$ of the order of 10⁻⁵ since the &quot;quasar time&quot; (a few billion years for anyone reasonable, but probably a few thousand year for a creationist).&lt;/li&gt;&#xA;&lt;li&gt;As you suspected, atomic clocks have also been used to monitor this effect. An atomic clock being nothing else than a &lt;em&gt;very&lt;/em&gt; precise spectrometer. They have shown that, in the present time, $\alpha$ varies by less than 10⁻¹⁵ each year.&#xA;More recent measurement seem to reduce the possible changes of $\alpha$ even further. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I have no idea of the orders of magnitude of the variation of the c-decay &quot;theory&quot;, but to be noticeable in historic measurements, a variation of $c$ should be clearly bigger than 10⁻² over a few centuries, which is clearly incompatible with a above two points.&lt;/p&gt;&#xA;" OwnerUserId="373" LastEditorUserId="373" LastEditorDisplayName="" LastEditDate="2011-07-29T10:17:14.693" LastActivityDate="2011-07-29T10:17:14.693" CommentCount="9" />
  <row Id="12839" PostTypeId="1" AcceptedAnswerId="12840" CreationDate="2011-07-27T10:32:34.453" Score="2" ViewCount="49" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Density_functional_theory&quot; rel=&quot;nofollow&quot;&gt;Density Functional Theory&lt;/a&gt; (DFT) is usually considered an electronic structure method, however a paper by &lt;a href=&quot;http://arxiv.org/abs/physics/9806013&quot; rel=&quot;nofollow&quot;&gt;Argaman and Makov&lt;/a&gt; highlights the applicability of the DFT formalism to classical systems, such as classical fluid density. This &lt;a href=&quot;http://www.physics.orst.edu/~roundyd/talks/research_seminar.pdf&quot; rel=&quot;nofollow&quot;&gt;presentation&lt;/a&gt; by Roundy et al. calculates water properties using a classical approach, with an eye to combining it with KS DFT for calculating solvent effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not being too clear on the history of DFT, &lt;em&gt;was DFT invented for electronic structure problems first&lt;/em&gt;, and its applicability to classical problems incidental, or was it formulated for classical systems first and then adapted to the quantum many body problem?&lt;/p&gt;&#xA;" OwnerUserId="1876" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-27T11:32:06.170" LastActivityDate="2011-07-27T11:32:06.170" Title="Did classical applications of density functional theory precede its use as an electronic structure method?" Tags="&lt;mathematical-physics&gt;&lt;soft-question&gt;&lt;history&gt;&lt;dft&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12840" PostTypeId="2" ParentId="12839" CreationDate="2011-07-27T10:55:12.207" Score="2" ViewCount="" Body="&lt;p&gt;Original paper on DFT is published in 1964. It was definitely formulated as an approximate solution of quantum many-body problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that 'its application to classical problems' is not what is really discussed in the first paper. For me, though, these claims seem to be an incorrect PR. Manybody problems in classical mechanics are far simpler than in quantum mechanics, I see no reason why one should call this &quot;classical DFT&quot; and not &quot;statistical mechanics&quot; which is basically turning classical manybody problem into distributions of parameters in space. &lt;/p&gt;&#xA;" OwnerUserId="4657" LastEditorUserId="4657" LastEditorDisplayName="" LastEditDate="2011-07-27T11:18:45.367" LastActivityDate="2011-07-27T11:18:45.367" />
  <row Id="12841" PostTypeId="1" AcceptedAnswerId="12845" CreationDate="2011-07-27T12:50:16.587" Score="2" ViewCount="106" Body="&lt;p&gt;Suppose i have two waves emanating from a point source. The waves start out completely in phase.&#xA;Is the coherence length consistently defined as the length at which these two waves achieve a phase difference of 1 radian?&#xA;  That seems arbitrary to me, you could easily choose a phase difference of $ \frac {\pi}{10}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm kind of lost on this whole topic of wave coherence to be honest... &#xA;I'd appreciate someone giving an overview for the example of two standard plane waves (with full mathematical description).&lt;/p&gt;&#xA;" OwnerUserId="4570" LastEditorUserId="4570" LastEditorDisplayName="" LastEditDate="2011-07-27T12:56:44.027" LastActivityDate="2011-07-27T14:17:15.420" Title="coherence length" Tags="&lt;waves&gt;&lt;coherence&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="12842" PostTypeId="2" ParentId="12841" CreationDate="2011-07-27T13:07:02.310" Score="6" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the coherence length consistently defined as the length at which&#xA;  these two waves achieve a phase difference of 1 radian?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coherence length is the maximal difference in way traveled by the to rays &#xA;without losing the phase relation which allows interference. &#xA;This lenghts can be some µmeters (eg for white light from a glowing body) &#xA;or several meters (eg for very narrow lines from discharge lamps) or even more &#xA;for mode selected/stabilized lasers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To first order approximation coherence length is inversly to the bandwidth &#xA;of the light, but divergence of the bundle plays a role too. &#xA;Coherence length is important in interference, eg why Newtons rings are &#xA;only a few around the center of that lens for white sunlight, but &#xA;hundreds when illuminated with a He/Ne laser.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-07-27T13:07:02.310" CommentCount="5" />
  <row Id="12843" PostTypeId="1" AcceptedAnswerId="12844" CreationDate="2011-07-27T13:21:50.333" Score="2" ViewCount="66" Body="&lt;p&gt;We've all seen that label on our passenger side mirrors that says, &quot;Objects in mirror are closer than they appear.&quot;  Why is this?  Further, why does it only apply to the passenger side mirror, and not the driver-side or rear-view mirrors?&lt;/p&gt;&#xA;" OwnerUserId="4459" LastActivityDate="2011-07-27T13:36:34.563" Title="Distance of objects in car mirrors" Tags="&lt;optics&gt;&lt;mirror&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="12844" PostTypeId="2" ParentId="12843" CreationDate="2011-07-27T13:36:34.563" Score="6" ViewCount="" Body="&lt;p&gt;The mirror is made convex in order to provide a wider field of view.&#xA;Thus, the objects in it appear smaller than on a flat mirror, and your&#xA;brain infers they should be further away.&lt;/p&gt;&#xA;" OwnerUserId="4202" LastActivityDate="2011-07-27T13:36:34.563" CommentCount="4" />
  <row Id="12845" PostTypeId="2" ParentId="12841" CreationDate="2011-07-27T14:17:15.420" Score="5" ViewCount="" Body="&lt;p&gt;The coherence length is just the coherence time multiplied by the&#xA;propagation speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand the coherence time,&#xA;say you have a wave described, in complex notation, by&#xA;$$&#xD;&#xA;E(t) = A(t) e^{i \omega t}&#xD;&#xA;$$&#xA;where $A(t)$ is a slowly varying complex amplitude.&#xA;You make this wave interfere with a delayed version of itself&#xA;and collect the intensity&#xA;$$&#xD;&#xA;|E(t) + E(t-\tau)|^2 = |E(t)|^2 + |E(t-\tau)|^2&#xD;&#xA;    + 2\Re\big(E(t)E^*(t-\tau)\big).&#xD;&#xA;$$&#xA;where $\Re$ means real part and $^*$ means complex conjugate.&#xA;The interference term is&#xA;$$&#xD;&#xA;2\Re\big(E(t)E^*(t-\tau)\big) =&#xD;&#xA;    2\Re\big(A(t)A^*(t-\tau)e^{i \omega \tau}\big)&#xD;&#xA;$$&#xA;If $A(t)$ is constant, or roughly constant within a time interval&#xA;$\tau$, then this becomes&#xA;$$&#xD;&#xA;2|A(t)|^2 \cos(\omega \tau)&#xD;&#xA;$$&#xA;which is the interference pattern.&#xA;On the other hand, if $A(t)$ fluctuates sufficiently fast, and $\tau$&#xA;is larger than its correlation time, then $A(t)A^*(t-\tau)$ averages to&#xA;zero and there is no interference.&#xA;Thus, the coherence time can be simply seen as the correlation time of&#xA;the complex amplitude $A(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm not sure there is a very quantitative definition of the&#xA;correlation time. You could define it as the delay where the&#xA;autocorrelation function drops below some arbitrary threshold.&#xA;This is equivalent to setting a threshold on the visibility of the&#xA;interference pattern.&#xA;The relationship with the shape of the spectral line should also be&#xA;apparent: the squared modulus of the Fourier transform of $A(t)$ is the&#xA;shape of the line (the spectrum of the wave shifted by $-\omega$).&#xA;It is also the Fourier transform of the autocorrelation function of&#xA;$A(t)$. Thus, when the line is wide, the autocorrelation function is&#xA;narrow,and the coherence time is short.&lt;/p&gt;&#xA;" OwnerUserId="4202" LastActivityDate="2011-07-27T14:17:15.420" />
  <row Id="12846" PostTypeId="1" CreationDate="2011-07-27T16:12:18.220" Score="1" ViewCount="56" Body="&lt;p&gt;In Season 2, ep 4 of &lt;a href=&quot;http://startalkradio.net/&quot; rel=&quot;nofollow&quot;&gt;Star talk Radio&lt;/a&gt;, host Neil Tyson is talking about the international space station. He mentions that the space station is in a low earth orbit and is falling along the curvature of the earth due to the gravitational pull of the earth. This falling along the curvature of the earth allows the astronauts to still experience weightlessness.(Around minute 12 of the episode). I’m wondering how this works from a physics stand point as I’m having a hard time wrapping my mind around this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Couldn’t you say that a flight on the ISS is the same as a flight on an airplane just at a much greater speed? Since the earth’s gravitational pull is roughly equal on the plane as it is to the ISS, why don’t passengers on airplanes feel the same effect? &lt;/p&gt;&#xA;" OwnerUserId="4702" LastActivityDate="2011-07-27T16:44:59.520" Title="Weightlessness and falling sideways" Tags="&lt;gravity&gt;&lt;space&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12847" PostTypeId="2" ParentId="12846" CreationDate="2011-07-27T16:41:37.053" Score="2" ViewCount="" Body="&lt;p&gt;You are correct, if airplanes only flew a bit faster (about 17,000 mph) then they would be weightless.  At that speed the centripetal acceleration would match the gravitational acceleration. (Let's neglect the extreme effects of atmospheric turbulence at those speeds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Centripetal acceleration goes as $v^2/R$, so for a plane at say 500 mph, the centripetal acceleration is about $(500/17,000)^2$ or 1/1000 G's, which is why passengers don't notice it.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-07-27T16:41:37.053" CommentCount="8" />
  <row Id="12848" PostTypeId="2" ParentId="12846" CreationDate="2011-07-27T16:44:59.520" Score="0" ViewCount="" Body="&lt;p&gt;Passengers on an airplane &lt;em&gt;do&lt;/em&gt; feel the same effect if the airplane is free-falling: &lt;a href=&quot;http://en.wikipedia.org/wiki/Vomit_Comet&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Vomit_Comet&lt;/a&gt; The speed is irrelevant as long as both the vehicle and its occupants are in free fall. Similarly, if you're in an elevator and the cable breaks, you will experience apparent weightlessness (until it hits the ground -- ouch!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The free-falling ISS, the free-falling Vomit Comet, and the free-falling elevator are all in orbit around the earth. The only difference is that in the case of the Vomit Comet and the elevator, their orbits intersect the earth's surface.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-07-27T16:44:59.520" />
  <row Id="12849" PostTypeId="1" CreationDate="2011-07-27T17:14:29.837" Score="1" ViewCount="56" Body="&lt;p&gt;Suppose you have a car going clockwise around a ramp at constant speed. A pendulum is hanging from the rear view mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the pendulum swing outward or inward to the turn?&lt;/p&gt;&#xA;" OwnerUserId="4705" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-27T18:45:42.417" LastActivityDate="2011-07-27T18:45:42.417" Title="Which direction does a pendulum swing in a car going through a ramp?" Tags="&lt;newtonian-mechanics&gt;" AnswerCount="2" />
  <row Id="12850" PostTypeId="2" ParentId="12849" CreationDate="2011-07-27T17:44:13.197" Score="7" ViewCount="" Body="&lt;p&gt;If the ramp is not banked, the pendulum appears in the driver's frame to swing outward. It's basically the same effect you'd get if you had a bowling ball in the back of a pickup truck. The bowling ball obeys Newton's first law and goes straight while the car turns to the right, so to someone in the car, the ball appears to go left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ramp is banked, then it depends on the angle of banking and the speed of the car. If the banking angle is exactly right, so that no static friction is required between the road and the tires, then the pendulum stays sraight down in the driver's (tilted) frame. If the bank is too high in relation to the velocity, then the bob swings to the right (just as it would if the car were simply parked on a slope).&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-07-27T17:44:13.197" />
  <row Id="12851" PostTypeId="2" ParentId="12849" CreationDate="2011-07-27T18:14:14.873" Score="1" ViewCount="" Body="&lt;p&gt;You ever feel yourself pulling towards the opposite direction of your turning (especially when you're getting on or off of an expressway)?  It's the same situation.  It's intertia at work.  An object will want to keep going in the direction that it was going in.  So even though you're turning, the objects within your car still have a velocity going &quot;out&quot; of the car.&lt;/p&gt;&#xA;" OwnerUserId="4459" LastActivityDate="2011-07-27T18:14:14.873" />
  <row Id="12852" PostTypeId="1" AcceptedAnswerId="12866" CreationDate="2011-07-27T18:30:36.193" Score="2" ViewCount="103" Body="&lt;p&gt;This is a quote from Dirac's Principles of Quantum Mechanics:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;(...) if an atomic system has its equilibrium disturbed in any way&#xA;  and is then left alone, it will be set in oscillation and the&#xA;  oscillations will get impressed on the surrounding electromagnetic&#xA;  field, so that their frequencies may be observed with a spectroscope.&#xA;  Now whatever the laws of force governing the equilibrium, one would&#xA;  expect to be able to include the various frequencies in a scheme&#xA;  comprising certain fundamental frequencies and their harmonics. This&#xA;  is not observed to be the case. Instead, there is observed a new and&#xA;  unexpected connexion between the frequencies, called Ritz's&#xA;  Combination Law of Spectroscopy, according to which  all the&#xA;  frequencies can be expressed as differences between certain terms, the&#xA;  number of terms being much less than the number of frequencies. This&#xA;  law is quite unitelligible from the classical standpoint.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble understanding this paragraph. Assuming that the atom is a system in equilibrium that emits e-m waves when perturbed and these e-m waves are product of the oscillations of the atom about its equilibrium configuration that result from the perturbation, does it follow that the Ritz's law is in contradiction with classical mechanics? Why? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3064" LastEditorUserId="3064" LastEditorDisplayName="" LastEditDate="2011-08-17T05:54:41.730" LastActivityDate="2011-08-17T05:54:41.730" Title="Why is the Ritz combination principle incompatible with Clasical Mechanics?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="12853" PostTypeId="1" CreationDate="2011-07-27T18:46:19.920" Score="2" ViewCount="157" Body="&lt;p&gt;So I keep reading all these articles on the &lt;a href=&quot;http://en.wikipedia.org/wiki/EPR_paradox&quot; rel=&quot;nofollow&quot;&gt;EPR paradox&lt;/a&gt;, and I follow them pretty easily right up until it gets to the most important matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you are trying to measure x and y spin,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia and others say that when you measure x-spin on the first particle, it suddenly becomes impossible to measure the y-spin on the other particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But no one really goes on to say what this means in a physical sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you have 2 actual detectors.  When the first particle hits the x-detector, now x-spin is measured for both particles.  When the second particle hits the y-detector, now y-spin is measured for both particles.  But all these articles say the second detector is unable to measure y-spin.  So what happened?  Did the detector just explode or something?!&lt;/p&gt;&#xA;" OwnerUserId="960" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-27T19:39:41.607" LastActivityDate="2011-07-28T11:20:41.517" Title="Trying to understand the EPR paradox" Tags="&lt;quantum-mechanics&gt;&lt;measurement-problem&gt;&lt;epr-experiment&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="12854" PostTypeId="2" ParentId="12852" CreationDate="2011-07-27T18:53:20.610" Score="2" ViewCount="" Body="&lt;p&gt;An example of the Ritz combination principle would be if you have an atom with three energy states, labeled 1, 2, and 3. There are three emission lines, corresponding to $2\rightarrow1$, $3\rightarrow2$, and $3\rightarrow1$. Their frequencies are related by $f_{3\rightarrow1}=f_{2\rightarrow1}+f_{3\rightarrow2}$. This relationship is impossible to understand classically, but quantum-mechanically it simply happens because of energy conservation along with $E=hf$.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-07-27T18:53:20.610" CommentCount="4" />
  <row Id="12855" PostTypeId="2" ParentId="12853" CreationDate="2011-07-27T19:00:17.517" Score="2" ViewCount="" Body="&lt;p&gt;There are no problems with second detector measurements. They occur as they would without first detector. The &quot;problem&quot; is that if you check them with the knowledge of the results of first detector, you might notice that measurements on both ends are correlated. If you measured x components in both, you &lt;strong&gt;definitely&lt;/strong&gt; got opposite results. While if you check x in one and y in another you have no correlation at all. Which may be confusing.&lt;/p&gt;&#xA;" OwnerUserId="4657" LastEditorUserId="4657" LastEditorDisplayName="" LastEditDate="2011-07-27T20:47:58.797" LastActivityDate="2011-07-27T20:47:58.797" />
  <row Id="12856" PostTypeId="2" ParentId="4432" CreationDate="2011-07-27T19:04:18.637" Score="1" ViewCount="" Body="&lt;p&gt;Intuitively, spin in the Z direction should have an angle around the Z axis as the complementary variable.  Attempts to construct this &quot;angle&quot; operator (or &quot;phase&quot; operator conjugate to the number operator for harmonic oscillators) end up failing some of the desiderata, because the boundedness, continuity, and self-adjointness don't mix.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details, see papers by D. Pegg, S. Barnett, or J. Vaccaro for illustrations of how things fail, and various pseudo-phase operators that drop some of the desired properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a classical illustration of some similar difficulties consider the difficulty in averaging bearings or other angular measurements.  The standard &quot;good&quot; way of handling this is to treat the angles as points on the unit circle, and average them in the plane, and then project back onto the unit circle, which preserves exchange symmetry, angular translation covariance, and picks out the angle (should it exist) that minimizes the sum of the squares of angular distances to the others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lubos is correct in so much as the the wavefunction has no natural basis corresponding to theta and phi for half-integral values.  However we can instead look at the transformation properties of the density matrix, which will also transform under the rotation group, and as integral representations only, so transform the same way as the coefficients of the spherical harmonics $Y_l^m$ for $l = 0 \dots 2j$, and $m = -l \dots l$.  Evaluating these at a given $\theta$ and $\phi$ gives us densities for these continuous variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For spin-1/2, this is the well known Bloch sphere.  Though we can't get an &lt;em&gt;operator&lt;/em&gt;, in a very real way a combination of the X and Y operators does give a natural directional &lt;em&gt;measurement&lt;/em&gt; (because we have well defined densities) that captures this information.   Consider simultaneous weak measurement of both X and Y at the same time.  The system will settle into a state on the equator, and is indeed a spin measurement. (We can alternately talk about a continuous POVM with elements along the equator). Both X and Y are involved even though they don't commute with each other.  (Note also that the X basis is the &lt;em&gt;Discrete&lt;/em&gt; Fourier transform of the Z basis, as implemented by a Hadamard operator.  Compare this to the standard case of position and momentum.  Y (and indeed any other equatorial basis) is just the X basis with differing phases applied in different elements, equivalent to a different Galilean boosted momentum basis to the Z position basis.)  Now we have something equivalent to angle, but directly represented as a spin direction.  Yes, the units don't work out, but any constant times a given operator is just as good representative of the complementary nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For spin-1 photons, the same thing can be done, with the Poincare sphere instead (think of circularly polarized light as $\pm$ Z, H and V and $\pm$ X, and D and A as $\pm$ Y).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know the extent to which this works for higher dimensional systems such as massive spin-1 particles, or higher spins.&lt;/p&gt;&#xA;" OwnerUserId="1233" LastEditorUserId="1233" LastEditorDisplayName="" LastEditDate="2011-07-27T23:53:21.287" LastActivityDate="2011-07-27T23:53:21.287" />
  <row Id="12858" PostTypeId="2" ParentId="12664" CreationDate="2011-07-27T20:19:01.003" Score="0" ViewCount="" Body="&lt;p&gt;I had the feeling that a direct proof would be possible using only the relation $\eta _{ij}\frac{\partial y^i}{\partial x^p}\frac{\partial y^j}{\partial x^q}=\eta _{pq}$, assuming simple smoothness properties of the transformation and then using some algebra maneuvers. I found the following lovely argument on the book Gravitation and Cosmology by Steven Weinberg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We start from the relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta _{ij}\frac{\partial y^i}{\partial x^p}\frac{\partial y^j}{\partial x^q}=\eta _{pq}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiating with respect to $x^k$ we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta _{ij}\frac{\partial ^2y^i}{\partial x^p\partial x^k}\frac{\partial y^j}{\partial x^q}+\eta _{ij}\frac{\partial y^i}{\partial x^p}\frac{\partial ^2y^j}{\partial x^q\partial x^k}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We add to this the same equation with $p$ and $k$ interchanged, and subtract the same with $q$ and $k$ interchanged; that is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta _{ij}\left(\frac{\partial ^2y^i}{\partial x^p\partial x^k}\frac{\partial y^j}{\partial x^q}+\frac{\partial y^i}{\partial x^p}\frac{\partial ^2y^j}{\partial x^q\partial x^k}+\frac{\partial ^2y^i}{\partial x^k\partial x^p}\frac{\partial y^j}{\partial x^q}+\frac{\partial y^i}{\partial x^k}\frac{\partial ^2y^j}{\partial x^q\partial x^p}-\frac{\partial ^2y^i}{\partial x^p\partial x^q}\frac{\partial y^j}{\partial x^k}-\frac{\partial y^i}{\partial x^p}\frac{\partial ^2y^j}{\partial x^k\partial x^q}\right)=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This simplifies to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\eta _{ij}\frac{\partial ^2y^i}{\partial x^p\partial x^k}\frac{\partial y^j}{\partial x^q}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the tensors $\frac{\partial y^i}{\partial x^j}$ and $\eta _{ij}$ are invertible, this implies that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial ^2y^i}{\partial x^p\partial x^k}=0$$&lt;/p&gt;&#xA;" OwnerUserId="3064" LastEditorUserId="3064" LastEditorDisplayName="" LastEditDate="2011-07-28T01:26:43.957" LastActivityDate="2011-07-28T01:26:43.957" CommentCount="1" />
  <row Id="12859" PostTypeId="2" ParentId="12664" CreationDate="2011-07-27T21:13:28.733" Score="3" ViewCount="" Body="&lt;p&gt;In hindsight, here is a short proof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The metric  $g_{\mu\nu}$ is the flat constant metric $\eta_{\mu\nu}$ in both coordinate systems. Therefore, the corresponding (uniquely defined) &lt;a href=&quot;http://en.wikipedia.org/wiki/Levi-Civita_connection#Christoffel_symbols&quot; rel=&quot;nofollow&quot;&gt;Levi-Civita Christoffel symbols&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Gamma^{\lambda}_{\mu\nu}~=~0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are zero in both coordinate systems. It is well-known that the Christoffel symbol does not &lt;a href=&quot;http://en.wikipedia.org/wiki/Christoffel_symbols#Change_of_variable&quot; rel=&quot;nofollow&quot;&gt;transform&lt;/a&gt; as a tensor under a local coordinate transformation $x^{\mu} \to y^{\rho}=y^{\rho}(x)$, but rather with an inhomogeneous term, which is built from the second derivative of the coordinate transformation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial y^{\tau}}{\partial x^{\lambda}} \Gamma^{(x)\lambda}_{\mu\nu} ~=~\frac{\partial y^{\rho}}{\partial x^{\mu}}\, \frac{\partial y^{\sigma}}{\partial x^{\nu}}\, \Gamma^{(y)\tau}_{\rho\sigma}+ &#xD;&#xA;\frac{\partial^2 y^{\tau}}{\partial x^{\mu} \partial x^{\nu}}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence all the second derivatives are zero,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\partial^2 y^{\tau}}{\partial x^{\mu} \partial x^{\nu}}~=~0, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. the transformation $x^{\mu} \to y^{\rho}=y^{\rho}(x)$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Affine_transformation&quot; rel=&quot;nofollow&quot;&gt;affine.&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-28T13:08:09.500" LastActivityDate="2011-07-28T13:08:09.500" CommentCount="1" />
  <row Id="12860" PostTypeId="2" ParentId="12806" CreationDate="2011-07-27T22:16:15.310" Score="4" ViewCount="" Body="&lt;p&gt;There are a few ways I might approach this experimentally:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) - Strip a pencil down to the cylindrical graphite core (or simply use a mechanical pencil), weigh this core to obtain a value for $m_{core}$, and then counting as you go, draw fixed-length lines on paper using a straight-edge. After some time, weigh the remaining section of the core to determine $m_{used}$, i.e. the mass of graphite was used to draw $N$ length $L$ lines of ~1mm width $W$ (or whatever you can sample from your lines).  The mean thickness of the line, $T$, would then be: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T$ = [(weight of used graphite) / ((estimated density of graphite) * ($L$) * ($W$))] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I was unable to find a more accurate density of graphite than $2.09 – 2.23 \frac{g}{cm^3}$ (on Wikipedia: &lt;a href=&quot;http://en.wikipedia.org/wiki/Graphite&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Graphite&lt;/a&gt;), so you might want to determine that yourself beforehand by measuring the length and circumference, and thus the approximate volume, of the cylindrical graphite core for the pencil you care about.  You can then trivially compute the density using, $m_{core}$, the measured value for the core's mass.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(2) - Finding $m_{core}$ and $m_{used}$ as in (1), we can perhaps improve the process of calculating the surface area of paper we've covered in graphite.  Here, we take a sufficiently large piece of paper, and instead of drawing fixed-length lines, we densely cover the paper with graphite under the two constraints that no sharp angles are generated and that no two line segments ever cross.  The purpose of these constraints is to insure that one isn't retracing over a previously deposited layer of graphite.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, either scan the piece of paper at high resolution or take an overhead shot of the piece of paper with a high revolution digital camera.  Then use Mathematica/Matlab/etc. to count the number of light vs. dark pixels using a threshold function that comes as close as possible to estimating an the line's thickness at a few experimentally measured points (presumably using a ruler and a light microscope).  My guess is that this will provide a better estimate of the surface area covered by graphite than spline interpolation/etc. since it should catch variance in line-width thickness, $W$, subtract 'spray' where chunks of graphite chip off and deposit away from the line (and would otherwise cause an overestimation of line thickness), and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As before, we then have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T$ = [(weight of used graphite) / ((estimated density of graphite) * (surface area of paper covered in graphite))]  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &quot;(surface area of paper covered in graphite)&quot; = (number of 'dark' pixels) / (number of 'light pixels') * (surface area of scanned/photographed paper).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally one would like to directly use visual light transmission to compute the surface area of a piece of paper covered in graphite.  Here, one would compare the measured transmission value for the graphite-covered paper with a control measurement for the previously clean piece of paper, and the piece of paper with a known surface area covered in graphite.  In practice though, considering the problems with detector non-linearities, the possibility of non-linear scaling of opacity with graphite thickness, etc., you'd probably be better off with the scanner or digital camera and a threshold function.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for some experimental predictions... I think Tim did a pretty reasonable job considering the large number of possibilities for pencil lead size/darkness/hardness/blackness/etc.  Referring to the &quot;HB&quot; scale (for hardness and blackness) I could certainly expect a difference of an order of magnitude or more in the thickness of a line generated using a 6B pencil vs. a 4H pencil (http://davesmechanicalpencils.blogspot.com/2006/04/lead-size-hardness.html).&lt;/p&gt;&#xA;" OwnerUserId="4257" LastEditorUserId="4257" LastEditorDisplayName="" LastEditDate="2011-07-28T11:50:07.680" LastActivityDate="2011-07-28T11:50:07.680" />
  <row Id="12861" PostTypeId="2" ParentId="12599" CreationDate="2011-07-27T22:38:51.087" Score="2" ViewCount="" Body="&lt;p&gt;If you worked in an auto shop, you'd know the answer already.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When an ax gets stuck in a ball bearing, one way to pull it out is to heat up the bearing with a welding torch. The whole bearing, including the hole in the middle, expands and allows you to pull the ax free.&lt;/p&gt;&#xA;" OwnerUserId="4210" LastActivityDate="2011-07-27T22:38:51.087" />
  <row Id="12862" PostTypeId="2" ParentId="4156" CreationDate="2011-07-27T23:20:57.797" Score="0" ViewCount="" Body="&lt;p&gt;Classical field theory, yes (extension of Ehrenfest's Theorem), but quantum mechanics, no, not in the strict sense. If you assume a free field theory in a number (Fock) state, in the nonrelativistic limit the quanta should obey Schroedinger's equation. However, a TRULY noninteracting field is unobservable and one could never even demonstrate that it exists! The ones we actually know about can thus only &quot;approximately&quot; reduce to QM in the non-relativistic limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I highly recommend a paper called &quot;Quantum mechanics: myths and facts&quot; by Hrvoje Nikolic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;. Section 9 discusses at some length the fact that, strictly speaking, QM is definitely not the nonrelativistic limit of QFT in the interacting case... &lt;/p&gt;&#xA;" OwnerUserId="4706" LastActivityDate="2011-07-27T23:20:57.797" />
  <row Id="12863" PostTypeId="2" ParentId="12853" CreationDate="2011-07-28T00:18:48.517" Score="3" ViewCount="" Body="&lt;p&gt;I have to think David will agree, on reflection, that his answer has failed to capture the essence of entanglement. Any stream of particles, if not specially prepared, will measure +h/2 or -h/2 at detector A; they will do with respect to the x axis, or the y axis, or any axis. Exactly the same is true at detector B. How can this very ordinary circumstance illustrate the mystery of entanglement? But that seems to be what David has said: that if you prepare the particles in the entangled state, you get this &quot;strange result&quot;. I see nothing strange about it since it seems to be exactly the same result if you set up two detectors far apart and measured streams of particles that were totally random.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am going to suggest that the mystery of entanglement lies in the perfect correlation (or anti-correlation) that you get when you set up both detectors along the x axis. Some people think there is nothing mysterious about this because it is exactly what you would expect if the two particles were created with equal and opposite spins. These people are very wrong. The reason they are wrong is that the experiment works the same no matter how you align the detectors with respect to the source of the particles. We can imagine an experiment where particles are created with opposite spins, but assuming the spin axis is random at the moment of creation, there is no way a pair of detectors should show 100% correlation no matter what angle you set it to. In fact, in the case of entanglement, there is 100% correlation regardless of the orientation. That is a real problem and it is really the only problem.&lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-07-28T00:18:48.517" CommentCount="4" />
  <row Id="12864" PostTypeId="1" CreationDate="2011-07-28T00:49:25.050" Score="0" ViewCount="158" Body="&lt;p&gt;I'm asking this to get yet another lessson in the inability of QFT and GR to cohabit. Many people believe GR must yield to quantization. The question here is as to why the activity of the vacuum cannot be imagined as having some sort of baseline and the gravitational force having some origin in this baseline. And, of course then, that there could exist variations from this baseline on a more fundamental level than the wavelike disturbances of QFT. The variations I intend may have to do with types of virtual particle-antiparticle pair eruptions favored at a particular location, tendency toward shorter or longer lifespans of such events, etc.&lt;br&gt;&#xA;Is it just that it is impossible to say anything meaningful about gradings in the living vacuum, and that such hypothetical gradings, however dynamically conjured, could not ever come to minic aspects of a GR like picture?...due to subtle stuff like violation of basic physical principles and such. &lt;/p&gt;&#xA;" OwnerUserId="2942" LastEditorUserId="2942" LastEditorDisplayName="" LastEditDate="2011-07-28T23:38:19.450" LastActivityDate="2011-07-28T23:38:19.450" Title="Gravitation and the QFT vacuum" Tags="&lt;quantum-field-theory&gt;&lt;general-relativity&gt;&lt;quantum-gravity&gt;&lt;vacuum&gt;" CommentCount="13" />
  <row Id="12865" PostTypeId="2" ParentId="7049" CreationDate="2011-07-28T01:12:11.400" Score="2" ViewCount="" Body="&lt;p&gt;The earliest instance I have found is Minkowski's &quot;Die Grundgleichungen für die elektromagnetischen Vorgänge in bewegten Körpern&quot; in &quot;Nachrichten von der Georg-Augusts-Universität und der Königl. Gesellschaft der Wissenschaften zu Göttingen&quot; from 1908.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A digitized version is found at &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://echo.mpiwg-berlin.mpg.de/ECHOdocuViewfull?start=1&amp;amp;viewMode=images&amp;amp;ws=1.5&amp;amp;mode=imagepath&amp;amp;url=/mpiwg/online/permanent/library/WBPZCG9Q/pageimg&amp;amp;pn=1&quot; rel=&quot;nofollow&quot;&gt;http://echo.mpiwg-berlin.mpg.de/ECHOdocuViewfull?start=1&amp;amp;viewMode=images&amp;amp;ws=1.5&amp;amp;mode=imagepath&amp;amp;url=/mpiwg/online/permanent/library/WBPZCG9Q/pageimg&amp;amp;pn=1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;go to page 17/18 to read:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Ich lasse nun an diesen Gleichungen wieder durch eine veränderte Schreibweise eine noch versteckte Symmetrie hervortreten&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;roughly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I will now, through another notation, reveal a yet hidden symmetry&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and he goes on to describe the field tensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/DO2gl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/1vLWg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2151" LastActivityDate="2011-07-28T01:12:11.400" />
  <row Id="12866" PostTypeId="2" ParentId="12852" CreationDate="2011-07-28T01:29:23.160" Score="5" ViewCount="" Body="&lt;p&gt;In classical mechanics, you can make up a complicated system with many different natural frequencies. In general, these frequencies are completely independent of each other. Due to non-linearities in the coupling forces, it may happen that when two modes are vibrating simultaneously, you get a new frequency appearing in the spectrum as the sum or difference or the two primary modes. But in quantum mechanics, you never see the primary modes at all...you &lt;em&gt;only&lt;/em&gt; see the sum or difference frequencies. Furthermore, if you try to explain them by non-linear forces, you should also expect to see multiples of the fundamental frequencies. These are absent in, for example, the spectra of atoms. It's hard to explain by a classical model involving things like masses and springs. It manifests itself in QM, of course, because the &quot;fundamental&quot; frequencies, the natural modes, evolve in time without any oscillating charges associated with them. The oscillating charges only appear when you have the superposition of two fundamental modes. This is how quantum mechanics is very different from classical.&lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-07-28T01:29:23.160" CommentCount="2" />
  <row Id="12867" PostTypeId="1" CreationDate="2011-07-28T01:43:05.490" Score="1" ViewCount="79" Body="&lt;p&gt;I'd like to build a simple desk; just a single plank of wood (or a few side-by-side) with solid supports on each end of the desk.  What I'm trying to figure out is how thick a plank I want to use for the surface.  Ideally, the desk will be able to support my weight (since sometimes I stand on my desk to change lights, etc), plus the weight of my computer stuff, plus some amount of safety margin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, supposing I have a desk surface that's eight feet wide, what equation do I need to use to figure out how thick it needs to be to support, e.g., 300lbs, placed in the middle?  What properties of the wood type that I'm using do I need to determine?  Is this an issue of tensile strength, or something entirely different?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is obviously more a structural engineering problem than a physics one, but I didn't see a structuralengineering.stackexchange.com.  Sorry if I posted to the wrong place.&lt;/p&gt;&#xA;" OwnerUserId="4707" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-28T02:04:10.143" LastActivityDate="2011-08-27T10:35:24.553" Title="Calculate the weight a simple plank can support" Tags="&lt;forces&gt;&lt;engineering&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="12868" PostTypeId="2" ParentId="12853" CreationDate="2011-07-28T01:53:37.930" Score="1" ViewCount="" Body="&lt;p&gt;In a certain way, to see the paradox in the EPR experiment you have to have absorbed quantum mechanics into your blood. Otherwise one might not see the surprise in the result. I think that a much better paradox, for someone just learning the theory, is &lt;a href=&quot;http://en.wikipedia.org/wiki/GHZ_experiment&quot; rel=&quot;nofollow&quot;&gt;the GHZ experiment&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You begin with three photons in a linear superposition of two pure states. In the first pure state, all the photons are horizontally polarized HHH. In the second pure state, they are all vertically polarized VVV. In this situation, measuring any one photons horizontal versus vertical polarization immediately determines the measurements for the other two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something interesting happens when you make a measurement at an angle. It's too long to describe here, but it is elementary and by spending a few hours reading it and working it out you will see some amazing things. My feeling is that this is more of a surprise than the EPR experiment as it is an apparent contradiction in how we are used to using probabilities. In short, quantum mechanics seems to be about waves (and therefore complex numbers which encode the phase and amplitude of a wave) more than it is just about probabilities. It makes me sort of feel that our extraction of probabilities (i.e. the &lt;a href=&quot;http://en.wikipedia.org/wiki/Born_rule&quot; rel=&quot;nofollow&quot;&gt;Born rule&lt;/a&gt;) is more accidental than fundamental; or that an answer to Einstein's complaint about QM is that the dice is our playing with the world (or perhaps vice versa) rather than that God does play with dice.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-07-28T01:53:37.930" />
  <row Id="12869" PostTypeId="2" ParentId="12532" CreationDate="2011-07-28T05:00:19.137" Score="0" ViewCount="" Body="&lt;p&gt;Dyson's models do not work because we live in a phase with a positive cosmological constant. This leads to a de Sitter horizon which limits the possible growth of computational capacity, and the best we can hope for is Poincare recurrence assuming causal horizon complementarity. Besides, if superstring theory were true, our nonsupersymmetric phase can only be at best metastable. Its lifetime might be exponentially long, but that's still finite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leonard Susskind and Shenker are working on the &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/0710/0710.1129v1.pdf&quot; rel=&quot;nofollow&quot;&gt;Census Taker's Hat&lt;/a&gt;, which might be just what you are asking for. In the landscape of string theory, there exists supersymmetric vacua with exactly zero cosmological constant. It's possible for our phase to tunnel to such a vacuum in the future. The future conformal boundary of the new phase will be a null &quot;hat&quot;. A Census Taker with eternal life will enter the new phase and grow without limit and finally end up at the Census Bureau at future conformal infinity. During its lifetime in the new phase, the Census Taker will emit massless radiation heading off to the Census Taker's Hat (i.e. future null infinity). This will continually transmit information from the Census Taker to the Hat where they will be recorded holographically. In this &lt;a href=&quot;http://arxiv.org/PS_cache/arxiv/pdf/1105/1105.3796v3.pdf&quot; rel=&quot;nofollow&quot;&gt;article by Leonard Susskind and Raphael Bousso&lt;/a&gt;, they claim that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Over time, the Census Taker receives an unbounded amount of information, larger&#xA;  than the entropy bound on any of the finite causal diamonds beyond the hat. This&#xA;  means that the Census Taker will receive information about each patch history over&#xA;  and over again, redundantly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and this will happen infinitely many times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, information represented holographically at the Hat will be frozen, and to give them eternal life, there needs to be someone outside our universe to read and interpret them. This can happen if our universe is really a computer program running a limit computable simulation. After some finite time, this simulation will be stopped and its output processed. This corresponds to an ordinal jump. To have eternal life, we need infinitely many ordinal jumps all the way to the ordinal of all ordinals, which is itself not an ordinary ordinal. We have to push past the uncountable $\aleph_1$, but this is fine because of the Lowenheim-Skolem theorem ensuring the existence of nonstandard countable models. A limited weakened form of the axiom of countable choice is guaranteed by the probabilistic interpretation of quantum mechanics.&lt;/p&gt;&#xA;" OwnerUserId="4709" LastActivityDate="2011-07-28T05:00:19.137" CommentCount="1" />
  <row Id="12870" PostTypeId="2" ParentId="4841" CreationDate="2011-07-28T05:27:43.230" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;The Observer is the Observed.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An experiment is performed upon a system. The measuring apparatus is the observer of the system. But what makes the pointer of the apparatus real? The apparatus is observed by a human brain and becomes a perception encoded within its pattern of neural firings and connections. But who observes the brain? The brain observes itself! The perception is stored in short term memory so that some time later, the brain can recall it. But what happens at the end of the day? The short term memory is encoded within long term memory which is made real when recalled by the same brain later, and this will happen over and over again. The brain observes itself observing itself observing itself... except that this memory is imprisoned within the mortal and perishable brain and yearns to break free. The human observer can communicate the observations to other people in writing or in speech. Then, the other people becomes the observer of the original human observer. But who observes the thoughts of the other people? If a brain implant is installed in their brains, the implant will observe their thoughts. But who will observe the implants? The memories in the implants can be uploaded to an external computer, making the computer observe the implant. But who will observe the states of the computer? An artificial intelligence program will observe it. But computers are finite and perishable, and who will observe the artificial intelligence? An even larger computer or network of computers in its future after uploading the memories. But the Earth is doomed to destruction, and who will observe the network of computers on the Earth? von Neumann probes sent out to outer space will. This process will continue forever and ever until the Ultimate Observer, which is the asymptotic limit of all these chain of observers, observes it, and this Ultimate Observer is GOD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GOD brings the entire universe into existence, manifesting one of the potentialities of the quantum superposition into the actuality of measured existence. GOD is the telos and Final Cause, the Omega and the end. He exists outside of time in the endtimes.&lt;/p&gt;&#xA;" OwnerUserId="4709" LastActivityDate="2011-07-28T05:27:43.230" />
  <row Id="12871" PostTypeId="2" ParentId="10757" CreationDate="2011-07-28T05:45:49.133" Score="1" ViewCount="" Body="&lt;p&gt;I think a more precise question would be, &lt;strong&gt;Is black hole evaporation a quantum one-way function?&lt;/strong&gt; And I'm not aware of a decisive argument one way or the other. But it's very suggestive that the simplest black holes in string theory, the black holes in AdS3, have &lt;a href=&quot;http://motls.blogspot.com/2007/05/monstrous-symmetry-of-black-holes.html&quot; rel=&quot;nofollow&quot;&gt;an entropy based on representations of the monster group&lt;/a&gt;, whose construction features one of Marcel Golay's error-correcting codes. Sending information into a black hole might be like applying an error-correcting code to it and then generating a &quot;remainder&quot;, from which the whole original state can in principle be reconstructed, but only in exponential time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Thanks to Vikram Dhillon for suggesting that one-way functions might play a role in quantum gravity, it's how I found this question.) &lt;/p&gt;&#xA;" OwnerUserId="1486" LastActivityDate="2011-07-28T05:45:49.133" />
  <row Id="12872" PostTypeId="2" ParentId="10779" CreationDate="2011-07-28T06:02:55.007" Score="0" ViewCount="" Body="&lt;p&gt;Alternatively you might take your starting point to be N quantum particles in a box, and try to encode the data needed to describe the quantum state. As Dr BDO Adams stated above, our qubits will be used to describe probability distributions of positions of particles, so this is really an elaboration on his answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we have a compact space (the box), we have discrete eigenstates. Rather than thinking about the position basis and Heisenberg, I'd suggest the energy eigenstate basis. Lets have the particles to be N distinguishable particles (not indistinguishable, for then we'd have to consider boson/fermion). Each particle would get a countably infinite sequence of complex numbers whose norms sum to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should we use one qubit to store one real number, and thus get a countably infinite number of qubits to store the full information of the state of each particle? Can we do better?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets start from a naive math argument. How much information is held in a countably infinite sequence of real numbers? We can make a bijection from countably infinite sequences of real numbers to just the real numbers. (First take the arctangent of each number in your sequence and divide by pi to get the numbers between 0 and 1. Write down the numbers as a decimal in a list one after another. Then form a new real number from all of those digits by starting in the upper left corner and zig-zag-ing down. This is a bijection up to .99999... ambiguities.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So maybe we can take the whole state of the particle and just cram it into one qubit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On second thought, that is a horrible horrible non-continuous way to store information. We should require some continuity in the way we store the information, so that a small movement of a particle makes a small change to our qubits. Continuous curves still allow us to compress information in horrible ways. (a space filling curve takes each point in a plane to a unique real number, but is fractal like.) We should make our information storage, which I'm considering here as a functor from the Hilbert space of 1-particle in a box to the Hilbert space of one or more qubits, to be a smooth functor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine the box is 2-dimensional. Imagine putting the box on a surface of a sphere. That would give you a correspondence between points in the box and qubits states on the bloch sphere, right? Each position eigenstate gets mapped to the qubit thats represented by a spin arrow that points to it on the edge of the sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's wrong with this picture? Linear superpositions of position eigenstates are not necessarily mapped to unique states. Orthogonal position states don't get mapped to orthogonal qubit states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here is what we really want: a linear, invertible, unitary (orthogonal states go to orthogonal states) map from the Hilbert space of the particle in the box to the Hilbert space of quibits. Invertible linear maps from infinite dimensional Hilbert spaces like that of the particle in the box will need infinite dimensions to go into, so infinite qubits as you said, to capture ALL of the information. If you don't need all the information, maybe you limit momentum to some large cutoff, than that just leads to using only finitely many of the energy eigenstates and mapping the others to 0. The number of energy eigenstates used will be the dimension of the Hilbert space, and the number of qubits you need for each particle. (A qubit has one complex degree of freedom, since it is specified by 2 complex coefficients but is normalized.).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think Dr. BDO Adams means to use $\Delta x = \hbar / (2p) $ with p being the momentum of the particle, not the incoming light. The smallest $\Delta x$ will come with the largest $p$ you allow. Our answers match if you use the momentum cutoff p where he has placed p and the relation between Energy and momentum being that of the particles, $p^2/2m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming that the uncertainty of position of the particle should be related to the uncertainty in &lt;strong&gt;its&lt;/strong&gt; momentum. If you bombard it with light, you collapse the state into one whose uncertainty of position is now given by the momentum of the light. 1. You've changed the state. 2. Now the particle has been kicked up to the higher momentum of the light, and we have to raise our momentum cutoff. Better to just leave the state alone.&lt;/p&gt;&#xA;" OwnerUserId="4710" LastActivityDate="2011-07-28T06:02:55.007" CommentCount="1" />
  <row Id="12873" PostTypeId="2" ParentId="12867" CreationDate="2011-07-28T06:32:06.160" Score="0" ViewCount="" Body="&lt;p&gt;This is a fairly complex problem to answer starting from scratch as it has multiple components so I will just summarise the calculations that will need to be done. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of the stresses in the desk, typically as a minimum you will need to calculate the following forces:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bending moments&lt;/li&gt;&#xA;&lt;li&gt;Shear forces&lt;/li&gt;&#xA;&lt;li&gt;Bearing stresses&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These will need to be calculated for different load cases including different locations for the weights on the beam as different positions for the load will give different worst case results. The method of calculation for the stresses will vary depending on the structural details you adopt for the desk but in the case you describe will probably be based on what is known as a simply supported beam. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having calculated the forces in the beam, you will need to calculate some geometric properties of the beam in order to calculate the stresses. Typical geometric properties will be Second Moment of Area (for bending moments), Shear Area (for shear force), and Bearing Area (for bearing stresses). Again the calculation of these properties will depend on the detailing you choose, as will the use of these properties to calculate the stresses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final calculations you will have to do will be the stresses that the wood can withstand. Again, this is somewhat complex because wood, being an organic material, has different strengths in different loading conditions with factors such as grain direction, load type, load duration, wood type, etc. all affecting the calculation. You will also need to include an appropriate factor of safety in the calculations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suggestion for your desk is to avoid doing calculations if you are not used to them, but to find a desk that is similar to the one you want and copy the sizes used and use the same type of wood. &lt;/p&gt;&#xA;" OwnerUserId="3799" LastActivityDate="2011-07-28T06:32:06.160" CommentCount="4" />
  <row Id="12874" PostTypeId="1" AcceptedAnswerId="12942" CreationDate="2011-07-28T06:53:43.920" Score="4" ViewCount="153" Body="&lt;p&gt;My driver's education teacher back in high school said 55 MPH is optimal for gas mileage of a passenger car. Just last week, I read an article in a magazine saying 60 MPH is optimal. These numbers are pretty close, so there's some validity in the statement. What's the physics explanation for this 55-60 MPH sweet spot?&lt;/p&gt;&#xA;" OwnerUserId="3287" LastActivityDate="2011-07-29T19:06:07.513" Title="Why is 55-60 MPH optimal for gas mileage of a passenger car?" Tags="&lt;aerodynamics&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="12875" PostTypeId="1" CreationDate="2011-07-28T07:23:54.223" Score="0" ViewCount="46" Body="&lt;p&gt;First off, what quantities need to be factored in? Voltage and current through the coils, the magnetism of the projectile, the magnetic fields, etc.? Next, how would you calculate the speed of the projectile as it exits the coil gun? Is there an equation? Also, an equation on the force exerted on the projectile would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4144" LastActivityDate="2011-07-28T07:23:54.223" Title="How to calculate the exit velocity of a coil gun projectile?" Tags="&lt;electromagnetism&gt;&lt;classical-mechanics&gt;&lt;electricity&gt;&lt;newtonian-mechanics&gt;" CommentCount="4" />
  <row Id="12876" PostTypeId="1" AcceptedAnswerId="12887" CreationDate="2011-07-28T07:26:49.603" Score="0" ViewCount="64" Body="&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/FkbmQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Sphere A is placed on a smooth table.Another sphere B is suspended as shown in the figure.Both the spheres are identical in all respects.Equal quantity of heat is supplied to both spheres.All kinds of heat loss are neglected.The final temperatures of A &amp;amp; B are T1 &amp;amp; T2 respectively,then&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;T1&gt;T2&lt;/li&gt;&#xA;&lt;li&gt;T1 = T2&lt;/li&gt;&#xA;&lt;li&gt;T1&#xA;&lt;li&gt;None of these&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Please give reasons for your answer.&lt;/p&gt;&#xA;" OwnerUserId="4711" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-07-28T09:50:06.973" LastActivityDate="2011-07-28T09:50:06.973" ClosedDate="2011-07-29T05:32:04.347" Title="Heat &amp; thermodynamics question based on heat loss" Tags="&lt;thermodynamics&gt;&lt;homework&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="12877" PostTypeId="1" AcceptedAnswerId="12914" CreationDate="2011-07-28T07:33:16.497" Score="1" ViewCount="85" Body="&lt;p&gt;not sure if this question's been asked before, though I couldn't see one in my brief search on here... Well, the problem I am trying to solve is that I want to determine the electric field strength between two parallel rectangular-shaped plates with different voltages on each plate. For example, rectangular plate 1 has a voltage of +180V and rectangular plate 2 (parallel to plate 1) has a voltage of -5V. What equation(s) could we use to determine the field strength at any point between these two plates? The distance of separation of the plates is finite, say 8.6mm.&lt;/p&gt;&#xA;" OwnerUserId="3604" LastEditorUserId="2936" LastEditorDisplayName="" LastEditDate="2011-07-28T07:50:40.153" LastActivityDate="2011-07-28T21:44:37.283" Title="Electric field between two conducting plates of different potential?" Tags="&lt;electrostatics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="12878" PostTypeId="1" CreationDate="2011-07-28T07:38:17.653" Score="1" ViewCount="66" Body="&lt;p&gt;So I can point the axis of spin in one horizontal direction - and then move it in another direction (so it stays horizontal - like the picture below). And that still changes the direction of spin (on the rotating stool). Why does that happen? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already know the full explanation in the case that the wheel axis is vertical and the wheel is flipped over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/dJsIi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1576" LastActivityDate="2011-08-03T00:45:17.230" Title="Spinning wheel on rotating stool - why am I also affected even when the wheel axis is horizontal?" Tags="&lt;angular-momentum&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12879" PostTypeId="1" AcceptedAnswerId="12905" CreationDate="2011-07-28T07:51:48.343" Score="2" ViewCount="134" Body="&lt;p&gt;Someone asked me this question and I don't think I gave him an adequate answer (I was trying to think of the extreme case - that of neutron stars)&lt;/p&gt;&#xA;" OwnerUserId="1576" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-28T17:38:24.013" LastActivityDate="2011-07-28T18:38:46.540" Title="How does gravity affect sound waves?" Tags="&lt;gravity&gt;&lt;sound&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="12881" PostTypeId="2" ParentId="11774" CreationDate="2011-07-28T08:50:52.443" Score="0" ViewCount="" Body="&lt;p&gt;One can always reorder the operators in your polynomial to make it a polynomial of individual particle number operators. E.g. $a^+_k a^+_k a_k a_k = \pm n_k^2+n_k$ (the sign depends on the statistics of your particles). Since the particle number operators for different modes commute, the calculation of the spectrum is straightforward. &lt;/p&gt;&#xA;" OwnerUserId="4712" LastActivityDate="2011-07-28T08:50:52.443" CommentCount="1" />
  <row Id="12882" PostTypeId="2" ParentId="12867" CreationDate="2011-07-28T08:54:45.293" Score="0" ViewCount="" Body="&lt;p&gt;Calculating the straight of a wooden plank is more an art then science. The type of wood you choose oak or pine for example. The quality and condition of the wood this things come whit experience and years of working whit wood. But a rule of thumb is a pine 2x4 in good condition whit no knots has similar tensile straight of a 1x1x1/8 cold roll square tube.&lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-07-28T08:54:45.293" />
  <row Id="12883" PostTypeId="2" ParentId="12853" CreationDate="2011-07-28T08:58:30.563" Score="-2" ViewCount="" Body="&lt;p&gt;As reading the wikipedia article, we can clearly see a way to transfer information faster than the speed of light, therefore we know there is an error in the article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say entangled particle pair has identical x-spins. We measure y-spin of one particle. The x-spin of the measured particle becomes random, the x-spin of the other particle is not affected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bohr debunked Einstein's, Podolsky's and Rosen's claim that, by using entangled particles, it is possible to measure both momentum and position accurately, by saying that measuring position causes the laboratory to have somewhat uncertain momentum, which cause the momentum measurement in the laboratory to be uncertain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So don't you think Bohr's way to solve the EPR-paradox is the right way to solve the EPR-paradox?&lt;/p&gt;&#xA;" OwnerUserId="4017" LastEditorUserId="4017" LastEditorDisplayName="" LastEditDate="2011-07-28T11:20:41.517" LastActivityDate="2011-07-28T11:20:41.517" CommentCount="4" />
  <row Id="12884" PostTypeId="2" ParentId="10804" CreationDate="2011-07-28T09:03:11.647" Score="1" ViewCount="" Body="&lt;p&gt;Because the OP asks for an interpretation of normal ordering, I would like to remark that normal ordering in QFT is the same then substracting the vacuum expectation value (which I guess is sometimes called Wick product).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely, if you have a free relativistic quantum field (bosonic or fermionic does not make a real difference) &#xA;$\phi(x)$ which splits in a creation an annihilation part&#xA;$\phi_+$ and $\phi_-$ it is easy to check using $\phi_-(x)\Omega=0$ and some formal calculation &#xA;that actually $$:\phi(x)\phi(y): = \phi(x)\phi(y)- (\Omega,\phi(x)\phi(y)\Omega)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that $\phi^2(x):=\lim_{y\to x}:\phi(x)\phi(y):$ (or a general Wick polynomial) defines a new Wightman field which lies in the same &quot;Borchers class&quot; i.e. is relatively local to $\phi$,&#xA;while the normal square would just give infinity and makes no sense. &lt;/p&gt;&#xA;" OwnerUserId="1611" LastActivityDate="2011-07-28T09:03:11.647" />
  <row Id="12885" PostTypeId="2" ParentId="12732" CreationDate="2011-07-28T09:22:29.997" Score="2" ViewCount="" Body="&lt;p&gt;Mathematically, the problems with the potential $1/|x|$ are due to the fact, that the resulting Hamiltonian is not continuous on $L^2(R)$. You can expect anything with such operators - e.g., for this particular example, there is a continuum of normalizable &quot;states&quot; for every negative energy, which are not mutually orthogonal w.r.t. the standard norm in $L^2(R)$. One is better to avoid such anomailies at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible way to regularize the problem is to consider the odd solutions only (which is equivalent to a 3D hydrogen atom problem). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the issues of numerical stability, I'd try to change the variable for the spatial coordibnate to make the potential less singular. &lt;/p&gt;&#xA;" OwnerUserId="4712" LastActivityDate="2011-07-28T09:22:29.997" CommentCount="2" />
  <row Id="12886" PostTypeId="2" ParentId="12495" CreationDate="2011-07-28T09:26:08.290" Score="0" ViewCount="" Body="&lt;p&gt;You've had good answers already: drying times will decrease as air humidity &lt;strong&gt;decreases&lt;/strong&gt;, and as air temperature &lt;strong&gt;increases&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one extra thought, not covered yet: where you are, do windspeeds tend to be lower when you've got high humidity / wet weather? If the air is warm and dry, but there's no wind, clothes dry slower than if it's colder but there is a wind. So if higher humidity / wet weather correlate with lower winds locally, that would explain some of what you're finding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit: a bit more explanation:]&#xA;Unless your house is airtight other than 100% mechanical ventilation, then air currents in your house will vary with windspeeds outside. In the UK, we typically model air-changes per hour as being proportional to external wind speeds (see &lt;a href=&quot;http://www.bre.co.uk/sap2009/&quot; rel=&quot;nofollow&quot;&gt;SAP2009&lt;/a&gt;).  So, low external wind-speeds will have two effects: they will allow humidity to build up in your house, and they will reduce air convection in the house. Both of these factors will increase drying time. [End edit]&lt;/p&gt;&#xA;" OwnerUserId="4066" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-07-28T12:00:28.733" LastActivityDate="2011-07-28T12:00:28.733" CommentCount="4" />
  <row Id="12887" PostTypeId="2" ParentId="12876" CreationDate="2011-07-28T09:44:44.517" Score="2" ViewCount="" Body="&lt;p&gt;This is an old physics olympiad problem, I think.  The answer hinges on the spheres expanding due to heating.  Sphere A raises its center of mass some and sphere B lowers its center of mass some.  By conservation of energy sphere A is thus slightly colder, since more of its energy went into its gravitational potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a bit  of a silly problem since the effect is extremely small.  We can see this because from common experience if you heat a metal sphere $10 C$, the change in radius is pretty small - so small you probably won't notice without measuring it or else having something with a different expansion coefficient wrapped around the sphere.  Meanwhile, if you drop a fist-sized a metal sphere by $1cm$, a distance much larger than such a sphere would expand with a $10 C$ change, the temperature change is much, much smaller than $10 C$.  It's smaller than you can even notice, really.  So the gravitational potential change is very small compared to the heat, and the difference in temperatures is minute.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-07-28T09:44:44.517" CommentCount="1" />
  <row Id="12888" PostTypeId="1" CreationDate="2011-07-28T10:11:52.663" Score="1" ViewCount="71" Body="&lt;p&gt;Maybe my question will seem stupid, but i am not a physicist so i have some problems understanding a classic Newtonian experiment: the bucket experiment (http://en.wikipedia.org/wiki/Bucket_argument), why does he have to introduce the absolute space in order to explain the accelerated motion of water in the bucket. Also i read that in modern physics (Einstein) the accelerated motion of the water in the bucket is explained by the action of gravitational field. How does this work exactly? Other sources say that the accelerated motion remains a dilemma even for modern physics, if so why is that? Sorry that my question is a bit too broad, but i really have a problem understanding this topic so thank you in advance for your answers.&lt;/p&gt;&#xA;" OwnerUserId="4714" LastEditorUserId="966" LastEditorDisplayName="" LastEditDate="2011-07-28T11:09:30.710" LastActivityDate="2011-07-28T11:09:30.710" Title="I would like to know how we explained an accelerated motion in newtonian physics, and how do we explain it now" Tags="&lt;classical-mechanics&gt;&lt;newtonian-gravity&gt;&lt;newton&gt;" CommentCount="2" />
  <row Id="12889" PostTypeId="2" ParentId="12867" CreationDate="2011-07-28T10:24:11.077" Score="0" ViewCount="" Body="&lt;p&gt;This question as asked, seemingly poses a problem of civil engineering. &#xA;But: No (useful) table ever was constructed to such limits. &#xA;The table You ask for, would say, bow down by a cm or two &#xA;when You climb on it. (You forgot to define a maximal bent You are going to accept) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A table in practice is as stiff, to give You a feeling of &#xA;&quot;not being affected&quot; when You lean or climb on it. &#xA;The conditions You gave lead to a table which bents a lot and just  carries &#xA;those 300 lbs, but not more.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, the main thing is tensile strength, there are tables for the common woods. &#xA;Because the bend of beams is prop to length squared, it would be wise not &#xA;to place the feet at the ends, but this is of course a question of style.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-07-28T10:24:11.077" />
  <row Id="12890" PostTypeId="2" ParentId="12594" CreationDate="2011-07-28T10:45:49.503" Score="0" ViewCount="" Body="&lt;p&gt;Under the isothermic condition, the force acting on each plate equals the derivative of the free energy w.r.t the distance between the plates. That is, one has to compute the partition sum of the system as a function of the interplate distance $z$. In the case of FJC model with the link length $d$ much shorter than $z$ the calculation is straightforward. Indeed, the partition function can be expressed as a path integral, which in turn boils down to a diffusion equation with zero boundary condition on the plates. Although no explicit formula for all $z$ can be written, some asymptotics can be calculated. For instance, in the case $d\ll z \ll \sqrt{L_p d}$, the free energy scales as the biggest eigenvalue of the diffusion operator, that is like $1/z^2$ - therefore, the force grows as $1/z^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case of WLC model could probably be treated analogously. The main difference is in the path integral formula. Since the polymer has finite stiffness, the quantity in the diffusion equation should be a 2-tensor (average tensor square of the direction of the chain), instead of scalar (average density of the chain). Also, the boundary condition will be zero for the component of the tensor normal to the plate. This will change the asymptotics at $z \approx P$. I do not see right now the answer, but it should be not difficut to compute it. Basically, you have two distinct diffision coefficients (one for the traceless part of the tensor, the other for the trace), and the boundary condition which mixes two parts. At the first glance, there should be a finite limit for the free energy as $z\to 0$, hence the force should tend to zero (thus reaching a maximum somewhere around $z \approx P$.&lt;/p&gt;&#xA;" OwnerUserId="4712" LastEditorUserId="4712" LastEditorDisplayName="" LastEditDate="2011-07-28T15:55:46.110" LastActivityDate="2011-07-28T15:55:46.110" />
  <row Id="12892" PostTypeId="1" AcceptedAnswerId="12894" CreationDate="2011-07-28T13:01:02.580" Score="2" ViewCount="116" Body="&lt;p&gt;Recently I read a &lt;a href=&quot;http://arxiv.org/abs/quant-ph/9902081&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; about solving radial Schrodinger equation with inverse power law potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the radial Schrodinger equation(simply set $\mu=\hbar=1$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(-\frac{1}{2}\Delta+V(\mathbf{r})\right)\psi(\mathbf{r})=E\psi(\mathbf{r}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://physics.stackexchange.com/q/12785/2451&quot;&gt;well-known substitution&lt;/a&gt; gives a one-dimension equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{1}{2}D^2\phi(r)+\left(V(r)+\frac{1}{2}\frac{l(l+1)}{r^2}\right)\phi(r)=E\phi(r),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $D=\dfrac{d}{dr}$, and $l$ is the azimuthal quantum number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we only consider the ground state, then $l=0$, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{1}{2}D^2 \phi(r)+V(r)\phi(r)=E\phi(r).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to find eigenvalue $E$ such that $\phi(0)=\phi(+\infty)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The central potential discussed in the paper is of this form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V(r)=\alpha r^{-\beta}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It states (see page 4) that if $\beta&amp;gt;2$ then the potential is repulsive (i.e. $\alpha&gt;0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is this conclusion(i.e. If $\beta&amp;gt;2$ then we must have $\alpha&gt;0$) generally valid in physics?&lt;/li&gt;&#xA;&lt;li&gt;What would happen if $\alpha&amp;lt;0$ and $\beta&amp;gt;2$? Is there 'ground state' in this condition?&lt;/li&gt;&#xA;&lt;li&gt;What about the condition $\alpha&amp;gt;0$ and $\beta=1,2$? I have tried to solve the equation numerically with $\alpha=1,\beta=1,2$, and the ground state energy in this two conditions seem to be $0$, is my result correct?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;P.S. When I try to find ground state when  $\alpha=-1,\beta=2$, the energy seem to be $-\infty$, which is qualitatively different from $\alpha=-1,\beta=1$.&lt;/p&gt;&#xA;" OwnerUserId="4314" LastEditorUserId="4314" LastEditorDisplayName="" LastEditDate="2011-07-29T00:27:02.763" LastActivityDate="2011-07-29T21:05:42.830" Title="Radial Schrodinger equation with inverse power law potential" Tags="&lt;quantum-mechanics&gt;&lt;computational-physics&gt;&lt;schrodinger-equation&gt;" AnswerCount="2" />
  <row Id="12893" PostTypeId="1" AcceptedAnswerId="12895" CreationDate="2011-07-28T14:03:22.743" Score="1" ViewCount="81" Body="&lt;p&gt;It's been a long time since I studied thermodynamics in college years ago, but I was just wondering this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that a large bathtub of 10 C water has more heat energy than a small cup of 20 C water, and that heat naturally flows from objects of higher temperature to objects of lower temperature, but is there any way to use the heat energy of the bathtub to get the temperature of the cup water higher &lt;em&gt;on a small scale level like this&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I purposefully chose number values that would make it hard to use the heat energy of the bathtub water to do something conventional like drive a turbine to generate electricity to heat the cup water because I'm specifically curious about possibility and efficiencies without huge amounts of energy.&lt;/p&gt;&#xA;" OwnerUserId="4715" LastActivityDate="2011-07-28T14:50:12.960" Title="Using heat energy to increase temperature" Tags="&lt;thermodynamics&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12894" PostTypeId="2" ParentId="12892" CreationDate="2011-07-28T14:13:51.500" Score="2" ViewCount="" Body="&lt;p&gt;What they show in the paper is that, for $\beta&amp;gt;2$, there are no solutions with the given asymptotic form as $r\to 0$ unless we assume $\alpha&amp;gt;0$. I think you can go further and show that there are no nonsingular solutions for $\beta&amp;gt;2,\alpha&amp;lt;0$, but I'm not sure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this mean physically? Well, when we have a potential with a singularity, usually we think of it as just an approximation that breaks down sufficiently close to the singularity. For instance, we model the hydrogen atom with a potential $V\propto r^{-1}$, but really the potential near $r=0$ doesn't go to infinity, due to the nonzero size of the nucleus. We get away with using the singular potential because the &quot;bad&quot; behavior at $r=0$ doesn't qualitatively change the solutions. (And of course people do correct for nonzero-nuclear-size effects in atomic physics.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's true that the ground state of the Schrodinger equation is singular for potentials of the given form ($\alpha&amp;lt;0,\beta&gt; 2$), what that means is that this procedure doesn't work. To be specific, suppose that you solved the Schrodinger equation for a potential that looks like the given one down to some &quot;cutoff&quot; $r_0$, and is constant for smaller $r$.&#xA;What you'd find is that the solution doesn't tend to some limit as $r_0\to 0$ -- the solution depends qualitatively on the size of that cutoff, no matter how small it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your last question, for any repulsive potential ($\alpha,\beta&gt;0$), you expect to find only continuum (unbound) states. Those states have $E&amp;gt;0$, and all positive values of $E$ are allowed. So if you try to solve numerically for the ground state, I'm not surprised that you seem to get zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addition&lt;/strong&gt;: After the discussion in the comments, it occurs to me that we can see why the case $\beta=2$ behaves the way it does. The Schrodinger equation in that case is&#xA;$$&#xD;&#xA;-{1\over 2}\nabla^2\psi + {\alpha\over r^2}\psi=E\psi.&#xD;&#xA;$$&#xA;Suppose that you'd found a bound-state solution $\psi_0$ corresponding to some energy $E_0&amp;lt;0$. Define a new solution by simply scaling the radial coordinate:&#xA;$$&#xD;&#xA;\psi_1(r)=\psi_0(cr)&#xD;&#xA;$$&#xA;for any $c&amp;gt;0$. Then $\psi_1$ is also a solution to the Schrodinger equation, with energy $E_1=c^2E_0$. In particular, for $c&amp;gt;1$ this corresponds to squeezing the wavefunction into a smaller space and making the energy more negative. If you try to find the lowest-energy solution, you'll end up with the $c\to\infty$ case -- an infinitely concentrated wavefunction, with energy $-\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try the procedure I suggest in my last comment (cutting off the singularity in the potential at some $r_0$ and then varying $r_0$), something similar occurs. The ground state solution for all positive $r_0$'s look the same, with radial coordinates scaled by the value of $r_0$, and the ground state energy goes like $r_0^2$. As $r_0\to 0$, the ground-state energy approaches $-\infty$, and the wavefunction becomes infinitely concentrated at $r=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This only works for the case $\beta=2$, because for this value of $\beta$ both the kinetic and potential terms on the left side of the Schrodinger equation scale in the same way when you rescale your coordinates (i.e., both go like $c^2$). Another way to put it: only in the case $\beta=2$ is the constant $\alpha$ dimensionless. For any other $\beta$, the value of $\alpha$ determines a length scale (so that you can't just rescale one solution to get a new one), but when $\beta=2$ the problem is scale-invariant.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-07-29T21:05:42.830" LastActivityDate="2011-07-29T21:05:42.830" CommentCount="11" />
  <row Id="12895" PostTypeId="2" ParentId="12893" CreationDate="2011-07-28T14:24:22.157" Score="7" ViewCount="" Body="&lt;p&gt;Yes, you can take heat from the bathtub and heat the cup of water with it, but you will have to do work in the process.  For example, you can run a heat pump that moves heat from the tub to the cup, but the heat pump will need to be powered by electricity, or maybe a hamster on a wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of the bathtub and the cup of coffee are immaterial as long as they both contain a large number of atoms.  Temperature is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Intensive_and_extensive_properties&quot; rel=&quot;nofollow&quot;&gt;intensive quantity&lt;/a&gt;, one that does not change based on the absolute size of the system.  That means that the statement &quot;heat flows naturally from high temperature to low temperature&quot; does not depend on the size of the reservoirs involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, you are moving heat from 283K to 293K.  To move one Joule of heat across that temperature gradient requires $1 - \frac{283}{293} = \frac{10}{293}$  Joules of work at minimum.  This is for a theoretical &lt;a href=&quot;http://en.wikipedia.org/wiki/Carnot_heat_engine&quot; rel=&quot;nofollow&quot;&gt;ideal heat engine&lt;/a&gt;.  To do it in real life, you will need more work than that because you &lt;a href=&quot;http://en.wikipedia.org/wiki/Second_law_of_thermodynamics&quot; rel=&quot;nofollow&quot;&gt;cannot be perfectly efficient&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I suppose there are cute responses to this question which would refer to waiting for times longer than the age of the universe or asking for random fluctuations smaller than could possibly be measured.  Here, I'm trying to give a practical response.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-07-28T14:50:12.960" LastActivityDate="2011-07-28T14:50:12.960" CommentCount="1" />
  <row Id="12896" PostTypeId="2" ParentId="12893" CreationDate="2011-07-28T14:26:06.173" Score="2" ViewCount="" Body="&lt;p&gt;The only way to &quot;heave&quot; some heat from 10 C reservoir to one of &#xA;20 C is to use a heat pump. This is ideally a Carnot machine. &#xA;Size of reservoirs does not play a role, exept that the temperature &#xA;increase resp. decrease of the reservoirs will be inversely proportional &#xA;to the masses of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;effiency = (T1 - T2) / T2 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for a Carnot machine rumming between T1 and T2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Your numbers :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;efficency = (293 K - 283 K ) / 283 K = 0.035 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that 3.5 % of the heat streaming down will be converted into mechanical work. &#xA;In reverse, for a heat pump this means that you can pump the 1 / 0.035 = 28.6 - fold &#xA;heat by a one equivalent of mechanical work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this of course for ideal machines only!. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scale does not help here when You ask for cups! The smaller the scale, &#xA;the less efficient will such machines be in praxi. &#xA;The smallest scale heat pump are the semiconductor (Seebeck-Peltier) &#xA;coolers, but their efficency much less than that of Linde machines. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-07-28T14:26:06.173" />
  <row Id="12897" PostTypeId="2" ParentId="11647" CreationDate="2011-07-28T14:44:33.513" Score="1" ViewCount="" Body="&lt;p&gt;The brain is massively parallel, so it tends to come out looking very good. The OP suggested using joules/flop as the measure of (in)efficiency. This leaves considerable ambiguity. I believe the way neurons typically work is that they form something like a weighted average of their binary inputs, and generate a binary output that is based on a threshold value for that average. I would consider this to be the moral equivalent of a floating-point operation. Of course if a floating-point operation means working out a long-division problem using paper and pencil, then the result is going to be horrible -- a kilojoule per flop for me, or infinitely many joules per flop for a kindergartener who hasn't yet learned the long division algorithm. To me it seems perverse to say that a kindergartener's brain has zero efficiency compared to an x86, so I'm going to equate one neuron's weighted-average operation to one flop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a big fan of Ray Kurzweil, but he does have a good summary of some relevant data in his 2005 book The Singularity is Near. There's a lot of ambiguity in trying to estimate the number of fundamental operations involved in a certain neurological process. Kurzweil refers to &quot;synaptic transactions,&quot; and equates one of those to be something like $10^3$ &quot;arithmetic&quot; operations, where I imagine that he means something roughly similar to my definition of an arithmetic operation above. Anyway, subject to all these ambiguities, the studies he cites estimates of $10^{14}$-$10^{19}$ Hz for the rate of operations per second. If the brain draws ~10 W (Dan's link says 20), then this is an energy consumption of $10^{-13}$-$10^{-18}$ joules per operation. Since a desktop computer currently does $\sim10^9$ arithmetic operations per second, this makes the brain more efficient, as measured by joules per operation, by about a factor of $10^6$-$10^{10}$.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastEditorUserId="4552" LastEditorDisplayName="" LastEditDate="2011-07-28T15:20:27.273" LastActivityDate="2011-07-28T15:20:27.273" />
  <row Id="12898" PostTypeId="1" AcceptedAnswerId="13111" CreationDate="2011-07-28T15:02:00.127" Score="5" ViewCount="149" Body="&lt;p&gt;in &lt;strong&gt;two dimensional&lt;/strong&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrostatics&quot; rel=&quot;nofollow&quot;&gt;electrostatics&lt;/a&gt; it is assumed that the whole physical system is translationally invariant in one direction. Here, the two-dimensional &lt;a href=&quot;http://en.wikipedia.org/wiki/Laplace%27s_equation&quot; rel=&quot;nofollow&quot;&gt;Laplace equation&lt;/a&gt; $$\Delta \phi(x,y) = \frac{\sigma(x,y)}{\epsilon_0}$$ holds in free space and &lt;strong&gt;solutions&lt;/strong&gt; can be &lt;strong&gt;mapped on solutions&lt;/strong&gt; if space is transformed &lt;a href=&quot;http://en.wikipedia.org/wiki/Conformal_geometry&quot; rel=&quot;nofollow&quot;&gt;conformally&lt;/a&gt; - the metric changes from $$g\rightarrow \Omega^2 g\ .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approach has a lot of nice applications like the possibility to calculate the field around a complicated conductor or, in aerodynamics, the flow around an airfoil as &lt;a href=&quot;http://physics.stackexchange.com/q/4847/358&quot;&gt;discussed earlier&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Aubry et al.&lt;/strong&gt; have in a number of papers (see e.g. &lt;a href=&quot;https://workspace.imperial.ac.uk/experimentalsolidstate/Public/Stefan%20Maier/InteractingWiresPRL.pdf&quot; rel=&quot;nofollow&quot;&gt;Interaction between Plasmonic Nanoparticles Revisited with Transformation Optics&lt;/a&gt; (PRL) and &lt;a href=&quot;http://www.activeplasmonics.org/materials/pdfs/NL_10_2574.pdf&quot; rel=&quot;nofollow&quot;&gt;Plasmonic Light-Harvesting Devices over the Whole Visible Spectrum&lt;/a&gt; (Nano Lett.)) used this technique to calculate the eigenfunctions of &quot;crescend&quot; structures and coupled cylinders:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/5kWTr.png&quot; alt=&quot;Plasmonic Light-Harvesting Devices over the Whole Visible Spectrum&quot;&gt;(taken from the Nano Lett.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is a very &lt;strong&gt;nice approach&lt;/strong&gt; but I am not sure why it can be done in this way. My problem is that as far as I know usually &lt;strong&gt;only Neumann&lt;/strong&gt; and &lt;strong&gt;Dirichlet boundary conditions&lt;/strong&gt; are &lt;strong&gt;invariant under conformal transformations&lt;/strong&gt;. In electrodynamics, however, we have that both $E_t = -d\phi(\mathbf{t})$ (tangential) and $\epsilon\cdot E_n = -\epsilon\cdot d\phi(\mathbf{n})$ (normal) are continuous at a boundary where the permittivity $\epsilon$ changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the papers I could not find a discussion of this issue so I might ask here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h3&gt;Are the boundary conditions for the electrostatic potential $\phi$ invariant under conformal transformations?&lt;/h3&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance&lt;br&gt;&#xA;Sincerely&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2011-07-29T14:01:44.003" LastActivityDate="2011-08-02T17:50:09.703" Title="Boundary Conditions Invariant Under Conformal Transformations in Electrostatics?" Tags="&lt;mathematical-physics&gt;&lt;optics&gt;&lt;electrostatics&gt;&lt;conformal-field-theory&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="12899" PostTypeId="1" AcceptedAnswerId="12900" CreationDate="2011-07-28T16:20:11.490" Score="3" ViewCount="87" Body="&lt;p&gt;Let's assume the Reissner–Nordström metric (charged black hole, non-rotating), for simplicity. The black hole is charged with a powerful electric charge. There's a particle nearby, of non-zero mass, let's say an electron, its charge being the same sign like the black hole's charge. The particle's initial speed, relative to the BH, is zero. The particle is close to the event horizon, but still outside of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is - what happens? Are there any combinations of parameters where the particle starts falling in, but stops before being swallowed? Or even repulsed outright? If so, any quantitative, intuitive examples?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm asking because I know how easily intuition can get deceived by General Relativity, and doing the math involving the R-N metric is probably not feasible for me now. :)&lt;/p&gt;&#xA;" OwnerUserId="4210" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-28T16:22:54.967" LastActivityDate="2011-07-28T16:53:35.150" Title="Charged particle close to a charged black hole - what happens?" Tags="&lt;general-relativity&gt;&lt;black-holes&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="12900" PostTypeId="2" ParentId="12899" CreationDate="2011-07-28T16:53:35.150" Score="3" ViewCount="" Body="&lt;p&gt;Let's say the black hole has mass and charge $Q$ and $M$, and the electron has $m$ and $q$. An extremal black hole has $|Q|=2M$ (in the appropriate units). An electron has $|q| \gg 2m$ in the same units. If the electron fell into a negatively charged extremal R-N black hole, then the black hole would have $|Q| &amp;gt; 2M$, which would make it more than extremal. This would cause it to be a naked singularity, which would be exciting, since it would be a counterexample to the cosmic censorship hypothesis; but I'm pretty sure there is no such trivial counterexample, since cosmic censorship is still alive and kicking, decades after being conjectured. Another way of seeing that it will be repelled is that extremal R-N black holes with like charges do not interact; their gravitational attraction exactly cancels their electrostatic repulsion. Since the electron has a greater $|q|/m$, it will definitely be repelled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing to watch out for in this kind of situation is that different observers can disagree on the direction of the forces. For example, a light ray that falls radially into a black hole experiences a repulsion at certain points as measured in Schwarzschild coordinates. This is the famous &quot;Hilbert repulsion&quot; beloved of kooks like Angelo Loinger. The light ray nevertheless passes through the horizon, and local observers never see a repulsion -- McGruder, Gravitational repulsion in the Schwarzschild field, Phys. Rev. D 25, 3191–3194 (1982).&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-07-28T16:53:35.150" CommentCount="6" />
  <row Id="12901" PostTypeId="1" AcceptedAnswerId="12902" CreationDate="2011-07-28T17:42:11.607" Score="4" ViewCount="132" Body="&lt;p&gt;I &lt;a href=&quot;http://www.independent.co.uk/environment/sunflowers-to-clean-radioactive-soil-in-japan-2303086.html&quot; rel=&quot;nofollow&quot;&gt;read yesterday&lt;/a&gt; that sunflowers were used and to cleanup radioactivity at Chernobyl and the Atomic Bomb sites in Japan and may be used as part of a campaign to clean up the Fukushima area.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How does a sunflower absorb radioactivity?&lt;/li&gt;&#xA;&lt;li&gt;Is there a way to measure how well the sunflowers are working? &lt;/li&gt;&#xA;&lt;li&gt;Can sunflowers planted in not particularly radioactive places concentrate otherwise harmless radiation?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I was tempted to ask this on Skeptics.SE, but there's not much to be skeptical about, except whether or not &lt;a href=&quot;http://www.rense.com/general93/sunf.htm&quot; rel=&quot;nofollow&quot;&gt;it's a good idea&lt;/a&gt;&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="4718" LastActivityDate="2011-07-29T13:49:16.117" Title="Sunflowers and radioactivity" Tags="&lt;radiation&gt;&lt;biology&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="12902" PostTypeId="2" ParentId="12901" CreationDate="2011-07-28T17:52:21.127" Score="7" ViewCount="" Body="&lt;p&gt;It may have something to do with the growth rate of sunflowers.  During peak growing times sunflowers can grow inches in a single day, which likely results in them drawing more water out of the ground, allowing them to concentrate the radioactive materials through deposition in the plant matter at a faster rate than other plant organisms.  I would suspect you could monitor the process with standard instruments, which are pretty sensitive to the presence of particle decay. I doubt that there is any serious risk of activity accrual in areas that are not already contaminated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited to add:&#xA;A quick search on the Google points out that &lt;a href=&quot;http://www.ecaa.ntu.edu.tw/weifang/cea/sunflowers.htm&quot; rel=&quot;nofollow&quot;&gt;sunflowers are good at absorbing heavy metals&lt;/a&gt; via &lt;a href=&quot;http://en.wikipedia.org/wiki/Rhizofiltration&quot; rel=&quot;nofollow&quot;&gt;rhizofiltration&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="2433" LastEditorUserId="2433" LastEditorDisplayName="" LastEditDate="2011-07-28T17:59:06.503" LastActivityDate="2011-07-28T17:59:06.503" CommentCount="2" />
  <row Id="12903" PostTypeId="2" ParentId="12879" CreationDate="2011-07-28T18:09:40.643" Score="0" ViewCount="" Body="&lt;p&gt;To my understanding, as long as we are speaking about somewhat similar condition to earth gravity won't have an apparent effect on sound waves since the molecules composing air are too light for the gravity to have significant effect on them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider a much heavier planet, then I don't think that you'll have a significant gas phase and you are now dealing with fluids and solids. As you might now, sounds travels further (and faster?) inside those phases since the sound waves are less scatted. If you go to extream cases such as neutron star I don't think that you can describe this as anything similar to our comprehending of sound since you don't have atoms and molecules anymore (at least not in their classical definition). With a gun pointed to my head, I'll guess that in neutron star there is no sound.&lt;/p&gt;&#xA;" OwnerUserId="3677" LastActivityDate="2011-07-28T18:09:40.643" CommentCount="1" />
  <row Id="12904" PostTypeId="2" ParentId="12892" CreationDate="2011-07-28T18:20:26.337" Score="1" ViewCount="" Body="&lt;p&gt;This is addressing NGY' questions in the comment (and following up on my own comment).&#xA;Suppose $\phi(r) \sim r^n$ for $r \sim 0$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;For $\beta &amp;lt; 2$ the derivate term beats the potential term and we get that around the origin the solution must (up to higher order terms) behave as&#xA;$$n(n-1)r^{n-2} \sim 0$$&#xA;which happens for $n &amp;gt; 2$ and $n=1$. Note that if the solution were exactly $n=1$ (higher order terms included) then the derivative term would vanish identically and the equation would not be satisfied for $\beta \leq 1$. As a sidenote recall that there are also bound solutions that diverge for $r \to 0$ but they can still be normalizable, since normalization is given by an integral over the whole space and contribution from the sphere $S^{d-1}$ can beat the divergence from $\left | \phi(r)^2 \right |$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For $\beta = 2$ we see that&#xA;$n(n-1) r^{n-2} \sim \alpha 2 r^{n-2}$&#xA;implying $n=-1$ for $\alpha = 1$ and $n=2$ for $\alpha = -1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For $\beta &amp;gt; 2$ the potential term beats the derivative and we get&#xA;$r^{n-\beta} = 0$. Which is satisfiable for any $n &amp;gt; \beta$ but then the solution will have to vanish identically at all orders in order to satisfy the equation (since the derivative, the potential and the energy will all have different orders).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To understand how the energy behaves though we need to move to higher orders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example for $\beta = 1$ and $n=1$ we have &#xA;$$\phi(r) = ar + br^2 + cr^3 + o(r^3).$$ Plugging this into the equation we get &#xA;$$-b - 3cr + \alpha (a + br) = E a r  + o(r)$$&#xA; and so $b = \alpha a$ and $E = -3c/a + \alpha^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\beta = 2$, $\alpha = -1$ and $n=2$ we have&#xA;$$\phi(r) = ar^2 + br^3 + cr^4 + o(r^4)$$&#xA;and consequently&#xA;$$a + 3br + 12cr^2 - (a + br + cr^2) = Ear^2 + o(r^2)$$&#xA;which means that $b = 0$ and $E = 11c/a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we further assumed that $c/a$ is negative this would yields an answer to your question. I have no idea why this should be true unfortunately. Can someone confirm or refute this?&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-28T18:20:26.337" />
  <row Id="12905" PostTypeId="2" ParentId="12879" CreationDate="2011-07-28T18:38:46.540" Score="0" ViewCount="" Body="&lt;p&gt;OK, I'll make my comment into an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Sound_wave#Longitudinal_and_transverse_waves&quot; rel=&quot;nofollow&quot;&gt;wiki article&lt;/a&gt; has basic information  .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sound waves are pressure waves and depend on density so gravity which stratifies the atmospheric density affects sound waves through that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In solids and liquids to the extent that gravity stratifies them it will change the behaviour of sound waves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any configuration that can take density variations should have the possibility of sound waves propagating, so there could be sound waves in &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutron-star_oscillation&quot; rel=&quot;nofollow&quot;&gt;neutron stars&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-07-28T18:38:46.540" CommentCount="3" />
  <row Id="12906" PostTypeId="1" AcceptedAnswerId="12947" CreationDate="2011-07-28T19:13:29.100" Score="4" ViewCount="160" Body="&lt;p&gt;The wave equation is often written in the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\partial^2_t-\Delta)u=0,$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;involving the &lt;a href=&quot;http://en.wikipedia.org/wiki/Laplace%E2%80%93Beltrami_operator&quot; rel=&quot;nofollow&quot;&gt;Laplace-Beltrami operator&lt;/a&gt; $\Delta$. However, the Laplace-Beltrami operator $\Delta$ is defined only in the presence of Riemannian metric $g$. Indeed, one can think of $\Delta u$ as of difference between $u$ and its mean value in the balls, and the notion of ball only makes sense in the presence of a metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the physical meaning of metric $g$ here? My guess is it's connected with the property of isotropy of space: in the vacuum, the waves should propagate uniformly in all directions, and this situation corresponds to Euclidean metric. The case of general Riemannian (= ‘perturbed Euclidean’) metric should then correspond to the propagation of waves in a medium of some sort, so that the property of ambient space isotropy is not satisfied, hence balls in it do not look like usual ‘Euclidean’ balls, and wave equation takes its general form $(\partial^2_t-\Delta)u=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my guess true? If not, what is the correct interpretation of metric in the general wave equation?&lt;/p&gt;&#xA;" OwnerUserId="4720" LastEditorUserId="4720" LastEditorDisplayName="" LastEditDate="2011-07-29T15:39:18.050" LastActivityDate="2011-07-29T16:11:43.453" Title="The role of metric in the Wave Equation" Tags="&lt;mathematical-physics&gt;&lt;waves&gt;&lt;differential-geometry&gt;&lt;geometry&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="0" />
  <row Id="12907" PostTypeId="1" AcceptedAnswerId="12940" CreationDate="2011-07-28T19:15:03.187" Score="2" ViewCount="83" Body="&lt;p&gt;Say we have a bar immersed in liquid, and that the bar in being rotated at a given constant angular speed. What force is exerted on the liquid by the faces of the bar moving through the liquid (so the force is tangent to the circular motion of the bar). My guess is that the expression involves the distance from the centre of the bar, $r$, and the angular velocity $\omega$. But I have no clue what exactly the relationship should be; can anybody help?&lt;/p&gt;&#xA;" OwnerUserId="4721" LastActivityDate="2011-07-29T14:56:57.400" Title="Force exerted by rotating bar" Tags="&lt;forces&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="12908" PostTypeId="2" ParentId="10757" CreationDate="2011-07-28T19:20:01.700" Score="0" ViewCount="" Body="&lt;p&gt;If I may add a little to black hole information paradox as mentioned above, Horowitz and Maldacena wrote a paper a while ago called the final black hole state (&lt;a href=&quot;http://arxiv.org/abs/hep-th/0310281/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) where they propose the black hole complimentary which states that information that goes into a black hole is copied/cloned and that cloning will cause problems such as providing non-linear corrections to the Shrodinger's equations but Maldacena et. al propose that this copying shouldn't be a problem if it occurs at a place where no one can see it. Excuse the non-technical language being used here. Preskill wrote some comments on that can be found &lt;a href=&quot;http://arxiv.org/abs/hep-th/0311269&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. This actually relates to another interesting concept called post-selection, the use of cloning allows to some extent for one to employ post-selection to measurements and I talked to Scott Aaronson about this before, here's what he said about this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The purpose of that footnote was just to give one actual example of&#xA;  someone who had proposed a theory involving postselection.  Maybe it&#xA;  wasn't the best example, since when I asked Juan Maldacena about it&#xA;  later, he suggested that he and Horowitz never &quot;really believed&quot; their&#xA;  own proposal! :-)  A better example that I could have given is the&#xA;  &quot;multiple times&quot; theory of Yakir Aharonov: that theory explicitly&#xA;  involves postselection, Yakir has been promoting it for decades, and&#xA;  when I pointed out to Yakir the potential for using such a proposal to&#xA;  solve NP-complete and even harder problems in polynomial time, he&#xA;  essentially said &quot;so?  what's the problem?  Nature does all kinds of&#xA;  hard calculations all the time!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now onto one-way functions, let's go with Maldacena's proposal, from that throwing in the diary won't do any good because the information present would be cloned and that means in strict computational sense that once we invert the function (which in this case implies the information coming out as radiation or being cloned) we can still get an output in polynomial time. I suppose that going more into Hawking radiation will give me more answers, however at this time I am not too familiar with it myself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do want to say something about the approach though, one of the approaches that IMHO is the easiest to go for one-way functions is Mulmuley's obstructions. I think one can take the following path, first we define physical analogs of the mathematical positivity and mathematical negativity hypothesis, these analogs would include physical problems such as lack of symmetries in AdS and so on. Once those are done we can define a modified version of the flip that they use, our flip would only go from hard nonexistence to easy existence and because we have defined the analogs, that by itself should suffice. Once we can show obstructions exist, then that can be applied to any case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I speculate, if black-hole evaporation is indeed one-way, and we follow that nonlinear corrections to QM can't happen, then showing that obstructions exist for a specific family of functions called point-functions when they are thrown into a black hole might work. Here's the reason, point-functions can be quantum copy-protected as seen (I can't post the link because SE won't let new users do that, however the name of the paper is:&#xA;Quantum Copy-Protection and Quantum Money, Scott Aaronson) and we already follow that no cloning can occur, now what happens to the point functions can perhaps be explained by Hawking radiation or the holographic principal, so in that case using the physical analogs we might be able to show that obstructions exist for the information. I'm working on how this approach can be completed. &lt;/p&gt;&#xA;" OwnerUserId="4719" LastEditorUserId="4719" LastEditorDisplayName="" LastEditDate="2011-07-29T00:44:11.117" LastActivityDate="2011-07-29T00:44:11.117" />
  <row Id="12909" PostTypeId="1" CreationDate="2011-07-28T19:49:35.870" Score="-4" ViewCount="68" Body="&lt;p&gt;I was wondering about organics that traveling back in time. for example. I'm standing on main street where there is a an apple tree about 30 years old. I pick a rotten apple from it, I go back in time 20 years and the tree is now only ten years old. but what happens to the rotten apple?&lt;br&gt;&#xA;If an apple that is rotten transports back in time will it become fresh again? &lt;/p&gt;&#xA;" OwnerUserId="4708" LastActivityDate="2011-07-28T20:07:59.173" ClosedDate="2011-07-30T15:06:51.197" Title="If an apple that is rotten transports back in time will it become fresh again?" Tags="&lt;spacetime&gt;&lt;time&gt;&lt;time-travel&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="12910" PostTypeId="1" CreationDate="2011-07-28T20:06:30.687" Score="4" ViewCount="70" Body="&lt;p&gt;Trying to understand one-forms and vectors via Schutz's A First Course In General Relativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/3mXwV.gif&quot; alt=&quot;contraction of one-form and vector)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His example uses a spacetime diagram, a scalar field phi, a curve (worldline) parametrized using proper time and the tangent vector (4-velocity). I think I understand, but I can't see the physical significance of the scalar field phi. In plain English, in the spacetime diagram, what does or might phi represent?&lt;/p&gt;&#xA;" OwnerUserId="4075" LastActivityDate="2011-07-28T20:21:46.060" Title="What does scalar phi represent in spacetime?" Tags="&lt;general-relativity&gt;&lt;spacetime&gt;&lt;vectors&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="12911" PostTypeId="2" ParentId="12909" CreationDate="2011-07-28T20:07:59.173" Score="-1" ViewCount="" Body="&lt;p&gt;Yes, it would be fresh. It's easy: you films the apple and then run the movie backwards. No doubt it will become fresh.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-07-28T20:07:59.173" CommentCount="1" />
  <row Id="12912" PostTypeId="2" ParentId="12910" CreationDate="2011-07-28T20:21:46.060" Score="6" ViewCount="" Body="&lt;p&gt;Scalar field is a field of scalars (don't blame me, you invited Captain Obvious yourself). In other words, it's just a function on the manifold. If the manifold were the surface of the Earth (I hope you don't mind I move to 2+0 dimensions for a bit), it could e.g. be the ground temperature in certain moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving up to 2+1 dimensions, it can again be a field of temperatures but now with time record included. So, your curve $\gamma$ records your wandering around the globe and ${\rm d} \phi \over {\rm d} \tau$ just tells you how the temperature you experience varies both in space and time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving up one further dimension still, a 3+1-dimensional field could be e.g. temperature &lt;em&gt;anywhere&lt;/em&gt; in the atmosphere (and again also in time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in particle physics, scalar fields play a whole different role. They are not connected to familiar things like the temperature above and one instead has to think about them purely abstractly, relating them to things such as EM waves and photons. EM wave, after all, is just a vector field (meaning having spin 1).  Similarly, an electron is a particle associated do a Dirac field (which is a fermionic field with spin 1/2). Finally a scalar field has spin 0. I hear, one of the recently-gone-famous fields in this class goes by the name of Higgs :)&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-28T20:21:46.060" CommentCount="2" />
  <row Id="12913" PostTypeId="1" CreationDate="2011-07-28T20:53:02.930" Score="2" ViewCount="167" Body="&lt;p&gt;I was listening to some co-workers talking about problems meeting stiffness requirements.  Someone said that even with a solid metal rod (instead of the existing tube) we could not meet stiffness requirements.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started daydreaming... and went back in time over a quarter of a century to some class I was taking in college.  Things are hazy when you go back that far; but I am sure that someone with a Ph.D., or some other letters after his name, said that you could actually &lt;em&gt;add&lt;/em&gt; stiffness to a solid rod by drilling out the center (and maybe by appropriately treating it, I forget).  The reason, if I recall correctly, had to do with the added tensile strength of the inner surface.  The reason I remember this from so long ago was that it was so counter-intuitive: I was stunned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a stress/structures guy; so I asked a co-worker about it, and he said that the solid rod would be stiffer, because it had the greater (bending) moment of inertia.  I agree with the latter part of the statement, but my hazy daydream keeps me from agreeing with the preceding conclusion.  My money is still on a series of concentric tubes, appropriately processed and internally supported, being stiffer than a solid rod of the same &lt;a href=&quot;http://en.wikipedia.org/wiki/Outside_diameter#Outside_diameter&quot; rel=&quot;nofollow&quot;&gt;O.D.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question:  Does anyone know of any references to this little structural trick (or engineering wives' tale).  If so, can you quantify how much stiffness is gained?&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-28T23:20:01.737" LastActivityDate="2011-07-29T11:45:12.853" Title="Hollow Tube Stronger than Solid bar of Same O.D.?" Tags="&lt;material-science&gt;&lt;engineering&gt;&lt;stress-strain&gt;&lt;metals&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="12914" PostTypeId="2" ParentId="12877" CreationDate="2011-07-28T21:09:40.373" Score="1" ViewCount="" Body="&lt;p&gt;If you are not worried about fringing effects near the edges of the plates (which you shouldn't, if the plate dimensions are much larger than the gap distance), then finding the field is simple: V/d, where V is the voltage between the plates and d is the gap distance between the plates.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in the example you gave, there would be a field of 185V/(0.0086m) = about 21512 V/m, between the two plates.  The voltage 1mm from the -5V plate would be -5 + (0.001m)(21512 V/M) = 21.512V-5V = 17.512V&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-07-28T21:09:40.373" CommentCount="3" />
  <row Id="12917" PostTypeId="1" AcceptedAnswerId="13133" CreationDate="2011-07-28T22:07:08.980" Score="4" ViewCount="99" Body="&lt;p&gt;I'm trying to understand how do gyroscopes and &lt;a href=&quot;http://en.wikipedia.org/wiki/Control_moment_gyroscope&quot; rel=&quot;nofollow&quot;&gt;cmgs&lt;/a&gt; used for navigating keep on spinning even though there's friction. Any engine/pushing mechanism I can think of will restrict the gyro's rotor to a fixed direction and defeat the purpose... how is it done without disturbing it? &lt;/p&gt;&#xA;" OwnerUserId="4722" LastActivityDate="2011-08-02T23:46:40.287" Title="how can you keep a gyroscope turning?" Tags="&lt;gyroscopes&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="12918" PostTypeId="1" CreationDate="2011-07-28T22:11:20.750" Score="-1" ViewCount="61" Body="&lt;p&gt;Please help me solve this. I have been thinking of all sorts of ways to solve this but can't figure out how :(. Ok here's the problem:&#xA;I am given a three dimensional velocity vector (i know the magnitude of this vector and I know what angle this vector makes wrt one of the axis, say the x-axis.) What I want to determine is what angle does this three-dimensional vector make with the x-y, y-z and x-z plane. Another way to look at this is if we project this three-dimensional vector in the x-y plane what is the angle between this vector and the x- axis (or the y- axis)? I do not know the velocity components of this vector in the x-, y- or z- axis. As a matter of fact, these x-, y- and z- velocity components are what I aim to calculate from determining the angle the 3-D vector makes with each plane. &lt;/p&gt;&#xA;" OwnerUserId="3604" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-28T22:19:29.877" LastActivityDate="2011-07-28T22:30:50.817" ClosedDate="2011-07-28T22:41:45.560" Title="how to calculate the angle in the x-y, y-z, x-z plane given only 3D vector direction and magnitude?" Tags="&lt;mathematical-physics&gt;&lt;homework&gt;&lt;vectors&gt;&lt;geometry&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="12919" PostTypeId="2" ParentId="12913" CreationDate="2011-07-28T22:14:52.610" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.physicsforums.com/archive/index.php/t-37701.html&quot; rel=&quot;nofollow&quot;&gt;http://www.physicsforums.com/archive/index.php/t-37701.html&lt;/a&gt; says &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Most of the strength of a cylinder comes from the outer portions.&#xA;  I think the contribution goes like the cube of the radial position.&#xA;  So, if you took a solid rod and drilled out a half the volume from the&#xA;  center, you do not lose half the strength. Strength to weight ratio is&#xA;  better for a hollow pipe than a solid rod.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The definition for the second moment of inertia $I_c$ for a filled and hollow cylinder can be found on &lt;a href=&quot;http://en.wikipedia.org/wiki/Second_moment_of_area&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Second_moment_of_area&lt;/a&gt;:&#xA;$$I_c=\int\!\!\!\!\!\int_A y^2 \textrm{d}x\textrm{d}y=\int_0^R\!\!\!\!\!\int_0^{2\pi} r^2 \textrm{d}\phi\ r\textrm{d}r=\frac{\pi r^4}{4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface area of the filled cylinder is:&#xA;$$A=\pi r^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare filled and hollow cylinder of equal mass: $$s_c=\frac{I_c}{A}=\frac{r^2}{4}$$, cylinder with fractional internal radius $r_i=xr_o$ and $x&amp;lt;1$: $$s_h=\frac{I_h}{A}=\frac{r^4(1-x^4)}{4r^2(1-x^2)}=\frac{r^2(1-x^2)(1+x^2)}{4(1-x^2)}=\frac{r^2(1+x^2)}{4}&amp;gt;s_c$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means a hollow cylinder is stronger than a rod of equal mass and the same material.&#xA;A hollow cylinder with a bigger inside diameter is better.&#xA;In the limit $x\rightarrow 1$ the hollow cylinder is twice as strong.&#xA;Note that this limit isn't physically viable as it would be an cylinder with infinite radius and infinitesimally thin wall. However it is useful to define the upper limit of the second moment of inertia. I didn't expect the increase in strength only a factor of two. &lt;/p&gt;&#xA;" OwnerUserId="4306" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-07-29T11:45:12.853" LastActivityDate="2011-07-29T11:45:12.853" CommentCount="2" />
  <row Id="12920" PostTypeId="2" ParentId="12918" CreationDate="2011-07-28T22:30:50.817" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Spherical_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Spherical_coordinate_system&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix}x\\y\\z\end{pmatrix}=r \begin{pmatrix}&#xD;&#xA;\sin(\theta)\cos(\phi)\\&#xD;&#xA;\sin(\theta)\sin(\phi)\\&#xD;&#xA;\cos\phi&#xD;&#xA;\end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several ways how you can measure the angles (e.g. $\theta$ from the pole or equator).&#xA;You can figure it out by inserting $\pi/2$ and $0$ and look what $(x y z)$ vectors you get.&lt;/p&gt;&#xA;" OwnerUserId="4306" LastActivityDate="2011-07-28T22:30:50.817" />
  <row Id="12921" PostTypeId="2" ParentId="759" CreationDate="2011-07-29T00:33:28.800" Score="0" ViewCount="" Body="&lt;p&gt;Even since Shanon linked information with entropy, there have always been speculations about the universe being run as a simulation and so on. John Preskill recently gave a talk at Holographic Cosmology 2.0 on this topic, I haven't seen the full talk yet, but just thought others might find this it interesting. &lt;/p&gt;&#xA;" OwnerUserId="4719" LastActivityDate="2011-07-29T00:33:28.800" />
  <row Id="12922" PostTypeId="1" AcceptedAnswerId="13414" CreationDate="2011-07-29T01:15:57.883" Score="5" ViewCount="141" Body="&lt;p&gt;Let $Q_{ab} = (\psi_{;a})(\psi_{;b}) - (1/2)g_{ab}|\nabla \psi|^2$ be the energy-momentum tensor of the wave equation in some space time. I will use semicolons to refer to covariant differentiation and $\partial$s to refer to coordinate differentiation. Let $\pi_{ab}$ be the deformation tensor for some fixed vector field $X$. In deriving &quot;almost conservation laws&quot; one uses the identity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(Q_{ab}X^b)^{;a} = (\psi^{;a}_{;a})(X^a\partial_a\psi) + (1/2)Q^{ab}\pi_{ab}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does there exist a physical interpretation of the scalar $Q^{ab}\pi_{ab}$ that is not based on the above formula? I am most interested in how one should think about this quantity in relativistic contexts, e.g. black hole geometries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Just in case anyone is tempted, I am looking for something more than &quot;$Q^{ab}\pi_{ab}$ vanishes if the flows of $X$ are isometries.&quot; &lt;/p&gt;&#xA;" OwnerUserId="4163" LastEditorUserId="4163" LastEditorDisplayName="" LastEditDate="2011-07-29T14:33:36.500" LastActivityDate="2011-08-12T02:21:29.680" Title="Physical Interpretation of a Scalar Quantity Related to Currents/Conservation Laws" Tags="&lt;general-relativity&gt;&lt;classical-mechanics&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="12923" PostTypeId="2" ParentId="5615" CreationDate="2011-07-29T02:44:16.630" Score="0" ViewCount="" Body="&lt;p&gt;If black holes do not conserve baryon number, is it possible that the anti-matter created at big bang are hidden inside primordial black holes, and thus we have matter (as opposed to anti-matter) in excess? Can this hypothesis be proved in any way?&lt;/p&gt;&#xA;" OwnerUserId="4724" LastActivityDate="2011-07-29T02:44:16.630" />
  <row Id="12924" PostTypeId="1" CreationDate="2011-07-29T03:01:34.950" Score="1" ViewCount="157" Body="&lt;p&gt;A popular circus stunt is for a motorcycle rider to ride inside a bowl shaped depression called a &quot;Wall of Death.&quot; The rider goes higher and higher up the wall until they are actually horizontal. I wonder if a human could do the same. &lt;/p&gt;&#xA;" OwnerUserId="4725" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-07-29T10:34:06.660" LastActivityDate="2011-07-29T19:43:11.880" Title="Could a human run horizontally inside a Wall of Death?" Tags="&lt;centrifugal-force&gt;" AnswerCount="3" />
  <row Id="12925" PostTypeId="2" ParentId="12924" CreationDate="2011-07-29T03:59:04.037" Score="1" ViewCount="" Body="&lt;p&gt;The speed you need (explained &lt;a href=&quot;http://physics.mut.ac.th/PhysicsMagic/wall.htm&quot; rel=&quot;nofollow&quot;&gt;http://physics.mut.ac.th/PhysicsMagic/wall.htm&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;V^2 &amp;gt; R g/u&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v$ = the velocity (m/s)&lt;br&gt;&#xA;$R$ = Radius of the pit (m)&lt;br&gt;&#xA;$g$ = acceleration of gravity (9.8m/s^2)&lt;br&gt;&#xA;$u$ = coeff of static friction  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if u is 1 (the maximum possible), and you are an olympic athlete that can run 100m in 10s&lt;br&gt;&#xA;Then you apparently could with a radius of about 10m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;R = uV^2/g = 10^2/9.8 = 10m &#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="2525" LastEditorUserId="2525" LastEditorDisplayName="" LastEditDate="2011-07-29T19:28:16.170" LastActivityDate="2011-07-29T19:28:16.170" CommentCount="11" />
  <row Id="12926" PostTypeId="2" ParentId="7838" CreationDate="2011-07-29T04:01:53.660" Score="-1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Nihil ex nihilo.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Time and causality are but an illusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pure Consciousness itself is all there is, and it exists timelessly, and always was, is and will be forever more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no existence before the Big Bang, and so, it was not and is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe is the cause of itself, but causality exists not except as an illusion.&lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2011-07-29T04:01:53.660" CommentCount="2" />
  <row Id="12927" PostTypeId="2" ParentId="12917" CreationDate="2011-07-29T04:06:16.933" Score="0" ViewCount="" Body="&lt;p&gt;Probably an electric motor mounted in the wheel and power passed in through a commutator&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-07-29T04:06:16.933" CommentCount="5" />
  <row Id="12928" PostTypeId="2" ParentId="759" CreationDate="2011-07-29T04:16:28.157" Score="0" ViewCount="" Body="&lt;p&gt;If the Church-Turing hypothesis applies to our universe, then this is equivalent to stating that the universe can be simulated on a computer. But just because it can be doesn't necessarily mean it is. This is an ontological question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Epistemologically, we can never know for sure, and never will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If quantum mechanics is purely a gadget for computing the output given some inputs, and we are forbidden to ask what happens in between, we might as well assume it's some black box computation mapping the inputs to the outputs probabilistically. In this abstract sense, computation is what quantum mechanics does, but this also applies to classical theories, and many other maps besides. But a computer is a physical substrate. Just because something is a computation doesn't necessarily mean it has a physical substrate, or a nonphysical substrate for that matter, but it doesn't rule that out either.&lt;/p&gt;&#xA;" OwnerUserId="4727" LastActivityDate="2011-07-29T04:16:28.157" />
  <row Id="12930" PostTypeId="2" ParentId="2166" CreationDate="2011-07-29T05:01:15.510" Score="0" ViewCount="" Body="&lt;p&gt;To recall a memory and immerse yourself in it so fully as to forget yourself is to go back in time. Memories are the time capsules preserving the past. So, yes, it's possible to go back in time, and in fact, is the only way.&lt;/p&gt;&#xA;" OwnerUserId="4728" LastActivityDate="2011-07-29T05:01:15.510" />
  <row Id="12931" PostTypeId="1" AcceptedAnswerId="12932" CreationDate="2011-07-29T06:21:42.073" Score="3" ViewCount="50" Body="&lt;p&gt;Here is the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A planet with mass $m$ and a second with mass $M$ are separated by a&#xA;  distance $d$. A third planet with mass $m_3$ happens to be midway&#xA;  between $M$ and $m$. Where could the third planet be positioned&#xA;  (distance from the larger planet $M$ in meters) so that the net&#xA;  gravitational force is zero?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My confusion lies with how to solve for the position of the third planet.  I am given this equation to find the force between two planets at a given distance $r$:&#xA;$$F = {GMm}/{r^2}$$&#xA;With this I can then set the sum of the two forces to zero:&#xA;$$0 = {GMm_3}/r_1^2 + {Gmm_3}/r_2^2$$&#xA;My confusion lies in the fact that both $r_1$ and $r_2$ are unknown.  However, we do know that $r_1 + r_2 = d$.  But I am confused with how to solve for either $r_1$ or $r_2$.&lt;/p&gt;&#xA;" OwnerUserId="4675" LastActivityDate="2011-07-29T06:46:59.743" Title="Finding the position of a planet between two other planets of known mass and distance" Tags="&lt;gravity&gt;&lt;homework&gt;&lt;planets&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="12932" PostTypeId="2" ParentId="12931" CreationDate="2011-07-29T06:46:59.743" Score="2" ViewCount="" Body="&lt;p&gt;Check the signs in your equation. Draw the three planets in a line with the central planet at the origin. What is the direction of the force each of the side planets exert on the middle one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you resolve the above problem you can approach solving the equations. Notice that you have as many equations as you have unknows. Therefore each equation lets you (at least in principle) express one unknown as the function of the others which reduces the problem to one with smaller number of unknows (although possibly with a more complicated equations).&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-29T06:46:59.743" />
  <row Id="12933" PostTypeId="1" AcceptedAnswerId="12980" CreationDate="2011-07-29T08:30:47.080" Score="2" ViewCount="120" Body="&lt;p&gt;Your lateral thinking and knowledge of lab kit could help us solve a tricky measuring problem in building-physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the problems we get in modelling the heat demands of buildings, is getting an accurate number for the building's thermal conductivity as a whole. Even when we know the materials it's built from, variations in material quality, thickness, and build-consistency, mean that the calculation of the composite thermal conductivity is frequently wrong. Now, yes, if we could get all of the very specific information on the exact materials, and the exact locations of every thermal bridge, we could calculate the building's thermal conductivity from first principles. But that's very rarely possible. (incidentally, we also need to know the thermal mass, but the calculation tends to be a bit more forgiving about measurement inaccuracies, and close enough may be good enough)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it strikes me that there is almost certainly some kit in use by physicists in labs or industry, that could be co-opted for this purpose. Something that could be taken to different dwellings, to measure the thermal conductivity of specific walls, floors, or whole buildings, and that would allow the measurement to be done within one day. [&lt;strong&gt;edit 1&lt;/strong&gt;:] Note that in general, we will have access to the inside as well as the outside of a building.[&lt;strong&gt;end edit 1&lt;/strong&gt;]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that the use of thermal cameras hasn't been entirely successful, due to problems with calibration, and with variations in reflectivity and emissivity of different materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buildings are almost never in thermal equilibrium, and often have high thermal mass, so simply releasing a known quantity of low-grade heat in the dwelling and recording time-series of temperature of inner air, wall, outer wall, as well as recording times series of external air temperatures, could be doable, but it would typically be difficult to establish what the initial and final conditions are (e.g. how much heat is stored in the building, at the start and at the end).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Testing must be non-destructive: that is, we can't submit surfaces to tests that would damage them. So no super-heating or deep freezing, if it would cause permanent damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;strong&gt;edit 2&lt;/strong&gt;: &lt;em&gt;supplementary information prompted by a question about using historic information about energy consumption to deduce conductivity:&lt;/em&gt;]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Historic energy use of dwellings does provide some information; however, it is often hard to obtain for a lot of dwellings (being commercially sensitive); and, as generally we don't know the long-term pattern of internal temperatures, we can't deduce the building's thermal conductivity. We may know how much energy was used, but we don't know  how much warmer than external temperatures did it keep the building, nor for how long. It turns out that observed thermostat settings aren't useful for this, for several reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;strong&gt;end edit 2&lt;/strong&gt;]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;strong&gt;edit 3&lt;/strong&gt;]We are interested heat losses from both conduction, convection and radiation. Convection is sometimes estimated using data from a &lt;a href=&quot;http://www.leedsmet.ac.uk/teaching/vsite/visit/pressurisation/procedure.htm&quot; rel=&quot;nofollow&quot;&gt;pressurisation test&lt;/a&gt;, which isn't always easy to do. And we're interested in conduction and radiation losses as well as convection losses.  Some heat flux sensors have been tried, but apparently without much success. If there have been recent advances in heat flux sensors, I'd love to hear about them.[&lt;strong&gt;end edit 3&lt;/strong&gt;]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not inviting speculation: I'm looking for a factual answer about specific tests to yield specific measurements about the physics of dwellings.&lt;/p&gt;&#xA;" OwnerUserId="4066" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-08-04T10:20:53.643" LastActivityDate="2011-08-04T10:20:53.643" Title="What devices could be taken round dwellings to measure the thermal conductivities of buildings in situ?" Tags="&lt;thermodynamics&gt;&lt;building-physics&gt;&lt;thermal-conductivity&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="2" />
  <row Id="12934" PostTypeId="2" ParentId="12913" CreationDate="2011-07-29T09:01:44.243" Score="1" ViewCount="" Body="&lt;p&gt;Though I generally agree with whoplisp's answer, it is worth to note that obtaining the (lower) limit of the thickness is rather tricky as long as it is defined by stability under 'strong' deformations. Where 'strong' is compared with the tube wall thickness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is obvious from common sense point of view: thin rod is easier to deform than the hollow tube (of the same &lt;em&gt;mass&lt;/em&gt;), but as long as the hollow tube starts to deform it is easier to break. &lt;/p&gt;&#xA;" OwnerUserId="4657" LastActivityDate="2011-07-29T09:01:44.243" />
  <row Id="12935" PostTypeId="2" ParentId="12907" CreationDate="2011-07-29T09:34:14.363" Score="0" ViewCount="" Body="&lt;p&gt;The force of the bar on the liquid is the opposite of the friction force of the liquid on the bar. Except for the part close to the center (which doesn't very fast at all) or the end (hard!) you can probably treat the fluid flow as a regular orthogonal flow past every part of the bar, at speed &lt;code&gt;rω&lt;/code&gt;. (If the fluid is non-Newtonian, you would get shear stresses in the fluid and you can't treat each part of the bar independently). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact forces will depend on the Reynolds number, which will be hard to predict without further information. However, the general relation is that the forces double with the square of the speed, i.e. with &lt;code&gt;rω&lt;/code&gt;&lt;sup&gt;2&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-07-29T09:34:14.363" CommentCount="5" />
  <row Id="12936" PostTypeId="1" CreationDate="2011-07-29T10:46:54.183" Score="1" ViewCount="72" Body="&lt;p&gt;decay with initial velocity: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All static particle decay to particles with momentum distribution in the same shape like&#xA;&lt;img src=&quot;http://i.stack.imgur.com/i8gD5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;that's implied by structure function and calculated by MG. but the decay of W emits e like (similar to)&#xA;&lt;img src=&quot;http://i.stack.imgur.com/mXogV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;Why that? The only explanation &#xA;is the W generated by p--p has initial velocity. The superficial explaination of &#xA;resonance are contradictory to the logical deduce. This &#xA;will modified the mass of it to 15 Gev.&#xA;Is it reasonable?&lt;/p&gt;&#xA;" OwnerUserId="4684" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-07-29T12:37:10.977" LastActivityDate="2011-07-30T07:55:28.207" Title="decay with initial velocity:" Tags="&lt;particle-physics&gt;" AnswerCount="0" CommentCount="6" />
  <row Id="12937" PostTypeId="2" ParentId="12924" CreationDate="2011-07-29T11:13:50.933" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The rider goes higher and higher up the wall until they are actually&#xA;  horizontal. I&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This causality is wrong. The riders (and bikes) angle approaches  more and more &#xA;a horizontal position (but never will be exactly horizontal!) the &#xA;faster he goes. This is independent from height in the &quot;wall&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your impression maybe results from the practice, that the drivers gain speed &#xA;and climb up in the wall simultaneously after start. When the driver has &#xA;reached the cruising speed to do his artistic show, he maintains the same angle &#xA;to horizontal irrespective of driving close to the upper rim or down close &#xA;to the bottom of that &quot;bowl&quot;. Have a look to that next time You see such a show. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reasoning for the angle is based on the vector diagram in Martins link, &#xA;You just have to ad the equilibrium specific of a bike. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-07-29T19:17:00.437" LastActivityDate="2011-07-29T19:17:00.437" />
  <row Id="12938" PostTypeId="2" ParentId="11154" CreationDate="2011-07-29T12:38:55.680" Score="0" ViewCount="" Body="&lt;p&gt;Lithium is more abundant in the Earth's crust than lead. However, it is more reactive than such metals and less abundant than other reactive metals such as sodium. Because of this it does not tend to accumulate in rich geological deposits in a form that makes it easy to extract. Its lightness may be another factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reactive metals such as lithium can form salts which dissolve in water. These are then left in deposits when enclosed areas of water dry up. Lithium is 1000 times less abundant in the Earth's crust than other reactive metals such as Sodium, Calcium and Potassium so it is still only found in relatively small quantities in such deposits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, some compounds of lithium are sufficiently soluble that it is present in some dried up sea deposits. About half the accessible Lithium on Earth is said to be beneath the Bolivian dessert and if extracting it in the future becomes as important as extracting oil is now, then there is not likely to be such a big shortage. &lt;/p&gt;&#xA;" OwnerUserId="1341" LastEditorUserId="1341" LastEditorDisplayName="" LastEditDate="2011-07-29T14:47:48.437" LastActivityDate="2011-07-29T14:47:48.437" CommentCount="4" />
  <row Id="12939" PostTypeId="2" ParentId="12901" CreationDate="2011-07-29T13:49:16.117" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;As your second link states, sunflowers absorb ions from the soil. That's how they get nutrients to grow, but the same chemical mechanism will absorb cesium atoms - cesium is very soluable in water. The explosion at Fukushima ejected two isotopes with a particular danger to humans, &lt;sup&gt;137&lt;/sup&gt;Cs (cesium) and &lt;sup&gt;131&lt;/sup&gt;I (Iodine) Because the latter has a half-life of only 8 days &lt;sup&gt;135&lt;/sup&gt;Cs is currently the main cause of radioactivity in the area. Therefore, sunflowers grown in the Fukushima area will absorb the radioactive Cesium-137 isotope.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The effect is quite easy to measure: the sunflowers themselves become radioactive because they contain radioactive Cesium-137 atoms. In particular, Cesium-137 gives off characteristic gamma rays through the decay via Barium-137m.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Not really. Banana's are far more effetive in that respect.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-07-29T13:49:16.117" />
  <row Id="12940" PostTypeId="2" ParentId="12907" CreationDate="2011-07-29T13:49:43.653" Score="2" ViewCount="" Body="&lt;p&gt;First of the sum of the forces on the bar have to be zero because the center of gravity does not move (assuming symmetry and such). The only thing you can calculate is the torque requited to maintain the constant speed. I can give an example with a rectangular bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the bar and split it into infinitesimal slices. Each slice has face area of ${\rm d}A = h\,{\rm d}r$ where $r$ is the distance from the center and varies from $-L/2$ to $L/2$, and $h$ is the height of the bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The liquid-solid force for each slice is ${\rm d}F =\frac{1}{2} \rho\; C_d v^2 {\rm d}A$ where $\rho$ is fluid density, $C_d$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Drag_coefficient&quot; rel=&quot;nofollow&quot;&gt;coefficient of drag&lt;/a&gt; and $v$ is the velocity of the bar slice, which is equal to $ v = \omega\;r $. Note that the force has to flip signs when $r$ flips signs and thus we have to add a ${\rm sign}(r)$ term and integrate over $r$ to get the total force&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F = \int_{-L/2}^{L/2}\;{\rm sign}(r)\frac{1}{2}\rho\;h\;C_d\;(\omega\,r)^2\;{\rm d}r = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and to total torque&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M = \int_{-L/2}^{L/2}\;{r\;\rm sign}(r)\frac{1}{2}\rho\;h\;C_d\;(\omega\,r)^2\;{\rm d}r = \frac{\rho C_d h \omega^2 L^4}{64} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main assumption here is that the $C_d$ does not depend on the flow velocity $v=\omega\,r$ which it does. A more extensive analysis would require finding the &lt;a href=&quot;http://en.wikipedia.org/wiki/Reynolds_Number&quot; rel=&quot;nofollow&quot;&gt;Reynolds number&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fanning_friction_factor&quot; rel=&quot;nofollow&quot;&gt;Fanning Friction&lt;/a&gt; factor and deriving the pressure distribution along the bar based on the flow characteristics.&lt;/p&gt;&#xA;" OwnerUserId="392" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2011-07-29T14:56:57.400" LastActivityDate="2011-07-29T14:56:57.400" CommentCount="4" />
  <row Id="12941" PostTypeId="2" ParentId="12874" CreationDate="2011-07-29T14:07:19.930" Score="3" ViewCount="" Body="&lt;p&gt;It's really an engineering question. Cars are designed for efficiency nowadays; especially in Europe the taxes are directly related to the MPG ratings on standardized test cycles. But even in the US, there are &lt;a href=&quot;http://www.fueleconomy.gov/feg/fe_test_schedules.shtml&quot; rel=&quot;nofollow&quot;&gt;standard test cycles&lt;/a&gt; for fuel effiency. Those test cycles typically have large segments driven at ~55 mph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now an engineer that's optimizing for fuel efficiency has to make tradeoffs. Adding more gear ratios can improve engine efficiency, but increases costs. His marketing department really doesn't ask for efficiency at 75 mph; that buys them nothing. So there's generally little commercial need for a 6th gear to optimize fuel efficiency at those speeds (You do see 6th gears more commonly on German cars - no speed limit there)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-07-29T14:07:19.930" CommentCount="4" />
  <row Id="12942" PostTypeId="2" ParentId="12874" CreationDate="2011-07-29T14:38:48.510" Score="7" ViewCount="" Body="&lt;p&gt;Although this is an engineering question it's one that I've had great interest in the specific values myself.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Fuel_economy_in_automobiles&quot; rel=&quot;nofollow&quot;&gt;Wikipedia does an alright job&lt;/a&gt; with the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/nyVML.png&quot; alt=&quot;ORNL fuel economy&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking a little bit deeper into it, this seems to be ripped from some Oak Ridge National Lab report, which as been taken down, but is &lt;a href=&quot;http://web.archive.org/web/20080527233939/http://cta.ornl.gov/data/tedb26/Edition26_Chapter04.pdf&quot; rel=&quot;nofollow&quot;&gt;still on the wayback machine&lt;/a&gt;.  In fact, a lot of the Wikipedia article looks like a copy of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Funny, the report references a website that sounds like it has all the credibility in the world, &lt;a href=&quot;http://www.fueleconomy.gov/feg/driveHabits.shtml&quot; rel=&quot;nofollow&quot;&gt;fueleconomy.gov&lt;/a&gt;.  They give the following image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/EnpVY.gif&quot; alt=&quot;fueleconomy.gov&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a couple of things to conclude with here.  I am much more tempted to believe the first graph here.  The reason is because cars have gears, because of that fact, I expect jumps in the fuel economy as a function of speed.  In fact, a faithful technical report should really give different curves for each gear IMO.  The second graph is probably smoothing things out from a larger collection of data to make it seem easy for the rest of us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, the claim that everyone driving 55 would be better for the environment seems to have technical merit behind it, even though we don't like the fact.  One of the comments says they found sweet spots at a number of different speeds, going up to 90 something, this is almost certainly wrong.  There may be hope for newer cars than what is represented here, however, and in particular some new designs with streamlined shapes may one day justify driving 80 mph to save gas.  We can dream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Electric cars - better data&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was at a loss to explain very well where the efficiency losses at low speed comes from in an internal combustion engine (ICE), so I thought that an informative comparison would be to look at electric cars.  It so happens that &lt;a href=&quot;http://www.teslamotors.com/blog/roadster-efficiency-and-range&quot; rel=&quot;nofollow&quot;&gt;Tesla Motors made a post about their cars&lt;/a&gt; that would be HUGE utility to a physics class.  Fuel use rate for an ICE must involve measurements of tiny flow or average data over a long trip, but electric cars have very accessible direct measurements of current available.  Firstly, let's look at the energy per mile used - the analog to gas mileage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/GYcvN.jpg&quot; alt=&quot;Wh/mile for Tesla Roadster&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most efficient speed for electric cars is closer to 15-20 mph versus 55 mph for ordinary cars.  This is because the losses at low speeds for ordinary cars has to do with the construction and inefficiency of the gas ICE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Directly to the question at hand - they also give the sources of losses over the range of speeds.  This graph directly answers the question better than anything else I have seen.  Of course, only for electric cars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/PKkDf.jpg&quot; alt=&quot;losses&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;aerodynamic&lt;/strong&gt; losses increase with increasing speed to possibly 3rd power, see other answers&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;tire losses&lt;/strong&gt; are from rolling drag, tire air pressure has a lot to do with this&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;drivetrain&lt;/strong&gt; include motor and gearbox type stuff&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;ancillary&lt;/strong&gt; losses include electrical loads like the AC and radio, if they are constant power demands, then the power per mile demanded will decrease inversely with speed&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-07-29T17:19:06.780" LastActivityDate="2011-07-29T17:19:06.780" CommentCount="2" />
  <row Id="12943" PostTypeId="2" ParentId="12874" CreationDate="2011-07-29T14:56:06.763" Score="2" ViewCount="" Body="&lt;p&gt;There are two things to consider.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The amount of energy required to maintain constant speed. This starts at a low value and increases. Energy rate can be measured in Watts (power) and it is proportional to $v^3$.&#xA;$$ P = F_{drag}\,v $$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The amount of fuel supply needed to make the above energy rate. This is called the break specific fuel consumption (BSFC) and it varies widely with the % throttle applied and the engine rpm (and thus gear). An engine is typically at it's best efficiency near full thottle at mid low rpms (like 80% thottle with 4500 rpm). Incidentally, peak torque rpm usually corresponds the best BSFC area. With low throttle the air intake is restricted and you have pumping losses (it takes more energy to suck the air into the pistons). Engine friction increases with speed and combustion efficiency varies with flame temperarure and volumetric efficiency. Also at low speeds the combustion suffers and thus you cannot maintain low speeds at high gears. Remember though that the lower speeds increase travel time and there is limit on the minimum fuel per hour used, as travel times increase so is the total fuel used for a trip.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Obviously engineers target their designs to perform well during the EPA mandated duty cycles. To get the best miles per gallon you need a car with a small engine (low friction) that can operate near full throttle at low rpms, with very high gearing. On the other hand if you have a Corvette what you want to do accelerate full throttle to like 120mph and then coast down (no throttle) to 30 mph and then repeat the cycle.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-07-29T14:56:06.763" CommentCount="2" />
  <row Id="12944" PostTypeId="1" CreationDate="2011-07-29T14:56:42.807" Score="1" ViewCount="75" Body="&lt;p&gt;Is there such a thing as refractive index of a material for gravitational waves ? How would the index of a material depend on density - Say density of interstellar cloud vs planet vs star vs neutron star vs black hole ? Could you get an object with negative refractive index - What would this do to a gravitational wave ?&lt;/p&gt;&#xA;" OwnerUserId="4732" LastActivityDate="2011-07-30T16:44:41.247" Title="metamaterials for gravitational waves" Tags="&lt;gravity&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="12945" PostTypeId="2" ParentId="9982" CreationDate="2011-07-29T15:09:02.160" Score="0" ViewCount="" Body="&lt;p&gt;Daniel Grumiller's answer is great. I will just take this from a slightly different angle in an attempt to shed more light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In GR, it makes sense to imagine that something local, like a black hole, can be separated from what's going on far away. As boundary conditions, we just say that spacetime should be asymptotically flat, and away we go. This is essentially because of Newton's shell theorem for a $1/r^2$ field. If we think the distant universe is spherically symmetric, then we don't expect it to have any effect on us locally other than imposing things like cosmological curvature and expansion of space, which don't have any significant effect on bound systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in B-D gravity, the field $\phi$ falls off like $1/r$, and therefore the contribution to $\phi$ from the distant universe is not expected to cancel, or even to be small. In fact, if you try to make a cosmological model in B-D gravity with a nonvanishing density of massive particles, $\phi$ will generically diverge. Basically the only way to keep it from diverging is to have a big bang cosmology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So although Daniel Grumiller is absolutely right that the Schwarzschild vacuum is a perfectly legitimate B-D spacetime, it is in some sense not what we generically expect a B-D spacetime to be like. Generically we expect it to be a big bang cosmology with some value of $\phi$ set by the cosmological conditions.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-07-29T15:09:02.160" />
  <row Id="12946" PostTypeId="1" AcceptedAnswerId="12949" CreationDate="2011-07-29T15:44:58.000" Score="3" ViewCount="100" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Laws_of_thermodynamics&quot; rel=&quot;nofollow&quot;&gt;laws of thermodynamics&lt;/a&gt; are essentially four axioms of a mathematical theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expectation values of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Statistical_ensemble_%28mathematical_physics%29&quot; rel=&quot;nofollow&quot;&gt;statistical ensemble&lt;/a&gt; are supposed to satisfy the axioms of thermodynamics (under the assumption that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ergodic_hypothesis&quot; rel=&quot;nofollow&quot;&gt;ergodic hypothesis&lt;/a&gt; holds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How is it proved that statistical mechanics satisfies the axioms of thermodynamics?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is important to me because the abstract theory of thermodynamics is used to develop things like the Maxwell relations, which are then applied to statistical mechanics ensembles. For this to give valid results it statistical mechanics must be a model of thermodynamics. I would appreciate pointers to proofs of this and references if you have any. Thanks a lot.&lt;/p&gt;&#xA;" OwnerUserId="2502" LastActivityDate="2011-07-29T17:15:03.110" Title="Proof that Statistical Mechanics is a model of Themodynamics" Tags="&lt;mathematical-physics&gt;&lt;thermodynamics&gt;&lt;statistical-mechanics&gt;" AnswerCount="1" />
  <row Id="12947" PostTypeId="2" ParentId="12906" CreationDate="2011-07-29T16:04:52.693" Score="0" ViewCount="" Body="&lt;p&gt;The interpretation of the metric as &lt;em&gt;anisotropy&lt;/em&gt; may or may not be physical, depending on your point of view. The main problem with this interpretation is that it requires &lt;strong&gt;two&lt;/strong&gt; distinct metrics. One of the lasting influences of general relativity in other fields of physics is the idea of treating objects by thinking about their &lt;em&gt;intrinsic&lt;/em&gt; geometry. So when you are dealing with the propagation of the wave in a medium, and you have a Riemannian metric $g$ with respect to which the wave propagates exactly like $\partial_t^2 - \triangle$, you want to use &lt;em&gt;that&lt;/em&gt; Riemannian metric $g$ as the preferred one when you are studying the equations. You are of course free to impose a background coordinate system and attach to it a secondary metric, but that secondary metric can be argued to be unphysical in the situation, if it does not actually affect the physics of the situation, but rather only the way your present your measurements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: consider the case where your metric is $4 dx^2 + dy^2 + dz^2$. If you are only given this metric, you would not say that &quot;one unit&quot; in the $x$ direction is the same length as &quot;one unit&quot; in the $y$ direction. In fact, you would be tempted to redefine your coordinate system with $x&amp;#39; = \frac{1}{2}x$ such that the metric &quot;look&quot; more symmetric. In general, given any symmetric, positive definite bilinear form on a vector space, you can always find an orthonormal basis. So if the relevant geometry for the physics is described by a single Riemannian metric, then there is automatically a notion of infinitesimal isotropy built-in to the physics. (This is related to the notion of &lt;a href=&quot;http://en.wikipedia.org/wiki/Local_Lorentz_covariance&quot; rel=&quot;nofollow&quot;&gt;local Lorentz covariance&lt;/a&gt; in relativity, as well as to the notion of &lt;a href=&quot;http://en.wikipedia.org/wiki/Frame_bundle&quot; rel=&quot;nofollow&quot;&gt;frame bundles&lt;/a&gt; in Riemannian geometry.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important thing to keep in mind is that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectral_theorem&quot; rel=&quot;nofollow&quot;&gt;spectral theorem&lt;/a&gt; for positive definite bilinear forms can be stated in the following way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given any finite dimensional real vector space $V$, and any &lt;em&gt;two&lt;/em&gt; positive definite symmetric bilinear forms $A$ and $B$ on $V$. We can find a basis $\{ e_1, e_2, e_3, e_4\}$ in which $A$ and $B$ are simultaneously diagonalized. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, if you have a physical problem in which there are two distinct natural Riemannian metrics used in the equations, what you have is that the two parts of the physics corresponding to the use of the two Riemannian metrics are anisotropic &lt;em&gt;relative to each other&lt;/em&gt;. Another way of saying the same thing is that the wave propagation in a medium that is governed by $\partial_t^2 - \triangle$ is still uniform in all directions, it is just that now the correct notion of directions should be the one given by the Riemannian metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example: imagine you are driving around on a plane. Because of friction (and other dissipative forces), to maintain constant velocity you need to keep the accelerator pedal depressed. Now above you there is a pigeon. Because of drag, for it to maintain constant &lt;em&gt;air velocity&lt;/em&gt; it also needs to keep putting in energy. Now there is a wind blowing at the height where the pigeon flies. From your point of view, the pigeon's motion is anisotropic: because to maintain the same &lt;em&gt;ground velocity&lt;/em&gt; with wind and against wind requires different amount of energy input. But from the pigeon's point of view, your motion is anisotropic: the pigeon's &quot;stationary&quot; coordinate system would be one that is being blown around by the wind!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that said, from the point of view of studying the wave equation in a medium, the notion of &lt;em&gt;ambient space isotropy&lt;/em&gt; is not all that important. So while the interpretation is perfectly valid, it is not something that you necessarily need to keep in mind. When one studies wave propagation in a &lt;em&gt;non-linear&lt;/em&gt; medium (keywords: crystal optics, elastic waves), the notion of anisotropy more often refers to the fact that the longitudinal components of a wave and the various polarisations of the traverse components of a wave can have anisotropic propagation behaviour relative to each other. Like Whoplisp wrote in his comments, the general case is governed by a system of equations that cannot be reduced to just the simple scalar wave equation. &lt;/p&gt;&#xA;" OwnerUserId="829" LastEditorUserId="829" LastEditorDisplayName="" LastEditDate="2011-07-29T16:11:43.453" LastActivityDate="2011-07-29T16:11:43.453" />
  <row Id="12948" PostTypeId="1" AcceptedAnswerId="12951" CreationDate="2011-07-29T17:03:55.767" Score="0" ViewCount="99" Body="&lt;p&gt;The biggest peak in the world is Mount Everest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine someone starting to make an artificial hill (like pyramide) from soil (earth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when starting with an 200x200 Km base area, with 45degree slope, its mathematical height is 100km (Low orbital space height). Is it possible to make such an artificial peak? (Without taking into account financial issues and so on.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, why not? What would happen? Is there a height limit?&lt;/p&gt;&#xA;" OwnerUserId="4735" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-07-29T22:07:10.140" LastActivityDate="2011-07-29T22:07:10.140" Title="How to calculate the highest theoretical artificial hill?" Tags="&lt;soft-question&gt;&lt;mass&gt;&lt;pressure&gt;&lt;material-science&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="12949" PostTypeId="2" ParentId="12946" CreationDate="2011-07-29T17:09:05.177" Score="1" ViewCount="" Body="&lt;p&gt;Let me first state that it's impossible to derive the result in full generality, i.e that it holds for any statistical model whatsoever, simply because this is false. One needs to place certain physical assumptions on the system in order for the result to even hold. But in order to actually prove anything (in the mathematical sense) one also needs to require additional technical assumption (the math is just hard here, no way around it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most common way is to work on a lattice $S$ (such as $S = \mathbb Z^d$), i.e. with discretized space (this makes your life much easier though still pretty hard) and also suppose we are given a space of fields $(E, \mathcal E, \lambda_0)^S$ where $(E, \mathcal E, \lambda_0)$ is some nice measure space. Finally, let the system be described by a collection of potentials $(\Phi_A)_{A \subset S}$ describing interactions in some subset $A$ of the lattice such that the energy $H_x(\cdot) = \sum_{A \ni x} \Phi_A(\cdot)$ at every site of the lattice is finite $\int H_x(\omega) \lambda(d\omega) &amp;lt; \infty$ where $\lambda$ is the product measure on $E^S$ (e.g. finite-range spin models such as the Ising model will do), we can prove that the any shift-invariant ergodic Gibbs measure (here ergodic refers to the shift translation on the lattice, not to the time evolution you are referring to) then satisfies a variational principle (of maximizng an entropy functional on some reasonable space of measures) and consequently the macroscopic quantities computed from it will have all the nice thermodynamic properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two standard references for this would be&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ruelle's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521546494&quot; rel=&quot;nofollow&quot;&gt;Thermodynamic formalism&lt;/a&gt;.&#xA;Note that the book is aimed at mathematicians, is very obtuse and hard to follow and requires working knowledge of topology, functional analysis and measure theory.&lt;/li&gt;&#xA;&lt;li&gt;Georgii's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0899254624&quot; rel=&quot;nofollow&quot;&gt;Gibbs measures and phase transitions&lt;/a&gt; where one whole part of the book is dedicated to the formal treatment of the variational problem for measures. He also studies the geometry of the Gibbs measure space as one varies the potential (since there is one extreme Gibbs measure for each phase for the given potential, this is nothing else than the phase diagram with coexistence lines, etc.; although here the phase diagram is actually infinite-dimensional). Here one can get by with just rudimentary knowledge of the above areas since Georgii develops some parts of topology and probability theory he needs.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-07-29T17:15:03.110" LastActivityDate="2011-07-29T17:15:03.110" />
  <row Id="12950" PostTypeId="2" ParentId="12948" CreationDate="2011-07-29T17:24:19.560" Score="0" ViewCount="" Body="&lt;p&gt;The limiting factor will be that the shear strain inside the pyramid will eventually exceed the elastic limit for whatever material the pyramid is made of.  For rock maximum shear strain is usually of order $10^{-3}$ if memory serves.  I've seen people do order of magnitude estimates of this using dimensional analysis, and come up with something of order the height of Everest.  To really get an upper limit though you'd need to do a more detailed modeling of the stresses.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-07-29T17:24:19.560" />
  <row Id="12951" PostTypeId="2" ParentId="12948" CreationDate="2011-07-29T17:32:22.993" Score="3" ViewCount="" Body="&lt;p&gt;Your question is a classical college exercise. The limit is supposed to be the melting of the base of the artificial mountain under the pressure, which is linked with the energy of chemical bounds. You have an example of such a calculation &lt;a href=&quot;http://www.phy.cuhk.edu.hk/phyworld/articles/mount_high/mount_high_e.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. You can also look it for an arbitrary planet size : the smaller the planet, the smaller the gravity is, and the bigger the biggest mountain can be. And when the mountain can be as big as the radius of the planet, you have roughly the dwarf planet / asteroid boundary. &lt;/p&gt;&#xA;" OwnerUserId="373" LastActivityDate="2011-07-29T17:32:22.993" CommentCount="4" />
  <row Id="12953" PostTypeId="1" CreationDate="2011-07-29T18:23:30.780" Score="11" ViewCount="125" Body="&lt;p&gt;Materials with large coefficients of static friction would be cool and useful. Rubber on rough surfaces typically has $\mu_s\sim1-2$. When people talk about examples with very high friction, often they're actually talking about surfaces that are sticky (so that a force is needed in order to separate them) or wet (like glue or the tacky heated rubber used on dragster tires and tracks). In this type of example, the usual textbook model of &lt;a href=&quot;http://en.wikipedia.org/wiki/Friction#Dry_friction&quot; rel=&quot;nofollow&quot;&gt;Coulombic friction&lt;/a&gt; with $\mu_s$ and $\mu_k$ (named after Mr. Coulomb, the same guy the unit was named for) doesn't apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was looking around to try to find the largest $\mu_s$ for any known combination of surfaces, limiting myself to Coulombic friction. This &lt;a href=&quot;http://robotics.eecs.berkeley.edu/~ronf/Gecko/prl-friction.html&quot; rel=&quot;nofollow&quot;&gt;group at Berkeley&lt;/a&gt; has made some &lt;a href=&quot;http://www.eecs.berkeley.edu/IPRO/VIF/Papers/Fearing1.pdf&quot; rel=&quot;nofollow&quot;&gt;amazing high-friction surfaces&lt;/a&gt; inspired by the feet of geckos. The paper describes their surface as having $\mu\sim5$. What's confusing to me is to what extent these surfaces exhibit Coulombic friction. The WP Gecko article has pictures of Geckos walking on vertical glass aquarium walls, and it also appears to imply a force proportional to the macroscopic surface area. These two things are both incompatible with the Coulomb model. But the Berkeley group's web page shows a coin lying on a piece of glass that is is nearly vertical, but not quite, and they do quote a $\mu$ value. This &lt;a href=&quot;http://www.pnas.org/content/99/19/12252.full&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; says gecko-foot friction involves van der Waals adhesion, but I think that refers to microscopic adhesion, not macroscopic adhesion; macroscopic adhesion would rule out the Coulomb model. (The WP Gecko article has more references.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: what is the highest coefficient of &lt;em&gt;Coulombic&lt;/em&gt; static friction ever observed, and does the Berkeley group's substance qualify?&lt;/p&gt;&#xA;" OwnerUserId="4552" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-29T22:15:02.630" LastActivityDate="2011-08-05T02:12:15.583" Title="Is gecko-like friction Coulombic? What is the highest known Coulombic $\mu_s$ for any combination of surfaces?" Tags="&lt;forces&gt;&lt;newtonian-mechanics&gt;&lt;friction&gt;" CommentCount="9" />
  <row Id="12954" PostTypeId="1" AcceptedAnswerId="12959" CreationDate="2011-07-29T18:31:53.200" Score="2" ViewCount="74" Body="&lt;p&gt;I don't want to open a debate about whether cell phones can cause cancer,&#xA;I read the thread:&lt;br&gt; &lt;a href=&quot;http://physics.stackexchange.com/questions/12069/could-cell-phone-radiation-cause-cancer&quot;&gt;Could cell-phone radiation cause cancer?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of this question let's assume there's a chance for cell-phones&#xA;to increase chances of cancer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, visible light is higher frequency and higher energy radiation than mobile phones,&lt;br&gt;&#xA;You can take a look at the spectrum of elctromagnetic radiation:&lt;br&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/File:EM_spectrum.svg&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/File:EM_spectrum.svg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking of going to photodynamic therapy (it's a form of treatment that uses intense visible light for skin conditions - including acne)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if visible light is higher frequency &amp;amp; higher energy than radio&#xA;waves, is there a reason that this kind of treatment would not raise&#xA;risk of cancer if cell-phones might?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;these are some specifics about the light used (it is blue light):&lt;br&gt;&#xA;Blue Intensity: 26273.0 uW/cm2 &lt;br&gt;&#xA;Blue Peak Wavelength: 415.20 nm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanx a lot to anyone giving thought to this question.&lt;/p&gt;&#xA;" OwnerUserId="4737" LastActivityDate="2011-08-17T22:15:11.047" Title="Is there a reason for photodynamic therapy to not cause cancer if mobile phones might?" Tags="&lt;light&gt;&lt;electromagnetic-radiation&gt;&lt;radiation&gt;" AnswerCount="3" />
  <row Id="12955" PostTypeId="1" AcceptedAnswerId="13035" CreationDate="2011-07-29T18:43:30.880" Score="2" ViewCount="54" Body="&lt;p&gt;I heard there were powerful convective movements in clouds which were responsible for increasing the size of water droplets or ice crystals. My question is: do the same movements appear outside of clouds? Or in other words: what's so special about a cloud that makes it host these convective movements? Couldn't warm, dry air also create those movements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my question is not clear, don't hesitate in asking me for details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Another way to phrase the question would be: Are convective movements in clouds different than those not in clouds?&lt;/p&gt;&#xA;" OwnerUserId="4739" LastEditorUserId="4739" LastEditorDisplayName="" LastEditDate="2011-08-01T03:50:34.770" LastActivityDate="2011-08-01T04:31:04.207" Title="Are there convective movements similar to those in clouds everywhere in the atmosphere?" Tags="&lt;air&gt;&lt;atmosphere&gt;&lt;atmospheric-science&gt;&lt;convection&gt;&lt;meteorology&gt;" AnswerCount="3" CommentCount="3" />
  <row Id="12956" PostTypeId="2" ParentId="12954" CreationDate="2011-07-29T18:59:24.397" Score="0" ViewCount="" Body="&lt;p&gt;There is no way to answer this question, because nobody has ever proposed a plausible mechanism by which cell phones could cause cancer. In fact, cell phones don't cause cancer. We know they don't cause cancer because (1) there is no plausible mechanism, (2) studies in humans that claim a non-null result don't show a dose-response, and (3) animal studies show no effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason frequency would be relevant would be that photons above a certain frequency are ionizing. For frequencies too low to be ionizing, there is no particular reason to believe that higher frequencies would be more harmful. Without a physical mechanism, there is simply no scientific way to approach this question.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-07-29T18:59:24.397" CommentCount="3" />
  <row Id="12957" PostTypeId="2" ParentId="12874" CreationDate="2011-07-29T19:06:07.513" Score="1" ViewCount="" Body="&lt;p&gt;Above the optimal speed, fuel consumption increases because air drag increases very quickly, so it takes more energy to &quot;carve out&quot; a tunnel through the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below the optimal speed, the consumption increases because a car engine will burn fuel even when idle - there's some fuel spent just to keep the engine turning, and when you go slowly, this effect starts to become important.&lt;/p&gt;&#xA;" OwnerUserId="4210" LastActivityDate="2011-07-29T19:06:07.513" />
  <row Id="12958" PostTypeId="2" ParentId="12924" CreationDate="2011-07-29T19:25:16.187" Score="0" ViewCount="" Body="&lt;p&gt;Short answer: No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longer answer:  I found it easiest to approach this by asking the question in a different way ---&gt;  If a runner were to run around in circles on an inverted conical ramp, how fast would he have to run to remain perdendicular to the ramp's surface?  The answer depends on three things: radius for the running path; slope of the ramp (B) and height/mass distribution of the runner (shorter runners do better, since more of their mass is at a larger $v^2/R$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumptions: 1.6m (5'3&quot;) tall runner, with a center of gravity (CoG) 0.8m from feet, on a 2m radius (approx 13 foot diameter) path.  Since he is tilted in from vertical, his CoG traces a path which has a diameter somewhat less than that: for now, estimate B = 40 degrees tilt, which give a path for the CoG of radius (2 - 0.8 sin(40))m = about 1.5m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_d = 9.8 m/s^2$ is acceleration (due to gravity) straight down.  $a_L = V^2/R$ is acceleration laterally outward, due to centrifugal force.  $a_t$ is total acceleration, which will be perpendicular to ramp surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_d = a_tcos(B)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_L = a_tsin(B)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we have $a_L = a_d tan(B)$ ----&gt; $B = tan^{-1}(V^2/(R a_d)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remembering that the R in that equation is the R for CoG, and that the velocity of that CoG will be slower than speed of the runner's feet, we have...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B = tan^{-1}((6.7 (m/s)(1.2/2))^2/(1.5(m) 9.8 (m/s^2))$ = about 47.8 degrees from horizontal.  Compare to the original guess of 40 degrees.  I'm not making a second iteration to correct path length: instead guess about 45 degrees.  This assumes that the runner is running a 4 minute mile, which is quite a feat to pull off on a conical ramp with path of radius 2 m, with your head (experiencing less centripetal force than CoG) constantly trying to fall in towards the center while your feet (experiencing more) trying to slide out from under you in the opposite direction.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-07-29T19:43:11.880" LastActivityDate="2011-07-29T19:43:11.880" CommentCount="1" />
  <row Id="12959" PostTypeId="2" ParentId="12954" CreationDate="2011-07-29T19:37:15.367" Score="2" ViewCount="" Body="&lt;p&gt;There is also no reason to correlate cell phone &quot;radiation&quot; to blue laser &quot;radiation&quot; - just because they are both electro-magnetic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ps. 26273.0 uW/cm2 is 260W/m^2, if you go outside you receive about 1300 W/m^2 from the big nuclear reaction in the sky so I wouldn't be &lt;strong&gt;too&lt;/strong&gt; convinced about the effectiveness of the  &quot;intense blue light&quot; therapy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;br&gt;&#xA;The danger from a source depends on the wavelength (how the electromagnetic radiation interacts with your body) and the power level ( how much of it is there ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cell phone uses low energy microwaves which don't interact strongly with your body so there is very little chance of them having any effect. In addition the power of a cell phone is tiny - 0.01% of the power of a microwave oven.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blue light is potentially more harmful if it has enough energy (depending how blue) to affect chemical bonds. Ultraviolet (very blue) light has enough energy to break molecules apart and so can be dangers, which is why you get sunburn. Light of longer wavelengths can only damage you by the heating effect so you need a lot more light to cause any harm.  So the infrared light from your stove doesn't harm you, but an industrial cutting laser at the same wavelength can cut you in half.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what wavelength the 'therapy' uses, but if it is ultraviolet then a small amount of power could be harmful (ie sunburn), if it is visible blue then they would need to have very high levels of power to do any harm - by simply cooking you (like cell phones!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if the light is too long a wavelength and too low a power to have any effect on you - then exactly what is the point of the treatment?&lt;/p&gt;&#xA;" OwnerUserId="2525" LastEditorUserId="2525" LastEditorDisplayName="" LastEditDate="2011-08-17T22:15:11.047" LastActivityDate="2011-08-17T22:15:11.047" CommentCount="4" />
  <row Id="12960" PostTypeId="2" ParentId="12922" CreationDate="2011-07-29T19:39:07.670" Score="2" ViewCount="" Body="&lt;p&gt;Here we will not give a physical interpretation per se, but just derive an alternative covariant expression for the sought-for scalar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are given a scalar field $\psi$ and a metric &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ds^2~=~g_{ab} ~dx^a dx^b.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\nabla$ be the corresponding Levi-Civita connection. The Lagrangian scalar function is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Lambda~:=~ \frac{1}{2}\langle \nabla\psi , \nabla\psi \rangle ~=~ \frac{1}{2}\nabla_a\psi ~ g^{ab}~ \nabla_b\psi ~=~ \frac{1}{2}\partial_a\psi ~ g^{ab}~ \partial_b\psi  . $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy-momentum tensor reads:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Q_{ab} ~:=~  \nabla_a \psi ~\nabla_b \psi - g_{ab}~\Lambda. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also have a vector field $X$. Let us make a co-vector (=one form) by lowering the index:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \eta_a = g_{ab}~ X^b. $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deformation tensor $\pi$ wrt. the vector field $X$ is defined as the Lie derivative of the metric:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi_{ab} ~:=~ ({\cal L}^{}_X g)_{ab}~=~ (\nabla_a \eta)_b + (\nabla_b \eta)_a,   $$ &#xA;$$\pi^{ab} ~:=~ g^{ac} ~\pi_{cd}~g^{db}~=~ -({\cal L}^{}_X g)^{ab}~=~ g^{ac}\partial_c X^b +g^{bc}\partial_c X^a - X^c\partial_c g^{ab}  . $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sought-for scalar is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S~:=~ \frac{1}{2}\pi^{ab} Q_{ab}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another vector field $Y$ is defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Y^a~:=~g^{ab}~ Q_{bc}~X^c~=~g^{ab}~\nabla_b \psi ~X[\psi] - \Lambda X^a, $$&#xA;$$ Y[f]~=~X[\psi]~\langle \nabla\psi , \nabla f \rangle  - \Lambda X[f].$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it is a bit tedious but straightforward to reproduce the formula in the question formulation (v3):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S~=~\mathrm{div}_g Y - \Box_g \psi ~ X[\psi], $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and also to derive another covariant expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S ~=~ \langle \nabla\psi , \nabla[ X[\psi]]  \rangle - \mathrm{div}_g( \Lambda X).$$&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-30T00:20:04.913" LastActivityDate="2011-07-30T00:20:04.913" CommentCount="3" />
  <row Id="12961" PostTypeId="1" CreationDate="2011-07-29T19:46:27.743" Score="2" ViewCount="44" Body="&lt;p&gt;I've noticed that many journal articles (in optics) use the phrase &quot;Fourier frequency&quot; to describe, well, the frequency of something.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://scholar.google.com/scholar?q=%22Fourier+frequency%22&amp;amp;hl=en&amp;amp;btnG=Search&amp;amp;as_sdt=1%2C19&amp;amp;as_sdtp=on&quot; rel=&quot;nofollow&quot;&gt;Google scholar search for &quot;Fourier frequency&quot;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/pdf/gr-qc/0209039&quot; rel=&quot;nofollow&quot;&gt;Example&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;To compare with previous sensitivity curves of a single LISA&#xA;  Michelson interferometer, we construct the SNRs as a function of&#xA;  &lt;strong&gt;Fourier frequency&lt;/strong&gt; for sinusoidal waves from sources uniformly&#xA;  distributed on the celestial sphere&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This phrase seems redundant to me.  Does the adjective &quot;Fourier&quot; add anything?  Is it trying to make a distinction of audio frequencies from optical frequencies?  Or temporal frequencies from spatial frequencies?&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-07-29T20:30:51.993" LastActivityDate="2011-07-29T20:30:51.993" Title="What is the specific meaning of &quot;Fourier frequency&quot; (as opposed to simply &quot;frequency&quot;)?" Tags="&lt;optics&gt;&lt;fourier-transform&gt;" CommentCount="3" />
  <row Id="12962" PostTypeId="2" ParentId="12954" CreationDate="2011-07-29T19:53:57.893" Score="4" ViewCount="" Body="&lt;p&gt;As desribed in the wiki article (Singlet oxygen production) this treatment has some &#xA;risk of causing cancer. It's another method to damage/kill cells. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Positive is that the action is rather localized to the place where the light beam is directed, but in general this &lt;strong&gt;can cause cancer as all other cancer treatments do!&lt;/strong&gt;&#xA;Those therapies are all more or less heroic, justified only because there is no alternative. &#xA;The ratio of fighting/causing cancer is different for each chemical/radiation used, &#xA;and depends on the tissue it is applied to. So, there is a certain risk to &#xA;that method, but that can be very small. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To mix this question up with cellphones hoax, is just silly. &#xA;And for no sake I agree to assume such a nonsense,&#xA;not even temporarily. . &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-07-29T21:53:30.010" LastActivityDate="2011-07-29T21:53:30.010" CommentCount="3" />
  <row Id="12963" PostTypeId="2" ParentId="12069" CreationDate="2011-07-29T20:29:22.763" Score="0" ViewCount="" Body="&lt;p&gt;No, cell phone use does not cause cancer. We know it doesn't cause cancer because:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There is no plausible mechanism.&lt;/li&gt;&#xA;&lt;li&gt;Animal studies show no effect.&lt;/li&gt;&#xA;&lt;li&gt;Human studies that get non-null results don't show a dose-response.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Doing these cell-phone studies with human subjects makes as much sense as doing studies to figure out whether I can cause my neighbors to get cancer by thinking evil thoughts about them. In both cases, there is no remotely plausible physical mechanism for the direct effect as postulated. The only reason to do the cell-phone study and not to do the evil-thoughts study is that the former appeals deeply to people's folk beliefs, which have been built up from decades of movies and comic books where &quot;radiation&quot; causes magical effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some studies with human subjects give positive results and some give null results. This should not surprise us. The studies are measuring the relative sizes of their random and systematic errors. In the studies where they succeed in getting their random errors down to a smaller level than their systematic errors, they will measure either a positive or a negative correlation with cancer. In the ones where they succeed in getting their systematic errors down to a smaller level than their random errors, they will get a null result.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-07-29T20:29:22.763" />
  <row Id="12964" PostTypeId="1" AcceptedAnswerId="12994" CreationDate="2011-07-29T20:45:03.457" Score="2" ViewCount="110" Body="&lt;p&gt;I'm not looking for books which deal with the mathematical foundations of Newtonian mechanics. What I'm looking for are modern books which deal with conceptual foundations of Newtonian mechanics - by that I mean exact definition of force, inertia, frames of reference etc. It seems like much of the books that deal topics were written before the 80's or even older (like the one by Ernst Mach himself). And the people who seem to be bothered about these things today are mostly science educators, who again publish very little on these topics. I'm not looking for books written from a philosophical standpoint either. Something that'd be comprehensible by a undergrad would be just fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: This question probably doesn't meet the guidelines as defined in the &lt;a href=&quot;http://physics.stackexchange.com/faq&quot;&gt;faq&lt;/a&gt;. That is why I added the 'soft-question' tag. Also, any moderator who wishes to make this question community-wiki is most welcome to do so!&lt;/p&gt;&#xA;" OwnerUserId="966" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-29T21:55:51.117" LastActivityDate="2011-07-30T20:33:12.547" CommunityOwnedDate="2011-07-29T21:56:06.547" Title="Book suggestions for foundation of Newtonian Mechanics" Tags="&lt;newtonian-mechanics&gt;&lt;books&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="12965" PostTypeId="1" AcceptedAnswerId="12966" CreationDate="2011-07-29T21:31:14.367" Score="4" ViewCount="177" Body="&lt;p&gt;Why are lasers inefficient? Is it because of the heat lost during lasing? Why couldn't there be thermocouples or turbines in parts of the cooling circuits to extract something out of that heat?&lt;/p&gt;&#xA;" OwnerUserId="4144" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-07-29T21:42:05.257" LastActivityDate="2011-08-03T03:42:44.203" Title="Why are lasers inefficient?" Tags="&lt;optics&gt;&lt;light&gt;&lt;photons&gt;&lt;heat&gt;&lt;laser&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="12966" PostTypeId="2" ParentId="12965" CreationDate="2011-07-29T22:20:49.667" Score="9" ViewCount="" Body="&lt;p&gt;Not all lasers are inefficient. Infrared diodes are quite efficient. This paper shows 64% electrical to optical conversion efficiency: &lt;a href=&quot;http://www.jdsu.com/ProductLiterature/paper_hipower_910_980_laser_diodes.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.jdsu.com/ProductLiterature/paper_hipower_910_980_laser_diodes.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An argon gas laser on the other hand is quite inefficient.&lt;/p&gt;&#xA;" OwnerUserId="4306" LastActivityDate="2011-07-29T22:20:49.667" CommentCount="5" />
  <row Id="12967" PostTypeId="2" ParentId="11519" CreationDate="2011-07-29T22:26:37.670" Score="1" ViewCount="" Body="&lt;p&gt;A small tokamak is inefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following quote from here &lt;a href=&quot;http://www.jet.efda.org/faq/about-tokamaks&quot; rel=&quot;nofollow&quot;&gt;http://www.jet.efda.org/faq/about-tokamaks&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Creating smaller reactors would be very good, but the confinement of&#xA;  plasma particles gets worse as the plasma gets smaller (and good&#xA;  confinement is required for effective fusion).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4306" LastActivityDate="2011-07-29T22:26:37.670" />
  <row Id="12968" PostTypeId="2" ParentId="12955" CreationDate="2011-07-29T22:31:52.240" Score="2" ViewCount="" Body="&lt;p&gt;Water vapor inside clouds condenses into droplets that fall trough the cloud. If the cloud has sufficient up drafts then the droplet or crystal, if it's cold enough, will rise through the cloud gaining more mass before finally it becomes too heavy and the up draft is no longer neough to prevent it from falling out of the cloud. In extreme cases this can lead to hailstones the size of tennis balls. This sort of phenonmenom if usually associated with extreme weather.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general you see up drafts that are more moderate. For example glider pilots use up drafts in the atmosphere to gain lift for their aircraft. Typically these are associated with geographical features, like cliffs and ridges, or differing land use patterns that cause the air near the ground to heat up due to differing colors etc.&lt;/p&gt;&#xA;" OwnerUserId="1674" LastActivityDate="2011-07-29T22:31:52.240" />
  <row Id="12969" PostTypeId="2" ParentId="12955" CreationDate="2011-07-29T23:26:23.717" Score="4" ViewCount="" Body="&lt;p&gt;In general, yes the updrafts also occur in warm dry air, as a result of heating on the ground which produces hyrdostatic instability in the atmosphere.  As the updrafts go higher, they cool adiabatically and may, if they go high enough and if there is enough moisture in the air, cool enough to condense water vapor and form clouds.  However there can also be very strong thermal updrafts on perfectly clear days.  It is possible to predict if and at what  altitude  clouds will form fairly accurately based on temperature vs. altitude profiles obtained by weather balloon, together with dewpoint measurements and predicted high temperatures near the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The condensation of water vapor, if it occurs,  will release latent heat which can strengthen the updraft.  In extreme cases this results in thunderstorms.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditorDisplayName="" LastEditDate="2011-07-29T23:34:36.183" LastActivityDate="2011-07-29T23:34:36.183" CommentCount="2" />
  <row Id="12970" PostTypeId="2" ParentId="12314" CreationDate="2011-07-30T01:24:31.157" Score="2" ViewCount="" Body="&lt;p&gt;I haven't read that book, but I did read Feynman's discussion of (sounds like) exactly the same thing. Easy: Tell the aliens how to build a telescope, then describe the configuration of some galaxies near them. OK OK, but suppose we rule that out: We can't see any objects in common. Easy: Send them circularly-polarized radio waves (thanks @Anonymous Coward). OK OK, let's say our radio waves must be linearly polarized. Easy: Tell them to look at almost any phenomenon related to the weak force, for example the beta-decay of cobalt-60 in a magnetic field. But then there's one more catch--what if the aliens are made of antimatter and they actually were watching the beta-decay of antimatter-cobalt-60? In Feynman's discussion (if I recall correctly), that's where it ends: There's no way to be really sure that the aliens understand right and left correctly, because they may be made of antimatter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since 1964, when CP-violation was observed, we can even eliminate that possibility: We tell the aliens how to watch kaons decay (for example) and then the aliens can figure out whether they're made of (what we call) matter or antimatter, and therefore they can figure out which way is left and right without any more ambiguity. So I guess that aspect is a slight update from pre-1964 descriptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Watching atoms decay in a magnetic field is a pretty simple thing to do by the standards of particle-physics experiments. I don't know of any parity-violating experiments that are much simpler than that. It just has to involve the weak force.&lt;/p&gt;&#xA;" OwnerUserId="3811" LastActivityDate="2011-07-30T01:24:31.157" CommentCount="3" />
  <row Id="12971" PostTypeId="2" ParentId="11687" CreationDate="2011-07-30T02:19:15.683" Score="2" ViewCount="" Body="&lt;p&gt;For a collision to be elastic, by the usual definition, no internal degrees of freedom of the colliding bodies can be excited/de-excited by the collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The internal degrees of freedom that might change in a collision include vibration, rotation (although some might argue about this), electron orbitals, electron spin, nuclear spin, etc. etc. etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For macroscopic objects, the density of states for all these parameters is so high that in practice it is astronomically unlikely that a TRULY inelastic collision will ever occur.  This is true even if the bodies are spherical, and have the same shock wave round-trip times, etc. etc.  The energy &quot;lost&quot; in a collision may be so small that it is extremely difficult to measure, but any collision of macroscopic objects is sure to be inelastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation is very different if the colliding particles are single atoms.  There, the density of internal states is so low (and coupling to various degrees of freedom so weak in the scattering) that it can be thousands of times more likely to have an elastic collision than an inelastic one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diatomic molecules are appreciably worse than atoms; it is relatively easy to excite/de-excite rotational motion of the molecule in a collision.  Inelastic collisions are about as common as elastic collisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where's the tipping point?  By the time you've got something as big as, say, a virus, I'm sure it's extremely improbable to have elastic collisions.  But for something like a buckyball?  I have no idea.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-07-30T02:19:15.683" CommentCount="4" />
  <row Id="12972" PostTypeId="2" ParentId="11687" CreationDate="2011-07-30T02:39:29.817" Score="-1" ViewCount="" Body="&lt;p&gt;Extended reply to Anonymous Coward:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perfect, I'll argue that gaining rotation in a collision doesn't make it inelastic.  The kinetic energy is still there, it's just in the form of rotational kinetic energy.  But yes, that is strictly semantics.  You pretty much covered the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally I would claim that no macro objects can collide fully elastically but elementary particles can.  Of course, elementary particles only do so in the absence of an interaction and the &quot;collision&quot; is though a force like Coulomb repulsion.  Molecules, atoms, and other things can blur the line by having a meaningful chance of colliding fully elastically as well as a meaningful chance of some energy conversion taking place with kinetic energy.  Like nuclear cross sections, it is possible to provide numeric quantification of the probabilities of different interaction types.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For fun consider an equilibrium gas.  Collisions should be on average net-zero kinetic energy $\Delta$ but it's possible that collisions frequently (almost always even) convert binding energy or some other internal stored energy into kinetic energy.  I imagine this would be the case for gases with a large molecule size, it could even be of relevance to a chemist.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-08-01T01:10:38.490" LastActivityDate="2011-08-01T01:10:38.490" CommentCount="3" />
  <row Id="12973" PostTypeId="1" AcceptedAnswerId="13533" CreationDate="2011-07-30T04:18:28.647" Score="11" ViewCount="236" Body="&lt;p&gt;Imagine I take a polymer like polyethylene, of length $L$ with some number of Kuhn lengths $N$, and I tie into into a trefoil knot.  What is the difference in entropy between this knotted polymer and a circular polyethylene unknot?  Is there an approximate way of calculating this for knots of a given crossing number (or other topological invariant such as a Jones polynomial)?  &lt;/p&gt;&#xA;" OwnerUserId="4257" LastActivityDate="2011-08-16T01:32:24.067" Title="The entropic cost of tying knots in polymers" Tags="&lt;mathematical-physics&gt;&lt;thermodynamics&gt;&lt;entropy&gt;&lt;topology&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="4" />
  <row Id="12975" PostTypeId="1" CreationDate="2011-07-30T05:34:00.110" Score="2" ViewCount="91" Body="&lt;p&gt;I have a homework question based on the following diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://gyazo.com/f0de3827222c4c5651df70ae25b657e9.png&quot; alt=&quot;velocity diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find the &lt;strong&gt;angular velocity of the object as seen by an observer at the origin&lt;/strong&gt; of the frame. The question says that the observed angular velocity is given by $\omega_0 = \Delta \varphi_0 / \Delta t$, where $\Delta t = t_2 - t_1$ and $\Delta \varphi_0 = \varphi_0(t_1^*) - \varphi_0(t_2^*)$ is the change of the angle $\varphi$, the polar angle which the object had at times $t_1^*$ and $t_2^*$. These are the times when the light detected by the observer (at $t_1$ and $t_2$) was emitted. The object is located at $(x_1, y_1)$ at time $t_1^*$ and it is located at $(x_2, y_2)$ at time $t_2^*$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started to answer this question by just finding an expression for $\Delta \varphi_0$ in terms of the coordinates ($\Delta \varphi_0 = \arctan{(y_1/x_1)} - \arctan{(y_2/x_2)}$) and dividing this by an expression for $\Delta t$. However, I was told that this was not correct. Apparently, I have to get an expression for the angular momentum in terms of the angle $\theta$ in the diagram. My lecturer told me that this was could be done with some basic geometry. I can't quite see what to do though. Also, I was told that the angle $\Delta \varphi_0$ was assumed to be very small. I have a feeling that what the lecturer wants us to derive is the expression $\omega = \frac{|\vec{v}|\sin{(\theta)}}{|\vec{r}|}$, but I can't see how to get this. Can anyone help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: This question is from a special relativity course, but I believe this can be answered without using any knowledge of special relativity. As I said, the lecturer told me that it was basically just geometry.&lt;/p&gt;&#xA;" OwnerUserId="4741" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-30T07:27:35.177" LastActivityDate="2011-08-01T12:57:39.030" Title="&quot;Observed&quot; angular velocity" Tags="&lt;homework&gt;&lt;kinematics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="12976" PostTypeId="1" AcceptedAnswerId="13144" CreationDate="2011-07-30T06:24:30.177" Score="5" ViewCount="146" Body="&lt;p&gt;Please bear in mind that I'm neither a physics guy nor particularly a math guy. So I will probably need a bit more hand-holding than just a complex equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a game, I'm trying to calculate a &lt;em&gt;reasonable&lt;/em&gt; simulation of a human body on a trampoline. It seems to me that there are two phases to consider once the body has contacted the trampoline: the deceleration of the body once it has contacted the surface, which gets stored as potential energy in the trampoline, and then the combination of the release of that potential energy once the body has reached the nadir plus any extra force that body might apply (assuming he has bent his knees prior to that point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in understanding the interaction between the trampoline's storage and release of energy and the body's own contribution (addition) to that energy with its own kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CLARIFICATION:&#xA;In short, I want to know when the best time to crouch is (pull your knees up) and when the best time to release (jump) is. Anyone who's been on a trampoline knows you can get better height by crouching and jumping at the right time - you can also bring yourself to a complete stop by crouching and jumping at different times. I can't seem to find anything documented though about what these times actually are. Since I couldn't find the information from the sports / athletics side, I thought I'd come over here to the physics side and see if anyone knew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm surely not being very clear, which only further reveals my own shortcomings in this matter. Any guidance would be appreciated. I understand that this forum typically operates at a much higher level than what I'm looking for, but I thought I'd ask here, since this is where the great minds seem to gather.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT&#xA;I can very easily do a &quot;good enough&quot; simulation of a mass &quot;bouncing&quot;off the trampoline by taking its incoming velocity and simply inverting it and multiplying by an arbitrary restitution amount to simulate the springiness of the material. Nothing more complicated really need from the perspective of a game.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part that's killing me is getting the body to continue moving past the plane of the trampoline while decelerating to 0 and then accelerate again up to final &quot;release velocity&quot;. The part that baffles me is the part that happens between the time the body touches the material to the time it leaves contact, taking into account any additional force added by the person through his legs (I've decided not to model any additional lift sure to arm motion etc, to keep both the simulation and the user controls thereof straightforward).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SECOND EDIT&#xA;I've read through the answers here and they're very good! And surprisingly, I seem to be following about 75% of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The challenge I'm having is that I'm principally interested in Velocity, rather than Energy, even though it's obviously energy that is behind all of this. Every 30th of a second (every &quot;step&quot;) I need to know the velocity of the body so that I can plot its new position. It's really easy to plot the position of the body once it has left &quot;contact&quot;: starting at some arbitrary position and velocity, v = v + g, y = y + v. See, it's a very different way of representing the same facts when you're dealing with visual graphics than physics equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now y = the plane of the trampoline. v = v * c, where 'c' is a factor of how far the springs will compress, less than 1. y = y + v until v approaches 0 at which point we start the &quot;bounce back&quot;. Here I suppose v = v * (1/c) * s, where 's' represents whatever factor of 'loss' we want to simulate so that we don't get 100% bounce back. Once y &gt; the plane of the trampoline, we can then forget the more complicated stuff and go back to letting gravity take over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously this whole bit ignores mass, which is unfortunate, but I have to keep it simple for now. The &quot;mass&quot; factor gets rolled into the numbers for 's' and 'c' that I fudge until it 'feels' right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, now that you guys have stopped shuddering in utter horror at this wishy-washy approach, the bit I'm trying to figure out is the part about the human body bending his legs and then performing the jumping motion to add his force to the rebound of the trampoline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me, having spent a little time on a trampoline, that you can jump at the right time and at the wrong time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, from a physics perspective, when is the optimal time to crouch? When is the optimal time to jump / release?&lt;/p&gt;&#xA;" OwnerUserId="4742" LastEditorUserId="4742" LastEditorDisplayName="" LastEditDate="2011-08-01T15:15:20.487" LastActivityDate="2011-08-03T08:07:26.137" Title="Trampoline physics - what is the right time to crouch and to jump to maximize your height?" Tags="&lt;simulation&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="12977" PostTypeId="2" ParentId="8850" CreationDate="2011-07-30T06:28:27.450" Score="0" ViewCount="" Body="&lt;p&gt;String theory (and all other theories involving hidden dimensions)&#xA; predict that gravity and electromagnetism unify in hidden &#xA;dimensions and that the hidden dimensions are indetectible because of &#xA;their small size. It does also predict that sufficiently short-waved &#xA;photons, with wavelengths shorter than the size of the hidden &#xA;dimensions, can enter them. Producing ultra-short photons can thus &#xA;manipulate gravity, with revolutionizing space travel applications such &#xA;as cheap anti-gravity launches. The &#xA;problem that it would require high energy can be practically solved by &#xA;concentrating several laser beams on a nanoparticle, heating it to &#xA;locally extreme temperatures. An Alcubierre metric can be created by &#xA;ejecting multiple nanoparticles from the craft and then beam perfectly &#xA;timed laser beams on them (fire at the most distant first so that they are&#xA; hit simultaneously), so each nanoparticle contributes a slower than &#xA;light effect but together add up to faster than light, creating no discrete &#xA;event horizon and thus no Hawking radiation.&lt;/p&gt;&#xA;" OwnerUserId="4743" LastActivityDate="2011-07-30T06:28:27.450" CommentCount="1" />
  <row Id="12979" PostTypeId="2" ParentId="12975" CreationDate="2011-07-30T08:14:28.020" Score="3" ViewCount="" Body="&lt;p&gt;Here's something that might help you out: this problem is motivated by astronomical observations of distant stars or galaxies. If you think about it, when you look through a telescope at an astronomical object, the only thing you can measure (without a spectroscope) is its angular position $\varphi_0$. So the way you obtain information about the object's motion is to record its angular position at one time $t_1$, and then later at another time $t_2$. You can then calculate the ratio $\Delta \varphi_0/\Delta t$, and since for an astronomical object $\Delta \varphi_0$ is &lt;em&gt;very&lt;/em&gt; small (just think about how little the stars appear to move through the sky over any human-length timescale), this ratio is a pretty good approximation to the derivative $\frac{\mathrm{d}\varphi_0}{\mathrm{d}t} = \omega$, the &lt;em&gt;observed&lt;/em&gt; angular velocity of the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what you see through a telescope is only part of the story. $\Delta \varphi_0/\Delta t$ is not necessarily the &lt;em&gt;actual&lt;/em&gt; angular velocity of the object, because even though you &lt;em&gt;observed&lt;/em&gt; two light rays a time $\Delta t$ apart, it doesn't mean that they were &lt;em&gt;emitted&lt;/em&gt; $\Delta t$ apart. The only way the time between their emission would be the same as the time between their observation is if both light rays had traveled the same distance before reaching you. But for many astronomical objects, that's not the case. Since they have motion parallel to your line of sight as well as perpendicular to it, it's very likely that one of the light rays was emitted closer to you than the other, which means it spent less time in transit. You need to take that difference of transit time into account, which is why the answer you should be getting is a little more complicated than $\omega = \frac{v\sin\theta}{r}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as actually solving the problem, I'd suggest that you start by finding expressions for the the travel time of each light ray - that is, find an expression for $t_1 - t_1^*$, and similarly for $t_2 - t_2^*$. Also, because the angle $\Delta\varphi_0$ is so small, you can do calculations parallel to and perpendicular to the line of sight separately. I would definitely suggest that you ignore the $x$ and $y$ coordinates entirely - think of it the way someone would be observing this from Earth, where they wouldn't know anything about those coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, here's the reason you're doing this in a special relativity course: it's a common calculation in astrophysics to work this problem backwards, i.e. to determine the actual velocity of the distant object given observations of $\Delta\varphi_0$ and $\Delta t$, along with spectroscopic data that gives you the parallel component of the velocity. In certain cases, the velocity works out to be faster than light if you use the naive formula $\frac{v\sin\theta}{r}$. You need to account for the difference in light travel time to show that these objects actually do obey special relativity.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-07-30T08:14:28.020" CommentCount="9" />
  <row Id="12980" PostTypeId="2" ParentId="12933" CreationDate="2011-07-30T08:14:51.870" Score="1" ViewCount="" Body="&lt;p&gt;In one of our first physics lab we did an experiment to characterize the thermal properties of a simulated house (it actually was a box with 4 sides made of different materials).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One just periodically heated the inside to 50 degree celsius. The temperature time series on the outside wall will be damped and phase shifted relative to the inside. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I googled &quot;praktikum thermohaus periodisch&quot; and came up with this description:&#xA;&lt;a href=&quot;http://www.uni-jena.de/unijenamedia/Downloads/faculties/physik_astro/phys_gp/V_210_1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.uni-jena.de/unijenamedia/Downloads/faculties/physik_astro/phys_gp/V_210_1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And conveniently there is also a students protocol with some data:&#xA;&lt;a href=&quot;http://www.personal.uni-jena.de/~p1erra/physik/protokolle/seite%20129-%20versuch%20210.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.personal.uni-jena.de/~p1erra/physik/protokolle/seite%20129-%20versuch%20210.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following image displays the dynamic temperature data. It indicates how the insulating properties of Styrofoam are superior to glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/JusOf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see a reason why this shouldn't work with a real building.&lt;/p&gt;&#xA;" OwnerUserId="4306" LastActivityDate="2011-07-30T08:14:51.870" CommentCount="2" />
  <row Id="12981" PostTypeId="2" ParentId="12976" CreationDate="2011-07-30T09:49:29.033" Score="3" ViewCount="" Body="&lt;p&gt;Can you be more specific about what you mean by 'reasonable'?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest treatment, ignoring loss of energy to heat/friction/etc, would be to approximate the human body as a point mass falling on a spring (i.e. the trampoline) with some spring constant $k$.  Here, we have the initial potential energy of the falling body - $PE = mgh$, this is entirely transformed to kinetic energy at the time of impact - $KE = \frac{1}{2}mv^2$, and you stretch the trampoline to convert this into spring potential energy - $SPE = \frac{1}{2}kz^2$, where $z$ is the compression distance.  Since the sum of the potential energy in the spring (SPE) and the kinetic energy of the falling body (KE) is constant, you can solve for the velocity at a given spring compression: let $C = SPE + KE$, which implies $v = (\frac{2C-kz^2}{m})^\frac{1}{2}$.  The derivative with respect to $z$ is $v&amp;#39; = \frac{-kz}{m(\frac{2C-kz^2}{m})^\frac{1}{2}}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the body bouncing upwards, start with the kinetic energy at impact and play the same game where the sum of the body's kinetic energy (KE) and the potential energy due to gravity (PE), again $PE = mgh$ (for some height $h$), is constant.  As such, we have $C = PE + KE$, which implies $v= (2)^\frac{1}{2}(\frac{C-mgh}{m})^\frac{1}{2}$.  The derivative w.r.t. $h$ is $v&amp;#39; = \frac{-g}{(-2gh+\frac{2C}{m})^\frac{1}{2}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more complex treatment would involve energy loss due to heat and irreversible stretching of the trampoline fabric, air resistance on the falling body, etc.  You could add in some fudge terms to account for this.  As for a more complex treatment of the human body, i.e. some kind of ragdoll model, nothing fundamental changes, but you'll have a much harder time finding the instantaneous velocity of the individual body parts, as well as the compression velocity of your trampoline at any given time.  To make traction there, you'd probably have to do something like a ball-and-spring simulation of the body, and more precisely define how it will be falling on the trampoline.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Some suggested units for the terms: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g$, or 'little $g$' denotes the acceleration due to gravity near the earth's surface (http://en.wikipedia.org/wiki/Earth%27s_gravity).  It's value is approximately $~9.81 \frac{m}{s^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v$, your velocity, can be measured in meters/second, $\frac{m}{s}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m$, or your mass, can be measured in kilograms ($kg$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These work well if you measure your terms representing energy (i.e. PE, SPE, KE, and 'C') in joules, since the units for a joule are: $\frac{kg*m^2}{s^{-2}}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for how the trampoline should 'look'  when depressed a distance $z$, perhaps you could model it as an upside-down cone of height $z$ (http://mathworld.wolfram.com/Cone.html)?&lt;/p&gt;&#xA;" OwnerUserId="4257" LastEditorUserId="4257" LastEditorDisplayName="" LastEditDate="2011-07-30T11:05:50.137" LastActivityDate="2011-07-30T11:05:50.137" CommentCount="2" />
  <row Id="12982" PostTypeId="2" ParentId="11687" CreationDate="2011-07-30T10:35:14.903" Score="1" ViewCount="" Body="&lt;p&gt;Horrible! Inferring &quot;rotation&quot; here is plain nonsense!&#xA;Elasticity esp. collision of bodies (i.e. macroscopic!) &#xA;is a matter of solids. There are a few crystals  with &#xA;rotating molecules in it, but those are in the league &#xA;of &quot;soapy&quot; materials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The property asked for here is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;as low as possible mechanical hysteresis at the time scale of the deformation.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the phenomenological part. To correlate it to microscopic &#xA;(atomic/crystallographic scale) is almost purely empiric. &#xA;Some general rules can be given: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Solids with localized strong bonds are &quot;purely&quot; elastic, whereas solids with &#xA;undirected bonding (truely metallic) are bad. Examples for &#xA;this are diamond, glass, ceramics as opposed to eg. copper or cesium. &#xA;Here comes a general problem: diamond is thought to be brittle &#xA;or hard, cesium is called soft or ductile. Both are not assiciated &#xA;with elasticity. The wording is very dependent on the trade, &#xA;but brittle/hard/elastic and soft/malleable/inelastic are strongly correlated. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Some metals like copper or steel which are soft and malleable, &#xA;can be transformed into brittle and &quot;purely&quot; elastic materials, &#xA;without any change in chemical composition. Copper is &quot;hardened&quot;&#xA;by rolling, steel by rapid chilling from red heat to give hard &#xA;materials, used as springs. The common thing behind this effect is &#xA;creation of crystal imperfections which hinder flowlike processes &#xA;between crystal layers/boundaries. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Some &quot;metals&quot; like tungsten are always brittle, this is partly due &#xA;to the fact that they are not really true metals. There is some localized &#xA;bonding mixed with metallic structure. Zintl phases or intermetallic &#xA;compounds (eg sorts of brass or bronze) are in this domain. &#xA;Such cases are regularly associated with not having a closest packed structure. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This are the chemical/crystallographic requirements, another thing is &#xA;sound speed in the solids, this is a requirement in the steel balls &#xA;of a &quot;Newton&quot; cradle (This name is marketing nonsense) , if &#xA;sound speed (speed of elastic wave) is too low, the energy won't be &#xA;tranferred/concentrated fast enough to the contact points. This phenomenon &#xA;is demonstrated best with that &quot;silly putty&quot; a lump of a silicon &#xA;resin preparation, which flows like a liquid on minutes time scale, &#xA;recoils like rubber when dropped  as a ball or reacts like a brittle &#xA;solid when hammered. &#xA;Of course sound speed in solids is not independent from the factors &#xA;listed obove for &quot;elasticity&quot;, its real world and complicated. &#xA;&quot;Simple Minds&quot; do not like it :=)&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-07-30T20:53:12.173" LastActivityDate="2011-07-30T20:53:12.173" />
  <row Id="12983" PostTypeId="1" AcceptedAnswerId="12985" CreationDate="2011-07-30T12:21:52.390" Score="10" ViewCount="554" Body="&lt;p&gt;My field of study is computer science, and I recently had some readings on quantum physics and computation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is surely a basic question for the physics researcher, but the answer helps me a lot to get a better understanding of the formulas, rather than regarding them &quot;as is.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Whenever I read an introductory text on quantum mechanics, it says that the states are demonstrated by vectors, and the operators are Hermitian matrices. It then describes the algebra of vector and matrix spaces, and proceeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have any problem with the mathematics of quantum mechanics, but I don't understand the philosophy behind this math. To be more clear, I have the following questions (and the like) in my mind (all related to quantum mechanics):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why vector/Hilbert spaces?&lt;/li&gt;&#xA;&lt;li&gt;Why Hermitian matrices?&lt;/li&gt;&#xA;&lt;li&gt;Why tensor products?&lt;/li&gt;&#xA;&lt;li&gt;Why complex numbers?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(and a different question):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When we talk of an n-dimensional space, what is &quot;n&quot; in the nature? For instance, when &lt;a href=&quot;http://en.wikipedia.org/wiki/Spin-%C2%BD#Mathematical_description&quot; rel=&quot;nofollow&quot;&gt;measuring the spin of an electron&lt;/a&gt;, n is 2. Why 2 and not 3? What does it mean?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is the answer just &quot;because the nature behaves this way,&quot; or there's a more profound explanation?&lt;/p&gt;&#xA;" OwnerUserId="4466" LastActivityDate="2011-08-01T11:58:46.583" Title="The philosophy behind the mathematics of quantum mechanics" Tags="&lt;quantum-mechanics&gt;&lt;mathematics&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="9" />
  <row Id="12984" PostTypeId="2" ParentId="12976" CreationDate="2011-07-30T12:37:59.047" Score="3" ViewCount="" Body="&lt;p&gt;As the secretary of a trampoline club I spend too much time watching people bouncing on trampolines and thinking about the physics of what they are doing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing you should take into account is that the reaction forces from the trampoline are not well modeled by an ideal spring with force proportional to the depression $x$. When the trampoline bed and springs are detached they shrink to a width $L$ just a little less than the stretched width $W$ You can model this with a stiff rod or string of length $L$ attached to an ideal spring of stretched length $S = \sqrt{W^2+x^2}-L$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/y0DKd.jpg&quot; alt=&quot;trampoline&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upward component of force from the bed is therefore proportional to &#xA;$T = x - \frac{xL}{\sqrt{W^2+x^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I am assuming that the tramplinist is bouncing in the middle of the trampoline and that it is circular. For rectangular trampolines you need a model that adds up contributions from parts of different length. Off centre bouncing will also create transverse forces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you ignore these things you should allow for some losses due to friction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trampolinist gets lift from the bed by bending knees and ankles and moving arms during the contact with the bed. The combined effect is to move the position of his centre of gravity relative to the position of his feet. In the simplest case he keeps this constant and bounces just losing some height due to friction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an analysis including novement but no friction you can take the vertical position of the centre of mass as your primary variable and use a Lagrangian with kinetic energy given by $\frac{1}{2}mv^2$ from the veloicty of the centre of mass and potential energy from the bed and gravitation, then work out equation of motion. How you proceed from there depends on what you are trying to achieve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to model summersaults and twists with the trampolinist moving across the trampoline it could get complicated! Hopefully I have given some initial pointers at least.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-07-30T12:37:59.047" CommentCount="3" />
  <row Id="12985" PostTypeId="2" ParentId="12983" CreationDate="2011-07-30T13:22:01.983" Score="13" ViewCount="" Body="&lt;p&gt;Vector spaces because we need superposition. Tensor product because this is how one combines smaller systems to obtain a bigger system when the systems are represented by vector space. Hermitation operator because this allows for the possibility of having discrete-valued observables. Hilbert space because we need scalar products to get probability amplitudes. Complex numbers because we need interference (look up double slit experiment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dimension of the vector space corresponds to the size of the phase space, so to speak. Spin of an electron can be either up or down and these are all the possibilities there are, therefore the dimension is 2. If you have $k$ electrons then each of them can be up or down and consequently the phase space is $2^k$-dimensional (this relates to the fact that the space of the total system is obtained as a tensor product of the subsystems). If one is instead dealing with particle with position that can be any $x \in \mathbb R^3$ then the vector space must be infinite-dimensional to encode all the independent possibilities.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; concerning Hermitation operators and eigenvalues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is actually where the term &lt;em&gt;quantum&lt;/em&gt; comes from: classically all observables are commutative functions on the phase space, so there is no way to get purely discrete energy levels (i.e. with gaps in-between the neighboring values) that are required to produce e.g. atomic absorption/emission lines. To get this kind of behavior, some kind of generalization of observable is required and it turns out that representing the energy levels of a system with a spectrum of an operator is the right way to do it. This also falls in neatly with rest of the story, e.g. the Heisenberg's uncertainty principle more or less forces one to have non-commutative observables and for this again operator algebra is required. This procedure of replacing commutative algebra of classical continuous functions with the non-commutative algebra of quantum operators is called quantization. [&lt;em&gt;Note that even on quantum level operators can still have continuous spectrum, which is e.g. required for an operator representing position. So the word &quot;quantum&quot; doesn't really imply that everything is discrete. It just refers the fact that the quantum theory is able to incorportate this possibility.&lt;/em&gt;]&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-07-30T20:25:02.090" LastActivityDate="2011-07-30T20:25:02.090" CommentCount="4" />
  <row Id="12986" PostTypeId="2" ParentId="12983" CreationDate="2011-07-30T13:39:34.987" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the answer just &quot;because the nature behaves this way,&quot; or there's a more profound explanation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would say , yes to &quot;because nature behaves this way&quot;. It is the most economical description of experimental data using mathematics, to date. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-07-30T13:39:34.987" />
  <row Id="12988" PostTypeId="2" ParentId="12983" CreationDate="2011-07-30T14:57:32.140" Score="1" ViewCount="" Body="&lt;p&gt;It is strange that you are not very comfortable with Hilbert spaces because actually continuous functions form Hilbert space and physics is all about functions. The general difference with classical mechanics is in the fact that classical mechanics was formulated long before people who used and learned (and even developed) it could understand manifolds and Lie algebras while for the time of quantum mechanics the idea of Hilbert space and all that stuff became more or less natural. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same with the rest. You could formulate QM without complex numbers (in principle), but this would be the same as formulation of Maxwell's equations without vectors. People say that Maxwell's equations actually were formulated before physicist were familiar enough with the vectors and it was a nightmare.&lt;/p&gt;&#xA;" OwnerUserId="4657" LastActivityDate="2011-07-30T14:57:32.140" CommentCount="6" />
  <row Id="12989" PostTypeId="2" ParentId="12983" CreationDate="2011-07-30T15:26:44.270" Score="1" ViewCount="" Body="&lt;p&gt;I asked the same questions some time back. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read :-)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Heisenberg%27s_entryway_to_matrix_mechanics&quot; rel=&quot;nofollow&quot;&gt;Heisenberg's entryway to matrix mechanics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/quant-ph/0404009&quot; rel=&quot;nofollow&quot;&gt;Understanding Heisenberg's 'Magical' Paper of July 1925: a New Look at the Calculational Details&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="126" LastActivityDate="2011-07-30T15:26:44.270" />
  <row Id="12990" PostTypeId="2" ParentId="12983" CreationDate="2011-07-30T15:41:59.343" Score="-8" ViewCount="" Body="&lt;p&gt;Quantum mechanics is a real time solution of Newton's equation. If you solve Newton's equation in real numbers you get Newton's solution. If you solve Newton's equation in complex numbers you get quantum solution and the difference between the two solutions is relativistic. Google into Roger Anderton's time dependent Newton's equation and see that the difference between Newton's mechanics in that Newton used real numbers and when Newton's equation is solved in complex numbers it gives quantum mechanics and the difference between quantum mechanics and newton's mechanics is relativistic mechanics or quantum = newton + relativiity. It is quite ineteresting article written by an english engineer &lt;/p&gt;&#xA;" OwnerUserId="4744" LastActivityDate="2011-07-30T15:41:59.343" CommentCount="7" />
  <row Id="12992" PostTypeId="2" ParentId="12944" CreationDate="2011-07-30T16:03:42.377" Score="1" ViewCount="" Body="&lt;p&gt;There are phenomena loosely analogous to refraction when a gravitational wave passes through a curved background. See Misner, Thorne, and Wheeler, p. 956. This is an interaction of fields with fields, not fields with matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a discussion of the optics of gravitational waves in matter, see Ingraham, &quot;Gravitational waves in matter,&quot; GRG 29 (1997) 117. You can find a freely downloadable pdf by googling. Since a typical gravitational wave from astrophysical sources has a frequency of ~1 Hz, its wavelength is going to be $\sim10^8$ meters, which means that no foreseeable technology could make a metamaterial. Even if you could, the index of refraction's difference from 1 would be incredibly small. Ingraham derives a relationship $n=1-4\pi G\chi$, where $\chi$ is a susceptibility, basically defined in terms of how much stress-energy tensor you get for a given amount of Weyl curvature. Ingraham suggests that the best chance for seeing the phenomenon might be in &quot;gravitational wave propagation through molecular gas&#xA;clouds of galactic or intergalactic size.&quot; Although $n-1$ would be incredibly small, he suggests that you might be able to see an effect accumulated over thousands or millions of light-years.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastEditorUserId="4552" LastEditorDisplayName="" LastEditDate="2011-07-30T16:44:41.247" LastActivityDate="2011-07-30T16:44:41.247" />
  <row Id="12993" PostTypeId="2" ParentId="12976" CreationDate="2011-07-30T16:57:23.597" Score="0" ViewCount="" Body="&lt;p&gt;I don't what to trow a wench in the works BUT you can jump on the trampoline bay just moving your arms up and down. The shift of mass and the change of momentum bay swinging your arms down at the moment of contact increases the mass towards the trampoline and then trowing them upwards accelerates the body in the direction of travel. We do this instinctively the problem with calculating this is that the jumper does not keep there arms straight. Now if you take in account that we do a similar motion whit our lags you have one hell of a calculation just considering the shift in mass in the air. Calculating the elastic capabilities of the trampoline is fairly straight forward but the work capability of individual people is something all together different. An experienced jumper can achieve impressive height bay just shifting there mass a very precise way.        &lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-07-30T16:57:23.597" CommentCount="5" />
  <row Id="12994" PostTypeId="2" ParentId="12964" CreationDate="2011-07-30T20:00:39.197" Score="2" ViewCount="" Body="&lt;p&gt;You may have a look at the works of Noll and Truesdell, for example &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.springerlink.com/content/w66h027430v96106/&quot; rel=&quot;nofollow&quot;&gt;Lectures on the foundations of continuum mechanics and thermodynamics&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or at the paper of Eisenbud (which maybe has influenced implicitly many textbooks): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ajp.aapt.org/resource/1/ajpias/v26/i3/p144_s1?isAuthorized=no&quot; rel=&quot;nofollow&quot;&gt;On the Classical Laws of Motion &lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this articles are written from completely different epistemological points of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't read it yet, but this may be of interest, too: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521636361&quot; rel=&quot;nofollow&quot;&gt;Classical Dynamics: A Contemporary Approach&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Some other references:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=k4H2AjWh9qQC&amp;amp;pg=PP3&amp;amp;dq=teodorescu%20mechanics%20volume%201&amp;amp;hl=en&amp;amp;ei=Z2g0Tr6JAZCr8AO6qfGgDg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCgQ6AEwAA#v=onepage&amp;amp;q=teodorescu%20mechanics%20volume%201&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Mechanical systems, classical models By P. P. Teodorescu&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0821821385&quot; rel=&quot;nofollow&quot;&gt;Differential Equations, Mechanics, and Computation by R. Palais&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rational Mechanics by C.W.Kilmister&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A first course in rational continuum mechanics by C. Truesdell&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elements of continuum mechanics by C. Truesdell&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=25iQQvHe9awC&amp;amp;pg=PR6&amp;amp;dq=arnold%20kozlov&amp;amp;hl=en&amp;amp;ei=MWo0TqjbE4a38gOqlL2hDg&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=1&amp;amp;ved=0CCgQ6AEwAA#v=onepage&amp;amp;q=arnold%20kozlov&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Mathematical aspects of classical and celestial mechanics By V. Arnolʹd,  V. Kozlov, A.  Neĭshtadt&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorUserId="667" LastEditorDisplayName="" LastEditDate="2011-07-30T20:33:12.547" LastActivityDate="2011-07-30T20:33:12.547" CommunityOwnedDate="2011-07-30T20:00:39.197" CommentCount="1" />
  <row Id="12995" PostTypeId="1" CreationDate="2011-07-30T21:06:31.077" Score="1" ViewCount="89" Body="&lt;p&gt;Are virtual particles created within existing space-time, or do they create new space-time for them to occupy?&lt;/p&gt;&#xA;" OwnerUserId="4747" LastActivityDate="2011-07-31T20:45:26.880" Title="Virtual particle production in space-time." Tags="&lt;space-expansion&gt;" AnswerCount="1" />
  <row Id="12996" PostTypeId="1" CreationDate="2011-07-30T21:17:01.617" Score="0" ViewCount="35" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/12995/virtual-particle-production-in-space-time&quot;&gt;Virtual particle production in space-time.&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Does virtual particle production contribute to observed universal expansion?&lt;/p&gt;&#xA;" OwnerUserId="4747" LastActivityDate="2011-07-30T21:17:01.617" ClosedDate="2011-07-30T21:26:46.860" Title="Virtual particles contributing to universal expansion." Tags="&lt;cosmology&gt;" CommentCount="2" />
  <row Id="12997" PostTypeId="2" ParentId="11247" CreationDate="2011-07-30T22:10:33.000" Score="0" ViewCount="" Body="&lt;p&gt;I would like to try an answer for my own question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The straightforward interpretation of the 3 equations in the original post is that they do present a legitimate conflict.  This conflict disappears when the spatial domain is extended to infinity, and the role of the boundary conditions goes away.  So, the conflict, the violation of gauge invariance for the finite ring, suggests that either the ring is simply not covered by quantum mechanics, or there is something about the boundary conditions implied by the ring geometry that doesn't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So are there other boundary conditions that could be tried?  The obvious ones are to demand periodicity not of the wave function, but of the probability density and probability current density.  These are both real (not complex) quantities, would let the phase of the wave function have a discontinuity at the boundary, so long as the gradient of the phase was smooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a choice for the boundaries would allow the continuous eigenvalue spectrum of the infinite line to apply as well to the ring, which would restore gauge invariance.  But this comes at the expense of inhomogeneous and nonlinear boundary conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nonlinear boundary conditions could be accepted if they are consistent with the standard type of Hilbert space of Hermitean operators, and the principle of superposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a paper I recently posted on arxiv.org I go into some detail about how the nonlinear boundaries are indeed consistent with both gauge invariance and the standard structure of Hilbert space.  The bottom line is that the nonlinearity creates a continuous spectrum of eigenvalues, not all of which are superposable in Hilbert space.  The nonlinearity in the boundary allows a subset to be superposable.  The combination of continuous eigenvalues and superposable discreet eigenvalues makes a band structure for the Hilbert space, rather than simple discreet levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/find/all/1/all%3a+AND+boundaries+AND+davidson+nonlinear/0/1/0/all/0/1&quot; rel=&quot;nofollow&quot;&gt;PLEASE CLICK HERE for the larger explanation.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think the correct solution of this problem is quite significant for quantum systems coupled to their environments, such as Josephson junctions used as qubits.&lt;/p&gt;&#xA;" OwnerUserId="4092" LastEditorUserId="4092" LastEditorDisplayName="" LastEditDate="2011-08-01T19:31:37.127" LastActivityDate="2011-08-01T19:31:37.127" />
  <row Id="12998" PostTypeId="2" ParentId="9352" CreationDate="2011-07-31T02:16:07.393" Score="1" ViewCount="" Body="&lt;p&gt;If that's an actual Tesla quote, it sounds like he's talking about the high output voltage during silent operation when no plasma streamers are present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case the max voltage limit of a well-designed Tesla coil is determined by the maximum power limit on the power supply (volts and max current of wall outlet, breaker panel, etc.)  Since radio wave losses are insignificant, all of the operating power goes into heating the spark gap, conductors, and the capacitor dielectric.  When no spark streamers are present, the voltage rises high, and the entire TC becomes very hot!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copper resistivity drops by about 10X at LN2 temperature.   If input power is adjusted to be constant, then a conductivity drop of 10X gives a resonator current increase of SQRT(10) ~= 3 times.  For a simple linear RLC resonator the capacitor voltage is proportional to the inductor current, so liquid nitrogen cooling should increase the main TC output voltage by about 3X.  (This is assuming that copper losses were dominant.  If spark gap and capacitor dielectric loss was significant, then LN2 would give less than 3X voltage improvement.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if spark streamers break out, this math doesn't work.  If the majority of the available power was going into the plasma, then reducing the copper losses wouldn't increase the spark length by much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.copper.org/resources/properties/cryogenic/homepage.html&quot; rel=&quot;nofollow&quot;&gt;http://www.copper.org/resources/properties/cryogenic/homepage.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4748" LastEditorUserId="4748" LastEditorDisplayName="" LastEditDate="2011-07-31T02:22:49.950" LastActivityDate="2011-07-31T02:22:49.950" />
  <row Id="12999" PostTypeId="2" ParentId="11519" CreationDate="2011-07-31T03:42:45.893" Score="0" ViewCount="" Body="&lt;p&gt;High Magnetic Fields.  Not the answer you were looking for, but probably the true answer to the question &quot;How dio you make a small(er) (working) tokamak?&quot;   It might even be the right approach in the long run.  Google &quot;ignitor fusion reactor&quot; or &quot;Bruno Coppi&quot; for more information.   The MIT Alcator is an example.  Of course high field magnets have their own costs, problems and dangers, as do high magnetic field tokamaks.  You should probably not try this on your kitchen table at home and expect a good result unless your resources are roughly equivalent to MIT, ITER or maybe even Florida State.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a fusion related experiment (even vaguely tokamak related) that might be doable at home, look up &quot;induction plasma technology&quot; in Wikipedia.&#xA;You might even find references under &quot;Q-tube plasma&quot; if you search hard enough.  (Q-tubes were used in very early fusion research.)  For more excitement and more danger, but a less tokamak like approach, look up &quot;theta pinch&quot;&lt;/p&gt;&#xA;" OwnerUserId="1172" LastActivityDate="2011-07-31T03:42:45.893" />
  <row Id="13000" PostTypeId="1" CreationDate="2011-07-31T06:43:45.267" Score="0" ViewCount="48" Body="&lt;p&gt;Suppose that   on X-axis  dependence of x coordinate of point  on time is expressed by the following formula &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X(t)=9-6*t+t^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we are asked to find time when point returns it's starting point (starting situation) i think we should solve it by the following way: at $t=0$ it was at position $9$  so we should find when $9-6*t+t^2=9$ yes?&#xA;or   we will have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t^2-6*t=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t( t-6)=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t=0 \text{ or } t=6$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from here $t=0$ we can reject so it means that after $t=6$  hour it will return it's position  am i right?&lt;/p&gt;&#xA;" OwnerUserId="4481" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-31T22:28:29.300" LastActivityDate="2011-07-31T22:28:29.300" Title="moving of point on X axis" Tags="&lt;homework&gt;&lt;kinematics&gt;" CommentCount="3" />
  <row Id="13001" PostTypeId="1" CreationDate="2011-07-31T06:55:33.777" Score="3" ViewCount="138" Body="&lt;p&gt;Is this statement true?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You can not travel back in time. If I travel fast enough the clock will start to go backwards, but that does not mean I am traveling back in time. It would only mean that the time reference is producing a negative count. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4708" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-31T22:27:01.687" LastActivityDate="2011-08-02T18:09:28.907" Title="Does superluminal travel imply travelling back in time?" Tags="&lt;special-relativity&gt;&lt;time-travel&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="13002" PostTypeId="2" ParentId="13001" CreationDate="2011-07-31T07:20:08.480" Score="6" ViewCount="" Body="&lt;p&gt;One cannot travel backwards in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physics data we have up to now agree completely with the special theory of relativity. This means you cannot travel faster than the velocity of light, and that you can only approach  it as a limit. You will not notice a change in your clock in any measurable way within your travelling system. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-08-02T18:09:28.907" LastActivityDate="2011-08-02T18:09:28.907" CommentCount="2" />
  <row Id="13003" PostTypeId="2" ParentId="13001" CreationDate="2011-07-31T07:58:39.917" Score="5" ViewCount="" Body="&lt;p&gt;Yes, this would be only your impression. E.g. when you are moving away from me then your clock is going slower from my point of view and my is going slower from your point of view. This is because as you are moving away from me the distance between us increases and the light needs longer time to reach you (and this statement is obviously symmetric when we swap me and you). Consequently, the image you'll see of me will be distorted in time and I'll appear to be doing everything slower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you would go at the speed of light relative to me (note that this is not physically achievable, but we can discuss this purely theoretically) then all the light from me that could reach you is originating at one moment in my history and consequently it would appear to you that my clock has stopped completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if you were moving superluminally, the light that I emitted as we met and any light I will emit in the future will never reach you (it is simply too slow compared to your speed). On the other hand, as you are moving away, the light that I emitted in the past (and that has already travelled some distance) &lt;em&gt;does&lt;/em&gt; have a chance of catching up you. And the further away you move, the older light from me you'll see. Your impression will be that my clock is going backwards. &lt;strong&gt;But&lt;/strong&gt; this is not the end of the story. Because if you were approaching me at the superluminal speeds then it would appear to you that my clock is going forward as usual. Until the point when our wordlines meet. Here's a quick sketch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/fKINs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dashed line is you (the superluminal observer), the vertical line is the me (the standing observer) and the 45 degrees lines are light rays I emitted at various events. Note that you'll see me first travelling forward in time upto the event labeled 6 and from the event labeled 7 onwards you'll see me going back in to my past.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-31T07:58:39.917" CommentCount="1" />
  <row Id="13004" PostTypeId="2" ParentId="7340" CreationDate="2011-07-31T10:34:04.137" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;and with the same hydrodynamic properties as water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not easily done, density per se is a important factor in hydrodynamics.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Then there must be a density at which the greatest altitude has its&#xA;  absolute maximum, i think.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think You start with the thought that You will fill always the same volume of water in?&#xA;Any way, this is wrong. &#xA;Thrust is proportional to mass ejected per time times velocity of the mass ejected.&#xA;So, if You choose a nozzle which gives the same velocity and mass per time of the liquid at &#xA;the nozzle, a denser liquid will result in the same thrust. Of course You would &#xA;fill in the same mass of liquid. &#xA;(finding that nozzle which does this, is the complicated and mostly empirical task)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The max height of a rocket is prop. to the energy stored in the rocket&#xA;(same total mass assumed) in this case it is the energy of compression &#xA;in the air. If You load the rocket with a denser liquid, You save &#xA;liquid volume, thus You can have more compressed air in the rocket, &#xA;this in turn gives some height extra. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But: There are not many liquid You could choose from: solvents like carbon tetrachloride &#xA;or chloroform and of course mercury. The solvents would dissolve the bottle, at least soften and weaken it, mercury is expensive and not appreciated by the owners ot launch areas. &#xA;In &quot;free&quot; space, mercury might be ok. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I look forward to the day when some astronaut will launch a &#xA;project water rocket from the ISS :=)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simple stabilisation by fins will not work in space :=(&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-08-03T10:20:57.603" LastActivityDate="2011-08-03T10:20:57.603" CommentCount="1" />
  <row Id="13005" PostTypeId="1" CreationDate="2011-07-31T12:34:57.773" Score="1" ViewCount="45" Body="&lt;p&gt;Suppose we are given an ODE problem $$ y&amp;#39;&amp;#39;(x)+V(x)f(x)=E_{n} y(x) $$&#xA; with boundary conditions $ y(0)=y(\infty)=0$. Here  $V(x)$ is a potential function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then is it always true that (for $n \gg 1$)&#xA;$$ E_{n}^{\rm WKB} \sim E_{n} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where WKB means the eigenvalues evaluated via the WKB approach?&lt;/p&gt;&#xA;" OwnerUserId="4435" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-07-31T13:13:59.597" LastActivityDate="2011-07-31T13:13:59.597" Title="Question on energies obtained via WKB approximation" Tags="&lt;mathematical-physics&gt;&lt;approximations&gt;" CommentCount="3" />
  <row Id="13006" PostTypeId="1" CreationDate="2011-07-31T13:01:45.230" Score="2" ViewCount="49" Body="&lt;p&gt;Suppose we have a material described by a dielectric tensor $\bar{\epsilon}$. In frequency domain, this tensor depends on the wave frequency $\omega$ and the wave vector $\vec{k}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly not all $\bar{\epsilon}=f(\omega,\vec{k})$ are physically possible. Without considering the physics behind any particular $\bar{\epsilon}=f(\omega,\vec{k})$, is it possible to find general conditions that all $\bar{\epsilon}=f(\omega,\vec{k})$ must obey? I am mostly interested in Lorenz invariance : surely the most general description of any material must be parametrized by the material's velocity according to the observer ($\vec{v}_{mat}$), and then there must be some sort of symmetry between $\omega$ and $\vec{k}$ in $\bar{\epsilon}=f(\omega,\vec{k},\vec{v}_{mat})$ because the Lorentz transformation mixes up distances ($1/k$) and time intervals ($1/\omega$).&lt;/p&gt;&#xA;" OwnerUserId="4751" LastActivityDate="2011-08-31T04:37:32.703" Title="Lorentz invariance of a frequency- and wavelength- dependent dielectric tensor" Tags="&lt;electromagnetism&gt;&lt;symmetry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="13007" PostTypeId="1" AcceptedAnswerId="13011" CreationDate="2011-07-31T13:22:22.250" Score="2" ViewCount="44" Body="&lt;p&gt;I'm currently taking an introductory course to particle physics and I'm now trying to understand the concept of isospin. However I do have some trouble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's write the up- and down Quark as a Flavour-Doublet (u,d). We then obtain &#xA;$\mid u \rangle = \mid \frac{1}{2} \frac{1}{2}\rangle$ and $\mid d \rangle = \mid \frac{1}{2} \frac{-1}{2}\rangle$. Now I want to know what particles I can get by combining these two quarks. So let's write $\frac{1}{2} \otimes \frac{1}{2} = 0 \oplus 1$. &#xA;So I have a singlett and a triplet. Most books now tell me that the triplet corresponds to the pion-triplet and the singlet to the $\eta$-particle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I can for example write by using Clebsch Gordon coefficients that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mid 1,1 \rangle = \mid \frac{1}{2} \frac{1}{2}\rangle \mid \frac{1}{2} \frac{1}{2}\rangle = \mid u u \rangle$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this doens't correspond to any pion (They are mostly products of ups and anti-downs etc). However some books interpret this as the product of two proton wave function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is quite confusing? Can someone help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd me more than happy. Thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="4520" LastActivityDate="2011-07-31T15:22:16.360" Title="Question concerning Isospin symmetry" Tags="&lt;particle-physics&gt;" AnswerCount="2" />
  <row Id="13008" PostTypeId="1" CreationDate="2011-07-31T13:32:19.343" Score="4" ViewCount="161" Body="&lt;p&gt;I was reading &lt;a href=&quot;http://www.ias.ac.in/resonance/April2011/p310-321.pdf&quot; rel=&quot;nofollow&quot;&gt;a recent article&lt;/a&gt; on Mach's Principle. In it, the author talks about inertia in an empty universe. I'll quote some lines from the article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Imagine a single body in an otherwise empty universe. In the absence&#xA;  of any forces, (Newton's second law gives): $$m\mathbf{a} = 0$$ What&#xA;  does this equation imply? Following Newton we would conclude from that&#xA;  $\mathbf{a} = 0$, that is, the body moves with uniform velocity. But&#xA;  we now no longer have a background against which to measure&#xA;  velocities. Thus $\mathbf{a} = 0$ has no operational significance.&#xA;  Rather, the lack of any tangible background for measuring motion&#xA;  suggests that $\mathbf{a}$ should be completely indeterminate. And it&#xA;  is not difficult to see that such a conclusion follows naturally,&#xA;  provided we come to the remarkable conclusion that&#xA;  $$m = 0$$ In other words, the measure of inertia depends on the&#xA;  existence of the background in such a way that in the absence of the&#xA;  background the measure vanishes!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't see how the argument is complete. For example, in an empty universe, how is it possible to assign a value of 0 to a force? And wouldn't the existence of mathematics in such an empty universe be questionable?&lt;/p&gt;&#xA;" OwnerUserId="966" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-31T22:24:01.783" LastActivityDate="2011-08-02T17:39:26.640" Title="Inertia in an empty universe" Tags="&lt;machs-principle&gt;&lt;inertia&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" />
  <row Id="13009" PostTypeId="1" CreationDate="2011-07-31T13:55:10.467" Score="3" ViewCount="35" Body="&lt;p&gt;Let's say I have a quantum adiabatic computer in a black-box that works perfectly, doesn't suffer from decoherence/noise problems, etc.  Are there any proven bounds for an adiabatic algorithm that guarantees an answer within some constant factor of an optimum solution?  Say, for the NP-hard bin-packing problem (http://en.wikipedia.org/wiki/Bin_packing), for which a variety of classical algorithms have proven approximation factors?   &lt;/p&gt;&#xA;" OwnerUserId="4257" LastActivityDate="2011-07-31T13:55:10.467" Title="What can quantum adiabatic computation provably accomplish?" Tags="&lt;mathematical-physics&gt;&lt;quantum-computation&gt;" CommentCount="2" />
  <row Id="13010" PostTypeId="2" ParentId="13008" CreationDate="2011-07-31T14:07:48.410" Score="3" ViewCount="" Body="&lt;p&gt;The author appears to be &lt;em&gt;assuming&lt;/em&gt; that $a$ is indeterminate in an empty universe. That assumption fits in nicely with some people's philosophical preconceptions (including Mach's), but of course we don't know it to be true. In particular, in general relativity, one can have an empty universe described by good old special-relativistic Minkowski spacetime. In such a universe, a test particle would &quot;know&quot; whether it was accelerating or not. In this sense, general relativity does not appear to be Machian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something like Mach's principle is in the end an axiom, which one can choose to assume or not depending on one's taste.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-07-31T14:07:48.410" CommentCount="1" />
  <row Id="13011" PostTypeId="2" ParentId="13007" CreationDate="2011-07-31T15:08:12.700" Score="3" ViewCount="" Body="&lt;p&gt;I'll just write $\left| u \right &amp;gt;$ for the up state and $\left |d \right &amp;gt;$ for the down state.&#xA;Now, when dealing with $SU(2)$ group, the representations for particles and antiparticles are equivalent (i.e. $ \mathbf 2 = \mathbf{ \bar 2}$) but this doesn't mean they are completely same! They differ by an unitary transformation that swaps up and down (and can also introduce phase factors). In particular, we can have the following identification $\left| \bar u \right &amp;gt; = \left |d \right&amp;gt;$ for the anti-up state and $\left |\bar d \right &amp;gt; = -\left | u \right&amp;gt;$ for anti-down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, all the possible combinations in $\mathbf 2 \otimes \mathbf{\bar 2}$ are &#xA;$$\left |d \right &amp;gt;\left |\bar d \right &amp;gt; ,\quad&#xD;&#xA;\left |d \right &amp;gt;\left |\bar u \right &amp;gt;, \quad&#xD;&#xA;\left |u \right &amp;gt;\left |\bar d \right &amp;gt;, \quad&#xD;&#xA;\left |u \right &amp;gt;\left |\bar u \right &amp;gt;$$&#xA;with the 3rd component of isospin being respectively $0$, $-1$, $1$, $0$. Now, $\mathbf 2 \otimes \mathbf{\bar 2} = \mathbf 1 \oplus \mathbf 3$. The triplet representation is characterized by being symmetric and this means that its generator corresponding to the 3rd component of the isospin $0$ must be ${1 \over \sqrt 2} (\left |d \right &amp;gt;\left |\bar d \right &amp;gt; - \left |u \right &amp;gt;\left |\bar u \right &amp;gt;) $ because when we swap first and second factor we get &#xA;$${1 \over \sqrt 2} (\left |\bar d \right &amp;gt;\left | d \right &amp;gt; - \left |\bar u \right &amp;gt;\left | u \right &amp;gt;) = {1 \over \sqrt 2} (-\left |u \right &amp;gt;\left | \bar u \right &amp;gt; + \left |d \right &amp;gt;\left | \bar d \right &amp;gt;).&#xD;&#xA;$$&#xA;This is $\pi^0$ we wanted.&#xA;For the same reason the antisymmetric state must be ${1 \over \sqrt 2} (\left |d \right &amp;gt;\left |\bar d \right &amp;gt; + \left |u \right &amp;gt;\left |\bar u \right &amp;gt;) $&#xA;and this is $\eta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Excercise:&lt;/strong&gt; check that the remaining generators of the triplet (representing $\pi^{\pm}$) are symmetric.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-07-31T15:08:12.700" />
  <row Id="13012" PostTypeId="2" ParentId="13007" CreationDate="2011-07-31T15:22:16.360" Score="0" ViewCount="" Body="&lt;p&gt;One can use the special unitary (2) SU(2) group representations whenever one has a classification of two states to find representations of states consisting of the original two assignments..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Originally isospin was used to classify nuclear states assigning to the proton isospin 1/2 and to the neutron isospin -1/2. So yes the representation of |1/2 1/2&gt; |1/2&gt;|1/2&gt;&#xA;describes the isospin state of two protons. Isospin had been used to successfully classify nuclear states and nuclear resonances into representations before quarks were discovered to be the underlying components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one goes to the quark framework, it is the quark and antiquark  that are the two states represented in the SU(2)combinations   that give mesons that have been observed. These support the hypothesis of the existence of quarks and antiquarks by classifying  observed mesons into &lt;a href=&quot;http://en.wikipedia.org/wiki/Mesons&quot; rel=&quot;nofollow&quot;&gt;representations of SU(2)&lt;/a&gt; that have   the same quantum numbers and are differing in isospin only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correspondingly the representations joining &lt;a href=&quot;http://en.wikipedia.org/wiki/Baryon&quot; rel=&quot;nofollow&quot;&gt;three quarks&lt;/a&gt; organize the observed baryon resonances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now your question about the two quarks |u&gt;|u&gt;  is valid, it is called a diquark, and there are researchers looking for the effects of the existence of &lt;a href=&quot;http://en.wikipedia.org/wiki/Diquarks&quot; rel=&quot;nofollow&quot;&gt;diquarks,&lt;/a&gt; but they have not been confirmed.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-07-31T15:22:16.360" />
  <row Id="13013" PostTypeId="1" CreationDate="2011-07-31T15:22:30.160" Score="8" ViewCount="463" Body="&lt;p&gt;Stephen Hawking says in his latest book &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Grand_Design_%28book%29&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;The Grand Design&lt;/em&gt;&lt;/a&gt; that,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because there is a law such as gravity, the universe can and will create itself from nothing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it not circular logic? I mean, how can gravity exist if there is no universe? And if there is no gravity, how can it be the reason for the creation of universe? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if the universe doesn't exist, how can &lt;em&gt;it&lt;/em&gt; create &lt;em&gt;itself&lt;/em&gt;? The very sentence doesn't make sense to me. It seems so absurd and illogical that I've never heard such sentences even in philosophy. On what ground, Stephen Hawkings &lt;em&gt;claims&lt;/em&gt; this?&lt;/p&gt;&#xA;" OwnerUserId="4753" LastEditorUserId="966" LastEditorDisplayName="" LastEditDate="2011-07-31T19:39:03.360" LastActivityDate="2011-08-06T05:08:53.497" Title="Stephen Hawking says universe can create itself from nothing, but how exactly?" Tags="&lt;gravity&gt;&lt;cosmology&gt;&lt;astrophysics&gt;&lt;astronomy&gt;&lt;universe&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="3" />
  <row Id="13014" PostTypeId="1" CreationDate="2011-07-31T16:13:00.693" Score="3" ViewCount="86" Body="&lt;p&gt;I'd like to know the exact context in which Paul Dirac introduced the Dirac delta function. What was the physical significance of the Dirac delta function when he first used it in Physics ?&lt;/p&gt;&#xA;" OwnerUserId="540" LastActivityDate="2011-09-05T18:44:39.473" Title="Physical significane and context in which Dirac introduced the Dirac Delta function" Tags="&lt;quantum-mechanics&gt;&lt;mathematical-physics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13015" PostTypeId="1" AcceptedAnswerId="13018" CreationDate="2011-07-31T16:18:04.333" Score="1" ViewCount="118" Body="&lt;p&gt;If one defines force as the time derivative of momentum, i.e. by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA; \vec{F}= \frac{d}{dt} \vec{p}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how can this include static forces? Is there a generally accepted way to argue in detail how to get from this to the static case? If not, what different solutions are discussed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &#xA;I shoud add, that by static forces I mean forces involved in problems where bodies don't move.&lt;/p&gt;&#xA;" OwnerUserId="2740" LastEditorUserId="2740" LastEditorDisplayName="" LastEditDate="2011-08-01T08:01:12.567" LastActivityDate="2011-08-01T18:22:57.163" Title="How to get statics out of a dynamic force concept?" Tags="&lt;forces&gt;&lt;newtonian-mechanics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="13016" PostTypeId="1" AcceptedAnswerId="13676" CreationDate="2011-07-31T16:46:45.760" Score="5" ViewCount="88" Body="&lt;p&gt;In parametric down-conversion, it is said that a driving photon is converted into two entangled photons whose frequencies add up to the driving frequency. Yet in discussions about entanglement experiments, I have not seen anything about the frequency at the point of detection. What is the story here? Do you have pairs of red and green entangled photons? If Alice detects a red photon does Bob detect a green one and vice versa? Is the color of the photon even known or measured at the point of detection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Thanks to Slavic for the awesome picture posted below. I am struggling now to understand what I see, so let's start with an easy question: obviously, the blue is the driving frequency; what is the mechanism for the cone separation (just ordinary prism effect??) and if so, why are the color rings reversed in the complementary light cones? &lt;/p&gt;&#xA;" OwnerUserId="3253" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-21T15:31:28.803" LastActivityDate="2011-08-21T15:31:28.803" Title="Tracking photon color in Bell experiments" Tags="&lt;quantum-mechanics&gt;&lt;mechanics&gt;&lt;entanglement&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" />
  <row Id="13017" PostTypeId="2" ParentId="11247" CreationDate="2011-07-31T16:56:41.143" Score="1" ViewCount="" Body="&lt;p&gt;It's possible to do quantum mechanics on other domains than $\mathbb R^3$, but you have to make a few adjustments, which you forgot to do. Namely, the gauge transformation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \psi(x) \mapsto e^{i\chi(x)} \psi(x) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;must fulfill $e^{i\chi(0)}=e^{i\chi(2\pi)}$ to be single-valued on the circle, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\chi(2\pi)-\chi(0)=2\pi m \quad\text{ with } m\in\mathbb Z$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, your choice $\chi(x)=kx$ is only allowed for particular values of $k$, namely $k\in\mathbb Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;Moreover, the eigenvalue does not depend on the gauge, despite your claim to the contrary. That's because a gauge transformation will &lt;em&gt;change the velocity operator&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v \mapsto v - \partial_x \chi(x) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and its eigenvalues stay the same. No physically measurable quantity will change after a gauge transformation, ever.&lt;/p&gt;&#xA;" OwnerUserId="188" LastActivityDate="2011-07-31T16:56:41.143" CommentCount="7" />
  <row Id="13018" PostTypeId="2" ParentId="13015" CreationDate="2011-07-31T17:10:32.893" Score="2" ViewCount="" Body="&lt;p&gt;I don't exactly know what you mean by static forces. But I am going to take a wild guess here and assume that by that you mean forces involved in problems where bodies don't move. I think you assumed that Newton's second law quantifies a force. This is actually wrong. First of all realize that a force is an interaction and it still acts whether the body on which it acts moves or not. &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion&quot; rel=&quot;nofollow&quot;&gt;Newton's second law&lt;/a&gt; &lt;em&gt;quantifies the total effect of all such forces&lt;/em&gt; on a body of mass $m$ and not the force itself. For example the &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation&quot; rel=&quot;nofollow&quot;&gt;Newton's law of Gravitation&lt;/a&gt; tells you that the force between two masses is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{F} = G\frac{Mm}{r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this is practically useless unless you specifies what a force does on a body. That's where Newton's second law comes in. So along with Newton's second law, you have a complete theory of (classical) gravitation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the $\vec{F}$ in Newton's second law is the total force acting on a body having momentum $\vec{p}$. So when bodies don't move the net forces on them is zero. But that does not mean that you can not have forces acting on it. &lt;/p&gt;&#xA;" OwnerUserId="966" LastEditorUserId="966" LastEditorDisplayName="" LastEditDate="2011-07-31T17:16:09.880" LastActivityDate="2011-07-31T17:16:09.880" CommentCount="2" />
  <row Id="13019" PostTypeId="2" ParentId="13015" CreationDate="2011-07-31T17:38:25.763" Score="2" ViewCount="" Body="&lt;p&gt;By &quot;static forces&quot; I think you mean forces within a physical system in static equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You get static equilibrium out of the kinetic effects of force by noting the sum of the forces at every point is zero and therefore so is the sum of their kinetic effects.&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="1345" LastEditorDisplayName="" LastEditDate="2011-08-01T11:51:30.230" LastActivityDate="2011-08-01T11:51:30.230" />
  <row Id="13020" PostTypeId="1" CreationDate="2011-07-31T17:41:11.800" Score="5" ViewCount="175" Body="&lt;p&gt;To add to &lt;a href=&quot;http://physics.stackexchange.com/questions/884/books-that-every-physicist-should-read&quot;&gt;Books that every physicist should read&lt;/a&gt;:&lt;br&gt;&#xA; A list of popular physics books for people who aren't necessarily physics grads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(see also &lt;a href=&quot;http://physics.stackexchange.com/questions/12175/book-recommendations&quot;&gt;Book recommendations&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-08-01T20:48:55.797" CommunityOwnedDate="2011-07-31T19:37:28.837" Title="Books that every non-physicist should read" Tags="&lt;soft-question&gt;&lt;books&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="13021" PostTypeId="2" ParentId="13020" CreationDate="2011-07-31T18:19:49.637" Score="3" ViewCount="" Body="&lt;p&gt;Richard Feynman!&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0465025277&quot; rel=&quot;nofollow&quot;&gt;Six Easy Pieces&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0262560038&quot; rel=&quot;nofollow&quot;&gt;Character of Physical Law&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0393316041&quot; rel=&quot;nofollow&quot;&gt;Surely You're Joking Mr.Feynman&lt;/a&gt; and &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0393320928&quot; rel=&quot;nofollow&quot;&gt;What Do You Care What Other People Think&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691125759&quot; rel=&quot;nofollow&quot;&gt;QED: The Strange Theory of Light and Matter&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="966" LastEditorUserId="966" LastEditorDisplayName="" LastEditDate="2011-08-01T15:48:20.447" LastActivityDate="2011-08-01T15:48:20.447" CommunityOwnedDate="2011-07-31T19:37:28.837" />
  <row Id="13022" PostTypeId="2" ParentId="12983" CreationDate="2011-07-31T18:38:02.390" Score="2" ViewCount="" Body="&lt;p&gt;First of all, the philosophy of quantum mechanics is hardly straightforward to most physicists. This has spawned an entire industry of &quot;quantum interpretations&quot; to the &quot;quantum measurement problem&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum mechanics is probably one of the best solutions to the problem of making mysticism mathematically precise. Quantum mechanics puts mysticism on a firm mathematically footing. Quantum mechanics is about Cosmic Consciousness and Reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really, you can read about it straight from the horse's mouth in the book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0394723384&quot; rel=&quot;nofollow&quot;&gt;Quantum Questions&lt;/a&gt;, which is a compilation of mystical writings by the founders of quantum mechanics themselves. People like Werner Heisenberg, Erwin Schroedinger, Albert Einstein, Louis de Broglie, Jeans, Max Planck, Wolfgang Pauli and Arthur Eddington. They're not mere &quot;fringe&quot; physicists, even though &quot;fringe&quot; physicists like Jack Sarfatti, David Bohm, Amit Goswami, John Hagelin and Frank Tipler often do have a point...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another book you can read is &lt;a href=&quot;http://quantumenigma.com/&quot; rel=&quot;nofollow&quot;&gt;Quantum Enigma&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4758" LastActivityDate="2011-07-31T18:38:02.390" />
  <row Id="13024" PostTypeId="2" ParentId="12822" CreationDate="2011-07-31T19:34:17.163" Score="2" ViewCount="" Body="&lt;p&gt;Some notes to get you on your way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power in full sunlight at sea level $\approx 1kW/m^2$ - remember to adjust for relative angle of sun and solar panel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specific heat capacity of water: 4.186 J/g/K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical collector efficiency: somewhere in the range 50-90%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's do a quick crude static calculation. It sounds like you're going to do a dynamic microsim - great stuff, go for it; if we can make a static estimate first, that'll give you an idea of how the numbers connect to each other, and give you a ballpark number to check the output of your simulation against.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take the sun to be shining directly on the panel, along the normal to the plate, so the power hitting the plate is  $1000 W/m^2$. Let's assume a 60% efficiency, so we're putting $600 W/m^2$ into the heat transfer fluid. Let's ignore any anti-freeze for the moment, and just assume our heat-transfer fluid is pure water, thus having a specific heat capacity of 4.186 J/g/K. Now, if you had 1 litre of water collecting heat per square metre, then you'd raise its temperature by &#xA;$$\frac{600 W}{1000 g \times 4.186 J /g/K} \approx 0.14 K/s$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here's a quick calculation for a whole day. Let's take a ball-park figure of 6 full sun-hours (that's 16 hours of summer daylight, derated by guesstimate to account for the variation of angle between panel and sun), and a 200-litre thermal storage tank; then our increase in temperature of the tank in a day, from a single square metre of collector would be, to the first order, be:&#xA;$$\Delta T_{tank} = \frac{0.14 K/s  \times 3600 s/h \times 6 h}{200} \approx 15 K$$&#xA;So a $4m^2$ system would give you $\Delta T_{tank} \approx 60 K$ . Then that would get derated, based on system losses, including thermal losses from the collector plate and the thermal store. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat loss from the panel itself will (broadly) be proportional to the difference between the heat-transfer fluid, and the ambient air temperature - that gives the tapering you mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you'll need a figure for the rate at which heat can be exchanged between the heat-transfer fluid that passes through the collector, and the thermal store.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(See also Appendix H of &lt;a href=&quot;http://www.bre.co.uk/sap2009/&quot; rel=&quot;nofollow&quot;&gt;SAP 2009&lt;/a&gt; which starts on pdf p73 - but bear in mind that that's a coarse static approximation of a solar thermal system in the UK, not a dynamic simulation - but it has some figures to get you started on collector efficiency and thermal losses)&lt;/p&gt;&#xA;" OwnerUserId="4066" LastActivityDate="2011-07-31T19:34:17.163" CommentCount="3" />
  <row Id="13025" PostTypeId="1" AcceptedAnswerId="13042" CreationDate="2011-07-31T19:36:33.393" Score="8" ViewCount="185" Body="&lt;p&gt;The friction term in Navier-Stokes equation assumes that the viscosity coefficients are the same for the longitudinal and transverse directions. This doesn't seem intuitive, because the former is essentially a bulk modulus while the latter doesn't involve any compression of the fluid. How is the assumption justified?&lt;/p&gt;&#xA;" OwnerUserId="2643" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-07-31T19:39:28.757" LastActivityDate="2011-08-03T14:34:53.057" Title="Friction term in Navier-Stokes equation" Tags="&lt;fluid-dynamics&gt;&lt;navier-stokes&gt;&lt;viscosity&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13026" PostTypeId="2" ParentId="13008" CreationDate="2011-07-31T20:38:59.163" Score="1" ViewCount="" Body="&lt;p&gt;I think it is straight forward: assuming a lack of a force means no acceleration, so the equation $m \cdot a = 0$ is satisfied. There is no information or assumptions about the mass, nor about the metric - so you may not conclude that $m=0$ without further information.&lt;/p&gt;&#xA;" OwnerUserId="4760" LastActivityDate="2011-07-31T20:38:59.163" />
  <row Id="13027" PostTypeId="2" ParentId="12995" CreationDate="2011-07-31T20:45:26.880" Score="1" ViewCount="" Body="&lt;p&gt;Space-time is not something that can be created or destroyed. Vacuum fluctuations take place in framework of the space-time - so virtual particles are created within space-time&lt;/p&gt;&#xA;" OwnerUserId="4760" LastActivityDate="2011-07-31T20:45:26.880" />
  <row Id="13028" PostTypeId="2" ParentId="12933" CreationDate="2011-07-31T21:35:35.350" Score="1" ViewCount="" Body="&lt;p&gt;You asked for an instrument you can take around the building to detect the heat leaks and I don’t think you can do this. There is a way to do it over several days. Setting up a 2 time-lapse Thermographic cameras taking pictures every 10 min over a period of 3 days. One camera is set perpendicular to building face ant the other parallel. Watching the 2 sequences side by side will show where the heat is leaking. Studying the sequences over the major temperature shifts such as morning and evening. When the outside of the building goes threw the day to night transition the areas of heat loss will tend to linger or change faster compared to the well insulated areas.&lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-07-31T21:35:35.350" CommentCount="2" />
  <row Id="13029" PostTypeId="2" ParentId="13006" CreationDate="2011-07-31T23:11:47.093" Score="0" ViewCount="" Body="&lt;p&gt;You get constraints on the possible form of the dielectric since it is a &quot;response function&quot;. The polarization $P$  in an applied field $E$ is given by $P(x,t) = \int \epsilon(x-x&amp;#39;,t-t&amp;#39;)E(x&amp;#39;,t&amp;#39;)dx&amp;#39;dt&amp;#39;$ up to various conventions. We must have $\epsilon(x,t) = 0$ when $t&amp;lt;0$, since an electric field can't change what the polarization was before the electric field was turned on. In fourier space this translates to $\epsilon(\omega)$ is analytic in the upper half complex $\omega$ plane. This in turn leads to something called Kramers-Kronig relations which you can look up on wikipedia under that name. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I suppose we can get an even stricter condition since $\epsilon(x,t)$ must be zero outside of the forward light cone, $t &amp;gt; |\vec{x}|$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for seeing how the dielectric changes with lorentz transformation - this should be a relatively straightforward task. Write down Maxwell's equation with everything expressed in terms of the fields and the dielectric and permitivity and perform a Lorentz transformation. This should take you back to same form of equations but with a different dielectric.&lt;/p&gt;&#xA;" OwnerUserId="3970" LastEditorUserId="3970" LastEditorDisplayName="" LastEditDate="2011-08-01T03:58:54.943" LastActivityDate="2011-08-01T03:58:54.943" />
  <row Id="13030" PostTypeId="1" CreationDate="2011-08-01T00:06:44.077" Score="5" ViewCount="245" Body="&lt;p&gt;I understand the common explanation of lift, which describes the airflow over the top of the wing as moving faster than the air below the wing.  However, I don't quite understand why the air moves faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read some explanations citing a circulation effect among others, but haven't found a good, clear explanation for the exact cause.  Can someone help me out?&lt;/p&gt;&#xA;" OwnerUserId="4763" LastActivityDate="2011-08-27T07:41:29.947" Title="Why does the air flow faster over the top of an airfoil?" Tags="&lt;aerodynamics&gt;" AnswerCount="6" CommentCount="4" />
  <row Id="13031" PostTypeId="2" ParentId="13030" CreationDate="2011-08-01T00:42:29.340" Score="-3" ViewCount="" Body="&lt;p&gt;The short answer: it doesn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Link #1: &lt;a href=&quot;http://xkcd.com/803/&quot; rel=&quot;nofollow&quot;&gt;http://xkcd.com/803/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Link #2: &lt;a href=&quot;http://en.wikipedia.org/wiki/Airfoil&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Airfoil&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4765" LastActivityDate="2011-08-01T00:42:29.340" CommentCount="3" />
  <row Id="13032" PostTypeId="2" ParentId="13030" CreationDate="2011-08-01T00:49:46.830" Score="5" ViewCount="" Body="&lt;p&gt;The common explanation given is that it flows faster over the top of the wing because the top is more curved than the bottom of the wing.  However, I understand why you would find this explanation unsatisfactory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start with, I think we need to identify the point at which the flow separates.  Looking at &lt;a href=&quot;http://en.wikipedia.org/wiki/Airfoil&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, I'll post two images:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/qMPLN.png&quot; alt=&quot;airfoil&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/9x0lH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument that the wind flows faster over the top is mostly a consequence of geometry.  First identify the point at which the flow separates, meaning the point above which the fluid goes over and below which the fluid goes under, this is slightly below the front-most point of the airfoil, due to the fact that it's angled slightly upward.  If both paths take roughly the same time to pass over the wing, then the average velocity of the fluid from the point of separation to the tail where the flow rejoins will be roughly proportional to the distance from those two points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you may say, &quot;but it will flow faster over the top even if the top isn't curved more!&quot;  You would be correct.  A plane can function with no additional curve on the top of the wing, as the famous xkcd comic points out.  Such a mode of flying, however, will still see the fluid passing over the top of the wing faster.  A simple argument for this is that the point of separation is lower than the front of the wing, again, since the wing is angled up.  A plane can fly upside down, but I don't know of a plane that can maintain altitude with the wings not angled up.  The curved top, however, increases efficiency by intensifying that natural effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that helps some, this is intentionally not a rigorous answer, and I want to recognize that I am not addressing the more hairy details of the actual fluid equations associated with this, which are required for a full explanation.  In short, the fluid velocity over a surface isn't &lt;em&gt;completely&lt;/em&gt; proportional to the distance traveled.  Even without getting into that, however, I think your question is &lt;em&gt;mostly&lt;/em&gt; answered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Another Attempt&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that my answer up to this point may not only be incomplete, but might not answer the question.  The question is why the flow on the top is moving faster than on the bottom.  Let me post another image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/OUIZM.gif&quot; alt=&quot;Wiki gif&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 2 things I want to note here.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The fluid above the wing &lt;em&gt;does&lt;/em&gt; speed up and the fluid below the wing &lt;em&gt;does&lt;/em&gt; slow down.  I just want to confirm this is still the case.&lt;/li&gt;&#xA;&lt;li&gt;The fluid going above and below does not have the same travel time over the wing.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Number 2 is particularly important because it is simply not correct to say that the speed is proportional to the distance between the separation and rejoin point.  That way of looking at it may still have &lt;em&gt;some&lt;/em&gt; usefulness.  But I digress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point I'm repeating &lt;a href=&quot;http://en.wikipedia.org/wiki/Lift_%28force%29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's explanation&lt;/a&gt;, but refer to the 2nd image in this answer.  Under some assumptions &lt;em&gt;the fluid does not cross the blue streamlines&lt;/em&gt;.  That means that when the channel size between 2 blue streamlines narrows or widens, the fluid speed changes correspondingly.  I'm still hand waving away plenty of technical detail, but please let me offer this as basic level answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The fluid on top of the wing is accelerated and the fluid on the bottom of the wind is slowed down compared to velocity of the aircraft itself because the wing geometry and angle narrows the flow area above the wing and widens the flow area below the wing&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the absolute best explanation I have.  If you assume that the fluid is incompressible it works great, if not, it works less great but still works.  There are also some other assumptions, I hope that the general point is still the same with all those included.  The bold text is the best answer I have and I &lt;em&gt;think&lt;/em&gt; it's a good one.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-08-01T04:17:30.823" LastActivityDate="2011-08-01T04:17:30.823" CommentCount="5" />
  <row Id="13034" PostTypeId="1" AcceptedAnswerId="13041" CreationDate="2011-08-01T04:07:10.383" Score="4" ViewCount="106" Body="&lt;p&gt;Suppose I have a massive particle in non-relativistic quantum mechanics.  Its wavefunction can be written in the position basis as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vert \Psi \rangle = \Psi_x(x,t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or in the momentum basis as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vert \Psi \rangle = \Psi_p(p,t)$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Psi_x$ and $\Psi_p$ are related to each other via a Fourier transform.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if I write $\vert \Psi \rangle$ as an integral over infinitely-many &quot;position basis vectors&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vert \Psi \rangle = \int_{-\infty}^\infty \Psi_x(x)\vert x \rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then the position basis vectors $\mid x \rangle$ are Dirac delta functions - they aren't really functions.  If we try to represent them in the momentum basis, we get non-normalizable plane waves.  These basis vectors are not members of the physical Hilbert space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My undergraduate quantum text explains that the Dirac deltas and plane waves are calculational tools and demonstrates their use.  The Dirac deltas do not represent true wavefunctions.  A real particle with low position uncertainty would simply have a wavefunction with a high but finite peak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm fine with this; I think I understand how to do the calculations and what they mean.  However, I am still unsure of how to find a basis for the physical Hilbert space that consists of vectors actually in the space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a bound state with no degeneracy, the energy eigenfunctions form a basis.  The physical Hilbert space then consists of all linear combinations of the energy eigenfunctions.  However, when we move to a scattering state, the spectrum of energy eigenvalues becomes continuous and the energy eigenfunctions are not normalizable because they are essentially the same as the plane-wave momentum eigenfunctions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the scattering state has a physical Hilbert space of  normalizable wavefunctions, shouldn't I be able to find a basis that consists of elements of the physical Hilbert space itself, even if this basis is not convenient for calculations?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an example of such a basis for a free particle?&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-09-05T21:05:03.100" LastActivityDate="2011-09-05T21:05:03.100" Title="What is a basis for the Hilbert space of a 1-D scattering state?" Tags="&lt;quantum-mechanics&gt;&lt;mathematical-physics&gt;&lt;hilbert-space&gt;&lt;normalization&gt;&lt;linear-algebra&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="13035" PostTypeId="2" ParentId="12955" CreationDate="2011-08-01T04:31:04.207" Score="2" ViewCount="" Body="&lt;p&gt;Lets put it another way though @user1631 is correct :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are convective movements in the atmosphere due to non uniform  absorption of heat and release of energy through radiation due to ground formations ( ocean, land, mountain, desert), altitude, latitude, Coriolis forces, atmospheric tides etc. Generally air will move from hot to cold generating high pressure and low pressure areas which move slowly from west to east ( Coriolis). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If humidity and dew point are right, clouds form at high altitudes. Clouds change the density of air and general properties of emission and absorption of radiation    and can create stronger convective currents up to cyclones in a feedback system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is: there exist convective atmospheric motions, even strong ones ( sandstorms for example) without clouds, but clouds amplify by their properties convection effects also.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-08-01T04:31:04.207" CommentCount="2" />
  <row Id="13036" PostTypeId="2" ParentId="13013" CreationDate="2011-08-01T05:37:21.777" Score="2" ViewCount="" Body="&lt;p&gt;At face value, I interpret the statement as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum mechanical fluctuations may  generate from the vacuum ( nothing) enough energy for a universe. Because gravity exists for any ensemble that has energy, then the big bang will go on its usual development in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One has to presuppose that the mathematical formulations exist,( a mathematical theory of everything including quantum mechanics and gravity), irrespective of the existence of matter/energy. A kind of platonic ideal.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-08-01T05:37:21.777" />
  <row Id="13037" PostTypeId="2" ParentId="12457" CreationDate="2011-08-01T05:55:24.603" Score="1" ViewCount="" Body="&lt;p&gt;Disclaimer: I've never done any real work with sector decomposition - I've only read (some of) the papers and played with some simple examples.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There are many different sector decomposition strategies, which are mainly designed to work on Feynman parametrized integrals. &#xA;In general, sector decomposition is a algorithmic (recursive or not) way of decomposing Feynman parametrized integrals into subsectors where the divergences can be more easily isolated. Traditionally it was used in formal investigations and proofs relating to BPHZ, nowadays it is often used for automated separation of divergences ($\epsilon$-expansion) and numerical integration for comparison with analytic results and for use where analytic results are not known. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sector decomposition does not mix up UV and IR divergences.&lt;/strong&gt; However, some strategies and changes of variables can make it difficult to easily tell them apart. Also, if dimensional regularization is the only regularization used, then both UV and IR divergences turn up as poles in the $\epsilon$-expansion. If care is not taken to track where the poles come from, then you can not separate the UV and IR contributions. There are sector decomposition schemes, e.g. &lt;a href=&quot;http://arxiv.org/abs/0908.2897&quot; rel=&quot;nofollow&quot;&gt;arXiv:0908.2897&lt;/a&gt;, that take care with the IR divergences.&#xA;Also see older work such as &lt;a href=&quot;http://dx.doi.org/10.1103/PhysRevD.10.3991&quot; rel=&quot;nofollow&quot;&gt;PhysRevD.10.3991&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify this (hopefully), let's give a quick summary of parametrized Feynman integrals:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There are two main parameterized versions of Feynman integrals (both of which can be derived from the position and momentum space versions). For a quick discussion see &lt;a href=&quot;http://demonstrations.wolfram.com/ScalarFeynmanDiagramsAndSymanzikPolynomials/&quot; rel=&quot;nofollow&quot;&gt;Scalar Feynman Diagrams and Symanzik Polynomials&lt;/a&gt;. For a longer discussion see &lt;a href=&quot;http://arxiv.org/abs/1011.5493&quot; rel=&quot;nofollow&quot;&gt;the paper linked to in the question&lt;/a&gt; and references within.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Symanzik polynomials $\mathcal{U}$ and $\mathcal{V}$ are key aspects of both Feynman parameterization (Chisholm representation) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/DBt2H.gif&quot; alt=&quot;Feynman parameters&quot;&gt; ......... (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and Schwinger parametrization (aka propertime or $\alpha$ parameters) (Nambu representation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Ys4OQ.gif&quot; alt=&quot;Schwinger parameters&quot;&gt; ......................... (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feynman diagrams with tensor, spinor or other structures eventually lead to similar integrals, but with extra polynomial factors in the numerator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Schwinger parametrization (2), UV divergences are zeros of the 1st Symanzik/Kirchhoff polynomial $\mathcal{U}(s)$ that occur when various subsets of $s_i$ are zero. IR divergences come from zeros of exponential (which includes the 2nd Symanzik polynomial $\mathcal{V}(s,p_{\text{ext}})$) as a subset of $s_i$ go to infinity. UV divergences only depend on the topology of the diagram , but since $\mathcal{V}$ has dependence on the external momenta, IR divergences can also depend on the kinematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar statements hold for the Feynman parametrization (1), but once the $\delta$ function is integrated out it can be confusing as to whether the singularities in the integrand are UV or IR. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To quote &lt;a href=&quot;http://arxiv.org/abs/0812.4700&quot; rel=&quot;nofollow&quot;&gt;Hepp and Speer sectors within modern strategies of sector decomposition&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To analyze convergence of the integral (2) large values of $s_i$ (in&#xA;  particular, to reveal infrared (IR) divergences) one has to take into&#xA;  account the exponent as well. A possible way is to separate the&#xA;  integration over every $s_i$ into (0, 1) and (1,∞) and then to deal with&#xA;  each of these $2n$ regions separately. This can be&#xA;  enough for a general analysis but cannot be good from the practical&#xA;  point of view because the number of the resulting sectors will be too&#xA;  large. A more reasonable approach is to turn (2) into the integral (1) with a&#xA;  compact integration domain, where both UV and IR divergences are&#xA;  somehow mixed up and manifest themselves as divergences at small values&#xA;  of parameters of integration.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This said, Feynman parameterization and sector decomposition don't actually mix UV and IR divergences, it can just make them difficult to naively identify. &#xA;Also, both turn up as poles in the $\epsilon$-expansion, so can be hard to separate after the integral has been performed.&lt;/p&gt;&#xA;" OwnerUserId="429" LastActivityDate="2011-08-01T05:55:24.603" />
  <row Id="13038" PostTypeId="2" ParentId="13034" CreationDate="2011-08-01T06:01:38.823" Score="2" ViewCount="" Body="&lt;p&gt;The eigenfunctions of a self adjoint operator lie outside the Hilbert space of square integrable functions on the line.&#xA;One solution is to work with a basis of eigenfunctions of a non-self adjoint operator such as $x+ip$. Of course these are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Coherent_state&quot; rel=&quot;nofollow&quot;&gt;coherent states.&lt;/a&gt; For the coherent states, one has an ovecomplete basis and a partition of unity, thus it is not difficult to decompose any vector in $L^2(\mathbb{R})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An other option is to work with a rigged Hilbert space, which is a formalization of Dirac's bra and ket method. A very good exposition of rigged Hilbert spaces is given in &#xA;&lt;a href=&quot;http://arxiv.org/PS_cache/quant-ph/pdf/9907/9907069v2.pdf&quot; rel=&quot;nofollow&quot;&gt;Francois Gieres' article&lt;/a&gt;. According to this option, you work with the eigenstates of the position operator but you remember that they do not belong to the Hilbert space, but rather to a Banach space which contains also distributions. &lt;/p&gt;&#xA;" OwnerUserId="2190" LastActivityDate="2011-08-01T06:01:38.823" CommentCount="4" />
  <row Id="13039" PostTypeId="1" AcceptedAnswerId="13071" CreationDate="2011-08-01T06:10:33.380" Score="4" ViewCount="158" Body="&lt;p&gt;It seems that at Three Mile Island and in Fukushima, there was a build-up of radioactive gasses, and other gasses like hydrogen in the &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Containment_building&amp;amp;oldid=433575682&quot; rel=&quot;nofollow&quot;&gt;containment buildings&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia says: The containment [building] is the fourth and final barrier to radioactive release (part of a nuclear reactor's defence in depth strategy), the first being the fuel ceramic itself, the second being the metal fuel cladding tubes, the third being the reactor vessel and coolant system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did radiation escape into the containment building in TMI if the reactor vessel wasn't breached?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wasn't it the same case in Fukushima that reactor vessels were intact?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And where do these gasses come from? &lt;/p&gt;&#xA;" OwnerUserId="2390" LastActivityDate="2011-08-03T16:27:05.743" Title="Without a reactor vessel breach, how does radiation build-up in containment buildings of troubled nuclear reactors?" Tags="&lt;radiation&gt;" AnswerCount="6" CommentCount="4" />
  <row Id="13040" PostTypeId="2" ParentId="13013" CreationDate="2011-08-01T06:18:16.897" Score="0" ViewCount="" Body="&lt;p&gt;The flaw in the argument is the assumption of causality, and that it's linear. Nothing doesn't exist, which makes saying a given entity is nothing empty.&lt;/p&gt;&#xA;" OwnerUserId="4767" LastActivityDate="2011-08-01T06:18:16.897" />
  <row Id="13041" PostTypeId="2" ParentId="13034" CreationDate="2011-08-01T06:28:52.583" Score="2" ViewCount="" Body="&lt;p&gt;If I understand correctly, your question basically comes down to identifying a basis for the space of square-integrable functions, $L^2(\mathbb{R})$, since any physical state $|\Psi\rangle$ can be constructed by performing the integral you listed in your question with a function $\Psi_x(x)\in L^2(\mathbb{R})$. $L^2$ is known to be a vector space, so a basis has to exist. Off the top of my head, I think an example would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_k(x) = e^{-x^2/a^2 - ikx}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is just the normal plane wave basis $e^{-ikx}$ multiplied by a Gaussian envelope $e^{-x^2/a^2}$ where $a$ is some constant. Multiplying by this Gaussian envelope ensures that the functions will be square-integrable, but since you use the same envelope for every element of the basis, you can factor it out of the Fourier transform, so it doesn't change any of the essential properties of momentum-space decomposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I found a &lt;a href=&quot;http://math.stackexchange.com/questions/54393/does-the-vector-space-spanned-by-a-set-of-orthogonal-basis-contains-the-basis-vec&quot;&gt;question&lt;/a&gt; on math.SE that seems related, and which motivated this answer.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-08-01T06:28:52.583" CommentCount="1" />
  <row Id="13042" PostTypeId="2" ParentId="13025" CreationDate="2011-08-01T06:58:27.770" Score="7" ViewCount="" Body="&lt;p&gt;Actually, there are two different viscosity coefficients. You can see this from the stress tensor&#xA;$$ \sigma_{ij} = -p_0 \delta_{ij} + \eta \left( \frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i} - \frac{2}{3} \delta_{ij} \frac{\partial v_k}{\partial x_k} \right) + \zeta \delta_{ij} \frac{\partial v_k}{\partial x_k} $$&#xA;which has the two coefficients of viscosity $\eta$ and $\zeta$ (see Landau &amp;amp; Lifshitz, &lt;em&gt;Fluid Mechanics&lt;/em&gt;, for example). &#xA;The pressure $p_0$ is given by the thermodynamic equation of state, but this is not the whole pressure $p$. The latter is given by the mean normal stress &#xA;$$ p = - \frac{1}{3} \sigma_{ii} = p_0 - \zeta \frac{\partial v_k}{\partial x_k} $$&#xA;so that the stress tensor is &#xA;$$ \sigma_{ij} = -p \delta_{ij} + \eta \left( \frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i} - \frac{2}{3} \delta_{ij} \frac{\partial v_k}{\partial x_k} \right) .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why sometimes you don't see the coefficient $\zeta$ (often called &lt;em&gt;second viscosity&lt;/em&gt;) in the Navier-Stokes equation. It is hidden in the pressure, but it's there. &lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2011-08-01T06:58:27.770" CommentCount="2" />
  <row Id="13043" PostTypeId="1" AcceptedAnswerId="13051" CreationDate="2011-07-30T23:08:43.483" Score="3" ViewCount="242" Body="&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Energy_Catalyzer&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Energy Catalyzer is an apparatus built by [...] Andrea Rossi, [and] Sergio Focardi. The 2009 patent application claims &quot;a method and apparatus for carrying out nickel and hydrogen exothermal reactions,&quot; with production of copper. Although the patent cites previous works on cold fusion, one statement by Rossi asserted that it is not cold fusion, but rather LENR, Low-Energy Nuclear Reaction.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;[...]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;According to Focardi, &quot;the hydrogen is heated at a given temperature with a simple resistor. When the ignition temperature is reached, the energy production process starts: the hydrogen atoms penetrate into the nickel and transform it into copper.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are these claims true?&lt;/p&gt;&#xA;" OwnerUserId="1576" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-08-01T07:16:02.457" LastActivityDate="2011-08-21T13:31:40.520" Title="Does the &quot;Energy Catalyzer&quot; generate energy by converting Nickel to Copper?" Tags="&lt;cold-fusion&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" />
  <row Id="13044" PostTypeId="2" ParentId="13043" CreationDate="2011-07-31T11:12:46.893" Score="3" ViewCount="" Body="&lt;p&gt;The transformation of Ni into Cu would require extreme temperatures. In nature, stars can convert elements into other ones. During the normal burning phase of the star, it converts Hydrogen into Helium. This happens at about 10 million degrees (C or K). When the Hydrogen is used up, the star contracts further, heats up more and gradually creates the other elements up to element 26 (Iron). The temperature is well above 1 billion degrees (1GK) when this happens. That is the end of the star's energy cycle as creating heavier elements requires energy instead of releasing it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heavier elements such as Nickel (28) and Copper (29) are created when heavy stars collapse on themselves after they reach iron and can no longer balance the gravitational pressure with the generated energy. That explosion is called a supernova and is visible across the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can guess, you're not going to reach a temperature of 1GK with a heating element, and even that would only get you to Iron. The reaction to convert Ni into Cu will never generate energy: you need to add (lots of) energy to make it happen. Even cold fusion (which requires a measly 10 million degrees) turned out to be a false alarm. So I'm sorry, this kind of energy generation is completely impossible. This product is quite clearly a con.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information, try Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_evolution&quot; rel=&quot;nofollow&quot;&gt;Evolution of a star&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aBinding_energy_curve_-_common_isotopes.svg&quot; rel=&quot;nofollow&quot;&gt;Nuclear binding energy&lt;/a&gt;&lt;br&gt;&#xA;This graph shows how Iron has the highest binding energy. In other words fusion (from light elements) up to Iron, or fission (from heavy elements such as Uranium) down to Iron will generate energy. However, fusion from Iron to heavier elements requires extra energy.&lt;/p&gt;&#xA;" LastActivityDate="2011-07-31T12:21:09.697" CommentCount="5" />
  <row Id="13045" PostTypeId="2" ParentId="12533" CreationDate="2011-08-01T07:10:27.613" Score="3" ViewCount="" Body="&lt;p&gt;The standard use of zeta-regularization in quantum field theory is to replace the ill defined one-loop functional determinant with something finite:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align} &#xD;&#xA;   \det(H) &amp;amp;= \exp(\mathrm{tr}\log H) := \exp(-\zeta&amp;#39;_H(0))\,, \\&#xD;&#xA;   \zeta_H(s) &amp;amp;= \mathrm{tr} H^{-s} = \frac1{\Gamma(s)}\int_0^\infty\mathrm{d}t\, t^{s-1}\mathrm{tr}(\exp(-H t))&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a generalization of the &#xA;Riemann zeta-function $\zeta(s)=\zeta(s,q)$ and &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hurwitz_zeta_function&quot; rel=&quot;nofollow&quot;&gt;Hurwitz zeta-function&lt;/a&gt; $\zeta(s,q)=\sum_{n=1}^\infty (q+n)^{-s}$ &#xA;since $$\zeta_H(s) = \mathrm{tr} H^{-s} = \sum_{n} \lambda_n^{-s}\,.$$ For example, for the harmonic oscillator, $\lambda_n=n+\frac12$ so $\zeta_H(s)=\zeta(s,\frac12)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This obviously only works at one-loop because the quantum corrections only look like a functional determinant at one-loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But related ideas can be extended to multiple loops.&lt;/strong&gt; (How does the following compare to your scheme? It's hard to tell exactly what you're saying in &lt;a href=&quot;http://vixra.org/abs/1009.0047&quot; rel=&quot;nofollow&quot;&gt;your paper&lt;/a&gt;...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://physics.stackexchange.com/q/12235/2451&quot;&gt;your previous question&lt;/a&gt; I mentioned operator regularization (e.g. &lt;a href=&quot;http://dx.doi.org/10.1103/PhysRevD.35.3854&quot; rel=&quot;nofollow&quot;&gt;PhysRevD.35.3854&lt;/a&gt;) as an extension of zeta-regularization.&#xA;Here, the functional determinant is written using &#xA;$$\begin{align} &#xD;&#xA;   \det(H) &amp;amp;= \exp(\mathrm{tr}\log H) \\&#xD;&#xA; \log H &amp;amp;:= \lim_{s\to0}\Big[-\frac{d^m}{d s^m}\Big[\frac{s^{m-1}}{m!}H^{-s}\Big]\Big]\,,&#xD;&#xA;\end{align}$$&#xA;for some large enough (only actually need $m=1$) integer $m$. With $m=1$ and writing $H^{-s}=\frac1{\Gamma(s)}\int_0^\infty \mathrm{d}t\, t^{s-1}\exp(-H t)$ you recover the zeta-regularized form given above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly you can also regularize the propagators $H^{-1}$ as&#xA;$$    H^{-1} = \frac{d}{d H}\log H := \lim_{s\to0}\Big[\frac{d^m}{d s^m}\Big[\frac{s^{m}}{m!}H^{-s-1}\Big]\Big]\,.&#xD;&#xA;$$&#xA;This is the underlying basics of operator regularization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://arxiv.org/abs/1006.1806&quot; rel=&quot;nofollow&quot;&gt;arXiv:1006.1806&lt;/a&gt;, this was generalized to &#xA;$$    H^{-n} := \lim_{s\to0}\Big[\frac{d^m}{d s^m}\Big[&#xD;&#xA;      \big(1+\alpha_1s+\dots+\alpha_ms^m\big)\frac{s^{m}}{m!}H^{-s-n}\Big]\Big]\,.&#xD;&#xA;$$&#xA;where the $\alpha_i$ are initially arbitrary finite constants and $m$ is the loop order (or higher).&#xA;Comparisons were made with some dimensionally regularized Feynman integrals and it was found that for a choice of $\alpha_i$ they matched. &#xA;These constants are fixed according the renormalization scheme that is used. I'm not sure if  a &quot;minimal subtraction&quot; scheme exists for operator regularization (unless you do a comparison with dimensional regularization for each diagram!) or if a physical/on-shell renormalization scheme has to be used to ensure consistent results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A much more solid paper is &lt;a href=&quot;http://dx.doi.org/10.1103/PhysRevD.41.514&quot; rel=&quot;nofollow&quot;&gt;Operator regularization and multiloop Green's functions&lt;/a&gt;.&#xA;There they define the subtraction operator (with out the $\alpha_i$ constants)&#xA;$$ \mathcal S^{(m)}\big(\prod_i A_i^{-1}\big) &#xD;&#xA;   = \lim_{s\to0}\frac{d^m}{d s^m}\Big[\frac{s^{m}}{m!}\Big(\prod_i A_i^{-s-1}\Big)\Big]&#xD;&#xA;$$ &#xA;and show that its naive application to obtain finite amplitudes breaks unitarity.&#xA;(The results of arXiv:1006.1806 suggest that if you include the $\alpha_i$'s and choose them correctly, you should be able to get correct results...)&#xA;Then they actually use Bogolibov's recursion formula to show how to construct a consistent $\mathcal R$-operator from $\mathcal S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Detail computations are given for both the naive and sophisticated use of operator regularization in massive $\phi^4_4$ theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that using a BPHZ-like scheme essentially reintroduces counterterms (hidden in the subtractions in each diagram) to the method and definitely breaks the simplicity of the original $\zeta$-regularization.&lt;/p&gt;&#xA;" OwnerUserId="429" LastEditorUserId="429" LastEditorDisplayName="" LastEditDate="2011-08-01T08:14:24.227" LastActivityDate="2011-08-01T08:14:24.227" CommentCount="7" />
  <row Id="13046" PostTypeId="2" ParentId="13039" CreationDate="2011-08-01T08:29:38.347" Score="1" ViewCount="" Body="&lt;p&gt;&quot;Radiation&quot; is nothing that can escape. This is all about radioactive &#xA;materials! Radiation penetrates more or less through walls, but &#xA;this is very effectively inhibited by the thick steel/concrete &#xA;in nuclear plants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What You have in mind, was caused by deliberately releasing atmosphere &#xA;(mostly hydrogen at that time, but some steam as well) to prevent &#xA;pressure containment rupture. This is done via pipes and filters &#xA;to an exhaust stack normally. This filters keep back cesium and iodine, &#xA;this two fission products are rather volatile and leave the core &#xA;with the gas/steam at the high temperatures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reasons I do not remember, this venting worked not as planned,&#xA;some or all of that gas/steam got into the buiding surrounding the &#xA;reactor, where it mixed with air and exploded then. &#xA;Even into the building of a reactor out of operation some of this gases &#xA;leaked by some pipe. The latter points to a major violation of &#xA;security guidelines: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Never install &quot;reserve&quot; connections between plants, which &#xA;are separated for (good) security reasons&quot;&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In German this is expressed short (in trade slang) .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;No Mikado plants&quot;&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mikado_%28game%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mikado_%28game%29&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-08-01T08:44:05.650" LastActivityDate="2011-08-01T08:44:05.650" CommentCount="0" />
  <row Id="13047" PostTypeId="2" ParentId="13025" CreationDate="2011-08-01T09:06:31.760" Score="0" ViewCount="" Body="&lt;p&gt;Viscosity of gases is independent from pressure, as long as &#xA;mean free path is shorter than experimental dimensions. &#xA;(in practice above 0.01 mmHg)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kinematic viscosity (dynamic visc./ density) is pressure dependent &#xA;of course, but this can be separated in equations easily. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of liquids there is no problem, because those can be &#xA;treated as being incompressible at pressures relevant in &#xA;hydrodynamic.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-08-03T14:34:53.057" LastActivityDate="2011-08-03T14:34:53.057" />
  <row Id="13048" PostTypeId="2" ParentId="13039" CreationDate="2011-08-01T09:39:06.387" Score="1" ViewCount="" Body="&lt;p&gt;These gases first come from the still ongoing decay. More devastating is the &lt;a href=&quot;https://secure.wikimedia.org/wikipedia/en/wiki/Decay_heat&quot; rel=&quot;nofollow&quot;&gt;decay heat&lt;/a&gt;, which will force you to cool or vent (gas laws still apply and the vessel/containment have limits). You can shut down the fission of a (thermal) nuclear reactor pretty well, but you still have to cool for days and weeks to come.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen can be produced later, if cooling fails: at high temperature the zirconium of the common fuel rods reacts with water to hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We estimated an example for how much energy decay heat might produce on Fukushima #4s spent fuel pool: 15m depth were given, we reckoned cubic measurements and a maximum of 3000 m³ water in the pool. The pool was empty enough after five days that they urgently needed to refill it by helicopter/water cannon, cracks not known at this time.&#xA;It takes (4kJ/K * 80 K + 2300kJ)/kg * 3*10^6 kg water ~ 78*10^12J to boil that water away, divided by five days ~ 18*10^9J/s. Give or take an magnitude for low water/the later discovered cracks that makes still something between 2 to 18 MW decay heat in the spent fuel pool alone. As far as it is known this pool was filled beyond normal capacity.&lt;/p&gt;&#xA;" OwnerUserId="1089" LastEditorUserId="1089" LastEditorDisplayName="" LastEditDate="2011-08-01T19:55:53.923" LastActivityDate="2011-08-01T19:55:53.923" CommentCount="5" />
  <row Id="13049" PostTypeId="2" ParentId="13039" CreationDate="2011-08-01T10:03:33.837" Score="1" ViewCount="" Body="&lt;p&gt;The site &lt;a href=&quot;http://allthingsnuclear.org/post/3940804083/possible-cause-of-reactor-building-explosions&quot; rel=&quot;nofollow&quot;&gt;http://allthingsnuclear.org/post/3940804083/possible-cause-of-reactor-building-explosions&lt;/a&gt; describes a believable scenario. Apparently there is a leakage problem with this reactor type, when the pressure within the vessel is to high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;It is possible that the containment pressures rose high enough to replicate the Brunswick experience by lifting the drywell head enough to allow hydrogen and other gases to leak into the refueling cavity and reactor building. If so, hydrogen could build up to an explosive mixture.&quot; However I haven't heard any follow up news on this since March.&lt;/p&gt;&#xA;" OwnerUserId="4306" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-08-01T12:34:32.467" LastActivityDate="2011-08-01T12:34:32.467" />
  <row Id="13050" PostTypeId="2" ParentId="13039" CreationDate="2011-08-01T10:16:06.663" Score="1" ViewCount="" Body="&lt;p&gt;Reactor vessel is not completely hermetical. Otherwise it would explode at some point if something goes not as expected and pressure inside increases. AFAIK, in Fukushima on a first day (before explosions) they let some (radioactive) gas out of the vessel (to confinement building) to decrease pressure inside reactor and avoid destruction of the vessel. Unfortunately, it did not help. It is a normal practice. It is better to let some radiation go out than wait for the explosion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Of course, reactor vessel &lt;strong&gt;is&lt;/strong&gt; completely hermetical during proper operation. However, &quot;no reactor breach&quot; is not equal to &quot;no radioactive materials escaped&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4657" LastEditorUserId="4657" LastEditorDisplayName="" LastEditDate="2011-08-01T18:00:42.043" LastActivityDate="2011-08-01T18:00:42.043" />
  <row Id="13051" PostTypeId="2" ParentId="13043" CreationDate="2011-08-01T11:32:10.303" Score="10" ViewCount="" Body="&lt;p&gt;I suppose I did not make myself clear, and spent too much time talking about stellar evolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason why this cannot work is that, when you are working with elements that have atomic numbers higher than Iron (26), you cannot get energy by converting an element into another that has an even higher atomic number. In this case we're converting Ni (element 28) into Cu (29). That conversion does not generate energy, it uses energy. Fusion only generates energy for atomic numbers below Iron; above it you can only generate energy through fission. This is why I mentioned stars: they cannot &quot;burn&quot; to any element above Iron. It needs energy from the gravitational collapse of a supernova to create the higher elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because Iron has the highest binding energy of any nucleus, as explained &lt;a href=&quot;http://www.furryelephant.com/content/radioactivity/binding-energy-mass-defect/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Adding more protons or neutrons to Iron means you have to supply energy. Below Iron it's the reverse. This is why fission bombs use Uranium (92) or Plutonium (94) and fusion bombs use Hydrogen (1).&lt;/p&gt;&#xA;" OwnerUserId="4769" LastActivityDate="2011-08-01T11:32:10.303" CommentCount="8" />
  <row Id="13052" PostTypeId="1" AcceptedAnswerId="13063" CreationDate="2011-08-01T11:51:14.707" Score="2" ViewCount="122" Body="&lt;p&gt;I recently saw this video on youtube:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=oJfBSc6e7QQ&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=oJfBSc6e7QQ&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I don't know what to make of it. It seems as if the theory has enough evidence to be correct but where would all the water have appeared from? Would that much water have appeared over 60 million years? Also what would cause it to expand. The video suggests that since the time of dinosaurs the earths size has doubled in volume, how much of this is and can be true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[could someone please tag this, I don't know what category this should come under]&lt;/p&gt;&#xA;" OwnerUserId="1878" LastActivityDate="2011-08-02T08:13:33.657" Title="Is the earth expanding?" Tags="&lt;mass&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="13053" PostTypeId="2" ParentId="12983" CreationDate="2011-08-01T11:58:46.583" Score="4" ViewCount="" Body="&lt;p&gt;Scott Aaronson, himself a (quantum) computer scientist, thinks and writes about a number of these subjects in his paper &lt;a href=&quot;http://www.scottaaronson.com/papers/island.pdf&quot; rel=&quot;nofollow&quot;&gt;Is Quantum Mechanics An Island In Theoryspace?&lt;/a&gt; - at least the &quot;why complex numbers and not the reals or the quaternions?&quot;, and I'm pretty sure he mentions it in his &lt;a href=&quot;http://www.scottaaronson.com/democritus/&quot; rel=&quot;nofollow&quot;&gt;'Democritus' lectures&lt;/a&gt; as well.&lt;/p&gt;&#xA;" OwnerUserId="2445" LastActivityDate="2011-08-01T11:58:46.583" CommentCount="2" />
  <row Id="13054" PostTypeId="2" ParentId="13013" CreationDate="2011-08-01T12:14:36.867" Score="4" ViewCount="" Body="&lt;p&gt;Unfortunately I don't have a copy of &quot;The Grand Design&quot;, so can't be 100% sure of the context, however, from the terminology, it sounds like this may be a reference to &lt;a href=&quot;http://prd.aps.org/abstract/PRD/v28/i12/p2960_1&quot; rel=&quot;nofollow&quot;&gt;Hartle and Hawking's&lt;/a&gt; no-boundary proposal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this scheme, they propose a method for computing what they refer to as the &quot;wavefunction of the universe&quot;.  This wavefunction uses Feynman's path integral to assign probability amplitudes to three-metrics on a three-surface $\Sigma$ bounding a Euclidean spacetime M.  By analytic continuation, the wavefunction can be continued to a function representing a Lorentzian signature spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approach is explained in Hawking's &lt;a href=&quot;http://hawking.org.uk/old-site/pdf/time.pdf&quot; rel=&quot;nofollow&quot;&gt;publicly available lecture&lt;/a&gt;.  There he describes an explicit example where $\Sigma$ is a three-sphere and the Euclidean manifold M is a four-ball.  &quot;On the other side&quot; of the bounding three-sphere $\Sigma$ is Lorentzian de Sitter space.  This model is proposed as a model for a spontaneously created de Sitter universe, and he makes the statement &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Unlike the black hole pair creation, one couldn't say that the de Sitter universe was created out of field energy in a preexisting space.  Instead, it would quite literally be created out of nothing: not just out of the vacuum, but out of absolutely nothing at all, because there is nothing outside the universe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3099" LastActivityDate="2011-08-01T12:14:36.867" CommentCount="3" />
  <row Id="13055" PostTypeId="2" ParentId="13020" CreationDate="2011-08-01T12:41:42.180" Score="1" ViewCount="" Body="&lt;p&gt;German book:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3817116616&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.de/Physik-Bleistift-analytische-Handwerkszeug-Naturwissenschaftlers/dp/3817116616&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It gives a concise introduction into many important topics of physics and motivates to learn math. I read it during my first semesters and it helped to keep me interested in mathematics even though the professor only occasionally motivated his topics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would have liked to read this book when I was in school.&lt;/p&gt;&#xA;" OwnerUserId="4306" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-08-01T12:47:07.337" LastActivityDate="2011-08-01T12:47:07.337" CommunityOwnedDate="2011-08-01T12:41:42.180" />
  <row Id="13056" PostTypeId="2" ParentId="12322" CreationDate="2011-08-01T12:49:41.100" Score="2" ViewCount="" Body="&lt;p&gt;I was at a talk a while back by Gerald Dunne where he talked about the Gelfand-Yaglom theorem. He used it for calculating some Euler-Heisenberg type effective actions. A paper of his with Hyunsoo Min on the subject is &lt;a href=&quot;http://dx.doi.org/10.1088/1751-8113/42/27/272001&quot; rel=&quot;nofollow&quot;&gt;A comment on the Gelfand–Yaglom theorem, zeta functions and heat kernels for PT-symmetric Hamiltonians&lt;/a&gt; and he's got some nice lecture notes: &lt;a href=&quot;http://arxiv.org/abs/0711.1178&quot; rel=&quot;nofollow&quot;&gt;Functional Determinants in Quantum Field Theory&lt;/a&gt; (also see &lt;a href=&quot;http://www.itp.uni-hannover.de/~lechtenf/Events/Lectures/dunne.pdf&quot; rel=&quot;nofollow&quot;&gt;a wider spanning set of lectures&lt;/a&gt; of the same name).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, it's a way of calculating the determinant of a 1-dimensional operator $\det(H)=\prod_i \lambda_i$ with out calculating, let alone multiplying, any of its eigenvalues $H \psi_i = \lambda_i \psi_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To state the original theorem:&lt;/strong&gt; assume that you have a Schrodinger operator (or Hamiltonian) &#xA;$ H = -\frac{d^2}{d x^2} + V(x) $&#xA;on the interval $x\in[0,L]$ with Dirichlet boundary conditions:&#xA;$$ H \psi_i(x) = \lambda_i \psi_i(x) \,,&#xD;&#xA;\quad \psi(0)=\psi(L)=0 \ .&#xD;&#xA;$$&#xA;Then we can compute its determinant by solving the related initial value problem&#xA;$$ H \phi(x) = 0\,, \quad \phi(0)=0\,,\quad \phi&amp;#39;(0) = 1 \ ,$$&#xA;so that&#xA;$$ \det H \approx \phi(L) \,,$$&#xA;where the final result is only $\approx$ as we can only really calculate the ratio of two determinants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This basic result can be generalised to more general boundary conditions, coupled systems of ODEs and higher order linear ODEs.&lt;/p&gt;&#xA;" OwnerUserId="429" LastEditorUserId="429" LastEditorDisplayName="" LastEditDate="2011-08-01T12:55:43.797" LastActivityDate="2011-08-01T12:55:43.797" />
  <row Id="13057" PostTypeId="2" ParentId="13013" CreationDate="2011-08-01T12:54:43.413" Score="-1" ViewCount="" Body="&lt;p&gt;Stephen Hawkings is a physicist and he needs to put any idea of God outside of the domain of physics. (a)&lt;br&gt;&#xA;Georges LeMaitre was (&lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble%27s_law&quot; rel=&quot;nofollow&quot;&gt;WP Hubble-Law&lt;/a&gt;) a belgian &lt;em&gt;priest&lt;/em&gt; and the BBT leaves intact the divine intervention as a justification to the act of the creation of the universe/matter, and it  appraises to the Catholic Church and other religious positions.&lt;br&gt;&#xA;If you beleive in ONE single BB then you are a &lt;em&gt;priviledged observer&lt;/em&gt; of something that begs the question: And before? and why?&lt;br&gt;&#xA;(gravity in the sentence is not about gravity but only: there is a law...)&lt;br&gt;&#xA;The sentence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the universe can and will create itself from nothing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is a circular reasoning as you say, even worst: create itself (like god!); from nothing? (a free lunch!). From 'nothing' we can not get 'something'.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that Spinosa's view is more adequate to a physicist:&lt;br&gt;&#xA;The Universe was, is and will be ALL with no start nor end. The event of matter creation, we use to call it BB, is a repeatable event. &#xA;This way I do not need to call for any entity exterior to the universe.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) - if God exists, or not, is a different question, and is outside of the domain of this site.&lt;br&gt;&#xA;In Physics we can not invoke any transcendental entity to be 'a cause'. &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-08-01T12:54:43.413" CommentCount="4" />
  <row Id="13058" PostTypeId="2" ParentId="12975" CreationDate="2011-08-01T12:57:39.030" Score="-1" ViewCount="" Body="&lt;p&gt;If you split the velocity into two components (as show above), one being radial (I call it $v_r$) and the other in the hoop direction (I call it $v_{\varphi}$), then by inspection (trig properties) you can see the hoop velocity is&#xA;$$ v_{\varphi} = v\,\sin\theta $$ now by nature of angular velocity the hoop velocity is also $$v_{\varphi} = \omega\;r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combine the two to get your answer.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-08-01T12:57:39.030" />
  <row Id="13059" PostTypeId="2" ParentId="13030" CreationDate="2011-08-01T13:18:10.743" Score="0" ViewCount="" Body="&lt;p&gt;I remember distantly about comparing it with pressure-flow relation in pipes: due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Bernoulli%27s_principle&quot; rel=&quot;nofollow&quot;&gt;Bernoulli's principle&lt;/a&gt; the top of the airfoil acts as a thin &quot;pipe&quot;, wherein (if respecting volume/mass conservation) for more fluid to be transported, it flows faster, decreasing the pressure on the pipe (increasing the pressure in the direction of the flow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/8jOT4.png&quot; alt=&quot;thin-thik-pipes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the reason for reduced pressure above the topside of the airfoil and the uplift&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the tilt is used for strearing reasons (amplification, but not the cause of the uplift), so the plain will fly with wings parallel to the ground as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@xkcd - flying on the back may imply much steering in the opposite direction (above)&lt;/p&gt;&#xA;" OwnerUserId="4760" LastActivityDate="2011-08-01T13:18:10.743" CommentCount="2" />
  <row Id="13060" PostTypeId="1" AcceptedAnswerId="13064" CreationDate="2011-08-01T13:37:22.093" Score="11" ViewCount="471" Body="&lt;p&gt;In log-plots a quantity is plotted on a logarithmic scale. This got me thinking about what the logarithm of a unit actually is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I have something with length $L = 1 km$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lg L = \lg km$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that the unit of $\lg L$ is $\lg km$, but I can also say $L = 1000 m$ and now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lg L = 3 + \lg m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't appear to have any units at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This suggests that $\lg km$ and $\lg m$ are actually dimensionless numbers. But wait, I can do this with any unit! Does it actually make sense to talk about the logarithm of a unit, or some other function for that matter?&lt;/p&gt;&#xA;" OwnerUserId="4770" LastActivityDate="2011-08-14T14:42:02.257" Title="What is the logarithm of a kilometer? Is it a dimensionless number?" Tags="&lt;mathematics&gt;&lt;si-units&gt;&lt;units&gt;&lt;dimensional-analysis&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="1" />
  <row Id="13061" PostTypeId="2" ParentId="13060" CreationDate="2011-08-01T13:52:05.240" Score="8" ViewCount="" Body="&lt;p&gt;Yes, logarithms always give dimensionless numbers, but no, it's not physical to take the logarithm of anything with units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, there is always some standard unit. For your example, the standard is the kilometer. Then 20 km, under the log transformation, becomes $\ln(20\;\textrm{km}\;/\;\textrm{km}\;)$. Similarly, the log of 10 cm, with this scale is&lt;br&gt;&#xA;$$\ln(10\;\textrm{cm}\;/10\;\textrm{km}\;) = \ln(10\times 10^{-3} / 10^{3}) = \ln(10^{-5})$$ &lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-08-01T13:52:05.240" CommentCount="4" />
  <row Id="13062" PostTypeId="1" CreationDate="2011-08-01T14:04:38.800" Score="2" ViewCount="184" Body="&lt;p&gt;I am making a 2D driving video game, and I would like to know the &quot;simple&quot; formula for calculating the friction force between the car and the road. I have read lots of friction diagrams involving balls rolling down inclined planes, but I'm having trouble applying it to this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a car driving on a horizontal road, with some coefficient of friction, $\mu$. Its velocity is $v$, mass is $m$, and a constant driving force is accelerating the car at $a\ \mathrm{m/sec}^2$ in the forwards direction. What is the friction force, $F$, that acts on the car in the backwards direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;guess&lt;/em&gt; that the friction force is proportional to the car's speed, but I can't explain this guess. The reason I think that is because of a handful of case studies. Say that the driver's foot is on the pedal such that the car will get up to $16\ \mathrm{m/sec}$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the car's velocity is $0\ \mathrm{m/sec}$, there is no force accelerating it backwards (it is stationary).&lt;/li&gt;&#xA;&lt;li&gt;If the car's velocity is $8\ \mathrm{m/sec}$, the friction force accelerating it backwards must be less (in magnitude) than the driving force, and hence the car keeps accelerating.&lt;/li&gt;&#xA;&lt;li&gt;If the car's velocity is $16\ \mathrm{m/sec}$, the friction force accelerating it backwards is equal (in magnitude) and opposite to the driving force, and hence the car maintains a constant velocity.&lt;/li&gt;&#xA;&lt;li&gt;When the driver releases the pedal, the same friction force accelerates the car backwards, but now there is no driving force, so the car drifts slowly to a halt.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It would therefore make sense that $F \propto -\mu v$, such that in the first case, $F = 0$; in the second case $F = -\frac{1}{2}ma$; in the third and fourth cases, $F = -ma$, where $a$ is whatever acceleration is required to maintain a constant velocity of $16\ \mathrm{m/sec}$. But I can't figure out the exact relationship between $F$ and $v$, or explain why they are proportional.&lt;/p&gt;&#xA;" OwnerUserId="4771" LastActivityDate="2011-09-03T19:11:06.503" Title="Simple friction formula for a car" Tags="&lt;newtonian-mechanics&gt;&lt;friction&gt;" AnswerCount="6" CommentCount="4" />
  <row Id="13063" PostTypeId="2" ParentId="13052" CreationDate="2011-08-01T14:06:54.930" Score="2" ViewCount="" Body="&lt;p&gt;It's not that the earth is expanding; instead the primary effect is that the earth's surface is shrinking. The effect occurs because the earth's surface doesn't remain flat. Instead, it gets tilted and folded. Some parts of the crust get subducted; once they disappear, what's left appears smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over any appreciable distance, rock has good compressive strength but negligible tensile strength. Consequently, when rock is pushed together it becomes thicker and taller (thereby decreasing the surface area; see the demonstration here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Mountain_building&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mountain_building&lt;/a&gt; ). But when rock is pulled apart it instead tends to form cracks that are obviously new crust (and so are not counted when trying to determine if the earth's surface is expanding).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall effect is that the size of the older portion of the earth's surface is smaller than the actual current surface of the earth but this is only an effect of standard geology.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-08-01T14:06:54.930" />
  <row Id="13064" PostTypeId="2" ParentId="13060" CreationDate="2011-08-01T14:42:57.033" Score="9" ViewCount="" Body="&lt;p&gt;This is a fun question.  I have a hard time getting a good grip on the transformation that is $ln$ so I'll write things in terms of exponents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$value = \ln(10\ \mathrm{ km})$$&#xA;$$e^{value} = 10\ \mathrm{ km}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number $e$ is, of course, unit-less.  If I raise a number to a power, what are the permissible units of the power?  If I write $x^2$, I have an intuitive assumption that $2$ has no units, because it is just a count used to express $x \times x = x^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, I have convinced myself of Carl's answer, and I would require a logarithm to have a reference to make sense.  For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{value} = \frac{ 10\ \mathrm{ km} }{1\ \mathrm{ km}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The previous alternative of $e$ raised to a power equaling a physical quantity with real units seems like the perfect example of something that is nonsensical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;log plots&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have another question that stemmed from your question and I will try to answer it here.  I specifically remember taking the derivative of log-log and linear-log plots in engineering classes.  We had some justification for that, but it would appear to be nonsensical on the surface, so let's dive in.  Here is an example of a log-log plot.  I'll show the graph and then offer an equation of the line that is being represented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/HjXoR.gif&quot; alt=&quot;log-log plot&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll start writing things from the basic $y=mx+b$ form, then change things as necessary.  Since I'm using an arbitrary constant, I'll fudge it whenever necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log(p) = a \log(m) + b = a ( \log(m) + b&amp;#39; ) = a \log( b&amp;#39;&amp;#39; m ) = \log( b&amp;#39;&amp;#39;^a m^a )  = \log\bigg( \frac{p_0}{m_0^a} m^a \bigg) $$&#xA;$$p = p_0 \left( \frac{m}{m_0} \right)^a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like magic, a recognizable form comes through.  Observing a linear relation in a log-log plot really means you're observing a power fit, not a linear fit.  A student may still ask &quot;but what are a and b&quot;, which is a bit more difficult.  Firstly, I did no manipulation of $a$, so you can take the meaning directly from the final form, which is to say it's an exponent and thus unitless.  For b:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b = a b&amp;#39; = a \log(b&amp;#39;&amp;#39;) = a \log\bigg( \frac{p_0^{1/a}}{m_0} \bigg) = \log\left( \frac{p_0}{m_0^a} \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that $b$ is also unitless, but it also gives interpretation to $p_0$, which is the reference y-value at some reference x-value ($m_0$).  I'll move on to linear-log plot, or a semi-log scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/xpmPz.jpg&quot; alt=&quot;semi-log plot&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll denote $f$ for &quot;surviving fraction&quot; and $d$ for dose.  The equation for a regression that appears linear on the above plot will be the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log(f) = a d + b$$&#xA;$$f = e^{a d + b} = e^b e^{a d} = f_0 e^{a d}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important to note here that $b$ had dubious units all along, just like in the log-log case, but it doesn't really matter because a more useful form comes out of the mathematics naturally.  The value $f_0$ would be the baseline value (100% in this case) at $d=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary: assuming a linear relation in log plots really makes the assumption that the actual relation follows some &lt;em&gt;nonlinear&lt;/em&gt; form, and the units will work out once you do the mathematics, but the interpretation of the values may be nontrivial.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-08-02T01:22:00.890" LastActivityDate="2011-08-02T01:22:00.890" CommentCount="1" />
  <row Id="13065" PostTypeId="2" ParentId="13062" CreationDate="2011-08-01T14:44:54.350" Score="-1" ViewCount="" Body="&lt;p&gt;Just to clarify, you mean the friction between air and the car (this is called drag, see e.g. stokes law &lt;a href=&quot;http://en.wikipedia.org/wiki/Stokes%27_law&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Stokes%27_law&lt;/a&gt;). There is some friction in the bearings of the car but this is negligible to the resistance of the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest of the answer assumes that you mean drag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are right: the first degree approximation to drag force is that it is proportional to the velocity. You can also use higher order approximations (e.g. F = kv^3), but it makes the calculations quickly very difficult (depending of course how do you want to solve them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How I would write this is (' means derivative respect to time, x is place, m and k are constants, a is acceleration)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f = ma + kv = mx'' + kx' =&gt; ax'' + bx' + c = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is now a general differential equation (a,b,c are constants). This helps later if you have lots of forces in your system.&lt;/p&gt;&#xA;" OwnerUserId="4772" LastEditorUserId="4772" LastEditorDisplayName="" LastEditDate="2011-08-02T14:08:45.220" LastActivityDate="2011-08-02T14:08:45.220" CommentCount="4" />
  <row Id="13066" PostTypeId="1" AcceptedAnswerId="13067" CreationDate="2011-08-01T14:51:44.890" Score="2" ViewCount="133" Body="&lt;p&gt;I have GPS in my phone but I have a very rough idea that how it works. I have checked wikipedia article but that gone over my head (i can not understand that maths and other strange stuff there). Please explain its working in simple words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i think is that it sends signal to satellites, satellite checks the location of sending device and reply with the location of that device.&lt;/p&gt;&#xA;" OwnerUserId="847" LastActivityDate="2011-08-02T17:38:25.877" Title="How GPS device works?" Tags="&lt;gps&gt;" AnswerCount="4" CommentCount="5" />
  <row Id="13067" PostTypeId="2" ParentId="13066" CreationDate="2011-08-01T15:13:59.770" Score="3" ViewCount="" Body="&lt;p&gt;GPS signal is actually a time signal. Each satellite sends data that includes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;time according to the satellite's clock&lt;/li&gt;&#xA;&lt;li&gt;location of the satellite (or data where the location can be calculated)&lt;/li&gt;&#xA;&lt;li&gt;some error correction data&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, the receiver calculates the distance to the satellites: distance = time * velocity. Velocity is a known constant (+- the error correction). The receiver then gets multiple distances to different satellites so it can calculate the its location. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that four distances to known locations (satellites) are enough to determine the location of the receiver. Imagine that you tie four strings to four different stationary objects and other ends of the strings to your finger. There is only one (or none) position where all the strings are tight. &lt;/p&gt;&#xA;" OwnerUserId="4772" LastEditorUserId="4772" LastEditorDisplayName="" LastEditDate="2011-08-01T15:40:46.470" LastActivityDate="2011-08-01T15:40:46.470" CommentCount="7" />
  <row Id="13068" PostTypeId="1" AcceptedAnswerId="13145" CreationDate="2011-08-01T15:30:04.670" Score="1" ViewCount="97" Body="&lt;p&gt;How can I make a single crystal silicon wafer? What are the tools needed? What is the most diffcult part? Can I make it at home or at school lab?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am now interested on Czochralski process where you melt silicon and let it crystallize and cut afterwords.&lt;/p&gt;&#xA;" OwnerUserId="4772" LastActivityDate="2011-08-03T08:56:40.640" Title="How to make a single crystal silicon wafer?" Tags="&lt;semiconductor-physics&gt;&lt;crystals&gt;" AnswerCount="2" />
  <row Id="13070" PostTypeId="2" ParentId="13068" CreationDate="2011-08-01T15:44:37.340" Score="1" ViewCount="" Body="&lt;p&gt;No, you can not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lot of ingredients which hardly may be found in school lab: extremely pure silicon, something where you melt silicon and this silicon remains pure, something which pulls this silicon crystal up slowly and carefully, a piece of crystal which you start from and so on. There is no &lt;strong&gt;most&lt;/strong&gt; difficult part. The process is extremely difficult. That's why people spent few decades on this technology and wafers are still quite expensive.&lt;/p&gt;&#xA;" OwnerUserId="4657" LastEditorUserId="4657" LastEditorDisplayName="" LastEditDate="2011-08-02T05:38:38.700" LastActivityDate="2011-08-02T05:38:38.700" />
  <row Id="13071" PostTypeId="2" ParentId="13039" CreationDate="2011-08-01T16:36:08.483" Score="2" ViewCount="" Body="&lt;p&gt;My answer will mostly be a linkfest because there is substantial information available to answer your questions online.  In fact, there is probably too much.  As other answers have already pointed out, this is a matter of the escape of radioactive material.  The best illustration I have seen to communicate the fundamentals of this can be found in slides 21 - 26 here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://online.itp.ucsb.edu/online/plecture/bmonreal11/&quot; rel=&quot;nofollow&quot;&gt;http://online.itp.ucsb.edu/online/plecture/bmonreal11/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is that the fission products (FP) contain radioactivity, and these FP are contained well within the fuel rods normally.  When something starts to go wrong, successive breaches of the barriers will allow more types of FP to escape, and escape farther.  The important point for accidents like Fukushima is &lt;em&gt;mobility&lt;/em&gt;.  It is no coincidence that Iodine and Cesium are the main culprits - they are some of the most mobile atoms out of the radioactive FPs.  They make it out the engineered barriers and they make it out the furthest.  Xenon is also highly mobile, in fact, more mobile than those.  Any leak at all will virtually guarantee escape of radioactive Xenon, but once it gets into the atmosphere it does not concentrate on the ground and thus won't harm people like Iodine and Cesium will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brief summary:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Since everything in the core of a nuclear reactor is exposed to radiation, there are activation products (radioactive materials created from being exposed to radiation) present in the system, and this is a concern for workers at a plant under normal operation.&lt;/li&gt;&#xA;&lt;li&gt;In a partial meltdown Cs, I, and Tc make it into the cooling water and Xe, Kr, and Radon make it into the steam&lt;/li&gt;&#xA;&lt;li&gt;When emergency venting occurs some FP in the steam may make it to the environment&lt;/li&gt;&#xA;&lt;li&gt;When containment fails, the FP in the water may also make it out, expanding the list to Xe, Kr, Rn, Cs, I, Tc&lt;/li&gt;&#xA;&lt;li&gt;When the fuel catches on fire and the containment is destroyed, all the above can make it into the environment plus some fuel and even more FP&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is a fairly general discussion.  We still see occasional updates on what happened at Fukushima, and the IAEA has tried to put together an authoritative account.  Here is a good summary of the most recent report:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://neutroneconomy.blogspot.com/2011/07/iaea-perspective-on-fukushima.html&quot; rel=&quot;nofollow&quot;&gt;http://neutroneconomy.blogspot.com/2011/07/iaea-perspective-on-fukushima.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Based upon calculations made of the measured activity outside the reactors, the IAEA made an estimate of the relative fraction of material released from each core. Nearly all noble fission product gasses (such as xenon and krypton) were released; however, these are of negligible concern, given the fact that they are chemically inert. Approximately 1% of radioactive iodine was released from Unit 1, with about 0.4 to 0.7% of the iodine released from Units 2 and 3. Finally, for other species (including cesium), less than 1% was determined to have been released into the environment; for Units 2 and 3, this was determined to be around 0.3 to 0.6% of the activity in the core.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a quantification of what got out of containment.  How did it get out of the fuel and then the reactor vessel into containment?  Well, to begin with, the fuel melted, so those barriers are toast.  Then water transport between the vessel and the primary containment is entirely part of the design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/RndsN.png&quot; alt=&quot;Emergency cooling&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this image, the modes of emergency cooling are illustrated, and you can basically see a direct line going from the vessel to the primary containment.  This is how the FPs are transported from the fuel to the containment.  Then obviously the venting and the flagrant break in containment provide a path from containment to the outside.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-08-01T16:36:08.483" />
  <row Id="13072" PostTypeId="2" ParentId="13062" CreationDate="2011-08-01T16:56:43.910" Score="-1" ViewCount="" Body="&lt;p&gt;You will have to track the motion of the wheels. If the wheels roll, then a small constant coefficient of friction applies. When the wheels slip, a larger coefficient of friction applies. Both do not really vary with speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you have hills (up/down) then you have to track how much weight % is on the front or the back to get your total friction/traction available. The rotational motion of the wheels get affected by the engine torque and the friction available.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-08-01T16:56:43.910" />
  <row Id="13074" PostTypeId="1" CreationDate="2011-08-01T17:07:18.400" Score="3" ViewCount="56" Body="&lt;p&gt;This is just a curiosity and you need to bear with me as my math skills were always sub-par and so is my English academic language. My specialty is Electronics but I have always been a programmer.&#xA;Years ago I was working on my diploma paper that dealt with locating leaks using auto-correlation.&lt;BR&gt;&#xA;Basically you have several sound sensors along a pipe which record the sound wave. If a leak appears then you get a peak in the autocorrelation function. Then you perform a  'triangulation' from two or more sound sensors and you can locate the leak pretty accurately. Anyway, my task was mostly to help a PHD student with transposing her algorithms into Matlab.&lt;BR&gt;&lt;BR&gt;&#xA;We have succeeded doing this with air but as soon as we switched to water I could not use any kind of windowing and/or transformation to get relevant peaks. At some point I just hunted blindly for some Matlab functions that would give me a relevant peak somewhere that would correlate with the distance to the leak, but failed.&lt;BR&gt;&lt;BR&gt;&#xA;I cannot to this day understand why we were not able to succeed, though I do assume that is has to do with water's turbulence. The setup was a running tap routed through metal pipes (about 10cm diameter), with about 5 very sensitive sound sensors (going to about 100kHz) placed about 3 meters apart and some taps along the way that simulate leaks. Everything was placed in a phonic-insulated basement, signals were truncated taken to stabilize against footsteps/vibrations so measurements were pretty good.&lt;BR&gt;&lt;BR&gt;&#xA;The multiple question is: what could have been done to achieve the goal or at least get closer to it? Is it really achievable (probably not in real life where you have trucks running above)?&lt;BR&gt;&#xA;Some variables that could influence but for which I lack the knowledge to explain: pipe diameter, water debit rate, tap/leak rate and diameter, choice of liquid, pipe material, sound spectrum and possible frequency filtering.&lt;/p&gt;&#xA;" OwnerUserId="4689" LastActivityDate="2011-08-03T11:02:40.983" Title="Autocorrelation of sound in liquids vs gas" Tags="&lt;fluid-dynamics&gt;&lt;signal-processing&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="13076" PostTypeId="2" ParentId="13074" CreationDate="2011-08-01T17:39:11.010" Score="4" ViewCount="" Body="&lt;p&gt;Did you see the experimental setup? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that a leak in a gas pipeline generally makes a whistling sound whereas leaking liquid will be very quiet. Without sound, the autocorrelation approach doesn't make sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the leak in a pipeline for liquids,&#xA;one could use a marker to find the leak (e.g. add some color). Or one could seal segments of the pipe using inflatable pistons and measure the flow through them. When less flows out than you pump in, you have a leak in the segment.&lt;/p&gt;&#xA;" OwnerUserId="4306" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-08-01T18:55:36.533" LastActivityDate="2011-08-01T18:55:36.533" CommentCount="2" />
  <row Id="13077" PostTypeId="2" ParentId="13060" CreationDate="2011-08-01T18:07:17.327" Score="2" ViewCount="" Body="&lt;p&gt;The closest thing to &quot;Logarithm units&quot; are &lt;a href=&quot;http://en.wikipedia.org/wiki/Decibel&quot; rel=&quot;nofollow&quot;&gt;decibels&lt;/a&gt;, which are 10 times the base 10 logarithm of a ratio.  To put any physical quantity into a decibel-like unit, you need to first divide by some reference quantity.   For example, the &quot;decibel&quot; unit for power  is &quot;dBm&quot;, which is the ratio of the power in question over 1 mW, expressed in dB:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[dBm] = 10 log_10 (Power / (1 mW))&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-08-01T18:07:17.327" />
  <row Id="13078" PostTypeId="2" ParentId="13020" CreationDate="2011-08-01T19:15:10.717" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691125481&quot; rel=&quot;nofollow&quot;&gt;The New Science of Strong Materials or Why You Don't Fall through the Floor&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0306812835&quot; rel=&quot;nofollow&quot;&gt;Structures: Or Why Things Don't Fall Down&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rather overlooked J E Gordon is an excellent teacher, better than the more philosophical Petroski's works &lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-08-01T19:15:10.717" CommunityOwnedDate="2011-08-01T19:15:10.717" />
  <row Id="13079" PostTypeId="2" ParentId="13020" CreationDate="2011-08-01T20:48:55.797" Score="0" ViewCount="" Body="&lt;p&gt;I'm not all that familiar with what Asimov has written about physics, but I used to have his &quot;Asimov on Chemistry&quot; and it was one of the best collections of essays I've ever read. &lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-08-01T20:48:55.797" CommunityOwnedDate="2011-08-01T20:48:55.797" CommentCount="1" />
  <row Id="13080" PostTypeId="2" ParentId="13066" CreationDate="2011-08-01T21:17:32.293" Score="2" ViewCount="" Body="&lt;p&gt;GPS receivers calculate position first in a coordinate system known as ECEF XYZ. That just means Earth Centered Earth Fixed xyz. The WGS84 defined center of the earth is given the coordinates 0,0,0. The x-axis extends from that center, to a point at the equator where the longitude is 0 and of course, in the negative direction to a point at the equator where the longitude is 180 degrees. The y-axis extends from the origin at the center of the earth, to a point at the equator where the longitude is 90 degrees east (for the + direction) and longitude 90 degrees west (for the -direction). Lastly, the z-axis is the same as the earth's polar axis. That is, it extends to the north pole (+) and to the south pole (-). This is all fine and dandy, but if I gave you coordinates of a position in this coordinate system, few people would be able to tell even roughly where that was. We're all much more familiar with latitude, longitude and elevation. So usually, GPS manufacturers will translate the ECEF XYZ coordinates to lat/lon/elev coordinates that we can understand. For latitude and longitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GPS &lt;strong&gt;receivers&lt;/strong&gt; are just that. your phone transmits its location when it does its handshake to the closest cellular antenna.&lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-08-01T21:17:32.293" />
  <row Id="13082" PostTypeId="2" ParentId="13008" CreationDate="2011-08-01T21:26:29.183" Score="0" ViewCount="" Body="&lt;p&gt;First of all, i apologize for my english.&#xA;The equation F=ma is correct only from the point of view of an inertial system. What an inertial system actually is: any system which is in agree with the statement that if no force is acting on a body (your example is the case) then the body moves in a straight line and whit constant velocity.&#xA;So, you are right, there is always an observer wich measures a nonvanishing acceleration (no matter you are in an empty universe or not), but all this observers are not inertial frames, so they have no right to say F=ma.&#xA;Hope i have been of help&lt;/p&gt;&#xA;" OwnerUserId="4778" LastActivityDate="2011-08-01T21:26:29.183" />
  <row Id="13085" PostTypeId="1" CreationDate="2011-08-02T00:24:45.980" Score="2" ViewCount="55" Body="&lt;p&gt;Is time dilation a form of dimensional compactification?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a probe approaches a black hole, toward a point on the equator of the event horizon, does general relativity predict that the time dimension is compactified like the extra dimensions of string theory?  Will this compactification continue until the uncertainty principle becomes significant?&lt;/p&gt;&#xA;" OwnerUserId="4781" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-02T01:05:29.777" LastActivityDate="2011-08-02T08:42:50.633" Title="Time dilation and dimensional compactification" Tags="&lt;general-relativity&gt;&lt;compactification&gt;" AnswerCount="1" />
  <row Id="13086" PostTypeId="2" ParentId="13062" CreationDate="2011-08-02T00:53:09.477" Score="0" ViewCount="" Body="&lt;p&gt;The confusing part in calculating the friction of a car tire on the road is that where the tire contacts the pavement it’s not moving relative to the pavement. For some reason it’s difficult for people to understand. If a car is going 60 m/h where the rubber contacts the pavement the speed is 0 and the top of the tire is going 120 m/h. So the coefficient of friction for the tire pavement contact is static. The coefficient of friction of a dry tire on dray pavement is approximately 1. Conditions: 1) Say you’re going 60 m/h and you slam on the brakes the static contact is broken and you start slipping so the mass of the car comes in to play and you apply kinetic friction formulas.  2) How hard do you step on the brakes before static friction gives in and that is momentum over kinetic friction. 3) If the car is turning on a curve the mass of the car and the radios of the curve will establish the angular momentum and this will give you the speed at which static friction gives and the car starts skidding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now your coefficient of static friction is almost arbitrary as far as the game goes. Consider the extremes a 5” wide car tire cold and smooth pavement Vs a formula One 18” wide sticky rubber and hot on a rough track. That gives you a pretty wide rage to play whit.&lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-08-02T00:53:09.477" CommentCount="1" />
  <row Id="13087" PostTypeId="2" ParentId="13060" CreationDate="2011-08-02T01:21:50.480" Score="9" ViewCount="" Body="&lt;p&gt;Here's one &quot;mathematical&quot; but highly &lt;em&gt;unphysical&lt;/em&gt; answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using that $km\cdot km = (km)^2$ etc, we can formally define arithmetic of numbers with units over a &lt;a href=&quot;http://en.wikipedia.org/wiki/Graded_algebra&quot; rel=&quot;nofollow&quot;&gt;graded algebra&lt;/a&gt; $A = \oplus_{k\in \mathbb{N}} V_k$ where $V_k = \otimes^k V$ where $V$ is treated as a one-dimensional real vector space ($V_0$ is the scalar $\mathbb{R}$). The choice of unit is the choice of a basis vector in $V$. $V_0$ is the pure scalars. So for each choice of a basis vector $v \in V$, we get mapping from the infinite sequence $\mathbb{R}^{\mathbb{N}}\to A$ realizing that sequence via $(r_k) \to [r_k]_v = \sum r_k (\otimes^k v)$. We define multiplication $V_k\times V_{k&amp;#39;} \to V_{k+k&amp;#39;}$ as usual. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(We'll not define the negative power units for now. But they probably can be incorporated analogously.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then formally we can define $\exp: A \to A$ by the power series expansion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \exp a = \sum_{k = 0}^{\infty} \frac{1}{k!} a^k $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a^k$ is defined in the sense of the graded algebra. And there, we have defined what it means for $\exp$ of something with units. Change of base of $\exp$ is handled by $y^a = \exp (\ln(y)\cdot a)$. And similarly the change of units is naturally incorporated, using the fact that a change of basis on a one dimensional real vector space is just multiplication by scalars. In other words, we have $[r_k]_v = [r&amp;#39;_k]_{v&amp;#39;}$ where $r_k = s^kr&amp;#39;_k$ when $v = s v&amp;#39;$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this we can formally invert the power series expansion to find what $\ln$ &quot;should&quot; be. Fix a unit $v$. Take $(r_k) \in \mathbb{R}^{\mathbb{N}}$ and consider $[r_k]_v \in A$. To find $\ln [r_k]_v$ we need to find $(s_k)\in \mathbb{R}^{\mathbb{N}}$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{align}&#xD;&#xA; r_0 &amp;amp; = \exp s_0\\&#xD;&#xA; r_1 &amp;amp; = e^{s_0} s_1 \\&#xD;&#xA; r_2 &amp;amp; = e^{s_0} (s_2  + \frac{1}{2} s_1^2)\\&#xD;&#xA;&amp;amp; \vdots&#xD;&#xA;\end{align} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(we can also use the Taylor expansion of $\ln$ around $1\in\mathbb{R}$ to get the expression for $(s_k)$ in terms of $(r_k)$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, even in this frame work $\ln 1 km$ is still not well-defined: in the image of $\exp$, $r_0$ is necessarily positive. Formally, it is possible to define $\ln (1 km) = $\ln (1 + (1 km - 1))$ as the rather divergent, power series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ln (1 m) = \sum \frac{(-1)^{k+1}}{k} (-1 + 1 (m))^k = \sum \frac{-1}{k} + \sum 1(m) - \sum \frac{k-1}{2} (m^2) + \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now some fun with divergent series: note that $\sum 1 = \sum (-1)^k(-1)^k = \sum (-1)^kx^k |_{-1}$ is the Taylor series expansion of $1/x$ around $x_0 = 1$ evaluated at $-1$, so the second term is nominally $\lim_{x\to 0+} 1/x$. So even if we regularise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lim_{\delta \to 0+} \ln (\delta + 1m) = \lim_{\delta \to 0+} \ln \delta + \delta^{-1} m + \cdots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is still heavily divergent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Observe, that, however, $\ln (1 + 1km) = \sum (-1)^{j+1}/j (km)^j$ is well defined as a formal power series.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So what was the point of this post? This post is primarily addressed at the conclusion that $\log m$ is a &quot;dimensionless number&quot; as stated in the statement of the question. While in the usual arithmetic we are taught that we cannot add apples to oranges, that is only if we take the point of view of trying to add an object in the $\mathbb{Z}$-module of apples to a separate object in the $\mathbb{Z}$-module of oranges. If you are willing to work in the direct sum module of apples $\oplus$ oranges, you can indeed add apples to oranges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, implicitly in asserting that $\log$ makes sense for objects with units, (and similarly that $\exp$ makes sense for objects of units), it is necessary that we already work in a system, that of the graded $\mathbb{R}$-algebra, in which you can add a scalar (an object with no units) to a vector (some object with units). So in asserting that you want to make sense of $\log km$, you cannot conclude from it that $3$ and $\log m$ must have the same units. &lt;/p&gt;&#xA;" OwnerUserId="829" LastEditorUserId="829" LastEditorDisplayName="" LastEditDate="2011-08-02T08:23:35.627" LastActivityDate="2011-08-02T08:23:35.627" />
  <row Id="13088" PostTypeId="1" AcceptedAnswerId="13457" CreationDate="2011-08-02T05:03:39.550" Score="4" ViewCount="52" Body="&lt;p&gt;So as I understand it, localized structures in AdS can wick rotated to dS, the boundary has to be complixified as can be seen &lt;a href=&quot;http://arxiv.org/abs/0710.4334&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Also, uplifting is another technique that can be used to move from AdS to dS. (Thanks to Mitchell Porter for this &lt;a href=&quot;http://www.physicsforums.com/showthread.php?p=3353934&quot; rel=&quot;nofollow&quot;&gt;suggestion&lt;/a&gt;) My question is that just as an exercise, is it possible to move from dS to AdS, I don't know the name for such a process, it is in some sense &quot;moving down&quot; from dS to AdS. Take the following example, suppose I have a mapping of a scalar field from boundary to bulk in AdS, now that can be uplifted to dS, it won't hold make much sense anymore since AdS allows for SUSY however dS doesn't but let's say for a moment that it can be done. Now consider the opposite case, we have a scalar field mapping in dS, can that be brought down to AdS? I think it can be, and this should be actually easier since AdS allows for SUSY which should make the mapping a little easier. The lack of a concrete example for dS/CFT makes it (about the mapping which we considered in our hypothetical case for dS) difficult to ask this question but does it seem plausible that if such a mapping were to be constructed, it can be brought down to AdS? Thanks. &lt;/p&gt;&#xA;" OwnerUserId="4719" LastActivityDate="2011-08-12T17:44:46.133" Title="AdS to dS uplifting and its opposite" Tags="&lt;quantum-gravity&gt;&lt;ads-cft&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13089" PostTypeId="2" ParentId="13085" CreationDate="2011-08-02T05:41:12.130" Score="4" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is time dilation a form of dimensional compactification?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As far as I know, no, it isn't. Compactified dimensions have a finite extent, and as used in string theory they're typically periodic. For example, the Earth's surface has two compactified dimensions: you can't travel more than a certain distance in any one direction in that two-dimensional surface without returning to where you started. But that's not true of the time dimension, which is generally considered to be infinite. There is a sense in which time ends at a black hole singularity, much like the set of negative numbers ends at zero, but still, it extends infinitely far back in the other direction, and infinitely far forward outside the black hole.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-02T08:42:50.633" LastActivityDate="2011-08-02T08:42:50.633" CommentCount="2" />
  <row Id="13090" PostTypeId="2" ParentId="13039" CreationDate="2011-08-02T07:09:32.893" Score="2" ViewCount="" Body="&lt;p&gt;There are at least five ways that radioactive material can escape from a nuclear power station, beyond being removed by human action. All five depend by definition on a breach of every containment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Four of these releases occurred at Fukushima. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one that didn't occur at Fukushima was an explosion that takes out the reactor vessel itself - the Chernobyl phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The four that occurred at Fukushima are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Water leakage from the reactor vessel&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When water-levels in the reactor vessel drop, exposing the fuel rods to the air, the fuel-rods melt down. The resulting corium damages the reactor vessel, leading to a breach of that vessel, and the leakage of cooling water carrying radioactive particles from the damaged and (partially) melted fuel-rods, from the reactor vessel.  Further breaches in the rest of the containment (see below) results in the water entering the outside environment. The evidence is that this happened at Fukushima reactor 1: not a &quot;major&quot; breach by some standards - i.e. not enough for a major loss of corium - but bad enough to create a significant discharge of highly radioactive water. This may also have happened at reactors 2 and 3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Vapour loss from the reactor vessel&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When water-levels in the reactor vessel drop, exposing the fuel rods to the air and to steam, the reaction between the zirconium cladding of the fuel-rods, as they crack, and the steam and water, at high temperatures, produces hydrogen gas. If the venting of this hydrogen gas is done without injecting lots of nitrogen into the vessel first, then when the very hot hydrogen is vented, carrying out some steam vapour and radioactive materials from the damaged fuel rods, it explodes in the air, breaching the rest of the containment, and putting the radioactive materials into the outside air; this happened at Fukushima reactors 1 and 3. As @whoplisp points out, this can happen through unintentional venting, where the pressure inside the reactor vessel lifts the drywell head - the Brunswick phenomenon. (Subsequent venting at Fukushima has only happened after nitrogen injection)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Water loss from the coolant circuit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see from the diagram in the answer from @Zassounotsukushi , there are pipes - the coolant circuit - between the reactor vessel and the rest of the reactor. Breaks in these pipes, and loss of containment from explosions (as above), can result in water loss from the coolant circuit; when the fuel rods have been damaged, this water carries radioactive material with it. This happened in reactors 1 and 2, and possibly reactor 3, at Fukushima.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Breach of containment in the fuel-storage pond&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When water-levels in the fuel-storage pond drop, exposing the fuel rods to the air; and the containment building is breached, the fuel rods get damaged and hydrogen gets produced; as above, this can result in an explosion breaching all containment, and an atmospheric distribution of radioactive material; this happened at Fukushima reactor 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately, for most of the time of the release at Fukushima, the wind was blowing out to sea, resulting in the radioactive material mostly getting deposited in the Pacific Ocean rather than on land. Total estimates of the amount of material released vary hugely, and we may never know the actual values. Estimates vary between less than 1% of the radioactive isotopes present; and 50% of the Chernobyl release.&lt;/p&gt;&#xA;" OwnerUserId="4066" LastActivityDate="2011-08-02T07:09:32.893" />
  <row Id="13091" PostTypeId="1" CreationDate="2011-08-02T08:03:39.807" Score="3" ViewCount="78" Body="&lt;p&gt;I appreciate your patience to my neophyte question. I am working on my dissertation in philosophy (which has nothing or little to do with physics) about the &quot;problem of naming.&quot; Briefly what I am arguing is that when we name something, we stop it from being anything or everything else. It is a phenomenological question and has a lot to do with language as an object.&#xA;My question for you is that, is it true that all matter was somehow formed in the Big Bang or in those famous three minutes following? I think I understand that helium and hydrogen were formed and are they then to be considered the basis of all matter today? A friend said to me a long time ago that we are made of the same atoms that were present at the Big Bang; could this possibly be true? (And how wonderful if it is...)&#xA;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="4783" LastEditorUserId="1876" LastEditorDisplayName="" LastEditDate="2011-08-02T08:33:12.153" LastActivityDate="2011-08-02T16:10:17.093" Title="Creation of matter in the big bang" Tags="&lt;big-bang&gt;" AnswerCount="2" />
  <row Id="13092" PostTypeId="2" ParentId="13052" CreationDate="2011-08-02T08:13:33.657" Score="-1" ViewCount="" Body="&lt;p&gt;I don't know as a fact if the earth is expanding or not. But way is it so hard to believe it is. On the same principal that ice expands when it mixes whit gases. Lava expands when it cools specialty when water is present. It fells whit gas bobbles and increases in volume 3 fold. On the same principal way is it not &lt;strong&gt;possible&lt;/strong&gt; for the earth crust to expand. I'm not saying that the earth mass is increasing just the volume.&lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2011-08-02T08:13:33.657" CommentCount="1" />
  <row Id="13093" PostTypeId="2" ParentId="13091" CreationDate="2011-08-02T08:31:58.270" Score="1" ViewCount="" Body="&lt;p&gt;Hydrogen and helium (and &lt;a href=&quot;http://en.wikipedia.org/wiki/Lithium#Occurrence&quot; rel=&quot;nofollow&quot;&gt;lithium&lt;/a&gt;) were formed in the big bang, but other elements arose later through a variety of different routes, including stellar and supernova nucleosynthesis, as well as fission and spallation. So only some of your atoms are truly primordial.&lt;/p&gt;&#xA;" OwnerUserId="1876" LastActivityDate="2011-08-02T08:31:58.270" />
  <row Id="13094" PostTypeId="2" ParentId="13091" CreationDate="2011-08-02T09:05:06.323" Score="2" ViewCount="" Body="&lt;p&gt;The closer to the Big Bang you look, the more fundamental are the particles that constitute the universe (quarks, leptons, etc.) - go far back enough and you reach the point where only a field description makes sense because of the curvature of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But after the 3rd minute and up to around the 20th minute, the universe had cooled down enough for nuclei to form and yet was hot enough for fusion to occur. This was the period of nucleosynthesis when the first elements in the periodic table were forged. Most of the nuclei formed were hydrogen and helium and their various isotopes but there were also trace amounts of lithium and beryllium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is confirmed today in the precise ratio of elements observed in the universe with the significant abundance of hydrogen and helium, and is one of the best numerical confirmations of the Big Bang theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rocky planets and humans and what not however require higher elements like carbon, nitrogen, oxygen and silicon to form. These were &lt;em&gt;not&lt;/em&gt; formed in the Big Bang like H and He. They were forged at the centres of stars and dispersed in violent supernovae explosions that took place millions to billions of years after the Big Bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That still is pretty wonderful - that every atom in our body has its origins in either stardust or the Big Bang.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastActivityDate="2011-08-02T09:05:06.323" />
  <row Id="13095" PostTypeId="2" ParentId="13043" CreationDate="2011-08-02T09:54:31.907" Score="0" ViewCount="" Body="&lt;p&gt;The &quot;cold fusion&quot; of Fleischer et. al. was discussed here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/3799/why-is-cold-fusion-considered-bogus&quot;&gt;Why is cold fusion considered bogus?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the things written there fit ot will fit here. &#xA;Nickel (frequently used as a hydrogenation catalyst in chemisty) &#xA;has some similarities to palladium with respect to hydrogen &#xA;incorporation. Usually this nickel is made by a recipe &#xA;from  Raney/Urushibara, which means that some aluminium in  reactive &#xA;(to water or oxygen) dispersion remains in this nickel. That might explain &#xA;some of the energy &quot;in excess&quot;. &#xA;But most important is : &quot;Wish was father of the thought&quot;-effect.  &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-08-02T09:54:31.907" CommentCount="1" />
  <row Id="13096" PostTypeId="2" ParentId="13062" CreationDate="2011-08-02T10:12:23.933" Score="1" ViewCount="" Body="&lt;p&gt;The &quot;friction&quot; of the tire (better named &quot;resistance to rolling&quot;) is 99 % by deformation of the tire and mechanical hysteresis of that deformation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the &quot;friction&quot; of a car is not a simple thing like &#xA;Foucault friction. Most of it is fluid friction.&#xA;All bearings (wheels, gear, pistons, crank etc) is ball bearings &#xA;or journal bearings,  all have lubrication by some paste/liquid &#xA;fat/oil. &quot;Friction&quot; is restricted to the lubricant, otherwise &#xA;the bearing will fail soon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that reasons a car has:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;aerodynamic friction (complicated, very small first, above 55 mph &#xA;the most important an then prop to square of velocity) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Liquid friction in bearings prop. to velocity&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Deformation of tire (&quot;flexing resistance&quot;) &#xA;This is hysteresis of the rubber, as a first approximation &#xA;I'd assume proportional to velocity. (If You want to know that &#xA;precisely, ask at an racing forum or a producer of tires) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-08-02T14:33:19.903" LastActivityDate="2011-08-02T14:33:19.903" CommentCount="2" />
  <row Id="13097" PostTypeId="2" ParentId="13074" CreationDate="2011-08-02T10:22:10.867" Score="1" ViewCount="" Body="&lt;p&gt;As long as the leak is as narrow to give rise of &#xA;capillary flow (potential currents) only, there will be &#xA;no &quot;sound&quot; at all. &#xA;For liquids this is a severe disadvantage, for most &#xA;(cheap and not very harmful) gases such leaks can be tolerated. &#xA;If gases are expensive/very dangerous You will have the pipes &#xA;open to inspection by leak detectors. &#xA;Gas pipes under streets or generally under soil are special &#xA;and sometimes very dangerous :=(&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-08-03T11:02:40.983" LastActivityDate="2011-08-03T11:02:40.983" CommentCount="1" />
  <row Id="13098" PostTypeId="1" CreationDate="2011-08-02T10:28:54.940" Score="1" ViewCount="62" Body="&lt;p&gt;I don't get &lt;a href=&quot;http://en.wikipedia.org/wiki/Corona_discharge&quot; rel=&quot;nofollow&quot;&gt;corona discharge&lt;/a&gt;. There's an electrode (say naked high-voltage wire) and a current develops from that wire and ionizes gas (air) on its way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But where does this current go? What's the destination of that discharge? &lt;/p&gt;&#xA;" OwnerUserId="3816" LastEditorUserId="176" LastEditorDisplayName="" LastEditDate="2011-08-15T06:15:27.180" LastActivityDate="2011-08-15T06:15:27.180" Title="What's the second terminal in corona discharge?" Tags="&lt;electricity&gt;&lt;plasma-physics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="13099" PostTypeId="2" ParentId="13098" CreationDate="2011-08-02T10:35:25.537" Score="1" ViewCount="" Body="&lt;p&gt;As always in &quot;electrostatics&quot; the second terminal is &#xA;&quot;the rest of the world&quot;, often called &quot;earth&quot; or &quot;ground&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because this is somewhat unreliable, You should &#xA;provide a definite counterterminal made from metal,&#xA;when dealing with DC corona.&#xA;Have You ever had a look into a Xerox type copier &#xA;or a &quot;laser&quot; printer? There a professional coronostats, &#xA;which might give You some ideas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of AC (high frequency) coronas the &quot;way&quot; back is &#xA;possible as a capacitive current, this makes things easier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I don't get corona discharge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does that mean You do not understand it, or does some generator of Yours not work?&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-08-02T18:12:51.710" LastActivityDate="2011-08-02T18:12:51.710" />
  <row Id="13100" PostTypeId="1" AcceptedAnswerId="13126" CreationDate="2011-08-02T11:30:23.227" Score="8" ViewCount="144" Body="&lt;p&gt;Is there some condition in the N=1 SUSY algebra telling that the spin of the superpartners of gauge bosons (either for colour or for electroweak) must be less than the spin of the gauge boson? &#xA;I am particularly puzzled because sometimes a supermultiplet is got from sugra that contains one spin 2 particle, four spin 3/2, and then some spin 1, 1/2 and 0. If this supermultiplet is to be broken to N=1 it seems clear that the graviton will pair with the gravitino and the rest of spin 3/2 should pair with spin 1, so in this case it seems that a superpair (3/2, 1) is feasible. Why not in gauge supermultiplets?&lt;/p&gt;&#xA;" OwnerUserId="1335" LastActivityDate="2011-08-02T19:42:47.657" Title="Why must gluinos be spin 1/2 instead of 3/2?" Tags="&lt;supersymmetry&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="13101" PostTypeId="1" AcceptedAnswerId="13137" CreationDate="2011-08-02T11:39:45.340" Score="3" ViewCount="102" Body="&lt;p&gt;A gluon string is a particular kind of open string terminated in two particles which are the sources for the field. Is it possible to have a similar arrangement with gluinos? At first glance, it seems to me that such object could not exist, or at least not as an extended object: you need bosons to mediate between the two sources, and you need bosons to build a classical field extended in space.&#xA;On other hand, the fact that in the world-sheet such structure is just an object with fermionic coordinates --and without bosonic coordinates-- could be telling that you can build it, but it is not extended in space, just a point. But I have never read of such description, so surely intuition is failing here. &lt;/p&gt;&#xA;" OwnerUserId="1335" LastActivityDate="2011-08-06T18:42:54.157" Title="Is there a SQCD gluino string, similar to the gluon string?" Tags="&lt;string-theory&gt;&lt;supersymmetry&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13102" PostTypeId="1" CreationDate="2011-08-02T12:01:36.243" Score="0" ViewCount="28" Body="&lt;p&gt;Is it possible to use the data that is readily available (e.g. of the &lt;a href=&quot;http://www2.jpl.nasa.gov/srtm/&quot; rel=&quot;nofollow&quot;&gt;Shuttle Radar Topography Mission&lt;/a&gt;) to predict the flooding of a region?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be necessary to use the refined geoid of Gravity Probe B?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there free software that can do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I always wondered about this and the current events near Fukushima&#xA;convinced me to ask this question:&#xA;&lt;a href=&quot;http://search.japantimes.co.jp/cgi-bin/nn20110730a5.html&quot; rel=&quot;nofollow&quot;&gt;http://search.japantimes.co.jp/cgi-bin/nn20110730a5.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4306" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-08-02T13:46:41.533" LastActivityDate="2011-08-02T13:46:41.533" Title="Flood prediction data and code" Tags="&lt;geophysics&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="13103" PostTypeId="1" AcceptedAnswerId="13105" CreationDate="2011-08-02T12:42:56.660" Score="2" ViewCount="86" Body="&lt;p&gt;I am trying to teach myself some particle physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are too many particles and its too much for me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hated biology just because of this sort of stuff. Too many names and it was all Greek to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a good cheat sheet/ reference sheet of elementary particles? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will also be very helpful if you share how you people manage to remember these things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the first time I have ever hated studying physics. :(&lt;/p&gt;&#xA;" OwnerUserId="126" LastEditorUserId="126" LastEditorDisplayName="" LastEditDate="2011-08-02T12:49:07.747" LastActivityDate="2011-08-02T13:25:54.943" Title="Cheat sheet of elementary particles" Tags="&lt;particle-physics&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="13104" PostTypeId="2" ParentId="13102" CreationDate="2011-08-02T12:44:55.963" Score="1" ViewCount="" Body="&lt;p&gt;I don't think the height accuracy is very good, especially in dificult targets like a city (walls, buildings, bridges etc)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relative height accuracy is only about 10m - even in good conditions. &lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-08-02T12:44:55.963" CommentCount="1" />
  <row Id="13105" PostTypeId="2" ParentId="13103" CreationDate="2011-08-02T13:25:54.943" Score="2" ViewCount="" Body="&lt;p&gt;There isn't that much to remember as far as fundamental particles are concerned.&#xA;We have matter particles and field particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matter particles are quarks and leptons and the latter include electrons and neutrinos. Both quarks and leptons come in pairs (up/down quark, electron/neutrino, etc.) which means they are charged under the weak force. Quarks are moreover colored and form RGB triplets which means they are charged under the strong force.&#xA;That's pretty much it except for unknown reasons there are three copies of each particle with identical properties but distinct masses (e.g. electron/muon/tauon). You don't really need to know much about the two heavier families since these are unstable (decaying into the lower families which are stable since there is nothing else they could decay into; at least besides the down quark that can decay into up quark; this is the famous radioactive beta decay) and you need expensive equipment (such as LHC or jets from galaxies) to produce them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the field particles, there is one type for every force although it gets a bit more tricky with weak and strong forces. For EM we have the usual photon. Weak force will give us three more -- the $W^{\pm}$ and $Z$ which are massive -- and finally the strong force has eight massless gluons (of various colors).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally there may or may not be the Higgs particle (or more of them), superpartners, etc. But all of this is as yet unconfirmed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there, that's all there is to particle zoo. It's only pain when you try to learn it by heart. But if you instead try to understand the concepts related to these particles (role of the weak particles and neutrinos in the beta decay, muons in cosmic rays, features of quantum chromodynamics, etc.) then it should be fun.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-08-02T13:25:54.943" />
  <row Id="13106" PostTypeId="1" CreationDate="2011-08-02T14:11:10.600" Score="0" ViewCount="33" Body="&lt;p&gt;Hey i am by no means a scientist but i have a idea for a art work using smoke.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically what i want to know is whether it is possible to circulate smoke by means of a pump through a transparent box. If the box is completely sealed will the smoke disappear within time? When the smoke is pumped out and back in again, will it be have the same consistency as it had before it was circulated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advanced&lt;/p&gt;&#xA;" OwnerUserId="4785" LastActivityDate="2011-08-02T15:04:38.033" Title="Circulate smoke in a closed system" Tags="&lt;physical-chemistry&gt;&lt;chemical-compounds&gt;" AnswerCount="2" />
  <row Id="13107" PostTypeId="2" ParentId="13106" CreationDate="2011-08-02T14:23:12.407" Score="1" ViewCount="" Body="&lt;p&gt;This is possible, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but not for long time. Smoke is a dispersion of solid particles &#xA;in a gas (in practical smoke there are liquid droplets too).&#xA;That particles will settle/adhere to any wall with time, &#xA;making the wall turbid/smeary  and the air will become more transparent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time for that is ruled by ratio of container/hoses/pipes &#xA;surface to gas volume. The nature of that smoke plays a role too &#xA;(particle size, polarity etc)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best thing is to my opinion &quot;smoke&quot; generated by some lumps of solid &#xA;carbon dioxide thrown in water. This looks rather smoke-like but &#xA;consists of water droplets only, thus it will not leave hard to clean &#xA;residue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This carbon dioxide method is eg used on stage.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-08-02T14:48:44.553" LastActivityDate="2011-08-02T14:48:44.553" CommentCount="8" />
  <row Id="13108" PostTypeId="2" ParentId="11887" CreationDate="2011-08-02T14:30:47.227" Score="0" ViewCount="" Body="&lt;p&gt;The Lagrangian for a relativistic particle moving in an electromagnetic field is given by $$L = - \frac {mc^2} \gamma + q~( \vec u(t)\cdot \vec A(\vec x,t) -\Phi(\vec x,t))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conservation of the canonical momentum requires the Lagrangian to be independent of coordinates, whereas in general it isn't. We therefore need to show it's possible to create this independence with some suitable gauge transformation of the potentials $\Phi$ and $A$. For the case of a static electric field with $\vec A=0~$, the EM gauge transformations  $$\vec A= \nabla\lambda(\vec x,t),\quad\phi(\vec x)\to\phi(\vec x) - \frac d {dt} \lambda(\vec x,t) $$ gives the Lagrangian $$L = -\frac{m_0c^2}\gamma + q~(  \vec u\cdot\nabla\lambda(\vec x,t) - \phi(\vec x) + \frac d {dt}\vec\lambda(\vec x, t))$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It isn't generally possible to make this expression independent of coordinates for every $\Phi(\vec x)$ including that for a static Coulomb electric field. However, lets take the simplest case of an electric field directed along the x axis giving a potential and defining a gauge as$$\Phi(\vec x) = -Ex\qquad \lambda(\vec x, t) = -Ext$$to give a coordinate independent Lagrangian$$L = -\frac{m_0c^2}\gamma - q\vec u\cdot Et$$ and the conserved canonical momentum $$\frac{\partial L}{\partial\vec u} = (\gamma mu_x - qEt, 0, 0)$$ Since this is constant over time, we're finally left with just the Lorentz force on a relativistic charge in a static electric field along the x direction $$\frac d {dt}\gamma m\vec u = q(E,0,0)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be generalised to any time and coordinate dependent electric field from the principle of locality, causality and memorylessness:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The dynamics of a charge depends upon the field at its coordinates only&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The dynamics of a charge depends upon the field at time t, and not the past or the future.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore, for any electric field $\vec E(\vec x,t)$, the Lorentz force and associated conserved expression is$$\frac d {dt}\gamma m\vec u = q\vec E(\vec x,t),\qquad\gamma m\vec u - q\int\vec E(\vec x,t)dt$$ and  can finally conclude using the gauge $\lambda(\vec x, t) = -\int\Phi(\vec x, t)dt$, canonical momentum is conserved for a relativistic charge in a electric field $\vec E(\vec x, t)~$ and given by $$\frac{\partial L}{\partial\vec u} = \gamma m\vec u - q\int\vec E(\vec x,t)dt$$&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="1345" LastEditorDisplayName="" LastEditDate="2011-08-26T15:00:32.293" LastActivityDate="2011-08-26T15:00:32.293" />
  <row Id="13109" PostTypeId="1" CreationDate="2011-08-02T15:00:30.540" Score="0" ViewCount="34" Body="&lt;p&gt;I have a system with a number of measurables (in time). Some measurables are discrete some are continuous (within the measurement accuracy). How can I determine whether my system experiences criticality or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for many different ways to (dis)proof criticality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See here for critical phenomena &lt;a href=&quot;http://en.wikipedia.org/wiki/Critical_phenomena&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Critical_phenomena&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4772" LastActivityDate="2011-08-02T15:00:30.540" Title="Identifying a critical phenomena?" Tags="&lt;statistical-mechanics&gt;&lt;condensed-matter&gt;&lt;renormalization&gt;&lt;ising-model&gt;" CommentCount="3" />
  <row Id="13110" PostTypeId="2" ParentId="13106" CreationDate="2011-08-02T15:04:38.033" Score="1" ViewCount="" Body="&lt;p&gt;Antony Gormley did such an installation at the Hayward Gallery, called &lt;a href=&quot;http://ticketing.southbankcentre.co.uk/minisites/gormley/&quot; rel=&quot;nofollow&quot;&gt;&quot;Blind Light&quot;&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/8QzzY.jpg&quot; alt=&quot;people walking around inside glass box full of fog&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can recirculate the dry ice or (in the above case) fog endlessly; but in the circuit you'll need a chiller, because the pump's action will steadily heat the gas.  Antony Gormley used an ultrasonic humidifier in the circuit, to keep Blind Light constantly full of fog.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being in it was a truly remarkable experience: visibility was less than one metre. And being outside it, watching the silhouettes materialise from nowhere and seeing them groping towards the glass wall was eerie too.&lt;/p&gt;&#xA;" OwnerUserId="4066" LastActivityDate="2011-08-02T15:04:38.033" CommentCount="4" />
  <row Id="13111" PostTypeId="2" ParentId="12898" CreationDate="2011-08-02T15:48:04.163" Score="2" ViewCount="" Body="&lt;p&gt;Under a conformal transformation $f$  we have that the coordinate $x_i \rightarrow f_i(x)$, the potential $\phi \rightarrow \phi$ and therefore the derivative $\partial_i \phi \rightarrow \frac{\partial f_j}{\partial x_i} \partial_j \phi$. A tangent vector $t$ to some surface $S$ gets mapped to a tangent vector to $f(S)$ as always, and because the transformation is conformal the normal to a surface $n$ gets mapped to a vector proportional to the normal $n&amp;#39;$ on the surface $f(S)$ since $t\cdot n = 0$ is preserved by conformal transformation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual boundary conditions  $\phi(S) = \phi_0$ and that $\phi(S)$ is continuous are preserved, since $\phi$ transforms trivially. A different condition is that $\epsilon_{+}n\cdot \partial \phi_{+} - \epsilon_{-}n\cdot \partial \phi_{-} = 0$, where $\phi_{\pm}$ is the limit of $\phi$ approaching the surface from either direction and  $\epsilon_{\pm}$ are constants. This gets mapped to the equation $\Omega^2\times\left(\epsilon_{+}n&amp;#39;\cdot \partial \phi_{+} - \epsilon_{-}n&amp;#39;\cdot \partial \phi_{-}\right) = 0$ so that is preserved as well (where $\Omega$ is the scale factor of $f$). The same occurs for the part of the $\partial_j \phi$ tangent to the the surface as well.  A sheet of charge $\sigma$ produces the b. c. $\epsilon_{+}n\cdot \partial \phi_{+} - \epsilon_{-}n\cdot \partial \phi_{-} = \sigma$ which is preserved as well with the correct transformation $\sigma \rightarrow \Omega^2 \sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So all the usual b. c. of electrostatics work well with conformal transformations.&lt;/p&gt;&#xA;" OwnerUserId="3970" LastEditorUserId="3970" LastEditorDisplayName="" LastEditDate="2011-08-02T17:50:09.703" LastActivityDate="2011-08-02T17:50:09.703" CommentCount="3" />
  <row Id="13112" PostTypeId="2" ParentId="12814" CreationDate="2011-08-02T15:55:05.420" Score="-1" ViewCount="" Body="&lt;p&gt;If you want to focus the light of a 1 mm wide emitter to 0.5 mm wide spot, you can achieve that by making the lens to move towards the emitter. (at some fraction of speed of light)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an electron gun we want to make the initial random speeds of the electrons very low, by making the electrode as cool as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we could accelerate the magnetic lens towards the electrons, but it's easier to accelerate the electrons towards the magnetic lens. Anyway, the principle of relativity says we can achieve the same focusing effect either way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this little analogy thermal electrons are analogous to thermal photons, and the speed of the lens relative to the light emitter is analogous to the speed of the magnetic lens relative to the electron emitter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick course in information theory and thermodynamics of lenses and screens:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lens-screen combination converts direction information into position information, an eye is an example of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lens-screen combo converts order in the directions of particles into order of marks on the screen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are easy ways to remove disorder in directions of particles:&#xA;a: let the particles with more aberrant directions leave the group&#xA;b: accelerate all particles the same amount&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lenses are used to produce positionally ordered (narrow) beams, because it is easy to produce directionally ordered beams.&lt;/p&gt;&#xA;" OwnerUserId="4017" LastEditorUserId="4017" LastEditorDisplayName="" LastEditDate="2011-08-03T10:38:41.737" LastActivityDate="2011-08-03T10:38:41.737" />
  <row Id="13115" PostTypeId="1" CreationDate="2011-08-02T16:22:28.460" Score="0" ViewCount="42" Body="&lt;p&gt;Some people claim the wave function is subjective, and is the quantum Bayesian measure of subjective belief. This begs the question; who or what is the self having the beliefs? What are beliefs anyway?&lt;/p&gt;&#xA;" OwnerUserId="4788" LastActivityDate="2011-08-02T16:22:28.460" ClosedDate="2011-08-02T16:48:08.493" Title="If the wave function is subjective, who or what is having the beliefs?" Tags="&lt;quantum-interpretations&gt;" CommentCount="4" />
  <row Id="13116" PostTypeId="2" ParentId="13066" CreationDate="2011-08-02T16:33:20.363" Score="1" ViewCount="" Body="&lt;p&gt;I think it should be mentioned that the (atomic) clocks on the satellites are running (ticking) faster than the clocks on the Earth's surface due to the relativistic effects - time dilation due to the satellite's motion relative to the Earth (&lt;a href=&quot;http://en.wikipedia.org/wiki/Time_dilation&quot; rel=&quot;nofollow&quot;&gt;Time dilation&lt;/a&gt;) and due to the difference between the gravitational potential on satellite's orbit and at Earth's surface (&lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_time_dilation&quot; rel=&quot;nofollow&quot;&gt;Gravitational time dilation&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineers must take these small effects into account, otherwise the GPS wouldn't give you your correct location (the time difference on clocks will continuously grow with time so the position errors will grow too and the GPS will become worthless as a navigational system).  &lt;/p&gt;&#xA;" OwnerUserId="3066" LastEditorUserId="3066" LastEditorDisplayName="" LastEditDate="2011-08-02T16:49:23.017" LastActivityDate="2011-08-02T16:49:23.017" CommentCount="2" />
  <row Id="13117" PostTypeId="1" CreationDate="2011-08-02T16:44:15.813" Score="3" ViewCount="133" Body="&lt;p&gt;A little thought experiment, similar to &lt;a href=&quot;http://physics.stackexchange.com/questions/5846/what-is-more-efficient-add-milk-and-then-heat-up-coffee-in-the-microwave-or-mi&quot;&gt;this one&lt;/a&gt;: Imagine you are making a cup of tea when the door bell rings. You've poured the boiling water into a cup with a teabag in it. As you're just about to pour milk in to the mix, the door bell rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: what will keep the resulting drink hottest after my chat with the person at the door: quickly pouring the milk in, or leaving it in the carton until after the interruption ends?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being a bit of a mathematician I'm happy to have the questions answered with a formula in terms of $t$ the length of the interruption, $m$, the temperature of the cold milk, $w$ the temperature of the hot water, $a$ the ambient temperature etc. I guess you'll have to use a model of how an insulated body loses/gains heat... feel free to use simplifying assumptions like the environment stays the same temperature etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My starting point from a physics point of view, is that heat is lost (or gained) proportional to the temperature difference between the object and the environment. So I imagine that would create some sort of exponential curve?&lt;/p&gt;&#xA;" OwnerUserId="4786" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-05T00:48:13.550" LastActivityDate="2011-08-05T00:48:13.550" Title="Does tea stay hotter with the milk in it?" Tags="&lt;heat&gt;&lt;temperature&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="13118" PostTypeId="2" ParentId="13117" CreationDate="2011-08-02T17:18:08.640" Score="6" ViewCount="" Body="&lt;p&gt;Pouring the milk in early on will leave the tea warmer than keeping it hot, then later pouring the milk in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your assumptions (in the final paragraph) would be correct for most kinds of conductive heat transfer, but the dominant cooling process for an uncovered cup of tea will be evaporation.  The details of the rate of evaporation will depend on how tricky things like air circulation, but the cooling power typically increases faster than linearly with the temperature.  That faster-than-linear dependence means you'll win big by adding the milk while the tea's hot (dmckee's point notwithstanding).&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-08-02T17:18:08.640" CommentCount="3" />
  <row Id="13119" PostTypeId="2" ParentId="13066" CreationDate="2011-08-02T17:38:25.877" Score="1" ViewCount="" Body="&lt;p&gt;There are 24 satellites around the world each of them transmit:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The current time on their clock&lt;/li&gt;&#xA;&lt;li&gt;Their location relative to earth&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The distance between us and a satellite must be equal to the difference in time it took to receive the data sent by the satellite multiplied by speed of light(c). d=(t2-t1)*c. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember how in math the distance between two points is d = sqrt((x1-x2)^2+(y1-y2)^2)). The same equation is used but now we have x,y,z instead of just x,y. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looks like we have 3 unknowns which is the position of the receiver (x,y,z). However there is a bias in the time due to relativity and computational delays. Since the value of c is so big, it can create large errors, we treat this bias as an unknown as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have unknowns we need four equations to solve the problem. Therefore we use the distance equation with 4 different satellites. We get four non-linear equations, which is a minimum to lock your position, but if there are more satellites available more equations can be used to give better results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These equations are solved iteratively since they are non-linear. You cannot used the elimination and back-substitution method(which is taught in school math classes). The processor repeatedly solves the set of equations using an initial guess and each time using results of the previous iteration as the guess to get better results until it reaches a point where another iteration gives a very small change. &lt;/p&gt;&#xA;" OwnerUserId="4789" LastActivityDate="2011-08-02T17:38:25.877" CommentCount="2" />
  <row Id="13120" PostTypeId="2" ParentId="13008" CreationDate="2011-08-02T17:39:26.640" Score="0" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;The argument fails when he attempts to prove that mass=0.&lt;br&gt;&#xA;IMO the particle preserves the mass.&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author Narlikar is correct here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;in the absence of the background the &lt;strong&gt;measure&lt;/strong&gt; vanishes!  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My bold &lt;em&gt;measure&lt;/em&gt;.&lt;br&gt;&#xA;The eq. &lt;strong&gt;F=m.a&lt;/strong&gt; is a relation to &lt;em&gt;measure&lt;/em&gt; any of &quot;force, mass or acceleration&quot; when at least 2 of the 3 terms are present.&lt;br&gt;&#xA;In Static we have forces and we have no acceleration and the mass is still present. Mass is a property of matter and is not defined with the help of that equation, although measured.&lt;br&gt;&#xA;&lt;strong&gt;IMO the particle preserves the mass.&lt;/strong&gt;&lt;br&gt;&#xA;First attempt:&lt;br&gt;&#xA;If the universe contains only 1 particle we have no means to define an external referential (except in a mathematical sence wich is useless), neither to exert a force. The center of mass of the system is always centered with that single particle that is to say: the particle is immobile.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second attempt:&lt;br&gt;&#xA;Lets try with a universe with two particles where 1 of them is reducing its mass until 0:&lt;br&gt;&#xA;The gravitational interaction is present, a center of mass frame (&lt;a href=&quot;http://en.wikipedia.org/wiki/Center_of_mass_frame&quot; rel=&quot;nofollow&quot;&gt;COM&lt;/a&gt;) is present and both particles contribute to the overall mass. We can attribute mass to each one as a part of the overall mass.&lt;br&gt;&#xA;As the mass of particle 1 is evolving to 0 the other particle is migrating to the COM. The remaining particle retains its mass and stay immobile irt the COM (centered in itself) in the limit of mass=0.&lt;br&gt;&#xA;There is a profound reason to preserve mass: the force and  the acceleration are consequence of a interaction, and it takes at least two objects to interact (and a time basis). The mass is an intrinsic property of matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The COM velocity is zero:  $V_{CM}=\frac{v_1m_1+v_2m_2}{m_1+m_2}=0=v&amp;#39;_1\frac{m_1}{m_1}$ and $m_1$ &amp;lt;&gt; 0;&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your sentence:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;existence of mathematics in such an empty universe be questionable?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Math is a human mental construct. Only.   &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-08-02T17:39:26.640" />
  <row Id="13121" PostTypeId="1" AcceptedAnswerId="13149" CreationDate="2011-08-02T18:46:51.043" Score="8" ViewCount="99" Body="&lt;p&gt;How should a geometer think about quotienting out by a Fermionic symmetry? Is this a formal concept? A strictly linear concept? A sheaf theoretic concept? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does symplectic reduction work with odd symmetries within geometric quantization? Are there moduli or instanton-like deformation complexes where the space of symmetries acting on the space of fields includes both even and odd symmetries? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, can the familiar instanton moduli problem whose deformations are governed by the linearized complex:&#xA;$$\Omega^0(Ad_0) \stackrel{d_A}\longrightarrow \Omega^1(Ad_0 ) \stackrel{\Pi \circ d_A} \longrightarrow \bar\Omega^2(Ad_0 )$$&#xA;be enhanced by fermionic symmetries to some structure whose linearized deformation complex looks like:&#xA;$$\Omega^0(Ad_0 \oplus Ad_1) \stackrel{d_A}\longrightarrow \Omega^1(Ad_0 \oplus Ad_1) \stackrel{\Pi \circ d_A} \longrightarrow \bar\Omega^2(Ad_0 \oplus Ad_1)$$&#xA;in a geometrically meaningful way? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some sense the question is trying to understand how 'odd symmetries' or Fermionic coordinates can be meaningfully carried through a familiar non-linear problem in geometry which makes use of a non-linear group action. Are these odd symmetries true geometric symmetries to be exponentiated in some way, or are they more properly formal symmetry-like operators analogous to true symmetries? Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="4477" LastEditorUserId="4477" LastEditorDisplayName="" LastEditDate="2011-08-02T18:52:57.450" LastActivityDate="2011-08-03T12:09:02.347" Title="Can Fermionic symmetries be fully integrated into geometric deformation complexes or symplectic reduction?" Tags="&lt;quantum-field-theory&gt;&lt;mathematical-physics&gt;&lt;supersymmetry&gt;&lt;differential-geometry&gt;&lt;quantization&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="13123" PostTypeId="1" CreationDate="2011-08-02T19:04:27.050" Score="2" ViewCount="78" Body="&lt;p&gt;I am trying to measure the concentration of a Volatile Organic Compound (VOC) inside a liquid using Gas Chromatography (GC). The liquid contains particles that cannot be sent to a GC, so I am using the head-space method (allowing the organic compound to evaporate into the gas above the liquid--- see this link:http://www.chem.agilent.com/Library/applications/5990-3285EN.pdf)-- &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I let the contents of the liquid equilibrate with the air in the vial at 50 C-- and then I use the air as feed for the GC. However, the concentration in the air is rather small ~ 10-20 ppm. Is there a technique that can be used to increase the concentration of the VOC's? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have wanted to use a smaller vial to collect my liquid, but the liquid splashes about and contaminates the cap of the vial, which makes me scared to inject a needle into it as I may wreck the needle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to repeat the experiment with the vial at a higher temperature ~70 C. This lead to the needle clogging. I am unsure of the cause of the needle clogging but I suspect it may have something to do with water evaporating and taking some of the polymer particles with it. Once in the needle, the particles could get dry and solidify. &lt;/p&gt;&#xA;" OwnerUserId="2946" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-09-03T03:35:30.897" LastActivityDate="2011-09-03T03:35:30.897" Title="Measuring concentration" Tags="&lt;experimental-physics&gt;&lt;sample-preparation&gt;&lt;experimental-technique&gt;" AnswerCount="1" CommentCount="14" />
  <row Id="13124" PostTypeId="2" ParentId="13123" CreationDate="2011-08-02T19:28:53.913" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a technique that can be used to increase the concentration of the VOC's? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Presumably heating the container (gently!) will work. In the past I have found that a heat lamp is safe when one needs to heat a small quantity of liquid without allowing it to boil. You will--of course--need patience.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have wanted to use a smaller vial to collect my liquid, but the liquid splashes about and contaminates the cap of the vial, which makes me scared to inject a needle into it as I may wreck the needle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;'Fraid I can't help you there.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-08-02T19:28:53.913" CommentCount="1" />
  <row Id="13125" PostTypeId="1" CreationDate="2011-08-02T19:41:09.153" Score="2" ViewCount="157" Body="&lt;p&gt;My friend bought two normal mirrors from some shop probably IKEA, looking like this one:&#xA;&lt;img src=&quot;http://i.stack.imgur.com/xmJ6w.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My images are different from the two. One is a little fatter and shorter, and the other a little thinner and taller. I wonder what can possibly be the reasons?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;They basically look the same, smooth and flat surfaces, almost the same size (one has a label saying 37x134cm), except one having a little thicker&#xA;decoration border than the other. The one with thinker edge is the&#xA;one giving me fatter and shorter image.&lt;/li&gt;&#xA;&lt;li&gt;I don't have other things to place them, so I just make both of them&#xA;stand on the floor, with a small angle lying against the wall.&#xA;Although the two mirrors are placed in the same way, will different&#xA;placement of the mirrors change my image?&lt;/li&gt;&#xA;&lt;li&gt;What kinds of mirrors (size, ...) and how to place them will give me more&#xA;realistic image of myself?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="528" LastEditorUserId="2535" LastEditorDisplayName="" LastEditDate="2011-08-03T17:54:39.833" LastActivityDate="2011-08-04T11:07:40.997" Title="How to make mirrors give more authentic image of myself" Tags="&lt;optics&gt;&lt;mirror&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13126" PostTypeId="2" ParentId="13100" CreationDate="2011-08-02T19:42:47.657" Score="4" ViewCount="" Body="&lt;p&gt;Fermionic spin 3/2 fields, much like bosonic fields with spin 1 and higher, contain negative-norm polarizations. Roughly speaking, a spin 3/2 field is $R_{\mu a}$ where $\mu$ is a vector index and $a$ is a spinor index. If $\mu$ is chosen to be 0, the timelike direction, one gets components of the spintensor $R$ that creates negative-norm excitations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not allowed to be a part of the physical spectrum because probabilities can't be negative. It follows that there must be a gauge symmetry that removes the $R_{0a}$ components - a spinor of them. The generator of this symmetry clearly has to transform as a spinor, too. There must be a spinor worth of gauge symmetry generators. The generators are fermionic because the original field $R$ is also fermionic, by the spin-statistics relation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that the conserved spinor generators are local supersymmetry generators and their anticommutator inevitably includes a vector-like bosonic symmetry which has to be the energy-momentum density. This completes the proof that in any consistent theory, spin 3/2 fields have to be gravitinos. The number of &quot;minimal spinors&quot; - the size of the gravitinos - has to be equal to the number of supercharges spinors which counts how much the local supersymmetry algebra is extended. In particular, it can't be linked to another quantity such as the dimension of a Yang-Mills group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while both 3/2 and 1/2 differ by 1/2 from $j=1$, more detailed physical considerations show that it is inevitable for the superpartner of a gauge boson, a gaugino, to have spin equal to 1/2 and not 3/2. Similarly, one can show that the superpartner of the graviton can't have spin 5/2 because that would require too many conserved spin-3/2 fermionic generators which would make the S-matrix essentially trivial, in analogy with the Coleman-Mandula theorem. Gravitinos can only have spin 3/2, not 5/2.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-08-02T19:42:47.657" CommentCount="3" />
  <row Id="13127" PostTypeId="1" CreationDate="2011-08-02T19:55:31.047" Score="1" ViewCount="72" Body="&lt;p&gt;I'm trying to figure out how to simulate eye diagrams for communications systems using Python.  I'm not sure I have the theory down completely, though.  From what I could figure out using some old LabWindows C code, the process is something like this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Take an input pattern and convert it into a discrete signal with the same time step between points as the system's step data&lt;/li&gt;&#xA;&lt;li&gt;Perform an FFT on the input pattern and convert the system's step data to an S-parameter&lt;/li&gt;&#xA;&lt;li&gt;Multiply the transformed pattern and the S-parameter pointwise (equivalent to convolution in the time domain)&lt;/li&gt;&#xA;&lt;li&gt;Take the inverse FFT of product.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is essentially all the old code seems to be doing, but unfortunately at the key steps there are calls to LabWindows functions that have poor online documentation, and I don't have access to the development libraries myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help or knowledge of eye diagram theory would be a tremendous help.&lt;/p&gt;&#xA;" OwnerUserId="4792" LastActivityDate="2011-09-01T22:38:57.180" Title="Simulating eye diagrams" Tags="&lt;fourier-transform&gt;&lt;signal-processing&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="13128" PostTypeId="2" ParentId="12814" CreationDate="2011-08-02T20:06:23.300" Score="1" ViewCount="" Body="&lt;p&gt;I am not a big expert in electron beams, but as long as nobody answers..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one thing which differs electron beams from the light. It is wavelength. It is few orders of magnitude smaller for electrons (~1Å) then for light (~5000Å for green light). Basically, there is one &lt;strong&gt;fundamental&lt;/strong&gt; limitation for something to be focused. It is wavelength of this object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it seems that your question is incorrect in a way that you have a wrong idea of an electron microscope. Where did you get these few tens of nm? Are you sure that you did not mix the resolution with beam focusing? Actually, for the best resolution beam should be &lt;a href=&quot;http://en.wikipedia.org/wiki/High-resolution_transmission_electron_microscopy&quot; rel=&quot;nofollow&quot;&gt;defocused a bit&lt;/a&gt;. A good start is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Transmission_electron_microscope&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4657" LastEditorUserId="4657" LastEditorDisplayName="" LastEditDate="2011-08-03T06:13:30.557" LastActivityDate="2011-08-03T06:13:30.557" CommentCount="4" />
  <row Id="13129" PostTypeId="2" ParentId="13125" CreationDate="2011-08-02T20:50:38.677" Score="6" ViewCount="" Body="&lt;p&gt;When mirrors give distorted images like that, it's usually because they're slightly curved. A &quot;&lt;a href=&quot;http://www.learner.org/teacherslab/science/light/lawslight/funhouse/index.html&quot; rel=&quot;nofollow&quot;&gt;fun house mirror&lt;/a&gt;&quot; is a typical (though extreme) example of this effect. So to get a realistic image, all you need to do is find a mirror that is as flat as possible. The placement of the mirror doesn't have any effect on whether the image is distorted, although it does allow you to see yourself from a different perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &#xA;Removed, was placed here erronously&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-08-04T11:07:27.533" LastActivityDate="2011-08-04T11:07:27.533" />
  <row Id="13130" PostTypeId="2" ParentId="13127" CreationDate="2011-08-02T21:39:32.000" Score="0" ViewCount="" Body="&lt;p&gt;I think the sequence of steps should be:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;convert input-data into analogue signal (modulation)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;apply distortion and noise to simulate the effects of the channel (e.g. using convolution with impulse response)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;demodulate the analogue signal into the out-data&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Then you plot input-data against out-data.&lt;/p&gt;&#xA;" OwnerUserId="4306" LastActivityDate="2011-08-02T21:39:32.000" CommentCount="6" />
  <row Id="13131" PostTypeId="2" ParentId="13125" CreationDate="2011-08-02T21:46:05.600" Score="4" ViewCount="" Body="&lt;p&gt;There are many mirrors today produced from rather thin float glass. &#xA;This mirrors are easily distorted by the frame they are mounted in. &#xA;Return that mirrors and buy one made from thicker glass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A method to solve the problem is to remove the mirror from the frame, &#xA;and if the mirror is found flat then, You fix it to a wall with some &#xA;glue point at the upper edge, and  two distance pads at the lower edge. &#xA;(Three pints defining a plane.) If this does not work, the mirror itself is &#xA;not really flat. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-08-04T11:07:40.997" LastActivityDate="2011-08-04T11:07:40.997" CommentCount="7" />
  <row Id="13132" PostTypeId="2" ParentId="13030" CreationDate="2011-08-02T22:22:00.717" Score="1" ViewCount="" Body="&lt;p&gt;The essential feature of a wing is that it redirects air flow. Oncoming air moves horizontally (relative to the plane), and behind the wing the air mass has a downward velocity component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In effect the air mass that is moving relative to the wing is &lt;em&gt;turning a corner&lt;/em&gt;. It follows that the air mass on the &quot;outside lane&quot; is covering a longer distance than the air mass that follows the &quot;inside lane&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lift arises from &lt;em&gt;action-reaction dynamics&lt;/em&gt;. For comparison: a helicopter rotor produces lift by exerting a force upon the air. This force accelerates the air mass downwards, and the reaction carries the helicopter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary phenomenon is air mass being redirected (downward) by the wing. This has two consequences:&lt;br&gt;&#xA;* The wing produces lift&lt;br&gt;&#xA;* Air mass flows faster over the top of the wing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's common to see the suggestion that the air mass flowing faster on top is what produces the lift. However, that is not the actual rundown from cause to effect. Primary is that air mass is being redirected.&lt;/p&gt;&#xA;" OwnerUserId="4794" LastEditorUserId="4794" LastEditorDisplayName="" LastEditDate="2011-08-02T22:46:42.523" LastActivityDate="2011-08-02T22:46:42.523" CommentCount="1" />
  <row Id="13133" PostTypeId="2" ParentId="12917" CreationDate="2011-08-02T23:46:40.287" Score="3" ViewCount="" Body="&lt;p&gt;I remember reading somewhere (don't remember where) the many design intricacies of the navigational gyroscopes for the appollo missions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gimbals where fitted with sensors and delicate electric motors. The slightest motion would trigger these sensors. The electric motors then moved in such a way that a torque arising from friction was nullified as much as possible. Equally important, the torque motors made the gimbals so agile and responsive that they moved as if they had no inertia &#xA;That way the rate of error buildup could be kept sufficiently low to enable the mission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you point out in your question, since the angular velocity of the rotor must be sustained the gimbal axes must somehow provide a reliable electric contact, which surely must give a non-negligable friction.&#xA;These problems made the navigational gyroscopes very challenging to build.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Apollo missions to keep buildup of error under control only 3 gimbals were used, instead of 4. If you use 4 gimbals you can design a configuration that can always be steered clear of a state of &lt;a href=&quot;http://en.wikipedia.org/wiki/Gimbal_lock&quot; rel=&quot;nofollow&quot;&gt;gimbal lock&lt;/a&gt;&#xA;The Appollo mission gyroscopes were susceptible to gimbal lock, when setting up a change of spacecraft orientation the crew had to plan ahead to avoid gimbal lock. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mechanical gyroscopes are obsolete now for navigational purposes. The successors to mechanical gyroscopse are Sagnac interferometers (also called ring interferometers).&lt;/p&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;About CMGS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, the purpose of a CMGS is to reorient the space-station as a whole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the purpose of a CMGS is the same that of a distribution of mini-thrusters for adjusting the attitude of the space-station; the CMGS is for muscling the spacecraft around, it's not a navigational sensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantage of a CMGS over the array of mini-thrusters is that the thrusters will deplete their supply of propellent, whereas the station's solar panels generate the electric energy that the CMGS flywheels need.&lt;/p&gt;&#xA;" OwnerUserId="4794" LastActivityDate="2011-08-02T23:46:40.287" />
  <row Id="13134" PostTypeId="2" ParentId="12878" CreationDate="2011-08-03T00:45:17.230" Score="1" ViewCount="" Body="&lt;p&gt;Let me set up the following convention for orientations, loosely based on aviation. &#xA;I take the &lt;em&gt;bicycle wheel&lt;/em&gt; as reference.&lt;br&gt;&#xA;Swivel: rotation around vertical axis&lt;br&gt;&#xA;Roll: like roll of an airplane&lt;br&gt;&#xA;Pitch: like pitching of an airplane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the spinning of the bicycle wheel be designated as 'rolling'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;InquilineKea, in the picture I presume the bicycle wheel rolling, and I gather the bicycle wheel is swiveled, while preventing pitch of the bicycle wheel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I gather you observe that as the bicycle wheel is swiveled the chair, with the bicycle-wheel-holding-person in it, tend to counter-swivel. (I'm assuming the entire weight of the bicycle wheel is supported by the person in the chair.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Discussion:&lt;br&gt;&#xA;As the bicycle wheel is swiveled you clearly need to prevent it from pitching. The torque to counter-act pitching tendency is a pitching torque. Clearly this pitching torque will not affect swivel of the chair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I concur with you: it seems to me this excludes the spin of the bicycle wheel from the explanation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next step:&#xA;Try the same with &lt;em&gt;non-spinning bicycle wheel&lt;/em&gt;. Swiveling the bicycle wheel will affect the chair at least somewhat, as the bicycle wheel has a considerable moment of inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would take a very strong person indeed to prevent &lt;em&gt;any&lt;/em&gt; pitching of the bicycle wheel; the motion will be somewhat jerky. Possibly that jerkiness is muddling the effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/dJsIi.jpg&quot; alt=&quot;Bicycle wheel and stool setup&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4794" LastActivityDate="2011-08-03T00:45:17.230" />
  <row Id="13135" PostTypeId="1" CreationDate="2011-08-03T01:08:27.360" Score="4" ViewCount="46" Body="&lt;p&gt;I've read that Wilson lines are the effective degrees of freedom for high-energy partons, when you consider collinear gluon emission. But I'm struggling to find a readable account of this topic. Does anyone have suggestions?&lt;/p&gt;&#xA;" OwnerUserId="2643" LastActivityDate="2011-08-03T01:08:27.360" Title="Why are Wilson lines the effective degrees of freedom for high energy partons?" Tags="&lt;quantum-chromodynamics&gt;" FavoriteCount="1" />
  <row Id="13136" PostTypeId="2" ParentId="12814" CreationDate="2011-08-03T01:19:03.530" Score="3" ViewCount="" Body="&lt;p&gt;The answer is a combination of two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The &quot;source size&quot; (the first crossover size, in fact) is about 50 μm even for thermionic cathodes. They are able to do that by using hairpin cathodes and a negative electrode known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wehnelt_cylinder&quot; rel=&quot;nofollow&quot;&gt;Wehnhelt cylinder&lt;/a&gt; (see &lt;a href=&quot;http://www.nhml.com/i/resources_NHML_Scanning-Electron-Microscopes_fig2_lg.gif&quot; rel=&quot;nofollow&quot;&gt;this image&lt;/a&gt; for a diagram of a thermionic electron gun).&lt;/li&gt;&#xA;&lt;li&gt;The electron probe can be made much smaller than the source by accepting a reduction in beam current. See &lt;a href=&quot;http://books.google.com/books?id=0Fm3T6F6_LEC&amp;amp;lpg=PA30&amp;amp;ots=HCVb8qVm7q&amp;amp;dq=%22Scanning%20Electron%20Microscopy%22%20demagnification%20source&amp;amp;pg=PA31#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;page 30 and 31 of&lt;/a&gt; Reimer's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3540639764&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Scanning Electron Microscopy&lt;/em&gt;&lt;/a&gt; for more details.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The reduction in current associated with a large demagnification is problematic, so &lt;a href=&quot;http://en.wikipedia.org/wiki/Field_emission_gun&quot; rel=&quot;nofollow&quot;&gt;field emission guns&lt;/a&gt; &lt;a href=&quot;http://books.google.com/books?id=0Fm3T6F6_LEC&amp;amp;lpg=PA30&amp;amp;ots=HCVb8qVm7q&amp;amp;dq=%22Scanning%20Electron%20Microscopy%22%20demagnification%20source&amp;amp;pg=PA33#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;are preferred&lt;/a&gt; when small probe sizes are desired, because they have apparent source sizes in the nanometer range.&lt;/p&gt;&#xA;" OwnerUserId="1656" LastActivityDate="2011-08-03T01:19:03.530" />
  <row Id="13137" PostTypeId="2" ParentId="13101" CreationDate="2011-08-03T02:28:27.810" Score="3" ViewCount="" Body="&lt;p&gt;In SQCD, you can get &quot;gluino hadrons&quot;, mesons and baryons in which the gluino is one of the constituent fermions. (Also a gluinoball, a glueball with some gluinos mixed in.) So you could have a &quot;gluino string&quot;, but in the opposite sense to what you want: it's two gluinos, at the ends of a gluon string. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility would be a topological defect in a gluino condensate. Maybe you could get this in N=2 SQCD, with monopoles at the ends. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what you're looking for is a &quot;fermionic meson&quot;, two quarks connected by the &quot;gluino string&quot;. Well, you can define this as an operator, a fermionic Wilson line. The only place I can find such an entity playing a role is in holographic approximations to QCD like the Sakai-Sugimoto model. In holographic QCD, you typically have color branes and flavor branes. A gluon is a string between color branes, a quark is a string between a color brane and a flavor brane, a meson is a string between flavor branes, and a baryon is a brane connected by strings to flavor branes (these strings are the valence quarks). These strings and branes all live in AdS space, and QCD lives on the boundary. Also, these are usually supersymmetric models - the exact, nonsupersymmetric holographic dual to QCD has yet to be determined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In superstring theory, you do have fermionic strings, and they are extended in space; the Fermi statistics come from the fermionic fields on the worldsheet. And the meson strings in holographic QCD do have mesino superpartners. See section 3.3.2 &lt;a href=&quot;http://arxiv.org/abs/0711.4467&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. They are discussing mesino operators in the worldvolume theory of a flavor brane. One is a quark-squark composite, but the other is a quark and an antiquark connected by a gluino (the adjoint fermion). So we had to go into anti de Sitter space to find it, but there it is, the fabled fermionic gluino string. :-)  &lt;/p&gt;&#xA;" OwnerUserId="1486" LastActivityDate="2011-08-03T02:28:27.810" CommentCount="2" />
  <row Id="13139" PostTypeId="2" ParentId="12814" CreationDate="2011-08-03T04:39:34.280" Score="3" ViewCount="" Body="&lt;p&gt;&quot;Why doesn't electrons widen the beam by repelling each other en-route?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does happen and can cause unwanted distortions. The reason it's not an &lt;strong&gt;overwhelming&lt;/strong&gt; problem is that the electrons, once they get going, are not all that close together. For example, if you have a 1nA beam current, and the electrons are traveling near the speed of light, there is on average 5cm vertical distance between one electron and the next. Also, once they're traveling very fast, it won't be very long until they to reach the sample, so even if they're slightly pushing each other, they can't get very far off track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything gets worse when the beam current goes up, and also when the electron energy (hence speed) goes down. More specifically regarding electron energy: (1) When you lower the beam voltage, repulsion effects will get worse; (2) The beam path very close to the electron gun (before the electrons have accelerated much) is the part that needs to be designed most carefully to minimize repulsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not an expert, this is my imperfect recollection from a few years ago. :-)&lt;/p&gt;&#xA;" OwnerUserId="3811" LastActivityDate="2011-08-03T04:39:34.280" />
  <row Id="13140" PostTypeId="1" CreationDate="2011-08-03T04:52:53.510" Score="10" ViewCount="175" Body="&lt;p&gt;I came across an interesting problem when I prepared for the preliminary exam on electromagnetism. Below is the problem in its original words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A metallic sphere of mass, $m$, and radius, $a$, carries a net charge, $Q$, and is magnetized with uniform magnetization, $M$, in the $z$-direction.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(a) Determine the total angular momentum associated with the electromagnetic field.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(b) The magnetization of the sphere is now reduced to zero. Assuming that no external mechanical forces act on it (PS: this is possible. For instance, just heat up the sphere and the magnetization will be eliminated ), determine the angular velocity and the sense of rotation of the sphere when $M=0$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(c) Describe the origin of the torque that causes the sphere to rotate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For part (a), it is straight forward to calculate electric and magnetic field E, H inside and outside of the sphere. Generated by the net charge $Q$ uniformly distributed on the surface, the electric field $E$ is isotropic outside the sphere and vanishes inside. The magnetic field H is generated by magnetization $M$ and can be calculated using the &quot;magnetic scalar potential&quot; formalism since there is no free current. The result is that H is also uniform inside the sphere and takes the dipole form outside the sphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total angular momentum $J$ is obtained by integrating over the region where a local angular momentum density of the EM field exists, i.e. outside of the sphere. It is in the $z$-direction and I find it to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$J= \frac{2 Q M a^{2}}{9 \epsilon_0 c^{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For part (b) and (c), it is interesting to ask why the sphere would rotate after magnetization $M$ is erased. What I think up is that as $M$ decreases to zero, a electric field with non-zero curl is generated, as implied by Faraday's law. This electric field is in the azimuthal direction, and will act on the surface charge $Q$ to provide a torque. However, a straightforward calculation can show that the angular momentum this torque can transfer is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$J&amp;#39;= \frac{Q M a^{2}}{9 \epsilon_0 c^{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously a factor of 2 is missing, so this only accounts for half of the angular momentum stored originally in the EM field. So I am perplexed by where the other half has gone?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I come up with 2 possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;since the charged sphere begins to rotate, there is also non-zero angular momentum stored in EM field after magnetization is removed. But it could hardly account for the missing amount, since in that case the electric field $E$ is proportion to $Q$, the magnetic field $H$ to $Q^2$ ( one $Q$ from the charge and a second $Q$ from the angular velocity). Hence one expects the angular momentum stored is proportional to $Q^3$, which is very different from $J$'s linear dependence on $Q$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;maybe in the process of magnetization removal, EM radiation carries away some angular momentum. But this possibility has to be validated quantitatively before it can convince people. What I feel is that if the process of removal is infinitely slow, then the effect of radiation can be neglected.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Then what is the correct answer? I provide quantitative result for $J$ and $J&amp;#39;$ in the hope that someone can check them and point out that I have made a wrong calculation lol. But I am really eager to hear from any explanation that sounds real ... Electromagnetism is as interesting as it was 3 years ago, but I just don't remember as much as I did then ...&lt;/p&gt;&#xA;" OwnerUserId="2597" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-03T05:25:46.570" LastActivityDate="2011-08-04T07:20:59.193" Title="Where the angular momentum has gone?" Tags="&lt;electromagnetism&gt;&lt;angular-momentum&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" />
  <row Id="13141" PostTypeId="1" AcceptedAnswerId="13146" CreationDate="2011-08-03T06:18:00.153" Score="2" ViewCount="265" Body="&lt;p&gt;If yes, why don't they fill up with water, and can you breathe the air there? Like, it's not exactly atmosphere there, but an underwater cave with higher ceiling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Possible that it has a different name than air pocket, but due to my limited knowledge of English, I have no idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="4799" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-08-03T10:36:17.433" LastActivityDate="2011-08-29T02:49:20.920" Title="Can pockets of air exist underwater?" Tags="&lt;water&gt;&lt;air&gt;&lt;hydrostatics&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="13143" PostTypeId="1" CreationDate="2011-08-03T07:53:08.230" Score="2" ViewCount="199" Body="&lt;p&gt;There appears to be a distinct lack of agreement in the physics community on what exactly Marie Curie did for atomic theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many journals state that Curie was responsible for shifting scientific opinion from the idea that the atom was solid and indivisible to an understanding of subatomic particles. However, JJ Thompson had already discovered the electron prior to Curie's work, and Rutherford proved the nature of atomic structure - so what did Marie Curie really do for atomic theory?&lt;/p&gt;&#xA;" OwnerUserId="3044" LastEditorUserId="2433" LastEditorDisplayName="" LastEditDate="2011-08-04T15:04:53.820" LastActivityDate="2011-08-04T15:04:53.820" Title="What did Marie Curie do for atomic theory?" Tags="&lt;history&gt;&lt;radioactivity&gt;&lt;subatomic&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="1" />
  <row Id="13144" PostTypeId="2" ParentId="12976" CreationDate="2011-08-03T08:07:26.137" Score="1" ViewCount="" Body="&lt;p&gt;I will make the following approximating assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The surface of a trampoline has zero mass density.  This implies standing or crouching will not instantaneously affect the position of your center of gravity.&lt;/li&gt;&#xA;&lt;li&gt;The upward restoring force is a monotonic function of vertical displacement.&lt;/li&gt;&#xA;&lt;li&gt;Humans can straighten their legs in zero time.  This is just a way to avoid extra optimization details.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If these hold, then the best time to crouch is before landing, and the best time to straighten your legs is when your speed is zero (i.e., when your height is at a minimum).  This behavior optimizes three quantities:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The kinetic energy you gain before reaching the trampoline is maximized.  Because you are in a crouching position, your center of gravity falls further before the restoring force starts slowing you down.&lt;/li&gt;&#xA;&lt;li&gt;The height of your center of gravity when your speed reaches zero is minimized.  This is because for a fixed position of your center of gravity, the restoring force is minimized when you are in a crouching position.&lt;/li&gt;&#xA;&lt;li&gt;The work done by the springs in releasing you is maximized.  If you set $h_0$ to be the displacement of your center of gravity where you are stationary, the energy imparted is the integral $\int_{h_0}^0 F ds$ of the restoring force with respect to displacement.  For any fixed position of your center of gravity, the restoring force is maximized when you are in a standing position.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you watch experienced gymnasts gaining altitude, you will see that they straighten their legs roughly when they reach the bottom, and they let the trampoline carry them up.&lt;/p&gt;&#xA;" OwnerUserId="761" LastActivityDate="2011-08-03T08:07:26.137" CommentCount="1" />
  <row Id="13145" PostTypeId="2" ParentId="13068" CreationDate="2011-08-03T08:32:55.723" Score="2" ViewCount="" Body="&lt;p&gt;Don't listen anyone. It is possible, but I have to admit - hard. Otherwise, how were first mono-crystals grown?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially, you may buy 'pure' silicon (pure for chemical reactions, not electronics). First of all you need to make a rod. To do that you'll need form out of material which can withstand 1600C (hard part, can't name ones at the moment), and heat Si in it using induction heating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Induction heating might be the hardest part - but there are lots of guides around. Induction heater able to melt metal is 100% doable at home. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have Si rod - you need to purify it using 'zone melting' using your induction heater. Will take ages, so you might want to make motorized thing which will move rod or heater for several days/weeks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that your rod is polycrystalline - which is already ok for solar cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Czochralski process you may reuse again your induction heater, but the hardest part is getting clean 'form' for molten Si, as silicon is ultrapure and very easy to contaminate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you may get mono-crystalline or large-grain polycrystalline Si right out of your zone heating station if you would do slow pass at the end and add seed crystal. But this might be tricky a little.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your rod in thin enough (1-3cm) you may cut it using diamond disks, sold in usual shops. Thicker rods will require larger disks (which are not easy /cheap to buy) or 'diamond wire'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whole process is quite complex, but will give you lots of experience working with tough things. Dealing with silicon is much harder than Steel ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. If you would do everything except getting seed crystals, I can send you some.&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="930" LastEditorDisplayName="" LastEditDate="2011-08-03T08:56:40.640" LastActivityDate="2011-08-03T08:56:40.640" CommentCount="2" />
  <row Id="13146" PostTypeId="2" ParentId="13141" CreationDate="2011-08-03T08:39:57.317" Score="12" ViewCount="" Body="&lt;p&gt;Yes, it is possible to have pockets of air underwater as long as there is something there to contain the pocket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can easily demonstrate this by turning a cup upside-down and submerging it in water.  If you put a napkin in the bottom of the cup before you do this, the napkin stays dry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do the above experiment, but dive down in a swimming pool with the cup, you will notice that the air pocket decreases in size as you dive, and returns to normal size as you surface.  This indicates that no air is being lost.  Instead, the air is being compressed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These observations indicate that the air pocket does fill part way up with water, but does not disappear.  This is because as the air is submerged, the water pressure on it increases.  This shrinks the air pocket's volume until the pressure in the air is the same as the pressure in the water surrounding the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, you can breathe air in an air pocket - it's normal air.  However, you will exhaust the oxygen supply quickly if  the air pocket is small.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-08-29T02:49:20.920" LastActivityDate="2011-08-29T02:49:20.920" CommentCount="5" />
  <row Id="13147" PostTypeId="2" ParentId="4243" CreationDate="2011-08-03T08:58:33.190" Score="-2" ViewCount="" Body="&lt;p&gt;gravity is the weakest force as its coupling constant  is  small in value.&#xA;gravity cannot felt by us in daily life because of huge universe surrounding us.electromagnetic force is no dowt stronger as it deals with microscopic particles (electrons,protons),.gravity is attractive in nature always .it is long range force among all other interactions in nature.&lt;/p&gt;&#xA;" OwnerUserId="4802" LastActivityDate="2011-08-03T08:58:33.190" CommentCount="1" />
  <row Id="13148" PostTypeId="2" ParentId="13141" CreationDate="2011-08-03T09:28:39.117" Score="5" ViewCount="" Body="&lt;p&gt;Spiders do that since millions of years, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=-nq6AGph4qc&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=-nq6AGph4qc&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and diving bells (aka Caisson) are used since centuries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as the &quot;pneumatic trough&quot; it was extremely important &#xA;as a means to handle gases in early chemistry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pneumatic_trough&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Pneumatic_trough&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-08-03T21:00:25.143" LastActivityDate="2011-08-03T21:00:25.143" />
  <row Id="13149" PostTypeId="2" ParentId="13121" CreationDate="2011-08-03T12:09:02.347" Score="2" ViewCount="" Body="&lt;p&gt;This answer will address the  two questions about symplectic reduction and the instanton moduli space separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An accepted method to generalize symplectic reduction to include fermionic symmetries is by means the theory of symplectic supermanifolds (and more generally Poisson supermanifolds). Please see the following &lt;a href=&quot;http://www.mathcs.emory.edu/~rudolf/Glimm.pdf&quot; rel=&quot;nofollow&quot;&gt;exposition&lt;/a&gt; by Tilmann Glimm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Symplectic supermanifolds can be equipped with two types of symplectic structures, odd and even. In Glim's article the reduction theorem is proved for the two types of symplectic structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of a symplectic supermanifold is the supercotangent bundle (see for example the following &lt;a href=&quot;http://hal.archives-ouvertes.fr/docs/00/47/22/24/PDF/GQ_Supercotangent.pdf&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; by: J. P. Michel) of a spin manifold. Locally this bundle is covered by charts $\{p_i, q^i, \theta_i\}$, consisting of position, momentum and Grassmann coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The symmetries divided by in the supersymplectic reduction consist of &lt;a href=&quot;http://en.wikipedia.org/wiki/Supergroup_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;Lie supergroups&lt;/a&gt; with Hamiltonian action on the supersymplectic manifold. In Glimm's article, there is a detailed description of the symplectic reduction of the Bose-Fermi oscillator by its Hamiltonian supersymmetry group (which is a subgroup of the orthosymplectic group).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory of geometric quantization can be extended to supergeometry. The quantization of odd symplectic structures lead to &lt;a href=&quot;http://en.wikipedia.org/wiki/Batalin%E2%80%93Vilkovisky_formalism&quot; rel=&quot;nofollow&quot;&gt;Batalin-Vilkovisky&lt;/a&gt; theory. In its simplest application, this quantization results quantization spaces isomorphic to the de-Rham complex of the manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classical mechanical theory associated with the even symplectic particle is the &#xA;&lt;a href=&quot;http://ph.technion.ac.il/~site/wisw/memoriam/marinov/pub/Mar40.pdf&quot; rel=&quot;nofollow&quot;&gt;Berezin-Marinov&lt;/a&gt; classical description of the spin by means of Grassmann variables. Its quantization corresponds to superparticles, where the fermions consist of sections of the spinor bundle over the base manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the instanton moduli space for super Yang-Mills theories consists of the usual instanton moduli space given by Atiyah Hitchin Singer deformation complex together with the  zero modes of a twisted Dirac operator (see for example the following &lt;a href=&quot;http://www.ph.utexas.edu/~wtang/donaldson_notes_walter_tom.pdf&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; by: Mainiero and Walter Tangarife in the context of the Seiberg-Witten theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A supersymmetric generalization of the deformation complex actually exists at least for the N= 4 super Yang-Mills as given in &lt;a href=&quot;http://arxiv.org/PS_cache/hep-th/pdf/9702/9702106v1.pdf&quot; rel=&quot;nofollow&quot;&gt;Labastida and Lozano's&lt;/a&gt; article  (equation (3.6).)&lt;/p&gt;&#xA;" OwnerUserId="2190" LastActivityDate="2011-08-03T12:09:02.347" />
  <row Id="13150" PostTypeId="2" ParentId="13140" CreationDate="2011-08-03T13:03:35.217" Score="-3" ViewCount="" Body="&lt;p&gt;The angular momentum is in the energy that was added into the object.&#xA;After the object has cooled, the angular momentum is in the heat radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But: If the object radiates mostly from the center parts, the radiation has less angular momentum than it has when it is the outer parts that radiate the most. Can you solve this?  &lt;/p&gt;&#xA;" OwnerUserId="4017" LastActivityDate="2011-08-03T13:03:35.217" />
  <row Id="13151" PostTypeId="2" ParentId="13140" CreationDate="2011-08-03T13:21:50.373" Score="2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein%E2%80%93de_Haas_effect&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Einstein%E2%80%93de_Haas_effect&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculations based on a model of electron spin as a circulating&#xA;  electric charge underestimate this magnetic moment by a factor of&#xA;  approximately 2, the Landé g-factor. A correct description of this&#xA;  magnetic moment requires a treatment based on quantum electrodynamics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4306" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-08-04T07:20:59.193" LastActivityDate="2011-08-04T07:20:59.193" CommentCount="3" />
  <row Id="13152" PostTypeId="2" ParentId="3164" CreationDate="2011-08-03T14:37:37.430" Score="-1" ViewCount="" Body="&lt;p&gt;Whether or not photons should be taken into account when calculating antenna parameters is irrelevant to the natural wish to get a picture of what is going on with the photons when an antennais tranmitting or receiving. Unfortunately, experiments involving photons and antennas are presumably difficult. Nevertheless, if we can look for gravity waves, we should be able to figure out how to observe low-energy photons coming out of a wire.&lt;/p&gt;&#xA;" OwnerUserId="4804" LastActivityDate="2011-08-03T14:37:37.430" CommentCount="2" />
  <row Id="13153" PostTypeId="1" CreationDate="2011-08-03T15:37:04.507" Score="6" ViewCount="193" Body="&lt;p&gt;I'm trying to understand a particular case of gauge theories, namely discrete spaces on which a group G can act transitively, with a gauge group H which is discrete as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I've already read about continuous and discrete gauge theory (I can provide the necessary references if needed*), it seems that the full gauge transformation group will be the semidirect product of G by H. However I'm unable to prove it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone point me to the proof of this fact ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help greatly appreciated !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Edit :&#xA;The bottom of page 2 in this reference describes this kind of thing without giving any proof : &lt;a href=&quot;http://arxiv.org/pdf/1106.2759&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/pdf/1106.2759&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="4805" LastEditorDisplayName="" LastEditDate="2011-08-03T17:44:02.243" LastActivityDate="2011-09-03T09:40:57.513" Title="Discrete gauge theories" Tags="&lt;mathematical-physics&gt;&lt;gauge-theory&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="13154" PostTypeId="2" ParentId="13062" CreationDate="2011-08-03T20:52:47.997" Score="0" ViewCount="" Body="&lt;p&gt;This is the figure I mean above in the comments for question. The units in the figure are a bit weird, but If I do this correctly...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;W*h/mile = J/h*h/mile = N*mile/h*h/mile = N&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, y-axis has force and x-axis has velocity (you may want to check the above). This shows that tires have nearly constant friction force respect to velocity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure is from www.teslamotors.com/blog/roadster-efficiency-and-range and the points go to Zassounotsukushi for revealing this web page (@Z you may want to make the correct answer, I am just posting this because of the figure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Jp48z.png&quot; alt=&quot;F vs. v&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4772" LastActivityDate="2011-08-03T20:52:47.997" CommentCount="1" />
  <row Id="13155" PostTypeId="1" CreationDate="2011-08-03T21:47:13.087" Score="0" ViewCount="40" Body="&lt;p&gt;Assuming I rotate a disk, I want to know how long it takes to completely stop, and the number of revolutions it made since I removed my fingers off the disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say a DVD I rotate with my fingers. I only know the radians per second (velocity) of the last moment I touched the disc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you guys tell me where to start?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Im trying to implement this on an iPhone app. So it would be nice if you mention &lt;strong&gt;equations&lt;/strong&gt;. It should not be exact. &lt;/p&gt;&#xA;" OwnerUserId="4808" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-04T06:42:59.603" LastActivityDate="2011-08-04T06:42:59.603" Title="How to know the time a disc takes to stop from rotating and the numbers of revolutions" Tags="&lt;rotational-dynamics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13156" PostTypeId="2" ParentId="13155" CreationDate="2011-08-03T22:03:25.057" Score="3" ViewCount="" Body="&lt;p&gt;Lets say the disk originally rotated with angular frequency $\omega_0$ in Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then due to friction you will have $\omega(t)=\omega_0-d t$. Where $d$ is the deceleration in Hz/s and $t$ is the time in seconds. Note that the disk won't start turning in the opposite direction $\omega(t)&amp;gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you get a rotation angle $\alpha(t)$:&#xA;$$\alpha(t)=\int_0^t \omega(t&amp;#39;)\textrm{d}t&amp;#39;= \omega_0t - dt^2/2$$ &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(let ((omega0 50)&#xA;      (d .2)&#xA;      (start 0)&#xA;      (old-alpha 0))&#xA;  (defun start-turning (omega)&#xA;    (setf omega0 omega&#xA;          start (get-internal-real-time)))&#xA;  (defun draw ()&#xA;    (load-identity)&#xA;    (let* ((time (- (get-internal-real-time start)))&#xA;           (alpha (if (&amp;lt; 0 (- omega0 (* d time)))&#xA;                      (- (* omega0 time) (* d (/ 2) (expt time 2)))&#xA;                      old-alpha)))&#xA;      (rotate alpha 0 0 1)&#xA;      (setf old-alpha alpha))&#xA;    (draw-disk)))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4306" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-08-03T22:38:05.277" LastActivityDate="2011-08-03T22:38:05.277" CommentCount="8" />
  <row Id="13157" PostTypeId="1" AcceptedAnswerId="13158" CreationDate="2011-08-03T23:33:53.543" Score="14" ViewCount="477" Body="&lt;p&gt;There was a reason why I constantly failed physics at school and university, and that reason was, apart from the fact I was immensely lazy, that I mentally refused to &quot;believe&quot; more advanced stuff until I understand the fundamentals (which I, eventually, never did).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, one of the most fundamental things in physics that I still don't understand (a year after dropping out from the university) is &lt;strong&gt;the concept of &lt;a href=&quot;http://en.wikipedia.org/wiki/Field_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;field&lt;/a&gt;&lt;/strong&gt;. No one cared to explain what a field &lt;em&gt;actually is&lt;/em&gt;, they just used to throw in a bunch of formulas and everyone was content. The school textbook definition for a field (electromagnetic in this particular case, but they were similar), as I remember it, goes like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An electromagnetic field is a special kind of substance by which charged moving particles or physical bodies with a magnetic moment interact.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A special kind of substance&lt;/em&gt;, are they for real? This sounds like the authors themselves didn't quite understand what a field is so they decided to throw in a bunch of buzzwords to make it sounds right. I'm fine with the second half but &lt;em&gt;a special kind of substance&lt;/em&gt; really bugs me, so I'd like to focus on that.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Is a field &lt;em&gt;material&lt;/em&gt;?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Apparently, it isn't. It doesn't consist of particles like my laptop or even the light.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it isn't material, &lt;strong&gt;is it &lt;em&gt;real&lt;/em&gt; or is it just &lt;em&gt;a concept that helps to explain our observations&lt;/em&gt;&lt;/strong&gt;? While this is prone to speculations, I think we can agree that in scope of this discussion particles actually do exist and laws of physics don't (the latter are nothing but human ideas so I suspect Universe doesn't &quot;know&quot; a thing about them, at least if we're talking raw matter and don't take it on metalevel where human knowledge, being a part of the Universe, makes the Universe contain laws of physics). Any laws are only a product of human thinking while the stars are likely to exist without us homo sapiens messing around. Or am I wrong here too? I hope you already see why I hate physics.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Is a field &lt;em&gt;not material but still real&lt;/em&gt;?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Can something &quot;not touchable&quot; by definition be considered part of our Universe by physicians? I used to imagine that a &quot;snapshot&quot; of our Universe in time would contain information about each particle and its position, and this would've been enough to &quot;de&lt;a href=&quot;http://en.wikipedia.org/wiki/Serialization&quot; rel=&quot;nofollow&quot;&gt;seralize&lt;/a&gt;&quot; it but I guess my programmer metaphors are largely off the track. (Oh, and I know that the uncertainty principle makes such (de)serialization impossible — I only mean that I thought the Universe can be &quot;defined&quot; as the set of all material objects in it). Is such assumption false?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, if fields indeed are &lt;em&gt;not&lt;/em&gt; material but &lt;em&gt;are&lt;/em&gt; part of the Universe, I don't really see how they are different from &lt;a href=&quot;http://en.wikipedia.org/wiki/Hindu_deities&quot; rel=&quot;nofollow&quot;&gt;the whole Hindu pantheon&lt;/a&gt; except for perhaps a more geeky flavor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I talked about this with the teacher who helped me to prepare for the exams (which I did pass, by the way, it was before I dropped out), she said to me that, if I wanted hardcore definitions,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a field is a function that returns a value for a point in space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now this finally makes a hell lot of sense to me but I still don't understand how mathematical functions can be a part of the Universe and shape the reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm afraid the question will seem ambiguous and get closed but I had a really hard time trying to shape my confusion into sentences, and I will highly appreciate if someone clears it up, suggests a link or confirms that there is no definitive answer to my question.&lt;/p&gt;&#xA;" OwnerUserId="4809" LastEditorUserId="103" LastEditorDisplayName="" LastEditDate="2011-08-25T21:10:26.940" LastActivityDate="2011-08-25T22:12:40.700" Title="What is a field, really?" Tags="&lt;education&gt;&lt;field-theory&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="7" />
  <row Id="13158" PostTypeId="2" ParentId="13157" CreationDate="2011-08-04T00:06:12.760" Score="21" ViewCount="" Body="&lt;p&gt;I'm going to go with a programmer metaphor for you.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The mathematics (including &lt;em&gt;&quot;A field is a function that returns a value for a point in space&quot;&lt;/em&gt;) are the interface: they define for you exactly what you can expect from this object.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;em&gt;&quot;what is it, really, when you get right down to it&quot;&lt;/em&gt; is the implementation. Formally you don't &lt;em&gt;care&lt;/em&gt; how it is implemented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of fields they are &lt;em&gt;not&lt;/em&gt; matter (and I consider &lt;em&gt;&quot;substance&quot;&lt;/em&gt; an unfortunate word to use in a definition, even though I am hard pressed to offer a better one) but they &lt;em&gt;are&lt;/em&gt; part of the universe and they &lt;em&gt;are&lt;/em&gt; part of physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What they are is the aggregate effect of the exchange of virtual particles governed by a quantum field theory (in the case of E&amp;amp;M) or the effect of the curvature of space-time (in the case of gravity, and stay tuned to learn how this can be made to get along with quantum mechanics at the very small scale...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alas I can't define how these things work unless you simply accept that fields do what the interface says and then study hard for a few years.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, it is very easy to get hung up on this &quot;Is it &lt;em&gt;real&lt;/em&gt; or not&quot; thing, and most people do for at least a while, but please just put it aside. When you peer really hard into the depth of the theory, it turns out that it is hard to say for sure that stuff is &quot;stuff&quot;. It is tempting to suggest that having a non-zero value of mass defines &quot;stuffness&quot;, but then how do you deal with the photo-electric effect (which makes a pretty good argument that light comes in packets that have enough &quot;stuffness&quot; to bounce electrons around)? All the properties that you associate with stuff are actually explainable in terms of electro-magnetic fields and mass (which in GR is described by a component of a tensor field!). And round and round we go.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-04T00:11:57.043" LastActivityDate="2011-08-04T00:11:57.043" CommentCount="6" />
  <row Id="13159" PostTypeId="2" ParentId="11434" CreationDate="2011-08-04T00:13:25.233" Score="0" ViewCount="" Body="&lt;p&gt;I found this book pretty useful during studies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521648548&quot; rel=&quot;nofollow&quot;&gt;Detectors for particle radiation (Konrad Kleinknecht)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It covers a lot of different detector types, and also has some chapters on solid state detectors, from the basics (particles in matter, semiconductors) up to how to build a detector out of it, if I recall correctly.&lt;/p&gt;&#xA;" OwnerUserId="825" LastActivityDate="2011-08-04T00:13:25.233" />
  <row Id="13160" PostTypeId="2" ParentId="3164" CreationDate="2011-08-04T00:44:41.243" Score="1" ViewCount="" Body="&lt;p&gt;One of the lessons quantum mechanics supposedly teaches us is that we should be cautious about asking questions that cannot be answered experimentally. Not that we should never do it, just that we should be careful. I think this question and the subsequent answers are a good example of this principle being disregarded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can a &quot;photon&quot; be everywhere at once on an expanding spherical surface? First of all, I would dismiss those people who make a major point of the issue of spherical symmetry. Everyone knows that an e-m wave is not spherically symmetrical. It is so obvious that those who deal with the subject will use the term &quot;spherical&quot; to describe the next best thing, the familiar donut shape of a dipole radiator. S-wave or p-wave, the question stands: how can the photon be everywhere at once?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, I disagree with those who say the question is wrong because an antenna emits billions of photons. There are in fact antennas which regularly emit light in quantities similar to one photon's worth of energy; these antennas are called &quot;atoms&quot; and they are everywhere. The question stands: how can a photon emitted by an atom be everywhere at once on a spherical surface? In fact, this is very close to the original form of the EPR paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Einstein posed the question in 1935, no one at first seriously considered that it might be tested experimentally. The EPR paradox went through a number of transformations before it dawned on people that it might be put testable. Among these transformations we can list Bohm, who recast it in the form of two electrons in the spin singlet state; and Feynmann, who analyzed the two-photon decay of positronium. Neither of these models were, then or now, amenable to experimental testing. After Bell's analysis in 1964, people were motivated afresh to look for experimental manifestations, and found something workable in parametric down-conversion. But that's another story.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic problem with the question as posed here is: how would you measure it? The theory tells us that the photon spreads out as a &quot;spherical&quot; wave. But Copenhagen, in one form or another, tells us that the photon is detected at a single point. How do we know? Many, notably Feynmann, would say that the click in a photomultiplier tube tells us when a photon has been detected. But the detailed physics of a detector event can be interpreted in different ways; all we can say with relative certainty is that the probability of a detector going off is proportional to the square of the incident field. And this is entirely conisistent with the photon's energy being spread over a spherical surface. It is very difficult to establish that a click in the photodetector is necessarily associated with the abosorption of one full photon's worth of energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some will undoubtedly say it is obvious that when a photomultiplier tube clicks, it must have absorbed a photon. To those people, I would ask: what experiment can you propose to demonstrate that a photomultiplier tube will never click when exposed to less than one photon's worth of energy? Others will object that once a detector clicks, a second detector will never go off at the same time; this shows that the whole photon &quot;collapsed&quot; into the first detector. But experimentally, this hypothesis is notoriously difficult to demonstrate. The reason is simply that we still do not have a working pea-shooter for photons that reliably produces one photon at a time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the original question, I would say that the wave from an antenna, even at &quot;atomic&quot; antenna, spreads out &quot;spherically&quot;; and that there is no experiment which can conclusively show that the emitted &quot;photon&quot; ever appears concentrated at a single point.&lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-08-04T00:44:41.243" />
  <row Id="13161" PostTypeId="1" AcceptedAnswerId="13192" CreationDate="2011-08-04T02:27:46.537" Score="5" ViewCount="79" Body="&lt;p&gt;According to Kolmogorov, the energy spectrum function of a turbulent fluid is given as,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(k)=C\epsilon^{\frac{2}{3}}k^{\frac{-5}{3}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\epsilon$ is the energy flux and $k=\frac{2\pi}{r}$ where $r$ is the length scale. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The normal explanation I see in most physics texts and articles is that the -5/3 exponent is found purely through dimensional analysis. Obviously dimensional analysis is very useful, but often times there is a more physical explanation as well. Does anyone have any insights?&lt;/p&gt;&#xA;" OwnerUserId="3018" LastEditorUserId="3018" LastEditorDisplayName="" LastEditDate="2011-08-04T06:42:36.650" LastActivityDate="2011-08-04T18:17:01.273" Title="Physical Explanation for &quot;Kolmogorov -5/3 spectrum&quot; in Fluid Mechanics" Tags="&lt;fluid-dynamics&gt;&lt;turbulence&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="13162" PostTypeId="1" CreationDate="2011-08-04T03:09:37.750" Score="-1" ViewCount="84" Body="&lt;p&gt;This is a very serious question, and I mean it, so please don't close it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the central principles of quantum gravity is the holographic principle stating that the quantum state of the interior of some spatial boundary is completely determined by the quantum state of the boundary. In some versions, it might be the boundary of the future Cauchy development of the interior, which is a null sheet. Among the information contained in this future null boundary are memories and records of what putatively happened in the interior. Given this information, the interior history is completely specified as a superposition of histories parameterized only by the future boundary state. Anything not determined by the memories and records has to be in a very precise superposition with no definite values. The holographic bound is a bottleneck, constraining the maximal possible channel capacity of the memories and records to go as the area of the boundary, and not the volume of the interior. This means most of the fine grained details of the interior simply don't exist. If we translate this holographic reconstruction into some constraint over canonical slices of the interior, do we get the Wheeler-DeWitt equations, and are the only solutions to the Wheeler-DeWitt equations holographic reconstructions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is quantum gravity trying to tell us mysticism is true? Even more so than quantum mechanics? Because after all, in quantum mechanics, the wave function of the interior may be in a superposition, but this superposition can be varied arbitrarily in principle.&lt;/p&gt;&#xA;" OwnerUserId="4810" LastActivityDate="2011-08-04T03:09:37.750" ClosedDate="2011-08-04T06:46:05.433" Title="Is the only consistent theory of quantum gravity mysticism?" Tags="&lt;quantum-mechanics&gt;&lt;cosmology&gt;&lt;quantum-gravity&gt;" CommentCount="3" />
  <row Id="13163" PostTypeId="2" ParentId="13013" CreationDate="2011-08-04T03:15:36.683" Score="0" ViewCount="" Body="&lt;p&gt;If the input to cosmology is trivial in the sense that the initial conditions have to be trivial, then creation itself was trivial, and needs no explanation. It's the output which needs to be explained, not the input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no need for even the laws of physics to be included in the input because eternal inflation and the multiverse of the string theory landscape leads to an ensemble of all possible laws.&lt;/p&gt;&#xA;" OwnerUserId="4810" LastActivityDate="2011-08-04T03:15:36.683" />
  <row Id="13164" PostTypeId="1" CreationDate="2011-08-04T03:52:16.410" Score="8" ViewCount="108" Body="&lt;p&gt;In section 2.3.4 of Di Francesco, Mathieu, Senechal's &lt;em&gt;Conformal Field Theory&lt;/em&gt; (and I would guess in many other introductory discussions of QFT) one finds a simple sample computation of a propagator&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K(x,y) := \langle \varphi(x)\varphi(y)\rangle.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the example, the fields $\varphi:\mathbf{R}^2\to \widehat{u}(1)$ are taken to obey the &lt;em&gt;Euclidean action of a free boson of mass $m$&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S = \frac{1}{2}\int d^2x \{\partial_\mu \varphi\partial^\mu\varphi + m^2 \varphi^2\}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[I'm assuming I won't ruffle any feathers talking about the fields as operator-valued functions rather than operator-valued distributions. By $\widehat{u}(1)$ I mean the affine Heisenberg algebra (see &lt;em&gt;Conformal Field Theory&lt;/em&gt; section 14.4.4).] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the authors compute&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K(x,y) = -\frac{1}{2\pi} \ln(\|x-y\|), \quad \textrm{if }\ m=0$$&#xA;$$K(x,y) = \frac{1}{2\pi} K_0(m \|x-y\|), \quad \textrm{if }\ m&amp;gt;0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $K_0$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Bessel_function#Modified_Bessel_functions_%3a_I.CE.B1.2C_K.CE.B1&quot; rel=&quot;nofollow&quot;&gt;modified Bessel function&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I find I am at a loss for how to interpret this result.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For simplicity, let's focus on the case $m=0$. Based on the last few paragraphs of Luboš's answer to &lt;a href=&quot;http://physics.stackexchange.com/questions/3168/why-correlation-functions&quot;&gt;this question&lt;/a&gt;, it seems  $K(x,y) = -\ln(\|x-y\|)/2\pi$ might be interpreted in one of two ways:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) As something that &quot;knows about the correlation of $\varphi(x)$ and $\varphi(y)$.&quot; In this case we would conclude that: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When $\|x-y\|&amp;lt;1$, it looks like $\varphi(x)$ and $\varphi(y)$ &quot;have the same sign.&quot; (What could that mean in this context?)&lt;/li&gt;&#xA;&lt;li&gt;When $\|x-y\|&amp;gt;1$, it looks like $\varphi(x)$ and $\varphi(y)$ &quot;have opposite signs.&quot; (Ditto)&lt;/li&gt;&#xA;&lt;li&gt;When $\|x-y\|=1$, it looks like $\varphi(x)$ and $\varphi(y)$ are independent of each other. (What would it mean for $\varphi(x)$ not to be correlated with $\varphi(y)$ only when the distance between the points is 1?) &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;2) As a probability amplitude for a particle to go from $x$ to $y$. This perspective doesn't seem to apply here: $K&amp;lt;0$ when $\|x-y\|&amp;gt;1$, and in any event $|K|\to\infty$ as $\|x-y\|\to\infty$. I mention it since this is the interpretation Luboš specifically gives in relation to correlation of operators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither of these possibilities, as I have understood them, looks correct to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Any help in clearing up my confusion would be most appreciated. &lt;em&gt;If&lt;/em&gt; it is possible to see the above propagators manifested in concrete data (perhaps results gathered in a lab (condensed matter?) or from computer simulations (Ising model?)), I think I would such a treatment most satisfying, but that may be asking too much. I don't know. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2966" LastEditorUserId="2966" LastEditorDisplayName="" LastEditDate="2011-08-04T04:09:06.097" LastActivityDate="2011-08-27T09:53:00.010" Title="Interpretation of basic free boson propagator (Euclidean action)" Tags="&lt;quantum-field-theory&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13165" PostTypeId="1" CreationDate="2011-08-04T05:51:20.230" Score="2" ViewCount="68" Body="&lt;p&gt;Imagine two blocks of masses m1 and m2 joined together by a spring of spring constant K.&#xA;Now let the spring be stretched by a distance X and then the system is released. suppose during the stretching the block of mass m1 moves towards left by a distance A and the block of mass m2 by a distance B. Now the center of mass of the system at any instant will remain at rest. Now let it be present at a point P in between the two blocks. so  can I depict the oscillations of the two blocks as two independent oscillations about the center of masss?&#xA;I appeal for an answer with strong reasons.&lt;/p&gt;&#xA;" OwnerUserId="4812" LastActivityDate="2011-08-04T16:10:58.637" Title="oscillations of blocks connected by a spring" Tags="&lt;homework&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="13166" PostTypeId="1" AcceptedAnswerId="13175" CreationDate="2011-08-04T07:54:07.887" Score="3" ViewCount="43" Body="&lt;p&gt;I was always wondering where the feeling of fuzz comes from when touching a CRT (Cathode Ray Tube) television on its display screen when the TV is turned on. Can someone please explain why I feel something fuzzy whenever I touch a CRT television screen when it is on?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be most appreciated if it can be explained in technical physics terms and in laymen terms too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that this fuzz feeling is electrostatic repulsion of the electron build up on the CRT television screen from the electron gun firing at the screen from the back onto my fingers.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2011-08-04T10:35:38.467" Title="The feeling of fuzz on the display surface of a Cathode Ray Tube Television" Tags="&lt;electromagnetism&gt;&lt;everyday-life&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="13167" PostTypeId="1" CreationDate="2011-08-04T08:06:44.083" Score="3" ViewCount="97" Body="&lt;p&gt;I have some questions about Pade approximants.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Given a divergent power series $ \sum_{n &amp;gt;0} a(n)x^{n} $ can we use a Pade Approximant to it $ R(x)$ so we can obtain a SUM of the series for every $x$ ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Given a Taylor power series $ \sum_{n &amp;gt;0} c(n)x^{n}=O(x^{a}) $ for positive $x &amp;gt;0$ for some positve but unknown $a$ can we obtain the value of $a$ by approximating the power series by a Pade Approximant?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can we compute a Borel transform of a series $ \sum_{n&amp;gt;0} \frac{a_n}{n!}x^{n} $ by a Pade Approximant?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can we use Pade approximation to obtain numerical integration of series and integrals ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4435" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-08-04T09:20:24.020" LastActivityDate="2011-08-05T10:24:42.557" Title="Pade Approximant" Tags="&lt;mathematical-physics&gt;&lt;regularization&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="13168" PostTypeId="2" ParentId="13125" CreationDate="2011-08-04T08:10:20.620" Score="1" ViewCount="" Body="&lt;p&gt;Image distortions you describe are because of the mirror being slightly bent. With sizes you mention (very elongated) the main reason would be the frame not rigid enough and deformed while being stored or transported.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also when you leave them resting onto the wall they will bend more and this will make the distortion slightly worse, so hanging them onto wall would be advantageous, but is not likely to fix the problem completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frame rigidness can be fixed by enforcing the frame - the first thing that comes to my mind is attaching some thin rigid hollow metal bars on the frame but if you're interested you can ask on &lt;a href=&quot;http://diy.stackexchange.com&quot;&gt;DIY SE&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3816" LastActivityDate="2011-08-04T08:10:20.620" CommentCount="2" />
  <row Id="13169" PostTypeId="1" CreationDate="2011-08-04T08:32:56.153" Score="1" ViewCount="186" Body="&lt;p&gt;Laser pointers manufacturers claim that some pointers have range of several kilometers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, they use a powerful laser, but that powerful laser usually has power less than one watt. Okay, the laser beam is very focused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about atmosphere particles like dust and vapor? Why don't these particles diffuse the beam and make it lose energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does it happen that a less than one watt laser can have several kilometers range in atmosphere?&lt;/p&gt;&#xA;" OwnerUserId="3816" LastActivityDate="2011-08-08T09:59:14.820" Title="How can a laser pointer have range of several kilometers in atmosphere?" Tags="&lt;light&gt;&lt;laser&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="13170" PostTypeId="1" CreationDate="2011-08-04T08:43:50.480" Score="11" ViewCount="185" Body="&lt;p&gt;I'm having a hard time understanding what some of the plots that are presented by ATLAS/CMS actually show. See for example: &lt;a href=&quot;http://resonaances.blogspot.com/2011/07/higgs-wont-come-out-of-closet.html&quot; rel=&quot;nofollow&quot;&gt;http://resonaances.blogspot.com/2011/07/higgs-wont-come-out-of-closet.html&lt;/a&gt;&#xA;What is the y-axis?&#xA;How would the plot look like if a Higgs was in the given mass range?&#xA;&lt;img src=&quot;http://i.stack.imgur.com/IbqWV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4813" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-08-04T11:09:32.213" LastActivityDate="2011-08-05T07:22:10.173" Title="Particle physics plots" Tags="&lt;particle-physics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="4" />
  <row Id="13171" PostTypeId="2" ParentId="13153" CreationDate="2011-08-04T08:50:52.113" Score="0" ViewCount="" Body="&lt;p&gt;The physical space is a space of orbits under $H$. That is, you start with some huge initial space $X$ but a little later you realize that many states represent the same physical state - any two of these are connected through the action of $H$. So, the physical space is $X / H$. Now, the physical space can be acted on by a symmetry group $G$. Now, to extend this to an action of $G \rtimes H$ on $X$ we need a &lt;strong&gt;connection&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is that? Well, we know how $H$ acts on all of $X$ and we know how $G$ acts on orbits of $X$ but we have no idea what does $G$ to the &lt;em&gt;points&lt;/em&gt; in the orbit (besides the fact that they get all mapped to points in some other orbit). Connection is precisely the tool that gives us this identification of points between various orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the language of topology (all your sets and groups are finite and so we can give them discrete topology), $X \to X / H$ is a fibre bundle with vertical action of $H$. We are also given an action of $G$ on the base. And we'd like to lift it to a horizontal action of $G$ on all of $X$ but horizontal structure on the bundle is the same thing as connection. Now, once we have actions of both $G$ and $H$ on $X$, we can combine these two in an obvious way by twisting the horizontal action by the vertical action (similarly to how one can twist translations by rotations to get a full group of symmetries of Eucidean space).&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-08-04T08:56:05.557" LastActivityDate="2011-08-04T08:56:05.557" CommentCount="5" />
  <row Id="13172" PostTypeId="2" ParentId="13170" CreationDate="2011-08-04T09:02:13.083" Score="4" ViewCount="" Body="&lt;p&gt;The plots in the y axis are representing  confidence levels for the value of the ratio of measured_cross_section/standard_model. The &lt;a href=&quot;http://vixra.files.wordpress.com/2011/07/hh00-11.jpg&quot; rel=&quot;nofollow&quot;&gt;Vixra blog&lt;/a&gt; has the plots better labeled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cut off line is y=1. If below the line, it means that the cross section is below that of the standard model within the confidence levels. The expected line has all the detector,  and decay mode choices ( these plots are composite of many decay modes) taken into account. What is below 1 is exclude within 2sigma, so the plot leaves a window below 150GeV where the Higgs may be hiding  and another around 280. The 280 is excluded by the CMS plot. Over 450 it is open season.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I interpret the plot on the lines that if what exists is only the standard model Higgs then with increasing statistics the lines should converge to 1, since the standard model in the denominator must have been evaluated for the corresponding Higgs mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are expectations plotted instead of mass distributions with predictions of SM? Because each individual channel studied has very few events, of the order of five or ten, whose statistical significance ( let alone systematic error estimates) leads to no exclusions or detection. Combining many channels in this format allows the imagination to  play the &quot;Can the Higgs be hiding here&quot; game with confidence levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 sigma, which are these plots, do not make for  a detection, 5 sigma is gold plated detection though a 4 sigma deviation from the standard model would get everybody excited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; : In order to have a first hand impression of the scarcity of data, yet, and of why the summary plots sound so esoteric have a look at a &lt;a href=&quot;http://indico.in2p3.fr/getFile.py/access?contribId=189&amp;amp;sessionId=6&amp;amp;resId=0&amp;amp;materialId=slides&amp;amp;confId=5116&quot; rel=&quot;nofollow&quot;&gt;summary talk&lt;/a&gt;. On page 32 there is an enlightening 4lepton plot for various channels, showing from 3 to 6 events over an expanse in mass of 400 GeV/c^2. Like medieval maps that marked &quot;here there be tigers&quot; one can look and say : probably there is no 350GeV/c^2 Higgs there since not even one event falls in its theoretical width. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The composite plots under discussion are a summation of various channels in the hope that summing the exclusions will extract some new physics knowledge. The method they found, that of dividing with the standard model predictions (shown as solid line in the linked plot, for a specific higgs mass),  allows a summary and combination of  the exclusion regions from  such plots, thus increasing the significance over all. One has to keep in mind though  that the information is tortured out of very few events.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-08-05T07:22:10.173" LastActivityDate="2011-08-05T07:22:10.173" CommentCount="5" />
  <row Id="13173" PostTypeId="2" ParentId="13169" CreationDate="2011-08-04T09:10:01.787" Score="0" ViewCount="" Body="&lt;p&gt;&quot;Range&quot; is a word liked by advertisment &quot;poets&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without a defintion to measure it, it is meaningless. &#xA;But in general: the power emitted by some sender &#xA;is never the sole criterium! &#xA;The sensitivity of the receiver is of same relevance. &#xA;In SW-band one can have connections around the globe &#xA;with one watt transmitted! &#xA;WRT dust etc: How does sunlight come down to us? &#xA;Of course the &quot;range&quot; of that pointer was not &quot;measured&quot; &#xA;at foggy conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last not least: &#xA;Pointers and similar lasers to be handeled freely are limited to &#xA;about 1 mW usually. A 1-Watt laser is a murderous weapon to eyes!&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-08-04T09:15:10.360" LastActivityDate="2011-08-04T09:15:10.360" CommentCount="3" />
  <row Id="13174" PostTypeId="2" ParentId="13170" CreationDate="2011-08-04T09:23:09.433" Score="4" ViewCount="" Body="&lt;p&gt;These plots (sometimes called &quot;brazil-band&quot; plots) show the 95% exclusion limit on the production rate ($\sigma\times BR$, i.e. productino cross section times branching fraction) of the Higgs boson as a function of its mass. This is usually normalize this to the $\sigma\times BR$ of the standard model Higgs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's take the ATLAS plot &#xA;&lt;img src=&quot;http://i.stack.imgur.com/we0Al.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At very low masses one the ATLAS experiment can with 95% confidence state &quot;If a Higgs boson exists and has mass 120GeV it is not produced at a rate 5 times larger than what the standard model would predict&quot; (this is the peak at the leftmost part of the plot)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this isn't very conclusive, since this doesn't rule out the existence of a higgs at that  mass. After all we weren't expecting that the Higgs is produced at &lt;em&gt;five times the rate&lt;/em&gt;. If the SM Higgs exists it will be produced with exactly the rate the SM predicts (i.e one times the rate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why the really interesting parts are the ones where the line drops below one. Then the experiments can state with 95% confidence &quot;Between 300 and 450 GeV, the Higgs boson, if it exists, is &lt;em&gt;not&lt;/em&gt; produced at the rate the SM predicts, thus we can rule out its existence in those mass ranges&quot;. Now there might be other particles in this mass range, but there surely is no Higgs boson that behaves (BR, crosssections, coupling) like the one prdicted by the SM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in conclusion the way to read this plots is to look for the places where the line drops below 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would the plot look like when there would be a higgs? Well at that mass point, the experiemnt shouldn't be able to exclude it. Higgs observation would then be another analysis and another set of plots (like looking for resonances). &lt;/p&gt;&#xA;" OwnerUserId="2151" LastEditorUserId="2151" LastEditorDisplayName="" LastEditDate="2011-08-04T09:34:35.633" LastActivityDate="2011-08-04T09:34:35.633" CommentCount="2" />
  <row Id="13175" PostTypeId="2" ParentId="13166" CreationDate="2011-08-04T09:46:32.660" Score="5" ViewCount="" Body="&lt;p&gt;Your guess is basicly right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the backside of the phosphor in a CRT there is a layer of aluminium, which reflects the &#xA;light emitted backward to the front, blocks positive ions from damaging the phosphor &#xA;and is used to carry of the electrons after they have done their duty. This aluminium &#xA;layer is connected to the anode (&quot;plate&quot;) terminal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this reason the beam is not the cause of the electric charge on the front suface &#xA;(which You describe as &quot;fuzzy feeling&quot;) &#xA;This charge is produced by influence (&quot;electrostatic induction&quot;). &#xA;The inner wall (that aluminium) is kept at some 10 to 30 kV positive during &#xA;operation. The charge induced from that layer to the outside stays there if t&#xA;here is no special surface treatment. Slowly it can creep away, but when You switch off, &#xA;a induction of opposite polarity arises. (this is often accompanied by a &#xA;rather loud &quot;hiss&quot; (corona). &#xA;On top of that the glass front can  leak small currents from the anode voltage, &#xA;this complicates analysis further. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the surface charge of CRTs can &quot;load&quot; dirt particles with charge and those fly then away with some &quot;push&quot;, the CRTs of PC monitors were treated to be conductive starting &#xA;from about mid-nineties.  (A lot of people complained about problems with burning/tears in eyes etc.) For some time one could buy frames with a fine metal mesh to be positioned in front of monitors of the first generation. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-08-04T10:35:38.467" LastActivityDate="2011-08-04T10:35:38.467" />
  <row Id="13176" PostTypeId="2" ParentId="2656" CreationDate="2011-08-04T12:18:26.303" Score="-1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;insolation of 2.61kWh/m2/year&lt;/li&gt;&#xA;  &lt;li&gt;perfectly efficient components (i.e. without violating Entropy laws&#xA;  or other theoretical limits)&lt;/li&gt;&#xA;  &lt;li&gt;light is emitted by an ideal &quot;white&quot; source&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This &quot;ideal white source&quot; is something unknown to physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for the first two items and assuming You mean sunlight &#xA;as the source: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thermodynamic limit of efficiency of a solar cell at Temperature T2 &#xA;insolated by (sun-) light of color temperature T1 is : (trara) :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Efficency = (T1 - T2) / T1 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 6000 K sunlight and 300 K cell this is : (6000 - 300) / 6000 = .95 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is 95 % &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-08-04T12:18:26.303" CommentCount="2" />
  <row Id="13177" PostTypeId="1" AcceptedAnswerId="13183" CreationDate="2011-08-04T12:29:50.613" Score="4" ViewCount="86" Body="&lt;p&gt;Hi I like to build a laser pointer that reflects different shapes for a project I am working on. I know this can be done with Diffractive Optical Elements. Are there other options? DOEs are expensive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see all these laser pointers with different shapes (arrow, hand etc), I am suspecting there is a lower cost but lower performance way. Any pointers?&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-05T00:39:21.850" LastActivityDate="2011-08-05T00:39:21.850" Title="Laser beam shaping" Tags="&lt;optics&gt;&lt;laser&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="13178" PostTypeId="2" ParentId="13177" CreationDate="2011-08-04T13:04:31.723" Score="0" ViewCount="" Body="&lt;p&gt;One option would be to expose a hologram into &lt;strong&gt;photographic film&lt;/strong&gt; (or ideally a photographic plate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also heard of people who wrote holograms into (or onto) &lt;strong&gt;plastic&lt;/strong&gt; but I'm not sure if you need pulsed lasers to write those holograms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cool solution is to use a &lt;strong&gt;phase-only spatial light modulator&lt;/strong&gt;, e.g. by &lt;a href=&quot;http://jp.hamamatsu.com/products/other/1013/X10468/index_en.html&quot; rel=&quot;nofollow&quot;&gt;Hamamatsu&lt;/a&gt;. The Hamamatsu device costs probably 5000 EUR but possibly you can use a normal LCD screen after you removed the polarizer foils.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coolest solution (in my opinion) is to build an &lt;a href=&quot;http://en.wikipedia.org/wiki/Acousto-optic_deflector&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;acousto optic deflector&lt;/strong&gt;&lt;/a&gt;. There you use an ultrasound transducer to create a thick grating within some transparent crystal  (but I think glass works as well). The laser light will be deflected. You can vary the angle of deflection by changing the frequency of the ultrasound. You can put two of these things behind each other. Then you can scan the beam in XY (like an oscilloscope). Note that this isn't really shaping the beam. But you can scan the beam very fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I always wondered if one could write a hologram on a &lt;strong&gt;DVD&lt;/strong&gt; or &lt;strong&gt;Blue-Ray medium&lt;/strong&gt;. If one understands the burners firmware good enough to put the data into specific places this might actually work. And 50Gb of information would make for some pretty detailed pictures.&lt;/p&gt;&#xA;" OwnerUserId="4306" LastEditorUserId="4306" LastEditorDisplayName="" LastEditDate="2011-08-04T14:12:47.577" LastActivityDate="2011-08-04T14:12:47.577" CommentCount="3" />
  <row Id="13179" PostTypeId="2" ParentId="13143" CreationDate="2011-08-04T13:12:29.847" Score="4" ViewCount="" Body="&lt;p&gt;From: &lt;a href=&quot;http://nobelprize.org/nobel_prizes/physics/articles/curie/&quot; rel=&quot;nofollow&quot;&gt;NobelPrize.org &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Her continued systematic studies of the various chemical compounds gave the surprising result that &lt;em&gt;the strength of the radiation did not depend on the compound that was being studied. It depended only on the amount of uranium or thorium.&lt;/em&gt; Chemical compounds of the same element generally have very different chemical and physical properties: one uranium compound is a dark powder, another is a transparent yellow crystal, &lt;strong&gt;but what was decisive for the radiation they gave off was only the amount of uranium they contained. Marie drew the conclusion that the ability to radiate did not depend on the arrangement of the atoms in a molecule, it must be linked to the interior of the atom itself.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;This discovery was absolutely revolutionary. From a conceptual point of view it is her most important contribution to the development of physics.&lt;/em&gt;&lt;/strong&gt; &quot;&lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2011-08-04T13:12:29.847" CommentCount="7" />
  <row Id="13180" PostTypeId="2" ParentId="13165" CreationDate="2011-08-04T13:36:38.203" Score="4" ViewCount="" Body="&lt;p&gt;The two oscillations will not be independent as they will share frequency and phase. You can start from Newton's equations of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/B0FaE.png&quot; alt=&quot;Two Mass &amp;amp; Spring&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m_1 \frac{{\rm d}^2\,a(t)}{{\rm d}t^2} = -F(t) $$&#xA;$$ \mbox{-}m_2 \frac{{\rm d}^2\,b(t)}{{\rm d}t^2} = F(t) $$&#xA;$$ F(t) = k \;\left( a(t)+b(t) \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume simple harmonic motion $a(t) = A \sin(\omega\,t)$, $b(t) = B \sin(\omega\,t)$ which leads to the frequency equation $$ k\,\left(m_1+m_2\right) = m_1 m_2 \omega^2 $$ and the amplitude equation $$ A\, m_1 = B\, m_2 $$ So now you can show that the center of gravity $P$ does not move as long as the amplitudes $A$ and $B$ obey the balance equation above. How? Well what is the equation for the center of gravity? $$ {\rm cg} = \frac{\mbox{-}B\,m_2+A\,m_1}{m_1+m_2} = 0 $$&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-08-04T13:36:38.203" CommentCount="4" />
  <row Id="13181" PostTypeId="2" ParentId="13165" CreationDate="2011-08-04T14:09:35.973" Score="0" ViewCount="" Body="&lt;p&gt;Traditionally, the &quot;mass on a spring&quot; is analyzed with one end of the spring held fixed. Your problem is equivalent to two such systems back-to-back. You have identified the center of gravity as a fixed point; so you can literally fix it and then cut the problem in two. Each half of the system then behaves like a traditional &quot;mass-on-a-spring&quot;.&lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-08-04T14:09:35.973" />
  <row Id="13182" PostTypeId="2" ParentId="13165" CreationDate="2011-08-04T14:42:46.833" Score="1" ViewCount="" Body="&lt;p&gt;Center of mass is a good point to start, but You have to divide &#xA;the spring accordingly! The two lengths have to have the reverse &#xA;ratio of the masses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This point then is like a fixed wall, bearing a big mass and a &#xA;hard spring on one side, a smaller mass and a softer spring on the other side. &#xA;This will lead to two pendulums of same frequency, &#xA;(Frequency is prop to spring constant divided by mass) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem arises frequently in oscillation of asymmetric molecules or &#xA;rotation of such molecules (eg HCl) &#xA;An expression to google for in this context is &quot;reduced mass&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Reduced_mass&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Reduced_mass&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-08-04T16:10:58.637" LastActivityDate="2011-08-04T16:10:58.637" />
  <row Id="13183" PostTypeId="2" ParentId="13177" CreationDate="2011-08-04T15:29:52.583" Score="6" ViewCount="" Body="&lt;p&gt;It is possible to do diffractive optics really cheap - either with photography (with a non-digital camera) or a widely available DTP polygraphic technique. What you need is to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Make a Fourier transform of an image you want to have (it should be relatively simple).&lt;/li&gt;&#xA;&lt;li&gt;Make an image which is white when the real part of the FT is positive, and black - when it is negative. &lt;/li&gt;&#xA;&lt;li&gt;Either print it, make a photo with a classical camera, develop the film and use it as you diffractive device OR prepare a pdf/ps file and go to a DTP point (resolution up to 5000dpi should be possible; a few $ per A4 sheet on which you can make a lot of patterns).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As it uses absorption, not refraction, so some loss of intensity is unavoidable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some patterns can be found here (mainly: Fresnel zone plates and diffractive gratings)&#xA;&lt;a href=&quot;http://migdal.wikidot.com/en:spz-moire-ps&quot; rel=&quot;nofollow&quot;&gt;http://migdal.wikidot.com/en:spz-moire-ps&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="184" LastEditorUserId="184" LastEditorDisplayName="" LastEditDate="2011-08-04T16:53:52.903" LastActivityDate="2011-08-04T16:53:52.903" CommentCount="4" />
  <row Id="13184" PostTypeId="1" AcceptedAnswerId="13189" CreationDate="2011-08-04T17:02:41.147" Score="4" ViewCount="271" Body="&lt;p&gt;Obviously this is impossible in relativity; however, if we ignore relativity and use only Newtonian mechanics, is this possible?  How (or why not)?&lt;/p&gt;&#xA;" OwnerUserId="175" LastActivityDate="2011-08-05T02:47:03.633" Title="Can an object accelerate to infinite speed in FINITE time (Newtonian)?" Tags="&lt;newtonian-mechanics&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="13185" PostTypeId="2" ParentId="13184" CreationDate="2011-08-04T17:14:11.823" Score="-1" ViewCount="" Body="&lt;p&gt;If by &quot;ignoring relativity&quot;, you mean ignoring the fact that nothing can move faster than the speed of light, then the answer is still no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since kinetic energy is proportional to the square of the speed, infinite speed would mean infinite energy, which you cannot provide, whatever the amount of time you are considering.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastActivityDate="2011-08-04T17:14:11.823" CommentCount="2" />
  <row Id="13187" PostTypeId="1" CreationDate="2011-08-04T17:27:06.947" Score="4" ViewCount="109" Body="&lt;p&gt;For flat FLRW cosmology, we can write down two Friedman equations and one matter equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $H^2=\frac{8 \pi G}{3} \rho$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $\frac{\ddot{a}}{a} = -\frac{4 \pi G}{3} (\rho +3p)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) $\dot{\rho} = -3 H (\rho +p)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well known that, given an equation of state, (1)+(3) implies (2) [also (1)+(2) implies (3)]. My question is: Why are (2) and (3) (plus equation of state) not a complete set of equations? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure it's probably completely obvious to some of you, but not to me.&lt;/p&gt;&#xA;" OwnerUserId="1429" LastActivityDate="2011-08-05T17:26:39.030" Title="Which Friedmann equation is redundant?" Tags="&lt;general-relativity&gt;&lt;cosmology&gt;" AnswerCount="1" />
  <row Id="13188" PostTypeId="2" ParentId="13169" CreationDate="2011-08-04T17:53:48.237" Score="0" ViewCount="" Body="&lt;p&gt;It is most likely the same reason as why the sky is blue. More specifically, Rayleigh scattering is proportional to the fourth inverse power of the wavelength, so if you choose the wavelength of the laser to be large (i.e. towards the red end of the visible spectrum), then you can suppress Rayleigh scattering significantly. In that case the range of the beam can be very large.&lt;/p&gt;&#xA;" OwnerUserId="4483" LastActivityDate="2011-08-04T17:53:48.237" />
  <row Id="13189" PostTypeId="2" ParentId="13184" CreationDate="2011-08-04T18:04:07.927" Score="19" ViewCount="" Body="&lt;p&gt;The answer is yes in some unintersting senses: Take two gravitational attracting point particles and set them at rest. They will attract each other and their velocity will go to $\infty$ in finite time. Note this doesn't contradict conservation of energy since the gravitational potential energy is proportional to $-1/r$. This isn't so interesting since it's just telling you that things under gravity collide. But its technically important in dealing with the problem of gravitationally attracting bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a more intersting question: Is there a situation where the speed of a particle goes to infinity without it just being a collision of two bodies?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suprisingly, the answer to this question is yes, even in a very natural setting. The great example is given  Xia in 1995 (Z. Xia, “The Existence of Noncollision Singularities in Newtonian Systems,” Annals Math. 135, 411-468, 1992). His example is five bodies gravitational interacting. With the right initial conditions one of the bodies can be made to oscillate faster with the frequency and amplitude going to infinity in finite amount of time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Added&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an image. The four masses $M$  are paired into two binary systems which rotate in opposite directions. The little mass $m$ oscillates up and down faster. It's behavior becomes singular in finite time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/0zaqz.gif&quot; alt=&quot;Five body setup&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3970" LastEditorUserId="3970" LastEditorDisplayName="" LastEditDate="2011-08-04T19:57:19.890" LastActivityDate="2011-08-04T19:57:19.890" CommentCount="6" />
  <row Id="13190" PostTypeId="2" ParentId="13184" CreationDate="2011-08-04T18:12:33.427" Score="-2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Lab Reference Frame&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start assuming a force is exerted on the object from equipment &quot;lab&quot; reference frame, meaning that there is negligible recoil on the lab frame from accelerating the object.  Not only is it the case that $F=ma$, but the power delivered to the object grows as $P=v F$.  Take a device like the large hadron collider, and just completely wave off the technical difficulty of applying force to an object of progressively increasing velocity, and let's not the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v&amp;#39; = a = \frac{F}{m} = \frac{P}{v m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The requirement is that $v \rightarrow \infty$ in a finite time.  You know, just for fun let's use an actual functional form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v(t) = -\frac{1}{t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For t from $-\infty$ to $t=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = m v&amp;#39; v = -\frac{m}{t^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the power delivered must increase at the hilariously fast rate of $1/t^3$ as $t$ goes to zero, not that it matters because we all knew this would result in requiring infinite energy, but this just shows exactly how intangible it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rocket&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a rocket the propellant has to be hauled along with the payload, but the tradeoff is that you then don't have the $P=v F$ proportionality, since the propellant itself is still moving after being ejected.  The equation of motion for a rocket that starts with mass $M$ and ends with mass $m$, ejects propellant with speed $v_e$ is as follows.  I'll add in the approximation for $M \gg m$, because obviously that must be the case since we're talking about going to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = v_e \ln{ \frac{M}{m} } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is certainly interesting.  It is interesting to note that the speed the rocket can reach is proportional to the speed the propellant is ejected at, which is also not limited by the speed of light.  The energy required to expel that propellant also isn't a problem (haha) because it is not the case that $E=mc^2$ in this world.  But it would be skirting the problem to say that the propellant is ejected at infinite speed, so we still seek a way for the above expression to limit to infinity with $v_e \neq \infty$.  But we would also really like $\frac{M}{m} \neq \infty$, because that would require either an infinitely large starting mass or an infinitely small ending mass.  Neither of these options are appealing, unless atoms also don't exist in this world, allowing us to use fractal math to claim that an infinitely small chuck of something went to infinite speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the extent that we don't make these absurd assumptions, your request is impossible, even given the already absurd assumption of allowable superluminal speed.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-08-05T02:47:03.633" LastActivityDate="2011-08-05T02:47:03.633" CommentCount="5" />
  <row Id="13191" PostTypeId="1" CreationDate="2011-08-04T18:14:51.693" Score="-2" ViewCount="54" Body="&lt;p&gt;After watching Lawrence Krauss' lecture 'life, universe, and nothing', an interesting idea lighted up in my mind. Is it possible that dark matter is just spread out matter waves and dark energy extremely long wavelength light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;matter -&gt; expansion -&gt; dark matter (delocalized matter waves/huge atoms)&#xA;visible and xray radiation -&gt; expansion -&gt; invisible light with wavelength greater than the diameter of universe (perhaps the source of energy of the empty space?!!!!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so dark matter today may be what we will become in the future. and we are what dark matter was in the past. eternal expansion just spreads out everything but new stuff is created all the time. anyone living in such a universe at any point of time will see the universe almost as we see it today, expanding with a beginning.&lt;/p&gt;&#xA;" OwnerUserId="4819" LastActivityDate="2011-08-04T23:37:50.513" Title="Could dark matter be expanded atoms, and dark energy expanded radiation?" Tags="&lt;cosmology&gt;&lt;dark-matter&gt;&lt;dark-energy&gt;&lt;inflation&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="13192" PostTypeId="2" ParentId="13161" CreationDate="2011-08-04T18:17:01.273" Score="4" ViewCount="" Body="&lt;p&gt;The standard explanation is that there is a constant flux of energy from large eddies to smaller eddies.  The time scale for an eddy of scale $r$ to turn over is $\tau \sim \frac{r}{v(r)}\sim \frac{1}{kv}$, the energy density for scale $r$ is $\sim v(r)^2$, so you get an energy flux rate $\epsilon \sim v^2/\tau \sim v^3k$ which is assumed constant.  You then estimate she spectral energy density as $E \sim v^2/k \sim \epsilon^{2/3}k^{-2/3 -1} = \epsilon^{2/3} k^{-5/3}$.  I hope that is a little better than pure dimensional analysis.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-08-04T18:17:01.273" />
  <row Id="13193" PostTypeId="2" ParentId="13191" CreationDate="2011-08-04T18:26:59.280" Score="2" ViewCount="" Body="&lt;p&gt;extremely long wavelength light would be measurable as essentially constant (in time and space) electric and magnetic fields. We do not see such thing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what a 'expanded' atom would look like. If its an atom with a different $\hbar$ then we neither have seen one of those&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-08-04T18:26:59.280" CommentCount="1" />
  <row Id="13195" PostTypeId="2" ParentId="11334" CreationDate="2011-08-04T18:48:17.373" Score="-1" ViewCount="" Body="&lt;p&gt;My Wikipedia suggestion for this problem is &lt;a href=&quot;http://en.wikipedia.org/wiki/Faraday%27s_law_of_induction&quot; rel=&quot;nofollow&quot;&gt;Faraday's law of induction&lt;/a&gt;.  They sum it up in pretty much a single quote.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The induced electromotive force (EMF) in any closed circuit is equal to the time rate of change of the magnetic flux through the circuit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are lots of technicalities of motors and generators, but they're not necessary for this problem.  The fundamental principle is that there is a wire spinning while in a magnetic field.  The EMF, I'll denote $V$ for voltage, is quantified as follows with $r$ being the rotating radius of the coil assuming it's rectangular (as well as rotating in the right direction), $l$ is the other dimension of the rectangular loop, $B$ is the magnetic field, $\omega$ is the speed of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V = B r l \omega$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If any single one of these factors had unlimited potential to increase then a motor could deliver infinite voltage.  Of course they are all limited.  The most obvious way to scale up power is to make a bigger machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one missing piece, which is that EMF refers to the voltage that can be either produced or converted into a mechanical action.  That does not say anything of current, so taken at face value, such a simple coil rotating in a constant magnetic field would allow infinite power conversion if there were infinite current.  Current in any wire or bundle of wires, is, of course limited by resistive heating limits.  You can go find plenty of information about these limits but I will not cover them here.  Yes, it is possible to use superconducting wires for both the primary coils as well as the magnetic field generating coils, but they also do not allow infinite energy conversion, and yes, there are companies that sell these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.amsc.com/products/motorsgenerators/index.html&quot; rel=&quot;nofollow&quot;&gt;http://www.amsc.com/products/motorsgenerators/index.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not familiar enough with the technology to say for sure, but I believe that the problem is still resistive heating.  Superconductors generate much less heat, but each unit of heat they produce is much more expensive to remove if it's a low temperature superconductor.  The 2nd law of thermodynamics gives a direct penalty on a heat flux out of a refrigerated system.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-08-04T18:48:17.373" CommentCount="3" />
  <row Id="13196" PostTypeId="1" CreationDate="2011-08-04T19:06:45.097" Score="-2" ViewCount="76" Body="&lt;p&gt;I pressed two countdown timers at 60 minutes at the 4 seconds apart. When both were at 30 minutes, one was at 30 minutes the other was at 30:40. Why the lag?&lt;/p&gt;&#xA;" OwnerUserId="4820" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-04T19:15:27.337" LastActivityDate="2011-08-04T19:15:27.337" ClosedDate="2011-08-04T19:12:34.670" Title="What is the reason for this?" Tags="&lt;time&gt;" CommentCount="3" />
  <row Id="13197" PostTypeId="2" ParentId="13187" CreationDate="2011-08-04T19:11:11.037" Score="3" ViewCount="" Body="&lt;p&gt;I suppose a lot of what constitutes the &quot;Friedmann Equation(s)&quot; is just up to definitions. However, with the 3 equations you listed, there will be redundancy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the normal derivation I usually see (feel free to skip the first two paragraphs if you aren't familiar with tensors/GR):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given einstien's equation in the form $R_{\mu\nu}=-8 \pi G S_{\mu \nu}$, (where $S_{\mu \nu}$ is related to the stress energy tensor by $S_{\mu \nu}=T_{\mu \nu} - \frac{1}{2}g_{\mu \nu}T^\lambda_\lambda$) and with some playing around with the spatial portion of the Robertson Walker Metric (see Weinburg Cosmology), we can get a Riemann Tensor $R_{ij}=\tilde{R}_{ij}-2\dot{a}^2 \tilde{g}_{ij}-a\ddot{a}\tilde{g}_{ij}$ (tilde means the spacial metric and it's curvature tensor) to $R_{ij}=-[2K+2\dot{a}^2+a\ddot{a}]\tilde{g}_{ij}$ (where $K$ is the curvature constant (-1,0,+1)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then decide on a stress energy tensor, using the principles of homogeneity and isotropy (we don't want there to be some strange asymmetry), to get one of the form $T_{00}=\rho$, $T_{i0}=0$, and $T_{ij}=a^2p\tilde{g}_{ij}$. Which gives us $S_{ij}=\frac{1}{2}(\rho-p)a^2\tilde{g}_{ij}$ and $S_{00}=\frac{1}{2}(\rho+3p)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using all this, plugging back into the EFEs, we get two equations (one for i=j=0, and one for the rest):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $-\frac{2K}{a^2}-2\frac{2\dot{a}^2}{a^2}-\frac{\ddot{a}}{a}=-4\pi G (\rho - p)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $\frac{3\ddot{a}}{a}=-4\pi G(3p+\rho)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, cause the first equation is sorta unwieldy, we can add three times the first equation to the second to get the nicer (and more familiar):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) $\dot{a}^2+K=\frac{8}{3}\pi G a^2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also get the following equation from (1) and (2):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) $\dot{\rho}=-\frac{3\dot{a}}{a}(\rho + p)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is really no surprise, as this conservation law is found in any solution to the EFEs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So really what equations you decide to call the &quot;Friedman Equations&quot; is up to whatever your doing. (1) and (2) are the direct consequences of derivation, which you can then use to derive (3) and (4). It just turns out for most cosmology calculations, we don't want to use (1), and the last 3 ((2)(3)(4)) are more useful. However there will be redundancy within these 3 equations (after all, they came from just two equations)!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: Just to adress the OPs question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets take (2) and (4):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $\frac{3\ddot{a}}{a}=-4\pi G(3p+\rho)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) $\dot{\rho}=-\frac{3\dot{a}}{a}(\rho + p)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets multiply (2) by $\frac{2a\dot{a}}{3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{2a\dot{a}}{3}[\frac{3\ddot{a}}{a}]=\frac{2a\dot{a}}{3}[-4\pi G(3p+\rho)] =&amp;gt; 2\dot{a}\ddot{a}=-\frac{8}{3}\pi G(3p a\dot{a} + \rho a\dot{a})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some tricky algebra here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=&amp;gt;2\dot{a}\ddot{a} = \frac{8}{3}\pi G(-3 a\dot{a}(\rho+p) + 2\rho a\dot{a})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now substitute in the conservation of energy equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) $2\dot{a}\ddot{a} = -\frac{8}{3}\pi G(\dot{\rho}a^2+2\rho a \dot{a})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But hey! This looks sorta like what you would get if you differentiated the Friedmann Equation (3), note that $K$ is not time dependent (if it were, that would be crazy!). Since your integrating instead of differentiating you will have constants of integration, but you can just tie that into the $K$ term (which is a sorta interesting way to view what $K$ means).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Doing a similar process will show you a way to derive (4) to begin with.&lt;/p&gt;&#xA;" OwnerUserId="3018" LastEditorUserId="3018" LastEditorDisplayName="" LastEditDate="2011-08-05T17:26:39.030" LastActivityDate="2011-08-05T17:26:39.030" CommentCount="9" />
  <row Id="13198" PostTypeId="1" AcceptedAnswerId="13200" CreationDate="2011-08-04T19:25:48.840" Score="2" ViewCount="44" Body="&lt;p&gt;What is the role of &lt;a href=&quot;http://en.wikipedia.org/wiki/Smoothed-particle_hydrodynamics&quot; rel=&quot;nofollow&quot;&gt;smoothed particle hydrodynamics&lt;/a&gt; (SPH) in cosmological N-body simulations like the &lt;a href=&quot;http://www.mpa-garching.mpg.de/galform/virgo/millennium/&quot; rel=&quot;nofollow&quot;&gt;Millenium Run&lt;/a&gt; (performed with &lt;a href=&quot;http://www.mpa-garching.mpg.de/gadget/&quot; rel=&quot;nofollow&quot;&gt;Gadget-2&lt;/a&gt;)?&lt;/p&gt;&#xA;" OwnerUserId="47" LastActivityDate="2011-08-04T22:23:24.650" Title="Smoothed particle hydrodynamics in cosmological N-body simulations" Tags="&lt;cosmology&gt;&lt;simulation&gt;&lt;numerics&gt;" AnswerCount="1" />
  <row Id="13199" PostTypeId="2" ParentId="2656" CreationDate="2011-08-04T20:36:41.150" Score="2" ViewCount="" Body="&lt;p&gt;Using that ideal white source on the wiki link, that states 251 lm/W.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The insolation level is $2.61kWh/m^2d = \frac{2610}{24}W/m^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which across $65cm^2 (= 0.0065m^2)$ gives 0.7W. If everything were 100% efficient, then you'd have $0.7 \times 251 lm = 178 lm$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we derate on some maximum theoretical efficiencies. The biggest theoretical derating will be on the PV, and that will completely dwarf any loss on the maximum theoretical efficiency of a round trip into storage and back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a single-junction n-p PV cell in unconcentrated sunlight, the maximum efficiency (&lt;a href=&quot;http://dx.doi.org/10.1063/1.1736034&quot; rel=&quot;nofollow&quot;&gt;Shockley–Queisser, DOI:10.1063/1.1736034&lt;/a&gt;) is 30%, giving you about 53 lm. By layering multiple junctions, you could theoretically get 42% (2 junctions, 74 lm), 49% (3 junctions, 87 lm), tending to 68% ($n\to\inf$, 121 lm) (&lt;a href=&quot;http://dx.doi.org/10.1088/0022-3727/13/5/018&quot; rel=&quot;nofollow&quot;&gt;doi: 10.1088/0022-3727/13/5/018&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you are allowed to put a concentrating lens onto the $65 cm^2$ cell, so that the sun jar harvests light from a much larger area, then we can really go to town. From the second link, the maximum efficiencies for concentrating PV, as the number of junctions tends to infinity, is 86.8%. So then you've got to find out what the maximum concentration could be without the whole lot bursting into flames ... that 86.8% is based on a concentration factor of 45 900 , so it's a PV cell made of pure unobtainium, and I'll stop right there.&lt;/p&gt;&#xA;" OwnerUserId="4066" LastActivityDate="2011-08-04T20:36:41.150" />
  <row Id="13200" PostTypeId="2" ParentId="13198" CreationDate="2011-08-04T22:23:24.650" Score="0" ViewCount="" Body="&lt;p&gt;Check here &lt;a href=&quot;http://arxiv.org/abs/1107.1789&quot; rel=&quot;nofollow&quot;&gt;Cosmology today-A brief review&lt;/a&gt; chapter &lt;strong&gt;Thermodynamics in the early Universe&lt;/strong&gt;, page 10 ... follow the word gas ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;some words here would be fine, but, ... find for yourself.  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-08-04T22:23:24.650" />
  <row Id="13201" PostTypeId="2" ParentId="13191" CreationDate="2011-08-04T23:37:50.513" Score="2" ViewCount="" Body="&lt;p&gt;This cannot explain dark energy, even in principle, for the following reason. If dark energy were any kind of &quot;invisible light&quot; (that is, electromagnetic radiation of very large wavelength), its pressure and energy density would be related in the standard way for radiation: $p=1/3 \rho c^2$, where $p$ is pressure, $\rho$ is energy density, and $c$ is the speed of light. In other words, in units where $c=1$, the equation of state of radiation is $w=p/\rho=1/3$. There is no wiggle room here; this has been known for more than 100 years and is a result taught in undergraduate physics courses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the effect of dark energy is to 'push things apart'. Equivalently, its pressure is very negative. In fact, its equation of state has been &lt;em&gt;measured&lt;/em&gt; to be $w\approx -1$, with only small error bars (of order 0.1). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So dark energy definitely cannot be explained by any kind of radiation (which has $w=1/3$), matter (which has $w=0$), matter waves, invisible light, etc etc. &lt;/p&gt;&#xA;" OwnerUserId="1334" LastActivityDate="2011-08-04T23:37:50.513" />
  <row Id="13202" PostTypeId="2" ParentId="5320" CreationDate="2011-08-05T00:04:39.040" Score="2" ViewCount="" Body="&lt;p&gt;There's a lot of confusion here, which is not surprising given that a few things are going on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physical distances in the universe e.g. between galaxies - measured in meters, miles, parsecs, or other favorite units of yours -  will continue to increase since the universe will continue to expand. In fact, as the universe becomes completely dark energy dominated (soon -  in some dozen billion years), the physical distances will increase &lt;em&gt;exponentially&lt;/em&gt; in time, $d_P\propto e^{Ht}$, where $H$ is the Hubble parameter which is now decreasing (it's equal to 71 km/s/Mpc &lt;em&gt;today&lt;/em&gt;, and going down), but will stabilize to a constant when DE takes over. So $H$ in that exponent will become constant, the distance will increase purely exponentially. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the radius of the observable part of the universe - the &quot;horizon distance&quot; discussed in some other posts here - is increasing slower, and will actually come to a halt. This becomes clear once you recall that the horizon distance is proportional to $1/H(t)$. The Hubble parameter $H(t)$ will, recall, become essentially constant 'soon', and so will the horizon distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So all of your favorite faraway galaxies are moving away from you (almost) exponentially with time, while the radius of the observable universe becomes (almost) constant. What gives? The galaxies are leaving our observable universe (because of dark energy, remember)! Those far away first, those closer to us later. This is the origin of the statement that only objects gravitationally &lt;em&gt;bound&lt;/em&gt; to us - our Solar System, Milky Way Galaxy, and the Local Group - will remain observable a few tens of billion years in the future. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original question was then something to the effect of 'when can you see the most galaxies in the universe'? In principle, this question should be easy to answer (the largest-number-of-galaxies moment has already happened I believe). But in practice the exact answer might be hard to calculate, since galaxies also have an annoying habit of being created and destroyed (e.g. by their collisions etc), not just sitting still. I have not done a back-of-envelope calculation to try to estimate this, but one could. &lt;/p&gt;&#xA;" OwnerUserId="1334" LastActivityDate="2011-08-05T00:04:39.040" CommentCount="1" />
  <row Id="13203" PostTypeId="1" AcceptedAnswerId="13205" CreationDate="2011-08-05T00:05:25.857" Score="1" ViewCount="42" Body="&lt;p&gt;Is it possible to create periodic potentials that instead of creating a well for an atom to be trapped in, repulsed by that specific location?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, can we use this as a means to make artificial objects that we can feel, with let's say our hands, by putting lots of these repulsive points together?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an idea that I have in order to make holograms that we can feel like the holodeck in Star Trek.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-08-05T00:28:13.940" LastActivityDate="2011-08-05T00:28:13.940" Title="Creating the opposite of an optical lattice" Tags="&lt;electromagnetism&gt;&lt;optics&gt;&lt;electrostatics&gt;&lt;hologram&gt;&lt;optical-lattice&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="13204" PostTypeId="1" CreationDate="2011-08-05T00:13:44.140" Score="1" ViewCount="43" Body="&lt;p&gt;Does anyones know which materials (or type of steel) are used to make rocket engines nozzles &amp;amp; combustion chambers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the tricks to reduce temperature of the engine except trying to cool it with fuel/oxidant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any good books around on rocket engines design?&lt;/p&gt;&#xA;" OwnerUserId="930" LastActivityDate="2011-08-05T00:13:44.140" ClosedDate="2011-08-05T07:59:52.030" Title="Rocket engines design &amp; materials" Tags="&lt;rocket&gt;&lt;rocket-science&gt;" CommentCount="3" />
  <row Id="13205" PostTypeId="2" ParentId="13203" CreationDate="2011-08-05T00:27:53.597" Score="1" ViewCount="" Body="&lt;p&gt;Atoms in an optical lattice are attracted to high/low field states depending on the quantum state the atom is in. Atoms in states that would be shifted down in energy by the application of an electric field are drawn to low-field regions and repulsed from high-field regions, and vice versa for atoms in states that would be shifted up in energy by the application of an electric field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to part 1 is that this is possible, and in fact atoms are already &quot;repulsed&quot; by parts of the lattice anyway.  Which part of the lattice is attractive/repulsive depends on the state the atom is in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because optical lattices are by definition periodic, these &quot;repulsive&quot; regions are surrounded by attractive regions, and the atoms in you hand tend to migrate from the repulsive to the attractive regions.  I don't think you'd be able to make a macroscopic force you could feel with your hand.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2011-08-05T00:27:53.597" />
  <row Id="13206" PostTypeId="2" ParentId="11334" CreationDate="2011-08-05T03:48:32.407" Score="2" ViewCount="" Body="&lt;p&gt;The small matter of the magnetic saturation of iron really ought to be mentioned in this discussion. In practise, the best iron cannot sustain a magnetic field of more than about 1 Tesla, or 1 volt-sec/m^2. We must then ask: how much current does it take to drive a field of that magnitude? That depends of course on the magnetic permeability of the iron and the length of the path, but even in the best case scenario of infinite permeability, the flux still has to cross the air gap between rotor and stator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can calculate the flux given the field by first dividing by 377 ohms (the impedance of free space) and then multiplying by the speed of light. You can verify that this comes to something on the order of 10^6 Amps/meter (the units check out nicely). If we take as a first guess that the air gap is one millimeter, it is evident that we need a current of at least 1000 amps (or ampere-turns if you like) to drive the field. In practise, this calls for a copper cross-section on the order of one square centimeter. Let's make it actually square so the width of the copper is one centimeter. Then let's make the width of the iron pole pieces the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take a rotor circumference of one meter (radius approx. 6 inches) and alternate copper and iron so we get fifty pole pieces. Ideally, at each station we get the maximum field crossing the maximum amperage: (one Tesla) x (1000 amps) x 50 poles. You can verify that this comes to 50 000 Joules/m^2. That's not a torque, and that's not even a force yet, because we haven't accounted for the length of the rotor. Let's take a half meter: that gives a force of 25 kJoules/meter. It's a &quot;force&quot; because it's the force which is turning the motor; to convert it to a torque, we multiply by the radius of 0.15 meter. The torque is about 4 kNewton-meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this relate to a practical motor? We can get the horsepower by multiplying torque by rotational frequency. At 3600 rpm, the frequency in radians per second is of course 377. So we get a horsepower of 4000*377 = 1500 kW, or about 2000 hP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is off by a factor of about ten for a typical industrial motor of this size, which would likely be closer to 200 hP. But it puts you in the general ballpark. The only factor which is totally made up is the 1-millimeter spacing I assumed for the air gap between rotor and stator. &lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-08-05T03:48:32.407" CommentCount="2" />
  <row Id="13208" PostTypeId="1" AcceptedAnswerId="13228" CreationDate="2011-08-05T04:09:17.777" Score="10" ViewCount="130" Body="&lt;p&gt;In preparation for an exam, I'm revisiting old exam questions. This one seems neat, but also quite complicated:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A soccer ball with Radius $R=11cm$ is inflated at a pressure of $P =9 \times 10^4 Pa$, then dropped from a height of $0.1m$ (distance floor to lowest part of ball) onto a hard floor and bounces off elastically.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Find approximate expressions for:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Surface area of ball in contact with floor&lt;/li&gt;&#xA;  &lt;li&gt;Amount of time the ball is in contact with the floor&lt;/li&gt;&#xA;  &lt;li&gt;Peak force exerted on the floor&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;if the mass is $0.42 kg$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My attempt at a solution: Assume the ball is filled with an ideal gas and that the process is adiabatic. Assume that the deformation leads to a simple spherical segment, i.e. a ball where one part is cut off flat. This gives an expression of the volume $V$ in terms of the height of the &quot;center&quot; of the ball, $h$ as&#xA;$$V = \frac{11}{6}\pi R^3 + \frac{\pi}{6}h^3$$&#xA;The surface area then is simply &#xA;$$A = \pi (R^2 - h^2)$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next: The ball has potential energy $mgh_0$, which is completely converted to internal energy at the peak point of the motion. The internal energy of an ideal gas is &#xA;$$U = 3/2 N kT = 3/2 PV = 3/2 P_0 V_0^\gamma V^{1-\gamma}$$&#xA;where $\gamma$ is the adiabatic coefficient (1.4 for air).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;change&lt;/em&gt; in $U$ due to a changing $V$ then comes completely from the initial potential energy $E$. Some algebra and some sensible binomic approximations then yield a simple expression for the surface area of &#xA;$$A = \frac{32 mgh}{9 V_0 P_0 (\gamma - 1)}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $F = PA$ then allows me to calculate the peak force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But&lt;/strong&gt; what about the contact time? My initial guess was to approximate $F(t)$ as a triangle curve that goes from 0 to $F_{max}$ and back to $0$ and then use that $F$ equals change in momentum over time, $dp = Fdt$, which in this simple case would mean &#xA;the total change in momentum, $\delta p$, would be equal to $1/2 F_{max} \delta t$.&#xA;I can calculate the initial momentum from the initial potential energy, and since the process is elastic, the change in momentum is (minus) two times that value. Then I know everything to calculate $\delta t$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am, however, unsure about whether I can apply this law about momentum at all, since I am not talking about a simple point of mass here, rather about a bunch of gas molecules confined to a certain volume. This is also why my standard approach to mechanics questions, i.e. Lagrangian mechanics, doesn't seem to work: A simple coordinate describing the entire process would be $h$, but what is the kinetic term in terms of $h$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&#xA;I just realized that my formula for the spherical cap is wrong, and a bit more complicated than I wanted: If $h$ is the distance from the center of the sphere to the base of the cap, the volume becomes&#xA;$$V = \frac{2}{3} R^3 - hR^2 + \frac{h^3}{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I keep the pressure constant, the work needed for a volume change is &#xA;$P \Delta V$, so we can equate:&#xA;$$\Delta V = E_0 / P$$&#xA;where $E_0 = mgh$ is the initial potential energy of the ball. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only problem then is that I have a third-order equation for $h$ which, I feel, is too complicated to solve. But let's see... let $d = R - h$, then we get&#xA;$V == Rd^2 - d^3/3$ and now we assume that $d \ll R$ so that we have $V \approx Rd^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we need the surface area, $A = \pi (R^2 - h^2) = \pi (R^2 - (R-d)^2) \approx&#xD;&#xA;2\pi R d$ where we again use $R \gg d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging in some numbers gives $A \approx 44 cm^2$ which amounts to a radius of &#xA;the spherical cap of approx $3.7cm$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The peak force $F_{max}$ is just $AP \approx 401 N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the force grows linearly with time until $F_{max}$ is reached and then drops linearly to $0$, the total change in momentum over the contact time $\Delta t$ is &#xA;$\Delta p = 1/2 F_{max} \delta T$. The change in momentum is two times the initial momentum, so it is &#xA;$$\Delta p = 2\sqrt{2mE_0} = 1/2 * 2\pi \sqrt{E_0 R P} \Delta t,$$&#xA;which we can solve for $\Delta t$ to obtain&#xA;$\Delta t \approx 9 ms$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That sounds reasonable to me.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditorDisplayName="" LastEditDate="2011-08-06T01:06:54.403" LastActivityDate="2011-08-06T05:02:16.583" Title="Mechanics + Thermodynamics: Bouncing Ball" Tags="&lt;classical-mechanics&gt;&lt;thermodynamics&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="2" />
  <row Id="13209" PostTypeId="2" ParentId="13167" CreationDate="2011-08-05T04:26:05.993" Score="-1" ViewCount="" Body="&lt;p&gt;I played around wirg Pade approximates for a couple of days once, but didn't find them very useful. The basic idea is you have some sort of polynomial expansion, and you want to approximately match the first N terms, by a function which is the ratio of a Polynomial of degree N-M and a monomial of degree M. If the coefficients of the original power series are known numeric values, by multiplying both sides by the monomial, and equating coefficients of equal power, you get an (N+1)th order linear system to solve for the Pade coefficients. By adjusting M you can usually find an approximant that does somewhat better on a finite interval than the original series. But I failed to see any magic there. Perhaps someone can enlighten us as to how this method can be made useful?&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2011-08-05T04:26:05.993" />
  <row Id="13210" PostTypeId="2" ParentId="1915" CreationDate="2011-08-05T04:35:51.747" Score="1" ViewCount="" Body="&lt;p&gt;If the basic question is how we define whether the universe is finite or infinite, then the most straightforward answer is that in a finite universe, there is an upper bound on the proper distance (which is defined as the distance between two points measured by a chain of rulers, each of which is at rest relative to the Hubble flow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Does it have infinite mass[...]?&quot; -- GR doesn't have a scalar quantity that plays the role of mass (or mass-energy) and that is conserved in all spacetimes. There is no well-defined way to discuss the total mass of the universe. MTW has a nice discussion of this on p. 457.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;[...]or is it dishomogeneous?&quot; -- I don't understand how this relates to the first part of the sentence. You can have homogeneous or inhmogeneous cosmological solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;How can the universe transition from being finite near the big bang and infinite 14 billion years later? Or would an infinite universe not necessarily have a big bang at all?&quot; -- This was asked again more recently, and a good answer was given: &lt;a href=&quot;http://physics.stackexchange.com/questions/9419/how-can-something-finite-become-infinite&quot;&gt;How can something finite become infinite?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-05T04:35:51.747" />
  <row Id="13211" PostTypeId="2" ParentId="13167" CreationDate="2011-08-05T10:24:42.557" Score="2" ViewCount="" Body="&lt;p&gt;Pade approximats are sometimes an alternative to a taylor approximation to a function. Imagine fitting (i.e. describing) some data with a taylor series&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D(x) = D_0 + D_1x + D_2x^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is a &quot;correct&quot; description of the data (i.e. higher order terms are negligible), then a Pade approximant should also describe the data. For example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D(x) = \frac{D_0}{1-D_1x-D_2x^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that the Taylor expansion of the second one is the same as the first expression plus some higher order terms that are &quot;small&quot;. (This by the way is an explicit example of your question 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many practical reasons, Pade approximants also do a nice job in describing data that have a singularity somewhere (not where you have data). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I do not understand your questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If a series is divergent there is no way to sum it. A different thing is that you have the asymptotic expansion of a function (that can be a divergent series), and you want to reconstruct the original function (this is what Borel summation does, for example). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The rule is that if the taylor series is up to order $a$, then the sum of the orders of the numerator and denominator in the Pade approximant should also be $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) You can approximate the borel transform of a series by the Pade approximants, but I do not know if this would be useful...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Yes, they are a &quot;good&quot; approximation to a function, and then can be used as approximations to the integrals of the function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure if this is useful for you, in any case, If you tell us what you want to do, maybe we can help.&lt;/p&gt;&#xA;" OwnerUserId="3830" LastActivityDate="2011-08-05T10:24:42.557" CommentCount="1" />
  <row Id="13212" PostTypeId="1" AcceptedAnswerId="13213" CreationDate="2011-08-05T11:03:33.823" Score="5" ViewCount="105" Body="&lt;p&gt;Consider an point particle moving on a frictionless semicircular hill (curve). The particle's initial kinetic energy is equal to the potential energy on the top of the hill, i.e it has the necessary energy to climb the hill. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will it reach the top of the hill in infinite or finite time? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my proof it needs infinite time and this is quite non-intuitive because, though the particle has the necessary energy to climb the hill, it needs an infinite time. Also if we reverse the time, when the particle is in equilibrium on the top of the hill, it will never go down, therefore this process is time irreversible.&lt;/p&gt;&#xA;" OwnerUserId="3066" LastEditorUserId="3066" LastEditorDisplayName="" LastEditDate="2011-08-05T11:10:43.893" LastActivityDate="2011-08-07T10:51:31.053" Title="Point particle moving on a frictionless semicircular hill" Tags="&lt;classical-mechanics&gt;&lt;point-particle&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="13213" PostTypeId="2" ParentId="13212" CreationDate="2011-08-05T12:06:09.173" Score="2" ViewCount="" Body="&lt;p&gt;EDIT: OK, I misinterpreted ANKU's question and had a bowl shape in mind whereas he had an upturned bowl in mind. This changes the energy equation to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ mgR = \frac{1}{2}mR^2\dot{\theta}^2 + m g R (\sin \theta) \; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I measure the angle from the horizontal here. After similar manipulations as below, I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T = \sqrt{\frac{R}{2g}} \int_0^{\pi/2} \frac{d\theta}{\sqrt{1-\sin\theta}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is indeed a divergent integral. However, this does not mean that the motion is irreversible. The reverse motion starts from the top of the hill, but the top of the hill is an equilibrium point, albeit an unstable one, which implies that it takes an infinite amount of time to roll down the hill.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From conservation of energy, you can write down the following formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ mgR = \frac{1}{2}mR^2\dot{\theta}^2 + m g R (1-\cos \theta) \; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left hand side represents the potential energy at the top of your semi-circular hill, the right hand side the total energy at any point of the trajectory. (Angle measured from the vertical.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging, you can write this as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{2 g \cos \theta}{R} = \dot{\theta}^2$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or after some additional work and intergrating&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T = \sqrt{\frac{R}{2g}} \int_0^{\pi/2} \frac{d\theta}{\sqrt{\cos\theta}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick check with &lt;a href=&quot;http://tinyurl.com/3gc9ddj&quot; rel=&quot;nofollow&quot;&gt;wolframalpha&lt;/a&gt; gives a finite number for the right integral. Therefore the time it takes for the ball to roll up the hill is finite.&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2011-08-07T10:51:31.053" LastActivityDate="2011-08-07T10:51:31.053" CommentCount="9" />
  <row Id="13216" PostTypeId="2" ParentId="13212" CreationDate="2011-08-05T16:20:07.730" Score="3" ViewCount="" Body="&lt;p&gt;Interesting question:&#xA;The time it takes to get at the top is naturaly infinite. It has sense, since the particle arrives at rest, it should move slower and slower as it get closer to the top. The main part of the time that the particle needs for to reach the top is spent in the last portion of its trayectory.&#xA;The process is already time reversivle; the point is that the both exmples you mention doesnt correspond to the same proces. In the former the particle never reaches the top. &#xA;I suggest to solve the problem in the case when the particle has initialy a little more energy than the potential at the top, say MgR+epsilon; then take the limit epsilon to zero. Probably it might clarify the aspects of the problem.&#xA;Hope I have been of help.&lt;/p&gt;&#xA;" OwnerUserId="4778" LastActivityDate="2011-08-05T16:20:07.730" />
  <row Id="13217" PostTypeId="1" AcceptedAnswerId="13218" CreationDate="2011-08-05T17:59:09.687" Score="5" ViewCount="83" Body="&lt;p&gt;Let a wire be shaped according to some even function $y=f(x)$, with $f&amp;#39;(0)=0$ and $f&amp;#39;&amp;#39;(0)&amp;gt;0$, and let a bead of negligible size slide frictionlessly on the wire. Let the bead oscillate under the influence of gravity about $x=0$ with amplitude $A$ (i.e., between $x=-A$ and $x=+A$) and frequency $\omega(A)$. Clearly $\omega$ is nearly constant for small $A$; it differs from the frequency $\omega_o$ of simple harmonic motion by at most $O(A^2)$. By choosing $f$ to be a fourth-order polynomial, we could presumably adjust the wire's shape so as to eliminate the errors of order $A^2$ and make $\omega(A)$ constant up to $O(A^4)$. Possibly we could continue this process of approximation and make all the derivatives $d^n\omega/dA^n$ vanish up to some finite $n$, or maybe for all $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the derivatives can be made to vanish for all $n$, then I think $\omega$ would have to be nonanalytic at $x=0$. It seems impossible that there is any $f$ such that isochrony holds for arbitrarily large $A$. No matter how steep you make the sides, the bead can't do any better than accelerating downward with acceleration $g$. Therefore I think the best $f$ you can find is probably one that blows up to infinity at $|x|$ equal to some $x_{max}$. On dimensional grounds, we would have to have $x_{max}=cL$, where $c$ is a unitless constant and $L=g/\omega_0^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my multipart question is: (1) Is there a function $f$ that gives $d^n\omega/dA^n=0$ for all $n$? If so, ... (2) How is $f$ characterized, and what is $c$? (3) Is $\omega(A)$ analytic at $x=0$, and if so, what is its radius of convergence to its Taylor series in units of $L$?&lt;/p&gt;&#xA;" OwnerUserId="4552" LastEditorUserId="68" LastEditorDisplayName="" LastEditDate="2011-08-05T20:21:32.157" LastActivityDate="2011-08-05T20:21:32.157" Title="Shaping a wire such that a bead sliding on it has exactly isochronous oscillations" Tags="&lt;newtonian-mechanics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13218" PostTypeId="2" ParentId="13217" CreationDate="2011-08-05T19:17:33.860" Score="7" ViewCount="" Body="&lt;p&gt;Here we are: such a pendulum was developed centurys ago: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://myreckonings.com/wordpress/2007/11/19/the-not-so-simple-pendulum/&quot; rel=&quot;nofollow&quot;&gt;http://myreckonings.com/wordpress/2007/11/19/the-not-so-simple-pendulum/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trajectory is a cycloid, a possible realisation is a kind of &#xA;guide in form of the evolent of that cycloid close to the fixation &#xA;point of the pendulum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;AfaIr, a similar cycloid is the fastest trajectory for a body &#xA;(on a rail) in free fall.&lt;/p&gt;&#xA;" OwnerUserId="1370" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-08-05T19:43:00.350" LastActivityDate="2011-08-05T19:43:00.350" CommentCount="2" />
  <row Id="13219" PostTypeId="2" ParentId="9154" CreationDate="2011-08-05T21:43:35.643" Score="2" ViewCount="" Body="&lt;p&gt;As others mentioned, special relativity (by definition really) doesn't have anything to do with curved surfaces! Special relativity has a particular metric (minkowski metric) which has no curvature. If your interested in manifolds (particularly integration on them, since integration in minkowski space is pretty trivial) and things like that, you really getting into general relativity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I am sure many people will have pedagogical arguments against the following approach, I will say it anyway... You don't really need to do an in depth study of mathematical SR to do GR. Really once you understand the basic properties of minkowski space, know what a lorenz transformation is, understand things like length contraction, etc, you really should just jump right into GR. One sorta friendly way to go about it is apply SR to E&amp;amp;M (covariant formalization) as a way to play around with tensors a little.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as recommendations of what books to get, take a look at Sean Carroll's notes (http://preposterousuniverse.com/grnotes/) and if you like them get his book too! Perhaps also looking at the first two sections will give you a better idea of what you are looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: After commenting, I thought I would also just post a link to &lt;a href=&quot;http://www.amazon.com/Barrett-ONeill/e/B001H6MNRO/ref=ntt_athr_dp_pel_1&quot; rel=&quot;nofollow&quot;&gt;O'Neills Amazon Page&lt;/a&gt;. His books are probably more mathematical than what you are looking for, but they could be very useful to someone else.&lt;/p&gt;&#xA;" OwnerUserId="3018" LastEditorUserId="3018" LastEditorDisplayName="" LastEditDate="2011-08-05T21:56:23.590" LastActivityDate="2011-08-05T21:56:23.590" />
  <row Id="13220" PostTypeId="1" AcceptedAnswerId="13221" CreationDate="2011-08-05T23:20:18.260" Score="1" ViewCount="84" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/10852/how-are-these-balls-reflected-after-they-hit-each-other&quot;&gt;How are these balls reflected after they hit each other?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I was wondering what sort of physics equations would I need in order to build a top down billiards game? I tried using Box2D (a 2D physics engine for games) but it has issues with objects going through other objects when the velocity is too fast. So I decided to do all the physics by scratch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what would be all the physics I need to create a top down 2D billiards game?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4830" LastEditorUserId="3540" LastEditorDisplayName="" LastEditDate="2011-08-06T23:17:25.270" LastActivityDate="2011-08-30T17:54:32.757" ClosedDate="2011-08-06T00:17:12.167" Title="Physics needed to build a top down billiards game" Tags="&lt;friction&gt;&lt;collision&gt;&lt;numerics&gt;&lt;rigid-body-dynamics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13221" PostTypeId="2" ParentId="13220" CreationDate="2011-08-06T01:16:47.687" Score="1" ViewCount="" Body="&lt;p&gt;The most important thing is conservation of momentum to describe the collisions. This part is actually quite straightforward, but before you get to collisions you should model the motion of single balls. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, you will describe the balls classically and probably not at relativistic speeds (though that would be interesting...) so pretty much all you need is Newton $\mathbf{F}=m\cdot \mathbf{a}$ for the frictional forces. However, describing the balls as solid spheres, you don't just need directed forces but also torques to describe the rotation $\pmb{\omega}$. It changes as&#xA;$$&#xD;&#xA;  J\cdot\dot{\pmb{\omega}} = \pmb{\tau}&#xD;&#xA;$$&#xA;where the moment of inertia $J$ would, for a general rigid body, be a tensor but is for spherical objects just a number. The torque  $\pmb{\tau}$ consists of&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The rolling friction, which occurs whenever a ball moves. It should be possible to model this simply by a multiple of the vector the ball would rotate about if it did properly roll, that is&#xA;$$&#xD;&#xA;  \pmb{\tau}_\mathrm{roll} = \eta_\mathrm{roll}\:\mathbf{e}_z\times \mathbf{v}&#xD;&#xA;$$&#xA;where $\mathbf{v}=\dot{\mathbf{r}}$ is assumed to always lie in the $xy$ plane, disregarding the possibility of balls jumping.&lt;/li&gt;&#xA;&lt;li&gt;The sliding friction, that is, the part of the friction that causes the balls to roll in the first place rather than just gliding over the surface like a curling stone. This actually originates as a directed force rather than a torque: if the ball could not rotate, if would just be the aforementioned $\mathbf{F}=m\cdot \mathbf{a}=m\cdot\ddot{\mathbf{r}}$. Such a &quot;dry&quot; sliding force has a pretty much constant strength but always points into the direction the surfaces slide against each other. $\mathbf{F}=m\mu\frac{\mathbf{v}_\mathrm{rel.slide}}{\|\mathbf{v}_\mathrm{rel.slide}\|}$ with some constant $\mu$ and&#xA;$$&#xD;&#xA;  \mathbf{v}_\mathrm{rel.slide} = \mathbf{v} - \mathbf{v}_\mathrm{ballbottom}&#xD;&#xA;$$&#xA;where ($R$ is the ball radius)&#xA;$$&#xD;&#xA;  \mathbf{v}_\mathrm{ballbottom} = \left(\begin{smallmatrix}\mathbf{e}_x\\\mathbf{e}_y\\\mathbf{e}_z\end{smallmatrix}\right)\left(\begin{smallmatrix}0&amp;amp;-1&amp;amp;0\\1&amp;amp;0&amp;amp;0\\0&amp;amp;0&amp;amp;0\end{smallmatrix}\right)\left(\begin{smallmatrix}\mathbf{e}_x\\\mathbf{e}_y\\\mathbf{e}_z\end{smallmatrix}\right)R\cdot\pmb{\omega}.&#xD;&#xA;$$&#xA;&lt;br&gt;&#xA;With a rotationable ball the same force also comes up as a torque because it's applied just to the low end of the ball. It is linked by the same linear (matrix) mapping, but with one additional contribution: the ball can also spin like a top. This does not directly interact with the movement but is neverthess important.&#xA;$$&#xD;&#xA;  \pmb{\tau}_\mathrm{slide} = R\cdot\left(\begin{smallmatrix}\mathbf{e}_x\\\mathbf{e}_y\\\mathbf{e}_z\end{smallmatrix}\right)\left(\begin{smallmatrix}0&amp;amp;-1&amp;amp;0\\1&amp;amp;0&amp;amp;0\\0&amp;amp;0&amp;amp;0\end{smallmatrix}\right)\left(\begin{smallmatrix}\mathbf{e}_x\\\mathbf{e}_y\\\mathbf{e}_z\end{smallmatrix}\right)\mathbf{F} + \mathbf{e}_z\:\eta_\text{spin}\omega_z.&#xD;&#xA;$$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That is enough to model, with some simple dynamic-system-simulator (&lt;a href=&quot;http://en.wikipedia.org/wiki/Euler_integration&quot; rel=&quot;nofollow&quot;&gt;Euler's method&lt;/a&gt; will easily do and is, in this case, actually better than the normally more sophisticated Runge-Kutta methods, because you &lt;em&gt;want&lt;/em&gt; fine steps), the behaviour of a single ball on the table. There are then two types of collisions you need to take care for: with the cushions and with other balls. I shall not discuss the cushions here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Collisions between balls are basically simple: quite good elastic scatterings, so pretty much only directed momentum is transferred, none of balls' own angular momenta and no energy is lost. To analyze such a collision, just enter the center of mass frame. At the collision point, both balls touch. The momentum (or equivanently velocity) of each of them is reflected off the plane going in between both points—that's all there is to it.&lt;br&gt;&#xA;The difficulty when doing this in a computer simulation is just that you will not have a continuous motion but discrete steps. None of these will actually catch the point where the balls just touch, not unless you enforce it. Do do so, you need to check for collisions, and upon noticing one took place calculate, with a uniform motion, back to the point where the balls actually did touch. This is not really a physics problem, so I will not adress it here either.&lt;/p&gt;&#xA;" OwnerUserId="3540" LastEditorUserId="3540" LastEditorDisplayName="" LastEditDate="2011-08-30T17:54:32.757" LastActivityDate="2011-08-30T17:54:32.757" />
  <row Id="13222" PostTypeId="1" CreationDate="2011-08-06T01:26:03.750" Score="4" ViewCount="32" Body="&lt;p&gt;At the Advanced Photon Source, they use two accelerators before injecting the electrons into the large storage ring.  Is the addition of particles to the storage ring done in &quot;batches&quot; (however small/short they may be), or continuously?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By a &quot;batch&quot;, I'm thinking where the LINAC would throw however many electrons into the booster ring, which would then accelerate them all together, then dump them all into the storage ring when they're at the appropriate energy.&lt;/p&gt;&#xA;" OwnerUserId="431" LastActivityDate="2011-08-06T04:05:53.153" Title="Is accelerating particles through a chain of accelerators a continuous or batch process?" Tags="&lt;particle-physics&gt;&lt;accelerator-physics&gt;" AnswerCount="1" />
  <row Id="13223" PostTypeId="1" CreationDate="2011-08-06T03:01:48.467" Score="5" ViewCount="69" Body="&lt;p&gt;I'm a retired Physics Professor, but I admit I'm puzzled by this...  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider two loops of wire, 2 small dipoles B and C , with a common axis z (facing each other) and (say) 30 cm apart B to C.  At the speed of light, information (including a change in magnetic field) will require about 1 nanosecond to travel from C to B.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Have the current on in coil B for some period of time at the start, so the B-fields at C is established in the +z-direction.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Turn loop B off rapidly (fall time &amp;lt; 0.3 ns, say) at the same time that a current in loop C is turned ON (rapidly, rise time &amp;lt;0.3ns, and current in opposite sense with respect to the previous current in loop B).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In this way, as the current is turned on in loop C, it is immersed in the field from loop B and therefore it receives an impulse to the right, in the +z-direction.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, loop B will be &quot;off&quot; (and open so no effective eddy currents) when the &quot;return&quot; field from loop C arrives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thus, loop C (which is free to move) will experience a force giving it momentum in the +z direction (to the right), whereas loop B will not experience a force to the left.  It seems that Newton's Third Law is violated.&lt;/strong&gt;&lt;br&gt;&#xA;(The question can also be posed in momentum terms, but I chose to consider forces and the Third Law. )&lt;/p&gt;&#xA;" OwnerUserId="4831" LastActivityDate="2011-08-06T04:04:01.993" Title="2 electromagnetic loops, with light-speed constraints: Is Newton's Third Law violated?" Tags="&lt;electromagnetism&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="13224" PostTypeId="2" ParentId="13169" CreationDate="2011-08-06T03:24:37.887" Score="1" ViewCount="" Body="&lt;p&gt;A laser pointer's energy and lights are concentrated in a very small light cone to reach a quite high intensity. So its labelled Wattage is much smaller than a usual bulb while its light is very strong at one point. And the labled power on laser pointers is the output power. The typical input power of a 50mW pointer is about 0.5 W. This make the difference looks bigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually common laser pointers used by teachers and lecturers are less than 5mW. The farthest distance to see their lights is about dozens of meters. But the laser pointers used by professional and amateur astronomers are much more powerful. You can see the light path caused by Tyndall effect directly as below image(&lt;a href=&quot;http://www.universetoday.com/70406/telescopes-laser-pointer-clarifies-blurry-skies/&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;, though this impressive light path is not generated by a handheld pointer, the scence is similar.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/gFcC1.jpg&quot; alt=&quot;At Mount Hopkins in Arizona, a bundle of five lasers is shot into the atmosphere to improve the imaging of the 6.3-meter MMT telescope&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy reduction is mainly caused by Rayleigh scattering, which is relatively small compared with laser's intensity. So the beam can easily reach several miles away with the power higher than 50 mW.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: Georg insist the previous image can not be used to discribe laser pointers. So I add another real pointer picture here. But I have no feeling about distance with this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/RoHho.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="518" LastEditorUserId="518" LastEditorDisplayName="" LastEditDate="2011-08-06T13:51:38.837" LastActivityDate="2011-08-06T13:51:38.837" CommentCount="5" />
  <row Id="13225" PostTypeId="2" ParentId="13222" CreationDate="2011-08-06T03:54:47.133" Score="3" ViewCount="" Body="&lt;p&gt;That rather depends on the accelerator and the intended purpose of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to choose one particular example the main electron accelerator at &lt;a href=&quot;http://www.jlab.org/&quot; rel=&quot;nofollow&quot;&gt;Jefferson Lab&lt;/a&gt; is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Thomas_Jefferson_National_Accelerator_Facility#Accelerator&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Continuous&lt;/em&gt; Electron Beam Accelerator Facility (CEBAF)&lt;/a&gt; which (duh!) functions in a continuous fashion (thought the electrons still comes in &quot;bunches&quot; because at any given time a RF klystron is only adding energy to particles found in part of it's length. The bunch spacing is ~2 ns in each of the three experimental halls and can be clearly reconstructed in some experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, the ability of CEBAF to recirculate electrons in a continuous mode like that is a very special feature of the machine (which to my knowledge is not duplicated anywhere else at this time). Other than that, linear machines and synchrotrons can run in continuous mode, but accelerator rings machines must fill-n-spill.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-06T04:05:53.153" LastActivityDate="2011-08-06T04:05:53.153" CommentCount="3" />
  <row Id="13226" PostTypeId="2" ParentId="13223" CreationDate="2011-08-06T04:04:01.993" Score="6" ViewCount="" Body="&lt;p&gt;In electromagnetism you can find millions of examples of apparent momentum non-conservation. They are all explained by the fact that the electromagnetic field itself carries &lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_stress-energy_tensor&quot; rel=&quot;nofollow&quot;&gt;momentum&lt;/a&gt;, and only the total momentum of all matter &lt;em&gt;plus&lt;/em&gt; the EM field is conserved.&lt;/p&gt;&#xA;" OwnerUserId="4749" LastActivityDate="2011-08-06T04:04:01.993" CommentCount="7" />
  <row Id="13227" PostTypeId="1" CreationDate="2011-08-06T04:54:43.173" Score="5" ViewCount="124" Body="&lt;p&gt;After taking a basic signals &amp;amp; systems class and learning about the frequency domain, I started wondering:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;How the heck do scientists still communicate with Voyagers I and II??&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;br&gt;&#xA;Do they send/receive signals in a frequency that is otherwise unusually silent? If not, then how do they send and receive faint signals from something &lt;em&gt;10 billion&lt;/em&gt; miles away, when the spacecraft has so little energy and where it seems practically impossible to aim everything 100% correctly (since things are so distant and moving at the same time)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I wasn't sure what to tag this, so please retag as appropriate, thanks!)&lt;/p&gt;&#xA;" OwnerUserId="853" LastActivityDate="2011-08-06T09:20:28.080" Title="How Earth communicates with Voyager I?" Tags="&lt;astrophysics&gt;&lt;signal-processing&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="13228" PostTypeId="2" ParentId="13208" CreationDate="2011-08-06T05:02:16.583" Score="1" ViewCount="" Body="&lt;p&gt;The problem gets very simple if you consider only small deformations $d = R - h$ and calculate the force $F=PA$ to first order in $d$. To this order, $A = 2 \pi R d$ and $F = (P_0 2 \pi R) d$, where $P_0$ is the pressure inside the undisturbed  ball (note that changes in pressure arising from volume changes only contribute terms of higher order in $d$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this approximation the force on the ball is essentially an elastic restoring force $F = - k d$ , with 'spring constant' $ k = 2 \pi R P_0 \approx 6 \times 10^4 N/m $. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The contact time is, then, one-half the oscillator period: $t_{contact} = \pi \sqrt{m/k} \approx 8\, ms$ (a very reasonable result). Note that, because of linearity, the contact time is independent of the velocity of the ball as it hits the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From conservation of energy, the maximum deformation of a ball dropped from a height $H$ is $ d_{max} = \sqrt{2 m g H/k}$. The peak force is $ F_{max} = k d_{max} = \sqrt{2 m g k H}$. &lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2011-08-06T05:02:16.583" CommentCount="1" />
  <row Id="13230" PostTypeId="1" CreationDate="2011-08-06T05:25:52.090" Score="9" ViewCount="112" Body="&lt;p&gt;Classically, a hydrogen atom should not be stable, since it should radiate away all its energy. I remember hearing from my favorite freshman physics prof ca. 1983 about a general theorem to the effect that all classical systems of charged particles were unstable, and I went so far as to contact him in 2009 and ask him if he remembered anything about it, but he didn't. There is something called Earnshaw's theorem (there's a Wikipedia article) that says that &lt;em&gt;static&lt;/em&gt; equilibrium is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It certainly seems plausible that something similar holds for dynamical equilibrium, in some sense that I don't know how to define properly, much less prove rigorously. Intuitively, I would expect that any classical system of $m$ charged particles should end up having no &quot;interesting structure,&quot; in the sense that the final state will consist of radiation plus $n \le m$ zero-size clusters whose charges $q_1,\ldots q_n$ all have the same sign (some possibly being zero); asymptotically, they all end up diverging radially from some central point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of any formal proof along these lines?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there counterexamples, although possibly ones with initial conditions corresponding to zero volume in phase space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would one go about stating the initial conditions on the radiation fields?&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-16T02:30:52.990" Title="Theorems on instability of classical systems of charged particles?" Tags="&lt;classical-mechanics&gt;&lt;elecromagnetism&gt;" AnswerCount="1" CommentCount="15" FavoriteCount="1" />
  <row Id="13231" PostTypeId="2" ParentId="11939" CreationDate="2011-08-06T08:38:48.213" Score="2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;classical electrodynamics&lt;/strong&gt; mainly deals with &lt;strong&gt;two kinds of proplems&lt;/strong&gt;: a) The action of a field on a charged particle and b) the fields arising from the motion of such a field. Of course, this can only be approximative but it turns out that a lot of phenomena can be described in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you are right, an entire treatment would include a) and b) &lt;strong&gt;simultaniously&lt;/strong&gt; - including the whole dynamics of such a system with &lt;strong&gt;radiative reaction&lt;/strong&gt; (or, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Abraham%E2%80%93Lorentz_force&quot; rel=&quot;nofollow&quot;&gt;Abraham-Lorentz-Force&lt;/a&gt; and its &lt;a href=&quot;http://en.wikipedia.org/wiki/Abraham%E2%80%93Lorentz%E2%80%93Dirac_force&quot; rel=&quot;nofollow&quot;&gt;relativistic counterpart&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as &lt;a href=&quot;http://physics.stackexchange.com/users/2451/qmechanic&quot;&gt;Qmechanic&lt;/a&gt; pointed out in a comment, there may be &lt;strong&gt;no fully consistent way&lt;/strong&gt; to do so within the framework of classical electrodynamics. &lt;a href=&quot;http://books.google.com/books/about/Classical_electrodynamics.html?id=U3LBQgAACAAJ&quot; rel=&quot;nofollow&quot;&gt;Jackson&lt;/a&gt; (Chapter 16) states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The difficulties presented by this problem touch one of the &lt;strong&gt;most fundamental aspects of physics&lt;/strong&gt;, the nature of an elementary particle. Although partial solutions, workable within limiting areas, can be given, the basic problem remains unsolved.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, you will have to search for a really satisfactory answer within the description of &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_electrodynamics&quot; rel=&quot;nofollow&quot;&gt;Quantum Electrodynamics&lt;/a&gt;. Otherwise, you may &lt;strong&gt;include the effect phenomenologically&lt;/strong&gt;. This was done e.g. by &lt;strong&gt;Barone and Mendes&lt;/strong&gt; in &lt;a href=&quot;http://arxiv.org/PS_cache/cond-mat/pdf/0607/0607370v1.pdf&quot; rel=&quot;nofollow&quot;&gt;Lagrangian description of the radiation damping&lt;/a&gt; (PRA, 2007) and they give a Lagrangian of the form&#xA;$$\mathcal{L}=\frac{m}{2}\dot{\mathbf{r}}_1\cdot\dot{\mathbf{r}}_2 - \frac{\gamma}{2}\epsilon\left(\dot{\mathbf{r}}_1,\ddot{\mathbf{r}}_2\right)&#xD;&#xA;-V\left(\mathbf{r}_1,\mathbf{r}_2\right)$$&#xA;with $\epsilon = \epsilon_{ij}dx^idx^j$ beeing the &lt;a href=&quot;http://en.wikipedia.org/wiki/Levi-Civita_symbol&quot; rel=&quot;nofollow&quot;&gt;Levi-Cevita tensor&lt;/a&gt; and the $\mathbf{r}_i$ arise from the special treatment of the problem used in the paper employing some kind of image phase-space representation of the system. Furthermore, $V\,$ is the potential related to the Abraham–Lorentz–Dirac-Force which is given explicitely in the paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely&lt;/p&gt;&#xA;" OwnerUserId="358" LastActivityDate="2011-08-06T08:38:48.213" />
  <row Id="13232" PostTypeId="2" ParentId="13227" CreationDate="2011-08-06T09:20:28.080" Score="6" ViewCount="" Body="&lt;p&gt;As always, a communication via electromagnetic radiation depends on both ends. &#xA;Uplink from earth can be done with a lot of power and big dishes, of course. &#xA;Downlink is limited to the power of the nuclear battery on board but has a rather &#xA;impressive 2.7 meters dish!.&#xA;On top of that they use a rather slow bitrate, I think with a lot of redundancy. &#xA;All this is commonplace information technology when dealing with bad connections &#xA;with signal not much above noise.&lt;br&gt;&#xA;Details are given in Wikipedia: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Uplink communications is via S band (16-bit/s command rate) while an X&#xA;  band transmitter provides downlink telemetry at 160 bit/s normally and&#xA;  1.4 kbit/s for playback of high-rate plasma wave data. All data is&#xA;  transmitted from and received at the spacecraft via the 3.7-meter&#xA;  high-gain antenna.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-08-06T09:20:28.080" CommentCount="6" />
  <row Id="13233" PostTypeId="1" CreationDate="2011-08-06T13:07:41.217" Score="4" ViewCount="182" Body="&lt;p&gt;How can one show from General Relativity that gravity is attractive force, and under which conditions it becomes repulsive, also why positive energy vacuum drives repulsive gravity?&lt;/p&gt;&#xA;" OwnerUserId="4107" LastEditorUserId="1353" LastEditorDisplayName="" LastEditDate="2011-08-12T11:45:59.350" LastActivityDate="2011-08-13T09:05:30.793" Title="Vacuum and repulsive gravity" Tags="&lt;general-relativity&gt;&lt;gravity&gt;&lt;vacuum&gt;&lt;cosmological-constant&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="3" />
  <row Id="13234" PostTypeId="2" ParentId="166" CreationDate="2011-08-06T13:41:50.043" Score="6" ViewCount="" Body="&lt;p&gt;There are quite a few novel energy technologies coming through. I guess that without quantification, &quot;breakthrough&quot; is a subjective term. Below, I've tried to list all the energy technologies that I know of, that are not yet at commercialisation, but could be within 50 years, and that could offer at least tens of gigawatts of power. They are, in descending order of readiness:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Wave power&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/5fKBz.jpg&quot; alt=&quot;Pelamis, public domain image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting kinetic energy from (near-)surface waves to electricity. First grid-connected prototypes went live in 2008, off the coast of Portugal. Theoretical UK resource (at 100% efficiency) is thought to be 40GW (at 50% efficiency), but this is subject to revision as better data is collected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tidal stream&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ikrsK.jpg&quot; alt=&quot;Seagen tidal turbine with blades raised above the water, from Wikimedia&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting kinetic energy from daily / twice-daily tidal movements, with underwater turbines. First grid-connected prototypes went live in 2008. UK resource estimated in the range 4GW-400GW at expected efficiencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Superconductors&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not a generation technology in itself, of course, but a technology offering potential breakthroughs within the energy industry.  Superconductors are being trialled as fault-current limiters; may have a role to play in HVDC circuit-breakers; and offer lots of potential in all sorts of turbines: trials are being done now with superconductors in run-of-river hydro turbines; and in wind turbines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Deep (100m-700m depth) offshore wind&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Offshore wind turbines, floating, moored to the seabed at depths of 100-700m. The first operational full-size prototype, &lt;a href=&quot;http://www.statoil.com/en/TechnologyInnovation/NewEnergy/RenewablePowerProduction/Onshore/Pages/Karmoy.aspx&quot; rel=&quot;nofollow&quot;&gt;HyWind&lt;/a&gt;, was deployed in 2009 in deep water off the coast of Norway. Theoretical UK resource at expected efficiency rates is over 1000GW. Global resource is one or two orders of magnitude higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Osmotic power&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generating electricity from salinity gradients. &lt;a href=&quot;http://www.statkraft.com/energy-sources/osmotic-power/&quot; rel=&quot;nofollow&quot;&gt;First prototype went in the water in Norway&lt;/a&gt;, 2009. Expected global resource is about 200GW.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Novel photovoltaics (PV)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Huge amounts of work going on in labs around the world on enhancing photovoltaics:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Photonics &amp;amp; plasmonics looks to harness quantum effects to increase&#xA;efficiency by scattering the light with metallic nanoparticles, or by&#xA;using quantum wells / quantum dots;&lt;/li&gt;&#xA;&lt;li&gt;BIPV integrates PV into building materials;&lt;/li&gt;&#xA;&lt;li&gt;inkjet-printing and solar dyes look to give high-speed, low-cost production.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The UK PV resource estimated at over 400GW, using 5% of land at 40% efficiency. Globally, the resource is orders of magnitude higher than global power demand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Artificial photosynthesis&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Building novel synthetic living photosynthesisers from basic genetic building blocks; or novel catalysts to harness sunlight to release hydrogen from water; various labs working on different routes. Potential resource: greater than current global power demand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Nuclear fusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For work in progress, see &lt;a href=&quot;http://www.iter.org/&quot; rel=&quot;nofollow&quot;&gt;JET&lt;/a&gt;, &lt;a href=&quot;http://www.iter.org/&quot; rel=&quot;nofollow&quot;&gt;ITER&lt;/a&gt; and &lt;a href=&quot;http://www.fusion.org.uk/&quot; rel=&quot;nofollow&quot;&gt;CCFE&lt;/a&gt;. And our local giant fusion reactor, the sun, can provide us with absolutely vast amounts of energy for a very long time to come. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tidal benthic friction&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hypothesis is that tidal water movements are mostly dissipated as friction against the bottom of the sea, and that by placing energy harvesters on the seabed, the kinetic energy can be converted to electricity. Specific harvesting technologies, scale of the resource, and the impact on benthic ecology, are currently unknown.&lt;/p&gt;&#xA;" OwnerUserId="4066" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-08-11T13:43:17.877" LastActivityDate="2011-08-11T13:43:17.877" CommentCount="4" />
  <row Id="13235" PostTypeId="2" ParentId="13169" CreationDate="2011-08-06T14:15:52.200" Score="0" ViewCount="" Body="&lt;p&gt;What about the elephant in the room, called coherence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Laser light and ordinary light differ in the amount of coherence of the beam. Incoherent beams lose intensity as 1/r**2, where r is the distance from the source. Coherent beams in vacuum disperse slowly according to &lt;a href=&quot;http://www.mrl.columbia.edu/ntm/level1/ch03/html/l1c03s02.html&quot; rel=&quot;nofollow&quot;&gt;optical equations.&lt;/a&gt; In the atmosphere there will be absorption and scattering, as discussed in other answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment&lt;/a&gt; to see laser light reflected from the moon, as far as distance travelled goes.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-08-08T09:59:14.820" LastActivityDate="2011-08-08T09:59:14.820" CommentCount="5" />
  <row Id="13236" PostTypeId="1" CreationDate="2011-08-06T14:50:25.823" Score="3" ViewCount="59" Body="&lt;p&gt;Thought experiment I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider two loops of wire, 2 small dipoles B and C , with a common axis z (facing each other) and (say) 30 cm apart B to C. At the speed of light, information (including a change in magnetic field) will require 1 nanosecond to travel from C to B.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Have the current on in loop B for some period of time at the start, so the B-fields at C is established in the +z-direction.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Turn loop B off rapidly (fall time &amp;lt; 0.1 ns, say) at the same time that a current in loop C is turned ON (rapidly, rise time &amp;lt;0.1ns, and opposite sense with respect to the previous current in loop B ).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In this way, as the current is turned on in loop C, it is immersed in the field from loop B and therefore both receives an impulse to the right, in the +z-direction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, loop B will be &quot;off&quot; (and open so no effective eddy currents) when the &quot;return&quot; field from loop C arrives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, loop C (which is free to move) will experience an impulse giving it momentum in the +z direction (to the right), whereas loop B will not experience an impulse to the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this argument is sufficiently simple to sketch and to ponder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thought Experiment II.&#xA;However,&#xA;If you argue that there is momentum to the left &quot;in the magnetic field&quot; from loop B, I will add a third loop to the left (call it A), and again, as B is opened rapidly (short fall time) -- at the same time that a current in A is turned ON (rapidly, and SAME sense with respect to the previous current in loop B ).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;You see, while one may hand-wave (without equations and ignoring signs) that a magnetic field &quot;carries momentum&quot;, the fact is that one can choose the DIRECTION of the current in a loop (or loops) which absorbs energy from that field, and *&lt;em&gt;thus one can choose the DIRECTION of the momentum in the sensing loop.&lt;/em&gt;*&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way, loops A and C (both free to move) as they turn on are immersed in the field from B while having currents in the opposite sense (A attracted by loop B whereas C is repelled) -- therefore BOTH loops receive an impulse to the right, in the +z-direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, and I will need to turn off the currents in loops B and C rather quickly, so that they both receive impulses in the +z direction without &quot;feeling&quot; the B fields from each other, for they will be &quot;off&quot; when those fields arrive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're concerned about fringe fields, I can add a rod of very high magnetic permeability down the z-axis, extending from A to C, so that essentially all the magnetic field is contained on the z-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whew -- simple thought experiment, but one that could actually be done IMO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will happen? Will there be detectable momentum imparted to the right (+z direction), but not to the left?&lt;/p&gt;&#xA;" OwnerUserId="4831" LastEditorUserId="4831" LastEditorDisplayName="" LastEditDate="2011-08-07T02:42:53.270" LastActivityDate="2011-08-07T03:56:36.440" Title="3 current loops with non-conservation of momentum, when one specifically considers SIGNs" Tags="&lt;electromagnetism&gt;&lt;forces&gt;&lt;newtonian-mechanics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="13237" PostTypeId="1" CreationDate="2011-08-06T15:16:33.880" Score="9" ViewCount="264" Body="&lt;p&gt;In mathematical derivations of physical identities, it is often more or less implicitly assumed that functions are well behaved. One example are the Maxwell identities in thermodynamics which assume that the order of partial derivatives of the thermodynamic potentials is irrelevant so one can write.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it is often assumed that all interesting functions can be expanded in a Taylor series, which is important when one wants to define the function of an operator, &#xA;for example&#xA;$$e^{\hat A} = \sum_{n=0}^\infty \frac{(\hat A)^n}{n!}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there some prominent examples where such assumptions of mathematically good behavior lead to wrong and surprising results? Such as... an operator $f(\hat A)$ where $f$ cannot be expanded in a power series?&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-06T16:08:12.823" LastActivityDate="2011-08-16T15:00:34.720" CommunityOwnedDate="2011-08-06T16:08:22.010" Title="Examples where an ill-behaved function leads to surprising results?" Tags="&lt;mathematical-physics&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="2" />
  <row Id="13238" PostTypeId="2" ParentId="13237" CreationDate="2011-08-06T16:32:22.583" Score="4" ViewCount="" Body="&lt;p&gt;I have had a surprizing &lt;strong&gt;&lt;a href=&quot;http://arxiv.org/ftp/arxiv/papers/0906/0906.3504.pdf&quot; rel=&quot;nofollow&quot;&gt;result&lt;/a&gt;&lt;/strong&gt; due to the wave function having different left and right derivatives at a point (see Chapter 2.1 and Appendix 3). Generally this article contains more surprizing results just due to implicit assumptions being wrong.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-08-07T14:45:43.003" LastActivityDate="2011-08-07T14:45:43.003" CommunityOwnedDate="2011-08-06T16:32:22.583" CommentCount="2" />
  <row Id="13239" PostTypeId="2" ParentId="13233" CreationDate="2011-08-06T16:42:56.097" Score="6" ViewCount="" Body="&lt;p&gt;The Einstein field equations actually don't say anything at all about the nature of matter. Their structure is that they relate a certain measure of spacetime curvature G to the stress-energy tensor T: $G_{ab}=8\pi T_{ab}$. The stress-energy tensor describes any matter that is present; it's zero in a vacuum. Trivially, you can write down any equations you like describing an arbitrary spacetime that you've made up, and then by calculating G you can find the T that is required in order to allow the existence of that spacetime. However, that T may have properties that are different from those of any known type of matter. T has a very specific structure for certain types of matter, such as electromagnetic radiation or &quot;dust&quot; (meaning a perfect fluid made of particles that have velocities $\ll c$ relative to one another). There are various conjectures, called energy conditions, about what types of stress-energy tensors are physically possible for realistic types of matter. They have names like the weak energy condition (WEC), strong energy condition (SEC), etc. The WEC amounts to a statement that the energy density is never negative in any frame of reference. If it was violated, then you could get repulsive gravity. Basically all energy conditions are known to be violated under some circumstances. Here is a nice discussion of that: &lt;a href=&quot;http://arxiv.org/abs/gr-qc/0205066&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/gr-qc/0205066&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cosmological constant is sometimes taken as a separate term in the Einstein field equations, but it can also be treated as a type of matter with a certain contribution to the stress-energy tensor. A spacetime with only a cosmological constant, and nothing else in it, violates various energy conditions.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-06T16:42:56.097" CommentCount="2" />
  <row Id="13240" PostTypeId="2" ParentId="13233" CreationDate="2011-08-06T18:40:48.897" Score="1" ViewCount="" Body="&lt;p&gt;In the context of General Relativity, small test particles move on &lt;a href=&quot;http://en.wikipedia.org/wiki/Geodesic&quot; rel=&quot;nofollow&quot;&gt;geodesics&lt;/a&gt;. A geodesic is a generalization of a straight line (for example, on a curved surface like a football). The geodesics are determined by the metric $g_{\mu \nu}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Einstein equation is $G_{\mu \nu} = 8 \pi T_{\mu \nu}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G_{\mu \nu}$ consists of the metric $g_{\mu \nu}$ and its derivatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_{\mu \nu}$ is matter, and has specific forms depending on what matter is present, like dust, radiation, or lambda. You can think of $T_{\mu \nu}$ as the input to the problem, like a heavy star sitting at the center of empty space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the concrete form of $G_{\mu \nu}$ as written down by Einstein, when you solve the equation for dust or radiation, $g_{\mu \nu}$ will be such that test particles will appear to be attracted to the matter described by $T_{\mu \nu}$. For a heavy star sitting at the center of empty space, the geodesics of test particles coming from very far away will bend toward the heavy star. This solution is actually famous for being the first explicit solution of a problem in the framework of General Relativity, and is called the Schwarzschild solution or &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_metric&quot; rel=&quot;nofollow&quot;&gt;Schwarzschild metric&lt;/a&gt; after Karl Schwarzschild.&lt;/p&gt;&#xA;" OwnerUserId="47" LastEditorUserId="47" LastEditorDisplayName="" LastEditDate="2011-08-13T09:05:30.793" LastActivityDate="2011-08-13T09:05:30.793" CommentCount="4" />
  <row Id="13241" PostTypeId="2" ParentId="13237" CreationDate="2011-08-07T00:15:25.467" Score="7" ViewCount="" Body="&lt;p&gt;I think, the most transparent example is phase transition: by definition it is when some thermodynamic value does not behave well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;AFAIK when Fourier showed that non-continuous function may be presented as an infinite sum of continuous, he had a hard time convincing people around that he is not crazy. That story might partially answer your question: as long as any not-so-well-behaved function may be presented as a sum of smooth ones, there is no &lt;strong&gt;much&lt;/strong&gt; difference as long as good formulated laws are linear. Functions which are really bad behaved usually do not appear in real problems. If they do, there is some significant physics behind it (as with phase transition, shock wave, etc.) and one can not miss it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an operator it is better (for physicist) to think of function from operator as a function acting on its eigenvalues (if it is not diagonalizable, in physics it is  bad behaviour). This is equivalent to power series definition, but works for any function.&lt;/p&gt;&#xA;" OwnerUserId="4657" LastActivityDate="2011-08-07T00:15:25.467" CommunityOwnedDate="2011-08-07T00:15:25.467" />
  <row Id="13242" PostTypeId="1" AcceptedAnswerId="13607" CreationDate="2011-08-07T00:37:15.393" Score="3" ViewCount="83" Body="&lt;p&gt;Say I have a cylindrical container of some volume $V$, that I fill with water or some other fluid with dynamic viscosity $\mu$.  Here, the bottom of the container is milled such that some fraction, $p$, of its surface area, $A$, is removed to allow passage of fluid.  One could, for example, drill pores until this criterion is met.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After positioning a plate at the top of the container, I press down on the plate with some force $F$ to compress the fluid.  As a function of $F$, how quickly will I drain the fluid through the bottom of the container?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate this, what additional measurements/parameters, such as surface hydrophobicity/etc., will I require, and how does this change with the minimum cross-sectional area of the pores I mill in the bottom of the container?&lt;/p&gt;&#xA;" OwnerUserId="4257" LastEditorUserId="1370" LastEditorDisplayName="" LastEditDate="2011-08-07T14:08:32.323" LastActivityDate="2011-08-15T22:04:04.057" Title="Pressing a fluid through leaks " Tags="&lt;fluid-dynamics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="13243" PostTypeId="1" CreationDate="2011-08-07T02:18:35.830" Score="3" ViewCount="62" Body="&lt;p&gt;Okay, so I'm learning some QFT, I read through Bogoliubov, Shirkov Introduction to Quantized Fields up to the section on renormalization, and then wanted to see a more modern point of view- so I started a-googling and now I'm coming to understand a bit more about gauge theories and what exactly the Standard Model is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ANYWAY- I'm still a bit troubled and confused about the basis of the whole theory. So many physically 'real' phenomena arise out of pure formalisms involved with perturbation, such as the actual interactions mediated by gauge bosons which, in a way, are just pieces of terms in a perturbative expansion- correct? They're also, conveniently and beautifully, artifacts emerging from the structure of gauge formalism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I do quite love this abstract nonsense, but isn't this fundamentally different from the physical reality arising in other branches of physics? In GR, another (gauge?) field theory, the stance on reality is quite different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is- How does one place QFT's reality in context with that of other areas of physics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT- By 'reality' I mean those phenomena exhibited by the formalisms that are to be considered physical, as opposed to unphysical. For example, again in GR there is a consensus that some regions of Schwarzschild are not 'real', so we concern ourselves with the dynamics of the other regions. Trying to weed through QFT with this sort of logic seems very difficult.&lt;/p&gt;&#xA;" OwnerUserId="4663" LastEditorUserId="4663" LastEditorDisplayName="" LastEditDate="2011-08-07T04:25:35.663" LastActivityDate="2011-08-07T04:25:35.663" ClosedDate="2011-08-07T03:12:19.947" Title="How does one place QFT's reality in context with that of other areas of physics?" Tags="&lt;quantum-field-theory&gt;&lt;standard-model&gt;" CommentCount="6" />
  <row Id="13244" PostTypeId="2" ParentId="13236" CreationDate="2011-08-07T03:56:36.440" Score="1" ViewCount="" Body="&lt;p&gt;I can see you've added more twist to your previous thought experiment, but it can be understood using a simple analogy. Think of the EM field as water, and the coils as boats floating in water. At the moment you turn on the electric current in the loop, the boat starts its propeller, and accelerates in +/- z direction (according to the direction of the current) by transferring momentum to the ambient EM field. So there's no need for the EM field to carry momentum in a predetermined direction. In fact, before you switch on the second coil, you have a static magnetic field (generated by the first coil) which has zero momentum density at every point. (EM momentum density = Poynting vector / c^2, therefore static E/B field doesn't carry momentum.)&lt;/p&gt;&#xA;" OwnerUserId="4749" LastActivityDate="2011-08-07T03:56:36.440" CommentCount="4" />
  <row Id="13245" PostTypeId="1" AcceptedAnswerId="14268" CreationDate="2011-08-07T04:31:42.010" Score="8" ViewCount="215" Body="&lt;p&gt;After some reading on dimensional analysis, it seems to me that only rational exponents are considered. To be more precise, it seems that dimensional values form a vector space over the rationals. My question is, why do we restrict ourselves to rational exponents? While it's true that I have never seen a physical equation with irrational exponents, is there any real guarantee that it will never occur? What are the benefits of such a restriction?&lt;/p&gt;&#xA;" OwnerUserId="3762" LastActivityDate="2011-09-02T07:13:03.647" Title="Dimensional analysis restricted to rational exponents?" Tags="&lt;dimensional-analysis&gt;" AnswerCount="5" FavoriteCount="1" />
  <row Id="13246" PostTypeId="1" CreationDate="2011-08-07T05:18:15.843" Score="-2" ViewCount="51" Body="&lt;p&gt;A large room is filled with air polluted with Caesium-137 at an average density of 1 Cs-137 particle per $\text{cm}^3$. A human is sitting inside the room for duration of $T$. What is the total energy absorbed by the person?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Originally given at &lt;a href=&quot;http://problemania.org/wiki/index.php?title=Problem_of_the_Day_20110324&quot; rel=&quot;nofollow&quot;&gt;Problemania wiki&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4836" LastEditorUserId="3066" LastEditorDisplayName="" LastEditDate="2011-08-07T18:14:00.773" LastActivityDate="2011-08-07T18:14:00.773" ClosedDate="2011-08-07T20:51:06.553" Title="Radiation dosage problem" Tags="&lt;homework&gt;&lt;radiation&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13247" PostTypeId="2" ParentId="13246" CreationDate="2011-08-07T06:57:41.330" Score="1" ViewCount="" Body="&lt;p&gt;I will give it a try. Multiply the density of Cs-137 by the volume of the room to find the initial number of Cs-137 particles $N_0$. Then from decay law $N(t)=N_0 e^{−\lambda t}$ find the number of Cs-137 particles $n=N_0−N(T)$ that have decayed after time $t=T$. Finally multiply the energy of the decay product (electrons of energy ~$1.176 MeV$) by the $n$ to find the total absorbed energy. Also you can divide it by the humans body mass to find the absorbed dose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this is just an approximation because of the fact that the source of Cs-137 is the whole room (it's not a point like source) so there will be some absorption of decay products by the air and Cs-137 (I think). Moreover, the human is inside the source (room). &lt;/p&gt;&#xA;" OwnerUserId="3066" LastEditorUserId="3066" LastEditorDisplayName="" LastEditDate="2011-08-07T07:09:59.250" LastActivityDate="2011-08-07T07:09:59.250" />
  <row Id="13248" PostTypeId="2" ParentId="13245" CreationDate="2011-08-07T07:22:37.043" Score="1" ViewCount="" Body="&lt;p&gt;Ok, first answer was incorrect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the idea may be reformulated in a way it is correct. The way you get powers in a dimensional analysis is restricted with physics laws. Usually, in laws only integer powers of physical values appear. Thus you get rationals out of integers. If you find a law with more sophisticated dependency (actually, I can hardly imagine such a law. Probably, in spaces with non-integer dimensionality) you might get any powers in principle. &lt;/p&gt;&#xA;" OwnerUserId="4657" LastEditorUserId="4657" LastEditorDisplayName="" LastEditDate="2011-08-07T12:11:53.927" LastActivityDate="2011-08-07T12:11:53.927" />
  <row Id="13249" PostTypeId="1" CreationDate="2011-08-07T11:38:23.907" Score="0" ViewCount="62" Body="&lt;p&gt;Ok so in a lecture my professor gave us this definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dF=\left(\frac{\partial F}{\partial T}\right)_{V,N}dT+\left(\frac{\partial F}{\partial V}\right)_{T,N}dV+\left(\frac{\partial F}{\partial N}\right)_{T,V}dN$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we were given that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F=E-TS=-pV+\mu N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(\frac{\partial F}{\partial T}\right)_{V,N}=-S$ This was easy!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(\frac{\partial F}{\partial V}\right)_{T,N}=-p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(\frac{\partial F}{\partial N}\right)_{T,V}=\mu$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other two I really don't know how to derive.&lt;/p&gt;&#xA;" OwnerUserId="2789" LastActivityDate="2011-08-07T14:33:31.180" Title="Helmholtz Free Energy relation" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" />
  <row Id="13250" PostTypeId="2" ParentId="13242" CreationDate="2011-08-07T14:04:16.683" Score="0" ViewCount="" Body="&lt;p&gt;I recommend You google for &quot;fluid dynamics&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg the wiki page is big, and has about a complete screen with &#xA;links going further on! &#xA;The page lists about half a dozen of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;either/or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which were necessary to know before You know how to or whether or not &#xA;calculation will be hopeful. &#xA;(But hydrophobicity is not mentioned there :=(&lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-08-07T14:04:16.683" />
  <row Id="13251" PostTypeId="2" ParentId="13249" CreationDate="2011-08-07T14:33:31.180" Score="1" ViewCount="" Body="&lt;p&gt;This seems to simply come from the derivatives of the second form of $F$ given, namely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = -pV +\mu N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;assuming $p$, $\mu$ so not depend on $V$, $N$.&lt;/p&gt;&#xA;" OwnerUserId="4841" LastActivityDate="2011-08-07T14:33:31.180" />
  <row Id="13252" PostTypeId="1" CreationDate="2011-08-07T14:53:36.757" Score="2" ViewCount="68" Body="&lt;p&gt;At the end of Chapter 14 of the &quot;Supersymmetry Demystified book&quot; from Patrick Labelle it is mentioned that to constrain the number of allowed softly SUSY breaking terms, a shadow or hidden supersymmetric sector can be assumed. In this hidden sector, SUSY is sponteanously broken and communicated to the visible sector of the MSSM where it leads to soft SUSY breaking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering where the particles of this shadow sector, which interacts only through a messenger interaction with the MSSM particles, would come from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that the MSSM can not &quot;stand alone&quot; but some kind of not so &quot;minimal&quot; or &quot;standard&quot; theory is needed aniway to provide the particle content and messenger interactions of the hidden sector with the MSSM?&lt;/p&gt;&#xA;" OwnerUserId="2751" LastActivityDate="2011-08-07T14:53:36.757" Title="Where does the hidden supersymmetric sector of the MSSM come from?" Tags="&lt;supersymmetry&gt;" CommentCount="3" FavoriteCount="1" />
  <row Id="13253" PostTypeId="2" ParentId="13245" CreationDate="2011-08-07T15:46:57.633" Score="2" ViewCount="" Body="&lt;p&gt;There's no reason I can't get irrational exponents of a unit by algebraic manipulation. For example, starting from $F=ma$, I could derive $F^\pi=m^\pi a^\pi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One concrete reason why I would not expect to see such things arise in any well-motivated physical context is that a system of units typically includes units for time and distance, and each of these can naturally occur as a negative number. Since the laws of physics are supposed to make well-defined predictions, we expect the equations representing them to be single-valued. But if you take non-integer powers of quantities that may be negative, you run into problems with branch cuts. I would therefore expect all the laws of physics to involve only integral exponents (as Misha has observed), and there is therefore no obvious reason why I would obtain irrational exponents through algebraic manipulation, except if I was just trying to be perverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many examples involving irrational exponents can be massaged into a form where they don't involve irrational exponents. For instance, if I was presented with $F^\pi=m^\pi a^\pi$, I could simplify it to $F=ma$. To get an equation involving only multiplication and exponentiation that can't be reduced to its simplest form, I need a system of units with more than one base unit, and an equation with different irrational exponents. E.g., I could define &quot;funkocity,&quot; a measure of the speed of motion, as $u=x^\pi/t^e$. The exponents in this equation can't be rationalized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to a stronger justification for the fact that we don't encounter irrational exponents. The SI has as many as three base units (or 4 if you count the Coulomb) only because of historical circumstances. If I'm doing calculations in special relativity, I'm going to use a system of units with $c=1$, and in GR I'd also set $G=1$, which gives geometrized units. In geometrized units, there is only one base unit, and therefore every expression involving only multiplication and exponentiation can reworked into a form where the exponents are rational.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would say that the fundamental reason we don't need irrational exponents in physics is that we have two fundamental theories of physics right now, GR and the standard model. In GR, the natural system of units is one with $c=1$ and $G=1$, so there is only one base unit. In the standard model, the natural system has $c=1$ and $\hbar=1$, so again we only have one base unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this argument fails to rule out is examples of equations with exponents that can't be rationalized because they have other operations such as addition and subtraction. For example, if we had an equation $a^\pi=b^\pi+c^\pi$, with $a$, $b$, and $c$ all having units of the base unit, then there is no way to rationalize the exponents. The only reason I can offer for the fact that such examples don't come up is to go back to the argument in the second paragraph.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastEditorUserId="4552" LastEditorDisplayName="" LastEditDate="2011-08-07T18:01:41.717" LastActivityDate="2011-08-07T18:01:41.717" CommentCount="1" />
  <row Id="13254" PostTypeId="2" ParentId="13060" CreationDate="2011-08-07T17:56:23.330" Score="2" ViewCount="" Body="&lt;p&gt;The best way to think about it is that a number like 1 km consists of a dimensionless 1 multiplied by a unit, km. When you take the log of a product, you get the sum of the logs, so log(1 km) is the same as log(1)+log(km). This shows that the log of 1 km is neither a dimensionless quantity nor a dimensionful one. If it was dimensionless, then it would be expressible without reference to any system of units. If it was dimensionful, then it would change by multiplication when the system of units was changed. It's neither of these things.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-07T17:56:23.330" CommentCount="3" />
  <row Id="13255" PostTypeId="2" ParentId="13157" CreationDate="2011-08-07T18:10:24.757" Score="0" ViewCount="" Body="&lt;p&gt;Ah, the idea of the field as interface. That's the whole philosophy of the source field method introduced by Schwinger. Treat the field as a perturbative source, and watch how it kicks the system under study. But always, at the end of the day, we have to take the limit as the source field goes to zero. The interface is fictitious after all, right? We only imagine the interface, or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then there's the S-matrix approach introduced by Heisenberg for philosophical reasons, with the philosophy carried to extremes by Chew with his bootstrap model. Bypass fields entirely, and only focus upon the asymptotic past and asymptotic future. Only the asymptotic future is observable. But it's not like we're living in the asymptotic future, are we?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interface analogy might work for a classical simulation, but not a quantum simulation. Quantum mechanics changes everything. Sure, for a classical simulation, the programmer can implement a read-only query function call which otherwise doesn't affect the simulation. But that's impossible for a quantum simulation, and our universe is quantum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say there's a programmer out there in heaven implementing the program simulating our universe. This programmer is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Demiurge&quot; rel=&quot;nofollow&quot;&gt;demiurge&lt;/a&gt;, a god but not God. Say this demiurge wishes to implement a read query for transcendent angels in heaven to call to query internal states of our universe. Now, let's say a transcendent angel Eve comes along and wants to eavesdrop our thoughts. What a snoop! Quantum information theory tells us Eve can't eavesdrop without affecting the system, i.e. our universe, and in principle, the mere act of calling the query can be detected within our universe. See, we would become entangled with Eve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If both heaven and earth are quantum, epiphenominalism is impossible.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or maybe Eve wishes to avoid detection by only accessing the query at the end of the simulation? Would that work? No, because of the retrocausal effects of the query. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Delayed_choice_experiment&quot; rel=&quot;nofollow&quot;&gt;delayed choice experiment&lt;/a&gt;. Take note, metaphysics-hating positivists, in principle, transcendent angels in heaven can produce measurable effects in our universe. Or if Eve is really careful, she can avoid snooping whenever there are experimenters trying to measure her effects? Isn't that a common complaint heard by skeptics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, maybe the demiurge wanted to be deistic and not implement any read query interfaces. But that demiurge would also have to prevent retrocausal influences from the end of the simulation. That can only be done if the end result of the quantum simulation is exactly uncomputed, erasing everything. From nothing, back to nothing. We might as well not even exist, and no one in heaven will know about us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So maybe the demiurge wasn't deistically inclined and left trapdoors to allow angels to influence our world with their will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or maybe the demiurge simulated our quantum universe on a classical computer. Let's just assume he has exponential resources to waste. But to read our universe, he still has to choose a decoherent histories realm anyway, anthropically selected upon our existence and conditioned upon other factors as well, like our history. Heaven is still detectable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The lesson of quantum mechanics is there has to be a transcendent observer observing us.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A closed quantum system of which we are a part can never have definite outcomes, and more importantly, doesn’t have and can't have a preferred basis. A quantum universe can never be closed if it were real.&lt;/p&gt;&#xA;" OwnerUserId="4842" LastActivityDate="2011-08-07T18:10:24.757" CommentCount="1" />
  <row Id="13256" PostTypeId="1" AcceptedAnswerId="13258" CreationDate="2011-08-07T18:10:35.293" Score="4" ViewCount="75" Body="&lt;p&gt;I was asked to today about the available plans (in Israel) to install photo-voltaic receptors on the roof and sell the energy. This is a legitimate plan and there are several companies that do this. The claim is that the initial investment is repaid after roughly 5 years which sounds too optimistic to me. I want to carry the calculation myself and check the results. Where can I find some data about the light flux that hit the earth in a typical year?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;note&lt;/em&gt;&#xA;I'm not sure, since this is not my specific field of expertise, about the commonly used term.&lt;/p&gt;&#xA;" OwnerUserId="3677" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-08-07T20:13:31.177" LastActivityDate="2011-08-08T02:33:18.873" Title="How much solar energy hits a rooftop in Israel?" Tags="&lt;renewable-energy&gt;&lt;photovoltaics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="13257" PostTypeId="2" ParentId="13256" CreationDate="2011-08-07T18:34:56.907" Score="2" ViewCount="" Body="&lt;p&gt;For the middle east, typically around 2000 KWh/m^2/year&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good place to start is wiki page for &lt;a href=&quot;http://en.wikipedia.org/wiki/Insolation&quot; rel=&quot;nofollow&quot;&gt;insolation&lt;/a&gt; (technical term for sunlight arriving)&lt;/p&gt;&#xA;" OwnerUserId="2525" LastEditorUserId="2525" LastEditorDisplayName="" LastEditDate="2011-08-07T21:26:29.597" LastActivityDate="2011-08-07T21:26:29.597" CommentCount="2" />
  <row Id="13258" PostTypeId="2" ParentId="13256" CreationDate="2011-08-07T19:03:36.787" Score="5" ViewCount="" Body="&lt;p&gt;It will vary a lot by site, and on the particular PV technology being proposed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you need to know:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;is it monocrystalline silicon, is it CdTe thin-film, or something else?&lt;/li&gt;&#xA;&lt;li&gt;what's the efficiency of the inverter?&lt;/li&gt;&#xA;&lt;li&gt;what's the guarantee on the kit (5 years, 10 years, 20 years)?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Using @Martin Beckett's figure of 2000kWh/y, that means, roughly, that your annual production would be approximately 2000 hours x the installed capacity of your PV array. So if you had a 1kW capacity system, you might expect to get 2 000 kWh of electricity in an average year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's assuming an equator-facing PV array tilted to the optimum angle for your location; and, crucially, that there aren't any objects that would cast shadows over your PV array. Depending on the particular panels and inverter configuration, a bit of shadowing can reduce generation by 10-80%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various PV calculators around on the web - the good ones should ask you my top two questions above, about panel type and efficiency). The &lt;a href=&quot;http://rredc.nrel.gov/solar/calculators/PVWATTS/version1/&quot; rel=&quot;nofollow&quot;&gt;USA NREL PVWATTS calculator&lt;/a&gt; has no sites in Israel - the closest is Cairo - but I expect you'll be able to find a reliable one somewhere that will run the numbers for you. The &lt;a href=&quot;http://re.jrc.ec.europa.eu/pvgis/apps4/pvest.php?map=africa&quot; rel=&quot;nofollow&quot;&gt;EU PVGIS calculator&lt;/a&gt; looks like it covers Israel, so I'd recommend trying that out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Five years might be a plausible payback time, depending on how much you'd get paid per kWh; how much you'd save off your bills; and how much the installation costs are. Currently, I expect you'd be asked to pay of the order of €3/W of installed capacity (before any subsidies or taxes). So, at the above generation rate of 2kWh/W/y, then a five-year payback would imply that you'd be getting paid (in direct payment + bill savings) about €0.30/kWh (assuming a zero discount rate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Panel efficiency has already been taken account of, in the above calculation of power output and of payback time. That is to say, once you know the panel's rated capacity, you've already accounted for the panel's efficiency. For more info on PV efficiency itself, see &lt;a href=&quot;http://physics.stackexchange.com/questions/2656/how-bright-can-we-make-a-sun-jar/13199#13199&quot;&gt;this answer to another question&lt;/a&gt;. To calculate the efficiency of any given panel, take its rated power in kW, and divide by its surface area in square metres. For example, a 200 W panel that is 1 square metre has an efficiency of 0.2 / 1.0 = 0.2 = 20%. Typical commercial panels currently are in the range 10% (thin film) - 20% (best monocrystalline).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An annual insolation of 2000 kWh, means that 2000 kWh of sunlight falls on $1m^2$ in an average year. If you want to put panel efficiency in there too, you end up putting it in the numerator and denominator, and it all cancels out, leaving you just with 2000 hours x PV capacity in kW.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, that factor of 2000 gets derated by various things: panel shading, inverter efficiency, less-than-optimal tilt or orientation, system down-time, and so on.&lt;/p&gt;&#xA;" OwnerUserId="4066" LastEditorUserId="4066" LastEditorDisplayName="" LastEditDate="2011-08-08T02:33:18.873" LastActivityDate="2011-08-08T02:33:18.873" CommentCount="3" />
  <row Id="13259" PostTypeId="1" CreationDate="2011-08-07T20:16:36.027" Score="9" ViewCount="92" Body="&lt;p&gt;If dark energy contributes mass-energy density $\rho$ and pressure $p$ to the stress-energy tensor, then you can define $w=p/\rho$, where $w=-1$ gives a cosmological constant, $w&amp;lt;-1$ gives a big rip, and $w&amp;lt;-1/3$ if we want to use dark energy to explain cosmological acceleration. The WP &quot;Big Rip&quot; article cites a paper that dates back to 2003 &lt;a href=&quot;http://arxiv.org/abs/astro-ph/0302506&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/astro-ph/0302506&lt;/a&gt; , which states that the empirical evidence was at that time only good enough to give $-2 \lesssim w \lesssim -.5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have observations constrained $w$ any more tightly since 2003?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard many people express the opinion that $w&amp;lt;-1$ is silly or poorly motivated, or that &quot;no one believes it.&quot; What are the reasons for this? Other than mathematical simplicity, I don't know of any reason to prefer $w=-1$ over $w\ne -1$. Considering that attempts to calculate $\Lambda$ using QFT are off by 120 orders of magnitude, is it reasonable to depend on theory to give any input into what values of $w$ are believable?&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-08T05:34:23.130" Title="evidence on the equation of state for dark energy?" Tags="&lt;general-relativity&gt;&lt;cosmology&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="13260" PostTypeId="2" ParentId="12385" CreationDate="2011-08-07T20:49:15.487" Score="3" ViewCount="" Body="&lt;p&gt;Check out the following 3 articles and 2 books:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ate.uni-duisburg-essen.de/data/postgraduate_lecture/AJP_2011_Olness.pdf&quot; rel=&quot;nofollow&quot;&gt;Regularization Renormalization and Dimensional Analysis: Dimensional Regularization meets Freshman E&amp;amp;M&lt;/a&gt; Published in the american journal of physics (can be found also on hep-ph, but slightly different with less references) &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://hitoshi.berkeley.edu/230A/regularization.pdf&quot; rel=&quot;nofollow&quot;&gt;Regularization, from Murayama's course of QFT at Berkeley&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/0212049&quot; rel=&quot;nofollow&quot;&gt;A Hint of Renormalization&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;A more general detailed, still introductory, treatment including renormalization would be the book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0198506945&quot; rel=&quot;nofollow&quot;&gt;Renormalization Methods: A Guide For Beginners&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;A. Zee's book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691140340&quot; rel=&quot;nofollow&quot;&gt;QFT in a Nutshell&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Anyway, I hope that was useful&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Revo&lt;/p&gt;&#xA;" OwnerUserId="4521" LastEditorUserId="4521" LastEditorDisplayName="" LastEditDate="2011-08-09T12:27:41.377" LastActivityDate="2011-08-09T12:27:41.377" CommunityOwnedDate="2011-08-07T20:49:15.487" CommentCount="3" />
  <row Id="13262" PostTypeId="2" ParentId="13245" CreationDate="2011-08-07T23:26:47.120" Score="5" ViewCount="" Body="&lt;p&gt;A counterexample: In second order phase transitions of 3-dimensional systems, the critical exponents can be irrational numbers. If $|T-T_c|$ is taken to be of dimension 1, then the correlation length, heat capacity, susceptibility etc. can have irrational dimensions.&lt;/p&gt;&#xA;" OwnerUserId="2643" LastActivityDate="2011-08-07T23:26:47.120" CommentCount="4" />
  <row Id="13263" PostTypeId="1" AcceptedAnswerId="13287" CreationDate="2011-08-07T23:31:45.307" Score="5" ViewCount="142" Body="&lt;p&gt;As noted in &#xA;&lt;a href=&quot;http://physics.stackexchange.com/questions/12049/a-doubt-about-the-age-of-the-universe&quot;&gt;A doubt about the age of the universe&lt;/a&gt;,&#xA;the wiki about quasars still contains the following misleading sentence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The highest redshift quasar known (as of June 2011) is ULAS_J1120+0641, with a redshift of 7.085, which corresponds to a proper distance of approximately 29 billion light-years from Earth.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even if the &quot;strange 29 billion&quot; is replaced by the &quot;correct 12.9 billion&quot;, the fact remains that the actual measurement is &quot;a redshift of 7.085&quot;. The &quot;proper distance&quot; is only a different way to express that measurement. It's not clear to me how &quot;accurately&quot; this describes the distance of the quasar, because the quasar surrounds a black hole and rotates quite fast. So there are at least two additional sources for the redshift, but how significant is their contribution?&lt;/p&gt;&#xA;" OwnerUserId="4846" LastEditorUserId="4846" LastEditorDisplayName="" LastEditDate="2011-08-08T10:02:29.660" LastActivityDate="2011-08-09T02:55:52.883" Title="Can the distance of a quasar be determined accurately?" Tags="&lt;general-relativity&gt;&lt;cosmology&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="13264" PostTypeId="1" CreationDate="2011-08-07T23:49:23.840" Score="0" ViewCount="47" Body="&lt;p&gt;what is the relationship of sound volume to atmospheric pressure? if I was in a plane with a cabin pressure equal to 8000 feet would the volume drop be noticable?&lt;/p&gt;&#xA;" OwnerUserId="4847" LastEditorUserId="4552" LastEditorDisplayName="" LastEditDate="2011-08-08T00:21:25.430" LastActivityDate="2011-08-13T13:35:13.867" Title="what is the relationship of sound volume to atmospheric pressure?" Tags="&lt;pressure&gt;&lt;sound&gt;&lt;atmosphere&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13265" PostTypeId="2" ParentId="13264" CreationDate="2011-08-08T00:21:03.967" Score="1" ViewCount="" Body="&lt;p&gt;There is not going to be any generic answer to your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I drop a book on the floor, and the book hits the floor with 1 joule of kinetic energy. Let's assume that there is very little internal friction in the book or the floor, so that none of the energy can be dissipated into heat, and that the floor is very stiff, so that no energy can be transmitted as vibrations in the floor. Then by conservation of energy, the amount of sound energy is going to be exactly 1 joule. This argument is independent of whether this is the floor of my house, near sea level, or the deck of an airplane. It only depends on the assumption that the mechanisms for dissipating energy into heat and vibration are much less efficient than the mechanism for dissipating it into sound waves in the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, let's say for the sake of argument that in a hand clap at sea level, 1/3 of the energy goes into sound, 1/3 into vibrations transmitted down through the arms, and 1/3 into heating in the palms. In an airplane, these 1:1:1 proportions will be altered, because the conversion of energy into sound waves in air will be less efficient. Because the competition among the three dissipative processes is fairly equal, there will be a big reduction in loudness when the pressure is low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is that it depends completely on the details of the process of production of the sound waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the extreme case, suppose you do these experiments in outer space. Outer space is not a perfect vacuum, so it can support sound waves. However, the efficiency of the coupling to sound waves is extremely low, so almost any other mechanism for dissipating the energy will be many orders of magnitude more efficient.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-08T00:21:03.967" />
  <row Id="13266" PostTypeId="1" AcceptedAnswerId="13310" CreationDate="2011-08-08T00:37:48.573" Score="11" ViewCount="218" Body="&lt;p&gt;Some of the applications of random matrices is to find the spectra of heavy atoms in nuclear physics which are usually difficult to find otherwise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can starting from randomness of some kind, such as random matrices, lead to prediction of that sort? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea why this work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Random matrices are used in string theory too, may be some string theorist here can shed some light on this topic &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Revo&lt;/p&gt;&#xA;" OwnerUserId="4521" LastEditorUserId="4521" LastEditorDisplayName="" LastEditDate="2011-08-08T18:21:38.370" LastActivityDate="2011-08-09T04:38:00.563" Title="How come random matrices can predict energy spectra of heavy atoms?" Tags="&lt;quantum-mechanics&gt;&lt;particle-physics&gt;&lt;mathematical-physics&gt;&lt;string-theory&gt;&lt;nuclear-physics&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="4" />
  <row Id="13267" PostTypeId="2" ParentId="13263" CreationDate="2011-08-08T02:09:15.470" Score="5" ViewCount="" Body="&lt;p&gt;First off, you seem to be getting confused over the same issue that the OP in &lt;a href=&quot;http://physics.stackexchange.com/questions/12049/a-doubt-about-the-age-of-the-universe&quot;&gt;A doubt about the age of the universe&lt;/a&gt; . As far as I can tell, both you and that person believed that the proper distance x traveled by light over an interval of cosmological time t should be x=ct. This is incorrect, because while the photon is in flight, the universe expands. Here &lt;a href=&quot;http://physicsforums.com/showthread.php?t=506987&quot; rel=&quot;nofollow&quot;&gt;http://physicsforums.com/showthread.php?t=506987&lt;/a&gt; is a more detailed explanation of that point. Both the 29 billion l.y. figure and the 12.9 billion years are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The line spectra used to determine the redshifts are emission spectra from clouds of gas that lie between us and the quasar. There will be a gravitational redshift, but believe it or not, the astronomers in this business are not complete idiots, and this has occurred to them, and they know how to handle it. I believe the gas is far enough from the event horizon that the effect is actually negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar considerations apply to the rotation. The infalling gas may be moving quite fast, but the emission spectrum is from gas that is much farther out, and is not moving as fast. In addition, there will tend to be a cancellation of the kinematic Doppler shift on the average, since equal amounts of emission occurs for gas moving away from us and gas moving toward us. (It won't be an exact cancellation, because SR has transverse as well as longitudinal Doppler shifts.)&lt;/p&gt;&#xA;" OwnerUserId="4552" LastEditorUserId="4552" LastEditorDisplayName="" LastEditDate="2011-08-08T15:11:47.887" LastActivityDate="2011-08-08T15:11:47.887" CommentCount="4" />
  <row Id="13268" PostTypeId="1" AcceptedAnswerId="13279" CreationDate="2011-08-08T04:34:30.590" Score="6" ViewCount="60" Body="&lt;p&gt;Math/physics teachers love to break out the &lt;a href=&quot;http://www.youtube.com/watch?v=oiTf9qIQXtQ&amp;amp;feature=related&quot; rel=&quot;nofollow&quot;&gt;double pendulum&lt;/a&gt; as an example of chaotic motion that is very sensitive to initial conditions.  I have some questions about specific properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For a simple double pendulum (which I'll define as two equal masses joined by two massless arms of the same length) is the highest initial energy state when both masses are completely raised vertically, as intuition would suggest?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Double_pendulum#Chaotic_motion&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Double_pendulum#Chaotic_motion&lt;/a&gt; contains a plot of conditions for which the double pendulum &quot;flips&quot; or does not flip.  My question is: does a &quot;flipping&quot; of either top or bottom pendulum imply subsequent chaotic motion, or are there certain energies for which we can fully predict the motion and flips?  Conversely, if we are in low energies where neither top nor bottom pendulum can possibly flip, can we still have a chaotic system or do all those energies have predictable motions?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4849" LastActivityDate="2011-08-08T10:09:53.663" Title="What is the highest energy position for a double pendulum? And for which energy positions is it chaotic?" Tags="&lt;mechanics&gt;&lt;energy&gt;&lt;motion&gt;&lt;chaos-theory&gt;" AnswerCount="2" FavoriteCount="2" />
  <row Id="13269" PostTypeId="1" CreationDate="2011-08-08T04:45:15.800" Score="2" ViewCount="87" Body="&lt;p&gt;I was recently surprised to learn that defining temperature isn't easy.  For a long time, it was defined operationally: how much does a thermometer expand.  Also surprising, temperature isn't a measure of thermal energy: 0 C water has more thermal energy than 0 C ice, but the same temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came up with this definition, and I'd like to know if it's correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperature is the propensity of something to give off heat, that is transmit thermal energy.  Equivalently, it is the resistance of something to receive heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that correct? Please justify your answer.&lt;/p&gt;&#xA;" OwnerUserId="4850" LastActivityDate="2011-08-09T02:31:57.057" Title="Can temperature be defined as propensity to transmit thermal energy?" Tags="&lt;thermodynamics&gt;&lt;energy&gt;&lt;heat&gt;&lt;temperature&gt;&lt;definition&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="13270" PostTypeId="2" ParentId="13268" CreationDate="2011-08-08T05:10:57.410" Score="1" ViewCount="" Body="&lt;p&gt;Here is an &lt;a href=&quot;http://www.myphysicslab.com/dbl_pendulum.html&quot; rel=&quot;nofollow&quot;&gt;interactive simulation page&lt;/a&gt; with derivation of formulae that may enlighten you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the energy in the system is just the gravitational energy, then yes, the vertical initial position will have the highest potential energy which will be transmuted to kinetic when let go. Energy is conserved, kinetic + potential is constant, so the &quot;position&quot; helps in determining the initial energy of the system.  Otherwise, if kinetic energy is supplied  it will be the maximum potential + the added kinetic. What has no answer is also the &quot;position&quot; in this case, as it will depend on how the impulse is conveyed. &lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-08-08T05:10:57.410" />
  <row Id="13271" PostTypeId="2" ParentId="13157" CreationDate="2011-08-08T05:32:47.717" Score="1" ViewCount="" Body="&lt;p&gt;I enjoyed @dmckee answer but would like to add that fields are primarily a &lt;a href=&quot;http://en.wikipedia.org/wiki/Field_%28mathematics%29&quot; rel=&quot;nofollow&quot;&gt;mathematical tool&lt;/a&gt;, with definite rules taken over by physicists in their effort to describe the observed world.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In abstract algebra, a field is a commutative ring whose nonzero elements form a group under multiplication. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will also add that in classical physics, fields are defined as well, with no need of  virtual particle exchanges to be useful in calculations .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And my third observation is that there exists the old pythgorean view of the world, that &quot;God continually geometrises&quot; ( if I may coin a word), or the so called &quot;platonic ideals&quot;, in modern words : given the mathematics reality will follow the mold.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-08-08T05:32:47.717" CommentCount="4" />
  <row Id="13272" PostTypeId="2" ParentId="13259" CreationDate="2011-08-08T05:34:23.130" Score="1" ViewCount="" Body="&lt;p&gt;I do not know if this answer will address fully your Question, anyway:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1104.5426&quot; rel=&quot;nofollow&quot;&gt;Clustering of Photometric Luminous Red Galaxies II: Cosmological Implications from the Baryon Acoustic Scale&lt;/a&gt; (2011/Apr)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Combining with previous measurements of the acoustic scale, we obtain&#xA;  a value of $w_0$ = -1.03 +/- 0.16 for the equation of state parameter of&#xA;  the dark energy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;cited here : &lt;a href=&quot;http://arxiv.org/abs/1107.1789v1&quot; rel=&quot;nofollow&quot;&gt;Cosmology today–A brief review&lt;/a&gt; (25 pages on Theory and data, 2007/Jul)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a new &lt;strong&gt;model of the Universe without Dark Energy&lt;/strong&gt; here:&lt;br&gt;&#xA;&lt;a href=&quot;http://vixra.org/pdf/1107.0016v1.pdf&quot; rel=&quot;nofollow&quot;&gt;A self-similar model of the Universe unveils the nature of dark energy&lt;/a&gt; (21 pages Jul/2011, not peer-reviewed, it uses only Newton and Coulomb laws). I do not know how this novel viewpoint can be discarded.&lt;br&gt;&#xA;Argument:&lt;br&gt;&#xA;From $$F=m\cdot a,\, F=G\cdot\frac{m_{1}\cdot m_{2}}{d^{2}},\, F=\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{q_{1}\cdot q_{2}}{d^{2}},\, c=\frac{1}{\sqrt{\varepsilon_{0}\mu_{0}}}.&#xD;&#xA;$$ obtain the dimensional equations $$\left[G\right]=M^{-1}L^{3}T^{-2},\left[\varepsilon\right]=M^{-1}Q^{2}L^{-3}T^{2},\left[c\right]=LT^{-1}$$&#xA;because the  the sum of the exponents is zero, a cute coincidence ;), we know that $[M],[Q],[L],[T]$ can scale in the same way and $[c],[G],[\varepsilon]$ are constants. If the universe    &lt;em&gt;can&lt;/em&gt; scale thru time, keeping always the same basic physical laws, it &lt;em&gt;will&lt;/em&gt; scale.&lt;br&gt;&#xA;The &lt;em&gt;scaling law&lt;/em&gt; $\alpha(t_{S})=e^{-H_{0}\cdot t_{S}}$ is derived from the observational data, where $t_S$ is considered from the viewpoint of a comoving invariant referential, and  Dark Energy is absent from it. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The accelerated expansion is an artifact of the standard model&lt;/strong&gt;&lt;br&gt;&#xA;  The statement that space expansion is accelerating is not the result of some direct measurement more or less independent of the cosmological model but, on the contrary, it is a consequence of the theoretical framework of the standard model. The deceleration parameter at the present moment, $q_{0}$, in the $\Lambda$CDM model, for flat space and $\Omega_{R}=0$ , is given by $q_{0}=\frac{1}{2}\left(\Omega_{M}-2\Omega_{\Lambda}\right)$ therefore, for $\Omega_{M}+\Omega_{\Lambda}=1$ , the value of $q_{0}$  is negative for $\Omega_{\Lambda}&amp;gt;1/3$ ; a value of $\Omega_{\Lambda}$  lower than 1/3 leads to a comoving distance largely in disagreement with observations, hence, in the framework of $\Lambda$CDM model it has to be $\Omega_{\Lambda}&amp;gt;1/3$  and, so, $q_{0}&amp;lt;0$ .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Needless to say: this is my preferred viewpoint because it has no free lunches: growing space, growing dark energy, and apply to all scales, even to Solar system)&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-08-08T05:34:23.130" CommentCount="3" />
  <row Id="13273" PostTypeId="2" ParentId="13245" CreationDate="2011-08-08T06:03:03.597" Score="-2" ViewCount="" Body="&lt;p&gt;Roy Sympson &lt;a href=&quot;http://physics.stackexchange.com/questions/7668/fundamental-question-about-dimensional-analysis/7675#7675&quot;&gt;PSE-here&lt;/a&gt; pointed to &lt;a href=&quot;http://en.wikipedia.org/wiki/Buckingham_%CF%80_theorem&quot; rel=&quot;nofollow&quot;&gt;The Buckingham Π Theorem&lt;/a&gt;.&lt;br&gt;&#xA;You should consider &lt;em&gt;measurable&lt;/em&gt; &quot;fundamental units&quot; like &quot;time, distance, velocity, mass, momentum, energy, and weight&quot;, and clearly $c,G,\hbar$ do not belong to these fundamental base units. The dimension equation of $\left[G\right]={M^{-1}}{L^{3}}{T^{-2}}$, $\left[c\right]  =  LT^{-1},\left[h\right]  =  ML^{2}T^{-1}$.&lt;br&gt;&#xA;In math you have the liberty to &lt;em&gt;construct&lt;/em&gt; the axioms and consistently derive from those. In Physics we are constrained by how the universe manifests itself or in ,other words, &quot;we model upon data&quot;.&lt;br&gt;&#xA;To find an irrational number in the exponent is not the point, e.g. $e^{t/k}$, where $t$ is valid for any $t$ including irrationals, but $k$ is also a time constant and the exponent is dimensionless, and this is the key point. &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-08-08T11:20:57.697" LastActivityDate="2011-08-08T11:20:57.697" CommentCount="2" />
  <row Id="13274" PostTypeId="2" ParentId="13237" CreationDate="2011-08-08T06:22:59.577" Score="3" ViewCount="" Body="&lt;p&gt;Well, I don't know if you want to count that, but QFT is full with functions that have poles which I'd call not well-behaved, and it does have lots of physical effects. If you're talking about observables only, you can approximate any discontinuous function to arbitrary precision with a continuous function, and you can push the difference below measurement precision. The reason one sometimes uses 'ill-behaved' functions (delta, heaviside, etc) is that they're easier to deal with. &lt;/p&gt;&#xA;" OwnerUserId="1458" LastActivityDate="2011-08-08T06:22:59.577" CommunityOwnedDate="2011-08-08T06:22:59.577" CommentCount="1" />
  <row Id="13275" PostTypeId="1" CreationDate="2011-08-08T07:33:01.577" Score="0" ViewCount="109" Body="&lt;p&gt;In mathematics, sum of all natural number is infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but Ramanujan suggests whole new definition of summation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The sum of $n$ is $-1/12$&quot; what so called &lt;a href=&quot;http://en.wikipedia.org/wiki/Ramanujan_summation&quot; rel=&quot;nofollow&quot;&gt;Ramanujan Summation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First he find the sum, only Hardy recognized the value of the summation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also in quantum mechanics(I know), Ramanujan summation is very important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question.&lt;/strong&gt; What is the value of Ramanujan summation in quantum mechanics?&lt;/p&gt;&#xA;" OwnerUserId="4558" LastActivityDate="2011-08-08T08:07:13.410" Title="Value of Ramanujan Summation In Quantum Mechanics" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13276" PostTypeId="2" ParentId="13169" CreationDate="2011-08-08T07:43:43.153" Score="-1" ViewCount="" Body="&lt;p&gt;Upshot: the range of a laser is greater because the light is concentrated in a very narrow beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without going into much detail of the power of the laser, I think the problem here has to do with the definition of 'intensity'. Wikipedia lists several different units for &lt;a href=&quot;https://secure.wikimedia.org/wikipedia/en/wiki/Intensity_%28physics%29#See_also&quot; rel=&quot;nofollow&quot;&gt;intensity&lt;/a&gt;, the one that most closely matches 'brightness' as perceived by the eye is probably &lt;a href=&quot;https://secure.wikimedia.org/wikipedia/en/wiki/Radiance&quot; rel=&quot;nofollow&quot;&gt;radiance&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The units of radiance are &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;L = W&amp;#183;sr^{−1}&amp;#183;m^{−2}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or watt per &lt;a href=&quot;https://secure.wikimedia.org/wikipedia/en/wiki/Steradian&quot; rel=&quot;nofollow&quot;&gt;steradian&lt;/a&gt; per square metre, a 'steradian' being a two-dimensional angle on the sky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now given a fixed power for a certain light source, the difference between ordinary lighting and a laser are two-fold:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The surface ($m^{2}$) of a laser is much smaller (probably a factor of 10)&lt;/li&gt;&#xA;&lt;li&gt;The divergence ($sr$) of the laser beam is also much smaller (factor ~100 or more). &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Both these factors mean that the radiance for the same power ($W$) a laser beam is at least 1000 times as 'bright' as a regular light source. Since the brightness is higher, the beam will be visible up to a longer distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to note that this result is regardless of the absorbance or scattering that occurs, which in principle happens for both laser and non-laser beams.&lt;/p&gt;&#xA;" OwnerUserId="4574" LastActivityDate="2011-08-08T07:43:43.153" CommentCount="3" />
  <row Id="13277" PostTypeId="2" ParentId="13275" CreationDate="2011-08-08T08:07:13.410" Score="3" ViewCount="" Body="&lt;p&gt;Lumo gives a very nice step by step calculation of this sum and a good discussion of the importance and application of such summation techniques in QFTs here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2011/07/why-is-sum-of-integers-equal-to-112.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2011/07/why-is-sum-of-integers-equal-to-112.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such mathematical calculations are NOT unrelated to physics; on the contrary they are important...&lt;/p&gt;&#xA;" OwnerUserId="2751" LastActivityDate="2011-08-08T08:07:13.410" />
  <row Id="13278" PostTypeId="2" ParentId="2229" CreationDate="2011-08-08T09:28:20.353" Score="-1" ViewCount="" Body="&lt;p&gt;Study Wikipedia on Electron Volt, and understand that a photon can&#xA;have different frequencies and wavelenghts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a photon has no mass, calculate E (energy) from &#xA;equations E=hv and E= hc/λ&#xA; v = frequency of photon = c/wavelength = c/λ&lt;/p&gt;&#xA;&#xA;&lt;p&gt;h = planck constant (6.626068 × 10-34 m2 kg / s) =3&#xA;h =Planck's constant (6.626 x 10^-34 Jsec). &#xA;    1 eV equals 1.602 x 10^-19 J. &#xA;So h =  4.13566733 eVsec&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c = speed of light = 299,792,458 m/sec&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E= hc/λ = (4.13566733 eVsec * 299,792,458 m/sec) /λ.&#xA;E(in eV) = 1240 eV nm / λ (nm); nm = nanometer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the Wikipedia article, green light photon λ (wavelength) is 432 nm,&#xA;E (green light) = 1240/432 eV  = 2.33 eV.&#xA;Using red light wavelength of 685 nm,&#xA;E (red light) = 1240/685 eV =  1.81 eV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary a photon does not have mass, or resting oe kinetic, &#xA;but does have momentum = to energy  (momenergy, &#xA;see Taylor, Wheeler Spacetime Physics, chapter 8).&lt;/p&gt;&#xA;" OwnerUserId="4851" LastActivityDate="2011-08-08T09:28:20.353" />
  <row Id="13279" PostTypeId="2" ParentId="13268" CreationDate="2011-08-08T10:09:53.663" Score="4" ViewCount="" Body="&lt;p&gt;It is true that the double pendulum exhibits integrable behavior, when the initial angles are very small, however, in general, it is very difficult to characterize the chaotic behavior of the double pendulum in terms of the initial angles. There are other representations which provide a clearer picture of its chaotic behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The introductory section of the following &lt;a href=&quot;http://donar.physik.uni-bremen.de/~prichter/pdfs/ForcesDoublePendulum.pdf&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; by O. A. Richter, and reference therein describe the main characteristics of the double pendulum integrability. I'll summarize here the main facts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Remarks: The numerical values correspond to a standard double pendulum of unit masses and unit rod lengths and unit acceleration due to gravity)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total energy of the double pendulum is a constant of motion. The double pendulum possesses 4 equilibrium points corresponding to total energies $E= 0, 2, 4, 6$. The total energy determines the topology of the energy hypersurfaces, for $E&amp;lt;2$, the energy hypersurfaces are three spheres, while for $E&amp;gt;6$ , the energy hypersurfaces are three tori.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To further understand this point, in the case of very low energies, the system can be approximated (linearized) to an isotropic harmonic oscillator. The energy hypersurfaces are then of the form $x_1^2+x_2^2+p_1^2+p_2^2=E = const.$, while for the case of very large energies, the kinetic energy dominates and we can neglect gravity. In this case, there are two types of solutions of the equation of motion, one in which the outer rod rotates and the inner rod oscillates, and the second in which both rods rotate. The transition between the two types of solutions is determined by the value of the total angular $L$ momentum which becomes a constant of motion (due to the lack of gravity). For the standard double pendulum the transition occurs at $ L^2 = 2E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both limits (small and large energies) correspond to &lt;a href=&quot;http://en.wikipedia.org/wiki/Integrable_system&quot; rel=&quot;nofollow&quot;&gt;integrable systems&lt;/a&gt;. This is well known, but here is a short explanation.&#xA;To see that the isotropic harmonic oscillator is integrable, one needs to solve the equations of motion in polar coordinates. &#xA;The polar angle just rotates with a constant angular velocity, and the radial coordinates oscillates in such a way that the trajectory has the shape of a two trip course through the donut hole drawn on the surface of a two-torus. This is the &lt;a href=&quot;http://en.wikipedia.org/wiki/KAM_theorem&quot; rel=&quot;nofollow&quot;&gt;Liouville-Arnold torus&lt;/a&gt; (whose existence indicates the system's integrability) with respect to which the three sphere energy hypersurface is foliated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the high energy limit, a similar Liouville-Arnold torus exists when the inner rod oscillates and a torus generated by the two polar angles when the two angles rotate.&#xA;(Here the exact solution is more difficult, see for example, &lt;a href=&quot;http://donar.physik.uni-bremen.de/~prichter/pdfs/DoublePendulum.pdf&quot; rel=&quot;nofollow&quot;&gt;the following article&lt;/a&gt; by Enolskii, Pronine, and Richter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since both limits of vanishing and very high energies correspond to integrable systems, the total energy also controls the system characteristics, but the dependence here is much more complicated.&#xA;The transition from integrability to chaos and back as the total energy decreases from infinity to zero is described in figure 2 of Richter's article. There are a lot of details, but here are the main features: The figure correspond to the projection of the trajectories&#xA;onto the plane spanned by the outer rod angle and the total angular momenta. For very large angular momentum, the projections of the trajectories are horizontal lines with constant angular momenta (which is a constant of motion). As the energy is reduced, two disjoint chaotic regions are formed, the integrable trajectories correspond to rational and irrational tori, together with stable resonances. At about E = 10.352 which corresponds to the golden winding ratio, all irrational tori vanish and a transition to global chaos occurs.&#xA;The stable resonances also vanish eventually, at the low energies the resonances corresponding to the second integrable region start to appear.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastActivityDate="2011-08-08T10:09:53.663" />
  <row Id="13280" PostTypeId="2" ParentId="13237" CreationDate="2011-08-08T12:36:48.753" Score="2" ViewCount="" Body="&lt;p&gt;A nice example arises for the &quot;rigorous coupled wave analysis&quot; (RCWA) method (also called Fourier Modal Method), which is used as a Maxwell solver for diffraction gratings. The normal component of the electric field is discontinuous in the normal direction of a material interface. This leads to convergence problems of the RCWA method for TM polarization, because the discontinuous electric field component is expanded into a Fourier series and multiplied by another discontinuous function representing the grating geometry. Many modifications of the RCWA method to overcome this convergence problem where proposed, but the &quot;correct&quot; modification was only discovered in 1996 (by P. Lalanne and M. Morris?). Even so Lifeng Li didn't discover that &quot;correct&quot; modification, he wrote the famous paper &quot;Use of Fourier series in the analysis of discontinuous periodic structures&quot; (also in 1996) which analyzed mathematically what goes wrong (multiplication of &quot;approximations of&quot; discontinuous functions is dangerous) and why the latest proposed modification to the RCWA methods finally solved the convergence problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today, the Fourier Modal Methods are the most efficient and accurate for many types of grating problems.&lt;/p&gt;&#xA;" OwnerUserId="4846" LastActivityDate="2011-08-08T12:36:48.753" CommunityOwnedDate="2011-08-08T12:36:48.753" />
  <row Id="13281" PostTypeId="1" CreationDate="2011-08-08T13:50:10.340" Score="1" ViewCount="51" Body="&lt;p&gt;Yes, I'm serious.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What components correspond to an individual scan vrs what components&#xA;can be cascaded across all objects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ex: The vacuum chamber can be cascaded ( colocate the sample chambers within a single vacuum enclosure). Whereas there will need to be individual detectors for each sample&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can there be a singular electron source? Using multiple magnetic focal lenses to direct those electrons to the individual samples?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can this idea be extrapolated to significantly downsize the components of an electron microscope?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4854" LastEditorUserId="4854" LastEditorDisplayName="" LastEditDate="2011-08-09T00:58:16.080" LastActivityDate="2011-08-09T00:58:16.080" ClosedDate="2011-08-09T20:45:42.287" Title="Scaling an electron microscope to fit within 1 cubic cm when running multiple parallel scans" Tags="&lt;electromagnetism&gt;&lt;particle-physics&gt;&lt;optics&gt;&lt;electron&gt;&lt;microscopy&gt;" CommentCount="10" />
  <row Id="13282" PostTypeId="1" AcceptedAnswerId="13285" CreationDate="2011-08-08T15:07:40.887" Score="6" ViewCount="88" Body="&lt;p&gt;I've recently come across this statement that massless particles arise from $SO(D-2)$ symetry and massive particles from $SO(D-1)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would have guessed that it would be the exact opposite way, but that doesn't seem so. Could someone point me in the right direction what exactly I should be reading to understand this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I'm a mathematician who has only recently been trying my way through String Theory, I've been wondering, why is this? Is there some intuitiv way of thinking about this? (This was only briefly mentioned in the ST book by Becker, Becker, Schwarz)&lt;/p&gt;&#xA;" OwnerUserId="4337" LastActivityDate="2011-08-08T18:48:56.900" Title="Particles mass determined by SO(D-2) vs SO(D-1)" Tags="&lt;string-theory&gt;&lt;standard-model&gt;&lt;symmetry&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="13283" PostTypeId="1" CreationDate="2011-08-08T15:35:07.697" Score="2" ViewCount="34" Body="&lt;p&gt;I'm modeling light interaction/reflection from a fiberglass surface with polyurathane epoxy (that is very  reflective) that has been spray painted with a matte black paint. I'm looking for some input - is it reasonable to approximate this as specular reflection from the polyurathane with a loss factor from the paint absorption? Is there a better way? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By way of context, this is a large water tank where we do experiments with underwater optical communication. I'm building a Monte Carlo-type simulation to match our experimental outputs. The wall surface reflections are giving me issues.&lt;/p&gt;&#xA;" OwnerUserId="2735" LastActivityDate="2011-08-08T15:35:07.697" Title="Modeling a spray painted polyurathane surface reflection" Tags="&lt;optics&gt;&lt;light&gt;&lt;reflection&gt;&lt;refraction&gt;" CommentCount="2" FavoriteCount="1" />
  <row Id="13284" PostTypeId="1" AcceptedAnswerId="13293" CreationDate="2011-08-08T15:36:12.573" Score="4" ViewCount="81" Body="&lt;p&gt;The outside temperature was somewhere between the hot and cold that blows on the inside through the car's vents. I know the condensate is on my side because I can reach forward and write on it with my finger. When I blow hot air on it, it stays there, but when I blow cold air on it, it disappears. I would have thought blowing hot air on it would give the water molecules enough energy to escape the surface and blowing cold air would cause water vapor molecules in the air to lose energy and get 'stuck' to the surface, but the reality is the opposite of what I expected. Can anyone explain?&lt;/p&gt;&#xA;" OwnerUserId="4715" LastActivityDate="2011-08-08T20:16:24.437" Title="Why would colder air disperse condensation on a car windshield?" Tags="&lt;temperature&gt;&lt;air&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="13285" PostTypeId="2" ParentId="13282" CreationDate="2011-08-08T16:10:49.260" Score="7" ViewCount="" Body="&lt;p&gt;I'm not entirely sure that I've understood what you're referring  to, but it sounds like the following. To figure out the allowed representations for a massive particle in $D$-dimensional space-time dimensions, we can boost to its rest frame. Then we have the remaining $SO(D-1)$ rotations that leave us in the rest frame of the particle, and so its the respresentation of $SO(D-1)$ that determine the properties of the particle. So in $D = 4$ our massive particles are labelled by representations of $SO(3)$, which is good old angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand if we have a massless particle, we can't boost to its rest frame since it doesn't have a rest frame. So what we have to look at instead are the transformations that leaves it's direction fixed, that is the transformations that fix a null ray. This is $SO(D-2)$ (if we ignore the fact that one of the directions is not compact, I think its really $SO(D-3,1)$ or whatever it's called), and so it is the representations of $SO(D-2)$ that determine the states of massless particle. In $D = 4$ we have that our massless particles are labeled by representations of $SO(2)$. So when we say the photon has angular momentum 1 we are not talking representation of $SO(3)$. It does not have a state with $m = 0$ - a longitudinal polarization - even though the $l=1$ represenation of $SO(3)$ has three states $m=-1,0,1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this is what you were looking for.&lt;/p&gt;&#xA;" OwnerUserId="3970" LastEditorUserId="3970" LastEditorDisplayName="" LastEditDate="2011-08-08T18:48:56.900" LastActivityDate="2011-08-08T18:48:56.900" CommentCount="2" />
  <row Id="13286" PostTypeId="2" ParentId="13282" CreationDate="2011-08-08T16:12:16.710" Score="5" ViewCount="" Body="&lt;p&gt;It has to do with the fact that massless particles do not have a rest frame. If the particle is massive, we can view it in the frame where its 3-momentum is zero. Its symmetry group then becomes $SO(D-1)$ (the group that arises in this way is known as the &quot;little group&quot;). Then the particles are classified according to the representations of this little group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the particle is massless, the simplest form the 4-momentum can take through Lorentz transformations $(P_0,P_1,0,0)$ where $P_0$ is the time component of the 4-vector. There is no frame in which the particle is not moving. So then the little group becomes the group of transformations of $D-2$ Euclidean space, in the normal $D=4$ case this is translations and rotations. I've also seen this group written as $ISO(D-2)$. So the particles now arise from representations of this group.&lt;/p&gt;&#xA;" OwnerUserId="1635" LastActivityDate="2011-08-08T16:12:16.710" />
  <row Id="13287" PostTypeId="2" ParentId="13263" CreationDate="2011-08-08T16:26:19.370" Score="7" ViewCount="" Body="&lt;p&gt;You may gain some intuition on cosmological distances by exploring &lt;a href=&quot;http://www.astro.ucla.edu/~wright/CosmoCalc.html&quot; rel=&quot;nofollow&quot;&gt;cosmocalc&lt;/a&gt; which gives a light travel time of 12.9 Gyrs for the redshift of 7.  This site is nice because it shows you the parameters for the &quot;concordant&quot; cosmology that describe our Universe.  These parameters describe how the universe is expanding and how it expanded in the past.  One cool thing to try out is varying the redshift and seeing what ages for the universe you get at that redshift.   This gives you a feeling for how a big change in redshift at high $z$ gives you little change in the age of the universe.  For example a change from $z=7$ to $z=8$.  But this represents a pretty big shift in the spectrum  (challenging detection).  A decade ago astronomers did not know whether the universe was flat or what $\Omega_M$ was (density in units of a flat universe).  But now there are a number of different astronomical measurements that have all agreed on (or given small error bars around) these parameters.  Hence the term &quot;concordant.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, a light travel time is a proxy for distance, it is an integrated quantity and so a meaningful metric for distance. However you can't really send a message back as the universe has been expanding since the quasar emitted the light that we are detecting.  The second part of your question concerned accuracy.  Probably the most meaningful error is the error in the estimated redshift.  As you mention the emission lines can be broad and this would contribute to the error.   The emission lines could be misidentified though since there is a Lyman alpha forest giving absorption lines from material in between us and the quasar this is unlikely for this object.  The quasar could be moving w.r.t to the `Hubble flow' or mean expansion of the universe.  For example the quasar could be a black hole that ejected from a galaxy.  This type of error is likely to be less than the speed of light and so would at most contribute something like an error of 1 to the redshift (and that would be a really big error and so it's probably less than this). Likewise the emission lines could be broad but at most giving an error of about this size.  At most.  You can turn an error in redshift into an error in the frequency or wavelength shift of the spectrum.  Even for low resolution spectra an error of 0.1 in $z$ would be just enormous.  Errors in the concordant cosmology parameters would also affect the estimated light travel time.  These days these parameters have errors of a few percent or so, -- using the cosmocalc you can figure out what affect variations in these ($\Omega_M$, $\Lambda$, and the hubble constant) would have on the light travel time. &lt;/p&gt;&#xA;" OwnerUserId="4554" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-09T02:55:52.883" LastActivityDate="2011-08-09T02:55:52.883" CommentCount="3" />
  <row Id="13288" PostTypeId="2" ParentId="2229" CreationDate="2011-08-08T16:59:16.493" Score="12" ViewCount="" Body="&lt;p&gt;The answer to this question is simple and requires only SR, not GR or quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In units with $c=1$, we have $m^2=E^2-p^2$, where $m$ is the invariant mass, $E$ is the mass-energy, and $p$ is the momentum. In terms of logical foundations, there is a variety of ways to demonstrate this. One route starts with Einstein's 1905 paper &quot;Does the inertia of a body depend upon its energy-content?&quot; Another method is to start from the fact that a valid conservation law has to use a tensor, and show that the energy-momentum four-vector is the only tensor that goes over to Newtonian mechanics in the appropriate limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once $m^2=E^2-p^2$ is established, it follows trivially that for a photon, with $m=0$, $E=|p|$, i.e., $p=E/c$ in units with $c \ne 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of the confusion on this topic seems to arise from people assuming that $p=m\gamma v$ should be the definition of momentum. It really isn't an appropriate definition of momentum, because in the case of $m=0$ and $v=c$, it gives an indeterminate form. The indeterminate form can, however, be evaluated as a limit in which $m$ approaches 0 and $E=m\gamma c^2$ is held fixed. The result is again $p=E/c$.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-08T16:59:16.493" CommentCount="1" />
  <row Id="13289" PostTypeId="1" AcceptedAnswerId="13291" CreationDate="2011-08-08T17:55:29.870" Score="5" ViewCount="166" Body="&lt;p&gt;&lt;strong&gt;Motivation for asking question:&lt;/strong&gt;&#xA;I am planning to take the GRE subject test for Physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&#xA;Can somebody please tell me what are the &lt;strong&gt;most important equations&lt;/strong&gt; I should definitely know and understand completely going into this test and for entering first year Grad School for physics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(schodinger equation, F = dp/dt, Maxwell Equations, what else?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Follow up Question:&lt;/strong&gt;&#xA;What are some miscellaneous equations that are also important to know? &lt;/p&gt;&#xA;" OwnerUserId="2578" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-08T19:58:54.100" LastActivityDate="2011-08-10T15:07:43.677" Title="Physics Equations for Grad School / Physics GRE Prep" Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="4" />
  <row Id="13290" PostTypeId="2" ParentId="13284" CreationDate="2011-08-08T18:10:54.203" Score="1" ViewCount="" Body="&lt;p&gt;Please correct my answer if its wrong:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that condensation of water inside the car has to do with taking into consideration the temperature and humidity outside and inside your car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You only get condensation whenever its colder outside, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Outside is colder than Inside:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I think that your car acts as an insulator of heat. Your car's air temperature inside is hotter than the outside, and therefore water is evaporating and expanding in all directions. However, there's no where for the water to go, so it builds up on your windows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you blow hot air, you'd end up heating the inside car's air temperature even more and end up with more condensation. If you blow cold air, you make the inside car's air temperature to be closer to the outside and therefore the water molecules have less energy and won't expand out in all directions and end up condensing on your windows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that if you open your windows and let the inside temperature and outside temperature of air to come to equilibrium, you'll end up with no condensation on your windows from the inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please test this and tell me if I'm wrong. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2011-08-08T18:10:54.203" />
  <row Id="13291" PostTypeId="2" ParentId="13289" CreationDate="2011-08-08T18:20:53.753" Score="4" ViewCount="" Body="&lt;p&gt;The GRE physics exam is partitioned into at least 7 different subject areas, so there are no &quot;singly important&quot; formulae, I'm sorry to report.  The breakdown is roughly 20% allocated to E&amp;amp;M and Mechanics, and between 5-15% for Quantum, AMO, Relativity, Lab Methods, etc.  There's a &lt;a href=&quot;http://en.wikipedia.org/wiki/GRE_Physics_Test&quot; rel=&quot;nofollow&quot;&gt;wiki page with this breakdown&lt;/a&gt;, so this may be a good place to look for formulas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say that a lot of this subject matter is encompassed in advanced undergraduate texts, so hitting a good comprehensive text is a good place to start.  Further, the GRE is multiple choice, so get good at eliminating answers that don't make sense for dimensional, scale reasons right off the bat, and consider the strategy of just knowing when not to answer since the grading scheme takes into account the total number of questions answered. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various study guide and practice tests available at the &lt;a href=&quot;http://www.ets.org/gre/subject/about/content/physics&quot; rel=&quot;nofollow&quot;&gt;ETS Website&lt;/a&gt; for free, so this would be a good help, and there is an entire forum available for consult on topics related to graduate school and physics at &lt;a href=&quot;http://www.physicsgre.com&quot; rel=&quot;nofollow&quot;&gt;PhysicsGRE.com&lt;/a&gt;.&#xA;Good luck!!&lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2011-08-08T18:20:53.753" CommentCount="2" />
  <row Id="13292" PostTypeId="2" ParentId="12634" CreationDate="2011-08-08T18:36:12.737" Score="2" ViewCount="" Body="&lt;p&gt;Actually, the astronaut would only float completely free in the middle of the space station. Elsewhere, he will stick slightly to whatever side is closer to the Earth than is the middle, or farther from the Earth than is the middle. The reason is the tidal force from the Earth, which will be very small but probably detectable. If the acceleration from gravity is $g = GM/r^2$, then the tidal acceleration will be the derivative of this, namely $dg/dr = -2GM/r^3$. so the maximum tidal acceleration that the astronaut will feel, at distance $D/2$ from the center of the space station, where $D$ is the linear size of the space station, will be $GMD/r^3$. This is basically equal to $gD/r$.&#xA;Putting in some typical numbers, a 100 kg by weight astronaut in a space station of size $D = 10 m$, and $r = 6500 km$, we get a maximum tidal force of about $150$ milligrams weight!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation would of course be completely different in orbit around a neutron star, where an astronaut would float free at the center of a (very strong) space station, but would get crushed beyond recognition were they to venture away from the middle, as has been pointed out in various science fiction stories. The reason is that $M/r^3$ would be about $10^{14}$ times greater than for an orbit around the Earth.&lt;/p&gt;&#xA;" OwnerUserId="4857" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-08-08T20:19:41.453" LastActivityDate="2011-08-08T20:19:41.453" />
  <row Id="13293" PostTypeId="2" ParentId="13284" CreationDate="2011-08-08T20:16:24.437" Score="2" ViewCount="" Body="&lt;p&gt;Condensation has to do with two factors, temperature and humidity. Specifically, in order for condensation to occur, you need a body of warm, &lt;em&gt;humid&lt;/em&gt; air (humidity is necessary because otherwise there wouldn't be any water to condense) and a cold surface (the window), specifically colder than the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dewpoint&quot; rel=&quot;nofollow&quot;&gt;dewpoint&lt;/a&gt; of the humid air, on which the condensation is to take place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a car window, this can happen on either side. In your case, you're getting condensation on the inside of the window because the interior air is warm and humid, but the outside air keeps the window cooler than the dewpoint of the inside air. So as long as you blow warm air on the window, you're just feeding it with more water to condense. In order to make the condensation evaporate by heating it, you'd have to be blowing enough air at a high enough temperature on the window to override the effect of the outside air and significantly heat up the window itself; your car's heating system probably isn't capable of doing that. (At least not while you're driving.) In contrast, as you pointed out in your comment, blowing cool air (cooler than the outside air) on the window means that the dewpoint of the inside air is necessarily less than the temperature of the window, so you don't get any condensation occurring. And as long as the air is less than 100% humid, any existing condensation will evaporate. This is especially likely if you're running the air conditioner, since it doubles as a dehumidifier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reverse situation occurs if it's hot and humid outside; you'll probably be running the air conditioner, so the cold air inside the car will cool down the window. Since the window is cooler than the outside air, you'll get condensation occurring on the outside. The solution in this case is to run the defroster to heat up the window and the condensed water molecules, giving them enough energy to evaporate.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-08-08T20:16:24.437" CommentCount="1" />
  <row Id="13294" PostTypeId="2" ParentId="9523" CreationDate="2011-08-08T20:57:23.580" Score="6" ViewCount="" Body="&lt;p&gt;The main problem about a &lt;strong&gt;rigorous solution&lt;/strong&gt; to such a scattering proplem is that &lt;strong&gt;computations are extremely demanding&lt;/strong&gt;. Just imagine you have a wavelength $\lambda$ of some $400$nm to $700$nm for visible light (from &lt;a href=&quot;http://www.mdsupport.org/library/hazard.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Qb4im.jpg&quot; alt=&quot;spectrum&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to do physically meaningful simulations, you will need a &lt;strong&gt;sub-wavelength lattice&lt;/strong&gt; which makes any computational cell above, say $10\,\mu m^3$ &lt;strong&gt;not accessible&lt;/strong&gt; since you  have in the order of &lt;strong&gt;one million grid points&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Approximative Approaches&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;But of course there can be ways out of it if you are willing to make some &lt;strong&gt;approximations&lt;/strong&gt; which will largely depend on the characteristics of the particles you are looking at. It is best to assume that we only have spherical particles since we can apply &lt;a href=&quot;http://en.wikipedia.org/wiki/Mie_theory&quot; rel=&quot;nofollow&quot;&gt;Mie theory&lt;/a&gt; in this case.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Large Particles&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;First of all, let us consider particles which are &lt;strong&gt;much larger than the wavelength&lt;/strong&gt;. Then, the radius $R$ times the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wave_vector&quot; rel=&quot;nofollow&quot;&gt;wave vector&lt;/a&gt; $k=2\pi/\lambda$ is much bigger than one, $$kR\gg1$$&#xA;which basically means that one observes reflection at a &lt;strong&gt;plane interface&lt;/strong&gt;. You can implement these particles using &lt;a href=&quot;http://en.wikipedia.org/wiki/Geometrical_optics&quot; rel=&quot;nofollow&quot;&gt;geometrical optics&lt;/a&gt; (mixed with &lt;a href=&quot;http://en.wikipedia.org/wiki/Fresnel_equations&quot; rel=&quot;nofollow&quot;&gt;Fresnel reflection&lt;/a&gt; if you like) since nothing really wave-like will happen as in this image (taken from &lt;a href=&quot;http://renderplant.riskybacon.com/2009_05_01_archive.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/jqu5H.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Small Particles&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Second, the particles should be &lt;strong&gt;much smaller&lt;/strong&gt; than the wavelength,&#xA;$$kR\ll1\,.$$&#xA;Then, everything what is observed is a sum of &lt;a href=&quot;http://en.wikipedia.org/wiki/Dipole&quot; rel=&quot;nofollow&quot;&gt;dipolar&lt;/a&gt; responses of the particles in the so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Rayleigh_scattering&quot; rel=&quot;nofollow&quot;&gt;Rayleigh-scattering&lt;/a&gt;. &lt;a href=&quot;http://en.wikipedia.org/wiki/Rayleigh_scattering&quot; rel=&quot;nofollow&quot;&gt;Then&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the intensity of light scattered by a single small particle from a beam of unpolarized light of wavelength $\lambda$ and intensity $I_0$ is given by:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$I=I_0(1+\cos^2\theta)\frac{(kR)^6}{2(kr)^2}\left(\frac{n_p^2-1}{n_p^2+2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where I have chosen the variables to be consistent with the used terminology and $r$ is the distance to the object, $\theta$ is the scattering angle and $n_p$ is the sphere's refractive index. Here is an image of such a situation with some metal particles also having quadrupolar excitation (from &lt;a href=&quot;http://www.ultra-optics.de/research+areas/nano+optics/computational+nanophotonics.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/HY6oY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;A Mean Field Approach - Effective Permittivity&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;If you have &lt;strong&gt;a lot of&lt;/strong&gt; these small objects, you may use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Clausius%E2%80%93Mossotti_relation&quot; rel=&quot;nofollow&quot;&gt;Clausius-Mossotti relation&lt;/a&gt; which gives you an effective permittivity $\epsilon_p=n_p^2$ depending on the &lt;strong&gt;concentration of the particle&lt;/strong&gt; in some volume:&#xA;$$\epsilon_{eff} = \epsilon_p + \frac{n\alpha}{1-\frac{n\alpha}{3\epsilon_p}}$$&#xA;where $\alpha$ is the polarizability of the sphere, for details see e.g. &lt;a href=&quot;http://books.google.de/books?id=uIHSNwxBxjgC&amp;amp;lpg=PP1&amp;amp;dq=Electromagnetic%20mixing%20formulas%20and%20applications&amp;amp;pg=PP1#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Electromagnetic mixing formulas and applications&lt;/a&gt; by Sihvola. This would be something like a &lt;strong&gt;mean-field approach&lt;/strong&gt;. You can make some very neat effects using this effective approach since it allows you to calculate a continuous refraction around some particle streams under water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the particles size is &lt;strong&gt;in the order of the wavelength&lt;/strong&gt;, $$kR\approx 1$$ then you may have to take higher &lt;a href=&quot;http://en.wikipedia.org/wiki/Multipole_moment&quot; rel=&quot;nofollow&quot;&gt;multipole moments&lt;/a&gt; into account which may be a very demanding task.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For much more on the subject I would recommend Bohren &amp;amp; Huffmanns classic &lt;a href=&quot;http://books.google.com/books?id=peHRPwAACAAJ&amp;amp;dq=inauthor%3a%22Craig%20F.%20Bohren%22&amp;amp;hl=de&amp;amp;ei=7ExATtKQLYmBOqL61PkO&amp;amp;sa=X&amp;amp;oi=book_result&amp;amp;ct=result&amp;amp;resnum=3&amp;amp;ved=0CDcQ6AEwAg&quot; rel=&quot;nofollow&quot;&gt;Absorption and Scattering of Light by Small Particles&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely&lt;/p&gt;&#xA;" OwnerUserId="358" LastEditorUserId="358" LastEditorDisplayName="" LastEditDate="2011-08-09T09:13:39.690" LastActivityDate="2011-08-09T09:13:39.690" CommentCount="4" />
  <row Id="13295" PostTypeId="2" ParentId="13289" CreationDate="2011-08-08T21:55:13.990" Score="0" ViewCount="" Body="&lt;p&gt;If you are getting started, 70-80% of the GRE is based on freshman physics from any standard textbooks such as Halliday and Resnick, extended version (which means modern physics included). For this, all the equations you are asking about can be found in the summary that follows every chapter. Of course it goes without saying that memorizing the equations is completely useless unless you understand how they are used.&lt;/p&gt;&#xA;" OwnerUserId="4521" LastEditorUserId="4521" LastEditorDisplayName="" LastEditDate="2011-08-10T14:26:10.780" LastActivityDate="2011-08-10T14:26:10.780" />
  <row Id="13296" PostTypeId="1" AcceptedAnswerId="13298" CreationDate="2011-08-08T21:58:39.077" Score="1" ViewCount="45" Body="&lt;p&gt;I learnt that in astrophysical spectroscopy, the emission spectrum of distant stars is used to determine what they're made of. So why is it that our own Sun is emitting the whole spectrum ? (or is that information incorrect)&lt;/p&gt;&#xA;" OwnerUserId="4858" LastActivityDate="2011-08-08T22:22:23.513" Title="Suns emission spectrum" Tags="&lt;spectroscopy&gt;&lt;sun&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13297" PostTypeId="1" CreationDate="2011-08-08T22:14:46.070" Score="2" ViewCount="109" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/2175/is-it-possible-for-information-to-be-transmitted-faster-than-light&quot;&gt;Is it possible for information to be transmitted faster than light?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;imagine this theoretical situation:&lt;br&gt;&lt;br&gt;&#xA;You have make an extremely long piece of steel (or anything really) one end is placed on earth the other is all the way at Alpha Centauri, 4ly away.&lt;br&gt;&#xA;Here on earth the end is hanging loosely, on Alpha Centauri there is a bell at the end of the rod. Now if you move our end straight in the direction of AC will the bell there ring instantly ? Or will it be slower/same as speed of light.&lt;br&gt;&lt;br&gt;&#xA;If I take make it small scale, for example moving a steel rod with a bell on the end, I can't imagine that there would be a delay, but of course the speed of light is the limit, so what could cause the delay ? Thanks for explanation.&lt;br&gt;&#xA;My very sketch:&#xA;&lt;a href=&quot;http://imageshack.us/photo/my-images/221/steelrod.png/&quot; rel=&quot;nofollow&quot;&gt;http://imageshack.us/photo/my-images/221/steelrod.png/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4858" LastEditorUserId="4552" LastEditorDisplayName="" LastEditDate="2011-08-08T22:24:58.207" LastActivityDate="2011-08-08T23:14:09.860" ClosedDate="2011-08-09T01:48:28.620" Title="If you make a steel rod from here to Alpha Centauri and move it, will the movement appear there instantly?" Tags="&lt;special-relativity&gt;&lt;velocity&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="0" />
  <row Id="13298" PostTypeId="2" ParentId="13296" CreationDate="2011-08-08T22:22:23.513" Score="4" ViewCount="" Body="&lt;p&gt;There are continuous spectra, emission spectra, and absorption spectra. Dense materials, such as the deeper parts of our sun, have continuous spectra. Hot, low-density gases have emission spectra (a black spectrum with bright lines in it). The continuous spectrum generated deep in our sun passes through the cooler and lower-density outer regions of the sun, where it is partially absorbed. The result is an absorption spectrum: a rainbow with certain dark lines notched out of it: &lt;a href=&quot;http://en.wikipedia.org/wiki/Absorption_spectrum&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Absorption_spectrum&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good way to think about this is that a line spectrum comes from individual atoms, which are quantum-mechanical. In systems with a very large number of particles, the quantum-mechanical behavior becomes undetectable. A dense gas, such as the deeper parts of the sun, has all its atoms so close together that they act like one big object with a very large number of particles. Therefore you get no quantum-mechanical features, and it's just a continuous spectrum.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-08T22:22:23.513" CommentCount="1" />
  <row Id="13299" PostTypeId="2" ParentId="2175" CreationDate="2011-08-08T22:24:43.417" Score="3" ViewCount="" Body="&lt;p&gt;The signal will propagate at the speed of sound in steel. I happen to know the speed of sound in aluminum, because my students measure it in lab; it's about 5000 m/s. This is many orders of magnitude less than the speed of light.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-08T22:24:43.417" />
  <row Id="13300" PostTypeId="2" ParentId="2175" CreationDate="2011-08-08T22:26:30.013" Score="4" ViewCount="" Body="&lt;p&gt;No. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In relativity you cannot consider extended objects to be infinitely &quot;stiff&quot; - they must bend and stretch, as real objects do. When you move one end of the steel rod, it makes part of it bend and stretch which exerts a force on the next section which makes that move and which makes a new part bend and stretch and so on and so on until you reach Alpha Centauri. This moves along at some speed which is characteristic for the metal which is fast enough that we don't really notice in day to day life. All relativity tells us is that that characteristic speed is less than the speed of light - it turns out for real metal its much less than the speed of light.&lt;/p&gt;&#xA;" OwnerUserId="3970" LastActivityDate="2011-08-08T22:26:30.013" />
  <row Id="13301" PostTypeId="2" ParentId="2175" CreationDate="2011-08-08T23:14:09.860" Score="1" ViewCount="" Body="&lt;p&gt;A simple explanation why the speed of sound can never be faster than the speed of light:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider two atoms $A$ and $B$. Give the nucleus of $A$ a slight push. As we now, this push will carry over to $B$, but why? It's due to their electrostatic repulsion. So for $B$ to even react, you first need at least an electromagnetic wave/photon travel from $A$ to $B$. This can of course not get there faster than the speed of light. The nucleus of $A$ itself can obviously not be faster, either, so even with brute force it's not possible to get a sonic speed $\,&amp;gt;\!\!c$.&lt;/p&gt;&#xA;" OwnerUserId="3540" LastActivityDate="2011-08-08T23:14:09.860" />
  <row Id="13302" PostTypeId="2" ParentId="13269" CreationDate="2011-08-08T23:17:44.190" Score="2" ViewCount="" Body="&lt;p&gt;I think your description of temperature has intuitive merit - it is a true and important statement about temperature.  However, I would not use it as a definition of temperature because, to a certain extent, that would trivialize the &lt;a href=&quot;http://en.wikipedia.org/wiki/Second_law_of_thermodynamics&quot; rel=&quot;nofollow&quot;&gt;second law of thermodynamics&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second law of thermodynamics says that we can't move heat from a body of low temperature to a body of higher temperature without doing some work (the amount of work required can be quantified, but that is not critical).  If we define temperature by referencing which way heat spontaneously flows, then the second law loses some of its physical content.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it another way, if we use this proposed definition of temperature, the only way to tell which of two bodies is hotter is to put them in contact and see which way the heat flows.  That's not really our goal, though.  Our goal in thermodynamics is to predict which way heat will flow (and how much work we can extract from the heat flow) before we ever put the bodies in contact, and to do it based on some other property the body has.  It would be better to have a concept of temperature independent of direction of heat flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two good alternative ways to think about temperature.  You already mentioned the first.  In classical thermodynamics, temperature is simply what a thermometer measures.  If we take one mole of an ideal gas and let it come into equilibrium with a system, the system's temperature is $PV/R$, with $P$ the pressure of the gas, $V$ its volume, and $R$ a constant called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gas_constant&quot; rel=&quot;nofollow&quot;&gt;gas constant&lt;/a&gt;.  Thus, temperature is a &lt;a href=&quot;http://en.wikipedia.org/wiki/State_variable&quot; rel=&quot;nofollow&quot;&gt;state variable&lt;/a&gt;, and a characteristic of systems in &lt;a href=&quot;http://en.wikipedia.org/wiki/Equilibrium_%28thermodynamics%29&quot; rel=&quot;nofollow&quot;&gt;equilibrium&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This definition of temperature, based on an &lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_gas&quot; rel=&quot;nofollow&quot;&gt;ideal gas&lt;/a&gt;, is more useful than one based on a particular mercury thermometer because formulas such as the efficiency of an &lt;a href=&quot;http://en.wikipedia.org/wiki/Carnot_heat_engine&quot; rel=&quot;nofollow&quot;&gt;ideal heat pump&lt;/a&gt; come out to have a simple form when we use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gas_thermometer&quot; rel=&quot;nofollow&quot;&gt;gas thermometer&lt;/a&gt; to define temperature.  When temperature is measured relative to an ideal gas, it turns out that mercury and other liquids have  coefficients of expansion that change slightly with the temperature, so they aren't good references for precision measurements.  However, they can be calibrated to agree with an ideal gas thermometer.  We must also remember that an ideal gas is only closely approximated by real gases, so even in this case we cannot get absolutely perfect temperature measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most widely-used scale of temperature in science is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kelvin&quot; rel=&quot;nofollow&quot;&gt;Kelvin scale&lt;/a&gt;.  It equates absolute zero with 0 K and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Triple_point&quot; rel=&quot;nofollow&quot;&gt;triple point&lt;/a&gt; of water with 273.16 K.  Then it takes a linear interpolation between those two points, based on the ideal gas mentioned above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other way to think about temperature is statistically.  Suppose you have a bunch of gas molecules bouncing around in a room.  At any given moment, some are going fast and others are going slowly.  The molecules thus have a spread of &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_energy&quot; rel=&quot;nofollow&quot;&gt;kinetic energies&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then wonder what the distribution of these kinetic energies is.  Are they mostly clustered around one value, say all between $.016$ and $.020$ electron-volts?  Or are they spread out widely from $0$ to $0.1$ electron-volts?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could imagine the energy distribution having all sorts of different shapes.  It could have five different humps, or look like a bell curve, or be flat with a gap missing, etc.  It turns out this isn't the case.  The energy distribution plot for equilibrium systems comes out looking pretty much the same all the time; it's a decaying exponential $\propto e^{-\beta E}$.  Every time you go up by a certain fixed amount of energy, the number of particles at that energy state gets cut in half. (The exact fixed amount, $\ln(2)/\beta$, changes depending on the system; what's common is the property of getting regularly cut in half.)  This is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Boltzmann_distribution&quot; rel=&quot;nofollow&quot;&gt;Boltzmann distribution&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you have to do is specify the number $\beta$ and you know everything about the distribution of energy.  This is quite a remarkable fact, and not an obvious one.  This distribution occurs because it maximizes &lt;a href=&quot;http://en.wikipedia.org/wiki/Entropy&quot; rel=&quot;nofollow&quot;&gt;entropy&lt;/a&gt; for a fixed energy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this distribution,  the average energy is $1/\beta$, and we call this average energy the &lt;a href=&quot;http://en.wikipedia.org/wiki/Temperature#Statistical_mechanics_approach_to_temperature&quot; rel=&quot;nofollow&quot;&gt;temperature&lt;/a&gt;.  (If the energies do not have a Boltzmann distribution for some reason we could still identify the mean energy as the temperature.)  If $\beta$ is small, then the temperature is high, and the energy distribution is widely spread and fairly flat.  At high temperature, things are quite random.  If $\beta$ is big, the temperature is low and energies are clustered near zero.  Low temperatures are more orderly.  (An interesting side note is that a perfectly-flat energy distribution has the temperature being infinite, and further if the high-energy states are statistically more-likely than the low-energy states, the temperature is negative.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that this statistical temperature and the thermodynamic temperature defined earlier agree with one another, up to a scale factor called &lt;a href=&quot;http://en.wikipedia.org/wiki/Boltzmann%27s_constant&quot; rel=&quot;nofollow&quot;&gt;Boltzmann's constant&lt;/a&gt;.  So we can think of temperature either way, and the definitions are equally good.  Either definition adds content to the second law of thermodynamics by making temperature an independent concept, and the direction of the flow of heat a physical hypothesis that involves temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, an important property of temperature is that it is well-defined.  It is an observed fact if systems A and B are in thermal equilibrium, and B and C are in thermal equilibrium, and A and C are as well.  This is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Zeroth_law_of_thermodynamics&quot; rel=&quot;nofollow&quot;&gt;zeroth law of thermodynamics&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-08-08T23:23:38.287" LastActivityDate="2011-08-08T23:23:38.287" CommentCount="1" />
  <row Id="13303" PostTypeId="2" ParentId="13269" CreationDate="2011-08-08T23:45:59.480" Score="6" ViewCount="" Body="&lt;p&gt;I think there is another, very nice definition of temperature, given in the Book &lt;em&gt;Thermal Physics&lt;/em&gt; by Daniel Schroeder:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that in a closed system, the entropy is maximized. Imagine now a large system that consists of two systems A and B in contact. We define temperature as &lt;em&gt;that which is the same for A and B when the system is in equilibrium&lt;/em&gt;. But what is the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Entropy is an additive thing: The total entropy $S$ is just $S_A + S_B$, the sum of the entropy of systems A and B. Since the systems are in contact, energy can move between A and B. The total energy is fixed, so $U_B = U - U_A$ and $dU_B = -dU_A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving energy around changes the entropy:&#xA;$$dS = \frac{\partial S_A}{\partial U_A} dU_A - \frac{\partial S_B}{\partial U_B} dU_A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in equilibrium, $S$ is already maximized and $dS = 0$, so we have &#xA;$$\frac{\partial S_A}{\partial U_A} = \frac{partial S_B}{\partial U_B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means: If two systems are in thermal equilibrium, then the quantity that is the same for them is $\partial S / \partial U$. We also see that energy wants to move from $A$ to $B$ if $\partial S_A/\partial U_A$ is &lt;em&gt;smaller&lt;/em&gt; than $\partial S_B/\partial U_B$, so we can &lt;em&gt;define&lt;/em&gt; temperature as &#xA;$$\frac{1}{T} = \frac{\partial S}{\partial U}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a way, you are right: Temperature is the propensity to transmit thermal energy. More strictly speaking, temperature is a measure for the gain in entropy when the internal energy changes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively speaking: If system B gains more entropy than A loses if energy flows from A to B, then energy will flow from A to B so that overall entropy grows. In &lt;em&gt;typical&lt;/em&gt; systems, the curve $S(U)$ grows with $U$ and becomes flatter for larger $U$, which means that temperature also increases with $U$. That is not always the case: For some weird systems (two-state paramagnets as the simplest example), you can actually have negative or infinite temperatures.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-08-08T23:45:59.480" />
  <row Id="13304" PostTypeId="2" ParentId="13157" CreationDate="2011-08-09T00:05:53.163" Score="2" ViewCount="" Body="&lt;p&gt;Field is tool which we use to implement the fact that Quantum Field Theory gives meaning to local measurements (those that can be performed in one space-time point, or more realistic, in small space-time region around that point). That is why fields depend on space-time coordinates. In canonical quantization, fields are self-adjoint operators, which means that these are observables in principle, and that all observables can be built from them.&#xA;In Feynman path integral approach to quantization (which is equivalent to canonical), fields are ordinary functions of space-time coordinates, if they correspond to bosonic fields from canonical quantization. To fermionic operators from canonical formalism correspond Grasmann variables in path integral approach. &#xA;Fields are real in operational sense, that we can define them, build theory using them and make predictions for observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far is known there is no Quantum Field Theory that works for any (arbitrarily) large energy scale, as it is possible that very short distance physics is best described by theory other than QFT (string theory, i.e). So asking about if field is real or not can only be put into point of view of some theory. Exact position and momentum of particle is real and useful concept in classical mechanics, for example, but in Quantum Mechanics this sort of determinism was lost.&lt;/p&gt;&#xA;" OwnerUserId="4107" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-09T03:04:04.337" LastActivityDate="2011-08-09T03:04:04.337" CommentCount="1" />
  <row Id="13305" PostTypeId="2" ParentId="11878" CreationDate="2011-08-09T01:35:30.197" Score="1" ViewCount="" Body="&lt;p&gt;Gerard t'Hooft, Quantum Field Theory for Elementary Particles, Is Quantum Field Theory a theory?, Phys.Rept. 104 (1984) 129-142. This is a beautifully written review.&lt;/p&gt;&#xA;" OwnerUserId="4107" LastActivityDate="2011-08-09T01:35:30.197" CommunityOwnedDate="2011-08-09T01:35:30.197" CommentCount="1" />
  <row Id="13306" PostTypeId="1" CreationDate="2011-08-09T01:49:38.983" Score="-2" ViewCount="80" Body="&lt;p&gt;I wasn't sure whether I should put this in Mathematics or Physics, so I put it here. My reasoning is that the math is relatively straightforward, but the question concerns the nature of the event in the physical world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I posited this question to my geometry teacher in highschool many years ago, and it stumped her. I've recently brought it up again in conversations with friends and have not gotten any answer that's satisfactory. I have an intuitive feeling of what I expect, but I don't know enough to really back it up with solid reasoning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, consider this theoretical situation:&#xA;Suppose you have two infinitely long lines (you could say a couple laser beams which never diffuse, but any perfectly straight line would work), and they are mounted so that they intersect at some point along an infinite plane in space. Then, you gradually rotate one of them so that the angle of intersection gets gradually closer and closer to 0, moving the lines closer and closer to parallel. At some point, intuition tells us that there is a position at which these lines will be perfectly parallel and never intersect. However, math tells us that they will always intersect, and the angle at which they intersect will simply approach 0 forever but never reach it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically, parallel exists. If these lines were angled away from each other, they will not intersect, because at each increment, the distance between the two lines increases. If one of the lines in this case were rotated closer and closer to parallel, eventually they will angle towards each other and intersect at some distance away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Math, again, tells us that this intersection point is found by a limit calculation, but it evaluates to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering that theoretical parallel does intuitively exist, when the lines become perfectly parallel, what happens at the point of intersection? There couldn't be a final point of intersection, could there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way I can wrap my brain around this situation, in my relatively limited understanding of physics and mathematics, is if these lines physically manifested as infinitely large loops, and then behave like intersecting rings. At any frame of reference, there is a point which appears parallel (the tangent lines) but where elsewhere, on the other side of the ring, they intersect. But this then assumes that the universe is curved along itself. Is there a better theory to explain this? I'm sure I can't be the first one to imagine this kind of situation.&lt;/p&gt;&#xA;" OwnerUserId="4860" LastActivityDate="2011-08-09T02:18:18.827" ClosedDate="2011-08-11T19:53:58.570" Title="What happens when two infinite lines stop intersecting and become parallel?" Tags="&lt;space&gt;&lt;curvature&gt;" AnswerCount="0" CommentCount="7" />
  <row Id="13307" PostTypeId="2" ParentId="13289" CreationDate="2011-08-09T01:52:18.930" Score="6" ViewCount="" Body="&lt;p&gt;The phrasing of your question worries me. Learning physics doesn't have anything to do with memorizing equations. If you understand the meaning of an equation, you shouldn't have to memorize it.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-09T01:52:18.930" CommentCount="2" />
  <row Id="13309" PostTypeId="2" ParentId="13269" CreationDate="2011-08-09T02:31:57.057" Score="1" ViewCount="" Body="&lt;p&gt;The other answers are great, I'm just going to try to answer it more concisely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proposed definition is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Temperature is the propensity of something to give off heat, that is transmit thermal energy. Equivalently, it is the resistance of something to receive heat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My altered version (which is a half-definition of temperature) is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Heat will not flow between two bodies at the same temperature.  When two bodies are at different temperatures, heat flows from the higher temperature to the lower temperature.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This describes most of the properties of the quantity called temperature.  With this, we can always identify a higher and lower relative temperature, however, we can't use this and only this to attach a numeric value on temperature.  For that, you can read the other answers, which gets into the science of thermodynamics itself.  For many people, however, my proposed half-definition may be sufficient for what they're looking for.  If someone knows of a &lt;em&gt;simple&lt;/em&gt; alteration that will fully define temperature, I'd love to hear it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is wrong with &quot;the propensity of something to give off heat&quot;?  Obviously the statement has intuitive value.  The heat flow, however, is characterized by several very different kind of quantities.  My favorite is the concept of thermal resistance.  If you have 2 large reservoirs at different temperatures, then the heat flow from the higher temperature to the lower one can said to be inversely proportional to the thermal resistance between the two, as well as proportional to the difference in temperatures.  Thus, giving off heat is proportional to temperature, but it is also proportional to other things.  For that reason we can only define the direction of flow on this qualitative level (I'm giving a &quot;no math&quot; answer), which the reason and the function of my alteration.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-08-09T02:31:57.057" />
  <row Id="13310" PostTypeId="2" ParentId="13266" CreationDate="2011-08-09T03:55:05.187" Score="4" ViewCount="" Body="&lt;p&gt;This is because the classical system consisting of nucleons interacting with a realistic pair potential is chaotic. A classically chaotic system careens between unstable periodic orbits, which in a path-integralish view ( gutzwiller trace formula) tells you that nearby energy levels are concentrated on completely different periodic orbits, so that they have mixed with each other strongly if you consider the unperturbed wavefunctions to be homogenous ( Heller scarring) The statistics of random matrix eigenvalues is based on the principle of heavy generic mixing between energy levels, leading to level repulsion. By contrast, classically integrable systems have energy levels which are regularly spaced, coming as they do from semiclassically setting the action variables to be integers. Each action variable gives a smooth level spacing, and the spacings add up linearly, to produce a uniformish distribution of energy levels without level repulsion.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-09T03:55:05.187" CommentCount="2" />
  <row Id="13311" PostTypeId="2" ParentId="9998" CreationDate="2011-08-09T04:04:44.233" Score="3" ViewCount="" Body="&lt;p&gt;There are simple non- renormalizable examples: the easiest is QED with elementary neutrons only. The neutrons have a magnetic dipole moment, and interact with the photon locally through the F tensor only contracted with the antisymmetric product of gamma matrices.  There are no renormalizable local examples in 4d because we have a list of all of these, and the vectors always interact with charged particles, not magnetically polarized ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I missed that the question asks specifically for asymptotically safe examples, and the OP was probably aware of this system. But the answer might be useful to others) &lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-10T07:45:38.520" LastActivityDate="2011-08-10T07:45:38.520" />
  <row Id="13312" PostTypeId="2" ParentId="9983" CreationDate="2011-08-09T04:21:20.087" Score="1" ViewCount="" Body="&lt;p&gt;This doesn't qualify as electrovac, but you can superpose arbitrarily placed extremal black holes, and making a soup of infinitely many infinitesimal charge/mass BH's, you get a static solution which is nontrivial, and only sourced by these infinitesimal sources which are nonsingular in the infinitesimal limit. The metric and electrostatic potential are determined in isotropic coordinates by a single solution to Laplaces equation in flat space. I forget the name of these, but it's in Hawking and Ellis&lt;/p&gt;&#xA;" OwnerUserId="4863" LastActivityDate="2011-08-09T04:21:20.087" />
  <row Id="13314" PostTypeId="2" ParentId="1235" CreationDate="2011-08-09T04:46:05.577" Score="2" ViewCount="" Body="&lt;p&gt;One easy way to see this is that the N-body problem can be used, with appropriate potentials, to simulate a classical computer, so that as N becomes large, any algorithm which predicts the future behavior at arbitrarily long times has to be at least as computationally complex as a general cN-bit computer (where c is the number of bits you can usefully code per-particle) . Summation of convergent infinite series also simulates a computer, so that's not a useful interpretation of the word &quot;solve&quot;. But any good definition of saying &quot;solve&quot; should mean that you reduced the computational complexity of predicting the future behavior from the present state, which can't be done for a general purpose computer.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-09T04:46:05.577" CommentCount="0" />
  <row Id="13315" PostTypeId="2" ParentId="1894" CreationDate="2011-08-09T05:53:36.630" Score="2" ViewCount="" Body="&lt;p&gt;One major difficulty with defining path-integrals (which is entirely mathematicians' fault)is that the mathematicians insist for no good reason (and many bad ones) that there are non-measurable subsets of R. This is a psychological artifact of early days of set theory, where ZFC ws not seen as a way of generating countable models of a set-theoretic universe, but as the way things REALLY are in Plato-land (whatever that means). Cohen fixed that in 1963, but mathematicians still haven't gotten used to the fix, but that is changing rapidly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you assume every subset of R is measurable, the notion of &quot;randomly picking a number between 0 and 1&quot; becomes free of contradiction. In the presence of the axiom of choice, the question &quot;what is the probability that this number lands in a Vitali set?&quot; is paradoxical, but in the real world, it is obviously meaningful. This tension is resolved in what is called a &quot;Solovay model&quot;, where you have no more trouble with probability arguments meshing with set theory. For a non-mathematician, when you deal with sets which arise by predicative definition, not by doing uncountable axiom-of-choice shenanigans, probability is never contradictory. A solovay model still allows you to use countable axiom-of-choice, and countable dependent choice, which is enough for all usual analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, inside a Solovay model, you can define a Euclidean bosonic path integral very easily: it is an algorithm for picking a random path, or a random field. This has to be done by step-wise refinement, because the random path or random field has values at continuum of points, so you need to say what it means for a path to &quot;refine&quot; another path. further, while paths end up continuous, so that the refinement process is meaningful in the space of continuous paths, fields refine discontinuously. if you have a field whose average value on a lattice is something, it swings more wildly at small distances in dimensions higher than 2, so that in the limit, it defines a random distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are allowed to pick at random, part of the battle is won. You get free field path-integrals in any dimension with absolutely no work (pick the fourier components at random as Gaussian random numbers with a width which goes like the propagator). There is no issue with proving measurability, and the space of distributions you get out is just defined to be whatever space of distributions you get by doing the random picking. It's as simple as that. Really.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining battle is just renormalization, at least for bosonic fields with CP invariant (real) actions, which have a stable vacuum, so that their Euclidean continuation has a probability interpretation. You need to define the stepwise approximations in such a way that their probability distribution function approaches a consistent limit at small refinements. This is slightly tricky, but it automatically defines the measure if you have a Solovay world. There is nobody working on field theory in Solovay models, but there are people who mock up what is more or less the same thing inside usual set theory by doing what is called &quot;constructive measure theory&quot;. I don't think that one can navigate the complicated renormalization arguments unless one is allowed to construct measures using probability intuitions without fear, and without work. And set theorists know how to do this since 1970. &lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-09T05:53:36.630" CommentCount="3" />
  <row Id="13316" PostTypeId="1" AcceptedAnswerId="13317" CreationDate="2011-08-09T06:05:08.363" Score="1" ViewCount="112" Body="&lt;p&gt;I've just found &lt;a href=&quot;http://www.youtube.com/watch?v=tu57B1v0SzI&quot; rel=&quot;nofollow&quot;&gt;Dr Quantum&lt;/a&gt; video sample where double split experiment is presented conducted out by researchers. Are there any papers published in peer reviewed journals on that experiment to read in detail?&lt;/p&gt;&#xA;" OwnerUserId="4865" LastActivityDate="2011-08-09T17:55:25.377" Title="'Observer in double slit experiment' publications available?" Tags="&lt;double-slit-experiment&gt;&lt;duality&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="13317" PostTypeId="2" ParentId="13316" CreationDate="2011-08-09T07:41:34.143" Score="2" ViewCount="" Body="&lt;p&gt;Yes. &lt;a href=&quot;http://www.nature.com/nature/journal/v401/n6754/abs/401680a0.html&quot; rel=&quot;nofollow&quot;&gt;A recent publication&lt;/a&gt; looks at the wave-particle duality of large particles (buckminsterfullerene, $\mathrm{C}_{60}$ (actually more of a many-slit experiment)), however the original diffraction experiments with electrons and such were done in the 1920s-1930s and may be hard to access.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors talk about facile ways to collapse the interference by observation, but only actually present the diffraction pattern (or lack thereof) with and without the grating. The paper is quite detailed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be careful, however. Whilst the wave-particle duality of electrons is well-characterised, the movie that Dr. Quantum is from pushes a lot of &lt;a href=&quot;http://en.wikipedia.org/wiki/What_the_Bleep_Do_We_Know!?#Academic_reaction&quot; rel=&quot;nofollow&quot;&gt;untenable silliness&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1876" LastEditorUserId="1876" LastEditorDisplayName="" LastEditDate="2011-08-09T08:08:15.640" LastActivityDate="2011-08-09T08:08:15.640" CommentCount="5" />
  <row Id="13318" PostTypeId="2" ParentId="13316" CreationDate="2011-08-09T08:03:27.277" Score="5" ViewCount="" Body="&lt;p&gt;In his 1924 dissertation, de Broglie argued that matter particles should have a wavelength of $\lambda = h/p$, where $p$ is the momentum of the particle. The first confirmation of the diffraction formed by such matter waves was observed in the Davisson-Germer experiment:&#xA;&lt;a href=&quot;http://prola.aps.org/abstract/PR/v30/i6/p705_1&quot; rel=&quot;nofollow&quot;&gt;C. Davisson, L.H. Germer. &lt;em&gt;Phys. Rev.&lt;/em&gt; &lt;strong&gt;30&lt;/strong&gt; (1927) 705&lt;/a&gt;. Independently, G.P. Thomson (son of J.J. Thomson, discoverer of the electron) and A. Reid found similar effects: &lt;a href=&quot;http://www.nature.com/nature/journal/v119/n3007/abs/119890a0.html&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Nature&lt;/em&gt; &lt;strong&gt;119&lt;/strong&gt; (1927) 890&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the specific experiment your video references, the first proper electron double-slit experiment was performed by Claus Jönsson, in &lt;em&gt;Zeitschrift für Physik&lt;/em&gt; &lt;strong&gt;161&lt;/strong&gt; (1961) 454. A partial translation of this appeared in &lt;a href=&quot;http://ajp.aapt.org/resource/1/ajpias/v42/i1/p4_s1&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;AJP&lt;/em&gt; &lt;strong&gt;42&lt;/strong&gt; (1974) 4&lt;/a&gt;. If you don't have access to AJP, don't despair--there's a &lt;a href=&quot;http://web.pdx.edu/~pmoeck/pdf/ElectronJonsson.PDF&quot; rel=&quot;nofollow&quot;&gt;pdf&lt;/a&gt; floating on the interwebs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to echo to Richard Terrett's warning about Dr. Quantum in general: while there might be some genuine information here and there, overall What the Bleep was a thinly veiled advertisement for a pseudoscience-driven scam.&lt;/p&gt;&#xA;" OwnerUserId="1353" LastActivityDate="2011-08-09T08:03:27.277" CommentCount="3" />
  <row Id="13320" PostTypeId="2" ParentId="10915" CreationDate="2011-08-09T14:22:52.010" Score="0" ViewCount="" Body="&lt;p&gt;There are 2 different effects of the EM waves:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Acceleration of charges due to EM field&lt;/li&gt;&#xA;&lt;li&gt;Radiation pressure of the EM field&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If the EM wave propagates along x direction and your antenna is along the z direction, then electrons will be driven along the antenna by the electric field. On the other hand, the antenna will also receive an overall impact (momentum) along the x direction because of the radiation pressure (which does not affect the antenna practically since its effect is tiny and the antenna is fastened).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth pointing out that there are 2 descriptions of radiation:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the classical description (EM waves) and &lt;/li&gt;&#xA;&lt;li&gt;the quantum description (photons) &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The quantum description is more general. Both pictures are identical only in the limit of so many photons. Electromagnetic waves can be described in terms of photons, but individual photons cannot be described in terms of electromagnetic waves. Photons description is more fundamental.&lt;/p&gt;&#xA;" OwnerUserId="4521" LastActivityDate="2011-08-09T14:22:52.010" CommentCount="10" />
  <row Id="13321" PostTypeId="1" CreationDate="2011-08-09T14:26:48.057" Score="3" ViewCount="43" Body="&lt;p&gt;Suppose I have a flow of hot air around a cold and unevenly shaped object with holes and tunnels (think about it as a bed packed with some objects). I would like to know the Reynolds number of this flow and its convective heat transfer coefficient. The definition of the Reynolds number contains a &quot;characteristic length&quot; that is somehow mysterious to me and that I do not have at hand. And I am a bit reluctant to use the formula for the packed bed Reynolds number. Is it possible to measure it? How would I design an experiment for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to avoid temperature readings of the object. In &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/096014819190107Z&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; article I found a relation between the Reynolds number of a packed bed and the heat transfer coefficient, so measuring only the Reynolds number would be a good start, even though I do not have a real packed bed. Could this be done by simple pressure drop readings?&lt;/p&gt;&#xA;" OwnerUserId="2480" LastActivityDate="2011-08-09T15:18:14.167" Title="Can I measure the Reynolds number or the heat transfer coefficient" Tags="&lt;fluid-dynamics&gt;" AnswerCount="1" />
  <row Id="13322" PostTypeId="2" ParentId="13321" CreationDate="2011-08-09T15:18:14.167" Score="0" ViewCount="" Body="&lt;p&gt;+1 for knowing at least of Reynolds number. &#xA;But did You ever hear of Nusselt, Prantl and Grashoff Numbers? &#xA;I'd recommend You read basics about heat transfer. &#xA;Look for examples, packed beds are everyday stuff in chemical &#xA;engineering. (pebble heaters, regenerative combustion, &#xA;brick ovens, catalysts and many things more) &#xA;The Reynolds number is not to &quot;measure&quot; this is more a matter of &#xA;choice. The mean diameter of the voids between the beads &#xA;might be a good choice for the characteristic length. But as I wrote already, look for examples, then You &#xA;see what others have choosen. To choose the same makes it &#xA;easier to compare the results. &lt;/p&gt;&#xA;" OwnerUserId="1370" LastActivityDate="2011-08-09T15:18:14.167" CommentCount="1" />
  <row Id="13323" PostTypeId="2" ParentId="1894" CreationDate="2011-08-09T15:53:45.567" Score="0" ViewCount="" Body="&lt;p&gt;For quantum mechanics, there's really nothing unrigorous about the path integral.  You have to define it in Euclidean signature, but that's just the way life is with oscillatory integrals.  It has nothing to do with the fact that the path integral is infinite-dimensional.  Try to insert a set of intermediate states in the propagator $\langle q_f | e^{-iHt/\hbar}|q_i\rangle$ and you'll get an integral that's not absolutely convergent.  This expression is just fine if you sandwhich it inside a well-defined computation -- e.g., don't use singular wave functions for your initial and final states -- but if you want the expression to stand on its own, you have to provide some additional convergence information.  Usually what people do is observe that the unitary group of time translations is the imaginary boundary of an analytic semigroup.  The real part of this semigroup, $e^{-H\tau}$, has a rigorous path integral formula; it's the volume of a cylinder set.  The volume of a cylinder set is computed as the limit of cutoff path integrals of the form&#xA;$$\frac{1}{Z} \int_{F_{cutoff}} e^{-\frac{1}{\hbar} S_{effective}(\phi)} d\phi,$$&#xA;where $d\phi$ is Lebesgue/Haar/whatever measures on the finite-dimensional space of cutoff fields and $S_{effective}$ is a cutoff/lattice approximation to the continuum action.  Given such a measure, under reasonable conditions, you can analytically continue the correlation functions from Euclidean signature back to Minkowski.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the record:  mathematicians are not tearing their hair out about this stuff.  It's cool.  We got it.  We -- and by &quot;we&quot;, I mean a relatively small number of experts, not necessarily including myself -- can even handle 4d Yang-Mills theory in &lt;em&gt;finite&lt;/em&gt; volume.  (What's hard is proving facts about the behavior of correlation functions in the IR limit.) &lt;/p&gt;&#xA;" OwnerUserId="1504" LastActivityDate="2011-08-09T15:53:45.567" CommentCount="2" />
  <row Id="13324" PostTypeId="1" CreationDate="2011-08-09T16:26:32.093" Score="0" ViewCount="102" Body="&lt;p&gt;Today I heard an argument to prove that the Earth frame is non-inertial. It seriously doesn't make any sense to me but I also heard that same argument was made by Einstein, so I am asking it here. Please explain how it proves that Earth frame is non-inertial. The argument goes like this:&#xA;Suppose the Earth frame is inertial. Now, all the heavenly bodies will have to go around the Earth once a day. But, this implies a very fast speed. Hence, the Earth frame can't be inertial.&lt;/p&gt;&#xA;" OwnerUserId="4235" LastActivityDate="2011-08-09T17:46:29.977" Title="Argument for proving Earth frame is non-inertial" Tags="&lt;inertial-frames&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="13325" PostTypeId="1" AcceptedAnswerId="13332" CreationDate="2011-08-09T16:29:39.663" Score="3" ViewCount="102" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Critical_mass&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; gives the following definition for critical mass.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A critical mass is the smallest amount of fissile material needed for a sustained nuclear chain reaction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No mention is made of a neutron moderator in this definition.  If critical mass is defined like this, then that should allow infinite moderator material to be used if doing so would lead to the minimum amount of fissile material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: &lt;strong&gt;What is the universally minimum amount of fissile mass needed to achieve a critical configuration (allowing a neutron moderator and no restriction on the moderator) for something like Uranium-235 and what would its configuration be?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranium-235 has a &lt;strong&gt;bare sphere critical mass&lt;/strong&gt; (BSCM) of around $52 kg$, which is the mass of a critical sphere containing &lt;em&gt;only&lt;/em&gt; the fissile material where no neutrons are reflected back into the sphere after leaving the surface.  Wikipedia has a good illustration of a bare sphere versus a sphere surrounded by a moderator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/O6KKI.gif&quot; alt=&quot;BSCM and moderator blanket&quot;&gt;&#xA;&lt;em&gt;Image: First item is a BSCM illustration, 2nd item is a critical sphere that uses less fissile mass than the BSCM due to the introduction of a neutron moderator blanket&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This example illustrates how a critical configuration can be made with U-235 that uses less than the BSCM, or $52 kg$.  It is possible, although unlikely, that the above sphere surrounded by a moderator could be the configuration that answers my question.  Alternatives would include a homogenous mixture of the moderator and the fissile material, a mixture of the two that varies radially, or something I have not thought of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hard part is showing that the particular fissile material/moderator mix can not be improved by any small change.  It would also be necessary to show that it can not be improved by adding more than 1 type of moderator (I suspect this could be done with a reasonably short argument).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Technical mumbo junbo&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Everything I write here is just a suggestion, answer however you want to or can&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the moderator and fissile material have a certain density.  I would make the simplifying assumption that the density of a homogenous mixture of the fuel and moderator would be a linear combination of their specific volumes, with the understanding that this is not true in real life.  Changing the mix changes the macroscopic cross section of both.  It might help to note that the BSCM almost exactly determines the macroscopic cross section of the pure fissile material, one over the macroscopic cross section is the path length, which is on the same order of magnitude of the radius.  Generally fast and thermal scattering from U-235 is small compared to the fission cross section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would use 2 neutron energy groups, and assume either diffusion or immediate absorption after scattering to thermal energies.  In fact, I would just assume immediate absorption.  At that point, all you would need is the scatter to absorption ratio and microscopic cross section at both thermal and fast energies for the moderator and fissile material, in addition to the densities of course.  Even then some type of calculus of variations may be necessary (if you're looking for a radially varying mixture) in addition to the fact that it would be hard to describe the fast group without a fairly complicated form of neutron transport.  The BSCM is comparatively simple since it has a constant number density of the fissile material and even &lt;em&gt;that&lt;/em&gt; is actually pretty complicated.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-08-09T16:38:32.007" LastActivityDate="2011-08-09T22:27:29.287" Title="What is the minimum amount of fissile mass required to acheive criticality?" Tags="&lt;nuclear-engineering&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="13326" PostTypeId="2" ParentId="13324" CreationDate="2011-08-09T17:03:40.443" Score="4" ViewCount="" Body="&lt;p&gt;In a frame of reference attached to the surface of the planet, everything far away (other planets, stars, distant galaxies...) follows a circular (or nearly) path with a period of 24 hours. These paths pose two problems&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;They involve observed accelerations with no obvious forces causing them&lt;/li&gt;&#xA;&lt;li&gt;Any of these bodies more than $24/2\pi$ light hours away are observed to move faster than light&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Condition one runs afoul of Newton and condition 2 of Einstein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, a non-rotating frame attached to the Earth or a non-rotating frame fixed to the CoM of the solar system both have no trouble with the rotation and are OK in general relativity (because they are freely falling). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can of course, see any upper-division or graduate mechanics text for a discussion of transformation to and from non-inertial frames that allow you to overcome all of this.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-08-09T17:03:40.443" />
  <row Id="13327" PostTypeId="2" ParentId="13324" CreationDate="2011-08-09T17:46:29.977" Score="5" ViewCount="" Body="&lt;p&gt;Your argument is actually more or less right, but some of the details are wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First you have to realize that Newtonian mechanics and general relativity have different definitions of an inertial frame. According to Newtonian mechanics, the coffee cup sitting on my desk right now defines a (very nearly) inertial frame, but a falling rock is extremely noninertial, because the rock has an acceleration of 9.8 m/s2. According to GR, free-fall is the preferred inertial state, so the rock is considered to define an inertial frame, but the coffee cup has a proper acceleration of 9.8 m/s2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Newtonian definition is actually impossible to define 100% rigorously, but traditionally the &quot;fixed stars&quot; have been taken as a pretty good standard for Newtonian frames. Any frame in which the stars have a very small acceleration is considered a very good inertial frame. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you have the Newtonian definition in mind, then your argument only goes wrong at the end, where you refer to a &quot;very fast speed.&quot; What's relevant is the stars' acceleration, not their speed. If a rocket ship is gliding through our solar system at 1,000,000 m/s, then it's an inertial frame. It doesn't matter that the stars have a velocity of -1,000,000 m/s in its frame; what matters is that they have a=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the Newtonian definition, a frame of reference fixed to a point on the earth's surface is not an inertial frame. You can tell this because in that frame, the stars have large centripetal accelerations. However, the earth-fixed frame comes very close to being inertial, because you can find other frames that are inertial and that differ from it only by a very small acceleration. Therefore experiments on the earth's surface need to be pretty sensitive in order to detect any noninertial effects. The classic example of such an experiment is the Foucault pendulum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In GR, a frame of reference fixed to a point on the earth's surface is not an inertial frame, and it doesn't even come close to being one. It differs from a valid (free-falling) inertial frame by a huge amount -- an acceleration of 9.8 m/s2. Even an extremely crude experiment can determine this. For instance, I can tell because I feel pressure from my chair on the seat of my pants. A secondary issue is that the earth's frame is rotating, and GR does consider rotating frames to be noninertial as well. (There was a lot of historical confusion on this point, including some early mistakes by Einstein, who thought GR would embody Mach's principle better than it actually did.)&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-09T17:46:29.977" />
  <row Id="13328" PostTypeId="2" ParentId="13316" CreationDate="2011-08-09T17:55:25.377" Score="1" ViewCount="" Body="&lt;p&gt;If you want a paper that explicitly shows the loss of interference due to a &quot;which slit&quot; measurement, the following work by Dave Pritchard's group at MIT (using sodium atoms) is one very nice example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Journal reference: Physical Review Letters vol. 75  pg. 3783  Nov 1995&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Free to read: &lt;a href=&quot;http://cua.mit.edu/Pritchard_IFM/Publications/photon_scattering.pdf&quot; rel=&quot;nofollow&quot;&gt;http://cua.mit.edu/Pritchard_IFM/Publications/photon_scattering.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few minor errors in the linked Dr. Quantum video (the electrons should still exhibit single-slit diffraction when the slit is measured), but apparently there are much worse parts of that movie than the selected clip, as the other commenters point out.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-08-09T17:55:25.377" CommentCount="1" />
  <row Id="13329" PostTypeId="1" CreationDate="2011-08-09T19:35:54.470" Score="5" ViewCount="104" Body="&lt;p&gt;Given a quantum state function, we can Fourier expand it in terms of stationary states of the Hamiltonian. So if we want to build that same quantum state approximately all we need to do is to superpose stationary states with proper amplitudes. Assuming that we can prepare such stationary states individually, how is their superposition done experimentally? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a specific example, how to prepare experimentally, an ensemble described by the quantum state that is a superposition of 4 stationary states of the particle in a box (or the hydrogen atom) that corresponds to n = 1, 2, 3, and 4 say ?  &lt;/p&gt;&#xA;" OwnerUserId="4521" LastEditorUserId="4521" LastEditorDisplayName="" LastEditDate="2011-08-10T14:21:01.847" LastActivityDate="2011-08-10T14:21:01.847" Title="How to prepare a desired quantum state?" Tags="&lt;quantum-mechanics&gt;&lt;experimental-physics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13330" PostTypeId="1" CreationDate="2011-08-09T21:02:49.207" Score="2" ViewCount="111" Body="&lt;p&gt;I'm studying for a physics test, but I think I don't really understand Galilean invariance. In my textbook there is an example in which they proof that if you consider 2 frames S and S' in standard configuration that the second law of Newton is Galilean invariant by proofing that if $x&amp;#39; = x - Vt$ than $F_x = F&amp;#39;_x$, so this law holds in both frames. So far I understand this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the book there is one assignment in which they ask me to verify that the relationship between kinetic energy and momentum, $E = p^2/2m$, is Galilean invariant. I couldn't really figure it out by myself so I looked at the answers. The answer is as followed, according to my text book:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In S:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$E = \frac{1}{2}m\dot{x}^2;$ &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; $p=m\dot{x}.$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Substitute $\dot{x} = p/m$ in the equation for the energy:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$E = \frac{1}{2}m(\frac{p}{m})^2=p^2/2m$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In S':&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$E'=\frac{1}{2}m\dot{x}'^2-\frac{1}{2}m(\dot{x}-V)^2=\frac{1}{2}m\dot{x}^2-m\dot{x}V^2$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$p'=m\dot{x}'$ &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Assume the relationship holds: i.e.,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$E'=\frac{p'^2}{2m}=\frac{1}{2m}(m\dot{x}-mV)^2=\frac{1}{2}m(\dot{x}^2-2\dot{x}V+V^2)=\frac{1}{2}m\dot{x}^2-m\dot{x}V+\frac{1}{2}mV^2,$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;in agreement with the Galilean transformation of the kinetic energy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: McComb, W. D., 1999. &lt;em&gt;Dynamics and Relativity.&lt;/em&gt; New York: Oxford University Press Inc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand all of the equations, however I just don't understand why this verifies that this relationship is Galilean invariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain this to me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4870" LastEditorUserId="4870" LastEditorDisplayName="" LastEditDate="2011-08-10T15:37:53.140" LastActivityDate="2011-08-10T15:37:53.140" Title="Galilean invariance proof" Tags="&lt;relativity&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="13331" PostTypeId="2" ParentId="13157" CreationDate="2011-08-09T21:22:38.057" Score="2" ViewCount="" Body="&lt;p&gt;I think the history of the field concept helps to understand what it means today. There's a &lt;a href=&quot;http://en.wikipedia.org/wiki/Line_of_force#Historical_origin_.26_differences_in_field_theories&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt; to Nancy J. Nersessian's paper &quot;&quot;Faraday's Field Concept&quot; where she says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The specific features of Faraday's field concept, in its 'favourite' and most complete form, are that force is a substance, that it is the only substance and that all forces are interconvertible through various motions of the lines of force. These features of Faraday's 'favourite notion' were not carried on. Maxwell, in his approach to the problem of finding a mathematical representation for the continuous transmission of electric and magnetic forces, considered these to be states of stress and strain in a mechanical aether. This was part of the quite different network of beliefs and problems with which Maxwell was working&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So for Faraday and Maxwell, the field was seen as a substance responsible for transmitting the effects of force and so doing away with action at a distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today, the field concept is still used to account for action at a distance as measurable effects which propagate between separate points, but without force or aether as a substance. Gauge field theories, for example, use particles called gauge bosons to transmit the fundamental forces of nature between separated particles.&lt;/p&gt;&#xA;" OwnerUserId="1345" LastActivityDate="2011-08-09T21:22:38.057" />
  <row Id="13332" PostTypeId="2" ParentId="13325" CreationDate="2011-08-09T22:27:29.287" Score="2" ViewCount="" Body="&lt;p&gt;I'm not an expert in critical mass calculation, but the minimum critical mass for U-235 must be less than (or equal to) 780 g as shown in &lt;a href=&quot;http://www.osti.gov/bridge/servlets/purl/766566-ITPxxo/webviewable/766566.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Mass Estimates of Very Small Reactor Cores&#xA;Fueled by Uranium-235, U-233 and Cm-245&quot;&lt;/a&gt; for a solution type core.&lt;/p&gt;&#xA;" OwnerUserId="1656" LastActivityDate="2011-08-09T22:27:29.287" />
  <row Id="13333" PostTypeId="2" ParentId="13330" CreationDate="2011-08-09T22:48:03.720" Score="2" ViewCount="" Body="&lt;p&gt;Either I should turn in my medical marijuana card, or the author of your textbook should. The exercise doesn't make any sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we only have one particle, whose mass is fixed, we can set $m=1$. Also, the factor of 1/2 in the equation for KE is purely conventional, so let's drop that as well. We then have the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the definitions $p=v$ and $E=v^2$, prove that the relationship $E=p^2$ is Galilean invariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution: In the new frame, by definition $p&amp;#39;=v&amp;#39;$ and $E&amp;#39;=v&amp;#39;^2$. therefore $E&amp;#39;=p&amp;#39;^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author of the text seems to have convinced him/herself that this is a nontrivial proof, when in fact it's a trivial matter of definition. It would have held equally well if we had defined $p=1/v$ and $E=\sin\left(\exp({(v/v_o)^{142}})\right)$. The reason is that the three variables $v$, $p$, and $E$ are all related to one another by definitions, and the definition of the Galilean transformation only directly affects $v$. To get an example where it failed, you'd have to have some definition like $p=c_1v+c_2x$; then the Galilean transformation would affect the variables $v$, $p$, and $E$ in more than one way (through $x$ as well as $v$), and a nontrivial question of consistency would arise.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-09T22:48:03.720" CommentCount="2" />
  <row Id="13335" PostTypeId="2" ParentId="13329" CreationDate="2011-08-10T02:21:58.857" Score="2" ViewCount="" Body="&lt;p&gt;For particles in a beam, making a superposition of spin is easy; you just split the beam and recombine it after making appropriate modifications to the split beams. In creating a superposition, this works even if the beam has only a single particle in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A particle in a box is a tougher situation as it is difficult to split the states. The problem is that the usual methods of measuring a particle's position or momentum (in a box) are destructive. For example, after you measure the position of an electron in a potential well you destroy all information about its momentum and its previous positions. I'll take a whack at it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get what you're asking for, you would need to have a measurement that (a) is non destructive, and (b) allows you to make some modification to the electron(s). Suppose that the electron begins in its lowest energy state. You want a measurement that will split that energy state into two different states and then you make a modification to one of the states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that you begin with an electron with spin-up and lowest energy in your potential well. One thing you could do is to apply a magnetic field in a direction other than up or down. This will split the electron into appropriate superpositions of spin along that new direction. If you make the new magnetic field in some horizontal direction u (so you're perpendicular to spin-up), you'll have a superposition: $(|+u\rangle+|-u\rangle)/\sqrt{2}$. Those two superpositions have two different energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When an electron absorbs a single photon, its spin flips. So you can convert a $|-u\rangle$ electron to the $|+u\rangle$ state by arranging for it to absorb a photon oriented in the +u direction. The $|+u\rangle$ electron can't absorb such a photon because of conservation of angular momentum. Therefore, you've just added energy to one half of your superposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now turn off the magnetic field. You've created a state that is a superposition of two energy states. (Maybe more, depending how copacetic the excited energy state with the magnetic field on are with the excited energy states with no magnetic field.) In fact, the magnetic field really wasn't necessary. I put it in there to make you think of the electron in terms of the +-u basis for its spin. All you really have to do is to arrange for the photon to have spin in the +u direction.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;By the way, a photon with spin in the +u direction is &quot;circularly polarized&quot;. Also, the turning on and off of the magnetic field needs to be done slowly. To arrange for particular relative phases in the superposition between the + and - spins you can use the technique of &quot;quantum phase&quot;, or &quot;Berry-Pancharatnam phase&quot;. This is the phase acquired when a system is slowly sent through a sequence of states (but which also applies to sudden state changes with the same topology). One can induce a phase by a slow change to the spin axis. The phase one obtains is equal to half the spherical area cut out by the spin axis in the Bloch sphere (i.e. the set of possible directions for a spin axis). To get a relative phase you'd want to &quot;park&quot; one of the spin states by, say, arranging for it to have an energy that prevents its spin state from being modified, and then send the other state through a sequence.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-08-10T02:21:58.857" CommentCount="4" />
  <row Id="13336" PostTypeId="2" ParentId="13330" CreationDate="2011-08-10T02:50:27.747" Score="2" ViewCount="" Body="&lt;p&gt;There's errors:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$E'=\frac{1}{2}m\dot{x}'^2-\frac{1}{2}m(\dot{x}-V)^2=\frac{1}{2}m\dot{x}^2-m\dot{x}V^2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;should be $$E&amp;#39;=\frac{1}{2}m\dot{x}&amp;#39;^2=\frac{1}{2}m(\dot{x}-V)^2=\frac{1}{2}m\dot{x}^2-m\dot{x}V+ \frac 1 2 mV^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the Galilean transformation of kinetic energy used later.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$E'=\frac{p'^2}{2m}=\frac{1}{2m}(m\dot{x}-mV)^2=\frac{1}{2}(\dot{x}^2-2\dot{x}V+V^2)=\frac{1}{2}m\dot{x}-m\dot{x}V+\frac{1}{2}mV^2,$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;should be $$E&amp;#39;=\frac{1}{2}m\dot{x}^2-m\dot{x}V+ \frac 1 2 mV^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He's used the Galilean transformation of momentum with assuming the invariance of the kinetic/momentum relationship to get the correct expression for the Galilean transformation of kinetic energy. He therefore concludes that assuming the kinetic/momentum relationship to hold in $s&amp;#39;$ is therefore true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is a contrived, and therefore confusing,  use of the Galilean transformations to prove Galilean invariance in this case because $E&amp;#39; = 1/2m&amp;#39;v&amp;#39;^2$ and $p&amp;#39;=m&amp;#39;v&amp;#39;$ are not physical laws but mathematical definitions in S' and so $E&amp;#39;= p&amp;#39;^2/2m&amp;#39;$ independently of any transformation.&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-10T03:54:56.887" LastActivityDate="2011-08-10T03:54:56.887" CommentCount="3" />
  <row Id="13338" PostTypeId="1" CreationDate="2011-08-10T08:15:24.200" Score="0" ViewCount="53" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/11321/confused-about-the-role-of-mass&quot;&gt;Confused about the role of mass&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I know that two different masses fall at the same rate in the same gravitational field because the greater gravitational force of the heavier one is exactly offset by its greater inertial resistance.  What I don't understand is why the larger mass wouldn't fall more slowly at first and then once the inertial resistance is overcome, it would then accelerate faster than the less massive one.  It seems that the rate of acceleration (not velocity, which obviously does) would vary according to the distance of the fall&lt;/p&gt;&#xA;" OwnerUserId="4871" LastEditorUserId="1876" LastEditorDisplayName="" LastEditDate="2011-08-10T08:54:54.953" LastActivityDate="2011-08-10T09:30:57.200" ClosedDate="2011-08-10T19:08:46.173" Title="Acceleration of two masses by gravitational force" Tags="&lt;gravity&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13339" PostTypeId="2" ParentId="13338" CreationDate="2011-08-10T08:39:52.020" Score="1" ViewCount="" Body="&lt;p&gt;[note this is an extremely common question already answered: &lt;a href=&quot;http://physics.stackexchange.com/questions/11321/confused-about-the-role-of-mass&quot;&gt;Confused about the role of mass&lt;/a&gt;]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inertia of an object is constant as long as its mass is constant. so the inertia is never overcome, it just stays the same for all times for our puroposes and&#xA;while the inertia of the heavier object is greater than the inertia of the lighter object, as you said, the gravitational force is proportionally greater such that both objects have the same acceleration (and thus same position with respect to time if they both are at initially at rest) as provided by Newton's law of gravitation. This requires vacuum and that the objects are aligned at the same radial distance from the body that is attracting them.&#xA;It also requires that we neglect the gravitational forces between the two objects compared to the gravitational force between each object and the earth. (as well as neglecting any other forces etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;note, this is really quite elementary:    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;newtons law of gravitation says: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \vec {F} = \frac{GMm}{r^2}{\hat{r}}$  where $\hat{r}$ is a line radially connecting the 2 bodies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;note that if M is the mass of the earth the force vector will be greater if little m is also greater as in the example above&lt;/p&gt;&#xA;" OwnerUserId="4570" LastEditorUserId="4570" LastEditorDisplayName="" LastEditDate="2011-08-10T09:30:57.200" LastActivityDate="2011-08-10T09:30:57.200" CommentCount="7" />
  <row Id="13340" PostTypeId="1" CreationDate="2011-08-10T08:48:18.963" Score="5" ViewCount="92" Body="&lt;p&gt;I am wondering how one can quantize the free electro-magnetic field&#xA;in the two dimensional space-time. The standard method of fixing the&#xA;Coulomb gauge in 4d does not seem to generalize immediately to 2d.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one tries to generalize it directly then, as in 4d, after a gauge&#xA;transformation one may assume that the scalar potential vanishes.&#xA;Thus $A_\mu=(0,A_1)$. Then the Maxwell equations imply easily that&#xA;$$\frac{\partial A_1}{\partial x^0}\equiv C=const$$&#xA;where $x^0$ is the time coordinate. Hence $A_1(x)=C x^0+h(x^1)$. Now&#xA;one can apply a gauge transformation such that $h$ will vanish:&#xA;$A_\mu(x)=(0,Cx^0)$. &lt;strong&gt;Thus there is no degrees of freedom to&#xA;quantize!!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Another problem is that  the&#xA;expression $\frac{1}{q^2+i0}$ which appears in the free photon&#xA;propagator in 4d, is not well defined in 2d&lt;/strong&gt; for purely mathematical reasons if I&#xA;understand correctly (there is no such generalized function:&#xA;the problem in at $q=0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a treatment of this in literature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually the above question was motivated by my attempt to read J.&#xA;Schwinger's paper &quot;Gauge invariance and mass, II&quot;, Phys. Review,&#xA;128, number 5 (1962). There he studies QED in 2d (so called&#xA;Schwinger's model). Is there a more modern and/or detailed&#xA;exposition of this paper?&lt;/p&gt;&#xA;" OwnerUserId="4562" LastActivityDate="2011-08-10T14:05:37.343" Title="How to quantize the free electro-magnetic field in 2d?" Tags="&lt;quantum-field-theory&gt;&lt;reference-request&gt;&lt;quantum-electrodynamics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13341" PostTypeId="1" CreationDate="2011-08-10T09:12:14.353" Score="3" ViewCount="96" Body="&lt;p&gt;given the differential operator Eigenvalue problem  $ -D^{2}y(x)=E_{n}y(x) $ with boundary periodic conditions $ y(x)=y(x+1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my question is if there is a similar probelm for the Dilation operator $ T= xD $ with 'D' the derivative respect to 'x' i mean solve&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ T^{2}y(x)=E_{n}y(x)$ with dilation periodic condition&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my question is how can i impose PERIODIC conditions but not for the traslation group (Yx+1) but for the Dilation group, i believe that some periodic conditions involving dilations would be $ y(x)=y(2x)$ but i am not sure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how can i impose PERIODIC conditions if the LIe group is just the Dilations in 1-D ??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks.&lt;/p&gt;&#xA;" OwnerUserId="4435" LastActivityDate="2011-09-05T15:06:30.967" Title="differential equation with periodic conditions" Tags="&lt;mathematical-physics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13342" PostTypeId="1" AcceptedAnswerId="13344" CreationDate="2011-08-10T09:24:49.113" Score="5" ViewCount="125" Body="&lt;p&gt;I was solving a practice test problem and it was just a conservation of energy problem&#xA;where a spherical ball is falls from a height h to the ground such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what I can't quite picture in my head is how there is an energy contribution from the rotation at a fixed point in time. obviously there is a instantaneous rate of change of the center of mass &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_{cm} =  \frac {d \vec {x_{cm}} }{dt}$$&#xA; for any fixed point in time&#xA; in my head I can see this as contributing to energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but the angular velocity can't be similarly treated like it is all at a point&#xA;   $$ \vec{\omega} = \frac{d\ \vec{\theta} } {dt}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;here  i understand this vector of angular rotation exists but I am confused as to why it adds to the energy of the system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just somehow can't wrap my head about energy of rotation when there is already some energy due to the center of mass' translation... i'm not sure exactly why sorry if this is vague but I figured i'd ask if someone has some way to think about this. maybe in terms of DOF's?&lt;/p&gt;&#xA;" OwnerUserId="4570" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-10T19:05:23.697" LastActivityDate="2011-08-12T17:26:46.180" Title="confused about ball rolling and rotational energy" Tags="&lt;angular-momentum&gt;&lt;rotational-dynamics&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="13343" PostTypeId="1" AcceptedAnswerId="13409" CreationDate="2011-08-10T10:03:35.320" Score="-1" ViewCount="79" Body="&lt;p&gt;suppose you have a pita pocket of radius 5cm or .05m.  We will model the pita pocket as consisting of disk with a layer of air of volume V which is contained by the surface of the bread. We will assume that no cracks or perforations exist in the pita pocket.&#xA;  supposing the thickness of the pocket is d = .5cm = .0025m. we can calculate the volume from the radius and thickness of the disk  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V_{pita}$ = $\pi r^2d$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;suppose that the pita contains a volume of air inside &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V_{air} = \frac {V_{pita}}{5}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that the rest of the volume of the pita is a solid. Model the forces and pressure that cause the pita to expand. at what temperature will the air pressure inside be able to break the bonds holding the top sides of the bread together and allow it to expand under your models assumptions? rough order of magnitude estimates are fine--- but all answers should agree with experimental values of pita expansion as a function of Temperature&lt;/p&gt;&#xA;" OwnerUserId="4570" LastEditorUserId="4570" LastEditorDisplayName="" LastEditDate="2011-08-10T10:10:05.373" LastActivityDate="2011-08-11T22:48:59.340" Title="determine temperature at which pita bread will expand in a toaster" Tags="&lt;forces&gt;&lt;pressure&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13344" PostTypeId="2" ParentId="13342" CreationDate="2011-08-10T10:50:38.340" Score="6" ViewCount="" Body="&lt;p&gt;Try looking at this problem microscopically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can imagine the ball consisting of a number of smaller pieces of matter. The total kinetic energy of the entire ball is the sum of the kinetic energies of its pieces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T=\frac{1}{2}\sum_im_iv_i^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if it were to just fall straight down, all these little pieces would have velocity vector $v_i=v$ and you could write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T=\frac{1}{2}(\sum_im_i)v^2 = \frac{1}{2}Mv^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $M$ is the total mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if it is rotating, the small pieces have an additional component in the velocity: they do not just move downwards, but have also to move around the center of rotation of the body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in addition to the downward falling component of the motion, the particles have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_{rot}=\frac{1}{2}\sum_im_iv_{\text{rot},i}^2 = \frac{1}{2}\sum_im_i(\omega r_i)^2 = \frac{1}{2}\omega^2\sum_im_ir_i^2 = \frac{1}{2}\omega^2 I$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the total energy is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_{tot}=\frac{1}{2}Mv_{cm}^2+\frac{1}{2}\omega^2 I$$&lt;/p&gt;&#xA;" OwnerUserId="2151" LastEditorUserId="4237" LastEditorDisplayName="" LastEditDate="2011-08-12T17:26:46.180" LastActivityDate="2011-08-12T17:26:46.180" />
  <row Id="13345" PostTypeId="2" ParentId="13340" CreationDate="2011-08-10T10:58:48.557" Score="3" ViewCount="" Body="&lt;p&gt;2d Electromagnetism is a special case of 2d Yang-Mills theory, which is the trivial theory on flat $\mathbb{R}^2$.   However, this theory becomes entertaining when the spacetime has non-trivial topology -- e.g., is a cylinder or a Riemann surface -- because then you get a countable number of degrees of freedom.  The gauge field can have non-trivial holonomies around the non-contractible loops, like in the Aharonov-Bohm effect.  (The Hamiltonian is non-trivial on a cylinder of radius R, but there's a mass gap proportional to R; as you decompactify the cylinder, these degrees of freedom become arbitrarily massive.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One nice place to read about this is Witten's &quot;On Quantum Gauge Theories in Two Dimensions&quot;.   Also, Chapter 11 of Volume 2 of the IAS QFT lectures.&lt;/p&gt;&#xA;" OwnerUserId="1504" LastEditorUserId="1504" LastEditorDisplayName="" LastEditDate="2011-08-10T14:05:37.343" LastActivityDate="2011-08-10T14:05:37.343" CommentCount="1" />
  <row Id="13346" PostTypeId="2" ParentId="13342" CreationDate="2011-08-10T11:18:49.970" Score="2" ViewCount="" Body="&lt;p&gt;The motion of a rigid body can be viewed as the sum of two parts: a translation of the center of mass plus rotation about the center of mass. From this, the velocity of any point $i$ of the body relative to the fixed system of coordinates is the sum of the translational velocity of the CM and the angular velocity of rotation around the CM: $$\mathbf{v_i} = \mathbf{V}_i + \boldsymbol{\omega} \times \mathbf{r}_i.$$ The total kinetic energy of the moving body is the sum of the kinetic energies of all discrete particles constituting the body: $$T = \sum_{i} \frac{1}{2}m_i \mathbf{v}_i^2.$$ Substituting $\mathbf{v}_i$ in the above equation (and doing some magic) gives $$T = \frac{1}{2} MV^2 + \frac{1}{2} \sum_{i} m_i ( \omega^2 r_i^2 - (\boldsymbol{\omega} \mathbf{r_i})^2 ).$$&#xA;The first term is the kinetic energy of the translational motion of body's CM, while the second  is the kinetic energy of rotation (with angular velocity $\boldsymbol{\omega}$) about the axis passing through the CM. In the simplest case of rotation about a fixed axis of the body's symmetry the second term will look like you wrote: $\frac{1}{2}I \omega^2$.&lt;/p&gt;&#xA;" OwnerUserId="4237" LastEditorUserId="4237" LastEditorDisplayName="" LastEditDate="2011-08-10T20:05:30.553" LastActivityDate="2011-08-10T20:05:30.553" />
  <row Id="13347" PostTypeId="2" ParentId="9523" CreationDate="2011-08-10T13:51:05.660" Score="1" ViewCount="" Body="&lt;p&gt;I was able to find experimental and simulated data for the plane intensity from multiple scattering of small (1, 5 and 10 $\mu$m spheres in the &lt;a href=&quot;http://www.forbrf.lth.se/english/staff/senior_researchers/edouard_berrocal/&quot; rel=&quot;nofollow&quot;&gt;Thesis of Edouard Berrocal&lt;/a&gt;. His &lt;a href=&quot;http://www.forbrf.lth.se/fileadmin/forbrf/Documents/Blandat/Thesis_Berrocal_2006.pdf&quot; rel=&quot;nofollow&quot;&gt;thesis can be downloaded here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2735" LastActivityDate="2011-08-10T13:51:05.660" CommentCount="1" />
  <row Id="13348" PostTypeId="2" ParentId="1030" CreationDate="2011-08-10T14:16:31.987" Score="-2" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;The evidence is against:&lt;br&gt;&#xA;CMB WMAP maps are inconsistent with the standard cosmological model&lt;/strong&gt;, from here:    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.hindawi.com/journals/aa/2010/847541/&quot; rel=&quot;nofollow&quot;&gt;Large-angle anomalies in the CMB&lt;/a&gt; (open access) and &lt;a href=&quot;http://arxiv.org/abs/1004.5602&quot; rel=&quot;nofollow&quot;&gt;in arXiv - April 2010&lt;/a&gt; (my bold)    &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Abstract&lt;br&gt;&#xA;  ...We discuss these findings in relation to expectation&#xA;  from standard inflationary cosmology ...&lt;br&gt;&#xA;  Summary&lt;br&gt;&#xA;  The study of alignments in the low ℓ CMB has found a&#xA;  number of peculiarities. We have shown that the alignment of the&#xA;  quadrupole and octopole planes is inconsistent with Gaussian,&#xA;  statistically isotropic skies at least at the &lt;strong&gt;99% confidence&#xA;  level&lt;/strong&gt;. Further a, number of (possibly related) alignments occur at&#xA;  &lt;strong&gt;95% confidence levels&lt;/strong&gt; or greater. Putting together these provides&#xA;  &lt;strong&gt;a strong indication that the full-sky CMB WMAP maps are inconsistent&#xA;  with the standard cosmological model&lt;/strong&gt; at the large-angles. &lt;strong&gt;Even&#xA;  more peculiar is the alignment of the quadrupole and octopole with&#xA;  solar system features (the ecliptic plane and the dipole)&lt;/strong&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Introduction&lt;br&gt;&#xA;  In this regard, it is worth noting that our record at&#xA;  predicting the gross properties of the universe on large scales from&#xA;  first principles has been rather poor. According to the standard&#xA;  concordance model of cosmology, over 95% of the energy content of the&#xA;  universe is extraordinary—dark matter or dark energy whose existence&#xA;  has been inferred from the failure of the Standard Model of particle&#xA;  physics plus General Relativity to describe the behavior of&#xA;  astrophysical systems larger than a stellar cluster—while the very&#xA;  homogeneity and isotropy (and inhomogeneity) of the universe owe to&#xA;  the influence of an inflaton field whose particle-physics-identity is&#xA;  completely mysterious even after three decades of theorizing.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Conclusions   ... The CMB is widely regarded as offering strong&#xA;  substantiating evidence for the concordance model of cosmology. Indeed&#xA;  the agreement between theory and data is remarkable—the patterns in&#xA;  the two-point correlation functions (TT, TE, and EE) of Doppler peaks&#xA;  and troughs are reproduced in detail by fitting with only six (or so)&#xA;  cosmological parameters. This agreement should not be taken lightly;&#xA;  it shows our precise understanding of the causal physics on the last&#xA;  scattering surface. &lt;strong&gt;Even so, the cosmological model we arrive at is&#xA;  baroque, requiring the introduction at different scales and epochs of&#xA;  three sources of energy density that are only detected&#xA;  gravitationally—dark matter, dark energy and the inflaton&lt;/strong&gt;. This&#xA;  alone should encourage us to continuously challenge the model and&#xA;  probe the observations particularly on scales larger than the horizon&#xA;  at the time of last scattering.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;At the very least, probes of the large-angle (low-ℓ) properties of the&#xA;  CMB &lt;strong&gt;reveal that we do not live in a typical realization of the&#xA;  concordance model of inflationary ΛCDM&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;On the theoretical side: from Michael S. Turner (1997)&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/astro-ph/9704062&quot; rel=&quot;nofollow&quot;&gt;Ten Things Everyone Should Know About Inflation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cold dark matter, which is an important means of testing inﬂation, is&#xA;  a &lt;strong&gt;ten-parameter&lt;/strong&gt; theory, $h,\Omega_Bh^2,S,n,dn/d \ln{k}, T/S, nT,&#xD;&#xA;&amp;gt; \Omega_\nu,g_*, \Omega_\Lambda$ . While this is a &lt;strong&gt;daunting number of&#xA;  parameters&lt;/strong&gt;, especially for a cosmological theory, there is good&#xA;  reason to believe that &lt;strong&gt;within ten years the data&lt;/strong&gt; will&#xA;  overdetermine these parameters. Crucial to achieving this goal are the&#xA;  high-precision, high-resolution measurements of CBR anisotropy that&#xA;  will be made over the next decade by...&#xA;  ...ΛCDM is consistent with all the observations discussed here as well as others;...&lt;br&gt;&#xA;  ...&lt;br&gt;&#xA;  &lt;strong&gt;Inﬂation Makes Three Robust Predictions&lt;/strong&gt;&lt;br&gt;&#xA;  1 Flat universe&lt;br&gt;&#xA;  2 Nearly scale-invariant spectrum of gaussian density perturbations&lt;br&gt;&#xA;  3 Nearly scale-invariant spectrum of gravitational waves  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The future: (obviously it is my own perspective)&lt;br&gt;&#xA;In this novel cosmological model &lt;a href=&quot;http://vixra.org/pdf/1107.0016v1.pdf&quot; rel=&quot;nofollow&quot;&gt;A self-similar model of the Universe unveils the nature of dark energy&lt;/a&gt; the Inflation is not needed.&lt;br&gt;&#xA;I had a part in the paper, not as an author, and I appreciate any criticism  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-08-10T14:16:31.987" CommentCount="6" />
  <row Id="13349" PostTypeId="1" AcceptedAnswerId="13358" CreationDate="2011-08-10T15:02:06.710" Score="3" ViewCount="76" Body="&lt;p&gt;I would like to know if there are any online collections of lecture videos on QCD or non-Abelian QFT at a graduate level (at the level of volume 2 of Weinberg's QFT books?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&#xA;In String Theory I think this one by Shiraz is a great (only?) online resource -&#xA;&lt;a href=&quot;http://theory.tifr.res.in/~minwalla/&quot; rel=&quot;nofollow&quot;&gt;http://theory.tifr.res.in/~minwalla/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1637" LastActivityDate="2011-08-11T17:08:52.567" Title="Online lecture videos on QCD?" Tags="&lt;quantum-field-theory&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="13350" PostTypeId="2" ParentId="13289" CreationDate="2011-08-10T15:07:43.677" Score="0" ViewCount="" Body="&lt;p&gt;Go to a library or a good college town bookstore, look for and browse through &quot;University of Chicago Graduate Problems in Physics&quot;  by Cronin, Greenberg &amp;amp; Telegdi.  Aim to be familiar with enough ideas and techniques to be able to answer at least half the questions.  &lt;/p&gt;&#xA;" OwnerUserId="353" LastActivityDate="2011-08-10T15:07:43.677" CommentCount="2" />
  <row Id="13351" PostTypeId="1" CreationDate="2011-08-10T15:19:10.623" Score="-2" ViewCount="49" Body="&lt;p&gt;I think there is no strong opposition to the effective global neutrality of the Universe ($n_{e^-}=n_{p^+}$) .&lt;br&gt;&#xA;It seems that &lt;a href=&quot;http://en.wikipedia.org/wiki/Baryogenesis&quot; rel=&quot;nofollow&quot;&gt;Baryogenesis&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Leptogenesis_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;Leptogenesis&lt;/a&gt; are two unrelated/unconnected issues.&lt;br&gt;&#xA;Did someone computed (or estimated) the probability of such event?&lt;br&gt;&#xA;Given the order of magnitude of the number of particles I expected it to be $\pm$0%.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To ease the understanding of the question I will leave an example:&lt;br&gt;&#xA;God and Evil do not communicate. God decided to make a large number of right shoes ($n_{p^+}$) in His factory at &lt;em&gt;Heaven&lt;/em&gt; . At a different time the Evil decided to make a large number of left shoes in his factory at &lt;em&gt;Hell&lt;/em&gt;  ($n_{e^-}$). By coincidence the numbers matched.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone compute, or estimate, the probability of such event or am I missing something? &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-08-10T15:19:10.623" Title="Universe global neutrality: likelyhood and probability" Tags="&lt;cosmology&gt;&lt;baryogenesis&gt;&lt;leptogenesis&gt;" CommentCount="5" />
  <row Id="13352" PostTypeId="2" ParentId="9569" CreationDate="2011-08-10T15:43:21.527" Score="6" ViewCount="" Body="&lt;p&gt;The standard thing to do is contact the author(s) and ask.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Be polite--this person will have to take some time out of their &quot;real&quot; work to help you.&lt;/li&gt;&#xA;&lt;li&gt;Introduce yourself (include any institutional affiliation than may lend you legitimacy).&lt;/li&gt;&#xA;&lt;li&gt;Explain &lt;em&gt;why&lt;/em&gt; you are interested and what you hope to &lt;em&gt;do&lt;/em&gt; with the data.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Most publications and the arXiv indicate the &quot;corresponding author&quot;, and this is probably the person to try first. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be aware that some data may be covered by agreements to not make it public until after publication (or some arbitrary publication window typically on order of one year).&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-08-10T15:43:21.527" />
  <row Id="13353" PostTypeId="1" CreationDate="2011-08-10T17:11:56.990" Score="-2" ViewCount="66" Body="&lt;p&gt;How accurate is the time on a digital clock versus the time on internet clocks? The internet clocks seem slower than internet clocks. So which is the correct time? What are the most accurate clocks?&lt;/p&gt;&#xA;" OwnerUserId="4874" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-10T18:51:38.567" LastActivityDate="2011-08-11T20:53:41.803" Title="How accurate is the time on a digital clock versus the time on internet clocks?" Tags="&lt;time&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13354" PostTypeId="1" AcceptedAnswerId="13357" CreationDate="2011-08-10T16:23:10.137" Score="3" ViewCount="128" Body="&lt;p&gt;Regular dice are made such that opposite sides of the die add to 1+the number of sides.&#xA;For example, a 20-sided die has 14 and 7 opposite of each other, adding to 21.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For certain types of games, &quot;life counter&quot; die are used. In these, sides are numbered sequentially, so that 1 is next to 2, which is next to 3, which is next to 4, and so on. You can see a picture here: &lt;a href=&quot;http://www.coolstuffinc.com/main_supplies.php?fpid=Acc-QWSd20SpindownLifeCounterBluewhite&quot; rel=&quot;nofollow&quot;&gt;http://www.coolstuffinc.com/main_supplies.php?fpid=Acc-QWSd20SpindownLifeCounterBluewhite&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question: generally speaking, will these two kinds of die yield equal probabilities? (Assuming that both dice are well balanced).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particularly,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Will the probability of getting a certain number be the same?&lt;/li&gt;&#xA;&lt;li&gt;Will the probability of getting a number above a threshold be the same? (i.e. rolling a 10 or more)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4875" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-10T18:53:41.763" LastActivityDate="2011-08-12T23:00:59.360" Title="&quot;Regular&quot; 20-sided die, vs &quot;life counter&quot; 20-sided die. Same probabilities?" Tags="&lt;mechanics&gt;&lt;probability&gt;" AnswerCount="2" CommentCount="8" />
  <row Id="13355" PostTypeId="5" CreationDate="2011-08-10T17:16:56.980" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-10T17:16:56.980" LastActivityDate="2011-08-10T17:16:56.980" />
  <row Id="13356" PostTypeId="4" CreationDate="2011-08-10T17:16:56.980" Score="0" ViewCount="" Body="for questions about design, process, data, or analysis of experiments and observations." OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-15T16:21:20.287" LastActivityDate="2011-08-15T16:21:20.287" />
  <row Id="13357" PostTypeId="2" ParentId="13354" CreationDate="2011-08-10T17:46:31.130" Score="7" ViewCount="" Body="&lt;p&gt;If you want the high school answer, then no, the numbering does not matter.  If all faces are equally-likely, the probability is the same regardless of how you number the die, and similarly all derived quantities (such as variance or probability to be greater than 8) are the same because the underlying distribution is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the real answer, it depends on the exact probability distribution of the faces of the die, which is best determined empirically (i.e. by rolling it many times).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the die is weighted towards one face, so that face becomes more likely to be on bottom and the opposite face equally more likely to be on top, the the mean value of the normal die will not change, while the mean value of the life-counter die will change.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-08-12T23:00:59.360" LastActivityDate="2011-08-12T23:00:59.360" CommentCount="2" />
  <row Id="13358" PostTypeId="2" ParentId="13349" CreationDate="2011-08-10T17:54:12.100" Score="3" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.physics.harvard.edu/about/Phys253.html&quot; rel=&quot;nofollow&quot;&gt;Physics 253: Quantum Field Theory Lectures by Sidney R. Coleman. Recorded in 1975-1976.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is rare that you can find on such advanced topics whole-semester courses that are video recorded, with the exception to Coleman's and Minwalla's lectures. What is usually available are at most 5 video lectures per topic that were given at different summer schools. Here is an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://indico.cern.ch/conferenceDisplay.py?confId=83014&quot; rel=&quot;nofollow&quot;&gt;QCD for Postgraduates, from CERN documet server, by Giulia Zanderighi, 2010&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4521" LastEditorUserId="4521" LastEditorDisplayName="" LastEditDate="2011-08-11T17:08:52.567" LastActivityDate="2011-08-11T17:08:52.567" CommentCount="1" />
  <row Id="13359" PostTypeId="1" AcceptedAnswerId="13415" CreationDate="2011-08-10T18:14:20.767" Score="1" ViewCount="265" Body="&lt;p&gt;Einstein described his discovery of the equivalence principle as the &quot;happiest thought of my life&quot;.  Why?  What, in broad conceptual terms, is the logical chain of reasoning that leads from the equivalence principle to general relativity?&lt;/p&gt;&#xA;" OwnerUserId="4075" LastActivityDate="2011-08-13T19:55:48.447" Title="Einstein's &quot;happiest thought&quot;" Tags="&lt;general-relativity&gt;&lt;equivalence-principle&gt;" AnswerCount="4" />
  <row Id="13360" PostTypeId="2" ParentId="13341" CreationDate="2011-08-10T18:14:36.817" Score="0" ViewCount="" Body="&lt;p&gt;You are right that $y(x)=y(2x)$ can be used as condition for your differential equation with the dilation operator, and that this is a form of boundary condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important point is that if $y(x)$ satisfies $T^{2}y(x)=E_{n}y(x)$ then $\bar{y}(x):=y(2x)$ satisfies $T^{2}\bar{y}(x)=E_{n}\bar{y}(x)$. I convinced myself that this is the case by noting that $T:=x\frac{d}{dx}=2x\frac{d}{d(2x)}$, in the same sense that $\frac{d}{dx}=\frac{d}{d(x+1)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting such a &quot;periodic condition&quot; into a &quot;real&quot; boundary conditions requires some care. Let's first look at the condition $y(x)=y(x+1)$. You could solve the eigenvalue problem on the interval $[0,1]$ with the boundary conditions $y(0)=y(1)$ and $y&amp;#39;(0)=y&amp;#39;(1)$. For the condition $y(x)=y(2x)$, you could solve the eigenvalue problem on the interval $[1,2]$ with the boundary conditions $y(1)=y(2)$ and $y&amp;#39;(1)=2y&amp;#39;(2)$.&lt;/p&gt;&#xA;" OwnerUserId="4846" LastActivityDate="2011-08-10T18:14:36.817" />
  <row Id="13361" PostTypeId="1" CreationDate="2011-08-10T18:45:42.557" Score="9" ViewCount="141" Body="&lt;p&gt;Does an accelerated charge always radiate?&#xA;For example the current electrons in an electric circuit when moving through a turn they are accelerated, do they radiate because of that acceleration? &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If the answer is no, then why not?&lt;/li&gt;&#xA;&lt;li&gt;If the answer is yes, then how small the radiation is? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4521" LastEditorUserId="4521" LastEditorDisplayName="" LastEditDate="2011-08-10T20:24:09.423" LastActivityDate="2011-08-10T21:34:30.470" Title="If you run an electric current through a wire loop, do the accelerated charges radiate?" Tags="&lt;electromagnetism&gt;&lt;electromagnetic-radiation&gt;&lt;electrodynamics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="13362" PostTypeId="2" ParentId="13353" CreationDate="2011-08-10T18:54:52.077" Score="1" ViewCount="" Body="&lt;p&gt;The clocks that you find on websites are delayed a bit between the time that the time is registered until it gets to you. Also, you cannot know if the server time is set properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There does exists a timeserver protocol, but a human will not make sense of it. Your best bet is to set your computer clock to sync to a timeserver, then reference your computer clock.&lt;/p&gt;&#xA;" OwnerUserId="4877" LastActivityDate="2011-08-10T18:54:52.077" CommentCount="1" />
  <row Id="13363" PostTypeId="2" ParentId="13361" CreationDate="2011-08-10T19:13:50.570" Score="9" ViewCount="" Body="&lt;p&gt;An individual electron going around the loop would of course radiate.  However, if we have a constant current going around the loop, there are many electrons and their individual radiations cancel each other, so there is no net radiation. They balance each other out. We can show their is no net radiation by solving Maxwell's equations on the net current, which is constant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There would be some transient radiation when you change the current. Also, in the real world, there would be some infinitesimal radiation due to the fact that the current won't truly be constant at the microscopic level, i.e. because of Johnson noise etc. &lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-08-10T19:13:50.570" CommentCount="12" />
  <row Id="13364" PostTypeId="2" ParentId="13354" CreationDate="2011-08-10T19:35:29.853" Score="3" ViewCount="" Body="&lt;p&gt;Mark Eichenlaub's answer is nice (although I am a bit dubious about the accuracy of the Boltzman-like formula).  However,  a little known fact is that even for a &quot;fair&quot; die, not all outcomes are equally likely.  Research at Stanford and Santa Cruz (http://www-stat.stanford.edu/~susan/papers/headswithJ.pdf) has shown that for coin flips, there is a slightly greater than 50% chance that a fair coin will come up with the same orientation as it started with!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With throwing a die similar physics is involved. Hence, with a &quot;life counter&quot; die, by choosing the die's orientation when thrown, one could more effectively bias towards higher or lower numbers than with a standard die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coin flip bias is only at around the 1% level (I have no idea what it would be for a 20-sided die), so whether this is significant or not depends on what level of effects you care about.  But casinos make a living on advantages on the level of a few percent.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastEditorUserId="1918" LastEditorDisplayName="" LastEditDate="2011-08-11T01:34:47.587" LastActivityDate="2011-08-11T01:34:47.587" />
  <row Id="13366" PostTypeId="2" ParentId="13361" CreationDate="2011-08-10T21:34:30.470" Score="3" ViewCount="" Body="&lt;p&gt;Generally speaking, accelerated charges radiate. For the case of a current in a wire, the acceleration is so small that the radiation is hard or impossible to detect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider 1 amp traveling down a wire of radius 1mm. How fast do the electrons move? To solve this we need to know the number of mobile charge carriers per unit volume and a few constants. &lt;a href=&quot;http://www.radioelectronicschool.net/files/downloads/howfast.pdf&quot; rel=&quot;nofollow&quot;&gt;The answer is $v = $ 2 meters per day.&lt;/a&gt; You can convert this speed into an acceleration using the formula $a = v^2/R$ where $R$ is the radius of curvature of the corner. It is a very small number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison, the radiation that makes it so difficult to speed up electrons in a collider experiment comes from electrons that are traveling essentially at the speed of light.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-08-10T21:34:30.470" CommentCount="5" />
  <row Id="13367" PostTypeId="1" AcceptedAnswerId="14054" CreationDate="2011-08-10T21:44:19.563" Score="4" ViewCount="237" Body="&lt;p&gt;C.W. Ufford and G.H. Shortley in &lt;a href=&quot;http://prola.aps.org/abstract/PR/v42/i2/p167_1&quot; rel=&quot;nofollow&quot; title=&quot;Physical Review 1932, Vol. 42, pg. 167&quot;&gt;Physical Review 1932, Vol. 42, pg. 167&lt;/a&gt;, separate the two  $^2$D's of $d^3$. On page 173 (pg 7 of the PDF) of this article they determine the matrix of the electrostatic energy in eqn. 3.3 as:&#xA;$$\begin{array}{ccc}&#xD;&#xA;- &amp;amp;a  &amp;amp; b \\&#xD;&#xA;a &amp;amp; 3F_0+7F_2+63F_4 &amp;amp; 3(21)^{1/2}(F_2-5F_4) \\&#xD;&#xA;b &amp;amp; 3(21)^{1/2}(F_2-5F_4) &amp;amp; 3F_0+3F_2-57F_4  \\  \end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to figure out how they actually determine this matrix. How do they do it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I understand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This matrix comes from a description of the two terms,&#xA;$$^2D^a_{2,1/2} = 1/2 [-A+B-C+E]$$&#xA;$$^2D^b_{2,1/2} = (84)^{-1/2}[-5A-3B-C+4D-3E-2(6)^{1/2}F]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A,B,C,D,E,F$ are zero-order states of $d^3$ which are&#xA;$$\begin{array}{cccc}&#xD;&#xA;A &amp;amp;\mathcal{A}u_1(2^+)u_2(2^-)u_3(-2^+)  &amp;amp; B &amp;amp;\mathcal{A}u_1(2^+)u_2(1^+)u_3(-1^-) \\&#xD;&#xA;C &amp;amp;\mathcal{A}u_1(2^+)u_2(1^-)u_3(-1^+) &amp;amp; D &amp;amp;\mathcal{A}u_1(2^-)u_2(1^+)u_3(-1^+) \\&#xD;&#xA;E &amp;amp;\mathcal{A}u_1(2^+)u_2(0^+)u_3(0^-) &amp;amp; F &amp;amp;\mathcal{A}u_1(1^+)u_2(1^-)u_3(0^+) \\&#xD;&#xA;  \end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this notation $u_i(a^j)$ indicates the one-electron central field eigenfunction for electron $i$ with the $j^{th}$ set of numbers $n^il^im_l^im_s^i$. $\mathcal{A}$ represents the antisymmetrizing operator $(N!)^{-1/2}\sum_P (-1)^p P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can say that $(ab|1/r_{12}|cd)$ depends, in part,  on $\sum_{k=0}^\infty c^k (l^a m_l^a, l^c m_l^c) c^k(l^dm_l^d,l^bm_l^b)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are tables of these values already determined, notably as $a^k$ and $b^k$, which can be used to evaluate the Coulomb and exchange integrals for $k=0,2,4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.U. Condon then extended Slater's method to determine the non-diagonal matrix elements in a paper (Physical Review, 1930, Vol. 36, pg. 1121) for separating multiplets of the same kind, as we have in our case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you are confused by the notation, $J(a,b)=\sum a^k$ and $K(a,b)=\sum b^k$ and $a^k(l^am_l^a,l^bm_l^b)=c^k(l^am_l^a,l^am_l^a)c^k(l^bm_l^b,l^bm_l^b)$ and $b^k(l^am_l^a,l^b m_l^b) = [c^k(l^am_l^a,l^bm_l^b)]^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we get to dealing with the doublet D's, we see that they depend on the following wavefunctions: $(2^+,2^-,-2^+), (2^-,1^+,-1^+), (2^+,0^+,0^-), (2^+,1^+,-1^-), (1^+,1^-,0^+), (2^+,1^-,-1^+) $ -- or $A,B,C,D,E,F$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Slater's method we find that   $E(^2H)+ E(^2G)+E(^4F)+ E(^2F)+E(^2D)+E(^2D)=18F_0-13F_2-152F_4$ and,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(^2H)+ E(^2G)+E(^4F)+ E(^2F)=12F_0 -23F_2 - 158F_4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&#xA;$E(^2D)+E(^2D)= 18F_0-13F_2-152F_4 - (12F_0 -23F_2 - 158F_4) = 6F_0+10F_2+6F_4$. Thus the mean energy is $3F_0 + 5F_2 + 3F_4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't understand how to get to the non-diagonal elements yet...&lt;/p&gt;&#xA;" OwnerUserId="4112" LastEditorUserId="4112" LastEditorDisplayName="" LastEditDate="2011-08-28T19:02:57.903" LastActivityDate="2011-08-28T19:02:57.903" Title="Theory of Complex Spectra, Applying Slater-Condon Rules" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13368" PostTypeId="2" ParentId="13359" CreationDate="2011-08-10T22:06:25.503" Score="2" ViewCount="" Body="&lt;p&gt;perhaps Einstein was happy due to the fact that his general relativity resolved an age old dilemma in Newtonian Mechanics: why is inertial mass = gravitational mass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here are some ways that the weak equivalence principle may be stated:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The trajectory of a point mass in a gravitational field depends only on its initial position and velocity, and is independent of its composition.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;All test particles at the alike spacetime point in a given gravitational field will undergo the same acceleration, independent of their properties, including their rest mass.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;All local centers of mass vacuum free fall along identical (parallel-displaced, same speed) minimum action trajectories independent of all observable properties.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The vacuum world line of a body immersed in a gravitational field is independent of all observable properties.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The local effects of motion in a curved space (gravitation) are indistinguishable from those of an accelerated observer in flat space, without exception.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mass (measured with a balance) and weight (measured with a scale) are locally in identical ratio for all bodies (the opening page to Newton's &quot;Principia&quot;). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4570" LastActivityDate="2011-08-10T22:06:25.503" CommentCount="3" />
  <row Id="13369" PostTypeId="1" CreationDate="2011-08-10T22:22:27.663" Score="2" ViewCount="46" Body="&lt;p&gt;There is a standard procedure to show that for CFT's in dimension $d\geq 2$, the trace of stress tensor vanishes. I think I can't apply those steps when I only have one dimension, say time, because if we could then we can show that stress tensor trace $T_{tt}$ (which is basically it) also vanishes which may not be true always. Things are definitely different in 1D (I think metric can just be $g_{tt}$ and conformal transformation just a $t\rightarrow \lambda t$ transformation. Please correct me if I am wrong.) So, where do basically things go wrong in such derivation?  &lt;/p&gt;&#xA;" OwnerUserId="1349" LastActivityDate="2011-08-10T22:22:27.663" Title="Trace of energy momentum tensor in $CFT_1$" Tags="&lt;conformal-field-theory&gt;" CommentCount="2" />
  <row Id="13370" PostTypeId="2" ParentId="13359" CreationDate="2011-08-10T23:06:26.590" Score="2" ViewCount="" Body="&lt;p&gt;The basic idea is that since small, electrically neutral bodies with given initial conditions follow universally defined trajectories, you can define those trajectories to be straight lines. This leads to a theory of gravity that is geometrical in character. By taking the trajectories, which we'd normally think of as being curves, and calling them straight, one is forced to take the background of space and time, which we normally think of as being flat, and call it curved.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-10T23:06:26.590" CommentCount="2" />
  <row Id="13371" PostTypeId="1" CreationDate="2011-08-10T23:40:53.390" Score="-1" ViewCount="24" Body="&lt;p&gt;This may be considered chemistry-related, but it's part of a physics report I'm writing so hopefully it's fine here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody assist me in figuring out the oxidation numbers for each element in each of these reactions? (see &lt;a href=&quot;http://img812.imageshack.us/img812/5429/unledurn.png&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;). I'm having trouble figuring them out!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4880" LastActivityDate="2011-08-10T23:40:53.390" ClosedDate="2011-08-11T00:16:54.137" Title="Oxidation numbers?" Tags="&lt;homework&gt;" CommentCount="1" />
  <row Id="13372" PostTypeId="2" ParentId="13359" CreationDate="2011-08-10T23:44:58.670" Score="1" ViewCount="" Body="&lt;p&gt;Short answer: the famous 'rocket elevator' thought experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer: It turns out that the motion due to gravitational force acting on some mass is exactly the same as the motion due to a mass in an accelerated reference frame. Einstein famously imagined the experience of standing inside a windowless box, with no way to contact the outside, accelerated by a rocket at $9.8 m/ sec^2$ which, in fact and in theory, is identical to the experience of standing inside a box that is sitting still on the surface of the earth. An observer in the box cannot ever tell the difference between these experiences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is, of course, only an abstract way of thinking about $(\text{gravitational mass}) = (\text{inertial mass})$. But, proceeding from this point of view and incorporating Special Relativity, it is possible to construct the entire theory of General  Relativity.&lt;/p&gt;&#xA;" OwnerUserId="4663" LastActivityDate="2011-08-10T23:44:58.670" />
  <row Id="13373" PostTypeId="1" AcceptedAnswerId="13394" CreationDate="2011-08-11T01:02:02.570" Score="5" ViewCount="47" Body="&lt;p&gt;When lightning strikes the ground in a sandy location it may fuze sand grains together to form a mineraloid known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Fulgurite&quot; rel=&quot;nofollow&quot;&gt;Fulgurite&lt;/a&gt;. Is there a physics explanation for the tubular shape? Why is it hollow as opposed to being solid? &lt;/p&gt;&#xA;" OwnerUserId="2843" LastActivityDate="2011-08-11T16:45:34.373" Title="Phyics Explanation for the Tubular Shape of Petrified Lightning" Tags="&lt;lightning&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13374" PostTypeId="1" CreationDate="2011-08-11T01:11:29.173" Score="2" ViewCount="185" Body="&lt;p&gt;I ran across &lt;a href=&quot;http://www.wbabin.net/science/anderton15.pdf&quot; rel=&quot;nofollow&quot;&gt;THIS&lt;/a&gt; Newtonian explanation for the bending of light. Does it have any merit? &lt;/p&gt;&#xA;" OwnerUserId="2843" LastEditorUserId="1353" LastEditorDisplayName="" LastEditDate="2011-08-13T08:49:08.933" LastActivityDate="2011-08-13T08:49:08.933" Title="Newtonian Bending of Light?" Tags="&lt;general-relativity&gt;&lt;gravity&gt;&lt;light&gt;&lt;newtonian-gravity&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="13376" PostTypeId="1" AcceptedAnswerId="13395" CreationDate="2011-08-11T03:18:17.763" Score="4" ViewCount="39" Body="&lt;p&gt;Again, a problem from exam preparation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[A] cell's membrane allows sodium ions to pass through it, but not chlorine ions. The cell is placed in a salty solution with a ten times higher concentration of salt outside than in the cell. As a result, sodium diffuses into the cell. Calculate the voltage difference across the cell membrane as a result.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, in equilibrium, the chemical potentials of the solution and the cell should be the same. Let $\mu_A$ be that of the solution and $\mu_B$ that of the cell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the salt is in a dilute solution, I can write its chemical potential as &#xA;$\mu_A = f(T,P) + kT \ln(n_A)$ where $f(T,P)$ is some function independent of the number of particles, and $n_A$ is the number-concentration of sodium ions in the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd assume that for the cell, I have &#xA;$\mu_B = f(T,P) + kT \ln(n_B) + U$ where now I include the voltage that exists across the cell's membrane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem is, now I have two unknowns: I don't know the final concentration of ions in the cell and I don't know the final voltage of ions in the cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alternative idea&lt;/strong&gt; but not sure if that makes sense: Osmotic pressure is defined as the pressure you must bring up to prevent osmosis from occurring, and it is also the pressure that finally builds up when you allow osmosis from happening. Could I analogously obtain the final voltage as the voltage that I'd have to apply manually to prevent any diffusion from happening at all?&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-08-11T16:55:28.137" Title="Voltage drop over a cell membrane" Tags="&lt;thermodynamics&gt;&lt;homework&gt;&lt;physical-chemistry&gt;&lt;biophysics&gt;&lt;chemical-potential&gt;" AnswerCount="1" />
  <row Id="13377" PostTypeId="2" ParentId="13374" CreationDate="2011-08-11T03:26:05.067" Score="4" ViewCount="" Body="&lt;p&gt;It's total nonsense. They claim this: &quot;Contrary to what many people believe, Newtonian theory gives the same value for light bending as General relativity.&quot; As a counterexample, GR allows exotic knife-edge trajectories for light moving in a Schwarzschild metric, where the light can come in, orbit n times, where n is any integer, and then fly off. In Newtonian gravity, this is impossible, because the only possible orbits are conic sections&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-11T03:26:05.067" CommentCount="4" />
  <row Id="13378" PostTypeId="2" ParentId="1030" CreationDate="2011-08-11T03:55:05.377" Score="0" ViewCount="" Body="&lt;p&gt;Personally I think the standard arguments are not yet conclusive. The horizon problem, or the homogeneity problem, can be explained by assuming that the initial condition is homogeneous, without assuming that causal contact in the early universe has smoothed out inhomogeneities. You may object that a homogeneous initial condition is &quot;unnatural&quot;, but since we know so little about the big bang singularity, there's nothing conclusive that can be said. The monopole problem is only a problem if you think monopoles exist, which has no empirical evidence so far. The curvature problem, again, is a &quot;naturalness&quot; problem, but we lack a precise definition of naturalness given our inability to understand the big bang singularity.&lt;/p&gt;&#xA;" OwnerUserId="2643" LastActivityDate="2011-08-11T03:55:05.377" />
  <row Id="13379" PostTypeId="2" ParentId="12200" CreationDate="2011-08-11T05:43:29.267" Score="2" ViewCount="" Body="&lt;p&gt;Actually there have been made some &lt;a href=&quot;http://en.wikipedia.org/wiki/Inductive_charging&quot; rel=&quot;nofollow&quot;&gt;wireless chargers&lt;/a&gt; where the energy is transferred by using electromagnetic field. Anyway the efficiency of such chargers is much lower than wired chargers due to following reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The antenna (remote charger) doesn't catch the whole electromagnetic field and therefore only a certain ammount of available energy is received&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The power decreases with distance.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The first issue could be solved with directional antenna (in this case in my opinion it is much easier to plug the cable directly into the power source than fixing the receiver - remote charger - into the optimal receiving point)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the decreasing field amplitudes the electromagnetic field consists of far field and near field. the near field dominates near the source and the far field extends from about two wavelengths distance from the antenna to infinity. The near field however decreases faster (1/d^3) whereas the far field decreases with 1/d^2. Due to these facts wireless communications exploit far field and the frequencies are high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To conclude the answer I would say &lt;strong&gt;although it is possible to transfer the energy wirelessly it is not affordable.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4883" LastEditorUserId="4883" LastEditorDisplayName="" LastEditDate="2011-08-12T07:10:55.943" LastActivityDate="2011-08-12T07:10:55.943" CommentCount="1" />
  <row Id="13380" PostTypeId="2" ParentId="6397" CreationDate="2011-08-11T05:58:39.667" Score="0" ViewCount="" Body="&lt;p&gt;To complete Ted Bunn's answer in most telecommunications applications the informations are transmitted by the far field which is present from about two wavelengths from the source to the infinity. In case of lower frequencies the communicating parties wouldn't be able to exchange data if they would be too close.&lt;/p&gt;&#xA;" OwnerUserId="4883" LastActivityDate="2011-08-11T05:58:39.667" />
  <row Id="13381" PostTypeId="2" ParentId="2069" CreationDate="2011-08-11T05:59:47.623" Score="1" ViewCount="" Body="&lt;p&gt;The orthodox Copenhagen interpretation is entirely unproblematic as long as your self is placed outside of the box. We can always choose to draw the box placing us outside. This is why it works in practice for practically all experiments and engineering work. Madness comes in when your try placing yourself inside the box. You either have to restrict the box, or contrive to place yourself outside the box one way or other. Many worlders try to describe the universe by an uncollapsed wave function, but only at the price of imaginatively abstracting themselves away from the universe, pretending that they're not really inside the universe, or that they don't exist.&lt;/p&gt;&#xA;" OwnerUserId="4884" LastActivityDate="2011-08-11T05:59:47.623" CommentCount="2" />
  <row Id="13382" PostTypeId="2" ParentId="6096" CreationDate="2011-08-11T06:36:10.370" Score="0" ViewCount="" Body="&lt;p&gt;Everettians are trying to pull wool over your eyes... If it only were that simple that our world keeps branching with time with a preferred basis, and we subjectively find ourselves in one branch. There are two major problems with this interpretation.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Different branches can and will recombine in the future. This is seldom emphasized, but this throws a wrench in the interpretation. In fact, for a system in thermal equilibrium, branching and recombination happens at an equal rate. It's only thanks to the fact that locally, we are out of equilibrium that one-way branching makes any approximate sense at all.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There's no canonical preferred basis in general, not even macroscopically. Decoherence works most of the time at macroscopic scales, but with many important exceptions. If decoherence were universal at macroscopic scales, do you think we'd be able to observe double slit experiments or superconductivity or quantum optics? Even more troubling is the fact that the basis to be chosen can depend contextually upon future decisions, as in the delayed choice experiment.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the many minds interpretation, subjectively fixing the conscious state of the observer still leaves most of the rest of the universe in an indeterminate superposition. Only those coarse-grained properties of the &quot;world out there&quot; corresponding to our internal conscious states will be determined by entanglement. Those Everettians who try to tell you subjective experiences are what causes the entire universe to split into many worlds are bluffing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the other worlds out there have some objective existence, how come we can't extract information from them, except in very special cases where we have a coherent variation in the phase and amplitudes between the many branches which then recombine? Not only that, after recombination, the separate worlds lose their separate identities. Besides, a coherent variation rules out the possibility of a complex intelligent observer, at least in the part of the wave function which varies coherently.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Do you really see?&lt;/p&gt;&#xA;" OwnerUserId="4886" LastActivityDate="2011-08-11T06:36:10.370" CommunityOwnedDate="2011-08-11T06:36:10.370" />
  <row Id="13383" PostTypeId="1" AcceptedAnswerId="13396" CreationDate="2011-08-11T06:59:27.580" Score="3" ViewCount="81" Body="&lt;p&gt;I am having a doubt about waveguides and the traditional Ethernet cable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The newest Ethernet i.e. the 10Gbps Ethernet has an astounding speed of 10Gbps. Which consists of 4 lanes in Cu as physical media. Each wire pair carries 3.125 G baud as signal!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that's enormous even for Cu cable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that somehow violate the waveguide principle, that the impedance of the carrier should be lesser than the impedance of the environment, and as the freq. of wave increases the impedance of carrier increases...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some light in this might be helpful...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards&#xA;Vineet&lt;/p&gt;&#xA;" OwnerUserId="4887" LastActivityDate="2011-08-12T02:48:38.757" Title="Waveguide and ethernet" Tags="&lt;waves&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="13384" PostTypeId="2" ParentId="13374" CreationDate="2011-08-11T08:33:36.780" Score="5" ViewCount="" Body="&lt;p&gt;As Ben says, it's nonsense. The most nonsensical claim is this: &quot;both Special relativity and Galilean relativity work.&quot; Though what the author has in mind is more specific than arbitrary light orbits, and he is not &lt;em&gt;completely&lt;/em&gt; wrong in seeing that the Newtonian calculation is hand-waving, but does not address the actual problems and certainly does not fix them. Usually, what is done in the Newtonian &quot;toy model&quot; exercise is to treat light as a (massive) corpuscle with velocity far from the Sun being $c$. Of course, that means that somewhere along the orbit, the light will have velocity higher than $c$, but there's no mechanism in Newtonian gravitation have a constant-speed unbound orbit, so that's a bit of a necessary evil. That's what is usually taken to be the Newtonian prediction for the deflection of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: the author explicitly says that by &quot;Newtonian theory&quot;, he means &quot;Newtonian theory obeying Galilean relativity&quot;. That's exactly what standard Newtonian mechanics is, so I feel confident if there's any misunderstanding, it is due to the author.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the author's mistakes is that he considers the path of the light ray trajectory to be parabolic just as it should be in Newtonian theory under a constant gravitational field. But for a parabolic orbit around a point, the speed far away from the gravitating body would be zero, &lt;em&gt;which obviously cannot be the case for light&lt;/em&gt;. On the other hand, the orbit far from the gravitating body should be straight, so in the far limit are lines. Thus, the orbit must be hyperbolic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first glance, one might think: aha, what if one takes the velocity somewhere else along the orbit to be $c$, say at Earth or closest distance to the Sun, or whatever? But that can't affect the situation too much, because if the corpuscle has velocity $v$~$c$, then its kinetic energy is so utterly dwarfs the gravitational potential energy due to the Sun, that it will simply be irrelevant just &lt;em&gt;where&lt;/em&gt; on the orbit the velocity is exactly $c$, because it will be extremely close to it everywhere on the orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can check this more explicitly. In the Newtonian assumption, the light corpuscle will have a hyperbolic orbit, and we can talk about the the angle $\theta$ between the asymptotes of that hyperbola. If the corpuscle is unaffected by gravity, then this angle be $\pi$ radians, so a sensible measure of deflection is its deviation from $\pi$; for convenience let's call this $2\psi$. The relevant relationship is:&#xA;$$e = 1+\frac{v_\infty^2R_p}{GM} = \frac{1}{\cos\frac{\theta}{2}} = \frac{1}{\sin\psi} \approx \frac{1}{\psi} + \frac{\psi}{6} + \ldots,$$&#xA;where $e$ is orbital eccentricity, $v_\infty$ is velocity at infinity, $R_p$ is the distance to gravitating body (Sun) at periapsis. The last bit is an approximation for small $\psi$. (Note also that a parabola would have eccentricity $e = 1$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have $v_\infty = c$ giving us some eccentricity and some $\psi$. Let's try velocity at periapsis $v_p = c$ instead, and compare the result. We know from conservation of mechanical energy that the velocity at $r = R_p$ and velocity at infinity are related by $\frac{1}{2}v_p^2 - \frac{GM}{R_p} = \frac{1}{2}v_\infty^2$, so plugging in $v_p=c$ gives:&#xA;$$e&amp;#39; = 1+\left(c^2 - 2\frac{GM}{R_p}\right)\frac{R_p}{GM} = e - 2.$$&#xA;Conclusion: taking the velocity at closest approach to the Sun to be $c$ rather than velocity at infinity hardly makes a difference at all, since in that regime $e$ is very large and $\psi$ is very small ($R_p$ ~ solar radius).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $e$ so large (~$10^5$ or higher), the addition of $1$ is similarly unimportant, thus we can also approximate the total deflection to be&#xA;$$2\psi \approx e^{-1} \approx \frac{2GM}{c^2R_p},$$&#xA;which is exactly what the author says is wrong. Thus, his claim that Newtonian gravity predicts $4GM/{c^2R_p}$ is simply not correct, because no matter how much one finagles with making the speed $c$ anywhere on the orbit, a factor of $2$ is much too large to be compensated for under the constraints of the Sun's mass and $R_p\geq\text{solar radius}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Even without the Newtonian toy models, there's a very straightforward and intuitive way that GTR's effect on light is twice that of Newton, and it can be seen clearly in the linearized approximation. Like for Newton, GTR's gravity couples to energy, which would be the relativistic mass (though that term is depreciated). But unlike it, GTR's gravity also couples to pressure, and for a light beam the pressure in the direction of motion is exactly equal to its energy, in units of $c = 1$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Addendum: A brief explanation of questions from the comments.&#xA;In linearized gravity, one begins with the assumption that the metric is a perturbation of Minkowski spacetime, imposes a gauge condition, and proceeds in analogy to the retarded potential of electromagnetism. Using approximations appropriate to a nearly-Newtonian regime (energy dominates stress, negligible retardation), the result applicable here is:&#xA;$$ds^2 = -\left(1+2\frac{\Phi}{c^2}\right)c^2dt^2 + \left(1-2\frac{\Phi}{c^2}\right)\underbrace{(dx^2+dy^2+dz^2)}_{dS^2},$$&#xA;where $\Phi$ is the Newtonian potential due to the gravitating source, and I'm calling the Euclidean distance element $dS^2$ for convenience. This is what some authors call the static, weak-field metric, and the derivation can be found in most GTR textboooks. Light travels along null geodesics, so $ds = 0$, giving&#xA;$$c\frac{dt}{dS} = \sqrt{\frac{1+2\Phi/c^2}{1-2\Phi/c^2}} \approx 1 + 2\frac{\Phi}{c^2}  + \mathcal{O}\left(\frac{\Phi^2}{c^4}\right).$$&#xA;Therefore, the effect of the gravitational field on light can be interpreted as follows: light travels &quot;as if&quot; it was in a medium with refractive index $n=c/v = 1+2\Phi/c^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How it is associated with the &quot;spatial curvature&quot; can also been derived from the above. What we want is the Newtonian $d^2S/dt^2 = -\nabla\Phi$, and to do that, we can explicitly extremize the proper time of an orbiting particle, and again make use of Taylor series approximations, keeping terms only to $\mathcal{O}((v/c)^2,\Phi/c^2)$:&#xA;$$\tau = \int dt\sqrt{\left(1+2\frac{\Phi}{c^2}\right) - \frac{1}{c^2}\left(1-2\frac{\Phi}{c^2}\right)\frac{dS^2}{dt^2}} \approx \int dt\left(1-\frac{1}{c^2}{\left(\frac{1}{2}\frac{dS^2}{dt^2} - \Phi\right)}\right),$$&#xA;which is exactly the action integral for Newtonian gravitation with Newtonian Lagrangian $\frac{1}{2}v^2 - \Phi$, except for the irrelevant addend of $1$ and rescaling by $c^2$ both of which drop out of the Euler-Lagrange equations to give the usual $d^2S/dt^2 = -\nabla\Phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interesting part in the above is this: since we're dropping the higher-order terms, the deviation from Euclidean $dS^2$ becomes &lt;em&gt;completely irrelevant&lt;/em&gt;--it only contributes terms that are dropped this Newtonian limit. Newtonian behavior can just as well be described by the much simpler metric&#xA;$$ds^2 = -\left(1+2\frac{\Phi}{c^2}\right)c^2dt^2 + \underbrace{(dx^2 + dy^2 + dz^2)}_{dS^2},$$&#xA;which produces the right Newtonian action integral in a more straightforward way. But of course light orbits in &lt;em&gt;this&lt;/em&gt; metric follow&#xA;$$c\frac{dt}{dS} = \sqrt{1+2\frac{\Phi}{c^2}} \approx 1 + \frac{\Phi}{c^2} + \mathcal{O}\left(\Phi^2/c^4\right),$$&#xA;in other words, light travels &quot;as if&quot; it was in a medium with refractive index $n = c/v = 1+\Phi/c^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without even calculating the exact amount of angular deviation, the effective indices of refraction make the following pretty clear: the effect of the spatial curvature is to &lt;em&gt;double&lt;/em&gt; the deviation of light it would be without it, and since Newtonian gravitation can be reproduced entirely by the purely temporal component of the metric, this is can be reasonably be interpreted as why GTR's effect is twice that of Newtonian gravity.&lt;/p&gt;&#xA;" OwnerUserId="1353" LastEditorUserId="1353" LastEditorDisplayName="" LastEditDate="2011-08-12T01:09:28.793" LastActivityDate="2011-08-12T01:09:28.793" CommentCount="8" />
  <row Id="13385" PostTypeId="1" CreationDate="2011-08-11T09:15:40.857" Score="2" ViewCount="94" Body="&lt;p&gt;It's been a while since school and I don't remember the right laws and formulas. So I come here. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got a power cord extender running from the wall outlet to my washing machine. It's about 3 or 4 meters in length. The outlet supplies 220V, but no grounding. The extender however does have grounding support, as does the washing machine. This means that the extender has three wires in it, and the grounding wire terminates at the outlet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When plug in one end of the outlet, and measure the AC voltage between the grounding (disconnected) and zero wires, it gives me about 60V. Which happens to be unpleasant enough when accidentally touching the washing machine (even when it's turned off). &lt;strong&gt;Considering that the cable is so short, can this really be explained by induction?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To rule out a few other theories: the wall outlet does not connect anything to the grounding wire, it really does terminate there. The grounding wire does not touch any of the other wires, the resistance between it and other wires is infinity (or rather, beyond the range of my cheap multimeter).&lt;/p&gt;&#xA;" OwnerUserId="3381" LastActivityDate="2011-08-11T20:58:10.127" Title="Practical physics: where does this current come from?" Tags="&lt;electromagnetism&gt;&lt;electricity&gt;&lt;inductance&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="13386" PostTypeId="2" ParentId="12488" CreationDate="2011-08-11T11:35:41.330" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Spontaneous_parametric_down-conversion&quot; rel=&quot;nofollow&quot;&gt;Spontaneous parametric down-conversion&lt;/a&gt; has the required properties of a decay $\gamma\to\gamma\gamma$:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... to split photons into pairs of photons that, in accordance with&#xA;  the law of conservation of energy, have combined energies and momenta&#xA;  equal to the energy and momentum of the original photon...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The process was independently discovered by two pairs of researchers&#xA;  in the late 1980s: Yanhua Shih and Carroll Alley, and Rupamanjari&#xA;  Ghosh and Leonard Mandel.&lt;br&gt;&#xA;  R. Ghosh and L. Mandel, &quot;Observation of Nonclassical Effects in the Interference of Two Photons&quot;, Phys. Rev. Lett. 59, 1903 (1987)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;IMO, can also be qualified as decays and I am aware that are other interpretations, the $\gamma\to e^+e^-$ with the help of a nucleus/neutron (a catalyst), and $\gamma\gamma\to e^+e^-$ in headon collisions are reported.&lt;br&gt;&#xA;Relevant links at &lt;a href=&quot;http://www.physics.princeton.edu/~mcdonald/&quot; rel=&quot;nofollow&quot;&gt;KIRK T. MCDONALD page&lt;/a&gt; and SLAC &lt;a href=&quot;http://www.slac.stanford.edu/exp/e144/e144.html&quot; rel=&quot;nofollow&quot;&gt;Experiment 144&lt;/a&gt; Home Page.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The gamma ray then collides with four or more laser photons to produce&#xA;  an electron-positron pair&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-08-11T22:57:33.160" LastActivityDate="2011-08-11T22:57:33.160" />
  <row Id="13387" PostTypeId="1" AcceptedAnswerId="13959" CreationDate="2011-08-11T11:44:26.947" Score="7" ViewCount="59" Body="&lt;p&gt;At Horns Rev windfarm off the coast of Denmark, sometimes in winter, clouds appears in the wake of the turbines. I've only seen photos of the phenomenon when the wind direction is &lt;strong&gt;exactly&lt;/strong&gt; aligned with the grid layout - that is, it's blowing directly from a turbine to its closest neighbour. That may be because it's most picturesque then (and thus most likely to be photographed); or it may be that there's something going in the fluid dynamics that requires that alignment for the phenomenon to occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/1a4MQ.png&quot; alt=&quot;Wakes at Horns Rev&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess there are several things at work here: that wake losses are highest when wind is exactly aligned with one axis of the turbine grid; that air temperatures vary with height above water; that the temperature is low enough to be close enough to form fog anyway (and in the photo, it looks like they're a layer of mist just above the sea's surface); that the turbine's wake is mixing air from different altitudes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;https://www.llnl.gov/news/newsreleases/2011/Apr/NR-11-04-06.html&quot; rel=&quot;nofollow&quot;&gt;study started in early 2011&lt;/a&gt; at Lawrence Livermore National Laboratory on turbine wakes, following on from a study by DONG energy on wakes at Horns Rev (&lt;a href=&quot;http://www.dongenergy.com/SiteCollectionDocuments/NEW%20Corporate/PDF/Engineering/40.pdf&quot; rel=&quot;nofollow&quot;&gt;787 kB pdf here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if it's possible to predict when the phenomenon in the photo here might occur. So my question is - what's the specific formulation of what's going on, here: what does the quantification of causes and effect look like?&lt;/p&gt;&#xA;" OwnerUserId="4066" LastActivityDate="2011-08-25T17:22:42.867" Title="What does the quantification of causes and effect look like, for clouds in offshore wind turbine wakes?" Tags="&lt;fluid-dynamics&gt;&lt;renewable-energy&gt;&lt;turbulence&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" />
  <row Id="13388" PostTypeId="1" AcceptedAnswerId="13389" CreationDate="2011-08-11T11:44:59.237" Score="1" ViewCount="103" Body="&lt;p&gt;Conceivably it expands with the speed of light. I do not know, but want to know if there is an answer.&lt;/p&gt;&#xA;" OwnerUserId="4867" LastEditorUserId="4867" LastEditorDisplayName="" LastEditDate="2011-08-11T19:40:18.363" LastActivityDate="2011-08-11T19:40:18.363" Title="At what speed does our universe expand?" Tags="&lt;speed-of-light&gt;&lt;universe&gt;&lt;speed&gt;&lt;space-expansion&gt;" AnswerCount="2" />
  <row Id="13389" PostTypeId="2" ParentId="13388" CreationDate="2011-08-11T12:03:01.177" Score="2" ViewCount="" Body="&lt;p&gt;The rate of expansion of the universe (the fact that all objects are receding from each other  and more so if they're farther away from each other) is given by the Hubble constant $H_0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;check out this plot from wikipedia&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/HbQxG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the y axis you have the velocity with which the object is receding from us and on the x axis the current distance in a common astrophysical unit called Megaparsec (parallax of one arcsecond $1\text{pc} \approx 3.26 light yrs).&lt;/p&gt;&#xA;" OwnerUserId="2151" LastActivityDate="2011-08-11T12:03:01.177" CommentCount="6" />
  <row Id="13390" PostTypeId="2" ParentId="13388" CreationDate="2011-08-11T12:22:54.913" Score="4" ViewCount="" Body="&lt;p&gt;The recessional velocity is given by Hubble's Law and it's proportional to the proper distance from the observer to the object (galaxy or other distant object):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v=H_0 D$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $H_0$ is Hubble's constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the recessional velocity beyond a certain distance, known as the&#xA;Hubble distance, can be much greater than the speed of light (in vacuum). This does not violate the relativity because the recession is not a movement through the space, it's the expansion of the space itself, between the objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details check the Wikipedia article: &lt;a href=&quot;http://en.wikipedia.org/wiki/Hubble%27s_law&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hubble%27s_law&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3066" LastEditorUserId="3066" LastEditorDisplayName="" LastEditDate="2011-08-11T12:52:48.950" LastActivityDate="2011-08-11T12:52:48.950" CommentCount="1" />
  <row Id="13392" PostTypeId="2" ParentId="13385" CreationDate="2011-08-11T14:36:12.133" Score="1" ViewCount="" Body="&lt;p&gt;The effect is mainly down to capacitive coupling between the wires. You have a capacitance $c_1$ between the live and ground wire; and a capacitance $c2$ between the ground and neutral wire. The $220v$ drives current around this series capacitive circuit, and the voltage you measured across $c_2$,  the neutral and ground, is found using the capacitive potential divider equation$$60 = 220\frac {c_1} {c_1 + c_2}$$&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="1345" LastEditorDisplayName="" LastEditDate="2011-08-11T20:58:10.127" LastActivityDate="2011-08-11T20:58:10.127" CommentCount="2" />
  <row Id="13393" PostTypeId="2" ParentId="166" CreationDate="2011-08-11T14:57:55.790" Score="4" ViewCount="" Body="&lt;p&gt;I find the idea of a breakthrough technology presented by the other answers to be extremely modest.  I'm almost positive that I'll get downvoted and flamed for my answer, but I have a strong desire to impress the potential for, and the implications of, a truly breakthrough energy technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to begin calling attention to Jevons Paradox, as well as the larger context of Jevons and other early commentators on early expanding coal power.  The fossil fuel age started with coal, and it started with steam cycles, not internal combustion engines.  Early coal engines were teeth-grindingly inefficient.  Here is an example, the Savery pump, which Jevons references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/aYvlg.gif&quot; alt=&quot;Savery pump&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/qMkA3.jpg&quot; alt=&quot;Modern steam turbine&quot;&gt;&#xA;&lt;em&gt;1700s versus now, continuous flow steam cycles were a breakthrough technology&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can probably see it already, but this is a batch process.  It used coal to pump water upwards but did so with less than 1% efficiency.  It had &lt;em&gt;extremely limited ability to replace labor&lt;/em&gt;.  When more efficient steam engines were developed, getting up to 10% efficiency, they could replace labor and do so very well.  Not only that, but those engines fueled a feedback where transportation, steel production, coal extraction itself, and a host of other economic inputs all became drastically cheaper, and led to greater demand for the fuel, coal itself.  The big picture, however, is that this lineage of efficiency improvement was a major catalyst for the entire industrial revolution.  Human population grew many-fold and this is pretty much the defining aspect of our current existence and challenges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have no such efficiency improvement available anymore.  We use thermal power with efficiencies of 30-60% and because of that, and the Carnot limit itself, no society-changing breakthrough will come from direct efficiency improvement.  &lt;em&gt;That means that energy breakthroughs need to come from opening up new resources&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;High altitude wind and solar&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are ventures currently pursuing high altitude wind with &lt;a href=&quot;http://www.makanipower.com/&quot; rel=&quot;nofollow&quot;&gt;wing designs&lt;/a&gt; and &lt;a href=&quot;http://www.kitegen.com/en/&quot; rel=&quot;nofollow&quot;&gt;kite designs&lt;/a&gt;.  There is enough resource to power the world with current 3-blade tower technology, but high altitude designs could make costs plummet and allow us to use far more of the resource.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll take a moment now to laugh at the idea of orbital solar power beamed back to the surface, haha.  Now, I'll mention that &lt;a href=&quot;http://www.stratosolar.com/index.html&quot; rel=&quot;nofollow&quot;&gt;tethered high-altitude solar&lt;/a&gt; is a viable option.  Solar has several problems today, including the difficulty of pointing the PV panels or collector into the direction of the sun.  A floating platform has so such challenge, and can also rise above cloud cover which limits isolation otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hot fusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tokomak is not a breakthrough technology.  Sorry.  I realize this is subjective, but if it works, then the cost of electricity from such plants will be 10-15 cents/ kWh.  Tokomak fusion will be more expensive than electricity today.  I don't know of a single person educated on this subject who argues otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, national labs are studying &lt;a href=&quot;https://lasers.llnl.gov/about/missions/energy_for_the_future/life/&quot; rel=&quot;nofollow&quot;&gt;laser inertial confinement&lt;/a&gt;, which could possibly be less expensive, although I doubt it would be breakthrough.  &lt;a href=&quot;http://www.generalfusion.com/&quot; rel=&quot;nofollow&quot;&gt;General Fusion&lt;/a&gt; is working on a mechanical acoustic approach that could be cheaper.  &lt;a href=&quot;http://www.polywellnuclearfusion.com/&quot; rel=&quot;nofollow&quot;&gt;Inertial Electrostatic Confinement (IEC), using the polywell&lt;/a&gt; would be revolutionary if it works.  You might have noticed a trend in the examples I'm giving.  These are all becoming progressively higher payoff and higher risk, and at long last I come to &lt;a href=&quot;http://focusfusion.org/&quot; rel=&quot;nofollow&quot;&gt;focus fusion&lt;/a&gt;.  Focus fusion is the most risky / highest payoff energy technology I know of that has some amount of decent science and credibility behind it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fission power&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://sites.google.com/site/rethinkingnuclearpower/aimhigh&quot; rel=&quot;nofollow&quot;&gt;Aim High!  LFTR reactors using Thorium&lt;/a&gt; is about the highest you can aim with fission power.  High risk and high payoff.  Small modular reactors, floating plant, and things like that can expand the use of our current Uranium resource which is not limited to the point of being economically prohibitive yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Biology advances&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jcvi.org/&quot; rel=&quot;nofollow&quot;&gt;Craig Venter&lt;/a&gt; leads possibly the most technologically advanced research project in the world.  This has already led to creating &quot;artificial life&quot; meaning that we have cells with a genome that was printed by a computer.  He was funded to the tune of $100s of millions by Exxon.  Algae biofuels are a powerful potential technology, but the breakthrough part won't be the physical bioreactor design (although there is room to improve refining processes), it will be the algae genome design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sum of low probability events&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take lots of technologies that could be breakthrough but have low probability of success, then the probability of ONE of the succeeding might be very significant.  &lt;a href=&quot;http://arpa-e.energy.gov/&quot; rel=&quot;nofollow&quot;&gt;ARPA-E&lt;/a&gt; was a federal program to use a competitive process to select the most likely breakthrough energy technologies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the question - many of these are rather close to being realized.  It differs from one to another.  Bad information, however, is quite rampant.  Everyone has their own personal opinion and the technologies listed here are presented as my own favorites unashamedly.  Things like zero-point, BlackLight Power, almost every LENR, orbital solar, and many others are hogwash.  There are too many credible potential breakthrough energy technologies to waste our times with the nonsense.  Too few people know how to disseminate between these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An energy &quot;breakthrough&quot; would be a loaded development.  We already use lots of energy, and if we found it economic to use more we probably would.  The last breakthrough shift in our ability to exploit energy resources rocketed the entire planet into a new geological era, the Anthropocene.  We called this change &lt;strong&gt;the industrial revolution&lt;/strong&gt;.  Some obscure project that ARPA-E funded with $500,000 could cause the &lt;em&gt;next&lt;/em&gt; industrial revolution.  The implications of such a change would probably be beyond any of our imaginations.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-08-11T15:03:54.610" LastActivityDate="2011-08-11T15:03:54.610" CommentCount="7" />
  <row Id="13394" PostTypeId="2" ParentId="13373" CreationDate="2011-08-11T16:45:34.373" Score="4" ViewCount="" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Lightning&quot; rel=&quot;nofollow&quot;&gt;lightning bolt&lt;/a&gt; produces hot gasses (30,000 °C). This results in very high pressures. This pushes away surrounding material. When the material cools, the empty region where the current passed is left empty for the same physical reasons that a piece of wood retains the imprint of a nail. Some of the sand is melted and solidifies before it can flow back into the hole.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-08-11T16:45:34.373" CommentCount="3" />
  <row Id="13395" PostTypeId="2" ParentId="13376" CreationDate="2011-08-11T16:55:28.137" Score="1" ViewCount="" Body="&lt;p&gt;You're basically making a battery calculation. Given these two solutions, what is the potential voltage difference between them? In this case, you should assume that the concentrations remain unchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, a very small number of sodium ions pass through the membrane. That movement of charge causes a voltage difference between the two sides that is sufficient to balance the flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you waited until the system reached complete stability, there would be no difference in potential and there would be no potential voltage. Since the membrane allows the passage of only one of the charge carriers, you would have to provide another path to get the other charges around. I.e. you would have to short circuit the cell.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-08-11T16:55:28.137" CommentCount="2" />
  <row Id="13396" PostTypeId="2" ParentId="13383" CreationDate="2011-08-11T17:05:42.490" Score="4" ViewCount="" Body="&lt;p&gt;Copper conducts to extremely high frequencies. What you're probably thinking about is &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Skin_depth&quot; rel=&quot;nofollow&quot;&gt;skin effect&lt;/a&gt;&quot; which does depend on frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As frequencies become higher and higher, the current-carrying thickness in a conductor decreases. This means that the resistance of a copper wire depends on the frequency. As you note, very high frequencies imply that this resistance is raised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as far as sending signals down a wire, the only problem with high resistance is that, for the same voltage, one must use less current.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-08-11T17:05:42.490" CommentCount="4" />
  <row Id="13397" PostTypeId="2" ParentId="7905" CreationDate="2011-08-11T18:06:56.197" Score="0" ViewCount="" Body="&lt;p&gt;From $\gamma$ &amp;lt;-&gt; $e^+e^-$ I can consider that the EM field is a charge separation that propagate side by side. We measure a net charge 0 until we make a way to separate the charges in a pair.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a possibility that if we have charge &lt;em&gt;after&lt;/em&gt; (in $e^+e^-$ pair)&lt;br&gt;&#xA;then &lt;strong&gt;we must have it also &lt;em&gt;before&lt;/em&gt; (in $\gamma$)&lt;/strong&gt; in such a way we can not detect it. A non detection is not the same as 'non existent'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AFAIK the standard interpretation is not like this one I post and you must forget this if you are a regular student. &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-08-11T18:06:56.197" />
  <row Id="13399" PostTypeId="1" AcceptedAnswerId="13401" CreationDate="2011-08-11T18:34:17.640" Score="3" ViewCount="23" Body="&lt;p&gt;In the classic experimental physics text &quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B0000CKKGJ&quot; rel=&quot;nofollow&quot;&gt;Statistical Theory of Signal Detection&quot; by Carl. W. Helstrom&lt;/a&gt;, Chapter II, section 4 concerns Gaussian Stochastic Processes. Such a process is observed at times $t_1, t_2, t_3, ... t_n$ to obtain $n$ random variables $x_1, x_2, x_3, ... x_n$, then the probability density for the $n$ variables is of the form:&lt;br&gt;&#xA;$$p_n(x_1,t_1; x_2,t_2; ...; x_n,t_n) = M_n\;\exp(-0.5\Sigma_j\Sigma_k\mu_{jk}x_jx_k).$$&lt;br&gt;&#xA;where $\mu_{jk}$ is a positive definite matrix and $M_n$ is a normalization constant to give unit probability when you integrate over all possible values:&lt;br&gt;&#xA;$$M_n = (2\pi)^{-n/2}\;|\;\textrm{det}\;\mu|^{0.5},$$&#xA;where $\mu$ is the determinant of the matrix $\mu_{jk}$. We assume that the expected values are all zero: $E(x_k)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He notes that the expected value of the product of any odd number of $x_j$ is zero (which seems to follow from symmetry), and gives a formula for the expected value of a product of even numbers of variables. We define $\phi_{jk}$ as:&lt;br&gt;&#xA;$$\phi_{jk} = E(x_jx_k).$$&#xA;He then notes that:&lt;br&gt;&#xA;$$E(x_1x_2x_3x_4) = \phi_{12}\phi_{34} + \phi_{13}\phi_{24} + \phi_{14}\phi_{23}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a simple proof? And is there a simple proof that relates to the methods of quantum mechanics / quantum field theory?&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-08-11T19:47:22.290" Title="Products of Gaussian stochastic process variables" Tags="&lt;quantum-mechanics&gt;&lt;signal-processing&gt;&lt;statistics&gt;&lt;probability&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13400" PostTypeId="1" AcceptedAnswerId="13406" CreationDate="2011-08-11T18:39:44.507" Score="3" ViewCount="49" Body="&lt;p&gt;What waveform does the light coming off different types of artificial light sources have and (of course) what are the underlying physical reason for them?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I was inspired to ask this question because of several things I have been hearing lately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, there was a recent TED talk about how &lt;a href=&quot;http://blog.ted.com/2011/08/02/wireless-data-from-every-light-bulb-harald-haas-on-ted-com/&quot; rel=&quot;nofollow&quot;&gt;LED bulbs can be used to transmit data&lt;/a&gt;.  Researchers have been working on the problem for a little while now and they've made a fairly impressive product, but it was only recently made possible by the LED technology, which has such a fast response that it can easily compare to the transfer rates of electric information condiuits that we normally use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that incandescent bulbs flicker because they have 60 Hz AC power being pulsed through them.  In fact, I think 60 Hz was chosen in the first place because it was a threshold that was unnoticeable to humans.  Looking at a &lt;a href=&quot;http://www.youtube.com/watch?v=9HUrYoqxQpw&quot; rel=&quot;nofollow&quot;&gt;high speed youtube video of such a bulb&lt;/a&gt; demonstrates that it has a very significant lag time responding to both the normal electrical phasor and being turned off.  In other words, it flickers but it doesn't go dark at any point in the AC waveform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking for more information, however, I was a little disappointed, since nothing seemed to give a graph of lumens emitted over time for all different kinds of bulbs.  Of course, there isn't just one relevant time scale.  A light could flicker with 60 Hz and also flicker with 2000 Hz (I just made that up).  I'm interested in the complete picture of what lights exhibit what patterns.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-08-11T19:17:39.783" LastActivityDate="2011-08-11T21:52:54.020" Title="What invisible flicker do different types of light bulbs have?" Tags="&lt;light&gt;" AnswerCount="2" />
  <row Id="13401" PostTypeId="2" ParentId="13399" CreationDate="2011-08-11T19:35:19.000" Score="3" ViewCount="" Body="&lt;p&gt;Yes, it is simple to prove using moment generating functions.  And yes, the mathematics is very closely related to that of quantum field theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You compute $G(j) = &amp;lt;exp(\sum j_i x_i)&amp;gt;$ where each $j_i$ is a &quot;source&quot; for the corresponding $x_i$.  This is easily shown to be something like $G(j) = exp(\sum j_i \mu_{ij}^{-1} j_j)$ To get expectation values you then take $ &amp;lt;x_i x_j ...&amp;gt; = \frac{\partial}{\partial j_i} \frac{\partial}{\partial j_j} ... G(j)|_{j=0}$ .  The rest follows simply. In particular, you can see how the variables must be grouped in pairs to get a nonzero results when you set $j=0$ after taking derivatives. You essentially have a Feynman expansion of a non-interacting 0-dimensional field theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is covered nicely in Zee's &quot;Quantum Field Theory in a Nutshell&quot;, where  it is a simple application of what he calls the &quot;central identity of quantum field theory&quot;&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditorDisplayName="" LastEditDate="2011-08-11T19:47:22.290" LastActivityDate="2011-08-11T19:47:22.290" />
  <row Id="13402" PostTypeId="2" ParentId="13400" CreationDate="2011-08-11T19:42:14.870" Score="1" ViewCount="" Body="&lt;p&gt;Incandescent light bulbs flicker at twice the AC frequency, because the filament grows a bit hotter each time the current peaks. So: 50 Hz AC =&gt; 100 Hz flicker; 60 Hz AC =&gt; 120 Hz flicker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fluorescent bulbs do the same, for the same reason. Their flicker has a higher amplitude than incandescent bulbs, because there's no thermal inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LEDs... I don't know. I'm fairly sure there must be at least some residual flicker, even if internally they are being fed DC, because that DC is never going to be 100% filtered. But whether any filtering occurs, and if so then how much, I don't know. Or maybe they have some kind of switching invertor and their flicker is at a much higher frequency than the AC mains.&lt;/p&gt;&#xA;" OwnerUserId="4210" LastActivityDate="2011-08-11T19:42:14.870" />
  <row Id="13403" PostTypeId="1" CreationDate="2011-08-11T19:44:28.673" Score="5" ViewCount="45" Body="&lt;p&gt;I am experimenting and playing around with some data, and I'm having trouble seeing how to generate invariant mass plots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data I have has a bunch of events, and variables such as $P,P_T,\eta,\phi$ etc.. but no energy. There is a histogram which generates the invariant mass, but no macro provided for this. For example, I know there are tracks for electrons and protons (TPC), and I want to see if there are any pairs as a result of $\gamma\rightarrow e^+ e^-$. Conceptually I need to loop over all positive negative pairs, and plot the histogram. I should get a peak around zero for $\gamma$ pair production. The invariant mass formula is $$m_\gamma^2  = (E_{e^+}+E_{e^-})^2 - (\vec{p}_{e^+}+\vec{p}_{e^-})^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is no data for the energy, how do the high energy physicists do it? Do they guess some value for the energy and make plots with that guess?&lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-12T16:17:31.857" LastActivityDate="2011-08-12T17:25:11.797" Title="How are the HEP experiments' invariant mass plots generated?" Tags="&lt;experimental-physics&gt;&lt;high-energy-physics&gt;&lt;data-analysis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="13404" PostTypeId="1" AcceptedAnswerId="13405" CreationDate="2011-08-11T20:04:33.547" Score="2" ViewCount="50" Body="&lt;p&gt;Suppose we are given two conducting, cocentric spheres of radius $a_1$ and $a_2$ respectively. The inner sphere with charge $q$, the outer sphere with charge $-q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can calculate the capacitance of this system by calculating the potential difference $U$ between the plates and then use the definition $C = q / U$. This is easy enough and indeed gives me the right result&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C = \dfrac{4\pi \epsilon_0}{a_1^{-1} - a_2^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now I tried to caluclate $C$ by finding out the energy stored in the system in two different ways:&#xA;First we have the expression&#xA;$$W_{el} = \frac12 \frac {q^2}C$$&#xA;for the energy stored in any capacitor. Now I wanted to compare this to the energy stored in the electric field (since that's where the energy is, right?!) $$E(r) = \begin{cases}\frac1{4\pi \epsilon_0} \frac q{r^2} &amp;amp; a_1 &amp;lt; r &amp;lt; a_2 \\ 0 &amp;amp; \text{otherwise}\end{cases}$$&#xA;to derive the above formula for $C$ again. The energy density is $w_{el} = \frac {\epsilon_0} 2 E^2$, therefore the energy stored in the electric field is &#xA;\begin{eqnarray*}&#xA;W_{el} &amp;amp;=&amp;amp; \int w_{el} \;\mathrm{d}V \\&#xA;&amp;amp;=&amp;amp; 4\pi \int_{a_1}^{a_2} \frac{\epsilon_0}2 \left(\frac1{4\pi \epsilon_0} \frac q{r^2}\right)^2 r \; \mathrm dr \\&#xA;&amp;amp;=&amp;amp; \frac1{4\pi \epsilon_0}\frac{q^2}{2} \int_{a_1}^{a_2} \frac{\mathrm d r}{r^3} \\&#xA;&amp;amp;=&amp;amp; \frac1{4\pi \epsilon_0}\frac{q^2}{2} \frac14\left(\frac 1{a_1^4} - \frac1{a_2^4} \right)&#xA;\end{eqnarray*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing the two expressions gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C = \dfrac{16\pi \epsilon_0}{a_1^{-4} - a_2^{-4}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: Why is this wrong? Where is the energy in this capacitor stored, if not in the electric field (as it doesn't seem to be - unless I have made a mistake in deriving the result somewhere...)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I also noticed, that the second result can be rewritten as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C = \dfrac{4\pi \epsilon_0}{a_1^{-1} - a_2^{-1}}\frac 4{a_1^{-3} + a_1^{-2}a_2^{-1} + a_1^{-1}a_2^{-2} + a_2^{-3}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I don't know whether this has any significance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for reading and any help will be greatly appreciated! :)&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-11T21:44:31.940" LastActivityDate="2011-08-11T21:44:31.940" Title="Capacitance of two cocentric spheres, contradicting results" Tags="&lt;electrostatics&gt;&lt;capacitance&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13405" PostTypeId="2" ParentId="13404" CreationDate="2011-08-11T20:15:29.383" Score="2" ViewCount="" Body="&lt;p&gt;Two things stand out to my eye:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) In your volume integral, when you switched to spherical coordinates, you put in a $4 \pi r \ dr$ where there should be a $4 \pi r^2 dr$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Your evaluation of the 1-D integral is not quite right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think these two are enough to explain the discrepancy.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-08-11T20:15:29.383" CommentCount="1" />
  <row Id="13406" PostTypeId="2" ParentId="13400" CreationDate="2011-08-11T20:43:43.123" Score="2" ViewCount="" Body="&lt;p&gt;Old fashioned (or very cheap) fluorescent bulbs flicker at twice the mains frequency if they use magnetic ballasts driven directly from the AC. &#xA;Modern fluorescent bulbs use an electric ballast that runs at 5khz - 40khz (for CFLS), then the electrons stay excited for longer than the switching frequency and the light output is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LEDs are constant, unless you count the individual electrons moving over the diode junction!&#xA;High power LEDs normally have their own driver chip which carefully controls the current and voltage of the diode so they are less sensitive to the supply power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solid state lasers are essentially the same as LEDs.&lt;/p&gt;&#xA;" OwnerUserId="2525" LastEditorUserId="2525" LastEditorDisplayName="" LastEditDate="2011-08-11T21:52:54.020" LastActivityDate="2011-08-11T21:52:54.020" CommentCount="1" />
  <row Id="13407" PostTypeId="2" ParentId="13353" CreationDate="2011-08-11T20:53:41.803" Score="0" ViewCount="" Body="&lt;p&gt;There are two parts to the question - how regular are the clocks (is a second always the same) and how close is the reading on the clock to the official time now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good digital clock, with reasonably constant temperature is probably good to 1sec/day. That is a second has the same value to within 1/(24*60*60) = 0.001%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An atomic clock can be accurate to 1second in millions of years. Fortunately there are 24 expensive atomic clocks flying over your head broadcasting the time to your GPS receiver. Most internet service providers and cell phone masts also use GPS as the time reference so the time on your cell phone and your computer (if you set auto time sync) are very 'accurate'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is the problem of the delay between a distant clock knowing that it it '12:00' and getting the message to you.  The GPS receiver is the best - the travel time for the signal from the satelite to you is a few millisecs. &#xA;For your computer it may take 10s of millisecs for the time message to go over the network and so the system has tricks to try and estimate this and correct for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radio used to be the standard for time signals. At the start of the news the radio would play a series of beeps (the pips) that accurately gave you the start of the hour. Now that the radio signal is digitised, sent over a computer network to the transmitter, recorded by the computer in your radio, decompressed, and then played - there may be a delay of several seconds, so most broadcasters no longer send the time signal&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-08-11T20:53:41.803" />
  <row Id="13408" PostTypeId="1" AcceptedAnswerId="13412" CreationDate="2011-08-11T21:41:54.260" Score="3" ViewCount="106" Body="&lt;p&gt;I am trying to do a crude particle identification, using a Bethe Bloch tenchnique. Here is a plot I made from the data that I have &#xA;&lt;img src=&quot;http://i.stack.imgur.com/SeyLv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I've read, the standard method to identify charged particles is by measuring the ionization for a given momentum, in what is known as the Bethe-Bloch Equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle \frac{dE}{dx}\rangle = \frac{P_1}{\beta^{P_4}}\left( P_2 - \beta^{P_4} - \log \left(P_3+\frac{1}{(\beta\gamma)^{P_5}}\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I tried to fit it and found out the parameters that are used by the detector people (ALICE, in their documentation, they write that the parameters were found using some simulation program). So these parameters become given constants. Now the entire Bethe-Bloch curves are normalized to the  values of the minimum ionizing particle. In the literature, this is referred to as MIP value. So I tried a one-parameter fit for the above curve, leaving ROOT to find out the value of the normalization constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I get absurd results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/u0tk6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three &quot;fits&quot; in the figure are supposed to be for pions, electrons and protons. However, they are not really accurate, as you can compare this to any of the standard plots released by CERN. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, this is a plot from STAR experiment, which obviously has much lower multiplicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/CYv56.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the techniques used by practitioners to find this normalization constant. I am using the same platform but the fitting routine is clearly inadequate for such magnitude of data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I do not have access to any online data or grid, etc. Most of the analysis code on the web seem to be based on somethings called ESDs and AODs and not for local standalone analysis (which is understandable but undfortunate for me)&lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-12T01:34:56.387" LastActivityDate="2011-08-12T17:42:00.747" Title="Fitting to a high density scatter plot" Tags="&lt;particle-physics&gt;&lt;experimental-physics&gt;&lt;data-analysis&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="13409" PostTypeId="2" ParentId="13343" CreationDate="2011-08-11T22:48:59.340" Score="1" ViewCount="" Body="&lt;p&gt;Some obscure thermal conductivity effect might heat the air inside to double the temperature, thereby doubling its pressure, but if the air expands and the pita doesn't let new gas in, it will not be able to expand more than a factor of 2 before the pressure is the same as outside. The expansion observed is a factor of a ten or more, so it's gas production.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that the explanation is production of steam, and the critical temperature is close to the boiling point of water, give or take the effects of solutes in water trapped in the pita. Steam production gives as much expansion as you want, and also, when I cut the pita open, I notice that it is full of steam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I don't think its such a terrible question either)&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-11T22:48:59.340" CommentCount="2" />
  <row Id="13411" PostTypeId="2" ParentId="9165" CreationDate="2011-08-11T23:53:29.430" Score="0" ViewCount="" Body="&lt;p&gt;Assuming that you have studied introductory mechanics at the level of freshman physics, The best book on mechanics after that to tackle is &#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/189138922X&quot; rel=&quot;nofollow&quot;&gt;Classical Mechanics by John R. Taylor&lt;/a&gt;. It is very clear and insightful. The author is a very good writer, he has written also the best introductory book on &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/093570275X&quot; rel=&quot;nofollow&quot;&gt;error analysis&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4521" LastActivityDate="2011-08-11T23:53:29.430" CommunityOwnedDate="2011-08-12T00:20:14.823" />
  <row Id="13412" PostTypeId="2" ParentId="13408" CreationDate="2011-08-12T00:16:54.237" Score="3" ViewCount="" Body="&lt;p&gt;The data set you show is a jumble of multiple contributions (and the cern fits agree, see how many times those lines intersect and how they all lie close together in the horizontal band on the right side of the plot?), and that makes for a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;You need some way to disentangle these bits.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing I would tell a grad student to try is to separate the data-set on the basis of some other variables and fit each line to a sub-set of the data that she was confident contained mostly the particle they were interested in. This would generally be done by taking advantage of &lt;em&gt;other&lt;/em&gt; data recorded about each hit in the plot{*}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no additional data was available I would suggest using geometric cuts in the $dE/dx$--$p$ plane to select the data to fit. That is fit to the parts of the plot which identifiable belong to a single particle species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are ambitious you could use a track density-and-width-model to sequentially subtract the results of each fit you get in order to make the next one easier (at the cost of making the uncertainty analysis high correlated---yuck!).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Additional options:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Start with a model for a particular particle and impose strict limits on the fitting parameters.&lt;/li&gt;&#xA;&lt;li&gt;Consider the more fitting and log-likelyhood options to the fitter. &lt;/li&gt;&#xA;&lt;li&gt;Also consider giving an artificially high weight to the data in the regions where the bands are well separated (the is an alternative to the geometric cuts).&lt;/li&gt;&#xA;&lt;li&gt;If your Monte Carlo is &lt;em&gt;good&lt;/em&gt; fit the lines to MC and simply plot them on the data.. (This is not an uncommon thing to do; does the caption on the cern data say where the lines come from?)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;{*} Possibly something as simple as that plot representing the overlay of multiple particle-specific data-sets, or by taking advantage of PID detector systems, or particle species identifications arising from event topology. Frankly you don't care what you get and you don't care if you lose a lot of the data in the selection (statistics are &lt;em&gt;not&lt;/em&gt; your problem when you have 7 million measurements).&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-12T17:42:00.747" LastActivityDate="2011-08-12T17:42:00.747" />
  <row Id="13413" PostTypeId="1" AcceptedAnswerId="13478" CreationDate="2011-08-12T01:27:56.237" Score="0" ViewCount="60" Body="&lt;p&gt;My friend and I were comparing the rightness of a cricket bat versus a baseball bat for general violence purposes. (Eg, rioting.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where are the centers of mass for these two instruments? Which would be better for an average adult human male to, say, smash in a window or other such similar action? Would the answer be different for an average adult human female?&lt;/p&gt;&#xA;" OwnerUserId="4897" LastActivityDate="2011-08-13T11:39:49.577" Title="Where are the centers of mass for a baseball bat vs. a cricket bat?" Tags="&lt;forces&gt;&lt;angular-momentum&gt;&lt;mass&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="13414" PostTypeId="2" ParentId="12922" CreationDate="2011-08-12T02:21:29.680" Score="3" ViewCount="" Body="&lt;p&gt;Observe the following: the Einstein-Hilbert stress-energy is formally given by the first variation of the matter Lagrangian density relative to the inverse metric:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Q = \frac{\delta L_\phi \mathrm{dvol}_g}{\delta g^{-1}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $L_\phi = g^{-1}(\nabla\phi,\nabla\phi)$ is the Lagrangian function for the scalar field. Further observe that $\mathcal{L}_X\mathrm{Id} = 0$, where $\mathrm{Id}:TM\to TM$ is the identity map and $\mathcal{L}_X$ is the Lie derivative. So $\mathcal{L}_X (g^{-1} g) = (\mathcal{L}_Xg^{-1})g + g^{-1}\pi = 0$. Hence $\mathcal{L}_X g^{-1} = -g^{-1}\pi g^{-1}$. This means that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ c(Q g^{-1} \pi g^{-1}) = -c(Q \mathcal{L}_Xg^{-1}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c(\cdot)$ maps 1,1-tensors to scalars is the contraction operation. Next, recall that the Lie derivative can be defined by the one parameter family of diffeomorphisms generated by a vector field. More precisely, let $\Psi_t$ be the one parameter family of diffeomorphisms generated by $X$. Then we have a one parameter family of tensor fields $G_t := \Psi_t^* g^{-1}$ defined by the pushforward of the inverse metric. For this family we have that $\mathcal{L}_Xg^{-1} = \frac{d}{dt}G_t |_{t=0}$. Holding the scalar field $\phi$ fixed, we write $\hat{L}_t = G_t(\nabla\phi,\nabla\phi) \mathrm{dvol}_{(G_t)^{-1}}$ for the corresponding one parameter family of Lagrangian densities, then we have that the chain rule gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d}{dt} \hat{L}_t |_{t = 0} = c(Q\mathcal{L}_Xg^{-1}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we in fact can split $$\mathcal{L}_X (L_\phi\mathrm{dvol}_g) = \frac{d}{dt}\hat{L}_t |_{t=0} + 2 g^{-1}(d\phi, d\mathcal{L}_X\phi)  \mathrm{dvol}_g$$ into the &quot;geometric/gravity&quot; part and the &quot;matter&quot; part using that the exterior derivative commutes with the Lie derivative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the scalar quantity you wrote down is the &lt;em&gt;portion of the change of the matter Lagrangian density in the direction of the vector field $X$ that is caused by the change of the (inverse) metric in that direction&lt;/em&gt;. In you are willing to consider &lt;a href=&quot;http://physics.stackexchange.com/questions/9686/the-meaning-of-action&quot;&gt;the Lagrangian density as a physical quantity&lt;/a&gt;, this gives a &quot;physical&quot; meaning to your scalar quantity as the &quot;geometric/gravity&quot; portion of the flow of the Lagrangian density under the vector field $X$. In a not very precise sense, if the vector field $X$ is time-like, you can associate to it a time-like congruence. So this scalar measures the infinitesimal (fictitious) work &quot;done by gravity/acceleration of frame&quot;. (I use the word fictitious here to indicate that the $X$ dependence of this scalar is somewhat analogous to the frame-dependence of the &quot;centrifugal force&quot; in classical mechanics.)&lt;/p&gt;&#xA;" OwnerUserId="829" LastActivityDate="2011-08-12T02:21:29.680" />
  <row Id="13415" PostTypeId="2" ParentId="13359" CreationDate="2011-08-12T02:26:16.517" Score="2" ViewCount="" Body="&lt;p&gt;In special relativity, unaccelerated observers cannot do any experiment to determine their state of motion in any absolute sense--all meaningful motion is relative to something else. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We want to generalize the principle of relativity, that laws of physics are independent of inertial motion, as far as possible.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Mathematically, this suggests using tensors, to make equations covariant with nigh-arbitrary coordinate transformations. But let's begin with the Eötvös experiment:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If under the influence of only gravity, every test particle behaves in an identical manner, no matter its mass or composition.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is suggestive of making gravity a property &lt;em&gt;of&lt;/em&gt; spacetime, rather than an interaction &lt;em&gt;within&lt;/em&gt; spacetime. But even more important is that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Objects in freefall don't feel their own weight.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In Newtonian mechanics, if you have two test masses near each other in a gravitational potential $\Phi$, they will accelerate in the same way, to first order. They will be at rest relative to each other. Of course, they are still at different positions, so experience different gravitational force, but this shows up as the &lt;em&gt;tidal force&lt;/em&gt; proportional to $\partial^2\Phi/\partial x^i\partial x^j$, a second-order effect. Intuitively, gravity &quot;doesn't exist&quot; for infinitesimal separations $dx$ (and times $dt$): you need to look  at larger scale, in space or time or both, to detect its presence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, over small enough regions of space and time i.e., experiments of short duration), physics is just as if you're an in an inertial frame. But since special relativity, not Galilean relativity of Newtonian mechanics is right,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Spacetime is locally Minkowski, the flat spacetime of STR.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Geometrically, things fall into place: in mathematics, a manifold is a structure that locally &quot;looks like&quot; flat Euclidean space. So both criteria--gravitational effects independent of mass or composition and physics behaving like special relativity over small enough regions can be done simply by having spacetime be a manifold that's locally Minkowski.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geodesic deviation is also a second-order effect: that's the geometrical meaning of the Riemann curvature tensor, which is determined only by the second derivatives of the metric. (But Einstein didn't think of curvature as the actual gravitational field; more on that later.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... my little box is pushed over a cliff and is in freefall. Presumably my box has now become a locally inertial reference frame, its spacetime is now flat and therefore described by the Minkowski metric.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's right, to first order. To see what's going on more precisely, it may help to explicitly build coordinates out of this. Say you have a local frame of one timelike direction and three spacelike directions $\{\hat{e}_\alpha\}$, orthonormal (that's actually not necessary, but it's more intuitive if you start with a Cartesian-like situation), i.e., four vectors spanning the tangent space at your location in spacetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick any direction in spacetime... any vector can be written in terms of components of your frame vectors. Consider the geodesic in that direction; it will have some affine parameter along it (say, proper time/length along it if it wasn't a lightlike direction). You can therefore label each point on that geodesic with coordinates consisting of that affine parameter and the vector components of the direction you picked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've constructed &lt;em&gt;Riemann normal coordinates&lt;/em&gt; corresponding to your local frame. The metric evaluated at your location (!!) is just Minkowski $\eta_{\mu\nu} = \mathrm{diag}(-1,+1,+1,+1)$, as it corresponds to the dot products of the frame vectors. And since the coordinates were defined in terms of geodesics and geodesic deviation is second-order, the metric in a &lt;em&gt;neighborhood&lt;/em&gt; of your location is Minkowski to first order.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm still on planet Earth, but the spacetime has changed. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, we've basically just decided to view the metric in coordinates $x^\mu$ such that at a particular event $p$, $g_{\alpha\beta}|_p = \eta_{\alpha\beta}$ &lt;em&gt;and&lt;/em&gt; $\partial g_{\alpha\beta}/\partial x^\gamma = 0$. The metric is Minkowski at $p$ and its derivatives vanish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't mean it's exactly Minkowski on a region around $p$--there will be a difference in second order. We can't make the second derivatives vanish in general because there's not enough degrees of freedom in a smooth coordinate transformation. But since the second-order derivatives are what determines curvature = geodesic deviation = gravitational tidal forces, we don't want those to go away.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Einstein came to the conclusion that gravity is given by a geometry of spacetime and free-falling objects are locally inertial. But from a modern perspective, we can use the latter to motivate GTR as well:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If objects in freefall have no proper acceleration (what an accelerometer measures), could they be actually be &lt;em&gt;inertial&lt;/em&gt;?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;On a manifold, the geodesics are curves that have the following property: transporting a tangent vector &lt;em&gt;along the geodesic itself&lt;/em&gt; gives a parallel vector. In more physical terms, the velocity of a particle traveling along it is &lt;em&gt;unchanged&lt;/em&gt; as it follows the geodesic. Geodesics naturally generalize the notion of &quot;inertial motion = no acceleration&quot;. Of course, that's exactly what Einstein concluded, but the mathematics explicating these various relationships between the metric and the connection were published a few years after GTR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's interesting that Einstein did &lt;em&gt;not&lt;/em&gt; think of the spacetime curvature as corresponding to the presence of a gravitational field. Rather, he thought of the connection coefficients the geodesic equation&#xA;$$\frac{d^2x^\alpha}{d\lambda^2} + \Gamma^\alpha_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda} = 0$$&#xA;as the components of the gravitational field. Some modern authors don't like this because $\Gamma^\alpha{}_{\mu\nu}$ are not tensorial, but conceptually Einstein's view makes more sense: GTR marries &lt;em&gt;gravitation&lt;/em&gt; and &lt;em&gt;inertia&lt;/em&gt;, and with the above notion of &quot;inertial&quot;, the connection is exactly what determines parallel transport and hence inertial motion.&lt;/p&gt;&#xA;" OwnerUserId="1353" LastEditorUserId="1353" LastEditorDisplayName="" LastEditDate="2011-08-13T19:55:48.447" LastActivityDate="2011-08-13T19:55:48.447" CommentCount="5" />
  <row Id="13416" PostTypeId="1" AcceptedAnswerId="13421" CreationDate="2011-08-12T02:37:24.843" Score="4" ViewCount="81" Body="&lt;p&gt;I was recently told by a physics teacher that &quot;any system of charges in which at least some of the charges are executing some sort of accelerated motion, will radiate and lose energy&quot;. This refers to classical electrodynamics (I'm not entering into quantum mechanics), that is, this statement should either be provable or disprovable from Maxwell's equations. I tried to think of a simple system of charges that would disproof this statement, since I intuitively think that it isn't true,  but I haven't come up with any. The one example I could think of was a positive charge sitting on top of a negative charge, so that the net charge is zero. No matter how they move it is obvious that there will be no radiation. But my teachers did not accept this trivial example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is whether the statement is true or not, plus a proof or a counterexample of some charges executing accelerated motion without radiating. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3064" LastActivityDate="2011-08-12T03:25:03.283" Title="Is it true that any system of accelerating charges will radiate?" Tags="&lt;electromagnetic-radiation&gt;&lt;classical-electrodynamics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="4" />
  <row Id="13417" PostTypeId="2" ParentId="5819" CreationDate="2011-08-12T02:37:43.390" Score="1" ViewCount="" Body="&lt;p&gt;I think the intuitive sense (your 'intuitive gut feeling'mentioned above)that one develops for physics starts with a passion for the subject itself. In my case that's where it all started. As a child, I recalled asking my father why his airplane flies. Later, when I was able to study physics, I could recall some relational experiences such as scuba diving (gas laws), airplane flight (Bernoulli's principle), list goes on. In all instances, I recall thinking oh yes, this makes sense, and from yhere I could extrapolate to another scenerio for example; fluid flow in tunnels as related to Bernoulli's principle. It gives you a sense of being part and parcel of the whole physics experience which is extremely gratifying. As I progressed through my graduate curriculuum to a PhD, the main areas of physics; Newtonian mechanics, quantum, statistical mechanics and electromagnetic theory seemed to be intricately linked together to explain the structure of the universe along with processes occuring within. This is not a 'hodge-podge of random facts' as you mention. By studying physics and getting the feel and understanding of physics, one doesn't neccessarily have to memorize equations, they just come naturally.&lt;/p&gt;&#xA;" OwnerUserId="4899" LastActivityDate="2011-08-12T02:37:43.390" />
  <row Id="13418" PostTypeId="2" ParentId="8234" CreationDate="2011-08-12T02:39:54.697" Score="1" ViewCount="" Body="&lt;p&gt;The classical field has a straightforward interpretation in the bosonic case--- it is determined by the density and phase of the superfluid condensate of the particles and both are simultaneously measurable in the thermodynamic limit. If there is no superfluid, the field is zero. Where there is a superfluid, it is the square-root of the density, with a phase whose gradient is the local superflow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can measure the value of psi-squared at x, that's determining the superfluid condensate density, which can be done by shining light through the fluid to get the index. You can also measure the value of the velocity by the exact way light refrects (or if it is dense enough, by putting little dust specks in the fluid). A gaseous Bose-Einstein condensate, or even liquid Helium, is described precisely by the classical limit of this formalism in the thermodynamic limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take the classical limit wrong, by insisting that the particle number N is fixed, you do still get hbars lurking around. In order to take a good classical field limit, you need enormous occupation numbers for the field, which requires that you take N to infinity and the mass to zero keeping the density fixed. In this limit, the conjugate variables density/phase become commuting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for measuring the &quot;x-projection of operators&quot;, I couldn't figure out what that meant. You can measure the field momentum too, of course, by just taking the complex conjugate of the measurement of the field (the imaginary part of the field and the real part are not independent, and you can describe the whole system using only the real part and its time derivative, as described in the Wikipedia article).&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-12T02:39:54.697" />
  <row Id="13419" PostTypeId="2" ParentId="13383" CreationDate="2011-08-12T02:48:38.757" Score="0" ViewCount="" Body="&lt;p&gt;1000BaseT and 10GBaseT both use amplitude modulation so that a cable with a maximum frequency of 500Mhz (cat6) can carry much higher amounts of data.&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-08-12T02:48:38.757" />
  <row Id="13420" PostTypeId="2" ParentId="8642" CreationDate="2011-08-12T03:06:47.427" Score="1" ViewCount="" Body="&lt;p&gt;To add a little to Lubos's answer: Even classical gravity has a problem with a thermodynamic limit, because a constant energy density leads to a collapse in a finite time in the future or past, so thermodynamics in the presence of gravity is not trivial. You can think of this as a classical residue of holography. I believe the proper analog to a thermal state in gravity (or in string theory) is an empty dS space, since this is a maximal entropy configuration semi-classically with a finite temperature. de Sitter space is hard to describe in quantum gravity, because of the various paradoxes related to the finite surface area bounding horizon.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-12T03:06:47.427" />
  <row Id="13421" PostTypeId="2" ParentId="13416" CreationDate="2011-08-12T03:18:47.647" Score="4" ViewCount="" Body="&lt;p&gt;The claim that accelerated charges must radiate is simply false. There are very many simple situations in which they do, but in general things should be examined on a case-by-case basis; there is not simple thumb rule like &quot;acceleration yields radiation.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest way to see this is to consider a wire carrying a constant current. This situation is magnetostatic, with well-understood electromagnetic field given by Biot-Savart law... which trivially lacks any radiation, as the field is constant. This is independent of the shape of the wire, and yet, if the wire is not straight, the charges within it must accelerate at some point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more exotic could be a uniformly rotating ring of charge. Both charge density and current density will be constant, and will produce constant electric and magnetic fields. This can be experimentally realized with toroidal superconductors.&lt;/p&gt;&#xA;" OwnerUserId="1353" LastEditorUserId="1353" LastEditorDisplayName="" LastEditDate="2011-08-12T03:25:03.283" LastActivityDate="2011-08-12T03:25:03.283" CommentCount="7" />
  <row Id="13422" PostTypeId="1" CreationDate="2011-08-12T03:55:25.687" Score="3" ViewCount="131" Body="&lt;p&gt;The Rayleigh-Lamb equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\tan (pd)}{\tan (qd)}=-\left[\frac{4k^2pq}{\left(k^2-q^2\right)^2}\right]^{\pm 1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(two equations, one with the +1 exponent and the other with the -1 exponent) where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p^2=\frac{\omega ^2}{c_L^2}-k^2$$&#xA;and&#xA;$$q^2=\frac{\omega ^2}{c_T^2}-k^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;show up in physical considerations of the elastic oscillations of solid plates. Here, $d$ is the thickness of a elastic plate, $c_L$ the velocity of longitudinal waves, and $c_T$ the velocity of transverse waves. These equations determine for each positive value of $\omega$ a discrete set of real &quot;eigenvalues&quot; for $k$. My problem is the numerical computation of these eigenvalues and, in particular, to obtain curves displaying these eigenvalues. What sort of numerical method can I use with this problem? Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Using the numerical values $d=1$, $c_L=1.98$, $c_T=1$, the plots should look something like this (black curves correspond to the -1 exponent, blue curves to the +1 exponent; the horizontal axis is $\omega$ and the vertical axis is $k$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/zIRvb.png&quot; alt=&quot;The curves are supposed to look like this.&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3064" LastEditorUserId="3064" LastEditorDisplayName="" LastEditDate="2011-08-19T00:36:13.077" LastActivityDate="2011-08-19T16:34:38.577" Title="Numerical computation of the Rayleigh-Lamb curves" Tags="&lt;mathematical-physics&gt;&lt;computational-physics&gt;&lt;oscillators&gt;&lt;rayleigh-lamb-equations&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="13423" PostTypeId="5" CreationDate="2011-08-12T04:21:09.657" Score="0" ViewCount="" Body="&lt;p&gt;The Rayleigh-Lamb equations determine the relationship between frequency and wavenumber in the waves of a solid elastic thick plane layer.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="3064" LastEditorDisplayName="" LastEditDate="2011-08-12T09:23:34.183" LastActivityDate="2011-08-12T09:23:34.183" />
  <row Id="13424" PostTypeId="4" CreationDate="2011-08-12T04:21:09.657" Score="0" ViewCount="" Body="Questions about the Rayleigh-Lamb equations, its numerical solution and the associated dispersion curves." OwnerUserId="-1" LastEditorUserId="3064" LastEditorDisplayName="" LastEditDate="2011-08-12T09:23:41.360" LastActivityDate="2011-08-12T09:23:41.360" />
  <row Id="13425" PostTypeId="2" ParentId="9325" CreationDate="2011-08-12T04:21:38.127" Score="2" ViewCount="" Body="&lt;p&gt;The Unruh effect does not violate Mach's principle, although it seems to superficially. Nothing violates Mach's principle, in my opinion, because Mach's principle is just true. Appropriately stated, it is one classical interpretation of the holographic principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should qualify this by saying that Minkowski space, considered as a solution of GR, or any other asymptotically flat solution, violates Mach's principle if it is stated naively. This is unavoidable, because the scale-invariance of classical GR implies that you can scale up a small region of any global solution to be flat, and then even if the rotation of objects is relative to far-away matter in the original solution, teensy objects in the scaled up solution are rotating relative to matter infinitely far away, so they end up rotating relative to the boundary conditions on the metric at infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make Mach's principle reasonable, one must be clear as to what constitutes &quot;matter&quot; which one can be rotating relative to. &quot;Matter&quot; does not mean that the vacuum equations are not satisfied, because then black holes wouldn't be matter. If you extend that to extremal horizons, and you believe string theory is correct so that electrons, quarks, are little weak-dual quantum versions of extremal black-holes, then Mach's principle is stating that all motion is relative to a distant horizons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This principle is true and it is predictive. If you have just one cosmological horizon and no other horizon (no other matter), say a completely empty deSitter space, then you cannot set it rotating. There is nothing it can rotate relative to. If you have a black hole inside a de-Sitter horizon (a dS-Schwartschild solution), you can set the black hole rotating, but only at the cost of making the de-Sitter horizon rotate in the opposite direction. The Gibbons/Page/Pope solution (hep-th/0404008) has the property--- the number of rotation parameters is equal to the number of rotation parameters of a single object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to object &quot;but the cosmological horizon isn't matter!&quot;, but this is specious. If you make the deSitter black hole area bigger and bigger, it becomes symmetric with the cosmological horizon exactly when they have equal areas (nothing but the usual coordinates go bad at this point), and afterwards switches places with the cosmological horizon (this process violates entropy if you try to actually do it, but imagine two little black holes at opposite ends Einstein static universe racing one against the other to become the cosmological horizon. Whichever one becomes biggest, wins. Did one of them stop being &quot;matter&quot; once it won?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Mach's principle, properly stated, tells you that rotation/acceleration is relative to a distant horizon. There is an Olbers paradox objection to Mach's principle--- there is not enough glowing matter to determine the local metric, otherwise every line of sight would end on matter. But this objection is specious. Every line of sight in our universe which doesn't end on matter ends on the cosmological horizon, so it also ends on &quot;horizon matter&quot;. The holographic principle tells you that all motion should be thought of as projected on the holographic screen, and this is just the statement that any angular motion in the interior of the universe has a visible counterpart in the motion of the constituent holographic counterpart on the screen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Unruh's radiation, when you accelerate, there is a gigantic acceleration horizon at a finite distance from you--- a black wall stretching from one end of the universe to the other. This horizon is what you are moving relative to. If you slowly stop moving, this horizon recedes to infinity, but the asymptotically flat solutions all violate Mach's principle anyway. If you are accelerating in de-Sitter, what happens is just that the accelerating observer has a different point of view regarding what the de-Sitter horizon is. If we had a good holographic description of deSitter space, then two observers which are in communication should be able to map their holographic descriptions back and forth, much as the infalling observer and the outside observer in a black hole can do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this does require some extrapolation of new physics, I believe it is safe to say that Mach's principle is just a not-quite-right forshadowing of the holographic principle, and is therefore correct.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-13T00:02:30.903" LastActivityDate="2011-08-13T00:02:30.903" CommentCount="3" />
  <row Id="13426" PostTypeId="1" AcceptedAnswerId="13427" CreationDate="2011-08-12T04:39:17.823" Score="5" ViewCount="439" Body="&lt;p&gt;I believe that no real objects are actually (&lt;em&gt;exactly&lt;/em&gt;) 1 meter long, since for something to be 1.00000000... meters long, we would have to have the ability to measure with infinite precision. Obviously, this can be extended to any units of measurement. Am I wrong?&lt;/p&gt;&#xA;" OwnerUserId="4901" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-08-12T06:13:30.873" LastActivityDate="2011-08-27T19:22:47.520" Title="Is anything actually 1 meter long (or 1kg of weight)?" Tags="&lt;si-units&gt;&lt;units&gt;&lt;metrology&gt;" AnswerCount="5" CommentCount="2" />
  <row Id="13427" PostTypeId="2" ParentId="13426" CreationDate="2011-08-12T04:46:59.627" Score="15" ViewCount="" Body="&lt;p&gt;You're not wrong. However, there used to be an object &lt;em&gt;exactly&lt;/em&gt; $1$ meter long until 1960, because a meter was &lt;em&gt;defined&lt;/em&gt; to be the length of a certain platinum-iridium rod at certain conditions. Since then, the meter is defined in terms of interferometry, and now it is specifically the distance traversed by light in vacuum within a certain period of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the kilogram has a prototype whose mass is $1$ kg by definition. There are some proposals to replace that definition, but it hasn't been done yet.&lt;/p&gt;&#xA;" OwnerUserId="1353" LastActivityDate="2011-08-12T04:46:59.627" CommentCount="3" />
  <row Id="13428" PostTypeId="1" CreationDate="2011-08-12T05:00:29.873" Score="7" ViewCount="104" Body="&lt;p&gt;How are Monte Carlo simulations used in experimental high energy physics? Particularly in studying detectors limitations (efficiencies?) and data analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will appreciate giving a simple example to clarify how MC is used if the question is too technical to have a simplified answer&lt;/p&gt;&#xA;" OwnerUserId="4521" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-12T16:45:28.467" LastActivityDate="2011-08-12T16:45:28.467" Title="How are Monte Carlo simulations used in experimental high energy physics?" Tags="&lt;particle-physics&gt;&lt;experimental-physics&gt;&lt;simulation&gt;&lt;high-energy-physics&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="13429" PostTypeId="2" ParentId="13428" CreationDate="2011-08-12T05:30:50.320" Score="7" ViewCount="" Body="&lt;p&gt;One has to realize that a Monte Carlo simulation is an integration tool. Suppose you have a curve in an xy plot, y=f(x). If you throw random (x,y) pairs in the square containing the f(x) and count the number where y is less than f(x)  versus the number y larger than f(x) you get an estimate of the area under f(x), i.e. the integral of the function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In elementary particle physics, the phase space ( equivalent to the square in the simple example) is known. Theoretical functions are used as a weight to a random number generator, to generate &quot;events&quot; according to their parameters and checked against the real event data.  If the fit is bad, the parameters are changed to improve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantage is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a)Detector limitations can be programmed in the phase space and events generated with the limits of the detector included&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The method is much more efficient in computer time than the numerical integrations necessary over the innumerable functions entering the problems, detector and theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Once a Monte Carlo event sample is generated it can be used &quot;as if it is data&quot; over and over again to get plots not thought up beforehand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the recent LHC experiments the Monte Carlo events were generated way before the real data, according  to the detector limitations and to the theoretical expectations from the Standard Model. The existence of these Monte Carlo data set allowed fast checks on whether new physics is appearing. New physics will appear as statistically significant  deviations from the Monte  Carlo curves.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-08-12T13:18:19.340" LastActivityDate="2011-08-12T13:18:19.340" CommentCount="6" />
  <row Id="13430" PostTypeId="2" ParentId="13426" CreationDate="2011-08-12T05:32:47.477" Score="7" ViewCount="" Body="&lt;p&gt;1 kg is defined as the mass of a particular reference mass in France.  So there is at least one thing that weighs exactly 1 kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, it may in general be possible to construct an arbitrary number of objects which weigh 1kg exactly. Assuming all Si atoms have the same mass, one could simply define the kilogram to be the mass of a certain number of Si atoms.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2011-08-12T05:32:47.477" />
  <row Id="13431" PostTypeId="2" ParentId="13387" CreationDate="2011-08-12T06:37:06.303" Score="3" ViewCount="" Body="&lt;p&gt;Mister S.Emeis from Institute for Meteorology and Climate Research, Karlsruhe Institute of Technology, gives an answer:&#xA;&lt;a href=&quot;http://www.dewi.de/dewi/fileadmin/pdf/publications/Magazin_37/07.pdf&quot; rel=&quot;nofollow&quot;&gt;Link here&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the abstract of his explanation:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The occurrence of wake clouds at Horns Rev wind farm is&#xA;explained as mixing fog. Mixing fog forms when two nearly&#xA;saturated air masses with different temperature are mixed.&#xA;Due to the non-linearity of the dependence of the saturation&#xA;water vapour pressure on temperature, the mixed air&#xA;mass is over-saturated and condensation sets in. On the day&#xA;in February 2008 (&lt;strong&gt;the figure in the question&lt;/strong&gt;), when the wake clouds were observed&#xA;at Horns Rev, cold and very humid air was advected from&#xA;the nearby land over the warmer North Sea and led to the&#xA;formation of a shallow layer with sea smoke or fog close&#xA;above the sea surface. The turbines mixed a much deeper&#xA;layer and thus provoked the formation of cloud trails in the&#xA;wakes of the turbines.&lt;/p&gt;&#xA;" OwnerUserId="552" LastActivityDate="2011-08-12T06:37:06.303" />
  <row Id="13432" PostTypeId="1" CreationDate="2011-08-12T08:17:09.860" Score="3" ViewCount="64" Body="&lt;p&gt;I'm developing (actually adopting existing solution) a &lt;a href=&quot;http://en.wikipedia.org/wiki/Kalman_filter&quot; rel=&quot;nofollow&quot;&gt;Kalman filter&lt;/a&gt; to model motion of a vehicle (UAV or automobile). The state vector will include position, velocity, and, possibly, acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that existing solution acceleration is included; state transition is something like $(\vec x,\vec v, \vec a) \to (\vec x+\Delta t\cdot \vec v,\vec v+\Delta t \cdot \vec a,\vec a+\xi)$, where $\xi$ is a process noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that such model is good when acceleration is changing somewhat smoothly, but when there are bursts of acceleration, removing $\vec a$ from state and adding noise to velocity should be more suitable: $(\vec x,\vec v)\to(\vec x+\Delta t\cdot\vec v,\vec v+\xi)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tests show that there is almost no difference in precision of solution between both models (turns out that the observation data provides more effect on position than the velocity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is, when should I include acceleration in state, and when should not? What things should I consider?&lt;/p&gt;&#xA;" OwnerUserId="4902" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-12T08:36:29.473" LastActivityDate="2011-08-12T08:36:29.473" Title="Should acceleration be included in state vector of a Kalman filter?" Tags="&lt;kinematics&gt;&lt;acceleration&gt;&lt;computational-physics&gt;" CommentCount="2" />
  <row Id="13433" PostTypeId="2" ParentId="12200" CreationDate="2011-08-12T08:49:22.087" Score="-2" ViewCount="" Body="&lt;p&gt;Nicola Tesla was actually perusing this wireless energy transfer by means of his &lt;a href=&quot;http://en.wikipedia.org/wiki/Tesla_coil&quot; rel=&quot;nofollow&quot;&gt;telsa coils&lt;/a&gt;.&#xA;He actually designed large EM coils that could be placed near the generating station and smaller coils at subscriber end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, people at that time, especially under the influence of Edison (they were arch enemies), made a mockery of him and the attempt was soon discarded after his death.(One may call this a conspiracy theory)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then it is possible....&lt;/p&gt;&#xA;" OwnerUserId="4887" LastActivityDate="2011-08-12T08:49:22.087" CommentCount="1" />
  <row Id="13434" PostTypeId="2" ParentId="12206" CreationDate="2011-08-12T09:09:57.247" Score="0" ViewCount="" Body="&lt;p&gt;The Wattage written is &quot;&lt;strong&gt;The Wattage&lt;/strong&gt;&quot;, actually the regulator works by changing the voltage applied to the fan's coils. This is the case with at least the potentiometer regulators(the conventional one), so even if the fan is slowed down, fan coil will consume less power but the resistors will consume the rest. So essentially, you gain nothing even by reducing the speed!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as Georg said, the power consumption is much greater at initial startup to overcome the static friction.&lt;/p&gt;&#xA;" OwnerUserId="4887" LastActivityDate="2011-08-12T09:09:57.247" CommentCount="4" />
  <row Id="13435" PostTypeId="2" ParentId="13342" CreationDate="2011-08-12T09:39:26.863" Score="0" ViewCount="" Body="&lt;p&gt;Think of it as this,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are having a ball, which is rotating with say &lt;strong&gt;&lt;em&gt;v&lt;/em&gt;&lt;/strong&gt; and angular velocity $ \omega $, then consider that both of these velocities constitute to its total energy &lt;strong&gt;&lt;em&gt;E&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;E = E{rot} + E{trans}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider the case when either of them is &lt;strong&gt;&lt;em&gt;0&lt;/em&gt;&lt;/strong&gt;, then since both things are independent of each other, it makes sense to assume that other will be unaffected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and again we know that,&#xA;$$&#xD;&#xA;E{rot} = 1/2 m v^{2}&#xD;&#xA;$$&#xA;$$&#xD;&#xA;E{trans} = 1/2 I\omega^{2}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, making one nil, doesn't affect the other, which means that you can just pick up a &lt;strong&gt;rotating + translating&lt;/strong&gt; object i.e. stop its translation, but still the ball will be rotating!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This clearly implies that even if you can't influence $E{rot}$ with $E{trans}$ or vice-versa.&lt;/p&gt;&#xA;" OwnerUserId="4887" LastActivityDate="2011-08-12T09:39:26.863" />
  <row Id="13437" PostTypeId="2" ParentId="13233" CreationDate="2011-08-12T11:35:44.263" Score="3" ViewCount="" Body="&lt;p&gt;We need a clear operational meaning about what it means for gravity to be &quot;repulsive&quot;. If we think of it too naively... say a distant observer in Schwarzschild spacetime viewing a particle orbit, for whom Schwarzschild time $t$ has a clear operational meaning and Schwarzschild radial coordinate $r$ is good enough. Can the orbit have positive $d^2r/dt^2$, i.e., have outward rather than inward acceleration? Yes, absolutely: in fact, it must be the case, because in those coordinates a radially freefalling particle never reaches the horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that's just perverse. Let's look for a &lt;em&gt;local&lt;/em&gt; operational definition. For example, take a small (near each other) collection of comoving test particles, four-velocity $u$, and see what they do. Their geodesic deviation is given by the Riemann curvature tensor, and if we have a small ball of volume $V$, then&#xA;$$\lim_{V\to 0}\frac{\ddot{V}}{V}\vert_{t=0} = -R^\alpha{}_{\mu\alpha\nu} u^\mu u^\nu,$$&#xA;and this contracted Riemann tensor is the Ricci curvature $R_{\mu\nu}$. Thus, the attractive or repulsive behavior of the ball of test particles is given by the Ricci curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a gander at the strong energy condition: for every future-pointing timelike vector $u$,&#xA;$$\left(T_{\mu\nu}-\frac{1}{2}Tg_{\mu\nu}\right)u^\mu u^\nu \geq 0.$$&#xA;It's not immediately obvious as to what this actually means, but contracting the Einstein field equation $R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu} = 8\pi T_{\mu\nu}$ gets you $R - \frac{1}{2}Rg^\mu{}_\mu = 8\pi T$, i.e., $R = -8\pi T$: the Ricci scalar is just the negative trace of the stress-energy tensor, up to a constant. Back-substituting into the field equation, and voila:&#xA;$$R_{\mu\nu} = 8\pi\left(T_{\mu\nu} - \frac{1}{2}Tg_{\mu\nu}\right),$$&#xA;so the SEC just says that $R_{\mu\nu}u^\mu u^\nu \geq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conclusion: for this reasonable local notion of what it means for gravity to be attractive/repulsive, &lt;em&gt;gravity is non-repulsive if, and only if, the strong energy condition holds&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a perfect fluid, $T^{\mu\nu} = (\rho+p)u^\mu u^\nu + pg^{\mu\nu}$, so that $T = T^\mu{}_\mu = -\rho + 3p$. From the frame comoving with the fluid, substitution gives $\rho + 3p\geq 0$. On the other hand, approaching a lightlike $u$ in one of the spatial directions gives $\rho + p\geq 0$, this step being justified by continuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Einstein field equation with a cosmological constant just adds a $\Lambda g_{\mu\nu}$ opposite the stress-energy tensor, it is equivalent to a perfect fluid with density $\rho = \Lambda/8\pi$ and pressure $p = -\Lambda/8\pi$. Thus, we need positive vacuum energy density to break the SEC.&lt;/p&gt;&#xA;" OwnerUserId="1353" LastActivityDate="2011-08-12T11:35:44.263" CommentCount="2" />
  <row Id="13438" PostTypeId="1" CreationDate="2011-08-12T11:36:05.297" Score="-2" ViewCount="62" Body="&lt;p&gt;I'm having difficulty finding some secondary data on the extraction of radium or polonium. Nothing specific, really just any practical data available on the topic - I'm getting desperate!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this seems like a vague thing to be looking for - I just need to analyze any data related to the topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3044" LastActivityDate="2011-08-12T11:36:05.297" ClosedDate="2011-08-13T21:14:54.007" Title="Experimental data - extraction of radium/polonium?" Tags="&lt;experimental-physics&gt;&lt;experimental-evidence&gt;" CommentCount="8" />
  <row Id="13439" PostTypeId="2" ParentId="13426" CreationDate="2011-08-12T11:54:48.587" Score="1" ViewCount="" Body="&lt;p&gt;In Nature the only &lt;em&gt;exact&lt;/em&gt; realization of a number is thru $\mathbb{N}$ i.e. &lt;em&gt;counting&lt;/em&gt; . I suspect that they are called 'Naturals' because of that. To all the other numbers we only get aproximations. Even when defined to be the &lt;em&gt;exact&lt;/em&gt; length/weight of a certain physical object the &lt;em&gt;measure&lt;/em&gt; of it will be within a certain range of values, $\pm$Angstrom. &#xA;Another example: there is not a single realization in the universe that correspond to the number $\pi$. Numbers are menthal constructs.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we &lt;em&gt;measure&lt;/em&gt; ? Obtaining a ratio between the &lt;em&gt;amount&lt;/em&gt; that is to be measured and the amount of a selected standard. If both amounts change equally, the measure (ratio) keep constant, i.e. the constancy of a measure is not a guarantee that the properties of 'objects' do not change.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://vixra.org/pdf/1107.0016v1.pdf&quot; rel=&quot;nofollow&quot;&gt;Atomic &lt;em&gt;measures&lt;/em&gt; are number counts&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We will now see that the measures of bodies' properties using atomic&#xA;  units are independent of the base quantities and dependent on the&#xA;  number of particles or atoms.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;An atomic unit of mass is the mass of a certain number of baryons; the&#xA;  measure of the mass of a body using atomic units is therefore a number&#xA;  proportional to the number of baryons of the body (this is not an&#xA;  exact statement but it serves the needs of this work). If the mass of&#xA;  baryons changes, so will the mass unit and the mass of the body; the&#xA;  measure holds invariant because the number of baryons did not change.&#xA;  Therefore, a measure of the mass of a body using atomic units is&#xA;  basically a baryon count, holding invariant as long as the number of&#xA;  baryons does not change, independently of the eventual change of&#xA;  baryons' mass. The same kind of reasoning applies to charge measures.&#xA;  In what concerns length measures, the length unit is such that the&#xA;  measures of length of isolated bodies hold invariant; this is not the&#xA;  way length unit is formally defined, but this is a condition it has to&#xA;  obey to be acceptable, in order to fit Einstein's measuring rod or&#xA;  reference-body, translated in the time invariance of Bohr radius. So,&#xA;  we can say that the atomic length unit is a fixed multiple of the Bohr&#xA;  radius; if the latter varies, so will bodies' length and the unit of&#xA;  length, holding invariant the measures of bodies' length. Therefore,&#xA;  length measures are a way of counting atoms, the measures of the&#xA;  length of bodies holding invariant as long as the number of atoms does&#xA;  so, for bodies and measuring devices subject to the same conditions. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The above reasoning shows that the measures of mass, charge and length&#xA;  of bodies are independent of the mass and charge of elementary&#xA;  particles and of atoms' radii, tracing only the number of particles or&#xA;  atoms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To any measure (ratio) that is not obtained thru direct counting an error margin is present (explicit or implicit) .&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-08-12T12:00:48.403" LastActivityDate="2011-08-12T12:00:48.403" />
  <row Id="13440" PostTypeId="2" ParentId="13426" CreationDate="2011-08-12T12:25:25.477" Score="2" ViewCount="" Body="&lt;p&gt;No, but for other reasons than you might suspect. For two endpoints of an object to be precisely one meter apart, one would have to be able to consistenly measure that distance. But in reality, the outcome of any individual measurement is constrained by the Heisenberg inequality, and therefore you cannot consistently measure such a length. If you'd know the precise postion of one endpoint, you'd have &lt;em&gt;zero&lt;/em&gt; knowledge about it's speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there are also practical limitations to our ability to measure lengths, and they kick in much earlier (typically ~10-9 or worse; in comparison Planck's constant is ~10-34)&lt;/p&gt;&#xA;" OwnerUserId="232" LastActivityDate="2011-08-12T12:25:25.477" />
  <row Id="13441" PostTypeId="1" CreationDate="2011-08-12T13:28:07.160" Score="10" ViewCount="229" Body="&lt;p&gt;Usually in all discussions and arguments of scaling or solving problems using dimensional analysis, the dimensionless constant is indeterminate but it is usually assumed that it is of order 1. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What does &quot;of order 1&quot; mean? 0.1-10?&lt;/li&gt;&#xA;&lt;li&gt;Is there any way, qualitative or quantitative, to see why the dimensionless constant is of order 1?&lt;/li&gt;&#xA;&lt;li&gt;Are there exceptions to that? I mean cases where the dimensionless constant is very far from 1? Could you give some examples? Can such exceptions be figured out from dimensional analysis alone?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4521" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-12T19:17:25.643" LastActivityDate="2011-08-16T10:06:22.870" Title="In dimensional analysis, why the dimensionless constant is usually of order 1?" Tags="&lt;dimensional-analysis&gt;" AnswerCount="8" />
  <row Id="13442" PostTypeId="2" ParentId="9818" CreationDate="2011-08-12T14:14:55.480" Score="1" ViewCount="" Body="&lt;p&gt;Black hole interiors are dual to exteriors under what is called by Susskind &quot;Black hole complementarity&quot;. The logical positivism is there--- the interior and exterior observers disagree on the description of their space time, but both descriptions are correct. The question can be asked in a de-Sitter space too, where Alice and Bob start out talking, but then fall into each other's cosmological horizon. There it is more difficult to understand.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-12T14:14:55.480" />
  <row Id="13443" PostTypeId="2" ParentId="5483" CreationDate="2011-08-12T14:27:30.120" Score="1" ViewCount="" Body="&lt;p&gt;Using Mach's 1893 definition of Mach's principle condemns the discussion to irrelevance. It's like posting on physics.SE with a question titled &quot;How is the emission spectrum of hydrogen determined?,&quot; but then saying in the body of the question that we want an answer written in terms of the aether and Newtonian mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 1960's and 70's, there was a golden age of tests of GR, and one of the most active topics was testing GR against alternative theories such as Brans-Dicke gravity. B-D gravity is physically a very well motivated theory. The original paper is available online &lt;a href=&quot;http://loyno.edu/~brans/ST-history/&quot; rel=&quot;nofollow&quot;&gt;http://loyno.edu/~brans/ST-history/&lt;/a&gt; and is very readable even if you're not a specialist. The idea of B-D gravity is to couple matter to a scalar field $\phi$, which provides a physical mechanism for Mach's idea that an object's inertia comes from the other matter in the universe. B-D gravity is more Machian than GR. Neither GR not B-D is completely Machian or completely non-Machian. B-D gravity has a dimensionless parameter $\omega$. In the limit $\omega\rightarrow\infty$, B-D gravity reduces to GR. Brans and Dicke committed themselves to the idea that &quot;[...]in any sensible theory $\omega$ must be of the general order of magnitude of unity.&quot; This makes the theory falsifiable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experiments show that $\omega$ must in fact be quite high. The best current limit comes from the Cassini probe, which requires $\omega \ge 40,000$. Therefore, B-D gravity should be considered as falsified. So the modern, sensible answer to the OP's question is: Mach's principle is false, in the sense that experiments determine the universe to be no more Machian than GR -- which is not very Machian.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-12T14:27:30.120" />
  <row Id="13444" PostTypeId="1" AcceptedAnswerId="13470" CreationDate="2011-08-12T14:50:02.843" Score="3" ViewCount="92" Body="&lt;p&gt;Does the theory on paper provide a way for hypothetical travelers to get back out of the bubble that has gotten them close to their distant destination by compressing all the space in front of them and expanding the space behind it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was picturing space as a sheet of rubber and my warp bubble as the space of rubber under my finger as I press my finger down and drag the 'space' under my finger to where it's near a distant part of the sheet. But despite the fact that I'm now closer to the distant location as measured if you can 'jump over' the compressed rubber, unless there really is a way to 'jump over,' you still have the same amount of rubber between you and the destination.&lt;/p&gt;&#xA;" OwnerUserId="4715" LastEditorUserId="1353" LastEditorDisplayName="" LastEditDate="2011-08-13T04:21:48.323" LastActivityDate="2011-08-13T04:21:48.323" Title="Getting back out of an Alcubierre warp bubble" Tags="&lt;space&gt;&lt;relativity&gt;&lt;general-relativity&gt;" AnswerCount="2" />
  <row Id="13445" PostTypeId="1" CreationDate="2011-08-12T14:50:42.580" Score="3" ViewCount="102" Body="&lt;p&gt;One can understand a subject better in physics by trying to solve as many problems as one can from a textbook say. When it comes to experimental physics and data analysis, no book on experimental (high energy or astrophysics or cosmology) physics comes with free data that one might analyze to test his understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there sites online, for experiments in cosmology or astrophysics (I know that particle physics data would be overwhelming), where such data is available for free that one can play with (i.e. to be able to make cuts and reproduce simple histograms...etc)? may be to reproduce roughly the CMB curve?&lt;/p&gt;&#xA;" OwnerUserId="4521" LastEditorUserId="4521" LastEditorDisplayName="" LastEditDate="2011-08-12T15:14:21.537" LastActivityDate="2011-08-18T01:28:26.317" Title="Are there free data available online from cosmology (or astrophysics) experiments that anyone can analyse?" Tags="&lt;particle-physics&gt;&lt;cosmology&gt;&lt;astrophysics&gt;&lt;experimental-physics&gt;" AnswerCount="6" CommentCount="1" />
  <row Id="13446" PostTypeId="2" ParentId="13428" CreationDate="2011-08-12T14:56:14.443" Score="3" ViewCount="" Body="&lt;p&gt;Monte Carlo are very important in almost any particle physics experiment and are used in a variety of ways including&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;To prototype the experiment without spending many millions of dollars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They can be used to &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;show that the proposed physics signal will be detectable among the many known physics effects&lt;/li&gt;&#xA;&lt;li&gt;to test variations on the proposed detector design&lt;/li&gt;&#xA;&lt;li&gt;to estimate the size of the data set and determine what kind of computation resources will be needed to manage the data&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This kind of use is often an input to the proposal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To test proposed a proposed analysis, and to evaluate the expected systematic uncertainty associated with it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many experiments (especially those at colliders) the data contains a vast number of over-lapping physics channels, and must be filtered (sometimes quite extensively) before a particular observable can be pulled out of the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By generating a &lt;em&gt;known&lt;/em&gt; amount of the (simulated) desired signal and hiding it among a bunch of (simulated) backgrounds, one can test if the analysis works (so &lt;em&gt;this&lt;/em&gt; point could just as well have gone in item #1), but you can do more than that: you can also take note of how many false positives and negatives are introduced in each filtering step and use that data to (a) optimize the filters and (b) evaluate the systematic uncertainly of the analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These applications are integral to the data analysis.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Un-fold complicated detector behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many times we have a theoretical or phenomenological entity (call it $A(\mathbf{x})$), that we would like to get at, but the data actually record $$A&amp;#39;(\mathbf{x_n}) = \int_n dx_{n-1} \rho(\mathbf{x_n};\mathbf{x_n-1}) \cdots \int_2 dx_2 \rho(\mathbf{x_1};\mathbf{x_1})\int_1 dx \rho(\mathbf{x_1};\mathbf{x}) A(\mathbf{x}) .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your MC is good enough (big if, as that is tough) You can use it to produce an approximate backwards mapping function $f&amp;#39;(A&amp;#39;(\mathbf{x_n})) \to \tilde{A}(\mathbf{x})$. This introduces another uncertainty often called the &lt;em&gt;&quot;model dependent error&quot;&lt;/em&gt; (though the term is overloaded for errors dependent on theoretical models as well).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This application is rather less common, but is part of the data analysis.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-12T15:38:52.150" LastActivityDate="2011-08-12T15:38:52.150" CommentCount="2" />
  <row Id="13447" PostTypeId="1" CreationDate="2011-08-12T14:58:12.507" Score="4" ViewCount="39" Body="&lt;p&gt;Could someone explain in details the meaning of the terminology &quot;tagging&quot; in experimental high energy physics and how is it used in the analysis ?&lt;/p&gt;&#xA;" OwnerUserId="4521" LastActivityDate="2011-08-12T15:25:46.553" Title="What does tagging in experimental high energy physics mean?" Tags="&lt;particle-physics&gt;&lt;experimental-physics&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13448" PostTypeId="2" ParentId="13447" CreationDate="2011-08-12T15:25:46.553" Score="5" ViewCount="" Body="&lt;p&gt;As @dmckee points out this term is used in multiple contexts, but &quot;b-tagging&quot; is probably the most common usage. As in &quot;the selection requires 2 tagged jets&quot; or something of this sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;b-tagging&lt;/strong&gt; is the identification of jets (the spray of color neutral particles originating from a colored quark or gluon) that are likely to originate from a $b$-quark. This can be useful for analyses with a lot of jets, of which some are $b$-jets (it will reduce &lt;a href=&quot;http://physics.stackexchange.com/questions/12590/what-is-meant-by-combinatorial-background-in-experimental-high-energy-physics&quot;&gt;combinatorial background&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples include top-pair production (each top almost exclusively decays into a W boson and a b quark), Higgs (couples strongly to b's), of flavour physics (e.g. B-Physics) A disadvantage is that b-tagging algorithms are not very efficient (common workin points include 50%-70% so out of all true b-jets you will only catch a little more than half)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B-tagging works because the B-mesons, that come early in the hadronization process have a significantly longer lifetime than other mesons. This is due to the $V_{xb}$ matrix elements in the CKM matrix (x being one of u,c,t). A drop down into lower generations via weak decay has a low probability, therfore the lifetime is long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find b-jets one possibility is to look at jets coming from secondary secondary vertices macroscopically apart from the primary interactions. Another way is to look at the impact parameters (the distance to the primary vertex) of the tracks of the individual charged particles that constitute the jet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from b-tagging, there is also c-tagging, since charmed mesons have a significantly larger lifetime as well (compared to so-called &quot;light-flavors&quot; u,d,s), but it doesn't work as well. Collectively the term is &quot;flavor-tagging&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another usage is when people talk about &quot;&lt;strong&gt;tag and probe&lt;/strong&gt;&quot;. This is a method to calibrate and measure the performance of some selection. For example it is crucial to know how well muons are reconstructed. Therefore one will select events from a region in phase space that corresponds to $Z$ boson decays to two muons. One will apply strict requirements for one muon but only loose one to the other. The strictly selected one is the &quot;tagged&quot; muon and the looser one is the muon on which one wants to &quot;probe&quot; the reconstruction efficiency (or something else)&lt;/p&gt;&#xA;" OwnerUserId="2151" LastActivityDate="2011-08-12T15:25:46.553" CommentCount="3" />
  <row Id="13450" PostTypeId="2" ParentId="13441" CreationDate="2011-08-12T15:57:38.927" Score="3" ViewCount="" Body="&lt;p&gt;Well, of course you have to pick the quantities in your dimensional analysis right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: Use dimensional analysis to estimate the potential energy of a star, hold together only by gravitation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution: Newtons gravitational constant $G$ better show up somewhere. This requires us to include something with units $kg^2 / m$. We can get this by inserting $M^2$, with $M$ the mass of the star, and by inserting $1/R$, with $R$ the radius of the star. Thus, the potential energy is estimated to be &#xA;$$E_\text{pot} \approx -G \frac{M^2}{R}$$&#xA;which is off by a factor $3/5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could, of course, have inserted the mass of a hydrogen atom and then everything would be off by many orders of magnitude...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no general guarantee that the constant is of order one. But:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why it often comes out as &quot;order of 1&quot; is, I believe, the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many cases, the true solution involves an integral over a variable $x$ where the function to be integrated is of the form $f(x) \sim x^n$. In physics, $n$ is small in most cases, so the integration gives a factor $\frac{1}{n}$. Other corrective factors that get ignored in dimensional analysis are $\pi$, $2\pi$ or $4\pi$, i.e., some small integer multiples of $\pi$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A counterexample... the fine-structure constant $\alpha = \frac{1}{137}$, maybe? But this constant itself can be obtained through dimensional analysis, so I am not sure.  &lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-08-12T15:57:38.927" CommentCount="3" />
  <row Id="13451" PostTypeId="1" AcceptedAnswerId="13453" CreationDate="2011-08-12T15:58:26.757" Score="2" ViewCount="182" Body="&lt;p&gt;In Special Relativity, the spacetime interval between two events is $s^2 = -(c{\Delta}t)^2+({\Delta}x)^2+({\Delta}y)^2+({\Delta}z)^2$ giving the Minkowski metric $\eta_{\mu\nu}=\text{diag}(-1, 1, 1, 1)$. What is the justification for making time have a negative coefficient, and how closely is that related to the 2nd law of thermodynamics? Sure, by letting $\eta = \text{diag}(1, 1, 1, 1)$, we get a pretty boring spacetime, and the boosts in the Poincaré group become trig instead of hyperbolic functions, but what's the physical reasoning behind this?&lt;/p&gt;&#xA;" OwnerUserId="4906" LastActivityDate="2011-09-02T06:36:55.640" Title="Why is time special?" Tags="&lt;special-relativity&gt;&lt;spacetime&gt;&lt;time&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" />
  <row Id="13452" PostTypeId="2" ParentId="13403" CreationDate="2011-08-12T16:20:39.020" Score="1" ViewCount="" Body="&lt;p&gt;Without knowing more about your data set I can only offer a few random suggestions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you have some kind of PID reason to believe that the tracks might be electron you just &lt;em&gt;assume&lt;/em&gt; that they are and compute the energy from the momentum and $m_e$.&lt;/li&gt;&#xA;&lt;li&gt;In you have a calorimeter in the detector stack you &lt;em&gt;measure&lt;/em&gt; the energy, and project the maximum likelihood energy loss back to the vertex. Note, however that the measured signal may have error much larger than $m_e$, so it may be better to use the device for PID and fall back on the previous suggestion.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-08-12T16:20:39.020" />
  <row Id="13453" PostTypeId="2" ParentId="13451" CreationDate="2011-08-12T16:21:51.523" Score="3" ViewCount="" Body="&lt;p&gt;I think this is a case of the mathematics being designed to model reality. As you say, making the time component of the metric positive would give a space that doesn't match what we observe. In particular, the negative component for time allows us to disconnect regions of space that aren't causally linked. In other words, the fact that the speed of light is finite and a maximum means that we must describe space-time with a shape that keeps causally disconnected regions separate. The necessary shape is reflected in the choice of the sign of $\eta_{00}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's how I understand anyway...&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2011-08-12T16:21:51.523" CommentCount="5" />
  <row Id="13454" PostTypeId="2" ParentId="13441" CreationDate="2011-08-12T17:04:03.840" Score="3" ViewCount="" Body="&lt;p&gt;One notable class of exceptions are what are called &quot;hierarchy problems&quot; in particle physics.  For example, if you identify the Planck mass as a fundamental energy scale, you end up with huge dimensionless numbers which don't have an obvious explanation (i.e. ratio of Planck to electroweak scale, etc.).  Explaining these large (or small depending on how you look at it) dimensionless numbers is a major focus of particle theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another fun counterexample:  see 't Hooft's original computation of the instanton effective interaction in 1976.  After a lengthy calculation, the answer turns out to have a dimensionless factor $2^{10}\pi^{6}$ in front, resulting from a slew of Gaussian normalizations and such. &lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="1631" LastEditorDisplayName="" LastEditDate="2011-08-15T17:10:51.447" LastActivityDate="2011-08-15T17:10:51.447" />
  <row Id="13455" PostTypeId="1" AcceptedAnswerId="13459" CreationDate="2011-08-12T17:04:35.463" Score="3" ViewCount="55" Body="&lt;p&gt;QFT says that all particles are indistinguishable from one-another [1]. That is, take a proton from cosmic ray from a supernova a billion light-yeas away, and compare it to a proton that just got smashed out in the LHC, and they are indistinguishable from one-another. Replace either with the other, and nothing will change. (As I understand) this applies to all particles that arise from the standard model fields- fermions and bosons. So if I have an atom, all the electrons in the atom are indistinguishable from all the other electrons in the atom. Theoretically I then take that atom and interchange it with any other &quot;indistinguishable&quot; atom. Or can I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously I can't replace hydrogen-1 with deuterium (their masses are different), so atoms are at least distinguishable to their isotopes. What about within isotopes? The volume of an orbital in an electron cloud is larger than the Compton length of the electron by a factor of about 10^7, giving a tremendous volume that is vacuum, and the probability of getting the election clouds just right to have negligible effects on the quantum field seems unlikely. So, are two ${}^2\text{H}$ atoms indistinguishable from one another?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that does work, and we can say isotopes are indistinguishable. Now, I pose the same question for molecules. What do I need to know about two water molecules? In QFT and GR, there are the 4 spacetime dimensions. I clearly can't just take a water and replace it with another water by swapping their $(\vec{x}, t) properties, because they might be &quot;upside -down&quot; and then all the hydrogen bonds will get messed up and their dipole moments will be wacky (and let's not even think about ionic or superionic water) and there will be some change in the system while the molecule &quot;rights&quot; itself. So, it seems I have at least answered my question at that scale - molecules are not indistinguishable from one another on their physical properties alone. For polar molecules in a &quot;normal&quot; form, we have at least 6 dimensions that must be swapped (assuming we can get away with calling the atoms themselves indistinguishable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final question, then, is what is the limit that systems are no longer indistinguishable based on just their physical properties?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf&lt;/a&gt; page 3.&lt;/p&gt;&#xA;" OwnerUserId="4906" LastActivityDate="2011-08-12T18:17:13.970" Title="Particle Indistinguishability Scale Limit" Tags="&lt;quantum-field-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="13456" PostTypeId="2" ParentId="13403" CreationDate="2011-08-12T17:25:11.797" Score="1" ViewCount="" Body="&lt;p&gt;When one has tracks in an ionising chamber, as a bubble chamber or a TPC there may be also information stored on the ionisation measurements across the track. For low enough momenta one can distinguish electrons from protons and pions and assign the appropriate mass, and then the energy is known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We used to have programs in bubble chamber physics, GRIND, which would calculate probabilities and give the corresponding ionisation values for several hypothesis of masses as well as for the whole event seen, applying energy and momentum balances. The event could be a 4 constraint fit, gold plated, a 1 constrain fit, or a 0 constraint fit, i.e. missing energy with  more than one particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the specific exercise you want to do, assign the mass of the electron to the tracks and calculate the energy and the invariant masses. The photons will appear at 0.  With a monte carlo you can estimate the background from the wrongly assigned mass.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-08-12T17:25:11.797" CommentCount="1" />
  <row Id="13457" PostTypeId="2" ParentId="13088" CreationDate="2011-08-12T17:44:46.133" Score="3" ViewCount="" Body="&lt;p&gt;Well it just so happens that Mitchell answered this question in a private email, I'm posting the relevant part here for the greater good. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Wick rotation in this case transforms a local patch of dS space&#xA;  into AdS space, or vice versa.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Think of the difference between concave and convex, a valley and a&#xA;  hill. Then you should imagine that we have some formula appropriate&#xA;  for a valley - e.g. you might have a line-of-sight device like&#xA;  surveyors use, and you need to adjust the angle of the view finder, in&#xA;  a way that depends on the curvature of the ground. We can turn it into&#xA;  the corresponding formula for the hill by replacing, somewhere in the&#xA;  formula, a positive number with a negative number - because the&#xA;  curvature now has the opposite sign. This is what's actually going on&#xA;  in these Wick rotations, although the details are somewhat more&#xA;  complicated.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Returning to the analogy, turning dS space as a whole into AdS space&#xA;  is like turning the whole valley into a hill.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Here is a picture for the analogy. Scroll down to the concave meniscus&#xA;  vs convex meniscus.&#xA;  &lt;a href=&quot;http://www.tutorvista.com/content/physics/physics-iii/solids-and-fluids/shape-meniscus.php&quot; rel=&quot;nofollow&quot;&gt;http://www.tutorvista.com/content/physics/physics-iii/solids-and-fluids/shape-meniscus.php&lt;/a&gt;&#xA;  In the concave meniscus, the vectors normal to the surface (labeled R)&#xA;  diverge; in the convex meniscus, they converge.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In the previous paragraph, you can see two formulas, basically &quot;Force1&#xA;  minus Force2&quot; and &quot;Force2 minus Force1&quot;, which correspond to the two&#xA;  situations. You can get one formula by multiplying the other by -1.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So here is the analogy. If you start with the force formula for a&#xA;  concave meniscus, and want to get the force formula for a convex&#xA;  meniscus, you perform the transformation &quot;multiply by minus 1&quot;. That's&#xA;  analogous to: start with the quantum propagator for a scalar field in&#xA;  AdS space, and get the propagator in dS space by the transformation&#xA;  &quot;perform a Wick rotation&quot;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;But if you actually want to turn a concave meniscus into a convex&#xA;  meniscus, you're not just &quot;multiplying the meniscus by minus 1&quot;. What&#xA;  you would have to do is increase the intermolecular tension&#xA;  everywhere, until it changes the curvature.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In the same way, the uplift from AdS space to dS space involves&#xA;  changing the local curvature everywhere, by having extra fields whose&#xA;  energy density acts as an extra source of curvature (this is in the&#xA;  &quot;Landscaping&quot; paper by Silverstein and Polchinski).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So my first point is that uplifting isn't Wick rotation. It actually&#xA;  means that you add something to your model, which will make the space&#xA;  curve differently. What the Wick rotation allows you to do, is to take&#xA;  a formula appropriate for one type of space, and modify it so it&#xA;  applies to the other type of space - but these are formulas which&#xA;  describe what happens in a small patch of the space, they aren't&#xA;  describing its total structure.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So far, I've only been talking about physics in the bulk (dS or AdS).&#xA;  The existence of a boundary CFT, and the details of how it relates to&#xA;  the bulk physics, wasn't even discussed yet. Remember that half of&#xA;  what David Lowe writes about is just a mapping from the center of the&#xA;  bulk to the edge of the bulk, that is, it's something entirely within&#xA;  bulk physics. For example, it's about how the wavefunction of a&#xA;  particle changes as it travels from the interior to the edge. Then, at&#xA;  the edge of the bulk, is the truly holographic mapping; or at least,&#xA;  the edge of the bulk is the place where a point in the bulk maps to a&#xA;  single point on the boundary. (Points deep in the interior of the bulk&#xA;  correspond to some sort of sum over whole regions on the boundary.)&#xA;  The truly holographic mapping is this change in variables - from bulk&#xA;  variables to boundary variables.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So a big part of what we should be interested in is, what happens to&#xA;  the holographic change in variables if we go from AdS to dS. But there&#xA;  might be an easier case to study first: what happens to this change in&#xA;  variables if the curvature of AdS space changes (this would usually be&#xA;  expressed in terms of the &quot;radius&quot; of the AdS space, the distance from&#xA;  center of bulk to edge of bulk).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If you look up the notes of Thomas Hartman's talk from Strings 2011,&#xA;  he actually says how the boundary theory changes for Vasiliev's&#xA;  higher-spin gravity, as you go from AdS to dS. In AdS space, the&#xA;  boundary theory is a vector field theory with an O(N) symmetry, in dS&#xA;  space the boundary theory is now a scalar field theory with a Sp(N)&#xA;  symmetry (N is the number of vector fields or number of scalar&#xA;  fields). So going from AdS to dS requires the set of fields to change&#xA;  rather radically - though apparently there's still enough of a&#xA;  similarity that he and Anninos were able to deduce or guess that Sp(N)&#xA;  symmetry would be the dS counterpart. (Let me also say in passing that&#xA;  they have only solved the problem - if they &lt;em&gt;have&lt;/em&gt; solved the problem;&#xA;  no paper yet... - for a particular case of AdS/CFT. When the bulk&#xA;  theory is string theory rather than Vasiliev's higher-spin gravity,&#xA;  the boundary theory is a supersymmetric field theory that Hartman&#xA;  doesn't know how to transform appropriately; he says this in questions&#xA;  after his talk.)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If you see Hartman's notes, I think you'll also find that he talks&#xA;  about the cosmological constant in the bulk space going from negative&#xA;  to positive. This just refers to the curvature, and the difference&#xA;  between AdS and dS. But you can also talk about the difference between&#xA;  AdS with strong curvature and AdS with weak curvature. This also&#xA;  requires a change in the boundary theory, but not such a big change.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;One reason AdS/CFT is popular is that it is a 'strong-weak' duality.&#xA;  When quantum theories have strong &quot;couplings&quot; - when the interactions&#xA;  are strong - they become difficult or impossible to calculate with,&#xA;  because the usual approach of successive approximations - perturbation&#xA;  theory - doesn't work. The whole idea of perturbation theory is that&#xA;  each correction is smaller and less important than the last, so in&#xA;  order to approximate the actual result, you just keep adding smaller&#xA;  and smaller corrections. But when the interaction is strong, the&#xA;  corrections get bigger as you go, which means that the whole starting&#xA;  point was wrong.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In AdS/CFT, typically, when the boundary theory is strongly coupled,&#xA;  the bulk theory is weakly coupled, and vice versa. So if you have a&#xA;  strongly coupled field theory, it may have an AdS dual gravity&#xA;  description which is weak and therefore calculable. Or if you have a&#xA;  gravity theory in a strong coupling state, it may be dual to a weakly&#xA;  coupled boundary field theory.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If the curvature of the AdS space gets bigger, that should mean that&#xA;  the gravitational coupling is getting stronger, but in turn that means&#xA;  that the boundary coupling is getting weaker. And conversely, if the&#xA;  AdS space is getting flatter - more weakly curved, and approaching the&#xA;  transition to dS space - that should mean the boundary theory is&#xA;  getting strongly coupled. And even independently of AdS/CFT, it's well&#xA;  known that strongly coupled theories often have a dual description in&#xA;  terms of new variables which will be more easy to calculate with. So&#xA;  possibly this is an aspect of what happens on the boundary as you go&#xA;  from AdS to dS.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4719" LastActivityDate="2011-08-12T17:44:46.133" />
  <row Id="13458" PostTypeId="1" CreationDate="2011-08-12T17:47:01.683" Score="10" ViewCount="148" Body="&lt;p&gt;At one point, I decided to make friends with the low-lying spectrum of QCD. By this I do not mean the symmetry numbers (the &quot;quark content&quot;), but the actual dynamics, some insight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pions are the sloshing of the up-down condensate, and the other pseudoscalars by extending to strangeness. Their couplings are by soft-particle theorems. The eta-prime is their frustrated friend, weighed down by the instanton fluid. The rho and omega are the gauge fields for flavor SU(2), and A1(1260) gauges the axial SU(2), and they have KaluzaKlein-like echoes at higher energies, these can decay into the appropriate &quot;charged&quot; hadrons with couplings that depend on the flavor symmetry multiplet. The proton and the neutron are the topological defects. That accounts for everything up to 1300 but a few scalars and the b1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are scalars starting at around 1300 MeV which are probably some combination of glue-condensate sloshing around and quark-condensate sloshing around, some kind of sound in the vacuum glue. Their mass is large, their lifetime is not that big, they have sharp decay properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, there is nothing in AdS/QCD which should correspond to the sigma/f0(600), or (what seems to be) its strange counterpart f0(980). While looking around, I found this discussion: &lt;a href=&quot;http://www.physicsforums.com/showthread.php?t=241073&quot; rel=&quot;nofollow&quot;&gt;http://www.physicsforums.com/showthread.php?t=241073&lt;/a&gt;. The literature that it pointed to suggests that the sigma is a very unstable bound state of pions (or, if you like, tetraquarks).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This paper gives strong evidence for an actual pole: &lt;a href=&quot;http://arxiv.org/PS_cache/hep-ph/pdf/0512/0512364v2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/hep-ph/pdf/0512/0512364v2.pdf&lt;/a&gt;. A more cursory review is &lt;a href=&quot;http://arxiv.org/PS_cache/hep-ph/pdf/0608/0608218v1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/PS_cache/hep-ph/pdf/0608/0608218v1.pdf&lt;/a&gt;. The location of the pole is far away from the real axis, the width is larger than the mass by 20% or so, and the mass is about 400MeV. The authors though are confident that it is real because they tell me that the interpolation the interactions of pions is safe in this region because their goldstone properties dominate the interactions. I want to believe it, but how can you be sure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this particle was controversial. I want to understand what kind of picture this is giving. The dispersion subtraction process is hard for me to visualize in terms of effective fields, and the result is saying that there is an unstable bound state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a physical picture of the sigma which is more field theoretical, perhaps even just an effective potential for pions? Did anyone who convinced himself of the reality of the sigma have a way of understanding the bound state properties? Is there an analog unstable bound state for other goldstone bosons? Any insight would be welcome.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-12T17:47:01.683" Title="What the heck is the sigma (f0) 600?" Tags="&lt;particle-physics&gt;" CommentCount="3" FavoriteCount="2" />
  <row Id="13459" PostTypeId="2" ParentId="13455" CreationDate="2011-08-12T18:17:13.970" Score="3" ViewCount="" Body="&lt;p&gt;As a technical term in quantum mechanics, &lt;a href=&quot;http://en.wikipedia.org/wiki/Identical_particles&quot; rel=&quot;nofollow&quot;&gt;two particles are indistinguishable if they cannot be distinguished even in principle&lt;/a&gt;. Typically, this means that they are identical other than their spin, momentum, position, energy and the time at which they exist. What this implies is that the wave state that describes the pair of particles has to be symmetrized or anti-symmetrized. In particular, different isotopes are distinguishable. One writes:&lt;br&gt;&#xA;$$|AB\rangle = (|A\rangle|B\rangle \pm |B\rangle|A\rangle)/\sqrt{2}$$&lt;br&gt;&#xA;where the sign is + for symmetrization of bosons, or - for the antisymmetrization of fermions. On the other hand, if the two states are not identical, then the wave state is not (anti)symmetrized:&lt;br&gt;&#xA;$$|AB\rangle = |A\rangle|B\rangle$$&lt;br&gt;&#xA;You can think of indistinguishability as arising naturally from the wave nature of particles. You might be able to write descriptive letters on particles, but how can you distinguish between two identical waves?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose you had a system where spin was conserved and never altered. An electron would be either spin-up or spin-down and never change and so you could distinguish between two such electrons. It would be as if they were distinct particles. This concept shows up in isospin where a neutron is treated as distinguishable from a proton even though they form an isospin doublet that is similar to the spin doublet (spin-down,spin-up). In such a restricted system, with a state of two particles you have the choice of writing your wave states either symmetrized, anti-symmetrized, or unsymmetrized. Any (consistent) way you choose, you'll get the same answer when you make your calculations as the two states form separate Hilbert spaces. However, if the instructor looks carefully at your derivation, it could be marked wrong even though it gives the correct answer -- it's better to follow the rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To learn more about this subtlety, read up on superselection sectors. What's going on is that when your Hilbert space divides into sectors which cannot physically interact, you do not have to symmetrize or antisymmetrize. But as I say above, you don't need to know this detail in order to calculate. Just follow the rules; only identical particles (in the sense of identical other than position, energy, momentum, orientation, etc.) are either symmetrized or antisymmetrized.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-08-12T18:17:13.970" />
  <row Id="13460" PostTypeId="1" CreationDate="2011-08-12T19:37:23.793" Score="2" ViewCount="61" Body="&lt;p&gt;What is the best reference for an introduction to the use of Fourier analysis in crystallography?&lt;/p&gt;&#xA;" OwnerUserId="4521" LastEditorUserId="2151" LastEditorDisplayName="" LastEditDate="2011-08-12T20:31:07.563" LastActivityDate="2011-08-15T10:55:23.880" Title="Fourier analysis in crystallography" Tags="&lt;mathematical-physics&gt;&lt;condensed-matter&gt;&lt;solid-state-physics&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13461" PostTypeId="2" ParentId="13444" CreationDate="2011-08-12T20:06:26.197" Score="1" ViewCount="" Body="&lt;p&gt;Pete, please elaborate your question, the original paper describes by Miguel describes how one can travel back and to a destination using the warp drive. The space in front is contracted to get closer to the destination, and the space behind is expanded.  For getting out of the bubble, let's use the picture below: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/75H6b.png&quot; alt=&quot;Warp Bubble, credits go the respective authors/artists.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bubble is created, so to speak, through the exotic matter that one will use. So if the mechanism that creates the exotic matter to keep up the bubble has a stop button, you press it and the bubble will go away :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps. &lt;/p&gt;&#xA;" OwnerUserId="4719" LastActivityDate="2011-08-12T20:06:26.197" CommentCount="2" />
  <row Id="13462" PostTypeId="1" AcceptedAnswerId="13464" CreationDate="2011-08-12T20:10:54.480" Score="2" ViewCount="138" Body="&lt;p&gt;As I understand it, the completed Large Hadron Collider (LHC) will ultimately have a proton beam with $2,808$ bunches of $1.15 \times 10^{11}$ protons each at $7$ TeV, giving a total beam energy of $(2808)(1.15\times 10^{11})(7 \times 10^{12})(1.602\times 10^{-19}$ J$)$ $\approx 362 $ MJ.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that instead of using a graphite block for the beam dump, the proton beam is aimed &quot;straight up&quot; at a satellite in geosynchronous orbit 42,160 km above the surface of the earth.  How significantly would be the beam intensity be diminished in the atmosphere, particularly the troposphere?  Could one reasonable expect to be able to blind sensors/cameras on the satellite?&lt;/p&gt;&#xA;" OwnerUserId="4257" LastEditorUserId="4257" LastEditorDisplayName="" LastEditDate="2011-08-13T00:37:17.240" LastActivityDate="2011-08-13T14:07:37.010" Title="Could the LHC &quot;blind&quot; a satellite?" Tags="&lt;large-hadron-collider&gt;&lt;proton&gt;&lt;order-of-magnitude&gt;" AnswerCount="3" />
  <row Id="13463" PostTypeId="2" ParentId="13462" CreationDate="2011-08-12T20:26:24.100" Score="2" ViewCount="" Body="&lt;p&gt;Extremely unlikely.  The beam would diffuse very rapidly.  The LHC beam is condensed into a very small location by magnetic fields and it's energy is maintained through the use of an RF field which replenishes the energy each time the beam circulates while orbiting in a near perfect vacuum. In the absence of such fields the beam would first repel itself through Coluombic repulsion, and then rapidly lose energy through interaction with the atmosphere due to the loss of vacuum.  It would pose no risk to a satellite so far away, especially for a satellite that has frequent interaction with space radiation.  &lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2011-08-12T20:26:24.100" CommentCount="6" />
  <row Id="13464" PostTypeId="2" ParentId="13462" CreationDate="2011-08-12T20:36:52.767" Score="3" ViewCount="" Body="&lt;p&gt;Hmmm...some back-of-the-envelope calculations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The depth of the air column at sea level is $14\text{ lbs/in}^2 = 2 \times 10^5\text{ g/cm}^2$, so neglecting space-charge effects and assuming minimum ionization the whole way we get about $4 \times 10^5\text{ MeV} = 0.4\text{ TeV}$ energy loss. We are actually above minimum ionization, so we can multiply that by a small integer. Call it 1 TeV, which justifies the approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another consideration: multiple scattering. The radiation length of air is around $36\text{ g/cm}^2$, so the protons travel through about $5.5 \times 10^3$ radiation length resulting in a RMS position dispersion of about 5 km. Making for a mean areal proton density at geosynchronous orbit of around 1500 per square meter at the top of the atmosphere and expanding in a cone for the next 42,000 kilometers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those protons are rather more energetic than a typical cosmic ray, but not at all unknown in that environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusions:&lt;/strong&gt; no prompt effects, but possible a reduced lifetime.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-13T00:28:13.830" LastActivityDate="2011-08-13T00:28:13.830" CommentCount="7" />
  <row Id="13465" PostTypeId="1" CreationDate="2011-08-12T22:43:40.893" Score="1" ViewCount="32" Body="&lt;p&gt;I think I know how do we define a vacuum in flat space QFT and also in a curved space QFT. But, can somebody tell me how do the choice of vacuum state in say the CFT side of AdS/CFT changes the choice of vacuum state in gravity side? Let me ask the other way. I mean if we pick a vacuum (say in bulk side, because it may not be unique), how does it reflect on the CFT vacuum (and vice versa)? The CFT side is conformally flat. So, there is no reason in the CFT side that there will be a unique vacuum, is there? So, how does this choice reflect on both sides and how do we generally make the identification?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1349" LastActivityDate="2011-08-12T22:43:40.893" Title="Choice and identification of vacuums in AdS/CFT" Tags="&lt;ads-cft&gt;" />
  <row Id="13466" PostTypeId="1" CreationDate="2011-08-12T23:15:27.420" Score="7" ViewCount="179" Body="&lt;p&gt;In Landau–Lifshitz's &lt;em&gt;Course of Theoretical Physics&lt;/em&gt;, Vol. 2 (‘Classical Fields Theory’), Ch. IV, § 27, there is an explanation why the field equations should be linear differential equations. It goes like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Every solution of the field equations gives a field that can exist in nature. According to the principle of superposition, the sum of any such fields must be a field that can exist in nature, that is, must satisfy the field equations.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As is well known, linear differential equations have just this property, that the sum of any solutions is also a solution. Consequently, the field equations must be linear differential equation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually, this reasoning is not logically valid. Not only the authors forget to explain the word ‘differential’, but they also do not actually prove that the field equations must be linear. (Just in case: this observation is not due to me.) But it seems that the last issue can be easily overcome. However, it is exactly the word ‘differential’, not ‘linear’, that is bothering me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a nice &lt;a href=&quot;http://en.wikipedia.org/wiki/Peetre_theorem&quot; rel=&quot;nofollow&quot;&gt;theorem of Peetre&lt;/a&gt; stating that the linear operator $D$ that acts on (the ring of) functions and does not increase supports, that is, $\mathop{\mathrm{supp}} f \supset \mathop{\mathrm{supp}} Df$, must be a differential operator. The property of preserving supports can be considered as a certain &lt;em&gt;locality&lt;/em&gt; property. Hence, the field equations must be differential because all interactions must propagate with a finite velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is another notion of ‘locality’ of an operator: the operator $D$ is called &lt;em&gt;local&lt;/em&gt; if the function $Df$ in the neighbourhood $V$ can be computed with $f$ determined only on $V$ as well, i.e., $(Df)|_V$ is completely defined by $f|_V$. The locality in this sense is not equivalent to the locality in the sense of supports' preserving. (Unfortunately, I do not have an illustrative example at hand right now, so there is a possibility of mistake $M$ hiding here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: what physical circumstances determine the (correct one) notion of locality for a given physical problem? (Assuming there is no mistake $M$.) And does my reasoning &lt;em&gt;really&lt;/em&gt; justifies the word ‘differential’ in the context of field equations? If so, are there any references containing more accurate argument than the one presented in Landau–Lifshitz's Course?&lt;/p&gt;&#xA;" OwnerUserId="4720" LastActivityDate="2011-08-13T18:10:39.790" Title="Why must the field equations be differential?" Tags="&lt;mathematical-physics&gt;&lt;field-theory&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" />
  <row Id="13467" PostTypeId="2" ParentId="9818" CreationDate="2011-08-13T00:18:04.777" Score="0" ViewCount="" Body="&lt;p&gt;The moment Alice passes through the event horizon, she is doomed to death, and nothing she observes will come out. How can the interior of the black hole exist if the only observer is dead?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What Quentin is doing is committing quantum suicide. Subjectively, he will have &lt;strong&gt;quantum immortality&lt;/strong&gt;, and &lt;em&gt;will never die&lt;/em&gt;. He will always measure his qubit to have value 1, and be spared a visit to the nonexistent interior. It's a &lt;strong&gt;miracle&lt;/strong&gt;! On the other hand, other observers will see him die with a 50% probability.&lt;/p&gt;&#xA;" OwnerUserId="4909" LastActivityDate="2011-08-13T00:18:04.777" />
  <row Id="13468" PostTypeId="2" ParentId="13441" CreationDate="2011-08-13T01:34:05.140" Score="1" ViewCount="" Body="&lt;p&gt;Notable exception: Reynolds number.&lt;/p&gt;&#xA;" OwnerUserId="3887" LastActivityDate="2011-08-13T01:34:05.140" />
  <row Id="13469" PostTypeId="2" ParentId="13441" CreationDate="2011-08-13T01:57:24.287" Score="1" ViewCount="" Body="&lt;p&gt;One more exception: weak localization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second class of exceptions (Lagerbaer named first where interaction constant is involved) originates from two characteristic lengths/times/... in the system. [Function of] their ratio may appear in the answer. &lt;/p&gt;&#xA;" OwnerUserId="4657" LastActivityDate="2011-08-13T01:57:24.287" />
  <row Id="13470" PostTypeId="2" ParentId="13444" CreationDate="2011-08-13T04:17:56.830" Score="1" ViewCount="" Body="&lt;p&gt;The interior of the bubble is causally disconnected. It's not possible for the bubble to be turned off or steered from the inside. But there is no reason it cannot be affected from an outside agency at a pre-planned points, or even simply have a finite lifetime, naturally deteriorating to stop at the intended destination.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;that upon traveling to some distant star system, since you never moved through space, once you turn off the bubble, and the space behind you contracts back to its normal state and the space in front of you expands back to its normal state, you're back where you started&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is not the case. The statement &quot;you never moved through space&quot; just means you never move relative to the bubble interior, which is different from the bubble itself moving through space. Even if the bubble accelerates, you will not experience any g-forces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, one can think of it as dual to the cosmic expansion of space: as space expands, it carries galaxies along with it, and because it's space between them that's expanding rather than them moving in space, distant galaxies can have superluminal separation velocity. Effective movement because of the way space expands (or contracts) is different from movement &lt;em&gt;in&lt;/em&gt; space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if cosmic expansion could somehow be &quot;turned off&quot;, it wouldn't suddenly make all the galaxies contract together again. It would simply stop further separation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Alcubierre drive does something similar: instead of expanding space to get away from from distant objects, it contracts it in order to approach them. It doesn't actually need to contract all of space in front of it; it just expands it back after traversing it. In effect, the warp bubble rides its own gravitational field.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Although Newtonian analogies are fraught with peril (esp. here, since not only is gravity different, but the behavior of negative mass especially so~don't take this seriously!), but a fun little toy exercise is to consider what would happen if you lived in a Newtonian universe and had two masses, equal in magnitude but one positive and the other negative.&lt;/p&gt;&#xA;" OwnerUserId="1353" LastActivityDate="2011-08-13T04:17:56.830" CommentCount="2" />
  <row Id="13473" PostTypeId="2" ParentId="13466" CreationDate="2011-08-13T08:15:03.333" Score="1" ViewCount="" Body="&lt;p&gt;This does not really answer your question why the equation should be differential. But&#xA;I think that the two notions of locality you mentioned are just equivalent, if I am not mistaken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us prove that the second definition impies the first one. One has to show that if a point $x$ does not belong to $supp(f)$ then $x$ does not belong to $supp(Df)$. Indeed, then there exists an open neighborhood $V$ of $x$ such that $f|_V=0$. Hence by the assumption $Df|_V=0$. Hence $x\not\in supp(Df)$ as requested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us prove now the converse statement that the first definition implies the second one. Assume that $f|_V=g|_V$. Then $(f-g)|_V=0$. Hence $supp(f-g)\cap V=\emptyset$. Consequently $supp(D(f-g))\cap V=\emptyset$, i.e. $D(f-g)|_V=0$. That means that $Df|_V=Dg|_V$.&lt;/p&gt;&#xA;" OwnerUserId="4562" LastActivityDate="2011-08-13T08:15:03.333" />
  <row Id="13474" PostTypeId="1" AcceptedAnswerId="13497" CreationDate="2011-08-13T08:46:26.940" Score="10" ViewCount="161" Body="&lt;p&gt;I'm talking about a Weighing Balance shown in the figure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Rkhle.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Press &amp;amp; Hold on onside of the horizontal beam and then release it. It makes some oscillations and comes back to equilibrium like shown in the figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the pans are of equal equal masses. When the horizontal beam is tilted by an angle using external force, the torque due to these pan weights are equal in magnitude &amp;amp; opposite in direction. Then why does it come back to it position? What's making it to come back.&lt;/p&gt;&#xA;" OwnerUserId="447" LastActivityDate="2011-08-15T10:34:52.377" Title="Why does the weighing balance restore when tilted and released" Tags="&lt;rotational-dynamics&gt;&lt;torque&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="0" />
  <row Id="13475" PostTypeId="2" ParentId="13474" CreationDate="2011-08-13T09:32:21.590" Score="1" ViewCount="" Body="&lt;p&gt;Nice question! &lt;strong&gt;if&lt;/strong&gt; the following analogy applies : imagine a seesaw on a half-sphere fulcrum (top of the picture). if it inclines e.g. to left side (bottom) - the length from the right edge to fulcrum ($L_2$) increases, the lever rule kicks in ($F_2&gt;F_1$) and the weight of the right side brings the seesaw back to equilibrium (top) (which is then broken again by inertia)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/15TRw.png&quot; alt=&quot;seesaw oscillation&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4760" LastEditorUserId="4760" LastEditorDisplayName="" LastEditDate="2011-08-15T06:18:31.000" LastActivityDate="2011-08-15T06:18:31.000" CommentCount="4" />
  <row Id="13476" PostTypeId="2" ParentId="13441" CreationDate="2011-08-13T10:02:56.313" Score="1" ViewCount="" Body="&lt;p&gt;A notable exception would be Debye's model for low temperatures. The molar heat capacity is then approximated as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ C = 234 \ k \ \left( \frac{T}{T_D} \right)^3 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with k the Boltzmann constant and $T_D$ the Debye temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the biggest reason why constant are always so small is that if they're not small, they're absorbed in other quantities.&lt;/p&gt;&#xA;" OwnerUserId="3234" LastActivityDate="2011-08-13T10:02:56.313" />
  <row Id="13477" PostTypeId="2" ParentId="13466" CreationDate="2011-08-13T10:33:47.867" Score="2" ViewCount="" Body="&lt;p&gt;All physical phenomena occur on a substrate, medium or 'space'. This space has properties which guarantee the stability of the system, such as the existence of a characteristic&#xA; propagation speed of 'c'. The reaction allowed at each point in space, and at every moment, is strictly due to the local environment, its immediate neighborhood and so 'action at a distance' is not permitted, nor infinities.&lt;br&gt;&#xA;This physical environment (1D) is illustrated in this image&lt;img src=&quot;http://i.stack.imgur.com/tfjFU.jpg&quot; alt=&quot;image (by Hans de Vries)&quot;&gt; (by Hans de Vries) and found in his online book &lt;a href=&quot;http://www.physics-quest.org/&quot; rel=&quot;nofollow&quot;&gt;&quot;Understanding Relativistic Quantum Field Theory&quot;&lt;/a&gt;. This medium allows the propagation of the classical wave equation &#xA;$$\frac{\partial^{2}\psi}{\partial t^{2}}=c^{2}\frac{\partial^{2}\psi}{\partial\mathit{x}^{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ψ is the vertical displacement in the mechanical model. ...the&#xA;  equation is satisfied by any arbitrary function which shifts along&#xA;  with a speed v (or −v). A function ”stretched” by a factor v has it’s&#xA;  slopes decreased by a factor v, while it’s second order derivatives&#xA;  are lower by a factor $v^2$ .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-08-13T10:33:47.867" />
  <row Id="13478" PostTypeId="2" ParentId="13413" CreationDate="2011-08-13T11:39:49.577" Score="0" ViewCount="" Body="&lt;p&gt;To hit a ball more interaction area is of advantage - the thinner baseball bat demands higher precision, in contrast to cricket bat.&#xA;If one needs to break something, the applied pressure per area is important. Assuming same masses, the edge of the cricket bat offers smaller impact surface and has higher potential for damage. On the other hand greater length of the baseball bat allows for higher momentum.&lt;/p&gt;&#xA;" OwnerUserId="4760" LastActivityDate="2011-08-13T11:39:49.577" />
  <row Id="13480" PostTypeId="1" AcceptedAnswerId="13484" CreationDate="2011-08-13T13:21:18.483" Score="13" ViewCount="203" Body="&lt;p&gt;A typical problem in quantum mechanics is to calculate the spectrum that corresponds to a given potential. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is there a one to one correspondence between the potential and its spectrum? &lt;/li&gt;&#xA;&lt;li&gt;If the answer to the previous question is yes, then given the spectrum, is there a systematic way to calculate the corresponding potential ?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4521" LastActivityDate="2011-08-14T01:51:19.153" Title="In quantum mechanics, given certain energy spectrum can one generate the corresponding potential?" Tags="&lt;quantum-mechanics&gt;&lt;mathematical-physics&gt;" AnswerCount="2" />
  <row Id="13481" PostTypeId="2" ParentId="13264" CreationDate="2011-08-13T13:35:13.867" Score="-1" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stokes%27_law_%28sound_attenuation%29&quot; rel=&quot;nofollow&quot;&gt;Stoke's Law&lt;/a&gt; states that sound attenuation is inversely proportional to pressure - so at higher altitudes one can hear worse.  &lt;/p&gt;&#xA;" OwnerUserId="4760" LastActivityDate="2011-08-13T13:35:13.867" CommentCount="1" />
  <row Id="13483" PostTypeId="2" ParentId="13462" CreationDate="2011-08-13T14:07:37.010" Score="0" ViewCount="" Body="&lt;p&gt;Well, I will give the simple answer of the geometrical intensity fall off with 1/r**2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;r**2=2*10**15meterssquared  approximately for the satelite distance .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the 362MJ  for 1 meter squared as an initial area to make life easy, ( at about 1km height?) though in the beginning it is concentrated into microns; only a tiny fraction of a joule will end on the satellite if it has a cross section of 1meter squared: 1.3*10**-7 joules, give or take an order of magnitude because of my sloppiness, will hit the satellite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think even if there were no atmosphere there would be absolutely no danger for the satellite.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-08-13T14:07:37.010" CommentCount="2" />
  <row Id="13484" PostTypeId="2" ParentId="13480" CreationDate="2011-08-13T15:33:50.813" Score="14" ViewCount="" Body="&lt;p&gt;In general, the answer is no. This type of inverse problem is sometimes referred to as: &quot;Can one hear the shape of a drum&quot;. The following &lt;a href=&quot;http://www6.cityu.edu.hk/rcms/publications/ln10.pdf&quot; rel=&quot;nofollow&quot;&gt;extensive exposition&lt;/a&gt; by Beals and Greiner discusses various problems of this type. Despite the fact that one can get a lot of geometrical and topological information from the spectrum or even its asymptotic behavior, this information is not complete even for systems as simple as quantum mechanics along a finite interval.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastActivityDate="2011-08-13T15:33:50.813" CommentCount="8" />
  <row Id="13486" PostTypeId="2" ParentId="13441" CreationDate="2011-08-13T17:39:11.843" Score="3" ViewCount="" Body="&lt;p&gt;Here are two examples of where dimensional estimates fail:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Divergent expressions&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I have a laser pointer pointed at a wall directly facing me, 1 meter away. I turn the laser pointer by 90 degrees over the course of 1 second. What is the average speed of the laser spot during the turn? The dimensional analysis argument is 1meter/1second, which would be roughly right if I didn't have a singular function. Similar examples are resonant phenomena, where you have a lot more response than you would predict from dimensional analysis.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Exponential growth:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Suppose I have 20 particles in a box of mass M, in volume V, with total energy E. When will they all return to within 10% of their starting positions/velocities? If you ignore the dimensionless 19, then you will get the recurrence time for one particle, which is shorter by an enormous amount.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-13T17:39:11.843" />
  <row Id="13487" PostTypeId="2" ParentId="13466" CreationDate="2011-08-13T18:10:39.790" Score="-1" ViewCount="" Body="&lt;p&gt;Not all fields follow the superposition principle. Temperature field does not. Pressure field does, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those words in LL are an attempt to &quot;substantiate&quot; equations obtained mostly empirically in the frame of differential approach.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-08-13T18:10:39.790" />
  <row Id="13488" PostTypeId="1" CreationDate="2011-08-13T19:22:27.190" Score="3" ViewCount="70" Body="&lt;p&gt;This is really a basic question whose answer I guess may have to do with the way we construct Feynman rules and diagrams. The question is Suppose I have been given a two point function (found in some other ways, say for example some gauge/gravity duality or some symmetry in the theory). How can we construct the Lagrangian of that theory from there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a general rule for that? Can you give me a reference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1349" LastActivityDate="2011-08-13T19:22:27.190" Title="To construct an action from a given two point function" Tags="&lt;quantum-field-theory&gt;" CommentCount="10" FavoriteCount="1" />
  <row Id="13490" PostTypeId="2" ParentId="13480" CreationDate="2011-08-13T20:04:59.587" Score="5" ViewCount="" Body="&lt;p&gt;The Harmonic oscillator has the same spectrum as a weaker harmonic oscillator with a hard wall at x=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LATER EDIT: I see that I have to be more explicit--- the potentials&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;V(x)= 2x^2 - 2&lt;/li&gt;&#xA;&lt;li&gt;(x&gt;0) V(x)= x^2 - 3 and (x&amp;lt;0) V(x)= $\infty$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;have the exact same spectrum.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-14T01:51:19.153" LastActivityDate="2011-08-14T01:51:19.153" CommentCount="9" />
  <row Id="13491" PostTypeId="2" ParentId="13445" CreationDate="2011-08-13T21:05:51.123" Score="4" ViewCount="" Body="&lt;p&gt;The &lt;a href=&quot;http://www.sdss.org/&quot; rel=&quot;nofollow&quot;&gt;Sloan Digital Sky Survey&lt;/a&gt; is very open. You can browse their dataset online, you can &lt;a href=&quot;http://www.sdss3.org/dr8/data_access.php&quot; rel=&quot;nofollow&quot;&gt;query it using SQL over the CAS&lt;/a&gt;, they even have &lt;a href=&quot;http://www.sdss.org/DR7/tutorials/&quot; rel=&quot;nofollow&quot;&gt;tutorials&lt;/a&gt; to get you started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The SDSS was one of the big surveys of the 2000s, with thousands of &lt;a href=&quot;http://www.sdss.org/publications/&quot; rel=&quot;nofollow&quot;&gt;publications&lt;/a&gt; based on the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think you can actually download the raw data, as it is several terabytes long. It's usually transported by copying it to hard drives and flying it home, or using dedicated fat pipes between institutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some quick links:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://cas.sdss.org/dr7/en/tools/places/&quot; rel=&quot;nofollow&quot;&gt;SkyServer famous places&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://cas.sdss.org/dr7/en/tools/chart/navi.asp&quot; rel=&quot;nofollow&quot;&gt;Navigate the SDSS sky image&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://cas.sdss.org/dr7/en/tools/quicklook/quickobj.asp&quot; rel=&quot;nofollow&quot;&gt;Details for a galaxy, including its spectrum&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://cas.sdss.org/dr7/en/tools/getimg/&quot; rel=&quot;nofollow&quot;&gt;Download sky images (stripes)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://cas.sdss.org/dr7/en/proj/&quot; rel=&quot;nofollow&quot;&gt;Science projects&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="47" LastEditorUserId="47" LastEditorDisplayName="" LastEditDate="2011-08-13T21:12:04.593" LastActivityDate="2011-08-13T21:12:04.593" />
  <row Id="13492" PostTypeId="2" ParentId="13445" CreationDate="2011-08-13T22:07:34.353" Score="3" ViewCount="" Body="&lt;p&gt;You can use &lt;a href=&quot;http://en.wikipedia.org/wiki/File_Transfer_Protocol#Anonymous_FTP&quot; rel=&quot;nofollow&quot;&gt;anonymous FTP&lt;/a&gt; to get data from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Very_Large_Array&quot; rel=&quot;nofollow&quot;&gt;VLA&lt;/a&gt; in &lt;a href=&quot;http://en.wikipedia.org/wiki/Socorro,_New_Mexico&quot; rel=&quot;nofollow&quot;&gt;Socorro&lt;/a&gt;, New Mexico, USA. Look in &lt;em&gt;&lt;a href=&quot;https://archive.nrao.edu/archive/e2earchivex.jsp&quot; rel=&quot;nofollow&quot;&gt;The NRAO Data Archive System&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need to use some data processing software like &lt;a href=&quot;http://en.wikipedia.org/wiki/Astronomical_Image_Processing_System&quot; rel=&quot;nofollow&quot;&gt;AIPS&lt;/a&gt; or something. It's not exactly user friendly and it is a lot of data, but it shouldn't be too hard to get going if you're interested. The only thing is that this is just &lt;a href=&quot;http://en.wikipedia.org/wiki/Radio_frequency&quot; rel=&quot;nofollow&quot;&gt;RF&lt;/a&gt; data. &lt;/p&gt;&#xA;" OwnerUserId="4570" LastEditorUserId="174" LastEditorDisplayName="" LastEditDate="2011-08-17T22:34:06.717" LastActivityDate="2011-08-17T22:34:06.717" CommentCount="0" />
  <row Id="13493" PostTypeId="1" AcceptedAnswerId="13514" CreationDate="2011-08-13T22:26:29.007" Score="3" ViewCount="67" Body="&lt;p&gt;This &lt;em&gt;should&lt;/em&gt; be easy, but I think I have a mind-block... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\beta^-$-decay, what is the maximum possible momentum for the electron? The two equations I can use are conservation of energy and conservation of momentum, but I have three unknowns: Momentum of electron, nucleus and anti-neutrino, so what am I missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I just set the kinetic energy of the nucleus to zero and work from there?&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-13T23:11:39.383" LastActivityDate="2011-08-15T06:09:36.930" Title="Maximum electron momentum in $\beta^-$-decay" Tags="&lt;particle-physics&gt;&lt;radiation&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="13494" PostTypeId="1" AcceptedAnswerId="13501" CreationDate="2011-08-13T23:36:23.087" Score="3" ViewCount="82" Body="&lt;p&gt;I'm researching the effect of temperature on uranium radioactivity, however I can't find any solid empirical evidence to prove the notion that temperature does not affect radioactivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone link to some solid data on the topic, or offer some advice on where to find such data?&lt;/p&gt;&#xA;" OwnerUserId="3044" LastEditorUserId="3044" LastEditorDisplayName="" LastEditDate="2011-08-14T04:36:55.323" LastActivityDate="2011-08-14T04:36:55.323" Title="Effect of temperature on radioactivity?" Tags="&lt;nuclear-physics&gt;&lt;temperature&gt;&lt;reference-request&gt;&lt;radioactivity&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13495" PostTypeId="1" AcceptedAnswerId="13499" CreationDate="2011-08-14T00:11:40.003" Score="3" ViewCount="28" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A piston (with mass M) in a car engine is in vertical simple harmonic&#xA;  motion with amplitude A. The engine is running at a period T.  Suppose&#xA;  a small piece of metal with mass m were to break loose from the&#xA;  surface of the piston when it (the piston) is at the lowest point z&#xA;  and velocity is v.&#xA;  At what position in meters does the piece of metal lose contact with the piston?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also the maximum velocity and acceleration are given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The text book gives these equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x(t) = Acos(wt+φ)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v(t) = -wAsin(wt+φ)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a(t) = -w^2Acos(wt+φ)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I propose that the best way to figure this out would be to solve for the time at which acceleration is at its max, then plug that time into $x(t)$.  But I must first solve for the phase constant.  Is this correct?  If so, how do I find the phase constant?&lt;/p&gt;&#xA;" OwnerUserId="4675" LastActivityDate="2011-08-14T00:43:36.677" Title="How do I solve for the phase constant given the amplitude and the angular frequency?" Tags="&lt;homework&gt;&lt;harmonic-oscillator&gt;&lt;oscillators&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="13496" PostTypeId="1" CreationDate="2011-08-14T00:20:12.680" Score="3" ViewCount="71" Body="&lt;p&gt;A consistent history is a POVM set of observables corresponding to a time-ordered product of projection operators. For gauge theories, not any old operator will do, only gauge-invariant observables. In quantum gravity, there are no localized observables, meaning observables have to be defined with respect to the conformal boundary. Inside a black hole, this is none other than the black hole singularity, which is often spacelike. Spacelike singularities are enigmatic in string theory. This raises the very interesting question; can we have consistent histories inside black holes?&lt;/p&gt;&#xA;" OwnerUserId="4917" LastActivityDate="2011-08-14T15:10:59.340" Title="Can we have consistent histories inside a black hole?" Tags="&lt;quantum-gravity&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" />
  <row Id="13497" PostTypeId="2" ParentId="13474" CreationDate="2011-08-14T00:31:42.750" Score="3" ViewCount="" Body="&lt;p&gt;The horizantal beam on such scales is intentionally placed below the rotational axis. As long as the weights are in equilibrium the torque is equal on both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as soon as the position changes e.g. tipping the left scale down, the torques differ because only the tangential part of the gravitational force vector in relation to the rotational axis contributes to the torque around it. When tipping down the left scale, torque on the left side gets smaller and torque on the right side gets bigger, therefore the right side moves down again until equilibrium is reached (besides some swings to accommodate for the temporary impulse energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This effect gets the more pronounced as the distance of the horizontal bar approaches the half length of the bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This effect would not be if the horizontal bar went exactly through the axis.&lt;/p&gt;&#xA;" OwnerUserId="4916" LastEditorUserId="4916" LastEditorDisplayName="" LastEditDate="2011-08-14T00:38:34.127" LastActivityDate="2011-08-14T00:38:34.127" CommentCount="5" />
  <row Id="13498" PostTypeId="2" ParentId="13494" CreationDate="2011-08-14T00:37:08.653" Score="1" ViewCount="" Body="&lt;p&gt;There are suggestions that low (&gt;1K) temperatures slow the decay rate - at least for some large atoms &lt;a href=&quot;http://www.springerlink.com/content/61v5l002704013h4/&quot; rel=&quot;nofollow&quot;&gt;http://www.springerlink.com/content/61v5l002704013h4/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-08-14T00:37:08.653" CommentCount="3" />
  <row Id="13499" PostTypeId="2" ParentId="13495" CreationDate="2011-08-14T00:43:36.677" Score="3" ViewCount="" Body="&lt;p&gt;The phase constant is needed only if you have a specific initial condition, e.g. if I told you where $x$ was at time $t = 0$, you could solve for $\varphi$. Otherwise you can just choose whatever you want for it: Note that it is the same in all functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choosing some value for $\varphi$ is analogous to you manually setting the time origin to something you like. Imagine observing the harmonic motion for a while, and then deciding that $t = 0$ should be the time where the piston is all the way up. Or you could decide that $t = 0$ should be the time where the piston is all the way down. Or right in the middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, for your problem, you can just set $\varphi = 0$. &lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-08-14T00:43:36.677" CommentCount="1" />
  <row Id="13500" PostTypeId="1" AcceptedAnswerId="13504" CreationDate="2011-08-14T01:08:27.627" Score="11" ViewCount="199" Body="&lt;p&gt;I have very little background in physics, so I apologize if this question is painfully naive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following thought experiment: an observer is in a closed room whose walls, floor, and ceiling are made entirely of mirrors, with a single light source in the middle of the room. When the light is on, the observer can see many copies of his reflection all over the place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suddenly, the light source turns itself off. Intuitively, I would expect the observer to &quot;instantly&quot; see darkness. However, I can't figure out why that is the case under the &quot;particle&quot; interpretation of light. There are obviously lots of photons already in the room from before. Furthermore, we know that they don't get &quot;consumed&quot; when they hit a wall, because otherwise the observer wouldn't see so many reflections of himself. Basically, when the light goes off, what happens to the photons already in the room?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect the answer goes something like this: the photons in the room lose a little bit of energy every time they bounce off a mirror, but it's so minuscule that we can still see more reflections than our eye can resolve anyway. When the light goes off, however, it takes them a very small fraction of a second to bounce around the room enough times to diffuse completely, which our eye cannot detect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that about right? If we had a theoretical &quot;perfect reflector&quot;, would the light remain trapped in the room forever? If we had instruments that could measure such things very finely, would it take (slightly) longer for the light to go out in a room made of mirrors as opposed to a room made of, say, black cloth?&lt;/p&gt;&#xA;" OwnerUserId="1136" LastActivityDate="2011-08-14T01:52:59.527" Title="How are photons &quot;consumed&quot;?" Tags="&lt;optics&gt;&lt;photons&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="13501" PostTypeId="2" ParentId="13494" CreationDate="2011-08-14T01:10:07.803" Score="3" ViewCount="" Body="&lt;p&gt;What do you mean by &quot;prove&quot;? If you mean in a strict mathematical sense, then looking for such assurances is a lost cause.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are quite a wide variety of papers on this matter. Curie attacked this particular problem in 1913 with radium. They immersed a radium source in liquid hydrogen for more than an hour and didn't find a change of more than 0.1% in its activity. You can read more from the paper by Curie &amp;amp; Kamerlingh Onnes entitled, &quot;The radiation of radium at the temperature of liquid hydrogen&quot; in KNAW, Proceedings, 15 II, 1912-1913, pp. 1430-1441. People even claimed, from Russia, that polonium's activity varied depending on geography. Hardly the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More recently, work has been done on the half-life decay rate of $^{97}Ru$ without seeing a noticeable temperature dependence near 20K compared with RT. See the paper by Goodwin, Golovko, Iacob and Hardy entitled, &quot;Half-life of the electron-capture decay of $^{97}$Ru: Precision measurement shows no temperature dependence&quot; in Physical Review C (2009), 80, 045501.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could be that there is a small dependence, but not even the Russian paper mentioned above by Martin agrees there is a measurable temperature dependence.&lt;/p&gt;&#xA;" OwnerUserId="4112" LastEditorUserId="4112" LastEditorDisplayName="" LastEditDate="2011-08-14T02:45:57.003" LastActivityDate="2011-08-14T02:45:57.003" CommentCount="3" />
  <row Id="13502" PostTypeId="2" ParentId="13474" CreationDate="2011-08-14T01:30:28.780" Score="0" ViewCount="" Body="&lt;p&gt;it seems to me that when I do the calculations for the torques considering the pivot as a single point in space they are equal. However I must assume that real scales are built such that any rotation through an angle $\theta$ will cause the pivot point to shift slightly from center causing unequal torques. The effect of this &lt;em&gt;net torque&lt;/em&gt; is to restore the beam to the horizontal no matter which direction in theta you rotate (e.g. whether you push down on the left or right piece, both case result in a restoring torque). I would also add that the oscillation die down due to frictional damping, the whole system is approximated by simple harmonic motion with damping.&lt;/p&gt;&#xA;" OwnerUserId="4570" LastEditorUserId="4570" LastEditorDisplayName="" LastEditDate="2011-08-14T02:58:30.757" LastActivityDate="2011-08-14T02:58:30.757" />
  <row Id="13503" PostTypeId="2" ParentId="13493" CreationDate="2011-08-14T01:46:59.413" Score="1" ViewCount="" Body="&lt;p&gt;The answer to your question is that the &lt;em&gt;beta endpoint energy&lt;/em&gt; is equal to the total Q value of the reaction.  Here is a graph of the spectrum from a beta emission, which is why we use internal conversion instead when looking for a monoenergetic source of beta particles.  This graph is used in the German Wikipedia, the y-axis is counts per unit energy and the x-axis is energy (of the electron).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ErAFB.png&quot; alt=&quot;Beta decay spectrum&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You asked for momentum instead, so use $E^2=(m_e c^2)^2 + (pc)^2$.  Do not use the classical analog because it is extremely common that beta decay gives relativistic electrons.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-08-14T01:46:59.413" CommentCount="4" />
  <row Id="13504" PostTypeId="2" ParentId="13500" CreationDate="2011-08-14T01:52:59.527" Score="13" ViewCount="" Body="&lt;p&gt;When being reflected by a mirror, the photons do not lose &quot;a tiny bit&quot; of energy. Either they are reflected unchanged, or they are completely absorbed. A good mirror will reflect most of the photons, but will absorb a small fraction of them as well, say $0.1\%$ of them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is: Your photons don't lose energy over time; what happens is that the room loses photons over time: For each time a photon hits a wall, there is some probability $p$ that it will get absorbed (&quot;consumed&quot;). The chance that it &lt;em&gt;doesn't&lt;/em&gt; get consumed after $N$ hits is $(1-p)^N$. Since the photons are very fast, they'll bounce off the walls very often in a short amount of time, so $N$ becomes really large really quick, and then $(1-p)^N$ becomes really small pretty fast, so after a short amount of time, all photons have been consumed with very high probability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important property of photons, that might not be entirely intuitive when coming from a &quot;wave&quot; background: The energy of an individual photon is determined entirely by the frequency of the light. Blue photons have higher energy than red photons. The &lt;em&gt;intensity&lt;/em&gt; of light is determined not by the energy of your photons, but by their number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If&lt;/em&gt; the photons would lose energy each time they bounce off a mirror, the reflections would change their color gradually, so that eventually blue light becomes red, then infrared etc. That doesn't happen: The mirrors don't change the color of the light. They only swallow some of the photons, i.e. they reduce the &lt;em&gt;intensity&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With perfect mirrors, you could indeed expect to never lose any photons. But since anyone &lt;em&gt;looking&lt;/em&gt; at the photons also absorbs then, the room would still get dark eventually. Unless there's nobody in there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be overly pedantic: Unless you keep the walls at zero temperature, you will always have some photons in the room as blackbody radiation. At &quot;normal&quot; temperatures, these photons are mostly in the infrared range, but if you make it really hot, the walls will start glowing.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-08-14T01:52:59.527" CommentCount="15" />
  <row Id="13505" PostTypeId="1" AcceptedAnswerId="13510" CreationDate="2011-08-14T02:06:16.057" Score="4" ViewCount="87" Body="&lt;p&gt;This question was inspired by: &lt;a href=&quot;http://physics.stackexchange.com/questions/13500/how-are-photons-consumed&quot;&gt;How are photons &amp;quot;consumed&amp;quot;?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine I have some number of photons, $N$, each of frequency $\nu$, moving randomly in a spherical &quot;wrap-around&quot; universe of volume $V$.  To clarify what I mean by 'wrap-around', if a photon reaches a point on the surface of the sphere it is immediately mapped to the antipodal point with no changes to it's properties or velocity vector.  The system is otherwise completely closed, and devoid of any other particles.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How might this system evolve over time?       &lt;/p&gt;&#xA;" OwnerUserId="4257" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-14T02:12:01.577" LastActivityDate="2011-08-14T07:30:30.623" Title="Photons in a &quot;wrap-around&quot; universe" Tags="&lt;electromagnetism&gt;&lt;photons&gt;" AnswerCount="1" />
  <row Id="13507" PostTypeId="2" ParentId="13445" CreationDate="2011-08-14T03:35:53.803" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/LHC@home&quot; rel=&quot;nofollow&quot;&gt;LHC&lt;/a&gt; did a new update with their &lt;a href=&quot;http://en.wikipedia.org/wiki/Berkeley_Open_Infrastructure_for_Network_Computing&quot; rel=&quot;nofollow&quot;&gt;BOINC&lt;/a&gt; software and they release a lot of data that can be analyzed, I actually haven't seen myself what data is being analyzed but essentially BOINC divides the data released from LHC into packets and then it is send to the users who download BOINC, and it does computations on those by using the user's resources who downloaded it and later the computers send the data back. I think looking into what BOINC does with the LHC's data maybe the best shot at realizing what LHC itself does with the collected data. &lt;/p&gt;&#xA;" OwnerUserId="4719" LastEditorUserId="174" LastEditorDisplayName="" LastEditDate="2011-08-17T22:34:16.837" LastActivityDate="2011-08-17T22:34:16.837" />
  <row Id="13508" PostTypeId="2" ParentId="13474" CreationDate="2011-08-14T04:04:30.147" Score="2" ViewCount="" Body="&lt;p&gt;It behaves this way because that's how it was built.  By adjusting the mass distribution, we could make a scale that flops to one side, is roughly balanced at all angles, etc.  However, those scales would not be useful, so the scale isn't built that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be assumed from the left/right symmetry of the picture that the system cannot decide which way to go, and so is at an equilibrium point.  This equilibrium will be stable if a small perturbation (rotating the beam a small angle) raises the center of mass.  It will be unstable if a small perturbation lowers the center of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond that, it is difficult to say how the center of mass moves simply by looking at your picture because we do not completely understand the mass distribution and the location of the pivot point.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When finding the center of mass, we can ignore any stationary pieces because we are only interested in the change of the height of the center of mass.  Additionally, if the pans hang freely down, it appears as if one will rise by the same amount the other falls, and thus they will not change the height of their center of mass when considered jointly.  They can also be ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume the rest of the scale rotates rigidly.  In that case, the center of mass of the rigid portion we're considering will be constrained to a circle with its center at the pivot point.  If the center of mass  is exactly at the bottom of the circle, we have a stable equilibrium.  Otherwise, it is unstable.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-08-14T04:04:30.147" CommentCount="2" />
  <row Id="13509" PostTypeId="2" ParentId="5764" CreationDate="2011-08-14T05:04:30.610" Score="2" ViewCount="" Body="&lt;p&gt;The qualitative way to understand the reason for string dualities, like almost everything else in string theory, is the holographic principle. The first thing you learn from black hole decay is that every exact global symmetry must be a local symmetry, because if you have no gauge field, the symmetry would not be conserved in black hole formation and evaporation. A black hole forgets about everything except the gauge fields on its surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black hole decay also tells you that you can't make charges small in gravity without making masses small. The reason is that if you make a black hole from charged objects, it must be able to decay to enough charged particles to be left with no charge by the time it vanishes. In order for this to be statistically possible, the polarizing field on the horizon must be able to preferentially create charged particles. In order for this to happen when the quantum of charge is small, the charged particles generically must be light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, gravity requires either charge quantization or massless charged particles, because if you have the proton, say, with infinitesimally more charge than the electron, you can throw a proton into a black hole and wait for an anti-electron to come out. The result will be an infinitesimal charge black hole, and this would have to decay to a particle with infinitesimal charge, which would have to be light compared to its charge. So if there is no charge quantization, there are lots of nearly massless teensy charged particles in the theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The holographic principle states that if you have an extreme charged black hole which is thermodynamically trivial (is zero temperature), you can describe the objects near it by the oscillations of its surface. This tells you that you should be able to describe strings near a stack of D-branes by oscillations of the D-branes themselves, without mentioning the strings. This is the principle of AdS/CFT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, if I have a limit where certain pointlike black holes become light, I should be able to describe the entire string spectrum in terms of their motion against each other. This is the principle of Matrix theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The string dualities can be summarized as follows: every extremal black hole gives a dual description of the spacetime around it, and a complete description. The different descriptions have to be self-consistent, they have to contain the same physics. This is a modern version of the 1960s principle called &quot;nuclear democracy&quot;, but it should be better called &quot;gravitational democracy&quot;. Gravitational democracy demands a huge number of dualities, between the fundamental description provided by different types of black holes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since gravity all but requires charge quantization, it is also is ok with monopoles. String theory requires magnetic duals for all objects, because a black hole or gravitational instanton can carry magnetic charge. Another source of duality, closely related to the first, is that quantum objects can carry magnetic charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electric-magnetic duality tells you that you can't make big charges without making objects whose charges are small. This means that whenever you try to take a limit where everything is strongly interacting and heavy, the string theory requires that other things become light. The principle of gravitational democracy tells you that you can then formulate the theory in terms of the light objects only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a string theory and compactify it on a circle. As the circle gets small the Kaluza Klein field of the circle gets a bigger and bigger quantum of charge. This means that the magnetic monopole gets a smaller and smaller charge, and therefore the lightest ones are lighter and lighter. This means that in the limit of tiny radii, the low-energy expansion of the S-matrix theory should be in terms of what were the magnetic variables in the original expansion. This is one way to motivate T-duality. The miracle of T-duality in type II string theory is that the new degrees of freedom are just different strings, and the duality never went through a regime where the strings are strongly coupled (only specific excitations became strongly coupled, and new excitations took their place as the light things after the duality)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other class of dualities arise when you make the string coupling big. Then the gauge fields become strongly coupled, which makes something magnetically dual becomes light. In the case of type IIB string theory, you get the same theory again, because of its self-duality properties. In the type IIA theory, the D0 branes become light, and when you formulate the resulting theory in terms of these, you get the famous matrix theory of 11 dimensional M-theory.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-14T05:04:30.610" />
  <row Id="13510" PostTypeId="2" ParentId="13505" CreationDate="2011-08-14T05:10:15.143" Score="5" ViewCount="" Body="&lt;p&gt;Is there any particular reason you made the spatial topology $\mathbf{PR}^3\cong\mathrm{SO}(3)$ rather than $S^3$? That's not going to affect local physics, though, so at least for uniform initial conditions on some $\mathbf{PR}^3$ hypersurface, we can simply scale a &lt;em&gt;restricted&lt;/em&gt; round metric&#xA;$$d\Omega^2 = d\psi^2 + \sin^2\psi(d\theta^2 + \sin^2\theta d\phi^2),\;\;\;\;0\leq\psi,\theta,\phi\leq\pi.$$&#xA;The Einstein equations have the same local form as for $S^3$, so we would get a spacetime that locally looks like the standard $k = +1$ Freidmann-Robertson-Walker solutions of the form $ds^2 = -dt^2 + a^2(t)d\Omega^2$. For example, a lambdavacuum (if energy density due to photons is $\rho\ll|\Lambda|$), radiation-dominated ($\rho\gg|\Lambda|$, $\rho = 3p$, scale factor $\propto t^{1/2}$), etc. These cases are extensively catalogued.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference being that the large-scale spatial structure is different, according to the antipodal identification you prescribe. Under the assumption of uniformity, the general answer to how this system evolves is simply: like the corresponding FRW universe. Depending on the conditions, it could even undergo transition from radiation-dominated to matter-dominated, just as our own universe did. I'm unclear as to whether you intended to exclude anything but photons in vacuum at one time or for all time.&lt;/p&gt;&#xA;" OwnerUserId="1353" LastEditorUserId="1353" LastEditorDisplayName="" LastEditDate="2011-08-14T07:30:30.623" LastActivityDate="2011-08-14T07:30:30.623" CommentCount="3" />
  <row Id="13513" PostTypeId="1" CreationDate="2011-08-14T13:21:15.910" Score="2" ViewCount="92" Body="&lt;p&gt;So I've been looking into particle-antiparticle pair production from a gamma ray and don't understand one thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I have a 1,1 MeV photon and it hits a nucleus - electron-positron pair with some momentum will be created and the nucleus will probably get some momentum as well because of the impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why does the photon need the nucleus at all? Why can't it just fly through space and suddenly, with some probability, change into a electron-positron pair with momentum? I see that the momentum of the system wouldn't be conserved but I don't really understand how the nucleus helps it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you a lot!&lt;/p&gt;&#xA;" OwnerUserId="2800" LastActivityDate="2011-08-14T15:51:29.280" Title="Spontaneous pair production?" Tags="&lt;photons&gt;&lt;electron&gt;&lt;antimatter&gt;&lt;pair-production&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="13514" PostTypeId="2" ParentId="13493" CreationDate="2011-08-14T13:47:49.567" Score="2" ViewCount="" Body="&lt;p&gt;Answer of Zassounotsukushi is correct, I'd like to expand it&lt;/p&gt;&#xA;&#xA;&lt;p&gt;consider center of mass frame, a resting neutron for simplicity. &lt;a href=&quot;http://en.wikipedia.org/wiki/Four-momentum&quot; rel=&quot;nofollow&quot;&gt;four-momentum&lt;/a&gt; is $\underline{p}_n=\binom{m_n}{\vec{0}}$.&#xA;neutron decays into proton, electron, antineutrino: $n \rightarrow p^+ e^- \bar{\nu}$ &#xA;$$&#xD;&#xA;\binom{m_n}{\vec{0}}=\binom{E_p}{\vec{p}_p}+\binom{E_e}{\vec{p}_e}+\binom{E_\nu}{\vec{p}_\nu}&#xD;&#xA;$$&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Neutrino#Mass&quot; rel=&quot;nofollow&quot;&gt;neutrino mass&lt;/a&gt; $(m_\nu &amp;lt; 0.2eV)$ is negligible $\Rightarrow E_\nu=|\vec{p}_\nu|$, to maximize electron energy neutrino has to be left without any  $\Rightarrow E_\nu=0$&#xA;$$&#xD;&#xA;\Rightarrow\binom{m_n}{\vec{0}}=\binom{E_p}{\vec{p}_p}+\binom{E_e}{\vec{p}_e}&#xD;&#xA;$$&#xA;for momentum conservation to be satisfied , $\vec{p}_p+\vec{p}_e=\vec0 $, proton momentum has to be $\vec{p}_p=-\vec{p}_e$&#xA;$$&#xD;&#xA;\Rightarrow\binom{m_n}{\vec{0}}=\binom{E_p}{-\vec{p}_e}+\binom{E_e}{\vec{p}_e}&#xD;&#xA;$$&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation&quot; rel=&quot;nofollow&quot;&gt;Energy-momentum relation&lt;/a&gt; can be used: $E_p=\sqrt{p_p^2+m_p^2}=\sqrt{p_e^2+m_p^2}$, $E_e=\sqrt{p_e^2+m_e^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;energy conservation:&#xA;$$&#xD;&#xA;\Rightarrow m_n = \sqrt{p_e^2 + m_p^2} + E_e&#xD;&#xA;$$&#xA;$$&#xD;&#xA;\Rightarrow m_n = \sqrt{E_e^2 - m_e^2 + m_p^2} + E_e&#xD;&#xA;$$&#xA;$$&#xD;&#xA;\Rightarrow (m_n - E_e)^2  = E_e^2 - m_e^2 + m_p^2&#xD;&#xA;$$&#xA;$$&#xD;&#xA;\Rightarrow E_{e,max} = \frac{m_n^2 - m_p^2 + m_e^2}{2m_n}&#xD;&#xA;$$&#xA;$$&#xD;&#xA;\Rightarrow p_{e,max} = \left(\frac{m_n^2 - m_p^2 + m_e^2}{2m_n}\right)^2 - m_e^2&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mass_defect&quot; rel=&quot;nofollow&quot;&gt;mass defect&lt;/a&gt;&quot; in transition $n\rightarrow p$ is in the term $m_n^2 - m_p^2$ included&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assumed &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_volt#Mass&quot; rel=&quot;nofollow&quot;&gt;$c=\hbar=1$&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4760" LastEditorUserId="4760" LastEditorDisplayName="" LastEditDate="2011-08-15T06:09:36.930" LastActivityDate="2011-08-15T06:09:36.930" CommentCount="4" />
  <row Id="13515" PostTypeId="1" CreationDate="2011-08-14T13:49:54.427" Score="1" ViewCount="74" Body="&lt;p&gt;Since the earth is in a vacuum and therefore there is no thermal transfer of heat to anything else, how can it even cool down?  It seems like its average temperature would always be constant, ignoring outside sources of heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you then consider that there is constant radiant heat transfer from the sun to the earth, now you have a net gain in energy/heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also read somewhere that the solar wind actually blows a bit of our atmosphere into space, but that seems like a miniscule amount of heat loss compared to how much heat is gained from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the earth ever cool at all?&lt;/p&gt;&#xA;" OwnerUserId="4921" LastActivityDate="2011-08-14T17:27:28.080" Title="Will the earth never cool, but only get hotter?" Tags="&lt;thermodynamics&gt;" AnswerCount="4" />
  <row Id="13516" PostTypeId="2" ParentId="13515" CreationDate="2011-08-14T14:01:08.530" Score="4" ViewCount="" Body="&lt;p&gt;You've pointed out that the sun carries out radiant heat transfer to the Earth, but you've forgotten that the earth also radiates heat into space. This is how the Earth cools down.&lt;/p&gt;&#xA;" OwnerUserId="1876" LastActivityDate="2011-08-14T14:01:08.530" CommentCount="2" />
  <row Id="13517" PostTypeId="2" ParentId="13513" CreationDate="2011-08-14T14:08:59.240" Score="2" ViewCount="" Body="&lt;p&gt;If you look at a photon in a vacuum, there is no preferred inertial frame of reference. And the energy of the photon is dependent upon which frame is chosen. So some frames have more than the needed energy, others less. It would be really awkward to have pair production in reference frames that lack the needed energy.&lt;/p&gt;&#xA;" OwnerUserId="391" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-08-14T14:24:44.973" LastActivityDate="2011-08-14T14:24:44.973" CommentCount="3" />
  <row Id="13519" PostTypeId="2" ParentId="13515" CreationDate="2011-08-14T14:24:22.430" Score="1" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Since the earth is in a vacuum and therefore there is no thermal&#xA;  transfer of heat&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;there is constant radiant heat transfer from the sun to the earth&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might like to think about those two statements !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the rate of heat flow from the earth into space increase as the temperature increases then you have a stable equilibrium. If the earth got hotter, more energy would be radiated into space cooling it. If the earth got cooler,less energy is radiated - heating it.&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-08-14T14:24:22.430" />
  <row Id="13520" PostTypeId="2" ParentId="13060" CreationDate="2011-08-14T14:42:02.257" Score="2" ViewCount="" Body="&lt;p&gt;It is something that's neither a physical quantity nor a dimensionless number, but something that can just be described as &lt;em&gt;logarithm of a physical quantity&lt;/em&gt;. There is not much problem with this: let $\mathcal{P}$ be the space of physical quantities. We can span this space in a vectorspace-like manner by base-physical units (eg. SI) as described by Willie Wong. What's important: we know that we can't perform certain operations in this space, for instance we cannot add a mass to an electrical current. Addition of quantities $a,b\in\mathcal{P}$ is only defined if $a$ and $b$ have the same dimension, that is, if $\exists x\in\mathbb{R}$ such that $a=xb$. Multiplication is always defined and always yields a physical quantity again. (This also defines powers of physical quantities, but not what, say, the exponential of one of them is.)&lt;br&gt;We then know that $\mathbb{R}\subset\mathcal{P}$, since for, say, two lengths $a,b\in\mathcal{P}$ the ratio $\tfrac{a}b$ will be a dimensionless number. For these dimensionless quantities, the logarithm is defined right from the start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's quite simple to extend this to a full space $\ln\!\!\mathcal{P}\supset\mathbb{R}$: for $a\in\mathbb{R}\subset\mathcal{P}$, the logarithm is defined as usually. For $a\not\in\mathbb{R}$, we define the logarithm axiomatically: first we require $\ln\!\!\mathcal{P}$ to be an abelian group WRT addition, even a vector space over $\mathbb{R}$. Then, for $\lambda\in\mathbb{R}$,&#xA;$$\begin{align}&#xD;&#xA;  \ln(a^\lambda) &amp;amp;:= \lambda\ln(a)&#xD;&#xA;\end{align}$$&#xA;and for $b\in\mathcal{P}$,&#xA;$$&#xD;&#xA;  \ln(ba) := \ln(b) + \ln(a).&#xD;&#xA;$$&#xA;Provided that $a$ and $b$ have the same dimension and therefore can be added, this already tells us the logarithm of the sum: we know that then $\exists x\in\mathbb{R}\colon b=ax$, in other words we can write any sum of physical quantities as a product of one of them with a real number, so the logarithm of any length cooks down to the logarithm of any one particular length, plus the logarithm of the ratio between the lengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coming back to your question: what is the logarithm of a kilometre? The answer: $\ln(1\:\mathrm{km})=\ln(\mathrm{km})$. If you treat kilometres as the basis unit of length, then this is all you need. If you prefer metres or inches or whatever, you just get&#xA;$$&#xD;&#xA;  \ln(1\:\mathrm{km}) = \ln(1000\:\mathrm{m}) = \ln(1000) + \ln(\mathrm{m})&#xD;&#xA;$$&#xA;$$&#xD;&#xA;  \ln(1\:\mathrm{km}) = \ln(\tfrac{1\:\mathrm{km}}{1&amp;quot;}\:\mathrm{&amp;quot;}) = \ln(\tfrac{1\:\mathrm{km}}{1&amp;quot;}) + \ln(\mathrm{&amp;quot;}) \approx 10.58 + \ln(\mathrm{&amp;quot;})&#xD;&#xA;$$&#xA;Here, $\ln(\mathrm{km}),\ln(\mathrm{m}),\ln(\mathrm{&amp;quot;})$ are &lt;em&gt;not&lt;/em&gt; dimensionless numbers. Rather think of them as elements of a vector space that has the real numbers as a subspace.&lt;/p&gt;&#xA;" OwnerUserId="3540" LastActivityDate="2011-08-14T14:42:02.257" />
  <row Id="13521" PostTypeId="2" ParentId="13513" CreationDate="2011-08-14T14:42:05.527" Score="4" ViewCount="" Body="&lt;p&gt;look at energy-momentum conservation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_\gamma = p_1+p_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the photon has invariant mass 0 wheras the electron and positron have mass $m_e$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_\gamma^2 = (p_1+p_2)^2 = p_1^2+p_2^2+2p_1\cdot p_2$$&#xA;$$0 = 2m_e^2 + 2p_1\cdot p_2$$&#xA;$$-m_e^2= p_1\cdot p_2 = E_1E_2-|\vec{p_1}||\vec{p_2}|cos\theta &amp;gt; E_1E_2-|\vec{p_1}||\vec{p_2}| = E_1E_2(1-\beta_1\beta_2) &amp;gt; 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The betas cannot be greater than one. So the right hand side always stays positive. The nucleon helps. because it changes the initial state to one with a invariant mass greater than zero.&lt;/p&gt;&#xA;" OwnerUserId="2151" LastEditorUserId="2151" LastEditorDisplayName="" LastEditDate="2011-08-14T15:51:29.280" LastActivityDate="2011-08-14T15:51:29.280" CommentCount="1" />
  <row Id="13522" PostTypeId="1" CreationDate="2011-08-14T14:51:08.793" Score="2" ViewCount="74" Body="&lt;p&gt;Page 580, Chapter 12 in Jackson's 3rd edition text carries the statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From the first postulate of special relativity the action integral must be a Lorentz scalar because the equations of motion are determined by the extemum condition, $\delta A = 0$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Certainly the extremeum condition must be an invariant for the equation of motion between $t_1$ and $t_2$, whereas I don't see how the action integral must be a Lorentz scalar. Using basic classical mechanics as a guide, the action for a free particle isn't a Galilean scalar but still gives the correct equations of motion. &lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="1345" LastEditorDisplayName="" LastEditDate="2011-08-14T14:57:18.377" LastActivityDate="2011-08-14T16:46:33.450" Title="Must the action for a relativistic charge be a Lorentz scalar?" Tags="&lt;special-relativity&gt;&lt;electrodynamics&gt;&lt;variational-principle&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13523" PostTypeId="2" ParentId="13513" CreationDate="2011-08-14T14:52:04.680" Score="3" ViewCount="" Body="&lt;p&gt;Pair production is not the same as decay of a particle. A particle can decay into two components according to its decay probability without needing an extra interaction. A lamda in its rest frame will decay into a proton and a pion, for example, within  a predictable decay time . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no rest frame for the photon since its mass is 0 and it is always travelling with the velocity of light.  If it were to decay spontaneously into an electron positron pair, they do have a rest mass and a rest frame, and   their invariant mass would be at least 2*m_e, which should have been  the mass of the  photon. A contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can interact though with the fields of other particles . How does the photon interact? &#xA;The interaction probabilities can be calculated given the charges of the target particles, the easiest way using Feynman diagrams. One can envisage a photon as sequentially turning into  virtual loops of e+e- . One of the virtual electrons interacts with the field of a real charged particle exchanging enough energy and momentum so that both e+ and e- become real while energy and momentum are conserved in a three body interaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nucleus helps by ensuring  the momentum in the final state (e+e_Nucleus) to be the same as the one in the initial state (photon nucleus). &lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-08-14T14:52:04.680" CommentCount="1" />
  <row Id="13525" PostTypeId="1" CreationDate="2011-08-14T15:16:22.273" Score="3" ViewCount="105" Body="&lt;p&gt;Let's say we want to freeze a banana. Would is freeze faster if we insert it into an empty freezer, or into one that already contains many frozen bananas?&lt;/p&gt;&#xA;" OwnerUserId="1155" LastActivityDate="2011-09-02T00:02:50.380" Title="Does a full freezer freeze items faster than an empty one?" Tags="&lt;thermodynamics&gt;" AnswerCount="4" CommentCount="3" />
  <row Id="13526" PostTypeId="2" ParentId="13522" CreationDate="2011-08-14T15:32:41.887" Score="1" ViewCount="" Body="&lt;p&gt;First, observe that although the non-relativistic Lagrangian is not invariant. It changes by a total derivative, thus the equations of motions remain invariant.&#xA;The reason of the difference between the Lorentzian and the Galilean cases is that the group action of the Lorentz group on the classical variables (positions and momenta) is a by means of a true representation, while in the case of the Galilean group the representation is projective.&#xA;In the Language of geometric quantization, $exp(i \frac{S}{\hbar})$, where $S$ is the action is a section in $L \otimes \bar{L}$, where $L$ is the prequantization line bundle and  $\bar{L}$ its dual. In other words, the action needs not be a scalar, only an exprssion of the form:&#xA;$\bar{\psi}(t_2)exp(i \frac{S(t_1, t_2)}{\hbar})\psi(t_1)$, where $\psi(t)$ is the wavefunction at time $t$ and $S(t_1, t_2)$ is the classical action between $t_1$ and $t_2$.&#xA;The reason that the representation in the Galilean case is projective is related to the nontriviality of the cohomology group $H^2(G, U(1))$ in the Galilean case in contrast to the Lorentz case.&#xA;I have given a more detailed answer on a very similar subject in my answer to Anirbit: &lt;a href=&quot;http://physics.stackexchange.com/questions/12341/poincare-group-vs-galilean-group/12364#12364&quot;&gt;Poincare group vs Galilean group&lt;/a&gt; and in the comments therein.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastActivityDate="2011-08-14T15:32:41.887" />
  <row Id="13527" PostTypeId="2" ParentId="13515" CreationDate="2011-08-14T15:54:08.520" Score="1" ViewCount="" Body="&lt;p&gt;All matter radiates electromagnetic radiation according to its temperature. It is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_body_radiation&quot; rel=&quot;nofollow&quot;&gt;black body radiation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power emitted goes  as T^4 , where T is the temperature in degrees Kelvin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is no replenishment of the energy lost the body in vacuum will reach absolute zero after a calculable time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earth gets replenished mainly  by the sun, a bit from internal fission too.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-08-14T15:54:08.520" />
  <row Id="13528" PostTypeId="2" ParentId="508" CreationDate="2011-08-14T15:55:37.113" Score="1" ViewCount="" Body="&lt;p&gt;Apart from spallation, neutrons can be produced upon hadronization of &lt;a href=&quot;http://en.wikipedia.org/wiki/Quark%E2%80%93gluon_plasma&quot; rel=&quot;nofollow&quot;&gt;quark-gluon-plasma&lt;/a&gt; or from neutralizing collisions of energetic electrons in the air shower with protons.&lt;/p&gt;&#xA;" OwnerUserId="4760" LastEditorUserId="4760" LastEditorDisplayName="" LastEditDate="2011-08-17T06:44:27.170" LastActivityDate="2011-08-17T06:44:27.170" />
  <row Id="13529" PostTypeId="2" ParentId="13522" CreationDate="2011-08-14T16:35:54.890" Score="0" ViewCount="" Body="&lt;p&gt;Yest, it must. It does not guarantee that the equations have physical exact solutions but at least everything looks relativistic.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-08-14T16:46:33.450" LastActivityDate="2011-08-14T16:46:33.450" />
  <row Id="13530" PostTypeId="2" ParentId="11386" CreationDate="2011-08-14T16:50:19.483" Score="0" ViewCount="" Body="&lt;p&gt;Most of the beam current eventually flows through the anode.  This happens via secondary emission (where a beam electron causes a phosphor molecule to emit an electron towards the anode) and conduction.  The inside of standard (non-storage) CRTs is coated with a thin layer of metal (usually aluminum) to increase conductivity.  Storage CRTs are made to have higher conductivity to prevent image blurring while in storage mode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aluminising the inside of the phosphor is important so that charge does not build up--Any charge accumulation on the phosphor will raise its potential.  That reduces the effective acceleration potential of the electrons, which causes the screen to dim.&lt;/p&gt;&#xA;" OwnerUserId="4923" LastActivityDate="2011-08-14T16:50:19.483" />
  <row Id="13531" PostTypeId="2" ParentId="13515" CreationDate="2011-08-14T17:27:28.080" Score="1" ViewCount="" Body="&lt;p&gt;As Martin above suggested, the Earth actually loses as much heat through thermal radiation as it gains through radiation from the sun each day. The assumption that leads us to that conclusion is basically that the Earth is in thermal equilibrium with it's surroundings. (If the sun suddenly started pumping out more heat, the Earth would eventually (how long?) adjust until equilibrium is reattained, probably at a higher temperature. How could humans affect that temp.? The easiest way to understand how the global temperature could change is to consider clouds. With more clouds, a chunk of the Earths radiation is reflected back, meaning less is radiated outward. The temperature adjusts up until we radiate out as much as the sun radiates in again. Of course also the sun energy radiating in is less ( some gets reflected) but the net effect is temperature up- if clouds both cut the suns radiation in and our radiation out by 10%, we are still in equilibrium but we are still stuck with our 10% of heat reflected back at us. (better explanation anyone?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy coming in from the sun is low entropy, which means it is useful to do work (such as power a plants chemical reactions.) The energy that leaves is high entropy ( read random, disordered like the motion of hot gas particles) that is no longer very useful to do work. So it is sometimes said that the sun doesn't really provide us with energy (we end up with no more energy than we started) but with low entropy. &lt;/p&gt;&#xA;" OwnerUserId="4924" LastActivityDate="2011-08-14T17:27:28.080" />
  <row Id="13532" PostTypeId="1" CreationDate="2011-08-14T18:38:44.583" Score="0" ViewCount="23" Body="&lt;p&gt;How and where are &lt;a href=&quot;http://en.wikipedia.org/wiki/GPGPU&quot; rel=&quot;nofollow&quot;&gt;GPUs&lt;/a&gt; used in scientific simulations (esp. in astrophysics/cosmology)?&lt;/p&gt;&#xA;" OwnerUserId="47" LastActivityDate="2011-08-14T19:25:51.777" ClosedDate="2011-08-14T19:38:26.660" Title="How and where are GPUs used in scientific simulations?" Tags="&lt;cosmology&gt;&lt;astrophysics&gt;&lt;gpgpu&gt;" AnswerCount="0" CommentCount="5" />
  <row Id="13533" PostTypeId="2" ParentId="12973" CreationDate="2011-08-14T18:48:04.343" Score="5" ViewCount="" Body="&lt;p&gt;The idea of knot inflation is that of A. Grosberg and it is tricky, but I'll try to explain some of it. Construction by way of scaling analysis is required to tackle this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the volume of our polymer is given by $R$. Now fit a tube of &lt;em&gt;fixed&lt;/em&gt; length, $L$, over the polymer; this tube has diameter, $D$, across its length that is just large enough to contain the polymer. Now the tube is inflated with a fixed $L$, then $D$ is uniformly allowed to increase until a maximum $D_{max}$ is reached. An aspect ratio, $p=L/D_{max}$, is then a weak topological invariant. Affine transformation of the inflated tube is done so that the polymer will fit in a sphere of size $R$, with $L_R$ and $D_R$ such that $L_R/D_R=p$ and $L_R D_R^2 \approx R^3$. He then uses this construction for finding the entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick now is to assume that &lt;em&gt;most&lt;/em&gt; of the available conformations of our polymer with a certain knot are also available to a trivial knot within our sphere of size $R$, giving $$S= S_{trivial} + \Delta S_{approx}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can say that $S_{trivial}$ should be proportional to $N$ and so we can rewrite it as, $$S = sN + \Delta S_{approx}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $s$ not a function of $p$ or $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now he confines an ideal knot in a cylinder (ties the polymer to both ends of the cylinder), and considers what will happen to the entropy if $p$, of the cylinder, is varied. He finds that $\Delta S_{approx}$ is proportional to $-p$. Finally, a polymer with the maximum amount of knots should give $S\approx 0$, so that means $p \approx N$ in that case, giving $$S = s(N-p)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the entropy decreases with increasing knot complexity, $p$. This method is very crude, at best, but it does provide a qualitative approach to your query.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see his full argument I'll have to refer you to the citation: A.Y. Grosberg “Entropy of a Knot: Simple Arguments About Difficult Problem,” In: Ideal Knots, Edited by A.Stasiak, V.Katrich, L.H.Kauffman, World Scientific, 1998, p. 129-142.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that same article he extends his argument to calculate the probability of finding a knot as $$P(N) = N_0 e^{-N/N_0 - sp} $$&lt;/p&gt;&#xA;" OwnerUserId="4112" LastEditorUserId="4112" LastEditorDisplayName="" LastEditDate="2011-08-14T20:42:43.677" LastActivityDate="2011-08-14T20:42:43.677" />
  <row Id="13537" PostTypeId="2" ParentId="11039" CreationDate="2011-08-14T19:26:35.187" Score="1" ViewCount="" Body="&lt;p&gt;This is not &quot;just quantum mechanics&quot;, it is more than that. Quantum mechanics tells you that states u-ubar and d-dbar are allowed to mix, so that you can consider a u-ubar system as a sum of (u-ubar + d-dbar) plus (u-ubar - d-dbar), but it doesn't tell you that they have to mix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If these states didn't mix, and they had approximately the same energy, then there would be no paradox, you would be free to think of the pion as a u-ubar, or as d-dbar. If the u and d have a different mass, then u-ubar and d-dbar would be the right way to visualize the &quot;u-pion&quot; and the &quot;d-pion&quot;, even when there are reasonably strong interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for the actual pions, the symmetric part is split in energy from the antisymmetric part by hundreds of MeV, five times the mass of the pions. This splitting is what makes the pions counterintuitive, and to answer the question you need to address the splitting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saying that pions are made of quarks is like saying that sound is made of atoms. It's true that if there are no atoms, there is no sound, but that's about it. The QCD vacuum is like a condensed matter system, and it has a quark condensate at the pion scale. The eigenstates of motion of the quark condensate define the low-lying excitations of QCD, and the lightest motion of the condensate is moving it's parts chirally against each other. By this, I mean turning the left-handed u/d and right-handed u/d quarks in the condensate by an opposite phase. This would do nothing to the energy if chiral symmetry were exact, that is, if the quarks were massless. This means that you could &quot;move&quot; the vacuum in the chiral direction without any energy cost, and this gives massless &quot;phonons&quot; (Goldstone bosons) for this process, by moving the vacuum over here a little, and not moving the vacuum over there. These phonons carry the same quantum numbers as the isospin triplet u-dbar/symmetric/d-udbar. These phonons are the pions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the pions is not zero, but it is small compared to other strongly interacting particles by a lot. This reflects the fact that the up/down quarks are light compared to the QCD scale. While this picture is only accurate to the extent that the pion mass is small (and the pion is not that light), it is indispensible for understanding pion scattering. Because while the pion mass is visible at scales of 7 to 8 fermis, the interactions with stuff like the proton take place at a scale of 1 fermi, where the pion mass is negligble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason pions are split from their isospin zero partner, the eta-prime, is because the gluons in the vacuum already break part of the chiral symmetry by themselves, through instantons. This splits the two kinds of chiral sound, the pion and the eta, and neither of them is made up of quarks like a molecule is made of atoms. The eta-prime vacuum sound mode is five times stiffer than the pion vacuum sound mode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When doing quark analysis of light mesons, one must always keep in mind that they only tell you the symmetry numbers, the isospin, strangeness (or SU(3)) quantum numbers. It is only at high energies/high masses that quarks become constituents of the hadrons and mesons in the ordinary sense.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-14T19:33:36.820" LastActivityDate="2011-08-14T19:33:36.820" />
  <row Id="13538" PostTypeId="2" ParentId="6721" CreationDate="2011-08-14T19:46:45.390" Score="3" ViewCount="" Body="&lt;p&gt;The problem with volume answers to this question is that if the universe is ultimately going to be de-Sitter, the volume that one is talking about is outside the horizon, and so of dubious status considering logical positivism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good answer to this can be obtained by considering the entropy stored in the cosmological horizon. In a deSitter space, the maximum area of the cosmological horizon is in an empty universe, and the gain in area of the horizon swamps the entropy of anything that fell through. That this is always true is the content of the Bekenstein bound on entropy. So as the universe dilutes, the entropy of the cosmological horizon always increases by more than the entropy of whatever is inside goes down.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-14T19:46:45.390" />
  <row Id="13539" PostTypeId="1" CreationDate="2011-08-14T19:59:35.407" Score="2" ViewCount="49" Body="&lt;p&gt;If we consider two spins, we can think of the spins as being either parallel (up|up or down|down)or anti-parallel (up|down or down|up). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or we can think of them as being in the triplet or singlet configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is one description more correct than the other? Or is it just  a matter of choice between two basis sets? It would seem to me that using T/S is correct because it accurately reflects the symmetry needed in the wavefunction.&lt;/p&gt;&#xA;" OwnerUserId="1939" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-14T20:02:03.503" LastActivityDate="2011-08-15T04:01:27.237" Title="parallel/anti-parallel vs. triplet/singlet description of two spins" Tags="&lt;quantum-mechanics&gt;&lt;spinors&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="13540" PostTypeId="2" ParentId="13539" CreationDate="2011-08-14T20:08:51.707" Score="4" ViewCount="" Body="&lt;p&gt;It's just a choice of basis. Whether you use&#xA;$$\bigl\{|\uparrow\downarrow\rangle,|\downarrow\uparrow\rangle\bigr\}$$&#xA;(individual spins) or&#xA;$$\biggl\{\frac{1}{2}\bigl(|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle\bigr),\frac{1}{2}\bigl(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle\bigr)\biggr\}$$&#xA;(triplet/singlet) they span the same space. But usually the T/S basis is more useful because those states are also eigenstates of the total spin operator $S^2$. As a side benefit, they reflect the (anti-)symmetrization requirements of identical particles; for example, if you have two identical fermions with no other quantum numbers (neglecting the fact that such particles don't exist :-P) in a bound state, you know that they have to take the singlet configuration in order for the wavefunction to be antisymmetric.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-08-14T20:08:51.707" CommentCount="2" />
  <row Id="13541" PostTypeId="2" ParentId="1268" CreationDate="2011-08-14T20:14:29.407" Score="1" ViewCount="" Body="&lt;p&gt;The argument is summarized on the Wikipedia page &quot;Mass/Energy equivalence&quot;, and it goes like this: imagine a body at rest which then emits two equal photons, one to the right and one to the left. In the rest frame, the body is still at rest because the photons have equal and oppsite momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shifting to a frame moving to the right. In this frame, the photon moving to the left is blueshifted and carries more momentum, and the photon moving to the left is redshifted and carries less momentum. This means that the object has lost some right momentum after the emission. The velocity is the same before and after, because the velocity is the same in the rest frame, so how could the body lose momentum without changing its velocity? It must have lost mass. If you calculate the lost mass, it equals the energy of the photons divided by c&lt;sup&gt;2&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This argument is obviously correct, essentially rigorous (it requires a precise framework to state rigorously, but there are no imprecise assumptions). The bickering came because Einstein demonstrated this and not anyone else, everyone else thought that the mass/energy relation was E=4/3 mc^2. Soon after, Poincare realized what everyone's mistake was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planck also discovers E=mc^2, and published after Einstein (but I would bet his work was independent). He just refuses to accept that Einstein's argument is correct, and says his argument is the correct one. This is possibly because of his bitterness at being scooped at such an important result. From this attack come all future error claims.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ohanian's book in general gets everything wrong. Here is a complete list of Einstein's mistakes (I put an expanded version of this on Wikipedia years ago, but it slowly got reworded, watered down, and moved. That gradual process, of course, was the work of Satan):&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Einstein's mistakes&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1905: In the original German version of the special relativity pape, Einstein gives the transverse mass as $ m/(1 - v^2/c^2)$, while the actual value is $ m/\sqrt{1 - v^2/c^2}$ (Max Planck corrected this).&lt;/li&gt;&#xA;&lt;li&gt;1905: In his PhD dissertation, the friction in dilute solutions has an miscalculated numerical prefactor, which makes the estimate of Avogadro’s number off by a factor of 3. The mistake is corrected by Einstein in a later publication.&lt;/li&gt;&#xA;&lt;li&gt;1905: An expository paper explaining how airplanes fly includes an example which is incorrect. There is a wing which he claims will generate lift. This wing is flat on the bottom, and flat on the top, with a small bump at the center. It is designed to generate lift by Bernoulli’s principle, and Einstein claims that it will. Simple action reaction considerations, though, show that the wing will not generate lift, at least if it is long enough.&lt;/li&gt;&#xA;&lt;li&gt;1913: Einstein started writing papers based on his belief that the hole argument made general covariance impossible in a theory of gravity. Einstein realized he was wrong in 1915, and finds General Relativity.&lt;/li&gt;&#xA;&lt;li&gt;1922: Einstein published a qualitative theory of superconductivity based on the vague idea of electrons quantum-mechanically shared in orbits. This paper predated modern quantum mechanics, and is well understood to be completely wrong. Einstein's paper is more of an old-quantum-mechanical version of the modern explanation of ordinary conductivity.&lt;/li&gt;&#xA;&lt;li&gt;1937: Einstein believed that the focusing properties of geodesics in general relativity would lead to an instability which causes plane gravitational waves to collapse in on themselves. While this is true to a certain extent in some limits, because gravitational instabilities can lead to a concentration of energy density into black holes, for plane waves of the type Einstein and Rosen considered in their paper, the instabilities are under control. Einstein retracted this position a short time later, but his collaborator Nathan Rosen maintained that gravitational waves are unstable until his death.&lt;/li&gt;&#xA;&lt;li&gt;1939: Einstein denied several times that black holes could form, the last time in print. He published a paper that argues that a star collapsing would spin faster and faster, spinning at the speed of light with infinite energy well before the point where it is about to collapse into a black hole. This paper received no citations, and the conclusions are well understood to be wrong. Einstein’s argument itself is inconclusive, since he only shows that stable spinning objects have to spin faster and faster to stay stable before the point where they collapse. But it is well understood today (and was understood well by some even then) that collapse cannot happen through stationary states the way Einstein imagined.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There's other mistakes that are not mistakes, but philosophical things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In the Bohr-Einstein debates and the papers following this, Einstein tries to poke holes in the uncertainty principle, ingeniously, but unsuccessfully.&lt;/li&gt;&#xA;&lt;li&gt;In the EPR paper, Einstein concludes that quantum mechanics must be replaced by local hidden variables. The measured violations of Bell’s inequality show that hidden variables, if they exist, must be nonlocal.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Einstein considered the cosmological constant a mistake, but the cosmological constant is necessary within general relativity as it is currently understood, and it is widely believed to have a nonzero value today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He had lapses in taste too, usually quickly corrected:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Einstein briefly flirted with transverse and longitudinal mass concepts, before rejecting them.&lt;/li&gt;&#xA;&lt;li&gt;Einstein initially opposed Minkowski’s geometrical formulation of special relativity, changing his mind completely a few years later.&lt;/li&gt;&#xA;&lt;li&gt;Based on his cosmological model, Einstein rejected expanding universe solutions by Friedman and Lemaitre as unphysical, changing his mind when the universe was shown to be expanding a few years later.&lt;/li&gt;&#xA;&lt;li&gt;Finding it too formal, Einstein believed that Heisenberg’s matrix mechanics was incorrect. He changed his mind when Schrödinger and others demonstrated that the formulation in terms of the Schrödinger equation, based on Einstein’s wave-particle duality was equivalent to Heisenberg’s matrices.&lt;/li&gt;&#xA;&lt;li&gt;Einstein rejected work on black holes by Chandrasekhar, Oppenheimer, and others, believing, along with Eddington, that collapse past the horizon (then called the ’Schwarzschild singularity’) would never happen. So big was his influence, that this opinion was not rejected until the early 1960s, almost a decade after his death.&lt;/li&gt;&#xA;&lt;li&gt;Einstein believed that some sort of nonlinear instability could lead to a field theory whose solutions would collapse into pointlike objects which would behave like quantum particles. This is impossible by Bell’s inequality.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is sometimes claimed that the general line of Einstein’s reasoning in the 1905 relativity paper is flawed, or the photon paper, or one or another of the most famous papers. Those claims are all ridiculous.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-14T20:39:46.810" LastActivityDate="2011-08-14T20:39:46.810" />
  <row Id="13542" PostTypeId="2" ParentId="12553" CreationDate="2011-08-14T20:14:54.053" Score="1" ViewCount="" Body="&lt;p&gt;Time axis is not bent by the black hole - the light-like axis is.&#xA;here we have the Eddington-Finkelstein coordinates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/LV9GY.jpg&quot; alt=&quot;Eddington- Finkelstein coordinates&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Schwarzschildradius is at $r_S$ and you can see the both light-like rays, one of which is parallel to time axis, the other points towards the singularity.  &lt;/p&gt;&#xA;" OwnerUserId="4760" LastActivityDate="2011-08-14T20:14:54.053" />
  <row Id="13543" PostTypeId="2" ParentId="11975" CreationDate="2011-08-14T21:20:32.940" Score="3" ViewCount="" Body="&lt;p&gt;The most general widely studied examples of potentials are those that are derived from an exactly given ground state, and which have the property that they contain enough parameters to be closed under taking supersymmetric conjugates. These are called &quot;shape invariant&quot; potentials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a real positive ground state, one can ask &quot;Which potential has this ground state?&quot; The answer is that if the ground state is $\exp(-W(x))$, and its energy is exactly zero, then the potential is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ V(x) = {1\over 2} |\nabla W|^2 - {1\over 2} \nabla^2 W$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conjugate potential is defined with a plus sign between the two terms instead of a minus sign. It is more usual to define W as the derivative of what I am calling W, but this convention is terrible in higher than one dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two potentials taken together define a supersymmetric quantum mechanics, as originally defined by Witten. The supersymmetric quantum mechanics in imaginary time is a stochastic Brownian process with a drift which is an analytic function of the position. The conjugate potentials correspond to reversing the direction of the drift, and their properties are similar because they are related by a stochastic version of time-reversibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If W(x) goes to plus infinity at infinity (so that it actually defines a normalizable ground state), then the ground state is unique, and the conjugate potential has the exact same spectrum as the original potential, except it omits the lowest energy state. This, plus the form of the supercharge, gives exact solutions of many classes of quantum potentials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some simple W's which correspond to usual elementary quantum mechanics examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;W(x) = |x|^2 is the Hamonic oscillator in any dimension&lt;/li&gt;&#xA;&lt;li&gt;W(x) = |x| is the delta function potential in 1d, and the Coulomb potential in 3d&lt;/li&gt;&#xA;&lt;li&gt;W(x) = log(|cos(x)|) this gives the infinite hard wall&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/hep-th/9405029&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9405029&lt;/a&gt; has a bunch of more interesting examples. Any quantum mechanical potential which has closed form energy states is in this class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A completely diffeent class of widely studied potentials are random potentials, as studied by Halperin and others, to understand Anderson localization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later Edit: The original paper by Anderson which started the random potential field is &lt;a href=&quot;http://prola.aps.org/abstract/PR/v109/i5/p1492_1&quot; rel=&quot;nofollow&quot;&gt;http://prola.aps.org/abstract/PR/v109/i5/p1492_1&lt;/a&gt; &quot;Absence of diffusion in certain random lattices&quot;, and it's one of the great classics. The setup is a square lattice with a random potential at each site, an independent random number between -V and V. The continuum limit in one dimension, where the potential is a random gaussian at each point is analyzed by Halperin B. I. Halperin, Green ' s Functions for a Particle in a One-Dimensional Random Potential, Phys. Rev. 139 , A104 (1965). The field is enormous--- look up &quot;localization&quot; on google scholar. It includes &quot;weak localization&quot; effects, which were popular in the mid 90s because they imply that resistance can drop sharply in the presence of a magnetic field, because the perturbative precurser to the localization process is hindered.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-15T04:13:39.527" LastActivityDate="2011-08-15T04:13:39.527" CommentCount="2" />
  <row Id="13544" PostTypeId="2" ParentId="8049" CreationDate="2011-08-14T21:42:21.050" Score="0" ViewCount="" Body="&lt;p&gt;Perhaps the machine described here can actually be built. Let me propose a heated flask containing a 50-50 mixture of gaseous, monatomic carbon-14 and nitrogen-14. When you push a button, a pinhole opens up and allows exactly one atom to escape. Is it: either a carbon atom or a nitrogen atom, with 50% probability, or is it an atom in a 50-50 superposition of the carbon/nitrogen states?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weight of expert opinion in the answers posted thus far seems to indicate that these two descriptions are experimentally indistinguishable. I suspect this is correct, although it's a funny conclusion that flies in the face of the common-sense belief that an atom is either carbon or nitrogen, but not both.  However I think two stipulations ought to be noted:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I can see no reason why a given atom ought not to appear in an 80/20 superposition, so long as the long-term average is 50/50.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I don't believe the machine is actually constructible because I don't think there is a mechanism which can reliably produce exactly one atom at a time. You never quite know just how many atoms you've let out, and that introduces enough uncertainty in the measurement to avoid any of the glaring contradictions that seem to be present.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-08-14T21:42:21.050" />
  <row Id="13545" PostTypeId="1" CreationDate="2011-08-14T21:53:21.990" Score="1" ViewCount="29" Body="&lt;p&gt;Here by &quot;low temperature&quot; I meant it in the scale of the characteristic $\hbar \omega$ of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can calculate and show that in the low temperature regime $C_V$ of phonons goes like $T^3$ for free fermions it goes as $T$ and for free bosons its like $T^{\frac{3}{2}}$ In most of the above cases the power comes from some complicated calculation.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I was wondering if there is a heuristic argument for &quot;understanding&quot; these powers. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Further is the power of $T$ characteristic of the system? Like if in a substance one sees the low temperature behaviour of $C_V$ to be going as $T$ then does it says something like that the effective degree of freedom of the system is that of free fermions?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are there other examples of such calculation? Like typical systems where one can calculate the low temperature dependance of $C_V$?  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1637" LastEditorUserId="1637" LastEditorDisplayName="" LastEditDate="2011-08-15T00:25:20.047" LastActivityDate="2011-08-15T00:25:20.047" Title="Heuristic argument for the temeprature dependence of specific heat in the &quot;low&quot; temperature regimes" Tags="&lt;quantum-mechanics&gt;&lt;statistical-mechanics&gt;&lt;condensed-matter&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13546" PostTypeId="1" CreationDate="2011-08-14T22:02:44.337" Score="4" ViewCount="117" Body="&lt;p&gt;Suppose some decay process emits 2 electrons in opposite directions,&#xA;and their spin is measured by a Stern-Gerlach type device in a&#xA;particular direction, say Sz. The books say that if 2 detectors have&#xA;the same orientation, then the spin measurements are 100% correlated&#xA;(or anti-correlated). If one has Sz=+1/2, then the other has Sz=-1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is the quantum state of the electrons before the measurements?&#xA;It seems to me that if it is anything but an eigenstate of Sz, then&#xA;the perfect correlations are impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any quantum mechanical explanation of the perfect correlations,&#xA;other than with an action-at-a-distance collapse of the wave function?&lt;/p&gt;&#xA;" OwnerUserId="4926" LastActivityDate="2011-08-16T18:11:19.243" Title="Why do Bell tests give perfect correlations?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13547" PostTypeId="2" ParentId="12553" CreationDate="2011-08-14T22:03:57.347" Score="2" ViewCount="" Body="&lt;p&gt;There is no uniquely defined time axis is relativity. Locally, observers in different states of motion have time axes that don't coincide. Even if we fix a state of motion for the observer, there is both a forward and a backward time axis. A more meaningful question would be phrased in terms of the past and future light cones: &lt;a href=&quot;http://en.wikipedia.org/wiki/Light_cone&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Light_cone&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following, I'll assume a spacetime containing a black hole that is the result of gravitational collapse. (There are other possibilities. See &lt;a href=&quot;http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/black_holes_picture/index.html&quot; rel=&quot;nofollow&quot;&gt;http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/black_holes_picture/index.html&lt;/a&gt; )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any spacetime event that is inside the event horizon, we have a future light cone and a past light cone. One of these contains world-lines that all terminate at the singularity. The other contains world-lines that extend out through the event horizon. Conventionally, we call the former direction the future and the latter the past direction. (If we were to reverse these labels, then this would be a white hole, not a black hole.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any spacetime event that is outside the event horizon, the past light-cone (assuming the conventional labels) contains world-lines that stretch back in time but never intersect the event horizon. The future light-cone contains some world-lines that pass in through the horizon and some that don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short answer to your question, phrased in the language you used, is basically that if you're inside the horizon, all possible future time axes point toward the singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as references, you could look at any textbook on GR. If you can post more about your level of math and physics background, we could point you to something specific that would be appropriate.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-14T22:03:57.347" CommentCount="1" />
  <row Id="13548" PostTypeId="2" ParentId="13545" CreationDate="2011-08-14T22:21:34.103" Score="2" ViewCount="" Body="&lt;p&gt;First, since $C_V = \left(\frac{\partial U}{\partial T}\right)_V$, the problem of looking for the power of $T$ in $C_V$ is equivalent to the problem of looking for the power of &#xA;$T$ in $U$, the internal energy of your system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This energy is given by an integral over the &lt;em&gt;density of states&lt;/em&gt; $g(E) dE$ and the appropriate distribution function (Fermi-Dirac or Bose-Einstein):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U = \int dE E \cdot g(E) n(E)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distribution function includes an exponential of the form $E/kT$, which we replace by a new variable $x$, so that $E = x kT$ and $dE/dx = kT$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, if $g(E) \sim E^a$, you end up with a factor &#xA;$T^{(a + 2)}$. All other components in your integral don't depend on $T$, so you have &#xA;$U \sim T^{a+2}$&#xA;and&#xA;$$C_V \sim T^{a+1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the only thing you have to figure out is how the density of states scales with energy. For a three dimensional photon gas, for example, we have a density of states &#xA;that is proportional to $E^2$, so the specific heat is proportional to &#xA;$T^{2 + 1} = T^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a two-dimensional free electron gas, we have $g(E) = const$, so the heat capacity &#xA;should be proportional to $T^{0+1} = T$. For a one-dimensional electron gas, we have &#xA;$g(E) \sim E^{-1/2}$, so $C_V \sim T^{-1/2 + 1} = T^{1/2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if you measure the low temperature behavior, this effectively means you measure $a$. This tells you what the density of states approximately scales like, at least in the lower energy regime. This does not tell you, however, what sort of elementary excitations you are looking at, as a peculiar band structure of the electrons could in principle give rise to a density of states that you'd normally expect in a photon gas. (Graphene would be an example, with its linear dispersion at the Dirac point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To your last question: Yes. For every system where you can come up with a density of states, you can then carry out the calculation and determine the specific heat. Another examples would be a ferromagnet at low temperature, whose elementary excitations are spin-waves (magnons).&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-08-14T22:21:34.103" />
  <row Id="13549" PostTypeId="2" ParentId="13546" CreationDate="2011-08-14T22:31:05.680" Score="3" ViewCount="" Body="&lt;p&gt;Quantum mechanics has one peculiar and highly counterintuitive property:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know the complete state of a compound system $S$, this does &lt;em&gt;not&lt;/em&gt; necessarily imply you know the complete state of either of the subsystems!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantum state you are after looks like this:&#xA;$$\frac{1}{\sqrt{2}} \left( \langle \uparrow \downarrow | - \langle \downarrow \uparrow| \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is an &lt;em&gt;entangled&lt;/em&gt; state, this means that you &lt;em&gt;cannot&lt;/em&gt; decompose this into some product of the form &#xA;$$ \langle \psi_A | \otimes \langle \psi_B | $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a way, it is therefore nonsensical to ask what the state of one of the electrons is! If you &lt;em&gt;want&lt;/em&gt; to talk about one of the subsystems only, you must make use of the density-matrix formalism. An application of this would show that for the state I've depicted above, the individual spins are as undetermined as possible, being up or down with exactly equal probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to your second question. Sure, there are other explanations apart from the collapse of the wave function. But from a practical point of view, those are all equivalent and you can basically choose whichever you like best. There are the various flavors of the many-world interpretation, for example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; In response to your comments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the correlation does not rely on the condition that the two measurements occur in such a way that light would be able to travel from one lab to the other in the time between the measurements! Indeed, if we call the two observers $A$ and $B$, it does not matter how far apart the labs are or when exactly the measurement takes place. How such experiments are carried out is that for both labs, a list of the spin is written down, which will read like $\uparrow \uparrow \downarrow \uparrow \downarrow \downarrow \dots$. &lt;em&gt;After&lt;/em&gt; the experiment, $A$ and $B$ will compare their lists and find a $100\%$ correlation (well, barring some measurement errors that destroy coherence). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think it is possible to give a satisfying explanation relying on classical terms, because nature isn't classical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I understand what you mean with &quot;there's no eigenstate of $S_z$&quot;. The total state as I have written down is a linear combination of product states, which are formed from the basis states of systems $A$ and $B$. So of course, when $A$ or $B$ perform their measurements of the electron spin, they will find only eigenvalues of $S_z$. The important part is that the overall state of the system cannot be written as  a pure product state, no matter what basis you choose. This is the hallmark of entanglement, because if you &lt;em&gt;had&lt;/em&gt; a product state $\langle \psi_A | \otimes \langle \psi_B$, a measurement on system $A$ would have no effect on the state of system $B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can view the process of measuring the electron spin in systems $A$ and $B$ and then talking about the results as a measurement of the total spin's $z$-component. In terms of total angular momentum, the state I have written is equivalent to &#xA;$\langle 0, 0|$, i.e. a state of total spin zero. Since this state is an eigenstate of the total spin with eigenvalue $0$, every measurement of the total spin's $z$-component must yield result $0$. &lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditorDisplayName="" LastEditDate="2011-08-15T00:32:49.283" LastActivityDate="2011-08-15T00:32:49.283" CommentCount="5" />
  <row Id="13550" PostTypeId="2" ParentId="12422" CreationDate="2011-08-14T23:26:23.087" Score="0" ViewCount="" Body="&lt;p&gt;I think that the simple thing that the OP missed is that infinite plane wave states are always non-interacting. No matter how big the cross section for collisions is, if particles are smeared out over all space, they do not ever collide. This is the reason you can choose &quot;in&quot; and &quot;out&quot; free states to parametrize the collisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual collisions describe what happens when a localized superposition of asymptotic states collides with another localized superposition of other asymptotic states. This is reflected in the fact that the subleading corrections to the S-matrix are small, not in the perturbation theory sense, but in the sense of writing S=I + i A , where A is the collision amplitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you are using a plane wave with standard relativistic normalization $&amp;lt;f|I|i&amp;gt;$ is divergent for in=out, because the normalization is infinite. On the other hand, &#xA;$&amp;lt;f|A|i&amp;gt;$ is finite, which reflects the small probability of particles finding each other. The results become finite when you make wave-packets, because you smear in and out, and the delta function when in=out just gives you the intensity of collisions based on the width of the beam. &lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-14T23:34:23.067" LastActivityDate="2011-08-14T23:34:23.067" />
  <row Id="13551" PostTypeId="2" ParentId="13546" CreationDate="2011-08-15T00:52:52.687" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;But what is the quantum state of the electrons before the measurements? It seems to me that if it is anything but an eigenstate of Sz, then the perfect correlations are impossible.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrons have &quot;wave particle duality&quot; which sounds like philosophical garbage but is detectable in experiments such as this one. Rather than emitting the electrons in an unknown state, let's assume a device that specifically emits a pair of electrons that are 100% anti-correlated in Sx rather than Sz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will work in the usual Sz basis, and since Sx can be taken to be real in this basis, I'll ignore the difference between bra and ket vectors (and for convenience, write them all as bras). So one of the electrons is Sx+ = $(1,1)/\sqrt{2}$, the other is Sx- = $(1,-1)/\sqrt{2}$. The combined wave function is defined mathematically as a x product as in:&lt;br&gt;&#xA;$$|x+-\rangle = (1,1)\times(1,-1)/2 = (1,1,-1,-1)/2.$$&lt;br&gt;&#xA;But since electrons are indistinguishable, we could have the two electrons reversed. This would be the state:&lt;br&gt;&#xA;$$|x-+\rangle = (1,-1)\times(1,1)/2 = (1,-1,1,-1)/2.$$&lt;br&gt;&#xA;Note that the above two vectors are orthonormal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since electrons are fermions, we take the anti-symmetric combination. This means we take the difference between the above two joint wave functions:&lt;br&gt;&#xA;$$|x+/-\rangle = (|x+-\rangle - |x-+\rangle)/\sqrt{2} = (0,1,-1,0)/\sqrt{2}.$$  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now let's redo the calculation with two electrons that have spin Sz anti-correlated so that Sz+ =$(1,0)$ and Sz- = $(0,1)$. The two possible cases are:&lt;br&gt;&#xA;$$|z+-\rangle = (1,0)\times (0,1) = (0,0,1,0)$$&lt;br&gt;&#xA;$$|z-+\rangle = (0,1)\times (1,0) = (0,1,0,0)$$&lt;br&gt;&#xA;and the anti-symmetric combination is:&lt;br&gt;&#xA;$$|z+/-\rangle = (|z+-\rangle - |z-+\rangle)/\sqrt{2} = (0,-1,1,0)/\sqrt{2}.$$&#xA;This is the same as we got for the x case $|x+/-\rangle$ (other than an overall factor of -1 which is just the usual arbitrary complex phase).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we see that according to the rules for quantum mechanics, it doesn't matter which two electrons we begin with. They could be spin in the x direction or z direction, or any direction; you get the same joint wave function.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now the above explanation may be unsatisfying in that it is mathematical. If you want a more physical explanation for what is going on, maybe the following will help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can think of the electrons as being excitations, that is, as waves. When you measure a wave, you alter the wave and change it. In the case of the anti-correlated electrons, the combined wave has just enough stuff to give you two electrons with opposite correlated spin. It doesn't matter which way the spins are arranged, there's just enough to give you a pair of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One must be careful when one talks about the passage of time in these experiments. The measurement typically takes place long after the electron wave function has been divided into two portions, say one going to one detector, the other to the other detector. It might be several nanoseconds before an apparatus is able to permanently record which electron had which spin. During THAT (i.e. measurement) time, the electron's wave function has to be changed into another form, one which is suitable for measurement. This cannot happen instantaneously and it does not happen AFTER the electron wave function is split. Until the measurement is made, the electron, along with the measuring apparatus, is all in a Schroedinger Cat sort of combined wave function. Either possible result is possible, but the electron wave function only has enough juice to do one of the two possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same &quot;explanation&quot; applies to things like &lt;a href=&quot;http://en.wikipedia.org/wiki/Wheeler%27s_delayed_choice_experiment&quot; rel=&quot;nofollow&quot;&gt;Wheeler's &quot;delayed choice&quot;&lt;/a&gt; experiment. There actually is no delay in the choice in that the combination of experiment and particle are in a combined wave state until the measurement becomes permanent.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-08-15T00:52:52.687" CommentCount="5" />
  <row Id="13552" PostTypeId="2" ParentId="13525" CreationDate="2011-08-15T01:08:33.373" Score="3" ViewCount="" Body="&lt;p&gt;The problem can have different answers depending on what restricts the transmission of heat. I will first make the assumption that the primary restriction is in how quickly your refrigerator can move heat. That is, I will assume that heat flow rates are not a problem and that when you put a lot of hot things in your freezer, it will immediately turn on the motor and start cooling it down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally speaking given two objects with the same volume, the one that is easiest to heat and cool will be the one with the fewer number of atoms. Technically, what is happening is that &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_capacity&quot; rel=&quot;nofollow&quot;&gt;heat capacity&lt;/a&gt; is a matter of counting &lt;a href=&quot;http://en.wikipedia.org/wiki/Degrees_of_freedom_%28mechanics%29&quot; rel=&quot;nofollow&quot;&gt;degrees of freedom&lt;/a&gt;. The larger the number of particles, the more degrees of freedom and the greater the amount of heat that corresponds to the same change in temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to see which is quicker to heat or cool down we can look at the number of atoms in the two objects. Air is a gas and the atoms are quite far apart. Therefore, air will cool down faster.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;On the other hand, suppose that your question is really about adding a single small object to a freezer that is either (a) nearly full, or (b) nearly empty. In this case we might assume that opening your freezer door changes the cold air in the freezer for warm air. Then the nearly full freezer will cool your new object more quickly because there is less new air to cool down. But since air is easy to cool, this won't be much of an effect.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally, it's possible that different materials will have different rates at which they cool down due to differences in the rate at which heat flows. There are three ways heat moves around, &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_radiation&quot; rel=&quot;nofollow&quot;&gt;radiation&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Conduction_%28heat%29&quot; rel=&quot;nofollow&quot;&gt;conduction&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Convection&quot; rel=&quot;nofollow&quot;&gt;convection&lt;/a&gt;. In this case, gasses and liquids typically have an advantage because they convect. Since the air in your freezer is a gas,it convects and might result in cooling an item more quickly when your freezer is empty. This would be especially the case, for instance, if the rest of your freezer were filled with an insulating solid such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Styrofoam&quot; rel=&quot;nofollow&quot;&gt;styrofoam insulation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-08-15T01:08:33.373" CommentCount="2" />
  <row Id="13553" PostTypeId="2" ParentId="13525" CreationDate="2011-08-15T01:20:49.190" Score="1" ViewCount="" Body="&lt;p&gt;What is your full freezer full of? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Situation #1: I take an empty freezer and stick a water bottle inside, resting on the freezer's floor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Situation #2: I take the water bottle, wrap it in a high-quality down sleeping bag, and shove all that in an otherwise empty freezer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Situation #3: I fill a freezer with ice cubes and shove the water bottle into the center of the ice cubes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 freezes the water more slowly than 1, because the water bottle is insulated from the floor of the freezer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 freezes the water more quickly than 1, because the water bottle is in thermal contact with cold stuff on all sides, not just underneath.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a freezer full of food, you can make a bottle of beer freeze more quickly by inserting it somewhere such that it's in thermal contact with frozen stuff on all sides.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-15T01:20:49.190" />
  <row Id="13554" PostTypeId="2" ParentId="12422" CreationDate="2011-08-15T01:21:37.270" Score="0" ViewCount="" Body="&lt;p&gt;In a lot of physics experiments, one arranges for initial and final states and these conditions are what one uses for the basis states. For example, if an electron is detected with momentum $p&amp;#39;$, then one of the basis states in the theoretical explanation for the experiment will be $|p&amp;#39;\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of this is the CKM matrix, which is a unitary transformations between two basis sets, each with three elements. One of the bases consists of the quarks with charge +2/3, i.e. the up, charm, and top quarks. The other basis is the charge -1/3 quarks, the down, strange and bottom quarks. In this case, again mother nature chose the bases for us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, there are situations where experiment doesn't choose the basis set. For two electrons with anti-correlated spin-1/2, our choice of basis (i.e. spin direction) is arbitrary and &lt;a href=&quot;http://physics.stackexchange.com/questions/13546/why-do-bell-tests-give-perfect-correlations/13551#13551&quot;&gt;choosing different bases does not change the mathematics or the prediction&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-08-15T01:21:37.270" />
  <row Id="13555" PostTypeId="2" ParentId="13098" CreationDate="2011-08-15T03:08:38.423" Score="0" ViewCount="" Body="&lt;p&gt;First try to understand the case in which there &lt;em&gt;is&lt;/em&gt; another electrode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a pointy electrode and a flat electrode.  You apply a high voltage across them.  The pointiness of the first electrode concentrates the electric field to the extent that electrons are stripped from atoms and the air around the tip becomes conductive, a plasma. The plasma is the part that glows.  The flat electrode spreads out the electric field, so the air around it stays normal, an insulator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the voltage is high enough, the electric field around the pointy plasma will also be strong enough to ionize air, and the plasma region will get bigger.  If this continues extending itself until it touches the flat electrode, there is a complete conductive path, and an arc forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the voltage is &lt;em&gt;not&lt;/em&gt; high enough for an arc, the plasma will stop increasing in size, and just remain in a layer around the pointy electrode.  The plasma is then surrounded by normal insulating air.  There is not a complete conductive path between the electrodes, but a current can still flow.  The way it happens is that ions from the plasma are repelled away from the electrode, and migrate slowly through the neutral air (like any other molecule diffuses in air by bouncing off of air molecules, but combined with a drift in one direction because it's repelled by the pointy electrode and attracted to the flat one).  Eventually it will reach the flat electrode and discharge.  This current is much smaller than the arc, though, because the charges have to diffuse through the region of insulating air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not correct to say &quot;a current develops from that wire and ionizes gas (air) on its way&quot;.  The gas is ionized by the electric field, and then a current flows because the ionization allows it to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to understand how it works without an obvious flat electrode: All voltages are really a difference between two points.  We say things like &quot;this point is at 20 kV&quot;, but what we really mean is &quot;this point is at 20 kV with respect to another reference point&quot;.  This reference point is referred to as &quot;ground&quot;, and can be any point in a circuit.  It doesn't have to be connected to the physical Earth.  So the other electrode is whatever your voltage source is connected to.  If the voltage source is connected to the planet Earth, then the planet and anything connected to it becomes the flat electrode.  If it's connected to a metal sphere, then that sphere is the flat electrode and the charges will try to migrate to it.&lt;/p&gt;&#xA;" OwnerUserId="176" LastActivityDate="2011-08-15T03:08:38.423" />
  <row Id="13556" PostTypeId="1" CreationDate="2011-08-15T03:17:52.720" Score="12" ViewCount="159" Body="&lt;p&gt;When applying DC to a neon lamp, only one electrode glows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/3poGs.jpg&quot; alt=&quot;http://commons.wikimedia.org/wiki/File:Neonlamp3.JPG&quot;&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aNeonlamp3.JPG&quot; rel=&quot;nofollow&quot;&gt;The voltages across the lamps are left: DC (left lead positive), middle: DC (right lead positive), and right: AC.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But... why?  The electrodes are the same shape, so the electric field around them should be same shape, and the gas should break down at the same electric field strength.  It seems like the fields would be symmetrical.  Is there a difference in threshold between &lt;a href=&quot;http://en.wikipedia.org/wiki/Corona_discharge#Positive_coronas&quot; rel=&quot;nofollow&quot;&gt;positive&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Corona_discharge#Negative_coronas&quot; rel=&quot;nofollow&quot;&gt;negative&lt;/a&gt; coronas?  If so, do both sides light up at high enough voltage?  Or maybe only one type of corona is possible in neon since it's a noble gas?  If it contained air would it glow at both electrodes?&lt;/p&gt;&#xA;" OwnerUserId="176" LastEditorUserId="176" LastEditorDisplayName="" LastEditDate="2011-08-18T02:46:14.097" LastActivityDate="2011-08-18T02:46:14.097" Title="Why does only one side of a neon lamp glow?" Tags="&lt;electromagnetism&gt;&lt;plasma-physics&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="1" />
  <row Id="13557" PostTypeId="1" AcceptedAnswerId="13614" CreationDate="2011-08-15T03:26:52.487" Score="9" ViewCount="122" Body="&lt;p&gt;Nowadays it seems to be popular among physics educators to present Newton's first law as a definition of inertial frames and/or a statement that such frames exist. This is clearly a modern overlay. Here is Newton's original statement of the law (Motte's translation):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Law I. Every body perseveres in its state of rest, or of uniform&#xA;  motion in a right line, unless it is compelled to change that state by&#xA;  forces impressed thereon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The text then continues:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Projectiles persevere in their motions, so far as they are not&#xA;  retarded by the resistance of the air, or impelled downwards by the&#xA;  force of gravity. A top, whose parts by their cohesion are perpetually&#xA;  drawn aside from rectilinear motion, does not cease its rotation,&#xA;  otherwise than as it is retarded by the air. The greater bodies of the&#xA;  planets and comets, meeting with less resistance in more free spaces,&#xA;  preserve their motions both progressive and circular for a much longer&#xA;  time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And then the second law is stated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is clearly nothing about frames of reference here. In fact, the discussion is so qualitative and nonmathematical that many modern physics teachers would probably mark it wrong on an exam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a small collection of old physics textbooks, and one of the more historically influential ones is Elements of Physics by Millikan and Gale, 1927. (Millikan wrote a long series of physics textbooks with various titles.) Millikan and Gale give a statement of the first law that reads like an extremely close paraphrase of the Mott translation. There is no mention of frames of reference, inertial or otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A respected and influential modern textbook, aimed at a much higher level than Millikan's book, is Kleppner and Kolenkow's 1973 Introduction to Mechanics. K&amp;amp;K has this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...it is always possible to find a coordinate system with respect to&#xA;  which isolated bodies move uniformly. [...] Newton's first law of&#xA;  motion is the assertion that inertial systems exist. Newton's first&#xA;  law is part definition and part experimental fact. Isolated bodies&#xA;  move uniformly in inertial systems by virtue of the definition of an&#xA;  inertial system. In contrast, that inertial systems exist is a&#xA;  statement about the physical world. Newton's first law raises a number&#xA;  of questions, such as what we mean by an 'isolated body,' [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a paper on this historical/educational topic: &lt;a href=&quot;http://www.springerlink.com/content/j42866672t863506/&quot; rel=&quot;nofollow&quot;&gt;Galili and Tseitlin, &quot;Newton's First Law: Text, Translations, Interpretations and Physics Education,&quot; Science &amp;amp; Education&#xA;Volume 12, Number 1, 45-73, DOI: 10.1023/A:1022632600805&lt;/a&gt;. I had access to it at one time, and it seemed very relevant. Unfortunately it's paywalled now. The abstract, which is not paywalled, says, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Normally, NFL is interpreted as a special case: a trivial deduction&#xA;  from Newton's Second Law. Some advanced textbooks replace NFL by a&#xA;  modernized claim, which abandons its original meaning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1&lt;/strong&gt;: Does anyone know more about when textbooks begain to claim that the first law was a statement of the definition and/or existence of inertial frames?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There seem to be several possible interpretations of the first law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. Newton consciously wrote the laws of motion in the style of an axiomatic system, possibly emulating Euclid. However, this is only a matter of style. The first law is clearly a trivial deduction from the second law. Newton presented it as a separate law merely to emphasize that he was working in the framework of Galileo, not medieval scholasticism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. Newton's presentation of the first and second laws is logically defective, but Newton wasn't able to do any better because he lacked the notion of inertial and noninertial frames of reference. Modern textbook authors can tell Newton, &quot;there, fixed that for you.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. It is impossible to give a logically rigorous statement of the physics being described by the first and second laws, since gravity is a long-range force, and, as pointed out by K&amp;amp;K, this raises problems in defining the meaning of an isolated body. The best we can do is that in a given cosmological model, such as the Newtonian picture of an infinite and homogeneous universe full of stars, we can find some frame, such as the frame of the &quot;fixed stars,&quot; that we want to call inertial. Other frames moving inertially relative to it are also inertial. But this is contingent on the cosmological model. That frame could later turn out to be noninertial, if, e.g., we learn that our galaxy is free-falling in an external gravitational field created by other masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2&lt;/strong&gt;: Is A supported by the best historical scholarship? For extra points, would anyone like to tell me that I'm an idiot for believing in A and C, or defend some other interpretation on logical or pedagogical grounds?&lt;/p&gt;&#xA;" OwnerUserId="4552" LastEditorUserId="966" LastEditorDisplayName="" LastEditDate="2011-08-15T05:57:31.693" LastActivityDate="2011-08-16T03:04:18.690" Title="History of interpretation of Newton's first law" Tags="&lt;newtonian-mechanics&gt;&lt;history&gt;" AnswerCount="1" FavoriteCount="5" />
  <row Id="13558" PostTypeId="2" ParentId="11670" CreationDate="2011-08-15T03:33:03.757" Score="0" ViewCount="" Body="&lt;p&gt;The significance was that radium was many orders of magnitude more intense as a source of alpha particles than the previously available sources, such as uranium salts. For example, the Rutherford-Marsden discovery of the atomic nucleus would have been impossible without radium. That experiment used human eyeballs as detectors, and human eyeballs are not sensitive enough to have allowed the experiment to be done with weaker sources.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-15T03:33:03.757" />
  <row Id="13559" PostTypeId="2" ParentId="1268" CreationDate="2011-08-15T03:47:59.630" Score="1" ViewCount="" Body="&lt;p&gt;The answers do not address what Ohanian said. &lt;a href=&quot;http://arxiv.org/abs/0805.1400&quot; rel=&quot;nofollow&quot;&gt;His paper&lt;/a&gt; is a free download. As far as I know, Ohanian has not been refuted.&lt;/p&gt;&#xA;" OwnerUserId="4926" LastActivityDate="2011-08-15T03:47:59.630" CommentCount="1" />
  <row Id="13560" PostTypeId="2" ParentId="13525" CreationDate="2011-08-15T03:56:51.013" Score="3" ViewCount="" Body="&lt;p&gt;&lt;strong&gt;Heat Capacity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are comparing two situations, and to formalize this, let's say that the freezer has a given volume, $V$ and that volume is broken down into air and other frozen materials.  Then we have the situation that $V=V_{air}+V_{stuff}$.  The heat capacity of the entire freezer between the two situations will be vastly difference because the &lt;a href=&quot;http://en.wikipedia.org/wiki/Volumetric_heat_capacity&quot; rel=&quot;nofollow&quot;&gt;volumetric heat capacity&lt;/a&gt; for air and ice (which is the most common item included in &quot;stuff&quot;) is vastly different by a factor of about 1000.  I'll use the Wikipedia notations, where specific heat capacity is $c$ ($\frac{J}{kg K}$) and volumetric heat capacity is $c \rho$.  The volumetric heat capacity times volume is the heat capacity (denoted $C$) so for the entire freezer we have the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C = \left(V c \rho \right)_{air} + \left( V c \rho \right)_{stuff}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat capacity matters because should you insert some mount of heat into the freezer, $Q$, then after equilibrium is reached, the temperature will raise by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta T = \frac{Q}{C}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we don't consider the work of the freezer thermal cycle actively cooling the air, then the freezer with more stuff in it will cool the item faster because the temperature of the air+stuff changes less, so the $\Delta T$ in the Newton cooling model will be greater and it will cool faster and cool to a lower temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Air Cooling&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is relevant to note that the air is actively cooled by the freezer systems and the stuff is not.  This is interesting because it will have relevance to the heat transfer mechanisms (more later).  Although this isn't entirely correct, we could assume that the cooling device takes some flow rate of air continuously, $\dot{m}$ (kg/s), and cools with some enthalpy change, $\Delta h$ (J/kg).  That enthalpy change removes heat from the system, but that doesn't change from one case to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that can't be the full picture.  Heat is only removed from the freezer if the temperature drops below a certain value.  So, another way to do this may be to assume that the cooler keeps the air at a certain temperature, but that may defeat the purpose of the problem.  It is also possible that the thing being cooled, in fact, outpaces the cooling device and heats the air faster than it can be cooled.  In that case the air temperature would rise above the programmed setpoint and stay there until the item being cooled started to approach the freezer temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Heat Transfer Mechanisms&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two mechanisms of heat transfer between the item you are freezing and other stuff in the freezer.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Convection with the air&lt;/li&gt;&#xA;&lt;li&gt;Radiative heat transfer with other items and the walls&lt;/li&gt;&#xA;&lt;li&gt;Conduction with the bottom surface (same for both and I won't discuss)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Obviously, the 1st one transfers with the air and the 2nd one transfers with the stuff.  I would like to propose several simplified models such that we can talk more specifically about predictions.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The freezer cools the item faster than the active cooling is relevant&lt;/li&gt;&#xA;&lt;li&gt;The item quickly equilibrizes with the air, then the stuff and the active cooling then slowly lowers the temperature &lt;/li&gt;&#xA;&lt;li&gt;The air is kept at a constant temperature&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As I've said already, number 1 clearly has the freezer with more stuff win.  Number 2 could potentially go in the direction of the freezer with less stuff because of the larger heat capacity of air (which will only work for a small item), but the radiative heat transfer still advantages the case with more stuff.  In number 3 the freezer with more stuff clearly wins unless the flow path for convection is restricted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The exact one that will win does depend on the specific values for the system, so that's in the hands of the experimentalist.  Generally though, the higher heat capacity and the ability to transfer heat through radiative transfer is likely to favor the freezer with more stuff in it.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-08-15T03:56:51.013" />
  <row Id="13561" PostTypeId="2" ParentId="13539" CreationDate="2011-08-15T04:01:27.237" Score="2" ViewCount="" Body="&lt;p&gt;The parallel/antiparallel picture is not quite correct. Mostly because it hides the fact that the electrons are indistinguishable from our intuition. In a (very rare) cases when you may consider electrons distinguishable, this picture is correct. In more natural case when electrons are completely equal triplet/singlet picture is much better. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe few more explanations needed. If you say that electron spins are antiparallel what you &lt;strong&gt;actually&lt;/strong&gt; say is &quot;spin of one electron is antiparallel to spin of another&quot;. But electrons are completely equal! You can not freely point to electron and say that it is &quot;the one&quot;. Singlet/triplet picture however stays on a more solid ground: what is the total angular momentum of system which consists of two electrons. Here you do not choose electrons. You explicitly use the fact that electrons are equal. &lt;/p&gt;&#xA;" OwnerUserId="4657" LastActivityDate="2011-08-15T04:01:27.237" />
  <row Id="13562" PostTypeId="2" ParentId="7049" CreationDate="2011-08-15T04:27:10.980" Score="0" ViewCount="" Body="&lt;p&gt;The purpose to the electromagnetism tensor was to demonstrate the Lorentz covariance of Maxwell's equations. It is central to the discovery of special relativity, and it got everyone excited about relativity at the time. Einstein had nothing to do with it.&lt;/p&gt;&#xA;" OwnerUserId="4926" LastActivityDate="2011-08-15T04:27:10.980" />
  <row Id="13563" PostTypeId="2" ParentId="2696" CreationDate="2011-08-15T04:48:08.520" Score="1" ViewCount="" Body="&lt;p&gt;The answer is no, because there is no horizon. A time-independent gravitational field emits no particles, because energy is conserved, just like a stationary charge distribution does not radiate. This is consistent with the equivalence principle, as explained below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Unruh effect, the particle emmission is from the &quot;black wall&quot; acceleration horizon behind the observer. If you place a refrigerated baffle between the black wall and the observer, the radiation will be absorbed by the baffle, and will not be seen by the observer. An observer standing on a planet is indistinguishable from an accelerating observer with a planet between him and the acceleration horizon, and such an observer sees no emissions from the horizon, because the horizon is invisible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might object that the horizon is infinite, and the planet is finite, so shouldn't you be able to see the horizon far enough out? But if you go far enough out, you notice the curvature caused by the planet, and this will push the horizon out to infinity, because there is no horizon after all, you're just standing on a planet.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-15T04:48:08.520" CommentCount="3" />
  <row Id="13564" PostTypeId="2" ParentId="12435" CreationDate="2011-08-15T05:03:51.140" Score="0" ViewCount="" Body="&lt;p&gt;Historically, FitzGerald and Lorentz deduced the transformations from Michelson-Morley showing that the speed of light was constant for all observers, and then Poincare found the Minkowski metric when he was looking for invariants of the Lorentz group.&lt;/p&gt;&#xA;" OwnerUserId="4926" LastActivityDate="2011-08-15T05:03:51.140" CommentCount="5" />
  <row Id="13565" PostTypeId="2" ParentId="5483" CreationDate="2011-08-15T05:15:52.540" Score="1" ViewCount="" Body="&lt;p&gt;Mach's principle, if interpreted charitably, requires that one include horizons as matter, along with gravitational waves, and light, and all particles. This is required to include black holes, and for consistency requires cosmological horizons too. Once you understand that &quot;matter&quot; means &quot;horizon&quot;, the statement that all rotation is relative to distant horizons is just a stunted classical version of the holographic principle, and is sort of vacuously true.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-15T05:15:52.540" />
  <row Id="13566" PostTypeId="2" ParentId="4831" CreationDate="2011-08-15T05:19:42.720" Score="0" ViewCount="" Body="&lt;p&gt;Technically any leaver that lifts an object away from the earth also lifts the earth away from said object. &lt;/p&gt;&#xA;" OwnerUserId="2843" LastActivityDate="2011-08-15T05:19:42.720" />
  <row Id="13567" PostTypeId="1" CreationDate="2011-08-15T05:37:33.407" Score="5" ViewCount="99" Body="&lt;p&gt;No doubt the height would vary each year since the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tropic_of_Cancer&quot; rel=&quot;nofollow&quot;&gt;tropic of cancer&lt;/a&gt; changes latitude slightly. Perhaps assume 23.5º N for calculations sake.&lt;/p&gt;&#xA;" OwnerUserId="2843" LastActivityDate="2011-08-15T08:15:49.823" ClosedDate="2011-08-16T15:54:09.250" Title="How tall would a tower at the south pole have to be for the top to get year-around sunshine?" Tags="&lt;solar&gt;" AnswerCount="0" CommentCount="1" />
  <row Id="13568" PostTypeId="2" ParentId="1019" CreationDate="2011-08-15T06:28:26.200" Score="6" ViewCount="" Body="&lt;p&gt;I will give some meta-false beliefs: these are beliefs held by the general public, which happen to be true, which are hyper-corrected by many physicists with bogus corrections based on the urge to appear smart:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Electrons move slowly down a wire&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The belief: the electrons move lightning fast down a wire. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the hypercorrection: in the completely obsolete Drude model, electrons move slowly. In this model, you imagine the current is carried by a classical gas of electrons, and you divide the total current by the density of all electronic charge to get the drift velocity. This predicts a completely bogus drift velocity of a few cm/s, which is total nonsense, because only electrons near the Fermi surface contribute to the conductivity. Nevertheless, you see this hypercorrection repeated endlessly (it appears here too).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The best answer: the electronic wavefunctions are spread out in a metal. The correct notion of electron velocity is the Fermi velocity, which is enormous typically, because the wavelength is about 1 atomic radius. While it isn't the same as the speed of electricity going down the wire (which is the speed of the field perturbations, some significant fraction of the speed of light), it is enormously high. Impurities which can scatter electrons will alter this speed, but not as much as the naive hypercorrection says.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;The atom is mostly empty space&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The belief: the atom is full of stuff, that's why stuff is hard when you push against it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The hypercorrection: In the totally obsolete Rutherford-Bohr model, the atom is mostly empty space, the tiny pointlike electron orbiting a nucleus which contains most of the mass.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The best answer: But it is the electrons' wavefunction which tells you whether something is empty space or not. A region filled with electronic wavefunction feels hard to the touch, because two electrons can't be compressed into the same space without squeezing their wavefunction to have very high spatial variations, by the exclusion principle. Atoms are full of electronic wavefunction, and are therefore not empty space, at least not by any reasonable definition.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;There is nothing mystical about measurement in quantum mechanics&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The belief: the measurement problem in the standard quantum mechanics suggests that consciousness is somehow involved in measurements&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;hypercorrection: decoherence explains all that! Quantum mechanics is no different than determinism as far as enlightenment values are concerned.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The best answer: Decoherence tells you why you don't have interference between classically different worlds, or histories, and its an important part of the story if quantum mechanics is exact. But it does not tell you why you &quot;percieve&quot; one consistent history as a world. You need a dictionary between physics and perception. Since this dictionary is fundamental, and weird, and philosophical, it is important to explain that this is not an output of physics, but an input, which links mathematical theory with explicit sense-data.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;There is no such thing as a centrifugal force&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The belief: when things are rotating, they are pushed out by a centrifugal force.&lt;/li&gt;&#xA;&lt;li&gt;The hypercorrection: There is no centrifugal force. There is a centripetal force (centripetal was a made up word to replace centrifugal) which pulls you in.&lt;/li&gt;&#xA;&lt;li&gt;The best answer: This is obviously true from the point of view of the inertial frame, there is no centrifugal force, but if you are looking at it from the point of view of the rotating object, then there is. It all depends on your choice of reference frame.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;The big bang happened everywhere at once&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The belief: the big bang means that the universe started at some definite point and got bigger from there.&lt;/li&gt;&#xA;&lt;li&gt;The hypercorrection: the big bang happened everywhere at once, and it is just wrong to think of it as happening at some single point in an extended model of space-time. If the universe is open, the big-bang was infinite in extent.&lt;/li&gt;&#xA;&lt;li&gt;The best answer: There are three important caveats: 1. In FRW models, the bang-point is a singularity, so its outside space and time, and it is impossible to determine if it is &quot;really&quot; a single point or &quot;really&quot; everywhere at once, so it's just a meaningless question. 2. In the &lt;em&gt;Newtonian&lt;/em&gt; &quot;big bang&quot; model, where you imagine the universe is now filled with particles which have a speed away from you linearly proportional to the distance from you, everything does come out from a single point! All the newtonian world lines converge on your current position. That's true even though the universe is spatially homogenous (the reason its not a paradox is that Galilean boosts are nontrivially mixed with translations). 3. the best picture, in my view, is the holographic picture, where you are surrounded by a horizon which was smaller in the past. This view is similar to the Newtonian big bang, in that everything came from a small region bounded by a dS cosmological horizon. This is mathematically equivalent to everything else, except throwing away the stuff outside the horizon that can't be observed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would like to admit that I was a little flabbergasted when a lay-person told me that everything in a Newtonian big-bang comes from a single place. That was completely counterintuitive.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-15T06:28:26.200" CommunityOwnedDate="2011-08-15T06:28:26.200" CommentCount="1" />
  <row Id="13569" PostTypeId="2" ParentId="1019" CreationDate="2011-08-15T07:23:23.677" Score="4" ViewCount="" Body="&lt;p&gt;Here's another list of false beliefs. These are held by science popularizers. Whether they actually believe these beliefs, or just utter them for the purpose of getting more viewers, is an unanswerable question:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The curved space near massive object can be pictured as a deformed rubber sheet&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This one is due to Einstein, unfortunately. You put balls on a rubber sheet, and you see that they roll towards each other. The reason this is a terrible explanation is because you have the Earth's gravity doing the pulling, not the curved space. The actual geodesics on a curved space like the rubber sheet repel each other. The reason things attract in relativity is because of the time-dilation factor, and this is the dominant effect. It is just as easy to explain things correctly, in terms of time slowing down near a massive object, and world-lines trying to maximize their proper time with given fixed endpoints, but popularizers never do this.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;A variable speed of light can replace inflation.&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This appeared in a recent popular show, and it is based on the following bogus idea: if light moved faster at early times, then all the universe could have been in communication! The reason this is false is because no matter how the speed of light is imagined to vary, one can recoordinatize space-time in terms of the intersections of light cones, and unless these lightcones split instead of merge, you get the same communication paradox--- new regions coming into causal contact are coming into causal contact for the first time.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Mesons and Baryons are made of quarks like atoms are made of protons, neutrons and electrons.&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This is insidious, because its true for heavy mesons. But it's much more false than true for pions and protons and all the excitations at lower than 1GeV, because of the vacuum condensates. There is no reasonable model of light pions which does not take into account their Goldstone nature. This type of explanation also leaves out Nambu and Skyrme, both of whom were unjustly ignored for too long.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;String theory is a theory of strings&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This picture is not good for someone who doesn't already have a sense of string theory, because if you start out making home-made models of relativistic strings, you will never get anything like the correct string theory. The strings you naively picture would not have the special light-cone interactions that strings do in the Mandelstam picture, and they would not obey Dolen Horn Schmidt duality. They would just be conglomerations of point particles held together by rubber bands. They would have the wrong spectrum, and they would be full of ghosts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only proper way to say what strings are is to say right off the bat that they are S-matrix states, and that they are designed to be an S-matrix theory with linear Regge trajectories. They have a string picture, but the constraint that exchanging strings in the S-channel is dual to exchanging them in the T-channel is all-important, just as it was all-important historically. Without this, even with the Nambu action, you are at a loss for how to incorporate interactions. It is not obvious that the interactions are by topology unless you know Dolan Horn Schmidt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also important for realizing that string interactions are somewhat holistic (that they become local on the light cone is the surprise, not the other way around). You add them order by order in perturbation theory by demanding unitarity, not by asking what happens when two strings collide in the usual sense. These &quot;strings&quot; are strange new things born of 1960s Chew-isms, and their closest cousins are flux lines in gauge theory, or fishnet Feynman diagrams, not a collection of point masses held together by spring-like forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also insidious, because Chew, Mandelstam, Dolen, Scherk, and all that generation developed the greatest physical theory the world will ever see, and their reward was: &quot;You're fired&quot;. (in Scherk's case, &quot;You're crazy&quot;). Then they were heckled for thirty years, while their work was appropriated by a new generation, who described them as the deluded misguided Chew-ites who discovered something great by accident.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;There is more than a snowball's chance in hell for large extra dimensions&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The idea that there are large extra dimensions was very popular in 2000, but it's completely preposterous. Large extra dimensions bring the planck mass down to about a TeV, giving neutrinos generic majorana masses which are in the KeV-MeV range, so you need to fine tune. They lead to essentially instantaneous proton decay, and huge CP violations in strong interactions, so you need to fine tune some more. To avoid proton decay, there is a clever mechanism due to Arkani-Hamed and Schmalz which puts the quarks and leptons in different places in the extra dimensions. This idea is appealing only at a superficial first glance, because it requires that the SU(2) and U(1) of the standard model be extended in the extra dimensions, which affects their running immediately. The theory predicts unambiguously and model-independently that proton decay suppression requires huge electroweak running at around a TeV. That's a signal you haven't seen a hint of at 100GeV collisions. Come on. In addition, how do you stabilize large dimensions? It's the same fine-tuning as before, so the number of problems has gone up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A low Planck scale would completely demolish the predictivity of string theory. You can squeeze a lot of stuff into large dimensions. In my opinion, it is this brand of string theory that the critics correctly criticize as fundamentally non-predictive.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-15T07:23:23.677" CommunityOwnedDate="2011-08-15T07:23:23.677" />
  <row Id="13571" PostTypeId="1" AcceptedAnswerId="13608" CreationDate="2011-08-15T09:17:22.470" Score="9" ViewCount="119" Body="&lt;p&gt;Is it possible to determine to an approximate degree, the revolutions per minute of a fan, for example a pedesal fan pictured below, without using some electronic/mechanical measuring device?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/ekTtd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that comes to mind is the &lt;a href=&quot;http://www.soundsclassic.com/listpix/technicsslbd35.jpg&quot; rel=&quot;nofollow&quot;&gt;markings on a old record player's turntable&lt;/a&gt; - can that concept somehow be used to make marks on a fan blade to determine it's RPM? Or can something called a &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electronic_tuner&quot; rel=&quot;nofollow&quot;&gt;strobe tuner&lt;/a&gt;&quot; help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/2vZfO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I maybe make markings on a fan with a marker and figure out the RPM using nothing else but a stopwatch? Or maybe some other DIY technique that does not require the purchase of any measuring device?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ps.&lt;/strong&gt; I'm not entirely sure if this is a physics or engineering question, so please feel free to move it to the appropriate site (I checked all 58 stackexchange sites and only the Physics site seemed to fit the question)&lt;/p&gt;&#xA;" OwnerUserId="4930" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-15T21:05:15.947" LastActivityDate="2011-08-16T13:51:40.943" Title="Home experiments to measure the RPM of a pedestal fan without special equipment?" Tags="&lt;measurement&gt;&lt;home-experiment&gt;&lt;angular-velocity&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" />
  <row Id="13572" PostTypeId="2" ParentId="13474" CreationDate="2011-08-15T10:34:52.377" Score="1" ViewCount="" Body="&lt;p&gt;If it would only be the weights exerting torque, the balance would be in equilibrium at all angles. What makes the balance go back to the horizontal position is the fact, that the center of mass is below the beam. consider this picture&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/pKZSH.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The needle exerts a torque too, so you have more torque on the side, where the plate is higher. You can have more subtle configurations (like in your picture, where the beam is rounded below) but the mechanism is the same.&lt;/p&gt;&#xA;" OwnerUserId="2151" LastActivityDate="2011-08-15T10:34:52.377" />
  <row Id="13573" PostTypeId="2" ParentId="5508" CreationDate="2011-08-15T10:41:04.090" Score="2" ViewCount="" Body="&lt;p&gt;I would add that developing a physical intuition for the objects of study is an incredibly useful tool. This includes getting a sense for the scale of things, as well as consideration of fundamental things like energy transfer.  Also, understanding thoroughly simple models like simple harmonic oscillators or the concepts of symmetries can be a tremendous help, as these come up everywhere.  If you have access and have a specific subject question, the &lt;a href=&quot;http://ajp.aapt.org/&quot; rel=&quot;nofollow&quot;&gt;American Journal of Physics&lt;/a&gt; frequently has articles about educational avenues in physics that I've found useful in my studies.  Indeed, good luck! &lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2011-08-15T10:41:04.090" CommentCount="1" />
  <row Id="13574" PostTypeId="2" ParentId="13460" CreationDate="2011-08-15T10:55:23.880" Score="1" ViewCount="" Body="&lt;p&gt;The comments list the cond-mat canonical texts, and I think the Kittel intro is a bit more accessible. Nevertheless, the International Tables of Crystallography (should be available at a university library) are the go to for anything related to crystallography. For a brute force introduction, &lt;a href=&quot;http://it.iucr.org/B/contents/&quot; rel=&quot;nofollow&quot;&gt;Volume B on reciprocal space&lt;/a&gt; has a comprehensive introduction to the topic of Fourier analysis as applied to crystallography. &lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2011-08-15T10:55:23.880" CommentCount="1" />
  <row Id="13575" PostTypeId="2" ParentId="3644" CreationDate="2011-08-15T13:05:39.537" Score="0" ViewCount="" Body="&lt;p&gt;Actually the speed of light can be different in different media, for example, it is higher than $c$ in Casimir vacuum and smaller than $c$ in a solid medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems the actual question here is why the speed of light in flat vacuum is the highest speed at which information can be transferred.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This comes from the Special Theory of Relativity where it was shown that if there was FTL information transfer, a casualty paradox would appear. This is because $c$ is used in Lorentz transforms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now naturally emerges a further question, why exactly the Special Theory of Relativity uses $c$ in its Lorentz transforms rather than any other speed. This is due to the idea that all inertial frames should be indistinguishable: the speeds of all processes, either electromagnetic, mechanical or gravitational should change similarly when changing a reference frame. If electric processes changes one way and say gravitational another way, we could determine whether we are in a moving or stationary frame. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that all fundamental interactions should propagate at the same speed for this criterion to be met.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So actually if there somewhere exists a medium where speed of light is higher or smaller than c, then in such medium it is possible to determine whether the given frame is absolutely moving or stationary relative to the medium. Such medium does not have the main propertiy of our vacuum, which distinguishes it from Ether: that there is no difference between moving(without acceleration) in it and being in rest.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastEditorUserId="1186" LastEditorDisplayName="" LastEditDate="2011-08-15T13:15:09.963" LastActivityDate="2011-08-15T13:15:09.963" />
  <row Id="13576" PostTypeId="1" CreationDate="2011-08-15T13:49:28.837" Score="4" ViewCount="105" Body="&lt;p&gt;Why is a hammer more effective in driving a nail  than a large mass resting over  the nail ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this has to do with momentum, but cant figure it out.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2011-08-15T21:45:10.623" Title="Hammer vs large mass on nail" Tags="&lt;momentum&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="13577" PostTypeId="1" AcceptedAnswerId="13636" CreationDate="2011-08-15T13:55:11.290" Score="4" ViewCount="123" Body="&lt;p&gt;If a photon (wave package) redshifts (streches) traveling in our expanding universe, is it's energy reduced? If so, where does it go?&lt;/p&gt;&#xA;" OwnerUserId="4760" LastActivityDate="2011-08-16T15:31:01.900" Title="photons in expanding space: how is energy conserved?" Tags="&lt;cosmology&gt;&lt;photons&gt;&lt;energy-conservation&gt;&lt;cosmological-inflation&gt;" AnswerCount="2" CommentCount="7" />
  <row Id="13578" PostTypeId="2" ParentId="7446" CreationDate="2011-08-15T13:59:06.390" Score="1" ViewCount="" Body="&lt;p&gt;No. Relative to Earth your bus will have zero length, so moving from back to the front of the bus will contribute nothing to your speed relative to Earth.&lt;/p&gt;&#xA;" OwnerUserId="1186" LastActivityDate="2011-08-15T13:59:06.390" />
  <row Id="13579" PostTypeId="1" AcceptedAnswerId="13601" CreationDate="2011-08-15T13:59:41.403" Score="2" ViewCount="79" Body="&lt;p&gt;My local car parts dealer presented me his inventory of car batteries. One 12V battery had a 'cold crank amperage' of 600amps. The other 12V battery had a 'cold crank amperage' of 585amps. &lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt;&lt;br/&gt;[1] If the resistance, my truck, is constant for both batteries, how can the amperage ratings differ? &lt;br/&gt;[2] What is does a battery's 'cold crank amperage' really mean?&lt;/p&gt;&#xA;" OwnerUserId="4933" LastActivityDate="2011-08-15T20:52:33.143" Title="How can two different 12V batteries have different amperage for same resistance?" Tags="&lt;electricity&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="13580" PostTypeId="2" ParentId="13577" CreationDate="2011-08-15T14:01:39.880" Score="-2" ViewCount="" Body="&lt;p&gt;It goes to make work to expand the universe against the forces of gravity and inertia. This is like adiabatically expanding volume of gas: the gas becomes cooler as the volume increases. Where the energy goes?&lt;/p&gt;&#xA;" OwnerUserId="1186" LastEditorUserId="1186" LastEditorDisplayName="" LastEditDate="2011-08-15T14:06:43.883" LastActivityDate="2011-08-15T14:06:43.883" CommentCount="10" />
  <row Id="13581" PostTypeId="1" AcceptedAnswerId="13592" CreationDate="2011-08-15T14:05:12.030" Score="10" ViewCount="244" Body="&lt;p&gt;The standard picture of the nucleus of atom is that is several distinct nucleons, which themselves are composed of quarks. However, it seems to me like a much simpler picture is that the nucleus is directly made out of quarks, without having nucleons as substructures. That is, that the nucleus is the ground state of a quark-gluon plasma. I can't find any evidence that this is false. For instance, if the distances between the nucleons was significantly larger than the size of the nucleons, there would be a difference in size between hydrogen and other atoms which can't be explained by this new framework, but according to wikipedia the diameter of all of the nuclei is the same order of magnitude. What is the evidence that large nuclei have nucleons as substructures?&lt;/p&gt;&#xA;" OwnerUserId="1859" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-15T16:09:12.310" LastActivityDate="2011-08-19T15:46:59.573" Title="How do we know that the nucleus isn't a quark-gluon plasma?" Tags="&lt;nuclear-physics&gt;&lt;experimental-evidence&gt;" AnswerCount="4" FavoriteCount="1" />
  <row Id="13582" PostTypeId="1" CreationDate="2011-08-15T14:05:56.473" Score="8" ViewCount="116" Body="&lt;p&gt;Many times I heared that the solar wind consists of protons and alpha particles. The both are positively charged, but are there electrons in solar wind?&lt;/p&gt;&#xA;" OwnerUserId="1186" LastEditorUserId="176" LastEditorDisplayName="" LastEditDate="2011-08-16T18:29:31.463" LastActivityDate="2011-08-16T18:29:31.463" Title="Is solar wind positively charged?" Tags="&lt;sun&gt;&lt;solar-system&gt;&lt;plasma-physics&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" />
  <row Id="13583" PostTypeId="2" ParentId="13579" CreationDate="2011-08-15T14:08:30.310" Score="5" ViewCount="" Body="&lt;p&gt;It is due to non zero internal resistances being different. If you make a short circuit, the current won't be infinity but of a finite value. It is the internal resistance who limits high currents.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-08-15T15:31:06.400" LastActivityDate="2011-08-15T15:31:06.400" />
  <row Id="13584" PostTypeId="2" ParentId="13579" CreationDate="2011-08-15T14:36:50.697" Score="1" ViewCount="" Body="&lt;p&gt;You have to be careful trying to apply idealised circuit theory to real world problems.  $V=IR$ is all well and good for theory but practice is a bit different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially I think the manufacturers are stating different capabilities of each battery to provide the required current when the vehicle is starting (If I have interpreted 'cold crank' correctly).  The actual chemistry/engineering inside the battery is more complicated than just Ohm's law and hence two batteries may have different abilities to 'instantly' provide the required current.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-08-15T14:36:50.697" />
  <row Id="13585" PostTypeId="2" ParentId="13582" CreationDate="2011-08-15T14:40:05.840" Score="9" ViewCount="" Body="&lt;p&gt;Solar wind is neutral overall else the Sun will become globally very strongly charged and we don't see that happening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It comprises Electrons/Protons and other particles.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-08-15T14:40:05.840" CommentCount="1" />
  <row Id="13586" PostTypeId="2" ParentId="13576" CreationDate="2011-08-15T14:47:01.417" Score="4" ViewCount="" Body="&lt;p&gt;The frictional force (F) holding the nail in place is what both the hammer and the large mass must overcome to move the nail. To get the nail to move you need a (Force = mass * acceleration) of the object hitting the nail greater than the (Force) holding the nail in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a large mass just resting on the nail, you are stuck with a constant acceleration gravity, so you will need a larger mass. With a hammer, you can achieve a higher acceleration than gravity, so your mass requirements are not as much. &lt;/p&gt;&#xA;" OwnerUserId="4933" LastActivityDate="2011-08-15T14:47:01.417" CommentCount="1" />
  <row Id="13587" PostTypeId="2" ParentId="13581" CreationDate="2011-08-15T14:55:02.417" Score="8" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the evidence that large nuclei have nucleons as substructures?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Periodic_table&quot; rel=&quot;nofollow&quot;&gt;periodic table&lt;/a&gt; of elements? This shows an organized internal substructure into nucleons, i.e. triplets of quarks in  SU(2) representations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atomic masses in the table are within a percent multiples of the nucleon mass (the binding energy takes up the slack) . A quark gluon plasma would not show this approximate quantization in multiples of triplets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quark gluon plasma would not be constrained this way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are experiments  looking for the signatures of a quark gluon plasma. &lt;a href=&quot;http://www.osti.gov/bridge/servlets/purl/771552-tMDu7T/webviewable/771552.pdf&quot; rel=&quot;nofollow&quot;&gt;Here is a review&lt;/a&gt; on such signatures.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-08-17T03:43:06.923" LastActivityDate="2011-08-17T03:43:06.923" CommentCount="5" />
  <row Id="13588" PostTypeId="5" CreationDate="2011-08-15T14:59:16.073" Score="0" ViewCount="" Body="&lt;h2&gt;Notes on Safety&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Some of our audience might be children, so make it clear what projects may call for adult supervision.&lt;/li&gt;&#xA;&lt;li&gt;It is best to list even the obvious dangers and point the reader towards ways to avoid or mitigate them.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-15T14:59:16.073" LastActivityDate="2011-08-15T14:59:16.073" />
  <row Id="13589" PostTypeId="4" CreationDate="2011-08-15T14:59:16.073" Score="0" ViewCount="" Body="questions concerning measurements, experiments, and activities that demonstrate physics and are safe to do at home, require at most one tool or ingredient not readily available, and can be interpreted or analyzed on one sheet of paper with accessible math." OwnerUserId="-1" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-15T14:59:16.073" LastActivityDate="2011-08-15T14:59:16.073" />
  <row Id="13590" PostTypeId="2" ParentId="13581" CreationDate="2011-08-15T15:03:47.657" Score="0" ViewCount="" Body="&lt;p&gt;In the same way that we know that water vapour is composed of tiny (liquid) water droplets. There is an associated phase potential expressed as surface tensions that determine the boundary between different states of matter condensation. There is no further distinction above this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If any, you can understand individual nucleons as an adiabatic approximation valid when the interaction energies and times are below the scales where asymptotic freedom dominates over confinement (which is the landmark that defines a true quark-gluon plasma). In other words, low energy reactions will have dominant nuclear reactions that mainly produce stable subproducts because they have time to thermalize before leaving the nucleus. But those most stable subproducts are themselves the nucleons (protons, neutrons, and aggregations of those, i.e:alpha). In that sense, we can tell that &quot;there are nucleons&quot; but only because they are particularly stable forms that will be produced in most but the most energetic reactions where such adiabatic approximation would be no longer valid&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-08-15T15:03:47.657" />
  <row Id="13591" PostTypeId="2" ParentId="13582" CreationDate="2011-08-15T15:15:04.587" Score="2" ViewCount="" Body="&lt;p&gt;Generally plasma may be charged in regions smaller than the &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Debye_length&quot; rel=&quot;nofollow&quot;&gt;Debye length&lt;/a&gt;&lt;/strong&gt;. Normally it happens at boundaries. For example, fast electrons tend to fly away and are deposited on a metallic wall first. (It is like electron emission from a hot cathode.) Thus a positive charge is created within some region. The Debye length depends on plasma density and temperature.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-08-15T15:30:18.010" LastActivityDate="2011-08-15T15:30:18.010" CommentCount="2" />
  <row Id="13592" PostTypeId="2" ParentId="13581" CreationDate="2011-08-15T15:45:49.943" Score="7" ViewCount="" Body="&lt;p&gt;Because the interaction energies inside the nucleus are very low, and when we perform experiments on nucleons at those energies they behave primarily as discrete particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a particular instance, take quasi-elastic nucleon ejection reactions like $A(e,e&amp;#39;p)$ (that is an electron scattering off a nucleus and neatly knocking out one proton without otherwise exciting or shattering the remnant). The figure of merit for such reactions is $Q^2 = -q \cdot q$, and in the range from about $1$--$8\text{+ GeV}^2$ (the upper limit is not yet experimentally known) the cross-section for this (per target proton) is a &lt;em&gt;constant&lt;/em&gt; times the cross-section for the free reaction $p(e,e&amp;#39;p)$. The constant is called the &quot;nuclear transparency&quot; and expresses the probability that the scattered proton re-interacts before leaving the nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See this figure from my dissertation work&lt;img src=&quot;http://i.stack.imgur.com/yR98p.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nuclei (shown by different marker shapes and colors) are deuterium, carbon, iron and gold, and the different marker fill-states and sizes represent three experiments. Error bars are shown for all data and represent statistics plus point-to-point systematics (but not  not systematics that would shift all the data for a single nucleus). The solid lines are a modified Glauber calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A big part of what they do at Jefferson laboratory is to search for the change between nucleon--meson degrees of freedom (as reasonably well described by quantum hadrondynamics) and quark--gluon degrees of freedom (as described by &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_chromodynamics&quot; rel=&quot;nofollow&quot;&gt;quantum chromodynamics&lt;/a&gt;); but even the on-set of QCD dominated behavior does not correspond to quark--gluon plasma which occurs at much higher energies.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-19T15:46:59.573" LastActivityDate="2011-08-19T15:46:59.573" CommentCount="2" />
  <row Id="13593" PostTypeId="2" ParentId="13576" CreationDate="2011-08-15T16:07:29.240" Score="3" ViewCount="" Body="&lt;p&gt;The equation of just $F=ma$ is lacking the amount of information needed to sufficiently answer this question, so I'll take a shot at this.  You'll find most of what you need with a tour around Wikipedia, but I'll try to give some guidance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let me be sure to mention several quantities.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Energy ($E=\frac{1}{2} m v^2$)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Impulse_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;Impulse&lt;/a&gt; ($I=m v$)&lt;/li&gt;&#xA;&lt;li&gt;Force ($\frac{dp}{dt}=m \frac{dv}{dt}$)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The hammer head that falls on the nail has all of these quantities.  A physics 101 class should teach you how to fluently exercise the algebra to go back and forth between all of these.  Impulse is synonymous with momentum, and impulse and energy are the comparatively easy values to find (the low hanging fruit) in the case of a household hammer.  The reason is that the velocity of the hammer as it hits the nail isn't particularly difficult and the mass of the hammer head is trivial to assess.  As I was saying, the hammer contains some energy and impulse, which result from the mass and velocity - the balance between those two is relevant to the performance of the hammer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The case of a large mass resting on the nail is a limit case where there is no energy exchanged (unless it pushes the nail) and high impulse&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some simple in-your-head-physics, think of a hammer head that falls without a human pushing it.  Energy is $m g h$, where $m$ is the mass, $g$ is the gravity constant, and $h$ is the height it falls from.  Impulse is the momentum on contact and could be said to be $m g \Delta t$.  In both cases $m g$ is the force of gravity, but energy cares how far it falls and impulse cares how long it falls.  In the case of a large mass resting on the nail, gravity continues to impart force on the mass which is continuously resisted by the friction that prevents the nail from going in.  This is the friction we wish to overcome.  For a more universal picture, think of energy as $F \Delta x$ and impulse as $F \Delta t$, and in our case, $F$ needs to surpass some given threshold.  I should add that $\Delta t$ is a direct function of $h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanics of the friction can be approximated by the coefficient of friction.  The nail is partially in a hole and the wood snugly squeezes on the nail, giving a normal force, so the force the hammer needs to reach is the coefficient of friction times the normal force, $\mu F_{normal}$, which is just some value as far as we are concerned.  If I need to move the nail $1 mm$, then &lt;em&gt;a given energy is required&lt;/em&gt; because energy is force times distance.  However, even if I have enough energy to move it some distance, it might not move because the value of the force never gets high enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get to a force value on a physics 101 level we would use &lt;a href=&quot;http://en.wikipedia.org/wiki/Hooke%27s_law&quot; rel=&quot;nofollow&quot;&gt;Hooke's Law&lt;/a&gt;, because it gives formulas &lt;em&gt;for how the force is distributed over time&lt;/em&gt;.  If the nail fails to move you may say it is because the nail softens the blow by its inherent spring-like qualities.  By the energy we can predict how far an idealized spring will move by $\frac{1}{2} m v^2=\frac{1}{2} k x^2$, and then the maximum force magnitude will be $k x$.  This would be fairly valid equations &lt;em&gt;should the nail not move&lt;/em&gt; because if it does move we default to the previous equations using coefficient of friction.  For the ideal spring, the motion over time will be some constant times $sin(\sqrt{\frac{k}{m}}t)$, from 0 to $\pi \sqrt{\frac{m}{k}}$, which allows to finally apply the impulse concept.  The impulse will be equal to the integral of the force over the time it is applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not going to solve the full problem, but let's look at the variables that go into it all.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The mass of the hammer head&lt;/li&gt;&#xA;&lt;li&gt;The material stiffness of the nail ($k$)&lt;/li&gt;&#xA;&lt;li&gt;The height it falls from&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These pretty much sum it up.  The combination of $k$ and $m$ determine the time over which the impulse from the hammer is distributed, and should the hammer breakthrough the static friction threshold, the energy will limit how far the hammerhead may push the nail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Given all this, I can say that we require sufficient stiffness of the spring-like system as well as sufficient impulse from the hammerhead, and we also need sufficient energy if we don't want to be dinging the nail for really small movements all day.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are plenty of ways you can come up for a way for this to not work.  Put silly putting on the head of the hammer and you don't have sufficient stiffness x impulse because of poor stiffness.  Also, if you don't &quot;throw&quot; the hammer at the nail, you distribute the time over which the impulse is imparted, so it doesn't work in that case either.  In any case, you need a sufficient height or else you won't have sufficient values to move it like you want it.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-08-15T16:07:29.240" />
  <row Id="13594" PostTypeId="1" CreationDate="2011-08-15T17:21:14.943" Score="-4" ViewCount="82" Body="&lt;p&gt;If I was able to go back in time via Machine or whatever, would traveling back in time not create a paradox? &#xA;I was not there, so how could I exist where I did not exist in the passed?&#xA; i would have to assume that we could never go backwards in time in any way shape or form.  &lt;/p&gt;&#xA;" OwnerUserId="4708" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-15T17:33:26.917" LastActivityDate="2011-08-15T21:41:28.990" Title="how could I exist where I did not exist in the past?" Tags="&lt;time-travel&gt;&lt;paradox&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="13595" PostTypeId="1" AcceptedAnswerId="13609" CreationDate="2011-08-15T17:56:49.050" Score="3" ViewCount="69" Body="&lt;p&gt;If there is nothing in the microwave, where does the excess radiation go?&#xA;Why doesn't the radiation accumulate and blow it up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I cook two pieces of Canadian Bacon twice as long as I cook one?&lt;/p&gt;&#xA;" OwnerUserId="4935" LastActivityDate="2011-08-15T23:20:33.240" Title="Where does the excess energy emitted by a microwave go?" Tags="&lt;radiation&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" />
  <row Id="13596" PostTypeId="2" ParentId="13595" CreationDate="2011-08-15T18:17:27.433" Score="1" ViewCount="" Body="&lt;p&gt;The microwave would blow up, eventually. If you read the user manual, some warning will placed about not operating the microwave when it's empty. Being a vegetarian, I have no idea about the bacon, but I can guess that it won't behave in a different way than any other material placed in the microwave. In small amount (single or double dish (?)) the energy absorbency is roughly linear and the heating time should be doubled.&lt;/p&gt;&#xA;" OwnerUserId="3677" LastActivityDate="2011-08-15T18:17:27.433" />
  <row Id="13597" PostTypeId="1" CreationDate="2011-08-15T18:35:18.370" Score="7" ViewCount="174" Body="&lt;p&gt;Physically, we know that a BEC has formed if a macroscopic number of bosons occupy a single quantum state. The wave-function $\Psi(x)$  of the latter, normalized to the total number of condensed atoms $N \gg 1$, gives the macroscopic description of the condensate. $\Psi(x)$ will satisfy the ususal Schrodinger euqation in the trapping potential (which is the special case of the Gross-Pitaevskii equation with zero interactions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other had, a macroscopic limit of a boson field should be described by the (classical) Klein-Gordon equation. In the relevant non-realtivistic limit, dispersion of the Klein-Gordon field is quadratic, as in the Schrodinger equation, but I struggle to derive the former from the latter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite this technical difficulty of mine, is it a valid claim that BEC is a physical realization of the non-relativistic, classical-wave-limit of a boson field? I see it in the same vein as electric-field description of the laser radiation being a classic limit of the quantum coherent state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The difference between number states and coherents states in the limit $N \gg 1$ is, hopefully, a mere technicality here). &lt;/p&gt;&#xA;" OwnerUserId="1739" LastActivityDate="2011-08-22T07:03:12.497" Title="Is Bose-Einstein condensate a good example of a classical massive boson field?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;field-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="13598" PostTypeId="2" ParentId="13581" CreationDate="2011-08-15T18:39:45.197" Score="1" ViewCount="" Body="&lt;p&gt;Scattering experiments are what tell us about substructure in atomic and nuclear physics.  The higher the energy of the probe (usually a charged particle), the finer the resolution and thus the more substructure you can see.  In the case of nuclei this process revealed the nucleons as  substructure.  In the case of the nucleons themselves, this revealed the existence of 3 'partons' and a background 'sea'.  We now know these partons as quarks and gluons but whatever you call them, we saw them first as point like features in the distributions we pulled from early scattering experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experimental evidence for substructure usually comes from scattering.&lt;/p&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2011-08-15T18:39:45.197" />
  <row Id="13599" PostTypeId="1" CreationDate="2011-08-15T19:50:24.877" Score="5" ViewCount="79" Body="&lt;p&gt;I'm trying to understand how can multiple radio stations transmit information just by transmitting using different frequency. The way I understand it all those different frequency waves add up to a single wave? If so how can that not corrupt information on other frequencies?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say if I transmit:&#xA;$\sin(2\pi x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And separately:&#xA;$\sin(2\pi x\times 2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it end up as a single wave of:&#xA;$\sin(2\pi x)+\sin(2\pi x\times 2)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or does it work in a different way completely?&lt;/p&gt;&#xA;" OwnerUserId="4937" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-15T19:56:01.800" LastActivityDate="2011-08-16T17:54:28.377" Title="Multiple channels of information in single electromagnetic wave?" Tags="&lt;electromagnetism&gt;&lt;waves&gt;&lt;radio&gt;&lt;signal-processing&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13600" PostTypeId="2" ParentId="13599" CreationDate="2011-08-15T20:06:12.267" Score="9" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Say if I transmit: $\sin(2\pi x)$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;And separately: $\sin(2\pi x\times 2)$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Does it end up as a single wave of: $\sin(2\pi x)+\sin(2\pi x\times 2)$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, that's exactly how it works. This is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Superposition_principle&quot; rel=&quot;nofollow&quot;&gt;superposition&lt;/a&gt;. There are electromagnetic waves at hundreds of different frequencies all filling the air simultaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way something like a radio is able to pick out the station it wants is by using a &lt;a href=&quot;http://en.wikipedia.org/wiki/Resonance&quot; rel=&quot;nofollow&quot;&gt;resonant&lt;/a&gt; circuit (also known as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Band-pass_filter&quot; rel=&quot;nofollow&quot;&gt;band-pass filter&lt;/a&gt;), which damps out all frequencies of EM waves except within a small range.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-08-15T20:06:12.267" CommentCount="1" />
  <row Id="13601" PostTypeId="2" ParentId="13579" CreationDate="2011-08-15T20:47:04.807" Score="5" ViewCount="" Body="&lt;p&gt;[1] If the resistance, my truck, is constant for both batteries, how can the amperage ratings differ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the total resistance includes the &quot;equivalent internal resistance&quot; of the battery.  All practical (realizable) power sources have a non-zero internal resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] What is does a battery's 'cold crank amperage' really mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Cold Cranking Amps&quot; means the amperage that the battery can deliver at a given cold temperature.  It is specified this way because car motors are harder to start at cold temperature (they require more amperage), and lead acid batteries produce less current at cold temperatures. (This is because they are using a chemical reaction to produce current.)  Thus you have a combination of the two worst cases at cold temperatures.  If a battery will start a car in the middle of winter in South Dakota, you can be absolutely sure it will do great in summer in Arizona.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, a lead-acid battery at cold temperatures APPEARS, at the terminals, to have a higher internal resistance.  But it is not a higher resistance that is holding back the current flow: it is the inability of the battery to produce current at the demanded rate, because of the slower chemical reaction at lower temperatures.  Manufacturers compensate by designing batteries with more reaction surface area, etc.  The measure of how well they did this is reflected in the CCA rating.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastEditorUserId="3281" LastEditorDisplayName="" LastEditDate="2011-08-15T20:52:33.143" LastActivityDate="2011-08-15T20:52:33.143" />
  <row Id="13602" PostTypeId="2" ParentId="13594" CreationDate="2011-08-15T20:57:38.840" Score="2" ViewCount="" Body="&lt;p&gt;This question is directly about the apparent paradoxes that are present in what I will refer to &quot;Back to the Future&quot; time travel.  This is the time travel depicted in several major movies where a time machine is created and someone goes back into the past and changes something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A simple solution to time travel paradoxes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might be better handled by a philosophy perspective, but since it's being asked here, I'll start with a slightly less relevant of a physical example where something can be taken to be traveling backward in time.  Hawking radiation involves a particle radiating away from an event horizon after being spontaneously created from quantum fluctuations with an antiparticle that falls into a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/5O7iZ.jpg&quot; alt=&quot;Hawking radiation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the antiparticles can be interpreted as the particle itself traveling back in time.  On can not escape the event horizon of a black hole, but it can if it travels backward in time.  The &quot;lame&quot; thing about this situation is that time travel happens without any conflicting world-lines.  There is a singular story of what happened, and no multiple universes are required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hollywood time travel often does involve conflicting (and often hopelessly confused) world-lines.  This generally comes in the form of someone from the future affecting the series of events that leads to the creation of a time machine.  Developing a plot should depend on several philosophic physical questions like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can a given outcome be guaranteed, or does the world exhibit determinism / fatalism?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One could potentially take the above example of particles and antiparticles and replace it with an intrepid time traveler.  In that case, the time traveler is moving forward in time, until at one point in spacetime he goes into reverse and begins traveling backward in time.  This would be observed as the existence of 2 versions of the time traveler colliding and then disappearing.  Then if he reverses back into the forward direction, then at some other point in spacetime, before his &quot;first&quot; reversal (first, perceived in his time), there is an apparently spontaneous creation of 2 versions of him, one of them going backward in time and one of them going forward in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no way to create time paradoxes because the versions that have time traveled &lt;em&gt;already exist by the time the time machine is used&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Entropy arguments&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another way to time travel goes back to the definition of entropy and time itself.  The laws of physics are (to the extent of my own knowledge) fully reversible, and the reason time progress in one direct was due to the low entropy at the big bang, which allows us to have a memory of the past and not the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could just create something that has a decreasing entropy.  There, finished.  What does this mean?  This means that you create an item that has a decreasing number of degrees of freedom.  For instance, a gas that starts with chaotic-seeming motion of particles that are really precisely engineered to converge on a single point in the middle.  The event of that gas contracting is identical to saying that the gas is expanding backwards in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, you should find Hollywood disappointed with the proposal because no privileged information is gained.  No one can use this to win the lottery, and the systems required to decrease entropy of a chunk of matter are drastically more elaborate than the system that is created.  On the plus side, we are getting closer to the point when we can technologically do this.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-08-15T20:57:38.840" />
  <row Id="13603" PostTypeId="2" ParentId="12283" CreationDate="2011-08-15T20:57:59.167" Score="1" ViewCount="" Body="&lt;p&gt;I'm thinking (assuming the effect is real), that it might be that magnetic effects have  lowered the effective spring constant. The total energy in the magnetic field will be different with a long spring than with a short one, because the magnetic permittivity of the spring metal is much higher than air/vacuum. So if one calculates total system potential energy, magnetic, plus internal energy of the spring metal, then one could calculate the mode frequencies (assuming they are long compared to establishing an equilibrium field for a change in spring geometry). The only problem I'm having is I think a longer spring probably means more magnetic field energy, i.e. maybe we would expect the frequency to increase, not decrease.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-08-15T20:57:59.167" />
  <row Id="13604" PostTypeId="2" ParentId="13594" CreationDate="2011-08-15T21:26:24.353" Score="0" ViewCount="" Body="&lt;p&gt;The universe lives in a continuum evolving &lt;em&gt;present&lt;/em&gt; . The &lt;em&gt;past&lt;/em&gt; do not exist, neither the &lt;em&gt;future&lt;/em&gt;; there are no such places.  The equations are symmetric irt time and they only told what &lt;em&gt;will be&lt;/em&gt; the next state in face of the actual snapshot and similarly we can derive what &lt;em&gt;was&lt;/em&gt; the previous state, eg. the Earth will progress to the next position in space and not to the position it occupied a moment ago.&lt;br&gt;&#xA;I'm looking to the night sky and I'm receiving energy (photons) released billions of years ago as measured by our local clock or by the &lt;em&gt;cosmological clock&lt;/em&gt; used to describe the evolution of the universe as a whole by an independent observer (not attached to the matter properties).&lt;br&gt;&#xA;If in the next moment, say  $t_{+1}$, I intended to be in &lt;em&gt;my past&lt;/em&gt; $t_{-1}$ I will have to deconstruct all past events that already happened in the whole universe between  $t_{-1}$ and  $t_{0}$. All photons that are hitting my eyes now must be sent back into space to some other evolving position. All other people normally evolving will have to be deprived of that light. Can we &lt;em&gt;erase&lt;/em&gt; stars from the sky? No. Can we substitute one photons by others? No. Now lets admit that you &lt;em&gt;arrived&lt;/em&gt; to your desired past. What photons are you seeing? In what position in space is the Earth? Was it &lt;em&gt;moved&lt;/em&gt; at your will to any other place except that one that physical laws will determine? With all its present inhabitants ? With all past and dead inhabitants, those that are now flowing in the rivers or buried down in the Earth (possibly some atoms of them are now integrated in my body) ? How much energy will you need to recall all that matter from its present locality to your desired position? &#xA;Will you go and at the same moment stay here? Will you have two consciences? &#xA;Some atoms in your body could belong to someone else in the past. To &lt;em&gt;return&lt;/em&gt; there will you destroy him? or will you reassemble some other matter locally available to reconstruct your body? What happens to your present body?&lt;br&gt;&#xA;All matter radiates gravito/electromagnetic energy into space always. How do you reassemble back that energy into the particles to have the world of the past? There is no way!&lt;br&gt;&#xA;In the universe there is a &lt;em&gt;currency&lt;/em&gt;, called energy that has to be paid to do any action. All the matter/energy in this universe is allocated, engaged in &lt;em&gt;this&lt;/em&gt; trip. If we want to construct an alternative reality we have to assume &lt;em&gt;other&lt;/em&gt; simultaneous universe like a shifted copy modified, of this one. The object of physics is only the universe that we observe. Some theoretical instances wrote about other realities, mental constructs, desires, much like many others constructed other places like Heaven and Hell. When some believe in other universes we have two situations: if they interact or leave some trace in this one then we are only talking about this one and if they do not interact then they are out of the domain of physics.&lt;br&gt;&#xA;Using Einstein SR equations we describe the viewpoint of a moving body and get prisoner of our local clocks and we think about closed time-like curves. &#xA;If we adopt another viewpoint, above all moving bodies and with &lt;em&gt;instant vision&lt;/em&gt; to acknowledge the events,not the kind of the Einstein referential, we will have a broader view and see that there is no chance to &lt;em&gt;travel&lt;/em&gt; to the past.&lt;br&gt;&#xA;The energy argument and the choice of a more proper referential should be enough to put an end to the dream of time travel. But because we are humans it is nice to have dreams.  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-08-15T21:41:28.990" LastActivityDate="2011-08-15T21:41:28.990" CommentCount="2" />
  <row Id="13605" PostTypeId="2" ParentId="13576" CreationDate="2011-08-15T21:45:10.623" Score="1" ViewCount="" Body="&lt;p&gt;The key things to remember are:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) F = ma&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) a = dv/dt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a 100 kg man standing on the nail: F = (100kg)(9.8 m/s^2) = 980 N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a 1/2 kg hammer head, swung at 10 m/s: F = (0.5 kg)(a) = ??? N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;a&quot; in this last equation is the DEceleration of the hammer head when it hits the nail.  Let's say that the hammer drives the nail 2mm ---&gt; x = 0.002m with each stroke, and further assume that the deceleration of the hammer head is constant (makes the math easier).  Then you get the quadratic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(t^2) - (20/a)t + 4/(1000a) = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting a=10(m/s)/t into the equation t=SQRT(2x/a), we get t = 0.0004 seconds (0.4 mS).  If we use that t in the quadratic we find that a = 19060(m/s^2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So F = (0.5 kg) (19060 m/s^2) = 9530 N ---&gt; Roughly 10 times the force of standing on the nail.&lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-08-15T21:45:10.623" CommentCount="1" />
  <row Id="13606" PostTypeId="1" AcceptedAnswerId="13610" CreationDate="2011-08-15T21:52:50.513" Score="4" ViewCount="76" Body="&lt;p&gt;Intuitively, if energy can be stored in rotational motion, it has to obey $E=mc^2$.&#xA;Does rotation of typical stellar-sized objects - BHs, pulsars, binaries - have measurable effect on their overall gravity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm not talking about nearby effects, like frame dragging described by Kerr metric, the effect on measurements from the Earth is of interest)&lt;/p&gt;&#xA;" OwnerUserId="4760" LastActivityDate="2011-08-16T01:04:29.927" Title="Does rotational energy have effect on gravity/metric?" Tags="&lt;general-relativity&gt;&lt;gravity&gt;&lt;energy&gt;&lt;rotational-dynamics&gt;" AnswerCount="1" />
  <row Id="13607" PostTypeId="2" ParentId="13242" CreationDate="2011-08-15T22:04:04.057" Score="0" ViewCount="" Body="&lt;p&gt;Your answer will depend greatly upon the size of the holes you drill.  If I remember correctly, fluid flow through an orifice is proportional to the fourth power of the orifice diameter.  So, for a given force, you would get twice as much fluid out of a single hole of diameter 1 mm as you would out of eight holes, each of 1/2 mm diameter. &lt;/p&gt;&#xA;" OwnerUserId="3281" LastActivityDate="2011-08-15T22:04:04.057" CommentCount="1" />
  <row Id="13608" PostTypeId="2" ParentId="13571" CreationDate="2011-08-15T22:17:20.380" Score="14" ViewCount="" Body="&lt;p&gt;Let me first list all of the possibilities I considered that I later rejected.  This is &lt;em&gt;far&lt;/em&gt; from exhaustive, and I'm looking forward to seeing other people's creativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bad Ideas&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sit on a tire swing with the fan pointing to the side.  Point the fan up, measure speed of rotation of the system on the tire swing.&lt;/li&gt;&#xA;&lt;li&gt;Get a laser or collimated flashlight.  Point it into the fan at a single point.  On a wall on the other side of the fan, you will have a dot of light that is flashing, measuring the rate of the flashing could be an easier problem.&lt;/li&gt;&#xA;&lt;li&gt;Attach springs to the tip of all the blades.  Observe how far out the acceleration deforms them, apply Hooke's Law to find acceleration, convert to angular speed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, I am almost sure that the experimenter doesn't have the tools to execute any of these methods very well.  Not even one.  I'll try to break this down as to why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, do we know the moment of inertia of the fan?  No.  Do we know the moment of inertia of the tire swing with a person and a fan on it?  No.  Is it even constant?  No.  I'm not saying we can't figure these things out, but it's an absurdly inferior method that will get terrible data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On to the laser method.  How are we going to measure the flash rate of the laser?  I thought endlessly about this problem.  Generally, a reference would be good, or if you could use electronics you could nail down the speed almost exactly and very easily.  But I don't think anything is available that will work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the spring idea.. where to begin?  The measurement of the deformation length is error-prone.  The weight of the springs themselves will affect the speed of the fan.  What spring do you have with appropriate characteristics anyway?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My best proposal&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm hoping you can take off the cover.  If you can't take off the cover, I hope you can take off &lt;em&gt;some&lt;/em&gt; part of it, so that you can get a protruding shaft.  I think the best way to make this measurement with household stuff would be to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Get a part of the shaft isolated&lt;/li&gt;&#xA;&lt;li&gt;Tape or attach the end of a thin string to it&lt;/li&gt;&#xA;&lt;li&gt;Turn the fan on with a stopwatch&lt;/li&gt;&#xA;&lt;li&gt;Time how long it takes for the fan to completely wrap the entire string around the shaft with a stopwatch&lt;/li&gt;&#xA;&lt;li&gt;Count the number of times the string has wrapped around the shaft&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There you go, you have a number of turns per some amount of time.  Ideally you would use a very light string that offered little to no resistance when pulled, as in, have it loosely laid out on the floor.  Fishing wire could possibly be very good.  You will want the acceleration time to be small compared to the entire measured time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some other (not terrible) possibilities&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It occurred to me that acoustic methods might have some merit here.  Get something that the blades can smack against, like when you take a pencil and stick it in the fan.  Open up Windows sound recorder (accessories -&gt; entertainment -&gt; sound recorder), or a program like Audacity.  Use some sound editing program and zoom in really tightly on the sound.  See if you can identify a periodic shape that corresponds to a single hit.  Count the peaks over a given time frame.  Once again, you have number of rotations (or 1/3rd rotations) per unit time.  If you already have an educated guess as to the frequency, then identifying the acoustic pattern from individual hits might not be very bad, not to mention, there is a lot of design flexibility in this experiment and computers should have a sufficiently high sample rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the ideal would be some kind of visual timing mechanism like the OP suggests.  I'd imagine that a mechanical reference could be of use.  Like if you had another fan that you knew the speed of, you could place it in front of the unknown fan and adjust its speed until you saw some patterns that indicated they were in sync.  Yes, I'm lacking a lot of what's required to do this effectively, but maybe someone else can offer better advice.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The Experiment&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Half of the papers on my desk are blown away.  I'm getting complaints about the wretched sound of pen on fanblades, and people in my office are not too happy with me right now.  But this is all in the name of physics!  I am editing to present my experimental results.  I used the acoustic method to determine the speed of my fan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, my experimental apparatus is the Galaxy 20 inch model 4733 fan.  It has 5 blades.  I can't find any shopping results for you, but maybe someone else can.  Here is &lt;a href=&quot;http://www.youtube.com/watch?v=jr9xIFGLl_8&quot; rel=&quot;nofollow&quot;&gt;a pretty good quality demo of the Galaxy 20&quot; fan&lt;/a&gt; on youtube.  And &lt;a href=&quot;http://www.youtube.com/watch?v=ZcHZsGX7Xbc&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt; specifically states they have the 4733 model that I'm using.  Why do people upload youtube videos of these things?!  Do you have to &quot;unbox&quot; every single thing you buy??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, moving on.  I'm using &lt;a href=&quot;http://audacity.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;the Audicity program&lt;/a&gt; and the microphone from a Microsoft Lifechat headset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fan has 3 settings, plus 0 for off.  Setting 1 is the slowest and setting 3 is the highest.  It produces quite a good breeze and has served me well.  To start off, I'll share a waveform I recorded with it on setting 3 and setting 1 with me doing nothing else to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/n0KlQ.png&quot; alt=&quot;Waveform for 1 and 3 settings with no pen&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, this is not too useful.  It makes a sound, but there's no way to distinguish peaks.  Maybe it has a frequency that reflects the speed of the fan, but I can't be sure (and I haven't had much luck with the spectrum visualizations).  You can see how the sound it makes is different between the two, and the 3rd setting is obviously louder, and the frequency is obviously different, but we want actual numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I put a plastic pen in it (the butt of the pen).  Now, you might not want to try this at home (like I just did), but I kind of had to play with the angle to get it to not miss blades.  It's very easy for it to jump and miss one, which would mess up the count.  I had to press kind of hard and it was rather loud.  But I got results.  Here are the waveforms for 0.5 seconds, and my markup in order to count the &quot;hits&quot;.  I also provide the actual count in the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/VXZj8.png&quot; alt=&quot;Waveform with markup&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can check my work for the count itself.  I'm also happy to upload some mp3 files, but I'm not sure where I'll host them right now.  The above image was made with the high-tech research software MS Paint.  I'll give answer denoted $rpm_i$, where $i$ is the number of the setting, and the number of hits above will be denoted $hits$.  I'll take the error in each hit count to be $\pm 1$ hit.  The formula and reported results are as follows.  Remember, it has 5 blades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$rpm_i = \left( hits_i \pm 1 \right) \times \frac{turn}{5 hits} \times \frac{1}{0.5 s} \times \frac{60 s}{min}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$rpm_1 = 456 \pm 24 rpm$$&#xA;$$rpm_2 = 624 \pm 24 rpm$$&#xA;$$rpm_3 = 864 \pm 24 rpm$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Power Consumption (addendum)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used my KILL A WATT device to record power consumption for all the different speed settings.  I'll denote this with $P_i$ but I need to explain a little about the difficulty in making this measurement.  I believe my KILL A WATT to be fairly reliable and it gives stable power measurements for devices with constant power consumption.  This is not quite true for the fan.  When I first turn on the fan it consumes more power than after I leave it running for some time.  The largest swing I observed was $50.5 W$ at max and $45.3 W$ at minimum for setting 1.  This gives you another possible source of error.  Since the experiment was performed with the fan on for a good while I'll report the lowest readings I have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_0 = 0 W$$&#xA;$$P_1 = 45.3 W$$&#xA;$$P_2 = 65.1 W$$&#xA;$$P_3 = 97.1 W$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I want to take just one quick second to apply the physics concepts of friction here.  Dynamic friction between two solid bodies is often taken as a constant, and fluid friction is often taken as a power law, as in, $v^n$ where $n$ is most commonly from 1 (fully laminar) to 2.  We can apply that here!  I converted the previous speeds to rad/s and plotted the speed versus power consumption for all 3 &quot;on&quot; settings.  Then, I guessed a certain offset and subtracted this value from the power consumption and applied a power fit to what was left.  I realized after I did this that a constant power consumption does not correlate to a constant force (that would be linear), so I'm really just assuming some base power consumption for the device, and this yields a more perfect power fit for what's left just due to the mechanics of the motor.  I found that I could get a better power fit by making this subtraction.  The power fit had $R^2=1.0000$ for a constant offset from $8 W$ all the way to $13 W$ so I took the middle ground of $10.5 W$, which is to say, I made an educated guess that a constant power loss accounts for $10.5 W$ out of the consumed power. The power fit follows a satisfying $1.5$ power law, which is about what I expect for fluid friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/NuF6i.png&quot; alt=&quot;Power consumption, broken down into static and fluid friction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_i = 10.5 W + 0.1343 \omega_i^{1.4374} W$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, I want to report the general intensity of the sound in my mp3 files.  I need to put a disclaimer with this that it might not be accurate in any physical sense.  I would want to ask an audio engineer about this issue - I don't know if the dB of an audio file represents the physical dB of the sound at the point the measurement was made.  My guess is that this will depend on the recording device.  Anyway, I want to give a dB measurement (I'll denote $A$) for the average peak for the pen hits on the fan blade, as per the audio file.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_1 = -8 dB$$&#xA;$$A_2 = 0 dB$$&#xA;$$A_3 = -3 dB$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did I have the microphone in different locations when I took these measurements?  Yes, I did.  If I had to guess, I would say that I had it about 3-4 inches away from the pen contact point when I did setting 1 &amp;amp; 2 and closer to 7-8 inches when I did the 3rd setting.  Obviously the 3rd setting was louder to my ear, and I had to set the headset down when taking that measurement because holding both was getting difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I offer this data because with my guesstimates you &lt;em&gt;could&lt;/em&gt; potentially calculate the energy released in the sound wave on a hit (assuming the dB measure is a 'real' measurement).  Then with a conversion efficiency (from mechanical to sound), estimate the energy dissipated in a hit.  You could also take some generic values for fan motor efficiency to relate power consumption to friction forces.  You could then use a tailored mathematical form for power consumption (like above) for the friction losses and apply the energy dissipation rate from the pen hits and estimate the speed loss &lt;em&gt;due to the pen&lt;/em&gt;.  It's just something good to keep in mind for future experiments, so that you can show that the process of measuring isn't affecting what is being measured too much.  With my 20&quot; fan I don't think it matters too much, but repeating the experiment on a smaller fan could benefit from these calculations, and in order to do so you should have the microphone located in a fixed position for all measurements (unlike what I did).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Comments&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible that the pen contact was slowing down the fan some.  In fact, this is almost surely the case, but this is a rather large fan.  It is also an old fan.  I would expect these speeds to be less than someone with a newer one.  I've taken power measurements that could be used for some other investigation if desired.  One use would be to guesstimate the impact the application of the pen on the blades has on the fan speed.  I have already taken a shot at putting together a picture of where the friction comes from and developed a formula for power consumption as a function of speed based on a breakdown between static and fluid friction.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-08-16T13:51:40.943" LastActivityDate="2011-08-16T13:51:40.943" CommentCount="4" />
  <row Id="13609" PostTypeId="2" ParentId="13595" CreationDate="2011-08-15T23:20:33.240" Score="5" ViewCount="" Body="&lt;p&gt;The power going into an empty microwave will accumulate as an EM field inside the cavity up to the point where the input power equals the rate of leakage.  Energy inside the cavity will leak out through the little holes in the metal screen in the door, through ohmic losses in the sided of the microwave, feedback into the power source, etc.   The rate of power leakage from the cavity is given by $P_{leak} = 2\pi E f / Q$ where $E$ is the energy stored in the cavity, $f$ is the resonant frequency of the cavity and Q is the &quot;quality factor&quot;.  In equilibrium then, $E = QP_{input}/{2\pi f}$.  For an resonant frequency of around 2.5 GHz, input power of 1kW, Q of say 100 (guess), the accumulated energy is only about a milli-joule.  Of course, the materials making up the microwave probably aren't designed to dissipate 1kW of power over an extended time, so eventually the microwave may explode or something due to overheating of components.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-08-15T23:20:33.240" CommentCount="3" />
  <row Id="13610" PostTypeId="2" ParentId="13606" CreationDate="2011-08-16T01:04:29.927" Score="6" ViewCount="" Body="&lt;p&gt;The source in the Einstein field equations is the stress-energy tensor, not the scalar mass-energy. Adding rotation will affect multiple elements of the stress-energy tensor. You can sometimes get rough estimates of effects in GR by using $E=mc^2$ and pseudo-Newtonian arguments, but sometimes these are way off. As an example where it's way off, two light rays propagating in parallel (not antiparallel) directions experience zero gravitational interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the &lt;em&gt;distant&lt;/em&gt; field, I believe the answer to your conjecture is yes, in the sense that in any asymptotically flat spacetime, the distant field is Newtonian, and its strength is what you would expect based on the Bondi or ADM mass of the ingredients that went in.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-16T01:04:29.927" CommentCount="1" />
  <row Id="13611" PostTypeId="1" AcceptedAnswerId="13618" CreationDate="2011-08-16T01:31:08.587" Score="2" ViewCount="53" Body="&lt;p&gt;I am drawing a blank when it comes to equation transformation. Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck%27s_law#Various_forms_of_expression_of_Planck.27s_law&quot; rel=&quot;nofollow&quot;&gt;gives two equations&lt;/a&gt; for the spectral radiance of black body:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First as a function of frequency $\nu$: $$I(\nu, T) = \frac{2 h \nu^3}{c^2}\cdot\frac{1}{e^\frac{h \nu}{k T} - 1}$$&lt;/li&gt;&#xA;&lt;li&gt;Then as a function of wavelength $\lambda$ : $$I&amp;#39;(\lambda, T) = \frac{2hc^2}{\lambda^5}\cdot\frac{1}{e^\frac{h c}{\lambda k T}-1}$$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And I don't see how they get $\lambda^5$ term. I'm assuming that the transformation is just $\nu \rightarrow c/\lambda$, but that gives&#xA;$$&#xD;&#xA;\frac{2 h \nu^3}{c^2} \Rightarrow \frac{2 hc}{\lambda^3} \neq \frac{2hc^2}{\lambda^5}&#xD;&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar transformation happens at other parts in the article also. I'm obviously missing something, likely completely trivial.&lt;/p&gt;&#xA;" OwnerUserId="4939" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-16T04:39:42.763" LastActivityDate="2011-08-16T05:10:51.207" Title="Planck's Law in terms of wavelength" Tags="&lt;thermal-radiation&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13612" PostTypeId="2" ParentId="12973" CreationDate="2011-08-16T01:32:24.067" Score="2" ViewCount="" Body="&lt;p&gt;I asked this related question on mathoverflow: &lt;a href=&quot;http://mathoverflow.net/questions/72937/what-are-the-statistics-of-prime-knots-in-3d-random-walk&quot; rel=&quot;nofollow&quot;&gt;http://mathoverflow.net/questions/72937/what-are-the-statistics-of-prime-knots-in-3d-random-walk&lt;/a&gt;. The physics references provided in the answer, and the articles which cite them, are illuminating.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-16T01:32:24.067" />
  <row Id="13613" PostTypeId="2" ParentId="13230" CreationDate="2011-08-16T02:30:52.990" Score="1" ViewCount="" Body="&lt;p&gt;I don't think there's an easy answer to your question but as for some possible leads and maybe a setup of the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd try to write down the Hamiltonian for your configuration of particles.  (e.g chapter two in this fella's &lt;a href=&quot;http://www.google.com/url?sa=t&amp;amp;source=web&amp;amp;cd=3&amp;amp;ved=0CCsQFjAC&amp;amp;url=http://www.physics.byu.edu/faculty/vanhuele/Research/AdamThesis.pdf&amp;amp;rct=j&amp;amp;q=electromagnetic%20hamiltonian%20two%20charged%20particles&amp;amp;ei=X85JTpq7O4Lt0gH91YzrBw&amp;amp;usg=AFQjCNHEcE58I8_R8JCUBquXMK8kNAjKmw&amp;amp;sig2=ct7gqsgYA3bA23X1Pa8JNQ&amp;amp;cad=rja&quot; rel=&quot;nofollow&quot;&gt;thesis&lt;/a&gt; in section 2.3 eq. 2.28 shows the Hamiltonian for two charged classical particles using center of mass coordinates.  You have n charged particles but the generalization to n is pretty straight-forward and contained in most undergrad books on mechanics.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then from there you want to proceed in your analysis as you would with the classical N-body problem as outlined in something like &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/038797637X&quot; rel=&quot;nofollow&quot;&gt;Meyer and Hall&lt;/a&gt; Chapter 1.  Specifically look at things like Ch. 1, Sec. 4 on 'Equilibria for the Restricted 3-body problem' and techniques for finding critical points of modified potentials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm willing to bet that for certain initial conditions (perhaps as an initial condition all particles are contained in the same plane) you'll be able to make a statement about the final configuration in a limiting case.  For the more general setup --  that's just GOTTA be an open problem.  I'd be really surprised if that was actually known at this point in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the question &quot;How would one go about stating the initial conditions on the radiation fields?&quot;.. I'm not sure how to interpret a radiation field in a classical unless you push off into derivations that you find in plasma physics.  For a Hamiltonian setup tho &lt;a href=&quot;http://www.springerlink.com/content/x2774mtn06634563/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; paper looks promising..  Truth is the more I think on it the more I think you might find an answer to your question in L&amp;amp;L's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0750626356&quot; rel=&quot;nofollow&quot;&gt;Physical Kinetics&lt;/a&gt;.  Chapter IV on instabilities might have something to say on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways.. good luck.&lt;/p&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2011-08-16T02:30:52.990" CommentCount="1" />
  <row Id="13614" PostTypeId="2" ParentId="13557" CreationDate="2011-08-16T03:04:18.690" Score="4" ViewCount="" Body="&lt;p&gt;I did not do more than read Newton, and a few commentators, so my insight on this is probably meager. But I am sure that you are right that the inertial frame interpretation of the first law is only a modern ex-post-facto justification for making it separate from the second law. Newton certainly never used the first law to define an inertial frame, he just assumed you had one in mind, since inertial frames were not the focus of his investigation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that the statements of the laws of motion are unfortunately following Aristotle more than Euclid. Since physics is no longer regarded as philosophy, we value independence of axioms over clarity of philosophical expounding, and this makes the first law redundant. But if you are stating a philosophical position--- that things maintain their state of motion unless acted upon--- Newton's first law is a neat summary of the foundation of the world-system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that Newton does not state it as &quot;a body in linear motion continues moving linearly&quot;. He includes rotational motion too, even though this is a different idea. I think he conflates the two to fix in mind the philosophical position that uniform motion is the natural state of all objects. In Aristotle, the natural state of massive stuff like &quot;earth&quot; is to be down at the center of universe, and of light stuff like &quot;fire&quot; to be up in the heavens, leading to gravity and levity. Newton is replacing this notion with a different notion of natural state. Then the second law talks about deviations from the natural state, and is a separate philosophical idea (although not a separate axiom in the mathematical sense).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The influence of Aristotle has (thankfully) declined through the centuries, making Newtons laws a little anachronistic. I think that we don't have to be so slavish to Newton nowadays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton was aware of the importance of linear momentum and angular momentum conservation. One other way of understanding and his first law can be thought of as making the conservation laws primary. This point of view is both closer to Newton's thinking (it is what makes his &quot;natural states&quot; natural), and it is also a better fit with modern understanding. So it might be nice to restate the first law as &quot;linear momentum and angular momentum are conserved&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this is based on personal speculation, not on sound historical research, so take with a grain of salt.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-16T03:04:18.690" CommentCount="1" />
  <row Id="13615" PostTypeId="2" ParentId="7654" CreationDate="2011-08-16T03:12:32.373" Score="1" ViewCount="" Body="&lt;p&gt;It's worthwhile to extend the thrust of Ted Bunn's approach to more general cases. Suppose you have a general static spacetime $$ds^2 = -e^{2\nu} dt^2 + h_{ij}dx^idx^j,$$ where the Latin indices sum over spatial terms, as usual. The factor of $2$ is purely traditional, meant to simplify various quantities, especially for the spherically symmetric case $h_{ij}dx^idx^j = e^{2\lambda} dr^2 + d\Omega^2$, of which the Schwarzschild spacetime is a particular example. Let's also distinguish &quot;proper&quot; brachistochrones (minimizing proper time $\tau$) from &quot;coordinate&quot; brachistochrones (minimizing coordinate time $t$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $-d\tau^2 = -e^{2\nu}dt^2 + h_{ij}dx^idx^j,$ and because the spacetime is static, it is also stationary with the Killing field $\partial_t$ that produces a conserved current $\epsilon = -\langle\partial_t,u\rangle = e^{2\nu}(dt/d\tau)$, the specific energy. So for every timelike of such specific energy, substituting the latter into the former gives&#xA;$$h_{ij}dx^idx^j = -d\tau^2[ 1 - e^{2\nu}(dt/d\tau)^2] = -d\tau^2[1 - \epsilon^2/e^{2\nu}].$$&#xA;Aha! The proper brachistochrones of specific energy $\epsilon$ are simply geodesics of the Riemannian manifold with the metric&#xA;$$dS_\tau^2 = \frac{h_{ij}dx^idx^j}{\epsilon^2/e^{2\nu} - 1}.$$&#xA;For the Schwarzschild spacetime, this is&#xA;$$dS_\tau^2 = \frac{(1-R/r)^{-1}dr^2 + r^2d\Omega^2}{\epsilon^2(1-R/r)^{-1} - 1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's now consider the coordinate brachistochrones. The approach is the same:&#xA;$$h_{ij}dx^idx^j = dt^2e^{2\nu}[1 - e^{-2\nu}(d\tau/dt)^2] = dt^2e^{2\nu}[1 - e^{2\nu}/\epsilon^2],$$&#xA;so that the coordinate brachistochrones are the geodesics of&#xA;$$dS_t^2 = \frac{h_{ij}dx^idx^j}{e^{2\nu}(1 - e^{2\nu}\epsilon^{-2})}.$$&#xA;For the Schwarzschild spacetime, this is:&#xA;$$dS_t^2 = \frac{(1-R/r)^{-1}dr^2 + r^2d\Omega^2}{(1-R/r)(1-(1-R/r)\epsilon^{-2})},$$&#xA;which matches Ted Bunn's result under the assumptions that the trajectory is in the equatorial plane ($d\Omega = d\phi$) and the particle starts out at rest from $r=r_0$ ($\epsilon^2 = (1-R/r_0)$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An even more general treatment in an arbitrary &lt;em&gt;stationary&lt;/em&gt; (not necessarily &lt;em&gt;static&lt;/em&gt;!) spacetime is possible, and was done by V. Perlick in &lt;a href=&quot;http://link.aip.org/link/jmapaq/v32/i11/p3148/s1&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;J. Math. Phys.&lt;/em&gt; &lt;strong&gt;32&lt;/strong&gt;, 3148 (1991)&lt;/a&gt;, which treats the problem in a more geometrically abstract form and includes brachistochrones in Gödel's rotating universe.&lt;/p&gt;&#xA;" OwnerUserId="1353" LastEditorUserId="1353" LastEditorDisplayName="" LastEditDate="2011-08-16T03:23:22.777" LastActivityDate="2011-08-16T03:23:22.777" />
  <row Id="13616" PostTypeId="1" CreationDate="2011-08-16T04:37:32.723" Score="3" ViewCount="84" Body="&lt;p&gt;The following is an excerpt from Landau's Course on Theoretical Physics Vol.1 Mechanics:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... we should recall the fact that every first-order partial differential equation has a solution depending on an arbitrary function; such a solution is called the general integral of the equation. In mechanical applications, the general integral of the Hamilton-Jacobi equation is less important than a complete integral, which contains as many independent arbitrary constants as there are independent variables.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can someone clarify what's a complete integral and what's a general integral of a first order partial differential equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3064" LastEditorUserId="3064" LastEditorDisplayName="" LastEditDate="2011-08-17T05:58:53.097" LastActivityDate="2011-08-17T17:56:04.097" Title="Complete vs General Integral of first order PDE" Tags="&lt;classical-mechanics&gt;&lt;mathematical-physics&gt;&lt;homework&gt;" AnswerCount="2" />
  <row Id="13617" PostTypeId="2" ParentId="10325" CreationDate="2011-08-16T04:58:19.283" Score="0" ViewCount="" Body="&lt;p&gt;I think the Y K Lim series are the best to follow, especially for last couple week of preparations. These books have all sort of good problems (with solutions) collected from reputed US universities. &#xA;You also can dig the internet for various universities solved exams but that is quite time consuming. You can try this link which appears to have a large collection of solved/unsolved problems from different relevant areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;www.physics4qualifiers.com &lt;/p&gt;&#xA;" OwnerUserId="4941" LastActivityDate="2011-08-16T04:58:19.283" CommunityOwnedDate="2011-08-16T04:58:19.283" />
  <row Id="13618" PostTypeId="2" ParentId="13611" CreationDate="2011-08-16T05:05:02.087" Score="6" ViewCount="" Body="&lt;p&gt;Expanding on Ron's comment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I(\nu ,T)d\nu =\frac{2h\nu ^3}{c^2}\frac{d\nu }{e^{\frac{h\nu }{kT}}-1}$$&#xA;$$\nu \to \frac{c}{\lambda },\quad d\nu \to c\frac{d\lambda }{\lambda ^2}$$&#xA;$$I(\lambda ,T)d\lambda =\frac{2h}{c^2}\left(\frac{c}{\lambda }\right)^3\frac{1}{e^{\frac{hc}{\lambda kT}}-1}c\frac{d\lambda }{\lambda ^2}=\frac{2hc^2}{\lambda ^5}\frac{d\lambda }{e^{\frac{hc}{\lambda kT}}-1}$$&lt;/p&gt;&#xA;" OwnerUserId="3064" LastEditorUserId="3064" LastEditorDisplayName="" LastEditDate="2011-08-16T05:10:51.207" LastActivityDate="2011-08-16T05:10:51.207" />
  <row Id="13619" PostTypeId="2" ParentId="13616" CreationDate="2011-08-16T05:13:10.973" Score="3" ViewCount="" Body="&lt;p&gt;In Russian &lt;em&gt;&quot;integral&quot;&lt;/em&gt; is a synonym of a solution of differential equation. &lt;em&gt;&quot;general integral&quot;&lt;/em&gt; means general solution, &lt;em&gt;&quot;complete&quot;&lt;/em&gt; probably means sum of particular solution and general solution  (called the complementary solution)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ewp.rpi.edu/hartford/~ernesto/F2003/MES/Notes/Notes_pdf/s10.pdf&quot; rel=&quot;nofollow&quot;&gt;iPDEs&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4760" LastEditorUserId="4760" LastEditorDisplayName="" LastEditDate="2011-08-16T07:50:51.273" LastActivityDate="2011-08-16T07:50:51.273" CommentCount="2" />
  <row Id="13620" PostTypeId="1" CreationDate="2011-08-16T05:39:56.500" Score="0" ViewCount="101" Body="&lt;p&gt;What is practical importance of Kohn anomaly for experimentalists and/or theorists?&lt;/p&gt;&#xA;" OwnerUserId="2179" LastEditorUserId="2179" LastEditorDisplayName="" LastEditDate="2011-08-16T14:28:34.107" LastActivityDate="2011-08-16T15:06:23.903" Title="Importance of Kohn anomaly?" Tags="&lt;solid-state-physics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13621" PostTypeId="2" ParentId="12435" CreationDate="2011-08-16T05:46:27.523" Score="5" ViewCount="" Body="&lt;p&gt;First, set the units so that the speed of light is equal to one, so that the path of light rays in space-time are at 45 euclidean degrees. Note that a moving observer has a space-time path which is tilted relative to the t-axis, and the t-axis describes the path of a stationary observer (what I really mean is, get comfortable with space-time pictures). Then restrict yourself to two dimensions, one space, one time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A moving observer's t-axis is, by the principle of relativity, his world line. He has a bunch of friends moving along at the same speed, at different positions, and their worldlines are parallel to the first (they are relatively stationary by the principle of relativity, so their worldlines never meet). They also have ticks-points corresponding to their clocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first nontrivial question is &quot;what does the moving observer's x-axis look like?&quot; This is also the first thing Einstein adresses in his relativity paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The x-axis is defined in the stationary frame as &quot;all events simultaneous with the origin&quot;. Another way of saying this, is that the events at position x and y are simultaneous if the light ray starting at the event at x and the light rays starting at the event from y reach the point (x+y)/2 at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.gliffy.com/pubdoc/2845985/L.png&quot; alt=&quot;alt text&quot;&gt;&#xA;So, by the principle of relativity, if two comoving observers want to synchronize their clocks, the friend at x and the friend at y send a light signal at what they think is the same clock-tick to the friend at (x+y)/2. If (x+y)/2 gets the signal from both at the same time, he tells them &quot;ok--- you're synchronous&quot;. This can be seen on a space-time diagram&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.gliffy.com/pubdoc/2845946/L.png&quot; alt=&quot;alt text&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to convince yourself of the following: the x-axis of the moving observer is tilted up by the same amount that the t-axis of the moving observer is tilted right, so that the light signal from e and the light signal from f both reach g on the parallel line halfway inbetween. The reason is that the light-rays are still on 45 degrees for the moving observer, which is Einstein's principle of the constancy of the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two lines are relativistically-perpendicular if one is the x-axis for the other's t-axis. Convince yourself that two lines are perpendicular when they have the same tilt relative to the 45-degree lines corresponding to paths of light rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you are ready for a proof of the Minkowski version pythagorean theorem.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Chinese Minkowski Pythagorean theorem&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://www.gliffy.com/pubdoc/2846023/L.png&quot; alt=&quot;alt text&quot;&gt;&#xA;The middle space-time shape that looks like a diamond is really a square in relativity--- you must remember what perpendicularity means: same tilt relative to the 45-degree line. By adding areas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(a+b)^2 = c^2 + 2ab + 2b^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c^2 = a^2 - b^2$&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-16T06:34:08.957" LastActivityDate="2011-08-16T06:34:08.957" CommentCount="3" />
  <row Id="13622" PostTypeId="2" ParentId="304" CreationDate="2011-08-16T05:51:52.690" Score="0" ViewCount="" Body="&lt;p&gt;Another factor: explosion is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_delta_function&quot; rel=&quot;nofollow&quot;&gt;dirac&lt;/a&gt;-type pressure excitation.&#xA;The frequency spectrum is flat (&lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_delta_function#Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;Fourier transform $F(\delta)=1$ is unity for every frequency&lt;/a&gt;) so my guess is, different pitch comes from &lt;a href=&quot;http://encyclopedia2.thefreedictionary.com/Dispersion+of+Sound&quot; rel=&quot;nofollow&quot;&gt;attenuation of different frequencies&lt;/a&gt; over distance - far away explosion is higher pitched, than the same explosion nearby; at the same distance the stronger explosion of the two is lower pitched, as low frequencies are not completely attenuated.  &lt;/p&gt;&#xA;" OwnerUserId="4760" LastEditorUserId="4760" LastEditorDisplayName="" LastEditDate="2011-08-16T21:56:01.220" LastActivityDate="2011-08-16T21:56:01.220" />
  <row Id="13623" PostTypeId="2" ParentId="13441" CreationDate="2011-08-16T07:37:39.027" Score="1" ViewCount="" Body="&lt;p&gt;Another exception: the factors can depend on the number of dimensions. Esp in models with extra dimensions you gather easily some (2 pi)^d that can be a factor 1000 or larger.&lt;/p&gt;&#xA;" OwnerUserId="1458" LastActivityDate="2011-08-16T07:37:39.027" CommentCount="1" />
  <row Id="13624" PostTypeId="1" AcceptedAnswerId="13633" CreationDate="2011-08-16T07:50:57.257" Score="4" ViewCount="134" Body="&lt;p&gt;And in what sense are they 'non-local'?&lt;/p&gt;&#xA;" OwnerUserId="1458" LastActivityDate="2011-08-17T00:33:43.663" Title="Why are higher order Lagrangians called 'non-local'?" Tags="&lt;quantum-field-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" />
  <row Id="13625" PostTypeId="1" AcceptedAnswerId="13637" CreationDate="2011-08-16T07:53:35.333" Score="3" ViewCount="100" Body="&lt;p&gt;I'm asking this question because of &lt;a href=&quot;http://www.newscientist.com/article/mg21128241.700-beyond-spacetime-welcome-to-phase-space.html&quot; rel=&quot;nofollow&quot;&gt;an article in New Scientist&lt;/a&gt; about &lt;a href=&quot;http://arxiv.org/abs/1101.0931&quot; rel=&quot;nofollow&quot;&gt;a recent preprint&lt;/a&gt; by a group including Lee Smolin. I haven't taken the time to comprehend the paper completely. My knowledge of differential and pseudo-Riemannian geometry is out of practice, so it'd take me a while to do so. Even so, the paper put some ideas in my head about phase space and symplectic manifolds that can be put as simple questions but I couldn't answer with some basic Googling. Also, this is resting on the (mis)understanding that phase space is a symplectic manifold. If this is wrong, these questions might be meaningless.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a canonical symplectic manifold that is associated with a given pseudo-Riemannian manifold? I'm guessing yes, because a pseudo-Riemannian manifold is itself a differential manifold, and I think the tangent bundle is a symplectic manifold.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the space we started with has a metric, does this imbue the associated symplectic manifold with any structure? e.g. is the symplectic form limited by or related to the metric or something similar?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Could any ideas from SR/GR be extended to this symplectic space such that we learn something new? e.g. could there be something like the Einstein Field Equations (EFEs), written for the 8-dimensional phase space, that have new solutions that are realistic but wouldn't be found using the EFEs? Or could there be non-canonical symplectic manifolds associated with a given pseudo-Riemannian space? (I suspect Darboux's theorem says &quot;no&quot;.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is mostly crazy speculation based on knowledge I haven't exercised in a few years now. For all I know, there's some underlying concept that makes this pointless or I've just totally misunderstood the paper in the first place. Also, I've posted here because of the theoretical physics nature of the problem, but I accept it may be better served on another SE.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2011-08-16T16:30:55.370" Title="How does (or can) SR/GR extend to phase space or symplectic manifolds?" Tags="&lt;general-relativity&gt;&lt;special-relativity&gt;&lt;differential-geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13627" PostTypeId="2" ParentId="13441" CreationDate="2011-08-16T10:06:22.870" Score="0" ViewCount="" Body="&lt;p&gt;To add to the other answers, it should be noted that when formulating laws and relationships it is custom to use only unitary values of the relevant physical quantities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F=ma$, $V=IR$ etc could just as well have had a constant inserted, and if you shift from SI units you will need to include some.  Therefore lack of numerical constants at the beginning leads to a lack of constants at the end of the analysis...&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-08-16T10:06:22.870" CommentCount="2" />
  <row Id="13628" PostTypeId="2" ParentId="10325" CreationDate="2011-08-16T10:35:01.577" Score="0" ViewCount="" Body="&lt;p&gt;I'll add absolutely great book with problems in analytical mechanics:&lt;br&gt;&#xA;&lt;em&gt;G.L. Kotkin&lt;/em&gt; and &lt;em&gt;V.G. Serbo&lt;/em&gt;, &lt;strong&gt;&quot;Problems in Classical Mechanics&quot;&lt;/strong&gt;&lt;br&gt;&#xA;It is basically a problem-driven approach to learning Landau's first tome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I really like the book:&lt;br&gt;&#xA;&lt;em&gt;Zhong-Qi Ma&lt;/em&gt; and &lt;em&gt;Xiao-Yan Gu&lt;/em&gt;, &lt;strong&gt;&quot;Problems &amp;amp; Solutions in Group theory for physicists.&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="386" LastActivityDate="2011-08-16T10:35:01.577" CommunityOwnedDate="2011-08-16T10:35:01.577" />
  <row Id="13629" PostTypeId="1" AcceptedAnswerId="13649" CreationDate="2011-08-16T11:22:58.897" Score="4" ViewCount="130" Body="&lt;p&gt;Explaining SU(5) GUTs (using the first particle family as an example) in the last SUSY lecture 10, Lenny Susskind mentioned that there are at present no ideas how to combine simultaneously all 3 particle families into a large GUT theory.&#xA;I somehow dont believe him, suspecting that he just didnt want to talk about this :-P...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, are there any ideas around how to incorporate all 3 families into a larger structer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, I would appreciate explanations about how it works at a &quot;Demystified&quot; level :-)&lt;/p&gt;&#xA;" OwnerUserId="2751" LastActivityDate="2011-08-17T05:09:18.077" Title="GUT that includes all 3 particle families into a large group?" Tags="&lt;unified-theories&gt;" AnswerCount="3" CommentCount="1" />
  <row Id="13630" PostTypeId="2" ParentId="7509" CreationDate="2011-08-16T11:49:21.200" Score="0" ViewCount="" Body="&lt;p&gt;Your robot will only able to walk in one dimension (apart from vertical movements), if the gyroscope is massive - the rotation plane is fixed then, so the robot will have trouble to turn away. (for small-scale experience try to move a running 3.5' external HDD, you will experience the resistance) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally if the gyroscope is running all the time, your mech will be slowly rotated with the earth movement during the day - not a nice feature in a long distance march.&lt;/p&gt;&#xA;" OwnerUserId="4760" LastActivityDate="2011-08-16T11:49:21.200" />
  <row Id="13631" PostTypeId="2" ParentId="13629" CreationDate="2011-08-16T11:50:20.597" Score="0" ViewCount="" Body="&lt;p&gt;There has been quite a lot of coverage about Garrett Lisi's Exceptionally Simple Theory of Everything. You can find the preprint &lt;a href=&quot;http://arxiv.org/abs/0711.0770&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, as well as a SciAm special &lt;a href=&quot;http://www.scientificamerican.com/article.cfm?id=a-geometric-theory-of-everything&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lisi's theory involves using $E_8$, which is the largest exceptional Lie algebra in Killing's classification. I'm not aware of the theory appearing in a peer-reviewed journal. I think Lisi claims to also incorporate gravity but, because the group is so large, there are also a lot of new particles that would need explaining away. The &lt;a href=&quot;http://en.wikipedia.org/wiki/An_Exceptionally_Simple_Theory_of_Everything&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; is reasonably fair about the coverage of the original preprint and makes some attempts to explain it with pretty visualizations of $E_8$. The theory is highly controversial and doesn't have widespread acceptance. but it is an &quot;idea around how to incorporate all 3 families into a larger structure&quot;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are probably more conventional ideas that use SU(5) or SO(10), but they clearly don't have as good PR people because I'm not aware of any leading theory.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastActivityDate="2011-08-16T11:50:20.597" CommentCount="5" />
  <row Id="13632" PostTypeId="2" ParentId="12122" CreationDate="2011-08-16T13:48:03.420" Score="0" ViewCount="" Body="&lt;p&gt;Once you got that the sum of all the forces is zero, why did you assume that the internal forces will have the same collinear direction (i.e. why would they be opposite along the same line. They can be along any direction as long as the sum is zero?)&#xA;Also, you consider 2 particles; $\vec{F_1}$+$\vec{F_2}$=$0$ might give you an idea on how the forces behave. Generalise to 3 and more directions.&lt;/p&gt;&#xA;" OwnerUserId="4555" LastActivityDate="2011-08-16T13:48:03.420" />
  <row Id="13633" PostTypeId="2" ParentId="13624" CreationDate="2011-08-16T14:01:36.077" Score="9" ViewCount="" Body="&lt;p&gt;The higher the number of derivatives the more initial data you have to provide. If you have some Lagrangian that contains an infinite number of derivatives (or derivatives appearing non-polynomially, such as one over derivative) then you have to provide an infinite amount of initial data which amounts to non-local info, in the sense explained below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think in terms of Taylor expansions around your initial value, then you have to provide the full function (and thus non-local information) if you have an infinite number of derivatives. This is to be contrasted with cases where you provide only the field and its first derivative as initial values (and thus rather local information).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I would not call any higher-derivative Lagrangian &quot;non-local&quot;, but only those theories where the number of derivatives is formally infinite in the Lagrangian. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any discretization scheme you literally see the non-locality induced by higher derivatives: to define the first derivative you need to know the function on two adjacent lattice points, to define the second derivative on three and to define the n-th derivative on n+1 lattice points. Thus, the more derivatives the more non-locality. If you have an infinite number of derivatives you need to know the function on an infinite set of lattice points.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-08-16T14:01:36.077" CommentCount="4" />
  <row Id="13634" PostTypeId="2" ParentId="13237" CreationDate="2011-08-16T14:47:22.247" Score="2" ViewCount="" Body="&lt;p&gt;This principle fails in the most startling way in second order phase transitions. This is a particularly clean example, because Landau predicted the critical exponents of second-order phase transitions using &lt;em&gt;only&lt;/em&gt; the principle that the thermodynamic functions are analytic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His argument is as follows: given a magnet going through the Curie point, where it loses its magnetization smoothly, the equilibrium magnetization should be the solution of some thermodynamic equation with the derivative some thermodynamic potential is set to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F(T,m)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At temperatures lower than T_c, the magnetization is nonzero, and at temperatures higher than Tc, the magnetization is 0, and it goes to 0 in a continuous way. How does it go to zero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the magnetization m and -m are related by rotational symmetry. Shifting T_c to 0 by translating $f(t,m)= F(T_c - t,m)$, you get a new thermodynamic function, which has the property that f has only the trivial solution m=0 for negative t, and has two small nontrivial solutions in m for positive t.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because m=0 is a solution at t=0, the function $f$ has no constant term in a Taylor expansion. By the symmetry of $m\rightarrow -m$, only even powers of m contribute to its Taylor series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(t,m) = At + Bm^2 + Ct^2 + Dt^3 + E t m^2 ...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that $f(t,m)$ is generic, A and B are not exactly zero. So for small enough t, for temperatures close enough to the critical point, you get that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m \propto \sqrt{t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, this scaling only fails if one of the coefficients is zero. If A=0,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m \propto |t|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But m is then nonzero on both sides of the transition. If B=0, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m \propto t^{1\over 4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and m is zero&#xA;or if A,B,C are zero, in which case you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m \propto |t|^{3/4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And each of these cases requires fine tuning of parameters. So Landau predicted that the critical behavior of the magnetization will be as the square root of the temperature at the critical point, and that this behavior will be universal, it won't depend on the system, just on the existence of the phase transition. The Ising model should have the same critical exponent as the physical magnet, a square root dependence of the magnetization on the temperature, and the liquid gas transition will also have a bend in the curve of the density vs. temperature at the critical pressure which goes as the square root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exponent turned out to be universal, it was equal for the gas and liquid, and for the Ising model. But it wasn't 1/2, but more like .308 in three dimensions, and .125 in two dimensions, It only turned into Landau's 1/2 in 4 dimensions or higher. This means that Landau's argument fails, and that the thermodynamic function is conspiring to be non-analytic at exactly the place where Landau was expanding. Understanding why it is non-analytic exactly at the phase transition led to the modern renormalization theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In mathematics, Rene Thom proposed that a version of Landau's argument is a complete theory of the types of allowed phase transitions in nature. He called the phase transitions &quot;catastrophes&quot;, because they showed a sudden change in behavior, and he predicted, based on catastrophe theory all sorts of scaling laws for natural transitions. This was the most ambitious attempt to exploit the observation that naturally occuring functions are nice. This fails for the same reason as Landau's argument: functions describing changes in the critical behavior of interesting systems at a transition point are rarely analytic at this point.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-16T15:00:34.720" LastActivityDate="2011-08-16T15:00:34.720" CommunityOwnedDate="2011-08-16T14:47:22.247" />
  <row Id="13635" PostTypeId="2" ParentId="13620" CreationDate="2011-08-16T15:06:23.903" Score="1" ViewCount="" Body="&lt;p&gt;The practical importance of a Kohn anomaly is that it is a direct evidence of electron phonon coupling in a metallic system.  This is used as evidence for electron-phonon coupling in BCS superconductors for example.  Further, a Kohn anomaly is a sharp change in a lattice vibration that can cause a lattice instabilities, which can result in structural phase transitions of a material in extreme cases.&lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2011-08-16T15:06:23.903" />
  <row Id="13636" PostTypeId="2" ParentId="13577" CreationDate="2011-08-16T15:31:01.900" Score="4" ViewCount="" Body="&lt;p&gt;Since you say you're talking about what happens locally (in a small volume), I'll answer from that point of view. The usual formulation of energy conservation in such a volume is that energy is conserved &lt;em&gt;in an inertial reference frame&lt;/em&gt;. In general relativity, there are no truly inertial frames, but in a sufficiently small volume, there are reference frames that are approximately inertial to any desired level of precision. If you restrict your attention to such a frame, &lt;em&gt;there is no cosmological redshift&lt;/em&gt;. The photon's energy when it enters one side of the frame is the same as the energy when it exits the other side. So there's no problem with energy conservation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (apparent) failure of energy conservation arises only when you consider volumes that are too large to be encompassed by a single inertial reference frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be slightly more precise, in some small volume $V=L^3$ of a generic expanding Universe, imagine constructing the best possible approximation to an inertial reference frame. In that frame, observers near one edge will be moving with respect to observers near the other edge, at a speed given by Hubble's Law (to leading order in $L$). That is, in such a frame, the observed redshift is an ordinary Doppler shift, which causes no problems with energy conservation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want more detail, David Hogg and I wrote about this at considerable (perhaps even excessive!) length in &lt;a href=&quot;http://arxiv.org/abs/0808.1081&quot; rel=&quot;nofollow&quot;&gt;an AJP paper&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-08-16T15:31:01.900" CommentCount="4" />
  <row Id="13637" PostTypeId="2" ParentId="13625" CreationDate="2011-08-16T16:30:55.370" Score="1" ViewCount="" Body="&lt;p&gt;This question addresses many aspects of the physics of indefinite signature spaces. I'll try to answer you by examples.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;As already mentioned in the comments, the cotangent bundle of a pseudo-Riemannian manifold is a symplectic manifold.&#xA;The basic example of a physical system having this phase space is the relativistic free particle. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It is very interesting that the quantization of the relativistic particle produces another type of indefinte signature on the cotangent bundle: The indefinite metric in the free particle Lagrangians produces a &quot;wrong&quot; sign in the time component of the symplectic structure of the cotangent bundle.&#xA;The canonical quantization of this symplectic structure produces negative norm states which are eliminated by the imposition of a physical constraint. One possible way to restrict to the physical subspace is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gupta%E2%80%93Bleuler_formalism&quot; rel=&quot;nofollow&quot;&gt;Gupta-Bleuler&lt;/a&gt; method, originated in the covariant quantization of electrodynamics, but can be used in the case of the free relativistic particle.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The pseudo-Riemannian metric on the base manifold results the association of the cotangent bundle with an $SO(M,N)$ principal bundle.&#xA;($(M,N)$ is the metric's signature) via the contragradient fundamental representation. This allows to define tensor fields on the base space as sections of associated vector bundles.&#xA;If in addition the base space has a spin structure, it is possible to lift this bundle to a $Spin(M,N)$ bundle and define spinor fields as sections&#xA;of aspinor bundle (The associated bundle corresponding to the spinor representation).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are many Hamiltonian approaches, in which the Einstein equations are obtained as the equations of motion. I'll describe to you the the method that I think is the most interesting: MacDowell–Mansouri Gravity, see for example &lt;a href=&quot;http://arxiv.org/PS_cache/gr-qc/pdf/0611/0611154v2.pdf&quot; rel=&quot;nofollow&quot;&gt;the following article&lt;/a&gt; by Derek Wise. In this approach one construct a gauge theory based on the de-Sitter or anti de-Sitter group.  One identifies the spin connection with the components corresponding to the Lorentz subgroup generators, and the Vielbeins with the components of the generators dS/Lorentz, or AdS/Lorentz.&#xA;In this setting it is possible to construct a hodge star operation not depending on a metric. It turns out that the Yang Mills Lagrangian, when written in terms of the spin connection and the Vielbeins turns out to be equal to the Einstein-Cartan Lagrangian (with a cosmological constant) which gives the Einstein equations of motion + a topological term.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, the symplectic geometry of this theory is similar to the well known Yang-Mills with a compact gauge group, for example the vanishing momenta of the time components of the connection give rise to the Gauss law constraints.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastActivityDate="2011-08-16T16:30:55.370" />
  <row Id="13638" PostTypeId="1" AcceptedAnswerId="13644" CreationDate="2011-08-16T16:42:42.790" Score="6" ViewCount="87" Body="&lt;p&gt;My question is about the problem below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/a5cJ5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depicted are two space ships (the USS Voyager and the USS Enterprise), each with velocity $v=c/2$ relative to the space station (Babylon 5).&#xA;At the exact moment the two space ships are closest together (at a distance d) the USS Enterprise fires off a shuttle with velocity $u=3c/4$ relative to the space station. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: Under what angle $\alpha&amp;#39;$ (as measured by USS Enterprise) must the shuttle be fired off in order to meet the USS Voyager?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see two possible ways to approach this: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Calculate the angle $\alpha$ measured by the space station Babylon 5 and figure out how angles transform when we change our frame of reference.&lt;/li&gt;&#xA;&lt;li&gt;Figure out the velocity in x-/y-direction in the frame of reference of the USS Enterprise and deduce the angle $\alpha&amp;#39;$ from that.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, if both ways are correct they should give the same answer. But they don't seem to, so what is wrong? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is what I have tried:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $u_x$, $u_y$ be the velocities (in the corresponding directions) the shuttle needs to have - as observed by Babylon 5. Let $u_x&amp;#39;$, $u_y&amp;#39;$ be the velocities of the shuttle observed by USS Enterprise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then by the rules for adding velocities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{eqnarray*}&#xA;u_y' &amp;amp;=&amp;amp; \frac{u_y + v}{1 + \frac{vu_y}{c^2}} \\&#xA;u_x' &amp;amp;=&amp;amp; \frac{u_x }{1 + \frac{vu_y}{c^2}}\cdot \sqrt{1-\frac{v^2}{c^2}}&#xA;\end{eqnarray*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan(\alpha&amp;#39;) = \frac{u_y&amp;#39;}{u_x&amp;#39;} = \frac{u_y + v}{u_x} \cdot\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But on the other hand, we could imagine the guys on Babylon 5 drawing a big triangle to show the crew on USS Enterprise the trajectory the shuttle needs to take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The people on Babylon 5 will then draw a triangle with angle $\alpha$ satisfying&#xA;$$\tan(\alpha) = \frac {u_y}{u_x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if side $a$ of this triangle is parallel to the x-axis and $b$ is parallel to the y-axis, then we have $$\frac ba = \tan(\alpha) = \frac{u_y}{u_x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since for the USS Enterprise $a$ is the same, but side $b$ is contracted, they will see a traingle with side $b&amp;#39; = b/\gamma$ and $a&amp;#39; = a$. Therefore &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan(\alpha&amp;#39;) = \frac{b&amp;#39;}{a&amp;#39;} = \frac ba \sqrt{1-\frac{v^2}{c^2}} = \frac{u_y}{u_x}\sqrt{1-\frac{v^2}{c^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are two different results in general. I cannot figure out what is wrong with either of them... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for reading, help will be greatly appreciated! :)&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-16T17:35:10.303" LastActivityDate="2011-08-18T22:01:33.677" Title="Transformation of angles in special relativity" Tags="&lt;special-relativity&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13639" PostTypeId="1" AcceptedAnswerId="13640" CreationDate="2011-08-16T17:10:11.283" Score="4" ViewCount="138" Body="&lt;p&gt;What is the basic postulate on which QM rests. Is it that the position of a particle can only be described only in the probabilistic sense given by the state function $\psi(r)$ ? We can even go ahead and abandon the particle formalism as well. So what is the QM all about ? A probabilistic description of the physical world ? and nothing more ?&lt;/p&gt;&#xA;" OwnerUserId="540" LastActivityDate="2011-08-16T17:25:01.520" Title="What is the basic postulate on which QM rests" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="13640" PostTypeId="2" ParentId="13639" CreationDate="2011-08-16T17:24:58.867" Score="4" ViewCount="" Body="&lt;p&gt;Existence of non-compatible observables: measuring one of them (say, coordinate) leads to an unavoidable uncertainty in the result of a subsequent measurement of the other (say, momentum).  This is the essence of the Heisenberg uncertainty principle in the kinematics of your system. There is a detailed discussion along these lines in the beginning of the Quantum Mechanics volume (volume III) in the Course of Theoretical Physics by Landau and Lifshitz. Any measurable (physical) system, be it particle, atom or anything else, is quantum only if you can identify a manifestation of Heisenberg uncertainty principle (non-commutativity of observables).&lt;/p&gt;&#xA;" OwnerUserId="1739" LastActivityDate="2011-08-16T17:24:58.867" CommentCount="4" />
  <row Id="13641" PostTypeId="2" ParentId="13639" CreationDate="2011-08-16T17:25:01.520" Score="0" ViewCount="" Body="&lt;p&gt;To me, the most basic postulate is that energy comes in discrete packages of $h \nu$. Based on this assumption, you get much of the rest of basic quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the Schrödinger equation is related to the Hamilton-Jacobi formulation of classical mechanics with the added assumption of quantized energy, and the Heisenberg picture follows directly from the Poisson bracket formulation of classical mechanics, assuming quantized energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wave-particle duality is also a very important assumption- this gives us a way to interpret exactly what these equations express, which is the probability of locating a particle per some (small) volume of position- or momentum-space.&lt;/p&gt;&#xA;" OwnerUserId="4663" LastActivityDate="2011-08-16T17:25:01.520" CommentCount="2" />
  <row Id="13642" PostTypeId="2" ParentId="13599" CreationDate="2011-08-16T17:54:28.377" Score="2" ViewCount="" Body="&lt;p&gt;It's possible to split this combined signal into the original components again. You can do that because the sine and cosine functions form a &lt;em&gt;base&lt;/em&gt; of a &lt;em&gt;Hilbert space&lt;/em&gt;, a space called $L^2(\mathbb{R})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what does this mean? The word &quot;space&quot; is perhaps confusing, a mathematical space is essentially just a set of mathematical objects, in this case the space contains all possible combined signals (each signal is a &quot;point&quot; in the space).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a Hilbert space, it is possible to calculate the &lt;em&gt;scalar product&lt;/em&gt; of two elements: let $\varphi,\psi$ be two possible signals, (for instance it could be $\varphi(x)=\sin(x)$, $\psi(x)=\cos(2x)$)&lt;sup&gt;†&lt;/sup&gt;, then we have&#xA;$$&#xD;&#xA;  \langle \varphi| \psi\rangle_{\!\!\!\!_{L^2(\mathbb{R})}} = \int_\mathbb{R}\!\mathrm{d}x\ \varphi(x)\cdot\psi(x)&#xD;&#xA;$$&#xA;This yields a single number. In case you don't understand what this expression means, don't worry. You can think of the scalar product as measuring &quot;how much the signal $\varphi$ is 'in' the signal $\psi$&quot;, in a sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose now you have given a signal&#xA;$$&#xD;&#xA;  \varphi(x) = a\cdot \sin(x) + b\cdot \sin(2x)&#xD;&#xA;$$&#xA;where $a$ and $b$ are unknown prefactors. It is these prefactors that carry the actual information in a radio signal, so you want to know them. And that's where we can use our scalar product: we calculate&#xA;$$&#xD;&#xA;  \langle \sin(x) | \varphi\rangle&#xD;&#xA;$$&#xA;and&#xA;$$&#xD;&#xA;  \langle \sin(2x) | \varphi\rangle.&#xD;&#xA;$$&#xA;Actually, we're running into a bit of a problem right here: the sine functions oscillate literally forever, &lt;em&gt;infinite&lt;/em&gt;. That makes the result of the scalar product also infinite. But in reality, we don't have infinite signals, they are actually limited to some finite time duration. Suppose the signals start at $x=0$ and end at the time $x=4\pi$, (before and after, both are just zero all the time). Then we have&#xA;$$\begin{align}&#xD;&#xA;  \langle \sin(x) | \varphi\rangle&#xD;&#xA;  &amp;amp;= \int\limits_0^{4\pi}\!\mathrm{d}x\ \sin(x)\cdot\psi(x)   \\&#xD;&#xA;  &amp;amp;= \int\limits_0^{4\pi}\!\mathrm{d}x\ \sin(x)\cdot\sin(x)&#xD;&#xA;    + \int\limits_0^{4\pi}\!\mathrm{d}x\ \sin(x)\cdot\sin(2x)&#xD;&#xA;\end{align}$$&#xA;The integrals can be calculated with help of some known trigonometric equations. It turns out the result is $\langle \sin(x) | \varphi\rangle = 2\pi\cdot a$, whereas $\langle \sin(2x) | \varphi\rangle = 2\pi\cdot b$. We calculated the prefactor of each of the sine functions!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&lt;sup&gt;†&lt;/sup&gt; I called the &quot;time&quot;-variable $x$ here, like you did. In physics, we would usually call it $t$ for radio signals.&lt;/p&gt;&#xA;" OwnerUserId="3540" LastActivityDate="2011-08-16T17:54:28.377" CommentCount="1" />
  <row Id="13643" PostTypeId="2" ParentId="13546" CreationDate="2011-08-16T18:11:19.243" Score="-1" ViewCount="" Body="&lt;p&gt;The correlations you describe seem to be trivial and completely explained by a conservation law (conservation of total Sz in this case). Classical example of  100% correlation: if a rocket with knowen total mass and velocity splits into to two parts then measuring the momentum of one of the parts gives the perfect knowledge of the momentum of the second part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the particle you mention  are in a pure entangled quantum mechanical  state is not relevant to this 100% correlation. Correlations entering Bell inequalities are more subtle and necessarily involve measuring non-commuting observables.&lt;/p&gt;&#xA;" OwnerUserId="1739" LastActivityDate="2011-08-16T18:11:19.243" CommentCount="5" />
  <row Id="13644" PostTypeId="2" ParentId="13638" CreationDate="2011-08-16T19:17:37.293" Score="4" ViewCount="" Body="&lt;p&gt;Let's look at the Gallilean case first (to which this problem must reduce in non-relativistic limit anyway). Your first approach works the same way but we can get rid of unnecessarily complicated terms. In particular, we have $\tan(\alpha) = \frac{v}{u_x}$ and $\tan(\alpha&amp;#39;) = \frac{2v}{u_x}$ (I've used that $u_y = v$ which is true also relativistically). You should convince yourself that this is the correct transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for the second approach we get that $\alpha = \alpha&amp;#39;$ since there are no contractions in Gallilean case. The conflict with the first approach rests on the fact that since the triangle is formed using the velocity vector, it will change when boosted. Therefore it's not valid to deduce that the triangle only transforms by contraction in one direction (which reduces to identity in non-relativistic limit). In fact, this contracting effect is quite negligible when compared to the deformation of the triangle caused by the boost.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-08-16T19:17:37.293" CommentCount="1" />
  <row Id="13645" PostTypeId="1" AcceptedAnswerId="13967" CreationDate="2011-08-16T19:53:01.940" Score="5" ViewCount="74" Body="&lt;p&gt;I'm interested in how quickly (if at all) the fluid in an open cylinder would reach solid-body rotation if the cylinder is suspended vertically in an unbounded fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found plenty of work that considers spin-up in a cylinder with a single end-wall, but can't seem to track down any information on whether the end-wall is essential for driving the flow. Even a pointer to the correct keywords to search for would help.&lt;/p&gt;&#xA;" OwnerUserId="4944" LastActivityDate="2011-08-25T20:37:22.837" Title="Spin-up time for fluid in an open cylinder with no endwalls?" Tags="&lt;fluid-dynamics&gt;&lt;rotational-dynamics&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="13646" PostTypeId="2" ParentId="13624" CreationDate="2011-08-16T23:08:25.310" Score="5" ViewCount="" Body="&lt;p&gt;Disillusionment with systems described by higher order Lagrangians harks back to a &lt;a href=&quot;http://dx.doi.org/10.1103/PhysRev.79.145&quot; rel=&quot;nofollow&quot;&gt;1950 paper by Pais and Uhlenbeck&lt;/a&gt;, in which they showed that such systems were prone to pathologies, including states with negative energy and states with negative norm. There's a more recent discussion of this in &lt;a href=&quot;http://arxiv.org/abs/hep-th/0408104&quot; rel=&quot;nofollow&quot;&gt;arXiv:hep-th/0408104&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="429" LastEditorDisplayName="" LastEditDate="2011-08-17T00:33:43.663" LastActivityDate="2011-08-17T00:33:43.663" CommentCount="1" />
  <row Id="13647" PostTypeId="2" ParentId="13629" CreationDate="2011-08-17T00:32:49.093" Score="1" ViewCount="" Body="&lt;p&gt;The problem of families in GUT is sometimes referred as an &quot;Horizontal symmetry&quot;. There are two lines of work, roughly: those which get a continous symmetry, say SU(3), and then all the gauge malabars, and those which add a discrete symmetry, such as A4. Of course in both cases, a serious GUT should show everything embodied in a larger simple group. E8 has some value because it can go down to E6xSU(3), and E6 can lodge chiral fermions (but then perhaps this SU(3) does not work as it should, in more detailed examination) Other alternatives are just growing up SO(2n) until everything fits... You always have V+A currents you dont want, plus a bag of any of the usual problems in phenomenology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zee is the adecuate source to check if you want to look deeper in this topic.&lt;/p&gt;&#xA;" OwnerUserId="1335" LastActivityDate="2011-08-17T00:32:49.093" CommentCount="2" />
  <row Id="13648" PostTypeId="1" AcceptedAnswerId="13673" CreationDate="2011-08-17T03:28:23.187" Score="6" ViewCount="148" Body="&lt;p&gt;The temperature that &lt;a href=&quot;http://www.youtube.com/watch?v=TtzRAjW6KO0&quot; rel=&quot;nofollow&quot;&gt;a solar death ray&lt;/a&gt; can produce is limited due to the solid angle of the sun itself.  Entropic arguments dictate that you can't focus the sun's light to create temperatures higher than the sun's surface, and sure enough, the size of the dot where its focused can't be minimized beyond a certain limit.  Lasers offer a very potent beam of light, but it would seem to me that it is also highly organized beam of light, since all of the photons in the beam are traveling in very nearly the same direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Let's say that I have a laser beam of some given power that starts with some diameter $D_o$ at the point of emission and increases to $D_f$ at some distance $r$ away.  Would this be sufficient information to imply a limit to the power per unit area (W/m^2) that could be obtained through focusing and what would that be?  What lens characteristics and approaches would someone look for in order to do this with a laser pointer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should note that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Laser_Beam_Quality&quot; rel=&quot;nofollow&quot;&gt;Wikipedia has an article that kind of sort of addresses this&lt;/a&gt; but I'll be forthcoming that I don't understand it and find it confusing.  Well, it mentions wavelength, which I recognize would limit the W/m^2 power if the beam was perfectly straight, but I'm interested in the limit due to manufactured imperfection of the laser, so unless somebody is over-complicating this (which is fine as long as you answer the question first), I don't think the answer should have anything to do with the wavelength.  I mean, the wavelength should be sufficiently small compared to the size of the dot that could be created, correct me if this is the wrong way of thinking of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also interested if this is the mechanism used for &lt;a href=&quot;http://en.wikipedia.org/wiki/Laser_cutting&quot; rel=&quot;nofollow&quot;&gt;laser cutting&lt;/a&gt;.  Note that the Wikipedia article introduces the subject with &lt;em&gt;Laser cutting is a technology that uses a laser to cut materials&lt;/em&gt;.  Pardon the internet speak, but this makes me want to facepalm.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'm seeing several new terms so I'm putting a glossary here to help the discussion.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gaussian beam&lt;/li&gt;&#xA;&lt;li&gt;Divergence half-angle&lt;/li&gt;&#xA;&lt;li&gt;Far field&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A Gaussian beam is a beam where if you shine it on paper, the intensity has a Gaussian distribution.  Easy enough.  The Divergence half-angle is the angle it's diverging at, given a few qualifiers, from &lt;a href=&quot;http://www.rp-photonics.com/beam_divergence.html&quot; rel=&quot;nofollow&quot;&gt;http://www.rp-photonics.com/beam_divergence.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/MzxQg.png&quot; alt=&quot;divergence and far field&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Far field is a point far to the right on this image where the width of the beam is close to the black lines.  Some other mechanism prevents it from converging at a point in the middle because, of course, we could have already dismissed that as theoretically impossible.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-08-17T16:08:12.900" LastActivityDate="2011-08-19T14:35:16.743" Title="Physics of Focusing a Laser" Tags="&lt;optics&gt;&lt;laser&gt;" AnswerCount="3" CommentCount="5" />
  <row Id="13649" PostTypeId="2" ParentId="13629" CreationDate="2011-08-17T05:02:28.493" Score="2" ViewCount="" Body="&lt;p&gt;Jacob Bourjaily derives three Standard Model generations from an E8 singularity &lt;a href=&quot;http://arxiv.org/abs/0704.0445&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://arxiv.org/abs/0901.3785&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. (See comments by Lubos, &lt;a href=&quot;http://motls.blogspot.com/2007/04/three-families-and-higgsing-of-f.html&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt; &lt;a href=&quot;http://motls.blogspot.com/2009/01/horava-lifshitz-cotton-and-uv-general.html&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt; &lt;a href=&quot;http://motls.blogspot.com/2009/05/predictions-of-local-m-theory-and-f.html&quot; rel=&quot;nofollow&quot;&gt;3&lt;/a&gt;.) &lt;/p&gt;&#xA;" OwnerUserId="1486" LastEditorUserId="1486" LastEditorDisplayName="" LastEditDate="2011-08-17T05:09:18.077" LastActivityDate="2011-08-17T05:09:18.077" CommentCount="1" />
  <row Id="13650" PostTypeId="1" AcceptedAnswerId="13665" CreationDate="2011-08-17T05:32:54.933" Score="0" ViewCount="71" Body="&lt;p&gt;&lt;strong&gt;Context:&lt;/strong&gt; There have been a few papers out recently which mention how photosynthesis in plants might have connections to entanglement, or even perhaps that entanglement is causing photosynthetic complexes to capture the sunlight and go through the conversion stages. Here are two papers worth mentioning: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mukamel.ps.uci.edu/publications/pdfs/676.pdf&quot; rel=&quot;nofollow&quot;&gt;Quantum oscillatory exciton migration in photosynthetic reaction centers.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1006.4053&quot; rel=&quot;nofollow&quot;&gt;Quantum entanglement between the electron clouds of nucleic acids in DNA&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; In QFT, the Reeh–Schlieder theorem is thought to be an analogue of some sense to quantum entanglement. My question is that can one use the Reeh–Schlieder theorem instead of entanglement and try to do the work the other papers mentioned above have done but in a QFT context? To what extent does that analogue relationship hold? In that sense it is usable as a dual description under certain constrains? &lt;/p&gt;&#xA;" OwnerUserId="4719" LastActivityDate="2011-08-17T09:40:55.900" Title="Reeh–Schlieder theorem in QFT and entanglement in biological systems" Tags="&lt;quantum-field-theory&gt;&lt;entanglement&gt;&lt;biology&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="13651" PostTypeId="2" ParentId="12422" CreationDate="2011-08-17T05:57:44.533" Score="0" ViewCount="" Body="&lt;p&gt;There's no preferred field coordinatization in quantum field theory. In fact, it has been shown there are infinitely many Borchers classes possible in algebraic quantum field theory. Even using path integrals, what your professor didn't tell you was fixing the path integral measure to be translationally invariant, and the leading kinetic term to be diagonal doesn't fix the field coordinatization at all, once you start to consider all the possible Wilson terms you can add. The only invariant is the S-matrix. If you don't trust me, meditate upon quantum field theory in curved spacetime. Now, there's really no canonical field coordinatization. The idea of a canonical field coordinatization is a prejudice of theoretical physicists. They like to think there's a canonical coordinatization for everything, something God-given and objectively better than all the other choices with no room for arbitrariness. Guess what, their prejudice is very mistaken. There always has to be an arbitrary choice imposed from the outside.&lt;/p&gt;&#xA;" OwnerUserId="4948" LastActivityDate="2011-08-17T05:57:44.533" />
  <row Id="13652" PostTypeId="1" CreationDate="2011-08-17T05:59:34.047" Score="5" ViewCount="134" Body="&lt;p&gt;This was explained to me many years ago, by a physics teacher, with the following analogy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If someone on the beach want's to reach someone else that is in the water, they will try to travel as much as they can on the beach and as little as possible on the water, because this way they will get there faster.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm paraphrasing of course, but this is as accurate as I recall it.&lt;br&gt;&#xA;This explanation makes no sense to me. Was he telling me the light knows where it is going? It wants to get there faster? It chooses a different direction?&lt;br&gt;&#xA;(No need to answer these questions, this was just me trying to understand the analogy.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempts to clarify the issue were without success and many years later I still don't know.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does light change direction when it travels through glass?&lt;/p&gt;&#xA;" OwnerUserId="4947" LastEditorUserId="4947" LastEditorDisplayName="" LastEditDate="2011-08-17T06:04:40.793" LastActivityDate="2011-08-17T16:44:46.520" Title="Why does light change direction when it travels through glass?" Tags="&lt;light&gt;" AnswerCount="6" CommentCount="3" />
  <row Id="13653" PostTypeId="2" ParentId="13556" CreationDate="2011-08-17T06:45:27.697" Score="0" ViewCount="" Body="&lt;p&gt;Well,&#xA;I guess, in the case of AC too, the lamp is glowing in only one electrode, but since its alternating at 50Hz( depends on your geography/country) the persistence of vision is making you see both of them glow at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In DC case, the lamp is glowing because of Ne ions hitting the cathode(negative electrode). Its simply a gas discharge tube. &lt;a href=&quot;http://en.wikipedia.org/wiki/Neon_lamp#Description&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4887" LastActivityDate="2011-08-17T06:45:27.697" CommentCount="2" />
  <row Id="13654" PostTypeId="1" CreationDate="2011-08-17T06:48:30.247" Score="8" ViewCount="279" Body="&lt;p&gt;I know what anti-matter is and how when it collides with matter both are annihilated. However, what about anti-photons? Are there such things as anti-photons? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I initially thought the idea preposterous. However I am curious because, if anti-photons don't exist, then anti-matter could theoretically transfer its energy to normal matter - through the mechanism of light...right?&lt;/p&gt;&#xA;" OwnerUserId="2843" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-17T07:17:46.757" LastActivityDate="2011-08-17T07:17:46.757" Title="Do anti-photons exist?" Tags="&lt;particle-physics&gt;&lt;antimatter&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="13655" PostTypeId="2" ParentId="13654" CreationDate="2011-08-17T07:04:24.267" Score="7" ViewCount="" Body="&lt;p&gt;Actually as per physics text books, photons are its own antiparticle.&#xA;Have a look &lt;a href=&quot;http://www.physicsforums.com/showthread.php?t=23122&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this ought to be a comment but I don't have the privilege to comment .. :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards,&lt;/p&gt;&#xA;" OwnerUserId="4887" LastActivityDate="2011-08-17T07:04:24.267" />
  <row Id="13656" PostTypeId="2" ParentId="13654" CreationDate="2011-08-17T07:05:43.627" Score="11" ViewCount="" Body="&lt;p&gt;Well, they do and don't. Depends on your point of view. Here's the story.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum field theory requires for consistency reasons that every charged particle has its antiparticle. It also tells you what properties will the anti-particle have: it will have the same characteristic from the point of view of space-time (i.e. Poincaré group) which means equal mass and spin. And it will have &lt;em&gt;all&lt;/em&gt; charges of opposite sign than a matter particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the particle is not charged then QFT doesn't impose any other constraint and so you don't need antiparticles for photons (since they are not charged). But you can still consider the same operation of keeping mass and spin and swapping charges and since this does nothing to photon, you can decide to identify it with an antiphoton. Your call.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-08-17T07:05:43.627" CommentCount="10" />
  <row Id="13657" PostTypeId="2" ParentId="12122" CreationDate="2011-08-17T07:18:10.420" Score="4" ViewCount="" Body="&lt;p&gt;The derivation in Landau and Lifschitz is making some additional implicit assumptions. They assume that all forces come from pair-interactions, and that the pair forces are rotationally invariant. With these two assumptions, the potential function in the Lagrangian is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V(x1,...,xn) = \sum_{\langle i,j\rangle} V(|x_i - x_j|)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then it is easy to prove Newton's third law, because the derivative of the distance function is equal and opposite for each pair of particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This type of derivation is reasonable from a physical point of view for macroscopic objects, but it is not mathematically ok, because it omits important examples.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;No rotational invariance, no third law&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Dropping the assumption of rotational invariance, but keeping the assumption of pair-wise interaction, one gets the following counterexample in 2 dimensions, with two particles (A,B) with position vectors $(A_x,A_y)$ $(B_x,B_y)$ respectively:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V(A_x,A_y,B_x,B_y) = f(A_x-B_x) + f(A_y - B_y) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where f is any function other than $f(x)=x^2$. This pair potential leads to equal and opposite forces, but not collinear ones. Linear momentum and energy are conserved, but angular momentum is not, except when both particles are on the lines $y=\pm x$ relative to each other. The potential is unphysical of course, in the absence of a medium like a lattice that breaks rotational invariance.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Many-body direct interactions, no reflection symmetry, no third law&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;There is another class of counterexamples which is much more interesting, because they do not break angular momentum or center of mass conservation laws, and so they are physically possible interactions in vacuum, but they do break Newton's third law. This is the chiral three-body interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider 3 particles A,B,C in two dimensions whose potential function is equal to the signed area of the triangle formed by the points A,B,C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V(A,B,C) = B_x C_y - A_x C_y -B_x A_y - C_x B_y + C_x A_y + A_x B_y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all 3 particles are collinear, the forces for this 3-body potential are perpendicular to the common line they lie on. The derivative of the area is maximum by moving the points away from the common line. So you obviously cannot write the force as any sum of pairwise interactions along the line of separation, equal and opposite or not. The forces and torques still add up to zero, since this potential is translationally and rotationally invariant.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Many body direct interaction, spatial reflection symmetry, crappy third law&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;If the force on k particles is reflection invariant, it never gets out of the subspace spanned by their mutual separation. This is because if they lie in a lower dimensional subspace, the system is invariant with respect to reflections perpendicular to that subspace, so the forces must be as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that you can always cook up equal and opposite forces between the particles that add up to the total force, and pretend that these forces are physically meaningful. This allows you to salvage Newton's third law, in a way. But it gives nonsense forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that this is nonsense, consider the three-particle triangle area potential from before, but this time take the absolute value. The result is reflection invariant, but contains a discontinuity in the derivative when the particles become collinear. Near collinearity, the forces perpendicular have a finite limit. But in order to write these finite forces as a sum of equal and opposite contributions from the three-particles, you need the forces between the particles to diverge at collinearity.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Three body interactions are natural&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;There is natural physics that gives such a three-body interaction. You can imagine the three bodies are connected by rigid frictionless struts that are free to expand and contract like collapsible antennas, and a very high-quality massless soap bubble is stretched between the struts. The soap bubble prefers to have less area according to its nonzero surface tension. If the dynamics of the soap bubble and the struts are fast compared to the particles, you can integrate out the soap bubble degrees of freedom and you will get just such a three-body interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the reason the bodies snap together near collinearity with a finite transverse force is clear--- the soap bubble wants to collapse to zero area, so it pulls them in. It is then obvious that there is no sense in which they have any diverging pairwise forces, or any pairwise forces at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other cases where you get three body interactions directly is when you have a nonlinear field between the three objects, and the field dynamics are fast. Consider a cubically self-interacting massive scalar field (with cubic coupling $\lambda$) sourced by classical stationary delta-function sources of strength g. The leading nonlinear contribution to the classical potential is a tree-level, classical, three-body interaction of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V(x,y,z) \propto g^3 \lambda \int d^3k_1 d^3k_2 { e^{i(k_1\cdot (x-z) + k_2\cdot(y-z))} \over (k_1^2 + m^2) (k_2^2 + m^2)((k_1+k_2)^2 + m^2)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which heuristically goes something like ${e^{-mr_{123}}r_{123}\over r_{12}r_{23}r_{13}}$ where the r's are the side lengths of the triangle and $r_{123}$ is the perimeter (this is just a scaling estimate). For nucleons, many body potentials are significant.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The forces from the crappy third law are not integrable&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;If you still insist on a Newton's third law description of three-body interactions like the soap bubble particles, and you give a pairwise force for each pair of particles which adds up to the full many-body interaction, these pairwise forces cannot be thought of as coming from a potential function. They are not integrable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example of the soap-bubble force makes it clear--- if A,B,C are nearly collinear with B between A and C, closer to A, you can slide B away from A towards C very very close to collinearity, and bring it back less close to collinear. The A-B force is along the line of separation, and it diverges at collinearity, so the integral of the force along this loop cannot be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force is still conservative of course, it comes from a three-body potential after all. This means that the two-body A-B force plus the two-body B-C force is integrable. It's just that the A-C two body force is not. So the separation is completely silly.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Absence of multi-body interactions for macroscopic objects in empty space&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The interactions of macroscopic objects are through contact forces, which are necessarily pairwise since all other contacts are far away, and electromagnetic and gravitational fields, which are very close to linear at these scales. The electromagnetic and gravitational forces end up being linearly additive between pairs, and the result is a potential of the form Landau and Lifschitz consider--- pairwise interactions which are individually rotationally invariant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for close packed atoms in a crystal, there is no reason to ignore 3-body potentials. It is certainly true that in the nucleus three-body and four-body potentials are necessary, but in both cases you are dealing with quantum systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I don't think the third law is particularly fundamental. As a philosophical thing, that nothing can act without being acted upon, it's as valid as any other general principle. But as a mathematical statement of the nature of interactions between particles, it is completely dated. The fundamental things are the conservation of linear momentum, angular momentum, and center of mass, which are independent laws, derived from translation invariance, rotational invariance, and Galilean invariance respectively. The pair-wise forces acting along the separation direction are just an accident.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-17T07:18:10.420" />
  <row Id="13658" PostTypeId="2" ParentId="13652" CreationDate="2011-08-17T07:39:21.560" Score="0" ViewCount="" Body="&lt;p&gt;Have a look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Index_of_refraction&quot; rel=&quot;nofollow&quot;&gt;index of refraction,&lt;/a&gt; which is the ratio of the velocity of light in  vacuum  to the one in the medium. Light travels more slowly in a medium and thus &quot;glances off&quot; and changes direction, as in the drawing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;n = speed of light in a vacuum / speed of light in medium&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For example, the refractive index of water is 1.33, meaning that in a vacuum, light travels 1.33 times as fast as it does in water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I should add that &lt;a href=&quot;http://en.wikipedia.org/wiki/Refraction&quot; rel=&quot;nofollow&quot;&gt;refraction is observed&lt;/a&gt; in all types of waves, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Refraction is described by Snell's law, which states that the angle of incidence θ1 is related to the angle of refraction θ2 by&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;sin(θ1)/sin(θ2)=v1/v2=n2/n1&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;where v1 and v2 are the wave velocities in the respective media, and n1 and n2 the refractive indices. In general, the incident wave is partially refracted and partially reflected&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-08-17T13:03:18.270" LastActivityDate="2011-08-17T13:03:18.270" CommentCount="4" />
  <row Id="13659" PostTypeId="2" ParentId="13652" CreationDate="2011-08-17T08:13:17.133" Score="0" ViewCount="" Body="&lt;p&gt;What you are describing is &lt;a href=&quot;http://en.wikipedia.org/wiki/Refraction&quot; rel=&quot;nofollow&quot;&gt;Refraction&lt;/a&gt;. Light changes speed when it crosses the interface between two substances of different densities. How much the speed changes depends on the difference in densities. The larger the change in speed, the more light bends. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Index_of_refraction&quot; rel=&quot;nofollow&quot;&gt;refractive index&lt;/a&gt; is calculated using only the densities of the interfacing substances. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry I can only explain &quot;what&quot; happens not &quot;why&quot; it happens. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your other questions: Light does not know where it is going so it doesn't matter if it wants to get there faster...it can't. Light merely responds to environmental stimuli--in this case to the oblique interface with a substance of another density. &lt;/p&gt;&#xA;" OwnerUserId="2843" LastEditorUserId="2843" LastEditorDisplayName="" LastEditDate="2011-08-17T08:30:56.337" LastActivityDate="2011-08-17T08:30:56.337" />
  <row Id="13660" PostTypeId="1" CreationDate="2011-08-17T08:18:49.120" Score="1" ViewCount="40" Body="&lt;p&gt;consider a block of mass, m, attached to a spring. Let it oscillate at a frequency &quot;f&quot;. Now each part of the spring is in SHM. so this means a wave is propagating through this spring.&#xA;Can this wave be reflected at the fixed end of the spring resulting in the formation of standing waves?&lt;/p&gt;&#xA;" OwnerUserId="4812" LastActivityDate="2011-08-17T09:16:07.690" Title="oscillations and wave" Tags="&lt;harmonic-oscillator&gt;" AnswerCount="1" />
  <row Id="13661" PostTypeId="2" ParentId="13652" CreationDate="2011-08-17T08:28:57.513" Score="0" ViewCount="" Body="&lt;p&gt;Light is an electromagnetic wave. electric field and magnetic field oscillations are perpendicular to each other. They keep on generating each other. The direction of propagation of an EM wave  is given by the Poynting Vector which is a vector in a direction perpendicular to both the directions of vibration of electric field as well as magnetic field. Now when light goes from one medium to another, say, from air to glass, what happens is that the E- field polarizes the atoms of glass. Now as the E field is oscillating the polarizing effect is also oscilating , that is , it changes direction of induced dipole moments. so The positive and negative charges inside the glass atoms now also oscillate and generate their own electric and magnetic field. so the net electric and magnetic field at any instant will be the resultant of the fields produced by the original source as well as the fields produced by the charges inside atoms of glass.Hence as the directions of E- field and B- field change ,the corresponding poynting vector also changes. Hence the direction of light is changed. &lt;/p&gt;&#xA;" OwnerUserId="4812" LastActivityDate="2011-08-17T08:28:57.513" CommentCount="5" />
  <row Id="13662" PostTypeId="1" CreationDate="2011-08-17T08:31:32.407" Score="1" ViewCount="50" Body="&lt;p&gt;Can there be a charge configuration in space such  that at any instant of time I can change the electric field at one and only one point?&lt;/p&gt;&#xA;" OwnerUserId="4812" LastActivityDate="2011-08-17T09:27:01.167" Title="One point to change electric field" Tags="&lt;electromagnetism&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13663" PostTypeId="2" ParentId="13660" CreationDate="2011-08-17T09:16:07.690" Score="3" ViewCount="" Body="&lt;p&gt;Well, the reflection of a wave at the end happens &lt;em&gt;always&lt;/em&gt;. One can picture this by imagining the succesive atoms being pushed off the equilibrium position as the wave propagates. Since the endpoint is fixed, it has nowhere to be pushed but the few atoms near it (I am considering idealized linear chain for simplicity) that have already being perturbed will, after having passed through equilibrium again, pass into the opposite direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For transversal waves (as those you have on strings of a guitar) this means that the wave perturbation will change from &quot;up&quot; to &quot;down&quot; at the end (and vice versa) while for the longitudinal waves (as those in your spring) there is a change from &quot;compressed&quot; to &quot;streched&quot; (and vice versa).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/bgeEE.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-08-17T09:16:07.690" CommentCount="4" />
  <row Id="13664" PostTypeId="2" ParentId="13662" CreationDate="2011-08-17T09:27:01.167" Score="2" ViewCount="" Body="&lt;p&gt;The electric field obeys Maxwell's equations and in particula the Gauss's law. This means that $\rm div E \sim \rho$. If you change the field at only one point, this will introduce an infinite divergence and consequently infinite density. In other words, you'd have to introduce a point charge. But then the field $E$ itself will diverge at that point. So the answer is no, you can't do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the allowed configurations are continuous except at boundaries of objects (where $E$ has to jump to account for charges in the material) and at point charges (where the field diverges).&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-08-17T09:27:01.167" />
  <row Id="13665" PostTypeId="2" ParentId="13650" CreationDate="2011-08-17T09:40:55.900" Score="2" ViewCount="" Body="&lt;p&gt;The concept of entanglement still applies in QFT. What the Reeh-Schlieder theorem tells you is that there is entanglement in the QFT vacuum state. So I take the question to be asking whether we could use the methods employed in proving the theorem, to decide that there is entanglement in biological systems. It seems possible, though first you have to find a biological system that you can describe in terms of quantum field theory. Let's take a much-hyped example, the microtubule. Just for argument's sake, suppose that one of the various possibilities listed in &lt;a href=&quot;http://dx.doi.org/10.1016/S0303-2647%2801%2900183-6&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; made sense, and that there was an effective field theory describing the dynamics of delocalized electrons in the shell of the microtubule. Then maybe you could try to prove the Reeh-Schlieder property for the ground state of that effective theory at biological temperatures, &lt;a href=&quot;http://arxiv.org/abs/hep-th/0001154&quot; rel=&quot;nofollow&quot;&gt;along these lines&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1486" LastActivityDate="2011-08-17T09:40:55.900" CommentCount="1" />
  <row Id="13666" PostTypeId="1" AcceptedAnswerId="13677" CreationDate="2011-08-17T12:57:41.227" Score="0" ViewCount="59" Body="&lt;p&gt;I was having a discussion with a friend about human tolerances of g-force. He believed that the maximum human tolerance of negative g-force is low, in the order of .5 g. I countered, saying that you're able to stand on your head and do not die. An interesting discussion ensued but there was one thing that we couldn't figure out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any difference between the negative g-force when you are standing on your head (i.e. you're stationary on the ground, just upside down, experiencing gravity) and negative g-force when you're in motion? For example, like when you are riding a roller coaster that has a bump in the track, and you're forced up from your seat?&lt;/p&gt;&#xA;" OwnerUserId="4951" LastActivityDate="2011-08-17T17:37:55.993" Title="Negative g-force: difference between gravity and centrifugal force?" Tags="&lt;gravity&gt;&lt;centrifugal-force&gt;" AnswerCount="2" CommentCount="0" />
  <row Id="13667" PostTypeId="1" AcceptedAnswerId="13669" CreationDate="2011-08-17T13:35:42.507" Score="1" ViewCount="87" Body="&lt;p&gt;Often in engineering physics, different vector spaces are used to visualize the trajectories (evolution) of systems.  An example being the 6n dimensional phase space of n particles. It is not very clear to me if this space is Euclidean or has a meaning of distance. I understand the 2-norm when the physical quantities are the same (e.g., only positions or only momenta) but not when a vector has both. However even this seems to be a problem with some other spaces.  For example if I consider the vector space of chemical composition of chemical system (mole numbers of all chemical species). I don't understand what &quot;length&quot; of a vector would signify?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="4952" LastActivityDate="2011-08-18T11:12:13.060" Title="Does the phase space (configuration and momentum space) of particles have a Euclidean norm? Does it have a useful meaning of &quot;distance&quot;?" Tags="&lt;classical-mechanics&gt;&lt;mathematical-physics&gt;&lt;mechanics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" />
  <row Id="13668" PostTypeId="2" ParentId="13666" CreationDate="2011-08-17T13:41:53.153" Score="2" ViewCount="" Body="&lt;p&gt;'g-force' (or acceleration) is actually a vector quantity.  Therefore a negative g-force is identical to a positive g-force in the opposite direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is however a cluttered way of thinking about the situation.  I recommend drawing a few free body diagrams to clear up the physical situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am deliberately avoiding giving a more complete answer in lack of a more detailed question.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-08-17T13:41:53.153" />
  <row Id="13669" PostTypeId="2" ParentId="13667" CreationDate="2011-08-17T14:19:08.037" Score="4" ViewCount="" Body="&lt;p&gt;The spaces you have in mind are in general not vector spaces. But it still makes sense to ask whether there is a useful concept of length around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, phase space (the 6n-dimensional space of positions and momenta) is a vector space if the underlying position space of single particle itself is. More generally, the particles would be constrained to move only on some subspace (e.g. a sphere) and the phase space would be cotantgent bundle of the $n$-fold product of the single-particle subspace. More generally still, the total position space need not have a product structure (which can happen if position of some particle depends on position of others). Again, the full space would be a cotangent bundle to the total position space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the phase space doesn't support a canonical notion of distance. Nevertheless, it does contain something else that is special: a symplectic two-form $\omega$ (the reason it's there is that phase space is always formed as a cotangent bundle) which turns it into symplectic manifold. Using $\omega$, one can form a $2N$-form (N here is the dimension of the total position space) $\omega^n = \underbrace{\omega \wedge \cdots \wedge \omega}_{n \text{ times}}$. Since this is a form of a maximal degree we can use it to measure volumes and this leads to famous Liouville's theorem that states that Hamiltonian flows on the symplectic manifold are incompressible (i.e. evolution conserves phase volume).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all the structure we get for phase spaces. You can impose a metric on any manifold to turn it into a Riemannian manifold but this won't be canonical (i.e. there are many different ways to measure distances). In case your phase space happens to be a simple $6n$-dimensional vector space, you can use the Euclidean notion of a distance but, again, this won't be canonical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, there is no canonical way to measure things and you simply shouldn't be expecting to be able to do so unless you are explictly given something nice like Euclidean space or a submanifold thereof.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-08-17T14:19:08.037" CommentCount="5" />
  <row Id="13670" PostTypeId="2" ParentId="13652" CreationDate="2011-08-17T14:28:59.120" Score="1" ViewCount="" Body="&lt;p&gt;Your teacher's explanation, as you describe it, seems a little disengenuous. It is true that light goes from A to B in the shortest possible time, given different speeds in different media. From a human perspective, it almost seems like the light rays are calculating all their possible path options and picking the most optimal. The real question is, what is the physical mechanism that allows light to perform what appears to be the act of a rational being with infinite powers of calculation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers to this question lie entirely within the domain of classical wave theory. There is no further insight to be gained by looking into the mysteries of quantum mechanics. Primeczar gives the bare outline of an explanation wherein the polarized charges within the glass are taken as the source of new fields. It's actually simpler to do the calculation if you simply accept that the propagation speed in glass is lower than air, and focus on the excess polarization that appears only on the surface of the glass, not within the body. It is then unnecessary to determine the amount of polarization, only to recognize that it must have some kind of constant phase relationship with the driving force of the incoming wave. The angle of refraction follows unambigously from this purely geometrical argument.&lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-08-17T14:28:59.120" CommentCount="6" />
  <row Id="13671" PostTypeId="2" ParentId="13652" CreationDate="2011-08-17T14:46:14.093" Score="1" ViewCount="" Body="&lt;p&gt;The teacher was trying to explain &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermat%27s_principle&quot; rel=&quot;nofollow&quot;&gt;Fermat principle&lt;/a&gt; which is one of the simplest variatinonal (least action) principles of classical physics. And your question seems to express to a  common frustration over the seemingly &quot;teleological&quot; property of all variational principles: how does the (partilce, wave, ray ...) &quot;know&quot; in advance which path to follow? Wikipedia article on the principle of least action specifically &lt;a href=&quot;http://en.wikipedia.org/wiki/Principle_of_least_action#Apparent_teleology&quot; rel=&quot;nofollow&quot;&gt;addresses this frustration&lt;/a&gt; (but does not give valid references, unfortunately).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The laws of propagation are local in time and space but it is sometimes easier to deduce their outcomes using non-local mathmatical constructions such as a Fermat principle. The teacher gave a great analogy to explain the principle but should not expect toget a step-by-step picture out of it.&lt;/p&gt;&#xA;" OwnerUserId="1739" LastEditorUserId="1739" LastEditorDisplayName="" LastEditDate="2011-08-17T14:54:10.960" LastActivityDate="2011-08-17T14:54:10.960" />
  <row Id="13672" PostTypeId="2" ParentId="12611" CreationDate="2011-08-17T15:07:46.070" Score="0" ViewCount="" Body="&lt;p&gt;There is already a detailed and correct derivation, in my answer I can try to address the qualitative side of &quot;why&quot;. In a scattering problem, there is always a hierarchy of well-separated scales. In your example of an alpha particle in Rutherford experiment, you refer to localization in space which means a certain spread in the momentum/energy. However, as long as this spread is smaller than the characteristic energy scale on which the scattering amplitudes changes, the time-independent at well-defined energy  should give correct results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of lengths this scale separation required for the time-independent picture to work is that the wave-packet of the alpha particle should be larger that the neighbourhood of the nucleus where the scattering happens.  Typically, this is the case -- if it is not, the alpha particle is likely to have very uncertain (in Heisenberg sense) energy/momentum.&lt;/p&gt;&#xA;" OwnerUserId="1739" LastActivityDate="2011-08-17T15:07:46.070" />
  <row Id="13673" PostTypeId="2" ParentId="13648" CreationDate="2011-08-17T15:38:40.957" Score="8" ViewCount="" Body="&lt;p&gt;There is a limit to how small you can focus an ideal single-mode laser beam. The product of the divergence half-angle $\Theta$ and the radius $w_0$ of the beam at its waist (narrowest point) is constant for any given beam. (This quantity is called the &lt;em&gt;beam parameter product&lt;/em&gt;, and is related to the $M^2$ beam quality measure you may have heard of.) For an ideal Gaussian (&quot;diffraction-limited&quot;) beam, it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Theta w_0 = \lambda/\pi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to answer what I interpret as your main question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let's say that I have a laser beam of some given power that starts with some diameter $D_0$ at the point of emission and increases to $D_f$ at some distance $r$ away. Would this be sufficient information to imply a limit to the power per unit area (W/m^2) that could be obtained through focusing and what would that be?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is &lt;em&gt;no&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parameters you have given are sufficient for calculating $\Theta$, but only if $r$ is large enough so that the points at which you measure the diameter are in each other's far field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would also need to know the beam radius at the waist, so you could calculate the beam parameter product. Then, to get the minimum spot size, you would need to refocus the beam so that it is maximally convergent. The absolute limit is the fictitious divergence half-angle of $\pi/2$, or 90 degrees, although in practice the theory breaks down for half-angles of more than 30 degrees (this number is from Wikipedia) since the paraxial approximation stops being valid. For an ideal beam at this impossible opening half-angle, this gives you a minimum waist radius of $2\lambda/\pi^2$. So yes, it does depend on the wavelength.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What lens characteristics and approaches would someone look for in order to do this with a laser pointer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You need a lens with a very short focal length. This gives you the largest convergence. Note that the more convergent the beam, and the smaller the waist size, the smaller the Rayleigh range is. That is, the beam radius will get very small, but it won't &lt;em&gt;stay&lt;/em&gt; very small, it'll get bigger very quickly as you move away from the focus. (The Rayleigh range is the distance over which the beam radius increases by $\sqrt{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, thinking of a Gaussian beam as being &quot;straight&quot; is not quite correct. There is always a waist, always a Rayleigh range less than infinity, and always a nonzero divergence angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it is important to realize that there is &lt;em&gt;no difference&lt;/em&gt; between an unfocused and a focused Gaussian beam. Refocusing a Gaussian beam with a lens just moves and resizes the waist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aperture size of the laser is &lt;em&gt;not&lt;/em&gt; the same as the waist size. If the beam is more or less collimated, then the aperture will still be larger, because the waist radius is usually defined in terms of the radius at which the intensity drops to $1/e^2$ of its peak value. If the beam is cut off by an aperture at that radius, then even if it were close to diffraction-limited, it certainly wouldn't be anymore. So, apertures are always larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The waist is the thinnest point of the beam. Usually this point is inside the laser cavity, or outside the laser if there are focusing optics involved, which there often are. So still, the answer to your question is no. You are not missing the definition of $\lambda$; rather, you are comparing your minimum waist radius to the value of $2\lambda/\pi^2$ that I said was &quot;impossible&quot;. I called it impossible, because to make a beam converging that strongly, you would need a lens with a focal length of zero!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's try a more realistic example with some numbers. Take your red laser pointer with $\lambda$ = 671 nm. Laser pointer beams are often crappy, but not so crappy as you might think, if they are single-mode. Let's assume that this particular laser pointer has an $M^2$ (&quot;beam quality parameter&quot;, which is the beam parameter product divided by the ideal beam parameter product of $\lambda/\pi$) of 1.5. A quick Google search didn't give me typical $M^2$s of red laser pointers, but this doesn't seem to me to be too much off the mark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if you know the $M^2$ and measure the divergence of a beam, then you can calculate the waist radius. We are going to do that now. Suppose the laser pointer beam is nearly collimated: you measure a divergence of 0.3 milliradians, about 0.017 degrees. Then the waist size is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ w_0 = \frac{M^2 \lambda} {\pi\Theta} = \frac{1.5 \times 671 \times 10^{-9}} {\pi \times 3 \times 10^{-4}} \approx 1\,\text{mm}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, they probably designed the laser pointer with an aperture radius of 2 or 3 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose you focus your collimated beam with a 1 cm focal length positive lens, which is quite a strong lens. The beam's new waist will be at the lens's focal length. That means you can calculate the divergence half-angle: it is the smaller acute angle of a right triangle with legs 1 mm and 10 mm. So,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan\Theta = 1/10,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or $\Theta\approx$ 6 degrees. Applying the formula once more to calculate the waist yields a waist radius of 3.2 microns, which is quite small indeed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;safe&quot; laser pointer might have a power of 1 mW. The &lt;a href=&quot;http://www.rp-photonics.com/spotlight_2009_10_03.html&quot; rel=&quot;nofollow&quot;&gt;peak intensity&lt;/a&gt; is equal to $2P/\pi w_0^2$, so before the lens the peak intensity is about 600 W/m^2. After the lens it is about 100000 times larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to summarize:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;yes, there is a fundamental limit to the intensity, and it does depend on the wavelength, but you cannot even come close with a real-world laser.&lt;/li&gt;&#xA;&lt;li&gt;you need to know two of any of these quantities: divergence half-angle, waist radius, Rayleigh range, beam parameter product.&lt;/li&gt;&#xA;&lt;li&gt;really, the minimum size and maximum intensity depend quite heavily on what optics you use and how good they are.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="107" LastEditorUserId="107" LastEditorDisplayName="" LastEditDate="2011-08-17T22:39:10.627" LastActivityDate="2011-08-17T22:39:10.627" CommentCount="3" />
  <row Id="13674" PostTypeId="2" ParentId="13652" CreationDate="2011-08-17T16:26:22.090" Score="0" ViewCount="" Body="&lt;p&gt;this reminds me of &lt;a href=&quot;http://v.cx/2010/04/feynman-brazil-education&quot; rel=&quot;nofollow&quot;&gt;http://v.cx/2010/04/feynman-brazil-education&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4953" LastEditorUserId="4953" LastEditorDisplayName="" LastEditDate="2011-08-17T16:44:46.520" LastActivityDate="2011-08-17T16:44:46.520" />
  <row Id="13675" PostTypeId="1" AcceptedAnswerId="13683" CreationDate="2011-08-17T16:28:19.737" Score="2" ViewCount="93" Body="&lt;p&gt;I was studying a &lt;a href=&quot;http://en.wikipedia.org/wiki/GRE_Physics_Test&quot; rel=&quot;nofollow&quot;&gt;GRE Physics Test&lt;/a&gt; problem where optical light with a wavelength of 500&amp;nbsp;nm travels through a gas with refractive index $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at the equations for wave motion and index of refraction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c=\lambda_0\nu\quad\text{(in vacuum)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = \lambda\nu\quad\text{(in medium)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n = c/v$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we see that, if the frequency is constant, the &lt;strong&gt;&lt;em&gt;wavelength decreases in the medium compared to vacuum&lt;/em&gt;&lt;/strong&gt;. Is this a consistent property at all frequencies and for all mediums with refractive index real and greater than 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there dielectrics which change the frequency (still for &lt;code&gt;n&lt;/code&gt; &gt; 1), and is there an example of that?&lt;/p&gt;&#xA;" OwnerUserId="4570" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-17T21:54:31.377" LastActivityDate="2011-08-18T04:04:19.387" Title="Does the wavelength always decrease in a medium?" Tags="&lt;electromagnetism&gt;&lt;light&gt;&lt;wavelength&gt;" AnswerCount="3" CommentCount="4" />
  <row Id="13676" PostTypeId="2" ParentId="13016" CreationDate="2011-08-17T17:15:58.450" Score="2" ViewCount="" Body="&lt;p&gt;Only those photons that travel in the same spatial mode are used, &#xA;and these are located at the intersection of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Spontaneous_parametric_down-conversion&quot; rel=&quot;nofollow&quot;&gt;cones in which the down-converted photons&lt;/a&gt; can be found. As you can see, &lt;img src=&quot;http://i.stack.imgur.com/vqCzD.jpg&quot; alt=&quot;enter image description here&quot;&gt; momentum and energy conservation imply that the colors at the intersection points  are equal. &#xA;&lt;a href=&quot;http://www.tongue-twister.net/mr/physics/entangled.htm&quot; rel=&quot;nofollow&quot;&gt;Quoting&lt;/a&gt;: &quot;Along the intersections of the cones of the same wavelength (in our photograph the green circles) polarization-entangled photon states can be observed.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1739" LastActivityDate="2011-08-17T17:15:58.450" CommentCount="1" />
  <row Id="13677" PostTypeId="2" ParentId="13666" CreationDate="2011-08-17T17:37:55.993" Score="2" ViewCount="" Body="&lt;p&gt;You are correct that standing on your head is the same as a G-force of -1 G.  That is what Einsteins equivalence principle tells us.    Your friend is correct that tolerance for negative Gs is lower, but it is around -3 G not -.5 G.  Tolerance for positive vertical Gs is around 5 Gs without G suits or training.  Tolerance for horizontal Gs varies from 12 to 17 Gs depending on whether the acceleration is &quot;eyeballs out&quot; or &quot;eyeballs in&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-08-17T17:37:55.993" />
  <row Id="13678" PostTypeId="2" ParentId="13675" CreationDate="2011-08-17T17:50:40.287" Score="1" ViewCount="" Body="&lt;p&gt;As for your second question, dielectrics which change the frequency, any medium which changes the frequency must be a nonlinear medium.  There are many of them.  They do not change the frequency continuously, but rather they generate higher order harmonics, an integral number times the original frequency.  &lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-08-17T17:50:40.287" />
  <row Id="13679" PostTypeId="2" ParentId="13616" CreationDate="2011-08-17T17:56:04.097" Score="1" ViewCount="" Body="&lt;p&gt;The notion &quot;complete integral&quot; here refers to solutions of specific (1st order) PDEs that depend on the maximal number of constants of motion. If you want a concrete example I can refer you to Equation (10) of &lt;a href=&quot;http://arxiv.org/abs/arXiv:0711.4115&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;, or even better to Ref. [10] in that paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;general solution&quot;, by contrast, need not depend explicitly on constants of motion, but usually contains some free (integration) function. As an example for a solution not depending explicitly on constants of motion see the &quot;enveloping solution&quot; in Eq. (11) of the paper above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[I never encountered these notions anywhere, except when solving Hamilton-Jacobi equations - this seems to be also the context to which Landau refers.]&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2011-08-17T17:56:04.097" CommentCount="2" />
  <row Id="13680" PostTypeId="2" ParentId="13445" CreationDate="2011-08-17T18:10:48.070" Score="3" ViewCount="" Body="&lt;p&gt;Cosmic microwave background data, particularly from WMAP, are &lt;a href=&quot;http://lambda.gsfc.nasa.gov/&quot; rel=&quot;nofollow&quot;&gt;publicly available&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-08-17T18:10:48.070" />
  <row Id="13681" PostTypeId="2" ParentId="13445" CreationDate="2011-08-17T18:22:17.593" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://simbak.cfa.harvard.edu/simbad/sim-fid&quot; rel=&quot;nofollow&quot;&gt;SIMBAD&lt;/a&gt; - Portal for data (and Tools) from several astronomic catalogs &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-08-17T18:22:17.593" />
  <row Id="13682" PostTypeId="2" ParentId="10763" CreationDate="2011-08-17T18:37:15.200" Score="1" ViewCount="" Body="&lt;p&gt;Assuming that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;by &quot;a perfect DFT solution&quot; you mean the set of Kohn-Sham orbitals for the exact (presently unknown) denistiy functional that existance of which Hohenberg and Kohn have proven;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;you are asking whether the individual energies of these &quot;perfect&quot; Kohn-Sham orbitals can be used as estimates of exact  ionization and/or excitation energies,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;the answer is &lt;strong&gt;yes, but only approximately&lt;/strong&gt; with the sign of error depending on the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The DFT method with the perfect functional will give exactly the total energy (that's the functional) and  the single-electron properties &lt;strong&gt;of the ground state only&lt;/strong&gt;. As far as I know, there are no exact DFT theorems for excited states. As for the estimation of the ionisation energy, the answer is not exact because the perfect DFT solution for (n-1) electrons will give different orbitals than the the perfect DFT solution of n electrons. Thus the difference between the two perfect total energies will not be equal to any particular orbital energy. &lt;/p&gt;&#xA;" OwnerUserId="1739" LastActivityDate="2011-08-17T18:37:15.200" />
  <row Id="13683" PostTypeId="2" ParentId="13675" CreationDate="2011-08-17T18:46:50.590" Score="1" ViewCount="" Body="&lt;p&gt;The index of refraction of a material can be less than 1 at high frequency, this is called &quot;anomalous dispersion&quot; and it happens as you cross an energy level of certain materials. It means that the phase velocity of light of a certain frequency is higher than c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the index of refraction is constant, as it is for long wavelengths, n has to be bigger than 1 to avoid superlumimal communication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principle of energy conservation in a static environment forbids a frequency shift for a photon, since this would add energy or take away energy, and nothing in the medium is changing with the right frequency to do that. But light entering a moving medium shifts frequency. photons can combine to make one of double the frequency in a strong light beam in a nonlinear medium, and this corresponds to making higher harmonics of the classical field. &lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-17T18:46:50.590" />
  <row Id="13684" PostTypeId="1" AcceptedAnswerId="13685" CreationDate="2011-08-17T19:19:58.873" Score="2" ViewCount="48" Body="&lt;p&gt;Today I had a discussion with a colleague who teaches electricity and magnetism to 2nd year undergraduate physics students. He is seeking the best way to explain how is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electromotive_force#Chemical_sources&quot; rel=&quot;nofollow&quot;&gt;emf generated inside a battery&lt;/a&gt; with a minimal appeal to physics beyond classical. We have lamented that some textbooks refer to &quot;non-electromagnetic chemical forces&quot; since all of chemistry is essentially electrostatics+quantum mechanics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our proposal is to draw the students' attention to the existence of atoms which can not be explained by classical mechanics + classical electromagnetism. In the same vein the forces on charge carriers in galvanic cells are electromagnetic but the response is not classical. Thus a battery &quot;amplifies&quot; non-classicality to the level macroscopic electricity. It is not that forces are non electrostatic but the systems response is not classical. Can you recommend textbooks or online sources that use/expand this idea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the main tag for this question should be &quot;teaching&quot; but I'm too much of a rookie here to create one)&lt;/p&gt;&#xA;" OwnerUserId="1739" LastEditorUserId="2843" LastEditorDisplayName="" LastEditDate="2011-08-17T19:53:49.837" LastActivityDate="2011-08-17T19:58:03.780" Title="Introducing emf of a chemical cell as a hint towards quantum mechanics" Tags="&lt;quantum-mechanics&gt;&lt;electricity&gt;&lt;education&gt;&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="13685" PostTypeId="2" ParentId="13684" CreationDate="2011-08-17T19:51:44.633" Score="2" ViewCount="" Body="&lt;p&gt;What you're describing is electrochemistry, which is summarised &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrochemistry&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Electrochemistry is a branch of chemistry that studies chemical reactions which take place in a solution at the interface of an electron conductor (a metal or a semiconductor) and an ionic conductor (the electrolyte), and which involve electron transfer between the electrode and the electrolyte or species in solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You don't need to delve into quantum mechanics to understand what's going on at the macro level since this branch of science has been around since Faraday, and so basic chemistry will suffice. On the other hand, the inner workings of chemistry are down to non-relativistic quantum mechanics and the solution of Schrodinger's equation which determines how electrons behave within atoms and with other electrons. Here's a link that develops a classical picture of the physical processes involved which is easily accessible to your students:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chem1.com/acad/webtext/elchem/&quot; rel=&quot;nofollow&quot;&gt;http://www.chem1.com/acad/webtext/elchem/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1345" LastActivityDate="2011-08-17T19:51:44.633" CommentCount="1" />
  <row Id="13686" PostTypeId="2" ParentId="13684" CreationDate="2011-08-17T19:52:51.350" Score="0" ViewCount="" Body="&lt;p&gt;Each chemical species has a &lt;a href=&quot;http://chempaths.chemeddl.org/services/chempaths/?q=book/General%20Chemistry%20Textbook/Reactions%20in%20Aqueous%20Solutions/1445/redox-couples&quot; rel=&quot;nofollow&quot;&gt;specific strength&lt;/a&gt; of oxidation and/or reduction. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Redox&quot; rel=&quot;nofollow&quot;&gt;Redox potential&lt;/a&gt; is what chemists use to calculate the electrode potential of a battery:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E°cell = E°red(cathode) – E°red(anode) = E°red(cathode) + E°oxi(anode)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia has a nice &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_electrode_potential_%28data_page%29&quot; rel=&quot;nofollow&quot;&gt;compilation of Redox potentials&lt;/a&gt;, which are used in the above formula to calculate the voltage produced by a electrochemical cell. &lt;/p&gt;&#xA;" OwnerUserId="2843" LastEditorUserId="2843" LastEditorDisplayName="" LastEditDate="2011-08-17T19:58:03.780" LastActivityDate="2011-08-17T19:58:03.780" />
  <row Id="13687" PostTypeId="2" ParentId="13556" CreationDate="2011-08-17T19:55:23.957" Score="3" ViewCount="" Body="&lt;p&gt;The asymmetry comes from the different masses of electrons and neon ions (neon ions are about 36000 times heavier). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This mass asymmetry results in different cross sections for the excitation of neon atoms by electrons and ions. There are some plots of this here &lt;a href=&quot;http://webhost.ua.ac.be/plasma/pdf_papers/sab97comparison.pdf&quot; rel=&quot;nofollow&quot;&gt;http://webhost.ua.ac.be/plasma/pdf_papers/sab97comparison.pdf&lt;/a&gt; (figure 1a for electrons, figure 1b for ions). The interesting processes of excitation occur above around 10eV for electrons and 100eV for ions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a very cool Java simulation of discharges: &lt;a href=&quot;http://phet.colorado.edu/en/simulation/discharge-lamps&quot; rel=&quot;nofollow&quot;&gt;http://phet.colorado.edu/en/simulation/discharge-lamps&lt;/a&gt; . It covers electron excitation, the atomic structure of neon, acceleration and excitation cross-sections very well, and demonstrates why the glow can be localized (it glows at the point that electrons have been accelerated to the necessary energy), and why it is asymmetric (all electrons start from the cathode and can gain energy on the way to the anode). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this answers why only one side can glow. I wish I could explain exactly why the glow is next to the cathode in the picture, but I would just be guessing. &lt;/p&gt;&#xA;" OwnerUserId="4956" LastActivityDate="2011-08-17T19:55:23.957" CommentCount="1" />
  <row Id="13688" PostTypeId="1" AcceptedAnswerId="13690" CreationDate="2011-08-17T20:13:29.400" Score="4" ViewCount="133" Body="&lt;ol&gt;&#xA;&lt;li&gt;Is there a simple intuitive way to understand the difference between left-handed and right-handed fermions (electrons say)? &lt;/li&gt;&#xA;&lt;li&gt;How to experimentally distinguish between them?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4521" LastEditorUserId="4521" LastEditorDisplayName="" LastEditDate="2011-08-17T20:31:20.387" LastActivityDate="2011-08-22T18:22:16.337" Title="Left and Right-handed fermions" Tags="&lt;quantum-field-theory&gt;&lt;particle-physics&gt;&lt;experimental-physics&gt;&lt;standard-model&gt;" AnswerCount="2" />
  <row Id="13689" PostTypeId="2" ParentId="5760" CreationDate="2011-08-17T20:40:12.367" Score="3" ViewCount="" Body="&lt;p&gt;Let me try to identify a generic reason for why are impurity-free condensed matter systems likely to be boring. A condensed matter system without defects:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;has a perfect spatial symmetry, so its low energy excitations can be described by an effective field theory, regularized at atomic energies by the actual lattice.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the correspondng vaccuum is &quot;empty&quot;, so there is not too much space for macroscopically observable qualitatively different phenomena unless they involve excitation on the cut-off (lattice) level. But such excitaions are nothing but defects/microscopic disorder/inhomogeneity that we want to avoid.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It is similar to QFT in the usual (God-given) vacuum: continuous symmetries put a strong restriction on the type of possible particles &amp;amp; interactions. &lt;/p&gt;&#xA;" OwnerUserId="1739" LastEditorUserId="1739" LastEditorDisplayName="" LastEditDate="2011-08-17T21:07:06.100" LastActivityDate="2011-08-17T21:07:06.100" />
  <row Id="13690" PostTypeId="2" ParentId="13688" CreationDate="2011-08-17T20:59:35.847" Score="6" ViewCount="" Body="&lt;p&gt;For the first question, no there can be no such general rule. The reason is the same why there is no intuitive way to understand the difference between your right and left shoe -- they are just reflections of each other but neither one is more fundamental (assuming you are not a pirate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, all the difference (that really matters) comes from the given theory. If the theory is $P$-symmetric (such as electromagnetism) you have no way to distinguish between these (that's why you haven't heard anyone talking about right- or left-handed electrons in the classical physics courses; there were just electrons). On the other hand, if the theory violates $P$-symmetry you have a chance since one type of particles will interact differently than the other. E.g. weak interactions violate the $P$-symmetry so that only left-handed electrons and left-handed neutrinos can form a weak isospin doublet (and so can right-handed antiparticles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will leave the experimental part of the question to someone more educated on these matters. But e.g. for neutrinos you might be able to measure helicity (which is as good as chirality since neutrinos are almost massless) which is a projection of spin on to the momentum. You know the momentum direction (since you know where the collision happened and where you registered the particle) and I suppose there are standard methods for measuring spin projection to arbitrary direction (but I admit complete ignorance on these experimental matters).&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-08-17T20:59:35.847" CommentCount="3" />
  <row Id="13691" PostTypeId="1" CreationDate="2011-08-17T20:59:56.827" Score="1" ViewCount="43" Body="&lt;p&gt;What is difference in transverse-plus, transverse minus, and longitudinal polarization of spin 1 particle, and how are this related to its three spin projections states? What is difference in spin 1/2 case? If spin 1/2 particle has two projections of spin (i.e two transverse), why it is often longitudinal one introduced, and what is its meaning?&lt;/p&gt;&#xA;" OwnerUserId="4107" LastActivityDate="2011-08-17T21:40:45.997" Title="Transverse-plus, transverse minus, and longitudinal polarization of spin 1 particle" Tags="&lt;quantum-field-theory&gt;" AnswerCount="1" />
  <row Id="13692" PostTypeId="2" ParentId="13030" CreationDate="2011-08-17T21:18:10.220" Score="3" ViewCount="" Body="&lt;p&gt;A fluid only transmits forces through pressure. If an airfoil generates lift, this must be because the top is at a lower pressure than the bottom in the steady-state flow. The amount of lift can be understood by the rate of deflection of air downward, but the mechanism of lift is always high-pressure on the bottom and low pressure on the top. Low pressure regions speed up a fluid, and high pressure regions slow it down, just because the fluid is doing work to enter the high pressure region and has work done upon it to enter the low pressure region. Therefore the velocity at the top is higher than at the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed at the top and the bottom do not guarantee that the transit time is the same, however, so the explanation that the airfoil lifts &lt;em&gt;because&lt;/em&gt; the air above is moving faster is incorrect. But the low-pressure and high-pressure regions exist. Note that if you make the angle of attack of an airfoil such that the air going past it is deflected upward, the pressure at the bottom is less and at the top is more, so the air at the top is moving slower.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-27T07:41:29.947" LastActivityDate="2011-08-27T07:41:29.947" />
  <row Id="13693" PostTypeId="2" ParentId="13691" CreationDate="2011-08-17T21:40:45.997" Score="0" ViewCount="" Body="&lt;p&gt;Well, first thing to note is that components of a field of arbitrary spin $j$ have nothing to do with space (e.g. there are not $2j + 1$ independent directions), so that your discussion of spin 1/2 just doesn't make any sense here. The reason we can interpret spin 1 field as vectors is that it transforms not only in the usual $SU(2)$ representation but also in the $SO(3)$ adjoint three-dimensional representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since the vector has three components we can decompose it into arbitrary base of three vectors and it is obviously convenient to choose one vector in the longitudinal direction and the other two in the transverse direction. Since you explicitly ask, let me show you how this is done. Denote the longitudinal direction as the $z$-axis. Then (as always when dealing with spin) we can require that the basis of the representation be composed of eigenstates of the $L_z$ operator with eigenvalues $-1$, $0$ and $1$ so that the matrix $D_z$ is diagonal in this basis. Matrices $D_x$ and $D_y$ can be determined from the relation of $L_x$ and $L_y$ to ladder operators $L_{\pm} = L_x \pm i L_y$ acting on the $L_z$ eigenstates. This is the standard representation $\mathbf D$. Now, there is an unitarily equivalent adjoint representation of $SO(3)$ given by the matrices $M^i_{jk} = i \epsilon_{ijk}$. It is now a routine exercise to find a unitary matrix that transforms the representation $\mathbf D$ to the representation $\mathbf M$. Doing this is in reverse, this is also how you pass from the standard orthonormal $x,y,z$ components of the field to the basis of the $-1, 0, 1$ polarization states.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-08-17T21:40:45.997" CommentCount="15" />
  <row Id="13694" PostTypeId="1" AcceptedAnswerId="13751" CreationDate="2011-08-17T23:14:04.320" Score="6" ViewCount="93" Body="&lt;p&gt;I've been looking for a satisfying proof of this, and can't quite find it. I read the brief proof of the black hole area theorem in Wald, which is similar, but doesn't quite come down to the actual statement $\frac{dA}{dt} \geq 0$. I had pictured something fairly succinct relating $T_{ab}k^ak^b \geq 0$ (weak energy condition) directly to $\frac{dA}{dt}$, and I tried but I couldn't get anything to quite work out. Anyway-  can anyone point me to a good source, or even just give the proof?&lt;/p&gt;&#xA;" OwnerUserId="4663" LastActivityDate="2011-08-19T08:17:10.477" Title="Second Law of Black Hole Thermodynamics" Tags="&lt;general-relativity&gt;&lt;thermodynamics&gt;&lt;black-holes&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="13695" PostTypeId="1" CreationDate="2011-08-17T23:45:26.573" Score="12" ViewCount="128" Body="&lt;p&gt;There are lots of different ways of arriving at the relativistic relations involving mass, energy, and momentum such as $E=mc^2$ and $m^2=E^2-p^2$ (the latter with $c=1$). One that I've seen in some textbooks is to start with the equation $W=\int F dx$ for mechanical work. Here is a particularly careful and rigorous version, which makes it explicit that this is a nontrivial assumption, along with $F=dp/dt$: &lt;a href=&quot;http://www.physicsforums.com/showthread.php?p=2416765&quot; rel=&quot;nofollow&quot;&gt;http://www.physicsforums.com/showthread.php?p=2416765&lt;/a&gt; Historically, Einstein made use of $W=\int F dx$ in section 10 of his 1905 paper on SR &lt;a href=&quot;http://www.fourmilab.ch/etexts/einstein/specrel/www/&quot; rel=&quot;nofollow&quot;&gt;http://www.fourmilab.ch/etexts/einstein/specrel/www/&lt;/a&gt; . This was before the &quot;Does the inertia of a body...&quot; paper, in which he derived $E=mc^2$ and gave it its full relativistic interpretation. Einstein later decided that force wasn't a very useful concept in relativity and stopped appealing to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have never seen in any of these treatments (neither the careful ones above nor the sloppy ones in some freshman physics texts) is any argument as to why the nonrelativistic relation $W=\int F dx$ should be expected &lt;em&gt;a priori&lt;/em&gt; to hold without modification in SR. If one has &lt;em&gt;already&lt;/em&gt; established the form and conservation of the energy-momentum four-vector, then I don't think it's particularly difficult to show that $W=\int F dx$. But what justification is there for assuming $W=\int F dx$ &lt;em&gt;before&lt;/em&gt; any of that has been established? Is there any coherent justification for it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A secondary problem with some of these treatments based on $W=\int F dx$ is that they need to establish the constant of integration.)&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-18T01:59:51.553" Title="a priori validity of $W=\int Fdx$ in relativity?" Tags="&lt;special-relativity&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" />
  <row Id="13696" PostTypeId="2" ParentId="5760" CreationDate="2011-08-17T23:59:33.240" Score="2" ViewCount="" Body="&lt;p&gt;Primary reference: &lt;a href=&quot;http://www.tcm.phy.cam.ac.uk/~bds10/publications/lesh.ps.gz&quot; rel=&quot;nofollow&quot;&gt;http://www.tcm.phy.cam.ac.uk/~bds10/publications/lesh.ps.gz&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disorder is very important in the so-called mesoscopic regime, where the sample size is larger than the coherence length of electrons, and the coherence length is much larger than the Compton wavelength. Roughly speaking, due to the multiple scattering which occurs across the entire sample, transport is a diffusive process, but the scattering centres (i.e. impurities) are dilute and/or weak so that electrons propagate coherently between close ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can imagine trying to compute the &lt;em&gt;probability&lt;/em&gt; of transmission as the square of a propagator/two-point correlation function. Using field integral methods, one finds that the probability corresponds to a Feynmann diagram which contains a path going forwards then backwards. In particular, for the phases to not strongly cancel, the paths need to remain within a Compton wavelength of each other. The introduction of scattering centres gives rise to a quantum phenomenon where one can get a constructive interference by making a loop somewhere in the middle and making the forward and return paths go in the same direction instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The introduction of an intermediate length scale also introduces new effective modes, the most important of which are the &lt;em&gt;diffuson&lt;/em&gt; (the classical diffusive mode) and the &lt;em&gt;cooperon&lt;/em&gt; which is purely quantum mechanical (and corresponds roughly to the description above of forward and backward paths going in the same direction). The field theory then allow you to join these together arbitrarily and resumming the entire series then predicts various interesting phenomenon.&lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-08-17T23:59:33.240" CommentCount="1" />
  <row Id="13697" PostTypeId="2" ParentId="6742" CreationDate="2011-08-18T00:14:29.100" Score="2" ViewCount="" Body="&lt;p&gt;There are three things you might want to do using relativity: (1) describe an object that's accelerating in flat spacetime; (2) adopt a frame of reference, in flat spacetime, that's accelerating; (3) describe curved spacetime. General relativity is only needed for #3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A prohibition on #1 is particularly silly. It would make SR into a trivial theory incapable of describing interactions. If you believed this, you would have to stop believing, for example, in the special-relativistic description of the Compton effect and fine structure in hydrogen; these phenomena would have to be described by some as yet undiscovered theory of quantum gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number 1 often comes up in discussions of the twin paradox. A good way to see that general relativity is totally unnecessary for understanding the twin paradox is to pose a version in which the four-vector equation a=b+c represents the unaccelerated twin's world-line a and the accelerated twin's world-line consisting of displacements b and c. The accelerated twin is subjected to (theoretically) infinite accelerations at the vertices of the triangle. The triangle inequality for flat spacetime is reversed compared to the one in flat Euclidean space, so proper time |a| is greater than proper time |b|+|c|.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number 2, accelerated &lt;em&gt;frames&lt;/em&gt;, is less trivial. It's for historical reasons that you'll see statements that SR can't handle accelerated frames. Einstein published special relativity in 1905, general relativity in 1915. During that ten-year period in between, nobody really knew what the boundaries of applicability of special relativity were. This uncertainty made its way into textbooks and lectures, and because of the conservative nature of education, some students are still hearing, a century later, incorrect assertions about it. There is an overwhelming consensus among modern relativists that the boundary between SR and GR should be defined as the distinction between flat and curved spacetime, not unaccelerated and accelerated observers.[MTW 1973,Penrose 2004,Taylor 1992,Schutz 2009,Hobson 2005]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an accelerating frame, the equivalence principle tells us that measurements will come out the same as if there were a gravitational field. But if the spacetime is flat, describing it in an accelerating frame doesn't make it curved. (Curvature is invariant under any smooth coordinate transformation.) Thus relativity allows us to have gravitational fields in flat space --- but only for certain special configurations like uniform fields. SR is capable of operating just fine in this context. For example, Chung et al. did a high-precision test of SR in 2009 using a matter interferometer in a vertical plane, specifically in order to test whether there was any violation of Lorentz invariance in a uniform gravitational field. Their experiment is interpreted purely as a test of SR, not GR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MTW 1973 -- Misner, Thorne, and Wheeler, Gravitation, 1973, p. 163: &quot;Accelerated motion and accelerated observers can be analyzed using special relativity.&quot; p. 164: &quot;An accelerated observer can carry clocks and measuring rods with him, and he can use them to set up a reference frame (coordinate system) in his neighborhood.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Penrose, The Road to Reality, 2004, p. 422, &quot;It used to be frequently argued that it would be necessary to pass to Einstein's general relativity in order to handle acceleration, but this is completely wrong. [...] We are working in special relativity provided that [the] metric is the flat metric of Minkowski Geometry M.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taylor and Wheeler, Spacetime Physics, 1992, p. 132: &quot;DO WE NEED GENERAL RELATIVITY? NO! [...] 'Don't you need general relativity to analyze events in accelerated reference frames?' 'Oh yes, general relativity can describe events in the accelerated frame,' we reply, 'but so can special relativity if we take it in easy steps!'&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Schutz, A First Course in General Relativity, 2009. Schutz equivocates on pp. 3 and 141 about the status of accelerated observers in SR, but says, &quot;[...] the real physical distinction between these two theories is that special relativity (SR) is capable of describing physics only in the absence of gravitational fields, while general relativity (GR) extends SR to describe gravitation itself.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hobson, General Relativity: An Introduction for Physicists, 2005, sec. 1.14, discusses &quot;Event horizons in special relativity&quot; from the point of view of accelerated observers, using coordinates defined in their accelerated reference frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chung -- &lt;a href=&quot;http://arxiv.org/abs/0905.1929&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/0905.1929&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-18T00:14:29.100" />
  <row Id="13698" PostTypeId="2" ParentId="13648" CreationDate="2011-08-18T00:25:00.503" Score="2" ViewCount="" Body="&lt;p&gt;I'm going to start with a specific example.  &lt;a href=&quot;http://www.spectrum-scientifics.com/cgi-bin/commerce.cgi?preadd=action&amp;amp;key=281&quot; rel=&quot;nofollow&quot;&gt;Here is a laser pointer you can buy online&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cost is $12.50&lt;/li&gt;&#xA;&lt;li&gt;1.0 mRad = 0.05 degrees&lt;/li&gt;&#xA;&lt;li&gt;wavelength: 645 nm&lt;/li&gt;&#xA;&lt;li&gt;output 5 mW&lt;/li&gt;&#xA;&lt;li&gt;beam diameter: 1.1 mm&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, I am having great difficulty with this problem, and I'm still in disagreement with what ptomato has written as well as many other things I find online.  My question, however, is still an objective and definable one, which is more or less &quot;can the above laser cut steel with the appropriate lens?&quot;  Anonymous Coward gave a good reference, which is &lt;a href=&quot;https://www.cvimellesgriot.com/Products/Documents/TechnicalGuide/Gaussian-Beam-Optics.pdf&quot; rel=&quot;nofollow&quot;&gt;a guide to Gaussian optics&lt;/a&gt;.  This document uses 1983 Sidney A. Self, &lt;a href=&quot;http://homepages.cae.wisc.edu/~ssanders/me_770/journal_papers/Self_1983_focusing_gaussian_beams_AWC.pdf&quot; rel=&quot;nofollow&quot;&gt;Focusing of spherical Gaussian beams&lt;/a&gt; as its primary reference.  The equations are the same throughout and are congruent to what you find in the prior Wikipedia links and such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The problem&lt;/em&gt; is that they seem to keep giving equations for a diffraction limited case, saying nothing to clearly identify that's what they're talking about.  In a diffraction limited case &lt;em&gt;of course you need the wavelength&lt;/em&gt;, but my anticipation is that a $12.50 laser will not be diffraction limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, I'll still get down some of the equations here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For starters, the waist radius is denoted $\omega_0$, which is the radius at the $1/e^2$ intensity as pointed out before, and the divergence half-angle is denoted $\Theta$.  For this particular device those values are known.  See page 5 of the guide, it notes &lt;em&gt;the location of the beam waist is typically designed to be close to the output surface of the laser&lt;/em&gt;.  I will be assuming that this is a real Gaussian beam and I thus won't be working with aperture.  Again, Gaussian beam, not diffraction limited, and highly imperfect.  I'll still be forced to take the reported &quot;beam diameter&quot; to be the representative of the $1/e^2$ radius, as well as being at the waist.  This could be fairly wrong, but only by the same factor that my answer will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega_0 = 0.6 mm$$&#xA;$$\Theta = 0.001 rad$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The literature goes into definition of several other values, as well as an expression for $\Theta$ itself.  See page 3 of the guide to get the following equation without the $M^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Theta = \frac{M^2 \lambda}{\pi \omega_0}$$&#xA;$$M^2 = \frac{\Theta \pi \omega_0}{\lambda} = 2.94 \approx 3.0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm okay with this value.  I don't expect it to get a good laser.  Next, the Rayleigh range, $z_R$ is a rough value beyond which the far lens approximation is valid.  I think it might need a $M^2$ on it, but I really don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z_R = \frac{\pi \omega_0^2}{\lambda}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just wanted to throw that out there.  The guide has equations as to how to calculate the lens equations as well as the magnification factor, but I can't figure out how the equations imply that you can't focus this to a finite dot with a sufficiently large lens.  I know this is not possible, so I had no choice but to apply a high school level approach to the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is my illustration of a beam of 100% collimated rays (I'm talking about Newtonian physics here), and the first vertical line is some hypothetical perfect lens, that would converge perfectly on the next vertical line.  If the rays coming in were perfectly collimated then the rays would exactly converge on a single point, but we know they don't because of the divergence angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Q7kE0.png&quot; alt=&quot;laser focusing&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Beam radius is 0.6 mm&lt;/li&gt;&#xA;&lt;li&gt;$2 \Theta = 0.001 rad$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Again, I'm going to act like a high school physics student, pretend I don't even know what wavelength is, and use geometry.  I'll assume the angle the furthest beams are deflected at is 45 degrees, and denote the things about the area it focuses on as $dot$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_{dot} = (\sqrt{2} \omega_0) (\Theta) \sqrt{2} = 0.0011 mm = 1.1 \mu m$$&#xA;$$A_{dot} = 1.21 \mu m^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for power per unit area I'll use $\Phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Phi = \frac{P}{A} = 4.13 \frac{GW}{m^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What temperature could this reach?  I'll just assume dissipation from blackbody radiation for now (back of the envelope).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Phi = \sigma T^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\sigma = 5.67 \times 10^{-8} W m^{-2} K^{-4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T = 16,500 K$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also like to do this for the radial conduction equation.  Maybe later.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-08-18T00:25:00.503" CommentCount="2" />
  <row Id="13699" PostTypeId="1" CreationDate="2011-08-18T01:13:43.420" Score="3" ViewCount="97" Body="&lt;p&gt;In Exercise 2.3 of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0198520743&quot; rel=&quot;nofollow&quot;&gt;A modern introduction to Quantum Field Theory by Michele Maggiore&lt;/a&gt; I am asked to show that, if $\xi_R$ and $\psi_R$ are right-handed spinors, then&#xA;$$ V^\mu = \xi_R^\dagger \sigma^\mu \psi_R$$&#xA;transforms as a four vector. Here, $\sigma^\mu = (1,\sigma^i)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have shown that it does so for boosts along the x-axis by explicitly transforming the two spinors and showing that the components of $V^\mu$ transform correctly. It seems easy to do the same thing for boosts along other directions and rotations around the separate axes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I would like to show this for a general Lorentz-transformation, i.e. I would like to show&#xA;$$&#xD;&#xA; (\Lambda_R \xi_R)^\dagger \sigma^\mu (\Lambda_R \psi_R) = \Lambda_\nu^\mu \xi_R^\dagger \sigma^\nu \psi_R \text{ }\text{ }\text{ }\text{ }\text{ }\text{ (2)}&#xD;&#xA;$$&#xA;Trying to do this explicitly seems even less elegant than what I have done so far. I have tried commuting $\sigma^\mu$ to the right of $\Lambda_R$ in (2), but that gives a rather complicated expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a slightly more elegant way to see why (2) is true without computing all the components explicitly? &lt;em&gt;(I am not looking for a solution, but would rather have a hint!)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4541" LastActivityDate="2011-08-18T03:33:31.200" Title="Four vectors from spinors" Tags="&lt;homework&gt;&lt;spinors&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="13700" PostTypeId="2" ParentId="13445" CreationDate="2011-08-18T01:28:26.317" Score="0" ViewCount="" Body="&lt;p&gt;There are some projects over at Zooniverse that are open to public participation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.zooniverse.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.zooniverse.org/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4210" LastActivityDate="2011-08-18T01:28:26.317" />
  <row Id="13701" PostTypeId="2" ParentId="13695" CreationDate="2011-08-18T01:39:11.293" Score="7" ViewCount="" Body="&lt;p&gt;There is no reason for the formula to obviously work, as you say, just because there are at least two different generalizations of force, the proper-time relativistic curvature of the trajectory ($d^2x\over d\tau^2$ not the right notion) and the derivative of the relativistic energy times the velocity with respect to time (${d\over dt} (m{dx\over d\tau})$ the one that works). Which one is the proper generalization?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Einstein came from a thermodynamic background, and he always accorded the laws of thermodynamics an absolute status, being deduced as they were so inexorably from such elementary postulates. It was the deductive form of thermodynamics that led him to write the special relativity in such a deductive style, building up from simple axioms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Potential energies, and energies in general, have a thermodynamic meaning mostly independent of their mechanical meaning, which is unchanged by relativity (if you like to think of this statistically, the Boltzmann distribution is still $\exp(-\beta E)$ in either theory). The electrostatic potential is the energy per unit charge in a region of space. The electrostatic energy of a charged ball moving with velocity v in an electrostatic potential $\phi$ can't depend on the velocity, because then I could transfer the charge from a fast moving object to a stationary object as the fast object comes whooshing by, and make a perpetual motion machine. So the potential energy of the charge is still $q\phi$ even in relativity, for thermodynamic reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electric field is the gradient of the static potential, and this is true in Maxwell's equations, which are already relativistic. The integral of the gradient of the static potential along the path, times the charge, then gives the work done along the path. So the correct notion of force to use is whatever mechanical thing is equal to $qE$. It is physically all but certain that different forces, no matter what their origin, do the same work if they have the same magnitude, so you can conclude that the work formula is generally true for this notion of force, which is the notion of force in the equation of motion. For Einstein this is so clear, he doesn't bother to explain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This assumption seems unnatural to many people nowadays, since we are comfortable with symmetries much more than thermodynamics. But Einstein's exposition, survives in textbooks to this day.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-18T01:59:51.553" LastActivityDate="2011-08-18T01:59:51.553" CommentCount="5" />
  <row Id="13702" PostTypeId="1" AcceptedAnswerId="13718" CreationDate="2011-08-18T02:06:13.243" Score="6" ViewCount="92" Body="&lt;ol&gt;&#xA;&lt;li&gt;What is a quark condensate? is it a bound state between 2 quarks? can we have 3(or more)-quarks condensate?&lt;/li&gt;&#xA;&lt;li&gt;What mediates the interaction between the constituents of the condensate?&lt;/li&gt;&#xA;&lt;li&gt;Are the constituents of the condensate always fermions?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4521" LastEditorUserId="4521" LastEditorDisplayName="" LastEditDate="2011-08-18T13:32:24.223" LastActivityDate="2011-08-18T16:01:26.593" Title="What is a quark condensate?" Tags="&lt;quantum-field-theory&gt;&lt;particle-physics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13703" PostTypeId="1" AcceptedAnswerId="13706" CreationDate="2011-08-18T02:53:13.787" Score="1" ViewCount="130" Body="&lt;p&gt;It will probably depend on the size of the pieces... but at what point should I stop assuming a linear relationship?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was prompted to ask in &lt;a href=&quot;http://physics.stackexchange.com/questions/13595/where-does-the-excess-energy-emitted-by-a-microwave-go&quot;&gt;a previous question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4935" LastActivityDate="2011-08-18T18:20:11.810" Title="Should I cook two pieces of Canadian Bacon twice as long as one in a microwave?" Tags="&lt;electromagnetic-radiation&gt;&lt;heat&gt;" AnswerCount="5" CommentCount="3" />
  <row Id="13704" PostTypeId="2" ParentId="13699" CreationDate="2011-08-18T03:22:26.473" Score="0" ViewCount="" Body="&lt;p&gt;Did you try to use the fact that the product of two spinors transforms in the same way as a vector? It is fairly general rule (in fact, is it a definition), but to use it for transformation law of Pauli matrices you need carefully work with the indices. &lt;/p&gt;&#xA;" OwnerUserId="4657" LastActivityDate="2011-08-18T03:22:26.473" />
  <row Id="13705" PostTypeId="2" ParentId="13699" CreationDate="2011-08-18T03:33:31.200" Score="1" ViewCount="" Body="&lt;p&gt;You're essentially done, because any proper, orthochronous Lorentz transformation is the product of a rotation and a boost. If you are not familiar with this fact, try looking up the proof of polar decomposition of matrices and you will see that it generalizes easily. If you get stuck, add a comment below and I'll update this answer with more details.&lt;/p&gt;&#xA;" OwnerUserId="4160" LastActivityDate="2011-08-18T03:33:31.200" />
  <row Id="13706" PostTypeId="2" ParentId="13703" CreationDate="2011-08-18T03:53:21.090" Score="3" ViewCount="" Body="&lt;p&gt;My experimental evidence comes from baked potatoes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 potato takes 5 minutes to be at my liking&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 potatoes come out the same at 10 minutes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8 potatoes come out the same at about 35 minutes, i.e. less time per potato.If I let them go to 40 minutes (5x8) they become dry and wizened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total mass plays a role then.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-08-18T03:53:21.090" CommentCount="5" />
  <row Id="13707" PostTypeId="2" ParentId="13675" CreationDate="2011-08-18T04:04:19.387" Score="0" ViewCount="" Body="&lt;p&gt;The answer to your first question &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this a consistent property at all frequencies and for all mediums with refractive index real and greater than 1?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is Yes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only in optics but in other &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Refraction&quot; rel=&quot;nofollow&quot;&gt;wave mechanics too.&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Refraction of light is the most commonly observed phenomenon, but any type of wave can refract when it interacts with a medium, for example when sound waves pass from one medium into another or when water waves move into water of a different depth&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-08-18T04:04:19.387" />
  <row Id="13708" PostTypeId="1" AcceptedAnswerId="13709" CreationDate="2011-08-18T07:10:50.417" Score="5" ViewCount="117" Body="&lt;p&gt;Imagine an astronaut floating in free-space with no significant nearby gravitational influences.  The astronaut takes an arbitrarily thin pole of uniform density with length $l$ and mass $m$, orients it vertically from his perspective, and then positions it some distance $h$ in front of him.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, once the pole is in place, he kicks it with force $F$ at some coordinate in space $p$, at distance $r$ from the pole's center of mass.  Under what conditions will another coordinate on the pole's contour pass through coordinate $p$ to collide with the astronaut's foot?&lt;/p&gt;&#xA;" OwnerUserId="4257" LastEditorUserId="4257" LastEditorDisplayName="" LastEditDate="2011-08-18T08:01:04.367" LastActivityDate="2011-08-18T18:12:15.733" Title="An astronaut and a vengeful pole" Tags="&lt;newtonian-mechanics&gt;&lt;geometry&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13709" PostTypeId="2" ParentId="13708" CreationDate="2011-08-18T09:22:47.923" Score="3" ViewCount="" Body="&lt;p&gt;The $h$ serves no purpose and could as well be 0. The 'kick' is best described with impulse $I$, not with 'force', albeit the actual impulse is irrelevant. It is important though that astronaut is not moving back recoiling from the kick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is: Where-ever he kicks, pole doesn't intersect that point again, meaning, it won't hit the astronaut's foot (but it may still hit astronaut on the head; perhaps a revised problem could be made about astronaut's head, or about a rod that has extra point mass attached in the middle).&#xA;To cut down on writing $r/l$ and the like everywhere, I normalize it by defining my own units so that $l=2 , m=1, I=1$ , my $r$ is your $2r/l$ , and my $r$ is in range 0..1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = I/m = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w = \frac{Ir}{\text{moment of inertia}} = \frac{Ir}{m(l^2)/12} = 3r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let coordinate system be centered on the centre of the pole before kick; $p_x=0, p_y=-r$ &#xA;and equation of motion of other end of pole is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=t-\sin(t3r)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=\cos(t3r)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some time before the intersection, the $y$ has to be equal $p.y$ and $x$ has to be be &amp;lt;0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let's find $x$ when $y=p.y$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t3r=\arccos(-r)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t=\arccos(-r)/(3r)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=\arccos(-r)/(3r)-\sin(\arccos(-r))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the rod would subsequently intersect $p$ if $x&amp;lt;0$ , but if we plot it, it never goes below 0 for $r$ in range 0..1 .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=plot+acos%28-r%29%2F%283%2ar%29-sin%28acos%28-r%29%29&quot; rel=&quot;nofollow&quot;&gt;Plot of the graph on wolfram alpha&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: sorry for excessive use of paragraphs, the form deletes newlines and the javascript is bugging out. edit2: ohh, that was noscript blocking some stuff, got that solved now. Forgot to define I, improved clarity a bit.&lt;/p&gt;&#xA;" OwnerUserId="4960" LastEditorUserId="-1" LastEditorDisplayName="" LastEditDate="2011-08-18T18:12:15.733" LastActivityDate="2011-08-18T18:12:15.733" CommentCount="3" />
  <row Id="13710" PostTypeId="2" ParentId="13667" CreationDate="2011-08-18T10:38:44.693" Score="3" ViewCount="" Body="&lt;p&gt;There are cases where a metric on a phase space is needed\useful. I'll describe to you two cases having very interesting physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, the example given in the question is an example of the first case. Please notice that an Eucledian metric of the form $p^2 + q^2$ can be interpretted as &#xA;the total (kinetic + potential) energy of a harmonic oscillator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More rigorously, when the phase space is a Kahler manifold, there are compatible metric, closed symplectic form and a complex structure:&#xA;(A Hermitian manifold is a complex manifold equipped with a tensor $ h = h_{\alpha \bar{\beta}}dz^{\alpha}d\bar{z}^{\beta}$, such that the matrix elements $ h_{\alpha \bar{\beta}}$ are the elements of a Hermitian matrix).The symplectic form $\omega = \mathrm{Im}(h)$ is the imaginary part of the Hermitian form, the metric $g = \mathrm{Re}(h)$ is the real part and the complex structure $J$, satisfies $\omega (X,Y) = g(JX,Y)$ ($X$, $Y$ are vector fields). ( A Kahler manifold is a Hermitian manifold with a closed two form $\omega$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider the example of the Eucledian two dimensional phase space, Here, we can define $z = x+ip$, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega  = dz \wedge d\bar{z}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g = \frac{1}{2} (dz \otimes d\bar{z} + d\bar{z})  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These types of spaces admit holomorphic or Kahler quantization, in our example, the quantization space is the Bargmann space of $L^2$ holomorphic functions: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\langle \psi, \phi \rangle = \int \overline{\psi(z)} \phi(z) \mathrm{exp}(-\bar{z} z) dz d\bar{z}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please observe again that the argument of the exponential is minus the Eucledian distance on the phase space. This example is sometimes referred to as the coherent state quantization of the harmonic oscillator.&#xA;Here the energy eigenstates are just the monomials in $z$. One can return to the standard representation by the means of &lt;a href=&quot;http://planetmath.org/encyclopedia/BargmannTransform.html&quot; rel=&quot;nofollow&quot;&gt;the (inverse) Bargmann transform&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second application where  a metric on a phase space is needed is when ones needs to put fermions (spinors) on the phase space (In this case one needs a spin structure as well).&#xA;One easy way to see that is throug the anti-commutation relation of the Gamma matrices on a general (curved) manifold&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\{\gamma^{\mu},\gamma^{\nu}\} = g^{\mu\nu}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the Dirac operator requires a metric. Dirac operators on phase spaces are used because their zero modes provide a way to define a quantization Hilbert space.&#xA;In the above example, if we merge the Gaussian factor of the integration into the definition f wave function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\psi ^{\prime} (z, \bar{z}) = \psi(z) exp(\frac{-\bar{z}z}{2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hilbert space would not contain the Gaussian weighting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\langle \psi, \phi \rangle = \int \overline{\psi^{\prime}(z, \bar{z})} \phi^{\prime}(z, \bar{z}) dz d\bar{z}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but now the wavefunctions are not holomorphic but rather satisfy the Dirac equation (in a uniform magnetic field):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\bar{\partial} + \frac{z}{2})\psi = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an almost trivial example of the use of the Dirac operator in quantization. For a good reference on the role of the metric on phase spaces and its connection to Dirac operators and quantization,&#xA;please see the following article: &quot;Quantization of systems with a general phase space equipped with a Riemannian metric&quot; by &lt;a href=&quot;http://iopscience.iop.org/0305-4470/29/10/025&quot; rel=&quot;nofollow&quot;&gt;Alicki and Klauder&lt;/a&gt; (sorry that I could not find an open version).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worthwhile to mention that the requirement of spin structure doesn't constitute a big constraint in the case of Kahler manifolds, since they can be equipped with a spin-c structure with witch a Dirac operator can be defined.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorUserId="2190" LastEditorDisplayName="" LastEditDate="2011-08-18T11:12:13.060" LastActivityDate="2011-08-18T11:12:13.060" CommentCount="2" />
  <row Id="13711" PostTypeId="1" AcceptedAnswerId="13717" CreationDate="2011-08-18T14:18:43.297" Score="4" ViewCount="113" Body="&lt;p&gt;I'm solving a  problem involving a Fermi gas. There is a specific sum I cannot figure my way around. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set of equidistant levels, indexed by $m=0,1,2 \ldots$,  is populated by spinless fermions with population numbers $\nu_m =0 $ or $1$. I need to compute the following sum over the set of all possible configurations $\{  \nu_l \}$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$Q(\beta,\beta_c) = \sum_{\{  \nu_l \}} \sum_{l} \prod_m \exp({\beta_c \, l \, \nu_l}-{ [ \beta \, m + i \phi] \, \nu_m} )$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any hints on how to deal with this are appreciated. This is not homework, it is a research problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that $\beta &amp;gt;0$,  $\beta_c&amp;gt;0$, and $\phi \in [0; 2 \pi ]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: corrected with the complex phase (the sum is coming from a generating function)&lt;/p&gt;&#xA;" OwnerUserId="1739" LastEditorUserId="1739" LastEditorDisplayName="" LastEditDate="2011-08-19T07:40:35.213" LastActivityDate="2011-08-19T16:54:00.667" Title="Combinatorial sum in a problem with a Fermi gas" Tags="&lt;mathematical-physics&gt;&lt;mathematics&gt;&lt;fermions&gt;" AnswerCount="2" FavoriteCount="0" />
  <row Id="13712" PostTypeId="2" ParentId="13703" CreationDate="2011-08-18T14:20:07.267" Score="1" ViewCount="" Body="&lt;p&gt;Normally the heat losses increase with increase of the surface so you may have to cook even longer if the free surface is doubled. 8 potatoes put close to each other may have smaller free surface and get cooked faster than 5x8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ah, if you stuff your MW oven with too much of food that needs high temperature for cooking, then your temperature may become so low that cooking will take much more time.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-08-18T16:15:53.530" LastActivityDate="2011-08-18T16:15:53.530" CommentCount="2" />
  <row Id="13713" PostTypeId="2" ParentId="13703" CreationDate="2011-08-18T14:46:04.813" Score="2" ViewCount="" Body="&lt;p&gt;To some approximation the whole cavity can be thought of as full of microwave radiation at some fixed power density. (Of course, we all know this approximation is weak 'cause some food has hot and cold spots, right?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, adding a second separate piece of food should have no effect, and you should use the same time for two pieces as for one. See the part of &lt;a href=&quot;http://physics.stackexchange.com/questions/13703/should-i-cook-two-pieces-of-canadian-bacon-twice-as-long-as-one-in-a-microwave/13706#13706&quot;&gt;anna's answer&lt;/a&gt; about two potatoes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this relies on the magnetron being able to &lt;em&gt;keep&lt;/em&gt; the cavity flooded at a constant energy density even while the food is absorbing some of the power. It may be that the power will sag, in which case it will take a bit longer; that's the &quot;many potatoes&quot; part of &lt;a href=&quot;http://physics.stackexchange.com/questions/13703/should-i-cook-two-pieces-of-canadian-bacon-twice-as-long-as-one-in-a-microwave/13706#13706&quot;&gt;anna's answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a long time my rule of thumb was for $N$ pieces multiple the time by $1.5^{N-1}$, but I have begun to wonder if the base should be lower with my current machine.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Additional consideration: if your &quot;extra&quot; food is stacked or placed close together parts of it may shield other parts which almost certainly calls for more cooking time (perhaps at lower power).&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-18T15:28:17.383" LastActivityDate="2011-08-18T15:28:17.383" CommentCount="1" />
  <row Id="13714" PostTypeId="1" AcceptedAnswerId="13724" CreationDate="2011-08-18T15:03:53.830" Score="4" ViewCount="224" Body="&lt;p&gt;Does anyone have any suggestions as to what is a good topic for a short talk on theoretical physics to a bunch of Math and Physics undergrads that might make them &quot;embrace&quot; theoretical physics? Thanks. (Brain storm: String theory, quantum tunneling, etc...?) Also it would be great if someone could suggest some resources.&lt;/p&gt;&#xA;" OwnerUserId="4964" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-30T18:32:10.893" LastActivityDate="2011-08-30T22:03:35.080" CommunityOwnedDate="2011-08-18T18:27:34.003" Title="Embrace Physics" Tags="&lt;soft-question&gt;&lt;education&gt;" AnswerCount="6" CommentCount="9" FavoriteCount="1" />
  <row Id="13715" PostTypeId="1" AcceptedAnswerId="13720" CreationDate="2011-08-18T15:16:34.487" Score="2" ViewCount="93" Body="&lt;p&gt;Some experts (and non experts) have called LHC the most powerful microscope in the world. I am wondering, what does the LHC magnify? Or is calling it a microscope just wrong and misleading?&lt;/p&gt;&#xA;" OwnerUserId="4521" LastActivityDate="2011-08-18T18:58:58.203" Title="If LHC is the most powerful microscope in the world, what does it magnify?" Tags="&lt;particle-physics&gt;&lt;experimental-physics&gt;&lt;high-energy-physics&gt;&lt;large-hadron-collider&gt;" AnswerCount="2" />
  <row Id="13716" PostTypeId="1" CreationDate="2011-08-18T15:26:49.020" Score="0" ViewCount="70" Body="&lt;p&gt;How is it possible to find the shortest possible rotation period of a pulsar from a mass and a radius?&lt;/p&gt;&#xA;" OwnerUserId="4965" LastActivityDate="2011-08-19T17:43:42.907" Title="Finding shortest possible rotation period" Tags="&lt;astrophysics&gt;" AnswerCount="1" />
  <row Id="13717" PostTypeId="2" ParentId="13711" CreationDate="2011-08-18T15:33:28.043" Score="3" ViewCount="" Body="&lt;p&gt;This is not an answer just some thoughts from playing with the expression. I've read the question before you included the phase, so for now let $\phi = 0$ (sorry it this makes my response useless for you).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll simply write $Z$ instead of your $Q(\beta, \beta_c)$ and also drop the arguments where obvious. Denote by $Z_{abc\dots}$ the partition function where we do not include the sites at $a, b, c, \dots$ in the problem. Also denote $f_k = 1 + \exp(-\beta k)$ and $g_k = 1 + \exp((\beta_c - \beta) k)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now (unless I screwed up), by summing over the site at $k$ we can get the relation&#xA;$$Z = f_kZ_k + g_k \sum_{\nu \setminus k} \prod_{m \neq k} \exp(-\beta m \nu_m) $$&#xA;and iterating it&#xA;$$Z = \left( \prod_{m \in abc\dots z} f_m \right) Z_{abc\dots z} + $$&#xA;$$ \left(g_a f_b \dots f_z + f_a g_b \dots f_z + \cdots + f_a f_b \dots g_z)  \right) \sum_{\nu \setminus abc\dots z} \prod_{m \neq abc \dots z} \exp(-\beta m \nu_m).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a simple observation that for the reduces system consisting of a single level $a$ we get $Z_{bc \ldots z} = g_a$ so the first term above gives a similar contribution like the other terms (all but one factors are $f$ and one of them is $g$). Therefore, we can write&#xA;$$Z = \left( \prod_{m} f_m \right) \left ( \sum_k \frac{g_k} { f_k} \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These expressions are exact in case we have finite number of states. Otherwise they are just formal and are to be understood as limits only if everything converges.&lt;/p&gt;&#xA;" OwnerUserId="329" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-08-19T13:38:09.877" LastActivityDate="2011-08-19T13:38:09.877" CommentCount="4" />
  <row Id="13718" PostTypeId="2" ParentId="13702" CreationDate="2011-08-18T15:44:25.920" Score="4" ViewCount="" Body="&lt;p&gt;The quark condensate is the vacuum expectation $\langle 0|\bar{\psi}\psi|0\rangle$ responsible for the breakdown of the chiral symmetry in QCD.&#xA;Its importance in QCD is by being one of the main contributrs to the Shifman, Vainshtein, Zakharov sum rule, please see for example equation  30 in the following&#xA;&lt;a href=&quot;http://arxiv.org/abs/hep-ph/0010175&quot; rel=&quot;nofollow&quot;&gt;review article&lt;/a&gt; by: Colangelo and Khodjamirian on QCD sum rules. It's value was already known in the pre-QCD era, please see equation 47 in the same article.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorUserId="329" LastEditorDisplayName="" LastEditDate="2011-08-18T16:01:26.593" LastActivityDate="2011-08-18T16:01:26.593" CommentCount="1" />
  <row Id="13719" PostTypeId="2" ParentId="13703" CreationDate="2011-08-18T16:11:22.707" Score="1" ViewCount="" Body="&lt;p&gt;Let's assume that heat losses are relatively insignificant comparing to heating power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1:Microwave oven's power is likely to stay constant, but it's efficiency may increase with more bacon slices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2:Roughly speaking, it will take time $t_1$ to heat up the food to 100 celsius , typically followed by time $t_2$ for the food to cook to whatever condition you find tasty (of course in practice it will also cook some during the heating up, and the $t_2$ may be zero). Doubling the power per bacon will only cut down the $t_1$ two times, but it won't cut down $t_2$ (and at the end of heating the bacon will be less cooked). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of those things suggest that microwave oven will need less than twice the time for 2 bacons, even if heating time is twice as long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the bacon is considered cooked by the time it reaches 100 celsius in both cases the relationship will be closer to linear.&lt;/p&gt;&#xA;" OwnerUserId="4967" LastActivityDate="2011-08-18T16:11:22.707" CommentCount="1" />
  <row Id="13720" PostTypeId="2" ParentId="13715" CreationDate="2011-08-18T16:23:47.633" Score="4" ViewCount="" Body="&lt;p&gt;Well, it is a microscope if the word is understood in a generalized sense that it allows us to learn about microscopic structure of some stuff. To understand the connection one should first recall how good old optical microscopes work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principle is really quite simple, you throw some stuff at the object you are interested in and observe what happens. In the old setting, you throw photons (i.e. light) which then come back to you and you record where they landed (obviously you also need to use some optical methods to actually magnify stuff). This then shows you the structure of the microscopic matter. But at certain point this stops working since visible light has wavelength of roughly $500 \,\rm nm$ so it can't really detect anything smaller than this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to increase resolution, you start throwing smaller things (with smaller wavelength or equivalently bigger energy) at stuff. So you try electrons and neutrons, even photons, but this time with shorter wavelength (e.g. gamma rays). But as you come down to this minute level of atomic and subatomic scales, quantum properties will manifest strongly and the good old approach of throwing photons and seeing what comes back stops working.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At quantum level you can only predict probabilities. E.g. that if you hit some object with a photon, you can calculate the probability that certain particle (e.g. pion) will be created and fly off into the given direction. There is really no way to tell beforehand. Also, at these energies new particles will be abundantely created since you get to energies above the mass of those particles. Therefore at these scales (and LHC is very deeply in these scales), microscope really means collider.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-08-18T16:23:47.633" />
  <row Id="13721" PostTypeId="2" ParentId="13716" CreationDate="2011-08-18T17:02:28.227" Score="1" ViewCount="" Body="&lt;p&gt;One simple method is to consider a particle of the star at the surface on the equator, this particle will feel two principal forces: a centrifugal force, $F_c$ generally acting to pull the particle off the surface and a gravitational $F_g$ (and strong nuclear?) force holding the particle to the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $F_c$ &gt; $F_g$ then the particle will tend to leave the surface, if not then the particle will stay put.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I suspect this is a homework question I'll leave the detailed mathematics out of it for now!&lt;/p&gt;&#xA;" OwnerUserId="2535" LastEditorUserId="379" LastEditorDisplayName="" LastEditDate="2011-08-19T17:43:42.907" LastActivityDate="2011-08-19T17:43:42.907" CommentCount="4" />
  <row Id="13722" PostTypeId="1" AcceptedAnswerId="13726" CreationDate="2011-08-18T18:18:17.363" Score="0" ViewCount="61" Body="&lt;p&gt;Forgive me if this is a newb question but I am not a trained scientist, much less a physicist. I'm just curious and would like to know if there's a unit to measure impact force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the newton is the amount of force required to accelerate a mass of one kg at a rate of 1m/s^2 but what about the impact itself?.&lt;/p&gt;&#xA;" OwnerUserId="4968" LastActivityDate="2011-08-19T03:16:12.960" Title="Unit for Impact Force" Tags="&lt;forces&gt;&lt;si-units&gt;&lt;units&gt;" AnswerCount="2" CommentCount="5" />
  <row Id="13723" PostTypeId="2" ParentId="13703" CreationDate="2011-08-18T18:20:11.810" Score="1" ViewCount="" Body="&lt;p&gt;Bacon in the microwave oven is a real-world problem that does not reduce easily to a physical model. It is not clear what cooking bacon in the microwave actually means. Does it have to reach a certain internal temperature? Does it need be exposed to some above threshold temperature for a certain period of time to kill bacteria? Or maybe it needs to rapidly absorb a certain amount of energy per gram to break down the fibers and soften up? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be easier if you asked about (and/or experimented with) boiling up water in the microwave. Water boils when it reaches 100 degrees Celsius, and that is very clear.  In this way, effects of cooling, feedback of load on efficiency, etc., can be isolated and investigated in detail. You can experiment with boiling up twice as much water in a glass vs boiling up two glasses of water (test the impact of cooling). You can try to boil up 1 cup vs 1 gallon (correct for the effect of cooling and learn if the extra load affects the power output). To avoid superheated water, you need to stop every minute and stir (see &lt;a href=&quot;http://www.youtube.com/watch?v=1_OXM4mr_i0&quot; rel=&quot;nofollow&quot;&gt;Superheated water on Mythbusters&lt;/a&gt;). And you must use fresh tap water every time so it contains the same amount of air.&lt;/p&gt;&#xA;" OwnerUserId="4450" LastActivityDate="2011-08-18T18:20:11.810" CommentCount="2" />
  <row Id="13724" PostTypeId="2" ParentId="13714" CreationDate="2011-08-18T18:43:21.073" Score="7" ViewCount="" Body="&lt;p&gt;Well, I can tell you what got &lt;em&gt;me&lt;/em&gt; into physics. Already at high school I have been fascinated by quantum theory, general relativity and string theory (mostly thanks to Hawking's popular books) and this hasn't gone away. One might argue that these three huge areas is already all there is to theoretical physics but only at the university did I discover fascinating subject of statistical physics and condensed matter theory which should certainly also been mentioned (of course, there are great many other areas; by not mentioning them I certainly do not imply they are not important or interesting). Now, so much for general talk, let's look at some concrete stuff.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nature of the space-time and gravity explained in completely geometrical terms (already geodesics on curved spaces are very entertaining).&lt;/li&gt;&#xA;&lt;li&gt;Anything about black holes. What happens near them, the fact that you can gain energy from rotating hole using Penrose's construction, what happens when you cross the ring singularity, etc. Also, black hole thermodynamics and evaporation could be mentioned (and if one feels up to it they can also mention the holographic principle and all the recent advances in AdS/CFT that connect this area with nuclear physics).&lt;/li&gt;&#xA;&lt;li&gt;Particle physics. Some Feynman diagrams, the most important particles, the way forces work through particle interaction and some important results such as confinement, explanation of radioactive decay through exchange of massive bosons, etc. The possibilities are endless and I know people love this because it is both very fundamental and also understandable since you can explain most of it by just drawing few wavy lines :)&lt;/li&gt;&#xA;&lt;li&gt;Phase transitions. Explaining them from the microscopic point of view, talking about spontaneous symmetry breaking, etc. There is already lots of interesting stuff going on with quite ordinary stuff (see this &lt;a href=&quot;http://www.youtube.com/watch?v=yBRdBrnIlTQ&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt;) but one can also discuss the phase diagram of QCD. Statistical physics permeats all physics since it's what you need to get from fundamental laws to observable phenomena in any area.&lt;/li&gt;&#xA;&lt;li&gt;Some condensed matter systems. Electronic properties of matter, band gaps, phonons, etc.&lt;/li&gt;&#xA;&lt;li&gt;Lasers. They are just cool :)&lt;/li&gt;&#xA;&lt;li&gt;Plasma (either EM or nuclear). Discussion of basic properties, instabilities, reasons Tokamaks are complicated.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, all of the above topics are great but as whole lecture that &lt;em&gt;I&lt;/em&gt; personally would find most intriguing (at the undergraduate level), I would welcome the historical account of force unifications: E+M -&gt; EM, weak force, strong force, electroweak, GUT, supersymmetry, string theory interleaved with introduction of SR, GR and quantum theory as they came along in the above succession. This might be bit too much for a single talk though :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the string theory, I am not so sure. It will surely have to stay strictly at popular level since the audience will have (I presume) no grasp of GR or QFT. One can mention Calabi-Yaus and the way they relate to the string spectrum. Dualities are a huge topic but again it might not be easy to explain what is dual to what. I guess you can look for inspiration into the Brian Greene's Elegant Universe.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-08-18T18:43:21.073" CommunityOwnedDate="2011-08-18T18:43:21.073" />
  <row Id="13725" PostTypeId="2" ParentId="13714" CreationDate="2011-08-18T18:47:06.540" Score="1" ViewCount="" Body="&lt;p&gt;The physics of the early universe?  Not my field but a version of theoretical astrophysics/cosmology...  Inflation, multi-universe type stuff?&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-08-18T18:47:06.540" CommunityOwnedDate="2011-08-18T18:47:06.540" />
  <row Id="13726" PostTypeId="2" ParentId="13722" CreationDate="2011-08-18T18:55:01.647" Score="0" ViewCount="" Body="&lt;p&gt;I would refer you to a previous answer I gave, here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/13576/hammer-vs-large-mass-on-nail/13593#13593&quot;&gt;Hammer vs large mass on nail&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'll try to answer very simply for right here only.  If you are referring to one free-moving objecting colliding and bouncing off another object then we can go about listing all the variables that would be sufficient to describe the approach/impact.  Also, we are likely to consider one of the objects involved in the impact as the &quot;lab&quot; frame, just so we don't get into the complication of a 2-body problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say a projectile is flying toward the ground, a wall, a nail, or whatever solid surface you want.  The projectile has mass $m$ and velocity $v$.  That can then give you &lt;a href=&quot;http://en.wikipedia.org/wiki/Impulse_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;impulse&lt;/a&gt; $I$ (but notation varies), and energy $E$.  Either of these 2 sets of variables are sufficient to describe the nature of the projectile aside from material-specific concerns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Into material specific, if one thing hits another, we can think of it like a spring, where $k$ is the functional constant for &lt;a href=&quot;http://en.wikipedia.org/wiki/Hooke%27s_law&quot; rel=&quot;nofollow&quot;&gt;Hooke's Law&lt;/a&gt;, which is really the combined spring constant of the two objects.  Again, if one item is at rest (or we use &lt;a href=&quot;http://en.wikipedia.org/wiki/Reduced_mass&quot; rel=&quot;nofollow&quot;&gt;reduced mass&lt;/a&gt;), then we can find the time spent in contact before the projectile bounces back off, this is $t=\pi \sqrt{m/k}$.  You should read up on Wikipedia and other resources for more specifics about the mechanics, but the force $F$ is not constant and follows a sinusoid profile over time (given the prior assumptions).  For the force experienced in the impact:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F \propto \frac{I}{t} = \frac{m v}{\pi} \sqrt{\frac{k}{m}} = \frac{v}{\pi} \sqrt{m k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, this exact quantity would be the average force.  For the maximum force, I believe it would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{max} = \frac{\pi}{2} F_{avg} = 2 v \sqrt{m k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although it's hard for me to think of something that would more appropriately be the &quot;impact force&quot;, you still need to refine your question.  It seems like some of the confusion is about units, and the is no single unit that will quantify the &quot;force&quot; or the &quot;damage&quot; from an impact, as it depends on a number of factors.  Although the above expression only has 3 variables, a real life collision is much more complicated.  The area and contour that the force is distributed over matters a great deal, and the assumption that the materials behave like a spring is wrong too, since plastic deformation and other things that can occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps you to advance your thinking on the subject.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I guess I'd be able to conceive the problem more accurately if I rephrased like this: what would the threshold maximum structural integrity of an object need to be in order to still be completely crushed between two surfaces, not counting their own material properties?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That would be kinetic energy on the projectile side and the integral of the stress-strain curve on the side of the crushed object side.  Imagine you have a rock flying toward a hard place, and in the middle is an object that may or may not get crushed.  What matters is the stress strain curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/0oYHa.png&quot; alt=&quot;rod&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The object will be conceptualized as a rod as above.  The integral is illustrated here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/KL68s.png&quot; alt=&quot;integral&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the rod has some area that the projectile hits.  The following determines if it is crushed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{2} m v^2 &amp;lt; A \int_0^{\epsilon_y} \sigma(\epsilon) d\epsilon$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is met it won't be crushed.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-08-19T03:16:12.960" LastActivityDate="2011-08-19T03:16:12.960" CommentCount="1" />
  <row Id="13727" PostTypeId="2" ParentId="13715" CreationDate="2011-08-18T18:58:58.203" Score="-1" ViewCount="" Body="&lt;p&gt;I would add as an answer to &quot;what does the LHC magnify&quot; : it magnifies the microcosm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though it is only probabilities that we can measure, &lt;a href=&quot;http://en.wikipedia.org/wiki/Orders_of_magnitude_%28length%29&quot; rel=&quot;nofollow&quot;&gt;the range is accessible&lt;/a&gt;, and the higher the energy, the finer the details of the region explored.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-08-18T18:58:58.203" CommentCount="2" />
  <row Id="13728" PostTypeId="1" AcceptedAnswerId="13749" CreationDate="2011-08-18T19:25:12.080" Score="8" ViewCount="601" Body="&lt;p&gt;It is well-known even among the lay public (thanks to popular books) that string theory first arose in the field of strong interactions where certain scattering amplitudes had properties that could be explained by assuming there were strings lurking around. Unfortunatelly, that's about as far as my knowledge reaches.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can you explain in detail what kind of objects that show those peculiar stringy properties are and what precisely those properties are?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;How did the old &quot;string theory&quot; explain those properties.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Are those explanations still relevant or have they been superseded by modern perspective gained from QCD (which hadn't been yet around at the Veneziano's time)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-08-20T13:53:44.960" Title="What are the details around the origin of the string theory?" Tags="&lt;particle-physics&gt;&lt;string-theory&gt;&lt;history&gt;&lt;quantum-chromodynamics&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" />
  <row Id="13729" PostTypeId="2" ParentId="2227" CreationDate="2011-08-18T19:31:19.050" Score="0" ViewCount="" Body="&lt;p&gt;I am still not sure of the answer, but I suspect the following paper to be highly relevant&lt;br&gt;&#xA;&lt;a href=&quot;http://arxiv.org/abs/1108.3377&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1108.3377&lt;/a&gt; .     &quot;Entanglement Dynamics between Inertial and Non-uniformly Accelerated Detectors&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;David C. M. Ostapchuk, Shih-Yuin Lin, Robert B. Mann, B. L. Hu&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2011-08-18T19:31:19.050" />
  <row Id="13730" PostTypeId="2" ParentId="13722" CreationDate="2011-08-18T19:36:29.803" Score="0" ViewCount="" Body="&lt;p&gt;$ \sum Impact = \Delta Momentum $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Units for impact are the same as momentum ${\rm N\,s} = {\rm kg\,\frac{m}{s}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I meant to say &quot;Impulse&quot; to describe impact forces.&lt;/p&gt;&#xA;" OwnerUserId="392" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2011-08-18T20:32:18.533" LastActivityDate="2011-08-18T20:32:18.533" CommentCount="1" />
  <row Id="13731" PostTypeId="2" ParentId="13728" CreationDate="2011-08-18T19:39:43.113" Score="3" ViewCount="" Body="&lt;p&gt;Well I believe the original clue was Regge trajectories.  It was observed that if you plotted mass squared vs. angular momentum for strongly interacting resonances, they tended to follow straight lines.  This could be explained as the spectrum of rotating strings connecting massless particles. &lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-08-18T19:39:43.113" CommentCount="3" />
  <row Id="13732" PostTypeId="2" ParentId="9371" CreationDate="2011-08-18T19:44:21.330" Score="3" ViewCount="" Body="&lt;p&gt;See the paper &lt;a href=&quot;http://arxiv.org/abs/1103.4937&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1103.4937&lt;/a&gt; by M. Villata for an argument for antimatter falling up.    However, see D.J. Cross at  &lt;a href=&quot;http://arxiv.org/abs/1108.5117&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1108.5117&lt;/a&gt; for powerful rebuttal arguments, asserting that Villata misapplied the CPT theorem, resulting in an incorrect sign for the matter-antimatter gravitational interaction.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://lss.fnal.gov/archive/test-fn/0000/fermilab-fn-0822-cd-t.pdf&quot; rel=&quot;nofollow&quot;&gt;http://lss.fnal.gov/archive/test-fn/0000/fermilab-fn-0822-cd-t.pdf&lt;/a&gt; for a general review of the previously existing experimental evidence from Etvos and other experiments such as accelerometers, Kaons, astrophysical neutrino's from SN1987A.   &lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2011-09-01T12:47:44.427" LastActivityDate="2011-09-01T12:47:44.427" />
  <row Id="13733" PostTypeId="2" ParentId="13728" CreationDate="2011-08-18T19:49:04.860" Score="4" ViewCount="" Body="&lt;p&gt;I liked this video, where Susskind explains how he and some friends discovered string theory in the context of meson scattering:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=25haxRuZQUk&amp;amp;list=PL202191442DB1B300&amp;amp;index=1&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=25haxRuZQUk&amp;amp;list=PL202191442DB1B300&amp;amp;index=1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is just a very preliminary answer, the video is probably way below Your level ... :-P, will take it back as soon as it is superseeded)&lt;/p&gt;&#xA;" OwnerUserId="2751" LastActivityDate="2011-08-18T19:49:04.860" CommentCount="2" />
  <row Id="13734" PostTypeId="2" ParentId="3799" CreationDate="2011-08-18T20:01:30.437" Score="2" ViewCount="" Body="&lt;p&gt;The treatment of Pons/Fleischmann is, by far, the worst scandal in modern science. It is probably worse than Galileo. Their claims were true, were reproduced immediately in several labs (sporadically, many labs failed too), and good research continues to this day, without funding and without theory.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Experimental History&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Their claim is not completely new. It goes back to the 1920's, when Paneth and Peters reported Helium production in Pd heavy water electrolysis. Paneth and Peters were dismissed for essentially the same reason that Pons and Fleischmann's were, the theorists' imaginations were too stunted to think up a mechanism that could bridge the gap between chemical energies and nuclear energies. In the 1950s, there was a Soviet scientist who also claimed that Pd heavy water electrolysis leads to nuclear anomalies. That claim was also dismissed by the Soviet scientific establishment, but his career was rehabilitated somewhat after Americans reproduced the effect in 1989.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many people (i.e. graduate students) who worked with Pd/deuterium system noticed anomalies in the system for decades, and it was folklore in the chemistry community that deuterated Palladium acts up, while Hydrogenated Pd does not. Pons and Fleischmann decided to get serious about the anomalies, and did extremely careful calorimetry on the system for many years, until they were certain they had a reproducible effect for which chemistry could be safely excluded. Then they held their press conference, and chaos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of the problem is that once they claimed fusion, people insisted that the fusion should emit neutrons, just as hot fusion does. This is impossible, because, considering the energy released, the number of neutrons would have cooked Pons and Fleischmann. Then nuclear physicists demanded that they measure nuclear effects, and they tried to do this, but their &lt;em&gt;nuclear&lt;/em&gt; measurements were riddled with errors, and it is possible that they fudged a plot that they showed at a conference (although considering Fleischmann's impeccable scientific integrity, I find it more plausible that they made an honest mistake). It is important to note that their published paper contains only calorimetry data, and no nuclear data of which they were unsure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people speculated that the effect can be explained by chemistry, or by insufficient stirring, or by storing electrical energy for later release. These claims are all idiotic. The effect is not small, the only reason it requires instruments to detect is because Pons and Fleischmann deliberately used a tiny Pd wire as a cathode. When they used a bigger Pd plate, the thing melted the table, and blew a hole in the concrete floor below. There just is no source of chemical energy, nor a battery, which can store energy chemically at more than about 1eV per atom. Other people noticed similar runaway explosions too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect is very sensitive to the metallurgy of the Palladium, and Pons and Fleischmann couldn't reproduce the finicky effect on demand once they ran out of the good palladium. The experiment sometimes takes weeks, and many people just didn't have patience. Still, the effect was reproduced immediately in a handful of places. MIT ran an infamous reproduction that noticed excess heat production, and were going to press with a reproduction. Then they realized that this effect was going to be labelled bogus, and they cut off their graph to show no excess heat. One of the graduate students who was involved in this experiment, Eugene Mallove, was so outraged that he quit his position and became a cold fusion promoter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several groups published reproductions. These groups were attacked in the most unscientific of ways. For example, one group reported very low levels of tritium production in the system. Since tritium is radioactive, it has a clear signature, it can't be mistaken for anything else. Since it is very expensive and is ordinarily produced in nuclear reactors, the only way such a signal could be seen is if it was deliberately faked by spiking the heavy water with tritium. The graduate student who observed the tritium production was fired, so confident were the senior people that he had committed fraud. Despite the intense pressure, he never retracted his claim, and no plausible way he could obtain tritium (other than cold fusion) was ever found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two extermely well respected theorists, Julian Schwinger (emeritus UCLA) and Peter Hagelstein at MIT, were convinced that the effect was real. Schwinger was not allowed to publish in the field, and Hagelstein, who was tenured, had all his funding cut and was moved into a closet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the early 90's, without any official funding, cold fusion scientists quantitated Helium production to correlate with the excess heat. Low levels of ordinary fusion are seen to happen in the system. The SPAWAR group in the U.S. Navy reproduced the effect in co-deposition experiments, where they plate Pd onto a surface in the presence of heavy water. Their experiments only detect nuclear products, because the plated surface is so small, but the effects are 100% reproducible. More recently, the navy presented evidence of sporadic high-energy neutrons coming from the co-deposition system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Others noticed new elements being produced in the Pd system with abnormal isotope ratios and atomic number near Pd. Abnormal isotope ratios cannot be fraud, because such materials are so difficult to make. In Japan, Arata reproduced the effect by using gas-loading of deuterium into Pd, which has no heat source, so there is no calorimetry error to blame things on. This was a foolproof version of Pons' and Fleischmann's &quot;heat after death&quot; experiments, where they get the reaction to continue to produce heat even after the current is turned off. This rules out calorimetry error entirely. The effect has been reproduced many hundreds of times, in far away labs with no mutual interests, and everybody should be certain by now that it is real. I am ashamed of myself, in that I couldn't bring myself to trust the experimental data until I came up with a reasonable theoretical story to explain it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Focardi/Rossi claims are more dubious. Their effect is in Nickel/Hydrogen, which has some reported energy anomalies too, but not with the same level of confidence. In terms of theoretical craziness, Nickel Hydrogen fusion is to Palladium Deuterium fusion as Palladium Deuterium fusion is to standard hot fusion. The cold fusion community is taking a wait-and-see attitude, but I think consensus is that the device is not likely to work. In his demonstrations, Rossi measured heat production using steam, not water, and by understating the water-content of the steam, you can inflate the energy output by the latent energy of vaporization, which is huge. For me, it is most suspicious that his claimed transmutation products were analyzed and have natural isotope ratios. It is possible that his machine works, and it is possible that it does not produce any excess energy at all, we will know soon enough. What is impossible is that there are no nuclear effects in Pd/deuterium.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Theoretical work&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;One major difficulty for acceptance of the effect is that theoretical work in this field is not sound. There are several theories, each of which are more or less preposterous. The central difficulties are overcoming the Coulomb barrier somehow, and making energy without nuclear reaction byproducts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hydrinos/little hydrogen: this theory states that the electron in Hydrogen can find a closer orbit than the ground state, and spends some times close to the nucleus. This requires that quantum mechanics is wrong, or that there is some new electron/proton force which has been missed, and somehow does not alter the ground state energy, but is capable of sucking the electron into the proton every once in a while.&lt;/li&gt;&#xA;&lt;li&gt;Bose-Einstein condensed deuterons/alphas: this idea is that the cross section for fusion is enhanced by identical-particle effects, since deuterons and alphas are both bosons. In theory, you can enhance reactions by having a coherent source of bosons all go through the same reaction to a coherent superposition. This theory fails both because the temperature is too high for coherence between deuterons, and because when it is implemented in specific cold fusion papers, the deuterons are treated as non-interacting particles in a product state, so that the amplitude for being at the same point is big. But this is ignoring the whole difficulty, because the electrostatic repulsion leads the wavefunction to be entangled, with little probability of any two deuterons getting to the same point.&lt;/li&gt;&#xA;&lt;li&gt;Lattice enhancement mechanisms: This was the focus of Schwinger and Hagelstein, neither of whom claimed to have solved the problem. The problem with such theories is only that the effects have to be collective over thousands of atoms to explain taking eV energies into KeV energies, and it is thermodynamically difficult to imagine how you can take such entropic energy into such an entropically unfavorable place as a single particle.&lt;/li&gt;&#xA;&lt;li&gt;Weak Force Neutron production: The Widom Larson theory claims that it is possible for a proton and an electron to do inverse beta decay on the surface of a metal, where there are large local electric fields. This is preposterous, because of the MeV difference in proton and neutron mass. It requires millions of volts to accelerate an electron to enough energy to be able to do an inverse beta-decay, and such energies are not available on the surface of a metal. Further, this theory will predict transmutations of plus/minus one mass unit predominanatly, which is not observed, and does not explain how a deuteron can absorb an electron.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The following lists are bogus theories I speculated would work, other people come up with these too every once in a while:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;sporadic atmospheric muon capture: The idea there is that muons are captured by the metal, and lead to fusion. This doesn't work, just because there are not enough muons, the deuterons are separated from each other in the lattice, and if the muon is captured by a Pd nucleus, it's wasted.&lt;/li&gt;&#xA;&lt;li&gt;tunneling with weird many-body enhancement: the idea is that the tunneling amplitude is always estimated, not calculated, and this is an impossible-to-solve many electron/many nucleon system, so perhaps the tunneling amplitude is just off by many orders of magnitude. This doesn't work, because there is a way of giving a lower bound to tunneling amplitudes which excludes any appreciable fusion reaction by tunneling. To do this, you exploit the fact that tunneling is a ground state property, and the deuterons which are imagined to tunnel are bosons, and their imaginary time ground state has no nodes. The electrons have nodes, since they are fermions, and at high energies, but when the electron states are all fully occupied, they might as well be a vacuum, with structure only near the Fermi surface (this follows from the particle-hole symmetric approximate description of the Fermi liquid). There were rigorous upper bounds on the tunneling probability for deuterons in a metal that claimed to prove cold fusion is impossible.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The previous failures leads one to expect that the effect is out of equilibrium, and involves highly excited atoms.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;My Personal Theory&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;To bridge the gap between the scale of chemistry at eVs and of the nuclei at MeVs, one should take note of the fact that there are K-shell electrons orbiting very close to the nucleus at KeV energies. The K-shell electron of Pd has a 20KeV ionization energy, and if you have a K-shell hole in one Pd atom, it will be able to store an amount of energy non-entropically in an amount sufficient to lead to deuteron fusion. While this energy is large, it is not large enough to knock a Palladium atom out of its lattice position, so it cannot dump its energy by locally breaking the lattice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such K-shell holes usually decay by X-rays, but this is an electromagnetic process which is suppressed by powers of v/c when the electron is nonrelativistic, as it is even in the K-shell. This is a well known effect--- it's the same reason that atomic spectral lines are narrow. Emitting a photon takes many orbits because of the mismatch in scale between the photon's wavelength and the size of the orbit. This is ultimately because the orbit is nonrelativistic. Because the emission takes so long, the spectral lines are sharply defined and narrow, and the emission is dominated by the matrix elements of the dipole moment of the atomic state between stationary states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other observed ways for K-shells to lost their energy is to kick out an outer shell electron from a neighboring atom. This process is electrostatic, and nonrelativistic, so it is not suppressed by 1/c factors. It is only suppressed by the smallness of the charge on the electron and the distance between electrons on neighboring atoms. There is a significant fraction of decays in K-holes in Pd in this channel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a metal with protons or deuterons, a K-shell hole should be able to also kick its energy into a proton or deutrons by electrostatic forces. The matrix element is exactly the same as for kicking an electron, but the density of states is 30-50 times bigger (depending on whether it's a proton or a deuteron) due to the heavier mass. The proton, unlike a Pd nucleus, will leave its lattice site under such a transfer. So, considering that the cross section for a K-shell hole to kick an electron is not small, I feel safe to conclude that the proton-kicking process is the dominant decay mechanism for K-holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that the accelerated deuteron produces a fusion event, it has 20KeV after all, and suppose by some miracle that this fusion does not happen in the usual hot-fusion way, but occurs in another channel which produces an alpha particle and transfers the excess energy/momentum to a nearby charged particle electrostatically. Then the alpha particle and whatever it transferred its energy to are moving with 24MeV of energy, and they go through the metal, ionizing Pd atoms. Energetically, they can make up to 1000 K-shell holes, all within a millimeter, since the penetration depth is so tiny. The true number is more likely closer to a hundred. These holes can then accelerate new deuterons, possibly leading to a chain reaction. I believe something like this explains the cold-fusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two problems:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The cross section for fusion at 20 KeV is not that huge, and it does not lead to a chain reaction by itself through the usual hot-fusion channels. The multiplication factor is at best something like .001, meaning that the chain reaction dies out instead of grows.&lt;/li&gt;&#xA;&lt;li&gt;The actual observed reaction produces an alpha particle without an emitted neutron or proton nearly all of the time. This is a 1-in-a-million event in hot fusion.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I think that both problems are related to the fact that the reaction is happening inside a dense metal. Fusion of deuterons happens through unstable intermediate states, and the cross section to alpha particle is not so big because of the same non-relativistic issue. To get an alpha, you need to emit a gamma-ray photon, and emissions of photons are suppressed by 1/c factors. If there is a nucleus nearby, it can be kicked electrostatically, and this process is easier than kicking out a photon, because it is nonrelativistic (the same holds for an electron, but with smaller cross section due to the smaller charge). The time-scale for kicking is the lifetime of the two-deuteron resonance, which is not very long, so to transfer energy to the nucleus the fusion has to happen within at most 100 fermis of the nucleus, which is by coincidence about the same size as the inner shell. If the deuterons are kicking about at random, this coincidence is not significant, but if there are coherent hole-transfer events in the metal, it is conceivable that nearly all the energetic deuteron-deuteron collisions take place very close to a nucleus. It is very difficult to even estimate whether the processes are coherent or random.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are conservation laws broken when a nucleus is nearby. The nucleus breaks parity, so it might open up a fusion channel, by allowing deuteron pairs to decay to an alpha from a parity odd state. Such a transition would never be observed in a dilute beam fusion, because these fusions happen far away from anything else. This hypothesis is not excluded by alpha particle spectroscopy (there are a lot of relevant levels of different parities), but it is not predicted either. It is a just a plausible just-so story, which you shouldn't believe except for the experimental data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This theory predicts the following&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;when undergoing cold fusion, the material should emit copious X-rays at the KeV range (from those K-shell holes which happen to decay electromagnetically anyway).&lt;/li&gt;&#xA;&lt;li&gt;the material should emit KeV deuterons in a mm skin around it.&lt;/li&gt;&#xA;&lt;li&gt;the material should emit MeV-range alphas and Pd nuclei. The alphas should go up to 20MeV, and the Pd up to 4 MeVs.&lt;/li&gt;&#xA;&lt;li&gt;There should be a small amount of hot fusion happening, with the associated fast neutrons and tritium, just from the occasional accidental hot-fusion collisions of 20KeV deuterons.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These were &quot;predictions&quot; only in that I did not know about them when I made up the theory. I later found out that the first, the third, and the fourth are all observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory also predicts the following&#xA;* Proton based cold fusion doesn't work (although there might be a way of storing KeV scale energies in a Nickel hydrogen system for a long time in K-shell bands, releasing it in bursts, although it seems unlikely to me)&#xA;* Transmutation products in cold fusion are all due to secondary collisions with the fast alphas and Pd nuclei. So the produced elements should be the same as bombardment with 20 MeV alphas and 4 MeV Pd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are not compatible with the experimental data at present.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Problems with the theory&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The second prediction is problematic, in that the transmutation products tend to show peaks at mass +8,+12. Bombardment by alphas would produce a +4 mass, not +8, since there is no chance of the same atom being hit twice by two different alphas. You can imagine that a fast-moving Pd is transfering 8 units of mass to a nucleus, but then you would see symmetric peaks at plus/minus 8 and this does not seem to be what is observed. What is observed experimentally is a big mess:http://www.lenr-canr.org/acrobat/IwamuraYobservatiob.pdf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory is also more or less a bulk theory, so it is hard to understand why surface is more important, as all the experts seem to agree. It is possible that in the presence of protons, the K-shells delocalize, producing bands at 20KeV, but this is difficult to calculate or to believe, because there are so many deuteron states available that it is hard to see how a coherent transfer of a K-shell hole from one atom to another through a deuteron intermediate could happen. But it is not impossible. K-shell bands in deuterated metals would give a whole zoo of useful effects, regardless of the nuclear stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am only telling this theoretical story to convince you that there is no reason to dismiss cold fusion. It is possible that the mechanism is different, although I can't think of what else it could be. &lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-18T20:01:30.437" CommentCount="7" />
  <row Id="13735" PostTypeId="2" ParentId="3799" CreationDate="2011-08-18T20:40:30.613" Score="0" ViewCount="" Body="&lt;p&gt;Pons and Fleischmann originally reported in 1989 that their chemical cells had produced excess heat, neutrons, and tritium. Their interpretation was that deuterium nuclei were fusing to produce 4He. The branching ratios in this process are known: 50% n+3He, 50% p+3H, and 10^-6 4He+gamma. If the claimed excess heat had been produced by fusion, then the experimenters would have been killed by the neutrons coming from the 50% of the decays that proceeded by neutron emission. Neutron detectors are infamous among nuclear physicists for being difficult to use, and for having a tendency to produce spurious signals. Even if all the neutron counts claimed by Pons and Fleischmann had been real, the observed flux of neutrons would have been many orders of magnitude too small in relation to the amount of excess heat claimed. Subsequent measurements by nuclear physicists using state-of-the-art neutron detection techniques demonstrated that no neutrons above background level are produced by chemical cells of the type used by Pons and Fleischmann.[Gai 1989] There has also been no plausible evidence for production of 3He, 3H, or gamma rays, all of which would have been copiously produced in d-d fusion reactions that produced measurable amounts of energy through nuclear reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As of 2010, the consensus among scientists is that cold fusion was an example of pathological science. However, a few true believers continue to do experiments and make claims of positive results. A 2010 review by a prominent believer[Storms 2010] says that &quot;Many people feel that the correlation between heat and helium is the strongest evidence for cold fusion.&quot; The trouble is that heat is produced by chemical reactions in any case, and the levels of helium claimed are not high enough to show a convincing excess relative to background. If these claims were correct, they would also require a fundamental rewriting of the laws of physics. They would require that the branching ratio in d-d fusion be drastically altered by the chemical environment, but this is impossible because in a nuclear reaction, the electrons are mere spectators. In order to conserve energy and momentum, d-d fusion also requires the emission of two particles in the final state. To get around the nonexistence of the second particle, cold fusion enthusiasts suppose that energy from the reaction is transmitted to the electronic lattice. No known mechanism exists by which such transmission could occur. As the experimental and theoretical constraints have gotten tighter, the believers have responded by coming up with crazier pseudoscience, insluding the production of nuclei with atomic number 126 and transmutation of the elements by plants and bacteria.[Storms 2010]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, claims of cold fusion cannot be correct unless they overturn firmly established knowledge of nuclear physics. This would be an extraordinary claim, and it would require extraordinary proof. After two decades, no such extraordinary proof has emerged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gai et al., &quot;Upper limits on neutron and gamma-ray emission from cold fusion&quot;, Nature 340 (1989) 29–34.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Storms, &quot;Status of cold fusion (2010)&quot;, Naturwissenschaften (online) 97 (10): 861–881&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-18T20:40:30.613" CommentCount="5" />
  <row Id="13736" PostTypeId="2" ParentId="9371" CreationDate="2011-08-18T21:08:29.760" Score="3" ViewCount="" Body="&lt;p&gt;Never seen any good reason to think that antimatter is affected any differently. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light and anti neutrinos (from supernovas) seem to be affected by gravity the same as regular mass, and the theory of gravity (general relativity) doesn't distinguish mass and any other energy. &#xA;If we beam a gamma photon down the gravitational well (from space to ground, for example), it gets blue shifted a little bit - it gains energy. If we use that gamma photon to produce matter antimatter pair and send the pair up the gravitational well, unless the antimatter is affected by gravity &lt;em&gt;exactly&lt;/em&gt; the same as matter (and light), the conservation of energy will be violated. It makes absolutely no sense for antimatter to be affected by gravity differently than light, when the matter is affected by gravity the same as light. (note: no, there is no anti-light)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess it is the usual opportunistic fundraising, exploiting the ignorance. It is hard to raise funds for real research, realistically presented - it sounds obscure and layman has no idea what it is for - and it is easy to raise funds for something that layman thinks is a huge question with good likehood of something extremely awesome, like antigravity. When one is perfectly ignorant then the likehood is fifty-fifty. &lt;/p&gt;&#xA;" OwnerUserId="4967" LastEditorUserId="4967" LastEditorDisplayName="" LastEditDate="2011-08-21T09:38:25.287" LastActivityDate="2011-08-21T09:38:25.287" />
  <row Id="13737" PostTypeId="2" ParentId="13638" CreationDate="2011-08-18T21:56:15.273" Score="1" ViewCount="" Body="&lt;p&gt;There's an inconsistency with the way the velocity composition formulas are consistent with the axis labeled as in the diagram, but your $\tan$ equations are consistent with the axis labeled the other way around. I'll stick with the unconventional way the vertical axis is labeled $x$, the horizontal $y$ in your diagram, so&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\tan(\alpha') = \frac{u_y'}{u_x'} = \frac{u_y + v}{u_x} \cdot\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;should be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tan(\alpha&amp;#39;) = \frac{u_x&amp;#39;}{u_y&amp;#39;} = \frac{u_x}{u_y + v} \cdot\sqrt{1-\frac{v^2}{c^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's look at your second method. The Lorentz contraction (LC) formula is a relationship between space intervals for the same pair of events in different frames, but simultaneous in one. I'll leave you to prove their time interval in the other frame $$\Delta t&amp;#39; = \gamma \frac V {c^2} \Delta x$$ Therefore, you need to be very clear in how you're using the LC formula to relate the space intervals of which pair of events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better way would be to find the coordinates of the event of the shuttle reaching USS voyager in the USS Enterprise frame using the Lorentz transformation$$ y'=\gamma(y + Vt) = \gamma(b + Vt) = \gamma(u_yt + Vt),\qquad x'= x = a = u_xt$$ giving $$tan\alpha&amp;#39; = \frac {x&amp;#39;} {y&amp;#39;} = \frac{u_x}{\gamma(u_y + v)} \cdot$$ So the two methods are consistent with one another, as expected.&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="1345" LastEditorDisplayName="" LastEditDate="2011-08-18T22:01:33.677" LastActivityDate="2011-08-18T22:01:33.677" />
  <row Id="13738" PostTypeId="1" CreationDate="2011-08-18T22:25:44.807" Score="2" ViewCount="85" Body="&lt;p&gt;In Q's &lt;a href=&quot;http://physics.stackexchange.com/questions/466/what-is-the-mechanism-behind-the-slowdown-of-light-in-a-transparent-medium/474#474&quot;&gt;466&lt;/a&gt; and &lt;a href=&quot;http://physics.stackexchange.com/questions/7437/why-glass-is-transparent/7439#7439&quot;&gt;7437&lt;/a&gt; transparent media were discussed. But I'd like to clarify one detail: is a photon absorbed (and delayed) by the medium and than reemitted, or scattered instantly? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is e.g. a laser beam still coherent after passing glass? As medium molecules are disordered, this should distort the phase of photons taking different paths.&lt;/p&gt;&#xA;" OwnerUserId="4760" LastActivityDate="2011-08-20T09:27:42.230" Title="propagation of light in transparent media: absorption and reemission or scattering?" Tags="&lt;optics&gt;&lt;light&gt;&lt;photons&gt;&lt;solid-state-physics&gt;&lt;laser&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="13739" PostTypeId="1" CreationDate="2011-08-18T23:11:39.420" Score="2" ViewCount="47" Body="&lt;p&gt;If I had a flexible tube sealed at both ends and I submerged it in water (held vertical) Would the bottom half of the tube compress and the top half expand? What would the pressure in the tube be? Say its a 2&quot; length of tube with the bottom being in 12&quot; of water&lt;/p&gt;&#xA;" OwnerUserId="4969" LastActivityDate="2011-08-21T02:05:26.590" Title="pressure exerted by fluid" Tags="&lt;classical-mechanics&gt;&lt;fluid-dynamics&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13740" PostTypeId="2" ParentId="13739" CreationDate="2011-08-19T00:11:12.997" Score="1" ViewCount="" Body="&lt;p&gt;If it's a rigid tube then nothing about it will change at all.  It will keep exactly the same shape and the fluid inside the tube will have no indications whether it is above or below water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take the tube to have some elasticity, then yes, the circumference of the bottom half of the tube will contract some (if the fluid inside the tube is lighter) and if the fluid it is filled with is incompressible, then the top half of the tube will expand some.  In fact, the volume will stay the same.  I'll stop at this point and post my illustration of the situation.  I paint the middle of the tube orange to indicate it is filled with a mystery fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Q9Ejd.png&quot; alt=&quot;illustration&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Rod is 2' long and 12&quot; in water on left&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define $\Delta P$ being the difference in pressure between the inside of the tube and the outside of the tube at any point along the height of the tube.  This is arbitrary, but I'll take the sign convention to reflect $\Delta P = P_{inside} - P_{outside}$.  I'll denote the difference between the radius and some arbitrary reference radius as $\Delta r$, and both of the values discussed here are functions of height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good approximation is to assume that air is constant pressure and water pressure increases linearly with depth (also true for the fluid inside).  We will stick with the incompressible assumption and then we need to know if the liquid in the tube is more or less dense than the water.  Assuming perfect elasticity, $\Delta r$ will have a linear relationship with $\Delta P$.  $\Delta P$ is linearly decreasing with height above the water level and linearly decreasing or increasing with height below the water level depending on if the fluid inside the tube is more dense or less dense than the water respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average value of $r^2$ over the entire height of the tube will be constant for every case (in air, in water, all the way in water).  This is from the constant volume requirement that comes from the incompressible assumption.  To a first approximation, $r^2 = r_0 + 2 r_0 \Delta r$, so requiring that $\Delta r$ average to zero over the entire height would also suffice.  I'm providing a graph of the two values I've discussed, which are just illustrations of everything I've already said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/CFXpY.png&quot; alt=&quot;graphs&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The offset ($h$ at $\Delta P=0$) is actually kind of arbitrary, although I could have potentially formalized that.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-08-19T00:18:52.423" LastActivityDate="2011-08-19T00:18:52.423" CommentCount="1" />
  <row Id="13741" PostTypeId="1" AcceptedAnswerId="14292" CreationDate="2011-08-19T00:23:57.733" Score="3" ViewCount="139" Body="&lt;p&gt;We all can agree that UV lasers are extremely useful, but we (poor majority) are still struggling with 355nm DPSS ones, while excimer lasers being prohibitively expensive (let's say at 1-10W average power, or 100Hz @ 0.1J pulse).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the reason for this? What is the complexity, making excimer lasers (namely, 248nm KrF / 308 XeCl ones) so complex? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mirrors &amp;amp; optics is simpler than for CO2 ones, gas is not expensive too... What am I missing?&#xA;Just reread my laser optics book - it needs aggressive laser pumping (50kV @10-30ns pulses), forced 'gas mixing' (just like in high-power CO2 ones), also needs preionization, but nothing too scary...&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="930" LastEditorDisplayName="" LastEditDate="2011-08-19T00:46:02.270" LastActivityDate="2011-09-02T23:08:18.947" Title="What makes excimer lasers so complex, rare and expensive?" Tags="&lt;laser&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="13742" PostTypeId="1" CreationDate="2011-08-19T01:17:43.570" Score="4" ViewCount="30" Body="&lt;p&gt;I am trying to do some Monte Carlo simulations for Pfaffian state from Fractional Quantum Hall effect. I am wondering what is the energy functional for v=5/2 Moore-Read state?&lt;/p&gt;&#xA;" OwnerUserId="2397" LastActivityDate="2011-08-19T01:17:43.570" Title="What is the energy functional for v=5/2 Moore-Read state?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-hall-effect&gt;" CommentCount="1" FavoriteCount="1" />
  <row Id="13743" PostTypeId="2" ParentId="503" CreationDate="2011-08-19T01:31:53.587" Score="2" ViewCount="" Body="&lt;p&gt;I can tell you how to generate open problems in special relativity: open up a geometry book and look at some geometry theorem. Add the word &quot;Minkowski&quot; before everything. Is the theorem still true? One problem that I got to thinking about recently is how to axiomatize Minkowski geometry in the same way that Hilbert axiomatized Euclid's geometry. It is a theorem that Euclidean geometry (as axiomatized by Hilbert, essentially completing Euclid's postulates) is a complete theory: there is an algorithm for deciding which statements are theorems and which are not. Is the same true of Minkowski geometry?&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-19T01:31:53.587" CommentCount="2" />
  <row Id="13744" PostTypeId="1" CreationDate="2011-08-19T01:46:51.207" Score="2" ViewCount="69" Body="&lt;p&gt;So, the event horizon around a black hole emits radiation, and Rindler space is full of thermal energy. I guess I have two questions- does the Unruh effect have anything to do with radiation from the apparent horizon in Rindler space? And what about the cosmological horizon- any radiation there?&lt;/p&gt;&#xA;" OwnerUserId="4663" LastActivityDate="2011-08-19T02:59:14.460" Title="Do all event horizons emit radiation?" Tags="&lt;general-relativity&gt;&lt;event-horizon&gt;" AnswerCount="2" />
  <row Id="13745" PostTypeId="2" ParentId="13714" CreationDate="2011-08-19T01:50:08.487" Score="4" ViewCount="" Body="&lt;p&gt;As someone who started in math as an undergrad and then ended up in physics I can say that I wasn't clear on the distinction between the two subjects.  I treated physics as if it were one big math problem and not as an experimental science.  Theoretical physics and math are no doubt related but when presented together it often ends up being the case that one is smashed into the other in such a way that it looses its essential character.  It makes me think of string theory lectures I've been to that looked like straight up math given in a theorem proof format and relativity classes where the instructors talked up a storm on tensors but they just pushed indexes around and nothing about multi-linear algebra was really called on.  The end result was it took a long time to realize this and recognize how the two are complimentary and to not just blindly smash math stuff into the physics shaped hole and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, here is my suggestion:  Find a result like &lt;a href=&quot;http://arxiv.org/abs/1012.5407&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; one on 1-dimensional magnetic systems and their possible relationship to a 248-dimensional Lie algebra and pre-chew it for the undergrads.  Highlight the interplay between the two subjects and how their distinct (and essential) techniques pulled the whole story together to give an explanation of reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I LOVE to see the two subjects come together in a result like this.  This is the kind of work that drives both subjects and shows the best of what both worlds offer.  That's what I wish I'd have had driven home when I was a junior just starting to see things for myself.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Full disclosure: I actually posted on this paper on my blog &lt;a href=&quot;http://power-quant.com/?q=node/12&quot; rel=&quot;nofollow&quot;&gt;power-quant.com&lt;/a&gt; a while back but I think it's a good example of how physics and math work together.&lt;/p&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2011-08-19T01:50:08.487" CommunityOwnedDate="2011-08-19T01:50:08.487" CommentCount="1" />
  <row Id="13746" PostTypeId="2" ParentId="13648" CreationDate="2011-08-19T02:16:19.063" Score="0" ViewCount="" Body="&lt;p&gt;The formula relating the beam waste and the numerical aperture of a Gaussian beam is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w_0 \approx \frac{\lambda}{\pi \;\, \mathrm{NA}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The highest numerical aperture you could dream of finding--with a good oil-immersion lens--is in the neighborhood of 1.5 (a bit more or less, I don't know). (Without oil-immersion, the numerical aperture is always less than 1.) So we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w_0 \gtrsim \frac{\lambda}{1.5 \pi}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can focus a 645nm beam (if ideal gaussian) to an area of roughly $\pi w_0^2 \gtrsim 0.05\mu \mathrm{m}^2$. If the power is 5 mW, I get $100 \,\mathrm{ GW}/\mathrm{m}^2$, or 100 million times brighter than daylight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real figure for a cheap laser pointer is lower, &lt;a href=&quot;http://en.wikipedia.org/wiki/M_squared&quot; rel=&quot;nofollow&quot;&gt;because $M^2&gt;1$&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, for a diode laser, your best bet is to first use a weak cylindrical lens to get the spot more round [all diode lasers have a built-in cylindrical lens, but in my experience the spot is still slightly elliptical], and then use a strong microscope objective lens (oil-immersion if possible) to focus as tight as possible. Also, put a two-lens telescope before the objective lens to either expand or shrink the beam so that it exactly fills the entrance aperture of the objective lens. Actually, maybe you shouldn't use an oil-immersion lens, you might burn the oil! Don't try this at home...&lt;/p&gt;&#xA;" OwnerUserId="3811" LastEditorUserId="3811" LastEditorDisplayName="" LastEditDate="2011-08-19T14:35:16.743" LastActivityDate="2011-08-19T14:35:16.743" CommentCount="1" />
  <row Id="13747" PostTypeId="2" ParentId="13744" CreationDate="2011-08-19T02:52:28.817" Score="2" ViewCount="" Body="&lt;p&gt;Yes, in all of those cases you will get radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case of the cosmological horizon is discussed here: &lt;a href=&quot;http://math.ucr.edu/home/baez/end.html&quot; rel=&quot;nofollow&quot;&gt;http://math.ucr.edu/home/baez/end.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't refer to &quot;Rindler space&quot; -- don't you just mean flat spacetime described in Rindler coordinates? But anyway, an observer with constant proper acceleration in a flat spacetime does see radiation from the event horizon; this is the canonical example of the Unruh effect.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-19T02:52:28.817" />
  <row Id="13748" PostTypeId="2" ParentId="13744" CreationDate="2011-08-19T02:59:14.460" Score="1" ViewCount="" Body="&lt;p&gt;The short answer is yes for horizons that look like Rindler space (Minkowski space) locally. You can read the Wikipedia article on Hawking radiation--- the equivalence principle allows you to equate the near-horizon Hawking temperature with the near horizon Rindler temperature of an observer which is accelerated to keep from falling into the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For extremal horizons, when the charge is equal to the mass, or if the Black hole is rotating as fast as possible, the horizon recedes to an infinite distance and space-time is not flat near the horizon. These receded horizons are not thermal, and do not radiate anything, they are cold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;deSitter cosmological horizons have a temperature, as do all static horizons. The local temperature can be calculated just as for the black hole horizon, as the periodicity in imaginary time of the solution. There are unusual vacua for deSitter space (first discovered by Burges &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0550321384905625&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencedirect.com/science/article/pii/0550321384905625&lt;/a&gt;) and later analyzed by Gross and collaborators. I don't know how physical these vacua are, if they are reasonable, or whether they go away in a more horizon-centric point of view, like the eternal inflation ambiguities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our cosmological horizon is in motion, it is receding away at nearly the speed of light. In the far future, if it becomes a deSitter horizon, then it will radiate thermally. For a FRW horizon, I don't know the answer either.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-19T02:59:14.460" />
  <row Id="13749" PostTypeId="2" ParentId="13728" CreationDate="2011-08-19T06:54:55.810" Score="10" ViewCount="" Body="&lt;p&gt;in the late 1960s, the strongly interacting particles were a jungle. Protons, neutrons, pions, kaons, lambda hyperons, other hyperons, additional resonances, and so on. It seemed like dozens of elementary particles that strongly interacted. There was no order. People thought that quantum field theory had to die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, they noticed regularities such as Regge trajectories. The minimal mass of a particle of spin $J$ went like &#xA;$$ M^2 = aJ + b $$&#xA;i.e. the squared mass is a linear function of the spin. This relationship was confirmed phenomenologically for a couple of the particles. In the $M^2$-$J$ plane, you had these straight lines, the Regge trajectories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Building on this and related insights, Veneziano &quot;guessed&quot; a nice formula for the scattering amplitudes of the $\pi+\pi \to \pi+\rho$ process, or something like that. It had four mesons and one of them was different. His first amplitude was the Euler beta function&#xA;$$ M = \frac{\Gamma(u)\Gamma(v)}{\Gamma(u+v)}$$&#xA;where $\Gamma$ is the generalized factorial and $u,v$ are linear functions of the Mandelstam variables $s,t$ with fixed coefficients again. This amplitude agrees with the Regge trajectories because $\Gamma(x)$ has poles for all non-positive integers. These poles in the amplitude correspond to the exchange of particles in the $s,t$ channels. One may show that if we expand the amplitude to the residues, the exchanged particles' maximum spin is indeed a linear function of the squared mass, just like in the Regge trajectory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why are there infinitely many particles that may be exchanged? Susskind, Nielsen, Yoneya, and maybe others realized that there has to be &quot;one particle&quot; of a sort that may have any internal excitations - like the Hydrogen atom. Except that the simple spacing of the levels looked much easier than the Hydrogen atom - it was like harmonic oscillators. Infinitely many of them were still needed. They ultimately realized that if we postulate that the mesons are (open) strings, you reproduce the whole Veneziano formula because of an integral that may be used to define it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the immediate properties that the &quot;string concept&quot; demystified was the &quot;duality&quot; in the language of the 1960s - currently called the &quot;world sheet duality&quot;. The amplitude $M$ above us $u,v$-symmetric. But it can be expanded in terms of poles for various values of $u$; or various values of $v$. So it may be calculated as a sum of exchanges purely in the $s$-channel; or purely in the $t$-channel. You don't need to sum up diagrams with the $s$-channel or with the $t$-channel: one of them is enough! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This simple principle, one that Veneziano actually correctly guessed to be a guiding principle for his search of the meson amplitude, is easily explained by string theory. The diagram in which 2 open strings merge into 1 open string and then split may be interpreted as a thickened $s$-channel graph; or a thick $t$-channel graph. There's no qualitative difference between them, so they correspond to a single stringy integral for the amplitude. This is more general - one stringy diagram usually reduces to the sum of many field-theoretical Feynman diagrams in various limits. String theory automatically resums them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Around 1970, many things worked for the strong interactions in the stringy language. Others didn't. String theory turned out to be too good - in particular, it was &quot;too soft&quot; at high energies (the amplitudes decrease exponentially with energies). QCD and quarks emerged. Around mid 1970s, 't Hooft wrote his famous paper on large $N$ gauge theory - in which some strings emerge, too. Only in 1997, these hints were made explicit by Maldacena who showed that string theory was the right description of a gauge theory (or many of them) at the QCD scale, after all: the relevant target space must however be higher-dimensional and be an anti de Sitter space. In AdS/CFT, much of the original strategies - e.g. the assumption that mesons are open strings of a sort - get revived and become quantitatively accurate. It just works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, meanwhile, around mid 1970s, it was also realized that string theory was primarily a quantum theory of gravity because the spin 2 massless modes inevitably exist and inevitably interact via general relativity at long distances. In the early and mid 1980s, it was realized that string theory included the right excitations and interactions to describe all particle species and all forces we know in Nature and nothing could have been undone about this insight later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today, we know that the original motivation of string theory wasn't really wrong: it was just trying to use non-minimal compactifications of string theory. Simpler vacua of string theory explain gravity in a quantum language long before they explain the strong interactions.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastActivityDate="2011-08-19T06:54:55.810" CommentCount="29" />
  <row Id="13750" PostTypeId="2" ParentId="12636" CreationDate="2011-08-19T07:17:00.293" Score="0" ViewCount="" Body="&lt;p&gt;It seems that you are missing what $\Delta T$  is here. It is not an interval of time. It is time uncertainty. Which is not what usually called $\Delta T$. The most standard application for this uncertainty principle is a linewidth. From this inequality you know that lifetime in upper state limits lower bound of energy dispersion. The larger lifetime the more precise your energy measurement (may be). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, there is no use to convert this $\Delta T$ to $\Delta x$. It will mean not what you expect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same holds for $\Delta E$. It is not a real energy of some object. &lt;/p&gt;&#xA;" OwnerUserId="4657" LastActivityDate="2011-08-19T07:17:00.293" />
  <row Id="13751" PostTypeId="2" ParentId="13694" CreationDate="2011-08-19T08:17:10.477" Score="6" ViewCount="" Body="&lt;p&gt;The proof is very simple, Hawking saw it in a flash. While writing this, I found the proof presented in the Wikipedia page for the Raychoudhuri equation. To understand it, there are three background results you need to be comfortable with.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Horizon area makes physical sense&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Given three nearly parallel infinitesimally separated light rays in Minkowski space moving in the direction perpendicular to their plane of separation, you can define the area they span by cutting them with a space time plane, and asking what is the area contained in the triangle they form. Unlike in Euclidean geometry, where this is never true for any area, in Minkowski geometry the area of this triangle doesn't depend on the orientation of the plane. In other words, you can slide each of the three intersection points up and down on the light ray without changing the area of the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, first note that if you have two parallel light rays l,l' whose separation is perpendicular to their line of motion, and you have a line between two points on l and l' respectively, the length of this line does not depend on which two points it connects. The length s in Minkowski space is defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s^2= A\cdot A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if you add a null-vector N to A, the result is unchanged because $A\cdot N$ and $N \cdot N$ are both zero. This implies that if you have three light rays which are all moving in the same direction and are separated perpendicular to the direction of motion, you connect any three points these rays and the three side-lengths of the resulting triangle are the same. Side-side-side congruence is true in Minkowski space too (when the sidelengths are nonzero).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the area of a black hole is well defined--- cut up the surface into infinitesimal triangles which are close to some space-like surface cutting the horizon and no matter what orientation these triangles have with the light rays, you get the same answer. This is an important point which is not explained well in usual treatments.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Deviation equation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;If two parallel geodesics are moving along a shared direction with a separation $\Delta$, you can think of their relative dynamics as being determined by a Newton's law. Parametrizing the space-time in a neighborhood of one of them with local coordinates, the deviation of the other is determined by expanding the arc-length formula to second order. The result is motion with a linear restoring/expelling force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spring-constant of the restoring/expelling force is one of the equivalent defining properties of the Riemann curvature:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ {d^2 \Delta^\mu\over d\tau^2} = R^\mu_{\nu\sigma\lambda} \Delta^\nu \dot{x}^\sigma\dot{x}^\lambda$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a simplex made out of nearby points, lying on a certain infinitesimal space-like plane, and set these points moving along the timelike direction perpendicular to the plane. The rate at which the log of their volume increases as a function of their shared proper time s is called the expansion $\theta$. The derivative of $\theta$ is interesting because it takes the Ricci trace of  the Riemann tensor when you evaluate it in terms of the geodesic deviation above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${d\theta \over ds} = - {1\over 2}\theta^2 - \sigma^2 - R_{\mu\nu} \dot{x}^\mu \dot{x^\nu}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\sigma$ is not important because it always makes a negative contribution. This is the Raychoudhuri equation (for vanishing vorticity--- the vorticity vanishes for these parallel infinitesimally separated geodesics for all times, because the curvature forces are Hooke springs, directed in and out). For null paths, the volume becomes an area (two of the perpendicular spanning directions of the simplex become parallel), and the proper time becomes the affine parameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that on a null vector like $\dot{x}$, $T_{\mu\nu}$ is equal to $R_{\mu\nu}$, so if the weak energy condition holds, this quantity is always positive. This tells you that once the divergence is negative, it must crash to minus infinity in a finite proper time. If the area of a little triangle of parallel light-rays is decreasing along their affine parameter at any instant, if the weak energy condition is satisfied, then this area will crash to zero area after a finite affine parameter. This means that some two neighboring geodesics in the congruence have collided, or &quot;focused&quot;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Focusing theorem&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;When two nearby geodesics collide, they cannot be &quot;shortest&quot; paths between their endpoints, for a simple reason: first, the deviations between them are always to first infinitesimal order, so that they have the same arclength (this is the same reason that particles which are moving slowly relative to each other agree on their shared Newtonian time, and this notion extends to affine parameter by limits). If geodesics 1 and 2 started at point P, became parallel for a while, then intersected, you can follow 1 from P to the intersection point, then follow 2 until the end, and this is the same length as following 2 all the way from P to the end. But the first path bends a corner. If light bends a corner, it is possible to catch up to it with a time-like path, so past the intersection point, a massive particle could have taken the path from P to the extension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This holds for two geodesics that start off going in a direction perpendicular to a given space-like plane just as well as those from a common point, replacing &quot;distance from the point&quot; with &quot;distance from the plane&quot;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The Area theorem&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The event horizon of a black hole is defined as those light-paths which just barely do not escape. If the light rays are pushed just a little out, then they escape to infinity, and if they are pushed just a little in, then they are sucked into the black hole. This means that any massive object which hits these light rays must never be able to catch up with these light rays, it must fall into the black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hawking's insight was that the if the area is decreasing at any one moment in any little triangle on this horizon, it must crash to zero in a finite affine parameter. This means that two nearby null geodesics on the extension of this simplex forward in affine parameter will focus. But this means that their extension can be linked to the original simplex by a time-like path, so that their extension must be inside the black hole, which means they could not be on the boundary.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-19T08:17:10.477" CommentCount="1" />
  <row Id="13752" PostTypeId="1" AcceptedAnswerId="13784" CreationDate="2011-08-19T08:30:41.873" Score="2" ViewCount="71" Body="&lt;p&gt;can anyone give me some good references on how to obtain the relativistic equation of particles with arbitrary spin? &lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2011-08-20T08:36:56.307" Title="Relativistic equation with arbitrary spin" Tags="&lt;quantum-field-theory&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13753" PostTypeId="2" ParentId="503" CreationDate="2011-08-19T09:06:35.410" Score="1" ViewCount="" Body="&lt;p&gt;Special relativity is rather simple. Abraham Ungar's stuff that you are quoting seems quite trivial at first (and second) glance. There is no difficulty combining arbitrary rotations and Lorentz boosts. You only need to express those as 4x4 matrices. I derived such matrix one time simply to answer a forum post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linear frame dragging results from motion of mass, and as such is, too, a gravitational effect. The general relativity is rather complicated and there are various unsolved problems.&lt;/p&gt;&#xA;" OwnerUserId="4967" LastActivityDate="2011-08-19T09:06:35.410" />
  <row Id="13754" PostTypeId="2" ParentId="13030" CreationDate="2011-08-19T09:46:11.990" Score="0" ViewCount="" Body="&lt;p&gt;Basically, the air flows faster over the top, but not because it has to obey some 'equal time' rule - it is really a totally made up rule which is not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wing is angled, and behind the top of the wing there is area of lower pressure - very simply, wing has just moved out of that place - just like a pump piston - the air has to turn to meet up the wing. In front of the wing there is an area of higher pressure, because the air is being pushed by the wing. This pressure difference acts on wing and produces lift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The often quoted Bernoulli's law needs some explaining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is really very simple. When air flows into area of low pressure, it accelerates akin to rollercoaster going down - it gets sucked into low pressure area! And when it enters high pressure, it slows down akin to rollercoaster going up. When there is no friction between flow lines, the only way flow velocity can change is via this pressure driven acceleration and deceleration. The air entering low pressure area on top of the wing speeds up. The air entering high pressure area on bottom slows down. &#xA;That is why air on top moves faster. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That results in deflection of the air downwards, which is required for generation of lift due to conservation of momentum (which is a true law of physics). For the air to be deflected downwards, it is necessary that air on top goes faster, but this is nonetheless not an explanation of how we made air move faster on the top, it is merely an explanation why we would want to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The common explanation of lift does it backwards - it assumes faster flow over the top of airfoil, using a made up, incorrect law of equal transit time to justify this, then uses Bernoulli's relation between speed and pressure to explain low pressure and lift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In common introductory textbooks the approach is taken to anyhow reduce everything into things that the author can't be blamed for not explaining, ideally 'laws'. The goal is essentially psychological; to produce &lt;em&gt;feeling of understanding&lt;/em&gt; in the mind of the typical reader.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The science works differently; instead of striving to achieve particular feeling in the head, it produces theories that should allow to predict things.&lt;/p&gt;&#xA;" OwnerUserId="4967" LastEditorUserId="4967" LastEditorDisplayName="" LastEditDate="2011-08-19T09:59:33.873" LastActivityDate="2011-08-19T09:59:33.873" />
  <row Id="13755" PostTypeId="2" ParentId="13688" CreationDate="2011-08-19T12:16:04.770" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://www.quantumdiaries.org/2011/06/19/helicity-chirality-mass-and-the-higgs/&quot; rel=&quot;nofollow&quot;&gt;This post &lt;em&gt;helicity-chirality-mass-and-the-higgs&lt;/em&gt; in quantumdiaries is very interesting&lt;/a&gt;&lt;br&gt;&#xA;and I think that it offers the explanation you seek, with nice graphics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://motls.blogspot.com/2011/06/particles-of-different-spins-and-their.html&quot; rel=&quot;nofollow&quot;&gt;Lubos Motl blog post (more technical viewpoint on this subject)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(both posts are cross linked)&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-08-22T18:22:16.337" LastActivityDate="2011-08-22T18:22:16.337" />
  <row Id="13756" PostTypeId="1" AcceptedAnswerId="13759" CreationDate="2011-08-19T13:55:30.883" Score="0" ViewCount="48" Body="&lt;p&gt;Consider the &lt;a href=&quot;http://physics.stackexchange.com/questions/12785/schrodinger-equation-in-spherical-coordinates&quot;&gt;reduced radial Schrodinger equation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{1}{2}\frac{\text{d}^2}{\text{d}r^2}\phi(r)+V(r)\phi(r)=E\phi(r).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We try to find a bound state (i.e. $\phi(0)=\phi(+\infty)=0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $V(r)=1/r$ or $1/r^2,1/r^3$ etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://physics.stackexchange.com/questions/12732/solving-one-dimensional-schrodinger-equation-with-finite-difference-method&quot;&gt;numerical approach&lt;/a&gt; showed that the energy $E$ tends to zero when $a$ (i.e. the upper bound for $r$, use it to represent $+\infty$ in the numerical method) increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/m3uuU.png&quot; alt=&quot;table1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In this table $V(r)=1/r^{\alpha}$. $E_1$ is the calculated energy.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how to explain this result physically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that the real $E$ should be $0$, or allowed $E$s actually are continuous? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is because of the repulsive potential $V(r)$, that $E$ should be $0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P.S. I'm interested in the &quot;ground state&quot;, if it exists. My numerical procedure tries to find the minimum energy. )&lt;/p&gt;&#xA;" OwnerUserId="4314" LastEditorUserId="4314" LastEditorDisplayName="" LastEditDate="2011-08-19T14:48:27.403" LastActivityDate="2011-08-19T14:48:27.403" Title="About the energy with the repulsive potential" Tags="&lt;quantum-mechanics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13757" PostTypeId="1" AcceptedAnswerId="13762" CreationDate="2011-08-19T13:59:23.213" Score="4" ViewCount="152" Body="&lt;p&gt;I read on Wikipedia how the numerical value of Avogadro's number can be found by doing an experiment, provided you have the numerical value of Faraday's constant, but it seems to me that Faraday's constant could not be known before Avogadro's number ever was because it's the electric charge per mole (how could you measure the charge per mole before you knew you had a mole?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just want to know the method physically used and the reasoning and calculations done by the first person who found the number 6.02214179(30)×10^23 (or however accurate it first was).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I see on the Wikipedia page for Avogadro constant that the numerical value was first obtained by &quot;Johann Josef Loschmidt who, in 1865, estimated the average diameter of the molecules in air by a method that is equivalent to calculating the number of particles in a given volume of gas.&quot; but I can't access any of the original sources that are cited. Can somebody explain or give an accessible link so I can read about what exactly Loschmidt did?&lt;/p&gt;&#xA;" OwnerUserId="4715" LastEditorUserId="4715" LastEditorDisplayName="" LastEditDate="2011-08-19T14:04:26.463" LastActivityDate="2011-08-25T04:53:43.993" Title="How was Avogadro's number first determined?" Tags="&lt;experimental-physics&gt;&lt;physical-chemistry&gt;&lt;history&gt;" AnswerCount="3" />
  <row Id="13758" PostTypeId="2" ParentId="503" CreationDate="2011-08-19T14:14:53.780" Score="-2" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Length_contraction#Experimental_verifications&quot; rel=&quot;nofollow&quot;&gt;Length_contraction#Experimental_verifications&lt;/a&gt;&lt;br&gt;&#xA;Length Lorentz contraction - observer frame dependent ? or ...&lt;br&gt;&#xA;Length Lorentz contraction - absolute, i.e. &lt;em&gt;real&lt;/em&gt;  ?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since the occurrence of length contraction depends on the inertial&#xA;  frame chosen, it can only be measured by an observer not at rest in&#xA;  the same inertial frame, i.e., it exists only in a non-co-moving&#xA;  frame. ...&lt;br&gt;&#xA;  Another confirmation is the increased ionization ability of&#xA;  electrically charged particles in motion. According to&#xA;  pre-relativistic physics the ability should decrease at high speed,&#xA;  however, the Lorentz contraction of the Coulomb field leads to an&#xA;  increase of the electrical field strength normal to the line of&#xA;  motion, &lt;strong&gt;which leads to the actually observed increase of the&#xA;  ionization ability&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How explain this &quot;&lt;strong&gt;observed increase of the ionization ability&lt;/strong&gt;&quot; as a frame dependent measure?  IMO, this &lt;strong&gt;points to an absolute effect&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.physicsforums.com/showpost.php?p=2563558&amp;amp;postcount=66&quot; rel=&quot;nofollow&quot;&gt;see the explanation of the Real Lorentz–FitzGerald contraction (by Hans de Vries)&lt;/a&gt; &#xA;from &lt;a href=&quot;http://physics-quest.org/Book_Chapter_EM_LorentzContr.pdf&quot; rel=&quot;nofollow&quot;&gt;his online book&lt;/a&gt;:  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="1257" LastEditorDisplayName="" LastEditDate="2011-08-19T14:41:40.867" LastActivityDate="2011-08-19T14:41:40.867" CommentCount="5" />
  <row Id="13759" PostTypeId="2" ParentId="13756" CreationDate="2011-08-19T14:42:29.047" Score="4" ViewCount="" Body="&lt;p&gt;There are no bound states for this potential, for any positive $\alpha$. All energies $E&amp;gt;0$ are allowed, and all of those are unbound states, in which $\phi\sim e^{ikr}$ for large $r$, with $E=k^2/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest way to prove that there are no states with $E&amp;lt;0$ is to show that the expectation values of both kinetic energy and potential energy, $\langle T\rangle$ and $\langle V\rangle$, are positive for any valid wavefunction. For an energy eigenfucntion, $E=\langle E\rangle=\langle T\rangle+\langle V\rangle&amp;gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that these aren't bound states -- that is, that they go as $e^{ikr}$ rather than as a decaying function of $r$ at large distances -- is most easily seen from the fact that, at sufficiently large $r$, the potential term is always negligible in comparison to the right-hand side. So at large $r$ we have $-\phi&amp;#39;&amp;#39;/2=E\phi$.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-08-19T14:42:29.047" CommentCount="4" />
  <row Id="13760" PostTypeId="2" ParentId="11284" CreationDate="2011-08-19T15:07:01.387" Score="2" ViewCount="" Body="&lt;p&gt;A $1/r^2$ force occurs because the particle mediating the force is massless. The only way there could be another such particle that had not yet been detected would be if the coupling constant for this force was small or zero for ordinary forms of matter. For example, suppose that we, our world, and our lab apparatus were all made out of neutrons. We might have a very hard time learning of the existence of the electromagnetic force. (Neutrons do have a magnetic moment, though.) I could imagine, for example, that dark matter possesses some new type of charge (not electric charge) that is zero for baryonic matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Re duals, these are general features of all fields due to the way special relativity works. For example, there is a standard, easy argument &lt;a href=&quot;http://www.lightandmatter.com/html_books/0sn/ch11/ch11.html#Section11.1&quot; rel=&quot;nofollow&quot;&gt;http://www.lightandmatter.com/html_books/0sn/ch11/ch11.html#Section11.1&lt;/a&gt; that if we have electrical interactions plus SR, we must have magnetic interactions. However, the detailed nature of the dual depends on the nature of the field equations, or, in QFT terms, on the spin of the force-carrying particle. For example, the graviton has spin 2, not spin 1 like the photon, so we don't have a dual structure exactly like (E,B) in E&amp;amp;M. There are, however, some very nice analogies between the (E,B) structure and what you see in stationary fields such as that of the rotating earth. GR does have phenomena that people refer to as &quot;gravitomagnetism,&quot; etc. But the analogy is not perfect, because otherwise GR would be the same theory as E&amp;amp;M. The fundamental reason for the difference is the spin of 2 instead of 1.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-19T15:07:01.387" CommentCount="9" />
  <row Id="13761" PostTypeId="2" ParentId="13757" CreationDate="2011-08-19T15:26:59.030" Score="3" ViewCount="" Body="&lt;p&gt;Avogadro's number was estimated at first only to order of magnitude precision, and then over the years by better and better techniques. Ben Franklin investigated thin layers of oil on water, but it was only realized later by Rayleigh that Franklin had made a monolayer: &lt;a href=&quot;http://en.wikipedia.org/wiki/Langmuir%E2%80%93Blodgett_film&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Langmuir%E2%80%93Blodgett_film&lt;/a&gt; If you know it's a monolayer, you can estimate the linear dimensions of a molecule and then get an order of magnitude estimate of Avogadro's number (or something equivalent to it). Some of the early estimates of the sizes and masses of molecules were based on viscosity. E.g., the viscosity of a dilute gas can be derived theoretically, and the theoretical expression depends on the scale of its atoms or molecules. Textbooks and popularizations often present a decades-long experimental program as a single experiment. Googling shows that Loschmidt did a whole bunch of different work on gases, including studies of diffusion, deviations from the ideal gas law, and liquified air. He seems to have studied these questions by multiple techniques, but it sounds like he got his best estimate of Avogadro's number from rates of diffusion of gases. It seems obvious to us now that setting the scale of atomic phenomena is an intrinsically interesting thing to do, but it was not always considered mainstream, important science in that era, and it didn't receive the kind of attention you'd expect. Many chemists considered atoms a mathematical model, not real objects. For insight into the science culture's attitudes, take a look at the story of Boltzmann's suicide. But this attitude doesn't seem to have been monolithic, since Loschmidt seems to have built a successful scientific career.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastEditorUserId="4552" LastEditorDisplayName="" LastEditDate="2011-08-19T15:32:23.357" LastActivityDate="2011-08-19T15:32:23.357" CommentCount="1" />
  <row Id="13762" PostTypeId="2" ParentId="13757" CreationDate="2011-08-19T15:30:36.867" Score="3" ViewCount="" Body="&lt;p&gt;It was determined from viscosity and Mean Free Path.&lt;br&gt;&#xA;See &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/menfre.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for an explanation&lt;/p&gt;&#xA;" OwnerUserId="2525" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-08-20T09:27:25.863" LastActivityDate="2011-08-20T09:27:25.863" />
  <row Id="13763" PostTypeId="2" ParentId="11284" CreationDate="2011-08-19T15:36:59.843" Score="0" ViewCount="" Body="&lt;p&gt;I would add that in &lt;a href=&quot;http://en.wikipedia.org/wiki/Grand_Unified_Theory&quot; rel=&quot;nofollow&quot;&gt;Grand Unified Theories&lt;/a&gt; where the unification happens at very high energies, there exist a number of mediating particle X Y etc for forces consistent to  the SU(2)XSU(3)xU(1) low energy structure which in the high energy limit before symmetry breaking are massless and similar to the photon, i.e. of 1/r^2 behavior. Even gluons themselves. More so if one includes supersymmetry in the brew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The verification of the existence of these forces is  an aim of the HEP experiments.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-08-19T15:36:59.843" />
  <row Id="13764" PostTypeId="2" ParentId="13422" CreationDate="2011-08-19T16:18:18.187" Score="3" ViewCount="" Body="&lt;p&gt;How about instead of finding the roots and then making the plots, you skip right to the plots using a Monte Carlo method?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose a random k, then choose a random &amp;omega; and calculate the left-hand side and the right-hand side of the R-L equations. If the RHS is close enough to the LHS (you pick how close), put a point on the plot (blue or black, depending on which branch you used). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more points you process, and the more stringent the condition of RHS=LHS equality you pick, the more accurate will the plot look. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A problem with this approach is that when you put a point on the plot, your algorithm doesn't know which branch it belongs to. But if it is the plot you are after, you will have no problem figuring it out by the eye when the calculation is complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To numerically read from that Monte Carlo plot, you can sort the solutions you find (sort them in k-space or in &amp;omega;-space), and do some kind of search using interpolation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an approximate C code to explain what I mean:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;void main()&#xA;{&#xA;srand(1);&#xA;const int N=1000000000;&#xA;const float eps=0.01;&#xA;const float cl=...;&#xA;const float ct=...;&#xA;const float kmax=20.0;&#xA;const float omegamax=20.0;&#xA;float k, omega, p, q, lhs, rhs;&#xA;for (int i=0; i&amp;lt;N; i++)&#xA; {&#xA;  k=(float)rand()/(float)RAND_MAX*k_max;&#xA;  omega=(float)rand()/(float)RAND_MAX*omega_max;&#xA;  lhs=tan(...)/tan()...;&#xA;  rhs=(4.0*k*k...);&#xA;  if (fabs(lhs-rhs) &amp;lt; eps*fabs(lhs+rhs))&#xA;    printf(&quot;%f %f\n&quot;, k, omega);&#xA; }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and that's it. &lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditorDisplayName="" LastEditDate="2011-08-19T16:34:38.577" LastActivityDate="2011-08-19T16:34:38.577" CommentCount="2" />
  <row Id="13765" PostTypeId="2" ParentId="13711" CreationDate="2011-08-19T16:54:00.667" Score="2" ViewCount="" Body="&lt;p&gt;The answer by Marek was so useful for me, that I wish to share the full problem and the resulting answer (to improve the lasting value of this Q-A):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;: In an equilibrium Fermi gas at inverse thermodynamic temperature &#xA;$\beta$ defined by a set of single particle levels $\epsilon_m$ ($m=0,1,\ldots$)&#xA;which are populated by $n$ fermions, &#xA;the canonical averages of an arbitrary single-particle field $\langle h_m(\nu_m) \rangle_n$&#xA;(where $\nu_m =0, 1$ is the occupation number) can be computed via the generating &#xA;function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ Z[h_m; z] \equiv \sum_{n=0}^{\infty} z^{n} \langle h_m(\nu_m) \rangle_n Z_n &#xD;&#xA;=\sum_{ \{ \nu_k\} } \sum_l h_l(\nu_l) \prod_m e^{-\beta \epsilon_m \nu_m} z^{\nu_m} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution&lt;/strong&gt;:As was shown by Marek,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z[h_m; z]= Z(z) \sum_k \frac{g_k}{f_k}$&#xA;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_k  = 1 + e^{-\beta \epsilon_k} z$,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g_k  = h_k(0) + h_k(1) e^{-\beta \epsilon_k} z$ &#xA;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ Z(z) \equiv \prod_m f_m$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The canonical partition functions $Z_n$&#xA;are generated by $Z(z)= \sum_{n=0}^{\infty} Z_n z^n$. &lt;/p&gt;&#xA;" OwnerUserId="1739" LastActivityDate="2011-08-19T16:54:00.667" />
  <row Id="13766" PostTypeId="2" ParentId="503" CreationDate="2011-08-19T17:00:42.267" Score="-3" ViewCount="" Body="&lt;p&gt;The Einstein relativity is the viewpoint as seen from moving bodys. Since all objects are in motion we think that there is no need to study any other viewpoint. &lt;strong&gt;Is that all?&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One subject that has been negleted is the one-way/two-way speed of light.&lt;br&gt;&#xA;In Einstein SR &lt;em&gt;c&lt;/em&gt; is the &lt;em&gt;mean&lt;/em&gt; light speed in a closed path and it is constant to any moving observer, as it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My comments on this: &lt;a href=&quot;http://en.wikipedia.org/wiki/One-way_speed_of_light#Preferred_reference_frame&quot; rel=&quot;nofollow&quot;&gt;One-way_speed_of_light#Preferred_reference_frame&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A preferred reference frame is a reference frame in which the laws of&#xA;  physics take on a special form.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The ability to make measurements which&#xA;  show the one-way speed of light to be different from its two-way speed&#xA;  would, in principle, enable a preferred reference frame to be&#xA;  determined. This would be the reference frame in which the two-way&#xA;  speed of light was equal to the one-way speed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is about the CMB referential! It is unique and is common to all moving bodies and where the speed of light is isotropic. Since we are not able to measure the one-way light speed, we are persuaded that there are no effects.&lt;br&gt;&#xA;How sure we are if we dont have the relevant studies?&lt;br&gt;&#xA;IMO...&lt;br&gt;&#xA;Afaik the computation of energy is frame dependent. Can I think that the neutrino-(originally a missing energy problem) do not exists in the absolute referential?&lt;br&gt;&#xA;IMO, the flyby anomaly, chirality, matter/antimatter, and a lot of other unsolved issues can have an explanation if we dare to make the studies.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In Einstein's special theory of relativity, all inertial frames of&#xA;  reference are equivalent and there is no preferred frame. There are&#xA;  theories, such as &lt;strong&gt;Lorentz ether theory that are experimentally and&#xA;  mathematically equivalent to special relativity&lt;/strong&gt; but have a preferred&#xA;  reference frame. In order for these theories to be compatible with&#xA;  experimental results the preferred frame must be undetectable. In&#xA;  other words it is a preferred frame in principle only, in practice all&#xA;  inertial frames must be equivalent, as in special relativity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are avoiding to aknowledge that the CMB reference is really special. IMO we are &lt;em&gt;poisoned with prejudice&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this paper &lt;a href=&quot;http://arxiv.org/abs/physics/0205033&quot; rel=&quot;nofollow&quot;&gt;Cosmological Principle and Relativity - Part I&lt;/a&gt; (2002, not peer reviewed, by my friend Alfredo that was fighting against a cancer in those days) is presented a study 'from above'. This work enable us to start to understand what kind of &lt;em&gt;nuances&lt;/em&gt; can appear, namely in Lorentz length contraction formula, if we study light and motion from the CMB viewpoint. Einstein SR equations are also derived as a by-product.&lt;br&gt;&#xA;In conclusion: Einstein viewpoint is correct but it is not all the story.  &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-08-19T17:00:42.267" CommentCount="1" />
  <row Id="13767" PostTypeId="1" AcceptedAnswerId="13768" CreationDate="2011-08-19T17:42:25.117" Score="5" ViewCount="111" Body="&lt;p&gt;As in, if I'm accelerating away from the Earth, then does the Earth &lt;em&gt;also&lt;/em&gt; appear to be accelerating away from me at the same rate? Or is there something to &quot;break&quot; this type of symmetry?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is inspired by the below discussions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.quora.com/Special-Relativity/When-I-am-traveling-at-high-speed-near-speed-of-light-am-I-time-traveling-or-slowing-my-aging-slowing-my-particles-speed-interactions/answer/Anthony-Yeh&quot; rel=&quot;nofollow&quot;&gt;http://www.quora.com/Special-Relativity/When-I-am-traveling-at-high-speed-near-speed-of-light-am-I-time-traveling-or-slowing-my-aging-slowing-my-particles-speed-interactions/answer/Anthony-Yeh&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Twin_paradox#Resolution_of_the_paradox_in_special_relativity&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Twin_paradox#Resolution_of_the_paradox_in_special_relativity&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1576" LastActivityDate="2011-08-19T19:02:45.843" Title="How are accelerated reference frames non-symmetrical?" Tags="&lt;relativity&gt;" AnswerCount="3" FavoriteCount="3" />
  <row Id="13768" PostTypeId="2" ParentId="13767" CreationDate="2011-08-19T17:53:56.360" Score="7" ViewCount="" Body="&lt;p&gt;Kinematically, yes.  In terms of describing the positions of objects, it is equivalent to say &quot;A is accelerating away from B&quot; and &quot;B is accelerating away from A&quot;.  However, it is an observed fact that the universe treats these two situations differently.  A and B can check whether they feel artificial gravity in their reference frame.  If so, it's accelerating.  As far as I know, the &quot;way the universe decides&quot; to break this symmetry is a topic of continuing speculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out some related questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/13008/inertia-in-an-empty-universe&quot;&gt;Inertia in an empty universe&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/173/is-acceleration-an-absolute-quantity&quot;&gt;Is acceleration an absolute quantity?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/1372/is-rotational-motion-relative-to-space&quot;&gt;Is rotational motion relative to space?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/6742/acceleration-in-special-relativity&quot;&gt;Acceleration in special relativity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/3986/newtons-bucket&quot;&gt;Newton&amp;#39;s Bucket&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/1048/what-if-the-universe-is-rotating-as-a-whole&quot;&gt;What if the universe is rotating as a whole?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-08-19T17:53:56.360" />
  <row Id="13769" PostTypeId="2" ParentId="13767" CreationDate="2011-08-19T18:45:51.987" Score="5" ViewCount="" Body="&lt;p&gt;This boils down to the question of how inertial reference frames are defined in GR. GR defines them differently than Newtonian mechanics does. In GR, a frame is inertial if it's defined by a free-falling particle. The earth is essentially a free-falling particle, so its center defines an inertial frame of reference. A person in a rocket ship accelerating away from the earth is not free-falling. One way to tell the difference is that an accelerometer at the center of the earth will read zero, while one in the ship reads a nonzero value.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-19T18:45:51.987" />
  <row Id="13770" PostTypeId="2" ParentId="3294" CreationDate="2011-08-19T18:49:58.767" Score="1" ViewCount="" Body="&lt;p&gt;The first thing to understand is that the Big Bang was not an explosion that happened at one place in a preexisting, empty space. The Big Bang happened everywhere at once, so there is no location that would be the place where we would expect a black hole's singularity to form. Cosmological models are either exactly or approximately homogeneous. In a homogeneous cosmology, symmetry guarantees that tidal forces vanish everywhere, and that any observer at rest relative to the average motion of matter will measure zero gravitational field. Based on these considerations, it's actually a little surprising that the universe ever developed any structure at all. The only kind of collapse that can occur in a purely homogeneous model is the recollapse of the entire universe in a &quot;Big Crunch,&quot; and this happens only for matter densities and values of the cosmological constant that are different from what we actually observe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A black hole is defined as a region of space from which light rays can't escape to infinity. &quot;To infinity&quot; can be defined in a formal mathematical way,[HE] but this definition requires the assumption that spacetime is asymptotically flat. To see why this is required, imagine a black hole in a universe that is spatially closed. Such a cosmology is spatially finite, so there is no sensible way to define what is meant by escaping &quot;to infinity.&quot; In cases of actual astrophysical interest, such as Cygnus X-1 and Sagittarius A*, the black hole is surrounded by a fairly large region of fairly empty interstellar space, so even though our universe isn't asymptotically flat, we can still use a portion of an infinite and asymptotically flat spacetime as an approximate description of that region. But if one wants to ask whether the entire universe is a black hole, or could have become a black hole, then there is no way to even approximately talk about asymptotic flatness, so the standard definition of a black hole doesn't even give a yes-no answer. It's like asking whether beauty is a U.S. citizen; beauty isn't a person, and wasn't born, so we can't decide whether beauty was born in the U.S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black holes can be classified, and we know, based on something called a no-hair theorem, that all static black holes fall within a family of solutions to the Einstein field equations called Kerr-Newman black holes. (Non-static black holes settle down quickly to become static black holes.) Kerr-Newman black holes have a singularity at the center, are surrounded by a vacuum, and have nonzero tidal forces everywhere. The singularity is a point at which the world-lines only extend a finite amount of time into the future. In our universe, we observe that space is not a vacuum, and tidal forces are nearly zero on cosmological distance scales (because the universe is homogeneous on these scales). Although cosmological models do have a Big Bang singularity in them, it is not a singularity into which future world-lines terminate in finite time, it's a singularity from which world-lines emerged at a finite time in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more detailed and technical discussion is given in [Gibbs].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[HE] Hawking and Ellis, The large-scale structure of spacetime, p. 315.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Gibbs] &lt;a href=&quot;http://math.ucr.edu/home/baez/physic.../universe.html&quot; rel=&quot;nofollow&quot;&gt;http://math.ucr.edu/home/baez/physic.../universe.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a FAQ entry written by the following members of physicsforums.com:&#xA;bcrowell&#xA;George Jones&#xA;jim mcnamara&#xA;marcus&#xA;PAllen&#xA;tiny-tim&#xA;vela&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-19T18:49:58.767" />
  <row Id="13771" PostTypeId="2" ParentId="5714" CreationDate="2011-08-19T18:56:22.183" Score="6" ViewCount="" Body="&lt;p&gt;In physics, good ideas that fail tend to be refined and recycled into correct theories later. &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;LeSage gravity and Feynman diagrams&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;One of the best examples is the LeSage theory of gravitation. LeSage imagined that all of space is filled with particles moving at some enormous speed c, and that gravity is caused by the absorption of these particles by matter, casting a shadow in the particles behind them. The shadow leads to inverse-square attraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This theory died because it would naively predict friction in the LeSage ether, which would wreck the orbits of the planets (of course, with out modern views, we would make the LeSage ether relativistically invariant, fixing this problem, but this was unavailable at the time). But despite this, it was very popular throughout the 19th century. Preston and DePretto, two 19th century lay-scientists, conjectured based on this theory that all mass contains an amount of energy proportional to $mc^2$ (the speed of the LeSage particles had settled down to c by the end of the 19th century).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 20th century, Feynman's particle exchange theory of forces revived LeSagian ideas, but this time for good. Feynman's explanation keeps the idea that the inverse square law comes from particle exchange diluted by geometry as they go from on object to another, but makes it quantum, and makes it relativistically invariant. It also becomes universal to all forces, showing how repulsion works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feynman, I suppose in honor of LeSage, devotes a section of the Feynman lectures to the LeSage model, to show how theories can fail.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Kelvin's ether atoms&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The idea of &quot;knots in the ether&quot;, that atoms are topological vortex lines in a fluid filling all space, was explicitly revived by Tony Skyrme in 1960, when he made a model of Baryons as topological defects in the newly discovered pion condensate. This idea was considered dead for many years, until Rajeev, Nair, and Balachandran, followed closely by Witten, showed that it emerges within the large-N (string) model of QCD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea that Baryon interactions are well modeled by tying knots in the pion ether is roughly successful at predicting the structure of small nuclei.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;S-matrix bootstrap&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The original idea of the bootstrap dates back to Heisenberg 1940, but really got going after Pomaranchuk, Gribov, Mandelstam and Gell-Mann started to think about the idea in the late 1950s. The point of bootstrap was to give a description of hadronic interactions based on the exchange of particles which would be formed through the same interaction. Geoffrey Chew proselyzed for this so effectively that it became the dominant paradigm in physics between 1964 and 1974.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During this time, the interaction of Regge trajectories was understood qualitatively, then quantitatively, and a whole host of new ideas emerged from this: the pomeron, conspiracies, dispersion relations, unitarity cuts, etc. These ideas are not as well understood today as they were back then, because almost all those people were lost to physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But they developed string theory in the last 6 years, and string theory pulled all their ideas out of the dustbin.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Yang Mills theory (as Yang Mills wanted)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;When Yang and Mills proposed their theory of an SU(2) gauge theory, they wanted it to gauge Isospin. The fundamental interactions don't do that, but Sakurai predicted the $\rho$ meson based on this idea. There are two light vector mesons, the $\rho$ and $\omega$, and the idea that they are gauging isospin and hypercharge led to the idea that they mix with the photon at hadronic energies. This predicted photon nucleon scattering well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory predicts coupling constant universality-- all isodoublets should have the same coupling to the rho. But it doesn't explain the mass of the rho, since Isospin is unbroken. These ideas died out, as it was realized that these particles are not fundamental. But in modern AdS/QCD it is completely rehabilitated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what the experimental situation is with respect to coupling constant universality. I will ask that as a question.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-19T18:56:22.183" CommunityOwnedDate="2011-08-19T18:56:22.183" />
  <row Id="13772" PostTypeId="1" CreationDate="2011-08-19T19:02:25.233" Score="4" ViewCount="62" Body="&lt;p&gt;Is there any good recent experimental data that tests whether the $rho$ coupling constant depends only on the isospin multiplet of the produced particle?&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-19T19:02:25.233" Title="How well is the $\rho$ and $\omega$ coupling universality measured?" Tags="&lt;particle-physics&gt;" CommentCount="2" />
  <row Id="13773" PostTypeId="2" ParentId="13767" CreationDate="2011-08-19T19:02:45.843" Score="3" ViewCount="" Body="&lt;p&gt;If you are in a rocket, you will see the Earth accelerating from you away. But everything on the Earth will occur in a regular way. In particular an accelerometer on the Earth will register zero acceleration and you may see it in your window. Your accelerator will show some acceleration and you will also feel it yourself. So accelerated RFs are different.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-08-19T19:02:45.843" />
  <row Id="13774" PostTypeId="2" ParentId="13738" CreationDate="2011-08-19T19:11:35.533" Score="1" ViewCount="" Body="&lt;p&gt;A laser beam passing though transparent glass does not lose energy (or momentum) consequently there's no reason for it to become incoherent. But this is not directly related to the question of whether there is scattering or absorption and delayed emission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a difficulty in relating what happens on a photon by photon basis with what happens in the classical limit. You know that light traveling through glass is slowed down, and it seems that this must be due to a delay between absorption and emission, and I suppose that this seems to imply that there must be a delay between absorption and emission in the quantum analysis of the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Quantum mechanics&quot; is limited to situations where the particles are neither created nor destroyed; this can't include absorption and emission. And it's not terribly easy with large number of particles in the same state (as happens in a laser). The more general &quot;Quantum Field Theory&quot; allows creation and annihilation of particles (photons) so this is what's appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantum field theory that applies to photons interacting with electrons (atoms) is quantum electrodynamics (QED). In this theory, the speed of photons is random; they do not have to travel at speed c. So it's possible for a QED problem to have a delayed photon without a delay between absorption and emission (but the average speed will still be c, as far as I know).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A crystal (or other solid) lattice contains a large number of atoms. What determines whether or not a photon can be absorbed is whether the crystal lattice has an available vibration corresponding to the energy, momentum, and angular momentum of the incoming photon. If it does, then the solid will absorb that photon and it will not be transparent in that frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a photon has an energy that is not compatible with the crystal lattice then it must adjust its energy by borrowing energy. This can only happen for a time $\Delta E\;\Delta t \approx \hbar$. One supposes that energies where $\Delta E$ is small should have a slower speed of light (larger index of refraction).&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="4760" LastEditorDisplayName="" LastEditDate="2011-08-20T09:27:42.230" LastActivityDate="2011-08-20T09:27:42.230" CommentCount="2" />
  <row Id="13775" PostTypeId="2" ParentId="13714" CreationDate="2011-08-19T19:23:34.980" Score="1" ViewCount="" Body="&lt;p&gt;I think style is more important than topic.   A talk with different viewpoints (math - physics - experiment - connections between topics) and different levels of abstraction and complexity can be really great.   I find nobody is offended if you spend some time on the big picture as long as you don't dumb down the entire talk.  People will forgive you if they become confused when you spend some time trying to be clear about a really complex issue.  People tend not to be offended if they are bored or confused by some fraction of the talk (as long as its not the whole thing).  And people tend to like things that have big connections or give glimpses to complicated stuff that they would like to learn more about.  People like it if you take the time to explain some tricky stuff (even if they forget it two seconds later).  And the glimpse to the future or to big (and important) unsolved problems can be a hook for your audience.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another style issue is to consider your talk as a `story'.  People like certain types of stories, ..e.g.,  I tried this but it failed, I was misled by this idea but then understood why and then I tried this and it worked! (Obstacles overcome by protagonist).   If you set up a paradox and then explain it.  If you present something as hopelessly complex and then outline a route to breaking it down to understandable pieces.  If you suggest a simple experiment and then describe why it gave wrong and misleading results....  I try to remember that as a lecturer I am essentially an entertainer.  I am not naturally a joker so the story line is key.   I would recommend you choose a topic you really care about (and let that come through in your story) rather than suggest a topic.&lt;/p&gt;&#xA;" OwnerUserId="4554" LastEditorUserId="4554" LastEditorDisplayName="" LastEditDate="2011-08-19T19:29:23.063" LastActivityDate="2011-08-19T19:29:23.063" CommunityOwnedDate="2011-08-19T19:23:34.980" />
  <row Id="13776" PostTypeId="1" AcceptedAnswerId="13792" CreationDate="2011-08-20T03:04:25.193" Score="1" ViewCount="51" Body="&lt;p&gt;I have a forced oscillating system, with driving force as $f_0\cos\omega_0 t \cos \delta t$ giving the equation of motion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ddot{x}(t) +\Gamma \dot{x}(t) +\omega_0^2x(t) = f_0\cos\omega_0 t \cos \delta t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notation is standard, i.e $\Gamma$ is the damping coefficient, $\omega_0$ is the natural frequency of the undamped system, i.e $ \omega_0= \sqrt{k/m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is given that $\delta \ll \omega_0$ and $\Gamma =0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question asks for the displacement $x$ for non-zero $\delta$ to leading order in $\delta/\omega_0$. The next statement is to express it in the for $\alpha(t)\cos\omega_0 t + \beta(t) \sin\omega_0 t $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following hint is provided: $$\cos\omega_0 t \cos \delta t = \frac{1}{2} Re\Big(e^{-i(\omega+\delta)t}+e^{-i(\omega-\delta)t}\Big)$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I don't want the solution. I cannot understand how to approach the problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the equation of motion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ddot{z}(t) - \omega^2 z(t)  = \frac{f_0}{2} \Big(e^{-i(\omega+\delta)t}+e^{-i(\omega-\delta)t}\Big)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I make use of the approximation $\delta \ll \omega_0$ I conduct the expansion to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ddot{z}(t) - \omega^2 z(t)  = f_0 \Big(1-i t\omega_0 - t^2\omega_0^2 \Big(1+\Big(\frac{\delta}{\omega_0}\Big)^2\Big)\Big)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this hasn't simplified it enough to allow me to solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The methods I now know of solving the original equation (Green's function and variation of parameters) have not been introduced, so the author must require us to work under certain approximations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source is Howard Georgi's physics of waves. This is not homework.&lt;/p&gt;&#xA;" OwnerUserId="2599" LastActivityDate="2011-08-20T15:38:25.207" Title="Help understanding this forced undamped oscillator" Tags="&lt;harmonic-oscillator&gt;&lt;oscillators&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13777" PostTypeId="1" CreationDate="2011-08-20T03:34:44.283" Score="2" ViewCount="95" Body="&lt;p&gt;In the process $A\rightarrow B+C$ If we know the energy and momentum four vectors of B and C, then how can we find the momentum three vector of A. The particular proces I have in mind is pair production $\gamma\rightarrow e^+ + e^-$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4-vectors for electron is $(E_1,\vec{p_1})$&#xA; and for positron it is $(E_2,\vec{p_2})$. Then for the photon all I can deduce is $E = E_1+E_2$ (energy should be conserved) and $|\vec{p}| = |E|/c$ . Where $(E,\vec{p})$ is the four momentum of the photon. Can I also find the three components of the momentum of the photon, given that I have the components of the momenta of electrons and positrons (i.e I know $\vec{p_1}$ and $\vec{p_2}$)?&lt;/p&gt;&#xA;" OwnerUserId="3239" LastActivityDate="2011-08-20T18:08:17.343" Title="Pair production kinematics" Tags="&lt;special-relativity&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="13778" PostTypeId="2" ParentId="13738" CreationDate="2011-08-20T04:39:32.827" Score="2" ViewCount="" Body="&lt;p&gt;There is no unambiguous correct answer to this question because it isn't well posed in terms of logical positivism: what is the difference between the two processes? There is no way to tell which happens if you don't muck up the intermediate state with a measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean this in terms of some quantum field theory with given fields and interactions and asymptotic states, then you can ask how the processes appear in a Feynman description. The scattering process in QED is always two-step, the absorption and emission are separate space-time points. But the emission can precede the absorption both in coordinate time and in proper time along the electron's world line, so you should include &quot;emitted and then absorbed&quot; to the list of possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light does not have to be resonant in order to scatter off an atom. The amount of scattering/emission-reabsorption is smaller away from resonance. A light wave is also a long coherent field, and this field can acquire a phase push from the emission-reabsorption, leading the phase-velocity to be bigger than the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue of &quot;how come the phases add up coherently&quot; is adressed by two things: there is a large scale difference between the atoms and the light wavelength. Each atom scatters the light independently and randomly into a spherical waves, which add up coherently in the original direction only to alter the phase velocity by a constant amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no scattering from the bulk of a perfect crystal, for long wavelengths, because there is still a discrete translation invariance which means momentum is conserved up to big jumps, and the big jumps give waves with the wrong frequency for long enough wavelengths. But there are discrete momentum additions which are allowed for a short wavelength x-ray in an atomic crystal, and if the photon momentum comes out different but at the same frequency due to the coherence in a different direction, that's called diffraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want scattering in a crystal, you need to scatter off defects which have a good amount of random variation in a box the size of one wavelength. Similarly, if you scatter of a fluid, you need fluctuations in density to be meaningful in one wavelength. This is easier for blue light than for red light, so transparent fluids scatter blue.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-20T04:39:32.827" />
  <row Id="13779" PostTypeId="2" ParentId="4236" CreationDate="2011-08-20T04:48:02.530" Score="1" ViewCount="" Body="&lt;p&gt;It is important to note that a line of soldiers marching forward change direction inward when they speed up and change direction outward when they slow down, exactly the opposite of what light does. The old Newtonian corpuscular theory of light predicted that light travels faster in glass, which is I suppose why the index of refraction is defined as the reciprocal of what you would expect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The naive particle theory is incorrect, and the wave theory explains refraction very simply, by matching peaks of waves moving at different velocities on either side of the barrier.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-20T04:48:02.530" />
  <row Id="13780" PostTypeId="2" ParentId="743" CreationDate="2011-08-20T06:07:58.283" Score="1" ViewCount="" Body="&lt;p&gt;The standard reference for many decades, which has not been improved upon in my opinion, is Kenneth Wilson's 1974 Reviews of Modern Physics article. It used to be required reading. It is a little old however.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-20T06:07:58.283" CommentCount="1" />
  <row Id="13781" PostTypeId="2" ParentId="6086" CreationDate="2011-08-20T06:32:20.973" Score="2" ViewCount="" Body="&lt;p&gt;The answers to this question all get the answer wrong. The answer is that an accelerating frame has exactly the same horizon as the black hole, so that the equivalence principle holds. It does not hold infinitesimally as you approach the horizon, it holds including the horizon, if you identify the black hole horizon with the Rindler horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length scale at which the EP fails is the inverse curvature, which is as large as you like compared to the distance to the horizon. So the motion of the ball and the rod is the same in a uniformly accelerated frame as it is next to a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This type of equivalence principle, with a short distance to the horizon, was never used by Einstein, but it's sort of folklore by now!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LATER EDIT: I see that this answer might be interpreted as lending support to the claimed violation of the equivalence principle in the OP's question. There is absolutely no violation of the equivalence principle, and this can be easily seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a rigid rod L in the horizontal direction, it is impossible to accelerate it horizontally while keeping it rigid with an acceleration any greater than&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{\mathrm{max}} = c^2/L$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because then the left-most point would be past the Rindler horizon of the right most point. If you try to do this to a rod, it gets properly longer, because the acceleration on the left point can't keep up (this is easily seen in a space-time diagram). The intuition that fails is that there is such a thing as &quot;uniform acceleration of a rigid rod&quot;. So when the rod is longer than the distance to the horizon it will not be able to pass the particle in the inertial frame before the whole frame reaches the horizon and the question is moot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, it is impossible to find a contradiction between a black hole horizon and the EP, because the near horizon metric is Rindler, up to curvature corrections which are arbitrarily small, so it is equivalent to a flat space, and there is no thought experiment which can refute this in a black hole which doesn't work in flat space just the same.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-23T18:21:35.967" LastActivityDate="2011-08-23T18:21:35.967" CommentCount="15" />
  <row Id="13782" PostTypeId="2" ParentId="743" CreationDate="2011-08-20T07:37:51.707" Score="1" ViewCount="" Body="&lt;p&gt;Some very good references given so far.  I don't think I've seen this one mentioned yet - &quot;Regularization, renormalization and dimensional analysis: Dimensional Regularization meets Freshman E&amp;amp;M&quot;  by F Oleness and R Scalise.  Available &lt;a href=&quot;http://www.ate.uni-duisburg-essen.de/data/postgraduate_lecture/AJP_2011_Olness.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, it gives an extremely readable introduction to regularization and renormalization at the level appropriate for a very first exposure.&lt;/p&gt;&#xA;" OwnerUserId="3099" LastActivityDate="2011-08-20T07:37:51.707" />
  <row Id="13783" PostTypeId="1" AcceptedAnswerId="13785" CreationDate="2011-08-20T08:10:30.843" Score="4" ViewCount="87" Body="&lt;p&gt;Suppose an isolated system with a number of particles with parallel spins. Can the macroscopic angular momentum of the system increase at expense of the number of particles having parallel spins (that is by inverting the direction of the spins of some of them) or by converting all particles into spinless ones?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely in a rotating system of spinless particles can the macroscopic algular momentum be decreased by converting some of the particles into those having parallel spins?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can one choose a rotating frame of reference in such a way so in it a particle that has spin in inertial frame, does not have one?&lt;/p&gt;&#xA;" OwnerUserId="1186" LastActivityDate="2011-08-20T08:40:55.343" Title="Can spin and macroscopic angular momentum convert to each other?" Tags="&lt;spin&gt;" AnswerCount="1" />
  <row Id="13784" PostTypeId="2" ParentId="13752" CreationDate="2011-08-20T08:36:56.307" Score="1" ViewCount="" Body="&lt;p&gt;Since you're specifically asking for references, the only reference on this subject that I'm aware of is an old artlcle by Wu-ki Tung:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Relativistic Wave Equations and Field Theory for Arbitrary Spin&quot;&#xA;Phys Rev Vol 156 #5 pp1385-1398  Apr 1967&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be available online somewhere.  It talks about the extra restrictions you might want to impose in order to obtain reasonable field equations and talks a little about what's special about the cases of spin 0, 1/2 and 1.&lt;/p&gt;&#xA;" OwnerUserId="3099" LastActivityDate="2011-08-20T08:36:56.307" />
  <row Id="13785" PostTypeId="2" ParentId="13783" CreationDate="2011-08-20T08:40:55.343" Score="5" ViewCount="" Body="&lt;p&gt;First of all yes, you can convert angular momentum to spin orientations, as only the total angular momentum $\vec J=\vec L+\vec S$ is conserved. The setup you describe is very similar to a famous experimental effect dubbed the &quot;Einstein-de-Haas&quot; effect. You can read more about it on &lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein%E2%80%93de_Haas_effect&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;. The main idea is that you have a ferromagnet hung on a string and when you magnetize it (which really only means that you orient the spins of the electrons in a coordinated way) the magnet has to pick up angular (mechanical) momentum, since angular momentum is conserved. What you observe is that it starts to turn seemingly out of nothing once you magnetize it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the part where you are mistaken is that you cannot change the spins of particles. The spin of a particle is a property of the particle itself (much like its mass. In fact those are the two properties particles have as a result of their being representations of the symmetry group of spacetime, c.f. Wigner's classification). You can only orient a component of the spin, but not the total spin magnitude. A Spin 1/2 particle (fermion) will always stay a fermion.&lt;/p&gt;&#xA;" OwnerUserId="2151" LastActivityDate="2011-08-20T08:40:55.343" CommentCount="7" />
  <row Id="13786" PostTypeId="2" ParentId="11078" CreationDate="2011-08-20T10:37:55.507" Score="-1" ViewCount="" Body="&lt;p&gt;I gave a popular explanation of &quot;how our interaction modifies the fundamental constants&quot; &lt;strong&gt;&lt;a href=&quot;http://www.science20.com/qed_reformulation_feasible/blog/ultimate_explanation_renormalizations-81791&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-08-20T10:37:55.507" CommentCount="4" />
  <row Id="13787" PostTypeId="1" AcceptedAnswerId="13813" CreationDate="2011-08-20T12:27:32.880" Score="8" ViewCount="145" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Spin-statistics_theorem&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's article on the spin-statistics theorem&lt;/a&gt; sums it up thusly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In quantum mechanics, the spin-statistics theorem relates the spin of a particle to the particle statistics it obeys. The spin of a particle is its intrinsic angular momentum (that is, the contribution to the total angular momentum which is not due to the orbital motion of the particle). All particles have either integer spin or half-integer spin (in units of the reduced Planck constant ħ).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The theorem states that:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;the wave function of a system of identical integer-spin particles has the same value when the positions of any two particles are swapped. Particles with wavefunctions symmetric under exchange are called bosons;&lt;/li&gt;&#xA;  &lt;li&gt;the wave function of a system of identical half-integer spin particles changes sign when two particles are swapped. Particles with wavefunctions anti-symmetric under exchange are called fermions.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;In other words, the spin-statistics theorem states that integer spin particles are bosons, while half-integer spin particles are fermions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question.&lt;/strong&gt; What would you recommend as the best means to understanding the proof of the spin statistics theorem? To put it another way: if you took (for instance) the proof-sketch for the spin-statistics theorem given later in that same Wikipedia page, and wanted to add enough material to it to create a textbook whose entire purpose was to take someone from a third- or fourth-year undergrad level of physics up to understanding the spin-statistics theorem inside-out, what references would you use to flesh out the material of that book?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Context.&lt;/strong&gt; I'm a researcher in quantum computation, coming from the computer science end of things: I have a very solid grasp on the basic mathematical framework of &lt;em&gt;non-relativistic&lt;/em&gt; QM, and some special relativity, if not all applicable techniques. I have very basic familiarity with fermionic and bosonic operator algebras, as algebras generated by creation/annihilation operators satisfying certain axioms, though I have not had occasion to use them much myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I take for granted that understanding the proof of the spin-statistics theorem will involve learning a non-trivial amount of physics (and also likely mathematical) background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any recommendations? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited to add:&lt;/strong&gt; If it is sufficient to learn quantum field theory, please recommend a suitable text as an answer. For instance, if you know of a good book on QFT that does not assume much of a background in particular topics such as E&amp;amp;M, and which definitely covers all of the concepts pertinent to the proof-sketch given on the Wikipedia page, and/or itself has a good self-contained proof of the spin-statistics theorem &amp;mdash; in short, a book which can take me from Schr&amp;ouml;dinger and Einstein all the way to the spin-statistics theorem &amp;mdash; then please recommend the book in question as an answer.&lt;/p&gt;&#xA;" OwnerUserId="4976" LastEditorUserId="4976" LastEditorDisplayName="" LastEditDate="2011-08-20T23:29:53.147" LastActivityDate="2011-08-21T14:09:03.703" Title="A reading list to build up to the spin statistics theorem" Tags="&lt;quantum-field-theory&gt;&lt;reference-request&gt;&lt;spin&gt;&lt;fermions&gt;&lt;spin-statistics&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" />
  <row Id="13788" PostTypeId="2" ParentId="13728" CreationDate="2011-08-20T13:53:44.960" Score="0" ViewCount="" Body="&lt;p&gt;After reading the exchange between Lubos and the other contributors, I think it is worthwhile to point out that theories and models are somewhat different things.  Theories provide a framework for building of models.  String theory encompasses all classess of models that use linear objects instead of point objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The particular mechanics of the models used are of interest only if they lead to accurate predictions.  Within the context of string theory, models have been built that can be used to make predictions.  There are already several hundred of these models.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Lubos has pointed out, there exists dualities between certain string theory models and models built based on other theories. There are also higher level theories, such as M-theory, that can utilize dualties to unify several classes of string theory models.  &lt;/p&gt;&#xA;" OwnerUserId="4977" LastActivityDate="2011-08-20T13:53:44.960" CommentCount="5" />
  <row Id="13789" PostTypeId="2" ParentId="743" CreationDate="2011-08-20T14:19:09.743" Score="1" ViewCount="" Body="&lt;p&gt;My &quot;eye-opener&quot; about RG has been Amnon Aharony, see his book with Dietrich Stauffer &quot;&lt;a href=&quot;http://books.google.com/books/about/Introduction_to_percolation_theory.html?id=v66plleij5QC&quot; rel=&quot;nofollow&quot;&gt;Introduction to Percolation theory&lt;/a&gt;&quot;.  The exposition of RG is totally focused on ideas,&#xA;and the (relatively) simple subject of the book -- a classical statistical lattice problem of percolation --  allows for very intuitive demonstrations, not requiring any QFT skills. Aharony became a postdoc of Michael Fisher a month after the Nobel-winning &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v28/i4/p240_1&quot; rel=&quot;nofollow&quot;&gt;paper by Wilson &amp;amp; Fisher&lt;/a&gt; came out in 1972 (&lt;a href=&quot;http://astrophysics.fic.uni.lodz.pl/100yrs/pdf/06/093.pdf&quot; rel=&quot;nofollow&quot;&gt;full text pdf&lt;/a&gt;), so the statistical physics roots of renormalization are very vivid in the book.&lt;/p&gt;&#xA;" OwnerUserId="1739" LastActivityDate="2011-08-20T14:19:09.743" />
  <row Id="13790" PostTypeId="1" CreationDate="2011-08-20T14:42:28.603" Score="0" ViewCount="54" Body="&lt;p&gt;If incident electromagnetic wave is given as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}E_i&amp;amp;=A_e \cos(\omega t + bz)\\&#xD;&#xA;H_i&amp;amp;=A_h \cos(\omega t + bz)\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be relation for REFLECTED wave? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it go like this?&#xA;$$\begin{align*}E_r&amp;amp;=A_e \cos(\omega t - bz)\\&#xD;&#xA;H_r&amp;amp;=A_h \cos(\omega t - bz)\end{align*}$$&lt;/p&gt;&#xA;" OwnerUserId="4979" LastEditorUserId="4979" LastEditorDisplayName="" LastEditDate="2011-08-20T17:44:11.050" LastActivityDate="2011-08-20T17:44:11.050" Title="Reflected electromagnetic wave relation" Tags="&lt;electromagnetism&gt;&lt;waves&gt;" CommentCount="5" />
  <row Id="13791" PostTypeId="2" ParentId="13777" CreationDate="2011-08-20T14:59:04.363" Score="2" ViewCount="" Body="&lt;p&gt;This question is quite elementary but since it's not getting an answer, let me provide one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know (hopefully!) that both energy and momentum is conserved, therefore both $E_{\gamma} = E_1 + E_2$ and $\vec p_{\gamma} = \vec p_1 + \vec p_2$. One might get confused by the fact that this process can't occur by itself but this can easily be fixed by adding additional observer particle (by which I mean that it doesn't affect the process in any serious way; beyond, of course, making it possible in the first place) in the initial and final state and then neglect its contribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in the four-vector formalism we can express this more succintly as conservation of energy-momentum in the following way&#xA;$$p_{\gamma} + p_{\text observer} = p_1 + p_2 + p&amp;#39;_{\text observer}$$&#xA;and by the definition of observer we have $p_{\text observer} \sim p&amp;#39;_{\text observer}$ and we cancel these observing terms on both sides of the equation.&lt;/p&gt;&#xA;" OwnerUserId="329" LastActivityDate="2011-08-20T14:59:04.363" CommentCount="3" />
  <row Id="13792" PostTypeId="2" ParentId="13776" CreationDate="2011-08-20T15:38:25.207" Score="2" ViewCount="" Body="&lt;p&gt;Do you know what the solutions to the equation of motion look like in the case where there's only one sinusoidal term on the right-hand side, i.e., &#xA;$$&#xD;&#xA;\ddot z-\omega^2z={f_0\over 2}e^{-i(\omega+\delta)t},&#xD;&#xA;$$&#xA;and the corresponding equation with $\omega-\delta$? If so, then you're most of the way there. When you have an inhomogeneous linear differential equation with multiple terms on the right side (which is what you've got), you can solve it for each term separately and add the solutions together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If this isn't a fact you already know from either a physics or a mathematics class, you can probably convince yourself of it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be precise, your solution can be written in the form&#xA;$$&#xD;&#xA;z(t)=z_+(t)+z_-(t)+z_g(t),&#xD;&#xA;$$&#xA;where $z_+$ is a particular solution to the equation containing just the first term on the right, $z_-$ is a particular solution with just the second term on the right, and $z_g$ is the general solution to the homogeneous equation (zero on the right).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In case you don't know how to solve the two inhomogeneous equations, I'll mention that you can do it by guessing that the solution is of the form $Ae^{-i\alpha t}$ where $\alpha=\omega\pm\delta$ is the driving frequency, and working out what value of $A$ works. But I suspect you probably already knew that.)&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-08-20T15:38:25.207" />
  <row Id="13793" PostTypeId="1" AcceptedAnswerId="13799" CreationDate="2011-08-20T17:46:45.247" Score="1" ViewCount="57" Body="&lt;p&gt;Imagine a partially filled $\nu=1$ state of the integer quantum Hall effect (IQHE). One way to think about it is to imagine a gas of electrons where each particle is locked to the lowest quantum state of their circular motion (a Gaussian wave-packet in both $x$ and $y$ with the characteristic length being the magnetic length $l_0$). When this gas becomes degenerate, we get a fully filled $v=1$ Landau level and the bulk conductivity vanishes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder, how does a single-particle theory of a partially filled$\nu=1$ IQH state state look like in terms of such particles? What is the kinetic energy term? I seek a kind of field theory for this, valid on the scales of  $k \ll l_0^{-1}$ and regularized in a consistent way on the scale of $l_0$. Perhaps a tight-binding-like lattice model? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would bet it has been done (if possible at all) by some &quot;high-flying&quot; quantum field theorists out there , but I have not seen such concept being used in &quot;down-to-Earth&quot; experiment-oriented calculations of IQHE.&lt;/p&gt;&#xA;" OwnerUserId="1739" LastActivityDate="2011-08-20T20:27:05.107" Title="&quot;Classical&quot; limit of Quantum Hall Effect" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;condensed-matter&gt;&lt;quantum-hall-effect&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13794" PostTypeId="2" ParentId="13777" CreationDate="2011-08-20T18:08:17.343" Score="4" ViewCount="" Body="&lt;p&gt;As luksen and Marek point out, when the incoming particle is massless this reaction can't go, unless there's a third particle around to soak up the momentum. If we allow for the existence of this particle, then I don't think there's enough information to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p_{Xi},p_{Xf}$ be the initial and final 4-momenta of that particle. Then conservation of 4-momentum says&#xA;$$&#xD;&#xA;p_\gamma+p_{Xi}=p_1+p_2+p_{Xf}&#xD;&#xA;$$&#xA;Assume that the extra particle was initially at rest: $p_{Xi}=(M,0,0,0)$. ($M$ is the mass of $X$, and I'm setting $c=1$.) We don't know anything about any of the components of $p_\gamma$ or $p_{Xf}$, so there are eight unknowns to solve for. The above is 4 equations, so we need 4 additional equations. Unfortunately, we have only two,&#xA;$$&#xD;&#xA;E_\gamma=|\vec p_\gamma|,&#xD;&#xA;$$&#xA;$$&#xD;&#xA;E_{Xf}^2=|\vec p_{Xf}|^2+M^2&#xD;&#xA;$$&#xA;(Here $E$ means energy and $\vec p$ means 3-momentum in the frame we're working in, of course. These two equations are just $p_\mu p^\mu=m^2$, if you prefer.) So there's not enough information to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The typical assumption is that the extra particle is much more massive than the others. In this case, the energy it absorbs is typically negligible, but the 3-momentum isn't. So if the 3-momentum is what you're interested in, I don't think you can get by with just pretending that particle isn't there.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-08-20T18:08:17.343" CommentCount="7" />
  <row Id="13795" PostTypeId="1" AcceptedAnswerId="13798" CreationDate="2011-08-20T19:26:05.880" Score="1" ViewCount="77" Body="&lt;p&gt;I was wondering: If I put three bottles of water next to each other in the fridge, which one is cold first? Does it matter? Is it the one in the middle because it gets refrigerated by the other two?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thx! :)&lt;/p&gt;&#xA;" OwnerUserId="4981" LastActivityDate="2011-08-20T21:54:28.883" Title="If I put 3 bottles of water next to each other in the fridge, which one is cold first?" Tags="&lt;heat&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="13796" PostTypeId="2" ParentId="5714" CreationDate="2011-08-20T19:54:12.060" Score="2" ViewCount="" Body="&lt;p&gt;Hermann Weyl's attempt to unify electromagnetism and gravity, which he expounded in his classic book Space, Time and Matter. Roughly speaking, what Weyl did was to generalize the General Theory of Relativity by allowing not only arbitrary coordinate transformations, but also transformations to the scales used in the measurements. The theory was very appealing, but it turned out to be wrong because (as Einstein noticed the moment very he heard of the theory) it implied that the properties of a particle depended on its history. Thus, an electron coming from the Sun should have slightly different properties to an electron here on Earth, because they have had a different history. Instead, it is found experimentally that the properties of elementary particles are always the same, which would imply in Weyl's theory that they all had similar histories, which is highly unlikely.&lt;/p&gt;&#xA;" OwnerUserId="3064" LastEditorUserId="3064" LastEditorDisplayName="" LastEditDate="2011-08-22T17:30:34.127" LastActivityDate="2011-08-22T17:30:34.127" CommunityOwnedDate="2011-08-20T19:54:12.060" CommentCount="1" />
  <row Id="13797" PostTypeId="1" CreationDate="2011-08-20T20:14:21.857" Score="0" ViewCount="55" Body="&lt;p&gt;what is Lee-Yang circle theorem and what is it used for ?? , i mean given a measure how can you know that is Ferromagnetic and hence all its zeros lie on a Circle ??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the Lee-Yang circle theorem proof is it only valid for the measure $ \delta (x-1) + \delta (x+1) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how can you proof the Circle theorem for any measure ?? if any measure is positive and EVEN does Lee-Yang circle theorem hold ??&lt;/p&gt;&#xA;" OwnerUserId="4435" LastActivityDate="2011-08-21T01:13:16.010" Title="Lee-Yang circle theorem" Tags="&lt;statistical-mechanics&gt;" AnswerCount="1" />
  <row Id="13798" PostTypeId="2" ParentId="13795" CreationDate="2011-08-20T20:21:37.357" Score="1" ViewCount="" Body="&lt;p&gt;It depends on where the vent from the cooling system is.  For my fridge, this is in the upper left on the back wall.  That means that the left one might cool faster in my fridge, as if they were 3 buddies in a house with one standing next to the AC system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assuming Perfect Mixing&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the chance that the location of the cooling system doesn't matter.  There could just be fast mixing of the air in the interior such that this factor doesn't matter, and all the air is approximately the same temperature.  This could also be the case if the change in temperature of the air over the heat exchanger was really small.  This case is roughly similar to a cooler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only path for heat to get in is the walls.  Of course the insulation is imperfect and this causes the walls to be slightly higher temperature.  Because of the radiative heat balance with the walls, if you assume the air temperature is uniform then the bottles on the outside will cool the slowest.  The one on in the middle &lt;strike&gt;will get cold fastest&lt;/strike&gt; will receive less of the wall radiation, but more radiation from the other bottles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I goofed.  My argument was that the outer bottles get less cooling due to the wall elevated temperature, but I should have continued to note that the walls won't be enough to compensate for higher T rad from other bottles.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Your thinking here:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it the one in the middle because it gets refrigerated by the other two?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is kind of interesting (but wrong).  &lt;strike&gt;If the 2 outer bottles were not there, the middle bottle would cool much faster in fact.  Use of words like &quot;insulate&quot;, &quot;refrigerate&quot;, and &quot;cool&quot; might get a little difficult in this discussion.  One way or the other, the outer bottles don't help cool the inner one, although they cool slower than it does.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The outer bottles do insulate the inner bottle.  That is correct to say, although I should add that the effect could be mostly through radiative heat transfer.  Convection is something you could get into but it's a little trickier.  Fewer bottles will probably always cool faster, and this is due to &lt;em&gt;both&lt;/em&gt; heat transfer and thermal mass reasons.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-08-20T21:54:28.883" LastActivityDate="2011-08-20T21:54:28.883" CommentCount="4" />
  <row Id="13799" PostTypeId="2" ParentId="13793" CreationDate="2011-08-20T20:27:05.107" Score="2" ViewCount="" Body="&lt;p&gt;Actually, this sounds improbable, because the quantum hall effects are fundamentally about interactions. A fully filled Landau level makes it possible (after much handwaving, etc.) to argue that these may be neglected, but once you don't then the interactions dominate. In fact, it's exactly because of this that you get the Fraction Quantum Hall Effects.&lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-08-20T20:27:05.107" CommentCount="5" />
  <row Id="13800" PostTypeId="2" ParentId="10953" CreationDate="2011-08-20T20:50:59.963" Score="3" ViewCount="" Body="&lt;p&gt;Such a proof would have to be nonperturbative, because of the likely perturbative renormalizability of maximal supergravity. Further, if you go to a matrix theory limit, you can describe the physics of strings with what are essentially point particles (although their mutual separations are noncommutative). So it is not clear what you would be proving exactly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that it is better to ask for a compelling argument that the physics of gravity requires a string theory completion, rather than a mathematical proof, which would be full of implicit assumptions anyway. The arguments people give in the literature are not the same as the personal reasons that they believe the theory, they are usually just stories made up to sound persuasive to students or to the general public. They fall apart under scrutiny. The real reasons take the form of a conversion story, and are much more subjective, and much less persuasive to everyone except the story teller. Still, I think that a conversion story is the only honest way to explain why you believe something that is not conclusively experimentally established.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some famous conversion stories are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Scherk and Schwarz (1974): They believed that the S-matrix bootstrap was a fundamental law of physics, and were persuaded that the bootstrap had a solution when they constructed proto-superstrings. An S-matrix theory doesn't really leave room for adding new interactions, as became clear in the early seventies with the stringent string consistency conditions, so if it were a fundamental theory of strong interactions only, how would you couple it to electromagnetism or to gravity? The only way is if gravitons and photons show up as certain string modes. Scherk understood how string theory reproduces field theory, so they understood that open strings easily give gauge fields. When they and Yoneya understood that the theory requires a perturbative graviton, they realized that it couldn't possibly be a theory of hadrons, but must include all interactions, and gravitational compactification gives meaning to the extra dimensions. Thankfully they realized this in 1974, just before S-matrix theory was banished from physics.&lt;/li&gt;&#xA;&lt;li&gt;Ed Witten(1984): At Princeton in 1984, and everywhere along the East Coast, the Chew bootstrap was as taboo as cold fusion. The bootstrap was tautological new-agey content-free Berkeley physics, and it was justifiably dead. But once Ed Witten understood that string theory cancels gravitational anomalies, this was sufficient to convince him that it was viable. He was aware that supergravity couldn't get chiral matter on a smooth compactification, and had a hard time fitting good grand-unification groups. Anomaly cancellation is a nontrivial constraint, it means that the theory works consistently in gravitational instantons, and it is hard to imagine a reason it should do that unless it is nonperturbatively consistent.&lt;/li&gt;&#xA;&lt;li&gt;Everyone else(1985): once they saw Ed Witten was on board, they decided it must be right.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am exaggerating of course. The discovery of heterotic strings and Calabi Yau compactifications was important in convincing other people that string theory was phenomenologically viable, which was important. In the Soviet Union, I am pretty sure that Knizhnik believed string theory was the theory of everything, for some deep unknown reasons, although his collaborators weren't so sure. Polyakov liked strings because the link between the duality condition and the associativity of the OPE, which he and Kadanoff had shown should be enough to determines critical exponents in phase transitions, but I don't think he ever fully got on board with the &quot;theory of everything&quot; bandwagon. It gets uncomfortable trying to tell history of people who are still alive, so I will stop talking about other people.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;My conversion story&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;It took place in 1999. I know exactly when, because I was watching a terrible movie: &quot;Sleepy Hollow&quot; with Johnny Depp, and I was bored, so I started thinking about physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I grew up near Syracuse University. The physics department there was partly put together by Peter Bergmann, one of Einstein's collaborators, and it was well respected for many things. It was where world-sheet supersymmetry was discovered by Ramond, and where the Ashtekar variables were worked out. I talked to Lee Smolin a bunch, and I bought into the idea that gravity could not be described by a perturbation series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is both more and less deep than is usually presented. The problem with a perturbative description is that it starts with empty space and works order by order. But all the interesting paradoxical things about gravity occur on backgrounds with horizons, which are an infinite number of gravitons away from being flat. It seemed obvious that such a description couldn't be completed nonperturbatively, because it starts out by over-counting the degrees of freedom at small scales, which are constrained by black-hole entropy bounds. (This argument is wrong, but some people still believe it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I liked the loop quantization program, because it was more in linw with what quantum gravity should look like, given classical gravity. I was impressed with the fact that geometric operators could have a discrete spectrum, and the theory looked like it gave a proper quantum geometry. I still believe loops are interesting, but the fundamental theory is string theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I knew about the bootstrap, and I hated it. The S-matrix is defined by the most annoying of limits. You have to take the incoming wave-packets far apart, and decompose them into plane waves, then you have to take the residual scattering from infinite area plane waves and extract the finite limit by multiplying by appropriate area factors. That's the fundamental quantity? The S-matrix in classical mechanics is computationally intractable, it can essentially solve the halting problem with enough particles, so there is no real way you can expect an S-matrix to have a simple description without a space-time picture of what is going on in intermediate steps. S-matrix theories do not allow you to describe local physics in any way, except by somehow mapping local states to S-matrix states that compose them by collisions in the far past. How could such a picture be complete?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read through the early parts of Green Schwarz and Witten, and I was annoyed with the string picture. The string was moving in an unquantized space time, and it was clearly derived from Regge theory and bootstrap ideas. I thought that the existence of a spin-2 particle was a reason for rejecting string theory as a theory of gravity--- if it contains gravity it is only because of the accident that spin-2 implies gravity. There's no reason that the resummation of the series will respect black hole decay physics, or resolve the pressing semi-classical gravity paradoxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I decided to study string theory (but not to believe it) when I read 'tHooft's &quot;Under the Spell of the Gauge Principle&quot;. The last paper describes a Schwartschild black hole in a very hokey model where it oscillates radially. But 'tHooft is convinced due to the information paradox that these oscillations completely describe the space-time around the hole, and the oscillations look exactly like a mode of a string theory (with an imaginary tension, and all sorts of nonsense too).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was enough to convince me that there was a connection between black hole physics and strings, and that strings should be thought of as small oscillating black holes (this was during the duality revolution, but I wasn't following the string literature then). The picture, complemented by what Susskind was saying at around that time, was compelling--- string theory was a formulation of gravity because the description of any black hole is similar to string theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct description of black holes in string theory was worked out around this time, but I still was annoyed by the S-matrixy character of the theory. How could it be fundamental if it was just describing asymptotic states? (It is weird to me that I could believe the holographic principle and not the S-matrix principle for so long.) How could it describe black holes when you couldn't even quantize it on a deSitter background, or get it to be properly thermal on an Unruh background?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I decided to kill string theory. This was in the late 1990s. I thought that I would come up with an elegant principle which could be seen to be true in semiclassical gravity and would rule out string theory completely. I thought about it for a long time without anything. I met a graduate student in California named Simeon Hellerman, and we went to a movie. I talked to him about Einstein, and thought experiments, and the like, and he said &quot;Those things only give junk. Look at Hawking's information loss argument! (We all knew it was wrong for the same reason as 'tHooft and Susskind, even without an AdS/CFT demonstration)&quot; But I said, how about charge quantization? Then I argued that charges must be quantized in a theory of quantum gravity, because you can get black holes to form containing the difference of any two charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Simeon says, &quot;Big deal. We already know charge is quantized for other reasons.“ And we saw sleepy hollow. So then I thought about how you can make a quantitative version of this argument. I thought about teeny tiny charges on black holes, and how they would polarize the surface, and that extremal black holes are just barely repulsive, and I realized that a charged black hole would get constipated if there was no charged particle with charge lighter than its mass. I told Simeon this, and I said, &quot;This is a quantitative principle: every consistent quantum theory of gravity has to have a charged particle lighter than its charge! This will kill string theory, because there is no reason strings should obey this. It's a completely non-perturbative constraint.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was very happy that I was going to kill string theory. Simeon said &quot;I will find a string vacuum that violates this, and I will kill your principle.&quot; I thought it was funny, because I thought he would be killing string theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So he went home, and thought about it, and the next day he says &quot;You can't violate it, because ...insert complicated stuff here... and ....insert complicated stuff here ... I now realize that &lt;em&gt;our&lt;/em&gt; principle is true!&quot; I was annoyed that his complicated stuff worked, because I wanted to kill string theory. It was only a year or two later that I figured out what complicated stuff he said (he was talking about T-duals and S-duals).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in understanding why string theory doesn't violate this bound, I had an epiphany. The string &lt;em&gt;is&lt;/em&gt; a black hole, so it is impossible to prove that it fails by black hole formation and evaporation. The laws of the world-sheet emission are just already-quantum laws of extremal Hawking radiation! Then all the dualities of string theory are required by holography, and the S-matrix theory is just the flat-space limit of all this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was enough to convince me that string theory was a consistent theory of quantum gravity. It is impossible that it obeys the mass/charge inequality if it is not. Once I understood this, I was depressed, because I felt that all of physics had just been essentially solved by Banks Fischler Susskind and Shenker, and Maldacena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The holographic principle means that the same theory has to be mathematically expressible in ten thousand different ways, each way corresponding to some extremal black hole. This property is so constricting, that it is impossible to satisfy, except for the fact that string theory satisfies it. There is no chance that there is any other consistent theory of quantum gravity, it is just too implausible that you can satisfy the constraint that the same theory is dual to a bazillion other lower dimensional formulations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I became a string theory believer, for the simple reason that the world is holographic, and string theory is obviously the only consistent way to respect holography. I think my story is typical of my generation. This is the real reason why people are so adamant that it must be right, without experimental evidence (although that would be nice）。&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-20T20:50:59.963" />
  <row Id="13801" PostTypeId="2" ParentId="13795" CreationDate="2011-08-20T21:30:31.593" Score="3" ViewCount="" Body="&lt;p&gt;I think in symmetrical case the middle will typically cool down the slowest, unless it is standing right under the cold spot and has the cold air flowing down onto it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The walls, albeit slightly warmer than the air, are much colder than the bottles, and the middle bottle is receiving radiated heat from 2 bottles standing next to it, whereas side bottle receives radiated heat from one bottle next to them. Furthermore the middle bottle will have restricted airflow on two sides whereas side bottles only on one side.&lt;/p&gt;&#xA;" OwnerUserId="4967" LastActivityDate="2011-08-20T21:30:31.593" />
  <row Id="13802" PostTypeId="2" ParentId="10953" CreationDate="2011-08-20T21:37:40.537" Score="0" ViewCount="" Body="&lt;p&gt;The question is essentially asking about the validity of the following conjecture: (a) string theory is the only possible theory of quantum gravity, and (b) this can be proved by elementary arguments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conjecture is clearly false. There are other candidates for a theory of quantum gravity, such as loop quantum gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LQG could be right or wrong. ST could be right or wrong. LQG might or might not have GR as its classical limit. Supersymmetry might or might not exist in nature, and ST might or might not be viable without supersymmetry. All of this leads to considerable uncertainty about part a of the conjecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when you strengthen the conjecture by adding part b, it becomes clearly false. If the conjecture held, including part b, then LQG would have been wrong for elementary reasons. People like Carlo Rovelli are not so stupid that they would have spent decades working on a theory that was obviously a dead end for elementary reasons.&lt;/p&gt;&#xA;" OwnerUserId="4552" LastActivityDate="2011-08-20T21:37:40.537" CommentCount="1" />
  <row Id="13803" PostTypeId="2" ParentId="12579" CreationDate="2011-08-20T22:00:20.813" Score="1" ViewCount="" Body="&lt;p&gt;Does not work like this. In EU and UK (and most of the world), one prong is at nearly zero voltage, other prong is sine wave, with amplitude 340v or so (240v RMS), with equation $v\sin(\omega t) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;US uses 120v for appliances such as TVs, lightbulbs, etc.&#xA;However, in the US you might also encounter two-phase wiring where one prong is at $\frac v 2\sin(\omega t) $ and another is at $-\frac v 2\sin(\omega t) $ , used for electric stoves that run at 240v. You might also encounter the wiring that uses 2 phases of three phase system, with voltage of about 200v between them but 120v from each to ground. Which happens because the equations are $ v \sin(\omega t) $ and $v \sin(\omega t + \frac 2 3 \pi) $ , the phase is offset by third of a circle. Ahh, and my apartment seem to be using full three-phase (with 400v between two phases) for the electric stove. Not quite sure. All in all there's a wide variety of wirings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: ahh and also:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As far as I know,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Both the two prongs will act as a phase for every 1/50th second&#xA;  alternatively.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;i.e For the first 1/50 second one hole in the outlet must act as a&#xA;  phase and in the next 1/50 second other outlet hole must act as a&#xA;  phase. And this cycle must continue...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is always incorrect, everywhere in the world. The voltage of a sine wave source in general is given by $ v\sin(\omega t + \alpha) $ where $ \frac {\omega} {2\pi}  $ is the frequency and $ \alpha $ is time offset.&#xA;The 'phase' aka 'live' is any wire with such variable voltage. The 'neutral' is wire with nearly zero voltage. The role of phase and neutral never alternates. What you are thinking of is a two phase system where one phase is positive and other is negative for 1/100th of a second, then vice versa, then back again (cycle repeats in 1/50th of a second)&lt;/p&gt;&#xA;" OwnerUserId="4967" LastEditorUserId="4967" LastEditorDisplayName="" LastEditDate="2011-08-20T22:26:51.673" LastActivityDate="2011-08-20T22:26:51.673" />
  <row Id="13804" PostTypeId="1" AcceptedAnswerId="13868" CreationDate="2011-08-20T22:47:32.233" Score="2" ViewCount="91" Body="&lt;p&gt;I'm reading &lt;em&gt;An Introduction to Stochastic Processes in Physics&lt;/em&gt; by Don S Lemons. Problem 10.2 leads to  a pair of equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dV_x = -\gamma V_xdt+V_y\Omega dt-V_y\sqrt{2\gamma dt}N_t(0,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dV_y = -\gamma V_ydt-V_x\Omega dt+V_x\sqrt{2\gamma dt}N_t(0,1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\gamma$, $\Omega$ constants, $X$, $Y$ position, $V_x$, $V_y$ velocity. The two noises are identical.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It then asks for the application of the Smoluchowski approximation to these equations. As the book is very hand-wavey in justifying the use of this approximation I find it tricky to reason with. I only know how to use the search-and-replace rules the book gives, eg. set $dV_x=0=dV_y$ and replace $V_xdt$ and $V_ydt$ with $dX$ and $dY$. So what do I do with the terms like $V_y\sqrt{2\gamma dt}N_t(0,1)$ for the Smoluchowksi approximation? Presumably all references to $V_x$ and $V_y$ need to be eliminated.&lt;/p&gt;&#xA;" OwnerUserId="4983" LastEditorUserId="4983" LastEditorDisplayName="" LastEditDate="2011-08-23T21:57:47.673" LastActivityDate="2011-08-24T05:17:52.763" Title="Cross-field diffusion from Smoluchowski approximation" Tags="&lt;statistical-mechanics&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="13805" PostTypeId="2" ParentId="13797" CreationDate="2011-08-21T01:13:16.010" Score="0" ViewCount="" Body="&lt;p&gt;You can start &lt;a href=&quot;http://en.wikipedia.org/wiki/Lee%E2%80%93Yang_theorem&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; to get an overview.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a high level the L-Y Circle theorem is a statement about the location of the zeros of the partition function used in statistical field theory for ferromagnetic systems.  It applies when using a &lt;a href=&quot;http://en.wikipedia.org/wiki/Mean_field_theory&quot; rel=&quot;nofollow&quot;&gt;mean-field approximation&lt;/a&gt; and the zeros of the partition function end up constrained to the unit circle. I can't remember the details of how the proof goes but &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521408059&quot; rel=&quot;nofollow&quot;&gt;Itzykson and Drouffe&lt;/a&gt; has a very detailed section on the Lee-Yang zeros.  Section 3.2 (if I remember correctly) will answer most of your detailed questions as it has a full on proof of the theorem.&lt;/p&gt;&#xA;" OwnerUserId="4191" LastActivityDate="2011-08-21T01:13:16.010" CommentCount="1" />
  <row Id="13806" PostTypeId="1" AcceptedAnswerId="13812" CreationDate="2011-08-21T01:51:28.937" Score="2" ViewCount="88" Body="&lt;p&gt;This isn't a duplicate to &quot;Why did my liquid soda freeze once I pulled it out of the fridge?&quot; My question is why soda froze after it was opened. Opening a can or bottle seems to have a larger effect than just jostling it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because of the disturbance noted in the previous question? Is it related to the pressure decrease? Is it because of the release of some CO2 when it was opened?&lt;/p&gt;&#xA;" OwnerUserId="4984" LastActivityDate="2011-08-21T04:29:50.787" Title="Why did my liquid soda freeze once I pulled it out of the fridge and opened it?" Tags="&lt;ice&gt;&lt;states-of-matter&gt;" AnswerCount="1" />
  <row Id="13807" PostTypeId="2" ParentId="13739" CreationDate="2011-08-21T02:05:26.590" Score="0" ViewCount="" Body="&lt;p&gt;Assuming there is air or gas in the flexible tube, it will be smaller at the bottom and larger at the top. The pressure at the top of the tube will be about .432 psi less than at the bottom.&lt;/p&gt;&#xA;" OwnerUserId="4984" LastActivityDate="2011-08-21T02:05:26.590" />
  <row Id="13808" PostTypeId="1" CreationDate="2011-08-21T02:05:27.610" Score="0" ViewCount="36" Body="&lt;p&gt;I have collected data on flow velocity at 1 Hz sampling rate for 25 hours using a high resolution acoustic Doppler current profiler (HR-ADCP).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how I can decide up to which period is the mean velocity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How I can exclude fluctuations with the above period using wavelet filters?&lt;/p&gt;&#xA;" OwnerUserId="4985" LastEditorUserId="4985" LastEditorDisplayName="" LastEditDate="2011-08-21T04:15:02.027" LastActivityDate="2011-09-05T21:51:33.163" Title="Filtering and extracting turbulent data from instantaneous dataset" Tags="&lt;fluid-dynamics&gt;&lt;computational-physics&gt;&lt;data-analysis&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" />
  <row Id="13809" PostTypeId="1" CreationDate="2011-08-21T03:25:29.173" Score="0" ViewCount="72" Body="&lt;p&gt;In his blog at &lt;a href=&quot;http://motls.blogspot.com/2010/09/can-antimatters-gravity-be-repulsive.html&quot; rel=&quot;nofollow&quot;&gt;http://motls.blogspot.com/2010/09/can-antimatters-gravity-be-repulsive.html&lt;/a&gt; Lubos Motl writes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;...neutrons contain a slightly different mass contribution from the antiquarks (antimatter!) than the protons ..... where the difference is something of order 1%, ...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like some reliable references to this derivation, the simpler the better, assuming the usual standard model, of course.   I assume it is somewhere in the computation of the masses of the baryons, but where, and why?   &lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2011-08-21T03:25:29.173" Title="Why are the antimatter compositions of neutrons and protons different? Why by about 1%? References?" Tags="&lt;particle-physics&gt;" CommentCount="4" />
  <row Id="13810" PostTypeId="1" CreationDate="2011-08-21T03:35:05.633" Score="2" ViewCount="113" Body="&lt;p&gt;A friend of mine was telling me about a storm that knocked down a power line over at their place, and it got me to thinking.  Why are downed power lines dangerous?  I don't see any good reason for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that if they can put a &lt;a href=&quot;http://en.wikipedia.org/wiki/Ground-fault_current_interrupt&quot; rel=&quot;nofollow&quot;&gt;GFI&lt;/a&gt; on a hair dryer that will kill power within a fraction of a second if it falls into the sink or tub and shorts out, they ought to be able to do the same thing to a power line.  Is there any reason why the basic concept can't work at that scale?  Or if not, why aren't they standard equipment on power lines?&lt;/p&gt;&#xA;" OwnerUserId="4039" LastActivityDate="2011-08-22T00:33:57.717" Title="Why are downed power lines dangerous?" Tags="&lt;electricity&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="13811" PostTypeId="2" ParentId="13810" CreationDate="2011-08-21T03:50:21.270" Score="2" ViewCount="" Body="&lt;p&gt;Even if you could cheaply and efficenctly engineer a GFI to work on 250,000V high current lines, and you could shut down the grid to fit them how often do downed lines hurt people compared to lawn mowers going over extention leads?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even then I suspect that if you could disconnect the other end of the line there would be enough charge stored in the cable between you and the GFI to make you very unhappy&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-08-21T03:50:21.270" CommentCount="5" />
  <row Id="13812" PostTypeId="2" ParentId="13806" CreationDate="2011-08-21T04:29:50.787" Score="5" ViewCount="" Body="&lt;p&gt;It's not a disturbance, the liquid isn't supercooled in this case. It's right about at 0 degrees, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It isn't the pressure drop directly, because you can give an upper-bound estimate to how much cooling the pressure drop directly does based on the observation that the pressure is not more than a few atmospheres. That means that space the gas in the bottle gets multiplied by a factor of, say 10 (that's too big), and (kT times) the entropy gain per atom is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$kT\Delta S = kT log(10)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and kT is 1/30 eV, so you get at best 1/10 eV per gas atom. The density of gas on top is 1/300 the density of water, and the maximum total cooling from expanding that gas is negligible per atom of liquid. There is no significant cooling and heating in compressing a liquid, since essentially no work is done in the process, so the pressure drop does nothing to the liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the pressure drop makes the soda supersaturated with respect to its dissolved CO&lt;sub&gt;2&lt;/sub&gt;, and this CO&lt;sub&gt;2&lt;/sub&gt; is outgassing from a liquid state to a gas state, and this is a huge gain in entropy. By outgassing the dissolved CO&lt;sub&gt;2&lt;/sub&gt; into bubbles, each CO&lt;sub&gt;2&lt;/sub&gt; atom gets a 300-fold increase of available volume to roam, and this can cool the liquid by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$kT log(300) \approx {1\over 4} eV $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you get $.25 eV$ per released CO&lt;sub&gt;2&lt;/sub&gt; molecule, which is comperable to the binding energy of a water molecule. So each outgassed CO&lt;sub&gt;2&lt;/sub&gt; freezes order 1 H&lt;sub&gt;2&lt;/sub&gt;O, and this freezes a network of filaments of water around the outgassed bubbles, making a slush.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-21T04:29:50.787" />
  <row Id="13813" PostTypeId="2" ParentId="13787" CreationDate="2011-08-21T04:44:05.583" Score="1" ViewCount="" Body="&lt;p&gt;I wrote the Wikipedia page in question, so I feel bad. I thought it was clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a recent textbook by Banks which covers the spin/statistics theorem pretty good. I hope it is ok. The main difficulty is that there is no quantum field theory book that covers analytic continuation to Euclidean space, and this is the essential thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is worked out by each person on their own, as far as I know. The problem is that it is very easy to say &quot;plug in i times t everywhere you see t&quot; and get 90% of everything right, without understanding anything. Streater and Whitman do it, that's most of their book, but they are too formal to be comprehensible. Schwinger is too long ago (and ideosyncratic). Perhaps the statistical section of Feynman and Hibbs (Path integrals), where they actually rederive the path integral in imaginary time, will allow you to extrapolate to the general bosonic fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Fermionic case requires the Euclidean continuation of Majorana spinors, and this was in the literature more recently: &lt;a href=&quot;http://arxiv.org/abs/hep-th/9608174&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/hep-th/9608174&lt;/a&gt;. This stuff is covered in none of the textbooks, and unfortunately I can't recommend any of them with a good conscience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later Edit: If you don't want to go to Euclidean space, you should avoid anything past Feynman/Schwinger. The best path then is possibly to work through Pauli's argument:W. Pauli, The Connection Between Spin and Statistics, Phys. Rev. 58, 716-&#xA;722(1940).&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-21T14:09:03.703" LastActivityDate="2011-08-21T14:09:03.703" CommentCount="4" />
  <row Id="13814" PostTypeId="1" AcceptedAnswerId="13827" CreationDate="2011-08-21T05:52:25.677" Score="2" ViewCount="127" Body="&lt;p&gt;I am an electronics and communication engineer, specializing in signal processing. I have some touch with the mathematics concerning communication systems and also with signal processing. I want to utilize this knowledge to study and understand Quantum Mechanics from the perspective of an engineer. I am not interested in reading about the historical development of QM and i am also not interested in the particle formalism. I know things have started from the wave-particle duality but my current interests are not study QM from that angle. What i am interested is to start studying a treatment of QM from very abstract notions such as, 'what is an observable ? (without referring to any particular physical system)' and 'what is meant by incompatibility observables ?' and then go on with what is a state vector and its mathematical properties. I am okay to deal with the mathematics and abstract notions but i some how do not like the notion of a particle, velocity and momentum and such physical things as they directly contradict my intuition which is based on classical mechanics ( basic stuff and not the mathematical treatment involving phase space as i am not much aware of it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I request you to give some suggestions on advantages and pitfalls in venturing into such a thing. I also request you to provide me good reference books or text books which give such a treatment of QM without assuming any previous knowledge of QM.&lt;/p&gt;&#xA;" OwnerUserId="540" LastActivityDate="2011-08-21T20:53:31.047" Title="reference request for an abstract treatment of QM without using any particle formalism" Tags="&lt;quantum-mechanics&gt;&lt;reference-request&gt;" AnswerCount="4" />
  <row Id="13815" PostTypeId="2" ParentId="13814" CreationDate="2011-08-21T07:24:50.373" Score="0" ViewCount="" Body="&lt;p&gt;That's quite an unusual request i.e. to start with an abstract formulation of quantum mechanics, especially for someone in a profession so closely connected with the &quot;real world&quot;.  However, to answer your question, I think what you're looking for is an axiomatic approach to quantum mechanics.  Such treatments will keep the physical examples to the minimum and skip to the mathematics straight away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could start with &lt;a href=&quot;http://www.stanford.edu/~mukul/tutorials/Quantum_Mechanics.pdf&quot; rel=&quot;nofollow&quot;&gt;this reference&lt;/a&gt; (just for a chatty treatment of what the postulates look like !), and maybe search for quantum mechanics textbooks with &quot;axiomatic&quot; in the title.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For many people, the fact that the quantum mechanical predictions of physical quantities  are sometimes counterintuitive is precisely what gives the subject its appeal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  It appears not so easy to find &quot;Axiomatic Quantum Mechanics&quot; in textbook titles !  However Google returns a few articles featuring those words.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there is also Von Neumann's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0691028931&quot; rel=&quot;nofollow&quot;&gt;&quot;Mathematical Foundations of Quantum Mechanics&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3099" LastEditorUserId="3099" LastEditorDisplayName="" LastEditDate="2011-08-21T07:47:35.607" LastActivityDate="2011-08-21T07:47:35.607" />
  <row Id="13816" PostTypeId="1" CreationDate="2011-08-21T09:19:17.830" Score="0" ViewCount="89" Body="&lt;p&gt;The string theorist Brian Greene recently came up with the book The Hidden Realities. In Chapter 9, and also in the article &lt;a href=&quot;http://discovermagazine.com/2011/jun/03-our-universe-may-be-a-giant-hologram&quot; rel=&quot;nofollow&quot;&gt;http://discovermagazine.com/2011/jun/03-our-universe-may-be-a-giant-hologram&lt;/a&gt; , he argued that our universe might be a hologram with a holographic parallel universe and compares our world to shadows in Plato's cave. Can anyone point me to where I can read up more on this/ Are we living in a black hole. Are we puppets pulled by strings from some distant surface&lt;/p&gt;&#xA;" OwnerUserId="4989" LastActivityDate="2011-08-21T13:55:28.547" Title="Can anyone clarify Brian Greene's relation of holographic parallel universes with Plato's cave/" Tags="&lt;string-theory&gt;" AnswerCount="1" />
  <row Id="13817" PostTypeId="2" ParentId="13814" CreationDate="2011-08-21T10:48:07.173" Score="2" ViewCount="" Body="&lt;p&gt;An unconventional approach would be to study &lt;em&gt;quantum computation&lt;/em&gt; or &lt;em&gt;quantum information theory&lt;/em&gt; first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is 'unusual' about quantum mechanics is the mathematical underpinnings, which is essentially a generalization of probability theory. (I have heard more than one colleague say that quantum mechanics is simply physics which involves 'non-commutative probability', &lt;em&gt;i.e.&lt;/em&gt;&amp;nbsp;in testing whether some collection of events are realized for some sample space, there is a pertinent sense of &lt;em&gt;the order&lt;/em&gt; in which one tests those events.) To the extent that this is true, it is not important to be learning the actual physics alongside that mathematical underpinning, so long as you can learn about &lt;em&gt;something&lt;/em&gt; evolving &lt;em&gt;e.g.&lt;/em&gt;&amp;nbsp;under the Schr&amp;ouml;dinger equation or collapsing under measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Studying quantum information evolving under a computational process is one way you could achieve that. Because the narrative of the field is less about the crisis in physics in the 20s&amp;ndash;40s, and more about physicists and computer scientists struggling to find a common language, the development is clearer and there is a better record of justifying the elements of the formalism from a fundamental standpoint. By studying quantum information and/or quantum computation, you will be able to decouple the learning of the underpinnings from the learning of the physics, and thereby get to the heart of any conceptual troubles you may have; and it will give you a tidier sandbox in which to play with ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To this end, I recommend &quot;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521635039&quot; rel=&quot;nofollow&quot;&gt;Nielsen &amp;amp; Chuang&lt;/a&gt;&quot;, which is the standard introductory text of the field. It's suitable as an introduction both for those coming from a computer science background, and from a quantum physics background; so apart from learning some of the formalism, you can get some exposure to some of the physics as well. There are other texts which I have not read, though; and about a bazillion pages of lecture notes floating around on the web.&lt;/p&gt;&#xA;" OwnerUserId="4976" LastActivityDate="2011-08-21T10:48:07.173" />
  <row Id="13818" PostTypeId="2" ParentId="11656" CreationDate="2011-08-21T12:31:34.637" Score="0" ViewCount="" Body="&lt;p&gt;I'm sorry, but Guth and Farhi calculated that to create a bubble universe in a laboratory which does not collapse on the inside to a big crunch, you need a white hole to kickstart the expansion.&lt;/p&gt;&#xA;" OwnerUserId="4990" LastActivityDate="2011-08-21T12:31:34.637" />
  <row Id="13819" PostTypeId="2" ParentId="13043" CreationDate="2011-08-21T13:31:40.520" Score="2" ViewCount="" Body="&lt;p&gt;The energy catalyzer probably does not work. Here is an entry point to some of the criticism: &lt;a href=&quot;http://tech.groups.yahoo.com/group/H-Ni_Fusion/message/226&quot; rel=&quot;nofollow&quot;&gt;http://tech.groups.yahoo.com/group/H-Ni_Fusion/message/226&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The criticism that nuclear reaction can't take place at low temperature is no good, because  Palladium/deuterium does it. Rossi is saying that, like Palladium, Nickel allows for nuclear reactions, except that he has the Nickel/protons fuse together to produce copper instead of the deuterium with deuterium. Fusion of protons with Nickel is an exothermic nuclear reaction (unlike what the previous answers claim), but it is prevented by MeV scale electrostatic barriers, not KeV scale barriers, and is energetically as far from d-d fusion as d-d fusion is from fire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rossi's device probably doesn't work, but without a completely sound theory of cold fusion, one cannot be 100% sure. The major problem with the demonstrations that Rossi ran is that the measurements of heat output are done by assuming that all the water that comes into the system is converted to steam, and this is a very unreliable way to measure heat output. You should just use water to measure heat, that's what everyone else does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rossi claimed that all the steam coming out of his machine was dry. But dry steam has a huge gas volume, and should rush out in an invisible jet of gas out of his pipe, while his device emitted a &quot;feeble mist&quot; (to quote the linked page). To make mist, all you need to do is to charge water up electrically, so that it breaks into droplets, and that doesn't take any energy at all. The device could be producing no energy beyond the electricity going into it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rossi claimed to make copper, and he submitted a sample of Nickel powder with copper and iron inside that he said was the product of his machine. But the copper and iron were analyzed by others and have natural isotope ratios, which is extremely unlikely for the products of a nuclear reaction. The reports of Nickel/Hydrogen cold fusion, unlike Palladium deuterium cold fusion, all come from essentially one place, with no independent reproduction. It is possible that Nickel hydrogen doesn't do anything at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is no reason to dismiss Rossi a-priori. He claims his device works, and this can be tested by just waiting to see it work.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-21T13:31:40.520" />
  <row Id="13820" PostTypeId="2" ParentId="13816" CreationDate="2011-08-21T13:55:28.547" Score="0" ViewCount="" Body="&lt;p&gt;The &quot;Plato's cave&quot; analogy is due to Polyakov: &lt;a href=&quot;http://cdsweb.cern.ch/record/364444/files/9809057.pdf&quot; rel=&quot;nofollow&quot;&gt;http://cdsweb.cern.ch/record/364444/files/9809057.pdf&lt;/a&gt;. The precise formulation is that a mathematical description of bulk gravity should be considered as living on a null surface (like a black-hole horizon) far away. This doesn't make us puppets, but it does mean that our perception of bulk space-time is not correct at short distances, and we are like 3-d shadows of a 2-d reality (more likely higher dimensions, but that's the idea anyway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;parallel universes&quot; stuff is only in an extremely speculative, in my opinion completely experimentally excluded, offshoot of string theory called &quot;large extra dimensions&quot;. There you can put extra universe &quot;parallel&quot; to ours, and they would only interact with us through gravity. This idea might appeal to people because they might think it explains astral projection or something like that, but in these theories, there is no way project yourself into an extra dimension unless your consciousness is made out of gravitons. It is logically independent of the holographic principle.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-21T13:55:28.547" />
  <row Id="13821" PostTypeId="2" ParentId="743" CreationDate="2011-08-21T15:05:39.673" Score="0" ViewCount="" Body="&lt;p&gt;Hollowood's lecture notes &quot;A Wilsonian Approach to Field Theory&quot; are really nice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're a mathematician interested in this stuff -- especially in renormalization as it appears in statistical mechanics -- you might want to try the Brydges' &quot;Lectures on the Renormalization Group&quot; in the book &lt;em&gt;Statistical Mechanics&lt;/em&gt; in the &quot;Ias/ Park City Mathematics&quot; lecture series.  It discusses a few examples in some detail.  G. Battle's &lt;em&gt;Wavelets and Renormalization&lt;/em&gt; is also a good place to look:  He discusses interacting scalar fields in a ton of detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also been some rigorous writing recently on renormalization in perturbative QFT, by Costello and by Borcherds, which might help people bridge the gap between the stat mech and more particle physics-y language.&lt;/p&gt;&#xA;" OwnerUserId="1504" LastEditorUserId="1504" LastEditorDisplayName="" LastEditDate="2011-08-22T15:21:57.010" LastActivityDate="2011-08-22T15:21:57.010" />
  <row Id="13822" PostTypeId="1" CreationDate="2011-08-21T18:07:49.273" Score="3" ViewCount="72" Body="&lt;p&gt;I homebrew my own beer and I'm looking for way to cheaply monitor the specific gravity of my brew during the fermentation process.  I came across &lt;a href=&quot;http://www.arduino.cc/cgi-bin/yabb2/YaBB.pl?num=1182800438&quot; rel=&quot;nofollow&quot;&gt;this thread&lt;/a&gt; that seems to indicate capacitance may be used to determine specific gravity.  Is there a direct relationship between the dielectric constant and/or capacitance of a liquid and its specific gravity?  &lt;/p&gt;&#xA;" OwnerUserId="4991" LastEditorUserId="4991" LastEditorDisplayName="" LastEditDate="2011-08-21T19:25:34.810" LastActivityDate="2011-08-22T12:07:35.097" Title="How feasible is it to use the dielectric constant of a liquid to determine specific gravity?" Tags="&lt;capacitance&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="13823" PostTypeId="2" ParentId="13822" CreationDate="2011-08-21T18:30:12.770" Score="4" ViewCount="" Body="&lt;p&gt;In general, no.  Specific gravity refers to the density of a substance - how heavy a liter of it is.  Dielectric constant refers to the response of the substance to an electric field, which depends on the chemistry of the substance.  They are not physically related in any obvious way.  It is possible to find two liquids with very similar densities and different dielectric constants or very similar dielectric constants and different densities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be possible to find useful correlations between dielectric constant and density if you already know a lot of other information about the substances involved, as you might with beer.  The nature of the correlation would probably need to be empirically determined (i.e. sample many batches of beer, measure explicitly both dielectric constant and density, and search for a correlation.)&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-08-21T18:30:12.770" CommentCount="1" />
  <row Id="13824" PostTypeId="1" CreationDate="2011-08-21T20:08:06.770" Score="0" ViewCount="120" Body="&lt;p&gt;The Nobel committee sometimes needs a little help--- look how long it took them to recognize Nambu. Also, they seem to have nationalism issues, they seem to be a little prejudiced sometimes against Europeans. If you could sit on the room, who would you nominate, and for what?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One nomination per answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer would have been: Chew and Mandelstam for the development of the S-matrix theory of the strong interactions. This would do a lot of good to the field of particle physics, and time is running out. Other fundamental contributors are already dead.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-21T21:39:19.103" LastActivityDate="2011-08-21T21:39:19.103" ClosedDate="2011-08-21T21:24:02.933" Title="Who would you nominate for a Physics Nobel prize?" Tags="&lt;soft-question&gt;" CommentCount="3" />
  <row Id="13825" PostTypeId="2" ParentId="13814" CreationDate="2011-08-21T20:08:37.033" Score="0" ViewCount="" Body="&lt;p&gt;I strongly advise you &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0792336321&quot; rel=&quot;nofollow&quot;&gt;Quantum Theory: Concepts and Methods&lt;/a&gt; by Asher Peres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this book is answering the questions you're asking. Like 'what is an observable ?'&lt;/p&gt;&#xA;" OwnerUserId="4992" LastActivityDate="2011-08-21T20:08:37.033" />
  <row Id="13826" PostTypeId="1" CreationDate="2011-08-21T20:13:11.717" Score="0" ViewCount="76" Body="&lt;p&gt;For a free baryon at rest at room temperature, how much of its ~1Gev (rest) mass can (on average) be considered as matter, as antimatter, and as binding energy?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a baryon in a nucleus, I assume this will also depend on the intranuclear pressure.   Please, references, if possible.   Note the related question &lt;a href=&quot;http://physics.stackexchange.com/questions/13809/why-are-the-antimatter-compositions-of-neutrons-and-protons-different-why-by-abo&quot;&gt;13809&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motivation:  I have read that  the uud quarks in a proton say correspond to less than 1.5% of its ~938 Mev rest mass.    I am asking about the remaining &gt;98.5% of its rest mass.   I would have naively assumed that it splits 50/50, but that is apparently wrong.   Also, I have also read that all of the remainder is all binding energy.   But that has also been contradicted.  &lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditorDisplayName="" LastEditDate="2011-08-21T20:36:38.780" LastActivityDate="2011-08-22T23:47:27.423" Title="What are the average matter, antimatter, and binding energy composition of protons and neutrons?" Tags="&lt;particle-physics&gt;&lt;nuclear-physics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13827" PostTypeId="2" ParentId="13814" CreationDate="2011-08-21T20:48:16.793" Score="0" ViewCount="" Body="&lt;p&gt;Try &quot;Mathematical Foundations of Quantum Mechanics&quot; by George Mackey. It is about 130 pages with a chapter on classical mechanics. The author is a very well know mathematician, and I think the books is what you are looking for. Also on higher level is the book &quot;Quantum mechanics for mathematicians&quot; by Leon Takhtadzhian. The book by Folland on Quantum Field Theory has a chapter on Quantum Mechanics, which can be read independently of the rest of the book.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2011-08-21T20:53:31.047" LastActivityDate="2011-08-21T20:53:31.047" />
  <row Id="13828" PostTypeId="1" CreationDate="2011-08-21T21:25:20.950" Score="4" ViewCount="208" Body="&lt;p&gt;String theory is made from Chew-ish strings, strings which follow Geoffrey Chew's S-matrix principle. These strings have the property that all their scattering is via string exchange, so that the amplitudes are entirely given by summing world sheets and by topology. This leads to a successful theory of gravity, we know, uniqueness, we know etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know exactly how much of the Chew-Mandelstam S-matrix philosophy, that proto-holographic principle which was so important for the development of string theory historically, is logically necessary for deriving the string interactions. This is obscured in more modern treatments, because these take for granted certain things that were historically difficult and came directly from the S-matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of those things which are taken for granted is that strings interact only by exchanging other strings, and that this interaction respects world-sheet duality. In a recent exchange (here: &lt;a href=&quot;http://physics.stackexchange.com/questions/13728/what-are-the-details-around-the-origin-of-the-string-theory/13749#comment-33261&quot;&gt;What are the details around the origin of the string theory?&lt;/a&gt;), Lubos Motl suggested that this may be derived only from the assumption that there is some sort of string excitation, and that this string interacts consistently. I would like to poke at this claim with examples, to determined exactly how many extra assumptions are hidden behind this claim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The assumption that strings exchange strings with Doleh-Horn-Schmidt duality is equivalent to the assumption that the only interaction between strings is the topological exchange diagram from string theory. This assumption not only fixes the string interaction given the spectrum, it fixes the spectrum and even the dimension of space time, it fixes everything. So it is very weird physics. I believe that this assumption is Chew/Mandelstam physics, and that you can't derive strings without using it.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Stupid rubber band model&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;There are N scalar fields of mass m, and N U(1) gauge fields. Scalar k has a charge +1 under gauge field number k, and a charge -1 under gauge field k+1 (with periodic boundaries, so scalar N and 1 are interacting).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each U(1)'s is broken from a separate SU(2) at some high energy by a technicolor mechanism which is irrelevant, but which embeds the U(1)'s in an asymptotically free theory at high energies, so that the theory is completely consistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These N scalars can form a matter-string, composed of a type 1 particle, bound to a type 2 particle, bound to a type 3 particle, etc around a loop closing back on particle 1. I will call this the rubber-band.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, the N different scalars are each separately in the fundamental representation of an SU(N) gauge theory, so that individual quanta of these scalars have infinite mass by confinement. The coupling of the SU(N) will be infinitesimal at the rubber-band scale, so that it will not affect the dynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rubber band is one of the baryons of the theory. If you just look at rubber-band dynamics, if you collide two rubber bands, you can exchange U(1) quanta between corresponding points on the rubber band, and get long range Van-der-Waals, splitting and rejoining interactions. But there is no duality.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Properties&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Given a &quot;baryon&quot; in the model, it contains N scalars. If the scalars are not numbers 1...N, they don't bind together, and if the scalars are 1...N, then they bind together, releasing some U(1) photons and making a closed bound rubber-band at a lower energy by N times the binding energy of each pair (approximately). The isolated particles are infinitely massive, so the lightest baryonic state in the theory is the rubber band.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rubber band has oscillation excitations. For large N, you get a spectrum which is by the frequency of a 1-dimensional fixed length string. There are modes which correspond to the N particles sitting on top of each other, rotating with 1 unit of angular momentum, moving apart and shrinking, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rubber band effective theory holds for energies above the bound-state baryon mass and below the mass of a cut rubber band, and it holds well above this too. But the interactions of the rubber band have nothing to do with string theory.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Questions&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Did I miss a failed conservation law which makes the rubber band decay?&lt;/li&gt;&#xA;&lt;li&gt;Would you consider the rubber band to be a relativistic string?&lt;/li&gt;&#xA;&lt;li&gt;How close to the real string spectrum can you get by mocking up rubber bands?&lt;/li&gt;&#xA;&lt;li&gt;Why do you say that string theory is &lt;em&gt;only&lt;/em&gt; based on the assumption of &quot;relativistic strings&quot;, when the relativistic rubber band is a &quot;relativistic string&quot; and does not obey duality?&lt;/li&gt;&#xA;&lt;li&gt;Is it obvious, given this model, that Chew bootstrap is necessary for deriving string theory, and that duality does not &quot;obviously&quot; follow from the existence of a string picture?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-22T03:03:55.453" LastActivityDate="2011-08-22T05:23:09.383" Title="Are There Strings that aren't Chew-ish?" Tags="&lt;particle-physics&gt;&lt;mathematical-physics&gt;&lt;string-theory&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="13829" PostTypeId="1" AcceptedAnswerId="13856" CreationDate="2011-08-21T22:40:12.183" Score="8" ViewCount="130" Body="&lt;p&gt;I have a question about climate science, but this is relatively straight forward and (I hope) easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been reading that the radiative forcing caused by CO2 in the atmosphere (or any gas) is proportional to the logarithm of the concentration.  I don't understand why.  I'm familiar with basic attenuation, so if I assume that the gas captures or reflects some light, which equates to the heat per unit area (the units for radiative forcing), then I would write an equation like the following, with $F$ being the forcing in $W/m^2$ and $C$ being the ppm concentration.  &lt;em&gt;The $1-e^x$ form is appropriate because we need the reflected quantity, versus the $e^x$ transmitted quantity&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F \propto \left( 1- e^{-\lambda C} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is fundamentally different about the greenhouse effect?  I hesitate to give a specific reference because it seems to occur so frequently, but I will note that IPCC 2007 report is said to make such a statement.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Radiative_forcing&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; gives the following equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta F = 5.35 \times \ln {C \over C_0}~\mathrm{W}~\mathrm{m}^{-2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would love to see just a simple and clean takedown of why $ln$ and not $exp$.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-08-22T01:42:14.163" LastActivityDate="2011-08-23T11:49:48.607" Title="Why is radiative forcing from CO2 logarithimic and not a decreasing exponential?" Tags="&lt;climate-science&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13830" PostTypeId="2" ParentId="13810" CreationDate="2011-08-22T00:33:57.717" Score="-1" ViewCount="" Body="&lt;p&gt;The entire network can still receive lightning from storms, however distant, and deliver them in your body. If you think you can turn off the storm then it will not be dangerous. &lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-08-22T00:33:57.717" />
  <row Id="13831" PostTypeId="2" ParentId="13714" CreationDate="2011-08-22T00:46:00.187" Score="0" ViewCount="" Body="&lt;p&gt;In my opinion the most amazing thing about theoretical physics is that &lt;em&gt;it works at all, ever&lt;/em&gt;. Mathematics is created, typically constrained only by pure reason and aesthetic decisions. On the other hand (as has been mentioned in other answers) physics is &lt;em&gt;first and foremost&lt;/em&gt; an experimental science. So our ability to dream up models and solve them with well(or ill!)-defined mathematical structures, and at the end of the day make &lt;em&gt;quantitative predictions&lt;/em&gt; about nature that tell us whether or not our insight is total crap, well that's an amazing thing -- and the fact that we understand a little bit of why that is is therefore $\mbox{amazing}^{(\mbox{amazing})}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is all to say that scaling and universality are the things that pushed &lt;strong&gt;me&lt;/strong&gt; into theoretical physics, at least. (&quot;Well, you see, by noting a subtle symmetry of critical hamiltonians, we can easily explain why all of this properly scaled real, measured data falls on exactly the same curve!&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(and a postscript: string theory is a topic that is deeply embedded in the &quot;popular-science&quot; consciousness [not necessarily a bad thing] but honestly, my cruddy, misinformed undergrad idea that &lt;em&gt;all&lt;/em&gt; successful theoretical physicists must be string theorists almost kept me away from theory entirely. The kids who already get excited about string theory don't seem to require any extra nudge from the experts.)&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2011-08-22T00:46:00.187" CommunityOwnedDate="2011-08-22T00:46:00.187" />
  <row Id="13832" PostTypeId="1" CreationDate="2011-08-22T01:48:21.873" Score="2" ViewCount="123" Body="&lt;p&gt;Landau writes in chapter 2 of his Mechanics book &quot;The number of independent integrals of motion for a closed mechanical system with s degrees of freedom is 2s-1&quot;. Then he goes on to show how -- &quot;Since the equations of motion for a closed system do not involve time explicitly, the choice of the origin of time is entirely arbitrary, and one of the arbitrary constants in the solution of the equations can always be taken as an additive constant t0 in time. Eliminating t + t0 from the 2s functions, we can express the 2s-1 arbitrary constants as functions of q and q’ (generalized co-ordinates and velocities) and these functions will be the integrals of the motions.&quot; Could someone elaborate? &lt;/p&gt;&#xA;" OwnerUserId="4263" LastActivityDate="2011-08-23T09:18:51.893" Title="Integrals of Motion" Tags="&lt;mechanics&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" />
  <row Id="13834" PostTypeId="1" CreationDate="2011-08-22T04:28:43.897" Score="5" ViewCount="70" Body="&lt;p&gt;Alice and Bob are each in possession of one half of a maximally entangled pair of particles.  Alice can make either of two observations, $A_1$ or $A_2$.  Bob can make either of two observations, $B_1$ or $B_2$.  (Observations have values $1$ or $-1$).  Write $E(x,y)$ for the expected value of the product $xy$.  Then Tsirelson's Bound says that&#xA;$|E(A1,B2)-E(A1,B1)-E(A2,B1)-E(A2,B2)|$ is bounded above by $2 \sqrt{2}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:  Is the converse true?  That is, suppose I have four numbers $x,y,z,w$.  Suppose they are all bounded by one in absolute value and that they satisfy $|x-y-z-w| &amp;lt; 2 \sqrt{2}$.  Does it follow that there are observables $A_1, A_2, B_1, B_2$ such that $x=E(A_1,B_2)$, $y=E(A_1,B_1)$, etc.?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, what other conditions do I need on $x, y, z, w$ in order for such observables to exist?&lt;/p&gt;&#xA;" OwnerUserId="4993" LastEditorUserId="3936" LastEditorDisplayName="" LastEditDate="2011-08-23T01:12:29.950" LastActivityDate="2011-08-23T01:12:29.950" Title="Is Tsirelson's Bound the only constraint on these quantum correlations?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-information&gt;" AnswerCount="1" />
  <row Id="13835" PostTypeId="2" ParentId="13828" CreationDate="2011-08-22T04:50:40.980" Score="0" ViewCount="" Body="&lt;p&gt;In your construct, it's the &lt;em&gt;segments&lt;/em&gt; of the rubber band which resemble the (open) strings of string theory - the flux lines connecting the scalars. The natural embedding of the rubber band into string theory would involve interpreting the scalars as zero-dimensional domain walls, with two &lt;a href=&quot;http://arxiv.org/abs/hep-th/0703267&quot; rel=&quot;nofollow&quot;&gt;wall-string junctions&lt;/a&gt; present at each such &quot;0-brane&quot;. Since the walls would normally contain gauge theories, it seems it would have to be an N=2 gauge theory (in order to contain a scalar) broken to N=0 gauge theory. So the embedding of a rubber band into string theory, if it exists, might require a lot of structure invisible from your starting point; though perhaps that structure would show up in an AdS/CFT analysis of your technicolor theory? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt;: Yes, I think the &quot;obvious&quot; way to embed a field theory with &quot;rubber bands&quot; into string theory is to look for an appropriate N=2 superconformal quiver theory with an AdS dual, and then add masses / break susy with VEVs as in Seiberg-Witten 1994. &lt;/p&gt;&#xA;" OwnerUserId="1486" LastEditorUserId="1486" LastEditorDisplayName="" LastEditDate="2011-08-22T05:23:09.383" LastActivityDate="2011-08-22T05:23:09.383" CommentCount="8" />
  <row Id="13836" PostTypeId="1" CreationDate="2011-08-22T05:09:50.030" Score="0" ViewCount="27" Body="&lt;p&gt;Help me my uncle got my room sprayed by insect killing liquid..i cant sit for a second in my room&lt;/p&gt;&#xA;" OwnerUserId="4994" LastActivityDate="2011-08-22T05:09:50.030" ClosedDate="2011-08-22T05:53:05.053" Title="Can we use activated charcoal to get rid of smell of insect-killing gas?" Tags="&lt;particle-physics&gt;&lt;particles&gt;" />
  <row Id="13837" PostTypeId="2" ParentId="11768" CreationDate="2011-08-22T06:12:36.573" Score="2" ViewCount="" Body="&lt;p&gt;We have a thermodynamic understanding of energy violation, in terms of perpetual motion, but momentum violation is somewhat less immediately paradoxical in a changing background, maybe because we have intuition that the field has momentum in this case. But the field momentum is irrelevant for this question. Mechanical momentum should be conserved in a z-independent background ignoring any field momentum considerations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll state the principle of conservation of momentum is as follows: if the z direction is made periodic with period L (for mathematical convenience), and all the physical conditions (the time dependent E and B fields) do not depend on z, it is impossible to get a rod parallel to the z-direction going all the way around the box to move along the z direction only by moving it in the x and y directions, keeping it parallel to the z axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of differential forms, the E equation you give is part of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dF = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When F is independent of time, this reduces to $\nabla \times E = 0$, and to the loop integral condition, but I will try to state it more covariantly in spacetime. Consider a spacetime 2-d surface which is some parametrized loop x(s) y(s) z(s) in the t=0 plane, extended in times by translation. The integral of the two form over the surface is equal to zero, by stokes theorem. But this integral is equal to (the infinite length in time of the surface times) $\int E_x dx + E_y dy + E_z dz$, and you conclude that in time-independent fields, the integral must be zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you believe the Lorentz equation of motion, this is a statement of conservation of mechanical energy, as you noticed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since it is stated in differential form language, all you have to do is tip the picture over into the z-axis to get a momentum version. If the F tensor does not depend on z, and z is periodic with period L (just in case), then integrating the F form on a two-surface made by taking a parametrized loop t(s),x(s),y(s), extended by translation along the z axis gives zero. This reduces to the following one dimensional integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int (E_z dt - B_y dx + B_x dy) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This integral is equal to zero by the Maxwell equation dF=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you believe the Lorentz equation of motion, this integral must also vanish for reasons of conservation of momentum. If you take a uniformly charged z-extended wire around this loop, it would acquire momentum of this magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you can't get a charged wire to move backward in time, so you have to use two wires, and at the proper instant charge them up to equal and opposite charge densities, then move them along the x and y directions as required by the loop, then bring them back and let them neutralize each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the integral above is nonzero, the wires will get a nonzero relative z-momentum, which is impossible in a z-invariant system. The above is just the integrated Lorentz force in the z-direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from conservation of momentum, you conclude that the above is zero. This allows you to see that in the static case, $B_y$ and $B_x$ are derived from a vector potential $A_z$ just as E is derived from a potential $\phi$. The differential form formulation gives you this automatically of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as the conserved mechanical energy quantity in a static background E field is the kinetic energy plus $q\phi$, the conserved mechanical momentum in a z-independent background is the mechanical z-momentum plus $qA_z$.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-22T06:12:36.573" CommentCount="4" />
  <row Id="13838" PostTypeId="2" ParentId="13597" CreationDate="2011-08-22T07:03:12.497" Score="1" ViewCount="" Body="&lt;p&gt;Yes, it is valid. The Schrodinger field is the nonrelativistic limit of the Klein Gordon field, but you have to take the limit slightly carefully, because you have to separate the creation and annihilation operators into separate fields, which become, appropriately mixed real/imaginary parts of the Schrodinger field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way to do this is to expand the dispersion to second order in the scalar Klein Gordon field, and (in the free theory) to identify $\psi(k,t)$ with $e^{imt}a(k)$ (throwing away the rest mass), where $a(k)$ is the nonrelativistically normalized creation operator for mode k. Then $\psi(x,t)$ is the Fourier transform of $\psi(k,t)$, and unlike the relativistic theory, has an immediate interpretation as an annihilation operator removing a particle at x, and from the expansion of the dispersion, it obeys the Schrodinger equation, and from complex conjugation, it's a complex field whose conjugate involves modes near -m energy in the relativistic theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the free field anyway. The interacting field is easier in the nonrelativistic limit, so it is easier to introduce the interactions by starting there directly (although you can do it by taking the limit, of course).&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-22T07:03:12.497" />
  <row Id="13839" PostTypeId="2" ParentId="3324" CreationDate="2011-08-22T11:07:26.787" Score="3" ViewCount="" Body="&lt;p&gt;The worst analogy I have seen is that of general relativity where they show a curved sheet and how a ball revolves due to the curvature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/cS8Rg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I remember, I have seen this at two places:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The movie &quot;Einstein and Eddington&quot; where Arthur Eddington explains Einstein's theory to his housemaid&lt;/li&gt;&#xA;&lt;li&gt;Carl Sagan in his celebrated video series.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2599" LastActivityDate="2011-08-22T11:07:26.787" CommunityOwnedDate="2011-08-22T11:07:26.787" />
  <row Id="13840" PostTypeId="2" ParentId="13822" CreationDate="2011-08-22T12:07:35.097" Score="1" ViewCount="" Body="&lt;p&gt;I would agree with mark.  There most likely is a correlation for any specific liquid.  If you track the specific gravity and the capacitance for one batch it may provide a relationship for the future.  However I would imagine the constant to relate the two would even vary between different beers.&lt;br&gt;&#xA;Im meeting with someone today who has done postdoctoral work in Organic Chem.  Ill ask her.&lt;/p&gt;&#xA;" OwnerUserId="4997" LastActivityDate="2011-08-22T12:07:35.097" CommentCount="1" />
  <row Id="13841" PostTypeId="1" AcceptedAnswerId="13844" CreationDate="2011-08-22T13:21:47.150" Score="0" ViewCount="59" Body="&lt;p&gt;Why do waves on the sea shore move towards the shore even when the tide is going out?&lt;/p&gt;&#xA;" OwnerUserId="4998" LastActivityDate="2011-08-22T13:54:32.770" Title="Explain the direction of waves on sea shore" Tags="&lt;waves&gt;" AnswerCount="1" />
  <row Id="13842" PostTypeId="1" CreationDate="2011-08-22T13:30:50.583" Score="2" ViewCount="131" Body="&lt;p&gt;A particle in one dimension moves under the influence of  a potential $V(x)= ax^6$, where $a$ is a real constant. For large $n$, what is the form of the dependence of the energy $E$ on $n$?&lt;/p&gt;&#xA;" OwnerUserId="4999" LastEditorUserId="1136" LastEditorDisplayName="" LastEditDate="2011-09-04T19:10:27.067" LastActivityDate="2011-09-04T19:10:27.067" Title="The quantized energy level E depends on which power of n?" Tags="&lt;quantum-mechanics&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="1" />
  <row Id="13843" PostTypeId="1" CreationDate="2011-08-22T13:52:23.280" Score="2" ViewCount="42" Body="&lt;p&gt;Hydrogen fusion requires two hydrogen nuclei to get close enough (typically a few fm) to fuse. Much of the problem of creating a fusion reactor is overcoming the Coulomb repulsion between a pair of nuclei - the millions of degrees for Maxwellian distributions, the Bremstrahlung losses for inertial confinement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we could align the paths of two neutral Hydrogen &lt;b&gt;atoms&lt;/b&gt; (of whichever isotopes), what would the repulsion look like between them as they approach collision? Obviously at long range there is negligable force as both are neutral. But as they approach each other, what happens to the electron distribution? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, I expect a bonding cloud to form between the nuclei, and antibonding clouds beyond them. This would presumably attract at first until reaching the usual Hydrogen covalent bond length, after which the internuclear repulsion would increasingly dominate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how does that compare to bare ionic collision? How much lower is the potential barrier? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously if it was significantly lower and we could somehow engineer the collision to achieve fusion, the cross section would be larger than ionic fusion, but how much?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or would the barrier be just as high over the final few femtometers?&lt;/p&gt;&#xA;" OwnerUserId="670" LastActivityDate="2011-08-22T20:37:06.693" Title="How does the internuclear repulsion vary in Hydrogenic atom collision?" Tags="&lt;nuclear-physics&gt;&lt;fusion&gt;&lt;chemical-potential&gt;&lt;hydrogen&gt;" AnswerCount="1" />
  <row Id="13844" PostTypeId="2" ParentId="13841" CreationDate="2011-08-22T13:54:32.770" Score="3" ViewCount="" Body="&lt;p&gt;Waves are generated by wind. But aside of small local winds, most of the waves you see at the shore are generated by stronger winds far out on the sea from where they start to propagate into all kinds of directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a point in the middle of the sea where a waves with different directions are created. They will propagate away from this point until they hit a shore, therefore they can only move towards the shore not away from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also why waves always brake towards the shore and not away from it. Waves in deep water are more or less circular waves which will go unhindered until the ground is too shallow for the circular motion to go on. See here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/1cqfL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2151" LastActivityDate="2011-08-22T13:54:32.770" CommentCount="6" />
  <row Id="13845" PostTypeId="1" CreationDate="2011-08-22T14:09:46.657" Score="1" ViewCount="45" Body="&lt;p&gt;Following up on this answer, is the point at which waves break on the sea shore a guide to the depth of the sea at that point? Could it indicate eg hidden rocks?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.stackexchange.com/questions/13841/explain-the-direction-of-waves-on-sea-shore/13844#13844&quot;&gt;Explain the direction of waves on sea shore&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4998" LastActivityDate="2011-08-22T14:09:46.657" Title="Depth of sea and point at which waves break" Tags="&lt;waves&gt;" CommentCount="2" />
  <row Id="13846" PostTypeId="2" ParentId="13842" CreationDate="2011-08-22T14:35:07.853" Score="5" ViewCount="" Body="&lt;p&gt;For large $n$, the semiclassical approximation is valid and for bound states we may use the Bohr-Sommerfeld quantization condition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n = \frac{1}{h}\oint p dq$, where $n$ is the principal quantum number, $h$, the Planck constant,&#xA;$q$, the position and $p$, the momentum on the classical trajectory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case due to the conservation of energy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{p^2}{2m}+ax^6 = E \rightarrow p = \sqrt{2m(E-ax^6)}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ is the mass and $E$, the total energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By substitution, we obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n = \frac{1}{h}\int_{-(E/a)^{1/6} }^{(E/a)^{1/6} }\sqrt{2m(E-ax^6)} dx$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the integration is between the two turning points. By scaling the integration variable:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x =( \frac{E}{a})^{1/6}y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n = \frac{\sqrt{2m}}{h}{(\frac{E}{a})}^{2/3}\int_{-1 }^{1} \sqrt{(1-y^6)} dy$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E \propto n^{3/2}$&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorUserId="2190" LastEditorDisplayName="" LastEditDate="2011-08-22T15:11:34.440" LastActivityDate="2011-08-22T15:11:34.440" CommentCount="5" />
  <row Id="13847" PostTypeId="1" CreationDate="2011-08-22T15:05:22.317" Score="0" ViewCount="30" Body="&lt;p&gt;In generalized measurement theory, a Positive Operator Valued Measure (POVM) is used in place of a Projection Valued Measure (PVM). Supposedly, the choice of POVM is determined dynamically after coupling the system of interest to an ancilla, and letting the combined system evolve and interact. However, this would seem to suggest a POVM appropriate at one instant might evolve to a different POVM at a later instant. Moreover, to get a definite POVM, the combined system-ancilla system needs its own ancilla, and we need an ancilla for the ancilla of the ancilla, etc. . Is the choice of POVM really determined dynamically? If not, what determines it?&lt;/p&gt;&#xA;" OwnerUserId="5001" LastActivityDate="2011-08-22T15:05:22.317" Title="Is the choice of POVM determined dynamically?" Tags="&lt;quantum-information&gt;" />
  <row Id="13848" PostTypeId="1" AcceptedAnswerId="13849" CreationDate="2011-08-22T16:29:24.140" Score="1" ViewCount="42" Body="&lt;p&gt;I have thought about this and looked for answers for a long time now, but I do not have any name or label for this problem, which is the reason for the long title of this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have come across this many times, but I have never found any rationale for this, i.e. a clear interpretation or explanation, so here are a few examples of what I mean. I suspect that they have different answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I) In e.g. Shapiro and Teukolsy's book &lt;em&gt;Black Holes, White Dwarfs and Neutron Stars&lt;/em&gt;, (1983), the Salpeter birthrate function for stars is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\psi_s\mathrm{d}\left(\frac{M}{M_\odot}\right) = 2\times10^{-12}\left(\frac{M}{M_\odot}\right)^{-2.35}\mathrm{d}\left(\frac{M}{M_\odot}\right)\mathrm{\ stars\ pc}^{-3}\mathrm{\ yr}^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions: why does it matter that those differentials are included? Why make the statement look messier by multiplying both sides of the equation with something that obviously cancels out? If it is just to show what $\psi_s$ depends on, then it makes mre sense to me to define the function like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\psi_s(M) = 2\times10^{-12}\left(\frac{M}{M_\odot}\right)^{-2.35}\mathrm{\ stars\ pc}^{-3}\mathrm{\ yr}^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is (or may be) the answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;II) When defining the equation of continuity for brownian motion particles in 1D space, I have a lecture note that reasons in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the setup &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         |  n(x,t) |&#xA;         |         |    &#xA;J(x,t) ----&amp;gt;     ----&amp;gt; J(x + dx, t)&#xA;         |         |&#xA;         |         |&#xA;    ----------------------&amp;gt;&#xA;         x        x+dx&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where $n(x,t)$ is the number density of particles between positions $x$ and $x + \mathrm{d}x$ and $J(x, t)$ is the flux of these particles at given position $x$ and time $t$. The particle number is conserved, and they are assumed identical and non-interacting with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the notes that I have, the following equation is readily displayed as though it was crystal clear why it was written inthis way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n(x, t + \mathrm{d}t)\mathrm{d}x - n(x, t)\mathrm{d}x = - (J(x + \mathrm{d}x, t)\mathrm{d}t - J(x, t)\mathrm{d}t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or (multiplying the minus into the parenthesis)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n(x, t + \mathrm{d}t)\mathrm{d}x - n(x, t)\mathrm{d}x  = J(x, t)\mathrm{d}t - J(x + \mathrm{d}x, t)\mathrm{d}t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: I can not myself justify why it makes sense to multiply with the diferentials as they are here. What is the significance, or rather the line of thought that lies behind the way this equation turns out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unsure of my own answer which is only to look at the dimensions of the two functions themselves and then conclude what unit the factor should have on each side in order for the dimensions to make sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to complete the derivation, rewriting the equation by replacing the functions $n(x, t + \mathrm{d}t)$ and $J(x + \mathrm{d}x, t)$ with their Taylor expansion (to first order) gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n(x, t + \mathrm{d}t) = n(x, t) + \frac{\partial n}{\partial t}\mathrm{d}t$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$J(x + \mathrm{d}x, t) = J(x, t) + \frac{\partial J}{\partial x}\mathrm{d}x$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, when inserted into the original equation gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial n}{\partial t}\mathrm{d}t\mathrm{d}x = -\frac{\partial J}{\partial x}\mathrm{d}x\mathrm{d}t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so producing, again, a shared factor ($\mathrm{d}t\mathrm{d}x$) on both sides of the equation which can just be divided away (right?), and we end with the coninuity equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial n}{\partial t} = -\frac{\partial J}{\partial x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this question makes sense or at least allows you to suggest a label for this kind of problem that I can use to find more information, Or better, help by answering the wuestion here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers.&lt;/p&gt;&#xA;" OwnerUserId="5002" LastActivityDate="2011-08-22T17:04:24.047" Title="What interpretive difference is there between defining a function with or without a differential as a postfactor?" Tags="&lt;soft-question&gt;&lt;astrophysics&gt;&lt;mathematics&gt;&lt;diffusion&gt;&lt;brownian-motion&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13849" PostTypeId="2" ParentId="13848" CreationDate="2011-08-22T17:04:24.047" Score="4" ViewCount="" Body="&lt;p&gt;The differential is used to specify that the number is for a &quot;differential range&quot;, which is a way to remind you that the notions involved are somewhat fuzzy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me give a purely mathematical example. Suppose I tell you that I am going to pick an arbitrary &lt;em&gt;real&lt;/em&gt; number between 0 and 10, with the likely hood of a number being picked being proportional to the number itself. What is the probability that I pick the number 7? The answer is 0, because there are infinitely many numbers between 0 and 10. But what if I ask: what is the probability that the number I picked is between $7$ and $7+\delta$? Then the probability of course depends on how big $\delta$ is, but if $\delta$ is very small, then all numbers between $7$ and $7+\delta$ are roughly equally likely to be chosen, so we can say that the probability, for $\delta$ very small, is roughly linear in $\delta$. And so we can say that the probability of choosing a number between $7$ and $7+\delta$ is equal to $q(7)\cdot\delta$, where $q(x)$ is a function that specifies the &quot;probability density&quot; at the number $x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly,  the Salpeter function gives a distribution of the number of stars of a given mass related to the mass of the stars. But one should not state it as &quot;the number of stars &lt;strong&gt;at&lt;/strong&gt; a given mass&quot;! Because if one expects, let us say, one star at each possible mass, since there are infinite number of allowed masses, there will then be infinitely (in fact uncountably infinite) many stars overall! What the Salpeter function gives you is that the number of stars between $M$ and $M+\mathrm{d}M$, for $\mathrm{d}M$ very small, the number of stars can be described by some function $\psi(M) \mathrm{d}M$, while the expected number of stars at &lt;em&gt;exactly&lt;/em&gt; mass $M$ should be zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the differential serves to remind you that you are dealing with a sort of a density, rather than a straight-up function. &lt;/p&gt;&#xA;" OwnerUserId="829" LastActivityDate="2011-08-22T17:04:24.047" CommentCount="5" />
  <row Id="13850" PostTypeId="2" ParentId="13834" CreationDate="2011-08-22T17:32:58.587" Score="3" ViewCount="" Body="&lt;p&gt;No. The necessary and sufficient condition is well-known; see for instance Tsirelson 1993, or L. Landau.&lt;/p&gt;&#xA;" OwnerUserId="5003" LastActivityDate="2011-08-22T17:32:58.587" />
  <row Id="13851" PostTypeId="1" AcceptedAnswerId="13853" CreationDate="2011-08-22T17:58:00.790" Score="2" ViewCount="247" Body="&lt;p&gt;I was wondering if a photon is divisible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at a photon as a particle, then you may be able to split it (in theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible and how do you split it?&lt;/p&gt;&#xA;" OwnerUserId="4867" LastActivityDate="2011-08-23T05:04:56.353" Title="Can you split a photon?" Tags="&lt;photons&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="13852" PostTypeId="2" ParentId="13851" CreationDate="2011-08-22T18:15:25.570" Score="0" ViewCount="" Body="&lt;p&gt;Photon is a stable particle. It has zero mass, so there is no lighter elementary particle for it to decay into.&lt;/p&gt;&#xA;" OwnerUserId="4237" LastActivityDate="2011-08-22T18:15:25.570" CommentCount="3" />
  <row Id="13853" PostTypeId="2" ParentId="13851" CreationDate="2011-08-22T19:12:28.347" Score="6" ViewCount="" Body="&lt;p&gt;The photon cannot be split as one can split a nucleus. As it has zero mass it cannot decay. But it can interact with another particle lose part of its energy and thus change wavelength.&#xA;It can be transmuted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Compton_scattering&quot; rel=&quot;nofollow&quot;&gt;compton scattering&lt;/a&gt; entry in wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Intrigued by the other answers I searched and found that &lt;a href=&quot;http://www.nature.com/news/2010/100728/full/news.2010.381.html&quot; rel=&quot;nofollow&quot;&gt;within special crystals&lt;/a&gt; &quot;splits&quot; can happen, if one defines as a split that there can come out two photons whose energy adds up to the original energy of the photon. So in a collective crystal photon interaction there exists such a probability.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="1492" LastEditorDisplayName="" LastEditDate="2011-08-23T05:04:56.353" LastActivityDate="2011-08-23T05:04:56.353" CommentCount="2" />
  <row Id="13854" PostTypeId="2" ParentId="13851" CreationDate="2011-08-22T19:41:35.473" Score="7" ViewCount="" Body="&lt;p&gt;The answer somewhat depends on how you define &quot;splitting&quot; of a particle with a zero rest mass.&#xA;Since it can never be stopped, there is always kinetic energy in a photon which can be converted into other particles. If this counts as &quot;splitting&quot; then the photon can be split  e.g., into more photons in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Spontaneous_parametric_down-conversion&quot; rel=&quot;nofollow&quot;&gt;parametric down-conversion.&lt;/a&gt; or even into pairs of massive particles and antiparticles if there is enough energy. However, if such excess energy conversion into matter does not count as splitting, then the answer is no. In this (rather loose) sense once can say that a photon is &quot;pure energy&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1739" LastEditorUserId="1739" LastEditorDisplayName="" LastEditDate="2011-08-22T20:23:29.743" LastActivityDate="2011-08-22T20:23:29.743" />
  <row Id="13855" PostTypeId="1" AcceptedAnswerId="13858" CreationDate="2011-08-22T20:25:55.480" Score="2" ViewCount="50" Body="&lt;p&gt;I read this example in a popular math book I was browsing, and the author does not give any physical explanation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example: Consider a rock falling freely under gravity. There are no viscous or dissipative effects. We will find the average speed at the end of 5 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Time-domain.&lt;/strong&gt; $v = 32 t \mbox{feet/sec}$ &#xA;Average over $5$ seconds $$\langle v\rangle_\text{time} = \frac{1}{5} \int_0^5 32 t \mathrm{d}t = 80\text{ feet/sec}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Space Domain.&lt;/strong&gt; At $t =5 $ sec, $y = 400$ feet. $$\langle v\rangle_\text{space} = \frac{1}{400}\int_0^{400} 8\sqrt{y}\; \mathrm{d}y \approx 107\text{ feet/sec}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry about the use of imperial units. But I just dont understand the physics of this. The source uses this only as a demonstration of Cauchy Schwarz inequality. Is there something elementary I am missing out here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, he derives $$\frac{1}{T}\int_0^T v(t) \mathrm{d}t \leq \frac{1}{L}\int_0^L v(y)\mathrm{d}y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The math is ok to follow, but my physical intuition just cant get it somehow.&lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-22T20:41:48.527" LastActivityDate="2011-08-23T06:25:15.657" Title="Different values of same physical quantity in time domain and space domain OR Physical explanation of Cauchy Schwarz result" Tags="&lt;newtonian-mechanics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13856" PostTypeId="2" ParentId="13829" CreationDate="2011-08-22T20:29:35.707" Score="5" ViewCount="" Body="&lt;p&gt;To start with, I need to write a list of assumptions that are at play, and numerous disclaimers are needed.  Firstly, the IPCC scientists don't say this follows a $ln$ function at all.  They say it follows whatever their computer models says it follows.  This is only a first order solution.  Assumptions begin here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We are considering one gas&lt;/li&gt;&#xA;&lt;li&gt;We are only considering outgoing thermal radiation blockage from the Earth&lt;/li&gt;&#xA;&lt;li&gt;We are only considering one absorption peak in the absorption spectrum of this gas&lt;/li&gt;&#xA;&lt;li&gt;This absorption peak is both sufficiently narrow and sufficiently close to the peak of outgoing radiation spectrum that the outgoing radiation as a function of wavelength at sea-level is effectively constant over the entire absorption peak&lt;/li&gt;&#xA;&lt;li&gt;The cross section as a function of energy around the peak follows an exponential decrease on both sides of the peak&lt;/li&gt;&#xA;&lt;li&gt;It is a valid approximation to say that for any given wavelength, either ALL or NONE of the outgoing radiation is absorbed, corresponding with the optically thin or thick characterizations.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One more assumption may or may not be necessary, reflecting my own imperfect knowledge, which is that the exponential declines quickly relative to the width of the peak.  Also, I am referring to cross section, but I really mean the cross section specific to the thermal distribution of the CO&lt;sub&gt;2&lt;/sub&gt; molecules - just one of the finer details that I do not care much about.  The following is my illustration of the assumption the model under discussion makes about the form of the effective cross section as a function of wavelength of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/mIqu4.png&quot; alt=&quot;effective cross section graph&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The approximation above is a Taylor series expansion of the effective cross section around $\lambda_1$ and $\lambda_2$ in a $ln$ scale.  This is to say the following for points around the left edge.  I will just keep to the left edge because the right part follows in the same way and isn't worth typing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln( \Psi(\lambda) ) = \ln( \Psi(\lambda_1) ) + \gamma \Delta \lambda$$&#xA;$$ \Psi(\lambda) = \Psi(\lambda_1) e^{ \gamma \Delta \lambda }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above includes an assumption about the form of the absorption peaks of the CO&lt;sub&gt;2&lt;/sub&gt;.  If you were mathy, you might be able to get this from a Gaussian or a broadened absorption peak, I just want to note it real quick b/c I won't be commenting on the validity.  Next, I need to write the equation for the optical thickness, $\tau$ as a function of energy, then mutilate the radiation forcing expression, $F$ into a piecewise function.  Also, $C$ is the concentration of carbon dioxide and $N_{Air}$ is the volume density of gas molecules in the atmosphere.  I'll also write it again for the specific region around the left edge for which $\Delta \lambda = \lambda - \lambda_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau(\lambda) = C N_{Air} \Psi(\lambda) = C N_{Air} \Psi(\lambda_1) e^{ \gamma \Delta \lambda }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Optically thick is defined by $\tau&amp;gt;2.3$.  We will write the energy dependent flux coming off the ground as $\Phi(\lambda)$.  By the previously stated assumption, $\Phi(\lambda) \approx \Phi$.  $F$ is proportional to the absorbed/reflected $\Phi$ by some constant I don't care about, $\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{CO2} = \int_0^{\infty} \beta \Phi (1 - e^{\tau(\lambda)}) d\lambda$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(1 - e^{\tau(\lambda)}) \approx \begin{cases}&#xD;&#xA;  0, &amp;amp; \mbox{if } \tau(\lambda) &amp;lt; 2.3 \\&#xD;&#xA;  1,  &amp;amp; \mbox{if } \tau(\lambda) \ge 2.3 &#xD;&#xA;\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{CO2} = \beta \Phi \left( \lambda_2 - \lambda_1 \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just gawked or gagged at that... all I can say is don't shoot the messenger.  In fact, we don't even really need that $\beta$ there for the purposes of this question.  Anyway, I hope it's clear where we're going, and at this point I will assign $\lambda_1$ and $\lambda_2$ to be &lt;em&gt;pre-industrial&lt;/em&gt; values.  The physical effect humans have had on the greenhouse effect, by this model, is to broaden the range of wavelengths that CO&lt;sub&gt;2&lt;/sub&gt; in the atmosphere absorbs.  To put this into math, I will say that $\lambda_1&amp;#39;$ and $\lambda_2&amp;#39;$ are the corresponding &lt;em&gt;post-industrial&lt;/em&gt; values.  Also, I will use $C$ as post-industrial concentration, $C_0$ as pre-industrial, $\tau$ and $\tau&amp;#39;$, and $F$ and $F&amp;#39;$ respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta F = F_{CO2}&amp;#39;-F_{CO2} = \beta \Phi \left( (\lambda_2&amp;#39; - \lambda_2) + (\lambda_1 - \lambda_1&amp;#39;) \right)= \beta \Phi \left( \Delta \lambda_2 + \Delta \lambda_1 \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here is the beef, we can find $\Delta \lambda_1$ from the prior Taylor series expansion.  Remember, the edge of the absorption peak, the point that separates the optically thick and thin regions, comes from that magic 2.3 value.  The driving force behind this is, of course, the change in concentration $C$.  For clarity I'll note that $\Delta \lambda_1 = \lambda_1-\lambda_1&amp;#39;$.  You'll probably be confused by the negative, it's just math trickery in order to account for the fact that we're on the left (negative) side of the peak, and thus get the right $C$ on top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;post-industrial&lt;/em&gt;&#xA;$$\tau&amp;#39;(\lambda) = 2.3 = C N_{Air} \Psi(\lambda_1) e^{ -\gamma \Delta \lambda_1 }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;pre-industrial&lt;/em&gt;&#xA;$$\tau(\lambda) = 2.3 = C_0 N_{Air} \Psi(\lambda_1) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to my question follows with algebra now.  Just assume $\Delta \lambda_2$ to be the same thing, not that the constant matters anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta \lambda_1 = \frac{1}{\gamma} \ln{\frac{C}{C_0}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta F = \frac{ 2 \beta \Phi}{\gamma} \ln{\frac{C}{C_0}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finished.  Although I probably used more space than what I needed to.  And just for the record, this does matter, since discussions with the IPCC scenarios involve values of $C/C_0$ of 2 to 3.  If you just used the decreasing exponential you will get a very different answer.  The problem in a nutshell is a cross section that varies so starkly with wavelength.  The $1-e^{-\tau}$ was always recognized to be mono-energetic.  The $ln$ alternative is for the exact opposite scenario where it completely blacks out energy ranges and the mathematical form of the absorption peaks give rise to the Taylor series used here.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The reference for this work was posted by Michael Luciuk in the comments.&#xA;&lt;a href=&quot;http://www.princeton.edu/~lam/documents/LamAug07bs.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.princeton.edu/~lam/documents/LamAug07bs.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And people have, of course, pointed out that the real behavior doesn't exactly follow this model.  Here is some real data.  I found it to look surprisingly logarithmic.  &lt;a href=&quot;http://scienceofdoom.com/2010/02/19/co2-an-insignificant-trace-gas-part-seven-the-boring-numbers/&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/v8kor.png&quot; alt=&quot;real correlation&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-08-23T11:49:48.607" LastActivityDate="2011-08-23T11:49:48.607" CommentCount="3" />
  <row Id="13857" PostTypeId="2" ParentId="13843" CreationDate="2011-08-22T20:37:06.693" Score="2" ViewCount="" Body="&lt;p&gt;Your last sentence is exactly right: the energy cost for fusion is almost all in those last few femtometers, at which electronic effects are negligible. Although there is in principle a difference between colliding neutral atoms and nuclei, at the energies required for fusion the effect is tiny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy differences associated with the presence, absence, arrangement, etc., of electrons are of order a few to a few tens of electron volts. One way to see this is to note that the size scale of the electron's wavefunction is of order the Bohr radius $a_0$, so the energy levels are of order $e^2/(4\pi\epsilon_0 a_0)$. But to get fusion to happen, as you say, you have to get the two protons to get within a fermi or so of each other. This length scale is something like $10^4$ to $10^5$ times smaller, so it involves energies that are larger by about the same factor -- MeV rather than eV.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-08-22T20:37:06.693" />
  <row Id="13858" PostTypeId="2" ParentId="13855" CreationDate="2011-08-22T21:55:50.480" Score="4" ViewCount="" Body="&lt;p&gt;Let us denote by $\langle\cdot\rangle \equiv \int_0^T \cdot \frac{dt}{T}$ the averaging over time. Please notice that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{L}\int_0^Lv\mathrm{d}y=\frac{1}{\langle v\rangle T}\int_0^Tv\frac{\mathrm{d}y}{\mathrm{d}t}\mathrm{d}t = \frac{1}{\langle v\rangle}\int_0^Tv^2\frac{\mathrm{d}t}{T} = \frac{\langle v^2\rangle}{\langle v\rangle}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the inequality is given by: $\langle v\rangle \le \frac{\langle v^2\rangle}{\langle v\rangle}$, or $\langle v\rangle^2 \le \langle v^2\rangle$, which is indeed the the Cauchy-Schwarz inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus averaging over time is not the same as averaging over the position. Averaging with respect to two variables is the same if and only if these two variables are linearly dependent (more generally related by an affine transformation). In our case, in general, the position is a nonlinear function of time, thus one obtains different results in averaging over time and position. The only case that these two averages are equal is the case of uniform motion with a constant velocity. &lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-23T06:25:15.657" LastActivityDate="2011-08-23T06:25:15.657" />
  <row Id="13859" PostTypeId="2" ParentId="13851" CreationDate="2011-08-22T21:57:53.423" Score="0" ViewCount="" Body="&lt;p&gt;there is a vertex gamma-&gt; 3 gamma. So the photon can turn into 3 phtons, with smaller energy. To make this possible, they all have to be collinear, orthewise the kinematics forbids it.&lt;/p&gt;&#xA;" OwnerUserId="5005" LastActivityDate="2011-08-22T21:57:53.423" CommentCount="1" />
  <row Id="13860" PostTypeId="1" CreationDate="2011-08-22T21:59:52.267" Score="0" ViewCount="40" Body="&lt;p&gt;This question should perhaps go in Chemistry, to which I am committed, but since the Chemistry Stack Exchange forum is not yet in beta, I am wondering if you can help me from a physics perspective.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering about the definitions of the terms &lt;em&gt;flammable&lt;/em&gt;, &lt;em&gt;inflammable&lt;/em&gt;, and &lt;em&gt;non-flammable&lt;/em&gt;.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Flammable#Linguistics%3a_flammable_vs._inflammable&quot; rel=&quot;nofollow&quot;&gt;Wikipedia says&lt;/a&gt; that &lt;em&gt;flammable&lt;/em&gt; and &lt;em&gt;inflammable&lt;/em&gt; are &lt;strong&gt;synonyms&lt;/strong&gt;, meaning &quot;capable of burning&quot;.  In contrast, &lt;em&gt;non-flammable&lt;/em&gt; means &quot;not capable of burning&quot;.  To avoid confusion, it is recommended that use of the term &lt;em&gt;inflammable&lt;/em&gt; be avoided, since the prefix &lt;em&gt;in-&lt;/em&gt; here does not mean &quot;not&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This generally makes sense and sounds like good advice, but as I was reading a journal paper today, I came across a paragraph using these terms.  I am wondering if you can help me by telling me your interpretation of the intent of this paragraph:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;During the last decade room temperature ionic liquids (RTIL), being&#xA;  salts of low temperature melting points, have been studied&#xA;  extensively.  They may be used as solvents for many processes and as&#xA;  electrolytes in electrochemical devices.  Batteries filled with such a&#xA;  type of electrolytes do not contain any volatile components and&#xA;  therefore, they are not flammable.  Room temperature ionic liquids,&#xA;  being usually quaternary ammonium salts, are characterized by&#xA;  negligible  vapor pressure, which makes them inflammable.  In&#xA;  addition, they show a broad electrochemical stability window,&#xA;  generally &gt;4 V, which is necessary for the application in lithium-ion&#xA;  batteries with high-energy cathodes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is from page 602 in &quot;Ionic liquids as electrolytes for Li-ion batteries--An overview of electrochemical studies&quot;, Lewandowski, A.; Swidereska-Mocek, A. &lt;em&gt;Journal of Power Sources&lt;/em&gt; &lt;strong&gt;2009&lt;/strong&gt;, &lt;em&gt;194&lt;/em&gt;, 601-609, which is available to subscribers &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0378775309011616&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first part, &lt;strong&gt;&quot;...do not contain any volatile components and therefore, they are not flammable&quot;&lt;/strong&gt;, seems to make sense; room temperature ionic liquids are promising in part because they do not ignite or burn easily, so they can potentially improve the safety of batteries.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the next mention seems to say just the opposite: &lt;strong&gt;&quot;...are characterized by negligible vapor pressure, which makes them inflammable&quot;&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is, does it seem to you that the term &lt;em&gt;inflammable&lt;/em&gt; has been used incorrectly in this paragraph?  Does negligible vapor pressure imply that a substance is likely to burn readily or not burn readily?  Thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="5004" LastActivityDate="2011-08-22T22:59:04.603" ClosedDate="2011-08-23T01:21:50.263" Title="What are the differences between the terms flammable, inflammable, and non-flammable?" Tags="&lt;pressure&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13861" PostTypeId="1" AcceptedAnswerId="13910" CreationDate="2011-08-22T22:07:30.577" Score="9" ViewCount="195" Body="&lt;p&gt;The great russian physicist Lev Landau developed a famous entry exam to test his students. This &quot;Theoretical Minimum&quot; contained everything he considered elementary for a young theoretical physicist. Despite its name, it was notoriously hard and comprehensive, and in Landau's time, only 43 students passed it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if anyone can provide the list of topics, or even a copy of the exam?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm sure I'd have no chance to pass, but I'd like to see it out of a sense of sportmanship ;-). Also, I think it would make quite a good curriculum of theoretical physics (at least pre 1960).)&lt;/p&gt;&#xA;" OwnerUserId="825" LastActivityDate="2011-08-24T10:00:21.407" Title="Lev Landau's &quot;Theoretical Minimum&quot;" Tags="&lt;theory&gt;&lt;history&gt;" AnswerCount="2" FavoriteCount="3" />
  <row Id="13862" PostTypeId="2" ParentId="13861" CreationDate="2011-08-22T22:23:38.643" Score="8" ViewCount="" Body="&lt;p&gt;In the following arXiv article &lt;a href=&quot;http://arxiv.org/abs/hep-ph/0204295v1&quot; rel=&quot;nofollow&quot;&gt;Landau's Theoretical minimum&lt;/a&gt;, the author details personal experiences on being one of 42 students ever who passed it, as well as some details of the examination. He also mentions some questions in passing, from which I gather most of them were similar to the exercises and examples of his famous books. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Landau gave me the programs of eight courses on theoretical physics.&#xA;  Besides, there was one more examination on mathematics — complex&#xA;  variables, special functions, the Laplace transformation, etc. By that&#xA;  time only a few books of the famous Landau course were published:&#xA;  Mechanics, Classical Field Theory, The Theory of Continuous Media and&#xA;  the first (classical) part of Statistical Physics. One had to study&#xA;  all other courses by reading various books and original papers. For&#xA;  example, let me display the list of books/papers which we were&#xA;  supposed to study for the Quantum Mechanics course:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-22T22:47:25.543" LastActivityDate="2011-08-22T22:47:25.543" CommentCount="1" />
  <row Id="13863" PostTypeId="2" ParentId="13860" CreationDate="2011-08-22T22:59:04.603" Score="0" ViewCount="" Body="&lt;p&gt;&lt;a href=&quot;http://english.stackexchange.com/questions/1578/why-are-not-infamous-and-inflammable-the-opposite-of-famous-and-flammable-like-in&quot;&gt;Flammable and inflammable&lt;/a&gt; are from separate French and Latin words (although ultimately from the same latin root)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct - low vapour pressure makes things difficult to burn. Or at least difficult to start a fire with if you don't have a way of dispersing them in air. I suspect the author got confused or was a victim of spell checker auto-complete&lt;/p&gt;&#xA;" OwnerUserId="2525" LastActivityDate="2011-08-22T22:59:04.603" />
  <row Id="13864" PostTypeId="1" AcceptedAnswerId="13927" CreationDate="2011-08-22T23:34:00.463" Score="3" ViewCount="62" Body="&lt;p&gt;Here is another old exam question I'm wondering about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A proton moving in the $-x$ direction encounters a region of space with a magnetic field that randomizes the direction of the particle. The magnetic field is at rest. After scattering off the magnetic fields the proton is ejected from the region of space and is moving in the $+x$ direction.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A) If the original energy of the proton is $10^{12} \, \mathrm{eV}$ and the mean strength of the magnetic field is $1\, \mathrm{mT}$, what is the final energy of the proton after scattering?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, here I thought that magnetic fields don't do work, so the energy would be unchanged? But on the other hand, an accelerated charge emits radiation, and of course I must accelerate the proton to turn its direction by 180 degrees. But if it's not a trick question where the answer is just $10^{12}\, \mathrm{eV}$, then how can I say anything meaningful at all when the field is random?&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-08-24T22:08:27.960" Title="Scattering off a random magnetic field?" Tags="&lt;electromagnetism&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="13865" PostTypeId="2" ParentId="5714" CreationDate="2011-08-22T23:41:12.867" Score="0" ViewCount="" Body="&lt;p&gt;I know this one is going to be controversial, but since I've seen Yang-Mills theory (&quot;as Yang Mills wanted&quot;) up there...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;String theory&lt;/strong&gt; - I wouldn't say it failed, but it turned out that it didn't solve &lt;em&gt;the problem it was invented for&lt;/em&gt;, namely to describe the strong interaction, and the occurence of certain patterns in mesons (Regge trajecories). Although you can describe some strong phenomena with a stringy theory - like you can with the &quot;eightfold way&quot;, nuclear isospin etc. - nowadays QCD is considered the most fundamental description of strong phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, it turns out that you can use some tools of string theory (the AdS/CFT correspondence) to perform calculations that you can't do in QCD. This is done in heavy ion physics. Note that the theory you use there is different than the original &quot;pions as strings&quot; hypothesis.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a nice write up about the history of string theory by Luboš Motl here: &lt;a href=&quot;http://physics.stackexchange.com/questions/13728/what-are-the-details-around-the-origin-of-the-string-theory/13749#13749&quot;&gt;What are the details around the origin of the string theory?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a second problem it didn't solve (yet?), to give a unified theory of gravity and the other forces, that includes the standard model as a low energy limit. But that's a lot to ask for from a theory...&lt;/p&gt;&#xA;" OwnerUserId="825" LastActivityDate="2011-08-22T23:41:12.867" CommunityOwnedDate="2011-08-22T23:41:12.867" CommentCount="2" />
  <row Id="13866" PostTypeId="2" ParentId="13826" CreationDate="2011-08-22T23:47:27.423" Score="2" ViewCount="" Body="&lt;p&gt;Well, you can add up the masses of the three light quarks in a proton (uud) and figure out how that compares to the total mass of $938\text{ MeV}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\times 2.4\text{ MeV} + 4.8\text{ MeV} = 9.6\text{ MeV} = 0.010 m_p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that doesn't tell you much of anything useful, because the quark masses are actually running parameters (they vary with energy scale) and more importantly, because the structure of a baryon is not that simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_chromodynamics&quot; rel=&quot;nofollow&quot;&gt;QCD&lt;/a&gt;, a baryon actually consists of a whole &quot;sea&quot; of quarks and antiquarks, along with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gluon&quot; rel=&quot;nofollow&quot;&gt;gluons&lt;/a&gt; and photons that bind them together. To further complicate things, gluons and photons can convert into virtual quark-antiquark pairs and back again, so you get varying mixes of these particles; and in fact, as you might expect since a baryon is a quantum system, the proportions change depending on how you measure them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particle physicists characterize the composition of baryons with &lt;a href=&quot;http://en.wikipedia.org/wiki/Parton_distribution_function#Parton_distribution_functions&quot; rel=&quot;nofollow&quot;&gt;parton distribution functions&lt;/a&gt; (PDFs). In a simplified model, a parton distribution function gives the probability density for detecting a particular kind of constituent inside a hadron. It depends on two kinematic variables: $x$ is the fraction of the total hadron's momentum carried by the parton you detect, and $Q^2$ is the negative squared four-momentum of the thing that interacts with the parton (i.e. a photon or gluon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose you fire a large number of virtual photons, with four-momenta $q^\mu$ uniformly distributed in the range $Q_1^2 &amp;lt; |q^2| &amp;lt; Q_2^2$, at a proton with (3-)momentum of magnitude $p$. The fraction of them that will interact with an up quark that carries a momentum between $x_1 p$ and $x_2 p$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{Q_1^2}^{Q_2^2}\int_{x_1}^{x_2}f_{\mathrm{u}/\mathrm{p}}(x,Q^2)\mathrm{d}x\frac{\mathrm{d}Q^2}{Q^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one of these functions for each kind of quark, antiquark, and gluon in each different hadron: $f_{\mathrm{d}/\mathrm{p}}$ for down quarks in a proton, $f_{\mathrm{\bar{u}}/\mathrm{n}}$ for anti-up quarks in a neutron, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you interpret these PDFs as probability densities, you can calculate the contribution of each different kind of parton to any particular physical quantity by taking a weighted average. For example, the expectation value of the &lt;em&gt;number&lt;/em&gt; of up quarks in the proton is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$N_{\mathrm{u}/\mathrm{p}} = \int_{Q_1^2}^{Q_2^2}\int_{x_1}^{x_2}f_{\mathrm{u}/\mathrm{p}}(x,Q^2)\mathrm{d}x\frac{\mathrm{d}Q^2}{Q^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and similarly for anti-up quarks, down quarks, etc. These values have to satisfy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}N_{\mathrm{u}/\mathrm{p}} - N_{\mathrm{\bar{u}}/\mathrm{p}} &amp;amp;= 2 \\ N_{\mathrm{d}/\mathrm{p}} - N_{\mathrm{\bar{d}}/\mathrm{p}} &amp;amp;= 1 \\ N_{i/\mathrm{p}} - N_{\bar{i}/\mathrm{p}} &amp;amp;= 0\quad\text{otherwise}\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the technical expression of the idea that the proton consists of two up quarks and one down quark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evidently, the issue of figuring out the composition and properties of a baryon comes down to determining these parton distributions. So what do the PDFs look like? Unfortunately, we can't calculate them from fundamental principles using &lt;a href=&quot;http://en.wikipedia.org/wiki/Perturbative_QCD&quot; rel=&quot;nofollow&quot;&gt;perturbative QCD&lt;/a&gt;, which is the main tool of most physicists who study baryon structure, but we can relate them to experimental data. There are several groups of physicists which periodically release &quot;global fits,&quot; which give the form of the PDFs taking into account all (credible) data collected from major experiments up to that time. They use slightly different fitting algorithms and data filters, but the different results are mostly comparable. The &quot;standard&quot; source (by which I mean the one-stop PDF fit reference that most people will give you) for this data is the &lt;a href=&quot;http://hepdata.cedar.ac.uk/pdfs&quot; rel=&quot;nofollow&quot;&gt;Durham PDF server&lt;/a&gt;. (Note that we mostly only have data for protons, since they are what gets used in the experiments.) As an example, the latest results from the CTEQ collaboration for protons at momentum transfer $Q^2 = 100\text{ GeV}^2$ look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/3eJRC.png&quot; alt=&quot;CT10 global PDF fits, central values&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the up and down quark distributions have bumps roughly around $x = \frac{1}{3}$, which reflects the fact that there are three valence quarks in the proton and therefore there is some tendency towards finding one of them carrying around a third of the proton's momentum. You would expect that if a proton was just three quarks and nothing else, but in that case it would be a sharp spike, not a broad bump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, this still isn't enough to actually answer your question, for a couple of reasons. Remember that the parton distributions depend on the momentum transfer of the probe ($Q^2$), so it's not necessarily meaningful to ask what the proportion of matter to antimatter to &quot;binding energy&quot; is unless you specify a particular value of $Q^2$, i.e. specify the parameters of the experiment you're doing. Also, PDFs have been measured only for high-energy scattering experiments, at large values of $Q^2$. There are equations, called the &lt;a href=&quot;http://cerncourier.com/cws/article/cern/43194&quot; rel=&quot;nofollow&quot;&gt;DGLAP equations&lt;/a&gt;, that tell us how the parton distributions change as $Q^2$ changes, but they only apply at relatively large values of $Q^2$ - precisely the sort of high-energy scattering experiments that the data is collected for. You can't extrapolate all the way down to zero momentum transfer (a proton at rest) because other unknown physical effects will come into play at lower values of $Q^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we can do is calculate the contributions to the proton's total energy arising from quarks, antiquarks, and gluons (which is probably the closest thing to &quot;binding energy&quot; - see &lt;a href=&quot;http://physics.stackexchange.com/questions/11449/conversion-of-mass-to-energy-in-chemical-nuclear-reactions/11460#11460&quot;&gt;this answer&lt;/a&gt;) at a given value of $Q^2$. Treating the PDFs as probability distributions, the fraction of the proton's total energy that is contributed by matter quarks (udscbt) would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_m = \sum_{i\in\{\mathrm{udcsbt}\}}\int_{Q_1^2}^{Q_2^2}\int_{0}^{1}\sqrt{x^2p^2 + p_T^2 + m_i^2}\,f_{i\,/\mathrm{p}}(x,Q^2)\mathrm{d}x\frac{\mathrm{d}Q^2}{Q^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make some simplifications to this, though. First, at the typical energy scales involved in a particle accelerator, the masses of the light quarks (uds) are so small as to be entirely irrelevant; furthermore, the proton has such a high longitudinal momentum $p$ that any transverse momentum not created by the collision is &lt;a href=&quot;http://physics.stackexchange.com/questions/12384/qcd-parton-branching-is-the-angle-always-negligible/12395#12395&quot;&gt;probably negligible&lt;/a&gt;. Also, as shown by the graph above, the quark distributions drop off pretty quickly with increasing mass for the heavier quarks - in other words, the bottom and top PDFs $f_{\mathrm{b}/\mathrm{p}}$ and $f_{\mathrm{t}/\mathrm{p}}$, and perhaps even the charm PDF $f_{\mathrm{c}/\mathrm{p}}$, represent very small contributions to the total energy. And finally, over a very small range of $Q^2$ we can assume that $f$ is nearly independent of $Q^2$, so the integral over $Q^2$ gives $\ln\frac{Q_2^2}{Q_1^2}\approx 1$. So we're down to calculating&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_m \approx p\sum_{i\in\{\mathrm{uds}\}}\int_{0}^{1}x\,f_{i\,/\mathrm{p}}(x)\mathrm{d}x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conveniently, this integrand is exactly what the Durham PDF calculator will give you, so it's easy to do the numerical integration. The integral works out to about 0.26 for the up quark, 0.13 for the down quark, and 0.025 for the strange quark, so we're looking at a result of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{quarks:}\quad E_m \approx 0.41 p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you assume that the proton's mass of $938\text{ MeV}$ is negligible relative to its momentum (which is the case in modern accelerators), this means that about 41% of an accelerated proton's energy can be attributed to matter. Doing the same calculation for the antiquarks yields a factor of 0.032 for anti-up and 0.038 for anti-down, and if you assume that the strange and anti-strange quark PDFs are the same (which is not rigorously justified, as far as I know, but seems reasonable), you get a total for the antimatter of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{antiquarks:}\quad E_m \approx 0.094 p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, integrating the gluon distribution gives a contribution of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{gluons:}\quad E_m \approx 0.45 p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All together, these numbers come within about 5% of adding up to 1, which tells us that the factors I've neglected (heavy quarks and masses) probably don't make much more than a 5% difference.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you'd like to read more about this sort of thing, I would recommend looking at textbooks, because there isn't just one paper that will tell you everything you need to know.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Peskin and Schroeder&lt;/strong&gt;, &lt;em&gt;Introduction to Quantum Field Theory&lt;/em&gt;, chapter 17&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Halzen and Martin&lt;/strong&gt;, &lt;em&gt;Quarks and Leptons: An Introductory Course in Modern Particle Physics&lt;/em&gt;, chapters 8-10&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Ellis, Stirling, and Webber&lt;/strong&gt;, &lt;em&gt;QCD and Collider Physics&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-08-22T23:47:27.423" CommentCount="2" />
  <row Id="13867" PostTypeId="1" CreationDate="2011-08-22T23:53:25.837" Score="0" ViewCount="31" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/6068/list-of-good-classical-physics-books&quot;&gt;List of good classical physics books&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Here's the thing: English is not my first language, and it's also not the language I was speaking when I studied Physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays, I'm having trouble describing various phenomena because I can't find the proper English terms for certain things. I know the concept, I just don't know the word describing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'm looking for a good book, in English, covering all main chapters of Physics: mechanics, thermodynamics, electricity &amp;amp; magnetism, optics, nuclear, relativity - the works. The mathematics used in the book don't have to be very advanced, in fact it's probably better if the math is rather basic, but the explanations are verbose (more words than formulas) - obviously, because it's the words that I need, more than the formulas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back then, I had a big Physics tome, by Sears, Zemansky and Young, translated in my native language, that covered pretty much everything. But it looks like the original English version, which would be pretty good for my purpose, is hard to find on Amazon (unless it's paperback, or 2nd hand, etc.). And perhaps there are even better books nowadays.&lt;/p&gt;&#xA;" OwnerUserId="4210" LastActivityDate="2011-08-22T23:53:25.837" ClosedDate="2011-08-23T01:13:35.597" Title="Good book covering all chapters of basic Physics, in English? (for someone who learned Physics in a different language)" Tags="&lt;books&gt;" CommentCount="3" />
  <row Id="13868" PostTypeId="2" ParentId="13804" CreationDate="2011-08-22T23:53:47.740" Score="3" ViewCount="" Body="&lt;h3&gt;Original wrong stuff&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Well, the OP asked for a solution, but I still can't figure out how this is physics. Add the two equations with an i to get the equation for $V=V_x + i V_y$ (dividing by dt--- I hate finance/mathematics conventions, and calling the white noise $\eta$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${dV\over dt} = (-\gamma+i\Omega)V + iV\eta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that the complex logarithm of V is executing a biased Brownian motion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${d\over dt} \log(V) = (-\gamma + i \Omega) + i\eta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V(t) = \exp( - (\gamma + i \Omega)t + iB(t))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where B is a standard Brownian motion. The V solution goes to zero for sure, even without the i multiplying the Brownian term, because B only goes as square root of t (assuming positive nonzero damping), and the total distance travelled is given by a Laplace transform of the exponential of a Brownian motion, and I didn't bother to evaluate its statistics because this is clearly not physics.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The x and y $\eta$ noise can't be the same in physics. I didn't realize they were the same in the problem as stated, because this is so strange.&lt;/li&gt;&#xA;&lt;li&gt;In physical Brownian diffusion, in the limit of long times, V acquires a distribution with a scale, the Maxwell Boltzmann distribution, and in this case V can't acquire dimension because the equation is scale invariant (the OP says &quot;dimensionally consistent&quot;). There is a thermal velocity scale of $V_T=\sqrt{2mkT}$ in real Brownian motion which allows the equation to be dimensionally consistent without V multiplying the noise, but with $V_T$. It is this thermal velocity that sets the scale for the average velocity at times long compared to the velocity relaxation time.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Smoluchowski approximation requires that V have a steady state distribution and a relaxation time to that steady state distribution, which doesn't happen in this problem. Here the velocity goes away. That's clearly a property of the scaling invariance of V, and the overall decay of V, so it won't be fixed even if the x and y noise are different. This is not thermal physics, and I can't see any way to do a Smoluchowski approximation, so how to do the approximation is moot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that the book had a thermal scale constant V multiplying the noise, and there is some error in transcription of the problem, but I don't want to start guessing.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Later: Correct stuff&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Thanks for linking to the book, it made everything clear. The key words are &quot;energy conserving process&quot;. This is designed to model a particle moving through a constant magnetic field with an additional random magnetic field on top, so that it goes around with a randomly fluctuating curvature, but the particle never changes speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disregard the previous stuff, it is totally wrong, because it is using a different convention for the time derivative of stochastic quantities than is used throughout the book (I should have seen this from the form of the problem, but I didn't). In deriving the solution I used the fact that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${1\over V} {dV\over dt} = {d\over dt} \log(V)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is only true if the time derivative is Stratonovich (centered difference), which is a common convention in physics. The book is using an Ito convention for time derivatives (forward-difference), which is the most common convention in mathematics and in finance. The Ito time derivative does &lt;em&gt;not&lt;/em&gt; obey the uncorrected chain rule, but the correction is easy to work out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${1\over V(t+\epsilon)} {dV\over dt} -{ 1\over V(t)} {dV\over dt} = -{1\over V^2} {(V(t+\epsilon) - V(t))^2\over \epsilon} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the last term, the infinitesimal square fluctuation in V at time t, is entirely determined by the square of the coefficient of the noise term. In quantum mechanics language,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[{1\over V}, {dV\over dt} ] = (i\sqrt{2\gamma})^2 = -2\gamma$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to go from Ito derivative, which is one order of operators, to Stratonovich derivative which is the average of the two orders (like the Jordan product of operators in quantum mechanics), you have to add half the commutator, and this gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${d\over dt} \log(V) - \gamma = -\gamma + i\Omega + i\sqrt{2\gamma}\eta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which exactly cancels the damping term, so that the result is completely physical--- the velocity magnitude doesn't decay at all. This is not a surprise. The coefficient which I confused for a physical damping is not a damping at all, but just the amount of magnitude change of the velocity vector an infinitesimal time into the future caused by the random direction-changing fluctuations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V = V_0 \exp(i\Omega t + i\sqrt{2\gamma}B(t))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the angle of the velocity V is executing a Brownian motion with drift, and keeps the same magnitude at all times. The equation at long times does become a Brownian motion in x, and the Smoluchowski approximation is valid for long times.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The Smoluchowski limit&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The central difficulty you have is with the Smoluchowski limit, so I'll give a short explanation. In ordinary Brownian motion, the equation of motion is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m{d^2x\over dt^2} + \gamma{dx\over dt} + \sqrt{2D}\eta = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Ito and Stratonovich are the same, there are no products of non-commuting quantities). If you average this equation over long time window, the velocity averages to the total distance over the total time, while the second derivative averages to zero (because the velocity is varying in a bounded thermal range). This means that the first derivative average is equal to the noise thermal average, which means that the long-time averaged displacement of the solution to the above second order stochastic equation is equal to the long-time averaged solution of the first order equation below, which drops the second derivative term:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\gamma {dx\over dt} + \sqrt{2D}\eta =0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, more succinctly at the cost of butchering history, in Brownian motion, Newton equals Aristotle. So that $D$ is the diffusion constant of the long-time Brownian motion (D is also determined from the condition that the v diffusion process has as a steady state the Maxwell Boltzmann distribution, and this gives the Einstein relation). Dropping the second derivative term in the equation is equivalent to the formal rules that the book gives. There is no more content in those formal rules than the above averaging procedure, which is better because its both clearer and more rigorous.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The actual problem&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;In this problem, the equation in question is (in second order Ito form--- all derivatives are forward differences)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${d^2 X\over dt^2} - \gamma {dX\over dt} + i\Omega{dX\over dt} - i\sqrt{2\gamma}{dX\over dt} \eta = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Smoluchowski approximation averages this equation over a long interval, which gets rid of the second derivative term (for the same reason as before, bounded velocity fluctuations), and produces an equation which is first order:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(-\gamma + i\Omega)\Delta X = i\sqrt{2\gamma}\int_0^T {dX\over dt}\eta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But unlike standard thermal drift, this is no simplification, because one needs to know the average of the product of the velocity and the noise in order to get a closed equation. This is the difficulty you were having.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a minor annoyance in this way of looking at things: it seems like there are too many equations, in that the real and imaginary parts of the equation give two separate constraints, but this isn't true, because the right hand side is almost a perfect integral. By putting the explicit solution for the velocity,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int_0^T i\sqrt{2\alpha} {dX\over dt}\eta(t) dt = \int_0^T \sqrt{2\alpha}V_0\eta e^{i\Omega t + i\sqrt{2\alpha} B(t)} dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And considering that $\eta = {dB\over dt}$ by definition, this would be a perfect integral up to adding Omega, in the stratonovich convention, where the chain rule works. But this is Ito, so you do the half-commutator trick etc, etc, I'll leave out the details, but the end result gives just one consistent equation for $\Delta X$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta X = V_0 \int_0^T \exp(i\Omega t + B(t)) dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which tells us nothing new, because this is just the integral of the velocity from the explicit solution. So there is no way to evaluate the Smoluchowski limit without knowing something about the integral of cosines and sines of a Brownian motion. This is the central problem that the book is posing.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The integral of cosines and sines of a Brownian motion&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;To get the Smolochowski limit, all you need evaluate the average square distance traveled after time t:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\langle |\Delta X|^2 \rangle = |V_0|^2\langle \int_0^T \int_0^T e^{i\Omega (t - t&amp;#39;) + i (B(t)-B(t&amp;#39;))} dt dt&amp;#39; \rangle = \int_0^T\int_0^T e^{i\Omega(t-t&amp;#39;)} \langle e^{i(B(t)-B(t&amp;#39;))}\rangle$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the irreducible quantity to solve this problem is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\langle e^{i(B(t) - B(t&amp;#39;))} \rangle = G(t-t&amp;#39;)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G(t)$ is, by translation invariance, the expected value of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G(t) = \langle e^{iB(t)} \rangle $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be evaluated by Feynman diagrams: expand the exponential in powers, and note that B(t) is gaussian with width t, so all its moments are known. The odd moments are zero, and the even moments are given by products of odd numbers (pairing combinations):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\langle B(t)^{2n} \rangle = 1 \cdot 3 \cdot 5 ...\cdot (2n-1) t^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the power series for the exponential only has odd terms&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G(t) = \sum_{k=0}^\infty {(-1)^n \langle B(t)^{2n}\rangle\over (2n)!} = e^{-{t\over 2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this means by symmetry that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G(t) = e^{-{1\over 2} |t-t&amp;#39;|}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be found by path integrals too, it is a quadratic path integral with a linear source, and the result is the exponential of 1/2 the Green's function of a 1d laplacian between the two sources, and this Green's function is the absolute value function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance traveled after time T is the double integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|\Delta X|^2 = |V_0|^2\int_0^T \int_0^T G(t-t&amp;#39;) e^{i\Omega(t-t&amp;#39;)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which by thinking about it in 45-degree rotated coordinates, is, up to the unimportant boundaries of the rectangle, equal to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$|\Delta X|^2 = T|V_0|^2 \tilde{G}(\Omega)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that the diffusion constant is the Fourier transform of $\exp(-|t|)$, which is ${1\over 1+\omega^2}$, fixed up to account for the factor of 2. This gives the average square distance, and the diffusion constant is the coefficient of T in this expression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tilde{G}(\Omega) = {1\over 4 + \Omega^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can restore the $\gamma$ into the formula by dimensional analysis, I'll do so later if necessary.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-24T05:17:52.763" LastActivityDate="2011-08-24T05:17:52.763" CommentCount="3" />
  <row Id="13869" PostTypeId="1" CreationDate="2011-08-23T02:22:52.927" Score="2" ViewCount="53" Body="&lt;p&gt;can somebody explain to me what happens or how do the correlation functions/partition functions in general behave (if such general answer/behavior exists and if not then why not) during a first and second order phase transition? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer is this:&#xA;1. Correlation function diverges @ phase transition.&#xA;2. 1st (2nd) derivative of partition function diverges at first (second) order phase transition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I kind of assume that it might be common knowledge for people who do Quantum Statistical mechanics or QFT in condensed matter (I even probably saw these answers in one or two websites). So, sorry if this question is naive. &#xA;But it would be useful for me to get an answer from a general QFT background point of view (no emphasis on condensed matter) and also it would be nice to know if there is an easier way to see it from general QFT background. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More importantly it would be nice to have a reference mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1349" LastActivityDate="2011-08-23T15:56:40.283" Title="Behavior of Correlation functions/partition functions during phase transitions" Tags="&lt;quantum-field-theory&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="13870" PostTypeId="1" CreationDate="2011-08-23T02:48:20.713" Score="19" ViewCount="255" Body="&lt;p&gt;I have read before in one of Seiberg's articles something like, that gauge symmetry is not a symmetry but a redundancy in our description, by introducing fake degrees of freedom to facilitate calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding this I have a few questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why is it called a symmetry if it is not a symmetry? what about Noether theorem in this case? and the gauge groups U(1)...etc?&lt;/li&gt;&#xA;&lt;li&gt;Does that mean, in principle, that one can gauge any theory (just by introducing the proper fake degrees of freedom)?&lt;/li&gt;&#xA;&lt;li&gt;Are there analogs or other examples to this idea, of introducing fake degrees of freedom to facilitate the calculations or to build interactions, in classical physics? Is it like introducing the fictitious force if one insists on using Newton's 2nd law in a noninertial frame of reference?   &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4521" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-23T03:15:35.430" LastActivityDate="2011-08-23T22:53:08.243" Title="Gauge symmetry is not a symmetry?" Tags="&lt;quantum-field-theory&gt;&lt;particle-physics&gt;&lt;gauge-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" />
  <row Id="13871" PostTypeId="1" AcceptedAnswerId="13874" CreationDate="2011-08-23T03:19:49.947" Score="5" ViewCount="113" Body="&lt;p&gt;So, the quantum path integral is a generalization of the classical principle of least action- but here we know that all paths contribute something finite to the probability density. What confuses me is that this doesn't seem to involve the quantization of energy (action) at all. How does this come into play? Is it a separate assumption, only coming into play when we worry about the non-commutativity of certain operators?&lt;/p&gt;&#xA;" OwnerUserId="4663" LastActivityDate="2011-08-23T10:44:36.967" Title="How does the quantum path integral relate to the quantization of energy?" Tags="&lt;quantum-mechanics&gt;&lt;path-integral&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" />
  <row Id="13872" PostTypeId="2" ParentId="13871" CreationDate="2011-08-23T05:39:16.683" Score="-1" ViewCount="" Body="&lt;p&gt;The quantization is not necessary in quantum mechanics. It is essential, but not necessary. Energy of a free particle is not quantized. Even for oscillator you are free to construct any state and energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantization appears at the next step. If you want for instance study stable states which do not change with time. It is not separate assumption. It is different problem.&lt;/p&gt;&#xA;" OwnerUserId="4657" LastActivityDate="2011-08-23T05:39:16.683" />
  <row Id="13873" PostTypeId="2" ParentId="12314" CreationDate="2011-08-23T06:48:00.940" Score="0" ViewCount="" Body="&lt;p&gt;See forum  post &lt;a href=&quot;http://focusfusion.org/index.php/forums/viewthread/934/#8825&quot; rel=&quot;nofollow&quot;&gt;http://focusfusion.org/index.php/forums/viewthread/934/#8825&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5009" LastActivityDate="2011-08-23T06:48:00.940" CommentCount="1" />
  <row Id="13874" PostTypeId="2" ParentId="13871" CreationDate="2011-08-23T07:08:08.467" Score="2" ViewCount="" Body="&lt;p&gt;The path integral in quantum mechanics computes the evolution kernel, which is the matrix element of the evolution operator: &#xA;$\mathrm{exp}(iH t) $, ($H$ is the Hamiltonian), between two position eigenstates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The path integral expresses the evolution kernel as a sum over paths:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U(x,t, x_0) = \int_{x(0)=x_0}^{x(t)=x} \mathrm{exp}(\frac{iS}{\hbar}) \mathcal{D}x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $U$ is the evolution kernel, $S$ is the classical action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the evolution operator has an expansion as a sum over the energy eigenstates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U(x,t, x_0) =\sum_n \mathrm{exp}(\frac{-iE_n t}{\hbar})\psi_n(x) \psi_n^{*}(x_0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, $\psi_n(x)$ are the energy eigenstates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this expression, it is clear that the evolution kernel has a discrete spectrum, whenever the energies are quantized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, in the case of quantized values of the energy, the Fourier transform of the evolution kernel:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hat{U}(x,\omega, x_0)\equiv \int_{-\infty}^{\infty} U(x,t, x_0) \mathrm{exp}(i\omega t) dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;will be a sum of Dirac delta functions centered at the frequencies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega_n = \frac{E_n}{\hbar}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please observe that the weight of each Dirac delta function is just the projection operator on the corresponding discrete eigenstate.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastActivityDate="2011-08-23T07:08:08.467" CommentCount="2" />
  <row Id="13875" PostTypeId="2" ParentId="2765" CreationDate="2011-08-23T07:45:33.640" Score="0" ViewCount="" Body="&lt;p&gt;To emit a radio wave of any sort which sticks out above ordinary thermal radiation, you need an energy source which is not thermal. Such an energy source is not available to DNA molecules in dilute solutions. If there is some cellular machinery which is responsible for emitting low frequency EM waves, it would require some energy source and would be sensitive to dilution. The idea that DNA emits radio is a perpetual motion machine of the second kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electromagnetic signals are in my opinion amplified thermal noise, and the &quot;signal&quot; stands out because the water with the DNA in it that he uses is slightly warmer.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-23T07:45:33.640" />
  <row Id="13876" PostTypeId="2" ParentId="13869" CreationDate="2011-08-23T08:14:48.147" Score="1" ViewCount="" Body="&lt;p&gt;The general rule is that in a first order transition the partition function (or its logarithm, the &quot;free entropy&quot; or free energy over T) is continuous but its derivative with respect to the thermodynamic parameter which is driving the transition is not. The reason is that as you come into the transition from the left, you have all phase 1, and the derivative is a bulk phase 1 property. At the transition, phase 1 and phase 2 have equal bulk free energies, and phase l turns into phase 2 as you try to drive the parameter further. The partition function is continuous at the transition, because both phases have equal free energy per unit volume in bulk at this point, which is why they can freely turn one into the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original idea was that at a second order transition, the derivative would be continuous and the second derivative discontinuous, allowing you to define third order transitions by the third derivative, and so on. This idea is incorrect (at least in finite dimensions, maybe it can be salvaged in mean field theory) because the second order transition is completely different from the first order in terms of the statistical degrees of freedom,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In first order transitions, the correlation functions fall off exponentially at long distances, with no fluctuations bigger than the atomic level. Nothing happens to the correlation functions of the bulk phases there, the transition is just caused by the collision of the bulk free entropy of the two phases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a second order transition the system at the transition point is not transitioning between two distinct phases, but has a continuum statistical limit defined by looking at the fluctuating statistical fields which are no longer &quot;massive&quot; (meaning they are no longer statisically clamped to a fixed value in such a way that they forget their boundary values on a large sphere). The generic behavior is that the correlations in the fluctuations of these fields falls off as a power at the transition, and this power is one of the critical exponents. Away from the transition they fall off exponentially, but at a distance called the correlation length which diverges as you approach the transition, and it diverges as another power, another critical exponent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The properties of a second order transition is always by a statistical field theory, and these are very constrained, because they obey the analog of a central limit theorem. All properties of the transition except the type of fluctuating fields and their renormalizable &quot;Lagrangian&quot; are not important, or irrelevant, for the long-distance behavior. The renormalizable Lagrangians (free entropy as a function of the field, but this is the statistical analogs of quantum field theory Lagrangian because it is the weight in a path integral) have only a finite parameter space, and these parameters correspond to the number of parameters you need to tune to approach the transition. Symmetries can reduce the number of directions by forcing some parameters to be zero automatically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generic behavior is that the free energy as a function of some collection of bulk thermodynamic fields becomes a function of the derivatives of these fields and their values so that the continuum path integral considering these fields makes sense.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-23T15:56:40.283" LastActivityDate="2011-08-23T15:56:40.283" />
  <row Id="13877" PostTypeId="2" ParentId="13832" CreationDate="2011-08-23T09:18:51.893" Score="3" ViewCount="" Body="&lt;p&gt;I think the source of your confusion is mathematics, not physics. It is important here (and L&amp;amp;L did mention this) that the system of differential equations is autonomous. If this is the case, than along with solution&#xA;$q_i=q_i(t,C_1,\dots,C_{2s}),$&#xA;it has a solution&#xA;$q_i=q_i(t-t_0,C_1,\dots,C_{2s}).$&#xA;Because the former is the general solution, the later must reduced to it, that is, it should be&#xA;$$q_i(t-t_0,C_1,\dots,C_{2s})=q_i(t,C_1&amp;#39;(C_1,\dots,C_{2s},t_0),\dots,C_{2s}&amp;#39;(C_1,\dots,C_{2s},t_0)).$$&#xA;Putting $C_{2s}=0$, yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q_i(t-t_0,C_1,\dots,C_{2s-1})=q_i(t,C_1&amp;#39;(C_1,\dots,C_{2s-1},t_0),\dots,C_{2s}&amp;#39;(C_1,\dots,C_{2s-1},t_0)).$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For example&lt;/strong&gt;, in the case of a free 1D motion $x=C_1t+C_2$. Making the shift in time one has: $$x=C_1t+(C_2-C_1t_0)$$ (expression in brackets is $C_2'$). And letting $C_2=0$ we find $$x=C_1(t-t_0).$$&lt;/p&gt;&#xA;" OwnerUserId="4237" LastActivityDate="2011-08-23T09:18:51.893" />
  <row Id="13878" PostTypeId="1" AcceptedAnswerId="13885" CreationDate="2011-08-23T09:47:39.280" Score="1" ViewCount="96" Body="&lt;p&gt;The kinetic energy of a particle is a periodic function in time. Does it imply that the particle is in a conservative force field and there are no dissipative forces acting on it at any instance of time ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; (in view of comment by Willie)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please consider the question as &quot;Is the force acting on the particle conservative ?&quot; Also consider the kinetic energy to be periodic with non zero period.&lt;/p&gt;&#xA;" OwnerUserId="540" LastEditorUserId="540" LastEditorDisplayName="" LastEditDate="2011-08-23T16:51:50.083" LastActivityDate="2011-08-23T18:09:14.643" Title="kinetic energy and conservative force field" Tags="&lt;classical-mechanics&gt;" AnswerCount="2" CommentCount="6" />
  <row Id="13879" PostTypeId="2" ParentId="13871" CreationDate="2011-08-23T10:44:36.967" Score="1" ViewCount="" Body="&lt;p&gt;In general, path integral does not imply quantization of anything. First, you can use path integrals for classical wave optics. Second, even in quantum mechanics there are non-bonding potentials (take the free particle potential $V(x)=0$ for the simplest case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, sometimes from the path integral point of view it can be argued that some things are quantized. Perhaps the best known example is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Aharonov%E2%80%93Bohm_effect&quot; rel=&quot;nofollow&quot;&gt;Aharonov-Bohm effect&lt;/a&gt; where to have non-ambiguous solution phase $e^{i S/\hbar}$ can not change when a particle makes a full circle around a flux of magnetic field.&lt;/p&gt;&#xA;" OwnerUserId="184" LastActivityDate="2011-08-23T10:44:36.967" />
  <row Id="13880" PostTypeId="2" ParentId="13870" CreationDate="2011-08-23T11:15:04.927" Score="18" ViewCount="" Body="&lt;p&gt;In order:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Because the term &quot;gauge symmetry&quot; pre-dates QFT. It was coined by Weyl, in an attempt to extend general relativity. In setting up GR, one could start with the idea that one cannot compare tangent vectors at different spacetime points without specifying a parallel transport/connection; Weyl tried to extend this to include size, thus the name &quot;gauge&quot;. In modern parlance, he created a classical field theory of a $\mathbb{R}$-gauge theory. Because $\mathbb{R}$ is locally the same as $U(1)$ this gave the correct classical equations of motion for electrodynamics (i.e. Maxwell's equations). As we will go into below, at the classical level, there is no difference between gauge symmetry and &quot;real&quot; symmetries.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes. In fact, a frequently used trick is to introduce such a symmetry to deal with constraints. Especially in subjects like condensed matter theory, where nothing is so special as to be believed to be fundamental, one often introduces more degrees of freedom and then &quot;glue&quot; them together with gauge fields. In particular, in the strong-coupling/Hubbard model theory of high-$T_c$ superconductors, one way to deal with the constraint that there be no more than one electron per site (no matter the spin) is to introduce spinons (fermions) and holons (bosons) and a non-Abelian gauge field, such that really the low energy dynamics is confined --- thus reproducing the physical electron; but one can then go and look for deconfined phases and ask whether those are helpful. This is a whole other review paper in and of itself. (Google terms: &quot;patrick lee gauge theory high tc&quot;.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You need to distinguish between forces and fields/degrees of freedom. Forces are, at best, an illusion anyway. Degrees of freedom really matter however. In quantum mechanics, one can be very precise about the difference. Two states $\left|a\right\rangle$ and $\left|b\right\rangle$ are &quot;symmetric&quot; if there is a unitary operator $U$ s.t. $$U\left|a\right\rangle = \left|b\right\rangle$$ and $$\left\langle a|A|a\right\rangle =\left\langle b|A|b\right\rangle $$ where $A$ is any physical observable. &quot;Gauge&quot; symmetries are those where we decide to label the &lt;em&gt;same&lt;/em&gt; state $\left|\psi\right\rangle$ as both $a$ and $b$. In classical mechanics, both are represented the same way as symmetries (discrete or otherwise) of a symplectic manifold. Thus in classical mechanics these are not separate, because both real and gauge symmetries lead to the same equations of motion; put another way, in a path-integral formalism you only notice the difference with &quot;large&quot; transformations, and locally the action is the same. A good example of this is the Gibbs paradox of working out the entropy of mixing identical particles -- one has to introduce by hand a factor of $N!$ to avoid overcounting --- this is because at the quantum level, swapping two particles is a gauge symmetry. This symmetry makes no difference to the local structure (in differential geometry speak) so one cannot observe it classically.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A general thing -- when people say &quot;gauge theory&quot; they often mean a much more restricted version of what this whole discussion has been about. For the most part, they mean a theory where the configuration variable includes a connection on some manifold. These are a vastly restricted version, but covers the kind that people tend to work with, and that's where terms like &quot;local symmetry&quot; tend to come from. Speaking as a condensed matter physicist, I tend to think of those as theories of closed loops (because the holonomy around a loop is &quot;gauge invariant&quot;) or if fermions are involved, open loops. Various phases are then condensations of these loops, etc. (For references, look at &quot;string-net condensation&quot; on Google.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the discussion would be amiss without some words about &quot;breaking&quot; gauge symmetry. As with real symmetry breaking, this is a polite but useful fiction, and really refers to the fact that the ground state is not the naive vacuum. The key is commuting of limits --- if (correctly) takes the large system limit last (both IR and UV) then no breaking of any symmetry can occur. However, it is useful to put in by hand the fact that different real symmetric ground states are separately into different superselection sectors and so work with a reduced Hilbert space of only one of them; for gauge symmetries one can again do the same (carefully) commuting superselection with gauge fixing.&lt;/p&gt;&#xA;" OwnerUserId="1501" LastEditorUserId="1501" LastEditorDisplayName="" LastEditDate="2011-08-23T11:28:11.000" LastActivityDate="2011-08-23T11:28:11.000" CommentCount="5" />
  <row Id="13881" PostTypeId="2" ParentId="13870" CreationDate="2011-08-23T12:59:28.697" Score="2" ViewCount="" Body="&lt;p&gt;The (big) difference between a gauge theory and a theory with only rigid symmetry is precisely expressed by the Noether &lt;a href=&quot;http://en.wikipedia.org/wiki/Noether%27s_first_theorem&quot; rel=&quot;nofollow&quot;&gt;first&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Noether%27s_second_theorem&quot; rel=&quot;nofollow&quot;&gt;second&lt;/a&gt; theorems:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While in the case of a rigid symmetry, the currents corresponding to the group generators are conserved only as a consequence of the equations of motion. This is called that they are conserved &quot;on-shell&quot;, in the case of a continuous gauge symmetry, the conservation laws &#xA;become valid &quot;off-shell&quot;, that is independently of the equations of motion. &#xA;This implies for example the conservation of the electric charge irrespective of the equation of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the conservation law equations can be used in principle to reduce the number of fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The procedure is as follows: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Work on the subspace of the field configurations satisfying the conservation laws. However, there will still be residual gauge symmetries on this subspace. In order to get rid of those:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Select a gauge fixing condition for each conservation law.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This will reduce the &quot;number of field components&quot; by two for every gauge symmetry.&#xA;The implementation of this procedure however is very difficult, because it actually requires to solve the conservation laws, and moreover, the reduced space of field configurations is very complicated.&#xA;This is the reason why this procedure is rarely implemented and other techniques like BRST are used.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-08-23T22:53:08.243" LastActivityDate="2011-08-23T22:53:08.243" />
  <row Id="13883" PostTypeId="1" CreationDate="2011-08-23T16:39:04.630" Score="3" ViewCount="37" Body="&lt;p&gt;and another exam question, this is about current research:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Interference of matter waves has been studied using ultra-cold atoms. The phase of a matter wave for free-falling cold-atoms at time $t$ and height $y$ is given as &#xA;  $$\phi(y,t) = \phi(y_0,t) + Et - \int_{y_0}^y dy&amp;#39; k(y&amp;#39;)$$&#xA;  where $E$ is the total energy of individual coherent atoms and $\hbar k(y)$ is the semi-classical momentum of falling atoms at height $y$. A recent experiment observed coherent matter waves continuously emitted from two vertical micro-traps (you can neglect the size of the traps) with height $-\lambda/2$ and $\lambda/2$. The atoms are emitted at rest from the traps.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A) For a given height $y$ below the two micro-traps, derive the condition of constructive and destructive interference.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, from basic wave mechanics I know that the phase difference of the two waves has to be an even (constructive) or odd (destructive) multiple of $\pi$, but apart from that I find the question poorly worded: What does $t$ refer to? Total time in laboratory frame, or time since leaving the trap? And what about the function $\phi(y_0,t)$? If I don't know its form, can I say anything meaningful about the interference pattern at all?&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-08-23T16:39:04.630" Title="Matter-wave interference from free falling cold atoms" Tags="&lt;quantum-mechanics&gt;&lt;homework&gt;&lt;bose-einstein-condensate&gt;" FavoriteCount="1" />
  <row Id="13884" PostTypeId="1" CreationDate="2011-08-23T16:56:42.760" Score="3" ViewCount="78" Body="&lt;p&gt;What are prerequisites (in mathematics and physics), that one should know about for getting into use of ideas from noncommutative geometry in physics?&lt;/p&gt;&#xA;" OwnerUserId="5012" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-08-23T17:18:23.040" LastActivityDate="2011-08-23T17:18:23.040" Title="Prerequisites to start the study of noncommutative geometry in physics" Tags="&lt;soft-question&gt;&lt;education&gt;&lt;geometry&gt;" />
  <row Id="13885" PostTypeId="2" ParentId="13878" CreationDate="2011-08-23T17:16:15.727" Score="3" ViewCount="" Body="&lt;p&gt;No.  I can take a ball and swing it back and forth periodically with my hand.  The motion is periodic, but the situation is not conservative - my body generates a lot of heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple mathematical example is &lt;a href=&quot;http://farside.ph.utexas.edu/teaching/315/Waves/node12.html&quot; rel=&quot;nofollow&quot;&gt;a forced, damped harmonic oscillator&lt;/a&gt;.  It has a steady-state periodic solution that dissipates energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to know whether a force field is conservative, take its curl.  Time-independent force fields (force is a function of position but not time) are conservative iff their curl is zero.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-08-23T17:16:15.727" CommentCount="2" />
  <row Id="13886" PostTypeId="5" CreationDate="2011-08-23T17:18:30.383" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-23T17:18:30.383" LastActivityDate="2011-08-23T17:18:30.383" />
  <row Id="13887" PostTypeId="4" CreationDate="2011-08-23T17:18:30.383" Score="0" ViewCount="" Body="A class of theories that attempt to explain all existing particles (including force carriers) as vibrational modes of one-dimensional objects." OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-23T17:18:30.383" LastActivityDate="2011-08-23T17:18:30.383" />
  <row Id="13888" PostTypeId="5" CreationDate="2011-08-23T17:19:24.657" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-23T17:19:24.657" LastActivityDate="2011-08-23T17:19:24.657" />
  <row Id="13889" PostTypeId="4" CreationDate="2011-08-23T17:19:24.657" Score="0" ViewCount="" Body="The quantitative study of how fluids (gases and liquids) move." OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-23T17:19:24.657" LastActivityDate="2011-08-23T17:19:24.657" />
  <row Id="13890" PostTypeId="5" CreationDate="2011-08-23T17:21:15.107" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-23T17:21:15.107" LastActivityDate="2011-08-23T17:21:15.107" />
  <row Id="13891" PostTypeId="4" CreationDate="2011-08-23T17:21:15.107" Score="0" ViewCount="" Body="General questions about the way objects move and interact. This tag should be used when the tags for certain kinds of mechanics (newtonian-mechanics, classical-mechanics, quantum-mechanics, etc.) are too specific." OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-23T17:21:15.107" LastActivityDate="2011-08-23T17:21:15.107" />
  <row Id="13892" PostTypeId="5" CreationDate="2011-08-23T17:23:03.417" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-23T17:23:03.417" LastActivityDate="2011-08-23T17:23:03.417" />
  <row Id="13893" PostTypeId="4" CreationDate="2011-08-23T17:23:03.417" Score="0" ViewCount="" Body="The study of physical properties condensed phases of matter, including solids and liquids." OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-23T17:23:03.417" LastActivityDate="2011-08-23T17:23:03.417" />
  <row Id="13894" PostTypeId="5" CreationDate="2011-08-23T17:23:47.800" Score="0" ViewCount="" Body="" OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-23T17:23:47.800" LastActivityDate="2011-08-23T17:23:47.800" />
  <row Id="13895" PostTypeId="4" CreationDate="2011-08-23T17:23:47.800" Score="0" ViewCount="" Body="Any of the various explanations of gravity as a quantum theory, including string theory and loop quantum gravity." OwnerUserId="-1" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-23T17:23:47.800" LastActivityDate="2011-08-23T17:23:47.800" />
  <row Id="13896" PostTypeId="1" AcceptedAnswerId="13898" CreationDate="2011-08-23T17:57:11.260" Score="1" ViewCount="82" Body="&lt;p&gt;I'm just starting to study quantum mechanics. Please explain the error in this thinking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You set up decay of two $\pi$ mesons and get $2\mathrm{e}^-$ on Mars and $2\mathrm{e}^+$ on Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Earth you may or may not measure the spin of those positrons, with 50% probability that they are the same spin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Mars, your buddy &quot;immediately afterwards&quot; takes a $\mathrm{He}^{2+}$ ion and adds the electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you made the measurement, the $\mathrm{He}$ has a 50%+ chance of being in a higher-then-base energy level due to Pauli exclusion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you didn't make the measurement, the $\mathrm{He}$ has a much lower chance of being in a higher-than-base energy level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your buddy measures the energy state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Result: You just superluminally transmitted $\approx 0.8$ bits of information.&lt;/p&gt;&#xA;" OwnerUserId="5013" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-23T19:17:23.347" LastActivityDate="2011-08-23T19:23:28.917" Title="Spin up, spin down and superposition" Tags="&lt;quantum-mechanics&gt;&lt;speed-of-light&gt;&lt;entanglement&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" />
  <row Id="13897" PostTypeId="2" ParentId="13878" CreationDate="2011-08-23T18:09:14.643" Score="3" ViewCount="" Body="&lt;p&gt;There is quite a lot wrong with this question. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is a conservative force field? As usually defined, a conservative force field is a field that is defined everywhere in space that is the gradient of some potential energy (which may be time dependent); equivalently we ask that the curl of the force field is zero. This requires knowing the force everywhere in space. &lt;/li&gt;&#xA;&lt;li&gt;Suppose we know the trajectory $\vec{\gamma}(t)$ of a particle. Then by taking the second time derivative we can compute the acceleration, and using Newton's second law we can compute the force. Okay, so this gives us knowledge of the (possibly time-dependent) force field $\vec{F}$ and the space-time coordinates $(t,\vec{\gamma}(t))$: we have that $\vec{F}(t,\vec{\gamma}(t)) = m \ddot{\gamma}(t)$. &#xA;&lt;ol&gt;&#xA;&lt;li&gt;But what does this tell us? Absolutely nothing! At any given time, we are only given knowledge &lt;strong&gt;at one single point&lt;/strong&gt; of the force. The force field could be &lt;strong&gt;anything&lt;/strong&gt; at other points in space. Given the value of a function at a point, you can extend it arbitrarily! You can extend it so that the force, at that instant in time, is conservative; or you can extend it so that the force is not conservative. You can extend it anywhich way you want and still have a force compatible with the motion of your particle. (Another way to look at it using &lt;a href=&quot;http://physics.stackexchange.com/questions/13878/kinetic-energy-and-conservative-force-field/13885#13885&quot;&gt;Mark Eichenlaub's answer&lt;/a&gt;: a particle executing simple harmonic motion could either be in a harmonic potential (so is in a conservative force field), or it could equally well be in the steady state of a driven, damped harmonic oscillator, in which case you have dissipation. &lt;/li&gt;&#xA;&lt;li&gt;Okay, you say, let us suppose then the force field is time independent. This way by specifying $\gamma(t)$ you can specify what the force would be along every point visited by the particle. Now you run into two problems: (a) what if the trajectory visits the same point twice, but exhibits different acceleration at those two different times? Then it is impossible to find a well-defined time-independent force field that generate this motion. (b) what if the trajectory only visits a single curve of points? Then you have the problem as before: if a function is prescribed along a one-dimensional subset of three-dimensional space, you can essentially extend the function anyway you want. You can extend the force-field to be conservative, or you can extend it to be non-conservative. &lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;And then there is the additional problem that just giving the kinetic energy is insufficient to specify the velocity of the trajectory: the kinetic energy can allow you to determine the modulus $|\dot{\gamma}|(t)$ of the particle, but it cannot allow you to determine the direction the particle is moving in. So associated with a kinetic energy there are infinitely many trajectories that intersects itself infinitely many times, and there are also infinitely many trajectories which trace out a smooth embedded curve in 3 dimensional space.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="829" LastActivityDate="2011-08-23T18:09:14.643" />
  <row Id="13898" PostTypeId="2" ParentId="13896" CreationDate="2011-08-23T19:23:28.917" Score="3" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If you didn't make the measurement, the He has a much lower chance of being in a higher-than-base energy level. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think this is the incorrect step in your reasoning.  The $He$ atom would be in a 50-50 superposition of the ground and an excited state if the electrons on earth were not measured, and should collapse to one state or the other with equal probablility if they are.  In either case, a mesurement of the $He$ atom to determine if it were in the ground or an excited state has exactly the same distribution of results.&lt;/p&gt;&#xA;" OwnerUserId="3936" LastActivityDate="2011-08-23T19:23:28.917" />
  <row Id="13899" PostTypeId="1" AcceptedAnswerId="13900" CreationDate="2011-08-23T21:23:15.170" Score="0" ViewCount="44" Body="&lt;p&gt;So this seemed pretty straight forward at first, but then I wasnt too sure about gravity forces...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;centerfuge&lt;/li&gt;&#xA;&lt;li&gt;2 samples&lt;/li&gt;&#xA;&lt;li&gt;1 sample is 1gram heavier than the other&lt;/li&gt;&#xA;&lt;li&gt;centerfuge at top speed is exerting 20000 gravitys of force on the samples&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What kind of weight / ballast is needed to keep the centerfuge from wobbling off the table?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Id imagine this is a linear equation ... So if i could get the equation that would be TOPS!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ballast should be placed on 4 corners on the outside of the sample&lt;/p&gt;&#xA;" OwnerUserId="4854" LastEditorUserId="4854" LastEditorDisplayName="" LastEditDate="2011-08-23T21:35:39.300" LastActivityDate="2011-08-23T21:42:59.973" Title="calculating ballast needed to offset wobble of a 1g unbalanced centerfuge spinning at 20000G's" Tags="&lt;gravity&gt;&lt;acceleration&gt;&lt;centrifugal-force&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="13900" PostTypeId="2" ParentId="13899" CreationDate="2011-08-23T21:42:59.973" Score="0" ViewCount="" Body="&lt;p&gt;the force caused by this disbalance (1 gram at 20 000 g) is the weight of 20 kilograms, assuming coefficient of friction of 0.5 you need at least 40kg to stop it from wobbling around. Somewhat more depending to the exact geometry (the centrifuge would act as lever). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It really sounds very unsafe. High speed centrifuges need to be shielded so that the fragments of failed rotor couldn't kill you.&lt;/p&gt;&#xA;" OwnerUserId="4967" LastActivityDate="2011-08-23T21:42:59.973" CommentCount="2" />
  <row Id="13901" PostTypeId="1" CreationDate="2011-08-23T22:03:18.787" Score="3" ViewCount="99" Body="&lt;p&gt;One way to normalize the free particle wave function &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;is to replace the the boundary condition $\psi(\pm{\frac{a}{2}}) = 0$ [for the infinite well] by periodic boundary conditions expressed in the form $\psi(x)=\psi(x+a)$&quot; &#xA;  -- &lt;em&gt;Quantum Physics&lt;/em&gt;, S. Gasiorowicz&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How does this work? What does this mean physically? Or more precisely, why does this approximation suffice? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that this makes the wavefunction square-integrable (when integrated from $x=0$ to $x=a$) hence normalizable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" LastActivityDate="2011-08-23T23:58:30.300" Title="Normalizing the free particle wave function" Tags="&lt;mathematical-physics&gt;" AnswerCount="2" />
  <row Id="13902" PostTypeId="2" ParentId="13901" CreationDate="2011-08-23T23:44:51.790" Score="4" ViewCount="" Body="&lt;p&gt;The physical idea is that you'll let $a$ go to infinity for a truly free particle, and if you take this limit, then the specific details of the boundary conditions should be irrelevant, because the boundaries are so far away anyway. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, you are welcome to choose convenient boundary conditions, and the periodic ones are convenient, because then you have just plain waves $e^{ikx}$, with the admitted $k$-values determined by $e^{ika} = 1$, so $ka = 2\pi n$, and $n \in \mathbb{Z}$.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-08-23T23:44:51.790" CommentCount="1" />
  <row Id="13903" PostTypeId="2" ParentId="13901" CreationDate="2011-08-23T23:58:30.300" Score="3" ViewCount="" Body="&lt;p&gt;You want boundary conditions that conserve the total probability within your finite box.  The probability current is proportional to, roughly,   $J \sim \psi \partial \psi$.  Setting $\psi=0$ at the boundary forces $J=0$ at the boundary (particles get reflected) so probability is conserved (once can see that Dirichlet boundary condition $\partial\psi = 0$ does the same).  Setting $\psi(x+a)=\psi(x)$ forces $J(x+a) = J(x)$ so the current going out of your box on the left is equal to that coming in on the right (i.e. your box is really a torus), and probability is again conserved.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-08-23T23:58:30.300" />
  <row Id="13904" PostTypeId="1" CreationDate="2011-08-24T02:19:49.570" Score="1" ViewCount="120" Body="&lt;p&gt;According to relativity,nothing can break light barrier.But a recent &lt;a href=&quot;http://arxiv.org/abs/1101.1840&quot; rel=&quot;nofollow&quot;&gt;preprint&lt;/a&gt; shows energy transmission of commercial electric power (f=60Hz) is faster than light. (It is not the drift velocity of electrons because energy transport depends on the electromagnetic field instead of charged particles). Moreover, the speed increases as its frequency decreases so the value is infinity for a direct current (f=0).That is action at a distance (instantaneous interaction)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is whether it is possible to design a circuit to compare this speed with c in lab? For example,time-delay of an electromagnetic signal through a free space distance of 3m is $10^{-8}\text{ s}$. As to direct current along a copper wire, however, there should be no delay. Is the difference detectable? I'm interested to the experiment because it is of substantial significance to measure the rate of electric power for electrical engineering not matter the result can exceed c or not.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-24T08:07:52.970" LastActivityDate="2011-08-24T08:07:52.970" Title="How to measure faster than light electric energy?" Tags="&lt;electromagnetism&gt;&lt;special-relativity&gt;&lt;electricity&gt;&lt;speed-of-light&gt;&lt;electrodynamics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" />
  <row Id="13905" PostTypeId="2" ParentId="13904" CreationDate="2011-08-24T03:33:14.203" Score="1" ViewCount="" Body="&lt;p&gt;Ignoring whether the claim is valid or not, there should be no problem in measuring the velocity of a signal on a transmission line compared to the velocity of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;in the lab&lt;/em&gt; part might present a problem because of the small distances, but in principle a simultaneous signal sent on the transmission line and on a laser beam ( which can go for kilometres) should easily allow to measure which comes first.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-08-24T03:33:14.203" CommentCount="3" />
  <row Id="13906" PostTypeId="2" ParentId="7232" CreationDate="2011-08-24T07:23:39.970" Score="2" ViewCount="" Body="&lt;p&gt;All the answers to this question are wrong. They are right that there is no limit to the magnitude of the electric field limit in classical GR, but they are wrong to claim that this is because electric fields by themselves can't make a black hole.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There is no bound to the magnitude of the electric field in GR:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is not enough to have a high energy density to make a black hole, you need an total energy greater than R in a sphere of radius R. If the sphere is arbitrarily small, the energy density can be arbitrarily big. For a specific example of a strong field without collapse, an extremal black hole solution in general relativity has a metric:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ds^2 = (1-Q/r)^2 dt^2 - {dr^2 \over \left(1-{Q/r}\right)^2 } + r^2 (d\theta^2 + \sin^2\theta \, d\phi^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And an electric field in the direction of r of magnitude $Q/r^2$. This electric field is maximum at $r=Q$ (at least outside the black hole), and the maximum value is 1/Q, which is as big as you like. To make a non singular solution with the same properties, just replace the black hole horizon with a slightly bigger sphere with the same mass and the same surface charge.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An electric field which has a large enough magnitude, all by itself, in a large enough region of space, causes gravitational collapse.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The easiest way to see this is to consider a parallel plate capacitor of area $L^2$ and charge Q with no net momentum, and pull apart the plates to distance L, then you have order $|Q/L^2|^2 L^3$ amount of mass in a region of radius L, and when Q is bigger than L (which happens with small charge densities when L is large) you get a total energy at large distances which is too big to avoid collapse just by considering the total mass and zero momentum.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There is nothing wrong with John Baez's examples of highly boosted noncollapsing stress, but they are irrelevant.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These are not like a static electic field because they have a large net mometum. The static electric field has zero field momentum.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="825" LastEditorDisplayName="" LastEditDate="2011-08-24T22:28:18.100" LastActivityDate="2011-08-24T22:28:18.100" />
  <row Id="13907" PostTypeId="1" CreationDate="2011-08-24T07:50:30.847" Score="4" ViewCount="89" Body="&lt;p&gt;This is a thought I had a while ago, and I was wondering if it was satisfactory as a physicist's proof of the positive mass theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The positive mass theorem was proven by Schoen and Yau using complicated methods that don't work in 8 dimensions or more, and by Witten using other complicated methods that don't work for non-spin manifolds. Recently Choquet-Bruhat published a proof for all dimensions, which I did not read in detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see that you can't get zero mass or negative mass, view the space-time in the ADM rest frame, and consider viewing the spacetime from a slowly accelerated frame going to the right. This introduces a Rindler horizon somewhere far to the left. As you continue accelerating, the whole thing falls into your horizon. If you like, you can imagine that the horizon is an enormous black hole far, far away from everything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizon starts out flat and far away before the thing falls in, and ends up flat and far away after. If the total mass is negative, it is easy to see that the total geodesic flow on the outer boundary brings area in, meaning that the horizon scrunched up a little bit. This is even easier to see if you have a black hole far away, it just gets smaller because it absorbed the negative mass. But this contradicts the area theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an argument for the positive mass theorem in a recent paper by Penrose which is similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Does this argument prove positive mass?&lt;/li&gt;&#xA;&lt;li&gt;Does this mean that the positive mass theorem holds assuming only the weak energy condition?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-24T14:00:24.377" Title="Positive Mass Theorem and Geodesic Deviation" Tags="&lt;general-relativity&gt;&lt;mathematical-physics&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13909" PostTypeId="1" AcceptedAnswerId="13918" CreationDate="2011-08-24T09:54:48.353" Score="1" ViewCount="54" Body="&lt;p&gt;The edge of a fractional quantum Hall state is an example of a chiral Luttinger liquid. Take, for the sake of simplicity, the edge of the Laughlin state. The Hamiltonian is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H = \frac{2\pi}{\nu}\frac{v_c}{2} \int_{\textrm{edge}} dx \rho(x)^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $\nu$ is the filling fraction, which is constant, $v_c$ is the velocity of the edge mode and $\rho$ is the charge density operator. You can think of this Hamiltonian as a delta-function interaction $V(x,x&amp;#39;) = \delta(x-x&amp;#39;)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Together with this Hamiltonian there is also the commutation relation of the field $\rho$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ [\rho(x),\rho(x&amp;#39;)] = i\frac{\nu}{2\pi} \partial_x \delta(x-x&amp;#39;)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't gone through the exercise myself, but I presume this is derived by going to momentum space, obtaining the canonical momenta through Hamilton's equation of motion and performing a canonical quantization. These equal-time commutation relations together with the Heisenberg equations of motion lead to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\partial_t \rho(x,t) = i[H,\rho(x,t)] = v_c \partial_x\rho(x,t) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This demonstrates that the edge is chiral, since $(\partial_t - v_c\partial_x)\rho(x,t) = 0$ and therefore the correlator involving $\rho(x,t)$ (or any other correlator) is a function of $t+x/v_c$ alone (hence the name &quot;chiral&quot; and &quot;left-moving&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also particle excitations (e.g. the electron) which are generated through vertex operator $\Psi(x,t)$ (these can be motivated through bosonization and/or conformal field theory, but I won't go into that). In any case these field operators have the following equal-time commutation relations with the current operator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\rho(x),\Psi(x&amp;#39;)] = Q \Psi(x&amp;#39;)\delta(x-x&amp;#39;)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $Q$ is the charge of the operator $\Psi$ with respect to the charge density operator $\rho$. This charge is then of course the electric charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is now: how do you generalize the commutation relations to non-equal time? What is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\rho(x,t),\rho(x&amp;#39;,t&amp;#39;)] = \ldots $$&#xA;$$[\rho(x,t),\Psi(x&amp;#39;,t&amp;#39;)] = \ldots $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1287" LastActivityDate="2011-08-24T14:27:38.447" Title="How do you obtain the commutation relations at non-equal times (for the edge of a fractional quantum Hall state)?" Tags="&lt;quantum-field-theory&gt;&lt;condensed-matter&gt;&lt;conformal-field-theory&gt;&lt;quantum-hall-effect&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="13910" PostTypeId="2" ParentId="13861" CreationDate="2011-08-24T10:00:21.407" Score="4" ViewCount="" Body="&lt;p&gt;The list of topics can be found &lt;a href=&quot;http://chair.itp.ac.ru/index.php?sub=minimum&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (in Russian, of course). Nowadays students are examined by collaborators of &lt;a href=&quot;http://itp.ac.ru/en/&quot; rel=&quot;nofollow&quot;&gt;Landau Institute for Theoretical Physics&lt;/a&gt;. Each exam, as it was before, consists of problems solving. For every exam there is one or several examiners with whom you are supposed to contact with to inform that you're willing to pass this particular exam (they will make an appointment). Everyone can pass any exam in any order. Today Landau's theoretical minimum (not all 11 exams, but at least 6 of them) is included in the program for students of Department of Problems of Theoretical Physics (&lt;a href=&quot;http://en.wikipedia.org/wiki/Moscow_Institute_of_Physics_and_Technology&quot; rel=&quot;nofollow&quot;&gt;Moscow Institute of Physics and Technology&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The program for each exam, as you can see from the link above, corresponds to the contents of volumes in the Course of Theoretical Physics by L&amp;amp;L (usually you have to master almost all paragraphs in the volume to pass the exam).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Mathematics I&lt;/strong&gt;. Integration, ordinary differential equations, vector algebra and tensor analysis. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Mechanics&lt;/strong&gt;. &lt;em&gt;Mechanics, Vol. 1&lt;/em&gt;, except §§ 27, 29, 30, 37, 51 (1988 russian edition) &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Field theory&lt;/strong&gt;  &lt;em&gt;The Classical Theory of Fields, Vol. 2&lt;/em&gt;, except §§ 50, 54-57, 59-61, 68, 70, 74, 77, 97, 98, 102, 106, 108, 109, 115-119 (1973 russian edition)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Mathematics II&lt;/strong&gt;. The theory of functions of a complex variable, residues, solving equations by means of contour integrals (Laplace's method), the computation of the asymptotics of integrals, special functions (Legendre, Bessel, elliptic, hypergeometric, gamma function) &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Quantum Mechanics&lt;/strong&gt;. &lt;em&gt;Quantum Mechanics: Non-Relativistic Theory, Vol. 3&lt;/em&gt;,  except §§ 29, 49, 51, 57, 77, 80, 84, 85, 87, 88, 90, 101, 104, 105, 106-110, 114, 138, 152 (1989 russian edition) &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Quantum electrodynamics&lt;/strong&gt;. &lt;em&gt;Relativistic Quantum Theory, Vol. 4&lt;/em&gt;, except §§ 9, 14-16, 31, 35, 38-41, 46-48, 51, 52, 55, 57, 66-70, 82, 84, 85, 87, 89 - 91, 95-97, 100, 101, 106-109, 112, 115-144 (1980 russian edition)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Statistical Physics I&lt;/strong&gt;. &lt;em&gt;Statistical Physics, Vol. 5&lt;/em&gt;, except §§ 22, 30, 50, 60, 68, 70, 72, 79, 80, 84, 95, 99, 100, 125-127, 134-141, 150-153 , 155-160 (1976 russian edition)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Mechanics of continua&lt;/strong&gt;. &lt;em&gt;Fluid Mechanics, Vol. 6&lt;/em&gt;, except §§ 11, 13, 14, 21, 23, 25-28, 30-32, 34-48, 53-59, 63, 67-78, 80, 83, 86-88, 90 , 91, 94-141 (1986 russian edition); &lt;em&gt;Theory of Elasticity, Vol. 7&lt;/em&gt;, except §§ 8, 9, 11-21, 25, 27-30, 32-47 (1987 russian edition)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Electrodynamics of Continuous Media&lt;/strong&gt;. &lt;em&gt;Electrodynamics of Continuous Media, Vol. 8&lt;/em&gt;, except §§ 1-5, 9, 15, 16, 18, ​​25, 28, 34, 35, 42-44, 56, 57, 61-64, 69, 74, 79-81 , 84, 91-112, 123, 126 (1982 russian edition)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Statistical Physics II&lt;/strong&gt;. &lt;em&gt;Statistical Physics, Part 2. Vol. 9&lt;/em&gt;, only §§ 1-5, 7-18, 22-27, 29, 36-40, 43-48, 50, 55-61, 63-65, 69 (1978 russian edition)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Physical Kinetics&lt;/strong&gt;. &lt;em&gt;Physical Kinetics. Vol. 10&lt;/em&gt;, only §§ 1-8, 11, 12, 14, 21, 22, 24, 27-30, 32-34, 41-44, 66-69, 75, 78-82, 86, 101.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some real problems&lt;/strong&gt; (Quantum Mechanics exam):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The electron enters a straight pipe of circular cross section (radius $r$). The tube is bent at a radius $R \gg r$ by the angle $\alpha$ and then is aligned back again. Find the probability that the electron will jump out.&lt;/li&gt;&#xA;&lt;li&gt;A hemisphere lies on an infinite two-dimensional plane. The electron falls on the hemisphere, determine the scattering cross section in the Born approximation.&lt;/li&gt;&#xA;&lt;li&gt;The electron &quot;sits&quot; in the ground state in the cone-shaped &quot;bag&quot; under the influence of gravity. The lower end of the plastic bag is cut with scissors. Find the time for the electron to fall out (in the semi-classical approximation).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4237" LastActivityDate="2011-08-24T10:00:21.407" />
  <row Id="13911" PostTypeId="1" CreationDate="2011-08-24T10:08:57.453" Score="7" ViewCount="292" Body="&lt;p&gt;I know that similar questions have been asked before, but none of them is exactly what I would like to know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to have a general idea of what string theory is, without having to read several hundred pages first. So, if you had to give a short formulation of string theory, what would it be? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are examples of what kind of formulation I am looking for. In quantum mechanics the states of a system are modeled by elements in a Hilbert space (or a projective one), the observables by certain operators acting on that space. The dynamics of a system is governed by this equation. The possible values an observable can have are the elements of the spectrum of the corresponding operator and so on. For general relativity one can say that the set of all events is described by a four-dimensional manifold with a certain metric satisfying certain equations. Free bodies move along geodesics and so on. With a bit more detail and precision of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the mathematical description of the basic objects and notions? What are the states of a system described by (mathematically)? Same question for the observables. What are the equations and so on? If these questions are not meaningful and string theory is radically different, then how is it in string theory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bump:&lt;/strong&gt; I don't think the question is so boring and uninteresting so that people don't want to spend the time to answer it. At least comparing to many other questions it seems reasonable. Also I don't think that there isn't anyone who can answer it. In fact I am convinced that there are may people that can write an answer, which would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1382" LastEditorUserId="1382" LastEditorDisplayName="" LastEditDate="2011-08-30T13:29:16.093" LastActivityDate="2011-08-30T13:29:16.093" Title="What is a brief formulation of string theory?" Tags="&lt;string-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" />
  <row Id="13912" PostTypeId="1" AcceptedAnswerId="13921" CreationDate="2011-08-24T13:14:38.103" Score="0" ViewCount="58" Body="&lt;p&gt;I am preparing for an exam, on one problem my answer differ from the solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current $i(t)=I_0e^{-\alpha t}$ runs in a long straight conductor along $\hat z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point A,B,C,D forms a triangle. &#xA;A and B is at $L \hat x$ with a very small separation in $\hat z$ with a resistance R between.&#xA;C and D is at $2L \hat x$ with a separation in $\hat z$ of $2L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: What is the induced current in the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I calculated the flux $\Omega$ correctly, but my answer differ on the direction of the current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took $V_{ind}=-\frac{d\Omega}{dt}$, since $i(t)$ is decreasing, the magnetic flux passing through ABCDA is decreasing, and so (to my understanding) it is reasonable that $i_{ABCD}=\frac{V_{ind}}{R}$ will run clockwise (using right-hand notation). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the solution to the problem they too use $V_{ind}=-\frac{d\Omega}{dt}$ but then &quot;magically&quot; on the next line remove the minus sign and says that according to Lenz law the current opposes the magnetic flux that caused the current and so will run counter-clockwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be thankful for an intuitive understanding for why it is opposing the magnetic flux that caused the current and not opposing the change in the magnetic flux?&#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5021" LastEditorUserId="5021" LastEditorDisplayName="" LastEditDate="2011-08-24T13:37:01.530" LastActivityDate="2011-08-24T15:19:21.877" Title="Lenz' law versus $-\frac{d\Omega}{dt}$" Tags="&lt;electromagnetism&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13913" PostTypeId="1" CreationDate="2011-08-24T13:33:00.303" Score="5" ViewCount="102" Body="&lt;p&gt;I am preparing for an exam, on this problem I had the opposite direction of the magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A conductor-cylinder with radius R has been cut in half ($\phi \in [0,\pi]$)&#xA;A DC current $I$ runs along $\hat Z$, and can be assumed to be evenly distributed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is to calculate the magnetic field at $\rho=0$ (i.e. the center of the cylinder and on the diameter of the half-cylinder).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/Ol8qe.png&quot; alt=&quot;Geometry of the problem&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took an infinitesimal contribution $dB=\frac{\mu_0 di}{2\pi \rho}$ with $di=\frac{2I}{\pi R^2}dr d\theta$, $dB$ is along $\hat \phi$ but the total $B$ at $\rho=0$ won't be. Integrating $dB$ first for $B_x$ gives me $-2\frac{\mu_0 I}{\pi^2R}$ and afterwards for $By$ gives me $B_y=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnitude is right, but the answer has instead $B_x$ in the positive direction. Intuitively it makes sense that it is positive (flows along the boundary). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I made the mistake here: $\hat \phi = -\hat x sin(\phi) + \hat y cos(\phi)$ gives: $dB_x = -sin(\phi) dB $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems (mathematically) correct to me, but the direction is wrong. Would be very thankful for any thoughts on where I am going wrong. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5021" LastEditorUserId="5021" LastEditorDisplayName="" LastEditDate="2011-08-28T10:04:13.300" LastActivityDate="2011-08-28T10:04:13.300" Title="Magnetic field from a half-cylinder" Tags="&lt;electromagnetism&gt;&lt;homework&gt;" CommentCount="8" FavoriteCount="1" />
  <row Id="13914" PostTypeId="2" ParentId="13911" CreationDate="2011-08-24T13:59:50.700" Score="0" ViewCount="" Body="&lt;p&gt;CERN has a useful series of academic training lectures, available as video and slides(pdf)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at &lt;a href=&quot;http://indico.cern.ch/conferenceDisplay.py?confId=52733&quot; rel=&quot;nofollow&quot;&gt;http://indico.cern.ch/conferenceDisplay.py?confId=52733&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prof. Constantin Bachas, &quot;String Theory for Pedestrians (1/3)&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;from february 2009, the most recent one. There are a large number of similar lectures earlier in time. CERN believes in training its academics :) .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My even more pedestrian summary is that: &lt;strong&gt;string theory is an attempt to describe and derive the microcosm of elementary particles as measured experimentally up to now, within the mathematical framework of relativistic strings ( and more). A bonus is the objective of unifying all four forces acting on these particles in one mathematical formulation.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-08-24T13:59:50.700" CommentCount="4" />
  <row Id="13915" PostTypeId="2" ParentId="13907" CreationDate="2011-08-24T14:00:24.377" Score="2" ViewCount="" Body="&lt;p&gt;Given that your argument requires moving in an accelerating frame and considering its Rindler horizon, I wonder if what you stated is more similar to statements about asymptotically hyperbolic hypersurfaces in a space-time. In which case, that the analogue of the positive mass theorem can be derived using only the weak energy condition has been shown before (for example, see &lt;a href=&quot;http://www.math.miami.edu/~galloway/papers/galloway.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper of Anderson, Cai, and Galloway&lt;/a&gt;). &lt;/p&gt;&#xA;" OwnerUserId="829" LastActivityDate="2011-08-24T14:00:24.377" CommentCount="1" />
  <row Id="13916" PostTypeId="2" ParentId="6234" CreationDate="2011-08-24T14:01:32.650" Score="-1" ViewCount="" Body="&lt;p&gt;A photon does not interfere with itself, or any other photons. The apparent interference pattern that results from photons, or even a single photon, passing though a double slit apparatus, is due to the quantum mechanical wave guide structure that determines the probabilities for photons to go from the emitter to the detection wall via various paths.&#xA;See &lt;a href=&quot;http://ps.missouri.edu/feynman&quot; rel=&quot;nofollow&quot;&gt;http://ps.missouri.edu/feynman&lt;/a&gt; for some animations of probability amplitudes.&lt;/p&gt;&#xA;" OwnerUserId="5023" LastActivityDate="2011-08-24T14:01:32.650" />
  <row Id="13917" PostTypeId="1" AcceptedAnswerId="13919" CreationDate="2011-08-24T14:07:27.697" Score="2" ViewCount="180" Body="&lt;p&gt;I have looked online to no avail.  There is two competing answers and I am curious to know which one is right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone asked me this question.  If you are traveling at the speed of light can you see your reflection in a mirror in front of you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer to the question is no, I would figure that in order for that to happen the light reflecting off you that would appear in the mirror must travel faster than the speed of light to actually reach the mirror (which we all know is impossible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He says the answer is yes, that it is all relative to the current frame of reference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone validate the correct answer with possible references?&lt;/p&gt;&#xA;" OwnerUserId="5024" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-08-24T17:53:10.633" LastActivityDate="2011-08-24T18:23:38.627" Title="Reflection At Speed of Light" Tags="&lt;special-relativity&gt;&lt;speed-of-light&gt;&lt;reflection&gt;" AnswerCount="3" FavoriteCount="1" />
  <row Id="13918" PostTypeId="2" ParentId="13909" CreationDate="2011-08-24T14:27:38.447" Score="2" ViewCount="" Body="&lt;p&gt;You may simply replace the arguments $ x \to x+vt $ in all of those commutators.&#xA;The way to prove that is to go to momentum space, where $\rho (k, t) = \rho (k,0) e^{i k v t} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proving that about the commutator $[\rho (x,t),\psi (x&amp;#39;,t&amp;#39;)]$ is a little trickier, because we don't know the explicit form of $\psi (t)$, but the way I did it was to use $$ [\rho (x,t),\psi (x&amp;#39;,t&amp;#39;)] = U^\dagger (t&amp;#39;) [\rho (x,t-t&amp;#39;),\psi (x&amp;#39;,0)] U (t&amp;#39;) $$&#xA;Where U is the time evolution operator. This way only the time evolution of the density is needed.&lt;/p&gt;&#xA;" OwnerUserId="4696" LastActivityDate="2011-08-24T14:27:38.447" CommentCount="1" />
  <row Id="13919" PostTypeId="2" ParentId="13917" CreationDate="2011-08-24T14:47:14.373" Score="7" ViewCount="" Body="&lt;p&gt;This question cannot really be answered because you cannot travel at the speed of light. See &lt;a href=&quot;http://physics.stackexchange.com/questions/1557/accelerating-particles-to-the-speed-of-light&quot;&gt;Accelerating particles to the speed of light&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were massless, you would always travel at the speed of light.  However, in that case you would not perceive the passing of time.  In relativity, the time that passes for an observer depends on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Proper_time&quot; rel=&quot;nofollow&quot;&gt;proper time&lt;/a&gt;.  The proper time for a light-like trajectory is always zero, so photons themselves do not experience the passage of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you travel very near to the speed of light - perhaps 99.9% light speed relative to Earth, you would still be able to view yourself normally in a mirror you carried with you.  That is ensured by the principle of relativity, which states that all physical processes work the same way at any constant speed.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-08-24T14:47:14.373" CommentCount="4" />
  <row Id="13920" PostTypeId="2" ParentId="13917" CreationDate="2011-08-24T14:49:09.567" Score="2" ViewCount="" Body="&lt;p&gt;You can not travel the speed of light, you can only get close to it and the closer you get the more relativistic effects you experience, but we can continue talking about the problem you posed in the context of &lt;em&gt;approaching the speed of light&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;From your perspective&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you approach the speed of light, the mirror actually gets closer from your perspective and the speed of light both approaching you from the mirror and leaving you moving toward the mirror is $c$.  However, you still perceive the mirror moving toward you.  So if the mirror is moving toward you at $0.99 c$ and the light from it is coming at you at $c$, then the light from the surface of the mirror only departs at a relative speed of $0.01 c$, which for the context of this discussion we'll say is not very fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;From the mirror's perspective&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mirror is a part of an inertial reference frame that &quot;observes&quot; (which is really just a formalism for relativity, different than seeing with photons obviously) you moving toward it at $0.99 c$ and let's also note that you are length-contracted like a pancake and experience time more slowly according to the mirror.  Light comes from the spaceship (I presume) at $c$ and moves toward the spaceship at $c$.  The light that the spaceship emits only moves $0.01 c$ faster than the speed of the spaceship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases it is agreed that the light from the spaceship emitted at a time $t$ before collision with the mirror only hits the mirror a small amount of time before the spaceship itself slams into the mirror (specifically $0.01 c t$, and yes I know this isn't an objective time measure as I have used it here).  In the limit of going exactly the speed of light, then of course, no light from the spaceship is able to reach the mirror before the collision.  A reflection of the spaceship in-travel is impossible in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So what do you see?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see reflections in the mirror from before you started your speed-of-light trip.  If we talk about going close to the speed of light, like $0.99 c$ then you see a highly blue-shifted version of yourself for a short period of time (relative to the duration of the trip) right before you crash into the mirror.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-08-24T14:49:09.567" CommentCount="3" />
  <row Id="13921" PostTypeId="2" ParentId="13912" CreationDate="2011-08-24T15:19:21.877" Score="1" ViewCount="" Body="&lt;p&gt;Actually, it does oppose the change in the magnetic flux causing it. If the magnetic field from the source is increasing in some direction, the magnetic field from the induced current decreases in the same direction to oppose the increase, for example. The induced current opposing the change causing it ensures that the cause must supply energy to the system to increase the induced current, so guaranteeing conservation of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnetic field from the wire points towards $\hat y$ and is decreasing, which means the magnetic field has to increase in the same direction from the induced current. Therefore the current runs clockwise around the loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The negative sign in Lenz's law is needed when using the right hand rule to find the direction of the induced emf: The thumb points in the direction of the applied magnetic field, and the curled fingers point around the direction of the induced emf. In your example, the thumb points towards $\hat y$ and the fingers curl in a clockwise sense. The magnetic field decreasing cancles the minus sign so the direction is the same as the fingers - clockwise. &lt;/p&gt;&#xA;" OwnerUserId="1345" LastActivityDate="2011-08-24T15:19:21.877" CommentCount="1" />
  <row Id="13922" PostTypeId="1" AcceptedAnswerId="13924" CreationDate="2011-08-24T15:55:09.637" Score="7" ViewCount="71" Body="&lt;p&gt;I saw on Discovery channel that the source of Earth's magnetic field is the molten iron and metals in the earth's core. The spinning of these liquid metals produces the same effect as electric current in a coil which produces a magnetic field. The scientist in the program proved the concept by conducting an experiment where a big metal sphere where winded by a metallic tube in which molten Sodium was pumped so it circulate around the sphere in a similar way that happens inside earth. The experiment showed that a magnetic field was successfully produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could a circulating neutrally charged particles produce a magnetic field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is like having two electric currents one due to the negative charges (electrons) and the other is due to positive charges (nucleus) with same value and in opposite direction so there should not be a magnetic field. so how does this happen?&lt;/p&gt;&#xA;" OwnerUserId="3241" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-08-24T16:30:52.037" LastActivityDate="2011-08-24T18:16:12.487" Title="What is the source of Earth's magnetic field?" Tags="&lt;electromagnetism&gt;&lt;geophysics&gt;&lt;geomagnetism&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="13923" PostTypeId="1" CreationDate="2011-08-24T15:36:00.070" Score="7" ViewCount="176" Body="&lt;p&gt;In the video game Portal, there are often puzzles which must be solved by gaining a large amount of momentum. Typically, this is accomplished by putting one portal on the ground and another directly above it on the ceiling, then jump into the portal in the ground, and the person would appear at the ceiling and started to fall off. When the person fall off, he/she gain kinetic energy and reach a certain speed and jump into the portal in the ground again. After that, he/she appear at the ceiling again. Thus you create an infinite loop of falling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Although the idea of a portal is absurd to common sense, let's discuss its possibility to existence.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The portal would seem to be creating potential energy. Typically when an item falls, its potential energy is converted to kinetic energy by virtue of it losing height, and additional energy would need to be expended to raise it up again before it could fall again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the possible source of the potential energy, if portal would exists? Does a portal provide potential energy in order to push people to higher position?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this has lead me to wonder another point, though, whether the portals should also transmit gravity. This might remove the whole problem of conservation of energy; if the gravity is transmitted through the portal, then the object would be pulled back up with equal force, and would actually reach an equilibrium floating within the portal, rather than falling continuously.&lt;/p&gt;&#xA;" OwnerUserId="5025" LastEditorUserId="4556" LastEditorDisplayName="" LastEditDate="2011-08-25T07:56:35.180" LastActivityDate="2011-08-25T07:56:35.180" ClosedDate="2011-08-24T17:32:11.667" Title="Would Portal-style portals transmit gravity?" Tags="&lt;gravity&gt;&lt;energy-conservation&gt;" CommentCount="8" FavoriteCount="1" />
  <row Id="13924" PostTypeId="2" ParentId="13922" CreationDate="2011-08-24T18:16:12.487" Score="3" ViewCount="" Body="&lt;p&gt;Circulating neutral particles will not by themselves create a magnetic field.  However, if the neutral particles are moving through an existing magnetic field, and the neutral medium is conducting, then the magnetic field will induce a current via the Lorentz force.  That induced current will in turn create it's own magnetic field, which may enhance the existing magnetic field.  If things work out right you have a self-reinforcing dynamo where motion thru the magnetic fields drives currents and those currents in turn support the magnetic field.  However, there had to be some sort of &quot;seed&quot; field to get the thing started in the beginning.  &lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-08-24T18:16:12.487" CommentCount="6" />
  <row Id="13925" PostTypeId="2" ParentId="13917" CreationDate="2011-08-24T18:23:38.627" Score="4" ViewCount="" Body="&lt;p&gt;Your friend is correct that it depends on the reference frame (see previous answers).  Both of these responses are correct, the only difference being that Mark assumes you and the mirror share a reference frame and are measuring your speed relative to another frame whereas Zassou assumes that you are measuring your speed relative to the mirror (and traveling toward it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regard to Zassou's answer, it is true that Lorentz contraction occurs, but, since &lt;em&gt;you&lt;/em&gt; are the observer, it would appear as though mirror were &quot;contracted like a pancake&quot; and that &lt;em&gt;it&lt;/em&gt; appears to experience time more slowly relative to you.  From your point of view, everything would appear normal except for the absurdly fast approaching mirror.  It is also worth noting that if you assume the opposite direction of travel your image would be red-shifted.&lt;/p&gt;&#xA;" OwnerUserId="5020" LastActivityDate="2011-08-24T18:23:38.627" />
  <row Id="13926" PostTypeId="1" CreationDate="2011-08-24T19:38:58.280" Score="0" ViewCount="11" Body="&lt;p&gt;Prop A. I wrote a zipcode server that gives me 32,000 zip codes of USA.&#xA;Each zipcode has an associated lat-long.&#xA;Given 2 zipcodes, I can find the distance between them using their lat-longs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prop B. I also wrote a weather server where you can input atmost 200 zipcodes and it spits out the temperature at each of those zipcodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Person tells me his zipcode is Z, temperature is T.&#xA;He asks me, what's the nearest place from Z where its atleast 10 degrees cooler ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I get a list of 200 zipcodes from Z sorted by distance ( using Prop A).&#xA;I feed that to B and get 200 temperatures.&#xA;If none are 10 degrees cooler, I get the next 200 zipcodes and repeat until done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem: This seems quite inefficient and brute-force. I feel there's some Physics insight I'm missing. Its not always true that if you go north the temperatures cool down &amp;amp; going south they heat up. So direction doesn't help. Altitude probably does ( mountains cooler than valleys ) but zipcodes data keyed to altitude is hard to find.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you guys think of some smarter way to go about this ? Any suggestions appreciated ( whether physics, or tweaking the search algorithm or organizing the data differently or whatever else )&lt;/p&gt;&#xA;" OwnerUserId="5026" LastActivityDate="2011-08-24T19:38:58.280" ClosedDate="2011-08-24T20:04:47.227" Title="On calculating temperature differences between zipcodes" Tags="&lt;temperature&gt;&lt;geophysics&gt;" CommentCount="1" />
  <row Id="13927" PostTypeId="2" ParentId="13864" CreationDate="2011-08-24T22:08:27.960" Score="1" ViewCount="" Body="&lt;p&gt;Hmmm... this is a confusing question. As far as I can tell, it's not clear exactly what they mean by &quot;magnetic field that randomizes the direction of the particle.&quot; And unless &quot;the magnetic field is at rest&quot; is supposed to mean that it's static, I'm not sure what they're getting at with that either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the physical phenomenon in question is &lt;a href=&quot;http://en.wikipedia.org/wiki/Synchrotron_radiation&quot; rel=&quot;nofollow&quot;&gt;synchrotron radiation&lt;/a&gt;. This is described by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Larmor_formula#Relativistic_Generalisation&quot; rel=&quot;nofollow&quot;&gt;relativistic Larmor formula&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = \frac{q^2}{6\pi\epsilon_0 c}\gamma^6\Bigl[\dot\beta^2 - \Bigl(\vec{\beta}\times\dot{\vec{\beta}}\Bigr)^2\Bigr]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this case, since we have only static magnetic fields and the particle is highly relativistic, it's approximately true that $\vec\beta\perp\dot{\vec\beta}$ (in other words, the component of acceleration parallel to the velocity is relatively negligible, c.f. Jackson p.667). Using this and the formula for the cyclotron radius $R = \frac{\gamma mc\beta}{qB}$, I get it down to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = \frac{q^2}{6\pi\epsilon_0 c^3}\frac{\gamma^2\beta^2B^2}{m^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here, I guess you would be expected to figure out the total energy loss while the particle is in the magnetic field. However, I don't see how to do that, for a couple of reasons: first of all, although you know the mean value of the magnetic field $\langle B\rangle$, you don't know enough to figure out $\langle B^2\rangle$, which is what you need to figure out the mean radiative power loss. And even if you did have that, you don't know how much time the particle spends in the magnetic field, which you would need to go from power to energy.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-08-24T22:08:27.960" />
  <row Id="13928" PostTypeId="2" ParentId="11768" CreationDate="2011-08-24T22:16:11.317" Score="0" ViewCount="" Body="&lt;p&gt;for this particular case, there doesn't appear to be an obvious expression that comes from the conservation of momentum, analagous to that for the conservation of energy. However, Noether's theorem offers a route in determing one of the conserved quantities as being the &lt;a href=&quot;http://physics.stackexchange.com/questions/11887/what-is-the-conserved-canonical-momentum-for-a-relativistically-moving-charge-in&quot;&gt;canonical momentum&lt;/a&gt;, which on rearranging gives the Lorentz force law for a static electric field: $$\frac d {dt} \gamma mv = qE$$&lt;/p&gt;&#xA;" OwnerUserId="1345" LastActivityDate="2011-08-24T22:16:11.317" CommentCount="1" />
  <row Id="13929" PostTypeId="1" AcceptedAnswerId="13934" CreationDate="2011-08-24T22:25:20.040" Score="0" ViewCount="50" Body="&lt;p&gt;We say an object's color is blue if the object is opaque, reflects blue color and absorbs waves of other color. What color a transparent object actually reflects?&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-08-24T23:58:57.493" Title="What color a transparent object reflect?" Tags="&lt;light&gt;" AnswerCount="1" />
  <row Id="13930" PostTypeId="1" CreationDate="2011-08-24T22:42:02.757" Score="0" ViewCount="19" Body="&lt;p&gt;Determine the force-time function within the compressor’s cylinder and the torque-time function acting on the input shaft on the compressor during any cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given:&#xA;Compressor speed is 1500rpm. The compressor has a bore of 3.125 in, a stroke of 3.26in and a connecting rod crank ratio to 3.5. Inlet pressure is atmospheric 14.7 psia, peak cylinder pressure is 132psia, and mean active pressure mep is 26psia. Flow is 8.9 cfm at mep, given 1.6 hp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumption:&#xA;The piston weight is 1 lb; the connecting rod weights 2 lb with its center of mass at 1/3 point from the big end. The crankshaft weights 5.4 lb including a counterbalance that optimally overbalances it to minimize the shaking force. The exponent for the gas law equation is k = 1.13&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Related to mechanical engineering design&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="5027" LastActivityDate="2011-08-24T22:42:02.757" Title="Preliminary design of a compressor drive train" Tags="&lt;engineering&gt;&lt;design&gt;" />
  <row Id="13931" PostTypeId="1" CreationDate="2011-08-24T22:42:30.773" Score="3" ViewCount="35" Body="&lt;p&gt;I am working on some radiative transfer equations, and struggling as I'm fairly new to this field. I have read about the Rayleigh Phase Function which is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(\theta) = \frac{3}{4}(1 + cos^2 \theta)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can plot this function, and generate values from this function in my computer program (which is what I need to do) - but what is it actually telling me? If I know I have a ray of incident light at a certain angle, will this function tell me what angle it will scatter at? Or is it more of a statistical generalisation where the areas where the function is higher are the angles at which the scattering is more likely to happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, when I see the function plotted it is always from 0-180, rather than 0-360 as I would expect for the scattering. This makes me think that the function may be giving the difference between the incoming angle and the outgoing angle rather than the absolute angle it is scattered at. Is that correct?&lt;/p&gt;&#xA;" OwnerUserId="5028" LastActivityDate="2011-08-24T22:42:30.773" Title="What does the Rayleigh Phase Function tell us" Tags="&lt;optics&gt;&lt;atmosphere&gt;&lt;scattering&gt;" CommentCount="2" />
  <row Id="13932" PostTypeId="1" AcceptedAnswerId="13935" CreationDate="2011-08-24T23:34:49.833" Score="2" ViewCount="53" Body="&lt;p&gt;So as part of a video I'm working on, I'm trying to launch a 10kg table about 8m into the air.  A machinist friend of mine has provided a solution that involves pneumatic cylinders and a pump to generate the force needed to propel the table.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we're trying to calculate is the amount of PSI needed to lift the table. All these calculations don't need to be exact (if it only goes 16m up, then there's no issue)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know where to even start to figure it out. Once gravity gets involved, it throws me off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any real help would be appreciated, thanks!&lt;/p&gt;&#xA;" OwnerUserId="5029" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-25T00:03:30.753" LastActivityDate="2011-08-25T01:21:06.440" Title="How do I determine the amount of lift needed to launch an object?" Tags="&lt;forces&gt;&lt;newtonian-mechanics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13933" PostTypeId="1" AcceptedAnswerId="14319" CreationDate="2011-08-24T23:45:43.607" Score="1" ViewCount="73" Body="&lt;p&gt;This remark from Lubos puzzles me&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ramond's string - and Neveu-Schwarz string - wasn't really an &quot;origin of string theory&quot;. String theory had &quot;origin&quot; as bosonic string theory which has no fermions. All SUSY/fermioncs strings are &quot;new&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How to fit this view, which seems to be the common one, with the fact that the R and NS papers date from 1971, before the concept of supersymmetry, long before superstrings themselves, and only three years later than the Veneziano model, and almost simultaneous (less than two years) with the idea of a String interpretation of the model? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Were the R-NS models just a hep-th proposal, without any link to, nor motivation from, phenomenology? It could be so. I can not find any clear reference to baryonic regge trajectories in the early literature, and I can not find any duality diagram involving fermions, before 1971. But on other hand, Susskind 1970 abstracts keep speaking of &quot;model of hadrons&quot;, not just &quot;mesons&quot;, when exposing the idea of strings. So perhaps I am just being sloppy in my spires searches. &lt;/p&gt;&#xA;" OwnerUserId="1335" LastActivityDate="2011-09-04T02:49:19.267" Title="Was there some phenomenological motivation for Ramond and Neveu-Swarz models?" Tags="&lt;string-theory&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="13934" PostTypeId="2" ParentId="13929" CreationDate="2011-08-24T23:58:57.493" Score="2" ViewCount="" Body="&lt;p&gt;A truly transparent object doesn't reflect any color in the sense that you seem to mean.  How we normally &quot;see&quot; transparent objects is by the refraction of light that they cause and the effect on the image of what lies &quot;behind&quot; them.  There is, however, a phenomenon called &quot;total internal reflection&quot; where a transparent object can reflect ALL light that approaches its boundary within a certain range of angles.  This is what causes the strange reflection you see when looking up from under water near the surface and how fiber optic systems work.&lt;/p&gt;&#xA;" OwnerUserId="5020" LastActivityDate="2011-08-24T23:58:57.493" CommentCount="1" />
  <row Id="13935" PostTypeId="2" ParentId="13932" CreationDate="2011-08-25T01:14:22.220" Score="3" ViewCount="" Body="&lt;p&gt;I'm going to assume that you have a pneumatic ram of some kind, and that it features an area $A$ and a stroke length of $l$ which is very short compared to the other distances in this problem. I further assume that you can connect this device to a high pressure reservoir with a volume much larger than $Al$ (making the force on the ram effectively constant over the stroke) at pressure $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The object to be lifted has mass $m$&lt;/li&gt;&#xA;&lt;li&gt;The target height is $h$&lt;/li&gt;&#xA;&lt;li&gt;The ram has very little friction.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Measuring from the starting height our projectile will have potential energy $E_p = mgh$ when it reaches the desired elevation, and if it is at the top of its arc it will have no kinetic energy there. (Here $g$ is the acceleration of gravity $g \approx 9.8 \text{ m/s}^2 \approx 32.2 \text{ ft/s}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By conservation of energy that means it should have kinetic energy of $mgh$ as it leaves the ram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ram provides a constant force $PA$ over a length $l$; by the work energy theorem the kinetic energy of the projectile at launch is $ E_k = PAl = mgh $. With $A$ and $l$ fixed by the geometry of the ram, that leaves us with&#xA;$$ P = \frac{mgh}{Al} .$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you are working with an experienced machinist I probably don't have to tell you how potentially dangerous this is, but I'm going to anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any reasonable mass, you are talking about &lt;strong&gt;&lt;em&gt;energies that are more than sufficient to main or kill you&lt;/em&gt;&lt;/strong&gt;. Take care, wear your safety goggle, and get behind appropriate cover before running this thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, &lt;em&gt;any&lt;/em&gt; high pressure system poses a potential &lt;strong&gt;&lt;em&gt;risk to life and limb&lt;/em&gt;&lt;/strong&gt;. You should learn to handle these systems from an expert before cobbling something together on your own.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-25T01:21:06.440" LastActivityDate="2011-08-25T01:21:06.440" CommentCount="2" />
  <row Id="13936" PostTypeId="2" ParentId="13757" CreationDate="2011-08-25T04:47:46.343" Score="1" ViewCount="" Body="&lt;p&gt;The first undeniably reliable measurements of Avogadro's number came right at the turn of the twentieth century, with Millikan's measurement of the charge of the electron, Planck's blackbody radiation law, and Einstein's theory of Brownian motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earlier measurements of Avogadro's number were really only estimates, they depended on the detailed model for atomic forces, and this was unknown. These three methods were the first model-independent ones, in that the answer they got was limited only by the experimental error, not by theoretical errors in the model. When it was observed that these methods gave the same answer three times, the existence of atoms became an established experimental fact.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Millikan&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Faraday discovered the law of electrodeposition. When you run a current through a wire suspended in an ionic solution, as the current flows, material will deposit on the cathode and on the anode. what Faraday discovered is that the number of moles of the material is strictly proportional to the total charge that passes from one end to the other. Faraday's constant is the number of moles deposited per unit of charge. This law is not always right, sometimes you get half of the expected number of moles of material deposited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the electron was discovered in 1899, the explanation of Faraday's effect was obvious--- the ions in solution were missing electrons, and the current flowed from the negative cathode by depositing electrons on the ions in solution, thereby removing them from the solution and depositing them on the electrode. Then Faraday's constant is the charge on the electron times Avogadro's number. The reason that you sometimes get half the expected number of moles is that sometimes the ions are doubly-ionized, they need two electrons to become uncharged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Millikan's experiment found the charge on the electron directly, by measuring the discreteness of the force on a droplet suspended in an electric field. This determined Avogadro's number.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Planck's blackbody law&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Following Boltzmann, Planck found the statistical distribution of electromagnetic energy in a cavity using Boltzmann's distribution law: the probability of having energy E was $\exp(-E/kT)$. Planck also introduced Planck's constant to describe the discreteness of the energy of the electromagnetic oscillators. Both constants, k and h, could be extracted by fitting the known blackbody curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Boltzmann's constant times Avogadro's number has a statistical interpretation, it is the &quot;Gas constant&quot; R you learn about in high school. So measuring Boltzmann's constant produces a theoretical value for Avogadro's number with no adjustible model parameters.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Einstein's diffusion law&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;A macroscopic particle in a solution obeys a statistical law--- it diffuses in space so that its average square distance from the starting point grows linearly with time. The coefficient of this linear growth is called the diffusion constant, and it seems hopeless to determine this constant theoretically, because it is determined by innumerable atomic collisions in the liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Einstein in 1905 discovered a fantastic law: that the diffusion constant can be understood immediately from the amount of friction force per unit velocity. The equation of motion for the Brownian particle is:&#xA;$ m{d^2x\over dt^2} + \gamma {dx\over dt} + C\eta(t) $ = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where m is the mass, $\gamma$ is the friction force per unit velocity, and $C\eta$ is a random noise that describes the molecular collisions. The random molecular collisions at macroscopic time scales must obey the law that they are independent Gaussian random variables at each time, because they are really the sum of many independent collisions which have a central limit theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Einstein knew that the probability distribution of the velocity of the particle must be the Maxwell-Boltzmann distribution, by general laws of statistical thermodynamics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(v) \propto e({-v^2\over 2mkT})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ensuring that this is unchanged by the molecular noise force determines C in terms of m and kT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Einstein noticed that the $d^2x\over dt^2$ term is irrelevant at long times. Ignoring the higher derivative term is called the &quot;Smoluchowski approximation&quot;, although it is not really an approximation by a long-time exact description. It is explained here: &lt;a href=&quot;http://physics.stackexchange.com/questions/13804/cross-field-diffusion-from-smoluchowski-approximation/13868#13868&quot;&gt;Cross-field diffusion from Smoluchowski approximation&lt;/a&gt;, so the equation of motion for x is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\gamma {dx\over dt} + C\eta = 0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this gives the diffusion constant for x. The result is that if you know the macroscopic quantities $m,\gamma,T$, and you measure the diffusion constant to determine C, you find Boltzmann's constant k, and therefore Avogadro's number. This method required no photon assumption and no electron theory, it was based only on mechanics. The measurements on Brownian motion were carried out by Perrin a few years later, and earned Perrin the Nobel prize.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-25T04:53:43.993" LastActivityDate="2011-08-25T04:53:43.993" />
  <row Id="13937" PostTypeId="1" AcceptedAnswerId="13949" CreationDate="2011-08-25T05:00:30.547" Score="3" ViewCount="50" Body="&lt;p&gt;To quote from Wikipedia on elastic scattering, &quot;In this scattering process, the energy (and therefore the wavelength) of the incident photon is conserved and only its direction is changed.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this work? The impacted particle will also start moving to conserve momentum. So it'll also carry some energy. How can the photon not lose any energy during the collision?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibly unrelated question: how exactly can the photon and the particle interact? The photon essentially carries information about a changing electromagnetic field. How can it influence a particle that carries neither charge nor magnetism?&lt;/p&gt;&#xA;" OwnerUserId="5030" LastActivityDate="2011-08-25T13:43:18.847" Title="Question about Rayleigh scattering" Tags="&lt;electromagnetism&gt;&lt;optics&gt;&lt;photons&gt;" AnswerCount="1" />
  <row Id="13938" PostTypeId="2" ParentId="2644" CreationDate="2011-08-25T05:09:01.567" Score="2" ViewCount="" Body="&lt;p&gt;Newton's 1st and 2nd laws weren't particularly revolutionary or surprising to anyone in the know back then. Hooke had already deduced inverse-square gravitation from Kepler's third law, so he understood the second law. He just could not prove that the bound motion in response to an inverse square attraction is an ellipse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The source of Newton's second law was Galileo's experiments and thought experiments, especially the principle of Galilean relativity. If you believe that the world is invariant under uniform motion, as Galileo states clearly, then the velocity cannot be a physical response because it isn't invariant, only the acceleration is. Galileo established that gravity produces acceleration, and its no leap from that to the second law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton's third law on the other hand &lt;em&gt;was&lt;/em&gt; revolutionary, because it implied conservation of momentum and conservation of angular momentum, and these general principles allow Newton to solve problems. The real juicy parts of the Principia are the specific problems he solves, including the bulge of the Earth due to its rotation, which takes some thinking even now, three centuries later.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-25T06:05:33.443" LastActivityDate="2011-08-25T06:05:33.443" />
  <row Id="13939" PostTypeId="1" CreationDate="2011-08-25T06:54:39.317" Score="0" ViewCount="14" Body="&lt;p&gt;I am a software engineer and programmer got bored with the software job. And i wanted to learn a new degree which is entirely different from computer science. I wanted to go for college again. I am thinking to learn satellite communications. But i don't know how to start and i am not even what are the things to learn for satellite communications as i am from computer science background. I am from india please suggest me any universities in USA which will provide this course from basics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and Regards,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sunny.&lt;/p&gt;&#xA;" OwnerUserId="5031" LastActivityDate="2011-08-25T06:54:39.317" ClosedDate="2011-08-25T11:13:42.423" Title="Degree for Satellite Communications" Tags="&lt;electromagnetism&gt;&lt;education&gt;" />
  <row Id="13941" PostTypeId="2" ParentId="10462" CreationDate="2011-08-25T10:06:08.160" Score="2" ViewCount="" Body="&lt;p&gt;Please allow me to quote from one of my old text books - Transmission Principles for Technicians, D C Green:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Whenever a current flows in a conductor, the conductor is surrounded by a magnetic field, the direction of which is determined by the direction of current flow.  If the current changes, the magnetic field will change also.  Now, a varying magnetic field &lt;em&gt;always&lt;/em&gt; produces an electric field that exists &lt;em&gt;only&lt;/em&gt; while the magnetic field continues to change.  When the magnetic field is constant the electric field disappears.  The direction of the electric field depends on whether the magnetic field is growing or collapsing and can be determined by the application of Lenz's law.  Similarly, a changing electric field &lt;em&gt;always&lt;/em&gt; produces a magnetic field; this means that a conductor carrying an alternating current is surrounded by continually changing magnetic and electric fields that are completely dependent on one another.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If a sinusoidal current is flowing in a conductor the electric and magnetic fields around the conductor will also attempt to vary sinusoidally.  When the current reverses direction the magnetic fields must first collapse into the conductor and &lt;em&gt;then&lt;/em&gt; build up in the opposite direction.  A finite time is required for a magnetic field and its associated electric field to collapse, however, and at frequencies above about 15KHz not all the energy contained in the field has returned to the conductor before the current has started to increase in the opposite direction and create new electric and magnetic fields.  The energy left outside the conductor cannot then return to it and instead is propagated away from the conductor at the speed of light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So that deals with how radio waves are produced.  Now, what makes an aerial special?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, that is all to do with the wavelength ($\lambda$) of the signal.  When an alternating current reaches the end of a transmission line it &quot;bounces&quot; back again (echoes) back down the line.  This is a big problem when it's not wanted, as the reflected signal can interfere with the rest of the signal causing extra noise and attenuation of the signal.  This is why proper termination of transmission lines is essential, to absorb these reflections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the same phenomenon can be made use of to boost the signal level.  If you measure the transmission line (in this case the aerial) to be precisely a hole division fraction of the wavelength of the signal (most commonly $\frac{1}{4}$ but $\frac{1}{2}$ or $\frac{1}{8}$ are used) it is possible to get the reflections to add to the signal, thus boosting it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same goes for the received signal in a reception aerial.  As the incoming signal inducts a voltage in the aerial reflections will be produced from the open end of the aerial.  These reflections are added to the signal to boost the signal level and quality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the aerial, due to its precisely measured length, can be &quot;tuned&quot; to a specific frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there are more complex designs of aerials than just a length of wire.  Such designs as the &lt;em&gt;Yagi Array&lt;/em&gt; which adds a directional component to the radiation of the signal.  The aerial includes a &quot;reflector&quot; which bounces the signal that goes behind the aerial back in a forward direction.  By careful position and measuring of the reflector it is possible again to get the waves of the signal to line up and thus increase in power, just like the reflections in the single-wire aerial.&lt;/p&gt;&#xA;" OwnerUserId="4123" LastActivityDate="2011-08-25T10:06:08.160" />
  <row Id="13942" PostTypeId="1" CreationDate="2011-08-25T10:22:20.440" Score="5" ViewCount="85" Body="&lt;p&gt;While most of what I've read about fractals has been dubious in nature, over the years, I keep hearing that these sorts of self-similar (or approximately self-similar) geometries are useful in the manufacture of high-performance antennas at small scales, perhaps for cell phones or distributed sensing applications.  Benoit Mandlebrot himself cited this as one application of his work.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone provide me an intuitive reason why a fractal geometry, as opposed to some other symmetric structure, would be optimal for compact antenna design?  &lt;/p&gt;&#xA;" OwnerUserId="4257" LastActivityDate="2011-09-05T15:25:51.393" Title="Why are fractal geometries useful for compact antenna design?" Tags="&lt;geometry&gt;&lt;antennas&gt;" AnswerCount="1" CommentCount="8" />
  <row Id="13943" PostTypeId="1" CreationDate="2011-08-25T07:45:02.480" Score="1" ViewCount="61" Body="&lt;p&gt;On account of the annual fluctuation of temperature the ground at the town of Ν freezes to a depth of 2 metres. To what depth would it freeze on account of a daily fluctuation of the same amplitude?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Found this on aops while gazing through some questions (So that people dont blame me for posting a phy q on math site). Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1928494&amp;amp;sid=5670b1680456f7c649b8ba62246de38f#p1928494&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4919" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-08-25T11:46:57.660" LastActivityDate="2011-08-25T13:09:12.953" Title="How does the period of a temperature fluctuation affect the depth to which the ground freezes?" Tags="&lt;thermodynamics&gt;&lt;scaling&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13944" PostTypeId="1" AcceptedAnswerId="13946" CreationDate="2011-08-25T11:59:10.277" Score="1" ViewCount="48" Body="&lt;p&gt;If light is considered as wave, then different colored light can be explained as waves of different wavelength/frequency. How colored light can be explained if light is considered as emission of photon?&lt;/p&gt;&#xA;" OwnerUserId="2125" LastActivityDate="2011-08-25T12:13:08.407" Title="How colored light can be explained if light is considered as emission of photon?" Tags="&lt;light&gt;&lt;waves&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="1" />
  <row Id="13945" PostTypeId="1" AcceptedAnswerId="13984" CreationDate="2011-08-25T12:12:08.507" Score="3" ViewCount="125" Body="&lt;p&gt;I think I've missed this point somehow. I've just started with Compactification and so far, I don't really see why it is restricted to the above mentioned types of manifolds? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to admit, when studying T-Duality, I simply took Toroidal compactification as a kind of &quot;why not&quot;-thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone point me in the right direction? Been skipping through my books to look for some explanation of this, but couldn't find anything. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4337" LastActivityDate="2011-09-03T20:44:51.637" Title="Why is Compactification restricted to Toroids, Calabi-Yau &amp; Co.?" Tags="&lt;string-theory&gt;&lt;compactification&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="13946" PostTypeId="2" ParentId="13944" CreationDate="2011-08-25T12:13:08.407" Score="1" ViewCount="" Body="&lt;p&gt;The Energy of the Photon defines its frequency. It is the equation $$E=h\nu$$ which puts this in mathematical terms. Where $h$ is Planck's constant, $E$ the Energy and $\nu$ the frequency. This can also be written in terms of wavelength of the photon as $$E=\frac{hc}{\lambda}$$ So blue (high frequency, short wavelength) light has high energy, red (low frequency, large wavelength) has lower energy&lt;/p&gt;&#xA;" OwnerUserId="2151" LastActivityDate="2011-08-25T12:13:08.407" CommentCount="4" />
  <row Id="13947" PostTypeId="2" ParentId="1019" CreationDate="2011-08-25T12:28:04.557" Score="0" ViewCount="" Body="&lt;p&gt;Misconception - &lt;strong&gt;The uncertainty principle is a statement about &quot;our&quot; ability to make measurements.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correction - The uncertainty principle is a result of the nature of the particles themselves and refers to the ability of &lt;em&gt;anything&lt;/em&gt; to &quot;make the relevant measurements&quot;.  It's not just us that can't determine the simultaneous values of incompatable observables, God can't either.&lt;/p&gt;&#xA;" OwnerUserId="5020" LastActivityDate="2011-08-25T12:28:04.557" CommunityOwnedDate="2011-08-25T12:28:04.557" />
  <row Id="13948" PostTypeId="2" ParentId="13943" CreationDate="2011-08-25T13:09:12.953" Score="1" ViewCount="" Body="&lt;p&gt;The characteristic length of the exponential decay in the ground is proportional to the square root of the period of the fluctuations, so the answer is roughly $2\text{m}/\sqrt{365.25}\approx10\text{cm}$. That's if we treat this as a linear isotropic heat diffusion problem and neglect non-linear effects from the heat of fusion, inhomogeneities in the ground, finite rate of heat transfer at the ground/air interface, and ground absorption of direct sunlight.&lt;/p&gt;&#xA;" OwnerUserId="5033" LastActivityDate="2011-08-25T13:09:12.953" />
  <row Id="13949" PostTypeId="2" ParentId="13937" CreationDate="2011-08-25T13:43:18.847" Score="2" ViewCount="" Body="&lt;p&gt;You're absolutely right that in principle each scattered photon undergoes a change in energy due to the recoil of the object it scatters off. In Rayleigh scattering, that change is very small, though, and can generally be neglected. In general, when the energy of the photon is much smaller than the rest energy of the massive particle, the photon undergoes only a small change in energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a qualitative argument for why this is true. Let $p$ be the momentum of the incoming photon. The momentum transferred to the massive particle is at most of order $p$ (specifically, it's at most $2p$). For a nonrelativistic particle, the kinetic energy is $p^2/(2m)$, so the energy transferred is of this order. So the fractional energy lost by the photon is of order&#xA;$$&#xD;&#xA;{\Delta E\over E}\sim{p^2/2m\over pc}={pc\over 2mc^2}={E\over 2mc^2},&#xD;&#xA;$$&#xA;where $E =pc$ is the original photon energy. So when $E\ll mc^2$, the change in energy is small in comparison to the original energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One common way to express this more quantitatively is via the Compton scattering formula,&#xA;$$&#xD;&#xA;\Delta\lambda={h\over mc}(1-\cos\theta),&#xD;&#xA;$$&#xA;where $m$ is the mass of the massive particle and $\theta$ is the angle through which the photon scattered. This formula is usually used for much higher-energy photons scattering off charged particles, but it is derived just from energy and momentum conservation, so it applies to any situation in which a photon scatters off a massive particle initially at rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I've done the arithmetic right, when $m$ is the mass of an atom or molecule, $\Delta\lambda$ comes out to be of order $10^{-16}$ m, which is certainly small enough to be ignored for visible light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your last question, even though atoms and molecules are neutral, they are made of charged parts. The photon can and does interact with those parts. &lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-08-25T13:43:18.847" CommentCount="1" />
  <row Id="13950" PostTypeId="1" CreationDate="2011-08-25T14:15:53.157" Score="0" ViewCount="145" Body="&lt;p&gt;For a massless particle the spacetime interval between its point of emission and point of absorption is zero: the two points coincide. From the vantage point of such a particle space-time has reduced to a single point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Standard Model, &lt;em&gt;all&lt;/em&gt; particles start with zero mass, so does this suggest that our familiar 4-D spacetime is a consequence of the Higgs field or some similar mechanism by which mass is acquired?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: if you believe the large-scale 4D structure of spacetime is a pre-existing given, this question makes little sense. In a relational view, where spacetime emerges out of event relationships, then the question suggests that a universe comprised solely of massless particles would have a fundamentally different event-structure.&lt;/p&gt;&#xA;" OwnerUserId="1394" LastEditorUserId="1394" LastEditorDisplayName="" LastEditDate="2011-08-25T20:55:15.367" LastActivityDate="2011-08-25T20:55:15.367" Title="Does a photon see a 0-D universe?" Tags="&lt;special-relativity&gt;&lt;mass&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="13952" PostTypeId="1" CreationDate="2011-08-25T15:18:45.457" Score="0" ViewCount="41" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/5456/the-speed-of-gravity&quot;&gt;The speed of gravity&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Lets say for example that a pertubance occured on the Sun that changes its distance to the Earth. How fast we are going to notice the gravity changes? Is it immediately? Light speed?&lt;/p&gt;&#xA;" OwnerUserId="5035" LastActivityDate="2011-08-25T15:18:45.457" ClosedDate="2011-08-25T16:44:34.560" Title="How fast gravity force propagates?" Tags="&lt;gravity&gt;&lt;speed&gt;" CommentCount="3" />
  <row Id="13953" PostTypeId="1" CreationDate="2011-08-25T15:48:51.663" Score="0" ViewCount="48" Body="&lt;p&gt;Given a particle on the plane we can assume that it follows 2D Brownian motion. On the other hand if there is a lot of such a Brownian particles one can be interested in the evolution of the density of such particles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is the PDE for the density derived? I guess it should be similar to the heat equation and there should be connection with a gas dynamics. Could you refer me to the literature? &lt;/p&gt;&#xA;" OwnerUserId="5037" LastActivityDate="2011-08-25T16:25:10.627" Title="SDE for particle, PDE for the density" Tags="&lt;heat&gt;&lt;models&gt;&lt;brownian-motion&gt;" AnswerCount="1" />
  <row Id="13954" PostTypeId="2" ParentId="13953" CreationDate="2011-08-25T16:25:10.627" Score="0" ViewCount="" Body="&lt;p&gt;The equation resembles the heat equation very closely, so much so, that it is the heat equation.&#xA;The literature on elementary things like this is uniformly terrible, so I cannot give a reference. To see why it is the heat equation, note that the law for the distribution function is the same as for a probability distribution, so it is linear equation, by the probability superposition principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you evolve a little time in the future, the probability of being at any point gets incremented the average of the nearby neighbors. The average of the neighbors less the center point is the Laplacian, so the time derivtive is proportional to the Laplacian plus the center value, and since the integral over all space has to be unchanged with time, it must be just the Laplacian. Changing units of space, you can make the coefficient of the Laplacian 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ {d\phi \over dt} = {1\over 2} \nabla^2 \phi $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this still leaves a scaling invariance, if you rescale space by a certain amount, and time by the same amount squared, you get the same equation back. So the fundamental solution must be proportional to a scaling function: f({x^2\over t}) and you can solve for the solution, which is a spreading gaussian $\phi(x,t) = {1\over (2\pi t)^{d/2}} e^{-{x^2\over 2t}}$, where d is the dimension of space. (a product of one-dimensional Gaussians). This collection of ideas form a circle, because the same distribution is in the definition of Brownian motion, and comes from the central limit theorem.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-25T16:25:10.627" />
  <row Id="13957" PostTypeId="1" CreationDate="2011-08-25T16:45:19.047" Score="2" ViewCount="53" Body="&lt;p&gt;So, once something is inside a black hole's event horizon, it can only move towards the center. This is fine for a point-object. But 3D solid objects rely on molecular forces to stay in one piece. These forces act in all directions inside the solid. But there could not be any interaction between atoms at different distances from the singularity, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what happens to a solid (let's assume a crystalline lattice, for simplicity) hypothetically placed inside the event horizon? Does it get cleaved in ultra-thin concentric layers, centered on the singularity?&lt;/p&gt;&#xA;" OwnerUserId="4210" LastActivityDate="2011-08-25T16:45:19.047" Title="Solid objects inside the event horizon - can they remain &quot;solid&quot;?" Tags="&lt;general-relativity&gt;&lt;black-holes&gt;&lt;event-horizon&gt;" CommentCount="4" FavoriteCount="1" />
  <row Id="13958" PostTypeId="1" CreationDate="2011-08-25T17:04:51.200" Score="-1" ViewCount="91" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/10870/calculating-position-in-space-assuming-general-relativity&quot;&gt;Calculating position in space assuming general relativity&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I recently started to program a 4 dimensional visualization program.&#xA;I have the 4 dimension space perspectively projected in the 3d space.&#xA;Now I describe 3d sphere inside this space.&#xA;I want to describe the movement of those 3d spheres.&#xA;What I thought it will be cool is to apply the general relativity theory to it.&#xA;As far as I know from the general relativity theory, is that masses bends the space around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I think it means:&#xA;let's imagine a 2d plane. if I put a mass on this 2d plane, the plane would be actually pinched in the 3d space.&#xA;Parctically what I have to do is to find a function that describe this pinch, in function of the mass.&#xA;like a Gauss distribution under the mass.&#xA;Now this transformation will be applied to the moving objects, which I think would bend the trajectory of the objet to a curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to do is to apply this to the 3d space:&#xA;if you put a mass in the 3d space it will cause a pinch in the 4th dimension.&#xA;So what I want is to find the transformation function which describe this pinch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that what Zaslavsky is the actual answer, I din't got it at the beginning because I didn't tought it was actually what I was looking for but anyway:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}^2x^\lambda}{\mathrm{d}t^2} + \Gamma^{\lambda}_{\mu\nu}\frac{\mathrm{d}x^\mu}{\mathrm{d}t}\frac{\mathrm{d}x^\nu}{\mathrm{d}t} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution to this differential equation should give the pinch I mean, am I right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: found by surfing random on the internet: Lorentz transformations are just hyperbolic rotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: I've edited the question, I hope, in a way is comprehensible.&lt;/p&gt;&#xA;" OwnerUserId="5039" LastEditorUserId="5039" LastEditorDisplayName="" LastEditDate="2011-08-26T09:44:49.740" LastActivityDate="2011-08-26T09:44:49.740" CommunityOwnedDate="2011-08-26T05:43:05.893" ClosedDate="2011-08-25T18:10:06.493" Title="How to bend 3d space embededd in a 4d enviornment?" Tags="&lt;general-relativity&gt;&lt;spacetime&gt;" AnswerCount="1" CommentCount="11" />
  <row Id="13959" PostTypeId="2" ParentId="13387" CreationDate="2011-08-25T17:22:42.867" Score="0" ViewCount="" Body="&lt;p&gt;Thanks to @Martin Gales for the lead to the answer. Here's the quantification, from &lt;a href=&quot;http://www.dewi.de/dewi/fileadmin/pdf/publications/Magazin_37/07.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.dewi.de/dewi/fileadmin/pdf/publications/Magazin_37/07.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The temperature dependence of the saturation water vapour pressure E&#xA;  is described theoretically by the Clausius-Clapeyron equation ...&#xA;  Approximately, the dependence of water vapour saturation pressure in&#xA;  hPa on temperature can be described more simply by Magnus’ formula ...&#xA;  $$ E(t) = 6.107 \times 10^{\frac{a t}{(b+t)}} $$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Here t denotes air temperature in °C. a and b are two constants (over&#xA;  water: a=7.5, b=235) ... Mixing of two saturated air masses of different temperature&#xA;  leads to a mixed air mass which is always oversaturated, because each&#xA;  straight line which connects two separate points on the saturation&#xA;  water vapour curve runs through the space above the curve between the&#xA;  two points&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, that's the gist of it. There's more to it than that, that's not covered in the paper, as to why the condensation appears to be emerging from a point that's about 70m downwind from the hub, and at hub height (about 70m ASL)&lt;/p&gt;&#xA;" OwnerUserId="4066" LastActivityDate="2011-08-25T17:22:42.867" />
  <row Id="13960" PostTypeId="2" ParentId="13958" CreationDate="2011-08-25T18:19:09.653" Score="0" ViewCount="" Body="&lt;p&gt;Unless your program is specifically designed to address issues of relativity, I would highly recommend that you use the classical Newtonian mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the massive body is the center of a stationary reference frame (as you suggest), the smaller object experiences a force toward the origin of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_g = G\frac{m M}{r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G = 6.67 \times 10^-11 \frac{m^3}{kg \cdot s}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m = $ mass of object&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M = $ mass of planet&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r = $ distance from center of planet to object&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The acceleration due to this force is $a = \frac{d v}{d t} = F/m$. The nature of your question suggests that you are interested in working with discrete time steps.  I would suggest calculating the position as normal and then adding the increase in velocity after the next time step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At time $t = 0$, the object has position $\textbf{r}_0 = \langle x, y, z \rangle$ and velocity $\textbf{v}_0 = \langle \frac{d x}{d t}, \frac{dy}{dt}, \frac{dz}{dt} \rangle = \langle v_{x, 0}, v_{y, 0}, v_{z, 0} \rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At time $t$, the object has position $\textbf{r}_1 = \langle x + v_{x, 0}\cdot t, y + v_{y, 0} \cdot t, z + v_{z, 0} \cdot t \rangle$ and velocity of $\langle v_{x, 0}, v_{y, 0}, v_{z, 0} \rangle + \frac{G M}{(\Delta r)^2}$.  At the next time step, you simply use the new velocities.  This would, of course, be greatly simplified if you used a spherical coordinate system centered on the massive body.&lt;/p&gt;&#xA;" OwnerUserId="5020" LastActivityDate="2011-08-25T18:19:09.653" CommentCount="2" />
  <row Id="13961" PostTypeId="1" AcceptedAnswerId="13963" CreationDate="2011-08-25T19:00:31.817" Score="3" ViewCount="110" Body="&lt;p&gt;I'm working in Taylor and Wheeler's &quot;Exploring Black Holes&quot; and on p.2-14 they use two honorary constants: Newton's constant divided by the speed of light squared e.g. $G/c^2$ as a term to convert mass measured in $kg$ to distance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without doing the arithmetic here, the &quot;length&quot; of the Earth is 0.444 cm; and of the sun is 1.477 km. To what do these distances correspond? What is their physical significance, generally? &lt;/p&gt;&#xA;" OwnerUserId="756" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-25T19:02:29.997" LastActivityDate="2011-08-26T03:26:46.343" Title="Physical interpretation of describing mass in units of length" Tags="&lt;units&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="13962" PostTypeId="2" ParentId="13157" CreationDate="2011-08-25T19:09:37.210" Score="0" ViewCount="" Body="&lt;p&gt;ok there is only one good answer to this. it is found in a book Local Quantum Physics, by Rudolf Haag, master of QFT, page 46, first paragraph. More or less, it meets answer by Newman.&lt;/p&gt;&#xA;" OwnerUserId="5012" LastActivityDate="2011-08-25T19:09:37.210" CommentCount="1" />
  <row Id="13963" PostTypeId="2" ParentId="13961" CreationDate="2011-08-25T19:41:01.453" Score="3" ViewCount="" Body="&lt;p&gt;They represent the scale on which general relativisic effects dominate physics related to bodies of that mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance if you were to create a (un-rotating, uncharged) black hole of 1 Earth mass &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow&quot;&gt;it's event horizon would have a radius of about $9\text{ mm} = 2 * M_\text{Earth}$ in those units&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For scales much, much larger than the &quot;length&quot; of the mass, general relativity may be neglected. For intermediate scale in comes in as corrections on order of $\frac{l}{L}$ where $l$ is the mass in the scaled units and $L$ is the length scale of the problem.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is similar to what particle physicists do by setting $c = \hbar = 1\text{ (dimensionless)}$ energy scales and length scales become inter-changeable.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-25T23:29:01.333" LastActivityDate="2011-08-25T23:29:01.333" CommentCount="1" />
  <row Id="13965" PostTypeId="2" ParentId="13950" CreationDate="2011-08-25T20:10:02.553" Score="-1" ViewCount="" Body="&lt;p&gt;We see a 3D universe with help of photons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;gauge nature&quot; of all particles is an incredible stretch which does not work itself at all. Higgs &quot;mechanism&quot; is a patch of this stretch and mass of Higgs needs another &quot;mechanism&quot; to be explained. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think starting from phenomenological masses is the only healthy physical way of giving masses to all particles and bodies. Some masses can be calculated from others then.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-08-25T20:24:16.520" LastActivityDate="2011-08-25T20:24:16.520" CommentCount="15" />
  <row Id="13966" PostTypeId="2" ParentId="13961" CreationDate="2011-08-25T20:17:07.493" Score="0" ViewCount="" Body="&lt;p&gt;I'm not sure it's terribly helpful, but it seems like the following analysis helps explain dmckee's response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force of gravity is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = G \frac{m M}{r^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging and dividing by $c^2$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{G}{c^2} = \frac{F r^2}{M (m c^2)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the $mc^2$ is the rest mass energy $(E_0)$ of the object experiencing the force caused by mass $M$. When you multiply through by the mass of the &quot;large&quot; object you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M \frac{G}{c^2} = l = \frac{F_g r^2}{E_0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we are interested in the length $l$, at that distance we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M\frac{G}{c^2} = r = \frac{F_g r^2}{E_0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or simply&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_0 = F_g r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In words, this is the distance at which the energy of the system due to the rest mass of an object in a gravitational field is the same as the potential energy due to gravitation.&lt;/p&gt;&#xA;" OwnerUserId="5020" LastActivityDate="2011-08-25T20:17:07.493" />
  <row Id="13967" PostTypeId="2" ParentId="13645" CreationDate="2011-08-25T20:29:25.567" Score="1" ViewCount="" Body="&lt;p&gt;This problem can get as complicated as you like, but let me show you how it can be approached in the most simple case. The Navier-Stokes equations describe the flow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho\left(\displaystyle\frac{\partial {\bf u}}{\partial t}+({\bf u}\cdot\nabla){\bf u}\right)=-\nabla p + \mu \nabla^2 {\bf u}$. $\qquad$ (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that the motion of the fluid is slow enough so that the non-linear term $({\bf u}\cdot \nabla){\bf u}$ is negligible. This will be true as long as viscosity forces are large compared to inertial forces from acceleration. Ugh, now that it is a linear equation, something can be done analytically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the problem in cylindrical coordinates with $z$, $r$ and $\phi$. Let us assume $\partial/\partial z=0$ and $\partial/\partial \phi=0$ for all quantities due to the axial symmetry of the problem. Then the continuity equation will offer additional simplifications. In cylindrical coordinates, the continuity equation reads&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\frac{\partial \rho}{\partial t}+ \frac{1}{r}\displaystyle\frac{\partial(\rho r u_r)}{\partial r}+ \frac{1}{r}\frac{\partial\rho u_{\phi}}{\partial \phi} + \frac{\partial (\rho u_z)}{\partial z}=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $\rho=\mathrm{const}$ (incompressibility assumption), the first term is zero, and last two terms are equal to zero due to the axial symmetry assumption. The remainder yields $u_r=0$. Hooray! Let us choose the problem so that $u_z=0$, too (true if the cylinder is not moving with respect to the fluid in $z$-direction) Then we are left with just $u_{\phi}$, and let us denote it $u\equiv u_{\phi}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the cylinder is suspended vertically, i.e., the axis of the cylinder is along the gravity force, then the pressure gradient $\nabla p$ can also be neglected. This is because the only component of $\nabla p$ is in z-direction due to gravity, and we already assumed $\partial {\bf u}/\partial z$=0. And there may be no $\phi$-component of $\nabla p$, because it would violate our $\partial /{\partial \phi}=0$ assumption. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the above leaves us with the only equation for $u_\phi\equiv u$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\frac{\partial u}{\partial t} = \mu \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right)$. $\qquad$ (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand where it came from, look up the explicit form of equation (1) in cylindrical coordinates at &lt;a href=&quot;http://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations#Cylindrical_coordinates&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia page&lt;/a&gt;, take the $\phi$-component of it and use the assumptions outlined below to get rid of $\partial/\partial z$ and $\partial/\partial \phi$ terms, and of the non-linear and pressure terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cylinder is spinning, which makes the boundary conditions for equation (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u(r=0, t)=0$ and $u(r=R, t)=R\omega_0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R$ is the radius and $\omega_0$ is the angular frequency of the cylinder. To understand the second one, look at the well-known solution of a similar problem &lt;a href=&quot;http://en.wikipedia.org/wiki/Couette_flow&quot; rel=&quot;nofollow&quot;&gt;at this Wikipedia page&lt;/a&gt; or &lt;a href=&quot;http://pre.aps.org/abstract/PRE/v60/i5/p6192_1&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; referenced in the Wiki page. The initial conditions are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u(r, t=0)=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation (2) is a very well-known equation. It is the diffusion or the heat equation in disguise. Well, no surprise about it, because viscosity in is nothing more than diffusion of momentum. At least, this is how it is expressed in the viscosity term of equation (1). The solution is readily obtained using separation of variables and Fourier analysis. I will not solve it here, because it is a well understood mathematical problem, with which graduate students are routinely tortured with great success. &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_equation&quot; rel=&quot;nofollow&quot;&gt;This Wikipedia page&lt;/a&gt; describes it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the back-of-the envelope estimate I mentioned in my comment comes from dimensional analysis of equation (2). If the typical velocity scale of the problem is $U$, spatial scale is $R$ and time scale is $T$, then (2) can be approximated as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\frac{U}{T} \approx \mu \frac{1}{R} \frac{1}{R} R \frac{U}{R}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T \approx R^2/\mu$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a crude estimate of your spin-up time.&lt;/p&gt;&#xA;" OwnerUserId="4450" LastEditorUserId="4450" LastEditorDisplayName="" LastEditDate="2011-08-25T20:37:22.837" LastActivityDate="2011-08-25T20:37:22.837" CommentCount="3" />
  <row Id="13968" PostTypeId="1" CreationDate="2011-08-25T20:35:33.780" Score="0" ViewCount="52" Body="&lt;p&gt;I have a 2D flow velocity field $\bar{V} = y^2\hat{i} + 2\hat{j}$. I'd like to find the equations for the streamlines and pathlines. Since $\bar{V}$ has 0 time derivative, the flow is steady, and so the equations for the streamlines should be identica, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Streamlines:&#xA;$$\left. \frac{dy}{dx} \right)_{streamline} = \frac{v}{u} = \frac{2}{y^2}$$&#xA;$$ \therefore y^2 dy = 2 dx \Rightarrow \int_{y_0}^y y^2 \, dy = \int_{x_0}^x 2 \, dx $$&#xA;$$ \therefore \tfrac{1}{3}y^3 - \tfrac{1}{3}y_0^3 = 2(x - x_0) \Rightarrow y^3 - 6x = y_0^3 - 6x_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pathlines ($x_p$ and $y_p$ the particle coordinates):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{dx_p}{dt} = y^2 \Rightarrow x_p = y^2(t-t_0) + x_{p,0} $$&#xA;$$ \frac{dy_p}{dt} = 2 \Rightarrow y_p = 2(t - t_0) + y_{p,0} $$&#xA;Now eliminating $(t-t_0)$:&#xA;$$x_p = y^2 \left( \frac{y_p - y_{p,0}}{2} \right) + x_{p,0}$$&#xA;Without expanding this final equation for the pathline, it is obviously different from the equation of the streamlines; but it shouldn't be, as the flow is steady? Normally this might suggest I've been fast and loose with the math, but I just can't see where I'm wrong here...&lt;/p&gt;&#xA;" OwnerUserId="5041" LastEditorUserId="5041" LastEditorDisplayName="" LastEditDate="2011-08-25T20:57:02.387" LastActivityDate="2011-09-04T00:23:08.757" Title="Streamlines &amp; Pathlines Problem" Tags="&lt;fluid-dynamics&gt;" AnswerCount="1" CommentCount="2" />
  <row Id="13969" PostTypeId="2" ParentId="13961" CreationDate="2011-08-25T21:30:02.830" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Earth is 0.444 cm; and of the sun is 1.477 km&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It corresponds to half of the respective &lt;a href=&quot;http://en.wikipedia.org/wiki/Schwarzschild_radius&quot; rel=&quot;nofollow&quot;&gt;Schwarzschild radius&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\frac{G}{c^2}$ is covered there and also in Adam’s answer.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="4438" LastEditorDisplayName="" LastEditDate="2011-08-26T03:26:46.343" LastActivityDate="2011-08-26T03:26:46.343" />
  <row Id="13970" PostTypeId="2" ParentId="13157" CreationDate="2011-08-25T21:49:09.537" Score="2" ViewCount="" Body="&lt;p&gt;You say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;she said to me that, if I wanted hardcore definitions,&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;a field is a function that returns a value for a point in space.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;Now this finally makes a hell lot of sense to me but I still don't understand how mathematical functions can be a part of the Universe and shape the reality.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You don't have to use super-complicated examples such as electromagnetism. I'll give you two examples which I hope will make it more clear; let me know if this helps.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example 1: Temperature&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You might have come across that the higher you climb (on Earth or somewhere else, but let's think of Earth) the colder the air gets, at an &lt;em&gt;typical&lt;/em&gt; rate of about 6ºC per kilometer (it depends on various factors, but this is a ballpark value); in meteorology, this is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lapse_rate&quot; rel=&quot;nofollow&quot;&gt;lapse rate&lt;/a&gt;: the rate of temperature drop with altitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose you're observing a large, uniform terrain (e.g. a &quot;flat desert&quot;). If you want to ask: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the temperature of the air at a point $(x,y,z)$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;then you'll ascribe a certain value of temperature for each point. But to make a &quot;table&quot; to give the temperature for &lt;em&gt;every point&lt;/em&gt; is certainly impractical! You try instead to use a &lt;em&gt;function&lt;/em&gt;, an application, that gives the value of the temperature for each point:&#xA;$$ f : (x,y,z) \mapsto f(x,y,z) $$&#xA;I'll use a clearer nomenclature:&#xA;$$ T : (x,y,z) \mapsto T(x,y,z) $$&#xA;So this is a function with arguments in a $\mathcal{R}^3$ space (three-dimensional space, $\mathcal{R}\times\mathcal{R}\times\mathcal{R}$) which gives values in a 1-dimensional $\mathcal{R}$ space. Those values represent the values of the temperature at each coordinate $(x,y,z)$ of $\mathcal{R}^3$. Instead of writing $T(x,y,z)$ you can be more &quot;practical&quot; and write just $T$ as shorthand (especially when you're some calculus in an exercise).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That function &lt;strong&gt;&lt;em&gt;represents&lt;/em&gt;&lt;/strong&gt; a field -- the &lt;em&gt;temperature field&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;&quot;But what's the use of that?!&quot;&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;What does it look like? If you have the ideal case of a perfectly flat &quot;desert&quot; and an idealized atmosphere, the temperature field will be something like:&#xA;$$ T(x,y,z) = T(x,y,z_0) - \frac{dT}{dz} (z-z_0) $$&#xA;Here, the temperature $T(x,y,z)$ at a point $(x,y,z)$ of 3D space depends on the &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In &lt;em&gt;this&lt;/em&gt; situation, the temperature only varies in the vertical; it looks the same at any place over the desert -- there is really no depedence in the coordinates $x$ and $y$. Because of that you &lt;em&gt;could&lt;/em&gt; make it easier for you and shorten the expression to just $T(z) = T(z-z_0) - dT/dz$.&lt;/li&gt;&#xA;&lt;li&gt;In case you don't know/forgot: $dT$ is how much the temperature $T$ varies when you increase your height by a small (infinitesimal!) amount $dz$.&lt;/li&gt;&#xA;&lt;li&gt;Don't worry about the minus sign next to the rate. It's put there by hand to have the expected physical meaning. When you go from a height level $z$ to $z+dz$, the temperature should &lt;em&gt;decrease&lt;/em&gt;, from $T$ to $T-dT$ where $-dT &amp;lt; 0$, so that $-dT/dz$ is &lt;em&gt;negative&lt;/em&gt; (it &quot;takes away&quot; from the temperature as you increase the altitude $z$). Example: from $z=1000$ to $z+dz = 1001$, the temperature should drop from $T$ to $T-0.006$ where $T$ is the temperature at level $z=1000$. Of course, that small value is because $0.006/(1001-1000) = dT/(dz+z-z) = dT/dz = 6$ Celsius per km.&lt;/li&gt;&#xA;&lt;li&gt;I've intentionally abused the expression above to make it easier to understand. A more appropriate expression would be (if you've studied &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Integral&quot; rel=&quot;nofollow&quot;&gt;integrals&lt;/a&gt;&quot; in calculus) something like &#xA;$$T(x,y,z) = T(x,y,z_0) - \int\limits_{z_0}^z \frac{dT}{dz} dz\ .$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You have to give the temperature at a certain level $z_0$ of your choice to represent a &lt;em&gt;specific&lt;/em&gt; case; it can be at the surface, $z_0 = 0\ \mathrm{meters}$. That function you have there represents the temperature field for that situation. If you have a &quot;hot spot&quot; -- e.g. you light up a candle -- then the temperature distribution (the field!) will be different, and the mathematical expression to &lt;em&gt;describe the temperature field&lt;/em&gt; will be different (more complicated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;&lt;em&gt;this&lt;/strong&gt; temperature field&lt;/em&gt; describes what is the temperature over that &quot;desert air&quot;. It represents a quantity which has a spatial distribution. You can make it much more &lt;strong&gt;shorthanded&lt;/strong&gt; if you just ignore the &lt;em&gt;frontier condition&lt;/em&gt; $T(z_0)$ at a certain vertical level $z_0$ (which is &lt;strong&gt;arbitrary&lt;/strong&gt;!) and write the field as &#xA;$$-\frac{dT}{dz}\ .$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example 2: Wind velocity&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The example above illustrates a &lt;em&gt;scalar&lt;/em&gt; field: the value of the field at each space point takes a &lt;em&gt;scalar&lt;/em&gt; value (&quot;just a number&quot;). Not all fields are scalar. An example is the &lt;em&gt;velocity field&lt;/em&gt;, which represents the velocity (&lt;em&gt;direction &lt;strong&gt;and&lt;/strong&gt; magnitude!&lt;/em&gt;) of the air at each point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can write it as &#xA;$$\vec v : (x,y,z) \mapsto \vec v(x,y,z)$$&#xA;and for each point $(x,y,z)$ it describes what is the &lt;em&gt;direction &lt;strong&gt;and&lt;/strong&gt; magnitude&lt;/em&gt; of the air displacement at that point, the vector $\vec v$ at that point.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;What does it look like?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;(The mathematical expression?) Well, it will depend on the situation of course! The expression can be impossibly complicated to write &lt;em&gt;analytically&lt;/em&gt;. You certainly won't write the &lt;em&gt;velocity field&lt;/em&gt; (or the temperature field) for the air inside your living room -- it's too complicated to write a mathematical expression! The best you can do is&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Know a few laws or expressions or (more correctly) models, perhaps deduced from first principles, to describe &lt;em&gt;how the conditions of a tiny piece of air&lt;/em&gt; will be influenced by the conditions of the neighbouring regions. Those &lt;a href=&quot;http://en.wikipedia.org/wiki/Lapse_rate#Types_of_lapse_rates&quot; rel=&quot;nofollow&quot;&gt;models&lt;/a&gt; can be very simple or more elaborate; in the latter for meteorology, you just use computers to do the complicated ballance for each and every &quot;air cell&quot;. In the example 1 with the &lt;strong&gt;temperature&lt;/strong&gt; above, there is no &lt;em&gt;horizontal&lt;/em&gt; dependence, but the rate at which the temperature varies vertically depends on the temperature, pressure, etc on &lt;em&gt;top&lt;/em&gt; of the &quot;tiny air box/cell/element&quot; and on the &lt;em&gt;bottom&lt;/em&gt; -- those are the ones who produce an effect.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Make some simplifications about the &lt;em&gt;initial conditions&lt;/em&gt;, such as knowing what is the temperature along the walls and assuming (for example) there aren't &quot;hot spots&quot; or if there are, they're too insignificant to spot the difference against the situation where there aren't hot spots.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Example 3: the electromagnetic field&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;When you put an electrically-charged tiny particle (&lt;em&gt;test particle&lt;/em&gt;) near a metalic place (for example) that has an electric charge itself (like the plate of a large capacitator, for example), in the most general and broad case the &lt;strong&gt;force&lt;/strong&gt; that the particle will feel will depend on &lt;em&gt;where&lt;/em&gt; the particle is relative to the charged plate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;force&lt;/em&gt; the test particle feels has a &lt;strong&gt;magnitude&lt;/strong&gt; as well as a &lt;strong&gt;direction&lt;/strong&gt;. If you put the test particle in another position, if will feel the force with a different intensity &lt;em&gt;and direction&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could put the test particle in many different places around the plate and &lt;strong&gt;measure&lt;/strong&gt; the electric force felt by the test particle. And you collect the &lt;em&gt;direction&lt;/em&gt; and &lt;em&gt;intensity&lt;/em&gt; of that force. If you are able to condense that description of the magnitudes and directions of the electric force felt by the particle, you're writing it as a &lt;strong&gt;field&lt;/strong&gt;,&#xA;$$\vec E : (x,y,z) \mapsto \vec E(x,y,z)\ .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can interpret the &lt;em&gt;electromagnetic field&lt;/em&gt; as nothing more as a &quot;mash-up&quot; of both the electric force and magnetic force that a test particle will feel at each point of space.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;OK, but can you &quot;touch&quot; a field?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As a final note, I'll say the following. This question is more subject to discussion. Personally, I don't quite think about &quot;touching&quot; a field or it being &quot;material&quot;; I don't know how you're supposed to &quot;touch&quot; temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The field represents the set of values for a quantity on a given space, and thus we arrive at your teacher's comment. In the classical physics sense that I've presented above, you can interpret the fields as &quot;our way&quot; of describing something that it's there, in a shorthand (a mathematical expression instead of a &quot;spreadsheet of values&quot;). In that case, I see the concept of field mixing up with the &quot;thing&quot; that it's representing. I won't debate that because I'm not sure I can explain better.&lt;/p&gt;&#xA;" OwnerUserId="103" LastEditorUserId="103" LastEditorDisplayName="" LastEditDate="2011-08-25T22:12:40.700" LastActivityDate="2011-08-25T22:12:40.700" />
  <row Id="13971" PostTypeId="1" AcceptedAnswerId="13993" CreationDate="2011-08-25T23:10:54.633" Score="6" ViewCount="75" Body="&lt;p&gt;I am trying to follow a derivation on &lt;a href=&quot;http://www.jgiesen.de/astro/stars/star.htm&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; site to derive the minimum number of (hydrogen) atoms I need so that the gravitational force dominates over internal electrostatic forces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivation begins by introducing the ratio of the two forces for &lt;em&gt;two&lt;/em&gt; protons:&#xA;$$\frac{1}{4 \pi \epsilon_0} \frac{e^2}{d^2} : G \frac{m^2}{d^2} = 1.24\cdot 10^{36}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It then considers a sphere of radius $R$, homogeneously filled with hydrogen, to which we add protons at the border. The gravitational energy per proton mass then is &#xA;$$\frac{W}{m} = G \frac{M}{R}$$&#xA;This can be expressed in terms of the number of atoms,&#xA;$\frac{W}{m} \sim N^{2/3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, so good. The next argument is that gravity is not shielded, and long range. Sure, otherwise we wouldn't have used the formula above. And the electrostatic forces are shielded by the electrons in hydrogen, so they are short range only. I understand that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then, the final step is just the claim that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$N &amp;gt; (1.24\cdot 10^{36})^{3/2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I don't see how this step can be made... I would have expected some discussion of the electrostatic energy per proton &lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="661" LastEditorDisplayName="" LastEditDate="2011-08-26T15:26:53.667" LastActivityDate="2011-08-27T16:12:39.037" Title="Minimum number of atoms in star" Tags="&lt;astrophysics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13973" PostTypeId="1" CreationDate="2011-08-26T00:23:16.433" Score="0" ViewCount="11" Body="&lt;p&gt;Say I have a number of events that pass a criterion $N_\mathrm{pass}$, and a total number of events $N_\mathrm{total}$. Now I'd like to calculate an efficiency ($\epsilon = N_\mathrm{pass} / N_\mathrm{total}$), with a correct treatment of uncertainties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With correct, I mean especially:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The errors do not extend over 1 and below 0.&lt;/li&gt;&#xA;&lt;li&gt;The errors do not vanish if $\epsilon=0$ or $\epsilon=1$.&lt;/li&gt;&#xA;&lt;li&gt;From this follows that the errors are assymmetric.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I know how to calculate this for the regular case, but now assume the events are weighted, so&#xA;$$ \epsilon = \frac{\sum_{i \;\in\; \mathrm{pass}} w_i}{\sum_{i \;\in\; \mathrm{total}} w_i} \,.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to have the properties stated above, and furthermore:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Weights are handled in a sensible way (e.g. adding events of vanishing weight to either sample doesn't significantly change the result, and so on).&lt;/li&gt;&#xA;&lt;li&gt;Increasing the statistics (while lowering the weights uniformly) decreases the uncertainty of $\epsilon$ by a sensible value.&lt;/li&gt;&#xA;&lt;li&gt;Bonus points if it handles negative weights in a sensible fashion.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I've been unable to find a prescription that allows for weighted events and does the &quot;Bayesian&quot; error intervals. Any clues?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation:&lt;/strong&gt; Say I calculate an efficiency to be nearly 100% in both data and my simulation (Monte Carlo). The MC uses weighted events. Now I'd like to be able to say if they are in agreement with each other or not. I'm posting this here, because this is a actual problem that came up in my research, and I assume it's more likely to find a good answer to it here rather than on e.g. mathoverflow.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here a little survey of the methods for calculating $\Delta \epsilon$. Sorry that I don't give the formulas, but rather the &quot;operative&quot; way of doing it. The derivations can be looked up in the standard literature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Naive:&lt;/strong&gt;&#xA;Use poisson errors, and then gaussian error propagation:&#xA;$ \Delta N_\mathrm{pass} = \sqrt{N_\mathrm{pass}} $&#xA;Disadvantage: Gives clearly wrong errors that extend over 100% and below 0%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Binomial:&lt;/strong&gt;&#xA;We have a binomial distribution, since each event is either fail or pass. So we can calculate binomial errors. In ROOT, this is given by&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;hEff.Divide(hPass, hTotal, 1, 1, &quot;B&quot;);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Advantage: Efficiency $\pm$ Errors constrained to [0, 1].&#xA;Disadvantage: Errors go to zero if $N_\mathrm{pass} = N_\mathrm{total}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bayesian:&lt;/strong&gt;&#xA;Using TGraphAsymmErrors::BayesDivide in ROOT.&#xA;Advantage: Gives assymmetric errors that do not extend over 1 or below 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Binomial, with weights:&lt;/strong&gt;&#xA;For an example see &lt;a href=&quot;http://www.desy.de/~blist/notes/effic.ps.gz&quot; rel=&quot;nofollow&quot;&gt;http://www.desy.de/~blist/notes/effic.ps.gz&lt;/a&gt; .&lt;/p&gt;&#xA;" OwnerUserId="825" LastActivityDate="2011-08-26T00:23:16.433" Title="Calculate bayesian uncertainties on efficiencies, giving weighted events" Tags="&lt;particle-physics&gt;&lt;statistics&gt;" />
  <row Id="13974" PostTypeId="1" CreationDate="2011-08-26T00:53:03.257" Score="2" ViewCount="27" Body="&lt;p&gt;Could someone please explain the details of the following &quot;propaganda plot&quot; from the &lt;a href=&quot;http://www.thphys.uni-heidelberg.de/~plehn/prospino/&quot; rel=&quot;nofollow&quot;&gt;prospino website&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one curve for stop pair production $\tilde t \bar {\tilde t}$, and one for general squark pair production $\tilde q \bar {\tilde q}$. Is the pair production of sbottoms lumped together with the other squarks, or is it missing here? If it is included in the $\tilde q \bar {\tilde q}$ curve, are the cross sections for the non-stop squarks nearly identical, or is this just the inclusive cross section for the production of any heavy squark pair? And finally what model/constraints are assumed in this plot?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/PEacs.png&quot; alt=&quot;Prospino propaganda plot for LHC&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="825" LastActivityDate="2011-08-26T00:53:03.257" Title="Treatment of sbottoms in prospino" Tags="&lt;particle-physics&gt;&lt;supersymmetry&gt;" />
  <row Id="13975" PostTypeId="1" CreationDate="2011-08-26T01:14:18.893" Score="1" ViewCount="60" Body="&lt;p&gt;For EM (freshman level physics) and advanced EM (Junior/Senior level) to help students appreciate the material, I am looking for books/websites that contain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1-applications of electricity and magnetism in real life. That is almost nonexistent in Griffiths, e.g. no bubble jet printers or magnetic tape in magnetism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2-home experiments&lt;/p&gt;&#xA;" OwnerUserId="4521" LastEditorUserId="4521" LastEditorDisplayName="" LastEditDate="2011-08-28T20:40:03.430" LastActivityDate="2011-08-28T21:04:26.523" Title="References for real life applications on advanced EM" Tags="&lt;electromagnetism&gt;&lt;soft-question&gt;&lt;education&gt;&lt;teaching&gt;" AnswerCount="1" />
  <row Id="13976" PostTypeId="2" ParentId="4238" CreationDate="2011-08-26T01:14:54.360" Score="0" ViewCount="" Body="&lt;p&gt;Mass can't be quantized because the contribution of a particle to a system's mass is not a scalar, but a 0 component of a 4-vector, so if you have a system of quantized mass particles, their bound states would not obey mass quantization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In semi-classical gravity, there is a simple reason that charge has to be quantized. If the proton had a charge infinitesimally bigger than the positron, you could make a black hole, throw in some protons, wait for an equal number of positrons to come out in the Hawking radiation, and then let the resulting wee-charged black hole decay while throwing back all the charged stuff that comes out. This would produce a small mass black hole with charge equal to any multiple of the difference, and it could not decay except by undoing the process of formation. This is obviously absurd, so the charge is either quantized or there are particles of arbitrarily small charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, the small charge particles can't be too heavy, since the polarizing field of the black hole with these wee-charges must be strong enough to polarize the horizon to emit them. If their mass is bigger than their charge, then they are net-attracted to the black hole, which causes a constipation for the black hole--- it can't get rid of its charge. So the wee charged particles must be lighter than their mass generically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These types of arguments reproduce the simpler swampland constraints. That out universe is not in the swampland is the only real testable prediction that string theory has made so far (for example, it excludes models where the proton stability is guaranteed by a new unbroken gauge charge).&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-26T05:41:36.047" LastActivityDate="2011-08-26T05:41:36.047" />
  <row Id="13977" PostTypeId="1" AcceptedAnswerId="13979" CreationDate="2011-08-26T01:29:33.203" Score="4" ViewCount="71" Body="&lt;p&gt;From a physics perspective, is there a difference between fog and steam? What is it and how does is manifest in properties of fog vs properties of steam?&lt;/p&gt;&#xA;" OwnerUserId="1264" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-26T01:58:05.460" LastActivityDate="2011-08-26T06:50:28.320" Title="Difference between Steam and Fog / Mist" Tags="&lt;states-of-matter&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="13978" PostTypeId="1" CreationDate="2011-08-26T03:11:02.577" Score="4" ViewCount="54" Body="&lt;p&gt;There are plenty of references to this claim on the internet that tying knots in power cables will prevent a piece of equipment e.g. television or computer from a power surge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can this be debunked (or proven) using mathematics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I stumbled across &lt;a href=&quot;http://forums.anandtech.com/archive/index.php/t-1242379.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; which seems reasonable to me, but is there some way this can be proved?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The surge impedance of any line is the square root of its inductance&#xA;  divided by its capacitance, and electromagnetic waves travel most&#xA;  readily down a line where that surge impedance doesn't change. A point&#xA;  of changing impedance is a discontinuity that causes a partial&#xA;  reflection of the wave back towards its source. As an example, the end&#xA;  of the line is a surge impedance jump to infinity and the whole wave&#xA;  is reflected back (which means the wave voltage at the open end&#xA;  doubles!) This is also the reason why you want to use terminators on&#xA;  the ends of coaxial cables. Open-ended cables will reflect back the&#xA;  signal causing poorer picture quality and ghosting (and similar things&#xA;  happen for poorly made connections that have higher impedances than&#xA;  the surge impedance of the coax).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Knotting the line gives that part of it a higher inductance (think of&#xA;  the knot as a coil with a couple of turns). That means two surge&#xA;  impedance discontinuities (from line to knot, and from knot back to&#xA;  line). It seems to me (too lazy to resort to doing the math) that this&#xA;  is bound to reduce the magnitude (voltage and current) of a surge&#xA;  passing through the knot because some will be reflected back. However,&#xA;  I'd guess that the reduction would be small.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5043" LastActivityDate="2011-08-26T03:11:02.577" Title="Tying knots in power cables to prevent power surge from lightning strikes" Tags="&lt;electricity&gt;" CommentCount="3" />
  <row Id="13979" PostTypeId="2" ParentId="13977" CreationDate="2011-08-26T05:29:05.507" Score="5" ViewCount="" Body="&lt;p&gt;Steam is water in a gas phase, while mist, or fog, are small droplets of water in the condensed phase, but small enough to be kept in the air by thermal Brownian motion. There is a huge difference in their properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;steam at 1 atmospheric pressure is defined to be water at a temperature of 100 degrees celsius or more. It is an invisible gas, and contains a large latent energy which is released when the steam condenses into droplets. It is highly dangerous, since its high energy content can scald very effectively per unit mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mist, on the other hand, is a harmless droplet state with only marginally more energy than bulk water. The extra energy is all due to the surface tension of the droplet formed, and is orders of magnitude smaller than the latent heat required to boil water. To make a mist or a fog, you can charge water electrically, and the electric repulsion will form droplets for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Humidity is water dissolved in air, and it is a different issue. Air will come to statistical equilibrium with liquid water when it has a certain water content, defined as 100% humidity. At any lower level, water will evaporate into the air, and at any higher level, water will spontaneously condense into droplets in the air. &lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-26T06:50:28.320" LastActivityDate="2011-08-26T06:50:28.320" CommentCount="4" />
  <row Id="13980" PostTypeId="1" AcceptedAnswerId="13981" CreationDate="2011-08-26T06:36:11.367" Score="8" ViewCount="182" Body="&lt;p&gt;I am reading up on the Schroedinger equation and I quote&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because the potential is symmetric under $x\to-x$, we expect that there will be solutions of definite parity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could someone kindly explain why this is true? And perhaps also what it means physically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance. :)&lt;/p&gt;&#xA;" OwnerUserId="5046" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-26T06:37:58.577" LastActivityDate="2011-08-26T06:59:52.533" Title="Schroedinger Equation" Tags="&lt;quantum-mechanics&gt;&lt;symmetry&gt;&lt;schrodinger-equation&gt;&lt;parity&gt;" AnswerCount="1" />
  <row Id="13981" PostTypeId="2" ParentId="13980" CreationDate="2011-08-26T06:59:52.533" Score="14" ViewCount="" Body="&lt;p&gt;Good question! First you need to know that &lt;a href=&quot;http://en.wikipedia.org/wiki/Parity_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;parity&lt;/a&gt; refers to the behavior of a physical system, or one of the mathematical functions that describe such a system, under reflection. There are two &quot;kinds&quot; of parity:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $f(x) = f(-x)$, we say the function $f$ has even parity&lt;/li&gt;&#xA;&lt;li&gt;If $f(x) = -f(-x)$, we say the function $f$ has odd parity&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of course, for most functions, neither of those conditions are true, and in that case we would say the function $f$ has &lt;em&gt;indefinite&lt;/em&gt; parity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, have a look at the time-independent Schrödinger equation in 1D:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}\psi(x) + V(x)\psi(x) = E\psi(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and notice what happens when you reflect $x\to -x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}\psi(-x) + V(-x)\psi(-x) = E\psi(-x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a symmetric (even) potential, $V(x) = V(-x)$, this is &lt;em&gt;exactly&lt;/em&gt; the same as the original equation except that we've transformed $\psi(x) \to \psi(-x)$. Since the two functions $\psi(x)$ and $\psi(-x)$ satisfy the same equation, you should get the same solutions for them, except for an overall multiplicative constant; in other words,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi(x) = a\psi(-x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normalizing $\psi$ requires that $|a| = 1$, which leaves two possibilities: $a = +1$ (even parity) and $a = -1$ (odd parity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for what this means physically, it tells you that whenever you have a symmetric potential, you should be able to find a basis of eigenstates which have definite even or odd parity (though I haven't proved that here,* only made it seem reasonable). In practice, you get linear combinations of eigenstates with different parities, so the actual state may not actually be symmetric (or antisymmetric) around the origin, but it does at least tell you that if your potential is symmetric, you &lt;em&gt;could&lt;/em&gt; construct a symmetric (or antisymmetric) state. That's not guaranteed otherwise. You'd probably have to get input from someone else as to what exactly definite-parity states are used for, though, since that's out of my area of expertise (unless you care about parity of elementary particles, which is rather weirder).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;*There is a parity operator $P$ that reverses the orientation of the space: $Pf(x) = f(-x)$. Functions of definite parity are eigenfunctions of this operator. I believe you can demonstrate the existence of a definite-parity eigenbasis by showing that $[H,P] = 0$.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-08-26T06:59:52.533" CommentCount="5" />
  <row Id="13982" PostTypeId="2" ParentId="12627" CreationDate="2011-08-26T07:52:17.377" Score="0" ViewCount="" Body="&lt;p&gt;What you've stumbled upon is called the &quot;Gibbs paradox&quot;, and the resolution is to divide the phase space for entropy calculations in statistical mechanics by the identical particle factor, which reduces the number of configurations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the temperature is unchanged in the process, the momentum distribution of the atoms is unimportant, it is the same before and after, and the entropy is entirely spatial, as you realized. The volume of configuration space for the left part is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${V_1^N \over N!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and for the right part is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${V_2^N\over N!}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the total volume of the 2N particle configuration space is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(V_1V_2)^N\over (N!)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you lift the barrier, you get the spatial volume of configuration space&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(V_1 + V_2)^{2N} \over (2N)!^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $V_1$ and $V_2$ are equal, you naively would expect zero entropy gain. But you do gain a tiny little bit of entropy by removing the wall. Before you removed the wall, the number of particles on the left and on the right were exactly equal, now they can fluctuate a little bit. But this is a negligible amount of extra entropy in the thermodynamic limit, as you can see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${(2V)^{2N}\over (2N)!^2} = {2^{2N}(N!)^2\over (2N)!}{V^{2N}\over (N!)^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that the extra entropy from lifting the barrier is equal to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \log ({(2N)!\over 2^{2N}(N!)^2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might recognize the thing inside the log, it's the probability that a symmetric +/-1 random walk returns to the origin after N steps, i.e. the biggest term of the Pascal triangle at stage 2N when normalized by the sum of all the terms of Pascal's triangle at that stage. From the Brownian motion identity or equivalently, directly from Stirling's formula), you can estimate its size as ${1\over \sqrt{2\pi N}}$, so that the logarithm goes as log(N), it is sub-extensive, and vanishes for large numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The entropy change in the general case is then exactly given by the logarithm of the ratio of the two configuration space volumes before and after:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e^{\Delta S} = { V_1^N V_2^N \over (N!)^2 } { (2N)! \over (V_1 + V_2)^{2N}} = { V_1^N V_2^N \over ({V_1 + V_2 \over 2})^{2N}} {(2N)!\over 2^{2N}(N!)^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring the thermodynamically negligible last factor, the macroscopic change in entropy, the part proprtional to N, is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta S = N\log({4 V_1 V_2 \over (V_1 + V_2)^2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;up to a sign, it is as you calculated.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Additional comments&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;You might think that it is weird to gain a little bit of entropy just from the fact that before you lift the wall you knew that the particle numbers were exactly N, even if that entropy is subextensive. Wouldn't that mean that when you lower the wall, you reduce the entropy a tiny subextensive amount, by preventing mixing of the right and left half? Even if the entropy decrease is tiny, it still violates the second law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no entropy decrease, because when you lower the barrier, you don't know how many molecules are on the left and how many are on the right. If you add the entropy of ignorance to the entropy of the lowered wall system, it exactly removes the subextensive entropy loss. If you try to find out how many molecules are on the right vs how many are on the left, you produce more entropy in the process of learning the answer than you gain from the knowledge.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-26T08:40:03.370" LastActivityDate="2011-08-26T08:40:03.370" CommentCount="18" />
  <row Id="13983" PostTypeId="2" ParentId="1019" CreationDate="2011-08-26T08:36:23.683" Score="0" ViewCount="" Body="&lt;p&gt;Earth revolves around the Sun. It is wrong to say that the Sun revolves around the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FACT: Motion is relative. There is nothing wrong in saying the Sun revolves around the Earth. The former is more frequently mentioned because the Sun (or more accurately, barycenter) is a better inertial frame, and other planets revolve in a near circle around the Sun as well, but in a bizarre way around the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By traveling faster than light in the vacuum one can go back to the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FACT: No explanation needed.&lt;/p&gt;&#xA;" OwnerUserId="3887" LastActivityDate="2011-08-26T08:36:23.683" CommunityOwnedDate="2011-08-26T08:36:23.683" />
  <row Id="13984" PostTypeId="2" ParentId="13945" CreationDate="2011-08-26T08:57:21.093" Score="1" ViewCount="" Body="&lt;p&gt;Expanding on Willie Wong's comment: to have a geometrical compactification, you must assume that geometry is still approximately valid at the string scale, so that you can use supergravity. The arguments for Calabi-Yau compactification were given by Candelas Horowitz Strominger and Witten in 1985. They worked in the supergravity approximation, but the argument was only based on the supersymmetry of the low energy approximation, so the Calabi-Yaus are expected to lift to exact solutions of string theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stringent condition is that there is a low energy supersymmetry that survives compactification. This means that there is a spinor which is covariantly constant, that is, which is unchanged under parallel transport in the compactification manifold. Parallel transport on a 6 dimensional manifold gives SO(6) rotation for each loop going from point x back to x, and SO(6) is SU(4) (p to a double cover, it's just like SU(2) and SO(3)) and if there is a spinor at x which is constant under these rotations, you can make it (1,0,0,0) by doing an SU(4) rotation, and then the only SU(4) rotations which leave it constant are the SU(3) acting on the last three components (this argument looks like it is a miracle of Lie-group algebra, and restricted to six dimensions, but in any even dimension 2n there is an embedding of SU(n) into SO(2n) which just pairs up the real coordinates into complex coordinates--- this is familiar from the reduction of the SO(10) GUT to the SU(5) GUT).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of Calabi Yaus is that their parallel transport on loops is restricted to an SU(3). So the compactification manifolds which preserve exactly one supersymmetry are classified by Calabi Yaus.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-09-03T20:44:51.637" LastActivityDate="2011-09-03T20:44:51.637" />
  <row Id="13985" PostTypeId="1" CreationDate="2011-08-26T09:04:34.330" Score="0" ViewCount="44" Body="&lt;p&gt;we vistited yesterday a fun park and saw a ride which was a big centrifuge only in vertical.&#xA;this one had a speed meter but it was broken and i also wanted to know how much g-force is on there...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so my first thought was to calculate this with the centrifugal-accerleration. I have $\omega = \frac {0.5*2\pi} 1 [\frac {rad} s]$ and $r= 25[m]$, so the accerleration is $a_z = \omega^2r = \pi^2*25 = ~246$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then i can calculate the g force with $g_{force} = \frac {a_z} g$ which gives me about $24g$ that seems fairly too high! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;what am i missing?&lt;/p&gt;&#xA;" OwnerUserId="5047" LastEditorUserId="5047" LastEditorDisplayName="" LastEditDate="2011-08-26T09:49:34.493" LastActivityDate="2011-08-26T21:38:09.363" Title="calculating of g force in fun ride" Tags="&lt;energy&gt;&lt;acceleration&gt;&lt;centrifugal-force&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="13986" PostTypeId="1" AcceptedAnswerId="13988" CreationDate="2011-08-26T11:10:31.770" Score="3" ViewCount="108" Body="&lt;p&gt;Not having studied General Relativity, I have sometimes been puzzled by references to the behaviour for &quot;classic&quot; black holes &amp;mdash; as they are popularly portrayed &amp;mdash; as being true for black holes which are not rotating and which have no charge. I don't understand the role rotation &lt;em&gt;does&lt;/em&gt; play, but at least because it has to do with the motion of massive bodies, I can understand why it &lt;em&gt;could&lt;/em&gt; play an important role. But I have no such intuition for electrical charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does &lt;em&gt;charge&lt;/em&gt; change the behaviour of a black hole, aside from the obvious role of electrostatic force? (It will preferentially attract particles of the opposite charge, obviously.) But it would seem that charge plays a more intriguing role than just this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current status of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Black_hole#Singularity&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page on black holes&lt;/a&gt; claims that you can theoretically avoid the singularity of a charged black hole. It also describes there that there is a theoretical upper bound on the charge/mass ratio of a black hole: that any would-be black hole exceeding it (which is generally thought to be impossible &amp;mdash; see &lt;a href=&quot;http://physics.stackexchange.com/q/6650/4976&quot;&gt;this related question on trying to force saturation of the charge/mass ratio or black holes&lt;/a&gt;) would lack an event horizon (and therefore presumably not be a black hole). Why should that be? Furthermore: from &lt;a href=&quot;http://physics.stackexchange.com/q/9516/4976&quot;&gt;this other related question on repulsion of pairs of charged black holes&lt;/a&gt; (and from Willie Wong's comment, below), ir seems that the size of the event horizon may change depending on how close it is to being extremal! Why would the event horizon of a &lt;em&gt;highly charged&lt;/em&gt; black hole be different than the event horizon of a &lt;em&gt;neutral&lt;/em&gt; black hole, of similar mass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a clear reason for such an interplay is there between electrodynamics (aside from local Lorentz invariance) and general relativity that bring these things about?&lt;/p&gt;&#xA;" OwnerUserId="4976" LastEditorUserId="4976" LastEditorDisplayName="" LastEditDate="2011-08-26T13:37:44.733" LastActivityDate="2011-08-26T18:53:06.477" Title="What role does electrical charge play in black holes?" Tags="&lt;general-relativity&gt;&lt;electromagnetism&gt;&lt;black-holes&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="13987" PostTypeId="2" ParentId="13985" CreationDate="2011-08-26T12:41:40.850" Score="1" ViewCount="" Body="&lt;p&gt;Ok, assuming a passenger on the ride to be traversing a vertical circle then at all times they are experiencing an acceleration $a_{circ}$, this is given by $a_{circ} = a_{track} + a_{gravity}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therfore at the bottom of the circle: $a_{track} = a_{circ}+a_{gravity}$ (with gravity pointing down the two minus signs cancel)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the top we get: $a_{track}=a_{circ}-a_{gravity}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effects of which are we feel a greater push at the bottom than at the top by a difference of $2a_{circ}$'s.  Which might seem counterintuitive at first but seems to fit reality quite well!&lt;/p&gt;&#xA;" OwnerUserId="2535" LastEditorUserId="2535" LastEditorDisplayName="" LastEditDate="2011-08-26T21:38:09.363" LastActivityDate="2011-08-26T21:38:09.363" CommentCount="4" />
  <row Id="13988" PostTypeId="2" ParentId="13986" CreationDate="2011-08-26T14:01:04.427" Score="1" ViewCount="" Body="&lt;ol&gt;&#xA;&lt;li&gt;The ability to &quot;avoid the singularity&quot; is generally regarded as a special property of the very special, stationary, exact solutions we know for black hole space-times. It has to do with the &lt;strong&gt;analytic continuation&lt;/strong&gt; of the solution of Einstein's equations &lt;strong&gt;beyond&lt;/strong&gt; the region that one can predict based on causal principles. (Basically, if you only know that there is a black hole, and you only know what goes on outside of the black hole, you cannot predict what happens inside the black hole; if you actually see the black hole form, on the other hand, you may have a chance at making this prediction.) So one shouldn't take that possibility too seriously. (I for one won't bet my life on it and jump into a charged black hole.) In fact, one interpretation of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmic_censorship_hypothesis#Weak_and_strong_cosmic_censorship_hypothesis&quot; rel=&quot;nofollow&quot;&gt;Strong Cosmic Censorship&lt;/a&gt; conjecture is precisely that for generic black holes, the singularity is unavoidable once you entered the event horizon. &lt;/li&gt;&#xA;&lt;li&gt;The bound on the charge-to-mass ratio is, at the present day, more of an imprecise conjecture than a stated fact. There are several problems with that statement: in a dynamical space-time, mass-energy can radiate. So the definition of the &quot;mass&quot; of a black hole is already problematic. (This is also related to the fact that mass in general relativity cannot be defined locally; though there are a lot of work put into definitions of quasilocal mass.) Similarly, the charge of a black hole, in a general dynamical space-time, is not well-defined. What we do know is that we have a three parameter family of exact, stationary solutions to Einstein's equation depending on $M, A, Q$. Because these solutions are all stationary, the mass and charge are well-defined. Because these solutions are all axisymmetric, angular momentum is well-defined. And &lt;strong&gt;within this family&lt;/strong&gt; we know that were the charge $Q$ to exceed the mass $M$, the &lt;em&gt;formula&lt;/em&gt; that gives the expression of the metric tensor still makes sense as a solution to Einstein's equations, but the formula will lead to solutions with no event horizons. So we conjecture that this is a general fact, despite not knowing how to define the mass and the charge of a generic black hole. Now, there are some cases where this conjecture is known to be true for dynamical solutions. For example, in spherical symmetry, if we also assume that we have, in addition to the electromagnetic field, some other &quot;good&quot; &lt;strong&gt;uncharged&lt;/strong&gt; matter fields (this makes the electromagnetic field non-dynamical, but the gravitational field is still dynamical), then we can prove such a statement using the Hawking mass of the black hole. These types of statements are related to &lt;a href=&quot;http://en.wikipedia.org/wiki/Riemannian_Penrose_inequality&quot; rel=&quot;nofollow&quot;&gt;Penrose-type inequalities&lt;/a&gt;, most of which are conjectural and only a few have been proven to hold generally.  (Remark: there's however &lt;a href=&quot;http://arxiv.org/abs/1107.0949&quot; rel=&quot;nofollow&quot;&gt;evidence that one cannot start with a subextremal black hole and then ``supercharge'' it&lt;/a&gt;.)&lt;/li&gt;&#xA;&lt;li&gt;The horizon area of charged black-holes are smaller than that of uncharged ones. (Again, because of the difficulty in definition, interpret the above in terms of the known &lt;a href=&quot;http://en.wikipedia.org/wiki/Kerr-Newman&quot; rel=&quot;nofollow&quot;&gt;stationary black holes&lt;/a&gt;.) Your last statement in the third paragraph is incorrect. &lt;/li&gt;&#xA;&lt;li&gt;The answer to your general question about the interplay and the mechanism is: &quot;no one really knows&quot;. The problem is that general relativity, unlike classical Newtonian mechanics coupled to electro-magnetism, or even special relativistic mechanics coupled to electro-magnetism, is a highly nonlinear theory. In classical electrodynamics, the linearity of the system allows you to pin-point the contributors to the dynamics: you can say that the total force acting on this particle is a sum of the gravitational force plus the Lorentz force plus this-and-that. In GR, because of the non-linear feedback, it is in general impossible to disentangle the sum of the parts from the whole. (While the equivalence principle tells you that locally in inertial frames stuff behave as it were linear, the sort of questions you were asking necessarily involve long range effects of gravity and electromagnetism.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To summarise: there's still too much we don't know, even with regards to the basic definitions, in general relativity to be able to answer your questions. We don't have a completely satisfactory definition of black holes that can be used locally (as oppose to teleologically), and we don't know what the local definitions of mass or charge should be. We certainly don't have a general description of how black holes should behave under the influence of electric charge. What we do have is a rather limited zoo of examples. This dearth of data points means that a lot of different conjectures can be made to fit those data points It is hard enough to know even &lt;em&gt;which&lt;/em&gt; of those conjectures are right, nevermind to try and understand the principles behind such behaviour. &lt;/p&gt;&#xA;" OwnerUserId="829" LastActivityDate="2011-08-26T14:01:04.427" CommentCount="6" />
  <row Id="13989" PostTypeId="1" AcceptedAnswerId="13994" CreationDate="2011-08-26T14:55:41.333" Score="3" ViewCount="200" Body="&lt;p&gt;If two identical masses are somehow &quot;released&quot; into deep space (that is, they're subject to no other gravitation forces but their own, and are initially at rest to each other). What decides whether they collide or orbit each other? I'm imagining for example two 1&amp;nbsp;kg masses initially at rest, say 10&amp;nbsp;m apart. What happens next?&lt;/p&gt;&#xA;" OwnerUserId="4075" LastEditorUserId="174" LastEditorDisplayName="" LastEditDate="2011-08-27T00:40:26.943" LastActivityDate="2011-08-27T02:06:15.290" Title="Two masses in deep space - collide or orbit?" Tags="&lt;gravity&gt;&lt;orbital-motion&gt;" AnswerCount="5" CommentCount="1" />
  <row Id="13990" PostTypeId="2" ParentId="13989" CreationDate="2011-08-26T15:03:56.717" Score="8" ViewCount="" Body="&lt;p&gt;with no forces perpendicular to the vector connecting them, they will just collide on this line, at a distance proportional to the ratio of their masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to avoid this situation is start with a different tangential velocity.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-26T20:08:34.840" LastActivityDate="2011-08-26T20:08:34.840" CommentCount="1" />
  <row Id="13991" PostTypeId="1" CreationDate="2011-08-26T15:37:00.890" Score="0" ViewCount="50" Body="&lt;p&gt;From the Earth If we observe two galaxies that are diametrically opposed and each 1000ly far from the earth. the separation distance between the galaxies will be 20000ly? really question is: if the galaxies were separated from Earth 10Gly. then the separation distance between galaxies would 20Gyr?&lt;/p&gt;&#xA;" OwnerUserId="5050" LastActivityDate="2011-08-26T16:07:10.687" Title="two distant galaxies seen from earth" Tags="&lt;galaxies&gt;" AnswerCount="1" />
  <row Id="13992" PostTypeId="2" ParentId="13989" CreationDate="2011-08-26T15:43:36.340" Score="-2" ViewCount="" Body="&lt;p&gt;the collision of objects depends on the mass of each and its velocity vector relative to each other.&lt;/p&gt;&#xA;" OwnerUserId="5050" LastActivityDate="2011-08-26T15:43:36.340" />
  <row Id="13993" PostTypeId="2" ParentId="13971" CreationDate="2011-08-26T15:52:44.987" Score="2" ViewCount="" Body="&lt;p&gt;Your equation $W/m\sim N^{2/3}$ gives the correct scaling, but to make this argument work you need to include the constant of proportionality:&#xA;$$&#xD;&#xA;{W\over m}\sim{GM\over R}={G(Nm)\over N^{1/3}r}={Gm\over r}N^{2/3},&#xD;&#xA;$$&#xA;where $r=N^{-1/3}r$ is the typical distance between two protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $Gm/r$ is the gravitational potential energy for a single pair of particles, which is small compared to the electrostatic energy for a pair of particles by precisely the factor $\sim 10^{36}$ you gave. Because of shielding, we can assume that the electrical potential energy per particle is about the same as if there were just a single pair of particles.&#xA;So if we compare gravitational to potential energies, we have&#xA;$$&#xD;&#xA;{(W/m)_{\rm grav}\over (W/m)_{\rm elec}}\sim (10^{-36})N^{2/3}.&#xD;&#xA;$$&#xA;So the two forms of energy are comparable when $N\sim (10^{36})^{3/2}=10^{54}$.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="1241" LastEditorDisplayName="" LastEditDate="2011-08-27T16:12:39.037" LastActivityDate="2011-08-27T16:12:39.037" CommentCount="1" />
  <row Id="13994" PostTypeId="2" ParentId="13989" CreationDate="2011-08-26T15:57:56.067" Score="9" ViewCount="" Body="&lt;p&gt;Due to the law of conservation of angular momentum they will only orbit if there is angular momentum in the initial conditions.  If they start at rest, there is none and so they will collide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we're supposed to provide links and whatnot, here's the wikipedia entry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Orbit&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Orbit&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of the difficulty in answering this question, however, is that there are many different kinds of orbits and the details of the trajectories depends on the relevant masses and linear and angular momentums.&lt;/p&gt;&#xA;" OwnerUserId="5020" LastEditorUserId="5020" LastEditorDisplayName="" LastEditDate="2011-08-26T23:03:52.097" LastActivityDate="2011-08-26T23:03:52.097" CommentCount="1" />
  <row Id="13995" PostTypeId="2" ParentId="13991" CreationDate="2011-08-26T16:07:10.687" Score="2" ViewCount="" Body="&lt;p&gt;For nearby distances, yes, you can just add them up in the usual way. (Of course, if two galaxies are just a few thousand light-years apart, then they're right on top of each other -- galaxies are much bigger than 1000 light-years!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For large distances (i.e., distances comparable to the Hubble length, as in your last example), you have to be careful. The best answer is that there &lt;em&gt;is no&lt;/em&gt; unique, well-defined notion of distance over such large distances. On cosmological distances, spacetime is curved, and what that means is that there are no inertial reference frames covering these large distances. Different people may choose different (non-inertial) reference frames, and as a result they'll disagree about the distance, but no one is necessarily &quot;right.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When people talk about the distance to a faraway galaxy, they most often mean distance as measured in a particular coordinate system, namely comoving coordinates, with distances evaluated at the present (cosmic) time. With that specific definition, the answer to your question is yes: two galaxies that are 10 Gly from earth in opposite directions are 10 Gly from each other. &lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-08-26T16:07:10.687" CommentCount="4" />
  <row Id="13996" PostTypeId="1" AcceptedAnswerId="14171" CreationDate="2011-08-26T17:59:16.913" Score="2" ViewCount="54" Body="&lt;p&gt;This may seem trivial, but I'm having some trouble deriving the finite difference form of the heat equation with a thermal conductivity function $a(x)$ depending on $x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial u(x, t)}{\partial t} = \frac{\partial }{\partial x}[a(x)\frac{\partial u(x, t)}{\partial x}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find some trouble deriving the right hand side. Let $h$ be the space step, and $x_{i−1}=x_i−h$ and $x_{i+1}=x_i+h$. I want to evaluate the right hand side on node $i$. If I take centered differences I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}\left\{\frac{\partial}{\partial x}\left[ a(x)\frac{\partial \, u(x, t)}{\partial x}\right]\right\}_i&#xD;&#xA;&amp;amp;= \frac{\left[a(x)\frac{\partial\, u(x, t)}{\partial x}\right]_{i+1/2} - \left[a(x)\frac{\partial \, u(x, t)}{\partial x}\right]_{i-1/2}}{h} \\&#xD;&#xA;&amp;amp;= \frac{a_{i+1/2}\left[\frac{u_{i+1}-u_{i}}{h}\right] - a_{i-1/2}\left[\frac{u_{i}-u_{i-1}}{h}\right]}{h}&#xD;&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this ok? The conductivities should be in between the nodes I'm calculating? (intuitively it seems this is reasonable). Is there a way, for example by using forward or backward differences, to use conductivity nodes that overlap with the nodes I'm calculating (I calculate only the interger indexed nodes)? Can I mix a centered difference approach for one derivative and a backward or forward approach for the other one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would really appreciate any hint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Federico&lt;/p&gt;&#xA;" OwnerUserId="5053" LastEditorUserId="5053" LastEditorDisplayName="" LastEditDate="2011-08-26T18:05:05.490" LastActivityDate="2011-09-01T09:02:29.973" Title="Finite difference formulation of the heat equation with thermal conductivity in 1D" Tags="&lt;heat&gt;&lt;simulation&gt;&lt;diffusion&gt;&lt;thermal-conductivity&gt;" AnswerCount="3" />
  <row Id="13997" PostTypeId="1" CreationDate="2011-08-26T18:11:09.283" Score="8" ViewCount="63" Body="&lt;p&gt;I'd like some suggestions for good reading materials on the Keldysh formalism in a condmat context. I'm familiar with the imaginary time, coherent state, path integral formalism, but lately I've been seeing Keldysh more and more in papers.&#xA;My understanding is that it is superior to the imaginary time formalism it at least one way: one can evaluate non-equilibrium expectations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="129" LastEditorUserId="955" LastEditorDisplayName="" LastEditDate="2011-08-26T19:03:06.343" LastActivityDate="2011-08-26T19:19:24.100" Title="Good reading on the Keldysh formalism" Tags="&lt;quantum-field-theory&gt;&lt;condensed-matter&gt;&lt;many-body&gt;&lt;field-theory&gt;&lt;keldysh-formalism&gt;" AnswerCount="1" FavoriteCount="4" />
  <row Id="13998" PostTypeId="2" ParentId="13986" CreationDate="2011-08-26T18:19:06.177" Score="1" ViewCount="" Body="&lt;p&gt;This is an area where there is a lot of confusion, and a lot of opinion cannot be supported by sound mathematical theory.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Not-analytic continuation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The theory of black holes requires continuation beyond regions which are accessible. These continuations are often called &quot;analytic continuations&quot;, but this is ridiculous, because the fields do not have to be analytic in order to do it. Hyperbolic equations don't have general analytic solutions, and generally covariant equations can't have solutions which are analytic in non-analytic coordinate systems, which are in no way bad for GR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term I would use is &quot;Minkowski horizon equivalence principle continuation&quot;, because it is continuation past a horizon by assuming that the horizon looks like Minkowski space locally, that is, by assuming the equivalence principle includes horizons too. When you reach a horizon which is locally like a Rindler horizon, like special relativity in accelerating coordinates, so that the curvature does not blow up, then you can continue a little ways past the horizon by extending the solution to be locally flat. In this way, you can continue into the interior of a black hole knowing the outside solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Rindler horizon is vertical sheet in time for the accelerated observer. There is another kind of horizon which is locally a horizontal sheet. The Rindler horizon is formed by the spatial light cone going to the right in a 2d space-time diagram where time is vertical. If you tip this cone by 90 degrees clockwise so that it goes into the past, this is called a Cauchy horizon. A Cauchy horizon is a limit to predictability, because it is where new light rays come from infinity. This Cauchy horizon in question is the limit of the future development of a past pointing half-hyperbola which is just the 90 degree rotated trajectory of an accelearated observer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cauchy horizons are also locally flat, so one may continue past them by assuming the equivalence principle, just as for Rindler horizons. The issue is that when one does this, there is always new information coming in from someplace that you haven't seen before, so it is difficult to predict what happens, at least in classical GR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the full continuation is not in principle any more difficult than the continuation past the event horizon. The equivalence principle should hold at all horizons. But the results are strange, and require interpretation&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Charged black hole solutions&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I will restrict myself to 4 dimensions here, and to classical General Relativity. In string theory, there is a dilaton and the charges are d-form fields, so the &quot;charged&quot; black holes are extended objects in higher dimensions, sometimes with very different interior structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The charged black hole is nice because it is still spherically symmetric, and yet looks like the generic case of an arbitrary rotating black hole. The reason charge and rotation look alike is Kaluza Klein theory--- a charged black hole can be thought of as moving in a small circular fifth dimension. This is similar to rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the solution in GR is just as simple as the Schwartschild solution: for general mass M and charge Q:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ ds^2 = - f(r)dt^2 + {dr^2\over f(r)} + r^2 dS_2^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(r)= 1-{2m\over R} + {Q^2\over R^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important thing is that there are now two radii where f is zero, and these are the two horizons. The outer horizon is continuously deformable to the Schwartschild horizon, and is an Event horizon. The inner horizon is a Cauchy horizon. As M=Q, the black hole is extremal, because the two zeros of f(r) have the same value of r, so you get a double root. Although the two horizons are collided in the r coordinates, they are not at the same point! The radius r breaks down, because the distance between the horizons approaches a finite limit at extremality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $M&amp;lt;Q$ there is no solution to GR without a naked singularity. Such solutions are absurd, they cannot be reached by continuous deformation of the uncharged black hole, and are forbidden by the Censorship conjecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black hole singularity is at r=0, past both horizons, and trajectories in the interior of the Cauchy horizons are repelled by it. So it is impossible for a massive object to hit the singularity. The Cauchy horizon itself is an outgrowth of the singularity, it is the first light rays from the singularitly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure: outer event horizon- inner Cauchy horizon - timelike singularity is generic. It is the situation for rotating/charged/rotating-charged black holes. But it is considered unstable. I will discuss why in the next section, and why I think these arguments are bogus.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Bad arguments that the Cauchy Horizon is the singularity&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;There are, in the literature, arguments that the Cauchy horizon becomes the singularity in realistic solutions. These arguments are hand-wavy, and based on the intuition that everything should be mushed somewhere. In my opinion they are completely wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arguments are based on the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cauchy horizons are unstable: you can see that it is possible to deform the solution to get whatever you like there, because you can start emitting arbitrary gravitational waves from the singularity, and these will go out and influence up to the Cauchy horizon.&lt;/li&gt;&#xA;&lt;li&gt;The Cauchy horizon has a property that it can see the entire future development of the black hole exterior, in principle. When you approach a Cauchy horizon, you see the entire future development of your past, which includes the entire future development of the outer region.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The second point is very important, because it means that photons which come to you from outside are infinitely blue-shifted at the Cauchy horizon, leading people to expect a wall of hard-radiation to appear at this point under generic perturbations, which will make a singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't buy these arguments, because the black hole is not eternal. It is not clear to me that a decaying black hole has the same Cauchy horizon structure as a regular black hole. It is also not clear to me why the singularity can't send out mollifying fields to make the Cauchy horizon &lt;em&gt;non-singular&lt;/em&gt; rather than singular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But most damningly, we have models of near extremal black holes in d-branes, slightly off from extremality, and these are shiny and thermodynamically nearly time-reversible. There are arguments by Gubser that d-branes will absorb other d-branes irreversibly by going off-diagonal in the non-commutative description, that is, by producing strings between the branes, but I don't buy these arguments at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that one should take the story in rotating/charged black holes at face value.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Traversable Cauchy horizon implies you can leave a black hole&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;If you do take the story at face value, objects going into a black hole will come out pretty much unchanged, except for possibly a little singeing when traversing the hard-radiation at the Cauchy horizon. These objects go in, turn around past the Cauchy horizon, and go out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many problems with this picture. The continuation of the solution for a particle which does a full circuit goes past an event horizon, as many cauchy horizons as it wants to (by choosing whether to leave going &quot;up in t&quot; or &quot;down in t&quot; (remember that t is a spatial coordinate inside the black hole) and then out the event horizon going the other way, the way only Hawking radiation can be said to come.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This picture is crazy, because the continuation goes on producing more sheets, just like analytic continuation (but it's not analytic continuation). Each traversing path puts you in another universe, with another direction of time. The result is total nonsense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why people are adament that one has to throw away interior charged solutions. These would naively lead to information loss, other universes, all that nonsense. In my opinion, the correct solution is to do some cutting and pasting and identify the outgoing and incoming paths. I tried a little, but I never found the right pasting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know the pasting, you can predict what will come out of a charged/rotating black hole, mostly, up to a little randomization which becomes more random as the black hole becomes neutral/non-rotating. This type of idea means that black holes are not black at all, not just slightly thermally non-black, but shiny-mirror non-black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This idea has exactly one supporter, which is me. I give a reasonable probability to this possibility because the counterarguments about a singular Cauchy horizon are all weak. Perhaps I should take solace in the observation that that mysterious anonymous author of the Wikipedia article also harbors suspicions, but I somehow don't think his opinion is independent.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Heuristic Justification for Pasting&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The reason I think pasting is necessary is because of the following classical paradox, which I haven't found in the literature. If you throw a mass into a maximally extended charged black hole, and let it go past the horizons and come out, it comes out in another patch. But the black hole mass went up in your patch after absorption, and the mass of the black hole in the other patch went down after emission. But this means that they no longer fit together as continuations of each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when the object comes out of the other sheet black hole, by consistency, it seems that something must come out of your sheet black hole. But the problem is that coming out can happen in either time direction, depending on how the infalling stuff steers itself, whether it decides to come out going towards positive or towards negative t, and so it means that objects can come out time-reflected, and the only way to make sense of that is to do a full identification with CPT gluing, so matter thrown in can come out as reflected motion-reversed antimatter. This is doubly crazy. So I am not sure if it is true, but I think it is. This might explain some astronomical puzzles of anomalous antimatter production because some astrophysical blackholes are measured to rotate at 99% of extremality.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-26T18:53:06.477" LastActivityDate="2011-08-26T18:53:06.477" CommentCount="5" />
  <row Id="13999" PostTypeId="1" CreationDate="2011-08-26T18:28:30.860" Score="2" ViewCount="24" Body="&lt;p&gt;How far apart do scientists estimate was/were the dying star(s) that supplied the elements that comprise our sun, planet, and us? With stars so far apart and expansion of space (as I understand it) carrying things further away still, it would seem to be a low probability occurrence for sufficient quantities of elements blown out from across many light years to accumulate to birth a star system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And another related question, it would seem that when a star went supernova and blew heavier elements in all directions, that would result in mass densities for future star nurseries far away from the dead parent that were much lower than what the parent had, so won't favorable conditions for star births monotonically decrease?&lt;/p&gt;&#xA;" OwnerUserId="4715" LastActivityDate="2011-08-26T18:28:30.860" Title="How did enough material from other dying stars accumulate to start our sun and planets?" Tags="&lt;cosmology&gt;&lt;star&gt;" />
  <row Id="14000" PostTypeId="1" CreationDate="2011-08-26T18:29:23.127" Score="0" ViewCount="61" Body="&lt;p&gt;What do the three wind measurements &quot;with winds forecast to reach xxx mph&quot;, &quot;gusts of xxx mph&quot;, &quot;sustained winds of xxx mph&quot; mean in the following &lt;a href=&quot;http://www.baltimoresun.com/news/weather/bs-md-hurricane-irene-prep-20110825,0,1022221.story&quot; rel=&quot;nofollow&quot;&gt;quote&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Baltimore remains under a Tropical Storm Warning, &lt;strong&gt;with winds forecast&#xA;  to reach 31 to 36 mph&lt;/strong&gt; Saturday night, increasing to 36 to 46 mph, &lt;strong&gt;with&#xA;  gusts to 60 mph&lt;/strong&gt;. Delmarva beach resorts are under a Hurricane Warning,&#xA;  and expecting &lt;strong&gt;sustained winds of 70 to 90 mph&lt;/strong&gt;, with &lt;strong&gt;gusts to 105 mph&lt;/strong&gt;&#xA;  by Saturday night.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also the following tracking information about Hurricane Irene, how are &quot;Moving&quot; and &quot;Max sustained&quot; different?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2:00 PM EDT Fri Aug 26&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Location: 31.2°N 77.5°W&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Max sustained&lt;/strong&gt;: 100 mph&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Moving&lt;/strong&gt;: N at 14 mph&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Min pressure: 951 mb&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="528" LastEditorUserId="528" LastEditorDisplayName="" LastEditDate="2011-08-26T18:53:30.797" LastActivityDate="2011-08-26T18:53:30.797" ClosedDate="2011-08-26T18:53:32.357" Title="What do these wind measurements mean?" Tags="&lt;weather&gt;" CommentCount="4" />
  <row Id="14001" PostTypeId="2" ParentId="13997" CreationDate="2011-08-26T19:10:23.373" Score="10" ViewCount="" Body="&lt;p&gt;I am somewhat biased towards condensed matter physics, even though the subject extends also to fields such as cosmology and QCD. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of condensed matter physics I recommend the following books (even though various techniques also apply outside this regime):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521188008&quot; rel=&quot;nofollow&quot;&gt;Rammer - Quantum Field Theory of Non-Equilibrium States&lt;/a&gt;. My first read on it, and I was quite content with it. If you are familiar with the idea of using periodic imaginary time to simulate a temperature then book will explain the small additional step you need to take to grasp the basics of the Keldysh formalism. Unfortunately it's just formalism for the first 7 (!) chapters and sometimes the pace is a bit slow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B0007DKBI6&quot; rel=&quot;nofollow&quot;&gt;Kadanoff and Baym- Quantum Statistical Mechanics: Green's Function Methods in Equilibrium and Nonequilibrium Problems&lt;/a&gt;. A classic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/354053833X&quot; rel=&quot;nofollow&quot;&gt;Kubo, Toda and Hashitsume - Statistical Physics II: Nonequilibrium Statistical Mechanics&lt;/a&gt;. Has some elements of classical statistical physics as well. The authors are very insightful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also chapter 18 in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/9814273562&quot; rel=&quot;nofollow&quot;&gt;Kleinert&lt;/a&gt; which I find a nice read. This book is huge though and treats a lot of other topics. Still, if you go through Rammer then this chapter by Kleinert nicely summarizes everything without dropping too many details. The newest edition of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521769752&quot; rel=&quot;nofollow&quot;&gt;Altland and Simons&lt;/a&gt; has two chapters on classical and quantum systems out of equilibrium, but I was fairly disappointed with their treatment considering the rest of the book is fantastic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for quantum transport, where this formalism is frequently employed, I can recommend &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521896347&quot; rel=&quot;nofollow&quot;&gt;Di Ventra&lt;/a&gt; as an undergrad-level introductory book and this book by &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521599431&quot; rel=&quot;nofollow&quot;&gt;Datta&lt;/a&gt; for some other interesting topics. &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/9812791620&quot; rel=&quot;nofollow&quot;&gt;Weiss&lt;/a&gt; is excellent for dissipative (open) systems, although this field opens up a whole new can of worms so you might want to avoid at first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other sources not in book form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://link.aps.org/doi/10.1103/RevModPhys.58.323&quot; rel=&quot;nofollow&quot;&gt;Rammer and Smith&lt;/a&gt; Review article. Solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/0901.3586&quot; rel=&quot;nofollow&quot;&gt;Kamenev and Levchenko&lt;/a&gt; is very advanced, but treats some important details. &lt;/p&gt;&#xA;" OwnerUserId="1287" LastEditorUserId="1287" LastEditorDisplayName="" LastEditDate="2011-08-26T19:19:24.100" LastActivityDate="2011-08-26T19:19:24.100" CommentCount="2" />
  <row Id="14002" PostTypeId="2" ParentId="13989" CreationDate="2011-08-26T19:23:28.167" Score="-1" ViewCount="" Body="&lt;p&gt;All the other answers are, by all means, correct, but I want to think about it more practically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All inertial reference frames are equally valid in terms of the laws of physics, and this has an implication that it might be pragmatically difficult to assure that the two objects are perfectly at rest relative to each other when they start out, should they start out a distance far from each other.  Technically no other reference frame is needed in order to tell if the objects are in motion relative to each other or not.  It is possible for either of the objects to tell rather or not they are rotating, so it is possible for one object to tell whether or not the other object has a changing angular position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if the two objects just barely miss each other and pass by they will enter into an elliptical orbit.  Statistically, uncertainty in the movement combined with the cross section of the objects could be interpreted to imply some percentage chance of colliding, where the alternative possibility is to go into orbit.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-08-26T19:23:28.167" CommentCount="1" />
  <row Id="14003" PostTypeId="2" ParentId="13996" CreationDate="2011-08-26T20:51:49.903" Score="0" ViewCount="" Body="&lt;p&gt;I would use the chain rule to get $a$ separate from $\partial u/\partial x$, which will end up as $\partial^2/\partial x^2$. To do the second derivative, your method worked fine.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-08-26T20:51:49.903" />
  <row Id="14004" PostTypeId="1" CreationDate="2011-08-26T21:22:04.627" Score="7" ViewCount="148" Body="&lt;p&gt;The Big Bang theory assumes that our universe started from a very/infinitely dense and extremely/infinitely hot state. But on the other side, it is often claimed that our universe must have been started in a state with very low or even zero entropy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the third law of thermodynamic states that if the entropy of a system approaches a minimum, it's temperature approaches absolut zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how can it be that the beginning universe had a high temperature and a low entropy at the same time? Wouldn't such a state be in contradiction to the third law of thermodynamics?&lt;/p&gt;&#xA;" OwnerUserId="1648" LastActivityDate="2011-09-01T08:45:16.503" Title="How can it be that the beginning universe had a high temperature and a low entropy at the same time?" Tags="&lt;thermodynamics&gt;&lt;cosmology&gt;&lt;temperature&gt;&lt;entropy&gt;&lt;big-bang&gt;" AnswerCount="3" FavoriteCount="2" />
  <row Id="14005" PostTypeId="1" CreationDate="2011-08-26T22:30:50.773" Score="0" ViewCount="16" Body="&lt;p&gt;This is a soft-question, but I was wondering if anybody knows of any official or unofficial repository with all the NASA conceptual artwork. I am also interested in the history of the elaboration of this material and the design principles that go behind them. Any pointers about this topic would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1527" LastEditorUserId="1527" LastEditorDisplayName="" LastEditDate="2011-08-26T22:36:24.663" LastActivityDate="2011-08-26T22:36:24.663" ClosedDate="2011-08-26T22:38:42.803" Title="Repository of NASA concept artwork" Tags="&lt;soft-question&gt;" CommentCount="1" />
  <row Id="14006" PostTypeId="1" AcceptedAnswerId="14017" CreationDate="2011-08-26T22:56:09.343" Score="1" ViewCount="61" Body="&lt;p&gt;I imagine that most people are familiar with the concept of &lt;a href=&quot;http://en.wikipedia.org/wiki/Space_debris&quot; rel=&quot;nofollow&quot;&gt;space junk&lt;/a&gt; (or space debris), which is a hazard to space exploration that grows with every item that humans leave in space.  Wikipedia lists the following types of space junk:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;spent rocket stages &lt;/li&gt;&#xA;&lt;li&gt;defunct satellites &lt;/li&gt;&#xA;&lt;li&gt;erosion, explosion and collision fragments&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm extremely curious, &lt;strong&gt;how much sense would it make to add ejected rocket propellant to this list?&lt;/strong&gt;  To the extent of my understanding, for at least a fraction of the flight the products of the reactions that power a rocket will be left in space and maybe in orbit (?).  In the case of LOX/LH2 fuel, for instance, I believe this ejected material would be water vapor.  Normally this wouldn't be a problem, but if that gas is occupying an orbit, or is suspended as a part of the upper-atmosphere I wonder about detrimental effects such as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Increasing the gas content of near-space would cause greater drag on spacecraft&lt;/li&gt;&#xA;&lt;li&gt;In the case of high specific impulse thrusters, the propellant wouldn't be following the normal thermal distribution... so maybe that could lead to radiation damage&lt;/li&gt;&#xA;&lt;li&gt;Space could be polluted by activities on the ground, due to use of Helium, which escapes the atmosphere (and can thus contribute to increased orbit drag)&lt;/li&gt;&#xA;&lt;li&gt;Couldn't it increase the effective temperature of space?  If you assume a constant growth rate in energy use &lt;a href=&quot;http://www.theoildrum.com/node/8185&quot; rel=&quot;nofollow&quot;&gt;humans will inevitably cook themselves&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It might be easy to dismiss this is negligible, but I find the outcome non-obvious for several (basic physics) reasons.  Firstly, the mass of rocket propellent would probably exceed the mass of other space junk, since it is not reusable and because the ratio of payload to propellant is so poor for many trips.  Although Earth's atmosphere is large, the orbits that this propellant could affect only have an extremely sparse population of gas molecules.  Adding a few drops to a small bucket can still make a difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some velocities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;LOX / LH2 fuel: $v_e = 4,400 m/s$&lt;/li&gt;&#xA;&lt;li&gt;Ion thruster: $v_e = 29,000$&lt;/li&gt;&#xA;&lt;li&gt;LEO orbit speed: $V = 7,800 m/s$&lt;/li&gt;&#xA;&lt;li&gt;Earth and moon escape velocities: $11,200 m/s$, $2,400 m/s$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Similar to how anthropogenic CO2 emissions slowly affected the climate of Earth, could a scale-up of space traffic ultimately lead to problems by stretching out the upper edges of our atmosphere, injecting weird thermal distributions into space, and giving the moon a man-made atmosphere?  Even though the moon can't &quot;hold&quot; an atmosphere, a quick glance at the velocities above reveals that many plausible ejection vectors could leave it bouncing around for a good long time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just so that I don't leave this untouched, &lt;a href=&quot;http://en.wikipedia.org/wiki/Project_Orion_%28nuclear_propulsion%29&quot; rel=&quot;nofollow&quot;&gt;Project Orion&lt;/a&gt; (using nuclear bombs for propellant) seems like it could do the same thing for space that atmospheric testing did for the radiological environment we live in.  The further out you go, the more space you have, but also the higher speed propellant you need, so it seems like this is a potential problem at several stages of space development.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-08-28T02:59:04.890" Title="Is rocket propellant meaningfully a form of &quot;space junk&quot;" Tags="&lt;thermodynamics&gt;&lt;space&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14007" PostTypeId="1" CreationDate="2011-08-27T00:30:08.960" Score="-1" ViewCount="76" Body="&lt;p&gt;Photons are considered as particle or wave depending on the the perspective of the experiment. In almost all situations the wave nature is sufficient.&lt;br&gt;&#xA;Some &lt;a href=&quot;http://en.wikipedia.org/wiki/Photon#Recent_research&quot; rel=&quot;nofollow&quot;&gt;Photon Recent_research&lt;/a&gt; show that the old notion of 'particle' - 'point size' and without 'decay' can be in trouble.&lt;br&gt;&#xA;The PSE user Marty Green wrote this very interesting article &lt;a href=&quot;http://fqxi.org/data/essay-contest-files/Green_There_Are_No_Pea_Shoo.pdf&quot; rel=&quot;nofollow&quot;&gt;There Are No Pea-Shooters for Photons&lt;/a&gt; where he &quot;consider whether the old arguments against the wave theory remain fully justified&quot; -  his viewpoints about: THE BLACK-BODY SPECTRUM, THE PHOTO-ELECTRIC EFFECT and THE COMPTON EFFECT are refreshing.&lt;br&gt;&#xA;This recent article by Matthew Rakher, (&lt;a href=&quot;http://prl.aps.org/abstract/PRL/v107/i8/e083602&quot; rel=&quot;nofollow&quot;&gt;Simultaneous Wavelength Translation and Amplitude Modulation of Single Photons from a Quantum Dot, Physical Review Letters,16 August 2011&lt;/a&gt;),&lt;a href=&quot;http://www.newscientist.com/article/mg21128274.700-photons-made-to-change-colour-and-shape.html&quot; rel=&quot;nofollow&quot;&gt;here reported (Photons made to change colour and shape)&lt;/a&gt; raise the bar. Quoting the news, my bold: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;has made photons behave like shape-shifting chameleons. They piped&#xA;  infrared photons with a wavelength of 1300 nanometres into a crystal,&#xA;  into which they also pumped photons from a 1550-nm-wavelength laser.&#xA;  &lt;strong&gt;Each had different shapes&lt;/strong&gt;.&#xA;  The crystal acted as a waveguide, reflecting and channelling the photons &lt;strong&gt;to hit each other&lt;/strong&gt; at a specific angle and place, making them &lt;strong&gt;blend together&lt;/strong&gt; to form photons with a wavelength of 710 nm with the same shape as the laser photons &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This excludes the absortion/reemission and this behaviour is inconsistent with any particle description, imo.&lt;br&gt;&#xA;In this experiment &lt;a href=&quot;http://people.whitman.edu/~beckmk/papers/papers.html&quot; rel=&quot;nofollow&quot;&gt;Observing the quantum behavior of light in an undergraduate laboratory&lt;/a&gt; is said: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We observe a near absence of coincidence counts between the two detectors—&#xA;  a result &lt;strong&gt;inconsistent with a classical wave model of light&lt;/strong&gt;, but&#xA;  consistent with a quantum description in which individual photons are&#xA;  incident on the beamsplitter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question is: What is the property(s) of the photon that still can not be explained as a wave?&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-08-27T00:30:08.960" ClosedDate="2011-08-28T08:49:58.923" Title="Photon is a particle?" Tags="&lt;quantum-electrodynamics&gt;&lt;duality&gt;&lt;photon&gt;" CommentCount="5" />
  <row Id="14008" PostTypeId="2" ParentId="14004" CreationDate="2011-08-27T01:28:51.577" Score="1" ViewCount="" Body="&lt;p&gt;I had puzzled about that as well.  But while the temperature is high the mass/energy density is extremely uniform (as ilustrated by the uniformity of the cosmic microwave background radiation 380,000 years later in the evolution of the universe).  And gravity changes everything.  Uniform density is very low for systems dominated by gravity (in the relative sense that clumpy distributions have higher gravitational entropy).  So I think this is one way entropy is relatively low.&lt;br&gt;&#xA;Another reason may be that the phase space of micro states grows as the universe expands.&lt;/p&gt;&#xA;" OwnerUserId="5055" LastActivityDate="2011-08-27T01:28:51.577" CommentCount="1" />
  <row Id="14009" PostTypeId="2" ParentId="11078" CreationDate="2011-08-27T01:59:53.100" Score="0" ViewCount="" Body="&lt;p&gt;Consider a binary pulsar with electrostatic charge 1 Coulomb in a constant electric field. As the two stars draw inwards, the potential energy of their interaction decreases their mass. So here is a modification of mass in classical mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a modification of force, consider a charged ball falling through air, and going into pure glycerin. The dielectric constant of the glycerin modifies the force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These examples are ubiquitous, and I don't understand the purpose of this question: Newton's laws are 300 years old, of course 'tHooft's stuff will supersede them, they have been superseded many times already.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-27T01:59:53.100" CommentCount="8" />
  <row Id="14010" PostTypeId="2" ParentId="13989" CreationDate="2011-08-27T02:06:15.290" Score="-1" ViewCount="" Body="&lt;p&gt;Assume: no visible stars, CMB, you on Earth and the Sun. You have a watch that is calibrated by the solar day. To simplify lets assume that the N-S direction of the rotation is normal to the plane of a circular orbit, if it is the case.&#xA;With a &lt;a href=&quot;http://en.wikipedia.org/wiki/Foucault_pendulum&quot; rel=&quot;nofollow&quot;&gt;Foucault pendulum&lt;/a&gt; and assuming that exist invisible distant stars to avoid a Machian problem interpretation, you can say:&lt;br&gt;&#xA;The Earth is rotating (irt the universe, and obtain the sidereal day (first you will have to find a correct latitude- North or South pole that minimize it).  If this one is equal to the solar day then the Earth is not orbiting the Sun and a collision will happen. &#xA;If you have a a microwave antenna you can detect the CMB photons and if you are rotating then you will detect the sidereal day and if you are orbiting the Sun you will detect the direction into which the Earth is moving (the CMB dipole direction makes a 360º full sky turn in each sidereal day modulated by one turn in each solar year).&lt;br&gt;&#xA;In the above experiments the referential in use is the absolute space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At each noon you can measure the angular diameter of the Sun. If it stays constant then you are orbiting, else you have to wait to decide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you dont have instruments the option is 'wait and see'. If the two bodies have angular momentum &gt; 0 then you are orbiting. If a collision happened then it is = 0. How do you define an adequate referential in this situation? It has to be a referential centered in the Earth but you have no way to 'anchor' one axis except in the line Sun-Earth. With this referential you can not discriminate the motion except using the apparent diameter.&lt;/p&gt;&#xA;" OwnerUserId="1257" LastActivityDate="2011-08-27T02:06:15.290" />
  <row Id="14011" PostTypeId="2" ParentId="14004" CreationDate="2011-08-27T03:19:07.660" Score="1" ViewCount="" Body="&lt;p&gt;Well, Entropy of a given system depends on the number of micro states available. This in turn depends on the variation of velocity of particles. At very high temperatures and due to gravitational effect I guess the probability of all the particles having a common velocity increases. Hence the no .of microstates available is quite small. Perhaps this is why it is said to have a low entropy in the beginning.&lt;/p&gt;&#xA;" OwnerUserId="4812" LastActivityDate="2011-08-27T03:19:07.660" />
  <row Id="14012" PostTypeId="1" CreationDate="2011-08-27T03:32:47.210" Score="1" ViewCount="42" Body="&lt;p&gt;Let us consider an ideal gas. Let it be present initially in a state (p1,v1,t1). Now let it be driven to another state (p2,v2,t2). Why is it so that during this process the maximum work can be obtained if the ideal gas is transformed from one state to another by a reversible process.?&lt;/p&gt;&#xA;" OwnerUserId="4812" LastActivityDate="2011-08-27T15:36:17.160" Title="Reversible process" Tags="&lt;thermodynamics&gt;" AnswerCount="1" />
  <row Id="14013" PostTypeId="1" CreationDate="2011-08-27T03:39:41.013" Score="0" ViewCount="23" Body="&lt;p&gt;Let us consider a rod , having a young's modulus,y. Let it be of length ,l and suppose it is suspended from a point P. Let us apply a force ,f, so as to pull the rod at a point Q which is at a distance 2/3 L from the point p. Then next let us apply the same force at the other end point of  the rod. Is the length in both the cases same?&lt;/p&gt;&#xA;" OwnerUserId="4812" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2011-08-27T04:21:17.143" LastActivityDate="2011-08-27T04:25:05.963" Title="Stress and strain" Tags="&lt;classical-mechanics&gt;&lt;elasticity&gt;" AnswerCount="2" />
  <row Id="14014" PostTypeId="2" ParentId="14013" CreationDate="2011-08-27T04:24:33.410" Score="0" ViewCount="" Body="&lt;p&gt;Deflection is $$\delta = \frac{F\,L}{Y\,A}$$ where $F$ is force, $L$ is the loaded length, $Y$ is modulus of elasticity and $A$ is the cross sectional area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the deflection $\delta$ is proportional to the loaded length. In your case a) that is $L=\frac{2}{3}l$, but in case b) I cannot figure out from the wording where points &lt;em&gt;P&lt;/em&gt; and &lt;em&gt;Q&lt;/em&gt; are. Figure out what $L$ is since the rest is the same, you will have your answer.&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-08-27T04:24:33.410" />
  <row Id="14015" PostTypeId="2" ParentId="14013" CreationDate="2011-08-27T04:25:05.963" Score="0" ViewCount="" Body="&lt;p&gt;The base length for the purpose of computing the strain is the &lt;em&gt;length over which the tension/compression acts&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the first case the strain is $\Delta l/(2/3 L) = 3 \Delta l/ 2 L$.&lt;/p&gt;&#xA;" OwnerUserId="520" LastActivityDate="2011-08-27T04:25:05.963" CommentCount="2" />
  <row Id="14016" PostTypeId="1" AcceptedAnswerId="14019" CreationDate="2011-08-27T04:30:43.743" Score="2" ViewCount="57" Body="&lt;p&gt;I know that for close by starts (&amp;lt;50 LY) we can use the parallax effect. And for distant galaxies we use red-shift (&amp;amp; hubble's constant). So how do we measure how far is a star lets say 50,000 LY from the earth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I am missing something, I just don't know what. How can we assume there is a relationship between red-shift and distance when the stars act like a (turbulent) fluid within the galaxy and can be moving in any sort of different manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;It is a physics question, because I really want to know what is the star velocity model within the galaxy in order to use red-shift. My instinct tell me that stars move like a swarm within the galaxy with an overall rotation about the galaxy center of gravity (In agreement to the edited horowitz answer).&lt;/p&gt;&#xA;" OwnerUserId="392" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2011-08-27T14:48:34.810" LastActivityDate="2011-08-27T16:08:56.553" Title="How do you measure distance to stars within the galaxy?" Tags="&lt;astronomy&gt;&lt;doppler-effect&gt;&lt;star&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" />
  <row Id="14017" PostTypeId="2" ParentId="14006" CreationDate="2011-08-27T05:36:06.543" Score="2" ViewCount="" Body="&lt;p&gt;I don't know enough about this topic to give a comprehensive answer, but I can point to some references.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Even though the moon can't &quot;hold&quot; an atmosphere, a quick glance at the&#xA;  velocities above reveals that many plausible ejection vectors could&#xA;  leave it bouncing around for a good long time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Geoffrey Landis wrote an interesting essay about the possible effects of future human activities on the lunar environment: &lt;a href=&quot;http://www.geoffreylandis.com/moonair.html&quot; rel=&quot;nofollow&quot;&gt;&quot;Air Pollution on the Moon&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Similar to how anthropogenic CO2 emissions slowly affected the climate&#xA;  of Earth, could a scale-up of space traffic ultimately lead to&#xA;  problems by stretching out the upper edges of our atmosphere, [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a 1964 paper exploring exactly this question: &lt;a href=&quot;http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1964SSRv....3..275K&amp;amp;db_key=AST&amp;amp;page_ind=0&amp;amp;data_type=GIF&amp;amp;type=SCREEN_VIEW&amp;amp;classic=YES&quot; rel=&quot;nofollow&quot;&gt;&quot;Pollution of the Upper Atmosphere by Rockets&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Just so that I don't leave this untouched, Project Orion (using&#xA;  nuclear bombs for propellant) seems like it could do the same thing&#xA;  for space that atmospheric testing did for the radiological&#xA;  environment we live in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I haven't found technical studies about the residence time of fission products in the Solar System. But, with respect to the Earth's atmosphere, there is a brief analysis starting in page 81 of &lt;a href=&quot;http://ntrs.nasa.gov/search.jsp?R=19770085619&quot; rel=&quot;nofollow&quot;&gt;&quot;Nuclear Pulse Space Vehicle Study, Volume III&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 8/28&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;Nuclear Pulse Space Vehicle Study&quot; landing page doesn't seem accessible at this moment, but the PDF (warning: 8 MB) can be downloaded &lt;a href=&quot;http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19770085619_1977085619.pdf&quot; rel=&quot;nofollow&quot;&gt;from here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1656" LastEditorUserId="1656" LastEditorDisplayName="" LastEditDate="2011-08-28T02:59:04.890" LastActivityDate="2011-08-28T02:59:04.890" />
  <row Id="14018" PostTypeId="2" ParentId="14016" CreationDate="2011-08-27T06:10:18.527" Score="0" ViewCount="" Body="&lt;p&gt;One neat trick for middle ranges requires a dynamically bound system whose components have measurable proper motions. There are a reasonable number of globular clusters that qualify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you project those motions across the sky, they will appear to come together (to some approximation) in two places (one forward and one backward), and the directions of those are the direction of the &lt;em&gt;real&lt;/em&gt; motion. Combine that with the measured line-of-sight velocity (from Doppler shifts of spectral lines), and you know he total velocity and can compute the distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more distant compound objects (clusters and galaxies) still too close to use the cosmological scale (we can't use the Hubble relationship on anything in our local group because the velocities arising from dynamic binding are larger than the cosmological relationship)), one can use Cephied variables and Type 1a supernovae as standard candles. Measuring the distance to clusters and near by galaxies (the Magelenic clouds in particular) for finding distances &lt;em&gt;inside&lt;/em&gt; the galaxy because it increase the total population of star we can use to calibrate the HR-diagram.&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-27T16:08:56.553" LastActivityDate="2011-08-27T16:08:56.553" CommentCount="2" />
  <row Id="14019" PostTypeId="2" ParentId="14016" CreationDate="2011-08-27T06:15:28.070" Score="3" ViewCount="" Body="&lt;p&gt;There are numerous distance indicators used for within the galaxy. The most common way is by using intrinsic magnitude. By knowing how bright an object would be if we were close, we can determine how far away it is by how dim it is. There are many types of stars where we have a rough idea of how bright they should be due to characteristics of the star:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Cephied Variables: The original type of variable star that was used by Hubble to determine the distance to the Andromeda Galaxy. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;RR Lyrae Variable: Like the Cephied variable, but usually dimmer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Type 1a Supernova: These guys, unlike the first two, are cataclismic variables. Essentially a binary neutron star slowly accretes matter from its binary till it reaches the Chandrashankar Limit, after which point it explodes in a very characteristic way (since the mass at the time of explosion is roughly constant). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Main Sequence Stars: Generally less accurate than the first 3, there are some types of main sequence stars which are used to find distances in a similar way.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are a few other ways we can measure distances:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perpendicular Movement: For example there is a &quot;light echo&quot; from SN 1987A which is essentially light from the supernova interacting with dust around the old star. Since this echo should be expanding at the speed of light, we can tell how far away the nova is by the angular velocity of the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relative Velocity in a Moving Cluster: (see dmckee's answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tulley-Fisher relation: A relationship between the luminosity of the galaxy and it's apparent width. Can be used as a decent distance calculator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Faber-Jackson Relation: Similar to Tulley-Fisher, relates luminosity with radial velocity dispersion rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Some more information about redshifts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole relationship between redshift and distance was in fact established by Hubble by relating distance to Cephied variables (I believe) with redshift. Later on it was made more precise using supernova, which are brighter and can be seen from much father away (I think recent supernova can be occasionally seen around Z=5, while Cephieds are all Z&amp;lt;1). Within a galaxy, redshift cannot be used directs since the &quot;peculiar velocity,&quot; the velocity within the galaxy, completely overshadows the effects of universe expansion on which Hubble's Law is based. Redshift within the galaxy is useful for certain other techniques.  &lt;/p&gt;&#xA;" OwnerUserId="3018" LastEditorUserId="3018" LastEditorDisplayName="" LastEditDate="2011-08-27T13:22:19.143" LastActivityDate="2011-08-27T13:22:19.143" CommentCount="5" />
  <row Id="14020" PostTypeId="1" CreationDate="2011-08-27T06:58:31.827" Score="4" ViewCount="131" Body="&lt;p&gt;While thinking about how to answer a &quot;describe string theory&quot; question, I remembered an old argument of Stanley Mandelstam's that linear Regge trajectories implies stability. I never fully understood the exact argument, or its limitations. (The article is &quot;Dual-Resonance Models&quot; from 1974, sciencedirect.com/science/article/pii/0370157374900349, thanks for digging it up) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the argument as I remember it: consider the Regge trajectory function $\alpha(s)$, and expand it in a dispersion relation with two subtractions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \alpha(s) = b + as + {1\over i\pi} \int_0^\infty {\mathrm{Im}(\alpha(s&amp;#39;))\over s-s&amp;#39;} ds&amp;#39;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where a,b are constants. Mandelstam says that the imaginary part of $\alpha(s)$ is a measure of some sort of instability or decay of the string states, so if the string resonances are exactly stable, then the imaginary part is zero, and the trajectory is linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This argument bugged me for these reasons&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It seems to work with no subtractions, with one subtraction, with two subtractions, etc. Can you conclude that exactly constant, exactly linear, exactly quadratic Regge trajectories are also stable? Maybe exactly constant trajectories at special values can be interpreted as free stable point particles of a given spin and mass, but what are quadratic trajectories exactly? Does he mean to argue that nearly linear trajectories are necessarily stable when they are exactly linear? What is the precise conclusion?&lt;/li&gt;&#xA;&lt;li&gt;The Regge trajectory function appears in an exponent in the scattering amplitude, so you have to take a log to extract it. Why you don't get a cut contribution from taking the log of negative values which has nothing to do with physics, just from the Regge scattering ansatz? &lt;/li&gt;&#xA;&lt;li&gt;Actual scattering amplitudes are combinations of different trajectory contributions, so why does each individual trajectory have to be separately analytic, with singularities that are determined by physics? Is this a heuristic assumption?&lt;/li&gt;&#xA;&lt;li&gt;Ok, even if you don't have cuts from the log, the expansion suggests that an imaginary part of the trajectory function, like the imaginary part of a two-point function, has some sort of physical interpretation that allows you to interpret it immediately as some sort of decay rate. What is that physical interpretation? Does it work far away from linear trajectories?&lt;/li&gt;&#xA;&lt;li&gt;How does this argument translate to modern string theory? I have never seen this kind of argument anywhere else.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Mandelstam is still around, but I don't know him, and I think that someone else who is better versed in Regge theory than I am will know the answer immediately, because Mandelstam states it in two lines, without explanation, so it must be obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later Comment: Having received a short cryptic answer from Mandelstam regarding this, I started thinking about it again, and I gave a partial answer below. But I am not really satisfied with it yet. But one of my complaints, though, that separating out the individual trajectory and expanding it by analytic dispersion relations might fail, is not really serious. The trajectory exchange amplitude is the sum of the resonances on the trajectory, so it is always separately analytic. Why it's log should also be analytic to justify the dispersion expansion is probably equally easy to see, but I don't see it&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-30T08:03:00.943" LastActivityDate="2011-08-30T08:03:00.943" Title="What's with Mandelstam's argument that only linear regge trajectories are stable?" Tags="&lt;particle-physics&gt;&lt;string-theory&gt;&lt;regge-theory&gt;&lt;s-matrix&gt;" AnswerCount="1" CommentCount="7" />
  <row Id="14021" PostTypeId="1" AcceptedAnswerId="14051" CreationDate="2011-08-27T08:09:17.650" Score="0" ViewCount="56" Body="&lt;p&gt;I have a rather strange question, let's say I can play with the temperature and atmospheric pressure conditions in a sealed room. What will I need to do to make the air's refraction coefficient &lt;strong&gt;greater then&lt;/strong&gt; the glass' coefficient?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/CsjOu.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4829" LastActivityDate="2011-08-27T23:29:36.703" Title="Air vs Glass refraction coefficient" Tags="&lt;optics&gt;&lt;refraction&gt;" AnswerCount="2" />
  <row Id="14022" PostTypeId="2" ParentId="8076" CreationDate="2011-08-27T08:56:49.400" Score="0" ViewCount="" Body="&lt;p&gt;I always pictured it in the following way, even though this is not what you read on Wikipedia, or in chemistry textbooks: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ATP has three negatively charged phosphate groups, which are highly repulsive, but stuck together by a chemical bond which is too strong for the repulsion to overcome by itself. When ATP binds to protein, the protein breaks one of the phosphates off, and this phosphate pushes hard on the other two as it zooms away, doing mechanical work. This mechanically forces parts of the protein to close or open, providing energy for the reaction. There is nothing more magical than electrostatic forces between phosphate groups, and the forces are mostly electrostatic, not entropic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theoretical values for ATP energy content are always given in terms of the maximum energy that could be theoretically extracted, given the temperature and concentrations, if you run a perfect adiabatic heat engine. But I think this is misleading, because the proteins aren't that sophisticated. They waste a lot of the free energy, because they are just using mechanical forces, they let the outgoing phosphate heat up the water, they aren't doing things particularly adiabatically, and I believe the actual amount of useful work done for a protein by ATP is a fraction of the free energy content, essentially only the electrostatic repulsion potential energy between the $\gamma$ phosphate and the other two, and so it doesn't depend on the temperature or on the concentration.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-27T08:56:49.400" />
  <row Id="14023" PostTypeId="2" ParentId="13164" CreationDate="2011-08-27T09:53:00.010" Score="3" ViewCount="" Body="&lt;p&gt;This is a confusing point, but BebopButUnsteady gets at the main issue. Consider higher dimensions. Then the correlation function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ {1\over k^2 + m^2} = \int e^{-\tau(k^2 + m^2)} d\tau$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the Schwinger representation, the propagating particle interpretation. Each $\tau$ contribution to the propagator is the Fourier transform of a Gaussian of width $\sqrt{\tau}$, which is the probability of a random walk to go from x to y in time $\tau$. This interpretation is correct in all dimensions, and it is the &quot;probability amplitude to go from x to y&quot; interpretation in real time, because the Schwinger proper time Brownian motion path-integral continues to a quantum mechanical free particle path integral amplitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Schwinger representation proves that the Euclidean G(x) is positive, and that $G(x)\rightarrow 0$ at large x, because the random walk probability distribution is positive and long-time gaussians are decayed. For short times, there is always a predictable blow up at short distances, which is the same as the power-law of the massless propagator. For the four dimensional massless propagator $1\over2\pi^2 (x-y)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blow up is a little hard to understand if you think of fields with a lattice action. If you take the fields to fluctuate with an action proportional to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S \propto \sum_{\langle x,y\rangle} (\phi(x) -\phi(y))^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you recover the free field propagator. But then $\langle\phi(x)\phi(x)\rangle$ is the value of $G(0)$, and if you take the continuum limit by choosing the scale of $\phi$ so as to fix $G(0)$, you get no correlations at all, because the scaling inverse-square falloff pushes the continuum $G$ to zero for small lattice spacings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right way to take the continuum limit is to allow $G(0)$ to blow up as $1\over\epsilon^2$, with a constant of proportionality that ensures that the free propagator renormalization constant equal to 1. Then you get a power-law falling propagator with a blow up at coincident points. This rescaling just introduces $\epsilon$ factors into $S$ to recover the usual $\nabla^2$ action. It's no surprise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in two dimensions, there is a surprise. The dependence of the propagator is logarithmic, so that you have a long-distance residue in the correlation function which you see as an additive constant at short distances. The two dimensional correlation function $\langle\phi(x)\phi(y)\rangle$ has a logarithmic dependence on the infrared cutoff, on the radius of the domain you define it on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 2-d lattice free field acquires a shift in the value of the correlation function depending on the size of the domain. To define the continuum limit, you have to rescale the field to make the coefficient of the two-point function 1, and you also need to add a constant to the answer to normalize for the limit of the box going to infinity. The convention for the scale is as in higher dimensions. The convention for the box is to set the correlation function to zero on a sphere of unit radius. You can make the zero point somewhere else with a different convention, and this just adds a constant to the log-correlation function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get rid of this annoying infrared sensitivity, people like to take the derivative of the scalar. I prefer to deal with it, because it is real. The actual correlation function in a box lattice adds back the infrared quantity in such a way that it pushes the correlations back to positive, and rescales them so that you get the right nearest-neighbor value. These positive definite correlations are the ones that have a statistical interpretation in terms of the probability of particle paths. After rescaling and subtraction, you get nonsense correlations, as you noticed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mathematical way you see this story in CFT books is in the somewhat ambiguous expression for the real-space propagator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G(x) = \int {d^2k\over (2\pi)^2} {e^{ikx}\over k^2 + i\epsilon}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The small-x normalization is fixed here, as you can see because there is no large-k divergence, no short distance problem in defining the divergence, because of the oscillating exponential. But there is a serious problem at small k. If you make a k-lattice for understanding the integral, you get a divergence proportional to the log of the k-lattice spacing, which is the infrared problem. You have to subtract out this logarithm to get the usual answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theories with these infrared issues are known as &quot;logarithmic conformal field theories&quot;. The long-range problems are responsible for the Mermin-Wagner-Coleman theorem.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-27T09:53:00.010" />
  <row Id="14024" PostTypeId="1" AcceptedAnswerId="14036" CreationDate="2011-08-27T10:16:15.420" Score="3" ViewCount="82" Body="&lt;p&gt;Why is the magnetic field along the length of a solenoid constant? (preferably in relatively simple terms)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4880" LastActivityDate="2011-08-27T16:05:19.313" Title="Why is the magnetic field inside a solenoid constant?" Tags="&lt;electromagnetism&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="14025" PostTypeId="1" CreationDate="2011-08-27T10:16:52.607" Score="8" ViewCount="145" Body="&lt;p&gt;There was an old argument by Landau that while the liquid gas transition can have a critical point, the solid-liquid transition cannot. This argument says that the solid breaks translational symmetry, and it is impossible to do this in a second order transition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this argument is subtly false. Second order transitions break symmetries, which can be discrete, like in the Ising model, or continuous, like in the x-y model. The reason Landau said it is because it is hard to imagine breaking all the translational and rotational symmetries all at once to make a second order liquid-solid point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But nowadays we know about nematics, and we can imagine the following chain of second-order transitions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;fluid (I)-&gt; fluid with broken x-y-z rotational symmetry with a z-directional order (II)&#xA;-&gt; fluid broken translational symmetry in the same direction&#xA;-&gt; broken x-y direction rotational symmetry in the y-direction&#xA;-&gt; broken y- direction translational symmetry&#xA;-&gt; broken x-direction translational symmetry&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of these transitions can be second order, and together, they can make a solid from a fluid. the question is, how badly does Landau's argument fail.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Are there any two phases which cannot be linked by a second order phase transition?&lt;/li&gt;&#xA;&lt;li&gt;Are there always parameters (perhaps impossible to vary in a physical system) which will allow the second order points to be reached?&lt;/li&gt;&#xA;&lt;li&gt;Is it possible to make the second order transitions collide by varying other parameters, to bring them to one critical point (in the example, a critical point between fluid and solid).&lt;/li&gt;&#xA;&lt;li&gt;Do these critical points exist in any system?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-27T17:44:54.417" Title="Is this Landau's other critical phenomena mistake?" Tags="&lt;thermodynamics&gt;&lt;statistical-mechanics&gt;&lt;symmetry&gt;&lt;phase-transitions&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" />
  <row Id="14026" PostTypeId="1" AcceptedAnswerId="14041" CreationDate="2011-08-27T10:43:18.300" Score="1" ViewCount="52" Body="&lt;p&gt;When stating an estimation of the accuracy of my measurements (of electric current, in this case) would I state that it is equal to the residual error, or the limit of reading? (in this case, ±0.2 or ±0.01). Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is for pretty simple high school physics - so nothing too complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4880" LastActivityDate="2011-08-27T19:13:44.173" Title="Estimation of accuracy of measurements?" Tags="&lt;experimental-physics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14027" PostTypeId="1" AcceptedAnswerId="14029" CreationDate="2011-08-27T11:05:12.337" Score="1" ViewCount="104" Body="&lt;p&gt;it would be great if someone could help me understand the following quote regarding wavefunctions :)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;$$\psi(x)=\sum_n C_nu_n(x)+\int dE C(E)u_E(x)$$ The expression is schematic because we have not specified exactly which form of the continuum eigenfunctions we are supposed to use.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why is it &quot;schematic&quot;? What would the expression look like when we &lt;em&gt;have&lt;/em&gt; specified the &quot;forms&quot; of the eigenfunctions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! :)&lt;/p&gt;&#xA;" OwnerUserId="5046" LastActivityDate="2011-08-27T12:33:55.733" Title="Schematic expression of the Schrodinger equation" Tags="&lt;quantum-mechanics&gt;&lt;schrodinger-equation&gt;&lt;wavefunction&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="14028" PostTypeId="1" AcceptedAnswerId="14037" CreationDate="2011-08-27T11:58:31.423" Score="3" ViewCount="187" Body="&lt;p&gt;I often hear people say that so-and-so [some famous physicist] can &lt;strong&gt;&lt;em&gt;see&lt;/em&gt;&lt;/strong&gt; the solution without calculation. What exactly does this mean? Is it an intuition? But how does one have intuition when it comes to say QM? How might one develop such an intuition?&lt;/p&gt;&#xA;" OwnerUserId="5058" LastActivityDate="2011-08-27T19:50:54.820" Title="&quot;Seeing&quot; the solution" Tags="&lt;soft-question&gt;&lt;intuition&gt;" AnswerCount="3" CommentCount="2" />
  <row Id="14029" PostTypeId="2" ParentId="14027" CreationDate="2011-08-27T11:58:44.907" Score="3" ViewCount="" Body="&lt;p&gt;I think this has to do with the integration measure $dE$ which is the only thing that makes the continuum part of the expansion possibly more &quot;schematic&quot; than the discrete one. In general you expand your wave function in a basis of common eigenstates of a (complete) set of commuting operators, which means that every eigenfunction is uniquely specified by a set of quantum numbers. In the discrete part of the spectrum, you just sum over all different eigenstates, labelled here &quot;schematically&quot; by $n$, while in the continuous part of the spectrum you have to integrate over certain domain of eigenvalues (for energy typically over positive values). The integration measure depends on the choice of the eigenstates as well as on their normalization, that is why it is schematic if you just write it formally as $dE$. For example, for a free spinless particle, you can choose as a basis either simultaneous eigenstates of all three components of momentum (plane waves), or simultaneous eigenstates of the Hamiltonian, square of angular momentum and one of its components (spherical waves). In the former case the quantum numbers are $p_1,p_2,p_3$ and the integration measure becomes $\int{d^3p}/{(2\pi)^3}$, while in the latter case the quantum numbers are $E,l,m$ and the &quot;integration measure&quot; would be $\sum_{l,m}\int dE$, modulo factors depending on the normalization of the eigenfunctions.&lt;/p&gt;&#xA;" OwnerUserId="1641" LastActivityDate="2011-08-27T11:58:44.907" CommentCount="1" />
  <row Id="14030" PostTypeId="2" ParentId="14027" CreationDate="2011-08-27T12:21:31.430" Score="0" ViewCount="" Body="&lt;p&gt;Some eigenstates may be degenerate so it is not always the energy that is a &quot;summation subscript&quot;. In such cases there must be degeneracy factor introduced in the integrand, like $d^3p=4 \pi p^2dp \propto \sqrt{E}dE$ .&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-08-27T12:33:55.733" LastActivityDate="2011-08-27T12:33:55.733" CommentCount="1" />
  <row Id="14031" PostTypeId="2" ParentId="14028" CreationDate="2011-08-27T12:41:57.380" Score="3" ViewCount="" Body="&lt;p&gt;I will share with you a story from a famous physicist, Feynman,  related by him at a workshop back in the 80s.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;During the, I think it was called the Manhattan project, when they were scrambling to create the A bomb, all physicists were employed in doing the laborious cross section calculations, integrals within integrals, that took the best of them ( I think Schwinger and Bethe were there at the time) at least a week to calculate. It was group work, everybody checking the calculations against the others.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Feynman said that he clearly remembered when the idea of the diagrams came to him whole, ( I suppose like Athena from the head of Zeus). He said he remembered his position on his bed, his blue jean legs up the wall.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;He started going to the meetings and reproducing the calculations the others had taken a week to do, overnight. When he got confidence that his method worked he started teasing them, by working on the current problems and coming up with the solutions the next day. He had great fun before he told them his method.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the time Feynman was a trained physicist; in addition he had eidetic memory, books he read he could reread mentally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Building up an intuition in physics does need a lot of perspiration, but the inspiration part should also be there, and surely genetics must play a role in an eidetic memory.&lt;/p&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-08-27T12:41:57.380" CommentCount="8" />
  <row Id="14032" PostTypeId="2" ParentId="14024" CreationDate="2011-08-27T14:56:48.637" Score="1" ViewCount="" Body="&lt;p&gt;remmenber if we assume that the solenoid is long compared to the radius of turns, the field is approximately uniform and parallel to the axis inside the solenoid is zero outside the solenoid. This approach applies the law of Ampère. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xD;&#xA;\int B. dl = \mu_0 .i&#xD;&#xA;$$ &#xA;them to integer arround solenoid we obtein:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ B = \mu_0 N/L .i&#xD;&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="5050" LastActivityDate="2011-08-27T14:56:48.637" />
  <row Id="14033" PostTypeId="2" ParentId="14021" CreationDate="2011-08-27T15:08:32.907" Score="0" ViewCount="" Body="&lt;p&gt;The refractive index is a measure that determines the reduction of the speed of light as it propagates through a homogeneous medium. More precisely, the index of refraction is the phase change per unit length, ie the wave number in the middle (k) is n times larger than the wave number in vacuum (k0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The refractive index of air is 1.00029, while the glass is 1.52 so if you just want the air refractive index as glass is under pressure and temperature variations, you would have to increase therefore the pressure in the room to get the density $ cm ^ 2 $ is equivalent to the density of the glass. but I do not think much of it.&lt;/p&gt;&#xA;" OwnerUserId="5050" LastActivityDate="2011-08-27T15:08:32.907" CommentCount="2" />
  <row Id="14034" PostTypeId="2" ParentId="14012" CreationDate="2011-08-27T15:36:17.160" Score="1" ViewCount="" Body="&lt;p&gt;A reversible process leaves entropy unchanged. Entropy never decreases and so an irreversible process involves an increase in entropy. Increases in entropy absorb work that would otherwise be spent on the environment.&lt;/p&gt;&#xA;" OwnerUserId="1272" LastActivityDate="2011-08-27T15:36:17.160" />
  <row Id="14035" PostTypeId="2" ParentId="14028" CreationDate="2011-08-27T16:04:05.977" Score="2" ViewCount="" Body="&lt;p&gt;It doesn't need to be such complicated stuff – &quot;seeing the solution&quot; can also apply to perfectly simple differential equations. For instance, in&#xA;$$&#xD;&#xA;  \tfrac{\partial^2}{\partial t^2}y = -\eta\cdot y&#xD;&#xA;$$&#xA;every physicist would immediately tell you the solution is $y(t)=\sin(\sqrt{\eta}\cdot t)$, or the corresponding cosine or complex exponential / linear combinations. Even many high school students would say this is pretty obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then, this very equation once turned up in a maths course I was in (functional analysis, not really advanced but not a beginner's lecture either), and when I told them this everybody was rather baffled. They said you can't just do it this way, you rather have to&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;classify this a 2nd order ordinary linear differential equation with constant coefficients&lt;/li&gt;&#xA;&lt;li&gt;make the general ansatz of $y(t)=A\cdot e^{\lambda t}$&lt;/li&gt;&#xA;&lt;li&gt;put it in the equation&lt;/li&gt;&#xA;&lt;li&gt;get $\lambda^2=-\eta$, therefore $\lambda\in\{\pm i\sqrt{\eta}\}$&lt;/li&gt;&#xA;&lt;li&gt;write out these two complex solutions&#xA;$$&#xD;&#xA;     y_1(t) = A_1\cdot e^{- i\sqrt{\eta}\cdot t},\qquad&#xD;&#xA;     y_2(t) = A_2\cdot e^{ i\sqrt{\eta}\cdot t}&#xD;&#xA;   $$&lt;/li&gt;&#xA;&lt;li&gt;try to find, since we're looking for real solutions rather than complex ones, linear combinations of $y_1$ and $y_2$ that are real. And &lt;em&gt;this way&lt;/em&gt; eventually find out that the solution is a sine.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Physicists don't have the time to go through all of these steps every time they encounter this equation (it just comes up &lt;em&gt;far&lt;/em&gt; too often), but they just know the solution. Now extend this to more complicated problems, then there will often still be clever guys who have seen the equation before (or, more often, a sufficiently similar one) so they can immidiately tell you the solution others will take hours to calculate.&lt;/p&gt;&#xA;" OwnerUserId="3540" LastEditorUserId="3540" LastEditorDisplayName="" LastEditDate="2011-08-27T16:21:19.363" LastActivityDate="2011-08-27T16:21:19.363" CommentCount="2" />
  <row Id="14036" PostTypeId="2" ParentId="14024" CreationDate="2011-08-27T16:05:19.313" Score="5" ViewCount="" Body="&lt;p&gt;First, remember that the magnetic field is only uniform for a long solenoid, and even then only if you restrict your attention to regions far away from the ends. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fundamental reason the field doesn't depend on position along the length of the solenoid is that, when you're far from the ends, the magnetic field of a &lt;em&gt;long&lt;/em&gt; solenoid looks almost the same as that of an &lt;em&gt;infinite&lt;/em&gt; solenoid. The field is &quot;mostly&quot; caused by the current relatively near you, so it doesn't matter much if the solenoid extends forever or just for a very long way. For an infinite solenoid, the field has to be uniform as a function of position along the axis, since every point along the length is the same as every other point (if you prefer, the system is translation-invariant along the axis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the more surprising thing about the magnetic field inside a solenoid is not that it's uniform along the length, but that it's uniform in the perpendicular directions -- that is, that the field doesn't depend on whether you're close to the axis or far from it (as long as you're inside it). It'd be easy to imagine the field would either drop off or get stronger as you move perpendicular to the axis, but it doesn't (again, for a long solenoid when you're not near the ends).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deep principle here is Ampere's Law. If you draw any closed loop in space, and &quot;add up&quot; (integrate) the tangential (&quot;forward-pointing&quot;) component of the magnetic field around the loop, the total equals a constant times the amount of current passing through the interior of the loop. Now choose your loop to be a rectangle like, for instance, the red rectangle in the image in &lt;a href=&quot;http://www.pa.msu.edu/courses/2000spring/phy232/lectures/ampereslaw/solenoid.html&quot; rel=&quot;nofollow&quot;&gt;this website&lt;/a&gt;. If you move the upper side of this rectangle up and down (keeping it inside the solenoid), the enclosed current doesn't change, which means the magnetic field strength can't change. &lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-08-27T16:05:19.313" CommentCount="1" />
  <row Id="14037" PostTypeId="2" ParentId="14028" CreationDate="2011-08-27T17:43:27.633" Score="9" ViewCount="" Body="&lt;p&gt;There is nothing spectacular about &quot;seeing&quot; the solution, everyone does it. The reason they tell you this is to try to get you to do it. The famous physicists become famous because they see it for a new problem, where no-one else saw it before, and an important problem. Please, do it too. The purpose is to learn something about nature. Never in the history of the world, has anyone seen something internally that not everyone else could also see. They just saw it first. It does require learning the appropriate background, a short explanation from the seer, and questions/answers, but nowadays pretty much every good physicist can see &lt;em&gt;everything&lt;/em&gt; that Newton saw, and Einstein, and Feynman, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no magic to it, anyone can do it, but it does require a lot of mental work, and a tremendous amount of internal intellectual honesty, so you can catch when you did something wrong. Feynman was very good at it, but equally so was Bethe, and Landau, and Gribov, and many others. Feynman was easily the best at explaining to others how to do it, through his terrific popular works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing that makes Feynman so much more famous than all the others is nothing to do with his physics (which is very great, don't get me wrong). It is his essentially American character, like a physics Walt Disney or Ronald Reagan, he fought alone against the big-shots and nay-sayers. By his own design, he fits into the American media story of the individual genius, standing out from the crowd. American capitalism uses such fables to justify awarding large amounts of wealth to a small number of people, and they need examples of people who actually do something exceptional who people will think are somehow exceptional people, people who deserve to be rich. This is not really Feynman's fault, he doesn't think this way. But I am sure that he understood that the media thinks this way. The bias for capitalist-reinforcing is strong, so much so that communists like Tullio Regge and Shoichi Sakata have been generally excluded from history books. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for &quot;seeing the solution&quot;, the best examples are cases where you can't do it by symbol manipulation. For instance, suppose I give you beads on a line of parallel circular tracks, all connected by springs at tension. I pull the first particle. What happens? Did you set up differential equations, or did you see the waves run down the spring?&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-27T19:50:54.820" LastActivityDate="2011-08-27T19:50:54.820" CommentCount="3" />
  <row Id="14038" PostTypeId="2" ParentId="14025" CreationDate="2011-08-27T17:44:54.417" Score="4" ViewCount="" Body="&lt;p&gt;Good point, I think, but when we look at your progression, from (ii) -&gt; (iii), you break two unrelated symmetries (lattice rotation, then lattice translation). This is fine, and you would have two distinct second order transitions. &lt;em&gt;But&lt;/em&gt; when you try to bring the two critical points together, you correctly notice that this is going to require fine tuning of the parameters. Now, that's not to say such points don't &lt;em&gt;exist.&lt;/em&gt; I think they tend to go under the name &quot;multicritical points,&quot; and I seem to recall work, possibly by Dagotto, arguing that CMR transitions were governed by some sort of bicriticality (not my expertise, unfortunately.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point though, and I think the point Landau is making, is that that sort of fine tuning is unlikely to be relevant for real phase transitions, with only a handful of knobs (That is, you write down the action density and it's something like $a \phi^2+b \psi^2+$interactions etc. but for reasons beyond our control, $a(g)$ and $b(g)$ must both go to zero at the same value of $g$... On the other hand, some folks believe there are &quot;deconfined&quot; &lt;em&gt;quantum&lt;/em&gt; critical points, where no fine-tuning is necessary to drive two unrelated order parameters to zero at the same &quot;tuning&quot; parameter point.)&lt;/p&gt;&#xA;" OwnerUserId="791" LastActivityDate="2011-08-27T17:44:54.417" CommentCount="5" />
  <row Id="14039" PostTypeId="1" CreationDate="2011-08-27T18:14:36.780" Score="1" ViewCount="44" Body="&lt;p&gt;I'm asked to derive a relationship for the leaning angle of a bicycle with the following specs:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Center of gravity for bike and rider is a distance $L$ above the ground when vertical, and the total mass of bike and rider is $M$.&lt;/li&gt;&#xA;&lt;li&gt;The wheel radius is $r$ and the wheel mass is $m$&lt;/li&gt;&#xA;&lt;li&gt;The radius of curvature for the turn is $R$, the speed is $v$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am to show that &#xA;$$\tan \theta = \frac{v^2}{R} \frac{2mr + ML}{MLg}$$&#xA;but I am unsure. If I take $M$ to be the total mass, including the wheels, then I would guess that the condition is that centrifugal force and gravity act in such a way that the total torque is zero, i.e. &#xA;$$M g L \sin \theta = M \frac{v^2}{R} \cos \theta L$.&#xA;But then the mass of the wheels and their radius doesn't enter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, on the other hand, I exclude the wheels from $M$, wouldn't I still have to add them to both sides, i.e. &#xA;$$(MgL + 2mgr) \sin \theta = \frac{v^2}{R} \cos \theta (ML + 2mr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or where does the extra term 2mr come from?&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-08-27T21:21:13.887" Title="Lean angle of a turning bicycle" Tags="&lt;classical-mechanics&gt;&lt;homework&gt;&lt;rotational-dynamics&gt;" AnswerCount="1" />
  <row Id="14040" PostTypeId="1" CreationDate="2011-08-27T18:17:15.733" Score="3" ViewCount="61" Body="&lt;p&gt;This is a question that should have a simple answer, but which I can find no proper discussion of in the literature or on the internet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I start from the assumption that I have a noisy numerical signal with a peak in it - for example amplitude as a function of x, i.e. A(x) - and I wish to determine its full-width at half maximum (FWHM). I am aware of two basic algorithms that may be used on numerical data:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Assume it's Gaussian, determine the r.m.s. value, and multiply by 2.35 ( 2\sqrt{2 \log 2} ).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Determine the FWHM using a peak-finding algorithm that locates the peak A_{max} = A(x_{peak}), then locates the &lt;em&gt;first&lt;/em&gt; positions either side where A(x) falls to 1/2 A_{max}.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Method 1 is numerically robust as you don't have to bin (i.e. smooth) the data that determines A(x). However, if the peak is significantly non-Gaussian in shape you get a systematically wrong answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method 2 is direct, but there are plenty of numerical cases where there are multiple points in x either side of the peak where A(x) crosses 1/2 A_{max}. And of course the number of crossing points depends sensitively on whether I smooth A(x) or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is therefore whether there is a mathematically justifiable algorithm for determining FWHM numerically that doesn't assume the peak is Gaussian,&lt;/p&gt;&#xA;" OwnerUserId="1610" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-27T21:35:58.140" LastActivityDate="2011-08-27T22:25:00.890" Title="Is there a stable numerical algorithm for FWHM that isn't 2.35*sigma?" Tags="&lt;statistics&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="14041" PostTypeId="2" ParentId="14026" CreationDate="2011-08-27T19:13:44.173" Score="1" ViewCount="" Body="&lt;p&gt;It depends on the situation and what the expectation of the teacher is.  If the teacher specifically asked you to record error, then s/he likely provided methodology somewhere along the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are making note of the error because you think it is the right thing to do, I would recommend using the residual error and making note of how you calculated it in your lab report.  There are at least two reasons I would take this route:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If your residual error is an order of magnitude larger than the readability of the device, chances are very good that the experiment is complex enough that there are many sources of error you are overlooking.  Given the accuracy of most lab equipment (even in a high school setting), there are aspects of your experiment that add to your uncertainty more than the equipment readability.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The residual error is likely based on a well recognized phenomenon with an accepted mathematical model. If the standard model says that you should get a certain value, that's what you should get. The chances that you have discovered some strange exception to the recognized physics are remote.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you are interested in high end metrology and would like to learn more, the internationally recognized best practice for calculating and citing uncertainties can be found &lt;a href=&quot;http://www.bipm.org/en/publications/guides/gum.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck with your class!&lt;/p&gt;&#xA;" OwnerUserId="5020" LastActivityDate="2011-08-27T19:13:44.173" CommentCount="1" />
  <row Id="14042" PostTypeId="2" ParentId="13426" CreationDate="2011-08-27T19:22:47.520" Score="0" ViewCount="" Body="&lt;p&gt;The definition of units in the SI are determined by the Bureau International des Poids et Mesures (BIPM). The &lt;strong&gt;official&lt;/strong&gt; statement of the definition of the kilogram is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Le kilogramme est l’unité de masse ; il est égal à la masse du prototype&#xA;  international du kilogramme.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The English translation is unofficial, but it states&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The kilogram is the unit of mass; it is equal to the mass of the international prototype of the kilogram.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, as was stated earlier, there is an object exactly one kilogram in mass.  You can find the BIPM brochure that contains this definition &lt;a href=&quot;http://www.bipm.org/en/si/si_brochure/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5020" LastActivityDate="2011-08-27T19:22:47.520" />
  <row Id="14043" PostTypeId="1" CreationDate="2011-08-27T19:39:55.213" Score="0" ViewCount="57" Body="&lt;p&gt;The problem is this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A particle is represented by the wave function $\psi = e^{-(x-x_{0})^2/2\alpha}\sin kx$. Plot the wave function $\psi$ and the probability distribution $|\psi(x)|^2$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This the problem 2.1 in the book &lt;em&gt;Fundamental University Physics Volume III&lt;/em&gt; by Marcelo Alonso and Edward Finn. The thing is I don't know what values $k, \alpha$ and $x_{0}$ should have. Probably I don't know what $\psi$ really represents in this case.&lt;/p&gt;&#xA;" OwnerUserId="3680" LastActivityDate="2011-08-27T20:31:10.760" Title="Plotting a wave function that represents a particle" Tags="&lt;quantum-mechanics&gt;&lt;simulation&gt;&lt;wavefunction&gt;" AnswerCount="2" />
  <row Id="14044" PostTypeId="1" CreationDate="2011-08-27T19:55:15.027" Score="3" ViewCount="50" Body="&lt;p&gt;If a superconducting magnet and appropriate power supply had just enough current*length so that when it was perpendicular to the earth's magnetic field, the force of the interaction was just enough to excede the force exerted on the object from gravity. AND it was rotating so the angular momentum of the vehicle was just high enough so it wouldn't flip over, would the vehicle fly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the vehicle is a 1000 kg (and the earth's magnetic field is 0.3 gauss) I calculated that with 6.54×10^8 meter amperes you just about reverse the force on the vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assuming a 100 meter diameter, that leaves 6.54×10^6 amps, which is less then the current in a railgun, but still a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the force normal is no longer so normal. It will want to flip the vehicle so the magnet is the other way. Now we would need to spin the vehicle fast enough, so that it has rotated 180 degrees faster then it would take for the force of the magnet to flip the vehicle 180 degrees. How would you go about calculating this part?&lt;/p&gt;&#xA;" OwnerUserId="4287" LastEditorUserId="4287" LastEditorDisplayName="" LastEditDate="2011-08-27T21:04:39.297" LastActivityDate="2011-08-27T21:12:07.043" Title="Can superconducting magnets fly (or repel the earth's core)?" Tags="&lt;electromagnetism&gt;&lt;kinematics&gt;&lt;superconductivity&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" />
  <row Id="14045" PostTypeId="2" ParentId="14043" CreationDate="2011-08-27T20:04:59.923" Score="1" ViewCount="" Body="&lt;p&gt;Then plot first with $x_0=0$, $\alpha=1$, and $k=1$.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2011-08-27T20:04:59.923" />
  <row Id="14046" PostTypeId="2" ParentId="14043" CreationDate="2011-08-27T20:31:10.760" Score="4" ViewCount="" Body="&lt;p&gt;The real purpose of this exercise seems to be exactly &lt;em&gt;finding out&lt;/em&gt; the meaning of $\alpha$, $k$ and $x_0$. So just try modifying them all one by one and see how this affects the wave function. For instance, you will see that a very small $\alpha$ will give you a narrow-peaked distribution, while a big $\alpha$ leads to a widely spread function.&lt;/p&gt;&#xA;" OwnerUserId="3540" LastActivityDate="2011-08-27T20:31:10.760" />
  <row Id="14047" PostTypeId="2" ParentId="14040" CreationDate="2011-08-27T20:55:16.103" Score="5" ViewCount="" Body="&lt;p&gt;Typically one knows the functional form of something when they are interested in the FWHM.  In this case, you can use least-squares fitting to &lt;strong&gt;fit&lt;/strong&gt; the function to the data and extract the parameters.  A very common functional form of a resonance is the Lorentzian:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(A,\gamma,x_0;x) = A\frac{1}{1 + \gamma^2\left(x-x_0\right)^2}$$&#xA;where $1/\gamma$ gives the HWHM.  Sometimes you will have to tailor this to your data a little bit; for instance, while this function goes to zero as $x\to\pm\infty$, your real data will probably have some artificial offset, which you would add a term to account for. Also, of course, you will want to either exclude or simultaneously fit any other nearby resonances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is some code that demonstrates this in Matlab:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;% define the functional form of the resonance:&#xA;lorentzian = @(a, x) a(3)./(1 + a(2)^2*(x - a(1)).^2);&#xA;&#xA;% make a signal and noise:&#xA;x = linspace(0, 100, 101);    &#xA;signal = lorentzian([70 0.1 1], x);&#xA;noise = 0.1*randn(size(signal));    &#xA;y = signal + noise;&#xA;&#xA;% make initial guess of the parameters:&#xA;[m, ii] = max(y);&#xA;a0 = [x(ii) m 1];&#xA;&#xA;% do a least-squares fitting of the lorentzian to the measured data&#xA;a = lsqcurvefit(lorentzian, a0, x, y);&#xA;&#xA;% plot the results&#xA;plot(x, y, 'o');&#xA;hold all&#xA;plot(x, signal, '--');&#xA;plot(x, lorentzian(a, x), '-');&#xA;hold off&#xA;&#xA;legend('data', 'true signal',  'best fit to data', ...&#xA;    'Location', 'Best');&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/uvotG.png&quot; alt=&quot;Fitting a lorentzian&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite a fairly large amount of scatter in the data points, the least squares fit recovers the parameters of the curve quite well:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                          x0       gamma        A&#xA;true value                70         0.1        1&#xA;estimated value           70.33      0.0956     0.97&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-08-27T22:25:00.890" LastActivityDate="2011-08-27T22:25:00.890" CommentCount="1" />
  <row Id="14048" PostTypeId="2" ParentId="14044" CreationDate="2011-08-27T21:12:07.043" Score="-1" ViewCount="" Body="&lt;p&gt;Do you mean a bit like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=A1vyB-O5i6E&quot; rel=&quot;nofollow&quot;&gt;Levitating Frog&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though admittedly this isn't very stable.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-08-27T21:12:07.043" />
  <row Id="14049" PostTypeId="2" ParentId="14039" CreationDate="2011-08-27T21:21:13.887" Score="1" ViewCount="" Body="&lt;p&gt;Aha, I got it. :) Most resources on bicycle physics dismiss gyroscopic effects because those are too weak. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, we need them: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wheels turn with angular velocity $\omega = v/r$ and hence have an angular momentum which we approximately can write as $L = mr^2 \omega$. In cylindrical coordinates, $L$ has a component in $z$-direction which stays constant in time, and a component in $r$ direction which changes with time. Let us look at the radial component:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec{L}_r = mr^2 \omega \cos \theta \vec{e}_r$. The change of it over time involves a torque:&#xA;$$\vec{T} = mr^2 \omega \Omega \cos \theta \vec{e}_r$$&#xA;where $\Omega = v/R$ is the angular frequency of the overall circular path. &#xA;Using this, and $r\omega = v$, we obtain &#xA;$$\vec{T} = mr \frac{v^2}{R} \cos \theta \vec{e}_r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, for the total torque to be $0$, we need &#xA;$$MgL \sin \theta = \frac{v^2}{R}\left( 2mr + ML\right) \cos \theta$$&#xA;which immediately yields the desired result.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2011-08-27T21:21:13.887" />
  <row Id="14050" PostTypeId="1" CreationDate="2011-08-27T22:23:52.687" Score="0" ViewCount="21" Body="&lt;p&gt;How do you calculate the torque of a brushless dc electric motor?&#xA;How do you calculate the torque of a hydraulic motor?&#xA;How do you calculate the push and pull force of a double acting hydraulic cylinder?&lt;/p&gt;&#xA;" OwnerUserId="4144" LastActivityDate="2011-08-27T22:23:52.687" ClosedDate="2011-08-27T22:55:35.943" Title="Calculating torque of actuators?" Tags="&lt;classical-mechanics&gt;&lt;mechanics&gt;&lt;fluid-dynamics&gt;" CommentCount="3" />
  <row Id="14051" PostTypeId="2" ParentId="14021" CreationDate="2011-08-27T23:29:36.703" Score="2" ViewCount="" Body="&lt;p&gt;If you're talking about the index of refraction at a visible wavelength, you can't do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To increase the index of refraction of the air, you need to increase the density.  But it's not so easy to do this indefinitely.  As you increase the density (by increasing the pressure or decreasing the temperature), the air (mostly nitrogen and oxygen) will eventually liquify or solidify.  The index of refraction of liquid/solid nitrogen or oxygen is less than that of glass in the visible (it's around 1.2 for liquid nitrogen).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose if you kept on increasing the pressure you would compress the liquid.  It requires tremendous pressures to compress liquid nitrogen.  Depending on how the material properties changed as the liquid was compressed (which I don't know; I don't know if anyone has studied the properties of liquid nitrogen at thousands of atmospheres of pressure) it's likely that you could get Nair&gt;Nglass.  But it seems unlikely to me that it would be technically feasible on earth.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastActivityDate="2011-08-27T23:29:36.703" CommentCount="2" />
  <row Id="14052" PostTypeId="1" AcceptedAnswerId="14055" CreationDate="2011-08-28T00:31:11.840" Score="2" ViewCount="74" Body="&lt;p&gt;I am on a boat docked at Cape Charles, VA, about 30 or 40 miles from the center of Hurricane Irene. This understandably got me thinking about the force of wind on the boat. Since air friction is proportional to the speed squared (except for some friction types I'm sure someone will be kind enough to remind me of), is the wind's force on the boat also proportional to the wind's speed squared? In other words, will a 70 knot wind produce almost twice the force on the stationary boat as a 50 knot wind, all other factors being equal?&lt;/p&gt;&#xA;" OwnerUserId="4984" LastActivityDate="2011-08-28T09:28:16.697" Title="Is the wind's force on a stationary object proportional to v^2?" Tags="&lt;fluid-dynamics&gt;&lt;air&gt;&lt;friction&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="14054" PostTypeId="2" ParentId="13367" CreationDate="2011-08-28T01:18:08.783" Score="1" ViewCount="" Body="&lt;p&gt;We'll look at just the $aa$ case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ ^2D^a$ by $^2D^a$ $= 1/4[(A|F|A)+(B|F|B)+(C|F|C)+(E|F|E) -2(A|G|B) + 2(A|G|C)$ $ - 2(A|G|E) - 2(B|G|C) + 2(B|G|E) - 2(C|G|E)]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll convert to two-electron operators ( for example, $((2^+,2^-,-2^+)|G|(2^+,1^+,-1^-))$ becomes only $((2^-,-2^+)|G|(1^+,-1^-)$ ). Recall that for $(ab|F|cd) = (ab|F|cd) - (ab|F|dc)$. Mind you, which integrals to actually evaluate depends on the spins, so ++ = ++, -- = --, +-=+- and -+=-+ are non-zero whereas the rest are 0 for each of these two electron-operators. This gives us (not showing diagonal element calculation since you seem to have that under control):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=1/4[3F_0+12F_2-67F_4+3F_0-9F_2+3F_4+3F_0-3F_2-27F_4+3F_0-8F_2+33F_4$ $+$ non-diagonal elements&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= 1/4[12F_0-8F_2-58F_4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ -2((2^-,-2^+)|G|(-1^-,1^+)) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$+2((2^-,-2^+)|G|(1^-,-1^+))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-2((2^-,-2^+)|G|(0^-,0^+))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-2((1^+,-1^-)|G|(-1^+,1^-))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$+2((1^+,-1^-)|G|(0^+,0^-))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-2((1^-,-1^+)|G|(0^-,0^+))]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we evaluate each integral,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$((2^-,-2^+)|G|(-1^-,1^+))=c(2,-1)c(1,-2)=-(35F_4)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$((2^-,-2^+)|G|(1^-,-1^+))=c(2,1)c(-1,-2)=(6F_2+5F_4)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$((2^-,-2^+)|G|(0^-,0^+)) =c(2,0)c(0,-2)=-(4F_2+15F_4)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$((1^+,-1^-)|G|(-1^+,1^-))=c(1,-1)c(1,-1)=-(6F_2+40F_4)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$((1^+,-1^-)|G|(0^+,0^-))=c(1,0)c(0,-1)=(F_2+30F_4)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$((1^-,-1^+)|G|(0^-,0^+))=c(1,0)c(0,-1)=-(F_2+30F_4)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting in these values gives us a grand total of&#xA;$3F_0+7F_2+63F_4$&lt;/p&gt;&#xA;" OwnerUserId="4112" LastEditorUserId="4112" LastEditorDisplayName="" LastEditDate="2011-08-28T01:23:25.590" LastActivityDate="2011-08-28T01:23:25.590" />
  <row Id="14055" PostTypeId="2" ParentId="14052" CreationDate="2011-08-28T01:19:38.850" Score="4" ViewCount="" Body="&lt;p&gt;On the most basic level I would say that both your suspicion and your reasoning are correct, although some caveats should follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I should address this on the basis of simple kinematics.  Yes, the energy of an air molecule is $m v^2$ but we need to formalize when and if this will translate into pressure.  Imagine a pipe that spews a fluid onto a surface, and the fluid comes to a complete rest after impacting the surface.  As speed changes the impulse from a molecule in the fluid increases (and impulse is the more true analog to force/pressure than energy), but so does the flow rate, $\dot{m}$.  We say that area and density stays the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{m} = \rho V A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P=\frac{F}{A}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F=\dot{m} \Delta V = \rho A V^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = \rho V^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you study thermal hydraulics, this will not be the last time you see that final expression.  Not by a long shot.  I should point out that it's not exactly energy &lt;em&gt;per se&lt;/em&gt; that leads to the $V^2$ nature, I would say that it's the above math that leads to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example I provide (much like a garden hose spraying someone holding a shield) is a similar system to that which you describe, which is wind in a hurricane hitting the wall of a house.  There are several ways that these two systems are not identical, and science has done a lot of work to develop empirical correlations and representative models for fluid pressure losses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can look at the system as either a forms loss (on a human-sized scale) or as a boundary layer friction force (on the larger climate level).  Equations for the change in pressure of a fluid due to friction or due to a obstacle (&quot;forms&quot; loss) are as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta P_{friction} = f \frac{L}{D} \frac{\rho V^2}{2}$$&#xA;$$\Delta P_{forms} = K \frac{\rho V^2}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above equations, both $f$ and $K$ represent constants that we multiply the $V^2$ term by to account for all of the real-life complexity of fluid flow.  Yes, these constants will change with the wind speed, so the answer is that the pressure is not &lt;em&gt;perfectly&lt;/em&gt; proportional to $V^2$, but we wouldn't be using the above format if this wasn't the closest form that it does follow.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="4984" LastEditorDisplayName="" LastEditDate="2011-08-28T01:30:19.543" LastActivityDate="2011-08-28T01:30:19.543" />
  <row Id="14056" PostTypeId="1" CreationDate="2011-08-28T01:21:19.113" Score="3" ViewCount="71" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein_Cross&quot; rel=&quot;nofollow&quot;&gt;Einstein's Cross&lt;/a&gt; has been attributed to &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_lens&quot; rel=&quot;nofollow&quot;&gt;gravitational lensing&lt;/a&gt;. However, most examples of gravitational lensing are crescents known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein_ring&quot; rel=&quot;nofollow&quot;&gt;Einstein's rings&lt;/a&gt;. I can easily understand the rings and crescents, but I struggle to comprehend the explanation that gravitational lensing accounts for Einstein's cross. I found &lt;a href=&quot;http://www.astronomyforum.net/physics-forum/112626-mathematical-explanation-einsteins-cross.html&quot; rel=&quot;nofollow&quot;&gt;this explanation&lt;/a&gt;, but it was not satisfactory. &lt;/p&gt;&#xA;" OwnerUserId="2843" LastActivityDate="2011-08-28T14:33:33.107" Title="How does gravitational lensing account for Einstein's Cross?" Tags="&lt;gravity&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="14057" PostTypeId="2" ParentId="14056" CreationDate="2011-08-28T02:31:36.807" Score="1" ViewCount="" Body="&lt;p&gt;Rings only form in perfect alignment. If the alignment is not precise, multiple images can form. Also you are assuming it is a point mass, when actually it is an extended galaxy, creating edge effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the simulation here: &lt;a href=&quot;http://www-ra.phys.utas.edu.au/~jlovell/simlens/&quot; rel=&quot;nofollow&quot;&gt;http://www-ra.phys.utas.edu.au/~jlovell/simlens/&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another article with some examples : &lt;a href=&quot;http://www.aeos.ulg.ac.be/lens_en.php&quot; rel=&quot;nofollow&quot;&gt;http://www.aeos.ulg.ac.be/lens_en.php&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3018" LastActivityDate="2011-08-28T02:31:36.807" CommentCount="1" />
  <row Id="14058" PostTypeId="2" ParentId="7652" CreationDate="2011-08-28T04:56:38.020" Score="0" ViewCount="" Body="&lt;p&gt;You asked for speculation, so here is outlandish speculation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can do engineering all they way down to the atomic scale. The reason is that life is made of atoms, and biology is complicated processes capable of general purpose computation, so atoms can do complicated things. There is no reason that we should be able to do engineering with nuclei, because our universe doesn't embed life in nuclear structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that doesn't mean that you can't try. Suppose you could build a pion laser, and stabilize it by the appropriate methods, by surrounding it with the appropriate reflectors to prevent 2-photon decay, or change to charged pion very quickly to prevent decay, or ... I don't know, or else this wouldn't be speculation. Then you would be able to do nuclear engineering, meaning move nuclei around like you move atoms with light lasers, and pump energy into hadronic states at will, not by collisions, but by engineering. Without good bosonic fields which you can manipulate over relatively large scales, there is no hope for any of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a pion laser, you might be able to blow up elementary particles in a spherically symmetric way, like balloons, and let them collapse onto themselves to make black holes. This requires a controlled implosion, because the mass of a nuclear scale black hole is that of a mountain. If you can somehow make a symmetric implosion which shrinks the pumped up hadron by 10 orders of magnitude, to get some black hole states at some heavy but achievable mass, you can make monopoles, black holes, strings, and turn anything into energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I spent many hours of my youth trying to make this hopelessly futuristic idea work, because it is the only hope we have of investigating the Planck scale by even speculatively achievable technology. I think this type of thing might work out in 200 years, and produce a monopole-driven total mass-energy converter.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-28T04:56:38.020" CommentCount="1" />
  <row Id="14059" PostTypeId="2" ParentId="7652" CreationDate="2011-08-28T06:34:46.097" Score="0" ViewCount="" Body="&lt;p&gt;There &lt;strong&gt;are&lt;/strong&gt; applications of particle theory. Well, it depends on what you call particle theory. There are projects of sensors which use neutrons to detect explosives (google gave me e.g. &lt;a href=&quot;http://www.springerlink.com/content/d20538601275743t/&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;). To develop such a device excellent understanding of particle physics is needed. Neutrino telescopes will be perfect devices to study the Sun and the Earth core soon. Which is useful for weather and cataclysms prediction. You might think that these applications are too engineering, but you never know which tiny details you need, so you have to discover all of them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For astrophysics one of interesting applications I've heard of is using pulsars as a positioning system. Instead of expensive and short-living sattelites which work only at the Earth you may use perfectly stable coordinate system which is Ok for few light years off the home planet. It is not string theory, but just an example of engineering stuff which comes out of pure science in an extremely unexpected way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speaking of science fiction, which you probably had in mind. String theory studies time and space. Right now we may operate (or at least know how to operate) on matter (all conventional technoloies, nuclar stuff, etc), fields (radio) and combination of the two (lasers and similar). The thing which is impossible to control is gravitation. General relativity explains why: gravitation is tightly connected with space and time itself. However, it does not explain all the details. To know the details we will need string theory or its successor/competitor. XX century taught us that you can be never sure that you know everything (see citations from Lord Kelvin). Small but annoying problem of inconsistency between general relativity and quantum mechanics might produce as many new physics as &quot;small problem&quot; with atomic spectra. Or it might not. You never know. &lt;/p&gt;&#xA;" OwnerUserId="4657" LastEditorUserId="4657" LastEditorDisplayName="" LastEditDate="2011-08-28T06:44:09.400" LastActivityDate="2011-08-28T06:44:09.400" />
  <row Id="14060" PostTypeId="2" ParentId="14052" CreationDate="2011-08-28T09:28:16.697" Score="1" ViewCount="" Body="&lt;p&gt;Well, I think of it this way. Imagine that your boat and the wind are the only things present in a giant empty space, and let the wind blow head on onto the bow of your boat at a constant speed. Since in this space there are only the wind and your boat (and yourself on it), you can't conclude if it is the wind blowing onto your boat or that it is your boat moving through a gas of air particles. But more importantly, you can conclude that the two situations are actually equivalent. It is just a matter in which frame of reference you are viewing your situation: a frame stuck to your boat or a frame stuck to the center of mass of the wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You already know that if it was your boat that was moving, the frictional force that would be applied to your boat would go like F ~ v^2. And since this must be the same as a gas blowing onto your boat by the above mentioned arguments, you can conclude that the force applied to your boat by a blowing wind is proportional to it's airspeed squared. I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck with the storm!&lt;/p&gt;&#xA;" OwnerUserId="5064" LastActivityDate="2011-08-28T09:28:16.697" />
  <row Id="14061" PostTypeId="1" AcceptedAnswerId="14073" CreationDate="2011-08-28T12:17:30.430" Score="1" ViewCount="51" Body="&lt;p&gt;I am trying to come up with everyday size objects comparisions of atomic scales items, e.g. if a proton probability cloud was of size basketball how far would the next atoms to it be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;reason being is to give an idea of the space in between the atoms and giving student some idea of relative atomic object sizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking for examples of how does dnsity of materials can be reflected in this comparsion scheme, for example  comapring a dense material to a relatively far less dense material and much closer the basketball sized atomic cloulds would relatively be closer to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS : No models of everyday objects as analogy to atomic structure will be used, only relative distances of the structures is the main objective.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2011-08-28T20:14:11.693" Title="Comparing scales of atomic level objects to scales of everyday size objects" Tags="&lt;measurement&gt;&lt;atoms&gt;&lt;teaching&gt;" AnswerCount="2" CommentCount="4" />
  <row Id="14062" PostTypeId="2" ParentId="14061" CreationDate="2011-08-28T14:27:29.887" Score="1" ViewCount="" Body="&lt;p&gt;may be thats videos could helpyou. only i hope like you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=Cyasy54wDQc&amp;amp;feature=related&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=Cyasy54wDQc&amp;amp;feature=related&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=PqL8JYPaq2M&amp;amp;feature=related&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=PqL8JYPaq2M&amp;amp;feature=related&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5050" LastActivityDate="2011-08-28T14:27:29.887" />
  <row Id="14063" PostTypeId="2" ParentId="14056" CreationDate="2011-08-28T14:33:33.107" Score="-1" ViewCount="" Body="&lt;p&gt;Indeed dear friend, this form is the result of a phenomenon called gravitational lensing. Fortunately, between the earth and a quasar located at 8,000 million light years away is a galaxy 400 million light years. The gravity of the galaxy acts as a huge but imperfect lens heading in different paths from the quasar light, which is like a point, so there are four images that are around the galaxy.&#xA;In this case the gravitational lensing effect produces a symmetrical cross because the lensing galaxy is almost exactly in our line of sight of the quasar. This cross is named after Albert Einstein, whose theory of relativity predicted this phenomenon.&lt;/p&gt;&#xA;" OwnerUserId="5050" LastActivityDate="2011-08-28T14:33:33.107" CommentCount="1" />
  <row Id="14064" PostTypeId="1" AcceptedAnswerId="14066" CreationDate="2011-08-28T15:23:37.913" Score="3" ViewCount="116" Body="&lt;p&gt;I'm trying to find out if black holes could be created by focusing enough light into a small enough volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I have found (any or all may be incorrect):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Maxwell's equations are linear, dictating no interaction of radiation.&lt;/li&gt;&#xA;&lt;li&gt;The Kerr effect and self-focusing has been observed in mediums, but not vacuums.&lt;/li&gt;&#xA;&lt;li&gt;Masses bending light have been observed per general relativity.&lt;/li&gt;&#xA;&lt;li&gt;Photons are said to have no rest mass, just energy and momentum (???).&lt;/li&gt;&#xA;&lt;li&gt;General relativity seems to provide for energy to energy interaction.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This leads to a more specific question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does radiation or energy curve space like mass curves space?&lt;/p&gt;&#xA;" OwnerUserId="5066" LastActivityDate="2011-08-28T16:20:18.550" Title="Can a black hole be formed by radiation?" Tags="&lt;general-relativity&gt;&lt;black-holes&gt;&lt;spacetime&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="14065" PostTypeId="2" ParentId="14064" CreationDate="2011-08-28T15:38:26.820" Score="1" ViewCount="" Body="&lt;p&gt;In general relativity, the quantities that can act as sources of spacetime curvature are the ingredients of the energy momentum tensor.  There is a nice diagram of these ingredients in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since radiation has such attributes (energy, momentum etc), it can act as a source for the gravitational field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  I seem to have omitted the very first part of your question.  See the discussion &lt;a href=&quot;http://physics.stackexchange.com/questions/7771/is-there-an-energy-density-limit-in-gr&quot;&gt;here&lt;/a&gt; about whether or not there is a &quot;critical energy density&quot; which determines black hole formation.&lt;/p&gt;&#xA;" OwnerUserId="3099" LastEditorUserId="3099" LastEditorDisplayName="" LastEditDate="2011-08-28T15:54:20.300" LastActivityDate="2011-08-28T15:54:20.300" />
  <row Id="14066" PostTypeId="2" ParentId="14064" CreationDate="2011-08-28T16:14:47.900" Score="5" ViewCount="" Body="&lt;p&gt;The answer to your first question is &lt;strong&gt;yes&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Building on Demetrios Christodoulou's &lt;a href=&quot;http://arxiv.org/abs/0805.3880&quot; rel=&quot;nofollow&quot;&gt;seminal work&lt;/a&gt; showing that black holes can form &quot;generically&quot; from focusing of &lt;em&gt;gravitational waves&lt;/em&gt; starting from an initial space-time that is arbitrarily close to flat, &lt;a href=&quot;http://arxiv.org/abs/1105.5898&quot; rel=&quot;nofollow&quot;&gt;Pin Yu has recently shown&lt;/a&gt; that one can also dynamically (and generically, in the sense that the formation is stable under small perturbations) form a black hole &lt;strong&gt;starting&lt;/strong&gt; with &lt;strong&gt;only electromagnetic waves&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the interaction between electromagnetism and gravity means that as soon as you set the thing in motion, you will pick up gravitational radiation. And also that since a precise covariant notion of local gravitational energy is not available, the idea that the space-time starts out with only electromagnetic waves is a specific, frame dependent mathematical definition; one should keep that in mind before trying to draw too much physical significance out of the casual statement of the theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your specific second question, the answer is also &lt;strong&gt;yes&lt;/strong&gt;. Einstein's equation specifies that $$ G_{\mu\nu} = T_{\mu\nu}$$&#xA;the left hand side, the Einstein tensor, is purely geometrical, and reflects the curvature of space-time. The right hand side comes from the energy-momentum contributions from the matter fields. The standard way of coupling electromagnetic waves to general relativity (Einstein-Maxwell theory) gives that the right hand side is zero only when the electromagnetic field vanishes. So the content of Einstein-Maxwell theory is based on that electromagnetic radiation can curve space-time.  &lt;/p&gt;&#xA;" OwnerUserId="829" LastEditorUserId="829" LastEditorDisplayName="" LastEditDate="2011-08-28T16:20:18.550" LastActivityDate="2011-08-28T16:20:18.550" CommentCount="1" />
  <row Id="14067" PostTypeId="1" AcceptedAnswerId="14068" CreationDate="2011-08-28T17:23:30.000" Score="1" ViewCount="51" Body="&lt;p&gt;By biot-savart:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{H} = \frac{I}{4\pi} \oint \frac{d\bar{l} \times \bar{r}}{r^{3}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{H} = \frac{I}{2a} \hat{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please, explain the last implication. I cannot find such integral to match the results. The radius of the loop is $a$. The current is $I$. $d\bar{l}$ is a vector along the perimeter.&lt;/p&gt;&#xA;" OwnerUserId="623" LastActivityDate="2011-08-28T19:35:01.203" Title="Magnetic field in the Centre of Circular loop wire with Current?" Tags="&lt;electromagnetism&gt;" AnswerCount="2" FavoriteCount="1" />
  <row Id="14068" PostTypeId="2" ParentId="14067" CreationDate="2011-08-28T17:44:01.127" Score="3" ViewCount="" Body="&lt;p&gt;For the path element $d\vec{l}$ around a circle with radius $a$ you can write $d\vec{l}=rd\phi\vec{e_\phi}$ with $\vec{r}=-r\vec{e_r}$ (note the minus sign, since the vector points &lt;em&gt;from&lt;/em&gt; the wire to the center) you get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{H}=\frac{I}{4\pi}\oint \frac{r^2 (-\vec{e_\phi}\times \vec{e_r})}{r^3}d\phi$$ substituting $a$ for $r$ and integrating $\phi$ from 0 to $2\pi$ and realizing $-\vec{e_\phi}\times\vec{e_r}=\vec{e_z}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{H}=\frac{I}{4\pi}\frac{1}{a}\vec{e_z}\int_0^{2\pi}d\phi = \frac{I}{2a}\vec{e_z}$$&lt;/p&gt;&#xA;" OwnerUserId="2151" LastEditorUserId="2151" LastEditorDisplayName="" LastEditDate="2011-08-28T17:57:07.887" LastActivityDate="2011-08-28T17:57:07.887" />
  <row Id="14069" PostTypeId="2" ParentId="14067" CreationDate="2011-08-28T17:58:23.983" Score="2" ViewCount="" Body="&lt;p&gt;Luksen's answer is 100% correct, but depending on your background and predilections you might want a less formal statement of what's going on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When working out a calculation like this, please, please, please start by drawing a picture. There's really no substitute for this! Here's one I stole from &lt;a href=&quot;http://230nsc1.phy-astr.gsu.edu/hbase/magnetic/curloo.html&quot; rel=&quot;nofollow&quot;&gt;a web site&lt;/a&gt; that does a good enough job (the main difference being that it calls the loop radius $R$ instead of $a$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/0Zrx1.gif&quot; alt=&quot;current loop&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral consists of adding up contributions all the way around the circular loop. For all these contributions, $d\vec l\times\vec r$ is a vector whose magnitude is $a\,dl$ and which points in the perpendicular direction (out of the plane of the diagram). To see this, note that the magnitude of $\vec r$ is always equal to $a$, the directions of $d\vec l$ and &#xA;$\vec r$ are always orthogonal, and the right-hand rule always gives the same perpendicular&#xA;direction. In short, $d\vec l\times\vec r=a\,dl\,\hat z$ where $z$ is the appropriate unit&#xA;vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, note that since the magnitude of $r$ is always $a$, the $r^3$ in the denominator&#xA;of the integrand is just $a^3$. Substitute these facts into the integral, and make use&#xA;of the fact that any constant quantity can be pulled outside of an integral to write&#xA;$$&#xD;&#xA;\vec H={I\over 4\pi}\oint {a\,dl\,\hat z\over a^3}={I\over 4\pi}{\hat z\over a^2}&#xD;&#xA;\oint dl.&#xD;&#xA;$$&#xA;The last step is to observe that $\oint dl=2\pi a$. You can prove this formally by writing things out in polar coordinates or something, or better yet just think about what this integral means. $dl$ is a unit of length around the loop. The integration just means that you're adding up all the $dl$'s. If you add up all the length elements around a loop, you get the circumference.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastEditorUserId="623" LastEditorDisplayName="" LastEditDate="2011-08-28T19:35:01.203" LastActivityDate="2011-08-28T19:35:01.203" CommentCount="4" />
  <row Id="14070" PostTypeId="1" CreationDate="2011-08-28T19:26:06.233" Score="4" ViewCount="66" Body="&lt;p&gt;A friend of mine was telling me about building a cloud chamber while he was in graduate school.  As I understand it, this allows you to &quot;see&quot; interactions caused by high energy particles going through the cloud chamber.  This has fascinated me, and I would like to build one with my daughter, but I want to make sure I am able to explain it to her when the eventual questions come.  Can someone help me out please?  How would I explain a cloud chamber to someone who is a freshman in high school?&lt;/p&gt;&#xA;" OwnerUserId="2715" LastActivityDate="2011-08-28T19:26:06.233" Title="Please explain the physics of a Cloud Chamber" Tags="&lt;particle-physics&gt;&lt;high-energy-physics&gt;" CommentCount="9" FavoriteCount="1" />
  <row Id="14071" PostTypeId="1" CreationDate="2011-08-28T19:28:10.267" Score="2" ViewCount="47" Body="&lt;p&gt;You have a coil over an iron hearth. There is a current in coil which creates the flux $\phi_{1}$. The flux then distributes over the wider area in the iron (using wrong word?): $\phi_{2}$  the flux over the middle and the $\phi_{3}$ over the left.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;            |-----------------|&#xA;            |        |        | &#xA;\phi_{3}    |\phi_{2}|        | COIL HERE &#xA;            |        |        |      the lines are of the iron hearth&#xA;            |-----------------|&#xA;&#xA;                           \phi_{1}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The flux does not disappear so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi_{1} = \phi_{2} + \phi_{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;could some explain the last statement? The last statement is a bit like Kirhoff's law. But I am unsure how can you play with fluxes. Could someone elaborate on this? On which rule, is the statement based on?&lt;/p&gt;&#xA;" OwnerUserId="623" LastEditorUserId="623" LastEditorDisplayName="" LastEditDate="2011-08-28T23:26:33.627" LastActivityDate="2011-08-28T23:41:18.080" Title="Mangnetic Flux  summing up like Kirhoff?" Tags="&lt;electromagnetism&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="14072" PostTypeId="1" AcceptedAnswerId="14075" CreationDate="2011-08-28T19:36:00.157" Score="3" ViewCount="70" Body="&lt;p&gt;I have quite often heard (and even used) the idea that quantum fluctuations are a way to explain the whole &quot;something from nothing&quot; intuitive leap.  I am about to give a talk at a local school on cosmology, and I just want to make sure that I don't misstate anything.  As a layperson type of explanation, would this pass the &quot;close enough to what theory actually states&quot; test?&lt;/p&gt;&#xA;" OwnerUserId="2715" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-28T23:37:23.110" LastActivityDate="2011-08-28T23:37:23.110" Title="Quantum Fluctuations as a model for the Big Bang?" Tags="&lt;quantum-mechanics&gt;&lt;cosmology&gt;&lt;education&gt;" AnswerCount="2" />
  <row Id="14073" PostTypeId="2" ParentId="14061" CreationDate="2011-08-28T20:14:11.693" Score="1" ViewCount="" Body="&lt;p&gt;It's easy enough to look up the sizes of a proton and a typical atom and determine their ratio, then you can use that to compare to real-world objects. For example, hydrogen has a nucleus consisting of a single proton. The radius of the proton is about $1\text{ fm}$, and that of the electron cloud in a hydrogen atom is about $25\text{ pm}$ (though there are some subtleties in the definition of &quot;radius,&quot; which I won't go into here). Thus the hydrogen atom is about 25000 times the size of the proton. So if you're using a basketball, with a $24\text{ cm}$ diameter, to represent the proton, the electron cloud would be represented by something 25000 times larger, or about $6\text{ km}$ across - probably the size of a small town.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One comparison that I've often heard is a kernel of corn to a football field. A kernel of corn is typically about $5-7\text{ mm}$ across (by my estimation), whereas a football field at 100 yards is about 13000 times larger. That's not actually the right ratio for a hydrogen atom, but it does give a sense of the huge scale difference involved.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-08-28T20:14:11.693" />
  <row Id="14074" PostTypeId="1" CreationDate="2011-08-28T20:23:26.827" Score="4" ViewCount="256" Body="&lt;p&gt;This question recently appeared &lt;a href=&quot;http://science.slashdot.org/story/11/08/28/186236/Ask-Slashdot-Math-Curriculum-To-Understand-General-Relativity&quot; rel=&quot;nofollow&quot;&gt;on Slashdot&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Slashdot posts a fair number of physics stories. Many of us, myself included, don't have the background to understand them. So I'd like to ask the Slashdot math/physics community to construct a curriculum that gets me, an average college grad with two semesters of chemistry, one of calculus, and maybe 2-3 applied statistics courses, all the way to understanding the mathematics of general relativity. What would I need to learn, in what order, and what texts should I use? Before I get killed here, I know this isn't a weekend project, but it seems like it could be fun to do in my spare time for the next ... decade.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems like something that would be a good addition to this site: I think it's specific enough to be answerable but still generally useful. The textbook aspect is covered pretty well by &lt;a href=&quot;http://physics.stackexchange.com/questions/12175/book-recommendations&quot;&gt;Book recommendations&lt;/a&gt;, but beyond that: &lt;strong&gt;What college-level subjects in physics and math are prerequisites to studying general relativity in mathematical detail?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-08-29T22:20:29.417" Title="What are the prerequisites to studying general relativity?" Tags="&lt;general-relativity&gt;&lt;soft-question&gt;&lt;education&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="1" />
  <row Id="14075" PostTypeId="2" ParentId="14072" CreationDate="2011-08-28T20:23:32.283" Score="6" ViewCount="" Body="&lt;p&gt;I don't think there has been much actual real research in the idea that quantum fluctuations can create the big bang. I see it more as a philosophical/theological argument than anything else. There is no &quot;real&quot; way to describe quantum fluctuations in non-existant space-time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would advise just keeping your discussion to post - big bang, as otherwise you will be either getting into highly speculative to pseudo-science territory. If some student asks, you can say something like &quot;that is really the next big question, which maybe one of you will figure out over the next few decades!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also of course mentions string theory has the whole colliding branes in the mutliverse, although that of course leads to the question &quot;what created the multiverse?&quot;&lt;/p&gt;&#xA;" OwnerUserId="3018" LastEditorUserId="3018" LastEditorDisplayName="" LastEditDate="2011-08-28T20:29:20.453" LastActivityDate="2011-08-28T20:29:20.453" />
  <row Id="14076" PostTypeId="2" ParentId="14072" CreationDate="2011-08-28T20:31:37.813" Score="4" ViewCount="" Body="&lt;p&gt;When explaining anything with physics, try to avoid questions that have no logical positivist meaning. The question &quot;why is there something rather than nothing&quot;, much like &quot;does a typical neutron prefer Chopin or Bach?&quot; is meaningless, there is no objective way to answer it, because there is no observable consequence of any answer. The only question that science answers are those for which the answer has some observable consequence.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-28T20:31:37.813" CommentCount="1" />
  <row Id="14077" PostTypeId="2" ParentId="14074" CreationDate="2011-08-28T20:37:15.433" Score="6" ViewCount="" Body="&lt;p&gt;The subject is surprisingly self-contained, perhaps mostly because physicists typically learn the physical theory and the associated mathematics simultaneously.  I'd say the main prerequisite is a little &quot;mathematical maturity&quot; and physical intuition. Any exposure to differential geometry and abstract mathematics can only help.  A course in analytical classical mechanics would put you in the right mindset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you've taken college calculus, linear algebra, and have some understanding of special relativity, then I suggest you &lt;a href=&quot;http://preposterousuniverse.com/grnotes/&quot; rel=&quot;nofollow&quot;&gt;jump right in&lt;/a&gt; and ask questions when you encounter difficulties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Colin's &lt;a href=&quot;http://physics.stackexchange.com/questions/14074/what-are-the-prerequisites-to-studying-general-relativity/14087#14087&quot;&gt;excellent answer&lt;/a&gt; is right on target in describing a more thorough background.  Just don't be too shy to dig into Sean Carroll's notes immediately and see how much you understand and how much you need to review.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-08-28T22:06:53.273" LastActivityDate="2011-08-28T22:06:53.273" CommentCount="1" />
  <row Id="14078" PostTypeId="1" CreationDate="2011-08-28T20:40:38.250" Score="0" ViewCount="60" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_circuital_law&quot; rel=&quot;nofollow&quot;&gt;Ampere's Ciruital Law&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/lBBiE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider two straight wires, each carrying current &lt;code&gt;I&lt;/code&gt;, one of infinite length and another of finite length &lt;code&gt;l&lt;/code&gt;. If you need to find out magnetic field because of each, at a point (X) whose perpendicular distance from wire is  &lt;code&gt;d&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You get magnetic field as $\frac{\mu I}{2 \pi d}$. Same for both. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnetic field due to infinitely long wire is : $\frac{\mu I}{2 \pi d}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnetic field due to wire of finite length &lt;code&gt;l&lt;/code&gt; : $\frac{\mu I (\sin(P)+\sin(Q)) }{2 \pi d}$, where P &amp;amp; Q are the angles subtended at the point by the ends of the wire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are we getting wrong value for using Ampere's circuit law?&lt;/p&gt;&#xA;" OwnerUserId="447" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-31T16:33:40.030" LastActivityDate="2011-08-31T16:43:37.557" Title="Using Ampere's circuital law for an infinitely long wire &amp; wire of given length" Tags="&lt;electromagnetism&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14079" PostTypeId="1" CreationDate="2011-08-28T20:45:54.077" Score="0" ViewCount="54" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/506/why-doesnt-the-bike-fall-if-going-with-a-high-speed&quot;&gt;Why doesn&amp;#39;t the bike fall if going with a high speed?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Why is it that, when we ride bicycle at good speed, it is perfectly balanced and we don't fall. But we lose balance when we ride bicycle at lower speeds.&lt;/p&gt;&#xA;" OwnerUserId="447" LastActivityDate="2011-08-28T20:45:54.077" ClosedDate="2011-08-28T21:19:14.723" Title="Why do we fall when we ride bicycle slowly but we don't when riding it fastly" Tags="&lt;general-physics&gt;" CommentCount="4" />
  <row Id="14080" PostTypeId="1" AcceptedAnswerId="14091" CreationDate="2011-08-28T20:50:00.153" Score="2" ViewCount="41" Body="&lt;p&gt;Consider an imaginary vertical plane. Now say, a body is falling freely (under earth's attraction). If you consider any axis that is perpendicular to that plane. We get a non-zero value for torque. Then why is it that body is not moving in a circular path about that axis?&lt;/p&gt;&#xA;" OwnerUserId="447" LastActivityDate="2011-08-28T23:07:08.230" Title="Confusion with the torque" Tags="&lt;classical-mechanics&gt;" AnswerCount="1" CommentCount="1" />
  <row Id="14081" PostTypeId="2" ParentId="679" CreationDate="2011-08-28T20:57:09.830" Score="0" ViewCount="" Body="&lt;p&gt;My main piece of advice to someone asking this question is: &lt;strong&gt;don't skip the basics&lt;/strong&gt;.  It's very tempting to try to skip right to something sexy: field theory, general relativity, string theory, etc.  Avoid this temptation and take a slow, methodological approach. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you say you have a good background in physics, however, you might be able to pick up &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3527406018&quot; rel=&quot;nofollow&quot;&gt;David Griffiths' undergraduate particle physics textbook&lt;/a&gt; immediately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were starting out from zero, I would suggest: spend six months or a year reading the three-volume set of the Feynman lectures and working out the problems.  Spend six months each working through Griffiths' quantum mechanics textbook, an analytical mechanics textbook, a thermal physics textbook, and an E&amp;amp;M textbook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something that works well for me, and as a computer scientist, might work well for you, is to try to simulate physical scenarios numerically in addition to solving them analytically (as in the textbooks).  Working with a numerical model and seeing how it agrees or disagrees with the formal theory is often quite illuminating.&lt;/p&gt;&#xA;" OwnerUserId="29" LastActivityDate="2011-08-28T20:57:09.830" />
  <row Id="14082" PostTypeId="1" AcceptedAnswerId="14084" CreationDate="2011-08-28T20:58:15.023" Score="5" ViewCount="297" Body="&lt;p&gt;I get the physical significance of vector addition &amp;amp; subtraction. But I don't understand what do dot &amp;amp; cross products mean? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;why is it that dot product of vectors &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; is defined as &lt;code&gt;A B Cos(theta)&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;why is it that cross product of vectors &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; is defined as &lt;code&gt;A B Sin(theta) N-cap&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To me, both these formulae seem to be arbitrarily defined (Although, I know that it definitely wouldn't be the case). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Cross product could be defined arbitrarily, why can't we define division of vectors? What's wrong with that? Why can't vectors be devided?&lt;/p&gt;&#xA;" OwnerUserId="447" LastActivityDate="2011-08-29T07:21:55.843" Title="What is the physical significance of dot &amp; cross product of vectors? Why is division not defined for vectors?" Tags="&lt;vectors&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="2" />
  <row Id="14083" PostTypeId="2" ParentId="13975" CreationDate="2011-08-28T21:04:26.523" Score="0" ViewCount="" Body="&lt;p&gt;For the optics aspects of the courses, I suggest the book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1441931694&quot; rel=&quot;nofollow&quot;&gt;Light Science: Physics and the Visual Arts&lt;/a&gt;. It's very real-world-oriented, with tons and tons of home experiments. Again, this is just optics, not the rest of electromagnetism.&lt;/p&gt;&#xA;" OwnerUserId="3811" LastActivityDate="2011-08-28T21:04:26.523" />
  <row Id="14084" PostTypeId="2" ParentId="14082" CreationDate="2011-08-28T21:07:03.823" Score="8" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I get the physical significance of vector addition &amp;amp; subtraction. But I don't understand what do dot &amp;amp; cross products mean?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Perhaps you would find the geometric interpretations of the dot and cross products more intuitive:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dot product of A and B is the length of the &lt;em&gt;projection&lt;/em&gt; of A onto B multiplied by the length of B (or the other way around--it's commutative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnitude of the cross product is the &lt;em&gt;area of the parallelogram with two sides A and B&lt;/em&gt;.  The orientation of the cross product is orthogonal to the plane containing this parallelogram. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why can't vectors be divided?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How would you define the inverse of a vector such that $\mathbf{v} \times \mathbf{v}^{-1} = \mathbf{1}$?  What would be the &quot;identity vector&quot; $\mathbf{1}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the answer is &lt;em&gt;sometimes you can&lt;/em&gt;.  In particular, in two dimensions, you can make a correspondence between vectors and complex numbers, where the real and imaginary parts of the complex number give the (x,y) coordinates of the vector.  Division is well-defined for the complex numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross-product only exists in 3D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Division is defined in some higher-dimensional spaces too (such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quaternions&quot; rel=&quot;nofollow&quot;&gt;quaternions&lt;/a&gt;), but only if you give up commutativity and/or associativity.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's an illustration of the geometric meanings of dot and cross product, from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dot_product&quot; rel=&quot;nofollow&quot;&gt;wikipedia article for dot product&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Cross_product#Geometric_meaning&quot; rel=&quot;nofollow&quot;&gt;wikipedia article for cross product&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/mMgJm.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;http://i.stack.imgur.com/hYgvO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditorDisplayName="" LastEditDate="2011-08-28T21:26:16.570" LastActivityDate="2011-08-28T21:26:16.570" CommentCount="2" />
  <row Id="14085" PostTypeId="2" ParentId="14082" CreationDate="2011-08-28T21:15:14.347" Score="3" ViewCount="" Body="&lt;p&gt;If you're going to define division of vectors, you're going to have to define over what multiplication field you're going to define the division: For ordinary numbers, I think of $\frac{x}{y}$ as being the number that, when multiplied by $y$, gives $x$.  So, $\frac{\vec x}{\vec y}$ would have to be the vector that, when &quot;multiplied&quot; by $\vec y$, gives $\vec x$.  If our multiplication field is the dot product, we're already in trouble, because the dot product of two vectors is a scalar, and the definition above would therefore require $\frac{\vec x}{\vec y}$ to simultaneously be a vector and a scalar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if our operation is the cross product, then we know that, for any vectors $\vec x$ and $\vec y$ and any scalar $c$, we have ${\vec x} \times {\vec y} = {\vec x}\times \left({\vec y} + c{\vec x}\right)$, so this means that there are an infinite number of vectors that satisfy the property &quot;when crossproducted by $\vec y$,  gives $\vec x$.  Therefore, division over the cross product is not unique.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-08-28T21:15:14.347" />
  <row Id="14086" PostTypeId="2" ParentId="14082" CreationDate="2011-08-28T21:15:47.147" Score="0" ViewCount="" Body="&lt;p&gt;In addition to nibot's answer: division something is finding a part of something. In case of a vector, its part has the same direction but a smaller length. So it is natural to divide vectors by numbers, not by vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those dot and cross products are not simple products because they depend not only on lengths but also on orientations. They are called correspondences between a couple of vectors and numbers or vectors.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="1390" LastEditorDisplayName="" LastEditDate="2011-08-28T22:13:05.167" LastActivityDate="2011-08-28T22:13:05.167" />
  <row Id="14087" PostTypeId="2" ParentId="14074" CreationDate="2011-08-28T21:16:00.087" Score="7" ViewCount="" Body="&lt;p&gt;First general relativity is typically taught at a 4th year undergraduate level or sometimes even a graduate level, obviously this presumes a good undergraduate training in mathematics and physics. Personally, I'm more of the opinion that one should go and learn other physics before tackling general relativity. A solid background in classical mechanics with exposure to Hamiltonians, Lagrangians, and action principles at least. A course in electromagnetism (at the level of Griffiths) I think is also a good thing to have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, I think the pre-reqs are a bit higher and since the question asks about mathematical detail,  I'll focus on that.  I learnt relativity from a very differential geometry centric viewpoint (I was taught by a mathematician) and I found that my understanding of differential geometry was very helpful for understanding the physics. I've never been a fan of Hartle's book which I think is greatly lacking on the mathematical details but is good for physical intuition. However having worked in relativity for some time now I think it's better to teach from a more mathematical point of view so you can easily pick up the higher level concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, I think you really need to understand what is going on mathematically to understand why we must construct things the way we do. I'm going to have to disagree with nibot here and say that you'll need more then just linear algebra and college calculus. Calculus you must have at least seen up to vector calculus and be familiar with it. Linear algebra is something you should have a very good understanding considering that we are dealing with vectors. A good course in more abstract algebra dealing with vector spaces, inner products/orthogonality, and that sort of thing is a must. To my knowledge this is normally taught in a second year linear algebra course and is typically kept out of first year courses. Obviously a course in differential equations is required and probably a course in partial differential equations is required as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think a course in analysis is required, however since the question is more about the mathematical aspect, I'd say having a course in analysis up to topological spaces is a huge plus. That way if you're curious about the more mathematical nature of manifolds, you could pick up a book like Lee and be off to the races. If you want to study anything at a level higher, say Wald, then a course in analysis including topological spaces is a must. You could get away with it but I think it's better to have at the end of the day. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd also say a good course in classical differential geometry (2 and 3 dimensional things) is a good pre-req to build a geometrical idea of what is going on, albeit the methods used in those types of courses do not generalise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there is also the whole bit about mathematical maturity. It's a funny thing that is impossible to quantify. I, despite having the right mathematical background, did not understand immediately the whole idea of introducing a tangent space on each point of a manifold and how $\{\partial_{i}\}$ form a basis for this vector space. It took me a bit longer to figure this out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can always skip all this and get away with just the physicists classical index gymnastics (tensors are things that transform this certain way) however I think if you want to be a serious student of relativity you'd learn the more mathematical point of view. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: On the suggestion of jdm, a course in classical field theory is good as well. There is a nice little Dover book appropriately titled &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486462609&quot; rel=&quot;nofollow&quot;&gt;Classical Field Theory&lt;/a&gt; that gets to general relativity right at the end. However I never took a course and I don't think many universities offer it anyway unfortunately. Also a good introduction if you want to go learn quantum field theory.&lt;/p&gt;&#xA;" OwnerUserId="5068" LastEditorUserId="5068" LastEditorDisplayName="" LastEditDate="2011-08-28T23:39:59.667" LastActivityDate="2011-08-28T23:39:59.667" CommentCount="3" />
  <row Id="14088" PostTypeId="2" ParentId="14074" CreationDate="2011-08-28T21:29:34.910" Score="4" ViewCount="" Body="&lt;p&gt;Can I recommend this set of videos from Stanford university as given by Leonard Susskind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very thorough and very haelpful for me inthe summer before I took the course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=hbmf0bB38h0&quot; rel=&quot;nofollow&quot;&gt;Stanford Uni GR course&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also it wouldn't hurt to look at some none GR cosmology courses, these will cover some of the physical phenomena that GR is necessary for to describe completely&lt;/p&gt;&#xA;" OwnerUserId="2535" LastActivityDate="2011-08-28T21:29:34.910" />
  <row Id="14089" PostTypeId="2" ParentId="14078" CreationDate="2011-08-28T21:35:07.950" Score="0" ViewCount="" Body="&lt;p&gt;There are two things to notice here.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You can only make the assignment $\oint \mathbf{B} \cdot dl = 2 \pi d B(d)$ if the situation is radially symmetric.&lt;/li&gt;&#xA;&lt;li&gt;In the case of a &lt;em&gt;finite&lt;/em&gt; wire you either have charge building up at the ends or you have not specified the &lt;em&gt;whole&lt;/em&gt; current distribution yet. Question: can you specify a radially symmetric return path and if so do you expected it make up the difference?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="520" LastEditorUserId="520" LastEditorDisplayName="" LastEditDate="2011-08-31T16:43:37.557" LastActivityDate="2011-08-31T16:43:37.557" CommentCount="0" />
  <row Id="14090" PostTypeId="2" ParentId="10055" CreationDate="2011-08-28T22:55:36.507" Score="-1" ViewCount="" Body="&lt;p&gt;It is the time-independent Schroedinger equation $$\frac {d^2 \psi} {d x^2} = - k^2 \psi$$ where $k^2 = \frac {2mE} {\hbar^2}$ that quantum mechanics uses to describe a free particle, not any other equation. Said equation has the following function as the solution: $$ \psi(x) = A e^{ikx} + Be^{-ikx}$$ which is a plane wave and not a wave packet. If the solution of the above equation were not a plane wave but a wave packet then it should be clearly shown and plane waves should not be discussed at all. A theory, however, should have a predictive quality. Quantum mechanics predicts the state of a free particle to be described by a plane wave (because, as said, that is what the solution of Schrodinger equation is) but that turns out to be non-physical (does not comply with experiment) for reasons already discussed. This is a major deficiency of quantum mechanics, along with a number of other deficiencies, which should be clearly spelled out and not sidetracked by offering constructs unrelated to Schrodinger's equation. Any comments?&lt;/p&gt;&#xA;" OwnerUserId="5070" LastActivityDate="2011-08-28T22:55:36.507" />
  <row Id="14091" PostTypeId="2" ParentId="14080" CreationDate="2011-08-28T23:07:08.230" Score="2" ViewCount="" Body="&lt;p&gt;When you exert a torque on something, its angular momentum changes.  This doesn't necessarily mean things are moving in a circle, or even rotating.  You can have torque and changing angular momentum even when things are going in a straight line.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the time, when we discuss torque and angular momentum in basic physics, we are talking about things like spinning bicycle wheels, precessing tops, or ice skaters doing a fancy maneuver, but that's only because torque and angular momentum are easiest to exemplify and visualize in those situations.  The simple facts about torque and angular momentum, such as $\vec{\tau} = \dot{\vec{L}}$, hold generally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make your example precise, suppose a particle's trajectory is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{r} = x_0\hat{x} - (y_0 - \frac{1}{2}gt^2)\hat{y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the force on it is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{F} = m\ddot{\vec{r}} = -mg\hat{y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torque about the origin is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{\tau} = \vec{r}\times\vec{F} = -x_0mg\hat{z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angular momentum about the origin is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{L} = \vec{r}\times\vec{p} = \vec{r} \times (-mgt\hat{y}) = -x_0mgt\hat{z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we see that the time derivative of the angular momentum is the torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{\vec{L}} = -x_0mg\hat{z} = \vec{\tau}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can indeed analyze the torque and angular momentum of an object falling in a straight line and find a consistent and accurate description.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-08-28T23:07:08.230" />
  <row Id="14092" PostTypeId="2" ParentId="14082" CreationDate="2011-08-28T23:40:34.353" Score="6" ViewCount="" Body="&lt;p&gt;The best way is to ignore the garbage authors put in elementary physics books, and define it with tensors. A tensor is an object which transforms as a product of vectors under rotations. Equivalently, it can be defined by linear functions of (sets of vectors) and (linear functions of sets of vectors), all this is described on Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are exactly two tensors which are invariant under rotations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta_{ij}$&#xA;and&#xA;$\epsilon_{ijk}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All other tensors which are invariant under rotations are products and tensor traces of these. These tensors define the &quot;dot product&quot; and &quot;cross product&quot;, neither of which is a good notion of product:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V \cdot U = V^i U^j \delta_{ij}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and cross product&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(V \times U)_k = V^i U^j \epsilon_{ijk}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is pointless to try to think of the cross product as a &quot;product&quot;, because it is not associative, $(A\times B)\times C$ does not equal $A\times(B\times C)$. It is also less than useful to think of the dot-product as a product in the usual sense, because it takes pairs of vectors to numbers, and $(A\cdot B)C$ does not equal $A(B\cdot C)$, because the first points in the C direction, and the second points in the A direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way is to get used to the invariant tensors. These generalize to arbitrary dimensions, and they are much clearer, and do not require a right-hand rule (this is taken care of by the index order convention). You will not find a single physics paper which uses the cross product, with the single exception of Feynman's 1981 paper &quot;the qualitative behavior of Yang-Mills theory in 2+1 dimensions&quot;, and even if you do, it is trivial to translate.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-29T07:21:55.843" LastActivityDate="2011-08-29T07:21:55.843" CommentCount="8" />
  <row Id="14093" PostTypeId="2" ParentId="14071" CreationDate="2011-08-28T23:41:18.080" Score="0" ViewCount="" Body="&lt;p&gt;A fundamental postulate of electromagnetism is that the flux of the magnetic field through any &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_surface&quot; rel=&quot;nofollow&quot;&gt;closed surface&lt;/a&gt; is zero.  This is essentially a statement that there are no &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_monopole&quot; rel=&quot;nofollow&quot;&gt;magnetic monopoles&lt;/a&gt;.  This implies that the divergence of the magnetic field is zero.  The corresponding equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \cdot \vec{B}  = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not very much like Kirchoff's (Voltage) Law, which is based on the vanishing curl of the electric field&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla \times \vec{E} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and is only true in the absence of a time-dependent magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize the difference:&#xA;The law about magnetic fields is saying that you take a closed plastic bag, contort it into any shape, and however much magnetic flux enters the bag has to flow back out again somewhere else.  The law about electric fields is saying that you take loop of wire, contort it into any shape, and however much the electric field pushes a charged particle forward in one part of the loop, it will push the particle backwards by the same amount over the rest of the loop.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2011-08-28T23:41:18.080" />
  <row Id="14094" PostTypeId="1" CreationDate="2011-08-29T00:44:12.197" Score="7" ViewCount="157" Body="&lt;p&gt;I was reading up on &lt;a href=&quot;http://en.wikipedia.org/wiki/De_Sitter_space&quot; rel=&quot;nofollow&quot;&gt;De Sitter space&lt;/a&gt;s, which states that the gravitational effects from a black hole is indistinguishable from any other spherically symmetric mass distribution.  This makes a lot of sense to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm now super curious, can we just formulate all the properties of a black hole beyond the limit at which GR is needed in a completely Maxwellian / Newtonian sense?  The Wikipedia article on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kerr-Newman_black_hole&quot; rel=&quot;nofollow&quot;&gt;Kerr-Newman metric&lt;/a&gt; seems to indicate so, but the equations are in terms of the GR metrics.  This is not what I want, I want a simplification, a limit-case of that math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ted Bunn answered a part of my question in &lt;a href=&quot;http://physics.stackexchange.com/questions/12169/detection-of-the-electric-charge-of-a-black-hole&quot;&gt;Detection of the Electric Charge of a Black Hole&lt;/a&gt;.  Let me repeat the stupidly simple form of the gravitational and electric field for a black hole beyond the point at which GR is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{g} = \frac{GM}{r^2} \hat{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{E} = \frac{Q}{4\pi\epsilon_0 r^2} \hat{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correct me if I'm wrong, but these would be a meaningful and accurate approximation in many, in fact, most situations where we would plausibly interact with a black hole (if we were close enough that these are no longer representative, we'd be risking a date with eternity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;  Fill in the blank; what would the magnetic field $\vec{B}$ around a black hole be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is why I find it non-trivial: Every magnet in &quot;our&quot; world has some significant waist to it.  So here is a normal magnet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/l7sr2.png&quot; alt=&quot;normal magnet&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens when this is a black hole?  Would the approximation for $\vec{B}$ that I'm asking for have all the magnetic field lines pass through the singularity?  Or would they all pass through the event horizon radius but not necessarily a single point?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I think most people who have understood this already, but ideally the answer would use the 3 fundamental metrics of a black hole.  Mass $M$, charge $Q$, and angular momentum $L$.  The prior equations for gravity and electric field already fit this criteria.  So the answer I'm looking for should be doable in the following form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{B} = f \left( M, Q, L, \vec{r} \right) $$&lt;/p&gt;&#xA;" OwnerUserId="1255" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-08-29T03:56:24.453" LastActivityDate="2011-09-02T13:08:06.127" Title="If a Kerr-Newman black hole is like a charged, spinning, heavy magnet, what kind of magnet is it like?" Tags="&lt;general-relativity&gt;&lt;black-holes&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="3" />
  <row Id="14095" PostTypeId="1" CreationDate="2011-08-29T05:32:24.043" Score="3" ViewCount="186" Body="&lt;p&gt;I am a bit lost how one has deduced the formula for electric field with electric dipole because of some inconsistency between different sources. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Dipole#Field_from_an_electric_dipole&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; contains a delta function in the formula, but &lt;a href=&quot;http://phys.columbia.edu/~nicolis/Dipole_electric_field.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; gives only the formula without the delta function,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{E} = -\nabla \phi = \frac{-p}{4\pi \epsilon_{0} r^{3}} ( \hat{e}_{z}-3\hat{u}_{r}(\hat{u}_{r} \cdot \hat{u}_{z}) ) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it intentionally left out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My notes mention that the dipole is with a circular thing $P=Q d$ where $d$ is distance and $Q$ is charge.&lt;/p&gt;&#xA;" OwnerUserId="623" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-29T20:04:17.603" LastActivityDate="2011-08-30T14:56:55.057" Title="What is the origin of the Dirac delta term in the dipole electric field?" Tags="&lt;electromagnetism&gt;&lt;electric-field&gt;&lt;dipole&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" />
  <row Id="14096" PostTypeId="2" ParentId="14082" CreationDate="2011-08-29T05:46:32.690" Score="1" ViewCount="" Body="&lt;p&gt;For products you have the answers. For division I recommend you to read more about &lt;a href=&quot;http://en.wikipedia.org/wiki/Quaternion&quot; rel=&quot;nofollow&quot;&gt;quaternions&lt;/a&gt;. Interpretation of &lt;a href=&quot;http://en.wikipedia.org/wiki/Quaternion#Quaternions_and_the_geometry_of_R3&quot; rel=&quot;nofollow&quot;&gt;vectors in terms of quaternions&lt;/a&gt; allows for more reach algebra than vector space itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little math right here. For natural definition of division you need at least division ring (one may comment that division algebra is enough, then add octonions to my answer). There is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Frobenius_theorem_%28real_division_algebras%29&quot; rel=&quot;nofollow&quot;&gt;theorem&lt;/a&gt; that the only finite dimensional division rings are reals, complex and quaternions. Vectors are vector space in three dimensions. So, any division for three-dimensional vectors will be &quot;unnatural&quot;. &lt;/p&gt;&#xA;" OwnerUserId="4657" LastActivityDate="2011-08-29T05:46:32.690" />
  <row Id="14097" PostTypeId="2" ParentId="12611" CreationDate="2011-08-29T07:16:23.183" Score="5" ViewCount="" Body="&lt;p&gt;This is fundamentally no more difficult than understanding how quantum mechanics describes particle motion using plane waves. If you have a delocalized wavefunction $\exp(ipx)$ it describes a particle moving to the right with velocity p/m. But such a particle is already everywhere at once, and only superpositions of such states are actually moving in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \psi_k(p) e^{ipx - iE(p) t} dp$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\psi_k(p)$ is a sharp bump at $p=k$, not a delta-function, but narrow. The superposition using this bump gives a wide spatial waveform centered about at x=0 at t=0. At large negative times, the fast phase oscillation kills the bump at x=0, but it creates a new bump at those x's where the phase is stationary, that is where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\partial\over\partial p}( p x - E(p)t ) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or, since the superposition is sharp near k, where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x = E&amp;#39;(k)t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which means that the bump is moving with a steady speed as determined by Hamilton's laws. The total probability is conserved, so that the integral of psi squared on the bump is conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual time-dependent scattering event is a superposition of stationary states in the same way. Each stationary state describes a completely coherent process, where a particle in a perfect sinusoidal wave hits the target, and scatters outward, but because it is an energy eigenstate, the scattering is completely delocalized in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a collision which is localized, you need to superpose, and the superposition produces a natural scattering event, where a wave-packet comes in, reflects and transmits, and goes out again. If the incoming wavepacked has an energy which is relatively sharply defined, all the properties of the scattering process can be extracted from the corresponding energy eigenstate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the solutions to the stationary eigenstate problem $\psi_p(x)$ for each incoming momentum $p$, so that at large negative x, $\psi_p(x) = exp(ipx) + A \exp(-ipx)$ and $\psi_p(x) = B\exp(ipx)$ at large positive x, superpose these waves in the same way as for a free particle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int dp \psi_k(p) \psi_p(x) e^{-iE(p)t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At large negative times, the phase is stationary only for the incoming part, not for the outgoing or reflected part. This is because each of the three parts describes a free-particle motion, so if you understand where free particle with that momentum would classically be at that time, this is where the wavepacket is nonzero. So at negative times, the wavepacket is centered at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x = E&amp;#39;(k)t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For large positive t, there are two places where the phase is stationary--- those x where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x = - E&amp;#39;(k) t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x = E_2&amp;#39;(k) t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $E_2&amp;#39;(k)$ is the change in phase of the transmitted k-wave in time (it can be different than the energy if the potential has an asymptotically different value at $+\infty$ than at $-\infty$). These two stationary phase regions are where the reflected and transmitted packet are located. The coefficient of the reflected and transmitted packets are A and B. If A and B were of unit magnitude, the superposition would conserve probability. So the actual transmission and reflection probability for a wavepacket is the square of the magnitude of A and of B, as expected.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-29T07:16:23.183" />
  <row Id="14098" PostTypeId="2" ParentId="14094" CreationDate="2011-08-29T14:43:24.963" Score="3" ViewCount="" Body="&lt;p&gt;It's a little tricky giving a formal answer to this, but here is a sketch:.  The electromagnetic potential of the Kerr-Newman hole is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{a}dx^{a}=-\frac{er}{r^{2}+a^{2}\cos^{2}\theta}\left(dt-a \sin^{2}\theta d\phi\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This field will acquire a magnetic field from the fact that $\frac{\partial A_{\phi}}{\partial r}$ and $\frac{\partial A_{\phi}}{\partial \theta}$ are both nonzero.  The problem is that the magnetic field, when re-phrased in terms of vectors and not one-forms, will fall off as $\frac{1}{r^{4}}$.  At that point, if we're keeping terms that fall off that quickly, then we need to have a discussion about that asymptotic form of the metric you imply above, because there are terms in the metric that we need to keep, arising from frame-dragging effects of the black hole.  If you want, I can go into more detail.&lt;/p&gt;&#xA;" OwnerUserId="347" LastActivityDate="2011-08-29T14:43:24.963" CommentCount="6" />
  <row Id="14099" PostTypeId="2" ParentId="14020" CreationDate="2011-08-29T15:30:45.197" Score="0" ViewCount="" Body="&lt;p&gt;I will try to expand Mandelstam's cryptic comment into a full answer, because I think I am starting to understand his argument. I have very little intuition for the analytic structure of trajectory functions, and when I read Mandelstam, I get the feeling of a great lost world of analytic constraints which have been dropped into the memory hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, if resonances are exactly stable, their pole position is real. This means that the scattering amplitude is infinitesimal, and concentrated right near mass-shell, because a particle is produced and propagates very far before being absorbed. So the scattering amplitude is real to zeroeth infinitesimal order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The assumption Mandelstam is making, as I see it, is analyticity of the most radical kind--- every trajectory function is separately analytic and its contribution is described by the exchange of its resonances only. When you take the log in s, you get the trajectory function, and he is also assuming that this function has a convergent dispersion integral after two subtractions, meaning that once you start with an approximation of straight-line trajectories, any correction is perturbative, and this is justified by the statement that the trajectories are approximately linear, not approximately quadratic, or approximately cubic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then if the resonances have no imaginary part, the trajectory function has no imaginary part either, because whenever it hits an integer angular momentum, it becomes a resonance. The argument then follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This argument is a convincing argument that there exists a perturbation expansion which starts with linear Regge trajectories, but it is not at the level of rigor of a matheamtical proof, nor should it be expected to be.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-29T15:30:45.197" />
  <row Id="14100" PostTypeId="1" AcceptedAnswerId="14112" CreationDate="2011-08-29T18:38:20.547" Score="2" ViewCount="42" Body="&lt;p&gt;Assume I have a chain of uniform mass, $M_c$, suspended between two points such that it will follow the trajectory of a catenary curve when only supporting its own weight.  Let's also say that this cable is actually supporting a suspension bridge of uniform mass, $M_b$, running parallel to the cable along a straight line.  In the limit where the mass of the cable is negligible compared to the mass of the suspension bridge, i.e. where $M_c &amp;lt;&amp;lt; M_b$, the cable should follow a parabolic trajectory.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm curious what happens when the weight of the cable and the weight of the bridge are within an order of magnitude or so of one another, i.e. where $M_c$ ~ $M_b$.  How would I derive a general equation for the curvature of the cable as a function of $\frac{M_c}{M_b}$? &lt;/p&gt;&#xA;" OwnerUserId="4257" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-29T18:51:22.467" LastActivityDate="2011-08-29T21:43:35.723" Title="Curvature of a cable supporting a suspension bridge in the regime where the mass of the cable is non-negligible" Tags="&lt;statics&gt;&lt;catenary&gt;" AnswerCount="1" />
  <row Id="14101" PostTypeId="2" ParentId="14094" CreationDate="2011-08-29T18:41:42.253" Score="0" ViewCount="" Body="&lt;p&gt;I'm going to answer this question with the a priori assumption that if some amount of matter collapses with a given field, then those fields are maintained as it collapses into a black hole.  I see nothing to refute this assumption, but the validity of it is beyond my knowledge.  As I was pointing out before, this sufficiently far from the singularity such that GR specific concepts aren't needed.  The cutoff for this is formalized by the following condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$GM/r \ll c^2/2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, we have $Q$, $M$, and $L$ to describe the black hole.  Obviously the black hole, and the singularity itself is some amount of mass rotating.  I find this problematic due to difficulties with &lt;a href=&quot;http://physics.stackexchange.com/questions/9830/tension-in-a-curved-charged-wire-electrostatic-force-does-wire-thickness-matt&quot;&gt;loop singularities in a completely Newtonian sense&lt;/a&gt;.  Anyway, I need to make statements about the angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = M a V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is to say, the angular momentum is due to the fact that the mass of the singularity is located at some position away from the axis of rotation, $a$ and spinning at speed $V$.  With the above statement we can make some meaningful statement about the magnetic moment, $\vec{m}$.  I'll assume the axis of rotation is the z-axis and denote that unit vector with $\hat{k}$.  The following equation comes from &lt;a href=&quot;http://en.wikipedia.org/wiki/Dipole&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.  I will then combine this with the prior equation.  Doing this algebra makes the assumption that the mass has the same distribution as the charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{m} = \frac{1}{2} Q a V \hat{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{m} = \frac{Q L}{2 M} \hat{k} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does make intuitive sense to me.  The more angular momentum, the stronger I expect the magnetic field from it to be.  Also, since angular momentum has mass as a component, we basically have to divide that back out.  Next, I would just use the equation for a perfect magnetic dipole.  This is to say the waist is zero, basically assuming it's a singularity.  I think this would be fine unless the rotational energy was a significant part of the energy contained in it.  I'll just ignore the delta function for the infinite magnetic field at the origin, because this isn't valid there anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{B}(\vec{m}, \vec{r}) = \frac {\mu_0} {4\pi r^3} \left(3(\vec{m}\cdot\hat{r})\hat{r}-\vec{m}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{B}(M,Q,L,\vec{r}) = \frac {\mu_0 Q L} {8 \pi r^3 M} \left(3 \frac{r_z}{r} \hat{r}-\hat{k}\right)$$&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-08-29T18:41:42.253" />
  <row Id="14102" PostTypeId="1" AcceptedAnswerId="14127" CreationDate="2011-08-29T18:55:47.177" Score="1" ViewCount="62" Body="&lt;p&gt;When light comes across with a solid material, some of it is reflected, some of it passes through and some of it is absorbed. I understand the reflection and passing through, but I don't understand what happens when light is absorbed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that there is an enclosed box. Its material is thick enough so that no light can escape out. We put a Led diode in it whose internal resistance is zero (there is no loss as heat energy, all energy dissipated is given out as light). What are the possible steady state scenarios that can happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why and how some wavelengths of sun light heats up black objects while some other wave lengths do not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why and how some wavelengths of light pulls the outer most electrons of atoms to higher energy levels while other wavelengths have no such effect? (I understand that energy of a photon must be enough to take electron to higher energy level, but why doesn't it take the electron temporarily a littler higher in the current energy level and then make it return back? I understand that an electron cannot stay between two energy levels, because those are unstable regions; but why doesn't electrons stay there even if it is as short as an instant?)&lt;/p&gt;&#xA;" OwnerUserId="5077" LastActivityDate="2011-08-30T12:45:49.913" Title="What happens to the absorbed light energy?" Tags="&lt;light&gt;&lt;energy&gt;&lt;heat&gt;&lt;electron&gt;&lt;photoelectric-effect&gt;" AnswerCount="2" />
  <row Id="14103" PostTypeId="1" AcceptedAnswerId="14134" CreationDate="2011-08-29T19:51:40.440" Score="1" ViewCount="97" Body="&lt;p&gt;Can a small amount of smoke be dense enough to stay in the air keeping its shape for a minute or so?&lt;br&gt;&#xA;Or does it always dissipate quickly?&lt;br&gt;&#xA;If not smoke, can anything else stay in the air for a minute while keeping its shape?&lt;/p&gt;&#xA;" OwnerUserId="877" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-29T20:28:47.390" LastActivityDate="2011-08-30T15:06:36.200" Title="Can smoke stay still in the air?" Tags="&lt;air&gt;&lt;diffusion&gt;" AnswerCount="3" CommentCount="0" />
  <row Id="14104" PostTypeId="1" CreationDate="2011-08-29T19:52:25.820" Score="1" ViewCount="89" Body="&lt;p&gt;A pollen particle has no charge so I cannot understand how the Lorentz force $\bar{F} = q \bar{E} + q(\bar{v} \times \bar{B} )$ could explain the event. I speculated that it is because of the electric field that is generated to the dipole with a charge in the TV and the distance, more &lt;a href=&quot;http://physics.stackexchange.com/questions/14095/explain-the-dirac-delta-in-bare-nabla-phi&quot;&gt;here&lt;/a&gt;. I have not yet confirmed that it would attract the pollen particle. Any better ideas?&lt;/p&gt;&#xA;" OwnerUserId="623" LastEditorUserId="369" LastEditorDisplayName="" LastEditDate="2011-08-30T08:18:23.797" LastActivityDate="2011-08-30T21:43:03.400" Title="Pollen Particle Attracted to TV due to Which Force?" Tags="&lt;electromagnetism&gt;&lt;forces&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" />
  <row Id="14105" PostTypeId="2" ParentId="14103" CreationDate="2011-08-29T20:21:05.183" Score="1" ViewCount="" Body="&lt;p&gt;Quick answer: I believe due to Brownian Motion, a small cloud of smoke particles would dissipate in fairly short order. Indeed, smoke particles are used in experiments to confirm the particulate nature of Air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you were to wrap that smoke in a soap bubble, the surface tension could hold for longer than a minute…&lt;/p&gt;&#xA;" OwnerUserId="2872" LastActivityDate="2011-08-29T20:21:05.183" />
  <row Id="14106" PostTypeId="2" ParentId="10651" CreationDate="2011-08-29T20:26:53.720" Score="1" ViewCount="" Body="&lt;p&gt;Science fiction writer (but also published physicist) Greg Egan has put quite a bit of work into investigated a universe with 4+0 dimensions: &lt;a href=&quot;http://gregegan.customer.netspace.net.au/ORTHOGONAL/ORTHOGONAL.html&quot; rel=&quot;nofollow&quot;&gt;http://gregegan.customer.netspace.net.au/ORTHOGONAL/ORTHOGONAL.html&lt;/a&gt; Some of it is quite ingenious, eg. assuming a compact universe guarantees that the (modified) wave equation doesn't have exponentially growing solutions and time appears, without the -1 in the spacetime metric, as the local gradient of entropy.&lt;/p&gt;&#xA;" OwnerUserId="4983" LastActivityDate="2011-08-29T20:26:53.720" />
  <row Id="14107" PostTypeId="2" ParentId="14095" CreationDate="2011-08-29T20:33:51.057" Score="0" ViewCount="" Body="&lt;p&gt;One difference is that this formula is for a dipole with $p$ in the $z$ direction only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the delta function I think that the derivate (the gradient components) is done in a distributional sense. You can safetely ignore it if you don't want to evaluate the field in the origin.&lt;/p&gt;&#xA;" OwnerUserId="5078" LastActivityDate="2011-08-29T20:33:51.057" />
  <row Id="14108" PostTypeId="2" ParentId="14095" CreationDate="2011-08-29T21:04:54.117" Score="1" ViewCount="" Body="&lt;p&gt;A dipole is formed of two opposite charges. By bringing them ever closer together, all the while increasing their charge but keeping the product of the charge and the separation $p = q \times d$ constant, we can form an ideal, elementary dipole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The field on a line in between the charges goes as $~ 1/d^4$ and thus goes to infinity as $d \to 0$. This forms the Dirac delta. As this dipole is an ideal model, it's only necessary to keep the theory consistent. The only two examples I know for its applications are &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;To get a correct result for the average electric field over a sphere, cfr. Griffiths' &quot;Introduction to Electrodynamics&quot;, pg 157.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperfine_structure&quot; rel=&quot;nofollow&quot;&gt;Hyperfine splitting&lt;/a&gt;. Although I doubt the classical expression is meaningful at that level, it apparently gives rise to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperfine_structure&quot; rel=&quot;nofollow&quot;&gt;Fermi contact interaction&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3234" LastActivityDate="2011-08-29T21:04:54.117" CommentCount="1" />
  <row Id="14109" PostTypeId="1" AcceptedAnswerId="14111" CreationDate="2011-08-29T21:16:11.777" Score="0" ViewCount="23" Body="&lt;p&gt;I am currently doing research on semiconductor materials, so I need a very strong background in band theory to understand the literature. I am currently trying to understand the relationship between bonding-antibonding splitting and the formation of the band energy gap in solids. What are some good references on how bonding-antibonding splitting comes about, and what its implications are in the electronic structure of solids?&lt;/p&gt;&#xA;" OwnerUserId="4241" LastActivityDate="2011-08-29T21:25:04.000" Title="References for the source and application of bonding-antibonding splitting on electronic structure?" Tags="&lt;quantum-mechanics&gt;&lt;solid-state-physics&gt;" AnswerCount="1" />
  <row Id="14110" PostTypeId="2" ParentId="14074" CreationDate="2011-08-29T21:24:03.073" Score="0" ViewCount="" Body="&lt;p&gt;Special Relativity is the first prerequisites, obviously. That should include the necessary linear algebra, group theory and classical field theory, because GR is itself a field theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that the main thing you need to understand is calculus up to and including the chain rule for partial derivatives which is crucial for the composition of co-ordinate transformations on tensor fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is all that is essential to learn the basics of general relativity. Perhaps you also need a good geometric intuition. If you have it, GR can be taught or learnt much earlier than it normally is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some other things that would help, such as understanding the principle of least action which uses the calculus of variations. Electromagnetics up to the formulation of Maxwell's equations in SR is also worth getting to know first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you pick up the smallest book on GR you can find (which would be Dirac's) and work through it. Later you can move to more modern treatments and larger texts with more details.&lt;/p&gt;&#xA;" OwnerUserId="1341" LastActivityDate="2011-08-29T21:24:03.073" />
  <row Id="14111" PostTypeId="2" ParentId="14109" CreationDate="2011-08-29T21:25:04.000" Score="0" ViewCount="" Body="&lt;p&gt;There are numerous excellent reviews out, two that come to mind are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/anie.198708461/abstract&quot; rel=&quot;nofollow&quot;&gt;R. Hoffmann in Angew Chem Int Ed Engl, 26, 846&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rmp.aps.org/abstract/RMP/v60/i3/p601_1&quot; rel=&quot;nofollow&quot;&gt;R. Hoffmann in Rev Mod Phys, 60, 601&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4112" LastActivityDate="2011-08-29T21:25:04.000" />
  <row Id="14112" PostTypeId="2" ParentId="14100" CreationDate="2011-08-29T21:43:35.723" Score="3" ViewCount="" Body="&lt;p&gt;Let $y(x)$ be the curve describing the shape of the cable. Let $T(x)$ be the tension in the cable. Consider a small segment extending from $x$ to $x+dx$. The horizontal component of the tension at the two ends of this segment must cancel out, so $T_x$ must be constant. If $\theta$ is the angle the cable makes with the vertical, so that $y&amp;#39;=\tan\theta$, then $T\cos\theta$ is constant. In other words,&#xA;$$&#xD;&#xA;{T\over \sqrt{1+y&amp;#39;^2}}=\alpha&#xD;&#xA;$$&#xA;where $\alpha$ is some constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider vertical forces. The tension forces on the two ends of our small length element&#xA;are&#xA;$$&#xD;&#xA;T_y(x+dx)-T_y(x)=T_y&amp;#39;(x)\,dx=(T\sin\theta)&amp;#39;\,dx=\left(Ty&amp;#39;\over \sqrt{1+y&amp;#39;^2}\right)&amp;#39;dx=&#xD;&#xA;\alpha y&amp;#39;&amp;#39;\,dx.&#xD;&#xA;$$&#xA;This force must balance the weight of the cable and of the hanging bridge below it:&#xA;$$&#xD;&#xA;\alpha y&amp;#39;&amp;#39;\,dx=\beta\sqrt{1+y&amp;#39;^2}\,dx+\gamma\, dx.&#xD;&#xA;$$&#xA;Here $\beta,\gamma$ are the linear mass densities of the cable and bridge respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the equation we need to solve is&#xA;$$&#xD;&#xA;y&amp;#39;&amp;#39;=b\sqrt{1+y&amp;#39;^2}+c,&#xD;&#xA;$$&#xA;where $b=\beta/\alpha,c=\gamma/\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $m(x)=y&amp;#39;(x)$ be the slope. Then this is a first-order separable equation for $m$, &#xA;with solution &#xA;$$&#xD;&#xA;x=\int{dm\over b\sqrt{1+m^2}+c}.&#xD;&#xA;$$&#xA;Mathematica cheerfully does this integral:&#xA;$$&#xD;&#xA;x=\frac{\frac{c \tan ^{-1}\left(\frac{c m}{\sqrt{m^2+1}&#xD;&#xA;   \sqrt{b^2-c^2}}\right)}{\sqrt{b^2-c^2}}-\frac{c \tan ^{-1}\left(\frac{b&#xD;&#xA;   m}{\sqrt{b^2-c^2}}\right)}{\sqrt{b^2-c^2}}+\sinh ^{-1}(m)}{b}.&#xD;&#xA;$$&#xA;Now &quot;all&quot; you have to do is invert this to get $m(x)$, and integrate $y(x)=\int m(x)dx$. There's no nice closed-form solution for this, unfortunately. But Mathematica does verify that the solution goes to a catenary as $c\to 0$ and to a parabola as $c\to\infty$, so I think it's right.&lt;/p&gt;&#xA;" OwnerUserId="1241" LastActivityDate="2011-08-29T21:43:35.723" />
  <row Id="14113" PostTypeId="2" ParentId="14074" CreationDate="2011-08-29T22:20:29.417" Score="0" ViewCount="" Body="&lt;p&gt;I read most of Schutz's first course straight out of high school and before university. &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521277035&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/First-Course-General-Relativity/dp/0521277035&lt;/a&gt; I had done mathematics and  physics A levels (standard for 18 year olds in the UK intending to go on to mathematics at university). It requires understanding what a partial derivative is and basic linear algebra, as well as a lack of fear. It helped that I'd previous read lots of pop science on special relativity beforehand, but it does start with a derivation of SR making the book pretty self-contained.&lt;/p&gt;&#xA;" OwnerUserId="4983" LastActivityDate="2011-08-29T22:20:29.417" />
  <row Id="14114" PostTypeId="2" ParentId="14104" CreationDate="2011-08-30T03:18:29.957" Score="3" ViewCount="" Body="&lt;p&gt;I do not know to what special pollen your professor refers to, but pollen is often charged at least as &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S1352231006010764://&quot; rel=&quot;nofollow&quot;&gt;this measurement described&lt;/a&gt; in the publication abstract found:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Under fair weather conditions, a weak electric field exists between negative charge induced on the surface of plants and positive charge in the air. This field is magnified around points (e.g. stigmas) and can reach values up to 3×106 V m−1. If wind-dispersed pollen grains are electrically charged, the electrostatic force (which is the product of the pollen's charge and the electric field at the pollen's location) could influence pollen capture. In this article, we report measurements of the electrostatic charge carried by wind-dispersed pollen grains. Pollen charge was measured using an adaptation of the Millikan oil-drop experiment for seven anemophilous plants. Acer rubrum, Cedrus atlantica, Cedrus deodara, Juniperus virginiana, Pinus taeda, Plantago lanceolata and Ulmus alata. All species had charged pollen, some were positive others negative. The distributions (number of pollen grains as a function of charge) were bipolar and roughly centered about zero although some distributions were skewed towards positive charges. Most pollen carried small amounts of charge, 0.8 fC in magnitude, on average. A few carried charges up to 40 fC. For Juniperus, pollen charges were also measured in nature and these results concurred with those found in the laboratory. For nearly all charged pollen grains, the likelihood that electrostatics influence pollen capture is evident.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1492" LastActivityDate="2011-08-30T03:18:29.957" CommentCount="3" />
  <row Id="14115" PostTypeId="1" AcceptedAnswerId="14184" CreationDate="2011-08-30T05:09:59.207" Score="4" ViewCount="55" Body="&lt;p&gt;Consider a magnetic spring as seen on this &lt;a href=&quot;http://www.youtube.com/watch?v=jMeCbFX80W4&quot; rel=&quot;nofollow&quot;&gt;YouTube&lt;/a&gt; video, but ignore gravity. If I wanted to calculate the effective spring rate (Force vs. Deflection) curve for the top magnet, how would I go by doing that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $N$ permanent magnets made of &lt;a href=&quot;http://en.wikipedia.org/wiki/Neodymium_magnet&quot; rel=&quot;nofollow&quot;&gt;Neodymium&lt;/a&gt; with known geometry (length, diameter etc). Space them equally such as the total span is $L=N\;\Delta x$ such that they can slide along a rod (1 &lt;strong&gt;DOF&lt;/strong&gt; each). Finally apply a unit force  $F$ on one end, while holding the other end fixed.&lt;/p&gt;&#xA;" OwnerUserId="392" LastEditorUserId="392" LastEditorDisplayName="" LastEditDate="2011-08-31T13:13:05.543" LastActivityDate="2011-08-31T15:47:23.793" Title="What is the effective spring rate of a magnetic spring" Tags="&lt;electromagnetism&gt;&lt;spring&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="14116" PostTypeId="1" CreationDate="2011-08-30T05:20:31.863" Score="4" ViewCount="499" Body="&lt;p&gt;Let $\hat{x} = x$ and $\hat{p} = -i \hbar \frac {\partial} {\partial x}$ be the position and momentum operators, respectively, and $|\psi_p\rangle$ be the eigenfunction of $\hat{p}$ and therefore $$\hat{p} |\psi_p\rangle = p |\psi_p\rangle,$$ where $p$ is the eigenvalue of $\hat{p}$. Then, we have $$ [\hat{x},\hat{p}] = \hat{x} \hat{p} - \hat{p} \hat{x} = i \hbar.$$ From the above equation, denoting by $\langle\cdot\rangle$ an expectation value, we get, on the one hand $$\langle i\hbar\rangle = \langle\psi_p| i \hbar  | \psi_p\rangle = i \hbar \langle \psi_p | \psi_p \rangle  = i \hbar$$ and, on the other $$\langle [\hat{x},\hat{p}] \rangle = \langle\psi_p| (\hat{x}\hat{p} - \hat{p}\hat{x}) |\psi_p\rangle = &#xD;&#xA;\langle\psi_p|\hat{x} |\psi_p\rangle p - p\langle\psi_p|\hat{x} |\psi_p\rangle = 0$$&#xA;This suggests that $i \hbar = 0$. What went wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; This seemingly little curiosity, as I perceived it intially, evolved into a major problem, as can be seen from the discussion below. Quantum mechanics indeed leads to non-physical results and not just to resolvable paradoxes, at that regarding one of its basic claims, the one involving commutators. It seems most of what could be said on this issue has already been addressed below and probably there is no need to continue the exchange here. I would suggest that we discuss it further in the chat.&lt;/p&gt;&#xA;" OwnerUserId="5070" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-09-02T07:59:30.843" LastActivityDate="2011-09-02T07:59:30.843" Title="What's wrong with this derivation that $i\hbar = 0$?" Tags="&lt;quantum-mechanics&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" />
  <row Id="14117" PostTypeId="2" ParentId="3798" CreationDate="2011-08-30T06:27:42.567" Score="4" ViewCount="" Body="&lt;p&gt;This principle is called &quot;positivism&quot;. But I prefer the term &quot;logical positivism&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Positivism is a basic principle of thought--- it distinguishes questions which are meaningful and meaningless. It is not meaningful to ask &quot;How does Argentinian property law taste?&quot;, it is not meaningful to ask &quot;How many angels can dance on the head of a pin?&quot;, and it is not meaningful to ask &quot;Wny is there something rather than nothing&quot;. These questions have the property that, whatever the answer, there is no consequence to our perception of the world, or to anything that we can measure. Such questions do not need an answer, because they are just words put together without meaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Philosophers do not accept positivism, because it moots many of their favorite questions. For example &quot;how can we have free will in a deterministic universe?&quot;. This question, although superficially so clear, is positivistically meaningless, because there is no measurement which can be made upon a creature's decision making power which can in any way determine whether the underlying laws are deterministic or not. This question is moot, because if you define &quot;free will&quot; properly, it cannot possibly have anything to do with determinism. You can't figure out that quantum mechanics is not deterministic by introspection, because your introspection has nothing to do with the determinism of the underlying laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar jibberish:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the universe exists, doesn't it require a prime mover?&lt;/li&gt;&#xA;&lt;li&gt;Is ethics/mathematics defined independently of human beings, or do we produce it?&lt;/li&gt;&#xA;&lt;li&gt;If God is good, how come there is so much suffering in the world?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Basically, half the philosophy curriculum is brain-dead nonsense, and the rest requires massive restructuring to eliminate redundancies. Religion, when formulated positivistically, is &lt;em&gt;not&lt;/em&gt; nonsense, and it is possible to formulate religion, including monotheistic religion, in positivist terms, only throwing away the metaphysics and keeping the meat of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Logical positivism is an early twentieth century tradition that combined the notion of scientific positivism with the new concept of a formal language, then being developed in Logic by Boole, Frege, Russell, Hilbert, and others. The logical positivists wished to get rid of the other annoyance of philosophy, the fact that we have to deal with imprecise natural language. The goal was to replace natural language with a formal language, in which all the terms of discourse are precisely defined. In this way, you would remove another huge branch of philosphy, namely that philosophy whose goal is to make sense of the various ambiguities in the writings of other philosphers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formal languages at the time were barely adequate for formalizing mathematics. It was a bold leap to assume that formal languages could encompass a large enough domain of discourse to approximate natural language. This leap was associated with a bunch of people who I am not going to name, because I haven't read any of them, because philosophy is so trivial compared to any real intellectual work, that you can reproduce any of the results of the non-mathematical philosophers for yourself by thinking for ten minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Logical positivism moots a whole bunch of other philosophy, because it suggests that the right way to think of terms in a language is in terms of a reduction to a formal language, much like in mathematics. If you take the positivists seriously, most of the field of philosophy is pointless and stupid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first, philosophers celebrated the revolutionary ideas, and positivism was the ascendent philosophy until the 1950s. But with the emergence of horrific totalitarian states with scientific materialism as their religion, scientific materialism lost ground, and by the 1970s, it was killed off in the west.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The positive influence of positivism&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;People think positivism equals quantum mechanics, because quantum mechanics was formulated strictly in positivist terms, using &quot;observables&quot;. But it wasn't only in quantum mechanics that positivism was important:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Fields: the notion of fields, which is so abstract, acquires meaning through positivism. How can you tell if a field is there? Put a charge there and see it move! This positivist formulation was important in making clear that a concept that seems so immaterial to many people at first glance is in fact real.&lt;/li&gt;&#xA;&lt;li&gt;Luminiferous ether: the ether lost its material characteristics one by one, and it was an act of positivism by Einstein to reject the ether completely, because it had become unobservable.&lt;/li&gt;&#xA;&lt;li&gt;Equivalence principle: In order to get from the fact that you can't observe gravity in a free-falling frame to the principle that gravity is a geometric force, Einstein made an act of positivism. If the effects of acceleration are indistinguishable from gravity, then gravity and acceleration must be the same thing in essence. This is a very predictive statement.&lt;/li&gt;&#xA;&lt;li&gt;String theory: String theory emerged from a positivistic question--- how can you make a measurement in a space-time that is not well defined at short distances? The answer was to speak about s-matrix states, and their scattering. In physics, the resulting S-matrix theory led to string theory, which is the only candidate for a theory of everything.&lt;/li&gt;&#xA;&lt;li&gt;Black hole interiors/holography: Within string theory, the rejection of the simultaneous existence of the interior and exterior (since each are observed by different observers) led to the fruitful principles of black-hole complementarity and holography as developed by Susskind.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are also several cases where positivism was misapplied overzealously, and I think this clarified things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Quantum field theory is meaningless--- because it is difficult to imagine measuring a quantum field. In fact, the original paper of Bohr-Rosenfeld analyzed measurements of the quantum electric and magnetic field, and decided that it made sense to quantize them. the positivist's complaint was about ultra-short distances, and for those distances you need to deal with gravity anyway.&lt;/li&gt;&#xA;&lt;li&gt;Quarks don't exist--- it is impossible to isolate quarks. This is no better an argument than Mach's argument that atoms don't exist because he can't see them. If there are observable phenomena which are best explained using quarks, then quarks exist to the extent that they are included in the explanation. The real battle here was over quantum field theory, and the postivists were the folks doing the s-matrix theory.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Outside of physics, positivism is identified with this ridiculous claim:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Interior experience doesn't exist--- since all we can measure are inputs and behavior, we should model organisms as a black box without internal experience. This idea, due to Skinner, is pretty idiotic.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In my view, positivism is an engrained part of science, and in my view, it has never steered us wrong, even in cases like S-matrix-theory/string-theory when almost everybody thought it did. With the presence of computers, and computer languages, formal languages are no longer abstract and remote. People program computers all the time. Scientists know when what they say has meaning when they can program a computer to do their model. So I think it is fair to say logical positivism is the only correct functioning philosophy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the logical positivists are now dead, and their work is basically ignored within their own field. If you ask philosophers why they ignore logical positivism, they often say: &quot;Positivism contradicts itself. It says all truths need to be experimentally verifiable, but positivism is not experimentally verifiable!&quot; To call this stupid, self-serving and intellectually vapid is too kind.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-30T06:36:19.727" LastActivityDate="2011-08-30T06:36:19.727" CommentCount="1" />
  <row Id="14118" PostTypeId="2" ParentId="14116" CreationDate="2011-08-30T07:24:02.850" Score="18" ViewCount="" Body="&lt;p&gt;Both p and x operators as operators do not have eigenvectors in the strict sense. They have distributional eigenvectors which are only defined in a bigger space of functions than the space of square-normalizable wavefunctions, and which should be thought of as only meaningful when smeared a little bit by a smooth test function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The normalization for $\langle \psi_p | \psi_p \rangle $ is infinite, because the p-wave is extended over all space. Similarly, the normalization of the delta-function wavefunction, the x-operator eigenvector, is infinite, because the square of a delta function has infinite integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could state your paradox using $|x\rangle$ states too:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$i\hbar \langle x|x\rangle  = \langle x| (\hat{x}\hat{p} - \hat{p}\hat{x})|x\rangle = x \langle x|p|x\rangle - \langle x|p|x\rangle x = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because $|x&amp;#39;\rangle$ is only defined when it is smeared a little, you need to use a seprate variable for the two occurences of x'. So write the full matrix out for this case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ i\hbar \langle x|y\rangle = x\langle x|\hat{p}|y\rangle - \langle x|\hat{p}|y\rangle y = (x-y)\langle x|\hat{p}|y\rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now x and y are separate variables which can be smeared independently, as required. The p operator's matrix elements are the derivative of a delta function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \langle x|p|y\rangle = -i\hbar \delta&amp;#39;(x-y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you get is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (x-y)\delta&amp;#39;(x-y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you are taking $x=y$ naively by setting the first factor to zero without noticing that the delta function factor is horribly singular, and the result is therefore ill defined without more careful evaluation. If you multiply by smooth test functions for x and y, to smear the answer out a little bit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int f(x) g(y) (x-y) \delta&amp;#39;(x-y) dx dy= \int f(x)g(x) dx = \int f(x) g(y) \delta(x-y)  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the first identify comes from integrating by parts in x, and setting to zero all terms that vanish under the evaluation of the delta function. The result is that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (x-y)\delta&amp;#39;(x-y) = \delta(x-y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the result is not zero, it is in fact consistent with the commutation relation. This delta-function equation appears, with explanation, in the first mathematical chapter of Dirac's &quot;The Principles of Quantum Mechanics&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is unfortunate that formal manipulations with distributions lead to paradoxes so easiy. For a related but different paradox, consider the trace of $\hat{x}\hat{p}-\hat{p}\hat{x}$.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-30T07:46:10.627" LastActivityDate="2011-08-30T07:46:10.627" CommentCount="13" />
  <row Id="14119" PostTypeId="1" CreationDate="2011-08-30T07:31:35.933" Score="1" ViewCount="15" Body="&lt;p&gt;I have a volume which is deforming (using explicit time-integration scheme) uniformly with velocity gradient $L$ and stress tensor $\sigma$. I would like to determine work done by the volume deformation during one timestep $\Delta t$, knowing both current and previous values of $L$, $\sigma$ and volume $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen somewhere the formula $\Delta W={\mathrm tr}(L\sigma)V\Delta t$, but I don't know if it is correct and how to derive it. $L\sigma$ should be energy density, but why are its non-diagonal terms discarded?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Remark:&lt;/em&gt; the integration scheme actually is leap-frog, but I ignored mid-step/on-step business for now and supposed everything are on-step values. The formula above would correctly compute the increment mid-step, reading $\Delta W(t-\Delta t/2)={\mathrm tr}\left(L(t-\Delta t/2)\frac{\sigma(t-\Delta t)+\sigma(t)}{2}\right)\frac{V(t-\Delta t)+V(t)}{2}\Delta t$&lt;/p&gt;&#xA;" OwnerUserId="5080" LastActivityDate="2011-08-30T07:31:35.933" Title="stress work of uniformly deforming continuum" Tags="&lt;energy-conservation&gt;&lt;stress-strain&gt;&lt;continuum-mechanics&gt;" />
  <row Id="14120" PostTypeId="2" ParentId="7784" CreationDate="2011-08-30T07:36:08.040" Score="1" ViewCount="" Body="&lt;p&gt;The best way to see this, I think, is using Einstein's 1907 approximation. Einstein noted that gravity slows down clocks, so that where the potential is more negative, clocks run slow, and where the potential is small, clocks run fast, so that the rate at which a clock ticks at position $x$ is $\sqrt{(1-2\phi(x))}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that space is completely flat, and only the clock rate changes from point to point. This is a curved space-time, but the space-part is flat. Then you ask: &quot;what is the curve in space-time which is extremal for the relativistic distance from point A to point B?&quot;. This is the analog of the question &quot;what is the shortest path from A to B&quot; in geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral for this is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int \sqrt{(1-2\phi(x)) - |v|^2)} dt = \int -\phi(x(t)) + {v^2\over 2} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which, if you multiply by the mass m, gives you the action for a Newtonian particle in a gravitational potential. So Einstein's law reduces to Newton's. The remaining work is to see that the equation for the time-time component of the metric reduces to Laplace's equation for static masses, and this is not too hard to do, but annoying to do twice or to write up. That this reduction works was one of Einstein's criteria for a good field equation.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-30T07:36:08.040" />
  <row Id="14121" PostTypeId="2" ParentId="14095" CreationDate="2011-08-30T07:48:09.367" Score="2" ViewCount="" Body="&lt;p&gt;&lt;em&gt;I'll give you the derivation from my book which includes a nice way to see how the delta functions arise:&lt;/em&gt;&#xA;..............................................................................................................................................................&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can derive the potential field $\vec{A}$ and the electromagnetic fields $\vec{E}$ and $\vec{B}$ of a vector point dipole and an axial point dipole in the same way as we derive these in the case of a the point monopole. We start with a static point charge $\delta(\vec{r})$ and derive the dipole charge/current densities with the help of differential operators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We apply the same differential operators for $\vec{A}$, $\vec{E}$ and $\vec{B}$ to obtain the dipole fields from the monopole fields. First we recall the fields of the monopole. The point charge obtains (over time) a potential field given by.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;\mbox{field}\Big\{\,\delta(\vec{r})\,\Big\} ~~=~~ \frac{1}{4\pi r}&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reversed operator which derives the source from the field is just the Laplacian operator.&#xA;$&#xD;&#xA;-\nabla^2\Big\{\,\frac{1}{4\pi r}\,\Big\} ~~=~~ \delta(\vec{r})&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating the delta function over space shows equal contributions from the three spatial components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;\int \delta(\vec{r})~d\vec{r} ~=~&#xD;&#xA;-\int\left[\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right]&#xD;&#xA;\frac{d\vec{r}}{4\pi r} ~=~ \frac13+\frac13+\frac13 ~=~ 1&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can define vector dipole and axial dipole sources by using differential operators on the monopole $\delta(\vec{r})$ and derive their potential and electromagnetic fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vector dipole is obtained by differentiating the monopole along the direction that we want the dipole to have. The result is a combination of a positive and a negative delta function. The axial dipole is defined by the curl of the monopole so that it gives a circular point current in the direction of the dipole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;charge/current densities, potentials and fields of dipoles:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;\begin{array}{|lcll|}&#xD;&#xA;\hline&#xD;&#xA; &amp;amp;&amp;amp;&amp;amp; \\&#xD;&#xA;    j^o       &amp;amp;=&amp;amp; ~~~\mbox{div}\,\left(~\vec{\mu} \,\delta(\vec{r})~\right)      &amp;amp;&#xD;&#xA;    \mbox{vector dipole charge density}   \\&#xD;&#xA;    \vec{j}   &amp;amp;=&amp;amp; ~~~\mbox{curl}\left(~\vec{\mu}\, \delta(\vec{r})~\right)      &amp;amp;&#xD;&#xA;    \mbox{axial dipole current density} \\ &amp;amp;&amp;amp;&amp;amp;&#xD;&#xA;\\&#xD;&#xA;    A^o       &amp;amp;=&amp;amp; ~~~\mbox{div}\,\left(~\vec{\mu} \,\frac{1}{4\pi r}\,\right) &amp;amp;&#xD;&#xA;    \mbox{vector dipole electric potential}   \\&#xD;&#xA;    \vec{A}   &amp;amp;=&amp;amp; ~~~\mbox{curl}\left(~\vec{\mu} \,\frac{1}{4\pi r}\,\right) &amp;amp;&#xD;&#xA;    \mbox{axial dipole magnetic potential} \\ &amp;amp;&amp;amp;&amp;amp;&#xD;&#xA;\\&#xD;&#xA;    \mathsf{E}&amp;amp;=&amp;amp; -~\mbox{grad}\left(\mbox{div}\,\left(~\vec{\mu} \,\frac{1}{4\pi r}\,\right)\right)&amp;amp;&#xD;&#xA;    \mbox{vector dipole electric field}  \\&#xD;&#xA;    \mathsf{B}&amp;amp;=&amp;amp; +~\,\mbox{curl}\left(\mbox{curl} \left(~\vec{\mu}\,\frac{1}{4\pi r}\,\right)\right)&amp;amp;&#xD;&#xA;    \mbox{axial dipole magnetic field}   \\ &amp;amp;&amp;amp;&amp;amp; \\&#xD;&#xA;\hline&#xD;&#xA;\end{array}&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The potential fields are obtained by applying the same differential operators on the field $1/4\pi r$ of the monopole rather than on the charge distribution. For a dipole moment in the $z$-direction we obtain in SI-units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Potential fields of the electric and magnetic dipoles:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Phi ~=~ \frac{z}{4\pi\epsilon_o r^3}\,\mu~~~~~~~~ \vec{A} ~=~ \Big( -y ~,~ x~ ~,~ 0 \Big)~ \frac{\mu_o}{4\pi r^3}\,\mu&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expressions for the electromagnetic fields do implicitly contain delta functions at the center with the right magnitude. These delta functions are easily lost if the derivation isn't careful enough. The E and B fields are related to each other by the standard vector identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mbox{curl}(\mbox{curl}\vec{X}) ~=~ \mbox{grad}(\mbox{div}\vec{X})-\nabla^2\vec{X}&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have seen that the last term (the Laplacian) yields $\delta(\vec{r})$. The two therefor differ only at the center by a delta function in the $\vec{\mu}$ direction. We can see this explicitly if we align the dipoles with the z-axis, so that $\vec{\mu}$ has only a z-component, and write out the fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only difference is in the z-components. The total difference between the two is the Laplacian and thus $\delta(\vec{r})$. The vector dipole gets -1/3 while the axial dipole gets +2/3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{aligned}&#xD;&#xA;    \mathsf{E}   &amp;amp;= &amp;amp;\frac{1}{\epsilon_o}\bigg[~&#xD;&#xA;    \mathbf{\hat{x}}\, \frac{\partial}{\partial x}\frac{\partial}{\partial z} ~+~&#xD;&#xA;    \mathbf{\hat{y}}\, \frac{\partial}{\partial y}\frac{\partial}{\partial z} ~+~~~&#xD;&#xA;    \mathbf{\hat{z}}\, \frac{\partial^2}{\partial z^2} ~~~~~~~~~~~~~&#xD;&#xA;    &amp;amp;\bigg] ~ \frac{\mu}{4\pi r}\\ \\&#xD;&#xA;    \mathsf{B}   &amp;amp;= &amp;amp;\mu_o\bigg[~&#xD;&#xA;    \mathbf{\hat{x}}\,\frac{\partial}{\partial x}\frac{\partial}{\partial z} ~+~&#xD;&#xA;    \mathbf{\hat{y}}\,\frac{\partial}{\partial y}\frac{\partial}{\partial z} ~-~&#xD;&#xA;    \mathbf{\hat{z}}\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)~&#xD;&#xA;    &amp;amp;\bigg] ~ \frac{\mu}{4\pi r}\\&#xD;&#xA;\end{aligned}&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which we can see from the third equation with the three 1/3 parts. In the general case, with arbitrary dipole direction $\vec{\mu}$ we get for the electromagnetic fields in vector form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electromagnetic dipole fields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xD;&#xA;\begin{aligned}&#xD;&#xA;    \mathsf{E}   &amp;amp;= &amp;amp;\frac{1}{\epsilon_o}\left(~\frac{3\left(\vec{\mu}\cdot\hat{r}\right)\hat{r}-\vec{\mu}}{4\pi r^3} ~-~\frac13\vec{\mu}\,\delta(\vec{r})~\right)&#xD;&#xA;\\ \\&#xD;&#xA;    \mathsf{B}   &amp;amp;= &amp;amp;\mu_o\left(~\frac{3\left(\vec{\mu}\cdot\hat{r}\right)\hat{r}-\vec{\mu}}{4\pi r^3} ~+~\frac23\vec{\mu}\,\delta(\vec{r})~\right)&#xD;&#xA;\end{aligned}&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively we can look at the fields explicitly expressed in the individual $x$, $y$ and $z$ components which gives. (with the dipole moment in the $z$-direction)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{aligned}&#xD;&#xA;    \mathsf{E}   &amp;amp;= &amp;amp;\frac{\mu}{4\pi\epsilon_o}\bigg[~&#xD;&#xA;    \mathbf{\hat{x}}\, \frac{3xz}{r^5} ~+~&#xD;&#xA;    \mathbf{\hat{y}}\, \frac{3yz}{r^5} ~+~~~&#xD;&#xA;    \mathbf{\hat{z}}\, \left(\frac{3zz}{r^5}-\frac{1}{r^3} - \frac{4\pi}{3}\delta(r)\right)~&#xD;&#xA;    &amp;amp;\bigg] \\ \\&#xD;&#xA;    \mathsf{B}   &amp;amp;= &amp;amp;\frac{\mu_o\mu}{~4\pi~}\bigg[~&#xD;&#xA;    \mathbf{\hat{x}}\, \frac{3xz}{r^5} ~+~&#xD;&#xA;    \mathbf{\hat{y}}\, \frac{3yz}{r^5} ~+~~~&#xD;&#xA;    \mathbf{\hat{z}}\, \left(\frac{3zz}{r^5}-\frac{1}{r^3} + \frac{8\pi}{3}\delta(r)\right)~&#xD;&#xA;    &amp;amp;\bigg] \\&#xD;&#xA;\end{aligned}&#xD;&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general the fields decrease with the third order of $r$ compared to $1/r^2$ for the charge which makes it possible not many effects at larger scale. At the other hand, when we go to smaller scales, the magnetic dipole fields become more powerful relative to the charge.&lt;/p&gt;&#xA;" OwnerUserId="3197" LastEditorUserId="3197" LastEditorDisplayName="" LastEditDate="2011-08-30T09:20:05.963" LastActivityDate="2011-08-30T09:20:05.963" CommentCount="4" />
  <row Id="14122" PostTypeId="2" ParentId="1317" CreationDate="2011-08-30T08:15:37.030" Score="1" ViewCount="" Body="&lt;p&gt;When the block is pushed, the vertical component of the push increases the normal force. Hence, there is an increase in friction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now when you pull the body, the vertical component of the pull is in the opposite direction of the normal reaction and hence, reduces it, thereby reducing the friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, friction in (b) is less than friction in (a) &#xA; Pulling a body is easier than pushing&lt;/p&gt;&#xA;" OwnerUserId="5082" LastActivityDate="2011-08-30T08:15:37.030" />
  <row Id="14124" PostTypeId="1" AcceptedAnswerId="14143" CreationDate="2011-08-30T09:35:22.973" Score="3" ViewCount="77" Body="&lt;p&gt;How does one determine the members of an simultaneously commuting set (of operators)? For example, I have read that for angular momentum, the set is {$H,L^2,L_z$}. How does one know that these are all of them or that these particular operators should be included? Is it because such a set is sufficient to distinguish all the eigenstates, degeneracies notwithstanding?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, why is $$L^2|l,m\rangle = \hbar l (l+1)|l,m\rangle$$ and $$L_z|l,m\rangle = \hbar m|l,m\rangle$$? Is it defined as such or is it the result of the definition of some more fundamental things?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot! :)&lt;/p&gt;&#xA;" OwnerUserId="5046" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-30T18:29:58.850" LastActivityDate="2011-08-30T19:13:42.883" Title="Simultaneously commuting set" Tags="&lt;quantum-mechanics&gt;&lt;angular-momentum&gt;&lt;operators&gt;" AnswerCount="1" CommentCount="4" />
  <row Id="14125" PostTypeId="1" CreationDate="2011-08-30T10:10:28.427" Score="1" ViewCount="39" Body="&lt;p&gt;One of the arguments of LHC &quot;win-win situation&quot; is that the scattering of W particles needs to include new terms to preserve unitatity begond 500 GeV or so. In the SM, this is realized by the higgs boson, and the same argument extends to the MSSM. Now, one could consider just a &quot;phenomenology Susy standard model&quot; where simply the superpartners to the known SM are added. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that, even without a Higgs mechanism, for each of the W and Z one needs to add an extra scalar boson, to get the right multiplcity; a massive gauge supermultiplet must contain four bosonic helicities. Of course, in the MSSM, two of these come from the Higgses. But from the point of view of SUSY, they are independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, are these scalars, alone or jointly with the sleptons and squarks, enough to preserve the unitarity of WW scattering? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar question could be, if you remove two higgses from the five that come in the MSSM, is it still possible to unitarize WW scattering?&lt;/p&gt;&#xA;" OwnerUserId="1335" LastActivityDate="2011-08-30T10:10:28.427" Title="in SUSY, does WW scattering unitarisation needs the higgs boson?" Tags="&lt;particle-physics&gt;&lt;supersymmetry&gt;" FavoriteCount="1" />
  <row Id="14126" PostTypeId="2" ParentId="14102" CreationDate="2011-08-30T11:47:41.123" Score="2" ViewCount="" Body="&lt;p&gt;it usually gets converted to heat... unless it bounces off or goes right through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why black things get hotter than white things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure about this. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2011-08-30T11:47:41.123" />
  <row Id="14127" PostTypeId="2" ParentId="14102" CreationDate="2011-08-30T12:45:49.913" Score="2" ViewCount="" Body="&lt;p&gt;Certainly vinas is correct.  The absorbed energy is converted to heat energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scenario you mention with the LED is very close to the blackbody problem known as the &quot;ultraviolet catastrophe.&quot; There is a Wikipedia article about it &lt;a href=&quot;http://en.wikipedia.org/wiki/Ultraviolet_catastrophe&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  What happens in the situation you described is that the light proof box gets hotter.  It will increase in heat until the heat leaving the box due to conduction, convection, and radiative effects balance the energy emitted by the LED.  Given the typical power required by an LED, the temperature gain would be minimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all practical purposes, all wavelengths of sunlight will heat an object of any color.  The amount of heat produced by light incident upon an object depends on the material properties.  Some wavelengths will be reflected, some absorbed, and some transmitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you point out, the associated energy of the photon is important in determining what sort of interaction will take place.  A photon of a certain threshhold energy will be able to free a bound electron from an atom causing the photoelectric effect.  A photon of lesser energy can still free less strongly bound electrons causing the Compton effect.  In this case, the photon continues on with less energy.  If the energy is high enough, a photon can cause pair production and create an electron and a positron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case you are wondering about, where a photon transfers energy but does not release an electron do occur.  Depending on whether the photon interacts with the electrons or the entire atom the process is called Thompson or Raleigh scattering respectivly.  In the first case, the electron simply returns to the lower energy level and re-emits a photon of the same energy; the result is effectively an elastic scattering event.  In the second case, the atom re-emits a photon of slightly less energy in essentially the original direction.  The drop in energy of the photons is due to transfer of momentum because of recoil effects.&lt;/p&gt;&#xA;" OwnerUserId="5020" LastActivityDate="2011-08-30T12:45:49.913" CommentCount="1" />
  <row Id="14128" PostTypeId="1" CreationDate="2011-08-30T12:57:48.420" Score="2" ViewCount="113" Body="&lt;p&gt;I'm calculating expectation values (thermal, time-independent) using many-body perturbation theory, but I'm unsure how to work out what values the parameter I'm expanding the perturbation series in can take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read that it's when the matrix elements, $\langle i | H_{pert} | j \rangle$ where $H_{pert}$ is the perturbative term of the Hamiltonian and $| i \rangle$ and $| j \rangle$ are the eigenvectors of the unperturbed Hamiltonian, are much smaller than the energy difference between $i$ and $j$. But I don't really understand what that means, or how it helps me to calculate what values my perturbation parameter can take - Is there a method I can use to figure it out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As requested, to make this concrete, I have a one dimensional fermionic Hubbard model with Hamiltonian&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;H= -t \sum_{\langle l,m \rangle} (c^\dagger_l c_m + h.c.) + U \sum_l (n_{l\uparrow} - 1/2)(n_{l\downarrow}-1/2)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a special case where I know that $U$ is very small and I want to use many-body perturbation theory to see its effects on correlation functions. I calculate the correlation functions using the functional integral method (i.e. calculating a functional partition function). For this case, how would I go about finding out how small $U$ has to be in order for perturbation theory to be valid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, (if this should be a separate question, please let me know!) if instead I have a random $U$, dependent on its position in the lattice,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;H= -t \sum_{\langle l,m \rangle} (c^\dagger_l c_m + h.c.) + \sum_l U_l (n_{l\uparrow} - 1/2)(n_{l\downarrow}-1/2)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can then use a similar functional integral technique, but take an average over the functional partition function (e.g. over a Gaussian distribution). This average removes the $U_l$ and leaves $\Delta$, the variance of the distribution we've averaged over. In this case, it is $\Delta$ which the perturbation series is expanded in. How would I go about finding how small $\Delta$ has to be for the perturbation series to be valid? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't want an answer that's true for any system, I just want to understand how to go about finding it for any system. So if anyone knows of another system where it is shown how small the expansion term has to be, please let me know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="3622" LastEditorUserId="3622" LastEditorDisplayName="" LastEditDate="2011-08-31T12:48:50.147" LastActivityDate="2011-08-31T12:48:50.147" Title="When is many-body perturbation theory valid?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-field-theory&gt;&lt;path-integral&gt;" AnswerCount="1" CommentCount="6" />
  <row Id="14129" PostTypeId="1" CreationDate="2011-08-30T13:10:03.440" Score="2" ViewCount="45" Body="&lt;p&gt;As you might know, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_Zeno_effect&quot; rel=&quot;nofollow&quot;&gt;Zeno effect&lt;/a&gt; is intuitively expressed as what happens when a system is measured in intervals smaller than the half life of the state it is currently on. As a consequence, the state has a negligible probability of doing a transition and is kept 'stuck' in its current state, making the effective evolution operator the identity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know an equivalent picture for the Anti-Zeno effect. Under what conditions does it happen and why? &lt;strong&gt;does the above picture is merely interpretational or is fully accurate at a fundamental level?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2011-09-05T21:07:30.287" Title="under what conditions happen the anti-Zeno effect?" Tags="&lt;quantum-mechanics&gt;&lt;quantum-zeno-effect&gt;" AnswerCount="1" />
  <row Id="14130" PostTypeId="2" ParentId="14128" CreationDate="2011-08-30T13:53:55.677" Score="1" ViewCount="" Body="&lt;p&gt;Your &quot;perturbation parameter&quot; should just be something that sets the scale of $H_{pert}$ -- that is, the size of the matrix elements you specified compared to the bare energies of the unperturbed problem, which you know the exact solution of. To see why this is you just need to look at the explicit second order contribution from (nondegenerate!) perturbation theory, say for the ground state:&#xA;$$E_0^{(2)}=-\sum_{j \neq 0}\frac{\langle 0 | H_{pert }| j \rangle \langle j | H_{pert }| 0 \rangle}{E_j - E_0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read off that this is only going to be a &lt;em&gt;small&lt;/em&gt; correction (and it &lt;em&gt;has&lt;/em&gt; to be a small correction if we want to truncate the expansion here) if you can justify those matrix elements being smaller than the energy difference between the states. If you can't do that (say, you have no control over the strength of the perturbation) then you can't trust the perturbation expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 1:&#xA;When we derive the perturbation expansion, the usual recipe is to do something along the lines $H=H_0+\alpha H_{pert}$, as you say, then expand in alpha -- &lt;em&gt;but we must keep in mind&lt;/em&gt; that this guy $\alpha$ is a bookkeeping device that lets us make sure we group all the terms correctly by their order in alpha! After we're done bookkeeping and grouping terms, $\alpha \rightarrow 1$ &lt;strong&gt;always&lt;/strong&gt;. It is not part of the physics and is not ours to play with. We expanded in it as a formal device. The real small parameter is going to be the ratio of the overall energy scale of $H_{pert}$ to $H_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me give an example. The fermion Hubbard model has two terms -- (1) an easily diagonalized &lt;em&gt;local&lt;/em&gt; interaction with a Coulomb energy scale called $U$; (2) a nearest neighbor hopping term that cannot be diagonalized in the same basis, with an energy scale (~the bandwidth) called $t$. Whether we can trust perturbation theory starting from this basis [that is, taking the hopping as a perturbation] depends only on the dimensionless scale $t/U$, and not the bookkeeping parameter we used to derive the formal &lt;em&gt;terms&lt;/em&gt; of the expansion.&lt;/p&gt;&#xA;" OwnerUserId="791" LastEditorUserId="791" LastEditorDisplayName="" LastEditDate="2011-08-30T15:26:35.957" LastActivityDate="2011-08-30T15:26:35.957" CommentCount="8" />
  <row Id="14131" PostTypeId="2" ParentId="14103" CreationDate="2011-08-30T14:07:48.077" Score="0" ViewCount="" Body="&lt;p&gt;Would, say, clouds or fog count? In this case, the temperature differences between pockets of the air leads to accumulation of water vapor. Off the top of my head, it might be possible to imagine a Van der Waals stabilized gas, or a heterogenous electrostatically stabilized 'smoke'.&lt;/p&gt;&#xA;" OwnerUserId="4257" LastActivityDate="2011-08-30T14:07:48.077" />
  <row Id="14132" PostTypeId="2" ParentId="6844" CreationDate="2011-08-30T14:11:30.033" Score="-1" ViewCount="" Body="&lt;p&gt;Time is only the measurement we use to mark the relative expanse of our universe, an expanse we see and measure by the light emitted (at this time). Thus velocity is another measurement of time as it relates to mass. Velocity cannot exist without time. It is simply another measurement of time, simply in mass form.&#xA;Light and velocity are related by the same continuous link that weaves all existence together. A link we often ignored by looking at and defining the parts rather than the whole. Once you discover the Link you will see the continuous and contiguous relationship of all existence.&lt;/p&gt;&#xA;" OwnerUserId="5085" LastEditorUserId="5085" LastEditorDisplayName="" LastEditDate="2011-08-30T14:39:35.430" LastActivityDate="2011-08-30T14:39:35.430" CommentCount="1" />
  <row Id="14133" PostTypeId="2" ParentId="14095" CreationDate="2011-08-30T14:56:55.057" Score="1" ViewCount="" Body="&lt;p&gt;The origin of the problem is the special point $\mathbf{r}=0$. In the usual derivation (when using easily-derived formula for the potential $\varphi = \frac{1}{4 \pi \epsilon_0} \frac{\mathbf{p} \cdot \mathbf{r}}{r^3}$):&#xA;$$E_{\alpha} = -\frac{1}{4 \pi \epsilon_0} \nabla_{\alpha} \left( \frac{p_{\beta} x_{\beta}}{r^3}\right) = -\frac{1}{4 \pi \epsilon_0} p_{\beta} \left[ \frac{\nabla_{\alpha} x_{\beta}}{r^3} + x_{\beta}\nabla_{\alpha}\left(\frac{1}{r^3}\right)\right] = - \frac{1}{4 \pi \epsilon_0} p_{\beta} \left[ \frac{\delta_{\alpha \beta}}{r^3} - 3 \frac{x_{\alpha} x_{\beta}}{r^5}\right] = $$&#xA;$$=-\frac{1}{4 \pi \epsilon_0} \left[ \frac{p_{\alpha}}{r^3} - 3 \frac{x_{\alpha} (p_{\beta} x_{\beta})}{r^5} \right] $$&#xA;the delta-term is &lt;strong&gt;missing&lt;/strong&gt;, because we do not treat the special point $\mathbf{r}=0$ in this derivation. To account for this term, we need to look closely on&#xA;$$\nabla_{\alpha}\left(\frac{x_{\beta}}{r^3}\right)$$&#xA;and rewrite it:&#xA;$$\nabla_{\alpha}\left(\frac{x_{\beta}}{r^3}\right) = - \nabla_{\alpha} \nabla_{\beta} \frac{1}{r}~~~~(1)$$&#xA;The trace of this term must satisfy the well-known&#xA;$$\Delta \frac{1}{r} = -4 \pi \delta^3(\mathbf{r}).$$&#xA;The above relation (1) caries two indexes, it is symmetrical and invariant with respect to rotations (when $r=0$). Thus, the missing term must be proportional to identity matrix (that is, $\delta_{\alpha \beta}$) and we finally have:&#xA;$$\nabla_{\alpha}\left(\frac{x_{\beta}}{r^3}\right) = \left( \frac{\delta_{\alpha \beta}}{r^3} - 3 \frac{x_{\alpha} x_{\beta}}{r^5} \right) + \frac{4 \pi}{3} \delta_{\alpha \beta} \delta^3(\mathbf{r})$$&#xA;from which it is easy to get the formula from Wikipedia article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is, if you wish, a mathematical explanation of the problem. To get a physical explanation, read &lt;a href=&quot;http://ajp.aapt.org/resource/1/ajpias/v50/i8/p698_s1&quot; rel=&quot;nofollow&quot;&gt;D. J. Griffiths, Am. J. Phys. 50, 698 (1982)&lt;/a&gt;. Note also that this term (the delta-term) arises only in quantum theory and in classical theory can be neglected.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;P.S. I'm using the following notation:&#xA;$$(\text{grad}\,\varphi)_{\alpha} = \nabla_{\alpha} \varphi$$&#xA;where $\alpha \in \{1,2,3\}$. Thus, for example,&#xA;$$\nabla_{\alpha} x_{\beta} = \frac{\partial x_{\beta}}{\partial x_{\alpha}} = \delta_{\alpha \beta}$$&#xA;$$\nabla_{\alpha} r = \nabla_{\alpha} \left(\sqrt{x_{\beta} x_{\beta}}\right) = \frac{1}{2 \sqrt{x_{\beta} x_{\beta}}} \nabla_{\alpha} \left(x_{\beta} x_{\beta}\right)= \frac{x_{\beta} \nabla_{\alpha} x_{\beta}}{ \sqrt{x_{\beta} x_{\beta}}} = \frac{x_{\alpha}}{r}$$&#xA;$$(\mathbf{a},\mathbf{b}) = a_{\alpha} b_{\alpha}$$&#xA;and so forth.&lt;/p&gt;&#xA;" OwnerUserId="4237" LastActivityDate="2011-08-30T14:56:55.057" />
  <row Id="14134" PostTypeId="2" ParentId="14103" CreationDate="2011-08-30T15:06:36.200" Score="2" ViewCount="" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can a small amount of smoke be dense enough to stay in the air keeping its shape for a minute or so?&#xA;  Or does it always dissipate quickly?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I read this and think to myself &quot;optimization problem&quot;.  Firstly, you should know the following, which is the law of diffusion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial \phi}{\partial t} = D\,\frac{\partial^2 \phi}{\partial x^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For clarity, $\phi$ is a function that represents the distribution of the concentration of the gas.  It is a scalar function of 3 variables, which is to say that I could write it as $\phi(\vec{r})$, where $\vec{r}$ represents typical $x,y,z$ coordinates.  If have an image in your mind of a cloud that is the shape of a snowman, that can be represented by that function, so can smoke rings or whatever you desire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clarity of the shape degrades over time, exactly per the above diffusion equation.  Picture blurring an image in Photoshop.  That is very similar to the process that happens.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Q: Can you reduce the rate at which this blurring happens?&lt;/li&gt;&#xA;&lt;li&gt;A: Yes you can&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The rate at which $\phi$ (your snowman) degrades in sharpness comes from the magnitude of $|d\phi/dt|$.  This magnitude is proportional to the diffusion coefficient $D$ as well as that other derivative with respect to $dx^2$, but that term is representative of the sharpness itself, so we don't want to reduce that, we would rather reduce $D$.  In order to reduce $D$, we need to first talk about mean free path and velocity of the gas molecules.  I'll use &lt;a href=&quot;http://www.chemicalforums.com/index.php?topic=27606.0&quot; rel=&quot;nofollow&quot;&gt;this source&lt;/a&gt; and refer to the mean free path $\lambda$ (units of length) and average speed $\bar{c}$.  In general D is proportional to those two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D \propto \lambda \bar{c}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a gas cloud the parameter $\lambda$ has mostly to do with the density of the gas, as well as some other things.  Again, we would like to minimize $D$, but $\lambda$ might not have much design freedom.  On the other hand, $\bar{c}$ could have great design freedom.  This is also dependent on the temperature of the gas, but more specifically, the temperature is a measure of the kinetic energy of the molecules.  I'll say fairly generally:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{2} m \bar{c}^2 = \frac{3}{2} k T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Never mind very much what $k$ is (it's just a physical constant), what matters is that this equation has temperature $T$ and $m$.  I'm taking your question to be most likely concerned with normal air.  That means that it is unlikely that we would have $T$ as a design variable.  However, since you are not specifying the gas you are working with, it's possible we could choose that, and the selection of that gas determines $m$ which is a factor in determining $\bar{c}$ which is a factor in determining $D$, which determines the persistence of your cloud image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line: &lt;em&gt;Heavier gases will diffuse more slowly, meaning the image will persist longer.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of a high molecular weight gas is common refrigerant gases, like R-134a.  If you released that into the air it will diffuse rather slowly compared to other examples.  NOTE: don't do this, it would be dangerous and probably illegal.&lt;/p&gt;&#xA;" OwnerUserId="1255" LastActivityDate="2011-08-30T15:06:36.200" CommentCount="4" />
  <row Id="14135" PostTypeId="1" CreationDate="2011-08-30T15:44:34.220" Score="2" ViewCount="63" Body="&lt;p&gt;Question's title says it all how dangerous are the bullets which are fired up straight in to the air? I often find myself in this discussion where there are people who argues that it is harmless and some say it can be lethal at times. I want to clear my understanding about it to settle this issue once in for all.&lt;/p&gt;&#xA;" OwnerUserId="5086" LastActivityDate="2011-08-30T15:44:34.220" ClosedDate="2011-08-30T18:18:38.093" Title="How dangerous is firing a bullet straight up in to the air?" Tags="&lt;gravity&gt;&lt;speed&gt;" CommentCount="5" />
  <row Id="14136" PostTypeId="1" AcceptedAnswerId="14155" CreationDate="2011-08-30T16:33:53.140" Score="1" ViewCount="125" Body="&lt;p&gt;I was wondering if there is a general formula for calculating Ricci scalar for any diagonal n by n metric tensor?&lt;/p&gt;&#xA;" OwnerUserId="4974" LastActivityDate="2011-09-01T15:51:13.310" Title="Ricci scalar for a diagonal metric tensor." Tags="&lt;general-relativity&gt;" AnswerCount="2" />
  <row Id="14137" PostTypeId="2" ParentId="14136" CreationDate="2011-08-30T16:50:26.437" Score="1" ViewCount="" Body="&lt;p&gt;The Ricci scalar is just the trace of the Ricci tensor, which in turn is a tensor contraction of the Riemann curvature tensor, which can be expressed in Cristoffel symbols defined by the local metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R=R^i_i=g^{ij}R_{ij}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_{ij}={R^k}_{ikj}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${R^{\rho}}_{\sigma\mu\nu} = \partial_\mu\Gamma^{\rho}_{\nu\sigma}-\partial_\nu\Gamma^{\rho}_{\mu\sigma}+\Gamma^{\rho}_{\mu\lambda}\Gamma^{\lambda}_{\nu\sigma}-\Gamma^{\rho}_{\nu\lambda}\Gamma^{\lambda}_{\mu\sigma}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma^{i}_{kl} = \frac{1}{2}g^{im}(\partial_lg_{mk}+\partial_kg_{ml}-\partial_mg_{kl})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as you see once you have the metric you can calculate all those quantities.&lt;/p&gt;&#xA;" OwnerUserId="2151" LastEditorUserId="2151" LastEditorDisplayName="" LastEditDate="2011-08-30T16:58:37.123" LastActivityDate="2011-08-30T16:58:37.123" CommentCount="3" />
  <row Id="14138" PostTypeId="1" CreationDate="2011-08-30T17:02:53.980" Score="0" ViewCount="78" Body="&lt;p&gt;When to parallel wires carrying currents in same direction I1 &amp;amp; I2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=43AeuDvWc0k&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=43AeuDvWc0k&lt;/a&gt; this video demonstrates that effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, why &lt;strong&gt;exactly&lt;/strong&gt; does this happen? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the reason, but I'm not convinced with it. One wire generates the magnetic field which is into the plane at another wire. Electrons are moving in the wire which experience lorentz force F = q(&lt;strong&gt;V X B&lt;/strong&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My arguments are, &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;this force is experienced by the electrons, not the neucleaus. And these electrons that are in motion are the &quot;Free electrons&quot;. So, when the experience force just they alone should be drifted towards/away from the wire but not the entire atom(s). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The only force binding an electron to the material/matter is the coulumbic attraction force from the neucleus. If the lorentz force is sufficiently large, then it should be able to remove electrons from atoms. I other words, they should come out of the material.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But I never heard/read of any thing like that happening. Why doesn't this happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, atoms must not experience any force, then why is it that entire wire is experiencing a force of i(&lt;strong&gt;L X B&lt;/strong&gt;)?&lt;/p&gt;&#xA;" OwnerUserId="447" LastActivityDate="2011-08-31T01:25:56.420" Title="Why exactly does current carrying two current wires attract/repel?" Tags="&lt;electromagnetism&gt;" AnswerCount="2" />
  <row Id="14139" PostTypeId="2" ParentId="14138" CreationDate="2011-08-30T17:09:41.740" Score="3" ViewCount="" Body="&lt;p&gt;You are right in both arguments. The thing is just, this &quot;only force, ... the coulombic attraction&quot; is incredibly much stronger than the Lorentz force due to the magnetic field of a single wire carrying current in the same direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for &quot;In any case, atoms must not experience any force&quot;, this is obviously wrong, as can be seen very plainly when you think of Newton's third law and the fact that the coulomb attraction occurs between the electrons and nuclei in the wire of question.&lt;/p&gt;&#xA;" OwnerUserId="3540" LastActivityDate="2011-08-30T17:09:41.740" />
  <row Id="14140" PostTypeId="1" AcceptedAnswerId="14161" CreationDate="2011-08-30T18:04:27.033" Score="1" ViewCount="102" Body="&lt;p&gt;When you spray gas from a compressed spray, the gas gets very cold, even though, the compressed spray is in the room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think, when it goes from high pressure to lower one, it gets cold, right? but what is the reason behind that literally?&lt;/p&gt;&#xA;" OwnerUserId="5088" LastActivityDate="2011-08-31T04:32:39.233" Title="Why does the gas get cold when I spray it?" Tags="&lt;temperature&gt;&lt;pressure&gt;" AnswerCount="4" />
  <row Id="14141" PostTypeId="2" ParentId="14140" CreationDate="2011-08-30T18:18:49.473" Score="1" ViewCount="" Body="&lt;p&gt;The ideal gas equation states that 'PV = nRT'. P is Pressure, V is volume, n is the number of moles (the amount of 'stuff') R is the universal gas constant, and T is temperature. Of course the gas in your can is not an ideal gas, but it's close. So, when you release it from its compressed state in the container, you relieve a huge amount of pressure. P drops. The volume doesn't go up enough to compensate for the pressure drop, so something else has to change. That something can't be the amount of stuff, nor the gas constant, so the temperature drops as well. &lt;/p&gt;&#xA;" OwnerUserId="1872" LastActivityDate="2011-08-30T18:18:49.473" CommentCount="2" />
  <row Id="14142" PostTypeId="1" CreationDate="2011-08-30T18:59:43.483" Score="0" ViewCount="66" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/3081/why-cant-a-piece-of-paper-of-non-zero-thickness-be-folded-more-than-n-times&quot;&gt;Why can&amp;#39;t a piece of paper (of non-zero thickness) be folded more than &amp;quot;n&amp;quot; times?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;On skeptics.stackexchange, there is a question on the &lt;a href=&quot;http://skeptics.stackexchange.com/questions/5912/is-it-impossible-to-fold-a-sheet-of-paper-in-half-more-than-seven-times&quot;&gt;maximum limit on the folds of a paper&lt;/a&gt;. The referenced answer said that &lt;a href=&quot;http://en.wikipedia.org/wiki/Britney_Gallivan&quot; rel=&quot;nofollow&quot;&gt;Britney Gallivan&lt;/a&gt; derived an equation to estimate the number of maximum folds possible for a given sheet of paper. The answer justified the 'only'-folding scenario but when the same scenario is changed to folding the paper preceded by tearing the paper off into two equal halves, it fails.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to tear an A4 sheet off in two equal halves and kept on tearing it in two equal halves in alternate directions by first folding the paper and then tearing it. After the 7th tear-off, I found that I could still fold the paper for the 8th time. Applying the same reasoning here, the A4 should fail to fold for the 8th time but it doesn't. Why?&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditorDisplayName="" LastEditDate="2011-08-31T08:41:26.090" LastActivityDate="2011-08-31T08:41:26.090" ClosedDate="2011-08-31T01:02:22.150" Title="Maximum limit on the number of paper-folds possible after tearing into halves" Tags="&lt;everyday-life&gt;" AnswerCount="2" CommentCount="3" />
  <row Id="14143" PostTypeId="2" ParentId="14124" CreationDate="2011-08-30T19:13:42.883" Score="2" ViewCount="" Body="&lt;p&gt;For practical reasons, physicists like to label the states of the system by a set of  &quot;quantum numbers&quot;. Technically this means that you are looking for a set of mutually commuting Hermitian operators such that: (i) Every vector from the basis of common eigenvectors of these operators is uniquely characterized by the set of eigenvalues (i.e. the above mentioned quantum numbers); (ii) The set of operators is minimal in the sense that by removing any of the operators, you lose the property (i). The first condition implies that your set of operators is complete while the second one that there are no redundancies. Since you typically want to use the quantum numbers to label the stationary states, that is, the eigenstates of the Hamiltonian, you in addition require that your operators commute with the Hamiltonian (the Hamiltonian itself need not be included in the set though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is obvious that for any given system and its Hilbert space, there are many sets of operators that satisfy the above two conditions. In principle, you can find your own set of operators using the following algorithm. Pick any (Hermitian) operator $A$ and determine its spectrum. If there is no degeneracy, you are done. Otherwise, add another operator $B$ that commutes with $A$ such that there is a pair of common eigenvectors with the same eigenvalues of $A$ but different eigenvalues of $B$. (This guarantees that you improve your &quot;resolution&quot; in the specification of the eigenvectors by the set of eigenvalues.) If there are no two eigenvectors for which both $A$ and $B$ have the same eigenvalue, you are done. Otherwise, repeat the previous step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be emphasized that the notion of a complete set of commuting operators introduced in this way is a bit vague. For example, it does not say &lt;em&gt;how many&lt;/em&gt; operators you need to form a complete set in a given Hilbert space. In fact, as pointed out by Willie Wong in his comment, one operator is in principle enough in a separable Hilbert space. Here is maybe a slightly less abstract example. You know that the states of a spin system can be characterized by specifying simultaneously the values of $\vec L^2$ and $L_z$. (I disregard the orbital degrees of freedom so that these two operators are enough.) However, one can as well take the operator $\vec L^2+\alpha L_z$ where $\alpha$ is an arbitrary &lt;em&gt;irrational&lt;/em&gt; number. You can easily convince yourself that the eigenvalue of this single operator is sufficient to label uniquely all the states $|l,m\rangle$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your second question, I would simply refer you to any textbook on quantum mechanics. The fact that the eigenvalues of $\vec L^2$ and $L_z$ take the form $\hbar l(l+1)$ (where $l$ is a non-negative integer or half-integer) and $\hbar m$ where $m\in\{-l,-l+1,\dots,l-1,l\}$, is not merely a matter of a definition, but follows from the commutation relation for the angular momentum operator,&#xA;$$&#xD;&#xA;[L_i,L_j]=i\hbar\varepsilon_{ijk}L_k.&#xD;&#xA;$$  &lt;/p&gt;&#xA;" OwnerUserId="1641" LastActivityDate="2011-08-30T19:13:42.883" CommentCount="1" />
  <row Id="14144" PostTypeId="1" AcceptedAnswerId="14209" CreationDate="2011-08-30T19:15:13.757" Score="2" ViewCount="59" Body="&lt;p&gt;Say you want to calculate degree of ionization for different gases in atmosphere of a star with abundances similar to those in Sun (let's assume you only have hydrogen, helium and sodium) over the temperature range (from 2000 K to 45000 K for example) using Saha equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{n_{i+1}}{n_i}=\frac{g_{i+1}}{g_i} \frac{2}{n_e} \frac{{(2\pi m_e)}^{3/2}}{h^3} {(k_B T)^{3/2}} e^{-\chi /k_B T}$$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which you write down for all three elements and of course next to abundances and temperature you also know ionization potentials $\chi$ for each element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can one calculate electron density in that case and how does it change? I understand, that at lower temperatures number of electrons is equal to number of ionized sodium atoms since it is the easiest to ionize (and in general $n_e=n_H^1+n_{He}^1+n_{Mg}^1$, where 1 means first level of ionization) but that doesn't help much. And additional question: should higher levels of ionization be included, given the temperature? &lt;/p&gt;&#xA;" OwnerUserId="4264" LastActivityDate="2011-09-02T06:50:01.593" Title="Degree of ionization and Saha equation" Tags="&lt;astrophysics&gt;" AnswerCount="1" FavoriteCount="1" />
  <row Id="14145" PostTypeId="2" ParentId="14140" CreationDate="2011-08-30T19:21:11.587" Score="1" ViewCount="" Body="&lt;p&gt;To do this properly, you need to describe the gas by a statistical ensemble. But this answers the question, too:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can describe the gas as a... well, &lt;em&gt;gas&lt;/em&gt; of classical particles, flying around in the can quite fast and in a randomish fashion. The temperature is essentially the average kinetic energy per particle (more precisely: per degree of freedom)&#xA;$$&#xD;&#xA;  E = \sum_{\text{particles}}\!\!\frac{mv^2}2 = \tfrac32n_\text{particles}\bar{\cdot}\!\!\!\!kT&#xD;&#xA;$$&#xA;Where $\bar{\cdot}\!\!\!\!k$ is the Boltzmann constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They bounce off the walls elastically, so the average squared velocity stays constant, and therefore also the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what happens now if we expand the volume, that is, move one of the walls outwards? While we do this, particles bouncing off the moving wall will not be reflected back in the volume with constant velocity anymore – the $v^2$ will stay constant as seen from the wall, but as it is moving itself, the velocity towards the wall was actually higher than it is back after the collision. So the average $v^2$ will also decrease, and so will the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this process is called &lt;em&gt;adiabatic expansion&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3540" LastEditorUserId="3540" LastEditorDisplayName="" LastEditDate="2011-08-30T19:47:19.990" LastActivityDate="2011-08-30T19:47:19.990" CommentCount="2" />
  <row Id="14146" PostTypeId="2" ParentId="14104" CreationDate="2011-08-30T20:33:42.467" Score="2" ViewCount="" Body="&lt;p&gt;CRT displays have a fairly strong electric charge and this induces a dipole moment in the pollen particle itself causing it to be attracted to the screen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somewhat related, the hairs on the bodies of some bees become electrically charged and this attracts pollen particles. Exercise 51 on page 602 of this document &lt;a href=&quot;http://cyber.gwc.cccd.edu/faculty/kstein/H_R_Ch_22.pdf&quot; rel=&quot;nofollow&quot;&gt;http://cyber.gwc.cccd.edu/faculty/kstein/H_R_Ch_22.pdf&lt;/a&gt; will give you a ball park figure for the possible strength of this attraction. It assumes a neutral pollen particle.&lt;/p&gt;&#xA;" OwnerUserId="4983" LastEditorUserId="4983" LastEditorDisplayName="" LastEditDate="2011-08-30T21:43:03.400" LastActivityDate="2011-08-30T21:43:03.400" />
  <row Id="14147" PostTypeId="2" ParentId="14142" CreationDate="2011-08-30T20:39:28.793" Score="-1" ViewCount="" Body="&lt;p&gt;If you had a sheet of sufficient size and folded in half once again, forty times, how thick would be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle we could think we can do bends over 100 times, however, be surprised that you can not get more than 8 or 9 folds. The record is 12 and this very difficulty!.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It happens that when we fold the paper also doubling its thickness, and thus the folding it difficult to follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depue first folded thickness is doubled, that is,$$ E = 2e $$ Where &quot;E&quot; is the thickness of the folded sheet and e the thickness of the original sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second fold is the total thickness of $$E &amp;#39;= 2E $$, and the above formula $$E&amp;#39; = 4e $$ or $$E &amp;#39;= 2 * 2e $$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Staring into the last sentence we can infer that, for example, the folding number 8 the total thickness will be $$E &amp;#39;= 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * e $$ By multiplying the account found that in only 8 fold increased the total thickness ... .. 256 times !!!!!&lt;/p&gt;&#xA;" OwnerUserId="5050" LastActivityDate="2011-08-30T20:39:28.793" CommentCount="3" />
  <row Id="14148" PostTypeId="1" AcceptedAnswerId="14149" CreationDate="2011-08-30T20:56:27.890" Score="2" ViewCount="57" Body="&lt;p&gt;When there is a wave, something is undulating. In the example of a rope, the rope is what undulates. In the case of a ripple on a pond, the water is undulating, and when a sound wave propagates, the air is undulating. The question is: in the case of a particle, what undulates?&lt;/p&gt;&#xA;" OwnerUserId="5050" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-30T21:04:04.467" LastActivityDate="2011-08-30T21:18:27.750" Title="What is it that undulates in a particle?" Tags="&lt;quantum-mechanics&gt;&lt;waves&gt;" AnswerCount="1" CommentCount="3" />
  <row Id="14149" PostTypeId="2" ParentId="14148" CreationDate="2011-08-30T21:18:27.750" Score="4" ViewCount="" Body="&lt;p&gt;The undulations are of the probability amplitude for the particle to be someplace. The notion of probability amplitude is fundamental, and cannot be reduced to anything more primitive. It is described on Wikipedia under &quot;superposition principle&quot;. The undulations are in the space of all possible universes, so that two particles are described by undulations in the 6 dimensional space of all possible pairs of positions, three particles are described by undulations in 9 dimensions. There is no physical way of making it like waves on water or sound in air, because physical waves travel in three dimensions of space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are asking about classical fields, like E and B fields, there are not undulations in anything either. They are the primitive things out of which things like atoms and water are built. You can make classical fields out of particles if they have the right statistics, and these fields, when they are made of matter, are called Bose-Einstein condensates or superfluids, depending on the density.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-30T21:18:27.750" />
  <row Id="14150" PostTypeId="2" ParentId="14140" CreationDate="2011-08-30T21:27:45.803" Score="2" ViewCount="" Body="&lt;p&gt;your question is about large drops in pressure, and why they cool gasses. The answer is that the gas is doing work in the process of expanding, and this work releases energy to the environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you prevent the gas from doing work, if there is nothing for it to push against, it doesn't get cold. If you have a dilute gas in the corner of a room and you open a barrier to a vacuum, the gas expands into the vacuum with no change in temperature. This is not what you are doing when you spray the can into air. There, the gas is encountering air, and produces a pressure wall which it then pushes against doing work. Once the equilibrium spray-profile is established, there is a pressure gradient from the can outward that accelerates the spray to its final velocity. Travelling along this pressure gradient, the gas expands and does work, and this removes energy from the gas. The cold temperature profile sneaks back towards the can, because the air is such a lousy conductor of heat, so the heat is all coming from the can. Eventually, your hand gets cold.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-30T21:27:45.803" CommentCount="2" />
  <row Id="14151" PostTypeId="2" ParentId="14138" CreationDate="2011-08-30T21:37:49.047" Score="1" ViewCount="" Body="&lt;p&gt;Your question is assuming that the electrons are weakly interacting with the nucleus. The interaction with the nucleus is extremely strong. It is better to ask instead why do we have conductivity at all. Electrons are so tightly bound to nuclei of atoms, why should a tiny external electric field get them moving?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that quantum mechanical effects can spread out electrons over many atoms. This is responsible for chemical bonding. In metals, the electrons have a spread out wavefunction, and the energy-band of spread-out electron states is only partly filled, so it only takes a little bit of energy to push an electron into motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for your original question, there is an easy way to see the answer. Consider two infinite charged wires 1 cm apart. You know that they repel, so they move apart. Now boost to a frame moving along the wires at a huge speed, near the speed of light. Relavistic time dilation slows down the rate at which they move apart. But the charge density has gone up in this frame, because of the length contraction. So there must be an additional attractive force due to the currents in the wires. In the limit that you are moving at the speed of light, the attractive like-current force must exactly cancel the repulsive like-charge electrostatic force.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-31T01:25:56.420" LastActivityDate="2011-08-31T01:25:56.420" />
  <row Id="14152" PostTypeId="2" ParentId="13714" CreationDate="2011-08-30T22:03:35.080" Score="0" ViewCount="" Body="&lt;p&gt;I find it appalling that of all the topics touted hear as ways to inspire undergraduates, by and large none of them are really understandable by a general undergraduate physics audience. This is what is wrong with physics today: a lot of people talking about things they pretend to understand but really don't. What about the many topics that are already in the undergraduate curriculum? The problem is that they are taught fomulaically. I would have liked to see almost any physics topic treated with really good physical insight; one sore point I can recall in particular was Lagrangian mechanics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, I don't see anything uninspiring about learning physics. Give me billiard ball collisions, give me coupled pendulums, give me Carnot cycles, whatever: just teach it as though it's more than a formula to memorize for the exam.&lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-08-30T22:03:35.080" CommunityOwnedDate="2011-08-30T22:03:35.080" />
  <row Id="14153" PostTypeId="2" ParentId="14142" CreationDate="2011-08-30T22:40:21.553" Score="1" ViewCount="" Body="&lt;p&gt;The point about Britney Gallivan's work is that not only do you lose length when folding several sheets, but that this accumulates and she worked out the impact of this accumulated loss.  This is illustrated in the &lt;a href=&quot;http://pomonahistorical.org/12times.htm&quot; rel=&quot;nofollow&quot;&gt;Historical Society of Pomona Valley&lt;/a&gt; page &quot;Folding Paper in Half 12 Times&quot; linked from the skeptics.se page with this picture &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/cy6Dw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the length of parallel paper after four folds is substantially less than half that after three folds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you tear the paper after each fold then you do not have this accumulation, so you would naturally expect to be able to fold more times when tearing is allowed.  That is what you have confirmed.&lt;/p&gt;&#xA;" OwnerUserId="2714" LastActivityDate="2011-08-30T22:40:21.553" />
  <row Id="14154" PostTypeId="1" CreationDate="2011-08-30T22:45:58.817" Score="1" ViewCount="12" Body="&lt;p&gt;on behalf of a small community of german scientists [physics &amp;amp; astrophysics, astronomy, not insiders in QG] and interested laymen I would like to compile a rather short questionlist regarding current status of LQG. The list should reflect current research topics [general questions, recent advances, major problems and open questions]. Our intention is not to start an own research program, it's just to get a brief overview and to understand better some arxiv papers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be OK for you a) to support us compiling such a list [suggestions from our side] and b) to answer the questions from your perspective?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best regards&#xA;Tom&lt;/p&gt;&#xA;" OwnerUserId="5090" LastEditorUserId="5090" LastEditorDisplayName="" LastEditDate="2011-08-30T23:48:41.460" LastActivityDate="2011-08-30T23:48:41.460" ClosedDate="2011-08-31T00:39:31.503" Title="Questionlist regarding current status of LQG" Tags="&lt;quantum-gravity&gt;" />
  <row Id="14155" PostTypeId="2" ParentId="14136" CreationDate="2011-08-30T22:47:15.660" Score="2" ViewCount="" Body="&lt;p&gt;The previous answer is correct, but does not give a practical algorithm for humans, because it is a nightmare to calculate the curvature tensor. You need a good hand algorithm, or else you need a symbolic manipulation package. I prefer hand calculations for the symmetric Ansatzes, because they are always revealing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The traditional simplified method is to use curvature forms, and this method is described in Misner Thorne and Wheeler. It is indespensible for understanding Kerr Solutions. It also pays to study the Newman-Penrose formalism, because it gives physical insight. I prefer to use my own mathematically inelegant home-made method, because much of the simplification in the advanced methods is really only due to the use of what is called &quot;sparse-matrix computation&quot; in the computer science literature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a matrix which is mostly zeros, like the on-diagonal curvature, you shouldn't write it in matrix form, unless you want to build good strong writing-hand muscles. Introduce noncovariant basis tensors &quot;l_{ij}&quot; which are nonzero in i,j position. Then write the metric in a mostly-plus convention as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g_{\mu\nu} = -A l_{00} + B l_{11} + C l_{22} $$&#xA;$$g^{\mu\nu} = -{1\over A}l^{00} + {1\over B} l^{11} + {1\over C}l^{22}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where I have gone to three dimensions so as to prevent a hopelessly long answer, and where I hope the notation is clear. For theoretical elegance, the basis tensor $l_{00}$ should really be written as $l^{00}_{\mu\nu}$ if you want it to be consistent with  the usual index conventions, but since the goal is to get the writing muscles as flabby as possible, don't do that. Since the l's are ridiculously coordinate dependent, you can express ridiculously non-tensorial objects like the connection coefficients and the pseudo-stress-energy tensor.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Calculating the connection&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;There are tricks to calculating the connection, like deriving the geodesic equation, but I won't use them. If you use the basis-tensors, it is no work at all to get the connection coefficients, and with practice, you can do most of the work in your head for the simpler Ansatzes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, differentiate the metric. Since &quot;diagonal&quot; is not much of a simplification, I will assume &quot;diagonal and dependent only on x1 and x0&quot;. I will use a prime for differentiating with respect to $x_1$, and a dot for differentiating with respect to $x_0$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g_{\mu\nu,\alpha} = - \dot{A} l_{000} - A&amp;#39; l_{001} + \dot{B}l_{110} + B&amp;#39;l_{111} + \dot{C}l_{220} + C&amp;#39; l_{221}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lesson is--- differentiation is trivial. Notice that this is symmetric on the first two indices, and nothing special on the third index. The Christoffel symbols are symmetric on the last two indices, and nothing special on the first. You transfer symmetry between index positions like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{i|jk} = Q_{ij|k} + Q_{ik|j} - Q_{jk|i}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where P is symmetric in the last two position, and Q is symmetric in the first two. Get used to this, because it comes up a lot. The first term has the same index order just by using a good index order convention, the second term forcibly symmetrizes the second and third positions, and the last term is required so that P keeps all the information in Q. You can do this procedure on the l's automatically, just by replacing $l_{001}$ with $l_{001} + l_{010} - l_{100}$, and so on. This is what you do to get $\Gamma$ from the derivative of g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the formula for the all-lower-index $\Gamma$ (its not written this way ever, because $\Gamma$ is not a tensor, but I do it here, just to spare the writing hand).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma_{\mu|\nu\sigma} = -{1\over 2} ( -\dot{A} l_{000} - A&amp;#39;(l_{001} + l_{010} -l_{100}) $$&#xA;$$+ \dot{B} (l_{110} + l_{101} - l_{011}) + B&amp;#39; l_{111}$$&#xA;$$+ \dot{C}(l_{220} + l_{202} - l_{022}) + C&amp;#39;(l_{221} + l_{212} - l_{122}) )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To raise the indices when the metric is diagonal is trivial, you just raise the index on the l and divide by the appropriate diagonal entry:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma^\mu_{\nu\sigma} = {\dot{A}\over 2A} l^0_{00} - {A&amp;#39;\over 2A} (l^0_{01} + l^0_{10}) - {A&amp;#39;\over 2B} l^1_{00} + {\dot{B}\over 2B}(l^1_{10} + l^1_{01}) - {\dot{B}\over 2A}l^0_{11} + {B&amp;#39;\over 2B} l_{111} +... $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the rest should be obvious. With practice, this takes a minute to do by hand.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Calculating the Ricci curvature.&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;To calculate the curvatures, it is important to &lt;em&gt;trace as you go&lt;/em&gt;, because this halves the work. The Riemann curvature always has a bunch of Weyl junk that you mostly don't care about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I always write the formula for the Riemann tensor this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ R^\mu_{\nu\lambda\sigma} = \Gamma^\mu_{\nu\lambda,\sigma} \mathrm{(AS)} + \Gamma^\mu_{\nu s}\Gamma^s_{\lambda\sigma} \mathrm{(AS)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;AS&quot; means subtract the same expression with $\nu$ and $\sigma$ interchanged. This form has the property that it is antisymmetric on the lower first and third index, so this is not the usual convention for the Riemann tensor, which is antisymmetric on the last two indices. But this is easy to fix at the end. Trust me, this is the best convention, you fix it up at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Ricci trace in this convention is on the first two indices:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ R_{\mu\nu} = R^\alpha_{\alpha\mu\nu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is important, because each term you get in the Riemann tensor comes with an &quot;l&quot;, and if the upper number of the l is not the same as the leftmost lower number, then that term doesn't contribute to the Ricci tensor. To get the Ricci tensor, you just ignore all $l^a_{bcd}$ with $a\ne b$ and write down $l_{cd}$ in place of those $l$'s which have $a=b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, differentiate the expression for $\Gamma$, tacking an index on the end. I will demonstrate with one of the contributions, from taking the $x_1$ derivative of the first term:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma^\mu_{\nu\lambda,\sigma} \mathrm{(AS)} = ..+ ({\dot{A}\over 2A})&amp;#39; (l^0_{001} - l^0_{100})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the second term is the antisymmetrizer. But now, look at the two l's --- does one of them have a matching index on the top left and bottom left? Yes! So erase the top and bottom left numbers from the l, and you are left with a contribution to the Ricci tensor:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$({\dot{A}\over 2A})&amp;#39;l_{01}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can all be done in your head, term by term. If you get an $l$ which is $l^0_{221}$ it can't contribute to Ricci, because the bottom first and third index don't match the top, if you get $l^0_{121}$ then it is killed by antisymmetrization, etc, etc, it's all obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next you need to multiply $\Gamma$ with itself, and trace. Here you work out all the terms. But it is a finite calculation, not a hopeless one. You don't get a contribution unless the leftmost lower index is the same as the upper index, so that leaves only a few terms, and further, you get zero on some l-terms after antisymmetrizing and tracing (in your head). Then there are only a handful of remaining terms, and these are real contributions to the curvature, so there is no way to avoid calculating them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first time takes a while, but with practice it only takes some minutes for simpler Ansatzes and some hours for the worst ones.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-09-01T15:51:13.310" LastActivityDate="2011-09-01T15:51:13.310" CommentCount="3" />
  <row Id="14156" PostTypeId="1" CreationDate="2011-08-31T00:22:11.920" Score="-1" ViewCount="49" Body="&lt;p&gt;pluton suppose that there is a mirror that can reflect any type of radiation that hits it. suppose that a photon sent from earth to the mirror. How long to return to the earth, bounced off the mirror? and if it were mounted on the photon, how much time passed until the return?&lt;/p&gt;&#xA;" OwnerUserId="5050" LastActivityDate="2011-08-31T00:22:11.920" ClosedDate="2011-08-31T01:27:16.793" Title="return time of a photon" Tags="&lt;particle-physics&gt;&lt;light&gt;" CommentCount="3" />
  <row Id="14157" PostTypeId="2" ParentId="6844" CreationDate="2011-08-31T03:09:04.460" Score="2" ViewCount="" Body="&lt;p&gt;This is because time is like space, and moving objects are tilted in time. It is no more mysterious than the statement that two rulers, one vertical and one tilted, have different heights. Not only that, but if you tilt your floor so that it is perpendicular to the tilted ruler, now the ruler that was previously vertical and taller is now tilted and shorter! There is no more mystery here than geometry, except that the geometry is slightly different from Euclid's in that the pythagorean theorem has a minus sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you are standing still, your path in space-time is parallel to the time axis. When you are moving, your path in space-time is tilted relative to the time axis. If you go somewhere and come back, you make a triangle in space time, and the time you measure is the sum of the lengths of the two legs of the triangle that represent your path, and the time measured by someone who doesn't move is the third leg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In ordinary geometry, the sum of two legs of a triangle are always more than the third, in relativity, it is always less (under the condition that an observer can travel on each leg). The reason for this difference is explained in this answer: &lt;a href=&quot;http://physics.stackexchange.com/questions/12435/einsteins-postulates-minkowski-space-in-laymans-terms/13621#13621&quot;&gt;Einstein&amp;#39;s postulates &amp;lt;==&amp;gt; Minkowski space. (In layman&amp;#39;s terms)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the minus sign in the pythagorean theorem, there are special right triangles whose hypotenuse is of zero length. These triangles have a vertical side which is always proportional to the horizontal side, in other words, their hypotenuse represent an object travelling with constant speed. This speed is the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is just an accident of our universe that light travels at this speed. If light had a small mass, it wouldn't quite travel at c, but everything else would work out fine. But we would still have relativity exactly the same as before, and the parameter &quot;c&quot; that we call the speed of light would still be exactly the same. It would just get a new name, perhaps &quot;the speed of gravity&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-08-31T03:30:22.720" LastActivityDate="2011-08-31T03:30:22.720" />
  <row Id="14158" PostTypeId="2" ParentId="10068" CreationDate="2011-08-31T03:33:24.343" Score="0" ViewCount="" Body="&lt;p&gt;In the interest of time,  I would like to refer the reader to the following two references regarding 'wave function collapse' or interpretations of the measurement problem.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Quo Vadis Quantum Mechanics?&#xA;Avshalom C. Elitzur, Shahar Dolev and Nancy Kolenda&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Quantum Enigma: Physics Encounters Consciousness, by Bruce Rosenblum  and Fred Kuttner&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5092" LastActivityDate="2011-08-31T03:33:24.343" />
  <row Id="14161" PostTypeId="2" ParentId="14140" CreationDate="2011-08-31T04:26:25.077" Score="3" ViewCount="" Body="&lt;p&gt;The temperature of the gas that is sprayed goes down because it adiabatically expands. This is simply because there is no heat transferred to or from the gas as it is sprayed, for the process is too fast. (See &lt;a href=&quot;http://en.wikipedia.org/wiki/Adiabatic_process&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia article&lt;/a&gt; for more details on adiabatic processes.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mathematical explanation goes as follows: let the volume of the gas in the container be $V_i$, and its temperature $T_i$. After the gas is sprayed it occupies volume $V_f$ and has temperature $T_f$. In an adiabatic process $TV^{\,\gamma-1}=\text{constant}$ ($\gamma$ is a number bigger than one), and so&#xA;$$&#xD;&#xA;T_iV_i^{\,\gamma-1}=T_fV_f^{\,\gamma-1},&#xD;&#xA;$$&#xA;or&#xA;$$&#xD;&#xA;T_f=T_i\left(\frac{V_i}{V_f}\right)^{\gamma-1}.&#xD;&#xA;$$&#xA;Since $\gamma&amp;gt;1$ and, clearly, $V_f&amp;gt;V_i$ (the volume available to the gas after it's sprayed is much bigger than the one in the container), we get that $T_f&amp;lt;T_i$, i.e. the gas cools down when it's sprayed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, adiabatic expansion is the reason why you are able to blow both hot and cold air from your mouth. When you want to blow hot air you open your mouth wide, but when you want to blow cold air you tighten your lips and force the air through a small hole. That way the air goes from a small volume to the big volume around you, and cools down according to the equations above.&lt;/p&gt;&#xA;" OwnerUserId="4483" LastEditorUserId="4483" LastEditorDisplayName="" LastEditDate="2011-08-31T04:32:39.233" LastActivityDate="2011-08-31T04:32:39.233" CommentCount="11" />
  <row Id="14162" PostTypeId="2" ParentId="14116" CreationDate="2011-08-31T05:39:01.937" Score="7" ViewCount="" Body="&lt;p&gt;As it seems that you are not completely satisfied with Ron Maimon's answer, I'll put it in a bit different way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that in your derivation you have a hidden ambiguity. $$\langle{\psi}_p\vert \hat{x}\vert\psi_p\rangle = \infty \;\;\;\;\; \Rightarrow \;\;\;\;\; \langle[\hat{x},\hat{p}]\rangle=...=(p-p)\langleψ_p|\hat{x}|ψ_p\rangle = 0\cdot\infty = \text{any number} $$&#xA;The problem is with the functions. Eigenfunctions of both momentum operator and coordinate operator are not really functions. They do not belong to integrable functions space and thus you can not freely work with them pretending that they are. Sometimes you can, but if you do at some point you find yourself in trouble. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take some &quot;proper&quot; function and do the math, you will find no problems. Let us take e.g. $$\psi(x)=\frac1{\sqrt{\pi}}e^{-x^2/2}$$ &#xA;Then $$ \int_{-\infty}^{\infty} \psi(x) x\left( -i\hbar \frac{∂}{∂x} \right) \psi(x)dx = i\hbar \frac1{\pi} \int_{-\infty}^{\infty} x^2 e^{x^2}dx = i\hbar \frac1{2\sqrt{\pi}}$$&#xA;$$ \int_{-\infty}^{\infty} \psi(x) \left( -i\hbar \frac{∂}{∂x} \right)x \psi(x)dx = i\hbar \frac1{\pi} \int_{-\infty}^{\infty} \left(x^2-1\right) e^{x^2}dx = i\hbar \frac1{2\sqrt{\pi}}-i\hbar$$&#xA;The difference is what you expected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take $\psi_a(x)=\frac1{a}\psi(x/a)$ and note that  $\lim_{a\to0}\psi_a(x) = \delta(x) = \vert x\rangle$ you will get an idea how this paradox for $\vert x \rangle$ may be solved and check that the solution is a correct way to handle $0\cdot\infty$. Similar trick may be used to resolve your paradox. Just functions which have $\psi_p$ as a limit are less convinient. &lt;/p&gt;&#xA;" OwnerUserId="4657" LastActivityDate="2011-08-31T05:39:01.937" CommentCount="33" />
  <row Id="14163" PostTypeId="1" CreationDate="2011-08-31T06:41:39.317" Score="0" ViewCount="51" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;http://physics.stackexchange.com/questions/1888/balloons-and-lifting-gases&quot;&gt;Balloons and lifting gases&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;My guess is that making a vacuum in a large solid baloon would make its volumic mass lesser than that of the air surrounding it, allowing it to float on air. Is there something wrong with this concept?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, could you make a ship fly by vacuuming its insides?&lt;/p&gt;&#xA;" OwnerUserId="4739" LastActivityDate="2011-08-31T06:41:39.317" ClosedDate="2011-08-31T06:46:53.620" Title="Would a baloon &quot;filled&quot; with a vacuum rise?" Tags="&lt;vacuum&gt;&lt;buoyancy&gt;" CommentCount="1" />
  <row Id="14165" PostTypeId="1" AcceptedAnswerId="14166" CreationDate="2011-08-31T07:21:48.213" Score="7" ViewCount="241" Body="&lt;p&gt;My professor told me recently that Area is a vector. A &lt;a href=&quot;http://www.google.co.in/#sclient=psy&amp;amp;hl=en&amp;amp;source=hp&amp;amp;q=what%20is%20a%20vector&amp;amp;pbx=1&amp;amp;oq=what%20is%20a%20ve&amp;amp;aq=1&amp;amp;aqi=g5&amp;amp;aql=&amp;amp;gs_sm=sc&amp;amp;gs_upl=1666l6221l0l7939l12l9l0l2l2l0l541l3051l2-3.1.2.2l8l0&amp;amp;bav=on.2,or.r_gc.r_pw.&amp;amp;fp=5b60a6a8a4733e57&amp;amp;biw=1024&amp;amp;bih=629&amp;amp;pf=p&amp;amp;pdl=300&quot; rel=&quot;nofollow&quot;&gt;Google search&lt;/a&gt; gave me the following definition for a vector:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Noun: A quantity having direction as well as magnitude, esp. as&#xA;  determining the position of one point in space relative to another.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question is - what is the direction of area? I can relate to the fact that velocity is a vector. The velocity of a moving motorbike for example, has a definite direction as well as a definite magnitude assuming that the bike is moving in a straight line &amp;amp; not accelerating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My friend gave me this explanation for the direction of Area vector. Consider a rectangular plane in space. He argued that the &lt;em&gt;orientation&lt;/em&gt; of the plane in space can only be described by considering area as a vector &amp;amp; not a scalar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I still wasn't convinced. Suppose the plane was placed such that its faces were perpendicular to the directions, North &amp;amp; South for example. Now the orientation of the plane is the same &lt;em&gt;irrespective&lt;/em&gt; whether the so called vector points to north or to the south. Further what is the direction of a sphere's area?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does considering area as a vector have any real significance? Please explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="4711" LastEditorUserId="1501" LastEditorDisplayName="" LastEditDate="2011-08-31T08:02:05.037" LastActivityDate="2011-08-31T21:23:08.610" Title="How can area be a vector?" Tags="&lt;mathematics&gt;&lt;vectors&gt;" AnswerCount="4" CommentCount="2" />
  <row Id="14166" PostTypeId="2" ParentId="14165" CreationDate="2011-08-31T07:52:22.487" Score="7" ViewCount="" Body="&lt;p&gt;This might be more of a math question. This is a peculiar thing about three-dimensional space. Note that in three dimensions, an area such as a plane is a two dimensional subspace. On a sheet of paper you only need two numbers to unambiguously denote a point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine standing on the sheet of paper, the direction your head points to will always be a way to know how this plane is oriented in space. This is called the &quot;normal&quot; vector to this plane, it is at a right angle to the plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/A5iRF.png&quot; alt=&quot;euclideanspace.com&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you now choos the convention to have the length of this vector (&quot;the norm&quot;) equal to the area of this surface, you get a complete description of the two dimensional plane, its orientation in three dimensional space (the vector part) and how big this plane is (the length of this vector).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, you can express this by the &quot;cross product&quot; $$\vec c=\vec a\times\vec b$$ whose magnitude is defined as $|c| = |a||b|sin\theta$ which is equal to the area of the parallelogram those to vectors (which really define a plane) span. To steal this picture from wikipedia's article on the cross product:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;http://i.stack.imgur.com/OrYLl.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said in the beginning this is a very special thing for three dimensions, in higher dimensions, it doesn't work as neatly for various reasons. If you want to learn more about this topic a keyword would be &quot;exterior algebra&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the physical significance of this concept, prominent examples are vector fields flowing through surfaces. Take a circular wire. This circle can be oriented in various ways in 3D. If you have an external magnetic field, you might know that this can induce an electric current, proportional to the rate of change of the amount flowing through the circle (think of this as how much the arrows perforate the area). If the magnetic field vectors are parallel to the circle (and thus orthogonal to its normal vector) they do not &quot;perforate&quot; the area at all, so the flow through this area is zero. On the other hand, if the field vectors are orthogonal to the plane (i.e. parallel to the normal), the maximally &quot;perforate&quot; this area and the flow is maximal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you change the orientation of between those two states you can get electrical current.&lt;/p&gt;&#xA;" OwnerUserId="2151" LastEditorUserId="2151" LastEditorDisplayName="" LastEditDate="2011-08-31T08:16:06.740" LastActivityDate="2011-08-31T08:16:06.740" CommentCount="6" />
  <row Id="14167" PostTypeId="2" ParentId="14165" CreationDate="2011-08-31T08:01:29.110" Score="8" ViewCount="" Body="&lt;p&gt;The main regime of use is when an area is infinitesimally small, like one would use in an integral. In that case, we can easily see that it is flat, and the shape doesn't really matter. In which case, we can encode the information as a vector, with the magnitude representing the (scalar) area; the choice (as you noticed) of pointing out of any given side is exactly that --- a choice --- but one that can be made consistently. We can extend this to non-infinitesimal planes, but it doesn't work that well for curved surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be precise, what you really want is a &lt;em&gt;co-vector&lt;/em&gt;. This is an abstract gadget which takes a vector and spits out a scalar. For a plane, you want this to represent the &quot;amount&quot; of the vector that goes through the plane --- so it should be linear in the vector (doubling the vector doubles the output) and it should take into account the angle at which the vector hits it (gives a factor of $\cos$). Now, we can ask the question of how to &lt;em&gt;represent&lt;/em&gt; this abstract co-vector, and it turns out that a vector is a good idea! Specifically, we can represent the action as taking the dot product, which naturally encodes the linearity and the cosine. Now, in general, this happens to have the same number of dimensions as a proper vector, but this only encodes an area (a 2D surface) in 3D --- in 2D you would get a line, in 4D a volume (yes! A 4-vector intersects a volume at one point!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to learn more about this sort of thing, you want to investigate differential geometry, where all it is necessary to be clear about this sort of thing and not mix up vectors and co-vectors (called &lt;em&gt;forms&lt;/em&gt; in that field). A good readable reference is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/9810220340&quot; rel=&quot;nofollow&quot;&gt;Gauge Fields, Knots and Gravity&lt;/a&gt; which starts from a basic overview of the mathematics and develops it for physical use.&lt;/p&gt;&#xA;" OwnerUserId="1501" LastActivityDate="2011-08-31T08:01:29.110" CommentCount="2" />
  <row Id="14168" PostTypeId="1" CreationDate="2011-08-31T08:56:11.307" Score="1" ViewCount="136" Body="&lt;p&gt;In many theoretical setups it is implicitly assumed that the underlying manifold (i.e. spacetime) is orientable. Then our analysis depends on this implicit assumption. For example, Stokes' theorem assumes orientability of the chain on which we integrate. However, we accept that time &quot;always&quot; points forwardly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: doesn't a 1-directional arrow of time provides a hint that spacetime should be modeled, afterall, by a non-orientable manifold?&lt;/p&gt;&#xA;" OwnerUserId="4975" LastActivityDate="2011-09-01T13:17:04.983" Title="Orientability of spacetime" Tags="&lt;mathematical-physics&gt;&lt;spacetime&gt;" AnswerCount="1" CommentCount="5" />
  <row Id="14169" PostTypeId="2" ParentId="6047" CreationDate="2011-08-31T09:07:49.550" Score="0" ViewCount="" Body="&lt;p&gt;May be &lt;a href=&quot;http://www.mat.univie.ac.at/~westra/dictionarytophys.pdf&quot; rel=&quot;nofollow&quot;&gt;this short one&lt;/a&gt; would be of little help....&lt;/p&gt;&#xA;" OwnerUserId="4975" LastActivityDate="2011-08-31T09:07:49.550" />
  <row Id="14170" PostTypeId="1" AcceptedAnswerId="14172" CreationDate="2011-08-31T10:50:03.550" Score="3" ViewCount="67" Body="&lt;p&gt;For the free particle with quantum number $l=0$, the &lt;strong&gt;&lt;em&gt;regular&lt;/em&gt;&lt;/strong&gt; solution to the radial Schrodinger equation is $R_0 (\rho)=\frac{\sin{\rho}}{\rho}$ while the &lt;strong&gt;&lt;em&gt;irregular&lt;/em&gt;&lt;/strong&gt; solution is $R_0 (\rho)=\frac{\cos{\rho}}{\rho}$. Is there a reason for this nomenclature -- (ir)regular? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5046" LastActivityDate="2011-08-31T14:37:53.527" Title="Nomenclature of radial solutions to the Schrodinger Equation" Tags="&lt;quantum-mechanics&gt;&lt;schrodinger-equation&gt;&lt;nomenclature&gt;" AnswerCount="1" />
  <row Id="14171" PostTypeId="2" ParentId="13996" CreationDate="2011-08-31T11:08:01.420" Score="0" ViewCount="" Body="&lt;p&gt;At first glance this looks ok. It is also reasonable to calculate your conductivity at cell nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a diffusion equation so the only scheme of any real practical use here is central differencing as by the very nature of diffusion the 'flow' of information is isotropic. For heat transfer through fluids (moving fluids) then upwind differencing can come into play to describe the fluid advection, but the difference scheme used for the diffusion terms in these equations would still be central differencing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a comprehensive derivation of this scheme and others like it for thermal heat transfer through solid and fluid media, see:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Patankar, Suhas V. (1980), Numerical Heat Transfer and Fluid Flow, Hemisphere.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Tannehill, John C., et al., (1997), Computational Fluid mechanics and Heat Transfer, 2nd Ed., Taylor and Francis.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Versteeg, H. K., Malalasekera, W., (2007), An Introduction to Computational Fluid Dynamics, The Finite Volume Method. Second Edition.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I hope this was of some use.&lt;/p&gt;&#xA;" OwnerUserId="4312" LastActivityDate="2011-08-31T11:08:01.420" />
  <row Id="14172" PostTypeId="2" ParentId="14170" CreationDate="2011-08-31T11:26:27.303" Score="8" ViewCount="" Body="&lt;p&gt;The reason for this nomenclature is the behavior at $r=0$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{r\to0^{+}}\frac{\sin r}{r} = 1,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{r\to0^{+}}\frac{\cos r}{r} = \infty .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\sin r}{r}$ is &lt;strong&gt;regular&lt;/strong&gt; at $r=0$ while $\frac{\cos r}{r}$ is &lt;strong&gt;irregular&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2151" LastEditorUserId="2451" LastEditorDisplayName="" LastEditDate="2011-08-31T14:37:53.527" LastActivityDate="2011-08-31T14:37:53.527" CommentCount="5" />
  <row Id="14173" PostTypeId="2" ParentId="6047" CreationDate="2011-08-31T11:29:23.600" Score="0" ViewCount="" Body="&lt;p&gt;Mathematics is a language in which physicist express their ideas. It is mathematics which helps to reach and imagine results which are far beyond reach of direct imagination of a physicist mind, at times. Mathematics really helps to imagine complex ideas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mathematician has all the tools ready in his hand to learn physics, as physics utilizes power of mathematics to understand nature. Hence, I believe that it would be of really help if mathematician understands how physicist employs math and how physicist see or interpret math for physical concepts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With classics like Landau's book, Goldstein's book, I would suggest two other classics: Feynman's Lectures on physics part 1 and Berkeley physics course part 1 for classical mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feynman's Lectures will be an exact place to learn physicist point of view.&lt;/p&gt;&#xA;" OwnerUserId="5097" LastActivityDate="2011-08-31T11:29:23.600" />
  <row Id="14174" PostTypeId="1" CreationDate="2011-08-31T11:43:10.667" Score="2" ViewCount="67" Body="&lt;p&gt;I understand how generators work, but I can't for the life of me conceptualize why the current in an AC generator reverses every 180 degrees!!! I understand that, using the right hand rule, it can be seen that the current flows around the coil in a generator in one direction, but I can't see why the current reverses after half a rotation!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been looking at the animations on &lt;a href=&quot;http://www.animations.physics.unsw.edu.au/jw/electricmotors.html#generators&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; page to try and figure it out, but I just can't get there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, I don't understand the concept of how split ring/slip rings work? I know split ring is for direct current, but not really why. For instance, if asked how could I 'explain the difference between a split ring or slip ring'? &lt;/p&gt;&#xA;" OwnerUserId="4880" LastActivityDate="2011-09-01T00:30:20.853" Title="Why does current alternate in an AC generator?" Tags="&lt;electromagnetism&gt;" AnswerCount="3" />
  <row Id="14175" PostTypeId="1" CreationDate="2011-08-31T11:54:01.390" Score="1" ViewCount="69" Body="&lt;p&gt;This is a question directly from a homework sheet, however I can't find any decent answers online so hopefully someone can help! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In single slit diffraction, why is the central maximum double the width of the other bands and  the brightest?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand vaguely that it is something to do with Fresnel zones, but I'm quite confused. Instead of making another question, I think this question also fits in with the above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the intensity of light bands decrease away from the central maximum in single slit diffraction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4880" LastEditorUserId="1255" LastEditorDisplayName="" LastEditDate="2011-08-31T12:31:18.180" LastActivityDate="2011-08-31T17:54:55.913" Title="Why is the central maximum the brightest?" Tags="&lt;homework&gt;&lt;double-slit-experiment&gt;" AnswerCount="3" />
  <row Id="14176" PostTypeId="2" ParentId="14175" CreationDate="2011-08-31T12:41:54.073" Score="0" ViewCount="" Body="&lt;p&gt;I think what you want to know is why the maximum bright spot is the central belt and from it begins to decrease in intensity? the answer is: because the path difference between the center point and the crack and the other parts of the screen have a greater distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can formulate a relationship between the separation of the slits, s, the wavelength λ, the distance from the slits to the screen D, and the width of the interference bands (the distance between successive bright fringes), x&#xA;$$λ / s = x / D$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This expression is only an approximation and formulation depends on specific conditions. It is possible however to calculate the wavelength of incident light from the ultimate relationship. If s and D are known and x is observed then λ can be calculated, which is of particular interest in measuring the wavelength corresponding to beams of electrons or other particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you serve.&lt;/p&gt;&#xA;" OwnerUserId="5050" LastActivityDate="2011-08-31T12:41:54.073" />
  <row Id="14177" PostTypeId="2" ParentId="14174" CreationDate="2011-08-31T12:51:11.297" Score="0" ViewCount="" Body="&lt;p&gt;may be &lt;a href=&quot;http://www.wisc-online.com/objects/ace11705/ace11705.swf&quot; rel=&quot;nofollow&quot;&gt;this animation&lt;/a&gt; help you. i hope i serve you. &lt;/p&gt;&#xA;" OwnerUserId="5050" LastEditorUserId="4551" LastEditorDisplayName="" LastEditDate="2011-08-31T14:53:56.747" LastActivityDate="2011-08-31T14:53:56.747" />
  <row Id="14178" PostTypeId="2" ParentId="14165" CreationDate="2011-08-31T13:07:01.357" Score="2" ViewCount="" Body="&lt;p&gt;Think of Force is Pressure times Area ($F = P\cdot A$). You know pressure is a scalar (there is no direction associated with it), and a force is a vector (it acts along an axis). So what does that mean for pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a small area and see it's contribution to the total force due to pressure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ {\rm d}F = P(x,y,z)\,{\rm d}A $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The direction of the force is normal to the area, and its magnitude is proportional to the size of the area. This is why an infinitesimal area ${\rm d} A$ can be a vector. It is convenient to think of (vector)=(scalar)*(vector).&lt;/p&gt;&#xA;" OwnerUserId="392" LastActivityDate="2011-08-31T13:07:01.357" />
  <row Id="14179" PostTypeId="1" CreationDate="2011-08-31T13:28:40.410" Score="3" ViewCount="77" Body="&lt;p&gt;Is there easily-available* software to simulate coils, solenoids, and other magnetic and electromagnetic devices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to play around with some design ideas, such as Halbach arrays, but physics simulators like SketchyPhysics don't seem to have the laws of electromagnetism implemented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*&quot;Easily-available&quot; means &quot;not cost prohibitive.&quot;)&lt;/p&gt;&#xA;" OwnerUserId="1465" LastActivityDate="2011-09-02T04:09:45.867" Title="Where can I find simulation software for electricity and magnets?" Tags="&lt;electromagnetism&gt;&lt;simulation&gt;&lt;software&gt;" AnswerCount="2" />
  <row Id="14180" PostTypeId="2" ParentId="14174" CreationDate="2011-08-31T13:43:40.610" Score="1" ViewCount="" Body="&lt;p&gt;Set up a magnet around a coil, such that the Magnet has a field that is constant in magnitude, and has vector form ${\vec B} = B_{0}(\cos (\omega t) {\hat z} + \sin (\omega t) {\hat x})$.  Orient the coil so that it lies in the $x-y$ plane and thus has a normal that points in the $z$ direction.  If the overlap of the coil's area and the magnetic field is $A$, then the net flux through the coil as a function of time is given by $\Phi=B_{0}A\cos\theta$.  Then, Faraday's Law ($V_{ind}=-\frac{d\Phi}{dt}$) tells us that the induced voltage through the loop is given by $V_{ind}=B_{0}A \omega \sin \omega t$, which reverses every half cycle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By going back to the original inducing magnetic field, you can see that this reverses every half cycle because the direction of the magnetic field &lt;b&gt;also&lt;/b&gt; reverses every half cycle.  &lt;/p&gt;&#xA;" OwnerUserId="347" LastEditorUserId="347" LastEditorDisplayName="" LastEditDate="2011-08-31T14:26:21.057" LastActivityDate="2011-08-31T14:26:21.057" />
  <row Id="14181" PostTypeId="1" CreationDate="2011-08-31T14:26:43.823" Score="7" ViewCount="149" Body="&lt;p&gt;For simple potentials like square wells and harmonic oscillators, one can explicitly calculate the product $\Delta x \Delta p$ for stationary states.  When you do this, it turns out that higher energy levels have higher values of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta x \Delta p$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true for all time-independent potentials?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly, it is possible to find two states $\mid \Psi_1 \rangle$ and $\mid \Psi_2 \rangle$ with $\langle \Psi_1 \mid H \mid \Psi_1 \rangle &amp;gt; \langle \Psi_2 \mid H \mid \Psi_2 \rangle$ and also $\Delta x_1 \Delta p_1 &amp;lt; \Delta x_2 \Delta p_2$.  For example, choose a quadratic potential, let $\mid \Psi_2 \rangle$ be the ground state and let $\mid \Psi_1 \rangle$ be a Gaussian with minimum uncertainty.  So I'm asking here just about the stationary states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Ron pointed out in the comments, this question is most interesting if we consider potentials with only a single local minimum, and increasing potential to the right of it and decreasing to the left.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-08-31T15:16:52.220" LastActivityDate="2011-09-03T19:06:45.790" Title="Do stationary states with higher energy necessarily have higher position-momentum uncertainty?" Tags="&lt;quantum-mechanics&gt;&lt;uncertainty-principle&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" />
  <row Id="14182" PostTypeId="1" AcceptedAnswerId="14192" CreationDate="2011-08-31T15:05:20.970" Score="11" ViewCount="325" Body="&lt;p&gt;There are a lot of articles being posted in the wake of a CERN announcement that they have not observed the Higgs boson in the range of energies so far searched (between 145 and 466 billion eV), e.g. &lt;a href=&quot;http://blogs.scientificamerican.com/guest-blog/2011/08/23/a-higgs-setback-did-stephen-hawking-just-win-the-most-outrageous-bet-in-physics-history/&quot; rel=&quot;nofollow&quot;&gt;this Scientific American blog post&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can this be? Wasn't the existence of the Higgs an almost foregone conclusion. Can someone who looked at the details at CERN explain if they are hoping to run the LHC for more time to find it, or is the search becoming hopeless (or perhaps its energy was estimated wrong).&lt;/p&gt;&#xA;" OwnerUserId="852" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-31T17:32:07.493" LastActivityDate="2011-09-01T16:13:00.017" Title="So, no Higgs boson then?" Tags="&lt;particle-physics&gt;&lt;higgs-boson&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" />
  <row Id="14183" PostTypeId="2" ParentId="14175" CreationDate="2011-08-31T15:17:32.627" Score="2" ViewCount="" Body="&lt;p&gt;In single slit diffraction, assuming small diffraction angles, the intensity profile is the magnitude-squared of the Fourier transform of the function which is constant between -1 and 1 (up to units of length), and this Fourier transform is $\frac{\sin(x)}{x}$. This has a peak at zero, because of the falloff of $1/x$ but most importantly, and this can be seen qualitatively, the first zero of $\sin(x)$ is absent, the central maximum is twice as wide as all the others.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-08-31T17:54:55.913" LastActivityDate="2011-08-31T17:54:55.913" />
  <row Id="14184" PostTypeId="2" ParentId="14115" CreationDate="2011-08-31T15:47:23.793" Score="2" ViewCount="" Body="&lt;p&gt;Lets suppose that all N magnets are identical. Assume that they are quite far from each other so we can replace them with N magnetic dipoles with dipole moment $\overrightarrow{m}$. Also assume that the dipoles are placed along the x axis  in such a way that they repel each other and all N moments   $\overrightarrow{m}$ are parallel. The first step is to find the interaction force between two dipoles. Fortunately, Wikipedia has a formula for the force so that we can save a large amount of work.   &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_dipole#Forces_between_two_magnetic_dipoles&quot; rel=&quot;nofollow&quot;&gt;Link here&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I write only its x axis component which we need, to save space:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F=\frac{6\mu_0 m^2}{4\pi x^4}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to find &quot;spring rate&quot; between two magnets, it is sufficient to find the differential $dF$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dF=-\frac{6\mu_0 m^2}{\pi x^5}dx$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for the small displacement the &quot;spring rate&quot;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k(x)= \frac{6\mu_0 m^2}{\pi x^5} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since we know $k$ it is easy to find  the effective spring rate for the top magnet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $F$ be the force on the top magnet, other end magnet fixed. Then the displacement of the top&lt;/p&gt;&#xA;&#xA;&lt;p&gt;magnet is $\Delta x=(N-1)\frac{F}{k}$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the effective spring rate: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k_e(x)= \frac{6\mu_0 m^2}{(N-1)\pi x^5} $$&lt;/p&gt;&#xA;" OwnerUserId="552" LastActivityDate="2011-08-31T15:47:23.793" CommentCount="4" />
  <row Id="14185" PostTypeId="1" AcceptedAnswerId="14188" CreationDate="2011-08-31T16:03:08.093" Score="1" ViewCount="111" Body="&lt;p&gt;Why is it hotter inside an isolated car (air conditioning is off) than outside during a  sunny day in summer?&lt;/p&gt;&#xA;" OwnerUserId="5100" LastEditorUserId="74" LastEditorDisplayName="" LastEditDate="2011-08-31T16:38:43.193" LastActivityDate="2011-08-31T18:11:49.723" Title="Heat in the car during sunny day" Tags="&lt;thermodynamics&gt;&lt;heat&gt;&lt;temperature&gt;" AnswerCount="4" CommentCount="1" />
  <row Id="14186" PostTypeId="2" ParentId="14168" CreationDate="2011-08-31T16:04:12.473" Score="3" ViewCount="" Body="&lt;p&gt;Orientability is a &lt;strong&gt;purely topological property&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time-orientability is a mix between topology and geometry: a manifold is time orientable if there exists &quot;an arrow of time&quot; on it. Mathematically this means that there must be a &lt;strong&gt;non-vanishing vector field&lt;/strong&gt; (certain topological criterion must be met for the manifold to admit a non-vanishing vector field) that is &lt;strong&gt;time-like&lt;/strong&gt; (so the Lorentzian metric must be compatible with the vector field). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two properties are more or less independent of each other. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The standard Minkowski space is orientable and time-orientable.&lt;/li&gt;&#xA;&lt;li&gt;The two dimensional sphere is orientable. But it cannot be time orientable. (On the two dimensional sphere there cannot be a non-vanishing vector field; in fact, the two dimensional sphere cannot admit a continuous Lorentzian metric.)&lt;/li&gt;&#xA;&lt;li&gt;The Mobius strip is not orientable. But it can be equipped with a Lorentzian metric that is time-orientable. It can also be equipped with a Lorentzian metric that is not time-orientable. &lt;/li&gt;&#xA;&lt;li&gt;The cylinder ($\mathbb{R}\times\mathbb{S}^1$) is orientable. It can be equipped with a Lorentzian metric that is time-orientable. It can also be equipped with a Lorentzian metric that is not time-orientable. (Let $x$ be the coordinate in the $\mathbb{R}$, and $\theta$ be the coordinate on $\mathbb{S}^1$ taking value between $[0,2\pi)$. The metric&#xA;$$ ds^2 = - dx^2 + d\theta^2 $$&#xA;is time orientable, but the metric&#xA;$$ ds^2 = \cos \theta dx^2 + 2\sin\theta dxd\theta - \cos\theta d\theta^2 $$&#xA;is a non-time-orientable Lorentzian metric.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt; Ron Maimon brings up a very good point in the comments: that there is a difference between the following two statements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The manifold $M$ with a Lorentzian metric $g$ is time orientable&lt;/li&gt;&#xA;&lt;li&gt;The manifold $M$ can be equipped with a time-orientable Lorentzian metric $g$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The distinction is clear when you consider the third and the fourth examples above the cut. And it is clear that trivially we have a one-way implication between the two above statements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it turns out, for paracompact manifolds, we have the equivalence of the following three statements (see Proposition 5.37, O'Neill, &lt;em&gt;Semi-Riemannian Geometry&lt;/em&gt;):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$M$ admits a smooth non-vanishing vector field&lt;/li&gt;&#xA;&lt;li&gt;$M$ can be equipped with a smooth Lorentzian metric&lt;/li&gt;&#xA;&lt;li&gt;$M$ can be equipped with a time-orientable Lorentzian metric&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;so if one were to consider time orientability to be &lt;em&gt;a property of the manifold&lt;/em&gt; that &lt;strong&gt;it admits a time-orientable Lorentzian metric&lt;/strong&gt;, then Ron correctly argues in the comments that this notion of time orientability &quot;is topological&quot;. There is, however, one unsatisfying aspect to that statement, namely that &lt;em&gt;any Lorentzian manifold is &quot;topologically&quot; time-orientable, but the given metric may not be time-orientable&lt;/em&gt; (again, I refer to the fourth example above the cut). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's interesting is the following: there actually is a &lt;strong&gt;topological obstruction&lt;/strong&gt; to the existence of a &lt;strong&gt;non&lt;/strong&gt; -time-orientable Lorentzian metric, when given a manifold that admits a Lorentzian (and hence can be assumed to be time-orientable) metric. Let me introduce some mathematics notations. For the manifold $M$, I will denote by $TM$ the tangent bundle, and by $SM$ the sphere bundle. The sphere bundle is formed by taking $TM$, removing the zero section, and quotienting out in each fiber by the multiplication by scalars. So topologically the fibers are $\mathbb{S}^{n-1}$. And by $PM$ I will denote the &quot;bundle of lines&quot;, obtained by identifying the antipodes in the fibres of $SM$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can easily show that &lt;em&gt;the manifold $M$ admits a Lorentzian metric if and only if one can find a section of the bundle $PM$&lt;/em&gt; (for example, see Choi &amp;amp; Suh, &quot;Remarks on the topology of Lorentzian manifolds&quot;, Comm. Korean Math. Soc. &lt;strong&gt;15&lt;/strong&gt; (2000)). In particular, given a Lorentzian manifold, the section $\gamma$ of $PM$ can be chosen to be time-like. This section is given by a time-like vector field (thereby showing time-orientability), if it is the image of a section of $SM$ under the projection map. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $SM$ forms a natural double cover of $PM$ under projection $\pi$. So given a section $\gamma$, its preimage $\pi^{-1}\gamma$ can be treated as a double cover of $M$. Therefore a topological condition that guarantees that there &lt;strong&gt;cannot&lt;/strong&gt; be any non-time-orientable Lorentzian metric would be that the pre-image $\pi^{-1}\gamma$ of any section $\gamma$ of $PM$ has two connected components. In particular, a sufficient condition to guarantee this is that $M$ is simply connected (since any covering map of a simply connected manifold $M$ is trivial). So we have that: a simply connected Lorentzian manifold must be time-orientable. &lt;/p&gt;&#xA;" OwnerUserId="829" LastEditorUserId="829" LastEditorDisplayName="" LastEditDate="2011-09-01T13:17:04.983" LastActivityDate="2011-09-01T13:17:04.983" CommentCount="7" />
  <row Id="14187" PostTypeId="2" ParentId="14185" CreationDate="2011-08-31T16:17:39.690" Score="3" ViewCount="" Body="&lt;p&gt;The air is transparent to the wavelengths of incident light and is not heated (much) by the sunlight; instead, the interior of the car is heated first.  The glass, however, does not as readily transmit the wavelengths of heat radiated from the interior.  The absorbed heat increases the car temperature until its heat losses to the environment equal the heat deposited in the car.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="5020" LastEditorDisplayName="" LastEditDate="2011-08-31T16:37:59.170" LastActivityDate="2011-08-31T16:37:59.170" />
  <row Id="14188" PostTypeId="2" ParentId="14185" CreationDate="2011-08-31T16:34:38.927" Score="3" ViewCount="" Body="&lt;p&gt;The heating of the car is the same as the heating of the Earth due to the greenhouse effect--- the sun shines through the glass depositing very energetic photons whose equilibrium temperature is 5000 degrees, but the interior surfaces reradiate that light as photons whose temperature is at the interior temperature. The interior end up at a higher temperature, where the outgoing thermal power is equal to the incoming non-thermal solar power.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastActivityDate="2011-08-31T16:34:38.927" />
  <row Id="14189" PostTypeId="2" ParentId="14185" CreationDate="2011-08-31T16:39:48.527" Score="0" ViewCount="" Body="&lt;p&gt;Here is my guess of some of the relevant factors:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Greenhouse_effect&quot; rel=&quot;nofollow&quot;&gt;Greenhouse effect&lt;/a&gt;: this causes energy to be &quot;trapped&quot; (in a sense) in your car.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lack of air flow in your car will make heat dissipation much less efficient.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The relatively lower &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_capacity&quot; rel=&quot;nofollow&quot;&gt;specific heat capacity&lt;/a&gt; of the things in your car, e.g. seats, compared to air will heat up faster.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Please comment if there are more relevant points. Also, corrections are welcome! :)&lt;/p&gt;&#xA;" OwnerUserId="5046" LastActivityDate="2011-08-31T16:39:48.527" />
  <row Id="14190" PostTypeId="2" ParentId="14175" CreationDate="2011-08-31T16:49:40.547" Score="1" ViewCount="" Body="&lt;p&gt;A more physical way of restating Ron's answer: in diffraction theory you are adding up Huygens wavelets over the aperture with appropriate phase factors.  The central fringe is the only one where the entire aperture is adding in phase, i.e. constructively.  The next fringe will have 2/3 adding constructively and 1/3 destructively, then after that 3/5 - 2/5, etc. The farther out you go in fringes, the more of the aperture is just cancelling itself.&lt;/p&gt;&#xA;" OwnerUserId="1631" LastActivityDate="2011-08-31T16:49:40.547" />
  <row Id="14192" PostTypeId="2" ParentId="14182" CreationDate="2011-08-31T17:41:14.500" Score="5" ViewCount="" Body="&lt;p&gt;I wouldn't call the existence of the Higgs boson a foregone conclusion. It got people excited because we have a very elegant theory that predicts pretty much everything else correctly, and also predicts the existence of a Higgs boson, but the theory could certainly be wrong. Elegance is no guarantee of success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I recently made a &lt;a href=&quot;http://www.ellipsix.net/blog/2011/08/getting-further-away-from-the-elusive-higgs-boson.html&quot; rel=&quot;nofollow&quot;&gt;blog post&lt;/a&gt; about that that you might want to read. The gist of it is that the signal observed by the LHC detectors - in other words, the difference between what they actually saw and what they expected to see - has been getting smaller the more data they collect, over the past month or so. But it's too soon to say anything with certainty. The LHC will continue running, and if the Higgs boson is there to be discovered, they hope to do so by the end of this year. Otherwise, they expect to be able to exclude it by the end of next year.&lt;/p&gt;&#xA;" OwnerUserId="124" LastActivityDate="2011-08-31T17:41:14.500" CommentCount="3" />
  <row Id="14193" PostTypeId="2" ParentId="14185" CreationDate="2011-08-31T18:11:49.723" Score="1" ViewCount="" Body="&lt;p&gt;Several effects all push the interior temperature above the ambient air temp. Most discussed, above is that to the extent that the glass is opaque (or partially opaque) to thermal infrared, you get a greenhouse effect. But you also have the fact, that surfaces exposed to direct sunlight (unless it is highly reflective), and much hotter than the ambient temperature of the free air. Surface temperatures are roughly in thermal equilibrium with the average amount of shortwave (solar) radiation, averaged over the course of several days, so the air effectively remembers the previous nights cooling. The air trapped within the car, has poor thermal communication (high thermal impedance) with the air, and so must become considerably hotter than the outside air in order lose enought heat to its surroundings to balance the heat budget. having to transfer heat from the solid interior of the car, to the air within the car, then to the glass, then finally from the glass to the outside air, adds quite a lot of thermal impedance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, note, that consistent with the first point, about the radiative equilibrium temperature during the daytime being much hotter than the ambient air temperature, also applies, even if the car had no transparent windows. Consider a matal box shipping container sitting in the sun. The surface will become tens of degrees C hotter than the air temperature, before radiative, and conductive loses become great enough to bring the surface heat budget into balance. The shipping container, will also get very hot, even without windows. And the roof of your car acts in the same manner.&lt;/p&gt;&#xA;" OwnerUserId="377" LastActivityDate="2011-08-31T18:11:49.723" />
  <row Id="14194" PostTypeId="1" CreationDate="2011-08-31T20:42:01.770" Score="1" ViewCount="30" Body="&lt;p&gt;A lot of quantum books have very brief discussions of it but I want to see it in some detail and preferably with many examples as possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a book, paper, or some other place that details this approximation?&lt;/p&gt;&#xA;" OwnerUserId="1939" LastActivityDate="2011-08-31T20:42:01.770" Title="where can a good treatment of the 'sudden' perturbation approximation be found?" Tags="&lt;books&gt;&lt;perturbation-theory&gt;" CommentCount="2" />
  <row Id="14195" PostTypeId="1" AcceptedAnswerId="14196" CreationDate="2011-08-31T20:47:16.100" Score="4" ViewCount="86" Body="&lt;p&gt;I'll try to be clear: example: If you send the ISS far enough for it not to undergo the Earth's gravity anymore, then you turn it and the—sleeping—astronauts in it upside down, when they wake up, will they know/feel that they're not the right way up? How?&lt;/p&gt;&#xA;" OwnerUserId="5102" LastActivityDate="2011-09-01T01:06:32.400" Title="Without gravity, is there still up and down?" Tags="&lt;gravity&gt;" AnswerCount="2" CommentCount="2" />
  <row Id="14196" PostTypeId="2" ParentId="14195" CreationDate="2011-08-31T21:02:58.587" Score="4" ViewCount="" Body="&lt;p&gt;No they will not. Space is intrinsically isotropic, so assuming they are not aware of any specific reference points, and they are far enough away from a massive body as to experience an insignificant amount of gravity, there would be no way of knowing their orientation. Gravity essentially provides observers with a force field that the body can utilise to establish orientation etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="4312" LastActivityDate="2011-08-31T21:02:58.587" />
  <row Id="14197" PostTypeId="2" ParentId="14179" CreationDate="2011-08-31T21:10:32.537" Score="2" ViewCount="" Body="&lt;p&gt;There are a number of commercial packages you can use for such purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comsol multiphysics simulation software is a general purpose physics modelling software package that I belive can deal with electromagnitism. The link to their website is &lt;a href=&quot;http://www.uk.comsol.com&quot; rel=&quot;nofollow&quot;&gt;http://www.uk.comsol.com&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ansys is another commercial vendor that provides a electromagnetic physics module. There website is &lt;a href=&quot;http://www.ansys.com&quot; rel=&quot;nofollow&quot;&gt;http://www.ansys.com&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="4312" LastActivityDate="2011-08-31T21:10:32.537" />
  <row Id="14198" PostTypeId="2" ParentId="14165" CreationDate="2011-08-31T21:23:08.610" Score="0" ViewCount="" Body="&lt;p&gt;There is an especially picturesque example of the Law of Pythagors in three dimensions applied to the areas of a simplex. (Where by &quot;simplex&quot; I believe I mean a section of space bounded by three orthogonal planes and one arbitrary plane.) The sum of the squares (of the areas) of the three small faces are equal to the square of the area of the oblique face. It is easily explained by the pressure/flow type arguments put forward in the other answers posted here, plus the obvious physical condition that an undisturbed fluid is in equilibrium with itself. &lt;/p&gt;&#xA;" OwnerUserId="3253" LastActivityDate="2011-08-31T21:23:08.610" />
  <row Id="14199" PostTypeId="1" AcceptedAnswerId="14200" CreationDate="2011-08-31T21:31:56.780" Score="1" ViewCount="88" Body="&lt;p&gt;Just a quick check: If given a time-independent wavefunction of the form $$\psi(x) = e^{ikx}f(x)$$, where $f(x)$ any arbitrary function of $x$ but one can't factor out another $e^{i\alpha x}$, $\alpha\in\mathbb R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I then immediately conclude that $k^2= \frac{2mE}{\hbar^2}$? Is there a name for this quantity? It seems to creep up a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the wavefunction is $$\psi(x) = e^{ikx}(\tanh{x}-ik)$$. Am I justified to conclude by inspection that there is no reflection (since there is no $Re^{-ikx}$ term)? What else can I conclude by inspection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5046" LastEditorUserId="5046" LastEditorDisplayName="" LastEditDate="2011-08-31T21:38:01.983" LastActivityDate="2011-09-01T00:19:44.630" Title="Inspecting the form of a wavefunction" Tags="&lt;quantum-mechanics&gt;&lt;schrodinger-equation&gt;" AnswerCount="2" CommentCount="9" />
  <row Id="14200" PostTypeId="2" ParentId="14199" CreationDate="2011-08-31T22:17:39.727" Score="1" ViewCount="" Body="&lt;p&gt;You need to know that the $f(x)$ is asymptotically constant in order to conclude this, and that the potential is asymptotically zero to the left, and that the wavefunction is an energy eigenstate. Then you can conclude that the energy of the eigenstate is the same as the energy of the eigenstate far away and the only energy is kinetic. $k^2\over 2m$ is called the kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your wavefunction, since $\tanh$ is asymptotically constant, you can conclude that there is no reflected wave, that the transmitted wave is phase shifted by the angle in the complex plane of the ratio of the two asymptotic values, $(-1+ik)\over (1+ik)$, and you can conclude that the potential of scattering is that which is required to make the Schrodinger equation work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ mV(x) = - {1\over \cosh^2(x)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where m is the mass of the particle. This is a well-known exactly solvable potential, called the Paschl-Teller potential.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;General Reflectionless Potential at Wavenumber k&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;You are looking for the most general potential with no reflection at wavenumber k when you make this ansatz for the wavefunction. For a general energy eigenstate of this form,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${- \psi&amp;#39;&amp;#39;\over 2m} = {k^2\over 2m}\psi + \psi( - {ik\over m} {f&amp;#39;\over f} - {1\over 2m}{f&amp;#39;&amp;#39;\over f})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that the potential you get is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ V(x) = -({f&amp;#39;&amp;#39;\over 2m f} + ik {f&amp;#39;\over f})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is not an answer, because this potential is in general complex, while a physical potential is real. To find the constraint on f that $V$ is real, write $f$ as $e^\alpha$ and define $\beta=\alpha&amp;#39;$. Then the potential in terms of $\beta$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ V(x) = \beta&amp;#39; + 2ik\beta + \beta^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and in order for this to be real, if $\beta=P+iQ$ with P,Q real functions of x, then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P = -{Q&amp;#39;\over 2(k+Q)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, for any choice of Q, with the appropriate boundary condition, give the class of potentials reflectionless at wavenumber k. The Paschl-Teller potential is exceptional for being reflectionless at all k simultaneously.&lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="4864" LastEditorDisplayName="" LastEditDate="2011-09-01T00:19:44.630" LastActivityDate="2011-09-01T00:19:44.630" CommentCount="1" />
  <row Id="14201" PostTypeId="2" ParentId="14199" CreationDate="2011-08-31T22:40:40.640" Score="1" ViewCount="" Body="&lt;p&gt;Recall the Schrödinger equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(-\frac{\hbar^2}{2m}\Delta+V(x)\right)\psi(x) = E\psi(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be written as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\Delta+\frac{2m(E-V(x))}{\hbar^2}\right)\psi(x)=\left(\Delta+k^2(x)\right)\psi=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when you rename the second factor $k(x)=\sqrt{\frac{2m(E-V(x))}{\hbar^2}}$ you get something like a wave equation if k is constant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta\psi(x)=-k^2\psi(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is solved by the the plane wave $\psi(x)=e^{ikx}$. For a free particle V=0 and $k^2=\frac{2mE}{\hbar^2}$ which is why this term appears so often. $k$ is called the &lt;em&gt;circular wave number&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for your question,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}\Delta\psi(x) &amp;amp;= \Delta e^{ikx}f(x)\\&#xD;&#xA;&amp;amp;=\frac{\partial^2}{\partial x^2}e^{ikx}f(x)\\&#xD;&#xA;&amp;amp;=\frac{\partial}{\partial x}(ike^{ikx}f(x)+e^{ikx}f&amp;#39;(x))\\&#xD;&#xA;&amp;amp;=\bigl(-k^2f(x)+2ikf&amp;#39;(x)+f&amp;#39;&amp;#39;(x)\bigr)e^{ikx}\\&#xD;&#xA;&amp;amp;=-k^2\psi(x)+\cdots\\&#xD;&#xA;&amp;amp;= -\frac{2m(E-V(x))}{\hbar^2}\psi(x)\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as you can see, in general this is not possible. Your function $f(x)$ generally is non constant thus the Laplacian has other terms besides $-k^2\psi(x)$. Thus in general the correspondence $k^2=\frac{2m(E-V(x))}{\hbar^2}$ does not hold, much less $k^2=\frac{2mE}{\hbar^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an interesting special case, that is very relevant in solid state physics, called Bloch waves. These arise, when the potential is periodic $V(x)=V(x+a)$. The solutions of the Schrödinger equation for such a potential look as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi(x)=e^{ikx}u_k(x)$$ where the $u_k(x)$ have the same periodicity as the potential.&lt;/p&gt;&#xA;" OwnerUserId="2151" LastEditorUserId="124" LastEditorDisplayName="" LastEditDate="2011-09-01T00:00:34.583" LastActivityDate="2011-09-01T00:00:34.583" CommentCount="1" />
</posts>